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Riley, Hobson. Mathematical methods for physics and engineering (2ed., 2002)(1253s)_MPt_-1

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Published textbook by Riley and Hobson, not Phil's own work, kept in a folder of math methods book downloads. The contents list covers algebra, calculus, complex numbers, series, partial differentiation, multiple integrals, vectors and matrices, normal modes, vector calculus, Fourier series and transforms, ordinary and partial differential equations, and Sturm-Liouville eigenfunction methods. Later chapters beyond the first 19 are not shown in the extracted text.

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Contents Preface to the second edition xix Preface to the ﬁrst edition xxi 1 Preliminary algebra 1 1.1 Simple functions and equations 1 Polynomial equations; factorisation; properties of roots 1.2 Trigonometric identities 10 Single angle; compound-angles; double- and half-angle identities 1.3 Coordinate geometry 151.4 Partial fractions 18 Complications and special cases; complex roots; repeated roots 1.5 Binomial expansion 251.6 Properties of binomial coeﬃcients 271.7 Some particular methods of proof 30 Methods of proof; by induction; by contradiction; necessary and suﬃcient conditions 1.8 Exercises 36 1.9 Hints and answers 39 2 Preliminary calculus 42 2.1 Diﬀerentiation 42 Diﬀerentiation from ﬁrst principles; products; the chain rule; quotients; implicit diﬀerentiation; logarithmic diﬀerentiation; Leibniz’ theorem; specialpoints of a function; theorems of diﬀerentiation v CONTENTS 2.2 Integration 60 Integration from ﬁrst principles; the inverse of diﬀerentiation; integration by inspection; sinusoidal functions; logarithmic integration; integrationusing partial fractions; substitution method; integration by parts; reduction formulae; inﬁnite and improper integrals; plane polar coordinates; integral inequalities; applications of integration 2.3 Exercises 77 2.4 Hints and answers 82 3 Complex numbers and hyperbolic functions 86 3.1 The need for complex numbers 86 3.2 Manipulation of complex numbers 88 Addition and subtraction; modulus and argument; multiplication; complex conjugate; division 3.3 Polar representation of complex numbers 95 Multiplication and division in polar form 3.4 de Moivre’s theorem 98 trigonometric identities; ﬁnding the nth roots of unity; solving polynomial equations 3.5 Complex logarithms and complex powers 102 3.6 Applications to diﬀerentiation and integration 1043.7 Hyperbolic functions 105 Deﬁnitions; hyperbolic–trigonometric analogies; identities of hyperbolic functions; solving hyperbolic equations ; inverses of hyperbolic functions; calculus of hyperbolic functions 3.8 Exercises 112 3.9 Hints and answers 116 4 Series and limits 118 4.1 Series 1184.2 Summation of series 119 Arithmetic series; geometric series; arithmetico-geometric series; the diﬀerence method; series involving natural numbers; transformation of series 4.3 Convergence of inﬁnite series 127 Absolute and conditional convergence; convergence of a series containing only real positive terms; alternating series test 4.4 Operations with series 134 4.5 Power series 134 Convergence of power series; operations with power series 4.6 Taylor series 139 Taylor’s theorem; approximation errors in Taylor series; standard Maclaurin series vi CONTENTS 4.7 Evaluation of limits 144 4.8 Exercises 147 4.9 Hints and answers 152 5 Partial diﬀerentiation 154 5.1 Deﬁnition of the partial derivative 154 5.2 The total diﬀerential and total derivative 156 5.3 Exact and inexact diﬀerentials 158 5.4 Useful theorems of partial diﬀerentiation 160 5.5 The chain rule 160 5.6 Change of variables 161 5.7 Taylor’s theorem for many-variable functions 163 5.8 Stationary values of many-variable functions 165 5.9 Stationary values under constraints 170 5.10 Envelopes 1765.11 Thermodynamic relations 179 5.12 Diﬀerentiation of integrals 181 5.13 Exercises 182 5.14 Hints and answers 188 6 Multiple integrals 190 6.1 Double integrals 190 6.2 Triple integrals 193 6.3 Applications of multiple integrals 194 Areas and volumes; masses, centres of mass and centroids; Pappus’ theorems; moments of inertia; mean values of functions 6.4 Change of variables in multiple integrals 202 Change of variables in double integrals; evaluation of the integral I=R∞ −∞e−x2dx; change of variables in triple integrals; general properties of Jacobians 6.5 Exercises 210 6.6 Hints and answers 214 7 Vector algebra 216 7.1 Scalars and vectors 216 7.2 Addition and subtraction of vectors 217 7.3 Multiplication by a scalar 218 7.4 Basis vectors and components 221 7.5 Magnitude of a vector 222 7.6 Multiplication of vectors 223 Scalar product; vector product; scalar triple product; vector triple product vii CONTENTS 7.7 Equations of lines, planes and spheres 230 Equation of a line; equation of a plane 7.8 Using vectors to ﬁnd distances 233 Point to line; point to plane; line to line; line to plane 7.9 Reciprocal vectors 237 7.10 Exercises 2387.11 Hints and answers 244 8 Matrices and vector spaces 246 8.1 Vector spaces 247 Basis vectors; the inner product; some useful inequalities 8.2 Linear operators 252 Properties of linear operators 8.3 Matrices 254 Matrix addition and multiplication by a scalar; multiplication of matrices 8.4 Basic matrix algebra 2558.5 Functions of matrices 2608.6 The transpose of a matrix 2608.7 The complex and Hermitian conjugates of a matrix 2618.8 The trace of a matrix 2638.9 The determinant of a matrix 264 Properties of determinants 8.10 The inverse of a matrix 2688.11 The rank of a matrix 2728.12 Special types of square matrix 273 Diagonal; symmetric and antisymmetri c; orthogonal; Hermitian; unitary; normal 8.13 Eigenvectors and eigenvalues 277 Of a normal matrix; of Hermitian and anti-Hermitian matrices; of a unitarymatrix; of a general square matrix 8.14 Determination of eigenvalues and eigenvectors 285 Degenerate eigenvalues 8.15 Change of basis and similarity transformations 288 8.16 Diagonalisation of matrices 2908.17 Quadratic and Hermitian forms 293 The stationary properties of the eigenvectors; quadratic surfaces 8.18 Simultaneous linear equations 297 Nsimultaneous linear equations in Nunknowns 8.19 Exercises 312 8.20 Hints and answers 319 viii CONTENTS 9 Normal modes 322 9.1 Typical oscillatory systems 3239.2 Symmetry and normal modes 3289.3 Rayleigh–Ritz method 333 9.4 Exercises 335 9.5 Hints and answers 338 10 Vector calculus 340 10.1 Diﬀerentiation of vectors 340 Composite vector expressions; diﬀerential of a vector 10.2 Integration of vectors 345 10.3 Space curves 346 10.4 Vector functions of several arguments 35010.5 Surfaces 35110.6 Scalar and vector ﬁelds 35310.7 Vector operators 353 Gradient of a scalar ﬁeld; divergence of a vector ﬁeld; curlof a vector ﬁeld 10.8 Vector operator formulae 360 Vector operators acting on sums and products; combinations of grad, div andcurl 10.9 Cylindrical and spherical polar coordinates 363 Cylindrical polar coordinates; spherical polar coordinates 10.10 General curvilinear coordinates 370 10.11 Exercises 37510.12 Hints and answers 381 11 Line, surface and volume integrals 383 11.1 Line integrals 383 Evaluating line integrals; physical examples of line integrals; line integrals with respect to a scalar 11.2 Connectivity of regions 389 11.3 Green’s theorem in a plane 390 11.4 Conservative ﬁelds and potentials 393 11.5 Surface integrals 395 Evaluating surface integrals; vector areas of surfaces; physical examples of surface integrals 11.6 Volume integrals 402 Volumes of three-dimensional regions 11.7 Integral forms for grad, div and curl 404 11.8 Divergence theorem and related theorems 407 Green’s theorems; other related integral theorems; physical applications of the divergence theorem ix CONTENTS 11.9 Stokes’ theorem and related theorems 412 Related integral theorems; physical applications of Stokes’ theorem 11.10 Exercises 415 11.11 Hints and answers 420 12 Fourier series 421 12.1 The Dirichlet conditions 42112.2 The Fourier coeﬃcients 42312.3 Symmetry considerations 42512.4 Discontinuous functions 42612.5 Non-periodic functions 428 12.6 Integration and diﬀerentiation 430 12.7 Complex Fourier series 43012.8 Parseval’s theorem 43212.9 Exercises 43312.10 Hints and answers 437 13 Integral transforms 439 13.1 Fourier transforms 439 The uncertainty principle; Fraunhofer diﬀraction; the Dirac δ-function; relation of the δ-function to Fourier transforms; properties of Fourier transforms; odd and even functions; c onvolution and de convolution; correlation functions and energy spectra; Parseval’s theorem; Fouriertransforms in higher dimensions 13.2 Laplace transforms 459 Laplace transforms of derivatives and integrals; other properties of Laplacetransforms 13.3 Concluding remarks 465 13.4 Exercises 46613.5 Hints and answers 472 14 First-order ordinary diﬀerential equations 474 14.1 General form of solution 475 14.2 First-degree ﬁrst-order equations 476 Separable-variable equations; exact equations; inexact equations: integrat- ing factors; linear equations; homogene ous equations; isobaric equations; Bernoulli’s equation; mi scellaneous equations 14.3 Higher-degree ﬁrst-order equations 486 Equations soluble for p;f o r x;f o r y; Clairaut’s equation 14.4 Exercises 490 14.5 Hints and answers 494 x CONTENTS 15 Higher-order ordinary diﬀerential equations 496 15.1 Linear equations with constant coeﬃcients 498 Finding the complementary function yc(x); ﬁnding the particular integral yp(x); constructing the general solution yc(x)+yp(x); linear recurrence relations; Laplace transform method 15.2 Linear equations with variable coeﬃcients 509 The Legendre and Euler linear equations; exact equations; partiallyknown complementary function; variation of parameters; Green’s functions;canonical form for second-order equations 15.3 General ordinary diﬀerential equations 524 Dependent variable absent; independent variable absent; non-linear exactequations; isobaric or homogeneous equations; equations homogeneous in x oryalone; equations having y=Ae xas a solution 15.4 Exercises 529 15.5 Hints and answers 535 16 Series solutions of ordinary diﬀerential equations 537 16.1 Second-order linear ordinary diﬀerential equations 537 Ordinary and singular points 16.2 Series solutions about an ordinary point 54116.3 Series solutions about a regular singular point 544 Distinct roots not diﬀering by an integer; repeated root of the indicial equation; distinct roots diﬀering by an integer 16.4 Obtaining a second solution 549 The Wronskian method; the derivative method; series form of the secondsolution 16.5 Polynomial solutions 554 16.6 Legendre’s equation 555 General solution for integer /lscript; properties of Legendre polynomials 16.7 Bessel’s equation 564 General solution for non-integer ν; general solution for integer ν; properties of Bessel functions 16.8 General remarks 575 16.9 Exercises 575 16.10 Hints and answers 579 17 Eigenfunction methods for diﬀerential equations 581 17.1 Sets of functions 583 Some useful inequalities 17.2 Adjoint and Hermitian operators 587 xi CONTENTS 17.3 The properties of Hermitian operators 588 Reality of the eigenvalues; orthogonalit y of the eigenfunctions; construction of real eigenfunctions 17.4 Sturm–Liouville equations 591 Valid boundary conditions; putting an equation into Sturm–Liouville form 17.5 Examples of Sturm–Liouville equations 593 Legendre’s equation; the associated Legendre equation; Bessel’s equation; the simple harmonic equation; Hermite’s equation; Laguerre’s equation;Chebyshev’s equation 17.6 Superposition of eigenfunctions: Green’s functions 597 17.7 A useful generalisation 601 17.8 Exercises 602 17.9 Hints and answers 606 18 Partial diﬀerential equations: general and particular solutions 608 18.1 Important partial diﬀerential equations 609 The wave equation; the diﬀusion equation; Laplace’s equation; Poisson’s equation; Schr ¨odinger’s equation 18.2 General form of solution 613 18.3 General and particular solutions 614 First-order equations; inhomogeneous e quations and problems; second-order equations 18.4 The wave equation 626 18.5 The diﬀusion equation 62818.6 Characteristics and the existence of solutions 632 First-order equations; second-order equations 18.7 Uniqueness of solutions 63818.8 Exercises 64018.9 Hints and answers 644 19 Partial diﬀerential equations: separation of variables and other methods 646 19.1 Separation of variables: the general method 646 19.2 Superposition of separated solutions 65019.3 Separation of variables in polar coordinates 658 Laplace’s equation in polar coordinates; spherical harmonics; other equations in polar coordinates; solution by expansion; separation ofvariables in inhomogeneous equations 19.4 Integral transform methods 681 19.5 Inhomogeneous problems – Green’s functions 686 Similarities with Green’s function for ordinary diﬀerential equations; general boundary-value problems; Dirichlet problems; Neumann problems xii CONTENTS 19.6 Exercises 702 19.7 Hints and answers 708 20 Complex variables 710 20.1 Functions of a complex variable 711 20.2 The Cauchy–Riemann relations 71320.3 Power series in a complex variable 71620.4 Some elementary functions 71820.5 Multivalued functions and branch cuts 72120.6 Singularities and zeroes of complex functions 72320.7 Complex potentials 725 20.8 Conformal transformations 730 20.9 Applications of conformal transformations 73520.10 Complex integrals 73820.11 Cauchy’s theorem 74220.12 Cauchy’s integral formula 74520.13 Taylor and Laurent series 74720.14 Residue theorem 752 20.15 Location of zeroes 754 20.16 Integrals of sinusoidal functions 75820.17 Some inﬁnite integrals 75920.18 Integrals of multivalued functions 76220.19 Summation of series 76420.20 Inverse Laplace transform 76520.21 Exercises 768 20.22 Hints and answers 773 21 Tensors 776 21.1 Some notation 77721.2 Change of basis 77821.3 Cartesian tensors 77921.4 First- and zero-order Cartesian tensors 781 21.5 Second- and higher-order Cartesian tensors 784 21.6 The algebra of tensors 78721.7 The quotient law 78821.8 The tensors δ ijand/epsilon1ijk 790 21.9 Isotropic tensors 79321.10 Improper rotations and pseudotensors 79521.11 Dual tensors 798 21.12 Physical applications of tensors 799 21.13 Integral theorems for tensors 80321.14 Non-Cartesian coordinates 804 xiii CONTENTS 21.15 The metric tensor 806 21.16 General coordinate transformations and tensors 80921.17 Relative tensors 81221.18 Derivatives of basis vectors and Christoﬀel symbols 814 21.19 Covariant diﬀerentiation 817 21.20 Vector operators in tensor form 82021.21 Absolute derivatives along curves 82421.22 Geodesics 82521.23 Exercises 82621.24 Hints and answers 831 22 Calculus of variations 834 22.1 The Euler–Lagrange equation 83522.2 Special cases 836 Fdoes not contain yexplicitly; Fdoes not contain xexplicitly 22.3 Some extensions 840 Several dependent variables; several i ndependent variables; higher-order derivatives; variable end-points 22.4 Constrained variation 844 22.5 Physical variational principles 846 Fermat’s principle in optics; Hamilton’s principle in mechanics 22.6 General eigenvalue problems 84922.7 Estimation of eigenvalues and eigenfunctions 85122.8 Adjustment of parameters 85422.9 Exercises 856 22.10 Hints and answers 860 23 Integral equations 862 23.1 Obtaining an integral equation from a diﬀerential equation 86223.2 Types of integral equation 86323.3 Operator notation and the existence of solutions 86423.4 Closed-form solutions 865 Separable kernels; integral transform methods; diﬀerentiation 23.5 Neumann series 87223.6 Fredholm theory 87423.7 Schmidt–Hilbert theory 87523.8 Exercises 87823.9 Hints and answers 882 24 Group theory 883 24.1 Groups 883 Deﬁnition of a group; further examples of groups xiv CONTENTS 24.2 Finite groups 891 24.3 Non-Abelian groups 89424.4 Permutation groups 89824.5 Mappings between groups 901 24.6 Subgroups 903 24.7 Subdividing a group 905 Equivalence relations and classes; congruence and cosets; conjugates and classes 24.8 Exercises 912 24.9 Hints and answers 915 25 Representation theory 918 25.1 Dipole moments of molecules 91925.2 Choosing an appropriate formalism 92025.3 Equivalent representations 92625.4 Reducibility of a representation 92825.5 The orthogonality theorem for irreducible representations 93225.6 Characters 934 Orthogonality property of characters 25.7 Counting irreps using characters 937 Summation rules for irreps 25.8 Construction of a character table 94225.9 Group nomenclature 94425.10 Product representations 94525.11 Physical applications of group theory 947 Bonding in molecules; matrix elemen ts in quantum mechanics; degeneracy of normal modes; breaking of degeneracies 25.12 Exercises 955 25.13 Hints and answers 959 26 Probability 961 26.1 Venn diagrams 961 26.2 Probability 966 Axioms and theorems; conditional probability; Bayes’ theorem 26.3 Permutations and combinations 975 26.4 Random variables and distributions 981 Discrete random variables; continuous random variables 26.5 Properties of distributions 985 Mean; mode and median; variance; higher moments; higher central moments 26.6 Functions of random variables 99226.7 Generating functions 999 Probability generating functions; moment generating functions xv CONTENTS 26.8 Important discrete distributions 1009 Binomial; hypergeometric; Poisson; Poisson approximation to the binomial distribution; multiple Poisson distributions 26.9 Important continuous distributions 1021 Gaussian; Gaussian approximation to the binomial distribution; Gaussianapproximation to the Poisson distribution; multiple Gaussian; exponential;uniform 26.10 The central limit theorem 1036 26.11 Joint distributions 1038 Discrete bivariate; continuous bivariate; conditional; marginal 26.12 Properties of joint distributions 1041 Expectation values; variance; covariance and correlation 26.13 Generating functions for joint distributions 104726.14 Transformation of variables in joint distributions 104826.15 Important joint distributions 1049 Multinominal; multivariate Gaussian; transformation of variables in multi- variate distributions 26.16 Exercises 1053 26.17 Hints and answers 1061 27 Statistics 1064 27.1 Experiments, samples and populations 106427.2 Sample statistics 1065 Averages; variance and standard deviation; moments; covariance and correlation 27.3 Estimators and sampling distributions 1072 Consistency, bias and eﬃciency; Fisher’s inequality; standard errors;conﬁdence limits 27.4 Some basic estimators 1086 Mean; variance; standard deviation; moments; covariance and correlation 27.5 Maximum-likelihood method 1097 ML estimator; transformation invariance and bias; eﬃciency; errors and conﬁdence limits; Bayesian interpretation; large Nbehaviour; extended maximum-likelihood 27.6 The method of least squares 1113 Linear least squares; non-linear least squares 27.7 Hypothesis testing 1119 Simple and composite hypotheses; statistical tests; Neyman-Pearson;generalised likelihood-ratio; Student’s t;F i s h e r ’ s F; goodness-of-ﬁt 27.8 Exercises 1140 27.9 Hints and answers 1145 xvi CONTENTS 28 Numerical methods 1148 28.1 Algebraic and transcendental equations 1149 Rearrangement of the equation; linea r interpolation; binary chopping; Newton–Raphson method 28.2 Convergence of iteration schemes 1156 28.3 Simultaneous linear equations 1158 Gaussian elimination; Gauss–Seidel iteration; tridiagonal matrices 28.4 Numerical integration 1164 Trapezium rule; Simpson’s rule; Gaussian integration; Monte-Carlo methods 28.5 Finite diﬀerences 117928.6 Diﬀerential equations 1180 Diﬀerence equations; Taylor series solutions; prediction and correction; Runge–Kutta methods; isoclines 28.7 Higher-order equations 1188 28.8 Partial diﬀerential equations 119028.9 Exercises 119328.10 Hints and answers 1198 Appendix Gamma, beta and error functions 1201 A1.1 The gamma function 1201A1.2 The beta function 1203A1.3 The error function 1204 Index 1206 xvii Preface to the second edition Since the publication of the ﬁrst edition of this book, we have, both through teaching the material it covers and as a result of receiving helpful comments fromcolleagues, become aware of the desirability of changes in a number of areas.The most important of these is the fact that the mathematical preparation ofcurrent senior college and university entrants is now less than it used to be. To match this, we have decided to include a preliminary chapter covering areas such as polynomial equations, trigonometric identities, coordinate geometry, partialfractions, binomial expansions, necessary and suﬃcient conditions, and proof byinduction and contradiction. Whilst the general level of what is included in this second edition has not been raised, some areas have been expanded to take in topics we now feel were not adequately covered in the ﬁrst. In particular, increased attention has been given to non-square sets of simultaneous linear equations and their associatedmatrices. We hope that this more extended treatment, together with the inclusionof singular value matrix decomposition will make the material of more practicaluse to engineering students. In the same spirit, an elementary treatment of linearrecurrence relations has been included. The topic of normal modes has now beengiven a small chapter of its own, though the links to matrices on the one hand, and to representation theory on the other, have not been lost. Elsewhere, the presentation of probability and statistics has been reorganised to give the two aspects more nearly equal weights. The early part of the probabilitychapter has been rewritten in order to present a more coherent developmentbased on Boolean algebra, the fundamental axioms of probability theory andthe properties of intersections and unions. Whilst this is somewhat more formalthan previously, we think that it has not reduced the accessibility of these topics and hope that it has increased it. The scope of the chapter has been somewhat extended to include all physically important distributions and an introduction tocumulants. xix PREFACE TO THE SECOND EDITION Statistics now occupies a substantial chapter of its own, one that includes systematic discussions of estimators and their eﬃciency, sample distributions,andt-a n d F-tests for comparing means and variances. Other new topics are applications of the chi-squared distribution, maximum-likelihood parameter es- timation and least-squares ﬁtting. In other chapters we have added material on the following topics: curvature, envelopes, curve-sketching, more reﬁned numer-ical methods for diﬀerential equations, and the elements of integration usingmonte-carlo techniques. Over the last four years we have received somewhat mixed feedback about the number of exercises to include at the ends of the various chapters. Afterconsideration, we decided to increase it substantially, partly to correspond to the additional topics covered in the text, but mainly to give both students and their teachers a wider choice. There are now nearly eight hundred such exercises, manywith several parts. An even more vexed question is that of whether or not toprovide hints and answers to all of the exercises, or just to ‘the odd-numbered’ones, as is the normal practice for textbooks in the United States, thus makingthe remainder more suitable for setting as homework. In the end, we decided thathints and outline solutions should be provided for all the exercises, in order to facilitate independent study while leaving the details of the calculation as a task for the student. In conclusion we hope that this edition will be thought by its users to be ‘heading in the right direction’ and would like to place on record our thanks toall who have helped to bring about the changes and adjustments. Naturally, thosecolleagues who have noted errors or ambiguities in the ﬁrst edition and broughtthem to our attention ﬁgure high on the list, as do the staﬀ at The Cambridge University Press. In particular, we are grateful to Dave Green for continued L ATEX advice, Susan Parkinson for copy-editing the 2nd edition with her usual keen eyefor detail and ﬂair for crafting coherent prose, and Alison Woollatt for once againturning our basic L ATEX into a beautifully typeset book. Our thanks go to all of them, though of course we accept full responsibility for any remaining errors orambiguities, of which, as with any new publication, there are bound to be some. On a more personal note, KFR again wishes to thank his wife Penny for her unwavering support, not only in his academic and tutorial work, but also in their joint eﬀorts to convert time at the bridge table into ‘green points’ on their record.MPH is once more indebted to his wife, Becky, and his mother, Pat, for theirtireless support and encouragement above and beyond the call of duty. MPHdedicates his contribution to this book to the memory of his father, RonaldLeonard Hobson, whose gentle kindness, patient understanding and unbreakablespirit made all things seem possible. Ken Riley, Michael Hobson Cambridge, 2002 xx Preface to the ﬁrst edition A knowledge of mathematical methods is important for an increasing number of university and college courses, particularly in physics, engineering and chemistry,but also in more general science. Students embarking on such courses come fromdiverse mathematical backgrounds, and their core knowledge varies considerably.We have therefore decided to write a textbook that assumes knowledge only ofmaterial that can be expected to be familiar to all the current generation of students starting physical science courses at university. In the United Kingdom this corresponds to the standard of Mathematics A-level, whereas in the UnitedStates the material assumed is that which would normally be covered at juniorcollege. Starting from this level, the ﬁrst six chapters cover a collection of topics with which the reader may already be familiar, but which are here extended and applied to typical problems encountered by ﬁrst-year university students. They are aimed at providing a common base of general techniques used inthe development of the remaining chapters. Students who have had additionalpreparation, such as Further Mathematics at A-level, will ﬁnd much of thismaterial straightforward. Following these opening chapters, the remainder of the book is intended to cover at least that mathematical material which an undergraduate in the physical sciences might encounter up to the end of his or her course. The book is also appropriate for those beginning graduate study with a mathematical content, andnaturally much of the material forms parts of courses for mathematics students.Furthermore, the text should provide a useful reference for research workers. The general aim of the book is to present a topic in three stages. The ﬁrst stage is a qualitative introduction, wherever possible from a physical point of view. The second is a more formal presentation, although we have deliberately avoided strictly mathematical questions such as the existence of limits, uniformconvergence, the interchanging of integration and summation orders, etc. on the xxi PREFACE TO THE FIRST EDITION grounds that ‘this is the real world; it must behave reasonably’. Finally a worked example is presented, often drawn from familiar situations in physical scienceand engineering. These examples have generally been fully worked, since, inthe authors’ experience, partially worked examples are unpopular with students. Only in a few cases, where trivial algebraic manipulation is involved, or where repetition of the main text would result, has an example been left as an exercisefor the reader. Nevertheless, a number of exercises also appear at the end of eachchapter, and these should give the reader ample opportunity to test his or herunderstanding. Hints and answers to these exercises are also provided. With regard to the presentation of the mathematics, it has to be accepted that many equations (especially partial diﬀerential equations) can be written more compactly by using subscripts, e.g. u xyfor a second partial derivative, instead of the more familiar ∂2u/∂x∂y , and that this certainly saves typographical space. However, for many students, the labour of mentally unpacking such equationsis suﬃciently great that it is not possible to think of an equation’s physicalinterpretation at the same time. Consequently, wherever possible we have decidedto write out such expressions in their more obvious but longer form. During the writing of this book we have received much help and encouragement from various colleagues at the Cavendish Laboratory, Clare College, Trinity Hall and Peterhouse. In particular, we would like to thank Peter Scheuer, whosecomments and general enthusiasm proved invaluable in the early stages. Forreading sections of the manuscript, for pointing out misprints and for numeroususeful comments, we thank many of our students and colleagues at the Universityof Cambridge. We are especially grateful to Chris Doran, John Huber, GarthLeder, Tom K ¨orner and, not least, Mike Stobbs, who, sadly, died before the book was completed. We also extend our thanks to the University of Cambridge and the Cavendish teaching staﬀ, whose examination questions and lecture hand-outshave collectively provided the basis for some of the examples included. Of course,any errors and ambiguities remaining are entirely the responsibility of the authors,and we would be most grateful to have them brought to our attention. We are indebted to Dave Green for a great deal of advice concerning typesetting in L ATEX and to Andrew Lovatt for various other computing tips. Our thanks also go to Anja Visser and Grac ¸a Rocha for enduring many hours of (sometimes heated) debate. At Cambridge University Press, we are very grateful to our editorAdam Black for his help and patience and to Alison Woollatt for her experttypesetting of such a complicated text. We also thank our copy-editor SusanParkinson for many useful suggestions that have undoubtedly improved the styleof the book. Finally, on a personal note, KFR wishes to thank his wife Penny, not only for a long and happy marriage, but also for her support and understanding during his recent illness – and when things have not gone too well at the bridge table!MPH is indebted both to Rebecca Morris and to his parents for their tireless xxii PREFACE TO THE FIRST EDITION support and patience, and for their unending supplies of tea. SJB is grateful to Anthony Gritten for numerous relaxing discussions about J.S.Bach, to SusannahTicciati for her patience and understanding, and to Kate Isaak for her calminglate-night e-mails from the USA. Ken Riley, Michael Hobson and Stephen Bence Cambridge, 1997 xxiii 1 Preliminary algebra This opening chapter reviews the basic algebra of which a working knowledge is presumed in the rest of the book. Many students will be familiar with much, ifnot all, of it, but recent changes in what is studied during secondary educationmean that it cannot be taken for granted that they will already have a masteryof all the topics presented here. The reader may assess which areas need furtherstudy or revision by attempting the exercises at the end of the chapter. The mainareas covered are polynomial equations and the related topic of partial fractions, curve sketching, coordinate geometry, trigonometric identities and the notions of proof by induction or contradiction. 1.1 Simple functions and equations It is normal practice when starting the mathematical investigation of a physical problem to assign an algebraic symbol to the quantity whose value is sought, eithernumerically or as an explicit algebraic expression. For the sake of deﬁniteness, inthis chapter we will use xto denote this quantity most of the time. Subsequent steps in the analysis involve applying a combination of known laws, consistency conditions and (possibly) given constraints to derive one or more equationssatisﬁed by x. These equations may take many forms, ranging from a simple polynomial equation to, say, a partial diﬀerential equation with several boundaryconditions. Some of the more complicated possibilities are treated in the laterchapters of this book, but for the present we will be concerned with techniquesfor the solution of relatively straightforward algebraic equations. 1.1.1 Polynomials and polynomial equations Firstly we consider the simplest type of equation, a polynomial equation in which apolynomial expression in x, denoted by f(x), is set equal to zero and thereby 1 PRELIMINARY ALGEBRA forms an equation which is satisﬁed by particular values of x; these values are called the rootsof the equation. f(x)=anxn+an−1xn−1+···+a1x+a0=0. (1.1) Here nis an integer >0, called the degree of both the polynomial and the equation, and the known coeﬃcients a0,a1,...,a nare real quantities with an/negationslash=0 . Equations such as (1.1) arise frequently in physical problems, the coeﬃcients ai being determined by the physical properties of the system under study. What is needed is to ﬁnd some or all of the roots solutions of (1.1), i.e. the x-values, αk,, that satisfy f(αk)=0 ;h e r e kis an index that, as we shall see later, can take up to ndiﬀerent values, i.e. k=1,2,...,n. The roots of the polynomial equations can equally well be described as the zeroes of the polynomial. When they are real, they correspond to the points at which a graph of f(x)c r o s s e st h e x-axis. Roots that are complex (see chapter 3) do not have such a graphical interpretation. For polynomial equations containing powers of xgreater tha x4general meth- ods do not exist for obtaining explicit expressions for the roots αk.E v e nf o r n=3a n d n= 4 the prescriptions for obtaining the roots are suﬃciently compli- cated that it is usually preferable to obtain exact or approximate values by othermethods. Only for n=1a n d n= 2 can closed-form solutions be given. These results will be well known to the reader, but they are given here for the sake of completeness. For n= 1, (1.1) reduces to the linear equation a 1x+a0= 0; (1.2) the solution (root) is α1=−a0/a1.F o r n= 2, (1.1) reduces to the quadratic equation a2x2+a1x+a0= 0; (1.3) the two roots α1andα2are given by α1,2=−a1±radicalBig a2 1−4a2a0 2a2. (1.4) When discussing speciﬁcally quadratic equations, as opposed to more general polynomial equations, it is usual to write the equation in one of the two notations ax2+bx+c=0,a x2+2bx+c=0, (1.5) with respective explicit pairs of solutions α1,2=−b±√ b2−4ac 2a,α 1,2=−b±√ b2−ac a. (1.6) Of course, these two notations are entirely equivalent and the only important 2 1.1 SIMPLE FUNCTIONS AND EQUATIONS point is to associate each form of answer with the corresponding form of equation; most people keep to one form, to avoid any possible confusion. If the value of the quantity appearing under the square root sign is positive then both roots are real; if it is negative then the roots form a complex conjugate pair, i.e. they are of the form p±iqwith pandqreal (see chapter 3); if it has zero value then the two roots are equal and special considerations usually arise. Thus linear and quadratic equations can be dealt with in a cut-and-dried way. We now turn to methods for obtaining partial information about the roots ofhigher-degree polynomial equations. In some circumstances the knowledge thatan equation has a root lying in a certain range, or that it has no real roots at all,is all that is actually required. For example, in the design of electronic circuits it is necessary to know whether the current in a proposed circuit will break into spontaneous oscillation. To test this, it is suﬃcient to establish whether acertain polynomial equation, whose coeﬃcients are determined by the physicalparameters of the circuit, has a root with a positive real part (see chapter 3);complete determination of all the roots is not needed for this purpose. If thecomplete set of roots of a polynomial equation is required, it can usually beobtained to any desired accuracy by numerical methods such as those described in chapter 28. There is no explicit step-by-step approach to ﬁnding the roots of a general polynomial equation such as (1.1). In most cases analytic methods yield onlyinformation about the roots, rather than their exact values. To explain the relevant techniques we will consider a particular example, ‘thinking aloud’ on paper andexpanding on special points about methods and lines of reasoning. In moreroutine situations such comment would be absent and the whole process briefer and more tightly focussed. Example: the cubic case Let us investigate the roots of the equation g(x)=4 x 3+3x2−6x−1 = 0 (1.7) or, in an alternative phrasing, investigate the zeroes of g(x). We note ﬁrst of all that this is a cubic equation. It can be seen that for xlarge and positive g(x) will be large and positive and equally that for xlarge and negative g(x) will be large and negative. Therefore, intuitively (or, more formally, by continuity)g(x) must cross the x-axis at least once and so g(x) = 0 must have at least one real root. Furthermore, it can be shown that if f(x)i sa n nth-degree polynomial then the graph of f(x) must cross the x-axis an even or odd number of times asxvaries between −∞and +∞, according to whether nitself is even or odd. Thus a polynomial of odd degree always has at least one real root, but one of even degree may have no real root. A small complication, discussed later in thissection, occurs when repeated roots arise. 3 PRELIMINARY ALGEBRA Having established that g(x) = 0, equation(1.7), has at least one real root, we may ask how many real roots it could have. To answer this we need one of the fundamental theorems of algebra, mentioned above: Annth-degree polynomial equation has exactly nroots. It should be noted that this does not imply that there are nrealroots (only that there are not more than n); some of the roots may be of the form p+iq. To make the above theorem plausible and to see what is meant by repeated roots, let us suppose that the nth-degree polynomial equation f(x) = 0, (1.1), has rroots α1,α2,...,α rconsidered distinct for the moment. That is, we suppose that f(αk)=0f o r k=1,2,...,r,s ot h a t f(x) vanishes only when xis equal to one of thervalues αk. But the same can be said for the function F(x)=A(x−α1)(x−α2)···(x−αr), (1.8) in which Ais a non-zero constant; F(x) can clearly be multiplied out to form a polynomial expression. We now call upon a second fudamental result in algebra: that if two polynomial functions f(x)a n d F(x) have equal values for allvalues of x, then their coeﬃcients are equal on a term-by-term basis. In other words, we can equate the coeﬃcients of each and every power of xin the two expressions; in particular we can equate the coeﬃcients of the highest power of x. From this we have Axr≡anxnand thus that r=nandA=an.A sris both equal to nand to the number of roots off(x) = 0, we conclude that the nth-degree polynomial f(x)=0h a s nroots. (Although this line of reasoning may make the theorem plausible, it does notconstitute a proof since we have not shown that it is permissible to write f(x)i n the form of equation (1.8).) We next note that the condition f(α k)=0f o r k=1,2,...,r, could also be met if (1.8) were replaced by F(x)=A(x−α1)m1(x−α2)m2···(x−αr)mr, (1.9) with A=an. In (1.9) the mkare integers ≥1 and are known as the multiplicities of the roots, mkbeing the multiplicity of αk. Expanding the right-hand side (RHS) leads to a polynomial of degree m1+m2+···+mr. This sum must be equal to n. Thus, if any of the mkis greater than unity then the number of distinct roots, r, is less than n; the total number of roots remains at n, but one or more of the αk counts more than once. For example, the equation F(x)=A(x−α1)2(x−α2)3(x−α3)(x−α4)=0 has exactly seven roots, α1being a double root and α2a triple root, whilst α3and α4are unrepeated ( simple )r o o t s . We can now say that our particular equation (1.7) has either one or three real roots but in the latter case it may be that not all the roots are distinct. To decide 4 1.1 SIMPLE FUNCTIONS AND EQUATIONS xxφ1(x) φ2(x) β1 β1β2 β2 Figure 1.1 Two curves φ1(x)a n d φ2(x), both with zero derivatives at the same values of x, but with diﬀerent numbers of real solutions to φi(x)=0 . how many real roots the equation has, we need to anticipate two ideas from the next chapter. The ﬁrst of these is the notion of the derivative of a function, andthe second is a result known as Rolle’s theorem. Thederivative f /prime(x) of a function f(x) measures the slope of the tangent to the graph of f(x) at that value of x(see ﬁgure 2.1 in the next chapter). For the moment, the reader with no prior knowledge of calculus is asked to acceptthat the derivative of ax nisnaxn−1, so that the derivative g/prime(x)o ft h ec u r v e g(x)=4 x3+3x2−6x−1i sg i v e nb y g/prime(x)=1 2 x2+6x−6. Similar expressions for the derivatives of other polynomials are used later in this chapter. Rolle’s theorem states that, if f(x) has equal values at two diﬀerent values of xthen at some point between these two x-values its derivative is equal to zero; i.e. the tangent to its graph is parallel to the x-axis at that point (see ﬁgure 2.2). Having brieﬂy mentioned the derivative of a function and Rolle’s theorem, we now use them to etablish whether g(x) has one or three real zeroes. If g(x)=0 does have three real roots αk,i . e . g(αk)=0f o r k=1,2,3, then it follows from Rolle’s theorem that between any consecutive pair of them (say α1andα2)t h e r e must be some real value of xat which g/prime(x) = 0. Similarly, there must be a further zero of g/prime(x) lying between α2andα3. Thus a necessary condition for three real roots of g(x)=0i st h a t g/prime(x) = 0 itself has two real roots. However, this condition on the number of roots of g/prime(x) = 0, whilst necessary, is not suﬃcient to guarantee three real roots of g(x) = 0. This can be seen by inspecting the cubic curves in ﬁgure 1.1. For each of the two functions φ1(x)a n d φ2(x), the derivative is equal to zero at both x=β1andx=β2. Clearly, though, φ2(x) = 0 has three real roots whilst φ1(x) = 0 has only one. It is easy to see that the crucial diﬀerence is that φ1(β1)a n d φ1(β2) have the same sign, whilst φ2(β1) andφ2(β2) have opposite signs. 5 PRELIMINARY ALGEBRA It will be apparent that for some cubic equations, φ(x)=0s a y , φ/prime(x)e q u a l s zero at a value of xfor which φ(x) is also zero. Then the graph of φ(x)j u s t touches the x-axis and there may appear to be only two roots. However, when this happens the value of xso found is, in fact, a double real root of the cubic (corresponding to one of the mkin (1.9) having the value 2) and must be counted twice when determining the number of real roots. Finally, then, we are in a position to decide the number of real roots of the equation g(x)=4 x3+3x2−6x−1=0 . The equation g/prime(x)=0 ,w i t h g/prime(x)=1 2 x2+6x−6, is a quadratic equation with explicit solutions † β1,2=−3±√ 9+7 2 12, so that β1=−1a n d β2=1/2. The corresponding values of g(x)a r e g(β1)=4a n d g(β2)=−11/4, which are of opposite sign. This indicates that 4 x3+3x2−6x−1=0 has three real roots, one lying in the range −1<x<1 2and the others one on each side of that range. The techniques we have developed above have been used to tackle a cubic equation, but they can be applied to polynomial equations f(x)=0o fd e g r e e greater than 3. However, much of the analysis centres around the equation f/prime(x) = 0 and this, itself, being then a polynomial equation of degree 3 or more either has no closed-form general solution or one that is complicated to evaluate.Thus the amount of information that can be obtained about the roots of f(x)=0 is correspondingly reduced. A more general case To illustrate what can (and cannot) be done in the more general case we now investigate as far as possible the real roots of f(x)=x 7+5x6+x4−x3+x2−2=0 . The following points can be made. (i) This is a seventh-degree polynomial equation; therefore the number of r e a lr o o t si s1 ,3 ,5o r7 . (ii)f(0) is negative whilst f(∞)=+∞, so there must be at least one positive root. †The two roots β1,β2are written as β1,2. By convention β1refers to the upper symbol in ±,β2to the lower symbol. 6 1.1 SIMPLE FUNCTIONS AND EQUATIONS (iii) The equation f/prime(x) = 0 can be written as x(7x5+3 0x4+4x2−3x+2 )=0 and thus x= 0 is a solution. The derivative of f/prime(x), denoted by f/prime/prime(x), equals 42 x5+ 150 x4+1 2x2−6x+2 . T h a t f/prime(x) is zero whilst f/prime/prime(x)i s positive at x= 0 indicates (subsection 2.1.8 ) that f(x) has a minimum there. This, together with the facts that f(0) is negative and f(∞)=∞, implies that the total number of real roots to the right of x= 0 must be odd. Since the total number of real roots must be odd, the number to theleft must be even (0, 2, 4 or 6). This is about all that can be deduced by simple analytic methods in this case, although some further progress can be made in the ways indicated in exercise 1.3. There are, in fact, more sophisticated tests that examine the relative signs of successive terms in an equation such as (1.1), and in quantities derived from them, to place limits on the numbers and positions of roots. But they are not prerequisites for the remainder of this book and will not be pursued further here. We conclude this section with a worked example which demonstrates that the practical application of the ideas developed so far can be both short and decisive.IFor what values of k,i fa n y ,d o e s f(x)=x3−3x2+6x+k=0 have three real roots? Firstly study the equation f/prime(x)=0 ,i . e .3 x2−6x+ 6 = 0. This is a quadratic equation but, using (1.6), because 62<4×3×6, it can have no real roots. Therefore, it follows immediately that f(x) has no turning points, i.e. no maximum or minimum; consequently f(x) = 0 cannot have more than one real root, whatever the value of k. J 1.1.2 Factorising polynomials In the previous subsection we saw how a polynomial with rgiven distinct zeroes αkcould be constructed as the product of factors containing those zeroes, f(x)=an(x−α1)m1(x−α2)m2···(x−αr)mr =anxn+an−1xn−1+···+a1x+a0, (1.10) with m1+m2+···+mr=n, the degree of the polynomial. It will cause no loss of generality in what follows to suppose that all the zeroes are simple, i.e. all mk=1 andr=n, and this we will do. Sometimes it is desirable to be able to reverse this process, in particular when one exact zero has been found by some method and the remaining zeroes are tobe investigated. Suppose that we have located one zero, α; it is then possible to write (1.10) as f(x)=(x−α)f 1(x), (1.11) 7 PRELIMINARY ALGEBRA where f1(x) is a polynomial of degree n−1. How can we ﬁnd f1(x)? The procedure is much more complicated to describe in a general form than to carry out foran equation with given numerical coeﬃcients a i. If such manipulations are too complicated to be carried out mentally, they could be laid out along the lines of an algebraic ‘long division’ sum. However, a more compact form of calculation is as follows. Write f1(x)a s f1(x)=bn−1xn−1+bn−2xn−2+bn−3xn−3+···+b1x+b0. Substitution of this form into (1.11) and subsequent comparison of the coeﬃcients ofxpforp=n,n−1,..., 1, 0 with those in the second line of (1.10) generates the series of equations bn−1=an, bn−2−αbn−1=an−1, bn−3−αbn−2=an−2, ... b0−αb1=a1, −αb0=a0. These can be solved successively for the bj, starting either from the top or from the bottom of the series. In either case the ﬁnal equation used serves as a check;if it is not satisﬁed, at least one mistake has been made in the computation –orαis not a zero of f(x) = 0. We now illustrate this procedure with a worked example.IDetermine by inspection the simple roots of the equation f(x)=3 x4−x3−10x2−2x+4=0 and hence, by factorisation, ﬁnd the rest of its roots. From the pattern of coeﬃcients it can be seen that x=−1 is a solution to the equation. We therefore write f(x)=(x+1 ) ( b3x3+b2x2+b1x+b0), where b3=3, b2+b3=−1, b1+b2=−10, b0+b1=−2, b0=4. These equations give b3=3,b2=−4,b1=−6,b0= 4 (check) and so f(x)=(x+1 )f1(x)=(x+ 1)(3 x3−4x2−6x+4 ). 8 1.1 SIMPLE FUNCTIONS AND EQUATIONS We now note that f1(x)=0i f xis set equal to 2. Thus x−2i saf a c t o ro f f1(x), which therefore can be written as f1(x)=(x−2)f2(x)=(x−2)(c2x2+c1x+c0) with c2=3, c1−2c2=−4, c0−2c1=−6, −2c0=4. These equations determine f2(x)a s3 x2+2x−2. Since f2(x) = 0 is a quadratic equation, its solutions can be written explicitly as x=−1±√1+6 3. Thus the four roots of f(x)=0a r e−1,2,1 3(−1+√7) and1 3(−1−√7). J 1.1.3 Properties of roots From the fact that a polynomial equation can be written in any of the alternative forms f(x)=anxn+an−1xn−1+···+a1x+a0=0, f(x)=an(x−α1)m1(x−α2)m2···(x−αr)mr=0, f(x)=an(x−α1)(x−α2)···(x−αn)=0 , it follows that it must be possible to express the coeﬃcients aiin terms of the roots αk. To take the most obvious example, comparison of the constant terms (formally the coeﬃcient of x0) in the ﬁrst and third expressions shows that an(−α1)(−α2)···(−αn)=a0, or, using the product notation, nproductdisplay k=1αk=(−1)na0 an. (1.12) Only slightly less obvious is a result obtained by comparing the coeﬃcients of xn−1in the same two expressions of the polynomial: nsummationdisplay k=1αk=−an−1 an. (1.13) Comparing the coeﬃcients of other powers of xyields further results, though they are of less general use than the two just given. One such, which the readermay wish to derive, is nsummationdisplay j=1nsummationdisplay k>jαjαk=an−2 an. (1.14) 9 PRELIMINARY ALGEBRA In the case of a quadratic equation these root properties are used suﬃciently often that they are worth stating explicitly, as follows. If the roots of the quadraticequation ax 2+bx+c=0a r e α1andα2then α1+α2=−b a, α1α2=c a. If the alternative standard form for the quadratic is used, bis replaced by 2 bin both the equation and the ﬁrst of these results.IFind a cubic equation whose roots are −4,3and5. From results (1.12) – (1.14) we can compute that, arbitrarily setting a3=1 , −a2=3X k=1αk=4,a 1=3X j=13X k>jαjαk=−17,a 0=(−1)33Y k=1αk=6 0. Thus a possible cubic equation is x3+(−4)x2+(−17)x+(60) = 0. Of course, any multiple ofx3−4x2−17x+ 60 = 0 will do just as well. J 1.2 Trigonometric identities So many of the applications of mathematics to physics and engineering are concerned with periodic, and in particular sinusoidal, behaviour that a sure and ready handling of the corresponding mathematical functions is an essential skill.Even situations with no obvious periodicity are often expressed in terms ofperiodic functions for the purposes of analysis. Later in this book whole chaptersare devoted to developing the techniques involved, but as a necessary prerequisitewe here establish (or remind the reader of) some standard identities with which heor she should be fully familiar, so that the manipulation of expressions containing sinusoids becomes automatic and reliable. So as to emphasise the angular nature of the argument of a sinusoid we will denote it in this section by θrather than x. 1.2.1 Single-angle identities We give without proof the basic identity satisﬁed by the sinusoidal functions sin θ and cos θ,n a m e l y cos 2θ+s i n2θ=1. (1.15) If sin θand cos θhave been deﬁned geometrically in terms of the coordinates of a point on a circle, a reference to the name of Pythagoras will suﬃce to establish this result. If they have been deﬁned by means of series (with θexpressed in radians) then the reader should refer to Euler’s equation (3.23) on page 96, andnote that e iθhas unit modulus if θis real. 10 1.2 TRIGONOMETRIC IDENTITIES xy x/primey/prime OABP TNR M Figure 1.2 Illustration of the compound-angle identities. Refer to the main text for details. Other standard single-angle formulae derived from (1.15) by dividing through by various powers of sin θand cos θare 1+t a n2θ=s e c2θ. (1.16) cot2θ+1=c o s e c2θ. (1.17) 1.2.2 Compound-angle identities The basis for building expressions for the sinusoidal functions of compound angles are those for the sum and diﬀerence of just two angles, since all othercases can be built up from these, in principle. Later we will see that a study of complex numbers can provide a more eﬃcient approach in some cases. To prove the basic formulae for the sine and cosine of a compound angle A+Bin terms of the sines and cosines of AandB, we consider the construction shown in ﬁgure 1.2. It shows two sets of axes, OxyandOx /primey/prime, with a common origin but rotated with respect to each other through an angle A. The point Plies on the unit circle centred on the common origin Oand has coordinates cos(A+B),sin(A+B) with respect to the axes Oxyand coordinates cos B,sinB with respect to the axes Ox/primey/prime. Parallels to the axes Oxy(dotted lines) and Ox/primey/prime(broken lines) have been drawn through P. Further parallels ( MRandRN)t ot h e Ox/primey/primeaxes have been 11 PRELIMINARY ALGEBRA drawn through R, the point (0 ,sin(A+B)) in the Oxysystem. That all the angles marked with the symbol •are equal to Afollows from the simple geometry of right-angled triangles and crossing lines. We now determine the coordinates of Pin terms of lengths in the ﬁgure, expressing those lengths in terms of both sets of coordinates: (i) cos B=x/prime=TN+NP=MR+NP =ORsinA+RPcosA= sin( A+B)s i nA+c o s ( A+B)cosA; (ii) sin B=y/prime=OM−TM=OM−NR =ORcosA−RPsinA= sin( A+B)cosA−cos(A+B)s i nA. Now, if equation (i) is multiplied by sin Aand added to equation (ii) multiplied by cos A, the result is sinAcosB+c o s AsinB= sin( A+B)(sin2A+c o s2A)=s i n ( A+B). Similarly, if equation (ii) is multiplied by sin Aand subtracted from equation (i) multiplied by cos A, the result is cosAcosB−sinAsinB=c o s ( A+B)(cos2A+s i n2A)=c o s ( A+B). Corresponding graphically based results can be derived for the sines and cosines of the diﬀerence of two angles; however, they are more easily obtained by settingBto−Bin the previous results and remembering that sin Bbecomes−sinB whilst cos Bis unchanged. The four results may be summarised by sin(A±B)=s i n AcosB±cosAsinB (1.18) cos(A±B)=c o s AcosB∓sinAsinB. (1.19) Standard results can be deduced from these by setting one of the two angles equal to πor to π/2: sin(π−θ)=s i n θ, cos(π−θ)=−cosθ,sinparenleftbig 1 2π−θparenrightbig (1.20) =c o s θ, cosparenleftbig1 2π−θparenrightbig =s i n θ, (1.21) From these basic results many more can be derived. An immediate deduction, obtained by taking the ratio of the two equations (1.18) and (1.19) and thendividing both the numerator and denominator of this ratio by cos AcosB,i s tan(A±B)=tanA±tanB 1∓tanAtanB. (1.22) One application of this result is a test for whether two lines on a graph are orthogonal (perpendicular); more generally, it determines the angle between them. The standard notation for a straight-line graph is y=mx+c,i nw h i c h m is the slope of the graph and cis its intercept on the y-axis. It should be noted that the slope mis also the tangent of the angle the line makes with the x-axis. 12 1.2 TRIGONOMETRIC IDENTITIES Consequently the angle θ12between two such straight-line graphs is equal to the diﬀerence in the angles they individually make with the x-axis, and the tangent of that angle is given by (1.22): tanθ12=tanθ1−tanθ2 1+t a n θ1tanθ2=m1−m2 1+m1m2. (1.23) For the lines to be orthogonal we must have θ12=π/2, i.e. the ﬁnal fraction on the RHS of the above equation must equal ∞,a n ds o m1m2=−1. (1.24) A kind of inversion of equations (1.18) and (1.19) enables the sum or diﬀerence of two sines or cosines to be expressed as the product of two sinusoids; theprocedure is typiﬁed by the following. Adding together the expressions given by(1.18) for sin( A+B) and sin( A−B) yields sin(A+B)+s i n ( A−B)=2s i n AcosB. If we now write A+B=CandA−B=D, this becomes sinC+s i n D=2s i nparenleftbiggC+D 2parenrightbigg cosparenleftbiggC−D 2parenrightbigg . (1.25) In a similar way each of the following equations can be derived: sinC−sinD=2c o sparenleftbiggC+D 2parenrightbigg sinparenleftbiggC−D 2parenrightbigg , (1.26) cosC+c o s D=2c o sparenleftbiggC+D 2parenrightbigg cosparenleftbiggC−D 2parenrightbigg , (1.27) cosC−cosD=−2s i nparenleftbiggC+D 2parenrightbigg sinparenleftbiggC−D 2parenrightbigg . (1.28) The minus sign on the right of the last of these equations should be noted; it may help to avoid overlooking this ‘oddity’ to recall that if C>D then cos C<cosD. 1.2.3 Double- and half-angle identities Double-angle and half-angle identities are needed so often in practical calculations that they should be committed to memory by any physical scientist. They can beobtained by setting Bequal to Ain results (1.18) and (1.19). When this is done, 13 PRELIMINARY ALGEBRA and use made of equation (1.15), the following results are obtained: sin2θ=2s i n θcosθ, (1.29) cos2θ=c o s2θ−sin2θ =2c o s2θ−1 =1−2s i n2θ, (1.30) tan2θ=2t a n θ 1−tan2θ. (1.31) A further set of identities enables sinusoidal functions of θto be expressed as polynomial functions of a variable t=t a n ( θ/2). They are not used in their primary role until the next chapter, but we give a derivation of them here forreference. Ift=t a n ( θ/2), then it follows from (1.16) that 1+ t 2=s e c2(θ/2) and cos( θ/2) = (1 +t2)−1/2, whilst sin( θ/2) = t(1 +t2)−1/2. Now, using (1.29) and (1.30), we may write: sinθ=2s i nθ 2cosθ 2=2t 1+t2, (1.32) cosθ=c o s2θ 2−sin2θ 2=1−t2 1+t2, (1.33) tanθ=2t 1−t2. (1.34) It can be further shown that the derivative of θwith respect to ttakes the algebraic form 2 /(1 + t2). This completes a package of results that enables expressions involving sinusoids, particularly when they appear as integrands, to be cast in more convenient algebraic forms. The proof of the derivative propertyand examples of use of the above results are given in subsection (2.2.7). We conclude this section with a worked example which is of such a commonly occurring form that it might be considered a standard procedure.ISolve for θthe equation asinθ+bcosθ=k, where a, bandkare given real quantities. To solve this equation we make use of result (1.18) by setting a=Kcosφandb=Ksinφ for suitable values of Kandφ. We then have k=Kcosφsinθ+Ksinφcosθ=Ksin(θ+φ), with K2=a2+b2and φ=t a n−1b a. Whether φlies in 0≤φ≤πor in−π<φ< 0 has to be determined by the individual signs of aandb. The solution is thus θ=s i n−1 /k K / −φ, 14 1.3 COORDINATE GEOMETRY with Kandφas given above. Notice that there is no re al solution to the original equation if|k|>|K|=(a2+b2)1/2. J 1.3 Coordinate geometry We have already mentioned the standard form for a straight-line graph, namely y=mx+c, (1.35) representing a linear relationship between the independent variable xand the dependent variable y.T h es l o p e mis equal to the tangent of the angle the line makes with the x-axis whilst cis the intercept on the y-axis. An alternative form for the equation of a straight line is ax+by+k=0, (1.36) to which (1.35) is clearly connected by m=−a band c=−k b. This form treats xandyon a more symmetrical basis, the intercepts on the two axes being−k/aand−k/brespectively. A power relationship between two variables, i.e. one of the form y=Axn,c a n also be cast into straight-line form by taking the logarithms of both sides. Whilstit is normal in mathematical work to use natural logarithms (to base e, written lnx), for practical investigations logarithms to base 10 are often employed. In either case the form is the same, but it needs to be remembered which has beenused when recovering the value of Afrom ﬁtted data. In the mathematical (base e) form, the power relationship becomes lny=nlnx+l nA. (1.37) Now the slope gives the power n, whilst the intercept on the ln yaxis is ln A, which yields A, either by exponentiation or by taking antilogarithms. The other standard coordinate forms of two-dimensional curves that students should know and recognise are those concerned with the conic sections – so called because they can all be obtained by taking suitable sections across a (double)cone. Because the conic sections can take many diﬀerent orientations and scalingstheir general form is complex, Ax 2+By2+Cxy+Dx+Ey+F=0, (1.38) but each can be represented by one of four generic forms, an ellipse, a parabola, a hyperbola or, the degenerate form, a pair of straight lines. If they are reduced to 15 PRELIMINARY ALGEBRA their standard representations, in which axes of symmetry are made to coincide with the coordinate axes, the ﬁrst three take the forms (x−α)2 a2+(y−β)2 b2= 1 (ellipse), (1.39) (y−β)2=4a(x−α) (parabola), (1.40) (x−α)2 a2−(y−β)2 b2= 1 (hyperbola). (1.41) Here, ( α, β) gives the position of the ‘centre’ of the curve, usually taken as the origin (0 ,0) when this does not conﬂict with any imposed conditions. The parabola equation given is that for a curve symmetric about a line parallel tothex-axis. For one symmetrical about a parallel to the y-axis the equation would read ( x−α) 2=4a(y−β). Of course, the circle is the special case of an ellipse in which b=aand the equation takes the form (x−α)2+(y−β)2=a2. (1.42) The distinguishing characteristic of this equation is that when it is expressed in the form (1.38) the coeﬃcients of x2andy2are equal and that of xyis zero; this property is not changed by any reorientation or scaling and so acts to identify a general conic as a circle. Deﬁnitions of the conic sections in terms of geometrical properties are also available; for example, a parabola can be deﬁned as the locus of a point thatis always at the same distance from a given straight line (the directrix )a si ti s from a given point (the focus). When these properties are expressed in Cartesian coordinates the above equations are obtained. For a circle, the deﬁning propertyis that all points on the curve are a distance afrom ( α, β); (1.42) expresses this requirement very directly. In the following worked example we derive the equation for a parabola.IFind the equation of a parabola that has the line x=−aas its directrix and the point (a,0)as its focus. Figure 1.3 shows the situation in Cartesian coordinates. Expressing the deﬁning requirement thatPNandPFare equal in length gives (x+a)=[ ( x−a)2+y2]1/2⇒(x+a)2=(x−a)2+y2 which, on expansion of the squared terms, immediately gives y2=4ax. This is (1.40) with αandβboth set equal to zero. J Although the algebra is more complicated, the same method can be used to derive the equations for the ellipse and the hyperbola. In these cases the distance from the ﬁxed point is a deﬁnite fraction, e, known as the eccentricity ,o ft h e distance from the ﬁxed line. For an ellipse 0 <e< 1, for a circle e=0 ,a n df o ra hyperbola e>1. The parabola corresponds to the case e=1 . 16 1.3 COORDINATE GEOMETRY xy OP FN x=−a(a,0)(x, y) Figure 1.3 Construction of a parabola using the point ( a,0) as the focus and the line x=−aas the directrix. The values of aandb(with a≥b) in equation (1.39) for an ellipse are related toethrough e2=a2−b2 a2 and give the lengths of the semi-axes of the ellipse. If the ellipse is centred on the origin, i.e. α=β= 0, then the focus is ( −ae,0) and the directrix is the line x=−a/e. For each conic section curve, although we have two variables, xandy,t h e ya r e not independent, since if one is given then the other can be determined. However, determining ywhen xis given, say, involves solving a quadratic equation on each occasion, and so it is convenient to have parametric representations of the curves. A parametric representation allows each point on a curve to be associated witha unique value of a single parameter t. The simplest parametric representations for the conic sections are as given below, though that for the hyperbola useshyperbolic functions, not formally introduced until chapter 3. That they do givevalid parameterizations can be veriﬁed by substituting them into the standard forms (1.39) – (1.41); in each case the standard form is reduced to an algebraic or trigonometric identity. x=α+acosφ,y=β+bsinφ(ellipse), x=α+at 2, y=β+2at (parabola), x=α+acoshφ,y=β+bsinhφ(hyperbola). As a ﬁnal example illustrating several topics from this section we now prove 17 PRELIMINARY ALGEBRA the well-known result that the angle subtended by a diameter at any point on a circle is a right angle.ITaking the diameter to be the line joining Q=(−a,0)andR=(a,0)and the point Pto be any point on the circle x2+y2=a2, prove that angle QP R is a right angle. IfPis the point ( x, y), the slope of the line QPis m1=y−0 x−(−a)=y x+a. That of RPis m2=y−0 x−(a)=y x−a. Thus m1m2=y2 x2−a2. But, since Pis on the circle, y2=a2−x2and consequently m1m2=−1. From result (1.24) this implies that QPandRPare orthogonal and that QP Ris therefore a right angle. Note that this is true for anypoint Pon the circle. J 1.4 Partial fractions In subsequent chapters, and in particular when we come to study integration in chapter 2, we will need to express a function f(x) that is the ratio of two polynomials in a more manageable form. To remove some potential complexity from our discussion we will assume that all the coeﬃcients in the polynomialsare real, although this is not an essential simpliﬁcation. The behaviour of f(x) is crucially determined by the location of the zeroes of its denominator, i.e. if f(x) is written as f(x)=g(x)/h(x) where both g(x)a n d h(x) are polynomials †,t h e n f(x) changes extremely rapidly when xis close to those values α ithat are the roots of h(x) = 0. To make such behaviour explicit, we write f(x) as a sum of terms such as A/(x−α)n,i nw h i c h Ais a constant, αis one of the αithat satisfy h(αi)=0a n d nis a positive integer. Writing a function in this way is known as expressing it in partial fractions . Suppose, for the sake of deﬁniteness, that we wish to express the function f(x)=4x+2 x2+3x+2 †It is assumed that the ratio has been reduced so that g(x)a n d h(x) do not contain any common factors, i.e. there is no value of xthat makes both vanish at the same time. We may also assume without any loss of generality that the coeﬃcient of the highest power of xinh(x) has been made equal to unity, if necessary, by dividing both numerator and denominator by the coeﬃcient of this highest power. 18 1.4 PARTIAL FRACTIONS in partial fractions, i.e. to write it as f(x)=g(x) h(x)=4x+2 x2+3x+2=A1 (x−α1)n1+A2 (x−α2)n2+···. (1.43) The ﬁrst question that arises is that of how many terms there should be on the right-hand side (RHS). Although some complications occur when h(x)h a s repeated roots (these are considered below) it is clear that f(x) only becomes inﬁnite at the twovalues of x,α1andα2,t h a tm a k e h(x) = 0. Consequently the RHS can only become inﬁnite at the same two values of xand therefore contains only two partial fractions – these are the ones shown explicitly. This argumentcan be trivially extended (again temporarily ignoring the possibility of repeatedroots of h(x)) to show that if h(x) is a polynomial of degree nthen there should be nterms on the RHS, each containing a diﬀerent root α iof the equation h(αi)=0 . A second general question concerns the appropriate values of the ni.T h i si s answered by putting the RHS over a common denominator, which will clearly have to be the product ( x−α1)n1(x−α2)n2···. Comparison of the highest power ofxin this new RHS with the same power in h(x)s h o w st h a t n1+n2+···=n. This result holds whether or not h(x) = 0 has repeated roots and, although we do not give a rigorous proof, strongly suggests the correct conclusions that: •The number of terms on the RHS is equal to the number of distinct roots of h(x) = 0, each term having a diﬀerent root αiin its denominator ( x−αi)ni; •Ifαiis a multiple root of h(x) = 0 then the value to be assigned to niin (1.43) is that of miwhen h(x) is written in the product form (1.9). Further, as discussed on p. 23, Aihas to be replaced by a polynomial of degree mi−1.This is also formally true for non-repeated roots, since then both miandniare equal to unity. Returning to our speciﬁc example we note that the denominator h(x)h a sz e r o e s atx=α1=−1a n d x=α2=−2; these x-values are the simple (non-repeated) roots of h(x) = 0. Thus the partial fraction expansion will be of the form 4x+2 x2+3x+2=A1 x+1+A2 x+2. (1.44) We now list several methods available for determining the coeﬃcients A1and A2. We also remind the reader that, as with all the explicit examples and techniques described, these methods are to be considered as models for the handling of any ratio of polynomials, with or without characteristics which makes it a specialcase. (i) The RHS can be put over a common denominator, in this case ( x+1)(x+2), and then the coeﬃcients of the various powers of xcan be equated in the 19 PRELIMINARY ALGEBRA numerators on both sides of the equation. This leads to 4x+2= A1(x+2 )+ A2(x+1 ), 4=A1+A22=2 A1+A2. Solving the simultaneous equations for A1andA2gives A1=−2a n d A2=6. (ii) A second method is to substitute two (or more generally n) diﬀerent values of xinto each side of (1.44) and so obtain two (or n) simultaneous equations for the two (or n)c o n s t a n t s Ai. To justify this practical way of proceeding it is necessary, strictly speaking, to appeal to method (i) above, which establishes that there are unique values for A1andA2valid for all values of x. It is normally very convenient to take zero as one of the values of x, but of course any set will do. Suppose in the present case that we use the values x=0a n d x= 1 and substitute in (1.44). The resulting equations are 2 2=A1 1+A2 2, 6 6=A1 2+A2 3, which on solution give A1=−2a n d A2= 6, as before. The reader can easily verify that any other pair of values for x(except for a pair that includes α1orα2) gives the same values for A1andA2. (iii) The very reason why method (ii) fails if xis chosen as one of the roots αiofh(x) = 0 can be made the basis for determining the values of the Ai corresponding to non-multiple roots without having to solve simultaneous equations. The method is conceptually more diﬃcult than the other meth-ods presented here, and needs results from the theory of complex variables(chapter 20) to justify it. However, we give a practical ‘cookbook’ recipefor determining the coeﬃcients. (a) To determine the coeﬃcient A k, imagine the denominator h(x) written as the product ( x−α1)(x−α2)···(x−αn), with any m-fold repeated root giving rise to mfactors in parentheses. (b) Now set xequal to αkand evaluate the expression obtained after omitting the factor that reads αk−αk. (c) Divide the value so obtained into g(αk); the result is the required coeﬃcient Ak. For our speciﬁc example we ﬁnd that in step (a) that h(x)=(x+1 ) ( x+2 ) and that in evaluating A1step (b) yields −1 + 2 = 1. Since g(−1) = 4(−1) + 2 =−2, step (c) gives A1as (−2)/(1), i.e in agreement with our other evaluations. In a similar way A2is evaluated as ( −6)/(−1) = 6. 20 1.4 PARTIAL FRACTIONS Thus any one of the methods listed above shows that 4x+2 x2+3x+2=−2 x+1+6 x+2. The best method to use in any particular circumstance will depend on the complexity, in terms of the degrees of the polynomials and the multiplicities ofthe roots of the denominator, of the function being considered and, to someextent, on the individual inclinations of the student; some prefer lengthy butstraightforward solution of simultaneous equations, whilst others feel more athome carrying shorter but more abstract calculations in their heads. 1.4.1 Complications and special cases Having established the basic method for partial fractions, we now show, through further worked examples, how some complications are dealt with by extensions to the procedure. These extensions are introduced one at a time, but of course in any practical application more than one may be involved. The degree of the numerator is greater than or equal to that of the denominator Although we have not speciﬁcally mentioned the fact, it will be apparent from trying to apply method (i) of the previous subsection to such a case, that if the degree of the numerator ( m) is not less than that of the denominator ( n) then the ratio of two polynomials cannot be expressed in partial fractions. To get round this diﬃculty it is necessary to start by dividing the denominator h(x) into the numerator g(x) to obtain a further polynomial, which we will denote bys(x), together with a function t(x)t h a t isa ratio of two polynomials for which the degree of the numerator is less than that of the denominator. The functiont(x)cantherefore be expanded in partial fractions. As a formula, f(x)=g(x) h(x)=s(x)+t(x)≡s(x)+r(x) h(x). (1.45) It is apparent that the polynomial r(x)i st h e remainder obtained when g(x)i s divided by h(x), and, in general, will be a polynomial of degree n−1. It is also clear that the polynomial s(x) will be of degree m−n. Again, the actual division process can be set out as an algebraic long division sum but is probably moreeasily handled by writing (1.45) in the form g(x)=s(x)h(x)+r(x) (1.46) or, more explicitly, as g(x)=(s m−nxm−n+sm−n−1xm−n−1+···+s0)h(x)+(rn−1xn−1+rn−2xn−2+···+r0) (1.47) and then equating coeﬃcients. 21 PRELIMINARY ALGEBRA We illustrate this procedure with the following worked example.IFind the partial fraction decomposition of the function f(x)=x3+3x2+2x+1 x2−x−6. Since the degree of the numerator is 3 and that of the denominator is 2, a preliminary long division is necessary. The polynomial s(x) resulting from the division will have degree 3−2 = 1 and the remainder r(x) will be of degree 2 −1 = 1 (or less). Thus we write x3+3x2+2x+1=( s1x+s0)(x2−x−6) + ( r1x+r0). From equating the coeﬃcients of the various powers of xon the two sides of the equation, starting with the highest, we now obtain the simultaneous equations 1=s1, 3=s0−s1, 2=−s0−6s1+r1, 1=−6s0+r0. These are readily solved, in the given order, to yield s1=1 , s0=4 , r1=1 2a n d r0= 25. Thus f(x) can be written as f(x)=x+4+12x+2 5 x2−x−6. The last term can now be decomposed into partial fractions as previously. The zeroes of the denominator are at x=3a n d x=−2 and the application of any method from the previous subsection yields the respective constants as A1=1 21 5andA2=−1 5. Thus the ﬁnal partial fraction decomposition of f(x)i s x+4+61 5(x−3)−1 5(x+2 ). J Factors of the form a2+x2in the denominator We have so far assumed that the roots of h(x) = 0, needed for the factorisation of the denominator of f(x), can always be found. In principle they always can but in some cases they are not real. Consider, for example, attempting to express inpartial fractions a polynomial ratio whose denominator is h(x)=x 3−x2+2x−2. Clearly x= 1 gives a zero of h(x), and so a ﬁrst factorisation is ( x−1)(x2+2 ) . However we cannot make any further progress because the factor x2+ 2 cannot be expressed as ( x−α)(x−β)f o ra n yr e a l αandβ. Complex numbers are introduced later in this book (chapter 3) and, when the reader has studied them, he or she may wish to justify the procedure set outbelow. It can be shown to be equivalent to that already given, but the zeroes ofh(x) are now allowed to be complex and terms that are complex conjugates of each other are combined to leave only real terms. Since quadratic factors of the form a 2+x2that appear in h(x) cannot be reduced to the product of two linear factors, partial fraction expansions including themneed to have numerators in the corresponding terms that are not simply constants 22 1.4 PARTIAL FRACTIONS Aibut linear functions of x,i . e .o ft h ef o r m Bix+Ci. Thus, in the expansion, linear terms (ﬁrst-degree polynomials) in the denominator have constants (zero-degree polynomials) in their numerators, whilst quadratic terms (second-degreepolynomials) in the denominator have linear terms (ﬁrst-degree polynomials) in their numerators. As a symbolic formula, the partial fraction expansion of g(x) (x−α1)(x−α2)···(x−αp)(x2+a2 1)(x2+a2 2)···(x2+a2q) should take the form A1 x−α1+A2 x−α2+···+Ap x−αp+B1x+C1 x2+a2 1+B2x+C2 x2+a2 2+···+Bqx+Cq x2+a2q. Of course, the degree of g(x) must be less than p+2q; if it is not, an initial division must be carried out as demonstrated earlier. Repeated factors in the denominator Consider trying (incorrectly) to expand f(x)=x−4 (x+1 ) ( x−2)2 in partial fraction form as follows: x−4 (x+1 ) ( x−2)2=A1 x+1+A2 (x−2)2. Multiplying both sides of this supposed equality by ( x+1 ) ( x−2)2produces an equation whose LHS is linear in x, whilst its RHS is quadratic. This is clearly wrong and so an expansion in the above form cannot be valid. The correction wemust make is very similar to that needed in the previous subsection, namely that since ( x−2) 2is a quadratic polynomial the numerator of the term containing it must be a ﬁrst-degree polynomial, and not simply a constant. The correct form for the part of the expansion containing the doubly repeated root is therefore ( Bx+C)/(x−2)2. Using this form and either of methods (i) and (ii) for determining the constants gives the full partial fraction expansion as x−4 (x+1 ) ( x−2)2=−5 9(x+1 )+5x−16 9(x−2)2, as the reader may verify. Since any term of the form ( Bx+C)/(x−α)2can be written as B(x−α)+C+Bα (x−α)2=B x−α+C+Bα (x−α)2, and similarly for multiply repeated roots, an alternative form for the part of the partial fraction expansion containing a repeated root αis D1 x−α+D2 (x−α)2+···+Dp (x−α)p. (1.48) 23 PRELIMINARY ALGEBRA In this form, all x-dependence has disappeared from the numerators but at the expense of p−1 additional terms; the total number of constants to be determined remains unchanged, as it must. When describing possible methods of determining the constants in a partial fraction expansion, we noted that method (iii), p. 20, which avoids the need to solve simultaneous equations, is restricted to terms involving non-repeated roots. In fact, it can be applied in repeated-root situations, when the expansion is putin the form (1.48), but only to ﬁnd the constant in the term involving the largestinverse power of x−α,i . e .D pin (1.48). We conclude this section with a more protracted worked example that contains all three of the complications discussed.IResolve the following expression F(x)into partial fractions: F(x)=x5−2x4−x3+5x2−46x+ 100 (x2+6 ) ( x−2)2. We note that the degree of the denominator (4) is not greater than that of the numerator (5), and so we must start by dividing the latter by the former. It follows, from the diﬀerencein degrees and the coeﬃcients of the highest powers in each, that the result will be a linearexpression s 1x+s0with the coeﬃcient s1equal to 1. Thus the numerator of F(x)m u s tb e expressible as (x+s0)(x4−4x3+1 0x2−24x+ 24) + ( r3x3+r2x2+r1x+r0), where the second factor in parentheses is the denominator of F(x) written as a polynomial. Equating the coeﬃcients of x4gives−2=−4+s0and ﬁxes s0as 2. Equating the coeﬃcients of powers less than 4 gives equations involving the coeﬃcients rias follows: −1=−8+1 0+ r3, 5=−24 + 20 + r2, −46 = 24−48 + r1, 100 = 48 + r0. Thus the remainder polynomial r(x) can be constructed and F(x) written as F(x)=x+2+−3x3+9x2−22x+5 2 (x2+6 ) ( x−2)2≡x+2+ f(x). The polynomial ratio f(x) can now be expressed in partial fraction form, noting that its denominator contains both a term of the form x2+a2and a repeated root. Thus f(x)=Bx+C x2+6+D1 x−2+D2 (x−2)2. We could now put the RHS of this equation over the common denominator ( x2+6)(x−2)2 and ﬁnd B,C,D 1andD2by equating coeﬃcients of powers of x. It is quicker, however, to use methods (iii) and (ii). Method (iii) gives D2as (−24 + 36−44 + 52) /(4 + 6) = 2. We choose to evaluate the other coeﬃ cients by method (ii), and setting x=0 , x=1a n d 24 1.5 BINOMIAL EXPANSION x=−1 gives respectively 52 24=C 6−D1 2+2 4, 36 7=B+C 7−D1+2, 86 63=C−B 7−D1 3+2 9. These equations reduce to 4C−12D1=4 0, B+C−7D1=2 2, −9B+9C−21D1=7 2, with solution B=0 , C=1 , D1=−3. Thus, ﬁnally, we may re-write the original expression F(x) in partial fractions as F(x)=x+2+1 x2+6−3 x−2+2 (x−2)2. J 1.5 Binomial expansion Earlier in this chapter we were led to consider functions containing powers of the sum or diﬀerence of two terms, e.g. ( x−α)m. Later in this book we will ﬁnd numerous occasions on which we wish to write such a product of repeated factorsas a polynomial in xor, more generally, as a sum of terms each of which contains powers of xandαseparately, as opposed to a power of their sum or diﬀerence. To make the discussion general and the result applicable to a wide variety of situations, we will consider the general expansion of f(x)=(x+y) n,w h e r e xand ymay stand for constants, variables or functions and, for the time being, nis a positive integer. It may not be obvious what form the general expansion takesbut some idea can be obtained by carrying out the multiplication explicitly forsmall values of n. Thus we obtain successively (x+y) 1=x+y, (x+y)2=(x+y)(x+y)=x2+2xy+y2, (x+y)3=(x+y)(x2+2xy+y2)=x3+3x2y+3xy2+y3, (x+y)4=(x+y)(x3+3x2y+3xy2+y3)=x4+4x3y+6x2y2+4xy3+y4. This does not establish a general formula, but the regularity of the terms in the expansions and the suggestion of a pattern in the coeﬃcients indicate that a general formula for power nwill have n+ 1 terms, that the powers of xandyin every term will add up to nand that the coeﬃcients of the ﬁrst and last terms will be unity whilst those of the second and penultimate terms will be n. 25 PRELIMINARY ALGEBRA In fact, the general expression, the binomial expansion for power n,i sg i v e nb y (x+y)n=k=nsummationdisplay k=0nCkxn−kyk, (1.49) wherenCkis called the binomial coeﬃcient and is expressed in terms of factorial functions by n!/[k!(n−k)!]. Clearly, simply to make such a statement does not constitute proof of its validity, but, as we will see in subsection 1.5.2, (1.49) canbeproved using a method called induction. Before turning to that proof, we investigate some of the elementary properties of the binomial coeﬃcients. 1.5.1 Binomial coeﬃcients As stated above, the binomial coeﬃcients are deﬁned by nCk≡n! k!(n−k)!≡parenleftbiggn kparenrightbigg for 0≤k≤n, (1.50) where in the second identity we give a common alternative notation fornCk. Obvious properties include (i)nC0=nCn=1, (ii)nC1=nCn−1=n, (iii)nCk=nCn−k. We note that, for any given n, the largest coeﬃcient in the binomial expansion is the middle one ( k=n/2) if nis even; the middle two coﬀﬁcients ( k=1 2(n±1)) are equal largest if nis odd. Somewhat less obvious is the result nCk+nCk−1=n! k!(n−k)!+n! (k−1)!(n−k+1 ) ! =n![(n+1−k)+k] k!(n+1−k)! =(n+1 ) ! k!(n+1−k)!=n+1Ck. (1.51) An equivalent statement, in which khas been redeﬁned as k+1 ,i s nCk+nCk+1=n+1Ck+1. (1.52) 1.5.2 Proof of the binomial expansion We are now in a position to prove the binomial expansion (1.49). In doing so, we introduce the reader to a procedure applicable to certain types of problems and known as the method of induction . The method is discussed much more fully in subsection 1.7.1. We start by assuming that(1.49) is true for some positive integer n=N.W e 26 1.6 PROPERTIES OF BINOMIAL COEFFICIENTS now proceed to show that this implies that it must also be true for n=N+1 ,a s follows: (x+y)N+1=(x+y)Nsummationdisplay k=0NCkxN−kyk =Nsummationdisplay k=0NCkxN+1−kyk+Nsummationdisplay k=0NCkxN−kyk+1 =Nsummationdisplay k=0NCkxN+1−kyk+N+1summationdisplay j=1NCj−1x(N+1)−jyj, where in the ﬁrst line we have used the assumption and in the third line have moved the second summation index, by unity by writing k+1= j.W en o w separate oﬀ the ﬁrst term of the ﬁrst sum,NC0xN+1, and write it asN+1C0xN+1; we can do this since, as noted in (i) following (1.50),nC0=1f o re v e r y n. Similarly, the last term of the second summation can be replaced byN+1CN+1yN+1. The remaining terms of each of the two summations are now written together, with the summation index denoted by kin both terms. Thus (x+y)N+1=N+1C0xN+1+Nsummationdisplay k=1parenleftbigNCk+NCk−1parenrightbig x(N+1)−kyk+N+1CN+1yN+1 =N+1C0xN+1+Nsummationdisplay k=1N+1Ckx(N+1)−kyk+N+1CN+1yN+1 =N+1summationdisplay k=0N+1Ckx(N+1)−kyk. In going from the ﬁrst to the second line we have used result (1.51). Now we observe that the ﬁnal overall equation is just the original assumed result (1.49) but with n=N+ 1. Thus it has been shown that if the binomial expansion is assumed to be true for n=N,t h e ni tc a nb e proved to be true for n=N+1 .B u t it holds trivially for n= 1, and therefore for n= 2 also. By the same token it is valid for n=3,4,..., and hence is established for all positive integers n. 1.6 Properties of binomial coeﬃcients 1.6.1 Identities involving binomial coeﬃcients There are many identities involving the binomial coeﬃcients that can be derived directly from their deﬁnition, and yet more that follow from their appearance in the binomial expansion. Only the most elementary ones, given earlier, are worth committing to memory but, as illustrations, we now derive two results involvingsums of binomial coeﬃcients. 27 PRELIMINARY ALGEBRA The ﬁrst is a further application of the method of induction. Consider the proposal that, for any n≥1a n d k≥0, n−1summationdisplay s=0k+sCk=n+kCk+1. (1.53) Notice that here n, the number of terms in the sum, is the parameter that varies, kis a ﬁxed parameter, whilst sis a summation index and does not appear on the RHS of the equation. Now we suppose that this statement about the value of the sum of the binomial coeﬃcientskCk,k+1Ck,...,k+n−1Ckis true for n=N. We next write down a series with an extra term and determine the implications of the supposition for the newseries: N+1−1summationdisplay s=0k+sCk=N−1summationdisplay s=0k+sCk+k+NCk =N+kCk+1+N+kCk =N+k+1Ck+1. But this is just proposal (1.53) with nnow set equal to N+ 1. To obtain the last line, we have used (1.52), with nset equal to N+k. It only remains to consider the case n= 1, when the summation only contains one term and (1.53) reduces to kCk=1+kCk+1. This is trivially valid for any ksince both sides are equal to unity, thus completing the proof of (1.53) for all positive integers n. The second result, which gives a formula for combining terms from two sets of binomial coeﬃcients in a particular way (a kind of ‘convolution’, for readerswho are already familiar with this term), is derived by applying the binomialexpansion directly to the identity (x+y) p(x+y)q≡(x+y)p+q. Written in terms of binomial expansions, this reads psummationdisplay s=0pCsxp−sysqsummationdisplay t=0qCtxq−tyt=p+qsummationdisplay r=0p+qCrxp+q−ryr. We now equate coeﬃcients of xp+q−ryron the two sides of the equation, noting that on the LHS all combinations of sandtsuch that s+t=rcontribute. This gives as an identity that rsummationdisplay t=0pCr−tqCt=p+qCr=rsummationdisplay t=0pCtqCr−t. (1.54) 28 1.6 PROPERTIES OF BINOMIAL COEFFICIENTS We have speciﬁcally included the second equality to emphasise the symmetrical nature of the relationship with respect to pandq. Further identities involving the coeﬃcients can be obtained by giving xandy special values in the deﬁning equation (1.49) for the expansion. If both are setequal to unity then we obtain (using the alternative notation so as to producefamiliarity with it) parenleftbiggn 0parenrightbigg +parenleftbiggn 1parenrightbigg +parenleftbiggn 2parenrightbigg +···+parenleftbiggn nparenrightbigg =2 n, (1.55) whilst setting x=1a n d y=−1 yields parenleftbiggn 0parenrightbigg −parenleftbiggn 1parenrightbigg +parenleftbiggn 2parenrightbigg −···+(−1)nparenleftbiggn nparenrightbigg =0. (1.56) 1.6.2 Negative and non-integral values of n Up till now we have restricted nin the binomial expansion to be a positive integer. Negative values can be accommodated, but only at the cost of an inﬁniteseries of terms rather than the ﬁnite one represented by (1.49). For reasons thatare intuitively sensible and will be discussed in more detail in chapter 4, very often we require an expansion in which, at least ultimately, successive terms in the inﬁnite series decrease in magnitude. For this reason, if x>y we consider (x+y) −m,w h e r e mitself is a positive integer, in the form (x+y)n=(x+y)−m=x−mparenleftBig 1+y xparenrightBig−m . Since the ratio y/xis less than unity, terms containing higher powers of it will be small in magnitude, whilst raising the unit term to any power will not aﬀect itsmagnitude. If y>x the roles of the two must be interchanged. We can now state, but will not explicitly prove, the form of the binomial expansion appropriate to negative values of n(nequal to−m): (x+y) n=(x+y)−m=x−m∞summationdisplay k=0−mCkparenleftBigy xparenrightBigk , (1.57) where the hitherto undeﬁned quantity−mCk, which appears to involve factorials of negative numbers, is given by −mCk=(−1)km(m+1 )···(m+k−1) k!=(−1)k(m+k−1)! (m−1)!k!=(−1)km+k−1Ck. (1.58) The binomial coeﬃcient on the extreme right of this equation has its normal meaning and is well deﬁned since m+k−1≥k. Thus we have a deﬁnition of binomial coeﬃcients for negative integer values ofnin terms of those for positive n. The connection between the two may not 29 PRELIMINARY ALGEBRA be obvious, but they are both formed in the same way in terms of recurrence relations. Whatever the sign of n, the series of coeﬃcientsnCkcan be generated by starting withnC0= 1 and using the recurrence relation nCk+1=n−k k+1nCk. (1.59) The diﬀerence is that for positive integer nthe series terminates when k=n, whereas for negative nthere is no such termination – in line with the inﬁnite series of terms in the corresponding expansion. Finally we note that, in fact, equation (1.59) generates the appropriate coef- ﬁcients for all values of n, positive or negative, integer or non-integer, with the obvious exception of the case in which x=−yandnis negative. For non-integer nthe expansion does not terminate, even if nis positive. 1.7 Some particular methods of proof Much of the mathematics used by physicists and engineers is concerned with obtaining a particular value, formula or function from a given set of data andstated conditions. However, just as it is essential in physics to formulate the basiclaws and so be able to set boundaries on what can or cannot happen, so itis important in mathematics to be able to state general propositions about theoutcomes that are or are not possible. To this end one attempts to establishtheorems that state in as general a way as possible mathematical results that apply to particular types of situation. We conclude this introductory chapter by describing two methods that can sometimes be used to prove particular classesof theorems. The two general methods of proof are known as proof by induction (which has already been met in this chapter) and proof by contradiction. They share the common characteristic that at an early stage in the proof an assumptionis made that a particular (unproven) statement is true; the consequences ofthat assumption are then explored. In an inductive proof the conclusion isreached that the assumption is self-consistent and has other equally consistentbut broader implications, which are then applied to establish the general validityof the assumption. A proof by contradiction, however, establishes an internal inconsistency and thus shows that the assumption is unsustainable; the natural consequence of this is that the negative of the assumption is established as true. Later in this book use will be made of these methods of proof to explore new territory, e.g. to examine the properties of vector spaces, matrices and groups. However, at this stage we will draw our illustrative and test examples from earliersections of this chapter and other topics in elementary algebra and number theory. 30 1.7 SOME PARTICULAR METHODS OF PROOF 1.7.1 Proof by induction The proof of the binomial expansion given in subsection 1.5.2 and the identity established in subsection 1.6.1 have already shown the way in which an inductiveproof is carried through. They also indicated the main limitation of the method,namely that only an initially supposed result can be proved. Thus the methodof induction is of no use for deducing a previously unknown result; a putative equation or result has to be arrived at by some other means, usually by noticing patterns or by trial and error using simple values of the variables involved. It will also be clear that propositions that can be proved by induction are limitedto those containing a parameter that takes a range of integer values (usuallyinﬁnite). For a proposition involving a parameter n, the ﬁve steps in a proof using induction are as follows. (i) Formulate the supposed result for general n. (ii) Suppose (i) to be true for n=N(or more generally for all values of n≤N;s e eb e l o w ) ,w h e r e Nis restricted to lie in the stated range. (iii) Show, using only proven results and supposition (ii), that proposition (i) is true for n=N+1 . (iv) Demonstrate directly, and without any assumptions, that proposition (i) is true when ntakes the lowest value in its range. (v) It then follows from (iii) and (iv) that the proposition is valid for all values ofnin the stated range. (It should be noted that, although many proofs at stage (iii) require the validity of the proposition only for n=N, some require it for all nless than or equal to N – hence the form of inequality given in parentheses in the stage (ii) assumption.) To illustrate further the method of induction, we now apply it to two worked examples; the ﬁrst concerns the sum of the squares of the ﬁrst nnatural numbers.IProve that the sum of the squares of the ﬁrst nnatural numbers is given by nX r=1r2=1 6n(n+ 1)(2 n+1 ). (1.60) As previously we start by assuming the result is true for n=N. Then it follows that N+1X r=1r2=NX r=1r2+(N+1 )2 =1 6N(N+ 1)(2 N+1 )+( N+1 )2 =1 6(N+1 ) [ N(2N+1 )+6 N+6 ] =1 6(N+ 1)[(2 N+3 ) ( N+2 ) ] =1 6(N+1 ) [ ( N+1 )+1 ] [ 2 ( N+1 )+1 ] . 31 PRELIMINARY ALGEBRA This is precisely the original assumption, but with Nreplaced by N+ 1. To complete the proof we only have to verify (1.60) for n= 1. This is trivially done and establishes the result for all positive n. The same and related results are obtained by a diﬀerent method in subsection 4.2.5. J Our second example is somewhat more complex and involves two nested proofs by induction: whilst trying to establish the main result by induction, we ﬁnd that we are faced with a second proposition which itself requires an inductive proof.IShow that Q(n)=n4+2n3+2n2+nis divisible by 6 (without remainder) for all positive integer values of n. Again we start by assuming the result is true for some particular value Nofn, whilst noting that it is trivially true for n= 0. We next examine Q(N+ 1), writing each of its terms as a binomial expansion: Q(N+1 )=( N+1 )4+2 (N+1 )3+2 (N+1 )2+(N+1 ) =(N4+4N3+6N2+4N+1 )+2 ( N3+3N2+3N+1 ) +2 (N2+2N+1 )+( N+1 ) =(N4+2N3+2N2+N)+( 4 N3+1 2N2+1 4N+6 ). Now, by our assumption, the group of terms within the ﬁrst parentheses in the last line is divisible by 6 and clearly so are the terms 12 N2and 6 within the second parentheses. Thus it comes down to deciding whether 4 N3+1 4Ni sd i v i s i b l eb y6–o re q u i v a l e n t l y , whether R(N)=2 N3+7Nis divisible by 3. To settle this latter question we try using a second inductive proof and assume that R(N)isdivisible by 3 for N=M, whilst again noting that the proposition is trivially true forN=M= 0. This time we examine R(M+1 ) : R(M+1 )=2 ( M+1 )3+7 (M+1 ) =2 (M3+3M2+3M+1 )+7 ( M+1 ) =( 2M3+7M)+3 ( 2 M2+2M+3 ) By assumption, the ﬁrst group of terms in the last line is divisible by 3 and the second group is patently so. We thus conclude that R(N) is divisible by 3 for all N≥M,a n d taking M= 0 shows that it is divisible by 3 for all N. We can now return to the main proposition and conclude that since R(N)=2 N3+7N is divisible by 3, 4 N3+1 2N2+1 4N+ 6 is divisible by 6. This in turn establishes that the divisibility of Q(N+ 1) by 6 follows from the assumption that Q(N) divides by 6. Since Q(0) clearly divides by 6, the proposition in the question is established for all values of n. J 1.7.2 Proof by contradiction The second general line of proof, but again one that is normally only useful when the result is already suspected, is proof by contradiction. The questions it canattempt to answer are only those that can be expressed in a proposition that is either true or false. Clearly, it could be argued that any mathematical result can be so expressed but, if the proposition is no more than a guess, the chancesof success are negligible. Valid propositions containing even modest formulae 32 1.7 SOME PARTICULAR METHODS OF PROOF are either the result of true inspiration or, much more normally, yet another reworking of an old chestnut! The essence of the method is to exploit the fact that mathematics is required to be self-consistent, so that, for example, two calculations of the same quantity,starting from the same given data but proceeding by diﬀerent methods, must givethe same answer. Equally, it must not be possible to follow a line of reasoning anddraw a conclusion that contradicts either the input data or any other conclusionbased upon the same data. It is this requirement on which the method of proof by contradiction is based. The crux of the method is to assume that the proposition to be proved isnottrue, and then use this incorrect assumption and ‘watertight’ reasoning to draw a conclusion that contradicts the assumption. The only way out of the self-contradiction is then to conclude that the assumption was indeed false andtherefore that the proposition is true. It must be emphasised that once a (false) contrary assumption has been made, every subsequent conclusion in the argument mustfollow of necessity. Proof by contradiction fails if at any stage we have to admit ‘this may or may not bethe case’. That is, each step in the argument must be a necessary consequence of results that precede it (taken together with the assumption), rather than simply apossible consequence. It should also be added that if no contradiction can be found using sound reasoning based on the assumption then no conclusion can be drawn about eitherthe proposition or its negative and some other approach must be tried. We illustrate the general method with an example in which the mathematical reasoning is straightforward so that attention can be focussed on the structure ofthe proof.IA rational number ris a fraction r=p/qin which pandqare integers with qpositive. Further, ris expressed in its lowest terms, any integer common factor of pandqhaving been divided out. Prove that the square root of an integer mcannot be a rational number, unless the square root itself is an integer. We begin by supposing that the stated result is nottrue and that we canwrite an equation √m=r=p qfor integers m, p, q with q/negationslash=1. It then follows that p2=mq2.B u t ,s i n c e ris expressed in its lowest terms, pandq,a n d hence p2andq2, have no factors in common whilst mis an integer. This is only possible ifq=1a n d p2=m. This conclusion contradicts the requirement that q/negationslash=1a n ds ol e a d s to the conclusion that it was wrong to suppose that√mcan be expressed as a non-integer rational number. This completes the proof of the statement in the question. J Our second worked example, also taken from elementary number theory, involves slightly more complicated mathematical reasoning but again exhibits thestructure associated with this type of proof. 33 PRELIMINARY ALGEBRAIThe prime integers piare labelled in ascending order, thus p1=1,p2=2,p5=7,etc. Show that there is no largest prime number. Assume, on the contrary, that there is a largest prime and let it be pN.C o n s i d e rn o wt h e number qformed by multiplying together all the primes from p1topNand then adding one to the product, i.e. q=p1p2···pN+1. By our assumption pNis the largest prime, and so no number can have a prime factor greater than this. However, for every prime pi(i=1,2,...,N ) the quotient q/p ihas the form Mi+( 1/pi)w i t h Mian integer and 1 /pinon-integer. This means that q/p icannot be an integer and so picannot be a divisor of q. Since qis not divisible by any of the (assumed) ﬁnite set of primes, it must be itself ap r i m e .A s qis also clearly greater than pN, we have a contradiction. Thus it follows that our assumption that there is a largest prime integer must be false, and so it has beenproved that there is no largest prime integer. It should be noted that the given construction for qdoes not generate all the primes that actually exist (e.g. for N=3,q= 7 rather than the next actual prime value of 5, is found), but this does not matter for the purposes of our proof by contradiction.J 1.7.3 Necessary and suﬃcient conditions As the ﬁnal topic in this introductory chapter, we consider brieﬂy the notion of, and distinction between, necessary and suﬃcient conditions in the contextof proving a mathematical proposition. In ordinary English the distinction iswell deﬁned, and that distinction is maintained in mathematics. However, in the authors’ experience students tend to overlook it and assume (wrongly) that, having proved that the validity of proposition Aimplies the truth of proposition B, it follows by ‘reversing the argument’ that the validity of Bautomatically implies that of A. As an example, let proposition Abe that an integer Nis divisible without remainder by 6, and proposition Bbe that Nis divisible without remainder by 2. Clearly, if Ais true then it follows that Bis true, i.e. Ais a suﬃcient condition forB; it is not however a necessary condition, as is trivially shown by taking N as 8. Conversely, the same value of Nshows that whilst the validity of Bis a necessary condition for Ato hold, it is not suﬃcient. An alternative terminology to ‘necessary’ and ‘suﬃcient’ often employed by mathematicians is that of ‘if’ and ‘only if’, particularly in the combination ‘if andonly if’ which is usually written as IFF or denoted by a double-headed arrow⇐⇒. The equivalent statements can be summarised by AifBA is true if Bis true or B=⇒A, Bis a suﬃcient condition for AB =⇒A, Aonly if BA is true only if Bis true or A=⇒B, Bis a necessary consequence of AA =⇒B, 34 1.7 SOME PARTICULAR METHODS OF PROOF AIFFBA is true if and only if Bis true or B⇐⇒ A, AandBnecessarily imply each other B⇐⇒ A. Although at this stage in the book we are able to employ for illustrative purposes only simple and fairly obvious results, the following example is given as a modelof how necessary and suﬃcient conditions should be proved. The essential pointis that for the second part of the proof (whether it be the ‘necessary’ part or the ‘suﬃcient’ part) one needs to start again from scratch; more often than not, the lines of the second part of the proof will notbe simply those of the ﬁrst written in reverse order.IProve that ( A) a function f(x)is a quadratic polynomial with zeroes at x=2andx=3 if and only if ( B) the function f(x)has the form λ(x2−5x+6)withλa non-zero constant. (1) Assume A,i . e .t h a t f(x)isa quadratic polynomial with zeroes at x=2a n d x=3 .L e t its form be ax2+bx+cwith a/negationslash= 0. Then we have 4a+2b+c=0, 9a+3b+c=0, and subtraction shows that 5 a+b=0a n d b=−5a. Substitution of this into the ﬁrst of the above equations gives c=−4a−2b=−4a+1 0a=6a. Thus, it follows that f(x)=a(x2−5x+6 ) w i t h a/negationslash=0, and establishes the ‘ Aonly if B’ part of the stated result. (2) Now assume that f(x)hasthe form λ(x2−5x+6 )w i t h λa non-zero constant. Firstly we note that f(x) is a quadratic polynomial, and so it only remains to prove that its zeroes occur at x=2a n d x=3 .C o n s i d e r f(x) = 0, which, after dividing through by the non-zero constant λ,g i v e s x2−5x+6=0 . We proceed by using a technique known as completing the square , for the purposes of illustration, although the factorisation of the above equation should be clear to the reader.Thus we write x 2−5x+(5 2)2−(5 2)2+6=0 , (x−5 2)2=1 4, x−5 2=±1 2. The two roots of f(x) = 0 are therefore x=2a n d x=3 ;t h e s e x-values give the zeroes off(x). This establishes the second (‘ AifB’) part of the result. Thus we have shown that the assumption of either condition implies the validity of the other and the proof iscomplete.J It should be noted that the propositions have to be carefully and precisely formulated. If, for example, the word ‘quadratic’ were omitted from A, statement Bwould still be a suﬃcient condition for Abut not a necessary one, since f(x) could then be x3−4x2+x+6a n d Awould not require B. Omitting the constant λfrom the stated form of f(x)i nBhas the same eﬀect. Conversely, if Awere to state that f(x)=3 ( x−2)(x−3) then Bwould be a necessary condition for Abut not a suﬃcient one. 35 PRELIMINARY ALGEBRA 1.8 Exercises Polynomial equations 1.1 Continue the investigation of equation (1.7), namely g(x)=4 x3+3x2−6x−1, as follows. (a) Make a table of values of g(x) for integer values of xbetween−2a n d2 .U s e it and the information derived in the text to draw a graph and so determinethe roots of g(x) = 0 as accurately as possible. (b) Find one accurate root of g(x) = 0 by inspection and hence determine precise values for the other two roots. (c) Show that f(x)=4 x 3+3x2−6x−k= 0 has only one real root unless −5≤k≤7 4. 1.2 Determine how the number of real roots of the equation g(x)=4 x3−17x2+1 0x+k=0 depends upon k. Are there any cases for which the equation has exactly two distinct real roots? 1.3 Continue the analysis of the polynomial equation f(x)=x7+5x6+x4−x3+x2−2=0 , investigated in subsection 1.1.1, as follows. (a) By writing the ﬁfth-degree polynom ial appearing in the expression for f/prime(x) in the form 7 x5+3 0x4+a(x−b)2+c, show that there is in fact only one positive root of f(x)=0 . (b) By evaluating f(1),f(0) and f(−1), and by inspecting the form of f(x)f o r negative values of x, determine what you can about the positions of the real roots of f(x)=0 . 1.4 Given that x=2i so n er o o to f g(x)=2 x4+4x3−9x2−11x−6=0 , use factorisation to determine how many real roots it has. 1.5 Construct the quadratic equations that have the following pairs of roots: (a) −6,−3; (b) 0 ,4; (c) 2 ,2; (d) 3 + 2 i,3−2i,w h e r e i2=−1. 1.6 Use the results of (i) equation (1.13), (ii) equation (1.12) and (iii) equation (1.14) to prove that if the roots of 3 x3−x2−10x+8=0a r e α1,α2andα3then (a)α−1 1+α−1 2+α−1 3=5/4, (b)α2 1+α2 2+α2 3=6 1/9, (c)α3 1+α3 2+α3 3=−125/27. (d) Convince yourself that eliminating (say) α2andα3from (i), (ii) and (iii) does notgive a simple explicit way of ﬁnding α1. Trigonometric identities 1.7 Prove that cosπ 12=√3+1 2√2 by considering 36 1.8 EXERCISES (a) the sum of the sines of π/3a n d π/6, (b) the sine of the sum of π/3a n d π/4. 1.8 (a) Use the fact that sin( π/6) = 1 /2 to prove that tan( π/12) = 2−√3. (b) Use the result of (a) to show further that tan( π/24) = q(2−q)w h e r e q2=2+√3. 1.9 Find the real solutions of (a) 3 sin θ−4cos θ=2, (b) 4 sin θ+3c o s θ=6, (c) 12 sin θ−5c os θ=−6. 1.10 If s= sin( π/8), prove that 8s4−8s2+1=0 , and hence show that s=[ ( 2−√2)/4]1/2. 1.11 Find all the solutions of sinθ+s i n4 θ=s i n2 θ+s i n3 θ that lie in the range −π<θ≤π. What is the multiplicity of the solution θ=0 ? Coordinate geometry 1.12 Obtain in the form (1.38) the equations that describe the following: (a) a circle of radius 5 with its centre at (1 ,−1); (b) the line 2 x+3y+ 4 = 0 and the line orthogonal to it which passes through (1,1); (c) an ellipse of eccentricity 0 .6 with centre (1 ,1) and its major axis of length 10 parallel to the y-axis. 1.13 Determine the forms of the conic sections described by the following equations: (a)x2+y2+6x+8y=0 ; (b) 9 x2−4y2−54x−16y+2 9=0 ; (c) 2 x2+2y2+5xy−4x+y−6=0 ; (d)x2+y2+2xy−8x+8y=0. 1.14 For the ellipse x2 a2+y2 b2=1 with eccentricity e, the two points ( −ae,0) and ( ae,0) are known as its foci. Show that the sum of the distances from anypoint on the ellipse to the foci is 2 a.( T h e constancy of the sum of the distances from two ﬁxed points can be used as analternative deﬁning property of an ellipse.) Partial fractions 1.15 Resolve the following into partial fractions using the three methods given in section 1.4, verifying that the same decomposition is obtained by each method: (a)2x+1 x2+3x−10, (b)4 x2−3x. 1.16 Express the following in partial fraction form: (a)2x3−5x+1 x2−2x−8, (b)x2+x−1 x2+x−2. 37 PRELIMINARY ALGEBRA 1.17 Rearrange the following functions in partial fraction form: (a)x−6 x3−x2+4x−4, (b)x3+3x2+x+1 9 x4+1 0x2+9. 1.18 Resolve the following into partial fractions in such a way that xdoes not appear in any numerator: (a)2x2+x+1 (x−1)2(x+3 ),(b)x2−2 x3+8x2+1 6x,(c)x3−x−1 (x+3 )3(x+1 ). Binomial expansion 1.19 Evaluate those of the following that are deﬁned: (a)5C3,( b )3C5,( c )−5C3,( d ) −3C5. 1.20 Use a binomial expansion to evaluate 1 /√ 4.2 to ﬁve places of decimals, and compare it with the accurate answer obtained using a calculator. Proof by induction and contradiction 1.21 Prove by induction that nX r=1r=1 2n(n+1 ) a n dnX r=1r3=1 4n2(n+1 )2. 1.22 Prove by induction that 1+r+r2+···+rk+···+rn=1−rn+1 1−r. 1.23 Prove that 32n+7 ,w h e r e nis a non-negative integer, is divisible by 8. 1.24 If a sequence of terms unsatisﬁes the recurrence relation un+1=( 1−x)un+nx with u1= 0 then show, using induction, that for n≥1 un=1 x[nx−1+( 1−x)n]. 1.25 Prove by induction that nX r=11 2rtan /θ 2r / =1 2ncot /θ 2n / −cotθ. 1.26 The quantities aiin this exercise are all positive real numbers. (a) Show that a1a2≤ /a1+a2 2 /2 . (b) Hence prove by induction on mthat a1a2···ap≤ /a1+a2+···+ap p /p , where p=2mwith ma positive integer. Note that each increase of mby unity doubles the number of factors in the product. 1.27 Establish the values of kfor which the binomial coeﬃcientpCkis divisible by p when pis a prime number. Use your result and the method of induction to prove thatnp−nis divisible by pfor all integers nand all prime numbers p. Deduce thatn5−nis divisible by 30 for any integer n. 38 1.9 HINTS AND ANSWERS 1.28 An arithmetic progression of integers anis one in which an=a0+nd,w h e r e a0 anddare integers and ntakes successive values 0 ,1,2,.... (a) Show that if any one term of the progression is the cube of an integer then so are inﬁnitely many others. (b) Show that no cube of an integer can be expressed as 7 n+5 for some positive integer n. 1.29 Prove, by the method of contradiction, that the equation xn+an−1xn−1+···+a1x+a0=0, in which all the coeﬃcients aiare integers, cannot have a rational root, unless that root is an integer. Deduce that any integral root must be a divisor of a0and hence ﬁnd all rational roots of (a)x4+6x3+4x2+5x+4=0 , (b)x4+5x3+2x2−10x+6=0 . Necessary and suﬃcient conditions 1.30 Prove that the equation ax2+bx+c=0 ,i nw h i c h a,bandcare real and a>0, has two real distinct solutions IFF b2>4ac. 1.31 For the real variable x, show that a suﬃcient, but not necessary, condition for f(x)=x(x+ 1)(2 x+ 1) to be divisible by 6 is that xis an integer. 1.32 Given that at least one of aandb, and at least one of candd, are non-zero, show that ad=bcis both a necessary and suﬃcient condition for the equations ax+by=0, cx+dy=0, to have a solution in which at least one of xandyis non-zero. 1.33 The coeﬃcients aiin the polynomial Q(x)=a4x4+a3x3+a2x2+a1xare all integers. Show that Q(n) is divisible by 24 for all integers n≥0 if and only if all the following conditions are satisﬁed:(i) 2a 4+a3is divisible by 4; (ii)a4+a2is divisible by 12; (iii)a4+a3+a2+a1is divisible by 24. 1.9 Hints and answers 1.1 (b) The roots are 1 ,1 8(−7+√ 33) =−0.1569,1 8(−7−√ 33) =−1.593. (c)−5a n d 7 4are the values of kthat make f(−1) and f(1 2) equal to zero. 1.2 Three distinct roots if −43 27<k<75 4; two distinct roots, one at x=1 3,i fk=−43 27; two distinct roots, one at x=5 2,i fk=75 4. 1.3 (a) a=4,b=3 8andc=23 6are all positive. Therefore f/prime(x)>0f o ra l l x>0. (b)f(1) = 5, f(0) =−2a n d f(−1) = 5, and so there is at least one root in each of the ranges 0 <x< 1a n d−1<x< 0. (x7+5x6)+(x4−x3)+(x2−2) is positive deﬁnite for −5<x<−√2. There are therefore no roots in this range, but there must be one to the left of x=−5. 1.4 g(x)=( x−2)(x+ 3)(2 x2+2x+ 1). The quadratic has complex roots and so g(x)=0h a so n l yt w or e a lr o o t s . 1.5 (a) x2+9x+1 8=0 ;( b ) x2−4x=0 ;( c ) x2−4x+4=0 ;( d ) x2−6x+1 3=0 . 1.6 (a) Divide (iii) by (ii). (b) Consider (i)2−2(iii). (c) Consider (i)3−3(i)(iii)+3(ii). 1.7 (a) Use sin( π/4) = 1 /√2. (b) Use results (1.20) and (1.20). 1.8 (a) Use (1.32). (b) Use (1.34) and show that q4+1=4 q2. 39 PRELIMINARY ALGEBRA 1.9 (a) 1 .339. (b) No solution because 62>42+32.( c )−0.0849. 1.10 Use the formula for sin(2 π/8) and square both sides. sin2(π/4) = 1 /2. 1.11 Show that the equation is equivalent to sin(5 θ/2)sin( θ)sin(θ/2) = 0. Solutions are −4π/5,−2π/5,0,2π/5,4π/5,π. Its multiplicity is 3. 1.12 (a) x2+y2−2x+2y−23 = 0 . (b) The orthogonal line is 3 x−2y−1 = 0. The pair of lines has equation 6x2−6y2+5xy+1 0x−11y−4=0 . (c) The minor axis has length 8. The ellipse has equation 25x2+1 6y2−50x−32y−359 = 0. 1.13 (a) A circle of radius 5 centred on ( −3,−4). (b) A hyperbola with ‘centre’ (3 ,−2) and ‘semi-axes’ 2 and 3. (c) The expression factorises into two lines, x+2y−3=0a n d2 x+y+2=0 . (d) Write the expression as ( x+y)2=8 (x−y) to see that it represents a parabola passing through the origin with the line x+y= 0 as its axis of symmetry. 1.14 Show that y2can be replaced by a2−x2−a2e2+x2e2and that the two lengths area+exanda−ex. 1.15 (a)5 7(x−2)+9 7(x+5 ), (b)−4 3x+4 3(x−3). 1.16 (a) 2 x+4+109 6(x−4)+5 6(x+2 ), (b) 1−1 3(x+2 )+1 3(x−1). 1.17 (a)x+2 x2+4−1 x−1, (b)x+1 x2+9+2 x2+1. 1.18 (a)1 (x−1)2+1 (x−1)+1 (x+3 ). (b)−1 8x+9 8(x+4 )−7 2(x+4 )2. (c)1 8 / −1 (x+1 )+9 (x+3 )−54 (x+3 )2+100 (x+3 )3 / . 1.19 (a) 10, (b) not deﬁned, (c) −35, (d)−21. 1.20 Write it as1 2( 1+0 .05)−1/2and evaluate−1/2Ckup to k= 3. The approximate and accurate values agree to ﬁve places of decimals, both giving 0 .48795. 1.23 Write 32nas 8m−7. 1.25 Use the half-angle formulae of equations (1.32) to (1.34) to relate functions of θ/2kto those of θ/2k+1. 1.26 (a) Consider ( a1−a2)2≥0. (b) Write a1+···+ap=Aandap+1+···+ap+p=B and use result (a) to replace the product ABwith an expression involving the sumA+B. Note that 2 p=2m+1. 1.27 Divisible for k=1,2,...,p−1.Expand ( n+1 )pasnp+ Pp−1 1pCknk+ 1. Apply the stated result for p= 5. Note that n5−n=n(n−1)(n+1 ) ( n2+ 1); the product of any three consecutive integers must divide by both 2 and 3. 1.28 (a) Suppose aN=a0+Nd=m3is the largest cube; then consider ( m+d)3. (b) Suppose that 7 N+5= m3. Show that ( m−7)3diﬀers from this by a multiple of 7. Deduce that q3must have the form 7 n+5f o rs o m e qin 0≤q≤7. Show explicitly that this is not so. Note. It is not suﬃcient to carry out the explicit valuations and rely on the construct from part (a). 1.29 By assuming x=p/qwith q/negationslash= 1, show that a fraction −pn/qis equal to an integer an−1pn−1+···+a1pqn−2+a0qn−1. This is a contradiction and is only resolved if q= 1 and the root is an integer. (a) The only possible candidates are ±1,±2,±4. None is a root. (b) The only possible candidates are ±1,±2,±3,±6. Only−3i sar o o t . 40 1.9 HINTS AND ANSWERS 1.30 (i) Show that the equation can be reformulated as a / x+b 2a /2 =b2−4ac 4a. (ii) If the real distinct solutions are αandβ, show that b=−(α+β)a,a n d c=αβa. Then consider the inequality 0 <(α−β)2=(α+β)2−4αβ. 1.31 f(x) can be written as x(x+1 ) ( x+2 )+ x(x+1 ) ( x−1). Each term consists of the product of three consecutive integers of which one must therefore divide by 2 and (a diﬀerent) one by 3. Thus each term separately divides by 6 and sotherefore does f(x). Note that if xis the root of 2 x 3+3x2+x−24 = 0 that lies near the non-integer value x=1.826 then x(x+ 1)(2 x+ 1) = 24 and therefore divides by 6. 1.32 (i) If x/negationslash= 0, multiply the ﬁrst equation by dand the second by band subtract. Ify/negationslash= 0, multiply by cand arespectively instead. (ii) Suppose a/negationslash=0a n d c/negationslash= 0. Whilst ensuring that no possible division by zero occurs, deduce that the equations are consistent, with solution x=−(b/a)y=−(d/c)yfor arbitrary non-zero y. 1.33 Note that, e.g., the condition for 6 a4+a3to be divisible by 4 is the same as the condition for 2 a4+a3to be divisible by 4. For the necessary (only if) part of the proof set n=1,2,3 and take integer combinations of the resulting equations.For the suﬃcient (if) part of the proof use the stated conditions to prove theproposition by induction. Note that n 3−nis divisible by 6 and that n2+3nis even. 41 2 Preliminary calculus This chapter is concerned with the formalism of probably the most widely used mathematical technique in the physical sciences, namely the calculus. The chapter divides into two sections. The ﬁrst deals with the process of diﬀerentiation and the second with its inverse process, integration. The material covered is essential forthe remainder of the book and serves as a reference. Readers who have previouslystudied these topics should ensure familiarity by looking at the worked examplesin the main text and by attempting the exercises at the end of the chapter. 2.1 Diﬀerentiation Diﬀerentiation is the process of determining how quickly or slowly a function varies, as the quantity on which it depends, its argument , is changed. More speciﬁcally it is the procedure for obtaining an expression (numerical or algebraic)for the rate of change of the function with respect to its argument. Familiarexamples of rates of change include acceleration (the rate of change of velocity)and chemical reaction rate (the rate of change of chemical composition). Both acceleration and reaction rate give a measure of the change of a quantity with respect to time. However, diﬀerentiation may also be applied to changes withrespect to other quantities, for example the change in pressure with respect to achange in temperature. Although it will not be apparent from what we have said so far, diﬀerentiation is in fact a limiting process, that is, it deals only with the inﬁnitesimal change inone quantity resulting from an inﬁnitesimal change in another. 2.1.1 Diﬀerentiation from ﬁrst principles Let us consider a function f(x) that depends on only one variable x, together with numerical constants, for example, f(x)=3 x 2orf(x)=s i n xorf(x)=2+3 /x. 42 2.1 DIFFERENTIATION A P xf(x) x+∆xf(x+∆x) ∆f θ∆x Figure 2.1 The graph of a function f(x) showing that the gradient of the function at P,g i v e nb yt a n θ, is approximately equal to ∆ f/∆x. Figure 2.1 shows an example of such a function. Near any particular point, P, the value of the function changes by an amount ∆ f,s a y ,a s xchanges by a small amount ∆ x. The slope of the tangent to the graph of f(x)a t P is then approximately ∆ f/∆x, and the change in the value of the function is ∆f=f(x+∆x)−f(x). In order to calculate the true value of the gradient, or ﬁrst derivative , of the function at P, we must let ∆ xbecome inﬁnitesimally small. We therefore deﬁne the ﬁrst derivative of f(x)a s f/prime(x)≡df(x) dx≡lim ∆x→0f(x+∆x)−f(x) ∆x, (2.1) provided that the limit exists. The limit will depend in almost all cases on the value of x. If the limit does exist at a point x=athen the function is said to be diﬀerentiable at a; otherwise it is said to be non-diﬀerentiable at a.T h ef o r m a l concept of a limit and its existence or non-existence is discussed in chapter 4; forpresent purposes we will adopt an intuitive approach. In the deﬁnition (2.1), we allow ∆ xto tend to zero from either positive or negative values and require the same limit to be obtained in both cases. Afunction that is diﬀerentiable at ais necessarily continuous at a(there must be no jump in the value of the function at a), though the converse is not necessarily true. This latter assertion is illustrated in ﬁgure 2.1: the function is continuousat the ‘kink’ Abut the two limits of the gradient as ∆ xtends to zero from 43 PRELIMINARY CALCULUS positive or negative values are diﬀerent and so the function is not diﬀerentiable atA. It should be clear from the above discussion that near the point Pwe may approximate the change in the value of the function, ∆ f, that results from a small change ∆ xinxby ∆f≈df(x) dx∆x. (2.2) As one would expect, the approximation improves as the value of ∆ xis reduced. In the limit in which the change ∆ xbecomes inﬁnitesimally small, we denote it by the diﬀerential dx, and (2.2) reads df=df(x) dxdx. (2.3) Thisequality relates the inﬁnitesimal change in the function, df, to the inﬁnitesimal change dxthat causes it. So far we have discussed only the ﬁrst derivative of a function. However, we can also deﬁne the second derivative as the gradient of the gradient of a function. Again we use the deﬁnition (2.1) but now with f(x) replaced by f/prime(x). Hence the second derivative is deﬁned by f/prime/prime(x)≡lim ∆x→0f/prime(x+∆x)−f/prime(x) ∆x, (2.4) provided that the limit exists. A physical example of a second derivative is the second derivative of the distance travelled by a particle with respect to time. Sincethe ﬁrst derivative of distance travelled gives the particle’s velocity, the secondderivative gives its acceleration. We can continue in this manner, the nth derivative of the function f(x)b e i n g deﬁned by f (n)(x)≡lim ∆x→0f(n−1)(x+∆x)−f(n−1)(x) ∆x. (2.5) It should be noted that with this notation f/prime(x)≡f(1)(x),f/prime/prime(x)≡f(2)(x), etc., and that formally f(0)(x)≡f(x). All this should be familiar to the reader, though perhaps not with such formal deﬁnitions. The following example shows the diﬀerentiation of f(x)=x2from ﬁrst principles. In practice, however, it is desirable simply to remember the derivatives of standard functions; the techniques given in the remainder of this section canbe applied to ﬁnd more complicated derivatives. 44 2.1 DIFFERENTIATIONIFind from ﬁrst principles the derivative with respect to xoff(x)=x2. Using the deﬁnition (2.1), f/prime(x) = lim ∆x→0f(x+∆x)−f(x) ∆x = lim ∆x→0(x+∆x)2−x2 ∆x = lim ∆x→02x∆x+( ∆x)2 ∆x = lim ∆x→0(2x+∆x). As ∆ xtends to zero, 2 x+∆xtends towards 2 x, hence f/prime(x)=2 x. J Derivatives of other functions can be obtained in the same way. The derivatives of some simple functions are listed below (note that ais a constant): d dx(xn)=nxn−1,d dx(eax)=aeax,d dx(lnax)=1 x, d dx(sinax)=acosax,d dx(cosax)=−asinax,d dx(secax)=asecaxtanax, d dx(tanax)=asec2ax,d dx(cosec ax)=−acosec axcotax, d dx(cotax)=−acosec2ax,d dxparenleftBig sin−1x aparenrightBig =1√ a2−x2, d dxparenleftBig cos−1x aparenrightBig =−1√ a2−x2,d dxparenleftBig tan−1x aparenrightBig =a a2+x2. Diﬀerentiation from ﬁrst principles emphasises the deﬁnition of a derivative as the gradient of a function. However, for most practical purposes, returning to the deﬁnition (2.1) is time consuming and does not aid our understanding. Instead, as mentioned above, we employ a number of techniques, which use the derivativeslisted above as ‘building blocks’, to evaluate the derivatives of more complicatedfunctions than hitherto encountered. S ubsections 2.1.2–2.1.7 develop the methods required. 2.1.2 Diﬀerentiation of products As a ﬁrst example of the diﬀerentiation of a more complicated function, we consider ﬁnding the derivative of a function f(x) that can be written as the product of two other functions of x,n a m e l y f(x)=u(x)v(x). For example, if f(x)= x 3sinxthen we might take u(x)= x3and v(x)=s i n x. Clearly the 45 PRELIMINARY CALCULUS separation is not unique. (In the given example, possible alternative break-ups would be u(x)=x2,v(x)=xsinx,o re v e n u(x)=x4tanx,v(x)=x−1cosx.) The purpose of the separation is to split the function into two (or more) parts, of which we know the derivatives (or at least we can evaluate these derivatives more easily than that of the whole). We would gain little, however, if we did not know the relationship between the derivative of fand those of uandv. Fortunately, they are very simply related, as we shall now show. Since f(x) is written as the product u(x)v(x), it follows that f(x+∆x)−f(x)=u(x+∆x)v(x+∆x)−u(x)v(x) =u(x+∆x)[v(x+∆x)−v(x)] + [ u(x+∆x)−u(x)]v(x). From the deﬁnition of a derivative (2.1), df dx= lim ∆x→0f(x+∆x)−f(x) ∆x = lim ∆x→0braceleftbigg u(x+∆x)bracketleftbiggv(x+∆x)−v(x) ∆xbracketrightbigg +bracketleftbiggu(x+∆x)−u(x) ∆xbracketrightbigg v(x)bracerightbigg . In the limit ∆ x→0, the factors in square brackets become dv/dx anddu/dx (by the deﬁnitions of these quantities) and u(x+∆x) simply becomes u(x). Consequently we obtain df dx=d dx[u(x)v(x)] =u(x)dv(x) dx+du(x) dxv(x). (2.6) In primed notation and without writing the argument xexplicitly, (2.6) is stated concisely as f/prime=(uv)/prime=uv/prime+u/primev. (2.7) This is a general result obtained without making any assumptions about the speciﬁc forms f,uandv, other than that f(x)=u(x)v(x). In words, the result reads as follows. The derivative of the product of two functions is equal to the ﬁrst function times the derivative of the second plus the second function times thederivative of the ﬁrst .IFind the derivative with respect to xoff(x)=x3sinx. Using the product rule, (2.6), d dx(x3sinx)=x3d dx(sinx)+d dx(x3)si nx =x3cosx+3x2sinx. J The product rule may readily be extended to the product of three or more functions. Considering the function f(x)=u(x)v(x)w(x) (2.8) 46 2.1 DIFFERENTIATION and using (2.6), we obtain, as before omitting the argument, df dx=ud dx(vw)+du dxvw. Using (2.6) again to expand the ﬁrst term on the RHS gives the complete result d dx(uvw)=uvdw dx+udv dxw+du dxvw (2.9) or (uvw)/prime=uvw/prime+uv/primew+u/primevw. (2.10) It is readily apparent that this can be extended to products containing any number nof factors; the expression for the derivative will then consist of nterms with the prime appearing in successive terms on each of the nfactors in turn. This is probably the easiest way to recall the product rule. 2.1.3 The chain rule Products are just one type of complicated function that we may encounter in diﬀerentiation. Another is the function of a function, e.g. f(x)=( 3+ x2)3=u(x)3, where u(x)=3+ x2.I f∆ f,∆uand ∆ xare small ﬁnite quantities, it follows that ∆f ∆x=∆f ∆u∆u ∆x; As the quantities become inﬁnitesimally small we obtain df dx=df dudu dx. (2.11) This is the chain rule , which we must apply when diﬀerentiating a function of a function.IFind the derivative with respect to xoff(x)=( 3+ x2)3. Rewriting the function as f(x)=u3,w h e r e u(x)=3+ x2, and applying (2.11) we ﬁnd df dx=3u2du dx=3u2d dx(3 +x2)=3 u2×2x=6x(3 +x2)2. J Similarly, the derivative with respect to xoff(x)=1 /v(x)m a yb eo b t a i n e db y rewriting the function as f(x)=v−1and applying (2.11): df dx=−v−2dv dx=−1 v2dv dx. (2.12) The chain rule is also useful for calculating the derivative of a function fwith respect to xwhen both xandfare written in terms of a variable (or parameter), sayt. 47 PRELIMINARY CALCULUSIFind the derivative with respect to xoff(t)=2 at,w h e r e x=at2. We could of course substitute for ta n dt h e nd i ﬀ e r e n t i a t e fas a function of x, but in this case it is quicker to use df dx=df dtdt dx=2a1 2at=1 t, where we have used the fact that dt dx= /dx dt /−1 . J 2.1.4 Diﬀerentiation of quotients Applying (2.6) for the derivative of a product to a function f(x)=u(x)[1/v(x)], we may obtain the derivative of the quotient of two factors. Thus f/prime=parenleftBigu vparenrightBig/prime =uparenleftbigg1 vparenrightbigg/prime +u/primeparenleftbigg1 vparenrightbigg =uparenleftbigg −v/prime v2parenrightbigg +u/prime v, where (2.12) has been used to evaluate (1 /v)/prime. This can now be rearranged into the more convenient and memorisable form f/prime=parenleftBigu vparenrightBig/prime =vu/prime−uv/prime v2. (2.13) This can be expressed in words as the derivative of a quotient is equal to the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared .IFind the derivative with respect to xoff(x)=s i n x/x. Using (2.13) with u(x)=s i n x,v(x)=xand hence u/prime(x)=c o s x,v/prime(x) = 1, we ﬁnd f/prime(x)=xcosx−sinx x2=cosx x−sinx x2. J 2.1.5 Implicit diﬀerentiation So far we have only diﬀerentiated functions written in the form y=f(x). However, we may not always be presented with a relationship in this simpleform. As an example consider the relation x 3−3xy+y3=2 .I nt h i sc a s ei ti s not possible to rearrange the equation to give yas a function of x. Nevertheless, by diﬀerentiating term by term with respect to x(implicit diﬀerentiation ), we can ﬁ n dt h ed e r i v a t i v eo f y. 48 2.1 DIFFERENTIATIONIFind dy/dx ifx3−3xy+y3=2. Diﬀerentiating each term in the equation with respect to xwe obtain d dx(x3)−d dx(3xy)+d dx(y3)=d dx(2), ⇒3x2− / 3xdy dx+3y / +3y2dy dx=0, where the derivative of 3 xyhas been found using the product rule. Hence, rearranging for dy/dx , dy dx=y−x2 y2−x. Note that dy/dx is a function of both xandyand cannot be expressed as a function of x only. J 2.1.6 Logarithmic diﬀerentiation In circumstances in which the variable with respect to which we are diﬀerentiating is an exponent, taking logarithms and then diﬀerentiating implicitly is the simplestway to ﬁnd the derivative.IFind the derivative with respect to xofy=ax. To ﬁnd the required derivative we ﬁrst take logarithms and then diﬀerentiate implicitly: lny=l nax=xlna⇒1 ydy dx=l na. Now, rearranging and substituting for y, we ﬁnd dy dx=ylna=axlna. J 2.1.7 Leibniz’ theorem We have discussed already how to ﬁnd the derivative of a product of two or more functions. We now consider Leibniz’ theorem , which gives the corresponding results for the higher derivatives of products. Consider again the function f(x)=u(x)v(x). We know from the product rule thatf/prime=uv/prime+u/primev. Using the rule once more for each of the products, we obtain f/prime/prime=(uv/prime/prime+u/primev/prime)+(u/primev/prime+u/prime/primev) =uv/prime/prime+2u/primev/prime+u/prime/primev. Similarly, diﬀerentiating twice more gives f/prime/prime/prime=uv/prime/prime/prime+3u/primev/prime/prime+3u/prime/primev/prime+u/prime/prime/primev, f(4)=uv(4)+4u/primev/prime/prime/prime+6u/prime/primev/prime/prime+4u/prime/prime/primev/prime+u(4)v. 49 PRELIMINARY CALCULUS The pattern emerging is clear and strongly suggests that the results generalise to f(n)=nsummationdisplay r=0n! r!(n−r)!u(r)v(n−r)=nsummationdisplay r=0nCru(r)v(n−r), (2.14) where the fraction n!/[r!(n−r)!] is identiﬁed with the binomial coeﬃcientnCr (see chapter 1). To prove that this is so, we use the method of induction as follows. Assume that (2.14) is valid for nequal to some integer N.T h e n f(N+1)=Nsummationdisplay r=0NCrd dxparenleftbig u(r)v(N−r)parenrightbig =Nsummationdisplay r=0NCr[u(r)v(N−r+1)+u(r+1)v(N−r)] =Nsummationdisplay s=0NCsu(s)v(N+1−s)+N+1summationdisplay s=1NCs−1u(s)v(N+1−s), where we have substituted summation index sforrin the ﬁrst summation, and forr+ 1 in the second. Now, from our earlier discussion of binomial coeﬃcients, equation (1.51), we have NCs+NCs−1=N+1Cs and so, after separating out the ﬁrst term of the ﬁrst summation and the last term of the second, obtain f(N+1)=NC0u(0)v(N+1)+Nsummationdisplay s=1N+1Csu(s)v(N+1−s)+NCNu(N+1)v(0). ButNC0=1=N+1C0andNCN=1=N+1CN+1, and so we may write f(N+1)=N+1C0u(0)v(N+1)+Nsummationdisplay s=1N+1Csu(s)v(N+1−s)+N+1CN+1u(N+1)v(0) =N+1summationdisplay s=0N+1Csu(s)v(N+1−s). This is just (2.14) with nset equal to N+ 1. Thus, assuming the validity of (2.14) forn=Nimplies its validity for n=N+ 1. However, when n=1e q u a t i o n (2.14) is simply the product rule, and this we have already proved directly. These results taken together establish the validity of (2.14) for all nand prove Leibniz’ theorem. 50 2.1 DIFFERENTIATION Q A BCf(x) xS Figure 2.2 A graph of a function, f(x), showing how diﬀerentiation corre- sponds to ﬁnding the gradient of the function at a particular point. Points B, QandSare stationary points (see text).IFind the third derivative of the function f(x)=x3sinx. Using (2.14) we immediately ﬁnd f/prime/prime/prime(x)=6s i n x+3 ( 6 x)cosx+3 ( 3 x2)(−sinx)+x3(−cosx) =3 ( 2−3x2)si nx+x(18−x2)c osx. J 2.1.8 Special points of a function We have interpreted the derivative of a function as the gradient of the function at the relevant point (ﬁgure 2.1). If the gradient is zero at some point then thefunction is said to have a stationary point there. Clearly, in graphical terms, this corresponds to a horizontal tangent to the graph at that point. Stationary points may be divided into three categories and an example of each is shown in ﬁgure 2.2. Point Bis said to be a minimum since the function increases in value in both directions away from it. Point Qis said to be a maximum since the function decreases in both directions away from it. Note that Bis not the overall minimum value of the function and Qis not the overall maximum; rather, they are a local minimum and a local maximum. The third type of stationary point isthestationary point of inﬂection ,S. In this case the function falls in the positive x-direction and rises in the negative x-direction so that Sis neither a maximum nor a minimum. Nevertheless, the gradient of the function is zero at S,i . e .t h e graph of the function is ﬂat there, and this justiﬁes our calling it a stationary point. Of course, a point at which the gradient of the function is zero but the function rises in the positive x-direction and falls in the negative x-direction is also a stationary point of inﬂection. 51 PRELIMINARY CALCULUS The above distinction between the three types of stationary point has been made rather descriptively. However, it is possible to deﬁne and distinguish sta-tionary points mathematically. From their deﬁnition as points of zero gradient,all stationary points must be characterised by df/dx = 0. In the case of the minimum, B, the slope, i.e. df/dx , changes from negative at Ato positive at C through zero at B. Thus df/dx is increasing and so the second derivative d 2f/dx2 must be positive. Conversely, at the maximum, Q, we must have that d2f/dx2is negative. It is less obvious, but intuitively reasonable, that at S,d2f/dx2is zero. This may be inferred from the following observations. To the left of Sthe curve is concave upwards so that df/dx is increasing with xand hence d2f/dx2>0. To the right ofS, however, the curve is concave downwards so that df/dx is decreasing with xand hence d2f/dx2<0. In summary, at a stationary point df/dx =0a n d (i) for a minimum, d2f/dx2>0, (ii) for a maximum, d2f/dx2<0, (iii) for a stationary point of inﬂection, d2f/dx2=0a n d d2f/dx2changes sign through the point. In case (iii), a stationary point of inﬂection, in order that d2f/dx2changes sign through the point we normally require d3f/dx3/negationslash= 0 at that point. This simple rule can fail for some functions, however, and in general if the ﬁrst non-vanishingderivative of f(x) at the stationary point is f (n)then if nis even the point is a maximum or minimum and if nis odd the point is a stationary point of inﬂection. This may be seen from the Taylor expansion (see equation (4.17)) of the function about the stationary point, but it is not proved here.IFind the positions and natures of the stationary points of the function f(x)=2 x3−3x2−36x+2. The ﬁrst criterion for a stationary point is that df/dx = 0, and hence we set df dx=6x2−6x−36 = 0 , from which we obtain (x−3)(x+2 )=0 . Hence the stationary points are at x=3a n d x=−2. To determine the nature of the stationary point we must evaluate d2f/dx2: d2f dx2=1 2x−6. 52 2.1 DIFFERENTIATION Gf(x) x Figure 2.3 The graph of a function f(x) that has a general point of inﬂection at the point G. Now, we examine each stationary point in turn. For x=3 , d2f/dx2= 30. Since this is positive, we conclude that x= 3 is a minimum. Similarly, for x=−2,d2f/dx2=−30 and sox=−2 is a maximum. J So far we have concentrated on stationary points, which are deﬁned to have df/dx = 0. We have found that at a stationary point of inﬂection d2f/dx2is also zero and changes sign. This naturally leads us to consider points at whichd 2f/dx2is zero and changes sign but at which df/dx isnot, in general, zero. Such points are called general points of inﬂection or simply points of inﬂection . Clearly, a stationary point of inﬂection is a special case for which df/dx is also zero. At a general point of inﬂection the graph of the function changes from beingconcave upwards to concave downwards (or vice versa), but the tangent to thecurve at this point need not be horizontal. A typical example of a general pointof inﬂection is shown in ﬁgure 2.3. The determination of the stationary points of a function, together with the identiﬁcation of its zeroes, inﬁnities and possible asymptotes, is usually suﬃcientto enable a graph of the function showing most of its signiﬁcant features to besketched. Some examples for the reader to try are included in the exercises at the end of this chapter. 2.1.9 Curvature of a function In the previous section we saw that at a point of inﬂection of the function f(x), the second derivative d 2f/dx2changes sign and passes through zero. The corresponding graph of fshows an inversion of its curvature at the point of inﬂection. We now develop a more quantitative measure of the curvature of afunction (or its graph), which is applicable at general points and not just in theneighbourhood of a point of inﬂection. As in ﬁgure 2.1, let θbe the angle made with the x-axis by the tangent at a 53 PRELIMINARY CALCULUS C PQρ θ θ+∆θ∆θ xf(x) Figure 2.4 Two neighbouring tangents to the curve f(x) whose slopes diﬀer by ∆ θ. The angular separation of the corresponding radii of the circle of curvature is also ∆ θ. point Pon the curve f=f(x), with tan θ=df/dx evaluated at P. Now consider also the tangent at a neighbouring point Qon the curve, and suppose that it makes an angle θ+∆θwith the x-axis, as illustrated in ﬁgure 2.4. It follows that the corresponding normals at PandQ, which are perpendicular to the respective tangents, also intersect at an angle ∆ θ. Furthermore, their point of intersection, Cin the ﬁgure, will be the position of the centre of a circle that approximates the arc PQ, at least to the extent of having the same tangents at the extremities of the arc. This circle is called the circle of curvature . For a ﬁnite arc PQ, the lengths of CPandCQwill not, in general, be equal, as they would be if f=f(x)werein fact the equation of a circle. But, as Q is allowed to tend to P,i . e .a s∆ θ→0, they do become equal, their common value being ρ, the radius of the circle, known as the radius of curvature . It follows immediately that the curve and the circle of curvature have a common tangentatPand lie on the same side of it. The reciprocal of the radius of curvature, ρ −1, deﬁnes the curvature of the function f(x) at the point P. The radius of curvature can be deﬁned more mathematically as follows. The length ∆ sof arc PQis approximately equal to ρ∆θand, in the limit ∆ θ→0, this relationship deﬁnes ρas ρ= lim ∆θ→0∆s ∆θ=ds dθ. (2.15) It should be noted that, as sincreases, θmay increase or decrease according to whether the curve is locally concave upwards (i.e. shaped as if it were near a minimum in f(x)) or concave downwards. This is reﬂected in the sign of ρ,w h i c h therefore also indicates the position of the curve (and of the circle of curvature) 54 2.1 DIFFERENTIATION relative to the common tangent, above or below. Thus a negative value of ρ indicates that the curve is locally concave downwards and that the tangent liesabove the curve. We next obtain an expression for ρ, not in terms of sandθbut in terms ofxandf(x). The expression, though somewhat cumbersome, follows from the deﬁning equation (2.15), the deﬁning property of θthat tan θ=df/dx≡f /primeand the fact that the rate of change of arc length with xis given by ds dx=bracketleftBigg 1+parenleftbiggdf dxparenrightbigg2bracketrightBigg1/2 . (2.16) This last result, simply quoted here, is proved more formally in subsection 2.2.13. From the chain rule (2.11) it follows that ρ=ds dθ=ds dxdx dθ. (2.17) Diﬀerentiating both sides of tan θ=df/dx with respect to xgives sec2θdθ dx=d2f dx2≡f/prime/prime, from which, using sec2θ=1+t a n2θ=1+( f/prime)2,w ec a no b t a i n dx/dθ as dx dθ=1+t a n2θ f/prime/prime=1+(f/prime)2 f/prime/prime. (2.18) Substituting (2.16) and (2.18) into (2.17) then yields the ﬁnal expression for ρ, ρ=bracketleftbig 1+(f/prime)2bracketrightbig3/2 f/prime/prime. (2.19) It should be noted that the quantity in brackets is always positive and that its3 2th root is also taken as positive. The sign of ρis thus solely determined by that of d2f/dx2, in line with our previous discussion relating the sign to whether the curve is concave or convex upwards. If, as happens at a point of inﬂection,d 2f/dx2is zero then ρis formally inﬁnite and the curvature of f(x)i sz e r o .A s d2f/dx2changes sign on passing through zero, both the local tangent and the circle of curvature change from their initial positions to the opposite side of thecurve. 55 PRELIMINARY CALCULUSIShow that the radius of curvature at the point (x, y)on the ellipse x2 a2+y2 b2=1 has magnitude (a4y2+b4x2)3/2/(a4b4)and the opposite sign to y. Check the special case b=a, for which the ellipse becomes a circle. Diﬀerentiating the equation of the ellipse with respect to xgives 2x a2+2y b2dy dx=0 and so dy dx=−b2x a2y. A second diﬀerentiation, using (2.13), then yields d2y dx2=−b2 a2 /y−xy/prime y2 / =−b4 a2y3 /y2 b2+x2 a2 / =−b4 a2y3, where we have used the fact that ( x, y) lies on the ellipse. We note that d2y/dx2,a n d hence ρ, has the opposite sign to y3, and hence to y. Substituting in (2.19) gives for the magnitude of the radius of curvature |ρ|= ///// / 1+b4x2/(a4y2) /3/2 −b4/(a2y3) /////=(a4y2+b4x2)3/2 a4b4. For the special case b=a,|ρ|reduces to a−2(y2+x2)3/2and, since x2+y2=a2,t h i si n turn gives|ρ|=a, as expected. J The discussion in this section has been conﬁned to the behaviour of curves that lie in one plane; examples of the application of curvature to the bending ofloaded beams and to particle orbits under the inﬂuence of a central forces can befound in the exercises at the ends of later chapters. A more general treatment ofcurvature in three dimensions is given in section 10.3, where a vector approach isadopted. 2.1.10 Theorems of diﬀerentiation Rolle’s theorem Rolle’s theorem (ﬁgure 2.5) states that if a function f(x) is continuous in the range a≤x≤c, is diﬀerentiable in the range a<x<c and satisﬁes f(a)=f(c) then for at least one point x=b,w h e r e a<b<c ,f /prime(b) = 0. Thus Rolle’s theorem states that for a well-behaved (continuous and diﬀerentiable) functionthat has the same value at two points either there is at least one stationary pointbetween those points or the function is a constant between them. The validity of the theorem is immediately apparent from ﬁgure 2.5 and a full analytic proof will not be given. The theorem is used in deriving the mean value theorem, which wenow discuss. 56 2.1 DIFFERENTIATION a b cf(x) x Figure 2.5 The graph of a function f(x), showing that if f(a)=f(c)t h e na t one point at least between x=aandx=cthe graph has zero gradient. a b cC Af(a)f(x) xf(c) Figure 2.6 The graph of a function f(x); at some point x=bit has the same gradient as the line AC. Mean value theorem The mean value theorem (ﬁgure 2.6) states that if a function f(x) is continuous in the range a≤x≤cand diﬀerentiable in the range a<x<c then f/prime(b)=f(c)−f(a) c−a, (2.20) for at least one value bwhere a<b<c . Thus the mean value theorem states that for a well-behaved function the gradient of the line joining two points on thecurve is equal to the slope of the tangent to the curve for at least one interveningpoint. The proof of the mean value theorem is found by examination of ﬁgure 2.6, as follows. The equation of the line ACis g(x)=f(a)+(x−a)f(c)−f(a) c−a, 57 PRELIMINARY CALCULUS and hence the diﬀerence between the curve and the line is h(x)=f(x)−g(x)=f(x)−f(a)−(x−a)f(c)−f(a) c−a. Since the curve and the line intersect at AandC,h(x) = 0 at both of these points. Hence, by an application of Rolle’s theorem, h/prime(x) = 0 for at least one point b between AandC. Diﬀerentiating our expression for h(x), we ﬁnd h/prime(x)=f/prime(x)−f(c)−f(a) c−a, and hence at b,w h e r e h/prime(x)=0 , f/prime(b)=f(c)−f(a) c−a. Applications of Rolle’s theorem and the mean value theorem Since the validity of Rolle’s theorem is intuitively obvious, given the conditions imposed on f(x), it will not be surprising that the problems that can be solved by applications of the theorem alone are relatively simple ones. Nevertheless we will illustrate it with the following example.IWhat semi-quantitative results can be deduced by applying Rolle’s theorem to the follow- ing functions f(x),w i t h aandcchosen so that f(a)=f(c)=0?( i )sinx, (ii)cosx, (iii) x2−3x+2,( i v ) x2+7x+3,( v )2x3−9x2−24x+k. (i) If the consecutive values of xthat make sin x=0a r e α1,α2,...(actually x=nπ,f o r any integer n) then Rolle’s theorem implies that the derivative of sin x,n a m e l yc o s x,h a s at least one zero lying between each pair of values αiandαi+1. (ii) In an exactly similar way, we conclude that the derivative of cos x,n a m e l y−sinx, has at least one zero lying between consecutive pairs of zeroes of cos x.T h e s et w o results taken together (but neither separately) imply that sin xand cos xhave interleaving zeroes. (iii) For f(x)=x2−3x+2 ,f(a)=f(c)=0i f aandcare taken as 1 and 2 respectively. Rolle’s theorem then implies that f/prime(x)=2 x−3=0h a sas o l u t i o n x=bwith bin the range 1 <b< 2. This is obviously so, since b=3/2. (iv) With f(x)= x2+7x+ 3, the theorem tells us that if there are two roots of x2+7x+ 3 = 0 then they have the root of f/prime(x)=2 x+ 7 = 0 lying between them. Thus any (real) roots of x2+7x+ 3 = 0 lie on either side of x=−7/2. The actual roots are (−7±√ 37)/2. (v) If f(x)=2 x3−9x2−24x+kthen f/prime(x) = 0 is the equation 6 x2−18x−24 = 0, which has solutions x=−1a n d x=4 .C o n s e q u e n t l y ,i f α1andα2are two diﬀerent roots off(x)=0t h e na tl e a s to n eo f −1 and 4 must lie in the open interval α1toα2.I f ,a si s the case for a certain range of values of k,f(x) = 0 has three roots, α1,α2andα3,t h e n α1<−1<α2<4<α3. 58 2.1 DIFFERENTIATION In each case, as might be expected, the applic ation of Rolle’s theorem does no more than focus attention on particular ranges of values; it does not yield precise answers. J Direct veriﬁcation of the mean value theorem is straightforward when it is applied to simple functions. For example, if f(x)=x2, it states that there is a value bin the interval a<b<c such that c2−a2=f(c)−f(a)=(c−a)f/prime(b)=(c−a)2b. This is clearly so, since b=(a+c)/2 satisﬁes the relevant criteria. As a slightly more complicated example we may consider a cubic equation, say f(x)=x3+2x2+4x−6 = 0, between two speciﬁed values of x,s a y1a n d2 .I n this case we need to verify that there is a value of xlying in the range 1 <x< 2 that satisﬁes 18−1=f(2)−f(1) = (2−1)f/prime(x)=1 ( 3 x2+4x+4 ). This is easily done, either by evaluating 3 x2+4x+4−17 at x=1a n da t x=2a n d checking that the values have opposite signs or by solving 3 x2+4x+4−17 = 0 and showing that one of the roots lies in the stated interval. The following applications of the mean value theorem establish some general inequalities for two common functions.IDetermine inequalities satisﬁed by lnxandsinxfor suitable ranges of the real variable x. Since for positive values of its argument the derivative of ln xisx−1, the mean value theorem gives us lnc−lna c−a=1 b for some bin 0<a<b<c . Further, since a<b<c implies that c−1<b−1<a−1,w e have 1 c<lnc−lna c−a<1 a, or, multiplying through by c−aand writing c/a=xwhere x>1, 1−1 x<lnx<x−1. Applying the mean value theorem to sin xshows that sinc−sina c−a=c o s b for some blying between aandc.I faandcare restricted to lie in the range 0 ≤a<c≤π, in which the cosine function is monotonically decreasing (i.e. there are no turning points),we can deduce that cosc<sinc−sina c−a<cosa. J 59 PRELIMINARY CALCULUS abf(x) x Figure 2.7 An integral as the area under a curve. 2.2 Integration The notion of an integral as the area under a curve will be familiar to the reader. In ﬁgure 2.7, in which the solid line is a plot of a function f(x), the shaded area represents the quantity denoted by I=integraldisplayb af(x)dx. (2.21) This expression is known as the deﬁnite integral off(x)b e t w e e nt h e lower limit x=aand the upper limit x=b,a n d f(x) is called the integrand . 2.2.1 Integration from ﬁrst principles The deﬁnition of an integral as the area under a curve is not a formal deﬁnition, but one that can be readily visualised. The formal deﬁnition of Iinvolves subdividing the ﬁnite interval a≤x≤binto a large number of subintervals, by deﬁning intermediate points ξisuch that a=ξ0<ξ1<ξ2<···<ξ n=b,a n d then forming the sum S=nsummationdisplay i=1f(xi)(ξi−ξi−1), (2.22) where xiis an arbitrary point that lies in the range ξi−1≤xi≤ξi(see ﬁgure 2.8). If now nis allowed to tend to inﬁnity in any way whatsoever, subject only to the restriction that the length of every subinterval ξi−1toξitends to zero, then S might, or might not, tend to a unique limit, I. If it does then the deﬁnite integral off(x) between aandbis deﬁned as having the value I. If no unique limit exists the integral is undeﬁned. For continuous functions and a ﬁnite interval a≤x≤b the existence of a unique limit is assured and the integral is guaranteed to exist. 60 2.2 INTEGRATION abx1x2 x3 x4 x5 ξ1 ξ2 ξ3 ξ4f(x) x Figure 2.8 The evaluation of a deﬁnite integral by subdividing the interval a≤x≤binto subintervals.IEvaluate from ﬁrst principles the integral I= Rb 0x2dx. We ﬁrst approximate the area under the curve y=x2between 0 and bbynrectangles of equal width h. If we take the value at the lower end of each subinterval (in the limit of an inﬁnite number of subintervals we could equally well have chosen the value at the upperend) to give the height of the corresponding rectangle, then the area of the kth rectangle will be ( kh) 2h=k2h3. The total area is thus A=n−1X k=0k2h3=(h3)1 6n(n−1)(2n−1), where we have used the expression for the sum of the squares of the natural numbers derived in subsection 1.7.1. Now h=b/nand so A= /b3 n3 /n 6(n−1)(2n−1) =b3 6 / 1−1 n // 2−1 n / . Asn→∞,A→b3/3, which is thus the value Iof the integral. J Some straightforward properties of deﬁnite integrals that are almost self-evident are: integraldisplayb a0dx=0,integraldisplaya af(x)dx=0, (2.23) integraldisplayc af(x)dx=integraldisplayb af(x)dx+integraldisplayc bf(x)dx, (2.24) integraldisplayb a[f(x)+g(x)]dx=integraldisplayb af(x)dx+integraldisplayb ag(x)dx. (2.25) 61 PRELIMINARY CALCULUS Combining (2.23) and (2.24) with cset equal to ashows that integraldisplayb af(x)dx=−integraldisplaya bf(x)dx. (2.26) 2.2.2 Integration as the inverse of diﬀerentiation The deﬁnite integral has been deﬁned as the area under a curve between two ﬁxed limits. Let us now consider the integral F(x)=integraldisplayx af(u)du (2.27) in which the lower limit aremains ﬁxed but the upper limit xis now variable. It will be noticed that this is essentially a restatement of (2.21), but that the variable xin the integrand has been replaced by a new variable u. It is conventional to rename the dummy variable in the integrand in this way in order that the same variable does not appear in both the integrand and the integration limits. It is apparent from (2.27) that F(x) is a continuous function of x, but at ﬁrst glance the deﬁnition of an integral as the area under a curve does not connect withour assertion that integration is the inverse process to diﬀerentiation. However, by considering the integral (2.27) and using the elementary property (2.24), we obtain F(x+∆x)=integraldisplay x+∆x af(u)du =integraldisplayx af(u)du+integraldisplayx+∆x xf(u)du =F(x)+integraldisplayx+∆x xf(u)du. Rearranging and dividing through by ∆ xyields F(x+∆x)−F(x) ∆x=1 ∆xintegraldisplayx+∆x xf(u)du. Letting ∆ x→0 and using (2.1) we ﬁnd that the LHS becomes dF/dx ,w h e r e a s the RHS becomes f(x). The latter conclusion follows because when ∆ xis small the value of the integral on the RHS is approximately f(x)∆x, and in the limit ∆x→0 no approximation is involved. Thus dF(x) dx=f(x), (2.28) or, substituting for F(x) from (2.27), d dxbracketleftbiggintegraldisplayx af(u)dubracketrightbigg =f(x). 62 2.2 INTEGRATION From the last two equations it is clear that integration can be considered as the inverse of diﬀerentiation. However, we see from the above analysis that thelower limit ais arbitrary and so diﬀerentiation does not have a unique inverse. Any function F(x) obeying (2.28) is called an indeﬁnite integral off(x), though any two such functions can diﬀer by at most an arbitrary additive constant. Since the lower limit is arbitrary, it is usual to write F(x)=integraldisplay x f(u)du (2.29) and explicitly include the arbitrary constant only when evaluating F(x). The evaluation is conventionally written in the form integraldisplay f(x)dx=F(x)+c (2.30) where cis called the constant of integration . It will be noticed that, in the absence of any integration limits, we use the same symbol for the arguments of both f andF. This can be confusing, but is suﬃciently common practice that the reader needs to become familiar with it. We also note that the deﬁnite integral of f(x) between the ﬁxed limits x=a andx=bc a nb ew r i t t e ni nt e r m so f F(x). From (2.27) we have integraldisplayb af(x)dx=integraldisplayb x0f(x)dx−integraldisplaya x0f(x)dx =F(b)−F(a), (2.31) where x0isanythird ﬁxed point. Using the notation F/prime(x)=dF/dx ,w em a y rewrite (2.28) as F/prime(x)=f(x), and so express (2.31) as integraldisplayb aF/prime(x)dx=F(b)−F(a)≡[F]b a. In contrast to diﬀerentiation, where repeated applications of the product rule and/or the chain rule will always give the required derivative, it is not alwayspossible to ﬁnd the integral of an arbitrary function. Indeed, in most real phys-ical problems exact integration cannot be performed and we have to revert tonumerical approximations. Despite this cautionary note, it is in fact possible tointegrate many simple functions and the following subsections introduce the most common types. Many of the techniques will be familiar to the reader and so are summarised by example. 2.2.3 Integration by inspection The simplest method of integrating a function is by inspection. Some of the more elementary functions have well-known integrals that should be remembered. Thereader will notice that these integrals are precisely the inverses of the derivatives 63 PRELIMINARY CALCULUS found near the end of subsection 2.1.1. A few are presented below, using the form given in (2.30). integraldisplay ad x=ax+c,integraldisplay axndx=axn+1 n+1+c, integraldisplay eaxdx=eax a+c,integraldisplaya xdx=alnx+c, integraldisplay acosbx dx =asinbx b+c,integraldisplay asinbx dx =−acosbx b+c, integraldisplay atanbx dx=−aln(cos bx) b+c,integraldisplay acosbxsinnbx dx=asinn+1bx b(n+1 )+c, integraldisplaya a2+x2dx=t a n−1parenleftBigx aparenrightBig +c,integraldisplay asinbxcosnbx dx =−acosn+1bx b(n+1 )+c, integraldisplay−1√ a2−x2dx=c o s−1parenleftBigx aparenrightBig +c,integraldisplay1√ a2−x2dx=s i n−1parenleftBigx aparenrightBig +c, where the integrals that depend on nare valid for all n/negationslash=−1a n dw h e r e aandb are constants. In the two ﬁnal results |x|≤a. 2.2.4 Integration of sinusoidal functions Integrals of the typeintegraltext sinnxd xandintegraltext cosnxd xmay be found by using trigono- metric expansions. Two methods are applicable, one for odd nand the other for even n. They are best illustrated by example.IEvaluate the integral I= R sin5xd x. Rewriting the integral as a product of sin xa n da ne v e np o w e ro fs i n x, and then using the relation sin2x=1−cos2xyields I= Z sin4xsinxd x = Z (1−cos2x)2sinxd x = Z (1−2cos2x+c o s4x)sinxd x = Z (sinx−2si nxcos2x+s i n xcos4x)dx =−cosx+2 3cos3x−1 5cos5x+c, where the integration has been carried out using the results of subsection 2.2.3. J 64 2.2 INTEGRATIONIEvaluate the integral I= R cos4xd x. Rewriting the integral as a power of cos2xand then using the double-angle formula cos2x=1 2(1 + cos2 x) yields I= Z (cos2x)2dx= Z /1+c o s2 x 2 /2 dx = Z 1 4(1 + 2 cos2 x+c o s22x)dx. Using the double-angle formula again we may write cos22x=1 2(1 + cos 4 x), and hence I= Z/1 4+1 2cos 2x+1 8(1 + cos 4 x) / dx =1 4x+1 4sin 2x+1 8x+1 32sin4x+c =3 8x+1 4sin 2x+1 32sin 4x+c. J 2.2.5 Logarithmic integration Integrals for which the integrand may be written as a fraction in which the numerator is the derivative of the denominator may be evaluated using integraldisplayf/prime(x) f(x)dx=l nf(x)+c. (2.32) This follows directly from the diﬀerentiation of a logarithm as a function of a function (see subsection 2.1.3).IEvaluate the integral I= Z6x2+2c o s x x3+s i n xdx. We note ﬁrst that the numerator can be factorised to give 2(3 x2+c o s x) ,a n dt h e nt h a t the quantity in brackets is the derivative of the denominator. Hence I=2 Z3x2+c o s x x3+s i n xdx=2l n ( x3+s i n x)+c. J 2.2.6 Integration using partial fractions The method of partial fractions was discussed at some length in section 1.4, but in essence consists of the manipulation of a fraction (here the integrand) in such a way that it can be written as the sum of two or more simpler fractions. Againwe illustrate the method by an example. 65 PRELIMINARY CALCULUSIEvaluate the integral I= Z1 x2+xdx. We note that the denominator factorises to give x(x+1 ) .H e n c e I= Z1 x(x+1 )dx. We now separate the fraction into two partial fractions and integrate directly: I= Z /1 x−1 x+1 / dx=l nx−ln(x+1 )+ c=l n /x x+1 / +c. J 2.2.7 Integration by substitution Sometimes it is possible to make a substitution of variables that turns a com- plicated integral into a simpler one, which can then be integrated by a standard method. There are many useful substitutions and knowing which to use is a matter of experience. We now present a few examples of particularly useful substitutions.IEvaluate the integral I= Z1√ 1−x2dx. Making the substitution x=s i n u, we note that dx=c o s ud u, and hence I= Z1√ 1−sin2ucosud u= Z1√ cos2ucosud u= Z du=u+c. Now substituting back for u, I=s i n−1x+c. This corresponds to one of the results given in subsection 2.2.3. J Another particular example of integration by substitution is aﬀorded by inte- grals of the form I=integraldisplay1 a+bcosxdx or I=integraldisplay1 a+bsinxdx. (2.33) In these cases making the substitution t=t a n ( x/2) yields integrals that can be solved more easily than the originals. Formulae expressing sin xand cos xin terms of twere derived in equations (1.32) and (1.33) [see p. 14], but before we can use them we must relate dxtodtas follows. 66 2.2 INTEGRATION Since dt dx=1 2sec2x 2=1 2parenleftBig 1+t a n2x 2parenrightBig =1+t2 2, the required relationship is dx=2 1+t2dt. (2.34)IEvaluate the integral I= Z2 1+3c o s xdx. Rewriting cos xin terms of tand using (2.34) yields I= Z2 1+3 / (1−t2)(1 + t2)−1 / /2 1+t2 / dt = Z2(1 + t2) 1+t2+3 ( 1−t2) /2 1+t2 / dt = Z2 2−t2dt= Z2 (√ 2−t)(√ 2+t)dt = Z1√ 2 /1√ 2−t+1√ 2+t / dt =−1√ 2ln(√ 2−t)+1√ 2ln(√ 2+t)+c =1√ 2ln /"√ 2+t a n( x/2)√ 2−tan (x/2) /# +c. J Integrals of a similar form to (2.33), but involving sin 2 x,c o s2 x,t a n 2 x,s i n2x, cos2xor tan2xinstead of cos xand sin x, should be evaluated by using the substitution t=t a n x.I nt h i sc a s e sinx=t√ 1+t2,cosx=1√ 1+t2and dx=dt 1+t2.(2.35) A ﬁnal example of the evaluation of integrals using substitution is the method of completing the square (cf. subsection 1.7.3). 67 PRELIMINARY CALCULUSIEvaluate the integral I= Z1 x2+4x+7dx. We can write the integral in the form I= Z1 (x+2 )2+3dx. Substituting y=x+ 2, we ﬁnd dy=dxand hence I= Z1 y2+3dy, Hence, by comparison with the table of standard integrals (see subsection 2.2.3) I=√ 3 3tan−1 /y√ 3 / +c=√ 3 3tan−1 /x+2√ 3 / +c. J 2.2.8 Integration by parts Integration by parts is the integration analogy of product diﬀerentiation. The principle is to break down a complicated function into two functions, at least one of which can be integrated by inspection. The method in fact relies on the resultfor the diﬀerentiation of a product. Recalling from (2.6) that d dx(uv)=udv dx+du dxv, where uandvare functions of x, we now integrate to ﬁnd uv=integraldisplay udv dxdx+integraldisplaydu dxvd x . Rearranging into the standard form for integration by parts gives integraldisplay udv dxdx=uv−integraldisplaydu dxvd x . (2.36) Integration by parts is often remembered for practical purposes in the form the integral of a product of two functions is equal to {the ﬁrst times the integral of the second}minus the integral of {the derivative of the ﬁrst times the integral of the second}. Here, uis ‘the ﬁrst’ and dv/dx is ‘the second’; clearly the integral v of ‘the second’ must be determinable by inspection.IEvaluate the integral I= R xsinxdx. In the notation given above, we identify xwith uand sin xwith dv/dx. Hence v=−cosx anddu/dx = 1 and so using (2.36) I=x(−cosx)− Z (1)(−cosx)dx=−xcosx+s i n x+c. J 68 2.2 INTEGRATION The separation of the functions is not always so apparent, as is illustrated by the following example.IEvaluate the integral I= R x3e−x2dx. Firstly we rewrite the integral as I= Z x2 / xe−x2 / dx. Now, using the notation given above, we identify x2with uandxe−x2with dv/dx. Hence v=−1 2e−x2anddu/dx =2x,s ot h a t I=−1 2x2e−x2− Z (−x)e−x2dx=−1 2x2e−x2−1 2e−x2+c. J A trick that is sometimes useful is to take ‘1’ as one factor of the product, as is illustrated by the following example.IEvaluate the integral I= R lnxd x. Firstly we rewrite the integral as I= Z (lnx)1dx. Now, using the notation above, we identify ln xwith uand 1 with dv/dx. Hence we have v=xanddu/dx =1/x,a n ds o I=( l n x)(x)− Z /1 x / xd x=xlnx−x+c. J It is sometimes necessary to integrate by parts more than once. In doing so, we may occasionally re-encounter the original integral I.I ns u c hc a s e sw ec a n obtain a linear algebraic equation for Ithat can be solved to obtain its value.IEvaluate the integral I= R eaxcosbx dx. Integrating by parts, taking eaxas the ﬁrst function, we ﬁnd I=eax /sinbx b / − Z aeax /sinbx b / dx, where, for convenience, we have omitted the constant of integration. Integrating by parts a second time, I=eax /sinbx b / −aeax /−cosbx b2 / + Z a2eax /−cosbx b2 / dx. Notice that the integral on the RHS is just −a2/b2times the original integral I. Thus I=eax /1 bsinbx+a b2cosbx / −a2 b2I. 69 PRELIMINARY CALCULUS Rearranging this expression to obtain Iexplicitly and including the constant of integration we ﬁnd I=eax a2+b2(bsinbx+acosbx)+c. (2.37) Another method of evaluating this integral, using the exponential of a complex number, is given in section 3.6. J 2.2.9 Reduction formulae Integration using reduction formulae is a process that involves ﬁrst evaluating a simple integral and then, in stages, using it to ﬁnd a more complicated integral.IUsing integration by parts, ﬁnd a relationship between InandIn−1where In= Z1 0(1−x3)ndx andnis any positive integer. Hence evaluate I2= R1 0(1−x3)2dx. Writing the integrand as a product and separating the integral into two we ﬁnd In= Z1 0(1−x3)(1−x3)n−1dx = Z1 0(1−x3)n−1dx− Z1 0x3(1−x3)n−1dx. The ﬁrst term on the RHS is clearly In−1and so, writing the integrand in the second term on the RHS as a product, In=In−1− Z1 0(x)x2(1−x3)n−1dx. Integrating by parts we ﬁnd In=In−1+ hx 3n(1−x3)n i1 0− Z1 01 3n(1−x3)ndx =In−1+0−1 3nIn, which on rearranging gives In=3n 3n+1In−1. We now have a relation connecting successive integrals. Hence, if we can evaluate I0,w e can ﬁnd I1,I2etc. Evaluating I0is trivial: I0= Z1 0(1−x3)0dx= Z1 0dx=[x]1 0=1. Hence I1=(3×1) (3×1) + 1×1=3 4,I 2=(3×2) (3×2) + 1×3 4=9 14. Although the ﬁrst few Incould be evaluated by direct multiplication, this becomes tedious for integrals containing higher values of n; these are therefore best evaluated using the reduction formula. J 70 2.2 INTEGRATION 2.2.10 Inﬁnite and improper integrals The deﬁnition of an integral given previously does not allow for cases in which either of the limits of integration is inﬁnite (an inﬁnite integral )o rf o rc a s e s in which f(x) is inﬁnite in some part of the range (an improper integral ), e.g. f(x)=( 2−x)−1/4near the point x= 2. Nevertheless, modiﬁcation of the deﬁnition of an integral gives inﬁnite and improper integrals each a meaning. In the case of an integral I=integraltextb af(x)dx, the inﬁnite integral, in which btends to∞, is deﬁned by I=integraldisplay∞ af(x)dx= lim b→∞integraldisplayb af(x)dx= lim b→∞F(b)−F(a). As previously, F(x) is the indeﬁnite integral of f(x) and lim b→∞F(b)m e a n st h e limit (or value) that F(b) approaches as b→∞;i ti se v a l u a t e d aftercalculating the integral. The formal concept of a limit will be introduced in chapter 4.IEvaluate the integral I= Z∞ 0x (x2+a2)2dx. Integrating, we ﬁnd F(x)=−1 2(x2+a2)−1+cand so I= lim b→∞ /−1 2(b2+a2) / − /−1 2a2 / =1 2a2. J For the case of improper integrals, we adopt the approach of excluding the unbounded range from the integral. For example, if the integrand f(x) is inﬁnite atx=c(say), a≤c≤bthen integraldisplayb af(x)dx= lim δ→0integraldisplayc−δ af(x)dx+ lim /epsilon1→0integraldisplayb c+/epsilon1f(x)dx.IEvaluate the integral I= R2 0(2−x)−1/4dx. Integrating directly, I= lim /epsilon1→0 / −4 3(2−x)3/4 /2−/epsilon1 0= lim /epsilon1→0 / −4 3/epsilon13/4 / +4 323/4= /;4 3 / 23/4. J 2.2.11 Integration in plane polar coordinates In plane polar coordinates ρ, φ, a curve is deﬁned by its distance ρfrom the origin as a function of the angle φbetween the line joining a point on the curve to the origin and the x-axis, i.e. ρ=ρ(φ). The area of an element is given by 71 PRELIMINARY CALCULUS dA ρ(φ)ρ(φ+dφ)ρd φ xy OBC Figure 2.9 Finding the area of a sector OBC deﬁned by the curve ρ(φ)a n d the radii OB,OC, at angles to the x-axis φ1,φ2respectively. dA=1 2ρ2dφ, as illustrated in ﬁgure 2.9, and hence the total area between two angles φ1andφ2is given by A=integraldisplayφ2 φ11 2ρ2dφ. (2.38) An immediate observation is that the area of a circle of radius ais given by A=integraldisplay2π 01 2a2dφ=bracketleftbig1 2a2φbracketrightbig2π 0=πa2.IThe equation in polar coordinates of an ellipse with semi-axes aandbis 1 ρ2=cos2φ a2+sin2φ b2. Find the area Aof the ellipse. Using (2.38) and symmetry, we have A=1 2 Z2π 0a2b2 b2cos2φ+a2sin2φdφ=2a2b2 Zπ/2 01 b2cos2φ+a2sin2φdφ. To evaluate this integral we write t=t a n φand use (2.35): A=2a2b2 Z∞ 01 b2+a2t2dt=2b2 Z∞ 01 (b/a)2+t2dt. Finally, from the list of standard integrals (see subsection 2.2.3), A=2b2 /1 (b/a)tan−1t (b/a) /∞ 0=2ab /π 2−0 / =πab. J 72 2.2 INTEGRATION 2.2.12 Integral inequalities Consider the functions f(x),φ1(x)a n d φ2(x) such that φ1(x)≤f(x)≤φ2(x)f o r allxin the range a≤x≤b. It immediately follows that integraldisplayb aφ1(x)dx≤integraldisplayb af(x)dx≤integraldisplayb aφ2(x)dx, (2.39) which gives us a way of estimating an integral that is diﬃcult to evaluate explicitly.IShow that the value of the integral I= Z1 01 (1 +x2+x3)1/2dx lies between 0.810and0.882. We note that for xin the range 0 ≤x≤1, 0≤x3≤x2. Hence (1 +x2)1/2≤(1 +x2+x3)1/2≤(1 + 2 x2)1/2, and so 1 (1 +x2)1/2≥1 (1 +x2+x3)1/2≥1 (1 + 2 x2)1/2. Consequently,Z1 01 (1 +x2)1/2dx≥ Z1 01 (1 +x2+x3)1/2dx≥ Z1 01 (1 + 2 x2)1/2dx, from which we obtainh ln(x+ p 1+x2) i1 0≥I≥ / 1√ 2ln / x+ q 1 2+x2 //1 0 0.8814≥I≥0.8105 0.882≥I≥0.810. In the last line the calculated values have been rounded to three signiﬁcant ﬁgures, one rounded up and the other rounded down so that the proved inequality cannot beunknowingly made invalid.J 2.2.13 Applications of integration Mean value of a function The mean value mof a function between two limits aandbis deﬁned by m=1 b−aintegraldisplayb af(x)dx. (2.40) The mean value may be thought of as the height of the rectangle that has the same area (over the same interval) as the area under the curve f(x). This is illustrated in ﬁgure 2.10. 73 PRELIMINARY CALCULUS mf(x) b x a Figure 2.10 The mean value mof a function.IFind the mean value mof the function f(x)=x2between the limits x=2andx=4. Using (2.40), m=1 4−2 Z4 2x2dx=1 2 /x3 3 /4 2=1 2 /43 3−23 3 / =28 3. J Finding the length of a curve Finding the area between a curve and certain straight lines provides one example of the use of integration. Another is in ﬁnding the length of a curve. If a curveis deﬁned by y=f(x) then the distance along the curve, ∆ s, that corresponds to small changes ∆ xand ∆ yinxandyis given by ∆s≈radicalbig (∆x)2+( ∆y)2; (2.41) this follows directly from Pythagoras’ theorem (see ﬁgure 2.11). Dividing (2.41) through by ∆ xand letting ∆ x→0w eo b t a i n † ds dx=radicalBigg 1+parenleftbiggdy dxparenrightbigg2 . Clearly the total length sof the curve between the points x=aandx=bis then given by integrating both sides of the equation: s=integraldisplayb aradicalBigg 1+parenleftbiggdy dxparenrightbigg2 dx. (2.42) †Instead of considering small changes ∆ xand ∆ yand letting these tend to zero, we could have derived (2.41) by considering inﬁnitesimal changes dxanddyfrom the start. After writing ( ds)2= (dx)2+(dy)2, (2.41) may be deduced by using the formal device of dividing through by dx. Although not mathematically rigorous, this method is often used and generally leads to the correct result. 74 2.2 INTEGRATION ∆x∆y∆sy=f(x) xf(x) Figure 2.11 The distance moved along a curve, ∆ s, corresponding to the small changes ∆ xand ∆ y. In plane polar coordinates, ds=radicalbig (dr)2+(rd φ)2⇒ s=integraldisplayr2 r1radicalBigg 1+r2parenleftbiggdφ drparenrightbigg2 dr. (2.43)IFind the length of the curve y=x3/2from x=0tox=2. Using (2.42) and noting that dy/dx =3 2√x, the length sof the curve is given by s= Z2 0 q 1+9 4xd x = h 2 3 /;4 9 //; 1+9 4x /3/2 i2 0=8 27 h/; 1+9 4x /3/2 i2 0 =8 27 h/;11 2 /3/2−1 i . J Surfaces of revolution Consider the surface Sformed by rotating the curve y=f(x) about the x-axis (see ﬁgure 2.12). The surface area of the ‘collar’ formed by rotating an element of the curve, ds, about the x- a x i si s2 πy ds, and hence the total surface area is S=integraldisplayb a2πy ds. Since ( ds)2=(dx)2+(dy)2from (2.41), the total surface area between the planes x=aandx=bis S=integraldisplayb a2πyradicalBigg 1+parenleftbiggdy dxparenrightbigg2 dx. (2.44) 75 PRELIMINARY CALCULUS Sy bVf(x) dx x ads Figure 2.12 The surface and volume of revolution for the curve y=f(x).IFind the surface area of a cone formed by rotating about the x-axis the line y=2x between x=0andx=h. Using (2.44), the surface area is given by S= Zh 0(2π)2x s 1+ /d dx(2x) /2 dx = Zh 04πx /; 1+22 /1/2dx= Zh 04√ 5πxdx = h 2√ 5πx2 ih 0=2√ 5π(h2−0) = 2√ 5πh2. J We note that a surface of revolution may also be formed by rotating a line about the y-axis. In this case the surface area between y=aandy=bis S=integraldisplayb a2πxradicalBigg 1+parenleftbiggdx dyparenrightbigg2 dy. (2.45) Volumes of revolution The volume Venclosed by rotating the curve y=f(x) about the x-axis can also be found (see ﬁgure 2.12). The volume of the disc between xandx+dxis given bydV=πy2dx.Hence the total volume between x=aandx=bis V=integraldisplayb aπy2dx. (2.46) 76 2.3 EXERCISESIFind the volume of a cone enclosed by the surface formed by rotating about the x-axis the line y=2xbetween x=0andx=h. Using (2.46), the volume is given by V= Zh 0π(2x)2dx= Zh 04πx2dx = /4 3πx3 /h 0=4 3π(h3−0) =4 3πh3. J As before, it is also possible to form a volume of revolution by rotating a curve about the y-axis. In this case the volume enclosed between y=aandy=bis V=integraldisplayb aπx2dy. (2.47) 2.3 Exercises 2.1 Obtain the following derivatives from ﬁrst principles: (a) the ﬁrst derivative of 3 x+4 ; (b) the ﬁrst, second and third derivatives of x2+x; (c) the ﬁrst derivative of sin x. 2.2 Find from ﬁrst principles the ﬁrst derivative of ( x+3)2and compare your answer with that obtained using the chain rule. 2.3 Find the ﬁrst derivatives of (a)x2expx,( b )2s i n xcosx,( c )s i n2 x,( d )xsinax, (e) (exp ax)(sinax)tan−1ax,( f )l n ( xa+x−a), (g) ln( ax+a−x), (h) xx. 2.4 Find the ﬁrst derivatives of (a)x/(a+x)2,( b ) x/(1−x)1/2,( c )t a n x,a ss i n x/cosx, (d) (3 x2+2x+1 )/(8x2−4x+2 ) . 2.5 Use result (2.12) to ﬁnd the ﬁrst derivatives of (a) (2 x+3 )−3,( b )s e c2x, (c) cosech33x,( d )1 /lnx,( e )1 /[sin−1(x/a)]. 2.6 Show that the function y(x)=e x p (−|x|) deﬁned by y(x)= /8/>/</>/:expx forx<0, 1f o r x=0, exp(−x)f o r x>0, isnotdiﬀerentiable at x= 0. Consider the limiting process for both ∆ x>0a n d ∆x<0. 2.7 Find dy/dx ifx=(t−2)/(t+2 )a n d y=2t/(t+1 )f o r−∞<t<∞. Show that it is always non-negative, and make use of this result in sketching the curve of y as a function of x. 2.8 If 2 y+s i n y+5= x4+4x3+2π, show that dy/dx =1 6w h e n x=1 . 2.9 Find the second derivative of y(x)=c o s [ ( π/2)−ax]. Now set a=1a n dv e r i f y that the result is the same as that obtained by ﬁrst setting a= 1 and simplifying y(x) before diﬀerentiating. 77 PRELIMINARY CALCULUS 2.10 The function y(x) is deﬁned by y(x)=( 1+ xm)n. (a) Use the chain rule to show that the ﬁrst derivative of yisnmxm−1(1 +xm)n−1. (b) The binomial expansion (see section 1.5) of (1 + z)nis (1 +z)n=1+ nz+n(n−1) 2!z2+···+n(n−1)···(n−r+1 ) r!zr+···. Keeping only the terms of zeroth and ﬁrst order in dx, apply this result twice to derive result (a) from ﬁrst principles. (c) Expand yin a series of powers of xbefore diﬀerentiating term by term. Show that the result is the series obtained by expanding the answer givenfordy/dx in (a). 2.11 Show by diﬀerentiation and substi tution that the diﬀerential equation 4x 2d2y dx2−4xdy dx+( 4x2+3 )y=0 has a solution of the form y(x)=xnsinx, and ﬁnd the value of n. 2.12 Find the positions and natures of the stationary points of the following functions: (a)x3−3x+3 ;( b ) x3−3x2+3x;( c )x3+3x+3 ; (d) sin axwith a/negationslash=0 ;( e ) x5+x3;( f )x5−x3. 2.13 Show that the lowest value taken by the function 3 x4+4x3−12x2+6i s−26. 2.14 By ﬁnding their stationary points and examining their general forms, determine the range of values that each of the following functions y(x) can take. In each case make a sketch-graph incorporating the features you have identiﬁed. (a)y(x)=(x−1)/(x2+2x+6 ) . (b)y(x)=1 /(4 + 3 x−x2). (c)y(x)=( 8s i n x)/(15 + 8 tan2x). 2.15 Show that y(x)=xa2xexpx2has no stationary points other than x=0 ,i f exp(−√ 2)<a< exp(√ 2). 2.16 The curve 4 y3=a2(x+3y) can be parameterised as x=acos 3θ,y=acosθ. (a) Obtain expressions for dy/dx (i) by implicit diﬀerentiation and (ii) in param- eterised form. Verify that they are equivalent. (b) Show that the only point of inﬂection occurs at the origin. Is it a stationary point of inﬂection? (c) Use the information gained in (a) and (b) to sketch the curve, paying particular attention to its shape near the points ( −a, a/2) and ( a,−a/2) and to its slope at the ‘end points’ ( a, a)a n d(−a,−a). 2.17 The parametric equations for the motion of a charged particle released from rest in electric and magnetic ﬁelds at right angles to each other take the forms x=a(θ−sinθ),y =a(1−cosθ). Show that the tangent to the curve has slope cot( θ/2). Use this result at a few calculated values of xandyto sketch the form of the particle’s trajectory. 2.18 Show that the maximum curvature on the catenary y(x)=acosh( x/a)i s1/a.Y o u will need some of the results about hyperbolic functions stated in subsection 3.7.6. 2.19 The curve whose equation is x2/3+y2/3=a2/3for positive xandyand which is completed by its symmetric reﬂections in both axes is known as an astroid.Sketch it and show that its radius of curvature in the ﬁrst quadrant is 3( axy) 1/3. 2.20 A two-dimensional coordinate system useful for orbit problems is the tangential- polar coordinate system (ﬁgure 2.13). In this system a curve is deﬁned by r,t h e distance from a ﬁxed point Oto a general point Pof the curve, and p,t h e 78 2.3 EXERCISES OC PQρ ρ rr+∆rc pp+∆p Figure 2.13 The coordinate system described in exercise 2.20. perpendicular distance from Oto the tangent to the curve at P. By proceeding as indicated below, show that the radius of curvature at Pc a nb ew r i t t e ni nt h e form ρ=r dr/dp . Consider two neighbouring points PandQon the curve. The normals to the curve through those points meet at C, with (in the limit Q→P)CP=CQ=ρ. Apply the cosine rule to triangles OPC andOQC to obtain two expressions for c2,o n ei nt e r m so f randpand the other in terms of r+∆randp+∆p.B y equating them and letting Q→Pdeduce the stated result. 2.21 Use Leibniz’ theorem to ﬁnd (a) the second derivative of cos xsin2x, (b) the third derivative of sin xlnx, (c) the fourth derivative of (2 x3+3x2+x+2 )e x p2 x. 2.22 If y=e x p (−x2), show that dy/dx =−2xyand hence, by applying Leibniz’ theorem, prove that for n≥1 y(n+1)+2xy(n)+2ny(n−1)=0. 2.23 (a) By considering its properties near x= 1, show that f(x)=5 x4−11x3+ 26x2−44x+ 24 takes negative values for some range of x. (b) Show that f(x)=t a n x−xcannot be negative for 0 ≤x≤π/2, and deduce thatg(x)=x−1sinxdecreases monotonically in the same range. 2.24 Determine what can be learned from applying Rolle’s theorem to the following functions f(x): (a) ex;( b ) x2+6x;( c )2 x2+3x+1 ; ( d ) 2 x2+3x+2 ; ( e ) 2x3−21x2+6 0x+k.( f )I f k=−45 in (e), show that x=3i so n er o o to f f(x) = 0, ﬁnd the other roots, and verify that the conclusions from (e) are satisﬁed. 2.25 By applying Rolle’s theorem to xnsinnx,w h e r e nis an arbitrary positive integer, show that tan nx+x=0h a sas o l u t i o n α1with 0 <α 1<π / n . Apply the theorem a second time to obtain the non sensical result that there is a real α2in 0<α2<π / n , such that cos2(nα2)=−n2. Explain why this incorrect result arises. 2.26 Use the mean value theorem to establish bounds (a) for−ln(1−y), by considering ln xin the range 0 <1−y<x< 1, (b) for ey−1, by considering ex−1 in the range 0 <x<y . 79 PRELIMINARY CALCULUS 2.27 For the function y(x)=x2exp(−x) obtain a simple relationship between yand dy/dx and then, by applying Leibniz’ theorem, prove that xy(n+1)+(n+x−2)y(n)+ny(n−1)=0. 2.28 Use Rolle’s theorem to deduce that if the equation f(x) = 0 has a repeated root x1then x1is also a root of the equation f/prime(x)=0 . (a) Apply this result to the ‘standard’ quadratic equation ax2+bx+c=0 ,t o show that the condition for equal roots is b2=4ac. (b) Find all the roots of f(x)=x3+4x2−3x−18 = 0, given that one of them is a repeated root. (c) The equation f(x)=x4+4x3+7x2+6x+2 = 0 has a repeated integer root. How many real roots does it have altogether? 2.29 Show that the curve x3+y3−12x−8y−16 = 0 touches the x-axis. 2.30 Find the following indeﬁnite integrals: (a) R (4 +x2)−1dx;( b ) R (8 + 2 x−x2)−1/2dxfor 2≤x≤4; (c) R (1 + sin θ)−1dθ;( d ) R (x√ 1−x)−1dxfor 0 <x≤1. 2.31 Find the indeﬁnite integrals Jof the following ratios of polynomials: (a) ( x+3 )/(x2+x−2); (b) ( x3+5x2+8x+ 12) /(2x2+1 0x+ 12); (c) (3 x2+2 0x+ 28) /(x2+6x+9 ) ; (d)x3/(a8+x8). 2.32 Express x2(ax+b)−1as the sum of powers of xand another integrable term, and hence evaluateZb/a 0x2 ax+bdx. 2.33 Find the integral Jof (ax2+bx+c)−1,w i t h a/negationslash= 0, distinguishing between the cases (i) b2>4ac, (ii)b2<4ac, and (iii) b2=4ac. 2.34 Use logarithmic integration to ﬁnd the indeﬁnite integrals Jof the following: (a) sin2 x/(1 + 4 sin2x); (b)ex/(ex−e−x); (c) (1 + xlnx)/(xlnx); (d) [ x(xn+an)]−1. 2.35 Find the derivative of f(x)=( 1+s i n x)/cosxand hence determine the indeﬁnite integral Jof sec x. 2.36 Find the indeﬁnite integrals Jof the following functions involving sinusoids: (a) cos5x−cos3x; (b) (1−cosx)/(1 + cos x); (c) cos xsinx/(1 + cos x); (d) sec2x/(1−tan2x). 2.37 By making the substitution x=acos2θ+bsin2θ, evaluate the de ﬁnite integrals Jbetween limits aandb(>a) of the following functions: (a) [( x−a)(b−x)]−1/2; (b) [( x−a)(b−x)]1/2; (c) [( x−a)/(b−x)]1/2. 80 2.3 EXERCISES 2.38 Determine whether the following integrals exist and, where they do, evaluate them: (a) Z∞ 0exp(−λx)dx;( b ) Z∞ −∞x (x2+a2)2dx; (c) Z∞ 11 x+1dx;( d ) Z1 01 x2dx; (e) Zπ/2 0cotθd θ;( f ) Z1 0x (1−x2)1/2dx. 2.39 Use integration by parts to evaluate the following: (a) Zy 0x2sinxd x;( b ) Zy 1xlnxd x; (c) Zy 0sin−1xd x;( d ) Zy 1ln(a2+x2)/x2dx. 2.40 Show, by each of the following methods, that the indeﬁnite integral Jofx3/(x+ 1)1/2is J=2 35(5x3−6x2+8x−16)(x+1 )1/2+c. (a) by using repeated integration by parts. (b) by setting x+1= u2and determining dJ/du as (dJ/dx )(dx/du ). 2.41 The gamma function Γ( n) is deﬁned for all n>−1b y Γ(n+1 )= Z∞ 0xne−xdx. Find a recurrence relation connecting Γ( n+1 )a n dΓ ( n). (a) Deduce (i) the value of Γ( n+1)when nis a non-negative integer and (ii) the value of Γ /;7 2 / ,g i v e nt h a tΓ /;1 2 / =√π. (b) Now, taking factorial mforanymto be deﬁned by m!=Γ ( m+ 1), evaluate/; −3 2 / !. 2.42 Deﬁne J(m, n), for non-negative integers mandn, by the integral J(m, n)= Zπ/2 0cosmθsinnθd θ . (a) Evaluate J(0,0),J(0,1),J(1,0),J(1,1),J(m,1),J(1,n). (b) Using integration by parts prove that, for mandnboth >0, J(m, n)=m−1 m+nJ(m−2,n)a n d J(m, n)=n−1 m+nJ(m, n−2). (c) Evaluate (i) J(5,3), (ii) J(6,5), (iii) J(4,8). 2.43 By integrating by parts twice, prove that Inas deﬁned in the ﬁrst equality below for positive integers nhas the value given in the second equality. In= Zπ/2 0sinnθcosθd θ=n−sin(nπ/2) n2−1. 2.44 Evaluate the following deﬁnite integrals: (a) R∞ 0xe−xdx;( b ) R1 0 / (x3+1 )/(x4+4x+1 ) / dx; (c) Rπ/2 0[a+(a−1)cos θ]−1dθwith a>1 2;( d ) R∞ −∞(x2+6x+ 18)−1dx. 81 PRELIMINARY CALCULUS 2.45 If Jris the integralZ∞ 0xrexp(−x2)dx show that (a)J2r+1=(r!)/2, (b)J2r=2−r(2r−1)(2r−3)···(5)(3)(1) J0. 2.46 (a) Find positive constants a,bsuch that ax≤sinx≤bxfor 0≤x≤π/2. Use this inequality to ﬁnd (to two signiﬁcant ﬁgures) upper and lower boundsfor the integral I= Zπ/2 0(1 + sin x)1/2dx. (b) Use the substitution t=t a n ( x/2) to evaluate Iexactly. 2.47 By noting that for 0 ≤η≤1,η1/2≥η3/4≥η, prove that 2 3≤1 a5/2 Za 0(a2−x2)3/4dx≤π 4. 2.48 Show that the total length of the astroid x2/3+y2/3=a2/3,w h i c hc a nb e parameterised as x=acos3θ,y=asin3θ,i s6a. 2.49 By noting that sinh x<1 2ex<coshx,a n dt h a t1+ z2<(1 +z)2forz>0, show that for x>0, the length Lof the curve y=1 2exmeasured from the origin satisﬁes the inequalities sinh x<L<x +s i n h x. 2.50 The equation of a cardioid in plane polar coordinates is ρ=a(1−sinφ). Sketch the curve and ﬁnd (i) its area, (ii) its total length, (iii) the surface area of the solid formed by rotating the cardioi d about its axis of symmetry and (iv) the volume of the same solid. 2.4 Hints and answers 2.1 (a) 3; (b) 2 x+ 1, 2, 0; (c) cos x. 2.2 2 x+6 . 2.3 (a) ( x2+2x)exp x;( b )2 ( c o s2x−sin2x)=2 c o s 2 x;( c )2 c o s 2 x;( d )s i n ax+ axcosax; (e) (aexpax)[(sin ax+c o s ax)tan−1ax+( s i n ax)(1 + a2x2)−1]; (f) [a(xa−x−a)]/[x(xa+x−a)]; (g) [( ax−a−x)lna]/(ax+a−x); (h) (1 + ln x)xx. 2.4 (a) ( a−x)(a+x)−3;( b )( 1−x/2)(1−x)−3/2;( c )s e c2x; (d) (−7x2−x+ 2)(4 x2−2x+1 )−2. 2.5 (a) −6(2x+3 )−4;( b )2 s e c2xtanx;( c )−9cosec h33xcoth3 x; (d)−x−1(lnx)−2;( e )−(a2−x2)−1/2[sin−1(x/a)]−2. 2.6 The two limits are −1( f o r∆ x>0) and +1 (for ∆ x<0) and are not equal. 2.7 ( t+2 )2/[2(t+1 )2]. 2.8 y=πatx=1 . 2.9−sinxin both cases. 2.10 (b) Write 1 + ( x+∆x)mas 1 + xm(1 + ∆ x/x)m; (c) in the general terms of the two series, the indices randsare related by r=s±1. 2.11 The required conditions are 8 n−4=0a n d4 n2−8n+ 3 = 0; both are satisﬁed byn=1 2. 82 2.4 HINTS AND ANSWERS −15−10−5 5 10 15 −0.4−0.20.20.4 (a)−3−2−11 2 3456 −0.8−0.40.40.8 (b) 0 −0.20.2 π 2π 3π (c) Figure 2.14 The solutions to exercise 2.14. 2.12 (a) Minimum at x= 1, maximum at x=−1; (b) inﬂection at x=1 ;( c )n o stationary points; (d) x=(n+1 2)π/a, maximum for neven, minimum for nodd; (e) inﬂection at x= 0; (f) inﬂection at x= 0, maximum at x=−(3 5)1/2, minimum at (3 5)1/2. 2.13−26 at x=−2; other stationary values are 6 at x= 0 and 1 at x=1 . 2.14 See ﬁgure 2.14(a)–(c). (a)y(1) = 0; no inﬁnities; minimum y(−2) =−1 2, maximum y(4) =1 10;−1 2≤ y≤1 10. (b) No zeroes; y(−1) =±∞,y(4) =±∞; minimum y(3 2)=4 25;y<0o ry≥4 25. (c) Periodic with period 2 π.W i t h i n0 ≤x≤π, symmetry about x=π/2. Within 0≤x≤2π, antisymmetry about x=π; zeroes at x=nπand x=( 2m+1)π/2; no inﬁnities; other stationary points at x=c o s−1(±2/√ 7); |y|≤8/(7√ 21). 2.15 Use logarithmic diﬀerentiation. Set dy/dx = 0, obtaining 2 x2+2xlna+1=0 . 2.16 (a) (i) a2/(12y2−3a2), (ii) (12 cos2θ−3)−1.( b )N o , dy/dx =−1/3. (c) Vertical tangents when y=±a/2;dy/dx =1/9a ty=±a. 2.17 See ﬁgure 2.15.2.18 First show that ρ=y 2/a. 2.19dy dx=− /y x /1/3 ;d2y dx2=a2/3 3x4/3y1/3. 2.20 For example, OC2=ρ2+r2−2pρ, where use has been made of the fact that rcosOPC =p. 2.21 (a) 2(2 −9cos2x)sinx;( b )( 2 x−3−3x−1)sinx−(3x−2+l nx)c osx;( c )8 ( 4 x3+ 30x2+6 2x+ 38)exp2 x. 83 PRELIMINARY CALCULUS πa 2πa2a xy Figure 2.15 The solution to exercise 2.17. 2.23 (a) f(1) = 0 whilst f/prime(1)/negationslash=0a n ds o f(x)m u s tb en e g a t i v ei ns o m er e g i o nw i t h x= 1 as an endpoint. (b)f/prime(x)=t a n2x>0a n d f(0) = 0; g/prime(x)=(−cosx)(tan x−x)/x2,which is never positive in the range. 2.24 (a) Any two consecutive roots of ex= 0 have another root of ex= 0 lying between them; thus there is at most one root of ex= 0 (formally −∞). (b) The root of 2x+ 6 = 0 lies in the range −6<x< 0. (c) Any roots of f(x) = 0 (actually −1 and−1 2) lie on either side of x=−3 4. (d) As in (c), but there are no real roots. More generally, if there are two values of xthat give 2 x2+3x+kequal values then they lie one on each side of x=−3 4.( e )f/prime(x)=6 x2−42x+60=0hasroots 2 and 5. Therefore, if f(x) = 0 has three real roots αithen α1<2<α2<5<α3. (f) The other roots are1 4(15±√ 105). 2.25 The false result arises because tan nxis not diﬀerentiable at x=π/(2n), which lies in the range 0 <x<π / n , and so the conditions for applying Rolle’s theorem are not satisﬁed. 2.26 (a) y<−ln(1−y)<y /(1−y); (b) y<ey−1<y ey. 2.27 xd y/ d x =( 2−x)y. 2.28 (a) Show that x=−b/(2a). (b) Possible repeated roots are −3a n d1 3;o n l y−3s a t i s ﬁ e s f(x)=0 .F a c t o r i s e f(x)a s( x+3 )2(x−b), giving b=2a n d x= 2 as the third root. (c)f/prime(x) = 0 has the integer solution x=−1 (by inspection); f(x) factorises as the product ( x+1)2(x2+2x+2) and hence f(x) = 0 has only two (coincident) real roots. 2.29 By implicit diﬀerentiation, y/prime(x)=( 3 x2−12)/(8−3y2), giving y/prime(±2) = 0. Since y(2) = 4 and y(−2) = 0, the curve touches the x-axis at the point ( −2,0). 2.30 (a) [tan−1(x/2)]/2; (b) sin−1[(x−1)/3]; (c)−2[1 + tan( θ/2)]−1; (d) put y= (1−x)1/2,l n / [1−(1−x)1/2]/[1 + (1−x)1/2] / . 2.31 (a) Express in partial fractions; J=1 3ln[(x−1)4/(x+2 ) ]+ c. (b) Divide the numerator by the denominator and express the remainder in partial fractions; J=x2/4+4l n ( x+2 )−3l n(x+3 )+ c. (c) After division of the numerator by the denominator the remainder can be expressed as 2( x+3 )−1−5(x+3 )−2;J=3x+2l n ( x+3 )+5 ( x+3 )−1+c. (d) Set x4=u;J=( 4a4)−1tan−1(x4/a4)+c. 2.32 Express as ( x/a)−(b/a2)+(b/a)2(ax+b)−1;(b2/a3)(ln2−1 2). 84 2.4 HINTS AND ANSWERS 2.33 Writing b2−4acas ∆2>0, or 4 ac−b2as ∆/prime2>0: (i) ∆−1ln[(2ax+b−∆)/(2ax+b+∆ ) ]+ k; (ii) 2∆/prime−1tan−1[(2ax+b)/∆/prime]+k; (iii)−2(2ax+b)−1+k. 2.34 (a) J=1 4ln(1 + 4sin2x)+c. (b) Multiply numerator and denominator by ex;J=1 2ln(e2x−1) +c. (c) First divide the numerator by the denominator. J=x+l n ( l n x)+c. (d) Multiply numerator and denominator by xn−1,a n dt h e ns e t xn=u. J=(nan)−1ln[xn/(xn+an)] +c. 2.35 f/prime(x)=( 1+s i n x)/cos2x=f(x)se cx;J=l n ( f(x)) +c=l n ( s e c x+t a n x)+c. 2.36 (a) Show cos4x−cos2x=s i n4x−sin2x;J=1 5sin5x−1 3sin3x+c. (b) Either write the numerator and denominator in terms of sinusoidal functions ofx/2 or make the substitution t=t a n ( x/2);J=2t a n ( x/2)−x+c. (c) Substitute t=t a n ( x/2);J=2l n ( c o s ( x/2))−2c os2(x/2) +c. (d) Either set tan x=uor show that the integrand is sec2 xand use the result of exercise 2.35. J=1 2ln(sec2 x+t a n2 x)+c=1 2ln[(1 + tan x)/(1−tanx)] +c. 2.37 (a) π;( b ) π(b−a)2/8; (c) π(b−a)/2. 2.38 (a) Yes, for λ>0, value λ−1; (b) yes, value 0; (c) no, ln(1 + R)→∞asR→∞; (d) no, /epsilon1−1→∞as/epsilon1→0; (e) no, ln(sin θ)→−∞ asθ→0; (f) yes, value 1. 2.39 (a) (2 −y2)c osy+2ysiny−2; (b) [( y2lny)/2] + [(1−y2)/4]; (c)ysin−1y+( 1−y2)1/2−1; (d) ln( a2+1 )−(1/y)ln(a2+y2)+( 2 /a)[tan−1(y/a)−tan−1(1/a)]. 2.40 (b) dJ/du =2 (u2−1)3. 2.41 Γ( n+1 )= nΓ(n); (a) (i) n!, (ii) 15√π/8; (b)−2√π. 2.42 (a) π/2, 1, 1, 1 /2, 1/(m+1 ) ,1 /(n+1 ) . (b) Write the initial integrand as cosm−1θsinnθcosθ, and later rewrite sinn+2θ as sinnθ(1−cos2θ). (c) (i) 1 /24, (ii) 8 /693, (iii) 7 π/2048. 2.44 (a) 1; (b) (ln6) /4; (c) / 2t a n−1[(2a−1)−1/2] / / (2a−1)1/2;( d ) π/3. 2.46 (a) a=2/π,b=1 ;2 3[(1+π 2)3/2−1]>I>π 3(23/2−1), 2.08>I> 1.91; (b) I=2 . 2.47 Set η=1−(x/a)2. 2.49 L= Rx 0 /; 1+1 4exp 2 x /1/2dx. 2.50 Note that to avoid any possible double counting, integrals should be taken from π/2t o3 π/2 and symmetry used for scaling up. The integrands (and inﬁnitesimals) should be as indicated, with ρ/primedenoting dρ/dφ : (i) (ρ2/2)dφ,3πa2/2; (ii) 2( ρ/prime2+ρ2)1/2dφ,8a; (iii) 2 πρcosφ(ρ/prime2+ρ2)1/2dφ,3 2πa2/5; (iv)πρ2cos2φd(ρsinφ), 8πa3/3. 85 3 Complex numbers and hyperbolic functions This chapter is concerned with the representation and manipulation of complex numbers. Complex numbers pervade this book, underscoring their wide appli-cation in the mathematics of the physical sciences. The application of complexnumbers to the description of physical systems is left until later chapters and only the basic tools are presented here. 3.1 The need for complex numbers Although complex numbers occur in many branches of mathematics, they arise most directly out of solving polynomial equations. We examine a speciﬁc quadratic equation as an example. Consider the quadratic equation z 2−4z+5=0 . (3.1) Equation (3.1) has two solutions, z1andz2, such that (z−z1)(z−z2)=0 . (3.2) Using the familiar formula for the roots of a quadratic equation, (1.4), the solutions z1andz2, written in brief as z1,2,a r e z1,2=4±radicalbig (−4)2−4(1×5) 2 =2±√ −4 2. (3.3) Both solutions contain the square root of a negative number. However, it is not true to say that there are no solutions to the quadratic equation. The fundamental theorem of algebra states that a quadratic equation will always have two solutions and these are in fact given by (3.3). The second term on the RHS of (3.3) iscalled an imaginary term since it contains the square root of a negative number; 86 3.1 THE NEED FOR COMPLEX NUMBERS 11 22 33 445 zf(z) Figure 3.1 The function f(z)=z2−4z+5 . the ﬁrst term is called a realterm. The full solution is the sum of a real term and an imaginary term and is called a complex number . A plot of the function f(z)=z2−4z+ 5 is shown in ﬁgure 3.1. It will be seen that the plot does not intersect the z-axis, corresponding to the fact that the equation f(z)=0h a sn o purely real solutions. The choice of the symbol zfor the quadratic variable was not arbitrary; the conventional representation of a complex number is z,w h e r e zis the sum of a real part xanditimes an imaginary part y,i . e . z=x+iy, where iis used to denote the square root of −1. The real part xand the imaginary part yare usually denoted by Re zand Im zrespectively. We note at this point that some physical scientists, engineers in particular, use jinstead of i. However, for consistency, we will use ithroughout this book. I no u rp a r t i c u l a re x a m p l e ,√ −4=2√ −1=2 i, and hence the two solutions of (3.1) are z1,2=2±2i 2=2±i. Thus here x=2a n d y=±1. For compactness a complex number is sometimes written in the form z=(x, y), where the components of zmay be thought of as coordinates in an xy-plot. Such a plot is called an Argand diagram and is a common representation of complex numbers; an example is shown in ﬁgure 3.2. 87 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS RezImz z=x+iy xy Figure 3.2 The Argand diagram. Our particular example of a quadratic equation may be generalised readily to polynomials whose highest power (degree) is greater than 2, e.g. cubic equations(degree 3), quartic equations (degree 4) and so on. For a general polynomial f(z), of degree n, the fundamental theorem of algebra states that the equation f(z)=0 will have exactly nsolutions. We will examine cases of higher-degree equations in subsection 3.4.3. The remainder of this chapter deals with: the algebra and manipulation of complex numbers; their polar representation, which has advantages in manycircumstances; complex exponentials and logarithms; the use of complex numbersin ﬁnding the roots of polynomial equations; and hyperbolic functions. 3.2 Manipulation of complex numbers This section considers basic complex number manipulation. Some analogy may be drawn with vector manipulation (see chapter 7) but this section stands aloneas an introduction. 3.2.1 Addition and subtraction The addition of two complex numbers, z 1and z2, in general gives another complex number. The real components and the imaginary components are added separately and in a like manner to the familiar addition of real numbers: z1+z2=(x1+iy1)+(x2+iy2)=(x1+x2)+i(y1+y2), 88 3.2 MANIPULATION OF COMPLEX NUMBERS RezImz z1z2z1+z2 Figure 3.3 The addition of two complex numbers. or in component notation z1+z2=(x1,y1)+(x2,y2)=(x1+x2,y1+y2). The Argand representation of the addition of two complex numbers is shown in ﬁgure 3.3. By straightforward application of the commutativity and associativity of the real and imaginary parts separately, we can show that the addition of complexnumbers is itself commutative and associative, i.e. z 1+z2=z2+z1, z1+(z2+z3)=(z1+z2)+z3. Thus it is immaterial in what order complex numbers are added.ISum the complex numbers 1+2 i,3−4i,−2+i. Summing the real terms we obtain 1+3−2=2 , and summing the imaginary terms we obtain 2i−4i+i=−i. Hence (1 + 2 i)+( 3−4i)+(−2+i)=2−i. J The subtraction of complex numbers is very similar to their addition. As in the case of real numbers, if two identical complex numbers are subtracted then theresult is zero. 89 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS RezImz |z| xy argz Figure 3.4 The modulus and argument of a complex number. 3.2.2 Modulus and argument The modulus of the complex number zis denoted by |z|and is deﬁned as |z|=radicalbig x2+y2. (3.4) Hence the modulus of the complex number is the distance of the corresponding point from the origin in the Argand diagram, as may be seen in ﬁgure 3.4. The argument of the complex number zis denoted by arg zand is deﬁned as argz=t a n−1parenleftBigy xparenrightBig . (3.5) Thus arg zis the angle that the line joining the origin to zon the Argand diagram makes with the positive x-axis. The anticlockwise direction is taken to be positive by convention. The angle arg zis shown in ﬁgure 3.4. Account must be taken of the signs of xandyindividually in determining in which quadrant arg zlies. Thus, for example, if xandyare both negative then arg zlies in the range −π<argz<−π/2 rather than in the ﬁrst quadrant (0 <argz<π / 2), though both cases give the same value for the ratio of ytox.IFind the modulus and the argument of the complex number z=2−3i. Using (3.4), the modulus is given by |z|= p 22+(−3)2=√ 13. Using (3.5), the argument is given by argz=t a n−1 /; −3 2 / . The two angles whose tangents equal −1.5a r e−0.9828rad and 2 .1588rad. Since x=2a n d y=−3,zclearly lies in the fourth quadrant; therefore arg z=−0.9828 is the appropriate answer. J 90 3.2 MANIPULATION OF COMPLEX NUMBERS 3.2.3 Multiplication Complex numbers may be multiplied together and in general give a complex number as the result. The product of two complex numbers z1andz2is found by multiplying them out in full and remembering that i2=−1, i.e. z1z2=(x1+iy1)(x2+iy2) =x1x2+ix1y2+iy1x2+i2y1y2 =(x1x2−y1y2)+i(x1y2+y1x2). (3.6)IMultiply the complex numbers z1=3+2 iandz2=−1−4i. By direct multiplication we ﬁnd z1z2=( 3+2 i)(−1−4i) =−3−2i−12i−8i2 =5−14i. J (3.7) The multiplication of complex numbers is both commutative and associative, i.e. z1z2=z2z1, (3.8) (z1z2)z3=z1(z2z3). (3.9) The product of two complex numbers also has the simple properties |z1z2|=|z1||z2|, (3.10) arg(z1z2)=a r g z1+a r g z2. (3.11) These relations are derived in subsection 3.3.1.IVerify that (3.10) holds for the product of z1=3+2 iandz2=−1−4i. From (3.7) |z1z2|=|5−14i|= p 52+(−14)2=√ 221. We also ﬁnd |z1|= p 32+22=√ 13, |z2|= p (−1)2+(−4)2=√ 17, and hence |z1||z2|=√ 13√ 17 =√ 221 =|z1z2|. J We now examine the eﬀect on a complex number zof multiplying it by ±1 and±i. These four multipliers have modulus unity and we can see immediately from (3.10) that multiplying zby another complex number of unit modulus gives a product with the same modulus as z. We can also see from (3.11) that if 91 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS RezImz iz −izz −z Figure 3.5 Multiplication of a complex number by ±1a n d±i. we multiply zby a complex number, the argument of the product is the sum of the argument of zand the argument of the multiplier. Hence multiplying zby unity (which has argument zero) leaves zunchanged in both modulus and argument, i.e. zis completely unaltered by the operation. Multiplying by −1 (which has argument π) leads to rotation, through an angle π, of the line joining the origin to zin the Argand diagram. Similarly, multiplication by ior −ilead to corresponding rotations of π/2o r−π/2 respectively. This geometrical interpretation of multiplication is shown in ﬁgure 3.5.IUsing the geometrical interpretation of multiplication by i, ﬁnd the product i(1−i). The complex number 1 −ihas argument −π/4 and modulus√ 2. Thus, using (3.10) and (3.11), its product with ihas argument + π/4 and unchanged modulus√ 2. The complex number with modulus√ 2 and argument + π/4i s1+ iand so i(1−i)=1+ i, as is easily veriﬁed by direct multiplication. J The division of two complex numbers is similar to their multiplication but requires the notion of the complex conjugate (see the following subsection) andso discussion is postponed until subsection 3.2.5. 3.2.4 Complex conjugate Ifzhas the convenient form x+iythen the complex conjugate, denoted by z ∗, may be found simply by changing the sign of the imaginary part, i.e. if z=x+iy then z∗=x−iy. More generally, we may deﬁne the complex conjugate of zas the (complex) number having the same magnitude as zthat when multiplied by zleaves a real result, i.e. there is no imaginary component in the product. 92 3.2 MANIPULATION OF COMPLEX NUMBERS RezImz z=x+iy xy −yz∗=x−iy Figure 3.6 The complex conjugate as a mirror image in the real axis. I nt h ec a s ew h e r e zc a nb ew r i t t e ni nt h ef o r m x+iyit is easily veriﬁed, by direct multiplication of the components, that the product zz∗gives a real result: zz∗=(x+iy)(x−iy)=x2−ixy+ixy−i2y2=x2+y2=|z|2. Complex conjugation corresponds to a reﬂection of zin the real axis of the Argand diagram, as may be seen in ﬁgure 3.6.IFind the complex conjugate of z=a+2i+3ib. The complex number is written in the standard form z=a+i(2 + 3 b); then, replacing iby−i,w eo b t a i n z∗=a−i(2 + 3 b). J In some cases, however, it may not be simple to rearrange the expression for zinto the standard form x+iy. Nevertheless, given two complex numbers, z1 andz2, it is straightforward to show that the complex conjugate of their sum (or diﬀerence) is equal to the sum (or diﬀerence) of their complex conjugates, i.e.(z 1±z2)∗=z∗ 1±z∗ 2. Similarly, it may be shown that the complex conjugate of the product (or quotient) of z1andz2is equal to the product (or quotient) of their complex conjugates, i.e. ( z1z2)∗=z∗ 1z∗ 2and ( z1/z2)∗=z∗ 1/z∗ 2. Using these results, it can be deduced that, no matter how complicated the expression, its complex conjugate may always be found by replacing every iby −i. To apply this rule, however, we must always ensure that all complex parts are ﬁrst written out in full, so that no i’s are hidden. 93 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIFind the complex conjugate of the complex number z=w(3y+2ix)where w=x+5i. Although we do not discuss complex powers until section 3.5, the simple rule given above still enables us to ﬁnd the complex conjugate of z. In this case witself contains real and imaginary components and so must be written out in full, i.e. z=w3y+2ix=(x+5i)3y+2ix. Now we can replace each iby−ito obtain z∗=(x−5i)(3y−2ix). It can be shown that the product zz∗is real, as required. J The following properties of the complex conjugate are easily proved and others may be derived from them. If z=x+iythen (z∗)∗=z, (3.12) z+z∗=2R e z=2x, (3.13) z−z∗=2iImz=2iy, (3.14) z z∗=parenleftbiggx2−y2 x2+y2parenrightbigg +iparenleftbigg2xy x2+y2parenrightbigg . (3.15) The derivation of this last relation relies on the results of the following subsection. 3.2.5 Division The division of two complex numbers z1andz2bears some similarity to their multiplication. Writing the quotient in component form we obtain z1 z2=x1+iy1 x2+iy2. (3.16) In order to separate the real and imaginary components of the quotient, we multiply both numerator and denominator by the complex conjugate of the denominator. By deﬁnition, this process will leave the denominator as a realquantity. Equation (3.16) gives z 1 z2=(x1+iy1)(x2−iy2) (x2+iy2)(x2−iy2)=(x1x2+y1y2)+i(x2y1−x1y2) x2 2+y2 2 =x1x2+y1y2 x2 2+y2 2+ix2y1−x1y2 x2 2+y2 2. Hence we have separated the quotient into real and imaginary components, as required. In the special case where z2=z∗ 1,s ot h a t x2=x1andy2=−y1, the general result reduces to (3.15). 94 3.3 POLAR REPRESENTATION OF COMPLEX NUMBERSIExpress zin the form x+iy,w h e n z=3−2i −1+4 i. Multiplying numerator and denominator by the complex conjugate of the denominator we obtain z=(3−2i)(−1−4i) (−1+4 i)(−1−4i)=−11−10i 17 =−11 17−10 17i. J In analogy to (3.10) and (3.11), which describe the multiplication of two complex numbers, the following relations apply to division: vextendsinglevextendsinglevextendsinglevextendsinglez1 z2vextendsinglevextendsinglevextendsinglevextendsingle=|z1| |z2|, (3.17) argparenleftbiggz1 z2parenrightbigg =a r g z1−argz2. (3.18) The proof of these relations is left until subsection 3.3.1. 3.3 Polar representation of complex numbers Although considering a complex number as the sum of a real and an imaginary part is often useful, sometimes the polar representation proves easier to manipulate. This makes use of the complex exponential function, which is deﬁned by ez=e x p z≡1+z+z2 2!+z3 3!+···. (3.19) Strictly speaking it is the function exp zthat is deﬁned by (3.19). The number e is the value of exp(1), i.e. it is just a number. However, it may be shown that ez and exp zare equivalent when zis real and rational and mathematicians then deﬁne their equivalence for irrational and complex z. For the purposes of this book we will not concern ourselves further with this mathematical nicety but,rather, assume that (3.19) is valid for all z. We also note that, using (3.19), by multiplying together the appropriate series we may show that (see chapter 20) e z1ez2=ez1+z2, (3.20) which is analogous to the familiar result for exponentials of real numbers. 95 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS RezImz r θ xy z=reiθ Figure 3.7 The polar representation of a complex number. From (3.19), it immediately follows that for z=iθ,θreal, eiθ=1+ iθ−θ2 2!−iθ3 3!+··· (3.21) =1−θ2 2!+θ4 4!−···+iparenleftbigg θ−θ3 3!+θ5 5!−···parenrightbigg , (3.22) and hence that eiθ=c o s θ+isinθ, (3.23) where the last equality follows from the series expansions of trigonometric func- tions (see subsection 4.6.3). This last relationship is called Euler’s equation . It also follows from (3.23) that einθ=c o s nθ+isinnθ for all n. From Euler’s equation (3.23) and using ﬁgure 3.7 we deduce that reiθ=r(cosθ+isinθ) =x+iy. Thus a complex number may be represented in the polar form z=reiθ. (3.24) Referring again to ﬁgure 3.7, we can identify rwith|z|andθwith arg z.T h e simplicity of the representation of the modulus and argument is one of the mainreasons for using the polar representation. The angle θlies conventionally in the range−π<θ≤π, but, since rotation by θis the same as rotation by 2 nπ+θ, where nis any integer, re iθ≡rei(θ+2nπ). 96 3.3 POLAR REPRESENTATION OF COMPLEX NUMBERS RezImz r1r2ei(θ1+θ2) r2eiθ2 r1eiθ1 Figure 3.8 The multiplication of two complex numbers. In this case r1and r2are both greater than unity. The algebra of the polar representation is diﬀerent from that of the real and imaginary component representation, though, of course, the results are identical. Some operations prove much easier in the polar representation, others much more complicated. The best representation for a particular problem must be determinedby the manipulation required. 3.3.1 Multiplication and division in polar form Multiplication and division in polar form are particularly simple. The product of z 1=r1eiθ1andz2=r2eiθ2is given by z1z2=r1eiθ1r2eiθ2 =r1r2ei(θ1+θ2). (3.25) The relations |z1z2|=|z1||z2|and arg( z1z2)=a r g z1+a r g z2follow immediately. An example of the multiplication of two complex numbers is shown in ﬁgure 3.8. Division is equally simple in polar form; the quotient of z1andz2is given by z1 z2=r1eiθ1 r2eiθ2=r1 r2ei(θ1−θ2). (3.26) The relations |z1/z2|=|z1|/|z2|and arg( z1/z2)=a r g z1−argz2are again imme- 97 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS RezImz r1 r2ei(θ1−θ2)r2eiθ2r1eiθ1 Figure 3.9 The division of two complex numbers. As in the previous ﬁgure, r1andr2are both greater than unity. diately apparent. The division of two complex numbers in polar form is shown in ﬁgure 3.9. 3.4 de Moivre’s theorem We now derive an extremely important theorem. Sinceparenleftbig eiθparenrightbign=einθ, we have (cosθ+isinθ)n=c o s nθ+isinnθ, (3.27) where the identity einθ=c o s nθ+isinnθfollows from the series deﬁnition of einθ(see (3.21)). This result is called de Moivre’s theorem a n di so f t e nu s e di nt h e manipulation of complex numbers. The theorem is valid for all nwhether real, imaginary or complex. There are numerous applications of de Moivre’s theorem but this section examines just three: proofs of trigonometric identities; ﬁnding the nth roots of unity; and solving complex equations. 3.4.1 Trigonometric identities The use of de Moivre’s theorem in ﬁnding trigonometric identities is best illus- trated by example. We consider the expression of a multiple-angle function interms of a polynomial in the single-angle function, and its converse. 98 3.4 DE MOIVRE’S THEOREMIExpress sin3θandcos 3θi nt e r m so fp o w e r so f cosθandsinθ. Using de Moivre’s theorem, cos 3θ+isin 3θ=( c o s θ+isinθ)3 =( c o s3θ−3c os θsin2θ)+i(3 sin θcos2θ−sin3θ). (3.28) We can equate the real and imaginary coeﬃcients separately, i.e. cos 3θ=c o s3θ−3cos θsin2θ =4c o s3θ−3c os θ (3.29) and sin 3θ=3s i n θcos2θ−sin3θ =3s i n θ−4si n3θ. J This method can clearly be applied to ﬁnding power expansions of cos nθand sinnθfor any positive integer n. The converse process uses the following properties of z=eiθ, zn+1 zn=2c o s nθ, (3.30) zn−1 zn=2isinnθ. (3.31) These equalities follow from simple applications of de Moivre’s theorem, i.e. zn+1 zn=( c o s θ+isinθ)n+( c o s θ+isinθ)−n =c o s nθ+isinnθ+c o s (−nθ)+isin(−nθ) =c o s nθ+isinnθ+c o s nθ−isinnθ =2c o s nθ and zn−1 zn=( c o s θ+isinθ)n−(cosθ+isinθ)−n =c o s nθ+isinnθ−cosnθ+isinnθ =2isinnθ. In the particular case where n=1 , z+1 z=eiθ+e−iθ=2c o s θ, (3.32) z−1 z=eiθ−e−iθ=2isinθ. (3.33) 99 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIFind an expression for cos3θin terms of cos 3θandcosθ. Using (3.32), cos3θ=1 23 / z+1 z /3 =1 8 / z3+3z+3 z+1 z3 / =1 8 / z3+1 z3 / +3 8 / z+1 z / . Now using (3.30) and (3.32), we ﬁnd cos3θ=1 4cos 3θ+3 4cosθ. J This result happens to be a simple rearrangement of (3.29), but cases involving larger values of nare better handled using this direct method than by rearranging polynomial expansions of multiple-angle functions. 3.4.2 Finding the nth roots of unity The equation z2= 1 has the familiar solutions z=±1. However, now that we have introduced the concept of complex numbers we can solve the generalequation z n= 1. Recalling the fundamental theorem of algebra, we know that the equation has nsolutions. In order to proceed we rewrite the equation as zn=e2ikπ, where kis any integer. Now taking the nth root of each side of the equation we ﬁnd z=e2ikπ/n. Hence, the solutions of zn=1a r e z1,2,...,n=1,e2iπ/n, ..., e2i(n−1)π/n, corresponding to the values 0 ,1,2,...,n−1f o r k. Larger integer values of kdo not give new solutions, since the roots already listed are simply cyclically repeatedfork=n, n+1,n+2 ,e t c .IFind the solutions to the equation z3=1. By applying the above method we ﬁnd z=e2ikπ/3. Hence the three solutions are z1=e0i=1 ,z2=e2iπ/3,z3=e4iπ/3. We note that, as expected, the next solution, for which k=3 ,g i v e s z4=e6iπ/3=1= z1, so that there are only three separate solutions. J 100 3.4 DE MOIVRE’S THEOREM RezImz e−2iπ/3e2iπ/3 2π/32π/3 1 Figure 3.10 The solutions of z3=1 . Not surprisingly, given that |z3|=|z|3from (3.10), all the roots of unity have unit modulus, i.e. they all lie on a circle in the Argand diagram of unit radius.The three roots are shown in ﬁgure 3.10. The cube roots of unity are often written 1, ωandω 2. The properties ω3=1 and 1 + ω+ω2= 0 are easily proved. 3.4.3 Solving polynomial equations A third application of de Moivre’s theorem is to the solution of polynomial equations. Complex equations in the form of a polynomial relationship must ﬁrst be solved for zin a similar fashion to the method for ﬁnding the roots of real polynomial equations. Then the complex roots of zmay be found.ISolve the equation z6−z5+4z4−6z3+2z2−8z+8=0 . We ﬁrst factorise to give (z3−2)(z2+4 ) ( z−1) = 0 . Hence z3=2o r z2=−4o r z= 1. The solutions to the quadratic equation are z=±2i; to ﬁnd the complex cube roots, we ﬁrst write the equation in the form z3=2=2 e2ikπ, where kis any integer. If we now take the cube root, we get z=21/3e2ikπ/3. 101 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS To avoid the duplication of solutions, we use the fact that −π<argz≤πand ﬁnd z1=21/3, z2=21/3e2πi/3=21/3 / −1 2+√ 3 2i /! , z3=21/3e−2πi/3=21/3 / −1 2−√ 3 2i /! . The complex numbers z1,z2andz3,t o g e t h e rw i t h z4=2i,z5=−2iandz6=1a r et h e solutions to the original polynomial equation. As expected from the fundamental theorem of algebra, we ﬁnd that the total number of complex roots (six, in this case) is equal to the largest power of zin the polynomial. J A useful result is that the roots of a polynomial with real coeﬃcients occur in conjugate pairs (i.e. if z1is a root, then z∗ 1is a second distinct root, unless z1is real). This may be proved as follows. Let the polynomial equation of which zis ar o o tb e anzn+an−1zn−1+···+a1z+a0=0. Taking the complex conjugate of this equation, a∗ n(z∗)n+a∗ n−1(z∗)n−1+···+a∗ 1z∗+a∗ 0=0. But the anare real, and so z∗satisﬁes an(z∗)n+an−1(z∗)n−1+···+a1z∗+a0=0, and is also a root of the original equation. 3.5 Complex logarithms and complex powers The concept of a complex exponential has already been introduced in section 3.3, where it was assumed that the deﬁnition of an exponential as a series was validfor complex numbers as well as for real numbers. Similarly we can deﬁne thelogarithm of a complex number and we can use complex numbers as exponents. Let us denote the natural logarithm of a complex number zbyw=L n z,w h e r e the notation Ln will be explained shortly. Thus, wmust satisfy z=e w. Using (3.20), we see that z1z2=ew1ew2=ew1+w2, and taking logarithms of both sides we ﬁnd Ln(z1z2)=w1+w2=L n z1+L n z2, (3.34) which shows that the familiar rule for the logarithm of the product of two real numbers also holds for complex numbers. 102 3.5 COMPLEX LOGARITHMS AND COMPLEX POWERS We may use (3.34) to investigate further the properties of Ln z. We have already noted that the argument of a complex number is multivalued, i.e. arg z=θ+2nπ, where nis any integer. Thus, in polar form, the complex number zshould strictly be written as z=rei(θ+2nπ). Taking the logarithm of both sides, and using (3.34), we ﬁnd Lnz=l nr+i(θ+2nπ), (3.35) where ln ris the natural logarithm of the real positive quantity ra n ds oi s written normally. Thus from (3.35) we see that Ln zis itself multivalued. To avoid this multivalued behaviour it is conventional to deﬁne another function ln z,t h e principal value of Ln z, which is obtained from Ln zby restricting the argument ofzto lie in the range −π<θ≤π.IEvaluate Ln(−i). By rewriting −ias a complex exponential, we ﬁnd Ln(−i)=L n / ei(−π/2+2nπ) / =i(−π/2+2 nπ), where nis any integer. Hence Ln( −i)=−iπ/2,3iπ/2,. . .. We note that ln( −i), the principal value of Ln( −i), is given by ln( −i)=−iπ/2. J Ifzandtare both complex numbers then the zth power of tis deﬁned by tz=ezLnt. Since Ln tis multivalued, so too is this deﬁnition.ISimplify the expression z=i−2i. Firstly we take the logarithm of both sides of the equation to give Lnz=−2iLni. Now inverting the process we ﬁnd eLnz=z=e−2iLni. We can write i=ei(π/2+2nπ),where nis any integer, and hence Lni=L n h ei(π/2+2nπ) i =i /; π/2+2 nπ / . We can now simplify zto give i−2i=e−2i×i(π/2+2nπ) =e(π+4nπ), which, perhaps surprisingly, is a real quantity rather than a complex one. J Complex powers and the logarithms of complex numbers are discussed further in chapter 20. 103 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS 3.6 Applications to diﬀerentiation and integration We can use the exponential form of a complex number together with de Moivre’s theorem (see section 3.4) to simplify the diﬀerentiation of trigonometric functions.IFind the derivative with respect to xofe3xcos 4x. We could diﬀerentiate this function straightforwardly using the product rule (see subsec- tion 2.1.2). However, an alternative method in this case is to use a complex exponential.Let us consider the complex number z=e 3x(cos4 x+isin4x)=e3xe4ix=e(3+4 i)x, where we have used de Moivre’s theorem to rewrite the trigonometric functions as a com- plex exponential. This complex number has e3xcos 4xas its real part. Now, diﬀerentiating zwith respect to xwe obtain dz dx=( 3+4 i)e(3+4 i)x=( 3+4 i)e3x(cos4 x+isin 4x), (3.36) where we have again used de Moivre’s theorem. Equating real parts we then ﬁnd d dx /; e3xcos 4x / =e3x(3cos 4 x−4si n4 x). By equating the imaginary parts of (3.36), we also obtain, as a bonus, d dx /; e3xsin 4x / =e3x(4cos4 x+3s i n4 x). J In a similar way the complex exponential can be used to evaluate integrals containing trigonometric and exponential functions.IEvaluate the integral I= R eaxcosbx dx. Let us consider the integrand as the real part of the complex number eax(cosbx+isinbx)=eaxeibx=e(a+ib)x, where we use de Moivre’s theorem to rewrite the trigonometric functions as a complex exponential. Integrating we ﬁndZ e(a+ib)xdx=e(a+ib)x a+ib+c =(a−ib)e(a+ib)x (a−ib)(a+ib)+c =eax a2+b2 /; aeibx−ibeibx / +c, (3.37) where the constant of integration cis in general complex. Denoting this constant by c=c1+ic2and equating real parts in (3.37) we obtain I= Z eaxcosbx dx=eax a2+b2(acosbx+bsinbx)+c1, which agrees with result (2.37) found using integration by parts. Equating imaginary parts in (3.37) we obtain, as a bonus, J= Z eaxsinbx dx=eax a2+b2(asinbx−bcosbx)+c2. J 104 3.7 HYPERBOLIC FUNCTIONS 3.7 Hyperbolic functions Thehyperbolic functions are the complex analogues of the trigonometric functions. The analogy may not be immediately apparent and their deﬁnitions may appearat ﬁrst to be somewhat arbitrary. However, careful examination of their propertiesreveals the purpose of the deﬁnitions. For instance, their close relationship with the trigonometric functions, both in their identities and in their calculus, means that many of the familiar properties of trigonometric functions can also be appliedto the hyperbolic functions. Further, hyperbolic functions occur regularly, and sogiving them special names is a notational convenience. 3.7.1 Deﬁnitions The two fundamental hyperbolic functions are cosh xand sinh x, which, as their names suggest, are the hyperbolic equivalents of cos xand sin x. They are deﬁned by the following relations: coshx= 1 2(ex+e−x), (3.38) sinhx=1 2(ex−e−x). (3.39) Note that cosh xis an even function and sinh xis an odd function. By analogy with the trigonometric functions, the remaining hyperbolic functions are tanh x=sinhx coshx=ex−e−x ex+e−x, (3.40) sechx=1 coshx=2 ex+e−x, (3.41) cosech x=1 sinhx=2 ex−e−x, (3.42) cothx=1 tanh x=ex+e−x ex−e−x. (3.43) All the hyperbolic functions above have been deﬁned in terms of the real variable x. However, this was simply so that they may be plotted (see ﬁgures 3.11–3.13); the deﬁnitions are equally valid for any complex number z. 3.7.2 Hyperbolic–trigonometric analogies In the previous subsections we have alluded to the analogy between trigonometric and hyperbolic functions. Here, we discuss the close relationship between the twogroups of functions. Recalling (3.32) and (3.33) we ﬁnd cosix= 1 2(ex+e−x), sinix=1 2i(ex−e−x). 105 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS sech xcoshx x1234 −1 −21 2 Figure 3.11 Graphs of cosh xand sech x. cosech xcosech x sinhx x24 −2 −4−1 −21 2 Figure 3.12 Graphs of sinh xand cosech x. Hence, by the deﬁnitions given in the previous subsection, coshx=c o s ix, (3.44) isinhx=s i n ix, (3.45) cosx=c o s h ix, (3.46) isinx= sinh ix. (3.47) These useful equations make the relationship between hyperbolic and trigono- 106 3.7 HYPERBOLIC FUNCTIONS cothxcothx tanh x x24 −2 −4−1 −21 2 Figure 3.13 Graphs of tanh xand coth x. metric functions transparent. The similarity in their calculus is discussed further in subsection 3.7.6. 3.7.3 Identities of hyperbolic functions The analogies between trigonometric functions and hyperbolic functions having been established, we should not be surprised that all the trigonometric identitiesalso hold for hyperbolic functions, with the following modiﬁcation. Whereversin 2xoccurs it must be replaced by −sinh2x, and vice versa. Note that this replacement is necessary even if the sin2xis hidden, e.g. tan2x=s i n2x/cos2x and so must be replaced by ( −sinh2x/cosh2x)=−tanh2x.IFind the hyperbolic identity analogous to cos2x+s i n2x=1. Using the rules stated above cos2xmust be replaced by cosh2x,a n ds i n2xmust be replaced by−sinh2x, and so the identity becomes cosh2x−sinh2x=1. This can be veriﬁed by direct substitution, using the deﬁnitions of cosh xand sinh x;s e e (3.38) and (3.39). J Some other identities that can be proved in a similar way are sech2x=1−tanh2x, (3.48) cosech2x=c o t h2x−1, (3.49) sinh2 x=2s i n h xcoshx, (3.50) cosh2 x=c o s h2x+ sinh2x. (3.51) 107 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS 3.7.4 Solving hyperbolic equations When we are presented with a hyperbolic equation to solve, we may proceed by analogy with the solution of trigonometric equations. However, it is almostalways easier to express the equation directly in terms of exponentials.ISolve the hyperbolic equation coshx−5si nh x−5=0. Substituting the deﬁnitions of the hyperbolic functions we obtain 1 2(ex+e−x)−5 2(ex−e−x)−5=0 . Rearranging, and then multiplying through by −ex, gives in turn −2ex+3e−x−5=0 and 2e2x+5ex−3=0 . N o ww ec a nf a c t o r i s ea n ds o l v e : (2ex−1)(ex+3 )=0 . Thus ex=1/2o r ex=−3. Hence x=−ln 2 or x=l n (−3). The interpretation of the logarithm of a negative number has been discussed in section 3.5. J 3.7.5 Inverses of hyperbolic functions Just like trigonometric functions, hyperbolic functions have inverses. If y= coshxthen x=c o s h−1y, which serves as a deﬁnition of the inverse. By using the fundamental deﬁnitions of hyperbolic functions, we can ﬁnd closed-formexpressions for their inverses. This is best illustrated by example.IFind a closed-form expression fo r the inverse hyperbolic function y=s i n h−1x. First we write xas a function of y,i . e . y=s i n h−1x⇒x=s i n h y. Now, since cosh y=1 2(ey+e−y) and sinh y=1 2(ey−e−y), ey=c o s h y+s i n h y = q 1+s i n h2y+s i n h y ey= p 1+x2+x, and hence y=l n ( p 1+x2+x). J In a similar fashion it can be shown that cosh−1x=l n (√ x2−1+x). 108 3.7 HYPERBOLIC FUNCTIONS sech−1xsech−1x cosh−1xcosh−1x x24 −2 −43 412 Figure 3.14 Graphs of cosh−1xand sech−1x.IFind a closed-form expression fo r the inverse hyperbolic function y=t a n h−1x. First we write xas a function of y,i . e . y=t a n h−1x⇒ x=t a n h y. Now, using the deﬁnition of tanh yand rearranging, we ﬁnd x=ey−e−y ey+e−y⇒ (x+1 )e−y=( 1−x)ey. Thus, it follows that e2y=1+x 1−x⇒ ey= r 1+x 1−x, y=l n r 1+x 1−x, tanh−1x=1 2ln /1+x 1−x / . J Graphs of the inverse hyperbolic functions are given in ﬁgures 3.14–3.16. 3.7.6 Calculus of hyperbolic functions Just as the identities of hyperbolic functions closely follow those of their trigono- metric counterparts, so their calculus is similar. The derivatives of the two basic 109 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS sinh−1x cosech−1xcosech−1x x24 −2 −4−1 −2 12 Figure 3.15 Graphs of sinh−1xand cosech−1x. coth−1xcoth−1xtanh−1x x24 −2 −4−1 −21 2 Figure 3.16 Graphs of tanh−1xand coth−1x. hyperbolic functions are d dx(coshx)= sinh x, (3.52) d dx(sinhx)=c o s h x. (3.53) These may be deduced by considering the deﬁnitions. 110 3.7 HYPERBOLIC FUNCTIONSIVerify the relation (d/dx)c osh x=s i n h x. Using the deﬁnition of cosh x, coshx=1 2(ex+e−x), and diﬀerentiating directly, we ﬁnd d dx(coshx)=1 2(ex−e−x) =s i n h x. J Clearly the integrals of the fundamental hyperbolic functions are also deﬁned by these relations. The derivatives of the remaining hyperbolic functions can bederived by product diﬀerentiation and are presented below only for complete-ness. d dx(tanh x)=s e c h2x, (3.54) d dx(sech x)=−sech xtanh x, (3.55) d dx(cosech x)=−cosech xcothx, (3.56) d dx(cothx)=−cosech2x. (3.57) The inverse hyperbolic functions also have derivatives, which are given by the following: d dxparenleftBig cosh−1x aparenrightBig =1√ x2−a2, (3.58) d dxparenleftBig sinh−1x aparenrightBig =1√ x2+a2, (3.59) d dxparenleftBig tanh−1x aparenrightBig =a a2−x2,forx2<a2, (3.60) d dxparenleftBig coth−1x aparenrightBig =−a x2−a2,forx2>a2. (3.61) These may be derived from the logarithmic form of the inverse (see subsec- tion 3.7.5). 111 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONSIEvaluate (d/dx)si nh−1xusing the logarithmic form of the inverse. From the results of section 3.7.5, d dx /; sinh−1x / =d dx h ln / x+ p x2+1 /i =1 x+√ x2+1 / 1+x√ x2+1 / =1 x+√ x2+1 / √ x2+1+ x√ x2+1 /! =1√ x2+1. J 3.8 Exercises 3.1 Two complex numbers zandware given by z=3+4 iandw=2−i.O na n Argand diagram plot (a)z+w,( b ) w−z,( c )wz,( d ) z/w, (e)z∗w+w∗z,( f )w2,( g )l n z,( h )( 1+ z+w)1/2. 3.2 By considering the real and imaginary parts of the product eiθeiφprove the standard formulae for cos( θ+φ) and sin( θ+φ). 3.3 By writing π/12 = ( π/3)−(π/4) and considering eiπ/12, evaluate cot( π/12). 3.4 Find the locus in the complex z-plane of points that satisfy the following equa- tions. (a)z−c=ρ /1+it 1−it / ,where cis complex, ρis real and tis a real parameter that varies in the range −∞<t<∞. (b)z=a+bt+ct2,i nw h i c h tis a real parameter and a,b,a n d care complex numbers with b/creal. 3.5 Evaluate (a) Re(exp 2 iz), (b) Im(cosh2z), (c) (−1+√ 3i)1/2, (d)|exp(i1/2)|, (e) exp( i3), (f) Im(2i+3), (g) ii,( h )l n [ (√ 3+i)3]. 3.6 Find the equations in terms of xandyof the sets of points in the Argand diagram that satisfy the following: (a) Re z2=I m z2; (b) (Im z2)/z2=−i; (c) arg[ z/(z−1)] = π/2. 3.7 Show that the locus of all points z=x+iyin the complex plane that satisfy |z−ia|=λ|z+ia|,λ > 0, is a circle of radius |2λ/(1−λ2)|acentred on the point z=ia[(1 + λ2)/(1−λ2)]. Sketch the circles for a few typical values of λ, including λ<1,λ>1a n d λ=1 . 3.8 The two sets of points z=a,z=b,z=c,a n d z=A,z=B,z=Care the corners of two similar triangles in the Argand diagram. Express in terms ofa, b, . . . , C 112 3.8 EXERCISES (a) the equalities of corresponding angles, and (b) the constant ratio of corresponding sides, in the two triangles. By noting that any complex quantity can be expressed as z=|z|exp(iargz), deduce that a(B−C)+b(C−A)+c(A−B)=0 . 3.9 For the real constant aﬁnd the loci of all points z=x+iyin the complex plane that satisfy (a) Re / ln /z−ia z+ia // =c, c>0, (b) Im / ln /z−ia z+ia // =k,0 ≤k≤π/2. Identify the two families of curves and verify that in case (b) all curves pass through the two points ±ia. 3.10 The most general type of transformation between one Argand diagram, in the z-plane, and another, in the Z-plane, that gives one and only one value of Zfor each value of z(and conversely) is known as the general bilinear transformation and takes the form z=aZ+b cZ+d. (a) Conﬁrm that the transformation from the Z-plane to the z-plane is also a general bilinear transformation. (b) Recalling that the equation of a circle can be written in the form////z−z1 z−z2 ////=λ, λ /negationslash=1, show that the general bilinear transformation transforms circles into circles (or straight lines). What is the condition that z1,z2andλmust satisfy if the transformed circle is to be a straight line? 3.11 Sketch the parts of the Argand diagram in which (a) Re z2<0,|z1/2|≤2, (b) 0≤argz∗≤π/2, (c)|expz3|→0a s|z|→∞ . What is the area of the region in which all three conditions are satisﬁed? 3.12 Denote the nth roots of unity by 1, ωn,ω2 n,...,ωn−1 n. (a) Prove that (i)n−1X r=0ωr n=0,(ii)n−1Y r=0ωr n=(−1)n+1. (b) Express x2+y2+z2−yz−zx−xyas the product of two factors, each linear inx,yandz, with coeﬃcients dependent on the third roots of unity (and those of the xterms arbitrarily taken as real). 113 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS 3.13 Prove that x2m+1−a2m+1,w h e r e mis an integer ≥1 ,c a nb ew r i t t e na s x2m+1−a2m+1=(x−a)mY r=1 / x2−2axcos /2πr 2m+1 / +a2 / . 3.14 The complex position vectors of two parallel interacting equal ﬂuid vortices moving with their axes of rotation always perpendicular to the z-plane are z1 andz2. The equations governing their motions are dz∗ 1 dt=−i z1−z2,dz∗ 2 dt=−i z2−z1. Deduce that (a) z1+z2,( b )|z1−z2|and (c)|z1|2+|z2|2are all constant in time, and hence describe the motion geometrically. 3.15 Solve the equation z7−4z6+6z5−6z4+6z3−12z2+8z+4=0 , (a) by examining the eﬀect of setting z3e q u a lt o2 ,a n d (b) by factorising and using the binomial expansion of ( z+a)4. Plot the seven roots of the equation on an Argand plot, exemplifying that complex roots of a polynomial equation always occur in conjugate pairs if the polynomialhas real coeﬃcients. 3.16 The polynomial f(z) is deﬁned by f(z)=z 5−6z4+1 5z3−34z2+3 6z−48. (a) Show that the equation f(z) = 0 has roots of the form z=λiwhere λis real, and hence factorize f(z). (b) Show further that the cubic factor of f(z) can be written in the form (z+a)3+b,w h e r e aandbare real, and hence solve the equation f(z)=0 completely. 3.17 The binomial expansion of (1 + x)n, discussed in chapter 1, can be written for a positive integer nas (1 +x)n=nX r=0nCrxr, wherenCr=n!/[r!(n−r)!]. (a) Use de Moivre’s theorem to show that the sum S1(n)=nC0−nC2+nC4−···+(−1)mnC2m,n−1≤2m≤n, has the value 2n/2cos(nπ/4). (b) Derive a similar result for the sum S2(n)=nC1−nC3+nC5−···+(−1)mnC2m+1,n−1≤2m+1≤n, and verify it for the cases n=6 ,7a n d8 . 3.18 By considering (1 + exp iθ)n, prove that nX r=0nCrcosnθ=2ncosn(θ/2)cos( nθ/2), nX r=0nCrsinnθ=2ncosn(θ/2)sin( nθ/2), wherenCr=n!/[r!(n−r)!]. 114 3.8 EXERCISES 3.19 Use de Moivre’s theorem with n= 4 to prove that cos 4θ=8c o s4θ−8cos2θ+1, and deduce that cosπ 8= / 2+√ 2 4 /!1/2 . 3.20 Express sin4θentirely in terms of the trigonometric functions of multiple angles and deduce that its average value over a complete cycle is3 8. 3.21 Use de Moivre’s theorem to prove that tan5θ=t5−10t3+5t 5t4−10t2+1, where t=t a n θ. Deduce the values of tan( nπ/10) for n=1 ,2 ,3 ,4 . 3.22 (a) Prove that coshx−coshy=2s i n h /x+y 2 / sinh /x−y 2 / . (b) Prove that, if y=s i n h−1x, (x2+1 )d2y dx2+xdy dx=0. 3.23 Determine the conditions under which the equation acoshx+bsinhx=c, c > 0, has zero, one, or two real solutions for x. What is the solution if a2=c2+b2? 3.24 (a) Solve cosh x=s i n h x+2 s e c h x. (b) Show that the real solution xof tanh x=c o s e c h xc a nb ew r i t t e ni nt h e form x=l n ( u+√u). Find an explicit value for u. (c) Evaluate tanh xwhen xis the real solution of cosh2 x=2c o s h x. 3.25 Express sinh4xin terms of hyperbolic cosines of multiples of x, and hence solve 2cosh4 x−8cosh2 x+5=0 . 3.26 In the theory of special relativity, the relationship between the position and time coordinates of an event as measured in two frames of reference that have parallelx-axes can be expressed in terms of hyperbolic functions. If the coordinates are x andtin one frame and x /primeandt/primein the other then the relationship take the form x/prime=xcoshφ−ctsinhφ, ct/prime=−xsinhφ+ctcoshφ. Express xandctin terms of x/prime,ct/primeandφand show that x2−(ct)2=(x/prime)2−(ct/prime)2. 3.27 A closed barrel has as its curved surface that obtained by rotating about the x-axis the part of the curve y=a[2−cosh( x/a)] lying in the range −b≤x≤b. Show that the total surface area Aof the barrel is given by A=πa[9a−8aexp(−b/a)+aexp(−2b/a)−2b]. 115 COMPLEX NUMBERS AND HYPERBOLIC FUNCTIONS 3.28 The principal value of the logarithmic function of a complex variable is deﬁned to have its argument in the range −π<argz≤π.B yw r i t i n g z=t a n win terms of exponentials show that tan−1z=1 2iln /1+iz 1−iz / . Use this result to evaluate tan−1 / 2√ 3−3i 7 /! . 3.9 Hints and answers 3.1 (a) 5 + 3 i;( b )−1−5i;( c )1 0+5 i;( d )2 /5+1 1 i/5; (e) 4; (f) 3 −4i; (g) 5 + i[tan−1(4/3) + 2 nπ]; (h)±(2.521 + 0 .595i). 3.3 2 +√ 3. 3.4 (a) Set t=t a n θwith−π/2<θ<π / 2. The equation becomes z−c=ρe2iθ. The locus is a circle, centre c,r a d i u s ρ. (b) Eliminate the tterm between xand y. Note that the coeﬃcient of tis proportional to Im( b/c). The locus is a straight line (Im k)[x−Re(a)] = (Rek)[y−Im(a)], where k=borc. 3.5 (a) exp( −2y)cos2 x;( b )( s i n2 ysinh2 x)/2; (c)√ 2e x p ( πi/3) or√ 2e x p ( 4 πi/3); (d) exp(1 /√ 2) or exp(−1/√ 2); (e) 0 .540−0.841i;( f )8s i n ( l n2 )=5 .11; (g) exp(−π/2−2πn); (h) ln 8 + i(2n+1/2)π. 3.6 (a) y=(±√ 2−1)x;( b ) x=±y;( c )t h eh a l fo ft h ec i r c l e( x−1 2)2+y2=1 4that lies in y<0. 3.7 Starting from |x+iy−ia|=λ|x+iy+ia|, show that the coeﬃcients of xandy are equal, and write the equation in the form x2+(y−α)2=r2. 3.8 (a) arg[( b−a)/(c−a)] = arg[( B−A)/(C−A)]. (b)|(b−a)|/|(c−a)|=|(B−A)|/|(C−A)|. 3.9 (a) Circles enclosing z=−ia,w i t h λ=e x p c>1. (b) The condition is that arg[( z−ia)/(z+ia)] = k. This can be rearranged to givea(z+z∗)=k(a2−|z|2), which becomes in x, ycoordinates the equation of a circle with centre ( −a/k,0) and radius a(1 +k−2)1/2. 3.10 (a) Z=(−dz+b)/(cz−a). (b)|(Z−Z1)/(Z−Z2)|=Λ ,w i t h Z1,2given by setting z=z1,2in the result in (a);|a−cz1|=λ|a−cz2|. 3.11 All three conditions are satisﬁed in 3 π/2≤θ≤7π/4,|z|≤4; area = 2 π. 3.12 (a) Express ωn−1 as a product of factors like ( ω−ωr n) and examine the coeﬃcients of (i) ωn−1and (ii) ω0. (b) ( x+ω3y+ω2 3z)(x+ω2 3y+ω3z). 3.13 Denoting exp[2 πi/(2m+ 1)] by Ω, express x2m+1−a2m+1as a product of factors like ( x−aΩr) and then combine those containing Ωrand Ω2m+1−r.U s et h ef a c t that Ω2m+1=1 . 3.14 (b) Diﬀerentiate ( z1−z2)(z∗ 1−z∗ 2). (c) Write 2 |z1|2+2|z2|2as|z1+z2|2−|z1−z2|2. Circular motion about a ﬁxed point with the vortices at the opposite ends of adiameter. 3.15 The roots are 2 1/3exp(2 πni/3) for n=0,1,2; 1±31/4;1±31/4i. 3.16 (a) The vanishing of the real and imaginary parts of f(λi)r e q u i r e s( λ2=3o r8 3) and ( λ2= 0 or 3 or 12); hence λ2=3a n d f(z)=(z2+3)(z3−6z2+12z−16). (b)a=−2,b=−8. The roots are ±i√ 3, 4, 1±i√ 3. 116 3.9 HINTS AND ANSWERS 3.17 (b) S2(n)=2n/2sin(nπ/4).S2(6) =−8,S2(7) =−8,S2(8) = 0. 3.18 Write 1 + cos θand sin θin terms of θ/2. 3.20 (cos4 θ)/8−(cos 2 θ)/2+3 /8. 3.21 Show that cos5 θ=1 6 c5−20c3+5c,w h e r e c=c o s θ, and correspondingly for sin 5θ.U s ec o s−2θ=1+t a n2θ. The four required values are [(5−√ 20)/5]1/2,( 5−√ 20)1/2,[ ( 5+√ 20)/5]1/2,( 5+√ 20)1/2. 3.23 Reality of the root(s) requires c2+b2≥a2anda+b>0. With these conditions, there are two roots if a2>b2, but only one if b2>a2. Fora2=c2+b2,x=1 2ln[(a−b)/(a+b)]. 3.24 (a) ln(1 /√ 3); (b) (1 +√ 5)/2; (c)±(12)1/4/(√ 3+1 ) . 3.25 Reduce the equation to 16sinh4x= 1, yielding x=±0.481. 3.26 The same expressions but with φreplaced by −φare obtained. 3.27 Show that ds=( c o s h x/a)dx; curved surface area = πa2[8sinh( b/a)−sinh(2 b/a)]−2πab. 3.28 π/6−iln√ 2. 117 4 Series and limits 4.1 Series Many examples exist in the physical sciences of situations where we are presented with a sum of terms to evaluate. For example, we may wish to add the contributions from successive slits in a diﬀraction grating to ﬁnd the total light intensity at a particular point behind the grating. A series may have either a ﬁnite or inﬁnite number of terms. In either case, the sum of the ﬁrst Nterms of a series (often called a partial sum) is written SN=u1+u2+u3+···+uN, where the terms of the series un,n=1,2,3,...,N are numbers, that may in general be complex. If the terms are complex then SNwill in general be complex also, and we can write SN=XN+iYN,w h e r e XNandYNare the partial sums of the real and imaginary parts of each term separately and are therefore real. If a series has only Nterms then the partial sum SNis of course the sum of the series. Sometimes we may encounter series where each term depends on some variable,x, say. In this case the partial sum of the series will depend on the value assumed byx. For example, consider the inﬁnite series S(x)=1+ x+x 2 2!+x3 3!+···. This is an example of a power series; these are discussed in more detail in section 4.5. It is in fact the Maclaurin expansion of exp x(see subsection 4.6.3). Therefore S(x)=e x p xand, of course, varies according to the value of the variable x. A series might just as easily depend on a complex variable z. A general, random sequence of numbers can be described as a series and a sum of the terms found. However, for cases of practical interest, there will usually be 118 4.2 SUMMATION OF SERIES some sort of relationship between successive terms. For example, if the nth term o fas e r i e si sg i v e nb y un=1 2n, forn=1,2,3,...,N then the sum of the ﬁrst Nterms will be SN=Nsummationdisplay n=1un=1 2+1 4+1 8+···+1 2N. (4.1) It is clear that the sum of a ﬁnite number of terms is always ﬁnite, provided that each term is itself ﬁnite. It is often of practical interest, however, to consider the sum of a series with an inﬁnite number of ﬁnite terms. The sum of an inﬁnite number of terms is best deﬁned by ﬁrst considering the partial sumof the ﬁrst Nterms, S N. If the value of the partial sum SNtends to a ﬁnite limit, S,a sNtends to inﬁnity, then the series is said to converge and its sum is given by the limit S. In other words, the sum of an inﬁnite series is given by S= lim N→∞SN, provided the limit exists. For complex inﬁnite series, if SNapproaches a limit S=X+iYasN→∞, this means that XN→XandYN→Yseparately, i.e. the real and imaginary parts of the series are each convergent series with sumsXandYrespectively. However, not all inﬁnite series have ﬁnite sums. As N→∞, the value of the partial sum S Nmay diverge: it may approach + ∞or−∞, or oscillate ﬁnitely or inﬁnitely. Moreover, for a series where each term depends on some variable,its convergence can depend on the value assumed by the variable. Whether aninﬁnite series converges, diverges or oscillates has important implications whendescribing physical systems. Methods for d etermining whether a series converges are discussed in section 4.3. 4.2 Summation of series It is often necessary to ﬁnd the sum of a ﬁnite series or a convergent inﬁnite series. We now describe arithmetic, geometric and arithmetico-geometric series, which are particularly common and for which the sums are easily found. Other methods that can sometimes be used to sum more complicated series are discussedbelow. 119 SERIES AND LIMITS 4.2.1 Arithmetic series Anarithmetic series has the characteristic that the diﬀerence between successive terms is constant. The sum of a general arithmetic series is written SN=a+(a+d)+(a+2d)+···+[a+(N−1)d]=N−1summationdisplay n=0(a+nd). Rewriting the series in the opposite order and adding this term by term to the original expression for SN, we ﬁnd SN=N 2[a+a+(N−1)d]=N 2(ﬁrst term + last term) . (4.2) If an inﬁnite number of such terms are added the series will increase (or decrease) indeﬁnitely; that is to say, it diverges.ISum the integers between 1and1000inclusive. This is an arithmetic series with a=1 ,d=1a n d N= 1000. Therefore, using (4.2) we ﬁnd SN=1000 2(1 + 1000) = 500500 , which can be checked directly only with considerable eﬀort. J 4.2.2 Geometric series Equation (4.1) is a particular example of a geometric series , which has the characteristic that the ratio of successive terms is a constant (one-half in thiscase). The sum of a geometric series is in general written S N=a+ar+ar2+···+arN−1=N−1summationdisplay n=0arn, where ais a constant and ris the ratio of successive terms, the common ratio .T h e sum may be evaluated by considering SNandrSN: SN=a+ar+ar2+ar3+···+arN−1, rSN=ar+ar2+ar3+ar4+···+arN. If we now subtract the second equation from the ﬁrst we obtain (1−r)SN=a−arN, and hence SN=a(1−rN) 1−r. (4.3) 120 4.2 SUMMATION OF SERIES For a series with an inﬁnite number of terms and |r|<1, we have lim N→∞rN=0 , and the sum tends to the limit S=a 1−r. (4.4) In (4.1), r=1 2,a=1 2,a n ds o S=1 .F o r|r|≥1, however, the series either diverges or oscillates.IConsider a ball that drops from a height of 27 mand on each bounce retains only a third of its kinetic energy; thus after one bounce it will return to a height of 9m,a f t e rt w o bounces to 3m, and so on. Find the total distance travelled between the ﬁrst bounce and theMth bounce. The total distance travelled between the ﬁrst bounce and the Mth bounce is given by the sum of M−1t e r m s : SM−1=2(9+3+1+ ···)=2M−2X m=09 3m forM> 1, where the factor of 2 is included to allow for both the upward and the downward journey. Inside the parentheses we clearly have a geometric series with ﬁrstterm 9 and common ratio 1 /3 and hence the distance is given by (4.3), i.e. S M−1=2×9 h 1− /;1 3 /M−1 i 1−1 3=2 7 h 1− /;1 3 /M−1 i , where the number of terms Nin (4.3) has been replaced by M−1. J 4.2.3 Arithmetico-geometric series An arithmetico-geometric series, as its name suggests, is a combined arithmetic and geometric series. It has the general form SN=a+(a+d)r+(a+2d)r2+···+[a+(N−1)d]rN−1=N−1summationdisplay n=0(a+nd)rn, and can be summed, in a similar way to a pure geometric series, by multiplying byrand subtracting the result from the original series to obtain (1−r)SN=a+rd+r2d+···+rN−1d−[a+(N−1)d]rN. Using the expression for the sum of a geometric series (4.3) and rearranging, we ﬁnd SN=a−[a+(N−1)d]rN 1−r+rd(1−rN−1) (1−r)2. For an inﬁnite series with |r|<1, lim N→∞rN= 0 as in the previous subsection, and the sum tends to the limit S=a 1−r+rd (1−r)2. (4.5) As for a geometric series, if |r|≥1 then the series either diverges or oscillates. 121 SERIES AND LIMITSISum the series S=2+5 2+8 22+11 23+···. This is an inﬁnite arithmetico-geometric series with a=2 , d=3a n d r=1/2. Therefore, from (4.5), we obtain S= 10. J 4.2.4 The diﬀerence method The diﬀerence method is sometimes useful in summing series that are more complicated than the examples discussed above. Let us consider the general series Nsummationdisplay n=1un=u1+u2+···+uN. If the terms of the series, un, can be expressed in the form un=f(n)−f(n−1) for some function f(n) then its (partial) sum is given by SN=Nsummationdisplay n=1un=f(N)−f(0). This can be shown as follows. The sum is given by SN=u1+u2+···+uN and since un=f(n)−f(n−1), it may be rewritten SN=[f(1)−f(0)] + [ f(2)−f(1)] + ···+[f(N)−f(N−1)]. By cancelling terms we see that SN=f(N)−f(0).IEvaluate the sum NX n=11 n(n+1 ). Using partial fractions we ﬁnd un=− /1 n+1−1 n / . Hence un=f(n)−f(n−1) with f(n)=−1/(n+1 ) ,a n ds ot h es u mi sg i v e nb y SN=f(N)−f(0) =−1 N+1+1=N N+1. J 122 4.2 SUMMATION OF SERIES The diﬀerence method may be easily extended to evaluate sums in which each term can be expressed in the form un=f(n)−f(n−m), (4.6) where mis an integer. By writing out the sum to Nterms with each term expressed in this form, and cancelling terms in pairs as before, we ﬁnd SN=msummationdisplay k=1f(N−k+1 )−msummationdisplay k=1f(1−k).IEvaluate the sum NX n=11 n(n+2 ). Using partial fractions we ﬁnd un=− /1 2(n+2 )−1 2n / . Hence un=f(n)−f(n−2) with f(n)=−1/[2(n+ 2)], and so the sum is given by SN=f(N)+f(N−1)−f(0)−f(−1) =3 4−1 2 /1 N+2+1 N+1 / . J In fact the diﬀerence method is quite ﬂexible and may be used to evaluate sums even when each term cannot be expressed as in (4.6). The method still relies,however, on being able to write u nin terms of a single function such that most terms in the sum cancel, leaving only a few terms at the beginning and the end.This is best illustrated by an example.IEvaluate the sum NX n=11 n(n+1 ) ( n+2 ). Using partial fractions we ﬁnd un=1 2(n+2 )−1 n+1+1 2n. Hence un=f(n)−2f(n−1) +f(n−2) with f(n)=1 /[2(n+ 2)]. If we write out the sum, expressing each term unin this form, we ﬁnd that most terms cancel and the sum is given by SN=f(N)−f(N−1)−f(0) + f(−1) =1 4+1 2 /1 N+2−1 N+1 / . J 123 SERIES AND LIMITS 4.2.5 Series involving natural numbers Series consisting of the natural numbers 1, 2, 3, ..., or the square or cube of these numbers, occur frequently and deserve a special mention. Let us ﬁrst consider the sum of the ﬁrst Nnatural numbers, SN=1+2+3+ ···+N=Nsummationdisplay n=1n. This is clearly an arithmetic series with ﬁrst term a= 1 and common diﬀerence d= 1. Therefore, from (4.2), SN=1 2N(N+1 ) . Next, we consider the sum of the squares of the ﬁrst Nnatural numbers: SN=12+22+32+...+N2=Nsummationdisplay n=1n2, which may be evaluated using the diﬀerence method. The nth term in the series isun=n2, which we need to express in the form f(n)−f(n−1) for some function f(n). Consider the function f(n)=n(n+ 1)(2 n+1 )⇒ f(n−1) = ( n−1)n(2n−1). For this function f(n)−f(n−1) = 6 n2,a n ds ow ec a nw r i t e un=1 6[f(n)−f(n−1)]. Therefore, by the diﬀerence method, SN=1 6[f(N)−f(0)] =1 6N(N+ 1)(2 N+1 ). Finally, we calculate the sum of the cubes of the ﬁrst Nnatural numbers, SN=13+23+33+···+N3=Nsummationdisplay n=1n3, again using the diﬀerence method. Consider the function f(n)=[n(n+1 ) ]2⇒ f(n−1) = [( n−1)n]2, for which f(n)−f(n−1) = 4 n3. Therefore we can write the general nth term of the series as un=1 4[f(n)−f(n−1)], and using the diﬀerence method we ﬁnd SN=1 4[f(N)−f(0)] =1 4N2(N+1 )2. Note that this is the square of the sum of the natural numbers, i.e. Nsummationdisplay n=1n3=parenleftBiggNsummationdisplay n=1nparenrightBigg2 . 124 4.2 SUMMATION OF SERIESISum the series NX n=1(n+1 ) ( n+3 ). Thenth term in this series is un=(n+1 ) ( n+3 )= n2+4n+3, and therefore we can write NX n=1(n+1 ) ( n+3 )=NX n=1(n2+4n+3 ) =NX n=1n2+4NX n=1n+NX n=13 =1 6N(N+ 1)(2 N+1 )+4×1 2N(N+1 )+3 N =1 6N(2N2+1 5N+ 31) . J 4.2.6 Transformation of series A complicated series may sometimes be summed by transforming it into a familiar series for which we already know the sum, perhaps a geometric series or the Maclaurin expansion of a simple function (see subsection 4.6.3). Varioustechniques are useful, and deciding which one to use in any given case is a matterof experience. We now discuss a few of the more common methods. The diﬀerentiation or integration of a series is often useful in transforming an apparently intractable series into a more familiar one. If we wish to diﬀerentiateor integrate a series that already depends on some variable then we may do soin a straightforward manner.ISum the series S(x)=x4 3(0!)+x5 4(1!)+x6 5(2!)+···. Dividing both sides by xwe obtain S(x) x=x3 3(0!)+x4 4(1!)+x5 5(2!)+···, which is easily diﬀerentiated to give d dx /S(x) x / =x2 0!+x3 1!+x4 2!+x5 3!+···. Recalling the Maclaurin expansion of exp xgiven in subsection 4.6.3, we recognise that the RHS is equal to x2expx. Having done so, we can now integrate both sides to obtain S(x)/x= Z x2expxd x . 125 SERIES AND LIMITS Integrating the RHS by parts we ﬁnd S(x)/x=x2expx−2xexpx+2e x p x+c, where the value of the constant of integration ccan be ﬁxed by the requirement that S(x)/x=0a t x= 0. Thus we ﬁnd that c=−2, and that the sum is given by S(x)=x3expx−2x2expx+2xexpx−2x. J Often, however, we require the sum of a series that does not depend on a variable. In this case, in order that we may diﬀerentiate or integrate the series,we deﬁne a function of some variable xsuch that the value of this function is equal to the sum of the series for some particular value of x(usually at x=1 ) .ISum the series S=1+2 2+3 22+4 23+···. Let us begin by deﬁning the function f(x)=1+2 x+3x2+4x3+···, so that the sum S=f(1/2). Integrating this function we obtainZ f(x)dx=x+x2+x3+···, which we recognise as an inﬁnite geometric series with ﬁrst term a=xand common ratio r=x. Therefore, from (4.4), we ﬁnd that the sum of this series is x/(1−x). In other wordsZ f(x)dx=x 1−x, so that f(x)i sg i v e nb y f(x)=d dx /x 1−x / =1 (1−x)2. The sum of the original series is therefore S=f(1/2) = 4. J Aside from diﬀerentiation and integration, an appropriate substitution can sometimes transform a series into a more familiar form. In particular, series withterms that contain trigonometric functions can often be summed by the use ofcomplex exponentials.ISum the series S(θ)=1+c o s θ+cos 2θ 2!+cos 3θ 3!+···. Replacing the cosine terms with a complex exponential, we obtain S(θ)=R e / 1+e x p iθ+exp2 iθ 2!+exp 3 iθ 3!+··· / =R e / 1+e x p iθ+(expiθ)2 2!+(expiθ)3 3!+··· / . 126 4.3 CONVERGENCE OF INFINITE SERIES Again using the Maclaurin expansion of exp xgiven in subsection 4.6.3, we notice that S(θ) = Re [exp(exp iθ)] = Re [exp(cos θ+isinθ)] =R e|:{[exp(cos θ)][exp( isinθ)]}= [exp(cos θ)]Re [exp( isinθ)] = [exp(cos θ)][cos(sin θ)]. J 4.3 Convergence of inﬁnite series Although the sums of some commonly occurring inﬁnite series may be found, the sum of a general inﬁnite series is usually diﬃcult to calculate. Nevertheless,it is often useful to know whether the partial sum of such a series converges toa limit, even if the limit cannot be found explicitly. As mentioned at the end of section 4.1, if we allow Nto tend to inﬁnity, the partial sum S N=Nsummationdisplay n=1un of a series may tend to a deﬁnite limit (i.e. the sum Sof the series), or increase or decrease without limit, or oscillate ﬁnitely or inﬁnitely. To investigate the convergence of any given series, it is useful to have available a number of tests and theorems of general applicability. We discuss them below; some we will merely state, since once they have been stated they become almostself-evident, but are no less useful for that. 4.3.1 Absolute and conditional convergence Let us ﬁrst consider some general points concerning the convergence, or otherwise, of an inﬁnite series. In general an inﬁnite seriessummationtextu ncan have complex terms, and in the special case of a real series the terms can be positive or negative. Fromany such series, however, we can always construct another seriessummationtext|u n|in which each term is simply the modulus of the corresponding term in the original series.Then each term in the new series will be a positive real number. If the seriessummationtext|u n|converges thensummationtextunalso converges, andsummationtextunis said to be absolutely convergent , i.e. the series formed by the absolute values is convergent. For an absolutely convergent series, the terms may be reordered without aﬀectingthe convergence of the series. However, ifsummationtext|u n|diverges whilstsummationtextunconverges thensummationtextunis said to be conditionally convergent . For a conditionally convergent series, rearranging the order of the terms can aﬀect the behaviour of the sumand, hence, whether the series converges or diverges. In fact, a theorem dueto Riemann shows that, by a suitable rearrangement, a conditionally convergent series may be made to converge to any arbitrary limit, or to diverge, or to oscillate ﬁnitely or inﬁnitely! Of course, if the original seriessummationtextu nconsists only of positive real terms and converges then automatically it is absolutely convergent. 127 SERIES AND LIMITS 4.3.2 Convergence of a series containing only real positive terms As discussed above, in order to test for the absolute convergence of a seriessummationtextun, we ﬁrst construct the corresponding seriessummationtext|un|that consists only of real positive terms. Therefore in this subsection we will restrict our attention to seriesof this type. We discuss below some tests that may be used to investigate the convergence of such a series. Before doing so, however, we note the following crucial consideration . In all the tests for, or discussions of, the convergence of a series, it is not whathappens in the ﬁrst ten, or the ﬁrst thousand, or the ﬁrst million terms (or anyother ﬁnite number of terms) that matters, but what happens ultimately . Preliminary test A necessary but not suﬃcient condition for a series of real positive termssummationtextu n to be convergent is that the term untends to zero as ntends to inﬁnity, i.e. we require lim n→∞un=0. If this condition is not satisﬁed then the series must diverge. Even if it is satisﬁed, however, the series may still diverge, and further testing is required. Comparison test The comparison test is the most basic test for convergence. Let us consider two seriessummationtextunandsummationtextvnand suppose that we know the latter to be convergent (by some earlier analysis, for example). Then, if each term unin the ﬁrst series is less than or equal to the corresponding term vnin the second series, for all ngreater than some ﬁxed number Nwhich will vary from series to series, then the original seriessummationtextunis also convergent. In other words, ifsummationtextvnis convergent and un≤vnforn>N , thensummationtextunconverges. However, ifsummationtextvndiverges and un≥vnfor all ngreater than some ﬁxed number thensummationtextundiverges.IDetermine whether the following series converges: ∞X n=11 n!+1=1 2+1 3+1 7+1 25+···. (4.7) Let us compare this series with the series ∞X n=01 n!=1 0!+1 1!+1 2!+1 3!+···=2+1 2!+1 3!+···, (4.8) 128 4.3 CONVERGENCE OF INFINITE SERIES which is merely the series obtained by setting x= 1 in the Maclaurin expansion of exp x (see subsection 4.6.3), i.e. exp(1) = e=1+1 1!+1 2!+1 3!+···. Clearly this second series is convergent, since it consists of only positive terms and has a ﬁnite sum. Thus, since each term unin the series (4.7) is less than the corresponding term 1/n! in (4.8), we conclude from the comparison test that (4.7) is also convergent. J D’Alembert’s ratio test The ratio test determines whether a series converges by comparing the relative magnitude of successive terms. If we consider a seriessummationtextunand set ρ= lim n→∞parenleftbiggun+1 unparenrightbigg , (4.9) then if ρ<1 the series is convergent; if ρ>1 the series is divergent; if ρ=1 then the behaviour of the series is undetermined by this test. To prove this we observe that if the limit (4.9) is less than unity, i.e. ρ<1t h e n we can ﬁnd a value rin the range ρ<r< 1 and a value Nsuch that un+1 un<r , for all n>N . Now the terms unof the series that follow uNare uN+1,u N+2,u N+3, ..., and each of these is less than the corresponding term of ruN,r2uN,r3uN, ... . (4.10) However, the terms of (4.10) are those of a geometric series with a common ratio rthat is less than unity. This geometric series consequently converges and therefore, by the comparison test discussed above, so must the original seriessummationtextun. An analogous argument may be used to prove the divergent case when ρ>1.IDetermine whether the following series converges: ∞X n=01 n!=1 0!+1 1!+1 2!+1 3!+···=2+1 2!+1 3!+···. As mentioned in the previous example, this series may be obtained by setting x=1i nt h e Maclaurin expansion of exp x, and hence we know already that it converges and has the sum exp(1) = e. Nevertheless, we may use the ratio test to conﬁrm that it converges. Using (4.9), we have ρ= lim n→∞ /n! (n+1 ) ! / = lim n→∞ /1 n+1 / = 0 (4.11) and since ρ<1, the series converges, as expected. J 129 SERIES AND LIMITS Ratio comparison test As its name suggests, the ratio comparison test is a combination of the ratio and comparison tests. Let us consider the two seriessummationtextunandsummationtextvnand assume that we know the latter to be convergent. It may be shown that if un+1 un≤vn+1 vn for all ngreater than some ﬁxed value Nthensummationtextunis also convergent. Similarly, if un+1 un≥vn+1 vn for all suﬃciently large n,a n dsummationtextvndiverges thensummationtextunalso diverges.IDetermine whether the following series converges: ∞X n=11 (n!)2=1+1 22+1 62+···. In this case the ratio of successive terms, as ntends to inﬁnity, is given by R= lim n→∞ /n! (n+1 ) ! /2 = lim n→∞ /1 n+1 /2 , which is less than the ratio seen in (4.11). Hence, by the ratio comparison test, the series converges. (It is clear that this series could also be found to be convergent using the ratiotest.)J Quotient test The quotient test may also be considered as a combination of the ratio and comparison tests. Let us again consider the two seriessummationtextunandsummationtextvn, and deﬁne ρas the limit ρ= lim n→∞parenleftbiggun vnparenrightbigg . (4.12) Then, it can be shown that: (i) if ρ/negationslash= 0 but is ﬁnite thensummationtextunandsummationtextvneither both converge or both diverge; (ii) if ρ=0a n dsummationtextvnconverges thensummationtextunconverges; (iii) if ρ=∞andsummationtextvndiverges thensummationtextundiverges. 130 4.3 CONVERGENCE OF INFINITE SERIESIGiven that the series P∞ n=11/ndiverges, determine whether the following series converges: ∞X n=14n2−n−3 n3+2n. (4.13) If we set un=( 4n2−n−3)/(n3+2n)a n d vn=1/nthen the limit (4.12) becomes ρ= lim n→∞ /(4n2−n−3)/(n3+2n) 1/n / = lim n→∞ /4n3−n2−3n n3+2n / =4. Since ρis ﬁnite but non-zero and Pvndiverges, from (i) above Punmust also diverge. J Integral test The integral test is an extremely powerful means of investigating the convergence of a seriessummationtextun. Suppose that there exists a function f(x) which monotonically decreases for xgreater than some ﬁxed value x0and for which f(n)=un,i . e .t h e value of the function at integer values of xis equal to the corresponding term in the series under investigation. Then it can be shown that, if the limit of the integral lim N→∞integraldisplayN f(x)dx exists, the seriessummationtextunis convergent. Otherwise the series diverges. Note that the integral deﬁned here has no lower limit; the test is sometimes stated with lowerlimit of unity for the integral, but this can lead to unnecessary diﬃculties.IDetermine whether the following series converges: ∞X n=11 (n−3/2)2=4+4+4 9+4 25+···. Let us consider the function f(x)=(x−3/2)−2. Clearly f(n)=unandf(x) monotonically decreases for x>3/2. Applying the integral test, we consider lim N→∞ ZN1 (x−3/2)2dx= lim N→∞ /−1 N−3/2 / =0. Since the limit exists the series converges. Note, however, that if we had included a lower limit of unity in the integral then we would have run into problems, since the integranddiverges at x=3/2.J The integral test is also useful for examining the convergence of the Riemann zeta series. This is a special series that occurs regularly and is of the form ∞summationdisplay n=11 np. It converges for p>1 and diverges if p≤1. These convergence criteria may be derived as follows. 131 SERIES AND LIMITS Using the integral test, we consider lim N→∞integraldisplayN1 xpdx= lim N→∞parenleftbiggN1−p 1−pparenrightbigg , and it is obvious that the limit tends to zero for p>1a n dt o∞forp≤1. Cauchy’s root test Cauchy’s root test may be useful in testing for convergence, especially if the nth terms of the series contains an nth power. If we deﬁne the limit ρ= lim n→∞(un)1/n, then it may be proved that the seriessummationtextunconverges if ρ<1. If ρ>1 then the series diverges. Its behaviour is undetermined if ρ=1 .IDetermine whether the following series converges: ∞X n=1 /1 n /n =1+1 4+1 27+···. Using Cauchy’s root test, we ﬁnd ρ= lim n→∞ /1 n / =0, and hence the series converges. J Grouping terms We now consider the Riemann zeta series, mentioned above, with an alternative proof of its convergence that uses the method of grouping terms. In general thereare better ways of determining convergence, but the grouping method may beused if it is not immediately obvious how to approach a problem by a better method. First consider the case where p>1 and group the terms in the series as follows: S N=1 1p+parenleftbigg1 2p+1 3pparenrightbigg +parenleftbigg1 4p+···+1 7pparenrightbigg +···. Now we can see that each bracket of this series is less than each term of the geometric series SN=1 1p+2 2p+4 4p+···. This geometric series has common ratio r=parenleftbig1 2parenrightbigp−1; therefore r<1s i n c e p>1, and so the geometric series converges. Then the comparison test shows that theRiemann zeta series also converges for p>1. 132 4.3 CONVERGENCE OF INFINITE SERIES The divergence of the Riemann zeta series for p≤1 can be seen by ﬁrst considering the case p= 1. The series is SN=1+1 2+1 3+1 4+···, which does notconverge, as may be seen by bracketing the terms of the series in groups in the following way: SN=Nsummationdisplay n=1un=1+parenleftbigg1 2parenrightbigg +parenleftbigg1 3+1 4parenrightbigg +parenleftbigg1 5+1 6+1 7+1 8parenrightbigg +···. The sum of the terms in each bracket is ≥1 2and, since as many such groupings can be made as we wish, it is clear that SNincreases indeﬁnitely as Nis increased. Now returning to the case of the Riemann zeta series for p<1, we note that each term in the series is greater than the corresponding one in the series forwhich p=1 .I no t h e rw o r d s1 /n p>1/nforn>1,p<1. The comparison test then shows us that the Riemann zeta series will diverge for all p≤1. 4.3.3 Alternating series test The tests discussed in the last subsection have been concerned with determining whether the series of real positive termssummationtext|un|converges, and so whethersummationtextun is absolutely convergent. Nevertheless, it is sometimes useful to consider whether a series is merely convergent rather than absolutely convergent. This is especially true for series containing an inﬁnite number of both positive and negative terms. In particular, we will consider the convergence of series in which the positive andnegative terms alternate, i.e. an alternating series . An alternating series can be written as ∞summationdisplay n=1(−1)n+1un=u1−u2+u3−u4+u5−···, with all un≥0. Such a series can be shown to converge provided (i) un→0a s n→∞and (ii) un<u n−1for all n>N for some ﬁnite N. If these conditions are not met then the series oscillates. To prove this, suppose for deﬁniteness that Nis odd and consider the series starting at uN. The sum of its ﬁrst 2 mterms is S2m=(uN−uN+1)+(uN+2−uN+3)+···+(uN+2m−2−uN+2m−1). By condition (ii) above, all the parentheses are positive, and so S2mincreases as mincreases. We can also write, however, S2m=uN−(uN+1−uN+2)−···−(uN+2m−3−uN+2m−2)−uN+2m−1, and since each parenthesis is positive, we must have S2m<u N. Thus, since S2m 133 SERIES AND LIMITS is always less than uNfor all mandun→0a s n→∞, the alternating series converges. It is clear that an analogous proof can be constructed in the casewhere Nis even.IDetermine whether the following series converges: ∞X n=1(−1)n+11 n=1−1 2+1 3−···. This alternating series clearly satisﬁes conditions (i) and (ii) above and hence converges. However, as shown above by the method of grouping terms, the corresponding series withall positive terms is divergent.J 4.4 Operations with series Simple operations with series are fairly intuitive, and we discuss them here only for completeness. The following points apply to both ﬁnite and inﬁnite seriesunless otherwise stated. (i) Ifsummationtextu n=Sthensummationtextkun=kSwhere kis any constant. (ii) Ifsummationtextun=Sandsummationtextvn=Tthensummationtext(un+vn)=S+T. (iii) Ifsummationtextun=Sthen a+summationtextun=a+S. A simple extension of this trivial result shows that the removal or insertion of a ﬁnite number of terms anywherein a series does not aﬀect its convergence. (iv) If the inﬁnite seriessummationtextu nandsummationtextvnare both absolutely convergent then the seriessummationtextwn,w h e r e wn=u1vn+u2vn−1+···+unv1, is also absolutely convergent. The seriessummationtextwnis called the Cauchy product of the two original series. Furthermore, ifsummationtextunconverges to the sum S andsummationtextvnconverges to the sum Tthensummationtextwnconverges to the sum ST. (v) It is not true in general that term-by-term diﬀerentiation or integration of a series will result in a new series with the same convergence properties. 4.5 Power series A power series has the form P(x)=a0+a1x+a2x2+a3x3+···, where a0,a1,a2,a3etc. are constants. Such series regularly occur in physics and engineering and are useful because, for |x|<1, the later terms in the series may become very small and be discarded. For example the series P(x)=1+ x+x2+x3+···, 134 4.5 POWER SERIES although in principle inﬁnitely long, in practice may be simpliﬁed if xhappens to have a value small compared with unity. To see this note that P(x)f o r x=0.1 has the following values: 1, if just one term is taken into account; 1.1, for twoterms; 1.11, for three terms; 1.111, for four terms, etc. If the quantity that it represents can only be measured with an accuracy of two decimal places, then all but the ﬁrst three terms may be ignored, i.e. when x=0.1o rl e s s P(x)=1+ x+x 2+O ( x3)≈1+x+x2. This sort of approximation is often used to simplify equations into manageable forms. It may seem imprecise at ﬁrst but is perfectly acceptable insofar as itmatches the experimental accuracy that can be achieved. The symbols O and ≈used above need some further explanation. They are used to compare the behaviour of two functions when a variable upon which both functions depend tends to a particular limit, usually zero or inﬁnity (and obvious from the context). For two functions f(x)a n d g(x), with gpositive, the formal deﬁnitions of the above symbols are as follows: (i) If there exists a constant ksuch that|f|≤kgas the limit is approached then f=O ( g). (ii) If as the limit of xis approached f/gtends to a limit l,w h e r e l/negationslash=0 ,t h e n f≈lg. The statement f≈gmeans that the ratio of the two sides tends to unity. 4.5.1 Convergence of power series The convergence or otherwise of power series is a crucial consideration in practical terms. For example, if we are to use a power series as an approximation, it is clearly important that it tends to the precise answer as more and more terms ofthe approximation are taken. Consider the general power series P(x)=a 0+a1x+a2x2+···. Using d’Alembert’s ratio test (see subsection 4.3.2), we see that P(x)c o n v e r g e s absolutely if ρ= lim n→∞vextendsinglevextendsinglevextendsinglevextendsinglea n+1 anxvextendsinglevextendsinglevextendsinglevextendsingle=|x|lim n→∞vextendsinglevextendsinglevextendsinglevextendsinglea n+1 anvextendsinglevextendsinglevextendsinglevextendsingle<1. Thus the convergence of P(x) depends upon the value of x,i . e .t h e r ei s ,i ng e n e r a l , a range of values of xfor which P(x)c o n v e r g e s ,a n interval of convergence .N o t e that at the limits of this range ρ= 1, and so the series may converge or diverge. The convergence of the series at the end-points may be determined by substituting these values of xinto the power series P(x) and testing the resulting series using any applicable method (discussed in section 4.3). 135 SERIES AND LIMITSIDetermine the range of values of xfor which the following power series converges: P(x)=1+2 x+4x2+8x3+···. By using the interval-of-convergence method discussed above, ρ= lim n→∞ ////2n+1 2nx ////=|2x|, and hence the power series will converge for |x|<1/2. Examining the end-points of the interval separately, we ﬁnd P(1/2 )=1+1+1+ ···, P(−1/2) = 1−1+1−···. Obviously P(1/2) diverges, while P(−1/2) oscillates. Therefore P(x) is not convergent at either end-point of the region but is convergent for −1<x< 1. J The convergence of power series may be extended to the case where the parameter zis complex. For the power series P(z)=a0+a1z+a2z2+···, we ﬁnd that P(z)c o n v e r g e si f ρ= lim n→∞vextendsinglevextendsinglevextendsinglevextendsinglea n+1 anzvextendsinglevextendsinglevextendsinglevextendsingle=|z|lim n→∞vextendsinglevextendsinglevextendsinglevextendsinglea n+1 anvextendsinglevextendsinglevextendsinglevextendsingle<1. We therefore have a range in |z|for which P(z)c o n v e r g e s ,i . e . P(z)c o n v e r g e s for values of zlying within a circle in the Argand diagram (in this case centred on the origin of the Argand diagram). The radius of the circle is called the radius of convergence :i fzlies inside the circle, the series will converge whereas ifzlies outside the circle, the series will diverge; if, though, zlies on the circle then the convergence must be tested using another method. Clearly the radius ofconvergence Ris given by 1 /R= lim n→∞|an+1/an|.IDetermine the range of values of zfor which the following complex power series converges: P(z)=1−z 2+z2 4−z3 8+···. We ﬁnd that ρ=|z/2|, which shows that P(z)c o n v e r g e sf o r |z|<2. Therefore the circle of convergence in the Argand diagram is centred on the origin and has a radius R=2 . On this circle we must test the conve rgence by substituting the value of zintoP(z)a n d considering the resulting series. On the circle of convergence we can write z=2 e x p iθ. Substituting this into P(z), we obtain P(z)=1−2exp iθ 2+4e x p2 iθ 4−··· =1−expiθ+[ e x p iθ]2−···, which is a complex inﬁnite geometric series with ﬁrst term a= 1 and common ratio 136 4.5 POWER SERIES r=−expiθ. Therefore, on the the circle of convergence we have P(z)=1 1+e x p iθ. Unless θ=πthis is a ﬁnite complex number, and so P(z) converges at all points on the circle|z|= 2 except at θ=π(i.e.z=−2), where it diverges. Note that P(z) is just the binomial expansion of (1 + z/2)−1, for which it is obvious that z=−2 is a singular point. In general, for power series expansions of complex functions about a given point in thecomplex plane, the circle of convergence extends as far as the nearest singular point. This is discussed further in chapter 20.J Note that the centre of the circle of convergence does not necessarily lie at the origin. For example, applying the ratio test to the complex power series P(z)=1+z−1 2+(z−1)2 4+(z−1)3 8+···, we ﬁnd that for it to converge we require |(z−1)/2|<1. Thus the series converges forzlying within a circle of radius 2 centred on the point (1,0) in the Argand diagram. 4.5.2 Operations with power series The following rules are useful when manipulating power series; they apply to power series in a real or complex variable. ( i )I ft w op o w e rs e r i e s P(x)a n d Q(x) have regions of convergence that overlap to some extent then the series produced by taking the sum, the diﬀerence or theproduct of P(x)a n d Q(x) converges in the common region. (ii) If two power series P(x)a n d Q(x) converge for all values of xthen one series may be substituted into the other to give a third series, which also convergesf o ra l lv a l u e so f x. For example, consider the power series expansions of sin xand e xgiven below in subsection 4.6.3, sinx=x−x3 3!+x5 5!−x7 7!+··· ex=1+ x+x2 2!+x3 3!+x4 4!+···, both of which converge for all values of x. Substituting the series for sin xinto that for exwe obtain esinx=1+ x+x2 2!−3x4 4!−8x5 5!+···, which also converges for all values of x. If, however, either of the power series P(x)a n d Q(x) has only a limited region of convergence, or if they both do so, then further care must be taken when substituting one series into the other. For example, suppose Q(x)c o n v e r g e sf o r allx, but P(x)o n l yc o n v e r g e sf o r xwithin a ﬁnite range. We may substitute 137 SERIES AND LIMITS Q(x)i n t o P(x)t oo b t a i n P(Q(x)), but we must be careful since the value of Q(x) may lie outside the region of convergence for P(x), with the consequence that the resulting series P(Q(x)) does not converge. (iii) If a power series P(x) converges for a particular range of xthen the series obtained by diﬀerentiating every term and the series obtained by integrating every term also converge in this range. This is easily seen for the power series P(x)=a0+a1x+a2x2+···, which converges if |x|<limn→∞|an/an+1|≡k. The series obtained by diﬀerenti- ating P(x) with respect to xis given by dP dx=a1+2a2x+3a3x2+··· and converges if |x|<lim n→∞vextendsinglevextendsinglevextendsinglevextendsinglenan (n+1 )an+1vextendsinglevextendsinglevextendsinglevextendsingle=k. Similarly the series obtained by integrating P(x) term by term, integraldisplay P(x)dx=a0x+a1x2 2+a2x3 3+···, converges if |x|<lim n→∞vextendsinglevextendsinglevextendsinglevextendsingle(n+2 )an (n+1 )an+1vextendsinglevextendsinglevextendsinglevextendsingle=k. So, series resulting from diﬀerentiation or integration have the same interval of convergence as the original series. However, even if the original series convergesat either end-point of the interval, it is not necessarily the case that the new serieswill do so. These new series must be tested separately at the end-points in orderto determine whether they converge there. Note that although power series maybe integrated or diﬀerentiated without altering their interval of convergence, thisis not true for series in general. It is also worth noting that diﬀerentiating or integrating a power series term by term within its interval of convergence is equivalent to diﬀerentiating orintegrating the function it represents. For example, consider the power seriesexpansion of sin x, sinx=x−x 3 3!+x5 5!−x7 7!+···, (4.14) which converges for all values of x. If we diﬀerentiate term by term, the series becomes 1−x2 2!+x4 4!−x6 6!+···, which is the series expansion of cos x,a sw ee x p e c t . 138 4.6 TAYLOR SERIES 4.6 Taylor series Taylor’s theorem provides a way of expressing a function as a power series in x, known as a Taylor series , but it can be applied only to those functions that are continuous and diﬀerentiable within the x-range of interest. 4.6.1 Taylor’s theorem Suppose that we have a function f(x) that we wish to express as a power series inx−aabout the point x=a. We shall assume that, in a given x-range, f(x) is a continuous, single-valued function of xhaving continuous derivatives with respect to x, denoted by f/prime(x),f/prime/prime(x) and so on, up to and including f(n−1)(x). We shall also assume that f(n)(x) exists in this range. From the equation following (2.31) we may write integraldisplaya+h af/prime(x)dx=f(a+h)−f(a), where a,a+hare neighbouring values of x. Rearranging this equation, we may express the value of the function at x=a+hin terms of its value at aby f(a+h)=f(a)+integraldisplaya+h af/prime(x)dx. (4.15) Aﬁrst approximation forf(a+h) may be obtained by substituting f/prime(a)f o r f/prime(x) in (4.15), to obtain f(a+h)≈f(a)+hf/prime(a). This approximation is shown graphically in ﬁgure 4.1. We may write this ﬁrst approximation in terms of xandaas f(x)≈f(a)+(x−a)f/prime(a), and, in a similar way, f/prime(x)≈f/prime(a)+(x−a)f/prime/prime(a) f/prime/prime(x)≈f/prime/prime(a)+(x−a)f/prime/prime/prime(a), and so on. Substituting for f/prime(x) in (4.15), we obtain the second approximation : f(a+h)≈f(a)+integraldisplaya+h a[f/prime(a)+(x−a)f/prime/prime(a)]dx ≈f(a)+hf/prime(a)+h2 2f/prime/prime(a). We may repeat this procedure as often as we like (so long as the derivatives 139 SERIES AND LIMITS f(a)f(x) a a+hxPQ R θhf/prime(a) h Figure 4.1 The ﬁrst-order Taylor series approximation to a function f(x). The slope of the function at P,i . e .t a n θ,e q u a l s f/prime(a). Thus the value of the function at Q,f(a+h), is approximated by the ordinate of R,f(a)+hf/prime(a). off(x) exist) to obtain higher-order approximations to f(a+h); we ﬁnd the (n−1)th-order approximation †to be f(a+h)≈f(a)+hf/prime(a)+h2 2!f/prime/prime(a)+···+hn−1 (n−1)!f(n−1)(a). (4.16) As might have been anticipated, the error associated with approximating f(a+h) by this ( n−1)th-order power series is of the order of the next term in the series. This error or remainder can be shown to be given by Rn(h)=hn n!f(n)(ξ), for some ξthat lies in the range [ a, a+h]. Taylor’s theorem then states that we may write the equality f(a+h)=f(a)+hf/prime(a)+h2 2!f/prime/prime(a)+···+h(n−1) (n−1)!f(n−1)(a)+Rn(h). (4.17) The theorem may also be written in a form suitable for ﬁnding f(x)g i v e n the value of the function and its relevant derivatives at x=a, by substituting †The order of the approximation is simply the highest power of hin the series. Note, though, that the (n−1)th-order approximation contains nterms. 140 4.6 TAYLOR SERIES x=a+hin the above expression. It then reads f(x)=f(a)+(x−a)f/prime(a)+(x−a)2 2!f/prime/prime(a)+···+(x−a)n−1 (n−1)!f(n−1)(a)+Rn(x), (4.18) where the remainder now takes the form Rn(x)=(x−a)n n!f(n)(ξ), andξlies in the range [ a, x]. Each of the formulae (4.17), (4.18) gives us the Taylor expansion of the function about the point x=a. A special case occurs when a= 0. Such Taylor expansions, about x= 0, are called Maclaurin series . Taylor’s theorem is also valid without signiﬁcant modiﬁcation for functions of a complex variable (see chapter 20). The extension of Taylor’s theorem tofunctions of more than one variable is given in chapter 5. For a function to be expressible as an inﬁnite power series we require it to be inﬁnitely diﬀerentiable and the remainder term R nto tend to zero as ntends to inﬁnity, i.e. lim n→∞Rn= 0. In this case the inﬁnite power series will represent the function within the interval of convergence of the series.IExpand f(x)=s i n xas a Maclaurin series, i.e. about x=0. We must ﬁrst verify that sin xmay indeed be represented by an inﬁnite power series. It is easily shown that the nth derivative of f(x)i sg i v e nb y f(n)(x)=s i n / x+nπ 2 / . Therefore the remainder after expanding f(x)a sa n( n−1)th-order polynomial about x= 0 is given by Rn(x)=xn n!sin / ξ+nπ 2 / , where ξlies in the range [0 ,x]. Since the modulus of the sine term is always less than or equal to unity, we can write |Rn(x)|<|xn|/n!. For any particular value of x,s a y x=c, Rn(c)→0a s n→∞. Hence lim n→∞Rn(x) = 0, and so sin xcan be represented by an inﬁnite Maclaurin series. Evaluating the function and its derivatives at x=0w eo b t a i n f(0) = sin 0 = 0 , f/prime(0) = sin( π/2) = 1 , f/prime/prime(0) = sin π=0, f/prime/prime/prime(0) = sin(3 π/2) =−1, and so on. Therefore, the Maclaurin series expansion of sin xis given by sinx=x−x3 3!+x5 5!−···. Note that, as expected, since sin xis an odd function, its power series expansion contains only odd powers of x. J 141 SERIES AND LIMITS We may follow a similar procedure to obtain a Taylor series about an arbitrary point x=a.IExpand f(x)=c o s xas a Taylor series about x=π/3. As in the above example, it is easily shown that the nth derivative of f(x)i sg i v e nb y f(n)(x)=c o s / x+nπ 2 / . Therefore the remainder after expanding f(x)a sa n( n−1)th-order polynomial about x=π/3i sg i v e nb y Rn(x)=(x−π/3)n n!cos / ξ+nπ 2 / , where ξlies in the range [ π/3,x]. The modulus of the cosine term is always less than or equal to unity, and so |Rn(x)|<|(x−π/3)n|/n!. As in the previous example, lim n→∞Rn(x)= 0 for any particular value of x,a n ds oc o s xcan be represented by an inﬁnite Taylor series about x=π/3. Evaluating the function and its derivatives at x=π/3w eo b t a i n f(π/3) = cos( π/3) = 1 /2, f/prime(π/3) = cos(5 π/6) =−√ 3/2, f/prime/prime(π/3) = cos(4 π/3) =−1/2, and so on. Thus the Taylor series expansion of cos xabout x=π/3i sg i v e nb y cosx=1 2−√ 3 2 /; x−π/3 / −1 2 /; x−π/3 /2 2!+···. J 4.6.2 Approximation errors in Taylor series In the previous subsection we saw how to represent a function f(x) by an inﬁnite power series, which is exactly equal to f(x)f o ra l l xwithin the interval of convergence of the series. However, in physical problems we usually do not wantto have to sum an inﬁnite number of terms, but prefer to use only a ﬁnite numberof terms in the Taylor series to approximate the function in some given range ofx. In this case it is desirable to know what is the maximum possible error associated with the approximation. As given in (4.18), a function f(x) can be represented by a ﬁnite ( n−1)th-order power series together with a remainder term such that f(x)=f(a)+(x−a)f /prime(a)+(x−a)2 2!f/prime/prime(a)+···+(x−a)n−1 (n−1)!f(n−1)(a)+Rn(x), where Rn(x)=(x−a)n n!f(n)(ξ) andξlies in the range [ a, x].Rn(x) is the remainder term, and represents the error in approximating f(x)b yt h ea b o v e( n−1)th-order power series. Since the exact 142 4.6 TAYLOR SERIES value of ξthat satisﬁes the expression for Rn(x) is not known, an upper limit on the error may be found by diﬀerentiating Rn(x)w i t hr e s p e c tt o ξand equating the derivative to zero in the usual way for ﬁnding maxima.IExpand f(x)=c o s xas a Taylor series about x=0and ﬁnd the error associated with using the approximation to evaluate cos(0 .5)if only the ﬁrst two non-vanishing terms are taken. (Note that the Taylor expansions of trigonometrical functions are only valid forangles measured in radians.) Evaluating the function and its derivatives at x= 0, we ﬁnd f(0) = cos0 = 1 , f/prime(0) =−sin 0 = 0 , f/prime/prime(0) =−cos 0 =−1, f/prime/prime/prime(0) = sin0 = 0 . So, for small |x|, we ﬁnd from (4.18) cosx≈1−x2 2. Note that since cos xis an even function, its power series expansion contains only even powers of x. Therefore, in order to estimate the error in this approximation, we must consider the term in x4, which is the next in the series. The required derivative is f(4)(x) a n dt h i si s( b yc h a n c e )e q u a lt oc o s x. Thus, adding in the remainder term R4(x), we ﬁnd cosx=1−x2 2+x4 4!cosξ, where ξlies in the range [0 ,x]. Thus, the maximum possible error is x4/4!, since cos ξ cannot exceed unity. If x=0.5, taking just the ﬁrst two terms yields cos(0 .5)≈0.875 with a predicted error of less than 0 .00260. In fact cos(0 .5) = 0 .87758 to 5 decimal places. Thus, to this accuracy, the true error is 0.00258, an error of about 0.3%. J 4.6.3 Standard Maclaurin series It is often useful to have a readily available table of Maclaurin series for standard elementary functions, and therefore these are listed below. sinx=x−x3 3!+x5 5!−x7 7!+···for−∞<x<∞, cosx=1−x2 2!+x4 4!−x6 6!+···for−∞<x<∞, tan−1x=x−x3 3+x5 5−x7 7+···for−1<x< 1, ex=1+ x+x2 2!+x3 3!+x4 4!+···for−∞<x<∞, ln(1 + x)=x−x2 2+x3 3−x4 4+···for−1<x≤1, (1 +x)n=1+ nx+n(n−1)x2 2!+n(n−1)(n−2)x3 3!+···for−∞<x<∞. 143 SERIES AND LIMITS These can all be derived by straightforward application of Taylor’s theorem to the expansion of a function about x=0 . 4.7 Evaluation of limits The idea of the limit of a function f(x)a sxapproaches a value ais fairly intuitive, though a strict deﬁnition exists and is stated below. In many cases, the limit of the function as xapproaches awill be simply the value f(a), but in others this is not so. Firstly, the function may be undeﬁned at x=a, as, for example, when f(x)=sinx x, which takes the value 0 /0a t x= 0. However, the limit as xapproaches zero does exist and can be evaluated as unity using l’H ˆopital’s rule below. Another possibility is that even if f(x) is deﬁned at x=aits value may not be equal to the limiting value lim x→af(x). This can occur for a discontinuous function at a point of discontinuity. The strict deﬁnition of a limit is that iflimx→af(x)=lthen for any number /epsilon1however small, it must be possible to ﬁnd a number ηsuch that |f(x)−l|</epsilon1whenever|x−a|<η.In other words, as xbecomes arbitrarily close to a,f(x) becomes arbitrarily close to its limit, l. To remove any ambiguity, it should be stated that, in general, the number ηwill depend on both /epsilon1and the form of f(x). The following observations are often useful in ﬁnding the limit of a function. (i) A limit may be ±∞. For example as x→0, 1/x2→∞ . (ii) A limit may be approached from below or above and the value may be diﬀerent in each case. For example consider the function f(x)=t a n x.A sxtends toπ/2f r o mb e l o w f(x)→∞, but if the limit is approached from above then f(x)→−∞ . Another way of writing this is lim x→π 2−tanx=∞, lim x→π 2+tanx=−∞. (iii) It may ease the evaluation of limits if the function under consideration is split into a sum, product or quotient. Provided each of the limits exists, the rulesfor evaluating such limits are as follows. (a) lim x→a{f(x)+g(x)}= lim x→af(x) + lim x→ag(x). (b) lim x→a{f(x)g(x)}= lim x→af(x) lim x→ag(x). (c) lim x→af(x) g(x)=limx→af(x) limx→ag(x), provided that the numerator and denominator are not both equal to zero or inﬁnity. Examples of cases (a)–(c) are discussed below. 144 4.7 EVALUATION OF LIMITSIEvaluate the limits lim x→1(x2+2x3),lim x→0(xcosx), lim x→π/2sinx x. Using (a) above, lim x→1(x2+2x3) = lim x→1x2+ lim x→12x3=3. Using (b), lim x→0(xcosx) = lim x→0xlim x→0cosx=0×1=0 . Using (c), lim x→π/2sinx x=limx→π/2sinx limx→π/2x=1 π/2=2 π. J (iv) Limits of functions of xthat contain exponents that themselves depend on xcan often be found by taking logarithms.IEvaluate the limit lim x→∞ / 1−a2 x2 /x2 . Let us deﬁne y= / 1−a2 x2 /x2 and consider the logarithm of the required limit, i.e. lim x→∞lny= lim x→∞ / x2ln / 1−a2 x2 // . Using the Maclaurin series for ln(1 + x) given in subsection 4.6.3, we can expand the logarithm as a series and obtain lim x→∞lny= lim x→∞ / x2 / −a2 x2−a4 2x4+··· // =−a2. Therefore, since lim x→∞lny=−a2it follows that lim x→∞y=e x p (−a2). J (v) L’H ˆopital’s rule may be used; it is an extension of (iii)(c) above. In cases where both numerator and denominator are zero or both are inﬁnite, furtherconsideration of the limit must follow. Let us ﬁrst consider lim x→af(x)/g(x), where f(a)=g(a) = 0. Expanding the numerator and denominator as Taylor series we obtain f(x) g(x)=f(a)+(x−a)f/prime(a)+[ ( x−a)2/2!]f/prime/prime(a)+··· g(a)+(x−a)g/prime(a)+[ ( x−a)2/2!]g/prime/prime(a)+···. However, f(a)=g(a)=0s o f(x) g(x)=f/prime(a)+[ ( x−a)/2!]f/prime/prime(a)+··· g/prime(a)+[ ( x−a)/2!]g/prime/prime(a)+···. 145 SERIES AND LIMITS Therefore we ﬁnd lim x→af(x) g(x)=f/prime(a) g/prime(a), provided f/prime(a)a n d g/prime(a) are not themselves both equal to zero. If, however, f/prime(a)a n d g/prime(a)areboth zero then the same process can be applied to the ratio f/prime(x)/g/prime(x) to yield lim x→af(x) g(x)=f/prime/prime(a) g/prime/prime(a), provided that at least one of f/prime/prime(a)a n d g/prime/prime(a) is non-zero. If the original limit does exist then it can be found by repeating the process as many times as is necessaryfor the ratio of corresponding nth derivatives not to be of the indeterminate form 0/0, i.e. lim x→af(x) g(x)=f(n)(a) g(n)(a).IEvaluate the limit lim x→0sinx x. We ﬁrst note that if x= 0, both numerator and denominator are zero. Thus we apply l’Hˆopital’s rule: diﬀerentiating, we obtain lim x→0(sinx/x) = lim x→0(cosx/1) = 1 . J So far we have only considered the case where f(a)=g(a)=0 .F o rt h ec a s e where f(a)=g(a)=∞we may still apply l’H ˆopital’s rule by writing lim x→af(x) g(x)= lim x→a1/g(x) 1/f(x), w h i c hi sn o wo ft h ef o r m0 /0a t x=a. Note also that l’H ˆopital’s rule is still valid for ﬁnding limits as x→∞,i . e .w h e n a=∞. This is easily shown by letting y=1/xas follows: lim x→∞f(x) g(x)= lim y→0f(1/y) g(1/y) = lim y→0−f/prime(1/y)/y2 −g/prime(1/y)/y2 = lim y→0f/prime(1/y) g/prime(1/y) = lim x→∞f/prime(x) g/prime(x). 146 4.8 EXERCISES Summary of methods for evaluating limits To ﬁnd the limit of a continuous function f(x) at a point x=a, simply substitute the value ainto the function noting that0 ∞=0a n dt h a t∞ 0=∞.T h eo n l y diﬃculty occurs when either of the expressions0 0or∞ ∞results. In this case diﬀerentiate top and bottom and try again. Continue diﬀerentiating until the topand bottom limits are no longer both zero or both inﬁnity. If the undeterminedform 0×∞occurs then it can always be rewritten as 0 0or∞ ∞. 4.8 Exercises 4.1 Sum the even numbers between 1000 and 2000 inclusive. 4.2 If you invest £1000 on the ﬁrst day of each year, and interest is paid at 5% on your balance at the end of each year, how much money do you have after 25 years? 4.3 How does the convergence of the series ∞X n=r(n−r)! n! depend on the integer r? 4.4 Show that for testing the convergence of of the series x+y+x2+y2+x3+y3+···, where 0 <x<y< 1, the D’Alembert ratio test fails but the Cauchy root test is successful. 4.5 Find the sum SNof the ﬁrst Nterms of the following series, and hence determine whether the series are convergent, divergent, or oscillatory: (a)∞X n=1ln /n+1 n / ,(b)∞X n=0(−2)n,(c)∞X n=1(−1)n+1n 3n. 4.6 By grouping and rearranging terms of the absolutely convergent series S=∞X n=11 n2, show that So=∞X nodd1 n2=3S 4. 4.7 Use the diﬀerence method to sum the series NX n=22n−1 2n2(n−1)2. 147 SERIES AND LIMITS 4.8 The N+ 1 complex numbers ωmare given by ωm=e x p ( 2 πim/N )f o r m= 0,1,2,... ,N . (a) Evaluate the following: (i)NX m=0ωm,(ii)NX m=0ω2 m,(iii)NX m=0ωmxm. (b) Use these results to evaluate (i)NX m=0 / cos /2πm N / −cos /4πm N // ,(ii)3X m=02msin /2πm 3 / . 4.9 Prove that cosθ+c o s ( θ+α)+···+c o s ( θ+nα)=sin1 2(n+1 )α sin1 2αcos(θ+1 2nα). 4.10 Determine whether the following series converge ( θand pare positive real numbers): (a)∞X n=12si nnθ n(n+1 ),(b)∞X n=12 n2,(c)∞X n=11 2n1/2, (d)∞X n=2(−1)n(n2+1 )1/2 nlnn,(e)∞X n=1np n!. 4.11 Find the real values of xfor which the following are series convergent: (a)∞X n=1xn n+1,(b)∞X n=1(sinx)n,(c)∞X n=1nx, (d)∞X n=1enx,(e)∞X n=2(lnn)x. 4.12 Determine whether the following series are convergent: (a)∞X n=1n1/2 (n+1 )1/2,(b)∞X n=1n2 n!,(c)∞X n=1(lnn)n nn/2,(d)∞X n=1nn n!. 4.13 Determine whether the following series are absolutely convergent, convergent or oscillatory: (a)∞X n=1(−1)n n5/2,(b)∞X n=1(−1)n(2n+1 ) n,(c)∞X n=0(−1)n|x|n n!, (d)∞X n=0(−1)n n2+3n+2,(e)∞X n=1(−1)n2n n1/2. 4.14 Determine the positive values of xfor which the following series converges: ∞X n=1xn/2e−n n. 148 4.8 EXERCISES 4.15 Prove that ∞X n=2ln /nr+(−1)n nr / is absolutely convergent for r= 2, but only conditionally convergent for r=1 . 4.16 An extension to the proof of the integral test (subsection 4.3.2) shows that, if f(x) is positive, continuous and monotonically decreasing, for x≥1, and the series f(1) + f(2) +···is convergent, then its sum does not exceed f(1) + L,w h e r e L is the integralZ∞ 1f(x)dx. Use this result to show that the sum ζ(p) of the Riemann zeta series Pn−p,w i t h p>1, is not greater than p/(p−1). 4.17 Demonstrate that rearranging the order of its terms can make a condition- ally convergent series converge to a diﬀerent limit by considering the seriesP(−1)n+1n−1=l n2=0 .693. Rearrange the series as S=1 1+1 3−1 2+1 5+1 7−1 4+1 9+1 11−1 6+1 13+··· and group each set of three successive terms. Show that the series can then be written ∞X m=18m−3 2m(4m−3)(4m−1), which is convergent (by comparison with Pn−2) and contains only positive terms. Evaluate the ﬁrst of these and hence deduce that Sis not equal to ln 2. 4.18 Illustrate result (iv) of section 4.4 about Cauchy products by considering the double summation S=∞X n=1nX r=11 r2(n+1−r)3. By examining the points in the nr-plane over which the double summation is to be carried out, show that Scan be written as S=∞X n=r∞X r=11 r2(n+1−r)3. Deduce that S≤3. 4.19 A Fabry–P ´erot interferometer consists of two parallel heavily silvered glass plates; light enters normally to the plates, an d undergoes repeated reﬂections between them, with a small transmitted fraction emerging at each reﬂection. Find theintensity|B| 2of the emerging wave, where B=A(1−r)∞X n=0rneinφ, with randφreal. 149 SERIES AND LIMITS 4.20 Identify the series ∞X n=1(−1)n+1x2n (2n−1)!, and then by integration and diﬀerentiation deduce the values Sof the following series, (a)∞X n=1(−1)n+1n2 (2n)!,( b )∞X n=1(−1)n+1n (2n+1 ) !, (c)∞X n=1(−1)n+1nπ2n 4n(2n−1)!,( d )∞X n=0(−1)n(n+1 ) (2n)!. 4.21 Starting from the Maclaurin series for cos x, show that (cosx)−2=1+ x2+2x4 3+···. Deduce the ﬁrst three terms in the Maclaurin series for tan x. 4.22 Find the Maclaurin series for (a)l n /1+x 1−x / ,(b)(x2+4 )−1,(c)s i n2x. 4.23 If f(x)=s i n h−1x,a n di t s nth derivative f(n)(x) is written as Pn(x)/(1 +x2)n−1/2, where Pn(x) is a polynomial (of order n−1), show that the Pn(x)s a t i s f yt h e recurrence relation Pn+1(x)=( 1+ x2)P/prime n(x)−(2n−1)xPn(x). Hence generate the coeﬃcients necessary to express sinh−1xas a Maclaurin series up to terms in x5. 4.24 Find the ﬁrst three non-zero terms in the Maclaurin series for the following functions: (a) (x2+9 )−1/2,(b) ln[(2 + x)3], (c) exp(sin x), (d) ln(cos x), (e) exp[−(x−a)−2],(f) tan−1x. 4.25 By using the logarithmic series, prove that if aandbare positive and nearly equal then lna b/similarequal2(a−b) a+b. Show that the error in this approximation is about 2( a−b)3/[3(a+b)3]. 4.26 Determine whether the following functions f(x) are (i) continuous, and (ii) diﬀerentiable at x=0 : (a)f(x)=e x p (−|x|); (b)f(x)=( 1−cosx)/x2forx/negationslash=0 , f(0) =1 2; (c)f(x)=xsin(1/x)f o r x/negationslash=0 , f(0) = 0; (d)f(x)=[ 4−x2], where [ y] denotes the integer part of y. 4.27 Find the limit as x→0o f[√ 1+xm−√ 1−xm]/xn,i nw h i c h mandnare positive integers. 4.28 Evaluate the following limits: (a) lim x→0sin3x sinhx, (b) lim x→0tanx−tanh x sinhx−x, (c) lim x→0tanx−x cosx−1, (d) lim x→0 /cosec x x3−sinhx x5 / . 150 4.8 EXERCISES 4.29 Find the limits of the following functions: (a)x3+x2−5x−2 2x3−7x2+4x+4,a s x→0,x→∞andx→2; (b)sinx−xcoshx sinhx−x,a s x→0; (c) Zπ/2 x /ycosy−siny y2 / dy,a s x→0. 4.30 Use Taylor expansions to three terms to ﬁnd approximations to (a)4√ 17, and (b)3√ 26. 4.31 Using a ﬁrst-order Taylor expansion about x=x0, show that a better approxi- mation than x0to the solution of the equation f(x)=s i n x+t a n x=2 is given by x=x0+h,w h e r e h=2−f(x0) cosx0+s e c2x0. (a) Use this procedure twice to ﬁnd the solution of f(x) = 2 to six signiﬁcant ﬁgures, given that it is close to x=0.9. (b) Use the result in (a) to deduce, to the same degree of accuracy, one solution of the quartic equation y4−4y3+4y2+4y−4=0 . 4.32 Evaluate lim x→0 /1 x3 / cosec x−1 x−x 6 // . 4.33 In quantum theory, a system of osc illators, each of fundamental frequency ν, interacting at temperature Thas an average energy ¯Egiven by ¯E= P∞ n=0nhνe−nxP∞ n=0e−nx, where x=hν/kT ,handkbeing the Planck and Boltzmann constants respectively. Prove that both series converge, evaluate their sums, and show that at hightemperatures ¯E≈kTwhilst at low temperatures ¯E≈hνexp(−hν/kT ). 4.34 In a very simple model of a crystal, point-like atomic ions are regularly spaced along an inﬁnite one-dimensional row with spacing R. Alternate ions carry equal and opposite charges ±e. The potential energy of the ith ion in the electric ﬁeld due to the jth ion is q iqj 4π/epsilon10rij, where qkis the charge on the kth ion and rijis the distance between the ith and jth ions. Write down a series giving the total contribution Viof the ith ion to the overall potential energy . Show that the series converges, and, if Viis written as Vi=αe2 4π/epsilon10R, ﬁnd a closed-form expression for α, the Madelung constant for this (unrealistic) lattice. 151 SERIES AND LIMITS 4.35 One of the factors contributing to the hi gh relative permittivi ty of water to static electric ﬁelds is the permanent electric dipole moment pof the water molecule. In an external ﬁeld Ethe dipoles tend to line up with the ﬁeld, but they do not do so completely because of thermal agitation at the temperature Tof the water. A classical (non-quantum) calculation using the Boltzmann distribution shows thatthe average polarisability per molecule αis given by α=p E(coth x−x−1), where x=pE/kT andkis the Boltzmann constant. At ordinary temperatures, even with high ﬁeld strengths (104Vm−1or more), x/lessmuch1. By making suitable series expansions of the hyperbolic functions involved, show that α=p2/3kTto an accuracy of about one part in 15 x−2. 4.36 In quantum theory a certain method (the Born approximation) gives the (so- called) amplitude f(θ) for the scattering of a particle of mass mthrough an angle θby a uniform potential well of depth V0and radius b(i.e. the potential energy of the particle is −V0within a sphere of radius band zero elsewhere) as f(θ)=2mV0/~2K3(sinKb−KbcosKb). Here /~is the Planck constant divided by 2 π, the energy of the particle is /~2k2/2m andKis 2ksin(θ/2). Use l’H ˆopital’s rule to evaluate the amplitude at low energies, i.e. when kand hence Ktend to zero, and so determine the low-energy total cross-section. (Note: the diﬀerential cross-section is given by |f(θ)|2a n dt h et o t a lc r o s s - s e c t i o n by the integral of this over all solid angles, i.e. 2 π Rπ 0|f(θ)|2sinθd θ.) 4.9 Hints and answers 4.1 2 P1000 500n= 751500. 4.2 Ar(rn−1)/(r−1) =£50 113 . 4.3 Divergent for r≤1; convergent for r≥2. 4.4 The ratio of successive terms oscillates between 0 and ∞asn→∞;un≤ (yn/2)1/n<1. 4.5 (a) ln( N+ 1), divergent; (b) [1 −(−2)n]/3, oscillates inﬁnitely; (c) Add SN/3t o theSNseries;3 16[1−(−3)−N]+3 4N(−3)−N−1,c o n v e r g e n tt o3 16. 4.6 Write all terms of the form (2 m)−2as1 4m−2; their sum is clearly1 4S. 4.7 (1 −N−2)/2. 4.8 (a) (i) 2 for N= 1; 1 otherwise. (ii) 2 for N=1 ;3f o r N= 2; 1 otherwise. (iii) 1 + xforN=1 ;[ 1−xN+1exp(2 πi/N)]/[1−xexp(2 πi/N)] otherwise. (b) (i) Consider Re( ωm−ω2 m);−2f o r N= 2; 0 otherwise. (ii) Consider Im(2mωm); −√ 3. 4.9 Sum the geometric series with rth term exp[ i(θ+rα)]. Its real part is {cosθ−cos[(n+1 )α+θ]−cos(θ−α)+c o s ( θ+nα)}/4sin2(α/2), which can be reduced to the given answer. 4.10 (a) Convergent, compare with Pn−1(n+1 )−1; (b) convergent, ratio test; (c) divergent, compare with Pn−1; (d) convergent, alternating signs; (e) convergent, ratio test. 152 4.9 HINTS AND ANSWERS 4.11 (a) −1≤x<1; (b) all xexcept x=( 2n±1)π/2; (c) x<−1; (d) x<0; (e) always divergent. Clearly divergent for x>−1. For−X=x<−1, consider ∞X k=1MkX n=Mk−1+11 (lnMk)X, where ln Mk=kand note that Mk−Mk−1=e−1(e−1)Mk; hence show that the series diverges. 4.12 (a) Divergent, undoes not tend to 0. (b) Convergen t, ratio test. (c) Convergent, root test. (d) Divergent, ratio tends to e,o rundoes not tend to 0. 4.13 (a) Absolutely convergent, compare wit h exercise 4.10(b). (b) Oscillates inﬁnitely. (c) Absolutely convergent for all x. (d) Absolutely convergent; use partial frac- tions. (e) Oscillates inﬁnitely. 4.14 x<e2, by the root test. 4.15 Divide the series into two series, nodd and neven. For r= 2 both are absolutely convergent, by comparison with Pn−2.F o r r=1n e i t h e rs e r i e si sc o n v e r g e n t , by comparison with Pn−1. However, the sum of the two is convergent, by the alternating sign test or by showing that the terms cancel in pairs. 4.17 The ﬁrst term has value 0.833 and all other terms are positive.4.18 The original summation ran along lines parallel to the r-axis; replace it with one running along lines parallel to the n-axis. Write n+1−r=sand deduce thatS=ζ(2)ζ(3), where ζ(p) is the Riemann zeta functionPn−p.U s et h er e s u l t proved in exercise 4.16 to give the stated conclusion. 4.19|A|2(1−r)2/(1 +r2−2rcosφ). 4.20 xsinx. (a) Diﬀerentiate once; set x=1 . S=( s i n1+c o s1 ) /4=0 .345. (b) Integrate once; set x=1 . S=( s i n1−cos 1) /2=0 .151. (c) Diﬀerentiate once; set x=π/2.S=π/4=0 .785. (d) Diﬀerentiate twice; set s=n−1a n d x=1 . S=( 2c o s1−sin1) /2=0 .120. 4.21 Use the binomial expansion and collect terms up to x4. Integrate both sides of the displayed equation. tan x=x+x3/3+2 x5/15 +···. 4.22 (a) X nodd2xn n;( b )∞X n=0(−1)n 4 /x 2 /2n ;( c )∞X n=1(−1)n+1(2x)2n 2(2n)!. 4.23 For example, P5(x)=2 4 x4−72x2+9 .s i n h−1x=x−x3/6+3 x5/40−···. 4.24 (a) [1 −(x2/18)−(3x4/648)] /3. (b) ln 8 + 3 x/2−3x2/8. (c) 1 + x+x2/2. (d) −x2/2−x4/12−x6/45. (e) exp(−a−2){1−2x/a3−x2[(3/a4)−(2/a6)]}.( f )x− x3/3+x5/5. 4.25 Set a=D+δandb=D−δand use the expansion for ln(1 ±δ/D). 4.26 (i) (a), (b) and (c) are continuous. (ii) Only (b) is diﬀerentiable.4.27 The limit is 0 for m>n ,1f o r m=n,a n d∞form<n . 4.28 (a) 3, (b) 4, (c) 0, (d) 1 90. 4.29 (a) −1 2,1 2,∞;( b )−4; (c)−1+2 /π. 4.30 (a) Expand f(x)=x1/4about x0= 16; approximation 2.030518, actual 2.030543. (b) Expand f(x)=x1/3about x0= 27; approximation 2.962506, actual 2.962496. 4.31 (a) First approximation 0.886452; second approximation 0.886287. (b) Set y= sinxand re-express f(x) = 2 as a polynomial equation. y= sin(0 .886287) = 0.774730. 4.32 7 /360. 4.33 E=hν[exp( hν/kT )−1]−1. 4.34 α=−2ln2. 4.36 f(θ)=2 mV0b3/3 /~2(i.e. independent of θ); 4π(2mV0b3/3 /~2)2. 153 5 Partial diﬀerentiation In chapter 2, we discussed functions fof only one variable x, which were usually written f(x). Certain constants and parameters may also have appeared in the deﬁnition of f,e . g . f(x)=ax+2 contains the constant 2 and the parameter a, but only xwas considered as a variable and only the derivatives f(n)(x)=dnf/dxn were deﬁned. However, we may equally well consider functions that depend on more than one variable, e.g. the function f(x, y)=x2+3xy, which depends on the two variables xandy. For any pair of values x, y, the function f(x, y) has a well-deﬁned value, e.g.f(2,3) = 22. This notion can clearly be extended to functions dependent on more than two variables. For the n-variable case, we write f(x1,x2,...,x n)f o r a function that depends on the variables x1,x2,...,x n.W h e n n=2 , x1andx2 correspond to the variables xandyused above. Functions of one variable, like f(x), can be represented by a graph on a plane sheet of paper, and it is apparent that functions of two variables can,with little eﬀort, be represented by a surface in three-dimensional space. Thus, we may also picture f(x, y) as describing the variation of height with position in a mountainous landscape. Functions of many variables, however, are usuallyvery diﬃcult to visualise and so the preliminary discussion in this chapter willconcentrate on functions of just two variables. 5.1 Deﬁnition of the partial derivative It is clear that a function f(x, y) of two variables will have a gradient in all directions in the xy-plane. A general expression for this rate of change can be found and will be discussed in the next section. However, we ﬁrst consider the simpler case of ﬁnding the rate of change of f(x, y) in the positive x-a n d y- directions. These rates of change are called the partial derivatives with respect 154 5.1 DEFINITION OF THE PARTIAL DERIVATIVE toxandyrespectively, and they are extremely important in a wide range of physical applications. For a function of two variables f(x, y) we may deﬁne the derivative with respect tox, for example, by saying that it is that for a one-variable function when yis held ﬁxed and treated as a constant. To signify that a derivative is with respect tox, but at the same time to recognize that a derivative with respect to yalso exists, the former is denoted by ∂f/∂x and is the partial derivative of f(x, y)with respect to x. Similarly, the partial derivative of fwith respect to yis denoted by ∂f/∂y . To deﬁne formally the partial derivative of f(x, y) with respect to x, we have ∂f ∂x= lim ∆x→0f(x+∆x, y)−f(x, y) ∆x, (5.1) provided that the limit exists. This is much the same as for the derivative of a one-variable function. The other partial derivative of f(x, y) is similarly deﬁned as a limit (provided it exists): ∂f ∂y= lim ∆y→0f(x, y+∆y)−f(x, y) ∆y. (5.2) It is common practice in connection with partial derivatives of functions involving more than one variable to indicate those variables that are held constant by writing them as subscripts to the derivative symbol. Thus, the partial derivatives deﬁned in (5.1) and (5.2) would be written respectively as parenleftbigg∂f ∂xparenrightbigg yandparenleftbigg∂f ∂yparenrightbigg x. In this form, the subscript shows explicitly which variable is to be kept constant. A more compact notation for these partial derivatives is fxandfy. However, it is extremely important when using partial derivatives to remember which variablesare being held constant and it is wise to write out the partial derivative in explicit form if there is any possibility of confusion. The extension of the deﬁnitions (5.1), (5.2) to the general n-variable case is straightforward and can be formally written as ∂f(x 1,x2,...,x n) ∂xi= lim ∆xi→0[f(x1,x2,...,x i+∆xi,...,x n)−f(x1,x2,...,x i,...,x n)] ∆xi, provided that the limit exists. Just as for one-variable functions, second (and higher) partial derivatives may be deﬁned in a similar way. For a two-variable function f(x, y)t h e ya r e ∂ ∂xparenleftbigg∂f ∂xparenrightbigg =∂2f ∂x2=fxx,∂ ∂yparenleftbigg∂f ∂yparenrightbigg =∂2f ∂y2=fyy, ∂ ∂xparenleftbigg∂f ∂yparenrightbigg =∂2f ∂x∂y=fxy,∂ ∂yparenleftbigg∂f ∂xparenrightbigg =∂2f ∂y∂x=fyx. 155 PARTIAL DIFFERENTIATION Only three of the second derivatives are independent since the relation ∂2f ∂x∂y=∂2f ∂y∂x, is always obeyed, provided that the second partial derivatives are continuous at the point in question. This relation often proves useful as a labour-savingdevice when evaluating second partial derivatives. It can also be shown that fora function of nvariables, f(x 1,x2,...,x n), under the same conditions, ∂2f ∂xi∂xj=∂2f ∂xj∂xi.IFind the ﬁrst and second partial derivatives of the function f(x, y)=2 x3y2+y3. The ﬁrst partial derivatives are ∂f ∂x=6x2y2,∂f ∂y=4x3y+3y2, and the second partial derivatives are ∂2f ∂x2=1 2xy2,∂2f ∂y2=4x3+6y,∂2f ∂x∂y=1 2x2y,∂2f ∂y∂x=1 2x2y, the last two being equal, as expected. J 5.2 The total diﬀerential and total derivative Having deﬁned the (ﬁrst) partial derivatives of a function f(x, y), which give the rate of change of falong the positive x-a n d y-axes, we consider next the rate of change of f(x, y) in an arbitrary direction. Suppose that we make simultaneous small changes ∆ xinxand ∆ yinyand that, as a result, fchanges to f+∆f. Then we must have ∆f=f(x+∆x, y+∆y)−f(x, y) =f(x+∆x, y+∆y)−f(x, y+∆y)+f(x, y+∆y)−f(x, y) =bracketleftbiggf(x+∆x, y+∆y)−f(x, y+∆y) ∆xbracketrightbigg ∆x+bracketleftbiggf(x, y+∆y)−f(x, y) ∆ybracketrightbigg ∆y. (5.3) In the last line we note that the quantities in brackets are very similar to those involved in the deﬁnitions of partial derivatives (5.1), (5.2). For them to be strictlyequal to the partial derivatives, ∆ xand ∆ ywould need to be inﬁnitesimally small. But even for ﬁnite (but not too large) ∆ xand ∆ ythe approximate formula ∆f≈∂f(x, y) ∂x∆x+∂f(x, y) ∂y∆y, (5.4) 156 5.2 THE TOTAL DIFFERENTIAL AND TOTAL DERIVATIVE can be obtained. It will be noticed that the ﬁrst bracket in (5.3) actually approxi- mates to ∂f(x, y+∆y)/∂xbut that this has been replaced by ∂f(x, y)/∂xin (5.4). This approximation clearly has the same degree of validity as that which replacesthe bracket by the partial derivative. How valid an approximation (5.4) is to (5.3) depends not only on how small ∆xand ∆ yare but also on the magnitudes of higher partial derivatives; this is discussed further in section 5.7 in the context of Taylor series for functions of more than one variable. Nevertheless, letting the small changes ∆ xand ∆ yin (5.4) become inﬁnitesimal, we can deﬁne the total diﬀerential dfof the function f(x, y), without any approximation, as df=∂f ∂xdx+∂f ∂ydy. (5.5) Equation (5.5) can be extended to the case of a function of nvariables, f(x1,x2,...,x n); df=∂f ∂x1dx1+∂f ∂x2dx2+···+∂f ∂xndxn. (5.6)IFind the total diﬀerential of the function f(x, y)=yexp(x+y). Evaluating the ﬁrst partial derivatives, we ﬁnd ∂f ∂x=yexp(x+y),∂f ∂y=e x p ( x+y)+yexp(x+y). Applying (5.5), we then ﬁnd that the total diﬀerential is given by df=[yexp(x+y)]dx+[ ( 1+ y)e x p( x+y)]dy. J In some situations, despite the fact that several variables xi,i=1,2,...,n, appear to be involved, eﬀectively only one of them is. This occurs if there aresubsidiary relationships constraining all the x ito have values dependent on the value of one of them, say x1. These relationships may be represented by equations that are typically of the form xi=xi(x1),i =2,3,...,n . (5.7) In principle fcan then be expressed as a function of x1alone by substituting from (5.7) for x2,x3,...,x n, and then the total derivative (or simply the derivative) offwith respect to x1is obtained by ordinary diﬀerentiation. Alternatively, (5.6) can be used to give df dx1=∂f ∂x1+parenleftbigg∂f ∂x2parenrightbiggdx2 dx1+···+parenleftbigg∂f ∂xnparenrightbiggdxn dx1. (5.8) It should be noted that the LHS of this equation is the total derivative df/dx 1, whilst the partial derivative ∂f/∂x 1forms only a part of the RHS. In evaluating 157 PARTIAL DIFFERENTIATION this partial derivative account must be taken only of explicit appearances of x1in the function f,a n dnoallowance must be made for the knowledge that changing x1necessarily changes x2,x3,...,x n. The contribution from these latter changes is precisely that of the remaining terms on the RHS of (5.8). Naturally, what has been shown using x1in the above argument applies equally well to any other of thexi, with the appropriate consequent changes.IFind the total derivative of f(x, y)=x2+3xywith respect to x, given that y=s i n−1x. We can see immediately that ∂f ∂x=2x+3y,∂f ∂y=3x,dy dx=1 (1−x2)1/2 and so, using (5.8) with x1=xandx2=y, df dx=2x+3y+3x1 (1−x2)1/2 =2x+3s i n−1x+3x (1−x2)1/2. Obviously the same expression would have resulted if we had substituted for yfrom the start, but the above method often produces results with reduced calculation, particularly in more complicated examples. J 5.3 Exact and inexact diﬀerentials In the last section we discussed how to ﬁnd the total diﬀerential of a function, i.e. its inﬁnitesimal change in an arbitrary direction, in terms of its gradients ∂f/∂x and∂f/∂y in the x-a n d y- directions (see (5.5)). Sometimes, however, we wish to reverse the process and ﬁnd the function fthat diﬀerentiates to give a known diﬀerential. Usually, ﬁnding such functions relies on inspection and experience. As an example, it is easy to see that the function whose diﬀerential is df= xd y+yd xis simply f(x, y)=xy+c,w h e r e cis a constant. Diﬀerentials such as this, which integrate directly, are called exact diﬀerentials , whereas those that do not are inexact diﬀerentials . For example, xd y+3yd xis not the straightforward diﬀerential of any function (see below). Inexact diﬀerentials can be made exact, however, by multiplying through by a suitable function called an integratingfactor. This is discussed further in subsection 14.2.3.IShow that the diﬀerential xd y+3yd xis inexact. On the one hand, if we integrate with respect to xwe conclude that f(x, y)=3 xy+g(y), where g(y) is any function of y. On the other hand, if we integrate with respect to ywe conclude that f(x, y)=xy+h(x)w h e r e h(x) is any function of x. These conclusions are inconsistent for any and every choice of g(y)a n d h(x), and therefore the diﬀerential is inexact. J 158 5.3 EXACT AND INEXACT DIFFERENTIALS It is naturally of interest to investigate which properties of a diﬀerential make it exact. Consider the general diﬀerential containing two variables, df=A(x, y)dx+B(x, y)dy. We see that ∂f ∂x=A(x, y),∂f ∂y=B(x, y) and, using the property fxy=fyx, we therefore require ∂A ∂y=∂B ∂x. (5.9) This is in fact both a necessary and a suﬃcient condition for the diﬀerential to be exact.IUsing (5.9) show that xd y+3yd xis inexact. In the above notation, A(x, y)=3 yandB(x, y)=xand so ∂A ∂y=3,∂B ∂x=1. As these are not equal it follows that the diﬀerential is inexact. J Determining whether a diﬀerential containing many variable x1,x2,...,x nis exact is a simple extension of the above. A diﬀerential containing many variablescan be written in general as df= nsummationdisplay i=1gi(x1,x2,...,x n)dxi and will be exact if ∂gi ∂xj=∂gj ∂xifor all pairs i, j. (5.10) There will be1 2n(n−1) such relationships to be satisﬁed.IShow that (y+z)dx+xd y+xd z is an exact diﬀerential. In this case, g1(x, y, z)=y+z,g2(x, y, z)=x,g3(x, y, z)=xand hence ∂g1/∂y=1= ∂g2/∂x,∂g3/∂x=1= ∂g1/∂z,∂g2/∂z=0= ∂g3/∂y; therefore, from (5.10), the diﬀerential is exact. As mentioned above, it is sometimes possible to show that a diﬀerential is exactsimply by ﬁnding by inspection the function from which it originates. In this example, itcan be seen easily that f(x, y, z)=x(y+z)+c.J 159 PARTIAL DIFFERENTIATION 5.4 Useful theorems of partial diﬀerentiation So far our discussion has centred on a function f(x, y) dependent on two variables, xandy. Equally, however, we could have expressed xas a function of fandy, oryas a function of fandx. To emphasise the point that all the variables are of equal standing, we now replace fbyz. This does not imply that x,yandz are coordinate positions (though they might be). Since xis a function of yandz, it follows that dx=parenleftbigg∂x ∂yparenrightbigg zdy+parenleftbigg∂x ∂zparenrightbigg ydz (5.11) and similarly, since y=y(x, z), dy=parenleftbigg∂y ∂xparenrightbigg zdx+parenleftbigg∂y ∂zparenrightbigg xdz. (5.12) We may now substitute (5.12) into (5.11) to obtain dx=parenleftbigg∂x ∂yparenrightbigg zparenleftbigg∂y ∂xparenrightbigg zdx+bracketleftBiggparenleftbigg∂x ∂yparenrightbigg zparenleftbigg∂y ∂zparenrightbigg x+parenleftbigg∂x ∂zparenrightbigg ybracketrightBigg dz. (5.13) Now if we hold zconstant, so that dz= 0, we obtain the reciprocity relation parenleftbigg∂x ∂yparenrightbigg z=parenleftbigg∂y ∂xparenrightbigg−1 z, which holds provided both partial derivatives exist and neither is equal to zero. Note, further, that this relationship only holds when the variable being keptconstant, in this case z, is the same on both sides of the equation. Alternatively we can put dx= 0 in (5.13). Then the contents of the square brackets also equal zero, and we obtain the cyclic relation parenleftbigg∂y ∂zparenrightbigg xparenleftbigg∂z ∂xparenrightbigg yparenleftbigg∂x ∂yparenrightbigg z=−1, which holds unless any of the derivatives vanish. In deriving this result we have used the reciprocity relation to replace ( ∂x/∂z )−1 yby (∂z/∂x )y. 5.5 The chain rule So far we have discussed the diﬀerentiation of a function f(x, y) with respect to its variables xandy.W en o wc o n s i d e rt h ec a s ew h e r e xandyare themselves functions of another variable, say u. If we wish to ﬁnd the derivative df/du , we could simply substitute in f(x, y) the expressions for x(u)a n d y(u)a n dt h e n diﬀerentiate the resulting function of u. Such substitution will quickly give the desired answer in simple cases, but in more complicated examples it is easier tomake use of the total diﬀerentials described in the previous section. 160 5.6 CHANGE OF VARIABLES From equation (5.5) the total diﬀerential of f(x, y)i sg i v e nb y df=∂f ∂xdx+∂f ∂ydy, but we now note that by using the formal device of dividing through by duthis immediately implies df du=∂f ∂xdx du+∂f ∂ydy du, (5.14) which is called the chain rule for partial diﬀerentiation. This expression provides a direct method for calculating the total derivative of fwith respect to uand is particularly useful when an equation is expressed in a parametric form.IGiven that x(u)=1+ auandy(u)=bu3, ﬁnd the rate of change of f(x, y)=xe−ywith respect to u. As discussed above, this problem could be addressed by substituting for xandyto obtain fas a function only of uand then diﬀerentiating with respect to u. However, using (5.14) directly we obtain df du=(e−y)a+(−xe−y)3bu2, which on substituting for xandygives df du=e−bu3(a−3bu2−3bau3). J Equation (5.14) is an example of the chain rule for a function of two variables each of which depends on a single variable. The chain rule may be extended to functions of many variables, each of which is itself a function of a variable u,i . e . f(x1,x2,x3,...,x n), with xi=xi(u). In this case the chain rule gives df du=nsummationdisplay i=1∂f ∂xidxi du=∂f ∂x1dx1 du+∂f ∂x2dx2 du+···+∂f ∂xndxn du. (5.15) 5.6 Change of variables It is sometimes necessary or desirable to make a change of variables during the course of an analysis, and consequently to have to change an equation expressedin one set of variables into an equation using another set. The same situation arisesif a function fdepends on one set of variables x i,s ot h a t f=f(x1,x2,...,x n), but the xiare given in terms of a further set of variables ujby the equations xi=xi(u1,u2,...,u m). (5.16) Thexion the right of this equation is a function (of the uj) whilst the xion the left is the value of that function. For each diﬀerent value of i,t h e xion the right 161 PARTIAL DIFFERENTIATION xy ρ φ Figure 5.1 The relationship between Cartesian and plane cylindrical polar coordinates. will be a diﬀerent function of the uj. In this case the chain rule (5.15) becomes ∂f ∂uj=nsummationdisplay i=1∂f ∂xi∂xi ∂uj,j =1,2,...,m , (5.17) and is said to express a change of variables . In general the number of variables in each set need not be equal, i.e. mneed not equal n, but if both the xiand the uiare sets of independent variables then m=n.IPlane polar coordinates, ρandφ, and Cartesian coordinates, xandy, are related by the expressions x=ρcosφ, y =ρsinφ, as can be seen from ﬁgure 5.1. An arbitrary function f(x, y)can be re-expressed as a function g(ρ, φ). Transform the expression ∂2f ∂x2+∂2f ∂y2 into one in ρandφ. We ﬁrst note that ρ2=x2+y2,φ=t a n−1(y/x). We can now write down the four partial derivatives ∂ρ ∂x=x (x2+y2)1/2=c o s φ,∂φ ∂x=−(y/x2) 1+(y/x)2=−sinφ ρ, ∂ρ ∂y=y (x2+y2)1/2=s i n φ,∂φ ∂y=1/x 1+(y/x)2=cosφ ρ. Thus, from (5.17), we may write ∂ ∂x=c o s φ∂ ∂ρ−sinφ ρ∂ ∂φ,∂ ∂y=s i n φ∂ ∂ρ+cosφ ρ∂ ∂φ. 162 5.7 TAYLOR’S THEOREM FOR MANY-VARIABLE FUNCTIONS Now it is only a matter of writing ∂2f ∂x2=∂ ∂x /∂f ∂x / =∂ ∂x /∂ ∂x / f = / cosφ∂ ∂ρ−sinφ ρ∂ ∂φ // cosφ∂ ∂ρ−sinφ ρ∂ ∂φ / g = / cosφ∂ ∂ρ−sinφ ρ∂ ∂φ // cosφ∂g ∂ρ−sinφ ρ∂g ∂φ / =c o s2φ∂2g ∂ρ2+2c os φsinφ ρ2∂g ∂φ−2c os φsinφ ρ∂2g ∂φ∂ρ +sin2φ ρ∂g ∂ρ+sin2φ ρ2∂2g ∂φ2 and a similar expression for ∂2f/∂y2, ∂2f ∂y2= / sinφ∂ ∂ρ+cosφ ρ∂ ∂φ // sinφ∂ ∂ρ+cosφ ρ∂ ∂φ / g =s i n2φ∂2g ∂ρ2−2cos φsinφ ρ2∂g ∂φ+2cos φsinφ ρ∂2g ∂φ∂ρ +cos2φ ρ∂g ∂ρ+cos2φ ρ2∂2g ∂φ2. When these two expressions are added together the change of variables is complete and we obtain ∂2f ∂x2+∂2f ∂y2=∂2g ∂ρ2+1 ρ∂g ∂ρ+1 ρ2∂2g ∂φ2. J 5.7 Taylor’s theorem for many-variable functions We have already introduced Taylor’s theorem for a function f(x)o fo n ev a r i a b l e , in section 4.6. In an analogous way, the Taylor expansion of a function f(x, y)o f two variables is given by f(x, y)=f(x0,y0)+∂f ∂x∆x+∂f ∂y∆y +1 2!bracketleftbigg∂2f ∂x2(∆x)2+2∂2f ∂x∂y∆x∆y+∂2f ∂y2(∆y)2bracketrightbigg +···, (5.18) where ∆ x=x−x0and ∆ y=y−y0, and all the derivatives are to be evaluated at (x0,y0). 163 PARTIAL DIFFERENTIATIONIFind the Taylor expansion, up to quadratic terms in x−2andy−3,o ff(x, y)=yexpxy about the point x=2,y=3. We ﬁrst evaluate the required partial derivatives of the function, i.e. ∂f ∂x=y2expxy,∂f ∂y=e x p xy+xyexpxy, ∂2f ∂x2=y3expxy,∂2f ∂y2=2xexpxy+x2yexpxy, ∂2f ∂x∂y=2yexpxy+xy2expxy. Using (5.18), the Taylor expansion of a two-variable function, we ﬁnd f(x, y)≈e6 n 3+9 ( x−2) + 7( y−3) +(2!)−1 / 27(x−2)2+ 48( x−2)(y−3) + 16( y−3)2 / o . J It will be noticed that the terms in (5.18) containing ﬁrst derivatives can be written as ∂f ∂x∆x+∂f ∂y∆y=parenleftbigg ∆x∂ ∂x+∆y∂ ∂yparenrightbigg f(x, y), where both sides of this relation should be evaluated at the point ( x0,y0). Similarly the terms in (5.18) containing second derivatives can be written as 1 2!bracketleftbigg∂2f ∂x2(∆x)2+2∂2f ∂x∂y∆x∆y+∂2f ∂y2(∆y)2bracketrightbigg =1 2!parenleftbigg ∆x∂ ∂x+∆y∂ ∂yparenrightbigg2 f(x, y), (5.19) where it is understood that the partial derivatives resulting from squaring the expression in parentheses act only on f(x, y) and its derivatives, and not on ∆ x or ∆y; again both sides of (5.19) should be evaluated at ( x0,y0). It can be shown that the higher-order terms of the Taylor expansion of f(x, y) can be written in an analogous way, and that we may write the full Taylor series as f(x, y)=∞summationdisplay n=01 n!bracketleftbiggparenleftbigg ∆x∂ ∂x+∆y∂ ∂yparenrightbiggn f(x, y)bracketrightbigg x0,y0 where, as indicated, all the terms on the RHS are to be evaluated at ( x0,y0). The most general form of Taylor’s theorem, for a function f(x1,x2,...,x n)o fn variables, is a simple extension of the above. Although it is not necessary to do so, we may think of the xias coordinates in n-dimensional space and write the function as f(x), where xis a vector from the origin to ( x1,x2,...,x n). Taylor’s 164 5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONS theorem then becomes f(x)=f(x0)+summationdisplay i∂f ∂xi∆xi+1 2!summationdisplay isummationdisplay j∂2f ∂xi∂xj∆xi∆xj+···, (5.20) where ∆ xi=xi−xi0and the partial derivatives are evaluated at ( x10,x20,...,x n0). For completeness, we note that in this case the full Taylor series can be writtenin the form f(x)=∞summationdisplay n=01 n!bracketleftbig (∆x·∇)nf(x)bracketrightbig x=x0, where∇is the vector diﬀerential operator del, to be discussed in chapter 10. 5.8 Stationary values of many-variable functions The idea of the stationary points of a function of just one variable has already been discussed in subsection 2.1.8. We recall that the function f(x) has a stationary point at x=x0if its gradient df/dx is zero at that point. A function may have any number of stationary points, and their nature, i.e. whether they are maxima,minima or stationary points of inﬂection, is determined by the value of the secondderivative at the point. A stationary point is (i) a minimum if d 2f/dx2>0; (ii) a maximum if d2f/dx2<0; (iii) a stationary point of inﬂection if d2f/dx2= 0 and changes sign through the point. We now consider the stationary points of functions of more than one variable; we will see that partial diﬀerential analysis is ideally suited to the determination of the position and nature of such points. It is helpful to consider ﬁrst the case of a function of just two variables but, even in this case, the general situationis more complex than that for a function of one variable, as can be seen fromﬁgure 5.2. This ﬁgure shows part of a three-dimensional model of a function f(x, y). At positions PandBthere are a peak and a bowl respectively or, more mathemati- cally, a local maximum and a local minimum. At position Sthe gradient in any direction is zero but the situation is complicated, since a section parallel to theplane x= 0 would show a maximum, but one parallel to the plane y=0w o u l d show a minimum. A point such as Sis known as a saddle point . The orientation of the ‘saddle’ in the xy-plane is irrelevant; it is as shown in the ﬁgure solely for ease of discussion. For any saddle point the function increases in some directionsaway from the point but decreases in other directions. 165 PARTIAL DIFFERENTIATION PS y xB Figure 5.2 Stationary points of a function of two variables. A minimum occurs at B, a maximum at Pand a saddle point at S. For functions of two variables, such as the one shown, it should be clear that a necessary condition for a stationary point (maximum, minimum or saddle point) to occur is that ∂f ∂x=0 a n d∂f ∂y=0. (5.21) The vanishing of the partial derivatives in directions parallel to the axes is enough to ensure that the partial derivative in any arbitrary direction is also zero. The latter can be considered as the superposition of two contributions, one alongeach axis; since both contributions are zero, so is the partial derivative in thearbitrary direction. This may be made more precise by considering the totaldiﬀerential df=∂f ∂xdx+∂f ∂ydy. Using (5.21) we see that although the inﬁnitesimal changes dxanddycan be chosen independently the change in the value of the inﬁnitesimal function dfis always zero at a stationary point. We now turn our attention to determining the nature of a stationary point of a function of two variables, i.e. whether it is a maximum, a minimum or a saddlepoint. By analogy with the one-variable case we see that ∂ 2f/∂x2and∂2f/∂y2 must both be positive for a minimum and both be negative for a maximum. However these are not suﬃcient conditions since they could also be obeyed at complicated saddle points. What is important for a minimum (or maximum) is that the second partial derivative must be positive (or negative) in alldirections, not just the x-a n d y- directions. 166 5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONS To establish just what constitutes suﬃcient conditions we ﬁrst note that, since fis a function of two variables and ∂f/∂x =∂f/∂y = 0, a Taylor expansion of the type (5.18) about the stationary point yields f(x, y)−f(x0,y0)≈1 2!bracketleftbig (∆x)2fxx+2 ∆ x∆yfxy+( ∆y)2fyybracketrightbig , where ∆ x=x−x0and ∆ y=y−y0and where the partial derivatives have been written in more compact notation. Rearranging the contents of the bracket as the weighted sum of two squares, we ﬁnd f(x, y)−f(x0,y0)≈1 2bracketleftBigg fxxparenleftbigg ∆x+fxy∆y fxxparenrightbigg2 +( ∆y)2parenleftBigg fyy−f2 xy fxxparenrightBiggbracketrightBigg . (5.22) For a minimum, we require (5.22) to be positive for all ∆ xand ∆ y, and hence fxx>0a n d fyy−(f2 xy/fxx)>0. Given the ﬁrst constraint, the second can be written fxxfyy>f2 xy. Similarly for a maximum we require (5.22) to be negative, and hence fxx<0a n d fxxfyy>f2 xy. For minima and maxima, symmetry requires thatfyyobeys the same criteria as fxx. When (5.22) is negative (or zero) for some values of ∆ xand ∆ ybut positive (or zero) for others, we have a saddle point. In this case fxxfyy<f2 xy. In summary, all stationary points have fx=fy=0a n d they may be classiﬁed further as (i) minima if both fxxandfyyare positive andf2 xy<f xxfyy, (ii) maxima if both fxxandfyyare negative andf2 xy<f xxfyy, (iii) saddle points if fxxandfyyhave opposite signs orf2 xy>f xxfyy. Note, however, that if f2 xy=fxxfyythen f(x, y)−f(x0,y0) can be written in one of the four forms ±1 2parenleftBig ∆x|fxx|1/2±∆y|fyy|1/2parenrightBig2 . For some choice of the ratio ∆ y/∆xthis expression has zero value showing that, for a displacement from the stationary point in this particular direction,f(x 0+∆x, y0+∆y) does not diﬀer from f(x0,y0) to second order in ∆ xand ∆y; in such situations further investigation is required. In particular, if fxx,fyy and fxyare all zero then the Taylor expansion has to be taken to a higher order. As examples, such extended investigations would show that the function f(x, y)=x4+y4has a mimimum at the origin but that g(x, y)=x4+y3has a saddle point there. 167 PARTIAL DIFFERENTIATIONIShow that the function f(x, y)=x3exp(−x2−y2)has a maximum at the point ( p 3/2,0), a minimum at (− p 3/2,0)and a stationary point at the origin whose nature cannot be determined by the above procedures. Setting the ﬁrst two partial derivatives to zero to locate the stationary points, we ﬁnd ∂f ∂x=( 3x2−2x4)e x p(−x2−y2)=0 , (5.23) ∂f ∂y=−2yx3exp(−x2−y2)=0 . (5.24) For (5.24) to be satisﬁed we require x=0o r y= 0 and for (5.23) to be satisﬁed we require x=0o r x=± p 3/2. Hence the stationary points are at (0 ,0), ( p 3/2,0) and (− p 3/2,0). We now ﬁnd the second partial derivatives: fxx=( 4x5−14x3+6x)exp(−x2−y2) fyy=x3(4y2−2)exp(−x2−y2) fxy=2x2y(2x2−3) exp(−x2−y2). We then substitute the pairs of values of xandyfor each stationary point and ﬁnd that at (0,0) fxx=0,f yy=0,f xy=0 and at (± p 3/2,0) fxx=∓6 p 3/2exp(−3/2),f yy=∓3 p 3/2exp(−3/2),f xy=0. Hence, applying criteria (i)–(iii) above, we ﬁnd that (0 ,0) is an undetermined stationary point, ( p 3/2,0) is a maximum and ( − p 3/2,0) is a minimum. The function is shown in ﬁgure 5.3. J Determining the nature of stationary points for functions of a general number of variables is considerably more diﬃcult and requires a knowledge of the eigenvectors and eigenvalues of matrices. Although these are not discussed until chapter 8, we present the analysis here for completeness. The remainder of thissection can therefore be omitted on a ﬁrst reading. For a function of nreal variables, f(x 1,x2,...,x n), we require that, at all stationary points, ∂f ∂xi=0 f o ra l l xi. In order to determine the nature of a stationary point, we must expand the function as a Taylor series about the point. Recalling the Taylor expansion (5.20)for a function of nvariables, we see that ∆f=f(x)−f(x 0)≈1 2summationdisplay isummationdisplay j∂2f ∂xi∂xj∆xi∆xj. (5.25) 168 5.8 STATIONARY VALUES OF MANY-VARIABLE FUNCTIONS minimum xy −1 1 2 3−22 −2 −3−0.2 −0.40.20.4 0 00maximum Figure 5.3 The function f(x, y)=x3exp(−x2−y2). If we deﬁne the matrix Mto have elements given by Mij=∂2f ∂xi∂xj, then we can rewrite (5.25) as ∆f=1 2∆xTM∆x, (5.26) where ∆ xis the column vector with the ∆ xias its components and ∆ xTis its transpose. Since Mis real and symmetric it has nreal eigenvalues λrand n orthogonal eigenvectors er, which after suitable normalisation satisfy Mer=λrer, eT res=δrs, where the Kronecker delta , written δrs, equals unity for r=sand equals zero otherwise. These eigenvectors form a basis set for the n-dimensional space and we can therefore expand ∆ xin terms of them, obtaining ∆x=summationdisplay rarer, 169 PARTIAL DIFFERENTIATION where the arare coeﬃcients dependent upon ∆ x. Substituting this into (5.26), we ﬁnd ∆f=1 2∆xTM∆x=1 2summationdisplay rλra2 r. Now, for the stationary point to be a minimum, we require ∆ f=1 2summationtext rλra2 r>0 for all sets of values of the ar, and therefore all the eigenvalues of Mto be greater than zero. Conversely, for a maximum we require ∆ f=1 2summationtext rλra2 r<0, and therefore all the eigenvalues of Mto be less than zero. If the eigenvalues have mixed signs, then we have a saddle point. Note that the test may fail if some or all of the eigenvalues are equal to zero and all the non-zero ones have the samesign.IDerive the conditions for maxima, minima and saddle points for a function of two real variables, using the above analysis. For a two-variable function the matrix Mis given by M= / fxxfxy fyxfyy / . Therefore its eigenvalues satisfy the equation////fxx−λf xy fxy fyy−λ ////=0. Hence (fxx−λ)(fyy−λ)−f2 xy=0 ⇒ fxxfyy−(fxx+fyy)λ+λ2−f2 xy=0 ⇒ 2λ=(fxx+fyy)± q (fxx+fyy)2−4(fxxfyy−f2 xy), which by rearrangement of the terms under the square root gives 2λ=(fxx+fyy)± q (fxx−fyy)2+4f2 xy. Now, that Mis real and symmetric implies that its eigenvalues are real, and so for both eigenvalues to be positive (corresponding to a minimum), we require fxxandfyypositive and also fxx+fyy> q (fxx+fyy)2−4(fxxfyy−f2 xy), ⇒ fxxfyy−f2 xy>0. A similar procedure will ﬁnd the criteria for maxima and saddle points. J 5.9 Stationary values under constraints In the previous section we looked at the problem of ﬁnding stationary values of a function of two or more variables when all the variables may be independently 170 5.9 STATIONARY VALUES UNDER CONSTRAINTS varied. However, it is often the case in physical problems that not all the vari- ables used to describe a situation are in fact independent, i.e. some relationshipbetween the variables must be satisﬁed. For example, if we walk through a hillylandscape and we are constrained to walk along a path, we will never reach the highest peak on the landscape, unless the path happens to take us to it. Nevertheless, we can still ﬁnd the highest point that we have reached during ourjourney. We ﬁrst discuss the case of a function of just two variables. Let us consider ﬁnding the maximum value of the diﬀerentiable function f(x, y) subject to the constraint g(x, y)=c,w h e r e cis a constant. In the above analogy, f(x, y)m i g h t represent the height of the land above sea-level in some hilly region, whilst g(x, y)=cis the equation of the path along which we walk. We could, of course, use the constraint g(x, y)=cto substitute for xoryin f(x, y), thereby obtaining a new function of only one variable whose stationary points could be found using the methods discussed in subsection 2.1.8. However, such a procedure can involve a lot of algebra and becomes very tedious for func- tions of more than two variables. A more direct method for solving such problemsis the method of Lagrange undetermined multipliers , which we now discuss. To maximise fwe require df=∂f ∂xdx+∂f ∂ydy=0. Ifdxanddywere independent, we could conclude fx=0= fy. However, here they are not independent, but constrained because gis constant: dg=∂g ∂xdx+∂g ∂ydy=0. Multiplying dgby an as yet unknown number λand adding it to dfwe obtain d(f+λg)=parenleftbigg∂f ∂x+λ∂g ∂xparenrightbigg dx+parenleftbigg∂f ∂y+λ∂g ∂yparenrightbigg dy=0, where λis called a Lagrange undetermined multiplier . In this equation dxanddy are to be independent and arbitrary; we must therefore choose λsuch that ∂f ∂x+λ∂g ∂x=0, (5.27) ∂f ∂y+λ∂g ∂y=0. (5.28) These equations, together with the constraint g(x, y)=c, are suﬃcient to ﬁnd the three unknowns, i.e. λand the values of xandyat the stationary point. 171 PARTIAL DIFFERENTIATIONIThe temperature of a point (x, y)on a unit circle is given by T(x, y)=1+ xy.F i n dt h e temperature of the two hottest points on the circle. We need to maximise T(x, y) subject to the constraint x2+y2= 1. Applying (5.27) and (5.28), we obtain y+2λx=0, (5.29) x+2λy=0. (5.30) These results, together with the original constraint x2+y2= 1, provide three simultaneous equations that may be solved for λ,xandy. From (5.29) and (5.30) we ﬁnd λ=±1/2, which in turn implies that y=∓x. Remem- bering that x2+y2= 1, we ﬁnd that y=x⇒x=±1√ 2,y =±1√ 2 y=−x⇒x=∓1√ 2,y =±1√ 2. We have not yet determined which of these stationary points are maxima and which are minima. In this simple case, we need only substitute the four pairs of x-a n d y- values into T(x, y)=1+ xyto ﬁnd that the maximum temperature on the unit circle is Tmax=3/2a t the points y=x=±1/√ 2. J The method of Lagrange multipliers can be used to ﬁnd the stationary points of functions of more than two variables, subject to several constraints, provided thatthe number of constraints is smaller than the number of variables. For example,if we wish to ﬁnd the stationary points of f(x, y, z) subject to the constraints g(x, y, z)=c 1andh(x, y, z)=c2,w h e r e c1andc2are constants, then we proceed as above, obtaining ∂ ∂x(f+λg+µh)=∂f ∂x+λ∂g ∂x+µ∂h ∂x=0, ∂ ∂y(f+λg+µh)=∂f ∂y+λ∂g ∂y+µ∂h ∂y=0, (5.31) ∂ ∂z(f+λg+µh)=∂f ∂z+λ∂g ∂z+µ∂h ∂z=0. We may now solve these three equations, together with the two constraints, to giveλ,µ,x,yandz. 172 5.9 STATIONARY VALUES UNDER CONSTRAINTSIFind the stationary points of f(x, y, z)=x3+y3+z3subject to the following constraints: (i)g(x, y, z)=x2+y2+z2=1; (ii)g(x, y, z)=x2+y2+z2=1andh(x, y, z)=x+y+z=0. Case (i). Since there is only one constraint in this case, we need only introduce a single Lagrange multiplier to obtain ∂ ∂x(f+λg)=3 x2+2λx=0, ∂ ∂y(f+λg)=3 y2+2λy=0, (5.32) ∂ ∂z(f+λg)=3 z2+2λz=0. These equations are highly symmetrical and clearly have the solution x=y=z=−2λ/3. Using the constraint x2+y2+z2= 1 we ﬁnd λ=±√ 3/2 and so stationary points occur at x=y=z=±1√ 3. (5.33) In solving the three equations (5.32) in this way, however, we have implicitly assumed thatx,yandzare non-zero. However, it is clear from (5.32) that any of these values can equal zero, with the exception of the case x=y=z= 0 since this is prohibited by the constraint x2+y2+z2= 1. We must consider the other cases separately. Ifx= 0, for example, we require 3y2+2λy=0, 3z2+2λz=0, y2+z2=1. Clearly, we require λ/negationslash= 0, otherwise these equations are inconsistent. If neither ynor zis zero we ﬁnd y=−2λ/3= za n df r o mt h et h i r de q u a t i o nw er e q u i r e y=z= ±1/√ 2. If y= 0, however, then z=±1 and, similarly, if z=0t h e n y=±1. Thus the stationary points having x=0a r e( 0 ,0,±1), (0 ,±1,0) and (0 ,±1/√ 2,±1/√ 2). A similar procedure can be followed for the cases y=0a n d z= 0 respectively and, in addition to those already obtained, we ﬁnd the stationary points ( ±1,0,0), (±1/√ 2,0,±1/√ 2) and (±1/√ 2,±1/√ 2,0). Case (ii). We now have two constraints and must therefore introduce two Lagrange multipliers to obtain (cf. (5.31)) ∂ ∂x(f+λg+µh)=3 x2+2λx+µ=0, (5.34) ∂ ∂y(f+λg+µh)=3 y2+2λy+µ=0, (5.35) ∂ ∂z(f+λg+µh)=3 z2+2λz+µ=0. (5.36) These equations are again highly symmetr ical and the simplest way to proceed is to subtract (5.35) from (5.34) to obtain 3(x2−y2)+2λ(x−y)=0 ⇒ 3(x+y)(x−y)+2λ(x−y)=0 . (5.37) This equation is clearly satisﬁed if x=y; then, from the second constraint, x+y+z=0 , 173 PARTIAL DIFFERENTIATION we ﬁnd z=−2x. Substituting these values into the ﬁrst constraint, x2+y2+z2=1 ,w e obtain x=±1√ 6,y =±1√ 6,z =∓2√ 6. (5.38) Because of the high degree of symmetry amongst the equations (5.34)–(5.36), we may obtain by inspection two further relations analogous to (5.37), one containing the variables y,z and the other the variables x, z. Assuming y=zin the ﬁrst relation and x=zin the second, we ﬁnd the stationary points x=±1√ 6,y =∓2√ 6,z =±1√ 6(5.39) and x=∓2√ 6,y =±1√ 6,z =±1√ 6. (5.40) We note that in ﬁnding the stationary points (5.38)–(5.40) we did not need to evaluate the Lagrange multipliers λandµexplicitly. This is not always the case, however, and in some problems it may be simpler to begin by ﬁnding the values of these multipliers. Returning to (5.37) we must now consider the case where x/negationslash=y; then we ﬁnd 3(x+y)+2λ=0. (5.41) However, in obtaining the stationary points (5.39), (5.40), we did notassume x=ybut only required y=zandx=zrespectively. It is clear that x/negationslash=yat these stationary points, and it can be shown that they do indeed satisfy (5.41). Similarly, several stationary pointsfor which x/negationslash=zory/negationslash=zhave already been found. Thus we need to consider further only two cases: ( a)x=y=z,a n d( b)x,yandzare all diﬀerent. The ﬁrst is clearly prohibited by the constraint x+y+z= 0. For the second case, (5.41) must be satisﬁed, together with the analogous equations containing y,zand x, zrespectively, i.e. 3(x+y)+2λ=0, 3(y+z)+2λ=0, 3(x+z)+2λ=0. Adding these three equations together and using the constraint x+y+z= 0 we ﬁnd λ=0 . However, for λ= 0 the equations are inconsistent for non-zero x,yandz. Therefore all the stationary points have already been found and are given by (5.38)–(5.40).J The method may be extended to functions of any number nof variables subject to any smaller number mof constraints. This means that eﬀectively there aren−mindependent variables and, as mentioned above, we could solve by substitution and then by the methods of the previous section. However, for large nthis becomes cumbersome and the use of Lagrange undetermined multipliers is a useful simpliﬁcation. 174 5.9 STATIONARY VALUES UNDER CONSTRAINTSIA system contains a very large number Nof particles, each of which can be in any of R energy levels with a corresponding energy Ei,i=1,2,...,R . The number of particles in the ith level is niand the total energy of the system is a constant, E. Find the distribution of particles amongst the energy level s that maximises the expression P=N! n1!n2!···nR!, subject to the constraints that both the number of particles and the total energy remain constant, i.e. g=N−RX i=1ni=0 a n d h=E−RX i=1niEi=0. The way in which we proceed is as follows. In order to maximise P, we must minimise its denominator (since the numerator is ﬁxed). Minimising the denominator is the same asminimising the logarithm of the denominator, i.e. f=l n(n 1!n2!···nR!)=l n(n1!)+l n(n2!)+···+l n(nR!). Using Stirling’s approximation, ln (n!)≈nlnn−n, we ﬁnd that f=n1lnn1+n2lnn2+···+nRlnnR−(n1+n2+···+nR) = / RX i=1nilnni /! −N. It has been assumed here that, for the desired distribution, all the niare large. Thus, we now have a function fsubject to two constraints, g=0a n d h= 0, and we can apply the Lagrange method, obtaining (cf. (5.31)) ∂f ∂n1+λ∂g ∂n1+µ∂h ∂n1=0, ∂f ∂n2+λ∂g ∂n2+µ∂h ∂n2=0, ... ∂f ∂nR+λ∂g ∂nR+µ∂h ∂nR=0. Since all these equations are alike, we consider the general case ∂f ∂nk+λ∂g ∂nk+µ∂h ∂nk=0, fork=1,2,...,R . Substituting the functions f,gandhinto this relation we ﬁnd nk nk+l nnk+λ(−1) +µ(−Ek)=0 , which can be rearranged to give lnnk=µEk+λ−1, and hence nk=CexpµEk. 175 PARTIAL DIFFERENTIATION We now have the general form for the distribution of particles amongst energy levels, but in order to determine the two constants µ,Cwe recall that RX k=1CexpµEk=N and RX k=1CE kexpµEk=E. This is known as the Boltzmann distribution and is a well-known result from statistical mechanics. J 5.10 Envelopes As noted at the start of this chapter, many of the functions with which the physicists, chemists and engineers have to deal contain, in addition to constantsand one or more variables, quantities that are normally considered as parametersof the system under study. Such parameters may, for example, represent thecapacitance of a capacitor, the length of a rod, or the mass of a particle – quantities that are normally taken as ﬁxed for any particular physical set-up. The corresponding variables may well be time, currents, charges, positions andvelocities. However, the parameters could be varied and in this section we study the eﬀects of doing so; in particular we study how the form of dependence ofone variable on another, typically y=y(x), is aﬀected when the value of a parameter is changed in a smooth and continuous way. In eﬀect, we are makingthe parameter into an additional variable. As a particular parameter, which we denote by α, is varied over its permitted range, the shape of the plot of yagainst xwill change, usually, but not always, in a smooth and continuous way. For example, if the muzzle speed vof a shell ﬁred from a gun is increased through a range of values then its height–distancetrajectories will be a series of curves with a common starting point that areessentially just magniﬁed copies of the original; furthermore the curves do notcross each other. However, if the muzzle speed is kept constant but θ, the angle of elevation of the gun, is increased through a series of values, the corresponding trajectories do not vary in a monotonic way. When θhas been increased beyond 45 ◦the resulting trajectory does cross some of the trajectories corresponding toθ<45◦. The trajectories all lie within a curve that touches each individual trajectory at one point. Such a curve is called the envelope to the set of trajectory solutions; it is to the study of such envelopes that this section is devoted. For our general discussion of envelopes we will consider an equation of the form f=f(x, y, α ) = 0. A function of three Cartesian variables, f=f(x, y, α ), is deﬁned at all points in xyα-space, whereas f=f(x, y, α )=0i sa surface in this space. A plane of constant α, which is parallel to the xy-plane, cuts such 176 5.10 ENVELOPES PP1 f(x, y, α 1)=0 f(x, y, α 1+h)=0 Figure 5.4 Two neighbouring curves in the xy-plane of the family f(x, y, α)= 0 intersecting at P.F o rﬁ x e d α1, the point P1is the limiting position of Pas h→0. As α1is varied, P1delineates the envelope of the family (broken line). a surface in a curve. Thus diﬀerent values of the parameter αcorrespond to diﬀerent curves, which can be plotted in the xy-plane. We now investigate how theenvelope equation for such a family of curves is obtained. 5.10.1 Envelope equations Suppose f(x, y, α 1)=0a n d f(x, y, α 1+h) = 0 are two neighbouring curves of a family for which the parameter αdiﬀers by a small amount h. Let them intersect at the point Pwith coordinates x, y, as shown in ﬁgure 5.4. Then the envelope, indicated by the broken line in the ﬁgure, touches f(x, y, α 1) = 0 at the point P1, which is deﬁned as the limiting position of Pwhen α1is ﬁxed but h→0. The full envelope is the curve traced out by P1asα1changes to generate successive members of the family of curves. Of course, for any ﬁnite h,f(x, y, α 1+h)=0i s one of these curves and the envelope touches it at the point P2. We are now going to apply Rolle’s theorem, see subsection 2.1.10, with the parameter αas the independent variable and xandyﬁxed as constants. In this context, the two curves in ﬁgure 5.4 can be thought of as the projections onto thexy-plane of the planar curves in which the surface f=f(x, y, α ) = 0 meets the planes α=α 1andα=α1+h. Along the normal to the page that passes through P,a sαchanges from α1 toα1+hthe value of f=f(x, y, α ) will depart from zero, because the normal meets the surface f=f(x, y, α )=0o n l ya t α=α1and at α=α1+h. However, at these end points the values of f=f(x, y, α ) will both be zero, and therefore equal. This allows us to apply Rolle’s theorem and so to conclude that for someθin the range 0 ≤θ≤1 the partial derivative ∂f(x, y, α 1+θh)/∂αis zero. When 177 PARTIAL DIFFERENTIATION his made arbitrarily small, so that P→P1, the three deﬁning equations reduce to two and deﬁne the envelope point P1: f(x, y, α 1)=0 a n d∂f(x, y, α 1) ∂α=0, (5.42) In (5.42) both the function and the gradient are evaluated at α=α1. The equation of the envelope g(x, y) = 0 is found by eliminating α1between the two equations. As a simple example we will now solve the problem which when posed mathe- matically reads ‘calculate the envelope appropriate to the family of straight lines in the xy-plane whose points of intersection with the coordinate axes are a ﬁxed distance apart’. In more ordinary language, the problem is about a ladder leaningagainst a wall.IA ladder of length Lcan be stood on level ground and leant at any angle against a vertical wall. Find the equation of the curve bounding the vertical area that can be accessed fromthe ladder. We take the ground and the wall as the x-a n d y-axes respectively. If the foot of the ladder isafrom the foot of the wall and the top is babove the ground, the straight-line equation of the ladder is x a+y b=1, where aandbare connected by a2+b2=L2. Expressed in standard form with only one independent parameter, a, the equation becomes f(x, y, a)=x a+y (L2−a2)1/2−1=0 . (5.43) Now, diﬀerentiating (5.43) with respect to aand setting the derivative ∂f/∂a equal to zero gives −x a2+ay (L2−a2)3/2=0 ; from which it follows that a=Lx1/3 (x2/3+y2/3)1/2and ( L2−a2)1/2=Ly1/3 (x2/3+y2/3)1/2. Eliminating aby substituting these values into (5.43) gives, for the equation of the envelope of all possible positions on the ladder, x2/3+y2/3=L2/3. This is the equation of an astroid (mentioned in exercise 2.19), and, together with the wall and the ground, marks the boundary of the vertical area that can be accessed by (theshoes of) a person standing on the ladder.J Other examples, drawn from both geometry and and the physical sciences, are considered in the exercises at the end of this chapter. The shell trajectory problem discussed earlier in this section is solved there, but in the guise of a questionabout the water bell of an ornamental fountain. 178 5.11 THERMODYNAMIC RELATIONS 5.11 Thermodynamic relations Thermodynamic relations provide a useful set of physical examples of partial diﬀerentiation. The relations we will derive are called Maxwell’s thermodynamic relations . They express relationships between four thermodynamic quantities de- scribing a unit mass of a substance. The quantities are the pressure P, the volume V, the thermodynamic temperature Tand the entropy Sof the substance. These four quantities are not independent; any two of them can be varied indepen-dently, but the other two are then determined. The ﬁrst law of thermodynamicsmay be expressed as dU=Td S−Pd V , (5.44) where Uis the internal energy of the substance. Essentially this is a conservation of energy equation, but we shall concern ourselves, not with the physics, but ratherwith the use of partial diﬀerentials to relate the four basic quantities discussedabove. The method involves writing a total diﬀerential, dUsay, in terms of the diﬀerentials of two variables, say XandY, thus dU=parenleftbigg∂U ∂Xparenrightbigg YdX+parenleftbigg∂U ∂Yparenrightbigg XdY , (5.45) and then using the relationship ∂2U ∂X∂Y=∂2U ∂Y ∂X to obtain the required Maxwell relation. The variables XandYare to be chosen from P,V,TandS.IShow that (∂T/∂V )S=−(∂P/∂S )V. Here the two variables that have to be held constant, in turn, happen to be those whose diﬀerentials appear on the RHS of (5.44). And so, taking XasSandYasVin (5.45), we have Td S−Pd V=dU= /∂U ∂S / VdS+ /∂U ∂V / SdV, and ﬁnd directly that/∂U ∂S / V=T and /∂U ∂V / S=−P. Diﬀerentiating the ﬁrst expression with respect to Vand the second with respect to S,a n d using ∂2U ∂V∂S=∂2U ∂S∂V, we ﬁnd the Maxwell relation/∂T ∂V / S=− /∂P ∂S / V. J 179 PARTIAL DIFFERENTIATIONIShow that (∂S/∂V )T=(∂P/∂T )V. Applying (5.45) to dS, with independent variables VandT, we ﬁnd dU=Td S−Pd V=T //∂S ∂V / TdV+ /∂S ∂T / VdT / −PdV. Similarly applying (5.45) to dU, we ﬁnd dU= /∂U ∂V / TdV+ /∂U ∂T / VdT. Thus, equating partial derivatives,/∂U ∂V / T=T /∂S ∂V / T−Pand /∂U ∂T / V=T /∂S ∂T / V. But, since ∂2U ∂T∂V=∂2U ∂V∂T,i.e.∂ ∂T /∂U ∂V / T=∂ ∂V /∂U ∂T / V, it follows that/∂S ∂V / T+T∂2S ∂T∂V− /∂P ∂T / V=∂ ∂V / T /∂S ∂T / V / T=T∂2S ∂V∂T. Thus ﬁnally we get the Maxwell relation/∂S ∂V / T= /∂P ∂T / V. J The above derivation is rather cumbersome, however, and a useful trick that can simplify the working is to deﬁne a new function, called a potential .T h e internal energy Udiscussed above is one example of a potential but three others are commonly deﬁned and they are described below.IShow that (∂S/∂V )T=(∂P/∂T )Vby considering the potential U−ST. We ﬁrst consider the diﬀerential d(U−ST). From (5.5), we obtain d(U−ST)=dU−SdT−TdS=−SdT−PdV when use is made of (5.44). We rewrite U−STasFfor convenience of notation; Fis called the Helmholtz potential . Thus dF=−SdT−PdV, and it follows that/∂F ∂T / V=−S and /∂F ∂V / T=−P. Using these results together with ∂2F ∂T∂V=∂2F ∂V∂T, we can see immediately that/∂S ∂V / T= /∂P ∂T / V, which is the same Maxwell relation as before. J 180 5.12 DIFFERENTIATION OF INTEGRALS Although the Helmholtz potential has other uses, in this context it has simply provided a means for a quick derivation of the Maxwell relation. The otherMaxwell relations can be derived similarly by using two other potentials, theenthalpy ,H=U+PV,a n dt h e Gibbs free energy ,G=U+PV−ST(see exercise 5.25). 5.12 Diﬀerentiation of integrals We conclude this chapter with a discussion of the diﬀerentiation of integrals. Let us consider the indeﬁnite integral (cf. equation (2.30)) F(x, t)=integraldisplay f(x, t)dt, from which it follows immediately that ∂F(x, t) ∂t=f(x, t). Assuming that the second partial derivatives of F(x, t) are continuous, we have ∂2F(x, t) ∂t∂x=∂2F(x, t) ∂x∂t, a n ds ow ec a nw r i t e ∂ ∂tbracketleftbigg∂F(x, t) ∂xbracketrightbigg =∂ ∂xbracketleftbigg∂F(x, t) ∂tbracketrightbigg =∂f(x, t) ∂x. Integrating this equation with respect to tthen gives ∂F(x, t) ∂x=integraldisplay∂f(x, t) ∂xdt. (5.46) Now consider the deﬁnite integral I(x)=integraldisplayt=v t=uf(x, t)dt =F(x, v)−F(x, u), where uandvare constants. Diﬀerentiating this integral with respect to x,a n d using (5.46), we see that dI(x) dx=∂F(x, v) ∂x−∂F(x, u) ∂x =integraldisplayv∂f(x, t) ∂xdt−integraldisplayu∂f(x, t) ∂xdt =integraldisplayv u∂f(x, t) ∂xdt. This is Leibnitz’ rule for diﬀerentiating integrals, and basically it states that for 181 PARTIAL DIFFERENTIATION constant limits of integration the order of integration and diﬀerentiation can be reversed. In the more general case where the limits of the integral are themselves functions ofx, it follows immediately that I(x)=integraldisplayt=v(x) t=u(x)f(x, t)dt =F(x, v(x))−F(x, u(x)), which yields the partial derivatives ∂I ∂v=f(x, v(x)),∂I ∂u=−f(x, u(x)). Consequently dI dx=parenleftbigg∂I ∂vparenrightbiggdv dx+parenleftbigg∂I ∂uparenrightbiggdu dx+∂I ∂x =f(x, v(x))dv dx−f(x, u(x))du dx+∂ ∂xintegraldisplayv(x) u(x)f(x, t)dt =f(x, v(x))dv dx−f(x, u(x))du dx+integraldisplayv(x) u(x)∂f(x, t) ∂xdt, (5.47) where the partial derivative with respect to xin the last term has been taken inside the integral sign using (5.46). This procedure is valid because u(x)a n d v(x) are being held constant in this term.IFind the derivative with respect to xof the integral I(x)= Zx2 xsinxt tdt. Applying (5.47), we see that dI dx=sinx3 x2(2x)−sinx2 x(1) + Zx2 xtcosxt tdt =2si nx3 x−sinx2 x+ /sinxt x /x2 x =3sinx3 x−2sinx2 x =1 x(3sin x3−2si nx2). J 5.13 Exercises 5.1 (a) Find all the ﬁrst partial derivatives of the following functions f(x, y): (i) x2y, (ii)x2+y2+ 4, (iii) sin( x/y), (iv) tan−1(y/x), (v) r(x, y, z)=(x2+y2+z2)1/2. 182 5.13 EXERCISES (b) For (i), (ii) and (v), ﬁnd ∂2f/∂x2,∂2f/∂y2,∂2f/∂x∂y. (c) For (iv) verify that ∂2f/∂x∂y =∂2f/∂y∂x. 5.2 Determine which of the following are exact diﬀerentials: (a) (3 x+2 )yd x+x(x+1 )dy, (b)ytanxd x+xtanyd y, (c)y2(lnx+1 )dx+2xylnxd y, (d)y2(lnx+1 )dy+2xylnxd x, (e) [ x/(x2+y2)]dy−[y/(x2+y2)]dx. 5.3 Show that the diﬀerential df=x2dy−(y2+xy)dx is not exact, but that dg=(xy2)−1dfis exact. 5.4 (a) Show that df=y(1 +x−x2)dx+x(x+1 )dy is not an exact diﬀerential. (b) Find the diﬀerential equation that a function g(x) must satisfy if dφ=g(x)df is to be an exact diﬀerential. Verify that g(x)=e−xis a solution of this equation and deduce the form of φ(x, y). 5.5 The equation 3 y=z3+3xzdeﬁnes zimplicitly as a function of xandy. Evaluate all three second partial derivatives of zwith respect to xand/or y.V e r i f yt h a t z is a solution of x∂2z ∂y2+∂2z ∂x2=0. 5.6 A possible equation of state for a gas takes the form pV=RTexp / −α VRT / , in which αandRare constants. Calculate expressions for/∂p ∂V / T, /∂V ∂T / p, /∂T ∂p / V, and show that their product is −1, as stated in section 5.4. 5.7 The function G(t) is deﬁned by G(t)=F(x, y)=x2+y2+3xy, where x(t)=at2andy(t)=2 at. Use the chain rule to ﬁnd the values of ( x, y)a t which G(t) has stationary values as a function of t. Do any of them correspond to the stationary points of F(x, y) as a function of xandy? 5.8 In the xy-plane, new coordinates sandtare deﬁned by s=1 2(x+y),t =1 2(x−y). Transform the equation ∂2φ ∂x2−∂2φ ∂y2=0 into the new coordinates and deduce that its general solution can be written φ(x, y)=f(x+y)+g(x−y), where f(u)a n d g(v) are arbitrary functions of uandvrespectively. 183 PARTIAL DIFFERENTIATION 5.9 The function f(x, y) satisﬁes the diﬀerential equation y∂f ∂x+x∂f ∂y=0. By changing to new variables u=x2−y2andv=2xy, show that fis, in fact, a function of x2−y2only. 5.10 If x=eucosθandy=eusinθ, show that ∂2φ ∂u2+∂2φ ∂θ2=(x2+y2) /∂2f ∂x2+∂2f ∂y2 / , where f(x, y)=φ(u, θ). 5.11 Find and evaluate the maxima, minima and saddle points of the function f(x, y)=xy(x2+y2−1). 5.12 Show that f(x, y)=x3−12xy+4 8x+by2,b /negationslash=0, has two, one, or zero stationary points according to whether |b|is less than, equal to, or greater than 3. 5.13 Locate the stationary points of the function f(x, y)=(x2−2y2)exp[−(x2+y2)/a2], where ais a non-zero constant. Sketch the function along the x-a n d y- axes and hence identify the nature and values of the stationary points. 5.14 Find the stationary points of the function f(x, y)=x3+xy2−12x−y2 and identify their nature. 5.15 Find the stationary values of f(x, y)=4 x2+4y2+x4−6x2y2+y4 and classify them as maxima, minima or saddle points. Make a rough sketch of the contours of fin the quarter plane x, y≥0. 5.16 The temperature of a point ( x, y, z) on the unit sphere is given by T(x, y, z)=1+ xy+yz. By using the method of Lagrange multipliers ﬁnd the temperature of the hottest point on the sphere. 5.17 A rectangular parallelepiped has all eight vertices on the ellipsoid x2+3y2+3z2=1. Using the symmetry of the parallelepiped about each of the planes x=0 , y=0 , z= 0, write down the surface area of the parallelepiped in terms of the coordinates of the vertex that lies in the octant x, y, z≥0. Hence ﬁnd the maximum value of the surface area of such a parallelepiped. 5.18 Two horizontal corridors, 0 ≤x≤awith y≥0, and 0≤y≤bwith x≥0, meet at right angles. Find the length Lof the longest ladder (considered as a stick) that may be carried horizontally around the corner. 5.19 A barn is to be constructed with a uniform cross-sectional area Athroughout its length. The cross-section is to be a rectangle of wall height h(ﬁxed) and width w, surmounted by an isosceles triangular roof that makes an angle θwith 184 5.13 EXERCISES the horizontal. The cost of construction is αper unit height of wall and βper unit (slope) length of roof. Show that, irrespective of the values of αandβ,t o minimise costs wshould be chosen to satisfy the equation w4=1 6A(A−wh), andθmade such that 2tan2 θ=w/h. 5.20 Show that the envelope of all concentric ellipses that have their axes along the x-a n d y-coordinate axes and that have the sum of their semi-axes equal to a constant Lis the same curve (an astroid) as that found in the worked example in section 5.10. 5.21 Find the area of the region covered by points on the lines x a+y b=1, where the sum of any line’s intercepts on the coordinate axes is ﬁxed and equal toc. 5.22 Prove that the envelope of the circles whose diameters are those chords of a given circle that pass through a ﬁxed point on its circumference, is the cardioid r=a(1 + cos θ). Here ais the radius of the given circle and ( r,θ) are the polar coordinates of the envelope. Take as the system parameter the angle φbetween a chord and the polar axis from which θis measured. 5.23 A water feature contains a spray head at water level at the centre of a round basin. The head is in the form of a hemisphere with many evenly distributedsmall holes in it, and through which water spurts out at the same speed v 0in all directions. (a) What is the shape of the ‘water bell’ so formed? (b) What must be the minimum diameter of the bowl if no water is to be lost? 5.24 In order to make a focussing mirror that concentrates parallel axial rays to one spot (or conversely forms a parallel beam from a point source) a parabolic shapeshould be adopted. If a mirror that is part of a circular cylinder or sphere wereused, the light would be spread out along a curve. This curve is known as acaustic and is the envelope of the rays reﬂected from the mirror. Denoting by θ the angle which a typical incident axial ray makes with the normal to the mirrorat the place where it is reﬂected, the geometry of reﬂection (the angle of incidenceequals the angle of reﬂection) is shown in ﬁgure 5.5. Show that a parametric speciﬁcation of the caustic is x=Rcosθ/;1 2+s i n2θ / ,y =Rsin3θ, where Ris the radius of curvature of the mirror. The curve is, in fact, part of an epicycloid. 5.25 By considering the diﬀerential dG=d(U+PV−ST), where Gis the Gibbs free energy, Pthe pressure, Vthe volume, Sthe entropy andTthe temperature of a system, and given further that dU=TdS−PdV, derive a Maxwell relation connecting ( ∂V/∂T )Pand ( ∂S/∂P )T. 185 PARTIAL DIFFERENTIATION OR xy θθ 2θ Figure 5.5 The reﬂecting mirror discussed in exercise 5.24. 5.26 Functions P(V,T),U(V,T)a n d S(V,T) are related by TdS=dU+PdV, where the symbols have the same meaning as in the previous question. Pis known from experiment to have the form P=T4 3+T V, in appropriate units. If U=αVT4+βT, where α,β, are constants (or at least do not depend on T,V), deduce that α must have a speciﬁc value but βmay have any value. Find the corresponding form of S. 5.27 As in the previous two exercises on the thermodynamics of a simple gas, the quantity dS=T−1(dU+PdV) is an exact diﬀerential. Use this to prove that/∂U ∂V / T=T /∂P ∂T / V−P. In the van der Waals model of a gas, Pobeys the equation P=RT V−b−a V2, where R,aandbare constants. Further, in the limit V→∞, the form of U becomes U=cT ,where cis another constant. Find the complete expression for U(V,T). 5.28 The entropy S(H,T), the magnetisation M(H,T) and the internal energy U(H,T) of a magnetic salt placed in a magnetic ﬁeld of strength Hat temperature Tare connected by the equation TdS=dU−HdM. 186 5.13 EXERCISES By considering d(U−TS−HM), or otherwise, prove that/∂M ∂T / H= /∂S ∂H / T. For a particular salt M(H,T)=M0[1−exp(−αH/T )]. Show that, at a ﬁxed temperature, if the applied ﬁeld is increased from zero to a strength such that the magnetization of the salt is3 4M0then the salt’s entropy decreases by an amount M0 4α(3−ln 4). 5.29 Using the results of section 5.12, evaluate the integral I(y)= Z∞ 0e−xysinx xdx. Hence show that J= Z∞ 0sinx xdx=π 2. 5.30 The integralZ∞ −∞e−αx2dx has the value ( π/α)1/2. Use this result to evaluate J(n)= Z∞ −∞x2ne−x2dx, where nis a positive integer. Express your answer in terms of factorials. 5.31 The function f(x) is diﬀerentiable and f(0) = 0. A second function g(y) is deﬁned by g(y)= Zy 0f(x)dx√y−x. Prove that dg dy= Zy 0df dxdx√y−x. For the case f(x)=xn, prove that dng dyn=2 (n!)√y. 5.32 The functions f(x, t)a n d F(x) are deﬁned by f(x, t)=e−xt, F(x)= Zx 0f(x, t)dt. Verify by explicit calculation that dF dx=f(x, x)+ Zx 0∂f(x, t) ∂xdt. 187 PARTIAL DIFFERENTIATION 5.33 If I(α)= Z1 0xα−1 lnxdx, α > −1, what is the value of I(0)? Show that d dαxα=xαlnx, and deduce that d dαI(α)=1 α+1. Hence prove that I(α)=l n ( 1+ α). 5.34 Find the derivative with respect to xof the integral I(x)= Z3x xexpxt dt. 5.35 The function G(t, ξ) is deﬁned for 0 ≤t≤πby G(t, ξ)= /( −costsinξ forξ≤t, −sintcosξ forξ>t . Show that the function x(t) deﬁned by x(t)= Zπ 0G(t, ξ)f(ξ)dξ satisﬁes the equation d2x dt2+x=f(t) foranyarbitrary (continuous) function f(t). Show further that x(0) = [dx/dt]x=π= 0, again for any f(t), but that the value of x(π) does depend upon the form of f(t). (The function G(t, ξ) is an example of a Green’s function, an important concept in the solution of diﬀerential equations and one studied extensively inlater chapters.) 5.14 Hints and answers 5.1 (a) (i) 2 xy, x2; (ii) 2 x,2y; (iii) y−1cos(x/y),(−x/y2)cos( x/y); (iv)−y/(x2+y2),x /(x2+y2); (v) x/r,y/r,z/r. (b) (i) 2 y,0,2x; (ii) 2 ,2,0; (v) ( y2+z2)r−3,(x2+z2)r−3,−xyr−3. (c) Both second derivatives are equal to ( y2−x2)(x2+y2)−2. 5.2 Only (c) and (e).5.3 2 x/negationslash=−2y−x.Forgboth sides of equation (5.9) equal y −2. 5.4 (a) 1 + x−x2/negationslash=2x+1 .( b ) g/prime=−g.φ(x, y)=x(x+1 )ye−x+k. 5.5 ∂2z/∂x2=2xz(z2+x)−3,∂2z/∂x∂y =(z2−x)(z2+x)−3,∂2z/∂y2=−2z(z2+x)−3. 5.6 The equation is most easily diﬀerentiated in the form ln p+l nV−lnR−lnT= −α/(VRT).p(α−VRT)/(V2RT);V(α+VRT)/[T(VRT−α)];VRT2/[p(α+ VRT)]. 5.7 (0 ,0),(a/4,−a)a n d( 1 6 a,−8a). Only the saddle point at (0 ,0). 5.8 The transformed equation is ∂2ψ/∂t∂s = 0 where ψ(s, t)=φ(x, y). 5.9 The transformed equation is 2( x2+y2)∂f/∂v = 0; hence fdoes not depend on v. 5.10 Write ∂/∂uand∂/∂θin terms of x,y,∂/∂x and∂/∂y using (5.17). The terms that cancel when ∂2φ/∂u2and∂2φ/∂θ2are added together are ±[x(∂f/∂x )+ y(∂f/∂y )+2xy(∂2f/∂x∂y )]. 188 5.14 HINTS AND ANSWERS 5.11 Maxima equal to 1 /8a t±(1/2,−1/2), minima equal to −1/8a t±(1/2,1/2), saddle points equalling 0 at (0 ,0), (0 ,±1), (±1,0). 5.12 From ∂f/∂y =0,y=6x/b. Substitute this into ∂f/∂x = 0 to obtain a quadratic equation for x. 5.13 Maxima equal to a2e−1at (±a,0), minima equal to −2a2e−1at (0 ,±a), saddle point equalling 0 at (0 ,0). 5.14 Maximum equal to 16 at ( −2,0), minimum equal to −16 at (2 ,0), saddle points equalling−11 at (1 ,±3). 5.15 Minimum at (0 ,0); saddle points at ( ±1,±1). 5.161+√ 2√ 2at± / 1 2,√ 2 2,1 2 / . 5.17 Lagrange multiplier method gives z=y=x/2f o rm a x i m a la r e ao f4 . 5.18 Put the ends of the ladder at ( a+ξ,0) and (0 ,b+η)a n dr e q u i r e( a, b)t ob eo n the ladder. L=(a2/3+b2/3)3/2. 5.19 The cost always includes 2 αhwhich can be ignored in the optimisation. With Lagrange multiplier λ,s i nθ=λw/(4β)a n d βsecθ−1 2λwtanθ=λh, leading to the stated results. 5.20 If the semi-axis in the x-direction is a,t h e n x2/y2=a3/(L−a)3for the envelope. 5.21 The envelope of lines x/a+y/(c−a)−1 = 0, as avaries, is√x+√y=√c.A r e a =c2/6. 5.22 The equation of a typical circle is r=2acosφcos(θ−φ). The envelope condition gives φ=θ/2. 5.23 (a) Using α=c o t θ,w h e r e θis the initial angle a jet makes with the vertical, the equation is f(z,ρ,α)=z−ρα+[gρ2(1+α2)/(2v2 0)], and setting ∂f/∂α = 0 gives α=v2 0/(gρ). The water bell has a parabolic proﬁle z=v2 0/(2g)−gρ2/(2v2 0). (b) Setting z= 0 gives the minimum diameter as 2 v2 0/g. 5.24 The reﬂected ray has equation y=t a n 2 θ(x−Rsinθ/sin 2θ). Put this into the standard form f(x, y, θ) = 0 and eliminate yorxfrom this equation and ∂f/∂θ =0 . 5.25 Show that ( ∂G/∂P )T=Vand ( ∂G/∂T )P=−S. From each result obtain an expression for ∂2G/∂T∂P and equate these, giving ( ∂V/∂T )P=−(∂S/∂P )T. 5.26 Establish that ( ∂U/∂V )T=T(∂S/∂V )T−Pand that ( ∂U/∂T )V=T(∂S/∂T )V. Equate expressions for ∂2S/∂T∂V and hence show α=1 .I n t e g r a t e( ∂S/∂V )T and ( ∂S/∂T )Vto show that S=4T3V/3+l n V+βlnT+c. 5.27 Find expressions for ( ∂S/∂V )Tand ( ∂S/∂T )V, and equate ∂2S/∂V∂T with ∂2S/∂T∂V .U(V,T)=cT−aV−1. 5.28 Show that dF=d(U−TS−HM)=−Sd T−Md H and ﬁnd two expressions for ∂2F/∂H∂T . Establish that ( ∂S/∂H )T=−M0αHT−2exp(−αH/T ) and integrate with respect to H. 5.29 dI/dy =−Im[ R∞ 0exp(−xy+ix)dx]=−1/(1 +y2). Integrate dI/dy from 0 to∞. I(∞)=0a n d I(0) = J. 5.30 Diﬀerentiate both the integral and its value ntimes with respect to αand then setα= 1. Note that 1 3 5 ···(2n−1) = (2 n)!/(2nn!).J(n)=( 2 n)!√π/(4nn!). 5.31 Integrate the RHS of the equation by parts before diﬀerentiating with respect toy. Repeated application of the method establishes the result for all orders of derivative. 5.32 Both sides of the equation equal e−x2+x−2(e−x2−1). 5.33 I(0) = 0; use Leibniz’ rule. 5.34 (6 −x−2)e x p( 3 x2)−(2−x−2)e x p x2. 5.35 Write x(t)=−cost Rt 0sinξf(ξ)dξ−sint Rπ tcosξf(ξ)dξand diﬀerentiate each term as a product to obtain dx/dt.O b t a i n d2x/dt2in a similar way. Note that integrals that have equal lower and upper limits have value zero. x(π)=Rπ 0sinξf(ξ)dξ. 189 6 Multiple integrals For functions of several variables, just as we may consider derivatives with respect to two or more of them, so may the integral of the function with respect to morethan one variable be formed. The formal deﬁnitions of such multiple integrals areextensions of that for a single variable, discussed in chapter 2. We ﬁrst discussdouble and triple integrals and illustrate some of their applications. We thenconsider changing the variables in multiple integrals and discuss some generalproperties of Jacobians. 6.1 Double integrals For an integral involving two variables – a double integral – we have a function, f(x, y) say, to be integrated with respect to xandybetween certain limits. These limits can usually be represented by a closed curve Cbounding a region Rin the xy-plane. Following the discussion of single integrals given in chapter 2, let us divide the region RintoNsubregions ∆ R pof area ∆ Ap,p=1,2,...,N , and let (xp,yp) be any point in subregion ∆ Rp. Now consider the sum S=Nsummationdisplay p=1f(xp,yp)∆Ap, and let N→∞ as each of the areas ∆ Ap→0. If the sum Stends to a unique limit, I, then this is called the double integral of f(x, y)over the region Rand is written I=integraldisplay Rf(x, y)dA, (6.1) where dAstands for the element of area in the xy-plane. By choosing the subregions to be small rectangles each of area ∆ A=∆x∆y, and letting both ∆ x 190 6.1 DOUBLE INTEGRALS V U C TSdxdy RdA=dxdyy d c a b x Figure 6.1 A simple curve Cin the xy-plane, enclosing a region R. and ∆ y→0 ,w ec a na l s ow r i t et h ei n t e g r a la s I=integraldisplayintegraldisplay Rf(x, y)dx dy, (6.2) where we have written out the element of area explicitly as the product of the two coordinate diﬀerentials (see ﬁgure 6.1). Some authors use a single integration symbol whatever the dimension of the integral; others use as many symbols as the dimension. In diﬀerent circumstancesboth have their advantages. We will adopt the convention used in (6.1) and (6.2),that as many integration symbols will be used as diﬀerentials explicitly written. The form (6.2) gives us a clue as to how we may proceed in the evaluation of a double integral. Referring to ﬁgure 6.1, the limits on the integration may be written as an equation c(x, y) = 0 giving the boundary curve C. However, an explicit statement of the limits can be written in two distinct ways. One way of evaluating the integral is ﬁrst to sum up the contributions from the small rectangular elemental areas into horizontal strips of width dy(as shown in the ﬁgure) and then to combine these horizontal strips to cover the region R. In this case, we write I=integraldisplay y=d y=cbraceleftbiggintegraldisplayx=x2(y) x=x1(y)f(x, y)dxbracerightbigg dy, (6.3) where x=x1(y)a n d x=x2(y) are the equations of the curves TSV andTUV respectively. This expression indicates that ﬁrst f(x, y) is to be integrated with respect to x(treating yas a constant) between the values x=x1(y)a n d x=x2(y) and then the result, considered as a function of y,i st ob ei n t e g r a t e db e t w e e nt h e limits y=candy=d. Thus the double integral is evaluated by expressing it in terms of two single integrals called iterated (orrepeated ) integrals. An alternative way of evaluating the integral, however, is ﬁrst to sum up the 191 MULTIPLE INTEGRALS contributions from the elemental rectangles arranged into vertical strips and then to combine these vertical strips to cover the region R.W et h e nw r i t e I=integraldisplayx=b x=abraceleftbiggintegraldisplayy=y2(x) y=y1(x)f(x, y)dybracerightbigg dx, (6.4) where y=y1(x)a n d y=y2(x) are the equations of the curves STU andSVU respectively. In going to (6.4) from (6.3), we have essentially interchanged the order of integration. In the discussion above we assumed that the curve Cwas such that any line parallel to either the x-o r y-axis intersected Cat most twice. In general, provided f(x, y) is continuous everywhere in Rand the boundary curve Chas this simple shape, the same result is obtained irrespective of the order of integration.In cases where the region Rhas a more complicated shape, it can usually be subdivided into smaller simpler regions R 1,R2etc. that satisfy this criterion. The double integral over Ris then merely the sum of the double integrals over the subregions.IEvaluate the double integral I= ZZ Rx2yd xd y , where Ris the triangular area bounded by the lines x=0,y=0andx+y=1. Reverse the order of integration and demonstrate that the same result is obtained. The area of integration is shown in ﬁgure 6.2. Suppose we choose to carry out theintegration with respect to yﬁrst. With xﬁxed, the range of yis 0 to 1−x.W ec a n therefore write I= Zx=1 x=0 /Zy=1−x y=0x2yd y / dx = Zx=1 x=0 /x2y2 2 /y=1−x y=0dx= Z1 0x2(1−x)2 2dx=1 60. Alternatively, we may choose to perform the integration with respect to xﬁrst. With y ﬁxed, the range of xis 0 to 1−y, so we have I= Zy=1 y=0 /Zx=1−y x=0x2yd x / dy = Zy=1 y=0 /x3y 3 /x=1−y x=0dx= Z1 0(1−y)3y 3dy=1 60. As expected, we obtain the same result irrespective of the order of integration. J We may avoid the use of braces in expressions such as (6.3) and (6.4) by writing (6.4), for example, as I=integraldisplayb adxintegraldisplayy2(x) y1(x)dy f(x, y), where it is understood that each integral symbol acts on everything to its right, 192 6.2 TRIPLE INTEGRALS y 11 dy 00dx xx+y=1 R Figure 6.2 The triangular region whose sides are the axes x=0 , y=0a n d the line x+y=1 . and that the order of integration is from right to left. So, in this example, the integrand f(x, y) is ﬁrst to be integrated with respect to ya n dt h e nw i t hr e s p e c t tox. With the double integral expressed in this way, we will no longer write the independent variables explicitly in the limits of integration, since the diﬀerentialof the variable with respect to which we are integrating is always adjacent to therelevant integral sign. Using the order of integration in (6.3), we could also write the double integral as I=integraldisplay d cdyintegraldisplayx2(y) x1(y)dx f(x, y). Occasionally, however, the interchange of the order of integration in a double integral is not permissible, as it yields a diﬀerent result. For example, diﬃcultiesmight arise if the region Rwere unbounded with some of the limits are inﬁ- nite, though, in many cases involving inﬁnite limits the same result is obtainedwhichever order of integration is used. Diﬃculties can also occur if the integrandf(x, y) has any discontinuities in the region Ror on its boundary C. 6.2 Triple integrals The above discussion for double integrals can easily be extended to triple integrals. Consider the function f(x, y, z) deﬁned in a closed three-dimensional region R. Proceeding as we did for double integrals, let us divide the region Rinto N subregions ∆ R pof volume ∆ Vp,p=1,2,...,N ,a n dl e t( xp,yp,zp) be any point in the subregion ∆ Rp. Now we form the sum S=Nsummationdisplay p=1f(xp,yp,zp)∆Vp, 193 MULTIPLE INTEGRALS and let N→∞as each of the volumes ∆ Vp→0. If the sum Stends to a unique limit, I, then this is called the triple integral of f(x, y, z)over the region Rand is written I=integraldisplay Rf(x, y, z)dV, (6.5) where dVstands for the element of volume. By choosing the subregions to be small cuboids, each of volume ∆ V=∆x∆y∆z, and proceeding to the limit, we c a na l s ow r i t et h ei n t e g r a la s I=integraldisplayintegraldisplayintegraldisplay Rf(x, y, z)dx dy dz, (6.6) where we have written out the element of volume explicitly as the product of the three coordinate diﬀerentials. Extending the discussion of double integrals, wemay write triple integrals as three iterated integrals, for example, I=integraldisplay x2 x1dxintegraldisplayy2(x) y1(x)dyintegraldisplayz2(x,y) z1(x,y)dz f(x, y, z), where the limits on each of the integrals describe the values that x,yandztake on the boundary of the region R. As for double integrals, in most cases the order of integration does not aﬀect the value of the integral. We can extend these ideas to deﬁne multiple integrals of higher dimensionality in a similar way. 6.3 Applications of multiple integrals Multiple integrals have many uses in the physical sciences, since there are numer- ous physical quantities which can be written in terms of them. We now discuss a few of the more common examples. 6.3.1 Areas and volumes Multiple integrals are often used in ﬁnding areas and volumes. For example, the integral A=integraldisplay RdA=integraldisplayintegraldisplay Rdx dy is simply equal to the area of the region R. Similarly, if we consider the surface z=f(x, y) in three-dimensional Cartesian coordinates then the volume under this surface that stands vertically above the region Ris given by the integral V=integraldisplay Rzd A=integraldisplayintegraldisplay Rf(x, y)dx dy, where volumes above the xy-plane are counted as positive, and those below as negative. 194 6.3 APPLICATIONS OF MULTIPLE INTEGRALS z c dx a xdz dyb ydV=dx dy dz Figure 6.3 The tetrahedron bounded by the coordinate surfaces and the plane x/a+y/b+z/c= 1 is divided up into vertical slabs, the slabs into columns and the columns into small boxes.IFind the volume of the tetrahedron bounded by the three coordinate surfaces x=0,y=0 andz=0and the plane x/a+y/b+z/c=1. Referring to ﬁgure 6.3, the elemental volume of the shaded region is given by dV=zd xd y , and we must integrate over the triangular region Rin the xy-plane whose sides are x=0 , y=0a n d y=b−bx/a. The total volume of the tetrahedron is therefore given by V= ZZ Rzd xd y = Za 0dx Zb−bx/a 0dy c / 1−y b−x a / =c Za 0dx / y−y2 2b−xy a /y=b−bx/a y=0 =c Za 0dx /bx2 2a2−bx a+b 2 / =abc 6. J Alternatively, we can write the volume of a three-dimensional region Ras V=integraldisplay RdV=integraldisplayintegraldisplayintegraldisplay Rdx dy dz, (6.7) where the only diﬃculty occurs in setting the correct limits on each of the integrals. For the above example, writing the volume in this way corresponds to dividing the tetrahedron into elemental boxes of volume dx dy dz (as shown in ﬁgure 6.3); integration over zthen adds up the boxes to form the shaded column in the ﬁgure. The limits of integration are z=0t o z=cparenleftbig 1−y/b−x/aparenrightbig ,a n d 195 MULTIPLE INTEGRALS the total volume of the tetrahedron is given by V=integraldisplaya 0dxintegraldisplayb−bx/a 0dyintegraldisplayc(1−y/b−x/a) 0dz, (6.8) which clearly gives the same result as above. This method is illustrated further in the following example.IFind the volume of the region bounded by the paraboloid z=x2+y2and the plane z=2y. The required region is shown in ﬁgure 6.4. In order to write the volume of the region in the form (6.7), we must deduce the limits on each of the integrals. Since the integrationscan be performed in any order, let us ﬁrst divide the region into vertical slabs of thicknessdyperpendicular to the y-axis, and then as shown in the ﬁgure we cut each slab into horizontal strips of height dz, and each strip into elemental boxes of volume dV=dx dy dz . Integrating ﬁrst with respect to x(adding up the elemental boxes to get a horizontal strip), the limits on xarex=−p z−y2tox= p z−y2. Now integrating with respect to z (adding up the strips to form a vertical slab) the limits on zarez=y2toz=2y. Finally, integrating with respect to y(adding up the slabs to obtain the required region), the limits onyarey=0a n d y= 2, the solutions of the simultaneous equations z=02+y2and z=2y. So the volume of the region is V= Z2 0dy Z2y y2dz Z√ z−y2 −√ z−y2dx= Z2 0dy Z2y y2dz2 p z−y2 = Z2 0dy /4 3(z−y2)3/2 /z=2y z=y2= Z2 0dy4 3(2y−y2)3/2. The integral over ymay be evaluated straightforwa rdly by making the substitution y= 1+s i n u, and gives V=π/2. J In general, when calculating the volume (area) of a region, the volume (area) elements need not be small boxes as in the previous example, but may be of anyconvenient shape. They are usually chosen to make the evaluation of the integralas simple as possible. 6.3.2 Masses, centres of mass and centroids It is sometimes necessary to calculate the mass of a given object having a non- uniform density. Symbolically, this mass is given simply by M=integraldisplay dM, where dMis the element of mass and the integral is taken over the extent of the object. For a solid three-dimensional body the element of mass is just dM=ρd V, where dVis an element of volume and ρis the variable density. For a laminar body (i.e. a uniform sheet of material) the element of mass is dM=σd A,w h e r e σis the mass per unit area of the body and dAis an area element. Finally, for a body in the form of a thin wire we have dM=λd s,w h e r e λis the mass per 196 6.3 APPLICATIONS OF MULTIPLE INTEGRALS z 02 y dV=dx dy dzz=2y z=x2+y2 x Figure 6.4 The region bounded by the paraboloid z=x2+y2and the plane z=2yis divided into vertical slabs, the slabs into horizontal strips and the strips into boxes. unit length and dsis an element of arc length along the wire. When evaluating the required integral, we are free to divide up the body into mass elements in the most convenient way, provided that over each mass element the density isapproximately constant.IFind the mass of the tetrahedron bounded by the three coordinate surfaces and the plane x/a+y/b+z/c=1, if its density is given by ρ(x, y, z)=ρ0(1 +x/a). From (6.8), we can immediately write down the mass of the tetrahedron as M= Z Rρ0 / 1+x a / dV= Za 0dx ρ 0 / 1+x a / Zb−bx/a 0dy Zc(1−y/b−x/a) 0dz, where we have taken the density outside the integrations with respect to zandysince it depends only on x. Therefore the integrations with respect to zandyproceed exactly as they did when ﬁnding the volume of the tetrahedron, and we have M=cρ0 Za 0dx / 1+x a / /bx2 2a2−bx a+b 2 / . (6.9) We could have arrived at (6.9) more directly by dividing the tetrahedron into triangular slabs of thickness dxperpendicular to the x-axis (see ﬁgure 6.3), each of which is of constant density, since ρdepends on xalone. A slab at a position xhas volume dV= 1 2c(1−x/a)(b−bx/a)dxand mass dM=ρd V=ρ0(1 + x/a)dV. Integrating over xwe again obtain (6.9). This integral is easily evaluated and gives M=5 24abcρ 0. J 197 MULTIPLE INTEGRALS The coordinates of the centre of mass of a solid or laminar body may also be written as multiple integrals. The centre of mass of a body has coordinates ¯x,¯y, ¯z) given by the three equations ¯xintegraldisplay dM=integraldisplay xd M ¯yintegraldisplay dM=integraldisplay yd M ¯zintegraldisplay dM=integraldisplay zd M , where again dMis an element of mass as described above, x,y,zare the coordinates of the centre of mass of the element dMand the integrals are taken over the entire body. Obviously, for any body that lies entirely in, or is symmetricalabout, the xy-plane (say), we immediately have ¯z= 0. For completeness, we note that the three equations above can be written as the single vector equation (seechapter 7) ¯r=1 Mintegraldisplay rdM, where ¯ris the position vector of the body’s centre of mass with respect to the origin, ris the position vector of the centre of mass of the element dMand M=integraltext dMis the total mass of the body. As previously, we may divide the body into the most convenient mass elements for evaluating the necessary integrals,provided each mass element is of constant density. We further note that the coordinates of the centroid of a body are deﬁned as those that its centre of mass would have if the body had uniform density.IFind the centre of mass of the solid hemisphere bounded by the surfaces x2+y2+z2=a2 and the xy-plane, assuming that it has a uniform density ρ. Referring to ﬁgure 6.5, we know from symmetry that the centre of mass must lie on thez-axis. Let us divide the hemisphere into volume elements that are circular slabs of thickness dzparallel to the xy-plane. For a slab at a height z, the mass of the element is dM=ρd V=ρπ(a2−z2)dz. Integrating over z, we ﬁnd that the z-coordinate of the centre of mass of the hemisphere is given by ¯z Za 0ρπ(a2−z2)dz= Za 0zρπ(a2−z2)dz. The integrals are easily evaluated and give ¯z=3a/8. Since the hemisphere is of uniform density, this is also the position of its centroid. J 6.3.3 Pappus’ theorems The theorems of Pappus (which are about seventeen centuries old) relate centroids to volumes of revolution and areas of surfaces, discussed in chapter 2, and can beuseful for ﬁnding one quantity given another that may be calculated more easily. 198 6.3 APPLICATIONS OF MULTIPLE INTEGRALS z xya a a√ a2−z2 dz Figure 6.5 The solid hemisphere bounded by the surfaces x2+y2+z2=a2 and the xy-plane. A yy xdA ¯y Figure 6.6 An area Ain the xy-plane, which may be rotated about the x-axis to form a volume of revolution. If a plane area is rotated about an axis that does not intersect it then the solid so generated is called a volume of revolution .Pappus’ ﬁrst theorem states that the volume of such a solid is given by the plane area Amultiplied by the distance moved by its centroid (see ﬁgure 6.6). This may be proved by considering thedeﬁnition of the centroid of the plane area as the position of the centre of massif the density is uniform, so that ¯y=1 Aintegraldisplay yd A . Now the volume generated by rotating the plane area about the x- a x i si sg i v e nb y V=integraldisplay 2πy dA =2π¯yA, which is the area multiplied by the distance moved by the centroid. 199 MULTIPLE INTEGRALS y yds ¯y x Figure 6.7 A curve in the xy-plane, which may be rotated about the x-axis to form a surface of revolution. Pappus’ second theorem states that if a plane curve is rotated about a coplanar axis that does not intersect it then the area of the surface of revolution so generated is given by the length of the curve Lmultiplied by the distance moved by its centroid (see ﬁgure 6.7). This may be proved in a similar manner to the ﬁrst theorem by considering the deﬁnition of the centroid of a plane curve, ¯y=1 Lintegraldisplay yd s , and noting that the surface area generated is given by S=integraldisplay 2πy ds=2π¯yL, which is equal to the length of the curve multiplied by the distance moved by its centroid.IA semicircular uniform lamina is freely suspended from one of its corners. Show that its straight edge makes an angle of 23.0◦with the vertical. Referring to ﬁgure 6.8, the suspended lamina will have its centre of gravity Cvertically below the suspension point and its straight edge will make an angle θ=t a n−1(d/a)w i t h the vertical, where 2 ais the diameter of the semicircle and dis the distance of its centre of mass from the diameter. Since rotating the lamina about the diameter generates a sphere of volume4 3πa3, Pappus’ ﬁrst theorem requires that 4 3πa3=2π×d×1 2πa2. Hence d=4 3a/πandθ=t a n−1(4 3π)=2 3 .0◦. J 200 6.3 APPLICATIONS OF MULTIPLE INTEGRALS aθ dC Figure 6.8 Suspending a semicircular lamina from one of its corners. 6.3.4 Moments of inertia For problems in rotational mechanics it is often necessary to calculate the moment of inertia of a body about a given axis. This is deﬁned by the multiple integral I=integraldisplay l2dM, where lis the distance of a mass element dMfrom the axis. We may again choose mass elements convenient for evaluating the integral. In this case, however, inaddition to elements of constant density we require all parts of each element tobe at approximately the same distance from the axis about which the moment of inertia is required.IFind the moment of inertia of a uniform rectangular lamina of mass Mwith sides aand babout one of the sides of length b. Referring to ﬁgure 6.9, we wish to calculate the moment of inertia about the y-axis. We therefore divide the rectangular lamina into elemental strips parallel to the y-axis of width dx. The mass of such a strip is dM=σbdx,w h e r e σis the mass per unit area of the lamina. The moment of inertia of a strip at a distance xfrom the y-axis is simply dI=x2dM=σbx2dx. The total moment of inertia of the lamina about the y-axis is therefore I= Za 0σbx2dx=σba3 3. Since the total mass of the lamina is M=σab,w ec a nw r i t e I=1 3Ma2. J 201 MULTIPLE INTEGRALS y xb dx adM=σbdx Figure 6.9 A uniform rectangular lamina of mass Mwith sides aandbcan be divided into vertical strips. 6.3.5 Mean values of functions In chapter 2 we discussed average values for functions of a single variable. This is easily extended to functions of several variables. Let us consider, for example,a function f(x, y) deﬁned in some region Rof the xy-plane. Then the average value ¯fof the function is given by ¯fintegraldisplay RdA=integraldisplay Rf(x, y)dA. (6.10) This deﬁnition is easily extended to three (and higher) dimensions; if a function f(x, y, z) is deﬁned in some three-dimensional region of space Rthen the average value ¯fof the function is given by ¯fintegraldisplay RdV=integraldisplay Rf(x, y, z)dV. (6.11)IA tetrahedron is bounded by the three coordinate surfaces and the plane x/a+y/b+z/c= 1and has density ρ(x, y, z)=ρ0(1 +x/a). Find the average value of the density. From (6.11), the average value of the density is given by ¯ρ Z RdV= Z Rρ(x, y, z)dV. Now the integral on the LHS is just the volume of the tetrahedron, which we found in subsection 6.3.1 to be V=1 6abc, and the integral on the RHS is its mass M=5 24abcρ 0, calculated in subsection 6.3.2. Therefore ¯ρ=M/V =5 4ρ0. J 6.4 Change of variables in multiple integrals It often happens that, either because of the form of the integrand involved or because of the boundary shape of the region of integration, it is desirable to 202 6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS y xu=c o n s t a n t v=c o n s t a n t NM L KR C Figure 6.10 A region of integration Roverlaid with a grid formed by the family of curves u=c o n s t a n ta n d v= constant. The parallelogram KLMN deﬁnes the area element dAuv. express a multiple integral in terms of a new set of variables. We now consider h o wt od ot h i s . 6.4.1 Change of variables in double integrals Let us begin by examining the change of variables in a double integral. Suppose that we require to change an integral I=integraldisplayintegraldisplay Rf(x, y)dx dy, in terms of coordinates xandy, into one expressed in new coordinates uandv, given in terms of xandyby diﬀerentiable equations u=u(x, y)a n d v=v(x, y) with inverses x=x(u, v)a n d y=y(u, v). The region Rin the xy-plane and the curve Cthat bounds it will become a new region R/primeand a new boundary C/primein theuv-plane, and so we must change the limits of integration accordingly. Also, the function f(x, y) becomes a new function g(u, v)o ft h en e wc o o r d i n a t e s . Now the part of the integral that requires most consideration is the area element. In the xy-plane the element is the rectangular area dAxy=dx dy generated by constructing a grid of straight lines parallel to the x-a n d y- axes respectively. Our task is to determine the corresponding area element in the uv-coordinates. In general the corresponding element dAuvwill not be the same shape as dAxy, but this does not matter since all elements are inﬁnitesimally small and the value ofthe integrand is considered constant over them. Since the sides of the area element are inﬁnitesimal, dA uvwill in general have the shape of a parallelogram. We can ﬁnd the connection between dAxyanddAuvby considering the grid formed by the family of curves u=c o n s t a n ta n d v= constant, as shown in ﬁgure 6.10. Since v 203 MULTIPLE INTEGRALS is constant along the line element KL, the latter has components ( ∂x/∂u )duand (∂y/∂u )duin the directions of the x-a n d y-axes respectively. Similarly, since u is constant along the line element KN, the latter has corresponding components (∂x/∂v )dvand ( ∂y/∂v )dv. Using the result for the area of a parallelogram given in chapter 7, we ﬁnd that the area of the parallelogram KLMN is given by dAuv=vextendsinglevextendsinglevextendsinglevextendsingle∂x ∂udu∂y ∂vdv−∂x ∂vdv∂y ∂uduvextendsinglevextendsinglevextendsinglevextendsingle =vextendsinglevextendsinglevextendsinglevextendsingle∂x ∂u∂y ∂v−∂x ∂v∂y ∂uvextendsinglevextendsinglevextendsinglevextendsingledu dv. Deﬁning the Jacobian ofx,ywith respect to u,vas J=∂(x, y) ∂(u, v)≡∂x ∂u∂y ∂v−∂x ∂v∂y ∂u, we have dAuv=vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y) ∂(u, v)vextendsinglevextendsinglevextendsinglevextendsingledu dv. The reader acquainted with determinants will notice that the Jacobian can also be written as the 2 ×2 determinant J=∂(x, y) ∂(u, v)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂x ∂u∂y ∂u ∂x ∂v∂y ∂vvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. Such determinants can in general be evaluated using the methods of chapter 8. So, in summary, the relationship between the size of the area element generated bydx,dyand the size of the corresponding area element generated by du,dvis dx dy =vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y) ∂(u, v)vextendsinglevextendsinglevextendsinglevextendsingledu dv. This equality should be taken as meaning that when transforming from coordi- nates x, yto coordinates u, v, the area element dx dy should be replaced by the expression on the RHS of the above equality. Of course, the Jacobian can, andin general will, vary over the region of integration. We may express the doubleintegral in either coordinate system as I=integraldisplayintegraldisplay Rf(x, y)dx dy =integraldisplayintegraldisplay R/primeg(u, v)vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y) ∂(u, v)vextendsinglevextendsinglevextendsinglevextendsingledu dv. (6.12) When evaluating the integral in the new coordinate system, it is usually advisable to sketch the region of integration R/primein the uv-plane. 204 6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALSIEvaluate the double integral I= ZZ R / a+ p x2+y2 / dx dy, where Ris the region bounded by the circle x2+y2=a2. In Cartesian coordinates, the integral may be written I= Za −adx Z√ a2−x2 −√ a2−x2dy / a+ p x2+y2 / , and can be calculated directly. However, because of the circular boundary of the integration region, a change of variables to plane polar coordinates ρ,φis indicated. The relationship between Cartesian and plane polar coordinates is given by x=ρcosφandy=ρsinφ. Using (6.12) we can therefore write I= ZZ R/prime(a+ρ) ////∂(x, y) ∂(ρ, φ) ////dρ dφ, where R/primeis the rectangular region in the ρφ-plane whose sides are ρ=0 , ρ=a,φ=0 andφ=2π. The Jacobian is easily calculated, and we obtain J=∂(x, y) ∂(ρ, φ)= ////cosφ sinφ −ρsinφρcosφ ////=ρ(cos2φ+s i n2φ)=ρ. So the relationship between the area elements in Cartesian and in plane polar coordinates is dx dy =ρd ρd φ . Therefore, when expressed in plane polar coordinates, the integral is given by I= ZZ R/prime(a+ρ)ρd ρd φ = Z2π 0dφ Za 0dρ(a+ρ)ρ=2π /aρ2 2+ρ3 3 /a 0=5πa3 3. J 6.4.2 Evaluation of the integral I=integraltext∞ −∞e−x2dx By making a judicious change of variables, it is sometimes possible to evaluate an integral that would be intractable otherwise. An important example of this method is provided by the evaluation of the integral I=integraldisplay∞ −∞e−x2dx. Its value may be found by ﬁrst constructing I2, as follows: I2=integraldisplay∞ −∞e−x2dxintegraldisplay∞ −∞e−y2dy=integraldisplay∞ −∞dxintegraldisplay∞ −∞dy e−(x2+y2) =integraldisplayintegraldisplay Re−(x2+y2)dx dy, 205 MULTIPLE INTEGRALS a a −a−ay x Figure 6.11 The regions used to illustrate the convergence properties of the integral I(a)= Ra −ae−x2dxasa→∞. where the region Ris the whole xy-plane. Then, transforming to plane polar coordinates, we ﬁnd I2=integraldisplayintegraldisplay R/primee−ρ2ρd ρd φ =integraldisplay2π 0dφintegraldisplay∞ 0dρ ρe−ρ2=2πbracketleftBig −1 2e−ρ2bracketrightBig∞ 0=π. Therefore the original integral is given by I=√π. Because the integrand is an even function of x, it follows that the value of the integral from 0 to ∞is simply√π/2. We note, however, that unlike in all the previous examples, the regions of integration RandR/primeare both inﬁnite in extent (i.e. unbounded). It is therefore prudent to derive this result more rigorously; this we do by considering theintegral I(a)=integraldisplay a −ae−x2dx. We then have I2(a)=integraldisplayintegraldisplay Re−(x2+y2)dx dy, where Ri st h es q u a r eo fs i d e2 acentred on the origin. Referring to ﬁgure 6.11, since the integrand is always positive the value of the integral taken over the square lies between the value of the integral taken over the region bounded by the inner circle of radius aand the value of the integral taken over the outer circle of radius√ 2a. Transforming to plane polar coordinates as above, we may 206 6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS z xyCR T SPQu=c1v=c2 w=c3 Figure 6.12 A three-dimensional region of integration R, showing an el- ement of volume in u, v, w coordinates formed by the coordinate surfaces u=c o n s t a n t , v=c o n s t a n t , w=c o n s t a n t . evaluate the integrals over the inner and outer circles respectively, and we ﬁnd πparenleftBig 1−e−a2parenrightBig <I2(a)<πparenleftBig 1−e−2a2parenrightBig . Taking the limit a→∞, we ﬁnd I2(a)→π.T h e r e f o r e I=√πas we found previously. We use this result in the discussion of the normal distribution inchapter 26. 6.4.3 Change of variables in triple integrals A change of variable in a triple integral follows the same general lines as that for a double integral. Suppose we wish to change variables from x, y, z tou, v, w . In the x, y, z coordinates the element of volume is a cuboid of sides dx, dy, dz and volume dV xyz=dx dy dz . If, however, we divide up the total volume into inﬁnitesimal elements by constructing a grid formed from the coordinate surfacesu=c o n s t a n t , v=c o n s t a n ta n d w= constant, then the element of volume dV uvw in the new coordinates will have the shape of a parallelepiped whose faces are the coordinate surfaces and whose edges are the curves formed by the intersections of these surfaces (see ﬁgure 6.12). Along the line element PQthe coordinates v andware constant, and so PQhas components of ( ∂x/∂u )du,(∂y/∂u )duand 207 MULTIPLE INTEGRALS (∂z/∂u )duin the direction of the x-,y-a n d z- axes respectively. The components of the line elements PSandSTare found by replacing ubyvandwrespectively. The expression for the volume of a parallelepiped in terms of the components of its edges with respect to the x-,y-a n d z-axes is given in chapter 7. Using this, we ﬁnd that the element of volume in u, v, w coordinates is given by dVuvw=vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y, z) ∂(u, v, w)vextendsinglevextendsinglevextendsinglevextendsingledu dv dw, where the Jacobian of x, y, z with respect to u, v, w is a short-hand for a 3 ×3 determinant: ∂(x, y, z) ∂(u, v, w)≡vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂x ∂u∂y ∂u∂z ∂u ∂x ∂v∂y ∂v∂z ∂v ∂x ∂w∂y ∂w∂z ∂wvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. So, in summary, the relationship between the elemental volumes in multiple integrals formulated in the two coordinate systems is given in Jacobian form by dx dy dz =vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y, z) ∂(u, v, w)vextendsinglevextendsinglevextendsinglevextendsingledu dv dw, and we can write a triple integral in either set of coordinates as I=integraldisplayintegraldisplayintegraldisplay Rf(x, y, z)dx dy dz =integraldisplayintegraldisplayintegraldisplay R/primeg(u, v, w)vextendsinglevextendsinglevextendsinglevextendsingle∂(x, y, z) ∂(u, v, w)vextendsinglevextendsinglevextendsinglevextendsingledu dv dw.IFind an expression for a volume element in sphe rical polar coordinates, and hence calcu- late the moment of inertia about a diameter of a uniform sphere of radius aand mass M. Spherical polar coordinates r,θ,φ are deﬁned by x=rsinθcosφ, y =rsinθsinφ, z =rcosθ (and are discussed fully in chapter 10). The required Jacobian is therefore J=∂(x, y, z) ∂(r,θ,φ)= //////sinθcosφ sinθsinφ cosθ rcosθcosφr cosθsinφ−rsinθ −rsinθsinφrsinθcosφ 0 //////. The determinant is most easily evaluated by expanding it with respect to the last column (see chapter 8), which gives J=c o s θ(r2sinθcosθ)+rsinθ(rsin2θ) =r2sinθ(cos2θ+s i n2θ)=r2sinθ. Therefore the volume element in spherical polar coordinates is given by dV=∂(x, y, z) ∂(r,θ,φ)dr dθ dφ =r2sinθd rd θd φ , which agrees with the result given in chapter 10. 208 6.4 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS If we place the sphere with its centre at the origin of an x, y, z coordinate system then its moment of inertia about the z-axis (which is, of course, a diameter of the sphere) is I= Z/; x2+y2 / dM=ρ Z/; x2+y2 / dV, where the integral is taken over the sphere, and ρis the density. Using spherical polar coordinates, we can write this as I=ρ ZZZ V /; r2sin2θ / r2sinθd rd θd φ =ρ Z2π 0dφ Zπ 0dθsin3θ Za 0dr r4 =ρ×2π×4 3×1 5a5=8 15πa5ρ. Since the mass of the sphere is M=4 3πa3ρ, the moment of inertia can also be written as I=2 5Ma2. J 6.4.4 General properties of Jacobians Although we will not prove it, the general result for a change of coordinates in ann-dimensional integral from a set xito a set yj(where iandjboth run from 1t on)i s dx1dx2···dxn=vextendsinglevextendsinglevextendsinglevextendsingle∂(x 1,x2,...,x n) ∂(y1,y2,...,y n)vextendsinglevextendsinglevextendsinglevextendsingledy 1dy2···dyn, where the n-dimensional Jacobian can be written as an n×ndeterminant (see chapter 8) in an analogous way to the two- and three-dimensional cases. For readers who already have suﬃcient familiarity with matrices (see chapter 8) and their properties, a fairly compact proof of some useful general properties of Jacobians can be given as follows. Other readers should turn straight to theresults (6.16) and (6.17) and return to the proof at some later time. Consider three sets of variables x i,yiandzi, with irunning from 1 to nfor each set. From the chain rule in partial diﬀerentiation (see (5.17)), we know that ∂xi ∂zj=nsummationdisplay k=1∂xi ∂yk∂yk ∂zj. (6.13) Now let A,Band Cbe the matrices whose ijth elements are ∂xi/∂y j,∂yi/∂z jand ∂xi/∂z jrespectively. We can then write (6.13) as the matrix product cij=nsummationdisplay k=1aikbkj or C=AB. (6.14) We may now use the general result for the determinant of the product of two matrices, namely |AB|=|A||B|, and recall that the Jacobian Jxy=∂(x1,...,x n) ∂(y1,...,y n)=|A|, (6.15) 209 MULTIPLE INTEGRALS and similarly for JyzandJxz. On taking the determinant of (6.14), we therefore obtain Jxz=JxyJyz or, in the usual notation, ∂(x1,...,x n) ∂(z1,...,z n)=∂(x1,...,x n) ∂(y1,...,y n)∂(y1,...,y n) ∂(z1,...,z n). (6.16) As a special case, if the set zii st a k e nt ob ei d e n t i c a lt ot h es e t xi,a n dt h e obvious result Jxx= 1 is used, we obtain JxyJyx=1 or, in the usual notation, ∂(x1,...,x n) ∂(y1,...,y n)=bracketleftbigg∂(y1,...,y n) ∂(x1,...,x n)bracketrightbigg−1 . (6.17) The similarity between the properties of Jacobians and those of derivatives is apparent, and to some extent is suggested by the notation. We further note from(6.15) that since |A|=|A T|,w h e r e ATis the transpose of A, we can interchange the rows and columns in the determinantal form of the Jacobian without changingits value. 6.5 Exercises 6.1 Sketch the curved wedge bounded by the surfaces y2=4ax,x+z=aandz=0 , and hence calculate its volume V. 6.2 Evaluate the volume integral of x2+y2+z2over the rectangular parallelepiped bounded by the six surfaces x=±a,y=±b,z=±c. 6.3 Find the volume integral of x2yover the tetrahedral volume bounded by the planes x=0 , y=0 , z=0 ,a n d x+y+z=1 . 6.4 Evaluate the surface integral of f(x, y) over the rectangle 0 ≤x≤a,0≤y≤b for the functions (a)f(x, y)=x x2+y2, (b)f(x, y)=(b−y+x)−3/2. 6.5 (a) Prove that the area of the ellipse x2 a2+y2 b2=1 isπab. (b) Use this result to obtain an expression for the volume of a slice of thickness dzof the ellipsoid x2 a2+y2 b2+z2 c2=1. Hence show that the volume of the ellipsoid is 4 πabc/ 3. 210 6.5 EXERCISES 6.6 The function Ψ(r)=A / 2−Zr a / e−Zr/2a gives the form of the quantum mechanical wavefunction representing the electron in a hydrogen-like atom of atomic number Zwhen the electron is in its ﬁrst allowed spherically symmetric excited state. Here ris the usual spherical polar coordinate, but, because of the spherical symmetry, the coordinates θandφdo not appear explicitly in Ψ. Determine the value that A(assumed real) must have if the wavefunction is to be correctly normalised, i.e. the volume integral of |Ψ|2 over all space is equal to unity. 6.7 In quantum mechanics the electron in a hydrogen atom in some particular state is described by a wavefunction Ψ, which is such that |Ψ|2dVis the probability of ﬁnding the electron in the inﬁnitesimal volume dV. In spherical polar coordinates Ψ=Ψ ( r,θ,φ)a n d dV=r2sinθd rd θd φ . Two such states are described by Ψ1= /1 4π /1/2 /1 a0 /3/2 2e−r/a0, Ψ2=− /3 8π /1/2 sinθeiφ /1 2a0 /3/2re−r/2a0 a0√ 3. (a) Show that each Ψ iis normalised, i.e. the integral over all space R |Ψ|2dVis equal to unity – physically, this means that the electron must be somewhere. (b) The (so-called) dipole matrix element between the states 1 and 2 is given by the integral px= Z Ψ∗ 1qrsinθcosφΨ2dV, where qis the charge on the electron. Prove that pxhas the value −27qa0/35. 6.8 A planar ﬁgure is formed from uniform wire and consists of two semicircular arcs, each with its own closing diameter, joined so as to form a letter ‘B’. Theﬁgure is freely suspended from its top left-hand corner. Show that the straightedge of the ﬁgure makes an angle θwith the vertical given by tan θ=( 2+ π) −1. 6.9 A certain torus has a circular vertical cross-section of radius acentred on a horizontal circle of radius c(>a). (a) Find the volume Vand surface area Aof the torus, and show that they can be written as V=π2 4(r2 o−r2 i)(ro−ri),A =π2(r2 o−r2 i), where roandroare respectively the outer and inner radii of the torus. (b) Show that a vertical circular cylinder of radius c, coaxial with the torus, divides Ain the ratio πc+2a:πc−2a. 6.10 A thin uniform circular disc has mass Mand radius a. (a) Prove that its moment of inertia about an axis perpendicular to its plane and passing through its centre is1 2Ma2. (b) Prove that the moment of inertia of the same disc about a diameter is1 4Ma2. This is an example of the general result for planar bodies that the moment of inertia of the body about an axis perpendicular to the plane is equal to the sum 211 MULTIPLE INTEGRALS of the moments of inertia about two perpendicular axes lying in the plane: in an obvious notation Iz= Z r2dm= Z (x2+y2)dm= Z x2dm+ Z y2dm=Iy+Ix. 6.11 In some applications in mechanics the moment of inertia of a body about a single point (as opposed to about an axis) is needed. The moment of inertia I about the origin of a uniform solid body of density ρis given by the volume integral I= Z V(x2+y2+z2)ρd V. Show that the moment of inertia of a right circular cylinder of radius a,l e n g t h 2b,a n dm a s s Mabout its centre is M /a2 2+b2 3 / . 6.12 The shape of an axially symmetric hard-boiled egg, of uniform density ρ0,i s given in spherical polar coordinates by r=a(2−cosθ), where θis measured from the axis of symmetry. (a) Prove that the mass Mof the egg is M=40 3πρ0a3. (b) Prove that the egg’s moment of inertia about its axis of symmetry is342 175Ma2. 6.13 In spherical polar coordinates r, θ, φ the element of volume for a body that is symmetrical about the polar axis is dV=2πr2sinθd rd θ , whilst its element of surface area is 2 πrsinθ[(dr)2+r2(dθ)2]1/2. A particular surface is deﬁned by r=2acosθ,w h e r e ais a constant, and 0 ≤θ≤π/2. Find its total surface area and the volume it encloses, and hence identify the surface. 6.14 By expressing both the integrand and the surface element in spherical polar coordinates, show that the surface integralZx2 x2+y2dS over the surface x2+y2=z2,0≤z≤1, has the value π/√2. 6.15 By transforming to cylindrical polar coordinates, evaluate the integral I= Z Z Z ln(x2+y2)dx dy dz over the interior of the conical region x2+y2≤z2,0≤z≤1. 6.16 Sketch the two families of curves y2=4u(u−x),y2=4v(v+x), where uandvare parameters. By transforming to the uv-plane evaluate the integral of y/(x2+y2)1/2over that part of the quadrant x>0,y>0 bounded by the lines x=0 , y=0a n d the curve y2=4a(a−x). 6.17 By making two successive simple changes of variables, evaluate I= Z Z Z x2dx dy dz over the ellipsoidal region x2 a2+y2 b2+z2 c2≤1. 212 6.5 EXERCISES 6.18 Sketch the domain of integration for the integral I= Z1 0 Z1/y x=yy3 xexp[y2(x2+x−2)]dx dy and characterise its boundaries in terms of new variables u=xyandv=y/x. Show that the Jacobian for the change from ( x, y)t o( u, v)i se q u a lt o( 2 v)−1,a n d hence evaluate I. 6.19 Sketch that part of the region 0 ≤x,0≤y≤π/2 which is bounded by the curves x=0 , y=0 ,s i n h xcosy=1a n dc o s h xsiny= 1. By making a suitable change of variables, evaluate the integral I= Z Z (sinh2x+c o s2y)si nh2 xsin 2yd xd y over the bounded sub-region. 6.20 Deﬁne a coordinate system u, vwhose origin coincides with that of the usual x, ysystem and whose u-axis coincides with the x-axis, whilst the v-axis makes an angle αwith it. By considering the integral I= R exp(−r2)dA,w h e r e ris the radial distance from the origin, over the area deﬁned by 0 ≤u<∞,0≤v<∞, prove thatZ∞ 0 Z∞ 0exp(−u2−v2−2uvcosα)du dv=α 2si nα. 6.21 As stated in section 5.11, the ﬁrst law of thermodynamics can be expressed as dU=TdS−PdV. By calculating and equating ∂2U/∂Y ∂X and∂2U/∂X∂Y ,w h e r e XandYare an unspeciﬁed pair of variables (drawn from P,V,T andS), prove that ∂(S,T) ∂(X,Y)=∂(V,P) ∂(X,Y). Using the properties of Jacobians, deduce that ∂(S,T) ∂(V,P)=1. 6.22 The distances of the variable point P, which has coordinates x, y, z, from the ﬁxed points (0 ,0,1) and (0 ,0,−1) are denoted by uandvrespectively. New variables ξ,η,φ are deﬁned by ξ=1 2(u+v),η =1 2(u−v), andφis the angle between the plane y= 0 and the plane containing the three points. Prove that the Jacobian ∂(ξ,η,φ )/∂(x, y, z) has the value ( ξ2−η2)−1and thatZ Z Z all space(u−v)2 uvexp / −u+v 2 / dx dy dz =32π 3e. 6.23 This is a more diﬃcult question about ‘volumes’ in an increasing number of dimensions. (a) Let Rbe a real positive number and deﬁne Kmby Km= ZR −R /; R2−x2 /mdx. Show, using integration by parts, that Kmsatisﬁes the recurrence relation (2m+1 )Km=2mR2Km−1. 213 MULTIPLE INTEGRALS (b) For integer n, deﬁne In=KnandJn=Kn+1/2. Evaluate I0andJ0directly and hence prove that In=22n+1(n!)2R2n+1 (2n+1 ) !and Jn=π(2n+1 ) ! R2n+2 22n+1n!(n+1 ) !. (c) A sequence of functions Vn(R) is deﬁned by V0(R)=1 , Vn(R)= ZR −RVn−1 /√ R2−x2 / dx, n≥1. Prove by induction that V2n(R)=πnR2n n!,V 2n+1(R)=πn22n+1n!R2n+1 (2n+1 ) !. (d) For interest, (i) show that V2n+2(1)<V 2n(1) and V2n+1(1)<V 2n−1(1) for all n≥3; (ii) hence, by explicitly writing out Vk(R)f o r1≤k≤8 (say), show that the ‘volume’ of the totally symmetric solid of unit radius is a maximum in ﬁve dimensions. 6.6 Hints and answers 6.1 For integration in the order z,y,x the limits are (0 ,a−x),(−√ 4ax,√ 4ax),(0,a). For integration in the order y,x,z the limits are ( −√ 4ax,√ 4ax),(0,a−z),(0,a). V=1 6a3/15. 6.2 8 abc(a2+b2+c2)/3. 6.3 1 /360. 6.4 (a) Integrate by parts to obtain ( b/2)ln[1 + ( a/b)2]+atan−1(b/a); (b) 4[ a1/2+b1/2−(a+b)1/2]. 6.5 (a) Evaluate R 2b[1−(x/a)2]1/2dxby setting x=acosφ; (b)dV=π×a[1−(z/c)2]1/2×b[1−(z/c)2]1/2dz. 6.6 A=±(Z/a)3/2/√ 32π. 6.8 If one of the semicircles has radius a, Pappus’ second theorem shows that its centre of gravity /angbracketleftx/angbracketrightis 2a/πfrom the centre of the circle of which it is half. For the whole ﬁgure, /angbracketleftx/angbracketright=4a/(2π+4 ) . 6.9 (a) V=2πc×πa2andA=2πa×2πc. Setting ro=c+aandri=c−agives the stated results. (b) See hint for previous exercise. 6.10 (b) Evaluate R 2(a2−x2)1/2x2(M/πa2)dxby setting x=acosφ. 6.11 Transform to cylindrical polar coordinates.6.12 (a) Show that dz=2asinθ(cosθ−1)dθ.W r i t i n gc o s θascto save space, the integrand is 2 πρ 0a3(1−c2)(1−c)(2−c)2dcover the range −1≤c≤1. (b) The integrand is πρ0a5(1−c2)2(1−c)(2−c)4dc. 6.13 4 πa2,4πa3/3, a sphere. 6.14 The coordinate ranges are 0 ≤r≤√2a n d0≤φ≤2π,w i t h θ=π/4. The integrand for the randφintegrations is ( rcos2φ)/√2. 6.15 The volume element is ρd φd ρd z . The integrand for the ﬁnal z-integration is given by 2 π[(z2lnz)−(z2/2)];I=−5π/9. 6.16 Jacobian = ( u/v)1/2+(v/u)1/2;a r e ai n uv-plane is the triangle bounded by v=0 , u=v,u=a;i n t e g r a l= a2. 6.17 Set ξ=x/a,η=y/b,ζ=z/cto map the ellipsoid onto the unit sphere, and then change from ( ξ,η,ζ) coordinates to spherical polar coordinates; I=4πa3bc/15. 214 6.6 HINTS AND ANSWERS 6.18 The boundaries of the three-sided region are u=v=0,v=1a n d u=1 . I=(e−1)2/8. 6.19 Set u=s i n h xcosy,v=c o s h xsiny;Jxy,uv=( s i n h2x+cos2y)−1and the integrand reduces to 4 uvover the region 0 ≤u≤1, 0≤v≤1;I=1. 6.20 x=vcosα+u,y=vsinα.J a c o b i a n=s i n α. I=(α/2π) R exp(−r2)dAover all space. 6.21 Terms such as T∂2S/∂Y ∂X cancel in pairs. Use equations (6.17) and (6.16). 6.22 Note that uv=(ξ2−η2). The ranges for the new variables are 1 ≤ξ<∞, −1≤η≤1, 0≤φ≤2π. 6.23 (d)(ii) 2, π,4π/3,π2/2, 8π2/15,π3/6, 16π3/105,π4/24. 215 7 Vector algebra This chapter introduces space vectors and their manipulation. Firstly we deal with the description and algebra of vectors and then we consider how vectorsmay be used to describe lines and planes and ﬁnally we look at the practical useof vectors in ﬁnding distances. Much use of vectors will be made in subsequentchapters; this chapter gives only some basic rules. 7.1 Scalars and vectors The simplest kind of physical quantity is one that can be completely speciﬁed by its magnitude, a single number, together with the units in which it is measured.Such a quantity is called a scalar and examples include temperature, time and density. Avector is a quantity that requires both a magnitude ( ≥0) and a direction in space to specify it completely; we may think of it as an arrow in space. A familiarexample is force, which has a magnitude (strength) measured in newtons and adirection of application. The large number of vectors that are used to describethe physical world include velocity, displacement, momentum and electric ﬁeld.Vectors are also used to describe quantities such as angular momentum andsurface elements (a surface element has an area and a direction deﬁned by the normal to its tangent plane); in such cases their deﬁnitions may seem somewhat arbitrary (though in fact they are standard) and not as physically intuitive as forvectors such as force. A vector is denoted by bold type, the convention of thisbook, or by underlining, the latter being much used in handwritten work. This chapter considers basic vector algebra and illustrates just how powerful vector analysis can be. All the techniques are presented for three-dimensionalspace but most can be readily extended to more dimensions. Throughout the book we will represent vectors in diagrams as a line together with an arrowhead. We will make no distinction between an arrowhead at the 216 7.2 ADDITION AND SUBTRACTION OF VECTORS aa b b a+bb+a Figure 7.1 Addition of two vectors showing the commutation relation. We make no distinction between an arrowhead at the end of the line and one along the line’s length, but rather use that which gives the clearer diagram. end of the line or one along the line’s length but, rather, use that which gives the clearer diagram. Furthermore, even though we are considering three-dimensionalvectors, we have to draw them in the plane of the paper. It should not be assumedthat vectors drawn thus are coplanar, unless this is explicitly stated. 7.2 Addition and subtraction of vectors Theresultant orvector sum of two displacement vectors is the displacement vector that results from performing ﬁrst one and then the other displacement, as shownin ﬁgure 7.1; this process is known as vector addition. However, the principleof addition has physical meaning for vector quantities other than displacements; for example, if two forces act on the same body then the resultant force acting on the body is the vector sum of the two. The addition of vectors only makesphysical sense if they are of a like kind, for example if they are both forcesacting in three dimensions. It may be seen from ﬁgure 7.1 that vector addition iscommutative, i.e. a+b=b+a. (7.1) The generalisation of this procedure to the addition of three (or more) vectors is clear and leads to the associativity property of addition (see ﬁgure 7.2), e.g. a+(b+c)=(a+b)+c. (7.2) Thus, it is immaterial in what order any number of vectors are added. The subtraction of two vectors is very similar to their addition (see ﬁgure 7.3), that is, a−b=a+(−b) where−bis a vector of equal magnitude but exactly opposite direction to vector b. 217 VECTOR ALGEBRA aa ab bb ccc a+(b+c) (a+b)+cb+cb+c a+b a+b Figure 7.2 Addition of three vectors showing the associativity relation. −b ba aa−b Figure 7.3 Subtraction of two vectors. The subtraction of two equal vectors yields the zero vector, 0, which has zero magnitude and no associated direction. 7.3 Multiplication by a scalar Multiplication of a vector by a scalar (not to be confused with the ‘scalar product’, to be discussed in subsection 7.6.1) gives a vector in the same direction as the original but of a proportional magnitude. This can be seen in ﬁgure 7.4.The scalar may be positive, negative or zero. It can also be complex in someapplications. Clearly, when the scalar is negative we obtain a vector pointingin the opposite direction to the original vector. Multiplication by a scalar isassociative, commutative and distributiv e over addition. These properties may be summarised for arbitrary vectors aandband arbitrary scalars λandµby (λµ)a=λ(µa)=µ(λa), (7.3) λ(a+b)=λa+λb, (7.4) (λ+µ)a=λa+µa. (7.5) 218 7.3 MULTIPLICATION BY A SCALAR a aλ Figure 7.4 Scalar multiplication of a vector (for λ>1). OAB P ab pµ λ Figure 7.5 An illustration of the ratio theorem. The point Pdivides the line segment ABin the ratio λ:µ. Having deﬁned the operations of addition, subtraction and multiplication by a scalar, we can now use vectors to solve simple problems in geometry.IA point Pdivides a line segment ABin the ratio λ:µ(see ﬁgure 7.5). If the position vectors of the points AandBareaandbrespectively, ﬁnd the position vector of the point P. As is conventional for vector geometry problems, we denote the vector from the point A to the point BbyAB. If the position vectors of the points AandB, relative to some origin O,a r eaandb, it should be clear that AB=b−a. Now, from ﬁgure 7.5 we see that one possible way of reaching the point Pfrom Ois ﬁrst to go from OtoAand to go along the line ABfor a distance equal to the the fraction λ/(λ+µ) of its total length. We may express this in terms of vectors as OP=p=a+λ λ+µAB =a+λ λ+µ(b−a) = / 1−λ λ+µ / a+λ λ+µb =µ λ+µa+λ λ+µb, (7.6) which expresses the position vector of the point Pin terms of those of AandB.W ew o u l d , of course, obtain the same result by considering the path from OtoBand then to P. J 219 VECTOR ALGEBRA OA BC DE F G a bc Figure 7.6 The centroid of a triangle. The triangle is deﬁned by the points A, BandCthat have position vectors a,bandc. The broken lines CD,BE,AF connect the vertices of the triangle to the mid-points of the opposite sides;these lines intersect at the centroid Gof the triangle. The result (7.6) is a version of the ratio theorem and we may use it in solving more complicated problems.IThe vertices of triangle ABC have position vectors a,bandcrelative to some origin O (see ﬁgure 7.6). Find the position vector of the centroid Gof the triangle. From ﬁgure 7.6, the points DandEbisect the lines ABandACrespectively. Thus from the ratio theorem (7.6), with λ=µ=1/2, the position vectors of DandErelative to the origin are d=1 2a+1 2b, e=1 2a+1 2c. Using the ratio theorem again, we may write the position vector of a general point on the lineCDthat divides the line in the ratio λ:( 1−λ)a s r=( 1−λ)c+λd, =( 1−λ)c+1 2λ(a+b), (7.7) where we have expressed din terms of aandb. Similarly, the position vector of a general point on the line BEcan be expressed as r=( 1−µ)b+µe, =( 1−µ)b+1 2µ(a+c). (7.8) Thus, at the intersection of the lines CDandBEwe require, from (7.7), (7.8), (1−λ)c+1 2λ(a+b)=( 1−µ)b+1 2µ(a+c). By equating the coeﬃcents of the vectors a,b,cwe ﬁnd λ=µ,1 2λ=1−µ, 1−λ=1 2µ. 220 7.4 BASIS VECTORS AND COMPONENTS These equations are consistent and have the solution λ=µ=2/3. Substituting these values into either (7.7) or (7.8) we ﬁnd that the position vector of the centroid Gis given by g=1 3(a+b+c). J 7.4 Basis vectors and components Given any three diﬀerent vectors e1,e2ande3, which do not all lie in a plane, it is possible, in three-dimensional space, to write any other vector in terms ofscalar multiples of them: a=a 1e1+a2e2+a3e3. (7.9) The three vectors e1,e2ande3are said to form a basis(for the three-dimensional space); the scalars a1,a2anda3, which may be positive, negative or zero, are called the components of the vector awith respect to this basis. We say that the vector has been resolved into components. Most often we shall use basis vectors that are mutually perpendicular, for ease of manipulation, though this is not necessary. In general, a basis set must (i) have as many basis vectors as the number of dimensions (in more formal language, the basis vectors must span the space) and (ii) be such that no basis vector may be described as a sum of the others, or, more formally, the basis vectors must be linearly independent . Putting this mathematically, in Ndimensions, we require c1e1+c2e2+···+cNeN/negationslash=0, for any set of coeﬃcients c1,c2,...,c Nexcept c1=c2=···=cN=0 . In this chapter we will only consider vectors in three dimensions; higher dimen- sionality can be achieved by simple extension. If we wish to label points in space using a Cartesian coordinate system ( x, y, z), we may introduce the unit vectors i,jandk, which point along the positive x-, y-a n d z- axes respectively. A vector amay then be written as a sum of three vectors, each parallel to a diﬀerent coordinate axis: a=axi+ayj+azk. (7.10) A vector in three-dimensional space thus requires three components to describe fully both its direction and its magnitude. A displacement in space may be t h o u g h to fa st h es u mo fd i s p l a c e m e n t sa l o n gt h e x-,y-a n d z- directions (see ﬁgure 7.7). For brevity, the components of a vector awith respect to a particular coordinate system are sometimes written in the form ( ax,ay,az). Note that the 221 VECTOR ALGEBRA ijk axiayj azka Figure 7.7 A Cartesian basis set. The vector ais the sum of axi,ayjandazk. basis vectors i,jandkmay themselves be represented by (1 ,0,0), (0 ,1,0) and (0,0,1) respectively. We can consider the addition and subtraction of vectors in terms of their components. The sum of two vectors aandbis found by simply adding their components, i.e. a+b=axi+ayj+azk+bxi+byj+bzk =(ax+bx)i+(ay+by)j+(az+bz)k (7.11) and their diﬀerence by subtracting them, a−b=axi+ayj+azk−(bxi+byj+bzk) =(ax−bx)i+(ay−by)j+(az−bz)k. (7.12)ITwo particles have velocities v1=i+3j+6kandv2=i−2krespectively. Find the velocity uof the second particle relative to the ﬁrst. The required relative velocity is given by u=v2−v1=( 1−1)i+( 0−3)j+(−2−6)k =−3j−8k. J 7.5 Magnitude of a vector The magnitude of the vector ais denoted by |a|ora. In terms of its components in three-dimensional Cartesian coordinates, the magnitude of ais given by a≡|a|=radicalBig a2x+a2y+a2z. (7.13) Hence, the magnitude of a vector is a measure of its length. Such an analogy is useful for displacement vectors but magnitude is better described, for example, by 222 7.6 MULTIPLICATION OF VECTORS ‘strength’ for vectors such as force or by ‘speed’ for velocity vectors. For instance, in the previous example, the speed of the second particle relative to the ﬁrst isgiven by u=|u|=radicalbig (−3)2+(−8)2=√ 73. A vector whose magnitude equals unity is called a unit vector. The unit vector in the direction ais usually notated ˆaand may be evaluated as ˆa=a |a|. (7.14) The unit vector is a useful concept because a vector written as λˆathen has mag- nitude λand direction ˆa. Thus magnitude and direction are explicitly separated. 7.6 Multiplication of vectors We have already considered multiplying a vector by a scalar. Now we consider the concept of multiplying one vector by another vector. It is not immediatelyobvious what the product of two vectors represents and in fact two productsare commonly deﬁned, the scalar product and the vector product . As their names imply, the scalar product of two vectors is just a number, whereas the vectorproduct is itself a vector. Although neither the scalar nor the vector product is what we might normally think of as a product, their use is widespread and numerous examples will be described elsewhere in this book. 7.6.1 Scalar product The scalar product (or dot product) of two vectors aandbis denoted by a·b and is given by a·b≡|a||b|cosθ,0≤θ≤π, (7.15) where θis the angle between the two vectors, placed ‘tail to tail’ or ‘head to head’. Thus, the value of the scalar product a·bequals the magnitude of amultiplied by the projection of bontoa(see ﬁgure 7.8). From (7.15) we see that the scalar product has the particularly useful property that a·b= 0 (7.16) is a necessary and suﬃcient condition for ato be perpendicular to b(unless either of them is zero). It should be noted in particular that the Cartesian basis vectorsi,jandk, being mutually orthogonal unit vectors, satisfy the equations i·i=j·j=k·k=1, (7.17) i·j=j·k=k·i=0. (7.18) 223 VECTOR ALGEBRA ab Oθ bcosθ Figure 7.8 The projection of bonto the direction of aisbcosθ. The scalar product of aandbisabcosθ. Examples of scalar products arise naturally throughout physics and in partic- ular in connection with energy. Perhaps the simplest is the work done F·rin moving the point of application of a constant force Fthrough a displacement r; notice that, as expected, if the displacement is perpendicular to the direction of the force then F·r= 0 and no work is done. A second simple example is aﬀorded by the potential energy −m·Bof a magnetic dipole, represented in strength and orientation by a vector m, placed in an external magnetic ﬁeld B. As the name implies, the scalar product has a magnitude but no direction. The scalar product is commutative and distributive over addition: a·b=b·a (7.19) a·(b+c)=a·b+a·c. (7.20)IFour points A, B, C, D are positioned such that the line ADis perpendicular to BCand BDis perpendicular to AC. Show that CDis perpendicular to AB. Let us denote the position vectors of the points A, B, C, D bya,b,c,drespectively. As the four points are not coplanar it is diﬃcult to draw a helpful diagram of the situation,but this is not a drawback when vector methods are used. We start by noting that, sinceAD⊥BC, we have from (7.16) that (d−a)·(c−b)=0 . Similarly, since BD⊥AC, (d−b)·(c−a)=0 . Combining these two equations we ﬁnd (d−a)·(c−b)=(d−b)·(c−a), which, on mutliplying out the parentheses, gives d·c−a·c−d·b+a·b=d·c−b·c−d·a+b·a. Cancelling terms that appear on both sides and rearranging yields d·b−d·a−c·b+c·a=0, which simpliﬁes to give (d−c)·(b−a)=0 . From (7.16), we see that this implies that CDis perpendicular to AB.J 224 7.6 MULTIPLICATION OF VECTORS If we introduce a set of basis vectors that are mutually orthogonal, such as i,j, k, we can write the components of a vector a, with respect to that basis, in terms of the scalar product of awith each of the basis vectors, i.e. ax=a·i,ay=a·jand az=a·k. In terms of the components ax,ayandazthe scalar product is given by a·b=(axi+ayj+azk)·(bxi+byj+bzk)=axbx+ayby+azbz, (7.21) where the cross terms such as axi·byjare zero because the basis vectors are mutually perpendicular; see equation (7.18). It should be clear from (7.15) thatthe value of a·bhas a geometrical deﬁnition and that this value is independent of the actual basis vectors used.IFind the angle between the vectors a=i+2j+3kandb=2i+3j+4k. From (7.15) the cosine of the angle θbetween aandbis given by cosθ=a·b |a||b|. From (7.21) the scalar product a·bhas the value a·b=1×2+2×3+3×4=2 0 , and from (7.13)the lengths of the vectors are |a|= p 12+22+32=√ 14 and |b|= p 22+32+42=√ 29. Thus, cosθ=20√ 14√ 29≈0.9926⇒ θ=0.12 rad . J We can see from the expressions (7.15), (7.21) for the scalar product that if θ is the angle between aandbthen cosθ=ax abx b+ay aby b+az abz b where ax/a,ay/aandaz/aare called the direction cosines ofa, since they give the cosine of the angle made by awith each of the basis vectors. Similarly bx/b,by/b andbz/bare the direction cosines of b. If we take the scalar product of any vector awith itself then clearly θ=0a n d from (7.15) we have a·a=|a|2. Thus the magnitude of acan be written in a coordinate-independent form as |a|=√a·a. Finally, we note that the scalar product may be extended to vectors with complex components if it is redeﬁned as a·b=a∗ xbx+a∗ yby+a∗ zbz, where the asterisk represents the operation of complex conjugation. To accom- 225 VECTOR ALGEBRA θa×b ab Figure 7.9 The vector product. The vectors a,banda×bform a right-handed set. modate this extension the commutation property (7.19) must be modiﬁed to read a·b=(b·a)∗. (7.22) In particular it should be noted that ( λa)·b=λ∗a·b,whereas a·(λb)=λa·b. However, the magnitude of a complex vector is still given by |a|=√a·a,s i n c e a·ais always real. 7.6.2 Vector product The vector product (or cross product) of two vectors aandbis denoted by a×b and is deﬁned to be a vector of magnitude |a||b|sinθin a direction perpendicular to both aandb; |a×b|=|a||b|sinθ. The direction is found by ‘rotating’ aintobthrough the smallest possible angle. The sense of rotation is that of a right-handed screw which moves forward inthe direction a×b(see ﬁgure 7.9). Again, θis the angle between the two vectors placed ‘tail to tail’ or ‘head to head’. With this deﬁnition a,banda×bform a right-handed set. A more directly usable description of the relative directions ina vector product is provided by a right hand whose ﬁrst two ﬁngers and thumbare held to be as nearly mutually perpendicular as possible. If the ﬁrst ﬁnger is pointed in the direction of the ﬁrst vector and the second ﬁnger in the direction of the second vector, then the thumb gives the direction of the vector product. The vector product is distributive over addition, but anticommutative andnon- associative : (a+b)×c=(a×c)+(b×c), (7.23) b×a=−(a×b), (7.24) (a×b)×c/negationslash=a×(b×c). (7.25) 226 7.6 MULTIPLICATION OF VECTORS θ ORP F r Figure 7.10 The moment of the force Fabout Oisr×F. The cross represents the direction of r×F, which is perpendicularly into the plane of the paper. From its deﬁnition, we see that the vector product has the very useful property that if a×b=0thenais parallel or antiparallel to b(unless either of them is zero). We also note that a×a=0. (7.26)IShow that if a=b+λc,for some scalar λ,t h e n a×c=b×c. From (7.23) we have a×c=(b+λc)×c=b×c+λc×c. However, from (7.26), c×c=0and so a×c=b×c. (7.27) We note in passing that the fact that (7.27) is satisﬁed does notimply that a=b. J An example of the use of the vector product is that of ﬁnding the area, A,o f a parallelogram with sides aandb, using the formula A=|a×b|. (7.28) Another example is aﬀorded by considering a force Facting through a point R, whose vector position relative to the origin Oisr(see ﬁgure 7.10). Its moment ortorque about Ois the strength of the force times the perpendicular distance OP, which numerically is just Frsinθ, i.e. the magnitude of r×F.Furthermore, the sense of the moment is clockwise about an axis through Othat points perpendicularly into the plane of the paper (the axis is represented by a crossin the ﬁgure). Thus the moment is completely represented by the vector r×F, in both magnitude and spatial sense. It should be noted that the same vectorproduct is obtained wherever the point Ris chosen, so long as it lies on the line of action of F. Similarly, if a solid body is rotating about some axis that passes through the origin, with an angular velocity ωthen we can describe this rotation by a vector ωthat has magnitude ωand points along the axis of rotation. The direction of ω 227 VECTOR ALGEBRA is the forward direction of a right-handed screw rotating in the same sense as the body. The velocity of any point in the body with position vector ris then given byv=ω×r. Since the basis vectors i,j,kare mutually perpendicular unit vectors, forming a right-handed set, their vector products are easily seen to be i×i=j×j=k×k=0, (7.29) i×j=−j×i=k, (7.30) j×k=−k×j=i, (7.31) k×i=−i×k=j. (7.32) Using these relations, it is straightforward to show that the vector product of two general vectors aandbis given in terms of their components with respect to the basis set i,j,k,b y a×b=(aybz−azby)i+(azbx−axbz)j+(axby−aybx)k. (7.33) For the reader who is familiar with determinants (see chapter 8), we record that this can also be written as a×b=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleijk a xayaz bxbybzvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. That the cross product a×bis perpendicular to both aandbcan be veriﬁed in component form by forming its dot products with each of the two vectors and showing that it is zero in both cases.IFind the area Aof the parallelogram with sides a=i+2j+3kandb=4i+5j+6k. The vector product a×bis given in component form by a×b=( 2×6−3×5)i+( 3×4−1×6)j+( 1×5−2×4)k =−3i+6j−3k. Thus the area of the parallelogram is A=|a×b|= p (−3)2+62+(−3)2=√ 54. J 7.6.3 Scalar triple product Now that we have deﬁned the scalar and vector products, we can extend our discussion to deﬁne products of three vectors. Again, there are two possibilities,thescalar triple product and the vector triple product . 228 7.6 MULTIPLICATION OF VECTORS θOφP abcv Figure 7.11 The triple scalar product gives the volume of a parallelepiped. The scalar triple product is denoted by [a,b,c]≡a·(b×c) and, as its name suggests, it is just a number. It is most simply interpreted as the volume of a parallelepiped whose edges are given by a,bandc(see ﬁgure 7.11). The vector v=a×bis perpendicular to the base of the solid and has magnitude v=absinθ, i.e. the area of the base. Further, v·c=vccosφ. Thus, since ccosφ =OPis the vertical height of the parallelepiped, it is clear that ( a×b)·c=a r e a of the base ×perpendicular height = volume. It follows that, if the vectors a,b andcare coplanar, a·(b×c)=0 . Expressed in terms of the components of each vector with respect to the Cartesian basis set i,j,kthe scalar triple product is a·(b×c)=ax(bycz−bzcy)+ay(bzcx−bxcz)+az(bxcy−bycx), (7.34) which can also be written as a determinant: a·(b×c)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglea xayaz bxbybz cxcyczvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. By writing the vectors in component form, it can be shown that a·(b×c)=(a×b)·c, so that the dot and cross symbols can be interchanged without changing the result. More generally, the triple scalar product is unchanged under cyclic permutation of the vectors a,b,c. Other permutations simply give the negative of the original triple scalar product. These results can be summarised by [a,b,c]=[b,c,a]=[c,a,b]=−[a,c,b]=−[b,a,c]=−[c,b,a]. (7.35) 229 VECTOR ALGEBRAIFind the volume Vof the parallelepiped with sides a=i+2j+3k,b=4i+5j+6kand c=7i+8j+1 0k. We have already found that a×b=−3i+6j−3k, in subsection 7.6.2. Hence the volume of the parallelepiped is given by V=|a·(b×c)|=|(a×b)·c| =|(−3i+6j−3k)·(7i+8j+1 0k)| =|(−3)(7) + (6)(8) + ( −3)(10)|=3. J Another useful formula involving both the scalar and vector products is La- grange’s identity (see exercise 7.9), i.e. (a×b)·(c×d)≡(a·c)(b·d)−(a·d)(b·c). (7.36) 7.6.4 Vector triple product By the vector triple product of three vectors a,b,cwe mean the vector a×(b×c). Clearly, a×(b×c) is perpendicular to aand lies in the plane of bandcand so can be expressed in terms of them (see (7.37) below). We note, from (7.25), thatthe vector triple product is not associative, i.e. a×(b×c)/negationslash=(a×b)×c. Two useful formulae involving the vector triple product are a×(b×c)=(a·c)b−(a·b)c, (7.37) (a×b)×c=(a·c)b−(b·c)a, (7.38) which may be derived by writing each vector in component form (see exercise 7.8). It can also be shown that for any three vectors a,b,c, a×(b×c)+b×(c×a)+c×(a×b)=0. 7.7 Equations of lines, planes and spheres Now that we have described the basic algebra of vectors, we can apply the results to a variety of problems, the ﬁrst of which is to ﬁnd the equation of a line in vector form. 7.7.1 Equation of a line Consider the line passing through the ﬁxed point Awith position vector aand having a direction b(see ﬁgure 7.12). It is clear that the position vector rof a general point Ron the line can be written as r=a+λb, (7.39) 230 7.7 EQUATIONS OF LINES, PLANES AND SPHERES OAR ab r Figure 7.12 The equation of a line. The vector bis in the direction ARand λbis the vector from AtoR. since Rcan be reached by starting from O, going along the translation vector ato the point Aon the line and then adding some multiple λbof the vector b. Diﬀerent values of λgive diﬀerent points Ron the line. Taking the components of (7.39), we see that the equation of the line can also b ew r i t t e ni nt h ef o r m x−ax bx=y−ay by=z−az bz=c o n s t a n t . (7.40) Taking the vector product of (7.39) with band remembering that b×b=0,g i v e s an alternative equation for the line (r−a)×b=0. We may also ﬁnd the equation of the line that passes through two ﬁxed points AandCwith position vectors aandc.S i n c e ACis given by c−a, the position vector of a general point on the line is r=a+λ(c−a). 7.7.2 Equation of a plane The equation of a plane through a point Awith position vector aand perpendic- ular to a unit position vector ˆn(see ﬁgure 7.13) is (r−a)·ˆn= 0; (7.41) this follows since the vector joining Ato a general point Rwith position vector r isr−a;rwill lie in the plane if this vector is perpendicular to the normal to the plane. Rewriting (7.41) as r·ˆn=a·ˆn, we see that the equation of the plane may also be expressed in the form r·ˆn=d, or in component form as lx+my+nz=d, (7.42) 231 VECTOR ALGEBRA Od aˆn rAR Figure 7.13 The equation of the plane is ( r−a)·ˆn=0 . where the unit normal to the plane is ˆn=li+mj+nkandd=a·ˆnis the perpendicular distance of the plane from the origin. The equation of a plane containing points a,bandcis r=a+λ(b−a)+µ(c−a). This is apparent because starting from the point ain the plane, all other points may be reached by moving a distance along each of two (non-parallel) directionsin the plane. Two such directions are given by b−aandc−a.I tc a nb es h o w n that the equation of this plane may also be written in the more symmetrical form r=αa+βb+γc, where α+β+γ=1 .IFind the direction of the line of intersection of the planes x+3y−z=5 and 2x−2y+4z=3. The two planes have normal vectors n1=i+3j−kandn2=2i−2j+4k.I ti sc l e a r that these are not parallel vectors and so the planes must intersect along some line. Thedirection pof this line must be parallel to both planes and hence perpendicular to both normals. Therefore p=n 1×n2 = [(3)(4)−(−2)(−1)]i+[ (−1)(2)−(1)(4)] j+ [(1)(−2)−(3)(2)] k =1 0i−6j−8k. J 7.7.3 Equation of a sphere Clearly, the deﬁning property of a sphere is that all points on it are equidistant from a ﬁxed point in space and that the common distance is equal to the radius 232 7.8 USING VECTORS TO FIND DISTANCES of the sphere. This is easily expressed in vector notation as |r−c|2=(r−c)·(r−c)=a2, (7.43) where cis the position vector of the centre of the sphere and ais its radius.IFind the radius ρof the circle that is the intersection of the plane ˆn·r=pand the sphere of radius acentred on the point with position vector c. The equation of the sphere is |r−c|2=a2, (7.44) and that of the circle of intersection is |r−b|2=ρ2, (7.45) where ris restricted to lie in the plane and bis the position of the circle’s centre. Asblies on the plane whose normal is ˆn, the vector b−cmust be parallel to ˆn,i . e . b−c=λˆnfor some λ. Further, by Pythagoras, we must have ρ2+|b−c|2=a2. Thus λ2=a2−ρ2. Writing b=c+ p a2−ρ2ˆnand substituting in (7.45) gives r2−2r· / c+ p a2−ρ2ˆn / +c2+2 (c·ˆn) p a2−ρ2+a2−ρ2=ρ2, whilst, on expansion, (7.44) becomes r2−2r·c+c2=a2. Subtracting these last two equations, using ˆn·r=pand simplifying yields p−c·ˆn= p a2−ρ2. On rearrangement, this gives ρas p a2−(p−c·ˆn)2, which places obvious geometrical constraints on the values a,c,ˆnandpcan take if a real intersection between the sphere and the plane is to occur. J 7.8 Using vectors to ﬁnd distances This section deals with the practical application of vectors to ﬁnding distances. Some of these problems are extremely cumbersome in component form, but theyall reduce to neat solutions when general vectors, with no explicit basis set,are used. These examples show the power of vectors in simplifying geometricalproblems. 7.8.1 Distance from a point to a line Figure 7.14 shows a line having direction bthat passes through a point Awhose position vector is a. To ﬁnd the minimum distance dof the line from a point P whose position vector is p, we must solve the right-angled triangle shown. We see thatd=|p−a|sinθ; so, from the deﬁnition of the vector product, it follows that d=|(p−a)×ˆb|. 233 VECTOR ALGEBRA OAP θp−a d abp Figure 7.14 The minimum distance from a point to a line.IFind the minimum distance from the point Pwith coordinates (1,2,1)to the line r=a+λb, where a=i+j+kandb=2i−j+3k. Comparison with (7.39) shows that the line passes through the point (1 ,1,1) and has direction 2 i−j+3k. The unit vector in this direction is ˆb=1√ 14(2i−j+3k). The position vector of Pisp=i+2j+kand we ﬁnd (p−a)×ˆb=1√ 14[j×(2i−3j+3k)] =1√ 14(3i−2k). Thus the minimum distance from the line to the point Pisd= p 13/14. J 7.8.2 Distance from a point to a plane The minimum distance dfrom a point Pwhose position vector is pto the plane deﬁned by ( r−a)·ˆn= 0 may be deduced by ﬁnding any vector from Pto the plane and then determining its component in the normal direction. This is shownin ﬁgure 7.15. Consider the vector a−p, which is a particular vector from Pto the plane. Its component normal to the plane, and hence its distance from the plane, is given by d=(a−p)·ˆn, (7.46) where the sign of ddepends on which side of the plane Pis situated. 234 7.8 USING VECTORS TO FIND DISTANCES OP d apˆn Figure 7.15 The minimum distance dfrom a point to a plane.IFind the distance from the point Pwith coordinates (1,2,3)to the plane that contains the points A,BandChaving coordinates (0,1,0),(2,3,1)and(5,7,2). Let us denote the position vectors of the points A, B, C bya,b,c. Two vectors in the plane are b−a=2i+2j+k and c−a=5i+6j+2k, and hence a vector normal to the plane is n=( 2i+2j+k)×(5i+6j+2k)=−2i+j+2k, and its unit normal is ˆn=n |n|=1 3(−2i+j+2k). Denoting the position vector of Pbyp, the minimum distance from the plane to Pis given by d=(a−p)·ˆn =(−i−j−3k)·1 3(−2i+j+2k) =2 3−1 3−2=−5 3. If we take Pto be the origin O, then we ﬁnd d=1 3, i.e. a positive quantity. It follows from this that the original point Pwith coordinates (1 ,2,3), for which dwas negative, is on the opposite side of the plane from the origin. J 7.8.3 Distance from a line to a line Consider two lines in the directions aandb, as shown in ﬁgure 7.16. Since a×b is by deﬁnition perpendicular to both aandb, the unit vector normal to both these lines is ˆn=a×b |a×b|. 235 VECTOR ALGEBRA O Q Pˆnq ab p Figure 7.16 The minimum distance from one line to another. Ifpandqare the position vectors of any two points PandQon diﬀerent lines then the vector connecting them is p−q. Thus, the minimum distance dbetween the lines is this vector’s component along the unit normal, i.e. d=|(p−q)·ˆn|.IA line is inclined at equal angles to the x-,y- and z- axes and passes through the origin. Another line passes through the points (1,2,4)and(0,0,1). Find the minimum distance between the two lines. The ﬁrst line is given by r1=λ(i+j+k), and the second by r2=k+µ(i+2j+3k). Hence a vector normal to both lines is n=(i+j+k)×(i+2j+3k)=i−2j+k, and the unit normal is ˆn=1√ 6(i−2j+k). A vector between the two lines is, for example, the one connecting the points (0 ,0,0) and (0 ,0,1), which is simply k. Thus it follows that the minimum distance between the two lines is d=1√ 6|k·(i−2j+k)|=1√ 6. J 7.8.4 Distance from a line to a plane Let us consider the line r=a+λb. This line will intersect any plane to which it is not parallel. Thus, if a plane has a normal ˆnthen the minimum distance from 236 7.9 RECIPROCAL VECTORS the line to the plane is zero unless b·ˆn=0, in which case the distance, d, will be d=|(a−r)·ˆn|, where ris any point in the plane.IA line is given by r=a+λb,w h e r e a=i+2j+3kandb=4i+5j+6k.F i n dt h e coordinates of the point Pat which the line intersects the plane x+2y+3z=6. A vector normal to the plane is n=i+2j+3k, from which we ﬁnd that b·n/negationslash= 0. Thus the line does indeed intersect the plane. To ﬁnd the point of intersection we merely substitute the x-,y-a n d z- values of a general point on the line into the equation of the plane obtaining 1+4 λ+2 ( 2+5 λ)+3 ( 3+6 λ)=6⇒ 14 + 32 λ=6. This gives λ=−1 4, which we may substitute into the equation for the line to obtain x=1−1 4(4) = 0, y=2−1 4(5) =3 4andz=3−1 4(6) =3 2. Thus the point of intersection is (0,3 4,3 2). J 7.9 Reciprocal vectors The ﬁnal section of this chapter introduces the concept of reciprocal vectors, which have particular uses in crystallography. The two sets of vectors a,b,canda/prime,b/prime,c/primeare called reciprocal sets if a·a/prime=b·b/prime=c·c/prime= 1 (7.47) and a/prime·b=a/prime·c=b/prime·a=b/prime·c=c/prime·a=c/prime·b=0. (7.48) It can be veriﬁed (see exercise 7.19) that the reciprocal vectors of a,bandcare given by a/prime=b×c a·(b×c), (7.49) b/prime=c×a a·(b×c), (7.50) c/prime=a×b a·(b×c), (7.51) where a·(b×c)/negationslash= 0. In other words, reciprocal vectors only exist if a,bandcare 237 VECTOR ALGEBRA not coplanar. Moreover, if a,bandcare mutually orthogonal unit vectors then a/prime=a,b/prime=bandc/prime=c, so that the two systems of vectors are identical.IConstruct the reciprocal vectors of a=2i,b=j+k,c=i+k. First we evaluate the triple scalar product: a·(b×c)=2i·[(j+k)×(i+k)] =2i·(i+j−k)=2 . Now we ﬁnd the reciprocal vectors: a/prime=1 2(j+k)×(i+k)=1 2(i+j−k), b/prime=1 2(i+k)×2i=j, c/prime=1 2(2i)×(j+k)=−j+k. It is easily veriﬁed that these reciprocal vectors satisfy their deﬁning properties (7.47), (7.48). J We may also use the concept of reciprocal vectors to deﬁne the components of a vector awith respect to basis vectors e1,e2,e3that are not mutually orthogonal. If the basis vectors are of unit length and mutually orthogonal, such as the Cartesian basis vectors i,j,k, then (see the text preceeding (7.21)) acan be w r i t t e ni nt h ef o r m a=(a·i)i+(a·j)j+(a·k)k. If the basis is not orthonormal, however, then this is no longer true. Nevertheless, we may write the components of awith respect to a non-orthonormal basis e1,e2,e3in terms of its reciprocal basis vectors e/prime 1,e/prime 2,e/prime 3, which are deﬁned as in (7.49)–(7.51). If we let a=a1e1+a2e2+a3e3, then the scalar product a·e/prime 1is given by a·e/prime 1=a1e1·e/prime 1+a2e2·e/prime 1+a3e3·e/prime 1=a1, where we have used the relations (7.48). Similarly, a2=a·e/prime 2anda3=a·e/prime 3;s o now a=(a·e/prime 1)e1+(a·e/prime 2)e2+(a·e/prime 3)e3. (7.52) 7.10 Exercises 7.1 Which of the following statements about general vectors a,bandcare true? (a)c·(a×b)=(b×a)·c. (b)a×(b×c)=(a×b)×c. (c)a×(b×c)=(a·c)b−(a·b)c. (d)d=λa+µbimplies ( a×b)·d=0. (e)a×c=b×cimplies c·a−c·b=c|a−b|. (f) (a×b)×(c×b)=b[b·(c×a)]. 238 7.10 EXERCISES 7.2 A unit cell of diamond is a cube of side Awith carbon atoms at each corner, at the centre of each face and, in addition, displaced by1 4A(i+j+k)f r o me a c ho f the previously mentioned ones, where i,j,kare unit vectors along the cube axes. One corner of the cube is taken as the origin of coordinates. What are the vectorsjoining the atom at 1 4A(i+j+k) to its four nearest neighbours? Determine the angle between the carbon bonds in diamond. 7.3 Identify the following surfaces: (a)|r|=k;( b )r·u=l;( c )r·u=m|r|for−1≤m≤+1; (d)|r−(r·u)u|=n. Here k,l,mandnare ﬁxed scalars and uis a ﬁxed unit vector. 7.4 Find the angle between the position vectors to the points (3 ,−4,0) and (−2,1,0) and ﬁnd the direction cosines of a vector perpendicular to both. 7.5 A, B, C andDare the four corners, in order, of one face of a cube of side 2 units. The opposite face has corners E,F,G andH,w i t h AE, BF, CG andDHas parallel edges of the cube. The centre Oo ft h ec u b ei st a k e na st h eo r i g i na n dt h e x-,y-a n d z-axes are parallel to AD,AEandABrespectively. Find the following: (a) the angle between the face diagonal AFand the body diagonal AG; (b) the equation of the plane through Bthat is parallel to the plane CGE; (c) the perpendicular distance from the centre Jof the face BCGF to the plane OCG; (d) the volume of the tetrahedron JOCG . 7.6 Use vector methods to prove that the lines joining the mid-points of the opposite edges of a tetrahedron OABC meet at a point and that this point bisects each of the lines. 7.7 The edges OP,OQand ORof a tetrahedron OPQR are vectors p,qandr respectively, where p=2i+4j,q=2i−j+3kandr=4i−2j+5k. Show that OPis perpendicular to the plane containing OQR. Express the volume of the tetrahedron in terms of p,qandrand hence calculate the volume. 7.8 Prove, by writing it out in component form, that (a×b)×c=(a·c)b−(b·c)a, and deduce the result, stated in (7.25), that the operation of forming the vector product is non-associative. 7.9 Prove Lagrange’s identity, i.e. (a×b)·(c×d)=(a·c)(b·d)−(a·d)(b·c). 7.10 For four arbitrary vectors a,b,candd, evaluate (a×b)×(c×d) in two diﬀerent ways and so prove that a[b,c,d]−b[c,d,a]+c[d,a,b]−d[a,b,c]=0 . Show that this reduces to the normal Cartesian representation of the vector d, i.e.dxi+dyj+dzkifa,bandcare taken as i,jandk, the Cartesian base vectors. 7.11 Show that the points (1 ,0,1), (1 ,1,0) and (1 ,−3,4) lie on a straight line. Give the equation of the line in the form r=a+λb. 7.12 The plane P1contains the points A,Band C, which have position vectors a=−3i+2j,b=7i+2jandc=2i+3j+2krespectively. Plane P2passes through Aand is orthogonal to the line BC, whilst plane P3passes through Band is orthogonal to the line AC. Find the coordinates of r, the point of intersection of the three planes. 239 VECTOR ALGEBRA 7.13 Two planes have non-parallel unit normals ˆnandˆmand their closest distances from the origin are λandµrespectively. Find the vector equation of their line of intersection in the form r=νp+a. 7.14 Two ﬁxed points, AandB, in three-dimensional space have position vectors a andb. Identify the plane Pgiven by (a−b)·r=1 2(a2−b2), where aandbare the magnitudes of aandb. Show also that the equation (a−r)·(b−r)=0 describes a sphere Sof radius|a−b|/2. Deduce that the intersection of Pand Sis also the intersection of two spheres, centred on AandBand each of radius |a−b|/√2. 7.15 Let O,A,BandCbe four points with position vectors 0,a,bandc, and denote byg=λa+µb+νcthe position of the centre of the sphere on which they all lie. (a) Prove that λ,µandνsimultaneously satisfy (a·a)λ+(a·b)µ+(a·c)ν=1 2a2 and two other similar equations. (b) By making a change of origin, ﬁnd the centre and radius of the sphere on which the points p=3i+j−2k,q=4i+3j−3k,r=7i−3kands=6i+j−k all lie. 7.16 The vectors a,bandcare coplanar and related by λa+µb+νc=0, where λ,µ,νare not all zero. Show that the condition for the points with position vectors αa,βbandγcto be collinear is λ α+µ β+ν γ=0. 7.17 (a) Show that the line of intersection of the planes x+2y+3z=0a n d 3x+2y+z= 0 is equally inclined to the x-a n d z-a x e sa n dm a k e sa na n g l e cos−1(−2/√ 6) with the y-axis. (b) Find the perpendicular distance between one corner of a unit cube and the major diagonal not passing through it. 7.18 Four points Xi(i=1,2,3,4), taken for simplicity as all lying within the octant x, y, z≥0, have position vectors xi. Convince yourself that vector xnlies within the sector of space deﬁned by the other three vectors if max over i / min over j/negationslash=i /xi·xj |xi||xj| // =n, i.e. if nequals that value of ifor which the largest of the set of angles which xi makes with the other vectors is the lowest. Determine whether any of the four points with coordinates X1=( 3,2,2),X 2=( 2,3,1),X 3=( 2,1,3),X 4=( 3,0,3) lies within the tetrahedron deﬁned by the origin and the other three points. 7.19 The vectors a,bandcare not coplanar. The vectors a/prime,b/primeandc/primeare the 240 7.10 EXERCISES d aa b c Figure 7.17 A face-centred cubic crystal. associated reciprocal vectors. Verify that the expressions (7.49)–(7.51) deﬁne a set of reciprocal vectors a/prime,b/primeandc/primewith the following properties: (a)a/prime·a=b/prime·b=c/prime·c=1 ; (b)a/prime·b=a/prime·c=b/prime·aetc = 0; (c) [a/prime,b/prime,c/prime]=1 /[a,b,c]; (d)a=(b/prime×c/prime)/[a/prime,b/prime,c/prime]. 7.20 Three non-coplanar vectors a,bandc, have as their respective reciprocal vectors the set a/prime,b/primeandc/prime. Show that the normal to the plane containing the points k−1a,l−1bandm−1cis in the direction of the vector ka/prime+lb/prime+mc/prime. 7.21 In a crystal with a face-centred cubic structure, the basic cell can be taken as a cube of edge awith its centre at the origin of coordinates and its edges parallel to the Cartesian coordinate axes; atoms are sited at the eight corners and at thecentre of each face. However, other basic cells are possible. One is the rhomboidshown in ﬁgure 7.17, which has the three vectors b,canddas edges. (a) Show that the volume of the rhomboid is one-quarter that of the cube. (b) Show that the angles between pairs of edges of the rhomboid are 60 ◦and that the corresponding angles between pairs of edges of the rhomboid deﬁned bythe reciprocal vectors to b,c,dare each 109 .5 ◦. (This rhomboid can be used as the basic cell of a body-centred cubic structure, more easily visualised asa cube with an atom at each corner and one at its centre.) (c) In order to use the Bragg formula, 2 dsinθ=nλ, for the scattering of X-rays by a crystal, it is necessary to know the perpendicular distance dbetween successive planes of atoms; for a given crystal structure, dhas a particular value for each set of planes considered. For the face-centred cubic structureﬁnd the distance between successive planes with normals in the k,i+jand i+j+kdirections. 7.22 In subsection 7.6.2 we showed how the moment or torque of a force about an axis could be represented by a vector in the direction of the axis. The magnitude ofthe vector gives the size of the moment and the sign of the vector gives the sense.Similar representations can be used for angular velocities and angular momenta. (a) The magnitude of the angular momentum about the origin of a particle of mass mmoving with velocity von a path that is a perpendicular distance d 241 VECTOR ALGEBRA from the origin is given by m|v|d. Show that if ris the position of the particle then the vector J=r×mvrepresents the angular momentum. (b) Now consider a rigid collection of particles (or a solid body) rotating about an axis through the origin, the angular velocity of the collection beingrepresented by ω. (i) Show that the velocity of the ith particle is v i=ω×ri and that the total angular momentum Jis J= X imi[r2 iω−(ri·ω)ri]. (ii) Show further that the component of Jalong the axis of rotation can be written as Iω,w h e r e I, the moment of inertia of the collection about the axis or rotation, is given by I= X imiρ2 i. Interpret ρigeometrically. (iii) Prove that the total kinetic energy of the particles is1 2Iω2. 7.23 By proceeding as indicated below, prove the parallel axis theorem ,w h i c hs t a t e s that, for a body of mass M, the moment of inertia Iabout any axis is related to the corresponding moment of inertia I0about a parallel axis that passes through the centre of mass of the body by I=I0+Ma2 ⊥, where a⊥is the perpendicular distance between the two axes. Note that I0can be written asZ (ˆn×r)·(ˆn×r)dm, where ris the vector position, relative to the centre of mass, of the inﬁnitesimal mass dmandˆnis a unit vector in the direction of the axis of rotation. Write a similar expression for Iin which ris replaced by r/prime=r−a,w h e r e ais the vector position of any point on the axis to which Irefers. Use Lagrange’s identity and the fact that R rdm=0(by the deﬁnition of the centre of mass) to establish the result. 7.24 Without carrying out any further integration, use the results of the previous exercise, the worked example in subsection 6.3.4 and exercise 6.10 to prove thatthe moment of inertia of a uniform rectangular lamina, of mass Mand sides a andb, about an axis perpendicular to its plane and passing through the point (αa/2,βb /2), with−1≤α, β≤1,is M 12[a2(1 + 3 α2)+b2(1 + 3 β2)]. 7.25 Deﬁne a set of (non-orthogonal) base vectors a=j+k,b=i+kandc=i+j. (a) Establish their reciprocal vectors and hence express the vectors p=3i−2j+k, q=i+4jandr=−2i+j+kin terms of the base vectors a,bandc. (b) Verify that the scalar product p·qhas the same value, −5, when evaluated using either set of components. 242 7.10 EXERCISES f= 200 Hz ω= 400 πs−1V0cosωtV1V2 V3V4 R1=5 0ΩR2 I1I2 I3L C=1 0 µF Figure 7.18 An oscillatory electric c ircuit. The power supply has angular frequency ω=2πf= 400 πs−1. 7.26 Systems that can be modelled as damped harmonic oscillators are widespread; pendulum clocks, car shock absorbers, tuning circuits in television sets and radios,and collective electron motions in plasmas and metals are just a few examples. In all these cases, one or more variables describing the system obey(s) an equation of the form ¨x+2γ˙x+ω 2 0x=Pcosωt, where ˙x=dx/dt, etc. and the inclusion of the factor 2 is conventional. In the steady state (i.e. after the eﬀects of any in itial displacement or velocity have been damped out) the solution of the equation takes the form x(t)=Acos(ωt+φ). By expressing each term in the form Bcos(ωt+/epsilon1) and representing it by a vector of magnitude Bmaking an angle /epsilon1with the x-axis, draw a closed vector diagram, att= 0, say, that is equivalent to the equation. (a) Convince yourself that whatever the value of ω(>0)φmust be negative (−π<φ≤0) and that φ=t a n−1 /−2γω ω2 0−ω2 / . (b) Obtain an expression for Ain terms of P,ω0andω. 7.27 According to alternating current theory, the currents and voltages in the compo- nents of the circuit shown in ﬁgure 7.18 are determined by Kirchhoﬀ’s laws andthe relationships I 1=V1 R1,I 2=V2 R2,I 3=iω CV 3,V 4=iω LI 2. The factor i=√ −1 in the expression for I3indicates that the phase of I3is 90◦ ahead of V3. Similarly the phase of V4is 90◦ahead of I2. Measurement shows that V3has an amplitude of 0 .661V0and a phase of +13.4◦relative to that of the power supply. Taking V0= 1 V and using a series of vector plots for voltages and currents (they could all be on the same plot ifsuitable scales were chosen), determine all unknown currents and voltages andﬁnd values for the inductance of Land the resistance of R 2. (Scales of 1 cm = 0.1 V for voltages and 1 cm = 1 mA for currents are convenient.) 243 VECTOR ALGEBRA 7.11 Hints and answers 7.1 (c), (d) and (e). 7.2 In units of1 4Athe vectors are −i−j−k,i+j−k,i−j+k,−i+j+k; cos−1(−1 3) = 109 .5◦. 7.3 (a) A sphere of radius kcentred on the origin; (b) a plane with its normal in the direction of uand a distance lfrom the origin; (c) a cone with its axis parallel to uand semiangle cos−1m; (d) a circular cylinder of radius nwith its axis parallel tou. 7.4 cos−1(−2/√ 5) = 153 .4◦;0 ,0 ,1 . 7.5 (a) cos−1 p 2/3; (b) z−x=2 ;( c )1 /√2; (d)1 31 2(c×g)·j=1 3. 7.6 With an obvious notation, the mid-points of OAandBCarea/2a n d( b+c)/2; the mid-point of the line joining them is ( a+b+c)/2. The same result is obtained forOBandAC,a n df o r OCandAB. 7.7 Show that q×ris parallel to p; volume =1 3 /1 2(q×r)·p / =5 3. 7.9 Note that ( a×b)·(c×d)=d·[(a×b)×c] and use the result from the previous question. 7.10 Consider ( a×b)×[(c×d)] as λa+µband [( a×b)]×(c×d)a sλ/primec+µ/primedusing t h er e s u l to fe x e r c i s e7 . 8 . 7.11 Show that the position vectors of the points are linearly dependent; r=a+λb where a=i+kandb=−j+k. 7.12 The conditions are ( r−a)·[(b−a)×(c−a)] = 0, ( r−a)·(b−c)=0a n d (r−b)·(c−a) = 0; the point of intersection is r=2i+7j+1 0k. 7.13 Show that pmust have the direction ˆn×ˆmand write aasxˆn+yˆm. By obtaining a pair of simultaneous equations for xandy, prove that x=(λ−µˆn·ˆm)/[1−(ˆn·ˆm)2] and that y=(µ−λˆn·ˆm)/[1−(ˆn·ˆm)2]. 7.14 Pis the plane orthogonal to the line joining AandBand equidistant from them. Sis|r−c|2=(|a−b|/2)2,w h e r e c=(a+b)/2. Add and subtract the equations forPandSand arrange the resulting equations in the form |r−d|2=R2. 7.15 (a) Note that |a−g|2=R2=|0−g|2, leading to a·a=2a·g. (b) Make pthe new origin and solve the three simultaneous linear equations to obtain λ=5/18,µ=1 0 /18,ν=−3/18, giving g=2i−kand a sphere of radius√5 centred on (5 ,1,−3). 7.16 For collinearity, γc=θαa+( 1−θ)βbfor some θ. 7.17 (a) Find two points on both planes, say (0 ,0,0) and (1 ,−2,1), and hence determine the direction cosines of the line of intersection; (b) (2 3)1/2. 7.18 The scalar products sij(i=1,2,3;j>i) between pairs of unit vectors are 0.907, 0.907, 0.857; 0.714, 0.567; 0.945. Thus i= 1 has the highest minimum ( s14=0.857) and so only X1could meet the condition. The plane containing X2,X3andX4 isx+y+z−6=0 ; s i n c e3+2+2 −6>0,X1lies outside the tetrahedron OX2X3X4. None of the points meets the condition. 7.19 For (c) and (d), use the result of exercise 7.8 to evaluate ( c×a)×(a×b). 7.20 The normal is in the direction ( l−1b−k−1a)×(m−1c−k−1a). 7.21 (b) b/prime=a−1(−i+j+k),c/prime=a−1(i−j+k),d/prime=a−1(i+j−k); (c) a/2 for direction k; successive planes through (0 ,0,0) and ( a/2,0,0) give a spacing of a/√ 8f o r direction i+j; successive planes through ( −a/2,0,0) and ( a/2,0,0) give a spacing ofa/√ 3 for direction i+j+k. 7.22 (a) Check both magnitude and rotational sense. (b)(i) Use the result of exercise 7.8 to evaluate ri×mi(ω×ri). (ii) Form ( J·ω)/ω;ρiis the distance of the ith particle from the axis of rotation. (iii) use Lagrange’s identity to evaluate (ω×ri)·(ω×ri). 7.23 Note that a2−(ˆn·a)2=a2 ⊥. 7.24 The moment of inertia about an axis through the centre of the rectangle and perpendicular to its plane is1 12M(a2+b2). 244 7.11 HINTS AND ANSWERS 7.26φ2φ12γωA 2γωAω2A ω2Aω2 0A ω2 0AP Figure 7.19 The vector diagram for the equation in exercise 7.26. 7.25 p=−2a+3b,q=3 2a−3 2b+5 2candr=2a−b−c. Remember that a·a=b·b= c·c=2a n d a·b=a·c=b·c=1 . See ﬁgure 7.19 and recall that −cosθ=c o s ( θ+π)a n d−sinθ=c o s ( θ+π/2). (a) With φ1>0, no matter what value ωtakes, the possible resultants (broken arrows) can never equal P.W i t h φ2<0, closure of the quadrilateral is possible. (b)A=P[(ω2 0−ω2)2+4γ2ω2]−1/2. 7.27 With currents in units of mA/ |V0|. Voltages in units of V0: I1=( 7.76,−23.2◦),I2=( 1 4 .36,−50.8◦),I3=( 8.30,103.4◦); V1=( 0.388,−23.2◦),V2=( 0.287,−50.8◦),V4=( 0.596,39.2◦); L= 33 mH, R2=2 0Ω . 245 8 Matrices and vector spaces In the previous chapter we deﬁned a vector as a geometrical object which has both a magnitude and a direction and which may be thought of as an arrow ﬁxedin our familiar three-dimensional space, a space which, if we need to, we deﬁneby reference to, say, the ﬁxed stars. This geometrical deﬁnition of a vector is bothuseful and important since it is independent of any coordinate system with which we choose to label points in space. In most speciﬁc applications, however, it is necessary at some stage to choose a coordinate system and to break down a vector into its component vectors in the directions of increasing coordinate values. Thus for a particular Cartesiancoordinate system (for example) the component vectors of a vector awill be a xi, ayjandazkand the complete vector will be a=axi+ayj+azk. (8.1) Although we have so far considered only real three-dimensional space, we may extend our notion of a vector to more abstract spaces, which in general canhave an arbitrary number of dimensions N. We may still think of such a vector as an ‘arrow’ in this abstract space, so that it is again independent of any ( N- dimensional) coordinate system with which we choose to label the space. As an example of such a space, which, though abstract, has very practical applications, we may consider the description of a mechanical or electrical system. If the stateof a system is uniquely speciﬁed by assigning values to a set of Nvariables, which may be angles or currents, for example, then that state can be representedby a vector in an N-dimensional space, the vector having those values as its components. In this chapter we ﬁrst discuss general vector spaces and their properties. We then go on to discuss the transformation of one vector into another by a linear operator. This leads naturally to the concept of a matrix , a two-dimensional array of numbers. The properties of matrices are then discussed and we conclude with 246 8.1 VECTOR SPACES a discussion of how to use these properties to solve systems of linear equations. The application of matrices to the study of oscillations in physical systems ist a k e nu pi nc h a p t e r9 . 8.1 Vector spaces A set of objects (vectors) a,b,c,...is said to form a linear vector space Vif: (i) the set is closed under commutative and associative addition, so that a+b=b+a, (8.2) (a+b)+c=a+(b+c); (8.3) (ii) the set is closed under multiplication by a scalar (any complex number) to form a new vector λa, the operation being both distributive and associative so that λ(a+b)=λa+λb, (8.4) (λ+µ)a=λa+µa, (8.5) λ(µa)=(λµ)a, (8.6) where λandµare arbitrary scalars; (iii) there exists a null vector 0such that a+0=afor all a; (iv) multiplication by unity leaves any vector unchanged, i.e. 1 ×a=a; (v) all vectors have a corresponding negative vector −asuch that a+(−a)=0. It follows from (8.5) with λ=1a n d µ=−1t h a t−ais the same vector as (−1)×a. We note that if we restrict all scalars to be real then we obtain a real vector space (an example of which is our familiar three-dimensional space); otherwise, in general, we obtain a complex vector space . We note that it is common to use the terms ‘vector space’ and ‘space’, instead of the more formal ‘linear vector space’. Thespanof a set of vectors a,b,...,sis deﬁned as the set of all vectors that may be written as a linear sum of the original set, i.e. all vectors x=αa+βb+···+σs (8.7) that result from the inﬁnite number of possible values of the (in general complex) scalars α ,β,...,σ .I fxin (8.7) is equal to 0for some choice of α ,β,...,σ (notall zero), i.e. if αa+βb+···+σs=0, (8.8) then the set of vectors a,b,...,s,i ss a i dt ob e linearly dependent .I ns u c has e t at least one vector is redundant, since it can be expressed as a linear sum of 247 MATRICES AND VECTOR SPACES the others. If, however, (8.8) is not satisﬁed by anyset of coeﬃcients (other than the trivial case in which all the coeﬃcients are zero) then the vectors are linearly independent , and no vector in the set can be expressed as a linear sum of the others. If, in a given vector space, there exist sets of Nlinearly independent vectors, but no set of N+ 1 linearly independent vectors, then the vector space is said to beN-dimensional. (In this chapter we will limit our discussion to vector spaces of ﬁnite dimensionality; spaces of inﬁnite dimensionality are discussed in chapter 17.) 8.1.1 Basis vectors IfVis an N-dimensional vector space then anyset of Nlinearly independent vectors e1,e2,...,eNforms a basisforV.I fxis an arbitrary vector lying in Vthen the set of N+ 1 vectors x,e1,e2,...,eN, must be linearly dependent and therefore such that αe1+βe2+···+σeN+χx=0, (8.9) where the coeﬃcients α ,β,...,χ are not all equal to 0, and in particular χ/negationslash=0 . Rearranging (8.9) we may write xas a linear sum of the vectors eias follows: x=x1e1+x2e2+···+xNeN=Nsummationdisplay i=1xiei, (8.10) for some set of coeﬃcients xithat are simply related to the original coeﬃcients, e.g.x1=−α/χ,x2=−β/χ,e t c .S i n c ea n y xlying in the span of Vcan be expressed in terms of the basis orbase vectors ei, the latter are said to form acomplete set. The coeﬃcients xiare the components ofxwith respect to the ei-basis. These components are unique , since if both x=Nsummationdisplay i=1xieiand x=Nsummationdisplay i=1yiei, then Nsummationdisplay i=1(xi−yi)ei=0, (8.11) which, since the eiare linearly independent, has only the solution xi=yifor all i=1,2,...,N . From the above discussion we see that anyset of Nlinearly independent vectors can form a basis for an N-dimensional space. If we choose a diﬀerent set e/prime i,i=1,...,N then we can write xas x=x/prime 1e/prime1+x/prime 2e/prime2+···+x/prime Ne/primeN=Nsummationdisplay i=1x/prime ie/primei. (8.12) 248 8.1 VECTOR SPACES We reiterate that the vector x(a geometrical entity) is independent of the basis – it is only the components of xthat depend on the basis. We note, however, that given a set of vectors u1,u2,...,uM,w h e r e M/negationslash=N,i na n N-dimensional vector space, then either there exists a vector that cannot be expressed as a linear combination of the uior, for some vector that can be so expressed, the components are not unique. 8.1.2 The inner product We may usefully add to the description of vectors in a vector space by deﬁning theinner product of two vectors, denoted in general by /angbracketlefta|b/angbracketright, which is a scalar function of aandb. The scalar or dot product, a·b≡|a||b|cosθ,o fv e c t o r s in real three-dimensional space (where θis the angle between the vectors), was introduced in the last chapter and is an example of an inner product. In eﬀect thenotion of an inner product /angbracketlefta|b/angbracketrightis a generalisation of the dot product to more abstract vector spaces. Alternative notations for /angbracketlefta|b/angbracketrightare (a,b), or simply a·b. The inner product has the following properties: (i)/angbracketlefta|b/angbracketright=/angbracketleftb|a/angbracketright ∗, (ii)/angbracketlefta|λb+µc/angbracketright=λ/angbracketlefta|b/angbracketright+µ/angbracketlefta|c/angbracketright. We note that in general, for a complex vector space, (i) and (ii) imply that /angbracketleftλa+µb|c/angbracketright=λ∗/angbracketlefta|c/angbracketright+µ∗/angbracketleftb|c/angbracketright, (8.13) /angbracketleftλa|µb/angbracketright=λ∗µ/angbracketlefta|b/angbracketright. (8.14) Following the analogy with the dot product in three-dimensional real space, two vectors in a general vector space are deﬁned to be orthogonal if/angbracketlefta|b/angbracketright=0 . Similarly, the norm of a vector ais given by /bardbla/bardbl=/angbracketlefta|a/angbracketright1/2and is clearly a generalisation of the length or modulus |a|of a vector ain three-dimensional space. In a general vector space /angbracketlefta|a/angbracketrightcan be positive or negative; however, we shall be primarily concerned with spaces in which /angbracketlefta|a/angbracketright≥0 and which are thus said to have a positive semi-deﬁnite norm .I ns u c has p a c e /angbracketlefta|a/angbracketright= 0 implies a=0. Let us now introduce into our N-dimensional vector space a basis ˆe1,ˆe2,...,ˆeN that has the desirable property of being orthonormal (the basis vectors are mutually orthogonal and each has unit norm), i.e. a basis that has the property /angbracketleftˆei|ˆej/angbracketright=δij. (8.15) Here δijis the Kronecker delta symbol (of which we say more in chapter 21) and has the properties δij=braceleftBigg 1f o r i=j, 0f o r i/negationslash=j. 249 MATRICES AND VECTOR SPACES In the above basis we may express any two vectors aandbas a=Nsummationdisplay i=1aiˆeiand b=Nsummationdisplay i=1biˆei. Furthermore, in such an orthonormal basis we have, for any a, /angbracketleftˆej|a/angbracketright=Nsummationdisplay i=1/angbracketleftˆej|aiˆei/angbracketright=Nsummationdisplay i=1ai/angbracketleftˆej|ˆei/angbracketright=aj. (8.16) Thus the components of aare given by ai=/angbracketleftˆei|a/angbracketright. Note that this is nottrue unless the basis is orthonormal. We can write the inner product of aandbin terms of their components in an orthonormal basis as /angbracketlefta|b/angbracketright=/angbracketlefta1ˆe1+a2ˆe2+···+aNˆeN|b1ˆe1+b2ˆe2+···+bNˆeN/angbracketright =Nsummationdisplay i=1a∗ ibi/angbracketleftˆei|ˆei/angbracketright+Nsummationdisplay i=1Nsummationdisplay j/negationslash=ia∗ ibj/angbracketleftˆei|ˆej/angbracketright =Nsummationdisplay i=1a∗ ibi, where the second equality follows from (8.14) and the third from (8.15). This is clearly a generalisation of the expression (7.21) for the dot product of vectors inthree-dimensional space. We may generalise the above to the case where the base vectors e 1,e2,...,eN arenotorthonormal (or orthogonal). In general we can deﬁne the N2numbers Gij=/angbracketleftei|ej/angbracketright. (8.17) Then, if a=summationtextN i=1aieiandb=summationtextN i=1biei, the inner product of aandbis given by /angbracketlefta|b/angbracketright=angbracketleftBiggNsummationdisplay i=1aieivextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleNsummationdisplay j=1bjejangbracketrightBigg =Nsummationdisplay i=1Nsummationdisplay j=1a∗ ibj/angbracketleftei|ej/angbracketright =Nsummationdisplay i=1Nsummationdisplay j=1a∗ iGijbj. (8.18) We further note that from (8.17) and the properties of the inner product we require Gij=G∗ ji. This in turn ensures that /bardbla/bardbl=/angbracketlefta|a/angbracketrightis real, since then /angbracketlefta|a/angbracketright∗=Nsummationdisplay i=1Nsummationdisplay j=1aiG∗ ija∗j=Nsummationdisplay j=1Nsummationdisplay i=1a∗ jGjiai=/angbracketlefta|a/angbracketright. 250 8.1 VECTOR SPACES 8.1.3 Some useful inequalities For a set of objects (vectors) forming a linear vector space in which /angbracketlefta|a/angbracketright≥0f o r alla, the following inequalities are often useful. (i)Schwarz’s inequality is the most basic result and states that |/angbracketlefta|b/angbracketright|≤/bardbla/bardbl/bardblb/bardbl, (8.19) where the equality holds when ais a scalar multiple of b,i . e .w h e n a=λb. It is important here to distinguish between the absolute value of a scalar, |λ|,a n dt h e normof a vector, /bardbla/bardbl. Schwarz’s inequality may be proved by considering /bardbla+λb/bardbl2=/angbracketlefta+λb|a+λb/angbracketright =/angbracketlefta|a/angbracketright+λ/angbracketlefta|b/angbracketright+λ∗/angbracketleftb|a/angbracketright+λλ∗/angbracketleftb|b/angbracketright. If we write /angbracketlefta|b/angbracketrightas|/angbracketlefta|b/angbracketright|eiαthen /bardbla+λb/bardbl2=/bardbla/bardbl2+|λ|2/bardblb/bardbl2+λ|/angbracketlefta|b/angbracketright|eiα+λ∗|/angbracketlefta|b/angbracketright|e−iα. However,/bardbla+λb/bardbl2≥0f o ra l l λ,s ow em a yc h o o s e λ=re−iαand require that, for all r, 0≤/bardbla+λb/bardbl2=/bardbla/bardbl2+r2/bardblb/bardbl2+2r|/angbracketlefta|b/angbracketright|. This means that the quadratic equation in rformed by setting the RHS equal to zero must have no real roots. This, in turn, implies that 4|/angbracketlefta|b/angbracketright|2≤4/bardbla/bardbl2/bardblb/bardbl2, which, on taking the square root (all factors are necessarily positive) of both sides, gives Schwarz’s inequality. (ii) The triangle inequality states that /bardbla+b/bardbl≤/bardbla/bardbl+/bardblb/bardbl (8.20) and may be derived from the properties of the inner product and Schwarz’s inequality as follows. Let us ﬁrst consider /bardbla+b/bardbl2=/bardbla/bardbl2+/bardblb/bardbl2+2R e/angbracketlefta|b/angbracketright≤/bardbla/bardbl2+/bardblb/bardbl2+2|/angbracketlefta|b/angbracketright|. Using Schwarz’s inequality we then have /bardbla+b/bardbl2≤/bardbla/bardbl2+/bardblb/bardbl2+2/bardbla/bardbl/bardblb/bardbl=(/bardbla/bardbl+/bardblb/bardbl)2, which, on taking the square root, gives the triangle inequality (8.20). (iii)Bessel’s inequality requires the introduction of an orthonormal basis ˆei, i=1,2,...,N into the N-dimensional vector space; it states that /bardbla/bardbl2≥summationdisplay i|/angbracketleftˆei|a/angbracketright|2, (8.21) 251 MATRICES AND VECTOR SPACES where the equality holds if the sum includes all Nbasis vectors. If not all the basis vectors are included in the sum then the inequality results(though of course the equality remains if those basis vectors omitted allhave a i= 0). Bessel’s inequality can also be written /angbracketlefta|a/angbracketright≥summationdisplay i|ai|2, where the aiare the components of ain the orthonormal basis. From (8.16) these are given by ai=/angbracketleftˆei|a/angbracketright. The above may be proved by considering vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglea−summationdisplay i/angbracketleftˆei|a/angbracketrightˆeivextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle2 =angbracketleftBig a−summationdisplay i/angbracketleftˆei|a/angbracketrightˆeivextendsinglevextendsinglevextendsinglea−summationdisplay j/angbracketleftˆej|a/angbracketrightˆejangbracketrightBig . Expanding out the inner product and using /angbracketleftˆei|a/angbracketright∗=/angbracketlefta|ˆei/angbracketright,w eo b t a i n vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglea−summationdisplay i/angbracketleftˆei|a/angbracketrightˆeivextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle2 =/angbracketlefta|a/angbracketright−2summationdisplay i/angbracketlefta|ˆei/angbracketright/angbracketleftˆei|a/angbracketright+summationdisplay isummationdisplay j/angbracketlefta|ˆei/angbracketright/angbracketleftˆej|a/angbracketright/angbracketleftˆei|ˆej/angbracketright. Now/angbracketleftˆei|ˆej/angbracketright=δij, since the basis is orthonormal, and so we ﬁnd 0≤vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglea−summationdisplay i/angbracketleftˆei|a/angbracketrightˆeivextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle2 =/bardbla/bardbl2−summationdisplay i|/angbracketleftˆei|a/angbracketright|2, which is Bessel’s inequality. We take this opportunity to mention also (iv) the parallelogram equality /bardbla+b/bardbl2+/bardbla−b/bardbl2=2parenleftbig /bardbla/bardbl2+/bardblb/bardbl2parenrightbig , (8.22) which may be proved straightforwardly from the properties of the inner product. 8.2 Linear operators We now discuss the action of linear operators on vectors in a vector space. A linear operator Aassociates with every vector xanother vector y=Ax, in such a way that, for two vectors aandb, A(λa+µb)=λAa+µAb, where λ,µare scalars. We say that A‘operates’ on xto give the vector y.W e note that the action of Aisindependent of any basis or coordinate system and 252 8.2 LINEAR OPERATORS may be thought of as ‘transforming’ one geometrical entity (i.e. a vector) into another. If we now introduce a basis ei,i=1,2,...,N , into our vector space then the action of Aon each of the basis vectors is to produce a linear combination of the latter; this may be written as Aej=Nsummationdisplay i=1Aijei, (8.23) where Aijis the ith component of the vector Aejin this basis; collectively the numbers Aijare called the components of the linear operator in the ei-basis. In this basis we can express the relation y=Axin component form as y=Nsummationdisplay i=1yiei=A Nsummationdisplay j=1xjej =Nsummationdisplay j=1xjNsummationdisplay i=1Aijei, and hence, in purely component form, in this basis we have yi=Nsummationdisplay j=1Aijxj. (8.24) If we had chosen a diﬀerent basis e/prime i, in which the components of x,yandA arex/prime i,y/prime iandA/prime ijrespectively then the geometrical relationship y=Axwould be represented in this new basis by y/prime i=Nsummationdisplay j=1A/prime ijx/primej. We have so far assumed that the vector yis in the same vector space as x.I f ,h o w e v e r , ybelongs to a diﬀerent vector space, which may in general be M-dimensional ( M/negationslash=N) then the above analysis needs a slight modiﬁcation. By introducing a basis set fi,i=1,2,...,M , into the vector space to which ybelongs we may generalise (8.23) as Aej=Msummationdisplay i=1Aijfi, where the components Aijof the linear operator Arelate to both of the bases ej andfi. 253 MATRICES AND VECTOR SPACES 8.2.1 Properties of linear operators Ifxis a vector and AandBare two linear operators then it follows that (A+B)x=Ax+Bx, (λA)x=λ(Ax), (AB)x=A(Bx), where in the last equality we see that the action of two linear operators in succession is associative. The product of two linear operators is not in generalcommutative, however, so that in general ABx/negationslash=BAx. In an obvious way we deﬁne the null (or zero) and identity operators by Ox=0and Ix=x, for any vector xin our vector space. Two operators AandBare equal if Ax=Bxfor all vectors x. Finally, if there exists an operator A −1such that AA−1=A−1A=I thenA−1is the inverse ofA. Some linear operators do not possess an inverse and are called singular , whilst those operators that do have an inverse are termed non-singular . 8.3 Matrices We have seen that in a particular basis eiboth vectors and linear operators can be described in terms of their components with respect to the basis. Thesecomponents may be displayed as an array of numbers called a matrix .I ng e n e r a l , if a linear operator Atransforms vectors from an N-dimensional vector space, for which we choose a basis e j,j=1,2,...,N , into vectors belonging to an M-dimensional vector space, with basis fi,i=1,2,...,M , then we may represent the operator Aby the matrix A= A11 A12... A 1N A21 A22... A 2N ............ A M1AM2... A MN . (8.25) Thematrix elements Aijare the components of the linear operator with respect to the bases ejandfi; the component Aijof the linear operator appears in the ith row and jth column of the matrix. The array has Mrows and Ncolumns a n di st h u sc a l l e da n M×Nmatrix. If the dimensions of the two vector spaces are the same, i.e. M=N(for example, if they are the same vector space) then we may represent Aby an N×Norsquare matrix of order N. The component Aij, which in general may be complex, is also denoted by ( A)ij. 254 8.4 BASIC MATRIX ALGEBRA In a similar way we may denote a vector xin terms of its components xiin a basisei,i=1,2,...,N , by the array x= x1 x2 ... xN , which is a special case of (8.25) and is called a column matrix (or conventionally, and slightly confusingly, a column vector or even just a vector – strictly speaking the term ‘vector’ refers to the geometrical entity x). The column matrix xcan also be written as x=(x1x2··· xN)T, which is the transpose of arow matrix (see section 8.6). We note that in a diﬀerent basis e/prime ithe vector xwould be represented by a diﬀerent column matrix containing the components x/prime iin the new basis, i.e. x/prime= x/prime 1 x/prime2 ... x/prime N . Thus, we use xand x/primeto denote diﬀerent column matrices which, in diﬀerent bases eiande/prime i, represent the samevector x.I nm a n yt e x t s ,h o w e v e r ,t h i sd i s t i n c t i o ni s not made and x(rather than x) is equated to the corresponding column matrix; if we regard xas the geometrical entity, however, this can be misleading and so we explicitly make the distinction. A similar argument follows for linear operators;the same linear operator Ais described in diﬀerent bases by diﬀerent matrices A and A /prime, containing diﬀerent matrix elements. 8.4 Basic matrix algebra The basic algebra of matrices may be deduced from the properties of the linear operators that they represent. In a given basis the action of two linear operatorsAandBon an arbitrary vector x(see the beginning of subsection 8.2.1), when written in terms of components using (8.24), is given by summationdisplay j(A+B)ijxj=summationdisplay jAijxj+summationdisplay jBijxj, summationdisplay j(λA)ijxj=λsummationdisplay jAijxj, summationdisplay j(AB)ijxj=summationdisplay kAik(Bx)k=summationdisplay jsummationdisplay kAikBkjxj. 255 MATRICES AND VECTOR SPACES Now, since xis arbitrary, we can immediately deduce the way in which matrices are added or multiplied, i.e. (A+B)ij=Aij+Bij, (8.26) (λA)ij=λAij, (8.27) (AB)ij=summationdisplay kAikBkj. (8.28) We note that a matrix element may, in general, be complex. We now discuss matrix addition and multiplication in more detail. 8.4.1 Matrix addition and multiplication by a scalar From (8.26) we see that the sum of two matrices, S=A+B, is the matrix whose elements are given by Sij=Aij+Bij for every pair of subscripts i, j, with i=1,2,...,M and j=1,2,...,N .F o r example, if Aand Bare 2×3 matrices then S=A+Bis given by parenleftbiggS11S12S13 S21S22S23parenrightbigg =parenleftbiggA11A12A13 A21A22A23parenrightbigg +parenleftbiggB11B12B13 B21B22B23parenrightbigg =parenleftbiggA11+B11A12+B12A13+B13 A21+B21A22+B22A23+B23parenrightbigg . (8.29) Clearly, for the sum of two matrices to have any meaning, the matrices must have the same dimensions, i.e. both be M×Nmatrices. From deﬁnition (8.29) it follows that A+B=B+Aand that the sum of a number of matrices can be written unambiguously without bracketting, i.e. matrixaddition is commutative andassociative . The diﬀerence of two matrices is deﬁned by direct analogy with addition. The matrix D=A−Bhas elements D ij=Aij−Bij,fori=1,2,...,M ,j=1,2,...,N . (8.30) From (8.27) the product of a matrix Awith a scalar λis the matrix with elements λAij, for example λparenleftbiggA11A12A13 A21A22A23parenrightbigg =parenleftbiggλA11λA12λA13 λA21λA22λA23parenrightbigg . (8.31) Multiplication by a scalar is distributive and associative. 256 8.4 BASIC MATRIX ALGEBRAIThe matrices A,Band Care given by A= / 2−1 31 / , B= / 10 0−2 / , C= / −21 −11 / . Find the matrix D=A+2B−C. D= / 2−1 31 / +2 / 100−2 / − / −21 −11 / = / 2+2×1−(−2)−1+2×0−1 3+2×0−(−1) 1 + 2×(−2)−1 / = / 6−2 4−4 / . J From the above considerations we see that the set of all, in general complex, M×Nmatrices (with ﬁxed MandN) forms a linear vector space of dimension MN. One basis for the space is the set of M×Nmatrices E(p,q)with the property that E(p,q) ij=1i f i=pandj=qwhilst E(p,q) ij= 0 for all other values of iand j, i.e. each matrix has only one non-zero entry, which equals unity. Here the pair (p, q) is simply a label that picks out a particular one of the matrices E(p,q),t h e total number of which is MN. 8.4.2 Multiplication of matrices Let us consider again the ‘transformation’ of one vector into another, y=Ax, which, from (8.24), may be described in terms of components with respect to aparticular basis as y i=Nsummationdisplay j=1Aijxjfori=1,2,...,M. (8.32) Writing this in matrix form as y=Axwe have  y 1 y2 ... yM = A 11 A12... A 1N A21 A22... A 2N ............ AM1AM2... A MN  x1 x2 ... xN (8.33) where we have highlighted with boxes the components used to calculate the element y 2: using (8.32) for i=2 , y2=A21x1+A22x2+···+A2NxN. All the other components yiare calculated similarly. If instead we operate with Aon a basis vector ejhaving all components zero 257 MATRICES AND VECTOR SPACES except for the jth, which equals unity, then we ﬁnd Aej= A 11 A12... A 1N A21 A22... A 2N ............ A M1AM2... A MN  0 0 ... 1 ... 0 = A 1j A2j ... AMj , and so conﬁrm our identiﬁcation of the matrix element A ijas the ith component ofAejin this basis. From (8.28) we can extend our discussion to the product of two matrices P=AB,w h e r e Pis the matrix of the quantities formed by the operation of the rows of Aon the columns of B, treating each column of Bin turn as the vector xrepresented in component form in (8.32). It is clear that, for this to be a meaningful deﬁnition, the number of columns in Amust equal the number of rows in B. Thus the product ABof an M×Nmatrix Awith an N×Rmatrix B is itself an M×Rmatrix P,w h e r e Pij=Nsummationdisplay k=1AikBkjfori=1,2,...,M ,j=1,2,...,R. For example, P=ABmay be written in matrix form parenleftBigg P11 P12 P21 P22parenrightBigg =parenleftbiggA11A12A13 A21A22A23parenrightbigg B11B12 B21B22 B31B32 where P 11=A11B11+A12B21+A13B31, P21=A21B11+A22B21+A23B31, P12=A11B12+A12B22+A13B32, P22=A21B12+A22B22+A23B32. Multiplication of more than two matrices follows naturally and is associative. So, for example, A(BC)≡(AB)C, (8.34) provided, of course, that all the products are deﬁned. As mentioned above, if Ais an M×Nmatrix and Bis an N×Mmatrix then two product matrices are possible, i.e. P=AB and Q=BA. 258 8.4 BASIC MATRIX ALGEBRA These are clearly not the same, since Pis an M×Mmatrix whilst Qis an N×Nmatrix. Thus, particular care must be taken to write matrix products in the intended order; P=ABbut Q=BA. We note in passing that A2means AA, A3means A(AA)=( AA)Aetc. Even if both Aand Bare square, in general AB/negationslash=BA, (8.35) i.e. the multiplication of matrices is not, in general, commutative.IEvaluate P=ABand Q=BAwhere A= /0/@32−1 03 21−34 /1A, B= /0/@2−23 110321 /1A. As we saw for the 2 ×2 case above, the element Pijof the matrix P=ABis found by mentally taking the ‘scalar product’ of the ith row of Awith the jth column of B.F o r example, P11=3×2+2×1+(−1)×3=5 , P12=3×(−2) + 2×1+(−1)×2=−6, etc. Thus P=AB= /0/@32−1 03 2 1−34 /1A /0/@2−23 110 321 /1A= /0 /@ 5−68 97 2 1 137 /1A, and, similarly, Q=BA= /0/@2−23 110 321 /1A /0/@32−1 03 2 1−34 /1A= /0 /@ 9−11 6 35 1 10 9 5 /1A. These results illustrate that, in general, two matrices do not commute. J The property that matrix multiplication is distributive over addition, i.e. that (A+B)C=AC+BC (8.36) and C(A+B)=CA+CB, (8.37) follows directly from its deﬁnition. 8.4.3 The null and identity matrices Both the null matrix and the identity matrix are frequently encountered, and we take this opportunity to introduce them brieﬂy, leaving their uses until later. Thenullorzeromatrix 0has all elements equal to zero, and so its properties are A0=0=0A, A+0=0+A=A. 259 MATRICES AND VECTOR SPACES Theidentity matrix Ihas the property AI=IA=A. It is clear that, in order for the above products to be deﬁned, the identity matrix must be square. The N×Nidentity matrix (often denoted by IN)h a st h ef o r m IN= 10 ···0 01... ......0 0··· 01 . 8.5 Functions of matrices If a matrix Aissquare then, as mentioned above, one can deﬁne powers ofAis a straightforward way. For example A2=AA,A3=AAA, or in the general case An=AA···A (ntimes) , where nis a positive integer. Having deﬁned powers of a square matrix A,w e may construct functions ofAof the form S=summationdisplay nanAn, where the akare simple scalars and the number of terms in the summation may be ﬁnite or inﬁnite. In the case where the sum has an inﬁnite number of terms,the sum has meaning only if it converges. A common example of such a functionis the exponential of a matrix, which is deﬁned by exp A= ∞summationdisplay n=0An n!. (8.38) This deﬁnition can, in turn, be used to deﬁne other functions such as sin Aand cosA. 8.6 The transpose of a matrix We have seen that the components of a linear operator in a given coordinate sys- tem can be written in the form of a matrix A. We will also ﬁnd it useful, however, to consider the diﬀerent (but clearly related) matrix formed by interchanging the rows and columns of A. The matrix is called the transpose ofAand is denoted byAT. 260 8.7 THE COMPLEX AND HERMITIAN CONJUGATES OF A MATRIXIFind the transpose of the matrix A= / 312 041 / . By interchanging the rows and columns of Awe immediately obtain AT= /0/@30 1421 /1A. J It is obvious that if Ais an M×Nmatrix then its transpose ATis aN×M matrix. As mentioned in section 8.3, the transpose of a column matrix is a row matrix and vice versa. An important use of column and row matrices isin the representation of the inner product of two real vectors in terms of theircomponents in a given basis. This notion is discussed fully in the next section,where it is extended to complex vectors. The transpose of the product of two matrices, ( AB) T, is given by the product of their transposes taken in the reverse order, i.e. (AB)T=BTAT. (8.39) This is proved as follows: (AB)T ij=(AB)ji=summationdisplay kAjkBki =summationdisplay k(AT)kj(BT)ik=summationdisplay k(BT)ik(AT)kj=(BTAT)ij, and the proof can be extended to the product of several matrices to give (ABC···G)T=GT···CTBTAT. 8.7 The complex and Hermitian conjugates of a matrix Two further matrices that can be derived from a given general M×Nmatrix are the complex conjugate , denoted by A∗,a n dt h e Hermitian conjugate , denoted byA†. The complex conjugate of a matrix Ais the matrix obtained by taking the complex conjugate of each of the elements of A,i . e . (A∗)ij=(Aij)∗. Obviously if a matrix is real(i.e. it contains only real elements) then A∗=A. 261 MATRICES AND VECTOR SPACESIFind the complex conjugate of the matrix A= / 12 3 i 1+i10 / . By taking the complex conjugate of each element we obtain immediately A∗= / 12−3i 1−i10 / . J The Hermitian conjugate, or adjoint ,o fam a t r i x Ais the transpose of its complex conjugate, or equivalently, the complex conjugate of its transpose, i.e. A†=(A∗)T=(AT)∗. We note that if Ais real (and so A∗=A)t h e n A†=AT, and taking the Hermitian conjugate is equivalent to taking the transpose. Following the previous line ofargument for the transpose of the product of several matrices, the Hermitianconjugate of such a product can be shown to be given by (AB···G) †=G†···B†A†. (8.40)IFind the Hermitian conjugate of the matrix A= / 12 3 i 1+i10 / . Taking the complex conjugate of Aand then forming the transpose we ﬁnd A†= /0/@11−i 21 −3i0 /1A. We obtain the same result, of course, if we ﬁrst take the transpose of Aand then take the complex conjugate. J An important use of the Hermitian conjugate (or transpose in the real case) is in connection with the inner product of two vectors. Suppose that in a givenorthonormal basis the vectors aandbmay be represented by the column matrices a= a 1 a2 ... aN and b= b1 b2 ... bN . (8.41) Taking the Hermitian conjugate of a, to give a row matrix, and multiplying (on 262 8 . 8T H ET R A C EO FAM A T R I X the right) by bwe obtain a†b=(a∗ 1a∗2···a∗ N) b1 b2 ... bN =Nsummationdisplay i=1a∗ ibi, (8.42) which is the expression for the inner product /angbracketlefta|b/angbracketrightin that basis. We note that for real vectors (8.42) reduces to aTb=summationtextN i=1aibi. If the basis eiisnotorthonormal, so that, in general, /angbracketleftei|ej/angbracketright=Gij/negationslash=δij, then, from (8.18), the scalar product of aandbin terms of their components with respect to this basis is given by /angbracketlefta|b/angbracketright=Nsummationdisplay i=1Nsummationdisplay j=1a∗ iGijbj=a†Gb, where Gis the N×Nmatrix with elements Gij. 8.8 The trace of a matrix For a given matrix A, in the previous two sections we have considered various other matrices that can be derived from it. However, sometimes one wishes toderive a single number from a matrix. The simplest example is the trace(orspur) of a square matrix, which is denoted by Tr A. This quantity is deﬁned as the sum of the diagonal elements of the matrix, TrA=A 11+A22+···+ANN=Nsummationdisplay i=1Aii. (8.43) It clear that the trace is a linear operation so that, for example, Tr(A±B)=T r A±TrB. A very useful property of traces is that the trace of the product of two matrices is independent of the order of their multiplication; this results holds whether ornot the matrices commute and is proved as follows: TrAB= Nsummationdisplay i=1(AB)ii=Nsummationdisplay i=1Nsummationdisplay j=1AijBji=Nsummationdisplay i=1Nsummationdisplay j=1BjiAij=Nsummationdisplay j=1(BA)jj=T r BA. (8.44) The result can be extended to the product of several matrices. For example, from (8.44), we immediately ﬁnd TrABC=T r BCA=T r CAB, 263 MATRICES AND VECTOR SPACES which shows that the trace of a product is invariant under cyclic permutations of the matrices in the product. Other easily derived properties of the trace are, forexample, Tr A T=T r Aand Tr A†=( T r A)∗. 8.9 The determinant of a matrix For a given matrix A, the determinant det A(like the trace) is a single number (or algebraic expression) that depends upon the elements of A. Also like the trace, the determinant is deﬁned only for square matrices. If, for example, Ais a 3×3 matrix then its determinant, of order 3, is denoted by detA=|A|=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleA 11A12A13 A21A22A23 A31A32A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. (8.45) In order to calculate the value of a determinant, we ﬁrst need to introduce the notions of the minor and the cofactor of an element of a matrix. (We shall see that we can use the cofactors to write an order-3 determinant as theweighted sum of three order-2 determinants, thereby simplifying its evaluation.)The minor M ijof the element Aijof an N×Nmatrix Ais the determinant of the (N−1)×(N−1) matrix obtained by removing all the elements of the ith row and jth column of A; the associated cofactor, Cij, is found by multiplying the minor by ( −1)i+j.IFind the cofactor of the element A23of the matrix A= /0/@A11A12A13 A21A22A23 A31A32A33 /1 A . Removing all the elements of the second row and third column of Aand forming the determinant of the remaining terms gives the minor M23= // // A11A12 A31A32 // // . Multiplying the minor by ( −1)2+3=(−1)5=−1g i v e s C23=− ////A11A12 A31A32 / /// . J We now deﬁne a determinant as the sum of the products of the elements of any row or column and their corresponding cofactors ,e . g . A21C21+A22C22+A23C23or A13C13+A23C23+A33C33.S u c has u mi sc a l l e da Laplace expansion . For example, in the ﬁrst of these expansions, using the elements of the second row of the 264 8.9 THE DETERMINANT OF A MATRIX determinant deﬁned by (8.45) and their corresponding cofactors, we write |A|as the Laplace expansion |A|=A21(−1)(2+1)M21+A22(−1)(2+2)M22+A23(−1)(2+3)M23 =−A21vextendsinglevextendsinglevextendsinglevextendsingleA 12A13 A32A33vextendsinglevextendsinglevextendsinglevextendsingle+A 22vextendsinglevextendsinglevextendsinglevextendsingleA 11A13 A31A33vextendsinglevextendsinglevextendsinglevextendsingle−A 23vextendsinglevextendsinglevextendsinglevextendsingleA 11A12 A31A32vextendsinglevextendsinglevextendsinglevextendsingle. We will see later that the value of the determinant is independent of the row or column chosen. Of course, we have not yet determined the value of |A|but, rather, written it as the weighted sum of three determinants of order 2. However,applying again the deﬁnition of a determinant, we can evaluate each of theorder-2 determinants.IEvaluate the determinant////A12A13 A32A33 ////. By considering the products of the elements of the ﬁrst row in the determinant, and their corresponding cofactors, we ﬁnd/// / A12A13 A32A33 /// / =A12(−1)(1+1)|A33|+A13(−1)(1+2)|A32| =A12A33−A13A32, where the values of the order-1 determinants |A33|and|A32|are deﬁned to be A33andA32 respectively. It must be remembered that the determinant is notthe same as the modulus, e.g. det (−2) =|−2|=−2, not 2. J We can now combine all the above results to show that the value of the determinant (8.45) is given by |A|=−A21(A12A33−A13A32)+A22(A11A33−A13A31) −A23(A11A32−A12A31) (8.46) =A11(A22A33−A23A32)+A12(A23A31−A21A33) +A13(A21A32−A22A31), (8.47) where the ﬁnal expression gives the form in which the determinant is usually remembered and is the form that is obtained immediately by considering theLaplace expansion using the ﬁrst row of the determinant. The last equality, which essentially rearranges a Laplace expansion using the second row into one using the ﬁrst row, supports our assertion that the value of the determinant is unaﬀectedby which row or column is chosen for the expansion. 265 MATRICES AND VECTOR SPACESISuppose the rows of a real 3×3matrix Aare interpreted as the components in a given basis of three (three-component) vectors a,bandc. Show that one can write the determinant ofAas |A|=a·(b×c). If one writes the rows of Aas the components in a given basis of three vectors a,bandc, we have from (8.47) that |A|= //////a1a2a3 b1b2b3 c1c2c3 //////=a1(b2c3−b3c2)+a2(b3c1−b1c3)+a3(b1c2−b2c1). From expression (7.34) for the scalar triple product given in subsection 7.6.3, it follows that we may write the determinant as |A|=a·(b×c). (8.48) In other words, |A|is the volume of the parallelepiped deﬁned by the vectors a,band c. (One could equally well interpret the columns of the matrix Aas the components of three vectors, and result (8.48) would still hold.) This result provides a more memorable(and more meaningful) expression than (8.47) for the value of a 3 ×3 determinant. Indeed, using this geometrical inte rpretation, we see immediately that, if the vectors a 1,a2,a3are not linearly independent then the value of the determinant vanishes: |A|=0 . J The evaluation of determinants of order greater than 3 follows the same general method as that presented above, in that it relies on successively reducing the orderof the determinant by writing it as a Laplace expansion. Thus, a determinantof order 4 is ﬁrst written as a sum of four determinants of order 3, which are then evaluated using the above method. For higher-order determinants, one cannot write down directly a simple geometrical expression for |A|analogous to that given in (8.48). Nevertheless, it is still true that if the rows or columns oftheN×Nmatrix Aare interpreted as the components in a given basis of N (N-component) vectors a 1,a2,...,aN, then the determinant |A|vanishes if these vectors are not all linearly independent. 8.9.1 Properties of determinants A number of properties of determinants follow straightforwardly from the deﬁni- tion of det A; their use will often reduce the labour of evaluating a determinant. We present them here without speciﬁc proofs, though they all follow readily fromthe alternative form for a determinant, given in equation (21.28) on page 791,and expressed in terms of the Levi–Civita symbol /epsilon1 ijk(see exercise 21.9). (i)Determinant of the transpose . The transpose matrix AT(which, we recall, is obtained by interchanging the rows and columns of A) has the same determinant as Aitself, i.e. |AT|=|A|. (8.49) 266 8.9 THE DETERMINANT OF A MATRIX It follows that anytheorem established for the rows of Awill apply to the columns as well, and vice versa. (ii)Determinant of the complex and Hermitian conjugate . It is clear that the matrix A∗obtained by taking the complex conjugate of each element of A has the determinant |A∗|=|A|∗. Combining this result with (8.49), we ﬁnd that |A†|=|(A∗)T|=|A∗|=|A|∗. (8.50) (iii)Interchanging two rows or two columns . If two rows (columns) of Aare interchanged, its determinant changes sign but is unaltered in magnitude. (iv)Removing factors . If all the elements of a single row (column) of Ahave a common factor, λ, then this factor may be removed; the value of the determinant is given by the product of the remaining determinant and λ. Clearly this implies that if all the elements of any row (column) are zerothen|A|= 0. It also follows that if every element of the N×Nmatrix A is multiplied by a constant factor λthen |λA|=λ N|A|. (8.51) (v)Identical rows or columns . If any two rows (columns) of Aare identical or are multiples of one another, then it can be shown that |A|=0 . (vi)Adding a constant multiple of one row (column) to another . The determinant of a matrix is unchanged in value by adding to the elements of one row(column) any ﬁxed multiple of the elements of another row (column). (vii)Determinant of a product .I fAand Bare square matrices of the same order then |AB|=|A||B|=|BA|. (8.52) A simple extension of this property gives, for example, |AB···G|=|A||B|···|G|=|A||G|···|B|=|A···GB|, which shows that the determinant is invariant to cyclic permutations of the matrices in the product. There is no explicit procedure for using the above results in the evaluation of any given determinant, and judging the quickest route to an answer is a matterof experience. A general guide is to try to reduce all terms but one in a row orcolumn to zero and hence in eﬀect to obtain a determinant of smaller size. The steps taken in evaluating the determinant in the example below are certainly not the fastest, but they have been chosen in order to illustrate the use of most of theproperties listed above. 267 MATRICES AND VECTOR SPACESIEvaluate the determinant |A|= // /// // 1023 01−21 3−34−2 −21−2−1 // /// // . Taking a factor 2 out of the third column and then adding the second column to the third gives |A|=2 ///////1013 01−11 3−32−2 −21−1−1 ///////=2 ///////1013 01013−3−1−2 −21 0 −1 ///////. Subtracting the second column from the fourth gives |A|=2 ///////1013 01003−3−11 −21 0 −2 ///////. We now note that the second row has only one non-zero element and so the determinant may conveniently be written as a Laplace expansion, i.e. |A|=2×1×(−1)2+2 /// /// 113 3−11 −20−2 /// /// =2 /// /// 404 3−11 −20−2 /// /// , where the last equality follows by adding the second row to the ﬁrst. It can now be seen that the ﬁrst row is minus twice the third, and so the value of the determinant is zero, byproperty (v) above.J 8.10 The inverse of a matrix Our ﬁrst use of determinants will be in deﬁning the inverse of a matrix. If we were dealing with ordinary numbers we would consider the relation P=ABas equivalent to B=P/A, provided that A/negationslash= 0. However, if A,Band Pare matrices then this notation does not have an obvious meaning. What we really want toknow is whether an explicit formula for Bcan be obtained in terms of Aand P. It will be shown that this is possible for those cases in which |A|/negationslash=0 .A square matrix whose determinant is zero is called a singular matrix; otherwise it isnon-singular . We will show that if Ais non-singular we can deﬁne a matrix, denoted by A −1and called the inverse ofA, which has the property that if AB=P then B=A−1P.I nw o r d s , Bcan be obtained by multiplying Pfrom the left by A−1. Analogously, if Bis non-singular then, by multiplication from the right, A=PB−1. It is clear that AI=A⇒ I=A−1A, (8.53) 268 8.10 THE INVERSE OF A MATRIX where Iis the unit matrix, and so A−1A=I=AA−1. These statements are equivalent to saying that if we ﬁrst multiply a matrix, Bsay, by Aand then multiply by the inverse A−1, we end up with the matrix we started with, i.e. A−1AB=B. (8.54) This justiﬁes our use of the term inverse. It is also clear that the inverse is only deﬁned for square matrices. So far we have only deﬁned what we mean by the inverse of a matrix. Actually ﬁnding the inverse of a matrix Amay be carried out in a number of ways. We will show that one method is to construct ﬁrst the matrix Ccontaining the cofactors of the elements of A, as discussed in the last subsection. Then the required inverse A−1can be found by forming the transpose of Cand dividing by the determinant ofA. Thus the elements of the inverse A−1are given by (A−1)ik=(C)T ik |A|=Cki |A|. (8.55) That this procedure does indeed result in the inverse may be seen by considering the components of A−1A,i . e . (A−1A)ij=summationdisplay k(A−1)ik(A)kj=summationdisplay kCki |A|Akj=|A| |A|δij. (8.56) The last equality in (8.56) relies on the property summationdisplay kCkiAkj=|A|δij; (8.57) this can be proved by considering the matrix A/primeobtained from the original matrix Awhen the ith column of Ais replaced by one of the other columns, say the jth. Thus A/primeis a matrix with two identical columns and so has zero determinant. However, replacing the ith column by another does not change the cofactors Cki of the elements in the ith column, which are therefore the same in Aand A/prime. Recalling the Laplace expansion of a determinant, i.e. |A|=summationdisplay kAkiCki, we obtain 0=|A/prime|=summationdisplay kA/prime kiC/prime ki=summationdisplay kAkjCki,i/negationslash=j, which together with the Laplace expansion itself may be summarised by (8.57). It is immediately obvious from (8.55) that the inverse of a matrix is not deﬁned if the matrix is singular (i.e. if |A|=0 ) . 269 MATRICES AND VECTOR SPACESIFind the inverse of the matrix A= /0/@243 1−2−2 −33 2 /1A. We ﬁrst determine |A|: |A|=2 [−2(2)−(−2)3] + 4[(−2)(−3)−(1)(2)] + 3[(1)(3) −(−2)(−3)] =1 1. (8.58) This is non-zero and so an inverse matrix can be constructed. To do this we need the matrix of the cofactors, C, and hence CT. We ﬁnd C= /0/@24−3 11 3−18 −27−8 /1A and CT= /0/@21−2 41 37 −3−18−8 /1A, and hence A−1=CT |A|=1 11 /0/@21−2 41 37 −3−18−8 /1A. J (8.59) For a 2×2 matrix, the inverse has a particularly simple form. If the matrix is A=parenleftbiggA11A12 A21A22parenrightbigg then its determinant |A|is given by |A|=A11A22−A12A21,a n dt h em a t r i xo f cofactors is C=parenleftbiggA22−A21 −A12 A11parenrightbigg . Thus the inverse of Ais given by A−1=CT |A|=1 A11A22−A12A21parenleftbiggA22−A12 −A21 A11parenrightbigg . (8.60) It can be seen that the transposed matrix of cofactors for a 2 ×2m a t r i xi st h e same as the matrix formed by swapping the elements on the leading diagonal (A11andA22) and changing the signs of the other two elements ( A12andA21). This is completely general for a 2 ×2 matrix and is easy to remember. The following are some further useful properties related to the inverse matrix 270 8.10 THE INVERSE OF A MATRIX and may be straightforwardly derived. (i) ( A−1)−1=A. (ii) ( AT)−1=(A−1)T. (iii) ( A†)−1=(A−1)†. (iv) ( AB)−1=B−1A−1. (v) ( AB···G)−1=G−1···B−1A−1.IProve the properties (i)–(v) stated above. We begin by writing down the fundamental expression deﬁning the inverse of a non- singular square matrix A: AA−1=I=A−1A. (8.61) Property (i): This follows immediately from the expression (8.61). Property (ii): Taking the transpose of each expression in (8.61) gives (AA−1)T=IT=(A−1A)T. Using the result (8.39) for the transpose of a product of matrices and noting that IT=I, we ﬁnd (A−1)TAT=I=AT(A−1)T. However, from (8.61), this implies ( A−1)T=(AT)−1and hence proves result (ii) above. Property (iii): This may be proved in an analogous way to property (ii), by replacing the transposes in (ii) by Hermitian conjugates and using the result (8.40) for the Hermitianconjugate of a product of matrices. Property (iv): Using (8.61), we may write (AB)(AB) −1=I=(AB)−1(AB), From the left-hand equality it follo ws, by multiplying on the left by A−1,t h a t A−1AB(AB)−1=A−1Iand hence B(AB)−1=A−1. Now multiplying on the left by B−1gives B−1B(AB)−1=B−1A−1, and hence the stated result. Property (v): Finally, result (iv) may extended to case (iv) in a straightforward manner. For example, using result (iv) twice we ﬁnd (ABC)−1=(BC)−1A−1=C−1B−1A−1. J We conclude this section by noting that the determinant |A−1|of the inverse matrix can be expressed very simply in terms of the determinant |A|of the matrix itself. Again we start with the fundamental expression (8.61). Then, using the property (8.52) for the determinant of a product, we ﬁnd |AA−1|=|A||A−1|=|I|. It is straightforward to show by Laplace expansion that |I|= 1, and so we arrive at the useful result |A−1|=1 |A|. (8.62) 271 MATRICES AND VECTOR SPACES 8.11 The rank of a matrix Therankof a general M×Nmatrix is an important concept, particularly in the solution of sets of simultaneous linear equations, to be discussed in the nextsection, and we now discuss it in some detail. Like the trace and determinant, the rank of matrix Ais a single number (or algebraic expression) that depends on the elements of A. Unlike the trace and determinant, however, the rank of a matrix can be deﬁned even when Ais not square. As we shall see, there are two equivalent deﬁnitions of the rank of a general matrix. Firstly, the rank of a matrix may be deﬁned in terms of the linear independence of vectors. Suppose that the columns of an M×Nmatrix are interpreted as the components in a given basis of N(M-component) vectors v 1,v2,...,vN,a s follows: A= ↑↑ ↑ v1v2... vN ↓↓ ↓ . Then the rankofA, denoted by rank Aor by R(A), is deﬁned as the number oflinearly independent vectors in the set v1,v2,...,vN, and equals the dimension of the vector space spanned by those vectors. Alternatively, we may consider therows of Ato contain the components in a given basis of the M(N-component) vectors w 1,w2,...,wMas follows: A= ← w1→ ← w2→ ... ← wM→ . It may then be shown †that the rank of Ais also equal to the number of linearly independent vectors in the set w1,w2,...,wM. From this deﬁnition it is should be clear that the rank of Ais unaﬀected by the exchange of two rows (or two columns) or by the multiplication of a row (or column) by a constant. Furthermore, suppose that a constant multiple of one row (column) is added to another row (column): for example, we might replace the row wibywi+cwj. This also has no eﬀect on the number of linearly independent rows and so leavesthe rank of Aunchanged. We may use these properties to evaluate the rank of a given matrix. A second (equivalent) deﬁnition may be given of the rank of a matrix and uses the concept of submatrices . A submatrix of Ais any matrix that can be formed from the elements of Aby ignoring one. or more than one, row or column. It †For a fuller discussion, see, for example, Modern Mathematical Methods for Physicists and Engineers , chapter 6, C. D. Cantrell (Cambridge University Press). 272 8.12 SPECIAL TYPES OF SQUARE MATRIX may be shown that the rank of a general M×Nmatrix is equal to the size of the largest square submatrix of Awhose determinant is non-zero. Therefore, if a matrix Ahas an r×rsubmatrix Swith|S|/negationslash= 0, but no ( r+1)×(r+1) submatrix with non-zero determinant then the rank of the matrix is r. From either deﬁnition it is clear that the rank of Ais less than or equal to the smaller of MandN.IDetermine the rank of the matrix A= /0/@110−2 202 2 413 1 /1A. The largest possible square submatrices of Amust be of dimension 3 ×3. Clearly, A possesses four such submatrices, the determinants of which are given by//////110 202413 //////=0, //////11−2 20 241 1 //////=0,//////10−2 22 243 1 //////=0, //////10−2 02 213 1 //////=0. (In each case the determinant may be evaluated as described in subsection 8.9.1.) The next largest square submatrices of Aare of dimension 2 ×2. Consider, for example, the 2×2 submatrix formed by ignoring the third row and the third and fourth columns ofA; this has determinant////11 20 ////=1×0−2×1=−2. Since its determinant is non-zero, Ais of rank 2 and we need not consider any other 2 ×2 submatrices. J In the special case in which the matrix Ais asquare N×Nmatrix, by comparing either of the above deﬁnitions of rank with our discussion of determinants insection 8.9, we see that |A|= 0 unless the rank of AisN. In other words, Ais singular unless R(A)=N. 8.12 Special types of square matrix Matrices that are square, i.e. N×N, are very common in physical applications. We now consider some special forms of square matrix that are of particularimportance. 8.12.1 Diagonal matrices The unit matrix, which we have already encountered, is an example of a diagonal matrix. Such matrices are characterised by having non-zero elements only on the 273 MATRICES AND VECTOR SPACES leading diagonal , i.e. only elements Aijwith i=jmay be non-zero. For example, A= 10 0 02 000−3 , is a 3×3 diagonal matrix. Such a matrix is often denoted by A= diag (1 ,2,−3). By performing a Laplace expansion, it is easily shown that the determinant of anN×Ndiagonal matrix is equal to the product of the diagonal elements. Thus, if the matrix has the form A=d i a g ( A 11,A22,...,A NN)t h e n |A|=A11A22···ANN. (8.63) Moreover, it is also straightforward to show that the inverse of Ais also a diagonal matrix given by A−1= diagparenleftbigg1 A11,1 A22,...,1 ANNparenrightbigg . Finally, we note that, if two matrices Aand Barebothdiagonal then they have the useful property that their product is commutative: AB=BA. This is nottrue for matrices in general. 8.12.2 Lower and upper triangular matrices A square matrix Ais called lower triangular if all the elements above the principal diagonal are zero. For example, the general form for a 3 ×3 lower triangular matrix is A= A1100 A21A220 A31A32A33 , where the elements Aijmay be zero or non-zero. Similarly an upper triangular square matrix is one for which all the elements below the principal diagonal are zero. The general 3 ×3 form is thus A= A11A12A13 0A22A23 00 A33 . By performing a Laplace expansion, it is straightforward to show that, in the general N×Ncase, the determinant of an upper or lower triangular matrix is equal to the product of its diagonal elements, |A|=A11A22···ANN. (8.64) 274 8.12 SPECIAL TYPES OF SQUARE MATRIX Clearly result (8.63) for diagonal matrices is a special case of this result. Moreover, it may be shown that the inverse of a non-singular lower (upper) triangular matrixis also lower (upper) triangular. 8.12.3 Symmetric and antisymmetric matrices A square matrix Aof order Nwith the property A=A Tis said to be symmetric . Similarly a matrix for which A=−ATis said to be anti-orskew-symmetric and its diagonal elements a11,a22,...,a NNare necessarily zero. Moreover, if Ais (anti-)symmetric then so too is its inverse A−1. This is easily proved by noting that if A=±ATthen (A−1)T=(AT)−1=±A−1. Any N×Nmatrix Acan be written as the sum of a symmetric and an antisymmetric matrix, since we may write A=1 2(A+AT)+1 2(A−AT)=B+C, where clearly B=BTand C=−CT. The matrix Bis therefore called the symmetric part of A,a n d Cis the antisymmetric part.IIfAis an N×Nantisymmetric matrix, show that |A|=0ifNis odd. IfAis antisymmetric then AT=−A. Using the properties of determinants (8.49) and (8.51), we have |A|=|AT|=|−A|=(−1)N|A|. Thus, if Nis odd then |A|=−|A|, which implies that |A|=0 . J 8.12.4 Orthogonal matrices A non-singular matrix with the property that its transpose is also its inverse, AT=A−1, (8.65) is called an orthogonal matrix . It follows immediately that the inverse of an orthogonal matrix is also orthogonal, since (A−1)T=(AT)−1=(A−1)−1. Moreover, since for an orthogonal matrix ATA=I, we have |ATA|=|AT||A|=|A|2=|I|=1. Thus the determinant of an orthogonal matrix must be |A|=±1. An orthogonal matrix represents, in a particular basis, a linear operator that leaves the norms (lengths) of real vectors unchanged, as we will now show. 275 MATRICES AND VECTOR SPACES Suppose that y=Axis represented in some coordinate system by the matrix equation y=Ax;t h e n/angbracketlefty|y/angbracketrightis given in this coordinate system by yTy=xTATAx=xTx. Hence/angbracketlefty|y/angbracketright=/angbracketleftx|x/angbracketright, showing that the action of a linear operator represented by an orthogonal matrix does not change the norm of a real vector. 8.12.5 Hermitian and anti-Hermitian matrices AnHermitian matrix is one that satisﬁes A=A†,w h e r e A†is the Hermitian conjugate discussed in section 8.7. Similarly if A†=−A,t h e n Ais called anti- Hermitian . A real (anti-)symmetric matrix is a special case of an (anti-)Hermitian matrix, in which all the elements of the matrix are real. Also, if Ais an (anti- )Hermitian matrix then so too is its inverse A−1,s i n c e (A−1)†=(A†)−1=±A−1. Any N×Nmatrix Acan be written as the sum of an Hermitian matrix and an anti-Hermitian matrix, since A=1 2(A+A†)+1 2(A−A†)=B+C, where clearly B=B†and C=−C†. The matrix Bis called the Hermitian part of A,a n d Cis called the anti-Hermitian part. 8.12.6 Unitary matrices Aunitary matrix Ais deﬁned as one for which A†=A−1. (8.66) Clearly, if Ais real then A†=AT, showing that a real orthogonal matrix is a special case of a unitary matrix, one in which all the elements are real. We note that the inverse A−1of a unitary is also unitary, since (A−1)†=(A†)−1=(A−1)−1. Moreover, since for a unitary matrix A†A=I, we have |A†A|=|A†||A|=|A|∗|A|=|I|=1. Thus the determinant of a unitary matrix has unit modulus. A unitary matrix represents, in a particular basis, a linear operator that leaves the norms (lengths) of complex vectors unchanged. If y=Axis represented in some coordinate system by the matrix equation y=Axthen/angbracketlefty|y/angbracketrightis given in this coordinate system by y†y=x†A†Ax=x†x. 276 8.13 EIGENVECTORS AND EIGENVALUES Hence/angbracketlefty|y/angbracketright=/angbracketleftx|x/angbracketright, showing that the action of the linear operator represented by a unitary matrix does not change the norm of a complex vector. The action of aunitary matrix on a complex column matrix thus parallels that of an orthogonalmatrix acting on a real column matrix. 8.12.7 Normal matrices A ﬁnal important set of special matrices consists of the normal matrices, for which AA †=A†A, i.e. a normal matrix is one that commutes with its Hermitian conjugate. We can easily show that Hermitian matrices and unitary matrices (or symmetric matrices and orthogonal matrices in the real case) are examples of normalmatrices. For an Hermitian matrix, A=A †and so AA†=AA=A†A. Similarly, for a unitary matrix, A−1=A†and so AA†=AA−1=A−1A=A†A. Finally, we note that, if Ais normal then so too is its inverse A−1,s i n c e A−1(A−1)†=A−1(A†)−1=(A†A)−1=(AA†)−1=(A†)−1A−1=(A−1)†A−1. This broad class of matrices is important in the discussion of eigenvectors and eigenvalues in the next section. 8.13 Eigenvectors and eigenvalues Suppose that a linear operator Atransforms vectors xin an N-dimensional vector space into other vectors Axin the same space. The possibility then arises that there exist vectors xeach of which is transformed by Ainto a multiple of itself. Such vectors would have to satisfy Ax=λx. (8.67) Any non-zero vector xthat satisﬁes (8.67) for some value of λis called an eigenvector of the linear operator A,a n d λis called the corresponding eigenvalue . As will be discussed below, in general the operator AhasNindependent eigenvectors xi, with eigenvalues λi.T h e λiare not necessarily all distinct. If we choose a particular basis in the vector space, we can write (8.67) in terms of the components of Aandxwith respect to this basis as the matrix equation Ax=λx, (8.68) where Ais an N×Nmatrix. The column matrices xthat satisfy (8.68) obviously 277 MATRICES AND VECTOR SPACES represent the eigenvectors xofAin our chosen coordinate system. Convention- ally, these column matrices are also referred to as the eigenvectors of the matrix A.†Clearly, if xis an eigenvector of A(with some eigenvalue λ) then any scalar multiple µxis also an eigenvector with the same eigenvalue. We therefore often usenormalised eigenvectors, for which x†x=1 (note that x†xcorresponds to the inner product /angbracketleftx|x/angbracketrightin our basis). Any eigen- vector xcan be normalised by dividing all its components by the scalar ( x†x)1/2. As will be seen, the problem of ﬁnding the eigenvalues and corresponding eigenvectors of a square matrix Aplays an important role in many physical investigations. Throughout this chapter we denote the ith eigenvector of a square matrix Abyxiand the corresponding eigenvalue by λi. This superscript notation for eigenvectors is used to avoid any confusion with components.IA non-singular matrix Ahas eigenvalues λiand eigenvectors xi. Find the eigenvalues and eigenvectors of the inverse matrix A−1. The eigenvalues and eigenvectors of Asatisfy Axi=λixi. Left-multiplying both sid es of this equation by A−1, we ﬁnd A−1Axi=λiA−1xi. Since A−1A=I, on rearranging we obtain A−1xi=1 λixi. Thus, we see that A−1has the same eigenvectors xias does A, but the corresponding eigenvalues are 1 /λi. J In the remainder of this section we will discuss some useful results concerning the eigenvectors and eigenvalues of certain special (though commonly occurring)square matrices. The results will be established for matrices whose elements maybe complex; the corresponding properties for real matrices may be obtained asspecial cases. 8.13.1 Eigenvectors and eigenvalues of a normal matrix In subsection 8.12.7 we deﬁned a normal matrix Aas one that commutes with its Hermitian conjugate, so that A †A=AA†. †In this context, when referring to linear combinations of eigenvectors xwe will normally use the term ‘vector’. 278 8.13 EIGENVECTORS AND EIGENVALUES We also showed that both Hermitian and unitary matrices (or symmetric and orthogonal matrices in the real case) are examples of normal matrices. We nowdiscuss the properties of the eigenvectors and eigenvalues of a normal matrix. Ifxis an eigenvector of a normal matrix Awith corresponding eigenvalue λ then Ax=λx, or equivalently, (A−λI)x=0. (8.69) Denoting B=A−λI, (8.69) becomes Bx=0and, taking the Hermitian conjugate, we also have (Bx) †=x†B†=0. (8.70) From (8.69) and (8.70) we then have x†B†Bx=0. (8.71) However, the product B†Bis given by B†B=(A−λI)†(A−λI)=( A†−λ∗I)(A−λI)=A†A−λ∗A−λA†+λλ∗. Now since Ais normal, AA†=A†A(see subsection 8.12.7) and so B†B=AA†−λ∗A−λA†+λλ∗=(A−λI)(A−λI)†=BB†, and hence Bis also normal. From (8.71) we then ﬁnd x†B†Bx=x†BB†x=(B†x)†B†x=0, from which we obtain B†x=(A†−λ∗I)x=0. Therefore, for a normal matrix A,the eigenvalues of A†are the complex conjugates of the eigenvalues of A. Let us now consider two eigenvectors xiand xjof a normal matrix Acorre- sponding to two diﬀerent eigenvalues λiandλj. We then have Axi=λixi, (8.72) Axj=λjxj. (8.73) Multiplying (8.73) on the left by ( xi)†we obtain (xi)†Axj=λj(xi)†xj. (8.74) However, on the LHS of (8.74) we have (xi)†A=(A†xi)†=(λ∗ ixi)†=λi(xi)†, (8.75) where we have used (8.40) and the property just proved for a normal matrix to 279 MATRICES AND VECTOR SPACES write A†xi=λ∗ ixi. From (8.74) and (8.75) we have (λi−λj)(xi)†xj=0. (8.76) Thus, ifλi/negationslash=λjthe eigenvectors xiand xjmust be orthogonal ,i . e .( xi)†xj=0. It follows immediately from (8.76) that if all Neigenvalues of a normal matrix Aare distinct then all Neigenvectors of Aare mutually orthogonal. If, however, two or more eigenvalues are the same then further consideration is required. Aneigenvalue corresponding to two or more diﬀerent eigenvectors (i.e. they are notsimply multiples of one another) is said to be degenerate . Suppose that λ 1isk-fold degenerate, i.e. Axi=λ1xifori=1,2,...,k, (8.77) but that it is diﬀerent from any of λk+1,λk+2, etc. Then any linear combination of these xiis also an eigenvector with eigenvalue λ1,s i n c e ,f o r z=summationtextk i=1cixi, Az≡Aksummationdisplay i=1cixi=ksummationdisplay i=1ciAxi=ksummationdisplay i=1ciλ1xi=λ1z. (8.78) If the xideﬁned in (8.77) are not already mutually orthogonal then we can construct new eigenvectors zithat are orthogonal by the following procedure: z1=x1, z2=x2−bracketleftBig (ˆz1)†x2bracketrightBig ˆz1, z3=x3−bracketleftBig (ˆz2)†x3bracketrightBig ˆz2−bracketleftBig (ˆz1)†x3bracketrightBig ˆz1, ... zk=xk−bracketleftBig (ˆzk−1)†xkbracketrightBig ˆzk−1−···−bracketleftBig (ˆz1)†xkbracketrightBig ˆz1. In this procedure, known as Gram–Schmidt orthogonalisation , each new eigen- vector ziis normalised to give the unit vector ˆzibefore proceeding to the construc- tion of the next one (the normalisation is carried out by dividing each element ofthe vector z iby [( zi)†zi]1/2). Note that each factor in brackets ( ˆzm)†xnis a scalar product and thus only a number. It follows that, as shown in (8.78), each vectorz iso constructed is an eigenvector of Awith eigenvalue λ1and will remain so on normalisation. It is straightforward to check that, provided the previous neweigenvectors have been normalised as prescribed, each z iis orthogonal to all its predecessors. (In practice, however, the method is laborious and the example in subsection 8.14.1 gives a less rigorous but considerably quicker way.) Therefore, even if Ahas some degenerate eigenvalues we can by construction obtain a set of Nmutually orthogonal eigenvectors. Moreover, it may be shown (although the proof is beyond the scope of this book) that these eigenvectorsarecomplete in that they form a basis for the N-dimensional vector space. As 280 8.13 EIGENVECTORS AND EIGENVALUES a result any arbitrary vector ycan be expressed as a linear combination of the eigenvectors xi: y=Nsummationdisplay i=1aixi, (8.79) where ai=(xi)†y. Thus, the eigenvectors form an orthogonal basis for the vector space. By normalising the eigenvectors so that ( xi)†xi=1t h i sb a s i si sm a d e orthonormal.IShow that a normal matrix Acan be written in terms of its eigenvalues λiand orthogonal eigenvectors xias A=NX i=1λixi(xi)†. (8.80) The key to proving the validity of (8.80) is to show that both sides of the expression give the same result when acting on an arbitary vector y.S i n c e Ais normal, we may expand y in terms of the eigenvectors xi, as shown in (8.79). Thus, we have Ay=ANX i=1aixi=NX i=1aiλixi. Alternatively, the action of the RHS of (8.80) on yis given by NX i=1λixi(xi)†y=NX i=1aiλixi, since ai=(xi)†y. We see that the two expressions for the action of each side of (8.80) on y are identical, which implies that this relationship is indeed correct. J 8.13.2 Eigenvectors and eigenvalues of Hermitian and anti-Hermitian matrices For a normal matrix we showed that if Ax=λxthen A†x=λ∗x. However, if Ais also Hermitian, A=A†, it follows necessarily that λ=λ∗. Thus, the eigenvalues of an Hermitian matrix are real, a result which may be proved directly.IProve that the eigenvalues of an Hermitian matrix are real. For any particular eigenvector xi, we take the Hermitian conjugate of Axi=λixito give (xi)†A†=λ∗ i(xi)†. (8.81) Using A†=A,s i n c e Ais Hermitian, and multiplying on the right by xi,w eo b t a i n (xi)†Axi=λ∗ i(xi)†xi. (8.82) But multiplying Axi=λixithrough on the left by ( xi)†gives (xi)†Axi=λi(xi)†xi. 281 MATRICES AND VECTOR SPACES Subtracting this from (8.82) yields 0=(λ∗ i−λi)(xi)†xi. But ( xi)†xiis the modulus squared of the non-zero vector xiand is thus non-zero. Hence λ∗ imust equal λiand thus be real. The same argument can be used to show that the eigenvalues of a real symmetric matrix are themselves real. J The importance of the above result will be apparent to any student of quantum mechanics. In quantum mechanics the eigenvalues of operators correspond to measured values of observable quantities, e.g. energy, angular momentum, parity and so on, and these clearly must be real. If we use Hermitian operators toformulate the theories of quantum mechanics, the above property guaranteesphysically meaningful results. Since an Hermitian matrix is also a normal matrix, its eigenvectors are orthog- onal (or can be made so using the Gram–Schmidt orthogonalisation procedure).Alternatively we can prove the orthogonality of the eigenvectors directly.IProve that the eigenvectors corresponding to di ﬀerent eigenvalues of an Hermitian matrix are orthogonal. Consider two unequal eigenvalues λiandλjand their corresponding eigenvectors satisfying Axi=λixi, (8.83) Axj=λjxj. (8.84) Taking the Hermitian conjugate of (8.83) we ﬁnd ( xi)†A†=λ∗ i(xi)†. Multiplying this on the right by xjwe obtain (xi)†A†xj=λ∗ i(xi)†xj, and similarly multiplying (8.84) through on the left by ( xi)†we ﬁnd (xi)†Axj=λj(xi)†xj. Then, since A†=A, the two left-hand sides are equal and, because the λiare real, on subtraction we obtain 0=(λi−λj)(xi)†xj. Finally we note that λi/negationslash=λjand so ( xi)†xj=0, i.e. the eigenvectors xiand xjare orthogonal. J In the case where some of the eigenvalues are equal, further justiﬁcation of the orthogonality of the eigenvectors is needed. The Gram–Schmidt orthogonalisa-tion procedure discussed above provides a proof of, and a means of achieving,orthogonality. The general method has already been described and we will notrepeat it here. We may also consider the properties of the eigenvalues and eigenvectors of an anti-Hermitian matrix, for which A †=−Aand thus AA†=A(−A)=(−A)A=A†A. Therefore matrices that are anti-Hermitian are also normal and so have mutu- ally orthogonal eigenvectors. The properties of the eigenvalues are also simply 282 8.13 EIGENVECTORS AND EIGENVALUES deduced, since if Ax=λxthen λ∗x=A†x=−Ax=−λx. Hence λ∗=−λand so λmust be pure imaginary (orzero). In a similar manner to that used for Hermitian matrices, these properties may be proved directly. 8.13.3 Eigenvectors and eigenvalues of a unitary matrix A unitary matrix satisﬁes A†=A−1and is also a normal matrix, with mutually orthogonal eigenvectors. To investigate the eigenvalues of a unitary matrix, we note that if Ax=λxthen x†x=x†A†Ax=λ∗λx†x, and we deduce that λλ∗=|λ|2= 1. Thus, the eigenvalues of a unitary matrix have unit modulus. 8.13.4 Eigenvectors and eigenvalues of a general square matrix When an N×Nmatrix is not normal there are no general properties of its eigenvalues and eigenvectors; in general it is not possible to ﬁnd any orthogonalset of Neigenvectors or even to ﬁnd pairsof orthogonal eigenvectors (except by chance in some cases). While the Nnon-orthogonal eigenvectors are usually linearly independent and hence form a basis for the N-dimensional vector space, this is not necessarily so. It may be shown (although we will not prove it) that anyN×Nmatrix with distinct eigenvalues has Nlinearly independent eigenvectors, which therefore form a basis for the N-dimensional vector space. If a general square matrix has degenerate eigenvalues, however, then it may or may not haveNlinearly independent eigenvectors. A matrix whose eigenvectors are not linearly independent is said to be defective . 8.13.5 Simultaneous eigenvectors We may now ask under what conditions two diﬀerent normal matrices can have a common set of eigenvectors. The result – that they do so if, and only if, theycommute – has profound signiﬁcance for the foundations of quantum mechanics. To prove this important result let Aand Bbe two N×Nnormal matrices and x ibe the ith eigenvector of Acorresponding to eigenvalue λi,i . e . Axi=λixifor i=1,2,... ,N. For the present we assume that the eigenvalues are all diﬀerent. (i) First suppose that Aand Bcommute. Now consider ABxi=BAxi=Bλixi=λiBxi, 283 MATRICES AND VECTOR SPACES where we have used the commutativity for the ﬁrst equality and the eigenvector property for the second. It follows that A(Bxi)=λi(Bxi) and thus that Bxiis an eigenvector of Acorresponding to eigenvalue λi. But the eigenvector solutions of (A−λiI)xi=0are unique to within a scale factor, and we therefore conclude that Bxi=µixi for some scale factor µi. However, this is just an eigenvector equation for Band shows that xiis an eigenvector of B, in addition to being an eigenvector of A.B y reversing the roles of Aand B, it also follows that every eigenvector of Bis an eigenvector of A. Thus the two sets of eigenvectors are identical. (ii) Now suppose that Aand Bhave all their eigenvectors in common, a typical one xisatisfying both Axi=λixiand Bxi=µixi. As the eigenvectors span the N-dimensional vector space, any arbitrary vector x in the space can be written as a linear combination of the eigenvectors, x=Nsummationdisplay i=1cixi. Now consider both ABx=ABNsummationdisplay i=1cixi=ANsummationdisplay i=1ciµixi=Nsummationdisplay i=1ciλiµixi, and BAx=BANsummationdisplay i=1cixi=BNsummationdisplay i=1ciλixi=Nsummationdisplay i=1ciµiλixi. It follows that ABxand BAxare the same for any arbitrary xand hence that (AB−BA)x=0 for all x.T h a ti s , Aand Bcommute . This completes the proof that a necessary and suﬃcient condition for two normal matrices to have a set of eigenvectors in common is that they commute.It should be noted that if an eigenvalue of A, say, is degenerate then not all of its possible sets of eigenvectors will also constitute a set of eigenvectors of B. However, provided that by taking linear combinations one set of joint eigenvectorscan be found, the proof is still valid and the result still holds. When extended to the case of Hermitian operators and continuous eigenfunc- tions (sections 17.2 and 17.3 the conn ection between commuting matrices and a set of common eigenvectors plays a fundamental role in the postulatory basis 284 8.14 DETERMINATION OF EIGENVALUES AND EIGENVECTORS of quatum mechanics. It draws the distinction between commuting and non- commuting observables and sets limits on how much information about a systemcan be known, even in principle, at any one time. 8.14 Determination of eigenvalues and eigenvectors The next step is to show how the eigenvalues and eigenvectors of a given N×N matrix Aare found. To do this we refer to (8.68) and as in (8.69) rewrite it as Ax−λIx=(A−λI)x=0. (8.85) The slight rearrangement used here is to write xasIx,w h e r e Iis the unit matrix of order N. The point of doing this is immediate since (8.85) now has the form of a homogeneous set of simultaneous equations, the theory of which will bedeveloped in section 8.18. What will be proved there is that the equation Bx=0 only has a non-trivial solution xif|B|= 0. Correspondingly, therefore, we must have in the present case that |A−λI|=0, (8.86) if there are to be non-zero solutions xto (8.85). Equation (8.86) is known as the characteristic equation forAa n di t sL H Sa s thecharacteristic orsecular determinant ofA. The equation is a polynomial of degree Nin the quantity λ.T h e Nroots of this equation λ i,i=1,2,...,N , give the eigenvalues of A. Corresponding to each λithere will be a column vector xi, which is the ith eigenvector of Aand can be found by using (8.68). It will be observed that when (8.86) is written out as a polynomial equation in λ, the coeﬃcient of −λN−1in the equation will be simply A11+A22+···+ANN relative to the coeﬃcient of λN. As discussed in section 8.8, the quantitysummationtextN i=1Aii is the traceofAand, from the ordinary theory of polynomial equations, will be equal to the sum of the roots of (8.86): Nsummationdisplay i=1λi=T r A. (8.87) This can be used as one check that a computation of the eigenvalues λihas been done correctly. Unless equation (8.87) is satisﬁed by a computed set of eigenvalues, they have not been calculated correctly. However, that equation (8.87) is satisﬁed is a necessary, but not suﬃcient, condition for a correct computation. An alternativeproof of (8.87) is given in section 8.16. 285 MATRICES AND VECTOR SPACESIFind the eigenvalues and normalised eige nvectors of the real symmetric matrix A= /0/@11 3 11−3 3−3−3 /1A. Using (8.86),//////1−λ13 11−λ−3 3−3−3−λ //////=0. Expanding out this determinant gives (1−λ)[(1−λ)(−3−λ)−(−3)(−3)]+1[(−3)(3)−1(−3−λ)] +3[1(−3)−(1−λ)(3)]=0, which simpliﬁes to give (1−λ)(λ2+2λ−12) + ( λ−6) + 3(3 λ−6) = 0 , ⇒ (λ−2)(λ−3)(λ+6 )=0 . Hence the roots of the characteristic equation, which are the eigenvalues of A,a r e λ1=2 , λ2=3 , λ3=−6. We note that, as expected, λ1+λ2+λ3=−1=1+1−3=A11+A22+A33=T r A. For the ﬁrst root, λ1= 2, a suitable eigenvector x1, with elements x1,x2,x3,m u s ts a t i s f y Ax1=2x1or, equivalently, x1+x2+3x3=2x1, x1+x2−3x3=2x2, (8.88) 3x1−3x2−3x3=2x3. These three equations are consistent (to ensure this was the purpose in ﬁnding the particular values of λ)a n dy i e l d x3=0 , x1=x2=k,w h e r e kis any non-zero number. A suitable eigenvector would thus be x1=(kk 0)T. If we apply the normalisation condition, we require k2+k2+02=1o r k=1/√ 2. Hence x1= /1√ 21√ 20 /T =1√ 2(110 )T. Repeating the last paragraph, but with the factor 2 on the RHS of (8.88) replaced successively by λ2=3a n d λ3=−6, gives two further normalised eigenvectors x2=1√ 3(1−11)T, x3=1√ 6(1−1−2)T. J In the above example, the three values of λare all diﬀerent and Ais a real symmetric matrix. Thus we expect, and it is easily checked, that the threeeigenvectors are mutually orthogonal, i.e. parenleftbig x 1parenrightbigTx2=parenleftbig x1parenrightbigTx3=parenleftbig x2parenrightbigTx3=0. It will be apparent also that, as expected, the normalisation of the eigenvectors has no eﬀect on their orthogonality. 286 8.14 DETERMINATION OF EIGENVALUES AND EIGENVECTORS 8.14.1 Degenerate eigenvalues We return now to the case of degenerate eigenvalues, i.e. those that have two or more associated eigenvectors. We have shown already that it is always possibleto construct an orthogonal set of eigenvectors for a normal matrix, see subsec-tion 8.13.1, and the following example illustrates one method for constructingsuch a set.IConstruct an orthonormal set of eigenvectors for the matrix A= /0/@103 0−20 301 /1A. We ﬁrst determine the eigenvalues using |A−λI|=0 : 0= //////1−λ 03 0−2−λ0 30 1 −λ //////=−(1−λ)2(2 +λ) + 3(3)(2 + λ) =( 4−λ)(λ+2 )2. Thus λ1=4 , λ2=−2=λ3. The eigenvector x1=(x1x2x3)Tis found from/0/@103 0−20 301 /1A /0/@x1 x2 x3 /1 A =4 /0 /@ x1 x2 x3 /1 A ⇒ x1=1√ 2 /0/@1 01 /1A. A general column vector that is orthogonal to x1is x=(ab−a)T, (8.89) and it is easily shown that Ax= /0/@103 0−20 301 /1A /0/@a b −a /1A=−2 /0/@a b −a /1A=−2x. Thus xis a eigenvector of Awith associated eigenvalue −2. It is clear, however, that there is an inﬁnite set of eigenvectors xall possessing the required property; the geometrical analogue is that there are an inﬁnite number of corresponding vectors xlying in the plane that has x1as its normal. We do require that the two remaining eigenvectors are orthogonal to one another, but this still le aves an inﬁnite number of possibilities. For x2, therefore, let us choose a simple form of (8.89), suitably normalised, say, x2=(010 )T. The third eigenvector is then speciﬁed (to within an arbitrary multiplicative constant) by the requirement that it must be orthogonal to x1and x2; thus x3may be found by evaluating the vector product of x1and x2and normalising the result. This gives x3=1√ 2(−101 )T, to complete the construction of an orthonormal set of eigenvectors. J 287 MATRICES AND VECTOR SPACES 8.15 Change of basis and similarity transformations Throughout this chapter we have considered the vector xas a geometrical quantity that is independent of any basis (or coordinate system). If we introduce a basise i,i=1,2,...,N , into our N-dimensional vector space then we may write x=x1e1+x2e2+···+xNeN, and represent xin this basis by the column matrix x=(x1x2···xn)T, having components xi. We now consider how these components change as a result of a prescribed change of basis. Let us introduce a new basis e/prime i,i=1,2,...,N , which is related to the old basis by e/prime j=Nsummationdisplay i=1Sijei, (8.90) the coeﬃcient Sijbeing the ith component of e/prime jwith respect to the old (unprimed) basis. For an arbitrary vector xit follows that x=Nsummationdisplay i=1xiei=Nsummationdisplay j=1x/prime je/primej=Nsummationdisplay j=1x/prime jNsummationdisplay i=1Sijei. From this we derive the relationship between the components of xin the two coordinate systems as xi=Nsummationdisplay j=1Sijx/prime j, w h i c hw ec a nw r i t ei nm a t r i xf o r ma s x=Sx/prime(8.91) where Sis the transformation matrix associated with the change of basis. Furthermore, since the vectors e/prime jare linearly independent, the matrix Sis non-singular and so possesses an inverse S−1. Multiplying (8.91) on the left by S−1we ﬁnd x/prime=S−1x, (8.92) which relates the components of xin the new basis to those in the old basis. Comparing (8.92) and (8.90) we note that the components of xtransform inversely to the way in which the basis vectors eithemselves transform. This has to be so, as the vector xitself must remain unchanged. We may also ﬁnd the transformation law for the components of a linear operator under the same change of basis. Now, the operator equation y=Ax 288 8.15 CHANGE OF BASIS AND SIMILARITY TRANSFORMATIONS (which is basis independent) can be written as a matrix equation in each of the two bases as y=Ax, y/prime=A/primex/prime. (8.93) But, using (8.91), we may rewrite the ﬁrst equation as Sy/prime=ASx/prime⇒ y/prime=S−1ASx/prime. Comparing this with the second equation in (8.93) we ﬁnd that the components of the linear operator Atransform as A/prime=S−1AS. (8.94) Equation (8.94) is an example of a similarity transformation – a transformation that can be particularly useful in converting matrices into convenient forms forcomputation. Given a square matrix A, we may interpret it as representing a linear operator Ain a given basis e i. From (8.94), however, we may also consider the matrix A/prime=S−1AS, for any non-singular matrix S, as representing the same linear operator Abut in a new basis e/prime j, related to the old basis by e/prime j=summationdisplay iSijei. Therefore we would expect that any property of the matrix Athat represents some (basis-independent) property of the linear operator Awill also be shared by the matrix A/prime. We list these properties below. (i) If A=Ithen A/prime=I, since, from (8.94), A/prime=S−1IS=S−1S=I. (8.95) (ii) The value of the determinant is unchanged: |A/prime|=|S−1AS|=|S−1||A||S|=|A||S−1||S|=|A||S−1S|=|A|.(8.96) (iii) The characteristic determinant and hence the eigenvalues of A/primeare the same as those of A: from (8.86), |A/prime−λI|=|S−1AS−λI|=|S−1(A−λI)S| =|S−1||S||A−λI|=|A−λI|. (8.97) (iv) The value of the trace is unchanged: from (8.87), TrA/prime=summationdisplay iA/prime ii=summationdisplay isummationdisplay jsummationdisplay k(S−1)ijAjkSki =summationdisplay isummationdisplay jsummationdisplay kSki(S−1)ijAjk=summationdisplay jsummationdisplay kδkjAjk=summationdisplay jAjj =T r A. (8.98) 289 MATRICES AND VECTOR SPACES An important class of similarity transformations is that for which Sis a uni- tary matrix; in this case A/prime=S−1AS=S†AS. Unitary transformation matrices are particularly important, for the following reason. If the original basis eiis orthonormal and the transformation matrix Sis unitary then /angbracketlefte/prime i|e/prime j/angbracketright=angbracketleftBigsummationdisplay kSkiekvextendsinglevextendsinglevextendsinglesummationdisplay rSrjerangbracketrightBig =summationdisplay kS∗ kisummationdisplay rSrj/angbracketleftek|er/angbracketright =summationdisplay kS∗ kisummationdisplay rSrjδkr=summationdisplay kS∗ kiSkj=(S†S)ij=δij, showing that the new basis is also orthonormal. Furthermore, in addition to the properties of general similarity transformations, for unitary transformations the following hold. (i) If Ais Hermitian (anti-Hermitian) then A/primeis Hermitian (anti-Hermitian), i.e. if A†=±Athen (A/prime)†=(S†AS)†=S†A†S=±S†AS=±A/prime. (8.99) (ii) If Ais unitary (so that A†=A−1)t h e n A/primeis unitary, since (A/prime)†A/prime=(S†AS)†(S†AS)=S†A†SS†AS=S†A†AS =S†IS=I. (8.100) 8.16 Diagonalisation of matrices Suppose that a linear operator Ais represented in some basis ei,i=1,2,...,N , by the matrix A. Consider a new basis xjgiven by xj=Nsummationdisplay i=1Sijei, where the xjare chosen to be the eigenvectors of the linear operator A,i . e . Axj=λjxj. (8.101) In the new basis, Ais represented by the matrix A/prime=S−1AS, which has a particularly simple form, as we shall see shortly. The element SijofSis the ith component, in the old (unprimed) basis, of the jth eigenvector xjofA,i . e .t h e columns of Sare the eigenvectors of the matrix A: S= ↑↑ ↑ x1x2··· xN ↓↓ ↓ , 290 8.16 DIAGONALISATION OF MATRICES That is Sij=(xj)i. Therefore A/primeis given by (S−1AS)ij=summationdisplay ksummationdisplay l(S−1)ikAklSlj =summationdisplay ksummationdisplay l(S−1)ikAkl(xj)l =summationdisplay k(S−1)ikλj(xj)k =summationdisplay kλj(S−1)ikSkj=λjδij. So the matrix A/primeis diagonal with the eigenvalues of Aas the diagonal elements, i.e. A/prime= λ10··· 0 0λ2... ......0 0··· 0λN . Therefore, given a matrix A,i fw ec o n s t r u c tt h em a t r i x Sthat has the eigen- vectors of Aas its columns then the matrix A/prime=S−1ASis diagonal and has the eigenvalues of Aas its diagonal elements. Since we require Sto be non-singular (|S|/negationslash= 0), the Neigenvectors of Amust be linearly independent and form a basis for the N-dimensional vector space. It may be shown that any matrix with distinct eigenvalues can be diagonalised by this procedure. If, however, a general square matrix has degenerate eigenvalues then it may, or may not, have Nlinearly independent eigenvectors. If it does not then it cannot be diagonalised. For normal matrices (which include Hermitian, anti-Hermitian and unitary matrices) the Neigenvectors are indeed linearly independent. Moreover, when normalised, these eigenvectors form an orthonormal s e t( o rc a nb em a d et od o so). Therefore the matrix Swith these normalised eigenvectors as columns, i.e. whose elements are Sij=(xj)i,h a st h ep r o p e r t y (S†S)ij=summationdisplay k(S†)ik(S)kj=summationdisplay kS∗ kiSkj=summationdisplay k(xi)∗ k(xj)k=(xi)†xj=δij. Hence Sis unitary ( S−1=S†) and the original matrix Acan be diagonalised by A/prime=S−1AS=S†AS. Therefore, any normal matrix Acan be diagonalised by a similarity transformation using a unitary transformation matrix S. 291 MATRICES AND VECTOR SPACESIDiagonalise the matrix A= /0/@103 0−20 301 /1A. The matrix Ais symmetric and so may be diagonalised by a transformation of the form A/prime=S†AS,w h e r e Shas the normalised eigenvectors of Aas its columns. We have already found these eigenvectors in subsection 8. 14.1, and so we can write straightaway S=1√ 2 /0/@10−1 0√ 20 10 1 /1A. We note that although the eigenvalues of Aare degenerate, its three eigenvectors are linearly independent and so Acan still be diagonalised. Thus, calculating S†ASwe obtain S†AS=1 2 /0/@10 1 0√ 20 −101 /1A /0/@103 0−20 301 /1A /0/@10−1 0√ 20 10 1 /1A = /0/@40 0 0−20 00−2 /1A, which is diagonal, as required, and has as its diagonal elements the eigenvalues of A. J If a matrix Ais diagonalised by the similarity transformation A/prime=S−1AS,s o that A/prime=d i a g ( λ1,λ2...,λ N), then we have immediately TrA/prime=T r A=Nsummationdisplay i=1λi, (8.102) |A/prime|=|A|=Nproductdisplay i=1λi, (8.103) since the eigenvalues of the matrix are unchanged by the transformation. More- over, these results may be used to prove the rather useful trace formula |exp A|=e x p ( T r A), (8.104) where the exponential of a matrix is as deﬁned in (8.38).IProve the trace formula (8.104). At the outset, we note that for the similarity transformation A/prime=S−1AS, we have (A/prime)n=(S−1AS)(S−1AS)···(S−1AS)=S−1AnS. Thus, from (8.38), we obtain exp A/prime=S−1(exp A)S, from which it follows that |exp A/prime|= 292 8.17 QUADRATIC AND HERMITIAN FORMS |exp A|. Moreover, by choosing the similarity transformation so that it diagonalises A,w e have A/prime= diag( λ1,λ2,...,λ N), and so |expA|=|expA/prime|=|exp[diag( λ1,λ2,...,λ N)]|=|diag(exp λ1,expλ2,...,expλN)|=NY i=1expλi. Rewriting the ﬁnal product of exponentials of the eigenvalues as the exponential of the sum of the eigenvalues, we ﬁnd |exp A|=NY i=1expλi=e x p / NX i=1λi /! =e x p ( T r A), which gives the trace formula (8.104). J 8.17 Quadratic and Hermitian forms Let us now introduce the concept of quadratic forms (and their complex ana- logues, Hermitian forms). A quadratic form Qis a scalar function of a real vector xgiven by Q(x)=/angbracketleftx|Ax/angbracketright, (8.105) for some real linear operator A. In any given basis (coordinate system) we can write (8.105) in matrix form as Q(x)=xTAx, (8.106) where Ais a real matrix. In fact, as will be explained below, we need only consider t h ec a s ew h e r e Ais symmetric, i.e. A=AT. As an example in a three-dimensional space, Q=xTAx=parenleftBig x1x2x3parenrightBig 11 3 11−3 3−3−3  x1 x2 x3  =x2 1+x2 2−3x2 3+2x1x2+6x1x3−6x2x3. (8.107) It is reasonable to ask whether a quadratic form Q=xTMx,w h e r e Mis any (possibly non-symmetric) real square matrix, is a more general deﬁnition. Thatthis not the case may be seen by expressing Min terms of a symmetric matrix A= 1 2(M+MT) and an antisymmetric matrix B=1 2(M−MT) such that M=A+B. We then have Q=xTMx=xTAx+xTBx. (8.108) However, Qis a scalar quantity and so Q=QT=(xTAx)T+(xTBx)T=xTATx+xTBTx=xTAx−xTBx. (8.109) 293 MATRICES AND VECTOR SPACES Comparing (8.108) and (8.109) shows that xTBx= 0, and hence xTMx=xTAx, i.e.Qis unchanged by considering only the symmetric part of M. Hence, with no loss of generality, we may assume A=ATin (8.106). From its deﬁnition (8.105), Qis clearly a basis- (i.e. coordinate-) independent quantity. Let us therefore consider a new basis related to the old one by an orthogonal transformation matrix S, the components in the two bases of any vector xbeing related (as in (8.91)) by x=Sx/primeor, equivalently, by x/prime=S−1x= STx. We then have Q=xTAx=(x/prime)TSTASx/prime=(x/prime)TA/primex/prime, where (as expected) the matrix describing the linear operator Ain the new basis is given by A/prime=STAS(since ST=S−1). But, from the last section, if we choose as Sthe matrix whose columns are the normalised eigenvectors of Athen A/prime=STASis diagonal with the eigenvalues of Aas the diagonal elements. (Since Ais symmetric, its normalised eigenvectors are orthogonal, or can be made so, and hence Sis orthogonal with S−1=ST.) In the new basis Q=xTAx=(x/prime)TΛx/prime=λ1x/prime 12+λ2x/prime 22+···+λNx/prime N2, (8.110) where Λ = diag( λ1,λ2,...,λ N)a n dt h e λiare the eigenvalues of A. It should be noted that Qcontains no cross-terms of the form x/prime 1x/prime2.IFind an orthogonal transformation that takes the quadratic form (8.107) into the form λ1x/prime 12+λ2x/prime 22+λ3x/prime 32. The required transformation matrix Shas the normalised eigenvectors of Aas its columns. We have already found these in section 8.14, and so we can write immediately S=1√ 6 /0/@√ 3√ 21√ 3−√ 2−1 0√ 2−2 /1A, which is easily veriﬁed as being orthogonal. Since the eigenvalues of Aareλ=2 ,3 ,a n d −6, the general result already proved shows that the transformation x=Sx/primewill carry (8.107) into the form 2 x/prime 12+3x/prime 22−6x/prime 32.This may be veriﬁed most easily by writing out the inverse transformation x/prime=S−1x=STxand substituting. The inverse equations are x/prime 1=(x1+x2)/√ 2, x/prime 2=(x1−x2+x3)/√ 3, (8.111) x/prime 3=(x1−x2−2x3)/√ 6. If these are substituted into the form Q=2x/prime 12+3x/prime 22−6x/prime 32then the original expression (8.107) is recovered. J In the deﬁnition of Qit was assumed that the components x1,x2,x3and the matrix Awere real. It is clear that in this case the quadratic form Q≡xTAxis real 294 8.17 QUADRATIC AND HERMITIAN FORMS also. Another, rather more general, expression that is also real is the Hermitian form H(x)≡x†Ax, (8.112) where Ais Hermitian (i.e. A†=A) and the components of xmay now be complex. It is straightforward to show that His real, since H∗=(HT)∗=x†A†x=x†Ax=H. With suitable generalisation, the properties of quadratic forms apply also to Her- mitian forms, but to keep the presentation simple we will restrict our discussionto quadratic forms. A special case of a quadratic (Hermitian) form is one for which Q=x TAx is greater than zero for all column matrices x. By choosing as the basis the eigenvectors of Awe have Qin the form Q=λ1x2 1+λ2x2 2+λ3x2 3. The requirement that Q>0f o ra l l xmeans that all the eigenvalues λiofAmust be positive. A symmetric (Hermitian) matrix Awith this property is called positive deﬁnite .I f ,i n s t e a d , Q≥0f o ra l l xthen it is possible that some of the eigenvalues are zero, and Ais called positive semi-deﬁnite . 8.17.1 The stationary properties of the eigenvectors Consider a quadratic form, such as Q(x)=/angbracketleftx|Ax/angbracketrightgiven in (8.105), in a ﬁxed basis. As the vector xis varied, through changes in its three components x1,x2 andx3, the value of the quantity Qalso varies. Because of the homogeneous form of Qwe may restrict any investigation of these variations to vectors of unit length (since multiplying any vector xby any scalar ksimply multiplies the value ofQby a factor k2). Of particular interest are any vectors xthat make the value of the quadratic form a maximum or minimum. A necessary, but not suﬃcient, condition for this is that Qis stationary with respect to small variations ∆ xinx, whilst/angbracketleftx|x/angbracketrightis maintained at a constant value (unity). In the chosen basis the quadratic form is given by Q=xTAxand, using Lagrange undetermined multipliers to incorporate the variational constraints, weare led to seek solutions of ∆[x TAx−λ(xTx−1)] = 0 . (8.113) This may be used directly, together with the fact that (∆ xT)Ax=xTA∆x,s i n c e A is symmetric, to obtain Ax=λx, (8.114) 295 MATRICES AND VECTOR SPACES as the necessary condition that xmust satisfy. If (8.114) is satisﬁed for some eigenvector xthen the value of Q(x)i sg i v e nb y Q=xTAx=xTλx=λ. (8.115) However, if xand yare eigenvectors corresponding to diﬀerent eigenvalues then they are (or can be chosen to be) orthogonal. Consequently the expression yTAx is necessarily zero, since yTAx=yTλx=λyTx=0. (8.116) Summarising, those column matrices xof unit magnitude that make the quadratic form Qstationary are eigenvectors of the matrix A, and the stationary value of Qis then equal to the corresponding eigenvalue. It is straightforward to see from the proof of (8.114) that, conversely, any eigenvector of Amakes Q stationary. Instead of maximising or minimising Q=xTAxsubject to the constraint xTx= 1, an equivalent procedure is to extremise the function λ(x)=xTAx xTx.IShow that if λ(x)is stationary then xis an eigenvector of Aandλ(x)is equal to the corresponding eigenvalue. We require ∆ λ(x) = 0 with respect to small variations in x.N o w ∆λ=1 (xTx)2 / (xTx) /; ∆xTAx+xTA∆x / −xTAx /; ∆xTx+xT∆x // =2∆xTAx xTx−2 /xTAx xTx /∆xTx xTx, since xTA∆x=( ∆ xT)Axand xT∆x=( ∆ xT)x. Thus ∆λ=2 xTx∆xT[Ax−λ(x)x]. Hence, if ∆ λ=0t h e n Ax=λ(x)x,i . e . xis an eigenvector of Awith eigenvalue λ(x). J Thus the eigenvalues of a symmetric matrix Aare the values of the function λ(x)=xTAx xTx at its stationary points. The eigenvectors of Alie along those directions in space for which the quadratic form Q=xTAxhas stationary values, given a ﬁxed magnitude for the vector x. Similar results hold for Hermitian matrices. 296 8.18 SIMULTANEOUS LINEAR EQUATIONS 8.17.2 Quadratic surfaces The results of the previous subsection may be turned round to state that the surface given by xTAx= constant = 1 (say) (8.117) and called a quadratic surface , has stationary values of its radius (i.e. origin– surface distance) in those directions that are along the eigenvectors of A.M o r e speciﬁcally, in three dimensions the quadratic surface xTAx= 1 has its principal axes along the three mutually perpendicular eigenvectors of A,a n dt h es q u a r e s of the corresponding principal radii are given by λ−1 i,i=1,2,3. As well as having this stationary property of the radius, a principal axis is characterised bythe fact that any section of the surface perpendicular to it has some degree ofsymmetry about it. If the eigenvalues corresponding to any two principal axes aredegenerate then the quadratic surface has rotational symmetry about the thirdprincipal axis and the choice of a pair of axes perpendicular to that axis is notuniquely deﬁned.IFind the shape of the quadratic surface x2 1+x2 2−3x2 3+2x1x2+6x1x3−6x2x3=1. If, instead of expressing the quadratic surface in terms of x1,x2,x3, as in (8.107), we were to use the new variables x/prime 1,x/prime 2,x/prime 3deﬁned in (8.111), for which the coordinate axes are along the three mutually perpendicular eigenvector directions (1 ,1,0), (1 ,−1,1) and (1,−1,−2), then the equation of the surface would take the form (see (8.110)) x/prime 12 (1/√ 2)2+x/prime 22 (1/√ 3)2−x/prime 32 (1/√ 6)2=1. Thus, for example, a section of the quadratic surface in the plane x/prime 3=0 ,i . e . x1−x2− 2x3= 0, is an ellipse, with semi-axes 1 /√ 2a n d1 /√ 3. Similarly a section in the plane x/prime 1=x1+x2= 0 is a hyperbola. J Clearly the simplest three-dimensional situation to visualise is that in which all the eigenvalues are positive, since then the quadratic surface is an ellipsoid. 8.18 Simultaneous linear equations In physical applications we often encounter sets of simultaneous linear equations. In general we may have Mequations in Nunknowns x1,x2,...,x Nof the form A11x1+A12x2+···+A1NxN=b1, A21x1+A22x2+···+A2NxN=b2, ... AM1x1+AM2x2+···+AMNxN=bM,(8.118) 297 MATRICES AND VECTOR SPACES where the Aijandbihave known values. If all the biare zero then the system of equations is called homogeneous , otherwise it is inhomogeneous . Depending on the given values, this set of equations for the Nunknowns x1,x2,...,xNmay have either a unique solution, no solution or inﬁnitely many solutions. Matrix analysis may be used to distinguish between the possibilities. The set of equations may be expressed as a single matrix equation Ax=b, or, written out in full, as  A11 A12... A 1N A21 A22... A 2N ............ AM1AM2... A MN  x1 x2 ... xN = b 1 b2 ... bM . 8.18.1 The range and null space of a matrix As we discussed in section 8.2, we may interpret the matrix equation Ax=bas representing, in some basis, the linear transformation Ax=bof a vector xin an N-dimensional vector space Vinto a vector bin some other (in general diﬀerent) M-dimensional vector space W. In general the operator Awill map anyvector in Vinto some particular subspace ofW, which may be the entire space. This subspace is called the range ofA(orA) and its dimension is equal to the rankofA.M o r e o v e r ,i f A(and hence A)i ssingular then there exists some subspace of Vthat is mapped onto the zero vector 0inW; that is, any vector ythat lies in the subspace satisﬁes Ay=0. This subspace is called the null space ofAand the dimension of this null space is called the nullity ofA. We note that the matrix Amustbe singular ifM/negationslash=Nandmaybe singular even if M=N. The dimensions of the range and the null space of a matrix are related through the fundamental relationship rank A+ nullity A=N, (8.119) where Nis the number of original unknowns x 1,x2,...,x N.IProve the relationship (8.119). As discussed in section 8.11, if the columns of an M×Nmatrix Aare interpreted as the components, in a given basis, of N(M-component) vectors v1,v2,...,vNthen rank Ais equal to the number of linearly independent vectors in this set (this number is also equalto the dimension of the vector space spanned by these vectors). Writing (8.118) in termsof the vectors v 1,v2,...,vN, we have x1v1+x2v2+···+xNvN=b. (8.120) From this expression, we immediately deduce that the range of Ais merely the span of the vectors v1,v2,...,vNand hence has dimension r=r a n k A. 298 8.18 SIMULTANEOUS LINEAR EQUATIONS If a vector ylies in the null space of AthenAy=0,w h i c hw em a yw r i t ea s y1v1+y2v2+···+yNvN=0. (8.121) As just shown above, however, only r(≤N) of these vectors are linearly independent. By renumbering, if necessary, we may assume that v1,v2,...,vrform a linearly independent set; the remaining vectors, vr+1,vr+2,...,vN, can then be written as a linear superposition ofv1,v2,...,vr.W ea r et h e r e f o r ef r e et oc h o o s et h e N−rcoeﬃcients yr+1,yr+2,...,y N arbitrarily and (8.121) will still be satisﬁed for some set of rcoeﬃcients y1,y2,...,y r(which are not all zero). The dimension of the null space is therefore N−r, and this completes the proof of (8.119). J Equation (8.119) has far-reaching consequences for the existence of solutions to sets of simultaneous linear equations such as (8.118). As mentioned previously,these equations may have no solution ,aunique solution orinﬁnitely many solutions . We now discuss these three cases in turn. No solution The system of equations possesses no solution unless blies in the range of A;i n this case (8.120) will be satisﬁed for some x 1,x2,...,x N. This in turn requires the s e to fv e c t o r s b,v1,v2,...,vNto have the same span (see (8.8)) as v1,v2,...,vN.I n terms of matrices, this is equivalent to the requirement that the matrix Aand the augmented matrix M= A 11 A12... A 1Nb1 A21 A22... A 2Nb1 ......... AM1AM2... A MN bM  have the samerank r. If this condition is satisﬁed then bdoes lie in the range of A, and the set of equations (8.118) will have either a unique solution or inﬁnitely many solutions. If, however, Aand Mhave diﬀerent ranks then there will be no solution. A unique solution Ifblies in the range of Aand if r=Nthen all the vectors v 1,v2,...,vNin (8.120) are linearly independent and the equation has a unique solution x1,x2,...,x N. Inﬁnitely many solutions Ifblies in the range of Aand if r<N then only rof the vectors v1,v2,...,vN in (8.120) are linearly independent. We may therefore choose the coeﬃcients of n−rvectors in an arbitrary way, while still satisfying (8.120) for some set of coeﬃcients x1,x2,...,x N. There are therefore inﬁnitely many solutions ,w h i c hs p a n an (n−r) dimensional vector space. We may also consider this space of solutions in terms of the null space of A:i fxis some vector satisfying Ax=bandyis 299 MATRICES AND VECTOR SPACES anyvector in the null space of A(i.e.Ay=0)t h e n A(x+y)=Ax+Ay=Ax+0=b, and so x+yis also a solution. Since the null space is ( n−r)-dimensional, so too is the space of solutions. We may use the above results to investigate the special case of the solution of ahomogeneous set of linear equations, for which b=0. Clearly the set always has the trivial solution x1=x2=···=xn=0 ,a n di f r=Nthis will be the only solution. If r<N , however, there are inﬁnitely many solutions; they form the null space of A, which has dimension n−r. In particular, we note that if M<N (i.e. there are fewer equations than unknowns) then r<N automatically. Hence a set of homogeneous linear equations with fewer equations than unknowns always has inﬁnitely many solutions. 8.18.2 Nsimultaneous linear equations in Nunknowns A special case of (8.118) occurs when M=N. In this case the matrix Aissquare and we have the same number of equations as unknowns. Since Ais square, the condition r=Ncorresponds to |A|/negationslash= 0 and the matrix Aisnon-singular .T h e caser<N corresponds to |A|=0 ,i nw h i c hc a s e Aissingular . As mentioned above, the equations will have a solution provided blies in the range of A. If this is true then the equations will possess a unique solution when |A|/negationslash= 0 or inﬁnitely many solutions when |A|= 0. There exist several methods for obtaining the solution(s). Perhaps the most elementary method is Gaussian elimination ; this method is discussed in subsection 28.3.1, where we also address numerical subtleties such as equation interchange (pivoting). In this subsection,we will outline three further methods for solving a square set of simultaneouslinear equations. Direct inversion Since Ais square it will possess an inverse, provided |A|/negationslash= 0. Thus, if Ais non-singular, we immediately obtain x=A −1b (8.122) as the unique solution to the set of equations. However, if b=0,t h e nw es e e immediately that the set of equations possesses only the trivial solution x=0.T h e direct inversion method has the advantage that, once A−1has been calculated, one may obtain the solutions xcorresponding to diﬀerent vectors b1,b2, ...on the RHS, with little further work. 300 8.18 SIMULTANEOUS LINEAR EQUATIONSIShow that the set of simultaneous equations 2x1+4x2+3x3=4, x1−2x2−2x3=0, (8.123) −3x1+3x2+2x3=−7, has a unique solution, and ﬁnd that solution. The simultaneous equations can be represented by the matrix equation Ax=b,i . e ./0/@243 1−2−2 −33 2 /1A /0/@x1 x2 x3 /1 A = /0 /@ 4 0 −7 /1A. As we have already shown that A−1exists and have calculated it, see (8.59), it follows that x=A−1bor, more explicitly, that/0/@x1 x2 x3 /1 A =1 11 /0/@21−2 41 37 −3−18−8 /1A /0/@4 0 −7 /1A= /0 /@ 2 −3 4 /1A. (8.124) Thus the unique solution is x1=2 , x2=−3,x3=4 . J LU decomposition Although conceptually simple, ﬁnding the solution by calculating A−1can be computationally demanding, especially when Nis large. In fact, as we shall now show, it is not necessary to perform the full inversion of Ain order to solve the simultaneous equations Ax=b. Rather, we can perform a decomposition of the matrix into the product of a square lower triangular matrix Land a square upper triangular matrix U, which are such that A=LU, (8.125) and then use the fact that triangular systems of equations can be solved very simply. We must begin, therefore, by ﬁnding the matrices Land Usuch that (8.125) is satisﬁed. This may be achieved straightforwardly by writing out (8.125) incomponent form. For illustration, let us consider the 3 ×3 case. It is, in fact, always possible, and convenient, to take the diagonal elements of Las unity, so we have A= 10 0 L 2110 L31L321  U11U12U13 0U22U23 00 U33  = U11 U12 U13 L21U11 L21U12+U22 L21U13+U23 L31U11L31U12+L32U22L31U13+L32U23+U33 (8.126) The nine unknown elements of Land Ucan now be determined by equating 301 MATRICES AND VECTOR SPACES the nine elements of (8.126) to those of the 3 ×3m a t r i x A. This is done in the particular order illustrated in the example below. Once the matrices Land Uhave been determined, one can use the decomposition to solve the set of equations Ax=bin the following way. From (8.125), we have LUx=b, but this can be written as twotriangular sets of equations Ly=b and Ux=y, where yis another column matrix to be determined. One may easily solve the ﬁrst triangular set of equations for y, which is then substituted into the second set. The required solution xis then obtained readily from the second triangular set of equations. We note that, as with direct inversion, once the LUdecomposition has been determined, one can solve for various RHS column matrices b1,b2,..., with little extra work.IUseLUdecomposition to solve the set of simultaneous equations (8.123). We begin the determination of the matrices Land Uby equating the elements of the matrix in (8.126) with those of the matrix A= /0/@243 1−2−2 −33 2 /1A. This is performed in the following order: 1st row: U11=2 , U12=4 , U13=3 1st column: L21U11=1 , L31U11=−3⇒L21=1 2,L31=−3 2 2nd row: L21U12+U22=−2 L21U13+U23=−2⇒U22=−4,U23=−7 2 2nd column: L31U12+L32U22=3 ⇒L32=−9 4 3rd row: L31U13+L32U23+U33=2 ⇒U33=−11 8 Thus we may write the matrix Aas A=LU= /0/@10 0 1 210 −3 2−9 41 /1A /0/@24 3 0−4−7 2 00−11 8 /1A. We must now solve the set of equations Ly=b,w h i c hr e a d/0/@10 0 1 210 −3 2−9 41 /1A /0/@y1 y2 y3 /1 A = /0 /@ 4 0 −7 /1A. Since this set of equations is triangular, we quickly ﬁnd y1=4,y 2=0−(1 2)(4) =−2,y 3=−7−(−3 2)(4)−(−9 4)(−2) =−11 2. These values must then be substituted into the equations Ux=y,w h i c hr e a d/0/@24 3 0−4−7 2 00−11 8 /1A /0/@x1 x2 x3 /1 A = /0 /@ 4 −2 −11 2 /1A. 302 8.18 SIMULTANEOUS LINEAR EQUATIONS This set of equations is also triangular, and we easily ﬁnd the solution x1=2,x 2=−3,x 3=4, which agrees with the result found above by direct inversion. J We note, in passing, that one can calculate both the inverse and the determinant ofAfrom its LUdecomposition. To ﬁnd the inverse A−1, one solves the system of equations Ax=brepeatedly for the Ndiﬀerent RHS column matrices b=ei (i=1,2,...,N ), where eiis the column matrix with its ith element equal to unity and the others equal to zero. The solution xin each case gives the corresponding column of A−1. Evaluation of the determinant |A|is much simpler. From (8.125), we have |A|=|LU|=|L||U|. (8.127) Since Land Uare triangular, however, we see from (8.64) that their determinants are equal to the products of their diagonal elements. Since Lii=1f o ra l l i,w e thus ﬁnd |A|=U11U22···UNN=Nproductdisplay i=1Uii. As an illustration, in the above example we ﬁnd |A|=( 2 ) (−4)(−11/8) = 11, which, as it must, agrees with our earlier calculation (8.58). Finally, we note that if the matrix Ais symmetric and positive semi-deﬁnite then we can decompose it as A=LL†, (8.128) where Lis a lower triangular matrix whose diagonal elements are not,i ng e n e r a l , equal to unity. This is known as a Cholesky decomposition (in the special case where Ais real, the decomposition becomes A=LLT). The reason that we cannot set the diagonal elements of Lequal to unity in this case is that we require the same number of independent elements in Las in A. The requirement that the matrix be positive semi-deﬁnite is easily derived by considering the Hermitian form (or quadratic form in the real case) x†Ax=x†LL†x=(L†x)†(L†x). Denoting the column matrix L†xbyy, we see that the last term on the RHS isy†y, which must be greater than or equal to zero. Thus, we require x†Ax≥0 for any arbitrary column matrix x,a n ds o Amust be positive semi-deﬁnite (see section 8.17). We recall that the requirement that a matrix be positive semi-deﬁnite is equiv- alent to demanding that all the eigenvalues of Aare positive or zero. If one of the eigenvalues of Ais zero, however, then from (8.103) we have |A|=0a n ds o A issingular . Thus, if Ais a non-singular matrix, it must be positive deﬁnite (rather 303 MATRICES AND VECTOR SPACES than just positive semi-deﬁnite) in order to perform the Cholesky decomposition (8.128). In fact, in this case, the inability to ﬁnd a matrix Lthat satisﬁes (8.128) implies that Acannot be positive deﬁnite. The Cholesky decomposition can be applied in an analogous way to the LU decomposition discussed above, but we shall not explore it further. Cramer’s rule An alternative method of solution is to use Cramer’s rule , which also provides some insight into the nature of the solutions in the various cases. To illustratethis method let us consider a set of three equations in three unknowns, A 11x1+A12x2+A13x3=b1, A21x1+A22x2+A23x3=b2, (8.129) A31x1+A32x2+A33x3=b3, which may be represented by the matrix equation Ax=b. We wish either to ﬁnd the solution(s) xto these equations or to establish that there are no solutions. From result (vi) of subsection 8.9.1, the determinant |A|is unchanged by adding to its ﬁrst column the combination x2 x1×(second column of |A|)+x3 x1×(third column of |A|). We thus obtain |A|=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleA 11A12A13 A21A22A23 A31A32A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleA 11+(x2/x1)A12+(x3/x1)A13A12A13 A21+(x2/x1)A22+(x3/x1)A23A22A23 A31+(x2/x1)A32+(x3/x1)A33A32A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle, which, on substituting b i/x1for the ith entry in the ﬁrst column, yields |A|=1 x1vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleb 1A12A13 b2A22A23 b3A32A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=1 x1∆1. The determinant ∆ 1is known as a Cramer determinant . Similar manipulations of the second and third columns of |A|yield x2andx3, and so the full set of results reads x1=∆1 |A|,x 2=∆2 |A|,x 3=∆3 |A|, (8.130) where ∆1=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleb 1A12A13 b2A22A23 b3A32A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle,∆ 2=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleA 11b1A13 A21b2A23 A31b3A33vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle,∆ 3=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleA 11A12b1 A21A22b2 A31A32b3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. It can be seen that each Cramer determinant ∆ iis simply|A|but with column i replaced by the RHS of the original set of equations. If |A|/negationslash= 0 then (8.130) gives 304 8.18 SIMULTANEOUS LINEAR EQUATIONS the unique solution. The proof given here appears to fail if any of the solutions xiis zero, but it can be shown that result (8.130) is valid even in such a case.IUse Cramer’s rule to solve the set of simultaneous equations (8.123). Let us again represent these simultaneous equations by the matrix equation Ax=b,i . e ./0/@243 1−2−2 −33 2 /1A /0/@x1 x2 x3 /1 A = /0 /@ 4 0 −7 /1A. From (8.58), the determinant of Ais given by |A|= 11. Following the discussion given above, the three Cramer determinants are ∆1= //////443 0−2−2 −73 2 //////,∆2= //////243 10−2 −3−72 //////,∆3= //////244 1−20 −33−7 //////. These may be evaluated using the properties of determinants listed in subsection 8.9.1 and we ﬁnd ∆ 1= 22, ∆ 2=−33 and ∆ 3= 44. From (8.130) the solution to the equations (8.123) is given by x1=22 11=2,x 2=−33 11=−3,x 3=44 11=4, which agrees with the solution found in the previous example. J At this point it is useful to consider each of the three equations (8.129) as rep- resenting a plane in three-dimensional Cartesian coordinates. Using result (7.42) of chapter 7, the sets of components of the vectors normal to the planes are(A 11,A12,A13), (A21,A22,A23)a n d( A31,A32,A33), and using (7.46) the perpendic- ular distances of the planes from the origin are given by di=biparenleftbig A2 i1+A2 i2+A2 i3parenrightbig1/2fori=1,2,3. Finding the solution(s) to the simultaneous equations above corresponds to ﬁnding the point(s) of intersection of the planes. If there is a unique solution the planes intersect at only a single point. This happens if their normals are linearly independent vectors. Since the rows of A represent the directions of these normals, this requirement is equivalent to |A|/negationslash=0 . Ifb=(000 )T=0then all the planes pass through the origin and, since there is only a single solution to the equations, the origin is that solution. Let us now turn to the cases where |A|= 0. The simplest such case is that in which all three planes are parallel; this implies that the normals are all paralleland so Ais of rank 1. Two possibilities exist: (i) the planes are coincident, i.e. d 1=d2=d3,i nw h i c hc a s et h e r ei sa n inﬁnity of solutions; (ii) the planes are not all coincident, i.e. d1/negationslash=d2and/or d1/negationslash=d3and/or d2/negationslash=d3, in which case there are no solutions. 305 MATRICES AND VECTOR SPACES (a) (b) Figure 8.1 The two possible cases when Ais of rank 2. In both cases all the normals lie in a horizontal plane but in ( a) the planes all intersect on a single line (corresponding to an inﬁnite number of solutions) whilst in ( b)t h e r ea r e no common intersection points (no solutions). It is apparent from (8.130) that case (i) occurs when all the Cramer determinants are zero and case (ii) occurs when at least one Cramer determinant is non-zero. The most complicated cases with |A|= 0 are those in which the normals to the planes themselves lie in a plane but are not parallel. In this case Ahas rank 2. Again two possibilities exist and these are shown in ﬁgure 8.1. Just as in therank-1 case, if all the Cramer determinants are zero then we get an inﬁnity ofsolutions (this time on a line). Of course, in the special case in which b=0(and the system of equations is homogeneous), the planes all pass through the originand so they must intersect on a line through it. If at least one of the Cramer determinants is non-zero, we get no solution. These rules may be summarised as follows. (i)|A|/negationslash=0 , b/negationslash=0: The three planes intersect at a single point that is not the origin, and so there is only one solution, given by both (8.122) and (8.130). (ii)|A|/negationslash=0 , b=0: The three planes intersect at the origin only and there is only the trivial solution, x=0 . (iii)|A|=0 , b/negationslash=0, Cramer determinants all zero: There is an inﬁnity of solutions either on a line if Ais rank 2, i.e. the cofactors are not all zero, or on a plane if Ais rank 1, i.e. the cofactors are all zero. (iv)|A|=0 , b/negationslash=0, Cramer determinants not all zero: No solutions. (v)|A|=0 , b=0: The three planes intersect on a line through the origin giving an inﬁnity of solutions. 8.18.3 Singular value decomposition There exists a very powerful technique for dealing with a simultaneous set of linear equations Ax=b, such as (8.118), which may be applied whether or not 306 8.18 SIMULTANEOUS LINEAR EQUATIONS the number of simultaneous equations Mis equal to the number of unknowns N. This technique is known as singular value decomposition (SVD) and is the method of choice in analysing anyset of simultaneous linear equations. We will consider the general case, in which Ais an M×N(complex) matrix. Let us suppose we can write Aas the product § A=USV†, (8.131) where the matrices U,Sand Vhave the following properties. (i) The square matrix Uhas dimensions M×Mand is unitary . (ii) The matrix Shas dimensions M×N( t h es a m ed i m e n s i o n sa st h o s eo f A) and is diagonal in the sense that Sij=0i f i/negationslash=j. We denote its diagonal elements by sifori=1,2,...,p,w h e r e p= min( M,N); these elements are termed the singular values ofA. (iii) The square matrix Vhas dimensions N×Nand is unitary . We must now determine the elements of these matrices in terms of the elements of A. From the matrix A, we can construct two square matrices: A†Awith dimensions N×Nand AA†with dimensions M×M. Both are clearly Hermitian . From (8.131), and using the fact that Uand Vare unitary, we ﬁnd A†A=VS†U†USV†=VS†SV†(8.132) AA†=USV†VS†U†=USS†U†, (8.133) where S†Sand SS†are diagonal matrices with dimensions N×NandM×M respectively. The ﬁrst pelements of each diagonal matrix are s2 i,i=1,2,...,p, where p= min( M,N), and the rest (where they exist) are zero. These two equations imply that both V−1A†AVparenleftbig =V−1A†A(V†)−1parenrightbig and, by a similar argument, U−1AA†U, must be diagonal. From our discussion of the diagonalisation of Hermitian matrices in section 8.16, we see that the columns of Vmust therefore be the normalised eigenvectors vi,i=1,2,...,N , of the matrix A†Aand the columns of Umust be the normalised eigenvectors uj,j=1,2,...,M , of the matrix AA†. Moreover, the singular values simust satisfy s2 i=λi,w h e r e theλiare the eigenvalues of the smaller of A†Aand AA†. Clearly, the λiare also some of the eigenvalues of the larger of these two matrices, the remainingones being equal to zero. Since each matrix is Hermitian, the λ iare real and the singular values simay be taken as real and non-negative. Finally, to make the decomposition (8.131) unique, it is customary to arrange the singular values in decreasing order of their values, so that s1≥s2≥···≥sp. §The proof that such a decomposition always exists is beyond the scope of this book. For a full account of SVD one might consult, for example, Golub & Van Loan, Matrix Computations , second edition (Johns Hopkins University Press). 307 MATRICES AND VECTOR SPACESIShow that, for i=1,2,...,p,Avi=siuiand A†ui=sivi,w h e r e p=m i n ( M,N). Post-multiplying both sides of (8.131) by V, and using the fact that Vis unitary, we obtain AV=US. Since the columns of Vand Uconsist of the vectors viand ujrespectively and Shas only diagonal non-zero elements, we ﬁnd immediately that, for i=1,2,...,p, Avi=siui. (8.134) Moreover, we note that Avi=0f o r i=p+1,p+2,...,N . Taking the Hermitian conjugate of both sides of (8.131) and post-multiplying by U,w e obtain A†U=VS†=VST, where we have used the fact that Uis unitary and Sis real. We then see immediately that, fori=1,2,...,p, A†ui=sivi. (8.135) We also note that A†ui=0f o r i=p+1,p+2,...,M . Results (8.134) and (8.135) are useful for investigating the properties of the SVD. J The decomposition (8.131) has some advantageous features for the analysis of sets of simultaneous linear equations. These are best illustrated by writing the decomposition (8.131) in terms of the vectors uiand vias A=psummationdisplay i=1siui(vi)†, where p= min( M,N). It may be, however, that some of the singular values si arezero, as a result of degeneracies in the set of Mlinear equations Ax=b. Let us suppose that there are rnon-zero singular values. Since our convention is to arrange the singular values in order of decreasing size, the non-zero singular values are si,i=1,2,...,r, and the zero singular values are sr+1,sr+2,...,s p. Therefore we can write Aas A=rsummationdisplay i=1siui(vi)†. (8.136) Let us consider the action of (8.136) on an arbitrary vector x.T h i si sg i v e nb y Ax=rsummationdisplay i=1siui(vi)†x. Since ( vi)†xis just a number, we see immediately that the vectors ui,i=1,2,...,r, must span the range of the matrix A; moreover, these vectors form an orthonor- mal basis for the range. Further, since this subspace is r-dimensional, we have rank A=r, i.e. the rank of Ais equal to the number of non-zero singular values. The SVD is also useful in characterising the null space of A. From (8.119), we already know that the null space must have dimension N−r;s oi f Ahasr 308 8.18 SIMULTANEOUS LINEAR EQUATIONS non-zero singular values si,i=1,2,...,r, then from the worked example above we have Avi=0 f o r i=r+1,r+2,...,N. Thus, the N−rvectors vi,i=r+1,r+2,...,N , form an orthonormal basis for the null space of A.IFind the singular value decompostion of the matrix A= /0B/@22 2 2 17 101 10−17 10−1 10 3 59 5−3 5−9 5 /1CA. (8.137) The matrix Ahas dimension 3 ×4( i . e . M=3 , N= 4), and so we may construct from it the 3×3m a t r i x AA†and the 4×4m a t r i x A†A(in fact, since Ais real, the Hermitian conjugates are just transposes). We begin by ﬁnding the eigenvalues λiand eigenvectors ui of the smaller matrix AA†. This matrix is easily found to be given by AA†= /0/@16 0 0 029 512 5 012 536 5 /1A, and its characteristic equation reads//// // 16−λ 00 029 5−λ12 5 012 536 5−λ //// // =( 1 6−λ)(36−13λ+λ2)=0 . Thus, the eigenvalues are λ1= 16, λ2=9 , λ3= 4. Since the singular values of Aare given bysi=√λiand the matrix Sin (8.131) has the same dimensions as A, we have S= /0/@4000 03000020 /1A, (8.138) where we have arranged the singular values in order of decreasing size. Now the matrix U has as its columns the normalised eigenvectors uiof the 3×3m a t r i x AA†. These normalised eigenvectors correspond to the eigenvalues of AA†as follows: λ1=1 6⇒ u1= ( 100 )T λ2=9⇒ u2=( 03 54 5)T λ3=4⇒ u3=( 0−4 53 5)T, a n ds ow eo b t a i nt h em a t r i x U= /0/@10 0 03 5−4 5 04 53 5 /1A. (8.139) The columns of the matrix Vin (8.131) are the normalised eigenvectors of the 4 ×4 matrix A†A, which is given by A†A=1 4 /0B/@29 21 3 11 21 29 11 3 31 1 2 9 2 1 1 132 1 2 9 /1CA. 309 MATRICES AND VECTOR SPACES We already know from the above discussion, however, that the non-zero eigenvalues of this matrix are equal to those of AA†found above, and that the remaining eigenvalue is zero. The corresponding normalised eigenvectors are easily found: λ1=1 6⇒ v1=1 2( 1111 )T λ2=9⇒ v2=1 2(1 1−1−1)T λ3=4⇒ v3=1 2(−111−1)T λ4=0⇒ v4=1 2(1−11−1)T and so the matrix Vis given by V=1 2 /0B/@11−11 11 1 −1 1−11 1 1−1−1−1 /1CA. (8.140) Alternatively, we could have found the ﬁrst three columns of Vby using the relation (8.135) to obtain vi=1 siA†uifori=1,2,3. The fourth eigenvector could then be found using the Gram–Schmidt orthogonalisation procedure. We note that if there were more than one eigenvector corresponding to a zeroeigenvalue then we would need to use this procedure to orthogonalise these eigenvectorsbefore constructing the matrix V. Collecting our results together, we ﬁnd the SVD of the matrix A: A=USV †= /0/@10 0 03 5−4 5 04 53 5 /1A /0/@4000 03000020 /1A /0BBB/@1 21 21 21 2 1 21 2−1 2−1 2 −1 21 21 2−1 2 1 2−1 21 2−1 2 /1CCCA; this can be veriﬁed by direct multiplication. J Let us now consider the use of SVD in solving a set of Msimultaneous linear equations in Nunknowns, which we write again as Ax=b. Firstly, consider the solution of a homogeneous set of equations, for which b=0. As mentioned previously, if Ais square and non-singular (and so possesses no zero singular values) then the equations have the unique trivial solution x=0.O t h e r w i s e , any of the vectors vi,i=r+1,r+2,...,N , or any linear combination of them, will be a solution. In the inhomogeneous case, where bis not a zero vector, the set of equations will possess solutions if blies in the range of A. To investigate these solutions, it is convenient to introduce the N×Mmatrix S, which is constructed by taking the transpose of Sin (8.131) and replacing each non-zero singular value sion the diagonal by 1 /si. It is clear that, with this construction, SS=I. (8.141) We note, however, that the matrix Sisnotthe inverse of Ssince SS/negationslash=I. 310 8.18 SIMULTANEOUS LINEAR EQUATIONS Nevertheless, using property (8.141) and the unitarity of the matrices Uand V,a solution to the equations Ax=bis given by x=VSU†b. (8.142) We may, however, add to this solution anylinear combination of the p−rvectors vi,i=r+1,r+2,...,p, that form an orthonormal basis for the null space of A; thus, in general, there exists an inﬁnity of solutions (although it is straightforward to show that (8.142) is the solution vector of shortest length). The only way inwhich the solution (8.142) can be unique is if the rank requals N, so that the matrix Adoes not possess a null space; this only occurs if Ais square and non-singular. Ifbdoes not lie in the range of Athen the set of equations Ax=bdoes not have a solution. Nevertheless, the vector (8.142) provides the closest possible‘solution’ in a least-squares sense. In other words, although the vector (8.142)does not exactly solve Ax=b, it is the vector that minimises the residual /epsilon1=|Ax−b|, where here the vertical lines denote the absolute value of the quantity they contain, not the determinant. This is proved as follows. Suppose we were to add some arbitrary vector x /primeto the vector xin (8.142). This would result in the addition of the vector b/prime=Ax/primetoAx−b;b/primeis clearly in the range of Asince any part of x/primebelonging to the null space of Acontributes nothing to Ax/prime. We would then have |Ax−b+b/prime|=|(USV†)(VSU†b)−b+b/prime| =|(USSU†−I)b+b/prime| =|U[(SS−I)U†b+U†b/prime]| =|(SS−I)U†b+U†b/prime|; (8.143) in the last line we have made use of the fact that the length of a vector is left unchanged under the action of the unitary matrix U. Now, the diagonal square matrix SS, with dimensions M×M, will have non-zero entries (actually all equal to unity ) only for those values of jfor which sj/negationslash= 0. Thus, the jth component of the vector ( SS−I)U†bwill only be non-zero when sj= 0. However, the jth element of the vector U†b/primeis given by the scalar product ( uj)†b/prime, which is non-zero only if sj/negationslash=0 ,s i n c e b/primelies in the range of A. Thus, as these two terms only contribute to (8.143) for two disjoint sets of j-values, its minimum value, as x/primeis varied, occurs when b/prime=0; this requires x/prime=0. 311 MATRICES AND VECTOR SPACESIFind the solution(s) to the set of simultaneous linear equations Ax=b,w h e r e Ais given by (8.137) and b= ( 100 )T. To solve the set of equations, we begin by calculating the vector given in (8.142), x=VSU†b, where Uand Vare given by (8.139) and (8.140) respectively and Sis obtained by taking the transpose of Sin (8.138) and replacing all the non-zero singular values siby 1/si. Thus, Sreads S= /0BBB/@1 400 01 30 001 2 000 /1CCCA. Substituting the appropriate matrices into the expression for xwe ﬁnd x=1 8( 1111 )T. (8.144) It is straightforward to show that this solves the set of equations Ax=bexactly, and so the vector b=( 1 0 0 )Tmust lie in the range of A. This is, in fact, immediately clear, since b=u1. The solution (8.144) is not, however, unique. There are three non-zero singular values, but N= 4. Thus, the matrix Ahas a one-dimensional null space, which is ‘spanned’ by v4, the fourth column of V, given in (8.140). The solutions to our set of equations, consisting of the sum of the exact solution and anyvector in the null space of A, therefore lie along the line x=1 8( 1111 )T+α(1−11−1)T, where the parameter αcan take any real value. We note that (8.144) is the point on this line that is closest to the origin. J 8.19 Exercises 8.1 Which of the following statements about linear vector spaces are true? Where a statement is false, give a counter-example to demonstrate this. (a) Non-singular N×Nmatrices form a vector space of dimension N2. (b) Singular N×Nmatrices form a vector space of dimension N2. (c) Complex numbers form a vector space of dimension 2. (d) Polynomial functions of xform an inﬁnite-dimensional vector space. (e) Series{a0,a1,a2,...,a N}for which PN n=0|an|2=1f o r ma n N-dimensional vector space. (f) Absolutely convergent series form an inﬁnite-dimensional vector space. (g) Convergent series with terms of alternating sign form an inﬁnite-dimensional vector space. 8.2 Evaluate the determinants (a) //////ahg hbf gfc //////, (b) /// /// / 1023 01−21 3−34−2 −21−21 /// /// / , 312 8.19 EXERCISES and (c) ////// / gc ge a +ge gb +ge 0bb b ce e b +e abb +fb +d ////// / . 8.3 Using the properties of determinants, solve with a minimum of calculation the following equations for x: (a) ///////xaa 1 axb 1 abx 1 abc 1 ///////=0, (b) //////x+2 x+4 x−3 x+3 xx +5 x−2x−1x+1 //////=0. 8.4 Consider the matrices (a) B= /0/@0−ii i0−i −ii 0 /1A,(b) C=1√ 8 /0/@√ 3−√ 2−√ 3 1√ 6−1 20 2 /1A. Are they (i) real, (ii) diagonal, (iii) symmetric, (iv) antisymmetric, (v) singular, (vi) orthogonal, (vii) Hermitian, (viii) anti-Hermitian, (ix) unitary, (x) normal? 8.5 By considering the matrices A= / 10 00 / , B= / 00 34 / show that AB=0doesnotimply that either AorBis the zero matrix but that it does imply that at least one of them is singular. 8.6 (a) The basis vectors of the unit cell of a crystal, with the origin Oat one corner, are denoted by e1,e2,e3.T h em a t r i x Ghas elements Gij,w h e r e Gij=ei·ej and Hijare the elements of the matrix H≡G−1. Show that the vectors fi= P jHijejare the reciprocal vectors and that Hij=fi·fj. (b) If the vectors uandvare given by u= X iuiei,v= X ivifi, obtain expressions for |u|,|v|,a n du·v. (c) If the basis vectors are each of length aand the angle between each pair is π/3, write down Gand hence obtain H. (d) Calculate (i) the length of the normal from Oonto the plane containing the points p−1e1,q−1e2,r−1e3, and (ii) the angle between this normal and e1. 8.7 (a) Show that if Ais Hermitian and Uis unitary then U−1AUis Hermitian. (b) Show that if Ais anti-Hermitian then iAis Hermitian. (c) Prove that the product of two Hermitian matrices Aand Bis Hermitian if and only if Aand Bcommute. (d) Prove that if Sis a real antisymmetric matrix then A=(I−S)(I+S)−1is orthogonal. If Ais given by A= / cosθsinθ −sinθcosθ / then ﬁnd the matrix Sthat is needed to express Ain the above form. (e) If Kis skew-hermitian, i.e. K†=−K, prove that V=(I+K)(I−K)−1is unitary. 313 MATRICES AND VECTOR SPACES 8.8 Aand Bare real non-zero 3 ×3 matrices and satisfy the equation (AB)T+B−1A=0. (a) Prove that if Bis orthogonal then Ais antisymmetric. (b) Without assuming that Bis orthogonal, prove that Ais singular. 8.9 The commutator [X,Y] of two matrices is deﬁned by the equation [X,Y]=XY−YX. Two anti-commuting matrices Aand Bsatisfy A2=I, B2=I,[A,B]=2 iC. (a) Prove that C2=Iand that [ B,C]=2 iA. (b) Evaluate [[[ A,B],[B,C]],[A,B]]. 8.10 The four matrices Sx,Sy,Szand Iare deﬁned by Sx= / 01 10 / , Sy= / 0−i i0 / , Sz= / 10 0−1 / ,I= / 10 01 / , where i2=−1. Show that S2 x=Iand SxSy=iSz, and obtain similar results by permutting x,yandz.G i v e nt h a t vis a vector with Cartesian components (vx,vy,vz), the matrix S(v) is deﬁned as S(v)=vxSx+vySy+vzSz. Prove that, for general non-zero vectors aandb, S(a)S(b)=a·bI+iS(a×b). Without further calculation, deduce that S(a)a n d S(b) commute if and only if a andbare parallel vectors. 8.11 A general triangle has angles α,βandγand corresponding opposite sides a, bandc. Express the length of each side in terms of the lengths of the other two sides and the relevant cosines, writing the relationships in matrix and vectorform using the vectors having components a, b, c and cos α,cosβ,cosγ. Invert the matrix and hence deduce the cosine-law expressions involving α,βandγ. 8.12 Given a matrix A= /0/@1α0 β10 001 /1A, where αandβare non-zero complex numbers, ﬁnd its eigenvalues and eigenvec- tors. Find the respective conditions for (a) the eigenvalues to be real and (b) theeigenvectors to be orthogonal. Show that the conditions are jointly satisﬁed ifand only if Ais Hermitian. 8.13 Using the Gram–Schmidt procedure: (a) construct an orthonormal set of vectors from the following: x 1= ( 0011 )T, x2=( 1 0−10 )T, x3= ( 1202 )T, x4= ( 2111 )T; 314 8.19 EXERCISES (b) ﬁnd an orthonormal basis, within a four-dimensional Euclidean space, for t h e s u b s p a c e s p a n n e d b y t h e t h r e e v e c t o r s ( 1200 )T,( 3−120 )T a n d ( 0021 )T. 8.14 If a unitary matrix Uis written as A+iB,w h e r e Aand Bare Hermitian with non-degenerate eigenvalues, show the following: (a) Aand Bcommute; (b) A2+B2=I; (c) The eigenvectors of Aare also eigenvectors of B; (d) The eigenvalues of Uhave unit modulus (as is necessary for any unitary matrix). 8.15 Determine which of the matrices below are mutually commuting, and, for those that are, demonstrate that they have a complete set of eigenfunctions in common: A= / 6−2 −29 / , B= / 18 8−11 / , C= / −9−10 −10 5 / ,D= / 14 2 21 1 / . 8.16 Find the eigenvalues and a set of eigenvectors of the matrix/0/@13−1 34−2 −1−22 /1A. Verify that its eigenvectors are mutually orthogonal. 8.17 Find three real orthogonal column matrices, each of which is a simultaneous eigenvector of A= /0/@001 010100 /1A and B= /0/@011 101110 /1A. 8.18 Use the results of the ﬁrst worked example in section 8.14 to evaluate, without repeated matrix multiplication, the expression A6x,w h e r e x=( 2 4−1)Tand Ais the matrix given in the example. 8.19 Given that Ais a real symmetric matrix with normalised eigenvectors eiobtain the coeﬃcients αiinvolved when column matrix x, which is the solution of Ax−µx=v, (∗) is expanded as x= P iαiei.H e r e µis a given constant and vis a given column matrix. (a) Solve (*) when A= /0/@210 120003 /1A, µ=2a n d v= ( 123 )T. (b) Would (*) have a solution if µ=1a n d( i ) v= ( 123 )T, (ii) v= (2 2 3)T? 315 MATRICES AND VECTOR SPACES 8.20 Demonstrate that the matrix A= /0/@20 0 −644 3−10 /1A, is defective, i.e. does not have three linearly independent eigenvectors, by showing the following: (a) its eigenvalues are degenerate and, in fact, all equal; (b) any eigenvector has the form ( µ(3µ−2ν)ν)T. (c) if two pairs of values, µ1,ν1andµ2,ν2, deﬁne two independent eigenvectors v1and v2thenanythird similarly deﬁned eigenvector v3c a nb ew r i t t e na sa linear combination of v1and v2,i . e . v3=av1+bv2 where a=µ3ν2−µ2ν3 µ1ν2−µ2ν1and b=µ1ν3−µ3ν1 µ1ν2−µ2ν1. Illustrate (c) using the example ( µ1,ν1)=( 1 ,1),(µ2,ν2)=( 1 ,2) and ( µ3,ν3)= (0,1). Show further that any matrix of the form/0/@200 6n−64−2n4−4n 3−3nn−12 n /1A is defective, with the same eigenvalues and eigenvectors as A. 8.21 By ﬁnding the eigenvectors of the Hermitian matrix H= / 10 3 i −3i2 / , construct a unitary matrix Usuch that U†HU= Λ, where Λ is a real diagonal matrix. 8.22 Use the stationary properties of quadratic forms to determine the maximum and minimum values taken by the expression Q=5x2+4y2+4z2+2xz+2xy on the unit sphere x2+y2+z2= 1. For what values of x, y, z do they occur? 8.23 Given that the matrix A= /0/@2−10 −12−1 0−12 /1A has two eigenvectors of the form (1 y1)T, use the stationary property of the expression J(x)= xTAx/(xTx) to obtain the corresponding eigenvalues. Deduce the third eigenvalue. 8.24 Find the lengths of the semi-axes of the ellipse 73x2+7 2xy+5 2y2= 100 , and determine its orientation. 8.25 The equation of a particular conic section is Q≡8x2 1+8x2 2−6x1x2= 110 . Determine the type of conic section this represents, the orientation of its principal axes, and relevant lengths in the directions of these axes. 316 8.19 EXERCISES 8.26 Show that the quadratic surface 5x2+1 1y2+5z2−10yz+2xz−10xy=4 is an ellipsoid with semi-axes of lengths 2, 1 and 0 .5. Find the direction of its longest axis. 8.27 Find the direction of the axis of symmetry of the quadratic surface 7x2+7y2+7z2−20yz−20xz+2 0xy=3. 8.28 Find the eigenvalues, and suﬃcient of the eigenvectors, of the following matrices to be able to describe the quadratic surfaces associated with them. (a) /0/@51−1 151 −11 5 /1A,(b) /0/@122 212221 /1A.(c) /0/@12 1 24 2 −121 /1A. 8.29 (a) Rearrange the result A/prime=S−1ASof section 8.16 to express the original matrix Ain terms of the unitary matrix Sand the diagonal matrix A/prime. Hence show how to construct a matrix Athat has given eigenvalues and given (orthogonal) column matrices as its eigenvectors. (b) Find the matrix with eigenvectors (1 2 1)T,(1−11 )Tand (1 0 −1)T, and corresponding eigenvalues λ,µandν. (c) Try a particular case, say λ=3 , µ=−2a n d ν= 1, and verify by explicit solution that the matrix so found does have these eigenvalues. 8.30 Find an orthogonal transformation that takes the quadratic form Q≡−x2 1−2x2 2−x2 3+8x2x3+6x1x3+8x1x2 into the form µ1y2 1+µ2y2 2−4y2 3, and determine µ1andµ2(see section 8.17). 8.31 One method of determining the nullity (and hence the rank) of an M×Nmatrix Ais as follows. •Write down an augmented transpose of A, by adding on the right an N×N unit matrix and thus producing an N×(M+N) array B. •Subtract a suitable multiple of the ﬁrst row of Bfrom each of the other lower rows so as to make Bi1=0f o r i>1. •Subtract a suitable multiple of the second row (or the uppermost row that does not start with Mzero values) from each of the other lower rows so as to make Bi2=0f o r i>2. •Continue in this way until all remaining rows have zeroes in the ﬁrst Mplaces. The number of such rows is equal to the nullity of Aand the Nrightmost entries of these rows are the components of vectors that span the null space.They can be made orthogonal if they are not so already. Use this method to show that the nullity of A= /0BBB/@−13 2 7 31 0−61 7 −1−22−3 23−44 40−8−4 /1CCCA is 2 and that an orthogonal base for the null space of Ais provided by any two column matrices of the form (2 + αi−2αi1αi)Tfor which the αi(i=1,2) are real and satisfy 6 α1α2+2 (α1+α2)+5=0 . 317 MATRICES AND VECTOR SPACES 8.32 Do the following sets of equations have non-zero solutions? If so, ﬁnd them. (a) 3 x+2y+z=0 , x−3y+2z=0 , 2 x+y+3z=0 . (b) 2 x=b(y+z), x=2a(y−z), x=( 6a−b)y−(6a+b)z. 8.33 Solve the simultaneous equations 2x+3y+z=1 1, x+y+z=6, 5x−y+1 0z=3 4. 8.34 Solve the following simultaneous equations for x1,x2and x3,u s i n gm a t r i x methods: x1+2x2+3x3=1, 3x1+4x2+5x3=2, x1+3x2+4x3=3. 8.35 Show that the following equations have solutions only if η= 1 or 2, and ﬁnd them in these cases: x+y+z=1, x+2y+4z=η, x+4y+1 0z=η2. 8.36 Find the condition(s) on αsuch that the simultaneous equations x1+αx2=1, x1−x2+3x3=−1, 2x1−2x2+αx3=−2 have (a) exactly one solution, (b) no solutions, or (c) an inﬁnite number of solutions; give all solutions where they exist. 8.37 Make an LUdecomposition of the matrix A= /0/@36 9 10 5 2−21 6 /1A and hence solve Ax=b,w h e r e( i ) b= (21 9 28)T, (ii) b= (21 7 22)T. 8.38 Make an LUdecomposition of the matrix A= /0B/@2−31 3 14−3−3 53−1−1 3−6−31 /1CA. Hence solve Ax=bfor (i) b=(−418−5)T, (ii) b=(−10 0−3−24)T. Deduce that det A=−160 and conﬁrm this by direct calculation. 8.39 Use the Cholesky separation method to determine whether the following matrices are positive deﬁnite. For each that is, determine the corresponding lower diagonalmatrix L: A= /0/@21 3 13−1 3−11 /1A, B= /0/@50√ 3 030√ 30 3 /1A. 318 8.20 HINTS AND ANSWERS 8.40 Find the equation satisﬁed by the squares of the singular values of the matrix associated with the following over-determined set of equations: 2x+3y+z=0 x−y−z=1 2x+y=0 2y+z=−2. Show that one of the singular values is close to zero. Determine the two larger singular values by an appropriate iteration process and the smallest by indirectcalculation. 8.41 Find the SVD of/0/@0−1 11 −10 /1A, showing that the singular values are√ 3a n d1 . 8.42 Find the SVD form of the matrix A= /0B/@22 28−22 1−2−19 19−2−1 −61 2 6 /1CA. Hence ﬁnd the best solution xto the equation Ax=bwhen (i) b=( 6− 39 15 18)T, (ii) b=( 9−42 15 15)T, showing that (i) has an exact solution, but that the best solution to (ii) has a residual of√ 18. 8.43 Four experimental measurements of particular combinations of three physical variables, x,yandz, gave the following inconsistent results: 13x+2 2y−13z=4, 10x−8y−10z=4 4, 10x−8y−10z=4 7, 9x−18y−9z=7 2. Find the SVD best values for x,yandz. Identify the null space of Aand hence obtain the general SVD solution. 8.20 Hints and answers 8.1 (a) False. ON,t h e N×Nnull matrix, is notnon-singular. (b) False. Consider the sum of / 10 00 / and / 0001 / . (c) True. (d) True. (e) False. Consider bn=an+anfor which PN n=0|bn|2=4/negationslash= 1, or note that there is no zero vector with unit norm. (f) True.(g) False. Consider the two series deﬁned by a 0=1 2,a n=2 (−1 2)nfor n≥1; bn=−(−1 2)nfor n≥0. T h es e r i e st h a ti st h es u mo f {an}and{bn}does not have alternating signs and so closure does not hold. 8.2 (a) abc+2fgh−af2−bg2−ch2,( b )0 ,( c ) ab(ab−cd). 319 MATRICES AND VECTOR SPACES 8.3 (a) x=a,borc;( b ) x=−1, equation is linear in x. 8.4 (a) iv, v, vii, x; (b) i, vi, ix, x.8.6 (b) (P ijuiGijuj)1/2,( P ijviHijvj)1/2, P iuivi; (c) H=1 a2 /0/@3/2−1/2−1/2 −1/23 /2−1/2 −1/2−1/23 /2 /1A.(d) (i) M−1, (ii) cos−1(p/Ma)w h e r e M=a−1[3(p2+q2+r2)/2−qr−pr−pq]1/2. 8.7 (d) S= / 0−tan(θ/2) tan(θ/2) 0 / .(e) Note that ( I+K)(I−K)= I−K2= (I−K)(I+K). 8.8 (b) Note that |−A|=(−1)3|A|. 8.9 (b) 32 iA. 8.10 S(a)S(b)−S(b)S(a)=2 iS(a×b) and equals zero only if a×b=0. 8.11 a=bcosγ+ccosβ, and cyclic permutations; a2=b2+c2−2bccosα, and cyclic permutations. 8.12 λ=1 ,(001 )T; λ=1+( αβ)1/2,(α1/2β1/20)T; λ=1−(αβ)1/2,(α1/2−β1/20)T; (a)αβreal and >0; (b)|α|=|β|. 8.13 (a) 2−1/2( 0011 )T,6−1/2(2 0−11 )T, 39−1/2(−16−11 )T,1 3−1/2( 212 −2)T. (b) 5−1/2( 1200 )T,(345)−1/2(14−71 00 )T, (18285)−1/2(−56 28 98 69)T. 8.14 (a) Use UU†=U†U; (b) use UU†=I; (c) apply the result of subsection 8.13.5 to give the eigenvalue for Uasλ+iµ; (d) apply result (b) to eigenvector uofUto deduce that λ2+µ2=1. 8.15 Cdoes not commute with the others; A,Band Dhave (1−2)Tand (2 1)Tas common eigenvectors. 8.16 λ=1 ,(113 )T; λ=3±√ 15, (5±√ 15 7±2√ 15−4∓√ 15)T; 8.17 For A:( 1 0−1)T,(1α11)T,(1α21)T. For B: ( 111 )T,(β1γ1−β1−γ1)T,(β2γ2−β2−γ2)T. Theαi,βiandγiare arbitrary. Simultaneous and orthogonal: (1 0 −1)T,( 111 )T,(1−21 )T. 8.18 Express xas a linear combination of the eigenvectors of Aand use the fact that Anx=λnxfor an eigenvector; x=3x(1)−x(2);A6x=(−537 921 729)T. 8.19 αj=(v·ej∗)/(λj−µ), where λjis the eigenvalue corresponding to ej. (a) x= ( 213 )T. (b) Since µis equal to one of A’s eigenvalues λj, the equation only has a solution ifv·ej∗= 0; (i) no solution; (ii) x= ( 113 /2)T. 8.20 (a) All eigenvalues equal 2; (c) a=−1,b=1 . 8.21 U= (10)−1/2(1,3i;3i,1), Λ = (1 ,0;0,11). 8.22 Maximum equal to 6 at ±(2,1,1)/√ 6; minimum equal to 3 at ±(1,−1,−1)/√ 3. 8.23 J=( 2y2−4y+4)/(y2+2) with stationary values at y=±√2 and corresponding eigenvalues 2 ∓√2. From the trace property of A, the third eigenvalue equals 2. 8.24 The eigenvalues, after making the RHS unity, are 1 /4 and 1, corresponding to semi-axis lengths of 2 and 1. The major axis makes an angle tan−1(−4/3) with the positive x-axis. 8.25 Ellipse; θ=π/4,a=√ 22;θ=3π/4,b=√ 10. 320 8.20 HINTS AND ANSWERS 8.26 The eigenvector corresponding to the smallest eigenvalue is in the direction (1,1,1)/√ 3. 8.27 The direction of the eigenvector having the non-repeated eigenvalue is (1,1,−1)/√ 3. 8.28 (a) Eigenvalues 6, 6, 3; an ellipsoid with circular cross-section of radius r, say, perpendicular to the direction (1 ,−1,1)/√3, and with semi-axis in that direction of√2r. (b) Eigenvalues 5, −1,−1; a hyperboloid of revolution about an axis in the direction (1 ,1,1)/√3, the two halves of the hyperboloid being asymptotic to that cone of semi-angle tan−1√5 that passes through the origin and also has its axis in that direction. (c) Eigenvalues 6, 0, 0; a pair of parallel planes, equidistant from the origin and with their normals in the directions ±(1,2,1)/√6. 8.29 (a) A=SA/primeS†,w h e r e Sis the matrix whose columns are the eigenvectors of the matrix Ato be constructed, and A/prime=d i a g( λ, µ, ν). (b) A=(λ+2µ+3ν,2λ−2µ, λ+2µ−3ν;2λ−2µ,4λ+2µ,2λ−2µ; λ+2µ−3ν,2λ−2µ, λ+2µ+3ν). (c)1 3(1,5,−2;5,4,5;−2,5,1). 8.30 y1=(x1+x2+x3)/√ 3,y2=(x1−2x2+x3)/√ 6,y3=(−x1+x3)/√ 2; µ1=6 , µ2=−6. 8.31 The null space is spanned by (2 0 1 0)Tand (1−201 )T. 8.32 (a) No, |A|=−24/negationslash=0 ;y e s , x:y:z=4ab:4a+b:4a−b. 8.33 x=3 , y=1 , z=2. 8.34 x1=−3/2,x2=7/2,x3=−3/2. 8.35 η=1 , x=1+2 z,y=−3z;η=2 , x=2z,y=1−3z. 8.36 (a) α/negationslash=6,α/negationslash=1 ; x1=( 1−α)/(1 +α),x2=2/(1−α),x3=0 . (b)α=1 .( c ) α=6 ; x1=1−6β, x 2=β, x 3=( 7β−2)/3f o ra n y β. 8.37 L=( 1,0,0;1 3,1,0;2 3,3,1),U =( 3,6,9;0,−2,2; 0,0,4). (i)x=(−112 )T. (ii) x=(−322 )T. 8.38 L=( 1,0,0,0;1 2,1,0,0;5 2,21 11,1,0;3 2,−3 11,−12 7,1); U=( 2,−3,1,3; 0,11 2−7 2,−9 2;0,0,35 11,1 11;0,0,0,−32 7). (i)x=( 2−14−5)T. (ii) x=(−114−3)T. 8.39 Ais not positive deﬁnite as L33is calculated to be√ −6. B=LLT, where the non-zero elements of Lare L11=√ 5,L31= p 3/5,L22=√ 3,L33= p 12/5. 8.40 λ3−27λ2+121 λ−3=0 .Find the two larger roots for λusing the rearrangement method described in subsection 28.1.1 and the smallest one using the property ofthe product of the roots. The singular values are 4.6190, 2.3748 and 0.1579. 8.41 A †A= / 21 12 / ,U=1√ 6 /0/@−1√ 3√ 2 20√ 2 −1−√ 3√ 2 /1A,V= / 11 1−1 / . 8.42 The singular values are 18√ 6,−18,−12√ 3. (i)x= ( 112 )Twith all four equations exactly satisﬁed. (ii)x=1 36(40 37 74)T, giving a residual column matrix ( −1223 )T. 8.43 The singular values are 12√ 6,0,−18√ 3 and the calculated best solution is x= 1.71,y=−1.94,z=−1.71. The null space is the line x=z,y= 0 and the general SVD solution is x=1.71 + λ, y=−1.94,z=−1.71 + λ. 321 9 Normal modes Any student of the physical sciences will encounter the subject of oscillations on many occasions and in a wide variety of circumstances, for example the voltageand current oscillations in an electric circuit, the vibrations of a mechanicalstructure and the internal motions of molecules. The matrices studied in the previous chapter provide a particularly simple way to approach what may appear, at ﬁrst glance, to be diﬃcult physical problems. We will consider only systems for which a position-dependent potential exists, i.e., the potential energy of the system in any particular conﬁguration dependsupon the coordinates of the conﬁguration, which need not be be lengths however;the potential must notdepend upon the time derivatives (generalised velocities) of these coordinates. So, for example, the potential −qv·Aused in the Lagrangian description of a charged particle in an electromagnetic ﬁeld is excluded. A further restriction that we place is that the potential has a local minimum atthe equilibrium point; physically, this is a necessary and suﬃcient condition forstable equilibrium. By suitably deﬁning the origin of the potential, we may takeits value at the equilibrium point as zero. We denote the coordinates chosen to describe a conﬁguration of the system byq i,i=1,2,...,N .T h e qineed not be distances; some could be angles, for example. For convenience we can deﬁne the qiso that they are all zero at the equilibrium point. The instantaneous veloc ities of various parts of the system will depend upon the time derivatives of the qi, denoted by ˙qi. For small oscillations the velocities will be linear in the ˙qiand consequently the total kinetic energy T will be quadratic in them – and will include cross terms of the form ˙qi˙qjwith i/negationslash=j. The general expression for Tcan be written as the quadratic form T=summationdisplay isummationdisplay jaij˙qi˙qj=˙qTA˙q, (9.1) where ˙qis the column vector ( ˙q1˙q2···˙qN)Tand the N×Nmatrix A is real and may be chosen to be symmetric. Furthermore, A, like any matrix 322 9.1 TYPICAL OSCILLATORY SYSTEMS corresponding to a kinetic energy, is positive deﬁnite (more strictly positive semi- deﬁnite); that is, whatever real values the ˙qitake, the quadratic form (9.1) has a value≥0. Turning now to the potential energy, we may write its value for a conﬁguration qby means of a Taylor expansion about the origin q=0, V(q)=V(0)+summationdisplay i∂V(0) ∂qiqi+1 2summationdisplay isummationdisplay j∂2V(0) ∂qi∂qjqiqj+···. However, we have chosen V(0) = 0 and, since the origin is an equilibrium point, there is no force there and ∂V(0)/∂q i= 0. Consequently, to second order in the qiwe also have a quadratic form, but in the coordinates rather than in their time derivatives: V=summationdisplay isummationdisplay jbijqiqj=qTBq, (9.2) where Bis, or can be made, symmetric. In this case, and in general, the requirement that the potential is a minimum means that the potential matrix B, like the kinetic energy matrix A, is real and positive deﬁnite. 9.1 Typical oscillatory systems We now introduce particular examples, although the results of this section are general, given the above restrictions and the reader will ﬁnd it easy to apply theresults to many other instances. Consider ﬁrst a uniform rod of mass Mand length l, attached by a light string also of length lto a ﬁxed point Pand executing small oscillations in a vertical plane. We choose as coordinates the angles θ 1andθ2shown, with exaggerated magnitude, in ﬁgure 9.1. In terms of these coordinates the centre of gravity of the rod has, to ﬁrst order in the θi, a velocity component in the x-direction equal to l˙θ1+1 2l˙θ2a n di nt h e y-direction equal to zero. Adding in the rotational kinetic energy of the rod about its centre of gravity we obtain, to second order in the ˙θi, T≈1 2Ml2(˙θ2 1+1 4˙θ2 2+˙θ1˙θ2)+1 24Ml2˙θ2 2 =1 6Ml2parenleftbig 3˙θ2 1+3˙θ1˙θ2+˙θ2 2parenrightbig =1 12Ml2˙qTparenleftbigg63 32parenrightbigg ˙q, (9.3) where ˙qT=(˙θ1˙θ2).The potential energy is given by V=Mlgbracketleftbig (1−cosθ1)+1 2(1−cosθ2)bracketrightbig (9.4) ≈1 4Mlg(2θ2 1+θ2 2)=1 12Mlg qTparenleftbigg60 03parenrightbigg q, (9.5) where gis the acceleration due to gravity and q=(θ1θ2)T; (9.5) is valid to second order in the θi. 323 NORMAL MODES P P P l lθ1θ1 θ1 θ2θ2θ2 (a) (b) (c) Figure 9.1 A uniform rod of length lattached to the ﬁxed point Pby a light string of the same length: ( a) the general coordinate system; ( b) approximation to the normal mode with lower frequency; ( c) approximation to the mode with higher frequency. With these expressions for TandVwe now apply the conservation of energy, d dt(T+V)=0 , (9.6) assuming that there are no external forces other than gravity. In matrix form (9.6) becomes d dt(˙qTA˙q+qTBq)=¨qTA˙q+˙qTA¨q+˙qTBq+qTB˙q=0, which, using A=ATand B=BT,g i v e s 2˙qT(A¨q+Bq)=0 . We will assume, although it is not clear that this gives the only possible solution, that the above equation implies that the coeﬃcient of each ˙qiis separately zero. Hence A¨q+Bq=0. (9.7) For a rigorous derivation Lagrange’s equations should be used, as in chapter 22. N o ww es e a r c hf o rs e t so fc o o r d i n a t e s qthatalloscillate with the same period, i.e. the total motion repeats itself exactly after a ﬁniteinterval. Solutions of this form will satisfy q=xcosωt; (9.8) the relative values of the elements of xin such a solution will indicate how each 324 9.1 TYPICAL OSCILLATORY SYSTEMS coordinate is involved in this special motion. In general there will be Nvalues ofωif the matrices Aand BareN×Nand these values are known as normal frequencies oreigenfrequencies . Putting (9.8) into (9.7) yields −ω2Ax+Bx=(B−ω2A)x=0. (9.9) Our work in section 8.18 showed that this can have non-trivial solutions only if |B−ω2A|=0. (9.10) This is a form of characteristic equation for B, except that the unit matrix Ihas been replaced by A. It has the more familiar form if a choice of coordinates is made in which the kinetic energy Tis a simple sum of squared terms, i.e. it has been diagonalised, and the scale of the new coordinates is then chosen to make each diagonal element unity. However, even in the present case, (9.10) can be solved to yield ω2 kfork= 1,2,...,N ,w h e r e Nis the order of Aand B. The values of ωkcan be used with (9.9) to ﬁnd the corresponding column vector xkand the initial (stationary) physical conﬁguration that, on release, will execute motion with period 2 π/ω k. In equation (8.76) we showed that the eigenvectors of a real symmetric matrix were, except in the case of degeneracy of the eigenvalues, mutually orthogonal. In the present situation an analogous, but not identical, result holds. It is shown in section 9.3 that if x1and x2are two eigenvectors satisfying (9.9) for diﬀerent values of ω2then they are orthogonal in the sense that (x2)TAx1=0 a n d ( x2)TBx1=0. The direct ‘scalar product’ ( x2)Tx1, formally equal to ( x2)TIx1,i sn o t ,i ng e n e r a l , equal to zero. Returning to the suspended rod, we ﬁnd from (9.10) vextendsinglevextendsinglevextendsinglevextendsingleMlg 12parenleftbigg60 03parenrightbigg −ω2Ml2 12parenleftbigg63 32parenrightbiggvextendsinglevextendsinglevextendsinglevextendsingle=0. Writing ω2l/g=λ, this becomes vextendsinglevextendsinglevextendsinglevextendsingle6−6λ−3λ −3λ3−2λvextendsinglevextendsinglevextendsinglevextendsingle=0⇒ λ2−10λ+6=0 , which has roots λ=5±√ 19. Thus we ﬁnd that the two normal frequencies are given by ω1=( 0.641g/l)1/2andω2=( 9.359g/l)1/2. Putting the lower of the two values for ω2,n a m e l y( 5 −√ 19)g/l, into (9.9) shows that for this mode x1:x2=3 ( 5−√ 19) : 6(√ 19−4) = 1 .923 : 2 .153. This corresponds to the case where the rod and string are almost straight out, i.e. they almost form a simple pendulum. Similarly it may be shown that the higher 325 NORMAL MODES frequency corresponds to a solution where the string and rod are moving with opposite phase and x1:x2=9.359 :−16.718. The two situations are shown in ﬁgure 9.1. In connection with quadratic forms it was shown in section 8.17 how to make a change of coordinates such that the matrix for a particular form becomesdiagonal. In exercise 9.6 a method is developed for diagonalising simultaneouslytwo quadratic forms (though the transformation matrix may not be orthogonal). If this process is carried out for Aand Bin a general system undergoing stable oscillations, the kinetic and potential energies in the new variables η itake the forms T=summationdisplay iµi˙η2 i=˙ηTM˙η, M=d i a g( µ1,µ2,...,µ N), (9.11) V=summationdisplay iνiη2 i=ηTNη, N=d i a g( ν1,ν2...,ν N), (9.12) and the equations of motion are the uncoupled equations µi¨ηi+νiηi=0,i=1,2,...,N. (9.13) Clearly a simple renormalisation of the ηican be made that reduces all the µi in (9.11) to unity. When this is done the variables so formed are called normal coordinates and equations (9.13) the normal equations . When a system is executing one of these simple harmonic motions it is said to be in a normal mode , and once started in such a mode it will repeat its motion exactly after each interval of 2 π/ω i. Any arbitrary motion of the system may be written as a superposition of the normal modes, and each component modewill execute harmonic motion with the corresponding eigenfrequency; however,unless by chance the eigenfrequencies are in integer relationship, the system will never return to its initial conﬁguration after any ﬁnite time interval. As a second example we will consider a number of masses coupled together by springs. For this type of situation the potential and kinetic energies are automat- ically quadratic functions of the coordinates and their derivatives, provided theelastic limits of the springs are not exceeded, and the oscillations do not have tobe vanishingly small for the analysis to be valid.IFind the normal frequencies and modes of oscillation of three particles of masses m,µm, mconnected in that order in a straight line by two equal light springs of force constant k. (This arrangement could serve as a model for some linear molecules, e.g. CO2.) The situation is shown in ﬁgure 9.2; the coordinates of the particles, x1,x2,x3,a r e measured from their equilibrium positions, at which the springs are neither extended norcompressed. The kinetic energy of the system is simply T= 1 2m /;˙x2 1+µ˙x2 2+˙x2 3 / , 326 9.1 TYPICAL OSCILLATORY SYSTEMS m m µm x1 x2 x3k k Figure 9.2 Three masses m,µmandmconnected by two equal light springs of force constant k. (a) (b) (c) Figure 9.3 The normal modes of the masses and springs of a linear molecule such as CO 2.(a)ω2=0 ;( b)ω2=k/m;(c)ω2=[ (µ+2 )/µ](k/m). whilst the potential energy stored in the springs is V=1 2k / (x2−x1)2+(x3−x2)2 / . The kinetic- and potential-energy symmetric matrices are thus A=m 2 /0/@100 0µ0 001 /1A, B=k 2 /0/@1−10 −12−1 0−11 /1A. From (9.10), to ﬁnd the normal frequencies we have to solve |B−ω2A|=0.Thus, writing mω2/k=λ, we have//////1−λ−10 −12−µλ−1 0−11−λ //////=0, which leads to λ=0 ,1o r1+2 /µ. The corresponding eigenvectors are respectively x1=1√ 3 /0/@1 11 /1A, x2=1√ 2 /0/@1 0 −1 /1A, x3=1p 2+( 4 /µ2) /0/@1 −2/µ 1 /1A. The physical motions associated with these normal modes are illustrated in ﬁgure 9.3. The ﬁrst, with λ=ω=0a n da l lt h e xiequal, merely describes bodily translation of the whole system, with no (i.e. zero-frequency) internal oscillations. In the second solution the central particle remains stationary, x2= 0, whilst the other two oscillate with equal amplitudes in antiphase with each other. This motion, which has frequency ω=(k/m)1/2, is illustrated in ﬁgure 9.3( b). The ﬁnal and most complicated of the three modes has frequency ω={[(µ+ 327 NORMAL MODES 2)/µ](k/m)}1/2, and involves a motion of the central particle which is in antiphase with that of the two outer ones and which has an amplitude 2 /µt i m e sa sg r e a t .I nt h i sm o t i o n (see ﬁgure 9.3( c)) the two springs are compressed and extended in turn. We also note that in the second and third normal modes the centre of mass of the molecule remainsstationary.J 9.2 Symmetry and normal modes It will have been noticed that the system in the above example has an obvious symmetry under the interchange of coordinates 1 and 3: the matrices Aand B, the equations of motion and the normal modes illustrated in ﬁgure 9.3 are all unaltered by the interchange of x1and−x3. This reﬂects the more general result that for each physical symmetry possessed by a system, there is at least onenormal mode with the same symmetry. The general question of the relationship between the symmetries possessed by a physical system and those of its normal modes will be taken up more formallyin chapter 25 where the representation theory of groups is considered. However,we can show here how an appreciation of a system’s symmetry properties will sometimes allow its normal modes to be guessed (and then veriﬁed), something that is particularly helpful if the number of coordinates involved is greater thantwo and the corresponding eigenvalue equation (9.10) is a cubic or higher-degreepolynomial equation. Consider the problem of determining the normal modes of a system consist- ing of four equal masses Mat the corners of a square of side 2 L, each pair of masses being connected by a light spring of modulus kthat is unstretched in the equilibrium situation. As shown in ﬁgure 9.4, we introduce Cartesiancoordinates x n,yn, with n=1,2,3,4, for the positions of the masses and de- note their displacements from their equilibrium positions Rnbyqn=xni+ynj. Thus rn=Rn+qnwith Rn=±Li±Lj. The coordinates for the system are thus x1,y1,x2,...,y 4and the kinetic en- ergy matrix Ais given trivially by MI8,w h e r e I8is the 8×8 identity ma- trix. The potential energy matrix Bis much more diﬃcult to calculate and involves, for each pair of values m, n, evaluating the quadratic approximation to the expression bmn=1 2kparenleftbig |rm−rn|−|Rm−Rn|parenrightbig2. Expressing each riin terms of qiandRiand remembering that |Rm−Rn|/greatermuch 328 9.2 SYMMETRY AND NORMAL MODES M MM M k k kk kk x1y1 x2y2 x3y3 x4y4 Figure 9.4 The arrangement of four equal masses and six equal springs discussed in the text. The coordinate systems xn,ynforn=1,2,3,4 measure the displacements of the masses from their equilibrium positions. |qm−qn|,w eo b t a i n bmn(=bnm): bmn=1 2kbracketleftbig |(Rm−Rn)+(qm−qn)|−|Rm−Rn|bracketrightbig2 =1 2kbraceleftBigbracketleftbig |Rm−Rn|2+2 (qm−qn)·(RM−Rn)+|qm−qn)|2bracketrightbig1/2−|Rm−Rn|bracerightBig2 =1 2k|Rm−Rn|2braceleftBiggbracketleftbigg 1+2(qm−qn)·(RM−Rn) |Rm−Rn|2+···bracketrightbigg1/2 −1bracerightBigg2 ≈1 2kbraceleftbigg(qm−qn)·(RM−Rn) |Rm−Rn|bracerightbigg2 . This ﬁnal expression is readily interpretable as the potential energy stored in the spring when it is extended by an amount equal to the component, along theequilibrium direction of the spring, of the relative displacement of its two ends. Applying this result to each spring in turn gives the following expressions for the elements of the potential matrix. mn 2b mn/k 12 ( x1−x2)2 13 ( y1−y3)2 141 2(−x1+x4+y1−y4)2 231 2(x2−x3+y2−y3)2 24 ( y2−y4)2 34 ( x3−x4)2. 329 NORMAL MODES The potential matrix is thus constructed as B=k 4 3−1−20 0 0 −11 −1 3000 −21−1 −2 031 −1−10 0 0013 −1−10−2 00−1−13 1 −20 0−2−1−1 1300 −1 100 −20 3 −1 1−10−20 0 −13 . To solve the eigenvalue equation |B−λA|= 0 directly would mean solving an eigth-degree polynomial equation. Fortunately, we can exploit intuition andthe symmetries of the system to obtain the eigenvectors and correspondingeigenvalues without such labour. Firstly, we know that bodily translation of the whole system, without any internal vibration, must be possible and that there will be two independentsolutions of this form, corresponding to translations in the x-a n d y- directions. The eigenvector for the ﬁrst of these (written in row form to save space) is x (1)= ( 10101010 )T. Evaluation of Bx(1)gives Bx(1)= ( 00000000 )T, showing that x(1)is a solution of ( B−ω2A)x=0corresponding to the eigenvalue ω2=0 ,w h a t e v e rf o r m Axmay take. Similarly, x(2)= ( 01010101 )T is a second eigenvector corresponding to the eigenvalue ω2=0 . The next intuitive solution, again involving no internal vibrations, and, there- fore, expected to correspond to ω2= 0, is pure rotation of the whole system about its centre. In this mode each mass moves perpendicularly to the line joining its position to the centre, and so the relevant eigenvector is x(3)=1√ 2( 111 −1−11−1−1)T. It is easily veriﬁed that Bx(3)=0thus conﬁrming both the eigenvector and the corresponding eigenvalue. The three non-oscillatory normal modes are illustrated in diagrams ( a)–(c) of ﬁgure 9.5. We now come to solutions that do involve real internal oscillations, and, because of the four-fold symmetry of the system, we expect one of them to be amode in which all the masses move along radial lines – the so-called ‘breathing 330 9.2 SYMMETRY AND NORMAL MODES (a)ω2=0 ( b)ω2=0 ( c)ω2=0 (d)ω2=2k/M (e)ω2=k/M (f)ω2=k/M (g)ω2=k/M (h)ω2=k/M Figure 9.5 The displacements and frequencies of the eight normal modes of the system shown in ﬁgure 9.4. Modes ( a), (b)a n d( c) are not true oscillations: (a)a n d( b) are purely translational whilst ( c) is one of bodily rotation. Mode ( d), the ‘breathing mode’, has the highest frequency and the remaining four, ( e)– (h), of lower frequency, are degenerate. mode’. Expressing this motion in coordinate form gives as the fourth eigenvector x(4)=1√ 2(−1111 −1−11−1)T. Evaluation of Bx(4)yields Bx(4)=k 4√ 2(−8888 −8−88−8)T=2kx(4), i.e. a multiple of x(4), conﬁrming that it is indeed an eigenvector. Further, since Ax(4)=Mx(4), it follows from ( B−ω2A)x=0that ω2=2k/Mf o rt h i sn o r m a l mode. Diagram (d) of the ﬁgure illustrates the corresponding motions of the fourmasses. As the next step in exploiting of the symmetry properties of the system we note that, because of its reﬂection symmetry in the x-axis, the system is invariant under the double interchange of y 1with−y3andy2with−y4. This leads us to try an eigenvector of the form x(5)=( 0 α0β0−α0−β)T. Substituting this trial vector into ( B−ω2A)x=0gives, of course, eight simulta- 331 NORMAL MODES neous equations for αandβ, but they are all equivalent to just two, namely α+β=0, 5α+β=4Mω2 kα; these have the solution α=−βandω2=k/M. The latter thus gives the frequency of the mode with eigenvector x(5)=( 0 1 0 −10−101 )T. Note that, in this mode, when the spring joining masses 1 and 3 is most stretched, the one joining masses 2 and 4 is at its most compressed. Similarly, based onreﬂection symmetry in the y-axis, x (6)=( 1 0−10−1010 )T can be shown to be an eigenvector corresponding to the same frequency. These two modes are sketched in diagrams ( e)a n d( f) of ﬁgure 9.5. This accounts for six of the expected eight modes, and the other two could be found by considering motions that are symmetric about both diagonals of the square or are invariant under successive reﬂections in the x-a n d y- axes. However, since Ais a multiple of the unit matrix, and since we know that ( x(j))TAx(i)=0i f i/negationslash=j, we can ﬁnd the two remaining eigenvectors more easily by requiring them to be orthogonal to each of those found so far. Let us take the next (seventh) eigenvector, x(7),t ob eg i v e nb y x(7)=(abcdefgh )T. Then orthogonality with each of the x(n)forn=1,2,...,6 yields six equations satisﬁed by the unknowns a ,b ,...,h . As the reader may verify, they can be reduced to the six simple equations a+g=0,d+f=0,a+f=d+g, b+h=0,c+e=0,b+c=e+h. With six homogeneous equations for eight unknowns, eﬀectively separated into two groups of four, we may pick one in each group arbitrarily. Taking a=b=1 gives d=e=1a n d c=f=g=h=−1 as a solution. Substitution of x(7)=( 1 1−111−1−1−1)T. into the eigenvalue equation checks that it is an eigenvector and shows that the corresponding eigenfrequency is given by ω2=k/M. We now have the eigenvectors for seven of the eight normal modeshand the eighth can be found by making it simultaneously orthogonal to each of the otherseven. It is left to the reader to show (or verify) that the ﬁnal solution is x (8)=( 1−111−1−1−11 )T 332 9.3 RAYLEIGH–RITZ METHOD and that this mode has the same frequency as three of the other modes. The general topic of the degeneracy of normal modes is discussed in chapter 25. Themovements associated with the ﬁnal two modes are shown in diagrams ( g)a n d (h) of ﬁgure 9.5; this ﬁgure summarises all eight normal modes and frequencies. Although this example has been lengthy to write out, we have seen that the actual calculations are quite simple and provide the full solution to what isformally a matrix eigenvalue equation involving 8 ×8 matrices. It should be noted that our exploitation of the intrinsic symmetries of the system played acrucial part in ﬁnding the correct eigenvectors for the various normal modes. 9.3 Rayleigh–Ritz method We conclude this chapter with a discussion of the Rayleigh–Ritz method for estimating the eigenfrequencies of an oscillating system. We recall from theintroduction to this chapter that for a system undergoing small oscillations thepotential and kinetic energy are given by V=q TBq and T=˙qTA˙q, where the components of qare the coordinates chosen to represent the conﬁgura- tion of the system and Aand Bare symmetric matrices (or may be chosen to be such). We also recall from (9.9) that the normal modes xiand the eigenfrequencies ωiare given by (B−ω2 iA)xi=0. (9.14) It may be shown that the eigenvectors xicorresponding to diﬀerent normal modes are linearly independent and so form a complete set. Thus, any coordinate vectorqcan be written q=summationtext jcjxj. We now consider the value of the generalised quadratic form λ(x)=xTBx xTAx=summationtext m(xm)Tc∗ mBsummationtext icixi summationtext j(xj)Tc∗ jAsummationtext kckxk, which, since both numerator and denominator are positive deﬁnite, is itself non- negative. Equation (9.14) can be used to replace Bxi, with the result that λ(x)=summationtext m(xm)Tc∗ mAsummationtext iω2 icixi summationtext j(xj)Tc∗ jAsummationtext kckxk =summationtext m(xm)Tc∗ msummationtext iω2 iciAxi summationtext j(xj)Tc∗ jAsummationtext kckxk. (9.15) Now the eigenvectors xiobtained by solving ( B−ω2A)x=0are not mutually orthogonal unless either AorBis a multiple of the unit matrix. However, it may 333 NORMAL MODES be shown that they do possess the desirable properties (xj)TAxi=0 a n d ( xj)TBxi=0 i f i/negationslash=j. (9.16) This result is proved as follows. From (9.14) it is clear that, for general iandj, (xj)T(B−ω2 iA)xi=0. (9.17) But, by taking the transpose of (9.14) with ireplaced by jand recalling that A and Bare real and symmetric, we obtain (xj)T(B−ω2 jA)=0. Forming the scalar product of this with xiand subtracting the result from (9.17) gives (ω2 j−ω2 i)(xj)TAxi=0. Thus, for i/negationslash=jand non-degenerate eigenvalues ω2 iand ω2 j, we have that (xj)TAxi= 0, and substituting this into (9.17) immediately establishes the corre- sponding result for ( xj)TBxi. Clearly, if either AorBis a multiple of the unit matrix then the eigenvectors are mutually orthogonal in the normal sense. Theorthogonality relations (9.16) are re-derived and extended in exercise 9.6. Using the ﬁrst of the relationships (9.16) to simplify (9.15), we ﬁnd that λ(x)=summationtext i|ci|2ω2 i(xi)TAxi summationtext k|ck|2(xk)TAxk. (9.18) Now, if ω2 0is the lowest eigenfrequency then ω2 i≥ω2 0for all iand, further, since (xi)TAxi≥0f o ra l l ithe numerator of (9.18) is ≥ω2 0summationtext i|ci|2(xi)TAxi.H e n c e λ(x)≡xTBx xTAx≥ω2 0, (9.19) for any xwhatsoever (whether xis an eigenvector or not). Thus we are able to estimate the lowest eigenfrequency of the system by evaluating λfor a variety of vectors x, the components of which, it will be recalled, give the ratios of the coordinate amplitudes. This is sometimes a useful approach if many coordinates are involved and direct solution for the eigenvalues is not possible. An additional result is that the maximum eigenfrequency ω2 mmay also be estimated. It is obvious that if we replace the statement ‘ ω2 i≥ω2 0for all i’b y ‘ω2 i≤ω2 mfor all i’, then λ(x)≤ω2 mfor any x. Thus λ(x) always lies between the lowest and highest eigenfrequencies of the system. Furthermore, λ(x)h a sa stationary value, equal to ω2 k,w h e n xis the kth eigenvector (see subsection 8.17.1). 334 9.4 EXERCISESIEstimate the eigenfrequencies of the oscillating rod of section 9.1. Firstly we recall that A=Ml2 12 / 63 32 / and B=Mlg 12 / 60 03 / . Physical intuition suggests that the slower mode will have a conﬁguration approximating that of a simple pendulum (ﬁgure 9.1), in which θ1=θ2, and so we use this as a trial vector.T a k i n g x=(θθ)T, λ(x)=xTBx xTAx=3Mlgθ2/4 7Ml2θ2/6=9g 14l=0.643g l, and we conclude from (9.19) that the lower (angular) frequency is ≤(0.643g/l)1/2.W e have already seen on p. 325 that the true answer is (0 .641g/l)1/2and so we have come very close to it. Next we turn to the higher frequency. Here, a typical pattern of oscillation is not so obvious but, rather preempting the answer, we try θ2=−2θ1; we then obtain λ=9g/l and so conclude that the higher eigenfrequency ≥(9g/l)1/2. We have already seen that the exact answer is (9 .359g/l)1/2and so again we have come close to it. J A simpliﬁed version of the Rayleigh–Ritz method may be used to estimate the eigenvalues of a symmetric (or in general Hermitian) matrix B, the eigenvectors of which will be mutually orthogonal. By repeating the calculations leading to(9.18), Abeing replaced by the unit matrix I, it is easily veriﬁed that if λ(x)=x TBx xTx is evaluated for anyvector xthen λ1≤λ(x)≤λm, where λ1,λ2...,λ mare the eigenvalues of Bin order of increasing size. A similar result holds for Hermitian matrices. 9.4 Exercises 9.1 Three coupled pendulums swing perpendicularly to the horizontal line containing their points of suspension, and the following equations of motion are satisﬁed: −m¨x1=cmx 1+d(x1−x2), −M¨x2=cMx 2+d(x2−x1)+d(x2−x3), −m¨x3=cmx 3+d(x3−x2), where x1,x2andx3are measured from the equilibrium points, m,Mandm are the masses of the pendulum bobs and canddare positive constants. Find the normal frequencies of the system and sketch the corresponding patterns ofoscillation. What happens as d→0o rd→∞? 9.2 A double pendulum, smoothly pivoted at A, consists of two light rigid rods, AB andBC, each of length l, which are smoothly jointed at Band carry masses mand αmatBandCrespectively. The pendulum makes s mall oscillations in one plane 335 NORMAL MODES under gravity; at time t,ABandBCmake angles θ(t)a n d φ(t) respectively with the downward vertical. Find quadratic e xpressions for the kinetic and potential energies of the system and hence show that the normal modes have angularfrequencies given by ω 2=g l h 1+α± p α(1 +α) i . Forα=1/3, show that in one of the normal modes the mid-point of BCdoes not move during the motion. 9.3 Continue the worked example modelling a linear molecule discussed at the end o fs e c t i o n9 . 1 ,f o rt h ec a s ei nw h i c h µ=2 . (a) Show that the eigenvectors derived there have the expected orthogonality properties with respect to both Aand B. (b) For the situation in which the atoms are released from rest with initial displacements x1=2/epsilon1,x2=−/epsilon1andx3= 0, determine their subsequent motions and maximum displacements. 9.4 Consider the circuit consisting of three equal capacitors and two diﬀerent induc- tors shown in the ﬁgure. For charges Qion the capacitors and currents Iithrough L1 L2C C C I1 I2Q1 Q2 Q3 the components, write down Kirchhoﬀ’s law for the total voltage change around each of two complete circuit loops. Note that, to within an unimportant constant,the conservation of current implies that Q 3=Q1−Q2and hence express the loop equations in the form given in (9.7), namely A¨Q+BQ=0. Use this to show that the normal frequencies of the circuit are given by ω2=1 CL1L2 / L1+L2±(L2 1+L2 2−L1L2)1/2 / . Obtain the same matrices and result by ﬁnding the total energy stored in the various capacitors (typically Q2/(2C)) and in the inductors (typically LI2/2). For the special case L1=L2=Ldetermine the relevant eigenvectors and so describe the patterns of current ﬂow in the circuit. 9.5 It is shown in physics and engineering textbooks that circuits containing capaci- tors and inductors can be analysed by replacing a capacitor of capacitance Cby a ‘complex impedance’ 1 /(iωC) and an inductor of inductance Lby an impedance iωL,w h e r e ωis the angular frequency of the currents ﬂowing and i2=−1. Use this approach and Kirchhoﬀ’s circuit laws to analyse the circuit shown in 336 9.4 EXERCISES the ﬁgure and obtain three linear equations governing the currents I1,I2andI3. Show that the only possible frequencies of self-sustaining currents satisfy either L L C CCI1 I2 I3P Q R STU (a)ω2LC=1o r( b )3 ω2LC= 1. Find the corresponding current patterns and, in each case, by identifying parts of the circuit in which no current ﬂows, draw an equivalent circuit that contains only one capacitor and one inductor. 9.6 The simultaneous reduction to diagonal form of two real symmetric quadratic forms. Consider the two real symmetric quadratic forms uTAuand uTBu,w h e r e uT stands for the row matrix ( xyz ), and denote by unthose column matrices that satisfy Bun=λnAun, (E9.1) in which nis a label and the λnare real, non-zero and all diﬀerent. (a) By multiplying (E9.1) on the left by ( um)Tand the transpose of the corre- sponding equation for umon the right by un, show that ( um)TAun=0f o r n/negationslash=m. (b) By noting that Aun=(λn)−1Bun, deduce that ( um)TBun=0f o r m/negationslash=n. It can be shown that the unare linearly independent; the next step is to construct a matrix Pwhose columns are the vectors un. (c) Make a change of variables u=Pvsuch that uTAubecomes vTCv,a n d uTBu becomes vTDv. Show that Cand Dare diagonal by showing that cij=0i f i/negationslash=jand similarly for dij. Thus u=Pvorv=P−1ureduces both quadratics to diagonal form. To summarise, the method is as follows: (a) ﬁnd the λnthat allow (E9.1) a non-zero solution, by solving |B−λA|=0 ; (b) for each λnconstruct un; (c) construct the non-singular matrix Pwhose columns are the vectors un; (d) make the change of variable u=Pv. 9.7 ( It is recommended that the reader does not attempt this question until exercise 9.6 has been studied .) If, in the pendulum system studied in section 9.1, the string is replaced by a second rod identical to the ﬁrst then the expressions for the kinetic energy Tand the potential energy Vbecome (to second order in the θi) T≈Ml2 /;8 3˙θ2 1+2˙θ1˙θ2+2 3˙θ2 2 / , V≈Mgl /;3 2θ2 1+1 2θ2 2 / . Determine the normal frequencies of the system and ﬁnd new variables ξandη that will reduce these two expressions to diagonal form, i.e. to a1˙ξ2+a2˙η2and b1ξ2+b2η2. 337 NORMAL MODES 9.8 ( It is recommended that the reader does not attempt this question until exercise 9.6 has been studied .) Find a real linear transformation that simultaneously reduces the quadratic forms 3x2+5y2+5z2+2yz+6zx−2xy, 5x2+1 2y2+8yz+4zx to diagonal form. 9.9 Three particles of mass mare attached to a light horizontal string having ﬁxed ends, the string being thus divided into four equal portions of length aeach under a tension T. Show that for small transverse vibrations the amplitudes xi of the normal modes satisfy Bx=(maω2/T)x,w h e r e Bis the matrix/0/@2−10 −12−1 0−12 /1A. Estimate the lowest and highest eigenfrequencies using trial vectors (343 )T and(3−43)T. Use also the exact vectors / 1√ 21 /T and / 1−√ 21 /T and compare the results. 9.10 Use the Rayleigh–Ritz method to estimate the lowest oscillation frequency of a heavy chain of Nlinks, each of length a(=L/N), which hangs freely from one end. (Try simple calculable conﬁgurations such as all links but one vertical, orall links collinear, etc.) 9.5 Hints and answers 9.1 See ﬁgure 9.6. 9.2 K.E. = (1 /2)ml2[(1 + α)˙θ2+α˙φ2+2α˙θ˙φ]; P.E. = (1 /2)mgl[(1 + α)θ2+αφ2]. For α=1/3a n d ω= p 2g/l,φ =−2θand the mid-point of BCremains vertically below A. 9.3 (b) x1=/epsilon1(cosωt+c o s√2ωt),x2=−/epsilon1cos√2ωt,x3=/epsilon1(−cosωt+c o s√2ωt). At various times the three displacements will reach 2 /epsilon1, /epsilon1,2/epsilon1respectively. For exam- ple,x1c a nb ew r i t t e na s2 /epsilon1cos[(√ 2−1)ωt/2]cos[(√ 2+1) ωt/2], i.e. an oscillation of angular frequency (√ 2+1) ω/2 and modulated amplitude 2 /epsilon1cos[(√ 2−1)ω/2]; the amplitude will reach 2 /epsilon1after a time ≈4π/[ω(√ 2−1)]. 9.4 Taking separate loops in the left-h and and right-hand sides of the diagram the relevant matrices are A=(L1,0; 0,L2)a n d B=( 2C−1,−C−1;−C−1,2C−1). Whatever the loop choice, ω2must satisfy L1L2C2ω4−2(L1+L2)Cω2+3=0 , which leads to the stated result. The energy stored in the central capacitor is(Q 1−Q2)2/(2C). IfL1=L2=Lthen one mode has ω2=(LC)−1and no current ﬂows through the central capacitor. The other mode has ω2=3 (LC)−1;i nt h i s mode equal currents I(one clockwise, one anticlockwise) ﬂow in the two loops and therefore the current through the central capacitor is 2 I. 9.5 As the circuit loops contain no voltage sources the equations are homogeneous and so for a non-trivial solution the determinant of coeﬃcients must vanish.(a)I 1=0 , I2=−I3; no current in PQ; capacitance C/2 and inductance 2 L. (b)I1=−2I2=−2I3; no current in TU; capacitance 3 C/2 and inductance 2 L. 9.6 (a) Obtain ( λ(n)−λ(m))(u(m))TAu(n)=0 ;( c ) cij=(PTAP)ij=(PT)ikAklPlj= u(i) kAklu(j) l=(u(i))TAu(j)=0f o r i/negationslash=j. 338 9.5 HINTS AND ANSWERS 1 2 3 m m M 2kmkM kM(a)ω2=c+d m (b)ω2=c (c)ω2=c+2d M+d m Figure 9.6 The normal modes, as viewed from above, of the coupled pendu- lums in example 9.1. 9.7 ω=( 2.634g/l)1/2or (0 .3661g/l)1/2;θ1=ξ+η,θ2=1.431ξ−2.097η. 9.8 λ=−1,2,4;x=2ξ−2η+2χ,y=ξ+η+χ,z=−3ξ+η−χ. 9.9 Estimated, 10 /17<M a ω2/T < 58/17; exact, 2 −√ 2≤Maω2/T≤2+√ 2. 9.10 The collinear case gives the best estimate, ω2≤6n2g/(4n3a)≈3g/(2l). 339 10 Vector calculus In chapter 7 we discussed the algebra of vectors, and in chapter 8 we considered how to transform one vector into another using a linear operator. In this chapterand the next we discuss the calculus of vectors, i.e. the diﬀerentiation and integration both of vectors describing particular bodies, such as the velocity of a particle, and of vector ﬁelds, in which a vector is deﬁned as a function of thecoordinates throughout some volume (one-, two- or three-dimensional). Since theaim of this chapter is to develop methods for handling multi-dimensional physicalsituations, we will assume throughout that the functions with which we have todeal have suﬃciently amenable mathematical properties, in particular that theyare continuous and diﬀerentiable. 10.1 Diﬀerentiation of vectors L e tu sc o n s i d e rav e c t o r athat is a function of a scalar variable u.B yt h i s we mean that with each value of uwe associate a vector a(u). For example, in Cartesian coordinates a(u)=a x(u)i+ay(u)j+az(u)k,w h e r e ax(u),ay(u)a n d az(u) a r es c a l a rf u n c t i o n so f uand are the components of the vector a(u)i nt h e x-,y- andz- directions respectively. We note that if a(u) is continuous at some point u=u0then this implies that each of the Cartesian components ax(u),ay(u)a n d az(u) is also continuous there. Let us consider the derivative of the vector function a(u) with respect to u. The derivative of a vector function is deﬁned in a similar manner to the ordinaryderivative of a scalar function f(x) given in chapter 2. The small change in the vector a(u) resulting from a small change ∆ uin the value of uis given by ∆a=a(u+∆u)−a(u) (see ﬁgure 10.1). The derivative of a(u)w i t hr e s p e c tt o uis deﬁned to be da du= lim ∆u→0a(u+∆u)−a(u) ∆u, (10.1) 340 10.1 DIFFERENTIATION OF VECTORS a(u)a(u+∆u)∆a=a(u+∆u)−a(u) Figure 10.1 A small change in a vector a(u) resulting from a small change inu. assuming that the limit exists, in which case a(u) is said to be diﬀerentiable at that point. Note that da/duis also a vector, which is not, in general, parallel to a(u). In Cartesian coordinates, the derivative of the vector a(u)=axi+ayj+azk is given by da du=dax dui+day duj+daz duk. Perhaps the simplest application of the above is to ﬁnding the velocity and acceleration of a particle in classical mechanics. If the time-dependent position vector of the particle with respect to the origin in Cartesian coordinates is given byr(t)=x(t)i+y(t)j+z(t)kthen the velocity of the particle is given by the vector v(t)=dr dt=dx dti+dy dtj+dz dtk. The direction of the velocity vector is along the tangent to the path r(t)a tt h e instantaneous position of the particle, and its magnitude |v(t)|is equal to the speed of the particle. The acceleration of the particle is given in a similar mannerby a(t)=dv dt=d2x dt2i+d2y dt2j+d2z dt2k.IThe position vector of a particle at time tin Cartesian coordinates is given by r(t)= 2t2i+( 3t−2)j+( 3t2−1)k. Find the speed of the particle at t=1and the component of its acceleration in the direction s=i+2j+k. The velocity and acceleration of the particle are given by v(t)=dr dt=4ti+3j+6tk, a(t)=dv dt=4i+6k. 341 VECTOR CALCULUS y xφˆeφ ˆeρ ρij Figure 10.2 Unit basis vectors for two-dimensional Cartesian and plane polar coordinates. The speed of the particle at t= 1 is simply |v(1)|= p 42+32+62=√ 61. The acceleration of the particle is constant (i.e. independent of t), and its component in the direction sis given by a·ˆs=(4i+6k)·(i+2j+k)√ 12+22+12=5√ 6 3. J Note that in the case discussed above i,jandkare ﬁxed, time-independent basis vectors. This may not be true of basis vectors in general; when we arenot using Cartesian coordinates the basis vectors themselves must also be dif-ferentiated. We discuss basis vectors for non-Cartesian coordinate systems indetail in section 10.10. Nevertheless, as a simple example, let us now considertwo-dimensional plane polar coordinates ρ, φ. Referring to ﬁgure 10.2, imagine holding φﬁxed and moving radially outwards, i.e. in the direction of increasing ρ. Let us denote the unit vector in this direction byˆe ρ. Similarly, imagine keeping ρﬁxed and moving around a circle of ﬁxed radius in the direction of increasing φ. Let us denote the unit vector tangent to the circle byˆeφ. The two vectors ˆeρandˆeφare the basis vectors for this two-dimensional coordinate system, just as iandjare basis vectors for two-dimensional Cartesian coordinates. All these basis vectors are shown in ﬁgure 10.2. An important diﬀerence between the two sets of basis vectors is that, while iandjare constant in magnitude and direction , the vectors ˆeρandˆeφhave constant magnitudes but their directions change as ρand φvary. Therefore, when calculating the derivative of a vector written in polar coordinates we mustalso diﬀerentiate the basis vectors. One way of doing this is to express ˆe ρandˆeφ 342 10.1 DIFFERENTIATION OF VECTORS in terms of iandj. From ﬁgure 10.2, we see that ˆeρ=c o s φi+s i n φj, ˆeφ=−sinφi+c o s φj. Since iandjare constant vectors, we ﬁnd that the derivatives of the basis vectors ˆeρandˆeφwith respect to tare given by dˆeρ dt=−sinφdφ dti+c o s φdφ dtj=˙φˆeφ, (10.2) dˆeφ dt=−cosφdφ dti−sinφdφ dtj=−˙φˆeρ, (10.3) where the overdot is the conventional notation for diﬀerentiation with respect to time.IThe position vector of a particle in plane polar coordinates is r(t)=ρ(t)ˆeρ.F i n de x p r e s - sions for the velocity and acceleration of the particle in these coordinates. Using result (10.4) below, the velocity of the particle is given by v(t)=˙r(t)=˙ρˆeρ+ρ˙ˆeρ=˙ρˆeρ+ρ˙φˆeφ, where we have used (10.2). In a similar way its acceleration is given by a(t)=d dt(˙ρˆeρ+ρ˙φˆeφ) =¨ρˆeρ+˙ρ˙ˆeρ+ρ˙φ˙ˆeφ+ρ¨φˆeφ+˙ρ˙φˆeφ =¨ρˆeρ+˙ρ(˙φˆeφ)+ρ˙φ(−˙φˆeρ)+ρ¨φˆeφ+˙ρ˙φˆeφ =(¨ρ−ρ˙φ2)ˆeρ+(ρ¨φ+2˙ρ˙φ)ˆeφ. J Here we have used (10.2) and (10.3). 10.1.1 Diﬀerentiation of composite vector expressions In composite vector expressions each of the vectors or scalars involved may be a function of some scalar variable u, as we have seen. The derivatives of such expressions are easily found using the deﬁnition (10.1) and the rules of ordinarydiﬀerential calculus. They may be summarised by the following, in which we assume that aandbare diﬀerentiable vector functions of a scalar uand that φ is a diﬀerentiable scalar function of u: d du(φa)=φda du+dφ dua, (10.4) d du(a·b)=a·db du+da du·b, (10.5) d du(a×b)=a×db du+da du×b; (10.6) 343 VECTOR CALCULUS the order of the factors in the terms on the RHS of (10.6) is, of course, just as important as it is in the original vector product.IAp a r t i c l eo fm a s s mwith position vector rrelative to some origin Oexperiences a force F, which produces a torque (moment) T=r×Fabout O. The angular momentum of the particle about Ois given by L=r×mv,w h e r e vis the particle’s velocity. Show that the rate of change of angular momentum is equal to the applied torque. The rate of change of angular momentum is given by dL dt=d dt(r×mv). Using (10.6) we obtain dL dt=dr dt×mv+r×d dt(mv) =v×mv+r×d dt(mv) =0+r×F=T, where in the last line we use Newton’s second law, namely F=d(mv)/dt. J If a vector ais a function of a scalar variable sthat is itself a function of u,s o thats=s(u), then the chain rule (see subsection 2.1.3) gives da(s) du=ds duda ds. (10.7) The derivatives of more complicated vector expressions may be found by repeated application of the above equations. One further useful result can be derived by considering the derivative d du(a·a)=2a·da du; sincea·a=a2,w h e r e a=|a|,w es e et h a t a·da du=0 i f ais constant. (10.8) In other words, if a vector a(u) has a constant magnitude as uvaries then it is perpendicular to the vector da/du. 10.1.2 Diﬀerential of a vector As a ﬁnal note on the diﬀerentiation of vectors, we can also deﬁne the diﬀerential of a vector, in a similar way to that of a scalar in ordinary diﬀerential calculus. In the deﬁnition of the vector derivative (10.1), we used the notion of a small change ∆ ain a vector a(u) resulting from a small change ∆ uin its argument. In the limit ∆ u→0, the change in abecomes inﬁnitesimally small, and we denote it by the diﬀerential da. From (10.1) we see that the diﬀerential is given by da=da dudu. (10.9) 344 10.2 INTEGRATION OF VECTORS Note that the diﬀerential of a vector is also a vector. As an example, the inﬁnitesimal change in the position vector of a particle in an inﬁnitesimal time dtis dr=dr dtdt=vdt, where vis the particle’s velocity. 10.2 Integration of vectors The integration of a vector (or of an expression involving vectors that may itself be either a vector or scalar) with respect to a scalar uc a nb er e g a r d e da st h e inverse of diﬀerentiation. We must remember, however, that (i) the integral has the same nature (vector or scalar) as the integrand, (ii) the constant of integration for indeﬁnite integrals must be of the same nature as the integral. For example, if a(u)=d[A(u)]/duthen the indeﬁnite integral of a(u)i sg i v e nb y integraldisplay a(u)du=A(u)+b, where bis a constant vector. The deﬁnite integral of a(u)f r o m u=u1tou=u2 is given by integraldisplayu2 u1a(u)du=A(u2)−A(u1).IA small particle of mass morbits a much larger mass Mcentred at the origin O. According to Newton’s law of gravitation, the position vector rof the small mass obeys the diﬀerential equation md2r dt2=−GMm r2ˆr. Show that the vector r×dr/dtis a constant of the motion. Forming the vector product of the diﬀerential equation with r,w eo b t a i n r×d2r dt2=−GM r2r×ˆr. Since randˆrare collinear, r×ˆr=0and therefore we have r×d2r dt2=0. (10.10) However, d dt / r×dr dt / =r×d2r dt2+dr dt×dr dt=0, 345 VECTOR CALCULUS xz yC Oˆbˆtˆn P r(u) Figure 10.3 The unit tangent ˆt,n o r m a l ˆnand binormal ˆbto the space curve C at a particular point P. since the ﬁrst term is zero by (10.10), and the second is zero because it is the vector product of two parallel (in this case identical) vectors. Integrating, we obtain the required result r×dr dt=c, (10.11) where cis a constant vector. As a further point of interest we may note that in an inﬁnitesimal time dtthe change in the position vector of the small mass is drand the element of area swept out by the position vector of the particle is simply dA=1 2|r×dr|. Dividing both sides of this equation bydt, we conclude that dA dt=1 2 ////r×dr dt ////=|c| 2, and that the physical interpretation of the above result (10.11) is that the position vector r of the small mass sweeps out equal areas in equal times. This result is in fact valid formotion under any force that acts along the line joining the two particles.J 10.3 Space curves In the previous section we mentioned that the velocity vector of a particle is a tangent to the curve in space along which the particle moves. We now give a morecomplete discussion of curves in space and also of the geometrical interpretationof the vector derivative. Ac u r v e Cin space can be described by the vector r(u) joining the origin Oof a coordinate system to a point on the curve (see ﬁgure 10.3). As the parameter u varies, the end-point of the vector moves along the curve. In Cartesian coordinates, r(u)=x(u)i+y(u)j+z(u)k, where x=x(u),y=y(u)a n d z=z(u)a r et h e parametric equations of the curve. 346 10.3 SPACE CURVES This parametric representation can be very useful, particularly in mechanics when the parameter may be the time t. We can, however, also represent a space curve byy=f(x),z=g(x), which can be easily converted into the above parametric form by setting u=x,s ot h a t r(u)=ui+f(u)j+g(u)k. Alternatively, a space curve can be represented in the form F(x, y, z)=0 , G(x, y, z) = 0, where each equation represents a surface and the curve is the intersection of the two surfaces. A curve may sometimes be described in parametric form by the vector r(s), where the parameter sis the arc length along the curve measured from a ﬁxed point. Even when the curve is expressed in terms of some other parameter, it isstraightforward to ﬁnd the arc length between any two points on the curve. Forthe curve described by r(u), let us consider an inﬁnitesimal vector displacement dr=dxi+dyj+dzk along the curve. The square of the inﬁnitesimal distance moved is then given by (ds) 2=dr·dr=(dx)2+(dy)2+(dz)2, f r o mw h i c hi tc a nb es h o w nt h a t parenleftbiggds duparenrightbigg2 =dr du·dr du. Therefore, the arc length between two points on the curve r(u), given by u=u1 andu=u2,i s s=integraldisplayu2 u1radicalbigg dr du·dr dudu. (10.12)IAc u r v el y i n gi nt h e xy-plane is given by y=y(x),z=0. Using (10.12), show that the arc length along the curve between x=aandx=bis given by s= Rb a p 1+y/prime2dx,w h e r e y/prime=dy/dx . Let us ﬁrst represent the curve in parametric form by setting u=x,s ot h a t r(u)=ui+y(u)j. Diﬀerentiating with respect to u, we ﬁnd dr du=i+dy duj, from which we obtain dr du·dr du=1+ /dy du /2 . 347 VECTOR CALCULUS Therefore, remembering that u=x, from (10.12) the arc length between x=aandx=b is given by s= Zb a r dr du·dr dudu= Zb a s 1+ /dy dx /2 dx. This result was derived using more elementary methods in chapter 2. J If a curve Cis described by r(u) then, by considering ﬁgures 10.1 and 10.3, we see that, at any given point on the curve, dr/duis a vector tangent to Cat that point, in the direction of increasing u. In the special case where the parameter u is the arc length salong the curve then dr/dsis aunittangent vector to Cand is denoted by ˆt. The rate at which the unit tangent ˆtchanges with respect to sis given by dˆt/ds, and its magnitude is deﬁned as the curvature κof the curve Cat a given point, κ=vextendsinglevextendsinglevextendsinglevextendsingledˆt dsvextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsingled 2ˆr ds2vextendsinglevextendsinglevextendsinglevextendsingle. We can also deﬁne the quantity ρ=1/κ, which is called the radius of curvature . Since ˆtis of constant (unit) magnitude, it follows from (10.8) that it is perpen- dicular to dˆt/ds. The unit vector in the direction perpendicular to ˆtis denoted byˆnand is called the principal normal at the point. We therefore have dˆt ds=κˆn. (10.13) The unit vector ˆb=ˆt×ˆn, which is perpendicular to the plane containing ˆt andˆn, is called the binormal toC. The vectors ˆt,ˆnandˆbform a right-handed rectangular cooordinate system (or triad) at any given point on C(see ﬁgure 10.3). Asschanges so that the point of interest moves along C, the triad of vectors also changes. The rate at which ˆbchanges with respect to sis given by dˆb/dsand is a measure of the torsion τof the curve at any given point. Since ˆbis of constant magnitude, from (10.8) it is perpendicular to dˆb/ds. We may further show that dˆb/dsis also perpendicular to ˆt, as follows. By deﬁnition ˆb·ˆt=0 ,w h i c ho n diﬀerentiating yields 0=d dsparenleftBig ˆb·ˆtparenrightBig =dˆb ds·ˆt+ˆb·dˆt ds =dˆb ds·ˆt+ˆb·κˆn =dˆb ds·ˆt, where we have used the fact that ˆb·ˆn= 0. Hence, since dˆb/dsis perpendicular to both ˆbandˆt, we must have dˆb/ds∝ˆn. The constant of proportionality is −τ, 348 10.3 SPACE CURVES so we ﬁnally obtain dˆb ds=−τˆn. (10.14) Taking the dot product of each side with ˆn, we see that the torsion of a curve is given by τ=−ˆn·dˆb ds. We may also deﬁne the quantity σ=1/τ, which is called the radius of torsion . Finally, we consider the derivative dˆn/ds.S i n c e ˆn=ˆb×ˆtwe have dˆn ds=dˆb ds×ˆt+ˆb×dˆt ds =−τˆn×ˆt+ˆb×κˆn =τˆb−κˆt. (10.15) In summary, ˆt,ˆnandˆband their derivatives with respect to sare related to one another by the relations (10.13), (10.14) and (10.15), the Frenet–Serret formulae , dˆt ds=κˆn,dˆn ds=τˆb−κˆt,dˆb ds=−τˆn. (10.16)IShow that the acceleration of a particle travelling along a trajectory r(t)is given by a(t)=dv dtˆt+v2 ρˆn, where vis the speed of the particle, ˆtis the unit tangent to the trajectory, ˆnis its principal normal and ρis its radius of curvature. The velocity of the particle is given by v(t)=dr dt=dr dsds dt=ds dtˆt, where ds/dt is the speed of the particle, which we denote by v,a n d ˆtis the unit vector tangent to the trajectory. Writing the velocity as v=vˆt, and diﬀerentiating once more with respect to time t,w eo b t a i n a(t)=dv dt=dv dtˆt+vdˆt dt; but we note that dˆt dt=ds dtdˆt ds=vκˆn=v ρˆn. Therefore, we have a(t)=dv dtˆt+v2 ρˆn. This shows that in addition to an acceleration dv/dt along the tangent to the particle’s trajectory, there is also an acceleration v2/ρin the direction of the principal normal. The latter is often called the centripetal acceleration. J 349 VECTOR CALCULUS Finally, we note that a curve r(u) representing the trajectory of a particle may sometimes be given in terms of some parameter uthat is not necessarily equal to the time tbut is functionally related to it in some way. In this case the velocity of the particle is given by v=dr dt=dr dudu dt. Diﬀerentiating again with respect to time gives the acceleration as a=dv dt=d dtparenleftbiggdr dudu dtparenrightbigg =d2r du2parenleftbiggdu dtparenrightbigg2 +dr dud2u dt2. 10.4 Vector functions of several arguments The concept of the derivative of a vector is easily extended to cases where the vectors (or scalars) are functions of more than one independent scalar variable,u 1,u2,...,u n. In this case, the results of subsection 10.1.1 are still valid, except that the derivatives become partial derivatives ∂a/∂u ideﬁned as in ordinary diﬀerential calculus. For example, in Cartesian coordinates, ∂a ∂u=∂ax ∂ui+∂ay ∂uj+∂az ∂uk. In particular, (10.7) generalises to the chain rule of partial diﬀerentiation discussed in section 5.5. If a=a(u1,u2,...,u n) and each of the uiis also a function ui(v1,v2,...,v n) of the variables vithen, generalising (5.17), ∂a ∂vi=∂a ∂u1∂u1 ∂vi+∂a ∂u2∂u2 ∂vi+···+∂a ∂un∂un ∂vi=nsummationdisplay j=1∂a ∂uj∂uj ∂vi. (10.17) A special case of this rule arises when ais an explicit function of some variable v,a sw e l la so fs c a l a r s u1,u2,...,u nthat are themselves functions of v;t h e nw e have da dv=∂a ∂v+nsummationdisplay j=1∂a ∂uj∂uj ∂v. (10.18) We may also extend the concept of the diﬀerential of a vector given in (10.9) to vectors dependent on several variables u1,u2,...,u n: da=∂a ∂u1du1+∂a ∂u2du2+···+∂a ∂undun=nsummationdisplay j=1∂a ∂ujduj. (10.19) As an example, the inﬁnitesimal change in an electric ﬁeld Ein moving from a position rto a neighbouring one r+dris given by dE=∂E ∂xdx+∂E ∂ydy+∂E ∂zdz. (10.20) 350 10.5 SURFACES xyz S OT P r(u, v)∂r ∂u ∂r ∂v v=c2u=c1 Figure 10.4 The tangent plane Tto a surface Sat a particular point P; u=c1andv=c2are the coordinate curves, shown by dotted lines, that pass through P. The broken line shows some particular parametric curve r=r(λ) lying in the surface. 10.5 Surfaces As u r f a c e Sin space can be described by the vector r(u, v) joining the origin Oof a coordinate system to a point on the surface (see ﬁgure 10.4). As the parametersuandvvary, the end-point of the vector moves over the surface. This is very similar to the parametric representation r(u) of a curve, discussed in section 10.3, but with the important diﬀerence that we require twoparameters to describe a surface, whereas we need only one to describe a curve. In Cartesian coordinates the surface is given by r(u, v)=x(u, v)i+y(u, v)j+z(u, v)k, where x=x(u, v),y=y(u, v)a n d z=z(u, v) are the parametric equations of the surface. We can also represent a surface by z=f(x, y)o rg(x, y, z)=0 .E i t h e r of these representations can be converted into the parametric form in a similarmanner to that used for equations of curves. For example, if z=f(x, y)t h e nb y setting u=xandv=ythe surface can be represented in parametric form by r(u, v)=ui+vj+f(u, v)k. Any curve r(λ), where λis a parameter, on the surface Scan be represented by a pair of equations relating the parameters uandv, for example u=f(λ) andv=g(λ). A parametric representation of the curve can easily be found by straightforward substitution, i.e. r(λ)=r(u(λ),v(λ)). Using (10.17) for the case where the vector is a function of a single variable λso that the LHS becomes a 351 VECTOR CALCULUS total derivative, the tangent to the curve r(λ) at any point is given by dr dλ=∂r ∂udu dλ+∂r ∂vdv dλ. (10.21) The two curves u=c o n s t a n ta n d v= constant passing through any point P onSare called coordinate curves .F o rt h ec u r v e u= constant, for example, we have du/dλ = 0, and so from (10.21) its tangent vector is in the direction ∂r/∂v. Similarly, the tangent vector to the curve v= constant is in the direction ∂r/∂u. If the surface is smooth then at any point PonSthe vectors ∂r/∂uand ∂r/∂vare linearly independent and deﬁne the tangent plane Tat the point P(see ﬁgure 10.4). A vector normal to the surface at Pis given by n=∂r ∂u×∂r ∂v. (10.22) In the neighbourhood of P, an inﬁnitesimal vector displacement dris written dr=∂r ∂udu+∂r ∂vdv. Theelement of area atP, an inﬁnitesimal parallelogram whose sides are the coordinate curves, has magnitude dS=vextendsinglevextendsinglevextendsinglevextendsingle∂r ∂udu×∂r ∂vdvvextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsingle∂r ∂u×∂r ∂vvextendsinglevextendsinglevextendsinglevextendsingledu dv=|n|du dv. (10.23) Thus the total area of the surface is A=integraldisplayintegraldisplay Rvextendsinglevextendsinglevextendsinglevextendsingle∂r ∂u×∂r ∂vvextendsinglevextendsinglevextendsinglevextendsingledu dv=integraldisplayintegraldisplay R|n|du dv, (10.24) where Ris the region in the uv-plane corresponding to the range of parameter values that deﬁne the surface.IFind the element of area on the surface of a sphere of radius a, and hence calculate the total surface area of the sphere. We can represent a point ron the surface of the sphere in terms of the two parameters θ andφ: r(θ,φ)=asinθcosφi+asinθsinφj+acosθk, where θandφare the polar and azimuthal angles respectively. At any point P, vectors tangent to the coordinate curves θ=c o n s t a n ta n d φ= constant are ∂r ∂θ=acosθcosφi+acosθsinφj−asinθk, ∂r ∂φ=−asinθsinφi+asinθcosφj. 352 10.6 SCALAR AND VECTOR FIELDS An o r m a l nto the surface at this point is then given by n=∂r ∂θ×∂r ∂φ= ///////ij k acosθcosφa cosθsinφ−asinθ −asinθsinφasinθcosφ 0 /////// =a2sinθ(sinθcosφi+s i n θsinφj+c o s θk), which has a magnitude of a2sinθ. Therefore, the element of area at Pis, from (10.23), dS=a2sinθd θd φ , and the total surface area of the sphere is given by A= Zπ 0dθ Z2π 0dφ a2sinθ=4πa2. This familiar result can, of course, be proved by much simpler methods! J 10.6 Scalar and vector ﬁelds We now turn to the case where a particular scalar or vector quantity is deﬁned not just at a point in space but continuously as a ﬁeldthroughout some region of space R(which is often the whole space). Although the concept of a ﬁeld is valid for spaces with an arbitrary number of dimensions, in the remainder of thischapter we will restrict our attention to the familiar three-dimensional case. Ascalar ﬁeld φ(x, y, z) associates a scalar with each point in R, while a vector ﬁeld a(x, y, z) associates a vector with each point. In what follows, we will assume that the variation in the scalar or vector ﬁeld from point to point is both continuousand diﬀerentiable in R. Simple examples of scalar ﬁelds include the pressure at each point in a ﬂuid and the electrostatic potential at each point in space in the presence of an electric charge. Vector ﬁelds relating to the same physical systems are the velocity vector in a ﬂuid (giving the local speed and direction of the ﬂow) and the electric ﬁeld. With the study of continuously varying scalar and vector ﬁelds there arises the need to consider their derivatives and also the integration of ﬁeld quantities alonglines, over surfaces and throughout volumes in the ﬁeld. We defer the discussionof line, surface and volume integrals until the next chapter, and in the remainderof this chapter we concentrate on the deﬁnition of vector diﬀerential operatorsand their properties. 10.7 Vector operators Certain diﬀerential operations may be performed on scalar and vector ﬁelds and have wide-ranging applications in the physical sciences. The most important operations are those of ﬁnding the gradient of a scalar ﬁeld and the divergence andcurlof a vector ﬁeld. It is usual to deﬁne these operators from a strictly 353 VECTOR CALCULUS mathematical point of view, as we do below. In the following chapter, however, we will discuss their geometrical deﬁnitions, which rely on the concept of integratingvector quantities along lines and over surfaces. Central to all these diﬀerential operations is the vector operator ∇,w h i c hi s called del(or sometimes nabla) and in Cartesian coordinates is deﬁned by ∇≡i∂ ∂x+j∂ ∂y+k∂ ∂z. (10.25) The form of this operator in non-Cartesian coordinate systems is discussed in sections 10.9 and 10.10. 10.7.1 Gradient of a scalar ﬁeld Thegradient of a scalar ﬁeld φ(x, y, z) is deﬁned by grad φ=∇φ=i∂φ ∂x+j∂φ ∂y+k∂φ ∂z. (10.26) Clearly,∇φis a vector ﬁeld whose x-,y-a n d z- components are the ﬁrst partial derivatives of φ(x, y, z) with respect to x,yandzrespectively. Also note that the vector ﬁeld ∇φshould not be confused with the vector operator φ∇, which has components ( φ∂ / ∂ x ,φ∂ / ∂ y,φ∂ / ∂ z ).IFind the gradient of the scalar ﬁeld φ=xy2z3. From (10.26) the gradient of φis given by ∇φ=y2z3i+2xyz3j+3xy2z2k. J The gradient of a scalar ﬁeld φhas some interesting geometrical properties. Let us ﬁrst consider the problem of calculating the rate of change of φin some particular direction . For an inﬁnitesimal vector displacement dr, forming its scalar product with ∇φwe obtain ∇φ·dr=parenleftbigg i∂φ ∂x+j∂φ ∂y+k∂φ ∂zparenrightbigg ·(idx+jdy+kdx), =∂φ ∂xdx+∂φ ∂ydy+∂φ ∂zdz, =dφ, (10.27) which is the inﬁnitesimal change in φin going from position rtor+dr.I n particular, if rdepends on some parameter usuch that r(u) deﬁnes a space curve 354 10.7 VECTOR OPERATORS φ=c o n s t a n t∇φ a PQ dφ dsin the direction aθ Figure 10.5 Geometrical properties of ∇φ.PQgives the value of dφ/ds in the direction a. then the total derivative of φwith respect to ualong the curve is simply dφ du=∇φ·dr du; (10.28) in the particular case where the parameter uis the arc length salong the curve, the total derivative of φwith respect to salong the curve is given by dφ ds=∇φ·ˆt, (10.29) where ˆtis the unit tangent to the curve at the given point, as discussed in section 10.3. In general, the rate of change of φwith respect to the distance sin a particular direction ais given by dφ ds=∇φ·ˆa (10.30) and is called the directional derivative. Since ˆais a unit vector we have dφ ds=|∇φ|cosθ where θis the angle between ˆaand∇φas shown in ﬁgure 10.5. Clearly ∇φlies in the direction of the fastest increase in φ,a n d|∇φ|is the largest possible value ofdφ/ds . Similarly, the largest rate of decrease of φisdφ/ds =−|∇φ|in the direction of −∇φ. 355 VECTOR CALCULUSIFor the function φ=x2y+yzat the point (1,2,−1), ﬁnd its rate of change with distance in the direction a=i+2j+3k. At this same point, what is the greatest possible rate of change with distance and in which direction does it occur? The gradient of φis given by (10.26): ∇φ=2xyi+(x2+z)j+yk, =4i+2kat the point (1 ,2,−1). The unit vector in the direction of aisˆa=1√ 14(i+2j+3k), so the rate of change of φ with distance sin this direction is, using (10.30), dφ ds=∇φ·ˆa=1√ 14(4 + 6) =10√ 14. From the above discussion, at the point (1 ,2,−1)dφ/ds will be greatest in the direction of∇φ=4i+2kand has the value |∇φ|=√ 20 in this direction. J We can extend the above analysis to ﬁnd the rate of change of a vector ﬁeld (rather than a scalar ﬁeld as above) in a particular direction. The scalardiﬀerential operator ˆa·∇can be shown to give the rate of change with distance in the direction ˆaof the quantity (vector or scalar) on which it acts. In Cartesian coordinates it may be written as ˆa·∇=a x∂ ∂x+ay∂ ∂y+az∂ ∂z. (10.31) Thus we can write the inﬁnitesimal change in an electric ﬁeld in moving from r tor+drgiven in (10.20) as dE=(dr·∇)E. A second interesting geometrical property of ∇φmay be found by considering the surface deﬁned by φ(x, y, z)=c,w h e r e cis some constant. If ˆtis a unit tangent to this surface at some point then clearly dφ/ds = 0 in this direction and from (10.29) we have ∇φ·ˆt= 0. In other words, ∇φis a vector normal to the surface φ(x, y, z)=cat every point , as shown in ﬁgure 10.5. If ˆnis a unit normal to the surface in the direction of increasing φ(x, y, z), then the gradient is sometimes written ∇φ≡∂φ ∂nˆn, (10.32) where ∂φ/∂n≡| ∇φ|is the rate of change of φin the direction ˆnand is called thenormal derivative .IFind expressions for the equations of the tangent plane and the line normal to the surface φ(x, y, z)=cat the point Pwith coordinates x0,y0,z0. Use the results to ﬁnd the equations of the tangent plane and the line normal to the surface of the sphere φ=x2+y2+z2=a2 at the point (0,0,a). A vector normal to the surface φ(x, y, z)=cat the point Pis simply∇φevaluated at that point; we denote it by n0.I fr0is the position vector of the point Prelative to the origin, 356 10.7 VECTOR OPERATORS xyz ˆn0 (0,0,a) O a φ=x2+y2+z2=a2z=a Figure 10.6 The tangent plane and the normal to the surface of the sphere φ=x2+y2+z2=a2at the point r0with coordinates (0 ,0,a). andris the position vector of any point on the tangent plane, then the vector equation of the tangent plane is, from (7.41), (r−r0)·n0=0. Similarly, if ris the position vector of any point on the straight line passing through P (with position vector r0) in the direction of the normal n0then the vector equation of this line is, from subsection 7.7.1, (r−r0)×n0=0. For the surface of the sphere φ=x2+y2+z2=a2, ∇φ=2xi+2yj+2zk =2akat the point (0 ,0,a). Therefore the equation of the tangent plane to the sphere at this point is (r−r0)·2ak=0. This gives 2 a(z−a)=0o r z=a, as expected. The equation of the line normal to the sphere at the point (0 ,0,a)i s (r−r0)×2ak=0, which gives 2 ayi−2axj=0orx=y=0 ,i . e .t h e z-axis, as expected. The tangent plane and normal to the surface of the sphere at this point are shown in ﬁgure 10.6. J Further properties of the gradient operation, which are analogous to those of the ordinary derivative, are listed in subsection 10.8.1 and may be easily proved. 357 VECTOR CALCULUS In addition to these, we note that the gradient operation also obeys the chain rule as in ordinary diﬀerential calculus, i.e. if φandψare scalar ﬁelds in some region Rthen ∇[φ(ψ)]=∂φ ∂ψ∇ψ. 10.7.2 Divergence of a vector ﬁeld Thedivergence of a vector ﬁeld a(x, y, z) is deﬁned by diva=∇·a=∂ax ∂x+∂ay ∂y+∂az ∂z, (10.33) where ax,ayandazare the x-,y-a n d z- components of a. Clearly,∇·ais a scalar ﬁeld. Any vector ﬁeld afor which∇·a= 0 is said to be solenoidal .IFind the divergence of the vector ﬁeld a=x2y2i+y2z2j+x2z2k. From (10.33) the divergence of ais given by ∇·a=2xy2+2yz2+2x2z=2 (xy2+yz2+x2z). J We will discuss fully the geometric deﬁnition of divergence and its physical meaning in the next chapter. For the moment, we merely note that the divergence can be considered as a quantitative measure of how much a vector ﬁeld diverges(spreads out) or converges at any given point. For example, if we consider thevector ﬁeld v(x, y, z) describing the local velocity at any point in a ﬂuid then ∇·v is equal to the net rate of outﬂow of ﬂuid per unit volume, evaluated at a point(by letting a small volume at that point tend to zero). Now if some vector ﬁeld ais itself derived from a scalar ﬁeld via a=∇φthen ∇·ahas the form ∇·∇φo r ,a si ti su s u a l l yw r i t t e n , ∇ 2φ,w h e r e∇2(del squared) is the scalar diﬀerential operator ∇2≡∂2 ∂x2+∂2 ∂y2+∂2 ∂z2. (10.34) ∇2φis called the Laplacian ofφand appears in several important partial diﬀer- ential equations of mathematical physics, discussed in chapters 18 and 19.IFind the Laplacian of the scalar ﬁeld φ=xy2z3. From (10.34) the Laplacian of φis given by ∇2φ=∂2φ ∂x2+∂2φ ∂y2+∂2φ ∂z2=2xz3+6xy2z. J 358 10.7 VECTOR OPERATORS 10.7.3 Curl of a vector ﬁeld Thecurlof a vector ﬁeld a(x, y, z) is deﬁned by curla=∇×a=parenleftbigg∂az ∂y−∂ay ∂zparenrightbigg i+parenleftbigg∂ax ∂z−∂az ∂xparenrightbigg j+parenleftbigg∂ay ∂x−∂ax ∂yparenrightbigg k, where ax,ayandazare the x-,y-a n d z- components of a. The RHS can be written in a more memorable form as a determinant: ∇×a=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleij k ∂ ∂x∂ ∂y∂ ∂z axayazvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle, (10.35) where it is understood that, on expanding the determinant, the partial derivatives in the second row act on the components of ain the third row. Clearly, ∇×a is itself a vector ﬁeld. Any vector ﬁeld afor which∇×a=0is said to be irrotational .IFind the curl of the vector ﬁeld a=x2y2z2i+y2z2j+x2z2k. The curl of ais given by ∇φ= ////// // ij k ∂ ∂x∂ ∂y∂ ∂z x2y2z2y2z2x2z2 ////// // =−2 / y2zi+(xz2−x2y2z)j+x2yz2k / . J For a vector ﬁeld v(x, y, z) describing the local velocity at any point in a ﬂuid, ∇×vis a measure of the angular velocity of the ﬂuid in the neighbourhood of that point. If a small paddle wheel were placed at various points in the ﬂuid thenit would tend to rotate in regions where ∇×v/negationslash=0, while it would not rotate in regions where ∇×v=0. Another insight into the physical interpretation of the curl operator is gained by considering the vector ﬁeld vdescribing the velocity at any point in a rigid body rotating about some axis with angular velocity ω.I fris the position vector of the point with respect to some origin on the axis of rotation then the velocityof the point is given by v=ω×r. Without any loss of generality, we may take ωto lie along the z-axis of our coordinate system, so that ω=ωk.T h ev e l o c i t y ﬁeld is then v=−ωyi+ωxj. The curl of this vector ﬁeld is easily found to be ∇×v=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleij k ∂ ∂x∂ ∂y∂ ∂z −ωy ωx 0vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=2ωk=2ω. (10.36) 359 VECTOR CALCULUS ∇(φ+ψ)=∇φ+∇ψ ∇·(a+b)=∇·a+∇·b ∇×(a+b)=∇×a+∇×b ∇(φψ)=φ∇ψ+ψ∇φ ∇(a·b)=a×(∇×b)+b×(∇×a)+(a·∇)b+(b·∇)a ∇·(φa)=φ∇·a+a·∇φ ∇·(a×b)=b·(∇×a)−a·(∇×b) ∇×(φa)=∇φ×a+φ∇×a ∇×(a×b)=a(∇·b)−b(∇·a)+(b·∇)a−(a·∇)b Table 10.1 Vector operators acting on sums and products. The operator ∇is deﬁned in (10.25); φandψare scalar ﬁelds, aandbare vector ﬁelds. Therefore the curl of the velocity ﬁeld is a vector equal to twice the angular velocity vector of the rigid body about its axis of rotation. We give a fullgeometrical discussion of the curl of a vector in the next chapter. 10.8 Vector operator formulae In the same way as for ordinary vectors (chapter 7), for vector operators certain identities exist. In addition, we must consider various relations involving theaction of vector operators on sums and products of scalar and vector ﬁelds. Some of these relations have been mentioned above, but we list all the most important ones here for convenience. The validity of these relations may be easily veriﬁedby direct calculation (a quick method of deriving them using tensor notation isgiven in chapter 21). Although some of the following vector relations are expressed in Cartesian coordinates, it may be proved that they are all independent of the choice ofcoordinate system. This is to be expected since grad, div and curl all have cleargeometrical deﬁnitions, which are discussed more fully in the next chapter andwhich do not rely on any particular choice of coordinate system. 10.8.1 Vector operators acting on sums and products Letφandψbe scalar ﬁelds and aandbbe vector ﬁelds. Assuming these ﬁelds are diﬀerentiable, the action of grad, div and curl on various sums and productsof them is presented in table 10.1. These relations can be proved by direct calculation. 360 10.8 VECTOR OPERATOR FORMULAEIShow that ∇×(φa)=∇φ×a+φ∇×a. Thex-component of the LHS is ∂ ∂y(φaz)−∂ ∂z(φay)=φ∂az ∂y+∂φ ∂yaz−φ∂ay ∂z−∂φ ∂zay, =φ /∂az ∂y−∂ay ∂z / + /∂φ ∂yaz−∂φ ∂zay / , =φ(∇×a)x+(∇φ×a)x, where, for example, ( ∇φ×a)xdenotes the x-component of the vector ∇φ×a. Incorporating they-a n d z- components, which can be similarly found, we obtain the stated result. J Some useful special cases of the relations in table 10.1 are worth noting. If ris the position vector relative to some origin and r=|r|,t h e n ∇φ(r)=dφ drˆr, ∇·[φ(r)r]=3φ(r)+rdφ(r) dr, ∇2φ(r)=d2φ(r) dr2+2 rdφ(r) dr, ∇×[φ(r)r]=0. These results may be proved straightforwardly using Cartesian coordinates but far more simply using spherical polar coordinates, which are discussed in subsec-tion 10.9.2. Particular cases of these results are ∇r=ˆr,∇·r=3,∇×r=0, together with ∇parenleftbigg1 rparenrightbigg =−ˆr r2, ∇·parenleftbiggˆr r2parenrightbigg =−∇2parenleftbigg1 rparenrightbigg =4πδ(r), where δ(r) is the Dirac delta function, discussed in chapter 13. The last equation is important in the solution of certain partial diﬀerential equations and is discussedfurther in chapter 18. 10.8.2 Combinations of grad, div and curl We now consider the action of two vector operators in succession on a scalar or vector ﬁeld. We can immediately discard four of the nine obvious combinations ofgrad, div and curl, since they clearly do not make sense. If φis a scalar ﬁeld and 361 VECTOR CALCULUS ais a vector ﬁeld, these four combinations are grad(grad φ), div(div a), curl(div a) and grad(curl a). In each case the second (outer) vector operator is acting on the wrong type of ﬁeld, i.e. scalar instead of vector or vice versa. In grad(grad φ), for example, grad acts on grad φ, which is a vector ﬁeld, but we know that grad only acts on scalar ﬁelds (although in fact we will see in chapter 21 that we can form the outer product of the del operator with a vector to form a tensor, but that need not concern us here). Of the ﬁve valid combinations of grad, div and curl, two are identically zero, namely curl grad φ=∇×∇ φ=0, (10.37) div curl a=∇·(∇×a)=0 . (10.38) From (10.37), we see that if ais derived from the gradient of some scalar function such that a=∇φthen it is necessarily irrotational ( ∇×a= 0). We also note that if ais an irrotational vector ﬁeld then another irrotational vector ﬁeld is a+∇φ+c,w h e r e φis any scalar ﬁeld and cis a constant vector. This follows since ∇×(a+∇φ+c)=∇×a+∇×∇ φ=0. Similarly, from (10.38) we may infer that if bis the curl of some vector ﬁeld a such that b=∇×athenbis solenoidal ( ∇·b= 0). Obviously, if bis solenoidal andcis any constant vector then b+cis also solenoidal. The three remaining combinations of grad, div and curl are div grad φ=∇·∇φ=∇2φ=∂2φ ∂x2+∂2φ ∂y2+∂2φ ∂z2, (10.39) grad div a=∇(∇·a), =parenleftbigg∂2ax ∂x2+∂2ay ∂x∂y+∂2az ∂x∂zparenrightbigg i+parenleftbigg∂2ax ∂y∂x+∂2ay ∂y2+∂2az ∂y∂zparenrightbigg j +parenleftbigg∂2ax ∂z∂x+∂2ay ∂z∂y+∂2az ∂z2parenrightbigg k, (10.40) curl curl a=∇×(∇×a)=∇(∇·a)−∇2a, (10.41) where (10.39) and (10.40) are expressed in Cartesian coordinates. In (10.41), the term∇2ahas the linear diﬀerential operator ∇2acting on a vector (as opposed to a scalar as in (10.39)), which of course consists of a sum of unit vectors multipliedby components. Two cases arise. (i) If the unit vectors are constants (i.e. they are independent of the values of the coordinates) then the diﬀerential operator gives a non-zero contribution only when acting upon the coordinates, the unit vectors being merelymultipliers. 362 10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES (ii) If the unit vectors vary as the values of the coordinates change (i.e. are not constant in direction throughout the whole space) then the derivativesof these vectors appear as contributions to ∇ 2a. Cartesian coordinates are an example of the ﬁrst case in which each component satisﬁes (∇2a)i=∇2ai. In this case (10.41) can be applied to each component separately: [∇×(∇×a)]i=[∇(∇·a)]i−∇2ai. (10.42) However, cylindrical and spherical polar coordinates come in the second class. For them (10.41) is still true, but the further step to (10.42) cannot be made. More complicated vector operator relations may be proved using the relations given above.IShow that ∇·(∇φ×∇ψ)=0 , where φandψare scalar ﬁelds. From the previous section we have ∇·(a×b)=b·(∇×a)−a·(∇×b). If we let a=∇φandb=∇ψthen we obtain ∇·(∇φ×∇ψ)=∇ψ·(∇×∇ φ)−∇φ·(∇×∇ ψ)=0 , (10.43) since∇×∇ φ=0=∇×∇ ψ, from (10.37). J 10.9 Cylindrical and spherical polar coordinates The operators we have discussed in this chapter, i.e. grad, div, curl and ∇2, have all been deﬁned in terms of Cartesian coordinates, but for many physicalsituations other coordinate systems are more natural. For example, many systems,such as an isolated charge in space, have spherical symmetry and spherical polarcoordinates would be the obvious choice. For axisymmetric systems, such as ﬂuidﬂow in a pipe, cylindrical polar coordinates are the natural choice. The physical laws governing the behaviour of the systems are often expressed in terms of the vector operators we have been discussing, and so it is necessary to be ableto express these operators in these other, non-Cartesian, coordinates. We ﬁrstconsider the two most common non-Cartesian coordinate systems, i.e. cylindricaland spherical polars, and go on to discuss general curvilinear coordinates in thenext section. 10.9.1 Cylindrical polar coordinates As shown in ﬁgure 10.7, the position of a point in space Phaving Cartesian coordinates x, y, z may be expressed in terms of cylindrical polar coordinates 363 VECTOR CALCULUS ρ, φ, z,w h e r e x=ρcosφ, y =ρsinφ, z =z, (10.44) andρ≥0, 0≤φ<2πand−∞<z<∞. The position vector of Pmay therefore be written r=ρcosφi+ρsinφj+zk. (10.45) If we take the partial derivatives of rwith respect to ρ,φandzrespectively then we obtain the three vectors eρ=∂r ∂ρ=c o s φi+s i n φj, (10.46) eφ=∂r ∂φ=−ρsinφi+ρcosφj, (10.47) ez=∂r ∂z=k. (10.48) These vectors lie in the directions of increasing ρ,φandzrespectively but are not all of unit length. Although eρ,eφandezform a useful set of basis vectors in their own right (we will see in section 10.10 that such a basis is sometimes themostuseful), it is usual to work with the corresponding unitvectors, which are obtained by dividing each vector by its modulus to give ˆe ρ=eρ=c o s φi+s i n φj, (10.49) ˆeφ=1 ρeφ=−sinφi+c o s φj, (10.50) ˆez=ez=k. (10.51) These three unit vectors, like the Cartesian unit vectors i,jandk,f o r ma n orthonormal triad at each point in space, i.e. the basis vectors are mutuallyorthogonal and of unit length (see ﬁgure 10.7). Unlike the ﬁxed vectors i,jandk, however, ˆe ρandˆeφchange direction as Pmoves. The expression for a general inﬁnitesimal vector displacement drin the position ofPis given, from (10.19), by dr=∂r ∂ρdρ+∂r ∂φdφ+∂r ∂zdz =dρeρ+dφeφ+dzez =dρˆeρ+ρd φˆeφ+dzˆez. (10.52) This expression illustrates an important diﬀerence between Cartesian and cylin- drical polar coordinates (or non-Cartesian coordinates in general). In Cartesian coordinates, the distance moved in going from xtox+dx, with yandzheld constant, is simply ds=dx. However, in cylindrical polars, if φchanges by dφ, with ρandzheld constant, then the distance moved is notdφ, but ds=ρd φ. 364 10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES xyz z ρr ijk OPˆez ˆeφ ˆeρ φ Figure 10.7 Cylindrical polar coordinates ρ, φ, z. xyz ρdzρd φ ρd φφ dφdρ Figure 10.8 The element of volume in cylindrical polar coordinates is given byρd ρd φd z . Factors, such as the ρinρd φ, that multiply the coordinate diﬀerentials (in the orthonormal basis) to get distances are known as scale factors . From (10.52), the scale factors for the ρ-,φ-a n d z- coordinates are therefore 1, ρand 1 respectively. The magnitude dsof the displacement dris given in cylindrical polar coordinates by (ds)2=dr·dr=(dρ)2+ρ2(dφ)2+(dz)2, where in the second equality we have used the fact that the basis vectors are orthonormal. We can also ﬁnd the volume element in a cylindrical polar system(see ﬁgure 10.8) by calculating the volume of the inﬁnitesimal parallelepiped 365 VECTOR CALCULUS ∇Φ=∂Φ ∂ρˆeρ+1 ρ∂Φ ∂φˆeφ+∂Φ ∂zˆez ∇·a=1 ρ∂ ∂ρ(ρaρ)+1 ρ∂aφ ∂φ+∂az ∂z ∇×a=1 ρ ////////ˆeρρˆeφˆez ∂ ∂ρ∂ ∂φ∂ ∂z aρρaφaz //////// ∇2Φ=1 ρ∂ ∂ρ / ρ∂Φ ∂ρ / +1 ρ2∂2Φ ∂φ2+∂2Φ ∂z2 Table 10.2 Vector operators in cylindrical polar coordinates; Φ is a scalar ﬁeld and ais a vector ﬁeld. deﬁned by the vectors dρˆeρ,ρd φˆeφanddzˆez;t h i si sg i v e nb y dV=|dρˆeρ·(ρd φˆeφ×dzˆez)|=ρd ρd φd z, which again uses the fact that the basis vectors are orthonormal. For a simple coordinate system such as cylindrical polars the expressions for ( ds)2anddVare obvious from the geometry. We will now express the vector operators discussed in this chapter in terms of cylindrical polar coordinates. Let us consider a scalar ﬁeld Φ( ρ, φ, z), where we use Φ for the scalar ﬁeld to avoid confusion with the azimuthal angle φ,a n da vector ﬁeld a(ρ, φ, z). We must ﬁrst write the vector ﬁeld in terms of the basis vectors of the cylindrical polar coordinate system, i.e. a=aρˆeρ+aφˆeφ+azˆez, where aρ,aφandazare the components of ain the ρ-,φ-a n d z- directions respectively. The expressions for grad, div, curl and ∇2c a nt h e nb ec a l c u l a t e d and are given in table 10.2. Since the derivations of these expressions are rathercomplicated we leave them until our discussion of general curvilinear coordinatesin the next section; the reader could well postpone examination of these formalproofs until some experience of using the expressions has been gained.IExpress the vector ﬁeld a=yzi−yj+xz2kin cylindrical polar coordinates, and hence calculate its divergence. Show that the same result is obtained by evaluating the divergencein Cartesian coordinates. The basis vectors of the cylindrical polar coordinate system are given in (10.49)–(10.51).Solving these equations simultaneously for i,jandkwe obtain i=c o s φˆe ρ−sinφˆeφ j=s i n φˆeρ+c o s φˆeφ k=ˆez. 366 10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES xyz r ijk OθPˆer ˆeφ ˆeθ φ Figure 10.9 Spherical polar coordinates r,θ,φ. Substituting these relations and (10.44) into the expression for awe ﬁnd a=zρsinφ(cosφˆeρ−sinφˆeφ)−ρsinφ(sinφˆeρ+c o s φˆeφ)+z2ρcosφˆez =(zρsinφcosφ−ρsin2φ)ˆeρ−(zρsin2φ+ρsinφcosφ)ˆeφ+z2ρcosφˆez. Substituting into the expression for ∇·agiven in table 10.2, ∇·a=2zsinφcosφ−2si n2φ−2zsinφcosφ−cos2φ+s i n2φ+2zρcosφ =2zρcosφ−1. Alternatively, and much more quickly in this case, we can calculate the divergence directly in Cartesian coordinates. We obtain ∇·a=∂ax ∂x+∂ay ∂y+∂az ∂z=2zx−1, which on substituting x=ρcosφyields the same result as the calculation in cylindrical polars. J Finally, we note that similar results can be obtained for (two-dimensional) polar coordinates in a plane by omitting the z-dependence. For example, ( ds)2= (dρ)2+ρ2(dφ)2, while the element of volume is replaced by the element of area dA=ρd ρd φ . 10.9.2 Spherical polar coordinates As shown in ﬁgure 10.9, the position of a point in space P, with Cartesian coordinates x, y, z, may be expressed in terms of spherical polar coordinates r,θ,φ,w h e r e x=rsinθcosφ, y =rsinθsinφ, z =rcosθ, (10.53) 367 VECTOR CALCULUS andr≥0, 0≤θ≤πand 0≤φ<2π. The position vector of Pmay therefore be written as r=rsinθcosφi+rsinθsinφj+rcosθk. If, in a similar manner to that used in the previous section for cylindrical polars, we ﬁnd the partial derivatives of rwith respect to r,θandφrespectively and divide each of the resulting vectors by its modulus then we obtain the unit basis vectors ˆer=s i n θcosφi+s i n θsinφj+c o s θk, ˆeθ=c o s θcosφi+c o s θsinφj−sinθk, ˆeφ=−sinφi+c o s φj. These unit vectors are in the directions of increasing r,θandφrespectively and are the orthonormal basis set for spherical polar coordinates, as shown inﬁgure 10.9. A general inﬁnitesimal vector displacement in spherical polars is, from (10.19), dr=drˆe r+rd θˆeθ+rsinθd φˆeφ; (10.54) thus the scale factors for the r-,θ-a n d φ- coordinates are 1, rand rsinθ respectively. The magnitude dsof the displacement dris therefore given by (ds)2=dr·dr=(dr)2+r2(dθ)2+r2sin2θ(dφ)2, since the basis vectors form an orthonormal set. The element of volume in spherical polar coordinates (see ﬁgure 10.10) is the volume of the inﬁnitesimalparallelepiped deﬁned by the vectors drˆe r,rd θˆeθandrsinθd φˆeφand is given by dV=|drˆer·(rd θˆeθ×rsinθd φˆeφ)|=r2sinθd rd θd φ , where again we use the fact that the basis vectors are orthonormal. The expres- sions for ( ds)2anddVin spherical polars can be obtained from the geometry of this coordinate system. We will now express the standard vector operators in spherical polar coordi- nates, using the same techniques as for cylindrical polar coordinates. We considera scalar ﬁeld Φ( r,θ,φ) and a vector ﬁeld a(r,θ,φ). The latter may be written in terms of the basis vectors of the spherical polar coordinate system as a=a rˆer+aθˆeθ+aφˆeφ, where ar,aθandaφare the components of ain the r-,θ-a n d φ- directions respectively. The expressions for grad, div, curl and ∇2are given in table 10.3. The derivations of these results are given in the next section. 368 10.9 CYLINDRICAL AND SPHERICAL POLAR COORDINATES ∇Φ=∂Φ ∂rˆer+1 r∂Φ ∂θˆeθ+1 rsinθ∂Φ ∂φˆeφ ∇·a=1 r2∂ ∂r(r2ar)+1 rsinθ∂ ∂θ(sinθaθ)+1 rsinθ∂aφ ∂φ ∇×a=1 r2sinθ ////////ˆerrˆeθrsinθˆeφ ∂ ∂r∂ ∂θ∂ ∂φ arraθrsinθaφ //////// ∇2Φ=1 r2∂ ∂r / r2∂Φ ∂r / +1 r2sinθ∂ ∂θ / sinθ∂Φ ∂θ / +1 r2sin2θ∂2Φ ∂φ2 Table 10.3 Vector operators in spherical polar coordinates. Φ is a scalar ﬁeld andais a vector ﬁeld. xyz rrd θ φdφdφ dr rsinθ rsinθd φrsinθd φ θ dθ Figure 10.10 The element of volume in spherical polar coordinates is given byr2sinθd rd θd φ . As a ﬁnal note we mention that in the expression for ∇2Φ given in table 10.3 we can rewrite the ﬁrst term on the RHS as follows: 1 r2∂ ∂rparenleftbigg r2∂Φ ∂rparenrightbigg =1 r∂2 ∂r2(rΦ), w h i c hc a no f t e nb eu s e f u li ns h o r t e n i n gc a l c u l a t i o n s . 369 VECTOR CALCULUS 10.10 General curvilinear coordinates As indicated earlier, the contents of this section are more formal and technically complicated than hitherto. The section could be omitted until the reader has had some experience of using its results. Cylindrical and spherical polars are just two examples of what are called general curvilinear coordinates . In the general case, the position of a point P having Cartesian coordinates x, y, z may be expressed in terms of the three curvilinear coordinates u1,u2,u3,w h e r e x=x(u1,u2,u3),y =y(u1,u2,u3),z =z(u1,u2,u3), and similarly u1=u1(x, y, z),u 2=u2(x, y, z),u 3=u3(x, y, z). We assume that all these functions are continuous, diﬀerentiable and have a single-valued inverse, except perhaps at or on certain isolated points or lines, so that there is a one-to-one correspondence between the x, y, z andu1,u2,u3 systems. The u1-,u2-a n d u3- coordinate curves of a general curvilinear system are analogous to the x-,y-a n d z- axes of Cartesian coordinates. The surfaces u1=c1,u2=c2and u3=c3,w h e r e c1,c2,c3are constants, are called the coordinate surfaces and each pair of these surfaces has its intersection in a curve called a coordinate curve orline(see ﬁgure 10.11). If at each point in space the three coordinate surfaces passing through the point meet at right angles then the curvilinear coordinate system is called orthogonal . For example, in spherical polars u1=r,u2=θ,u3=φand the three coordinate surfaces passing through the point ( R,Θ,Φ) are the sphere r=R, the circular cone θ= Θ and the plane φ= Φ, which intersect at right angles at that point. Therefore spherical polars (and cylindrical polars) form an orthogonal coordinate system. Ifr(u1,u2,u3) is the position vector of the point Pthene1=∂r/∂u1is a vector tangent to the u1-curve at P(for which u2andu3are constants) in the direction of increasing u1. Similarly, e2=∂r/∂u2ande3=∂r/∂u3are vectors tangent to theu2-a n d u3-c u r v e sa t Pin the direction of increasing u2andu3respectively. Denoting the lengths of these vectors by h1,h2andh3,t h eunitvectors in each of these directions are given by ˆe1=1 h1∂r ∂u1, ˆe2=1 h2∂r ∂u2, ˆe3=1 h3∂r ∂u3, where h1=|∂r/∂u1|,h2=|∂r/∂u2|andh3=|∂r/∂u3|. The quantities h1,h2,h3are called the scale factors of the curvilinear coordinate 370 10.10 GENERAL CURVILINEAR COORDINATES z xy ijk OPu2=c2u1=c1 u3=c3u1u2u3 ˆ/epsilon11ˆ/epsilon12ˆ/epsilon13 ˆe1 ˆe2ˆe3 Figure 10.11 General curvilinear coordinates. system. The element of distance associated with an inﬁnitesimal change duiin one of the coordinates is hidui. In the previous section we found that the scale factors for cylindrical and spherical polar coordinates were for cylindrical polars hρ=1 , hφ=ρ,hz=1 , for spherical polars hr=1 , hθ=r,hφ=rsinθ. Although the vectors e1,e2,e3form a perfectly good basis for the curvilinear coordinate system, it is usual to work with the corresponding unit vectors ˆe1,ˆe2, ˆe3. For an orthogonal curvilinear coordinate system these unit vectors form an orthonormal basis. An inﬁnitesimal vector displacement in general curvilinear coordinates is given by, from (10.19), dr=∂r ∂u1du1+∂r ∂u2du2+∂r ∂u3du3 (10.55) =du1e1+du2e2+du3e3 (10.56) =h1du1ˆe1+h2du2ˆe2+h3du3ˆe3. (10.57) I nt h ec a s eo f orthogonal curvilinear coordinates, where the ˆeiare mutually perpendicular, the element of arc length is given by (ds)2=dr·dr=h2 1(du1)2+h2 2(du2)2+h2 3(du3)2. (10.58) The volume element for the coordinate system is the volume of the inﬁnitesimal parallelepiped deﬁned by the vectors ( ∂r/∂u i)dui=duiei=hiduiˆei,f o r i=1,2,3. 371 VECTOR CALCULUS For orthogonal coordinates this is given by dV=|du1e1·(du2e2×du3e3)| =|h1ˆe1·(h2ˆe2×h3ˆe3)|du1du2du3 =h1h2h3du1du2du3. Now, in addition to the set {ˆei},i=1,2,3, there exists another useful set of three unit basis vectors at P.S i n c e∇u1is a vector normal to the surface u1=c1, a unit vector in this direction is ˆ/epsilon11=∇u1/|∇u1|. Similarly, ˆ/epsilon12=∇u2/|∇u2|and ˆ/epsilon13=∇u3/|∇u3|are unit vectors normal to the surfaces u2=c2andu3=c3 respectively. Therefore at each point Pin a curvilinear coordinate system, there exist, in general, two sets of unit vectors: {ˆei}, tangent to the coordinate curves, and {ˆ/epsilon1i}, normal to the coordinate surfaces. A vector acan be written in terms of either set of unit vectors: a=a1ˆe1+a2ˆe2+a3ˆe3=A1ˆ/epsilon11+A2ˆ/epsilon12+A3ˆ/epsilon13, where a1,a2,a3andA1,A2,A3are the components of ain the two systems. It may be shown that the two bases become identical if the coordinate system isorthogonal. Instead of the unitvectors discussed above, we could instead work directly with the two sets of vectors {e i=∂r/∂u i}and{/epsilon1i=∇ui},w h i c ha r en o t ,i ng e n e r a l ,o f unit length. We can then write a vector aas a=α1e1+α2e2+α3e3=β1/epsilon11+β2/epsilon12+β3/epsilon13, or more explicitly as a=α1∂r ∂u1+α2∂r ∂u2+α3∂r ∂u3=β1∇u1+β2∇u2+β3∇u3, where α1,α2,α3andβ1,β2,β3are called the contravariant andcovariant com- ponents of arespectively. A more detailed discussion of these components, in the context of tensor analysis, is given in chapter 21. The (in general) non-unit bases{ei}and{/epsilon1i}are often the most natural bases in which to express vector quantities.IShow that{ei}and{/epsilon1i}are reciprocal systems of vectors. Let us consider the scalar product ei·/epsilon1j; using the Cartesian expressions for rand∇,w e obtain ei·/epsilon1j=∂r ∂ui·∇uj = /∂x ∂uii+∂y ∂uij+∂z ∂uik / · /∂uj ∂xi+∂uj ∂yj+∂uj ∂zk / =∂x ∂ui∂uj ∂x+∂y ∂ui∂uj ∂y+∂z ∂ui∂uj ∂z=∂uj ∂ui, 372 10.10 GENERAL CURVILINEAR COORDINATES in the last step we have used the chain rule for partial diﬀerentiation. Therefore ei·/epsilon1j=1 ifi=j,a n dei·/epsilon1j= 0 otherwise. Hence {ei}and{/epsilon1j}are reciprocal systems of vectors. J We now derive expressions for the standard vector operators in orthogonal curvilinear coordinates. Despite the useful properties of the non-unit bases dis-cussed above, the remainder of our discussion in this section will be in terms ofthe unit basis vectors {ˆe i}. The expressions for the vector operators in cylindrical and spherical polar coordinates given in tables 10.2 and 10.3 respectively can be found from those derived below by inserting the appropriate scale factors. Gradient The change dΦ in a scalar ﬁeld Φ resulting from changes du1,d u2,d u3in the coordinates u1,u2,u3is given by, from (5.5), dΦ=∂Φ ∂u1du1+∂Φ ∂u2du2+∂Φ ∂u3du3. For orthogonal curvilinear coordinates u1,u2,u3we ﬁnd from (10.57), and com- parison with (10.27), that we can write this as dΦ=∇Φ·dr, (10.59) where∇Φi sg i v e nb y ∇Φ=1 h1∂Φ ∂u1ˆe1+1 h2∂Φ ∂u2ˆe2+1 h3∂Φ ∂u3ˆe3. (10.60) This implies that the del operator can be written ∇=ˆe1 h1∂ ∂u1+ˆe2 h2∂ ∂u2+ˆe3 h3∂ ∂u3.IShow that for orthogonal curvilinear coordinates ∇ui=ˆei/hi. Hence show that the two sets of vectors {ˆei}and{ˆ/epsilon1i}are identical in this case. Letting Φ = uiin (10.60) we ﬁnd immediately that ∇ui=ˆei/hi. Therefore |∇ui|=1/hi,a n d soˆ/epsilon1i=∇ui/|∇ui|=hi∇ui=ˆei. J Divergence In order to derive the expression for the divergence of a vector ﬁeld in orthogonal curvilinear coordinates, we must ﬁrst write the vector ﬁeld in terms of the basisvectors of the coordinate system: a=a 1ˆe1+a2ˆe2+a3ˆe3. The divergence is then given by ∇·a=1 h1h2h3bracketleftbigg∂ ∂u1(h2h3a1)+∂ ∂u2(h3h1a2)+∂ ∂u3(h1h2a3)bracketrightbigg . (10.61) 373 VECTOR CALCULUSIProve the expression for ∇·ain orthogonal curvilinear coordinates. Let us consider the sub-expression ∇·(a1ˆe1). Now ˆe1=ˆe2×ˆe3=h2∇u2×h3∇u3. Therefore ∇·(a1ˆe1)=∇·(a1h2h3∇u2×∇u3), =∇(a1h2h3)·(∇u2×∇u3)+a1h2h3∇·(∇u2×∇u3). However,∇·(∇u2×∇u3) = 0, from (10.43), so we obtain ∇·(a1ˆe1)=∇(a1h2h3)· /ˆe2 h2×ˆe3 h3 / =∇(a1h2h3)·ˆe1 h2h3; letting Φ = a1h2h3in (10.60) and substituting into the above equation, we ﬁnd ∇·(a1ˆe1)=1 h1h2h3∂ ∂u1(a1h2h3). Repeating the analysis for ∇·(a2ˆe2)a n d∇·(a3ˆe3), and adding the results we obtain (10.61), as required. J Laplacian In the expression for the divergence (10.61), let a=∇Φ=1 h1∂Φ ∂u1ˆe1+1 h2∂Φ ∂u2ˆe2+1 h3∂Φ ∂u3ˆe3, where we have used (10.60). We then obtain ∇2Φ=1 h1h2h3bracketleftbigg∂ ∂u1parenleftbiggh2h3 h1∂Φ ∂u1parenrightbigg +∂ ∂u2parenleftbiggh3h1 h2∂Φ ∂u2parenrightbigg +∂ ∂u3parenleftbiggh1h2 h3∂Φ ∂u3parenrightbiggbracketrightbigg , which is the expression for the Laplacian in orthogonal curvilinear coordinates. Curl The curl of a vector ﬁeld a=a1ˆe1+a2ˆe2+a3ˆe3in orthogonal curvilinear coordinates is given by ∇×a=1 h1h2h3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleh 1ˆe1h2ˆe2h3ˆe3 ∂ ∂u1∂ ∂u2∂ ∂u3 h1a1h2a2h3a3vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle. (10.62)IProve the expression for ∇×ain orthogonal curvilinear coordinates. Let us consider the sub-expression ∇×(a1ˆe1). Since ˆe1=h1∇u1we have ∇×(a1ˆe1)=∇×(a1h1∇u1), =∇(a1h1)×∇u1+a1h1∇×∇ u1. But∇×∇ u1=0 ,s ow eo b t a i n ∇×(a1ˆe1)=∇(a1h1)×ˆe1 h1. 374 10.11 EXERCISES ∇Φ=1 h1∂Φ ∂u1ˆe1+1 h2∂Φ ∂u2ˆe2+1 h3∂Φ ∂u3ˆe3 ∇·a=1 h1h2h3 /∂ ∂u1(h2h3a1)+∂ ∂u2(h3h1a2)+∂ ∂u3(h1h2a3) / ∇×a=1 h1h2h3 ///////h1ˆe1h2ˆe2h3ˆe3 ∂ ∂u1∂ ∂u2∂ ∂u3 h1a1h2a2h3a3 /////// ∇2Φ=1 h1h2h3 /∂ ∂u1 /h2h3 h1∂Φ ∂u1 / +∂ ∂u2 /h3h1 h2∂Φ ∂u2 / +∂ ∂u3 /h1h2 h3∂Φ ∂u3 // Table 10.4 Vector operators in orthogonal curvilinear coordinates u1,u2,u3. Φ is a scalar ﬁeld and ais a vector ﬁeld. Letting Φ = a1h1in (10.60) and substituting into the above equation, we ﬁnd ∇×(a1ˆe1)=ˆe2 h3h1∂ ∂u3(a1h1)−ˆe3 h1h2∂ ∂u2(a1h1). The corresponding analysis of ∇×(a2ˆe2) produces terms in ˆe3andˆe1, whilst that of ∇×(a3ˆe3) produces terms in ˆe1andˆe2. When the three results are added together, the coeﬃcients multiplying ˆe1,ˆe2andˆe3are the same as those obtained by writing out (10.62) explicitly, thus proving the stated result. J The general expressions for the vector operators in orthogonal curvilinear coordinates are shown for reference in table 10.4. The explicit results for cylindricaland spherical polar coordinates, given in tables 10.2 and 10.3 respectively, areobtained by substituting the appropriate set of scale factors in each case. A discussion of the expressions for vector operators in tensor form, which are valid even for non-orthogonal curvilinear coordinate systems, is given inchapter 21. 10.11 Exercises 10.1 Evaluate the integralZ/ a(˙b·a+b·˙a)+˙a(b·a)−2(˙a·a)b−˙b|a|2 / dt in which ˙a,˙bare the derivatives of a,bwith respect to t. 10.2 At time t= 0, the vectors EandBare given by E=E0andB=B0,w h e r et h e ﬁxed unit vectors E0andB0are orthogonal. The equations of motion are dE dt=E0+B×E0, dB dt=B0+E×B0. FindEandBat a general time t, showing that after a long time the directions ofEandBhave almost interchanged. 375 VECTOR CALCULUS 10.3 The general equation of motion of a (non-relativistic) particle of mass mand charge qwhen it is placed in a region where there is a magnetic ﬁeld Band an electric ﬁeld Eis m¨r=q(E+˙r×B); hereris the position of the particle at time tand˙r=dr/dtetc. Write this as three separate equations in terms of the Cartesian components of the vectors involved. For the simple case of crossed uniform ﬁelds E=Ei,B=Bjin which the particle starts from the origin at t=0w i t h ˙r=v0k, ﬁnd the equations of motion and show the following: (a) if v0=E/Bthen the particle continues its initial motion; (b) if v0= 0 then the particle follows the space curve given in terms of the parameter ξby x=mE B2q(1−cosξ),y =0,z =mE B2q(ξ−sinξ). Interpret this curve geometrically and relate ξtot. Show that the total distance travelled by the particle after time tis 2E B Zt 0 /// / sinBqt/prime 2m /// / dt/prime. 10.4 Use vector methods to ﬁnd the maximum angle to the horizontal at which a stone may be thrown so as to ensure that it is always moving away from the thrower. 10.5 If two systems of coordinates with a common origin Oare rotating with respect to each other, the measured accelerations diﬀer in the two systems. Denotingbyrandr /primeposition vectors in frames OXY Z andOX/primeY/primeZ/primerespectively, the connection between the two is ¨r/prime=¨r+˙ω×r+2ω×˙r+ω×(ω×r), where ωis the angular velocity vector of the rotation of OXY Z with respect to OX/primeY/primeZ/prime(taken as ﬁxed). The third term on the RHS is known as the Coriolis acceleration, whilst the ﬁnal term gives rise to a centrifugal force. Consider the application of this result to the ﬁring of a shell of mass mfrom a stationary ship on the steadily rotating earth, working to the ﬁrst order inω(= 7.3×10 −5rad s−1). If the shell is ﬁred with velocity vat time t=0a n do n l y reaches a height that is small compared to the radius of the earth, show that itsacceleration, as recorded on the ship, is given approximately by ¨r=g−2ω×(v+gt), where mgis the weight of the shell measured on the ship’s deck. The shell is ﬁred at another stationary ship (a distance saway) and vis such that the shell would have hit its target had there been no Coriolis eﬀect. (a) Show that without the Coriolis eﬀect the time of ﬂight of the shell would have been τ=−2g·v/g 2. (b) Show further that when the shell ac tually hits the sea it is oﬀ target by approximately 2τ g2[(g×ω)·v](gτ+v)−(ω×v)τ2−1 3(ω×g)τ3. (c) Estimate the order of magnitude ∆ of this miss for a shell for which v= 300 ms−1, ﬁring close to its maximum range ( vmakes an angle of π/4w i t ht h e vertical) in a northerly direction, whilst the ship is stationed at latitude 45◦ North. 376 10.11 EXERCISES 10.6 Prove that for a space curve r=r(s), where sis the arc length measured along the curve from a ﬁxed point, the triple scalar product/dr ds×d2r ds2 / ·d3r ds3 at any point on the curve has the value κ2τ,w h e r e κis the curvature and τthe torsion at that point. 10.7 For the twisted space curve y3+2 7axz−81a2y= 0, given parametrically by x=au(3−u2),y =3au2,z =au(3 +u2), show the following: (a) that ds/du =3√2a(1 +u2), where sis the distance along the curve measured from the origin; (b) that the length of the curve from the origin to the Cartesian point (2 a,3a,4a) is 4√2a; (c) that the radius of curvature at the point with parameter uis 3a(1 +u2)2; (d) that the torsion τand curvature κat a general point are equal; (e) that any of the Frenet–Serret formulae that you have not already used directly are satisﬁed. 10.8 The shape of the curving slip road joining two motorways that cross at right angles and are at vertical heights z=0a n d z=hcan be approximated by the space curve r=√ 2h πln cos /zπ 2h / i+√ 2h πln sin /zπ 2h / j+zk. Show that the radius of curvature ρof the curve is (2 h/π)cosec ( zπ/h)a th e i g h t zand that the torsion τ=−1/ρ. (To shorten the algebra, set z=2hθ/πand use θas the parameter.) 10.9 In a magnetic ﬁeld, ﬁeld lines are curves to which the magnetic induction Bis everywhere tangential. By evaluating dB/ds,w h e r e sis the distance measured along a ﬁeld line, prove that the radius of curvature at any point on a line isgiven by ρ=B 3 |B×(B·∇)B|. 10.10 (a) Using the parameterization x=ucosφ,y=usinφ,z=ucotΩ, ﬁnd the sloping surface area of a right circular cone of semi-angle Ω whose base hasradius a.V e r i f yt h a ti ti se q u a lt o 1 2×perimeter of the base ×slope height. (b) Using the same parameterization as in (a) for xandy, and an appropriate choice for z, ﬁnd the surface area between the planes z=0a n d z=Zof the paraboloid of revolution z=α(x2+y2). 10.11 (a) Parameterising the hyperboloid x2 a2+y2 b2−z2 c2=1 byx=acosθsecφ,y=bsinθsecφ,z=ctanφ, show that an area element on its surface is dS=s e c2φ / c2sec2φ /; b2cos2θ+a2sin2θ / +a2b2tan2φ /1/2dθ dφ. (b) Use this formula to show that the area of the curved surface x2+y2−z2=a2 between the planes z=0a n d z=2ais πa2 / 6+1√ 2sinh−12√ 2 / . 377 VECTOR CALCULUS 10.12 For the function z(x, y)=(x2−y2)e−x2−y2, ﬁnd the location(s) at which the steepest gradient occurs. What are the magnitude and direction of that gradient? (The algebra involved is easier if plane polarcoordinates are used.) 10.13 Verify by direct calculation that ∇·(a×b)=b·(∇×a)−a·(∇×b). 10.14 (a) Simplify ∇×a(∇·a)+ a×[∇×(∇×a)]+a×∇ 2a. (b) By explicitly writing out the terms in Cartesian coordinates prove that [c·(b·∇)−b·(c·∇)]a=(∇×a)·(b×c). (c) Prove that a×(∇×a)=∇(1 2a2)−(a·∇)a. 10.15 Evaluate the Laplacian of the function ψ(x, y, z)=zx2 x2+y2+z2 (a) directly in Cartesian coordinates, and (b) after changing to a spherical polar coordinate system. Verify that, as they must, the two methods give the sameresult. 10.16 Verify that (10.42) is valid for each component separately when ais the Cartesian vector x 2yi+xyzj+z2yk, by showing that each side of the equation is equal to zi+( 2x+2z)j+xk. 10.17 The (Maxwell) relationship between a time-independent magnetic ﬁeld Band the current density J(measured in S.I. units in A m−2) producing it, ∇×B=µ0J, can be applied to a long cylinder of conducting ionised gas which, in cylindrical polar coordinates, occupies the region ρ<a. (a) Show that a uniform current density (0 ,C,0) and a magnetic ﬁeld (0 ,0,B), with Bconstant (= B0)f o r ρ>a andB=B(ρ)f o r ρ<a , are consistent with this equation. Obtain expressions for CandB(ρ)i nt e r m so f B0anda, given that Bis continuous at ρ=a. (b) The magnetic ﬁeld can be expressed as B=∇×A,w h e r e Ais known as the vector potential. Show that a suitable Acan be found which has only one non-vanishing component, Aφ(ρ), and obtain explicit expressions for Aφ(ρ) for both ρ<a andρ>a.L i k e B, the vector potential is continuous at ρ=a. (c) The gas pressure p(ρ) satisﬁes the hydrostatic equation ∇p=J×Band vanishes at the outer wall of the cylinder. Find a general expression for p. 10.18 (a) For cylindrical polar coordinates ρ, φ, z evaluate the derivatives of the three unit vectors with respect to each of the coordinates, showing that only ∂ˆeρ/∂φ and∂ˆeφ/∂φare non-zero. (i) Hence evaluate ∇2awhenais the vector ˆeρ, i.e. a vector of unit magnitude everywhere directed radially outwards from the z-axis. (ii) Note that it is trivially obvious that ∇×a=0and hence that equation (10.41) requires that ∇(∇·a)=∇2a. (iii) Evaluate ∇(∇·a) and show that the latter equation holds, but that [∇(∇·a)]ρ/negationslash=∇2aρ. 378 10.11 EXERCISES (b) Rework the same problem in Cartesian coordinates (where, as it happens, the algebra is more complicated). 10.19 Maxwell’s equations for electromagnetism in free space (i.e. in the absence of charges, currents and dielectric or magnetic media) can be written (i)∇·B=0, (ii)∇·E=0, (iii)∇×E+∂B ∂t=0,(iv)∇×B−1 c2∂E ∂t=0. A vector Ais deﬁned by B=∇×A,a n das c a l a r φbyE=−∇φ−∂A/∂t. Show that if the condition (v)∇·A+1 c2∂φ ∂t=0 is imposed (this is known as choosing the Lorenz gauge), then both Aandφ satisfy the wave equations (vi)∇2φ−1 c2∂2φ ∂t2=0, (vii)∇2A−1 c2∂2A ∂t2=0. The reader is invited to proceed as follows. (a) Verify that the expressions for BandEin terms of Aandφare consistent with (i) and (iii). (b) Substitute for Ein (ii) and use the derivative with respect to time of (v) to eliminate Afrom the resulting expression. Hence obtain (vi). (c) Substitute for BandEin (iv) in terms of Aandφ. Then use the divergence of (v) to simplify the resulting equation and so obtain (vii). 10.20 For a description in spherical polar coordinates with axial symmetry of the ﬂow of a very viscous ﬂuid, the components of the velocity ﬁeld uare given in terms of the stream function ψby ur=1 r2sinθ∂ψ ∂θ,u θ=−1 rsinθ∂ψ ∂r. Find an explicit expression for the diﬀerential operator Edeﬁned by Eψ=−(rsinθ)(∇×u)φ. The stream function satisﬁes the equation of motion E2ψ= 0 and, for the ﬂow of a ﬂuid past a sphere, takes the form ψ(r,θ)=f(r)sin2θ. Show that f(r)s a t i s ﬁ e s the (ordinary) diﬀerential equation r4f(4)−4r2f/prime/prime+8rf/prime−8f=0. 10.21 Paraboloidal coordinates u, v, φ are deﬁned in terms of Cartesian coordinates by x=uvcosφ, y =uvsinφ, z =1 2(u2−v2). Identify the coordinate surfaces in the u, v, φ system. Verify that each coordinate surface ( u= constant, say) intersects every coordinate surface on which one of the other two coordinates ( v, say) is constant. Show further that the system of coordinates is an orthogonal one and determine its scale factors. Prove that theu-component of ∇×ais given by 1 (u2+v2)1/2 /aφ v+∂aφ ∂v / −1 uv∂av ∂φ. 379 VECTOR CALCULUS 10.22 Non-orthogonal curvilinear coordinates are diﬃcult to work with and should be avoided if at all possible, but the following example is provided to illustrate thecontent of section 10.10. In a new coordinate system for the region of space in which the Cartesian coordinate zsatisﬁes z≥0, the position of a point ris given by ( α 1,α2,R), where α1andα2are respectively the cosines of the angles made by rwith the x-a n d y- coordinate axes of a Cartesian system and R=|r|. The ranges are −1≤αi≤1, 0≤R<∞. (a) Express rin terms of α1,α2,Rand the unit Cartesian vectors i,j,k. (b) Obtain expressions for the vectors ei(=∂r/∂α1,...) and hence show that the scale factors hiare given by h1=R(1−α2 2)1/2 (1−α2 1−α2 2)1/2,h 2=R(1−α2 1)1/2 (1−α2 1−α2 2)1/2,h 3=1. (c) Verify formally that the system is not an orthogonal one. (d) Show that the volume element of the coordinate system is dV=R2dα1dα2dR (1−α2 1−α2 2)1/2, and demonstrate that this is always less than or equal to the corresponding expression for an orthogonal curvilinear system. (e) Calculate the expression for ( ds)2for the system, and show that it diﬀers from that for the corresponding orthogonal system by 2α1α2R2 1−α2 1−α2 2dα1dα2. 10.23 Hyperbolic coordinates u, v, φ are deﬁned in terms of Cartesian coordinates by x=c o s h ucosvcosφ, y =c o s h ucosvsinφ, z =s i n h usinv. Sketch the coordinate curves in the φ= 0 plane, showing that far from the origin they become concentric circles and radial lines. In particular, identify the curvesu=0,v=0,v=π/2a n d v=π. Calculate the tangent vectors at a general point, show that they are mutually orthogonal and deduce that the appropriatescale factors are h u=hv=( c o s h2u−cos2v)1/2,h φ=c o s h ucosv. Find the most general function ψ(u)o fuonly that satisﬁes Laplace’s equation ∇2ψ=0 . 10.24 In a Cartesian system, AandBare the points (0 ,0,−1) and (0 ,0,1) respectively. In a new coordinate system a general point Pis given by ( u1,u2,u3)w i t h u1=1 2(r1+r2),u2=1 2(r1−r2),u3=φ;h e r e r1andr2are the distances APand BPandφis the angle between the plane ABPandy=0 . (a) Express zand the perpendicular distance ρfrom Pto the z-axis in terms of u1,u2,u3. (b) Evaluate ∂x/∂u i,∂y/∂u i,∂z/∂u i,f o r i=1,2,3. (c) Find the Cartesian components of ˆujand hence show that the new coordi- nates are mutually orthogonal. Evaluate the scale factors and the inﬁnitesimalvolume element in the new coordinate system. (d) Determine and sketch the forms of the surfaces u i=c o n s t a n t . (e) Find the most general function fofu1only that satisﬁes ∇2f=0 . 380 10.12 HINTS AND ANSWERS 10.12 Hints and answers 10.1 a×(a×b)+h. 10.2 Taking E0=iandB0=j,E=( 1+ t)i+(t2/2+t3/6)j−(t+t2/2)k, B=(t2/2+t3/6)i+( 1+ t)j+(t+t2/2)k. 10.3 For crossed uniform ﬁelds ¨x+(Bq/m)2x=q(E−Bv0)/m,¨y=0 ,m˙z=qBx+mv0; (b)ξ=Bqt/m ; the path is a cycloid in the plane y=0 ; ds=[ (dx/dt)2+ (dz/dt)2]1/2dt. 10.4 Prove that the vector equation of the stone is r=v0t+gt2/2. Impose the condition r·˙r>0f o ra l l t,i . e .r·˙r=0h a sn or e a lr o o t sf o r t;8v2 0g2>9(v0·g)2. Maximum angle is 70 .5◦. 10.5 g=¨r/prime−ω×(ω×r), where ¨r/primeis the shell’s acceleration measured by an observer ﬁxed in space. To ﬁrst order in ω, the direction of gis radial, i.e. parallel to ¨r/prime. (a) Note that sis orthogonal to g. (b) If the actual time of ﬂight is T,u s e( s+∆ )·g=0t os h o wt h a t T≈τ(1 + 2 g−2(g×ω)·v+···). In the Coriolis terms it is suﬃcient to put T≈τ. (c) For this situation ( g×ω)·v=0a n d ω×v=0;τ≈43 s and ∆ = 10–15 m to the East. 10.6 Diﬀerentiate ˆb=ˆt×ˆnwith respect to s; express the result in terms of the derivatives of r; take the scalar product with d2r/ds2. 10.7 (a) Evaluate ( dr/du)·(dr/du). (b) Integrate the previous result between u=0a n d u=1 . (c)ˆt=[√2(1 + u2)]−1[(1−u2)i+2uj+( 1+ u2)k]. Use dˆt/ds=(dˆt/du)/(ds/du); ρ−1=|dˆt/ds|. (d)ˆn=( 1+ u2)−1[−2ui+(1−u2)j].ˆb=[√2(1+ u2)]−1[(u2−1)i−2uj+(1+ u2)k]. Usedˆb ds=dˆb du /ds duand show that this equals −[3a(1 +u2)2]−1ˆn. (e) Show that dˆn/ds=τ(ˆb−ˆt)=−2[3√2a(1 +u2)3]−1[(1−u2)i+2uj]. 10.8 ds/dθ =√ 2h/(πsinθcosθ);ˆt=−sin2θi+c o s2θj+√ 2si nθcosθk; ˆb=c o s2θi−sin2θj+√ 2si nθcosθk. 10.9 Note that dB=(dr·∇)Band that B=Bˆt,w i t h ˆt=dr/ds.O b t a i n( B·∇)B/B= ˆt(dB/ds )+ˆn(B/ρ) and then take the vector product of ˆtwith this equation. 10.10 (a) dS=|(−ucosφcot Ω ,−usinφcot Ω ,u)|dφ du;S=πa2cosec Ω . (b)z=αu2;dS=u(1 + 4 α2u2)1/2dφ du;S=(π/6)[(1 + 4 αZ)3/2−1]. 10.11 (b) Put tan φ=2−1/2sinhψ. 10.12|∇z|2=4ρ2e−2ρ2[(1−ρ2)2cos22φ+ρ2sin22φ], which is extremal when φ=nπ/4 and 1−ρ2= 0. Maximum slope = 2 e−1atx=±1,y=±1, along azimuthal directions x±y=±2a n d x±y=∓2. 10.14 (a) ( ∇·a)(∇×a); (b) terms of the form bxcx(∂ax/∂x) cancel; (c) for the x- component, add and subtract ax(∂ax/∂x) and regroup. 10.15 (a) 2 z(x2+y2+z2)−3[(y2+z2)(y2+z2−3x2)−4x4]. (b) 2 r−1cosθ(1−5sin2θcos2φ); both are equal to 2 zr−4(r2−5x2). 10.17 Use the formulae given in table 10.2. (a)C=−B0/(µ0a);B(ρ)=B0ρ/a. (b)B0ρ2/(3a)f o r ρ<a,a n d B0[ρ/2−a2/(6ρ)] for ρ>a. (c) [ B2 0/(2µ0)][1−(ρ/a)2]. 10.18 (a) ∂ˆeρ/∂φ=ˆeφ,∂ˆeφ/∂φ=−ˆeρ;( i )−ρ−2ˆeρ.( b )∇2a=−(x2+y2)−3/2(xi+yj). 381 VECTOR CALCULUS 10.20 E=∂2 ∂r2+sinθ r2∂ ∂θ /1 sinθ∂ ∂θ / . 10.21 Two sets of paraboloids of revolution about the z-axis and the sheaf of planes containing the z-axis. For constant u,−∞<z<u2/2; for constant v,−v2/2< z<∞. The scale factors are hu=hv=(u2+v2)1/2,hφ=uv. 10.22 (c) e1·e2=R2α1α2/(1−α2 1−α2 2)/negationslash=0 . 10.23 The tangent vectors are as follows: for u= 0, the line joining (1 ,0,0) and (−1,0,0); for v= 0, the line joining (1 ,0,0) and (∞,0,0). For v=π/2, the line (0,0,z); for v=π, the line joining ( −1,0,0) and (−∞,0,0).ψ(u)=2t a n−1eu+c, derived from ∂[cosh u(∂ψ/∂u )]/∂u=0 . 10.24 (a) z=u1u2,ρ=u2 1+u2 2−u2 1u22−1. (b)u1(1−u2 2)c osu3/ρ,u1(1−u2 2)si nu3/ρ,u2;u2(1−u2 1)c osu3/ρ,u2(1−u2 1)si nu3/ρ, u1;−ρsinu3,ρcosu3,0 . (c) [( u2 1−u2 2)/(u2 1−1)]1/2,[ (u2 2−u2 1)/(u2 2−1)]1/2,ρ;|u2 1−u2 2|du1du2du3. (d) Confocal ellipsoids, hyperboloids, half-planes containing the z-axis. (e)Bln[(u1−1)/(u1+ 1)]. 382 11 Line, surface and volume integrals In the previous chapter we encountered continuously varying scalar and vector ﬁelds and discussed the action of various diﬀerential operators on them. Inaddition to these diﬀerential operations, the need often arises to consider theintegration of ﬁeld quantities along lines, over surfaces and throughout volumes. In general the integrand may be scalar or vector in nature, but the evaluation of such integrals involves their reduction to one or more scalar integrals, whichare then evaluated. In the case of surface and volume integrals this requires theevaluation of double and triple integrals (see chapter 6). 11.1 Line integrals In this section we discuss lineorpath integrals , in which some quantity related to the ﬁeld is integrated between two given points in space, AandB, along a prescribed curve Cthat joins them. In general, we may encounter line integrals of the forms integraldisplay Cφdr,integraldisplay Ca·dr,integraldisplay Ca×dr, (11.1) where φis a scalar ﬁeld and ais a vector ﬁeld. The three integrals themselves are respectively vector, scalar and vector in nature. As we will see below, in physical applications line integrals of the second type are by far the most common. The formal deﬁnition of a line integral closely follows that of ordinary integrals and can be considered as the limit of a sum. We may divide the path Cjoining the points AandBintoNsmall line elements ∆ rp,p=1,...,N .I f(xp,yp,zp)i s any point on the line element ∆ rpthen the second type of line integral in (11.1), for example, is deﬁned as integraldisplay Ca·dr= lim N→∞Nsummationdisplay p=1a(xp,yp,zp)·∆rp, where it is assumed that all |∆rp|→0a sN→∞. 383 LINE, SURFACE AND VOLUME INTEGRALS Each of the line integrals in (11.1) is evaluated over some curve Cthat may be either open ( AandBbeing distinct points) or closed (the curve Cforms a loop, so that AandBare coincident). In the case where Cis closed, the line integral is writtencontintegraltext Cto indicate this. The curve may be given either parametrically by r(u)=x(u)i+y(u)j+z(u)kor by means of simultaneous equations relating x, y, z for the given path (in Cartesian coordinates). A full discussion of the diﬀerentrepresentations of space curves was given in section 10.3. In general, the value of the line integral depends not only on the end-points AandBbut also on the path Cjoining them. For a closed curve we must also specify the direction around the loop in which the integral is taken. It is usuallytaken to be such that a person walking around the loop Cin this direction always has the region Ron his/her left; this is equivalent to traversing Cin the anticlockwise direction (as viewed from above). 11.1.1 Evaluating line integrals The method of evaluating a line integral is to reduce it to a set of scalar integrals. It is usual to work in Cartesian coordinates, in which case dr=dxi+dyj+dzk. The ﬁrst type of line integral in (11.1) then becomes simply integraldisplay Cφdr=iintegraldisplay Cφ(x, y, z)dx+jintegraldisplay Cφ(x, y, z)dy+kintegraldisplay Cφ(x, y, z)dz. The three integrals on the RHS are ordinary scalar integrals that can be evaluated in the usual way once the path of integration Chas been speciﬁed. Note that in the above we have used relations of the formintegraldisplay φidx=iintegraldisplay φd x , which is allowable since the Cartesian unit vectors are of constant magnitude and direction and hence may be taken out of the integral. If we had been usinga diﬀerent coordinate system, such as spherical polars, then, as we saw in thelast chapter, the unit basis vectors would not be constant. In that case the basisvectors could not be factorised out of the integral. The second and third line integrals in (11.1) can also be reduced to a set of scalar integrals by writing the vector ﬁeld ain terms of its Cartesian components asa=a xi+ayj+azk,w h e r e ax,ay,azare each (in general) functions of x, y, z . The second line integral in (11.1), for example, can then be written as integraldisplay Ca·dr=integraldisplay C(axi+ayj+azk)·(dxi+dyj+dzk) =integraldisplay C(axdx+aydy+azdz) =integraldisplay Caxdx+integraldisplay Caydy+integraldisplay Cazdz. (11.2) 384 11.1 LINE INTEGRALS A similar procedure may be followed for the third type of line integral in (11.1). Line integrals have properties that are analogous to those of ordinary integrals. In particular, the following are useful properties (which we illustrate using the second form of line integral in (11.1) but which are valid for all three types). (i) Reversing the path of integration changes the sign of the integral. If the path Calong which the line integrals are evaluated has AandBas its end-points then integraldisplayB Aa·dr=−integraldisplayA Ba·dr. This implies that if the path Cis a loop then integrating around the loop in the opposite direction changes the sign of the integral. (ii) If the path of integration is subdivided into smaller segments then the sum of the separate line integrals along each segment is equal to the line integralalong the whole path. So, if Pis any point on the path of integration that lies between the path’s end-points AandBthen integraldisplay B Aa·dr=integraldisplayP Aa·dr+integraldisplayB Pa·dr.IEvaluate the line integral I= R Ca·dr,w h e r e a=(x+y)i+(y−x)j, along each of the paths in the xy-plane shown in ﬁgure 11.1, namely (i) the parabola y2=xfrom(1,1)to(4,2), (ii) the curve x=2u2+u+1,y=1+ u2from(1,1)to(4,2), (iii) the line y=1from(1,1)to(4,1), followed by the line x=4 from(4,1) to(4,2). Since each of the paths lies entirely in the xy-plane, we have dr=dxi+dyj.W ec a n therefore write the line integral as I= Z Ca·dr= Z C[(x+y)dx+(y−x)dy]. (11.3) We must now evaluate this line integral along each of the prescribed paths. Case (i) . Along the parabola y2=xwe have 2 yd y=dx. Substituting for xin (11.3) and using just the limits on y,w eo b t a i n I= Z(4,2) (1,1)[(x+y)dx+(y−x)dy]= Z2 1[(y2+y)2y+(y−y2)]dy=1 11 3. Note that we could just as easily have substituted for yand obtained an integral in x, which would have given the same result. Case (ii) . The second path is given in terms of a parameter u. We could eliminate u between the two equations to obtain a relationship between xandydirectly and proceed as above, but it is usually quicker to write the line integral in terms of the parameter u. Along the curve x=2u2+u+1 , y=1+ u2we have dx=( 4u+1 )duanddy=2ud u. 385 LINE, SURFACE AND VOLUME INTEGRALS y x(i) (ii) (iii) (1,1)(4,2) Figure 11.1 Diﬀerent possible paths between the points (1, 1) and (4, 2). Substituting for xandyin (11.3) and writing the correct limits on u,w eo b t a i n I= Z(4,2) (1,1)[(x+y)dx+(y−x)dy] = Z1 0[(3u2+u+ 2)(4 u+1 )−(u2+u)2u]du=1 02 3. Case (iii) . For the third path the line integral must be evaluated along the two line segments separately and the results added together. First, along the line y= 1 we have dy= 0. Substituting this into (11.3) and using just the limits on xf o rt h i ss e g m e n t ,w e obtainZ(4,1) (1,1)[(x+y)dx+(y−x)dy]= Z4 1(x+1 )dx=1 01 2. Next, along the line x= 4 we have dx= 0. Substituting this into (11.3) and using just the limits on yfor this segment, we obtainZ(4,2) (4,1)[(x+y)dx+(y−x)dy]= Z2 1(y−4)dy=−21 2. The value of the line integral along the whole path is just the sum of the values of the line integrals along each segment, and is given by I=1 01 2−21 2=8 . J When calculating a line integral along some curve C, which is given in terms ofx,yandz, we are sometimes faced with the problem that the curve Cis such that x,yandzare not single-valued functions of one another over the entire length of the curve. This is a particular problem for closed loops in the xy-plane (and also for some open curves). In such cases the path may be subdivided intoshorter line segments along which one coordinate is a single-valued function of the other two. The sum of the line integrals along these segments is then equal to the line integral along the entire curve C. A better solution, however, is to represent the curve in a parametric form r(u) that is valid for its entire length. 386 11.1 LINE INTEGRALSIEvaluate the line integral I= H Cxd y,w h e r e Cis the circle in the xy-plane deﬁned by x2+y2=a2,z=0. Adopting the usual convention mentioned above, the circle Cis to be traversed in the anticlockwise direction. Taking the circle as a whole means xis not a single-valued function of y. We must therefore divide the path into two parts with x=+ p a2−y2for the semicircle lying to the right of x=0 ,a n d x=− p a2−y2for the semicircle lying to the left of x= 0. The required line integral is then the sum of the integrals along the two semicircles. Substituting for x,i ti sg i v e nb y I= I Cxd y= Za −a p a2−y2dy+ Z−a a / − p a2−y2 / dy =4 Za 0 p a2−y2dy=πa2. Alternatively, we can represent the entire circle parametrically, in terms of the azimuthal angle φ,s ot h a t x=acosφandy=asinφwith φrunning from 0 to 2 π. The integral can therefore be evaluated over the whole circle at once. Noting that dy=acosφd φ,w ec a n rewrite the line integral completely in terms of the parameter φand obtain I= I Cxd y=a2 Z2π 0cos2φd φ=πa2. J 11.1.2 Physical examples of line integrals There are many physical examples of line integrals, but perhaps the most common is the expression for the total work done by a force Fwhen it moves its point of application from a point Ato a point Balong a given curve C. We allow the magnitude and direction of Fto vary along the curve. Let the force act at a point rand consider a small displacement dralong the curve; then the small amount of work done is dW=F·dr, as discussed in subsection 7.6.1 (note that dWcan be either positive or negative). Therefore, the total work done in traversing thepath Cis W C=integraldisplay CF·dr. Naturally, other physical quantities can be expressed in such a way. For example, the electrostatic potential energy gained by moving a charge qalong a path Cin an electric ﬁeld Eis−qintegraltext CE·dr. We may also note that Amp `ere’s law concerning the magnetic ﬁeld Bassociated with a current-carrying wire can be written as contintegraldisplay CB·dr=µ0I, where Iis the current enclosed by a closed path Ctraversed in a right-handed sense with respect to the current direction. Magnetostatics also provides a physical example of the third type of line 387 LINE, SURFACE AND VOLUME INTEGRALS integral in (11.1). If a loop of wire Ccarrying a current Iis placed in a magnetic ﬁeldBthen the force dFon a small length drof the wire is given by dF=Idr×B, and so the total (vector) force on the loop is F=Icontintegraldisplay Cdr×B. 11.1.3 Line integrals with respect to a scalar In addition to those listed in (11.1), we can form other types of line integral, which depend on a particular curve Cbut for which we integrate with respect to a scalar du, rather than the vector diﬀerential dr. This distinction is somewhat arbitrary, however, since we can always rewrite line integrals containing the vectordiﬀerential dras a line integral with respect to some scalar parameter. If the path Calong which the integral is taken is described parametrically by r(u)t h e n dr=dr dudu, and the second type of line integral in (11.1), for example, can be written as integraldisplay Ca·dr=integraldisplay Ca·dr dudu. A similar procedure can be followed for the other types of line integral in (11.1). Commonly occurring special cases of line integrals with respect to a scalar are integraldisplay Cφd s ,integraldisplay Cads, where sis the arc length along the curve C. We can always represent Cparamet- rically by r(u), and from section 10.3 we have ds=radicalbigg dr du·dr dudu. The line integrals can therefore be expressed entirely in terms of the parameter u and thence evaluated.IEvaluate the line integral I= R C(x−y)2ds,w h e r e Cis the semicircle of radius arunning from A=(a,0)toB=(−a,0)and for which y≥0. The semicircular path from AtoBcan be described in terms of the azimuthal angle φ (measured from the x-axis) by r(φ)=acosφi+asinφj, where φruns from 0 to π. Therefore the element of arc length is given, from section 10.3, by ds= s dr dφ·dr dφdφ=a(cos2φ+s i n2φ)dφ=ad φ . 388 11.2 CONNECTIVITY OF REGIONS (a)( b)( c) Figure 11.2 ( a) A simply connected region; ( b) a doubly connected region; (c) a triply connected region. Since ( x−y)2=a2(1−sin 2φ), the line integral becomes I= Z C(x−y)2ds= Zπ 0a3(1−sin2φ)dφ=πa3. J As discussed in the previous chapter, the expression (10.58) for the square of the element of arc length in three-dimensional orthogonal curvilinear coordinates u1,u2,u3is (ds)2=h2 1(du1)2+h2 2(du2)2+h2 3(du3)2, where h1,h2,h3are the scale factors of the coordinate system. If a curve Cin three dimensions is given parametrically by the equations ui=ui(λ)f o r i=1,2,3 then the element of arc length along the curve is ds=radicalBigg h2 1parenleftbiggdu1 dλparenrightbigg2 +h2 2parenleftbiggdu2 dλparenrightbigg2 +h2 3parenleftbiggdu3 dλparenrightbigg2 dλ. 11.2 Connectivity of regions In physical systems it is usual to deﬁne a scalar or vector ﬁeld in some region R. In the next and some later sections we will need the concept of the connectivity of such a region in both two and three dimensions. We begin by discussing planar regions. A plane region Ris said to be simply connected if every simple closed curve within Rcan be continuously shrunk to a point without leaving the region (see ﬁgure 11.2( a)). If, however, the region Rcontains a hole then there exist simple closed curves that cannot by shrunk to a point without leaving R(see ﬁgure 11.2( b) ) .S u c har e g i o ni ss a i dt ob e doubly connected, since its boundary has two distinct parts. Similarly, a regionwith n−1 holes is said to be n-fold connected ,o rmultiply connected (the region in ﬁgure 11.2( c) is triply connected). These ideas can be extended to regions that are not planar, such as general 389 LINE, SURFACE AND VOLUME INTEGRALS y d c a b xSR TCUV Figure 11.3 A simply connected region Rbounded by the curve C. three-dimensional surfaces and volumes. The same criteria concerning the shrink- ing of closed curves to a point also apply when deciding the connectivity of suchregions. In these cases, however, the curves must lie in the surface or volumein question. For example, the interior of a torus is not simply connected, sincethere exist closed curves in the interior that cannot be shrunk to a point without leaving the torus. On the other hand, the region between two concentric spheres of diﬀerent radii is simply connected. 11.3 Green’s theorem in a plane In subsection 11.1.1 we considered (amongst other things) the evaluation of line integrals for which the path Cis closed and lies entirely in the xy-plane. Since the path is closed it will enclose a region Rof the plane. We now discuss how to express the line integral around the loop as a double integral over the enclosed region R. Suppose the functions P(x, y),Q(x, y) and their partial derivatives are single- valued, ﬁnite and continuous inside and on the boundary Cof some simply connected region Rin the xy-plane. Green’s theorem in a plane then states contintegraldisplay C(Pd x+Qd y)=integraldisplayintegraldisplay Rparenleftbigg∂Q ∂x−∂P ∂yparenrightbigg dx dy, (11.4) and so relates the line integral around Cto a double integral over the enclosed region R. This theorem may be proved straightforwardly in the following way. Consider the simply connected region Rin ﬁgure 11.3, and let y=y1(x)a n d y=y2(x) be the equations of the curves STU andSVU respectively. We then 390 11.3 GREEN’S THEOREM IN A PLANE write integraldisplayintegraldisplay R∂P ∂ydx dy =integraldisplayb adxintegraldisplayy2(x) y1(x)dy∂P ∂y=integraldisplayb adxbracketleftBig P(x, y)bracketrightBigy=y2(x) y=y1(x) =integraldisplayb abracketleftBig P(x, y2(x))−P(x, y1(x))bracketrightBig dx =−integraldisplayb aP(x, y1(x))dx−integraldisplaya bP(x, y2(x))dx=−contintegraldisplay CPd x . If we now let x=x1(y)a n d x=x2(y) be the equations of the curves TSV and TUV respectively, we can similarly show that integraldisplayintegraldisplay R∂Q ∂xdx dy =integraldisplayd cdyintegraldisplayx2(y) x1(y)dx∂Q ∂x=integraldisplayd cdybracketleftBig Q(x, y)bracketrightBigx=x2(y) x=x1(y) =integraldisplayd cbracketleftBig Q(x2(y),y)−Q(x1(y),y)bracketrightBig dy =integraldisplayc dQ(x1,y)dy+integraldisplayd cQ(x2,y)dy=contintegraldisplay CQd y. Subtracting these two results gives Green’s theorem in a plane.IShow that the area of a region Renclosed by a simple closed curve Cis given by A= 1 2 H C(xd y−yd x)= H Cxd y=− H Cyd x. Hence calculate the area of the ellipse x=acosφ, y=bsinφ. In Green’s theorem (11.4) put P=−yandQ=x;t h e nI C(xd y−yd x)= ZZ R(1 + 1) dx dy =2 ZZ Rdx dy =2A. Therefore the area of the region is A=1 2 H C(xd y−yd x). Alternatively, we could put P=0 andQ=xand obtain A= H Cxd y, or put P=−yandQ= 0, which gives A=− H Cyd x. The area of the ellipse x=acosφ,y=bsinφis given by A=1 2 I C(xd y−yd x)=1 2 Z2π 0ab(cos2φ+s i n2φ)dφ =ab 2 Z2π 0dφ=πab. J It may further be shown that Green’s theorem in a plane is also valid for multiply connected regions. In this case, the line integral must be taken overall the distinct boundaries of the region. Furthermore, each boundary must be traversed in the positive direction, such that a person travelling along it in this direction always has the region Ron their left. In order to apply Green’s theorem to the region Rshown in ﬁgure 11.4, the line integrals must be taken over 391 LINE, SURFACE AND VOLUME INTEGRALS y xC1C2R Figure 11.4 A doubly connected region Rbounded by the curves C1andC2. both boundaries, C1andC2, in the directions indicated, and the results added together. We may also use Green’s theorem in a plane to investigate the path indepen- dence (or not) of line integrals when the paths lie in the xy-plane. Let us consider the line integral I=integraldisplayB A(Pd x+Qd y). For the line integral from AtoBto be independent of the path taken, it must have the same value along any two arbitrary paths C1andC2joining the points. Moreover, if we consider as the path the closed loop Cformed by C1−C2then the line integral around this loop must be zero. From Green’s theorem in a plane,(11.4), we see that a suﬃcient condition for I=0i st h a t ∂P ∂y=∂Q ∂x, (11.5) throughout some simply connected region Rcontaining the loop, where we assume that these partial derivatives are continuous in R. It may be shown that (11.5) is also a necessary condition for I=0a n di s equivalent to requiring Pd x+Qd yto be an exact diﬀerential of some function φ(x, y) such that Pd x+Qd y=dφ. It follows thatintegraltextB A(Pd x+Qd y)=φ(B)−φ(A) and thatcontintegraltext C(Pd x+Qd y) around any closed loop Cin the region Ris identically zero. These results are special cases of the general results for paths in threedimensions, which are discussed in the next section. 392 11.4 CONSERVATIVE FIELDS AND POTENTIALSIEvaluate the line integral I= I C[(exy+c o s xsiny)dx+(ex+s i n xcosy)dy], around the ellipse x2/a2+y2/b2=1. Clearly, it is not straightforward to calculate this line integral directly. However, if we let P=exy+c o s xsiny and Q=ex+s i n xcosy, then ∂P/∂y =ex+c o s xcosy=∂Q/∂x ,a n ds o Pd x+Qd yis an exact diﬀerential (it is actually the diﬀerential of the function f(x, y)=exy+s i n xsiny). From the above discussion, we therefore immediately conclude that I=0 . J 11.4 Conservative ﬁelds and potentials So far we have made the point that, in general, the value of a line integral between two points AandBdepends on the path Ctaken from AtoB.I nt h e previous section, however, we saw that, for paths in the xy-plane, line integrals whose integrands have certain properties are independent of the path taken. We now extend that discussion to the full three-dimensional case. For line integrals of the formintegraltext Ca·dr, there exists a class of vector ﬁelds for which the line integral between two points is independent of the path taken. Such vector ﬁelds are called conservative . A vector ﬁeld athat has continuous partial derivatives in a simply connected region Ris conservative if, and only if, any of the following is true. (i) The integralintegraltextB Aa·dr,w h e r e AandBlie in the region R, is independent of the path from AtoB. Hence the integralcontintegraltext Ca·draround any closed loop inRis zero. (ii) There exists a single-valued function φof position such that a=∇φ. (iii)∇×a=0. (iv)a·dris an exact diﬀerential. The validity or otherwise of any of these statements implies the same for the other three, which we will now show. First, let us assume that (i) above is true. If the line integral from AtoB is independent of the path taken between the points then its value must be afunction only of the positions of AandB. We may therefore write integraldisplay B Aa·dr=φ(B)−φ(A), (11.6) which deﬁnes a single-valued scalar function of position φ. If the points AandB are separated by an inﬁnitesimal displacement drthen (11.6) becomes a·dr=dφ, 393 LINE, SURFACE AND VOLUME INTEGRALS which shows that we require a·drto be an exact diﬀerential: condition (iv). From (10.27) we can write dφ=∇φ·dr, and so we have (a−∇φ)·dr=0. Since dris arbitrary, we ﬁnd that a=∇φ; this immediately implies ∇×a=0, condition (iii) (see (10.37)). Alternatively, if we suppose that there exists a single-valued function of position φsuch that a=∇φthen∇×a=0follows as before. The line integral around a closed loop then becomes contintegraldisplay Ca·dr=contintegraldisplay C∇φ·dr=contintegraldisplay dφ. Since we deﬁned φto be single-valued, this integral is zero as required. Now suppose ∇×a=0. From Stoke’s theorem, which is discussed in sec- tion 11.9, we immediately obtaincontintegraltext Ca·dr=0 ;t h e n a=∇φanda·dr=dφfollow as above. Finally, let us suppose a·dr=dφ. Then immediately we have a=∇φ,a n dt h e other results follow as above.IEvaluate the line integral I= RB Aa·dr,w h e r e a=(xy2+z)i+(x2y+2 )j+xk,Ais the point(c, c, h)andBis the point (2c, c/2,h), along the diﬀerent paths (i)C1, given by x=cu,y=c/u,z=h, (ii)C2, given by 2y=3c−x,z=h. Show that the vector ﬁeld ais in fact conservative, and ﬁnd φsuch that a=∇φ. Expanding out the integrand, we have I= Z(2c, c/2,h) (c, c, h) / (xy2+z)dx+(x2y+2 )dy+xd z / , (11.7) which we must evaluate along each of the paths C1andC2. (i) Along C1we have dx=cd u,dy=−(c/u2)du,dz= 0, and on substituting in (11.7) and ﬁnding the limits on u,w eo b t a i n I= Z2 1c / h−2 u2 / du=c(h−1). (ii) Along C2we have 2 dy=−dx,dz= 0 and, on substituting in (11.7) and using the limits on x,w eo b t a i n I= Z2c c /;1 2x3−9 4cx2+9 4c2x+h−1 / dx=c(h−1). Hence the line integral has the same value along paths C1andC2. Taking the curl of a, we have ∇×a=( 0−0)i+( 1−1)j+( 2xy−2xy)k=0, soais a conservative vector ﬁeld, and the line integral between two points must be 394 11.5 SURFACE INTEGRALS independent of the path taken. Since ais conservative, we can write a=∇φ. Therefore, φ must satisfy ∂φ ∂x=xy2+z, which implies that φ=1 2x2y2+zx+f(y,z) for some function f. Secondly, we require ∂φ ∂y=x2y+∂f ∂y=x2y+2, which implies f=2y+g(z). Finally, since ∂φ ∂z=x+∂g ∂z=x, we have g=c o n s t a n t= k. It can be seen that we have explicitly constructed the function φ=1 2x2y2+zx+2y+k. J The quantity φthat ﬁgures so prominently in this section is called the scalar potential function o ft h ec o n s e r v a t i v ev e c t o rﬁ e l d a(which satisﬁes ∇×a=0), and is unique up to an arbitrary additive constant. Scalar potentials that are multi-valued functions of position (but in simple ways) are also of value in describingsome physical situations, the most obvious example being the scalar magnetic potential associated with a current-carrying wire. When the integral of a ﬁeld quantity around a closed loop is considered, provided the loop does not enclosea net current, the potential is single-valued and all the above results still hold. Ifthe loop does enclose a net current, however, our analysis is no longer valid andextra care must be taken. If, instead of being conservative, a vector ﬁeld bsatisﬁes∇·b=0( i . e . b is solenoidal) then it is both possible and useful, for example in the theory ofelectromagnetism, to deﬁne a vector potential ﬁeld asuch that b=∇×a.I tm a y be shown that such a vector ﬁeld aalways exists. Further, if ais one such vector ﬁeld then a /prime=a+∇ψ+c,w h e r e ψis any scalar function and cis any constant vector, also satisﬁes the above relationship, i.e. b=∇×a/prime. This was discussed more fully in subsection 10.8.2. 11.5 Surface integrals As with line integrals, integrals over surfaces can involve vector and scalar ﬁelds and, equally, can result in either a vector or a scalar. The simplest case involvesentirely scalars and is of the form integraldisplay Sφd S. (11.8) As analogues of the line integrals listed in (11.1), we may also encounter surface integrals involving vectors, namely integraldisplay SφdS,integraldisplay Sa·dS,integraldisplay Sa×dS. (11.9) 395 LINE, SURFACE AND VOLUME INTEGRALS SS V CdS dS (a)( b) Figure 11.5 ( a) A closed surface and ( b) an open surface. In each case a normal to the surface is shown: dS=ˆndS. All the above integrals are taken over some surface S, which may be either open or closed, and are therefore, in general, double integrals. Following thenotation for line integrals, for surface integrals over a closed surfaceintegraltext Sis replaced bycontintegraltext S. The vector diﬀerential dSin (11.9) represents a vector area element of the surface S. It may also be written dS=ˆndS,w h e r e ˆnis a unit normal to the surface at the position of the element and dSis the scalar area of the element used in (11.8). The convention for the direction of the normal ˆnto a surface depends on whether the surface is open or closed. A closed surface, see ﬁgure 11.5( a), does not have to be simply connected (for example, the surface of a torus is not),but it does have to enclose a volume V, which may be of inﬁnite extent. The direction of ˆnis taken to point outwards from the enclosed volume as shown. An open surface, see ﬁgure 11.5( b), spans some perimeter curve C. The direction ofˆnis then given by the right-hand sense with respect to the direction in which the perimeter is traversed, i.e. follows the right-hand screw rule discussed in section 7.6.2. An open surface does not have to be simply connected but for our purposes it must be two-sided (a M ¨obius strip is an example of a one-sided surface). The formal deﬁnition of a surface integral is very similar to that of a line integral. We divide the surface SintoNelements of area ∆ S p,p=1,...,N ,e a c h with a unit normal ˆnp.I f(xp,yp,zp) is any point in ∆ Spthen the second type of surface integral in (11.9), for example, is deﬁned as integraldisplay Sa·dS= lim N→∞Nsummationdisplay p=1a(xp,yp,zp)·ˆnp∆Sp, where it is required that all ∆ Sp→0a sN→∞. 396 11.5 SURFACE INTEGRALS xz y RdAα SkdS Figure 11.6 A surface S(or part thereof) projected onto a region Rin the xy-plane; dSis the surface element at a point P. 11.5.1 Evaluating surface integrals We now consider how to evaluate surface integrals over some general surface. This involves writing the scalar area element dSin terms of the coordinate diﬀerentials of our chosen coordinate system. In some particularly simple cases this is verystraightforward. For example, if Sis the surface of a sphere of radius a(or some part thereof) then using spherical polar coordinates θ,φon the sphere we have dS=a 2sinθd θd φ . For a general surface, however, it is not usually possible to represent the surface in a simple way in any particular coordinate system. In suchcases, it is usual to work in Cartesian coordinates and consider the projections ofthe surface onto the coordinate planes. Consider a surface (or part of a surface) Sas in ﬁgure 11.6. The surface Sis projected onto a region Rof the xy-plane, so that an element of surface area dS at point Pprojects onto the area element dA. From the ﬁgure, we see that dA= |cosα|dS,w h e r e αis the angle between the unit vector kin the z-direction and the unit normal ˆnto the surface at P. So, at any given point of S, we have simply dS=dA |cosα|=dA |ˆn·k|. Now, if the surface Sis given by the equation f(x, y, z) = 0 then, as shown in subsection 10.7.1, the unit normal at any point of the surface is simply given by ˆn=∇f/|∇f|evaluated at that point, cf. (10.32). The scalar element of surface area then becomes dS=dA |ˆn·k|=|∇f|dA ∇f·k=|∇f|dA ∂f/∂z, (11.10) 397 LINE, SURFACE AND VOLUME INTEGRALS where|∇f|and∂f/∂z are evaluated on the surface S. We can therefore express any surface integral over Sas a double integral over the region Rin the xy-plane.IEvaluate the surface integral I= R Sa·dS,w h e r e a=xiandSis the surface of the hemisphere x2+y2+z2=a2with z≥0. The surface of the hemisphere is shown in ﬁgure 11.7. In this case dSmay be easily expressed in spherical polar coordinates as dS=a2sinθd θd φ , and the unit normal to the surface at any point is simply ˆr. On the surface of the hemisphere we have x=asinθcosφ and so a·dS=x(i·ˆr)dS=(asinθcosφ)(sinθcosφ)(a2sinθd θd φ ). Therefore, inserting the correct limits on θandφ, we have I= Z Sa·dS=a3 Zπ/2 0dθsin3θ Z2π 0dφcos2φ=2πa3 3. We could, however, follow the general prescription above and project the hemisphere S onto the region Rin the xy-plane that is a circle of radius acentred at the origin. Writing the equation of the surface of the hemisphere as f(x, y)=x2+y2+z2−a2=0a n du s i n g (11.10), we have I= Z Sa·dS= Z Sx(i·ˆr)dS= Z Rx(i·ˆr)|∇f|dA ∂f/∂z. Now∇f=2xi+2yj+2zk=2r,s oo nt h es u r f a c e Swe have|∇f|=2|r|=2a.O n Swe also have ∂f/∂z =2z=2 p a2−x2−y2andi·ˆr=x/a. Therefore, the integral becomes I= ZZ Rx2p a2−x2−y2dx dy. Although this integral may be evaluated directly, it is quicker to transform to plane polar coordinates: I= ZZ R/primeρ2cos2φp a2−ρ2ρd ρd φ = Z2π 0cos2φd φ Za 0ρ3dρp a2−ρ2. Making the substitution ρ=asinu, we ﬁnally obtain I= Z2π 0cos2φd φ Zπ/2 0a3sin3ud u=2πa3 3. J In the above discussion we assumed that any line parallel to the z-axis intersects Sonly once. If this is not the case, we must split up the surface into smaller surfaces S1,S2etc. that are of this type. The surface integral over Sis then the sum of the surface integrals over S1,S2and so on. This is always necessary for closed surfaces. We may also sometimes wish to project a surface S(or some part of it) onto thezx-o ryz-plane, rather than the xy-plane. In such cases, the above analysis is easily modiﬁed. 398 11.5 SURFACE INTEGRALS dS Sz Ca a a xy dA=dx dy Figure 11.7 The surface of the hemisphere x2+y2+z2=a2,z≥0. 11.5.2 Vector areas of surfaces The vector area of a surface Sis deﬁned simply as S=integraldisplay SdS, where the surface integral may be evaluated as above.IFind the vector area of the surface of the hemisphere x2+y2+z2=a2with z≥0. As in the previous example, dS=a2sinθd θd φ ˆrin spherical polar coordinates. Therefore the vector area is given by S= ZZ Sa2sinθˆrdθ dφ. Now, since ˆrvaries over the surface S, it also must be integrated. This is most easily achieved by writing ˆrin terms of the constant Cartesian basis vectors. On Swe have ˆr=s i n θcosφi+s i n θsinφj+c o s θk, so the expression for the vector area becomes S=i / a2 Z2π 0cosφd φ Zπ/2 0sin2θd θ /! +j / a2 Z2π 0sinφd φ Zπ/2 0sin2θd θ /! +k / a2 Z2π 0dφ Zπ/2 0sinθcosθd θ /! =0+0+πa2k=πa2k. Note that the magnitude of Sis the projected area, of the hemisphere onto the xy-plane, and not the surface area of the hemisphere. J 399 LINE, SURFACE AND VOLUME INTEGRALS dr r OC Figure 11.8 The conical surface spanning the perimeter Cand having its vertex at the origin. The hemispherical shell discussed above is an example of an open surface. For a closed surface, however, the vector area is always zero. This may be seen by projecting the surface down onto each Cartesian coordinate plane in turn. For each projection, every positive element of area on the upper surface is cancelledby the corresponding negative element on the lower surface. Therefore, eachcomponent of S=contintegraltext SdSvanishes. An important corollary of this result is that the vector area of an open surface depends only on its perimeter, or boundary curve, C. This may be proved as follows. If surfaces S1andS2have the same perimeter then S1−S2is a closed surface, for which contintegraldisplay dS=integraldisplay S1dS−integraldisplay S2dS=0. Hence S1=S2. Moreover, we may derive an expression for the vector area of an open surface Ssolely in terms of a line integral around its perimeter C. Since we may choose any surface with perimeter C, we will consider a cone with its vertex at the origin (see ﬁgure 11.8). The vector area of the elementarytriangular region shown in the ﬁgure is dS= 1 2r×dr. Therefore, the vector area of the cone, and hence of anyopen surface with perimeter C, is given by the line integral S=1 2contintegraldisplay Cr×dr. For a surface conﬁned to the xy-plane, r=xi+yjand dr=dxi+dyj, and we obtain for this special case that the area of the surface is given byA= 1 2contintegraltext C(xd y−yd x), as we found in section 11.3. 400 11.5 SURFACE INTEGRALSIFind the vector area of the surface of the hemisphere x2+y2+z2=a2,z≥0,b y evaluating the line integral S=1 2 H Cr×draround its perimeter. The perimeter Cof the hemisphere is the circle x2+y2=a2, on which we have r=acosφi+asinφj,d r=−asinφd φi+acosφd φj. Therefore the cross product r×dris given by r×dr= // //// ij k acosφa sinφ 0 −asinφd φ a cosφd φ 0 // //// =a2(cos2φ+s i n2φ)dφk=a2dφk, and the vector area becomes S=1 2a2k Z2π 0dφ=πa2k. J 11.5.3 Physical examples of surface integrals There are many examples of surface integrals in the physical sciences. Surface integrals of the form (11.8) occur in computing the total electric charge on a surface or the mass of a shell,integraltext Sρ(r)dS, when the charge or mass density ρ(r)is known. For surface integrals involving vectors, the second form in (11.9) is themost common. For a vector ﬁeld a, the surface integralintegraltext Sa·dSis called the ﬂux ofathrough S. Examples of physically important ﬂux integrals are numerous. For example, let us consider a surface Sin a ﬂuid with density ρ(r)t h a th a sa velocity ﬁeld v(r). The mass of ﬂuid crossing an element of surface area dSin time dtisdM=ρv·dSdt. Therefore the nettotal mass ﬂux of ﬂuid crossing S isM=integraltext Sρ(r)v(r)·dS. As a another example, the electromagnetic ﬂux of energy out of a given volume Vbounded by a surface Siscontintegraltext S(E×H)·dS. The solid angle, to be deﬁned below, subtended at a point Oby a surface (closed or otherwise) can also be represented by an integral of this form, although it isnot strictly a ﬂux integral (unless we imagine isotropic rays radiating from O). The integral Ω=integraldisplay Sr·dS r3=integraldisplay Sˆr·dS r2, (11.11) gives the solid angle Ωsubtended at Oby a surface Sifris the position vector measured from Oof an element of the surface. A little thought will show that (11.11) takes account of all three relevant factors: the size of the element ofsurface, its inclination to the line joining the element to Oand the distance from O. Such a general expression is often useful for computing solid angles when the three-dimensional geometry is complicated. Note that (11.11) remains valid when the surface Sis not convex and when a single ray from Oin certain directions would cut Sin more than one place (but we exclude multiply connected regions). 401 LINE, SURFACE AND VOLUME INTEGRALS In particular, when the surface is closed Ω = 0 if Ois outside Sand Ω = 4 πifO is an interior point. Surface integrals resulting in vectors occur less frequently. An example is aﬀorded, however, by the total resultant force experienced by a body immersed in a stationary ﬂuid in which the hydrostatic pressure is given by p(r). The pressure is everywhere inwardly directed and the resultant force is F=−contintegraltext SpdS,t a k e n over the whole surface. 11.6 Volume integrals Volume integrals are deﬁned in an obvious way and are generally simpler than line or surface integrals since the element of volume dVis a scalar quantity. We may encounter volume integrals of the form integraldisplay Vφd V,integraldisplay VadV. (11.12) Clearly, the ﬁrst form results in a scalar, whereas the second form yields a vector. Two closely related physical examples, one of each kind, are provided by the totalmass of a ﬂuid contained in a volume V,g i v e nb yintegraltext Vρ(r)dV, and the total linear momentum of that same ﬂuid, given byintegraltext Vρ(r)v(r)dV,w h e r e v(r)i st h ev e l o c i t y ﬁeld in the ﬂuid. As a slightly more complicated example of a volume integral wemay consider the following.IFind an expression for the angular momentum of a solid body rotating with angular velocity ωabout an axis through the origin. Consider a small volume element dVsituated at position r; its linear momentum is ρd V˙r, where ρ=ρ(r) is the density distribution, and its angular momentum about Oisr×ρ˙rdV. Thus for the whole body the angular momentum Lis L= Z V(r×˙r)ρd V. Putting ˙r=ω×ryields L= Z V[r×(ω×r)]ρd V= Z Vωr2ρd V− Z V(r·ω)rρd V. J The evaluation of the ﬁrst type of volume integral in (11.12) has already been considered in our discussion of multiple integrals in chapter 6. The evaluation of the second type of volume integral follows directly since we can write integraldisplay VadV=iintegraldisplay VaxdV+jintegraldisplay VaydV+kintegraldisplay VazdV, (11.13) where ax,ay,azare the Cartesian components of a. Of course, we could have written ain terms of the basis vectors of some other coordinate system (e.g. spherical polars) but, since such basis vectors are not, in general, constant, they 402 11.6 VOLUME INTEGRALS V OS rdS Figure 11.9 A general volume Vcontaining the origin and bounded by the closed surface S. cannot be taken out of the integral sign as in (11.13) and must be included as part of the integrand. 11.6.1 Volumes of three-dimensional regions As discussed in chapter 6, the volume of a three-dimensional region Vis simply V=integraltext VdV, which may be evaluated directly once the limits of integration have been found. However, the volume of the region obviously depends only on the surface Sthat bounds it. We should therefore be able to express the volume V in terms of a surface integral over S. This is indeed possible, and the appropriate expression may derived as follows. Referring to ﬁgure 11.9, let us suppose thatthe origin Ois contained within V. The volume of the small shaded cone is dV= 1 3r·dS; the total volume of the region is thus given by V=1 3contintegraldisplay Sr·dS. It may be shown that this expression is still valid even when Ois not contained inV. Although this surface integral form is available, in practice, in many cases it is simpler to evaluate the volume integral directly.IFind the volume enclosed between a sphere of radius acentred on the origin and a circular cone of half-angle αwith its vertex at the origin. The element of vector area dSon the surface of the sphere is given in spherical polar coordinates by a2sinθd θd φ ˆr. Now taking the axis of the cone to lie along the z-axis (from which θis measured) the required volume is given by V=1 3 I Sr·dS=1 3 Z2π 0dφ Zα 0a2sinθr·ˆrdθ =1 3 Z2π 0dφ Zα 0a3sinθd θ=2 3πa3(1−cosα). J 403 LINE, SURFACE AND VOLUME INTEGRALS 11.7 Integral forms for grad,divandcurl In the previous chapter we deﬁned the vector operators grad, div and curl in purely mathematical terms, which depended on the coordinate system in which they wereexpressed. An interesting application of line, surface and volume integrals is theexpression of grad, div and curl in coordinate-free, geometrical terms. If φis a scalar ﬁeld and ais a vector ﬁeld then it may be shown that at any point P ∇φ= lim V→0parenleftbigg1 Vcontintegraldisplay SφdSparenrightbigg (11.14) ∇·a= lim V→0parenleftbigg1 Vcontintegraldisplay Sa·dSparenrightbigg (11.15) ∇×a= lim V→0parenleftbigg1 Vcontintegraldisplay SdS×aparenrightbigg (11.16) where Vis a small volume enclosing PandSis its bounding surface. Indeed, we may consider these equations as the (geometrical) deﬁnitions of grad, div and curl. An alternative, but equivalent, geometrical deﬁnition of ∇×aat a point P, which is often easier to use than (11.16), is given by (∇×a)·ˆn= lim A→0parenleftbigg1 Acontintegraldisplay Ca·drparenrightbigg , (11.17) where Cis a plane contour of area Aenclosing the point Pandˆnis the unit normal to the enclosed planar area. It may be shown, in any coordinate system , that all the above equations are consistent with our deﬁnitions in the previous chapter although the diﬃculty of proof depends on the chosen coordinate system. The most general coordinate system encountered in that chapter was one with orthogonal curvilinear coordi-nates u 1,u2,u3, of which Cartesians, cylindrical polars and spherical polars are all special cases. Although it may be shown that (11.14) leads to the usual expressionfor grad in curvilinear coordinates, the proof requires complicated manipulationsof the derivatives of the basis vectors with respect to the coordinates and is notpresented here. In Cartesian coordinates, however, the proof is quite simple.IShow that the geometrical deﬁnition of gradleads to the usual expression for ∇φin Cartesian coordinates. Consider the surface Sof a small rectangular volume element ∆ V=∆x∆y∆zthat has its f a c e sp a r a l l e lt ot h e x,y,a n d zcoordinate surfaces and the point Pat one corner. We must calculate the surface integral (11.14) over each of its six faces. Remembering that thenormal to the surface points outwards from the volume on each face, the two faces withx= constant have areas ∆ S=−i∆y∆zand ∆ S=i∆y∆zrespectively. Furthermore, over each small surface element, we may take φto be constant, so that the net contribution to 404 11.7 INTEGRAL FORMS FOR grad, div AND curl the surface integral fr om these two faces is then [(φ+∆φ)−φ]∆y∆zi= / φ+∂φ ∂x∆x−φ / ∆y∆zi =∂φ ∂x∆x∆y∆zi. The surface integral over the pairs of faces with y=c o n s t a n ta n d z= constant respectively may be found in a similar way, and we obtainI SφdS= /∂φ ∂xi+∂φ ∂yj+∂φ ∂zk / ∆x∆y∆z. Therefore∇φat the point Pis given by ∇φ= lim ∆x,∆y,∆z→0 /1 ∆x∆y∆z /∂φ ∂xi+∂φ ∂yj+∂φ ∂zk / ∆x∆y∆z / =∂φ ∂xi+∂φ ∂yj+∂φ ∂zk. J We now turn to (11.15) and (11.17). These geometrical deﬁnitions may be shown straightforwardly to lead to the usual expressions for div and curl inorthogonal curvilinear coordinates.IBy considering the inﬁnitesimal volume element dV=h1h2h3∆u1∆u2∆u3shown in ﬁg- ure 11.10, show that (11.15) leads to the usual expression for ∇·ain orthogonal curvilinear coordinates. Let us write the vector ﬁeld in terms of its components with respect to the basis vectors of the curvilinear coordinate system as a=a1ˆe1+a2ˆe2+a3ˆe3. We consider ﬁrst the contribution to the RHS of () from the two faces with u1=c o n s t a n t , i . e . PQ RS and the face opposite it (see ﬁgure 11.10). Now, the volume element is formed from the orthogonalvectors h 1∆u1ˆe1,h2∆u2ˆe2andh3∆u3ˆe3, at the point Pand so for we have ∆S=h2h3∆u2∆u3ˆe3×ˆe2=−h2h3∆u2∆u3ˆe1. Reasoning along the same lines as in the previous example, we conclude that the contri- bution to the surface integral of a·dSover PQ RS and its opposite face taken together is given by ∂ ∂u1(a·∆S)∆u1=∂ ∂u1(a1h2h3)∆u1∆u2∆u3. The surface integrals over the pairs of faces with u2=c o n s t a n ta n d u3=c o n s t a n t respectively may be found in a similar way, and we obtainI Sa·dS= /∂ ∂u1(a1h2h3)+∂ ∂u2(a2h3h1)+∂ ∂u3(a3h1h2) / ∆u1∆u2∆u3. Therefore∇·aat the point Pis given by ∇·a= lim ∆u1,∆u2,∆u3→0 /1 h1h2h3∆u1∆u2∆u3 I Sa·dS / =1 h1h2h3 /∂ ∂u1(a1h2h3)+∂ ∂u2(a2h3h1)+∂ ∂u3(a3h1h2) / . J 405 LINE, SURFACE AND VOLUME INTEGRALS R Q PS xyz Th1∆u1ˆe1 h2∆u2ˆe2h3∆u3ˆe3 Figure 11.10 A general volume ∆ Vin orthogonal curvilinear coordinates u1,u2,u3.PTgives the vector h1∆u1ˆe1,PSgives h2∆u2ˆe2and PQgives h3∆u3ˆe3.IBy considering the inﬁnitesimal planar surface element PQ RS in ﬁgure 11.10, show that (11.17) leads to the usual expression for ∇×ain orthogonal curvilinear coordinates. The planar surface PQ RS is deﬁned by the orthogonal vectors h2∆u2ˆe2andh3∆u3ˆe3 at the point P. If we traverse the loop in the direction PSRQ then, by the right-hand convention, the unit normal to the plane is ˆe1.W r i t i n g a=a1ˆe1+a2ˆe2+a3ˆe3, the line integral around the loop in this direction is given byI PSRQa·dr=a2h2∆u2+ / a3h3+∂ ∂u2(a3h3)∆u2 / ∆u3 − / a2h2+∂ ∂u3(a2h2)∆u3 / ∆u2−a3h3∆u3 = /∂ ∂u2(a3h3)−∂ ∂u3(a2h2) / ∆u2∆u3. Therefore from (11.17) the component of ∇×ain the direction ˆe1atPis given by (∇×a)1= lim ∆u2,∆u3→0 /1 h2h3∆u2∆u3 I PSRQa·dr / =1 h2h3 /∂ ∂u2(h3a3)−∂ ∂u3(h2a2) / . The other two components are found by cyclically permuting the subscripts 1, 2, 3. J Finally, we note that we can also write the ∇2operator as a surface integral by setting a=∇φin (11.15), to obtain ∇2φ=∇·∇φ= lim V→0parenleftbigg1 Vcontintegraldisplay S∇φ·dSparenrightbigg . 406 11.8 DIVERGENCE THEOREM AND RELATED THEOREMS 11.8 Divergence theorem and related theorems The divergence theorem relates the total ﬂux of a vector ﬁeld out of a closed surface Sto the integral of the divergence of the vector ﬁeld over the enclosed volume V; it follows almost immediately from our geometrical deﬁnition of divergence (11.15). Imagine a volume V, in which a vector ﬁeld ais continuous and diﬀerentiable, to be divided up into a large number of small volumes Vi. Using (11.15), we have for each small volume (∇·a)Vi≈contintegraldisplay Sia·dS, where Siis the surface of the small volume Vi. Summing over iwe ﬁnd that contributions from surface elements interior to Scancel since each surface element appears in two terms with opposite signs, the outward normals in the two termsbeing equal and opposite. Only contributions from surface elements that are alsoparts of Ssurvive. If each V ii sa l l o w e dt ot e n dt oz e r ot h e nw eo b t a i nt h e divergence theorem , integraldisplay V∇·adV=contintegraldisplay Sa·dS. (11.18) We note that the divergence theorem holds for both simply and multiply con- nected surfaces, provided that they are closed and enclose some non-zero volumeV. The divergence theorem may also be extended to tensor ﬁelds (see chapter 21). The theorem ﬁnds most use as a tool in formal manipulations, but sometimes it is of value in evaluating surface integrals of the formintegraltext Sa·dSas volume integrals or vice versa. For example, setting a=rwe immediately obtain integraldisplay V∇·rdV=integraldisplay V3dV=3V=contintegraldisplay Sr·dS, which gives the expression for the volume of a region found in subsection 11.6.1. The use of the divergence theorem is further illustrated in the following example.IEvaluate the surface integral I= R Sa·dS,w h e r e a=(y−x)i+x2zj+(z+x2)kandS is the open surface of the hemisphere x2+y2+z2=a2,z≥0. We could evaluate this surface integral directly, but the algebra is somewhat lengthy. We will therefore evaluate it by use of the divergence theorem. Since the latter only holds for closed surfaces enclosing a non-zero volume V, let us ﬁrst consider the closed surface S/prime=S+S1,w h e r e S1is the circular area in the xy-plane given by x2+y2≤a2,z=0 ; S/prime then encloses a hemispherical volume V. By the divergence theorem we haveZ V∇·adV= I S/primea·dS= Z Sa·dS+ Z S1a·dS. Now∇·a=−1+0+1=0 ,s ow ec a nw r i t eZ Sa·dS=− Z S1a·dS. 407 LINE, SURFACE AND VOLUME INTEGRALS Ry C xdxdydr ˆnds Figure 11.11 A closed curve Cin the xy-plane bounding a region R. Vectors tangent and normal to the curve at a given point are also shown. The surface integral over S1is easily evaluated. Remembering that the normal to the surface points outward from the volume, a surface element on S1is simply dS=−kdx dy. OnS1we also have a=(y−x)i+x2k,s ot h a t I=− Z S1a·dS= ZZ Rx2dx dy, where Ris the circular region in the xy-plane given by x2+y2≤a2. Transforming to plane polar coordinates we have I= ZZ R/primeρ2cos2φ ρ dρ dφ = Z2π 0cos2φd φ Za 0ρ3dρ=πa4 4. J It is also interesting to consider the two-dimensional version of the divergence theorem. As an example, let us consider a two-dimensional planar region Rin thexy-plane bounded by some closed curve C(see ﬁgure 11.11). At any point on the curve the vector dr=dxi+dyjis a tangent to the curve and the vector ˆnds=dyi−dxjis a normal pointing out of the region R. If the vector ﬁeld ais continuous and diﬀerentiable in Rthen the two-dimensional divergence theorem in Cartesian coordinates gives integraldisplayintegraldisplay Rparenleftbigg∂ax ∂x+∂ay ∂yparenrightbigg dx dy =contintegraldisplay a·ˆnds=contintegraldisplay C(axdy−aydx). Letting P=−ayandQ=ax, we recover Green’s theorem in a plane, which was discussed in section 11.3. 11.8.1 Green’s theorems Consider two scalar functions φandψthat are continuous and diﬀerentiable in some volume Vbounded by a surface S. Applying the divergence theorem to the 408 11.8 DIVERGENCE THEOREM AND RELATED THEOREMS vector ﬁeld φ∇ψwe obtain contintegraldisplay Sφ∇ψ·dS=integraldisplay V∇·(φ∇ψ)dV =integraldisplay Vbracketleftbig φ∇2ψ+(∇φ)·(∇ψ)bracketrightbig dV. (11.19) Reversing the roles of φandψin (11.19) and subtracting the two equations gives contintegraldisplay S(φ∇ψ−ψ∇φ)·dS=integraldisplay V(φ∇2ψ−ψ∇2φ)dV. (11.20) Equation (11.19) is usually known as Green’s ﬁrst theorem and (11.20) as his second. Green’s second theorem is useful in the development of the Green’sfunctions used in the solution of partial diﬀerential equations (see chapter 19). 11.8.2 Other related integral theorems There exist two other integral theorems which are closely related to the divergence theorem and which are of some use in physical applications. If φis a scalar ﬁeld andbis a vector ﬁeld and both φandbsatisfy our usual diﬀerentiability conditions in some volume Vbounded by a closed surface Sthen integraldisplay V∇φd V=contintegraldisplay SφdS, (11.21) integraldisplay V∇×bdV=contintegraldisplay SdS×b. (11.22)IUse the divergence theorem to prove (11.21). In the divergence theorem (11.18) let a=φc,w h e r e cis a constant vector. We then haveZ V∇·(φc)dV= I Sφc·dS. Expanding out the integrand on the LHS we have ∇·(φc)=φ∇·c+c·∇φ=c·∇φ, sincecis constant. Also, φc·dS=c·φdS,s ow eo b t a i nZ Vc·(∇φ)dV= I Sc·φdS. Since cis constant we may take it out of both integrals to give c· Z V∇φd V=c· I SφdS, and since cis arbitrary we obtain the stated result (11.21). J Equation (11.22) may be proved in a similar way by letting a=b×cin the divergence theorem, where cis again a constant vector. 409 LINE, SURFACE AND VOLUME INTEGRALS 11.8.3 Physical applications of the divergence theorem The divergence theorem is useful in deriving many of the most important partial diﬀerential equations in physics (see chapter 18). The basic idea is to use the divergence theorem to convert an integral form, often derived from observation,into an equivalent diﬀerential form (used in theoretical statements).IFor a compressible ﬂuid with time-varying position-dependent density ρ(r,t)and velocity ﬁeld v(r,t), in which ﬂuid is neither being created nor destroyed, show that ∂ρ ∂t+∇·(ρv)=0 . For an arbitrary volume Vin the ﬂuid, the conservation of mass tells us that the rate of increase or decrease of the mass Mof ﬂuid in the volume must equal the net rate at which ﬂuid is entering or leaving the volume, i.e. dM dt=− I Sρv·dS, where Sis the surface bounding V. But the mass of ﬂuid in Vis simply M= R Vρd V,s o we have d dt Z Vρd V+ I Sρv·dS=0. Taking the derivative inside the ﬁrst integral on the RHS and using the divergence theorem to rewrite the second integral, we obtainZ V∂ρ ∂tdV+ Z V∇·(ρv)dV= Z V /∂ρ ∂t+∇·(ρv) / dV=0. Since the volume Vis arbitrary, the integrand (which is assumed continuous) must be identically zero, so we obtain ∂ρ ∂t+∇·(ρv)=0 . This is known as the continuity equation . It can also be applied to other systems, for example those in which ρis the density of electric charge or the heat content, etc. In the ﬂow of an incompressible ﬂuid ρ= constant and the continuity equation becomes simply ∇·v=0 . J In the previous example, we assumed that there were no sources or sinks in the volume V, i.e. that there was no part of Vin which ﬂuid was being created or destroyed. We now consider the case where a ﬁnite number of point sources and/or sinks are present in an incompressible ﬂuid. Let us ﬁrst consider thesimple case where a single source is located at the origin, out of which a quantityof ﬂuid ﬂows radially at a rate Q(m 3s−1). The velocity ﬁeld is given by v=Qr 4πr3=Qˆr 4πr2. Now, for a sphere S1of radius rcentred on the source, the ﬂux across S1is contintegraldisplay S1v·dS=|v|4πr2=Q. 410 11.8 DIVERGENCE THEOREM AND RELATED THEOREMS Since vhas a singularity at the origin it is not diﬀerentiable there, i.e. ∇·vis not deﬁned there, but at all other points ∇·v= 0, as required for an incompressible ﬂuid. Therefore, from the divergence theorem, for any closed surface S2that does not enclose the origin we have contintegraldisplay S2v·dS=integraldisplay V∇·vdV=0. Thus we see that the surface integralcontintegraltext Sv·dShas value Qor zero depending on whether or not Sencloses the source at the origin. In order that the divergence theorem is valid for allsurfaces S, irrespective of whether they enclose the source, we write ∇·v=Qδ(r), where δ(r) is the three-dimensional Dirac delta function. The properties of this function are discussed fully in chapter 13, but for the moment we note that it isdeﬁned in such a way that δ(r−a)=0 f o r r/negationslash=a, integraldisplay Vf(r)δ(r−a)dV=braceleftBigg f(a)i falies in V 0o t h e r w i s e for any well-behaved function f(r). Therefore, for any volume Vcontaining the source at the origin, we have integraldisplay V∇·vdV=Qintegraldisplay Vδ(r)dV=Q, which is consistent withcontintegraltext Sv·dS=Qfor a closed surface enclosing the source. Hence, by introducing the Dirac delta function the divergence theorem can bemade valid even for non-diﬀerentiable point sources. The generalisation to several sources and sinks is straightforward. For example, if a source is located at r=aand a sink at r=bthen the velocity ﬁeld is v=(r−a)Q 4π|r−a|3−(r−b)Q 4π|r−b|3 and its divergence is given by ∇·v=Qδ(r−a)−Qδ(r−b). Therefore, the integralcontintegraltext Sv·dShas the value QifSencloses the source, −Qif Sencloses the sink and 0 if Sencloses neither the source nor sink or encloses them both. This analysis also applies to other physical systems – for example, in electrostatics we can regard the sources and sinks as positive and negative pointcharges respectively and replace vby the electric ﬁeld E. 411 LINE, SURFACE AND VOLUME INTEGRALS 11.9 Stokes’ theorem and related theorems Stokes’ theorem is the ‘curl analogue’ of the divergence theorem and relates the integral of the curl of a vector ﬁeld over an open surface Sto the line integral of the vector ﬁeld around the perimeter Cbounding the surface. Following the same lines as for the derivation of the divergence theorem, we can divide the surface Sinto many small areas Siwith boundaries Ciand unit normals ˆni. Using (11.17), we have for each small area (∇×a)·ˆniSi≈contintegraldisplay Cia·dr. Summing over iwe ﬁnd that on the RHS all parts of all interior boundaries that are not part of Care included twice, being traversed in opposite directions on each occasion and thus contributing nothing. Only contributions from line elements that are also parts of Csurvive. If each Sii sa l l o w e dt ot e n dt oz e r o then we obtain Stokes’ theorem, integraldisplay S(∇×a)·dS=contintegraldisplay Ca·dr. (11.23) We note that Stokes’ theorem holds for both simply and multiply connected open surfaces, provided that they are two-sided. Stokes’ theorem may also be extendedto tensor ﬁelds (see chapter 21). Just as the divergence theorem (11.18) can be used to relate volume and surface integrals for certain types of integrand, Stokes’ theorem can be used in evaluatingsurface integrals of the formcontintegraltext S(∇×a)·dSas line integrals or vice versa.IGiven the vector ﬁeld a=yi−xj+zk, verify Stokes’ theorem for the hemispherical surface x2+y2+z2=a2,z≥0. Let us ﬁrst evaluate the surface integralZ S(∇×a)·dS over the hemisphere. It is easily shown that ∇×a=−2k, and the surface element is dS=a2sinθd θd φ ˆrin spherical polar coordinates. ThereforeZ S(∇×a)·dS= Z2π 0dφ Zπ/2 0dθ /; −2a2sinθ /ˆr·k =−2a2 Z2π 0dφ Zπ/2 0sinθ /z a / dθ =−2a2 Z2π 0dφ Zπ/2 0sinθcosθd θ=−2πa2. We now evaluate the line integral around the perimeter curve Cof the surface, which 412 11.9 STOKES’ THEOREM AND RELATED THEOREMS is the circle x2+y2=a2in the xy-plane. This is given byI Ca·dr= I C(yi−xj+zk)·(dxi+dyj+dzk) = I C(yd x−xd y). Using plane polar coordinates, on Cwe have x=acosφ,y=asinφso that dx= −asinφd φ,dy=acosφd φ, and the line integral becomesI C(yd x−xd y)=−a2 Z2π 0(sin2φ+c o s2φ)dφ=−a2 Z2π 0dφ=−2πa2. Since the surface and line integrals have the s ame value, we have veriﬁed Stokes’ theorem in this case. J The two-dimensional version of Stokes’ theorem also yields Green’s theorem in a plane. Consider the region Rin the xy-plane shown in ﬁgure 11.11, in which a vector ﬁeld ais deﬁned. Since a=axi+ayj, we have∇×a=(∂ay/∂x−∂ax/∂y)k, and Stokes’ theorem becomes integraldisplayintegraldisplay Rparenleftbigg∂ay ∂x−∂ax ∂yparenrightbigg dx dy =contintegraldisplay C(axdx+aydy). Letting P=axandQ=aywe recover Green’s theorem in a plane, (11.4). 11.9.1 Related integral theorems As for the divergence theorem, there exist two other integral theorems that are closely related to Stokes’ theorem. If φis a scalar ﬁeld and bis a vector ﬁeld, and both φandbsatisfy our usual diﬀerentiability conditions on some two-sided open surface Sbounded by a closed perimeter curve C,t h e n integraldisplay SdS×∇φ=contintegraldisplay Cφdr, (11.24) integraldisplay S(dS×∇)×b=contintegraldisplay Cdr×b. (11.25)IUse Stokes’ theorem to prove (11.24). In Stokes’ theorem, (11.23), let a=φc,w h e r e cis a constant vector. We then haveZ S[∇×(φc)]·dS= I Cφc·dr. (11.26) Expanding out the integrand on the LHS we have ∇×(φc)=∇φ×c+φ∇×c=∇φ×c, sincecis constant, and the triple scalar product on the LHS of (11.26) can therefore be written [∇×(φc)]·dS=(∇φ×c)·dS=c·(dS×∇φ). 413 LINE, SURFACE AND VOLUME INTEGRALS Substituting this into (11.26) and taking cout of both integrals because it is constant, we ﬁnd c· Z SdS×∇φ=c· I Cφdr. Since cis an arbitrary constant vector we therefore obtain the stated result (11.24). J Equation (11.25) may be proved in a similar way, by letting a=b×cin Stokes’ theorem, where cis again a constant vector. We also note that by setting b=r in (11.25) we ﬁndintegraldisplay S(dS×∇)×r=contintegraldisplay Cdr×r. Expanding out the integrand on the LHS we ﬁnd (dS×∇)×r=dS−dS(∇·r)=dS−3dS=−2dS. Therefore, as we found in subsection 11.5.2, the vector area of an open surface S is given by S=integraldisplay SdS=1 2contintegraldisplay Cr×dr. 11.9.2 Physical applications of Stokes’ theorem Like the divergence theorem, Stokes’ theorem is useful in converting integral equations into diﬀerential equations.IFrom Amp `ere’s law derive Maxwell’s equation in the case where the currents are steady, i.e.∇×B−µ0J=0. Amp`ere’s rule for a distributed current with current density JisI CB·dr=µ0 Z SJ·dS, for any circuit Cbounding a surface S. Using Stokes’ theorem, the LHS can be transformed into R S(∇×B)·dS; henceZ S(∇×B−µ0J)·dS=0 foranysurface S. This can only be so if ∇×B−µ0J=0, which is the required relation. Similarly, from Faraday’s law of electromagnetic induction we can derive Maxwell’sequation∇×E=−∂B/∂t.J In subsection 11.8.3 we discussed the ﬂow of an incompressible ﬂuid in the presence of several sources and sinks. Let us now consider vortex ﬂow in an incompressible ﬂuid with a velocity ﬁeld v=1 ρˆeφ, in cylindrical polar coordinates ρ, φ, z . For this velocity ﬁeld ∇×vequals zero 414 11.10 EXERCISES everywhere except on the axis ρ=0 ,w h e r e vhas a singularity. Thereforecontintegraltext Cv·dr equals zero for any path Cthat does not enclose the vortex line on the axis and 2πifCdoes enclose the axis. In order for Stokes’ theorem to be valid for all paths C, we therefore set ∇×v=2πδ(ρ), where δ(ρ) is the Dirac delta function, to be discussed in subsection 13.1.3. Now, since∇×v=0, except on the axis ρ= 0, there exists a scalar potential ψsuch thatv=∇ψ. It may easily be shown that ψ=φ, the polar angle. Therefore, if C does not enclose the axis thencontintegraldisplay Cv·dr=contintegraldisplay dφ=0, and if Cdoes enclose the axis, contintegraldisplay Cv·dr=∆φ=2πn, where nis the number of times we traverse C. Thus φis a multivalued potential. A similar analysis is valid for other physical systems – for example, in magneto- statics we may replace the vortex lines by current-carrying wires and the velocityﬁeldvby the magnetic ﬁeld B. 11.10 Exercises 11.1 The vector ﬁeld Fis deﬁned by F=2xzi+2yz2j+(x2+2y2z−1)k. Calculate∇×Fand deduce that Fcan be written F=∇φ.Determine the form ofφ. 11.2 The vector ﬁeld Qis deﬁned by Q= / 3x2(y+z)+y3+z3 / i+ / 3y2(z+x)+z3+x3 / j+ / 3z2(x+y)+x3+y3 / k. Show that Qis a conservative ﬁeld, construct its potential function and hence evaluate the integral J= R Q·dralong any line connecting the point Aat (1,−1,1) to Bat (2,1,2). 11.3 Fis a vector ﬁeld xy2i+2j+xk,a n d Lis a path parameterised by x=ct,y=c/t, z=dfor the range 1 ≤t≤2. Evaluate (a) R LFdt,( b ) R LFdyand (c) R LF·dr. 11.4 By making an appropriate choice for the functions P(x, y)a n d Q(x, y) that appear in Green’s theorem in a plane, show that the integral of x−yover the upper half of the unit circle centred on the origin has the value −2 3. Show the same result by direct integration in Cartesian coordinates. 11.5 Determine the point of intersection P, in the ﬁrst quadrant, of the two ellipses x2 a2+y2 b2=1 a n dx2 b2+y2 a2=1. Taking b<a, consider the contour Lthat bounds that area in the ﬁrst quadrant which is common to the two ellipses. Show that the parts of Lthat lie along the coordinate axes contribute nothing to the line integral around Lofxd y−yd x, and that this line integral can be written as the sum of two such integrals, I1 415 LINE, SURFACE AND VOLUME INTEGRALS andI2, around closed contours. Using a parameterisation of each ellipse similar to that employed in the example in section 11.3, evaluate these two integrals andhence ﬁnd the total area common to the two ellipses. 11.6 By using parameterisations of the form x=acos nθandy=asinnθfor suitable values of n, ﬁnd the area bounded by the curves x2/5+y2/5=a2/5and x2/3+y2/3=a2/3 . 11.7 Evaluate the line integral I= I C / y(4x2+y2)dx+x(2x2+3y2)dy / around the ellipse x2/a2+y2/b2=1 . 11.8 Criticise the following ‘proof’ that π=0 . (a) Apply Green’s theorem in a plane to the functions P(x, y)=t a n−1(y/x)a n d Q(x, y)=t a n−1(x/y), taking the region Rto be the unit circle centred on the origin. (b) The RHS of the equality so produced isZ Z Ry−x x2+y2dx dy which, either by symmetry considerations or by changing to plane polar coordinates, can be shown to have zero value. (c) In the LHS of the equality set x=c o s θandy=s i n θ, yielding P(θ)=θ andQ(θ)=π/2−θ. The line integral becomesZ2 0π h/π 2−θ / cosθ−θsinθ i dθ, which has value 2 π. (d) Thus 2 π= 0 and the stated result follows. 11.9 A single-turn coil Cof arbitrary shape is placed in a magnetic ﬁeld Band carries a current I. Show that the couple acting upon the coil can be written as M=I Z C(B·r)dr−I Z CB(r·dr). For a planar rectangular coil of sides 2 aand 2 bplaced with its plane vertical and at an angle φto a uniform horizontal ﬁeld B, show that Mis, as expected, 4abBIcosφk. 11.10 Find the vector area Sof the curved surface of the hyperboloid of revolution x2 a2−y2+z2 b2=1 which lies in the region z≥0a n d a≤x≤λa. 11.11 An axially symmetric solid body with its axis ABvertical is immersed in an incompressible ﬂuid of density ρ0. Use the following method to show that, whatever the shape of the body, for ρ=ρ(z) in cylindrical polars the Archimedean upthrust is, as expected, ρ0gV,w h e r e Vis the volume of the body. Express the vertical component of the resultant force ( − R pdS,w h e r e pis the pressure) on the body in terms of an integral; note that p=−ρ0gzand that for an annular surface element of width dl,n·nzdl=−dr. Integrate by parts and use the fact that ρ(zA)=ρ(zB)=0 . 416 11.10 EXERCISES 11.12 Show that the expression below is equal to the solid angle subtended by a rectangular aperture of sides 2 aand 2 bat a point a distance cfrom the aperture along the normal to its centre: Ω=4 Zb 0ac (y2+c2)(y2+c2+a2)1/2dy. By setting y=(a2+c2)1/2tanφ, change this integral into the formZφ1 04accosφ c2+a2sin2φdφ, where tan φ1=b/(a2+c2)1/2, and hence show that Ω=4t a n−1 /ab c(a2+b2+c2)1/2 / . 11.13 A vector ﬁeld ais given by−zxr−3i−zyr−3j+(x2+y2)r−3k,w h e r e r2=x2+y2+z2. Establish that the ﬁeld is conservative (a) by showing ∇×a=0and (b) by constructing its potential function φ. 11.14 A vector ﬁeld ais given by ( z2+2xy)i+(x2+2yz)j+(y2+2zx)k. Show that ais conservative and that the line integral R a·dralong any line joining (1 ,1,1) and (1 ,2,2) has the value 11. 11.15 A force F(r) acts on a particle at r. In which of the following cases can Fbe represented in terms of a potential? Where it can, ﬁnd the potential. (a)F=F0 / i−j−2(x−y) a2r / exp / −r2 a2 / ; (b)F=F0 a h zk+(x2+y2−a2) a2r i exp / −r2 a2 / ; (c)F=F0 / k+a(r×k) r2 / . 11.16 One of Maxwell’s electromagnetic equations states that all magnetic ﬁelds B are solenoidal (i.e. ∇·B= 0). Determine whether each of the following vectors could represent a real magnetic ﬁeld; where it could, try to ﬁnd a suitable vectorpotential A, i.e. such that B=∇×A. (Hint: seek a vector potential that is parallel to∇×B.): (a)B 0b r3[(x−y)zi+(x−y)zj+x2−y2k] in Cartesians with r2=x2+y2+z2; (b)B0b r3[cosθcosφˆer−sinθcosφˆeθ+s i n2 θsinφˆeφ] in spherical polars; (c)B0b2 /zr (b2+z2)2ˆeρ+1 b2+z2ˆez / in cylindrical polars. 11.17 The vector ﬁeld fhas components yi−xj+kandγis a curve given parametrically by r=(a−c+ccosθ)i+(b+csinθ)j+c2θk,0≤θ≤2π. Describe the shape of the path γand show that the line integral R γf·drvanishes. Does this result imply that fis a conservative ﬁeld? 11.18 A vector ﬁeld a=f(r)ris spherically symmetric and everywhere directed away from the origin. Show that ais irrotational but that it is also solenoidal only if f(r)i so ft h ef o r m Ar−3. 417 LINE, SURFACE AND VOLUME INTEGRALS 11.19 Evaluate the surface integral R r·dS,w h e r e ris the position vector, over that part of the surface z=a2−x2−y2for which z≥0, by each of the following methods: (a) parameterize the surface as x=asinθcosφ,y=asinθsinφ,z=a2cos2θ, and show that r·dS=a4(2 sin3θcosθ+c o s3θsinθ)dθ dφ. (b) apply the divergence theorem to the volume bounded by the surface and the plane z=0 . 11.20 Obtain an expression for the value φPat a point Pof a scalar function φthat satisﬁes∇2φ= 0 in terms of its value and normal derivative on a surface Sthat encloses it, by proceeding as follows. (a) In Green’s second theorem take ψat any particular point Qas 1/r,w h e r e r is the distance of Qfrom P. Show that ∇2ψ= 0 except at r=0 . (b) Apply the result to the doubly connected region bounded by Sand a small sphere Σ of radius δcentred on P. (c) Apply the divergence theorem to show that the surface integral over Σ involving 1 /δvanishes, and prove that the term involving 1 /δ2has the value 4πφP. (d) Conclude that φP=−1 4π Z Sφ∂ ∂n /1 r / dS+1 4π Z S1 r∂φ ∂ndS. This important result shows that the value at a point Pof a function φ that satisﬁes ∇2φ= 0 everywhere within a closed surface Sthat encloses P may be expressed entirely in terms of its value and normal derivative on S. This matter is taken up more generally in connection with Green’s functionsin chapter 19 and in connection with functions of a complex variable insection 20.12. 11.21 Use result (11.21), together with an appropriately chosen scalar function φto prove that the position vector ¯rof the centre of mass of an arbitrarily-shaped body of volume Vand uniform density can be written ¯r=1 V I S1 2r2dS . 11.22 A rigid body of volume Vand surface Srotates with angular velocity ω. Show that ω=−1 2V I Su×dS, where u(x) is the velocity of the point xon the surface S. 11.23 Demonstrate the validity of the divergence theorem: (a) by calculating the ﬂux of the vector F=αr (r2+a2)3/2 through the spherical surface |r|=√ 3a; (b) by showing that ∇·F=3αa2 (r2+a2)5/2 418 11.10 EXERCISES and evaluating the volume integral of ∇·Fover the interior of the sphere |r|=√ 3a. (The substitution r=atanθwill prove useful in carrying out the integration.) 11.24 Prove equation (11.22) and, by taking b=zx2i+zy2j+(x2−y2)k, show that the two integrals I= Z x2dVand J= Z cos2θsin3θcos 2φd θd φ , both taken over the unit sphere, must have the same value. Evaluate both directly to show that the common value is 4 π/15. 11.25 In a uniform, non-dielectric, conducting medium with unit relative permittivity, charge density ρ, current density J, electric ﬁeld Eand magnetic ﬁeld B, Maxwell’s electromagnetic equations take the form (with µ0/epsilon10=c−2) (i)∇·B= 0, (ii) ∇·E=ρ//epsilon10, (iii)∇×E+˙B=0,( i v )∇×B−(˙E/c2)=µ0J, The density of stored energy in the medium is given by1 2(/epsilon10E2+µ−1 0B2). Show that the rate of change of the total stored energy in a volume Vis equal to − Z VJ·EdV−1 µ0 I S(E×B)·dS, where Sis the surface bounding V. (The ﬁrst integral gives the ohmic heating loss, whilst the second gives the electromagnetic energy ﬂux out of the boundingsurface. The vector µ −1 0(E×B) is known as the Poynting vector.) 11.26 A vector ﬁeld Fis deﬁned in cylindrical polar coordinates ρ, θ, z by F=F0 /xcosλz ai+ycosλz aj+( s i n λz)k / ≡ρ a(cosλz)eρ+( s i n λz)k, where i,jandkare the unit vectors along the Cartesian axes and eρis the unit vector ( x/ρ)i+(y/ρ)j. (a) Calculate, as a surface integral, the ﬂux of Fthrough the closed surface bounded by the cylinders ρ=aandρ=2aand the planes z=±aπ/2. (b) Evaluate the same integral using the divergence theorem. 11.27 The vector ﬁeld Fis given by F=( 3x2yz+y3z+xe−x)i+( 3xy2z+x3z+yex)j+(x3y+y3x+xy2z2)k. Calculate (a) directly and (b) by using Stokes’ theorem the value of the line integral R LF·dr,w h e r e Lis the (three-dimensional) closed contour OABCDEO deﬁned by the successive vertices (0 ,0,0), (1 ,0,0), (1 ,0,1), (1 ,1,1), (1 ,1,0), (0 ,1,0), (0,0,0). 11.28 A vector force ﬁeld Fis deﬁned in Cartesian coordinates by F=F0 //y3 3a3+y aexy/a2+1 / i+ /xy2 a3+x+y aexy/a2 / j+z aexy/a2k / . Use Stokes’ theorem to calculateI LF·dr, where Lis the perimeter of the rectangle ABCD given by A=( 0,1,0),B=( 1,1,0), C=( 1,3,0) and D=( 0,3,0). 419 LINE, SURFACE AND VOLUME INTEGRALS 11.11 Hints and answers 11.1 Show that ∇×F=0. The potential φF(r)=x2z+y2z2−z. 11.2 Show that one component of ∇×Qis zero and apply symmetry. The potential φQ(r)=xy(x2+y2)+yz(y2+z2)+zx(z2+x2);J=φQ(B)−φQ(A) = 54. 11.3 (a) c3ln 2i+2j+( 3c/2)k;( b )(−3c4/8)i−cj−(c2ln 2)k;( c ) c4ln 2−c. 11.4 Take P=y2andQ=x2. Show that the line integral along the x-axis from (−1,0) to (1 ,0) contributes nothing. 11.5 For P,x=y=ab/(a2+b2)1/2. Note that the integral along the straight line joining Pto the origin is traversed in opposite directions in I1and I2.T h e relevant limits are 0 ≤θ1≤tan−1(b/a)a n dt a n−1(a/b)≤θ2≤π/2. As required by symmetry, I1=I2; the total common area is 4 abtan−1(b/a). 11.6 Use the result of the worked example in section 11.3 and the reduction formulae derived in exercise 2.42. Bounded area = 33 πa2/128. 11.7 Show that, in the notation of section 11.3, ∂Q/∂x−∂P/∂y =2x2;I=πa3b/2. 11.8 The conditions for Green’s theorem are not met as PandQare not continuous (or diﬀerentiable) at the origin. 11.9 M=I R Cr×(dr×B). 11.10 Since the vector area of a closed surface vanishes, S=−S1i+S2kwhere S1is the area of the semicircular intersection with the plane x=λaandS2is the area of the hyperbolic intersection with the plane z=0 ; S1=1 2πb2(λ2−1); S2=ab[λ√(λ2−1)−cosh−1λ]. 11.13 (b) φ=c+z/r. 11.14 The appropriate potential function is f(x, y, z)=z2x+x2y+y2z. 11.15 (a) Yes, F0(x−y)exp(−r2/a2); (b) yes,−F0[(x2+y2)/2a]exp(−r2/a2); (c) no,∇×F/negationslash=0. 11.16 Only (c) has zero divergence. A possible vector potential is1 2B0b2ρ(b2+z2)−1ˆeφ; to this could be added the gradient of anyscalar function. 11.17 A spiral of radius cwith its axis parallel to the z-direction and passing through (a, b). The pitch of the spiral is 2 πc2. No, because (i) γis not a closed loop and (ii) the line integral must be zero for every closed loop, not just for a particular one. In fact ∇×f=−2k/negationslash=0shows that fis not conservative. 11.18∇×a=0;∇·a=3f(r)+rf/prime(r)=0i f f(r)=Ar−3. 11.19 (a) dS=( 2a3cosθsin2θcosφi+2a3cosθsin2θsinφj+a2cosθsinθk)dθ dφ. (b)∇·r= 3; over the plane z=0 ,r·dS= 0; The necessarily common value is 3πa4/2. 11.20 (d) Remember that the outward normal to the region is the inward normal to Σ.11.21 Write ras∇( 1 2r2). 11.22 Use result (11.22) and the expression for ∇×(a×b) and note that ( ω·∇)x=ω. 11.23 T heansweris 3√ 3πα/2i ne a c hc a s e . 11.24 Follow the method indicated in subsection 11.8.2, using an identity given in table 10.1. Use Cartesian coordinates for the LHS of equation (11.22) and sphericalpolars for the RHS. Employ (anti)symmetry and periodicity arguments to setseveral integrals to zero without explicit calculation. 11.25 Identify the expression for ∇·(E×B) and use the divergence theorem. 11.26 6 πF 0(a2+2a/λ)sin(λaπ/2). 11.27 (a) The successive contributions to the integral are 1 ,0,2+1 2e,−7 3,−1,−1 2. (b)∇×F=2xyz2i−y2z2j+yexk. Show that the contour is equivalent to the sum of two plane square contours in the planes z=0a n d x= 1, the latter being traversed in the negative sense. Integral =1 6(3e−5). 11.28 Ra 0dx R3a ady F 0(y/a)2exy/a2=F0a(2e3−4). 420 12 Fourier series We have already discussed, in chapter 4, how complicated functions may be expressed as power series. However, this is not the only way in which a functionmay be represented as a series, and the subject of this chapter is the expressionof functions as a sum of sine and cosine terms. Such a representation is called aFourier series . Unlike Taylor series, a Fourier series can describe functions that are not everywhere continuous and/or diﬀerent iable. There are also other advantages in using trigonometrical terms. They are easy to diﬀerentiate and integrate, their moduli are easily taken and each term contains only one characteristic frequency.This last point is important because, as we shall see later, Fourier series are oftenused to represent the response of a system to a periodic input, and this responseoften depends directly on the frequency content of the input. Fourier series areused in a wide variety of such physical situations, including the vibrations of aﬁnite string, the scattering of light by a diﬀraction grating and the transmission of an input signal by an electronic circuit. 12.1 The Dirichlet conditions We have already mentioned that Fourier series may be used to represent some functions for which a Taylor series expansion is not possible. The particularconditions that a function f(x) must fulﬁl in order that it may be expanded as a Fourier series are known as the Dirichlet conditions , and may be summarised by the following four points: (i) the function must be periodic; (ii) it must be single-valued and continuous, except possibly at a ﬁnite number of ﬁnite discontinuities; (iii) it must have only a ﬁnite number of maxima and minima within one period; (iv) the integral over one period of |f(x)|must converge. 421 FOURIER SERIES L Lf(x) x Figure 12.1 An example of a function that may be represented as a Fourier series without modiﬁcation. If the above conditions are satisﬁed then the Fourier series converges to f(x) at all points where f(x) is continuous. The convergence of the Fourier series at points of discontinuity is discussed in section 12.4. The last three Dirichletconditions are almost always met in real applications, but not all functions areperiodic and hence do not fulﬁl the ﬁrst condition. It may be possible, however,to represent a non-periodic function as a Fourier series by manipulation of the function into a periodic form. This is discussed in section 12.5. An example of a function that may, without modiﬁcation, be represented as a Fourier series isshown in ﬁgure 12.1. We have stated without proof that any function that satisﬁes the Dirichlet conditions may be represented as a Fourier series. Let us now show why this is a plausible statement. We require that any reasonable function (one that satisﬁesthe Dirichlet conditions) can be expressed as a linear sum of sine and cosineterms. We ﬁrst note that we cannot use just a sum of sine terms since sine, beingan odd function (i.e. a function for which f(−x)=−f(x)), cannot represent even functions (i.e. functions for which f(−x)=f(x)). This is obvious when we try to express a function f(x) that takes a non-zero value at x= 0. Clearly, since sinnx= 0 for all values of n, we cannot represent f(x)a tx= 0 by a sine series. Similarly odd functions cannot be represented by a cosine series since cosine isan even function. Nevertheless, it is possible to represent allodd functions by a sine series and alleven functions by a cosine series. Now, since all functions may b ew r i t t e na st h es u mo fa no d da n da ne v e np a r t , f(x)= 1 2[f(x)+f(−x)] +1 2[f(x)−f(−x)] =feven(x)+fodd(x), 422 12.2 THE FOURIER COEFFICIENTS we can write any function as the sum of a sine series and a cosine series. All the terms of a Fourier series are mutually orthogonal, that is, the integrals, over one period, of the product of any two terms have the following properties: integraldisplayx0+L x0sinparenleftbigg2πrx Lparenrightbigg cosparenleftbigg2πpx Lparenrightbigg dx=0 f o ra l l randp, (12.1) integraldisplayx0+L x0cosparenleftbigg2πrx Lparenrightbigg cosparenleftbigg2πpx Lparenrightbigg dx=  Lforr=p=0, 1 2Lforr=p>0, 0f o r r/negationslash=p,(12.2) integraldisplayx0+L x0sinparenleftbigg2πrx Lparenrightbigg sinparenleftbigg2πpx Lparenrightbigg dx=  0f o r r=p=0, 1 2Lforr=p>0, 0f o r r/negationslash=p,(12.3) where randpare integers greater than or equal to zero; these formulae are easily derived. A full discussion of why it is possible to expand a function as a sum ofmutually orthogonal functions is given in chapter 17. The Fourier series expansion of the function f(x) is conventionally written f(x)=a 0 2+∞summationdisplay r=1bracketleftbigg arcosparenleftbigg2πrx Lparenrightbigg +brsinparenleftbigg2πrx Lparenrightbiggbracketrightbigg , (12.4) where a0,ar,brare constants called the Fourier coeﬃcients . These coeﬃcients are analogous to those in a power series expansion and the determination of their numerical values is the essential step in writing a function as a Fourier series. This chapter continues with a discussion of how to ﬁnd the Fourier coeﬃcients for particular functions. We then discuss simpliﬁcations to the general Fourier series that may save considerable eﬀort in calculations. This is followed by the alternative representation of a function as a complex Fourier series, and weconclude with a discussion of Parseval’s theorem. 12.2 The Fourier coeﬃcients We have indicated that a series that satisﬁes the Dirichlet conditions may be written in the form (12.4). We now consider how to ﬁnd the Fourier coeﬃcientsfor any particular function. For a periodic function f(x)o fp e r i o d Lwe will ﬁnd that the Fourier coeﬃcients are given by a r=2 Lintegraldisplayx0+L x0f(x)cosparenleftbigg2πrx Lparenrightbigg dx, (12.5) br=2 Lintegraldisplayx0+L x0f(x)s i nparenleftbigg2πrx Lparenrightbigg dx, (12.6) where x0is arbitrary but is often taken as 0 or −L/2. The apparently arbitrary factor1 2which appears in the a0term in (12.4) is included so that (12.5) may 423 FOURIER SERIES apply for r= 0 as well as r>0. The relations (12.5) and (12.6) may be derived as follows. Suppose the Fourier series expansion of f(x) can be written as in (12.4), f(x)=a0 2+∞summationdisplay r=1bracketleftbigg arcosparenleftbigg2πrx Lparenrightbigg +brsinparenleftbigg2πrx Lparenrightbiggbracketrightbigg . Then, multiplying by cos(2 πpx/L ), integrating over one full period in xand changing the order of the summation and integration, we get integraldisplayx0+L x0f(x)cosparenleftbigg2πpx Lparenrightbigg dx=a0 2integraldisplayx0+L x0cosparenleftbigg2πpx Lparenrightbigg dx +∞summationdisplay r=1arintegraldisplayx0+L x0cosparenleftbigg2πrx Lparenrightbigg cosparenleftbigg2πpx Lparenrightbigg dx +∞summationdisplay r=1brintegraldisplayx0+L x0sinparenleftbigg2πrx Lparenrightbigg cosparenleftbigg2πpx Lparenrightbigg dx. (12.7) We can now ﬁnd the Fourier coeﬃcients by considering (12.7) as ptakes diﬀerent values. Using the orthogonality conditions (12.1)–(12.3) of the previous section,we ﬁnd that when p= 0 (12.7) becomes integraldisplay x0+L x0f(x)dx=a0 2L. When p/negationslash= 0 the only non-vanishing term on the RHS of (12.7) occurs when r=p,a n ds o integraldisplayx0+L x0f(x)cosparenleftbigg2πrx Lparenrightbigg dx=ar 2L. The other Fourier coeﬃcients brmay be found by repeating the above process but multiplying by sin(2 πpx/L ) instead of cos(2 πpx/L ) (see exercise 12.2).IExpress the square-wave function illustrated in ﬁgure 12.2 as a Fourier series. Physically this might represent the input to a el ectrical circuit that sw itches between a high and a low state with time period T. The square wave may be represented by f(t)= /( −1f o r−1 2T≤t<0, +1 for 0≤t<1 2T. In deriving the Fourier coeﬃcients, we note ﬁrstly that the function is an odd function and so the series will contain only sine terms (this simpliﬁcation is discussed further in the 424 12.3 SYMMETRY CONSIDERATIONS 1 −10T 2−T 2 tf(t) Figure 12.2 A square-wave function. following section). To evaluate the coeﬃcients in the sine series we use (12.6). Hence br=2 T ZT/2 −T/2f(t)sin /2πrt T / dt =4 T ZT/2 0sin /2πrt T / dt =2 πr[1−(−1)r]. Thus the sine coeﬃcients are zero if ris even and equal to 4 /(πr)i fris odd. Hence the Fourier series for the square- wave function may be written as f(t)=4 π / sinωt+sin3ωt 3+sin5ωt 5+··· / , (12.8) where ω=2π/Tis called the angular frequency . J 12.3 Symmetry considerations The example in the previous section employed the useful property that since the function to be represented was odd, all the cosine terms of the Fourier series were zero. It is often the case that the function we wish to express as a Fourier serieshas a particular symmetry, which we can exploit to reduce the calculational labourof evaluating Fourier coeﬃcients. Functions that are symmetric or antisymmetricabout the origin (i.e. even and odd functions respectively) admit particularlyuseful simpliﬁcations. Functions that are odd in xhave no cosine terms (see section 12.1) and all the a-coeﬃcients are equal to zero. Similarly, functions that are even in xhave no sine terms and all the b-coeﬃcients are zero. Since the Fourier series of odd or even functions contain only half the coeﬃcients requiredfor a general periodic function, there is a considerable reduction in the algebraneeded to ﬁnd a Fourier series. The consequences of symmetry or antisymmetry of the function about the quarter period (i.e. about L/4) are a little less obvious. Furthermore, the results 425 FOURIER SERIES are not used as often as those above and the remainder of this section can be omitted on a ﬁrst reading without loss of continuity. The following argumentgives the required results. Suppose that f(x) has even or odd symmetry about L/4, i.e. f(L/4−x)= ±f(x−L/4). For convenience, we make the substitution s=x−L/4 and hence f(−s)=±f(s). We can now see that b r=2 Lintegraldisplayx0+L x0f(s)s i nparenleftbigg2πrs L+πr 2parenrightbigg ds, where the limits of integration have been left unaltered since fis, of course, periodic in sas well as in x. If we use the expansion sinparenleftbigg2πrs L+πr 2parenrightbigg =s i nparenleftbigg2πrs Lparenrightbigg cosparenleftBigπr 2parenrightBig +c o sparenleftbigg2πrs Lparenrightbigg sinparenleftBigπr 2parenrightBig , we can immediately see that the trigonometrical part of the integrand is an odd function of sifris even and an even function of sifris odd. Hence if f(s)i s even and ris even then the integral is zero, and if f(s) is odd and ris odd then the integral is zero. Similar results can be derived for the Fourier a-coeﬃcients and we conclude that (i) if f(x) is even about L/4t h e n a2r+1=0a n d b2r=0 , (ii) if f(x) is odd about L/4t h e n a2r=0a n d b2r+1=0 . All the above results follow automatically when the Fourier coeﬃcients are evaluated in any particular case, but prior knowledge of them will often enable some coeﬃcients to be set equal to zero on inspection and so substantially reducethe computational labour. As an example, the square-wave function shown inﬁgure 12.2 is (i) an odd function of t,s ot h a ta l l a r= 0, and (ii) even about the point t=T/4, so that b2r= 0. Thus we can say immediately that only sine terms of odd harmonics will be present and therefore will need to be calculated; this isconﬁrmed in the expansion (12.8). 12.4 Discontinuous functions The Fourier series expansion usually works well for functions that are discon- tinuous in the required range. However, the series itself does not produce adiscontinuous function and we state without proof that the value of the ex- panded f(x) at a discontinuity will be half-way between the upper and lower values. Expressing this more mathematically, at a point of ﬁnite discontinuity, x d, the Fourier series converges to 1 2lim /epsilon1→0[f(xd+/epsilon1)+f(xd−/epsilon1)]. At a discontinuity, the Fourier series representation of the function will overshoot its value. Although as more terms are included the overshoot moves in position 426 12.4 DISCONTINUOUS FUNCTIONS (a)( b) (c)( d)−1 −1−1 −11 11 1−T 2 −T 2−T 2 −T 2T 2 T 2T 2 T 2δ Figure 12.3 The convergence of a Fourier series expansion of a square-wave function, including ( a)o n et e r m ,( b) two terms, ( c) three terms and ( d)2 0 terms. The overshoot δis shown in ( d). arbitrarily close to the discontinuity, it never disappears even in the limit of an inﬁnite number of terms. This behaviour is known as Gibbs’ phenomenon .Af u l l discussion is not pursued here but suﬃce it to say that the size of the overshoot is proportional to the magnitude of the discontinuity.IFind the value to which the Fourier series of the square-wave function discussed in sec- tion 12.2 converges at t=0. It can be seen that the function is discontinuous at t= 0 and, by the above rule, we expect the series to converge to a value half-way between the upper and lower values, in otherwords to converge to zero in this case. Considering the Fourier series of this function,(12.8), we see that all the terms are zero and hence the Fourier series converges to zero asexpected. The Gibbs phenomenon for the square-wave function is shown in ﬁgure 12.3.J 427 FOURIER SERIES (a) (b) (c) (d) 0000 LLLL 2L2L2L Figure 12.4 Possible periodic extensions of a function.12.5 Non-periodic functions We have already mentioned that a Fourier representation may sometimes be used for non-periodic functions. If we wish to ﬁnd the Fourier series of a non-periodicfunction only within a ﬁxed range then we may continue this function outside the range so as to make it periodic. The Fourier series of this periodic function wouldthen correctly represent the non-periodic function in the desired range. Since we are often at liberty to extend the function in a number of ways, we can sometimes make it odd or even and so reduce the calculation required. Figure 12.4( b)s h o w s the simplest extension to the function shown in ﬁgure 12.4( a). However, this extension has no particular symmetry. Figures 12.4( c), (d) show extensions as odd and even functions respectively with the beneﬁt that only sine or cosine termsappear in the resulting Fourier series. We note that these last two extensions givea function of period 2 L. In view of the result of section 12.4, it must be added that the continuation must not be discontinuous at the end-points of the interval of interest; if it isthe series will not converge to the required value there. This requirement thatthe series converges appropriately may reduce the choice of continuations. Thisis discussed further at the end of the following example.IFind the Fourier series of f(x)=x2for0<x≤2. We must ﬁrst make the function periodic. We do this by extending the range of interest to −2<x≤2i ns u c haw a yt h a t f(x)=f(−x) and then letting f(x+4k)=f(x), where kis any integer. This is shown in ﬁgure 12.5. Now we have an even function of period 4. TheFourier series will faithfully represent f(x) in the range, −2<x≤2, although not outside it. Firstly we note that since we have made the speciﬁed function even in xby extending 428 12.5 NON-PERIODIC FUNCTIONS −22 0 Lxf(x)=x2 Figure 12.5 f(x)=x2,0<x≤2, with the range extended to give periodicity. the range, all the coeﬃcients brwill be zero. Now we apply (12.5) and (12.6) with L=4 to determine the remaining coeﬃcients: ar=2 4 Z2 −2x2cos /2πrx 4 / dx=4 4 Z2 0x2cos /πrx 2 / dx, where the second equality holds because the function is even in x. Thus ar= /2 πrx2sin /πrx 2 / /2 0−4 πr Z2 0xsin /πrx 2 / dx =8 π2r2 h xcos /πrx 2 /i2 0−8 π2r2 Z2 0cos /πrx 2 / dx =16 π2r2cosπr =16 π2r2(−1)r. Since this expression for arhasr2in its denominator, to evaluate a0we must return to the original deﬁnition, ar=2 4 Z2 −2f(x)cos /πrx 2 / dx. From this we obtain a0=2 4 Z2 −2x2dx=4 4 Z2 0x2dx=8 3. The ﬁnal expression for f(x)i st h e n x2=4 3+1 6∞X r=1(−1)r π2r2cos /πrx 2 / for 0 <x≤2. J We note that in the above example we could have extended the range so as to make the function odd. In other words we could have set f(x)=−f(−x)a n d then made f(x) periodic in such a way that f(x+4 )= f(x). In this case the resulting Fourier series would be a series of just sine terms. However, althoughthis will faithfully represent the function inside the required range, it does not 429 FOURIER SERIES converge to the correct values of f(x)=±4a tx=±2; it converges, instead, to zero, the average of the values at the two ends of the range. 12.6 Integration and diﬀerentiation It is sometimes possible to ﬁnd the Fourier series of a function by integration or diﬀerentiation of another Fourier series. If the Fourier series of f(x)i si n t e g r a t e d term by term then the resulting Fourier series converges to the integral of f(x). Clearly, when integrating in such a way there is a constant of integration that must be found. If f(x) is a continuous function of xfor all xandf(x) is also periodic then the Fourier series that results from diﬀerentiating term by term converges tof /prime(x), provided that f/prime(x) itself satisﬁes the Dirichlet conditions. These properties of Fourier series may be useful in calculating complicated Fourier series, sincesimple Fourier series may easily be evaluated (or found from standard tables)and often the more complicated series can then be built up by integration and/ordiﬀerentiation.IFind the Fourier series of f(x)=x3for0<x≤2. In the example discussed in the previous section we found the Fourier series for f(x)=x2 in the required range. So, if we integrate this term by term, we obtain x3 3=4 3x+3 2∞X r=1(−1)r π3r3sin /πrx 2 / +c, where cis, so far, an arbitrary constant. We have not yet found the Fourier series for x3 because the term4 3xappears in the expansion. However, by now diﬀerentiating the same initial expression for x2we obtain 2x=−8∞X r=1(−1)r πrsin /πrx 2 / . We can now write the full Fourier expansion of x3as x3=−16∞X r=1(−1)r πrsin /πrx 2 / +9 6∞X r=1(−1)r π3r3sin /πrx 2 / +c. Finally, we can ﬁnd the constant, c, by considering f(0). At x= 0, our Fourier expansion gives x3=csince all the sine terms are zero, and hence c=0 . J 12.7 Complex Fourier series As a Fourier series expansion in general contains both sine and cosine parts, it may be written more compactly using a complex exponential expansion. Thissimpliﬁcation makes use of the property that exp( irx)=c o s rx+isinrx.T h e 430 12.7 COMPLEX FOURIER SERIES complex Fourier series expansion is written f(x)=∞summationdisplay r=−∞crexpparenleftbigg2πirx Lparenrightbigg , (12.9) where the Fourier coeﬃcients are given by cr=1 Lintegraldisplayx0+L x0f(x)ex pparenleftbigg −2πirx Lparenrightbigg dx. (12.10) This relation can be derived, in a similar manner to that of section 12.2, by mul- tiplying (12.9) by exp( −2πipx/L ) before integrating and using the orthogonality relation integraldisplayx0+L x0expparenleftbigg −2πipx Lparenrightbigg expparenleftbigg2πirx Lparenrightbigg dx=braceleftBigg Lforr=p, 0f o r r/negationslash=p. The complex Fourier coeﬃcients in (12.9) have the following relations to the real Fourier coeﬃcients: cr=1 2(ar−ibr), c−r=1 2(ar+ibr).(12.11) Note that if f(x)i sr e a lt h e n c−r=c∗ r, where the asterisk represents complex conjugation.IFind a complex Fourier series for f(x)=xin the range −2<x< 2. Using (12.10), for r/negationslash=0 , cr=1 4 Z2 −2xexp / −πirx 2 / dx = / −x 2πirexp / −πirx 2 //2 −2+ Z2 −21 2πirexp / −πirx 2 / dx =−1 πir[exp(−πir)+e x p ( πir)]+ /1 r2π2exp / −πirx 2 //2 −2 =2i πrcosπr−2i r2π2sinπr=2i πr(−1)r. (12.12) Forr= 0, we ﬁnd c0= 0 and hence x=∞X r=−∞ r/negationslash=02i(−1)r rπexp /πirx 2 / . We note that the Fourier series derived for xin section 12.6 gives ar=0f o ra l l rand br=−4(−1)r πr, and so, using (12.11), we conﬁrm that crandc−rhave the forms derived above. It is also apparent that the relationship c∗ r=c−rholds, as we expect, since f(x)i sr e a l . J 431 FOURIER SERIES 12.8 Parseval’s theorem Parseval’s theorem gives a useful way of relating the Fourier coeﬃcients to the function that they describe. Essentially a conservation law, it states that 1 Lintegraldisplayx0+L x0|f(x)|2dx=∞summationdisplay r=−∞|cr|2 =parenleftbig1 2a0parenrightbig2+1 2∞summationdisplay r=1(a2 r+b2 r). (12.13) In a more memorable form, this says that the sum of the moduli squared of the complex Fourier coeﬃcients is equal to the average value of |f(x)|2over one period. Parseval’s theorem can be proved straightforwardly by writing f(x)a s a Fourier series and evaluating the required integral, but the algebra is messy. Therefore, we shall use an alternative method, for which the algebra is simple and which in fact leads to a more general form of the theorem. Let us consider two functions f(x)a n d g(x), which are (or can be made) periodic with period Land which have Fourier series (expressed in complex form) f(x)=∞summationdisplay r=−∞crexpparenleftbigg2πirx Lparenrightbigg , g(x)=∞summationdisplay r=−∞γrexpparenleftbigg2πirx Lparenrightbigg , where crandγrare the complex Fourier coeﬃcients of f(x)a n d g(x) respectively. Thus f(x)g∗(x)=∞summationdisplay r=−∞crg∗(x)ex pparenleftbigg2πirx Lparenrightbigg . Integrating this equation with respect to xover the interval ( x0,x0+L)a n d dividing by L, we ﬁnd 1 Lintegraldisplayx0+L x0f(x)g∗(x)dx=∞summationdisplay r=−∞cr1 Lintegraldisplayx0+L x0g∗(x)ex pparenleftbigg2πirx Lparenrightbigg dx =∞summationdisplay r=−∞crbracketleftbigg1 Lintegraldisplayx0+L x0g(x)ex pparenleftbigg−2πirx Lparenrightbigg dxbracketrightbigg∗ =∞summationdisplay r=−∞crγ∗ r, where the last equality uses (12.10). Finally, if we let g(x)=f(x) then we obtain Parseval’s theorem (12.13). This proof can be performed in a similar manner 432 12.9 EXERCISES using the sine and cosine form of the Fourier series, but the algebra is slightly more complicated. Parseval’s theorem is sometimes used to sum series. However, if one is presented with a series to sum, it is not usually possible to decide which Fourier series should be used to evaluate it. Instead, useful summations are sometimes found serendipitously. The following example shows the evaluation of a sum by aFourier series method.IUsing Parseval’s theorem and the Fourier series for f(x)=x2found in section 12.5, calculate the sum P∞ r=1r−4. Firstly we ﬁnd the average value of [ f(x)]2over the interval −2<x≤2: 1 4 Z2 −2x4dx=16 5. Now we evaluate the right-hand side of (12.13):/;1 2a0 /2+1 2∞X 1a2 r+1 2∞X 1b2 n= /;4 3 /2+1 2∞X r=1162 π4r4. Equating the two expression we ﬁnd ∞X r=11 r4=π4 90. J 12.9 Exercises 12.1 Prove the orthogonality relations stated in section 12.1. 12.2 Derive the Fourier coeﬃcients brin a similar manner to the derivation of the ar in section 12.2. 12.3 Which of the following functions of xcould be represented by a Fourier series over the range indicated? (a) tanh−1(x),−∞<x<∞. (b) tan x, −∞<x<∞. (c)|sinx|−1/2,−∞<x<∞. (d) cos−1(sin2 x),−∞<x<∞. (e)xsin(1/x)−π−1<x≤π−1, cyclically repeated. 12.4 By moving the origin of tto the centre of an interval in which f(t) = +1, i.e. by changing to a new independent variable t/prime=t−1 4T, express the square-wave function in the example in section 12.2 as a cosine series. Calculate the Fouriercoeﬃcients involved (a) directly and (b) by changing the variable in result (12.8). 12.5 Find the Fourier series of the function f(x)=xin the range −π<x≤π. Hence show that 1−1 3+1 5−1 7+···=π 4. 12.6 For the function f(x)=1−x, 0≤x≤1, ﬁnd (a) the Fourier sine series and (b) the Fourier cosine series. Which would be better for numerical evaluation? Relate your answer to the relevant periodiccontinuations. 433 FOURIER SERIES 12.7 For the continued functions used in exercise 12.6 and the derived corresponding series, consider (i) their derivatives and (ii) their integrals. Do they give meaningfulequations? You will probably ﬁnd it helpful to sketch all the functions involved. 12.8 The function y(x)=xsinxfor 0≤x≤πis to be represented by a Fourier series of period 2 πthat is either even or odd. By sketching the function and considering its derivative, determine which series will have the more rapid convergence. Findthe full expression for the better of these two series, showing that the convergence∼n −3and that alternate terms are missing. 12.9 Find the Fourier coeﬃcients in the expansion of f(x)=e x p xover the range −1<x< 1. What value will the expansion have when x=2 ? 12.10 By integrating term by term the Fourier series found in the previous question and using the Fourier series for f(x)= xfound in section 12.6, show thatR expxd x=e x p x+c. Why is it not possible to show that d(expx)/dx=e x p x by diﬀerentiating the Fourier series of f(x)=e x p xin a similar manner? 12.11 Consider the function f(x)=e x p (−x2) in the range 0 ≤x≤1. Show how it should be continued to give as its Fourier series a series (the actual form is notwanted) (a) with only cosine terms, (b) with only sine terms, (c) with period 1and (d) with period 2. Would there be any diﬀerence between the values of the last two series at (i) x= 0, (ii) x=1 ? 12.12 Find, without calculation, which terms will be present in the Fourier series for the periodic functions f(t), of period T, that are given in the range −T/2t oT/2 by: (a)f(t)=2f o r0≤|t|<T/4,f=1f o r T/4≤|t|<T/2; (b)f(t)=e x p [−(t−T/4) 2]; (c)f(t)=−1f o r−T/2≤t<−3T/8a n d3 T/8≤t<T / 2,f(t)=1f o r −T/8≤t<−T/8; the graph of fis completed by two straight lines in the remaining ranges so as to form a continuous function. 12.13 Consider the representation as a Fourier series of the displacement of a string lying in the interval 0 ≤x≤Land ﬁxed at its ends, when it is pulled aside by y0 at the point x=L/4. Sketch the continuations for the region outside the interval that will (a) produce a series of period L, (b) produce a series that is antisymmetric about x=0 ,a n d (c) produce a series that will contain only cosine terms. (d) What are (i) the periods of the series in (b) and (c) and (ii) the value of the ‘a0-term’ in (c)? (e) Show that a typical term of the series obtained in (b) is 32y0 3n2π2sinnπ 4sinnπx L. 12.14 Show that the Fourier series for the function y(x)=|x|in the range −π≤x<π is y(x)=π 2−4 π∞X m=0cos(2 m+1 )x (2m+1 )2. By integrating this equation term by term from 0 to x, ﬁnd the function g(x) whose Fourier series is 4 π∞X m=0sin(2m+1 )x (2m+1 )3. 434 12.9 EXERCISES Deduce the value of the sum Sof the series 1−1 33+1 53−1 73+···. 12.15 Using the result of exercise 12.14, determine, as far as possible by inspection, the form of the functions of which the following are the Fourier series: (a) cosθ+1 9cos 3θ+1 25cos 5θ+···; (b) sinθ+1 27sin 3θ+1 125sin 5θ+···; (c) L2 3−4L2 π2 / cosπx L−1 4cos2πx L+1 9cos3πx L−··· / . (You may ﬁnd it helpful to ﬁrst set x= 0 in the quoted result and so obtain values for S0= P(2m+1 )−2and other sums derivable from it.) 12.16 By ﬁnding a cosine Fourier series of period 2 for the function f(t)t h a tt a k e st h e form f(t)=c o s h ( t−1) in the range 0 ≤t≤1, prove that ∞X n=11 n2π2+1=1 e2−1. Deduce values for the sums P(n2π2+1 )−1over odd nand even nseparately. 12.17 Find the (real) Fourier series of period 2 for f(x)=c o s h xandg(x)=x2in the range−1≤x≤1. By integrating the series for f(x) twice, prove that ∞X n=1(−1)n+1 n2π2(n2π2+1 )=1 2 /1 sinh 1−5 6 / . 12.18 Express the function f(x)=x2as a Fourier sine series in the range 0 <x≤2 and show that it converges to zero at x=±2. 12.19 Demonstrate explicitly for the square-w ave function discussed in section 12.2 that Parseval’s theorem (12.13) is valid. You will need to use the relationship ∞X m=01 (2m+1 )2=π2 8. Show that a ﬁlter that transmits frequencies only up to 8 π/Twill still transmit more than 90 per cent of the power in such a square-wave voltage signal. 12.20 Show that the Fourier series for |sinθ|in the range −π≤θ≤πis given by |sinθ|=2 π−4 π∞X m=1cos 2mθ 4m2−1. By setting θ=0a n d θ=π/2, deduce values for ∞X m=11 4m2−1and∞X m=11 16m2−1. 435 FOURIER SERIES 12.21 Find the complex Fourier series for the periodic function of period 2 πdeﬁned in the range−π≤x≤πbyy(x)=c o s h x. By setting t= 0 prove that ∞X n=1(−1)n n2+1=1 2 /π sinhπ−1 / . 12.22 The repeating output from an electronic oscillator takes the form of a sine wave f(t)=s i n tfor 0≤t≤π/2; it then drops instantaneously to zero and starts again. The output is to be represented by a complex Fourier series of the form ∞X n=−∞cne4nti. Sketch the function and ﬁnd an expression for cn.V e r i f yt h a t c−n=c∗ n.D e m o n - strate that setting t=0a n d t=π/2 produces diﬀering values for the sum ∞X n=11 16n2−1. Determine the correct value and check it using the quoted result of exercise 12.5. 12.23 Apply Parseval’s theorem to the series found in the previous exercise and so derive a value for the sum of the series 17 (15)2+65 (63)2+145 (143)2+···+16n2+1 (16n2−1)2+···. 12.24 A string, anchored at x=±L/2, has a fundamental vibration frequency of 2 L/c, where cis the speed of transverse waves on the string. It is pulled aside at its centre point by a distance y0and released at time t= 0. Its subsequent motion can be described by the series y(x, t)=∞X n=1ancosnπx Lcosnπct L. Find a general expression for anand show that only odd harmonics of the fundamental frequency are present in the sound generated by the released string.By applying Parseval’s theorem, ﬁnd the sum Sof the seriesP∞ 0(2m+1 )−4. 12.25 Show that Parseval’s theorem for two functions whose Fourier expansions have cosine and sine coeﬃcients an,bnandαn,βntakes the form 1 L ZL 0f(x)g∗(x)dx=1 4a0α0+1 2∞X n=1(anαn+bnβn). (a) Demonstrate that for g(x)=s i n mxor cos mxthis reduces to the deﬁnition of the Fourier coeﬃcients. (b) Explicitly verify the above result for the case in which f(x)=xandg(x)i s the square-wave function, both in the interval −1≤x≤1. 12.26 An odd function f(x)o fp e r i o d2 πis to be approximated by a Fourier sine series having only mterms. The error in this approximation is measured by the square deviation Em= Zπ −π /" f(x)−mX n=1bnsinnx /#2 dx. By diﬀerentiating Emwith respect to the coeﬃcients bn, ﬁnd the values of bnthat minimise Em. 436 12.10 HINTS AND ANSWERS Sketch the graph of the function f(x), where f(x)= / −x(π+x)f o r−π≤x<0, x(x−π)f o r 0≤x<π . f(x) is to be approximated by the ﬁrst three terms of a Fourier sine series. What coeﬃcients minimise E3? What is the resulting value of E3? 12.10 Hints and answers 12.3 Only (c). In terms of the Dirichlet conditions (section 12.1), the others fail as follows: (a) (i); (b) (ii); (d) (ii); (e) (iii). 12.4 an=[ ( 4 /(nπ)](−1)(n−1)/2fornodd and an=0f o r neven. In (b) use the expansion of sin( A+B). 12.5 f(x)=2 P∞ 1(−1)n+1n−1sinnx;s e t x=π/2. 12.6 (a) P[(2/(nπ)]sin nπx,a l ln;( b ) P[(4/(n2π2)] cos nπxfor odd nonly. The cosine series, with n−2convergence and alternate terms missing; the sine continuation contains a discontinuity. 12.7 (i) Series (a) from exercise 12.6 does not converge and cannot represent the function y(x)=−1. Series (b) reproduces the square-wave function of equation (12.8).(ii) Series (a) gives the series for y(x)=−x− 1 2x2−1 2in the range −1≤x≤0 and for y(x)=x−1 2x2−1 2in the range 0 ≤x≤1. Series (b) gives the series for y(x)=x+1 2x2+1 2in the range −1≤x≤0a n df o r y(x)=x−1 2x2+1 2in the range 0≤x≤1. 12.8 The even continuation has a discontinuity in its derivative at x=π, whilst the odd continuation does not; thus the sine series will have better convergence.b 1=π/2;b2m+1=0f o r m>0;b2m=−16m/[π(4m2−1)2]. 12.9 f(x)=( s i n h1 ) / 1+2 P∞ 1(−1)n(1 +n2π2)−1[cos(nπx)−nπsin(nπx)] / ; f(2) = f(0) = 1. 12.10 Combine the coeﬃcients of the sin( nπx) terms from the Fourier series for xand (part of) R expxd x; the partial series obtained by diﬀerentiating the sin( nπx) terms does not converge, having coeﬃcients of the form ( nπ)2/[1 + ( nπ)2]. 12.11 See ﬁgure 12.6. (c) (i) (1 + e−1)/2, (ii) (1 + e−1)/2; (d) (i) (1 + e−4)/2, (ii) e−1. (a) (b) (c) (d)00 0 0 11 12 4 Figure 12.6 Continuations of exp( −x2)i n0≤x≤1t og i v e :( a)c o s i n et e r m s only; ( b) sine terms only; ( c)p e r i o d1 ;( d)p e r i o d2 . 12.12 (a) a0and odd cosines; (b) all, there is no symmetry about T/4 for the periodic function; (c) Odd cosines. 12.13 (d) (i) The periods are both 2 L; (ii) y0/2. 12.14 g(x)=1 2x(π−x)f o r x≥0a n d=1 2x(π+x)f o r x≤0. Set x=π/2;S=π3/32. 437 FOURIER SERIES 12.15 So=π2/8. If Se= P(2m)−2then Se=1 4(Se+So), yielding So−Se=π2/12 and (Se+So)=π2/6. (a) (π/4)(π/2−|θ|); (b) ( πθ/4)(π/2−|θ|/2) from integrating (a). (c) Even function; average value L2/3;y(0) = 0; y(L)=L2; probably y(x)=x2. Compare with the worked example in section 12.5. 12.16 cosh( t−1) = (sinh 1)[1+2 P∞ n=1(cosnπt)/(n2π2+1)]; set t= 0 to obtain the stated result; set t= 1 to evaluate P(−1)n/(n2π2+1) and add and subtract this quantity from P(n2π2+1 )−1. P odd=(e−1)/[4(e+ 1)]; P even=( 3−e)/[4(e−1)]. 12.17 cosh x=( s i n h1 ) [ 1+2 P∞ n=1(−1)n(cosnπx)/(n2π2+1)] and after integrating twice this form must be recovered. Use x2=1 3+4 P(−1)n(cosnπx)/(n2π2)] to eliminate the quadratic term arising from the constants of integration; there is no linearterm. 12.18 Consider f(x)=−x 2for−2<x≤0, to ensure a sine series;P nbnsin(nπx/2), with bn=(−1)n+18/(nπ)f o r neven and ( −1)n+18/(nπ)− 32/(nπ)3fornodd. 12.19 C±(2m+1)=∓2i/[(2m+1 )π]; P|Cn|2=( 4/π2)×2×(π2/8); the values n=±1, ±3 contribute >90% of the total. 12.20 Write sin θcosnθas1 2[sin(n+1 )θ−sin(n−1)θ]; obtain P∞ 1(4m2−1)−1=1 2as well as P∞ 1(−1)m(4m2−1)−1=1 2−π 4and add the two equations;1 2,1 2−π 8. 12.21 cn=(−1)n[sinh π+in(cosh π−1)]/[π(1 +n2)]. 12.22 cn=( 2 /π)[(4ni−1)/(16n2−1)].The correct value is the mean of the two incorrect ones, i.e. (4 −π)/8.Write (16 n2−1)−1in partial fractions and compare with exercise 12.5. 12.23 ( π2−8)/16. 12.24 an=8y0/(n2π2) for odd n, an= 0 otherwise; S=π4/96. 12.25 (b) All anandαnare zero; bn=2 (−1)n+1/(nπ)a n d βn=4/(nπ). You will need the result quoted in exercise 12.19. 12.26 Show that the minimising value bkis given by 0 = Rπ −πf(x)si nkx dx− Pm n=1bnπδkn and hence that bkis equal to the normal Fourier coeﬃcient; b1=−8/π,b2=0 , b3=−8/(27π);E3=( 6 4 /π) P∞ 2(2m+1 )−6. 438 13 Integral transforms In the previous chapter we encountered the Fourier series representation of a periodic function in a ﬁxed interval as a superposition of sinusoidal functions. It isoften desirable, however, to obtain such a representation even for functions deﬁnedover an inﬁnite interval and with no particular periodicity. Such a representationis called a Fourier transform and is one of a class of representations called integral transforms . We begin by considering Fourier transforms as a generalisation of Fourier series. We then go on to discuss the properties of the Fourier transform and its applications. In the second part of the chapter we present an analogous discussionof the closely related Laplace transform . 13.1 Fourier transforms The Fourier transform provides a representation of functions deﬁned over an inﬁnite interval and having no particular periodicity, in terms of a superpositionof sinusoidal functions. It may thus be considered as a generalisation of the Fourier series representation of periodic functions. Since Fourier transforms are often used to represent time-varying functions, we shall present much of ourdiscussion in terms of f(t), rather than f(x), although in some spatial examples f(x) will be the more natural notation and we shall use it as appropriate. Our only requirement on f(t) will be thatintegraltext ∞ −∞|f(t)|dtis ﬁnite. In order to develop the transition from Fourier series to Fourier transforms, we ﬁrst recall that a function of period Tmay be represented as a complex Fourier series, cf. (12.9), f(t)=∞summationdisplay r=−∞cre2πirt/T=∞summationdisplay r=−∞creiωrt, (13.1) where ωr=2πr/T. As the period Ttends to inﬁnity, the ‘frequency quantum’ 439 INTEGRAL TRANSFORMS c(ω)e x p iωt −100 12 r−2π T2π T4π T ωr Figure 13.1 The relationship between the Fourier terms for a function of period Tand the Fourier integral (the area below the solid line) of the function. ∆ω=2π/Tbecomes vanishingly small and the spectrum of allowed frequencies ωrbecomes a continuum. Thus, the inﬁnite sum of terms in the Fourier series becomes an integral, and the coeﬃcients crbecome functions of the continuous variable ω, as follows. We recall, cf. (12.10), that the coeﬃcients crin (13.1) are given by cr=1 TintegraldisplayT/2 −T/2f(t)e−2πirt/Tdt=∆ω 2πintegraldisplayT/2 −T/2f(t)e−iωrtdt, (13.2) where we have written the integral in two alternative forms and, for convenience, made one period run from −T/2t o+ T/2 rather than from 0 to T. Substituting from (13.2) into (13.1) gives f(t)=∞summationdisplay r=−∞∆ω 2πintegraldisplayT/2 −T/2f(u)e−iωrudu eiωrt. (13.3) At this stage ωris still a discrete function of requal to 2 πr/T. The solid points in ﬁgure 13.1 are a plot of (say, the real part of) creiωrtas a function of r(or equivalently of ωr) and it is clear that (2 π/T)creiωrtgives the area of the rth broken-line rectangle. If Ttends to∞then ∆ ω(= 2π/T) becomes inﬁnitesimal, the width of the rectangles tends to zero and, from the mathematical deﬁnition of an integral, ∞summationdisplay r=−∞∆ω 2πg(ωr)eiωrt→1 2πintegraldisplay∞ −∞g(ω)eiωtdω. In this particular case g(ωr)=integraldisplayT/2 −T/2f(u)e−iωrudu, 440 13.1 FOURIER TRANSFORMS and (13.3) becomes f(t)=1 2πintegraldisplay∞ −∞dω eiωtintegraldisplay∞ −∞du f(u)e−iωu. (13.4) This result is known as Fourier’s inversion theorem . From it we may deﬁne the Fourier transform off(t)b y tildewidef(ω)=1√ 2πintegraldisplay∞ −∞f(t)e−iωtdt, (13.5) and its inverse by f(t)=1√ 2πintegraldisplay∞ −∞tildewidef(ω)eiωtdω. (13.6) Including the constant 1 /√ 2πin the deﬁnition of tildewidef(ω) (whose mathematical existence as T→∞is assumed here without proof) is clearly arbitrary, the only requirement being that the product of the constants in (13.5) and (13.6) shouldequal 1 /(2π). Our deﬁnition is chosen to be as symmetric as possible.IFind the Fourier transform of the exponential decay function f(t)=0 fort<0and f(t)=Ae−λtfort≥0(λ>0). Using the deﬁnition (13.5) and separating the integral into two parts,ef(ω)=1√ 2π Z0 −∞(0)e−iωtdt+A√ 2π Z∞ 0e−λte−iωtdt =0+A√ 2π / −e−(λ+iω)t λ+iω /∞ 0 =A√ 2π(λ+iω), which is the required transform. It is clear that the multiplicative constant Adoes not aﬀect the form of the transform, merely its amplitude. This transform may be veriﬁed by re-substitution of the above re sult into (13.6) to recover f(t), but evaluation of the integral requires the use of complex-variable contour integration (chapter 20). J 13.1.1 The uncertainty principle An important function that appears in many areas of physical science, either precisely or as an approximation to a physical situation, is the Gaussian or normal distribution. Its Fourier transform is of importance both in itself and also because, when interpreted statistically, it readily illustrates a form of uncertainty principle . 441 INTEGRAL TRANSFORMSIFind the Fourier transform of the normalised Gaussian distribution f(t)=1 τ√ 2πexp / −t2 2τ2 / ,−∞<t<∞. This Gaussian distribution is centred on t= 0 and has a root mean square deviation ∆t=τ. (Any reader who is unfamiliar with this interpretation of the distribution should refer to chapter 26.) Using the deﬁnition (13.5), the Fourier transform of f(t)i sg i v e nb yef(ω)=1√ 2π Z∞ −∞1 τ√ 2πexp / −t2 2τ2 / exp(−iωt)dt =1√ 2π Z∞ −∞1 τ√ 2πexp / −1 2τ2 / t2+2τ2iωt+(τ2iω)2−(τ2iω)2 / / dt, where the quantity −(τ2iω)2/(2τ2) has been both added and subtracted in the exponent in order to allow the factors involving the variable of integration tto be expressed as a complete square. Hence the expression can be writtenef(ω)=exp(−1 2τ2ω2)√ 2π /1 τ√ 2π Z∞ −∞exp / −(t+iτ2ω)2 2τ2 / dt / . The quantity inside the braces is the normalisation integral for the Gaussian and equals unity, although to show this strictly needs results from complex variable theory (chapter 20).That it is equal to unity can be made plausible by changing the variable to s=t+iτ 2ω and assuming that the imaginary parts introduced into the integration path and limits (where the integrand goes rapidly to zero anyway) make no diﬀerence. We are left with the result thatef(ω)=1√ 2πexp /−τ2ω2 2 / , (13.7) which is another Gaussian distribution, centred on zero and with a root mean square deviation ∆ ω=1/τ. It is interesting to note, and an important property, that the Fourier transform of a Gaussian is another Gaussian. J In the above example the root mean square deviation in twasτ, and so it is seen that the deviations or ‘spreads’ in tand in ωare inversely related: ∆ω∆t=1, independently of the value of τ. In physical terms, the narrower in time is, say, an electrical impulse the greater the spread of frequency components it must contain.Similar physical statements are valid for other pairs of Fourier-related variables,such as spatial position and wave number. In an obvious notation, ∆ k∆x=1f o r a Gaussian wave packet. The uncertainty relations as usually expressed in quantum mechanics can be related to this if the de Broglie and Einstein relationships for momentum and energy are introduced; they are p=/planckover2pi1kand E=/planckover2pi1ω. Here/planckover2pi1is Planck’s constant hdivided by 2 π. In a quantum mechanics setting f(t) 442 13.1 FOURIER TRANSFORMS is a wavefunction and the distribution of the wave intensity in time is given by |f|2(also a Gaussian). Similarly, the intensity distribution in frequency is given by|tildewidef|2. These two distributions have respective root mean square deviations of τ/√ 2a n d1 /(√ 2τ), giving, after incorporation of the above relations, ∆E∆t=/planckover2pi1/2a n d∆ p∆x=/planckover2pi1/2. The factors of 1 /2 that appear are speciﬁc to the Gaussian form, but any distribution f(t) produces for the product ∆ E∆ta quantity λ/planckover2pi1in which λis strictly positive (in fact the value 1 /2 for a Gaussian is the minimum possible). 13.1.2 Fraunhofer diﬀraction We take our ﬁnal example of the Fourier transform from the ﬁeld of optics. The pattern of transmitted light produced by a partially opaque (or phase-changing)object upon which a coherent beam of radiation falls is called a diﬀraction pattern and, in particular, when the cross-section of the object is small compared with the distance at which the light is observed the pattern is known as a Fraunhofer diﬀraction pattern. We will consider only the case in which the light is monochromatic with wavelength λ. The direction of the incident beam of light can then be described by the wave vector k; the magnitude of this vector is given by the wave number k=2π/λof the light. The essential quantity in a Fraunhofer diﬀraction pattern is the dependence of the observed amplitude (and hence intensity) on the angle θ between the viewing direction k /primeand the direction kof the incident beam. This is entirely determined by the spatial distribution of the amplitude and phase ofthe light at the object, the transmitted intensity in a particular direction k /primebeing determined by the corresponding Fourier component of this spatial distribution. As an example, we take as an object a simple two-dimensional screen of width 2Yon which light of wave number kis incident normally; see ﬁgure 13.2. We suppose that at the position (0 ,y) the amplitude of the transmitted light is f(y) per unit length in the y-direction ( f(y) may be complex). The function f(y)i s called an aperture function . Both the screen and beam are assumed inﬁnite in the z-direction. Denoting the unit vectors in the x-a n d y- directions by iandjrespectively, the total light amplitude at a position r0=x0i+y0j, with x0>0, will be the superposition of all the (Huyghens’) wavelets originating from the various parts of the screen. For large r0(=|r0|), these can be treated as plane waves to give † A(r0)=integraldisplayY −Yf(y)ex p[ ik/prime·(r0−yj)] |r0−yj|dy. (13.8) †This is the approach ﬁrst used by Fresnel. For simplicity we have omitted from the integral a multiplicative inclination factor that depends on angle θand decreases as θincreases. 443 INTEGRAL TRANSFORMS −YYy xkk/prime 0θ Figure 13.2 Diﬀraction grating of width 2 Ywith light of wavelength 2 π/k being diﬀracted through an angle θ. The factor exp[ ik/prime·(r0−yj)] represents the phase change undergone by the light in travelling from the point yjon the screen to the point r0, and the denominator represents the reduction in amplitude with distance. (Recall that the system isinﬁnite in the z-direction and so the ‘spreading’ is eﬀectively in two dimensions only.) If the medium is the same on both sides of the screen then k /prime=kcosθi+ksinθj, and if r0/greatermuchYthen expression (13.8) can be approximated by A(r0)=exp(ik/prime·r0) r0integraldisplay∞ −∞f(y)ex p(−ikysinθ)dy. (13.9) We have used that f(y)=0f o r|y|>Y, to extend the integral to inﬁnite limits. The intensity in the direction θis then given by I(θ)=|A|2=2π r02|tildewidef(q)|2, (13.10) where q=ksinθ.IEvaluate I(θ)for an aperture consisting of two long slits each of width 2bwhose centres are separated by a distance 2a,a>b; the slits are illuminated by light of wavelength λ. The aperture function is plotted in ﬁgure 13.3. We ﬁrst need to ﬁnd ef(q):ef(q)=1√ 2π Z−a+b −a−be−iqxdx+1√ 2π Za+b a−be−iqxdx =1√ 2π / −e−iqx iq /−a+b −a−b+1√ 2π / −e−iqx iq /a+b a−b =−1 iq√ 2π / e−iq(−a+b)−e−iq(−a−b)+e−iq(a+b)−e−iq(a−b) / . 444 13.1 FOURIER TRANSFORMS f(y) 1 a−b −a−b a+b −a+ba −a x Figure 13.3 The aperture function f(y) for two wide slits. After some manipulation we obtainef(q)=4c os qasinqb q√ 2π. Now applying (13.10), and remembering that q=( 2πsinθ)/λ, we ﬁnd I(θ)=16cos2qasin2qb q2r02, where r0is the distance from the centre of the aperture. J 13.1.3 The Dirac δ-function Before going on to consider further properties of Fourier transforms we make a digression to discuss the Dirac δ-function and its relation to Fourier transforms. The δ-function is diﬀerent from most functions encountered in the physical sciences but we will see that a rigorous mathematical deﬁnition exists and theutility of the δ-function will be demonstrated throughout the remainder of this chapter. It can be visualised as a very sharp narrow pulse (in space, time, density,etc.) which produces an integrated eﬀect having a deﬁnite magnitude. The formal properties of the δ-function may be summarised as follows. The Dirac δ-function has the property that δ(t)=0 f o r t/negationslash=0, (13.11) but its fundamental deﬁning property is integraldisplay f(t)δ(t−a)dt=f(a), (13.12) provided the range of integration includes the point t=a; otherwise the integral 445 INTEGRAL TRANSFORMS equals zero. This leads immediately to two further useful results: integraldisplayb −aδ(t)dt=1 f o ra l l a, b > 0 (13.13) andintegraldisplay δ(t−a)dt=1, (13.14) provided the range of integration includes t=a. Equation (13.12) can be used to derive further useful properties of the Dirac δ-function: δ(t)=δ(−t), (13.15) δ(at)=1 |a|δ(t), (13.16) tδ(t)=0 . (13.17)IProve that δ(bt)=δ(t)/|b|. Let us ﬁrst consider the case where b>0. It follows thatZ∞ −∞f(t)δ(bt)dt= Z∞ −∞f /t/prime b / δ(t/prime)dt/prime b=1 bf(0) =1 b Z∞ −∞f(t)δ(t)dt, where we have made the substitution t/prime=bt.B u t f(t) is arbitrary and so we immediately see that δ(bt)=δ(t)/b=δ(t)/|b|forb>0. Now consider the case where b=−c<0. It follows thatZ∞ −∞f(t)δ(bt)dt= Z−∞ ∞f /t/prime −c / δ(t/prime) /dt/prime −c / = Z∞ −∞1 cf /t/prime −c / δ(t/prime)dt/prime =1 cf(0) =1 |b|f(0) =1 |b| Z∞ −∞f(t)δ(t)dt, where we have made the substitution t/prime=bt=−ct.B u t f(t) is arbitrary and so δ(bt)=1 |b|δ(t), for all b, which establishes the result. J Furthermore, by considering an integral of the form integraldisplay f(t)δ(h(t))dt, and making a change of variables to z=h(t), we may show that δ(h(t)) =summationdisplay iδ(t−ti) |h/prime(ti)|, (13.18) where the tiare those values of tfor which h(t)=0a n d h/prime(t) stands for dh/dt. 446 13.1 FOURIER TRANSFORMS The derivative of the delta function, δ/prime(t), is deﬁned by integraldisplay∞ −∞f(t)δ/prime(t)dt=bracketleftBig f(t)δ(t)bracketrightBig∞ −∞−integraldisplay∞ −∞f/prime(t)δ(t)dt =−f/prime(0), (13.19) and similarly for higher derivatives. For many practical purposes, eﬀects that are not strictly described by a δ- function may be analysed as such, if they take place in an interval much shorter than the response interval of the system on which they act. For example, the idealised notion of an impulse of magnitude Japplied at time t0can be represented by j(t)=Jδ(t−t0). (13.20) Many physical situations are described by a δ-function in space rather than in time. Moreover, we often require the δ-function to be deﬁned in more than one dimension. For example, the charge density of a point charge qat a point r0may be expressed as a three-dimensional δ-function ρ(r)=qδ(r−r0)=qδ(x−x0)δ(y−y0)δ(z−z0), (13.21) so that a discrete ‘quantum’ is expressed as if it were a continuous distribution. From (13.21) we see that (as expected) the total charge enclosed in a volume V is given by integraldisplay Vρ(r)dV=integraldisplay Vqδ(r−r0)dV=braceleftBigg qifr0lies in V, 0o t h e r w i s e . Closely related to the Dirac δ-function is the Heaviside orunit step function H(t), for which H(t)=braceleftBigg 1f o r t>0, 0f o r t<0.(13.22) This function is clearly discontinuous at t= 0 and it is usual to take H(0) = 1 /2. The Heaviside function is related to the delta function by H/prime(t)=δ(t). (13.23) 447 INTEGRAL TRANSFORMSIProve relation (13.23). Considering the integralZ∞ −∞f(t)H/prime(t)dt= / f(t)H(t) /∞ −∞− Z∞ −∞f/prime(t)H(t)dt =f(∞)− Z∞ 0f/prime(t)dt =f(∞)− / f(t) /∞ 0=f(0), and comparing it with (13.12) when a= 0 immediately shows that H/prime(t)=δ(t). J 13.1.4 Relation of the δ-function to Fourier transforms In the previous section we introduced the Dirac δ-function as a way of repre- senting very sharp narrow pulses, but in no way related it to Fourier transforms.We now show that the δ-function can equally well be deﬁned in a way that more naturally relates it to the Fourier transform. Referring back to the Fourier inversion theorem (13.4), we have f(t)=1 2πintegraldisplay∞ −∞dω eiωtintegraldisplay∞ −∞du f(u)e−iωu =integraldisplay∞ −∞du f(u)braceleftbigg1 2πintegraldisplay∞ −∞eiω(t−u)dωbracerightbigg . Comparison of this with (13.12) shows that we may write the δ-function as δ(t−u)=1 2πintegraldisplay∞ −∞eiω(t−u)dω. (13.24) Considered as a Fourier transform, this representation shows that a very narrow time peak at t=uresults from the superposition of a complete spectrum of harmonic waves, all frequencies having the same amplitude and all waves beingin phase at t=u. This suggests that the δ-function may also be represented as the limit of the transform of a uniform distribution of unit height as the width of this distribution becomes inﬁnite. Consider the rectangular distribution of frequencies shown in ﬁgure 13.4( a). From (13.6), taking the inverse Fourier transform, f Ω(t)=1√ 2πintegraldisplayΩ −Ω1×eiωtdω =2Ω√ 2πsin Ωt Ωt. (13.25) This function is illustrated in ﬁgure 13.4( b) and it is apparent that, for large Ω, it becomes very large at t= 0 and also very narrow about t= 0, as we qualitatively 448 13.1 FOURIER TRANSFORMS ω (a) (b)Ω −Ω t π Ω1 efΩfΩ(t) 2Ω (2π)1/2 Figure 13.4 ( a) A Fourier transform showing a rectangular distribution of frequencies between ±Ω; (b) the function of which it is the transform, which is proportional to t−1sin Ωt. expect and require. We also note that, in the limit Ω →∞,fΩ(t), as deﬁned by the inverse Fourier transform, tends to (2 π)1/2δ(t) by virtue of (13.24). Hence we may conclude that the δ-function can also be represented by δ(t) = lim Ω→∞parenleftbiggsinΩt πtparenrightbigg . (13.26) Several other function representations are equally valid, e.g. the limiting cases of rectangular, triangular or Gaussian distributions; the only essential requirementsare a knowledge of the area under such a curve and that undeﬁned operationssuch as dividing by zero are not inadvertently carried out on the δ-function whilst some non-explicit representation is being employed. We also note that the Fourier transform deﬁnition of the delta function, (13.24), shows that the latter is real since δ ∗(t)=1 2πintegraldisplay∞ −∞e−iωtdω=δ(−t)=δ(t). Finally, the Fourier transform of a δ-function is simply tildewideδ(ω)=1√ 2πintegraldisplay∞ −∞δ(t)e−iωtdt=1√ 2π. (13.27) 13.1.5 Properties of Fourier transforms Having considered the Dirac δ-function, we now return to our discussion of the properties of Fourier transforms. As we would expect, Fourier transforms have many properties analogous to those of Fourier series in respect of the connectionbetween the transforms of related functions. Here we list these properties without 449 INTEGRAL TRANSFORMS proof; they can be veriﬁed by working from the deﬁnition of the transform. As previously, we denote the Fourier transform of f(t)b ytildewidef(ω)o r F[f(t)]. (i) Diﬀerentiation:Fbracketleftbig f/prime(t)bracketrightbig =iωtildewidef(ω). (13.28) This may be extended to higher derivatives, so thatFbracketleftbig f/prime/prime(t)bracketrightbig =iω Fbracketleftbig f/prime(t)bracketrightbig =−ω2tildewidef(ω), a n ds oo n . (ii) Integration:Fbracketleftbiggintegraldisplayt f(s)dsbracketrightbigg =1 iωtildewidef(ω)+2πcδ(ω), (13.29) where the term 2 πcδ(ω) represents the Fourier transform of the constant of integration associated with the indeﬁnite integral. (iii) Scaling:F[f(at)]=1 atildewidefparenleftBigω aparenrightBig . (13.30) (iv) Translation:F[f(t+a)]=eiaωtildewidef(ω). (13.31) (v) Exponential multiplication:Fbracketleftbig eαtf(t)bracketrightbig =tildewidef(ω+iα), (13.32) where αmay be real, imaginary or complex.IProve relation (13.28). Calculating the Fourier transform of f/prime(t) directly, we obtainF / f/prime(t) / =1√ 2π Z∞ −∞f/prime(t)e−iωtdt =1√ 2π / e−iωtf(t) /∞ −∞+1√ 2π Z∞ −∞iω e−iωtf(t)dt =iω ef(ω), iff(t)→0a tt=±∞,a si tm u s ts i n c e R∞ −∞|f(t)|dtis ﬁnite. J To illustrate a use and also a proof of (13.32), let us consider an amplitude- modulated radio wave. Suppose a message to be broadcast is represented by f(t). The message can be added electronically to a constant signal aof magnitude such that a+f(t) is never negative, and then the sum can be used to modulate 450 13.1 FOURIER TRANSFORMS the amplitude of a carrier signal of frequency ωc. Using a complex exponential notation, the transmitted amplitude is now g(t)=A[a+f(t)]eiωct. (13.33) Ignoring in the present context the eﬀect of the term Aaexp(iωct), which gives a contribution to the transmitted spectrum only at ω=ωc,w eo b t a i nf o rt h en e w spectrum tildewideg(ω)=1√ 2πAintegraldisplay∞ −∞f(t)eiωcte−iωtdt =1√ 2πAintegraldisplay∞ −∞f(t)e−i(ω−ωc)tdt =Atildewidef(ω−ωc), (13.34) which is simply a shift of the whole spectrum by the carrier frequency. The use of diﬀerent carrier frequencies enables signals to be separated. 13.1.6 Odd and even functions Iff(t) is odd or even then we may derive alternative forms of Fourier’s inversion theorem, which lead to the deﬁnition of diﬀerent transform pairs. Let us ﬁrstconsider an odd function f(t)=−f(−t), whose Fourier transform is given by tildewidef(ω)=1 √ 2πintegraldisplay∞ −∞f(t)e−iωtdt =1√ 2πintegraldisplay∞ −∞f(t)(cos ωt−isinωt)dt =−2i√ 2πintegraldisplay∞ 0f(t)s i nωtdt, where in the last line we use the fact that f(t)a n ds i n ωtare odd, whereas cos ωt is even. We note that tildewidef(−ω)=−tildewidef(ω), i.e.tildewidef(ω) is an odd function of ω. Hence f(t)=1√ 2πintegraldisplay∞ −∞tildewidef(ω)eiωtdω=2i√ 2πintegraldisplay∞ 0tildewidef(ω)s i nωtdω =2 πintegraldisplay∞ 0dωsinωtbraceleftbiggintegraldisplay∞ 0f(u)s i nωudubracerightbigg . Thus we may deﬁne the Fourier sine transform pair for odd functions: tildewidefs(ω)=radicalbigg 2 πintegraldisplay∞ 0f(t)s i n ωtdt, (13.35) f(t)=radicalbigg 2 πintegraldisplay∞ 0tildewidefs(ω)s i n ωtdω. (13.36) 451 INTEGRAL TRANSFORMS g(y) (a) (b) (c) (d) y 0 Figure 13.5 Resolution functions: ( a)i d e a l δ-function; ( b) typical unbiased resolution; ( c)a n d( d) biases tending to shift observations to higher values than the true one. Note that although the Fourier sine transform pair was derived by considering an odd function f(t) deﬁned over all t, the deﬁnitions (13.35) and (13.36) only require f(t)a n dtildewidefs(ω) to be deﬁned for positive tandωrespectively. For an even function, i.e. one for which f(t)=f(−t), we can deﬁne the Fourier cosine transform pair in a similar way, but with sin ωtreplaced by cos ωt. 13.1.7 Convolution and deconvolution It is apparent that any attempt to measure the value of a physical quantity is limited, to some extent, by the ﬁnite resolution of the measuring apparatus used. On the one hand, the physical quantity we wish to measure will be in general a function of an independent variable, xsay, i.e. the true function to be measured takes the form f(x). On the other hand, the apparatus we are using does not give the true output value of the function; a resolution function g(y) is involved. By this we mean that the probability that an output value y= 0 will be recorded instead as being between yandy+dyis given by g(y)dy. Some possible resolution functions of this sort are shown in ﬁgure 13.5. To obtain good results we wish the resolution function to be as close to a δ-function as possible (case ( a)). A typical piece of apparatus has a resolution function of ﬁnite width, although ifit is accurate the mean is centred on the true value (case ( b)). However, some apparatus may show a bias that tends to shift observations to higher or lowervalues than the true ones (cases ( c)a n d( d)), thereby exhibiting systematic error. Given that the true distribution is f(x) and the resolution function of our 452 13.1 FOURIER TRANSFORMS −a −a aax y zf(x) b−b2b 2bg(y) h(z) ∗ = 1 Figure 13.6 The convolution of two functions f(x)a n d g(y). measuring apparatus is g(y), we wish to calculate what the observed distribution h(z) will be. The symbols x,yandzall refer to the same physical variable (e.g. length or angle), but are denoted diﬀerently because the variable appears in theanalysis in three diﬀerent roles. The probability that a true reading lying between xandx+dx, and so having probability f(x)dxof being selected by the experiment, will be moved by the instrumental resolution by an amount z−xinto a small interval of width dzis g(z−x)dz. Hence the combined probability that the interval dxwill give rise to an observation appearing in the interval dzisf(x)dx g(z−x)dz. Adding together the contributions from all values of xt h a tc a nl e a dt oa no b s e r v a t i o ni nt h er a n g e ztoz+dz, we ﬁnd that the observed distribution is given by h(z)=integraldisplay ∞ −∞f(x)g(z−x)dx. (13.37) The integral in (13.37) is called the convolution of the functions fandgand is often written f∗g. The convolution deﬁned above is commutative ( f∗g=g∗f), associative and distributive. The observed distribution is thus the convolution ofthe true distribution and the experimental resolution function. The result will bethat the observed distribution is broader and smoother than the true one and, if g(y) has a bias, the maxima will normally be displaced from their true positions. It is also obvious from (13.37) that if the resolution is the ideal δ-function, g(y)=δ(y)t h e n h(z)=f(z) and the observed distribution is the true one. It is interesting to note, and a very important property, that the convolution of any function g(y) with a number of delta functions leaves a copy of g(y)a tt h e position of each of the delta functions.IFind the convolution of the function f(x)=δ(x+a)+δ(x−a)with the function g(y) plotted in ﬁgure 13.6. Using the convolution integral (13.37) h(z)= Z∞ −∞f(x)g(z−x)dx= Z∞ −∞[δ(x+a)+δ(x−a)]g(z−x)dx =g(z+a)+g(z−a). This convolution h(z) is plotted in ﬁgure 13.6. J 453 INTEGRAL TRANSFORMS Let us now consider the Fourier transform of the convolution (13.37); this is given by tildewideh(k)=1√ 2πintegraldisplay∞ −∞dz e−ikzbraceleftbiggintegraldisplay∞ −∞f(x)g(z−x)dxbracerightbigg =1√ 2πintegraldisplay∞ −∞dx f(x)braceleftbiggintegraldisplay∞ −∞g(z−x)e−ikzdzbracerightbigg . If we let u=z−xin the second integral we have tildewideh(k)=1√ 2πintegraldisplay∞ −∞dx f(x)braceleftbiggintegraldisplay∞ −∞g(u)e−ik(u+x)dubracerightbigg =1√ 2πintegraldisplay∞ −∞f(x)e−ikxdxintegraldisplay∞ −∞g(u)e−ikudu =1√ 2π×√ 2πtildewidef(k)×√ 2πtildewideg(k)=√ 2πtildewidef(k)tildewideg(k). (13.38) Hence the Fourier transform of a convolution f∗gis equal to the product of the separate Fourier transforms multiplied by√ 2π; this result is called the convolution theorem . It may be proved similarly that the converse is also true, namely that the Fourier transform of the product f(x)g(x)i sg i v e nb yF[f(x)g(x)]=1√ 2πtildewidef(k)∗tildewideg(k). (13.39)IFind the Fourier transform of the function in ﬁgure 13.3 representing two wide slits by considering the Fourier transforms of (i) two δ-functions, at x=±a, (ii) a rectangular function of height 1and width 2bcentred on x=0. (i) The Fourier transform of the two δ-functions is given byef(q)=1√ 2π Z∞ −∞δ(x−a)e−iqxdx+1√ 2π Z∞ −∞δ(x+a)e−iqxdx =1√ 2π /; e−iqa+eiqa / =2c os qa√ 2π. (ii) The Fourier transform of the broad slit iseg(q)=1√ 2π Zb −be−iqxdx=1√ 2π /e−iqx −iq /b −b =−1 iq√ 2π(e−iqb−eiqb)=2si nqb q√ 2π. We have already seen that the convolution of these functions is the required function representing two wide slits (see ﬁgure 13.6). So, using the convolution theorem, the Fourier transform of the convolution is√ 2πtimes the product of the individual transforms, i.e. 4cos qasinqb/(q√ 2π). This is, of course, the same result as that obtained in the example in subsection 13.1.2. J 454 13.1 FOURIER TRANSFORMS The inverse of convolution, called deconvolution , allows us to ﬁnd a true distribution f(x) given an observed distribution h(z) and a resolution function g(y).IAn experimental quantity f(x)is measured using apparatus with a known resolution func- tiong(y)to give an observed distribution h(z).H o wm a y f(x)be extracted from the mea- sured distribution? From the convolution theorem (13.38), the Fourier transform of the measured distributioniseh(k)=√ 2π ef(k) eg(k), from which we obtainef(k)=1√ 2π eh(k)eg(k). Then on inverse Fourier transforming we ﬁnd f(x)=1√ 2π F−1 /"eh(k)eg(k) /# . In words, to extract the true distribution, we divide the Fourier transform of the observed distribution by that of the resolution function for each value of kand then take the inverse Fourier transform of the function so generated. J This explicit method of extracting true distributions is straightforward for exact functions but, in practice, because of experimental and statistical uncertainties inthe experimental data or because data over only a limited range are available, itis often not very precise, involving as it does three (numerical) transforms eachrequiring in principle an integral over an inﬁnite range. 13.1.8 Correlation functions and energy spectra Thecross-correlation of two functions fandgis deﬁned by C(z)=integraldisplay ∞ −∞f∗(x)g(x+z)dx. (13.40) Despite the formal similarity between (13.40) and the deﬁnition of the convolution in (13.37), the use and interpretation of the cross-correlation and of the convo-lution are very diﬀerent; the cross-correlation provides a quantitative measure ofthe similarity of two functions fandgas one is displaced through a distance z relative to the other. The cross-correlation is often notated as C=f⊗g, and, like convolution, it is both associative and distributive. Unlike convolution, however,it isnotcommutative, in fact [f⊗g](z)=[g⊗f] ∗(−z). (13.41) 455 INTEGRAL TRANSFORMSIProve the Wiener–Kinchin theorem,eC(k)=√ 2π[ ef(k)]∗eg(k). (13.42) Following a method similar to that for the convolution of fandg, let us consider the Fourier transform of (13.40):eC(k)=1√ 2π Z∞ −∞dz e−ikz /Z∞ −∞f∗(x)g(x+z)dx / =1√ 2π Z∞ −∞dx f∗(x) /Z∞ −∞g(x+z)e−ikzdz / . Making the substitution u=x+zin the second integral we obtaineC(k)=1√ 2π Z∞ −∞dx f∗(x) /Z∞ −∞g(u)e−ik(u−x)du / =1√ 2π Z∞ −∞f∗(x)eikxdx Z∞ −∞g(u)e−ikudu =1√ 2π×√ 2π[ ef(k)]∗×√ 2π eg(k)=√ 2π[ ef(k)]∗eg(k). J Thus the Fourier transform of the cross-correlation of fandgis equal to the product of [ tildewidef(k)]∗andtildewideg(k) multiplied by√ 2π.T h i sas t a t e m e n to ft h e Wiener–Kinchin theorem . Similarly we can derive the converse theoremFbracketleftbig f∗(x)g(x)bracketrightbig =1√ 2πtildewidef⊗tildewideg. If we now consider the special case where gis taken to be equal to fin (13.40) then, writing the LHS as a(z), we have a(z)=integraldisplay∞ −∞f∗(x)f(x+z)dx; (13.43) this is called the auto-correlation function off(x). Using the Wiener–Kinchin theorem (13.42) we see that a(z)=1√ 2πintegraldisplay∞ −∞tildewidea(k)eikzdk =1√ 2πintegraldisplay∞ −∞√ 2π[tildewidef(k)]∗tildewidef(k)eikzdk, so that a(z) is the inverse Fourier transform of√ 2π|tildewidef(k)|2, which is in turn called theenergy spectrum off. 13.1.9 Parseval’s theorem Using the results of the previous section we can immediately obtain Parseval’s theorem . The most general form of this (also called the multiplication theorem )i s 456 13.1 FOURIER TRANSFORMS obtained simply by noting from (13.42) that the cross-correlation (13.40) of two functions fandgc a nb ew r i t t e na s C(z)=integraldisplay∞ −∞f∗(x)g(x+z)dx=integraldisplay∞ −∞[tildewidef(k)]∗tildewideg(k)eikzdk. (13.44) Then, setting z= 0 gives the multiplication theorem integraldisplay∞ −∞f∗(x)g(x)dx=integraldisplay [tildewidef(k)]∗tildewideg(k)dk. (13.45) Specialising further, by letting g=f, we derive the most common form of Parseval’s theorem, integraldisplay∞ −∞|f(x)|2dx=integraldisplay∞ −∞|tildewidef(k)|2dk. (13.46) When fis a physical amplitude these integrals relate to the total intensity involved in some physical process. We have already met a form of Parseval’s theorem for Fourier series in chapter 12; it is in fact a special case of (13.46).IThe displacement of a damped harmonic oscillator as a function of time is given by f(t)= /( 0 fort<0, e−t/τsinω0tfort≥0. Find the Fourier transform of this function and so give a physical interpretation of Parseval’s theorem. Using the usual deﬁnition for the Fourier transform we ﬁndef(ω)= Z0 −∞0×e−iωtdt+ Z∞ 0e−t/τsinω0te−iωtdt. Writing sin ω0tas (eiω0t−e−iω0t)/2iwe obtainef(ω)=0+1 2i Z∞ 0 / e−it(ω−ω0−i/τ)−e−it(ω+ω0−i/τ) / dt =1 2 /1 ω+ω0−i/τ−1 ω−ω0−i/τ / , which is the required Fourier transform. The physical interpretation of | ef(ω)|2is the energy content per unit frequency interval (i.e. the energy spectrum ) whilst|f(t)|2is proportional to the sum of the kinetic and potential energies of the oscillator. Hence (to within a constant)Parseval’s theorem shows the equivalence of these two alternative speciﬁcations for the total energy.J 13.1.10 Fourier transforms in higher dimensions The concept of the Fourier transform can be extended naturally to more than one dimension. For instance we may wish to ﬁnd the spatial Fourier transform of 457 INTEGRAL TRANSFORMS two- or three-dimensional functions of position. For example, in three dimensions we can deﬁne the Fourier transform of f(x, y, z)a s tildewidef(kx,ky,kz)=1 (2π)3/2integraldisplayintegraldisplayintegraldisplay f(x, y, z)e−ikxxe−ikyye−ikzzdx dy dz, (13.47) and its inverse by as f(x, y, z)=1 (2π)3/2integraldisplayintegraldisplayintegraldisplay tildewidef(kx,ky,kz)eikxxeikyyeikzzdkxdkydkz. (13.48) Denoting the vector with components kx,ky,kzbykand that with components x, y, z byr, we can write the Fourier transform pair (13.47), (13.48) as tildewidef(k)=1 (2π)3/2integraldisplay f(r)e−ik·rd3r, (13.49) f(r)=1 (2π)3/2integraldisplay tildewidef(k)eik·rd3k. (13.50) From these relations we may deduce that the three-dimensional Dirac δ-function c a nb ew r i t t e na s δ(r)=1 (2π)3integraldisplay eik·rd3k. (13.51) Similar relations to (13.49), (13.50) and (13.51) exist for spaces of other dimen- sionalities.IIn three-dimensional space a function f(r)possesses spherical symmetry, so that f(r)= f(r). Find the Fourier transform of f(r)as a one-dimensional integral. Let us choose spherical polar coordinates in which the vector kof the Fourier transform lies along the polar axis ( θ= 0). This we can do since f(r) is spherically symmetric. We then have d3r=r2sinθd rd θd φ and k·r=krcosθ, where k=|k|. The Fourier transform is then given byef(k)=1 (2π)3/2 Z f(r)e−ik·rd3r =1 (2π)3/2 Z∞ 0dr Zπ 0dθ Z2π 0dφ f(r)r2sinθe−ikrcosθ =1 (2π)3/2 Z∞ 0dr2πf(r)r2 Zπ 0dθsinθe−ikrcosθ. The integral over θmay be straightforwardly evaluated by noting that d dθ(e−ikrcosθ)=ikrsinθe−ikrcosθ. Thereforeef(k)=1 (2π)3/2 Z∞ 0dr2πf(r)r2 /e−ikrcosθ ikr /θ=π θ=0 =1 (2π)3/2 Z∞ 04πr2f(r) /sinkr kr / dr. J 458 13.2 LAPLACE TRANSFORMS A similar result may be obtained for two-dimensional Fourier transforms in which f(r)=f(ρ), i.e. f(r) is independent of azimuthal angle φ. In this case, using the integral representation of the Bessel function J0(x) given at the very end of subsection 16.7.3, we ﬁnd tildewidef(k)=1 2πintegraldisplay∞ 02πρf(ρ)J0(kρ)dρ. (13.52) 13.2 Laplace transforms Often we are interested in functions f(t) for which the Fourier transform does not exist because f/negationslash→0a s t→∞, and so the integral deﬁning tildewidefdoes not converge. For example, the function f(t)=tdoes not possess a Fourier transform. Furthermore, often we are interested in a given function only for t>0, for example when we are given the value at t= 0 in an initial-value problem. This leads us to consider the Laplace transform, ¯f(s)o r L[f(t)],o ff(t), which is deﬁned by ¯f(s)≡integraldisplay∞ 0f(t)e−stdt, (13.53) provided that the integral exists. We assume here that sis real, but complex values would have to be considered in a more detailed study. In practice, for a givenfunction f(t) there will be some real number s 0such that the integral in (13.53) exists for s>s 0but diverges for s≤s0. Through (13.53) we deﬁne a linear transformation L[]that converts functions of the variable tto functions of a new variable s:L[af1(t)+bf2(t)]=a L[f1(t)]+b L[f2(t)]=a¯f1(s)+b¯f2(s). (13.54)IFind the Laplace transforms of the functions (i) f(t)=1, (ii) f(t)=eat, (iii) f(t)=tn, forn=0,1,2,.... (i) By direct application of the deﬁnition of a Laplace transform (13.53), we ﬁndL[1]= Z∞ 0e−stdt= /−1 se−st /∞ 0=1 s,ifs>0, where the restriction s>0 is required for the integral to exist. (ii) Again using (13.53) directly, we ﬁnd ¯f(s)= Z∞ 0eate−stdt= Z∞ 0e(a−s)tdt = /e(a−s)t a−s /∞ 0=1 s−aifs>a . 459 INTEGRAL TRANSFORMS (iii) Once again using the deﬁnition (13.53) we have ¯fn(s)= Z∞ 0tne−stdt. Integrating by parts we ﬁnd ¯fn(s)= /−tne−st s /∞ 0+n s Z∞ 0tn−1e−stdt =0+n s¯fn−1(s),ifs>0. We now have a recursion relation between successive transforms and by calculating ¯f0we can infer ¯f1,¯f2,e t c .S i n c e t0= 1, (i) above gives ¯f0=1 s,ifs>0, (13.55) and ¯f1(s)=1 s2,¯f2(s)=2! s3,. . . , ¯fn(s)=n! sn+1ifs>0. Thus, in each case (i)–(iii), direct application of the deﬁnition of the Laplace transform (13.53) yields the required result. J Unlike that for the Fourier transform, the inversion of the Laplace transform is not an easy operation to perform, since an explicit formula for f(t), given ¯f(s), is not straightforwardly obtained from ( 13.53). The general method for obtaining an inverse Laplace transform makes use of complex variable theory and is notdiscussed until chapter 20. However, progress can be made without having to ﬁndanexplicit inverse, since we can prepare from (13.53) a ‘dictionary’ of the Laplace transforms of common functions and, when faced with an inversion to carry out, hope to ﬁnd the given transform (together with its parent function) in the listing. Such a list is given in table 13.1. When ﬁnding inverse Laplace transforms using table 13.1, it is useful to note that for all practical purposes the inverse Laplace transform is unique †and linear so thatL−1bracketleftbig a¯f1(s)+b¯f2(s)bracketrightbig =af1(t)+bf2(t). (13.56) In many practical problems the method of partial fractions can be useful in producing an expression from which the inverse Laplace transform can be found.IUsing table 13.1 ﬁnd f(t)if ¯f(s)=s+3 s(s+1 ). Using partial fractions ¯f(s) may be written ¯f(s)=3 s−2 s+1. †This is not strictly true, since two functions can diﬀer from one another at a ﬁnite number of isolated points but have the sameLaplace transform. 460 13.2 LAPLACE TRANSFORMS f(t) ¯f(s) s0 cc / s 0 ctncn!/sn+10 sinbt b/ (s2+b2)0 cosbt s/ (s2+b2)0 eat1/(s−a) a tneatn!/(s−a)n+1a sinhat a/ (s2−a2) |a| coshat s/ (s2−a2) |a| eatsinbt a/ [(s−a)2+b2] a eatcosbt (s−a)/[(s−a)2+b2] a t1/2 1 2(π/s3)1/20 t−1/2(π/s)1/20 δ(t−t0) e−st0 0 H(t−t0)= /( 1f o r t≥t0 0f o r t<t 0e−st0/s 0 Table 13.1 Standard Laplace transforms. The transforms are valid for s>s 0. Comparing this with the standard Laplace transforms in table 13.1, we ﬁnd that the inverse transform of 3 /sis 3 for s>0 and the inverse transform of 2 /(s+1 )i s2 e−tfors>−1, and so f(t)=3−2e−t,ifs>0. J 13.2.1 Laplace transforms of derivatives and integrals One of the main uses of Laplace transforms is in solving diﬀerential equations. Diﬀerential equations are the subject of the next six chapters and we will returnto the application of Laplace transforms to their solution in chapter 15. Inthe meantime we will derive the required results, i.e. the Laplace transforms ofderivatives. The Laplace transform of the ﬁrst derivative of f(t)i sg i v e nb yLbracketleftbiggdf dtbracketrightbigg =integraldisplay∞ 0df dte−stdt =bracketleftbig f(t)e−stbracketrightbig∞ 0+sintegraldisplay∞ 0f(t)e−stdt =−f(0) + s¯f(s),fors>0. (13.57) The evaluation relies on integration by parts and higher-order derivatives may be found in a similar manner. 461 INTEGRAL TRANSFORMSIFind the Laplace transform of d2f/dt2. Using the deﬁnition of the Laplace transf orm and integrating by parts we obtainL /d2f dt2 / = Z∞ 0d2f dt2e−stdt = /df dte−st /∞ 0+s Z∞ 0df dte−stdt =−df dt(0) + s[s¯f(s)−f(0)],fors>0, where (13.57) has been substituted for the integral. This can be written more neatly asL /d2f dt2 / =s2¯f(s)−sf(0)−df dt(0),fors>0. J In general the Laplace transform of the nth derivative is given byLbracketleftbiggdnf dtnbracketrightbigg =sn¯f−sn−1f(0)−sn−2df dt(0)−···−dn−1f dtn−1(0),fors>0. (13.58) We now turn to integration, which is much more straightforward. From the deﬁnition (13.53),Lbracketleftbiggintegraldisplayt 0f(u)dubracketrightbigg =integraldisplay∞ 0dt e−stintegraldisplayt 0f(u)du =bracketleftbigg −1 se−stintegraldisplayt 0f(u)dubracketrightbigg∞ 0+integraldisplay∞ 01 se−stf(t)dt. The ﬁrst term on the RHS vanishes at both limits, and soLbracketleftbiggintegraldisplayt 0f(u)dubracketrightbigg =1 s L[f]. (13.59) 13.2.2 Other properties of Laplace transforms From table 13.1 it will be apparent that multiplying a function f(t)b yeathas the eﬀect on its transform that sis replaced by s−a. This is easily proved generally:Lbracketleftbig eatf(t)bracketrightbig =integraldisplay∞ 0f(t)eate−stdt =integraldisplay∞ 0f(t)e−(s−a)tdt =¯f(s−a). (13.60) As it were, multiplying f(t)b yeatmoves the origin of sby an amount a. 462 13.2 LAPLACE TRANSFORMS We may now consider the eﬀect of multiplying the Laplace transform ¯f(s)b y e−bs(b>0). From the deﬁnition (13.53), e−bs¯f(s)=integraldisplay∞ 0e−s(t+b)f(t)dt =integraldisplay∞ 0e−szf(z−b)dz, on putting t+b=z. Thus e−bs¯f(s) is the Laplace transform of a function g(t) deﬁned by g(t)=braceleftBigg 0f o r 0 <t≤b, f(t−b)f o r t>b . In other words, the function fhas been translated to ‘later’ t(larger values of t) by an amount b. Further properties of Laplace transforms can be proved in similar ways and are listed below. (i) L[f(at)]=1 a¯fparenleftBigs aparenrightBig , (13.61) (ii) L[tnf(t)]=(−1)ndn¯f(s) dsn,forn=1,2,3,..., (13.62) (iii) Lbracketleftbiggf(t) tbracketrightbigg =integraldisplay∞ s¯f(u)du, (13.63) provided lim t→0[f(t)/t] exists. Related results may be easily proved.IFind an expression for the Laplace transform of td2f/dt2. From the deﬁnition of the Laplace transform we haveL / td2f dt2 / = Z∞ 0e−sttd2f dt2dt =−d ds Z∞ 0e−std2f dt2dt =−d ds[s2¯f(s)−sf(0)−f/prime(0)] =−s2d¯f ds−2s¯f+f(0). J Finally we mention the convolution theorem for Laplace transforms (which is analogous to that for Fourier transforms discussed in subsection 13.1.7). If thefunctions fandghave Laplace transforms ¯f(s)a n d ¯g(s)t h e nLbracketleftbiggintegraldisplayt 0f(u)g(t−u)dubracketrightbigg =¯f(s)¯g(s), (13.64) 463 INTEGRAL TRANSFORMS t tt=u u=t u u(a) (b) Figure 13.7 Two representations of the Laplace transform convolution (see text). where the integral in the brackets on the LHS is the convolution offandg, denoted by f∗g. As in the case of Fourier transforms, the convolution deﬁned above is commutative, i.e. f∗g=g∗f, and is associative and distributive. From (13.64) we also see thatL−1bracketleftbig¯f(s)¯g(s)bracketrightbig =integraldisplayt 0f(u)g(t−u)du=f∗g.IProve the convolution theorem (13.64) for Laplace transforms. From the deﬁnition (13.64), ¯f(s)¯g(s)= Z∞ 0e−suf(u)du Z∞ 0e−svg(v)dv = Z∞ 0du Z∞ 0dv e−s(u+v)f(u)g(v). Now letting u+v=tchanges the limits on the integrals, with the result that ¯f(s)¯g(s)= Z∞ 0du f(u) Z∞ udt g(t−u)e−st. As shown in ﬁgure 13.7( a) the shaded area of integration may be considered as the sum of vertical strips. However, we may instead integrate over this area by summing overhorizontal strips as shown in ﬁgure 13.7( b). Then the integral can be written as ¯f(s)¯g(s)= Zt 0du f(u) Z∞ 0dt g(t−u)e−st = Z∞ 0dt e−st /Zt 0f(u)g(t−u)du / = L /Zt 0f(u)g(t−u)du / . J 464 13.3 CONCLUDING REMARKS The properties of the Laplace transform derived in this section can sometimes be useful in ﬁnding the Laplace transforms of particular functions.IFind the Laplace transform of f(t)=tsinbt. Although we could calculate the Laplace transform directly, we can use (13.62) to give ¯f(s)=(−1)d ds L[sinbt]=−d ds /b s2+b2 / =2bs (s2+b2)2,fors>0. J 13.3 Concluding remarks In this chapter we have discussed Fourier and Laplace transforms in some detail. Both are examples of integral transforms , which can be considered in a more general context. A general integral transform of a function f(t)t a k e st h ef o r m F(α)=integraldisplayb aK(α, t)f(t)dt, (13.65) where F(α)i st h et r a n s f o r mo f f(t) with respect to the kernel K(α, t), and αis the transform variable. For example, in the Laplace transform case K(s, t)=e−st, a=0 , b=∞. Very often the inverse transform can also be written straightforwardly and we obtain a transform pair similar to that encountered in Fourier transforms.Examples of such pairs are (i) the Hankel transform F(k)=integraldisplay ∞ 0f(x)Jn(kx)xd x , f(x)=integraldisplay∞ 0F(k)Jn(kx)kd k , where the Jnare Bessel functions of order n,a n d (ii) the Mellin transform F(z)=integraldisplay∞ 0tz−1f(t)dt, f(t)=1 2πiintegraldisplayi∞ −i∞t−zF(z)dz. Although we do not have the space to discuss their general properties, the reader should at least be aware of this wider class of integral transforms. 465 INTEGRAL TRANSFORMS 13.4 Exercises 13.1 Find the Fourier transform of the function f(t)=e x p (−|t|). (a) By applying Fourier’s inversion theorem prove that π 2exp(−|t|)= Z∞ 0cosωt 1+ω2dω. (b) By making the substitution ω=t a n θ, demonstrate the validity of Parseval’s theorem for this function. 13.2 Use the general deﬁnition and properties of Fourier transforms to show the following. (a) If f(x) is periodic with period athen˜f(k) = 0 unless ka=2πnfor integer n. (b) The Fourier transform of tf(t)i sid˜f(ω)/dω. (c) The Fourier transform of f(mt+c)i s eiωc/m m˜f /ω m / . 13.3 Find the Fourier transform of H(x−a)e−bx,w h e r e H(x) is the Heaviside function. 13.4 Prove that the Fourier transform of the function f(t) deﬁned in the tf-plane by straight-line segments joining ( −T,0) to (0 ,1) to ( T,0), with f(t) = 0 outside |t|<T,i s ˜f(ω)=T√ 2πsinc2 /ωT 2 / , where sinc xis deﬁned as (sin x)/x. Use the general properties of Fourier transforms to determine the transforms of the following functions, graphically deﬁned by straight-line segments and equalto zero outside the ranges speciﬁed: (a) (0 ,0) to (0 .5,1) to (1 ,0) to (2 ,2) to (3 ,0) to (4 .5,3) to (6 ,0); (b) (−2,0) to (−1,2) to (1 ,2) to (2 ,0); (c) (0 ,0) to (0 ,1) to (1 ,2) to (1 ,0) to (2 ,−1) to (2 ,0). 13.5 By taking the Fourier transform of the equation d 2φ dx2−K2φ=f(x) show that its solution φ(x) can be written as φ(x)=−1√ 2π Z∞ −∞eikxef(k) k2+K2dk, where ef(k) is the Fourier transform of f(x). 13.6 By diﬀerentiating the deﬁnition of the Fourier sine transform ˜fs(ω) of the function f(t)=t−1/2with respect to ω, and then integrating the resulting expression by parts, ﬁnd an elementary diﬀerential equation satisﬁed by ˜fs(ω). Hence show that this function is its own Fourier sine transform, i.e. ˜fs(ω)=Af(ω), where Ais a constant. Show that it is also its own Fourier cosine transform. (Assume that thelimit as x→∞ofx 1/2sinαxcan be taken as zero.) 13.7 (a) Find the Fourier transform of the unit rectangular distribution f(t)= /( 1|t|<1 0 otherwise . 466 13.4 EXERCISES (b) Determine the convolution of fwith itself and, without further integration, deduce its transform. (c) Deduce thatZ∞ −∞sin2ω ω2dω=π,Z∞ −∞sin4ω ω4dω=2π 3. 13.8 Calculate the Fraunhofer spectrum produced by a diﬀraction grating, uniformly illuminated by light of wavelength 2 π/k, as follows. Consider a grating with 4 N equal strips each of width aand alternately opaque and transparent. The aperture function is then f(y)= /( Afor (2 n+1 )a≤y≤(2n+2 )a,−N≤n<N , 0 otherwise. (a) Show, for diﬀraction at angle θto the normal to the grating, that the required Fourier transform can be writtenef(q)=( 2 π)−1/2N−1X r=−Nexp(−2iarq) Z2a aAexp(−iqu)du, where q=ksinθ. (b) Evaluate the integral and sum to show thatef(q)=( 2 π)−1/2exp(−iqa/2)Asin(2qaN) qcos(qa/2), and hence that the intensity distribution I(θ) in the spectrum is proportional to sin2(2qaN) q2cos2(qa/2). (c) For large values of N, the numerator in the above expression has very closely spaced maxima and minima as a function of θand eﬀectively takes its mean value, 1 /2, giving a low-intensity background. Much more signiﬁcant peaks inI(θ) occur when θ= 0 or the cosine term in the denominator vanishes. Show that the corresponding values of | ef(q)|are 2aNA (2π)1/2and4aNA (2π)1/2(2m+1 )πwith mintegral . Note that the constructive interference makes the maxima in I(θ)∝N2, not N. Of course, observable maxima only occur for 0 ≤θ≤π/2. 13.9 By ﬁnding the complex Fourier series for its LHS show that either side of the equation ∞X n=−∞δ(t+nT)=1 T∞X n=−∞e−2πnit/T can represent a periodic train of impulses. By expressing the function f(t+nX), in which Xis a constant, in terms of the Fourier transform ˜f(ω)o ff(t), show that ∞X n=−∞f(t+nX)=√ 2π X∞X n=−∞˜f /2nπ X / e2πnit/X, This result is known as the Poisson summation formula . 467 INTEGRAL TRANSFORMS 13.10 In many applications in which the frequency spectrum of an analogue signal is required, the best that can be done is to sample the signal f(t) a ﬁnite number of times at ﬁxed intervals and then use a discrete Fourier transform Fkto estimate discrete points on the (true) frequency spectrum ˜f(ω). (a) By an argument that is essentially the converse of that given in section 13.1, show that, if Nsamples fn, beginning at t= 0 and spaced τapart, are taken, then˜f(2πk/(Nτ))≈Fkτwhere Fk=1√ 2πN−1X n=0fne−2πnki/N. (b) For the function f(t) deﬁned by f(t)= /( 1f o r 0≤t<1 0 otherwise, from which eight samples are drawn at intervals of τ=0.25, ﬁnd a formula for|Fk|and evaluate it for k=0,1,...,7. (c) Find the exact frequency spectrum of f(t) and compare the actual and estimated values of√ 2π|˜f(ω)|atω=kπfork=0,1,...,7.Note the relatively good agreement for k<4 and the lack of agreement for larger values of k. 13.11 For a function f(t) that is non-zero only in the range |t|<T/2, the full frequency spectrum ˜f(ω) can be constructed, in principle exactly, from values at discrete sample points ω=n(2π/T) .P r o v et h i sa sf o l l o w s . (a) Show that the coeﬃcients of a complex Fourier series representation of f(t) with period Tcan be written as cn=√ 2π T˜f /2πn T / . (b) Use this result to represent f(t) as an inﬁnite sum in the deﬁning integral for ˜f(ω), and hence show that ˜f(ω)=∞X n=−∞˜f /2πn T / sinc / nπ−ωT 2 / , where sinc xis deﬁned as (sin x)/x. 13.12 A signal obtained by sampling a function x(t) at regular intervals Tis passed through an electronic ﬁlter, whose response g(t) to a unit δ-function input is represented in a tg-plot by straight lines joining (0 ,0) to ( T,1/T)t o( 2 T,0) and is zero for all other values of t. The output of the ﬁlter is the convolution of the input, P∞ −∞x(t)δ(t−nT), with g(t). Using the convolution theorem, and the result given in exercise 13.4, show thatthe output of the ﬁlter can be written y(t)=1 2π∞X n=−∞x(nT) Z∞ −∞sinc2 /ωT 2 / e−iω[(n+1)T−t]dω. 13.13 (a) Find the Fourier transform of f(γ,p,t)= /( e−γtsinpt t > 0 0 t<0, where γ(>0) and pare constant parameters. 468 13.4 EXERCISES (b) The current I(t) ﬂowing through a certain system is related to the applied voltage V(t) by the equation I(t)= Z∞ −∞K(t−u)V(u)du, where K(τ)=a1f(γ1,p1,τ)+a2f(γ2,p2,τ). The function f(γ,p,t) is as given in (a) and all the ai,γi(>0) and piare ﬁxed parameters. By considering the Fourier transform of I(t), ﬁnd the relationship that must hold between a1anda2if the total net charge Qpassed through the system (over a very long time) is to be zero for an arbitrary appliedvoltage. 13.14 Prove the equalityZ∞ 0e−2atsin2at dt=1 π Z∞ 0a2 4a4+ω4dω. 13.15 A linear ampliﬁer produces an output that is the convolution of its input and its response function. The Fourier transform of the response function for a particularampliﬁer is ˜K(ω)=iω √ 2π(α+iω)2. Determine the time variation of its output g(t) when its input is the Heaviside step function. (Consider the Fourier transform of a decaying exponential function and the result of exercise 13.2(b).) 13.16 In quantum mechanics, two equal-mass particles having momenta pj= /~kjand energies Ej= /~ωjand represented by plane wavefunctions φj=e x p [ i(kj·rj−ωjt)], j=1,2, interact through a potential V=V(|r1−r2|). In ﬁrst-order perturbation theory the probability of scattering to a state with momenta and energies p/prime j,E/prime j is determined by the modulus squared of the quantity M= ZZ Z ψ∗ fVψ idr1dr2dt. The initial state ψiisφ1φ2and the ﬁnal state ψfisφ/prime 1φ/prime2. (a) By writing r1+r2=2Randr1−r2=rand assuming that dr1dr2=dRdr, show that Mcan be written as the product of three one-dimensional integrals. (b) From two of the integrals deduce energy and momentum conservation in the form of δ-functions. (c) Show that Mis proportional to the Fourier transform of V,i . e . eV(k)w h e r e 2 /~k=(p2−p1)−(p/prime 2−p/prime 1). 13.17 For some ion–atom scattering processes, the potential Vof the previous example may be approximated by V=|r1−r2|−1exp(−µ|r1−r2|). Show, using the result of the worked example in subsection 13.1.10, that the probability that the ionwill scatter from, say, p 1top/prime 1is proportional to ( µ2+k2)−2where k=|k|andk is as given in part (c) of exercise 13.16. 13.18 The equivalent duration and bandwidth, TeandBe, of a signal x(t) are deﬁned in terms of the latter and its Fourier transform ˜x(ω): Te=1 x(0) Z∞ −∞x(t)dt, Be=1 ˜x(0) Z∞ −∞˜x(ω)dω, 469 INTEGRAL TRANSFORMS where neither x(0) nor ˜x(0) is zero. Show that the product TeBe=2π(this is a form of uncertainty principle), and ﬁnd the equivalent bandwidth of the signal x(t)=e x p (−|t|/T). For this signal, determine the fraction of the total energy that lies in the frequency range|ω|<B e/4. You will need the indeﬁnite integral with respect to xof (a2+x2)−2,w h i c hi s x 2a2(a2+x2)+1 2a3tan−1x a. 13.19 Calculate directly the auto-correlation function a(z) for the product of the expo- nential decay distribution and the Heaviside step function f(t)=1 λe−λtH(t). Use the Fourier transform and energy spectrum of f(t) to deduce thatZ∞ −∞eiωz λ2+ω2dω=π λe−λ|z|. 13.20 Prove that the cross-correlation C(z) of the Gaussian and Lorentzian distributions f(t)=1 τ√ 2πexp / −t2 2τ2 / ,g (t)= /a π /1 t2+a2, has as its Fourier transform the function 1√ 2πexp / −τ2ω2 2 / exp(−a|ω|). Hence show that C(z)=1 τ√ 2πexp /a2−z2 2τ2 / cos /az τ2 / . 13.21 Prove the expressions given in table 13.1 for the Laplace transforms of t−1/2and t1/2, by setting x2=tsin the resultZ∞ 0exp(−x2)dx=1 2√π. 13.22 Find the functions y(t) whose Laplace transforms are the following, (a) 1 /(s2−s−2), (b) 2 s/[(s+1 ) ( s2+4 ) ] , (c)e−(γ+s)t0/[(s+γ)2+b2]. 13.23 Use the properties of Laplace transforms to prove the following without evaluat- ing any Laplace integrals explicitly: (a) L / t5/2 / =15 8√πs−7/2. (b) L / (sinh at)/t / =1 2ln / (s+a)/(s−a) / ,s >|a|. (c) L[sinhatcosbt]=a(s2−a2+b2)[(s−a)2+b2]−1[(s+a)2+b2]−1. 13.24 Find the solution (the so-called impulse response orGreen’s function )o ft h e equation Tdx dt+x=δ(t) by proceeding as follows. 470 13.4 EXERCISES (a) Show by substitution that x(t)=A(1−e−t/T)H(t) is a solution, for which x(0) = 0, of Tdx dt+x=AH(t), (*) where H(t) is the Heaviside step function. (b) Construct the solution when the RHS of (*) is replaced by AH(t−τ)w i t h dx/dt =x=0f o r t<τ, and hence ﬁnd the solution when the RHS is a rectangular pulse of duration τ. (c) By setting A=1/τand taking the limit when τ→0, show that the impulse response is x(t)=T−1e−t/T. (d) Obtain the same result much more directly by taking the Laplace transform of each term in the original equation, solving the resulting algebraic equationand then using the entries in table 13.1. 13.25 (a) If f(t)=A+g(t), where Ais a constant and the indeﬁnite integral of g(t)i s bounded as its upper limit tends to ∞, show that lim s→0s¯f(s)=A. (b) For t>0 the function y(t) obeys the diﬀerential equation d2y dt2+ady dt+by=ccos2ωt, where a,bandcare positive constants. Find ¯y(s)a n ds h o wt h a t s¯y(s)→c/2b ass→0. Interpret the result in the t-domain. 13.26 By writing f(x) as an integral involving the δ-function δ(ξ−x) and taking the Laplace transforms of both sides, show that the transform of the solution of the equation d4y dx4−y=f(x) for which yand its ﬁrst three derivatives vanish at x= 0 can be written as ¯y(s)= Z∞ 0f(ξ)e−sξ s4−1dξ. Use the properties of Laplace transforms and the entries in table 13.1 to show that y(x)=1 2 Zx 0f(ξ)[sinh(x−ξ)−sin(x−ξ)]dξ. 13.27 The function fa(x) is deﬁned as unity for 0 <x<a and zero otherwise. Find its Laplace transform ¯fa(s) and deduce that the transform of xfa(x)i s 1 s2 / 1−(1 +as)e−sa / . Write fa(x) in terms of Heaviside functions and hence obtain an explicit expres- sion for ga(x)= Zx 0fa(y)fa(x−y)dy. Use the expression to write ¯ga(s) in terms of the functions ¯fa(s), and ¯f2a(s)a n d their derivatives, and hence show that ¯ga(s) is equal to the square of ¯fa(s), in accordance with the convolution theorem. 471 INTEGRAL TRANSFORMS 13.28 (a) Show that the Laplace transform of f(t−a)H(t−a), where a≥0, ise−as¯f(s). (b) If g(t) is a periodic function of period T, show that ¯g(s) can be written as 1 1−e−sT ZT 0e−stg(t)dt. (c) Sketch the periodic function deﬁned in 0 ≤t≤Tby g(t)= /( 2t/T 0≤t<T/ 2 2(1−t/T)T/2≤t≤T, and, using the result in (b), ﬁnd its Laplace transform. (d) Show, by sketching it, that 2 T[tH(t)+2∞X n=1(−1)n(t−1 2nT)H(t−1 2nT)] is another representation of g(t) and hence derive the relationship tanh x=1+2∞X n=1(−1)ne−2nx. 13.5 Hints and answers 13.1 (2 /π)1/2(1 +ω2)−1. 13.2 (a) Show ˜f(k)(1−e±ika)=0 . 13.3 (1 /√ 2π)[(b−ik)/(b2+k2)]e−a(b+ik). 13.4 (a) [8 /(√ 2πω2)][e−iω/2sin2(ω/4) +e−i2ωsin2(ω/2) +e−i9ω/2sin2(3ω/4)]. (b) Consider the superposition of a ‘triangle’ of height 2 with T=2a n dt w o triangles, each of unit height with T= 1, displaced by ±1; [8 sin2(ω/2) (1 + 2cos ω)]/(√ 2πω2). (c) Consider the superposition of a triangle and its derivative. [(1 + iω)e−iω/√ 2π]sinc2(ω/2). 13.6 d˜fs(ω)/dω=−˜fs(ω)/(2ω). 13.7 (a) (2 /√ 2π)(sinω/ω). (b) 2−|t|for|t|<2, zero otherwise. Use convolution theorem; (4 /√ 2π)(sin2ω/ω2). (c) Apply Parseval’s theorem to fand to f∗f. 13.8 (c) Use l’H ˆopital’s rule to evaluate the expressions of the form 0 /0. 13.10 (b) |Fk|=c o s e c( kπ/8)/√ 2πforkodd;|Fk|=0f o r keven, except |F0|=4/√ 2π. (c)√ 2π˜f(ω)=e−iω/2[sin(ω/2)/(ω/2)]. Actual (estimated) values at ω=kπfor k=0,1,...,7: 1( 1 ) ;0 .637 (0 .653); 0 (0); 0 .212 (0 .271); 0 (0); 0 .127 (0 .271); 0 (0); 0 .091 (0 .653). 13.11 (b) Recall that the inﬁnite integral involved in deﬁning ˜f(ω)o n l yh a san o n - z e r o integrand in |t|<T/2. 13.12 The Fourier transform of g(t) is found by moving the time origin by Tand then applying (13.31). It is (1 /√ 2π)si nc2(ωT/2)e−iωT. 13.13 (a) (1 /√ 2π){p/[(γ+iω)2+p2]}. (b) Show that Q=√ 2π˜I(0) and use the convolution theorem. The required relationship is a1p1/(γ2 1+p2 1)+a2p2/(γ2 2+p2 2)=0 . 13.14 Set p=γ=ain part (a) of exercise 13.13 and then apply Parseval’s theorem. 13.15 ˜g(ω)=1 /[√ 2π(α+iω)2], leading to g(t)=te−αt. 13.16 (b) The t-integral is R exp[i(E/prime 1+E/prime 2−E1−E2)]dt∝δ(E/prime 1+E/prime 2−(E1+E2)); similarly the R-integral yields δ(p/prime 1+p/prime 2−(p1+p2)). 472 13.5 HINTS AND ANSWERS 13.17 eV(k)∝[−2π/(ik)] R {exp[−(µ−ik)r]−exp[−(µ+ik)r]}dr. 13.18 By setting t=0a n d ω= 0 in the Fourier deﬁnitions, obtain two equations connecting x(0) and ˜x(0).Be=π/T;˜x(ω), proportional to the Fourier cosine transform of exp( −t/T), is equal to [2 T/√ 2π](1+ ω2T2)−1. The energy spectrum is proportional to |˜x(ω)|2. Fraction = 0.733. 13.19 Note that the lower limit in the calculation of a(z)i s0f o r z>0a n d|z|for z<0. Auto-correlation a(z)=[ ( 1 /(2λ3)]exp(−λ|z|). 13.20 Use the result of exercise 13.18 to deduce that ˜g(ω)=( 1 /√ 2π)e x p (−a|ω|). Apply the Wiener–Kinchin theorem. Note that, because of the presence of |ω|,t h e inverse transform giving C(z) is a cosine transform. 13.21 Prove the result for t1/2by integrating that for t−1/2by parts. 13.22 (a) y(t)=1 3(e2t−e−t). (b)y(t)=1 5(4 sin2 t+2c o s2 t−2e−t). (c) Note the factor e−st0and write y(t) as a function of ( t−t0);y(t)= b−1e−γtsinb(t−t0)H(t−t0). 13.23 (a) Use (13.62) with n=2o n L /√ t / ; (b) use (13.63); (c) consider L[exp(±at)cosbt]and use the translation property, subsection 13.2.2. 13.24 (b) Superimpose solutions with equal amplitudes but opposite signs. x(t)= A(1−e−t/T)H(t)−A(1−e−(t−τ)/T)H(t−τ). (c) Write e−(t−τ)/Tase−t/T[1 +τ/T+O(τ2)] and note that, with 0 <t<τ ,τ (1− e−t/T)/τ→0a sτ→0. (d) The algebraic equation is ¯x=( 1+ sT)−1. 13.25 (a) Note that |lim R g(t)e−stdt|≤|lim R g(t)dt|. (b) (s2+as+b)¯y(s)={c(s2+2ω2)/[s(s2+4ω2)]}+(a+s)y(0) + y/prime(0). For this damped system, at large t(corresponding to s→0) rates of change are negligible and the equation reduces to by=ccos2ωt,w i t hc o s2ωthaving an average value1 2. 13.26 Factorise ( s4−1)−1as1 2[(s2−1)−1−(s2+1 )−1]. 13.27 s−1[1−exp(−sa)];ga(x)=xfor 0 <x<a ,ga(x)=2 a−xfora≤x≤2a, ga(x) = 0 otherwise. 13.28 (a) Note that R∞ TH(t)···= R∞ 0H(t−T)···and that H(t−T)g(t)=H(t− T)g(t−T). (c)¯g(s)=[ 2 /(Ts2)] tanh( sT/4). (d) Use the result from (a) and L[tH(t)] =s−2;s e t sT=4x. 473 14 First-order ordinary diﬀerential equations Diﬀerential equations are the group of equations that contain derivatives. Chap- ters 14–19 discuss a variety of diﬀerential equations, starting in this chapter andthe next with those ordinary diﬀerential equations (ODEs) that have closed-formsolutions. As its name suggests, an ODE contains only ordinary derivatives (andnot partial derivatives) and describes the relationship between these derivatives ofthedependent variable , usually called y, with respect to the independent variable , usually called x. The solution to such an ODE is therefore a function of xand is written y(x). For an ODE to have a closed-form solution, it must be possible to express y(x) in terms of the standard elementary functions such as exp x,l nx, sinxetc. The solutions of some diﬀerential equations cannot, however, be written in closed form, but only as an inﬁnite series; these are discussed in chapter 16. Ordinary diﬀerential equations may be separated conveniently into diﬀer- ent categories according to their general characteristics. The primary groupingadopted here is by the order of the equation. The order of an ODE is simply the order of the highest derivative it contains. Thus equations containing dy/dx , but no higher derivatives, are called ﬁrst order, those containing d 2y/dx2are called second order and so on. In this chapter we consider ﬁrst-order equations, and inthe next, second- and higher-order equations. Ordinary diﬀerential equations may be classiﬁed further according to degree . The degree of an ODE is the power to which the highest-order derivative israised, after the equation has been rationalised to contain only integer powers ofderivatives. Hence the ODE d 3y dx3+xparenleftbiggdy dxparenrightbigg3/2 +x2y=0, is of third order and second degree, since after rationalisation it contains the term (d3y/dx3)2. Thegeneral solution to an ODE is the most general function y(x) that satisﬁes the equation; it will contain constants of integration which may be determined by 474 14.1 GENERAL FORM OF SOLUTION the application of some suitable boundary conditions . For example, we may be told that for a certain ﬁrst-order diﬀerential equation, the solution y(x)i se q u a lt o zero when the parameter xis equal to unity; this allows us to determine the value of the constant of integration. The general solutions tonth-order ODEs, which are considered in detail in the next chapter, will contain n(essential) arbitrary constants of integration and therefore we will need nboundary conditions if these constants are to be determined (see section 14.1). When the boundary conditionshave been applied, and the constants found, we are left with a particular solution to the ODE, which obeys the given boundary conditions. Some ODEs of degreegreater than unity also possess singular solutions , which are solutions that contain no arbitrary constants and cannot be found from the general solution; singular solutions are discussed in more detail in section 14.3. When any solution to an ODE has been found, it is always possible to check its validity by substitutioninto the original equation and veriﬁcation that any given boundary conditionsare met. In this chapter, ﬁrstly we discuss various types of ﬁrst-degree ODE, and then go on to examine those higher-degree equations that can be solved in closed form.At the outset, however, we discuss the general form of the solutions of ODEs; this discussion is relevant to both ﬁrst- and higher-order ODEs. 14.1 General form of solution It is helpful when considering the general form of the solution of an ODE to consider the inverse process, namely that of obtaining an ODE from a givengroup of functions, each one of which is a solution of the ODE. Suppose the members of the group can be written as y=f(x, a 1,a2,...,a n), (14.1) each member being speciﬁed by a diﬀerent set of values of the parameters ai.F o r example, consider the group of functions y=a1sinx+a2cosx; (14.2) here n=2 . Since an ODE is required for which anyof the group is a solution, it clearly must not contain any of the ai.A st h e r ea r e nof the aiin expression (14.1), we must obtain n+ 1 equations involving them in order that, by elimination, we can obtain one ﬁnal equation without them. Initially we have only (14.1), but if this is diﬀerentiated ntimes, a total of n+1 equations is obtained from which (in principle) all the aican be eliminated, to give one ODE satisﬁed by all the group. As a result of the ndiﬀerentiations, dny/dxnwill be present in one of the n+ 1 equations and hence in the ﬁnal equation, which will therefore be of nth order. 475 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS In the case of (14.2), we have dy dx=a1cosx−a2sinx, d2y dx2=−a1sinx−a2cosx. Here the elimination of a1anda2is trivial (because of the similarity of the forms ofyandd2y/dx2), resulting in d2y dx2+y=0, a second-order equation. Thus, to summarise, a group of functions (14.1) with nparameters satisﬁes an nth-order ODE in general (although in some degenerate cases an ODE of less than nth order is obtained). The intuitive converse of this is that the general solution of an nth-order ODE contains narbitrary parameters (constants); for our purposes, this will be assumed to be valid although a totally general proof isdiﬃcult. As mentioned earlier, external factors aﬀect a system described by an ODE, by ﬁxing the values of the dependent variables for particular values of theindependent ones. These externally imposed (or boundary ) conditions on the solution are thus the means of determining the parameters and so of specifying precisely which function is the required solution. It is apparent that the numberof boundary conditions should match the number of parameters and hence theorder of the equation, if a unique solution is to be obtained. Fewer independentboundary conditions than this will lead to a number of undetermined parametersin the solution, whilst an excess will usually mean that no acceptable solution ispossible. For an nth-order equation the required nboundary conditions can take many forms, for example the value of yatndiﬀerent values of x, or the value of any n−1o ft h e nderivatives dy/dx ,d 2y/dx2,...,dny/dxntogether with that of y,a l l for the same value of x, or many intermediate combinations. 14.2 First-degree ﬁrst-order equations First-degree ﬁrst-order ODEs contain only dy/dx equated to some function of x andy, and can be written in either of two equivalent standard forms, dy dx=F(x, y),A(x, y)dx+B(x, y)dy=0, where F(x, y)=−A(x, y)/B(x, y), and F(x, y),A(x, y)a n d B(x, y) are in general functions of both xandy. Which of the two above forms is the more useful for ﬁnding a solution depends on the type of equation being considered. There 476 14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS are several diﬀerent types of ﬁrst-degree ﬁrst-order ODEs that are of interest in the physical sciences. These equations and their respective solutions are discussedbelow. 14.2.1 Separable-variable equations A separable-variable equation is one which may be written in the conventional form dy dx=f(x)g(y), (14.3) where f(x)a n d g(y) are functions of xandyrespectively, including cases in which f(x)o rg(y) is simply a constant. Rearranging this equation so that the terms depending on xand on yappear on opposite sides (i.e. are separated), and integrating, we obtain integraldisplaydy g(y)=integraldisplay f(x)dx. Finding the solution y(x) that satisﬁes (14.3) then depends only on the ease with which the integrals in the above equation can be evaluated. It is also worth noting that ODEs that at ﬁrst sight do not appear to be of the form (14.3) can sometimes be made separable by an appropriate factorisation.ISolve dy dx=x+xy. Since the RHS of this equation can be factorised to give x(1 +y), the equation becomes separable and we obtainZdy 1+y= Z xd x . Now integrating both sides separately, we ﬁnd ln(1 + y)=x2 2+c, and so 1+y=e x p /x2 2+c / =Aexp /x2 2 / , where cand hence Ais an arbitrary constant. J Solution method. Factorise the equation so that it becomes separable. After rear- ranging it so that the terms depending on xand those depending on yappear on opposite sides, integrate directly. Remember the constant of integration, which canbe evaluated if further information is given. 477 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS 14.2.2 Exact equations Anexact ﬁrst-degree ﬁrst-order ODE is one of the form A(x, y)dx+B(x, y)dy= 0 and for which∂A ∂y=∂B ∂x. (14.4) In this case A(x, y)dx+B(x, y)dyis an exact diﬀerential, dU(x, y)s a y( s e e section 5.3). In other words Ad x+Bd y=dU=∂U ∂xdx+∂U ∂ydy, from which we obtain A(x, y)=∂U ∂x, (14.5) B(x, y)=∂U ∂y. (14.6) Since ∂2U/∂x∂y =∂2U/∂y∂x we therefore require ∂A ∂y=∂B ∂x. (14.7) If (14.7) holds then (14.4) can be written dU(x, y) = 0, which has the solution U(x, y)=c,w h e r e cis a constant and from (14.5) U(x, y)i sg i v e nb y U(x, y)=integraldisplay A(x, y)dx+F(y). (14.8) The function F(y) can be found from (14.6) by diﬀerentiating (14.8) with respect toyand equating to B(x, y).ISolve xdy dx+3x+y=0. Rearranging into the form (14.4) we have (3x+y)dx+xd y=0, i.e.A(x, y)=3 x+yandB(x, y)=x.S i n c e ∂A/∂y =1= ∂B/∂x , the equation is exact, and by (14.8) the solution is given by U(x, y)= Z (3x+y)dx+F(y)=c1⇒3x2 2+yx+F(y)=c1. Diﬀerentiating U(x, y) with respect to yand equating it to B(x, y)=xwe obtain dF/dy =0 , which integrates immediately to give F(y)=c2. Therefore, letting c=c1−c2,t h es o l u t i o n to the original ODE is 3x2 2+xy=c. J 478 14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS Solution method. Check that the equation is an exact diﬀerential using (14.7) then solve using (14.8). Find the function F(y)by diﬀerentiating (14.8) with respect to yand using (14.6). 14.2.3 Inexact equations: integrating factors Equations that may be written in the form A(x, y)dx+B(x, y)dy= 0 but for which∂A ∂y/negationslash=∂B ∂x(14.9) are known as inexact equations. However, the diﬀerential Ad x+Bd ycan always be made exact by multiplying by an integrating factor µ(x, y), which obeys ∂(µA) ∂y=∂(µB) ∂x. (14.10) For an integrating factor that is a function of both xandy,i . e .µ=µ(x, y), there exists no general method for ﬁnding it; in such cases it may sometimes be foundby inspection. If, however, an integrating factor exists that is a function of either xoryalone then (14.10) can be solved to ﬁnd it. For example, if we assume that the integrating factor is a function of xalone, i.e. µ=µ(x), then (14.10) reads µ∂A ∂y=µ∂B ∂x+Bdµ dx. Rearranging this expression we ﬁnd dµ µ=1 Bparenleftbigg∂A ∂y−∂B ∂xparenrightbigg dx=f(x)dx, where we require f(x) also to be a function of xonly; indeed this provides a general method of determining whether the integrating factor µis a function of xalone. This integrating factor is then given by µ(x)=e x pbraceleftbiggintegraldisplay f(x)dxbracerightbigg where f(x)=1 Bparenleftbigg∂A ∂y−∂B ∂xparenrightbigg . (14.11) Similarly, if µ=µ(y)t h e n µ(y)=e x pbraceleftbiggintegraldisplay g(y)dybracerightbigg where g(y)=1 Aparenleftbigg∂B ∂x−∂A ∂yparenrightbigg . (14.12) 479 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve dy dx=−2 y−3y 2x. Rearranging into the form (14.9), we have (4x+3y2)dx+2xy dy=0, (14.13) i.e.A(x, y)=4 x+3y2andB(x, y)=2 xy.N o w ∂A ∂y=6y,∂B ∂x=2y, so the ODE is not exact in its present form. However, we see that 1 B /∂A ∂y−∂B ∂x / =2 x, a function of xalone. Therefore an integrating factor exists that is also a function of x alone and, ignoring the arbitrary constant of integration, is given by µ(x)=e x p / 2 Zdx x / =e x p ( 2l n x)=x2. Multiplying (14.13) through by µ(x)=x2we obtain (4x3+3x2y2)dx+2x3yd y=4x3dx+( 3x2y2dx+2x3yd y)=0 . By inspection this integrates immediately to give the solution x4+y2x3=c,w h e r e cis a constant. J Solution method. Examine whether f(x)andg(y)are functions of only xory respectively. If so, then the required integrating factor is a function of either xor yonly, and is given by (14.11) or (14.12) respectively. If the integrating factor is a function of both xandy, then sometimes it may be found by inspection or by trial and error. In any case, the integrating factor µmust satisfy (14.10). Once the equation has been made exact, solve by the method of subsection 14.2.2. 14.2.4 Linear equations Linear ﬁrst-order ODEs are a special case of inexact ODEs (discussed in the previous subsection) and can be written in the conventional form dy dx+P(x)y=Q(x). (14.14) Such equations can be made exact by multiplying through by an appropriate integrating factor in a similar manner to that discussed above. In this case, however, the integrating factor is always a function of xalone and may be expressed in a particularly simple form. An integrating factor µ(x) must be such that µ(x)dy dx+µ(x)P(x)y=d dx[µ(x)y]=µ(x)Q(x), (14.15) 480 14.2 FIRST-DEGREE FIRST-ORDER EQUATIONS which may then be integrated directly to give µ(x)y=integraldisplay µ(x)Q(x)dx. (14.16) The required integrating factor µ(x) is determined by the ﬁrst equality in (14.15), i.e. d dx(µy)=µdy dx+dµ dxy=µdy dx+µPy, which immediately gives the simple relation dµ dx=µ(x)P(x)⇒ µ(x)=e x pbraceleftbiggintegraldisplay P(x)dxbracerightbigg . (14.17)ISolve dy dx+2xy=4x. The integrating factor is given immediately by µ(x)=e x p /Z 2xd x / =e x p x2. Multiplying through the ODE by µ(x)=e x p x2and integrating, we have yexpx2=4 Z xexpx2dx=2e x p x2+c. The solution to the ODE is therefore given by y=2+ cexp(−x2). J Solution method. Rearrange the equation into the form (14.14) and multiply by the integrating factor µ(x)given by (14.17). The left- and right-hand sides can then be integrated directly, giving y from (14.16). 14.2.5 Homogeneous equations Homogeneous equation are ODEs that may be written in the form dy dx=A(x, y) B(x, y)=FparenleftBigy xparenrightBig , (14.18) where A(x, y)a n d B(x, y) are homogeneous functions of the same degree. A function f(x, y) is homogeneous of degree nif, for any λ,i to b e y s f(λx, λy)=λnf(x, y). For example, if A=x2y−xy2andB=x3+y3then we see that AandBare both homogeneous functions of degree 3. In general, for functions of the form of AandB, we see that for both to be homogeneous, and of the same degree, we require the sum of the powers in xandyin each term of AandBto be the same 481 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS (in this example equal to 3). The RHS of a homogeneous ODE can be written as a function of y/x. The equation may then be solved by making the substitution y=vx,s ot h a t dy dx=v+xdv dx=F(v). This is now a separable equation and can be integrated directly to give integraldisplaydv F(v)−v=integraldisplaydx x. (14.19)ISolve dy dx=y x+t a n /y x / . Substituting y=vxwe obtain v+xdv dx=v+t a n v. Cancelling von both sides, rearranging and integrating givesZ cotvd v= Zdx x=l nx+c1. ButZ cotvd v= Zcosv sinvdv=l n ( s i n v)+c2, so the solution to the ODE is y=xsin−1Ax,w h e r e Ais a constant. J Solution method. Check to see whether the equation is homogeneous. If so, make the substitution y=vx, separate variables as in (14.19) and then integrate directly. Finally replace vbyy/xto obtain the solution. 14.2.6 Isobaric equations An isobaric ODE is a generalisation of the homogeneous ODE discussed in the previous section, and is of the form dy dx=A(x, y) B(x, y), (14.20) where the equation is dimensionally consistent if yanddyare each given a weight mrelative to xanddx, i.e. if the substitution y=vxmmakes it separable. 482 14.2 FIRST-DEGREE FIRST-ORDER EQUATIONSISolve dy dx=−1 2yx / y2+2 x / . Rearranging we have/ y2+2 x / dx+2yx dy =0. Giving yanddythe weight mandxanddxthe weight 1, the sums of the powers in each term on the LHS are 2 m+1 ,0a n d2 m+ 1 respectively. These are equal if 2 m+1=0 ,i . e . ifm=−1 2. Substituting y=vxm=vx−1/2, with the result that dy=x−1/2dv−1 2vx−3/2dx, we obtain vd v+dx x=0, which is separable and may be integrated directly to give1 2v2+l nx=c. Replacing vby y√xwe obtain the solution1 2y2x+l nx=c. J Solution method. Write the equation in the form Ad x+Bd y=0. Giving yand dyeach a weight mandxanddxeach a weight 1, write down the sum of powers in each term. Then, if a value of mthat makes all these sums equal can be found, substitute y=vxminto the original equation to make it separable. Integrate the separated equation directly, and then replace vbyyx−mto obtain the solution. 14.2.7 Bernoulli’s equation Bernoulli’s equation has the form dy dx+P(x)y=Q(x)ynwhere n/negationslash=0o r1 . (14.21) This equation is very similar in form to the linear equation (14.14), but is in fact non-linear due to the extra ynfactor on the RHS. However, the equation can be made linear by substituting v=y1−n, and correspondingly dy dx=parenleftbiggyn 1−nparenrightbiggdv dx. Substituting this into (14.21) and dividing through by yn, we ﬁnd dv dx+( 1−n)P(x)v=( 1−n)Q(x), which is a linear equation, and may be solved by the method described in subsection 14.2.4. 483 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve dy dx+y x=2x3y4. If we let v=y1−4=y−3then dy dx=−y4 3dv dx. Substituting this into the ODE and rearranging, we obtain dv dx−3v x=−6x3, which is linear and may be solved by multiplying through by the integrating factor (see subsection 14.2.4) exp / −3 Zdx x / =e x p (−3lnx)=1 x3. This yields the solution v x3=−6x+c. Remembering that v=y−3,w eo b t a i n y−3=−6x4+cx3. J Solution method. Rearrange the equation into the form (14.21) and make the sub- stitution v=y1−n. This leads to a linear equation in v, which can be solved by the method of subsection 14.2.4. Then replace vbyy1−nto obtain the solution. 14.2.8 Miscellaneous equations There are two further types of ﬁrst-degree ﬁrst-order equation that occur fairly regularly but do not fall into any of the above categories. They may be reducedto one of the above equations, however, by a suitable change of variable. Firstly, we consider dy dx=F(ax+by+c), (14.22) where a,bandcare constants, i.e. xandyonlyappear on the RHS in the particular combination ax+by+cand not in any other combination or by themselves. This equation can be solved by making the substitution v=ax+by+c,i nw h i c hc a s e dv dx=a+bdy dx=a+bF(v), (14.23) which is separable and may be integrated directly. 484 14.2 FIRST-DEGREE FIRST-ORDER EQUATIONSISolve dy dx=(x+y+1 )2. Making the substitution v=x+y+ 1, we obtain, as in (14.23), dv dx=v2+1, which is separable and integrates directly to giveZdv 1+v2= Z dx⇒tan−1v=x+c1. So the solution to the original ODE is tan−1(x+y+1 )= x+c1,w h e r e c1is a constant of integration. J Solution method. In an equation such as (14.22), substitute v=ax+by+cto obtain a separable equation that can be integrated directly. Then replace vbyax+by+c to obtain the solution. Secondly, we discuss dy dx=ax+by+c ex+fy+g, (14.24) where a,b,c,e,fandgare all constants. This equation may be solved by letting x=X+αandy=Y+β,w h e r e αandβare constants found from aα+bβ+c= 0 (14.25) eα+fβ+g=0. (14.26) Then (14.24) can be written as dY dX=aX+bY eX+fY, which is homogeneous and can be solved by the method of subsection 14.2.5. Note, however, that if a/e=b/fthen (14.25) and (14.26) are not independent and so cannot be solved uniquely for αandβ. However, in this case, (14.24) reduces to an equation of the form (14.22), which was discussed above.ISolve dy dx=2x−5y+3 2x+4y−6. Letx=X+αandy=Y+β,w h e r e αandβobey the relations 2α−5β+3=0 2α+4β−6=0 , which solve to give α=β= 1. Making these substitutions we ﬁnd dY dX=2X−5Y 2X+4Y, 485 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS which is a homogeneous ODE and can be solved by substituting Y=vX(see subsec- tion 14.2.5) to obtain dv dX=2−7v−4v2 X(2 + 4 v). This equation is separable, and using partial fractions we ﬁndZ2+4 v 2−7v−4v2dv=−4 3 Zdv 4v−1−2 3 Zdv v+2= ZdX X, which integrates to give lnX+1 3ln(4v−1) +2 3ln(v+2 )= c1, or X3(4v−1)(v+2 )2=e x p ( 3 c1). Remembering that Y=vX,x=X+1a n d y=Y+ 1, the solution to the original ODE is given by (4 y−x−3)(y+2x−3)2=c2,w h e r e c2=e x p ( 3 c1). J Solution method. If in (14.24) a/e/negationslash=b/fthen make the substitution x=X+α, y=Y+β,w h e r e αandβare given by (14.25) and (14.26); the resulting equation is homogeneous and can be solved as in subsection 14.2.5. Substitute v=Y/ X, X=x−α, and Y=y−βto obtain the solution. If a/e=b/fthen (14.24) is of the same form as (14.22) and may be solved accordingly. 14.3 Higher-degree ﬁrst-order equations First-order equations of degree higher than the ﬁrst do not occur often in the description of physical systems, since squared and higher powers of ﬁrst- order derivatives usually arise from resistive or driving mechanisms, when anacceleration or other higher-order derivative is also present. They do sometimesappear in connection with geometrical problems, however. Higher-degree ﬁrst-order equations can be written as F(x, y, dy/dx )=0 .T h e most general standard form is p n+an−1(x, y)pn−1+···+a1(x, y)p+a0(x, y)=0 , (14.27) where for ease of notation we write p=dy/dx . If the equation can be solved for one of x,yorpthen either an explicit or a parametric solution can sometimes be obtained. We discuss the main types of such equations below, including Clairaut’sequation, which is a special case of an equation explicitly soluble for y. 14.3.1 Equations soluble for p Sometimes the LHS of (14.27) can be factorised into the form (p−F 1)(p−F2)···(p−Fn)=0 , (14.28) 486 14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONS where Fi=Fi(x, y). We are then left with solving the nﬁrst-degree equations p=Fi(x, y). Writing the solutions to these ﬁrst-degree equations as Gi(x, y)=0 , the general solution to (14.28) is given by the product G1(x, y)G2(x, y)···Gn(x, y)=0 . (14.29)ISolve (x3+x2+x+1 )p2−(3x2+2x+1 )yp+2xy2=0. (14.30) This equation may be factorised to give [(x+1 )p−y][(x2+1 )p−2xy]=0 . Taking each bracket in turn we have (x+1 )dy dx−y=0, (x2+1 )dy dx−2xy=0, which have the solutions y−c(x+1 ) = 0 a n d y−c(x2+ 1) = 0 respectively (see section 14.2 on ﬁrst-degree ﬁrst-order equations). Note that the arbitrary constants inthese two solutions can be taken to be the same, since only one is required for a ﬁrst-orderequation. The general solution to (14.30) is then given by [y−c(x+1 )]/ y−c(x2+1 ) / =0. J Solution method. If the equation can be factorised into the form (14.28) then solve the ﬁrst-order ODE p−Fi=0in each factor and write the solution in the form Gi(x, y)=0 . The solution to the original equation is then given by the product (14.29). 14.3.2 Equations soluble for x Equations that can be solved for x, i.e. such that they may be written in the form x=F(y,p), (14.31) can be reduced to ﬁrst-degree ﬁrst-order equations in pby diﬀerentiating both sides with respect to y,s ot h a t dx dy=1 p=∂F ∂y+∂F ∂pdp dy. This results in an equation of the form G(y,p) = 0, which can be used together with (14.31) to eliminate pand give the general solution. Note that often a singular solution to the equation will be found at the same time (see the introduction tothis chapter). 487 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve 6y2p2+3xp−y=0. (14.32) This equation can be solved for xexplicitly to give 3 x=(y/p)−6y2p. Diﬀerentiating both sides with respect to y, we ﬁnd 3dx dy=3 p=1 p−y p2dp dy−6y2dp dy−12yp, which factorises to give/; 1+6 yp2 / / 2p+ydp dy / =0. (14.33) Setting the factor containing dp/dy equal to zero gives a ﬁrst-degree ﬁrst-order equation inp, which may be solved to give py2=c. Substituting for pin (14.32) then yields the general solution of (14.32): y3=3cx+6c2. (14.34) If we now consider the ﬁrst factor in (14.33), we ﬁnd 6 p2y=−1 as a possible solution. Substituting for pin (14.32) we ﬁnd the singular solution 8y3+3x2=0. Note that the singular solution contains no arbitrary constants and cannot be found from the general solution (14.34) by any choice of the constant c. J Solution method. Write the equation in the form (14.31) and diﬀerentiate both sides with respect to y. Rearrange the resulting equation into the form G(y,p)=0, which can be used together with the original ODE to eliminate pand so give the general solution. If G(y,p)can be factorised then the factor containing dp/dy should be used to eliminate pand give the general solution. Using the other factors in this fashion will instead lead to singular solutions. 14.3.3 Equations soluble for y Equations that can be solved for y, i.e. are such that they may be written in the form y=F(x, p), (14.35) can be reduced to ﬁrst-degree ﬁrst-order equations in pby diﬀerentiating both sides with respect to x,s ot h a t dy dx=p=∂F ∂x+∂F ∂pdp dx. This results in an equation of the form G(x, p) = 0, which can be used together with (14.35) to eliminate pand give the general solution. An additional (singular) solution to the equation is also often found. 488 14.3 HIGHER-DEGREE FIRST-ORDER EQUATIONSISolve xp2+2xp−y=0. (14.36) This equation can be solved for yexplicitly to give y=xp2+2xp. Diﬀerentiating both sides with respect to x, we ﬁnd dy dx=p=2xpdp dx+p2+2xdp dx+2p, which after factorising gives (p+1 ) / p+2xdp dx / =0. (14.37) To obtain the general solution of (14.36), we consider the factor containing dp/dx .T h i s ﬁrst-degree ﬁrst-order equation in phas the solution xp2=c(see subsection 14.3.1), which we then use to eliminate pfrom (14.36). Thus we ﬁnd that the general solution to (14.36) is (y−c)2=4cx. (14.38) If instead, we set the other factor in (14.37) equal to zero, we obtain the very simple solution p=−1. Substituting this into (14.36) then gives x+y=0, which is a singular solution to (14.36). J Solution method. Write the equation in the form (14.35) and diﬀerentiate both sides with respect to x. Rearrange the resulting equation into the form G(x, p)=0, which can be used together with the original ODE to eliminate pand so give the general solution. If G(x, p)can be factorised then the factor containing dp/dx should be used to eliminate pand give the general solution. Using the other factors in this fashion will instead lead to singular solutions. 14.3.4 Clairaut’s equation Finally, we consider Clairaut’s equation, which has the form y=px+F(p) (14.39) and is therefore a special case of equations soluble for y, as in (14.35). It may be solved by a similar method to that given in subsection 14.3.3, but for Clairaut’sequation the form of the general solution is particularly simple. Diﬀerentiating(14.39) with respect to x, we ﬁnd dy dx=p=p+xdp dx+dF dpdp dx⇒dp dxparenleftbiggdF dp+xparenrightbigg =0. (14.40) Considering ﬁrst the factor containing dp/dx , we ﬁnd dp dx=d2y dx2=0⇒ y=c1x+c2. (14.41) 489 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS Since p=dy/dx =c1, if we substitute (14.41) into (14.39) we ﬁnd c1x+c2= c1x+F(c1). Therefore the constant c2is given by F(c1), and the general solution to (14.39) is y=c1x+F(c1), (14.42) i.e. the general solution to Clairaut’s equation can be obtained by replacing p in the ODE by the arbitrary constant c1. Now, considering the second factor in (14.40), we also have dF dp+x=0, (14.43) which has the form G(x, p) = 0. This relation may be used to eliminate pfrom (14.39) to give a singular solution.ISolve y=px+p2. (14.44) From (14.42) the general solution is y=cx+c2. But from (14.43) we also have 2 p+x= 0⇒p=−x/2. Substituting this into (14.44) we ﬁnd the singular solution x2+4y=0 . J Solution method. Write the equation in the form (14.39), then the general solution is given by replacing pby some constant c, as shown in (14.42). Using the relation dF/dp +x=0to eliminate pfrom the original equation yields the singular solution. 14.4 Exercises 14.1 A radioactive isotope decays in such a way that the number of atoms present at ag i v e nt i m e , N(t), obeys the equation dN dt=−λN. If there are initially N0atoms present, ﬁnd N(t)a tl a t e rt i m e s . 14.2 Solve the following equations by separation of the variables: (a)y/prime−xy3=0 ; (b)y/primetan−1x−y(1 +x2)−1=0 ; (c)x2y/prime+xy2=4y2. 14.3 Show that the following equations are either exact or can be made exact, and solve them: (a)y(2x2y2+1 )y/prime+x(y4+1 )=0 ; (b) 2 xy/prime+3x+y=0 ; (c) (cos2x+ysin2x)y/prime+y2=0 . 14.4 Find the values of αandβthat make F(x, y)= /1 x2+2+α y / dx+(xyβ+1 )dy an exact diﬀerential. For these values solve F(x, y)=0 . 490 14.4 EXERCISES 14.5 By ﬁnding a suitable integrating factor, solve the following equations: (a) (1−x2)y/prime+2xy=( 1−x2)3/2; (b)y/prime−ycotx+c o s e c x=0 ; (c) ( x+y3)y/prime=y(treat yas the independent variable). 14.6 By ﬁnding an appropriate integrating factor, solve dy dx=−2x2+y2+x xy. 14.7 Find, in the form of an integral, the solution of the equation αdy dt+y=f(t) for a general function f(t). Find the speciﬁc solutions for (a)f(t)=H(t), (b)f(t)=δ(t), (c)f(t)=β−1e−t/βH(t)w i t h β<α . For case (c), what happens if β→0? 14.8 An electric circuit contains a resistance Rand a capacitor Cin series, and a battery supplying a time-varying electromotive force V(t). The charge qon the capacitor therefore obeys the equation Rdq dt+q C=V(t). Assuming that initially there is no charge on the capacitor, and given that V(t)=V0sinωt, ﬁnd the charge on the capacitor as a function of time. 14.9 Using tangential-polar coordinates (see exercise 2.20), consider a particle of mass mmoving under the inﬂuence of a force fdirected towards the origin O. By resolving forces along the instantaneous tangent and normal and making use ofthe result of exercise 2.20 for the instantaneous radius of curvature, prove that f=−mvdv drand mv2=fpdr dp. Show further that h=mpvis a constant of the motion and that the law of force can be deduced from f=h2 p3dp dr. 14.10 Use the result of the previous exercise to ﬁnd the law of force, acting towards the origin, under which a particle must move so as to describe the followingtrajectories: (a) A circle of radius awhich passes through the origin; (b) An equiangular spiral, which is deﬁned by the property that the angle α between the tangent and the radius vector is constant along the curve. 14.11 Solve (y−x)dy dx+2x+3y=0. 14.12 A mass mis accelerated by a time-varying force exp( −βt)v3,w h e r e vis its velocity. It also experiences a resistive force ηv,w h e r e ηis a constant, owing to its motion through the air. The equation of motion of the mass is therefore mdv dt=e x p (−βt)v3−ηv. 491 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS Find an expression for the velocity vof the mass as a function of time, given that it has an initial velocity v0. 14.13 Using the results about Laplace transforms given in chapter 13 for df/dt and tf(t), show, for a function y(t)t h a ts a t i s ﬁ e s tdy dt+(t−1)y=0 ( * ) with y(0) ﬁnite, that ¯y(s)=C(1 +s)−2for some constant C. Given that y(t)=t+∞X n=2antn, determine Cand show that an=(−1)n/n!. Compare this result with that obtained by integrating (*) directly. 14.14 Solve dy dx=1 x+2y+1. 14.15 Solve dy dx=−x+y 3x+3y−4. 14.16 If u=1+t a n y,c a l c u l a t e d(lnu)/dy; hence ﬁnd the general solution of dy dx=t a n xcosy(cosy+s i n y). 14.17 Solve x(1−2x2y)dy dx+y=3x2y2, given that y(1) = 1 /2. 14.18 A reﬂecting mirror is made in the shape of the surface of revolution generated by revolving the curve y(x) about the x-axis. In order that light rays emitted from a point source at the origin are reﬂected back parallel to the x-axis, the curve y(x) must obey y x=2p 1−p2, where p=dy/dx . By solving this equation for xﬁnd the curve y(x). 14.19 Find the curve such that at each point on it the sum of the intercepts on the x- andy-axes of the tangent to the curve (taking account of sign) is equal to 1. 14.20 Find a parametric solution of x /dy dx /2 +dy dx−y=0 as follows. (a) Write an equation for yin terms of p=dy/dx and show that p=p2+( 2px+1 )dp dx. (b) Using pas the independent variable, arrange this as a linear ﬁrst-order equation for x. 492 14.4 EXERCISES (c) Find an appropriate integrating factor to obtain x=lnp−p+c (1−p)2, which, together with the expression for yobtained in (a), gives a parameter- isation of the solution. (d) Reverse the roles of xandyin steps (a) to (c), putting dx/dy =p−1,a n d show that essentially the same parameterisation is obtained. 14.21 Using the substitutions u=x2andv=y2, reduce the equation xy /dy dx /2 −(x2+y2−1)dy dx+xy=0 to Clairaut’s form. Hence show that the equation represents a family of conics and the four sides of a square. 14.22 The action of the control mechanism on a particular system for an input f(t)i s described, for t≥0, by the coupled ﬁrst-order equations: ˙y+4z=f(t), ˙z−2z=˙y+1 2y. Use Laplace transforms to ﬁnd the response y(t) of the system to a unit step input f(t)=H(t), given that y(0) = 1 and z(0) = 0. Questions 23 to 31 are intended to give the reader practice in choosing an ap- propriate method. The level of diﬃculty varies within the set; if necessary, the hints may be consulted for an indication of the most appropriate approach . 14.23 Find the general solutions of the following: (a)dy dx+xy a2+x2=x;( b )dy dx=4y2 x2−y2. 14.24 Solve the following ﬁrst-order equations for the boundary conditions given: (a)y/prime−(y/x)=1 ,y (1) =−1; (b)y/prime−ytanx=1,y(π/4) = 3; (c)y/prime−y2/x2=1/4,y(1) = 1; (d)y/prime−y2/x2=1/4,y(1) = 1 /2. 14.25 An electronic system has two inputs, to each of which a constant unit signal is applied, but starting at diﬀerent times. The equations governing the system thustake the form ˙x+2y=H(t), ˙y−2x=H(t−3). Initially (at t=0 ) , x=1a n d y= 0; ﬁnd x(t)a tl a t e rt i m e s . 14.26 Solve the diﬀerential equation sinxdy dx+2ycosx=1 subject to the boundary condition y(π/2) = 1. 14.27 Find the complete solution of/dy dx /2 −y xdy dx+A x=0, where Ais a positive constant. 493 FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS 14.28 Find the solution of (5x+y−7)dy dx=3 (x+y+1 ). 14.29 Find the solution y=y(x)o f xdy dx+y−y2 x3/2=0, subject to y(1) = 1. 14.30 Find the solution of (2sin y−x)dy dx=t a n y, if (a) y(0) = 0, and (b) y(0) = π/2. 14.31 Find the family of solutions of d2y dx2+ /dy dx /2 +dy dx=0 that satisfy y(0) = 0. 14.5 Hints and answers 14.1 N(t)=N0exp(−λt). 14.2 (a) y=±(c−x2)−1/2;( b ) y=ctan−1(x); (c) y=( l n x+4x−1−c)−1. 14.3 (a) exact, x2y4+x2+y2=c;( b )I F= x−1/2,x1/2(x+y)= c;( c )I F= sec2x, y2tanx+y=c. 14.4 α=−1,β=−2; (1/√2) tan−1(x/√2)−(x/y)+y=c. 14.5 (a) IF = (1 −x2)−2,y=( 1−x2)(k+s i n−1x); (b) IF = cosec x, leading to y=ksinx+c o s x; (c) exact equation is y−1(dx/dy )−xy−2=y, leading to x=y(k+y2/2). 14.6 Integrating factor is x;3x4+2x3+3x2y2=c. 14.7 y(t)=e−t/α Rtα−1et/prime/αf(t/prime)dt/prime;( a ) y(t)=1−e−t/α;( b ) y(t)=α−1e−t/α;( c ) y(t)= (e−t/α−e−t/β)/(α−β). It becomes case (b). 14.8 q(t)=CV0[1 + ( ωCR)2]−1{sinωt+CRω[exp(−t/RC)−cosωt]}. 14.9 If the angle between the tangent and the radius vector is α, note that cos α=dr/ds and sin α=p/r. 14.10 (a) r2=2ap, f∝ar−5;( b ) p=rsinα, f∝(sinα)−2r−3. 14.11 Homogeneous equation, put y=vxto obtain (1 −v)(v2+2v+2 )−1dv=x−1dx; write 1−vas 2−(1 +v), and v2+2v+2a s1+( 1+ v)2; A[x2+(x+y)2]=e x p / 4t a n−1[(x+y)/x] / . 14.12 Bernoulli’s equation; set v=u−1/2to obtain m du/dt−2ηu=−2e x p (−βt); v−2=2 (mβ+2η)−1[exp(−βt)−exp(2 ηt/m)] +v−2 0exp(2 ηt/m). 14.13 (1 + s)(d¯y/ds)+2¯y=0 . C=1 ; y(t)=te−t. 14.14 Follow subsection 14.2.8; k+y=l n ( x+2y+3 ) . 14.15 Equation is of the form of (14.22), set v=x+y;x+3y+2l n ( x+y−2) = A. 14.16 y=t a n−1(ksecx−1). 14.17 Equation is isobaric with weight y=−2; setting y=vx−2gives v−1(1−v)−1(1−2v)dv=x−1dx;4xy(1−x2y)=1 . 14.18 Eliminate yto obtain, in turn, p(p2−1) = 2 x(dp/dx );p=±(1−Ax)−1/2; A2y2=4 ( 1−Ax), i.e. a parabola. 14.19 The curve must satisfy y=( 1−p−1)−1(1−x+px), which has solution x=(p−1)−2, leading to y=( 1±√x)2orx=( 1±√y)2; the singular solution p/prime= 0 gives straight lines joining ( θ,0) and (0 ,1−θ)f o ra n y θ. 494 14.5 HINTS AND ANSWERS 14.20 (a) y=p2x+p; (d) the constants of integration will diﬀer in the two cases. 14.21 v=qu+q/(q−1), where q=dv/du. General solution y2=cx2+c/(c−1), hyperbolae for c>0 and ellipses for c<0. Singular solution y=±(x±1). 14.22 ¯y(s2+2s+2 )= s(¯f+1 )+( 2−2¯f);y(t)=−1+e−t(2cos t+3s i n t). 14.23 (a) Integrating factor is ( a2+x2)1/2,y=(a2+x2)/3+A(a2+x2)−1/2; (b) separable, y=x(x2+Ax+4 )−1. 14.24 (a) y=xlnx−x;( b ) y=t a n x+√2se cx; (c) homogeneous, y=x(2−lnx)−1+ x/2; (d) singular solution y=x/2. 14.25 Use Laplace transforms; ¯xs(s2+4 )= s+s2−2e−3s; x(t)=1 2sin 2t+c o s2 t−1 2H(t−3) +1 2cos(2 t−6)H(t−3). 14.26 Integrating factor is sin x;y=( 1+c o s x)−1. 14.27 This is Clairaut’s equation with F(p)=A/p. General solution y=cx+A/c; singular solution, y=2√ Ax. 14.28 Follow the second method demonstrated in subsection 14.2.8; x=X+2,y= Y−3; X(dv/dX )=( 3−2v−v2)/(5 +v); (x−y−5)3=A(3x+y−3). 14.29 Either Bernoulli’s equation with n= 2 or an isobaric equation with m=3/2; y(x)=5 x3/2/(2 + 3 x5/2). 14.30 Treat yas the independent variable, giving the general solution xsiny= −(cos2 y)/2+k.( a )y=s i n−1x;( b ) x=−cosycoty. 14.31 Show that p=(Cex−1)−1,w h e r e p=dy/dx ;y=l n [ C−e−x)/(C−1)] or ln[D−(D−1)e−x]o rl n ( e−K+1−e−x)+K. 495 15 Higher-order ordinary diﬀerential equations Following on from the discussion of ﬁrst-order ordinary diﬀerential equations (ODEs) given in the previous chapter, we now examine equations of second andhigher order. Since a brief outline of the general properties of ODEs and theirsolutions was given at the beginning of the previous chapter, we will not repeat it here. Instead, we will begin with a discussion of various types of higher-order equation. This chapter is divided into three main parts. We ﬁrst discuss linearequations with constant coeﬃcients and then investigate linear equations withvariable coeﬃcients. Finally, we discuss a few methods that may be of use insolving general linear or non-linear ODEs. Let us start by considering somegeneral points relating to alllinear ODEs. Linear equations are of paramount importance in the description of physical processes. Moreover, it is an empirical fact that, when put into mathematicalform, many natural processes appear as higher-order linear ODEs, most oftenas second-order equations. Although we could restrict our attention to thesesecond-order equations, the generalisation to nth-order equations requires little extra work, and so we will consider this more general case. A linear ODE of general order nhas the form a n(x)dny dxn+an−1(x)dn−1y dxn−1+···+a1(x)dy dx+a0(x)y=f(x). (15.1) Iff(x) = 0 then the equation is called homogeneous ; otherwise it is inhomogeneous . The ﬁrst-order linear equation studied in subsection 14.2.4 is a special case of(15.1). As discussed at the beginning of the previous chapter, the general solutionto (15.1) will contain narbitrary constants, which may be determined if nboundary conditions are also provided. In order to solve any equation of the form (15.1), we must ﬁrst ﬁnd the general solution of the complementary equation , i.e. the equation formed by setting 496 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS f(x)=0 : an(x)dny dxn+an−1(x)dn−1y dxn−1+···+a1(x)dy dx+a0(x)y=0. (15.2) To determine the general solution of (15.2), we must ﬁnd nlinearly independent functions that satisfy it. Once we have found these solutions, the general solutionis given by a linear superposition of these nfunctions. In other words, if the n solutions of (15.2) are y 1(x),y2(x),...,y n(x), then the general solution is given by the linear superposition yc(x)=c1y1(x)+c2y2(x)+···+cnyn(x), (15.3) where the cmare arbitrary constants that may be determined if nboundary conditions are provided. The linear combination yc(x) is called the complementary function of (15.1). The question naturally arises how we establish that any nindividual solutions to (15.2) are indeed linearly independent. For nfunctions to be linearly independent over an interval, there must not exist anyset of constants c1,c2,...,c nsuch that c1y1(x)+c2y2(x)+···+cnyn(x) = 0 (15.4) over the interval in question, except for the trivial case c1=c2=···=cn=0 . A statement equivalent to (15.4), which is perhaps more useful for the practical determination of linear independence, can be found by repeatedly diﬀerentiating (15.4), n−1 times in all, to obtain nsimultaneous equations for c1,c2,...,c n: c1y1(x)+c2y2(x)+···+cnyn(x)=0 c1y1/prime(x)+c2y2/prime(x)+···+cnyn/prime(x)=0 ... c1y(n−1) 1(x)+c2y(n−1) 2+···+cny(n−1) n(x)=0 ,(15.5) where the primes denote diﬀerentiation with respect to x. Referring to the discussion of simultaneous linear equations given in chapter 8, if the determinantof the coeﬃcients of c 1,c2,...,c nis non-zero then the only solution to equations (15.5) is the trivial solution c1=c2=···=cn= 0. In other words, the nfunctions y1(x),y2(x),...,y n(x) are linearly independent over an interval if W(y1,y2,...,y n)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingley 1 y2... y n y1/primey2/prime... ......... y(n−1) 1 ... ... y(n−1) nvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle/negationslash= 0 (15.6) over that interval; W(y 1,y2,...,y n) is called the Wronskian of the set of functions. It should be noted, however, that the vanishing of the Wronskian does notguarantee that the functions are linearly dependent. 497 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS If the original equation (15.1) has f(x) = 0 (i.e. it is homogeneous) then of course the complementary function yc(x) in (15.3) is already the general solution. If, however, the equation has f(x)/negationslash= 0 (i.e. it is inhomogeneous) then yc(x)i so n l y one part of the solution. The general solution of (15.1) is then given by y(x)=yc(x)+yp(x), (15.7) where yp(x)i st h e particular integral ,w h i c hc a nb e anyfunction that satisﬁes (15.1) directly, provided it is linearly independent of yc(x). It should be emphasised for practical purposes that anysuch function, no matter how simple (or complicated), is equally valid in forming the general solution (15.7). It is important to realise that the above method for ﬁnding the general solution to an ODE by superposing particular solutions assumes crucially that the ODEis linear. For non-linear equations, discussed in section 15.3, this method cannotbe used, and indeed it is often impossible to ﬁnd closed-form solutions to suchequations. 15.1 Linear equations with constant coeﬃcients If the a min (15.1) are constants rather than functions of xthen we have andny dxn+an−1dn−1y dxn−1+···+a1dy dx+a0y=f(x). (15.8) Equations of this sort are very common throughout the physical sciences and engineering, and the method for their solution falls into two parts as discussed in the previous section, i.e. ﬁnding the complementary function yc(x) and ﬁnding the particular integral yp(x). If f(x) = 0 in (15.8) then we do not have to ﬁnd a particular integral, and the complementary function is by itself the generalsolution. 15.1.1 Finding the complementary function y c(x) The complementary function must satisfy andny dxn+an−1dn−1y dxn−1+···+a1dy dx+a0y= 0 (15.9) and contain narbitrary constants (see equation (15.3)). The standard method for ﬁnding yc(x) is to try a solution of the form y=Aeλx, substituting this into (15.9). After dividing the resulting equation through by Aeλx, we are left with a polynomial equation in λof order n;t h i si st h e auxiliary equation and reads anλn+an−1λn−1+···+a1λ+a0=0. (15.10) 498 15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS In general the auxiliary equation has nroots, say λ1,λ2,...,λ n. In certain cases, some of these roots may be repeated and some may be complex. The three maincases are as follows. (i)All roots real and distinct. In this case the nsolutions to (15.9) are exp λ mx form=1t o n. It is easily shown by calculating the Wronskian (15.6) of these functions that if all the λmare distinct then these solutions are linearly independent. We can therefore linearly superpose them, as in(15.3), to form the complementary function y c(x)=c1eλ1x+c2eλ2x+···+cneλnx. (15.11) (ii)Some roots complex. For the special (but usual) case that all the coeﬃcients amin (15.9) are real, if one of the roots of the auxiliary equation (15.10) is complex, say α+iβ, then its complex conjugate α−iβis also a root. In this case we can write c1e(α+iβ)x+c2e(α−iβ)x=eαx(d1cosβx+d2sinβx) =Aeαxbraceleftbiggsin cosbracerightbigg (βx+φ), (15.12) where Aandφare arbitrary constants. (iii)Some roots repeated. If, for example, λ1occurs ktimes ( k>1) as a root of the auxiliary equation, then we have not found nlinearly independent solutions of (15.9); formally the Wronskian (15.6) of these solutions, havingtwo or more identical columns, is equal to zero. We must therefore ﬁndk−1 further solutions that are linearly independent of those already found and also of each other. By direct substitution into (15.9) we ﬁnd that xe λ1x,x2eλ1x, ... , xk−1eλ1x are also solutions, and by calculating the Wronskian it is easily shown that they, together with the solutions already found, form a linearly independent set of nfunctions. Therefore the complementary function is given by yc(x)=(c1+c2x+···+ckxk−1)eλ1x+ck+1eλk+1x+ck+2eλk+2x+···+cneλnx. (15.13) If more than one root is repeated the above argument is easily extended. For example, suppose as before that λ1is ak-fold root of the auxiliary equation and, further, that λ2is an l-fold root (of course, k>1a n d l>1). Then, from the above argument, the complementary function reads yc(x)=(c1+c2x+···+ckxk−1)eλ1x +(ck+1+ck+2x+···+ck+lxl−1)eλ2x +ck+l+1eλk+l+1x+ck+l+2eλk+l+2x+···+cneλnx.(15.14) 499 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSIFind the complementary function of the equation d2y dx2−2dy dx+y=ex. (15.15) Setting the RHS to zero, substituting y=Aeλxand dividing through by Aeλxwe obtain the auxiliary equation λ2−2λ+1=0 . The root λ= 1 occurs twice and so, although exis a solution to (15.15), we must ﬁnd a further solution to the equation that is linearly independent of ex.F r o mt h ea b o v e discussion, we deduce that xexis such a solution, so that the full complementary function is given by the linear superposition yc(x)=(c1+c2x)ex. J Solution method. Set the RHS of the ODE to zero (if it is not already so), and substitute y=Aeλx. After dividing through the resulting equation by Aeλx, obtain annth-order polynomial equation in λ(the auxiliary equation, see (15.10)). Solve the auxiliary equation to ﬁnd the nroots, λ1,λ2,...,λ n, say. If all these roots are real and distinct then yc(x)is given by (15.11). If, however, some of the roots are complex or repeated then yc(x)is given by (15.12) or (15.13), or the extension (15.14) of the latter, respectively. 15.1.2 Finding the particular integral yp(x) There is no generally applicable method for ﬁnding the particular integral yp(x) but, for linear ODEs with constant coeﬃcients and a simple RHS, yp(x) can often be found by inspection or by assuming a parameterised form similar to f(x). The latter method is sometimes called the method of undetermined coeﬃcients .I ff(x) contains only polynomial, exponential, or sine and cosine terms then, by assuminga trial function for y p(x) of similar form but one which contains a number of undetermined parameters and substituting this trial function into (15.9), theparameters can be found and y p(x) deduced. Standard trial functions are as follows. (i) If f(x)=aerxthen try yp(x)=berx. (ii) If f(x)=a1sinrx+a2cosrx(a1ora2may be zero) then try yp(x)=b1sinrx+b2cosrx. (iii) If f(x)=a0+a1x+···+aNxN(some ammay be zero) then try yp(x)=b0+b1x+···+bNxN. 500 15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS (iv) If f(x) is the sum or product of any of the above then try yp(x)a st h e sum or product of the corresponding individual trial functions. It should be noted that this method fails if any term in the assumed trial function is also contained within the complementary function yc(x). In such a case the trial function should be multiplied by the smallest integer power of x such that it will then contain no term that already appears in the complementary function. The undetermined coeﬃcients in the trial function can now be foundby substitution into (15.8). Three further methods that are useful in ﬁnding the particular integral y p(x) are Green’s functions, the variation of parameters, and making a change in thedependent variable based on knowledge of the complementary function. However,since these methods are also applicable to equations with variable coeﬃcients, adiscussion of them is postponed until section 15.2.IFind a particular integral of the equation d2y dx2−2dy dx+y=ex. From the above discussion our ﬁrst guess at a trial particular integral would be yp(x)=bex. However, since the complementary function of this equation is yc(x)=( c1+c2x)ex(as in the previous subsection), we see that exis already contained in it, as indeed is xex. Multiplying our ﬁrst guess by the lowest integer power of xsuch that the result does not appear in yc(x), we therefore try yp(x)=bx2ex. Substituting this into the ODE, we ﬁnd thatb=1/2, so the particular integral is given by yp(x)=x2ex/2. J Solution method. If the RHS of an ODE contains only the functions mentioned at the start of this subsection then the appropriate trial function should be substitutedinto it, thereby ﬁxing the undetermined parameters. If, however, the RHS of the equation is not of this form then one of the more general methods outlined in sub- sections 15.2.3–15.2.5 should be used; perhaps the most straightforward of these isthe variation-of-parameters method. 15.1.3 Constructing the general solution y c(x)+yp(x) As stated earlier, the full solution to the ODE (15.8) is found by adding together the complementary function and any particular integral. In order to illustrate further the material discussed in the last two subsections, let us ﬁnd the generalsolution to a new example, starting from the beginning. 501 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve d2y dx2+4y=x2sin2x. (15.16) First we set the RHS to zero and assume the trial solution y=Aeλx. Substituting this into (15.16) leads to the auxiliary equation λ2+4=0 ⇒ λ=±2i. (15.17) Therefore the complementary function is given by yc(x)=c1e2ix+c2e−2ix=d1cos 2x+d2sin2x. (15.18) We must now turn our attention to the particular integral yp(x). Consulting the list of standard trial functions in the previous subsection, we ﬁnd that a ﬁrst guess at a suitabletrial function for this case should be (ax 2+bx+c)(dsin 2x+ecos 2x). (15.19) However, we see that this trial function contains terms in sin 2 xand cos 2 x, both of which already appear in the complementary function (15.18). We must therefore multiply (15.19) by the smallest integer power of xthat ensures that none of the resulting terms appears inyc(x). Since multiplying by xwill suﬃce, we ﬁnally assume the trial function (ax3+bx2+cx)(dsin2x+ecos 2x). (15.20) Substituting this into (15.16) to ﬁx the constants appearing in (15.20), we ﬁnd the particular integral to be yp(x)=−x3 12cos 2x+x2 16sin2x+x 32cos 2x. (15.21) The general solution to (15.16) then reads y(x)=yc(x)+yp(x) =d1cos 2x+d2sin2x−x3 12cos 2x+x2 16sin2x+x 32cos 2x. J 15.1.4 Linear recurrence relations The discrete analogues of diﬀerential equations are called recurrence relations (or sometimes diﬀerence equations ). Whereas a diﬀerential equation gives a prescrip- tion, in terms of current values, for the new value of an dependent variable at a point only inﬁnitesimally far away, a recurrence relation describes how the nextin a sequence of values u n, deﬁned only at (non-negative) integer values of the ‘independent variable’ n,i st ob ec a l c u l a t e d . In its most general form a recurrence relation expresses the way in which un+1 is to be calculated from all the preceding values u0,u1,... ,u n. Just as the most general diﬀerential equations are intractable, so are the most general recurrence relations, and we will limit ourselves to analogues of the types of diﬀerential equations studied earlier in this chapter, namely those that are linear, haveconstant coeﬃcients and possess simple functions on the RHS. Such equations 502 15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS occur over a broad range of engineering and statistical physics as well as in the realms of ﬁnance, business planning and gambling! They form the basis of manynumerical methods, particularly those concerned with the numerical solution ofordinary and partial diﬀerential equations. A general recurrence relation is exempliﬁed by the formula u n+1=N−1summationdisplay r=0arun−r+k, (15.22) where Nand the arare ﬁxed and kis a constant or a simple function of n. Such an equation, involving terms of the series whose indices diﬀer by up to N (ranging from n−N+1 to n), is called an Nth-order recurrence relation. It is clear that, given values for u0,u1,... ,u N−1, this is a deﬁnitive scheme for generating the series and therefore has a unique solution. Parallelling the nomenclature of diﬀerential equations, if the term not involving anyunis absent, i.e. k= 0, then the recurrence relation is called homogeneous . The parallel continues with the form of the general solution of (15.22). If vnis the general solution of the homogeneous relation, and wnisanysolution of the full relation, then un=vn+wn is the most general solution of the complete recurrence relation. This is straight- forwardly veriﬁed as follows: un+1=vn+1+wn+1 =N−1summationdisplay r=0arvn−r+N−1summationdisplay r=0arwn−r+k =N−1summationdisplay r=0ar(vn−r+wn−r)+k =N−1summationdisplay r=0arun−r+k. Of course, if k=0t h e n wn=0f o ra l l nis a trivial particular solution and the complementary solution, vn, is itself the most general solution. First-order recurrence relations First-order relations, for which N= 1, are exempliﬁed by un+1=aun+k, (15.23) with u0speciﬁed. The solution to the homogeneous relation is immediate, un=Can, 503 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS and, if kis a constant, the particular solution is equally straightforward: wn=K for all n,p r o v i d e d Kis chosen to satisfy K=aK+k, i.e.K=k(1−a)−1. This will be suﬃcient unless a=1 ,i nw h i c hc a s e un=u0+nk is obvious by inspection. Thus the general solution of (15.23) is un=braceleftBigg Can+k/(1−a)a/negationslash=1, u0+nk a =1.(15.24) Ifu0is speciﬁed for the case of a/negationslash=1t h e n Cmust be chosen as C=u0−k/(1−a), resulting in the equivalent form un=u0an+k1−an 1−a. (15.25) We now illustrate this method with a worked example.IA house-buyer borrows capital Bfrom a bank that charges a ﬁxed annual rate of interest R% .I ft h el o a ni st ob er e p a i do v e r Yyears, at what value should the ﬁxed annual payments P, made at the end of each year, be set? For a loan over 25 years at 6%, what percentage of the ﬁrst year’s payment goes towards paying oﬀ the capital? Letundenote the outstanding debt at the end of year n,a n dw r i t e R/100 = r. Then the relevant recurrence relation is un+1=un(1 +r)−P with u0=B. From (15.25) we have un=B(1 +r)n−P1−(1 +r)n 1−(1 +r). A st h el o a ni st ob er e p a i do v e r Yyears, uY= 0 and thus P=Br(1 +r)Y (1 +r)Y−1. The ﬁrst year’s interest is rBand so the fraction of the ﬁrst year’s payment going towards capital repayment is ( P−rB)/P, which, using the above expression for P,i se q u a l to (1 + r)−Y. With the given ﬁgures, this is (only) 23%. J With only small modiﬁcations, the method just described can be adapted to handle recurrence relations in which the constant kin (15.23) is replaced by kαn, i.e. the relation is un+1=aun+kαn. (15.26) As for an inhomogeneous linear diﬀerential equation (see subsection 15.1.2), we may try as a potential particular solution a form which resembles the term thatmakes the equation inhomogeneous. Here, the presence of the term kα nindicates 504 15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS that a particular solution of the form un=Aαnshould be tried. Substituting this into (15.26) gives Aαn+1=aAαn+kαn, from which it follows that A=k/(α−a) and that there is a particular solution having the form un=kαn/(α−a), provided α/negationslash=a. For the special case α=a,t h e reader can readily verify that a particular solution of the form un=Anαnis appro- priate. This mirrors the corresponding situation for linear diﬀerential equations when the RHS of the diﬀerential equation is contained in the complementary function of its LHS. In summary, the general solution to (15.26) is un=braceleftBigg C1an+kαn/(α−a)α/negationslash=a, C2an+knαn−1α=a,(15.27) with C1=u0−k/(α−a)a n d C2=u0. Second-order recurrence relations We consider next recurrence relations that involve un−1in the prescription for un+1and treat the general case in which the intervening term, un, is also present. A typical equation is thus un+1=aun+bun−1+k. (15.28) As previously, the general solution of this is un=vn+wn,w h e r e vnsatisﬁes vn+1=avn+bvn−1 (15.29) andwnisanyparticular solution of (15.28); the proof follows the same lines as that given earlier. We have already seen for a ﬁrst-order recurrence relation that the solution to the homogeneous equation is given by terms forming a geometric series, and weconsider a corresponding series of powers in the present case. Setting v n=Aλnin (15.29) for some λ, as yet undetermined, requires that λshould satisfy Aλn+1=aAλn+bAλn−1. Dividing through by Aλn−1(assumed non-zero) shows that λcould be either of the roots, λ1andλ2,o f λ2−aλ−b=0, (15.30) which is known as the characteristic equation of the recurrence relation. That there are two possible series of terms of the form Aλnis consistent with the fact that two initial values (boundary conditions) have to be provided before the series can be calculated by repeated use of (15.28). These two values are suﬃcientto determine the appropriate coeﬃcient Afor each of the series. Since (15.29) is 505 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS both linear and homogeneous, and is satisﬁed by both vn=Aλn 1andvn=Bλn 2,i t s general solution is vn=Aλn 1+Bλn 2. If the coeﬃcients aandbare such that (15.30) has two equal roots, i.e. a2=−4b, then, as in the analogous case of repeated roots for diﬀerential equations (seesubsection 15.1.1(iii)), the second term of the general solution is replaced by Bnλ n 1 to give vn=(A+Bn)λn 1. Finding a particular solution is straightforward if kis a constant: a trivial but adequate solution is wn=k(1−a−b)−1for all n. As with ﬁrst-order equations, particular solutions can be found for other simple forms of kby trying functions similar to kitself. Thus particular solutions for the cases k=Cnandk=Dαn can be found by trying wn=E+Fnandwn=Gαnrespectively.IFind the value of u16if the series unsatisﬁes un+1+4un+3un−1=n forn≥1,w i t h u0=1andu1=−1. We ﬁrst solve the characteristic equation, λ2+4λ+3=0 , to obtain the roots λ=−1a n d λ=−3. Thus the complementary function is vn=A(−1)n+B(−3)n. In view of the form of the RHS of the original relation, we try wn=E+Fn as a particular solution and obtain E+F(n+1 )+4 ( E+Fn)+3 [ E+F(n−1)] = n, yielding F=1/8a n d E=1/32. Thus the complete general solution is un=A(−1)n+B(−3)n+n 8+1 32, and now using the given values for u0andu1determines Aas 7/8a n d Bas 3/32. Thus un=1 32[28(−1)n+3 (−3)n+4n+1]. Finally, substituting n=1 6g i v e s u16= 4035633, a value the reader may (or may not) wish to verify by repeated application of the initial recurrence relation. J 506 15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS Higher-order recurrence relations It will be apparent that linear recurrence relations of order N>2 do not present any additional diﬃculty in principle, though two obvious practical diﬃculties are (i) that the characteristic equation is of order Nand in general will not have roots that can be written in closed form and (ii) that a correspondingly large numberof given values is required to determine the Notherwise arbitrary constants in the solution. The algebraic labour needed to solve the set of simultaneouslinear equations which determines them increases rapidly with N. We do not give speciﬁc examples here, but some are included in the exercises at the end of the chapter. 15.1.5 Laplace transform method The method of Laplace transforms is very useful for solving linear ODEs with constant coeﬃcients. Taking the Laplace transform of such an equation trans-forms it into a purely algebraic equation in terms of the Laplace transform of the required solution. Once the algebraic equation has been solved for thisLaplace transform, the general solution to the original ODE can be obtainedby performing an inverse Laplace transform. One advantage of this method isthat, for given boundary conditions, it provides the solution in just one step, instead of having to ﬁnd the complementary function and particular integral separately. In order to apply this method we need only two results from Laplace transform theory (see section 13.2). First, the Laplace transform of a function f(x) is deﬁned by ¯f(s)≡integraldisplay ∞ 0e−sxf(x)dx, (15.31) from which we can derive a second useful relation. This concerns the Laplace transform of derivatives of f(x): f(n)(s)=sn¯f(s)−sn−1f(0)−sn−2f/prime(0)−···−sf(n−2)(0)−f(n−1)(0), (15.32) where the primes and superscripts in parentheses denote diﬀerentiation with respect to x. Using these relations, along with the table 13.1, on p. 461, which gives Laplace transforms of standard functions, we are in a position to solve alinear ODE with constant coeﬃcients by this method. 507 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve d2y dx2−3dy dx+2y=2e−x, (15.33) subject to the boundary conditions y(0) = 2 ,y/prime(0) = 1 . Taking the Laplace transform of (15.33) and using the table of standard results we obtain s2¯y(s)−sy(0)−y/prime(0)−3[s¯y(s)−y(0)]+2¯y(s)=2 s+1, which reduces to (s2−3s+2 )¯y(s)−2s+5=2 s+1. (15.34) Solving this algebraic equation for ¯y(s), the Laplace transform of the required solution to (15.33), we obtain ¯y(s)=2s2−3s−3 (s+1 ) ( s−1)(s−2)=1 3(s+1 )+2 s−1−1 3(s−2), (15.35) where in the ﬁnal step we have used partial fractions. Taking the inverse Laplace transform of (15.35), again using table 13.1, we ﬁnd the speciﬁc solution to (15.33) to be y(x)=1 3e−x+2ex−1 3e2x. J Note that if the boundary conditions in a problem are given as symbols, rather than just numbers, then the step involving partial fractions can often involvea considerable amount of algebra. The Laplace transform method is also veryconvenient for solving sets of simultaneous linear ODEs with constant coeﬃcients.ITwo electrical circuits, both of negligible resistance, each consist of a coil having self- inductance Land a capacitor having capacitance C. The mutual inductance of the two circuits is M. There is no source of e.m.f. in either circuit. Initially the second capacitor is given a charge CV0, the ﬁrst capacitor being uncharged, and at time t=0as w i t c hi n the second circuit is closed to complete the circuit. Find the subsequent current in the ﬁrstcircuit. Subject to the initial conditions q1(0) = ˙q1(0) = ˙q2(0) = 0 and q2(0) = CV0=V0/G,s a y , we have to solve L¨q1+M¨q2+Gq1=0, M¨q1+L¨q2+Gq2=0. On taking the Laplace transform of the above equations, we obtain (Ls2+G)¯q1+Ms2¯q2=sMV 0C, Ms2¯q1+(Ls2+G)¯q2=sLV 0C. Eliminating ¯q2and rewriting as an equation for ¯q1, we ﬁnd ¯q1(s)=MV 0s [(L+M)s2+G][(L−M)s2+G] =V0 2G /(L+M)s (L+M)s2+G−(L−M)s (L−M)s2+G / . 508 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS Using table 13.1, q1(t)=1 2V0C(cosω1t−cosω2t), where ω2 1(L+M)=Gandω2 2(L−M)=G. Thus the current is given by i1(t)=1 2V0C(ω2sinω2t−ω1sinω1t). J Solution method. Perform a Laplace transform, as deﬁned in (15.31), on the entire equation, using (15.32) to calculate the transform of the derivatives. Then solve theresulting algebraic equation for ¯y(s), the Laplace transform of the required solution to the ODE. By using the method of partial fractions and consulting a table ofLaplace transforms of standard functions, calculate the inverse Laplace transform.The resulting function y(x)is the solution of the ODE that obeys the given boundary conditions. 15.2 Linear equations with variable coeﬃcients There is no generally applicable method of solving equations with coeﬃcients that are functions of x. Nevertheless, there are certain cases in which a solution is possible. Some of the methods discussed in this section are also useful in ﬁndingthe general solution or particular integral for equations with constant coeﬃcientsthat have proved impenetrable by the techniques discussed above. 15.2.1 The Legendre and Euler linear equations Legendre’s linear equation has the form a n(αx+β)ndny dxn+···+a1(αx+β)dy dx+a0y=f(x), (15.36) where α,βand the anare constants and may be solved by making the substitution αx+β=et. We then have dy dx=dt dxdy dt=α αx+βdy dt d2y dx2=d dxdy dx=α2 (αx+β)2parenleftbiggd2y dt2−dy dtparenrightbigg and so on for higher derivatives. Therefore we can write the terms of (15.36) as (αx+β)dy dx=αdy dt, (αx+β)2d2y dx2=α2d dtparenleftbiggd dt−1parenrightbigg y, ... (αx+β)ndny dxn=αnd dtparenleftbiggd dt−1parenrightbigg ···parenleftbiggd dt−n+1parenrightbigg y.(15.37) 509 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS Substituting equations (15.37) into the original equation (15.36), the latter becomes a linear ODE with constant coeﬃcients, i.e. anαnd dtparenleftbiggd dt−1parenrightbigg ···parenleftbiggd dt−n+1parenrightbigg y+···+a1αdy dt+a0y=fparenleftbigget−β αparenrightbigg , which can be solved by the methods of section 15.1. A special case of Legendre’s linear equation, for which α=1a n d β=0 ,i s Euler’s equation , anxndny dxn+···+a1xdy dx+a0y=f(x); (15.38) it may be solved in a similar manner to the above by substituting x=et.I f f(x) = 0 in (15.38), substituting y=xλleads to a simple algebraic equation in λ, which can be solved to yield the solution to (15.38). In the event that the algebraic equation for λhas repeated roots, extra care is needed. If λ1is ak-fold root ( k>1) then the klinearly independent solutions corresponding to this root arexλ1,xλ1lnx ,...,xλ1(lnx)k−1.ISolve x2d2y dx2+xdy dx−4y= 0 (15.39) by both of the methods discussed above. First we make the substitution x=et, which, after cancelling et, gives an equation with constant coeﬃcients, i.e. d dt /d dt−1 / y+dy dt−4y=0⇒d2y dt2−4y=0. (15.40) Using the methods of section 15.1, the general solution of (15.40), and therefore of (15.39), is given by y=c1e2t+c2e−2t=c1x2+c2x−2. Since the RHS of (15.39) is zero, we can reach the same solution by substituting y=xλ into (15.39). This gives λ(λ−1)xλ+λxλ−4xλ=0, which reduces to (λ2−4)xλ=0. This has the solutions λ=±2, so we obtain again the general solution y=c1x2+c2x−2. J Solution method. If the ODE is of the Legendre form (15.36) then substitute αx+ β=et. This results in an equation of the same order but with constant coeﬃcients, which can be solved by the methods of section 15.1. If the ODE is of the Euler form (15.38) with a non-zero RHS then substitute x=et; this again leads to an equation of the same order but with constant coeﬃcients. If, however, f(x)=0 in the Euler equation (15.38) then the equation may also be solved by substituting 510 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS y=xλ. This leads to an algebraic equation whose solution gives the allowed values ofλ; the general solution is then the linear superposition of these functions. 15.2.2 Exact equations Sometimes an ODE may be merely the derivative of another ODE of one order lower. If this is the case then the ODE is called exact. The nth-order linear ODE an(x)dny dxn+···+a1(x)dy dx+a0(x)y=f(x), (15.41) is exact if the LHS can be written as a simple derivative, i.e. if an(x)dny dxn+···+a0(x)y=d dxbracketleftbigg bn−1(x)dn−1y dxn−1+···+b0(x)ybracketrightbigg . (15.42) It may be shown that, for (15.42) to hold, we require a0(x)−a/prime 1(x)+a/prime/prime 2(x)−···+(−1)na(n) n(x)=0 , (15.43) where the prime again denotes diﬀerentiation with respect to x. If (15.43) is satisﬁed then straightforward integration leads to a new equation of one order lower. If this simpler equation can be solved then a solution to the original equation is obtained. Of course, if the above process leads to an equation that isitself exact then the analysis can be repeated to reduce the order still further.ISolve (1−x2)d2y dx2−3xdy dx−y=1. (15.44) Comparing with (15.41), we have a2=1−x2,a1=−3xanda0=−1. It is easily shown thata0−a/prime 1+a/prime/prime 2= 0, so (15.44) is exact and can the refore be written in the form d dx / b1(x)dy dx+b0(x)y / =1. (15.45) Expanding the LHS of (15.45) we ﬁnd d dx / b1dy dx+b0y / =b1d2y dx2+(b/prime 1+b0)dy dx+b/prime 0y. (15.46) Comparing (15.44) and (15.46) we ﬁnd b1=1−x2,b/prime 1+b0=−3x, b/prime 0=−1. These relations integrate consistently to give b1=1−x2andb0=−x, so (15.44) can be written asd dx / (1−x2)dy dx−xy / =1. (15.47) Integrating (15.47) gives us directly the ﬁrst-order linear ODE dy dx− /x 1−x2 / y=x+c1 1−x2, which can be solved by the method of subsection 14.2.4 and has the solution y=c1sin−1x+c2√ 1−x2−1. J 511 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS It is worth noting that, even if a higher-order ODE is not exact in its given form, it may sometimes be made exact by multiplying through by some suitable function,anintegrating factor , cf. subsection 14.2.3. Unfortunately, no straightforward method for ﬁnding an integrating factor exists and one often has to rely on inspection or experience.ISolve x(1−x2)d2y dx2−3x2dy dx−xy=x. (15.48) It is easily shown that (15.48) is not exact, but we also see immediately that by multiplying it through by 1 /xwe recover (15.44), which is exact and is solved above. J Another important point is that an ODE need not be linear to be exact, although no simple rule such as (15.43) exists if it is not linear. Nevertheless, it isoften worth exploring the possibility that a non-linear equation is exact, since itcould then be reduced in order by one and may lead to a soluble equation. This is discussed further in subsection 15.3.3. Solution method. For a linear ODE of the form (15.41) check whether it is exact using equation (15.43). If it is not then attempt to ﬁnd an integrating factor which when multiplying the equation makes it exact. Once the equation is exact write the LHS as a derivative as in (15.42) and, by expanding this derivative and comparingwith the LHS of the ODE, determine the functions b m(x)in (15.42). Integrate the resulting equation to yield another ODE, of one order lower. This may be solved orsimpliﬁed further if the new ODE is itself exact or can be made so. 15.2.3 Partially known complementary function Suppose we wish to solve the nth-order linear ODE a n(x)dny dxn+···+a1(x)dy dx+a0(x)y=f(x), (15.49) a n dw eh a p p e nt ok n o wt h a t u(x) is a solution of (15.49) when the RHS is set to zero, i.e. u(x) is one part of the complementary function. By making the substitution y(x)=u(x)v(x), we can transform (15.49) into an equation of order n−1i ndv/dx. This simpler equation may prove soluble. In particular, if the original equation is of second order then we obtain a ﬁrst-order equation in dv/dx, which may be soluble using the methods of section 14.2. In this way both the remaining term in the complementary function and the particular integral are found. This method therefore provides a useful way of calculating particular integrals for second-order equations with variable(or constant) coeﬃcients. 512 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSISolve d2y dx2+y=c o s e c x. (15.50) We see that the RHS does not fall into any of the categories listed in subsection 15.1.2, and so we are at an initial loss as to how to ﬁnd the particular integral. However, thecomplementary function of (15.50) is y c(x)=c1sinx+c2cosx, and so let us choose the solution u(x)=c o s x(we could equally well choose sin x)a n d make the substitution y(x)=v(x)u(x)=v(x)c osxinto (15.50). This gives cosxd2v dx2−2si nxdv dx=c o s e c x, (15.51) which is a ﬁrst-order linear ODE in dv/dx and may be solved by multiplying through by a suitable integrating factor, as discussed in subsection 14.2.4. Writing (15.51) as d2v dx2−2tan xdv dx=cosec x cosx, (15.52) we see that the required integrating factor is given by exp / −2 Z tanxd x / =e x p [2l n( c os x)]=c o s2x. Multiplying both sides of (15.52) by the integrating factor cos2xwe obtain d dx / cos2xdv dx / =c o t x, which integrates to give cos2xdv dx=l n ( s i n x)+c1. After rearranging and integrating again, this becomes v= Z sec2xln(sin x)dx+c1 Z sec2xd x =t a n xln(sin x)−x+c1tanx+c2. Therefore the general solution to (15.50) is given by y=uv=vcosx,i . e . y=c1sinx+c2cosx+s i n xln(sin x)−xcosx, which contains the full complementary function and the particular integral. J Solution method. Ifu(x)is a known solution of the nth-order equation (15.49) with f(x)=0, then make the substitution y(x)=u(x)v(x)in (15.49). This leads to an equation of order n−1indv/dx, which might be soluble. 513 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS 15.2.4 Variation of parameters The method of variation of parameters proves useful in ﬁnding particular integrals for linear ODEs with variable (and constant) coeﬃcients. However, it requiresknowledge of the entire complementary function, not just of one part of it as inthe previous subsection. Suppose we wish to ﬁnd a particular integral of the equation a n(x)dny dxn+···+a1(x)dy dx+a0(x)y=f(x), (15.53) and the complementary function yc(x) (the general solution of (15.53) with f(x)=0 )i s yc(x)=c1y1(x)+c2y2(x)+···+cnyn(x), where the functions ym(x) are known. We now assume that a particular integral of (15.53) can be expressed in a form similar to that of the complementary function,but with the constants c mreplaced by functions of x,i . e .w ea s s u m eap a r t i c u l a r integral of the form yp(x)=k1(x)y1(x)+k2(x)y2(x)+···+kn(x)yn(x). (15.54) This will no longer satisfy the complementary equation (i.e. (15.53) with the RHS set to zero) but might, with suitable choices of the functions ki(x), be made equal tof(x), thus producing not a complementary function but a particular integral. Since we have narbitrary functions k1(x),k2(x),...,k n(x), but only one restric- tion on them (namely the ODE), we may impose a further n−1 constraints. We can choose these constraints to be as convenient as possible, and the simplestchoice is given by k /prime 1(x)y1(x)+k/prime 2(x)y2(x)+···+k/prime n(x)yn(x)=0 k/prime 1(x)y/prime 1(x)+k/prime 2(x)y/prime 2(x)+···+k/prime n(x)y/prime n(x)=0 ... (15.55) k/prime 1(x)y(n−2) 1(x)+k/prime 2(x)y(n−2) 2(x)+···+k/prime n(x)y(n−2) n(x)=0 k/prime 1(x)y(n−1) 1(x)+k/prime 2(x)y(n−1) 2(x)+···+k/prime n(x)y(n−1) n(x)=f(x) an(x), where the primes denote diﬀerentiation with respect to x. The last of these equations is not a freely chosen constraint but must be satisﬁed given the previousn−1 constraints and the original ODE. This choice of constraints is easily justiﬁed (although the algebra is quite messy). Diﬀerentiating (15.54) with respect to x,w eo b t a i n y /prime p=k1y/prime 1+k2y/prime 2+···+kny/prime n+(k/prime 1y1+k/prime 2y2+···+k/prime nyn), where, for the moment, we drop the explicit x-dependence of these functions. Since 514 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS we are free to choose our constraints as we wish, let us deﬁne the expression in parentheses to be zero, giving the ﬁrst equation in (15.55). Diﬀerentiating againwe ﬁnd y /prime/prime p=k1y/prime/prime 1+k2y/prime/prime 2+···+kny/prime/prime n+(k/prime 1y/prime 1+k/prime 2y/prime 2+···+k/prime ny/prime n). Once more we can choose the expression in brackets to be zero, giving the second equation in (15.55). We can repeat this procedure, choosing the corresponding expression in each case to be zero. This yields the ﬁrst n−1 equations in (15.55). Themth derivative of ypform<n is then given by y(m) p=k1y(m) 1+k2y(m) 2+···+kny(m) n. Diﬀerentiating yponce more we ﬁnd that its nth derivative is given by y(n) p=k1y(n) 1+k2y(n) 2+···+kny(n) n+(k/prime 1y(n−1) 1+k/prime 2y(n−1) 2+···+k/prime ny(n−1) n). Substituting the expressions for y(m) p,m=0t o n, into the original ODE (15.53), we obtain nsummationdisplay m=0am(k1y(m) 1+k2y(m) 2+···+kny(m) n)+an(k/prime 1y(n−1) 1+k/prime 2y(n−1) 2+···+k/prime ny(n−1) n)=f(x). i.e. nsummationdisplay m=0amnsummationdisplay j=1kjy(m) j+an(k1/primey(n−1) 1+k2/primey(n−1) 2+···+kn/primey(n−1) n=f(x). Rearranging the order of summations on the LHS, we ﬁnd nsummationdisplay j=1kj(any(n) j+···+a1y/prime j+a0yj)+an(k/prime 1y(n−1) 1+k2/primey(n−1) 2+···+k/prime ny(n−1) n)=f(x). (15.56) But since the functions yjare solutions of the complementary equation of (15.53) we have (for all j) any(n) j+···+a1y/prime j+a0yj=0. Therefore (15.56) becomes an(k/prime 1y(n−1) 1+k2/primey(n−1) 2+···+k/prime ny(n−1) n)=f(x), which is the ﬁnal equation given in (15.55). Considering (15.55) to be a set of simultaneous equations in the set of unknowns k/prime 1(x),k/prime 2,...,k/prime n(x), we see that the determinant of the coeﬃcients of these functions is equal to the Wronskian W(y1,y2,...,y n), which is non-zero since the solutions ym(x) are linearly independent; see equation (15.6). Therefore (15.55) can be solved for the functions k/prime m(x), which in turn can be integrated, setting all constants of 515 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS integration equal to zero, to give km(x). The general solution to (15.53) is then given by y(x)=yc(x)+yp(x)=nsummationdisplay m=1[cm+km(x)]ym(x). Note that if the constants of integration are included in the km(x) then, as well as ﬁnding the particular integral, we introduce an addition to the complementaryfunction.IUse the variation of parameters method to solve d2y dx2+y=c o s e c x, (15.57) subject to the boundary conditions y(0) = y(π/2) = 0 . The complementary function of (15.57) is again yc(x)=c1sinx+c2cosx. We therefore assume a particular integral of the form yp(x)=k1(x)sinx+k2(x)cosx, and impose the additional constraints of (15.55), i.e. k/prime 1(x)sinx+k/prime 2(x)cosx=0, k/prime 1(x)cosx−k/prime 2(x)sinx=c o s e c x. Solving these equations for k/prime 1(x)a n d k/prime 2(x)g i v e s k/prime 1(x)=c o s xcosec x=c o t x, k/prime 2(x)=−sinxcosec x=−1. Hence, ignoring the constants of integration, k1(x)a n d k2(x) are given by k1(x) = ln(sin x), k2(x)=−x. The general solution to the ODE (15.57) is therefore y(x)=[c1+l n ( s i n x)]sinx+(c2−x)cosx, which is identical to the solution found in subsection 15.2.3. Applying the boundary conditions y(0) = y(π/2) = 0 we ﬁnd c1=c2=0a n ds o y(x) = ln(sin x)sinx−xcosx. J Solution method. If the complementary function of (15.53) is known then assume a particular integral of the same form but with the constants replaced by functions ofx. Impose the constraints in (15.55) and solve the resulting system of equations for the unknowns k/prime 1(x),k/prime 2,...,k/prime n(x). Integrate these functions, setting constants of integration equal to zero, to obtain k1(x),k2(x),...,k n(x)and hence the particular integral. 516 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS 15.2.5 Green’s functions The Green’s function method of solving linear ODEs bears a striking resemblance to the method of variation of parameters discussed in the previous subsection;it too requires knowledge of the entire complementary function in order to ﬁndthe particular integral and therefore the general solution. The Green’s functionapproach diﬀers, however, since once the Green’s function for a particular LHS of (15.1) and accompanying boundary conditions has been found, then the solution foranyRHS (i.e. any f(x)) can be written down immediately, albeit in the form of an integral. Although the Green’s function method can be approached by considering the superposition of eigenfunctions of the equation (see chapter 17) and is also applicable to the solution of partial diﬀerential equations (see chapter 19), thissection adopts a more utilitarian approach based on the properties of the Diracdelta function (see subsection 13.1.3) and deals only with the use of Green’sfunctions in solving ODEs. Let us again consider the equation a n(x)dny dxn+···+a1(x)dy dx+a0(x)y=f(x), (15.58) but for the sake of brevity we now denote the LHS by Ly(x), i.e. as a linear diﬀerential operator acting on y(x). Thus (15.58) now reads Ly(x)=f(x). (15.59) Let us suppose that a function G(x, z) exists (the Green’s function ) such that the general solution to (15.59), which obeys some set of imposed boundary conditions in the range a≤x≤b,i sg i v e nb y y(x)=integraldisplayb aG(x, z)f(z)dz, (15.60) where zis the integration variable. If we apply the linear diﬀerential operator L to both sides of (15.60) and use (15.59) then we obtain Ly(x)=integraldisplayb a[LG(x, z)]f(z)dz=f(x). (15.61) Comparison of (15.61) with a standard property of the Dirac delta function (see subsection 13.1.3), namely f(x)=integraldisplayb aδ(x−z)f(z)dz, fora≤x≤b, shows that for (15.61) to hold for any arbitrary function f(x), we require (for a≤x≤b)t h a t LG(x, z)=δ(x−z), (15.62) 517 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS i.e. the Green’s function G(x, z)must satisfy the original ODE with the RHS set equal to a delta function .G(x, z) may be thought of physically as the response of a system to a unit impulse at x=z. In addition to (15.62), we must impose two further sets of restrictions on G(x, z). The ﬁrst is the requirement that the general solution y(x) in (15.60) obeys the boundary conditions. For homogeneous boundary conditions, in which y(x) and/or its derivatives are required to be zeroat speciﬁed points, this is most simply arranged by demanding that G(x, z) itself obeys the boundary conditions when it is considered as a function of xalone; if, for example, we require y(a)=y(b) = 0 then we should also demand G(a, z)=G(b, z)=0 .P r o b l e m s having inhomogeneous boundary conditions are discussed at the end of thissubsection. The second set of restrictions concerns the continuity or discontinuity of G(x, z) and its derivatives at x=zand can be found by integrating (15.62) with respect toxover the small interval [ z−/epsilon1, z+/epsilon1] and taking the limit as /epsilon1→0. We then obtain lim /epsilon1→0nsummationdisplay m=0integraldisplayz+/epsilon1 z−/epsilon1am(x)dmG(x, z) dxmdx= lim /epsilon1→0integraldisplayz+/epsilon1 z−/epsilon1δ(x−z)dx=1. (15.63) Since dnG/dxnexists at x=zbut with value inﬁnity, the ( n−1)th-order derivative must have a ﬁnite discontinuity there, whereas all the lower-order derivatives,d mG/dxmform<n−1, must be continuous at this point. Therefore the terms containing these derivatives cannot co ntribute to the value of the integral on the LHS of (15.63). Noting that, apart from an arbitrary additive constant,integraltext (dmG/dxm)dx=dm−1G/dxm−1, and integrating the terms of (15.63) by parts we ﬁnd lim /epsilon1→0integraldisplayz+/epsilon1 z−/epsilon1am(x)dmG(x, z) dxmdx= 0 (15.64) form=0t o n−1. Thus, since only the term containing dnG/dxncontributes to the integral in (15.63), we conclude, after performing an integration by parts, that lim /epsilon1→0bracketleftbigg an(x)dn−1G(x, z) dxn−1bracketrightbiggz+/epsilon1 z−/epsilon1=1. (15.65) Thus we have the further nconstraints that G(x, z) and its derivatives up to order n−2 are continuous at x=zbut that dn−1G/dxn−1has a discontinuity of 1 /an(z) atx=z. Thus the properties of the Green’s function G(x, z)f o ra n nth-order linear ODE may be summarised by the following. (i)G(x, z) obeys the original ODE but with f(x) on the RHS set equal to a delta function δ(x−z). 518 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS (ii) When considered as a function of xalone G(x, z) obeys the speciﬁed (homogeneous) boundary conditions on y(x). (iii) The derivatives of G(x, z) with respect to xup to order n−2 are continuous atx=z, but the ( n−1)th-order derivative has a discontinuity of 1 /an(z) at this point.IUse Green’s functions to solve d2y dx2+y=c o s e c x, (15.66) subject to the boundary conditions y(0) = y(π/2) = 0 . From (15.62) we see that the Green’s function G(x, z)m u s ts a t i s f y d2G(x, z) dx2+G(x, z)=δ(x−z). (15.67) Now it is clear that for x/negationslash=zthe RHS of (15.67) is zero, and we are left with the task of ﬁnding the general solution to the homogeneous equation, i.e. the complementary function. The complementary function of (15.67) consists of a linear superposition of sin x and cos xandmustconsist of diﬀerent superpositions on either side of x=z, since its (n−1)th derivative (i.e. the ﬁrst derivative in this case) is required to have a discontinuity there. Therefore we assume the form of the Green’s function to be G(x, z)= /A(z)sinx+B(z)c osxforx<z , C(z)sinx+D(z)c osxforx>z . Note that we have performed a similar (but not identical) operation to that used in the variation of parameters method, i.e. we have replaced the constants in the complementaryfunction with func tions (this time of z). We must now impose the relevant restrictions on G(x, z) in order to determine the functions A(z),...,D (z). The ﬁrst of these is that G(x, z) should itself obey the homogeneous boundary conditions G(0,z)=G(π/2,z) = 0. This leads to the conclusion that B(z)= C(z) = 0, so we now have G(x, z)= /A(z)sinxforx<z , D(z)c osxforx>z . The second restriction is the continuity conditions given in equations (15.64), (15.65), namely that, for this second-order equation, G(x, z) is continuous at x=zanddG/dx has a discontinuity of 1 /a2(z) = 1 at this point. Applying these two constraints we have D(z)c osz−A(z)sinz=0 −D(z)sinz−A(z)cosz=1. Solving these equations for A(z)a n d D(z), we ﬁnd A(z)=−cosz, D (z)=−sinz. Thus we have G(x, z)= /−coszsinxforx<z , −sinzcosxforx>z . Therefore, from (15.60), the general solution to (15.66) that obeys the boundary conditions 519 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS y(0) = y(π/2) = 0 is given by y(x)= Zπ/2 0G(x, z)cosec zd z =−cosx Zx 0sinzcosec zd z−sinx Zπ/2 xcoszcosec zd z =−xcosx+s i n xln(sin x), which agrees with the result obtained in the previous subsections. J As mentioned earlier, once a Green’s function has been obtained for a given LHS and boundary conditions, it can be used to ﬁnd a general solution for anyRHS; thus, the solution of d 2y/dx2+y=f(x), with y(0) = y(π/2) = 0, is given immediately by y(x)=integraldisplayπ/2 0G(x, z)f(z)dz =−cosxintegraldisplayx 0sinzf(z)dz−sinxintegraldisplayπ/2 xcoszf(z)dz. (15.68) As an example, the reader may wish to verify that if f(x)=s i n2 xthen (15.68) gives y(x)=(−sin2x)/3, a solution easily veriﬁed by direct substitution. In general, analytic integration of (15.68) for arbitrary f(x) will, prove intractable; then the integrals must be evaluated numerically. Another important point is that although the Green’s function method above has provided a general solution, it is also useful for ﬁnding a particular integral if the complementary function is known. This is easily seen since in (15.68) theconstant integration limits 0 and π/2 lead merely to constant values by which the factors sin xand cos xare multiplied; thus the complementary function is reconstructed. The rest of the general solution, i.e. the particular integral, comesfrom the variable integration limit x. Therefore by changingintegraltext π/2 xto−integraltextx,a n ds o dropping the constant integration limits, we can ﬁnd just the particular integral. For example, a particular integral of d2y/dx2+y=f(x) that satisﬁes the above boundary conditions is given by yp(x)=−cosxintegraldisplayx sinzf(z)dz+s i n xintegraldisplayx coszf(z)dz. A very important point to realise about the Green’s function method is that a particular G(x, z) applies to a given LHS of an ODE andthe imposed boundary conditions, i.e. the same equation with diﬀerent boundary conditions will have a diﬀerent Green’s function . To illustrate this point, let us consider again the same ODE as solved above, but with diﬀerent boundary conditions. 520 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTSIUse Green’s functions to solve d2y dx2+y=f(x), (15.69) subject to the one-point boundary conditions y(0) = y/prime(0) = 0 . We again require (15.67) to hold and so again we assume a Green’s function of the form G(x, z)= /A(z)sinx+B(z)c osxforx<z , C(z)sinx+D(z)c osxforx>z . However, we now require G(x, z) to obey the boundary conditions G(0,z)=G/prime(0,z)=0 , which imply A(z)=B(z) = 0. Therefore we have G(x, z)= /0f o r x<z , C(z)sinx+D(z)c osxforx>z . Applying the continuity conditions on G(x, z) as before now gives C(z)sinz+D(z)c osz=0, C(z)cosz−D(z)s i nz=1, which are solved to give C(z)=c o s z, D (z)=−sinz. So ﬁnally the Green’s function is given by G(x, z)= /0f o r x<z , sin(x−z)f o r x>z , and the general solution to (15.69) that obeys the boundary conditions y(0) = y/prime(0) = 0 is y(x)= Z∞ 0G(x, z)f(z)dz = Zx 0sin(x−z)f(z)dz. J Finally, we consider how to deal with inhomogeneous boundary conditions such as y(a)=α,y(b)=βory(0) = y/prime(0) = γetc., where α, β, γ are non- zero. The simplest method of solution in this case is to make a change of variable such that the boundary conditions in the new variable, usay, are homogeneous, i.e. u(a)=u(b)=0o r u(0) = u/prime(0) = 0 etc. For nth-order equations we generally require nboundary conditions to ﬁx the solution, but these n boundary conditions can be of various types. For example we may have the n- point boundary conditions y(xm)=ymform=1t o n, or the one-point boundary conditions y(x0)=y/prime(x0)=···=y(n−1)(x0)=y0,o rs o m e t h i n gi nb e t w e e n .I na l l cases a suitable change of variable is u=y−h(x), where h(x)i sa n( n−1)th-order polynomial that obeys the boundary conditions. 521 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS For example, if we consider the second-order case with boundary conditions y(a)=α,y(b)=βthen a suitable change of variable is u=y−(mx+c), where y=mx+cis the straight line through the points ( a, α)a n d( b, β); this is given by m=(α−β)/(a−b)a n d c=(βa−αb)/(a−b). Alternatively, if the boundary conditions for our second-order equation are y(0) = y/prime(0) = γthen we would make the same change of variable, but this time y=mx+cwould be the straight line through (0 ,γ) with slope γ,i . e .m=c=γ. Solution method. Require that the Green’s function G(x, z)obeys the original ODE, but with the RHS set to a delta function δ(x−z). This is equivalent to assuming thatG(x, z)is given by the complementary function of the original ODE, with the constants replaced by functions of z; these functions are diﬀerent for x<z andx> z. Now require also that G(x, z)obeys the given homogeneous boundary conditions and impose the continuity conditions given in (15.64) and (15.65). The generalsolution to the original ODE is then given by (15.60). For inhomogeneous boundary conditions, make the change of dependent variable u=y−h(x),w h e r e h(x)is a polynomial obeying the given boundary conditions. 15.2.6 Canonical form for second-order equations In this section we specialise from nth-order linear ODEs with variable coeﬃcients to those of order 2. In particular we consider the equation d 2y dx2+a1(x)dy dx+a0(x)y=f(x), (15.70) which has been rearranged so that the coeﬃcient of d2y/dx2is unity. By making the substitution y(x)=u(x)v(x)w eo b t a i n v/prime/prime+parenleftbigg2u/prime u+a1parenrightbigg v/prime+parenleftbiggu/prime/prime+a1u/prime+a0u uparenrightbigg v=f u, (15.71) where the prime denotes diﬀerentiation with respect to x. Since (15.71) would be much simpliﬁed if there were no term in v/prime,l e tu sc h o o s e u(x) such that the ﬁrst factor in parentheses on the LHS of (15.71) is zero, i.e. 2u/prime u+a1=0⇒ u(x)=e x pbraceleftbigg −1 2integraldisplay a1(z)dzbracerightbigg . (15.72) We then obtain an equation of the form d2v dx2+g(x)v=h(x), (15.73) 522 15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS where g(x)=a0(x)−1 4[a1(x)]2−1 2a/prime 1(x) h(x)=f(x)ex pbraceleftbigg 1 2integraldisplay a1(z)dzbracerightbigg . Since (15.73) is of a simpler form than the original equation, (15.70), it may prove easier to solve.ISolve 4x2d2y dx2+4xdy dx+(x2−1)y=0. (15.74) Dividing (15.74) through by 4 x2, we see that it is of the form (15.70) with a1(x)=1 /x, a0(x)=(x2−1)/4x2andf(x) = 0. Therefore, making the substitution y=vu=vexp / − Z1 2xdx / =Av√x, we obtain d2v dx2+v 4=0. (15.75) Equation (15.75) is easily solved to give v=c1sin1 2x+c2cos1 2x, so the solution of (15.74) is y=v√x=c1sin1 2x+c2cos1 2x√x. J As an alternative to choosing u(x) such that the coeﬃcient of v/primein (15.71) is zero, we could choose a diﬀerent u(x) such that the coeﬃcient of vvanishes. For this to be the case, we see from (15.71) that we would require u/prime/prime+a1u/prime+a0u=0, sou(x) would have to be a solution of the original ODE with the RHS set to zero, i.e. part of the complementary function. If such a solution were known thenthe substitution y=uvwould yield an equation with no term in v, which could be solved by two straightforward integrations. This is a special (second-order) case of the method discussed in subsection 15.2.3. Solution method. Write the equation in the form (15.70), then substitute y=uv, where u(x)is given by (15.72). This leads to an equation of the form (15.73), in which there is no term in dv/dx and which may be easier to solve. Alternatively, if part of the complementary function is known then follow the method of subsec-tion 15.2.3. 523 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS 15.3 General ordinary diﬀerential equations In this section, we discuss miscellaneous methods for simplifying general ODEs. These methods are applicable to both linear and non-linear equations and in some cases may lead to a solution. More often than not, however, ﬁnding a closed-form solution to a general non-linear ODE proves impossible. 15.3.1 Dependent variable absent If an ODE does not contain the dependent variable yexplicitly, but only its derivatives, then the change of variable p=dy/dx l e a d st oa ne q u a t i o no fo n e order lower.ISolve d2y dx2+2dy dx=4x (15.76) This is transformed by the substitution p=dy/dx to the ﬁrst-order equation dp dx+2p=4x. (15.77) The solution to (15.77) is then found by the method of subsection 14.2.4 and reads p=dy dx=ae−2x+2x−1, where ais a constant. Thus by direct integration the solution to the original equation, (15.76), is y(x)=c1e−2x+x2−x+c2. J An extension to the above method is appropriate if an ODE contains only derivatives of ythat are of order mand greater. Then the substitution p=dmy/dxm reduces the order of the ODE by m. Solution method. If the ODE contains only derivatives of ythat are of order mand greater then the substitution p=dmy/dxmreduces the order of the equation by m. 15.3.2 Independent variable absent If an ODE does not contain the independent variable xexplicitly, except in d/dx, d2/dx2etc., then as in the previous subsection we make the substitution p=dy/dx 524 15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS but also write d2y dx2=dp dx=dy dxdp dy=pdp dy d3y dx3=d dxparenleftbigg pdp dyparenrightbigg =dy dxd dyparenleftbigg pdp dyparenrightbigg =p2d2p dy2+pparenleftbiggdp dyparenrightbigg2 , (15.78) and so on for higher-order derivatives. This leads to an equation of one order lower.ISolve 1+yd2y dx2+ /dy dx /2 =0. (15.79) Making the substitutions dy/dx =pandd2y/dx2=p(dp/dy ) we obtain the ﬁrst-order ODE 1+ypdp dy+p2=0, which is separable and may be solved as in subsection 14.2.1 to obtain (1 +p2)y2=c1. Using p=dy/dx we therefore have p=dy dx=± s c2 1−y2 y2, which may be integrated to give the general solution of (15.79); after squaring this reads (x+c2)2+y2=c2 1. J Solution method. If the ODE does not contain xexplicitly then substitute p= dy/dx , along with the relations for higher derivatives given in (15.78), to obtain an equation of one order lower, which may prove easier to solve. 15.3.3 Non-linear exact equations As discussed in subsection 15.2.2, an exact ODE is one that can be obtained by straightforward diﬀerentiation of an equation of one order lower. Moreover, thenotion of exact equations is useful for both linear and non-linear equations, sincean exact equation can be immediately integrated. It is possible, of course, thatthe resulting equation may itself be exact, so that the process can be repeated. In the non-linear case, however, there is no simple relation (such as (15.43) for the linear case) by which an equation can be shown to be exact. Nevertheless, ageneral procedure does exist and is illustrated in the following example. 525 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSISolve 2yd3y dx3+6dy dxd2y dx2=x. (15.80) Directing our attention to the term on the LHS of (15.80) that contains the highest-order derivative, i.e. 2 yd3y/dx3, we see that it can be obtained by diﬀerentiating 2 yd2y/dx2since d dx / 2yd2y dx2 / =2yd3y dx3+2dy dxd2y dx2. (15.81) Rewriting the LHS of (15.80) using (15.81), we are left with 4( dy/dx )(d2y/dy2), which may itself be written as a derivative, i.e. 4dy dxd2y dx2=d dx /" 2 /dy dx /2 /# . (15.82) Since, therefore, we can write the LHS of (15.80) as a sum of simple derivatives of other functions, (15.80) is exact. Integrating (15.80) with respect to x, and using (15.81) and (15.82), now gives 2yd2y dx2+2 /dy dx /2 = Z xd x=x2 2+c1. (15.83) Now we can repeat the process to ﬁnd whether ( 15.83) is itself exact. Considering the term on the LHS of (15.83) that contains the highest-order derivative, i.e. 2 yd2y/dx2, we note that we obtain this by diﬀerentiating 2 yd y / d x , as follows: d dx / 2ydy dx / =2yd2y dx2+2 /dy dx /2 . The above expression already contains all the terms on the LHS of (15.83), so we can integrate (15.83) to give 2ydy dx=x3 6+c1x+c2. Integrating once more we obtain the solution y2=x4 24+c1x2 2+c2x+c3. J It is worth noting that both linear equations (as discussed in subsection 15.2.2) and non-linear equations may sometimes be made exact by multiplying throughby an appropriate integrating factor. Although no general method exists forﬁnding such a factor, one may sometimes be found by inspection or inspiredguesswork. Solution method. Rearrange the equation so that all the terms containing yor its derivatives are on the LHS, then check to see whether the equation is exact byattempting to write the LHS as a simple derivative. If this is possible then theequation is exact and may be integrated directly to give an equation of one orderlower. If the new equation is itself exact the process can be repeated. 526 15.3 GENERAL ORDINARY DIFFERENTIAL EQUATIONS 15.3.4 Isobaric or homogeneous equations It is straightforward to generalise the discussion of ﬁrst-order isobaric equations given in subsection 14.2.6 to equations of general order n.A n nth-order isobaric equation is one in which every term can be made dimensionally consistent upongiving yanddyeach a weight m,a n d xanddxeach a weight 1. Then the nth derivative of ywith respect to x, for example, would have dimensions minyand −ninx. In the special case, with m= 1, for which the equation is dimensionally consistent the equation is called homogeneous (not to be confused with linearequations with a zero RHS). If an equation is isobaric or homogeneous then the change in dependent variable y=vx m(y=vxin the homogeneous case) followed by the change in independent variable x=etleads to an equation in which the new independent variable tis absent except in the form d/dt.ISolve x3d2y dx2−(x2+xy)dy dx+(y2+xy)=0 . (15.84) Assigning yanddythe weight m,a n d xanddxthe weight 1, the weights of the ﬁve terms on the LHS of (15.84) are, from left to right: m+1 , m+1 , 2 m,2m,m+1 . F o r t h e s e weights all to be equal we require m= 1; thus (15.84) is a homogeneous equation. Since it is homogeneous we now make the substitution y=vx, which, after dividing the resulting equation through by x3,g i v e s xd2v dx2+( 1−v)dv dx=0. (15.85) Now substituting x=etinto (15.85) we obtain (after some working) d2v dt2−vdv dt=0, (15.86) which can be integrated directly to give dv dt=1 2v2+c1. (15.87) Equation (15.87) is separable, and integrates to give 1 2t+d2= Zdv v2+d2 1 =1 d1tan−1 /v d1 / . Rearranging and using x=etandy=vxwe ﬁnally obtain the solution to (15.84) as y=d1xtan /;1 2d1lnx+d1d2 / . J Solution method. Assume that yanddyhave weight m, and xanddxweight 1, and write down the combined weights of each term in the ODE. If these weights canbe made equal by assuming a particular value for mthen the equation is isobaric (or homogeneous if m=1). Making the substitution y=vx mfollowed by x=et leads to an equation in which the new independent variable tis absent except in the form d/dt. 527 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS 15.3.5 Equations homogeneous in xoryalone It will be seen that the intermediate equation (15.85) in the example of the previous subsection was simpliﬁed by the substitution x=et, in that this led to an equation in which the new independent variable toccurred only in the form d/dt, see (15.86). A closer examination of (15.85) reveals that it is dimensionally consistent in the independent variable xtaken alone ; this is equivalent to giving the dependent variable and its diﬀerential a weight m= 0. For any equation that is homogeneous in xalone, the substitution x=etwill lead to an equation that does not contain the new independent variable texcept as d/dt. Note that the Euler equation of subsection 15.2.1 is a special, linear example of an equation homogeneous in xalone. Similarly, if an equation is homogeneous in yalone, then substituting y=evleads to an equation in which the new dependent variable, v, occurs only in the form d/dv.ISolve x2d2y dx2+xdy dx+2 y3=0. This equation is homogeneous in xalone, and on substituting x=etwe obtain d2y dt2+2 y3=0, which does not contain the new independent variable texcept as d/dt. Such equations may often be solved by the method of subsection 15.3.2, but in this case we can integratedirectly to obtain dy dt= p 2(c1+1/y2). This equation is separable, and we ﬁndZdyp 2(c1+1/y2)=t+c2. By multiplying the numerator and denominator of the integrand on the LHS by y, we ﬁnd the solutionp c1y2+1√ 2c1=t+c2. Remembering that t=l nxwe ﬁnally obtainp c1y2+1√ 2c1=l nx+c2. J Solution method. If the weight of xtaken alone is the same in every term in the ODE then the substitution x=etleads to an equation in which the new independent variable tis absent except in the form d/dt. If the weight of ytaken alone is the same in every term then the substitution y=evleads to an equation in which the new dependent variable vis absent except in the form d/dv. 528 15.4 EXERCISES 15.3.6 Equations having y=Aexas a solution Finally, we note that if any general (linear or non-linear) nth-order ODE is satisﬁed identically by assuming that y=dy dx=···=dny dxn(15.88) then y=Aexis a solution of that equation. This must be so because y=Aexis a non-zero function that satisﬁes (15.88).IFind a solution of (x2+x)dy dxd2y dx2−x2ydy dx−x /dy dx /2 =0. (15.89) Setting y=dy/dx =d2y/dx2in (15.89), we obtain (x2+x)y2−x2y2−xy2=0, which is satisﬁed identically. Therefore y=Aexis a solution of (15.89); this is easily veriﬁed by directly substituting y=Aexinto (15.89). J Solution method. If the equation is satisﬁed identically by assuming that y= dy/dx =···=dny/dxnthen y=Aexis a solution. 15.4 Exercises 15.1 A simple harmonic oscillator, with natural frequency ω0, experiences an oscillating driving force f(t)=c o s ωt. Therefore, its equation of motion is d2x dt2+ω2 0x=c o s ωt, where xis its position. Given that at t= 0 we have x=dx/dt = 0, ﬁnd the function x(t). Describe the solution if ωis approximately, but not exactly, equal toω0. 15.2 Find the roots of the auxiliary equation for the following. Hence solve them for the boundary conditions stated. (a)d2f dt2+2df dt+5f=0 w i t h f(0) = 1 ,f/prime(0) = 0 . (b)d2f dt2+2df dt+5f=e−tcos 3twith f(0) = 0 ,f/prime(0) = 0 . 15.3 The theory of bent beams shows that at any point in the beam the ‘bending moment’ is given by K/ρ,w h e r e Kis a constant (that depends upon the beam material and cross-sectional shape) and ρis the radius of curvature at that point. Consider a light beam of length Lwhose ends, x=0a n d x=L, are supported at the same vertical height and which has a weight Wsuspended from its centre. Verify that at any point x(0≤x≤L/2 for deﬁniteness) the net magnitude of the bending moments, (bending moment = force ×perpendicular distance) due to the weight and support reactions, evaluated on either side of x,i sWx/2. 529 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS If the beam is only slightly bent, so that ( dy/dx )2/lessmuch1, where y=y(x)i st h e downward displacement of the beam at x, show that the beam proﬁle satisﬁes the approximate equation d2y dx2=−Wx 2K. By integrating this equation twice and using physically imposed conditions on your solution at x=0a n d x=L/2, show that the downward displacement at the centre of the beam is WL3/(48K). 15.4 Solve the diﬀerential equation d2f dt2+6df dt+9f=e−t, subject to the conditions f=0a n d df/dt =λatt=0 . Find the equation satisﬁed by the positions of the turning points of f(t)a n d hence, by drawing suitable sketch graphs, determine the number of turning pointsthe solution has in the range t>0i f( a ) λ=1/4, and (b) λ=−1/4. 15.5 The function f(t) satisﬁes the diﬀerential equation d 2f dt2+8df dt+1 2f=1 2e−4t. For the following sets of boundary conditions determine whether it has solutions, and, if so, ﬁnd them: (a)f(0) = 0 ,f/prime(0) = 0 ,f(ln√2) = 0; (b)f(0) = 0 ,f/prime(0) =−2,f(ln√2) = 0 . 15.6 Determine the values of αandβfor which the following functions are linearly dependent: y1(x)=xcoshx+s i n h x, y2(x)=xsinhx+c o s h x, y3(x)=(x+α)ex, y4(x)=(x+β)e−x. You will ﬁnd it convenient to work with those linear combinations of the yi(x) that can be written the most compactly. 15.7 A solution of the diﬀerential equation d2y dx2+2dy dx+y=4e−x takes the value 1 when x= 0 and the value e−1when x= 1. What is its value when x=2 ? 15.8 The two functions x(t)a n d y(t) satisfy the simultaneous equations dx dt−2y=−sint, dy dt+2x=5c o s t. Find explicit expressions for x(t)a n d y(t), given that x(0) = 3 and y(0) = 2. Sketch the solution trajectory in the xy-plane for 0 ≤t<2π, showing that the trajectory crosses itself at (0 ,1/2) and passes through the points (0 ,−3) and (0,−1) in the negative x-direction. 530 15.4 EXERCISES 15.9 Find the general solutions of (a)d3y dx3−12dy dx+1 6y=3 2x−8, (b)d dx /1 ydy dx / +( 2acoth2 ax) /1 ydy dx / =2a2, where ais a constant. 15.10 Use the method of Laplace transforms to solve (a)d2f dt2+5df dt+6f=0,f (0) = 1 ,f/prime(0) =−4, (b)d2f dt2+2df dt+5f=0,f (0) = 1 ,f/prime(0) = 0 . 15.11 The quantities x(t),y(t) satisfy the simultaneous equations ¨x+2n˙x+n2x=0, ¨y+2n˙y+n2y=µ˙x, where x(0) = y(0) = ˙y(0) = 0 and ˙x(0) = λ. Show that y(t)=1 2µλt2 /; 1−1 3nt / exp(−nt). 15.12 Use Laplace transforms to solve, for t≥0, the diﬀerential equations ¨x+2x+y=c o s t, ¨y+2x+3y=2c o s t, which describe a coupled system that starts from rest at the equilibrium position. Show that the subsequent motion takes place along a straight line in the xy-plane. Verify that the frequency at which the system is driven is equal to one of theresonance frequencies of the system; explain why there is noresonant behaviour in the solution you have obtained. 15.13 Two unstable isotopes AandBand a stable isotope Chave the following decay rates per atom present: A→B,3 s −1;A→C,1 s−1;B→C,2 s−1. Initially a quantity x0ofAis present and none of the other two types. Using Laplace transforms, ﬁnd the amount of Cpresent at a later time t. 15.14 For a lightly damped ( γ<ω 0) harmonic oscillator driven at its undamped resonance frequency ω0, the displacement x(t)a tt i m e tsatisﬁes the equation d2x dt2+2γdx dt+ω2 0x=Fsinω0t. Use Laplace transforms to ﬁnd the displacement at a general time if the oscillator starts from rest at its equilibrium position. (a) Show that ultimately the oscillation has amplitude F/(2ω0γ) with a phase lag of π/2 relative to the driving force F. (b) By diﬀerentiating the original equation, conclude that if x(t) is expanded as a power series in tfor small tthen the ﬁrst non-vanishing term is Fω0t3/6. Conﬁrm this conclusion by expanding your explicit solution. 15.15 The ‘golden mean’, which is said to describe the most aesthetically pleasing proportions for the sides of a rectangle (e.g. the ideal picture frame), is givenby the limiting value of the ratio of successive terms of the Fibonacci series u n, which is generated by un+2=un+1+un, with u0=0a n d u1= 1. Find an expression for the general term of the series and 531 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS verify that the golden mean is equal to the larger root of the recurrence relation’s characteristic equation. 15.16 In a particular scheme for modelling numerically one-dimensional ﬂuid ﬂow, the successive values, un, of the solution are connected for n≥1 by the diﬀerence equation c(un+1−un−1)=d(un+1−2un+un−1), where canddare positive constants. The boundary conditions are u0=0a n d uM= 1. Find the solution to the equation and show that successive values of un will have alternating signs if c>d. 15.17 The ﬁrst few terms of a series un, starting with u0,a r e1 ,2,2,1,6,−3. The series is generated by a recurrence relation of the form un=Pun−2+Qun−4, where PandQare constants. Find an expression for the general term of the series and show that the series in fact consists of two other interleaved seriesgiven by u 2m=2 3+1 34m, u2m+1=7 3−1 34m, form=0,1,2,... . 15.18 Find an explicit expression for the unsatisfying un+1+5un+6un−1=2n, given that u0=u1= 1. Deduce that 2n−26(−3)nis divisible by 5 for all integer n. 15.19 Find the general expression for the unsatisfying un+1=2un−2−un−1 with u0=u1=0a n d u2= 1, and show that they can be written in the form un=1 5−2n/2 √5cos /3πn 4−φ / , where tan φ=2 . 15.20 Consider the seventh-order recurrence relation un+7−un+6−un+5+un+4−un+3+un+2+un+1−un=0. Find the most general form of its solution, and show that: (a) if only the four initial values u0=0 ,u1=2 ,u3=6a n d u3= 12, are speciﬁed, the relation has one solution which cycles repeatedly through this set of fournumbers. (b) but if, in addition, it is required that u 4= 20, u5=3 0a n d u6= 42 then the solution is unique, with un=n(n+1 ) . 15.21 Find the general solution of x2d2y dx2−xdy dx+y=x, given that y(1) = 1 and y(e)=2 e. 15.22 Find the general solution of (x+1 )2d2y dx2+3 (x+1 )dy dx+y=x2. 532 15.4 EXERCISES 15.23 Prove that the general solution of (x−2)d2y dx2+3dy dx+4y x2=0 is given by y(x)=1 (x−2)2 / k /2 3x−1 2 / +cx2 / . 15.24 Use the method of variation of parameters to ﬁnd the general solutions of (a)d2y dx2−y=xn,( b )d2y dx2−2dy dx+y=2xex. 15.25 Use the intermediate result of exercise 15.24(a) to ﬁnd the Green’s function which satisﬁes d2G(x, ξ) dx2−G(x, ξ)=δ(x−ξ)w i t h G(0,ξ)=G(1,ξ)=0 . 15.26 (a) Given that y1(x)=1 /xis a solution of F(x, y)=x(x+1 )d2y dx2+( 2−x2)dy dx−(2 +x)y=0, ﬁnd a second linearly independent solution, (i) by setting y2(x)=y1(x)u(x), (ii) by noting the sum of the coeﬃcients in the equation. (b) Hence, using the variation of parameters method, ﬁnd the general solution of F(x, y)=(x+1 )2. 15.27 Show generally that if y1(x)a n d y2(x) are linearly independent solutions of d2y dx2+p(x)dy dx+q(x)y=0, with y1(0) = 0 and y2(1) = 0, then the Green’s function G(x, ξ) for the interval 0≤x, ξ≤1a n dw i t h G(0,ξ)=G(1,ξ)=0c a nb ew r i t t e ni nt h ef o r m G(x, ξ)= /( y1(x)y2(ξ)/W(ξ)0<x<ξ y2(x)y1(ξ)/W(ξ)ξ<x< 1, where W(x)=W[y1(x),y2(x)] is the Wronskian of y1(x)a n d y2(x). 15.28 Use the result of the previous exercise to ﬁnd the Green’s function G(x, ξ)t h a t satisﬁes d2G dx2+3dG dx+2G=δ(x−x), in the interval 0 ≤x, ξ≤1w i t h G(0,ξ)=G(1,ξ) = 0. Hence obtain integral expressions for the solution of d2y dx2+3dy dx+2y= /( 00 <x<x 0, 1x0<x< 1, distinguishing between the cases (a) x<x 0,a n d( b ) x>x 0. 15.29 The equation of motion for a driven damped harmonic oscillator can be written ¨x+2˙x+( 1+ κ2)x=f(t), with κ/negationslash=0 .I fi ts t a r t sf r o mr e s tw i t h x(0) = 0 and ˙x(0) = 0, ﬁnd the corresponding Green’s function G(t, τ) and verify that it can b e written as a function of t−τ 533 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS only. Find the explicit solution when the driving force is the unit step function, i.e.f(t)=H(t). Conﬁrm your solution by taking the Laplace transforms of both it and the original equation. 15.30 Show that the Green’s function for the equation d2y dx2+y 4=f(x), subject to the boundary conditions y(0) = y(π) = 0, is given by G(x, z)= /( −2cos1 2xsin1 2z0≤z≤x, −2sin1 2xcos1 2zx≤z≤π. 15.31 Find the Green’s function x=G(t, t0) that solves d2x dt2+αdx dt=δ(t−t0) under the initial conditions x=dx/dt =0a t t= 0. Hence solve d2x dt2+αdx dt=f(t), where f(t)=0f o r t<0. Evaluate your answer explicitly for f(t)=Ae−at(t>0). 15.32 (a) By multiplying through by dy/dx , write down the solution to the equation d2y dx2+f(y)=0 , where f(y) can be any function. (b) A mass m, initially at rest at the point x= 0, is accelerated by a force f(x)=A(x0−x) / 1+2l n / 1−x x0 // . Its equation of motion is md2x/dt2=f(x). Find xas a function of time and show that ultimately the par ticle has travelled a distance x0. 15.33 Solve 2yd3y dx3+2 / y+3dy dx /d2y dx2+2 /dy dx /2 =s i n x. 15.34 Find the general solution of the equation xd3y dx3+2d2y dx2=Ax. 15.35 Express the equation d2y dx2+4xdy dx+( 4x2+6 )y=e−x2sin2x in canonical form and hence ﬁnd its general solution. 15.36 Find the form of the solutions of the equation dy dxd3y dx3−2 /d2y dx2 /2 + /dy dx /2 =0 which have y(0) =∞. (You will need the result Rzcosech ud u=−ln(cosech z+c o t h z).) 534 15.5 HINTS AND ANSWERS 15.37 Consider the equation xpy/prime/prime+n+3−2p n−1xp−1y/prime+ /p−2 n−1 /2 xp−2y=yn, in which p/negationslash=2a n d n>−1 but n/negationslash= 1. For the boundary conditions y(1) = 0 and y/prime(1) = λ, show that the solution is y(x)=v(x)x(p−2)/(n−1),w h e r e v(x)i sg i v e nb yZv(x) 0dz/ λ2+2zn+1/(n+1 ) /1/2=l nx. 15.5 Hints and answers 15.1 The function is ( ω2 0−ω2)−1(cosωt−cosω0t); for moderate t,x(t) is a sine wave of linearly increasing amplitude ( tsinω0t)/(2ω0); for large tit shows beats of maximum amplitude 2( ω2 0−ω2)−1. 15.2 m=−1±2i;( a ) f(t)=e−t(cos2 t+1 2sin2t); (b) f(t)=1 5e−t(cos2 t−cos 3t). 15.3 y=0a t x= 0. From symmetry, dy/dx =0a t x=L/2. 15.4 f(t)=1 4{e−t+[ ( 4 λ−2)t−1]e−3t}. For turning points, (4 λ+1 )+( 6−12λ)t=e2t. (a) 1, (b) 2. 15.5 General solution f(t)= Ae−6t+Be−2t−3e−4t. (a) No solution, inconsistent boundary conditions; (b) f(t)=2 e−6t+e−2t−3e−4t. 15.6 Set y5(x)=y1(x)+y2(x)a n d y6(x)=y1(x)−y2(x). Wronskian W(y3,y4,y5,y6)= −16(α−1)(β+ 1). Thus linear dependence if α=1 ,o r β=−1, or both. 15.7 The auxiliary equation has repeated roots and the RHS is contained in the complementary function. The solution is y(x)=(A+Bx)e−x+2x2e−x.y(2) = 5 e−2. 15.8 x=2s i n2 t+3c o s t, y=2c o s2 t−sint. The curve is symmetric about the y-axis and crosses each axis four times. Its outer perimeter is heart-shaped. 15.9 (a) The auxiliary equation has roots 2, 2, −4; (A+Bx)exp2 x+Cexp(−4x)+2x+1; (b) multiply through by sinh2 axand note thatR cosech 2 ax dx =( 2a)−1ln(|tanh ax|);y=B(sinh2 ax)1/2(|tanh ax|)A. 15.10 (a) f(t)=2 e−3t−e−2t,( b ) f(t)=e−t(cos 2 t+1 2sin 2t); compare with exercise 15.2(a). 15.11 Use Laplace transforms; write s(s+n)−4as (s+n)−3−n(s+n)−4. 15.12 y=2x=2 3(cost−cos 2t), i.e. yis always a ﬁxed multiple of x. There is no resonance because the driving forces form the components, cos t,2c os t,o fa vector that is a pure eigenvector corresponding to resonant frequency ω=2 ,a n d contains no component of the eigenvector (1 ,−1) corresponding to ω=1 ,t h e frequency of the forces. 15.13 L[C(t)]=x0(s+8 )/[s(s+2 ) ( s+ 4)], yielding C(t)=x0[1 +1 2exp(−4t)−3 2exp(−2t)]. 15.14 Write the numerator of the partial fraction with denominator ( s+γ)2+k2,w h e r e k2=ω2 0−γ2,i nt h ef o r m A(s+γ)+B. General solution is x(t)=(F/2ω0){γ−1[e−γtcoskt−cos(ω0t)] +k−1e−γtsinkt}. (b) Since x=dx/dt =s i n ω0t=0a t t=0 , d2x/dt2= 0 also. Diﬀerentiating and then setting t=0s h o w st h a t d3x/dt3has the initial value ω0F. 15.15 un=[ ( 1+√5)n−(1−√5)n]/(2n√5). 15.16 un=( 1−rn)/(1−rM)w h e r e r=(d+c)/(d−c). Ifc>d,t h e n r<−1. 15.17 P=5,Q=−4.un=3/2−5(−1)n/6+(−2)n/4+2n/12. 15.18 un= [35(−2)n−26(−3)n+2n]/10. Note that, with this recurrence relation and these intial values, all unmust be integers. 15.19 The general solution is A+B2n/2exp(i3πn/4)+C2n/2exp(i5πn/4). The initial values imply that A=1/5,B=(√5/10)exp[ i(π−φ)] and C=(√5/10)exp[ i(π+φ)]. 535 HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS 15.20 The general solution is un=(A+Bn+Cn2)1n+(D+En)(−1)n+Fin+G(−i)n. (a)B=C=E=0 , A=5 , D=−2,F=−3 2+5 2i,G=−3 2−5 2i; (b)B=C= 1 and all other coeﬃcients = 0. 15.21 This is Euler’s equation; setting x=e x p tproduces d2z/dt2−2dz/dt +z=e x p t with complementary function ( A+Bt)e x p tand particular integral t2(expt)/2; y(x)=x+[xlnx(1 + ln x)]/2. 15.22 This is Legendre’s linear equation with α=β= 1. Its reduced form is y/prime/prime+2y/prime+y= (et−1)2,w h e r e y=y(t)a n d x+1= et. A particular integral is y(t)=e2t/9−et/2 + 1 and the general solution is y(x)=(x+1 )−1[A+Bln(x+1 ) ]+ x2/9−5x/18 + 11 /18. 15.23 After multiplication through by x2the coeﬃcients are such that this is an exact equation. The resulting ﬁrst-order equation, in standard form, needs an integrating factor ( x−2)2/x2. 15.24 (a) The complementary function is Aex+Be−x; writing the particular integral in the form k1(x)ex+k2(x)e−xgives k/prime 1=xne−x/2a n d k/prime 2=−xnex/2. These lead to the particular integral −(n!/2) Pn m=0[1 + (−1)n+m]xm/m!. (b) Setting the particular integral equal to k1(x)ex+k2(x)xexgives the general solution y=(A+Bx+x3/3)ex. 15.25 Given the boundary conditions, it is better to work with sinh xand sinh(1−x) than with e±x;G(x, ξ)=−[sinh(1−ξ)sinh x]/sinh1 for x<ξ and−[sinh(1− x)sinh ξ]/sinh1 for x>ξ. 15.26 (a) (i) (1 + x)u/prime/prime=( 2+ x)u/prime, (ii) follow subsection 15.3.6. Both give y2(x)=ex. (b)y(x)=A/x+Bex−x/2−1. 15.27 Follow the method of subsection 15.2.5 but using general rather than speciﬁc functions. 15.28 The relevant independent solutions are y1(x)=A(e−x−e−2x)a n d y2(x)=B(e−x− e−2x+1) with Wronskian AB(e−1)e−3x.I fG1(x, ξ)=(e−1)−1(e−x−e−2x)(e2ξ−eξ+1) and G2(x, ξ)=( e−1)−1(e−x−e−2x+1)(e2ξ−eξ)t h e n( a )f o r x<x 0,y(x)=R1 x0G1(x, ξ)dξ,a n d( b )f o r x>x 0,y(x)= Rx x0G2(x, ξ)dξ+ R1 xG1(x, ξ)dξ. 15.29 G(t, τ)=0f o r t<τ,a n d κ−1e−(t−τ)sin[κ(t−τ)] for t>τ. For a unit step input, x(t)=( 1+ κ2)−1(1−e−tcosκt−κ−1e−tsinκt). Both transforms are equivalent to s[(s+1 )2+κ2)]¯x=1. 15.30 With y=A(x)sin(x/2)+B(x)cos( x/2), obtain A/prime(z)=2 f(z)cos( z/2) and B/prime(z)= −2f(z)sin(z/2) and hence identify G(x, z). 15.31 Use continuity and the step condition on ∂G/∂t att=t0to show that G(t, t0)=α−1{1−exp[α(t0−t)]}for 0≤t0≤t; x(t)=A(α−a)−1{a−1[1−exp(−at)]−α−1[1−exp(−αt)]}. 15.32 (a) B+x= Rydz[A−2 Rzf(u)du]−1/2; (b) show that the force is proportional to the derivative of ( x0−x)2ln[x0/(x0−x)];x=x0{1−exp[−At2/(2m)]}. 15.33 LHS of the equation is exact for two stages of integration and then needs an integrating factor exp x;2yd2y/dx2+2yd y / d x +2 (dy/dx )2;2yd y / d x +y2= d(y2)/dx+y2;y2=Aexp(−x)+Bx+C−(sinx−cosx)/2. 15.34 Set p=dy/dx ;y(x)=Ax3/18−Blnx+Cx+D. 15.35 Follow the method of subsection 15.2.6; u(x)=e−x2andv(x)s a t i s ﬁ e s v/prime/prime+4v= sin2x, for which a particular integral is ( −xcos 2x)/4. The general solution y(x)=[Asin 2x+(B−1 4x)cos2 x]e−x2. 15.36 Set p=dy/dx and follow subsection 15.3.2 to obtain pd2p/dy2+1=( dp/dy )2 and then set q=dp/dy to obtain ( q2−1)1/2=Ap.The substitution sinh θ=Ap gives ﬁnally that cosech ( Ay+B)+c o t h ( Ay+B)=e−x. 15.37 Equation is isobaric with yof weight m,w h e r e m+p−2=mn;v(x)s a t i s ﬁ e s x2v/prime/prime+xv/prime=vn.S e t x=etandv(x)=u(t), leading to u/prime/prime=unwith u(0) = 0,u/prime(0) = λ. Multiply both sides by u/primeto make the equation exact. 536 16 Series solutions of ordinary diﬀerential equations In the previous chapter the solution of both homogeneous and non-homogeneous linear ordinary diﬀerential equations (ODEs) of order ≥2 were discussed. In par- ticular we developed methods for solving some equations in which the coeﬃcientswere not constant but functions of the independent variable x. In each case we were able to write the solutions to such equations in terms of elementary func-tions, or as integrals. In general, however, the solutions of equations with variable coeﬃcients cannot be written in this way, and we must consider alternative approaches. In this chapter we discuss a method for obtaining solutions to linear ODEs in the form of convergent series. Such series can be evaluated numerically, andthose occurring most commonly are named and tabulated. There is in fact nodistinct borderline between this and the previous chapter, since solutions in termsof elementary functions may equally well be written as convergent series (i.e. therelevant Taylor series). Indeed, it is partly because some series occur so frequently that they are given special names such as sin x,c o sxor exp x. Since we shall be concerned principally with second-order linear ODEs in this chapter, we begin with a discussion of these equations, and obtain some general results that will prove useful when we come to discuss series solutions. 16.1 Second-order linear ordinary diﬀerential equations Any homogeneous second-order linear ODE can be written in the form y /prime/prime+p(x)y/prime+q(x)y=0, (16.1) where y/prime=dy/dx andp(x)a n d q(x) are given functions of x. From the previous chapter, we recall that the most general form of the solution to (16.1) is y(x)=c1y1(x)+c2y2(x), (16.2) 537 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS where y1(x)a n d y2(x)a r elinearly independent solutions of (16.1), and c1andc2 are constants that are ﬁxed by the boundary conditions (if supplied). A full discussion of the linear independence of sets of functions was given at the beginning of the previous chapter, but for just two functions y1andy2 to be linearly independent we simply require that y2is not a multiple of y1. Equivalently, y1andy2must be such that the equation c1y1(x)+c2y2(x)=0 isonlysatisﬁed for c1=c2= 0. Therefore the linear independence of y1(x)a n d y2(x) can usually be deduced by inspection but in any case can always be veriﬁed by the evaluation of the Wronskian of the two solutions, W(x)=vextendsinglevextendsinglevextendsinglevextendsingley1y2 y/prime 1y/prime 2vextendsinglevextendsinglevextendsinglevextendsingle=y1y/prime 2−y2y/prime 1. (16.3) IfW(x)/negationslash= 0 anywhere in a given interval then y1andy2are linearly independent in that interval. An alternative expression for W(x), of which we will make use later, may be derived by diﬀerentiating (16.3) with respect to xto give W/prime=y1y/prime/prime 2+y/prime 1y/prime 2−y2y/prime/prime 1−y/prime 2y/prime 1=y1y/prime/prime 2−y/prime/prime 1y2. Since both y1andy2satisfy (16.1), we may substitute for y/prime/prime 1andy/prime/prime 2to obtain W/prime=−y1(py/prime 2+qy2)+(py/prime 1+qy1)y2=−p(y1y/prime 2−y/prime 1y2)=−pW. Integrating, we ﬁnd W(x)=Cexpbraceleftbigg −integraldisplayx p(u)dubracerightbigg , (16.4) where Cis a constant. We note further that in the special case p(x)≡0w eo b t a i n W=c o n s t a n t .IThe functions y1=s i n xandy2=c o s xare both solutions of the equation y/prime/prime+y= 0. Evaluate the Wronskian of these two solutions, and hence show that they are linearly independent. The Wronskian of y1andy2is given by W=y1y/prime 2−y2y/prime 1=−sin2x−cos2x=−1. Since W/negationslash= 0 the two solutions are linearly independent. We also note that y/prime/prime+y=0i s a special case of (16.1) with p(x) = 0. We therefore expect, from (16.4), that Wwill be a constant, as is indeed the case. J From the previous chapter we recall that, once we have obtained the general solution to the homogeneous second-order ODE (16.1) in the form (16.2), thegeneral solution to the inhomogeneous equation y /prime/prime+p(x)y/prime+q(x)y=f(x) (16.5) 538 16.1 SECOND-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS can be written as the sum of the solution to the homogeneous equation yc(x) (the complementary function) and anyfunction yp(x) (the particular integral) that satisﬁes (16.5) and is linearly independent of yc(x). We have therefore y(x)=c1y1(x)+c2y2(x)+yp(x). (16.6) General methods for obtaining yp, that are applicable to equations with variable coeﬃcients, such as the variation of parameters or Green’s functions, were dis- cussed in the previous chapter. An alternative description of the Green’s function method for solving inhomogeneous equations is given in the next chapter. For thepresent, however, we will restrict our attention to the solutions of homogeneousODEs in the form of convergent series. 16.1.1 Ordinary and singular points of an ODE So far we have implicitly assumed that y(x)i sa realfunction of a realvariable x. However, this is not always the case, and in the remainder of this chapter we broaden our discussion by generalising to a complex function y(z)o fa complex variable z. Let us therefore consider the second-order linear homogeneous ODE y /prime/prime+p(z)y/prime+q(z)=0 , (16.7) where now y/prime=dy/dz ; this is a straightforward generalisation of (16.1). A full discussion of complex functions and diﬀerentiation with respect to a complexvariable zis given in chapter 20, but for the purposes of the present chapter we need not concern ourselves with many of the subtleties that exist. In particular,we may treat diﬀerentiation with respect to zin an way analogous to ordinary diﬀerentiation with respect to a real variable x. In (16.7), if at some point z=z 0the functions p(z)a n d q(z) are ﬁnite and can be expressed as complex power series (see section 4.5) p(z)=∞summationdisplay n=0pn(z−z0)n,q (z)=∞summationdisplay n=0qn(z−z0)n then p(z)a n d q(z) are said to be analytic atz=z0, and this point is called an ordinary point of the ODE. If, however, p(z)o rq(z), or both, diverge at z=z0 then it is called a singular point of the ODE. Even if an ODE is singular at a given point z=z0, it may still possess a non-singular (ﬁnite) solution at that point. In fact the necessary and suﬃcientcondition†for such a solution to exist is that ( z−z 0)p(z)a n d( z−z0)2q(z) are both analytic at z=z0. Singular points that have this property are regular singular †See, for example, Jeﬀreys and Jeﬀreys, Mathematical Methods of Physics ,3 r de d .( C a m b r i d g e University Press, 1966), p. 479. 539 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS points, whereas any singular point not satisfying both these criteria is termed an irregular oressential singularity.ILegendre’s equation has the form (1−z2)y/prime/prime−2zy/prime+/lscript(/lscript+1 )y=0, (16.8) where /lscriptis a constant. Show that z=0is an ordinary point and z=±1are regular singular points of this equation. Firstly, divide through by 1 −z2to put the equation into our standard form (16.7): y/prime/prime−2z 1−z2y/prime+/lscript(/lscript+1 ) 1−z2y=0. Comparing this with (16.7), we identify p(z)a n d q(z)a s p(z)=−2z 1−z2=−2z (1 +z)(1−z),q (z)=/lscript(/lscript+1 ) 1−z2=/lscript(/lscript+1 ) (1 +z)(1−z). By inspection, p(z)a n d q(z) are analytic at z= 0, which is therefore an ordinary point, but both diverge for z=±1, which are thus singular points. However, at z=1w es e e that both ( z−1)p(z)a n d( z−1)2q(z) are analytic and hence z= 1 is a regular singular point. Similarly, at z=−1 both ( z+1 )p(z)a n d( z+1 )2q(z) are analytic, and it too is a regular singular point. J So far we have assumed that z0is ﬁnite. However, we may sometimes wish to determine the nature of the point |z|→∞ . This may be achieved straightforwardly by substituting w=1/zinto the equation and investigating the behaviour at w=0 .IShow that Legendre’s equation has a regular singularity at |z|→∞ . Letting w=1/z, the derivatives with respect to zbecome dy dz=dy dwdw dz=−1 z2dy dw=−w2dy dw, d2y dz2=dw dzd dw /dy dz / =−w2 / −2wdy dw−w2d2y dw2 / =w3 / 2dy dw+wd2y dw2 / . If we substitute these derivatives into Legendre’s equation (16.8) we obtain/ 1−1 w2 / w3 / 2dy dw+wd2y dw2 / +21 ww2dy dw+/lscript(/lscript+1 )y=0, which simpliﬁes to give w2(w2−1)d2y dw2+2w3dy dw+/lscript(/lscript+1 )y=0. Dividing through by w2(w2−1) to put the equation into standard form, and comparing with (16.7), we identify p(w)a n d q(w)a s p(w)=2w w2−1,q (w)=/lscript(/lscript+1 ) w2(w2−1). Atw=0 , p(w) is analytic but q(w) diverges, and so the point |z|→∞ is a singular point of Legendre’s equation. However, since wpandw2qare both analytic at w=0 ,|z|→∞ is a regular singular point. J 540 16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINT Equation Regular singularities Essential singularities Legendre∗ (1−z2)y/prime/prime−2zy/prime+/lscript(/lscript+1 )y=0 −1,1,∞ — Chebyshev (1−z2)y/prime/prime−zy/prime+n2y=0 −1,1,∞ — Bessel z2y/prime/prime+zy+(z2−ν2)y=0 0 ∞ Laguerre∗ zy/prime/prime+( 1−z)y/prime+αy=0 0 ∞ Simple harmonic oscillator y/prime/prime+ω2y=0 — ∞ Hermite y/prime/prime−2zy/prime+2αy=0 — ∞ Table 16.1 Important ODEs in the physical sciences and engineering. The asterisks indicate that the corresponding associated equations (discussed in the next chapter) have the same singular points. Table 16.1 lists the singular points of several second-order linear ODEs that play important roles in the analysis of many physics and engineering problems.In sections 16.6 and 16.7 we consider the the solution of Legendre’s and Bessel’sequations in terms of convergent series and discuss some useful properties ofthese solutions. The solutions of the remaining equations in table 16.1 may alsobe found in the form of convergent series, but a discussion of these solutions andtheir properties is left until the next chapter, where they are considered in the context of Sturm–Liouville systems. We now discuss the methods by which series solutions may be obtained. 16.2 Series solutions about an ordinary point Ifz=z 0is an ordinary point of (16.7) then it may be shown that everysolution y(z) of the equation is also analytic at z=z0. In our subsequent discussion we will take z0as the origin, i.e. z0= 0. If this is not already the case, then a substitution Z=z−z0will make it so. Since every solution is analytic, y(z)c a n be represented by a power series of the form (see section 20.13) y(z)=∞summationdisplay n=0anzn. (16.9) Moreover, it may be shown that such a power series converges for |z|<R,w h e r e Ris the radius of convergence and is equal to the distance from z=0t ot h e nearest singular point of the ODE (see chapter 20). At the radius of convergence,however, the series may or may not converge (as shown in section 4.5). 541 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS Since every solution of (16.7) is analytic at an ordinary point, it is always possible to obtain two independent solutions (from which the general solution (16.2) can be constructed) of the form (16.9). The derivatives of ywith respect to zare given by y/prime=∞summationdisplay n=0nanzn−1=∞summationdisplay n=0(n+1 )an+1zn, (16.10) y/prime/prime=∞summationdisplay n=0n(n−1)anzn−2=∞summationdisplay n=0(n+2 ) ( n+1 )an+2zn. (16.11) Note that, in each case, in the ﬁrst equality the sum can still start at n=0s i n c e the ﬁrst term in (16.10) and the ﬁrst two terms in (16.11) are automatically zero.The second equality in each case is obtained by shifting the summation index sothat the sum can be written in terms of coeﬃcients of z n. By substituting (16.9)– (16.11) into the ODE (16.7), and requiring that the coeﬃcients of each power of zs u mt oz e r o ,w eo b t a i na recurrence relation expressing anas a function of the previous ar(0≤r≤n−1).IFind the series solutions, about z=0,o f y/prime/prime+y=0. By inspection z= 0 is an ordinary point of the equation, and so we may obtain two independent solutions by making the substitution y= P∞ n=0anzn. Using (16.9) and (16.11) we ﬁnd ∞X n=0(n+2 ) ( n+1 )an+2zn+∞X n=0anzn=0, which may be written as ∞X n=0[(n+2 ) ( n+1 )an+2+an]zn=0. For this equation to be satisﬁed we require that the coeﬃcient of each power of zvanishes separately , and so we obtain the two-term recurrence relation an+2=−an (n+2 ) ( n+1 )forn≥0. Using this relation, we can calculate, say, the even coeﬃcients a2,a4,a6a n ds oo n ,f o r ag i v e n a0. Alternatively, starting with a1, we obtain the odd coeﬃcients a3,a5etc. Two independent solutions of the ODE can be obtained by setting either a0=0o r a1=0 . Firstly if we set a1= 0 and choose a0= 1 then we obtain the solution y1(z)=1−z2 2!+z4 4!−···=∞X n=0(−1)n (2n)!z2n. Secondly, if we set a0= 0 and choose a1= 1 then we obtain a second, independent , solution y2(z)=z−z3 3!+z5 5!−···=∞X n=0(−1)n (2n+1 ) !z2n+1. 542 16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINT Recognising these two series as cos zand sin z, we can write the general solution as y(z)=c1cosz+c2sinz, where c1andc2are arbitrary constants that are ﬁxed by boundary conditions (if supplied). We note that both solutions converge for all z, as might be expected since the ODE possesses no singular points (except |z|→∞ ). J Solving the above example was quite straightforward and the resulting series were easily recognised and written in closed form (i.e. in terms of elementary functions); this is not usually the case . Another simplifying feature of the previous example was that we obtained a two-term recurrence relation relating an+2and an, so that the odd- and even-numbered coeﬃcients were independent of one another. In general the recurrence relation expresses anas a function of any number of the previous ar(0≤r≤n−1).IFind the series solutions, about z=0,o f y/prime/prime−2 (1−z)2y=0. By inspection z= 0 is an ordinary point, and therefore we may ﬁnd two independent solutions by substituting y= P∞ n=0anzn. Using (16.10) and (16.11), and multiplying through by (1−z)2, we ﬁnd (1−2z+z2)∞X n=0n(n−1)anzn−2−2∞X n=0anzn=0, which leads to ∞X n=0n(n−1)anzn−2−2∞X n=0n(n−1)anzn−1+∞X n=0n(n−1)anzn−2∞X n=0anzn=0. In order to write all these series in terms of the coeﬃcients of zn, we must shift the summation index in the ﬁrst two sums, obtaining ∞X n=0(n+2 ) ( n+1 )an+2zn−2∞X n=0(n+1 )nan+1zn+∞X n=0(n2−n−2)anzn=0, which can be written as ∞X n=0(n+1 ) [ ( n+2 )an+2−2nan+1+(n−2)an]zn=0. By demanding that the coeﬃcients of each power of zvanish separately, we obtain the three-term recurrence relation (n+2 )an+2−2nan+1+(n−2)an=0 f o r n≥0, which determines anforn≥2i nt e r m so f a0anda1. Three-term (or more) recurrence relations are a nuisance and, in general, can be diﬃcult to solve. This particular recurrencerelation, however, has two straightf orward solutions. One solution is a n=a0for all n,i n which case (choosing a0= 1) we ﬁnd y1(z)=1+ z+z2+z3+···=1 1−z. 543 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS The other solution to the recurrence relation is a1=−2a0,a2=a0andan=0f o r n>2, so that (again choosing a0=1 )w eo b t a i na polynomial solution to the ODE: y2(z)=1−2z+z2=( 1−z)2. The linear independence of y1andy2is obvious but can be checked by computing the Wronskian W=y1y/prime 2−y/prime 1y2=1 1−z[−2(1−z)]−1 (1−z)2(1−z)2=−3. Since W/negationslash= 0 the two solutions y1andy2are indeed linearly independent. The general solution of the ODE is therefore y(z)=c1 1−z+c2(1−z)2. We observe that y1(and hence the general solution) is singular at z=1 ,w h i c hi st h e singular point of the ODE nearest to z= 0, but the polynomial solution y2is valid for all ﬁnite z. J The above example illustrates the possibility that, in some cases, we may ﬁnd that the recurrence relation leads to an=0f o r n>N , for one or both of the two solutions; we then obtain a polynomial solution to the equation. Polynomial solutions are discussed more fully in section 16.5, but one obvious property ofsuch solutions is that they converge for all ﬁnite z. By contrast, as mentioned above, for solutions in the form of an inﬁnite series the circle of convergence extends only as far as the singular point nearest to that about which the solutionis being obtained. 16.3 Series solutions about a regular singular point From table 16.1 we see that several of the most important second-order linear ODEs in physics and engineering have regular singular points in the ﬁnite complex plane. We must extend our discussion, therefore, to obtaining series solutions toODEs about such points. In what follows we assume that the regular singularpoint about which the solution is required is at z= 0, since, as we have seen, if this is not already the case then a substitution of the form Z=z−z 0will make it so. Ifz= 0 is a regular singular point of the equation y/prime/prime+p(z)y/prime+q(z)y=0 then p(z)a n d q(z) are not analytic at z= 0, and in general we should not expect to ﬁnd a power series solution of the form (16.9). We must therefore extend the method to include a more general form for the solution. In fact it may be shown (Fuch’s theorem) that there exists at least one solution to the above equation, of the form y=zσ∞summationdisplay n=0anzn, (16.12) 544 16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT where the exponent σis a number that may be real or complex and where a0/negationslash=0 (since, if it were otherwise, σcould be redeﬁned as σ+1o r σ+2o r ···so as to make a0/negationslash= 0). Such a series is called a generalised power series or Frobenius series . As in the case of a simple power series solution, the radius of convergence of the Frobenius series is, in general, equal to the distance to the nearest singularity of the ODE. Since z= 0 is a regular singularity of the ODE, it follows that zp(z)a n d z2q(z) a r ea n a l y t i ca t z= 0, so that we may write zp(z)≡s(z)=∞summationdisplay n=0snzn z2q(z)≡t(z)=∞summationdisplay n=0tnzn, where we have deﬁned the analytic functions s(z)a n d t(z) for later convenience. The original ODE therefore becomes y/prime/prime+s(z) zy/prime+t(z) z2y=0. Let us substitute the Frobenius series (16.12) into this equation. The derivatives of (16.12) with respect to xare given by y/prime=∞summationdisplay n=0(n+σ)anzn+σ−1, (16.13) y/prime/prime=∞summationdisplay n=0(n+σ)(n+σ−1)anzn+σ−2, (16.14) and we obtain ∞summationdisplay n=0(n+σ)(n+σ−1)anzn+σ−2+s(z)∞summationdisplay n=0(n+σ)anzn+σ−2+t(z)∞summationdisplay n=0anzn+σ−2=0. Dividing this equation through by zσ−2we ﬁnd ∞summationdisplay n=0[(n+σ)(n+σ−1) +s(z)(n+σ)+t(z)]anzn=0. (16.15) Setting z= 0, all terms in the sum with n>0 vanish, implying that [σ(σ−1) +s(0)σ+t(0)]a0=0, which, since we require a0/negationslash= 0, yields the indicial equation σ(σ−1) +s(0)σ+t(0) = 0 . (16.16) This equation is a quadratic in σand in general has two roots, the nature of which determines the forms of possible series solutions. 545 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS The two roots of the indicial equation σ1andσ2are called the indices of the regular singular point. By substituting each of these roots into (16.15) in turn andrequiring that the coeﬃcients of each power of zvanish separately, we obtain a recurrence relation (for each root) expressing each a nas a function of the previous ar(0≤r≤n−1). Depending on the roots of the indicial equation σ1andσ2, there are three possible general cases, which we now discuss. 16.3.1 Distinct roots not diﬀering by an integer If the roots of the indicial equation σ1andσ2diﬀer by an amount that is not an integer then the recurrence relations corresponding to each root lead to twolinearly independent solutions of the ODE, y 1(z)=zσ1∞summationdisplay n=0anzn,y 2(z)=zσ2∞summationdisplay n=0bnzn. The linear independence of these two solutions follows from the fact that y2/y1 is not a constant since σ1−σ2is not an integer. Because y1andy2are linearly independent, we may use them to construct the general solution y=c1y1+c2y2. We also note that this case includes complex conjugate roots where σ2=σ∗ 1, since σ1−σ2=σ1−σ∗ 1=2iImσ1cannot be equal to a real integer.IFind the power series solutions about z=0of 4zy/prime/prime+2y/prime+y=0. Dividing through by 4 zto put the equation into standard form, we obtain y/prime/prime+1 2zy/prime+1 4zy=0, (16.17) and on comparing with (16.7) we identify p(z)=1 /(2z)a n d q(z)=1 /(4z). Clearly z=0 is a singular point of (16.17), but since zp(z)=1 /2a n d z2q(z)=z/4 are ﬁnite there, it is a regular singular point. We therefore substitute the Frobenius series y=zσ P∞ n=0anzn into (16.17). Using (16.13) and (16.14), we obtain ∞X n=0(n+σ)(n+σ−1)anzn+σ−2+1 2z∞X n=0(n+σ)anzn+σ−1+1 4z∞X n=0anzn+σ=0, which on dividing through by zσ−2gives ∞X n=0 / (n+σ)(n+σ−1) +1 2(n+σ)+1 4z / anzn=0. (16.18) If we set z= 0 then all terms in the sum with n>0 vanish, and we obtain the indicial equation σ(σ−1) +1 2σ=0, which has roots σ=1/2a n d σ= 0. Since these roots do not diﬀer by an integer we expect to ﬁnd two independent solutions to (16.17), in the form of Frobenius series. 546 16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINT Demanding that the coeﬃcients of znvanish separately in (16.18), we obtain the recurrence relation (n+σ)(n+σ−1)an+1 2(n+σ)an+1 4an−1=0. (16.19) If we choose the larger root, σ=1/2, of the indicial equation then (16.19) becomes (4n2+2n)an+an−1=0⇒ an=−an−1 2n(2n+1 ). Setting a0= 1 we ﬁnd an=(−1)n/(2n+ 1)! and so the solution to (16.17) is y1(z)=√z∞X n=0(−1)n (2n+1 ) !zn =√z−(√z)3 3!+(√z)5 5!−···=s i n√z. To obtain the second solution we set σ= 0 (the smaller root of the indicial equation) in (16.19), which gives (4n2−2n)an+an−1=0⇒ an=−an−1 2n(2n−1). Setting a0=1n o wg i v e s an=(−1)n/(2n)!, and so the second (independent) solution to (16.17) is y2(z)=∞X n=0(−1)n (2n)!zn=1−(√z)2 2!+(√ 4)4 4!−···=c o s√z. We may check that y1(z)a n d y2(z) are indeed linearly independent by computing the Wronskian W=y1y/prime 2−y2y/prime 1 =s i n√z / −1 2√zsin√z / −cos√z /1 2√zcos√z / =−1 2√z /; sin2√z+c o s2√z / =−1 2√z/negationslash=0. Since W/negationslash= 0 the solutions y1(z)a n d y2(z) are linearly independent. Hence the general solution to (16.17) is given by y(z)=c1sin√z+c2cos√z. J 16.3.2 Repeated root of the indicial equation If the indicial equation has a repeated root, so that σ1=σ2=σ, then obviously only one solution in the form of a Frobenius series (16.12) may be found asdescribed above, i.e. y 1(z)=zσ∞summationdisplay n=0anzn. Methods for obtaining a second, linearly independent, solution are discussed in section 16.4. 547 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 16.3.3 Distinct roots diﬀering by an integer Whatever the roots of the indicial equation, the recurrence relation corresponding to the larger of the two always leads to a solution of the ODE. However, if theroots of the indicial equation diﬀer by an integer then the recurrence relationcorresponding to the smaller root may or may not lead to a second linearlyindependent solution, depending on the ODE under consideration. Note that for complex roots of the indicial equation, the ‘larger’ root is taken to be the one with the larger real part.IFind the power series solutions about z=0of z(z−1)y/prime/prime+3zy/prime+y=0. (16.20) Dividing through by z(z−1) to put the equation into standard form, we obtain y/prime/prime+3 (z−1)y/prime+1 z(z−1)y=0, (16.21) and on comparing with (16.7) we identify p(z)=3 /(z−1) and q(z)=1 /[z(z−1)]. We immediately see that z= 0 is a singular point of (16.21), but since zp(z)=3 z/(z−1) and z2q(z)=z/(z−1) are ﬁnite there, it is a regular singular point and we expect to ﬁnd at least one solution in the form of a Frobenius series. We therefore substitute y=zσ P∞ n=0anzn into (16.21), and using (16.13) and (16.14), we obtain ∞X n=0(n+σ)(n+σ−1)anzn+σ−2+3 z−1∞X n=0(n+σ)anzn+σ−1 +1 z(z−1)∞X n=0anzn+σ=0, which on dividing through by zσ−2gives ∞X n=0 / (n+σ)(n+σ−1) +3z z−1(n+σ)+z z−1 / anzn=0. Although we could use this expression to ﬁnd the indicial equation and recurrence relations, the working is simpler if we now multiply through by z−1t og i v e ∞X n=0[(z−1)(n+σ)(n+σ−1) + 3 z(n+σ)+z]anzn=0. (16.22) If we set z= 0 then all terms in the sum with the exponent of zgreater than zero vanish, and we obtain the indicial equation σ(σ−1) = 0 , which has the roots σ=1a n d σ= 0. Since the roots diﬀer by an integer (unity), it may not be possible to ﬁnd two linearly independent solutions of (16.21) in the form of Frobeniusseries. We are guaranteed, however, to ﬁnd one such solution corresponding to the largerroot, σ=1 . Demanding that the coeﬃcients of z nvanish separately in (16.22), we obtain the recurrence relation (n−1+σ)(n−2+σ)an−1−(n+σ)(n+σ−1)an+3 (n−1+σ)an−1+an−1=0, 548 16.4 OBTAINING A SECOND SOLUTION which can be simpliﬁed to give (n+σ−1)an=(n+σ)an−1. (16.23) Substituting σ= 1 into this expression, we obtain an= /n+1 n / an−1, and setting a0= 1 we ﬁnd an=n+ 1; so one solution to (16.21) is y1(z)=z∞X n=0(n+1 )zn=z(1 + 2 z+3z2+···) =z (1−z)2. (16.24) If we attempt to ﬁnd a second solution (corresponding to the smaller root of the indicial equation) by setting σ= 0 in (16.23), we ﬁnd an= /n n−1 / an−1, but we require a0/negationslash=0 ,s o a1is formally inﬁnite and the method fails. We discuss how to ﬁnd a second linearly independent solution in the next section. J One particular case is also worth mentioning. If the point about which the solution is required, i.e. z= 0, is in fact an ordinary point of the ODE rather than a regular singular point, then substitution of the Frobenius series (16.12) leads toan indicial equation with roots σ=0a n d σ= 1. Although these roots diﬀer by an integer (unity), the recurrence relations corresponding to the two roots yieldtwo linearly independent power series solutions (one for each root), as expectedfrom section 16.2. It is always worth investigating whether a series found as a solution to a problem is summable in closed form or expressible in terms of known functions.Nevertheless, the reader should avoid gaining the impression that this is always so or that, if one worked hard enough, a closed-form solution could always be found without using the series method. As mentioned earlier, this is notthe case, and very often an inﬁnite series solution is the best one can do. 16.4 Obtaining a second solution Whilst attempting to ﬁnd a solution to an ODE in the form of a Frobenius series about a regular singular point, we found in the previous section that when theindicial equation has a repeated root, or roots diﬀering by an integer, we can (ingeneral) ﬁnd only one solution of this form. In order to construct the general solution to the ODE, however, we require two linearly independent solutions y 1 andy2. We now consider several methods for obtaining a second solution in this case. 549 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 16.4.1 The Wronskian method Ify1andy2are two linearly independent solutions of the standard equation y/prime/prime+p(z)y/prime+q(z)y=0 then the Wronskian of these two solutions is given by W(z)= y1y/prime 2−y2y/prime 1. Dividing the Wronskian by y2 1we obtain W y2 1=y/prime 2 y1−y/prime 1 y2 1y2=y/prime 2 y1+bracketleftbiggd dzparenleftbigg1 y1parenrightbiggbracketrightbigg y2=d dzparenleftbiggy2 y1parenrightbigg , which integrates to give y2(z)=y1(z)integraldisplayzW(u) y2 1(u)du. Now using the alternative expression for W(z) given in (16.4) with C=1( s i n c e we are not concerned with this normalising factor), we ﬁnd y2(z)=y1(z)integraldisplayz1 y2 1(u)expbraceleftbigg −integraldisplayu p(v)dvbracerightbigg du. (16.25) Hence, given y1, we can in principle compute y2. Note that the lower limits of integration have been omitted. If constant lower limits are included then theymerely lead to a constant times the ﬁrst solution.IFind a second solution to (16.21) using the Wronskian method. For the ODE (16.21) we have p(z)=3 /(z−1), and from (16.24) we see that one solution to (16.21) is y1=z/(1−z)2. Substituting for pandy1in (16.25) we have y2(z)=z (1−z)2 Zz(1−u)4 u2exp / − Zu3 v−1dv / du =z (1−z)2 Zz(1−u)4 u2exp[−3ln(u−1)]du =z (1−z)2 Zzu−1 u2du =z (1−z)2 / lnz+1 z / . By calculating the Wronskian of y1andy2it is easily shown that, as expected, the two solutions are linearly independent. In fact, as the Wronskian has already been evaluated asW(u)=e x p [−3ln(u−1)], i.e. W(z)=(z−1)−3, no calculation is needed. J An alternative (but equivalent) method of ﬁnding a second solution is simply to assume that the second solution has the form y2(z)=u(z)y1(z) for some function u(z) to be determined (this method was discussed more fully in subsection 15.2.3). From (16.25), we see that the second solution derived from the Wronskian is indeed of this form. Substituting y2(z)= u(z)y1(z) into the ODE leads to a ﬁrst-order ODE in which u/primeis the dependent variable; this may then be solved. 550 16.4 OBTAINING A SECOND SOLUTION 16.4.2 The derivative method The derivative method of ﬁnding a second solution begins with the derivation of a recurrence relation for the coeﬃcients anin a Frobenius series solution, as in the previous section. However, rather than putting σ=σ1in this recurrence relation to evaluate the ﬁrst series solution, we now keep σas a variable parameter. This means that the computed anare functions of σand the computed solution is now a function of zandσ: y(z,σ)=zσ∞summationdisplay n=0an(σ)zn. (16.26) Of course, if we put σ=σ1in this, we obtain immediately the ﬁrst series solution, but for the moment we leave σas a parameter. For brevity let us denote the diﬀerential operator on the LHS of our standard ODE (16.7) by L,s ot h a t L=d2 dz2+p(z)d dz+q(z), and examine the eﬀect of Lon the series y(z,σ) in (16.26). It is clear that the series Ly(z,σ) will contain only a term in zσ, since the recurrence relation deﬁning thean(σ) is such that these coeﬃcients vanish for higher powers of z.B u tt h e coeﬃcient of zσis simply the LHS of the indicial equation. Therefore, if the roots of the indicial equation are σ=σ1andσ=σ2then it follows that Ly(z,σ)=a0(σ−σ1)(σ−σ2)zσ. (16.27) Therefore, as in the previous section, we see that for y(z,σ)t ob eas o l u t i o no f the ODE Ly=0 , σmust equal σ1orσ2. For simplicity we shall set a0= 1 in the following discussion. Let us ﬁrst consider the case in which the two roots of the indicial equation are equal, i.e. σ2=σ1. From (16.27) we then have Ly(z,σ)=(σ−σ1)2zσ. Diﬀerentiating this equation with respect to σwe obtain ∂ ∂σ[Ly(z,σ)]=(σ−σ1)2zσlnz+2 (σ−σ1)zσ, which equals zero if σ=σ1. But since ∂/∂σandLare operators that diﬀerentiate with respect to diﬀerent variables we can reverse their order, implying that Lbracketleftbigg∂ ∂σy(z,σ)bracketrightbigg =0 a t σ=σ1. Hence the function in square brackets, evaluated at σ=σ1and denoted by bracketleftbigg∂ ∂σy(z,σ)bracketrightbigg σ=σ1, (16.28) 551 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS is also a solution of the original ODE Ly= 0, and is in fact the second linearly independent solution for which we were looking. The case in which the roots of the indicial equation diﬀer by an integer is slightly more complicated but can be treated in a similar way. In (16.27), since L diﬀerentiates with respect to zwe may multiply (16.27) by any function of σ,s a y σ−σ2, and take this function inside the operator Lon the LHS to obtain L[(σ−σ2)y(z,σ)]=(σ−σ1)(σ−σ2)2zσ. (16.29) Therefore the function [(σ−σ2)y(z,σ)]σ=σ2 is also a solution of the ODE Ly= 0. However, it can be proved †that this function is a simple multiple of the ﬁrst solution y(z,σ1), showing that it is not linearly independent and that we must ﬁnd another solution. To do this we diﬀerentiate (16.29) with respect to σand ﬁnd ∂ ∂σ{L[(σ−σ2)y(z,σ)]}=(σ−σ2)2zσ+2 (σ−σ1)(σ−σ2)zσ +(σ−σ1)(σ−σ2)2zσlnz, which is equal to zero if σ=σ2. As previously, since ∂/∂σ andLare operators that diﬀerentiate with respect to diﬀerent variables, we can reverse their order toobtain Lbraceleftbigg∂ ∂σ[(σ−σ2)y(z,σ)]bracerightbigg =0 a t σ=σ2, and so the function braceleftbigg∂ ∂σ[(σ−σ2)y(z,σ)]bracerightbigg σ=σ2(16.30) is also a solution of the original ODE Ly= 0, and is in fact the second linearly independent solution.IFind a second solution to (16.21) using the derivative method. From (16.23) the recurrence relation (with σas a parameter) is given by (n+σ−1)an=(n+σ)an−1. Setting a0= 1 we ﬁnd that the coﬃcients have the particularly simple form an(σ)= (σ+n)/σ. We therefore consider the function y(z,σ)=zσ∞X n=0an(σ)zn=zσ∞X n=0σ+n σzn. †For a fuller discussion see, for example, Riley, Mathematical Methods for the Physical Sciences , (Cambridge University Press, 1974), pp. 158–9. 552 16.4 OBTAINING A SECOND SOLUTION The smaller root of the indicial equation for (16.21) is σ2= 0, and so from (16.30) a second, linearly independent, solution to the ODE is/∂ ∂σ[σy(z,σ)] / σ=0= /( ∂ ∂σ /" zσ∞X n=0(σ+n)zn /# /) σ=0. The derivative with respect to σis given by ∂ ∂σ /" zσ∞X n=0(σ+n)zn /# =zσlnz∞X n=0(σ+n)zn+zσ∞X n=0zn, which on setting σ= 0 gives the second solution y2(z)=l n z∞X n=0nzn+∞X n=0zn =z (1−z)2lnz+1 1−z =z (1−z)2 / lnz+1 z−1 / . This second solution is the same as that obtained by the Wronskian method in the previous subsection except for the addition of some of the ﬁrst solution. J 16.4.3 Series form of the second solution Using any of the methods discussed above, we can ﬁnd the general form of the second solution to the ODE. This form is most easily found, however, using thederivative method. Let us ﬁrst consider the case where the two solutions of theindicial equation are equal. In this case a second solution is given by (16.28),which may be written as y 2(z)=bracketleftbigg∂y(z,σ) ∂σbracketrightbigg σ=σ1 =( l n z)zσ1∞summationdisplay n=0an(σ1)zn+zσ1∞summationdisplay n=1bracketleftbiggdan(σ) dσbracketrightbigg σ=σ1zn =y1(z)l nz+zσ1∞summationdisplay n=1bnzn, where bn=[dan(σ)/dσ]σ=σ1. In the case where the roots of the indicial equation diﬀer by an integer (not equal to zero), then from (16.30) a second solution is given by y2(z)=braceleftbigg∂ ∂σ[(σ−σ2)y(z,σ)]bracerightbigg σ=σ2 =l nzbracketleftBigg (σ−σ2)zσ∞summationdisplay n=0an(σ)znbracketrightBigg σ=σ2+zσ2∞summationdisplay n=0bracketleftbiggd dσ(σ−σ2)an(σ)bracketrightbigg σ=σ2zn. 553 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS But, as we mentioned in the previous section, [(σ−σ2)y(z,σ)]atσ=σ2is just a multiple of the ﬁrst solution y(z,σ1). Therefore the second solution is of the form y2(z)=cy1(z)l nz+zσ2∞summationdisplay n=0bnzn, where cis a constant. In some cases, however, cmight be zero and so the second solution would not contain the term in ln zand could be written simply as a Frobenius series. Clearly this corresponds to the case in which the substitution ofa Frobenius series into the original ODE yields two solutions automatically. 16.5 Polynomial solutions We have seen that the evaluation of successive terms of a series solution to a diﬀerential equation is carried out by means of a recurrence relation. The formof the relation for a ndepends upon n, the previous values of ar(r<n)a n dt h e parameters of the equation. It may happen, as a result of this, that for some value of n=N+ 1 the computed value aN+1is zero and that all higher aralso vanish. If this is so, and the corresponding solution of the indicial equation σ is a positive integer or zero, then we are left with a ﬁnite polynomial of degreeN /prime=N+σas a solution of the ODE: y(z)=Nsummationdisplay n=0anzn+σ. (16.31) In many applications in theoretical physics (particularly in quantum mechanics) the termination of a potentially inﬁnite series after a ﬁnite number of termsis of crucial importance in establishing physically acceptable descriptions andproperties of systems. The condition under which such a termination occurs istherefore of considerable importance.IFind power series solutions about z=0of y/prime/prime−2zy/prime+λy=0. (16.32) For what values of λdoes the equation possess a polynomial solution? Find such a solution forλ=4. Clearly z= 0 is an ordinary point of (16.32) and so we look for solutions of the form y= P∞ n=0anzn. Substituting this into the ODE and multiplying through by z2we ﬁnd ∞X n=0[n(n−1)−2z2n+λz2]anzn=0. By demanding that the coeﬃcients of each power of zvanish separately we derive the recurrence relation n(n−1)an−2(n−2)an−2+λan−2=0, 554 16.6 LEGENDRE’S EQUATION which may be rearranged to give an=2(n−2)−λ n(n−1)an−2forn≥2. (16.33) The odd and even coeﬃcients are therefore independent of one another, and two solutions to (16.32) may be derived. We either set a1=0a n d a0=1t oo b t a i n y1(z)=1−λz2 2!−λ(4−λ)z4 4!−λ(4−λ)(8−λ)z6 6!−··· (16.34) or set a0=0a n d a1=1t oo b t a i n y2(z)=z+( 2−λ)z3 3!+( 2−λ)(6−λ)z5 5!+( 2−λ)(6−λ)(10−λ)z7 7!+···. Now from the recurrence relation (16.33) (or in this case from the expressions for y1 andy2themselves) we see that for the ODE to possess a polynomial solution we require λ=2 (n−2) for n≥2o rm o r es i m p l y λ=2nforn≥0, i.e. λmust be an even positive integer. If λ= 4 then from (16.34) the ODE has the polynomial solution y1(z)=1−4z2 2!=1−2z2. J A simpler method of obtaining ﬁnite polynomial solutions is to assume a solution of the form (16.31), where aN/negationslash= 0. Instead of starting with the lowest power of z, as we have done up to now, this time we start by considering the coeﬃcient of the highest power zN; such a power now exists because of our assumed form of solution.IBy assuming a polynomial solution ﬁnd the values of λin (16.32) for which such a solution exists. We assume a polynomial solution to (16.32) of the form y= PN n=0anzn. Substituting this form into (16.32) we ﬁnd NX n=0 / n(n−1)anzn−2−2zna nzn−1+λanzn / =0. Now, instead of starting with the lowest power of z, we start with the highest. Thus, demanding that the coeﬃcient of zNvanishes, we require −2N+λ=0 ,i . e . λ=2N,a sw e found in the previous example. By demanding that the coeﬃcient of a general power of z is zero, the same recurrence relation as above may be derived and the solutions found. J 16.6 Legendre’s equation In previous sections we have discussed methods for obtaining series solutions of second-order linear ODEs. In this section and the next we apply some of thesemethods to ﬁnding the series solutions of the two most important equations listedin table 16.1, namely Legendre’s equation and Bessel’s equation. As mentioned earlier, the remaining equations in table 16.1 may also be solved by the methods discussed in this chapter. These equations, and the properties of their solutions,are discussed brieﬂy in the next chapter. 555 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS We now consider Legendre’s equation (1−z2)y/prime/prime−2zy/prime+/lscript(/lscript+1 )y=0, (16.35) which occurs in numerous physical applications and particularly in problems with axial symmetry when they are expressed in spherical polar coordinates. In normalusage the variable zin Legendre’s equation is the cosine of the polar angle in spherical polars, and thus −1≤z≤1. The parameter /lscriptis a given real number, and any solution of (16.35) is called a Legendre function . In subsection 16.1.1, we showed that z= 0 is an ordinary point of (16.35), and so we expect to ﬁnd two linearly independent solutions of the form y=summationtext ∞ n=0anzn. Substituting, we ﬁnd ∞summationdisplay n=0bracketleftbig n(n−1)anzn−2−n(n−1)anzn−2nanzn+/lscript(/lscript+1 )anznbracketrightbig =0, which on collecting terms gives ∞summationdisplay n=0{(n+2 ) ( n+1 )an+2−[n(n+1 )−/lscript(/lscript+1 ) ] an}zn=0. The recurrence relation is therefore an+2=[n(n+1 )−/lscript(/lscript+1 ) ] (n+1 ) ( n+2 )an, (16.36) forn=0,1,2,.... If we choose a0=1a n d a1= 0 then we obtain the solution y1(z)=1−/lscript(/lscript+1 )z2 2!+(/lscript−2)/lscript(/lscript+1 ) ( /lscript+3 )z4 4!−···, (16.37) whereas choosing a0=0a n d a1= 1 we ﬁnd a second solution y2(z)=z−(/lscript−1)(/lscript+2 )z3 3!+(/lscript−3)(/lscript−1)(/lscript+2 ) ( /lscript+4 )z5 5!−···.(16.38) By applying the ratio test to these series (see subsection 4.3.2), we ﬁnd that both series converge for |z|<1, and so their radius of convergence is unity, which (as expected) is the distance to the nearest singular point of the equation. Since(16.37) contains only even powers of zand (16.38) contains only odd powers, these two solutions cannot be proportional to one another, and are thereforelinearly independent. Hence y=c 1y1+c2y2is the general solution to (16.35) for |z|<1. 16.6.1 General solution for integer /lscript Now, if /lscriptis an integer in Legendre’s equation (16.35), i.e. /lscript=0,1,2,..., then the recurrence relation (16.36) gives a/lscript+2=[/lscript(/lscript+1 )−/lscript(/lscript+1 ) ] (/lscript+1 ) ( /lscript+2 )a/lscript=0, 556 16.6 LEGENDRE’S EQUATION P0 P1P2 P3−1 −1−0.5 0.5 11 z −22 Figure 16.1 The ﬁrst four Legendre polynomials. i.e. the series terminates and we obtain a polynomial solution of order /lscript.T h e s e solutions (suitably normalised) are called Legendre polynomials of order /lscript;t h e y are written P/lscript(z) and are valid for all ﬁnite z. It is conventional to normalise P/lscript(z)i ns u c haw a yt h a t P/lscript(1) = 1, and as a consequence P/lscript(−1) = (−1)/lscript.T h e ﬁrst few Legendre polynomials are easily constructed and are given by P0(z)=1 P1(z)=z P2(z)=1 2(3z2−1) P3(z)=1 2(5z3−3z) P4(z)=1 8(35z4−30z2+3 ) P5(z)=1 18(63z5−70z3+1 5z). The ﬁrst four Legendre polynomials are plotted in ﬁgure 16.1. According to whether /lscriptis an even or odd integer respectively, either y1(z) in (16.37) or y2(z) in (16.38) terminates to give a multiple of the corresponding Legendre polynomial P/lscript(z). In either case, however, the other series does not terminate and therefore converges only for |z|<1. According to whether /lscriptis even or odd we deﬁne Legendre functions of the second kind asQ/lscript(z)=α/lscripty2(z) orQ/lscript(z)=β/lscripty1(z) respectively, where the constants α/lscriptandβ/lscriptare conventionally taken to have the values α/lscript=(−1)/lscript/22/lscript[(/lscript/2)!]2 /lscript!for/lscripteven, (16.39) β/lscript=(−1)(/lscript+1)/22/lscript−1{[(/lscript−1)/2]!}2 /lscript!for/lscriptodd. (16.40) 557 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS These normalisation factors are chosen so that the Q/lscript(z) obey the same recurrence r e l a t i o n sa st h e P/lscript(z) (see subsection 16.6.2). The general solution of Legendre’s equation for integer /lscriptis therefore y(z)=c1P/lscript(z)+c2Q/lscript(z), (16.41) where P/lscript(z) is a polynomial of order /lscript, and so converges for all z,a n d Q/lscript(z)i s an inﬁnite series that converges only for |z|<1.† By using the Wronskian method, section 16.4, one may obtain closed forms for theQ/lscript(z).IUse the Wronskian method to ﬁnd a closed-form expression for Q0(z). From (16.25) a second solution to Legendre’s equation (16.35), with /lscript=0 ,i s y2(z)=P0(z) Zz1 [P0(u)]2exp /Zu2v 1−v2dv / du = Zz exp / −ln(1−u2) / du = Zzdu (1−u2)=1 2ln /1+z 1−z / , (16.42) where in the second line we have used the fact that P0(z)=1 . All that remains is to adjust the normalisation of this solution so that it agrees with (16.39). Expanding the logarithm in (16.42) as a Maclaurin series we obtain y2(z)=z+z3 3+z5 5+···. Comparing this with the expression for Q0(z), using (16.38) with /lscript= 0 and the normali- sation (16.39), we ﬁnd that y2(z) is already correctly normalised, and so Q0(z)=1 2ln /1+z 1−z / . Of course, we might have recognised the series (16.38) for /lscript= 0, but to do so for larger /lscript would prove progressively more diﬃcult. J Using the above method for /lscript= 1, we ﬁnd Q1(z)=1 2zlnparenleftbigg1+z 1−zparenrightbigg −1. Closed forms for higher-order Q/lscript(z) may now be found using the recurrence relation (16.55) derived in the next subsection. †It is possible, in fact, to ﬁnd a second solution in terms of an inﬁnite series of negative powers of zthat is ﬁnite for |z|>1. 558 16.6 LEGENDRE’S EQUATION 16.6.2 Properties of Legendre polynomials As stated earlier, when encountered in physical problems the variable zin Leg- endre’s equation is usually the cosine of the polar angle θin spherical polar coordinates, and we then require the solution y(z) to be regular at z=±1, which corresponds to θ=0o r θ=π. For this to occur we require the equation to have a polynomial solution, and so /lscriptmust be an integer. Furthermore, we also require the coeﬃcient c2of the function Q/lscript(z) in (16.41) to be zero, since Q/lscript(z) is singular atz=±1, with the result that the general solution is simply some multiple of the relevant Legendre polynomial P/lscript(z). In this section we will study the properties of the Legendre polynomials P/lscript(z) in some detail. Rodrigues’ formula As an aid to establishing further properties of the Legendre polynomials we now develop Rodrigues’ representation of these functions. Rodrigues’ formula for theP /lscript(z)i s P/lscript(z)=1 2/lscript/lscript!d/lscript dz/lscript(z2−1)/lscript. (16.43) To prove that this is a representation we let u=(z2−1)/lscript,s ot h a t u/prime=2/lscriptz(z2−1)/lscript−1 and (z2−1)u/prime−2/lscriptzu=0. If we diﬀerentiate this expression /lscript+ 1 times using Leibnitz’ theorem, we obtain bracketleftbig (z2−1)u(/lscript+2)+2z(/lscript+1 )u(/lscript+1)+/lscript(/lscript+1 )u(/lscript)bracketrightbig −2/lscriptbracketleftbig zu(/lscript+1)+(/lscript+1 )u(/lscript)bracketrightbig =0, which reduces to (z2−1)u(/lscript+2)+2zu(/lscript+1)−/lscript(/lscript+1 )u(/lscript)=0. Changing the sign all through and comparing the resulting expression with Legendre’s equation (16.35), we see that u(/lscript)satisﬁes the same equation as P/lscript(z), and so u(/lscript)(z)=c/lscriptP/lscript(z), (16.44) for some constant c/lscriptthat depends on /lscript. To establish the value of c/lscriptwe note that the only term in the expression for the /lscriptth derivative of ( z2−1)/lscriptthat does not contain a factor z2−1, and therefore does not vanish at z=1 ,i s (2z)/lscript/lscript!(z2−1)0. Putting z= 1 in (16.44) and recalling that P/lscript(1) = 1, therefore shows that c/lscript=2/lscript/lscript!, thus completing the proof of Rodrigues’ formula (16.43). 559 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSIUse Rodrigues’ formula to show that I/lscript= Z1 −1P/lscript(z)P/lscript(z)dz=2 2/lscript+1. (16.45) The result is trivially obvious for /lscript= 0 and so we assume /lscript≥1. Then, by Rodrigues’ formula, I/lscript=1 22/lscript(/lscript!)2 Z1 −1 /d/lscript(z2−1)/lscript dz/lscript //d/lscript(z2−1)/lscript dz/lscript / dz. Repeated integration by parts, with all boundary terms vanishing, reduces this to I/lscript=(−1)/lscript 22/lscript(/lscript!)2 Z1 −1(z2−1)/lscriptd2/lscript dz2/lscript(z2−1)/lscriptdz =(2/lscript)! 22/lscript(/lscript!)2 Z1 −1(1−z2)/lscriptdz. If we write K/lscript= Z1 −1(1−z2)/lscriptdz, then integration by parts (taking a factor 1 as the second part) gives K/lscript= Z1 −12/lscriptz2(1−z2)/lscript−1dz. Writing 2 /lscriptz2as 2/lscript−2/lscript(1−z2)w eo b t a i n K/lscript=2/lscript Z1 −1(1−z2)/lscript−1dz−2/lscript Z1 −1(1−z2)/lscriptdz =2/lscriptK/lscript−1−2/lscriptK/lscript and hence the recurrence relation (2 /lscript+1 )K/lscript=2/lscriptK/lscript−1. We therefore ﬁnd K/lscript=2/lscript 2/lscript+12/lscript−2 2/lscript−1···2 3K0=2/lscript/lscript!2/lscript/lscript! (2/lscript+1 ) !2=22/lscript+1(/lscript!)2 (2/lscript+1 ) !, which, when substituted into the expression for I/lscript, establishes the required result. J Mutual orthogonality of Legendre polynomials Another useful property of the P/lscript(z) is their mutual orthogonality, i.e. that integraldisplay1 −1P/lscript(z)Pk(z)dz=0 i f /lscript/negationslash=k. (16.46) More general considerations concerning the mutual orthogonality of solutions to various classes of second-order linear ODEs are discussed in the next chapter,but for the moment we concentrate on the speciﬁc proof of (16.46). Since the P /lscript(z) satisfy Legendre’s equation we may write bracketleftbig (1−z2)P/prime /lscriptbracketrightbig/prime+/lscript(/lscript+1 )P/lscript=0, 560 16.6 LEGENDRE’S EQUATION where P/prime /lscript=dP/lscript/dz. Multiplying through by Pkand integrating from z=−1t o z=1 ,w eo b t a i n integraldisplay1 −1Pkbracketleftbig (1−z2)P/prime /lscriptbracketrightbig/primedz+integraldisplay1 −1Pk/lscript(/lscript+1 )P/lscriptdz=0. Integrating the ﬁrst term by parts and noting that the boundary contribution vanishes at both limits because of the factor 1 −z2, we ﬁnd −integraldisplay1 −1P/prime k(1−z2)P/prime /lscriptdz+integraldisplay1 −1Pk/lscript(/lscript+1 )P/lscriptdz=0. Now, if we reverse the roles of /lscriptandkand subtract one expression from the other, we conclude that [k(k+1 )−/lscript(/lscript+1 ) ]integraldisplay1 −1PkP/lscriptdz=0, and therefore since k/negationslash=/lscriptwe must have the result (16.46). As a particular case we note that if we put k=0w eo b t a i n integraldisplay1 −1P/lscript(z)dz=0 f o r /lscript/negationslash=0. As will be discussed more fully in the next chapter, the mutual orthogonality of theP/lscript(z) means that any reasonable function f(z) (i.e. one obeying the Dirichlet conditions discussed at the start of chapter 12) can be expressed in the interval|z|<1 as an inﬁnite sum of Legendre polynomials, f(z)=∞summationdisplay /lscript=0a/lscriptP/lscript(z), (16.47) where the coeﬃcients a/lscriptare given by a/lscript=2/lscript+1 2integraldisplay1 −1f(z)P/lscript(z)dz. (16.48)IProve the expression (16.48) for the coeﬃcients in the Legendre polynomial expansion of a function f(z). If we multiply (16.47) by Pm(z) and integrate from z=−1t oz= 1 then we obtainZ1 −1Pm(z)f(z)dz=∞X /lscript=0a/lscript Z1 −1Pm(z)P/lscript(z)dz =am Z1 −1Pm(z)Pm(z)dz=2am 2m+1, where we have used the orthogonality property (16.46) and the normalisation property (16.45). J 561 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS Generating function for Legendre polynomials A useful device for manipulating and studying sequences of functions or quantities labelled by an integer variable (here, the Legendre polynomials P/lscript(z) labelled by /lscript)i sagenerating function . The generating function has perhaps its greatest utility in the area of probability theory (see chapter 26). However, it is also a greatconvenience in our present study. The generating function for, say, a series of functions f n(z)f o r n=0,1,2,...is a function G(z,h), containing as well as za dummy variable h, such that G(z,h)=∞summationdisplay n=0fn(z)hn, i.e.fn(z) is the coeﬃcient of hnin the expansion of Gin powers of h. The utility of the device lies in the fact that sometimes it is possible to ﬁnd a closed form forG(z,h). For our study of Legendre polynomials let us consider the functions Pn(z) deﬁned by the equation G(z,h)=( 1−2zh+h2)−1/2=∞summationdisplay n=0Pn(z)hn. (16.49) As we show below, the functions so deﬁned are identical to the Legendre poly- nomials and the function (1 −2zh+h2)−1/2is in fact the generating function for them. In the process we will also deduce several useful relationships between thevarious polynomials and their derivatives. In the following dP n(z)/dzwill be denoted by P/prime n. Firstly, we diﬀerentiate the deﬁning equation (16.49) with respect to zto get h(1−2zh+h2)−3/2=summationdisplay P/prime nhn. (16.50) Also, we diﬀerentiate (16.49) with respect to hto yield (z−h)(1−2zh+h2)−3/2=summationdisplay nPnhn−1; (16.51) equation (16.50) can then be written using (16.49) as hsummationdisplay Pnhn=( 1−2zh+h2)summationdisplay P/prime nhn, and thus equating coeﬃcients of hn+1we obtain the recurrence relation Pn=P/prime n+1−2zP/prime n+P/prime n−1. (16.52) Equations (16.50) and (16.51) can be combined as (z−h)summationdisplay P/prime nhn=hsummationdisplay nPnhn−1, from which the coeﬃcent of hnyields a second recurrence relation zP/prime n−P/prime n−1=nPn; (16.53) 562 16.6 LEGENDRE’S EQUATION eliminating P/prime n−1between (16.52) and (16.53) then gives the further result (n+1 )Pn=P/prime n+1−zP/prime n. (16.54) If we now take the result (16.54) with nreplaced by n−1a n da d d ztimes (16.53) to it then we obtain (1−z2)P/prime n=n(Pn−1−zPn); ﬁnally, diﬀerentiating both sides with respect to zand using (16.53) again, we ﬁnd (1−z2)P/prime/prime n−2zP/prime n=n[(P/prime n−1−zP/prime n)−Pn] =n(−nPn−Pn)=−n(n+1 )Pn, a n ds ot h e Pndeﬁned by (16.49) do indeed satisfy Legendre’s equation. It remains only to verify the normalisation. This is easily done at z=1 ,w h e n Gbecomes G(1,h)=[ ( 1−h)2]−1/2=1+ h+h2+···, and we can see that all the Pnso deﬁned have Pn(1) = 1 as required. Many other useful recurrence relations can be derived from those found above.IProve the recurrence relation (n+1 )Pn+1−(2n+1 )zPn+nPn−1=0. (16.55) Substituting from (16.49) into (16.51) we ﬁnd (z−h) X Pnhn=( 1−2zh+h2) X nPnhn−1. Equating coeﬃcients of hnwe obtain zPn−Pn−1=(n+1 )Pn+1−2znP n+(n−1)Pn−1, which on rearrangment gives the stated result. J Another use of the generating function (16.49) is in representing the inverse distance between two points in three-dimensional space in terms of Legendrepolynomials. If two points randr /primeare at distances randr/primerespectively from the origin, with r/prime<r,t h e n 1 |r−r/prime|=1 (r2+r/prime2−2rr/primecosθ)1/2 =1 r[1−2(r/prime/r)cosθ+(r/prime/r)2]1/2 =1 r∞summationdisplay /lscript=0parenleftbiggr/prime rparenrightbigg/lscript P/lscript(cosθ), (16.56) where θis the angle between the two position vectors randr/prime.I fr/prime>r,h o w e v e r , then randr/primemust be exchanged in (16.56) or the series would not converge. 563 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS To summarise the situation concerning Legendre polynomials, we now have three possible starting points, which have been shown to be equivalent: thedeﬁning equation (16.35) together with the condition P n(1) = 1; Rodrigues’ formula (16.43); and the generating function (16.49). In addition we have proved a variety of relationships and recurrence relations (not particularly memorable, but collectively useful) and, as will be apparent from the work of chapter 18,have developed a powerful tool for use in axially symmetric situations in whichthe∇ 2operator is involved and spherical polar coordinates are employed. 16.7 Bessel’s equation Bessel’s equation arises from physical situations similar to those involving Legen- dre’s equation but when cylindrical, rather than spherical, polar coordinates areemployed. It has the form z 2y/prime/prime+zy/prime+(z2−ν2)y=0, (16.57) where the parameter νis a given number, which we may take as ≥0w i t hn ol o s s of generality. In Bessel’s equation, zis usually a multiple of a radial distance and therefore ranges from 0 to ∞. Writing (16.57) in our standard form we have y/prime/prime+1 zy/prime+parenleftbigg 1−ν2 z2parenrightbigg y=0. (16.58) By inspection z= 0 is a regular singular point; hence we try a solution of the form y=zσsummationtext∞ n=0anzn. Substituting this into (16.58) and multiplying the resulting equation by z2−σ,w eo b t a i n ∞summationdisplay n=0bracketleftbig (σ+n)(σ+n−1) + ( σ+n)−ν2bracketrightbig anzn+∞summationdisplay n=0anzn+2=0, which simpliﬁes to ∞summationdisplay n=0bracketleftbig (σ+n)2−ν2bracketrightbig anzn+∞summationdisplay n=0anzn+2=0. Considering coeﬃcients of z0we obtain the indicial equation σ2−ν2=0, and so σ=±ν. For coeﬃcients of higher powers of zwe ﬁnd bracketleftbig (σ+1 )2−ν2bracketrightbig a1=0, (16.59)bracketleftbig (σ+n)2−ν2bracketrightbig an+an−2=0 f o r n≥2. (16.60) 564 16.7 BESSEL’S EQUATION Substituting σ=±νinto (16.59) and (16.60) we obtain the recurrence relations (1±2ν)a1=0, (16.61) n(n±2ν)an+an−2=0 f o r n≥2. (16.62) We consider now the form of the general solution to Bessel’s equation (16.57) for two cases, the case for which νis not an integer and that for which it is (including zero). 16.7.1 General solution for non-integer ν Ifνis a non-integer then in general the two roots of the indicial equation, σ1=ν and σ2=−ν, will not diﬀer by an integer, and we may obtain two linearly independent solutions in the form of Frobenius series. Special considerations do arise, however, when ν=m/2f o r m=1,3,5,...,a n d σ1−σ2=2ν=mis an (odd positive) integer. When this happens, we may always obtain a solution inthe form of a Frobenius series corresponding to the larger root σ 1=ν=m/2, as described above. For the smaller root σ2=−ν=−m/2, however, we must determine whether a second Frobenius series solution is possible by examiningthe recurrence relation (16.62), which reads n(n−m)a n+an−2=0 f o r n≥2. Since mis anoddpositive integer in this case, we can use this recurrence relation (starting with a0/negationslash=0 )t oc a l c u l a t e a2,a4,a6,...in the knowledge that all these terms will remain ﬁnite. It is possible in this case, therefore, to ﬁnd a secondsolution in the form of a Frobenius series corresponding to the smaller root σ 2. Thus, in general, for non-integer νwe have from (16.61) and (16.62) an=−1 n(n±2ν)an−2forn=2,4,6,..., =0 f o r n=1,3,5,.... Setting a0= 1 in each case, we obtain the two solutions y±ν(z)=z±νbracketleftbigg 1−z2 2(2±2ν)+z4 2×4(2±2ν)(4±2ν)−···bracketrightbigg . It is customary, however, to set a0=1 2±νΓ(1±ν), where Γ( x)i st h e gamma function , described in the appendix; it may be regarded as the generalisation of the factorial function to non-integer and/or negativearguments.†The two solutions of (16.57) are then written as J ν(z)a n d J−ν(z), †In particular, Γ( n+1 )= n!f o r n=0,1,2,...,and Γ( n) is inﬁnite if nis any integer ≤0. 565 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS where Jν(z)=1 Γ(ν+1 )parenleftBigz 2parenrightBigνbracketleftbigg 1−1 ν+1parenleftBigz 2parenrightBig2 +1 (ν+1 ) ( ν+2 )1 2!parenleftBigz 2parenrightBig4 −···bracketrightbigg =∞summationdisplay n=0(−1)n n!Γ(ν+n+1 )parenleftBigz 2parenrightBigν+2n ; (16.63) replacing νby−νgives J−ν(z). The functions Jν(z)a n d J−ν(z) are called Bessel functions of the ﬁrst kind, of order ν. Since the ﬁrst term of each series is a ﬁnite non-zero multiple of zνandz−νrespectively, if νis not an integer then Jν(z)a n d J−ν(z) are linearly independent. This may be conﬁrmed by calculating the Wronskian of these two functions. Therefore, for non-integer νthe general solution of Bessel’s equation (16.57) is y(z)=c1Jν(z)+c2J−ν(z). (16.64)IFind the general solution of z2y/prime/prime+zy/prime+(z2−1 4)y=0. This is Bessel’s equation with ν=1/2, so from (16.64) the general solution is simply y(z)=c1J1/2(z)+c2J−1/2(z). However, Bessel functions of half-integral order can be expressed in terms of trigonometric functions. To show this, we note from (16.63) that J±1/2(z)=z±1/2∞X n=0(−1)nz2n 22n±1/2n!Γ(1 + n±1 2). Using the fact that Γ( x+1 )= xΓ(x)a n dΓ (1 2)=√πwe ﬁnd that, for ν=1/2, J1/2(z)=(1 2z)1/2 Γ(3 2)−(1 2z)5/2 1!Γ(5 2)+(1 2z)9/2 2!Γ(7 2)−··· =(1 2z)1/2 (1 2)√π−(1 2z)5/2 1!(3 2)(1 2)√π+(1 2z)9/2 2!(5 2)(3 2)(1 2)√π−··· =(1 2z)1/2 (1 2)√π / 1−z2 3!+z4 5!−··· / =(1 2z)1/2 (1 2)√πsinz z= r 2 πzsinz, whereas for ν=−1/2w eo b t a i n J−1/2(z)=(1 2z)−1/2 Γ(1 2)−(1 2z)3/2 1!Γ(3 2)+(1 2z)7/2 2!Γ(5 2)−··· =(1 2z)−1/2 √π / 1−z2 2!+z4 4!−··· / = r 2 πzcosz. Therefore the general solution we require is y(z)=c1J1/2(z)+c2J−1/2(z)=c1 r 2 πzsinz+c2 r 2 πzcosz. J 566 16.7 BESSEL’S EQUATION Corresponding to the discussion in subsection 16.6.2 of the general solution of Legendre’s equation, we note that when Bessel’s equation is encountered inphysical situations the argument zis usually some multiple of a radial distance and so takes values in the range 0 ≤z≤∞. We often require that the solution is regular at z= 0 but, from (16.63), we see immediately that J −ν(z) is singular at the origin (remember that we restricted νto be non-negative). In such cases, the coeﬃcient c2in (16.64) must be set to zero, and the solution is simply some multiple of Jν(z). 16.7.2 General solution for integer ν The deﬁnition of the Bessel function Jν(z) given in (16.63) is, of course, valid for all values of νbut, as we shall see, in the case of integer νthe general solution of Bessel’s equation cannot be written in the form (16.64). Firstly let us consider thecase ν= 0, so that the two solutions to the indicial equation are equal, and we clearly obtain only one solution in the form of a Frobenius series. From (16.63),this is given by J 0(z)=∞summationdisplay n=0(−1)nz2n 22nn!Γ(1 + n) =1−z2 22+z4 2242−z6 224262+···. In general, however, if νis a positive integer then the solutions of the indicial equation diﬀer by an integer. For the larger root, σ1=ν, we may ﬁnd a solution Jν(z)f o r ν=1,2,3,..., in the form of a Frobenius series given by (16.63). Graphs ofJ0(z),J1(z)a n d J2(z) are plotted in ﬁgure 16.2 for real z. For the smaller root σ2=−ν, however, the recurrence relation (16.62) becomes n(n−m)an+an−2=0 f o r n≥2, where m=2νis now an evenpositive integer, i.e. m=2,4,6,.... Starting with a0/negationslash= 0 we may then calculate a2,a4,a6,..., but we see that when n=mthe coeﬃcient anis formally inﬁnite, and the method fails to produce a second solution in the form of a Frobenius series. In fact, by replacing νby−νin the deﬁnition of Jν(z) given in (16.63), it can be shown that, for integer ν, J−ν(z)=(−1)νJν(z) and hence that Jν(z)a n d J−ν(z) are linearly dependent. So, in this case, we cannot write the general solution to Bessel’s equation in the form (16.64). One therefore deﬁnes the function Yν(z)=Jν(z)cosνπ−J−ν(z) sinνπ, (16.65) 567 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS J0 J1 J2 24 6 81 0z 0 −0.4−0.20.20.40.60.81 Figure 16.2 The ﬁrst three integer-order Bessel functions. which is called a Bessel’s function of the second kind of order ν.A sB e s s e l ’ se q u a - tion is linear, Yν(z) is clearly a solution, since it is just the weighted sum of Bessel functions of the ﬁrst kind. Furthermore, for non-integer νit is clear that Yν(z)i s linearly independent of Jν(z). It may also be shown that the Wronskian of Jν(z) andYν(z) is non-zero for allvalues of ν. Hence Jν(z)a n d Yν(z) always constitute a pair of independent solutions. The expression (16.65) becomes an indeterminate form 0 /0w h e n νis an integer, however. This is so because for integer νwe have cosνπ=(−1)νandJ−ν(z)=(−1)νJν(z). Nevertheless, this indeterminate form can be evaluated using l’H ˆopital’s rule (see chapter 4). Thus for integer νwe set Yν(z) = lim µ→νbracketleftbiggJµ(z)cosµπ−J−µ(z) sinµπbracketrightbigg , (16.66) which gives a linearly independent second solution for integer ν. Therefore, we may write the general solution of Bessel’s equation, valid for allν,a s y(z)=c1Jν(z)+c2Yν(z). (16.67) As mentioned above for the case when νis not an integer, in physical situations we often require the solution of Bessel’s equation to be regular at z=0 .B u t , from its deﬁnition (16.65) or (16.66), it is clear that Yν(z) is singular at the origin, and so in such physical situations the coeﬃcient c2in (16.67) must be set to zero; the solution is then simply some multiple of Jν(z). 16.7.3 Properties of Bessel functions Bessel functions of the ﬁrst and second kind, Jν(z)a n d Yν(z), have various useful properties that are worthy of further discussion. 568 16.7 BESSEL’S EQUATION Recurrence relations The recurrence relations enjoyed by Bessel functions of the ﬁrst kind, Jν(z), can be derived directly from the power series deﬁnition (16.63).IProve the recurrence relation d dz[zνJν(z)] =zνJν−1(z). (16.68) From the power series deﬁnition (16.63) of Jν(z)w eo b t a i n d dz[zνJν(z)] =d dz∞X n=0(−1)nz2ν+2n 2ν+2nn!Γ(ν+n+1 ) =∞X n=0(−1)nz2ν+2n−1 2ν+2n−1n!Γ(ν+n) =zν∞X n=0(−1)nz(ν−1)+2 n 2(ν−1)+2 nn!Γ((ν−1) +n+1 )=zνJν−1(z). J It may similarly be shown that d dz[z−νJν(z)] =−z−νJν+1(z). (16.69) From (16.68) and (16.69) the remaining recurrence relations may be easily derived. Expanding out the derivative on the LHS of (16.68) and dividing through by zν−1 we obtain the relation zJ/prime ν(z)+νJν(z)=zJν−1(z). (16.70) Similarly, by expanding out the derivative on the LHS of (16.69), and multiplying through by zν+1, we ﬁnd zJ/prime ν(z)−νJν(z)=−zJν+1(z). (16.71) Adding (16.70) and (16.71) and dividing through by zgives Jν−1(z)−Jν+1(z)=2 J/prime ν(z). (16.72) Finally, subtracting (16.71) from (16.70) and dividing by zgives Jν−1(z)+Jν+1(z)=2ν zJν(z). (16.73) 569 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSIGiven that J1/2(z)=( 2 /πz)1/2sinzand that J−1/2(z)=( 2 /πz)1/2cosz,e x p r e s s J3/2(z) andJ−3/2(z)in terms of trigonometric functions. From (16.71) we have J3/2(z)=1 2zJ1/2(z)−J/prime 1/2(z) =1 2z /2 πz /1/2 sinz− /2 πz /1/2 cosz+1 2z /2 πz /1/2 sinz = /2 πz /1/2 /1 zsinz−cosz / . Similarly, from (16.70), we have J−3/2(z)=−1 2zJ−1/2(z)+J/prime −1/2(z) =−1 2z /2 πz /1/2 cosz− /2 πz /1/2 sinz−1 2z /2 πz /1/2 cosz = /2 πz /1/2 / −1 zcosz−sinz / . We shall see that, by repeated use of these recurrence relations, all Bessel functions Jν(z) of half-integer order may be expressed in terms of trigonometric functions. From their deﬁnition (16.65), Bessel functions of the second kind, Yν(z), of half-integer order can be similarly expressed. J Finally, we note that the relations (16.68) and (16.69) may be rewritten in integral form as integraldisplay zνJν−1(z)dz=zνJν(z) integraldisplay z−νJν+1(z)dz=−z−νJν(z). Ifνis an integer, the recurrence relations of this section may be proved using the generating function for Bessel functions discussed below. It may be shownthat Bessel functions of the second kind, Y ν(z), also satisfy the recurrence relations derived above. Mutual orthogonality of Bessel functions Bessel functions of the ﬁrst kind, Jν(z), possess an orthogonality relation analo- gous to that of the Legendre polynomials discussed in subsection 16.6.2. A moregeneral discussion of the mutual orthogonality of solutions to second-order linearODEs (such as Bessel’s equation) is given in chapter 17. By deﬁnition, the function J ν(z) satisﬁes Bessel’s equation (16.57), z2y/prime/prime+zy/prime+(z2−ν2)y=0. 570 16.7 BESSEL’S EQUATION Let us instead consider the functions f(z)=Jν(λz)a n d g(z)=Jν(µz), which, as will be proved below, respectively satisfy the equations z2f/prime/prime+zf/prime+(λ2z2−ν2)f=0, (16.74) z2g/prime/prime+zg/prime+(µ2z2−ν2)g=0. (16.75)IShow that f(z)=Jν(λz)satisﬁes (16.74). Iff(z)=Jν(λz) and we write w=λz,t h e n df dz=λdJν(w) dwandd2f dz2=λ2d2Jν(w) dw2. When these expressions are substi tuted, the LHS of (16.74) becomes z2λ2d2Jν(w) dw2+zλdJν(w) dw+(λ2z2−ν2)Jν(w) =w2d2Jν(w) dw2+wdJν(w) dw+(w2−ν2)Jν(w). But, from Bessel’s equation itself, this ﬁnal expression is equal to zero, thus verifying that f(z) does satisfy (16.74). J Now multiplying (16.75) by f(z) and (16.74) by g(z) and subtracting them gives d dz[z(fg/prime−gf/prime)] = ( λ2−µ2)zfg, (16.76) w h e r ew eh a v eu s e dt h ef a c tt h a t d dz[z(fg/prime−gf/prime)] =z(fg/prime/prime−gf/prime/prime)+(fg/prime−gf/prime). By integrating (16.76) over any given range z=atoz=bwe obtain integraldisplayb azf(z)g(z)dz=1 λ2−µ2bracketleftBig zf(z)g/prime(z)−zg(z)f/prime(z)bracketrightBigb a, which, on setting f(z)=Jν(λz)a n d g(z)=Jν(µz), becomes integraldisplayb azJν(λz)Jν(µz)dz=1 λ2−µ2bracketleftBig µzJ ν(λz)J/prime ν(µz)−λzJ ν(µz)J/prime ν(λz)bracketrightBigb a. (16.77) Ifλ/negationslash=µ, and the interval [ a, b] is such that the expression on the RHS of (16.77) equals zero then we obtain the orthogonality condition integraldisplayb azJν(λz)Jν(µz)dz=0. (16.78) This happens, for example, if Jν(λz)a n d Jν(µz) vanish at z=aandz=b,o ri f J/prime ν(λz)a n d J/prime ν(µz) vanish at z=aandz=b, or for many more general conditions. Ifλ=µ, however, then the RHS of (16.77) takes the indeterminant form 0 /0. This may be evaluated using l’H ˆopital’s rule, or alternatively we may calculate the relevant integral directly. 571 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSIEvaluate the integralZb aJ2 ν(λz)zd z . Ignoring the integration limits for the moment,Z J2 ν(λz)zd z=1 λ2 Z J2 ν(u)u du, where u=λz. Integrating by parts yields I= Z J2 ν(u)ud u=1 2u2J2 ν(u)− Z Jν(u)J/prime ν(u)u2du. Now Bessel’s equation (16.57) can be rearranged as u2Jν(u)=ν2Jν(u)−uJ/prime ν(u)−u2J/prime/prime ν(u), which, on substitution into the expression for I,g i v e s I=1 2u2J2 ν(u)− Z J/prime ν(u)[ν2Jν(u)−uJ/prime ν(u)−u2J/prime/prime ν(u)]du =1 2u2J2 ν(u)−1 2ν2J2 ν(u)+1 2u2[J/prime ν(u)]2+c. Since u=λzthe required integral is given byZb aJ2 ν(λz)zd z=1 2 // z2−ν2 λ2 / J2 ν(λz)+z2[J/prime ν(λz)]2 /b a, (16.79) which gives the normalisation condition for Bessel functions of the ﬁrst kind. J Since the Bessel functions Jν(z) possess the orthogonality property (16.78) we may expand any reasonable function f(z) (i.e. one obeying the Dirichlet conditions discussed in chapter 12) in the interval 0 ≤z≤aas a sum of Bessel functions of a given order ν, f(z)=∞summationdisplay n=0cnJν(λnz), (16.80) where the λnare chosen such that Jν(λna) = 0. The coeﬃcients cnare then given by cn=2 a2J2 ν+1(λna)integraldisplaya 0f(z)Jν(λnz)zd z . (16.81) 572 16.7 BESSEL’S EQUATIONIProve the expression (16.81) for the coeﬃ cients in a Bessel function expansion of a function f(z). If we multiply (16.80) by zJν(λmz) and integrate from z=0t o z=athen we obtainZa 0zJν(λmz)f(z)dz=∞X n=0cn Za 0zJν(λmz)Jν(λnz)dz =cm Za 0J2 ν(λmz)zd z =1 2cma2J/prime2 ν(λma)=1 2cma2J2 ν+1(λma), where in the last two lines we have used (16.77), (16.79), the fact that Jν(λma)=0a n d (16.71). J Generating function for Bessel functions The Bessel functions Jν(z), where νis an integer, can be described by a gener- ating function in a similar way to that discussed for Legendre polynomials insubsection 16.6.2. The generating function for Bessel functions of integer order isgiven by G(z,h)=e x pbracketleftbiggz 2parenleftbigg h−1 hparenrightbiggbracketrightbigg =∞summationdisplay n=−∞Jn(z)hn. (16.82) By expanding the exponential as a power series, it is straightfoward to verify that the functions Jn(z) deﬁned by (16.82) are indeed Bessel functions of the ﬁrst kind. The generating function (16.82) is useful for ﬁnding, for Bessel functions of integer order, properties which can often be extended to the non-integer case. Inparticular, the Bessel function recurrence relations may be derived.IUse the generating function (16.82) to prove, for integer ν, the recurrence relation (16.73), i.e. Jν−1(z)+Jν+1(z)=2ν zJν(z). Diﬀerentiating G(z,h) with respect to hwe obtain ∂G(z,h) ∂h=z 2 / 1+1 h2 / G(z,h)=∞X n=−∞nJn(z)hn−1, which can be written using (16.82) again as z 2 / 1+1 h2 /∞X n=−∞Jn(z)hn=∞X n=−∞nJn(z)hn−1. Equating coeﬃcients of hnwe obtain z 2[Jn(z)+Jn+2(z)] = ( n+1 )Jn+1(z), which on replacing nbyν−1 gives the required recurrence relation. J 573 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS The generating function (16.82) is also useful in deriving the integral represen- tation of Bessel functions of integer order.IShow that for integer nthe Bessel function Jn(z)is given by Jn(z)=1 π Zπ 0cos(nθ−zsinθ)dθ. (16.83) By expanding out the cosine term in the integrand in (16.83) we obtain the integral I=1 π Zπ 0[cos(zsinθ)c osnθ+s i n ( zsinθ)sinnθ]dθ. (16.84) Now, we may express cos( zsinθ) and sin( zsinθ) in terms of Bessel functions by setting h=e x p iθin (16.82) to give exp hz 2(expiθ−exp(−iθ)) i =e x p (izsinθ)=∞X m=−∞Jm(z)exp imθ. Using de Moivre’s theorem exp iθ=c o s θ+isinθwe then obtain exp(izsinθ)=c o s ( zsinθ)+isin(zsinθ)=∞X m=−∞Jm(z)(cos mθ+isinmθ). Equating the real and imaginary parts of this expression we ﬁnd cos(zsinθ)=∞X m=−∞Jm(z)cosmθ, sin(zsinθ)=∞X m=−∞Jm(z)sinmθ. Substituting these expressions into (16.84) we ﬁnd I=1 π∞X m=−∞ Zπ 0[Jm(z)c osmθcosnθ+Jm(z)si nmθsinnθ]dθ. However, using the orthogonality of the trigonometric functions, see equations (12.1)– (12.3), we obtain I=1 ππ 2[Jn(z)+Jn(z)] =Jn(z), which proves the integral representation (16.83). J Finally, we mention the special case of the integral representation (16.83) for n=0 , J0(z)=1 πintegraldisplayπ 0cos(zsinθ)dθ=1 2πintegraldisplay2π 0cos(zsinθ)dθ, since cos( zsinθ) repeats itself in the range θ=πtoθ=2π. However, sin( zsinθ) changes sign in this range and so 1 2πintegraldisplay2π 0sin(zsinθ)dθ=0. 574 16.8 GENERAL REMARKS Using de Moivre’s theorem, we can therefore write J0(z)=1 2πintegraldisplay2π 0exp(izsinθ)dθ=1 2πintegraldisplay2π 0exp(izcosθ)dθ. There are in fact many other integral representations of Bessel functions, which can be derived from those given. 16.8 General remarks As was our intention, in respect of inﬁnite series solutions we have concentrated to a very marked degree on Bessel’s equation and, in respect of ﬁnite polynomialsolutions, on Legendre’s equation. The techniques used are, however, applicable to many equations other than these, but since the procedures are in all essentials the same, we do not need to treat them explicitly. The solutions of the remainingequations in table 16.1 are discussed brieﬂy in the next chapter in connectionwith Sturm–Liouville systems. 16.9 Exercises 16.1 Find two power series solutions about z= 0 of the diﬀerential equation (1−z2)y/prime/prime−3zy/prime+λy=0. Deduce that the value of λfor which the corresponding power series becomes an Nth-degree polynomial UN(z)i sN(N+ 2). Construct U2(z)a n d U3(z). 16.2 Find solutions, as power series in z, of the equation 4zy/prime/prime+2 ( 1−z)y/prime−y=0. Identify one of the solutions and verify it by direct substitution. 16.3 Find power series solutions in zof the diﬀerential equation zy/prime/prime−2y/prime+9z5y=0. Identify closed forms for the two series, calculate their Wronskian, and verify that they are linearly independent. Compare the Wronskian with that calculatedfrom the diﬀerential equation. 16.4 Change the independent variable in the equation d 2f dz2+2 (z−a)df dz+4f=0 ( * ) from ztox=z−α, and ﬁnd two independent series solutions, expanded about x= 0, of the resulting equation. Deduce that the general solution of (*) is f(z,α)=A(z−α)e−(z−α)2+B∞X m=0(−4)mm! (2m)!(z−α)2m, with AandBarbitrary constants. 16.5 (a) Verify that z= 1 is a regular singular point of Legendre’s equation and that the indicial equation for a series solution in powers of ( z−1) has roots 0 and 3. (b) Obtain the corresponding recurrence relation and show that σ= 0 does not give a valid series solution. 575 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS (c) Determine the radius of convergence Rof the σ= 3 series and relate it to the positions of the singularities of Legendre’s equation. 16.6 Verify that z= 0 is a regular singular point of the equation z2y/prime/prime−3 2zy/prime+( 1+ z)y=0, and that the indicial equation has roots 2 and 1 /2. Show that the general solution is y(z)=6 a0z2∞X n=0(−1)n(n+1 ) 22nzn (2n+3 ) ! +b0 / z1/2+2z3/2−z1/2 4∞X n=2(−1)n22nzn n(n−1)(2n−3)! /! . 16.7 Use the derivative method to obtain as a second solution of Bessel’s equation for the case when ν= 0 the following expression: J0(z)lnz−∞X n=1(−1)n (n!)2 / nX r=11 r /!/z 2 /2n , given that the ﬁrst solution is J0(z) as speciﬁed by (16.63). 16.8 By initially writing y(x)a s x1/2f(x) and then making subsequent changes of variable, reduce d2y dx2+λxy=0 to Bessel’s equation. Hence show that a solution that is ﬁnite at x=0i sa multiple of x1/2J1/3(2 3√ λx3). 16.9 (a) Show that the indicial equation for zy/prime/prime−2y/prime+yz=0 has roots that diﬀer by an integer but that the two roots nevertheless generate linearly independent solutions y1(z)=3 a0∞X n=1(−1)n+12nz2n+1 (2n+1 ) !, y2(z)=a0∞X n=0(−1)n+1(2n−1)z2n (2n)!. (b) Show that y1(z)i se q u a lt o3 a0(sinz−zcosz) by expanding the sinusoidal functions. Then, using the Wronskian method, ﬁnd an expression for y2(z) in terms of sinusoids. (You will need to write z2as (z/sinz)(zsinz)a n d integrate by parts to evaluate the integral involved.) (c) Conﬁrm that the two solutions are linearly independent by showing that their Wronskian is equal to −z2, in accordance with (16.4). 16.10 Find series solutions of the equation y/prime/prime−2zy/prime−2y= 0. Identify one of the series asy1(z)=e x p z2and verify this by direct substitution. By setting y2(z)=u(z)y1(z) and solving the resulting equation for u(z), ﬁnd an explicit form for y2(z)a n d deduce thatZx 0e−v2dv=e−x2∞X n=0n! 2(2n+1 ) !(2x)2n+1. 576 16.9 EXERCISES 16.11 (a) Identify and classify the singular points of the equation z(1−z)d2y dz2+( 1−z)dy dz+λy=0, and determine their indices. (b) Find one series solution in powers of z. Give a formal expression for a second linearly independent solution. (c) Deduce the values of λfor which there is a polynomial solution PN(z)o f degree N. Evaluate the ﬁrst four polynomials, normalised in such a way that PN(0) = 1 . 16.12 Find the general power series solution about z=0o ft h ee q u a t i o n zd2y dz2+( 2z−3)dy dz+4 zy=0. 16.13 Find the radius of convergence of a series solution about the origin for the equation ( z2+az+b)y/prime/prime+2y= 0 in the following cases: (a)a=5 , b=6 ;( b ) a=5 , b=7 . Show that if aandbare real and 4 b>a2then the radius of convergence is always given by b1/2. 16.14 For the equation y/prime/prime+z−3y= 0, show that the origin becomes a regular singular point if the independent variable is changed from ztox=1/z. Hence ﬁnd a series solution of the form y1(z)= P∞ 0anz−n. By setting y2(z)=u(z)y1(z)a n d expanding the resulting expression for du/dz in powers of z−1, show that y2(z) has the asymptotic form y2(z)=c / z+l nz−1 2+O /lnz z // , where cis an arbitrary constant. 16.15 Prove that the Laguerre equation zd2y dz2+( 1−z)dy dz+λy=0 has polynomial solutions LN(z)i fλis a non-negative integer N, and determine the recurrence relationship for the polynomial coeﬃcients. Hence show that anexpression for L N(z), normalised in such a way that LN(0) = N!, is LN(z)=NX n=0(−1)n(N!)2 (N−n)!(n!)2zn. Evaluate L3(z) explicitly. [The Laguerre generating function is discussed in exer- cise 17.9.] 16.16 (a) Use Leibniz’ theorem to show that the Rodrigues’ formula for the Laguerre polynomials LN(z) of the previous question is LN(z)=ezdN dzN(zNe−z). (b) Use the Rodrigue formulation to prove that zL/prime N(z)=LN+1(z)−(N+1−z)LN(z). (c) Deduce the recurrence relation for the Laguerre polynomials, namely LN+1(z)+(z−2N−1)LN(z)+N2LN−1(z)=0 . 577 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 16.17 Equation (16.32) was shown to have a polynomial solution provided that λ=2n with nan integer≥0. The polynomials are known as Hermite polynomials Hn(x) and are of importance in the quantum mechanical treatment of the harmonicoscillator problem. They may also be deﬁned by Φ(x, h)=e x p ( 2 xh−h 2)=∞X n=01 n!Hn(x)hn. Show that ∂2Φ ∂x2−2x∂Φ ∂x+2h∂Φ ∂h=0, and hence that the Hn(x) satisfy (16.32). Use Φ to prove that (a)H/prime n(x)=2 nHn−1(x), (b)Hn+1(x)−2xHn(x)+2nHn−1(x)=0 . 16.18 By writing Φ( x, h) of the previous exercise as a function of h−xrather than of h, show that an alternative representation of the nth Hermite polynomial is Hn(x)=(−1)n /; expx2 /dn dxn[exp(−x2)]. (Note that Hn(x)=∂nΦ/∂hnath=0 . ) 16.19 Obtain the recurrence relations for the solution of Legendre’s equation (16.35) ininverse powers of z,i . e .s e t y(z)= Panzσ−n,w i t h a0/negationslash= 0. Deduce that if /lscriptis an integer then the series with σ=/lscriptwill terminate and hence converge for all z whilst that with σ=−(/lscript+ 1) does not terminate and hence converges only for |z|>1. 16.20 Carry through the following procedure as an alternative proof of result (16.45). (a) Square both sides of (16.49), giving the generating-function deﬁnition of the Legendre polynomials. (b) Express the RHS as a sum of powers of h, obtaining expressions for the coeﬃcients. (c) Integrate the RHS from −1 to 1 and use the orthogonality results (16.46). (d) Similarly integrate the LHS and expand the result in powers of h. (e) Compare coeﬃcients. 16.21 A charge +2 qis situated at the origin and charges of −qare situated at distances ±afrom it along the polar axis. By relating it to the generating function for the Legendre polynomials, show that the electrostatic potential Φ at a point ( r,θ,φ) with r>a is given by Φ(r,θ,φ)=2q 4π/epsilon10r∞X s=1 /a r /2s P2s(cosθ). 16.22 The origin is an ordinary point of the Chebyshev equation, (1−z2)y/prime/prime−zy/prime+m2y=0, which therefore has series solutions of the form zσ P∞ 0anznforσ=0a n d σ=1 . (a) Find the recurrence relationships for the anin the two cases and show that there exist polynomial solutions Tm(z): (i) for σ=0 ,w h e n mis an even integer, the polynomial having1 2(m+2 ) terms; (ii) for σ=1 ,w h e n mis an odd integer, the polynomial having1 2(m+1 ) terms. 578 16.10 HINTS AND ANSWERS (b)Tm(z) is normalised so as to have Tm(1) = 1. Find explicit forms for Tm(z) form=0,1,2,3. (c) Show that the corresponding non-terminating series solutions Sm(z) have as their ﬁrst few terms S0(z)=a0 / z+1 3!z3+9 5!z5+··· / , S1(z)=a0 / 1−1 2!z2−3 4!z4−··· / , S2(z)=a0 / z−3 3!z3−15 5!z5−··· / , S3(z)=a0 / 1−9 2!z2+45 4!z4+··· / . 16.23 By choosing a suitable form for hin (16.82), show that further integral repe- sentations of the Bessel functions of the ﬁrst kind are given, for integral m, by J2m(z)=(−1)m π Z2π 0cos(zcosθ)c o s 2 mθ dθ m ≥1, J2m+1(z)=(−1)m+1 π Z2π 0cos(zcosθ)s i n ( 2 m+1 )θd θ m≥0. 16.24 Show from the deﬁnition given in (16. 66) that the Bessel function of the second kind of order νcan be written as Yν(z)=1 π /∂Jµ(z) ∂µ−(−1)ν∂J−µ(z) ∂µ / µ=ν. Using the explicit expression (16.63) for Jµ(z), show that ∂Jµ(z)/∂µcan be written as Jν(z)ln /z 2 / +g(ν,z), and deduce that Yν(z) can be expressed as Yν(z)=2 πJν(z)ln /z 2 / +h(ν,z), h(ν,z), like g(ν,z), being a power series in z. 16.10 Hints and answers 16.1 Note that z= 0 is an ordinary point of the equation. Forσ=0,an+2/an=[n(n+2)−λ]/[(n+1)(n+2)] and correspondingly for σ=1 ; U2(z)=a0(1−4z2)a n d U3(z)=a0(z−2z3). 16.2 a0exp(z/2);b0z1/2 P∞ n=0(2z)nn!/(2n+1 ) ! . 16.3 σ=0a n d3 ; a6m/a0=(−1)m/(2m)! and a6m/a0=(−1)m/(2m+ 1)! respectively. y1(z)=a0cosz3andy2(z)=a0sinz3.T h eW r o n s k i a ni s ±3a2 0z2/negationslash=0. 16.4 x= 0 is an ordinary point of the transformed equation and so σ=0a n d1 . Forσ=1,an+2=−2an/(n+2 )a n ds o a2m/a0=(−1)m/m!.Forσ=0,an+2= −2an/(n+1 )a n ds o a2m/a0=(−2)m/ Qm r=1(2r−1). 16.5 (b) an+1/an=−[(σ+n)(σ+n−3) +/lscript(/lscript+1 ) ] /[2(σ+n)2−2]. For σ=0,a2=∞. (c)R= 2, equal to the distance between z= 1 and the closest singularity at z=−1. 579 SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 16.8 x2f/prime/prime+xf/prime+(λx3−1 4)f= 0. Then, in turn, set x3/2=u,a n d2 λ1/2u/3=v;t h e n v satisﬁes Bessel’s equation with ν=1/3. 16.9 (b) cos z+zsinz. 16.10 y2(z)=( e x p z2) Rz 0exp(−x2)dx. 16.11 (a) Regular singular points at z= 0 (indices 0, 0) and at z= 1 (indices 0, 1). (b)y1(z)=a0+a0 P∞ n=1(n!)−2zn Qn−1 r=0(r2−λ). y2(z)=y1(z)lnz+ P∞ n=1zn / (∂/∂σ) nQn−1 r=0[(r+σ)2−λ]/(n+σ+1 )2 o/ σ=0. (c)λ=N2; polynomials are 1, 1 −z,( 1−z)(1−3z), (1−z)(1−8z+1 0z2). 16.12 Repeated roots σ=2 . y(z)=az2+∞X n=1(n+1 ) (−2z)n+2 n! na 4+b[lnz+g(n)] o , where g(n)=1 n+1−1 n−1 n−1−···−1 2−2. 16.13 (a) 2; (b)√ 7. 16.14 Transformed equation is xy/prime/prime+2y/prime+y=0 ; an=(−1)n(n+1 )−1(n!)−2a0;du/dz = A[y1(z)]−2. 16.15 an+1=−(N−n)an/(n+1 )2;L3(z)=6−18z+9z2−z3. 16.16 (b) Calculate LN+1(z), considering zN+1e−zaszzNe−z. Later write dN/dzN(zNe−z) ase−zLN(z).(c) Use (b) to calculate L/prime N+1(z), substituting for L/prime/prime N(z)f r o mt h e Laguerre equation. Substitute from (b) for the ﬁrst derivatives, and ﬁnally change n+1t o n. 16.17 Consider ∂Φ/∂x; (b) diﬀerentiate result (a) and then use (a) again to replace the derivatives. 16.19 σ=/lscript;an+2=[ (/lscript−n)(/lscript−n−1)an]/[(n+2)(n−2/lscript+1)]. Note that ( n−2/lscript+1)/negationslash=0 forn≤/lscript+1a n d neven. σ=−(/lscript+1 ) ; an+2=[ (/lscript+n+1 ) ( /lscript+n+2 )an]/[(n+2 ) ( n+2/lscript+3 ) ] . 16.20 At step (d) 1 hln1+h 1−h=∞X h=0h2n Z1 −1P2 n(x)dx. 16.21 Using the cosine law, the distances from the charges −qare of the form r / 1±2(a/r)cosθ+(a/r)2 /1/2. 16.22 (a) (i) an+2=[an(n2−m2)]/[(n+2 ) ( n+ 1)], (ii)an+2={an[(n+1 )2−m2]}/[(n+3 ) ( n+ 2)]; (b) 1, z,2z2−1, 4z3−3z. 16.23 Set h=iexpiθand obtain an expression for cos( zcosθ). 16.24 Recall that J−ν(z)=(−1)νJν(z)f o ri n t e g e r ν. 580 17 Eigenfunction methods for diﬀerential equations In the previous three chapters we dealt with the solution of diﬀerential equations of order nby two methods. In one method, we found nindependent solutions of the equation and then combined them, weighted with coeﬃcients determinedby the boundary conditions; in the other we found solutions in terms of serieswhose coeﬃcients were related by (in general) an n-term recurrence relation and thence ﬁxed by the boundary conditions. For both approaches the linearity of theequation was an important or essential factor in the utility of the method, and in this chapter our aim will be to exploit the superposition properties of linear diﬀerential equations even further. We will be concerned with the solution of equations of the inhomogeneous form Ly(x)=f(x), (17.1) where f(x) is a prescribed or general function and the boundary conditions to be satisﬁed by the solution y=y(x), for example at the limits x=aandx=b, are given. The expression Ly(x) stands for a linear diﬀerential operator Lacting upon the function y(x). In general, unless f(x) is both known and simple, it will not be possible to ﬁnd particular integrals of (17.1), even if complementary functions can be found thatsatisfy Ly= 0. The idea is therefore to exploit the linearity of Lby building up the required solution as a superposition , generally containing an inﬁnite number of terms, of some set of functions that each individually satisfy the boundaryconditions. Clearly this brings in a quite considerable complication but since,within reason, we may select the set of functions to suit ourselves, we can obtain sizeable compensation for this complication. Indeed, if the set chosen is one containing functions that, when acted upon by L, produce particularly simple results then we can ‘show a proﬁt’ on the operation. In particular, if the set 581 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS consists of those functions yifor which Lyi(x)=λiyi(x), (17.2) where λiis a constant, then a distinct advantage may be obtained from the manoeuvre because all the diﬀerentiation will have disappeared from (17.1). Equation (17.2) is clearly reminiscent of the equation satisﬁed by the eigenvec- torsxiof a linear operator A,n a m e l y Axi=λixi, (17.3) where λiis a constant and is called the eigenvalue associated with xi. By analogy, in the context of diﬀerential equations a function yi(x) satisfying (17.2) is called aneigenfunction of the operator Landλiis then called the eigenvalue associated with the eigenfunction yi(x). Probably the most familiar equation of the form (17.2) is that which describes a simple harmonic oscillator, i.e. Ly≡−d2y dt2=ω2y,where L≡−d2/dt2. (17.4) In this case the eigenfunctions are given by yn(t)=Aneiωnt,w h e r e ωn=2πn/T, Tis the period of oscillation, n=0,±1,±2,...and the Anare constants. The eigenvalues are ω2 n=n2ω2 1=n2(2π/T)2. (Sometimes ωnis referred to as the eigenvalue of this equation but we will avoid this confusing terminology here.) Another equation of the form (17.2) is Legendre’s equation Ly≡−(1−x2)d2y dx2+2xdy dx=/lscript(/lscript+1 )y, (17.5) where L=−(1−x2)d2 dx2+2xd dx. (17.6) We found the eigenfunctions of Lby a series method in chapter 16, and for solutions to Legendre’s equation that are regular at x=±1 these are the Legendre polynomials, given by y/lscript(x)=P/lscript(x)=1 2/lscript/lscript!d/lscript dx/lscript(x2−1)/lscript(17.7) for/lscript=0,1,2,...; they have associated eigenvalues /lscript(/lscript+1). (Again, /lscriptis sometimes, confusingly, referred to as the eigenvalue of this equation.) We may discuss a somewhat wider class of diﬀerential equations by considering a slightly more general form of (17.2), namely Ly(x)=λρ(x)y(x), (17.8) where ρ(x)i sa weight function . In many applications ρ(x) is unity for all x,i n which case (17.2) is recovered; in general, though, it is a function determined by 582 17.1 SETS OF FUNCTIONS the choice of coordinate system used in describing a particular physical situation. The only requirement on ρ(x) is that it is real and does not change sign in the range a≤x≤b, so that it can, without loss of generality, be taken to be non- negative throughout. A function y(x) that satisﬁes (17.8) is called an eigenfunction of the operator Lwith respect to the weight function ρ(x). This chapter will not cover methods used to determine the eigenfunctions of (17.2) or (17.8), since we have discussed these in previous chapters, but, rather,will use the properties of the eigenfunctions to solve inhomogeneous equationsof the form (17.1). We shall see later that the sets of eigenfunctions y i(x)o f a particular class of operators called Hermitian operators (the operators in the simple harmonic oscillator equation and in Legendre’s equation are examples) have particularly useful properties and these will be studied in detail. I turns out that many of the interesting operators met with in the physical sciences areHermitian. Before continuing our discussion of the eigenfunctions of Hermitianoperators, however, we will consider the properties of general sets of functions. 17.1 Sets of functions In chapter 8 we discussed the deﬁnition of a vector space but concentrated on spaces of ﬁnite dimensionality. We consider now the inﬁnite -dimensional space of all reasonably well-behaved functions f(x),g(x),h(x),...on the interval a≤x≤b. That these functions form a linear vector space can be veriﬁed since the set is closed under (i) addition, which is commutative and associative, i.e. f(x)+g(x)=g(x)+f(x), [f(x)+g(x)]+h(x)=f(x)+[g(x)+h(x)], (ii) multiplication by a scalar, which is distributive and associative, i.e. λ[f(x)+g(x)]=λf(x)+λg(x), λ[µf(x)]=(λµ)f(x), (λ+µ)f(x)=λf(x)+µf(x). Furthermore, in such a space (iii) there exists a ‘null vector’ 0 such that f(x)+0= f(x), (iv) multiplication by unity leaves any function unchanged, i.e. 1 ×f(x)=f(x), (v) each function has an associated negative function −f(x) that is such that f(x)+[−f(x)] = 0. By analogy with ﬁnite-dimensional vector spaces we now introduce a set of linearly independent basis functions y n(x),n=0,1,...,∞, such that any 583 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS ‘reasonable’ function in the interval a≤x≤b(i.e. it obeys the Dirichlet conditions discussed in chapter 12) can be expressed as the linear sum of these functions: f(x)=∞summationdisplay n=0cnyn(x). Clearly if a diﬀerent set of linearly independent basis functions zn(x) is chosen then the function can be expressed in terms of the new basis, f(x)=∞summationdisplay n=0dnzn(x), where the dnare a diﬀerent set of coeﬃcients. In each case, provided the basis functions are linearly independent, the coeﬃcients are unique. We may also deﬁne an inner product on our function space by /angbracketleftf|g/angbracketright=integraldisplayb af∗(x)g(x)ρ(x)dx, (17.9) where ρ(x) is the weight function, which we require to be real and non-negative in the interval a≤x≤b. As mentioned above, ρ(x) is often unity for all x.T w o functions are said to be orthogonal on the interval [ a, b]i f /angbracketleftf|g/angbracketright=integraldisplayb af∗(x)g(x)ρ(x)dx=0, (17.10) and the normof a function is deﬁned as /bardblf/bardbl=/angbracketleftf|f/angbracketright1/2=bracketleftbiggintegraldisplayb af∗(x)f(x)ρ(x)dxbracketrightbigg1/2 =bracketleftbiggintegraldisplayb a|f(x)|2ρ(x)dxbracketrightbigg1/2 .(17.11) An inﬁnite-dimensional vector space of functions, for which an inner product is deﬁned, is called a Hilbert space . Using the concept of the inner product we can choose a basis of linearly independent functions φn(x),n=0,1,2,...,t h a t are orthonormal, i.e. such that /angbracketleftφi|φj/angbracketright=integraldisplayb aφ∗ i(x)φj(x)ρ(x)dx=δij. (17.12) Ifyn(x),n=0,1,2,..., are a linearly independent, but not orthonormal, basis for the Hilbert space then an orthonormal set of basis functions φnmay be produced (in a similar manner to that used in the construction of a set of orthogonal eigenvectors of an Hermitian matrix, see chapter 8) by the followingprocedure, in which each of the new functions ψ nis to be normalised, giving 584 17.1 SETS OF FUNCTIONS φn=ψn/angbracketleftψn|ψn/angbracketright−1/2, before proceeding to the construction of the next one: ψ0=y0, ψ1=y1−φ0/angbracketleftφ0|y1/angbracketright, ψ2=y2−φ1/angbracketleftφ1|y2/angbracketright−φ0/angbracketleftφ0|y2/angbracketright, ... ψn=yn−φn−1/angbracketleftφn−1|yn/angbracketright−···−φ0/angbracketleftφ0|yn/angbracketright ... It is straightforward to check that each φn=ψn/angbracketleftψn|ψn/angbracketright−1/2is orthogonal to all its predecessors φi,i=0,1,2,...,n−1. This method is called Gram–Schmidt orthogonalisation . Clearly the functions ψnalso form an orthogonal set, but in general they do not have unit norms.IStarting from the linearly independent functions yn(x)=xn,n=0,1,..., construct the ﬁrst three orthonormal functions over the range −1<x< 1. The ﬁrst unnormalised function ψ0is simply equal to the ﬁrst of the original functions, i.e. ψ0=1. The normalisation is carried out by dividing by /angbracketleftψ0|ψ0/angbracketright1/2= /Z1 −11×1du /1/2 =√ 2, with the result that the ﬁrst normalised function φ0is given by φ0=ψ0√ 2= q 1 2. The second unnormalised function is found by applying the above Gram–Schmidt orthog- onalisation procedure, i.e. ψ1=y1−φ0/angbracketleftφ0|y1/angbracketright. It can easily be shown that /angbracketleftφ0|y1/angbracketright=0 ,a n ds o ψ1=x. Normalising then gives φ1=ψ1 /Z1 −1u×ud u /−1/2 = q 3 2x. The third unnormalised function is similarly given by ψ2=y2−φ1/angbracketleftφ1|y2/angbracketright−φ0/angbracketleftφ0|y2/angbracketright =x2−0−1 3, which, on normalising, gives φ2=ψ2 /Z1 −1 /; u2−1 3 /2du /−1/2 =1 2 q 5 2(3x2−1). By comparing the functions φ0,φ1andφ2, with the list in subsection 16.6.1, we see that this procedure has generated (multiples of) the ﬁrst three Legendre polynomials. J 585 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS If a function is expressed in terms of an orthonormal basis φn(x)a s f(x)=∞summationdisplay n=0anφn(x) (17.13) then the coeﬃcients anare given by an=/angbracketleftφn|f/angbracketright=integraldisplayb aφ∗ n(x)f(x)ρ(x)dx. (17.14) Note that this is true only if the basis is orthonormal. 17.1.1 Some useful inequalities Since for a Hilbert space /angbracketleftf|f/angbracketright≥0, the inequalities discussed in subsection 8.1.3 hold. The proofs are not repeated here, but the relationships are listed forcompleteness. (i) The Schwarz inequality states that |/angbracketleftf|g/angbracketright|≤/angbracketleft f|f/angbracketright 1/2/angbracketleftg|g/angbracketright1/2, (17.15) where the equality holds when f(x) is a scalar multiple of g(x), i.e. when they are linearly dependent. (ii) The triangle inequality states that /bardblf+g/bardbl≤/bardbl f/bardbl+/bardblg/bardbl, (17.16) where again equality holds when f(x) is a scalar multiple of g(x). (iii) Bessel’s inequality requires the introduction of an orthonormal basis φn(x) so that any function f(x) can be written as f(x)=∞summationdisplay n=0cnφn(x), where cn=/angbracketleftφn|f/angbracketright. Bessel’s inequality then states that /angbracketleftf|f/angbracketright≥summationdisplay n|cn|2. (17.17) The equality holds if the summation is over all the basis functions. If some values of nare omitted from the sum then the inequality results (unless, of course, the cnhappen to be zero for all values of nomitted, in which case the equality remains). 586 17.2 ADJOINT AND HERMITIAN OPERATORS 17.2 Adjoint and Hermitian operators Having discussed general sets of functions we now return to the discussion of eigenfunctions of linear operators. The adjoint of an operator L, denoted by L†, is deﬁned by integraldisplayb af(x)∗[Lg(x)]ρ(x)dx=braceleftbiggintegraldisplayb ag∗(x)bracketleftbig L†f(x)bracketrightbig ρ(x)dxbracerightbigg∗ , (17.18) or, in inner product notation, /angbracketleftf|Lg/angbracketright=/angbracketleftg|L†f/angbracketright∗.A no p e r a t o ri st h e ns a i dt ob e self-adjoint orHermitian ifL†=L,i . e .i f integraldisplayb af∗(x)[Lg(x)]ρ(x)dx=braceleftbiggintegraldisplayb ag∗(x)[Lf(x)]ρ(x)dxbracerightbigg∗ , (17.19) or, in inner product notation, /angbracketleftf|Lg/angbracketright=/angbracketleftg|Lf/angbracketright∗. From (17.19) we note that, when applied to an Hermitian operator, the general property /angbracketleftb|a/angbracketright∗=/angbracketlefta|b/angbracketrighttakes the form /angbracketleftg|Lf/angbracketright∗=/angbracketleftLf|g/angbracketright⇒/angbracketleft Lf|g/angbracketright=/angbracketleftf|Lg/angbracketright=/angbracketleftf|L|g/angbracketright, where the notation of the ﬁnal equality emphasises that Lcan act on either forg without changing the value of the inner product. A little careful study will revealthe similarity between the deﬁnition of an Hermitian operator and the deﬁnitionof an Hermitian matrix given in chapter 8. In general, however, an operator L is Hermitian over an interval a≤x≤bonly if certain boundary conditions are met by the functions fandgon which it acts.IFind the required boundary conditions for the linear operator L=d2/dt2to be Hermitian over the interval t0tot0+T. Substituting into the LHS of the deﬁnition of an Hermitian operator (17.19) and integrating by parts givesZt0+T t0f∗d2g dt2dt= / f∗dg dt /t0+T t0− Zt0+T t0df∗ dtdg dtdt, where we have taken the weight function ρ(x) to be unity. Integrating the second term on the RHS by parts yieldsZt0+T t0f∗d2g dt2dt= / f∗dg dt /t0+T t0+ / −df∗ dtg /t0+T t0+ Zt0+T t0gd2f∗ dt2dt. Remembering that the operator is real and taking the complex conjugate outside the integral givesZt0+T t0f∗d2g dt2dt= / f∗dg dt /t0+T t0− /df∗ dtg /t0+T t0+ /Zt0+T t0g∗d2f dt2dt /∗ , which, by comparison with (17.19), proves that Lis Hermitian provided/ f∗dg dt /t0+T t0= /df∗ dtg /t0+T t0. J 587 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS We showed in chapter 8 that the eigenvalues of Hermitian matrices are real and that their eigenvectors can be chosen to be orthogonal. Similarly, the eigenvaluesof Hermitian operators are real and their eigenfunctions can be chosen to beorthogonal (we will prove these properties in the following section). Hermitian operators (or matrices) are often used in the formulation of quantum mechanics. The eigenvalues then give the possible measured values of an observable quantitysuch as energy or angular momentum, and the physical requirement that suchquantities must be real is ensured by the reality of these eigenvalues. Furthermore,the inﬁnite set of eigenfunctions of an Hermitian operator form a complete basisset, so that it is possible to expand in an eigenfunction series any function y(x) obeying the appropriate conditions: y(x)= ∞summationdisplay n=0cnyn(x), (17.20) where the choice of suitable values for the cnwill make the sum arbitrarily close toy(x).†These useful properties provide the motivation for a detailed study of Hermitian operators. 17.3 The properties of Hermitian operators We now provide proofs of some of the useful properties of Hermitian operators. Again much of the analysis is similar to that for Hermitian matrices in chapter 8, although the present section stands alone. (Here, and throughout the remainderof this chapter, we will write out inner products in full. We note, however,that the inner product notation often provides a neat form in which to expressresults.) 17.3.1 Reality of the eigenvalues Consider an Hermitian operator for which (17.8) is satisﬁed by at least two eigenfunctions y i(x)a n d yj(x), which have eigenvalues λiandλjrespectively, so that Lyi=λiρ(x)yi, (17.21) Lyj=λjρ(x)yj, (17.22) where ρ(x) is the weight function. Multiplying (17.21) by y∗ jand (17.22) by y∗ i †The proof of the completeness of the eigenfunctions of an Hermitian operator is beyond the scope of this book. The reader should refer to e.g. Courant and Hilbert, Methods of Mathematical Physics (Interscience Publishers, 1953). 588 17.3 THE PROPERTIES OF HERMITIAN OPERATORS and then integrating gives integraldisplayb ay∗ jLyidx=λiintegraldisplayb ay∗ jyiρd x , (17.23) integraldisplayb ay∗ iLyjdx=λjintegraldisplayb ay∗ iyjρd x . (17.24) Remembering that we have required ρ(x) to be real, the complex conjugate of (17.23) becomes bracketleftbiggintegraldisplayb ay∗ jLyidxbracketrightbigg∗ =λ∗ iintegraldisplayb ay∗ iyjρd x , (17.25) and using the deﬁnition of an Hermitian operator (17.19) it follows that the LHS of (17.25) is equal to the LHS of (17.24). Thus (λ∗ i−λj)integraldisplayb ay∗ iyjρd x=0. (17.26) Ifi=jthen λi=λ∗ i(sinceintegraltextb ay∗ iyiρd x/negationslash= 0), which is a statement that the eigenvalue λiis real. 17.3.2 Orthogonality of the eigenfunctions From (17.26), it is immediately apparent that two eigenfunctions yiandyjthat correspond to diﬀerent eigenvalues, i.e. such that λi/negationslash=λj,s a t i s f y integraldisplayb ay∗ iyjρd x=0, (17.27) which is a statement of the orthogonality of yiandyj. Because Lis linear, the normalisation of the eigenfunctions yi(x) is arbitrary and we shall assume for deﬁniteness that they are normalised so thatintegraltextb ay∗ iyiρd x= 1. Thus we can write (17.27) in the form integraldisplayb ay∗ iyjρd x=δij, (17.28) which is valid for all pairs of values i, j. If one (or more) of the eigenvalues is degenerate, however, we have diﬀerent eigenfunctions corresponding to the same eigenvalue, and the proof of orthogo-nality is not so straightforward. Nevertheless, an orthogonal set of eigenfunctionsmay be constructed using the Gram–Schmidt orthogonalisation method mentioned earlier in this chapter and used in chapter 8 to construct a set of orthogonal eigenvectors of an Hermitian matrix. We repeat the analysis here for complete-ness. 589 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS Suppose, for the sake of our proof, that λ0isk-fold degenerate, i.e. Lyi=λ0ρyifori=0,1,...,k−1, (17.29) but that λ0is diﬀerent from any of λk,λk+1, etc. Then any linear combination of these yiis also an eigenfunction with eigenvalue λ0since Lz≡Lk−1summationdisplay i=0ciyi=k−1summationdisplay i=0ciLyi=k−1summationdisplay i=0ciλ0ρyi=λ0ρz. (17.30) If the yideﬁned in (17.29) are not already mutually orthogonal then consider the new eigenfunctions ziconstructed by the following procedure, in which each of the new functions wiis to be normalised, to give zi, before proceeding to the construction of the next one (the normalisation can be carried out by dividing the eigenfunction wiby (integraltextb aw∗ iwiρd x)1/2): w0=y0, w1=y1−parenleftbigg z0integraldisplayb az∗ 0y1ρd xparenrightbigg , w2=y2−parenleftbigg z1integraldisplayb az∗ 1y2ρd xparenrightbigg −parenleftbigg z0integraldisplayb az∗ 0y2ρd xparenrightbigg , ... wk−1=yk−1−parenleftbigg zk−2integraldisplayb az∗ k−2yk−1ρd xparenrightbigg −···−parenleftbigg z0integraldisplayb az∗ 0yk−1ρd xparenrightbigg . Each of the integrals is just a number and thus each new function zi= wi(integraltextb aw∗ iwiρd x)−1/2is, as can be shown from (17.30), an eigenvector of Lwith eigenvalue λ0. It is straightforward to check that each ziis orthogonal to all its predecessors. Thus, by this explicit construction we have shown that an orthog-onal set of eigenfunctions of an Hermitian operator Lcan be obtained. Clearly the orthonormal set obtained, z i, is not unique. 17.3.3 Construction of real eigenfunctions Recall that the eigenfunction yisatisﬁes Lyi=λiρyi (17.31) and that the complex conjugate of this gives Ly∗ i=λ∗ iρy∗ i=λiρy∗ i, (17.32) where the last equality follows because the eigenvalues are real, i.e. λi=λ∗ i. Thus, yiandy∗ iare eigenfunctions corresponding to the same eigenvalue and hence, because of the linearity of L,a tl e a s to n eo f y∗ i+yiandi(y∗ i−yi) (which are both 590 17.4 STURM–LIOUVILLE EQUATIONS real) is a non-zero eigenfunction corresponding to that eigenvalue. Therefore the eigenfunctions can always be made real by taking suitable linear combinations.Such linear combinations will only be necessary in cases where a particular λis degenerate, i.e. corresponds to more than one linearly independent eigenfunction. 17.4 Sturm–Liouville equations One of the most important applications of our discussion of Hermitian operators is to the study of Sturm–Liouville equations , which take the general form p(x)d 2y dx2+r(x)dy dx+q(x)y+λρ(x)y=0,where r(x)=dp(x) dx(17.33) andp,qandrare real functions of x. (We note that sign conventions vary in this expression for the general Sturm–Liouville equation; some authors use −λρ(x)y on the LHS of (17.33).) A variational approach to the Sturm–Liouville equation,which is useful in estimating the eigenvalues λof the equation, is discussed in chapter 22. For now, however, we concentrate on a demonstration that the Sturm–Liouville equation can be solved by superposition methods. It is clear that (17.33) can be written Ly=λρ(x)ywhere L=−bracketleftbigg p(x)d 2 dx2+r(x)d dx+q(x)bracketrightbigg . (17.34) An example is Legendre’s equation (17.5), which is a Sturm–Liouville equation with p(x)=1−x2,r(x)=−2x=p/prime(x),q(x)=0 , ρ(x) = 1 and eigenvalues /lscript(/lscript+1 ) . It will be seen that the general Sturm–Liouville equation (17.33) can be rewritten (py/prime)/prime+qy+λρy=0, (17.35) where primes denote diﬀerentiation with respect to x. Using (17.34) this may also be written Ly=−(py/prime)/prime−qy=λρy. We will show in the next section that, under certain boundary conditions on the solutions y(x), linear operators that can be w r i t t e ni nt h i sf o r ma r e self-adjoint . Whilst it is true that Sturm–Liouville equations represent only a small fraction of the diﬀerential equations encountered in practice, as we shall demonstrate insubsection 17.4.2 anysecond-order diﬀerential equation of the form p(x)y /prime/prime+r(x)y/prime+q(x)y+λρ(x)y= 0 (17.36) can be converted into Sturm–Liouville form by multiplying through by a suitable factor; this is discussed in subsection 17.4.2. 591 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS 17.4.1 Valid boundary conditions For the linear operator of the Sturm–Liouville equation (17.34) to be Hermitian over the range [ a, b] requires certain boundary conditions to be met, namely, that any two eigenfunctions yiandyjof (17.34) must satisfy bracketleftbig y∗ ipy/prime jbracketrightbig x=a=bracketleftbig y∗ ipy/prime jbracketrightbig x=bfor all i, j. (17.37) Rearranging (17.37) we ﬁnd that bracketleftBig y∗ ipy/prime jbracketrightBigx=b x=a=0, (17.38) is an equivalent statement of the required boundary conditions. These boundary conditions are in fact not too restrictive and are met, for instance, by the sets y(a)=y(b)=0 ; y(a)=y/prime(b)=0 ; p(a)=p(b) = 0 and by many other sets. It is important to note that in order to satisfy (17.37) and (17.38) one boundarycondition must be speciﬁed at each end of the range.IProve that the Sturm–Liouville operator is Hermitian over the range [a, b]and under the boundary conditions (17.38). Putting the Sturm–Liouville form Ly=−(py/prime)/prime−qyinto the deﬁnition (17.19) of an Hermitian operator, the LHS may be written as a sum of two terms, i.e. − Zb a / y∗ i(py/prime j)/prime+y∗ iqyj / dx=− Zb ay∗ i(py/prime j)/primedx− Zb ay∗ iqyjdx. The ﬁrst term may be integrated by parts to give − / y∗ ipy/prime j /b a+ Zb a(y∗ i)/primepy/prime jdx. The ﬁrst term is zero because of the boundary conditions, and thus, integrating by parts again yields/ (y∗ i)/primepyj /b a− Zb a((y∗ i)/primep)/primeyjdx. The ﬁrst term is once again zero. Thus − Zb a / y∗ i(py/prime j)/prime+y∗ iqyj / dx= Zb a / −((y∗ i)/primep)/primeyj−y∗ iqyj / dx, = / − Zb a / y∗ j(py/prime i)/prime+y∗ jqyi / dx /∗ , which proves that the Sturm–Liouville operator is Hermitian over the prescribed interval. J 17.4.2 Putting an equation into Sturm–Liouville form The Sturm–Liouville equation (17.33) requires that r(x)=p/prime(x). However, any equation of the form p(x)y/prime/prime+r(x)y/prime+q(x)y+λρ(x)y=0, (17.39) 592 17.5 EXAMPLES OF STURM–LIOUVILLE EQUATIONS can be put into self-adjoint form by multiplying through by the integrating factor F(x)=e x pbraceleftbiggintegraldisplayxr(z)−p/prime(z) p(z)dzbracerightbigg . (17.40) It is easily veriﬁed that (17.39) then takes the Sturm–Liouville form [F(x)p(x)y/prime]/prime+F(x)q(x)y+λF(x)ρ(x)y=0, (17.41) with a diﬀerent, but still non-negative, weight function F(x)ρ(x).IPut the Hermite equation y/prime/prime−2xy/prime+2αy=0 into Sturm–Liouville form. Using (17.40), with p(z)=1 , p/prime(z)=0a n d r(z)=−2zgives the integrating factor F(x)=e x p /Zx −2zd z / =e x p /; −x2 / . Thus, the Hermite equation becomes e−x2y/prime/prime−2xe−x2y/prime+2αe−x2y=(e−x2y/prime)/prime+2αe−x2y=0, which is clearly in Sturm–Liouville form with p(x)=e−x2,q(x)=0 , ρ(x)=e−x2and λ=2α. J 17.5 Examples of Sturm–Liouville equations In order to illustrate the wide applicability of Sturm–Liouville theory, in this section we present a short catalogue of some common equations of Sturm–Liouville form. Many of them have already been discussed in chapter 16. Inparticular the reader should note the orthogonality properties of the varioussolutions, which, in each case, follow because the diﬀerential operator is self- adjoint. For completeness we also quote the associated generating functions. 17.5.1 Legendre’s equation We have already met Legendre’s equation , (1−x 2)y/prime/prime−2xy/prime+/lscript(/lscript+1 )y=[ ( 1−x2)y/prime]/prime+/lscript(/lscript+1 )y= 0 (17.42) and shown that it is a Sturm–Liouville equation with p(x)=1−x2,q(x)=0 , ρ(x) = 1 and eigenvalues /lscript(/lscript+1). In the previous chapter we found the solutions of Legendre’s equation that are regular for all ﬁnite x. These are the Legendre polynomials P/lscript(x), which are given by a Rodrigues’ formula: P/lscript(x)=1 2/lscript/lscript!d/lscript dx/lscript(x2−1)/lscript. 593 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS The orthogonality and normalisation of the functions in the interval −1≤x≤1 is expressed by integraldisplay1 −1P/lscript(x)Pk(x)dx=2 2/lscript+1δ/lscriptk. The generating function is G(x, h)=( 1−2xh+h2)−1/2=∞summationdisplay n=0Pn(x)hn. Legendre’s equations appear in the analysis of physical situations involving the operator∇2and axial symmetry, since the linear diﬀerential operator involved has the form of the polar-angle part of ∇2, when the latter is expressed in spherical polar coordinates. Examples include the solution of Laplace’s equation in axially symmetric situations and the solution of the Schr ¨odinger equation for a quantum mechanical system involving a central potential. 17.5.2 The associated Legendre equation Very closely related to the Legendre equation is the associated Legendre equation [(1−x2)y/prime]/prime+bracketleftbigg /lscript(/lscript+1 )−m2 1−x2bracketrightbigg y=0, (17.43) which reduces to Legendre’s equation when m= 0. In physical applications −/lscript≤m≤/lscriptand mis restricted to integer values. If y(x)i sas o l u t i o no f Legendre’s equation then w(x)=( 1−x2)|m|/2d|m|y dx|m| is a solution of the associated equation. The solutions of the associated Legendre equation that are regular for all ﬁnite xare called the associated Legendre functions and are therefore given by Pm /lscript(x)=( 1−x2)|m|/2d|m|P/lscript dx|m|. Note also that Pm /lscript(x)=0f o r m>/lscript . Like the Legendre polynomials, the associated Legendre functions Pm /lscript(x) are orthogonal in the range −1≤x≤1. This property, and their normalisation, is expressed by integraldisplay1 −1Pm /lscript(x)Pm k(x)dx=2 2/lscript+1(/lscript+m)! (/lscript−m)!δ/lscriptk. They have the generating function G(x, h)=(2m)!(1−x2)m/2 2mm!(1−2hx+h2)m+1/2=∞summationdisplay n=0Pm n+m(x)hn. 594 17.5 EXAMPLES OF STURM–LIOUVILLE EQUATIONS The associated Legendre equation arises in physical situations in which there is a dependence on azimuthal angle φof the form eimφor cos mφ. 17.5.3 Bessel’s equation Physical situations that when described in spherical polar coordinates give rise to Legendre and associated Legendre equations lead to Bessel’s equation when cylindrical polar coordinates are used. Bessel’s equation has the form x2y/prime/prime+xy/prime+(x2−n2)y=0, (17.44) but on dividing by xand changing variables to ξ=x/a,†it takes on the Sturm-Liouville form (ξy/prime)/prime+a2ξy+−n2 ξy=0, (17.45) where a prime now indicates diﬀerentiation with respect to ξ. We met Bessel’s equation in chapter 16, where we saw that those of its solutions that are regular for ﬁnite xare the Bessel functions, given by Jn(x)=∞summationdisplay r=0(−1)r(1 2x)n+2r r!Γ(n+r+1 ), (17.46) where Γ is the gamma function discussed in the Appendix. Their orthogonality and normalisation over the range 0 ≤x<∞have been discussed in detail in chapter 16. The generating function for the Bessel functions is G(x, h)=e x pbracketleftbiggx 2parenleftbigg h−1 hparenrightbiggbracketrightbigg =∞summationdisplay n=−∞Jn(x)hn. (17.47) 17.5.4 The simple harmonic equation The most trivial of Sturm–Liouville equations is the simple harmonic motion equation y/prime/prime+ω2y=0, (17.48) which has p(x)=1 , q(x)=0 , ρ(x) = 1 and eigenvalue ω2. We have already met the solutions of this equation in the Fourier analysis of chapter 12, and the properties of orthogonality and normalisation of the eigenfunctions given there can now be seen in the wider context of general Sturm–Liouville equations. †This change of scale is required to give the conventional normalisation, but is not needed for the transformation into Sturm–Liouville form. 595 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS 17.5.5 Hermite’s equation The Hermite equation appears in the description of the wavefunction of a harmonic oscillator and is given by y/prime/prime−2xy/prime+2αy=0. (17.49) We have already seen that it can be converted to Sturm–Liouville form by multiplying by the integrating factor exp( −x2), which yields e−x2y/prime/prime−2xe−x2y/prime+2αe−x2y=(e−x2y/prime)/prime+2αe−x2y=0. (17.50) The solutions, the Hermite polynomials Hn(x), are given by a Rodrigues’ formula: Hn(x)=(−1)nex2dn dxnparenleftBig e−x2parenrightBig . (17.51) Their orthogonality over the range −∞<x<∞and their normalisation are summarised by integraldisplay∞ −∞e−x2Hm(x)Hn(x)dx=2nn!√πδmn, (17.52) and their generating function is G(x, h)=e2hx−h2=∞summationdisplay n=0Hn(x) n!hn. (17.53) 17.5.6 Laguerre’s equation The Laguerre equation appears in the description of the wavefunction of the hydrogen atom and is given by xy/prime/prime+( 1−x)y/prime+ny=0. (17.54) It can be converted to Sturm–Liouville form by multiplying by the integrating factor exp(−x), which yields xe−xy/prime/prime+( 1−x)e−xy/prime+ne−xy=(xe−xy/prime)/prime+ne−xy=0. (17.55) The solutions, the Laguerre polynomials Ln(x), are again given by a Rodrigues’ formula: Ln(x)=exdn dxnparenleftbig xne−xparenrightbig . (17.56) Their orthogonality over the range 0 ≤x<∞and their normalisation are expressed by integraldisplay∞ 0e−xLm(x)Ln(x)dx=(n!)2δmn, (17.57) 596 17.6 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS and their generating function is G(x, h)=e−xh/(1−h) 1−h=∞summationdisplay n=0Ln(x) n!hn. (17.58) 17.5.7 Chebyshev’s equation The Chebyshev equation (1−x2)y/prime/prime−xy/prime+n2y= 0 (17.59) can be converted to an equation of Sturm–Liouville form by multiplying by the integrating factor (1 −x2)−1/2. Simplifying, this yields bracketleftBig (1−x2)1/2y/primebracketrightBig/prime +n2(1−x2)−1/2y=0. (17.60) The solutions, the Chebyshev polynomials Tn(x), are once again given by a Rodrigues’ formula: Tn(x)=(−2)nn!(1−x2)1/2 (2n)!dn dxn(1−x2)n−1/2. (17.61) Their orthogonality over the range −1≤x≤1 and their normalisation are given by integraldisplay1 −1(1−x2)−1/2Tm(x)Tn(x)dx=  0f o r m/negationslash=n, π/2f o r n=m/negationslash=0, π forn=m=0,(17.62) and their generating function is G(x, h)=1−xh 1−2xh+h2=∞summationdisplay n=0Tn(x)hn. (17.63) 17.6 Superposition of eigenfunctions: Green’s functions We have already seen that if Lyn(x)=λnρ(x)yn(x), (17.64) where Lis an Hermitian operator, then the eigenvalues λnare real and the eigenfunctions yn(x) are orthogonal (or can be made so). Let us assume that we know the eigenfunctions yn(x)o fLthat individually satisfy (17.64) and some imposed boundary conditions (for which Lis Hermitian). Now let us suppose we wish to solve the inhomogeneous diﬀerential equation Ly(x)=f(x), (17.65) 597 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS subject to the same boundary conditions. Since the eigenfunctions of Lform a complete set, the full solution, y(x), to (17.65) may be written as a superposition of eigenfunctions, i.e. y(x)=∞summationdisplay n=0cnyn(x), (17.66) for some choice of the constants cn. Making full use of the linearity of L, we have f(x)=Ly(x)=LparenleftBigg∞summationdisplay n=0cnyn(x)parenrightBigg =∞summationdisplay n=0cnLyn(x)=∞summationdisplay n=0cnλnρ(x)yn(x). (17.67) Multiplying the ﬁrst and last terms of (17.67) by y∗ jand integrating, we obtain integraldisplayb ay∗ j(z)f(z)dz=∞summationdisplay n=0integraldisplayb acnλny∗ j(z)yn(z)ρ(z)dz, (17.68) w h e r ew eh a v eu s e d zas the integration variable for later convenience. Finally, using the orthogonality condition (17.28), we see that the integrals on the RHSare zero unless n=j,a n ds oo b t a i n c n=1 λnintegraltextb ay∗ n(z)f(z)dz integraltextb ay∗n(z)yn(z)ρ(z)dz. (17.69) Thus, if we can ﬁnd all the eigenfunctions of a diﬀerential operator then (17.69) can be used to ﬁnd the weighting coeﬃcients for the superposition, to give as the full solution y(x)=∞summationdisplay n=01 λnintegraltextb ay∗ n(z)f(z)dz integraltextb ay∗n(z)yn(z)ρ(z)dzyn(x). (17.70) If the eigenfunctions have already been normalised, so that integraldisplayb ay∗ n(z)yn(z)ρ(z)dz=1 f o ra l l n, and we assume that we may interchange the order of summation and integration, then (17.70) can be written as y(x)=integraldisplayb abraceleftBigg∞summationdisplay n=0bracketleftbigg1 λnyn(x)y∗ n(z)bracketrightbiggbracerightBigg f(z)dz. The quantity in braces, which is a function of xandzonly, is usually written G(x, z), and is the Green’s function for the problem. With this notation, y(x)=integraldisplayb aG(x, z)f(z)dz, (17.71) 598 17.6 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS where G(x, z)=∞summationdisplay n=01 λnyn(x)y∗ n(z). (17.72) We note that G(x, z) is determined entirely by the boundary conditions and the eigenfunctions yn, and hence by Litself, and that f(z) depends purely on the RHS of the inhomogeneous equation (17.65). Thus, for a given Land boundary conditions we can establish, once and for all, a function G(x, z) that will enable us to solve the inhomogeneous equation for anyRHS. From (17.72) we also note that G(x, z)=G∗(z,x). (17.73) We have already met the Green’s function in the solution of second-order dif- ferential equations in chapter 15, as the function that satisﬁes the equationL[G(x, z)] = δ(x−z) (and the boundary conditions). The formulation given above is an alternative, though equivalent, one.IFind an appropriate Green’s function for the equation y/prime/prime+1 4y=f(x), with boundary conditions y(0) = y(π)=0. Hence, solve for (i) f(x)=s i n 2 xand (ii) f(x)=x/2. One approach to solving this problem is to use the methods of chapter 15 and ﬁnd a complementary function and particular integral. However, in order to illustrate the techniques developed in the present chapter we will use the superposition of eigenfunctions,which, as may easily be checked, produces the same solution. The operator on the LHS of this equation is already self-adjoint under the given boundary conditions, and so we seek its eigenfunctions. These satisfy the equation y /prime/prime+1 4y=λy. This equation has the familiar solution y(x)=Asin /q 1 4−λ / x+Bcos /q 1 4−λ / x. Now, the boundary conditions require that B=0a n ds i n / q 1 4−λ / π=0 ,a n ds oq 1 4−λ=n,where n=0,±1,±2,.... Therefore, the independent eigenfunctions that satisfy the boundary conditions are yn(x)=Ansinnx, where nis any non-negative integer. The normalisation condition further requiresZπ 0A2 nsin2nx dx =1⇒ An= /2 π /1/2 . 599 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS Comparison with (17.72) shows that the appropriate Green’s function is therefore given by G(x, z)=2 π∞X n=0sinnxsinnz 1 4−n2. Case (i). Using (17.71), the solution with f(x)=s i n2 xis given by y(x)=2 π Zπ 0 / ∞X n=0sinnxsinnz 1 4−n2 /! sin 2zd z=2 π∞X n=0sinnx 1 4−n2 Zπ 0sinnzsin2zd z . Now the integral is zero unless n= 2, in which case it isZπ 0sin22zd z=π 2. Thus y(x)=−2 πsin 2x 15/4π 2=−4 15sin 2x is the full solution for f(x)=s i n2 x. This is, of course, exactly the solution found by using the methods of chapter 15. Case (ii). The solution with f(x)=x/2i sg i v e nb y y(x)= Zπ 0 / 2 π∞X n=0sinnxsinnz 1 4−n2 /! z 2dz=1 π∞X n=0sinnx 1 4−n2 Zπ 0zsinnz dz. The integral may be evaluated by integrating by parts, i.e.Zπ 0zsinnz dz= /" −zcosnz n /#π 0+ Zπ 0cosnz ndz =−πcosnπ n+ /sinnz n2 /π 0 =−π(−1)n n. Forn= 0 the integral is zero, and thus y(x)=∞X n=1(−1)n+1sinnx n /;1 4−n2 /, is the full solution for f(x)=x/2. Using the methods of subsection 15.1.2 the solution is found to be y(x)=2 x−2πsin(x/2), which may be shown to be equal to the above solution by expanding 2 x−2πsin(x/2) as a Fourier sine series. J A useful relation between the eigenfunctions of Lis given by writing f(x)=summationdisplay nyn(x)integraldisplayb ay∗ n(z)f(z)ρ(z)dz =integraldisplayb af(z)ρ(z)summationdisplay nyn(x)y∗ n(z)dz, and hence ρ(z)summationdisplay nyn(x)y∗ n(z)=δ(x−z). (17.74) 600 17.7 A USEFUL GENERALISATION This is called the completeness orclosure property of the eigenfunctions. It deﬁnes a complete set. If the spectrum of eigenvalues of Lis anywhere continuous then the eigenfunction yn(x) must be treated as y(n, x) and an integration carried out over n. We also note that the RHS of (17.74) is a δ-function and so is only non-zero when z=x; thus ρ(z) on the LHS can be replaced by ρ(x) if required, i.e. ρ(z)summationdisplay nyn(x)y∗ n(z)=ρ(x)summationdisplay nyn(x)y∗ n(z). (17.75) 17.7 A useful generalisation Sometimes we encounter inhomogeneous equations of a form slightly more gen- eral than (17.1), given by Ly(x)−λρ(x)y(x)=f(x) (17.76) for some self-adjoint operator L, with ysubject to the appropriate boundary conditions and λa given (i.e. ﬁxed) constant. To solve this equation we expand y(x)a n d f(x) in terms of the eigenfunctions yn(x) of the operator L, which satisfy Lyn(x)=λnρ(x)yn(x). Firstly, we expand f(x) as follows: f(x)=∞summationdisplay n=0yn(x)integraldisplayb ay∗ n(z)f(z)ρ(z)dz =integraldisplayb aρ(z)∞summationdisplay n=0yn(x)y∗ n(z)f(z)dz. (17.77) Using (17.75) this becomes f(x)=integraldisplayb aρ(x)∞summationdisplay n=0yn(x)y∗ n(z)f(z)dz =ρ(x)∞summationdisplay n=0yn(x)integraldisplayb ay∗ n(z)f(z)dz. (17.78) Next, we expand y(x)a sy=summationtext∞ n=0cnyn(x) and seek the coeﬃcients cn. Substi- tuting this and (17.78) in (17.76) we have ρ(x)∞summationdisplay n=0(λn−λ)cnyn(x)=ρ(x)∞summationdisplay n=0yn(x)integraldisplayb ay∗ n(z)f(z)dz, 601 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS from which we ﬁnd that cn=∞summationdisplay n=0integraltextb ay∗ n(z)f(z)dz λn−λ. Hence the solution of (17.76) is given by y=∞summationdisplay n=0cnyn(x)=∞summationdisplay n=0yn(x) λn−λintegraldisplayb ay∗ n(z)f(z)dz=integraldisplayb a∞summationdisplay n=0yn(x)y∗ n(z) λn−λf(z)dz. From this we may identify the Green’s function G(x, z)=∞summationdisplay n=0yn(x)y∗ n(z) λn−λ. We note that if λ=λn,i . e .i f λequals one of the eigenvalues of L,t h e n G(x, z) becomes inﬁnite and this method runs into diﬃculty. No solution then existsunless the RHS of (17.76) satisﬁes the relation integraldisplay b ay∗ n(x)f(x)dx=0. If the spectrum of eigenvalues of the operator Lis anywhere continuous, the orthogonality and closure relationships of the eigenfunctions become integraldisplayb ay∗ n(x)ym(x)ρ(x)dx=δ(n−m), integraldisplay∞ 0y∗ n(z)yn(x)ρ(x)dn=δ(x−z). Repeating the above analysis we then ﬁnd that the Green’s function is given by G(x, z)=integraldisplay∞ 0yn(x)y∗ n(z) λn−λdn. 17.8 Exercises 17.1 By considering /angbracketlefth|h/angbracketright,w h e r e h=f+λgwith λreal, prove that, for two functions fandg, /angbracketleftf|f/angbracketright/angbracketleftg|g/angbracketright≥1 4[/angbracketleftf|g/angbracketright+/angbracketleftg|f/angbracketright]2. The function y(x) is real and positive for all x. Its Fourier cosine transform ˜yc(k) is deﬁned by ˜yc(k)= Z∞ −∞y(x)cos( kx)dx, and it is given that ˜yc(0) = 1. Prove that ˜yc(2k)≥2[˜yc(k)]2−1. 602 17.8 EXERCISES 17.2 (a) Write the homogeneous Sturm-Liouville eigenvalue equation for which y(a)=y(b)=0a s L(y;λ)≡(py/prime)/prime+qy+λρy=0, where p(x),q(x)a n d ρ(x) are continuously diﬀerentiable functions. Show that ifz(x)a n d F(x)s a t i s f y L(z;λ)=F(x)w i t h z(a)=z(b)=0t h e nZb ay(x)F(x)dx=0. (b) Demonstrate the validity of result (a) by direct calculation for the case in which p(x)=ρ(x)=1 , q(x)=0 , a=−1,b=1a n d z(x)=1−x2. 17.3 Consider the real eigenfunctions yn(x) of a Sturm–Liouville equation (py/prime)/prime+qy+λρy=0,a≤x≤b in which p(x),q(x)a n d ρ(x) are continuously diﬀerentiable real functions and p(x) does not change sign in a≤x≤b.T a k e p(x) as positive throughout the interval, if necessary by changing the signs of all eigenvalues. For a≤x1≤x2≤b, establish the identity (λn−λm) Zx2 x1ρynymdx= / ynpy/prime m−ympy/prime n /x2 x1. Deduce that if λn>λ mthen yn(x) must change sign between two successive zeroes ofym(x). (The reader may ﬁnd it helpful to illustrate this result by sketching the ﬁrst few eigenfunctions of the system y/prime/prime+λy=0 ,w i t h y(0) = y(π) = 0, and the Legendre polynomials Pn(z) given in subsection 16.6.1 for n=2,3,4,5.) 17.4 (a) Show that the equation y/prime/prime+aδ(x)y+λy=0, with y(±π)=0a n d areal, has a set of eigenvalues λsatisfying tan(π√λ)=2√λ a. (b) Investigate the conditions under which negative eigenvalues, λ=−µ2with µ real, are possible. 17.5 Express the hypergeometric equation (x2−x)y/prime/prime+[ ( 1+ α+β)x−γ]y/prime+αβy=0 in Sturm–Liouville form, determining the conditions imposed on xand on the parameters α,βandγby the boundary conditions and the allowed forms of weight function. 17.6 (a) Find the solution of (1 −x2)y/prime/prime−2xy/prime+by=f(x) valid in the range −1≤x≤1 and ﬁnite at x= 0, in terms of Legendre polynomials. (b) If b=1 4a n d f(x)=5 x3, ﬁnd the explicit solution and verify it by direct substitution. 17.7 Use the generating function for the Legendre polynomials Pn(x) to show thatZ1 0P2n+1(x)dx=(−1)n (2n)! 22n+1n!(n+1 ) ! and that, except for the case n=0 ,Z1 0P2n(x)dx=0. 603 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS 17.8 The quantum mechanical wavefunction for a one-dimensional simple harmonic oscillator in its nth energy level is of the form ψ(x)=e x p (−x2/2)Hn(x), where Hn(x)i st h e nth Hermite polynomial. The generating function for the polynomials (17.53) is G(x, h)=e2hx−h2=∞X n=0Hn(x) n!hn. (a) Find Hi(x)f o r i=1,2,3,4. (b) Evaluate by direct calculationZ∞ −∞e−x2Hp(x)Hq(x)dx, (i) for p=2 , q= 3; (ii) for p=2 , q= 4; (iii) for p=q= 3. Check your answers against equation (17.52). (You will ﬁnd it convenient to useZ∞ −∞x2ne−x2dx=(2n)!√π 22nn! for integer n≥0.) 17.9 The Laguerre polynomials, which are required for the quantum mechanical description of the hydrogen atom, can be deﬁned by the generating function(equation (17.58)) G(x, h)=e −hx/(1−h) 1−h=∞X n=0Ln(x) n!hn. By diﬀerentiating the equation separately with respect to xand h,a n dr e - substituting for G(x, h), prove that LnandL/prime n(=dLn(x)/dx) satisfy the recurrence relations L/prime n−nL/prime n−1+nLn−1=0, Ln+1−(2n+1−x)Ln+n2Ln−1=0. From these two equations and others derived from them, show that Ln(x)s a t i s ﬁ e s the Laguerre equation xL/prime/prime n+( 1−x)L/prime n+nLn=0. 17.10 Starting from the linearly independent functions 1, x,x2,x3,..., in the range 0≤x<∞, ﬁnd the ﬁrst three orthogonal functions φ0,φ1andφ2, with respect to the weight function ρ(x)=e−x. By comparing your answers with the Laguerre polynomials generated by the recurrence relation derived in exercise 17.9, deducethe form of φ 3(x). 17.11 Consider the set of functions {f(x)}of the real variable x, deﬁned in the interval −∞<x<∞,t h a t→0a tl e a s ta sq u i c k l ya s x−1asx→±∞ . For unit weight function, determine whether each of the following linear operators is Hermitianwhen acting upon {f(x)}: (a)d dx+x;( b )−id dx+x2;( c ) ixd dx;( d ) id3 dx3. 17.12 The Chebyshev polynomials Tn(x) can be written as Tn(x)=c o s ( ncos−1x). 604 17.8 EXERCISES (a) Verify that these functions do satisfy the Chebyshev equation. (b) Use de Moivre’s theorem to show that an alternative expression is Tn(x)=nX reven(−1)r/2n! (n−r)!r!xn−r(1−x2)r/2. 17.13 A particle moves in a parabolic potential in which its natural angular frequency of oscillation is 1 /2. At time t= 0 it passes through the origin with velocity v and is suddenly subjected to an addi tional acceleration of +1 for 0 ≤t≤π/2, and then−1f o r π/2<t≤π. At the end of this period it is at the origin again. Apply the results of the worked example in section 17.6 to show that v=−8 π∞X m=01 (4m+2 )2−1 4≈−0.81. 17.14 Find an eigenfunction expansion for the solution with boundary conditions y(0) = y(π) = 0 of the inhomogeneous equation d2y dx2+κy=f(x), where κis a constant and f(x)= /( x, 0≤x≤π/2, π−x, π/ 2<x≤π. 17.15 (a) Find those eigenfunctions yn(x) of the self-adjoint linear diﬀerential operator d2/dx2that satisfy the boundary conditions yn(0) = yn(π) = 0, and hence construct its Green’s function G(x, z). (b) Construct the same Green’s function using the methods of subsection 15.2.5, showing that it is G(x, z)= /( x(z−π)/π,0≤x≤z, z(x−π)/π, z≤x≤π. (c) By expanding the function given in (b) in terms of the eigenfunctions yn(x), verify that it is the same function as that derived in (a). 17.16 (a) The diﬀerential operator Lis deﬁned by Ly=−d dx / exdy dx / −exy 4. Determine the eigenvalues λnof the problem Lyn=λnexyn0<x< 1, with boundary conditions y(0) = 0 ,dy dx+y 2=0 a t x=1. (b) Find the corresponding unnormalised yn, and also a weight function ρ(x)w i t h respect to which the ynare orthogonal. Hence, select a suitable normalisation for the yn. (c) By making an eigenfunction expansion, solve the equation Ly=−ex/2,0<x< 1, subject to the same boundary conditions as previously. 605 EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS 17.17 Show that the linear operator L≡1 4(1 +x2)2d2 dx2+1 2x(1 +x2)d dx+a, acting upon functions deﬁned in −1≤x≤1 and vanishing at the endpoints of the interval, is Hermitian with respect to the weight function (1 + x2)−1. By making the change of variable x=t a n ( θ/2), ﬁnd two even eigenfunctions, f1(x)a n d f2(x), of the diﬀerential equation Lu=λu. 17.18 By substituting x=e x p tﬁnd the normalized eigenfunctions yn(x)a n dt h e eigenvalues λnof the operator Ldeﬁned by Ly=x2y/prime/prime+2xy/prime+1 4y, 1≤x≤e, with y(1) = y(e) = 0. Find, as a series Panyn(x), the solution of Ly=x−1/2. 17.19 Express the solution of Poisson’s equation in electrostatics, ∇2φ(r)=−ρ(r)//epsilon10, where ρis the non-zero charge density over a ﬁnite part of space, in the form of an integral and hence identify the Green’s function for the ∇2operator. 17.20 In the quantum mechanical study of the scattering of a particle by a potential, a Born-approximation solution can be obtained in terms of a function y(r)t h a t satisﬁes an equation of the form (−∇2−K2)y(r)=F(r). Assuming that yk(r)=( 2 π)−3/2exp(ik·r) is a suitably normalised eigenfunction of −∇2corresponding to eigenvalue −k2, ﬁnd a suitable Green’s function GK(r,r/prime). By taking the direction of the vector r−r/primeas the polar axis for a k-space integration, show that GK(r,r/prime) can be reduced to 1 4π|r−r/prime| Z∞ −∞wsinw w2−w2 0dw, where w0=K|r−r/prime|. (This integral can be evaluated using a contour integration (chapter 20) to give(4π|r−r /prime|)−1exp(iK|r−r/prime|).) 17.9 Hints and answers 17.1 Express the condition /angbracketlefth|h/angbracketright≥0 as a quadratic equation in λand then apply the condition for no real roots, noting that /angbracketleftf|g/angbracketright+/angbracketleftg|f/angbracketrightis real. To put a limit onR ycos2kx dx,s e tf=y1/2coskxandg=y1/2in the inequality. 17.2 (a) By twice integrating by parts the term containing p, show thatRb ayL(z;λ)dx= Rb azL(y;λ)dx. (b)y(x)=Acos(√λx)w i t h λ=n2π2/4, and F(x)=λ−2−λx2. 17.3 Follow an argument similar to that in subsection 17.3.1, but integrate from x1to x2, rather than from atob.T a k e x1andx2as two successive zeroes of ym(x)a n d note that, if the sign of ymisαthen the sign of y/prime m(x1)i sαwhilst that of y/prime m(x2) is−α. Now assume that yn(x) does not change sign in the interval and has a constant sign β; show that this leads to a contradiction between the signs of the two sides of the identity. 17.4 (a) Diﬀerent combinations of sinusoids are needed for negative and positive ranges of x.( b ) µmust satisfy tanh µπ=2µ/a,w h i c hr e q u i r e s a>2/π. 606 17.9 HINTS AND ANSWERS 17.5 [ xγ(1−x)α+β−γ+1y/prime]/prime=αβxγ−1(1−x)α+β−γy;0≤x≤1,α+β>γ> 1. 17.6 (a) y= PanPn(x)w i t h an=n+1/2 b−n(n+1 ) Z1 −1f(z)Pn(z)dz; (b) 5x3=2P3(x)+3P1(x), giving a1=1/4a n d a3= 1, leading to y=5 ( 2 x3−x)/4. 17.8 (a) 2 x,4x2−2, 8x3−12x,1 6x4−48x2+ 12; (b) (i) 0, (ii) 0, (iii) 48√π. 17.10 φ0(x)=1 ,φ1(x)=x−1,φ2(x)=(x2−4x+2 )/2;n!φn(x)=(−1)nLn(x); φ3(x)=(x3−9x2+1 8x−6)/6. 17.11 (a) No, R gf∗/primedx/negationslash= 0; (b) yes; (c) no, i R f∗gdx/negationslash=0 ;( d )y e s . 17.14 The normalised eigenfunctions are (2 /π)1/2sinnx,w i t h nan integer. y(x)=( 4 /π) P nodd[(−1)(n−1)/2sinnx]/[n2(κ−n2)]. 17.15 (a) The normalised eigenfunctions are (2 /π)1/2sinnx,w i t h nan integer. G(x, z)=(−2/π) P∞ n=0[sin(nz)sin(nx)]/n2. 17.16 (a) λn=(n+1/2)2π2,n=0,1,2,... . (b) Since yn(1)y/prime m(1)/negationslash= 0, the Sturm–Liouville boundary conditions are not sat- isﬁed and the appropriate weight function has to be justiﬁed by inspection. Thenormalised eigenfunctions are√2e −x/2sin[(n+1/2)πx], with ρ(x)=ex. (c)y(x)=(−2/π3) P∞ n=0e−x/2sin[(n+1/2)πx]/(n+1/2)3. 17.17 In terms of θ,Lisd2/dθ2+aand has eigenfunctions u(θ)=c o s (√ a−λθ), where√ a−λ=2n+1 ; f1(x)=( 1−x2)/(1 +x2);f2(x)=4 [ ( 1−x2)/(1 +x2)]3−3[(1−x2)/(1 +x2)]. 17.18 yn(x)=√ 2x−1/2sin(nπlnx)w i t h λn=−n2π2; an= /( −(nπ)−2 Re 1√ 2x−1sin(nπlnx)dx=−√ 8(nπ)−3fornodd, 0f o r neven. 17.19 G(r,r/prime)=( 4 π|r−r/prime|)−1. 607 18 Partial diﬀerential equations: general and particular solutions In this chapter and the next the solution of diﬀerential equations of types typically encountered in the physical sciences and engineering is extended tosituations involving more than one independent variable. A partial diﬀerentialequation (PDE) is an equation relating an unknown function (the dependentvariable) of two or more variables to its partial derivatives with respect tothose variables. The most commonly occurring independent variables are those describing position and time, and so we will couch our discussion and examples in notation appropriate to them. As in other chapters we will focus our attention on the equations that arise most often in physical situations. We will restrict our discussion, therefore, tolinear PDEs, i.e. those of ﬁrst degree in the dependent variable. Furthermore, wewill discuss primarily second-order equations. The solution of ﬁrst-order PDEs will necessarily be involved in treating these, and some of the methods discussed can be extended without diﬃculty to third- and higher-order equations. We shallalso see that many ideas developed for ordinary diﬀerential equations (ODEs)can be carried over directly into the study of PDEs. In this chapter we will concentrate on general solutions of PDEs in terms of arbitrary functions and the particular solutions that may be derived fromthem in the presence of boundary conditions. We also discuss the existence and uniqueness of the solutions to PDEs under given boundary conditions. In the next chapter the methods most commonly used in practice for obtaining solutions to PDEs subject to given boundary conditions will be considered. Thesemethods include the separation of variables, integral transforms and Green’sfunctions. This division of material is rather arbitrary and really has been madeonly to emphasise the general usefulness of the latter methods. In particular, it will be readily apparent that some of the results of the present chapter are in fact solutions in the form of separated variables, but arrived at by a diﬀerentapproach. 608 18.1 IMPORTANT PARTIAL DIFFERENTIAL EQUATIONS 18.1 Important partial diﬀerential equations Most of the important PDEs of physics are second-order and linear. In order to gain familiarity with their general form, some of the more important ones willnow be brieﬂy discussed. These equations apply to a wide variety of diﬀerentphysical systems. Since, in general, the PDEs listed below describe three-dimensional situations, the independent variables are randt,w h e r e ris the position vector and tis time. The actual variables used to specify the position vector rare dictated by the coordinate system in use. For example, in Cartesian coordinates the independentvariables of position are x,yandz, whereas in spherical polar coordinates they arer,θandφ. The equations may be written in a coordinate-independent manner, however, by the use of the Laplacian operator ∇ 2. 18.1.1 The wave equation The wave equation ∇2u=1 c2∂2u ∂t2(18.1) describes as a function of position and time the displacement from equilibrium, u(r,t), of a vibrating string or membrane or a vibrating solid, gas or liquid. The equation also occurs in electromagnetism, where umay be a component of the electric or magnetic ﬁeld in an elecromagnetic wave or the current or voltagealong a transmission line. The quantity cis the speed of propagation of the waves.IFind the equation satisﬁed by small transverse displacements u(x, t)of a uniform string of mass per unit length ρheld under a uniform tension T, assuming that the string is initially located along the x-axis in a Cartesian coordinate system. Figure 18.1 shows the forces acting on an elemental length ∆ sof the string. If the tension Tin the string is uniform along its length then the net upward vertical force on the element is ∆F=Tsinθ2−Tsinθ1. Assuming that the angles θ1andθ2are both small, we may make the approximation sinθ≈tanθ. Since at any point on the string the slope tan θ=∂u/∂x ,t h ef o r c ec a nb e written ∆F=T /∂u(x+∆x, t) ∂x−∂u(x, t) ∂x / ≈T∂2u(x, t) ∂x2∆x, where we have used the deﬁnition of the partial derivative to simplify the RHS. This upward force may be equated, by Newton’s second law, to the product of the mass of the element and its upward acceleration. The element has a mass ρ∆s,w h i c hi s approximately equal to ρ∆xif the vibrations of the string are small, and so we have ρ∆x∂2u(x, t) ∂t2=T∂2u(x, t) ∂x2∆x. 609 PDES: GENERAL AND PARTICULAR SOLUTIONS u x xT T∆s x+∆xθ1θ2 Figure 18.1 The forces acting on an element of a string under uniform tension T. Dividing both sides by ∆ xwe obtain, for the vibrations of the string, the one-dimensional wave equation ∂2u ∂x2=1 c2∂2u ∂t2, where c2=T/ρ. J The longitudinal vibrations of an elastic rod obey a very similar equation to that derived in the above example, namely ∂2u ∂x2=ρ E∂2u ∂t2; here ρis the mass per unit volume and Eis Young’s modulus. The wave equation can be generalised slightly. For example, in the case of the vibrating string, there could also be an external upward vertical force f(x, t)p e r unit length acting on the string at time t. The transverse vibrations would then satisfy the equation T∂2u ∂x2+f(x, t)=ρ∂2u ∂t2, which is clearly of the form ‘upward force per unit length = mass per unit length ×upward acceleration’. Similar examples, but involving two or three spatial dimensions rather than one, are provided by the equation governing the transverse vibrations of a stretchedmembrane subject to an external vertical force density f(x, y, t), Tparenleftbigg∂ 2u ∂x2+∂2u ∂y2parenrightbigg +f(x, y, t)=ρ(x, y)∂2u ∂t2, where ρis the mass per unit area of the membrane and Tis the tension. 610 18.1 IMPORTANT PARTIAL DIFFERENTIAL EQUATIONS 18.1.2 The diﬀusion equation The diﬀusion equation κ∇2u=∂u ∂t(18.2) describes the temperature uin a region containing no heat sources or sinks; it also applies to the diﬀusion of a chemical that has a concentration u(r,t). The constant κis called the diﬀusivity. The equation is clearly second-order in the three spatial variables, but ﬁrst order in time.IDerive the equation satisﬁed by the temperature u(r,t)at time tfor a material of uniform thermal conductivity k, speciﬁc heat capacity sand density ρ. Express the equation in Cartesian coordinates. Let us consider an arbitrary volume Vlying within the solid and bounded by a surface S (this may coincide with the surface of the solid if so desired). At any point in the solidthe rate of heat ﬂow per unit area in any given direction ˆris proportional to minus the component of the temperature gradient in that direction and so is given by ( −k∇u)·ˆr.T h e total ﬂux of heat outof the volume Vper unit time is given by −dQ dt= ZZ S(−k∇u)·ˆndS = ZZ Z V∇·(−k∇u)dV, (18.3) where Qis the total heat energy in Vat time tandˆnis the outward-pointing unit normal toS; note that we have used the divergence theorem to convert the surface integral into a volume integral. W ec a na l s oe x p r e s s Qas a volume integral over V, Q= ZZZ Vsρu dV, and its rate of change is then given by dQ dt= ZZ Z Vsρ∂u ∂tdV, (18.4) where we have taken the derivative with respect to time inside the integral (see section 5.12). Comparing (18.3) and (18.4), and remembering that the volume Vis arbitrary, we obtain the three-dimensional diﬀusion equation κ∇2u=∂u ∂t, where the diﬀusion coeﬃcient κ=k/(sρ). To express this equation in Cartesian coordinates, we simply write ∇2in terms of x,yandzto obtain κ /∂2u ∂x2+∂2u ∂y2+∂2u ∂z2 / =∂u ∂t. J The diﬀusion equation just derived can be generalised to k∇2u+f(r,t)=sρ∂u ∂t. 611 PDES: GENERAL AND PARTICULAR SOLUTIONS The second term, f(r,t), represents a varying density of heat sources throughout the material but is often not required in physical applications. In the most generalcase, k,sandρmay depend on position r, in which case the ﬁrst term becomes ∇·(k∇u). However, in the simplest application the heat ﬂow is one-dimensional with no heat sources, and the equation becomes (in Cartesian coordinates) ∂ 2u ∂x2=sρ k∂u ∂t. 18.1.3 Laplace’s equation Laplace’s equation, ∇2u=0, (18.5) may be obtained by setting ∂u/∂t = 0 in the diﬀusion equation (18.2), and describes (for example) the steady-state temperature distribution in a solid in which there are no heat sources – i.e. the temperature distribution after a longtime has elapsed. Laplace’s equation also describes the gravitational potential in a region con- taining no matter or the electrostatic potential in a charge-free region. Further, it applies to the ﬂow of an incompressible ﬂuid with no sources, sinks or vortices;in this case uis the velocity potential, from which the velocity is given by v=∇u. 18.1.4 Poisson’s equation Poisson’s equation, ∇ 2u=ρ(r), (18.6) describes the same physical situations as Laplace’s equation, but in regions containing matter, charges or sources of heat or ﬂuid. The function ρ(r)i s called the source density and in physical applications usually contains somemultiplicative physical constants. For example, if uis the electrostatic potential in some region of space, in which case ρis the density of electric charge, then ∇ 2u=−ρ(r)//epsilon10,w h e r e /epsilon10is the permittivity of free space. Alternatively, umight represent the gravitational potential in some region where the matter density isgiven by ρ;t h e n∇ 2u=4πGρ(r), where Gis the gravitational constant. 18.1.5 Schr ¨odinger’s equation The Schr ¨odinger equation −/planckover2pi12 2m∇2u+V(r)u=i/planckover2pi1∂u ∂t, (18.7) 612 18.2 GENERAL FORM OF SOLUTION describes the quantum mechanical wavefunction u(r,t) of a non-relativistic particle of mass m;/planckover2pi1is Planck’s constant divided by 2 π. Like the diﬀusion equation it is second order in the three spatial variables and ﬁrst order in time. 18.2 General form of solution Before turning to the methods by which we may hope to solve PDEs such as those listed in the previous section, it is instructive, as for ODEs in chapter 14, to study how PDEs may be formed from a set of possible solutions. Such a studycan provide an indication of how equations obtained not from possible solutionsbut from physical arguments might be solved. For deﬁniteness let us suppose we have a set of functions involving two independent variables xandy. Without further speciﬁcation this is of course a very wide set of functions, and we could not expect to ﬁnd a useful equation thatthey all satisfy. However, let us consider a type of function u i(x, y)i nw h i c h xand yappear in a particular way, such that uican be written as a function (however complicated) of a single variable p, itself a simple function of xandy. Let us illustrate this by considering the three functions u1(x, y)=x4+4 (x2y+y2+1 ), u2(x, y)=s i n x2cos2y+c o s x2sin 2y, u3(x, y)=x2+2y+2 3x2+6y+5. These are all fairly complicated functions of xandyand a single diﬀerential equation of which each one is a solution is not obvious. However, if we observe that in fact each can be expressed as a function of the variable p=x2+2yalone (with no other xoryinvolved) then a great simpliﬁcation takes place. Written in terms of pthe above equations become u1(x, y)=(x2+2y)2+4= p2+4= f1(p), u2(x, y)=s i n ( x2+2y)=s i n p=f2(p), u3(x, y)=(x2+2y)+2 3(x2+2y)+5=p+2 3p+5=f3(p). Let us now form, for each ui, the partial derivatives ∂ui/∂xand∂ui/∂y.I ne a c h case these are (writing both the form for general pand the one appropriate to o u rp a r t i c u l a rc a s e , p=x2+2y) ∂ui ∂x=dfi(p) dp∂p ∂x=2xf/prime i, ∂ui ∂y=dfi(p) dp∂p ∂y=2f/prime i, fori= 1, 2, 3. All reference to the form of fican be eliminated from these 613 PDES: GENERAL AND PARTICULAR SOLUTIONS equations by cross-multiplication, obtaining ∂p ∂y∂ui ∂x=∂p ∂x∂ui ∂y, or, for our speciﬁc form, p=x2+2y, ∂ui ∂x=x∂ui ∂y. (18.8) It is thus apparent that not only are the three functions u1,u2u3solutions of the PDE (18.8) but so also is any arbitrary function f(p) of which the argument phas the form x2+2y. 18.3 General and particular solutions In the last section we found that the ﬁrst-order PDE (18.8) has as a solution any function of the variable x2+2y. This points the way for the solution of PDEs of other orders, as follows. It is notgenerally true that an nth-order PDE can always be considered as resulting from the elimination of narbitrary functions from its solution (as opposed to the elimination of narbitrary constants for an nth-order ODE, see section 14.1). However, given speciﬁc PDEs we can try to solve them by seeking combinations of variables in terms of which the solutions may be expressed as arbitrary functions. Where this is possible we may expect n combinations to be involved in the solution. Naturally, the exact functional form of the solution for any particular situation must be determined by some set of boundary conditions. For instance, if the PDE contains two independent variables xandythen for complete determination of its solution the boundary conditions will take a form equivalent to specifyingu(x, y) along a suitable continuum of points in the xy-plane (usually along a line). We now discuss the general and particular solutions of ﬁrst- and second- order PDEs. In order to simplify the algebra, we will restrict our discussionto equations containing just two independent variables xandy. Nevertheless, the method presented below may be extended to equations containing severalindependent variables. 18.3.1 First-order equations Although most of the PDEs encountered in physical contexts are second order (i.e. they contain ∂ 2u/∂x2or∂2u/∂x∂y , etc.), we now discuss ﬁrst-order equations to illustrate the general considerations involved in the form of the solution andin satisfying any boundary conditions on the solution. The most general ﬁrst-order linear PDE (containing two independent variables) 614 18.3 GENERAL AND PARTICULAR SOLUTIONS is of the form A(x, y)∂u ∂x+B(x, y)∂u ∂y+C(x, y)u=R(x, y), (18.9) where A(x, y),B(x, y),C(x, y)a n d R(x, y) are given functions. Clearly, if either A(x, y)o r B(x, y) is zero then the PDE may be solved straightforwardly as a ﬁrst-order linear ODE (as discussed in chapter 14), the only modiﬁcation being that the arbitrary constant of integration becomes an arbitrary function ofxory respectively.IFind the general solution u(x, y)of x∂u ∂x+3u=x2. Dividing through by xwe obtain ∂u ∂x+3u x=x, which is a linear equation with integra ting factor (see subsection 14.2.4) exp /Z3 xdx / =e x p ( 3l n x)=x3. Multiplying through by this factor we ﬁnd ∂ ∂x(x3u)=x4, which, on integrating with respect to x,g i v e s x3u=x5 5+f(y), where f(y)i sa n arbitrary function ofy. Finally, dividing through by x3, we obtain the solution u(x, y)=x2 5+f(y) x3. J When the PDE contains partial derivatives with respect to both independent variables then, of course, we cannot employ the above procedure but must seekan alternative method. Let us for the moment restrict our attention to the specialcase in which C(x, y)=R(x, y) = 0 and, following the discussion of the previous section, look for solutions of the form u(x, y)=f(p)w h e r e pis some, at present unknown, combination of xandy. We then have ∂u ∂x=df(p) dp∂p ∂x, ∂u ∂y=df(p) dp∂p ∂y, 615 PDES: GENERAL AND PARTICULAR SOLUTIONS which, when substituted into the PDE (18.9), give bracketleftbigg A(x, y)∂p ∂x+B(x, y)∂p ∂ybracketrightbiggdf(p) dp=0. This removes all reference to the actual form of the function f(p) since for non-trivial pwe must have A(x, y)∂p ∂x+B(x, y)∂p ∂y=0. (18.10) Let us now consider the necessary condition for f(p) to remain constant as x andyvary; this is that pitself remains constant. Thus for fto remain constant implies that xandymust vary in such a way that dp=∂p ∂xdx+∂p ∂ydy=0. (18.11) The forms of (18.10) and (18.11) are very alike, and become the same if we require that dx A(x, y)=dy B(x, y). (18.12) By integrating this expression the form of pcan be found.IFor x∂u ∂x−2y∂u ∂y=0, (18.13) ﬁnd (i) the solution that takes the value 2y+1on the line x=1, and (ii) a solution that has the value 4at the point (1,1). If we seek a solution of the form u(x, y)=f(p), we deduce from (18.12) that u(x, y) will be constant along lines of ( x, y)t h a ts a t i s f y dx x=dy −2y, w h i c ho ni n t e g r a t i n gg i v e s x=cy−1/2. Identifying the constant of integration cwith p1/2 (to avoid fractional powers), we conclude that p=x2y. Thus the general solution of the PDE (18.13) is u(x, y)=f(x2y), where fis an arbitrary function. We must now ﬁnd the particular solutions that obey each of the imposed boundary conditions. For boundary condition (i) a little thought shows that the particular solutionrequired is u(x, y)=2 ( x 2y)+1=2 x2y+1. (18.14) For boundary condition (ii) some ob viously acceptable solutions are u(x, y)=x2y+3, u(x, y)=4 x2y, u(x, y)=4 . 616 18.3 GENERAL AND PARTICULAR SOLUTIONS Each is a valid solution (the freedom of choice of form arises from the fact that u is speciﬁed at only one point (1 ,1), and not along a continuum (say), as in boundary condition (i)). All three are particular exam ples of the general solution, which may be written, for example, as u(x, y)=x2y+3+ g(x2y), where g=g(x2y)=g(p) is an arbitrary function subject only to g(1) = 0. For this example, the forms of gcorresponding to the particular solutions listed above are g(p)=0 , g(p)=3 p−3,g(p)=1−p. J As mentioned above, in order to ﬁnd a solution of the form u(x, y)=f(p)w e require that the original PDE contains no term in u, but only terms containing its partial derivatives. If a term in uis present, so that C(x, y)/negationslash= 0 in (18.9), then the procedure needs some modiﬁcation, since we cannot simply divide outthe dependence on f(p) to obtain (18.10). In such cases we look instead for a solution of the form u(x, y)=h(x, y)f(p). We illustrate this method in the following example.IFind the general solution of x∂u ∂x+2∂u ∂y−2u=0. (18.15) We seek a solution of the form u(x, y)=h(x, y)f(p), with the consequence that ∂u ∂x=∂h ∂xf(p)+hdf(p) dp∂p ∂x, ∂u ∂y=∂h ∂yf(p)+hdf(p) dp∂p ∂y. Substituting these expressions into the PDE (18.15) and rearranging, we obtain/ x∂h ∂x+2∂h ∂y−2h / f(p)+ / x∂p ∂x+2∂p ∂y / hdf(p) dp=0. The ﬁrst factor in parentheses is just the original PDE with ureplaced by h. Therefore, if hisanysolution of the PDE, however simple , this term will vanish, to leave/ x∂p ∂x+2∂p ∂y / hdf(p) dp=0, from which, as in the previous case, we obtain x∂p ∂x+2∂p ∂y=0. From (18.11) and (18.12) we see that u(x, y) will be constant along lines of ( x, y)t h a t satisfy dx x=dy 2, which integrates to give x=cexp(y/2). Identifying the constant of integration cwith pwe ﬁndp=xexp(−y/2). Thus the general solution of (18.15) is u(x, y)=h(x, y)f(xexp(−1 2y)), where f(p) is any arbitrary function of pandh(x, y) is any solution of (18.15). 617 PDES: GENERAL AND PARTICULAR SOLUTIONS If we take, for example, h(x, y)=e x p y, which clearly satisﬁes (18.15), then the general solution is u(x, y)=( e x p y)f(xexp(−1 2y)). Alternatively, h(x, y)=x2also satisﬁes (18.15) and so the general solution to the equation c a na l s ob ew r i t t e n u(x, y)=x2g(xexp(−1 2y)), where gis an arbitrary function of p; clearly g(p)=f(p)/p2. J 18.3.2 Inhomogeneous equations and problems Let us discuss in a more general form the particular solutions of (18.13) found in the second example of the previous subsection. It is clear that, so far as thisequation is concerned, if u(x, y) is a solution then so is any multiple of u(x, y)o r any linear sum of separate solutions u 1(x, y)+u2(x, y). However, when it comes to ﬁtting the boundary conditions this is not so. For example, although u(x, y) in (18.14) satisﬁes the PDE and the boundary condition u(1,y)=2 y+ 1, the function u1(x, y)=4 u(x, y)=8 xy+ 4, whilst satisfying the PDE, takes the value 8 y+4 on the line x= 1 and so does not satisfy the required boundary condition. Likewise the function u2(x, y)=u(x, y)+f1(x2y), for arbitrary f1, satisﬁes (18.13) but takes the value u2(1,y)=2 y+1+ f1(y)o n the line x= 1, and so is not of the required form unless f1is identically zero. Thus we see that when treating the superposition of solutions of PDEs two considerations arise, one concerning the equation itself and the other connectedto the boundary conditions. The equation is said to be homogeneous if the fact thatu(x, y) is a solution implies that λu(x, y), for any constant λ, is also a solution. However, the problem is said to be homogeneous if, in addition, the boundary conditions are such that if they are satisﬁed by u(x, y) then they are also satisﬁed byλu(x, y). The last requirement itself is referred to as that of homogeneous boundary conditions . For example, the PDE (18.13) is homogeneous but the general ﬁrst-order equation (18.9) would not be homogeneous unless R(x, y) = 0. Furthermore, the boundary condition (i) imposed on the solution of (18.13) in the previous subsection is not homogeneous though, in this case, the boundary condition u(x, y) = 0 on the line y=4x −2 would be, since u(x, y)=λ(x2y−4) satisﬁes this condition for any λand, being a function of x2y, satisﬁes (18.13). The reason for discussing the homogeneity of PDEs and their boundary condi- tions is that in linear PDEs there is a close parallel to the complementary-function and particular-integral property of ODEs. The general solution of an inhomo-geneous problem can be written as the sum of anyparticular solution of the 618 18.3 GENERAL AND PARTICULAR SOLUTIONS problem and the general solution of the corresponding homogeneous problem (as for ODEs, we require that the particular solution is not already contained in thegeneral solution of the homogeneous problem). Thus, for example, the generalsolution of ∂u ∂x−x∂u ∂y+au=f(x, y), (18.16) subject to, say, the boundary condition u(0,y)=g(y), is given by u(x, y)=v(x, y)+w(x, y), where v(x, y) is any solution (however simple) of (18.16) such that v(0,y)=g(y) andw(x, y) is the general solution of ∂w ∂x−x∂w ∂y+aw=0, (18.17) with w(0,y) = 0. If the boundary conditions are suﬃciently speciﬁed then the only possible solution of (18.17) will be w(x, y)≡0a n d v(x, y) will be the complete solution by itself. Alternatively, we may begin by ﬁnding the general solution of the inhomoge- neous equation (18.16) without regard for any boundary conditions; it is just the sum of the general solution to the homogeneous equation and a particular inte- gral of (18.16), both without reference to the boundary conditions. The boundary conditions can then be used to ﬁnd the appropriate particular solution from thegeneral solution. We will not discuss at length general methods of obtaining particular integrals of PDEs but merely note that some of those methods available for ordinarydiﬀerential equations can be suitably extended. †IFind the general solution of y∂u ∂x−x∂u ∂y=3x. (18.18) Hence ﬁnd the most general particular solution (i) which satisﬁes u(x,0) = x2and (ii) which has the value u(x, y)=2 at the point (1,0). This equation is inhomogeneous, and so let us ﬁrst ﬁnd the general solution of (18.18) without regard for any boundary conditions. We begin by looking for the solution of thecorresponding homogeneous equation ((18.18) with the RHS equal to zero) of the formu(x, y)=f(p). Following the same procedure as that used in the solution of (18.13) we ﬁnd that u(x, y) will be constant along lines of ( x, y)t h a ts a t i s f y dx y=dy −x⇒x2 2+y2 2=c. Identifying the constant of integration cwith p/2, we ﬁnd that the general solution of the †See for example Piaggio, Diﬀerential Equations (Bell, 1954), p. 175 et seq. 619 PDES: GENERAL AND PARTICULAR SOLUTIONS homogeneous equation is u(x, y)=f(x2+y2) for arbitrary function f. Now by inspection a particular integral of (18.18) is u(x, y)=−3y, and so the general solution to (18.18) is u(x, y)=f(x2+y2)−3y. Boundary condition (i) requires u(x,0) = f(x2)=x2,i . e .f(z)=z, and so the particular solution in this case is u(x, y)=x2+y2−3y. Similarly, boundary condition (ii) requires u(1,0) = f(1) = 2. One possibility is f(z)=2 z, and if we make this choice, then one way of writing the most general particular solutionis u(x, y)=2 x 2+2y2−3y+g(x2+y2), where gis any arbitrary function for which g(1) = 0. Alternatively, a simpler choice would bef(z) = 2, leading to u(x, y)=2−3y+g(x2+y2). J Although we have discussed the solution of inhomogeneous problems only for ﬁrst-order equations, the general considerations hold true for linear PDEs ofhigher order. 18.3.3 Second-order equations As noted in section 18.1, second-order linear PDEs are of great importance in describing the behaviour of many physical systems. As in our discussion of ﬁrst- order equations, for the moment we shall restrict our discussion to equations withjust two independent variables; extensions to a greater number of independentvariables are straightforward. The most general second-order linear PDE (containing two independent vari- ables) has the form A∂ 2u ∂x2+B∂2u ∂x∂y+C∂2u ∂y2+D∂u ∂x+E∂u ∂y+Fu=R(x, y), (18.19) where A ,B,...,F andR(x, y) are given functions of xandy. Because of the nature of the solutions to such equations, they are usually divided into three classes, adivision of which we will make further use in subsection 18.6.2. The equation (18.19) is called hyperbolic ifB 2>4AC,parabolic ifB2=4ACandelliptic if B2<4AC. Clearly, if A,BandCare functions of xandy(rather than just constants) then the equation might be of diﬀerent types in diﬀerent parts of thexy-plane. Equation (18.19) obviously represents a very large class of PDEs, and it is usually impossible to ﬁnd closed-form solutions to most of these equations.Therefore, for the moment we shall consider only homogeneous equations, with R(x, y) = 0, and make the further (greatly simplifying) restriction that, throughout the remainder of this section, A ,B,...,F are not functions of xandybut merely constants. 620 18.3 GENERAL AND PARTICULAR SOLUTIONS We now tackle the problem of solving some types of second-order PDE with constant coeﬃcients by seeking solutions that are arbitrary functions of particularcombinations of independent variables, just as we did for ﬁrst-order equations. Following the discussion of the previous section, we can hope to ﬁnd such solutions only if all the terms of the equation involve the same total number of diﬀerentiations, i.e. all terms are of the same order, although the numberof diﬀerentiations with respect to the individual independent variables may bediﬀerent. This means that in (18.19) we require the constants D,EandFto be identically zero (we have, of course, already assumed that R(x, y) is zero), so that we are now considering only equations of the form A∂ 2u ∂x2+B∂2u ∂x∂y+C∂2u ∂y2=0, (18.20) where A,BandCare constants. We note that both the one-dimensional wave equation, ∂2u ∂x2−1 c2∂2u ∂t2=0, and the two-dimensional Laplace equation, ∂2u ∂x2+∂2u ∂y2=0, are of this form, but that the diﬀusion equation, κ∂2u ∂x2−∂u ∂t=0, is not, since it contains a ﬁrst-order derivative. Since all the terms in (18.20) involve two diﬀerentiations, by assuming a solution of the form u(x, y)=f(p), where pis some unknown function of xandy(ort), we may be able to obtain a common factor d2f(p)/dp2as the only appearance of fon the LHS. Then, because of the zero RHS, all reference to the form of fcan be cancelled out. We can gain some guidance on suitable forms for the combination p=p(x, y) by considering ∂u/∂x when uis given by u(x, y)=f(p), for then ∂u ∂x=df(p) dp∂p ∂x. Clearly diﬀerentiation of this equation with respect to x(ory) will not lead to a single term on the RHS, containing fonly as d2f(p)/dp2, unless the factor ∂p/∂x is a constant so that ∂2p/∂x2and∂2p/∂x∂y are necessarily zero. This shows that pmust be a linear function of x. In an exactly similar way pmust also be a linear function of y,i . e .p=ax+by. If we assume a solution of (18.20) of the form u(x, y)=f(ax+by), and evaluate 621 PDES: GENERAL AND PARTICULAR SOLUTIONS the terms ready for substitution into (18.20), we obtain ∂u ∂x=adf(p) dp,∂u ∂y=bdf(p) dp, ∂2u ∂x2=a2d2f(p) dp2,∂2u ∂x∂y=abd2f(p) dp2,∂2u ∂y2=b2d2f(p) dp2, w h i c ho ns u b s t i t u t i o ng i v e parenleftbig Aa2+Bab+Cb2parenrightbigd2f(p) dp2=0. (18.21) This is the form we have been seeking, since now a solution independent of the form of fcan be obtained if we require that aandbsatisfy Aa2+Bab+Cb2=0. From this quadratic, two values for the ratio of the two constants aandbare obtained, b/a=[−B±(B2−4AC)1/2]/2C. If we denote these two ratios by λ1andλ2thenanyfunctions of the two variables p1=x+λ1y, p 2=x+λ2y will be solutions of the original equation (18.20). The omission of the constant factor afrom p1andp2is of no consequence since this can always be absorbed into the particular form of any chosen function; only the relative weighting of x andyinpis important. Since p1andp2are in general diﬀerent, we can thus write the general solution of (18.20) as u(x, y)=f(x+λ1y)+g(x+λ2y), (18.22) where fandgare arbitrary functions. Finally, we note that the alternative solution d2f(p)/dp2= 0 to (18.21) leads only to the trivial solution u(x, y)=kx+ly+m, for which all second derivatives are individually zero.IFind the general solution of the one-dimensional wave equation ∂2u ∂x2−1 c2∂2u ∂t2=0. This equation is (18.20) with A=1 ,B=0a n d C=−1/c2, and so the values of λ1andλ2 are the solutions of 1−λ2 c2=0, namely λ1=−candλ2=c. This means that arbitrary functions of the quantities p1=x−ct, p 2=x+ct 622 18.3 GENERAL AND PARTICULAR SOLUTIONS will be satisfactory solutions of the equation and that the general solution will be u(x, t)=f(x−ct)+g(x+ct), (18.23) where fandgare arbitrary functions. This solution is discussed further in section 18.4. J The method used to obtain the general solution of the wave equation may also be applied straightforwardly to Laplace’s equation.IFind the general solution of the two-dimensional Laplace equation ∂2u ∂x2+∂2u ∂y2=0. (18.24) Following the established procedure, we look for a solution that is a function f(p)o f p=x+λy, where from (18.24) λsatisﬁes 1+λ2=0. This requires that λ=±i, and satisfactory variables parep=x±iy. The general solution required is therefore, in terms of arbitrary functions fandg, u(x, y)=f(x+iy)+g(x−iy). J It will be apparent from the last two examples that the nature of the appropriate linear combination of xandydepends upon whether B2>4ACorB2<4AC. This is exactly the same criterion as determines whether the PDE is hyperbolic or elliptic. Hence as a general result, hyperbolic and elliptic equations of the form (18.20), given the restriction that the constants A,BandCare real, have as solutions functions whose arguments have the form x+αyandx+iβyrespectively, where αandβthemselves are real. The one case not covered by this result is that in which B2=4AC,i . e .a parabolic equation. In this case λ1andλ2are not diﬀerent and only one suitable combination of xandyresults, namely u(x, y)=f(x−(B/2C)y). To ﬁnd the second part of the general solution we try, in analogy with the corresponding situation for ordinary diﬀerential equations, a solution of the form u(x, y)=h(x, y)g(x−(B/2C)y). Substituting this into (18.20) and using A=B2/4Cresults in parenleftbigg A∂2h ∂x2+B∂2h ∂x∂y+C∂2h ∂y2parenrightbigg g=0. Therefore we require h(x, y) to be any solution of the original PDE. There are several simple solutions of this equation, but as only one is required we take thesimplest non-trivial one, h(x, y)=x, to give the general solution of the parabolic equation u(x, y)=f(x−(B/2C)y)+xg(x−(B/2C)y). (18.25) 623 PDES: GENERAL AND PARTICULAR SOLUTIONS We could, of course, have taken h(x, y)=y, but this only leads to a solution that is already contained in (18.25).ISolve ∂2u ∂x2+2∂2u ∂x∂y+∂2u ∂y2=0, subject to the boundary conditions u(0,y)=0 andu(x,1) = x2. From our general result, functions of p=x+λywill be solutions provided 1+2 λ+λ2=0, i.e.λ=−1 and the equation is parabolic. The general solution is therefore u(x, y)=f(x−y)+xg(x−y). The boundary condition u(0,y) = 0 implies f(p)≡0, whilst u(x,1) = x2yields xg(x−1) = x2, which gives g(p)=p+ 1, Therefore the particular solution required is u(x, y)=x(p+1 )= x(x−y+1 ). J To reinforce the material discussed above we will now give alternative deriva- tions of the general solutions (18.22) and (18.25) by expressing the original PDEin terms of new variables before solving it. The actual solution will then become almost trivial; but, of course, it will be recognised that suitable new variables could hardly have been guessed if it were not for the work already done. Thisdoes not detract from the validity of the derivation to be described, only fromthe likelihood that it would be discovered by inspection. We start again with (18.20) and change to new variables ζ=x+λ 1y, η =x+λ2y. With this change of variables, we have from the chain rule that ∂ ∂x=∂ ∂ζ+∂ ∂η, ∂ ∂y=λ1∂ ∂ζ+λ2∂ ∂η. Using these and the fact that A+Bλi+Cλ2 i=0 f o r i=1,2, equation (18.20) becomes [2A+B(λ1+λ2)+2Cλ1λ2]∂2u ∂ζ∂η=0. 624 18.3 GENERAL AND PARTICULAR SOLUTIONS Then, providing the factor in brackets does not vanish, for which the required condition is easily shown to be B2/negationslash=4AC,w eo b t a i n ∂2u ∂ζ∂η=0, which has the successive integrals ∂u ∂η=F(η),u(ζ,η)=f(η)+g(ζ). This solution is just the same as (18.22), u(x, y)=f(x+λ2y)+g(x+λ1y). If the equation is parabolic (i.e. B2=4AC), we instead use the new variables ζ=x+λy, η =x, and recalling that λ=−(B/2C) we can reduce (18.20) to A∂2u ∂η2=0. Two straightforward integrations give as the general solution u(ζ,η)=ηg(ζ)+f(ζ), w h i c hi nt e r m so f xandyhas exactly the form of (18.25), u(x, y)=xg(x+λy)+f(x+λy). Finally, as hinted at in subsection 18.3.2 with reference to ﬁrst-order linear PDEs, some of the methods used to ﬁnd particular integrals of linear ODEscan be suitably modiﬁed to ﬁnd particular integrals of PDEs of higher order. Insimple cases, however, an appropriate solution may often be found by inspection.IFind the general solution of ∂2u ∂x2+∂2u ∂y2=6 (x+y). Following our previous methods and results, the complementary function is u(x, y)=f(x+iy)+g(x−iy), and only a particular integral remains to be found. By inspection a particular integral of the equation is u(x, y)=x3+y3, and so the general solution can be written u(x, y)=f(x+iy)+g(x−iy)+x3+y3. J 625 PDES: GENERAL AND PARTICULAR SOLUTIONS 18.4 The wave equation We have already found that the general solution of the one-dimensional wave equation is u(x, t)=f(x−ct)+g(x+ct), (18.26) where fandgare arbitrary functions. However, the equation is of such general importance that further discussion will not be out of place. Let us imagine that u(x, t)=f(x−ct) represents the displacement of a string at time tand position x. It is clear that all positions xand times tfor which x−ct= constant will have the same instantaneous displacement. But x−ct=c o n s t a n t is exactly the relation between the time and position of an observer travelling with speed calong the positive x-direction. Consequently this moving observer sees a constant displacement of the string, whereas to a stationary observer, theinitial proﬁle u(x,0) moves with speed calong the x-axis as if it were a rigid system. Thus f(x−ct) represents a wave form of constant shape travelling along the positive x-axis with speed c, the actual form of the wave depending upon the function f. Similarly, the term g(x+ct) is a constant wave form travelling with speed cin the negative x-direction. The general solution (18.23) represents a superposition of these. If the functions fand gare the same then the complete solution (18.23) represents identical progressive waves going in opposite directions. This may result in a wave pattern whose proﬁle does not progress, described as a standing wave. As a simple example, suppose both f(p)a n d g(p) have the form † f(p)=g(p)=Acos(kp+/epsilon1). Then (18.23) can be written as u(x, t)=A[cos(kx−kct+/epsilon1)+c o s ( kx+kct+/epsilon1)] =2Acos(kct)cos ( kx+/epsilon1). The important thing to notice is that the shape of the wave pattern, given by the factor in x, is the same at all times but that its amplitude 2 Acos(kct) depends upon time. At some points xthat satisfy cos(kx+/epsilon1)=0 there is no displacement at any time; such points are called nodes. So far we have not imposed any boundary conditions on the solution (18.26). The problem of ﬁnding a solution to the wave equation that satisﬁes given bound-ary conditions is normally treated using the method of separation of variables †In the usual notation, kis the wave number (= 2 π/wavelength) and kc=ω, the angular frequency of the wave. 626 18.4 THE WAVE EQUATION discussed in the next chapter. Nevertheless, we now consider D’Alembert’s solution u(x, t) of the wave equation subject to initial conditions (boundary conditions) in the following general form: initial displacement, u(x,0) = φ(x); initial velocity,∂u(x,0) ∂t=ψ(x). The functions φ(x)a n d ψ(x) are given and describe the displacement and velocity of each part of the string at the (arbitrary) time t=0 . It is clear that what we need are the particular forms of the functions fandg in (18.26) that lead to the required values at t= 0. This means that φ(x)=u(x,0) = f(x−0) +g(x+0 ), (18.27) ψ(x)=∂u(x,0) ∂t=−cf/prime(x−0) +cg/prime(x+0 ), (18.28) where it should be noted that f/prime(x−0) stands for df(p)/dpevaluated, after the diﬀerentiation, at p=x−c×0; likewise for g/prime(x+0 ) . Looking on the above two left-hand sides as functions of p=x±ct, but everywhere evaluated at t= 0, we may integrate (18.28) between an arbitrary (and irrelevant) lower limit p0and an indeﬁnite upper limit pto obtain 1 cintegraldisplayp p0ψ(q)dq+K=−f(p)+g(p), t h ec o n s t a n to fi n t e g r a t i o n Kdepending on p0. Comparing this equation with (18.27), with xreplaced by p, we can establish the forms of the functions fand gas f(p)=φ(p) 2−1 2cintegraldisplayp p0ψ(q)dq−K 2, (18.29) g(p)=φ(p) 2+1 2cintegraldisplayp p0ψ(q)dq+K 2. (18.30) Adding (18.29) with p=x−ctto (18.30) with p=x+ctgives as the solution to the original problem u(x, t)=1 2[φ(x−ct)+φ(x+ct)]+1 2cintegraldisplayx+ct x−ctψ(q)dq, (18.31) in which we notice that all dependence on p0has disappeared. Each of the terms in (18.31) has a fairly straightforward physical interpretation. In each case the factor 1 /2 represents the fact that only half a displacement proﬁle that starts at any particular point on the string travels towards any other positionx, the other half travelling away from it. The ﬁrst term 1 2φ(x−ct) arises from the initial displacement at a distance ctto the left of x; this travels forward arriving at xat time t. Similarly, the second contribution is due to the initial displacement at a distance ctto the right of x. The interpretation of the ﬁnal 627 PDES: GENERAL AND PARTICULAR SOLUTIONS term is a little less obvious. It can be viewed as representing the accumulated transverse displacement at position xdue to the passage past xof all parts of the initial motion whose eﬀects can reach xwithin a time t, both backward and forward travelling. The extension to the three-dimensional wave equation of solutions of the type we have so far encountered presents no serious diﬃculty. In Cartesian coordinatesthe three-dimensional wave equation is ∂ 2u ∂x2+∂2u ∂y2+∂2u ∂z2−1 c2∂2u ∂t2=0. (18.32) In close analogy with the one-dimensional case we try solutions that are functions of linear combinations of all four variables, p=lx+my+nz+µt. It is clear that a solution u(x, y, z, t )=f(p) will be acceptable provided that parenleftbigg l2+m2+n2−µ2 c2parenrightbiggd2f(p) dp2=0. Thus, as in the one-dimensional case, fcan be arbitrary provided that l2+m2+n2=µ2/c2. Using an obvious normalisation, we take µ=±candl,m,nas three numbers such that l2+m2+n2=1. In other words ( l,m,n) are the Cartesian components of a unit vector ˆnthat points along the direction of propagation of the wave. The quantity pcan be written in terms of vectors as the scalar expression p=ˆn·r±ct, and the general solution of (18.32) is then u(x, y, z, t )=u(r,t)=f(ˆn·r−ct)+g(ˆn·r+ct), (18.33) where ˆnisanyunit vector. It would perhaps be more transparent to write ˆn explicitly as one of the arguments of u. 18.5 The diﬀusion equation One important class of second-order PDEs, which we have not yet considered in detail, is that in which the second derivative with respect to one variableappears, but only the ﬁrst derivative with respect to another (usually time). Thisis exempliﬁed by the one-dimensional diﬀusion equation κ∂ 2u(x, t) ∂x2=∂u ∂t, (18.34) 628 18.5 THE DIFFUSION EQUATION in which κis a constant with the dimensions length2×time−1. The physical constants that go to make up κin a particular case depend upon the nature of the process (e.g. solute diﬀusion, heat ﬂow, etc.) and the material being described. With (18.34) we cannot hope to repeat successfully the method of subsection 18.3.3, since now u(x, t) is diﬀerentiated a diﬀerent number of times on the two sides of the equation; any attempted solution in the form u(x, t)=f(p) with p=ax+btwill lead only to an equation in which the form of fcannot be cancelled out. Clearly we must try other methods. Solutions may be obtained by using the standard method of separation of variables discussed in the next chapter. Alternatively, a simple solution is also given if both sides of (18.34), as it stands, are separately set equal to a constant α(say), so that ∂2u ∂x2=α κ,∂u ∂t=α. These equations have the general solutions u(x, t)=α 2κx2+xg(t)+h(t)a n d u(x, t)=αt+m(x) respectively, and may be made compatible with each other if g(t) is taken as constant, g(t)=g(where gcould be zero), h(t)=αtandm(x)=(α/2κ)x2+gx. An acceptable solution is thus u(x, t)=α 2κx2+gx+αt+c o n s t a n t . (18.35) Let us now return to seeking solutions of equations by combining the inde- pendent variables in particular ways. Having seen that a linear combination ofxandtwill be of no value, we must search for other possible combinations. It has been noted already that κhas the dimensions length 2×time−1and so the combination of variables η=x2 κt will be dimensionless. Let us see if we can satisfy (18.34) with a solution of the form u(x, t)=f(η). Evaluating the necessary derivatives we have ∂u ∂x=df(η) dη∂η ∂x=2x κtdf(η) dη, ∂2u ∂x2=2 κtdf(η) dη+parenleftbigg2x κtparenrightbigg2d2f(η) dη2, ∂u ∂t=−x2 κt2df(η) dη. Substituting these expressions into (18.34) we ﬁnd that the new equation can be 629 PDES: GENERAL AND PARTICULAR SOLUTIONS written entirely in terms of η, 4ηd2f(η) dη2+( 2+ η)df(η) dη=0. This is a straightforward ODE, which can be solved (using a minimum of explanation) as follows. Writing f/prime(η)=df(η)/dη, etc., we have f/prime/prime(η) f/prime(η)=−1 2η−1 4 ⇒ ln[η1/2f/prime(η)] =−η 4+c ⇒ f/prime(η)=A η1/2expparenleftBig−η 4parenrightBig ⇒ f(η)=Aintegraldisplayη η0µ−1/2expparenleftBig−µ 4parenrightBig dµ. If we now write this in terms of a slightly diﬀerent variable ζ=η1/2 2=x 2(κt)1/2, then dζ=1 4η−1/2dη, and the solution to (18.34) is given by u(x, t)=f(η)=g(ζ)=Bintegraldisplayζ ζ0exp(−ν2)dν. (18.36) Here Bis a constant and it should be noticed that xandtappear on the RHS only in the indeﬁnite upper limit ζ, and then only in the combination xt−1/2.I f ζ0is chosen as zero then u(x, t) is, to within a constant factor, †the error function erf[x/2(κt)1/2], which is tabulated in many reference books. Only non-negative values of xandtare to be considered here, so that ζ≥ζ0. Let us try to determine what kind of (say) temperature distribution and ﬂow this represents. For deﬁniteness we take ζ0= 0. Firstly, since u(x, t) in (18.36) depends only upon the product xt−1/2, it is clear that all points xat times tsuch that xt−1/2has the same value have the same temperature. Put another way, at any speciﬁc time tthe region having a particular temperature has moved along the positive x-axis a distance proportional to the square root of t.T h i si sat y p i c a l diﬀusion process. Notice that, on the one hand, at t= 0, the variable ζ→∞ andubecomes quite independent of x(except perhaps at x= 0); the solution then represents a uniform spatial temperature distribution. On the other hand, at x=0 , u(x, t)i s identically zero for all t. †Take B=2π−1/2to give the usual error function normalised such that erf( ∞) = 1. See the Appendix. 630 18.5 THE DIFFUSION EQUATIONIAn infrared laser delivers a pulse of (heat) energy Eto a point Pon a large insulated sheet of thickness b, thermal conductivity k, speciﬁc heat sand density ρ. The sheet is initially at a uniform temperature. If u(r,t)is the excess temperature a time tlater, at a point that is a distance r(/greatermuchb)from P, then show that a suitable expression for uis u(r,t)=α texp / −r2 2βt / , (18.37) where αandβare constants. (Note that we use rinstead of ρto denote the radial coordinate in plane polars so as to avoid confusion with the density.) Further, (i) show that β=2k/(sρ); (ii) show that the excess heat energy in the sheet is independent of t, and hence evaluate α; and (iii) show that the total heat ﬂow past any circle of radius risE. The equation to be solved is the heat diﬀusion equation k∇2u(r,t)=sρ∂u(r,t) ∂t. Since we only require the solution for r/greatermuchbwe can treat the problem as two-dimensional with obvious circular symmetry. Thus only the r-derivative term in the expression for ∇2u is non-zero, giving k r∂ ∂r / r∂u ∂r / =sρ∂u ∂t, (18.38) where now u(r,t)=u(r,t). (i) Substituting the given expression (18.37) into (18.38) we obtain 2kα βt2 /r2 2βt−1 / exp / −r2 2βt / =sρα t2 /r2 2βt−1 / exp / −r2 2βt / , from which we ﬁnd that (18.37) is a solution, provided β=2k/(sρ). (ii) The excess heat in the system at any time tis bρs Z∞ 0u(r,t)2πr dr=2πbρsα Z∞ 0r texp / −r2 2βt / dr =2πbρsαβ. The excess heat is therefore independent of tand must be equal to the total heat input E, implying that α=E 2πbρsβ=E 4πbk. (iii) The total heat ﬂow past a circle of radius ris −2πrbk Z∞ 0∂u(r,t) ∂rdt=−2πrbk Z∞ 0E 4πbkt /−r βt / exp / −r2 2βt / dt =E / exp / −r2 2βt //∞ 0=Efor all r. As we would expect, all the heat energy Edeposited by the laser will eventually ﬂow past a circle of any given radius r. J 631 PDES: GENERAL AND PARTICULAR SOLUTIONS 18.6 Characteristics and the existence of solutions So far in this chapter we have discussed how to ﬁnd general solutions to various types of ﬁrst- and second-order linear PDE. Moreover, given a set of boundary conditions we have shown how to ﬁnd the particular solution (or class of solutions)that satisﬁes them. For ﬁrst-order equations, for example, we found that if thevalue of u(x, y) is speciﬁed along some curve in the xy-plane then the solution to the PDE is in general unique, but that if u(x, y) is speciﬁed at only a single point then the solution is not unique: there exists a class of particular solutions all ofwhich satisfy the boundary condition. In this section and the next we make more rigorous the notion of the types of boundary condition that cause a PDE to have a unique solution, a class of solutions, or no solution at all. 18.6.1 First-order equations Let us consider the general ﬁrst-order PDE (18.9) but now write it as A(x, y)∂u ∂x+B(x, y)∂u ∂y=F(x, y, u). (18.39) Suppose we wish to solve this PDE subject to the boundary condition that u(x, y)=φ(s) is speciﬁed along some curve Cin the xy-plane that is described parametrically by the equations x=x(s)a n d y=y(s), where sis the arc length along C. The variation of ualong Cis therefore given by du ds=∂u ∂xdx ds+∂u ∂ydy ds=dφ ds. (18.40) We may then solve the two (inhomogeneous) simultaneous linear equations (18.39) and (18.40) for ∂u/∂x and∂u/∂y ,unless the determinant of the coeﬃcients vanishes (see section 8.18), i.e. unless vextendsinglevextendsinglevextendsinglevextendsingledx/ds dy/ds ABvextendsinglevextendsinglevextendsinglevextendsingle=0. At each point in the xy-plane this equation determines a set of curves called characteristic curves (or just characteristics ), which thus satisfy Bdx ds−Ady ds=0, or, multiplying through by ds/dx and dividing through by A, dy dx=B(x, y) A(x, y). (18.41) However, we have, already met (18.41) in subsection 18.3.1 on ﬁrst-order PDEs, where solutions of the form u(x, y)=f(p), where pis some combination of xandy, 632 18.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONS were discussed. Comparing (18.41) with (18.12) we see that the characteristics are merely those curves along which pis constant. Since the partial derivatives ∂u/∂x and∂u/∂y may be evaluated provided the boundary curve Cdoesnotlie along a characteristic, deﬁning u(x, y)=φ(s) along Cis suﬃcient to specify the solution to the original problem (equation plus boundary conditions) near the curve C, in terms of a Taylor expansion about C. Therefore the characteristics can be considered as the curves along which information about the solution u(x, y) ‘propagates’. This is best understood by using an example.IFind the general solution of x∂u ∂x−2y∂u ∂y= 0 (18.42) that takes the value 2y+1on the line x= 1 between y=0a n d y=1. We solved this problem in subsection 18.3.1 for the case where u(x, y) takes the value 2y+1a l o ngt he entire linex= 1. We found then that the general solution to the equation (ignoring boundary conditions) is of the form u(x, y)=f(p)=f(x2y), for some arbitrary function f. Hence the characteristics of (18.42) are given by x2y=c where cis a constant; some of these curves are plotted in ﬁgure 18.2 for various values of c. Furthermore, we found that the particular solution for which u(1,y)=2 y+1for all y was given by u(x, y)=2 x2y+1. In the present case the value of x2yis ﬁxed by the boundary conditions only between y=0a n d y= 1. However, since the characteristics are curves along which x2y, and hence f(x2y), remains constant, the solution is determined everywhere along any characteristic that intersects the line segment denoting the boundary conditions. Thus u(x, y)=2 x2y+1 is the particular solution that holds in the shaded region in ﬁgure 18.2 (corresponding to0≤c≤1). Outside this region, however, the solution is not precisely speciﬁed, and any function of the form u(x, y)=2 x 2y+1+ g(x2y) will satisfy both the equation and the boundary condition, provided g(p)=0f o r 0≤p≤1. J In the above example the boundary curve was not itself a characteristic and furthermore it crossed each characteristic once only . For a general boundary curve Cthis may not be the case. Firstly, if Cis itself a characteristic (or is just a single point) then information about the solution cannot ‘propagate’ away from C,a n d so the solution remains unspeciﬁed everywhere except on C. The second possibility is that C(although not a characteristic itself) crosses some characteristics more than once, as in ﬁgure 18.3. In this case specifying the value of u(x, y) along the curve PQdetermines the solution along all the character- istics that intersect it. Therefore, also specifying u(x, y)a l o n g QRcanoverdetermine the problem solution and generally results in there being no solution. 633 PDES: GENERAL AND PARTICULAR SOLUTIONS 112y x−1 x=1c=1 y=c/x2 Figure 18.2 The characteristics of equation (18.42). The shaded region shows where the solution to the equation is deﬁned, given the imposed boundary condition at x= 1 between y=0a n d y= 1, shown as a bold vertical line. Py QCR x Figure 18.3 A boundary curve Cthat crosses characteristics more than once. 18.6.2 Second-order equations The concept of characteristics can be extended naturally to second- (and higher-) order equations. In this case let us write the general second-order linear PDE (18.19) as A(x, y)∂2u ∂x2+B(x, y)∂2u ∂x∂y+C(x, y)∂2u ∂y2=Fparenleftbigg x, y, u,∂u ∂x,∂u ∂yparenrightbigg . (18.43) 634 18.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONS Cy xdxdy ˆndsdr Figure 18.4 A boundary curve Cand its tangent and unit normal at a given point. For second-order equations we might expect that relevant boundary conditions would involve specifying u, or some of its ﬁrst derivatives, or both, along a suitable set of boundaries bordering or enclosing the region over which a solution is sought. Three common types of boundary condition occur and are associatedwith the names of Dirichlet, Neumann and Cauchy. They are as follows. (i)Dirichlet : The value of uis speciﬁed at each point of the boundary. (ii)Neumann : The value of ∂u/∂n ,t h enormal derivative ofu,i ss p e c i ﬁ e da t each point of the boundary. Note that ∂u/∂n =∇u·ˆn,w h e r e ˆnis the normal to the boundary at each point. (iii)Cauchy :B o t h uand∂u/∂n are speciﬁed at each point of the boundary. Let us consider for the moment the solution of (18.43) subject to the Cauchy boundary conditions, i.e. uand∂u/∂n are speciﬁed along some boundary curve Cin the xy-plane deﬁned by the parametric equations x=x(s),y=y(s),sbeing the arc length along C(see ﬁgure 18.4). Let us suppose that along Cwe have u(x, y)=φ(s)a n d ∂u/∂n =ψ(s). At any point on Cthe vector dr=dxi+dyjis a tangent to the curve and ˆnds=dyi−dxjis a vector normal to the curve. Thus onCwe have ∂u ∂s≡∇u·dr ds=∂u ∂xdx ds+∂u ∂ydy ds=dφ(s) ds, ∂u ∂n≡∇u·ˆn=∂u ∂xdy ds−∂u ∂ydx ds=ψ(s). These two equations may then be solved straightforwardly for the ﬁrst partial derivatives ∂u/∂x and∂u/∂y along C. Using the chain rule to write d ds=dx ds∂ ∂x+dy ds∂ ∂y, 635 PDES: GENERAL AND PARTICULAR SOLUTIONS we may diﬀerentiate the two ﬁrst derivatives ∂u/∂x and∂u/∂y along the boundary to obtain the pair of equations d dsparenleftbigg∂u ∂xparenrightbigg =dx ds∂2u ∂x2+dy ds∂2u ∂x∂y, d dsparenleftbigg∂u ∂yparenrightbigg =dx ds∂2u ∂x∂y+dy ds∂2u ∂y2. We may now solve these two equations, together with the original PDE (18.43), for the second partial derivatives of u,except where the determinant of their coeﬃcients equals zero,vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleABC dx dsdy ds0 0dx dsdy dsvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=0. Expanding out the determinant, Aparenleftbiggdy dsparenrightbigg2 −Bparenleftbiggdx dsparenrightbiggparenleftbiggdy dsparenrightbigg +Cparenleftbiggdx dsparenrightbigg2 =0. Multiplying through by ( ds/dx)2we obtain Aparenleftbiggdy dxparenrightbigg2 −Bdy dx+C=0, (18.44) which is the ODE for the curves in the xy-plane along which the second partial derivatives of ucannot be found. As for the ﬁrst-order case, the curves satisfying (18.44) are called characteristics of the original PDE. These characteristics have tangents at each point given by(when A/negationslash=0 ) dy dx=B±√ B2−4AC 2A. (18.45) Clearly, when the original PDE is hyperbolic ( B2>4AC), equation (18.45) deﬁnes two families of real curves in the xy-plane; when the equation is parabolic (B2=4AC) it deﬁnes one family of real curves; and when the equation is elliptic (B2<4AC) it deﬁnes two families of complex curves. Furthermore, when A, BandCare constants, rather than functions of xandy, the equations of the characteristics will be of the form x+λy= constant, which is reminiscent of the form of solution discussed in subsection 18.3.3. 636 18.6 CHARACTERISTICS AND THE EXISTENCE OF SOLUTIONS 0ct x+ct=c o n s t a n tx−ct=c o n s t a n t x L Figure 18.5 The characteristics for the one-dimensional wave equation. The shaded region indicates the region over which the solution is determined by specifying Cauchy boundary conditions at t= 0 on the line segment x=0t o x=L.IFind the characteristics of the one-dimensional wave equation ∂2u ∂x2−1 c2∂2u ∂t2=0. This is a hyperbolic equation with A=1 , B=0a n d C=−1/c2. Therefore from (18.44) the characteristics are given by/dx dt /2 =c2, and so the characteristics are the straight lines x−ct=c o n s t a n ta n d x+ct=c o n s t a n t . J The characteristics of second-order PDEs can be considered as the curves along which partial information about the solution u(x, y) ‘propagates’. Consider a point in the space that has the independent variables as its coordinates; if either or both of the two characteristics which pass through the point does not intersectthe curve along which the boudary conditions are speciﬁed then the solution willnot be determined at that point.In particular, if the equation is hyperbolic, sothat we obtain two families of real characteristics in the xy-plane, then Cauchy boundary conditions propagate partial information concerning the solution alongthe characteristics, belonging to each family, that intersect the boundary curve C. The solution uis then speciﬁed in the region common to these two families of characteristics. For instance, the characteristics of the hyperbolic one-dimensionalwave equation in the last example are shown in ﬁgure 18.5. By specifying Cauchy 637 PDES: GENERAL AND PARTICULAR SOLUTIONS Equation type Boundary Conditions hyperbolic open Cauchy parabolic open Dirichlet or Neumannelliptic closed Dirichlet or Neumann Table 18.1 The appropriate boundary conditions for diﬀerent types of partialdiﬀerential equation. boundary conditions uand∂u/∂t on the line segment t=0 , x=0t o L,t h e solution is speciﬁed in the shaded region. As in the case of ﬁrst-order PDEs, however, problems can arise. For example, if for a hyperbolic equation the boundary curve intersects any characteristic more than once then Cauchy conditions along Ccan overdetermine the problem, resulting in there being no solution. In this case either the boundary curve C must be altered, or the boundary conditions on the oﬀending parts of Cmust be relaxed to Dirichlet or Neumann conditions. The general considerations involved in deciding which boundary conditions are appropriate for a particular problem are complex, and we do not discuss themany further here. †We merely note that whether the various types of boundary condition are appropriate (in that they give a solution that is unique, sometimesto within a constant, and is well deﬁned) depends upon the type of second-orderequation under consideration and on whether the region of solution is boundedby a closed or an open curve (or a surface if there are more than two independent variables). Note that part of a closed boundary may be at inﬁnity if conditions are imposed on uor∂u/∂n there. It may be shown that the appropriate boundary-condition and equation-type pairings are as given in table 18.1. For example, Laplace’s equation ∇ 2u= 0 is elliptic and thus requires either Dirichlet or Neumann boundary conditions on a closed boundary which, as wehave already noted, may be at inﬁnity if the behaviour of uis speciﬁed there (most often uor∂u/∂n→0 at inﬁnity). 18.7 Uniqueness of solutions Although we have merely stated the appropriate boundary types and conditions for which, in the general case, a PDE has a unique, well-deﬁned solution, some-times to within an additive constant, it is often important to be able to prove that a unique solution is obtained. †For a discussion the reader is referred, for example, to Morse and Feshbach, Methods of Theoretical Physics, Part I (McGraw-Hill, 1953) chapter 6. 638 18.7 UNIQUENESS OF SOLUTIONS As an extremely important example let us consider Poisson’s equation in three dimensions, ∇2u(r)=ρ(r), (18.46) with either Dirichlet or Neumann conditions on a closed boundary appropriate to such an elliptic equation; for brevity, in (18.46), we have absorbed any physicalconstants into ρ. We aim to show that, to within an unimportant constant, the solution of (18.46) is unique if either the potential uor its normal derivative ∂u/∂n is speciﬁed on all surfaces bounding a given region of space (including, if necessary, a hypothetical spherical surface of indeﬁnitely large radius on which u or∂u/∂n is prescribed to have an arbitrarily small value). Stated more formally this is as follows. Uniqueness theorem. Ifuis real and its ﬁrst and second partial derivatives are continuous in a region Vand on its boundary S, and∇ 2u=ρinVand either u=for∂u/∂n =gonS,w h e r e ρ,fandgare prescribed functions, then uis unique (at least to within an additive constant).IProve the uniqueness theorem for Poisson’s equation. Let us suppose on the contrary that two solutions u1(r)a n d u2(r) both satisfy the conditions given above, and denote their diﬀerence by the function w=u1−u2. We then have ∇2w=∇2u1−∇2u2=ρ−ρ=0, so that wsatisﬁes Laplace’s equation in V. Furthermore, since either u1=f=u2or ∂u1/∂n=g=∂u2/∂nonS, we must have either w=0o r ∂w/∂n =0o n S. If we now use Green’s ﬁrst theorem, (11.19), for the case where both scalar functions are taken as wwe haveZ V / w∇2w+(∇w)·(∇w) / dV= Z Sw∂w ∂ndS. However, either condition, w=0o r ∂w/∂n = 0, makes the RHS vanish whilst the ﬁrst term on the LHS vanishes since ∇2w=0i n V. Thus we are left withZ V|∇w|2dV=0. Since|∇w|2can never be negative, this can only be satisﬁed if ∇w=0, i.e. if w, and hence u1−u2, is a constant in V. If Dirichlet conditions are given then u1≡u2on (some part of) Sand hence u1=u2 everywhere in V. For Neumann conditions, however, u1andu2can diﬀer throughout V by an arbitrary (but unimportant) constant. J The importance of this uniqueness theorem lies in the fact that if a solution to Poisson’s (or Laplace’s) equation that ﬁts the given set of Dirichlet or Neumann conditions can be found by any means whatever, then that solution is the correct one, since only one exists. This result is the mathematical justiﬁcation for themethod of images , which is discussed more fully in the next chapter. 639 PDES: GENERAL AND PARTICULAR SOLUTIONS We also note that often the same general method, used in the above example for proving the uniqueness theorem for Poisson’s equation, can be employed toprove the uniqueness (or otherwise) of solutions to other equations and boundaryconditions. 18.8 Exercises 18.1 Determine whether the following can be written as functions of p=x2+2yonly, and hence whether they are solutions of (18.8): (a)x2(x2−4) + 4 y(x2−2) + 4( y2−1); (b)x4+2x2y+y2; (c) [ x4+4x2y+4y2+4 ]/[2x4+x2(8y+1 )+8 y2+2y]. 18.2 Find partial diﬀerential equations satisﬁed by the following functions u(x, y)f o r all arbitrary functions fand all arbitrary constants aandb: (a)u(x, y)=f(x2−y2); (b)u(x, y)=(x−a)2+(y−b)2; (c)u(x, y)=ynf(y/x); (d)u(x, y)=f(x+ay). 18.3 Solve the following partial diﬀerential equations for u(x, y) with the boundary conditions given: (a)x∂u ∂x+xy=u,u=2yon the line x=1 ; (b) 1 + x∂u ∂y=xu,u(x,0) = x. 18.4 Find the most general solutions u(x, y) of the following equations consistent with the boundary conditions stated: (a)y∂u ∂x−x∂u ∂y=0 , u(x,0) = 1 + sin x; (b)i∂u ∂x=3∂u ∂y,u=( 4+3 i)x2on the line x=y; (c) sin xsiny∂u ∂x+c o s xcosy∂u ∂y=0 , u=c o s2 yonx+y=π/2; (d)∂u ∂x+2x∂u ∂y=0 , u= 2 on the parabola y=x2. 18.5 Find solutions of 1 x∂u ∂x+1 y∂u ∂y=0 for which (a) u(0,y)=y,( b ) u(1,1) = 1. 18.6 Find the most general solutions u(x, y) of the following equations consistent with the boundary conditions stated: (a)y∂u ∂x−x∂u ∂y=3x,u=x2on the line y=0 ; 640 18.8 EXERCISES (b)y∂u ∂x−x∂u ∂y=3x,u(1,0) = 2; (c)y2∂u ∂x+x2∂u ∂y=x2y2(x3+y3), no boundary conditions. 18.7 Solve sinx∂u ∂x+c o s x∂u ∂y=c o s x subject to (a) u(π/2,y)=0 ,( b ) u(π/2,y)=y(y+1 ). 18.8 A function u(x, y)s a t i s ﬁ e s 2∂u ∂x+3∂u ∂y=1 0, and takes the value 3 on the line y=4x. Evaluate u(2,4). 18.9 If u(x, y)s a t i s ﬁ e s ∂2u ∂x2−3∂2u ∂x∂y+2∂2u ∂y2=0 andu=−x2and∂u/∂y =0f o r y=0a n da l l x, ﬁnd the value of u(0,1). 18.10 (a) Solve the previous question if the boundary condition is u=∂u/∂y =1 when y=0f o ra l l x. (b) In which region of the xy-plane would ube determined if the boundary condition were u=∂u/∂y = 1 when y=0f o ra l l x>0? 18.11 In those cases in which it is possible to do so, evaluate u(2,2), where u(x, y)i s the solution of 2y∂u ∂x−x∂u ∂y=2xy(2y2−x2) that satisﬁes the (separate) boundary conditions given below. (a)u(x,1) = x2for all x. (b)u(x,1) = x2forx≥0. (c)u(x,1) = x2for 0≤x≤3. (d)u(x,0) = xforx≥0. (e)u(x,0) = xfor all x. (f)u(1,√10) = 5 . (g)u(√10,1) = 5 . 18.12 Solve 6∂2u ∂x2−5∂2u ∂x∂y+∂2u ∂y2=1 4, subject to u=2x+1a n d ∂u/∂y =4−6x, both on the line y=0 . 18.13 By changing the independent variables in the previous question to ξ=x+2y and η=x+3y, show that it must be possible to write 14( x2+5xy+6y2)i nt h ef o r m f1(x+2y)+f2(x+3y)−(x2+y2), and determine the forms of f1(z)a n d f2(z). 18.14 Solve ∂2u ∂x∂y+3∂2u ∂y2=x(2y+3x). 18.15 Find the most general solution of ∂2u/∂x2+∂2u/∂y2=x2y2. 641 PDES: GENERAL AND PARTICULAR SOLUTIONS 18.16 An inﬁnitely long string on which waves travel at speed chas an initial displace- ment y(x)= /( sin(πx/a),−a≤x≤a, 0,|x|>a . It is released from rest at time t= 0, and its subsequent displacement is described byy(x, t). By expressing the initial displacement as one explicit function incorporating Heaviside step functions, ﬁnd an expression for y(x, t)a tag e n e r a lt i m e t>0. In particular, determine the displacement as a function of time (a) at x=0 ,( b )a t x=a,a n d( c )a t x=a/2. 18.17 The non-relativistic Schr ¨odinger equation (18.7) is similar to the diﬀusion equa- tion in having diﬀerent orders of derivatives in its various terms; this precludessolutions that are arbitrary functions of particular linear combinations of vari-ables. However, since exponential functions do not change their forms underdiﬀerentiation, solutions in the form of exponential functions of combinations ofthe variables may still be possible. Consider the Schr ¨odinger equation for the case of a constant potential, i.e. for a free particle, and show that it has solutions of the form Aexp(lx+my+nz+λt) where the only requirement is that −/~2 2m /; l2+m2+n2 / =i /~λ. In particular, identify the equation and wavefunction obtained by taking λas −iE/ /~,a n d l,mand nasipx/ /~,i py/ /~and ipz/ /~respectively, where Eis the energy and pthe momentum of the particle; these identiﬁcations are essentially the content of the de Broglie and Einstein relationships. 18.18 Like the Schr ¨odinger equation of the previous question, the equation describing the transverse vibrations of a rod, a4∂4u ∂x4+∂2u ∂t2=0, has diﬀerent orders of derivatives in its various terms. Show, however, that it has solutions of exponential form u(x, t)=Aexp(λx+iωt) provided that the relation a4λ4=ω2is satisﬁed. Use a linear combination of such allowed solutions, expressed as the sum of sinusoids and hyperbolic sinusoids of λx, to describe the transverse vibrations of a rod of length Lclamped at both ends. At a clamped point both uand∂u/∂x must vanish; show that this implies that cos( λL)cosh( λL) = 1, thus determining the frequencies ωat which the rod can vibrate. 18.19 An incompressible ﬂuid of density ρand negligible viscosity ﬂows with velocity valong a thin straight tube, perfectly light and ﬂexible, of cross-section Aand held under tension T. Assume that small transverse displacements uof the tube are governed by ∂2u ∂t2+2v∂2u ∂x∂t+ / v2−T ρA /∂2u ∂x2=0. (a) Show that the general solution consists of a superposition of two waveforms travelling with diﬀerent speeds. (b) The tube initially has a small transverse displacement u=acoskxand is suddenly released from rest. Find its subsequent motion. 18.20 A sheet of material of thickness w, speciﬁc heat capacity cand thermal con- ductivity kis isolated in a vacuum, but its two sides are exposed to ﬂuxes of 642 18.8 EXERCISES r a d i a n th e a to fs t r e n g t h s J1andJ2. Ignoring short-term transients, show that the temperature diﬀerence between its two surfaces is steady at ( J2−J1)w/2k, whilst their average temperature increases at a rate ( J2+J1)/cw. 18.21 In an electrical cable of resistance Rand capacitance Cper unit length, voltage signals obey the equation ∂2V/∂x2=RC∂V/∂t . This has solutions of the form given in (18.36) and also of the form V=Ax+D. (a) Find a combination of these that re presents the situation after a steady voltage V0is applied at x=0a tt i m e t=0 . (b) Obtain a solution describing the propagation of the voltage signal resulting from application of the signal V=V0for 0 <t<T ,V= 0 otherwise, to the end x= 0 of an inﬁnite cable. (c) Show that for t/greatermuchTthe maximum signal occurs at a value of xproportional tot1/2and has a magnitude proportional to t−1. 18.22 The daily and annual variations of temperature at the surface of the earth may be represented by sine-wave oscillations with equal amplitudes and periods of1 day and 365 days respectively. Assume that for (angular) frequency ωthe temperature at depth xin the earth is given by u(x, t)=Asin(ωt+µx)exp(−λx), where λandµare constants. (a) Use the diﬀusion equation to ﬁnd the values of λandµ. (b) Find the ratio of the depths below the surface at which the amplitudes have dropped to 1 /20 of their surface values. (c) At what time of year is the soil coldest at the greater of these depths, assuming that the smoothed annual variation in temperature at the surfacehas a minimum on February 1st? 18.23 Consider each of the following situations in a qualitative way and determine the equation type, the nature of the boundary curve and the type of boundaryconditions involved. (a) a conducting bar given an initial temperature distribution and then thermally isolated; (b) two long conducting concentric cylinders on each of which the voltage distribution is speciﬁed; (c) two long conducting concentric cylinders on each of which the charge dis- tribution is speciﬁed; (d) a semi-inﬁnite string the end of which is made to move in a prescribed way. 18.24 This example gives a formal demonstration that the type of a second-order PDE (elliptic, parabolic or hyperbolic) cannot be changed by a new choi ce of independent variable. The algebra is somewhat lengthy, but straightforward. If a change of variable ξ=ξ(x, y),η=η(x, y) is made in (18.19), so that it reads A /prime∂2u ∂ξ2+B/prime∂2u ∂ξ∂η+C/prime∂2u ∂η2+D/prime∂u ∂ξ+E/prime∂u ∂η+F/primeu=R/prime(ξ,η), show that B/prime2−4A/primeC/prime=(B2−4AC) /∂(ξ,η) ∂(x, y) /2 . Hence deduce the conclusion stated above. 18.25 The Klein–Gordon equation (which is satisﬁed by the quantum-mechanical wave- function Φ( r) of a relativistic spinless particle of non-zero mass m)i s ∇2Φ−m2Φ=0 . 643 PDES: GENERAL AND PARTICULAR SOLUTIONS Show that the solution for the scalar ﬁeld Φ( r)i na n yv o l u m e Vbounded by as u r f a c e Sis unique if either Dirichlet or Neumann boundary conditions are speciﬁed on S. 18.9 Hints and answers 18.1 (a) Yes, p2−4p−4; (b) no, ( p−y)2;( c )y e s ,( p2+4 )/(2p2+p). 18.2 (a) y(∂u/∂x )+x(∂u/∂y )=0 ;( b )( ∂u/∂x )2+(∂u/∂y )2=4u;( c ) x(∂u/∂x )+ y(∂u/∂y )= nu;( d )( ∂u/∂y )(∂2u/∂x2)=( ∂u/∂x )(∂2u/∂x∂y ), or with xand y reversed. 18.3 Each equation is eﬀectively an ordinary diﬀerential equation but with a function of the non-integrated variable as the constant of integration;(a)u=xy(2−lnx); (b) u=x −1(1−ey)+xey. 18.4 (a) p=x2+y2,u=s i n ( x2+y2)1/2+1 ;( b ) p=3x+iy,u=( 3x+iy)1/2/2; (c)p=s i n xcosy,u=2s i n xcosy−1; (d) p=y−x2,u=y−x2+2 . 18.5 (a) ( y2−x2)1/2;( b )1+ f(y2−x2)w h e r e f(0) = 0. 18.6 (a) p=x2+y2, particular integral u=−3y,u=x2+y2−3y; (b)u=x2+y2−3y+1+ g(x2+y2)w h e r e g(1) = 0; (c) (x6+y6)/6+g(x3−y3). 18.7 u=y+f(y−ln(sin x)); (a) u=l n ( s i n x); (b) u=y+[y−ln(sin x)]2. 18.8 u=f(3x−2y)+2 ( x+y);f(p)=3+2 p;u=8x−2y+3a n d u(2,4) = 11. 18.9 General solution is u(x, y)=f(x+y)+g(x+y/2). Show that 2 p=−g/prime(p)/2, and hence g(p)=k−2p2, whilst f(p)=p2−k, leading to u(x, y)=−x2+y2/2; u(0,1) = 1 /2. 18.10 (a) u(x, y)=2 ( x+y)−2(x+y/2) + 1 = y+1 ; u(0,1) = 2; (b) in the sector −π/4≤θ≤π/2+φ,w h e r et a n φ=1/2a n d θis measured from the positive x-axis. 18.11 p=x2+2y2;u(x, y)=f(p)+x2y2/2. (a)u(x, y)=( x2+2y2+x2y2−2)/2.u(2,2) = 13. The line y= 1 cuts each characteristic in zero or two distinct points, but this causes no diﬃculty withthe given boundary conditions. (b) As in (a). (c) The solution is deﬁned over the space between the ellipses p=2a n d p= 11; (2,2) lies on p= 12, and so u(2,2) is undetermined. (d)u(x, y)=(x 2+2y2)1/2+x2y2/2;u(2,2) = 8 +√12. (e) The line y= 0, cuts each characteristic in two distinct points. No diﬀer- entiable form of f(p)g i v e s f(±a)=±arespectively, and so there is no solution. (f) The solution is only speciﬁed on p= 21, and so u(2,2) is undetermined. (g) The solution is speciﬁed on p= 12, and so u(2,2) = 5 +1 2(4)(4) = 13 . 18.12 u(x, y)=f(x+2y)+g(x+3y)+x2+y2, leading to u=1+2 x+4y−6xy−8y2. 18.13 The equation becomes ∂2f/∂ξ∂η =−14, with solution f(ξ,η)=f(ξ)+g(η)−14ξη, which can be compared with the answer from the previous question; f1(z)=1 0 z2 andf2(z)=5 z2. 18.14 u=f(y−3x)+g(x)+x2y2/2. 18.15 u(x, y)=f(x+iy)+g(x−iy)+( 1 /12)x4(y2−(1/15)x2). In the last term, xandy may be interchanged. 18.16 y(x, t)=1 2sin[π(x−ct)/a][H(x−ct+a)−H(x−ct−a)] +1 2sin[π(x+ct)/a][H(x+ct+a)−H(x+ct−a)]. 644 18.9 HINTS AND ANSWERS (a) zero at all times; (b)1 2sin(πct/a)f o r0≤t≤2a/c, and 0 otherwise; (c) cos( πct/a)f o r0≤t≤a/2c,1 2cos(πct/a)f o r a/2c≤t≤3a/2c,a n d0 otherwise. 18.17 E=p2/(2m), the relationship between energy and momentum for a non- relativistic particle; u(r,t)=Aexp[i(p·r−Et)/ /~], a plane wave of wave number k=p/ /~and angular frequency ω=E/ /~travelling in the direction p/p. 18.18 λ=±ω1/2/aor±iω1/2/a;u(x, t)=e x p ( iωt)[Asinλx+Bcosλx+Csinhλx+ Dcoshλx], with C=−AandD=−B. The conditions at x=Land consistency establish the quoted result. 18.19 (a) c=v±αwhere α2=T/ρA ; (b)u(x, t)=acos[k(x−vt)] cos( kαt)−(va/α)sin[k(x−vt)] sin( kαt). 18.20 Use the ﬁrst form of solution given in (18.35). 18.21 (a) V0 / 1−(2/√π) R1 2x(CR/t)1/2exp(−ν2)dν / ; (b) consider as V0applied at t=0 and continued and −V0att=Tand continued; V(x, t)=2V0√π Z1 2x[CR/(t−T)]1/2 1 2x(CR/t)1/2exp /; −ν2 / dν; (c) For t/greatermuchT, maximum at x=[ 2t/(CR)]1/2with value V0Texp(−1/2) (2π)1/2t. 18.22 (a) λ=−µ=[ω/(2κ)]1/2,w h e r e κis the diﬀusion constant; (b) xa= (365)1/2xd; (c) only the annual variation is signiﬁcant at this depth and has a phase µaxa= ln 20 behind the surface. Thus the coldest day is 1 February + (365ln 20) /(2π) days≈23 July. 18.23 (a) Parabolic, open, Dirichlet u(x,0) given, Neumann ∂u/∂x =0a t x=±L/2 for all t; (b) elliptic, closed, Dirichlet;(c) elliptic, closed, Neumann ∂u/∂n =σ//epsilon1 0; (d) hyperbolic, open, Cauchy. 18.24 A/prime=A /∂ξ ∂x /2 +B∂ξ ∂x∂ξ ∂y+C /∂ξ ∂y /2 , B/prime=2A∂ξ ∂x∂η ∂x+B /∂ξ ∂x∂η ∂y+∂ξ ∂y∂η ∂x / +2C∂ξ ∂y∂η ∂y,etc. 18.25 Follow an argument similar to that in section 18.7 and argue that the additional term R m2|w|2dVmust be zero, and hence that w= 0 everywhere. 645 19 Partial diﬀerential equations: separation of variables and other methods In the previous chapter we demonstrated the methods by which general solutions of some partial diﬀerential equations (PDEs) may be obtained in terms ofarbitrary functions. In particular, solutions containing the independent variablesin deﬁnite combinations were sought, thus reducing the eﬀective number of them. In the present chapter we begin by taking the opposite approach, namely that of trying to keep the independent variables as separate as possible, using the method of separation of variables. We then consider integral transform methods by which one of the independent variables may be eliminated, at least fromdiﬀerential coeﬃcients. Finally, we discuss the use of Green’s functions in solvinginhomogeneous problems. 19.1 Separation of variables: the general method Suppose we seek a solution u(x, y, z, t ) to some PDE (expressed in Cartesian coordinates). Let us attempt to obtain one that has the product form † u(x, y, z, t )=X(x)Y(y)Z(z)T(t). (19.1) A solution that has this form is said to be separable inx,y,zandt, and seeking solutions of this form is called the method of separation of variables . As simple examples we may observe that, of the functions (i)xyz 2sinbt, (ii)xy+zt, (iii) ( x2+y2)zcosωt, (i) is completely separable, (ii) is inseparable in that no single variable can be separated out from it and written as a multiplicative factor, whilst (iii) is separableinzandtbut not in xandy. †It should be noted that the conventional use here of upper-case (capital) letters to denote the functions of the corresponding lower-case variable is intended to enable an easy correspondence between a function and its argument to be made. 646 19.1 SEPARATION OF VARIABLES: THE GENERAL METHOD When seeking PDE solutions of the form (19.1), we are requiring not that there is no connection at all between the functions X,Y,ZandT(for example, certain parameters may appear in two or more of them), but only that the Xdoes not depend upon y,z,t,t h a t Ydoes not depend on x,z,ta n ds oo n . For a general PDE it is likely that a separable solution is impossible, but certainly some common and important equations do have useful solutions ofthis form and we will illustrate the method of solution by studying the three-dimensional wave equation ∇ 2u(r)=1 c2∂2u(r) ∂t2. (19.2) We will work in Cartesian coordinates for the present and assume a solution of the form (19.1); the solutions in alternative coordinate systems, e.g. sphericalor cylindrical polars, are considered in section 19.3. Expressed in Cartesiancoordinates (19.2) takes the form ∂ 2u ∂x2+∂2u ∂y2+∂2u ∂z2=1 c2∂2u ∂t2; (19.3) substituting (19.1) gives d2X dx2YZT +Xd2Y dy2ZT+XYd2Z dz2T=1 c2XY Zd2T dt2, which can also be written as X/prime/primeYZT +XY/prime/primeZT+XY Z/prime/primeT=1 c2XY ZT/prime/prime, (19.4) w h e r ei ne a c hc a s et h ep r i m e sr e f e rt ot h e ordinary derivative with respect to the independent variable upon which the function depends. This emphasises the factthat each of the functions X,Y,ZandThas only one independent variable and thus its only derivative is its total derivative. For the same reason, in each termin (19.4) three of the four functions are unaltered by the partial diﬀerentiation and behave exactly as constant multipliers. If we now divide (19.4) throughout by u=XY ZT we obtain X /prime/prime X+Y/prime/prime Y+Z/prime/prime Z=1 c2T/prime/prime T. (19.5) This form shows the particular characteristic that is the basis of the method of separation of variables, namely that of the four terms the ﬁrst is a function of x only, the second of yonly, the third of zonly and the RHS a function of tonly and yet there is an equation connecting them. This can only be so for all x,y,z andtifeachof the terms does not in fact, despite appearances, depend upon the corresponding independent variable but is equal to a constant , the four constants being such that (19.5) is satisﬁed. Since there is only one equation to be satisﬁed and four constants involved, 647 PDES: SEPARATION OF VARIABLES AND OTHER METHODS there is considerable freedom in the values they may take. For the purposes of our illustrative example let us make the choice of −l2,−m2,−n2,f o rt h eﬁ r s t three constants. The constant associated with c−2T/prime/prime/Tmust then have the value −µ2=−(l2+m2+n2). Having recognised that each term of (19.5) is individually equal to a constant (or parameter), we can now replace (19.5) by four separate ordinary diﬀerentialequations (ODEs), X /prime/prime X=−l2,Y/prime/prime Y=−m2,Z/prime/prime Z=−n2,1 c2T/prime/prime T=−µ2. (19.6) The important point to notice is not the simplicity of the equations (19.6) (the corresponding ones for a general PDE are usually far from simple) but that, bythe device of assuming a separable solution, a partial diﬀerential equation (19.3), containing derivatives with respect to the four independent variables all in oneequation, has been reduced to four separate ordinary diﬀerential equations (19.6). The ordinary equations are connected through four constant parameters that satisfy an algebraic relation. These constants are called separation constants . The general solutions of the equations (19.6) can be deduced straightforwardly and are X(x)=Aexp(ilx)+Bexp(−ilx) Y(y)=Cexp(imy)+Dexp(−imy) Z(z)=Eexp(inz)+Fexp(−inz) T(t)=Gexp(icµt)+Hexp(−icµt),(19.7) where A , B,...,H are constants, which may be determined if boundary condtions are imposed on the solution. Depending on the geometry of the problem andany boundary conditions, it is sometimes more appropriate to write the solutions (19.7) in the alternative form X(x)=A /primecoslx+B/primesinlx Y(y)=C/primecosmy+D/primesinmy Z(z)=E/primecosnz+F/primesinnz T(t)=G/primecos(cµt)+H/primesin(cµt),(19.8) for some diﬀerent set of constants A/prime,B/prime,...,H/prime. Clearly the choice of how best to represent the solution depends on the problem being considered. As an example, suppose that we take as particular solutions the four functions X(x)=e x p ( ilx),Y (y)=e x p ( imy), Z(z)=e x p ( inz),T (t)=e x p (−icµt). 648 19.1 SEPARATION OF VARIABLES: THE GENERAL METHOD This gives a particular solution of the original PDE (19.3) u(x, y, z, t )=e x p ( ilx)ex p ( imy)ex p( inz)ex p (−icµt) =e x p [ i(lx+my+nz−cµt)], which is a special case of the solution (18.33) obtained in the previous chapter and represents a plane wave of unit amplitude propagating in a direction givenby the vector with components l,m,n in a Cartesian coordinate system. In the conventional notation of wave theory, l,mandnare the components of the wave-number vector k, whose magnitude is given by k=2π/λ,w h e r e λis the wavelength of the wave; cµis the angular frequency ωof the wave. This gives the equation in the form u(x, y, z, t )=e x p [ i(k xx+kyy+kzz−ωt)] =e x p [ i(k·r−ωt)], and makes the exponent dimensionless. The method of separation of variables can be applied to many commonly occurring PDEs encountered in physical applications.IUse the method of separation of variables to obtain for the one-dimensional diﬀusion equation κ∂2u ∂x2=∂u ∂t, (19.9) a solution that tends to zero as t→∞for all x. Here we have only two independent variables xandtand we therefore assume a solution of the form u(x, t)=X(x)T(t). Substituting this expression into (19.9) and dividing through by u=XT(and also by κ) we obtain X/prime/prime X=T/prime κT. Now, arguing exactly as above that the LHS is a function of xonly and the RHS is a function of tonly, we conclude that each side must equal a constant, which, anticipating the result and noting the imposed boundary condition, we will take as −λ2. This gives us two ordinary equations, X/prime/prime+λ2X=0, (19.10) T/prime+λ2κT=0, (19.11) which have the solutions X(x)=Acosλx+Bsinλx, T(t)=Cexp(−λ2κt). Combining these to give the assumed solution u=XTyields (absorbing the constant C intoAandB) u(x, t)=(Acosλx+Bsinλx)exp(−λ2κt). (19.12) 649 PDES: SEPARATION OF VARIABLES AND OTHER METHODS In order to satisfy the boundary condition u→0a s t→∞,λ2κmust be >0. Since κ is real and >0, this implies that λis a real non-zero number and that the solution is sinusoidal in xand is not a disguised hyperbolic function; this was our reason for choosing the separation constant as −λ2. J As a ﬁnal example we consider Laplace’s equation in Cartesian coordinates; this may be treated in a similar manner.IUse the method of separation of variables to obtain a solution for the two-dimensional Laplace equation, ∂2u ∂x2+∂2u ∂y2=0. (19.13) If we assume a solution of the form u(x, y)=X(x)Y(y) then, following the above method, and taking the separation constant as λ2, we ﬁnd X/prime/prime=λ2X, Y/prime/prime=−λ2Y. Taking λ2as>0, the general solution becomes u(x, y)=(Acoshλx+Bsinhλx)(Ccosλy+Dsinλy), (19.14) An alternative form, in which the exponentials are written explicitly, may be useful for other geometries or boundary conditions: u(x, y)=[Aexpλx+Bexp(−λx)](Ccosλy+Dsinλy), (19.15) with diﬀerent constants AandB. Ifλ2<0 then the roles of xandyinterchange. The particular combination of sinusoidal and hyperbolic functions and the values of λallowed will be determined by the geometrical properties of any speciﬁc problem, together with any prescribed or necessary boundaryconditions.J We note here that a particular case of the solution (19.14) links up with the ‘combination’ result u(x, y)=f(x+iy) of the previous chapter (equations (18.24) and following), namely that if A=B,a n d D=iCthen the solution is the same asf(p)=ACexpλpwith p=x+iy. 19.2 Superposition of separated solutions It will be noticed in the previous two examples that there is considerable freedom in the values of the separation constant λ, the only essential requirement being that λhas the samevalue in both parts of the solution, i.e. the part depending onxand the part depending on y(ort). This is a general feature for solutions in separated form, which, if the original PDE has nindependent variables, will contain n−1 separation constants. All that is required in general is that we associate the correct function of one independent variable with the appropriatefunctions of the others, the correct function being the one with the same values of the separation constants. If the original PDE is linear (as are the Laplace, Schr ¨odinger, diﬀusion and wave equations) then mathematically acceptable solutions can be formed by 650 19.2 SUPERPOSITION OF SEPARATED SOLUTIONS superposing solutions corresponding to diﬀerent allowed values of the separation constants. To take a two-variable example: if uλ1(x, y)=Xλ1(x)Yλ1(y) is a solution of a linear PDE obtained by giving the separation constant the value λ1then the superposition u(x, y)=a1Xλ1(x)Yλ1(y)+a2Xλ2(x)Yλ2(y)+···=summationdisplay iaiXλi(x)Yλi(y), (19.16) is also a solution for any constants ai, provided that the λiare the allowed values of the separation constant λgiven the imposed boundary conditions. Note that if the boundary conditions allow any of the separation constants to be zero then the form of the general solution is normally diﬀerent and must be deduced byreturning to the separated ordinary diﬀerential equations. We will encounter thisbehaviour in section 19.3. The value of the superposition approach is that a boundary condition, say that u(x, y) takes a particular form f(x)w h e n y= 0, might be met by choosing the constants a isuch that f(x)=summationdisplay iaiXλi(x)Yλi(0). In general, this will be possible provided that the functions Xλi(x) form a complete set – as do the sinusoidal functions of Fourier series or the spherical harmonicsthat we shall discuss in subsection 19.3.2.IA semi-inﬁnite rectangular metal plate occupies the region 0≤x≤∞and0≤y≤bin thexy-plane. The temperature at the far end of the plate and along its two long sides is ﬁxed at 0◦C. If the temperature of the plate at x=0is also ﬁxed and is given by f(y),ﬁ n d the steady-state temperature distribution u(x,y) of the plate. Hence ﬁnd the temperaturedistribution if f(y)=u 0,w h e r e u0is a constant. The physical situation is illustrated in ﬁgure 19.1. With the notation we have used several times before, the two-dimensional heat diﬀu sion equation satisﬁed by the temperature u(x, y, t)i s κ /∂2u ∂x2+∂2u ∂y2 / =∂u ∂t, with κ=k/(sρ). In this case, however, we are asked to ﬁnd the steady-state temperature, which corresponds to ∂u/∂t = 0, and so we are led to consider the (two-dimensional) Laplace equation ∂2u ∂x2+∂2u ∂y2=0. We saw that assuming a separable solution of the form u(x, y)=X(x)Y(y)l e dt o solutions such as (19.14) or (19.15), or equivalent forms with xandyinterchanged. In the current problem we have to satisfy the boundary conditions u(x,0) = 0 = u(x, b)a n d so a solution that is sinusoidal in yseems appropriate. Furthermore, since we require u(∞,y) = 0 it is best to write the x-dependence of the solution explicitly in terms of 651 PDES: SEPARATION OF VARIABLES AND OTHER METHODS y b 0 xu=0 u=0u→0 u=f(y) Figure 19.1 A semi-inﬁnite metal plate whose edges are kept at ﬁxed tem- peratures. exponentials rather than of hyperbolic functions. We therefore write the separable solution in the form (19.15) as u(x, y)=[Aexpλx+Bexp(−λx)](Ccosλy+Dsinλy). Applying the boundary conditions, we see ﬁrstly that u(∞,y) = 0 implies A=0i fw e take λ>0. Secondly, since u(x,0) = 0 we may set C= 0, which, if we absorb the constant DintoB, leaves us with u(x, y)=Bexp(−λx)sinλy. But, using the condition u(x, b) = 0, we require sin λb= 0 and so the constant λis constrained to equal nπ/b,w h e r e nis any positive integer. Using the principle of superposition (19.16), the general solution satisfying the given boundary conditions can therefore be written u(x, y)=∞X n=1Bnexp(−nπx/b )sin(nπy/b ), (19.17) for some constants Bn. Notice that in the sum in (19.17) we have omitted negative values of nsince they would lead to exponential terms that diverge as x→∞.T h e n= 0 term is also omitted since it is identically zero. Using the remaining boundary condition u(0,y)=f(y) we see that the constants Bnmust satisfy f(y)=∞X n=1Bnsin(nπy/b ). (19.18) This is clearly a Fourier sine series expansion of f(y) (see chapter 12). For (19.18) to hold, however, the continuation of f(y) outside the region 0 ≤y≤bmust be an odd periodic function with period 2 b(see ﬁgure 19.2). We also see from ﬁgure 19.2 that if the original function f(y) does not equal zero at either of y=0a n d y=bthen its continuation has a discontinuity at the corresponding point(s); nevertheless, as discussedin chapter 12, the Fourier series will converge to the mid-points of these jumps and hencetend to zero in this case. If, however, the top and bottom edges of the plate were held notat 0 ◦C but at some other non-zero temperature, then, in general, the ﬁnal solution would possess discontinuities at the corners x=0 , y=0a n d x=0 , y=b. Bearing in mind these technicalities, the coeﬃcients Bnin (19.18) are given by Bn=2 b Zb 0f(y)sin /nπy b / dy. (19.19) 652 19.2 SUPERPOSITION OF SEPARATED SOLUTIONS f(y) −b 0 b y Figure 19.2 The continuation of f(y) for a Fourier sine series. Therefore, if f(y)=u0(i.e. the temperature of the side at x= 0 is constant along its length), (19.19) becomes Bn=2 b Zb 0u0sin /nπy b / dy = / −2u0 bb nπcos /nπy b / /b 0 =−2u0 nπ[(−1)n−1] = / 4u0/nπ fornodd 0f o r neven. Therefore the required solution is u(x, y)= X nodd4u0 nπexp / −nπx b / sin /nπy b / . J In the above example the boundary conditions meant that one term in each part of the separable solution could be immediately discarded, making the prob- lem much easier to solve. Sometimes, however, a little ingenuity is required in writing the separable solution in such a way that certain parts can be neglectedimmediately.ISuppose that the semi-inﬁnite rectangular metal plate in the previous example is replaced by one that in the x-direction has ﬁnite length a. The temperature of the right-hand edge is ﬁxed at 0◦Cand all other boundary conditions remain as before. Find the steady-state temperature in the plate. As in the previous example, the boundary conditions u(x,0) = 0 = u(x, b) suggest a solution that is sinusoidal in y. In this case, however, we require u=0o n x=a(rather than at inﬁnity) and so a solution in which the x-dependence is written in terms of hyperbolic functions, such as (19.14), rather than exponentials is more appropriate. Moreover, sincethe constants in front of the hyperbolic functions are, at this stage, arbitrary, we maywrite the separable solution in the most convenient way that ensures that the conditionu(a, y) = 0 is straightforwardly satisﬁed. We therefore write u(x, y)=[Acoshλ(a−x)+Bsinhλ(a−x)](Ccosλy+Dsinλy). Now the condition u(a, y) = 0 is easily satisﬁed by setting A=0 .A sb e f o r et h e conditions u(x,0) = 0 = u(x, b)i m p l y C=0a n d λ=nπ/bfor integer n. Superposing the 653 PDES: SEPARATION OF VARIABLES AND OTHER METHODS solutions for diﬀerent nwe then obtain u(x, y)=∞X n=1Bnsinh[ nπ(a−x)/b] sin( nπy/b ), (19.20) for some constants Bn. We have omitted negative values of nin the sum (19.20) since the relevant terms are already included in those obtained for positive n. Again the n=0t e r m is identically zero. Using the ﬁnal boundary condition u(0,y)=f(y) as above we ﬁnd that the constants Bnmust satisfy f(y)=∞X n=1Bnsinh(nπa/b )sin(nπy/b ), and, remembering the caveats discussed in the previous example, the Bnare therefore given by Bn=2 bsinh(nπa/b ) Zb 0f(y) sin(nπy/b )dy. (19.21) For the case where f(y)=u0, following the working of the previous example gives (19.21) as Bn=4u0 nπsinh(nπa/b )fornodd,B n=0 f o r neven. (19.22) The required solution is thus u(x, y)= X nodd4u0 nπsinh(nπa/b )sinh[ nπ(a−x)/b]sin /; nπy/b / . We note that, as required, in the limit a→∞ this solution tends to the solution of the previous example. J Often the principle of superposition can be used to write the solution to problems with more complicated boundary conditions as the sum of solutions toproblems that each satisfy only some part of the boundary condition but when added togther satisfy all the conditions.IFind the steady-state temperature in the (ﬁnite) rectangular plate of the previous example, subject to the boundary conditions u(x, b)=0,u(a, y)=0 andu(0,y)=f(y)as before, but now in addition u(x,0) = g(x). Figure 19.3( c) shows the imposed boundary conditions for the metal plate. Although we could ﬁnd a solution to this problem using the methods presented above, we can arrive at the answer almost immediately by using the principle of superposition and the result of the previous example. Let us suppose the required solution u(x, y) is made up of two parts: u(x, y)=v(x, y)+w(x, y), where v(x, y) is the solution satisfying the boundary conditions shown in ﬁgure 19.3( a), 654 19.2 SUPERPOSITION OF SEPARATED SOLUTIONS yy y b bb f(y)f(y) 0 00 00 0 00 aa a xx x g(x)g(x) (a)( b) (c) Figure 19.3 Superposition of boundary conditions for a metal plate. whilst w(x, y) is the solution satisfying the boundary conditions in ﬁgure 19.3( b). It is clear thatv(x, y) is simply given by the solution to the previous example, v(x, y)= X noddBnsinh /nπ(a−x) b / sin /nπy b / , where Bnis given by (19.21). Moreover, by symmetry, w(x, y)m u s tb eo ft h es a m ef o r ma s v(x, y) but with xandainterchanged with yandbrespectively, and with f(y) in (19.21) replaced by g(x). Therefore the required solution can be written down immediately without further calculation as u(x, y)= X noddBnsinh /nπ(a−x) b / sin /nπy b / + X noddCnsinh /nπ(b−y) a / sin /nπx a / , theBnbeing given by (19.21) and Cnby Cn=2 asinh(nπb/a ) Za 0g(x)sin(nπx/a )dx. Clearly, this method may be extended to cases in which three or four sides of the plate have non-zero boundary conditions. J As a ﬁnal example of the usefulness of the principle of superposition we now consider a problem that illustrates how to deal with inhomogeneous boundaryconditions by a suitable change of variables. 655 PDES: SEPARATION OF VARIABLES AND OTHER METHODSIA bar of length Lis initially at a temperature of 0◦C. One end of the bar (x=0 )is held at0◦Cand the other is supplied with heat at a constant rate per unit area of H.F i n dt h e temperature distribution within the bar after a time t. With our usual notation, the heat diﬀusion equation satisﬁed by the temperature u(x, t)i s κ∂2u ∂x2=∂u ∂t, with κ=k/(sρ), where kis the thermal conductivity of the bar, sis its speciﬁc heat capacity and ρits density. The boundary conditions can be written as u(x,0) = 0 ,u (0,t)=0 ,∂u(L, t) ∂x=H k, the last of which is inhomogeneous. In general, inhomogeneous boundary conditions can cause diﬃculties and it is usual to attempt a transformation of the problem into anequivalent homogeneous one. To this end, let us assume that the solution to our problemtakes the form u(x, t)=v(x, t)+w(x), where the function w(x) is to be suitably determined. In terms of vandwthe problem becomes κ/∂2v ∂x2+d2w dx2 / =∂v ∂t, v(x,0) +w(x)=0 , v(0,t)+w(0) = 0 , ∂v(L, t) ∂x+dw(L) dx=H k. There are several ways of choosing w(x) so as to make the new problem straightforward. Using some physical insight, however, it is clear that ultimately (at t=∞), when all transients have died away, the end x=Lwill attain a temperature u0such that ku0/L=H and there will be a constant temperature gradient u(x,∞)=u0x/L. We therefore choose w(x)=Hx k. Since the second derivative of w(x) is zero, vsatisﬁes the diﬀusion equation and the boundary conditions on vare now v(x,0) =−Hx k,v (0,t)=0 ,∂v(L, t) ∂x=0, which are homogeneous in x. From (19.12) a separated solution for the one-dimensional diﬀusion equation is v(x, t)=(Acosλx+Bsinλx)exp(−λ2κt), corresponding to a separation constant −λ2.I fw er e s t r i c t λto be real then all these solutions are transient ones decaying to zero as t→∞. These are just what is needed for adding to w(x) to give the correct solution as t→∞. In order to satisfy v(0,t)=0 , however, we require A= 0. Furthermore, since ∂v ∂x=Bexp(−λ2κt)λcosλx, 656 19.2 SUPERPOSITION OF SEPARATED SOLUTIONS L −Lf(x) x 0 −HL/k Figure 19.4 The appropriate continuation for a Fourier series containing only sine terms. in order to satisfy ∂v(L, t)/∂x=0w er e q u i r ec o s λL=0 ,a n ds o λis restricted to take the values λ=nπ 2L, where nis an odd non-negative integer, i.e. n=1,3,5,.... Thus, to satisfy the boundary condition v(x,0) =−Hx/k, we must haveX noddBnsin /nπx 2L / =−Hx k, in the range x=0t o x=L. In this case we must be more careful about the continuation of the function −Hx/k for which the Fourier sine series is needed. We want a series that is odd in x(sine terms only) and continuous as x=0a n d x=L(no discontinuities, since the series must converge at the end-points). This leads to a continuation of the functionas shown in ﬁgure 19.4, with a period of L /prime=4L. Following the discussion of section 12.3, since this continuation is odd about x= 0 and even about x=L/prime/4=Lit can indeed be expressed as a Fourier sine series containing only odd-numbered terms. The corresponding Fourier series coeﬃcients are found to be Bn=−8HL kπ2(−1)(n−1)/2 n2fornodd, and thus the ﬁnal formula for u(x, t)i s u(x, t)=Hx k−8HL kπ2 X nodd(−1)(n−1)/2 n2sin /nπx 2L / exp / −kn2π2t 4L2sρ / , giving the temperature for all positions 0 ≤x≤Land for all times t≥0. J We note that in all the above examples the boundary conditions restricted the separation constant(s) to an inﬁnite number of discrete values, usually integers. If, however, the boundary conditions allow the separation constant(s) λto take acontinuum of values then the summation in (19.16) is replaced by an integral over λ. This is discussed further in connection with integral transform methods in section 19.4. 657 PDES: SEPARATION OF VARIABLES AND OTHER METHODS 19.3 Separation of variables in polar coordinates So far we have considered the solution of PDEs only in Cartesian coordinates, but many systems in two and three dimensions are more naturally expressed in some form of polar coordinates, in which full advantage can be taken ofany inherent symmetries. For example, the potential associated with an isolatedpoint charge has a very simple expression, q/(4π/epsilon1 0r), when polar coordinates are used, but involves all three coordinates and square roots, when Cartesians areemployed. For these reasons we now turn to the separation of variables in planepolar, cylindrical polar and spherical polar coordinates. Most of the PDEs we have considered so far have involved the operator ∇ 2,e . g . the wave equation, the diﬀusion equation, Schr ¨odinger’s equation and Poisson’s equation (and of course Laplace’s equation). It is therefore appropriate that werecall the expressions for ∇ 2when expressed in polar coordinate systems. From chapter 10, in plane polars, cylindrical polars and spherical polars respectivelywe have ∇ 2=1 ρ∂ ∂ρparenleftbigg ρ∂ ∂ρparenrightbigg +1 ρ2∂2 ∂φ2, (19.23) ∇2=1 ρ∂ ∂ρparenleftbigg ρ∂ ∂ρparenrightbigg +1 ρ2∂2 ∂φ2+∂2 ∂z2, (19.24) ∇2=1 r2∂ ∂rparenleftbigg r2∂ ∂rparenrightbigg +1 r2sinθ∂ ∂θparenleftbigg sinθ∂ ∂θparenrightbigg +1 r2sin2θ∂2 ∂φ2. (19.25) Of course the ﬁrst of these may be obtained from the second by taking zto be identically zero. 19.3.1 Laplace’s equation in polar coordinates The simplest of the equations containing ∇2is Laplace’s equation, ∇2u(r)=0 . (19.26) Since it contains most of the essential features of the other more complicated equations we will consider its solution ﬁrst. Laplace’s equation in plane polars Suppose that we need to ﬁnd a solution of (19.26) that has a prescribed behaviour on the circle ρ=a(e.g. if we are ﬁnding the shape taken up by a circular drumskin when its rim is slightly deformed from being planar). Then we may seek solutions of (19.26) that are separable in ρandφ(measured from some arbitrary radius asφ= 0) and hope to accommodate the boundary condition by examining the solution for ρ=a. 658 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES Thus, writing u(ρ, φ)=P(ρ)Φ(φ) and using the expression (19.23), Laplace’s equation (19.26) becomes Φ ρ∂ ∂ρparenleftbigg ρ∂P ∂ρparenrightbigg +P ρ2∂2Φ ∂φ2=0. Now, employing the same device as previously, that of dividing through by u=PΦ and multiplying through by ρ2, results in the separated equation ρ P∂ ∂ρparenleftbigg ρ∂P ∂ρparenrightbigg +1 Φ∂2Φ ∂φ2=0. Following our earlier argument, since the ﬁrst term on the RHS is a function of ρonly, whilst the second term depends only on φ, we obtain the two ordinary equations ρ Pd dρparenleftbigg ρdP dρparenrightbigg =n2(19.27) 1 Φd2Φ dφ2=−n2, (19.28) where we have taken the separation constant to have the form n2for later convenience; for the present nis a general (complex) number. Let us ﬁrst consider the case in which n/negationslash= 0. The second equation, (19.28), then has the general solution Φ(φ)=Aexp(inφ)+Bexp(−inφ). (19.29) Equation (19.27), on the other hand, is the homogeneous equation ρ2P/prime/prime+ρP/prime−n2P=0, which must be solved either by trying a power solution in ρor by making the substitution ρ=e x p tas described in subsection 15.2.1 and so reducing it to an equation with constant coeﬃcients. Carrying out this procedure we ﬁnd P(ρ)=Cρn+Dρ−n. (19.30) Returning to the solution (19.29) of the azimuthal equation (19.28), we can see that if Φ, and hence u, is to be single-valued and so not change when φ increases by 2 πthen nmust be an integer. Mathematically, other values of nare permissible, but for the description of real physical situations it is clear that thislimitation must be imposed. Having thus restricted the possible values of nin one part of the solution, the same limitations must be carried over into the radial part(19.30). Thus we may write a particular solution of the two-dimensional Laplace equation as u(ρ, φ)=(Acosnφ+Bsinnφ)(Cρ n+Dρ−n), where A, B, C, D are arbitrary constants and nis any integer. 659 PDES: SEPARATION OF VARIABLES AND OTHER METHODS We have not yet, however, considered the solution when n=0 .I nt h i sc a s e , the solutions of the separated ordinary equations (19.28) and (19.27) respectivelyare easily shown to be Φ(φ)=Aφ+B, P(ρ)=Clnρ+D. But, in order that u=PΦ is single-valued, we require A=0a n ds ot h es o l u t i o n forn= 0 is simply (absorbing BintoCandD) u(ρ, φ)=Clnρ+D. Superposing the solutions for the diﬀerent allowed values of n, we can write the general solution to Laplace’s equation in plane polars as u(ρ, φ)=(C 0lnρ+D0)+∞summationdisplay n=1(Ancosnφ+Bnsinnφ)(Cnρn+Dnρ−n), (19.31) where ncan take only integer values. Negative values of nhave been omitted from the sum since they are already included in the terms obtained for positive n. We note that, since ln ρis singular at ρ= 0, whenever we solve Laplace’s equation in a region containing the origin, C0must be identically zero.IA circular drumskin has a supporting rim at ρ=a. If the rim is twisted so that it is displaced vertically by a small amount /epsilon1(sinφ+2 s i n2 φ),w h e r e φis the azimuthal angle with respect to a given radius , ﬁnd the resulting displacement u(ρ, φ)over the entire drumskin. The transverse displacement of a circular drumskin is usually described by the two-dimensional wave equation. In this case, however, there is no time dependence and sou(ρ, φ) solves the two-dimensional Laplace equation, subject to the imposed boundary condition. Referring to (19.31), since we wish to ﬁnd a solution that is ﬁnite everywhere inside ρ=a,w er e q u i r e C 0=0a n d Dn=0f o ra l l n>0. Now the boundary condition at the rim requires u(a, φ)=D0+∞X n=1Cnan(Ancosnφ+Bnsinnφ)=/epsilon1(sinφ+2s i n2 φ). Firstly we see that we require D0=0a n d An=0f o ra l l n. Furthermore, we must have C1B1a=/epsilon1,C2B2a2=2/epsilon1andBn=0f o r n>2. Hence the appropriate shape for the drumskin (valid over the whole skin, not just the rim) is u(ρ, φ)=/epsilon1ρ asinφ+2/epsilon1ρ2 a2sin2φ=/epsilon1ρ a / sinφ+2ρ asin2φ / . J 660 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES Laplace’s equation in cylindrical polars Passing to three dimensions, we now consider the solution of Laplace’s equation in cylindrical polar coordinates, 1 ρ∂ ∂ρparenleftbigg ρ∂u ∂ρparenrightbigg +1 ρ2∂2u ∂φ2+∂2u ∂z2=0. (19.32) We note here that, even when considering a cylindrical physical system, if there is no dependence of the physical variables on z(i.e. along the length of the cylinder) then the problem may be treated using two-dimensional plane polars, as discussed above. For the more general case, however, we proceed as previously by trying a solution of the form u(ρ, φ, z)=P(ρ)Φ(φ)Z(z), which on substitution into (19.32) and division through by u=PΦZgives 1 Pρd dρparenleftbigg ρdP dρparenrightbigg +1 Φρ2d2Φ dφ2+1 Zd2Z dz2=0. The last term depends only on zand the ﬁrst and second (taken together) only onρandφ. Taking the separation constant to be k2, we ﬁnd 1 Zd2Z dz2=k2, 1 Pρd dρparenleftbigg ρdP dρparenrightbigg +1 Φρ2d2Φ dφ2+k2=0. The ﬁrst of these equations has the straightforward solution Z(z)=Eexp(−kz)+Fexpkz. Multiplying the second equation through by ρ2,w eo b t a i n ρ Pd dρparenleftbigg ρdP dρparenrightbigg +1 Φd2Φ dφ2+k2ρ2=0, in which the second term depends only on Φ and the other terms only on ρ. Taking the second separation constant to be m2, we ﬁnd 1 Φd2Φ dφ2=−m2, (19.33) ρd dρparenleftbigg ρdP dρparenrightbigg +(k2ρ2−m2)P=0. (19.34) The equation in the azimuthal angle φhas the very familiar solution Φ(φ)=Ccosmφ+Dsinmφ. 661 PDES: SEPARATION OF VARIABLES AND OTHER METHODS As in the two-dimensional case, single-valuedness of urequires that mis an integer. However, in the particular case m= 0 the solution is Φ(φ)=Cφ+D. This form is appropriate to a solution with axial symmetry ( C=0 )o ro n et h a ti s multivalued, but manageably so, such as the magnetic scalar potential associatedwith a current I(in which case C=I/(2π)a n d Dis arbitrary). Finally the ρ-equation (19.34) may be transformed into Bessel’s equation of order mby writing µ=kρ. This has the solution P(ρ)=AJ m(kρ)+BYm(kρ). The properties of these functions were investigated in chapter 16 and will not be pursued here. We merely note that Ym(kρ) is singular at ρ=0 ,a n ds ow h e n seeking solutions to Laplace’s equation in cylindrical coordinates within someregion containing the ρ= 0 axis, we require B=0 . The complete separated-variable solution in cylindrical polars of Laplace’s equation∇ 2u= 0 is thus u(ρ, φ, z)=[AJm(kρ)+BYm(kρ)][Ccosmφ+Dsinmφ][Eexp(−kz)+Fexpkz]. (19.35) Of course we may use the principle of superposition to build up more general solutions by adding together solutions of the form (19.35) for all allowed valuesof the separation constants kandm.IA semi-inﬁnite solid cylinder of radius ahas its curved surface held at 0◦Cand its base held at a temperature T0. Find the steady-state temperature distribution in the cylinder. The physical situation is shown in ﬁgure 19.5. The steady-state temperature distribution u(ρ, φ, z) must satisfy Laplace’s equation subject to the imposed boundary conditions. Let us take the cylinder to have its base in the z= 0 plane and to extend along the positive z-axis. From (19.35), in order that uis ﬁnite everywhere in the cylinder we immediately require B=0a n d F= 0. Furthermore, since the boundary conditions, and hence the temperature distribution, are axially symmetric we require m= 0, and so the general solution must be a superposition of solutions of the form J0(kρ)exp(−kz) for all allowed values of the separation constant k. The boundary condition u(a, φ, z) = 0 restricts the allowed values of ksince we must have J0(ka) = 0. The zeroes of Bessel functions are given in most books of mathematical tables, and we ﬁnd that, to two decimal places, J0(x)=0 f o r x=2.40,5.52,8.65,. . . . Writing the allowed values of kasknforn=1,2,3,...(so, for example, k1=2.40/a), the required solution takes the form u(ρ, φ, z)=∞X n=1AnJ0(knρ)e x p (−knz). 662 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES z xy au=0 u=0 u=T0 Figure 19.5 A uniform metal cylinder whose curved surface is kept at 0◦C and whose base is held at a temperature T0. By imposing the remaining boundary condition u(ρ, φ,0) = T0, the coeﬃcients Ancan be found in a similar way to Fourier coeﬃcients but this time by exploiting the orthogonalityof the Bessel functions, as discussed in chapter 16. From this boundary condition werequire u(ρ, φ,0) = ∞X n=1AnJ0(knρ)=T0. If we multiply this expression by ρJ0(krρ) and integrate from ρ=0t o ρ=a, and use the orthogonality of the Bessel functions J0(knρ), then the coeﬃcients are given by (16.81) as An=2T0 a2J2 1(kna) Za 0J0(knρ)ρd ρ . (19.36) The integral on the RHS can be evaluated using the recurrence relation (16.68) of chapter 16, d dz[zJ1(z)] =zJ0(z), which on setting z=knρyields 1 knd dρ[knρJ1(knρ)] =knρJ0(knρ). Therefore the integral in (19.36) is given byZa 0J0(knρ)ρd ρ= /1 knρJ1(knρ) /a 0=1 knaJ1(kna), 663 PDES: SEPARATION OF VARIABLES AND OTHER METHODS and the coeﬃcients Anmay be expressed as An=2T0 a2J2 1(kna) /aJ1(kna) kn / =2T0 knaJ1(kna). The steady-state temperature in the cylinder is then given by u(ρ, φ, z)=∞X n=12T0 knaJ1(kna)J0(knρ)e x p (−knz). J We note that if, in the above example, the base of the cylinder were not kept at a uniform temperature T0, but instead had some ﬁxed temperature distribution T(ρ, φ), then the solution of the problem would become more complicated. In such a case, the required temperature distribution u(ρ, φ, z) is in general notaxially symmetric, and so the separation constant mis not restricted to be zero but may take any integer value. The solution will then take the form u(ρ, φ, z)=∞summationdisplay m=0∞summationdisplay n=1Jm(knmρ)(Cnmcosmφ+Dnmsinmφ)ex p (−knmz), where the separation constants knmare such that Jm(knma)=0 ,i . e . knmais the nth zero of the mth-order Bessel function. At the base of the cylinder we would then require u(ρ, φ,0) =∞summationdisplay m=0∞summationdisplay n=1Jm(knmρ)(Cnmcosmφ+Dnmsinmφ)=T(ρ, φ). (19.37) The coeﬃcients Cnmcould be found by multiplying (19.37) by Jq(krqρ)cosqφ, integrating with respect to ρandφover the base of the cylinder and exploiting the orthogonality of the Bessel functions and of the trigonometric functions. The Dnmcould be found in a similar way by multiplying (19.37) by Jq(krqρ)s i nqφ. Laplace’s equation in spherical polars We now come to an eequation that is very widely applicable in physical science, namely∇2u= 0 in spherical polar coordinates: 1 r2∂ ∂rparenleftbigg r2∂u ∂rparenrightbigg +1 r2sinθ∂ ∂θparenleftbigg sinθ∂u ∂θparenrightbigg +1 r2sin2θ∂2u ∂φ2=0. (19.38) Our method of procedure will be as before; we try a solution of the form u(r,θ,φ)=R(r)Θ(θ)Φ(φ). Substituting this in (19.38), dividing through by u=RΘΦ and multiplying by r2, we obtain 1 Rd drparenleftbigg r2dR drparenrightbigg +1 Θs i n θd dθparenleftbigg sinθdΘ dθparenrightbigg +1 Φsin2θd2Φ dφ2=0. (19.39) 664 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES The ﬁrst term depends only on rand the second and third terms (taken together) only on θandφ. Thus (19.39) is equivalent to the two equations 1 Rd drparenleftbigg r2dR drparenrightbigg =λ, (19.40) 1 Θs i n θd dθparenleftbigg sinθdΘ dθparenrightbigg +1 Φsin2θd2Φ dφ2=−λ. (19.41) Equation (19.40) is a homogeneous equation, r2d2R dr2+2rdR dr−λR=0, which can be reduced by the substitution r=e x p t(and writing R(r)=S(t)) to d2S dt2+dS dt−λS=0. This has the straightforward solution S(t)=Aexpλ1t+Bexpλ2t, and so the solution to the radial equation is R(r)=Arλ1+Brλ2, where λ1+λ2=−1a n d λ1λ2=−λ. We can thus take λ1andλ2as given by /lscript and−(/lscript+1 ) ; λthen has the form /lscript(/lscript+ 1). (It should be noted that at this stage nothing has been either assumed or proved about whether /lscriptis an integer.) Hence we have obtained some information about the ﬁrst factor in the separated-variable solution, which will now have the form u(r,θ,φ)=bracketleftbig Ar/lscript+Br−(/lscript+1)bracketrightbig Θ(θ)Φ(φ), (19.42) where Θ and Φ must satisfy (19.41) with λ=/lscript(/lscript+1 ) . The next step is to take (19.41) further. Multiplying through by sin2θand substituting for λ, it too takes a separated form: bracketleftbiggsinθ Θd dθparenleftbigg sinθdΘ dθparenrightbigg +/lscript(/lscript+1 )s i n2θbracketrightbigg +1 Φd2Φ dφ2=0. (19.43) Taking the separation constant as m2, the equation in the azimuthal angle φ has the same solution as in cylindrical polars, namely Φ(φ)=Ccosmφ+Dsinmφ. As before, single-valuedness of urequires that mis an integer; for m= 0 we again have Φ( φ)=Cφ+D. 665 PDES: SEPARATION OF VARIABLES AND OTHER METHODS Having settled the form of Φ( φ), we are left only with the equation satisﬁed by Θ(θ), which is sinθ Θd dθparenleftbigg sinθdΘ dθparenrightbigg +/lscript(/lscript+1 )s i n2θ=m2. (19.44) A change of independent variable from θtoµ=c o s θwill reduce this to a form for which solutions are known, and of which some study has been made inchapter 16. Putting µ=c o s θ,dµ dθ=−sinθ,d dθ=−(1−µ2)1/2d dµ, the equation for M(µ)≡Θ(θ)r e a d s d dµbracketleftbigg (1−µ2)dM dµbracketrightbigg +bracketleftbigg /lscript(/lscript+1 )−m2 1−µ2bracketrightbigg M=0. (19.45) This equation is the associated Legendre equation , which was mentioned in sub- section 17.5.2 in the context of Sturm–Liouville equations. We recall that for the case m= 0, (19.45) reduces to Legendre’s equation, which was studied at length in chapter 16, and has the solution M(µ)=EP/lscript(µ)+FQ/lscript(µ). (19.46) We have not solved (19.45) explicitly for general m, but the solutions were given in subsection 17.5.2 and are the associated Legendre functions Pm /lscript(µ)a n d Qm /lscript(µ), where Pm /lscript(µ)=( 1−µ2)|m|/2d|m| dµ|m|P/lscript(µ), (19.47) and similarly for Qm /lscript(µ). We then have M(µ)=EPm /lscript(µ)+FQm /lscript(µ); (19.48) here mmust be an integer, 0 ≤|m|≤/lscript. We note that if we require solutions to Laplace’s equation that are ﬁnite when µ=c o s θ=±1 (i.e. on the polar axis where θ=0,π), then we must have F= 0 in (19.46) and (19.48) since Qm /lscript(µ) diverges at µ=±1. It will be remembered that one of the important conditions for obtaining ﬁnite polynomial solutions of Legendre’s equation is that /lscriptis an integer ≥0. This condition therefore applies also to the solutions (19.46) and (19.48) and isreﬂected back into the radial part of the general solution given in (19.42). Now that the solutions of each of the three ordinary diﬀerential equations governing R, Θ and Φ have been obtained, we may assemble a complete separated- 666 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES variable solution of Laplace’s equation in spherical polars. It is u(r,θ,φ)=(Ar/lscript+Br−(/lscript+1))(Ccosmφ+Dsinmφ)[EPm /lscript(cosθ)+FQm /lscript(cosθ)], (19.49) where the three bracketted factors are connected only through the integer pa- rameters /lscriptandm,0≤|m|≤/lscript. As before, a general solution may be obtained by superposing solutions of this form for the allowed values of the separationconstants /lscriptandm. As mentioned above, if the solution is required to be ﬁnite on the polar axis then F=0f o ra l l /lscriptandm.IAn uncharged conducting sphere of radius ais placed at the origin in an initially uniform electrostatic ﬁeld E. Show that it behaves as an electric dipole. The uniform ﬁeld, taken in the direction of the polar axis, has a electrostatic potential u=−Ez=−Ercosθ, where uis arbitrarily taken as zero at z=0 .T h i ss a t i s ﬁ e sL a p l a c e ’ se q u a t i o n ∇2u=0 ,a s must the potential vwhen the sphere is present; for large rthe asymptotic form of vmust still be−Ercosθ. Since the problem is clearly axially symmetric we have immediately that m=0 ,a n d since we require vto be ﬁnite on the polar axis we must have F= 0 in (19.49). Therefore the solution must be of the form v(r,θ,φ)=∞X /lscript=0(A/lscriptr/lscript+B/lscriptr−(/lscript+1))P/lscript(cosθ). Now the cos θ-dependence of vfor large rindicates that the ( θ,φ)-dependence of v(r,θ,φ) is given by P0 1(cosθ)=c o s θ. Thus the r-dependence of vmust also correspond to an /lscript= 1 solution, and the most general such solution (outside the sphere, i.e. for r≥a)i s v(r,θ,φ)=(A1r+B1r−2)P1(cosθ). The asymptotic form of vfor large rimmediately gives A1=−Eand so yields the solution v(r,θ,φ)= / −Er+B1 r2 / cosθ. Since the sphere is conducting, it is an equipotential region and so vmust not depend on θforr=a. This can only be the case if B1/a2=Ea, thus ﬁxing B1. The ﬁnal solution is therefore v(r,θ,φ)=−Er / 1−a3 r3 / cosθ. Since a dipole of moment pgives rise to a potential p/(4π/epsilon10r2), this result shows that the sphere behaves as a dipole of moment 4 π/epsilon10a3E, because of the charge distribution induced on its surface; see ﬁgure 19.6. J Often the boundary conditions are not so easily met, and it is necessary to use the mutual orthogonality of the associated Legendre functions (and thetrigonometric functions) to obtain the coeﬃcients in the general solution. 667 PDES: SEPARATION OF VARIABLES AND OTHER METHODS + + + + + + + +++ ++ −−−− − −− −−−−− θ a Figure 19.6 Induced charge and ﬁeld lines associated with a conducting sphere placed in an initially uniform electrostatic ﬁeld.IA hollow split conducting sphere of radius ais placed at the origin. If one half of its surface is charged to a potential v0and the other half is kept at zero potential, ﬁnd the potential vinside and outside the sphere. Let us choose the top hemisphere to be charged to v0and the bottom hemisphere to be at zero potential, with the plane in which the two hemispheres meet perpendicular to thepolar axis; this is shown in ﬁgure 19.7. The boundary condition then becomes v(a, θ, φ)= / v0for 0 <θ<π / 2( 0 <cosθ<1), 0f o r π/2<θ<π (−1<cosθ<0).(19.50) The problem is clearly axially symmetric and so we may set m= 0. Also, we require the solution to be ﬁnite on the polar axis and so it cannot contain Q/lscript(cosθ). Therefore the general form of the solution to (19.38) is v(r,θ,φ)=∞X /lscript=0(A/lscriptr/lscript+B/lscriptr−(/lscript+1))P/lscript(cosθ). (19.51) Inside the sphere (for r<a) we require the solution to be ﬁnite at the origin and so B/lscript=0f o ra l l /lscriptin (19.51). Imposing the boundary condition at r=awe must then have v(a, θ, φ)=∞X /lscript=0A/lscripta/lscriptP/lscript(cosθ), where v(a, θ, φ) is also given by (19.50). Exploiting the mutual orthogonality of the Legendre polynomials, the coeﬃcients in the Legendre polynomial expansion are given by (16.48)as (writing µ=c o s θ) A /lscripta/lscript=2/lscript+1 2 Z1 −1v(a, θ, φ)P/lscript(µ)dµ =2/lscript+1 2v0 Z1 0P/lscript(µ)dµ, 668 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES xz ya −aφθ v=0v=v0 r Figure 19.7 A hollow split conducting sphere with its top half charged to a potential v0and its bottom half at zero potential. where in the last line we have used (19.50). The integrals of the Legendre polynomials are easily evaluated (see exercise 17.7) and we ﬁnd A0=v0 2,A 1=3v0 4a,A 2=0,A 3=−7v0 16a3,···, so that the required solution inside the sphere is v(r,θ,φ)=v0 2 / 1+3r 2aP1(cosθ)−7r3 8a3P3(cosθ)+··· / . Outside the sphere (for r>a ) we require the solution to be bounded as rtends to inﬁnity and so in (19.51) we must have A/lscript=0f o ra l l /lscript. In this case, by imposing the boundary condition at r=awe require v(a, θ, φ)=∞X /lscript=0B/lscripta−(/lscript+1)P/lscript(cosθ), where v(a, θ, φ) is given by (19.50). Following the above argument the coeﬃcients in the expansion are given by B/lscripta−(/lscript+1)=2/lscript+1 2v0 Z1 0P/lscript(µ)dµ, so that the required solution outside the sphere is v(r,θ,φ)=v0a 2r / 1+3a 2rP1(cosθ)−7a3 8r3P3(cosθ)+··· / . J 669 PDES: SEPARATION OF VARIABLES AND OTHER METHODS In the above example, on the equator of the sphere (i.e. at r=aandθ=π/2) the potential is given by v(a, π/2,φ)=v0/2, i.e. mid-way between the potentials of the top and bottom hemispheres. This is so because a Legendre polynomial expansion of a function behaves in the sameway as a Fourier series expansion, in that it converges to the average of the twovalues at any discontinuities present in the original function. If the potential on the surface of the sphere had been given as a function of θ andφ, then we would have had to consider a double series summed over /lscriptand m(for−/lscript≤m≤/lscript), since, in general, the solution would not have been axially symmetric. 19.3.2 Spherical harmonics When obtaining solutions in spherical polar coordinates of ∇ 2u= 0, we found that, for solutions that are ﬁnite on the polar axis, the angular part of the solutionwas given by Θ(θ)Φ(φ)=P m /lscript(cosθ)(Ccosmφ+Dsinmφ). This general form is suﬃciently common that particular functions of θandφ called spherical harmonics are deﬁned and tabulated. The spherical harmonics Ym /lscript(θ,φ) are deﬁned for m≥0b y Ym /lscript(θ,φ)=(−1)mbracketleftbigg2/lscript+1 4π(/lscript−m)! (/lscript+m)!bracketrightbigg1/2 Pm /lscript(cosθ)ex p ( imφ). (19.52) For values of m<0t h er e l a t i o n Y−|m| /lscript(θ,φ)=(−1)|m|bracketleftBig Y|m| /lscript(θ,φ)bracketrightBig∗ deﬁnes the spherical harmonic, the asterisk denoting complex conjugation. Since they contain as their θ-dependent part the solution Pm /lscriptto the associated Legendre equation, which is a Sturm–Liouville equation (see chapter 17), the Ym /lscriptare mutually orthogonal when integrated from −1t o+ 1o v e r d(cosθ). Their mutual orthogonality with respect to φ(0≤φ≤2π) is even more obvious. The numerical factor in (19.52) is chosen to make the Ym /lscriptan orthonormal set, that is integraldisplay1 −1integraldisplay2π 0bracketleftbig Ym /lscript(θ,φ)bracketrightbig∗Ym/prime /lscript/prime(θ,φ)dφ d(cosθ)=δ/lscript/lscript/primeδmm/prime. In addition, the spherical harmonics form a complete set in that any reasonable function (i.e. one that is likely to be met in a physical situation) of θandφcan 670 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES be expanded as a sum of such functions, f(θ,φ)=∞summationdisplay /lscript=0/lscriptsummationdisplay m=−/lscripta/lscriptmYm /lscript(θ,φ), (19.53) the constants a/lscriptmbeing given by a/lscriptm=integraldisplay1 −1integraldisplay2π 0bracketleftbig Ym /lscript(θ,φ)bracketrightbig∗f(θ,φ)dφ d(cosθ). (19.54) This is in exact analogy with a Fourier series and is a particular example of the general property of Sturm–Liouville solutions. The ﬁrst few spherical harmonics Ym /lscript(θ,φ)≡Ym /lscriptare as follows: Y0 0=radicalBig 1 4π,Y0 1=radicalBig 3 4πcosθ, Y±1 1=∓radicalBig 3 8πsinθexp(±iφ),Y0 2=radicalBig 5 16π(3 cos2θ−1), Y±1 2=∓radicalBig 15 8πsinθcosθexp(±iφ),Y±2 2=radicalBig 15 32πsin2θexp(±2iφ). 19.3.3 Other equations in polar coordinates The development of the solutions of ∇2u= 0 carried out in the previous subsection can be employed to solve other equations in which the ∇2operator appears. Since we have discussed the general method in some depth already, only an outline ofthe solutions will be given here. Let us ﬁrst consider the wave equation ∇ 2u=1 c2∂2u ∂t2, (19.55) and look for a separated solution of the form u=F(r)T(t), so that initially we are separating only the spatial and time dependences. Substituting this form into (19.55) and taking the separation constant as k2we obtain ∇2F+k2F=0,d2T dt2+k2c2T=0. (19.56) The second equation has the simple solution T(t)=Aexp(iωt)+Bexp(−iωt), (19.57) where ω=kc; this may also be expressed in terms of sines and cosines, of course. The ﬁrst equation in (19.56) is referred to as Helmholtz’s equation ; we discuss it below. We may treat the diﬀusion equation κ∇2u=∂u ∂t 671 PDES: SEPARATION OF VARIABLES AND OTHER METHODS in a similar way. Separating the spatial and time dependences by assuming a solution of the form u=F(r)T(t), and taking the separation constant as k2,w e ﬁnd ∇2F+k2F=0,dT dt+k2κT=0. Just as in the case of the wave equation, the spatial part of the solution satisﬁes Helmholtz’s equation. It only remains to consider the time dependence, whichhas the simple solution T(t)=Aexp(−k 2κt). Helmholtz’s equation is clearly of central importance in the solutions of the wave and diﬀusion equations. It can be solved in polar coordinates in much the same way as Laplace’s equation, and indeed reduces to Laplace’s equation when k= 0. Therefore, we will merely sketch the method of its solution in each of the three polar coordinate systems. Helmholtz’s equation in plane polars In two-dimensional plane polar cooordinates Helmholtz’s equation takes the form 1 ρ∂ ∂ρparenleftbigg ρ∂F ∂ρparenrightbigg +1 ρ2∂2F ∂φ2+k2F=0. If we try a separated solution of the form F(r)= P(ρ)Φ(φ), and take the separation constant as m2, we ﬁnd d2Φ dφ2+m2φ=0, d2P dρ2+1 ρdP dρ+parenleftbigg k2−m2 ρ2parenrightbigg P=0. As for Laplace’s equation, the angular part has the familiar solution (if m/negationslash=0 ) Φ(φ)=Acosmφ+Bsinmφ, or an equivalent form in terms of complex exponentials. The radial equation diﬀers from that found in the solution of Laplace’s equation, but by making thesubstitution µ=kρit is easily transformed into Bessel’s equation of order m (discussed in chapter 16), and has the solution P(ρ)=CJ m(kρ)+DYm(kρ), where Ymis a Bessel function of the second kind, which is inﬁnite at the origin and is not to be confused with a spherical harmonic (these are written with asuperscript as well as a subscript). Putting the two parts of the solution together we have F(ρ, φ)=[Acosmφ+Bsinmφ][CJ m(kρ)+DYm(kρ)]. (19.58) 672 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES Clearly, for solutions of Helmholtz’s equation that are required to be ﬁnite at the origin, we must set D=0 .IFind the four lowest frequency modes of oscillation of a circular drumskin of radius a whose circumference is held ﬁxed in a plane. The transverse displacement u(r,t) of the drumskin satisﬁes the two-dimensional wave equation ∇2u=1 c2∂2u ∂t2, with c2=T/σ,w h e r e Tis the tension of the drumskin and σis its mass per unit area. From (19.57) and (19.58) a separated solution of this equation, in plane polar coordinates,that is ﬁnite at the origin is u(ρ, φ, t)=J m(kρ)(Acosmφ+Bsinmφ)e x p(±iωt), where ω=kc. Since we require the solution to be single-valued we must have mas an integer. Furthermore, if the drumskin is clamped at its outer edge ρ=athen we also require u(a, φ, t) = 0. Thus we need Jm(ka)=0 , which in turn restricts the allowed values of k. The zeroes of Bessel functions can be obtained from most books of tables, and the ﬁrst few are J0(x)=0 f o r x≈2.40,5.52,8.65,..., J1(x)=0 f o r x≈3.83,7.02,10.17,..., J2(x)=0 f o r x≈5.14,8.42,11.62.... The smallest value of xfor which any of the Bessel functions is zero is x≈2.40, which occurs for J0(x). Thus the lowest-frequency mode has k=2.40/aand angular frequency ω=2.40c/a.S i n c e m= 0 for this mode, the shape of the drumskin is u∝J0 / 2.40ρ a / ; this is illustrated in ﬁgure 19.8. Continuing in the same way the next three modes are given by ω=3.83c a,u∝J1 / 3.83ρ a / cosφ, J 1 / 3.83ρ a / sinφ; ω=5.14c a,u∝J2 / 5.14ρ a / cos 2φ, J 2 / 5.14ρ a / sin 2φ; ω=5.52c a,u∝J0 / 5.52ρ a / . These modes are also shown in ﬁgure 19.8. We note that the second and third frequencies have twocorresponding modes of oscillation; these frequencies are therefore two-fold degenerate. J Helmholtz’s equation in cylindrical polars Generalising the above method to three-dimensional cylindrical polars is straight- forward, and following a similar procedure to that used for Laplace’s equation 673 PDES: SEPARATION OF VARIABLES AND OTHER METHODS ω=2.40c/a ω=3.83c/a ω=5.14c/a ω=5.52c/aa Figure 19.8 For a circular drumskin of radius a, the modes of oscillation with the four lowest frequencies. The dotted lines indicate the nodes, wherethe displacement of the drumskin is always zero. we ﬁnd the separated solution of Helmholtz’s equation takes the form F(ρ, φ, z)=bracketleftBig AJmparenleftBig√ k2−α2ρparenrightBig +BYmparenleftBig√ k2−α2ρparenrightBigbracketrightBig ×(Ccosmφ+Dsinmφ)[Eexp(iαz)+Fexp(−iαz)], where αandmare separation constants. We note that the angular part of the solution is the same as for Laplace’s equation in cylindrical polars. Helmholtz’s equation in spherical polars In spherical polars, we ﬁnd again that the angular parts of the solution Θ( θ)Φ(φ) are identical to those of Laplace’s equation in this coordinate system, i.e. they are the spherical harmonics Ym /lscript(θ,φ), and so we shall not discuss them further. The radial equation in this case is given by r2R/prime/prime+2rR/prime+[k2r2−/lscript(/lscript+1 ) ] R=0, (19.59) which has an additional term k2r2Rcompared with the radial equation for the Laplace solution. The equation (19.59) looks very much like Bessel’s equation and can in fact be reduced to it by writing R(r)=r−1/2S(r). The function S(r)t h e n satisﬁes r2S/prime/prime+rS/prime+bracketleftBig k2r2−parenleftbig /lscript+1 2parenrightbig2bracketrightBig S=0, which, after changing the variable to µ=kr,i sB e s s e l ’ se q u a t i o no fo r d e r /lscript+1 2 and has as its solutions S(µ)=J/lscript+1/2(µ)a n d Y/lscript+1/2(µ). The separated solution to 674 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES Helmholtz’s equation in spherical polars is thus F(r,θ,φ)=r−1/2[AJ/lscript+1/2(kr)+BY/lscript+1/2(kr)](Ccosmφ+Dsinmφ) ×[EPm /lscript(cosθ)+FQm /lscript(cosθ)]. (19.60) For solutions that are ﬁnite at the origin we require B= 0, and for solutions that are ﬁnite on the polar axis we require F=0 . It is worth mentioning that the solutions proportional to r−1/2J/lscript+1/2(kr)w h e n suitably normalised are called spherical Bessel functions and are denoted by j/lscript(kr): j/lscript(µ)=radicalbiggπ 2µJ/lscript+1/2(µ). They are trigonometric functions of µ(as discussed in chapter 16), and for /lscript=0 and/lscript=1a r eg i v e nb y j0(µ)=s i n µ, j1(µ)=sinµ µ−cosµ. The second, linearly-independent, solution of (19.59), n/lscript(µ), is derived from Y/lscript+1/2(µ) in a similar way. As mentioned at the beginning of this subsection, the separated solution of the wave equation in spherical polars is the product of the time-dependent part(19.57) and a spatial part (19.60). It will be noticed that, although this solutioncorresponds to a solution of deﬁnite frequency ω=kc, the zeroes of the radial function j /lscript(kr) are not equally spaced in r, except for the case /lscript= 0 involving j0(kr), and so there is no precise wavelength associated with the solution. To conclude this subsection, let us mention brieﬂy the Schr ¨odinger equation for the electron in a hydrogen atom, the nucleus of which is taken at the originand is assumed massive compared with the electron. Under these circumstancesthe Schr ¨odinger equation is −/planckover2pi1 2 2m∇2u−e2 4π/epsilon10u r=i/planckover2pi1∂u ∂t. For a ‘stationary-state’ solution, for which the energy is a constant Eand the time- dependent factor Tinuis given by T(t)=Aexp(−iEt//planckover2pi1), the above equation is similar to, but not quite the same as, the Helmholtz equation. †However, as with the wave equation, the angular parts of the solution are identical to thosefor Laplace’s equation and are expressed in terms of spherical harmonics. The important point to note is that for anyequation involving ∇ 2,p r o v i d e d θ andφdo not appear in the equation other than as part of ∇2, a separated-variable †For the solution by series of the r-equation in this case the reader may consult, e.g., Schiﬀ, Quantum Mechanics (McGraw-Hill, 1955) p. 82. 675 PDES: SEPARATION OF VARIABLES AND OTHER METHODS solution in spherical polars will always lead to spherical harmonic solutions. This is the case for the Schr ¨odinger equation describing an atomic electron whenever the potential is central, i.e. whenever V(r)i si nf a c t V(r). 19.3.4 Solution by expansion It is sometimes possible to use the uniqueness theorem discussed in the last chapter, together with the results of the last few subsections, in which Laplace’s equation (and other equations) were considered in polar coordinates, to obtainsolutions of such equations appropriate to particular physical situations. We will illustrate the method for Laplace’s equation in spherical polars and ﬁrst assume that the required solution of ∇ 2u= 0 can be written as a superposition in the normal way: u(r,θ,φ)=∞summationdisplay /lscript=0/lscriptsummationdisplay m=−/lscript(Ar/lscript+Br−(/lscript+1))Pm /lscript(cosθ)(Ccosmφ+Dsinmφ). (19.61) Here, all the constants A, B, C, D may depend upon /lscriptand m, and we have assumed that the required solution is ﬁnite on the polar axis. As usual, boundaryconditions of a physical nature will then ﬁx or eliminate some of the constants;for example, uﬁnite at the origin implies all B= 0, or axial symmetry implies that only m= 0 terms are present. The essence of the method is then to ﬁnd the remaining constants by determin- inguat values of r,θ,φ for which it can be evaluated by other means , e.g. by direct calculation on an axis of symmetry. Once the remaining constants have been ﬁxed by these special considerations to have particular values, the uniqueness theoremcan be invoked to establish that they must have these values in general.ICalculate the gravitational potential at a general point in space due to a uniform ring of matter of radius aand total mass M. Everywhere except on the ring the potential u(r) satisﬁes the Laplace equation, and so if we use polar coordinates with the normal to the ring as polar axis, as in ﬁgure 19.9, asolution of the form (19.61) can be assumed. We expect the potential u(r,θ,φ) to tend to zero as r→∞, and also to be ﬁnite at r=0 . At ﬁrst sight this might seem to imply that all AandB, and hence u, must be identically zero, an unacceptable result. In fact, what it means is that diﬀerent expressions must applyto diﬀerent regions of space. On the ring itself we no longer have ∇ 2u= 0 and so it is not surprising that the form of the expression for uchanges there. Let us therefore take two separate regions. In the region r>a (i) we must have u→0a sr→∞, implying that all A=0 ,a n d (ii) the system is axially symmetric and so only m= 0 terms appear. With these restrictions we can write as a trial form u(r,θ,φ)=∞X /lscript=0B/lscriptr−(/lscript+1)P0 /lscript(cosθ). (19.62) 676 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES yP aθ O−arz x Figure 19.9 The polar axis Ozis taken as normal to the plane of the ring of matter and passing through its centre. The constants B/lscriptare still to be determined; this we do by calculating directly the potential where this can be done simply – in this case, on the polar axis. Considering a point Pon the polar axis at a distance z(>a) from the plane of the ring (taken as θ=π/2), all parts of the ring are at a distance ( z2+a2)1/2from it. The potential atPis thus straightforwardly u(z,0,φ)=−GM (z2+a2)1/2, (19.63) where Gis the gravitational constant. This must be the same as (19.62) for the particular values r=z,θ=0 ,a n d φundeﬁned. Since P0 /lscript(cosθ)=P/lscript(cosθ)w i t h P/lscript(1) = 1, putting r=zin (19.62) gives u(z,0,φ)=∞X /lscript=0B/lscript z/lscript+1. (19.64) However, expanding (19.63) for z>a (as it applies to this region of space) we obtain u(z,0,φ)=−GM z / 1−1 2 /a z /2 +3 8 /a z /4 −··· / , which on comparison with (19.64) gives † B0=−GM, B2/lscript=−GMa2/lscript(−1)/lscript(2/lscript−1)!! 2/lscript/lscript!for/lscript≥1, (19.65) B2/lscript+1=0. We now conclude the argument by saying that if a solution for a general point ( r,θ,φ) exists at all, which of course we very much expect on physical grounds, then it must be(19.62) with the B /lscriptgiven by (19.65). This is so because thus deﬁned it is a function with no arbitrary constants and which satisﬁes all the boundary conditions, and the uniqueness †(2/lscript−1)!! = 1×3×···×(2/lscript−1). 677 PDES: SEPARATION OF VARIABLES AND OTHER METHODS theorem states that there is only one such function. The expression for the potential in the region r>a is therefore u(r,θ,φ)=−GM r /" 1+∞X /lscript=1(−1)/lscript(2/lscript−1)!! 2/lscript/lscript! /a r /2/lscript P2/lscript(cosθ) /# . The expression for r<a can be found in a similar way. The ﬁniteness of uatr=0a n d the axial symmetry give u(r,θ,φ)=∞X /lscript=0A/lscriptr/lscriptP0 /lscript(cosθ). Comparing this expression for r=z,θ= 0 with the z<a expansion of (19.63), which is valid for any z, establishes A2/lscript+1=0 , A0=−GM/a and A2/lscript=−GM a2/lscript+1(−1)/lscript(2/lscript−1)!! 2/lscript/lscript!, so that the ﬁnal expression valid, and convergent, for r<a is thus u(r,θ,φ)=−GM a /" 1+∞X /lscript=1(−1)/lscript(2/lscript−1)!! 2/lscript/lscript! /r a /2/lscript P2/lscript(cosθ) /# . It is easy to check that the solution obtained has the expected physical value for large r and for r= 0 and is continuous at r=a. J 19.3.5 Separation of variables for inhomogeneous equations So far our discussion of the method of separation of variables has been limited to the solution of homogeneous equations such as the Laplace equation and thewave equation. The solutions of inhomogeneous PDEs are usually obtained usingthe Green’s function methods to be discussed below in section 19.5. However, as aﬁnal illustration of the usefulness of the separation of variables, we now consider its application to the solution of inhomogeneous equations. Because of the added complexity in dealing with inhomogeneous equations, we shall restrict our discussion to the solution of Poisson’s equation, ∇ 2u=ρ(r), (19.66) in spherical polar coordinates, although the general method can accommodate other coordinate systems and equations. In physical problems the RHS of (19.66)usually contains some multiplicative constant(s). If uis the electrostatic potential in some region of space in which ρis the density of electric charge then ∇ 2u= −ρ(r)//epsilon10. Alternatively, umight represent the gravitational potential in some region where the matter density is given by ρ,s ot h a t∇2u=4πGρ(r). We will simplify our discussion by assuming that the required solution uis ﬁnite on the polar axis and also that the system possesses axial symmetry about that axis – in which case ρdoes not depend on the azimuthal angle φ.T h ek e y to the method is then to assume a separated form for both the solution uandthe density term ρ. 678 19.3 SEPARATION OF VARIABLES IN POLAR COORDINATES From the discussion of Laplace’s equation, for systems with axial symmetry only m= 0 terms appear, and so the angular part of the solution can be expressed in terms of Legendre polynomials P/lscript(cosθ). Since these functions form an orthogonal set let us expand both uandρin terms of them: u=∞summationdisplay /lscript=0R/lscript(r)P/lscript(cosθ), (19.67) ρ=∞summationdisplay /lscript=0F/lscript(r)P/lscript(cosθ), (19.68) where the coeﬃcients R/lscript(r)a n d F/lscript(r) in the Legendre polynomial expansions are functions of r. Since in any particular problem ρis given, we can ﬁnd the coeﬃcients F/lscript(r) in the expansion in the usual way (see subsection 16.6.2). It then only remains to ﬁnd the coeﬃcients R/lscript(r) in the expansion of the solution u. Writing∇2in spherical polars and substituting (19.67) and (19.68) into (19.66) we obtain ∞summationdisplay /lscript=0bracketleftbiggP/lscript(cosθ) r2d drparenleftbigg r2dR/lscript drparenrightbigg +R/lscript r2sinθd dθparenleftbigg sinθdP/lscript(cosθ) dθparenrightbiggbracketrightbigg =∞summationdisplay /lscript=0F/lscript(r)P/lscript(cosθ). (19.69) However, if, in equation (19.44) of our discussion of the angular part of the solution to Laplace’s equation, we set m= 0 we conclude that 1 sinθd dθparenleftbigg sinθdP/lscript(cosθ) dθparenrightbigg =−/lscript(/lscript+1 )P/lscript(cosθ). Substituting this into (19.69), we ﬁnd that the LHS is greatly simpliﬁed and we obtain ∞summationdisplay /lscript=0bracketleftbigg1 r2d drparenleftbigg r2dR/lscript drparenrightbigg −/lscript(/lscript+1 )R/lscript r2bracketrightbigg P/lscript(cosθ)=∞summationdisplay /lscript=0F/lscript(r)P/lscript(cosθ). This relation is most easily satisﬁed by equating terms on both sides for each value of /lscriptseparately, so that for /lscript=0,1,2,...we have 1 r2d drparenleftbigg r2dR/lscript drparenrightbigg −/lscript(/lscript+1 )R/lscript r2=F/lscript(r). (19.70) This is an ODE in which F/lscript(r) is given, and it can therefore be solved for R/lscript(r). The solution to Poisson’s equation, u, is then obtained by making the superposition (19.67). 679 PDES: SEPARATION OF VARIABLES AND OTHER METHODSIIn a certain system, the electric charge density ρis distributed as follows: ρ= / Arcosθfor0≤r<a , 0 forr≥a. Find the electrostatic potential inside and outside the charge distribution, given that both the potential and its radial derivative are continuous everywhere. The electrostatic potential usatisﬁes ∇2u= / −(A//epsilon10)rcosθfor 0≤r<a , 0f o r r≥a. Forr<a the RHS can be written −(A//epsilon10)rP1(cosθ), and the coeﬃcients in (19.68) are simply F1(r)=−(Ar//epsilon1 0)a n d F/lscript(r)=0f o r /lscript/negationslash= 1. Therefore we need only calculate R1(r), which satisﬁes (19.70) for /lscript=1 : 1 r2d dr / r2dR1 dr / −2R1 r2=−Ar /epsilon10. This can be rearranged to give r2R/prime/prime 1+2rR/prime 1−2R1=−Ar3 /epsilon10, where the prime denotes diﬀerentiation with respect to r. The LHS is homogeneous and the equation can be reduced by the substitution r=e x p t, and writing R1(r)=S(t), to ¨S+˙S−2S=−A /epsilon10exp 3 t, (19.71) where the dots indicate diﬀerentiation with respect to t. This is an inhomogeneous second-order ODE with constant coeﬃcients and can be straightforwardly solved by the methods of subsection 15.2.1 to give S(t)=c1expt+c2exp(−2t)−A 10/epsilon10exp 3 t. Recalling that r=e x p twe ﬁnd R1(r)=c1r+c2r−2−A 10/epsilon10r3. Since we are interested in the region r<a we must have c2= 0 for the solution to remain ﬁnite. Thus inside the charge distribution the electrostatic potential has the form u1(r,θ,φ)= / c1r−A 10/epsilon10r3 / P1(cosθ). (19.72) Outside the charge distribution (for r≥a), however, the electrostatic potential obeys Laplace’s equation, ∇2u= 0, and so given the symmetry of the problem and the requirement thatu→∞asr→∞the solution must take the form u2(r,θ,φ)=∞X /lscript=0B/lscript r/lscript+1P/lscript(cosθ). (19.73) We can now use the boundary conditions at r=ato ﬁx the constants in (19.72) and (19.73). The requirement of continuity of the potential and its radial derivative at r=a imply that u1(a, θ, φ)=u2(a, θ, φ), ∂u1 ∂r(a, θ, φ)=∂u2 ∂r(a, θ, φ). 680 19.4 INTEGRAL TRANSFORM METHODS Clearly B/lscript=0f o r /lscript/negationslash= 1; carrying out the necessary diﬀerentiations and setting r=ain (19.72) and (19.73) we obtain the simultaneous equations c1a−A 10/epsilon10a3=B1 a2, c1−3A 10/epsilon10a2=−2B1 a3 which may be solved to give c1=Aa2/(6/epsilon10)a n d B1=Aa5/(15/epsilon10). Since P1(cosθ)=c o s θ, the electrostatic potentials inside and outside the charge distribution are given respectivelyby u 1(r,θ,φ)=A /epsilon10 /a2r 6−r3 10 / cosθ, u 2(r,θ,φ)=Aa5 15/epsilon10cosθ r2. J 19.4 Integral transform methods In the method of separation of variables our aim was to keep the independent variables in a PDE as separate as possible. We now discuss the use of integraltransforms in solving PDEs, a method by which one of the independent variablescan be eliminated from the diﬀerential coeﬃcients. It will be assumed that thereader is familiar with Laplace and Fourier transforms and their properties, asdiscussed in chapter 13. The method consists simply of transforming the PDE into one containing derivatives with respect to a smaller number of variables. Thus, if the originalequation has just two independent variables, it may be possible to reduce thePDE into a soluble ODE. The solution obtained can then (where possible) betransformed back to give the solution of the original PDE. As we shall see, boundary conditions can usually be incorporated in a natural way. Which sort of transform to use, and the choice of the variable(s) with respect to which the transform is to be taken, is a matter of experience; we illustrate thisin the example below. In practice, transforms can be taken with respect to eachvariable in turn, and the transformation that aﬀords the greatest simpliﬁcationcan be pursued further.IA semi-inﬁnite tube of constant cross-section contains initially pure water. At time t=0, one end of the tube is put into contact with a salt solution and maintained at a concentrationu 0. Find the total amount of salt that has diﬀused into the tube after time t, if the diﬀusion constant is κ. The concentration u(x, t)a tt i m e tand distance xfrom the end of the tube satisﬁes the diﬀusion equation κ∂2u ∂x2=∂u ∂t, (19.74) which has to be solved subject to the boundary conditions u(0,t)=u0for all tand u(x,0) = 0 for all x>0. Since we are interested only in t>0, the use of the Laplace transform is suggested. 681 PDES: SEPARATION OF VARIABLES AND OTHER METHODS Furthermore, it will be recalled from chapter 13 that one of the major virtues of Laplace transformations is the possibility they aﬀord of replacing derivatives of functions by simplemultiplication by a scalar. If the derivative with respect to time were so removed, equation(19.74), would contain only diﬀerentiation with respect to a single variable. Let us thereforetake the Laplace transform of (19.74) with respect to t:Z∞ 0κ∂2u ∂x2exp(−st)dt= Z∞ 0∂u ∂texp(−st)dt. On the LHS the (double) diﬀerentiation is with respect to x, whereas the integration is with respect to the independent variable t. Therefore the derivative can be taken outside the integral. Denoting the Laplace transform of u(x, t)b y¯u(x, s) and using result (13.57) to rewrite the transform of the derivative on the RHS (or by integrating directly by parts),we obtain κ∂ 2¯u ∂x2=s¯u(x, s)−u(x,0). But from the boundary condition u(x,0) = 0 the last term on the RHS vanishes, and the solution is immediate: ¯u(x, s)=Aexp / rs κx / +Bexp / − rs κx / , where the constants AandBmay depend on s. We require u(x, t)→0a s x→∞ and so we must also have ¯u(∞,s) = 0; consequently we require that A= 0. The value of Bis determined by the need for u(0,t)=u0and hence that ¯u(0,s)= Z∞ 0u0exp(−st)dt=u0 s. We thus conclude that the appropriate expression for the Laplace transform of u(x, t)i s ¯u(x, s)=u0 sexp / − rs κx / . (19.75) To obtain u(x, t) from this result requires the inversion of this transform – a task that is generally diﬃcult and requires a contour integration. This is discussed in chapter 20, butfor completeness we note that the solution is u(x, t)=u 0 / 1−erf /x√ 4κt // , where erf( x) is the error function discussed in the Appendix. (The more complete sets of mathematical tables list this inverse Laplace transform.) In the present problem, however, an alternative method is available. Let w(t)b et h e amount of salt that has diﬀused into the tube in time t;t h e n w(t)= Z∞ 0u(x, t)dx, and its transform is given by ¯w(s)= Z∞ 0dtexp(−st) Z∞ 0u(x, t)dx = Z∞ 0dx Z∞ 0u(x, t)e x p(−st)dt = Z∞ 0¯u(x, s)dx. 682 19.4 INTEGRAL TRANSFORM METHODS Substituting for ¯u(x, s) from (19.75) into the last integral and integrating, we obtain ¯w(s)=u0κ1/2s−3/2. This expression is much simpler to invert, and referring to the table of standard Laplace transforms (table 13.1) we ﬁnd w(t)=2 ( κ/π)1/2u0t1/2, which is thus the required expression for the amount of diﬀused salt at time t. J The above example shows that in some circumstances the use of a Laplace transformation can greatly simplify the solution of a PDE. However, it will havebeen observed that (as with ODEs) the easy elimination of some derivatives isusually paid for by the introduction of a diﬃcult inverse transformation. This problem, although still present, is less severe for Fourier transformations.IAn inﬁnite metal bar has an initial temperature distribution f(x)along its length. Find the temperature distribution at a later time t. We are interested in values of xfrom−∞to∞, which suggests Fourier transformation with respect to x. Assuming that the solution obeys the boundary conditions u(x, t)→0 and∂u/∂x→0a s|x|→∞ , we may Fourier-transform the one-dimensional diﬀusion equation (19.74) to obtain κ√ 2π Z∞ −∞∂2u(x, t) ∂x2exp(−ikx)dx=1√ 2π∂ ∂t Z∞ −∞u(x, t)exp(−ikx)dx, where on the RHS we have taken the partial derivative with respect to toutside the integral. Denoting the Fourier transform of u(x, t)b y eu(k,t), and using equation (13.28) to rewrite the Fourier transform of the second derivative on the LHS, we then have −κk2eu(k,t)=∂ eu(k,t) ∂t. This ﬁrst-order equation has the simple solutioneu(k,t)= eu(k,0)exp(−κk2t), where the initial conditions giveeu(k,0) =1√ 2π Z∞ −∞u(x,0)exp(−ikx)dx =1√ 2π Z∞ −∞f(x)exp(−ikx)dx= ef(k). Thus we may write the Fourier transform of the solution aseu(k,t)= ef(k)exp(−κk2t)=√ 2π ef(k) eG(k,t), (19.76) where we have deﬁned the function eG(k,t)=(√ 2π)−1exp(−κk2t). Since eu(k,t)c a nb e written as the product of two Fourier transforms, we can use the convolution theorem,subsection 13.1.7, to write the solution as u(x, t)= Z∞ −∞G(x−x/prime,t)f(x/prime)dx/prime, 683 PDES: SEPARATION OF VARIABLES AND OTHER METHODS where G(x, t) is the Green’s function for this problem (see subsection 15.2.5). This function is the inverse Fourier transform of eG(k,t) and is thus given by G(x, t)=1 2π Z∞ −∞exp(−κk2t)exp( ikx)dk =1 2π Z∞ −∞exp / −κt / k2−ix κtk // dk. Completing the square in the integrand we ﬁnd G(x, t)=1 2πexp / −x2 4κt /Z∞ −∞exp /" −κt / k−ix 2κt /2 /# dk =1 2πexp / −x2 4κt /Z∞ −∞exp / −κtk/prime2 / dk/prime =1√ 4πκtexp / −x2 4κt / , where in the second line we have made the substitution k/prime=k−ix/(2κt) ,a n di nt h el a s t line we have used the standard result for the integral of a Gaussian, given in subsection6.4.2. (Strictly speaking the change of variable from ktok /primeshifts the path of integration oﬀ the real axis, since k/primeis complex for real k, and so results in a complex integral, as will be discussed in chapter 20. Nevertheless, in this case the path of integration can be shiftedback to the real axis without aﬀecting the value of the integral.) Thus the temperature in the bar at a later time tis given by u(x, t)=1 √ 4πκt Z∞ −∞exp / −(x−x/prime)2 4κt / f(x/prime)dx/prime, (19.77) which may be evaluated (numerically if necessary) when the form of f(x)i sg i v e n . J As we might expect from our discussion of Green’s functions in chapter 15, we see from (19.77) that, if the initial temperature distribution is f(x)=δ(x−a), i.e. a ‘point’ source at x=a, then the temperature distribution at later times is simply given by u(x, t)=G(x−a, t)=1√ 4πκtexpbracketleftbigg −(x−a)2 4κtbracketrightbigg . The temperature at several later times is illustrated in ﬁgure 19.10, which shows that the heat diﬀuses out from its initial position; the width of the Gaussianincreases as√ t, a dependence on time which is characteristic of diﬀusion processes. The reader may have noticed that in both examples using integral transforms the solutions have been obtained in closed form – albeit in one case in the formof an integral. This diﬀers from the inﬁnite series solutions usually obtained viathe separation of variables. It should be noted that this behaviour is a result ofthe inﬁnite range in xrather than of the transform method itself. In fact the method of separation of variables would yield the same solutions, since in the inﬁnite-range case the separation constant is not restricted to take on an inﬁnite set of discrete values but may have any real value, with the result that the sumover λbecomes an integral, as mentioned at the end of section 19.2. 684 19.4 INTEGRAL TRANSFORM METHODS xu x=at1 t2 t3 Figure 19.10 Diﬀusion of heat from a point source in a metal bar: the curves show the temperature uat position xfor various diﬀerent times t1<t2<t3. The area under the curves remains constant, since the total heat energy is conserved.IAn inﬁnite metal bar has an initial temperature distribution f(x)along its length. Find the temperature distribution at a later time tusing the method of separation of variables. This is the same problem as in the previous example, but we now seek a solution by separating variables. From (19.12) a separated solution for the one-dimensional diﬀusionequation is u(x, t)=[Aexp(iλx)+Bexp(−iλx)]exp(−κλ 2t), where−λ2is the separation constant. Since the bar is inﬁnite we do not require the solution to take a given form at any ﬁnite value of x(for instance at x=0 )a n ds ot h e r e is no restriction on λother than its being real. Therefore instead of the superposition of such solutions in the form of a sum over allowed values of λwe have an integral over allλ, u(x, t)=1√ 2π Z∞ −∞A(λ)exp(−κλ2t)exp( iλx)dλ, (19.78) where in taking λfrom−∞to∞we need include only one of the complex exponentials; we have taken a factor 1 /√ 2πout of A(λ) for convenience. We can see from (19.78) that the expression for u(x, t) has the form of an inverse Fourier transform (where λis the transform variable). Therefore, Fourier-transforming both sides and using the Fourierinversion theorem, we ﬁndeu(λ, t)=A(λ)exp(−κλ2t). Now the initial boundary condition requires u(x,0) =1√ 2π Z∞ −∞A(λ)exp( iλx)dλ=f(x), 685 PDES: SEPARATION OF VARIABLES AND OTHER METHODS from which, using the Fourier inversion theorem once more, we see that A(λ)= ef(λ). Therefore we haveeu(λ, t)= ef(λ)exp(−κλ2t), which is identical to (19.76) in the previous example (but with kreplaced by λ), and hence leads to the same result. J 19.5 Inhomogeneous problems – Green’s functions In chapters 15 and 17 we encountered Green’s functions and found them a useful tool for solving inhomogeneous linear ODEs. We now discuss their usefulness in solving inhomogeneous linear PDEs. For the sake of brevity we shall again denote a linear PDE by Lu(r)=ρ(r), (19.79) where Lis a linear partial diﬀerential operator. For example, in Laplace’s equation we have L=∇2, whereas for Helmholtz’s equation L=∇2+k2. Note that we have not speciﬁed the dimensionality of the problem, and (19.79) may, for example,represent Poisson’s equation in two or three (or more) dimensions. The readerwill also notice that for the sake of simplicity we have not included any timedependence in (19.79). Nevertheless, the following discussion can be generalisedto include it. As we discussed in subsection 18.3.2, a problem is inhomogeneous if the fact that u(r) is a solution does notimply that any constant multiple λu(r)i sa l s oa solution. This inhomogeneity may derive from either the PDE itself or from theboundary conditions imposed on the solution. In our discussion of Green’s function solutions of inhomogeneous ODEs (see subsection 15.2.5) we dealt with inhomogeneous boundary conditions by making asuitable change of variable such that in the new variable the boundary conditionswere homogeneous. In an analogous way, as illustrated in the ﬁnal exampleof section 19.2, it is usually possible to make a change of variables in PDEs totransform between inhomogeneity of the boundary conditions and inhomogeneityof the equation. Therefore let us assume for the moment that the boundary conditions imposed on the solution u(r) of (19.79) are homogeneous. This most commonly means that if we seek a solution to (19.79) in some region Vthen on the surface Sthat bounds Vthe solution obeys the conditions u(r)=0o r ∂u/∂n =0 ,w h e r e ∂u/∂n is the normal derivative of uat the surface S. We shall discuss the extension of the Green’s function method to the direct so- lution of problems with inhomogeneous boundary conditions in subsection 19.5.2, but we ﬁrst highlight how the Green’s function approach to solving ODEs canbe simply extended to PDEs for homogeneous boundary conditions. 686 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS 19.5.1 Similarities with Green’s functions for ODEs As in the discussion of ODEs in chapter 15, we may consider the Green’s function for a system described by a PDE as the response of the system to a ‘unitimpulse’ or ‘point source’. Thus if we seek a solution to (19.79) that satisﬁes somehomogeneous boundary conditions on u(r) then the Green’s function G(r,r 0)f o r the problem is a solution of LG(r,r0)=δ(r−r0), (19.80) where r0lies in V. The Green’s function G(r,r0) must also satisfy the imposed (homogeneous) boundary conditions. It is understood that in (19.80) the Loperator expresses diﬀerentiation with respect to ras opposed to r0.A l s o , δ(r−r0) is the Dirac delta function (see chapter 13) of dimension appropriate for the problem; it may be thought of asrepresenting a unit-strength point source at r=r 0. Following an analogous argument to that given in subsection 15.2.5 for ODEs, if the boundary conditions on u(r) are homogeneous then a solution to (19.79) that satisﬁes the imposed boundary conditions is given by u(r)=integraldisplay G(r,r0)ρ(r0)dV(r0), (19.81) where the integral on r0is over some appropriate ‘volume’. In two or more dimensions, however, the task of ﬁnding directly a solution to (19.80) that satisﬁesthe imposed boundary conditions on Scan be a diﬃcult one, and we return to this in the next subsection. An alternative approach is to follow a similar argument to that presented in chapter 17 for ODEs and so to construct the Green’s function for (19.79) as a superposition of eigenfunctions of the operator L,p r o v i d e d Lis Hermitian. By analogy with an ordinary diﬀerential operator, a partial diﬀerential operator isHermitian if it satisﬁes integraldisplay Vv∗(r)Lw(r)dV=bracketleftbiggintegraldisplay Vw∗(r)Lv(r)dVbracketrightbigg∗ , where the asterisk denotes complex conjugation and vandware arbitrary func- tions obeying the imposed (homogeneous) boundary condition on the solution ofLu(r)=0 . The eigenfunctions u n(r),n=0,1,2,...,o fLsatisfy Lun(r)=λnun(r), where λnare the corresponding eigenvalues, which are all real for an Hermitian operator L. Furthermore, each eigenfunction must obey any imposed (homo- geneous) boundary conditions. Using an argument analogous to that given in 687 PDES: SEPARATION OF VARIABLES AND OTHER METHODS chapter 17, the Green’s function for the problem is given by G(r,r0)=∞summationdisplay n=0un(r)u∗ n(r0) λn. (19.82) From (19.82) we see immediately that the Green’s function (irrespective of how it is found) enjoys the property G(r,r0)=G∗(r0,r). Thus, if the Green’s function is real then it is symmetric in its two arguments. Once the Green’s function has been obtained, the solution to (19.79) is again given by (19.81). For PDEs this approach can become very cumbersome, however,and so we shall not pursue it further here. 19.5.2 General boundary-value problems As mentioned above, often inhomogeneous boundary conditions can be dealt with by making an appropriate change of variables, such that the boundaryconditions in the new variables are homogeneous although the equation itself isgenerally inhomogeneous. In this section, however, we extend the use of Green’sfunctions to problems with inhomogeneous boundary conditions (and equations). This provides a more consistent and intuitive approach to the solution of such boundary-value problems . For deﬁniteness we shall consider Poisson’s equation ∇ 2u(r)=ρ(r), (19.83) but the material of this section may be extended to other linear PDEs of the form (19.79). Clearly, Poisson’s equation reduces to Laplace’s equation for ρ(r)=0a n d so our discussion is equally applicable to this case. We wish to solve (19.83) in some region Vbounded by a surface S, which may consist of several disconnected parts. As stated above, we shall allow the possibilitythat the boundary conditions on the solution u(r) may be inhomogeneous on S, although as we shall see this method reduces to those discussed above in thespecial case that the boundary conditions are in fact homogeneous. The two common types of inhomogeneous boundary condition for Poisson’s equation are (as discussed in subsection 18.6.2): (i) Dirichlet conditions, in which u(r)i ss p e c i ﬁ e do n S,a n d (ii) Neumann conditions, in which ∂u/∂n is speciﬁed on S. In general, specifying bothDirichlet andNeumann conditions on Soverdetermines the problem and leads to there being no solution. The speciﬁcation of the surface Srequires some further comment, since S may have several disconnected parts. If we wish to solve Poisson’s equation 688 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS ˆnˆn ˆnV VS S1 S2 (a) (b) Figure 19.11 Surfaces used for solving Poisson’s equation in diﬀerent regions V. inside some closed surface Sthen the situation is straightforward and is shown in ﬁgure 19.11( a). If, however, we wish to solve Poisson’s equation in the gap between two closed surfaces (for example in the gap between two concentricconducting cylinders) then the volume Vis bounded by a surface Sthat has two disconnected parts S 1andS2, as shown in ﬁgure 19.11( b); the direction of the normal to the surface is always taken as pointing outof the volume V. A similar situation arises when we wish to solve Poisson’s equation outside some closed surface S1. In this case the volume Vis inﬁnite but is treated formally by taking the surface S2as a large sphere of radius Rand letting Rtend to inﬁnity. In order to solve (19.83) subject to either Dirichlet or Neumann boundary conditions on S, we ﬁrst remind ourselves of Green’s second theorem, equation (11.20), which states that for two scalar functions φ(r)a n d ψ(r) deﬁned in some volume Vbounded by a surface S integraldisplay V(φ∇2ψ−ψ∇2φ)dV=integraldisplay S(φ∇ψ−ψ∇φ)·ˆndS, (19.84) where on the RHS it is common to write, for example, ∇ψ·ˆndSas (∂ψ/∂n )dS. The expression ∂ψ/∂n stands for∇ψ·ˆn, the rate of change of ψin the direction of the unit outward normal ˆnto the surface S. The Green’s function for Poisson’s equation (19.83) must satisfy ∇2G(r,r0)=δ(r−r0), (19.85) where r0lies in V. (As mentioned above, we may think of G(r,r0)a st h es o l u t i o n to Poisson’s equation for a unit-strength point source located at r=r0.) Let us for the moment impose no boundary conditions on G(r,r0). If we now let φ=u(r)a n d ψ=G(r,r0) in Green’s theorem (19.84) then we 689 PDES: SEPARATION OF VARIABLES AND OTHER METHODS obtain integraldisplay Vbracketleftbig u(r)∇2G(r,r0)−G(r,r0)∇2u(r)bracketrightbig dV(r) =integraldisplay Sbracketleftbigg u(r)∂G(r,r0) ∂n−G(r,r0)∂u(r) ∂nbracketrightbigg dS(r), where we have made explicit that the volume and surface integrals are with respect to r. Using (19.83) and (19.85) the LHS can be simpliﬁed to give integraldisplay V[u(r)δ(r−r0)−G(r,r0)ρ(r)]dV(r) =integraldisplay Sbracketleftbigg u(r)∂G(r,r0) ∂n−G(r,r0)∂u(r) ∂nbracketrightbigg dS(r),(19.86) Since r0lies within the volume V, integraldisplay Vu(r)δ(r−r0)dV(r)=u(r0), and thus rearranging (19.86) the solution to Poisson’s equation (19.83) can be written u(r0)=integraldisplay VG(r,r0)ρ(r)dV(r)+integraldisplay Sbracketleftbigg u(r)∂G(r,r0) ∂n−G(r,r0)∂u(r) ∂nbracketrightbigg dS(r). (19.87) Clearly, we can interchange the roles of randr0in (19.87) if we wish. (Remember also that, for a real Green’s function, G(r,r0)=G(r0,r).) Equation (19.87) is central to the extension of the Green’s function method to problems with inhomogeneous boundary conditions, and we next discuss itsapplication to both Dirichlet and Neumann boundary-value problems. But, beforedoing so, we also note that if the boundary condition on Sis in fact homogeneous, so that u(r)=0o r ∂u(r)/∂n=0o n S, then demanding that the Green’s function G(r,r 0) also obeys the same boundary condition causes the surface integral in (19.87) to vanish, and we are left with the familiar form of solution given in (19.81). The extension of (19.87) to a PDE other than Poisson’s equation isdiscussed in exercise 19.30. 19.5.3 Dirichlet problems In a Dirichlet problem we require the solution u(r) of Poisson’s equation (19.83) to take speciﬁc values on some surface Sthat bounds V,i . e .w er e q u i r et h a t u(r)=f(r)o nSwhere fis a given function. If we seek a Green’s function G(r,r 0) for this problem it must clearly satisfy (19.85), but we are free to choose the boundary conditions satisﬁed by G(r,r0)i n 690 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS such a way as to make the solution (19.87) as simple as possible. From (19.87), we see that by choosing G(r,r0)=0 f o r ronS (19.88) the second term in the surface integral vanishes. Since u(r)=f(r)o n S, (19.87) then becomes u(r0)=integraldisplay VG(r,r0)ρ(r)dV(r)+integraldisplay Sf(r)∂G(r,r0) ∂ndS(r). (19.89) Thus we wish to ﬁnd the Dirichlet Green’s function that (i) satisﬁes (19.85) and hence is singular at r=r0,a n d (ii) obeys the boundary condition G(r,r0)=0f o r ronS. In general, it is, diﬃcult to obtain this function directly, and so it is useful to separate these two requirements. We therefore look for a solution of the form G(r,r0)=F(r,r0)+H(r,r0), where F(r,r0) satisﬁes (19.85) and has the required singular character at r=r0but does not necessarily obey the boundary condition on S, whilst H(r,r0) satisﬁes the corresponding homogeneous equation (i.e. Laplace’s equation) inside Vbut is adjusted in such a way that the sum G(r,r0) equals zero on S. The Green’s function G(r,r0) is still a solution of (19.85) since ∇2G(r,r0)=∇2F(r,r0)+∇2H(r,r0)=∇2F(r,r0)+0= δ(r−r0). The function F(r,r0) is called the fundamental solution and will clearly take diﬀerent forms depending on the dimensionality of the problem. Let us ﬁrst consider the fundamental solution to (19.85) in three dimensions.IFind the fundamental solution to Poisson’s equation in three dimensions that tends to zero as|r|→∞ . We wish to solve ∇2F(r,r0)=δ(r−r0) (19.90) in three dimensions, subject to the boundary condition F(r,r0)→0a s|r|→∞ .S i n c et h e problem is spherically symmetric about r0, let us consider a large sphere Sof radius R centred on r0, and integrate (19.90) over the enclosed volume V. We then obtainZ V∇2F(r,r0)dV= Z Vδ(r−r0)dV=1, (19.91) since Vencloses the point r0. However, using the divergence theorem,Z V∇2F(r,r0)dV= Z S∇F(r,r0)·ˆndS, (19.92) where ˆnis the unit normal to the large sphere Sat any point. 691 PDES: SEPARATION OF VARIABLES AND OTHER METHODS Since the problem is spherically symmetric about r0, we expect that F(r,r0)=F(|r−r0|)=F(r), i.e.Fhas the same value everywhere on S. Thus, evaluating the surface integral in (19.92) and equating it to unity from (19.91), we have † 4πr2dF dr //// r=R=1. Integrating this expression we obtain F(r)=−1 4πr+c o n s t a n t , but, since we require F(r,r0)→0a s|r|→∞ , the constant must be zero. The fundamental solution in three dimensions is consequently given by F(r,r0)=−1 4π|r−r0|. (19.93) This is clearly also the full Green’s function for Poisson’s equation subject to the boundary condition u(r)→0a s|r|→∞ . J Using (19.93) we can write down the solution of Poisson’s equation to ﬁnd, for example, the electrostatic potential u(r) due to some distribution of electric charge ρ(r). The electrostatic potential satisﬁes ∇2u(r)=−ρ /epsilon10, where u(r)→0a s|r|→∞ . Since the boundary condition on the surface at inﬁnity is homogeneous the surface integral in (19.89) vanishes, and using (19.93)we recover the familiar solution u(r 0)=integraldisplayρ(r) 4π/epsilon10|r−r0|dV(r), (19.94) where the volume integral is over all space. We can develop an analogous theory in two dimensions. As before the funda- mental solution satisﬁes ∇2F(r,r0)=δ(r−r0), (19.95) where δ(r−r0) is now the two-dimensional delta function. Following an analogous method to that used in the previous example, we ﬁnd the fundamental solutionin two dimensions to be given by F(r,r 0)=1 2πln|r−r0|+c o n s t a n t . (19.96) †A vertical bar to the right of an expression is a common alternative notation to enclosing the expression in square brackets; as usual, the subscript shows the value of the variable at which the expression is to be evaluated. 692 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS From the form of the solution we see that in two dimensions we cannot apply the condition F(r,r0)→0a s|r|→∞ , and in this case the constant does not necessarily vanish. We now return to the task of constructing the full Dirichlet Green’s function. To do so we wish to add to the fundamental solution a solution of the homogeneous equation (in this case Laplace’s equation) such that G(r,r0)=0o n S,a sr e q u i r e d by (19.89) and its attendant conditions. The appropriate Green’s function isconstructed by adding to the fundamental solution ‘copies’ of itself that represent‘image’ sources at diﬀerent locations outside V. Hence this approach is called the method of images . In summary, if we wish to solve Poisson’s equation in some region Vsubject to Dirichlet boundary conditions on its surface Sthen the procedure and argument are as follows. (i) To the single source δ(r−r 0) inside Vadd image sources outside V Nsummationdisplay n=1qnδ(r−rn) with rnoutside V, where the positions rnand the strengths qnof the image sources are to be determined as described in step (iii) below. (ii) Since all the image sources lie outside V, the fundamental solution cor- responding to each source satisﬁes Laplace’s equation inside V. Thus we may add the fundamental solutions F(r,rn) corresponding to each image source to that corresponding to the single source inside V, obtaining the Green’s function G(r,r0)=F(r,r0)+Nsummationdisplay n=1qnF(r,rn). (iii) Now adjust the positions rnand strengths qnof the image sources so that the required boundary conditions are satisﬁed on S. For a Dirichlet Green’s function we require G(r,r0)=0f o r ronS. (iv) The solution to Poisson’s equation subject to the Dirichlet boundary condition u(r)=f(r)o n Sis then given by (19.89). In general it is very diﬃcult to ﬁnd the correct positions and strengths for the images, i.e. to make them such that the boundary conditions on Sare satisﬁed. Nevertheless, it is possible to do so for certain problems that have simple geometry.In particular, for problems in which the boundary Sconsists of straight lines (in two dimensions) or planes (in three dimensions), positions of the image points can be deduced simply by imagining the boundary lines or planes to be mirrorsin which the single source in V(atr 0) is reﬂected. 693 PDES: SEPARATION OF VARIABLES AND OTHER METHODS yz xVr0 r1+ − Figure 19.12 The arrangement of images for solving Laplace’s equation in the half-space z>0.ISolve Laplace’s equation ∇2u=0in three dimensions in the half-space z>0, given that u(r)=f(r)on the plane z=0. The surface Sbounding Vconsists of the xy-plane and the surface at inﬁnity. Therefore, the Dirichlet Green’s function for this problem must satisfy G(r,r0)=0o n z=0a n d G(r,r0)→0a s|r|→∞ . Thus it is clear in this case that we require one image source at a position r1that is the reﬂection of r0in the plane z= 0, as shown in ﬁgure 19.12 (so that r1lies in z<0, outside the region in which we wish to obtain a solution). It is also clear that the strength of this image should be −1. Therefore by adding the fundamental solutions corresponding to the original source and its image we obtain the Green’s function G(r,r0)=−1 4π|r−r0|+1 4π|r−r1|, (19.97) where r1is the reﬂection of r0in the plane z=0 ,i . e .i f r0=(x0,y0,z0)t h e n r1=(x0,y0,−z0). Clearly G(r,r0)→0a s|r|→∞ as required. Also G(r,r0)=0o n z= 0, and so (19.97) is the desired Dirichlet Green’s function. The solution to Laplace’s equation is then given by (19.89) with ρ(r)=0 , u(r0)= Z Sf(r)∂G(r,r0) ∂ndS(r). (19.98) Clearly the surface at inﬁnity makes no contribution to this integral. The outward-pointing unit vector normal to the xy-plane is simply ˆn=−k(where kis the unit vector in the z-direction), and so ∂G(r,r0) ∂n=−∂G(r,r0) ∂z=−k·∇G(r,r0). We may evaluate this normal derivative by writing the Green’s function (19.97) explicitly in terms of x,yandz(and x0,y0andz0) and calculating the partial derivative with respect 694 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS tozdirectly. It is usually quicker, however, to use the fact that † ∇|r−r0|=r−r0 |r−r0|; (19.99) thus ∇G(r,r0)=r−r0 4π|r−r0|3−r−r1 4π|r−r1|3. Since r0=(x0,y0,z0)a n d r1=(x0,y0,−z0) the normal derivative is given by −∂G(r,r0) ∂z=−k·∇G(r,r0) =−z−z0 4π|r−r0|3+z+z0 4π|r−r1|3. Therefore on the surface z= 0, writing out the dependence on x,yandzexplicitly, we have −∂G(r,r0) ∂z //// z=0=2z0 4π[(x−x0)2+(y−y0)2+z2 0]3/2. Inserting this expression into (19.98) we obtain the solution u(x0,y0,z0)=z0 2π Z∞ −∞ Z∞ −∞f(x, y) [(x−x0)2+(y−y0)2+z2 0]3/2dx dy. J An analogous procedure may be applied in two-dimensional problems. For example, in solving Poisson’s equation in two dimensions in the half-space x>0 we again require just one image charge, of strength q1=−1, at a position r1that is the reﬂection of r0in the line x= 0. Since we require G(r,r0)=0w h e n rlies onx= 0, the constant in (19.96) must equal zero, and so the Dirichlet Green’s function is G(r,r0)=1 2πparenleftbig ln|r−r0|−ln|r−r1|parenrightbig . Clearly G(r,r0) tends to zero as |r|→∞ . If, however, we wish to solve the two- dimensional Poisson equation in the quarter space x>0,y>0, then more image points are required. †Since|r−r0|2=(r−r0)·(r−r0) we have∇|r−r0|2=2 (r−r0),from which we obtain ∇(|r−r0|2)1/2=1 22(r−r0) (|r−r0|2)1/2=r−r0 |r−r0|. Note that this result holds in two andthree dimensions. 695 PDES: SEPARATION OF VARIABLES AND OTHER METHODS −λ−λ +λ+λ x0 y0r0 r1 r2r3 C xy V Figure 19.13 The arrangement of images for ﬁnding the force on a line charge situated in the (two-dimensional) quarter-space x>0,y>0, when the planes x=0a n d y= 0 are earthed.IA line charge in the z-direction of charge density λis placed at some position r0in the quarter-space x>0,y>0. Calculate the force per unit length on the line charge due to the presence of thin earthed plates along x=0andy=0. Here we wish to solve Poisson’s equation ∇2u=−λ /epsilon10δ(r−r0) in the quarter space x>0,y>0. It is clear that we require three image line charges with positions and strengths as shown in ﬁgure 19.13 (all of which lie outside the regionin which we seek a solution). The boundary condition that the electrostatic potential uis zero on x=0a n d y= 0 (shown as the ‘curve’ Cin ﬁgure 19.13) is then automatically satisﬁed, and so this system of image charges is directly equivalent to the original situationof a single line charge in the presence of the earthed plates along x=0a n d y= 0. Thus the electrostatic potential is simply equal to the Dirichlet Green’s function u(r)=G(r,r 0)=−λ 2π/epsilon10 /; ln|r−r0|−ln|r−r1|+l n|r−r2|−ln|r−r3| / , which equals zero on Cand on the ‘surface’ at inﬁnity. The force on the line charge at r0, therefore, is simply that due to the three line charges atr1,r2andr3. The elecrostatic potential due to a line charge at ri,i= 1, 2 or 3, is given by the fundamental solution ui(r)=∓λ 2π/epsilon10ln|r−ri|+c, the upper or lower sign being taken according to whether the line charge is positive or negative respectively. Therefore the force per unit length on the line charge at r0, due to the one at ri,i sg i v e nb y −λ∇ui(r) //// r=r0=±λ2 2π/epsilon10r0−ri |r0−ri|2. 696 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS Adding the contributions from the three image charges shown in ﬁgure 19.13, the total force experienced by the line charge at r0is F=λ2 2π/epsilon10 / −r0−r1 |r0−r1|2+r0−r2 |r0−r2|2−r0−r3 |r0−r3|2 / , where, from the ﬁgure, r0−r1=2y0j,r0−r2=2x0i+2y0jandr0−r3=2x0i. Thus, in terms of x0andy0, the total force on the line charge due to the charge induced on the plates is given by F=λ2 2π/epsilon10 / −1 2y0j+2x0i+2y0j 4x2 0+4y2 0−1 2x0i / =−λ2 4π/epsilon10(x2 0+y2 0) /y2 0 x0i+x2 0 y0j / . J Further generalisations are possible. For instance, solving Poisson’s equation in the two-dimensional strip −∞<x<∞,0<y<b requires an inﬁnite series of image points. So far we have considered problems in which the boundary Sconsists of straight lines (in two dimensions) or planes (in three dimensions), in which simple reﬂection of the source at r0in these boundaries ﬁxes the positions of the image points. For more complicated (curved) boundaries this is no longer possible, andﬁnding the appropriate position(s) and strength(s) of the image source(s) requiresfurther work.IUse the method of images to ﬁnd the Dirichlet Green’s function for solving Poisson’s equation outside a sphere of radius acentred at the origin. We need to ﬁnd a solution of Poisson’s equation valid outside the sphere of radius a. Since an image point r1cannot lie in this region, it must be located within the sphere. The Green’s function for this problem is therefore G(r,r0)=−1 4π|r−r0|−q 4π|r−r1|, where|r0|>a,|r1|<aandqis the strength of the image which we have yet to determine. Clearly, G(r,r0)→0 on the surface at inﬁnity. By symmetry we expect the image point r1to lie on the same radial line as the original source, r0, as shown in ﬁgure 19.14, and so r1=kr0where k<1. However, for a Dirichlet Green’s function we require G(r−r0)=0o n|r|=a, and the form of the Green’s function suggests that we need |r−r0|∝|r−r1|for all|r|=a. (19.100) Referring to ﬁgure 19.14, if this relationship is to hold over the whole surface of the sphere, then it must certainly hold for the points AandB. We thus require |r0|−a a−|r1|=|r0|+a a+|r1|, which reduces to |r1|=a2/|r0|. Therefore the image point must be located at the position r1=a2 |r0|2r0. 697 PDES: SEPARATION OF VARIABLES AND OTHER METHODS yz a −aBAV x−a |r0|r1r0 +1 Figure 19.14 The arrangement of images for solving Poisson’s equation outside a sphere of radius acentred at the origin. For a charge +1 at r0,t h e image point r1is given by ( a/|r0|)2r0and the strength of the image charge is −a/|r0|. It may now be checked that, for this location of the image point, (19.100) is satisﬁed over the whole sphere. Using the geometrical result |r−r1|2=|r|2−2a2 |r0|2r·r0+a4 |r0|2 =a2 |r0|2 /; |r0|2−2r·r0+a2 / for|r|=a, (19.101) we see that, on the surface of the sphere, |r−r1|=a |r0||r−r0|for|r|=a. (19.102) Therefore, in order that G=0a t|r|=a, the strength of the image charge must be −a/|r0|. Consequently, the Dirichlet Green’s function for the exterior of the sphere is G(r,r0)=−1 4π|r−r0|+a/|r0| 4π|r−(a2/|r0|2)r0|. For a less formal treatment of the same problem see exercise 19.24. J If we seek solutions to Poisson’s equation in the interior of a sphere then the above analysis still holds, but randr0are now inside the sphere and the image r1lies outside it. For two-dimensional Dirichlet problems outside the circle |r|=a, we are led by arguments similar to those employed previously to use the same image pointas in the three-dimensional case, namely r 1=a2 |r0|2r0. (19.103) 698 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS As illustrated below, however, it is usually necessary to take the image strength as−1 in two-dimensional problems.ISolve Laplace’s equation in the two-dimensional region |r|≤a, subject to the boundary condition u=f(φ)on|r|=a. In this case we wish to ﬁnd the Dirichlet Green’s function in the interior of a disc of radius a, so the image charge must lie outside the disc. Taking the strength of the image to be−1, we have G(r,r0)=1 2πln|r−r0|−1 2πln|r−r1|+c, where r1=(a2/|r0|2)r0lies outside the disc, and cis a constant that includes the strength of the image charge and does not necessarily equal zero. Since we require G(r,r0)=0w h e n |r|=a, the value of the constant cis determined, and the Dirichlet Green’s function for this problem is given by G(r,r0)=1 2π / ln|r−r0|−ln /// / r−a2 |r0|2r0 /// / −ln|r0| a / . (19.104) Using plane polar coordinates, the solution to the boundary-value problem can be written as a line integral around the circle ρ=a: u(r0)= Z Cf(r)∂G(r,r0) ∂ndl = Z2π 0f(r)∂G(r,r0) ∂ρ //// ρ=aad φ . (19.105) The normal derivative of the Green’s function (19.104) is given by ∂G(r,r0) ∂ρ=r |r|·∇G(r,r0) =r 2π|r|· /r−r0 |r−r0|2−r−r1 |r−r1|2 / . (19.106) Using the fact that r1=(a2/|r0|2)r0and the geometrical result (19.102), we ﬁnd that ∂G(r,r0) ∂ρ //// ρ=a=a2−|r0|2 2πa|r−r0|2. In plane polar coordinates, r=ρcosφi+ρsinφjandr0=ρ0cosφ0i+ρ0sinφ0j,a n d so ∂G(r,r0) ∂ρ //// ρ=a= /1 2πa /a2−ρ2 0 a2+ρ2 0−2aρ0cos(φ−φ0). On substituting into (19.105), we obtain u(ρ0,φ0)=1 2π Z2π 0(a2−ρ2 0)f(φ)dφ a2+ρ2 0−2aρ0cos(φ−φ0), (19.107) which is the solution to the problem. J 699 PDES: SEPARATION OF VARIABLES AND OTHER METHODS 19.5.4 Neumann problems In a Neumann problem we require the normal derivative of the solution of Poisson’s equation to take on speciﬁc values on some surface Sthat bounds V, i.e. we require ∂u(r)/∂n=f(r)o nS,w h e r e fis a given function. As we shall see, much of our discussion of Dirichlet problems can be immediately taken over intothe solution of Neumann problems. As we proved in section 18.7 of the previous chapter, specifying Neumann boundary conditions determines the relevant solution of Poisson’s equation towithin an (unimportant) additive constant. Unlike Dirichlet conditions, Neumannconditions impose a self-consistency requirement. In order for a solution uto exist, it is necessary that the following consistency condition holds: integraldisplay Sfd S=integraldisplay S∇u·ˆndS=integraldisplay V∇2ud V=integraldisplay Vρd V, (19.108) where we have used the divergence theorem to convert the surface integral into a volume integral. As a physical example, the integral of the normal componentof an electric ﬁeld over a surface bounding a given volume cannot be chosenarbitrarily when the charge inside the volume has already been speciﬁed (Gauss’stheorem). Let us again consider (19.87), which is central to our discussion of Green’s functions in inhomogeneous problems. It reads u(r 0)=integraldisplay VG(r,r0)ρ(r)dV(r)+integraldisplay Sbracketleftbigg u(r)∂G(r,r0) ∂n−G(r,r0)∂u(r) ∂nbracketrightbigg dS(r). As always, the Green’s function must obey ∇2G(r,r0)=δ(r−r0), where r0lies in V. In the solution of Dirichlet problems in the previous subsection, we chose the Green’s function to obey the boundary condition G(r,r0)=0o n S and, in a similar way, we might wish to choose ∂G(r,r0)/∂n= 0 in the solution of Neumann problems. However, in general this is notpermitted since the Green’s function must obey the consistency condition integraldisplay S∂G(r,r0) ∂ndS=integraldisplay S∇G(r,r0)·ˆndS=integraldisplay V∇2G(r,r0)dV=1. The simplest permitted boundary condition is therefore ∂G(r,r0) ∂n=1 AforronS, where Ais the area of the surface S; this deﬁnes a Neumann Green’s function . If we require ∂u(r)/∂n=f(r)o n S, the solution to Poisson’s equation is given 700 19.5 INHOMOGENEOUS PROBLEMS – GREEN’S FUNCTIONS by u(r0)=integraldisplay VG(r,r0)ρ(r)dV(r)+1 Aintegraldisplay Su(r)dS(r)−integraldisplay SG(r,r0)f(r)dS(r) =integraldisplay VG(r,r0)ρ(r)dV(r)+/angbracketleftu(r)/angbracketrightS−integraldisplay SG(r,r0)f(r)dS(r), (19.109) where/angbracketleftu(r)/angbracketrightSis the average of uover the surface Sand is a freely speciﬁable constant. For Neumann problems in which the volume Vis bounded by a surface Sat inﬁnity, we do not need the /angbracketleftu(r)/angbracketrightSterm. For example, if we wish to solve a Neumann problem outside the unit sphere centred at the origin then r>a is the region Vthroughout which we require the solution; this region may be considered as being bounded by two disconnected surfaces, the surface of thesphere and a surface at inﬁnity. By requiring that u(r)→0a s|r|→∞ ,t h et e r m /angbracketleftu(r)/angbracketright Sbecomes zero. As mentioned above, much of our discussion of Dirichlet problems can be taken over into the solution of Neumann problems. In particular, we may use the method of images to ﬁnd the appropriate Neumann Green’s function.ISolve Laplace’s equation in the two-dimensional region |r|≤asubject to the boundary condition ∂u/∂n =f(φ)on|r|=a,w i t h R2π 0f(φ)dφ=0as required by the consistency condition (19.108). Let us assume, as in Dirichlet problems with this geometry, that a single image charge isplaced outside the circle at r 1=a2 |r0|2r0, where r0is the position of the source inside the circle (see equation (19.103)). Then, from (19.102), we have the useful geometrical result |r−r1|=a |r0||r−r0|for|r|=a. (19.110) Leaving the strength qof the image as a parameter, the Green’s function has the form G(r,r0)=1 2π /; ln|r−r0|+qln|r−r1|+c / . (19.111) Using plane polar coordinates, the radial (i.e. normal) derivative of this function is given by ∂G(r,r0) ∂ρ=r |r|·∇G(r,r0) =r 2π|r|· /r−r0 |r−r0|2+q(r−r1) |r−r1|2 / . Using (19.110), on the circumference of the circle ρ=athe radial derivative is ∂G(r,r0) ∂ρ //// ρ=a=1 2π|r| /|r|2−r·r0 |r−r0|2+q|r|2−q(a2/|r0|2)r·r0 (a2/|r0|2)|r−r0|2 / =1 2πa1 |r−r0|2 / |r|2+q|r0|2−(1 +q)r·r0 / , 701 PDES: SEPARATION OF VARIABLES AND OTHER METHODS where we have set |r|2=a2in the second term on the RHS, but not in the ﬁrst. If we take q= 1, the radial derivative simpliﬁes to ∂G(r,r0) ∂ρ //// ρ=a=1 2πa, or 1/Lwhere Lis the length of the circumference, and so (19.111) with q=1i st h e required Neumann Green’s function. Since ρ(r) = 0, the solution to our boundary-value problem is now given by (19.109) as u(r0)=/angbracketleftu(r)/angbracketrightC− Z CG(r,r0)f(r)dl(r), where the integral is around the circumference of the circle C. In plane polar coordinates r=ρcosφi+ρsinφjandr0=ρ0cosφ0i+ρ0sinφ0j, and again using (19.110) we ﬁnd that on Cthe Green’s function is given by G(r,r0)|ρ=a=1 2π / ln|r−r0|+l n /a |r0||r−r0| / +c / =1 2π / ln|r−r0|2+l na |r0|+c / =1 2π / ln / a2+ρ2 0−2aρ0cos(φ−φ0) / +l na ρ0+c / . (19.112) Since dl=ad φonC, the solution to the problem is given by u(ρ0,φ0)=/angbracketleftu/angbracketrightC−a 2π Z2π 0f(φ)ln[a2+ρ2 0−2aρ0cos(φ−φ0)]dφ. The contributions of the ﬁnal two terms terms in the Green’s function (19.112) vanish because R2π 0f(φ)dφ= 0. The average value of uaround the circumference, /angbracketleftu/angbracketrightC, is a freely speciﬁable constant as we would expect for a Neumann problem. This result should becompared with the result (19.107) for the corresponding Dirichlet problem, but it shouldbe remembered that in the one case f(φ) is a potential, and in the other the gradient of a potential.J 19.6 Exercises 19.1 Solve the following ﬁrst-order partial diﬀerential equations by separating the variables: (a)∂u ∂x−x∂u ∂y=0 ; ( b ) x∂u ∂x−2y∂u ∂y=0. 19.2 A conducting cube has as its six faces the planes x=±a,y=±aandz=±a, and contains no internal heat sources. Verify that the temperature distribution u(x, y, z, t )=Acosπx asinπz aexp / −2κπ2t a2 / obeys the appropriate diﬀusion equation. Across which faces is there heat ﬂow? What is the direction and rate of heat ﬂow at the point (3 a/4,a /4,a)a tt i m e t=a2/(κπ2)? 19.3 The wave equation describing the transverse vibrations of a stretched membrane under tension Tand having a uniform surface density ρis T /∂2u ∂x2+∂2u ∂y2 / =ρ∂2u ∂t2. 702 19.6 EXERCISES Find a separable solution appropriate to a membrane stretched on a frame of length aand width b, showing that the natural angular frequencies of such a membrane are ω2=π2T ρ /n2 a2+m2 b2 / , where nandmare any positive integers. 19.4 Schr ¨odinger’s equation for a non-relativistic particle in a constant potential region can be taken as − /~2 2m /∂2u ∂x2+∂2u ∂y2+∂2u ∂z2 / =i /~∂u ∂t. (a) Find a solution, separable in the four independent variables, that can be written in the form of a plane wave, ψ(x, y, z, t )=Aexp[i(k·r−ωt)]. Using the relationships associated with de Broglie ( p= /~k) and Einstein (E= /~ω), show that the separation constants must be such that p2 x+p2 y+p2 z=2mE. (b) Obtain a diﬀerent separable solution describing a particle conﬁned to a box of side a(ψmust vanish at the walls of the box). Show that the energy of the particle can only take the quantised values E= /~2π2 2ma2(n2 x+n2 y+n2 z), where nx,ny,nzare integers. 19.5 Denoting the three terms of ∇2in spherical polars by ∇2 r,∇2 θ,∇2 φin an obvious way, evaluate ∇2 ru, etc. for the two functions given below and verify that, in each case, although the individual terms are not necessarily zero their sum ∇2uis zero. Identify the corresponding values of /lscriptandm. (a)u(r,θ,φ)= / Ar2+B r3 /3c os2θ−1 2. (b)u(r,θ,φ)= / Ar+B r2 / sinθexpiφ. 19.6 Prove that the expression given in equation (19.47) for the associated Legendre function Pm /lscript(µ) satisﬁes the appropriate equation, (19.45), as follows. (a) Evaluate dPm /lscript(µ)/dµandd2Pm /lscript(µ)/dµ2using the forms given in (19.47) and substitute them into (19.45). (b) Diﬀerentiate Legendre’s equation mtimes using Leibniz’ theorem. (c) Show that the equations obtained in (a) and (b) are multiples of each other, and hence that the validity of (b) implies that of (a). 19.7 Use the expressions at the end of subsection 19.3.2 to verify for /lscript=0,1,2t h a t /lscriptX m=−/lscript|Ym /lscript(θ,φ)|2=2/lscript+1 4π and so is independent of the values of θandφ.T h i si st r u ef o ra n y /lscript, but a general proof is more involved. This result helps to reconcile intuition withthe apparently arbitrary choice of polar axis in a general quantum mechanicalsystem. 703 PDES: SEPARATION OF VARIABLES AND OTHER METHODS 19.8 Express the function f(θ,φ)=s i n θ[sin2(θ/2)cos φ+icos2(θ/2)sin φ]+s i n2(θ/2) as a sum of spherical harmonics. 19.9 Continue the analysis of exercise 10.20, concerned with the ﬂow of a very viscous ﬂuid past a sphere, to ﬁnd the full expression for the stream function ψ(r,θ). At the surface of the sphere r=athe velocity ﬁeld u=0, whilst far from the sphere ψ/similarequal(Ur2sin2θ)/2. Show that f(r) can be expressed as a superposition of powers of r,a n d determine which powers give acceptable solutions. Hence show that ψ(r,θ)=U 4 / 2r2−3ar+a3 r / sin2θ. 19.10 The motion of a very viscous ﬂuid in the two-dimensional (wedge) region −α< φ<α can be described in ( ρ, φ) coordinates by the (biharmonic) equation ∇2∇2ψ≡∇4ψ=0, together with the boundary conditions ∂ψ/∂φ =0a t φ=±α, which represents the fact that there is no radial ﬂuid velocity close to either of the boundingwalls because of the viscosity, and ∂ψ/∂ρ =±ρatφ=±α, which imposes the condition that azimuthal ﬂow increases linearly with ralong any radial line. Assuming a solution in separated-variable form, show that the full expression forψis ψ(ρ, φ)=ρ 2 2sin 2φ−2φcos 2α sin 2α−2αcos 2α. 19.11 A circular disk of radius ais such a way that its perimeter ρ=ais maintained with a temperature distribution A+Bcos2φ,w h e r e ρandφare plane polar coordinates and AandBare constants. Find the temperature T(ρ, φ) everywhere in the region ρ<a . 19.12 (a) Find the form of the solution of Laplace’s equation in plane polar coordinates ρ, φthat takes the value +1 for 0 <φ<π and the value −1f o r−π<φ< 0, when ρ=a. (b) For a point ( x, y) on or inside the circle x2+y2=a2, identify the angles α andβdeﬁned by α=t a n−1y a+xand β=t a n−1y a−x. Show that u(x, y)=( 2 /π)(α+β) is a solution of Laplace’s equation that satisﬁes the boundary conditions given in (a). (c) Deduce a Fourier series expansion for the function tan−1sinφ 1+c o s φ+t a n−1sinφ 1−cosφ. 19.13 The free transverse vibrations of a thick rod satisfy the equation a4∂4u ∂x4+∂2u ∂t2=0. Obtain a solution in separated-variable form and, for a rod clamped at one end, x= 0, and free at the other, x=L, show that the angular frequency of vibration ωsatisﬁes cosh /ω1/2L a / =−sec /ω1/2L a / . 704 19.6 EXERCISES (At a clamped end both uand∂u/∂x vanish, whilst at a free end, where there is no bending moment, ∂2u/∂x2and∂3u/∂x3are both zero.) 19.14 A membrane is stretched between two concentric rings of radii aandb(b>a). If the smaller ring is transversely distorted from the planar conﬁguration by anamount c|φ|,−π≤φ≤π, show that the membrane then has a shape given by u(ρ, φ)=cπ 2ln(b/ρ) ln(b/a)−4c π X moddam m2(b2m−a2m) /b2m ρm−ρm / cosmφ. 19.15 A string of length L, ﬁxed at its two ends, is plucked at its mid-point by an amount Aand then released. Prove that the subsequent displacement is given by u(x, t)=∞X n=08A π2(2n+1 )2sin /(2n+1 )πx L / cos /(2n+1 )πct L / , where, in the usual notation, c2=T/ρ. Find the total kinetic energy of the string when it passes through its unplucked position, by calculating it in each mode (each n) and summing, using the result ∞X 01 (2n+1 )2=π2 8. Conﬁrm that the total energy is equal to the work done in plucking the string initially. 19.16 Prove that the potential for ρ<a associated with a vertical split cylinder of radius a, the two halves of which (cos φ>0a n dc o s φ<0) are maintained at equal and opposite potentials ±V,i sg i v e nb y u(ρ, φ)=4V π∞X n=0(−1)n 2n+1 /ρ a /2n+1 cos(2 n+1 )φ. 19.17 A conducting spherical shell of radius ais cut round its equator and the two halves connected to voltages of + Vand−V. Show that an expression for the potential at the point ( r,θ,φ) anywhere inside the two hemispheres is u(r,θ,φ)=V∞X n=0(−1)n(2n)!(4n+3 ) 22n+1n!(n+1 ) ! /r a /2n+1 P2n+1(cosθ). (This is the spherical polar analogue of the previous question.) 19.18 A slice of biological material of thickness Lis placed into a solution of a radioactive isotope of constant concentration C0at time t=0 .F o ral a t e rt i m e t ﬁnd the concentration of radioactive ions at a depth xinside one of its surfaces if the diﬀusion constant is κ. 19.19 Two identical copper bars are each of length a. Initially, one is at 0◦Ca n dt h e other at 100◦C; they are then joined together end to end and thermally isolated. Obtain in the form of a Fourier series an expression u(x, t) for the temperature at any point a distance xfrom the join at a later time t.( B e a ri nm i n dt h eh e a t ﬂow conditions at the free ends of the bars.) Taking a=0.5m estimate the time it takes for one of the free ends to attain a temperature of 55◦C. The thermal conductivity of copper is 3 .8× 102Jm−1K−1s−1, and its speciﬁc heat capacity is 3 .4×106Jm−3K−1. 19.20 A sphere of radius aand thermal conductivity k1is surrounded by an inﬁnite medium of conductivity k2in which, far away, the temperature tends to T∞. A distribution of heat sources q(θ) embedded in the sphere’s surface establish steady temperature ﬁelds T1(r,θ) inside the sphere and T2(r,θ) outside it. It can 705 PDES: SEPARATION OF VARIABLES AND OTHER METHODS be shown, by considering the heat ﬂow through a small volume that includes part of the sphere’s surface, that k1∂T1 ∂r−k2∂T2 ∂r=q(θ)o n r=a. Given that q(θ)=1 a∞X n=0qnPn(cosθ), ﬁnd complete expressions for T1(r,θ)a n d T2(r,θ). What is the temperature at the centre of the sphere? 19.21 Using result (19.77) from the worked example in the text, ﬁnd the general expression for the temperature u(x, t) in the bar, given that the temperature distribution at time t=0i s u(x,0) = exp(−x2/a2). 19.22 (a) Show that the gravitational potential due to a uniform disc of radius aand mass M, centred at the origin, is given for r<a by 2GM a / 1−r aP1(cosθ)+1 2 /r a /2 P2(cosθ)−1 8 /r a /4 P4(cosθ)+··· / , and for r>a by GM r / 1−1 4 /a r /2 P2(cosθ)+1 8 /a r /4 P4(cosθ)−··· / , where the polar axis is normal to the plane of the disc. (b) Reconcile the presence of a term P1(cosθ), which is odd under θ→π−θ, with the symmetry with respect to the plane of the disc of the physicalsystem. (c) Deduce that the gravitational ﬁeld near an inﬁnite sheet of matter of constant density ρper unit area is 2 πGρ. 19.23 In the region −∞<x,y<∞and−t≤z≤t, a charge-density wave ρ(r)= Acosqx,i nt h e x-direction, is represented by ρ(r)=e iqx √ 2π Z∞ −∞˜ρ(α)eiαzdα. The resulting potential is represented by V(r)=eiqx √ 2π Z∞ −∞˜V(α)eiαzdα. Determine the relationship between ˜V(α)a n d ˜ρ(α), and hence show that the potential at the point ( x,0,0) is A π Z∞ −∞sinkt k(k2+q2)dk. 19.24 Point charges qand−qa/b(with a<b) are placed respectively at a point P,a distance bfrom the origin O, and a point Qbetween OandP,ad i s t a n c e a2/b from O. Show, by considering similar triangles QOSandSOP,w h e r e Sis any point on the surface of the sphere centred at Oand of radius a, that the net potential anywhere on the sphere due to the two charges is zero. Use this result (backed up by the uniqueness theorem) to ﬁnd the force with which a point charge qplaced a distance bfrom the centre of a spherical conductor of radius a(<b) is attracted to the sphere (i) if the sphere is earthed, and (ii) if the sphere is uncharged and insulated. 706 19.6 EXERCISES 19.25 Find the Green’s function G(r,r0) in the half-space z>0 for the solution of ∇2Φ = 0 with Φ speciﬁed in cylindrical polar coordinates ( ρ, φ, z) on the plane z=0b y Φ(ρ, φ, z)= /( 1f o r ρ≤1, 1/ρforρ>1. Determine the variation of Φ(0 ,0,z)a l o n gt h e z-axis. 19.26 Electrostatic charge is distributed in a sphere of radius Rcentred on the origin. Determine the form of the resultant potential φ(r) at distances much greater than R, as follows. (a) express in the form of an integral over all space the solution of ∇2φ=−ρ(r) /epsilon10; (b) show that, for r/greatermuchr/prime, |r−r/prime|=r−r·r/prime r+O /1 r / . (c) use results (a) and (b) to show that φ(r)h a st h ef o r m φ(r)=M r+d·r r3+O /1 r3 / ; Find expressions for Mandd, and identify them physically. 19.27 Find, in the form of an inﬁnite series the Green’s function of the ∇2operator for the Dirichlet problem in the region −∞<x<∞,−∞<y<∞,−c≤z≤c. 19.28 Find the Green’s function for the three-dimensional Neumann problem ∇2φ=0 f o r z>0a n d∂φ ∂z=f(x, y)o n z=0. Determine φ(x, y, z)i f f(x, y)= /( δ(y)f o r|x|<a , 0f o r|x|≥a. 19.29 (a) By applying the divergence theorem to the volume integralZ V / φ(∇2−m2)ψ−ψ(∇2−m2)φ / dV obtain a Green’s function expression, as the sum of a volume integral and a surface integral, for φ(r/prime)t h a ts a t i s ﬁ e s ∇2φ−m2φ=ρ inVand takes the speciﬁed form φ=fonS, the boundary of V.T h e Green’s function G(r,r/prime)t ob eu s e ds a t i s ﬁ e s ∇2G−m2G=δ(r−r/prime) and vanishes when ris on S. (b) When Vis all space, G(r,r/prime) can be written as G(t)=g(t)/twhere t=|r−r/prime| andg(t) is bounded as t→∞. Find the form of G(t). (c) Find φ(r) in the half space x>0i fρ(r)=δ(r−r1)a n d φ= 0 both on x=0 and as r→∞. 707 PDES: SEPARATION OF VARIABLES AND OTHER METHODS 19.30 Consider the PDE Lu(r)=ρ(r), for which the diﬀerential operator Lis given by L=∇·[p(r)∇]+q(r), where p(r)a n d q(r) are functions of position. By proving the generalised form of Green’s theorem,Z V(φLψ−ψLφ)dV= I Sp(φ∇ψ−ψ∇φ)·ˆndS, show that the solution of the PDE is given by u(r0)= Z VG(r,r0)ρ(r)dV(r)+ I Sp(r) / u(r)∂G(r,r0) ∂n−G(r,r0)∂u(r) ∂n / dS(r), where G(r,r0) is the Green’s function satisfying LG(r,r0)=δ(r−r0). 19.7 Hints and answers 19.1 (a) Cexp[λ(x2+2y)]; (b) C(x2y)λ. 19.2 There is heat ﬂow only across z=±a. It is into the cube at a rate of κAe−2/√2. 19.3 u(x, y, t)=s i n ( nπx/a )sin(mπy/b )(Asinωt+Bcosωt). 19.4 (a) − /~2 2mX/prime/prime X=p2 x 2m,etc.,i /~T/prime T=E; (b) As in (a), but with solutions X=Asin(pxx/ /~), etc. with pxa/ /~=nxπ. 19.5 (a) 6 u/r2,−6u/r2,0 ,/lscript=2 , m=0 ; (b) 2u/r2,( c o t2θ−1)u/r2;−u/(r2sin2θ),/lscript=1 , m=1 . 19.8 The ﬁrst term can contain only /lscript=1,2a n d m=±1, the second only /lscript=0,1,2 andm=0 ; f(θ,φ)=(π)1/2[Y0 0−3−1/2Y0 1−(2/3)1/2Y1 1−(2/15)1/2Y−1 2]. 19.9 Solutions of the form r/lscriptgive/lscriptas−1,1,2,4. Because of the asymptotic form of ψ,a nr4term cannot be present. The coeﬃcients of the three remaining terms are determined by the two boundary conditions u=0on the sphere and the form of ψfor large r. 19.10 If ψ(ρ, φ)=R(ρ)Φ(φ), show that Φ(4)+4 Φ/prime/prime= 0 and hence that Φ = A+Bφ+ Ccos 2φ+Dsin 2φ. 19.11 Express cos2φin terms of cos 2 φ;T(ρ, φ)=A+B/2+(Bρ2/2a2)c os2 φ. 19.12 (a) u(ρ, φ)=( 4 /π) P noddn−1(ρ/a)nsinnφ. (b)∇2α=0 ,a n d∇2β= 0 separately. On ρ=a, α+β+π/2=π. (c) Equate the two forms (uniqueness theorem) and then set ρ=a. The Fourier series is 2 P noddn−1sinnφ. 19.13 ( Acosmx+Bsinmx+Ccoshmx+Dsinhmx)cos( ωt+/epsilon1), with m4a4=ω2. 19.15 En=1 6ρA2c2/[(2n+1 )2π2L];E=2ρc2A2/L= RA 0[2Tv/(1 2L)]dv. 19.17 You will need the result from exercise 17.7.19.18 Write C(x, t)=C 0+ P∞ 1Ansin(nπx/L )fn(t)w h e r e fn(t)→0a st→∞; An=−4C0/(nπ)a n d fn(t)=e x p [−(κn2π2/L2)t]f o r nodd, and An=0f o r neven. 19.19 Since there is no heat ﬂow at x=±a, use a series of period 4 a,u(x,0) = 100 for 0<x≤2a,u(x,0) = 0 for−2a≤x<0. u(x, t)=5 0+200 π∞X n=01 2n+1sin /(2n+1 )πx 2a / exp / −k(2n+1 )2π2t 4a2s / . Taking only the n= 0 term gives t≈2300 s. 19.20 T1(r,θ)= P∞ 1bm(r/a)mPm(cosθ)+q0/k2+T∞, T2(r,θ)= P∞ 1bm(a/r)m+1Pm(cosθ)+aq0/(k2r)+T∞, where in both cases bm=qm/[mk1+(m+1 )k2];T(0,θ)=q0/k2+T∞. 708 19.7 HINTS AND ANSWERS 19.21 u(x, t)=[a/(a2+4κt)1/2]e x p [−x2/(a2+4κt)]. 19.22 (a) u(r=z,0) = 2 MGa−2[(a2+z2)1/2−z]. (b) For θ>π / 2, the factor in the square brackets is ( a2+z2)1/2+z.( c )F i n d ∂u/∂r atθ=0f o r r<a,a n dl e t a→∞. 19.23 Fourier-transform Poisson’s equation to show that ˜ρ(α)=/epsilon10(α2+q2)˜V(α). 19.24 (i) q2ab/[4π/epsilon10(b2−a2)2]; (ii) [ q2ab/(4π/epsilon10)][(b2−a2)−2−b−4]. Obtain (ii) from (i) by adding a further image charge + qa/batO, to give a net zero electrostatic ﬂux from the sphere while maintaining its equipotential property. 19.25 Follow the worked example that includes result (19.98). For part of the explicit integration, substitute ρ=ztanα. Φ(0,0,z)=z(1 +z2)1/2−z2+( 1+ z2)1/2−1 z(1 +z2)1/2. 19.26 (a) See equation (19.94); (c) M=( 4 π/epsilon10)−1 R ρ(r/prime)dV/prime= total charge on the sphere. d=( 4π/epsilon10)−1 R ρ(r/prime)r/primedV/prime= dipole moment of the sphere. 19.27 G(r,r0)=1 4π∞X n=2(−1)n /" 1p (x−x0)2+(y−y0)2+(z+(−1)nz0−nc)2 +1p (x−x0)2+(y−y0)2+(z+(−1)nz0+nc)2 /# . 19.28 G(r,r0)=−1 4π /" 1p (x−x0)2+(y−y0)2+(z−z0)2 +1p (x−x0)2+(y−y0)2+(z+z0)2 /# . φ(x, y, z)=1 2π / sinh−1a+xp y2+z2+s i n h−1a−xp y2+z2 /! . 19.29 (a) As given in equation (19.89), but with r0replaced by r/prime. (b) Move the origin to r/primeand integrate the deﬁning Green’s equation to obtain 4πt2dG dt−m2 Zt 0G(t/prime)4πt/prime2dt/prime=1, leading to G(t)=[−1/(4πt)]e−mt. (c)φ(r)=[−1/(4π)](p−1e−mp−q−1e−mq), where p=|r−r1|andq=|r−r2|with r1=(x1,y1,z1)a n d r2=(−x1,y1,z1). 709 20 Complex variables Throughout this book references have been made to results derived from the theory of complex variables. This theory thus becomes an integral part of themathematics appropriate to physical applications. The diﬃculty with it, from thepoint of view of a book such as the present one, is that although it has many practical applications its underlying basis has a distinctly pure mathematics ﬂavour. Thus, to adopt a comprehensive rigorous approach would involve a large amount of groundwork in analysis, for example formulating precise deﬁnitionsof continuity and diﬀerentiability, developing the theory of sets and making adetailed study of boundedness. Instead, we will be selective and pursue only thoseparts of the formal theory that are needed to establish the results used elsewherein this book and some others of general utility. In this spirit, the proofs that have been adopted for some of the standard results of complex variable theory have been chosen with an eye to simplicity rather than sophistication. This means that in some cases the imposed conditionsare more stringent than would be strictly necessary if more sophisticated proofswere used; where this happens the less restrictive results are usually stated aswell. The reader who is interested in a fuller treatment should consult one of themany excellent textbooks on this fascinating subject. † One further concession to ‘hand-waving’ has been made in the interests of keeping the treatment to a moderate length. In several places phrases such as ‘can be made as small as we like’ are used, rather than a careful treatment in terms of ‘given /epsilon1>0, there exists a δ>0 such that’. In the authors’ experience, some students are more at ease with the former type of statement despite its lack of †For example, Knopp, Theory of Functions, Part I (Dover, 1945); Phillips, Functions of a Complex Variable (Oliver and Boyd, 1954); Titchmarsh, The Theory of Functions (Oxford, 1952). 710 20.1 FUNCTIONS OF A COMPLEX VARIABLE precision whilst others, those who would contemplate only the latter, are usually well able to supply it for themselves. 20.1 Functions of a complex variable The quantity f(z) is said to be a function of the complex variable zif to every value ofzin a certain domain R(a region of the Argand diagram) there corresponds one or more values of f(z). Stated like this f(z) could be any function consisting of a real and an imaginary part, each of which is, in general, itself a function of x andy. If we denote the real and imaginary parts of f(z)b yuandvrespectively, then f(z)=u(x, y)+iv(x, y). In this chapter, however, we will be primarily concerned with functions that are single-valued, so that for each value of zthere corresponds just one value of f(z), and diﬀerentiable in a particular sense, which we now discuss. A function f(z) that is single-valued in some domain Risdiﬀerentiable at the point zinRif the derivative f/prime(z) = lim ∆z→0bracketleftbiggf(z+∆z)−f(z) ∆zbracketrightbigg (20.1) exists and is unique, in that its value does not depend upon the direction in the Argand diagram from which ∆ ztends to zero.IShow that the function f(z)=x2−y2+i2xyis diﬀerentiable for all values of z. Considering the deﬁnition (20.1), and taking ∆ z=∆x+i∆y, we have f(z+∆z)−f(z) ∆z =(x+∆x)2−(y+∆y)2+2i(x+∆x)(y+∆y)−x2+y2−2ixy ∆x+i∆y =2x∆x+( ∆x)2−2y∆y−(∆y)2+2i(x∆y+y∆x+∆x∆y) ∆x+i∆y =2x+i2y+(∆x)2−(∆y)2+2i∆x∆y ∆x+i∆y. Now, in whatever way ∆ xand ∆ yare allowed to tend to zero (e.g. taking ∆ y=0a n d letting ∆ x→0 or vice versa), the last term on the right will tend to zero and the unique limit 2 x+i2ywill be obtained. Since zwas arbitrary, f(z)w i t h u=x2−y2andv=2xy is diﬀerentiable at all points in the (ﬁnite) complex plane. J We note that the above working can be considerably reduced by recognising that, since z=x+iy, we can write f(z)a s f(z)=x2−y2+2ixy=(x+iy)2=z2. 711 COMPLEX VARIABLES We then ﬁnd that f/prime(z) = lim ∆z→0bracketleftbigg(z+∆z)2−z2 ∆zbracketrightbigg = lim ∆z→0bracketleftbigg(∆z)2+2z∆z ∆zbracketrightbigg =parenleftBig lim ∆z→0∆zparenrightBig +2z=2z, from which we see immediately that the limit both exists and is independent of the way in which ∆ z→0. Thus we have veriﬁed that f(z)=z2is diﬀerentiable for all (ﬁnite) z. We also note that the derivative is analogous to that found for real variables. Although the deﬁnition of a diﬀerentiable function clearly includes a wide class of functions, the concept of diﬀerentiability is restrictive and, indeed, some functions are not diﬀerentiable at any point in the complex plane.IShow that the function f(z)=2 y+ixis not diﬀerentiable anywhere in the complex plane. In this case f(z) cannot be written simply in terms of z, and so we must consider the limit (20.1) in terms of xandyexplicitly. Following the same procedure as in the previous example we ﬁnd f(z+∆z)−f(z) ∆z=2y+2 ∆ y+ix+i∆x−2y−ix ∆x+i∆y =2∆y+i∆x ∆x+i∆y. In this case the limit will clearly depend on the direction from which ∆ z→0. Suppose ∆z→0 along a line through zof slope m,s ot h a t∆ y=m∆x,t h e n lim ∆z→0 /f(z+∆z)−f(z) ∆z / = lim ∆x,∆y→0 /2∆y+i∆x ∆x+i∆y / =2m+i 1+im. This limit is dependent on mand hence on the direction from which ∆ z→0. Since this conclusion is independent of the value of z, and hence true for all z,f(z)=2 y+ixis nowhere diﬀerentiable. J A function that is single-valued and diﬀerentiable at all points of a domain R is said to be analytic (orregular )i nR. A function may be analytic in a domain except at a ﬁnite number of points (or an inﬁnite number if the domain is inﬁnite); in this case it is said to be analytic except at these points, which are called the singularities off(z). (In our treatment we will not consider cases in which an inﬁnite number of singularities occur in a ﬁnite domain.) 712 20.2 THE CAUCHY–RIEMANN RELATIONSIShow that the function f(z)=1 /(1−z)is analytic everywhere except at z=1. Since f(z) is given explicitly as a function of z, evaluation of the limit (20.1) is somewhat easier. We ﬁnd f/prime(z) = lim ∆z→0 /f(z+∆z)−f(z) ∆z / = lim ∆z→0 /1 ∆z /1 1−z−∆z−1 1−z // = lim ∆z→0 /1 (1−z−∆z)(1−z) / =1 (1−z)2, independently of the way in which ∆ z→0, provided z/negationslash= 1. Hence f(z)i sa n a l y t i c everywhere except at the singularity z=1 . J 20.2 The Cauchy–Riemann relations From examining the previous examples, it is apparent that for a function f(z) to be diﬀerentiable and hence analytic there must be some particular connectionbetween its real and imaginary parts uand v. We next establish what this connection must be, by considering a general function. If the limit L= lim ∆z→0bracketleftbiggf(z+∆z)−f(z) ∆zbracketrightbigg (20.2) is to exist and be unique, in the way required for diﬀerentiability, then any two speciﬁc ways of letting ∆ z→0 must produce the same limit. In particular, moving parallel to the real axis and moving parallel to the imaginary axis must do so. This is certainly a necessary condition, although it may not be suﬃcient. If we let f(z)=u(x, y)+iv(x, y)a n d∆ z=∆x+i∆ythen we have f(z+∆z)=u(x+∆x, y+∆y)+iv(x+∆x, y+∆y), and the limit (20.2) is given by L= lim ∆x,∆y→0bracketleftbiggu(x+∆x, y+∆y)+iv(x+∆x, y+∆y)−u(x, y)−iv(x, y) ∆x+i∆ybracketrightbigg . If we ﬁrst suppose that ∆ zis purely real, so that ∆ y=0 ,w eo b t a i n L= lim ∆x→0bracketleftbiggu(x+∆x, y)−u(x, y) ∆x+iv(x+∆x, y)−v(x, y) ∆xbracketrightbigg =∂u ∂x+i∂v ∂x, (20.3) provided each limit exists at the point z. Similarly, if ∆ zis taken as purely imaginary, so that ∆ x= 0, we ﬁnd L= lim ∆y→0bracketleftbiggu(x, y+∆y)−u(x, y) i∆y+iv(x, y+∆y)−v(x, y) i∆ybracketrightbigg =1 i∂u ∂y+∂v ∂y. (20.4) 713 COMPLEX VARIABLES Forfto be diﬀerentiable at the point z, expressions (20.3) and (20.4) must be identical. It follows from equating real and imaginary parts that necessary conditions for this are ∂u ∂x=∂v ∂yand∂v ∂x=−∂u ∂y. (20.5) These two equations are known as the Cauchy–Riemann relations . We can now see why for the earlier examples (i) f(z)=x2−y2+i2xymight be diﬀerentiable and (ii) f(z)=2 y+ixcould not be. (i)u=x2−y2,v=2xy: ∂u ∂x=2x=∂v ∂yand∂v ∂x=2y=−∂u ∂y, (ii)u=2y,v=x: ∂u ∂x=0=∂v ∂ybut∂v ∂x=1/negationslash=−2=−∂u ∂y. It is apparent that for f(z) to be analytic something more than the existence of the partial derivatives of uandvwith respect to xandyis required; this something is that they satisfy the Cauchy–Riemann relations. We may enquire also as to the suﬃcient conditions for f(z) to be analytic in R.I tc a nb es h o w n †that a suﬃcient condition is that the four partial derivatives exist, are continuous and satisfy the Cauchy–Riemann relations. It is the addi- tional requirement of continuity that makes the diﬀerence between the necessaryconditions and the suﬃcient conditions.IIn which domain(s) of the complex plane is f(z)=|x|−i|y|an analytic function? Writing f=u+ivit is clear that both ∂u/∂y and∂v/∂x a r ez e r oi na l lf o u rq u a d r a n t s and hence that the second Cauchy–Riemann relation in (20.5) is satisﬁed everywhere. Turning to the ﬁrst Cauchy–Riemann relation, in the ﬁrst quadrant ( x>0,y>0) we have f(z)=x−iyso that ∂u ∂x=1,∂v ∂y=−1, which clearly violates the ﬁrst relation in (20.5). Thus f(z) is not analytic in the ﬁrst quadrant. Following a similiar argument for the other quadrants, we ﬁnd ∂u ∂x=−1o r + 1 f o r x<0a n d x>0 respectively, ∂v ∂y=−1o r + 1 f o r y>0a n d y<0 respectively. Therefore ∂u/∂x and∂v/∂y are equal, and hence f(z) is analytic, only in the second and fourth quadrants. J †See for example any of the references given earlier. 714 20.2 THE CAUCHY–RIEMANN RELATIONS Since xandyare related to zand its complex conjugate z∗by x=1 2(z+z∗)a n d y=1 2i(z−z∗), (20.6) we may formally regard any function f=u+ivas a function of zandz∗,r a t h e r than xandy. If we do this and examine ∂f/∂z∗we obtain ∂f ∂z∗=∂f ∂x∂x ∂z∗+∂f ∂y∂y ∂z∗ =parenleftbigg∂u ∂x+i∂v ∂xparenrightbiggparenleftbigg1 2parenrightbigg +parenleftbigg∂u ∂y+i∂v ∂yparenrightbiggparenleftbigg −1 2iparenrightbigg =1 2parenleftbigg∂u ∂x−∂v ∂yparenrightbigg +i 2parenleftbigg∂v ∂x+∂u ∂yparenrightbigg . (20.7) Now, if fis analytic then the Cauchy–Riemann relations (20.5) must be satisﬁed, and these immediately give that ∂f/∂z∗is identically zero. Thus we conclude that iffis analytic then fcannot be a function of z∗and any expression representing an analytic function of zcan contain xandyonly in the combination x+iy,not in the combination x−iy. We conclude this section by discussing some properties of analytic functions that are of great practical importance in theoretical physics. These can be obtainedsimply from the requirement that the Cauchy–Riemann relations must be satisﬁedby the real and imaginary parts of an analytic function. The most important of these results can be obtained by diﬀerentiating the ﬁrst Cauchy–Riemann relation with respect to one independent variable, and thesecond with respect to the other independent variable, to obtain the two chains of equalities: ∂ ∂xparenleftbigg∂u ∂xparenrightbigg =∂ ∂xparenleftbigg∂v ∂yparenrightbigg =∂ ∂yparenleftbigg∂v ∂xparenrightbigg =−∂ ∂yparenleftbigg∂u ∂yparenrightbigg ; ∂ ∂xparenleftbigg∂v ∂xparenrightbigg =−∂ ∂xparenleftbigg∂u ∂yparenrightbigg =−∂ ∂yparenleftbigg∂u ∂xparenrightbigg =−∂ ∂yparenleftbigg∂v ∂yparenrightbigg . Thus both uandvareseparately solutions of Laplace’s equation in two dimen- sions, i.e. ∂2u ∂x2+∂2u ∂y2=0 a n d∂2v ∂x2+∂2v ∂y2=0. (20.8) We shall make use of this result in section 20.9. A further useful result concerns the two families of curves u(x, y) = constant andv(x, y) = constant, where uandvare the real and imaginary parts of any analytic function f=u+iv. As discussed in chapter 10, the vector normal to the curve u(x, y) = constant is given by ∇u=∂u ∂xi+∂u ∂yj, (20.9) 715 COMPLEX VARIABLES where iandjare the unit vectors along the x-a n d y- axes respectively. A similar expression exists for ∇v, the normal to the curve v(x, y) = constant. Taking the scalar product of these two normal vectors we obtain ∇u·∇v=∂u ∂x∂v ∂x+∂u ∂y∂v ∂y =−∂u ∂x∂u ∂y+∂u ∂y∂u ∂x=0, where in the last line we have used the Cauchy–Riemann relations to rewrite the partial derivatives of vas partial derivatives of u. Since the scalar product of the normal vectors is zero, they must be orthogonal and the curves u(x, y) = constant andv(x, y) = constant must therefore intersect at right angles .IUse the Cauchy–Riemann relations to show that, for any analytic function f=u+iv,t h e relation|∇u|=|∇v|must hold. From (20.9) we have |∇u|2=∇u·∇u= /∂u ∂x /2 + /∂u ∂y /2 . Using the Cauchy–Riemann relations to write the partial derivatives of uin terms of those ofv,w eo b t a i n |∇u|2= /∂v ∂y /2 + /∂v ∂x /2 =|∇v|2, from which the result |∇u|=|∇v|follows immediately. J 20.3 Power series in a complex variable The theory of power series in a real variable was considered in chapter 4, which also contained a brief discussion of the natural extension of this theory to a seriessuch as f(z)=∞summationdisplay n=0anzn, (20.10) where zis a complex variable and the anare in general complex. We now consider complex power series in more detail. Expression (20.10) is a power series about the origin and may be used for general discussion since a power series about any other point z0can be obtained by a change of variable from ztoz−z0.I fzwere written in its modulus and argument form z=rexpiθ, expression (20.10) would become f(z)=∞summationdisplay n=0anrnexp(inθ). (20.11) 716 20.3 POWER SERIES IN A COMPLEX VARIABLE This series is absolutely convergent if ∞summationdisplay n=0|an|rn, (20.12) which is a series of positive real terms, is convergent. Thus tests for the absolute convergence of real series can be used in the present context, and of these themost appropriate form is based on the Cauchy root test. The radius of convergence Ris deﬁned by 1 R= lim n→∞|an|1/n; (20.13) the series (20.10) is absolutely convergent if |z|<Rand divergent if |z|>R.I f |z|=Rno particular conclusion may be drawn, and this case must be considered separately, as discussed in subsection 4.5.1. A circle of radius Rcentred on the origin is called the circle of convergence of the seriessummationtextanzn.T h ec a s e s R=0a n d R=∞correspond respectively to convergence at the origin only and convergence everywhere. For Rﬁnite the convergence occurs in a restricted part of the z-plane (the Argand diagram). For a power series about a general point z0, the circle of convergence is of course centred on that point.IFind the parts of the z-plane for which the following series are convergent: (i)∞X n=0zn n!, (ii)∞X n=0n!zn,(iii)∞X n=1zn n. (i) Since ( n!)1/nbehaves like nasn→∞ we ﬁnd lim(1 /n!)1/n= 0. Hence R=∞and the series is convergent for all z. (ii) Correspondingly, lim( n!)1/n=∞. Thus R=0a n dt h e series converges only at z= 0. (iii) As n→∞,(n)1/nhas a lower limit of 1 and hence lim(1 /n)1/n=1/1 = 1. Thus the series is absolutely convergent if |z|<1. J Case (iii) in the above example provides a good illustration of the fact that on its circle of convergence a power series may or may not converge. For this particular series the circle of convergence is |z|= 1, so let us consider the convergence of the series at two diﬀerent points on this circle. Taking z= 1, the series becomes ∞summationdisplay n=11 n=1+1 2+1 3+1 4+···, which is easily shown to diverge (by, for example, grouping terms, as discussed in subsection 4.3.2). Taking z=−1, however, the series is given by ∞summationdisplay n=1(−1)n n=−1+1 2−1 3+1 4−···, 717 COMPLEX VARIABLES which is an alternating series whose terms decrease in magnitude and which therefore converges. The ratio test discussed in subsection 4.3.2 may also be employed to investi- gate the absolute convergence of a complex power series. A series is absolutely convergent if lim n→∞|an+1||z|n+1 |an||z|n= lim n→∞|an+1||z| |an|<1 (20.14) and hence the radius of convergence Ro ft h es e r i e si sg i v e nb y 1 R= lim n→∞|an+1| |an|. For instance, in case (i) of the previous example, we have 1 R= lim n→∞n! (n+1 ) != lim n→∞1 n+1=0. Thus the series is absolutely convergent for all (ﬁnite) z, conﬁrming the previous result. Before turning to particular power series, we conclude this section by stating the important result †thatthe power seriessummationtext∞ 0anznhas a sum that is an analytic function of zinside its circle of convergence. As a corollary to the above theorem, it may further be shown that if f(z)=summationtextanznthen, inside the circle of convergence of the series, f/prime(z)=∞summationdisplay n=0nanzn−1. Repeated application of this result demonstrates that any power series can be diﬀerentiated any number of times inside its circle of convergence. 20.4 Some elementary functions In the example at the end of the previous section it was shown that the function expzdeﬁned by expz=∞summationdisplay n=0zn n!(20.15) is convergent for all zof ﬁnite modulus and is thus, by the discussion of the previous section, an analytic function over the whole z-plane.‡Like its †For a proof see, for example, Riley, Mathematical Methods for the Physical Sciences (CUP, 1974), p. 446. ‡Functions that are analytic in the whole z-plane are usually called integral orentire functions. 718 20.4 SOME ELEMENTARY FUNCTIONS real-variable counterpart it is called the exponential function ; also like its real counterpart it is equal to its own derivative. The multiplication of two exponential functions results in a further exponential function, in accordance with the corresponding result for real variables.IShow that expz1expz2=e x p ( z1+z2). From the series expansion (20.15) of exp z1and a similar expansion for exp z2,i ti sc l e a r that the coeﬃcient of zr 1zs 2in the corresponding series expansion of exp z1expz2is simply 1/(r!s!). But, from (20.15) we also have exp(z1+z2)=∞X n=0(z1+z2)n n!. In order to ﬁnd the coeﬃcient of zr 1zs 2in this expansion, we clearly have to consider the term in which n=r+s,n a m e l y (z1+z2)r+s (r+s)!=1 (r+s)! /;r+sC0zr+s 1+···+r+sCszr 1zs 2+···+r+sCr+szr+s 2 / . The coeﬃcient of zr 1zs 2in this is given by r+sCs1 (r+s)!=(r+s)! s!r!1 (r+s)!=1 r!s!. Thus, since the corresponding coeﬃcients on the two sides are equal and all the series involved are absolutely convergent for all z, we can conclude that exp z1expz2=e x p ( z1+ z2). J As an extension of (20.15) we may also deﬁne the complex exponent of a real number a>0 by the equation az=e x p ( zlna), (20.16) where ln ais the natural logarithm of a. The particular case a=eand the fact that ln e= 1 enable us to write exp zinterchangeably with ez.I fzis real then the deﬁnition agrees with the familiar one. The result for z=iy, expiy=c o s y+isiny, (20.17) has been met already in equation (3.23). Its immediate extension is expz=( e x p x)(cos y+isiny). (20.18) Aszvaries over the complex plane the modulus of exp ztakes all real positive values, except that of 0. However, two values of zthat diﬀer by 2 πni,f o ra n y integer n, produce the same value of exp z, as given by (20.18), and so exp zis periodic with period 2 πi. If we denote exp zbytthen the strip −π<y≤πin thez-plane corresponds to the whole of the t-plane, except for the point t=0 . 719 COMPLEX VARIABLES The sine, cosine, sinh and cosh functions of a complex variable are deﬁned from the exponential function exactly as are those for real variables. The functionsderived from them (e.g. tan and tanh), the identities they satisfy, and theirderivative properties, are also just as for real variables. In view of this we will not give them further attention here. The inverse function of exp zis given by w, the solution of expw=z. (20.19) This inverse function was discussed in chapter 3, but we mention it again here for completeness. By virtue of the discussion following (20.18), wis not uniquely deﬁned and is indeterminate to the extent of any integer multiple of 2 πi.I fw e express zas z=rexpiθ, where ris the (real) modulus of z,a n d θis its argument ( −π<θ≤π), then multiplying zby exp(2 inπ), where nis an integer, will result in the same complex number z. Thus we may write z=rexp[i(θ+2nπ)], where nis an integer. If we denote win (20.19) by w=L n z=l nr+i(θ+2nπ), (20.20) where ln ris the natural logarithm (to base e) of the real positive quantity r,t h e n Lnzis an inﬁnitely multivalued function of z.I t sprincipal value , denoted by ln z, is obtained by taking n= 0 so that its argument lies in the range −πtoπ. Thus lnz=l nr+iθ, with−π<θ≤π. (20.21) Now that the logarithm of a complex variable has been deﬁned, deﬁnition (20.16) of a general power can be extended to cases other than those in which a is real and positive. If t(/negationslash=0 )a n d zare both complex then the zth power of tis deﬁned by t z=e x p ( zLnt). (20.22) Since Ln tis multivalued, so is this deﬁnition. Its principal value is obtained by giving Ln tits principal value, ln t. Ift(/negationslash= 0) is complex but zis real and equal to 1 /n, then (20.22) provides a deﬁnition of the nth root of t. Because of the multivaluedness of Ln t, there will be more than one nth root of any given t. 720 20.5 MULTIVALUED FUNCTIONS AND BRANCH CUTSIShow that there are exactly ndistinct nth roots of t. From (20.22) the nth roots of tare given by t1/n=e x p /1 nLnt / . On the RHS let us write tas follows: t=rexp[i(θ+2kπ)], where kis an integer. We then obtain t1/n=e x p /1 nlnr+i(θ+2kπ) n / =r1/nexp / i(θ+2kπ) n / , where k=0,1,...,n−1; for other values of kwe simply recover the roots already found. Thus thasndistinct nth roots. J 20.5 Multivalued functions and branch cuts In the deﬁnition of an analytic function, one of the conditions imposed was that the function is single-valued. However, as shown in the previous section, thelogarithmic function, a complex power and a complex root are all multivalued. Nevertheless, it happens that the properties of analytic functions can still be applied to these and other multivalued functions of a complex variable providedthat suitable care is taken. This care amounts to identifying the branch points of the multivalued function f(z) in question. If zis varied in such a way that its path in the Argand diagram forms a closed curve that encloses a branch point,then, in general, f(z) will not return to its original value. For deﬁniteness let us consider the multivalued function f(z)=z 1/2and express zasz=rexpiθ. From ﬁgure 20.1( a), it is clear that, as the point ztraverses any closed contour Cthat does not enclose the origin, θwill return to its original value after one complete circuit. However, for any closed contour C/primethat does enclose the origin, after one circuit θ→θ+2π(see ﬁgure 20.1( b)). Thus, for the function f(z)=z1/2, after one circuit r1/2exp(iθ/2)→r1/2exp[i(θ+2π)/2] =−r1/2exp(iθ/2). In other words, the value of f(z) changes around any closed loop enclosing the origin; in this case f(z)→− f(z). Thus z= 0 is a branch point of the function f(z)=z1/2. We note in this case that if any closed contour enclosing the origin is traversed twicethen f(z)=z1/2returns to its original value. The number of loops around a branch point required for any given function f(z) to return to its original value 721 COMPLEX VARIABLES y y y x x xC θ θr r (a)( b)( c)C/prime Figure 20.1 ( a) A closed contour not enclosing the origin; ( b) a closed contour enclosing the origin; ( c) a possible branch cut for f(z)=z1/2. depends on the function in question, and for some functions (e.g. Ln z,w h i c ha l s o has a branch point at the origin) the original value is never recovered. In order that f(z) may be treated as single-valued we may deﬁne a branch cut in the Argand diagram. A branch cut is a line (or curve) in the complex plane and may be regarded as an artiﬁcial barrier that we must not cross. Branch cuts are positioned in such a way that we are prevented from making a completecircuit around any one branch point, and so the function in question remainssingle-valued. For the function f(z)=z 1/2, we may take as a branch cut any curve starting at the origin z= 0 and extending out to |z|=∞in any direction, since all such curves would equally well prevent us from making a closed loop around the branch point at the origin. It is usual, however, to take the cut along the real or imaginary axis. For example, in ﬁgure 20.1( c), we take the cut as the positive real axis. By agreeing not to cross this cut, we restrict θto lie in the range 0 ≤θ<2π, a n ds ok e e p f(z) single-valued. These ideas are easily extended to functions with more than one branch point.IFind the branch points of f(z)=√ z2+1, and hence sketch suitable arrangements of branch cuts. We begin by writing f(z)a s f(z)= p z2+1= p (z−i)(z+i). As shown above the function g(z)=z1/2has a branch point at z= 0. Thus we might expect f(z) to have branch points at values of zthat make the expression under the square root equal to zero, i.e. at z=iandz=−i. As shown in ﬁgure 20.2( a), we use the notation z−i=r1expiθ1 and z+i=r2expiθ2. 722 20.6 SINGULARITIES AND ZEROES OF COMPLEX FUNCTIONS (a)( b)( c)−i −i −ii i iy y y x x xr1 r2θ1 θ2z Figure 20.2 ( a) Coordinates used in the analysis of the branch points of f(z)=( z2+1 )1/2;(b) one possible arrangement of branch cuts; ( c) another possible branch cut, which is ﬁnite. We can therefore write f(z)a s f(z)=√r1r2exp(iθ1/2)exp( iθ2/2) =√r1r2exp / i(θ1+θ2)/2 / . Let us now consider how f(z) changes as we make one complete circuit around various closed loops Cin the Argand diagram. If Cencloses (i) neither branch point, then θ1→θ1,θ2→θ2and so f(z)→f(z); (ii)z=ibut not z=−i,t h e n θ1→θ1+2π,θ2→θ2and so f(z)→−f(z); (iii)z=−ibut not z=i,t h e n θ1→θ1,θ2→θ2+2πand so f(z)→−f(z); (iv) both branch points, then θ1→θ1+2π,θ2→θ2+2πand so f(z)→f(z). Thus, as expected, f(z) changes value around loops containing either z=iorz=−i (but not both). We must therefore choose branch cuts that prevent us from making acomplete loop around either branch point; one suitable choice is shown in ﬁgure 20.2( b). For this f(z), however, we have noted that after traversing a loop containing bothbranch points the function returns to its original value. Thus we may choose an alternative, ﬁnite, branch cut that allows this possibility but still prevents us from making a complete looparound just one of the points. A suitable cut is shown in ﬁgure 20.2( c).J 20.6 Singularities and zeroes of complex functions A singular point of a complex function f(z) is any point in the Argand diagram at which f(z) fails to be analytic. We have already met one sort of singularity, the branch point, and in this section we shall consider other types of singularityas well as discuss the zeroes of complex functions. Iff(z) has a singular point at z=z 0but is analytic at all points in some neighbourhood containing z0but no other singularities then z=z0is called an isolated singularity . (Clearly branch points are not isolated singularities.) 723 COMPLEX VARIABLES The most important type of isolated singularity is the pole.I ff(z)h a st h ef o r m f(z)=g(z) (z−z0)n, (20.23) where nis a positive integer, g(z) is analytic at all points in some neighbourhood containing z=z0andg(z0)/negationslash=0 ,t h e n f(z)h a sa pole of order natz=z0.A n alternative (though equivalent) deﬁnition is that lim z→z0[(z−z0)nf(z)]=a, (20.24) where ais a ﬁnite, non-zero complex number. (If the limit equals zero then z=z0 is a pole of order less than n,o rf(z) is analytic there; if the limit is inﬁnite then the pole is of order greater than n.) It may also be shown that if f(z) has a pole atz=z0,t h e n|f(z)|→∞ asz→z0from any direction in the Argand diagram. † If no ﬁnite value of ncan be found such that (20.24) is satisﬁed then z=z0is called an essential singularity .IFind the singularities of the functions (i)f(z)=1 1−z−1 1+z, (ii)f(z)=t a n h z. (i) If we write f(z)a s f(z)=1 1−z−1 1+z=2z (1−z)(1 + z), we see immediately from either (20.23) or (20.24) that f(z) has poles of order 1 (or simple poles)a tz=1a n d z=−1. (ii) In this case we write f(z)=t a n h z=sinhz coshz=expz−exp(−z) expz+e x p (−z). Thus f(z) has a singularity when exp z=−exp(−z) or, equivalently, when expz=e x p [ i(2n+1 )π]e x p (−z), where nis any integer. Equating the arguments of the exponentials we ﬁnd z=(n+1 2)πi, for integer n. Furthermore, using l’H ˆopital’s rule (see chapter 4) we have lim z→(n+1 2)πi /( [z−(n+1 2)πi]si nh z coshz /) = lim z→(n+1 2)πi /( [z−(n+1 2)πi]cosh z+s i n h z sinhz /) =1. Therefore, from (20.24), each singularity is a simple pole. J Another type of singularity exists at points for which the value of f(z)t a k e s an indeterminate form such as 0 /0 but lim z→z0f(z) exists and is independent †Although perhaps intuitively obvious this result really requires formal demonstration by analysis. 724 20.7 COMPLEX POTENTIALS of the direction from which z0is approached. Such points are called removable singularities .IShow that f(z)=( s i n z)/zhas a removable singularity at z=0. It is clear that f(z) takes the indeterminate form 0 /0a t z= 0. However, by expanding sinzas a power series in z, we ﬁnd f(z)=1 z / z−z3 3!+z5 5!−··· / =1−z2 3!+z4 5!−···. Thus lim z→0f(z) = 1 independently of the way in which z→0, and so f(z) has a removable singularity at z=0 . J An expression common in mathematics, but which we have so far avoided using explicitly in this chapter, is ‘ ztends to inﬁnity’. For a real variable such as|z|orR, ‘tending to inﬁnity’ has a reasonably well-deﬁned meaning. For a complex variable needing a two-dimensional plane to represent it, the meaning is not intrinsically well deﬁned. However, it is convenient to have a unique meaningand this is provided by the following deﬁnition : the behaviour of f(z)at inﬁnity is given by that of f(1/ξ)a tξ=0 ,w h e r e ξ=1/z.IFind the behaviour at inﬁnity of (i) f(z)=a+bz−2, (ii) f(z)=z(1 + z2)and (iii) f(z)=e x p z. (i)f(z)=a+bz−2: on putting z=1/ξ,f(1/ξ)=a+bξ2, which is analytic at ξ= 0; thus f is analytic at z=∞. (ii)f(z)=z(1 +z2):f(1/ξ)=1 /ξ+1/ξ3; thus fhas a pole of order 3a tz=∞. (iii) f(z)=e x p z:f(1/ξ)= P∞ 0(n!)−1ξ−n; thus fhas an essential singularity atz=∞. J We conclude this section by brieﬂy mentioning the zeroes of a complex function. As the name suggests, if f(z0)=0t h e n z=z0is called a zero of the function f(z). Zeroes are classiﬁed in a similar way to poles, in that if f(z)=(z−z0)ng(z), where nis a positive integer and g(z0)/negationslash=0 ,t h e n z=z0is called a zero of order noff(z). Ifn=1t h e n z=z0is called a simple zero . It may further be shown that if z=z0is a zero of order noff(z) then it is also a pole of order nof the function 1 /f(z). We will return in section 20.13 to the classiﬁcation of zeroes and poles in terms of their series expansions. 20.7 Complex potentials Towards the end of section 20.2 it was shown that the real and the imaginary parts of an analytic function of zare separately solutions of Laplace’s equation in two dimensions. Analytic functions thus oﬀer a possible way of solving some 725 COMPLEX VARIABLES y x Figure 20.3 The equipotentials (broken) and ﬁeld lines (solid) for a line charge perpendicular to the z-plane. two-dimensional physical problems describable by a potential satisfying ∇2φ=0 . The general method is known as that of complex potentials . We found also that if f=u+ivis an analytic function of zthen any curve u= constant intersects any curve v= constant at right angles. In the context of solutions of Laplace’s equation, this result implies that the real and imaginary parts of f(z) have an additional connection between them, for if the set of contours on which one of them is a constant represents the equipotentials of asystem then the contours on which the other is constant, being orthogonal toeach of the ﬁrst set, must represent the corresponding ﬁeld lines or stream lines, depending on the context. The analytic function fis the complex potential. It is conventional to use φandψ(rather than uandv) to denote the real and imaginary parts of a complex potential, so that f=φ+iψ. As an example consider the function f(z)=−q 2π/epsilon10lnz, (20.25) in connection with the physical situation of a line charge of strength qper unit length passing through the origin, perpendicular to the z-plane (ﬁgure 20.3). Its real and imaginary parts are φ=−q 2π/epsilon10ln|z|,ψ =−q 2π/epsilon10argz. (20.26) The contours in the z-plane of φ= constant are concentric circles and of ψ= constant are radial lines. As expected these are orthogonal sets, but in additionthey are respectively the equipotentials and electric ﬁeld lines appropriate to the 726 20.7 COMPLEX POTENTIALS ﬁeld produced by the line charge (the minus sign is needed in (20.25) because the value of φmust decrease with increasing distance from the origin). Suppose we make the choice that the real part φof the analytic function f gives the conventional potential function; ψcould equally well be selected. Then we may consider how the direction and magnitude of the ﬁeld are related to f.IShow that for any complex (electrostatic) potential f(z)the strength of the electric ﬁeld is given by E=|f/prime(z)|and that its direction makes an angle of π−arg[f/prime(z)]with the x-axis. Because φ= constant is an equipotential, the ﬁeld has components Ex=−∂φ ∂xand Ey=−∂φ ∂y. (20.27) Since fis analytic, (i) we may use the Cauchy–Riemann relations (20.5) to change the second of these, obtaining Ex=−∂φ ∂xand Ey=∂ψ ∂x; (20.28) (ii) the direction of diﬀerentiation at a point is immaterial and so df dz=∂f ∂x=∂φ ∂x+i∂ψ ∂x=−Ex+iEy. (20.29) From these it can be seen that the ﬁeld at a point is given in magnitude by E=|f/prime(z)| and that it makes an angle with the x-axis given by π−arg[f/prime(z)]. J It will be apparent from the above that much of physical interest can be calculated by working directly in terms of fandz. In particular, the electric ﬁeld vector Emay be represented, using (20.29) above, by the quantity E=Ex+iEy=−[f/prime(z)]∗. Complex potentials can be used in two-dimensional ﬂuid mechanics problems in a similar way. If the ﬂow is stationary (i.e. the velocity of the ﬂuid does notdepend on time) and irrotational, and the ﬂuid is both incompressible and non- viscous, then the velocity of the ﬂuid can be described by V=∇φ,w h e r e φis the velocity potential and satisﬁes ∇ 2φ= 0. If, for a complex potential f=φ+iψ, the real part φis taken to represent the velocity potential then the curves ψ= constant will be the streamlines of the ﬂow. In a direct parallel with the electricﬁeld, the velocity may be represented in terms of the complex potential by V=V x+iVy=[f/prime(z)]∗, the diﬀerence of a minus sign reﬂecting the same diﬀerence between the deﬁnitions ofEandV. The speed of the ﬂow is equal to |f/prime(z)|. Points where f/prime(z) = 0, and so the velocity is zero, are called stagnation points of the ﬂow. Analogously to the electrostatic case, a line source of ﬂuid at z=z0, perpendic- ular to the z-plane (i.e. a point from which ﬂuid is emerging at a constant rate) 727 COMPLEX VARIABLES is described by the complex potential f(z)=kln(z−z0), where kis the strength of the source. A sink is similarly represented, but with k replaced by −k. Other simple examples are as follows. (i) The ﬂow of a ﬂuid at a constant speed V0and at an angle αto the x-axis is described by f(z)=V0(expiα)z. (ii) Vortex ﬂow, in which ﬂuid ﬂows azimuthally in an anticlockwise direction around some point z0, the speed of the ﬂow being inversely proportional to the distance from z0, is described by f(z)=−ikln(z−z0), where kis the strength of the vortex. For a clockwise vortex kis replaced by −k.IVerify that the complex potential f(z)=V0 / z+a2 z / , is appropriate to a circular cylinder of radius aplaced so that it is perpendicular to a uniform ﬂuid ﬂow of speed V0parallel to the x-axis. Firstly, since f(z)i sa n a l y t i ce x c e p ta t z= 0, both its real and imaginary parts satisfy Laplace’s equation in the region exterior to the cylinder. Also f(z)→V0zasz→∞,s o that Re f(z)→V0x, which is appropriate to a uniform ﬂow of speed V0in the x-direction far from the cylinder. Writing z=rexpiθand using de Moivre’s theorem we have f(z)=V0 / rexpiθ+a2 rexp(−iθ) / =V0 / r+a2 r / cosθ+iV0 / r−a2 r / sinθ. Thus we see that the streamlines of the ﬂow described by f(z) are given by ψ=V0 / r−a2 r / sinθ=c o n s t a n t . In particular, ψ=0o n r=a, independently of the value of θ,a n ds o r=amust be a streamline. Since there can be no ﬂow of ﬂuid across streamlines, r=amust correspond to a boundary along which the ﬂuid ﬂows tangentially. Thus f(z) is a solution of Laplace’s equation that satisﬁes all the physical boundary conditions of the problem, and so it is theappropriate complex potential.J By a similar argument the complex potential f(z)=−E(z−a2/z) (note the minus signs) is appropriate to a conducting circular cylinder of radius aplaced perpendicular to a uniform electric ﬁeld Ein the x-direction. The real and imaginary parts of a complex potential f=φ+iψhave another interesting relationship in the context of Laplace’s equation in electrostatics or ﬂuid mechanics. Let us choose φas the conventional potential, so that ψrepresents the stream function (or electric ﬁeld, depending on the application), and consider 728 20.7 COMPLEX POTENTIALS PQy x ˆn Figure 20.4 A curve joining the points PandQ. Also shown is ˆn, the unit vector normal to the curve. the diﬀerence in the values of ψat any two points PandQconnected by some path C, as shown in ﬁgure 20.4. This diﬀerence is given by ψ(Q)−ψ(P)=integraldisplayQ Pdψ=integraldisplayQ Pparenleftbigg∂ψ ∂xdx+∂ψ ∂ydyparenrightbigg , which, on using the Cauchy–Riemann relations, becomes ψ(Q)−ψ(P)=integraldisplayQ Pparenleftbigg −∂φ ∂ydx+∂φ ∂xdyparenrightbigg =integraldisplayQ P∇φ·ˆnds=integraldisplayQ P∂φ ∂nds, where ˆnis the vector unit normal to the path Candsis the arc length along the path; the last equality is written in terms of the normal derivative ∂φ/∂n≡∇φ·ˆn. Now suppose that in an electrostatics application, the path Cis the surface of a conductor; then ∂φ ∂n=−σ /epsilon10, where σis the surface charge density per unit length normal to the xy-plane. Therefore−/epsilon10[ψ(Q)−ψ(P)] is equal to the charge per unit length normal to the xy-plane on the surface of the conductor between the points PandQ. Similarly, in ﬂuid mechanics applications, if the density of the ﬂuid is ρand its velocity V then ρ[ψ(Q)−ψ(P)] =ρintegraldisplayQ P∇φ·ˆnds=ρintegraldisplayQ PV·ˆnds is equal to the mass ﬂux between PandQper unit length perpendicular to the xy-plane. 729 COMPLEX VARIABLESIA conducting circular cylinder of radius ais placed with its centre line passing through the origin and perpendicular to a uniform electric ﬁeld Ein the x-direction. Find the charge per unit length induced on the half of the cylinder that lies in the region x<0. As mentioned after the previous example, the appropriate complex potential for this problem is f(z)=−E(z−a2/z). Writing z=rexpiθthis becomes f(z)=−E / rexpiθ−a2 rexp(−iθ) / =−E / r−a2 r / cosθ−iE / r+a2 r / sinθ, so that on r=athe imaginary part of fis given by ψ=−2Easinθ. Therefore the induced charge qper unit length on the left half of the cylinder, between θ=π/2a n d θ=3π/2, is given by q=2/epsilon10Ea[sin(3 π/2)−sin(π/2)] =−4/epsilon10Ea. J 20.8 Conformal transformations We now turn our attention to the subject of transformations, by which we mean a change of coordinates from the complex variable z=x+iyto another, say w=r+is, by means of a prescribed formula: w=g(z)=r(x, y)+is(x, y). Under such a transformation, or mapping , the Argand diagram for the z-variable is transformed into one for the w-variable, although the complete z-plane might be mapped onto only a part of the w-plane, or onto the whole of the w-plane, or onto some or all of the w-plane covered more than once. We shall consider only those mappings for which wandzare related by a function w=g(z) and its inverse z=h(w) that are analytic, except possibly at a few isolated points; such mappings are called conformal . Their important properties are that, except at points at which g/prime(z), and hence h/prime(z), is zero or inﬁnite: (i) continuous lines in the z-plane transform into continuous lines in the w-plane; (ii) the angle between two intersecting curves in the z-plane equals the angle between the corresponding curves in the w-plane; (iii) the magniﬁcation, as between the z-a n d w-plane, of a small line element in the neighbourhood of any particular point is independent of the directionof the element; (iv) any analytic function of ztransforms to an analytic function of wand vice versa. 730 20.8 CONFORMAL TRANSFORMATIONS y θ1 θ2C1 C2z1 z2 z0 w=g(z)s rC/prime 1 C/prime 2 φ1 φ2w0w1 w2 x Figure 20.5 Two curves C1andC2in the z-plane, which are mapped onto C/prime 1andC/prime 2in the w-plane. Result (i) is immediate, and results (ii) and (iii) can be justiﬁed by the following argument. Let two curves C1andC2pass through the point z0in the z-plane and z1andz2be two points on their respective tangents at z0, each a distance ρfrom z0. The same prescription with wreplacing zdescribes the transformed situation; however, the transformed tangents may not be straight lines and the distancesofw 1andw2from w0have not yet been shown to be equal. This situation is illustrated in ﬁgure 20.5. In the z-plane z1andz2are given by z1−z0=ρexpiθ1and z2−z0=ρexpiθ2. The corresponding descriptions in the w-plane are w1−w0=ρ1expiφ1and w2−w0=ρ2expiφ2. The angles θiandφiare clear from ﬁgure 20.5. Now since w=g(z), where gis analytic, we have lim z1→z0parenleftbiggw1−w0 z1−z0parenrightbigg = lim z2→z0parenleftbiggw2−w0 z2−z0parenrightbigg =dg dzvextendsinglevextendsinglevextendsinglevextendsingle z=z0, which may be written as lim ρ→0braceleftbiggρ1 ρexp[i(φ1−θ1)]bracerightbigg = lim ρ→0braceleftbiggρ2 ρexp[i(φ2−θ2)]bracerightbigg =g/prime(z0). (20.30) Comparing magnitudes and phases (i.e. arguments) in the equalities (20.30) gives the stated results (ii) and (iii) and adds quantitative information to them, 731 COMPLEX VARIABLES namely that for smallline elements ρ1 ρ≈ρ2 ρ≈|g/prime(z0)|, (20.31) φ1−θ1≈φ2−θ2≈argg/prime(z0). (20.32) For strict comparison with result (ii), (20.32) must be written as θ1−θ2=φ1−φ2, with an ordinary equality sign, since the angles are only deﬁned in the limit ρ→0 when (20.32) becomes a true identity. We also see from (20.31) that the linear magniﬁcation factor is |g/prime(z0)|; similarly, small areas are magniﬁed by |g/prime(z0)|2. Since in the neighbourhoods of corresponding points in a transformation angles are preserved and magniﬁcations are independent of direction, it follows that smallplane ﬁgures are transformed into ﬁgures of the same shape, but, in general, onesthat are magniﬁed and rotated (though not distorted). However, we also notethat at any point where g /prime(z) = 0, the angle arg g/prime(z) through which line elements are rotated is undeﬁned; these are called critical points of the transformation. The ﬁnal result (iv) is perhaps the most important property of conformal transformations. If f(z) is an analytic function of zandz=h(w) is also analytic, then F(w)=f(h(w)) is analytic in w. Its importance lies in the further conclusions it allows us to draw from the fact that, since fis analytic, the real and imaginary parts of f=φ+iψare necessarily solutions of ∂2φ ∂x2+∂2φ ∂y2=0 a n d∂2ψ ∂x2+∂2ψ ∂y2=0. (20.33) Since the transformation property ensures that F=Φ+ iΨ is also analytic, we can conclude that its real and imaginary parts must themselves satisfy Laplace’sequation in the w-plane, ∂ 2Φ ∂r2+∂2Φ ∂s2=0 a n d∂2Ψ ∂r2+∂2Ψ ∂s2=0. (20.34) Further, suppose that (say) Re f(z)=φis constant over a boundary Cin the z-plane; then Re F(w) = Φ is constant over Cin the z-plane. But this is the same as saying that Re F(w) is constant over the boundary C/primein the w-plane, C/primebeing the curve into which Cis transformed by the conformal transformation w=g(z). This is discussed further in the next section. Examples of useful conformal transformations are numerous. For instance, w=z+b,w=( e x p iφ)zandw=azcorrespond respectively to a translation by b, a rotation through an angle φand a stretching (or contraction) in the radial direction (for areal). These three examples can be combined into the general linear transformation w=az+b, where in general aandbare complex. Another example is the inversion mapping w=1/z, which maps the interior of the unit circle to the exterior and vice versa. Other, more complicated, examples also exist. 732 20.8 CONFORMAL TRANSFORMATIONS y x Q R S TPi w=g(z)R/prime P/prime S/primeQ/prime T/primes r Figure 20.6 Transforming the upper half of the z-plane into the interior of the unit circle in the w-plane, in such a way that z=iis mapped onto w=0 and the points x=±∞are mapped onto w=1 .IShow that, if the point z0lies in the upper half of the z-plane then the transformation w=( e x p iφ)z−z0 z−z∗ 0 maps the upper half of the z-plane into the interior of the unit circle in the w-plane. Hence ﬁnd a similar transformation that maps the point z=ionto w=0and the points x=±∞ onto w=1. Taking the modulus of w, we have |w|= ////(expiφ)z−z0 z−z∗ 0 ////= ////z−z0 z−z∗ 0 ////. However, since the complex conjugate z∗ 0is the reﬂection of z0in the real axis, if zandz0 both lie in the upper half of the z-plane then |z−z0|≤|z−z∗ 0|; thus|w|≤1 as required. We also note that (i) the equality holds only when zlies on the real axis, and so this axis is mapped onto the boundary of the unit circle in the w-plane; (ii) the point z0is mapped onto w= 0, the origin of the w-plane. By ﬁxing the images of two points in the z-plane, the constants z0andφc a na l s ob e ﬁxed. Since we require the point z=ito be mapped onto w= 0, we have immediately z0=i. By further requiring z=±∞to be mapped onto w=1 ,w eﬁ n d1= w=e x p iφ and so φ= 0. The required transformation is therefore w=z−i z+i, and is illustrated in ﬁgure 20.6. J We conclude this section by mentioning the rather curious Schwarz–Christoﬀel transformation. †Suppose, as shown in ﬁgure 20.7, that we are interested in a (ﬁnite) number of points x1,x2,...,x non the real axis in the z-plane. Then by means of the transformation w=braceleftbigg Aintegraldisplayz 0(ξ−x1)(φ1/π)−1(ξ−x2)(φ2/π)−1···(ξ−xn)(φn/π)−1dξbracerightbigg +B,(20.35) †Strictly speaking the use of this transformation requires an understanding of complex integrals, which are discussed in section 20.10 below. 733 COMPLEX VARIABLES x1x2x3 x4x5w1 w2 w3w4w5 φ1 φ2 φ3φ4φ5y xs rw=g(z) Figure 20.7 Transforming the upper half of the z-plane into the interior of a polygon in the w-plane, in such a way that the points x1,x2,...,x nare mapped onto the vertices w1,w2,...,w nof the polygon with interior angles φ1,φ2,...,φ n. we may map the upper half of the z-plane onto the interior of a closed polygon in thew-plane having nvertices w1,w2,...,w n(which are the images of x1,x2,...,x n) with corresponding interior angles φ1,φ2,...,φ n, as shown in ﬁgure 20.7. The real axis in the z-plane is transformed into the boundary of the polygon itself. The constants Aand Bare complex in general and determine the position, size and orientation of the polygon. It is clear from (20.35) that dw/dz =0a t x=x1,x2,...,x n, and so the transformation is not conformal at these points. There are various subtleties associated with the use of the Schwarz–Christoﬀel transformation. For example, if one of the points on the real axis in the z-plane (usually xn) is taken at inﬁnity then the corresponding factor in (20.35) (i.e. the one involving xn) is not present. In this case, the point(s) x=±∞are considered as one point, since they transform to a single vertex of the polygon in the w-plane. We can also map the upper half of the z-plane onto an inﬁnite openpolygon by considering it as the limiting case of some closed polygon.IFind a transformation that maps the upper half of the z-plane onto the triangular region shown in ﬁgure 20.8 in such a way that the points x1=−1andx2=1are mapped onto the points w=−aandw=arespectively, and the point x3=±∞is mapped onto w=ib. Hence ﬁnd a transformation that maps the upper half of the z-plane into the region −a<r<a ,s>0of the w-plane, as shown in ﬁgure 20.9. Let us denote the angles at w1andw2in the w-plane by φ1=φ2=φ,w h e r e φ=t a n−1(b/a). Since x3is taken at inﬁnity we may omit the corresponding factor in (20.35) to obtain w= / A Zz 0(ξ+1 )(φ/π)−1(ξ−1)(φ/π)−1dξ / +B = / A Zz 0(ξ2−1)(φ/π)−1dξ / +B. (20.36) The required transformation may then be found by ﬁxing the constants AandBas follows. Since the point z= 0 lies on the line segment x1x2it will be mapped onto the line 734 20.9 APPLICATIONS OF CONFORMAL TRANSFORMATIONS φ1 φ2φ3 x1 x2 −11y xw1 w2w3 ib −aas rw=g(z) Figure 20.8 Transforming the upper half of the z-plane into the interior of a triangle in the w-plane. φ1 φ2 x1 x2 −11y xw1 w2w3 w3 −aas rw=g(z) Figure 20.9 Transforming the upper half of the z-plane into the interior of the region−a<r<a ,s>0i nt h e w-plane. segment w1w2in the w-plane, and by symmetry must be mapped onto the point w=0 . Thus setting z=0a n d w= 0 in (20.36) we obtain B= 0. An expression for Acan be found in the form of an integral by setting (for example) z=1a n d w=ain (20.36). We may consider the region in the w-plane in ﬁgure 20.9 to be the limiting case of the triangular region in ﬁgure 20.8 with the vertex w3at inﬁnity. Thus we may use the above, but with the angles at w1andw2set to φ=π/2. From (20.36), we obtain w=A Zz 0dξp ξ2−1=iAsin−1z. By setting z=1a n d w=a, we ﬁnd iA=2a/π, so the required transformation is w=2a πsin−1z. J 20.9 Applications of conformal transformations In the previous section it was shown that, under a conformal transformation w=g(z)f r o m z=x+iyto a new variable w=r+is, if a solution of Laplace’s equation in some region Rof the xy-plane can be found as the real or imaginary 735 COMPLEX VARIABLES part of an analytic function †ofzthen the same expression put in terms of rand swill be a solution of Laplace’s equation in the corresponding region R/primeof the w-plane, and vice versa. In addition, if the solution is constant over the boundary Cof the region Rin the xy-plane then the solution in the w-plane will take the same constant value over the corresponding curve C/primethat bounds R/prime. Thus, from any two-dimensional solution of Laplace’s equation for a particular geometry, further solutions for other geometries can be obtained by makingconformal transformations. From the physical point of view the given geometryis usually complicated and so the solution is sought by transforming to a simplerone. However, working from simpler to more complicated situations can provideuseful experience, and make it more likely that the reverse procedure can be tackled successfully.IFind the complex electrostatic potential associated with an inﬁnite charged conducting plate y=0, and thus obtain those associated with (i) a semi-inﬁnite charged conducting plate ( r>0,s=0), (ii) the inside of a right-angled charged conducting wedge ( r> 0,s=0 and r=0,s>0). Figure 20.10( a) shows the equipotentials (broken lines) and ﬁeld lines (solid) for the inﬁnite charged conducting plane y= 0. Suppose that we elect to make the real part of the complex potential coincide with the conventional electrostatic potential. If the plate isc h a r g e dt oap o t e n t i a l Vthen clearly φ(x, y)=V−ky, (20.37) where kis related to the charge density σbyk=σ//epsilon1 0, since physically the electric ﬁeld E has components (0 ,σ/ /epsilon1 0)a n d E=−∇φ. Thus what is needed is an analytic function of zof which the real part is V−ky.T h i s can be obtained by inspection, but we may proceed formally and use the Cauchy–Riemannrelations to obtain the imaginary part ψ(x, y) thus: ∂ψ ∂y=∂φ ∂x=0 a n d∂ψ ∂x=−∂φ ∂y=k. Hence ψ=kx+cand, absorbing cintoV, the required complex potential is f(z)=V−ky+ikx=V+ikz. (20.38) (i) Now consider the transformation w=g(z)=z2. (20.39) This satisﬁes the criteria for a conformal mapping (except at z= 0) and carries the upper half of the z-plane into the entire w-plane; the equipotential plane y=0g o e si n t ot h e half-plane r>0,s=0 . By the general results proved, f(z) when expressed in terms of randswill give a complex potential of which the real part will be constant on the half-plane in question; †In fact, the original solution in the xy-plane need not be given explicitly as the real or imaginary part of an analytic function. Any solution of ∇2φ= 0 in the xy-plane is carried over into another solution of ∇2φ= 0 in the new variables by a conformal transformation, and vice versa. 736 20.9 APPLICATIONS OF CONFORMAL TRANSFORMATIONS y s s r r x (a)z-plane (b)w-plane ( c)w-plane Figure 20.10 ( a) The equipotential lines (broken) and ﬁeld lines (solid) for an inﬁnite charged conducting plane at y=0 ,w h e r e z=x+iy;(b), (c) after the transformations w=z2,w=z1/2of the situation shown in ( a). we deduce that F(w)=f(z)=V+ikz=V+ikw1/2(20.40) is the required potential. Expressed in terms of r,sandρ=(r2+s2)1/2,w1/2is given by w1/2=ρ1/2 /"/ρ+r 2ρ /1/2 +i /ρ−r 2ρ /1/2 /# (20.41) and, in particular, the electrostatic potential is given by Φ(r,s)=R e F(w)=V−k√ 2 / (r2+s2)1/2−r /1/2. (20.42) The corresponding equipotentials and ﬁeld lines are shown in ﬁgure 20.10( b). Using results (20.27)–(20.29), the magnitude of the electric ﬁeld is |E|=|F/prime(w)|=|1 2ikw−1/2|=1 2k(r2+s2)−1/4. (ii) A transformation ‘converse’ to that used in (i), w=g(z)=z1/2, has the eﬀect of mapping the upper half of the z-plane into the ﬁrst quadrant of the w-plane and the conducting plane y= 0 into the wedge r>0,s=0a n d r=0 , s>0. The complex potential now becomes F(w)=V+ikw2 =V+ik[(r2−s2)+2irs], (20.43) showing that the electrostatic potential is V−2krsand the electric ﬁeld has components E=( 2ks,2kr). (20.44) Figure 20.10( c) indicates the approximate equipotentials and ﬁeld lines. (Note that, in both transformations, g/prime(z)i se i t h e r0o r ∞at the origin and so neither transformation is conformal there. Consequently there is no violation of result (ii), given at the start ofsection 20.8, concerning the angles between intersecting lines.)J 737 COMPLEX VARIABLES φ=0φ=0 Φ=0 Φ=0y xπ/az0w=zαs rw0 w∗ 0 (a) (b) Figure 20.11 ( a) An inﬁnite conducting wedge with interior angle π/αand a line charge at z=z0;(b) after the transformation w=zα,w i t ha na d d i t i o n a l image charge placed at w=w∗ 0. Themethod of images discussed in section 19.5 can also be used in conjunction with conformal transformations to solve Laplace’s equation in two dimensions.IA wedge of angle π/αwith its vertex at z=0is formed by two semi-inﬁnite conducting plates, as shown in ﬁgure 20.11(a). A line charge of strength qper unit length is positioned atz=z0, perpendicular to the z-plane. By considering the transformation w=zα,ﬁ n dt h e complex electrostatic potential for this situation. Let us consider the action of the transformation w=zαon the lines deﬁning the positions of the conducting plates. The plate that lies along the positive x-axis is mapped onto the positive r-axis in the w-plane, whereas the plate that lies along the direction exp( iπ/α)i s mapped into the negative r-axis, as shown in ﬁgure 20.11( b). Similarly the line charge at z0is mapped onto the point w0=zα 0. From ﬁgure 20.11( b), we see that in the w-plane the problem can be solved by introducing a second line charge of opposite sign at the point w∗ 0, so that the potential Φ = 0 along ther-axis. The complex potential for such an arrangement is simply F(w)=−q 2π/epsilon10ln(w−w0)+q 2π/epsilon10ln(w−w∗ 0). Substituting w=zαinto the above shows that the required complex potential in the original z-plane is f(z)=q 2π/epsilon10ln /zα−z∗α 0 zα−zα 0 / . J 20.10 Complex integrals Corresponding to integration with respect to a real variable, it is possible to deﬁne integration with respect to a complex variable between two complex limits.Since the z-plane is two-dimensional there is clearly greater freedom and hence ambiguity in what is meant by a complex integral. If a complex function f(z)i s single-valued and continuous in some region Rin the complex plane, then we can deﬁne the complex integral of f(z) between two points AandBalong some curve 738 20.10 COMPLEX INTEGRALS AB C1C2 C3xy Figure 20.12 Alternative paths for the integral of a function f(z) between A andB. inR; its value will depend, in general, upon the path taken between AandB (see ﬁgure 20.12). However, we will ﬁnd that for some paths that are diﬀerentbut bear a particular relationship to each other, the value of the integral does not depend upon which of the paths is adopted. Let a particular path Cbe described by a continuous (real) parameter t (α≤t≤β) that gives successive positions on Cby means of the equations x=x(t),y =y(t), (20.45) with t=αandt=βcorresponding to the points AandBrespectively. Then the integral along path Cof a continuous function f(z) is written integraldisplay Cf(z)dz (20.46) and can be given explicitly as a sum of real integrals as follows: integraldisplay Cf(z)dz=integraldisplay C(u+iv)(dx+idy) =integraldisplay Cud x−integraldisplay Cvd y+iintegraldisplay Cud y+iintegraldisplay Cvd x =integraldisplayβ αudx dtdt−integraldisplayβ αvdy dtdt+iintegraldisplayβ αudy dtdt+iintegraldisplayβ αvdx dtdt. (20.47) The question of when such an integral exists will not be pursued, except to state that a suﬃcient condition is that dx/dt anddy/dt are continuous. 739 COMPLEX VARIABLES (a)( b)( c)y y y x x x RRR R t t −R −Rs=1iR t=0C3b C3aC1 C2 Figure 20.13 Diﬀerent paths for an integral of f(z)=z−1. See the text for details.IEvaluate the complex integral of f(z)=z−1along the circle |z|=R, starting and ﬁnishing atz=R. The path C1is parameterised as follows (ﬁgure 20.13( a)): z(t)=Rcost+iRsint, 0≤t≤2π, whilst f(z)i sg i v e nb y f(z)=1 x+iy=x−iy x2+y2. Thus the real and imaginary parts of f(z)a r e u=x x2+y2=Rcost R2and v=−y x2+y2=−Rsint R2. Hence, using expression (20.47),Z C11 zdz= Z2π 0cost R(−Rsint)dt− Z2π 0 /−sint R / Rcostd t +i Z2π 0cost RRcostd t+i Z2π 0 /−sint R / (−Rsint)dt (20.48) =0+0+ iπ+iπ=2πi. J With a bit of experience, the reader may be able to evaluate integrals like the LHS of (20.48) directly without having to write them as four separate real integrals. In the present case, integraldisplay C1dz z=integraldisplay2π 0−Rsint+iRcost Rcost+iRsintdt=integraldisplay2π 0id t=2πi. (20.49) This very important result will be used many times later, and the following should be carefully noted: (i) its value, (ii) that this value is independent of R. In the above example the contour was closed, and so it began and ended at 740 20.10 COMPLEX INTEGRALS the same point in the Argand diagram. We can evaluate complex integrals along open paths in a similar way.IEvaluate the complex integral of f(z)=z−1along (i) the contour C2consisting of the semicircle |z|=Rin the half-plane y≥0, (see ﬁgure 20.13(b)), (ii) the contour C3made up of the two straight lines C3aand C3b(see ﬁgure 20.13(c)). (i) This is just as in the previous example, except that now 0 ≤t≤π. With this change we have from (20.48) or (20.49) thatZ C2dz z=πi. (20.50) (ii) The straight lines that make up the countour C3may be parameterised as follows: C3a,z =( 1−t)R+itR for 0≤t≤1; C3b,z =−sR+i(1−s)Rfor 0≤s≤1. With these parameterisations the required integrals may be writtenZ C3dz z= Z1 0−R+iR R+t(−R+iR)dt+ Z1 0−R−iR iR+s(−R−iR)ds. (20.51) If we could take over from real-variable theory that, for real t, R (a+bt)−1dt=b−1ln(a+bt) even if aandbare complex, then these integrals could be evaluated immediately. However, to do this would be presuming to some extent what we wish to show, and so the evaluationmust be made in terms of entirely real integrals. For example, the ﬁrst isZ1 0−R+iR R(1−t)+itRdt= Z1 0(−1+i)(1−t−it) (1−t)2+t2dt = Z1 02t−1 1−2t+2t2dt+i Z1 01 1−2t+2t2dt =1 2 / ln(1−2t+2t2) /1 0+i 2 /" 2tan−1 / t−1 2 1 2 /!/#1 0 =0+i 2 hπ 2− / −π 2 /i =πi 2. The second integral on the right of (20.51) can also be shown to have the value πi/2. ThusZ C3dz z=πi. J Considering the results of the last two examples, which have common inte- grands and limits, some interesting observations are possible. Firstly, the twointegrals from z=Rtoz=−R,a l o n g C 2andC3respectively, have the same value even though the paths taken are diﬀerent. It also follows that if we took aclosed path C 4,g i v e nb y C2from Rto−RandC3traversed backwards from −R toR, then the integral round C4ofz−1would be zero (both parts contributing equal and opposite amounts). This is to be compared with result (20.49), in whichclosed path C 1, beginning and ending at the same place as C4, yields a value 2 πi. 741 COMPLEX VARIABLES It is not true, however, that the integrals along the paths C2andC3are equal for any function f(z), or, indeed, that their values are independent of Rin general.IEvaluate the complex integral of f(z)=R e zalong the paths C1,C2andC3shown in ﬁgure 20.13. (i) If we take f(z)=R e zand the contour C1thenZ C1Rezd z= Z2π 0Rcost(−Rsint+iRcost)dt=iπR2. (ii) Using C2as the contour,Z C2Rezd z= Zπ 0Rcost(−Rsint+iRcost)dt=1 2iπR2. (iii) Finally the integral along C3=C3a+C3bis given byZ C3Rezd z= Z1 0(1−t)R(−R+iR)dt+ Z1 0(−sR)(−R−iR)ds =1 2R2(−1+i)+1 2R2(1 +i)=iR2. J The results of this section demonstrate that the value of an integral between the same two points may depend upon the path that is taken between them but,at the same time, suggest that under some circumstances it is independent of the path. The general situation is summarised in the result of the next section, namely Cauchy’s theorem, which is the cornerstone of the integral calculus of complexvariables. Before discussing Cauchy’s theorem, however, we note an important result concerning complex integrals that will be of some use later. Let us consider theintegral of a function f(z) along some path C.I fMis an upper bound on the value of|f(z)|on the path, i.e. |f(z)|≤MonC,a n d Lis the length of the path C, thenvextendsinglevextendsinglevextendsinglevextendsingleintegraldisplay Cf(z)dzvextendsinglevextendsinglevextendsinglevextendsingle≤integraldisplay c|f(z)||dz|≤Mintegraldisplay Cdl=ML. (20.52) It is straightforward to verify that this result does indeed hold for the complex integrals considered earlier in this section. 20.11 Cauchy’s theorem Cauchy’s theorem states that if f(z) is an analytic function, and f/prime(z) is continuous at each point within and on a closed contour C,t h e n contintegraldisplay Cf(z)dz=0. (20.53) In this statement and from now on we denote an integral around a closed contour bycontintegraltext C. 742 20.11 CAUCHY’S THEOREM To prove this theorem we will need the two-dimensional form of the divergence theorem, known as Green’s theorem in a plane (see section 11.3). This says thatifpandqare two functions with continuous ﬁrst derivatives within and on a closed contour C(bounding a domain R)i nt h e xy-plane, then integraldisplayintegraldisplay Rparenleftbigg∂p ∂x+∂q ∂yparenrightbigg dxdy=contintegraldisplay C(pd y−qd x). (20.54) With f(z)=u+ivanddz=dx+id y, this can be applied to I=contintegraldisplay Cf(z)dz=contintegraldisplay C(ud x−vd y)+icontintegraldisplay C(vd x+ud y) to give I=integraldisplayintegraldisplay Rbracketleftbigg∂(−u) ∂y+∂(−v) ∂xbracketrightbigg dx dy+iintegraldisplayintegraldisplay Rbracketleftbigg∂(−v) ∂y+∂u ∂xbracketrightbigg dx dy. (20.55) Now, recalling that f(z) is analytic and therefore that the Cauchy–Riemann relations (20.5) apply, we see that each integrand is identically zero and thus Iis also zero; this proves Cauchy’s theorem. In fact the conditions of the above proof are more stringent than are needed. The continuity of f/prime(z) is not necessary for the proof of Cauchy’s theorem, analyticity of f(z) within and on Cbeing suﬃcient. However, the proof then becomes more complicated and is too long to be given here. † The connection between Cauchy’s theorem and the zero value of the integral ofz−1around the composite path C4discussed towards the end of the previous section is apparent: the function z−1is analytic in the two regions of the z-plane enclosed by contours ( C2andC3a)a n d( C2andC3b).ISuppose two points AandBin the complex plane are joined by two diﬀerent paths C1 andC2. Show that if f(z)is an analytic function on each path and in the region enclosed by the two paths then the integral of f(z)is the same along C1andC2. The situation is shown in ﬁgure 20.14. Since f(z)i sa n a l y t i ci n Rit follows from Cauchy’s theorem that we haveZ C1f(z)dz− Z C2f(z)dz= I C1−C2f(z)dz=0, since C1−C2forms a closed contour enclosing R. Thus we immediately obtainZ C1f(z)dz= Z C2f(z)dz, and so the values of the integrals along C1andC2are equal. J An important application of Cauchy’s theorem is in proving that in some cases it †The reader may refer to almost any book devoted to complex variables and the theory of functions. 743 COMPLEX VARIABLES AB Ry xC1 C2 Figure 20.14 Two paths C1andC2enclosing a region R. is possible to deform a closed contour Cinto another contour γ,i ns u c haw a yt h a t the integrals of a function f(z) around each of the contours have the same value.IConsider two closed contours Candγin the Argand diagram, γbeing suﬃciently small that it lies completely within C. Show that if the function f(z)is analytic in the region between the two contours thenI Cf(z)dz= I γf(z)dz. (20.56) To prove this result we consider a contour as shown in ﬁgure 20.15. The two close parallel lines C1andC2joinγandC, which are ‘cut’ to accommodate them. The new contour Γ so formed consists of C,C1,γandC2. Within the area bounded by Γ the function f(z) is analytic and therefore, by Cauchy’s theorem (20.53),I Γf(z)dz=0. (20.57) Now the parts C1andC2of Γ are traversed in opposite directions, and in the limit lie on top of each other, and so their contributions to (20.57) cancel. ThusI Cf(z)dz+ I γf(z)dz=0. (20.58) The sense of the integral round γis opposite to the conventional (anticlockwise) one, and so by traversing γin the usual sense, we establish the result (20.56). J A sort of converse of Cauchy’s theorem is known as Morera’s theorem ,w h i c h states that if f(z) is a continuous function of zin a closed domain Rbounded by ac u r v e Cand, further,contintegraltext Cf(z)dz=0 ,t h e n f(z) is analytic in R. 744 20.12 CAUCHY’S INTEGRAL FORMULA y xγ C1 C2C Figure 20.15 The contour used to prove the result (20.56). 20.12 Cauchy’s integral formula Another very important theorem in the theory of complex variables is Cauchy’s integral formula , which states that if f(z) is analytic within and on a closed contour Candz0is a point within Cthen f(z0)=1 2πicontintegraldisplay Cf(z) z−z0dz. (20.59) This formula is saying that the value of an analytic function anywhere inside a closed contour is uniquely determined by its values on the contour †and that the speciﬁc expression (20.59) can be given for the value at the interior point. We may prove Cauchy’s integral formula by using (20.56) and taking γto be a circle centred on the point z=z0, of small enough radius ρthat it all lies inside C. Then, since f(z) is analytic inside C, the integrand f(z)/(z−z0) is analytic in the space between Candγ. Thus, from (20.56), the integral around γhas the same value as that around C. We then use the fact that any point zonγis given by z=z0+ρexpiθ(and sodz=iρexpiθ dθ). Thus the value of the integral around γis given by I=contintegraldisplay γf(z) z−z0dz=integraldisplay2π 0f(z0+ρexpiθ) ρexpiθiρexpiθ dθ =iintegraldisplay2π 0f(z0+ρexpiθ)dθ. †The similarity between this and the uniqueness theorem for Dirichlet boundary conditions (see chapter 18) is apparent. 745 COMPLEX VARIABLES If the radius of the circle γis now shrunk to zero, i.e. ρ→0, then I→2πif(z0), thus establishing the result (20.59). An extension to Cauchy’s integral formula can be made, yielding an integral expression for f/prime(z0): f/prime(z0)=1 2πiintegraldisplay Cf(z) (z−z0)2dz, (20.60) under the same conditions as previously stated.IProve Cauchy’s integral formula for f/prime(z0)given in (20.60). To show this, we use the deﬁnition of a derivative and (20.59) itself to evaluate f/prime(z0) = lim h→0f(z0+h)−f(z0) h = lim h→0 /1 2πi I Cf(z) h /1 z−z0−h−1 z−z0 / dz / = lim h→0 /1 2πi I Cf(z) (z−z0−h)(z−z0)dz / =1 2πi I Cf(z) (z−z0)2dz, which establishes the result (20.60). J Further, it may be proved by induction that the nth derivative of f(z) is also given by a Cauchy integral, f(n)(z0)=n! 2πicontintegraldisplay Cf(z)dz (z−z0)n+1. (20.61) Thus, if the value of the analytic function is known on Cthen not only may the value of the function at any interior point be calculated, but also the values ofallits derivatives. The observant reader will notice that (20.61) may also be obtained by the formal device of diﬀerentiating under the integral sign with respect to z 0in Cauchy’s integral formula (20.59), f(n)(z0)=1 2πicontintegraldisplay C∂n ∂zn 0bracketleftbiggf(z) (z−z0)bracketrightbigg dz =n! 2πicontintegraldisplay Cf(z)dz (z−z0)n+1. 746 20.13 TAYLOR AND LAURENT SERIESISuppose that f(z)is analytic inside and on a circle Cof radius Rcentred on the point z=z0.I f|f(z)|≤Mon the circle, where Mis some constant, show that |f(n)(z0)|≤Mn! Rn. (20.62) From (20.61) we have |f(n)(z0)|=n! 2π //// I Cf(z)dz (z−z0)n+1 //// and using (20.52) this becomes |f(n)(z0)|≤n! 2πM Rn+12πR=Mn! Rn. This result is known as Cauchy’s inequality . J We may use Cauchy’s inequality to prove Liouville’s theorem , which states that iff(z) is analytic and bounded for all zthen fis a constant. Setting n=1i n (20.62) and letting R→∞we ﬁnd|f/prime(z0)|= 0 and hence f/prime(z0) = 0. Since f(z)i s analytic for all zwe may take z0as any point in the z-plane and thus f/prime(z)=0 for all z; this implies f(z) = constant. Liouville’s theorem may be used in turn to prove the fundamental theorem of algebra (see exercise 20.12). 20.13 Taylor and Laurent series Following on from (20.61), we may establish Taylor’s theorem f o rf u n c t i o n so fa complex variable. If f(z) is analytic inside and on a circle Cof radius Rcentred on the point z=z0,a n d zis a point inside C,t h e n f(z)=∞summationdisplay n=0an(z−z0)n, (20.63) where anis given by f(n)(z0)/n!. The Taylor expansion is valid inside the region of analyticity and, for any particular z0,c a nb es h o w nt ob eu n i q u e . To prove Taylor’s theorem (20.63), we note that, since f(z) is analytic inside and on C, we may use Cauchy’s formula to write f(z)a s f(z)=1 2πicontintegraldisplay Cf(ξ) ξ−zdξ, (20.64) where ξlies on C. Now we may expand the factor ( ξ−z)−1as a geometric series in (z−z0)/(ξ−z0), 1 ξ−z=1 ξ−z0∞summationdisplay n=0parenleftbiggz−z0 ξ−z0parenrightbiggn , 747 COMPLEX VARIABLES so (20.64) becomes f(z)=1 2πicontintegraldisplay Cf(ξ) ξ−z0∞summationdisplay n=0parenleftbiggz−z0 ξ−z0parenrightbiggn dξ =1 2πi∞summationdisplay n=0(z−z0)ncontintegraldisplay Cf(ξ) (ξ−z0)n+1dξ =1 2πi∞summationdisplay n=0(z−z0)n2πif(n)(z0) n!, (20.65) where we have used Cauchy’s integral formula (20.61) for the derivatives of f(z). Cancelling the factors of 2 πi, we thus establish the result (20.63) with an=f(n)(z0)/n!.IShow that if f(z)andg(z)are analytic in some region R, and f(z)=g(z)within some subregion SofR,t h e n f(z)=g(z)throughout R. It is simpler to consider the (analytic) function h(z)=f(z)−g(z), and to show that because h(z)=0i n Sit follows that h(z) = 0 throughout R. If we choose a point z=z0inSthen we can expand h(z) in a Taylor series about z0, h(z)=h(z0)+h/prime(z0)(z−z0)+1 2h/prime/prime(z0)(z−z0)2+···, which will converge inside some circle Cthat extends at least as far as the nearest part of the boundary of R,s i n c e h(z)i sa n a l y t i ci n R. But since z0lies in S, we have h(z0)=h/prime(z0)=h/prime/prime(z0)=···=0, and so h(z)=0i n s i d e C. We may now expand about a new point, which can lie anywhere within C, and repeat the process. By continuing this procedure we may show that h(z)=0 throughout R. This result is called the identity theorem and, in fact, the equality of f(z)a n d g(z) throughout Rfollows from their equality along as little as some curve in R,o re v e na ta countably inﬁnite number of points in R. J So far we have assumed that f(z) is analytic inside and on the (circular) contour C.I f ,h o w e v e r , f(z) has a singularity inside Cat the point z=z0,t h e ni t cannot be expanded in a Taylor series. Nevertheless, suppose that f(z) has a pole of order patz=z0but is analytic at every other point inside and on C.T h e n the function g(z)=( z−z0)pf(z) is analytic at z=z0, and so may be expanded as a Taylor series about z=z0, g(z)=∞summationdisplay n=0bn(z−z0)n. (20.66) Thus, for all zinside C,f(z) will have a power series representation of the form f(z)=a−p (z−z0)p+···+a−1 z−z0+a0+a1(z−z0)+a2(z−z0)2+···, (20.67) with a−p/negationslash= 0. Such a series, which is an extension of the Taylor expansion, is 748 20.13 TAYLOR AND LAURENT SERIES xy R z0C1C2 Figure 20.16 The region of convergence Rfor a Laurent series of f(z) about a point z=z0where f(z) has a singularity. called a Laurent series . By comparing the coeﬃcients in (20.66) and (20.67), we see that an=bn+p. Now, the coeﬃcients bnin the Taylor expansion of g(z)a r e seen from (20.65) to be given by bn=g(n)(z0) n!=1 2πicontintegraldisplayg(z) (z−z0)n+1dz, and so for the coeﬃcients anin (20.67) we have an=1 2πicontintegraldisplayg(z) (z−z0)n+1+pdz=1 2πicontintegraldisplayf(z) (z−z0)n+1dz, an expression that is valid for both positive and negative n. The terms in the Laurent series with n≥0 are collectively called the analytic part, whilst the remainder of the series, consisting of terms in inverse powers of z−z0, is called the principal part . Depending on the nature of the point z=z0, the principal part may contain an inﬁnite number of terms, so that f(z)=+∞summationdisplay n=−∞an(z−z0)n. (20.68) In this case we would expect the principal part to converge only for |(z−z0)−1| less than some constant, i.e. outside some circle centred on z0. However, the analytic part will converge inside some (diﬀerent) circle also centred on z0.I ft h e latter circle has the greater radius then the Laurent series will converge in theregion Rbetween the two circles (see ﬁgure 20.16); otherwise it does not converge at all. In fact, it may be shown that any function f(z) that is analytic in a region Rbetween two such circles C 1andC2centred on z=z0can be expressed as 749 COMPLEX VARIABLES a Laurent series about z0that converges in R. We note that, depending on the nature of the point z=z0, the inner circle may be a point (when the principal part contains only a ﬁnite number of terms) and the outer circle may have aninﬁnite radius. We may use the Laurent series of a function f(z) about any point z=z 0to classify the nature of that point. If f(z) is actually analytic at z=z0then in (20.68) all anforn<0 must be zero. It may happen that not only are all an zero for n<0 but a0,a1,...,am−1are all zero as well. In this case the ﬁrst non-vanishing term in (20.68) is am(z−z0)mwith m>0, and f(z)i st h e ns a i dt o have a zero of order matz=z0. Iff(z) is not analytic at z=z0then two cases arise, as discussed above ( pis here taken as positive): (i) It is possible to ﬁnd an integer psuch that a−p/negationslash= 0 but a−p−k=0f o ra l l integers k>0; (ii) it is not possible to ﬁnd such a lowest value of −p. In case (i), f(z) is of the form (20.67) and is described as having a pole of order patz=z0; the value of a−1(not a−p) is called the residue off(z) at the pole z=z0, and will play an important part in later applications. For case (ii), in which the negatively decreasing powers of z−z0do not terminate, f(z) is said to have an essential singularity . These deﬁnitions should be compared with those given in section 20.6.IFind the Laurent series of f(z)=1 z(z−2)3 about the singularities z=0andz=2(separately). Hence verify that z=0is a pole of order 1andz=2is a pole of order 3, and ﬁnd the residue of f(z)at each pole. To obtain the Laurent series about z=0 ,w es i m p l yw r i t e f(z)=−1 8z(1−z/2)3 =−1 8z / 1+(−3) / −z 2 / +(−3)(−4) 2! / −z 2 /2 +(−3)(−4)(−5) 3! / −z 2 /3 +··· / =−1 8z−3 16−3z 16−5z2 32−···. Since the lowest power of zis−1, the point z= 0 is a pole of order 1. The residue of f(z) atz= 0 is simply the coeﬃcient of z−1in the Laurent expansion about that point and is equal to−1/8. The Laurent series about z= 2 is most easily found by letting z=2+ ξ(orz−2=ξ) 750 20.13 TAYLOR AND LAURENT SERIES and substituting into the expression for f(z)t oo b t a i n f(z)=1 (2 +ξ)ξ3=1 2ξ3(1 +ξ/2) =1 2ξ3 /" 1− /ξ 2 / + /ξ 2 /2 − /ξ 2 /3 + /ξ 2 /4 −··· /# =1 2ξ3−1 4ξ2+1 8ξ−1 16+ξ 32−··· =1 2(z−2)3−1 4(z−2)2+1 8(z−2)−1 16+z−2 32−···. From this series we see that z= 2 is a pole of order 3 and that the residue of f(z)a tz=2 is 1/8. J As we shall see in the next few sections, ﬁnding the residue of a function at a singularity is of crucial importance in the evaluation of complex integrals.Speciﬁcally, formulae exist for calculating the residue of a function at a particular(singular) point z=z 0without having to expand the function explicitly as a Laurent series about z0and identify the coeﬃcient of ( z−z0)−1. The type of formula generally depends on the nature of the singularity at which the residue is required.ISuppose that f(z)has a pole of order mat the point z=z0. By considering the Laurent series of f(z)about z0, derive a general expression for the residue R(z0)off(z)atz=z0. Hence evaluate the residue of the function f(z)=expiz (z2+1 )2 at the point z=i. Iff(z)h a sap o l eo fo r d e r matz=z0then its Laurent series about this point has the form f(z)=a−m (z−z0)m+···+a−1 (z−z0)+a0+a1(z−z0)+a2(z−z0)2+···, which, on multiplying both sides of the equation by ( z−z0)m,g i v e s (z−z0)mf(z)=a−m+a−m+1(z−z0)+···+a−1(z−z0)m−1+···. Diﬀerentiating both sides m−1t i m e s ,w eo b t a i n dm−1 dzm−1[(z−z0)mf(z)] = ( m−1)!a−1+∞X n=1bn(z−z0)n, for some coeﬃcients bn. In the limit z→z0, however, the terms in the sum disappear and after rearranging we obtain the formula R(z0)=a−1= lim z→z0 /1 (m−1)!dm−1 dzm−1[(z−z0)mf(z)] / , (20.69) which gives the value of the residue of f(z) at the point z=z0. If we now consider the function f(z)=expiz (z2+1 )2=expiz (z+i)2(z−i)2, 751 COMPLEX VARIABLES we see immediately that it has poles of order 2 ( double poles) at z=iandz=−i.T o calculate the residue at (for example) z=i, we may apply the formula (20.69) with m=2 . Performing the required diﬀerentiation we obtain d dz[(z−i)2f(z)] =d dz /expiz (z+i)2 / =1 (z+i)4[(z+i)2iexpiz−2(exp iz)(z+i)]. Setting z=iwe ﬁnd the residue is given by R(i)=1 1!1 16 /; −4ie−1−4ie−1 / =−i 2e. J An important special case of (20.69) occurs when f(z)h a sa simple pole (a pole of order 1) at z=z0. Then the residue at z0is given by R(z0) = lim z→z0[(z−z0)f(z)]. (20.70) Iff(z) has a simple pole at z=z0and, as is often the case, has the form g(z)/h(z), where g(z) is analytic and non-zero at z0andh(z0) = 0, then (20.70) becomes R(z0) = lim z→z0(z−z0)g(z) h(z)=g(z0) lim z→z0(z−z0) h(z) =g(z0) lim z→z01 h/prime(z)=g(z0) h/prime(z0), (20.71) where we have used l’H ˆopital’s rule. This result often provides the simplest way of determining the residue at a simple pole. 20.14 Residue theorem Having seen from Cauchy’s theorem that the value of an integral round a closed contour Cis zero if the integrand is analytic inside the contour, it is natural to ask what value it takes when the integrand is not analytic inside C.T h ea n s w e r to this is contained in the residue theorem, which we now discuss. Suppose the function f(z) has a pole of order mat the point z=z0,a n ds o can be written as a Laurent series about z0of the form f(z)=∞summationdisplay n=−man(z−z0)n. (20.72) Now consider the integral Ioff(z) around a closed contour Cthat encloses z=z0, but no other singular points. Using Cauchy’s theorem this integral has the same value as the integral around a circle γof radius ρcentred on z=z0, since f(z) is analytic in the region between Cand γ. On the circle we have 752 20.14 RESIDUE THEOREM z=z0+ρexpiθ(and dz=iρexpiθ dθ), and so I=contintegraldisplay γf(z)dz =∞summationdisplay n=−mancontintegraldisplay (z−z0)ndz =∞summationdisplay n=−manintegraldisplay2π 0iρn+1exp[i(n+1 )θ]dθ. For every term in the series with n/negationslash=−1, we have integraldisplay2π 0iρn+1exp[i(n+1 )θ]dθ=bracketleftbiggiρn+1exp[i(n+1 )θ] i(n+1 )bracketrightbigg2π 0=0, but for the n=−1t e r mw eo b t a i n integraldisplay2π 0id θ=2πi. Therefore only the term in ( z−z0)−1contributes to the value of the integral around γ(and therefore C), and Itakes the value I=contintegraldisplay Cf(z)dz=2πia−1. (20.73) Thus the integral around any closed contour containing a single pole of general order m(or, by extension, an essential singularity) is equal to 2 πitimes the residue off(z)a tz=z0. If we extend the above argument to the case where f(z) is continuous within and on a closed contour Cand analytic, except for a ﬁnite number of poles, within C, then we arrive at the residue theorem contintegraldisplay Cf(z)dz=2πisummationdisplay jRj, (20.74) wheresummationtext jRjis the sum of the residues of f(z) at its poles within C. The method of proof is indicated by ﬁgure 20.17, in which ( a) shows the original contour Creferred to in (20.74) and ( b) shows a contour C/primegiving the same value to the integral, because fis analytic between CandC/prime. Now the contribution to theC/primeintegral from the polygon (a triangle for the case illustrated) joining the small circles is zero, since fis also analytic inside C/prime. Hence the whole value of the integral comes from the circles and, by result (20.73), each of these contributes2πitimes the residue at the pole it encloses. All the circles are traversed in their positive sense if Cis thus traversed and so the residue theorem follows. Formally, Cauchy’s theorem (20.53) is a particular case of (20.74) in which Cencloses no poles. Finally we mention another important result, which we will use later. Suppose 753 COMPLEX VARIABLES CCC/prime (a) (b) Figure 20.17 The contours used to prove the residue theorem: ( a) the original contour; ( b) the contracted contour encircling each of the poles. that f(z) has a simple pole at z=z0and so may be expanded as the Laurent series f(z)=φ(z)+a−1(z−z0)−1, where φ(z) is analytic within some neighbourhood surrounding z0. We wish to ﬁnd an expression for the integral Ioff(z)a l o n ga n opencontour C,w h i c hi s the arc of a circle of radius ρcentred on z=z0given by |z−z0|=ρ, θ 1≤arg(z−z0)≤θ2, (20.75) where ρis chosen small enough that no singularity of f, other than z=z0, lies within the circle. Then Iis given by I=integraldisplay Cf(z)dz=integraldisplay Cφ(z)dz+a−1integraldisplay C(z−z0)−1dz. If the radius of the arc Cis now allowed to tend to zero then the ﬁrst integral tends to zero, since the path becomes of zero length and φis analytic and therefore continuous along it. On C,z=ρeiθand hence the required expression forIis I= lim ρ→0integraldisplay Cf(z)dz= lim ρ→0parenleftbigg a−1integraldisplayθ2 θ11 ρeiθiρeiθdθparenrightbigg =ia−1(θ2−θ1).(20.76) We note that result (20.73) is a special case of (20.76) in which θ2is equal to θ1+2π. 20.15 Location of zeroes An important use of the residue theorem is to locate the zeroes of functions of a complex variable. The location of such zeroes has a particular applicationin electrical network and general oscillation theory, since the complex zeroes of 754 20.15 LOCATION OF ZEROES certain functions give the system parameters (usually frequencies) at which system instabilities occur. As the basis of a method for locating these zeroes we nextprove three important theorems. (i) If f(z) has poles as its only singularities inside a closed contour Cand is not zero at any point on Cthen contintegraldisplay Cf/prime(z) f(z)dz=2πisummationdisplay j(Nj−Pj). (20.77) Here Njis the order of the jth zero of f(z) enclosed by C. Similarly Pjis the order of the jth pole of f(z) inside C. To prove this we note that, at each position zj,f(z) can be written as f(z)=(z−zj)mjφ(z), (20.78) where φ(z) is analytic and non-zero at z=zjandmjis positive for a zero and negative for a pole. Then the integrand f/prime(z)/f(z)t a k e st h ef o r m f/prime(z) f(z)=mj z−zj+φ/prime(z) φ(z). (20.79) Since φ(zj)/negationslash= 0, the second term on the right is analytic; thus the integrand has a simple pole at z=zj, with residue mj. For zeroes mj=Njand for poles mj=−Pj, and thus by the residue theorem (20.77) follows. (ii) If f(z) is analytic inside Cand not zero at any point on it then 2πsummationdisplay jNj=∆ C[argf(z)], (20.80) where ∆ C[x] denotes the variation in xaround the contour C. Since fis analytic there are no Pj; further, since f/prime(z) f(z)=d dz[Lnf(z)], (20.81) equation (20.77) can be written 2πisummationdisplay Nj=contintegraldisplay Cf/prime(z) f(z)dz=∆ C[Lnf(z)]. (20.82) However, ∆C[Lnf(z)] = ∆ C[ln|f(z)|]+i∆C[argf(z)], (20.83) and, since Cis a closed contour, ln |f(z)|must return to its original value and so the real term on the RHS is zero. Comparison of (20.82) and (20.83) then establishes (20.80), which is known as the principle of the argument . (iii) If f(z)a n d g(z) are analytic within and on a closed contour Cand |g(z)|<|f(z)|onCthen f(z)a n d f(z)+g(z) have the same number of zeroes inside C;t h i si s Rouch ´e’s theorem . 755 COMPLEX VARIABLES With the conditions given, neither f(z)n o r f(z)+g(z) can have a zero on C. So, applying theorem (ii) with an obvious notation, 2πsummationtext jNj(f+g)=∆ C[arg(f+g)] =∆ C[argf]+∆ C[arg(1 + g/f)] =2πsummationtext kNk(f)+∆ C[arg(1 + g/f)]. (20.84) Further, since |g|<|f|onC,1+ g/falways lies within a unit circle centred onz= 1; thus its argument always lies in the range −π/2<arg(1 + g/f)<π /2 and cannot change by any multiple of 2 π. It must therefore return to its original value when zreturns to its starting point having traversed C.H e n c e the second term on the right of (20.84) is zero and the theorem is estab- lished. The importance of Rouch ´e’s theorem is that for some functions, in particular polynomials, only the behaviour of a single term in the function need be con-sidered if the contour is chosen appropriately. For example, for a polynomial,treated as f(z)+g(z), only the properties of its largest- (smallest-) power, taken as f(z), need be investigated, if a circular contour is chosen with radius Rsuﬃciently large (small) that, on the contour, the magnitude of the largest (smallest) power term is greater than the sum of the magnitudes of all other terms. Further, if the zeroes of f(z)+g(z)=summationtext N 0bnznare considered as the roots of f(z)+g(z)=0 , w r i t t e ni nt h ef o r m 1+g(z) f(z)=0, (20.85) then it is apparent that no roots can lie outside (inside) |z|=Rand also that f(z)=bNzN(orb0)h a s N(or 0) zeroes inside |z|=R;f+gconsequently has the same number of zeroes inside the same circle. Aw e a kf o r mo ft h e maximum-modulus theorem may also be deduced. This states that if f(z) is analytic within and on a simple closed contour Cthen|f(z)| attains its maximum value on the boundary of C. Let|f(z)|≤MonCwith equality at at least one point of C. Now suppose that there is a point z=ainside Csuch that|f(a)|>M. Then the function h(z)≡f(a) is such that |h(z)|>|−f(z)|onC, and thus h(z)a n d h(z)−f(z) have the same number of zeroes inside C.B u t h(z)(≡f(a)) has no zeroes inside C and by Rouch ´e’s theorem this would imply that f(a)−f(z) has no zeroes in C. However, f(a)−f(z) clearly has a zero at z=a, and so we have a contradiction; the assumption of a point z=ainside Csuch that|f(a)|>Mmust be invalid. This establishes the theorem. The stronger form of the maximum-modulus theorem, which we do not prove, states in addition that the maximum value of f(z) is not attained at any interior point except for the case where f(z) is a constant. 756 20.15 LOCATION OF ZEROES Yy XR x O Figure 20.18 A contour for locating the zeroes of a polynomial that occur in the ﬁrst quadrant of the Argand diagram.IShow that the four zeroes of h(z)=z4+z+1occur one in each quadrant of the Argand diagram and that all four lie between the circles |z|=2/3and|z|=3/2. Putting z=xandz=iyshows that no zeroes occur on the real or imaginary axes. They must therefore occur in conjugate pairs (as is shown by taking the complex conjugate ofh(z)=0 ) . Now take Cas the contour OXY O shown in ﬁgure 20.18 and consider the changes ∆[arg h] in the argument of h(z)a sztraverses C. (i)OX:a r g his everywhere zero, since his real, and thus ∆ OX[argh]=0 . (ii)XY:z=Rexpiθa n ds oa r g hchanges by an amount ∆XY[argh]=∆ XY[argz4]+∆ XY[arg(1 + z−3+z−4)] =∆ XY[argR4e4iθ]+∆ XY / arg[1 + O( R−3)] / =2π+O ( R−3). (20.86) (iii)YO:z=iyand so arg h=y/(y4+ 1), which starts at O( R−3) and ﬁnishes at 0 as y goes from large Rto 0. It never reaches π/2 because y4+1 = 0 has no real positive root. Thus ∆ YO[argh]=0 . Hence for the complete contour ∆ C[argh]=0+2 π+0+O ( R−3) and, if Ris allowed to tend to inﬁnity, we deduce from (20.80) that h(z) has one zero in the ﬁrst quadrant. Furthermore, since the roots occur in conjugate pairs, a second root must lie in the fourthquadrant and the other pair in the second and third quadrants. To show that the zeroes lie within a given annulus in the z-plane we must apply Rouch ´e’s theorem, as follows. (i) With Cas|z|=3/2,f=z 4,g=z+1 .N o w|f|=8 1/16 on Cand|g|≤1+|z|< 5/2<81/16. Thus since z4=0h a sf o u rr o o t si n s i d e |z|=3/2, so also does z4+z+1=0 . (ii) With Cas|z|=2/3,f=1 , g=z4+z.N o w f=1o n Cand|g|≤|z4|+|z|= 16/81 + 2 /3=7 0 /81<1. Thus since f= 0 has no roots inside |z|=2/3, neither does 1 + z+z4=0 . Hence the four zeroes of h(z)=z4+z+ 1 occur one in each quadrant and all lie between the circles|z|=2/3a n d|z|=3/2. J 757 COMPLEX VARIABLES A further technique useful in locating function zeroes is explained in exer- cise 20.16. 20.16 Integrals of sinusoidal functions The remainder of this chapter is devoted to methods of applying contour inte- gration and the residue theorem to various types of deﬁnite integral. In each case not much preamble is given since, for this material, the simplest explanation is felt to be via a series of worked examples that can be used as models. Suppose that an integral of the form integraldisplay2π 0F(cosθ,sinθ)dθ (20.87) is to be evaluated. It can be made into a contour integral around the unit circle Cby writing z=e x p iθand hence cosθ=1 2(z+z−1),sinθ=−1 2i(z−z−1),d θ =−iz−1dz. (20.88) This contour integral can then be evaluated using the residue theorem, provided the transformed integrand has only a ﬁnite number of poles inside the unit circlea n dn o n eo ni t .IEvaluate I= Z2π 0cos 2θ a2+b2−2abcosθdθ, b > a > 0. (20.89) By de Moivre’s theorem (section 3.4), cosnθ=1 2(zn+z−n). (20.90) Using n= 2 in (20.90) and straightforward substitution for the other functions of θin (20.89) gives I=i 2ab I Cz4+1 z2(z−a/b)(z−b/a)dz. Thus there are two poles inside C, a double pole at z= 0 and a simple pole at z=a/b (recall that b>a). We could ﬁnd the residue of the integrand at z= 0 by expanding the integrand as a Laurent series in zand identifying the coeﬃcient of z−1. Alternatively, we may use the formula (20.69) with m= 2. Denoting the integrand by f(z) we have d dz[z2f(z)] =d dz /z4+1 (z−a/b)(z−b/a) / =(z−a/b)(z−b/a)4z3−(z4+1 ) [ ( z−a/b)+(z−b/a)] (z−a/b)2(z−b/a)2. Setting z= 0 and applying (20.69), we ﬁnd R(0) =a b+b a. 758 20.17 SOME INFINITE INTEGRALS y xR −R OΓ Figure 20.19 A semicircular contour in the upper half-plane. For the simple pole at z=a/b, equation (20.70) gives the residue as R(a/b) = lim z→(a/b) / (z−a/b)f(z) / =(a/b)4+1 (a/b)2(a/b−b/a) =−a4+b4 ab(b2−a2). Therefore by the residue theorem I=2πi×i 2ab /a2+b2 ab−a4+b4 ab(b2−a2) / =2πa2 b2(b2−a2). J 20.17 Some inﬁnite integrals Suppose we wish to evaluate an integral of the form integraldisplay∞ −∞f(x)dx, where f(z) has the following properties. (i)f(z) is analytic in the upper half-plane, Im z≥0, except for a ﬁnite number of poles, none of which are on the real axis. (ii) on a semicircle Γ of radius R(ﬁgure 20.19), Rtimes the maximum of |f| on Γ tends to zero as R→∞ (a suﬃcient condition is that zf(z)→0a s |z|→∞ ). (iii)integraltext0 −∞f(x)dxandintegraltext∞ 0f(x)dxboth exist. The required integral is then given by integraldisplay∞ −∞f(x)dx=2πi×(sum of the residues at poles with Im z>0). (20.91) 759 COMPLEX VARIABLES Sincevextendsinglevextendsinglevextendsinglevextendsingleintegraldisplay Γf(z)dzvextendsinglevextendsinglevextendsinglevextendsingle≤2πR×(maximum of |f|on Γ) , condition (ii) ensures that the integral along Γ tends to zero as R→∞, after which (20.91) is obvious from the residue theorem.IEvaluate I= Z∞ 0dx (x2+a2)4,where ais real . The complex function ( z2+a2)−4has poles of order 4 at z=±aiof which only z=ai is in the upper half-plane. Conditions (ii) and (iii) are clearly satisﬁed. For higher-orderpoles, formula (20.69) for evaluating residues can be tiresome to apply. So ,instead, we putz=ai+ξand expand for small ξto obtain† 1 (z2+a2)4=1 (2aiξ+ξ2)4=1 (2aiξ)4 / 1−iξ 2a /−4 . The coeﬃcient of ξ−1is 1 (2a)4(−4)(−5)(−6) 3! /−i 2a /3 =−5i 32a7, and hence by the residue theoremZ∞ −∞dx (x2+a2)4=10π 32a7, and so I=5π/(32a7). J Condition (i) of the previous method required there to be no poles of the integrand on the real axis, but in fact simple poles on the real axis can beaccommodated by indenting the contour as shown in ﬁgure 20.20. The indentationat the pole z=z 0is in the form of a semicircle γof radius ρin the upper half- plane, thus excluding the pole from the interior of the contour. What is then obtained from a contour integration, apart from the contributions for Γ and γ, is called the principal value of the integral, deﬁned as ρ→0b y : PintegraldisplayR −Rf(x)dx≡integraldisplayz0−ρ −Rf(x)dx+integraldisplayR z0+ρf(x)dx. The remainder of the calculation goes through as before, but the contribution from the semicircle γmust be included. Result (20.76) of section 20.14 shows that since only a simple pole is involved its contribution is −ia−1π, (20.92) where a−1is the residue at the pole and the minus sign arises because γis traversed in the clockwise (negative) sense. †This illustrates another useful technique for determining residues. 760 20.17 SOME INFINITE INTEGRALS y xR −R OΓ γ Figure 20.20 An indented contour used when the integrand has a simple pole on the real axis. We defer giving an example of an indented contour until we have established Jordan’s lemma ; we will then work through an example illustrating both. Jordan’s lemma enables inﬁnite integrals involving sinusoidal functions to be evaluated. Jordan’s lemma. For a function f(z)of a complex variable z,i f (i)f(z)is analytic in the upper half-plane except for a ﬁnite number of poles inImz>0, (ii)the maximum of |f(z)|→0as|z|→∞ in the upper half-plane, (iii)m>0, then IΓ=integraldisplay Γeimzf(z)dz→0asR→∞ , (20.93) where Γis the same semicircular contour as in ﬁgure 20.19. Notice that this condition (ii) is less stringent than the earlier condition (ii) (see the start of this section), since we now only require M(R)→0 and not RM(R)→0, where Mis the maximum †of|f(z)|on|z|=R. The proof of the lemma is straightforward, once it has been observed that, for 0≤θ≤π/2, 1≥sinθ θ≥2 π. (20.94) Then, since on Γ we have |exp(imz)|=|exp(−mRsinθ)|, IΓ≤integraldisplay Γ|eimzf(z)||dz|≤MRintegraldisplayπ 0e−mRsinθdθ=2MRintegraldisplayπ/2 0e−mRsinθdθ. †More strictly the least upper bound. 761 COMPLEX VARIABLES Thus, using (20.94), IΓ<2MRintegraldisplayπ/2 0e−mR(2θ/π)dθ=πM mparenleftbig 1−e−mRparenrightbig <πM m; hence,as R→∞,IΓtends to zero since Mdoes.IFind the principal value ofZ∞ −∞cosmx x−adx, forareal, m>0. Consider the function ( z−a)−1exp(imz); although it has no poles in the upper half-plane it does have a simple pole at z=a, and further |(z−a)−1|→0a s|z|→∞ . We will use a contour like that shown in ﬁgure 20.20 and apply the residue theorem. Symbolically,Za−ρ −R+ Z γ+ ZR a+ρ+ Z Γ=0. (20.95) Now as R→∞ andρ→0 we have R Γ→0, by Jordan’s lemma, and from (20.91) and (20.92) we obtain P Z∞ −∞eimx x−adx−iπa−1=0, (20.96) where a−1is the residue of ( z−a)−1exp(imz)a tz=a, which is exp( ima). Then taking the real and imaginary parts of (20.96) gives P Z∞ −∞cosmx x−adx=−πsinma, as required , P Z∞ −∞sinmx x−adx=πcosma, as a bonus. J 20.18 Integrals of multivalued functions We have discussed brieﬂy some of the properties and diﬃculties associated with certain multivalued functions such as z1/2or Ln z. It was mentioned that one method of managing such functions is by means of a ‘cut plane’. A similartechnique can be used with advantage to evaluate some kinds of inﬁnite integral involving real functions for which the corresponding complex functions are multi- valued. A typical contour employed for functions with a single branch pointlocated at the origin is shown in ﬁgure 20.21. Here Γ is a large circle of radius R andγa small one of radius ρ, both centred on the origin. Eventually we will let R→∞andρ→0. The success of the method is due to the fact that because the integrand is multivalued, its values along the two lines ABandCDjoining z=ρtoz=R arenotequal and opposite although both are related to the corresponding real integral. Again an example gives the best explanation. 762 20.18 INTEGRALS OF MULTIVALUED FUNCTIONS y xA B C DΓ γ Figure 20.21 A typical cut-plane contour for use with multivalued functions that have a single branch point located at the origin.IEvaluate I= Z∞ 0dx (x+a)3x1/2,a > 0. We consider the integrand f(z)=( z+a)−3z−1/2and note that |zf(z)|→0o nt h et w o circles as ρ→0a n d R→∞. Thus the two circles make no contribution to the contour integral. The only pole of the integrand inside the contour is at z=−a(and is of order 3). To determine its residue we put z=−a+ξand expand (noting that ( −a)1/2equals a1/2exp(iπ/2) = ia1/2): 1 (z+a)3z1/2=1 ξ3ia1/2(1−ξ/a)1/2 =1 iξ3a1/2 / 1+1 2ξ a+3 8ξ2 a2+··· / . The residue is thus −3i/(8a5/2). The residue theorem (20.74) now givesZ AB+ Z Γ+ Z DC+ Z γ=2πi /−3i 8a5/2 / . We have seen that R Γand R γvanish and if we denote zbyxalong the line ABthen it has the value z=xexp2 πialong the line DC(note that exp2 πimust not be set equal to 1 until after the substitution for zhas been made in R DC). Substituting these expressions,Z∞ 0dx (x+a)3x1/2+ Z0 ∞dx [xexp2 πi+a]3x1/2exp(1 22πi)=3π 4a5/2. 763 COMPLEX VARIABLES Thus/ 1−1 expπi /Z∞ 0dx (x+a)3x1/2=3π 4a5/2, and I=1 2×3π 4a5/2. J 20.19 Summation of series Sometimes a real inﬁnite series may be summed if a suitable complex function can be found that has poles on the real axis at positions corresponding to thevalues of the dummy variable in the summation, and whose residues at thesepoles are equal to the values of the terms of the series there.IBy consideringI Cπcotπz (a+z)2dz, where ais not an integer and Cis a circle of large radius, evaluate ∞X n=−∞1 (a+n)2. The integrand has (i) simple poles at z= integer n,f o r−∞<n<∞, (ii) a double pole at z=−a. (i) To ﬁnd the residue of cot πz, put z=n+ξfor small ξ: cotπz=cos(nπ+ξπ) sin(nπ+ξπ)≈cosnπ (cosnπ)ξπ=1 ξπ. The residue of the integrand at z=nis thus π(a+n)−2π−1. (ii) Putting z=−a+ξfor small ξand determining the coeﬃcient of ξ−1,† πcotπz (a+z)2=π ξ2cot(−aπ+ξπ) =π ξ2 / cot(−aπ)+ξ /d dz(cotπz) / z=−a+··· / , so that the residue at the double pole z=−ais π[−πcosec2πz]z=−a=−π2cosec2πa. Collecting together these results to express the residue theorem gives I= I Cπcotπz (a+z)2dz=2πi /"NX n=−N1 (a+n)2−π2cosec2πa /# , (20.97) where Nequals the integer part of R. But as the radius RofCtends to∞,c o tπz→∓i (depending on whether Im zis greater or less than zero respectively). Thus I<k Zdz (a+z)2, †This again illustrates one of the techniques for determining residues. 764 20.20 INVERSE LAPLACE TRANSFORM which tends to 0 as R→∞. Thus I→0a sR(and hence N)→∞and (20.97) establishes the result ∞X n=−∞1 (a+n)2=π2 sin2πa. J Series with alternating signs in the terms, i.e. ( −1)n, can also be attempted in this way but using cosec πzinstead of cot πz, since the former has residue (−1)nπ−1atz=n(see exercise 20.30). 20.20 Inverse Laplace transform As a ﬁnal example of contour integration we mention a method whereby the process of Laplace transformation, discussed in chapter 13, can be inverted. It will be recalled that the Laplace transform ¯f(s) of a function f(x),x≥0, is given by ¯f(s)=integraldisplay∞ 0e−sxf(x)dx, Res>s 0. (20.98) In chapter 13, functions f(x) were deduced from the transforms by means of a prepared dictionary. However, an explicit formula for an unknown inverse may be written in the form of an integral. It is known as the Bromwich integral and is given by f(x)=1 2πiintegraldisplayλ+i∞ λ−i∞esx¯f(s)ds, λ > 0, (20.99) where sis treated as a complex variable and the integration is along the line L indicated in ﬁgure 20.22. The position of the line is dictated by the requirementsthatλis positive and that all singularities of ¯f(s) lie to the left of the line. That (20.99) really is the unique inverse of (20.98) is diﬃcult to show for general functions and transforms, but the following veriﬁcation should at least make itplausible: f(x)=1 2πiintegraldisplayλ+i∞ λ−i∞ds esxintegraldisplay∞ 0e−suf(u)du, Re(s)>0,i.e.λ>0, =1 2πiintegraldisplay∞ 0du f(u)integraldisplayλ+i∞ λ−i∞es(x−u)ds =1 2πiintegraldisplay∞ 0du f(u)integraldisplay∞ −∞eλ(x−u)eip(x−u)i dp, putting s=λ+ip, =1 2πintegraldisplay∞ 0f(u)eλ(x−u)2πδ(x−u)du =braceleftBigg f(x)x≥0, 0 x<0.(20.100) 765 COMPLEX VARIABLES Ims Res L λ Figure 20.22 The integration path of the inverse Laplace transform is along the inﬁnite line L.T h eq u a n t i t y λmust be positive and large enough for all poles of the integrand to lie to the left of L. Our main interest here is in the use of contour integration. To employ it to evaluate the line integral in (20.99), the path Lmust be made into a closed contour in such a way that the contribution from the completion either vanishes or is simply calculable. A typical completion is shown in ﬁgure 20.23( a) and would be appropriate if ¯f(s) had a ﬁnite number of poles. For more complicated cases, in which ¯f(s)h a s an inﬁnite sequence of poles but all to the left of Las in ﬁgure 20.23( b), a sequence of circular-arc completions that pass between the poles must be used and f(x)i s obtained as a series. If ¯f(s) is a multivalued function then a cut plane is needed and a contour such as that shown in ﬁgure 20.23( c) might be appropriate. We consider here only the simple case in which the contour in ﬁgure 20.23( a) is used; we refer the reader to the exercises at the end of the chapter for others.Ideally, we would like the contribution to the integral from the circular arc Γ to tend to zero as its radius R→∞. Using a modiﬁed version of Jordan’s lemma, it may be shown that this is indeed the case if there exist constants M>0a n d α>0 such that on Γ |¯f(s)|≤M Rα. Moreover, this condition always holds when ¯f(s)h a st h ef o r m ¯f(s)=P(s) Q(s), where P(s)a n d Q(s) are polynomials and the degree of Q(s) is greater than that ofP(s). 766 20.20 INVERSE LAPLACE TRANSFORM ΓΓ ΓR R R L L L (a)( b)( c) Figure 20.23 Some contour completions for the integration path Lof the inverse Laplace transform. For details of when each is appropriate see the main text. When the contribution from the part-circle Γ tends to zero as R→∞,w e have from the residue theorem that the inverse Laplace transform (20.99) is givensimply by f(t)=summationdisplayparenleftbig residues of ¯f(s)e sxat all polesparenrightbig . (20.101)IFind the function f(x)whose Laplace transform is ¯f(s)=s s2−k2, where kis a constant. It is clear that ¯f(s) is of the form required for the integral over the circular arc Γ to tend to zero as R→∞, and so we may use the result (20.101). Now ¯f(s)esx=sesx (s−k)(s+k) and thus has simple poles at s=kands=−k. Using (20.70) the residues at each pole can be easily calculated as R(k)=kekx 2kand R(−k)=ke−kx 2k. Thus the inverse Laplace transform is given by f(x)=1 2 /; ekx+e−kx / =c o s h kx. This result may be checked by computing the forward transform of cosh kx. J Sometimes a little more care is required when deciding in which half-plane to close the contour C. 767 COMPLEX VARIABLESIFind the function f(x)whose Laplace transform is ¯f(s)=1 s(e−as−e−bs), where aandbare ﬁxed and positive, with b>a. From (20.99) we have the integral f(x)=1 2πi Zλ+i∞ λ−i∞e(x−a)s−e(x−b)s sds. (20.102) Now, despite appearances to the contrary, the integrand has no poles, as may be conﬁrmed by expanding the exponentials as Taylor series about s= 0. Depending on the value of x, several cases arise. (i) For x<a both exponentials in the integrand will tend to zero as Re s→∞. Thus we may close Lwith a circular arc Γ in the righthalf-plane ( λcan be as small as desired), and we observe that s×integrand tends to zero everywhere on Γ as R→∞.W i t hn o poles enclosed and no contribution from Γ, the integral along Lmust also be zero. Thus f(x)=0 f o r x<a . (20.103) (ii) For x>b the exponentials in the integrand will tend to zero as Re s→−∞ ,a n ds o we may close Lin the left half-plane, as in ﬁgure 20.23( a). Again the integral around Γ vanishes for inﬁnite Rand so, by the residue theorem, f(x)=0 f o r x>b . (20.104) (iii) For a<x<b the two parts of the integrand behave in diﬀerent ways and have to be treated separately: I1−I2≡1 2πi Z Le(x−a)s sds−1 2πi Z Le(x−b)s sds. The integrand of I1then vanishes in the far left-han d half-plane, but does now have a (simple) pole at s=0 .C l o s i n g Lin the left half-plane, and using the residue theorem, we obtain I1= residue at s=0o f s−1e(x−a)s=1. (20.105) The integrand of I2, however, vanishes in the far right-hand half-plane (and also has a simple pole at s= 0) and is evaluated by a circular-arc completion in that half-plane. Such a contour encloses no poles and leads to I2=0 . Thus, collecting together results (20.103)–(20.105) we obtain f(x)= /8/>/</>/:0f o r x<a, 1f o r a<x<b , 0f o r x>b, as shown in ﬁgure 20.24. J 20.21 Exercises 20.1 Find an analytic function of z=x+iywhose imaginary part is (ycosy+xsiny)e x p x. 768 20.21 EXERCISES a b1f(x) x Figure 20.24 The result of the Laplace inversion of ¯f(s)=s−1(e−as−e−bs) with b>a. 20.2 Find a function f(z), analytic in a suitable part of the Argand diagram, for which Ref=sin2x cosh 2 y−cos 2x. W h e r ea r et h es i n g u l a r i t i e so f f(z)? 20.3 Find the radii of convergence of the following Taylor series: (a)∞X n=2zn lnn,(b)∞X n=1n!zn nn, (c)∞X n=1znnlnn,(d)∞X n=1 /n+p n /n2 zn,with preal. 20.4 Find the Taylor series expansion about the origin of the function f(z) deﬁned by f(z)=∞X r=1(−1)r+1sin /pz r / where pis a constant. Hence verify that f(z) is a convergent series for all z. 20.5 Determine the types of singularities (if any) possessed by the following functions atz=0a n d z=∞: (a) (z−2)−1,(b) (1 + z3)/z2, (c) sinh(1 /z), (d)ez/z3, (e)z1/2/(1 +z2)1/2. 20.6 Identify the zeroes, poles and essential singularities of the following functions: (a) tan z, (b) [( z−2)/z2] sin[1 /(1−z)],(c) exp(1 /z), (d) tan(1 /z),(e)z2/3. 20.7 Find the real and imaginary parts of the functions (i) z2, (ii)ez, and (iii) cosh πz. By considering the values taken by these parts on the boundaries of the region0≤x,y≤1, determine the solution of Laplace’s equation in that region that satisﬁes the boundary conditions φ(x,0) = 0 ,φ (0,y)=0 , φ(x,1) = x, φ (1,y)=y+s i n πy. 769 COMPLEX VARIABLES 20.8 For the function f(z)=l n /z+c z−c / where cis real, show that the real part uoffis constant on a circle of radius ccosech ucentred on the point z=ccothu. Use this result to show that the electrical capacitance per unit length of two parallel cylinders of radii a, placed with their axes 2 dapart, is proportional to [cosh−1(d/a)]−1. 20.9 Find a complex potential in the z-plane appropriate to a physical situation in which the half-plane x>0,y= 0 has zero potential and the half-plane x<0, y= 0 has potential V. By making the transformation w=a(z+z−1)/2, with areal and positive, ﬁnd the electrostatic potential associated with the half-plane r>a ,s=0a n dt h e half-plane r<−a,s= 0 at potentials 0 and Vrespectively. 20.10 By considering in turn the transformations z=1 2c(w+w−1),w =e x p ζ, where z=x+iy,w=rexpiθ,ζ=ξ+iηandcis a real positive constant, show thatz=ccoshζmaps the strip ξ≥0, 0≤η≤2π, onto the whole z-plane. Which curves in the z-plane correspond to the lines ξ=c o n s t a n ta n d η=c o n s t a n t ? Identify those corresponding respectively to ξ=0 , η=0a n d η=2π. The electric potential φof a charged conducting strip −c≤x≤c,y=0 , satisﬁes φ∼−kln(x2+y2)1/2for large ( x2+y2)1/2, with φconstant on the strip. Show that φ=R e [−kcosh−1(z/c)] and that the magnitude of the electric ﬁeld near the strip is k(c2−x2)−1/2. 20.11 Show that the transformation w= Zz 01 (ζ3−ζ)1/2dζ transforms the upper half-plane into the interior of a square that has one corner at the origin of the w-plane and sides of length L,w h e r e L= Zπ/2 0cosec1/2θd θ . 20.12 The fundamental theorem of algebra states that a complex polynomial pn(z)o f degree nhas precisely ncomplex roots. By applying Liouville’s theorem (see the end of section 20.12) to f(z)=1 /pn(z) prove that pn(z) has at least one complex root. Factor out that root to obtain pn−1(z) and, by repeating the process, prove the above theorem. 20.13 Show that, if ais a positive real constant, the function exp( iaz2) is analytic and →0a s|z|→∞ for 0 <argz≤π/4. By applying Cauchy’s theorem to a suitable contour prove thatZ∞ 0cos(ax2)dx= rπ 8a. 20.14 For the equation 8 z3+z+1=0 : (a) show that all three roots lie between the circles |z|=3/8a n d|z|=5/8; (b) ﬁnd the approximate location of the real root, and hence deduce that the complex ones lie in the ﬁrst and fourth quadrants and have moduli greaterthan 0 .5. 770 20.21 EXERCISES 20.15 (a) Prove that z8+3z3+7z+ 5 has two zeroes in the ﬁrst quadrant. (b) Find in which quadrants the zeroes of 2 z3+7z2+1 0z+ 6 lie. Try to locate them. 20.16 The following is a method of determining the number of zeroes of an nth-degree polynomial f(z) inside the contour Cgiven by|z|=R: (a) put z=R(1 +it)/(1−it)w i t h t=t a n ( θ/2) in−∞≤ t≤∞; (b) obtain f(z)a s A(t)+iB(t) (1−it)n(1 +it)n (1 +it)n; (c) show that arg f(z)=t a n−1(B/A)+ntan−1t; (d) show that ∆ C[argf(z)] = ∆ C[tan−1(B/A)] +nπ; (e) using inspection or a sketch graph, determine ∆ C[tan−1(B/A)] by ﬁnding the discontinuities in B/Aand evaluating tan−1(B/A)a tt=±∞. Use this method, together with the results of the worked example in section 20.15, to show that the zeroes of z4+z+ 1 in the second and third quadrants have |z|<1. 20.17 By considering the real part ofZ−izn−1dz 1−a(z+z−1)+a2, where z=e x p iθandnis a non-negative integer, evaluateZπ 0cosnθ 1−2acosθ+a2dθ, forareal and >1. 20.18 Prove that if f(z) has a simple pole at z0then 1 /f(z) has residue 1 /f/prime(z0)t h e r e . Hence evaluateZπ −πsinθ a−sinθdθ, where ais real and >1. 20.19 The equation of an ellipse in plane polar coordinates r,θ, with one of its foci at the origin, is l r=1−/epsilon1cosθ, where lis a length (that of the latus rectum) and /epsilon1(0</epsilon1< 1) is the eccentricity of the ellipse. Express the area of the ellipse as an integral around the unit circlein the complex plane, and show that the only singularity of the integrand inside the circle is a double pole at z 0=/epsilon1−1−(/epsilon1−2−1)1/2. By setting z=z0+ξand expanding the integrand in powers of ξ, ﬁnd the residue at z0and hence show that the area is equal to πl2(1−/epsilon12)−3/2. (In terms of the semi-axes aandbof the ellipse, l=b2/aand/epsilon12=(a2−b2)/a2.) 20.20 Prove that, for α>0, the integralZ∞ 0tsinαt 1+t2dt has the value ( π/2)exp(−α). 20.21 Prove thatZ∞ 0cosmx 4x4+5x2+1dx=π 6 /; 4e−m/2−e−m / form>0. 771 COMPLEX VARIABLES 20.22 Show that the principal value of the integralZ∞ −∞cos(x/a) x2−a2dx is−(π/a)sin1. 20.23 (a) Prove that the integral of [exp( iπz2)]cosec πzaround the parallelogram with corners±1/2±Rexp(iπ/4) has the value 2 i. (b) Show that the parts of the contour parallel to the real axis give no contribu- tion when R→∞. (c) Evaluate the integrals along the other two sides by putting z/prime=rexp(iπ/4) and working in terms of z/prime+1 2andz/prime−1 2. Hence by letting R→∞show thatZ∞ −∞e−πr2dr=1. 20.24 By applying the residue theorem around a wedge-shaped contour of angle 2 π/n, with one side along the real axis, prove that the integralZ∞ 0dx 1+xn, where nis real and≥2, has the value ( π/n)cosec( π/n). 20.25 Using a suitable cut plane, prove that if αis real and 0 <α< 1t h e nZ∞ 0x−α 1+xdx has the value πcosec πα. 20.26 Show thatZ∞ 0lnx x3/4(1 +x)dx=−√ 2π2. 20.27 By integrating a suitable function around a large semicircle in the upper half plane and a small semicircle centred on the origin, determine the value of I= Z∞ 0(lnx)2 1+x2dx and deduce, as a by-product of your calculation, thatZ∞ 0lnx 1+x2dx=0. 20.28 Prove that ∞X −∞1 n2+3 4n+1 8=4π. Carry out the summation numerically, say between −4 and 4, and note how much of the sum comes from values near the poles of the contour integration. 20.29 (a) Determine the residues at all the poles of the function f(z)=πcotπz a2+z2, where ais a positive real constant. (b) By evaluating, in two diﬀerent ways, the integral Ioff(z) along the straight line joining −∞− ia/2a n d+∞−ia/2, show that ∞X n=11 a2+n2=πcothπa 2a−1 2a2. (c) Deduce the value of P∞ 1n−2. 772 20.22 HINTS AND ANSWERS 20.30 By considering the integral of/sinαz αz /2π sinπz,α <π 2 around a circle of large radius, prove that ∞X m=1(−1)m−1sin2mα (mα)2=1 2. 20.31 Use the Bromwich inversion, and contours such as those shown in ﬁgure 20.23( a), to ﬁnd the functions of which the following are the Laplace transforms: (a)s(s2+b2)−1; (b)n!(s−a)−(n+1),w i t h na positive integer and s>a; (c)a(s2−a2)−1,w i t h s>|a|. (Change variable to t=s−|a|.) Compare your answers with those given in table 13.1. 20.32 Find the function f(t) whose Laplace transform is ¯f(s)=e−s−1+s s2. 20.33 A function f(t) has the Laplace transform F(s)=1 2iln /s+i s−i / the complex logarithm being deﬁned by a ﬁnite branch cut running along the imaginary axis from −itoi. (a) Convince yourself that, for t>0,f(t) can be expressed as a closed contour integral that encloses only the branch cut. (b) Calculate F(s) on either side of the branch cut, evaluate the integral and hence determine f(t). (c) Conﬁrm that the derivative with respect to sof the Laplace transform integral of your answer is the same as that given by dF/ds . 20.34 Use the contour in ﬁgure 20.23( c) to show that the function with Laplace transform s−1/2is (πx)−1/2. (For an integrand of the form r−1/2exp(−rx) change variable to t=r1/2.) 20.22 Hints and answers 20.1 ∂u/∂y =−(expx)(ycosy+xsiny+s i n y);zexpz. 20.2 f=( s i n 2 x−isinh2 y)/(cosh2 y−cos 2x); the special case of zreal shows that f(z)=c o t z; poles at z=nπ. 20.3 (a) 1; (b) 1; (c) 1; (d) e−p. 20.4 The series is given by a2n+1=(−1)n+1p2n+1 (2n+1 ) !∞X r=1(−1)r r2n+1, for integer n≥0,a2n=0 ; R−1= lim [ p×1×(2n+1 )−1] = 0, and so the series is convergent by the root test. 20.5 (a) analytic, analytic; (b) double pole, single pole; (c) essential singularity, ana- lytic; (d) triple pole, essential singularity; (e) branch point, branch point. 773 COMPLEX VARIABLES 20.6 (a) Zeroes at z=nπ, simple poles at z=nπ+π/2, essential singularity at z=∞; (b) zeroes at z=∞,2a n d1−(nπ)−1, double pole at z= 0, essential singularity atz=1 ; (c) zero at∞, essential singularity at z=0 ; (d) zeroes at z=∞and ( nπ)−1, simple poles at z=(nπ+π/2)−1, essential singularity at z=0 ; (e) zero and branch point at the origin, essential singularity at z=∞. 20.7 (i) x2−y2,2xy; (ii) excosy,exsiny; (iii) cosh πxcosπy,sinhπxsinπy; φ(x, y)=xy+( s i n h πxsinπy)/sinhπ. 20.8 Set ccothu1=−d,ccothu2=+d,|ccosech u|=aand note that the capacitance is proportional to ( u2−u1)−1. 20.9 f(z)=−i(V/π)lnz;−i(V/π)ln / (z/a)±[(z/a)2−1]1/2 / . 20.10 ξ= constant, ellipses x2(a+1)−2+y2(a−1)−2=c2/(4a2);η= constant, hyperbolae x2(cosα)−2−y2(sinα)−2=c2. The curves are the cuts −c≤x≤c,y=0 ;a n d |x|≥c,y= 0. The curves for η=2πare the same as those for η=0 . 20.13 Use a contour bounding the sector 0 ≤argz≤π/4 to establish the relationship between the required integral and that with exp( −au2) as the integrand. 20.14 (a) |z|=3/8,|8z3+z|≤51/64<1;|z|=5/8,|8z3|= 125 /64>104/64≥|z+1|; (b) write as 8( z−γ)(z−α−iβ)(z−α+iβ)=0 , γ<0, and then the zero coeﬃcient ofz2shows α>0. Show−3/8>γ>−1/2 and use−8γ(α2+β2)=1 . 20.15 (a) For a quarter-circular contour enclosing the ﬁrst quadrant, the change in the argument of the function is 0 + 8( π/2) + 0 (since y8+ 5 = 0 has no real roots); (b) one negative real zero; a conjugate pair in the second and third quadrants,− 3 2,−1±i. 20.16 A=3−12t2+t4,B=−2t−2t3,∆C[tan−1(B/A)] = 0, ∆ C[argf(z)] = 4 π; hence there are two zeroes inside |z|=1 . 20.17 Pole at z=1/a;πa−n(a2−1)−1. 20.18 The only pole inside the unit circle is at z=ia−i(a2−1)1/2; the residue is given by−(i/2)(a2−1)−1/2; the integral has value 2 π[a(a2−1)−1/2−1]. 20.19 The integrand is 2 l2z(2z−/epsilon1z2−/epsilon1)−2; residue = (4 /epsilon1)−1(/epsilon1−2−1)−3/2. 20.20 Follow the ﬁrst example in section 20.17 and use Jordan’s lemma, pole at z=i. 20.21 Factorise the denominator, showing that the relevant simple poles are at i/2a n d i. 20.22 Use Jordan’s lemma and a semicircular contour indented at z=±a. 20.23 (a) The only pole is at the origin with residue π−1; (b) each is O[exp( −πR2− πR/√ 2)]; (c) the sum of the integrals is 2 i RR −Rexp(−πr2)dr. 20.24 The residue at the only pole inside the contour, z=e x p ( iπ/n)i s−n−1exp(iπ/n). The values of the integrals along the two radii diﬀer by a factor −exp(2 πi/n). 20.25 Use a contour like that shown in ﬁgure 20.21.20.26 See the previous example.20.27 Note that ρln nρ→0a s ρ→0f o ra l l n.W h e n zis on the negative real axis, (lnz)2contains three terms; one of the corresponding integrals is a standard form. The residue at z=iisiπ2/8;I=π3/8. 20.28 EvaluateZπcotπz/;1 2+z //;1 4+z /dz around a large circle centred on the origin; residue at z=−1/2 is 0; residue at z=−1/4i s4 πcot(−π/4). 20.29 (a) ( a2+n2)−1atz=n(integer);−πcoth( πa)/(2a)a tz=±ia. (b) Complete the contour separately in the upper half-plane (including all the poles on the real axis and the one at z=ia), and in the lower half plane (including only the pole at z=−ia). Equate the two expressions for I. (c) Take the limit as a→0, using l’H ˆopital’s rule to give π2/6. 774 20.22 HINTS AND ANSWERS 20.30 The behaviour of the integrand for large |z|is|z|−2exp[(2 α−π)|z|]. The residue atz=±m,f o re a c hi n t e g e r m,i ss i n2(mα)(−1)m/(mα)2. The contour contributes nothing. Required summation = [total sum −(m= 0 term)]/2. 20.31 Poles at (a) ±ib;( b ) t=s−a=0 ,o fo r d e r n+1 ,( c ) t=0a n d t=−2|a|.S e e table 13.1. 20.32 Note that ¯f(s) has no pole at s=0 .F o r t<0 close the Bromwich contour in the right half-plane, and for t>1 in the left half-plane. For 0 <t< 1 the integrand has to be split into separate terms containing e−sands−1 and the completions made in the right and left half-planes respectively. The last of these completedcontours now contains a second-order pole at s=0 . f(t)=1−tfor 0 <t< 1 but is 0 otherwise. 20.33 (a) Note that F(s) has no singularities in Re s<0 and apply Cauchy’s theorem to reshape the Bromwich contour. (b) The real parts of F(s)d i ﬀ e rb y πon either side of the branch cut; f(t)=s i n t/t. (c) Both are −1/(1 +s 2). 20.34 R Γand R γtend to 0 as R→∞andρ→0. Put s=rexpiπands=rexp(−iπ)o n the two sides of the cut and use R∞ 0exp(−t2x)dt=1 2(π/x)1/2. There are no poles inside the contour. 775 21 Tensors It may seem obvious that the quantitative description of physical processes cannot depend on the coordinate system in which they are represented. However, we mayturn this argument around: since physical results must indeed be independent of the choice of coordinate system, what does this imply about the nature of the quantities involved in the description of physical processes? The study of theseimplications and of the classiﬁcation of physical quantities by means of themforms the content of the present chapter. Although the concepts presented here may be applied, with little modiﬁ- cation, to more abstract spaces (most notably the four-dimensional space–time ofspecial or general relativity), we shall restrict our attention to our familiar three-dimensional Euclidean space. This removes the need to discuss the properties ofdiﬀerentiable manifolds and their tangent and dual spaces. The reader who isinterested in these more technical aspects of tensor calculus in general spaces,and in particular their application to general relativity, should consult one of the many excellent textbooks on the subject. † Before the presentation of the main development of the subject, we begin by introducing the summation convention, which will prove very useful in writingtensor equations in a more compact form. We then review the eﬀects of a change of basis in a vector space; such spaces were discussed in chapter 8. This is followed by an investigation of the rotation of Cartesian coordinate systems, andﬁnally we broaden our discussion to include more general coordinate systems andtransformations. †For example, D’Inverno, Introducing Einstein’s Relativity (Oxford, 1992); Foster and Nightingale, A Short Course in General Relativity (Springer-Verlag, 1994); Schutz, A First Course in General Relativity (Cambridge, 1990). 776 21.1 SOME NOTATION 21.1 Some notation Before proceeding further, we introduce the summation convention for subscripts, since its use looms large in the work of this chapter. The convention is that anylower-case alphabetic subscript that appears exactly twice in any term of an expression is understood to be summed over all the values that a subscript inthat position can take (unless the contrary is speciﬁcally stated). The subscriptedquantities may appear in the numerator and/or the denominator of a term in anexpression. This naturally implies that any such pair of repeated subscripts mustoccur only in subscript positions that have the same range of values. Sometimesthe ranges of values have to be speciﬁed but usually they are apparent from the context. The following simple examples illustrate what is meant (in the three-dimensional case): (i)a ixistands for a1x1+a2x2+a3x3; (ii)aijbjkstands for ai1b1k+ai2b2k+ai3b3k; (iii)aijbjkckstands forsummationtext3 j=1summationtext3 k=1aijbjkck; (iv)∂vi ∂xistands for∂v1 ∂x1+∂v2 ∂x2+∂v3 ∂x3; (v)∂2φ ∂xi∂xistands for∂2φ ∂x2 1+∂2φ ∂x2 2+∂2φ ∂x2 3. Subscripts that are summed over are called dummy subscripts and the others free subscripts . It is worth remarking that when introducing a dummy subscript into an expression, care should be taken not to use one that is already present,either as a free or as a dummy subscript. For example, a ijbjkcklcannot, and must not, be replaced by aijbjjcjlor by ailblkckl, but could be replaced by aimbmkckl or by aimbmncnl. Naturally, free subscripts must not be changed at all unless the working calls for it. Furthermore, as we have done throughout this book, we will make frequent use of the Kronecker delta δij, which is deﬁned by δij=braceleftBigg 1i f i=j, 0o t h e r w i s e . When the summation convention has been adopted, the main use of δijis to replace one subscript by another in certain expressions. Examples might include bjδij=bi, and aijδjk=aijδkj=aik. (21.1) 777 TENSORS In the second of these the dummy index shared by both terms on the left-hand side (namely j) has been replaced by the free index carried by the Kronecker delta (namely k), and the delta symbol has disappeared. In matrix language, (21.1) can be written as AI=A,w h e r e Ais the matrix with elements aijand Iis the unit matrix having the same dimensions as A. In some expressions we may use the Kronecker delta to replace indices in a number of diﬀerent ways, e.g. aijbjkδki=aijbjior akjbjk, where the two expressions on the RHS are totally equivalent to one another. 21.2 Change of basis In chapter 8 some attention was given to the subject of changing the basis set (or coordinate system) in a vector space and it was shown that, under such a change,diﬀerent types of quantity behave in diﬀerent ways. These results are given insection 8.15, but are summarised below for convenience, using the summationconvention. Although throughout this section we will remind the reader that weare using this convention, it will simply be assumed in the remainder of the chapter. If we introduce a set of basis vectors e 1,e2,e3into our familiar three-dimensional (vector) space, then we can describe any vector xin terms of its components x1,x2,x3with respect to this basis: x=x1e1+x2e2+x3e3=xiei, where we have used the summation convention to write the sum in a more compact form. If we now introduce a new basis e/prime 1,e/prime 2,e/prime 3related to the old one by e/prime j=Sijei(sum over i), (21.2) where the coeﬃcient Sijis the ith component of the vector e/prime jwith respect to the unprimed basis, then we may write xwith respect to the new basis as x=x/prime 1e/prime1+x/prime 2e/prime2+x/prime 3e/prime3=x/prime ie/primei(sum over i). If we denote the matrix with elements SijbyS, then the components x/prime iandxi in the two bases are related by x/prime i=(S−1)ijxj (sum over j), where, using the summation convention, there is an implicit sum over jfrom j=1t o j= 3. In the special case where the transformation is a rotation of the coordinate axes, the transformation matrix Sis orthogonal and we have x/prime i=(ST)ijxj=Sjixj(sum over j). (21.3) 778 21.3 CARTESIAN TENSORS Scalars behave diﬀerently under transformations, however, since they remain unchanged. For example, the value of the scalar product of two vectors x·y (which is just a number) is unaﬀected by the transformation from the unprimedto the primed basis. Diﬀerent again is the behaviour of linear operators. If a linear operator Ais represented by some matrix Ain a given coordinate system then in the new (primed) coordinate system it is represented by a new matrixA /prime=S−1AS. In this chapter we develop a general formulation to describe and classify these diﬀerent types of behaviour under a change of basis (or coordinate transfor-mation). In the development, the generic name tensor is introduced, and certain scalars, vectors and linear operators are described respectively as tensors of ze- roth, ﬁrst and second order (the order –o rrank– corresponds to the number of subscripts needed to specify a particular element of the tensor). Tensors of thirdand fourth order will also occupy some of our attention. 21.3 Cartesian tensors We begin our discussion of tensors by considering a particular class of coordinate transformation – namely rotations – and we shall conﬁne our attention strictly to the rotation of Cartesian coordinate systems. Our object is to study the prop-erties of various types of mathematical quantities, and their associated physicalinterpretations, when they are described in terms of Cartesian coordinates andthe axes of the coordinate system are rigidly rotated from a basis e 1,e2,e3(lying along the Ox1,Ox2andOx3axes) to a new one e/prime 1,e/prime 2,e/prime 3(lying along the Ox/prime 1, Ox/prime 2andOx/prime 3axes). Since we shall be more interested in how the components of a vector or linear operator are changed by a rotation of the axes than in the relationship betweenthe two sets of basis vectors e iande/prime i, let us deﬁne the transformation matrix L as the inverse of the matrix Sin (21.2). Thus, from (21.2), the components of a position vector x, in the old and new bases respectively, are related by x/prime i=Lijxj. (21.4) Because we are considering only rigid rotations of the coordinate axes, the transformation matrix Lwill be orthogonal, i.e. such that L−1=LT. Therefore the inverse transformation is given by xi=Ljix/prime j. (21.5) The orthogonality of Lalso implies relations among the elements of Lthat express the fact that LLT=LTL=I. In subscript notation they are given by LikLjk=δij and LkiLkj=δij. (21.6) Furthermore, in terms of the basis vectors of the primed and unprimed Cartesian 779 TENSORS O x1x2 x/prime 1x/prime 2 θ θ θ Figure 21.1 Rotation of Cartesian axes by an angle θabout the x3-axis. The three angles marked θand the parallels (broken lines) to the primed axes show how the ﬁrst two equations of (21.7) are constructed. coordinate systems, the transformation matrix is given by Lij=e/prime i·ej. We note that the product of two rotations is also a rotation. For example, suppose that x/prime i=Lijxjandx/prime/prime i=Mijx/prime j; then the composite rotation is described by x/prime/prime i=Mijx/prime j=MijLjkxk=(ML)ikxk, corresponding to the matrix ML.IFind the transformation matrix Lcorresponding to a rotation of the coordinate axes through an angle θabout the e3-axis (or x3-axis), as shown in ﬁgure 21.1. Taking xas a position vector – the most obvious choice – we see from the ﬁgure that the components of xwith respect to the new (primed) basis are given in terms of the components in the old (unprimed) basis by x/prime 1=x1cosθ+x2sinθ, x/prime 2=−x1sinθ+x2cosθ, (21.7) x/prime 3=x3. The (orthogonal) transformation matrix is thus L= /0/@cosθsinθ0 −sinθcosθ0 00 1 /1A. The inverse equations are x1=x/prime 1cosθ−x/prime 2sinθ, x2=x/prime 1sinθ+x/prime 2cosθ, (21.8) x3=x/prime 3, in line with (21.5). J 780 21.4 FIRST- AND ZERO-ORDER CARTESIAN TENSORS 21.4 First- and zero-order Cartesian tensors Using the above example as a guide, we may consider any set of three quantities vi, which are directly or indirectly functions of the coordinates xiand possibly involve some constants, and ask how their values are changed by any rotation ofthe Cartesian axes. The speciﬁc question to be answered is whether the speciﬁcforms v /prime iin the new variables can be obtained from the old ones viusing (21.4), v/prime i=Lijvj. (21.9) If so, the viare said to form the components of a vector orﬁrst-order Cartesian tensor v. The ﬁrst-order tensor vdoes not change under the rotation of the coordinate axes; nevertheless, since the basis set does change, from e1,e2,e3to e/prime 1,e/prime 2,v/prime 3, the components of vmust also change. The changes must be such that vgiven by v=viei=v/prime ie/primei(21.10) is unchanged. By deﬁnition, the position coordinates are themselves the compo- nents of such a tensor. Since the transformation (21.9) is orthogonal, the components of any such ﬁrst-order Cartesian tensor obey a relation that is the inverse of (21.9), vi=Ljiv/prime j. (21.11) We now consider explicit examples. In order to keep the equations to reasonable proportions, the examples will be restricted to the x1x2-plane, i.e. there are no components in the x3-direction. Three-dimensional cases are no diﬀerent in principle – but much longer to write out.IWhich of the following pairs (v1,v2)form the components of a ﬁrst-order Cartesian tensor in two dimensions?: (i)(x2,−x1), (ii)(x2,x1), (iii)(x2 1,x22). We shall consider the rotation discussed in the previous example, and to save space we denote cos θbycand sin θbys. (i) Here v1=x2andv2=−x1, referred to the old axes. In terms of the new coordinates they will be v/prime 1=x/prime 2andv/prime 2=−x/prime 1,i . e . v/prime 1=x/prime 2=−sx1+cx2 v/prime 2=−x/prime 1=−cx1−sx2.(21.12) Now if we start again and evaluate v/prime 1andv/prime 2as given by (21.9) we ﬁnd that v/prime 1=L11v1+L12v2=cx2+s(−x1) v/prime 2=L21v1+L22v2=−s(x2)+c(−x1).(21.13) The expressions for v/prime 1andv/prime 2in (21.12) and (21.13) are the same whatever the values ofθ(i.e. for allrotations) and thus by de ﬁnition (21.9) the pair ( x2,−x1)isa ﬁrst-order Cartesian tensor. 781 TENSORS (ii) Here v1=x2andv2=x1. Following the same procedure, v/prime 1=x/prime 2=−sx1+cx2 v/prime 2=x/prime 1=cx1+sx2. But, by (21.9), for a Cartesian tensor we must have v/prime 1=cv1+sv2=cx2+sx1 v/prime 2=(−s)v1+cv2=−sx2+cx1. These two sets of expressions do not agree and thus the pair ( x2,x1) is not a ﬁrst-order Cartesian tensor. (iii)v1=x2 1andv2=x2 2. As in (ii) above, considering the ﬁrst component alone is suﬃcient to show that this pair is nota ﬁrst-order tensor. Evaluating v/prime 1directly gives v/prime 1=x/prime 12=c2x2 1+2csx1x2+s2x2 2, whilst (21.9) requires that v/prime 1=cv1+sv2=cx2 1+sx2 2, which is quite diﬀerent. J There are many physical examples of ﬁrst-order tensors (i.e. vectors) that will be familiar to the reader. As a straightforward one, we may take the set of Cartesian components of the momentum of a particle of mass m,(m˙x1,m˙x2,m˙x3). This set transforms in all essentials as ( x1,x2,x3), since the other operations involved, multiplication by a number and diﬀerentiation with respect to time, are quiteunaﬀected by any orthogonal transformation of the axes. Similarly, accelerationand force are represented by the components of ﬁrst-order tensors. Other more complicated vectors involving the position coordinates more than once, such as the angular momentum of a particle of mass m,n a m e l y J= x×p=m(x×˙x), are also ﬁrst-order tensors. That this is so is less obvious in component form than for the earlier examples, but may be veriﬁed by writing out the components of Jexplicitly or by appealing to the quotient law to be discussed in section 21.7 and using the Cartesian tensor /epsilon1 ijkfrom section 21.8. Having considered the eﬀects of rotatio ns on vector-like sets of quantities we may consider quantities that are unchanged by a rotation of axes. In our previousnomenclature these have been called scalars but we may also describe them as tensors of zero order . They contain only one element (formally, the number of subscripts needed to identify a particular element is zero); the most obvious non-trivial example associated with a rotation of axes is the square of the distance of a point from the origin, r 2=x2 1+x2 2+x2 3. In the new coordinate system it will have the form r/prime2=x/prime 12+x/prime 22+x/prime 32, which for any rotation has the same value as x2 1+x2 2+x2 3. 782 21.4 FIRST- AND ZERO-ORDER CARTESIAN TENSORS In fact any scalar product of two ﬁrst-order tensors (vectors) is a zero-order tensor (scalar), as might be expected since it can be written in a coordinate-freeway as u·v.IBy considering the components of the vectors uandvwith respect to two Cartesian coordinate systems (related by a rotation), show that the scalar product u·vis invariant under rotation. In the original (unprimed) system the scalar product is given in terms of components byu ivi(summed over i), and in the rotated (primed) system by u/prime iv/prime i=LijujLikvk=LijLikujvk=δjkujvk=ujvj, where we have used the orthogonality relation (21.6). Since the resulting expression in the rotated system is the same as that in the original system, the scalar product is indeedinvariant under rotations.J The above result leads directly to the identiﬁcation of many physically im- portant quantities as zero-order tensors. Perhaps the most immediate of these isenergy, either as potential energy or as an energy density (e.g. F·dr,eE·dr,D·E, B·H,µ·B), but others, such as the angle between two directed quantities, are important. As mentioned in the ﬁrst paragraph of this chapter, in most analyses of physical situations it is a scalar quantity (such as energy) that is to be determined. Such quantities are invariant under a rotation of axes and so it is possible to work with the most convenient set of axes and still have conﬁdence in the results. Complementing the way in which a zero-order tensor was obtained from two ﬁrst-order tensors, so a ﬁrst-order tensor can be obtained from a zero-order tensor. We show this by taking a speciﬁc example, that of the electric ﬁeld E=−∇φ; this is derived from a scalar, the electrostatic potential φand has components E i=−∂φ ∂xi. (21.14) Clearly, Eisa ﬁrst-order tensor, but we may prove this more formally by considering the behaviour of its components (21.14) under a rotation of thecoordinate axes, since the components of the electric ﬁeld E /prime iare then given by E/prime i=parenleftbigg −∂φ ∂xiparenrightbigg/prime =−∂φ/prime ∂x/prime i=−∂xj ∂x/prime i∂φ ∂xj=LijEj, (21.15) where (21.5) has been used to evaluate ∂xj/∂x/prime i. Now (21.15) is in the form (21.9), thus conﬁrming that the components of the electric ﬁeld do behave as thecomponents of a ﬁrst-order tensor. 783 TENSORSIIfviare the components of a ﬁrst-order tensor, show that ∇·v=∂vi/∂x iis a zero-order tensor. In the rotated coordinate system ∇·vis given by/∂vi ∂xi //prime =∂v/prime i ∂x/primei=∂xj ∂x/prime i∂ ∂xj(Likvk)=LijLik∂vk ∂xj, since the elements Lijare not functions of position. Using the orthogonality relation (21.6) we then ﬁnd ∂v/prime i ∂x/primei=LijLik∂vk ∂xj=δjk∂vk ∂xj=∂vj ∂xj. Hence ∂vi/∂x iis invariant under rotation of the axes and is thus a zero-order tensor; this was to be expected since it can be written in a coordinate-free way as ∇·v. J 21.5 Second- and higher-order Cartesian tensors Following on from scalars with no subscripts and vectors with one subscript, we turn to sets of quantities that require two subscripts to identify a particularelement of the set. Let these quantities by denoted by T ij. Taking (21.9) as a guide we deﬁne a second-order Cartesian tensor as follows: theTijform the components of such a tensor if, under the same conditions as for (21.9), T/prime ij=LikLjlTkl (21.16) and Tij=LkiLljT/prime kl. (21.17) At the same time we may deﬁne a Cartesian tensor of general order as follows. The set of expressions Tij···kform the components of a Cartesian tensor if, for all rotations of the axes of coordinates given by (21.4) and (21.5), subject to (21.6),the expressions using the new coordinates, T /prime ij···kare given by T/prime ij···k=LipLjq···LkrTpq···r (21.18) and Tij···k=LpiLqj···LrkT/prime pq···r. (21.19) It is apparent that in three dimensions, an Nth-order Cartesian tensor has 3N components. Since a second-order tensor has two subscripts, it is natural to display its components in matrix form. The notation [ Tij] is used, as well as T, to denote the matrix having Tijas the element in the ith row and jth column.† We may think of a second-order tensor Tas a geometrical entity in a similar way to that in which we viewed linear operators (which transform one vector into †We can also denote the column matrix containing the elements viof a vector by [ vi]. 784 21.5 SECOND- AND HIGHER-ORDER CARTESIAN TENSORS another, without reference to any coordinate system) and consider the matrix containing its components as a representation of the tensor with respect to aparticular coordinate system. Moreover, the matrix T=[T ij], containing the components of a second-order tensor, behaves in the same way under orthogonal transformations T/prime=LTLTas a linear operator. However, not all linear operators are second-order tensors. More speciﬁcally, we require the two subscripts in a second-order tensor to refer to the same coordinatesystem, so that only linear operators that transform a vector to another vector inthe same vector space could be second-order tensors. Thus, although the elementsL ijof the transformation matrix are written with two subscripts, they cannot be the components of a tensor since the two subscripts each refer to a diﬀerent coordinate system. As examples of sets of quantities that are readily shown to be second-order tensors we consider the following. (i)The outer product of two vectors .L e t uiandvi,i=1,2,3, be the components of two vectors uandv, and consider the set of quantities Tijdeﬁned by Tij=uivj. (21.20) The set Tijare called the components of the the outer product ofuandv. Under rotations the components Tijbecome T/prime ij=u/prime iv/prime j=LikukLjlvl=LikLjlukvl=LikLjlTkl, (21.21) which shows that they do transform as the components of a second-order tensor. Use has been made in (21.21) of the fact that uiandviare the components of ﬁrst-order tensors. The outer product of two vectors is often denoted, without reference to any coordinate system, as T=u⊗v. (21.22) (This is not to be confused with the vector product of two vectors, which is itself a vector and is discussed in chapter 7.) The expression (21.22) gives the basis towhich the components T ijof the second-order tensor refer. Since u=uieiand v=viei, we may write the tensor Tas T=uiei⊗vjej=uivjei⊗ej=Tijei⊗ej. (21.23) Moreover, as for the case of ﬁrst-order tensors (see equation(21.10)) we note that the quantities T/prime ijare the components of the sametensor T, but referred to a diﬀerent coordinate system, i.e. T=Tijei⊗ej=T/prime ije/primei⊗e/prime j. These concepts can be extended to higher-order tensors. (ii)The gradient of a vector. Suppose virepresents the components of a vector; 785 TENSORS let us consider the quantities generated by forming the derivatives of each vi, i=1,2,3, with respect to each xj,j=1,2,3, i.e. Tij=∂vi ∂xj. These nine quantities form the components of a second-order tensor, as can be seen from the fact that T/prime ij=∂v/prime i ∂x/primej=∂(Likvk) ∂xl∂xl ∂x/prime j=Lik∂vk ∂xlLjl=LikLjlTkl. In coordinate-free language the tensor Tmay be written as T=∇vand hence gives meaning to the concept of the gradient of a vector, a quantity that was not discussed in the chapter on vector calculus (chapter 10). A test of whether any given set of quantities forms the components of a second- order tensor can always be made by direct substitution of the x/prime iin terms of the xi, followed by comparison with the right-hand side of (21.16). This procedure is extremely laborious, however, and it is almost always better to try to recognisethe set as being expressible in one of the forms just considered, or to make alternative tests based on the quotient law of section 21.7 below.IShow that the Tijgiven by T=[Tij]= / x2 2−x1x2 −x1x2 x2 1 / (21.24) are the components of a second-order tensor. Again we consider a rotation θabout the e3-axis. Carrying out the direct evaluation ﬁrst we obtain, using (21.7), T/prime 11=x/prime 22=s2x2 1−2scx1x2+c2x2 2, T/prime 12=−x/prime 1x/prime2=scx2 1+(s2−c2)x1x2−scx2 2, T/prime 21=−x/prime 1x/prime2=scx2 1+(s2−c2)x1x2−scx2 2, T/prime 22=x/prime 12=c2x2 1+2scx1x2+s2x2 2. Now, evaluating the right-hand side of (21.16), T/prime 11=ccx2 2+cs(−x1x2)+sc(−x1x2)+ssx2 1, T/prime 12=c(−s)x2 2+cc(−x1x2)+s(−s)(−x1x2)+scx2 1, T/prime 21=(−s)cx2 2+(−s)s(−x1x2)+cc(−x1x2)+csx2 1, T/prime 22=(−s)(−s)x2 2+(−s)c(−x1x2)+c(−s)(−x1x2)+ccx2 1. After reorganisation, the corresponding expressions are seen to be the same, showing, as required, that the Tijare the components of a second-order tensor. The same result could be inferred much more easily, however, by noting that the Tij are in fact the components of the outer product of the vector ( x2,−x1) with itself. That (x2,−x1) is indeed a vector was established by (21.12) and (21.13). J Physical examples involving second-order tensors will be discussed in the later sections of this chapter, but we might note here that, for example, the magnetic 786 21.6 THE ALGEBRA OF TENSORS susceptibility and electrical conductivity of materials are described by second- order tensors. 21.6 The algebra of tensors Because of the similarity of ﬁrst- and second-order tensors to column vectors and matrices, it would be expected that similar types of algebraic operation can be carried out with them and so provide ways of constructing new tensors from oldones. In the remainder of this chapter, instead of referring to the T ij(say) as the components of a second-order tensor T, we may sometimes simply refer to Tij as the tensor. It should always be remembered, however, that the Tijare in fact just the components of Tin a given coordinate system and that T/prime ijrefers to the components of the sametensor Tin a diﬀerent coordinate system. The addition and subtraction of tensors follows an obvious deﬁnition; namely that if Vij···kandWij···kare (the components of) tensors of the same order, then their sum and diﬀerence, Sij···kandDij···krespectively, are given by Sij···k=Vij···k+Wij···k, Dij···k=Vij···k−Wij···k, for each set of values i ,j,...,k .T h a t Sij···kandDij···kare the components of tensors follows immediately from the linearity of a rotation of coordinates. It is equally straightforward to show that if the Tij···kare the components of a tensor, then so is the set of quantities formed by interchanging the order of (apair of) indices, e.g. T ji···k. IfTji···kis found to be identical with Tij···kthen Tij···kis said to be symmetric with respect to its ﬁrst two subscripts (or simply ‘symmetric’, for second-ordertensors). If, however, T ji···k=−Tij···kfor every element then it is an antisymmetric tensor. An arbitrary tensor is neither symmetric nor antisymmetric but can alwaysbe written as the sum of a symmetric tensor S ij···kand an antisymmetric tensor Aij···k: Tij···k=1 2(Tij···k+Tji···k)+1 2(Tij···k−Tji···k) =Sij···k+Aij···k. Of course these properties are valid for any pair of subscripts. In (21.20) in the previous section we had an example of a kind of ‘multiplication’ of two tensors, thereby producing a tensor of higher order – in that case two ﬁrst-order tensors were multiplied to give a second-order tensor. Inspection of (21.21) shows that there is nothing particular about the orders of the tensorsinvolved and it follows as a general result that the outer product of an Nth-order tensor with an Mth-order tensor will produce an ( M+N)th-order tensor. An operation that produces the opposite eﬀect – namely, generates a tensor 787 TENSORS of smaller rather than larger order – is known as contraction and consists of making two of the subscripts equal and summing over all values of the equalisedsubscripts.IShow that the process of contraction of a tensor produces another tensor, but with an order reduced by 2. LetTij···l···m···kbe the components of an Nth-order tensor, then T/prime ij···l···m···k=LipLjq···Llr···Lms···Lkn/| /{z /} NfactorsTpq···r···s···n. Thus if, for example, we make the two subscripts landmequal and sum over all values of these subscripts, we obtain T/prime ij···l···l···k=LipLjq···Llr···Lls···LknTpq···r···s···n =LipLjq···δrs···LknTpq···r···s···n =LipLjq···Lkn/| /{z /} (N−2) factorsTpq···r···r···n, showing that Tij···l···l···kare the components of a (diﬀerent) Cartesian tensor of order N−2. J For a second-rank tensor, the process of contraction is the same as taking the trace of the corresponding matrix. The trace Tiiitself is thus a zero-order tensor (or scalar) and hence invariant under rotations, as was noted in chapter 8. The process of taking the scalar product of two vectors can be recast into tensor language as forming the outer product Tij=uivjof two ﬁrst-order tensors uand vand then contracting the second-order tensor Tso formed, to give Tii=uivi,a scalar (invariant under a rotation of axes). As yet another example of a familiar operation that is a particular case of a contraction, we may note that the multiplication of a column vector [ ui]b ya matrix [ Bij] to produce another column vector [ vi], Bijuj=vi, can be looked upon as the contraction Tijjof the third-order tensor Tijkformed from the outer product of Bijanduk. 21.7 The quotient law The previous paragraph appears to give a heavy-handed way of describing a familiar operation, but it leads us to ask whether it has a converse. To put thequestion in more general terms: if we know that BandCare tensors and also that A pq···k···mBij···k···n=Cpq···mij···n, (21.25) 788 21.7 THE QUOTIENT LAW does this imply that the Apq···k···malso form the components of a tensor A?H e r e A,BandCare respectively of Mth,Nth and ( M+N−2)th order and it should be noted that the subscript kthat has been contracted may be any of the subscripts inAandBindependently. Thequotient law for tensors states that if (21.25) holds in all rotated coordinate frames then the Apq···k···mdo indeed form the components of a tensor A.T op r o v e it for general MandNis no more diﬃcult regarding the ideas involved than to show it for speciﬁc MandN, but this does involve the introduction of a large number of subscript symbols. We will therefore take the case M=N= 2, but it will be readily apparent that the principle of the proof holds for general M andN. We thus start with (say) ApkBik=Cpi, (21.26) where BikandCpiare arbitrary second-order tensors. Under a rotation of coor- dinates the set Apk(tensor or not) transforms into a new set of quantities that we will denote by A/prime pk. We thus obtain in succession the following steps, using (21.16), (21.17) and (21.6): A/prime pkB/prime ik=C/prime pi (transforming (21.26)), =LpqLijCqj (since Cis a tensor), =LpqLijAqlBjl (from (21.26)), =LpqLijAqlLmjLnlB/prime mn(since Bis a tensor), =LpqLnlAqlB/prime in (since LijLmj=δim). Now kon the left and non the right are dummy subscripts and thus we may write (A/prime pk−LpqLklAql)B/prime ik=0. (21.27) Since Bik, and hence B/prime ik, is an arbitrary tensor, we must have A/prime pk=LpqLklAql, showing that the A/prime pkare given by the general formula (21.18) and hence that theApkare the components of a second-order tensor. By following an analogous argument, the same result (21.27) and deduction could be obtained if (21.26) were replaced by ApkBki=Cpi, i.e. the contraction being now with respect to a diﬀerent pair of indices. Use of the quotient law to test whether a given set of quantities is a tensor is generally much more convenient than making a direct substitution. A particularway in which it is applied is by contracting the given set of quantities, having 789 TENSORS Nsubscripts, with an arbitrary Nth-order tensor (i.e. one having independently variable components) and determining whether the result is a scalar.IUse the quotient law to show that the elements of T, equation (21.24), are the components of a second-order tensor. The outer product xixjis a second-order tensor. Contracting this with the Tijgiven in (21.24) we obtain Tijxixj=x2 2x21−x1x2x1x2−x1x2x2x1+x2 1x22=0, which is clearly invariant (a zeroth-order tensor). Hence by the quotient theorem Tijmust also be a tensor. J 21.8 The tensors δijand/epsilon1ijk In many places throughout this book we have encountered and used the two- subscript quantity δijdeﬁned by δij=braceleftBigg 1i f i=j, 0o t h e r w i s e . Let us now also introduce the three-subscript Levi–Civita symbol /epsilon1ijk, the value of which is given by /epsilon1ijk=  +1 if i, j, kis an even permutation of 1 ,2,3, −1i f i, j, kis an odd permutation of 1 ,2,3, 0o t h e r w i s e . We will now show that δ ijand/epsilon1ijkare respectively the components of a second- and a third-order Cartesian tensor. Notice that the coordinates xido not appear explicitly in the components of these tensors, their components consisting entirely of 0 and 1. In passing, we also note that /epsilon1ijkis totally antisymmetric, i.e. it changes sign under the interchange of any pair of subscripts. In fact /epsilon1ijk, or any scalar multiple of it, is the onlythree-subscript quantity with this property. Treating δijﬁrst, the proof that it is a second-order tensor is straightforward since, if, from (21.16), we consider the equation δ/prime kl=LkiLljδij=LkiLli=δkl, we see that the transformation generates the same expression (a pattern of 0’s and 1’s) as does the deﬁnition of δ/prime ijin the transformed coordinates. Thus δijtransforms according to the appropriate tensor transformation law and is therefore a second-order tensor. Turning now to /epsilon1ijk, we have to consider the quantity /epsilon1/prime lmn=LliLmjLnk/epsilon1ijk. 790 21.8 THE TENSORS δijAND /epsilon1ijk Let us begin, however, by noting that we may use the Levi–Civita symbol to write an expression for the determinant of a 3 ×3m a t r i x A, |A|/epsilon1lmn=AliAmjAnk/epsilon1ijk, (21.28) which may be shown to be equivalent to the Laplace expansion (see chapter 8). † Indeed many of the properties of determinants discussed in chapter 8 can be proved very eﬃciently using this expression (see exercise 21.9).IEvaluate the determinant of the matrix A= /0/@21−3 34 01−21 /1A. Setting l=1 , m=2a n d n= 3 in (21.28) we ﬁnd |A|=/epsilon1ijkA1iA2jA3k = (2)(4)(1)−(2)(0)(−2)−(1)(3)(1) + ( −3)(3)(−2) + (1)(0)(1)−(−3)(4)(1) = 35 , which may be veriﬁed using the Laplace expansion method. J We can now show that the /epsilon1ijkare in fact the components of a third-order tensor. Using (21.28) with the general matrix Areplaced by the speciﬁc transformation matrix L, we can rewrite the RHS of (21.8)in terms of |L| /epsilon1/prime lmn=LliLmjLnk/epsilon1ijk=|L|/epsilon1lmn. Since Lis orthogonal its determinant has the value unity, and so /epsilon1/prime lmn=/epsilon1lmn. Thus we see that /epsilon1/prime lmnhas exactly the properties of /epsilon1ijkbut with i, j, kreplaced by l,m,n, i.e. it is the same as the expression /epsilon1ijkwritten using the new coordinates. This shows that /epsilon1ijkis a third-order Cartesian tensor. In addition to providing a convenient notation for the determinant of a matrix, δijand /epsilon1ijkcan be used to write many of the familiar expressions of vector algebra and calculus as contracted tensors. For example, provided we are using right-handed Cartesian coordinates, the vector product a=b×chas as its ith component ai=/epsilon1ijkbjck; this should be contrasted with the outer product T=b⊗c, which is a second-order tensor having the components Tij=bicj. †This may be readily extended to an N×Nmatrix A,i . e . |A|/epsilon1i1i2···iN=Ai1j1Ai2j2···AiNjN/epsilon1j1j2···jN, where /epsilon1i1i2···iNequals 1 if i1i2···iNis an even permutation of 1 ,2,...,N and equals−1i fi ti sa n odd permutation; otherwise it equals zero. 791 TENSORSIWrite the following as contracted Cartesian tensors: a·b;∇2φ;∇×v;∇(∇·v);∇×(∇×v); (a×b)·c. The corresponding (contracted) tensor expressions are readily seen to be as follows: a·b=aibi=δijaibj, ∇2φ=∂2φ ∂xi∂xi=δij∂2φ ∂xi∂xj, (∇×v)i=/epsilon1ijk∂vk ∂xj, [∇(∇·v)]i=∂ ∂xi /∂vj ∂xj / =δjk∂2vj ∂xi∂xk, [∇×(∇×v)]i=/epsilon1ijk∂ ∂xj / /epsilon1klm∂vm ∂xl / =/epsilon1ijk/epsilon1klm∂2vm ∂xj∂xl, (a×b)·c=δijci/epsilon1jklakbl=/epsilon1iklciakbl. J An important relationship between the /epsilon1-a n d δ- tensors is expressed by the identity /epsilon1ijk/epsilon1klm=δilδjm−δimδjl. (21.29) To establish the validity of this identity between two fourth-order tensors (the LHS is a once-contracted sixth-order tensor) we consider the various possible cases. The RHS of (21.29) has the values +1 if i=landj=m/negationslash=i, (21.30) −1i fi=mandj=l/negationslash=i, (21.31) 0 for any other set of subscript values i, j, l, m . (21.32) In each product on the LHS khas the same value in both factors and for a non-zero contribution none of i, l, j, m can have the same value as k. Since there are only three values, 1, 2 and 3, that any of the subscripts may take, the onlynon-zero possibilities are i=landj=mor vice versa but not all four subscripts equal (since then each /epsilon1factor is zero, as it would be if i=jorl=m). This reproduces (21.32) for the LHS of (21.29) and also the conditions (21.30) and (21.31). The values in (21.30) and (21.31) are also reproduced in the LHS of(21.29) since (i) if i=landj=m,/epsilon1 ijk=/epsilon1lmk=/epsilon1klmand, whether /epsilon1ijkis +1 or−1, the product of the two factors is +1; and (ii) if i=mandj=l,/epsilon1ijk=/epsilon1mlk=−/epsilon1klmand thus the product /epsilon1ijk/epsilon1klm(no summation) has the value −1. This concludes the establishment of identity (21.29). 792 21.9 ISOTROPIC TENSORS A useful application of (21.29) is in obtaining alternative expressions for vector quantities that arise from the vector product of a vector product.IObtain an alternative expression for ∇×(∇×v). As shown in the previous example, ∇×(∇×v) can be expressed in tensor form as [∇×(∇×v)]i=/epsilon1ijk/epsilon1klm∂2vm ∂xj∂xl =(δilδjm−δimδjl)∂2vm ∂xj∂xl =∂ ∂xi /∂vj ∂xj / −∂2vi ∂xj∂xj =[∇(∇·v)]i−∇2vi, where in the second line we have used the identity (21.29). This result has already been mentioned in chapter 10 and the reader is referred there for a discussion of itsapplicability.J By examining the various possibilities, it is straightforward to verify that, more generally, /epsilon1ijk/epsilon1pqr=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleδ ipδiqδir δjpδjqδjr δkpδkqδkrvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle(21.33) and it is easily seen that (21.29) is a special case of this result. From (21.33) we can derive alternative forms of (21.29), for example, /epsilon1 ijk/epsilon1ilm=δjlδkm−δjmδkl. (21.34) The pattern of subscripts in these identities is most easily remembered by noting that the subscripts on the ﬁrst δon the RHS are those that immediately follow (cyclically, if necessary) the common subscript, here i,i ne a c h /epsilon1-term on the LHS; the remaining combinations of j,k,l,m as subscripts in the other δ-terms on the RHS can then be ﬁlled in automatically. Contracting (21.34) by setting j=l(say) we obtain, since δkk= 3 when using the summation convention, /epsilon1ijk/epsilon1ijm=3δkm−δkm=2δkm, and by contracting once more, setting k=m, we further ﬁnd that /epsilon1ijk/epsilon1ijk=6. (21.35) 21.9 Isotropic tensors It will have been noticed that, unlike most of the tensors discussed (except for scalars), δijand/epsilon1ijkhave the property that all their components have values that are the same whatever rotation of axes is made, i.e. the component values 793 TENSORS are independent of the transformation Lij. Speciﬁcally, δ11has the value 1 in all coordinate frames, whereas for a general second-order tensor Tall we know is that if T11=f11(x1,x2,x3)t h e n T/prime 11=f11(x/prime 1,x/prime2,x/prime3). Tensors with the former property are called isotropic (orinvariant ) tensors. It is important to know the most general form that an isotropic tensor can take, since the description of the physical properties, e.g. the conductivity, magneticsusceptibility or tensile strength, of an isotropic medium (i.e. a medium havingthe same properties whichever way it is orientated) involves an isotropic tensor.In the previous section it was shown that δ ijand/epsilon1ijkare second- and third-order isotropic tensors; we will now show that, to within a scalar multiple, they are theonly such isotropic tensors. Let us begin with isotropic second-order tensors. Suppose T ijis an isotropic tensor; then, by deﬁnition, for anyrotation of the axes we must have that Tij=T/prime ij=LikLjlTkl (21.36) for each of the nine components. First consider a rotation of the axes by 2 π/3 about the (1 ,1,1) direction; this takes Ox1,Ox2,Ox3intoOx/prime 2,Ox/prime 3,Ox/prime 1respectively. For this rotation L13=1 , L21=1 , L32= 1 and all other Lij= 0. This requires that T11=T/prime 11=T33. Similarly T12=T/prime 12=T31. Continuing in this way, we ﬁnd: (a)T11=T22=T33; (b)T12=T23=T31; (c)T21=T32=T13. Next, consider a rotation of the axes (from their original position) by π/2 about the Ox3-axis. In this case L12=−1,L21=1 ,L33= 1 and all other Lij=0 . Amongst other relationships, we must have from (21.36) that: T13=(−1)×1×T23; T23=1×1×T13. Hence T13=T23= 0 and therefore, by parts (b) and (c) above, each element Tij= 0 except for T11,T22andT33, which are all the same. This shows that Tij=λδij.IShow that λ/epsilon1ijkis the only isotropic third-order Cartesian tensor. The general line of attack is as above and so only a minimum of explanation will be given. Tijk=T/prime ijk=LilLjmLknTlmn (in all, there are 27 elements) . Rotate about the (1 ,1,1) direction: this is equivalent to making subscript permutations 1→2→3→1. We ﬁnd (a)T111=T222=T333, (b)T112=T223=T331 (and two similar sets), (c)T123=T231=T312 (and a set involving odd permutations of 1 ,2,3). 794 21.10 IMPROPER ROTATIONS AND PSEUDOTENSORS Rotate by π/2 about the Ox3-axis: L12=−1,L21=1 , L33= 1, the other Lij=0 . (d)T111=(−1)×(−1)×(−1)×T222=−T222, (e)T112=(−1)×(−1)×1×T221, (f)T221=1×1×(−1)×T112, (g)T123=(−1)×1×1×T213. Relations (a) and (d) show that elements with all subscripts the same are zero. Relations (e), (f) and (b) show that all elements with repe ated subscripts are zero. Relations (g) and (c) show that T123=T231=T312=−T213=−T321=−T132. In total, Tijkdiﬀers from /epsilon1ijkby at most a scalar factor, but since /epsilon1ijk(and hence λ/epsilon1ijk) has already been shown to be an isotropic tensor, Tijkmust be the most general third-order isotropic Cartesian tensor. J Using exactly the same procedures as those employed for δijand/epsilon1ijk,i tm a yb e shown that the only isotropic ﬁrst-order tensor is the trivial one with all elementszero. 21.10 Improper rotations and pseudotensors So far we have considered rigid rotations of the coordinate axes described by an orthogonal matrix Lwith|L|= +1, (21.4). Strictly speaking such transfor- mations are called proper rotations . We now broaden our discussion to include transformations that are still described by an orthogonal matrix Lbut for which |L|=−1; these are called improper rotations . This kind of transformation can always be considered as an inversion of the coordinate axes through the origin represented by the equation x /prime i=−xi, (21.37) combined with a proper rotation. The transformation may be looked upon alternatively as one that changes an initially right-handed coordinate system intoa left-handed one; any prior or subsequent proper rotation will not change thisstate of aﬀairs. The most obvious example of a transformation with |L|=−1i s the matrix corresponding to (21.37) itself; in this case L ij=−δij. As we have emphasised in earlier chapters, any real physical vector vmay be considered as a geometrical object (i.e. an arrow in space), which can be referred to independently of any coordinate system and whose direction and magnitude cannot be altered merely by describing it in terms of a diﬀerent coordinate system.Thus the components of vtransform as v /prime i=Lijvjunder allrotations (proper and improper). We can deﬁne another type of object, however, whose components may also be labelled by a single subscript but which transforms as v/prime i=Lijvjunder proper rotations and as v/prime i=−Lijvj(note the minus sign) under improper rotations. In this case, the viare not strictly the components of a true ﬁrst-order Cartesian tensor but instead are said to form the components of a ﬁrst-order Cartesianpseudotensor orpseudovector . 795 TENSORS OOpv x1x2x3 x/prime 1 x/prime2 x/prime 3v/prime p/prime Figure 21.2 The behaviour of a vector vand a pseudovector punder a reﬂection through the origin of the coordinate system x1,x2,x3giving the new system x/prime 1,x/prime2,x/prime3. It is important to realise that a pseudovector (as its name suggests) is not a geometrical object in the usual sense. In particular, it should notbe considered as a real physical arrow in space, since its direction is reversed by an improper transformation of the coordinate axes (such as an inversion through the origin).This is illustrated in ﬁgure 21.2, in which the pseudovector pis shown as a broken line to indicate that it is not a real physical vector. Corresponding to vectors and pseudovectors, zeroth-order objects may be divided into scalars and pseudoscalars – the latter being invariant under rotationbut changing sign on reﬂection. We may also extend the notion of scalars and pseudoscalars, vectors and pseu- dovectors, to objects with two or more subscripts. For two subcripts, as deﬁnedpreviously, any quantity with components that transform as T /prime ij=LikLjlTklun- derallrotations (proper and improper) is called a second-order Cartesian tensor. If, however, T/prime ij=LikLjlTklunder proper rotations but T/prime ij=−LikLjlTklunder improper ones (which include reﬂections), then the Tijare the components of a second-order Cartesian pseudotensor. In general the components of Cartesian pseudotensors of arbitary order transform as T/prime ij···k=|L|LilLjm···LknTlm···n, (21.38) where|L|is the determinant of the transformation matrix. For example, from (21.28) we have that |L|/epsilon1ijk=LilLjmLkn/epsilon1lmn, 796 21.10 IMPROPER ROTATIONS AND PSEUDOTENSORS but since|L|=±1 we may rewrite this as /epsilon1ijk=|L|LilLjmLkn/epsilon1lmn. From this expression, we see that although /epsilon1ijkbehaves as a tensor under proper rotations, as discussed in section 21.8, it should properly be regarded as a third-order Cartesian pseudo tensor.IIfbjandckare the components of vectors, show that the quantities ai=/epsilon1ijkbjckform the components of a pseudovector. In a new coordinate system we have a/prime i=/epsilon1/prime ijkb/primejc/primek =|L|LilLjmLkn/epsilon1lmnLjpbpLkqcq =|L|Lil/epsilon1lmnδmpδnqbpcq =|L|Lil/epsilon1lmnbmcn =|L|Lilal, from which we see immediately that the quantities aiform the components of a pseu- dovector. J The above example is worth some further comment. If we denote the vec- tors with components bjandckbybandcrespectively then, as mentioned in section 21.8, the quantities ai=/epsilon1ijkbjckare the components of the real vector a=b×c,provided that we are using a right-handed Cartesian coordinate system . However, in a coordinate system that is left-handed the quantitites a/prime i=/epsilon1/prime ijkb/prime jc/prime k arenotthe components of the physical vector a=b×c, which has, instead, the components −a/prime i. It is therefore important to note the handedness of a coordinate system before attempting to write in component form the vector relation a=b×c (which is true without reference to any coordinate system). It is worth noting that, although pseudotensors can be useful mathematical objects, the description of the real physical world must usually be in terms oftensors (i.e. scalars, vectors, etc.). †For example, the temperature or density of a gas must be a scalar quantity (rather than a pseudoscalar), since its value doesnot change when the coordinate system used to describe it is inverted through the origin. Similarly, velocity, magnetic ﬁeld strength or angular momentum can only be described by a vector, and not by a pseudovector. At this point, it may be useful to make a brief comment on the distinction between active andpassive transformations of a physical system, as this diﬀerence often causes confusion. In this chapter, we are concerned solely with passive trans- †In fact the quantum-mechanical description of elementary particles, such as electrons, protons and neutrons, requires the introduction of a new kind of mathematical object called a spinor ,w h i c hi s not a scalar, vector, or more general tensor. The study of spinors, however, falls beyond the scope of this book. 797 TENSORS formations, for which the physical system of interest is left unaltered, and only the coordinate system used to describe it is changed. In an active transformation,however, the system itself is altered. As an example, let us consider a particle of mass mthat is located at a position xrelative to the origin Oand hence has velocity ˙x. The angular momentum of the particle about Ois thus J=m(x×˙x). If we merely invert the Cartesian coordinates used to describe this system through O, neither the magnitude nor direction of any these vectors will be changed, since they may be consideredsimply as arrows in space that are independent of the coordinates used to de-scribe them. If, however, we perform the analogous active transformation on the system, by inverting the position vector of the particle through O,t h e ni t is clear that the direction of particle’s velocity will also be reversed, since itis simply the time derivative of the position vector, but that the direction ofits angular momentum vector remains unaltered. This suggests that vectors canbe divided into two categories, as follows: polar vectors (such as position and velocity) which reverse direction under an active inversion of the physical sys-tem through the origin and axial vectors (such as angular momentum), which remain unchanged. It should be emphasised that at no point in this discus- sion have we used the concept of a pseudovector to describe a real physicalquantity.† 21.11 Dual tensors Although pseudotensors are not themselves appropriate for the description of physical phenomena, they are sometimes needed; for example, we may use thepseudotensor /epsilon1 ijkto associate with every antisymmetric second-order tensor Aij (in three dimensions) a pseudovector pigiven by pi=1 2/epsilon1ijkAjk; (21.39) piis called the dualofAij. Thus if we denote the antisymmetric tensor Aby the matrix A=[Aij]= 0 A12−A31 −A12 0 A23 A31−A23 0  then the components of its dual pseudovector are ( p1,p2,p3)=(A23,A31,A12). †The scalar product of a polar vector and an axial vector is a pseudoscalar. It was the experimental detection of the dependence of the angular distribution of electrons of (polar vector) momentum peemitted by polarised nuclei of (axial vector) spin JNupon the pseudoscalar quantity JN·pethat established the existence of the non-conservation of parity in β-decay. 798 21.12 PHYSICAL APPLICATIONS OF TENSORSIUsing (21.39) show that Aij=/epsilon1ijkpk. By contracting both sides of (21.39) with /epsilon1ijk, we ﬁnd /epsilon1ijkpk=1 2/epsilon1ijk/epsilon1klmAlm. Using the identity (21.29) then gives /epsilon1ijkpk=1 2(δilδjm−δimδjl)Alm =1 2(Aij−Aji)=1 2(Aij+Aij)=Aij, where in the last line we use the fact that Aij=−Aji. J By a simple extension, we may associate a dual pseudoscalar swith every totally antisymmetric third-rank tensor Aijk,i.e. one that is antisymmetric with respect to the interchange of every possible pair of subscripts; sis given by s=1 3!/epsilon1ijkAijk. (21.40) Since Aijkis a totally antisymmetric three-subscript quantity, we expect it to equal some multiple of /epsilon1ijk(since this is the only such quantity). In fact Aijk=s/epsilon1ijk, as can be proved by substituting this expression into (21.40) and using (21.35). 21.12 Physical applications of tensors In this section some physical applications of tensors will be given. First-order tensors are familiar as vectors and so we will concentrate on second-order tensors,starting with an example taken from mechanics. Consider a collection of rigidly connected point particles of which the αth, which has mass m (α)and is positioned at r(α)with respect to an origin O,i s typical. Suppose that the rigid assembly is rotating about an axis through Owith angular velocity ω. The angular momentum Jabout Oof the assembly is given by J=summationdisplay αparenleftbig r(α)×p(α)parenrightbig . Butp(α)=m(α)˙r(α)and˙r(α)=ω×r(α),f o ra n y α, and so in subscript form the components of Jare given by Ji=summationdisplay αm(α)/epsilon1ijkx(α) j˙x(α) k =summationdisplay αm(α)/epsilon1ijkx(α) j/epsilon1klmωlx(α) m =summationdisplay αm(α)(δilδjm−δimδjl)x(α) jx(α) mωl =summationdisplay αm(α)bracketleftBigparenleftbig r(α)parenrightbig2δil−x(α) ix(α) lbracketrightBig ωl≡Iilωl, (21.41) where Iilis a symmetric second-order Cartesian tensor (by the quotient rule, see 799 TENSORS section 21.7, since Jand ωare vectors). The tensor is called the inertia tensor atO of the assembly and depends only on the distribution of masses in the assemblyand not upon the direction or magnitude of ω. A more realistic situation obtains if a continuous rigid body is considered. In this case, m (α)must be replaced everywhere by ρ(r)dx dy dz and all summations by integrations over the volume of the body. Written out in full in Cartesians,the inertia tensor for a continuous body would have the form I=[I ij]= integraltext (y2+z2)ρd V−integraltext xyρ dV −integraltext xzρ dV −integraltext xyρ dVintegraltext (z2+x2)ρd V−integraltext yzρdV −integraltext xzρ dV −integraltext yzρdVintegraltext (x2+y2)ρd V , where ρ=ρ(x, y, z) is the mass distribution and dVstands for dx dy dz ;t h e integrals are to be taken over the whole body. The diagonal elements of thistensor are called the moments of inertia and the oﬀ-diagonal elements without the minus signs are known as the products of inertia .IShow that the kinetic energy of the rotating system is given by T=1 2Ijlωjωl. By an argument parallel to that already made for J, the kinetic energy is given by T=1 2 X αm(α) /;˙r(α)·˙r(α) / =1 2 X αm(α)/epsilon1ijkωjx(α) k/epsilon1ilmωlx(α) m =1 2 X αm(α)(δjlδkm−δjmδkl)x(α) kx(α) mωjωl =1 2 X αm(α) h δjl /; r(α) /2−x(α) jx(α) l i ωjωl =1 2Ijlωjωl. Alternatively, since Jj=Ijlωlwe may write the kinetic energy of the rotating system as T=1 2Jjωj. J The above example shows that the kinetic energy of the rotating body can be expressed as a scalar obtained by twice contracting ωwith the inertia tensor. It also shows that the moment of inertia of the body about a line given by the unitvector ˆnisI jlˆnjˆnl(orˆnTIˆnin matrix form). Since I(≡Ijl) is a real symmetric second-order tensor, it has associated with it three mutually perpendicular directions that are its principal axes and have the following properties (proved in chapter 8): (i) with each axis is associated a principal moment of inertia λµ,µ=1,2,3; (ii) when the rotation of the body is about one of these axes, the angular velocity and the angular momentum are parallel and given by J=Iω=λµω, i.e.ωis an eigenvector of Iwith eigenvalue λµ; 800 21.12 PHYSICAL APPLICATIONS OF TENSORS (iii) referred to these axes as coordinate axes, the inertia tensor is diagonal with diagonal entries λ1,λ2,λ3. Two further examples of physical quantities represented by second-order tensors are magnetic susceptibility and electrical conductivity. In the ﬁrst case we have(in standard notation) M i=χijHj, (21.42) and in the second case ji=σijEj. (21.43) Here Mis the magnetic moment per unit volume and jthe current density (current per unit perpendicular area). In both cases we have on the left-hand sidea vector and on the right-hand side the contraction of a set of quantities withanother vector. Each set of quantities must therefore form the components of a second-order tensor. For isotropic media M∝Handj∝E, but for anisotropic materials such as crystals the susceptibility and conductivity may be diﬀerent along diﬀerent crystalaxes, making χ ijandσijgeneral second-order tensors, although they are usually symmetric.IThe electrical conductivity σin a crystal is measured by an observer to have components as shown [σij]= /0/@1√ 20√ 231 01 1 /1A. (21.44) Show that there is one direction in the crystal along which no current can ﬂow. Does the current ﬂow equally easily in the two perpendicular directions? The current density in the crystal is given by ji=σijEj,w h e r e σij,r e l a t i v et ot h e observer’s coordinate system, is given by (21.44). Since [ σij] is a symmetric matrix, it possess three mutually perpendicular eigenvectors (or principal axes) with respect to which the conductivity tensor is diagonal, with diagonal entries λ1,λ2,λ3, the eigenvalues of [σij]. As discussed in chapter 8, the eigenvalues of [ σij] are given by |σ−λI|= 0. Thus we require//////1−λ√ 20√ 23−λ1 01 1 −λ //////=0, from which we ﬁnd (1−λ)[(3−λ)(1−λ)−1]−2(1−λ)=0 . This simpliﬁes to give λ=0,1,4 so that, with respect to its principal axes, the conductivity tensor has components σ/prime ijgiven by [σ/prime ij]= /0/@400 010000 /1A. Since j/prime i=σ/prime ijE/prime j, we see immediately that along one of the principal axes there is no current ﬂow and along the two perpendicular directions the current ﬂows are not equal.J 801 TENSORS We can extend the idea of a second-order tensor that relates two vectors to a situation where two physical second-order tensors are related by a fourth-ordertensor. The most common occurrence of such relationships is in the theory ofelasticity. This is not the place to give a detailed account of elasticity theory, but suﬃce it to say that the local deformation of an elastic body at any interior point Pcan be described by a second-order symmetric tensor e ijcalled the strain tensor .I ti sg i v e nb y eij=1 2parenleftbigg∂ui ∂xj+∂uj ∂xiparenrightbigg , where uis the displacement vector describing the strain of a small volume element whose unstrained position relative to the origin is x. Similarly we can describe the stress in the body at Pby the second-order symmetric stress tensor pij;t h e quantity pijis the xj-component of the stress vector acting across a plane through Pwhose normal lies in the xi-direction. A generalisation of Hooke’s law then relates the stress and strain tensors by pij=cijklekl (21.45) where cijklis a fourth-order Cartesian tensor.IAssuming that the most general fourth-order isotropic tensor is cijkl=λδijδkl+ηδikδjl+νδilδjk, (21.46) ﬁnd the form of (21.45) for an isotropic medium having Young’s modulus Eand Poisson’s ratio σ. For an isotropic medium we must have an isotropic tensor for cijkl, and so we assume the form (21.46). Substituting this into (21.45) yields pij=λδijekk+ηeij+νeji. Buteijis symmetric, and if we write η+ν=2µ, then this takes the form pij=λekkδij+2µeij, in which λandµare known as Lam´e constants . It will be noted that if eij=0f o r i/negationslash=j t h e nt h es a m ei st r u eo f pij, i.e. the principal axes of the stress and strain tensors coincide. Now consider a simple tension in the x1-direction, i.e. p11=Sbut all other pij=0 . Then denoting ekk(summed over k)b yθwe have, in addition to eij=0f o r i/negationslash=j,t h et h r e e equations S=λθ+2µe11, 0=λθ+2µe22, 0=λθ+2µe33. Adding them gives S=θ(3λ+2µ). Substituting for θfrom this into the ﬁrst of the three, and recalling that Young’s modulus is deﬁned by S=Ee11,g i v e s Eas E=µ(3λ+2µ) λ+µ. (21.47) 802 21.13 INTEGRAL THEOREMS FOR TENSORS Further, Poisson’s ratio is deﬁned as σ=−e22/e11(or−e33/e11) and is thus σ= /1 e11 /λθ 2µ= /1 e11 //λ 2µ /Ee11 3λ+2µ=λ 2(λ+µ). (21.48) Solving (21.47) and (21.48) for λandµgives ﬁnally pij=σE (1 +σ)(1−2σ)ekkδij+E (1 +σ)eij. J 21.13 Integral theorems for tensors In chapter 11, we discussed various integral theorems involving vector and scalar ﬁelds. Most notably, we considered the divergence theorem, which states that, forany vector ﬁeld a, integraldisplay V∇·adV=contintegraldisplay Sa·ˆndS, (21.49) where Sis the surface enclosing the volume Vandˆnis the outward-pointing unit normal to Sat each point. Writing (21.49) in subscript notation, we have integraldisplay V∂ak ∂xkdV=contintegraldisplay SakˆnkdS. (21.50) Although we shall not prove it rigorously, (21.50) can be extended in an obvious manner to relate integrals of tensor ﬁelds , rather than just vector ﬁelds, over volumes and surfaces, with the result integraldisplay V∂Tij···k···m ∂xkdV=contintegraldisplay STij···k···mˆnkdS. This form of the divergence theorem for general tensors can be very useful in vector calculus manipulations.IA vector ﬁeld asatisﬁes∇·a=0inside some volume Vanda·ˆn=0on the bound- ary surface S. By considering the divergence theorem applied to Tij=xiaj, show thatR VadV=0. Applying the divergence theorem to Tij=xiajwe ﬁndZ V∂Tij ∂xjdV= Z V∂(xiaj) ∂xjdV= I SxiajˆnjdS=0, since ajˆnj= 0. By expanding the volume integral we obtainZ V∂(xiaj) ∂xjdV= Z V∂xi ∂xjajdV+ Z Vxi∂aj ∂xjdV = Z VδijajdV = Z VaidV=0, where in going from the ﬁrst to the second line we used the fact that ∂xi/∂x j=δijand ∂aj/∂x j=0 . J 803 TENSORS The other integral theorems discussed in chapter 11 can be extended in a similar way. For example, written in tensor notation Stokes’ theorem states that,for a vector ﬁeld a i,integraldisplay S/epsilon1ijk∂ak ∂xjˆnidS=contintegraldisplay Cakdxk. For a general tensor ﬁeld this has the straightforward extension integraldisplay S/epsilon1ijk∂Tlm···k···n ∂xjˆnidS=contintegraldisplay CTlm···k···ndxk. 21.14 Non-Cartesian coordinates So far we have restricted our attention to the study of tensors when they are described in terms of Cartesian coordinates and the axes of coordinates are rigidly rotated, sometimes together with an inversion of axes through the origin. In theremainder of this chapter we shall extend the concepts discussed in the previoussections by considering arbitrary coordinate transformations from one generalcoordinate system to another. Although this generalisation brings with it severalcomplications, we shall ﬁnd that many of the properties of Cartesian tensorsare still valid for more general tensors. Before considering general coordinate transformations, however, we begin by reminding ourselves of some properties of general curvilinear coordinates, as discussed in chapter 10. The position of an arbitrary point Pin space may be expressed in terms of the three curvilinear coordinates u 1,u2,u3. We saw in chapter 10 that if r(u1,u2,u3)i s the position vector of the point Pthen at Pthere exist two sets of basis vectors ei=∂r ∂uiand /epsilon1i=∇ui, (21.51) where i=1,2,3. In general, the vectors in each set neither are of unit length nor form an orthogonal basis. However, the sets eiand /epsilon1iare reciprocal systems of vectors and so ei·/epsilon1j=δij. (21.52) In the context of general tensor analysis, it is more usual to denote the second set of vectors /epsilon1iin (21.51) by ei, the index being placed as a superscript to distinguish it from the (diﬀerent) vector ei, which is a member of the ﬁrst set in (21.51). Although this positioning of the index may seem odd (not least becauseof the possibility of confusion with powers) it forms part of a slight modiﬁcationto the summation convention that we will adopt for the remainder of this chapter. This is as follows: any lower-case alphabetic index that appears exactly twice in any term of an expression, once as a subscript and once as a superscript ,i st ob e summed over all the values that an index in that position can take (unless the 804 21.14 NON-CARTESIAN COORDINATES contrary is speciﬁcally stated). All other aspects of the summation convention remain unchanged. With the introduction of superscripts, the reciprocity relation (21.52) should be rewritten so that both sides of (21.53) have one subscript and one superscript, i.e. as ei·ej=δj i. (21.53) The alternative form of the Kronecker delta is deﬁned in a similar way to previously, i.e. it equals unity if i=jand is zero otherwise. For similar reasons it is usual to denote the curvilinear coordinates themselves byu1,u2,u3, with the index raised, so that ei=∂r ∂uiand ei=∇ui. (21.54) From the ﬁrst equality we see that we may consider a superscript that appears in the denominator of a partial derivative as a subscript. Given the two bases eiandei,w em a yw r i t eag e n e r a lv e c t o r aequally well in terms of either basis as follows: a=a1e1+a2e2+a3e3=aiei; a=a1e1+a2e2+a3e3=aiei. The aiare called the contravariant components of the vector aand the ai thecovariant components, the position of the index (either as a subscript or superscript) serving to distinguish between them. Similarly, we may call the eithe covariant basis vectors and the eithe contravariant ones.IShow that the contravariant and covariant components of a vector aare given by ai=a·ei andai=a·eirespectively. For the contravariant components, we ﬁnd a·ei=ajej·ei=ajδi j=ai, where we have used the reciprocity relation (21.53). Similarly, for the covariant components, a·ei=ajej·ei=ajδj i=ai. J The reason that the notion of contravariant and covariant components of a vector (and the resulting superscript notation) was not introduced earlier isthat for Cartesian coordinate systems the two sets of basis vectors e iandeiare identical and, hence, so are the components of a vector with respect to eitherbasis. Thus, for Cartesian coordinates, we may speak simply of the componentsof the vector and there is no need to diﬀerentiate between contravariance and covariance, or to introduce superscripts to make a distinction between them. If we consider the components of higher-order tensors in non-Cartesian co- ordinates, there are even more possibilities. As an example, let us consider a 805 TENSORS second-order tensor T. Using the outer product notation in (21.23), we may write Tin three diﬀerent ways: T=Tijei⊗ej=Ti jei⊗ej=Tijei⊗ej, where Tij,Ti jandTijare called the contravariant, mixed andcovariant com- ponents of Trespectively. It is important to remember that these three sets of quantities form the components of the sametensor Tbut refer to diﬀerent (tensor) bases made up from the basis vectors of the coordinate system. Again, if we are using Cartesian coordinates then all three sets of components are identical. We may generalise the above equation to higher-order tensors; then com- ponents carrying only superscripts or only subscripts are referred to as thecontravariant and covariant components respectively and all others are calledmixed components. 21.15 The metric tensor Any particular curvilinear coordinate system is completely characterised at each point in space by the nine quantities g ij=ei·ej, (21.55) which, as we will show, are the covariant components of a symmetric second-order tensor gcalled the metric tensor . Since an inﬁnitesimal vector displacement can be written as dr=duiei, we ﬁnd that the square of the inﬁnitesimal arc length ( ds)2c a nb ew r i t t e ni nt e r m so ft h e metric tensor as (ds)2=dr·dr=duiei·dujej=gijduiduj. (21.56) It may further be shown that the volume element dVis given by dV=√gd u1du2du3, (21.57) where gis the determinant of the matrix [ gij], which has the covariant components of the metric tensor as its elements. If we compare equations (21.56) and (21.57) with the analogous ones in section 10.10 then we see that in the special case where the coordinate system is orthogonal(so that e i·ej=0f o r i/negationslash=j) the metric tensor can be written in terms of the coordinate-system scale factors hi,i=1,2,3a s gij=braceleftBigg h2 ii=j, 0i/negationslash=j. Its determinant is then given by g=h2 1h22h23. 806 21.15 THE METRIC TENSORICalculate the elements gijof the metric tensor for cylindrical polar coordinates. Hence ﬁnd the square of the inﬁnitesimal arc length (ds)2and the volume dVfor this coordinate system. As discussed in section 10.9, in cylindrical polar coordinates ( u1,u2,u3)=( ρ, φ, z)a n ds o the position vector rof any point Pmay be written r=ρcosφi+ρsinφj+zk. From this we obtain the (covariant) basis vectors: e1=∂r ∂ρ=c o s φi+s i n φj; e2=∂r ∂φ=−ρsinφi+ρcosφj; e3=∂r ∂z=k. (21.58) Thus the components of the metric tensor [ gij]=[ei·ej] are found to be G=[gij]= /0/@100 0ρ20 001 /1A, (21.59) from which we see that, as expected for an orthogonal coordinate system, the metric tensor is diagonal, the diagonal elements being equal to the squares of the scale factors of thecoordinate system. From (21.56), the square of the inﬁnitesimal arc length in this coordinate system is given by (ds) 2=gijduiduj=(dρ)2+ρ2(dφ)2+(dz)2, and, using (21.57), the volume element is found to be dV=√gd u1du2du3=ρd ρd φd z. These expressions are identical to those derived in section 10.9. J We may also express the scalar product of two vectors in terms of the metric tensor: a·b=aiei·bjej=gijaibj, (21.60) where we have used the contravariant components of the two vectors. Similarly, using the covariant components, we can write the same scalar product as a·b=aiei·bjej=gijaibj, (21.61) where we have deﬁned the nine quantities gij=ei·ej. As we shall show, they form the contravariant components of the metric tensor gand are, in general, diﬀerent from the quantities gij. Finally, we could express the scalar product in terms of the contravariant components of one vector and the covariant components of the other, a·b=aiei·bjej=aibjδi j=aibi, (21.62) 807 TENSORS where we have used the reciprocity relation (21.53). Similarly, we could write a·b=aiei·bjej=aibjδj i=aibi. (21.63) By comparing the four alternative expressions (21.60)–(21.63) for the scalar product of two vectors we can deduce one of the most useful properties of the quantities gijand gij.S i n c e gijaibj=aibiholds for any arbitrary vector components ai, it follows that gijbj=bi, which illustrates the fact that the covariant components gijof the metric tensor can be used to lower an index . In other words, it provides a means of obtaining the covariant components of a vector from its contravariant components. By asimilar argument, we have g ijbj=bi, so that the contravariant components gijcan be used to perform the reverse operation of raising an index . It is straightforward to show that the contravariant and covariant basis vectors, eiandeirespectively, are related in the same way as other vectors, i.e. by ei=gijejand ei=gijej. We also note that, since eiandeiare reciprocal systems of vectors in three- dimensional space (see chapter 7), we may write ei=ej×ek ei·(ej×ek), for the combination of subscripts i, j, k=1,2,3 and its cyclic permutations. A similar expression holds for eiin terms of the ei-basis. Moreover, it may be shown that the triple scalar product |e1·(e2×e3)|=√g.IShow that the matrix [gij]is the inverse of the matrix [gij]. Hence calculate the con- travariant components gijof the metric tensor in cylindrical polar coordinates. Using the index-lowering and index-raising properties of gijandgijon an arbitrary vector a, we ﬁnd δi kak=ai=gijaj=gijgjkak. But, since ais arbitrary, we must have gijgjk=δi k. (21.64) Denoting the matrix [ gij]b y Gand [ gij]b yˆG, equation (21.64) can be written in matrix form as ˆGG=I,w h e r e Iis the unit matrix. Hence GandˆGare inverse matrices of each other. 808 21.16 GENERAL COORDINATE TRANSFORMATIONS AND TENSORS Thus, by inverting the matrix Gin (21.59), we ﬁnd that the elements gijare given in cylindrical polar coordinates by ˆG=[gij]= /0/@100 01 /ρ20 001 /1A. J So far we have not considered the components of the metric tensor gi jwith one subscript and one superscript. By analogy with (21.55), these mixed componentsare given by g i j=ei·ej=δj i, and so the components of gi jare identical to those of δi j.W em a yt h e r e f o r e consider the δi jto be the mixed components of the metric tensor g. 21.16 General coordinate transformations and tensors We now discuss the concept of general transformations from one coordinate system, u1,u2,u3, to another, u/prime1,u/prime2,u/prime3. We can describe the coordinate transform using the three equations u/primei=u/primei(u1,u2,u3), fori=1,2,3, in which the new coordinates u/primeican be arbitrary functions of the old ones uirather than just represent linear orthogonal transformations (rotations) of the coordinate axes. We shall assume also that the transformation can beinverted, so that we can write the old coordinates in terms of the new ones as u i=ui(u/prime1,u/prime2,u/prime3), As an example, we may consider the transformation from spherical polar to Cartesian coordinates, given by x=rsinθcosφ, y=rsinθsinφ, z=rcosθ, which is clearly not a linear transformation. The two sets of basis vectors in the new coordinate system, u/prime1,u/prime2,u/prime3, are given as in (21.54) by e/prime i=∂r ∂u/primeiand e/primei=∇u/primei. (21.65) Considering the ﬁrst set, we have from the chain rule that ∂r ∂uj=∂u/primei ∂uj∂r ∂u/primei, 809 TENSORS so that the basis vectors in the old and new coordinate systems are related by ej=∂u/primei ∂uje/prime i. (21.66) Now, since we can write any arbitrary vector ain terms of either basis as a=a/primeie/prime i=ajej=aj∂u/primei ∂uje/prime i, it follows that the contravariant components of a vector must transform as a/primei=∂u/primei ∂ujaj. (21.67) In fact, we use this relation as the deﬁning property for a set of quantities aito form the contravariant components of a vector.IFind an expression analogous to (21.66) relating the basis vectors eiande/primeiin the two coordinate systems. Hence deduce the way i n which the covariant components of a vector change under a coordinate transformation. If we consider the second set of basis vectors in (21.65), e/primei=∇u/primei, we have from the chain rule that ∂uj ∂x=∂uj ∂u/primei∂u/primei ∂x and similarly for ∂uj/∂yand∂uj/∂z. So the basis vectors in the old and new coordinate systems are related by ej=∂uj ∂u/primeie/primei. (21.68) For any arbitrary vector a, a=a/prime ie/primei=ajej=aj∂uj ∂u/primeie/primei and so the covariant components of a vector must transform as a/prime i=∂uj ∂u/primeiaj. (21.69) Analogously to the contravariant case (21.67), we take this result as the deﬁning property of the covariant components of a vector. J We may compare the transformation laws (21.67) and (21.69) with those for a ﬁrst-order Cartesian tensor under a rigid rotation of axes. Let us consider a rotation of Cartesian axes xithrough an angle θabout the 3-axis to a new setx/primei,i=1,2,3, as given by (21.7) and the inverse transformation (21.8). It is straightforward to show that ∂xj ∂x/primei=∂x/primei ∂xj=Lij, 810 21.16 GENERAL COORDINATE TRANSFORMATIONS AND TENSORS where the elements Lijare given by L= cosθsinθ0 −sinθcosθ0 00 1 . Thus (21.67) and (21.69) agree with our earlier deﬁnition in the special case of a rigid rotation of Cartesian axes. Following on from (21.67) and (21.69), we proceed in a similar way to de- ﬁne general tensors of higher rank. For example, the contravariant, mixed and covariant components, respectively, of a second-order tensor must transform as follows: contravariant components, T/primeij=∂u/primei ∂uk∂u/primej ∂ulTkl; mixed components, T/primei j=∂u/primei ∂uk∂ul ∂u/primejTk l; covariant components, T/prime ij=∂uk ∂u/primei∂ul ∂u/primejTkl. It is important to remember that these quantities form the components of the sametensor Tbut refer to diﬀerent tensor bases made up from the basis vectors of the diﬀerent coordinate systems. For example, in terms of the contravariant components we may write T=Tijei⊗ej=T/primeije/prime i⊗e/prime j. We can clearly go on to deﬁne tensors of higher order, with arbitrary numbers of covariant (subscript) and contravariant (superscript) indices, by demandingthat their components transform as follows: T /primeij···k lm···n=∂u/primei ∂ua∂u/primej ∂ub···∂u/primek ∂uc∂ud ∂u/primel∂ue ∂u/primem···∂uf ∂u/primenTab···c de···f.(21.70) Using the revised summation convention described in section 21.14, the algebra of general tensors is completely analogous to that of the Cartesian tensorsdiscussed earlier. For example, as with Cartesian coordinates, the Kroneckerdelta is a tensor provided it is written as the mixed tensor δ i jsince δ/primei j=∂u/primei ∂uk∂ul ∂u/primejδk l=∂u/primei ∂uk∂uk ∂u/primej=∂u/primei ∂u/primej=δi j, where we have used the chain rule to justify the third equality. This also shows that δi jis isotropic. As discussed at the end of section 21.15, the δi jcan be considered as the mixed components of the metric tensor g. 811 TENSORSIShow that the quantities gij=ei·ejform the covariant com ponents of a second-order tensor. In the new (primed) coordinate system we have g/prime ij=e/prime i·e/prime j, but using (21.66) for the inverse transformation, we have e/prime i=∂uk ∂u/primeiek, and similarly for e/prime j. Thus we may write g/prime ij=∂uk ∂u/primei∂ul ∂u/primejek·el=∂uk ∂u/primei∂ul ∂u/primejgkl, which shows that the gijare indeed the covariant components of a second-order tensor (the metric tensor g). J A similar argument to that used in the above example shows that the quantities gijform the contravariant components of a second-order tensor which transforms according to g/primeij=∂u/primei ∂uk∂u/primej ∂ulgkl. In the previous section we discussed the use of the components gijandgijin the raising and lowering of indices in contravariant and covariant vectors. This can be extended to tensors of arbitrary rank. In general, contraction of a tensorwith g ijwill convert the contracted index fro m being contravariant (superscript) to covariant (subscript), i.e. it is lowered. This can be repeated for as many indicesare required. For example, T ij=gikTk j=gikgjlTkl. (21.71) Similarly contraction with gijraises an index, i.e. Tij=gikTj k=gikgjlTkl. (21.72) That (21.71) and (21.72) are mutually consistent may be shown by using the fact thatgikgkj=δi j. 21.17 Relative tensors In section 21.10 we introduced the concept of pseudotensors in the context of the rotation (proper or improper) of a set of Cartesian axes. Generalising to arbitrarycoordinate transformations leads to the notion of a relative tensor . For an arbitrary coordinate transformation from one general coordinate system 812 21.17 RELATIVE TENSORS uito another u/primei, we may deﬁne the Jacobian of the transformation (see chapter 6) as the determinant of the transformation matrix [ ∂u/primei/∂uj]: this is usually denoted by J=vextendsinglevextendsinglevextendsinglevextendsingle∂u/prime ∂uvextendsinglevextendsinglevextendsinglevextendsingle. Alternatively, we may interchange the primed and unprimed coordinates to obtain|∂u/∂u/prime|=1/J: unfortunately this also is often called the Jacobian of the transformation. Using the Jacobian J, we deﬁne a relative tensor of weight was one whose components transform as follows: T/primeij···k lm···n=∂u/primei ∂ua∂u/primej ∂ub···∂u/primek ∂uc∂ud ∂u/primel∂ue ∂u/primem···∂uf ∂u/primenTab···c de···fvextendsinglevextendsinglevextendsinglevextendsingle∂u ∂u/primevextendsinglevextendsinglevextendsinglevextendsinglew . (21.73) Comparing this expression with (21.70), we see that a true (or absolute )g e n e r a l tensor may be considered as a relative tensor of weight w=0 .I f w=−1, on the other hand, the relative tensor is known as a general pseudotensor ,a n di f w=1 as atensor density . It is worth comparing (21.73) with the deﬁnition (21.38) of a Cartesian pseu- dotensor. For the latter, we are concerned only with its behaviour under a rotation(proper or improper) of Cartesian axes, for which the Jacobian J=±1. Thus, general relative tensors of weight w=−1a n d w= 1 would both satisfy the deﬁnition (21.38) of a Cartesian pseudotensor.IIf the gijare the covariant components of the metric tensor, show that the determinant g of the matrix [gij]is a relative scalar of weight w=2. The components gijtransform as g/prime ij=∂uk ∂u/primei∂ul ∂u/primejgkl. Deﬁning the matrices U=[∂ui/∂u/primej],G=[gij]a n d G/prime=[g/prime ij], we may write this expression as G/prime=UTGU. Taking the determinant of both sides, we obtain g/prime=|U|2g= ////∂u ∂u/prime ////2 g, which shows that gis a relative scalar of weight w=2 . J From the discussion in section 21.8, it can be seen that /epsilon1ijkis a covariant relative tensor of weight −1. We may also deﬁne the contravariant tensor /epsilon1ijk, which is numerically equal to /epsilon1ijkbut is a relative tensor of weight +1. If two relative tensors have weights w1andw2respectively then, from (21.73), 813 TENSORS the outer product of the two tensors, or any contraction of them, is a relative tensor of weight w1+w2. As a special case, we may use /epsilon1ijkand/epsilon1ijkto construct pseudovectors from antisymmetric tensors and vice versa, in an analogous wayto that discussed in section 21.11. For example, if the A ijare the contravariant components of an antisymmetric tensor ( w=0 )t h e n pi=1 2/epsilon1ijkAjk are the covariant components of a pseudovector ( w=−1), since /epsilon1ijkhas weight w=−1. Similarly, we may show that Aij=/epsilon1ijkpk. 21.18 Derivatives of basis vectors and Christoﬀel symbols In Cartesian coordinates, the basis vectors eiare constant and so their derivatives with respect to the coordinates vanish. In a general coordinate system, however, the basis vectors eiandeiare functions of the coordinates. Therefore, in order that we may diﬀerentiate general tensors we must consider the derivatives of thebasis vectors. First consider the derivative ∂e i/∂uj. Since this is itself a vector, it can be written as a linear combination of the basis vectors ek,k=1,2,3. If we introduce the symbol Γk ijto denote the coeﬃcients in this combination, we have ∂ei ∂uj=Γk ijek. (21.74) The coeﬃcient Γk ijis the kth component of the vector ∂ei/∂uj.U s i n gt h er e c i - procity relation ei·ej=δi j, these 27 numbers are given (at each point in space) by Γk ij=ek·∂ei ∂uj. (21.75) Furthermore, by diﬀerentiating the reciprocity relation ei·ej=δi jwith respect to the coordinates, and using (21.75), it is straightforward to show that thederivatives of the contravariant basis vectors are given by ∂e i ∂uj=−Γi kjek. (21.76) The symbol Γk ijis called a Christoﬀel symbol (of the second kind), but, despite appearances to the contrary, these quantities do notform the components of a third-order tensor. It is clear from (21.75) that in Cartesian coordinates Γk ij=0 for all values of the indices i,jandk. 814 21.18 DERIVATIVES OF BASIS VECTORS AND CHRISTOFFEL SYMBOLSIUsing (21.75), deduce the way in which the quantities Γk ijtransform under a general coordinate transformation, and hence show that they do not form the components of athird-order tensor. In a new coordinate system Γ/primek ij=e/primek·∂e/prime i ∂u/primej, but from (21.68) and (21.66) respectively we have, on reversing primed and unprimed variables, e/primek=∂u/primek ∂unenand e/prime i=∂ul ∂u/primeiel. Therefore in the new coordinate system the quantities Γ/primek ijare given by Γ/primek ij=∂u/primek ∂unen·∂ ∂u/primej /∂ul ∂u/primeiel / =∂u/primek ∂unen· /∂2ul ∂u/primej∂u/primeiel+∂ul ∂u/primei∂el ∂u/primej / =∂u/primek ∂un∂2ul ∂u/primej∂u/primeien·el+∂u/primek ∂un∂ul ∂u/primei∂um ∂u/primejen·∂el ∂um =∂u/primek ∂ul∂2ul ∂u/primej∂u/primei+∂u/primek ∂un∂ul ∂u/primei∂um ∂u/primejΓn lm, (21.77) where in the last line we have used (21.75) and the reciprocity relation en·el=δn l.F r o m (21.77), because of the presence of the ﬁrst term on the right-hand side, we concludeimmediately that the Γ k ijdo not form the components of a third-order tensor. J In a given coordinate system, in principle, we may calculate the Γk ijusing (21.75). In practice, however, it is often quicker to use an alternative expression, which we now derive, for the Christoﬀel symbol in terms of the metric tensor gij and its derivatives with respect to the coordinates. Firstly we note that the Christoﬀel symbol Γk ijis symmetric with respect to the interchange of its two subscripts iandj. This is easily shown: since ∂ei ∂uj=∂2r ∂uj∂ui=∂2r ∂ui∂uj=∂ej ∂ui, it follows from (21.74) that Γk ijek=Γk jiek. Taking the scalar product with eland using the reciprocity relation ek·el=δl kgives immediately that Γl ij=Γl ji. To obtain an expression for Γk ijwe then use gij=ei·ejand consider the derivative ∂gij ∂uk=∂ei ∂uk·ej+ei·∂ej ∂uk =Γl ikel·ej+ei·Γl jkel =Γl ikglj+Γl jkgil, (21.78) 815 TENSORS where we have used the deﬁnition (21.74). By cyclically permuting the free indices i, j, kin (21.78), we obtain two further equivalent relations, ∂gjk ∂ui=Γl jiglk+Γl kigjl (21.79) and ∂gki ∂uj=Γl kjgli+Γl ijgkl. (21.80) If we now add (21.79) and (21.80) together and subtract (21.78) from the result, we ﬁnd ∂gjk ∂ui+∂gki ∂uj−∂gij ∂uk=Γl jiglk+Γl kigjl+Γl kjgli+Γl ijgkl−Γl ikglj−Γl jkgil =2 Γl ijgkl, where we have used the symmetry properties of both Γl ijandgij. Contracting both sides with gmkleads to the required expression for the Christoﬀel symbol in terms of the metric tensor and its derivatives, namely Γm ij=1 2gmkparenleftbigg∂gjk ∂ui+∂gki ∂uj−∂gij ∂ukparenrightbigg . (21.81)ICalculate the Christoﬀel symbols Γm ijfor cylindrical polar coordinates. We may use either (21.74) or (21.81) to calculate the Γm ijfor this simple coordinate system. In cylindrical polar coordinates ( u1,u2,u3)=( ρ, φ, z), the basis vectors eiare given by (21.58). It is straightforward to show that the only derivatives of these vectors with respectto the coordinates that are non-zero are ∂e ρ ∂φ=1 ρeφ,∂eφ ∂ρ=1 ρeφ,∂eφ ∂φ=−ρeρ. Thus, from (21.74), we have immediately that Γ2 12=Γ2 21=1 ρand Γ1 22=−ρ. (21.82) Alternatively, using (21.81) and the fact that g11=1 , g22=ρ2,g33= 1 and the other components are zero, we see that the only three non-zero Christoﬀel symbols are indeedΓ 2 12=Γ2 21and Γ1 22.T h e s ea r eg i v e nb y Γ2 12=Γ2 21=1 2g22∂g22 ∂u1=1 2ρ2∂ ∂ρ(ρ2)=1 ρ, Γ1 22=−1 2g11∂g22 ∂u1=−1 2∂ ∂ρ(ρ2)=−ρ, which agree with the expressions found directly from (21.74) and given in (21.82 ). J 816 21.19 COVARIANT DIFFERENTIATION 21.19 Covariant diﬀerentiation For Cartesian tensors we noted that the derivative of a scalar is a (covariant) vector. This is also true for general tensors, as may be shown by considering thediﬀerential of a scalar dφ=∂φ ∂uidui. Since the duiare the components of a contravariant vector and dφis a scalar, we have by the quotient law, discussed in section 21.7, that the quantities ∂φ/∂ui must form the components of a covariant vector. As a second example, in Cartesian coordinates, if the viare the contravariant components of a vector then the quantites ∂vi/∂xjform the components of a second-order tensor. It is straightforward, however, to show that (in contrast to what happens in Cartesian coordinates) the diﬀerentiation of the components ofa general tensor, other than a scalar, with respect to the coordinates does notin general result in the components of another tensor.IShow that, in general coordinates, the quantities ∂vi/∂ujdo not form the components of a tensor. We may show this directly by considering/∂vi ∂uj //prime =∂v/primei ∂u/primej=∂uk ∂u/primej∂v/primei ∂uk =∂uk ∂u/primej∂ ∂uk / ∂u/primei ∂ulvl /! =∂uk ∂u/primej∂u/primei ∂ul∂vl ∂uk+∂uk ∂u/primej∂2u/primei ∂uk∂ulvl. (21.83) The presence of the second term on the right-hand side of (21.83) shows that the ∂vi/∂xj do not form the components of a second-order tensor. This term arises because the ‘transformation matrix’ [ ∂u/primei/∂uj] changes as the position in space at which it is evaluated is changed. This is not true in Cartesian coordinates, for which the second term vanishes,and∂v i/∂xjis a second-order tensor. J We may, however, use the Christoﬀel symbols discussed in the previous section to deﬁne a new covariant derivative of the components of a tensor that does result in the components of another tensor. Let us ﬁrst consider the derivative of a vector vwith respect to the coordinates. Writing the vector in terms of its contravariant components v=viei, we ﬁnd ∂v ∂uj=∂vi ∂ujei+vi∂ei ∂uj, (21.84) where the second term arises because, in general, the basis vectors eiare not constant (this term vanishes in Cartesian coordinates). Using (21.74) we may 817 TENSORS write ∂v ∂uj=∂vi ∂ujei+viΓk ijek. Since iandkare dummy indices in the last term on the right-hand side, we may interchange them to obtain ∂v ∂uj=∂vi ∂ujei+vkΓi kjei=parenleftbigg∂vi ∂uj+vkΓi kjparenrightbigg ei. (21.85) The reason for the interchanging the dummy indices, as shown in (21.85), is that we may now factor out ei. The quantity in parentheses is called the covariant derivative , for which the standard notation is vi ;j≡∂vi ∂uj+Γi kjvk, (21.86) the semicolon subscript denoting covariant diﬀerentiation. A similar short-hand notation also exists for the partial derivatives, a comma being used for these instead of a semicolon; for example, ∂vi/∂ujis denoted by vi ,j. In Cartesian coordinates all the Γi kjare zero, and so the covariant derivative reduces to the simple partial derivative ∂vi/∂uj. Using the short-hand semicolon notation, the derivative of a vector may be written in the very compact form ∂v ∂uj=vi ;jei and, by the quotient rule (section 21.7), it is clear that the vi ;jare the (mixed) components of a second-order tensor. This may also be veriﬁed directly, usingthe transformation properties of ∂v i/∂ujand Γi kjgiven in (21.83) and (21.77) respectively. In general, we may regard the vi ;jas the mixed components of a second- order tensor called the covariant derivative of vand denoted by ∇v. In Cartesian coordinates, the components of this tensor are just ∂vi/∂xj.ICalculate vi ;iin cylindrical polar coordinates. Contracting (21.86) we obtain vi ;i=∂vi ∂ui+Γi kivk. Now from (21.82) we have Γi 1i=Γ1 11+Γ2 12+Γ3 13=1/ρ, Γi 2i=Γ1 21+Γ2 22+Γ3 23=0, Γi 3i=Γ1 31+Γ2 32+Γ3 33=0, 818 21.19 COVARIANT DIFFERENTIATION and so vi ;i=∂vρ ∂ρ+∂vφ ∂φ+∂vz ∂z+1 ρvρ =1 ρ∂ ∂ρ(ρvρ)+∂vφ ∂φ+∂vz ∂z. This result is identical to the expression for the divergence of a vector ﬁeld in cylindrical polar coordinates given in section 10.9. This is discussed further in section 21.20. J So far we have considered only the covariant derivative of the contravariant components viof a vector. The corresponding result for the covariant components vimay be found in a similar way, by considering the derivative of v=vieiand using (21.76) to obtain vi;j=∂vi ∂uj−Γk ijvk. (21.87) Comparing the expressions (21.86) and (21.87) for the covariant derivative of the contravariant and covariant components of a vector respectively, we seethat there are some similarities and some diﬀerences. It may help to rememberthat the index with respect to which the covariant derivative is taken ( jin this case), is also the last subscript on the Christoﬀel symbol; the remaining indicescan then be arranged in only one way without raising or lowering them. It remains to remember the sign diﬀerence, i.e. that for a covariant index (subscript) the Christoﬀel symbol carries a minus sign, whereas for a contravariant index(superscript) the sign is positive. Following a similar procedure to that which led to equation (21.86), we may obtain expressions for the covariant derivatives of higher-order tensors.IBy considering the derivative of the second-order tensor Twith respect to the coordinate uk, ﬁnd an expression for the covariant derivative Tij ;kof its contravariant components. Expressing Tin terms of its contravariant components, we have ∂T ∂uk=∂ ∂uk(Tijei⊗ej) =∂Tij ∂ukei⊗ej+Tij∂ei ∂uk⊗ej+Tijei⊗∂ej ∂uk. Using (21.74), we can rewrite the derivatives of the basis vectors in terms of Christoﬀel symbols to obtain ∂T ∂uk=∂Tij ∂ukei⊗ej+TijΓl ikel⊗ej+Tijei⊗Γl jkel. Interchanging the dummy indices iandlin the second term and jandlin the third term on the right-hand s ide, this becomes ∂T ∂uk= /∂Tij ∂uk+Γi lkTlj+Γj lkTil / ei⊗ej, 819 TENSORS where the expression in brackets is the required covariant derivative Tij ;k=∂Tij ∂uk+Γi lkTlj+Γj lkTil. (21.88) Using (21.88), the derivative of the tensor Twith respect to ukcan now be written in terms of its contravariant components as ∂T ∂uk=Tij ;kei⊗ej. J Results similar to (21.88) may be obtained for the the covariant derivatives of the mixed and covariant components of a second-order tensor. Collecting these results together, we have Tij ;k=Tij ,k+Γi lkTlj+Γj lkTil, Ti j;k=Ti j,k+Γi lkTl j−Γl jkTi l, Tij;k=Tij, k−Γl ikTlj−Γl jkTil, where we have used the comma notation for partial derivatives. The position of the indices in these expressions is very systematic: for each contravariant index(superscript) on the LHS we add a term on the RHS containing a Christoﬀelsymbol with a plus sign, and for every covariant index (subscript) we add acorresponding term with a minus sign. This is extended straightforwardly totensors with an arbitrary number of contravariant and covariant indices. We note that the quantities T ij ;k,Ti j;kandTij;kare the components of the samethird-order tensor ∇Twith respect to diﬀerent tensor bases, i.e. ∇T=Tij ;kei⊗ej⊗ek=Ti j;kei⊗ej⊗ek=Tij;kei⊗ej⊗ek. We conclude this section by considering brieﬂy the covariant derivative of a scalar. The covariant derivative diﬀers from the simple partial derivative withrespect to the coordinates only because the basis vectors of the coordinate system change with position in space (hence for Cartesian coordinates there is no diﬀerence). However, a scalar φdoes not depend on the basis vectors at all and so its covariant derivative must be the same as its partial derivative, i.e. φ ;j=∂φ ∂uj=φ,j. (21.89) 21.20 Vector operators in tensor form In section 10.10 we used vector calculus methods to ﬁnd expressions for vector diﬀerential operators, such as grad, div, curl and the Laplacian, in general orthog- onalcurvilinear coordinates, taking cylindrical and spherical polars as particular examples. In this section we use the framework of general tensors that we have developed to obtain, in tensor form, expressions for these operators that are validinallcoordinate systems, whether orthogonal or not. 820 21.20 VECTOR OPERATORS IN TENSOR FORM In order to compare the results obtained here with those given in section 10.10 for orthogonal coordinates, it is necessary to remember that here we areworking with the (in general) non-unit basis vectors e i=∂r/∂uiorei=∇ui. Thus the components of a vector v=vieiare not the same as the components ˆvi appropriate to the corresponding unit basis ˆei. In fact, if the scale factors of the coordinate system are hi,i=1,2,3, then vi=ˆvi/hi(no summation over i). As mentioned in section 21.15, for an orthogonal coordinate system with scale factors hiwe have gij=braceleftBigg h2 iifi=j, 0o t h e r w i s eand gij=braceleftBigg 1/h2 iifi=j, 0o t h e r w i s e , a n ds ot h ed e t e r m i n a n t gof the matrix [ gij]i sg i v e nb y g=h2 1h22h23. Gradient The gradient of a scalar φis given by ∇φ=φ;iei=∂φ ∂uiei, (21.90) since the covariant derivative of a scalar is the same as its partial derivative. Divergence Replacing the partial derivatives that occur in Cartesian coordinates with covari- ant derivatives, the divergence of a vector ﬁeld vin a general coordinate system is given by ∇·v=vi ;i=∂vi ∂ui+Γi kivk. Using the expression (21.81) for the Christoﬀel symbol in terms of the metric tensor, we ﬁnd Γi ki=1 2gilparenleftbigg∂gil ∂uk+∂gkl ∂ui−∂gki ∂ulparenrightbigg =1 2gil∂gil ∂uk. (21.91) The last two terms have cancelled because gil∂gkl ∂ui=gli∂gki ∂ul=gil∂gki ∂ul, where in the ﬁrst equality we have interchanged the dummy indices iandl,a n d in the second equality have used the symmetry of the metric tensor. We may simplify (21.91) still further by using a result concerning the derivative of the determinant of a matrix whose elements are functions of the coordinates. 821 TENSORSISuppose A=[aij],B=[bij]and that B=A−1. By considering the determinant a=|A|, show that ∂a ∂uk=abji∂aij ∂uk. If we denote the cofactor of the element aijby ∆ijthen the elements of the inverse matrix are given by (see chapter 8) bij=1 a∆ji. (21.92) However, the determinant of Ais given by a= X jaij∆ij, in which we have ﬁxed iand written the sum over jexplicitly, for clarity. Partially diﬀerentiating both sides with respect to aij, we then obtain ∂a ∂aij=∆ij, (21.93) since aijdoes not occur in any of the cofactors ∆ij. Now, if the aijdepend on the coordinates then so will the determinant aand, by the chain rule, we have ∂a ∂uk=∂a ∂aij∂aij ∂uk=∆ij∂aij ∂uk=abji∂aij ∂uk, (21.94) in which we have used (21.92) and (21.93). J Applying the result (21.94) to the determinant gof the metric tensor, and remembering both that gikgkj=δi jand that gijis symmetric, we obtain ∂g ∂uk=ggij∂gij ∂uk. (21.95) Substituting (21.95) into (21.91) we ﬁnd that the expression for the Christoﬀel symbol can be much simpliﬁed to give Γi ki=1 2g∂g ∂uk=1√g∂√g ∂uk. Thus ﬁnally we obtain the expression for the divergence of a vector ﬁeld in a general coordinate system as ∇·v=vi ;i=1√g∂ ∂uj(√gvj). (21.96) Laplacian If we replace vby∇φin∇·vthen we obtain the Laplacian ∇2φ. From (21.90), we have viei=v=∇φ=∂φ ∂uiei, 822 21.20 VECTOR OPERATORS IN TENSOR FORM and so the covariant components of vare given by vi=∂φ/∂ui. In (21.96), however, we require the contravariant components vi. These may be obtained by raising the index using the metric tensor, to give vj=gjkvk=gjk∂φ ∂uk. Substituting this into (21.96) we obtain ∇2φ=1√g∂ ∂ujparenleftbigg√ggjk∂φ ∂ukparenrightbigg . (21.97)IUse (21.97) to ﬁnd the expression for ∇2φin an orthogonal coordinate system with scale factors hi,i=1,2,3. For an orthogonal coordinate system√g=h1h2h3andgij=1/h2 iifi=jandgij=0 otherwise. Therefore, from (21.97) we have ∇2φ=1 h1h2h3∂ ∂uj / h1h2h3 h2 j∂φ ∂uj /! , which agrees with the results of section 10.10. J Curl The special vector form of the curl of a vector ﬁeld exists only in three dimensions. We therefore consider a more general form valid in higher-dimensional spaces aswell. In a general space the operation curl vis deﬁned by (curlv) ij=vi;j−vj;i, which is an antisymmetric covariant tensor. In fact the diﬀerence of derivatives can be simpliﬁed, since vi;j−vj;i=∂vi ∂uj−Γl ijvl−∂vj ∂ui+Γl jivl =∂vi ∂uj−∂vj ∂ui, where the Christoﬀel symbols have cancelled because of their symmetry properties. Thus curl vcan be written in terms of partial derivatives as (curlv)ij=∂vi ∂uj−∂vj ∂ui. Generalising slightly the discussion of section 21.17, in three dimensions we may associate with this antisymmetric second-order tensor a vector with contravariant 823 TENSORS components, (∇×v)i=−1 2√g/epsilon1ijk(curlv)jk =−1 2√g/epsilon1ijkparenleftbigg∂vj ∂uk−∂vk ∂ujparenrightbigg =1√g/epsilon1ijk∂vk ∂uj; this is the analogue of the expression in Cartesian coordinates discussed in section 21.8. 21.21 Absolute derivatives along curves In section 21.19 we discussed how to diﬀerentiate a general tensor with respect to the coordinates and introduced the covariant derivative. In this section weconsider the slightly diﬀerent problem of calculating the derivative of a tensor along a curve r(t) that is parameterised by some variable t. Let us begin by considering the derivative of a vector valong the curve. If we introduce an arbitrary coordinate system u iwith basis vectors ei,i=1,2,3, then we may write v=vieia n ds oo b t a i n dv dt=dvi dtei+videi dt =dvi dtei+vi∂ei ∂ukduk dt; here the chain rule has been used to rewrite the last term on the right-hand side. Using (21.74) to write the derivatives of the basis vectors in terms of Christoﬀelsymbols, we obtain dv dt=dvi dtei+Γj ikviduk dtej. Interchanging the dummy indices iandjin the last term, we may factor out the basis vector, and we ﬁnd dv dt=parenleftbiggdvi dt+Γi jkvjduk dtparenrightbigg ei. The term in parentheses is called the absolute (orintrinsic ) derivative of the components vialong the curve r(t)and is usually denoted by δvi δt≡dvi dt+Γi jkvjduk dt=vi ;kduk dt. With this notation, we may write dv dt=δvi δtei=vi ;kduk dtei. (21.98) 824 21.22 GEODESICS Using the same method, the absolute derivative of the covariant components viof a vector is given by δvi δt≡vi;kduk dt. Similarly, the absolute derivatives of the contravariant, mixed and covariant components of a second-order tensor Tare δTij δt≡Tij ;kduk dt, δTi j δt≡Ti j;kduk dt, δTij δt≡Tij;kduk dt. The derivative of Talong the curve r(t) may then be written in terms of, for example, its contravariant components as dT dt=δTij δtei⊗ej=Tij ;kduk dtei⊗ej. 21.22 Geodesics As an example of the use of the absolute derivative, we conclude this chapter with a brief discussion of geodesics. A geodesic in real three-dimensional spaceis a straight line, which has two equivalent deﬁning properties. Firstly, it is thecurve of shortest length between two points and, secondly, it is the curve whosetangent vector always points in the same direction (along the line). Althoughin this chapter we have considered explicitly only our familiar three-dimensional space, much of the mathematical formalism developed can be generalised to more abstract spaces of higher dimensionality in which the familiar ideas of Euclideangeometry are no longer valid. It is often of interest to ﬁnd geodesic curves insuch spaces by using the deﬁning properties of straight lines in Euclidean space. We shall not consider these more complicated spaces explicitly but will de- termine the equation that a geodesic in Euclidean three-dimensional space (i.e.a straight line) must satisfy, deriving it in a suﬃciently general way that our method may be applied with little modiﬁcation to ﬁnding the equations satisﬁed by geodesics in more abstract spaces. Let us consider a curve r(s), parameterised by the arc length sfrom some point on the curve, and choose as our deﬁning property for a geodesic that its tangentvector t=dr/dsalways points in the same direction everywhere on the curve, i.e. dt ds=0. (21.99) (We could alternatively exploit the property that the distance between two points 825 TENSORS is a minimum along a geodesic and use the calculus of variations (see chapter 22); this would lead to the same ﬁnal result (21.100).) If we now introduce an arbitrary coordinate system uiwith basis vectors ei, i=1,2,3, then we may write t=tieiand from (21.98) we ﬁnd dt ds=ti ;kduk dsei=0. Writing out the covariant derivative, we obtain parenleftbiggdti ds+Γi jktjduk dsparenrightbigg ei=0. But, since tj=duj/ds, it follows that the equation satisﬁed by a geodesic is d2ui ds2+Γi jkduj dsduk ds=0. (21.100)IFind the equations satisﬁed by a geodesic (straight line) in cylindrical polar coordinates. From (21.82), the only non-zero Christoﬀel symbols are Γ1 22=−ρand Γ2 12=Γ2 21=1/ρ. Thus the required geodesic equations are d2u1 ds2+Γ1 22du2 dsdu2 ds=0⇒d2ρ ds2−ρ /dφ ds /2 =0, d2u2 ds2+2 Γ2 12du1 dsdu2 ds=0⇒d2φ ds2+2 ρdρ dsdφ ds=0, d2u3 ds2=0⇒d2z ds2=0. J 21.23 Exercises 21.1 (a) Show that for any general, but ﬁxed, φ, (u1,u2)=(x1cosφ−x2sinφ, x 1sinφ+x2cosφ) are the components of a ﬁrst-order tensor in two dimensions. (b) Show that/ x2 2x1x2 x1x2x21 / is not a (Cartesian) tensor of order 2. To establish that a single element does not transform correctly is suﬃcient. 21.2 The components of two vectors AandBand a second-order tensor Tare given in one coordinate system by A= /0/@1 00 /1A,B= /0/@0 10 /1A,T= /0/@2√30√340 00 2 /1A. 826 21.23 EXERCISES In a second coordinate system, obtained from the ﬁrst by rotation, the components ofAandBare A/prime=1 2 /0/@√3 01 /1A,B/prime=1 2 /0/@−1 0√3 /1A. Find the components of Tin this new coordinate system and hence evaluate, with a minimum of calculation, TijTji,T kiTjkTij,T ikTmnTniTkm. 21.3 In section 21.3 the transformation matrix for a rotation of the coordinate axes was derived, and this approach is used in the rest of the chapter. An alternative view is that of taking the coordinate axes as ﬁxed and rotating the componentsof the system; this is equivalent to reversing the signs of all rotation angles. Using this alternative view, determine the matrices representing (a) a positive rotation of π/4 about the x-axis, and (b) a rotation of −π/4 about the y-axis. Determine the initial vector rwhich, when subjected to (a) followed by (b), ﬁnishes at (3 ,2,1). 21.4 Show how to decompose the tensor T ijinto three tensors, Tij=Uij+Vij+Sij, where Uijis symmetric and has zero trace, Vijis isotropic and Sijhas only three independent components. 21.5 Use the quotient law discussed in section 21.7 to show that the array/0/@y2+z2−x2−2xy −2xz −2yx x2+z2−y2−2yz −2zx −2zy x2+y2−z2 /1A forms a second-order tensor. 21.6 Use tensor methods to establish the following vector identities: (a) (u×v)×w=(u·w)v−(v·w)u; (b) curl( φu)=φcurlu+ (grad φ)×u; (c) div ( u×v)=v·curlu−u·curlv; (d) curl( u×v)=(v·grad)u−(u·grad)v+udivv−vdivu; (e) grad1 2(u·u)=u×curlu+(u·grad)u. 21.7 Use result (e) of the previous question and the general divergence theorem for tensors to show thatZ S / A(A·dS)−1 2A2dS / = Z V[AdivA−A×curlA]dV. 21.8 A column matrix ahas components ax,ay,azand Ais the matrix with elements Aij=−/epsilon1ijkak. (a) What is the relationship between column matrices band cifAb=c? (b) Find the eigenvalues of Aand show that ais one of its eigenvectors. Explain why this must be so. 21.9 Equation (21.28), |A|/epsilon1lmn=AliAmjAnk/epsilon1ijk, is a more general form of the expression (8.47) for the determinant of a 3 ×3 matrix A. The latter could have been written as |A|=/epsilon1ijkAi1Aj2Ak3, 827 TENSORS whilst the former removes the explicit mention of 1 ,2,3 at the expense of an additional Levi–Civita symbol. As stated in the footnote on p. 791, (21.28) canbe readily extended to cover a general N×Nmatrix. Use the form given in (21.28) to prove properties (i), (iii), (v), (vi) and (vii) of determinants stated in subsection 8.9.1. Property (iv) is obvious by inspection.For deﬁniteness take N= 3, but convince yourself that your methods of proof would be valid for any positive integer N. 21.10 A symmetric second-order Cartesian tensor is deﬁned by T ij=δij−3xixj. Evaluate the following surface integrals, each taken over the surface of the unit sphere: (a) Z TijdS;( b ) Z TikTkjdS;( c ) Z xiTjkdS. 21.11 Given a non-zero vector v, ﬁnd the value that should be assigned to αto make Pij=αvivjand Qij=δij−αvivj into parallel and orthogonal projection tensors respectively, i.e. tensors that satisfy respectively Pijvj=vi,Pijuj=0a n d Qijvj=0 , Qijuj=ui, for any vector uthat is orthogonal to v, Show, in particular, that Qijis unique, i.e. that if another tensor Tijhas the same properties as Qijthen ( Qij−Tij)wj=0f o r anyvector w. 21.12 In four dimensions deﬁne second-order antisymmetric tensors FijandQijand a ﬁrst-order tensor Sias follows: (a)F23=H1,Q23=B1and their cyclic permutations; (b)Fi4=−Di,Qi4=Eifori=1,2,3; (c)S4=ρ,Si=Jifori=1,2,3. Then, taking x4astand the other symbols to have their usual meanings in electromagnetic theory, show that the equations P j∂Fij/∂x j=Siand∂Qjk/∂x i+ ∂Qki/∂x j+∂Qij/∂x k= 0 reproduce Maxwell’s equations. Here i, j, kis any set of three subscripts selected from 1 ,2,3,4, but chosen in such a way that they are all diﬀerent. 21.13 In a certain crystal the unit cell can be taken as six identical atoms lying at the corners of a regular octahedron. Convince yourself that these atoms can also beconsidered as lying at the centres of the faces of a cube and hence that the crystalhas cubic symmetry. Use this result to prove that the conductivity tensor for the crystal, σ ij,m u s tb ei s o t r o p i c . 21.14 Assuming that the current density jand the electric ﬁeld Eappearing in equation (21.43) are ﬁrst-order Cartesian tensors, show explicitly that the electrical con-ductivity tensor σ ijtransforms according to the law appropriate to a second-order tensor. The rate Wat which energy is dissipated per unit volume, as a result of the current ﬂow, is given by E·j. Determine the limits between which Wmust lie for a given value of |E|as the direction of Eis varied. 21.15 In a certain system of units the electromagnetic stress tensor Mijis given by Mij=EiEj+BiBj−1 2δij(EkEk+BkBk), where the electric and magnetic ﬁelds, EandB, are ﬁrst-order tensors. Show that Mijis a second-order tensor. Consider a situation in which |E|=|B|but the directions of EandBare not parallel. Show that E±Bare principal axes of the stress tensor and ﬁnd 828 21.23 EXERCISES the corresponding principal values. Determine the third principal axis and its corresponding principal value. 21.16 A rigid body consists of four particles of masses m,2m,3m,4m, respectively situated at the points ( a, a, a), (a,−a,−a), (−a, a,−a), (−a,−a, a) and connected together by a light framework. (a) Find the inertia tensor at the origin and show that the principal moments of inertia are 20 ma2,a n d( 2 0±2√ 5)ma2. (b) Find the principal axes and verify that they are orthogonal. 21.17 A rigid body consists of eight particles, each of mass m, held together by light rods. In a certain coordinate frame the particles are at ±a(3,1,−1),±a(1,−1,3),a(1,3,−1),a(−1,1,3). Show that, when the body rotates about an axis through the origin, if the angular velocity and angular momentum vectors are parallel then their ratio must be40ma 2,6 4ma2or 72 ma2. 21.18 The paramagnetic tensor χijof a body placed in a magnetic ﬁeld, in which its energy density is −1 2µ0M·Hwith Mi= P jχijHj,i s/0/@2k00 03 kk 0 k3k /1A. Assuming depolarizing eﬀects are negligible, ﬁnd how the body will orientate itself if the ﬁeld is horizontal, in the following circumstances: (a) the body can rotate freely; (b) the body is suspended with the (1,0,0) axis vertical;(c) the body is suspended with the (0,1,0) axis vertical. 21.19 A block of wood contains a number of thin soft iron nails (of constant permeabil- ity). A unit magnetic ﬁeld directed eastwards induces a magnetic moment in theblock having components (3 ,1,−2) and similar ﬁelds directed northwards and vertically upwards induce moments (1 ,3,−2) and (−2,2,2) respectively. Show that all the nails lie in parallel planes. 21.20 For tin the conductivity tensor is diagonal, with entries a, a,andbwhen referred to its crystal axes. A single crystal is grown in the shape of a long wire of length L and radius r, the axis of the wire making polar angle θwith respect to the crystal’s 3-axis. Show that the resistance of the wire is L(πr 2ab)−1 /; acos2θ+bsin2θ / . 21.21 By considering an isotropic body subjected to a uniform hydrostatic pressure (no shearing stress), show that the bulk modulus k, deﬁned by the ratio of the pressure to the fractional decrease in volume, is given by k=E/[3(1−2σ)] where Eis Young’s modulus and σPoisson’s ratio. 21.22 For an isotropic elastic medium under dynamic stress, at time tthe displacement uiand the stress tensor pijsatisfy pij=cijkl /∂uk ∂xl+∂ul ∂xk / and∂pij ∂xj=ρ∂2ui ∂t2, where cijklis the isotropic tensor given in equation (21.46) and ρis a constant. Show that both ∇·uand∇×usatisfy wave equations and ﬁnd the corresponding wave speeds. 829 TENSORS 21.23 A fourth-order tensor Tijklhas the properties Tjikl=−Tijkl,T ijlk=−Tijkl. Prove that for any such tensor there exists a second-order tensor Kmnsuch that Tijkl=/epsilon1ijm/epsilon1klnKmn and give an explicit expression for Kmn. Consider two (separate) special cases, as follows. (a) Given that Tijklis isotropic and Tijji= 1, show that Tijklis uniquely deter- mined and express it in terms of Kronecker deltas. (b) If now Tijklhas the additional property Tklij=−Tijkl, show that Tijklhas only three linearly independent components and ﬁnd an expression for Tijklin terms of the vector Vi=−1 4/epsilon1jklTijkl. 21.24 Working in cylindrical polar coordinates ρ, φ, z, parameterise the straight line (geodesic) joining (1 ,0,0) to (1 ,π/2,1) in terms of s, the distance along the line. Show by substitution that the geodesic equations derived at the end of section21.22 are satisﬁed. 21.25 In a general coordinate system u i,i=1,2,3, in three-dimensional Euclidean space, a volume element is given by dV=|e1du1·(e2du2×e3du3)|. Show that an alternative form for this expression, written in terms of the deter- minant gof the metric tensor, is given by dV=√gd u1du2du3. Show that under a general coordinate transformation to a new coordinate system u/primeithe volume element dVremains unchanged, i.e. show that it is a scalar quantity. 21.26 By writing down the expression for the square of the inﬁnitesimal arc length ( ds)2 in spherical polar coordinates, ﬁnd the components gijof the metric tensor in this coordinate system. Hence, using (21.96), ﬁnd the expression for the divergenceof a vector ﬁeld vin spherical polars. Calculate the Christoﬀel symbols (of the second kind) Γ i jkin this coordinate system. 21.27 Find an expression for the second covariant derivative vi;jk≡(vi;j);kof a vector vi(see(21.86)). By interchanging the order of diﬀerentiation and then subtracting the two expressions, we deﬁne the components Rl ijkof the Riemann tensor as vi;jk−vi;kj≡Rl ijkvl. Show that in a general coordinate system uithese components are given by Rl ijk=∂Γl ik ∂uj−∂Γl ij ∂uk+Γm ikΓl mj−Γm ijΓl mk. By ﬁrst considering Cartesian coordinates, show that all the components Rl ijk≡0 foranycoordinate system in three-dimensional Euclidean space. In such a space, therefore, we may change the order of the covariant derivativeswithout changing the resulting expression. 830 21.24 HINTS AND ANSWERS 21.28 A curve r(t) is parameterised by a scalar variable t. Show that the length of the curve between two points, AandB,i sg i v e nb y L= ZB A r gijdui dtduj dtdt. Using the calculus of variations (see chapter 22), show that the curve r(t)t h a t minimises Lsatisﬁes the equation d2ui dt2+Γi jkduj dtduk dt=¨s ˙sdui dt, where sis the arc length along the curve, ˙s=ds/dt and¨s=d2s/dt2. Hence, show that if the parameter tis of the form t=as+b,w h e r e aandbare constants, then we recover the equation for a geodesic (21.100). (A parameter which, like t, is the sum of a linearly transformation of sand a translation is called an aﬃne parameter.) 21.29 We may deﬁne Christoﬀel symbols of the ﬁrst kind by Γijk=gilΓl jk. Show that these are given by Γijk=1 2 /∂gik ∂uj+∂gjk ∂ui−∂gij ∂uk / . By permuting indices, verify that ∂gij ∂uk=Γ ijk+Γ jik. Using the fact that Γl jk=Γl kj, show that gij;k≡0, i.e. that the covariant derivative of the metric tensor is identically zero in all coordinate systems. 21.24 Hints and answers 21.1 (a) u/prime 1=x1cos(φ−θ)−x2sin(φ−θ), etc.; (b)u/prime 11=s2x2 1−2scx1x2+c2x2 2/negationslash=c2x2 2+csx1x2+scx1x2+s2x2 1. 21.2 Determine entries for the third column of Lby requiring that it is orthogonal and has determinant +1. T=1 2(√3,−1,0;0,0,−2;1√3,0). They are all scalars with values 30, 134, 642. 21.3 (a) (1 /√2)(√2,0,0; 0,1,−1; 0,1,1). (b) (1 /√2)(1,0,−1;0,√2,0; 1,0,1). r=( 2√2,−1+√2,−1−√2)T. 21.4 If T0is Tr Tijthen Uij=1 2(Tij+Tji)−1 3T0δij,Vij=1 3T0δij,Sij=1 2(Tij−Tji). 21.5 Twice contract the array with the outer product of ( x, y, z)w i t hi t s e l ft oo b t a i n the expression −(x2+y2+z2)2, which is an invariant and therefore a scalar. 21.6 (a) /epsilon1ijk/epsilon1jlmulvmwkand use (21.29); (b) /epsilon1ijk∂(φuk)/∂x j;( c ) ∂(/epsilon1ijkujvk)/∂x i; (d)/epsilon1ijk/epsilon1klm∂(ulvm)/∂x jand use (21.29); (e) start with u×curluand obtain /epsilon1ijkuj/epsilon1klm∂um ∂xl=···=uj /∂uj ∂xi / −uj /∂ui ∂xj / . 21.7 Write Aj(∂Ai/∂x j)a s∂(AiAj)/∂x j−Ai(∂Aj/∂x j). 21.8 (a) c=a×b.( b )0 ,±i|a|.Aa=0asince a×a=0. 831 TENSORS 21.9 (i) Write out the expression for |AT|, contract both sides of the equation with /epsilon1lmn and pick out the expression for |A|o nt h eR H S .N o t et h a t /epsilon1lmn/epsilon1lmnis a numerical scalar.(iii) Each non-zero term on the RHS contains any particular row index once andonly once. The same can be said for the Levi–Civita symbol on the LHS. Thusinterchanging two rows is equivalent to interchanging two of the subscripts of/epsilon1 lmnand thereby reversing its sign. Consequently, the magnitude of |A|remains the same but its sign is changed.(v) If, say, A pi=λApj, for some particular pair of values iandjand all pthen, in the (multiple-) summation on the RHS, each Ankappears multiplied by (no summation over iandj) /epsilon1ijkAliAmj+/epsilon1jikAljAmi=/epsilon1ijkλAljAmj+/epsilon1jikAljλAmj=0, since /epsilon1ijk=−/epsilon1jik. Consequently, grouped in this way all terms are zero and |A|=0 . (vi) Replace AmjbyAmj+λAljand note that λAliAljAnk/epsilon1ijk= 0 by virtue of result (v).(vii) If C=AB, |C|/epsilon1 lmn=AlxBxiAmyByjAnzBzk/epsilon1ijk. Contract this with /epsilon1lmnand show that the RHS is equal to /epsilon1xyz|AT|/epsilon1xyz|B|.I tt h e n follows from result (i) that |C|=|A||B|. 21.10 Note that R xidS= R (xi)3dS=0a n dt h a t R (xi)2dS=4π/3. (a) 0 (the two contributions cancel when i=j); (b) 8 πδij; (c) 0 for all sets of i, j, k,w h e t h e ro r not some or all are equal. 21.11 α=|v|−2. Note that the most general vector has components wi=λvi+µu(1) i+νu(2) i, where both u(1)andu(2)are orthogonal to v. 21.12∇×H=J+˙D;∇·D=ρ;∇×E+˙B=0;∇·B=0 . 21.13 Construct the orthogonal transformation matrix Sfor the symmetry operation of (say) a rotation of 2 π/3 about a body diagonal and, setting L=S−1=ST, construct σ/prime=LσLTand require σ/prime=σ. Repeat the procedure for (say) a rotation ofπ/2 about the x3-axis. These together show that σ11=σ22=σ33and that all other σij= 0. Further symmetry requirements do not provide any additional constraints. 21.14 W=EiσijEjhas to be maximised or minimised subject to EiEibeing held constant. Extreme values are W±=λ±|E|2,w h e r e λ±are the maximum and minimum eigenvalues of the matrix σij. 21.15 The transformation of δijhas to be included; the principal values are ±E·B. The third axis is in the direction ±B×Ewith principal value −|E|2. 21.16 (b) xT 1=( 2−10 ) ,xT 2=( 1 2√ 5),xT 3=( 1 2−√ 5). 21.17 The principal moments give the required ratios.21.18 The principal susceptibilit ies and (unnormalised) axes are λ=4 ,±(0,1,1); λ=2 ,±(c i,1,−1) with c1c2=−2, leading to: (a) lowest energy when (0 ,1,1) axis is parallel to the ﬁeld; (b) permitted values o f orientation are (0 ,n2,n3), hence as in (a); (c) permitted values o f orientation are ( n1,0,n3), subject to n2 1+n2 3=1 . The energy = −1 2µ0kH2V(2n2 1+3n2 3), which is minimised when (0 ,0,1) is parallel to the ﬁeld. 21.19 The principal permeability, in direction (1 ,1,2), has value 0. Thus all the nails lie in planes to which this is the normal. 21.20 ji=σikEkgives Isinθcosφ=aπr2E1,Isinθsinφ=aπr2E2,Icosθ=bπr2E3. Also V/L=E1sinθcosφ+E2sinθsinφ+E3cosθ. The current must ﬂow along the wire; Eis not parallel to the wire. 832 21.24 HINTS AND ANSWERS 21.21 Take p11=p22=p33=−p,a n d pij=eij=0f o r i/negationslash=j, leading to −p= (λ+2µ/3)eii. The fractional volume change is eii;λandµare as deﬁned in (21.45) and the worked example that follows it. 21.22 Show that pij=2λδij∇·u+(η+ν)(∂ui/∂x j+∂uj/∂x i). Form the sum of the derivatives P j(∂/∂x j) for this equation, substitute for ∂pij/∂x jand then formP i(∂/∂x i) of the result. The wave speed for ∇·uis [2( λ+η+ν)/ρ]1/2.Show that ρ∂2(∇×u)k ∂t2=/epsilon1kji∂2pil ∂xj∂xl and then use the previous expression for pijand the identity /epsilon1kji∂2/∂x j∂xi=0 . The wave speed for ∇×uis [(η+ν)/ρ]1/2. 21.23 Consider Qpq=/epsilon1pij/epsilon1qklTijkland show that Kmn=Qmn/4 has the required property. (a) Argue from the isotropy of Tijkland/epsilon1ijkfor that of Kmnand hence that it must be a multiple of δmn. Show that the multiplier is uniquely determined and thatTijkl=(δilδjk−δikδjl)/6. (b) By relabelling dummy subscripts and using the stated antisymmetry property,show that K nm=−Kmn. Show that −2Vi=/epsilon1minKmnand hence that Kmn=/epsilon1imnVi. Tijkl=/epsilon1kliVj−/epsilon1kljVi. 21.24 ρ=( 1−2s/√ 3+2 s2/3)1/2,φ=t a n−1[s/(√ 3−s)],z=s/√ 3. 21.25 Use |e1·(e2×e3)|=√g. Recall that√g/prime=|∂u/∂u/prime|√ganddu/prime1du/prime2du/prime3=|∂u/prime/∂u|du1du2du3. 21.26 g=r4sin2θ; recall that, for each i,vi=ˆvi/hi,e . g . v3=vφ/(rsinθ). Γ1 22=−r;Γ1 33=−rsin2θ;Γ2 12=r−1;Γ2 32=−sinθcosθ;Γ3 13=r−1;Γ3 23= cotθ. 21.27 ( vi;j);k=(vi;j),k−Γl ikvl;j−Γl jkvi;landvi;j=vi, j−Γm ijvm. If all components of a tensor equal zero in one coordinate system then they are zero in all coordinatesystems. 21.28 Using ˙s=p gij˙ui˙uj, the Euler–Lagrange equation is d dt /gik˙ui ˙s / −1 2˙s∂gij ∂uk˙ui˙uj=0. Calculate the t-derivative, write ∂gik ∂uj=1 2 /∂gik ∂uj+∂gjk ∂ui / and multiply through by glk.I ft=as+bthen¨s=0 . 833 22 Calculus of variations In chapters 2 and 5 we discussed how to ﬁnd stationary values of functions of a single variable f(x), of several variables f(x ,y,... ) and of constrained variables, where x ,y,... are subject to the nconstraints gi(x ,y,... )=0 , i=1,2,...,n.I na l l these cases the forms of the functions fandgiwere known, and the problem was one of ﬁnding the appropriate values of the variables x,yetc. We now turn to a diﬀerent kind of problem in which we are interested in bringing about a particular condition for a given expression (usually maximising or minimising it) by varying the functions on which the expression depends. For instance, we might want to know in what shape a ﬁxed length of rope shouldbe arranged so as to enclose the largest possible area, or in what shape it willhang when suspended under gravity from two ﬁxed points. In each case we areconcerned with a general maximisation or minimisation criterion by which thefunction y(x) that satisﬁes the given problem may be found. The calculus of variations provides a method for ﬁnding the function y(x). The problem must ﬁrst be expressed in a mathematical form, and the formmost commonly applicable to such problems is an integral . In each of the above questions, the quantity that has to be maximised or minimised by an appropriatechoice of the function y(x) may be expressed as an integral involving y(x)a n d the variables describing the geometry of the situation. In our example of the rope hanging from two ﬁxed points, we need to ﬁnd the shape function y(x) that minimises the gravitational potential energy of the rope. Each elementary piece of the rope has a gravitational potential energyproportional both to its vertical height above an arbitrary zero level and to thelength of the piece. Therefore the total potential energy is given by an integralfor the whole rope of such elementary contributions. The particular function y(x) for which the value of this integral is a minimum will give the shape assumed by the hanging rope. So in general we are led by this type of question to study the value of an 834 22.1 THE EULER–LAGRANGE EQUATION y x a b Figure 22.1 Possible paths for the integral (22.1). The solid line is the curve along which the integral is assumed stationary. The broken curves represent small variations from this path. integral whose integrand has a speciﬁed form in terms of a certain function and its derivatives, and to study how that value changes when the form ofthe function is varied. Speciﬁcally, we aim to ﬁnd the function that makes theintegral stationary , i.e. the function that makes the value of the integral a local maximum or minimum. Note that, unless stated otherwise, y /primeis used to denote dy/dx throughout this chapter. We also assume that all the functions we need to deal with are suﬃciently smooth and diﬀerentiable. 22.1 The Euler–Lagrange equation Let us consider the integral I=integraldisplayb aF(y,y/prime,x)dx, (22.1) where a,band the form of the function Fare ﬁxed by given considerations, e.g. the physics of the problem, but the curve y(x) is to be chosen so as to make stationary the value of I, which is clearly a function (or more accurately a functional ) of this curve, i.e. I=I[y(x)]. Referring to ﬁgure 22.1, we wish to ﬁnd the function y(x) (given, say, by the solid line) such that ﬁrst-order small changes in it (for example the two broken lines) will make only second-order changes in the value of I. Writing this in a more mathematical form, let us suppose that y(x)i st h e function required to make Istationary and consider making the replacement y(x)→y(x)+αη(x), (22.2) where the parameter αis small and η(x) is an arbitrary function with suﬃciently amenable mathematical properties. For the value of Ito be stationary with respect 835 CALCULUS OF VARIATIONS to these variations, we require dI dαvextendsinglevextendsinglevextendsinglevextendsingle α=0=0 f o ra l l η(x). (22.3) Substituting (22.2) into (22.1) and expanding as a Taylor series in αwe obtain I(y,α)=integraldisplayb aF(y+αη, y/prime+αη/prime,x)dx =integraldisplayb aF(y,y/prime,x)dx+integraldisplayb aparenleftbigg∂F ∂yαη+∂F ∂y/primeαη/primeparenrightbigg dx+O ( α2). With this form for I(y,α) the condition (22.3) implies that for all η(x)w er e q u i r e δI=integraldisplayb aparenleftbigg∂F ∂yη+∂F ∂y/primeη/primeparenrightbigg dx=0, where δIdenotes the ﬁrst-order variation in the value of Idue to the variation (22.2) in the function y(x). Integrating the second term by parts this becomes bracketleftbigg η∂F ∂y/primebracketrightbiggb a+integraldisplayb abracketleftbigg∂F ∂y−d dxparenleftbigg∂F ∂y/primeparenrightbiggbracketrightbigg η(x)dx=0. (22.4) In order to simplify the result we will assume, for the moment, that the end-points are ﬁxed, i.e. not only aandbare given but also y(a)a n d y(b). This restriction means that we require η(a)=η(b) = 0, in which case the ﬁrst term on the LHS of (22.4) equals zero at both end-points. Since (22.4) must be satisﬁed for arbitrary η(x), it is easy to see that we require ∂F ∂y=d dxparenleftbigg∂F ∂y/primeparenrightbigg . (22.5) This is known as the Euler–Lagrange (EL) equation, and is a diﬀerential equation fory(x),since the function Fis known. 22.2 Special cases In certain special cases a ﬁrst integral of the EL equation can be obtained for a general form of F. 22.2.1 Fdoes not contain yexplicitly In this case ∂F/∂y = 0, and (22.5) can be integrated immediately giving ∂F ∂y/prime=c o n s t a n t . (22.6) 836 22.2 SPECIAL CASES A(a, y(a))dxdydsB(b, y(b)) Figure 22.2 An arbitrary path between two ﬁxed points.IShow that the shortest curve joining two points is a straight line. Let the two points be labelled Aand Band have coordinates ( a, y(a)) and ( b, y(b)) respectively (see ﬁgure 22.2). Whatever the shape of the curve joining AtoB, the length of an element of path dsis given by ds= / (dx)2+(dy)2 /1/2=( 1+ y/prime2)1/2dx, and hence the total path length along the curve is given by L= Zb a(1 +y/prime2)1/2dx. (22.7) We must now apply the results of the previous section to determine that path which makes Lstationary (clearly a minimum in this case). Since the integral does not contain y(or indeed x) explicitly, we may use (22.6) to obtain k=∂F ∂y/prime=y/prime (1 +y/prime2)1/2. where kis a constant. This is easily rearranged and integrated to give y=k (1−k2)1/2x+c, which, as expected, is the equation of a straight line in the form y=mx+c,w i t h m=k/(1−k2)1/2. The value of m(ork) can be found by demanding that the straight line passes through the points AandBand is given by m=[y(b)−y(a)]/(b−a). Substituting the equation of the straight line into (22.7) we ﬁnd that, again as expected, the total pathlength is given by L 2=[y(b)−y(a)]2+(b−a)2. J 837 CALCULUS OF VARIATIONS dxdydsy x Figure 22.3 A convex closed curve that is symmetrical about the x-axis. 22.2.2 Fdoes not contain xexplicitly In this case, multiplying the EL equation (22.5) by y/primeand using d dxparenleftbigg y/prime∂F ∂y/primeparenrightbigg =y/primed dxparenleftbigg∂F ∂y/primeparenrightbigg +y/prime/prime∂F ∂y/prime we obtain y/prime∂F ∂y+y/prime/prime∂F ∂y/prime=d dxparenleftbigg y/prime∂F ∂y/primeparenrightbigg . But since Fis a function of yandy/primeonly, and not explicitly of x,t h eL H So f this equation is just the total derivative of F,n a m e l y dF/dx . Hence, integrating we obtain F−y/prime∂F ∂y/prime=c o n s t a n t . (22.8)IFind the closed convex curve of length lthat encloses the greatest possible area. Without any loss of generality we can assume that the curve passes through the origin, and can further suppose that it is symmetric with respect to the x-axis; this assumption is not essential. Using the distance salong the curve, measured from the origin, as the independent variable and yas the dependent one, we have the boundary conditions y(0) = y(l/2) = 0. The element of area shown in ﬁgure 22.3 is then given by dA=yd x=y / (ds)2−(dy)2 /1/2, a n dt h et o t a la r e ab y A=2 Zl/2 0y(1−y/prime2)1/2ds; (22.9) herey/primestands for dy/ds rather than dy/dx . Since the integrand does not contain sexplicitly, 838 22.2 SPECIAL CASES we can use (22.8) to obtain a ﬁrst integral of the EL equation for y,n a m e l y y(1−y/prime2)1/2+yy/prime2(1−y/prime2)−1/2=k, where kis a constant. On rearranging this gives ky/prime=±(k2−y2)1/2, which, using y(0) = 0, integrates to y/k=s i n ( s/k). (22.10) The other end-point, y(l/2) = 0, ﬁxes the value of kasl/2πto yield y=l 2πsin2πs l. From this we obtain dy=c o s ( 2 πs/l)dsand since ( ds)2=(dx)2+(dy)2we ﬁnd also that dx=±sin(2πs/l)ds. This in turn can be integrated and, using x(0) = 0, gives xin terms ofsas x−l 2π=−l 2πcos2πs l. We thus obtain the expected result that xandylie on the circle of radius l/(2π)g i v e nb y/ x−l 2π /2 +y2=l2 4π2. Substituting the solution (22.10) into the expression for the total area (22.9), it is easily veriﬁed that A=l2/(4π). A much quicker derivation of this result is possible using plane polar coordinates. J The previous two examples have been carried out in some detail, even though the answers are more easily obtained in other ways, expressly so that the method is transparent and the way in which it works can be ﬁlled in mentally at almostevery step. The next example, however, does not have such an intuitively obvioussolution.ITwo rings, each of radius a, are placed parallel with their centres 2bapart and on a common normal. An axially symmetric soap ﬁlm is formed between them but does not coverthe ends of the rings (see ﬁgure 22.4). Find the shape assumed by the ﬁlm. Creating the soap ﬁlm requires an energy γper unit area (numerically equal to the surface tension of the soap solution). So the stable shape of the soap ﬁlm, i.e. the one thatminimises the energy, will also be the one that minimises the surface area (neglectinggravitational eﬀects). It is obvious that any convex surface, shaped such as that shown as the broken line in ﬁgure 22.4( a) cannot be a minimum but it is not clear whether some shape intermediate between the cylinder shown by solid lines in ( a), with area 4 πab(or twice this for the double surface of the ﬁlm), and the form shown in ( b), with area approximately 2 πa 2, will produce a lower total area than both of these extremes. If there is such a shape (e.g. that in ﬁgure 22.4( c)), then it will be that which best compromises between two requirements, the need to minimise the ring-to-ring distance measured on the ﬁlm surface ( a)a n dt h e need to minimise the average waist measurement of the surface ( b). We take cylindrical polar coordinates as in ﬁgure 22.4( c) and let the radius of the soap ﬁlm at height zbeρ(z)w i t h ρ(±b)=a. Counting only one side of the ﬁlm, the element of 839 CALCULUS OF VARIATIONS (a)( b)( c)b −bz ρ a Figure 22.4 Possible soap ﬁlms between two parallel circular rings. surface area between zandz+dzis dS=2πρ / (dz)2+(dρ)2 /1/2, so the total surface area is given by S=2π Zb −bρ(1 +ρ/prime2)1/2dz. (22.11) Since the integrand does not contain zexplicitly, we can use (22.8) to obtain an equation forρthat minimises S,i . e . ρ(1 +ρ/prime2)1/2−ρρ/prime2(1 +ρ/prime2)−1/2=k, where kis a constant. Multiplying through by (1 + ρ/prime2)1/2, rearranging to ﬁnd an explicit expression for ρ/primeand integrating we ﬁnd cosh−1ρ k=z k+c. where cis the constant of integration. Using the boundary conditions ρ(±b)=a,w e require c=0a n d ksuch that a/k=c o s h b/k(ifb/ais too large, no such kcan be found). Thus the curve that minimises the surface area is ρ/k=c o s h ( z/k), and in proﬁle the soap ﬁlm is a catenary (see section 22.4) with the minimum distance from the axis equal to k. J 22.3 Some extensions It is quite possible to relax many of the restrictions we have imposed so far. For example, we can allow end-points that are constrained to lie on given curves rather than being ﬁxed, or we can consider problems with several dependent and/or independent variables or higher-order derivatives of the dependent variable. Eachof these extensions is now discussed. 840 22.3 SOME EXTENSIONS 22.3.1 Several dependent variables Here we have F=F(y1,y/prime 1,y2,y/prime 2,...,y n,y/prime n,x)w h e r ee a c h yi=yi(x). The analysis in this case proceeds as before, leading to nseparate but simultaneous equations for the yi(x), ∂F ∂yi=d dxparenleftbigg∂F ∂y/prime iparenrightbigg ,i =1,2,...,n . (22.12) 22.3.2 Several independent variables With nindependent variables, we need to extremise multiple integrals of the form I=integraldisplayintegraldisplay ···integraldisplay Fparenleftbigg y,∂y ∂x1,∂y ∂x2,...,∂y ∂xn,x1,x2,...,x nparenrightbigg dx1dx2···dxn. Using the same kind of analysis as before, we ﬁnd that the extremising function y=y(x1,x2,...,x n) must satisfy ∂F ∂y=nsummationdisplay i=1∂ ∂xiparenleftbigg∂F ∂yxiparenrightbigg , (22.13) where yxistands for ∂y/∂x i. 22.3.3 Higher-order derivatives If in (22.1) F=F(y,y/prime,y/prime/prime,...,y(n),x) then using the same method as before and performing repeated integration by parts, it can be shown that the requiredextremising function y(x) satisﬁes ∂F ∂y−d dxparenleftbigg∂F ∂y/primeparenrightbigg +d2 dx2parenleftbigg∂F ∂y/prime/primeparenrightbigg −···+(−1)ndn dxnparenleftbigg∂F ∂y(n)parenrightbigg =0,(22.14) provided that y=y/prime=···=y(n−1)= 0 at both end-points. If y, or any of its derivatives, is not zero at the end-points then a corresponding contribution orcontributions will appear on the RHS of (22.14). 22.3.4 Variable end-points We now discuss the very important generalisation to variable end-points. Suppose, as before, we wish to ﬁnd the function y(x) that extremises the integral I=integraldisplay b aF(y,y/prime,x)dx, but this time we demand only that the lower end-point is ﬁxed, while we allow y(b) to be arbitrary. Repeating the analysis of section 22.1, we ﬁnd from (22.4) 841 CALCULUS OF VARIATIONS ∆x∆y y(x) h(x, y)=0y(x)+η(x) b Figure 22.5 Variation of the end-point balong the curve h(x, y)=0 . that we require bracketleftbigg η∂F ∂y/primebracketrightbiggb a+integraldisplayb abracketleftbigg∂F ∂y−d dxparenleftbigg∂F ∂y/primeparenrightbiggbracketrightbigg η(x)dx=0. (22.15) Obviously the EL equation (22.5) must still hold for the second term on the LHS to vanish. Also, since the lower end-point is ﬁxed, i.e. η(a)=0 ,t h eﬁ r s tt e r mo n the LHS automatically vanishes at the lower limit. However, in order that it alsovanishes at the upper limit, we require in addition that ∂F ∂y/primevextendsinglevextendsinglevextendsinglevextendsingle x=b=0. (22.16) Clearly if both end-points may vary then ∂F/∂y/primemust vanish at both ends. An interesting and more general case is where the lower end-point is again ﬁxed at x=a, but the upper end-point is free to lie anywhere on the curve h(x, y) = 0. Now in this case, the variation in the value of Idue to the arbitrary variation (22.2) is given to ﬁrst order by δI=bracketleftbigg∂F ∂y/primeηbracketrightbiggb a+integraldisplayb aparenleftbigg∂F ∂y−d dx∂F ∂y/primeparenrightbigg ηd x+F(b)∆x, (22.17) where ∆ xis the displacement in the x-direction of the upper end-point, as indicated in ﬁgure 22.5, and F(b) is the value of Fatx=b. In order for (22.17) to be valid, we of course require the displacement ∆ xto be small. From the ﬁgure we see that ∆ y=η(b)+y/prime(b)∆x. Since the upper end-point must lie on h(x, y) = 0 we also require that, at x=b, ∂h ∂x∆x+∂h ∂y∆y=0, which on substituting our expression for ∆ yand rearranging becomes parenleftbigg∂h ∂x+y/prime∂h ∂yparenrightbigg ∆x+∂h ∂yη=0. (22.18) 842 22.3 SOME EXTENSIONS A yx=x0 Bx Figure 22.6 A frictionless wire along which a small bead slides. We seek the shape of the wire that allows the bead to travel from the origin Oto the line x=x0in the least possible time. Now, from (22.17) the condition δI= 0 requires, besides the EL equation, that atx=b, the other two contributions cancel, i.e. F∆x+∂F ∂y/primeη=0. (22.19) Eliminating ∆ xandηbetween (22.18) and (22.19) leads to the condition that at the end-point parenleftbigg F−y/prime∂F ∂y/primeparenrightbigg∂h ∂y−∂F ∂y/prime∂h ∂x=0. (22.20) In the special case where the end-point is free to lie anywhere on the vertical line x=b, we have ∂h/∂x =1a n d ∂h/∂y = 0. Substituting these values into (22.20), we recover the end-point condition given in (22.16).IA frictionless wire in a vertical plane connects two points AandB,Abeing higher than B. Let the position of Abe ﬁxed at the origin of an xy-coordinate system, but allow Bto lie anywhere on the vertical line x=x0(see ﬁgure 22.6). Find the shape of the wire such that ab e a dp l a c e do ni ta t Awill slide under gravity to Bin the shortest possible time. This is a variant of the famous brachistochrone (shortest time) problem, which is often used to illustrate the calculus of variations. Conservation of energy tells us that the particle speed is given by v=ds dt= p 2gy, where sis the path length along the wire and gis the acceleration due to gravity. Since the element of path length is ds=( 1+ y/prime2)1/2dx, the total time taken to travel to the line x=x0is given by t= Zx=x0 x=0ds v=1√2g Zx0 0 s 1+y/prime2 ydx. Because the integrand does not contain xexplicitly, we can use (22.8) with the speciﬁc form F= p 1+y/prime2/√yto ﬁnd a ﬁrst integral; on simpliﬁcation this yieldsh y(1 +y/prime2) i1/2 =k, 843 CALCULUS OF VARIATIONS where kis a constant. Letting a=k2and solving for y/primewe ﬁnd y/prime=dy dx= ra−y y, which on substituting y=asin2θintegrates to give x=a 2(2θ−sin2θ)+c. Thus the parametric equations of the curve are given by x=b(φ−sinφ)+c, y =b(1−cosφ), where b=a/2a n d φ=2θ; they deﬁne a cycloid, the curve traced out by a point on the rim of a wheel of radius brolling along the x-axis. We must now use the end-point conditions to determine the constants bandc. Since the curve passes through the origin, we see immediately that c=0 .N o ws i n c e y(x0) is arbitrary, i.e. the upper end-point can lie anywhere on the curve x=x0, the condition (22.20) reduces to (22.16), so that we also require ∂F ∂y/prime //// x=x0=y/primep y(1 +y/prime2) ///// x=x0=0, which implies that y/prime=0a t x=x0. In words, that the tangent to the cycloid at Bmust b ep a r a l l e lt ot h e x-axis; this requires πb=x0. J 22.4 Constrained variation Just as the problem of ﬁnding thestationary values of a function f(x, y)s u b j e c tt o the constraint g(x, y) = constant is solved by means of Lagrange’s undetermined multipliers (see chapter 5), so the corresponding problem in the calculus ofvariations is solved by an analogous method. Suppose that we wish to ﬁnd the stationary values of I=integraldisplay b aF(y,y/prime,x)dx, subject to the constraint that the value of J=integraldisplayb aG(y,y/prime,x)dx is held constant. Following the method of Lagrange undetermined multipliers let us deﬁne a new functional K=I+λJ=integraldisplayb a(F+λG)dx, and ﬁnd its unconstrained stationary values. Repeating the analysis of section 22.1 we ﬁnd that we require ∂F ∂y−d dxparenleftbigg∂F ∂y/primeparenrightbigg +λbracketleftbigg∂G ∂y−d dxparenleftbigg∂G ∂y/primeparenrightbiggbracketrightbigg =0, 844 22.4 CONSTRAINED VARIATION −ay O a x Figure 22.7 A uniform rope with ﬁxed end-points suspended under gravity. which, together with the original constraint J= constant, will yield the required solution y(x). This method is easily generalised to cases with more than one constraint by the introduction of more Lagrange multipliers. If we wish to ﬁnd the stationary valuesof an integral Isubject to the multiple constraints that the values of the integrals J ibe held constant for i=1,2,...,n, then we simply ﬁnd the unconstrained stationary values of the new integral K=I+nsummationdisplay 1λiJi.IFind the shape assumed by a uniform rope when suspended by its ends from two points at equal heights. We will solve this problem using x(see ﬁgure 22.7) as the independent variable. Let the rope of length 2 Lbe suspended between the points x=±a,y=0( L>a )a n d have uniform linear density ρ. We then need to ﬁnd the stationary value of the rope’s gravitational potential energy, I=−ρg Z yd s=−ρg Za −ay(1 +y/prime2)1/2dx, with respect to small changes in the form of the rope but subject to the constraint that the total length of the rope remains constant, i.e. J= Z ds= Za −a(1 +y/prime2)1/2dx=2L. We thus deﬁne a new integral (omitting the factor −1f r o m Ifor brevity) K=I+λJ= Za −a(ρgy+λ)(1 + y/prime2)1/2dx and ﬁnd its stationary values. Since the integrand does not contain the independent variable xexplicitly, we can use (22.8) to ﬁnd the ﬁrst integral: (ρgy+λ) / 1+y/prime2 /1/2 −(ρgy+λ) / 1+y/prime2 /−1/2 y/prime2=k, 845 CALCULUS OF VARIATIONS where kis a constant; this reduces to y/prime2= /ρgy+λ k /2 −1. Making the substitution ρgy+λ=kcoshz, this can be integrated easily to give k ρgcosh−1 /ρgy+λ k / =x+c, where ci st h ec o n s t a n to fi n t e g r a t i o n . We now have three unknowns, λ,kandc, that must be evaluated using the two end conditions y(±a) = 0 and the constraint J=2L. The end conditions give coshρg(a+c) k=λ k=c o s hρg(−a+c) k, and since a/negationslash= 0, these imply c=0a n d λ/k=c o s h ( ρga/k ). Putting c=0i n t ot h e constraint, in which y/prime= sinh( ρgx/k ), we obtain 2L= Za −a h 1+s i n h2 /ρgx k / i1/2 dx =2k ρgsinh /ρga k / . Collecting together the values for the constants, the form adopted by the rope is therefore y(x)=k ρg h cosh /ρgx k / −cosh /ρga k / i , where kis the solution of sinh( ρga/k )=ρgL/k . This curve is known as a catenary. J 22.5 Physical variational principles Many results in both classical and quantum physics can be expressed as varia- tional principles, and it is often when expressed in this form that their physical meaning is most clearly understood. Moreover, once a physical phenomenon hasbeen written as a variational principle, we can use all the results derived in thischapter to investigate its behaviour. It is usually possible to identify conservedquantities, or symmetries of the system of interest, that otherwise might be foundonly with considerable eﬀort. From the wide range of physical variational princi-ples we will select two examples from familiar areas of classical physics, namely geometric optics and mechanics. 22.5.1 Fermat’s principle in optics Fermat’s principle in geometrical optics states that a ray of light travelling in a region of variable refractive index follows a path such that the total optical pathlength (physical length ×refractive index) is stationary. 846 22.5 PHYSICAL VARIATIONAL PRINCIPLES θ1θ2 n1n2 AB xy Figure 22.8 Path of a light ray at the plane interface between media with refractive indices n1andn2,w h e r e n2<n1.IFrom Fermat’s principle deduce Snell’s law of refraction at an interface. Let the interface be at y= constant (see ﬁgure 22.8) and let it separate two regions with refractive indices n1andn2respectively. On a ray the element of physical path length is ds=( 1+ y/prime2)1/2dx, and so for a ray that passes through the points AandB,t h et o t a l optical path length is P= ZB An(y)(1 + y/prime2)1/2dx. Since the integrand does not contain the independent variable xexplicitly, we use (22.8) to obtain a ﬁrst integral, which, a fter some rearrangement, reads n(y) / 1+y/prime2 /−1/2 =k, where kis a constant. Recalling that y/primeis the tangent of the angle φbetween the instantaneous direction of the ray and the x-axis, this general result, which is not dependent on the conﬁguration presently under consideration, can be put in the form ncosφ=c o n s t a n t along a ray, even though nandφvary individually. For our particular conﬁguration nis constant in each medium and therefore so is y/prime. Thus the rays travel in straight lines in each medium (as anticipated in ﬁgure 22.8, but not assumed in our analysis), and since kis constant along the whole path we have n1cosφ1=n2cosφ2, or in terms of the conventional angles in the ﬁgure n1sinθ1=n2sinθ2. J 22.5.2 Hamilton’s principle in mechanics Consider a mechanical system whose conﬁguration can be uniquely deﬁned by a number of coordinates qi(usually distances and angles) together with time tand which experiences only forces derivable from a potential. Hamilton’s principle 847 CALCULUS OF VARIATIONS y Odx lx Figure 22.9 Transverse displacement on a taut string that is ﬁxed at two points a distance lapart. states that in moving from one conﬁguration at time t0to another at time t1the motion of such a system is such as to make L=integraldisplayt1 t0L(q1,q2...,q n,˙q1,˙q2,...,˙qn,t)dt (22.21) stationary. The Lagrangian Lis deﬁned, in terms of the kinetic energy Tand the potential energy V(with respect to some reference situation), by L=T−V. Here Vis a function of the qionly, not of the ˙qi. Applying the EL equation to L we obtain Lagrange’s equations , ∂L ∂qi=d dtparenleftbigg∂L ∂˙qiparenrightbigg ,i =1,2,...,n .IUsing Hamilton’s principle derive the wave e quation for small transverse oscillations of a taut string. In this example we are in fact considering a generalisation of (22.21) to a case involvingone isolated independent coordinate t,t o g e t h e rw i t ha continuum in which the q ibecome the continuous variable x. The expressions for TandVtherefore become integrals over x rather than sums over the label i. Ifρandτare the local density and tension of the string, both of which may depend on x, then, referring to ﬁgure 22.9, the kinetic and potential energies of the string are given by T= Zl 0ρ 2 /∂y ∂t /2 dx, V = Zl 0τ 2 /∂y ∂x /2 dx and (22.21) becomes L=1 2 Zt1 t0dt Zl 0 /" ρ /∂y ∂t /2 −τ /∂y ∂x /2 /# dx. 848 22.6 GENERAL EIGENVALUE PROBLEMS Using (22.13) and the fact that ydoes not appear explicitly, we obtain ∂ ∂t / ρ∂y ∂t / −∂ ∂x / τ∂y ∂x / =0. If, in addition, ρandτdo not depend on xortthen ∂2y ∂x2=1 c2∂2y ∂t2, where c2=τ/ρ. This is the wave equation for small transverse oscillations of a taut uniform string. J 22.6 General eigenvalue problems We have seen in this chapter that the problem of ﬁnding a curve that makes the value of a given integral stationary when the integral is taken along the curveresults, in each case, in a diﬀerential equation for the curve. It is not a great extension to ask whether this may be used to solve diﬀerential equations, by setting up a suitable variational problem and then seeking ways other than theEuler equation of ﬁnding or estimating stationary solutions. We shall be concerned with diﬀerential equations of the form Ly=λρ(x)y, where the diﬀerential operator Lis self-adjoint, so that L=L †(with appropriate boundary conditions on the solution y)a n d ρ(x) is some weight function, as discussed in chapter 17. In particular, we will concentrate on the Sturm–Liouvilleequation as an explicit example, but much of what follows can be applied toother equations of this type. We have already discussed the solution of equations of the Sturm–Liouville type in chapter 17 and the same notation will be used here. In this section,however, we will adopt a variational approach to estimating the eigenvalues ofsuch equations. Suppose we search for stationary values of the integral I=integraldisplay b abracketleftBig p(x)y/prime2(x)−q(x)y2(x)bracketrightBig dx, (22.22) with y(a)=y(b)=0a n d pandqany suﬃciently smooth and diﬀerentiable functions of x. However, in addition we impose a normalisation condition J=integraldisplayb aρ(x)y2(x)dx=c o n s t a n t . (22.23) Here ρ(x) is a positive weight function deﬁned in the interval a≤x≤b, but which may in particular cases be a constant. Then, as in section 22.4, we use undetermined Lagrange multipliers, †and †We use−λ, rather than λ, so that the ﬁnal equation (22.24) appears in the conventional Sturm– Liouville form. 849 CALCULUS OF VARIATIONS consider K=I−λJgiven by K=integraldisplayb abracketleftBig py/prime2−(q+λρ)y2bracketrightBig dx. On application of the EL equation (22.5) this yields d dxparenleftbigg pdy dxparenrightbigg +qy+λρy=0, (22.24) which is exactly the Sturm–Liouville equation (17.35), with eigenvalue λ. Now, since both IandJare quadratic in yand its derivative, ﬁnding stationary values ofKis equivalent to ﬁnding stationary values of I/J. This may also be shown by considering the functional Λ = I/J, for which δΛ=( δI/J)−(I/J2)δJ =(δI−ΛδJ)/J =δK/J. Hence, extremising Λ is equivalent to extremising K. Thus we have the important result that ﬁnding functions ythat make I/Jstationary is equivalent to ﬁnding functions ythat are solutions of the Sturm–Liouville equation; the resulting value ofI/Jequals the corresponding eigenvalue of the equation. Of course this does not tell us how to ﬁnd such a function yand, naturally, to have to do this by solving (22.24) directly defeats the purpose of the exercise. Wewill see in the next section how some progress can be made. It is worth recallingthat the functions p(x),q(x)a n d ρ(x) can have many diﬀerent forms, and so (22.24) represents quite a wide variety of equations. We now recall some properties of the solutions of the Sturm–Liouville equation. The eigenvalues λ iof (22.24) are real and will be assumed non-degenerate (for simplicity). We also assume that the corresponding eigenfunctions have been made real, so that normalised eigenfunctions yi(x) satisfy the orthogonality relation (as in (17.27)) integraldisplayb ayiyjρd x=δij. (22.25) Further, we take the boundary condition in the form bracketleftBig yipy/prime jbracketrightBigx=b x=a= 0; (22.26) this can be satisﬁed by y(a)=y(b) = 0, but also by many other sets of boundary conditions. 850 22.7 ESTIMATION OF EIGENVALUES AND EIGENFUNCTIONSIShow thatZb a /; y/prime jpy/prime i−yjqyi / dx=λiδij. (22.27) Letyibe an eigenfunction of (22.24), corresponding to a particular eigenvalue λi,s ot h a t/; py/prime i //prime+(q+λiρ)yi=0. Multiplying this through by yjand integrating from atob(the ﬁrst term by parts) we obtain/ yj /; py/prime i / /b a− Zb ay/prime j(py/prime i)dx+ Zb ayj(q+λiρ)yidx=0. (22.28) The ﬁrst term vanishes by virtue of (22.26), and on rearranging the other terms and using (22.25), we ﬁnd the result (22.27). J We see at once that, if the function y(x) minimises I/J, i.e. satisﬁes the Sturm– Liouville equation, then putting yi=yj=yin (22.25) and (22.27) yields Jand Irespectively on the left-hand sides; thus, as mentioned above, the minimised value of I/Jis just the eigenvalue λ, introduced originally as the undetermined multiplier.IFor a function ysatisfying the Sturm–Liouville equat ion verify that, provided (22.26) is satisﬁed, λ=I/J. Firstly, we multiply (22.24) through by yto give y(py/prime)/prime+qy2+λρy2=0. Now integrating this expression by parts we have/ ypy/prime /b a− Zb a / py/prime2−qy2 / dx+λ Zb aρy2dx=0. The ﬁrst term on the LHS is zero, the second is simply −Ia n dt h et h i r di s λJ. Thus λ=I/J. J 22.7 Estimation of eigenvalues and eigenfunctions Since the eigenvalues λiof the Sturm–Liouville equation are the stationary values ofI/J(see above), it follows that any evaluation of I/Jmust yield a value that lies between the lowest and highest eigenvalues of the corresponding Sturm–Liouville equation, i.e. λmin≤I J≤λmax, where, depending on the equation under consideration, either λmin=−∞and 851 CALCULUS OF VARIATIONS λmaxis ﬁnite, or λmax=∞andλminis ﬁnite. Notice that here we have departed from direct consideration of the minimising problem and made a statement abouta calculation in which no actual minimisation is necessary. Thus, as an example, for an equation with a ﬁnite lowest eigenvalue λ 0any evaluation of I/Jprovides an upper bound on λ0. Further, we will now show that the estimate λobtained is a better estimate of λ0than the estimated (guessed) function yis of y0, the true eigenfunction corresponding to λ0. The sense in which ‘better’ is used here will be clear from the ﬁnal result. Firstly, we expand the estimated or trial function yin terms of the complete setyi: y=y0+c1y1+c2y2+···, where, if a good trial function has been guessed, the ciwill be small. Using (22.25) we have immediately that J=1+summationtext i|ci|2. The other required integral is I=integraldisplayb abracketleftBigg pparenleftbigg y/prime 0+summationdisplay iciy/prime iparenrightbigg2 −qparenleftbigg y0+summationdisplay iciyiparenrightbigg2bracketrightBigg dx. On multiplying out the squared terms, all the cross terms vanish because of (22.27) to leave λ=I J=λ0+summationtext i|ci|2λi 1+summationtext j|cj|2 =λ0+summationdisplay i|ci|2(λi−λ0)+O ( c4). Hence λdiﬀers from λ0by a term second order in the ci, even though ydiﬀered from y0by a term ﬁrst order in the ci; this is what we aimed to show. We notice incidentally that, since λ0<λ ifor all i,λis shown to be necessarily ≥λ0, with equality only if all ci=0 ,i . e .i f y≡y0. The method can be extended to the second and higher eigenvalues by imposing, in addition to the original constraints and boundary conditions, a restrictionof the trial functions to only those that are orthogonal to the eigenfunctionscorresponding to lower eigenvalues. (Of course, this requires complete or nearlycomplete knowledge of these latter eigenfunctions.) An example is given at theend of the chapter (exercise 22.26). We now illustrate the method we have discussed by considering a simple example, one for which, as on previous occasions, the answer is obvious. 852 22.7 ESTIMATION OF EIGENVALUES AND EIGENFUNCTIONS (a)(b)(c) (d) 0.20.2 0.40.4 0.60.6 0.80.8 11 xy(x) Figure 22.10 Trial solutions used to estimate the lowest eigenvalue λof −y/prime/prime=λywith y(0) = y/prime(1) = 0. They are: ( a)y=s i n ( πx/2), the exact result; (b)y=2x−x2;(c)y=x3−3x2+3x;(d)y=s i n2(πx/2).IEstimate the lowest eigenvalue of the equation −d2y dx2=λy, 0≤x≤1, (22.29) with boundary conditions y(0) = 0 ,y/prime(1) = 0 . (22.30) We need to ﬁnd the lowest value λ0ofλfor which (22.29) has a solution y(x) that satisﬁes (22.30). The exact answer is of course y=Asin(xπ/2) and λ0=π2/4≈2.47. Firstly we note that the Sturm–Liouville equation reduces to (22.29) if we take p(x)=1 , q(x)=0a n d ρ(x) = 1 and that the boundary conditions satisfy (22.26). Thus we are able to apply the previous theory. We will use three trial functions so that the eﬀect on the estimate of λ0of making better or worse ‘guesses’ can be seen. One further preliminary remark is relevant, namely that theestimate is independent of any constant multiplicative factor in the function used. Thisis easily veriﬁed by looking at the form of I/J. We normalise each trial function so that y(1) = 1, purely in order to facilitate comparison of the various function shapes. Figure 22.10 illustrates the trial functions used, curve ( a) being the exact solution y= sin( πx/2). The other curves are ( b)y(x)=2 x−x 2,(c)y(x)=x3−3x2+3x,a n d( d) y(x)=s i n2(πx/2). The choice of trial function is governed by the following considerations: (i) the boundary conditions (22.30) mustbe satisﬁed. (ii) a ‘good’ trial function ought to mimic the correct solution as far as possible, but it may not be easy to guess even the general shape of the correct solution in somecases. (iii) the evaluation of I/Jshould be as simple as possible. 853 CALCULUS OF VARIATIONS It is easily veriﬁed that functions ( b), (c)a n d( d) all satisfy (22.30) but, so far as mimicking the correct solution is concerned, we would expect from the ﬁgure that ( b)w o u l db e superior to the other two. All three evaluations are straightforward, using (22.22) and(22.23): λ b= R1 0(2−2x)2dxR1 0(2x−x2)2dx=4/3 8/15=2.50 λc= R1 0(3x2−6x+3 )2dxR1 0(x3−3x2+3x)2dx=9/5 9/14=2.80 λd= R1 0(π2/4)sin2(πx)dxR1 0sin4(πx/2)dx=π2/8 3/8=3.29. We expected all evaluations to yield estimates greater than the lowest eigenvalue, 2.47, and this is indeed so. From these trials alone we are able to say (only) that λ0≤2.50. As expected, the best approximation ( b) to the true eigenfunction yields the lowest, and therefore the best, upper bound on λ0. J We may generalise the work of this section to other diﬀerential equations of the form Ly=λρy,w h e r e L=L†. In particular, one ﬁnds λmin≤I J≤λmax, where IandJare now given by I=integraldisplayb ay∗(Ly)dx and J=integraldisplayb aρy∗yd x . (22.31) It is straightforward to show that, for the special case of the Sturm–Liouville equation, for which Ly=−(py/prime)/prime−qy, the expression for Iin (22.31) leads to (22.22). 22.8 Adjustment of parameters Instead of trying to estimate λ0by selecting a large number of diﬀerent trial functions, we may also use trial functions that include one or more parameters which themselves may be adjusted to give the lowest value to λ=I/Jand hence the best estimate of λ0. The justiﬁcation for this method comes from the knowledge that no matter what form of function is chosen, nor what values areassigned to the parameters, provided the boundary conditions are satisﬁed λcan never be less than the required λ 0. To illustrate this method an example from quantum mechanics will be used. The time-independent Schr ¨odinger equation is formally written as the eigenvalue equation Hψ=Eψ,w h e r e His a linear operator, ψthe wavefunction describing a quantum mechanical system and Ethe energy of the system. The energy 854 22.8 ADJUSTMENT OF PARAMETERS operator His called the Hamiltonian and for a particle of mass mmoving in a one-dimensional harmonic oscillator potential is given by H=−/planckover2pi12 2md2 dx2+kx2 2, (22.32) where/planckover2pi1is Planck’s constant divided by 2 π.IEstimate the ground-state energy of a quantum harmonic oscillator. Using (22.32) in Hψ=Eψ, the Schr ¨odinger equation is − /~2 2md2ψ dx2+kx2 2ψ=Eψ,−∞<x<∞. (22.33) The boundary conditions are that ψshould vanish as x→±∞ . Equation (22.33) is a form of the Sturm–Liouville equation in which p= /~2/(2m),q=−kx2/2,ρ=1a n d λ=E;i t can be solved by the methods developed previously, e.g. by writing the eigenfunction ψas a power series in x. However, our purpose here is to illustrate variational methods and so we take as a trial wavefunction ψ=e x p (−αx2), where αis a positive parameter whose value we will choose later. This function certainly →0a sx→±∞ and is convenient for calculations. Whether it approximates the true wave function is unknown, but if it does not our estimate willstill be valid, although the upper bound will be a poor one. With y=e x p (−αx 2) and therefore y/prime=−2αxexp(−αx2), the required estimate is E=λ= R∞ −∞[( /~2/2m)4α2x2+(k/2)x2]e−2αx2dxR∞ −∞e−2αx2dx= /~2α 2m+k 8α. (22.34) This evaluation is easily carried out using the reduction formula In=n−1 4αIn−2,for integrals of the form In= Z∞ −∞xne−2αx2dx. (22.35) So, we have obtained the estimate (22.34), involving the parameter α, for the oscillator’s ground-state energy, i.e. the lowest eigenvalue of H. In line with our previous discussion we now minimise λwith respect to α. Putting dλ/dα = 0 (clearly a minimum), yields α=(km)1/2/(2 /~), which in turn gives as the minimum value for λ E= /~ 2 /k m /1/2 = /~ω 2, (22.36) where we have put ( k/m)1/2equal to the classical angular frequency ω. The method thus leads to the conclusion that the ground-state energy E0is≤1 2 /~ω. In fact, as is well known, the equality sign holds,1 2 /~ωbeing just the zero-point energy of a quantum mechanical oscillator. Our estimate gives the exact value because ψ(x)= exp(−αx2) is the correct functional form for the ground state wavefunction and the particular value of αthat we have found is the that needed to make ψan eigenfunction of Hwith eigenvalue ≤1 2 /~ω. J An alternative but equivalent approach to this is developed in the exercises that follow, as is an extension of this particular problem to estimating the second-lowest eigenvalue (see exercise 22.26). 855 CALCULUS OF VARIATIONS 22.9 Exercises 22.1 A surface of revolution, whose equation in cylindrical polar coordinates is ρ= ρ(z), is bounded by the circles ρ=a,z=±c(a>c). Show that the function that makes the surface integral I= R ρ−1/2dSstationary with respect to small variations is given by ρ(z)=k+z2/(4k), where k=[a±(a2−c2)1/2]/2. 22.2 Show that the lowest value of the integralZB A(1 +y/prime2)1/2 ydx, where Ais (−1,1) and Bis (1,1), is 2ln(1+√ 2). Assume that the Euler–Lagrange equation gives a minimising curve. 22.3 The refractive index nof a medium is a function only of the distance rfrom a ﬁxed point O. Prove that the equation of a light ray, assumed to lie in a plane through O, travelling in the medium satisﬁes (in plane polar coordinates) 1 r2 /dr dφ /2 =r2 a2n2(r) n2(a)−1, where ais the distance of the ray from Oat the point at which dr/dφ =0 . Ifn=[ 1+( α2/r2)]1/2and the ray starts and ends far from O, ﬁnd its deviation (the angle through which the ray is tu rned) if its minimum distance from Oisa. 22.4 The Lagrangian for a π-meson is given by L(x,t)=1 2(˙φ2−|∇φ|2−µ2φ2), where µis the meson mass and φ(x,t) is its wavefunction. Assuming Hamilton’s principle ﬁnd the wave equation satisﬁed by φ. 22.5 (a) For a system described in terms of coordinates qiandt, show that if tdoes not appear explicitly in the expressions for x,yandz(x=x(qi,t), etc.) then the kinetic energy Tis a homogeneous quadratic function of the ˙qi(it may also involve the qi). Deduce that P i˙qi(∂T/∂˙qi)=2 T. (b) Assuming that the forces acting on th e system are derivable from a potential V, show, by expressing dT/dt in terms of qiand˙qi,t h a t d(T+V)/dt=0 . 22.6 For a system speciﬁed by the coordinates qandt, show that the equation of motion is unchanged if the Lagrangian L(q,˙q,t) is replaced by L1=L+dφ(q,t) dt, where φis an arbitrary function. Deduce that the equation of motion of a particle that moves in one dimension subject to a force −dV(x)/dx(xbeing measured from a point O) is unchanged if Ois forced to move with a constant velocity v (xstill being measured from O). 22.7 In cylindrical polar coordinates, the curve ( ρ(θ),θ,α ρ(θ)) lies on the surface of the cone z=αρ. Show that geodesics (curves of minimum length joining two points) on the cone satisfy ρ4=c2[β2ρ/prime2+ρ2], where cis an arbitrary constant, but βhas to have a particular value. Determine the form of ρ(θ) and hence ﬁnd the equation of the shortest path on the cone between the points ( R,−θ0,α R)a n d( R,θ0,α R). (You will ﬁnd it useful to determine the form of the derivative of cos−1(u−1).) 22.8 Derive the diﬀerential equations for the polar coordinates r,θof a particle of unit mass moving in a ﬁeld of potential V(r). Find the form of Vif the path of the particle is given by r=asinθ. 856 22.9 EXERCISES 22.9 You are provided with a line of length πa/2 and negligible mass and some lead shot of total mass M. Use a variational method to determine how the lead shot must be distributed along the line if the loaded line is to hang in a circular arc ofradius awhen its ends are attached to two points at the same height. (Measure the distance salong the line from its centre.) 22.10 Extend the result of subsection 22.2.2 to the case of several dependent variables y i(x), showing that, if xdoes not appear explicitly in the integrand, then a ﬁrst integral of the Euler–Lagrange equations is F−nX i=1y/prime i∂F ∂y/prime i=c o n s t a n t . 22.11 A general result is that light travels through a variable medium by a path which minimises the travel time (this is an alternative formulation of Fermat’s principle).With respect to a particular cylindrical polar coordinate system ( ρ, φ, z) the speed of light v(ρ, φ) is independent of z. If the path of the light is parameterised as ρ=ρ(z),φ=φ(z), use the result of the previous exercise to show that v 2(ρ/prime2+ρ2φ/prime2+1 ) is constant along the path. For the particular case when v=v(ρ)=b(a2+ρ2)1/2, show that the two Euler– Lagrange equations have a common solution in which the light travels along ahelical path given by φ=Az+B,ρ=C, provided that Ahas a particular value. 22.12 Light travels in a vertical xz-plane through a slab of material which lies between the planes z=z 0andz=2z0and in which the speed of light v(z)=c0z/z0.U s i n g the alternative formulation of Fermat’s principle given in the previous question,show that the ray paths are arcs of circles. Deduce that, if a ray enters the material at (0 ,z 0) at an angle to the vertical, π/2−θ,o fm o r et h a n3 0◦, it does not reach the far side of the slab. 22.13 A dam of capacity V(less than πb2h/2) is to be constructed on level ground next to a long straight wall which runs from ( −b,0) to ( b,0 ) .T h i si st ob ea c h i e v e db y joining the ends of a new wall, of height h, to those of the existing wall. Show that, in order to minimise the length Lof new wall to be built, it should form part of a circle, and that Lis then given byZb −bdx (1−λ2x2)1/2, where λis found from V hb2=sin−1µ µ2−(1−µ2)1/2 µ, andµ=λb. 22.14 The Schwarzchild metric for the static ﬁeld of a non-rotating spherically sym- metric black hole of mass Mis (ds)2=c2 / 1−2GM c2r / (dt)2−dr2 1−2GM/(c2r)−r2(dθ)2−r2sin2θ(dφ)2. Considering only motion conﬁned to the plane θ=π/2, and assuming that the path of a small test particle is such as to make R dsstationary, ﬁnd two ﬁrst integrals of the equations of motion. From their Newtonian limits, in whichGM/r ,˙r 2andr2˙φ2are all/lessmuchc2, identify the constants of integration. 22.15 In the brachistochrone problem of subsection 22.3.4 show that if the upper end- point can lie anywhere on the curve h(x, y) = 0 then the curve of quickest descent y(x) meets h(x, y) = 0 at right angles. 857 CALCULUS OF VARIATIONS 22.16 Use result (22.27) to evaluate J= Z1 −1(1−x2)P/prime m(x)P/prime n(x)dx, where Pm(x) is a Legendre polynomial of order m. 22.17 Determine the minimum value that the integral J= Z1 0[x4(y/prime/prime)2+4x2(y/prime)2]dx, can have, given that yis not singular at x=0a n dt h a t y(1) = y/prime(1) = 1. Assume that the Euler–Lagrange equation does give the lower limit, and verifyretrospectively that your solution makes the ﬁrst term on the LHS of equation(22.15) vanish. 22.18 Show that y /prime/prime−xy+λx2y=0h a sas o l u t i o nf o rw h i c h y(0) = y(1) = 0 and λ≤147/4. 22.19 Find an appropriate but simple trial function and use it to estimate the lowest eigenvalue λ0of Stokes’ equation d2y dx2+λxy=0,y (0) = y(π)=0 . Explain why your estimate must be strictly greater than λ0. 22.20 Estimate the lowest eigenvalue λ0of the equation d2y dx2−x2y+λy=0,y (−1) = y(1) = 0 , using a quadratic trial function. 22.21 A drumskin is stretched across a ﬁxed circular rim of radius a. Small transverse vibrations of the skin have an amplitude z(ρ, φ, t) that satisﬁes ∇2z=1 c2∂2z ∂t2 in plane polar coordinates. For a normal mode independent of azimuth, z= Z(ρ)cosωt, ﬁnd the diﬀerential equation satisﬁed by Z(ρ). By using a trial function of the form aν−ρν, obtain an estimate for the lowest normal mode frequency. (The exact answer is (5 .78)1/2c/a.) 22.22 (a) Recast the problem of ﬁnding the lowest eigenvalue λ0of the equation (1 +x2)d2y dx2+2xdy dx+λy=0,y (±1) = 0 , in variational form, and derive an approximation λ1toλ0by using the trial function y1(x)=1−x2. (b) Show that an improved estimate λ2is obtained by using y2(x)=c o s ( πx/2). (c) Prove that the estimate λ(γ) obtained by taking y1(x)+γy2(x)a st h et r i a l function is λ(γ)=64/15 + 16 γ/π+(π2/3+1 /2)γ2 16/15 + 64 γ/π3+γ2. Investigate λ(γ) numerically as γis varied, or, more simply, show that λ(1) = 3 .183, a signiﬁcant improvement on both λ1andλ2. 22.23 For the boundary conditions given below, obtain a functional Λ( y) whose sta- tionary values give the eigenvalues of the equation (1 +x)d2y dx2+( 2+ x)dy dx+λy=0,y (0) = 0 ,y/prime(2) = 0 . 858 22.9 EXERCISES Derive an approximation to the lowest eigenvalue λ0using the trial function y(x)=xe−x/2. For what value(s) of γwould y(x)=xe−x/2+βsinγx be a suitable trial function for attemp ting to obtain an improved estimate of λ0? 22.24 The upper and lower surfaces of a ﬁlm of liquid with surface energy per unit area (surface tension) equal to γand with density ρhave equations z=p(x)a n d z=q(x) respectively. The ﬁlm has a given volume V(per unit depth in the y- direction) and lies in the region −L<x<L ,w i t h p(0) = q(0) = p(L)=q(L)=0 . The total energy (per unit depth) of the ﬁlm consists of its surface energy and its gravitational energy, and is expressed by E=1 2 ZL −L(p2−q2)dx+γ ZL −L h (1 +p/prime2)1/2+( 1+ q/prime2)1/2 i dx. (a) Express Vin terms of pandq. (b) Show that, if the total energy is minimised, pandqmust satisfy p/prime2 (1 +p/prime2)1/2−q/prime2 (1 +q/prime2)1/2=c o n s t a n t . (c) As an approximate solution, consider the equations p=a(L−|x|),q =b(L−|x|), where aandbare suﬃciently small that a3andb3can be neglected compared to unity. Find the values of aandbthat minimise E. 22.25 This is an alternative approach to the example in section 22.8. Using the notation of that section, the expectation value of the energy of the state ψis given byR ψ∗Hψdv . Denote the eigenfunctions of Hbyψi,s ot h a t Hψ i=Eiψi, and, since His self-adjoint (Hermitian), R ψ∗ jψidv=δij. (a) By writing any function ψas Pcjψjand following an argument similar to that in section 22.7, show that E= R ψ∗HψdvR ψ∗ψd v≥E0, the energy of the lowest state. (This is the Rayleigh–Ritz principle.) (b) Using the same trial function as in section 22.8, ψ=e x p (−αx2),show that the same result is obtained. 22.26 This is an extension to section 22.8 and the previous question. With the ground- state (i.e. the lowest-energy) wavefunction as exp( −αx2), take as a trial function the orthogonal wave function x2n+1exp(−αx2), using the integer nas a variable parameter. Use either Sturm–Liouville theory or the Rayleigh–Ritz principle toshow that the energy of the second lowest state of a quantum harmonic oscillatoris≤3/~ω/2. 22.27 The Hamiltonian Hfor the hydrogen atom is − /~2 2m∇2−q 4π/epsilon10r. For a spherically symmetric state, as may be assumed for the ground state, the only relevant part of ∇2is that involving diﬀerentiation with respect to r. (a) Deﬁne the integrals Jnby Jn= Z∞ 0rne−2βrdr 859 CALCULUS OF VARIATIONS and show that, for a trial wavefunction of the form exp( −βr)w i t h β>0,R ψ∗Hψdv and R ψ∗ψd v(see exercise 22.25(a)) can be expressed as aJ1−bJ2 andcJ2respectively, where a, b, c are factors which you should determine. (b) Show that the estimate of Eis minimised when β=mq2/(4π/epsilon10 /~2). (c) Hence ﬁnd an upper limit for the ground-state energy of the hydrogen atom. In fact, exp( −βr) is the correct form for the wavefunction and the limit gives the actual value. 22.28 A particle of mass mmoves in a one-dimensional potential well of the form V(x)=−µ /~2α2 msech2αx, where µandαare positive constants. As in exercise 22.27, the expectation value /angbracketleftE/angbracketrightof the energy of the system is R ψ∗Hψdx , where the self-adjoint operator His given by −( /~2/2m)d2/dx2+V(x). Using trial wavefunctions of the form y=Asech βx, show the following: (a) for µ= 1 there is an exact eigenfunction of H, with a corresponding /angbracketleftE/angbracketrightof half of the maximum depth of the well; (b) for µ= 6 the ‘binding energy’ of the ground state is at least 10 /~2α2/(3m). (You will ﬁnd it useful to note that for u,v≥0, sech usech v≥sech ( u+v).) 22.29 The Sturm–Liouville equation can be extended to two independent variables, x andz, with little modiﬁcation. In equation (22.22) y/prime2is replaced by ( ∇y)2and the integrals of the various functions of y(x, z) become two-dimensional, i.e. the inﬁnitesimal is dx dz. The vibrations of a trampoline 4 units long and 1 unit wide satisfy the equation ∇2y+k2y=0. By taking the simplest possible permissible polynomial as a trial function, show that the lowest mode of vibration has k2≤10.63 and, by direct solution, that the actual value is 10.49. 22.10 Hints and answers 22.2 The minimising curve is x2+y2=2 . 22.3 I= R n(r)[r2+(dr/dφ )2]1/2dφ. Take axes such that φ= 0 when r=∞. Ifβ=(π−deviation angle) /2t h e n β=φatr=a, and the equation reduces to β (a2+α2)1/2= Z∞ −∞dr r(r2−a2)1/2, which can be evaluated by putting r=a(y+y−1)/2, or successively r=acoshψ, y=e x p ψto yield a deviation π[(a2+α2)1/2−a]/a. 22.4∇2φ−∂2φ/∂t2=µ2φ. 22.5 (a) ∂x/∂t =0a n ds o ˙x= P i˙qi∂x/∂q i; (b) useX i˙qid dt /∂T ∂˙qi / =d dt(2T)− X i¨qi∂T ∂˙qi. 22.6 φ(x, t)=m(vx+v2t/2). 22.7 Use result (22.8); β2=1+ α2.P u t ρ=ucto obtain dθ/du =β/[u(u2−1)1/2]. Remember that cos−1is a multivalued function; ρ(θ)=[Rcos(θ0/β)]/[cos(θ/β)]. 22.8 r2˙θ=k,¨r−r˙θ2+dV/dr =0 , V(r)=−k2a2/(2r4) + constant. 22.9−λy/prime(1−y/prime2)−1/2=2gP(s),y=y(s),P(s)= Rs 0ρ(s/prime)ds/prime.T h es o l u t i o n y= 860 22.10 HINTS AND ANSWERS −acos(s/a)a n d2 P(πa/4) = Mgive λ=−gM.T h er e q u i r e d ρ(s)i s [M/(2a)]sec2(s/a). 22.10 Note that dF/dx contains partial contributions from all yi(x)a n da l l y/prime i(x) but no∂F/∂x term. 22.11 A=1/a. 22.12 Circle is ( x−z0tanθ)2+z2=z2 0sec2θ. Consider the value of zwhen dz/dx =0 . 22.13 Circle is λ2x2+[λy+( 1−λ2b2)1/2]2=1 .U s et h ef a c tt h a t R yd x=V/hto determine the condition on λ. 22.14 Denoting ( ds)2/(dt)2byf2, the Euler–Lagrange equation for φgives r2˙φ=Af where Acorresponds to the angular momentum of the particle. Use the result of exercise 22.10 to obtain c2−(2GM/r )=Bf, where to ﬁrst order in small quantities cB=c2−GM r+1 2(˙r2+r2˙φ2), which reads ‘total energy = rest mass + gravitational energy + radial and azimuthal kinetic energy’. 22.16 Note that Legendre’s equation is a Sturm–Liouville equation with p(±1) = 0 andρ(x) = 1. For normalised eigenfunctions take ym(x)=[ ( 2 m+1 )/2]1/2Pm(x); J={[2m(m+1 ) ] /(2m+1 )}δmn. 22.17 Convert the equation to the usual form, by writing y/prime(x)=u(x), and obtain x2u/prime/prime+4xu/prime−4u= 0 with general solution Ax−4+Bx. Integrating a second time and using the boundary conditions gives y(x)=( 1+ x2)/2a n d J=1 ; η(1) = 0, since y/prime(1) is ﬁxed, and ∂F/∂u/prime=2x4u/prime=0a t x=0 . 22.18 The equation is of SL form with p=1 , q=−xand weight function x2.T r y y=x(1−x). The integrals have values 7 /20 and 1 /105. 22.19 Using y=s i n xas a trial function shows that λ0≤2/π. The estimate must be >λ0since the trial function does not satisfy the original equation. 22.20 Using y=1−x2as a trial function shows that λ0≤37/14. 22.21 Z/prime/prime+ρ−1Z/prime+(ω/c)2Z=0 ,w i t h Z(a)=0a n d Z/prime(0) = 0, an SL equation with p=ρ,q= 0 and weight function ρ/c2.E s t i m a t eo f ω2=[c2ν/(2a2)][0.5−2(ν+ 2)−1+( 2ν+2 )−1]−1, which minimises to c2(2 +√ 2)2/(2a2)=5 .83c2/a2when ν=√ 2. 22.22 (a) Follow the method of section 22.7 with p=1+ x2,q=0a n d ρ=1 ; λ1=4. (b)λ2=π2/3+1 /2≈3.79. (c)λ(γ) has a minimum value 3 .1768 at γ=1.1976. 22.23 Note that the original equation is not self-adjoint; it needs an integrating factor ofex.Λ (y)=[ R2 0(1 +x)exy/prime2dx]/[ R2 0exy2dx;λ0≤3/8. Since y/prime(2) must equal 0, γ=(π/2)(n+1 2)f o rs o m ei n t e g e r n. 22.24 (a) V= RL −L(p−q)dx.( c )U s e V=(a−b)L2to eliminate bfrom the expression forE; now the minimisation is with respect to aalone. The values for aandb are±V/(2L2)−Vρg/(6γ). 22.25 The estimate is /~2α/(2m)+k/(8α) and the minimum occurs at the value of αthat makes the two terms equal. 22.26 E1≤( /~ω/2)(8n2+1 2n+3 )/(4n+ 1), which has a minimum value 3 /~ω/2w h e n integer n=0 . 22.27 (a) a=4π /~2β/m−q2//epsilon10,b=2π /~2β2/m,c=4π;( c )−mq4/[2(4π/epsilon10 /~)2]. 22.28 (a) Hy=λyrequires that β=α. (b) R ψ∗[−( /~2/2m)d2/dx2]ψd x= /~2β2/(6m)a n d R ψ∗Vψd x≤−6 /~2α2β/[m(α+ β)]. The sum of the two integrals is minimised when β=2α, leading to the stated upper limit for /angbracketleftE/angbracketright. 22.29 The SL equation has p=1 , q=0 ,a n d ρ=1 . Useu(x, y)=x(4−x)y(1−y) as a trial function. Numerator = 1088 /90, denominator = 512 /450. Direct solution k2=1 7π2/16. 861 23 Integral equations It is not unusual in the analysis of a physical system to encounter an equation in which an unknown but required function y(x), say, appears under an integral sign. Such an equation is called an integral equation , and in this chapter we discuss several methods for solving the more straightforward examples of such equations. Before embarking on our discussion of methods for solving various integral equations, we begin with a warning that many of the integral equations met inpractice cannot be solved by the elementary methods presented here but mustinstead be solved numerically, usually on a computer. Nevertheless, the regularoccurrence of several simple types of integral equation that may be solved analytically is suﬃcient reason to explore these equations more fully. We shall begin this chapter by discussing how a diﬀerential equation can be transformed into an integral equation and by considering the most commontypes of linear integral equation. After introducing the operator notation andconsidering the existence of solutions for various types of equation, we go onto discuss elementary methods of obtaining closed-form solutions of simpleintegral equations. We then consider the solution of integral equations in terms of inﬁnite series and conclude by discussing the properties of integral equations with Hermitian kernels, i.e. those in which the integrands have particular symmetryproperties. 23.1 Obtaining an integral equation from a diﬀerential equation Integral equations occur in many situations, partly because we may always rewrite a diﬀerential equation as an integral equation. It is sometimes advantageous tomake this transformation, since questions concerning the existence of a solu- tion are more easily answered for integral equations (see section 23.3), and, furthermore, an integral equation can incorporate automatically any boundaryconditions on the solution. 862 23.2 TYPES OF INTEGRAL EQUATION We shall illustrate the principles involved by considering the diﬀerential equa- tion y/prime/prime(x)=f(x, y), (23.1) where f(x, y) can be any function of xandybut not of y/prime(x). Equation (23.1) thus represents a large class of linear and non-linear second-order diﬀerential equations. We can convert (23.1) into the corresponding integral equation by ﬁrst inte- grating with respect to xto obtain y/prime(x)=integraldisplayx 0f(z,y(z))dz+c1. Integrating once more, we ﬁnd y(x)=integraldisplayx 0duintegraldisplayu 0f(z,y(z))dz+c1x+c2. Provided we do not change the region in the uz-plane over which the double integral is taken, we can reverse the order of the two integrations. Changing the integration limits appropriately, we ﬁnd y(x)=integraldisplayx 0f(z,y(z))dzintegraldisplayx zdu+c1x+c2 (23.2) =integraldisplayx 0(x−z)f(z,y(z))dz+c1x+c2; (23.3) this is a non-linear (for general f(x, y))Volterra integral equation. It is straightforward to incorporate any boundary conditions on the solution y(x) by ﬁxing the constants c1andc2in (23.3). For example, we might have the one-point boundary condition y(0) = aandy/prime(0) = b, for which it is clear that we must set c1=bandc2=a. 23.2 Types of integral equation From (23.3), we can see that even a relatively simple diﬀerential equation such as (23.1) can lead to a corresponding integral equation that is non-linear. In this chapter, however, we will restrict our attention to linear integral equations, which have the general form g(x)y(x)=f(x)+λintegraldisplayb aK(x, z)y(z)dz. (23.4) In (23.4), y(x) is the unknown function, while the functions f(x),g(x)a n d K(x, z) are assumed known. K(x, z) is called the kernel of the integral equation. The integration limits aandbare also assumed known, and may be constants or functions of x,a n d λis a known constant or parameter. 863 INTEGRAL EQUATIONS In fact, we shall be concerned with various special cases of (23.4), which are known by particular names. Firstly, if g(x) = 0 then the unknown function y(x) appears only under the integral sign, and (23.4) is called a linear integral equationof the ﬁrst kind . Alternatively, if g(x) = 1, so that y(x) appears twice, once inside the integral and once outside, then (23.4) is called a linear integral equation of the second kind . In either case, if f(x) = 0 the equation is called homogeneous , otherwise inhomogeneous . We can distinguish further between diﬀerent types of integral equation by the form of the integration limits aandb. If these limits are ﬁxed constants then the equation is called a Fredholm equation. If, however, the upper limit b=x(i.e. it is variable) then the equation is called a Volterra equation; such an equation is analogous to one with ﬁxed limits but for which the kernel K(x, z)=0f o r z>x. Finally, we note that any equation for which either (or both) of the integrationlimits is inﬁnite, or for which K(x, z) becomes inﬁnite in the range of integration, is called a singular integral equation. 23.3 Operator notation and the existence of solutions There is a close correspondence between linear integral equations and the matrix equations discussed in chapter 8. However, the former involve linear, integral rela- tions between functions in an inﬁnite-dimensional function space (see chapter 17),whereas the latter specify linear relations among vectors in a ﬁnite-dimensionalvector space. Since we are restricting our attention to linear integral equations, it will be convenient to introduce the linear integral operator K, whose action on an arbitrary function yis given by Ky=integraldisplay b aK(x, z)y(z)dz. (23.5) This is analogous to the introduction in chapters 16 and 17 of the notation Lto describe a linear diﬀerential operator. Furthermore, we may deﬁne the Hermitianconjugate K †by K†y=integraldisplayb aK∗(z,x)y(z)dz, where the asterisk denotes complex conjugation and we have reversed the order of the arguments in the kernel. It is clear from (23.5) that Kis indeed linear. Moreover, since Koperates on the inﬁnite-dimensional space of (reasonable) functions, we may make an obviousanalogy with matrix equations and consider the action of Kon a function fas that of a matrix on a column vector (both of inﬁnite dimension). When written in operator form, the integral equations discussed in the pre- vious section resemble equations familiar from linear algebra. For example, the 864 23.4 CLOSED-FORM SOLUTIONS inhomogeneous Fredholm equation of the ﬁrst kind may be written as 0=f+λKy, which has the unique solution y=−K−1f/λ, provided that f/negationslash= 0 and the inverse operator K−1exists. Similarly, we may write the corresponding Fredholm equation of the second kind as y=f+λKy. (23.6) In the homogeneous case, where f= 0, this reduces to y=λKy,w h i c hi s reminiscent of an eigenvalue problem in linear algebra (except that λappears on the other side of the equation) and, similarly, only has solutions for at most acountably inﬁnite set of eigenvalues λ i. The corresponding solutions yiare called the eigenfunctions. In the inhomogeneous case ( f/negationslash= 0), the solution to (23.6) can be written symbolically as y=( 1−λK)−1f, again provided that the inverse operator exists. It may be shown that, in general, (23.6) does possess a unique solution if λ/negationslash=λi,i . e .w h e n λdoes not equal one of the eigenvalues of the corresponding homogeneous equation. When λdoes equal one of these eigenvalues, (23.6) may have either many solutions or no solution at all, depending on the form of f. If the function fis orthogonal to everyeigenfunction of the equation g=λ∗K†g (23.7) that belongs to the eigenvalue λ∗,i . e . /angbracketleftg|f/angbracketright=integraldisplayb ag∗(x)f(x)dx=0 for every function gobeying (23.7), then it can be shown that (23.6) has many solutions. Otherwise the equation has no solution. These statements are discussed further in section 23.7, for the special case of integral equations with Hermitiankernels, i.e. those for which K=K †. 23.4 Closed-form solutions In certain very special cases, it may be possible to obtain a closed-form solution of an integral equation. The reader should realise, however, when faced with an integral equation, that in general it will not be soluble by the simple methods presented in this section but must instead be solved using (numerical) iterativemethods, such as those outlined in section 23.5. 865 INTEGRAL EQUATIONS 23.4.1 Separable kernels The most straightforward integral equations to solve are Fredholm equations withseparable (ordegenerate ) kernels. A kernel is separable if it has the form K(x, z)=nsummationdisplay i=1φi(x)ψi(z), (23.8) where φi(x)a r e ψi(z) are respectively functions of xonly and of zonly and the number of terms in the sum, n, is ﬁnite. Let us consider the solution of the (inhomogeneous) Fredholm equation of the second kind, y(x)=f(x)+λintegraldisplayb aK(x, z)y(z)dz, (23.9) which has a separable kernel of the form (23.8). Writing the kernel in its separated form, the functions φi(x) may be taken outside the integral over zto obtain y(x)=f(x)+λnsummationdisplay i=1φi(x)integraldisplayb aψi(z)y(z)dz. Since the integration limits aandbare constant for a Fredholm equation, the integral over zin each term of the sum is just a constant. Denoting these constants by ci=integraldisplayb aψi(z)y(z)dz, (23.10) the solution to (23.9) is found to be y(x)=f(x)+λnsummationdisplay i=1ciφi(x), (23.11) where the constants cican be evalutated by substituting (23.11) into (23.10).ISolve the integral equation y(x)=x+λ Z1 0(xz+z2)y(z)dz. (23.12) The kernel for this equation is K(x, z)=xz+z2, which is clearly separable, and using the notation in (23.8) we have φ1(x)=x,φ2(x)=1 , ψ1(z)=zandψ2(z)=z2. From (23.11) the solution to (23.12) has the form y(x)=x+λ(c1x+c2), where the constants c1andc2are given by (23.10) as c1= Z1 0z[z+λ(c1z+c2)]dz=1 3+1 3λc1+1 2λc2, c2= Z1 0z2[z+λ(c1z+c2)]dz=1 4+1 4λc1+1 3λc2. 866 23.4 CLOSED-FORM SOLUTIONS These two simultaneous linear equations may be straightforwardly solved for c1andc2to give c1=24 + λ 72−48λ−λ2and c2=18 72−48λ−λ2, so that the solution to (23.12) is y(x)=(72−24λ)x+1 8λ 72−48λ−λ2. J In the above example, we see that (23.12) has a (ﬁnite) unique solution provided that λis not equal to either root of the quadratic in the denominator of y(x). The roots of this quadratic are in fact the eigenvalues of the corresponding homogeneous equation, as mentioned in the previous section. In general, if theseparable kernel contains nterms, as in (23.8), there will be nsuch eigenvalues, although they may not all be diﬀerent. Kernels consisting of trigonometric (or hyperbolic) functions of sums or diﬀer- ences of xandzare also often separable.IFind the eigenvalues and corresponding ei genfunctions of the homogeneous Fredholm equation y(x)=λ Zπ 0sin(x+z)y(z)dz. (23.13) The kernel of this integral equation can be written in separated form as K(x, z)=s i n ( x+z)=s i n xcosz+c o s xsinz, so, comparing with (23.8), we have φ1(x)=s i n x,φ2(x)=c o s x,ψ1(z)=c o s zand ψ2(z)=s i n z. Thus, from (23.11), the solution to (23.13) has the form y(x)=λ(c1sinx+c2cosx), where the constants c1andc2are given by c1=λ Zπ 0cosz(c1sinz+c2cosz)dz=λπ 2c2, (23.14) c2=λ Zπ 0sinz(c1sinz+c2cosz)dz=λπ 2c1. (23.15) Combining these two equations we ﬁnd c1=(λπ/2)2c1, and, assuming that c1/negationslash=0 ,t h i s gives λ=±2/π, the two eigenvalues of the integral equation (23.13). By substituting each of the eigenvalues back into (23.14) and (23.15), we ﬁnd that the eigenfunctions corresponding to the eigenvalues λ1=2/πandλ2=−2/πare given respectively by y1(x)=A(sinx+c o s x)a n d y2(x)=B(sinx−cosx), (23.16) where AandBare arbitrary constants. J 867 INTEGRAL EQUATIONS 23.4.2 Integral transform methods If the kernel of an integral equation can be written as a function of the diﬀerence x−zof its two arguments, then it is called a displacement kernel. An integral equation having such a kernel, and which also has the integration limits −∞to ∞, may be solved by the use of Fourier transforms (chapter 13). If we consider the following integral equation with a displacement kernel, y(x)=f(x)+λintegraldisplay∞ −∞K(x−z)y(z)dz, (23.17) the integral over zclearly takes the form of a convolution (see chapter 13). Therefore, Fourier-transforming (23.17) and using the convolution theorem, we obtain ˜y(k)=˜f(k)+√ 2πλ˜K(k)˜y(k), which may be rearranged to give ˜y(k)=˜f(k) 1−√ 2πλ˜K(k). (23.18) Taking the inverse Fourier transform, the solution to (23.17) is given by y(x)=1√ 2πintegraldisplay∞ −∞˜f(k)ex p( ikx) 1−√ 2πλ˜K(k)dk. If we can perform this inverse Fourier transformation then the solution can be found explicitly; otherwise it must be left in the form of an integral.IFind the Fourier transform of the function g(x)= /( 1if|x|≤a, 0if|x|>a. Hence ﬁnd an explicit expression for the solution of the integral equation y(x)=f(x)+λ Z∞ −∞sin(x−z) x−zy(z)dz. (23.19) Find the solution for the special case f(x)=( s i n x)/x. The Fourier transform of g(x) is given directly by ˜g(k)=1√ 2π Za −aexp(−ikx)dx= /1√ 2πexp(−ikx) (−ik) /a −a= r 2 πsinka k. (23.20) The kernel of the integral equation (23.19) is K(x−z)=[ s i n ( x−z)]/(x−z). Using (23.20), it is straightforward to show that the Fourier transform of the kernel is ˜K(k)= /(p π/2i f|k|≤1, 0i f|k|>1.(23.21) 868 23.4 CLOSED-FORM SOLUTIONS Thus, using (23.18), we ﬁnd the Fourier transform of the solution to be ˜y(k)= /(˜f(k)/(1−πλ)i f|k|≤1, ˜f(k)i f |k|>1.(23.22) Inverse Fourier-transforming, and writing the result in a slightly more convenient form, the solution to (23.19) is given by y(x)=f(x)+ /1 1−πλ−1 /1√ 2π Z1 −1˜f(k)e x p ( ikx)dk =f(x)+πλ 1−πλ1√ 2π Z1 −1˜f(k)e x p ( ikx)dk. (23.23) It is clear from (23.22) that when λ=1/π, which is the only eigenvalue of the corresponding homogeneous equation to (23.19), the solution becomes inﬁnite, as wewould expect. For the special case f(x)=( s i n x)/x, the Fourier transform ˜f(k) is identical to that in (23.21), and the solution (23.23) becomes y(x)=sinx x+ /πλ 1−πλ /1√ 2π Z1 −1 rπ 2exp(ikx)dk =sinx x+ /πλ 1−πλ /1 2 /exp(ikx) ix /k=1 k=−1 =sinx x+ /πλ 1−πλ /sinx x= /1 1−πλ /sinx x. J If, instead, the integral equation (23.17) had integration limits 0 and x(so making it a Volterra equation) then its solution could be found, in a similar way,by using the convolution theorem for Laplace transforms (see chapter 13). Wewould ﬁnd ¯y(s)=¯f(s) 1−λ¯K(s), where sis the Laplace transform variable. Often one may use the dictionary of Laplace transforms given in table 13.1 to invert this equation and ﬁnd the solution y(x). In general, however, the evaluation of inverse Laplace transform integrals is diﬃcult, since (in principle) it requires a contour integration; see chapter 20. As a ﬁnal example of the use of Fourier transforms in solving integral equations, we mention equations that have integration limits −∞and∞and a kernel of the form K(x, z)=e x p (−ixz). Consider, for example, the inhomogeneous Fredholm equation y(x)=f(x)+λintegraldisplay∞ −∞exp(−ixz)y(z)dz. (23.24) The integral over zis clearly just (a multiple of) the Fourier transform of y(z), 869 INTEGRAL EQUATIONS so we can write y(x)=f(x)+√ 2πλ˜y(x). (23.25) If we now take the Fourier transform of (23.25) but continue to denote the independent variable by x(i.e. rather than k, for example), we obtain ˜y(x)=˜f(x)+√ 2πλy(−x). (23.26) Substituting (23.26) into (23.25) we ﬁnd y(x)=f(x)+√ 2πλbracketleftBig ˜f(x)+√ 2πλy(−x)bracketrightBig , but on making the change x→−xand substituting back in for y(−x), this gives y(x)=f(x)+√ 2πλ˜f(x)+2πλ2bracketleftBig f(−x)+√ 2πλ˜f(−x)+2πλ2y(x)bracketrightBig . Thus the solution to (23.24) is given by y(x)=1 1−(2π)2λ4bracketleftBig f(x)+( 2 π)1/2λ˜f(x)+2πλ2f(−x)+( 2 π)3/2λ3˜f(−x)bracketrightBig . (23.27) Clearly, (23.24) possesses a unique solution provided λ/negationslash=±1/√ 2πor±i/√ 2π; these are easily shown to be the eigenvalues of the corresponding homogeneousequation (for which f(x)≡0).ISolve the integral equation y(x)=e x p / −x2 2 / +λ Z∞ −∞exp(−ixz)y(z)dz, (23.28) where λis a real constant. Show that the solution is unique unless λhas one of two particular values. Does a solution exist for either of these two values of λ? Following the argument given above, the solution to (23.28) is given by (23.27) with f(x)=e x p (−x2/2). In order to write the solution exp licitly, however, we must calculate the Fourier transform of f(x). Using equation (13.7), we ﬁnd ˜f(k)=e x p (−k2/2), from which we note that f(x) has the special property that its functional form is identical to that of its Fourier transform. Thus, the solution to (23.28) is given by y(x)=1 1−(2π)2λ4 / 1+( 2 π)1/2λ+2πλ2+( 2π)3/2λ3 / exp / −x2 2 / . (23.29) Since λis restricted to be real, the solution to (23.28) will be unique unless λ=±1/√ 2π, at which points (23.29) becomes inﬁnite. In order to ﬁnd whether solutions exist for eitherof these values of λwe must return to equations (23.25) and (23.26). Let us ﬁrst consider the case λ=+ 1 /√ 2π. Putting this value into (23.25) and (23.26), we obtain y(x)=f(x)+˜y(x), (23.30) ˜y(x)=˜f(x)+y(−x). (23.31) 870 23.4 CLOSED-FORM SOLUTIONS Substituting (23.31) into (23.30) we ﬁnd y(x)=f(x)+˜f(x)+y(−x), but on changing xto−xand substituting back in for y(−x), this gives y(x)=f(x)+˜f(x)+f(−x)+˜f(−x)+y(x). Thus, in order for a solution to exist, we require that the function f(x)o b e y s f(x)+˜f(x)+f(−x)+˜f(−x)=0 . This is satisﬁed if f(x)=−˜f(x), i.e. if the functional form of f(x)i sm i n u st h ef o r mo fi t s Fourier transform. We may repeat this analysis for the case λ=−1/√ 2π, and, in a similar way, we ﬁnd that this time we require f(x)=˜f(x). In our case f(x)=e x p (−x2/2), for which, as we mentioned above, f(x)=˜f(x). Therefore, (23.28) possesses no solution when λ=+ 1 /√ 2πbut has many solutions when λ=−1/√ 2π. J A similar approach to the above may be taken to solve equations with kernels of the form K(x, y)=c o s xyor sin xy, either by considering the integral over yin each case as the real or imaginary part of the corresponding Fourier transformor by using Fourier cosine or sine transforms directly. 23.4.3 Diﬀerentiation A closed-form solution to a Volterra equation may sometimes be obtained by diﬀerentiating the equation to obtain the corresponding diﬀerential equation,which may be easier to solve.ISolve the integral equation y(x)=x− Zx 0xz2y(z)dz. (23.32) Dividing through by x,w eo b t a i n y(x) x=1− Zx 0z2y(z)dz, which may be diﬀerentiated with respect to xto give d dx /y(x) x / =−x2y(x)=−x3 /y(x) x / . This equation may be integrated straightforwardly, and we ﬁnd ln /y(x) x / =−x4 4+c, where cis a constant of integration. Thus the solution to (23.32) has the form y(x)=Axexp / −x4 4 / , (23.33) where Ais an arbitrary constant. Since the original integral equation (23.32) contains no arbitrary constants, neither should its solution. We may calculate the value of the constant, A, by substituting the solution (23.33) back into (23.32), from which we ﬁnd A=1 . J 871 INTEGRAL EQUATIONS 23.5 Neumann series As mentioned above, most integral equations met in practice will not be of the simple forms discussed in the last section and so, in general, it is not possible toﬁnd closed-form solutions. In such cases, we might try to obtain a solution in theform of an inﬁnite series, as we did for diﬀerential equations (see chapter 16). Let us consider the equation y(x)=f(x)+λintegraldisplay b aK(x, z)y(z)dz, (23.34) where either both integration limits are constants (for a Fredholm equation) or the upper limit is variable (for a Volterra equation). Clearly, if λwere small then a crude (but reasonable) approximation to the solution would be y(x)≈y0(x)=f(x), where y0(x) stands for our ‘zeroth-order’ approximation to the solution (and is not to be confused with an eigenfunction). Substituting this crude guess under the integral sign in the original equation, we obtain what should be a better approximation: y1(x)=f(x)+λintegraldisplayb aK(x, z)y0(z)dz=f(x)+λintegraldisplayb aK(x, z)f(z)dz, which is ﬁrst order in λ. Repeating the procedure once more results in the second-order approximation y2(x)=f(x)+λintegraldisplayb aK(x, z)y1(z)dz =f(x)+λintegraldisplayb aK(x, z1)f(z1)dz1+λ2integraldisplayb adz1integraldisplayb aK(x, z1)K(z1,z2)f(z2)dz2. It is clear that we may continue this process to obtain progressively higher-order approximations to the solution. Introducing the functions K1(x, z)=K(x, z), K2(x, z)=integraldisplayb aK(x, z1)K(z1,z)dz1, K3(x, z)=integraldisplayb adz1integraldisplayb aK(x, z1)K(z1,z2)K(z2,z)dz2, a n ds oo n ,w h i c ho b e yt h er e c u r r e n c er e l a t i o n Kn(x, z)=integraldisplayb aK(x, z1)Kn−1(z1,z)dz1, 872 23.5 NEUMANN SERIES we may write the nth-order approximation as yn(x)=f(x)+nsummationdisplay m=1λmintegraldisplayb aKm(x, z)f(z)dz. (23.35) The solution to the original integral equation is then given by y(x)= limn→∞yn(x),provided the inﬁnite series converges . Using (23.35), this solution may be written as y(x)=f(x)+λintegraldisplayb aR(x, z;λ)f(z)dz, (23.36) where the resolvent kernel R(x, z;λ)i sg i v e nb y R(x, z;λ)=∞summationdisplay m=0λmKm+1(x, z). (23.37) Clearly, the resolvent kernel, and hence the series solution, will converge provided λis suﬃciently small. In fact, it may be shown that the series converges in some domain of |λ|provided the original kernel K(x, z) is bounded in such a way that |λ|2integraldisplayb adxintegraldisplayb a|K(x, z)|2dz <1. (23.38)IUse the Neumann series method to solve the integral equation y(x)=x+λ Z1 0xzy(z)dz. (23.39) Following the method outlined above, we begin with the crude approximation y(x)≈ y0(x)=x. Substituting this under the integral sign in (23.39), we obtain the next approxi- mation y1(x)=x+λ Z1 0xzy0(z)dz=x+λ Z1 0xz2dz=x+λx 3, Repeating the procedure once more, we obtain y2(x)=x+λ Z1 0xzy1(z)dz =x+λ Z1 0xz / z+λz 3 / dz=x+ /λ 3+λ2 9 / x. For this simple example, it is easy to see that by continuing this process the solution to (23.39) is obtained as y(x)=x+ /" λ 3+ /λ 3 /2 + /λ 3 /3 +··· /# x. Clearly the expression in brackets is an inﬁnite geometric series with ﬁrst term λ/3a n d 873 INTEGRAL EQUATIONS common ratio λ/3. Thus, provided|λ|<3, this inﬁnite series converges to the value λ/(3−λ), and the solution to (23.39) is y(x)=x+λx 3−λ=3x 3−λ. (23.40) Finally, we note that the requirement that |λ|<3 may also be derived very easily from the condition (23.38). J 23.6 Fredholm theory In the previous section, we found that a solution to the integral equation (23.34) can be obtained as a Neumann series of the form (23.36), where the resolventkernel R(x, z;λ) is written as an inﬁnite power series in λ. This solution is valid provided the inﬁnite series converges. A related, but more elegant, approach to the solution of integral equations using inﬁnite series was found by Fredholm. We will not reproduce Fredholm’s analysis here, but merely state the results we need. Essentially, Fredholm theory provides a formula for the resolvent kernel R(x, z;λ) in (23.36) in terms of the ratio of two inﬁnite series: R(x, z;λ)=D(x, z;λ) d(λ). (23.41) The numerator and denominator in (23.41) are given by D(x, z;λ)=∞summationdisplay n=0(−1)n n!Dn(x, z)λn, (23.42) d(λ)=∞summationdisplay n=0(−1)n n!dnλn, (23.43) where the functions Dn(x, z) and the constants dnare found from recurrence relations as follows. We start with D0(x, z)=K(x, z)a n d d0=1, (23.44) where K(x, z) is the kernel of the original integral equation (23.34). The higher- order coeﬃcients of λin (23.43) and (23.42) are then obtained from the two recurrence relations dn=integraldisplayb aDn−1(x, x)dx, (23.45) Dn(x, z)=K(x, z)dn−nintegraldisplayb aK(x, z1)Dn−1(z1,z)dz1. (23.46) Although the formulae for the resolvent kernel appear complicated, they are often simple to apply. Moreover, for the Fredholm solution the power series(23.42) and (23.43) are both guaranteed to converge for all values of λ, unlike 874 23.7 SCHMIDT–HILBERT THEORY Neumann series, which converge only if the condition (23.38) is satisﬁed. Thus the Fredholm method leads to a unique, non-singular solution, provided that d(λ)/negationslash=0 . In fact, as we might suspect, the solutions of d(λ) = 0 give the eigenvalues of the homogeneous equation corresponding to (23.34), i.e. with f(x)≡0.IUse Fredholm theory to solve the integral equation (23.39). Using (23.36) and (23.41), the solution to (23.39) can be written in the form y(x)=x+λ Z1 0R(x, z;λ)zd z=x+λ Z1 0D(x, z;λ) d(λ)zd z . (23.47) In order to ﬁnd the form of the resolvent kernel R(x, z;λ), we begin by setting D0(x, z)=K(x, z)=xz and d0=1 and use the recurrence relations (23.45) and (23.46) to obtain d1= Z1 0D0(x, x)dx= Z1 0x2dx=1 3, D1(x, z)=xz 3− Z1 0xz2 1zd z1=xz 3−xz /z3 1 3 /1 0=0. Applying the recurrence relations again we ﬁnd that dn=0a n d Dn(x, z)=0f o r n>1. Thus, from (23.42) and (23.43), the numerator and denominator of the resolvent respectively are given by D(x, z;λ)=xz and d(λ)=1−λ 3. Substituting these expressions into (23.47), we ﬁnd that the solution to (23.39) is given by y(x)=x+λ Z1 0xz2 1−λ/3dz =x+λ /x 1−λ/3z3 3 /1 0=x+λx 3−λ=3x 3−λ, which, as expected, is the same as the solution (23.40) found by constructing a Neumann series. J 23.7 Schmidt–Hilbert theory The Schmidt–Hilbert (SH) theory of integral equations may be considered as analogous to the Sturm–Liouville (SL) theory of diﬀerential equations, discussed in chapter 17, and is concerned with the properties of integral equations withHermitian kernels. An Hermitian kernel enjoys the property K(x, z)=K ∗(z,x), (23.48) and it is clear that a special case of (23.48) occurs for a real kernel that is also symmetric with respect to its two arguments. 875 INTEGRAL EQUATIONS Let us begin by considering the homogeneous integral equation y=λKy, where the integral operator Khas an Hermitian kernel. As discussed in sec- tion 23.3, in general, this equation will have solutions only for λ=λi,w h e r et h e λi are the eigenvalues of the integral equation, the corresponding solutions yibeing the eigenfunctions of the equation. By following similar arguments to those presented in chapter 17 for SL theory, it may be shown that the eigenvalues λiof an Hermitian kernel are real and that the corresponding eigenfunctions yibelonging to diﬀerent eigenvalues are orthogonal and form a complete set. If the eigenfunctions are suitably normalised, we have /angbracketleftyi|yj/angbracketright=integraldisplayb ay∗ i(x)yj(x)dx=δij. (23.49) If an eigenvalue is degenerate then the eigenfunctions corresponding to that eigenvalue can be made orthogonal by the Gram–Schmidt procedure, in a similarway to that discussed in chapter 17 in the context of SL theory. Like SL theory, SH theory does not provide a method of obtaining the eigen- values and eigenfunctions of any particular homogeneous integral equation withan Hermitian kernel; for this we have to turn to the methods discussed in the previous sections of this chapter. Rather, SH theory is concerned with the gen- eral properties of the solutions to such equations. Where SH theory becomesapplicable, however, is in the solution of inhomogeneous integral equations withHermitian kernels for which the eigenvalues and eigenfunctions of the corre-sponding homogeneous equation are already known. Let us consider the inhomogeneous equation y=f+λKy, (23.50) where K=K †and for which we know the eigenvalues λiand normalised eigenfunctions yiof the corresponding homogeneous problem. The function f may or may not be expressible solely in terms of the eigenfunctions yi,a n dt o accommodate this situation we write the unknown solution yasy=f+summationtext iaiyi, where the aiare expansion coeﬃcients to be determined. Substituting this into (23.50), we obtain f+summationdisplay iaiyi=f+λsummationdisplay iaiyi λi+λKf, (23.51) where we have used the fact that yi=λiKyi. Forming the inner product of both 876 23.7 SCHMIDT–HILBERT THEORY sides of (23.51) with yj, we ﬁnd summationdisplay iai/angbracketleftyj|yi/angbracketright=λsummationdisplay iai λi/angbracketleftyj|yi/angbracketright+λ/angbracketleftyj|Kf/angbracketright. (23.52) Since the eigenfunctions are orthonormal and Kis an Hermitian operator, we have that both /angbracketleftyj|yi/angbracketright=δijand/angbracketleftyj|Kf/angbracketright=/angbracketleftKyj|f/angbracketright=λ−1 j/angbracketleftyj|f/angbracketright. Thus the coeﬃcients ajare given by aj=λλ−1 j/angbracketleftyj|f/angbracketright 1−λλ−1 j=λ/angbracketleftyj|f/angbracketright λj−λ, (23.53) and the solution is y=f+summationdisplay iaiyi=f+λsummationdisplay i/angbracketleftyi|f/angbracketright λi−λyi. (23.54) This also shows, incidentally, that a formal representation for the resolvent kernel is R(x, z;λ)=summationdisplay iyi(x)y∗ i(z) λi−λ. (23.55) Iffcanbe expressed as a linear superposition of the yi,i . e .f=summationtext ibiyi,t h e n bi=/angbracketleftyi|f/angbracketrightand the solution can be written more brieﬂy as y=summationdisplay ibi 1−λλ−1 iyi. (23.56) We see from (23.54) that the inhomogeneous equation (23.50) has a unique solution provided λ/negationslash=λi,i . e .w h e n λis not equal to one of the eigenvalues of the corresponding homogeneous equation. However, if λdoes equal one of the eigenvalues λjthen, in general, the coeﬃcients ajbecome singular and no (ﬁnite) solution exists. Returning to (23.53) we notice that even if λ=λja non-singular solution to the integral equation is still possible provided that the function fis orthogonal to every eigenfunction corresponding to the eigenvalue λj,i . e . /angbracketleftyj|f/angbracketright=integraldisplayb ay∗ j(x)f(x)dx=0. The following worked example illustrates the case in which fcan be expressed in terms of the yi. One in which it cannot is considered in exercise 23.14. 877 INTEGRAL EQUATIONSIUse Schmidt–Hilbert theory to solve the integral equation y(x) = sin( x+α)+λ Zπ 0sin(x+z)y(z)dz. (23.57) It is clear that the kernel K(x, z)=s i n ( x+z) is real and symmetric in xandzand is thus Hermitian. In order to solve this inhomogeneous equation using SH theory, however,we must ﬁrst ﬁnd the eigenvalues and eigenfunctions of the corresponding homogeneousequation. In fact, we have considered the solution of the corresponding homogeneous equation (23.13) already in subsection 23.4.1, where we found that it has two eigenvalues λ 1=2/π andλ2=−2/π, with eigenfunctions given by (23.16). The normalised eigenfunctions are y1(x)=1√π(sinx+c o s x)a n d y2(x)=1√π(sinx−cosx) (23.58) and are easily shown to obey the orthonormality condition (23.49). Using (23.54), the solution to the inhomogeneous equation (23.57) has the form y(x)=a1y1(x)+a2y2(x), (23.59) where the coeﬃcients a1anda2are given by (23.53) with f(x) = sin( x+α). Therefore, using (23.58), a1=1 1−πλ/2 Zπ 01√π(sinz+c o s z)si n(z+α)dz=√π 2−πλ(cosα+s i n α), a2=1 1+πλ/2 Zπ 01√π(sinz−cosz)si n(z+α)dz=√π 2+πλ(cosα−sinα). Substituting these expressions for a1anda2into (23.59) and simplifying, we ﬁnd that the solution to (23.57) is given by y(x)=1 1−(πλ/2)2 / sin(x+α)+(πλ/2)cos( x−α) / . J 23.8 Exercises 23.1 Solve the integral equationZ∞ 0cos(xv)y(v)dv=e x p (−x2/2), for the function y=y(x)f o r x>0. Note that for x<0,y(x) can be chosen as is most convenient. 23.2 SolveZ∞ 0f(t)exp(−st)dt=a a2+s2. 23.3 Use the fact that its kernel is separable to solve for y(x) the integral equation y(x)=Acos(x+a)+λ Zπ 0sin(x+z)y(z)dz. (This equation is an inhomogeneous extension of the homogeneous Fredholm equation (23.13), and is similar to equation (23.57).) 878 23.8 EXERCISES 23.4 Convert f(x)=e x p x+ Zx 0(x−y)f(y)dy into a diﬀerential equation, and hence show that its solution is (α+βx)exp x+γexp(−x), where α, β, γ are constants that should be determined. 23.5 Solve for φ(x) the integral equation φ(x)=f(x)+λ Z1 0 //x y /n + /y x /n / φ(y)dy, where f(x) is bounded for 0 <x< 1a n d−1 2<n<1 2, expressing your answer in terms of the quantities Fm= R1 0f(y)ymdy. (a) Give the explicit solution when λ=1 . (b) For what values of λare there no solutions unless F±ntake particular values? What are these values? 23.6 (a) Consider the inhomogeneous integral equation f(x)=g(x)+λ Zb aK(x, y)f(y)dy; its kernel K(x, y) is real, symmetric and continuous in a≤x≤b,a≤y≤b. Ifλis one of the eigenvalues λiof the homogeneous equation fi(x)=λi Zb aK(x, y)fi(y)dy, prove that the inhomogeneous equation can only a have non-trivial solution ifg(x) is orthogonal to the corresponding eigenfunction fi(x). (b) Show that the only values of λfor which f(x)=λ Z1 0xy(x+y)f(y)dy has a non-trivial solution are the roots of the equation λ2+ 120 λ−240 = 0 . (c) Solve f(x)=µx2+ Z1 02xy(x+y)f(y)dy. 23.7 (a) If the kernel of the integral equation ψ(x)=λ Zb aK(x, y)ψ(y)dy has the form K(x, y)=∞X n=0hn(x)gn(y), where the hn(x) form a complete orthonormal set of functions over the interval [ a, b], show that the eigenvalues λiare given by |M−λ−1I|=0, 879 INTEGRAL EQUATIONS where Mis the matrix with elements Mkj= Zb agk(u)hj(u)du. If the corresponding solutions are ψ(i)(x)= P∞ n=0a(i) nhn(x), ﬁnd an expression fora(i) n. (b) Obtain the eigenvalues and eigenfunctions over the interval [0 ,2π]i f K(x, y)=∞X n=11 ncosnxcosny. 23.8 By taking its Laplace transform, and that of xne−ax, obtain the explicit solution of f(x)=e−x / x+ Zx 0(x−u)euf(u)du / . Verify your answer by substitution. 23.9 For f(t)=exp(−t2 2), use the relationships of the Fourier transforms of f/prime(t)a n d tf(t)t ot h a to f f(t) itself to ﬁnd a simple diﬀerential equation satisﬁed by ˜f(ω), the Fourier transform of f(t) and hence determine ˜f(ω) to within a constant. Use this result to solve the integral equationZ∞ −∞e−t(t−2x)/2h(t)dt=e3x2/8 forh(t). 23.10 Show that the equation f(x)=x−1/3+λ Z∞ 0f(y)e x p(−xy)dy has a solution of the form Axα+Bxβ. Determine the values of αandβand show that those of AandBare 1 1−λ2Γ(1 3)Γ(2 3)andλΓ(2 3) 1−λ2Γ(1 3)Γ(2 3), where Γ( z) is the gamma function, discussed in the appendix. 23.11 At an international ‘peace’ conference a large number of delegates are seated around a circular table with each delegation sitting near its allies and diametricallyopposite the delegation most bitterly opposed to it. The position of a delegate isdenoted by θ,w i t h0≤θ≤2π.T h ef u r y f(θ)f e l tb yt h ed e l e g a t ea t θis the sum of his own natural hostility h(θ) and the inﬂuences on him of each of the other delegates; a delegate at position φcontributes an amount K(θ−φ)f(φ). Thus f(θ)=h(θ)+Z2π 0K(θ−φ)f(φ)dφ. Show that if K(ψ)t a k e st h ef o r m K(ψ)=k0+k1cosψthen f(θ)=h(θ)+p+qcosθ+rsinθ and evaluate p,qandr. A positive value for k1implies that delegates tend to placate their opponents but upset their allies, whilst negative values imply thatthey calm their allies but infuriate their opponents. A walkout will occur if f(θ) exceeds a certain threshold value for some θ. Is this more likely to happen for positive or for negative values of k 1? 880 23.8 EXERCISES 23.12 By considering functions of the form h(x)= Rx 0(x−y)f(y)dy, show that the solution f(x)o ft h ei n t e g r a le q u a t i o n f(x)=x+1 2 Z1 0|x−y|f(y)dy satisﬁes the equation f/prime/prime(x)=f(x). By examining the special cases x=0a n d x= 1, show that f(x)=2 (e+3 ) ( e+1 )[(e+2 )ex−ee−x]. 23.13 The operator Mis deﬁned by Mf(x)≡ Z∞ −∞K(x, y)f(y)dy, where K(x, y) = 1 inside the square |x|<a ,|y|<a, and is equal to 0 elsewhere. Consider the possible eigenvalues of Mand the eigenfunctions that correspond to them; show that the only possible eigenvalues are 0 and 2 aand determine the corresponding eigenfunctions. Hence ﬁnd the general solution of f(x)=g(x)+λ Z∞ −∞K(x, y)f(y)dy. 23.14 For the integral equation y(x)=x−3+λ Zb ax2z2y(z)dz, show that the resolvent kernel is 5 x2z2/[5−λ(b5−a5)] and hence solve the equation. For what range of λis the solution valid? 23.15 Use Fredholm theory to show that, for the kernel K(x, z)=(x+z)e x p( x−z) over the interval [0 ,1], the resolvent kernel is R(x, z;λ)=exp(x−z)[(x+z)−λ(1 2x+1 2z−xz−1 3)] 1−λ−1 12λ2, and hence solve y(x)=x2+2 Z1 0(x+z)e x p ( x−z)y(z)dz, expressing your answer in terms of In,w h e r e In= R1 0unexp(−u)du. 23.16 (a) Determine the eigenvalues λ±of the kernel K(x, z)=(xz)1/2(x1/2+z1/2)a n d show that the corresponding eigenfunctions have the forms y±(x)=A±(√2x1/2±√3x), where A2 ±=5/(10±4√6). (b) Use Schmidt–Hilbert theory to solve y(x)=1+5 2 Z1 0K(x, z)y(z)dz. (c) As may be apparent, the algebra involved in the formal method used in (b) is long and error-prone, and it is in fact much more straightforward to usea trial function 1 + αx 1/2+βx. Check your answer by doing so. 881 INTEGRAL EQUATIONS 23.9 Hints and answers 23.1 Deﬁne y(−x)=y(x) and use the cosine Fourier transform inversion theorem; y(x)=( 2 /π)1/2exp(−x2/2). 23.2 Use the Laplace transform; f(t)=s i n at. 23.3 Set y(x)=c1sinx+c2cosx;y(x)=A[cos(x+a)+(λπ/2)sin( x−a)]/[1−(λ2π2/4)]. 23.4 f/prime/prime(x)−f(x)=e x p x;α=3/4,β=1/2,γ=1/4. 23.5 (a) φ(x)=f(x)−( 1+2 n)Fnxn−(1−2n)F−nx−n. (b) There are no solutions for λ=[ 1±(1−4n2)−1/2]−1unless F±n=0o r F−n/Fn=±[(1−2n)/(1 + 2 n)]1/2. 23.6 (b) Set f(x)=a1x2+a2xand obtain a1=(λ/4)a1+(λ/3)a2,a2=(λ/5)a1+(λ/4)a2; (c) set f(x)=(µ+a1)x2+a2x;f(x)=−6µx(5x+4 ) . 23.7 (a) a(i) n= Rb ahn(x)ψ(x)dx; (b) use (1 /√π)cosnxand (1 /√π)si nnx;Mis diagonal; eigenvalues λk=k/πwith ψ(k)(x)=( 1 /√π)c oskx. 23.8 Writing p(x)= exf(x)a n d q(x)= x, the integrand can be expressed as a convolution. Show that ¯p(s)=¯q(s)/[1−¯q(s)], leading to f(x)=( 1−e−2x)/2. 23.9 d˜f/dω =−ω˜f, leading to ˜f(ω)=Ae−ω2/2. Rearrange the integral as a convolution and deduce that ˜h(ω)= Be−3ω2/2;h(t)= Ce−t2/6, where re-substitution and Gaussian normalisation show that C=√ 6π/2. 23.10 Recall or prove that the Laplace transform of x−n,w h e r e n<1 but is not necessarily an integer, is Γ(1 −n)sn−1. For a possible solution α=−1/3a n d β=−2/3, or vice versa. 23.11 p=k0H/(1−2πk0),q=k1Hc/(1−1 2k1), and r=k1Hs/(1−1 2k1), where H= R2π 0h(z)dz,Hc= R2π 0h(z)coszd z,a n d Hs= R2π 0h(z)sinzd z. Positive values of k1(≈2) are most likely to cause a conference breakdown. 23.12 Write R1 0|x−y|f(y)dyas Rx 0(x−y)f(y)dy+ R1 x(y−x)f(y)dy. 23.13 For eigenvalue 0 : f(x)=0f o r|x|<aorf(x) is such that Ra −af(y)dy=0 .F o r eigenvalue 2 a:f(x)=µS(x, a)w i t h µac o n s t a n ta n d S(x, a)≡[H(a+x)−H(x− a)], where H(z) is the Heaviside step function. Take f(x)=g(x)+cGS(x, a), where G= Ra −ag(z)dz. Show that c=λ/(1−2aλ). 23.14 y(x)=x−3+[ 5x2λln(b/a)]/[5−λ(b5−a5)];|λ|<5/|b5−a5|. 23.15 y(x)=x2−(3I3x+I2)e x p x. 23.16 (a) 5√6/(2√6±5). (b)/angbracketlefty±|K1/angbracketright=A±[(31/60)√2±(19/45)√3]. [1−(5/2)/λ±]−1= ∓2√6/5. For (b) and (c) y(x)=1−4 3x1/2−3 2x. 882 24 Group theory For systems that have some degree of symmetry, full exploitation of that symmetry is desirable. Signiﬁcant physical results can sometimes be deduced simply by astudy of the symmetry properties of the system under investigation. Consequentlyit becomes important, for such a system, to identify all those operations (rotations,reﬂections, inversions) that carry the system into a physically indistinguishablecopy of itself. The study of the properties of the complete set of such operations forms one application of group theory . Though this is the aspect of most interest to the physical scientist, group theory itself is a much larger subject and of greatimportance in its own right. Consequently we leave until the next chapter any direct applications of group theoretical results and concentrate on building up the general mathematical properties of groups. 24.1 Groups As an example of symmetry properties, let us consider the sets of operations, such as rotations, reﬂections, and inversions, that transform physical objects, forexample molecules, into physically indistinguishable copies of themselves, so thatonly the labelling of identical components of the system (the atoms) changes in the process. For diﬀerently shaped molecules there are diﬀerent sets of operations, but in each case it is a well-deﬁned set, and with a little practice all members ofeach set can be identiﬁed. As simple examples, consider ( a) the hydrogen molecule, and ( b) the ammonia molecule illustrated in ﬁgure 24.1. The hydrogen molecule consists of two atoms H of hydrogen and is carried into itself by any of the following operations: (i) any rotation about its long axis; (ii) rotation through πabout an axis perpendicular to the long axis and passing through the point Mthat lies midway between the atoms; 883 GROUP THEORY (a)( b)H HH HHN M Figure 24.1 ( a) The hydrogen molecule, and ( b) the ammonia molecule. (iii) inversion through the point M; (iv) reﬂection in the plane that passes through Mand has its normal parallel to the long axis. These operations collectively form the set of symmetry operations for the hydro- gen molecule. The somewhat more complex ammonia molecule consists of a tetrahedron with an equilateral triangular base at the three corners of which lie hydrogen atomsH, whilst a nitrogen atom N is sited at the fourth vertex of the tetrahedron. Theset of symmetry operations on this molecule is limited to rotations of π/3a n d 2π/3 about the axis joining the centroid of the equilateral triangle to the nitrogen atom, and reﬂections in the three planes containing that axis and each of thehydrogen atoms in turn. However, if the nitrogen atom could be replaced by a fourth hydrogen atom, and all interatomic distances equalised in the process, the number of symmetry operations would be greatly increased. Once allthe possible operations in any particular set have been identiﬁed, it must follow that the result of applying two such operations in succession will beidentical to that obtained by the sole application of some third (usually diﬀerent)operation in the set – for if it were not, a new member of the set would havebeen found, contradicting the assumption that all members have been identiﬁed. Such observations introduce two of the main considerations relevant to decid- ing whether a set of objects, here the rotation, reﬂection and inversion operations,qualiﬁes as a group in the mathematically tightly deﬁned sense. These two consid- erations are (i) whether there is some law for combining two members of the set,and (ii) whether the result of the combination is also a member of the set. Theobvious rule of combination has to be that the second operation is carried outon the system that results from application of the ﬁrst operation, and we have already seen that the second requirement is satisﬁed by the inclusion of all such operations in the set. However, for a set to qualify as a group, more than thesetwo conditions have to be satisﬁed, as will now be made clear. 884 24.1 GROUPS 24.1.1 Deﬁnition of a group Ag r o u p Gis a set of elements {X,Y,...}, together with a rule for combining them that associates with each ordered pair X,Ya ‘product’ or combination law X•Yfor which the following conditions must be satisﬁed. (i) For everypair of elements X, Y that belongs to G, the product X•Yalso belongs to G. (This is known as the closure property of the group.) (ii) For all triples X, Y , Z theassociative law holds; in symbols, X•(Y•Z)=(X•Y)•Z. (24.1) (iii) There exists a unique element I,b e l o n g i n gt o G, with the property that I•X=X=X•I (24.2) forallXbelonging to G. This element Iis known as the identity element of the group. (iv) For every element XofG, there exists an element X−1,a l s ob e l o n g i n gt o G, such that X−1•X=I=X•X−1. (24.3) X−1is called the inverse ofX. An alternative notation in common use is to write the elements of a group G as the set{G1,G2,...}or, more brieﬂy, as {Gi}, a typical element being denoted byGi. It should be noticed that, as given, the nature of the operation •is not stated. It should also be noticed that the more general term element , rather than operation , has been used in this deﬁnition. We will see that the general deﬁnition of agroup allows as elements not only sets of operations on an object but also sets of numbers, of functions and of other objects, provided that the interpretation of • is appropriately deﬁned. In one of the simplest examples of a group, namely the group of all integers under addition, the operation •is taken to be ordinary addition. In this group the role of the identity Iis played by the integer 0, and the inverse of an integer Xis −X. That requirements (i) and (ii) are satisﬁed by the integers under addition is trivially obvious. A second simple group, under ordinary multiplication, is formedby the two numbers 1 and −1; in this group, closure is obvious, 1 is the identity element, and each element is its own inverse. It will be apparent from these two examples that the number of elements in a group can be either ﬁnite or inﬁnite. In the former case the group is called a ﬁnite group and the number of elements it contains is called the order of the group, which we will denote by g; an alternative notation is |G|but has obvious dangers 885 GROUP THEORY if matrices are involved. In the notation in which G={G1,G2,...,G n}the order of the group is clearly n. As we have noted, for the integers under addition zero is the identity. For the group of rotations and reﬂections, the operation of doing nothing, i.e. thenull operation, plays this role. This latter identiﬁcation may seem artiﬁcial, butit is an operation, albeit trivial, which does leave the system in a physicallyindistinguishable state, and needs to be included. One might add that without itthe set of operations would not form a group and none of the powerful resultswe will derive later in this and the next chapter could be justiﬁably applied to give deductions of physical signiﬁcance. In the examples of rotations and reﬂections mentioned earlier, •has been taken to mean that the left-hand operation is carried out on the system that resultsfrom application of the right-hand operation. Thus Z=X•Y (24.4) means that the eﬀect on the system of carrying out Zi st h es a m ea sw o u l d be obtained by ﬁrst carrying out Yand then carrying out X. The order of the operations should be noted; it is arbitrary in the ﬁrst instance but, once chosen, must be adhered to. The choice we have made is dictated by the fact that mostof our applications involve the eﬀect of rotations and reﬂections on functions ofspace coordinates, and it is usual, and our practice in the rest of this book, towrite operators acting on functions to the left of the functions. It will be apparent that for the above-mentioned group, integers under ordinary addition, it is true that Y•X=X•Y (24.5) for all pairs of integers X,Y. If any two particular elements of a group satisfy (24.5), they are said to commute under the operation •; if all pairs of elements in a group satisfy (24.5), then the group is said to be Abelian .T h es e to fa l li n t e g e r s forms an inﬁnite Abelian group under (ordinary) addition. As we show below, requirements (iii) and (iv) of the deﬁnition of a group are over-demanding (but self-consistent), since in each of equations (24.2) and(24.3) the second equality can be deduced from the ﬁrst by using the associativityrequired by (24.1). The mathematical steps in the following arguments are allvery simple, but care has to be taken to make sure that nothing that has notyet been proved is used to justify a step. For this reason, and to act as a modelin logical deduction, a reference in Roman numerals to the previous result, or to the group deﬁnition used, is given over each equality sign. Such explicit detailed referencing soon becomes tiresome, but it should always be available ifneeded. 886 24.1 GROUPSIUsing only the ﬁrst equalities in (24.2) and (24.3), deduce the second ones. Consider the expression X−1•(X•X−1); X−1•(X•X−1)(ii)=(X−1•X)•X−1(iv)=I•X−1 (iii)=X−1. (24.6) ButX−1belongs to G, and so from (iv) there is an element UinGsuch that U•X−1=I. (v) Form the product of Uwith the ﬁrst and last expressions in (24.6) to give U•(X−1•(X•X−1)) =U•X−1(v)=I. (24.7) Transforming the left-hand side of this equation gives U•(X−1•(X•X−1))(ii)=(U•X−1)•(X•X−1) (v)=I•(X•X−1) (iii)=X•X−1. (24.8) Comparing (24.7), (24.8) shows that X•X−1=I, (iv)/prime i.e. the second equality in group deﬁnition (iv). Similarly X•I(iv)=X•(X−1•X)(ii)=(X•X−1)•X (iv)/prime =I•X (iii)=X. (iii/prime) i.e. the second equality in group deﬁnition (iii). J The uniqueness of the identity element Ic a na l s ob ed e m o n s t r a t e dr a t h e rt h a n assumed. Suppose that I/prime,b e l o n g i n gt o G, also has the property I/prime•X=X=X•I/primefor all Xbelonging to G. Take XasI,t h e n I/prime•I=I. (24.9) Further, from (iii/prime), X=X•I for all Xbelonging to G, 887 GROUP THEORY and setting X=I/primegives I/prime=I/prime•I. (24.10) It then follows from (24.9), (24.10) that I=I/prime, showing that in any particular group the identity element is unique. In a similar way it can be shown that the inverse of any particular element is unique. If UandVare two postulated inverses of an element XofG,b y considering the product U•(X•V)=(U•X)•V, it can be shown that U=V. The proof is left to the reader. Given the uniqueness of the inverse of any particular group element, it follows that (U•V•···•Y•Z)•(Z−1•Y−1•···•V−1•U−1) =(U•V•···•Y)•(Z•Z−1)•(Y−1•···•V−1•U−1) =(U•V•···•Y)•(Y−1•···•V−1•U−1) ... =I, where use has been made of the associativity and of the two equations Z•Z−1=I andI•X=X. Thus the inverse of a product is the product of the inverses in reverse order, i.e. (U•V•···•Y•Z)−1=(Z−1•Y−1•···•V−1•U−1). (24.11) Further elementary results that can be obtained by arguments similar to those above are as follows. (i) Given any pair of elements X, Y belonging to G, there exist unique elements U, V,a l s ob e l o n g i n gt o G, such that X•U=Y and V•X=Y. Clearly U=X−1•Y,a n d V=Y•X−1, and they can be shown to be unique. This result is sometimes called the division axiom . (ii) The cancellation law can be stated as follows. If X•Y=X•Z for some Xbelonging to G,t h e n Y=Z.Similarly, Y•X=Z•X implies the same conclusion. 888 24.1 GROUPS L M K Figure 24.2 Reﬂections in the three perpendicular bisectors of the sides of an equilateral triangle take the triangle into itself. (iii) Forming the product of each element of Gwith a ﬁxed element XofG simply permutes the elements of G; this is often written symbolically as G•X=G. If this were not so, and X•YandX•Zwere not diﬀerent even though YandZwere, application of the cancellation law would lead to a contradiction. This result is called the permutation law . In any ﬁnite group of order g, any element Xwhen combined with itself to form successively X2=X•X,X3=X•X2,...will, after at most g−1s u c h combinations, produce the group identity I. Of course X2,X3,...are some of the original elements of the group, and not new ones. If the actual number of combinations needed is m−1, i.e. Xm=I,t h e n mis called the order of the element XinG. The order of the identity of a group is always 1, and that of any other element of a group that is its own inverse is always 2.IDetermine the order of the group of (two-dimensional) rotations and reﬂections that take a plane equilateral triangle into itself and the order of each of the elements. The group is usually known as 3m(to physicists and crystallographers) or C3v(to chemists). There are two (clockwise) rotations, by 2 π/3a n d4 π/3, about an axis perpendicular to the plane of the triangle. In addition, reﬂections in the perpendicular bisectors of the threesides (see ﬁgure 24.2) have the deﬁning property. To these must be added the identityoperation. Thus in total there are six distinct operations and so g= 6 for this group. To reproduce the identity operation either of the rotations has to be applied three times,whilst any of the reﬂections has to be applied just twice in order to recover the originalsituation. Thus each rotation element of the group has order 3, and each reﬂection elementhas order 2.J A so-called cyclic group is one for which all members of the group can be generated from just one element X(say). Thus a cyclic group of order gcan be written as G=braceleftbig I,X,X2,X3,...,Xg−1bracerightbig . 889 GROUP THEORY It is clear that cyclic groups are always Abelian and that each element, apart from the identity, has order g, the order of the group itself. 24.1.2 Further examples of groups In this section we consider some sets of objects, each set together with a law of combination, and investigate whether they qualify as groups and, if not, why not. We have already seen that the integers form a group under ordinary addition, but it is immediately apparent that (even if zero is excluded) they do notdo so under ordinary multiplication. Unity must be the identity of the set, but therequisite inverse of any integer n,n a m e l y1 /n, does not belong to the set of integers for any nother than unity. Other inﬁnite sets of quantities that do form groups are the sets of all real numbers, or of all complex numbers, under addition, and of the same two setsexcluding 0 under multiplication. All these groups are Abelian. Although subtraction and division are normally considered the obvious coun- terparts of the operations of (ordinary) addition and multiplication, they are notacceptable operations for use within groups since the associative law, (24.1), doesnot hold. Explicitly, X−(Y−Z)/negationslash=(X−Y)−Z, X÷(Y÷Z)/negationslash=(X÷Y)÷Z. From within the ﬁeld of all non-zero complex numbers we can select just those that have unit modulus, i.e. are of the form e iθwhere 0≤θ<2π,t of o r ma group under multiplication, as can easily be veriﬁed: eiθ1×eiθ2=ei(θ1+θ2)(closure) , ei0= 1 (identity) , ei(2π−θ)×eiθ=ei2π≡ei0= 1 (inverse) . Closely related to the above group is the set of 2 ×2 rotation matrices that take the form M(θ)=parenleftbiggcosθ−sinθ sinθcosθparenrightbigg where, as before, 0 ≤θ<2π. These form a group when the law of combination is that of matrix multiplication. The reader can easily verify that M(θ)M(φ)= M(θ+φ) (closure), M(0) = I2 (identity), M(2π−θ)= M−1(θ) (inverse). Here I2is the unit 2 ×2 matrix. 890 24.2 FINITE GROUPS 24.2 Finite groups Whilst many properties of physical systems (e.g. angular momentum) are related to the properties of inﬁnite, and, in particular, continuous groups, the symmetryproperties of crystals and molecules are more intimately connected with those ofﬁnite groups. We therefore concentrate in this section on ﬁnite sets of objects thatcan be combined in a way satisfying the group postulates. Although it is clear that the set of all integers does not form a group under ordinary multiplication, restricted sets can do so if the operation involved ismultiplication (mod N) for suitable values of N; this operation will be explained below. As a simple example of a group with only four members, consider the set S deﬁned as follows: S={1,3,5,7}under multiplication (mod 8) . To ﬁnd the product (mod 8) of any two elements, we multiply them together in the ordinary way, and then divide the answer by 8, treating the remainder after doing so as the product of the two elements. For example, 5 ×7 = 35, which on dividing by 8 gives a remainder of 3. Clearly, since Y×Z=Z×Y, the full set of diﬀerent products is 1×1=1 ,1×3=3 ,1×5=5 ,1×7=7 , 3×3=1 ,3×5=7 ,3×7=5 , 5×5=1 ,5×7=3 , 7×7=1 . The ﬁrst thing to notice is that each multiplication produces a member of the original set, i.e. the set is closed. Obviously the element 1 takes the role of theidentity, i.e. 1 ×Y=Yfor all members Yof the set. Further, for each element Y of the set there is an element Z(equal to Y, as it happens, in this case) such that Y×Z= 1, i.e. each element has an inverse. These observations, together with the associativity of multiplication (mod 8), show that the set Sis an Abelian group of order 4. It is convenient to present the results of combining any two elements of a group in the form of multiplication tables – akin to those which used to appear inelementary arithmetic books before electronic calculators were invented! Written in this much more compact form the above example is expressed by table 24.1. Although the order of the two elements being combined does not matter herebecause the group is Abelian, we adopt the convention that if the product in ageneral multiplication table is written X•Ythen Xis taken from the left-hand column and Yis taken from the top row. Thus the bold ‘ 7’ in the table is the result of 3×5, rather than of 5 ×3. Whilst it would make no diﬀerence to the basic information content in a table 891 GROUP THEORY 1357 11357 331 75 55713 77531 Table 24.1 The table of products for the elements of the group S={1,3,5,7} under multiplication (mod 8). to present the rows and columns with their headings in random orders, it is usual to list the elements in the same order in both the vertical and horizontalheadings in any one table. The actual order of the elements in the common list,whilst arbitrary, is normally chosen to make the table have as much symmetry as possible. This is initially a matter of convenience, but, as we shall see later, some of the more subtle properties of groups are revealed by putting next to each otherelements of the group that are alike in certain ways. Some simple general properties of group multiplication tables can be deduced immediately from the fact that each row or column constitutes the elements ofthe group. (i) Each element appears once and only once in each row or column of the table; this must be so since G•X=G(the permutation law) holds. (ii) The inverse of any element Ycan be found by looking along the row in which Yappears in the left-hand column (the Yth row), and noting the element Zat the head of the column (the Zth column) in which the identity appears as the table entry. An immediate corollary is that whenever the identity appears on the leading diagonal, it indicates thatthe corresponding header element is of order 2 (unless it happens to bethe identity itself). (iii) For any Abelian group the multiplication table is symmetric about the leading diagonal. To get used to the ideas involved in using group multiplication tables, we now consider two more sets of integers under multiplication (mod N): S /prime={1,5,7,11}under multiplication (mod 24), and S/prime/prime={1,2,3,4}under multiplication (mod 5) . These have group multiplication tables 24.2( a)a n d( b) respectively, as the reader should verify. If tables 24.1 and 24.2( a) for the groups SandS/primeare compared, it will be seen that they have essentially the same struc ture, i.e if the elements are written as {I,A,B,C}in both cases, then the two tables are each equivalent to table 24.3. ForS,I=1 ,A=3 ,B=5 ,C= 7 and the law of combination is multiplication 892 24.2 FINITE GROUPS (a)157 1 1 1157 1 1 551 1 1 7 771 11 5 111 1 751(b)1234 11234 22413 33142 44321 Table 24.2 ( a) The multiplication table for the group S/prime={1,5,7,11}under multiplication (mod 24). ( b) The multiplication table for the group S/prime/prime= {1,2,3,4}under multiplication (mod 5). IA B C IIA B C AAICB BBCI A CCBAI Table 24.3 The common structure exempliﬁed by tables 24.1 and 24.2( a). 1 i−1−i 1 1 i−1−i i i−1−i1 −1−1−i1 i −i−i1 i−1 Table 24.4 The group table for the set {1,i ,−1,−i}under ordinary multipli- cation of complex numbers. (mod 8), whilst for S/prime,I=1 , A=5 , B=7 , C= 11 and the law of combination is multiplication (mod 24). However, the really important point is that the twogroups SandS /primehave equivalent group multiplication tables – they are said to beisomorphic , a matter to which we will return more formally in section 24.5.IDetermine the behaviour of the set of four elements {1,i ,−1,−i} under the ordinary multiplication of complex numbers. Show that they form a group and determine whether the group is isomorphic to either of the groups S(itself isomorphic to S/prime) andS/prime/primedeﬁned above. That the elements form a group under the associative operation of complex multiplication is immediate; there is an identity (1), each possible product generates a member of the setand each element has an inverse (1 ,−i,−1,i, respectively). The group table has the form shown in table 24.4. We now ask whether this table can be made to look like table 24.3, which is the standardised form of the tables for SandS /prime. Since the identity element of the group (1) will have to be represented by I, and ‘1’ only appears on the leading diagonal twice whereas 893 GROUP THEORY 1 i−1−i 1 1 i−1−i i i−1−i1 −1−1−i1 i −i−i1 i−11243 11243 22431 44312 33124 Table 24.5 A comparison between tables 24.4 and 24.2( b), the latter with its columns reordered. IA B C IIA B C AABC I BBCI A CCIA B Table 24.6 The common structure exempliﬁed by tables 24.4 and 24.2( b), the latter with its columns reordered. Iappears on the leading diagonal four times in table 24.3, it is clear that no amount of relabelling (or, equivalently, no allocation of the symbols A, B, C ,a m o n g s t i,−1,−i)c a n bring table 24.4 into the form of table 24.3. We conclude that the group {1,i ,−1,−i}is not isomorphic to SorS/prime. An alternative way of stating the observation is to say that the group contains only one element of order 2 whilst a group corresponding to table 24.3contains three such elements. However, if the rows and columns of table 24.2( b) – in which the identity does appear twice on the diagonal and which therefore has the potential to beequivalent to table 24.4 – are rearranged by making the heading order 1 ,2,4,3 then the two tables can be compared in the forms sh own in table 24.5. They can thus be seen to have the same structure, namely that shown in table 24.6. We therefore conclude that the group of four elements {1,i ,−1,−i}under ordinary mul- tiplication of complex numbers is isomorphic to the group {1,2,3,4}under multiplication (mod 5).J What we have done does not prove it, but the two tables 24.3 and 24.6 are in fact the only possible tables for a group of order 4, i.e. a group containing exactlyfour elements. 24.3 Non-Abelian groups So far, all the groups for which we have constructed multiplication tables have been based on some form of arithmetic multiplication, a commutative operation,with the result that the groups have been Abelian and the tables symmetricabout the leading diagonal. We now turn to examples of groups in which some non-commutation occurs. It should be noted, in passing, that non-commutation cannot occur throughout a group, as the identity always commutes with any element in its group. 894 24.3 NON-ABELIAN GROUPS As the ﬁrst example we consider again as elements of a group the two- dimensional operations which transform an equilateral triangle into itself (seethe end of subsection 24.1.1). It has already been shown that there are sixsuch operations; the null operation, two rotations (by 2 π/3a n d4 π/3 about an axis perpendicular to the plane of the triangle) and three reﬂections in the perpendicular bisectors of the three sides. To abbreviate we will denote theseoperations by symbols as follows. (i)Iis the null operation. (ii)RandR /primeare (clockwise) rotations by 2 π/3a n d4 π/3 respectively. (iii)K,L,Mare reﬂections in the three lines indicated in ﬁgure 24.2. Some products of the operations of the form X•Y(where it will be recalled that the symbol •means that the second operation Xis carried out on the system resulting from the application of the ﬁrst operation Y) are easily calculated: R•R=R/prime,R/prime•R/prime=R, R•R/prime=I=R/prime•R(24.12) K•K=L•L=M•M=I. Others, such as K•M, are more diﬃcult, but can be found by a little thought, or by making a model triangle or drawing a sequence of diagrams such as thosefollowing. xx x xK•M = = =K R/prime showing that K•M=R/prime. In the same way, x x x xM•K = = =M R shows that M•K=R,a n d x x xxR•L = = =R K shows that R•L=K. Proceeding in this way we can build up the complete multiplication table (table 24.7). In fact, it is not necessary to draw any more diagrams, as allremaining products can be deduced algebraically from the three found above and 895 GROUP THEORY IR R/primeKL M I IR R/primeKL M R RR/primeIM KL R/primeR/primeIRL M K K KL MI RR/prime L LMKR/primeIR M MK L RR/primeI Table 24.7 The group table for the two-dimensional symmetry operations onan equilateral triangle. the more self-evident results given in (24.12). A number of things may be noticed about this table. (i) It is notsymmetric about the leading diagonal, indicating that some pairs of elements in the group do not commute. (ii) There is some symmetry within the 3 ×3 blocks that form the four quarters of the table. This occurs because we have elected to put similar operations close to each other when choosing the order of table headings – the tworotations (or three if Iis viewed as a rotation by 0 π/3) are next to each other, and the three reﬂections also occupy adjacent columns and rows.We will return to this later. That two groups of the same order may be isomorphic carries over to non- Abelian groups. The next two examples are each concerned with sets of sixobjects; they will be shown to form groups that, although very diﬀerent in naturefrom the rotation–reﬂection group just considered, are isomorphic to it. We consider ﬁrst the set Mof six orthogonal 2 ×2 matrices given by I=parenleftbigg10 01parenrightbigg A=parenleftBigg − 1 2√ 3 2 −√ 3 2−1 2parenrightBigg B=parenleftBigg −1 2−√ 3 2√ 3 2−1 2parenrightBigg C=parenleftbigg−10 01parenrightbigg D=parenleftBigg1 2−√ 3 2 −√ 3 2−1 2parenrightBigg E=parenleftBigg1 2√ 3 2√ 3 2−1 2parenrightBigg(24.13) the combination law being that of ordinary matrix multiplication. Here we use italic, rather than the sans serif used for matrices elsewhere, to emphasise thatthe matrices are group elements. Although it is tedious to do so, it can be checked that the product of any two of these matrices, in either order, is also in the set. However, the result isgenerally diﬀerent in the two cases, as matrix multiplication is non-commutative. The matrix Iclearly acts as the identity element of the set, and during the checking for closure it is found that the inverse of each matrix is contained in the set, I, C,DandEbeing their own inverses. The group table is shown in table 24.8. 896 24.3 NON-ABELIAN GROUPS IA B C D E IIA B C D E AAB I ECD BBIA D EC CCDEI AB DDECB I A EECDABI Table 24.8 The group table, under matrix multiplication, for the set Mof six orthogonal 2 ×2 matrices given by (24.13). The similarity to table 24.7 is striking. If {R,R/prime,K,L ,M}of that table are replaced by {A, B, C, D, E }respectively, the two tables are identical, without even the need to reshuﬄe the rows and columns. The two groups, one of reﬂectionsand rotations of an equilateral triangle, the other of matrices, are isomorphic. Our second example of a group isomorphic to the same rotation–reﬂection group is provided by a set of functions of an undetermined variable x.T h e functions are as follows: f 1(x)=x, f 2(x)=1 /(1−x),f 3(x)=(x−1)/x, f4(x)=1 /x, f 5(x)=1−x, f 6(x)=x/(x−1), and the law of combination is fi(x)•fj(x)=fi(fj(x)), i.e. the function on the right acts as the argument of the function on the left to produce a new function of x. It should be emphasised that it is the functions that are the elements of the group. The variable xis the ‘system’ on which they act, and plays much the same role as the triangle does in our ﬁrst example of a non-Abelian group. To show an explicit example, we calculate the product f6•f3. The product will be the function of xobtained by evaluating y/(y−1), when yis set equal to (x−1)/x. Explicitly f6(f3)=(x−1)/x (x−1)/x−1=1−x=f5(x). Thus f6•f3=f5. Further examples are f2•f2=1 1−1/(1−x)=x−1 x=f3, and f6•f6=x/(x−1) x/(x−1)−1=x=f1. (24.14) 897 GROUP THEORY The multiplication table for this set of six functions has all the necessary proper- ties to show that they form a group. Further, if the symbols f1,f2,f3,f4,f5,f6are replaced by I,A,B,C,D,E respectively the table becomes identical to table 24.8. This justiﬁes our earlier claim that this group of functions, with argument sub- stitution as the law of combination, is isomorphic to the group of reﬂections and rotations of an equilateral triangle. 24.4 Permutation groups The operation of rearranging ndistinct objects amongst themselves is called a permutation of degree n, and since many symmetry operations on physical systems can be viewed in that light, the properties of permutations are of interest. For example, the symmetry operations on an equilateral triangle, to which we have already given much attention, can be considered as the six possible rearrangementsof the marked corners of the triangle amongst three ﬁxed points in space, muchas in the diagrams used to compute table 24.7. In the same way, the symmetryoperations on a cube can be viewed as a rearrangement of its corners amongsteight points in space, albeit with many constraints, or, with fewer complications,as a rearrangement of its body diagonals in space. The details will be left until we review the possible ﬁnite groups more systematically. The notations and conventions used in the literature to describe permutations are very varied and can easily lead to confusion. We will try to avoid this by usingletters a ,b ,c ,... (rather than numbers) for the objects that are rearranged by a permutation and by adopting, before long, a ‘cycle notation’ for the permutationsthemselves. It is worth emphasising that it is the permutations ,i . e .t h ea c t so f rearranging, and not the objects themselves (represented by letters) that form the elements of permutation groups. The complete group of all permutations of degree nis usually denoted by S nor Σ n. The number of possible permutations of degree nisn!, and so this is the order of Sn. Suppose the ordered set of six distinct objects {abcdef}is rearranged by some process into {befadc}; then we can represent this mathematically as θ{abcdef}={befadc}, where θis a permutation of degree 6. The permutation θcan be denoted by [ 256143 ] ,s i n c et h eﬁ r s to b j e c t , a, is replaced by the second, b, the second object, b, is replaced by the ﬁfth, e, the third by the sixth, f,e t c .T h ee q u a t i o n can then be written more explicitly as θ{abcdef}=[ 256143 ] {abcdef}={befadc}. Ifφis a second permutation, also of degree 6, then the obvious interpretation of the product φ•θof the two permutations is φ•θ{abcdef}=φ(θ{abcdef}). 898 24.4 PERMUTATION GROUPS Suppose that φis the permutation [4 5 3 6 2 1]; then φ•θ{abcdef}=[ 453621 ] [ 256143 ] {abcdef} =[ 453621 ] {befadc} ={adfceb} =[ 146352 ] {abcdef}. Written in terms of the permutation notation this result is [ 453621 ] [ 256143 ]=[ 146352 ] . A concept that is very useful for working with permutations is that of decom- position into cycles. The cycle notation is most easily explained by example. For the permutation θgiven above: the 1st object, a, has been replaced by the 2nd, b; the 2nd object, b, has been replaced by the 5th, e; the 5th object, e, has been replaced by the 4th, d; the 4th object, d, has been replaced by the 1st, a. This brings us back to the beginning of a closed cycle, which is conveniently represented by the notation (1 2 5 4), in which the successive replacementp o s i t i o n sa r ee n c l o s e d ,i ns e q u e n c e ,i np a r e n t h e s e s .T h u s( 1254 )m e a n s2 n d →1st, 5th→2nd, 4th→5th, 1st→4th. It should be noted that the object initially in the ﬁrst listed position replaces that in the ﬁnal position indicated inthe bracket – here ‘ a’ is put into the fourth position by the permutation. Clearly the cycle (5 4 1 2), or any other which involved the same numbers in the samerelative order, would have exactly the same meaning and eﬀect. The remainingtwo objects, candf, are interchanged by θor, more formally, are rearranged according to a cycle of length 2, a transposition , represented by (3 6). Thus the complete representation (speciﬁcation) of θis θ=( 1254 ) ( 36 ) . The positions of objects that are unaltered by a permutation are either placed by themselves in a pair of parentheses or omitted altogether. The former is recom-mended as it helps to indicate how many objects are involved – important whenthe object in the last position is unchanged, or the permutation is the identity,which leaves all objects unaltered in position! Thus the identity permutation of degree 6 is I= (1)(2)(3)(4)(5)(6) , though in practice it is often shortened to (1). It will be clear that the cycle representation is unique, to within the internal 899 GROUP THEORY absolute ordering of the numbers in each bracket as already noted, and that each number appears once and only once in the representation of any particularpermutation. Theorder of any permutation of degree nwithin the group S ncan be read oﬀ from the cyclic representation and is given by the lowest common multiple (LCM) of the lengths of the cycles. Thus Ihas order 1, as it must, and the permutation θdiscussed above has order 4 (the LCM of 4 and 2). Expressed in cycle notation our second permutation φis (3)(1 4 6)(2 5), and the product φ•θis calculated as (3)(1 4 6)(2 5) •(1 2 5 4)(3 6) {abcdef}= (3)(1 4 6)(2 5) {befadc} ={adfceb} = (1)(5)(2 4 3 6) {abcdef}. i.e. expressed as a relationship amongst elements of the group of permutations of degree 6 (not yet proved as a group, but reasonably anticipated), this result reads (3)(1 4 6)(2 5) •(1 2 5 4)(3 6) = (1)(5)(2 4 3 6) . We note, for practice, that φhas order 6 (the LCM of 1, 3, and 2) and that the product φ•θhas order 4. The number of elements in the group Snof all permutations of degree nis n! and clearly increases very rapidly as nincreases. Fortunately, to illustrate the essential features of permutation groups it is suﬃcient to consider the case n=3 , which involves only six elements. They are as follows (with labelling which thereader will by now recognise as anticipatory): I= (1)(2)(3) A=( 123 ) B=( 132 ) C= (1)(2 3) D= (3)(1 2) E= (2)(1 3) It will be noted that AandBhave order 3, whilst C,DandEhave order 2. As perhaps anticipated, their combination products are exactly those correspondingto table 24.8, I,C,DandEbeing their own inverses. For example, putting in all steps explicitly, D•C{abc}= (3)(1 2)•(1)(2 3){abc} = (3)(12){acb} ={cab} =( 321 ){abc} =( 132 ){abc} =B{abc}. In brief, the six permutations belonging to S 3form yet another non-Abelian group isomorphic to the rotation–reﬂection symmetry group of an equilateral triangle. 900 24.5 MAPPINGS BETWEEN GROUPS 24.5 Mappings between groups Now that we have available a range of groups that can be used as examples, we return to the study of more general group properties. From here on, when there is no ambiguity we will write the product of two elements, X•Y, simply asXY, omitting the explicit combination symbol. We will also continue to use ‘multiplication’ as a loose generic name for the combination process betweenelements of a group. IfGandG /primeare two groups, we can study the eﬀect of a mapping Φ:G→G/prime ofGontoG/prime.I fXis an element of Gwe denote its image inG/primeunder the mapping Φb y X/prime=Φ ( X). A technical term that we have already used is isomorphic . We will now deﬁne it formally. Two groups G={X,Y,...}andG/prime={X/prime,Y/prime,...}are said to be isomorphic if there is a one-to-one correspondence X↔X/prime,Y↔Y/prime,··· between their elements such that XY=Z implies X/primeY/prime=Z/prime and vice versa. In other words, isomorphic groups have the same (multiplication) structure, although they may diﬀer in the nature of their elements, combination law andnotation. Clearly if groups GandG /primeare isomorphic, and GandG/prime/primeare isomorphic, then it follows that G/primeandG/prime/primeare isomorphic. We have already seen an example of four groups (of functions of x, of orthogonal matrices, of permutations and of the symmetries of an equilateral triangle) that are isomorphic, all having table 24.8 as their multiplication table. Although our main interest is in isomorphic relationships between groups, the wider question of mappings of one set of elements onto another is of someimportance, and we start with the more general notion of a homomorphism. LetGandG /primebe two groups and Φa mapping of G→G/prime. If for every pair of elements XandYinG (XY)/prime=X/primeY/prime thenΦis called a homomorphism, and G/primeis said to be a homomorphic image of G. The essential deﬁning relationship, expressed by ( XY)/prime=X/primeY/prime, is that the same result is obtained whether the product of two elements is formed ﬁrst andthe image then taken or the images are taken ﬁrst and the product then formed. Three immediate consequences of the above deﬁnition are proved as follows. 901 GROUP THEORY (i) If Iis the identity of Gthen IX=Xfor all XinG.C o n s e q u e n t l y X/prime=(IX)/prime=I/primeX/prime, for all X/primeinG/prime. Thus I/primeis the identity in G/prime. In words, the identity element ofGmaps into the identity element of G/prime. (ii) Further, I/prime=(XX−1)/prime=X/prime(X−1)/prime. That is, ( X−1)/prime=(X/prime)−1.In words, the image of an inverse is the same element in G/primeas the inverse of the image. (iii) If element XinGis of order m,i . e .I=Xm,t h e n I/prime=(Xm)/prime=(XXm−1)/prime=X/prime(Xm−1)/prime=···=X/primeX/prime···X/prime bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright mfactors. In words, the image of an element has the same order as the element. What distinguishes an isomorphism from the more general homomorphism are the requirements that in an isomorphism: (I) diﬀerent elements in Gmust map into diﬀerent elements in G/prime(whereas in a homomorphism several elements in Gmay have the same image in G/prime), that is, x/prime=y/primemust imply x=y; (II) any element in G/primemust be the image of some element in G. An immediate consequence of (I) and result (iii) for homomorphisms is that groups that are isomorphic each have the same number of elements of any given order. For a general homomorphism, the set of elements of Gwhose image in G/prime isI/primeis called the kernel of the homomorphism; this is discussed further in the next section. In an isomorphism the kernel consists of the identity Ialone. To illustrate both this point and the general notion of a homomorphism, considera mapping between the additive group of real numbers /Rfracturand the multiplicative group of complex numbers with unit modulus, U(1). Suppose that the mapping /Rfractur→ U(1) is Φ:x→e ix; then this is a homomorphism since (x+y)/prime→ei(x+y)=eixeiy=x/primey/prime. However, it is not an isomorphism because many (an inﬁnite number) of the elements of /Rfracturhave the same image in U(1). For example, π,3π,5π,... in/Rfracturall have the image −1i nU(1) and, furthermore, all elements of /Rfracturof the form 2 πn, where nis an integer, map onto the identity element in U(1). The latter set forms the kernel of the homomorphism. 902 24.6 SUBGROUPS (a)IA B C D E IIAB CDE AABI ECD BBIA DEC CCDEI AB DDECB I A EECDABI(b)IA B C IIA BC AAI CB BBCI A CCBAI Table 24.9 Reproduction of ( a) table 24.8 and ( b) table 24.3 with the relevant subgroups shown in bold. For the sake of completeness, we add that a homomorphism for which (I) above holds is said to be a monomorphism (or an isomorphism into), whilst a homomor- phism for which (II) holds is called an epimorphism (or an isomorphism onto). If, in either case, the other requirement is met as well then the monomorphism orepimorphism is also an isomorphism. Finally, if the initial and ﬁnal groups are the same, G=G /prime, then the isomorphism G→G/primeis termed an automorphism . 24.6 Subgroups More detailed inspection of tables 24.8 and 24.3 shows that not only do the complete tables have the properties associated with a group multiplication table(see section 24.2) but so do the upper left corners of each table taken on theirown. The relevant parts are shown in bold in the tables 24.9( a)a n d( b). This observation immediately prompts the notion of a subgroup . A subgroup of a group Gcan be formally deﬁned as any non-empty subset H={H i}of G, the elements of which themselves behave as a group under the same rule of combination as applies in Gitself. As for all groups, the order of the subgroup is equal to the number of elements it contains; we will denote it by hor|H|. All groups Gcontain two trivial subgroups: (i)Gitself; (ii) the set Iconsisting of the identity element alone. All other subgroups are termed proper subgroups . In a group with multiplication table 24.8 the elements {I,A,B}form a proper subgroup, as do {I,A}in a group with table 24.3 as its group table. Some groups have no proper subgroups. For example, the so-called cyclic groups , mentioned at the end of subsection 24.1.1, have no subgroups other than the whole group or the identity alone. Tables 24.10( a)a n d( b) show the multiplication tables for two of these groups. Table 24.6 is also the group tablefor a cyclic group, that of order 4. 903 GROUP THEORY (a)IA B IIA B AAB I BBIA(b)IA B C D IIA B C D AABCDI BBCDI A CCDI AB DDIABC Table 24.10 The group tables of two cyclic groups, of orders 3 and 5. They have no proper subgroups. It will be clear that for a cyclic group Grepeated combination of any element with itself generates all other elements of G, before ﬁnally reproducing itself. So, for example, in table 24.10( b), starting with (say) D, repeated combination with itself produces, in turn, C,B,A,Iand ﬁnally Dagain. As noted earlier, in any cyclic group Gevery element, apart from the identity, is of order g, the order of the group itself. The two tables shown are for groups of orders 3 and 5. It will be proved in subsection 24.7.2 that the order of any group is a multiple of the order of any ofits subgroups (Lagrange’s theorem), i.e. in our general notation, gis a multiple ofh. It thus follows that a group of order p,w h e r e pis any prime, must be cyclic and cannot have any proper subgroups. The groups for which tables 24.10( a)a n d (b) are the group tables are two such examples. Groups of non-prime order may (table 24.3) or may not (table 24.6) have proper subgroups. As we have seen, repeated multiplication of an element X(not the identity) by itself will generate a subgroup {X,X 2,X3,...}. The subgroup will clearly be Abelian, and if Xis of order m,i . e . Xm=I, the subgroup will have mdistinct members. If mis less than g– though, in view of Lagrange’s theorem, mmust be a factor of g– the subgroup will be a proper subgroup. We can deduce, in passing, that the order of any element of a group is an exact divisor of the orderof the group. Some obvious properties of the subgroups of a group G, which can be listed without formal proof, are as follows. (i) The identity element of Gbelongs to every subgroup H. (ii) If element Xbelongs to a subgroup H, so does X −1. (iii) The set of elements in Gthat belong to every subgroup of Gthemselves form a subgroup, though it may consist of the identity alone. Properties of subgroups that need more explicit proof are given in the follow- ing sections, though some need the development of new concepts before they can be established. However, we can begin with a theorem, applicable to allhomomorphisms, not just isomorphisms, that requires no new concepts. Let Φ : G→G /primebe a homomorphism of GintoG/prime;t h e n 904 24.7 SUBDIVIDING A GROUP (i) the set of elements H/primeinG/primethat are images of the elements of Gforms a subgroup of G/prime; (ii) the set of elements KinGthat are mapped onto the identity I/primeinG/primeforms a subgroup of G. As indicated in the previous section, the subgroup Kis called the kernel of the homomorphism. To prove (i), suppose ZandWbelong to H/prime, with Z=X/primeandW=Y/prime,w h e r e XandYbelong to G.T h e n ZW=X/primeY/prime=(XY)/prime and therefore belongs to H/prime,a n d Z−1=(X/prime)−1=(X−1)/prime and therefore belongs to H/prime. These two results, together with the fact that I/prime belongs to H/prime, are enough to establish result (i). To prove (ii), suppose XandYbelong to K;t h e n (XY)/prime=X/primeY/prime=I/primeI/prime=I/prime(closure), I/prime=(XX−1)/prime=X/prime(X−1)/prime=I/prime(X−1)/prime=(X−1)/prime and therefore X−1belongs to K. These two results, together with the fact that I belongs to K, are enough to establish (ii). An illustration of this result is provided by the mapping Φ of /Rfractur→ U(1) considered in the previous section. Its kernel consists of the set of real numbers of the form 2 πnwhere nis an integer; they form a subgroup of R, the additive group of real numbers. In fact the kernel Kof a homomorphism is a normal subgroup of G.T h e deﬁning property of such a subgroup is that for every element XinGand every element Yin the subgroup, XY X−1belongs to the subgroup. This property is easily veriﬁed for the kernel K,s i n c e (XY X−1)/prime=X/primeY/prime(X−1)/prime=X/primeI/prime(X−1)/prime=X/prime(X−1)/prime=I/prime. Anticipating the discussion of subsection 24.7.2, the cosets of a normal subgroup themselves form a group (see exercise 24.16). 24.7 Subdividing a group We have already noted, when looking at the (arbitrary) order of headings in a group table, that some choices appear to make the table more orderly than do others. In the following subsections we will identify ways in which the elements of a group can be divided up into sets with the property that the members of anyone set are more like the other members of the set, in some particular regard, 905 GROUP THEORY than they are like any element that does not belong to the set. We will ﬁnd that these divisions will be such that the group is partitioned , i.e. the elements will be divided into sets in such a way that each element of the group belongs to one,and only one, such set. We note in passing that the subgroups of a group do notform such a partition, not least because the identity element is in every subgroup, rather than being inprecisely one. In other words, despite the nomenclature, a group is not simply theaggregate of its proper subgroups. 24.7.1 Equivalence relations and classes We now specify in a more mathematical manner what it means for two elements of a group to be ‘more like’ one another than like a third element, as mentionedin section 24.2. Our introduction will apply to any set, whether a group or not, but our main interest will ultimately be in two particular applications to groups. We start with the formal deﬁnition of an equivalence relation. Anequivalence relation on a set Sis a relationship X∼Y,b e t w e e nt w o elements XandYbelonging to S, in which the deﬁnition of the symbol ∼must satisfy the requirements of (i) reﬂexivity, X∼X; (ii) symmetry, X∼Yimplies Y∼X; (iii) transitivity, X∼YandY∼Zimply X∼Z. Any particular two elements either satisfy or do not satisfy the relationship. The general notion of an equivalence relation is very straightforward, and the requirements on ∼seem undemanding; but not all relationships qualify. As an example within the topic of groups, if ∼meant ‘has the same order as’ then clearly all the requirements would be satisﬁed. However, if ∼meant ‘commutes with’ then it would not be an equivalence relation, since although Acommutes with I,a n d Icommutes with C, this does not necessarily imply that Acommutes with C, as is obvious from table 24.8. It may be shown that an equivalence relation on Sdivides up Sintoclasses C i such that: (i)XandYbelong to the same class if, and only if, X∼Y; (ii) every element WofSbelongs to exactly one class. This may be shown as follows. Let Xbelong to S, and deﬁne the subset SXof Sto be the set of all elements UofSsuch that X∼U. Clearly by reﬂexivity Xbelongs to SX. Suppose ﬁrst that X∼Y, and let Zbe any element of SY. Then Y∼Z, and hence by transitivity X∼Z, which means that Zbelongs to SX. Conversely, since the symmetry law gives Y∼X,i fZbelongs to SXthen 906 24.7 SUBDIVIDING A GROUP this implies that Zbelongs to SY. These two results together mean that the two subsets SXandSYhave the same members and hence are equal. Now suppose that SXequals SY.S i n c e Ybelongs to SYit also belongs to SX and hence X∼Y. This completes the proof of (i), once the distinct subsets of typeSXare identiﬁed as the classes Ci. Statement (ii) is an immediate corollary, the class in question being identiﬁed as SW. The most important property of an equivalence relation is as follows. Two diﬀerent subsets SXandSYcan have no element in common, and the collection of all the classes Ciis a ‘partition’ of S, i.e. every element in Sbelongs to one, and only one, of the classes. To prove this, suppose SXandSYhave an element Zin common; then X∼Z andY∼Zand so by the symmetry and transitivity laws X∼Y.B yt h ea b o v e theorem this implies SXequals SY. But this contradicts the fact that SXandSY are diﬀerent subsets. Hence SXandSYcan have no element in common. Finally, if the elements of Sare used in turn to deﬁne subsets and hence classes inS, every element Uis in the subset SUthat is either a class already found or constitutes a new one. It follows that the classes exhaust S,i . e .e v e r ye l e m e n ti s in some class. Having established the general properties of equivalence relations, we now turn to two speciﬁc examples of such relationships, in which the general set Shas the more specialised properties of a group Gand the equivalence relation ∼is chosen in such a way that the relatively transparent general results for equivalencerelations can be used to derive powerful, but less obvious, results about theproperties of groups. 24.7.2 Congruence and cosets As the ﬁrst application of equivalence relations we now prove Lagrange’s theorem which is stated as follows. IfGis a ﬁnite group of order gandHis a subgroup of Gof order h then gis a multiple of h. We take as the deﬁnition of ∼that, given XandYbelonging to G,X∼Yif X −1Ybelongs to H. This is the same as saying that Y=XH ifor some element Hibelonging to H; technically XandYare said to be left-congruent with respect toH. This deﬁnes an equivalence relation, since it has the following properties. (i) Reﬂexivity: X∼X,s i n c e X−1X=IandIbelongs to any subgroup. (ii) Symmetry: X∼Yimplies that X−1Ybelongs to Hand so, therefore, does its inverse, since His a group. But ( X−1Y)−1=Y−1Xand, as this belongs toH, it follows that Y∼X. 907 GROUP THEORY (iii) Transitivity: X∼YandY∼Zimply that X−1YandY−1Zbelong to H and so, therefore, does their product ( X−1Y)(Y−1Z)=X−1Z,f r o mw h i c h it follows that X∼Z. With∼proved as an equivalence relation, we can immediately deduce that it divides Ginto disjoint (non-overlapping) classes. For this particular equivalence relation the classes are called the left cosets ofH. Thus each element of Gis in one and only one left coset of H. The left coset containing any particular Xis usually written XH, and denotes the set of elements of the form XH i(one of which is Xitself since Hcontains the identity element); it must contain hdiﬀerent elements, since if it did not, and two elements were equal, XH i=XH j, we could deduce that Hi=Hjand that Hcontained fewer than helements. From our general results about equivalence relations it now follows that the left cosets of Hare a ‘partition’ of Ginto a number of sets each containing h members. Since there are gmembers of Gand each must be in just one of the sets, it follows that gis a multiple of h. This concludes the proof of Lagrange’s theorem. The number of left cosets of HinGis known as the index ofHinGand is written [ G:H]; numerically the index = g/h. For the record we note that, for the trivial subgroup I, which contains only the identity element, [ G:I]=gand that, for a subgroup Jof subgroup H,[G:H][H:J]=[G:J]. The validity of Lagrange’s theorem was established above using the far-reaching properties of equivalence relations. However, for this speciﬁc purpose there is amore direct and self-contained proof, which we now give. LetXbe some particular element of a ﬁnite group Gof order g,a n dHbe a subgroup of Gof order h, with typical element Y i. Consider the set of elements XH≡{XY1,XY 2,...,XY h}. This set contains hdistinct elements, since if any two were equal, i.e. XYi=XYj with i/negationslash=j, this would contradict the cancellation law. As we have already seen, the set is called a left coset of H. We now prove three simple results. •Two cosets are either disjoint or identical. Suppose cosets X1HandX2Hhave an element in common, i.e. X1Y1=X2Y2for some Y1,Y2inH.T h e n X1= X2Y2Y−1 1, and since Y1andY2both belong to Hso does Y2Y−1 1; thus X1 belongs to the left coset X2H. Similarly X2belongs to the left coset X1H. Consequently, either the two cosets are identical or it was wrong to assumethat they have an element in common. 908 24.7 SUBDIVIDING A GROUP •Two cosets X1HandX2Hare identical if, and only if, X−1 2X1belongs to H.If X−1 2X1belongs to Hthen X1=X2Yifor some i,a n d X1H=X2YiH=X2H, since by the permutation law YiH=H. Thus the two cosets are identical. Conversely, suppose X1H=X2H.T h e n X−1 2X1H=H. But one element of H(on the left of the equation) is I; thus X−1 2X1must also be an element of H (on the right). This proves the stated result. •Every element of Gis in some left coset XH.This follows trivially since H contains I, and so the element Xiis in the coset XiH. The ﬁnal step in establishing Lagrange’s theorem is, as previously, to note that each coset contains helements, that the cosets are disjoint and that every one of thegelements in Gappears in one and only one distinct coset. It follows that g=khfor some integer k. As noted earlier, Lagrange’s theorem justiﬁes our statement that any group of order p,w h e r e pis prime, must be cyclic and cannot have any proper subgroups: since any subgroup must have an order that divides p, this can only be 1 or p, corresponding to the two trivial subgroups Iand the whole group. It may be helpful to see an example worked through explicitly, and we again use the same six-element group.IFind the left cosets of the proper subgroup Hof the group Gthat has table 24.8 as its multiplication table. The subgroup consists of the set of elements H={I,A,B}. We note in passing that it has order 3, which, as required by Lagrange’s theorem, is a divisor of 6, the order of G.A si n all cases, Hitself provides the ﬁrst (left) coset, formally the coset IH={II,IA,IB}={I,A,B}. We continue by choosing an element not already selected, Csay, and form CH={CI,CA,CB}={C,D,E}. These two cosets of Hexhaust G, and are therefore the only cosets, the index of HinG being equal to 2. This completes the example, but it is useful to demonstrate that it would not have mattered if we had taken D, say, instead of Ito form a ﬁrst coset DH={DI, DA, DB}={D, E,C}, and then, from previously unselected elements, picked B, say: BH={BI,BA,BB}={B,I,A}. The same two cosets would have resulted. J It will be noticed that the cosets are the same groupings of the elements ofGwhich we earlier noted as being the choice of adjacent column and row headings that give the multiplication table its ‘neatest’ appearance. Furthermore, 909 GROUP THEORY ifHis anormal subgroup of Gthen its (left) cosets themselves form a group (see exercise 24.16). 24.7.3 Conjugates and classes Our second example of an equivalence relation is concerned with those elements XandYof a group Gthat can be connected by a transformation of the form Y=G−1 iXG i,w h e r e Giis an (appropriate) element of G. Thus X∼Yif there exists an element GiofGsuch that Y=G−1 iXG i. Diﬀerent pairs of elements X andYwill, in general, require diﬀerent group elements Gi. Elements connected in this way are said to be conjugates . We ﬁrst need to establish that this does indeed deﬁne an equivalence relation, as follows. (i) Reﬂexivity: X∼X,s i n c e X=I−1XIandIbelongs to the group. (ii) Symmetry: X∼Yimplies Y=G−1 iXG iand therefore X=(G−1 i)−1YG−1 i. Since Gibelongs to Gso does G−1 i, and it follows that Y∼X. (iii) Transitivity: X∼YandY∼Zimply Y=G−1 iXG iandZ=G−1 jYG j and therefore Z=G−1 jG−1 iXG iGj=(GiGj)−1X(GiGj). Since GiandGj belong to Gso does GiGj, from which it follows that X∼Z. These results establish conjugacy as an equivalence relation and hence show that it divides Ginto classes, two elements being in the same class if, and only if, they are conjugate. Immediate corollaries are: (i) If Zis in the class containing Ithen Z=G−1 iIGi=G−1 iGi=I. Thus, since any conjugate of Ican be shown to be I, the identity must be in a class by itself. (ii) If Xis in a class by itself then Y=G−1 iXG i must imply that Y=X.B u t X=GiG−1 iXG iG−1 i for any Gi,a n ds o X=Gi(G−1 iXG i)G−1 i=GiYG−1 i=GiXG−1 i, i.e.XG i=GiXfor all Gi. Thus commutation with all elements of the group is a necessary (and suﬃcient) condition for any particular group element to be in a class byitself. In an Abelian group each element is in a class by itself. 910 24.7 SUBDIVIDING A GROUP (iii) In any group Gthe set Sof elements in classes by themselves is an Abelian subgroup (known as the centre ofG). We have shown that Ibelongs to S, and so if, further, XG i=GiXandYG i=GiYfor all Gibelonging to G then: (a) (XY)Gi=XG iY=Gi(XY),i.e. the closure of S,a n d (b)XG i=GiXimplies X−1Gi=GiX−1, i.e. the inverse of Xbelongs toS. Hence Sis a group, and clearly Abelian. Yet again for illustration purposes, we use the six-element group that has table 24.8 as its group table.IFind the conjugacy classes of the group Ghaving table 24.8 as its multiplication table. As always, Iis in a class by itself, and we need consider it no further. Consider next the results of forming X−1AX,a sXruns through the elements of G. I−1AI A−1AA B−1AB C−1AC D−1AD E−1AE =IA =IA =AI =CE =DC =ED =A =A =A =B =B =B Only AandBare generated. It is clear that {A, B}is one of the conjugacy classes of G. This can be veriﬁed by forming all elements X−1BX; again only AandBappear. We now need to pick an element not in the two classes already found. Suppose we pick C.J u s ta sf o r A, we compute X−1CX,a sXruns through the elements of G.T h e calculations can be done directly using the table and give the following: X :IA B C D E X−1CX :CEDCED Thus C,DandEbelong to the same class. The group is now exhausted, and so the three conjugacy classes are {I},{A, B},{C,D,E}. J In the case of this small and simple, but non-Abelian, group, only the identity is in a class by itself (i.e. only Icommutes with all other elements). It is also the only member of the centre of the group. Other areas from which examples of conjugacy classes can be taken include permutations and rotations. Two permutations are in the same class if theircycle speciﬁcations have the same structure. For example, in S 5the permutations (1 3 5)(2)(4) and (2 5 3)(1)(4) are in the same class as each other but in a diﬀerentclass from that which contains (1 5)(2 4)(3). In the case of the continuous rotation group, rotations by the same angle θ about any two axes labelled iandjare in the same class, because the group contains a rotation that takes the ﬁrst axis into the second. Without going intomathematical details, a rotation about axis ican be represented by the operator R i(θ), and the two rotations are connected by a relationship of the form Rj(θ)=φ−1 ijRi(θ)φij, 911 GROUP THEORY in which φijis the member of the full continuous rotation group that takes axis iinto axis j. 24.8 Exercises 24.1 For each of the following sets, determine whether they form a group under the op- eration indicated (where it is relevant you may assume that matrix multiplicationis associative): (a) the integers (mod 10) under addition; (b) the integers (mod 10) under multiplication; (c) the integers 1,2,3, 4,5,6 under multiplication (mod 7);(d) the integers 1,2,3, 4,5 under multiplication (mod 6); (e) all matrices of the form/ aa−b 0 b / where aandbare integers (mod 5), and a/negationslash=0/negationslash=b, under matrix multiplica- tion; (f) those elements of the set in (e) that are of order 1 or 2 (taken together); (g) all matrices of the form/0/@100 a10 bc 1 /1A where a,b,care integers, under matrix multiplication. 24.2 Which of the following relationships between XandYare equivalence relations? Give a proof of your conclusions in each case: (a)XandYare integers and X−Yis odd; (b)XandYare integers and X−Yis even; (c)XandYare people and have the same postcode; (d)XandYare people and have a parent in common; (e)XandYare people and have the same mother; (f)XandYaren×nmatrices satisfying Y=PXQ,w h e r e PandQare elements of a group Gofn×nmatrices. 24.3 Deﬁne a binary operation •on the set of real numbers by x•y=x+y+rxy, where ris a non-zero real number. Show that the operation •is associative. Prove that x•y=−r−1if, and only if, x=−r−1ory=−r−1. Hence prove that the set of all real numbers excluding −r−1forms a group under the operation •. 24.4 Prove that the relationship X∼Y, deﬁned by X∼YifYcan be expressed in the form Y=aX+b cX+d, with a,b,canddas integers, is an equivalence relation on the set of real numbers /Rfractur. Identify the class that contains the real number 1. 24.5 The following is a ‘proof’ that reﬂexivity is an unnecessary axiom for an equiva- lence relation. 912 24.8 EXERCISES Because of symmetry X∼Yimplies Y∼X. Then by transitivity X∼Yand Y∼Ximply X∼X. Thus symmetry and transitivity imply reﬂexivity, which therefore need not be separately required. Demonstrate the ﬂaw in this proof using the set consisting of all real numbers plus the number i. Show by investigating the following speciﬁc cases that, whether or not reﬂexivity actually holds, it cannot be deduced from symmetry and transitivityalone. (a)X∼YifX+Yis real. (b)X∼YifXYis real. 24.6 Prove that the set Mof matrices A= / ab 0c / , where a, b, c are integers (mod 5) and a/negationslash=0/negationslash=c, forms a non-Abelian group under matrix multiplication. Show that the subset containing elements of Mthat are of order 1 or 2 does not form a proper subgroup of M (a) using Lagrange’s theorem, (b) by direct demonstration that the set is not closed. 24.7 Sis the set of all 2 ×2 matrices of the form A= / wx yz / where wz−xy=1 . Show that Sis a group under matrix multiplication. Which element(s) have order 2? Prove that an element Ahas order 3 if w+z+1=0 . 24.8 Show that, under matrix multiplication, matrices of the form M(a0,a)= / a0+a1i−a2+a3i a2+a3ia 0−a1i / , where a0and the components of column matrix a=(a1a2a3)Tare real num- bers satisfying a2 0+|a|2= 1, form a group. Deduce that, under the transformation z→Mz,w h e r e zis any column matrix, |z|2is invariant. 24.9 If Ais a group in which every element other than the identity, I,h a so r d e r2 , prove that Ais Abelian. Hence show that if XandYare distinct elements of A, neither being equal to the identity, then the set {I,X,Y ,XY }forms a subgroup ofA. Deduce that if Bis a group of order 2 p,w i t h pa prime greater than 2, then B must contain an element of order p. 24.10 The group of rotations (excluding reﬂections and inversions) in three dimensions that take a cube into itself is known as the group 432 (or Oin the usual chemical notation). Show by each of the following methods that this group has 24 elements. (a) Identify the distinct relevant axes and count the number of qualifying rota- tions about each. (b) The orientation of the cube is determined if the directions of two of its body diagonals are given. Consider the number of distinct ways in which one bodydiagonal can be chosen to be ‘vertical’ and a second diagonal made to liealong a particular direction. 24.11 Identify the eight symmetry operations on a square. Show that they form a group (known to crystallographers as 4 mmor to chemists as C 4v) having one element of order 1, ﬁve of order 2 and two of order 4. Find its proper subgroups and thecorresponding cosets. 913 GROUP THEORY 24.12 If AandBare two groups then their direct product, A×B,i sd e ﬁ n e dt ob e the set of ordered pairs ( X,Y), with Xan element of A,Yan element of B and multiplication given by ( X,Y)(X/prime,Y/prime)=(XX/prime,YY/prime).Prove that A×Bis a group. Denote the cyclic group of order nbyCnand the symmetry group of a regular n-sided ﬁgure (an n-gon) by Dn– thus D3is the symmetry group of an equilateral triangle, as discussed in the text. (a) By considering the orders of each of their elements, show (i) that C2×C3is isomorphic to C6, and (ii) that C2×D3is isomorphic to D6. (b) Are any of D4,C8,C2×C4,C2×C2×C2isomorphic? 24.13 Find the group Ggenerated under matrix multiplication by the matrices A= / 01 10 / , B= / 0i i0 / . Determine its proper subgroups, and verify for each of them that its cosets exhaust G. 24.14 Show that if pis prime then the set of rational number pairs ( a, b), excluding (0,0), with multiplication deﬁned by (a, b)•(c, d)=(e, f),where ( a+b√p)(c+d√p)=e+f√p, forms an Abelian group. Show further that the mapping ( a, b)→(a,−b)i sa n automorphism. 24.15 (a) Denote by Anthe subset of the permutation group Snthat contains all the even permutations. Show that Anis a subgroup of Sn. (b) List the elements of S3in cycle notation and identify the subgroup A3. (c) For each element XofS3,l e tp(X)=1i f Xbelongs to A3andp(X)=−1i fi t does not. Denote by C2the multiplicative cyclic group of order 2. Determine the images of each of the elements of S3for the following four mappings: Φ1:S3→C2 X→p(X) Φ2:S3→C2 X→−p(X) Φ3:S3→A3 X→X2 Φ4:S3→S3 X→X3 (d) For each mapping, determine whether the kernel Kis a subgroup of S3and, if so, whether the mapping is a homomorphism. 24.16 For the group Gwith multiplication table 24.8 and proper subgroup H={I,A,B}, denote the coset {I,A,B}byC1and the coset {C,D,E}byC2.F o r mt h es e to f all possible products of a member of C1with itself, and denote this by C1C1. Similarly compute C2C2,C1C2andC2C1. Show that each product coset is equal to C1or toC2and that a 2 ×2 multiplication table can be formed demonstrating thatC1andC2are themselves the elements of a group of order 2. A subgroup likeHwhose cosets themselves form a group is a normal subgroup . 24.17 The group of all non-singular n×nmatrices is known as the general linear group GL(n) and that with only real elements as GL(n,R). IfR∗denotes the multiplicative group of non-zero real numbers, prove that the mapping Φ :GL(n,R)→R ∗, deﬁned by Φ( M)=d e t M, is a homomorphism. Show that the kernel Kof Φ is a subgroup of GL(n,R). Determine its cosets and show that they themselves form a group. 24.18 The group of reﬂection–rotation symmetries of a square is known as D4;l e t Xbe one of its elements. Consider a mapping Φ : D4→S4, the permutation group on four objects, deﬁned by Φ( X) = the permutation induced by Xon 914 24.9 HINTS AND ANSWERS the set{x, y, d, d/prime},w h e r e xand yare the two principal axes and dand d/prime the two principal diagonals, of the square. For example, if Ris a rotation by π/2, Φ( R) = (12)(34). Show that D4is mapped onto a subgroup of S4and, by constructing the multiplication tables for D4and the subgroup, prove that the mapping is a homomorphism. 24.19 Given that matrix Mis a member of the multiplicative group GL(3,R), determine, for each of the following additional constraints on M(applied separately), whether the subset satisfying the constraint is a subgroup of GL(3,R): (a) MT=M; (b) MTM=I; (c)|M|=1 ; (d)Mij=0f o r j>iandMii/negationslash=0 . 24.20 In the quaternion group Qthe elements form the set {1,−1,i ,−i, j,−j,k,−k}, with i2=j2=k2=−1,ij=kand its cyclic permutations, and ji=−kand its cyclic permutations. Find the proper subgroups of Qand the corresponding cosets. Show that the subgroup of order 2 is a normal subgroup, but that theother subgroups are not. Show that Qcannot be isomorphic to the group 4 mm (C 4v) considered in exercise 24.11. 24.21 Show that D4, the group of symmetries of a square, has two isomorphic subgroups of order 4. Show further that there exists a two-to-one homomorphism from thequaternion group Qof exercise 24.20 onto one (and hence either) of these two subgroups, and determine its kernel. 24.22 Show that the matrices M(θ,x,y)= /0/@cosθ−sinθx sinθcosθy 00 1 /1A, where 0≤θ<2π,−∞<x<∞,−∞<y<∞, form a group under matrix multiplication. Show thatthose Mfor which θ= 0 form a subgroup and identify its cosets. Show that the cosets themselves form a group. 24.23 Find (a) all the proper subgroups and (b) all the conjugacy classes of the symmetry group of a regular pentagon. 24.9 Hints and answers 24.1†(a) Yes. (b) no, no inverse for 2. (c) yes. (d) no, 2 ×3 is not in the set. (e) yes. (f) yes, they form a subgroup of order 4, [1 ,0; 0,1] [4,0;0,4] [1,2; 0,4] [4,3;0,1]. (g) yes. 24.2 (a) No, not reﬂexive; (b) yes, partition of integers into odd and even; (c) yes. (d) no, not transitive, X→Y→ZifY’s parents both re-marry and XandZare children of the two second marriages; (e) yes; (f) yes. 24.3 x•(y•z)=x+y+z+r(xy+xz+yz)+r2xyz=(x•y)•z.Show that assuming x•y=−r−1leads to ( rx+1 ) ( ry+1 )=0 .The inverse of xisx−1=−x/(1 +rx); show that this is not equal to −r−1. †Where matrix elements are given as a list, the convention used is [row 1; row 2; ...], individual entries in each row being separated by commas. 915 GROUP THEORY 24.4 The relevant sets of values for [ a, b, c, d ]a r e[ 1 ,0,0,1],[−d, b, c,−a]a n d[ a/primea+ b/primec, a/primeb+b/primed, c/primea+d/primec, c/primeb+d/primed] for reﬂexivity, symmetry and transitivity respec- tively; the rational numbers. 24.5 (a) Consider both X=iandX/negationslash=i. Here, i/negationslash∼i. (b) In this case i∼i, but the conclusion cannot be deduced from the other axioms. In both cases iis in a class by itself and no Y, as used in the false proof, can be found. 24.6†Matrices [1 ,3;0,1] and [2 ,3;0,1] do not commute, so the group is non- Abelian. (a) 12 elements in the set {[1,0;0,1] [4,0; 0,4] [1,b;0,4] [4,b;0,1] with barbitrary}. The full group has order 4 ×4×5 = 80, which is not divisible by 12. (b) [1 ,0;0,4][1,3;0,4] = [1 ,3; 0,1], which has order >2. 24.7†Use|AB|=|A||B|=1×1 = 1 to prove closure. The inverse has w↔z, x↔−x,y↔−y, giving|A−1|= 1, i.e. it is in the set. The only element of order 2i s−I;A2can be simpliﬁed to [ −(w+1 ),−x;−y,−(z+ 1)]. 24.8 Note that if each matrix is written in the form N=(n1,−n∗ 2;n2,n∗ 1)w i t h|n1|2+ |n2|2=1t h e n NQ=P,w h e r e p1=n1q1−n∗ 2q2and p2=n2q1+n∗ 1q2with |p1|2+|p2|2= 1. The inverse of M(a0,a)i sM(a0,−a). Show that M∗TM=I. 24.9 If XY=Z, show that Y=XZandX=ZY,t h e nf o r m YX. Note that the elements of Bcan only have orders 1, 2 or p. Suppose they all have order 1 or 2; then using the earlier result, whilst noting that 4 does not divide 2 p,l e a d st o a contradiction. 24.10 (a) Identity = 1, three rotations of πabout face normals, six rotations of ±π/2 about face normals, six rotations of πabout edge diagonals, eight rotations of ±2π/3 about body diagonals. (b) The ‘vertical’ diagonal can be chosen in 4 ×2 ways (either end of each diagonal can be ‘up’). There are then three equivalentrotational positions about the vertical and thus 4 ×2×3 possibilities altogether. 24.11 Using the notation indicated in ﬁgure 24.3, Rbeing a rotation of π/2 about an axis perpendicular to the square, we have: Ihas order 1; R 2,m1,m2,m3,m4have order 2; R,R3have order 4. m1(π) m2(π) m3(π) m4(π) Figure 24.3 The notation for exercise 24.11. Subgroup{I,R,R2,R3}has cosets{I,R,R2,R3},{m1,m2,m3,m4}; subgroup{I,R2}has cosets{I,R2},{R,R3},{m1,m2},{m3,m4}; subgroup{I,m1}has cosets{I,m1},{R,m 3},{R2,m2},{R3,m4}; subgroup{I,m2}has cosets{I,m2},{R,m 4},{R2,m1},{R3,m3}; subgroup{I,m3}has cosets{I,m3},{R,m 2},{R2,m4},{R3,m1}; subgroup{I,m4}has cosets{I,m4},{R,m 1},{R2,m3},{R3,m2}. 24.12 (a) (i) Each has one element of order 1, one element of order 2 two elements of order 3 and two elements of order 6.(ii) Each has one element of order 1, seven of order 2, two of order 3 andtwo elements of order 6. 916 24.9 HINTS AND ANSWERS (b) No. C8contains elements of order 8; none of the others could. Every element ofC2×C2×C2is of order 1 or 2; the remaining two groups must each contain an element of order 4. D4has one, ﬁve and two elements of order 1, 2 and 4 respectively; C2×C4has correspondingly one, three and four elements. 24.13 G={I,A,B,B2,B3,AB,AB2,AB3}. The proper subgroups are as follows: {I,A};{I,B2},{I,AB2},{I,B,B2,B3},{I,B2,AB,AB3}. 24.14 ( a, b)−1=(a2−pb2)−1(a,−b), which has rational entries with a2/negationslash=pb2since pis prime. 24.15 (b) A3={(1),(123) ,(132)}. (d) For Φ 1,K={(1),(123) ,(132)}is a subgroup. For Φ 2,K={(23),(13),(12)}is not a subgroup because it has no identity element. For Φ 3,K={(1),(23),(13),(12)}is not a subgroup because it is not closed. For Φ 4,K={(1),(123) ,(132)}is a subgroup. Only Φ 1is a homomorphism; Φ 4fails because, for example, [(23)(13)]/prime/negationslash= (23)/prime(13)/prime. 24.16 C1C1=C2C2=C1,C1C2=C2C1=C2. 24.17 Recall that, for any pair of matrices Pand Q,|PQ|=|P||Q|.Kis the set of all matrices with unit determinant. The cosets of Kare the sets of matrices whose determinants are equal; Kitself is the identity in the group of cosets. 24.18 I,R2→(1);R,R3→(12)(34); mx,my→(34); md,md/prime→(12). The multiplication table for the subgroup is that given in table 24.3. 24.19 (a) No, because the set is not closed. (b) yes. (c) yes. (d) yes.24.20 The subgroup {1,−1}has cosets C 1={1,−1},Ci={i,−i},Cj={j,−j}, Ck={k,−k}. The subgroup {1,i ,−1,−i}has cosets Di={1,i ,−1,−i},D/prime i= {j,−j,k,−k}; corresponding pairs of cosets Dj,Dj/primeandDk,Dk/primeare obtained from subgroups {1,j,−1,−j}and{1,k,−1,−k}respectively. They can be written down by cyclically permuting i, j, kinDi,D/prime i. The cosets of {1,−1}form a group withC1as the identity and CiCj=Cketc. The cosets of {1,i ,−1,−i}do not form a group since, for example, the product DiD/prime iinvolves all elements of Q.I ti s suﬃcient to notice that 4 mmhas six elements of order 2, whilst Qhas only two. 24.21 Each subgroup contains the identity, a rotation by π, and two reﬂections. The homomorphism is ±1→I,±i→R2,±j→mx,±k→mywith kernel {1,−1}. 24.22 Closure is shown by M(θ,x,y)M(φ, x/prime,y/prime)=M(θ+φ, X, Y ), where X=x+x/primecosθ−y/primesinθandY=y+y/primecosθ+x/primesinθ. The inverse is given by M(θ, x,y)−1=M(−θ,−xcosθ−ysinθ, xsinθ−ycosθ). All members of any coset Cθhave the same value for θ. Cθ1×Cθ2=Cθ1+θ2(mod 2 π). The inverse coset is C−1 θ=C2π−θ. 24.23 There are 10 elements: I, rotations Ri(i=1,4) and reﬂections mj(j=1,5). (a) Five proper subgroups of order 2, {I,m j}and one of order 5, {I,R,R2,R3,R4}. (b) Four conjugacy classes, {I},{R,R4},{R2,R3},{m1,m2,m3,m4,m5}. 917 25 Representation theory As indicated at the start of the previous chapter, signiﬁcant conclusions can often be drawn about a physical system simply from the study of its symmetry properties. That chapter was devoted to setting up a formal mathematical basis,group theory, with which to describe and classify such properties; the currentchapter shows how to implement the consequences of the resulting classiﬁcationsand obtain concrete physical conclusions about the system under study. Theconnection between the two chapters is akin to that between working withcoordinate-free vectors, each denoted by a single symbol, and working with a coordinate system in which the same vectors are expressed in terms of components. The ‘coordinate systems’ that we will choose will be ones that are expressed in terms of matrices; it will be clear that ordinary numbers would not be suﬃcient,as they make no provision for any non-commutation amongst the elements of a group. Thus, in this chapter the group elements will be represented by matrices that have the same commutation relations as the members of the group,whatever the group’s original nature (symmetry operations, functional forms,matrices, permutations, etc.). For some abstract groups it is diﬃcult to give awritten description of the elements and their properties without recourse to suchrepresentations. Most of our applications will be concerned with representationsof the groups that consist of the symmetry operations on molecules containing two or more identical atoms. Firstly, in section 25.1, we use an elementary example to demonstrate the kind of conclusions that can be reached by arguing purely on symmetry grounds. Then in sections 25.2–25.10 we develop the formal side of representation theory and establish general procedures and results. Finally, these are used in section 25.11to tackle a variety of problems drawn from across the physical sciences. 918 25.1 DIPOLE MOMENTS OF MOLECULES (a)H C l (b)C O 2 (c)O3AB A/primeB/prime Figure 25.1 Three molecules, ( a)h y d r o g e nc h l o r i d e ,( b) carbon dioxide and (c) ozone, for which symmetry considerations impose varying degrees of constraint on their possible electric dipole moments. 25.1 Dipole moments of molecules Some simple consequences of symmetry can be demonstrated by considering whether a permanent electric dipole moment can exist in any particular molecule;three simple molecules, hydrogen chloride, carbon dioxide and ozone, are illus- trated in ﬁgure 25.1. Even if a molecule is electrically neutral, an electric dipole moment will exist in it if the centres of gravity of the positive charges (due toprotons in the atomic nuclei) and of the negative charges (due to the electrons)do not coincide. For hydrogen chloride there is no reason why they should coincide; indeed, the normal picture of the binding mechanism in this molecule is that the electron fromthe hydrogen atom moves its average position from that of its proton nucleus tosomewhere between the hydrogen and chlorine nuclei. There is no compensating movement of positive charge, and a net dipole moment is to be expected – and is found experimentally. For the linear molecule carbon dioxide it seems obvious that it cannot have a dipole moment, because of its symmetry. Putting this rather more rigorously,we note that any rotation about the long axis of the molecule leaves it totallyunchanged; consequently, any component of a permanent electric dipole perpen-dicular to that axis must be zero (a non-zero component would rotate althoughno physical change had taken place in the molecule). That only leaves the pos- sibility of a component parallel to the axis. However, a rotation of πradians about the axis AA /primeshown in ﬁgure 25.1( b) carries the molecule into itself, as does a reﬂection in a plane through the carbon atom and perpendicular to themolecular axis (i.e. one with its normal parallel to the axis). In both cases the twooxygen atoms change places but, as they are identical, the molecule is indistin-guishable from the original. Either ‘symmetry operation’ would reverse the signof any dipole component directed parallel to the molecular axis; this can only be compatible with the indistinguishability of the original and ﬁnal systems if the parallel component is zero. Thus on symmetry grounds carbon dioxide cannothave a permanent electric dipole moment. Finally, for ozone, which is angular rather than linear, symmetry does not 919 REPRESENTATION THEORY place such tight constraints. A dipole-moment component parallel to the axis BB/prime(ﬁgure 25.1( c)) is possible, since there is no symmetry operation that reverses the component in that direction and at the same time carries the molecule intoan indistinguishable copy of itself. However, a dipole moment perpendicular to BB /primeis not possible, since a rotation of πabout BB/primewould both reverse any such component and carry the ozone molecule into itself – two contradictoryconclusions unless the component is zero. In summary, symmetry requirements appear in the form that some or all components of permanent electric dipoles in molecules are forbidden; they do not show that the other components do exist, only that they may. The greaterthe symmetry of the molecule, the tighter the restrictions on potentially non-zerocomponents of its dipole moment. In section 23.11 other, more complicated, physical situations will be analysed using results derived from representation theory. In anticipation of these results,and since it may help the reader to understand where the developments in thenext nine sections are leading, we make here a broad, powerful, but rather formal,statement as follows. If a physical system is such that after the application of particular rotations or reﬂections (or a combination of the two) the ﬁnal system is indistinguishable fromthe original system then its behaviour, and hence the functions that describe itsbehaviour, must have the corresponding property of invariance when subjected tothe same rotations and reﬂections. 25.2 Choosing an appropriate formalism As mentioned in the introduction to this chapter, the elements of a ﬁnite group Gcan be represented by matrices; this is done in the following way. A suitable column matrix u, known as a basis vector ,†is chosen and is written in terms of its components u i,t h ebasis functions ,a s u=(u1u2···un)T.T h e uimay be of a variety of natures, e.g. numbers, coordinates, functions or even a set of labels,though for any one basis vector they will all be of the same kind. Once chosen, the basis vector can be used to generate an n-dimensional rep- resentation of the group as follows. An element Xof the group is selected and its eﬀect on each basis function u iis determined. If the action of Xonu1is to produce u/prime 1, etc. then the set of equations u/prime i=Xui (25.1) generates a new column matrix u/prime=(u/prime 1u/prime2···u/prime n)T. Having established uand u/prime †This usage of the term basis vector is not exactly the same as that introduced in subsection 8.1.1. 920 25.2 CHOOSING AN APPROPRIATE FORMALISM we can determine the n×nmatrix, M(X) say, that connects them by u/prime=M(X)u. (25.2) It may seem natural to use the matrix M(X) so generated as the representative matrix of the element X; in fact, because because we have already chosen the convention whereby Z=XYimplies that the eﬀect of applying element Zis the same as that of ﬁrst applying Yand then applying Xto the result, one further step has to be taken. So that the representative matrices D(X) may follow the same convention, i.e. D(Z)=D(X)D(Y), and at the same time respect the normal rules of matrix multiplication, it is necessary to take the transpose ofM(X) as the representative matrix D(X). Explicitly, D(X)=MT(X) (25.3) and (25.2) becomes u/prime=DT(X)u. (25.4) Thus the procedure for determining the matrix D(X) that represents the group element Xin a representation based on basis vector uis summarised by equations (25.1)–(25.4). † This procedure is then repeated for each element Xof the group, and the resulting set of n×nmatrices D={D(X)}is said to be the n-dimensional representation of Ghaving uas its basis. The need to take the transpose of each matrix M(X) is not of any fundamental signiﬁcance, since the only thing that really matters is whether the matrices D(X) have the appropriate multiplication properties – and, as deﬁned, they do. In cases in which the basis functions are labels, the actions of the group elements are such as to cause rearrangements of the labels. Correspondingly thematrices D(X) contain only ‘1’s and ‘0’s as entries; each row and each column contains a single ‘1’. †An alternative procedure in which a row vector is used as the basis vector is possible. Deﬁning equations of the form uTX=uTD(X) are used, and no additional transpositions are needed to deﬁne the representative matrices. However, row-matrix equations are cumbersome to write out and in all other parts of this book we have conventionally written operators (here the group element) to the left of the object on which they operate (here the basis vector). 921 REPRESENTATION THEORYIFor the group S3of permutations on three objects, which has group multiplication ta- ble 24.8 on p. 897, with (in cycle notation) I= (1)(2)(3) ,A=( 123 ) ,B =( 132 C= (1)(23) ,D = (3)(12) ,E= (2)(13) , use as the components of a basis vector the ordered letter triplets u1={PQR},u 2={QRP},u 3={RPQ}, u4={PRQ},u 5={QPR},u 6={RQP}. Generate a six-dimensional representation D={D(X)}of the group and conﬁrm that the representative matrices multiply according to table 24.8, e.g. D(C)D(B)=D(E). It is immediate that the identity permutation I= (1)(2)(3) leaves all uiunchanged, i.e. u/prime i=uifor all i. The representative matrix D(I) is thus I6,t h e6×6 unit matrix. We next take Xas the permutation A= (12 3) and, using (25.1), let it act on each of the components of the basis vector: u/prime 1=Au1=( 123 ){PQR}={QRP}=u2 u/prime 2=Au2=( 123 ){QRP}={RPQ}=u3 ...... u/prime 6=Au6=( 123 ){RQP}={QPR}=u5. The matrix M(A) has to be such that u/prime=M(A)u(here dots replace zeroes to aid readability): u/prime= /0BBBBB/@u2 u3 u1 u6 u4 u5 /1CCCCCA= /0BBBBB/@·1···· ·· 1··· 1····· ····· 1 ··· 1·· ···· 1· /1CCCCCA /0BBBBB/@u1 u2 u3 u4 u5 u6 /1CCCCCA≡M(A)u. D(A)i st h e ne q u a lt o MT(A). The other D(X) are determined in a similar way. In general, if Xui=uj, then[M(X)]ij= 1, leading to [D(X)]ji=1a n d [D(X)]jk=0f o r k/negationslash=i. For example, Cu3= (1)(23){RPQ}={RQP}=u6 implies that [D(C)]63=1a n d [D(C)]6k=0f o r k=1,2,4,5,6. When calculated in full D(C)= /0BBBBB/@··· 1·· ···· 1· ····· 1 1····· ·1···· ·· 1··· /1CCCCCA, D(B)= /0BBBBB/@·1···· ·· 1··· 1····· ····· 1 ··· 1·· ···· 1· /1CCCCCA, 922 25.2 CHOOSING AN APPROPRIATE FORMALISM (a) (b) (c)P P P Q Q Q R R R1 11 22 2 33 3 Figure 25.2 Diagram ( a) shows the deﬁnition of the basis vector, ( b)s h o w s the eﬀect of applying a clockwise rotation of 2 π/3a n d( c) shows the eﬀect of applying a reﬂection in the mirror axis through Q. D(E)= /0BBBBB/@····· 1 ··· 1·· ···· 1· ·1···· ·· 1··· 1····· /1CCCCCA, from which it can be veriﬁed that D(C)D(B)=D(E). J Whilst a representation obtained in this way necessarily has the same dimension as the order of the group it represents, there are, in general, square matrices ofboth smaller and larger dimensions that can be used to represent the group,though their existence may be less obvious. One possibility that arises when the group elements are symmetry opera- tions on an object whose position and orientation can be referred to a spacecoordinate system is called the natural representation . In it the representative matrices D(X) describe, in terms of a ﬁxed coordinate system, what happens to a coordinate system that moves with the object when Xis applied. There is usually some redundancy of the coordinates used in this type of represen- tation, since interparticle distances are ﬁxed and fewer than 3 Ncoordinates, where Nis the number of identical particles, are needed to specify uniquely the object’s position and orientation. Subsection 25.11.1 gives an example thatillustrates both the advantages and disadvantages of the natural representation.We continue here with an example of a natural representation that has no suchredundancy.IUse the fact that the group considered in the previous worked example is isomorphic to the group of two-dimensional symmetry operati ons on an equilateral triangle to generate a three-dimensional representation of the group. Label the triangle’s corners as 1, 2, 3 and three ﬁxed points in space as P, Q, R, so thatinitially corner 1 lies at point P, 2 lies at point Q, and 3 at point R. We take P, Q, R asthe components of the basis vector. In ﬁgure 25.2, ( a) shows the initial conﬁguration and also, formally, the result of applying the identity Ito the triangle; it is therefore described by the basis vector, (P Q R) T. 923 REPRESENTATION THEORY Diagram ( b) shows the the eﬀect of a clockwise rotation by 2 π/3, corresponding to element Ain the previous example; the new column matrix is (Q R P)T. Diagram ( c) shows the eﬀect of a typical mirror reﬂection – the one that leaves the corner at point Q unchanged (element Din table 24.8 and the previous example); the new column matrix is now (R Q P)T. In similar fashion it can be concluded that the column matrix corresponding to element B,r o t a t i o nb y4 π/3, is (R P Q)T, and that the other two reﬂections CandEresult in column matrices (P R Q)Tand (Q P R)Trespectively. The forms of the representative matrices Mnat(X), (25.2), are now determined by equations such as, for element E,/0/@Q P R /1A= /0 /@ 010 100001 /1A /0/@P QR /1A implying that Dnat(E)= /0/@010 100001 /1AT = /0 /@ 010 100001 /1A. In this way the complete repr esentation is obtained as Dnat(I)= /0/@100 010 001 /1A,Dnat(A)= /0/@001 100 010 /1A,Dnat(B)= /0/@010 001 100 /1A, Dnat(C)= /0/@100 001010 /1A,Dnat(D)= /0/@001 010100 /1A,Dnat(E)= /0/@010 100001 /1A. It should be emphasised that although the group contains six elements this representation is three-dimensional. J We will concentrate on matrix representations of ﬁnite groups, particularly rotation and reﬂection groups (the so-called crystal point groups). The general ideas carry over to inﬁnite groups, such as the continuous rotation groups, but ina book such as this, which aims to cover many areas of applicable mathematics,some topics can only be mentioned and not explored. We now give the formaldeﬁnition of a representation. Deﬁnition. A representation D={D(X)}of a group Gis an assignment of a non- singular square n×nmatrix D(X)to each element Xbelonging to G, such that (i)D(I)=I n,the unit n×nmatrix , (ii)D(X)D(Y)=D(XY)for any two elements XandYbelonging to G,i . e .t h e matrices multiply in the same way as the group elements they represent. As mentioned previously, a representation by n×nmatrices is said to be an n-dimensional representation ofG. The dimension nis not to be confused with g, the order of the group, which gives the number of matrices needed in the representation, though they might not all be diﬀerent. A consequence of the two deﬁning conditions for a representation is that the 924 25.2 CHOOSING AN APPROPRIATE FORMALISM matrix associated with the inverse of Xis the inverse of the matrix associated with X. This follows immediately from setting Y=X−1in (ii): D(X)D(X−1)=D(XX−1)=D(I)=In; hence D(X−1)=[D(X)]−1. As an example, the four-element Abelian group that consists of the set {1,i ,−1,−i} under ordinary multiplication has a two-dimensional representation based on thecolumn matrix (1i) T: D(1) =parenleftbigg10 01parenrightbigg , D(i)=parenleftbigg0−1 10parenrightbigg , D(−1) =parenleftbigg−10 0−1parenrightbigg ,D(−i)=parenleftbigg01 −10parenrightbigg . The reader should check that D(i)D(−i)= D(1), D(i)D(i)= D(−1) etc., i.e. that the matrices do have exactly the same multiplication properties as the elementsof the group. Having done so, the reader may also wonder why anybody wouldbother with the representative matrices, when the original elements are so muchsimpler to handle! As we will see later, once some general properties of matrixrepresentations have been established, the analysis of large groups, both Abelian and non-Abelian, can be reduced to routine, almost cookbook, procedures. Ann-dimensional representation of Gis a homomorphism of Ginto the set of invertible n×nmatrices (i.e. n×nmatrices that have inverses or, equivalently, have non-zero determinants); this set is usually known as the general linear groupand denoted by GL( n). In general the same matrix may represent more than one element of G; if, however, all the matrices representing the elements of Gare diﬀerent then the representation is said to be faithful , and the homomorphism becomes an isomorphism onto a subgroup of GL( n). A trivial but important representation is D(X)= I nfor all elements XofG. Clearly both of the deﬁning relationships are satisﬁed, and there is no restriction on the value of n. However, such a representation is not a faithful one. To sum up, in the context of a rotation–reﬂection group, the transposes of the set of n×nmatrices D(X)t h a tm a k eu par e p r e s e n t a t i o n Dmay be thought of as describing what happens to an n-component basis vector of coordinates, (xy ···)T, or of functions, (Ψ 1Ψ2···)T,t h eΨ ithemselves being functions of coordinates, when the group operation Xis carried out on each of the coordinates or functions. For example, to return to the symmetry operationson an equilateral triangle, the clockwise rotation by 2 π/3,R, carries the three- 925 REPRESENTATION THEORY dimensional basis vector ( xyz )Tinto the column matrix  −1 2x+√ 3 2y −√ 3 2x−1 2y z  whilst the two-dimensional basis vector of functions ( r23z2−r2)Tis unaltered, as neither rnorzis changed by the rotation. The fact that zis unchanged by any of the operations of the group shows that the components x,y,zactually divide (i.e. are ‘reducible’, to anticipate a more formal description) into two sets:one comprises z, which is unchanged by any of the operations, and the other comprises x,y, which change as a pair into linear combinations of themselves. This is an important observation to which we return in section 25.4. 25.3 Equivalent representations IfDis an n-dimensional representation of a group G,a n d Qis any ﬁxed invert- iblen×nmatrix (|Q|/negationslash= 0), then the set of matrices deﬁned by the similarity transformation D Q(X)=Q−1D(X)Q (25.5) also forms a representation DQofG,s a i dt ob e equivalent toD.W ec a ns e ef r o ma comparison with the deﬁnition in section 25.2 that they do form a representation: (i)DQ(I)=Q−1D(I)Q=Q−1InQ=In, (ii)DQ(X)DQ(Y)=Q−1D(X)QQ−1D(Y)Q=Q−1D(X)D(Y)Q =Q−1D(XY)Q=DQ(XY). Since we can always transform between equivalent representations using a non- singular matrix Q, we will consider such representations to be one and the same. Despite the similarity of words and mani pulations to those of subsection 24.7.1, that two representations are equivalent does not constitute an ‘equivalence re-lation’ – for example, the reﬂexive property does not hold for a general ﬁxedmatrix Q. However, if Qwere not ﬁxed, but simply restricted to belonging to a set of matrices that themselves form a group, then (25.5) would constitute anequivalence relation. The general invertible matrix Qthat appears in the deﬁnition (25.5) of equiv- alent matrices describes changes arising from a change in the coordinate system (i.e. in the set of basis functions). As before, suppose that the eﬀect of an opera- tionXon the basis functions is expressed by the action of M(X)( w h i c hi se q u a l toD T(X)) on the corresponding basis vector: u/prime=M(X)u=DT(X)u. (25.6) 926 25.3 EQUIVALENT REPRESENTATIONS A change of basis would be given by uQ=Quand u/prime Q=Qu/prime, and we may write u/prime Q=Qu/prime=QM(X)u=QDT(X)Q−1uQ. (25.7) This is of the same form as (25.6), i.e. u/prime Q=DT QT(X)uQ, (25.8) where DQT(X)=( QT)−1D(X)QTis related to D(X) by a similarity transforma- tion. Thus DQT(X) represents the same linear transformation as D(X), but with respect to a new basis vector uQ; this supports our contention that representa- tions connected by similarity transformations should be considered as the same representation.IFor the four-element Abelian group consisting of the set {1,i ,−1,−i}under ordinary multiplication, discussed near the end of section 25.2, change the basis vector from u= (1i)TtouQ=( 3−i2i−5)T. Find the real transformation matrix Q. Show that the transformed representative matrix for element i,DQT(i), is given by DQT(i)= / 17−29 10−17 / and verify that DT QT(i)uQ=iuQ. Firstly, we solve the matrix equation/ 3−i 2i−5 / = / ab cd // 1 i / , with a, b, c, d real. This gives Qand hence Q−1as Q= / 3−1 −52 / , Q−1= / 21 53 / . Following (25.7) we now ﬁnd the transpose of DQT(i)a s QDT(i)Q−1= / 3−1 −52 // 01 −10 // 21 53 / = / 17 10 −29−17 / and hence DQT(i) is as stated. Finally, DT QT(i)uQ= / 17 10 −29−17 // 3−i 2i−5 / = / 1+3 i −2−5i / =i / 3−i 2i−5 / =iuQ, as required. J Although we will not prove it, it can be shown that any ﬁnite representation of a ﬁnite group of linear transformations that preserve spatial length (or, inquantum mechanics, preserve the magnitude of a wavefunction) is equivalent to 927 REPRESENTATION THEORY a representation in which all the matrices are unitary (see chapter 8) and so from now on we will consider only unitary representations . 25.4 Reducibility of a representation We have seen already that it is possible to have more than one representation of any particular group. For example, the group {1,i ,−1,−i}under ordinary multiplication has been shown to have a set of 2 ×2 matrices, and a set of four unitn×nmatrices In, as two of its possible representations. Consider two or more representations, D(1),D(2), ..., D(N), which may be of diﬀerent dimensions, of a group G. Now combine the matrices D(1)(X), D(2)(X), ..., D(N)(X) that correspond to element XofGinto a larger block- diagonal matrix: D(2)(X) D(N)(X). . .D(X) = 00 D(1)(X) (25.9) Then D={D(X)}is the matrix representation of the group obtained by combining the basis vectors of D(1),D(2), ..., D(N)into one larger basis vector. If, knowingly or unknowingly, we had started with this larger basis vector and found the matricesof the representation Dto have the form shown in (25.9), or to have a form that can be transformed into this by a similarity transformation (25.5) (using,of course, the samematrix Qfor each of the matrices D(X)) then we would say that Disreducible and that each matrix D(X) can be written as the direct sum of smaller representations: D(X)=D (1)(X)⊕D(2)(X)⊕···⊕D(N)(X). It may be that some or all of the matrices D(1)(X),D(2)(X), ..., D(N)themselves can be further reduced – i.e. written in block diagonal form. For example,suppose that the representation D (1), say, has a basis vector ( xyz )T; then, for the symmetry group of an equilateral triangle, whilst xandyare mixed together for at least one of the operations X,zis never changed. In this case the 3 ×3 representative matrix D(1)(X) can itself be written in block diagonal form as a 928 25.4 REDUCIBILITY OF A REPRESENTATION 2×2m a t r i xa n da1 ×1 matrix. The direct-sum matrix D(X) can now be written a b cd 1 D(2)(X) D(N)(X). . .D (X) = 00 (25.10) but the ﬁrst two blocks can be reduced no further. When all the other representations D(2)(X),...have been similarly treated, w h a tr e m a i n si ss a i dt ob e irreducible and has the characteristic of being block diagonal, with blocks that individually cannot be reduced further. The blocks are known as the irreducible representations of G, often abbreviated to the irreps of G, and we denote them by ˆD(i). They form the building blocks of representation theory, and it is their properties that are used to analyse any given physicalsituation which is invariant under the operations that form the elements of G. Any representation can be written as a linear combination of irreps. If, however, the initial choice uof basis vector for the representation Dis arbitrary, as it is in general, then it is unlikely that the matrices D(X) will assume obviously block diagonal forms (it should be noted, though, that sincethe matrices are square, even a matrix with non-zero entries only in the extremetop right and bottom left positions is technically block diagonal). In general, itwill be possible to reduce them to block diagonal matrices with more than oneblock; this reduction corresponds to a transformation Qto a new basis vector u Q, as described in section 25.3. In any particular representation D, each constituent irrep ˆD(i)may appear any number of times, or not at all, subject to the obvious restriction that the sum ofall the irrep dimensions must add up to the dimension of Ditself. Let us say that ˆD(i)appears mitimes. The general expansion of Dis then written D=m1ˆD(1)⊕m2ˆD(2)⊕···⊕mNˆD(N), (25.11) where if Gis ﬁnite so is N. This is such an important result that we shall now restate the situation in somewhat diﬀerent language. When the set of matrices that forms a representation 929 REPRESENTATION THEORY of a particular group of symmetry operations has been brought to irreducible form, the implications are as follows. (i) Those components of the basis vector that correspond to rows in the representation matrices with a single-entry block, i.e. a 1 ×1b l o c k ,a r e unchanged by the operations of the group. Such a coordinate or functionis said to transform according to a one-dimensional irrep of G.I nt h e example given in (25.10), that the entry on the third row forms a 1 ×1 block implies that the third entry in the basis vector ( xyz ···) T, namely z, is invariant under the two-dimensional symmetry operations on an equilateral triangle in the xy-plane. (ii) If, in any of the gmatrices of the representation, the largest-sized block located on the row or column corresponding to a particular coordinate(or function) in the basis vector is n×n, then that coordinate (or function) is mixed by the symmetry operations with n−1 others and is said to transform according to an n-dimensional irrep of G. Thus in the matrix (25.10), xis the ﬁrst entry in the complete basis vector; the ﬁrst row of the matrix contains two non-zero entries, as does the ﬁrst column, and soxis part of a two-component basis vector whose components are mixed by the symmetry operations of G. The other component is y. The result (25.11) may also be formulated in terms of the more abstract notion of vector spaces (chapter 8). The set of gmatrices that forms an n-dimensional representation Dof the group Gcan be thought of as acting on column matrices corresponding to vectors in an n-dimensional vector space Vspanned by the basis functions of the representation. If there exists a proper subspace WofV,s u c h that if a vector whose column matrix is wbelongs to Wthen the vector whose column matrix is D(X)walso belongs to W,f o ra l l Xbelonging to G,t h e ni t follows that Dis reducible. We say that the subspace Wis invariant under the actions of the elements of G. With Dunitary, the orthogonal complement W ⊥of W,i . e .t h ev e c t o rs p a c e Vremaining when the subspace Whas been removed, is also invariant, and all the matrices D(X) split into two blocks acting separately onWandW⊥.B o t h WandW⊥may contain further invariant subspaces and be split still further. As a concrete example of this approach, consider in plane polar coordinates ρ, φ, the eﬀect of rotations about the polar axis on the inﬁnite-dimensional vector space Vof all functions of φthat satisfy the Dirichlet conditions for expansion as a Fourier series (see section 12.1). We take as our basis functions the set{sinmφ,cosmφ}for integer values m=0,1,2,...; this is an inﬁnite-dimensional representation ( n=∞) and, since a rotation about the polar axis can be through any angle α(0≤α<2π), the group Gis a subgroup of the continuous rotation group and has its order gformally equal to inﬁnity. 930 25.4 REDUCIBILITY OF A REPRESENTATION Now, for some k, consider a vector win the space Wkspanned by{sinkφ,coskφ}, say w=asinkφ+bcoskφ. Under a rotation by αabout the polar axis, asinkφ becomes asink(φ+α), which can be written as acoskαsinkφ+asinkαcoskφ,i . e as a linear combination of sin kφand cos kφ; similarly cos kφbecomes another linear combination of the same two functions. Thus w/prime=D(α)walso belongs toWkfor any αand we can conclude that Wkis an invariant irreducible two- dimensional subspace of V. It follows that D(α) is reducible and that, since the result holds for every k, in its reduced form D(α) has an inﬁnite series of identical 2×2 blocks on its leading diagonal; each block will have the form parenleftbiggcosα−sinα sinαcosαparenrightbigg . We note that the particular case k= 0 is special, in that then sin kφ=0a n d coskφ=1 ,f o ra l l φ; consequently the ﬁrst 2 ×2 block in D(α) is reducible further and becomes two single-entry blocks. A second illustration of the connection between the behaviour of vector spaces under the actions of the elements of a group and the form of the matrix repre-sentation of the group is provided by the vector space spanned by the sphericalharmonics Y /lscriptm(θ,φ). This contains subspaces, corresponding to the diﬀerent values of /lscript, that are invariant under the actions of the elements of the full three- dimensional rotation group; the corresponding matrices are block-diagonal, andthose entries that correspond to the part of the basis containing Y /lscriptm(θ,φ)f o r ma (2/lscript+1 )×(2/lscript+ 1) block. To illustrate further the irreps of a group, we return again to the group Gof two-dimensional rotation and reﬂection symmetries of an equilateral triangle, orequivalently the permutation group S 3; this may be shown, using the methods of section 25.7 below, to have three irreps. Firstly, we have already seen that the setMof six orthogonal 2 ×2 matrices given in section (24.3), equation (24.13), is isomorphic to G. These matrices therefore form not only a representation of G, but a faithful one. It should be noticed that, although Gcontains six elements, the matrices are only 2 ×2. However, they contain no invariant 1 ×1 sub-block (which for 2 ×2 matrices would require them all to be diagonal) and neither can allthe matrices be made block diagonal by the samesimilarity transformation; they therefore form a two-dimensional irrep of G. Secondly, as previously noted, every group has one (unfaithful) irrep in which every element is represented by the 1 ×1m a t r i x I 1, or, more simply, 1. Thirdly an (unfaithful) irrep of Gis given by assignment of the one-dimensional set of six ‘matrices’ {1,1,1,−1,−1,−1}to the symmetry operations {I,R,R/prime,K, L, M}respectively, or to the group elements {I,A,B,C,D,E }respectively; see section 24.3. In terms of the permutation group S3, 1 corresponds to even permutations and −1 to odd permutations, ‘odd’ or ‘even’ referring to the number of simple pair interchanges to which a permutation is equivalent. That these 931 REPRESENTATION THEORY assignments are in accord with the group multiplication table 24.8 should be checked. Thus the three irreps of the group G(i.e. the group 3 morC3vorS3), are, using the conventional notation A 1,A2, E (see section 25.8), as follows: Element IABCDE A11111 1 1 Irrep A 2111 −1−1−1 E MIMAMBMCMDME(25.12) where MI=parenleftBigg 10 01parenrightBigg , MA=parenleftBigg −1 2√ 3 2 −√ 3 2−1 2parenrightBigg , MB=parenleftBigg −1 2−√ 3 2√ 3 2−1 2parenrightBigg , MC=parenleftBigg −10 01parenrightBigg ,MD=parenleftBigg 1 2−√ 3 2 −√ 3 2−1 2parenrightBigg ,ME=parenleftBigg 1 2√ 3 2√ 3 2−1 2parenrightBigg . 25.5 The orthogonality theorem for irreducible representations We come now to the central theorem of representation theory, a theorem that justiﬁes the relatively routine application of certain procedures to determinethe restrictions that are inherent in physical systems that have some degree ofrotational or reﬂection symmetry. The development of the theorem is long andquite complex when presented in its entirety, and the reader will have to referelsewhere for the proof. † The theorem states that, in a certain sense, the irreps of a group Gare as orthogonal as possible, as follows. If, for each irrep, the elements in any one p o s i t i o ni ne a c ho ft h e gm a t r i c e sa r eu s e dt om a k eu p g-component column matrices then (i) any two such column matrices coming from diﬀerent irreps are orthogonal; (ii) any two such column matrices coming from diﬀerent positions in the matrices of the same irrep are orthogonal. This orthogonality is in addition to the irreps’ being in the form of orthogo- nal (unitary) matrices and thus each comprising mutually orthogonal rows andcolumns. †See, e.g., Groups, Representation and Physics , H.F. Jones (Institute of Physics), Group Theory in Quantum Mechanics , J. F. Cornwell (Academic Press), or Linear Representations of Finite Groups , J. P. Sore (Springer-Verlag). 932 25.5 ORTHOGONALITY THEOREM FOR IRREDUCIBLE REPRESENTATIONS More mathematically, if we denote the entry in the ith row and jth column of a matrix D(X)b y[D(X)]ij,a n d ˆD(λ)andˆD(µ)are two irreps of Ghaving dimensions nλandnµrespectively, then summationdisplay XbracketleftBig ˆD(λ)(X)bracketrightBig∗ ijbracketleftBig ˆD(µ)(X)bracketrightBig kl=g nλδikδjlδλµ. (25.13) This rather forbidding-looking equation needs some further explanation. Firstly, the asterisk indicates that the complex conjugate should be taken if necessary, though all our representations so far have involved only real matrixelements. Each Kronecker delta function on the right-hand side has the value 1if its two subscripts are equal and has the value 0 otherwise. Thus the right-handside is only non-zero if i=k,j=landλ=µ, all at the same time. Secondly, the summation over the group elements Xmeans that gcontributions have to be added together, each contribution being a product of entries drawn from the representative matrices in the two irreps ˆD (λ)={ˆD(λ)(X)}andˆD(µ)= {ˆD(µ)(X)}.T h e gcontributions arise as Xruns over the gelements of G. Thus, putting these remarks together, the summation will produce zero if either (i) the matrix elements are not taken from exactly the same position in every matrix, including cases in which it is not possible to do so because the irreps ˆD(λ)andˆD(µ)have diﬀerent dimensions, or (ii) even if ˆD(λ)andˆD(µ)do have the same dimensions and the matrix elements are from the same positions in every matrix, they are diﬀerent irreps, i.e.λ/negationslash=µ. Some numerical illustrations based on the irreps A 1,A2and E of the group 3 m (orC3vorS3) will probably provide the clearest explanation (see (25.12)). (a) Take i=j=k=l= 1, with ˆD(λ)=A 1andˆD(µ)=A 2. Equation (25.13) then reads 1(1) + 1(1) + 1(1) + 1( −1) + 1(−1) + 1(−1) = 0 , as expected, since λ/negationslash=µ. (b) Take ( i, j)a s( 1 ,2) and ( k,l)a s( 2 ,2), corresponding to diﬀerent matrix positions within the same irrep ˆD(λ)=ˆD(µ)= E. Substituting in (25.13) gives 0(1) +parenleftBig −√ 3 2parenrightBigparenleftbig −1 2parenrightbig +parenleftBig√ 3 2parenrightBigparenleftbig −1 2parenrightbig +0 ( 1 )+parenleftBig −√ 3 2parenrightBigparenleftbig −1 2parenrightbig +parenleftBig√ 3 2parenrightBigparenleftbig −1 2parenrightbig =0. (c) Take ( i, j)a s( 1 ,2), and ( k,l)a s( 1 ,2), corresponding to the same matrix positions within the same irrep ˆD(λ)=ˆD(µ)= E. Substituting in (25.13) gives 0(0)+parenleftBig −√ 3 2parenrightBigparenleftBig −√ 3 2parenrightBig +parenleftBig√ 3 2parenrightBigparenleftBig√ 3 2parenrightBig +0(0)+parenleftBig −√ 3 2parenrightBigparenleftBig −√ 3 2parenrightBig +parenleftBig√ 3 2parenrightBigparenleftBig√ 3 2parenrightBig =6 2. 933 REPRESENTATION THEORY (d) No explicit calculation is needed to see that if i=j=k=l= 1, with ˆD(λ)=ˆD(µ)=A 1(or A 2), then each term in the sum is either 12or (−1)2 and the total is 6, as predicted by the right-hand side of (25.13) since g=6 andnλ=1 . 25.6 Characters The actual matrices of general representations and irreps are cumbersome to work with, and they are not unique since there is always the freedom to change the coordinate system, i.e. the components of the basis vector (see section 25.3),and hence the entries in the matrices. However, one thing that does not changefor a matrix under such an equivalence (similarity) transformation – i.e. undera change of basis – is the trace of the matrix. This was shown in chapter 8,but is repeated here. The trace of a matrix Ais the sum of its diagonal ele- ments, TrA= nsummationdisplay i=1Aii or, using the summation convention (section 21.1), simply Aii. Under a similarity transformation, again using the summation convention, [DQ(X)]ii=[Q−1]ij[D(X)]jk[Q]ki =[D(X)]jk[Q]ki[Q−1]ij =[D(X)]jk[I]kj =[D(X)]jj, showing that the traces of equivalent matrices are equal. This fact can be used to greatly simplify work with representations, though with some partial loss of the information content of the full matrices. For example, using trace values alone it is not possible to distinguish between the two groupsknown as 4 mmand¯42m,o ra s C 4vandD2drespectively, even though the two groups are not isomorphic. To make use of these simpliﬁcations we now deﬁnethe characters of a representation. Deﬁnition. Thecharacters χ(D)of a representation Dof a group Gare deﬁned as the set of traces of the matrices D(X), one for each element XofG. At this stage there will be gcharacters, but, as we noted in subsection 24.7.3, elements A,BofGin the same conjugacy class are connected by equations of the form B=X −1AX. It follows that their matrix representations are connected by corresponding equations of the form D(B)=D(X−1)D(A)D(X) ,a n ds ob yt h e argument just given their representations will have equal traces and hence equalcharacters. Thus elements in the same conjugacy class have the same characters , 934 25.6 CHARACTERS 3mIA ,BC ,D,E A111 1 z;z2;x2+y2 A211 −1 Rz E2−10 (x, y); (xz, yz); (Rx,Ry); (x2−y2,2xy) Table 25.1 The character table for the irreps of group 3 m(C3vorS3). The right-hand column lists some common functions that transform according to the irrep against which each is shown (see text). though, in general, these will vary from one representation to another. However, it might also happen that two or more conjugacy classes have the same characters in a representation – indeed, in the trivial irrep A 1, see (25.12), every element inevitably has the character 1. For the irrep A 2of the group 3 m, the classes {I},{A, B}and{C,D,E}have characters 1, 1 and −1, respectively, whilst they have characters 2, −1a n d0 respectively in irrep E. We are thus able to draw up a character table for the group 3 mas shown in table 25.1. This table holds in compact form most of the important infor- mation on the behaviour of functions under the two-dimensional rotational andreﬂection symmetries of an equilateral triangle, i.e. under the elements of group3m. The entry under Ifor any irrep gives the dimension of the irrep, since it is equal to the trace of the unit matrix whose dimension is equal to that ofthe irrep. In other words, for the λth irrep χ (λ)(I)=nλ,w h e r e nλis its dimen- sion. In the extreme right-hand column we list some common functions of Cartesian coordinates that transform, under the group 3 m, according to the irrep on whose line they are listed. Thus, as we have seen, z,z2,a n d x2+y2are all unchanged by the group operations (though xandyindividually are aﬀected) and so are listed against the one-dimensional irrep A 1. Each of the pairs ( x, y), (xz, yz), and (x2−y2,2xy), however, is mixed as a pair by some of the operations, and so these pairs are listed against the two-dimensional irrep E: each pair forms a basis forthis irrep. The quantities R x,Ryand Rzrefer to rotations about the indicated axes; they transform in the same way as the corresponding components of angularmomentum J, and their behaviour can be established by examining how the components of J=r×ptransform under the operations of the group. To do this explicitly is beyond the scope of this book. However, it can be noted that R z, being listed opposite the one-dimensional A 2, is unchanged by Iand by the rotations AandBbut changes sign under the mirror reﬂections C,D,a n d E,a s would be expected. 935 REPRESENTATION THEORY 25.6.1 Orthogonality property of characters Some of the most important properties of characters can be deduced from the orthogonality theorem (25.13), summationdisplay XbracketleftBig ˆD(λ)(X)bracketrightBig∗ ijbracketleftBig ˆD(µ)(X)bracketrightBig kl=g nλδikδjlδλµ. If we set j=iandl=k,s ot h a tb o t hf a c t o r si na n yp a r t i c u l a rt e r mi nt h e summation refer to diagonal elements of the representative matrices, and then sum both sides over iandk,w eo b t a i n summationdisplay Xnλsummationdisplay i=1nµsummationdisplay k=1bracketleftBig ˆD(λ)(X)bracketrightBig∗ iibracketleftBig ˆD(µ)(X)bracketrightBig kk=g nλnλsummationdisplay i=1nµsummationdisplay k=1δikδikδλµ. Expressed in term of characters, this reads summationdisplay Xbracketleftbig χ(λ)(X)bracketrightbig∗χ(µ)(X)=g nλnλsummationdisplay i=1δ2 iiδλµ=g nλnλsummationdisplay i=11×δλµ=gδλµ. (25.14) In words, the ( g-component) ‘vectors’ formed from the characters of the various irreps of a group are mutually orthogonal, but each one has a squared magnitude(the sum of the squares of its components) equal to the order of the group. Since, as noted in the previous subsection, group elements in the same class have the same characters, (25.14) can be written as a sum over classes rather than elements. If c idenotes the number of elements in class CiandXiany element of Ci,t h e n summationdisplay icibracketleftbig χ(λ)(Xi)bracketrightbig∗χ(µ)(Xi)=gδλµ. (25.15) Although we do not prove it here, there also exists a ‘completeness’ relation for characters. It makes a statement about the products of characters for a ﬁxed pair of group elements, X1andX2, when the products are summed over all possible irreps of the group. This is the converse of the summation process deﬁned by(25.14). The completeness relation states that summationdisplay λbracketleftbig χ(λ)(X1)bracketrightbig∗χ(λ)(X2)=g c1δC1C2, (25.16) where element X1belongs to conjugacy class C1andX2belongs to C2. Thus the sum is zero unless X1andX2belong to the same class. For table 25.1 we can verify that these results are valid. (i) For ˆD(λ)=ˆD(µ)=A 1or A 2, (25.15) reads 1(1) + 2(1) + 3(1) = 6 , 936 25.7 COUNTING IRREPS USING CHARACTERS whilst for ˆD(λ)=ˆD(µ)=E ,i tg i v e s 1(22) + 2(1) + 3(0) = 6 . (ii) For ˆD(λ)=A 2andˆD(µ)= E, say, (25.15) reads 1(1)(2) + 2(1)( −1) + 3(−1)(0) = 0 . (iii) For X1=AandX2=D, say, (25.16) reads 1(1) + 1(−1) + (−1)(0) = 0 , whilst for X1=CandX2=E, both of which belong to class C3for which c3=3 , 1(1) + (−1)(−1) + (0)(0) = 2 =6 3. 25.7 Counting irreps using characters The expression of a general representation D={D(X)}in terms of irreps, as given in (25.11), can be simpliﬁed by going from the full matrix form to that ofcharacters. Thus D(X)=m 1ˆD(1)(X)⊕m2ˆD(2)(X)⊕···⊕mNˆD(N)(X) becomes, on taking the trace of both sides, χ(X)=Nsummationdisplay λ=1mλχ(λ)(X). (25.17) Given the characters of the irreps of the group Gto which the elements Xbelong, and the characters of the representation D={D(X)},t h e gequations (25.17) can be solved as simultaneous equations in the mλ, either by inspection or by multiplying both sides bybracketleftbig χ(µ)(X)bracketrightbig∗and summing over X, making use of (25.14) and (25.15), to obtain mµ=1 gsummationdisplay Xbracketleftbig χ(µ)(X)bracketrightbig∗χ(X)=1 gsummationdisplay icibracketleftbig χ(µ)(Xi)bracketrightbig∗χ(Xi). (25.18) That an unambiguous formula can be given for each mλ, once the character set(the set of characters of each of the group elements or, equivalently, of each of the conjugacy classes) of Dis known, shows that, for any particular group, two representations with the same characters are equivalent. This stronglysuggests something that can be shown, namely, the number of irreps = the number of conjugacy classes. The argument is as follows. Equation (25.17) is a set of simultaneous equations for Nunknowns, the m λ, some of which may be zero. The value of Nis equal to the number of irreps of G.T h e r ea r e gdiﬀerent values of X, but the number of diﬀerent equations is only equal to the number of distinct 937 REPRESENTATION THEORY conjugacy classes, since any two elements of Gin the same class have the same character set and therefore generate the same equation. For a unique solutionto simultaneous equations in Nunknowns, exactly Nindependent equations are needed. Thus Nis also the number of classes, establishing the stated result.IDetermine the irreps contained in the representation of the group 3min the vector space spanned by the functions x2,y2,xy. We ﬁrst note that although these functions are not orthogonal they form a basis set for a representation, since they are linearly independent quadratic forms in xandyand any other quadratic form can be written (uniquely) in terms of them. We must establish how they transform under the symmetry operations of group 3 m. We need to do so only for a repre- sentative element of each conjugacy class, and naturally we take the simplest in each case. T h eﬁ r s tc l a s sc o n t a i n so n l y I(as always) and clearly D(I)i st h e3×3 unit matrix. The second class contains the rotations, AandB, and we choose to ﬁnd D(A). Since, under A, x→−1 2x+√ 3 2y and y→−√ 3 2x−1 2y, it follows that x2→1 4x2−√ 3 2xy+3 4y2,y2→3 4x2+√ 3 2xy+1 4y2(25.19) and xy→√ 3 4x2−1 2xy−√ 3 4y2. (25.20) Hence D(A) can be deduced and is given below. The third and ﬁnal class contains the reﬂections, C,DandE;o ft h e s e Cis much the easiest to deal with. Under C,x→−xandy→y,c a u s i n g xyto change sign but leaving x2andy2unaltered. The three matrices needed are thus D(I)=I3,D(C)= /0/@10 0 01 0 00−1 /1A,D(A)= /0BB/@1 43 4−√ 3 2 3 41 4√ 3 2√ 3 4−√ 3 4−1 2 /1CCA; their traces are respectively 3, 1 and 0. It should be noticed that much more work has been done here than is necessary, since the traces can be computed immediately from the eﬀects of the symmetry operations on thebasis functions. All that is needed is the weight of each basis function in the transformedexpression for that function; these are clearly 1, 1, 1 for I,a n d 1 4,1 4,−1 2forA, from (25.19) and (25.20), and 1, 1, −1f o r C, from the observations made just above the displayed matrices. The traces are then the sums of these weights. The oﬀ-diagonal elements of thematrices need not be found, nor need the matrices be written out. From (25.17) we now need to ﬁnd a superposition of the characters of the irreps that gives representation Din the bottom line of table 25.2. By inspection it is obvious that D=A 1⊕E, but we can use (25.18) formally: mA1=1 6[1(1)(3) + 2(1)(0) + 3(1)(1)] = 1 , mA2=1 6[1(1)(3) + 2(1)(0) + 3( −1)(1)] = 0 , mE=1 6[1(2)(3) + 2( −1)(0) + 3(0)(1)] = 1 . Thus A 1and E appear once each in the reduction of D,a n dA 2not at all. Table 25.1 gives the further information, not needed here, that it is the combination x2+y2that transforms as a one-dimensional irrep and the pair ( x2−y2,2xy)t h a tf o r m sab a s i so f the two-dimensional irrep, E. J 938 25.7 COUNTING IRREPS USING CHARACTERS Classes Irrep IA BC D E A1 11 1 A2 11 −1 E 2−10 D 30 1 Table 25.2 The characters of the irreps of the group 3 mand of the represen- tation D, which must be a superposition of some of them. 25.7.1 Summation rules for irreps The ﬁrst summation rule for irreps is a simple restatement of (25.14), with µset equal to λ;i tt h e nr e a d s summationdisplay Xbracketleftbig χ(λ)(X)bracketrightbig∗χ(λ)(X)=g. In words, the sum of the squares (modulus squared if necessary) of the characters of an irrep taken over all elements of the group adds up to the order of thegroup. For group 3 m(table 25.1), this takes the following explicit forms: for A 1, 1(12)+2 ( 12)+3 ( 12)=6 ; for A 2, 1(12)+2 ( 12)+3 (−1)2=6 ; for E , 1(22)+2 (−1)2+3 ( 02)=6 . We next prove a theorem that is concerned not with a summation within an irrep but with a summation over irreps. Theorem. Ifnµis the dimension of the µth irrep of a group Gthen summationdisplay µn2 µ=g, where gis the order of the group. Proof. Deﬁne a representation of the group in the following way. Rearrange the rows of the multiplication table of the group so that whilst the elements ina particular order head the columns, their inverses in the same order head therows. In this arrangement of the g×gtable, the leading diagonal is entirely occupied by the identity element. Then, for each element Xof the group, take as representative matrix the multiplication-table array obtained by replacing Xby 1 and all other element symbols by 0. The matrices D reg(X) so obtained form the regular representation ofG;t h e ya r ee a c h g×g, have a single non-zero entry ‘1’ in each row and column and (as will be veriﬁed by a little experimentation) have 939 REPRESENTATION THEORY (a)IA B IIA B AAB I BBIA(b)IA B IIA B BBIA AAB I Table 25.3 ( a) The multiplication table of the cyclic group of order 3, and (b) its reordering used to generate the regular representation of the group. the same multiplication structure as the group Gitself, i.e. they form a faithful representation of G. Although not part of the proof, a simple example may help to make these ideas more transparent. Consider the cyclic group of order 3. Its multiplicationtable is shown in table 25.3( a) (a repeat of table 24.10( a) of the previous chapter), whilst table 25.3( b) shows the same table reordered so that the columns are still labelled in the order I,A,Bbut the rows are now labelled in the order I −1=I, A−1=B, B−1=A. The three matrices of the regular representation are then Dreg(I)= 100 010001 ,D reg(A)= 010 001100 ,D reg(B)= 001 100010 . An alternative, more mathematical, deﬁnition of the regular representation of a group is bracketleftbig D reg(Gk)bracketrightbig ij=braceleftBigg 1i f GkGj=Gi, 0o t h e r w i s e . We now return to the proof. With the construction given, the regular representa- t i o nh a sc h a r a c t e r sa sf o l l o w s : χreg(I)=g, χreg(X)=0 i f X/negationslash=I. We now apply (25.18) to Dregto obtain for the number mµof times that the irrep ˆD(µ)appears in Dreg(see 25.11)) mµ=1 gsummationdisplay Xbracketleftbig χ(µ)(X)bracketrightbig∗χreg(X)=1 gbracketleftbig χ(µ)(I)bracketrightbig∗χreg(I)=1 gnµg=nµ. Thus an irrep ˆD(µ)of dimension nµappears nµtimes in Dreg, and so by counting the total number of basis functions, or by considering χreg(I), we can conclude 940 25.7 COUNTING IRREPS USING CHARACTERS that summationdisplay µn2 µ=g. (25.21) This completes the proof. As before, our standard demonstration group 3 mprovides an illustration. In this case we have seen already that there are two one-dimensional irreps and onetwo-dimensional irrep. This is in accord with (25.21) since 1 2+12+22=6,which is the order gof the group. Another straightforward application of the relation (25.21), to the group with multiplication table 25.3( a), yields immediate results. Since g=3 ,n o n eo fi t s irreps can have dimension 2 or more, as 22= 4 is too large for (25.21) to be satisﬁed. Thus all irreps must be one-dimensional and there must be three ofthem (consistent with the fact that each element is in a class of its own, and thatthere are therefore three classes). The three irreps are the sets of 1 ×1 matrices (numbers) A 1={1,1,1}A2={1,ω,ω2}A∗ 2={1,ω2,ω}, where ω=e x p ( 2 πi/3); since the matrices are 1 ×1, the same set of nine numbers would be, of course, the entries in the character table for the irreps of the group. The fact that the numbers in each irrep are all cube roots of unity is discussed below. As will be noticed, two of these irreps are complex – an unusual occurrencein most applications – and form a complex conjugate pair of one-dimensionalirreps. In practice, they function much as a two-dimensional irrep, but this is tobe ignored for formal purposes such as theorems. A further property of characters can be derived from the fact that all elements in a conjugacy class have the same order. Suppose that the element Xhas order m,i . e .X m=I. This implies for a representation Dof dimension nthat [D(X)]m=In. (25.22) Representations equivalent to Dare generated as before by using similarity transformations of the form DQ(X)=Q−1D(X)Q. In particular, if we choose the columns of Qto be the eigenvectors of D(X) then, as discussed in chapter 8, DQ(X)= λ 10···0 0λ2... ......0 0···0 λn  941 REPRESENTATION THEORY where the λiare the eigenvalues of D(X). Therefore, from (25.22), we have that  λm 10···0 0λm 2... ......0 0···0 λm n = 10 ···0 01... ......0 0···01 . Hence all the eigenvalues λ iaremth roots of unity, and so χ(X), the trace of D(X), is the sum of nof these. In view of the implications of Lagrange’s theorem (section 24.6 and subsection 24.7.1), the only values of mallowed are the divisors of the order gof the group. 25.8 Construction of a character table In order to decompose representations into irreps on a routine basis using characters, it is necessary to have available a character table for the group inquestion. Such a table gives, for each irrep µof the group, the character χ (µ)(X) of the class to which group element Xbelongs. To construct such a table the following properties of a group, established earlier in this chapter, may be used: (i) the number of classes equals the number of irreps; (ii) the ‘vector’ formed by the characters from a given irrep is orthogonal to the ‘vector’ formed by the characters from a diﬀerent irrep; (iii)summationtext µn2 µ=g,w h e r e nµis the dimension of the µth irrep and gis the order of the group; (iv) the identity irrep (one-dimensional with all characters equal to 1) is present for every group; (v)summationtext Xvextendsinglevextendsingleχ(µ)(X)vextendsinglevextendsingle2=g. (vi)χ(µ)(X)i st h es u mo f nµmth roots of unity, where mis the order of X.IConstruct the character table for the group 4mm(orC4v) using the properties of classes, irreps and characters so far established. The group 4 mmis the group of two-dimensional symmetries of a square, namely rotations of 0, π/2,πand 3 π/2 and reﬂections in the mirror planes parallel to the coordinate axes and along the main diagonals. These are illustrated in ﬁgure 25.3. For this group there areeight elements: •the identity, I; •rotations by π/2a n d3 π/2,RandR /prime; •ar o t a t i o nb y π,Q; •four mirror reﬂections mx,my,mdandmd/prime. Requirements (i) to (iv) at the start of this section put tight constraints on the possible character sets, as the following argument shows. The group is non-Abelian (clearly Rm x/negationslash=mxR), and so there are fewer than eight classes, and hence fewer than eight irreps. But requirement (iii), with g= 8, then implies 942 25.8 CONSTRUCTION OF A CHARACTER TABLE mx mymd m/prime d Figure 25.3 The mirror planes associated with 4 mm,t h eg r o u po ft w o - dimensional symmetries of a square. that at least one irrep has dimension 2 or gr eater. However, there can be no irrep with dimension 3 or greater, since 32>8, nor can there be more than one two-dimensional irrep, since 22+22= 8 would rule out a contribution to the sum in (iii) of 12from the identity irrep, and this must be present. Thus the only possibility is one two-dimensionalirrep and, to make the sum in (iii) correct, four one-dimensional irreps. Therefore using (i) we can now deduce that there are ﬁve classes. This same conclusion can be reached by evaluating X −1YXfor every pair of elements in G, as in the description of conjugacy classes given in the previous chapter. However, it is tedious to do so andcertainly much longer than the above. The ﬁve classes are I,Q,{R,R /prime},{mx,my},{md,md/prime}. It is straightforward to show that only IandQcommute with every element of the group, so they are the only elements in classes of their own. Each other class must haveat least 2 members, but, as there are three classes to accommodate 8 −2=6e l e m e n t s , there must be exactly 2 in each class. This does not pair up the remaining 6 elements, butdoes say that the ﬁve classes have 1, 1, 2, 2, and 2 elements. Of course, if we had startedby dividing the group into classes, we would know the number of elements in each classdirectly. We cannot entirely ignore the group structure (though it sometimes happens that the results are independent of the group structure – for example, all non-Abelian groups oforder 8 have the same character table!); thus we need to note in the present case thatm 2 i=Ifori=x, y, d ord/primeand, as can be proved directly, Rm i=miR/primefor the same four values of label i. We also recall that for any pair of elements XandY,D(XY)=D(X)D(Y). We may conclude the following for the one-dimensional irreps. (a) In view of result (vi), χ(mi)=D(mi)=±1. (b) Since R4=I, result (vi) requires that χ(R) is one of 1, i,−1,−i.B u t ,s i n c e D(R)D(mi)=D(mi)D(R/prime), and the D(mi) are just numbers, D(R)=D(R/prime). Further D(R)D(R)=D(R)D(R/prime)=D(RR/prime)=D(I)=1 , and so D(R)=±1= D(R/prime). (c)D(Q)=D(RR)=D(R)D(R)=1 . If we add this to the fact that the characters of the identity irrep A 1are all unity then we can ﬁll in those entries in character table 25.4 shown in bold. Suppose now that the three missing entries in a one-dimensional irrep are p,qandr, where each can only be ±1. Then, allowing for the numbers in each class, orthogonality 943 REPRESENTATION THEORY 4mm IQR ,R/primemx,mymd,md/prime A1 11 1 1 1 A2 11 1−1−1 B1 11−11 −1 B2 11−1−11 E 2−20 0 0 Table 25.4 The character table deduced for the group 4 mm. For an explana- tion of the entries in bold see the text. with the characters of A 1requires that 1(1)(1) + 1(1)(1) + 2(1)( p) + 2(1)( q) + 2(1)( r)=0 . The only possibility is that two of p,q,a n d requal−1 and the other equals +1. This can be achieved in three diﬀerent ways, corresponding to the need to ﬁnd three furtherdiﬀerent one-dimensional irreps. Thus the ﬁrst four lines of entries in character table 25.4can be completed. The ﬁnal line can be completed by requiring it to be orthogonal to theother four. Property (v) has not been used here though it could have replaced part of theargument given.J 25.9 Group nomenclature The nomenclature of published character tables, as we have said before, is erratic and sometimes unfortunate; for example, often Eis used to represent, not only a two-dimensional irrep, but also the identity operation, where we have used I. Thus the symbol Emight appear in both the column and row headings of a table, though with quite diﬀerent meanings in the two cases. In this book we useroman capitals to denote irreps. One-dimensional irreps are regularly denoted by A and B, B being used if a rotation about the principal axis of 2 π/nhas character −1. Here nis the highest integer such that a rotation of 2 π/nis a symmetry operation of the system, and the principal axis is the one about which this occurs. For the group of operations on a square, n= 4, the axis is the perpendicular to the square and the rotation in question is R. The names for the group, 4 mmandC 4v,d e r i v ef r o mt h ef a c t that here nis equal to 4. Similarly, for the operations on an equilateral triangle, n= 3 and the group names are 3 mandC3v, but because the rotation by 2 π/3h a s character +1 in all its one-dimensional irreps (see table 25.1), only A appears inthe irrep list. Two-dimensional irreps are denoted by E, as we have already noted, and three- dimensional irreps by T, although in many cases the symbols are modiﬁed by primes and other alphabetic labels to denote variations in behaviour from one irrep to another in respect of mirror reﬂections and parity inversions. In the studyof molecules, alternative names based on molecular angular momentum properties 944 25.10 PRODUCT REPRESENTATIONS are common. It is beyond the scope of this book to list all these variations, or to give a large selection of character tables; our aim is to demonstrate and justifythe use of those found in the literature speciﬁcally dedicated to crystal physics ormolecular chemistry. Variations in notation are not restricted to the naming of groups and their irreps, but extend to the symbols used to identify a typical element, and henceall members, of a conjugacy class in a group. In physics these are usually of thetypes n z,¯nzormx. The ﬁrst of these denotes a rotation of 2 π/nabout the z-axis, and the second the same thing followed by parity inversion (all vectors rgo to −r), whilst the third indicates a mirror reﬂection in a plane, in this case the plane x=0 . Typical chemistry symbols for classes are NC n,NC2 n,NCx n,NSn,σv,σxy.H e r e the ﬁrst symbol N, where it appears, shows that there are Nelements in the class (a useful feature). The subscript nhas the same meaning as in the physics notation, but σrather than mis used for a mirror reﬂection, subscripts v,dorhor superscripts xy,xzoryzdenoting the various orientations of the relevant mirror planes. Symmetries involving parity inversions are denoted by S; thus Snis the chemistry analogue of ¯n. None of what is said in this and the previous paragraph should be taken as deﬁnitive, but merely as a warning of common variations innomenclature and as an initial guide to corresponding entities. Before using anyset of group character tables, the reader should ensure that he or she understands the precise notation being employed. 25.10 Product representations In quantum mechanical investigations we are often faced with the calculation of what are called matrix elements. These normally take the form of integrals over allspace of the product of two or more functions whose analytic forms depend on themicroscopic properties (usually angular momentum and its components) of the electrons or nuclei involved. For ‘bonding’ calculations involving ‘overlap integrals’ there are usually two functions involved, whilst for transition probabilities a thirdfunction, giving the spatial variation of the interaction Hamiltonian, also appearsunder the integral sign. If the environment of the microscopic system under investigation has some symmetry properties, then sometimes these can be used to establish, without detailed evaluation, that the multiple integral must have zero value. We nowexpress the essential content of these ideas in group theoretical language. Suppose we are given an integral of the form J=integraldisplay Ψφd τ or J=integraldisplay Ψξφdτ to be evaluated over all space in a situation in which the physical system is 945 REPRESENTATION THEORY invariant under a particular group Gof symmetry operations. For the integral to be non-zero the integrand must be invariant under each of these operations. Ingroup theoretical language, the integrand must transform as the identity, the one- dimensional representation A 1ofG; more accurately, some non-vanishing part of the integrand must do so. An alternative way of saying this is that if under the symmetry operations ofGthe integrand transforms according to a representation Dand Ddoes not contain A 1amongst its irreps then the integral Jis necessarily zero. It should be noted that the converse is not true; Jm a yb ez e r oe v e ni fA 1is present, since the integral, whilst showing the required invariance, may still have the value zero. It is evident that we need to establish how to ﬁnd the irreps that go to make up a representation of a double or triple product when we already know theirreps according to which the factors in the product transform. The method isestablished by the following theorem. Theorem. For each element of a group the character in a product representation is the product of the corresponding characters in the separate representations. Proof. Suppose that {u i}and{vj}are two sets of basis functions, that transform under the operations of a group Gaccording to representations D(λ)and D(µ) respectively. Denote by uand vthe corresponding basis vectors and let Xbe an element of the group. Then the functions generated from uiandvjby the action ofXare calculated as follows, using (25.1) and (25.4): u/prime i=Xui=bracketleftBigparenleftbig D(λ)(X)parenrightbigTubracketrightBig i=bracketleftbig D(λ)(X)bracketrightbig iiui+summationdisplay l/negationslash=ibracketleftBigparenleftbig D(λ)(X)parenrightbigTbracketrightBig ilul, v/prime j=Xvj=bracketleftBigparenleftbig D(µ)(X)parenrightbigTvbracketrightBig j=bracketleftbig D(µ)(X)bracketrightbig jjvj+summationdisplay m/negationslash=jbracketleftBigparenleftbig D(µ)(X)parenrightbigTbracketrightBig jmvm. Here[D(X)]ijis just a single element of the matrix D(X)a n d[ D(X)]kk=[DT(X)]kk is simply a diagonal element from the matrix – the repeated subscript does not indicate summation. Now, if we take as basis functions for a product represen-tation D prod(X) the products wk=uivj(where the nλnµvarious possible pairs of values i,jare labelled by k), we have also that w/prime k=Xw k=Xuivj=(Xui)(Xvj) =bracketleftbig D(λ)(X)bracketrightbig iibracketleftbig D(µ)(X)bracketrightbig jjuivj+ terms not involving the product uivj. This is to be compared with w/prime k=Xw k=bracketleftBigparenleftbig Dprod(X)parenrightbigTwbracketrightBig k=bracketleftbig Dprod(X)bracketrightbig kkwk+summationdisplay n/negationslash=kbracketleftBigparenleftbig Dprod(X)parenrightbigTbracketrightBig knwn, where Dprod(X) is the product representation matrix for element Xof the group. 946 25.11 PHYSICAL APPLICATIONS OF GROUP THEORY The comparison shows that bracketleftbig Dprod(X)bracketrightbig kk=bracketleftbig D(λ)(X)bracketrightbig iibracketleftbig D(µ)(X)bracketrightbig jj. It follows that χprod(X)=nλnµsummationdisplay k=1bracketleftbig Dprod(X)bracketrightbig kk =nλsummationdisplay i=1nµsummationdisplay j=1bracketleftbig D(λ)(X)bracketrightbig iibracketleftbig D(µ)(X)bracketrightbig jj =braceleftBiggnλsummationdisplay i=1bracketleftbig D(λ)(X)bracketrightbig iibracerightBiggbraceleftBiggnµsummationdisplay j=1bracketleftbig D(µ)(X)bracketrightbig jjbracerightBigg =χ(λ)(X)χ(µ)(X). (25.23) This proves the theorem, and a similar argument leads to the corresponding result for integrands in the form of a product of three or more factors. An immediate corollary is that an integral whose integrand is the product of two functions transforming according to two diﬀerent irreps is necessarily zero .T o see this, we use (25.18) to determine whether irrep A 1appears in the product character set χprod(X): mA1=1 gsummationdisplay Xbracketleftbig χ(A1)(X)bracketrightbig∗χprod(X)=1 gsummationdisplay Xχprod(X)=1 gsummationdisplay Xχ(λ)(X)χ(µ)(X). We have used the fact that χ(A1)(X)=1f o ra l l Xbut now note that, by virtue of (25.14), the expression on the right of this equation is equal to zero unless λ=µ. Any complications due to non-real characters have been ignored – in practice, they are handled automatically as it is usually Ψ∗φ, rather than Ψ φ, that appears in integrands, though many functions are real in any case, and nearly all charactersare. Equation (25.23) is a general result for integrands but, speciﬁcally in the context of chemical bonding, it implies that for the possibility of bonding to exist, the two quantum wavefunctions must transform according to the same irrep. This is discussed further in the next section. 25.11 Physical applications of group theory As we indicated at the start of chapter 24 and discussed in a little more detail at the beginning of the present chapter, some physical systems possess symmetries that allow the results of the present chapter to be used in their analysis. We consider now some of the more common sorts of problem in which these resultsﬁnd ready application. 947 REPRESENTATION THEORY 1 2 34xy Figure 25.4 A molecule consisting of four atoms of iodine and one of manganese. 25.11.1 Bonding in molecules We have just seen that whether chemical bonding can take place in a molecule is strongly dependent upon whether the wavefunctions of the two atoms forming a bond transform according to the same irrep. Thus it is sometimes useful to beable to ﬁnd a wavefunction that does transform according to a particular irrepof a group of transformations. This can be done if the characters of the irrep areknown and a sensible starting point can be guessed. We state without proof thatstarting from any n-dimensional basis vector Ψ ≡(Ψ 1Ψ2···Ψn)Twhere{Ψi}is a set of wavefunctions, the new vector Ψ(λ)≡(Ψ(λ) 1Ψ(λ) 2···Ψ(λ) n)Tgenerated by Ψ(λ) i=summationdisplay Xχ(λ)∗(X)XΨi (25.24) will transform according to the λth irrep. If the randomly chosen Ψ happens not to contain any component that transforms in the desired way then the Ψ(λ)so generated is found to be a zero vector and it is necessary to select a new startingvector. An illustration of the use of this ‘projection operator’ is given in the next example.IConsider a molecule made up of four iodine atoms lying at the corners of a square in the xy-plane, with a manganese atom at its centre, as shown in ﬁgure 25.4. Investigate whether the molecular orbital given by the superposition of p-state (angular momentum l=1) atomic orbitals Ψ1=Ψ y(r−R1)+Ψ x(r−R2)−Ψy(r−R3)−Ψx(r−R4) can bond to the d-state atomic orbitals of the mangane se atom described by either (a) φ1=( 3z2−r2)f(r)or (b) φ2=(x2−y2)f(r),w h e r e f(r)is a function of rand so is unchanged by any of the symmetry operations of th e molecule. Such linear combinations of atomic orbitals are known as ring orbitals. We have eight basis functions, the atomic orbitals Ψ x(N)a n dΨ y(N), where N=1,2,3,4 and indicates the position of an iodine atom. Since the wavefunctions are those of p-states they have the forms xf(r)o ryf(r) and lie in the directions of the x-a n d y-a x e ss h o w ni n the ﬁgure. Since ris not changed by any of the symmetry operations, f(r) can be treated as a constant. The symmetry group of the system is 4 mm, whose character table is table 25.4. Case (a). The manganese atomic orbital φ1=( 3z2−r2)f(r), lying at the centre of the 948 25.11 PHYSICAL APPLICATIONS OF GROUP THEORY molecule, is not aﬀected by any of the symmetry operations since zandrare unchanged by them. It clearly transforms according to the identity irrep A 1. We therefore need to know which combination of the iodine orbitals Ψ x(N)a n dΨ y(N), if any, also transforms according to A 1. We use the projection operator (25.24). If we choose Ψ x(1) as the arbitrary one- dimensional starting vector, we unfortunately obtain zero (as the reader may wish toverify), but Ψ y(1) does generate a new non-zero one-dimensional vector transforming according to A 1. The results of acting on Ψ y(1) with the various symmetry elements X can be written down by inspection (see the discussion in section 25.2). So, for example, theΨ y(1) orbital centred on iodine atom 1 and aligned along the positive y-axis, is changed by the anticlockwise rotation of π/2 produced by R/primeinto an orbital centred on atom 4 and aligned along the negative x-axis; thus R/primeΨy(1) =−Ψx(4). The complete set of group actions on Ψ y(1) is: I,Ψy(1); Q,−Ψy(3); R,Ψx(2); R/prime,−Ψx(4); mx,Ψy(1); my,−Ψy(3); md,Ψx(2); md/prime,−Ψx(4). Now χ(A1)(X)=1f o ra l l X, so (25.24) states that the sum of the above results for XΨy(1), all with weight 1, gives a vector (in this case of a one-dimensional irrep, just a wave-function) that transforms according to A 1and is therefore capable of forming a chemical bond with the manganese wavefunction φ1.I ti s Ψ(A1)=2 [ Ψ y(1)−Ψy(3) + Ψ x(2)−Ψx(4)], though, of course, the factor 2 is irrelevant. This is precisely the ring orbital Ψ 1given in the problem, but here it is generated rather than guessed beforehand. Case (b). The atomic orbital φ2=(x2−y2)f(r) behaves as follows under the action of typical conjugacy class members: I, φ 2;Q, φ 2;R,(y2−x2)f(r)=−φ2;mx,φ2;md,−φ2. From this we see that φ2transforms as a one-dimensional irrep, but, from table 25.4, that irrep is B 1not A 1(the irrep according to which Ψ 1transforms, as already shown). Thus φ2and Ψ 1cannot form a bond. J The original question did not ask for the the ring orbital to which φ2may bond, but it can be generated easily by using the values of XΨy(1) calculated in case (a) but now weighting them according to the characters of B1: Ψ(B1)=Ψ y(1)−Ψy(3) + (−1)Ψ x(2)−(−1)Ψ x(4) +Ψ y(1)−Ψy(3) + (−1)Ψ x(2)−(−1)Ψ x(4) =2 [ Ψ y(1)−Ψx(2)−Ψy(3) + Ψ x(4)]. Now we will ﬁnd the other irreps of 4 mmpresent in the space spanned by the basis functions Ψ x(N)a n dΨ y(N); at the same time this will illustrate the important point that since we are working with characters we are only interestedin the diagonal elements of the representative matrices. This means (section 25.2)that if we work in the natural representation D natwe need consider only those functions that transform, wholly or partially, into themselves. Since we have no need to write out the matrices explicitly, their size (8 ×8) is no drawback. All the irreps spanned by the basis functions Ψ x(N)a n dΨ y(N)c a nb ed e t e r m i n e db y considering the actions of the group elements upon them, as follows. 949 REPRESENTATION THEORY (i) Under Iall eight basis functions are unchanged, and χ(I)=8 . (ii) The rotations R,R/primeandQchange the value of Nin every case and so all diagonal elements of the natural representation are zero and χ(R)= χ(Q)=0 . (iii)mxtakes xinto−xandyintoyand, for N= 1 and 3, leaves Nunchanged, with the consequences (remember the forms of Ψ x(N)a n dΨ y(N)) that Ψx(1)→−Ψx(1),Ψx(3)→−Ψx(3), Ψy(1)→Ψy(1),Ψy(3)→Ψy(3). Thus χ(mx) has four non-zero contributions, −1,−1, 1 and 1, together with four zero contributions. The total is thus zero. (iv)mdandmd/primeleave no atom unchanged and so χ(md)=0 . The character set of the natural representation is thus 8, 0, 0, 0, 0, which, either by inspection or by applying formula (25.18), shows that Dnat=A 1⊕A2⊕B1⊕B2⊕2E, i.e. that all possible irreps are present. We have constructed previously the combinations of Ψ x(N)a n dΨ y(N) that transform according to A 1and B 1. The others can be found in the same way. 25.11.2 Matrix elements in quantum mechanics In section 25.10 we outlined the procedure for determining whether a matrix element that involves the product of three factors as an integrand is necessarily zero. We now illustrate this with a speciﬁc worked example.IDetermine whether a ‘dipole’ matrix element of the form J= Z Ψd1xΨd2dτ, where Ψd1andΨd2ared-state wavefunctions of the forms xyf(r)and(x2−y2)g(r)respec- tively, can be non-zero (i) in a molecule with symmetry C3v(or3m), such as ammonia, and (ii) in a molecule with symmetry C4v(or4mm), such as the MnI 4molecule considered in the previous example. We will need to make reference to the character tables of the two groups. The table forC 3vis table 25.1 (section 25.6); that for C4vis reproduced as table 25.5 from table 25.4 but with the addition of another column showing how some common functions transform. We make use of (25.23), extended to the product of three functions. No attention need be paid to f(r)a n d g(r) as they are unaﬀected by the group operations. Case (a). From the character table 25.1 for C3v, we see that each of xy,xandx2−y2 forms part of a basis set transforming according to the two-dimensional irrep E. Thus we may ﬁll in the array of characters (using chemical notation for the classes, except thatwe continue to use Irather than E) as shown in table 25.6. The last line is obtained by 950 25.11 PHYSICAL APPLICATIONS OF GROUP THEORY 4mm IQR ,R/primemx,mymd,md/prime A1 11 1 1 1 z;z2;x2+y2 A2 11 1 −1−1 Rz B1 11−11 −1 x2−y2 B2 11−1−11 xy E 2−20 0 0 (x, y); (xz, yz); (Rx,Ry) Table 25.5 The character table for the irreps of group 4 mm(orC4v). The right-hand column lists some common func tions, or, for the two-dimensional irrep E, pairs of functions, that transform according to the irrep against which they are shown. Function Irrep Classes I2C33σv xy E2−10 x E2−10 x2−y2E2−10 product 8 −10 Table 25.6 The character sets, for the group C3v(or 3mm), of three functions and of their product x2y(x2−y2). Function Irrep Classes IC 22C62σv2σd xy B2 11−1−11 x E2 −20 0 0 x2−y2B1 11−11−1 product 2 −20 0 0 Table 25.7 The character sets, for the group C4v(or 4mm), of three functions, and of their product x2y(x2−y2). multiplying together the corresponding characters for each of the three elements. Now, by inspection, or by applying (25.18), i.e. mA1=1 6[1(1)(8) + 2(1)( −1) + 3(1)(0)] = 1 , we see that irrep A 1does appear in the reduced representation of the product, and so J is not necessarily zero. Case (b). From table 25.5 we ﬁnd that, under the group C4v,xyandx2−y2transform as irreps B 2and B 1respectively and that xis part of a basis set transforming as E. Thus the calculation table takes the form of table 25.7 (again, chemical notation for the classeshas been used). Here inspection is suﬃcient, as the product is exactly that of irrep E and irrep A 1is certainly not present. Thus Jis necessarily zero and the dipole matrix element vanishes. J 951 REPRESENTATION THEORY x1x2x3 y1 y2y3 Figure 25.5 An equilateral array of masses and springs. 25.11.3 Degeneracy of normal modes As our ﬁnal area for illustrating the usefulness of group theoretical results we consider the normal modes of a vibrating system (see chapter 9). This analysishas far-reaching applications in physics, chemistry and engineering. For a givensystem, normal modes that are related by some symmetry operation have the samefrequency of vibration; the modes are said to be degenerate .I tc a nb es h o w nt h a t such modes span a vector space that transforms according to some irrep of thegroup Gof symmetry operations of the system. Moreover, the degeneracy of the modes equals the dimension of the irrep. As an illustration, we consider the following example.IInvestigate the possible vibrational modes o f the equilateral triangular arrangement of equal masses and springs shown in ﬁgure 25. 5. Demonstrate that two are degenerate. Clearly the symmetry group is that of the symme try operations on an equilateral triangle, namely 3 m(orC3v), whose character table is table 25.1. As on a previous occasion, it is most convenient to use the natural representation Dnatof this group (it almost always saves having to write out matrices explicitly) acting on the six-dimensional vector space(x 1,y1,x2,y2,x3,y3). In this example the natural and regular representations coincide, but this is not usually the case. We note that in table 25.1 the second class contains the rotations A(byπ/3) and B(by 2π/3), also known as RandR/prime. This class is known as 3 zin crystallographic notation, or C3in chemical notation, as explained in section 25.9. The third class contains C,D,E,t h e three mirror reﬂections. Clearly χ(I) = 6. Since all position labels are changed by a rotation, χ(3z)=0 .F o rt h e mirror reﬂections the simplest representative class member to choose is the reﬂection myin the plane containing the y3-axis, since then only label 3 is unchanged; under my,x3→−x3 andy3→y3, leading to the conclusion that χ(my) = 0. Thus the character set is 6, 0, 0. Using (25.18) and the character table 25.1 shows that Dnat=A 1⊕A2⊕2E. 952 25.11 PHYSICAL APPLICATIONS OF GROUP THEORY However, we have so far allowed xi,yito be completely general, and we must now identify and remove those irreps that do not correspond to vibrations. These will be the irrepscorresponding to bodily translations of the triangle and to its rotation without relativemotion of the three masses. Bodily translations are linear motions of the centre of mass, which has coordinates x=(x 1+x2+x3)/3a n d y=(y1+y2+y3)/3). Table 25.1 shows that such a coordinate pair ( x, y) transforms according to the two- dimensional irrep E; this accounts for one of the two such irreps found in the naturalrepresentation. It can be shown that, as stated in table 25.1, planar bodily rotations of the triangle – rotations about the z-axis, denoted by R z– transform as irrep A 2. Thus, when the linear motions of the centre of mass, and pure rotation about it, are removed from ourreduced representation, we are left with E ⊕A 1. These must be the irreps corresponding to the internal vibrations of the triangle. – one doubly degenerate mode and one non-degenerate mode. The physical interpretation of this is that two of the normal modes of thesystem have the same frequency and one normal mode has a diﬀerent frequency (barringaccidental coincidences for other reasons). It may be noted that in quantum mechanicsthe energy quantum of a normal mode is proportional to its frequency.J In general, group theory does not tell us what the frequencies are, since it is entirely concerned with the symmetry of the system and not with the values of masses and spring constants. However, using this type of reasoning, the results from representation theory can be used to predict the degeneracies of atomicenergy levels and, given a perturbation whose Hamiltonian (energy operator) hassome degree of symmetry, the extent to which the perturbation will resolve thedegeneracy. Some of these ideas are explored a little further in the next sectionand in the exercises. 25.11.4 Breaking of degeneracies If a physical system has a high degree of symmetry, invariant under a group Gof reﬂections and rotations, say, then, as implied above, it will normally be the casethat some of its eigenvalues (of energy, frequency, angular momentum etc.) are degenerate. However, if a perturbation that is invariant only under the operations of the elements of a smaller symmetry group (a subgroup of G)is added, some of the original degeneracies may be broken. The results derived from representationtheory can be used to decide the extent of the degeneracy-breaking. The normal procedure is to use an N-dimensional basis vector, consisting of theNdegenerate eigenfunctions, to generate an N-dimensional representation of the symmetry group of the perturbation. This representation is then decomposedinto irreps. In general, eigenfunctions that transform according to diﬀerent irrepsno longer share the same frequency of vibration. We illustrate this with the following example. 953 REPRESENTATION THEORY M MM Figure 25.6 A circular drumskin loaded with three symmetrically placed masses.IA circular drumskin has three equal masses placed on it at the vertices of an equilateral triangle, as shown in ﬁgure 25.6. Determine which degenerate normal modes of the drumskincan be split in frequency by this perturbation. When no masses are present the normal modes of the drum-skin are either non-degenerateor two-fold degenerate (see chapter 19). The degenerate eigenfunctions Ψ of the nth normal mode have the forms J n(kr)(cos nθ)e±iωtor Jn(kr)(sinnθ)e±iωt. Therefore, as explained above, we need to consider the two-dimensional vector space spanned by Ψ 1=s i n nθand Ψ 2=c o s nθ. This will generate a two-dimensional representa- tion of the group 3 m(orC3v), the symmetry group of the perturbation. Taking the easiest element from each of the three classes (identity, rotations, and reﬂections) of group 3 m, we have IΨ1=Ψ 1,IΨ2=Ψ 2, AΨ1=s i n / n /; θ−2 3π // = /; cos2 3nπ / Ψ1− /; sin2 3nπ / Ψ2, AΨ2=c o s / n /; θ−2 3π // = /; cos2 3nπ / Ψ2+ /; sin2 3nπ / Ψ1, CΨ1= sin[ n(π−θ)] =−(cosnπ)Ψ1, CΨ2=c o s [ n(π−θ)] = (cos nπ)Ψ2. The three representative matrices are therefore D(I)=I2,D(A)= / cos2 3nπ−sin2 3nπ sin2 3nπ cos2 3nπ /! ,D(C)= / −cosnπ 0 0c o s nπ /! . The characters of this representation are χ(I)=2 , χ(A)=2 c o s ( 2 nπ/3) and χ(C)=0 . Using (25.18) and table 25.1, we ﬁnd that mA1=1 6 /; 2+4c o s2 3nπ / =mA2 mE=1 6 /; 4−4c os2 3nπ / . Thus D= /( A1⊕A2ifn=3,6,9,. . ., E otherwise . Hence the normal modes n=3,6,9,. . .each transform under the operations of 3 m 954 25.12 EXERCISES as the sum of two one-dimensional irreps and, using the reasoning given in the previous example, are therefore split in frequency by the perturbation. For other values of nthe representation is irreducible and so the degeneracy cannot be split. J 25.12 Exercises 25.1 A group Gh a sf o u re l e m e n t s I,X,Y andZ, which satisfy X2=Y2=Z2= XY Z =I. Show that Gis Abelian and hence deduce the form of its character table. Show that the matrices D(I)= / 10 01 / , D(X)= / −10 0−1 / , D(Y)= / −1−p 01 / , D(Z)= / 1p 0−1 / , where pis a real number, form a representation DofG. Find its characters and decompose it into irreps. 25.2 Using a square whose corners lie at coordinates ( ±1,±1), form a natural rep- resentation of the dihedral group D4. Find the characters of the representation, and, using the information (and class order) in table 25.4 (p. 944), express therepresentation in terms of irreps. Now form a representation in terms of eight 2 ×2 orthogonal matrices, by considering the eﬀect of each of the elements of D 4on a general vector ( x, y). Conﬁrm that this representation is one of the irreps found using the naturalrepresentation. 25.3 The quaternion group Q(see exercise 24.20) has eight elements {±1,±i,±j,±k} obeying the relations i 2=j2=k2=−1,i j=k=−ji. Determine the conjugacy classes of Qand deduce the dimensions of its irreps. Show that Qis homomorphic to the four-element group V, which is generated by two distinct elements aandbwith a2=b2=(ab)2=I. Find the one-dimensional irreps of Vand use these to help determine the full character table for Q. 25.4 (a) By considering the possible forms of its cycle notation, determine the number of elements in each conjugacy class of the permutation group S4and show thatS4has ﬁve irreps. Give the logical reasoning that shows they must consist of two three-dimensional, one two-dimensional, and two one-dimensionalirreps. (b) By considering the odd and even permutations in the group S 4establish the characters for one of the one-dimensional irreps. (c) Form a natural matrix representation of 4 ×4 matrices based on a set of objects{a, b, c, d}, which may or may not be equal to each other, and, by selecting one example from each conjugacy class, show that this natural rep-resentation has characters 4, 2, 1, 0, 0. The one-dimensional vector subspace spanned by sets of the form {a, a, a, a}is invariant under the permutation group and hence transforms according to the invariant irrep A 1.T h er e m a i n - ing three-dimensional subspace is irreducible; use this and the charactersdeduced above to establish the characters for one of the three-dimensionalirreps, T 1. (d) Complete the character table using orthogonality properties, and check the summation rule for each irrep. You should obtain table 25.8. 955 REPRESENTATION THEORY Typical element and class size Irrep (1) (12) (123) (1234) (12)(34) 16 8 6 3 A1 11 1 1 1 A2 1−11 −11 E 20 −10 2 T1 31 0 −1−1 T2 3−10 1 −1 Table 25.8 The character table for the permutation group S4. 25.5 In exercise 24.10, the group of pure rotations taking a cube into itself was found to have 24 elements. The group is isomorphic to the permutation group S4, considered in the previous question, and hence has the same character table, oncecorresponding classes have been established. By counting the number of elementsin each class make the correspondences below (the ﬁnal two cannot be decidedpurely by counting, and should be taken as given). Permutation Symbol Action class type (physics) (1) I none (123) 3 rotations about a body diagonal(12)(34) 2 z rotation of πabout the normal to a face (1234) 4 z rotations of ±π/2 about the normal to a face (12) 2 d rotation of πabout an axis through the centres of opposite edges Reformulate the character table 25.8 in terms of the elements of the rotation symmetry group (432 or O) of a cube and use it when answering exercises 25.7 and 25.8. 25.6 Consider a regular hexagon orientated so that two of its vertices lie on the x-axis. Find matrix representations of a rotation Rthrough π/6 and a reﬂection myin they-axis by determining their eﬀects on vectors lying in the xy-plane . Show that a reﬂection mxin the x-axis can be written as mx=myR3and that the (12) elements of the symmetry group of the hexagon are given by RnorRnmy. Using the representations of Randmyas generators, ﬁnd a two-dimensional representation of the symmetry group, C6, of the regular hexagon. Is it a faithful representation? 25.7 In a certain crystalline compound, a thorium atom lies at the centre of a regular octahedron of six sulphur atoms at positions ( ±a,0,0), (0 ,±a,0), (0 ,0,±a). These can be considered as being positioned at the centres of the faces of a cube ofside 2 a. The sulphur atoms produce at the site of the thorium atom an electric ﬁeld that has the same symmetry group as a cube (432 or O). The ﬁve degenerate d-electron orbitals of the thorium atom can be expressed, relative to any arbitrary polar axis, as (3cos 2θ−1)f(r),e±iφsinθcosθf(r),e±2iφsin2θf(r). A rotation about that polar axis by an angle φ/primeeﬀectively changes φtoφ−φ/prime. Use this to show that the character of the rotation in a representation based onthe orbital wavefunctions is given by 1+2c o s φ /prime+2c o s2 φ/prime 956 25.12 EXERCISES and hence that the characters of the representation, in the order of the symbols given in exercise 25.5, is 5, −1, 1,−1, 1. Deduce that the ﬁve-fold degenerate level is split into two levels, a doublet and a triplet. 25.8 Sulphur hexaﬂuoride is a molecule with the same structure as the crystalline compound in exercise 25.7, except that a sulphur atom is now the central atom.The following are the forms of some of the electronic orbitals of the sulphuratom, together with the irreps according to which they transform under thesymmetry group 432 (or O). Ψ s=f(r)A 1 Ψp1=zf(r)T 1 Ψd1=( 3z2−r2)f(r)E Ψd2=(x2−y2)f(r)E Ψd3=xyf(r), T2 The function xtransforms according to the irrep T 1. Use the above data to determine whether dipole matrix elements of the form J= R φ1xφ2dτcan be non-zero for the following pairs of orbitals φ1,φ2in a sulphur hexaﬂuoride molecule: (a) Ψ d1,Ψs;( b )Ψ d1,Ψp1;( c )Ψ d2,Ψd1;( d )Ψ s,Ψd3;( e )Ψ p1,Ψs. 25.9 The hydrogen atoms in a methane molecule CH 4form a perfect tetrahedron with the carbon atom at its centre. The molecule is most conveniently describedmathematically by placing the hydrogen atoms at the points (1 ,1,1), (1 ,−1,−1), (−1,1,−1) and (−1,−1,1 ) .T h es y m m e t r yg r o u pt ow h i c hi tb e l o n g s ,t h et e t r a h e - dral group ( ¯43morT d) has classes typiﬁed by I,3 ,2 z,mdand¯4z, where the ﬁrst three are as in exercise 25.5, mdis a reﬂection in the mirror plane x−y=0a n d ¯4zis a rotation of π/2 about the z-axis followed by an inversion in the origin. A reﬂection in a mirror plane can be considered as a rotation of πabout an axis perpendicular to the plane, followed by an inversion in the origin. T h ec h a r a c t e rt a b l ef o rt h eg r o u p ¯43mis very similar to that for the group 432, and has the form shown in table 25.9. Typical element and class size Functions transforming Irreps I32 z¯4z md according to irrep 18 3 6 6 A1 11 1 1 1 x2+y2+z2 A2 11 1 −1−1 E 2−12 0 0 (x2−y2,3z2−r2) T1 30−11 −1 (Rx,Ry,Rz) T2 30−1−11 (x, y, z); (xy, yz, zx ) Table 25.9 The character table for group ¯43m. By following the steps given below, determine how many diﬀerent internal vibra- tion frequencies the CH 4molecule has. (a) Consider a representation based on the 12 coordinates xi,yi,zifori= 1,2,3,4. For those hydrogen atoms that transform into themselves, a rota- tion through an angle θabout an axis parallel to one of the coordinate axes gives rise in the natural representation to the diagonal elements 1 for thecorresponding coordinate and 2cos θfor the two orthogonal coordinates. If the rotation is followed by an inversion then these entries are multiplied by−1. Atoms not transforming into themselves give a zero diagonal contribu- tion. Show that the characters of the natural representation are 12, 0, 0, 0, 2 957 REPRESENTATION THEORY and hence that its expression in terms of irreps is A1⊕E⊕T1⊕2T2. (b) The irreps of the bodily translational and rotational motions are included in this expression and need to be identiﬁed and removed. Show that when thisis done it can be concluded that there are three diﬀerent internal vibrationfrequencies in the CH 4molecule. State their degeneracies and check that they are consistent with the expected number of normal coordinates neededto describe the internal motions of the molecule. 25.10 (a) The set of even permutations of four objects (a proper subgroup of S 4) is known as the alternating group A4. List its twelve members using cycle notation. (b) Assume that all permutations with the same cycle structure belong to the same conjugacy class. Show that this leads to a contradiction and hencedemonstrates that even if two permutations have the same cycle structurethey do not necessarily belong to the same class. (c) By evaluating the products p 1= (123)(4)•(12)(34)•(132)(4) and p2= (132)(4)•(12)(34)•(123)(4) deduce that the three elements of A4with structure of the form (12)(34) belong to the same class. (d) By evaluating products of the form (1 α)(βγ)•(123)(4)•(1α)(βγ), where α, β, γ are various combinations of 2, 3, 4, show that the class to which (123)(4)belongs contains at least four members. Show the same for (124)(3). (e) By combining results (b), (c) and (d) deduce that A 4has exactly four classes, and determine the dimensions of its irreps. (f) Using the orthogonality properties of characters and noting that elements of the form (124)(3) have order 3, ﬁnd the character table for A4. 25.11 Use the results of exercise 24.23 to ﬁnd the character table for the dihedral group D5, the symmetry group of a regular pentagon. 25.12 Demonstrate that equation (25.24) does indeed generate a set of vectors trans- forming according to an irrep λ, by sketching and superposing drawings of an equilateral triangle of springs and masses, based on that shown in ﬁgure 25.7. (a) (b) (c)A A A BB BC CC 30◦30◦ Figure 25.7 The three normal vibration modes of the equilateral array. Mode (a) is known as the ‘breathing mode’. Modes ( b)a n d( c) transform according to irrep E and have equal vibrational frequencies. (a) Make an initial sketch showing an arbitrary small mass displacement from, say, vertex C. Draw the results of operating on the initial sketch with each of the symmetry elements of the group 3 m(C3v). (b) Superimpose the results, weighting them according to the characters of irrep A1(table 25.1 in section 25.6) and verify that the resultant is a symmetrical arrangement in which all three masses move symmetrically towards (or awayfrom) the centroid of the triangle. The mode is illustrated in ﬁgure 25.7( a). 958 25.13 HINTS AND ANSWERS (c) Start again, now considering a displacement δofCparallel to the x-axis. Form a similar superposition of sketches weighted according to the charactersof irrep E (note that the reﬂections are not needed). The resultant containssome bodily displacement of the triangle, since this also transforms accordingto E. Show that the displacement of the centre of mass is ¯x=δ,¯y=0 . Subtract this out and verify that the remainder is of the form shown inﬁgure 25.7( c). (d) Using an initial displacement parallel to the y-axis, and an analogous proce- dure, generate the remaining normal mode, degenerate with that in ( c)a n d shown in ﬁgure 25.7( b). 25.13 Further investigation of the crystalline compound considered in exercise 25.7 shows that the octahedron is not quite perfect but is elongated along the (1 ,1,1) direction with the sulphur atoms at positions ±(a+δ,δ,δ),±(δ,a+δ,δ),±(δ,δ,a+ δ), where δ/lessmucha. This structure is invariant under the (crystallographic) symmetry group 32 with three two-fold axes along directions typiﬁed by (1 ,−1,0). The latter axes, which are perpendicular to the (1 ,1,1) direction, are axes of two- fold symmetry for the perfect octahedron. The group 32 is really the three-dimensional version of the group 3 mand has the same character table as table 25.1 (section 25.6). Use this to show that, when the distortion of the octahedron isincluded, the doublet found in exercise 25.7 is unsplit but the triplet breaks upinto a singlet and a doublet. 25.13 Hints and answers 25.1 There are four classes and hence four one-dimensional irreps, which must have e n t r i e sa sf o l l o w s :1 ,1 ,1 ,1 ; 1 ,1 , −1,−1; 1,−1, 1,−1; 1,−1,−1, 1. The characters of Dare 2,−2, 0, 0 and so the irreps present are the last two of these. 25.2 The characters are 4, 0, 0, 0, 2, and the irreps present are A 1+B 2+E .T h e characters of the classes are 2, −2, 0, 0, 0, showing that the representation is the irrep E. 25.3 There are ﬁve classes {1},{−1},{±i},{±j},{±k}; there are four one-dimensional irreps and one two-dimensional irrep. Show that ab=ba. The homomorphism is±1→I,±i→a,±j→b,±k→ab.Vis Abelian and hence has four one-dimensional irreps. In the class order given above, the characters for Qare as follows: ˆD(1),1,1,1,1,1; ˆD(2),1,1,1,−1,−1;ˆD(3),1,1,−1,1,−1;ˆD(4),1,1,−1,−1,1;ˆD(5),2,−2,0,0,0. 25.4 (a) One element of type (1)(2)(3)(4), six of type (12)(3)(4), eight of type (123)(4), six of type (1234), three of type (12)(34). Five classes implies ﬁve irreps. SincePn2 imust equal 24, at least one ni≥3. Assuming ni≥4 leads to a contradiction, and so n5(say) equals 3. The inequalities 12+3 ( 22)<15<4(22)i m p l yt h a t a second niequals 3. P3 1n2 i= 6 has only one integer solution. (b) D(2)[(12)] = D(2)[(1234)] =−1. (c) Characters for T 1are (4−1), (2−1), (1−1), (0−1), (0−1), i.e. 3, 1, 0,−1,−1. 25.6 The matrix representations are R=1 2[1,−√ 3;√ 3,1]; my=[−1,0;0,1] As examples, R4=1 2[−1,√ 3;−√ 3,−1] and R2my=1 2[1,−√ 3;−√ 3,−1]. The representation is faithful. 25.7 The ﬁve basis functions of the representation are multiplied by 1, e−iφ/prime,e+iφ/prime, e−2iφ/prime,e+2iφ/primeas a result of the rotation. The character is the sum of these for rotations of 0, 2 π/3,π,π/2,π;Drep=E+T 2. 25.8 (a) No; (b) yes; (c) no; (d) no; (e) yes. 959 REPRESENTATION THEORY 25.9 (b) The bodily translation has irrep T 2and the rotation has irrep T 1. The irreps of the internal vibrations are A 1,E ,T 2, with respective degeneracies 1, 2, 3, making six internal coordinates (12 in total minus three translational minus threerotational). 25.10 (a) The identity, eight elements of the form (124)(3) and three elements of the form (12)(34). (b) The assumption implies that there are three irreps, of which one must be the identity irrep. However, 1 + n 2 2+n2 3= 12 has no integer solutions. (c)p1= (13)(24) ,p2= (14)(23). (d) For example, (123)(4) generates (134)(2) ,(142)(3) ,(243)(1). (e) There are three one-dimensional irreps and one three-dimensional irrep. (f) The four sets of characters are: 1 ,1,1,1; 1,ω,ω2,1; 1,ω2,ω,1; 3,0,0,−1. Here ω=e x p ( 2 πi/3) and 1 + ω+ω2=0 . 25.11 There are four classes and hence four irreps, which can only be the identity irrep, one other one-dimensional irrep, and two two-dimensional irreps. In theclass order {I},{R,R 4},{R2,R3},{mi}the second one-dimensional irrep must (because of orthogonality) have characters 1, 1, 1, −1. The summation rules and orthogonality require the other two character sets to be 2 ,(−1+√5)/2,(−1−√5)/2,0a n d2 ,(−1−√5)/2,(−1+√5)/2,0. Note that Rhas order 5 and that, e.g., (−1+√5)/2=e x p ( 2 πi/5) + exp(8 πi/5). 25.12 (c) ¯x=1 3[2δ+(−1)(−1 2δ)+(−1)(−1 2δ)],¯y=1 3[0 + (−1)(−√ 3 2δ)+(−1)(√ 3 2δ)]. 25.13 The doublet irrep E (characters 2, −1, 0) appears in both 432 and 32 and so is unsplit. The triplet T 2(characters 3, 0, 1) splits under 32 into doublet E (characters 2, −1, 0) and singlet A 1(characters 1, 1, 1). 960 26 Probability All scientists will know the importance of experiment and observation and, equally, be aware that the results of some experiments depend to a degree onchance. For example, in an experiment to measure the heights of a random sampleof people, we would not be in the least surprised if all the heights were found to be diﬀerent; but, if the experiment were repeated often enough, we would expect to ﬁnd some sort of regularity in the results. Statistics, which is the subject of thenext chapter, is concerned with the analysis of real experimental data of this sort.First, however, we discuss probability. To a pure mathematician, probability is anentirely theoretical subject based on axioms. Although this axiomatic approach isimportant, and we discuss it brieﬂy, an approach to probability more in keepingwith its eventual applications in statistics is adopted here. We ﬁrst discuss the terminology required, with particular reference to the convenient graphical representation of experimental results as Venn diagrams.The concepts of random variables and distributions of random variables are thenintroduced. It is here that the connection with statistics is made; we assert thatthe results of many experiments are random variables and that those results havesome sort of regularity, which is represented by a distribution. Precise deﬁnitionsof a random variable and a distribution are then given, as are the deﬁning equations for some important distributions. We also derive some useful quantities associated with these distributions. 26.1 Venn diagrams We call a single performance of an experiment a trialand each possible result anoutcome .T h e sample space Sof the experiment is then the set of all possible outcomes of an individual trial. For example, if we throw a six-sided die then there are six possible outcomes that together form the sample space of the experiment.At this stage we are not concerned with how likely a particular outcome might 961 PROBABILITY ii iiii ivA B S Figure 26.1 A Venn diagram. be (we will return to the probability of an outcome in due course) but rather will concentrate on the classiﬁcation of possible outcomes. It is clear that some sample spaces are ﬁnite (e.g. the outcomes of throwing a die) whilst others are inﬁnite (e.g. the outcomes of measuring people’s heights). Most often, one is notinterested in individual outcomes but in whether an outcome belongs to a givensubset A(say) of the sample space S; these subsets are called events . For example, we might be interested in whether a person is taller or shorter than 180 cm, inwhich case we divide the sample space into just two events: namely, that theoutcome (height measured) is (i) greater than 180 cm or (ii) less than 180 cm. A common graphical representation of the outcomes of an experiment is the Venn diagram . A Venn diagram usually consists of a rectangle, the interior of which represents the sample space, together with one or more closed curves insideit. The interior of each closed curve then represents an event. Figure 26.1 showsa typical Venn diagram representing a sample space Sand two events Aand B. Every possible outcome is assigned to an appropriate region; in this example there are four regions to consider (marked i to iv in ﬁgure 26.1): (i) outcomes that belong to event Abut not to event B; (ii) outcomes that belong to event Bbut not to event A; (iii) outcomes that belong to both event Aand event B; (iv) outcomes that belong to neither event Anor event B.IA six-sided die is thrown. Let event Abe ‘the number obtained is divisible by 2’ and event Bbe ‘the number obtained is divisible by 3’. Draw a Venn diagram to represent these events. It is clear that the outcomes 2, 4, 6 belong to event Aand that the outcomes 3, 6 belong to event B. Of these, 6 belongs to both AandB. The remaining outcomes, 1, 5, belong to neither AnorB. The appropriate Venn diagram is shown in ﬁgure 26.2. J In the above example, one outcome, 6, is divisible by both 2 and 3 and so belongs to both AandB. This outcome is placed in region iii of ﬁgure 26.1, which is called the intersection ofAandBand is denoted by A∩B(see ﬁgure 26.3( a)). If no events lie in the region of intersection then AandBare said to be mutually exclusive ordisjoint . In this case, often the Venn diagram is drawn so that the closed curves representing the events AandBdo not overlap, so as to make 962 26.1 VENN DIAGRAMS A B S12 34 56 Figure 26.2 The Venn diagram for the outcomes of the die-throwing trials described in the worked example. AAA ABBB SSS S(a)( b) (c)( d)¯A Figure 26.3 Venn diagrams: the shaded regions show ( a)A∩B,t h ei n t e r - section of two events AandB,(b)A∪B, the union of events AandB,(c) the complement ¯Aof an event A,(d)A−B, those outcomes in Athat do not belong to B. graphically explicit the fact that AandBare disjoint. It is not necessary, however, to draw the diagram in this way, since we may simply assign zero outcomes to the shaded region in ﬁgure 26.3(a). An event that contains no outcomes is calledtheempty event and denoted by ∅. The event comprising all the elements that belong to either AorB, or to both, is called the union ofAandBand is denoted byA∪B(see ﬁgure 26.3( b)). In the previous example, A∪B={2,3,4,6}. It is sometimes convenient to talk about those outcomes that do notbelong to a particular event. The set of outcomes that do not belong to Ais called the complement ofAand is denoted by ¯A(see ﬁgure 26.3( c)); this can also be written as¯A=S−A. It is clear that A∪¯A=SandA∩¯A=∅. The above notation can be extended in an obvious way, so that A−Bdenotes the outcomes in Athat do not belong to B. It is clear from ﬁgure 26.3( d)t h a t A−Bc a na l s ob ew r i t t e na s A∩¯B. Finally, when allthe outcomes in event B (say) also belong to event A, but Amay contain, in addition, outcomes that do 963 PROBABILITY AB C S12 34 5 678 Figure 26.4 The general Venn diagram for three events is divided into eight regions. not belong to B,t h e n Bis called a subset ofA, a situation that is denoted by B⊂A; alternatively, one may write A⊃B, which states that Acontains B.I nt h i s case, the closed curve representing the event Bis often drawn lying completely within the closed curve representing the event A. The operations ∪and∩are extended straightforwardly to more than two events. If there exist nevents A1,A2,...,A n, in some sample space S, then the event consisting of all those outcomes that belong to one or more of the Aiis the union ofA1,A2,...,A nand is denoted by A1∪A2∪···∪An. (26.1) Similarly, the event consisting of all the outcomes that belong to every one of the Aiis called the intersection ofA1,A2,...,A nand is denoted by A1∩A2∩···∩An. (26.2) If, for anypair of values i, jwith i/negationslash=j, Ai∩Aj=∅ (26.3) then the events AiandAjare said to be mutually exclusive ordisjoint . Consider three events A,BandCwith a Venn diagram such as is shown in ﬁgure 26.4. It will be clear that, in general, the diagram will be divided into eight regions and they will be of four diﬀerent types. Three regions correspond to a single event; three regions are each the intersection of exactly two events; oneregion is the three-fold intersection of all three events; and ﬁnally one regioncorresponds to none of the events. Let us now consider the numbers of diﬀerentregions in a general n-event Venn diagram. For one-event Venn diagrams there are two regions, for the two-event case there are four regions and, as we have just seen, for the three-event case there areeight. In the general n-event case there are 2 nregions, as is clear from the fact that any particular region Rlies either inside or outside the closed curve of any particular event. With two choices (inside or outside) for each of nclosed curves, there are 2ndiﬀerent possible combinations with which to characterise R.O n c e n 964 26.1 VENN DIAGRAMS gets beyond three it becomes impossible to draw a simple two-dimensional Venn diagram, but this does not change the results. The 2nregions will break down into n+1 types, with the numbers of each type as follows† no events,nC0=1 ; one event but no intersections,nC1=n; two-fold intersections,nC2=1 2n(n−1); three-fold intersections,nC3=1 3!n(n−1)(n−2); ... ann-fold intersection,nCn=1 . That this makes a total of 2ncan be checked by considering the binomial expansion 2n=( 1+1 )n=1+ n+1 2n(n−1) +···+1. Using Venn diagrams, it is straightforward to show that the operations ∩and ∪obey the following algebraic laws: commutativity, A∩B=B∩A, A∪B=B∪A; associativity, ( A∩B)∩C=A∩(B∩C),(A∪B)∪C=A∪(B∪C); distributivity, A∩(B∪C)=(A∩B)∪(A∩C), A∪(B∩C)=(A∪B)∩(A∪C); idempotency, A∩A=A, A∪A=A.IShow that (i) A∪(A∩B)=A∩(A∪B)=A, (ii)(A−B)∪(A∩B)=A. (i) Using the distributivity and idempotency laws above, we see that A∪(A∩B)=(A∪A)∩(A∪B)=A∩(A∪B). By sketching a Venn diagram it is immediately clear that both expressions are equal to A. Nevertheless, we here proceed in a more formal manner in order to deduce this result algebraically. Let us begin by writing X=A∪(A∩B)=A∩(A∪B), (26.4) from which we want to deduce a simpler expression for the event X. Using the ﬁrst equality in (26.4) and the algebraic laws for ∩and∪,w em a yw r i t e A∩X=A∩[A∪(A∩B)] =(A∩A)∪[A∩(A∩B)] =A∪(A∩B)=X. †The symbolsnCi,f o r i=0,1,2,...,n, are a convenient notation for combinations; they and their properties are discussed in chapter 1. 965 PROBABILITY Since A∩X=Xwe must have X⊂A. Now, using the second equality in (26.4) in a similar way, we ﬁnd A∪X=A∪[A∩(A∪B)] =(A∪A)∩[A∪(A∪B)] =A∩(A∪B)=X, from which we deduce that A⊂X. Thus, since X⊂AandA⊂X, we must conclude that X=A. (ii) Since we do not know how to deal with compound expressions containing a minus sign, we begin by writing A−B=A∩¯Bas mentioned above. Then, using the distributivity law, we obtain (A−B)∪(A∩B)=(A∩¯B)∪(A∩B) =A∩(¯B∪B) =A∩S=A. In fact, this result, like the ﬁrst one, can be proved trivially by drawing a Venn diagram. J Further useful results may be derived from Venn diagrams. In particular, it is simple to show that the following rules hold: (i) if A⊂Bthen¯A⊃¯B; (ii)A∪B=¯A∩¯B; (iii)A∩B=¯A∪¯B. Statements (ii) and (iii) are known jointly as de Morgan’s laws and are sometimes useful in simplifying logical expressions.IThere exist two events AandBsuch that (X∪A)∪(X∪¯A)=B. Find an expression for the event Xin terms of AandB. We begin by taking the complement of both sides of the above expression: applying de Morgan’s laws we obtain ¯B=(X∪A)∩(X∪¯A). We may then use the algebraic laws obeyed by ∩and∪to yield ¯B=X∪(A∩¯A)=X∪∅=X. Thus, we ﬁnd that X=¯B. J 26.2 Probability In the previous section we discussed Venn diagrams, which are graphical repre- sentations of the possible outcomes of experiments. We did not, however, giveany indication of how likely each outcome or event might be when any particular experiment is performed. Most experiments show some regularity. By this we mean that the relative frequency of an event is approximately the same on eachoccasion that a set of trials is performed. For example, if we throw a die N 966 26.2 PROBABILITY times then we expect that a six will occur approximately N/6 times (assuming, of course, that the die is not biased). The regularity of outcomes allows us todeﬁne the probability ,P r (A), as the expected relative frequency of event Ain a large number of trials. More quantitatively, if an experiment has a total of n S outcomes in the sample space S,a n d nAof these outcomes correspond to the event A, then the probability that event Awill occur is Pr(A)=nA nS. (26.5) 26.2.1 Basic theorems From (26.5) we may deduce the following properties of the probability Pr( A). (i) For any event Ain a sample space S, 0≤Pr(A)≤1. (26.6) If Pr( A)=1t h e n Ais a certainty; if Pr( A)=0t h e n Ais an impossibility. (ii) For the entire sample space Swe have Pr(S)=nS nS=1, (26.7) which simply states that we are certain to obtain one of the possible outcomes. (iii) If AandBare two events in Sthen, from the Venn diagrams in ﬁgure 26.3, we see that nA∪B=nA+nB−nA∩B, (26.8) the ﬁnal subtraction arising because the outcomes in the intersection of AandBare counted twice when the outcomes of Aare added to those ofB. Dividing both sides of (26.8) by nS, we obtain the addition rule for probabilities Pr(A∪B)=P r ( A)+P r ( B)−Pr(A∩B). (26.9) However, if Aand Baremutually exclusive events ( A∩B=∅)t h e n Pr(A∩B) = 0 and we obtain the special case Pr(A∪B)=P r ( A)+P r ( B). (26.10) (iv) If ¯Ais the complement of Athen¯AandAare mutually exclusive events. Thus, from (26.7) and (26.10) we have 1=P r ( S)=P r ( A∪¯A)=P r ( A)+P r ( ¯A), from which we obtain the complement law Pr(¯A)=1−Pr(A). (26.11) 967 PROBABILITY This is particularly useful for problems in which evaluating the probability of the complement is easier than evaluating the probability of the eventitself.ICalculate the probability of drawing an ace or a spade from a pack of cards. LetAbe the event that an ace is drawn and Bthe event that a spade is drawn. It immediately follows that Pr( A)=4 52=1 13and Pr( B)=13 52=1 4. The intersection of Aand Bconsists of only the ace of spades and so Pr( A∩B)=1 52. Thus, from (26.9) Pr(A∪B)=1 13+1 4−1 52=4 13. In this case it is just as simple to recognise that there are 16 cards in the pack that satisfy the required condition (13 spades plus three other aces) and so the probability is16 52. J The above theorems can easily be extended to a greater number of events. For example, if A1,A2,...,A nare mutually exclusive events then (26.10) becomes Pr(A1∪A2∪···∪An)=P r ( A1)+P r ( A2)+···+P r ( An). (26.12) Furthermore, if A1,A2,...,A n(whether mutually exclusive or not) exhaust S,i . e . are such that A1∪A2∪···∪An=S,t h e n Pr(A1∪A2∪···∪An)=P r ( S)=1 . (26.13)IA biased six-sided die has probabilities1 2p,p,p,p,p,2pof showing 1, 2, 3, 4, 5, 6 respectively. Calculate p. Given that the individual events are mutually exclusive, (26.12) can be applied to give Pr(1∪2∪3∪4∪5∪6) =1 2p+p+p+p+p+2p=13 2p. The union of all possible outcomes on the LHS of this equation is clearly the sample space, S,a n ds o Pr(S)=13 2p. Now using (26.7), 13 2p=P r ( S)=1⇒ p=2 13. J When the possible outcomes of a trial correspond to more than two events, and those events are notmutually exclusive, the calculation of the probability of the union of a number of events is more complicated, and the generalisation ofthe addition law (26.9) requires further work. Let us begin by considering theunion of three events A 1,A2andA3, which need not be mutually exclusive. We ﬁrst deﬁne the event B=A2∪A3and, using the addition law (26.9), we obtain Pr(A1∪A2∪A3)=P r ( A1∪B)=P r ( A1)+P r ( B)−Pr(A1∩B). (26.14) 968 26.2 PROBABILITY However, we may write Pr( A1∩B)a s Pr(A1∩B)=P r [ A1∩(A2∪A3)] =P r [ ( A1∩A2)∪(A1∩A3)] =P r ( A1∩A2)+P r ( A1∩A3)−Pr(A1∩A2∩A3). Substituting this expression, and that for Pr( B) obtained from (26.9), into (26.14) we obtain the probability addition law for three general events, Pr(A1∪A2∪A3)=P r ( A1)+P r ( A2)+P r ( A3)−Pr(A2∩A3)−Pr(A1∩A3) −Pr(A1∩A2)+P r ( A1∩A2∩A3). (26.15)ICalculate the probability of drawing from a pack of cards one that is an ace or is a spade or shows an even number ( 2 ,4 ,6 ,8 ,1 0 ) . If, as previously, Ais the event that an ace is drawn, Pr( A)=4 52. Similarly the event B, that a spade is drawn, has Pr( B)=13 52. The further possibility C, that the card is even (but not a picture card) has Pr( C)=20 52. The two-fold intersections have probabilities Pr(A∩B)=1 52,Pr(A∩C)=0 ,Pr(B∩C)=5 52. There is no three-fold intersection as events AandCare mutually exclusive. Hence Pr(A∪B∪C)=1 52[(4 + 13 + 20) −( 1+0+5 )+( 0 ) ]=31 52. The reader should identify the 31 cards involved. J When the probabilities are combined to calculate the probability for the union of the ngeneral events, the result, which may be proved by induction upon n(see the answer to exercise 26.4), is Pr(A1∪A2∪···∪An)=summationdisplay iPr(Ai)−summationdisplay i,jPr(Ai∩Aj)+summationdisplay i,j,kPr(Ai∩Aj∩Ak) −···+(−1)n+1Pr(A1∩A2∩···∩An). (26.16) Each summation runs over all possible sets of subscripts, except those in which any two subscripts in a set are the same. The number of terms in the summationof probabilities of m-fold intersections of the nevents is given by nCm(as discussed in section 26.1). Equation (26.9) is a special case of (26.16) in which n=2a n d only the ﬁrst two terms on the RHS survive. We now illustrate this result with aworked example that has n= 4 and includes a four-fold intersection. 969 PROBABILITYIFind the probability of drawing from a pack a card that has at least one of the following properties: A,i ti sa na c e ; B, it is a spade; C, it is a black honour card (ace, king, queen, jack or 10); D, it is a black ace. Measuring all probabilities in units of1 52, the single-event probabilities are Pr(A)=4 , Pr(B)=1 3 , Pr(C)=1 0 , Pr(D)=2 . The two-fold intersection probabilities, measured in the same units, are Pr(A∩B)=1 , Pr(A∩C)=2 , Pr(A∩D)=2 , Pr(B∩C)=5 , Pr(B∩D)=1 , Pr(C∩D)=2 . The three-fold intersections have probabilities Pr(A∩B∩C)=1 ,Pr(A∩B∩D)=1 ,Pr(A∩C∩D)=2 ,Pr(B∩C∩D)=1 . Finally, the four-fold intersection, requiring all four conditions to hold, is satisﬁed only by the ace of spades, and hence (again in units of1 52) Pr(A∩B∩C∩D)=1 . Substituting in (26.16) gives P=1 52[( 4+1 3+1 0+2 ) −( 1+2+2+5+1+2 )+( 1+1+2+1 ) −(1)]=20 52. J We conclude this section on basic theorems by deriving a useful general expression for the probability Pr( A∩B)t h a tt w oe v e n t s AandBboth occur in the case where A(say) is the union of a set of nmutually exclusive events Ai.I n this case A∩B=(A1∩B)∪···∪(An∩B), where the events Ai∩Bare also mutually exclusive. Thus, from the addition law (26.12) for mutually exclusive events, we ﬁnd Pr(A∩B)=summationdisplay iPr(Ai∩B). (26.17) Moreover, in the special case where the events Aiexhaust the sample space S,w e have A∩B=S∩B=B, and we obtain the total probability law Pr(B)=summationdisplay iPr(Ai∩B). (26.18) 26.2.2 Conditional probability So far we have deﬁned only probabilities of the form ‘what is the probability that event Ahappens?’. In this section we turn to conditional probability , the probability that a particular event occurs given the occurrence of another, possibly related, event. For example, we may wish to know the probability of event B,d r a w i n ga n 970 26.2 PROBABILITY ace from a pack of cards from which one has already been removed, given that event A, the card already removed was itself an ace, has occurred. We denote this probability by Pr( B|A) and may obtain a formula for it by considering the total probability Pr( A∩B)=P r ( B∩A) that both AandBwill occur. This may be written in two ways, i.e. Pr(A∩B)=P r ( A)P r(B|A) =P r ( B)P r (A|B). From this we obtain Pr(A|B)=Pr(A∩B) Pr(B)(26.19) and Pr(B|A)=Pr(B∩A) Pr(A). (26.20) In terms of Venn diagrams, we may think of Pr( B|A) as the probability of Bin the reduced sample space deﬁned by A. Thus, if two events AandBare mutually exclusive then Pr(A|B)=0=P r ( B|A). (26.21) When an experiment consists of drawing objects at random from a given set of objects, it is termed sampling a population . We need to distinguish between two diﬀerent ways in which such a sampling experiment may be performed. After an object has been drawn at random from the set it may either be put asideor returned to the set before the next object is randomly drawn. The former istermed ‘sampling without replacement’, the latter ‘sampling with replacement’.IFind the probability of drawing two aces at random from a pack of cards (i) when the ﬁrst card drawn is replaced at random into the pack before the second card is drawn, and(ii) when the ﬁrst card is put aside after being drawn. LetAbe the event that the ﬁrst card is an ace, and Bthe event that the second card is an ace. Now Pr(A∩B)=P r ( A)P r(B|A), and for both (i) and (ii) we know that Pr( A)=4 52=1 13. (i) If the ﬁrst card is replaced in the pack before the next is drawn then Pr( B|A)= Pr(B)=4 52=1 13,s i n c e AandBare independent events. We then have Pr(A∩B)=P r ( A)P r(B)=1 13×1 13=1 169. (ii) If the ﬁrst card is put aside and the second then drawn, AandBare not independent and Pr( B|A)=3 51, with the result that Pr(A∩B)=P r ( A)P r(B|A)=1 13×3 51=1 221. J 971 PROBABILITY Two events AandBarestatistically independent if Pr( A|B)=P r ( A) (or equiva- lently if Pr( B|A)=P r ( B)). In words, the probability of Agiven Bis then the same as the probability of Aregardless of whether Boccurs. For example, if we throw a coin and a die at the same time, we would normally expect that the probability of throwing a six was independent of whether a head was thrown. If AandBare statistically independent then it follows that Pr(A∩B)=P r ( A)P r(B). (26.22) In fact, on the basis of intuition and experience, (26.22) may be regarded as the deﬁnition of the statistical independence of two events. The idea of statistical independence is easily extended to an arbitrary number of events A1,A2,...,A n. The events are said to be (mutually) independent if Pr(Ai∩Aj)=P r ( Ai)P r (Aj), Pr(Ai∩Aj∩Ak)=P r ( Ai)P r (Aj)P r (Ak), ... Pr(A1∩A2∩···∩An)=P r ( A1)P r(A2)···Pr(An), for all combinations of indices i,jandkfor which no two indices are the same. Even if all nevents are not mutually independent, any two events for which Pr(Ai∩Aj)=P r ( Ai)P r(Aj) are said to be pairwise independent . We now derive two results that often prove useful when working with condi- tional probabilities. Let us suppose that an event Ais the union of nmutually exclusive events Ai.I fBis some other event then from (26.17) we have Pr(A∩B)=summationdisplay iPr(Ai∩B). Dividing both sides of this equation by Pr( B), and using (26.19), we obtain Pr(A|B)=summationdisplay iPr(Ai|B), (26.23) which is the addition law for conditional probabilities . Furthermore, if the set of mutually exclusive events Aiexhausts the sample space Sthen, from the total probability law (26.18), the probability Pr( B) of some event BinScan be written as Pr(B)=summationdisplay iPr(Ai)P r (B|Ai). (26.24)IA collection of traﬃc islands connected by a system of one-way roads is shown in ﬁg- ure 26.5. At any given island a car driver chooses a direction at random from those available. What is the probability that a driver starting at Owill arrive at B? In order to leave Othe driver must pass through one of A1,A2,A3orA4, which thus form a complete set of mutually exclusive events. Since at each island (including O)t h e driver chooses a direction at random from those available, we have that Pr( Ai)=1 4for 972 26.2 PROBABILITY O BA1 A2A3A4 Figure 26.5 A collection of traﬃc islands connected by one-way roads. i=1,2,3,4. From ﬁgure 26.5, we see also that Pr(B|A1)=1 3,Pr(B|A2)=1 3,Pr(B|A3)=0 ,Pr(B|A4)=2 4=1 2. Thus, using the total probability law (26.24), we ﬁnd that the probability of arriving at B is given by Pr(B)= X iPr(Ai)P r(B|Ai)=1 4 /;1 3+1 3+0+1 2 / =7 24. J Finally, we note that the concept of conditional probability may be straightfor- wardly extended to several compound events. For example, in the case of three events A, B, C ,w em a yw r i t eP r ( A∩B∩C) in several ways, e.g. Pr(A∩B∩C)=P r ( C)P r(A∩B|C) =P r ( B∩C)P r(A|B∩C) =P r ( C)P r(B|C)P r(A|B∩C).ISuppose{Ai}is a set of mutually exclusive events that exhausts the sample space S.I fB andCare two other events in S, show that Pr(B|C)= X iPr(Ai|C)P r(B|Ai∩C). Using (26.19) and (26.17), we may write Pr(C)P r(B|C)=P r ( B∩C)= X iPr(Ai∩B∩C). (26.25) Each term in the sum on the RHS can be expanded as an appropriate product of conditional probabilities, Pr(Ai∩B∩C)=P r ( C)P r(Ai|C)P r(B|Ai∩C). Substituting this form into (26.25) and dividing through by Pr( C) gives the required result. J 973 PROBABILITY 26.2.3 Bayes’ theorem In the previous section we saw that the probability that both an event Aand a related event Bwill occur can be written either as Pr( A)P r (B|A)o rP r ( B)P r(A|B). Hence Pr(A)P r (B|A)=P r ( B)P r (A|B), from which we obtain Bayes’ theorem , Pr(A|B)=Pr(A) Pr(B)Pr(B|A). (26.26) This theorem clearly shows that Pr( B|A)/negationslash=P r ( A|B), unless Pr( A)=P r ( B). It is sometimes useful to rewrite Pr( B), if it is not known directly, as Pr(B)=P r ( A)P r(B|A)+P r ( ¯A)P r (B|¯A) so that Bayes’ theorem becomes Pr(A|B)=Pr(A)P r(B|A) Pr(A)P r(B|A)+P r ( ¯A)P r (B|¯A). (26.27)ISuppose that the blood test for some disease is reliable in the following sense: for people who are infected with the disease the test produces a positive result in 99.99% of cases; forpeople not infected a positive test result is obtained in only 0.02% of cases. Furthermore,assume that in the general population one person in 10000 people is infected. A person is selected at random and found to test positive for the disease. What is the probability thatthe individual is actually infected? LetAbe the event that the individual is infected and Bbe the event that the individual tests positive for the disease. Using Bayes’ theorem the probability that a person who tests positive is actually infected is Pr(A|B)=Pr(A)P r(B|A) Pr(A)P r(B|A)+Pr(¯A)Pr(B|¯A). Now Pr( A)=1 /10000 = 1−Pr(¯A), and we are told that Pr( B|A) = 9999 /10000 and Pr(B|¯A)=2 /10000. Thus we obtain Pr(A|B)=1/10000×9999/10000 (1/10000×9999/10000) + (9999 /10000×2/10000)=1 3. Thus, there is only a one in three chance that a person chosen at random, who tests positive for the disease, is actually infected. At a ﬁrst glance, this answer may seem a little surprising, but the reason for the counter- intuitive result is that the probability tha t a randomly selected person is not infected is 9999/10000, which is very high. Thus, the 0.02% chance of a match for an uninfected person becomes signiﬁcant. J We note that (26.27) may be written in a more general form if Sis not simply 974 26.3 PERMUTATIONS AND COMBINATIONS divided into Aand¯Abut, rather, into anyset of mutually exclusive events Aithat exhaust S. Using the total probability law (26.24), we may then write Pr(B)=summationdisplay iPr(Ai)P r (B|Ai), so that Bayes’ theorem takes the form Pr(A|B)=Pr(A)P r (B|A)summationtext iPr(Ai)P r (B|Ai), (26.28) where the event Aneed not coincide with any of the Ai. As a ﬁnal point, we comment that sometimes we are concerned only with the relative probabilities of two events AandC(say), given the occurrence of some other event B. From (26.26) we then obtain a diﬀerent form of Bayes’ theorem, Pr(A|B) Pr(C|B)=Pr(A)P r (B|A) Pr(C)P r (B|C), (26.29) which does not contain Pr( B)a ta l l . 26.3 Permutations and combinations In equation (26.5) we deﬁned the probability of an event Ain a sample space S as Pr(A)=nA nS, where nAis the number of outcomes belonging to event AandnSis the total number of possible outcomes. It is therefore necessary to be able to count thenumber of possible outcomes in various common situations. 26.3.1 Permutations Let us ﬁrst consider a set of nobjects that are all diﬀerent. We may ask in how many ways these nobjects may be arranged, i.e. how many permutations of these objects exist. This is straightforward to deduce, as follows: the object in theﬁrst position may be chosen in ndiﬀerent ways, that in the second position in n−1 ways, and so on until the ﬁnal object is positioned. The number of possible arrangements is therefore n(n−1)(n−2)···(1) = n! (26.30) Generalising (26.30) slightly, let us suppose we choose only k(<n)o b j e c t s from n. The number of possible permutations of these kobjects selected from n is given by n(n−1)(n−2)···(n−k+1 )bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright kfactors=n! (n−k)!≡nPk. (26.31) 975 PROBABILITY In calculating the number of permutations of the various objects we have so far assumed that the objects are sampled without replacement – i.e. once an object has been drawn from the set it is put aside. As mentioned previously, however,we may instead replace each object before the next is chosen. The number of permutations of kobjects from nwith replacement may be calculated very easily since the ﬁrst object can be chosen in ndiﬀerent ways, as can the second, the third, etc. Therefore the number of permutations is simply n k. This may also be viewed as the number of permutations of kobjects from nwhere repetitions are allowed, i.e. each object may be used as often as one likes.IFind the probability that in a group of kpeople, at least two have the same birthday (ignoring 29February). It is simplest to begin by calculating the probability that no two people share a birthday, as follows. Firstly, we imagine each of the kpeople in turn pointing to their birthday on a year planner. Thus, we are sampling the 365 days of the year ‘with replacement’ and so the total number of possible outcomes is (365)k. Now, (for the moment) we assume that no two people share a birthday and imagine the process being repeated, but as eachperson points out their birthday it is crossed oﬀ the planner. In this case, we are samplingthe days of the year ‘without replacement’, and so the possible number of outcomes forwhich all the birthdays are diﬀerent is 365Pk=365! (365−k)!. Hence the probability that all the birthdays are diﬀerent is p=365! (365−k)! 365k. Now using the complement rule (26.11), the probability qthat two or more people have the same birthday is simply q=1−p=1−365! (365−k)! 365k. This expression may be conveniently evalutated using Stirling’s approximation for n!w h e n nis large, namely n!∼√ 2πn /n e /n , to give q≈1−e−k /365 365−k /365−k+0.5 . It is interesting to note that if k= 23 the probability is a little greater than a half that at least two people have the same birthday, and if k= 50 the probability rises to 0.970. This can prove a good bet at a party of non-mathematicians! J So far we have assumed that all nobjects are diﬀerent (or distinguishable ). Let us now consider nobjects of which n1are identical and of type 1, n2are identical and of type 2, ...,n mare identical and of type m(clearly n=n1+n2+···+nm). From (26.30) the number of permutations of these nobjects is again n!. However, 976 26.3 PERMUTATIONS AND COMBINATIONS the number of distinguishable permutations is only n! n1!n2!···nm!, (26.32) since the ith group of identical objects can be rearranged in ni! ways without changing the distinguishable permutation.IA set of snooker balls consists of a white, a yellow, a green, a brown, a blue, a pink, a black and 15reds. How many distinguishable permutations of the balls are there? In total there are 22 balls, the 15 reds being indistinguishable. Thus from (26.32) the number of distinguishable permutations is 22! (1!)(1!)(1!)(1!)(1!)(1!)(15!)=22! 15!= 859541760 . J 26.3.2 Combinations We now consider the number of combinations of various objects when their order is immaterial. Assuming all the objects to be distinguishable, from (26.31) we seethat the number of permutations of kobjects chosen from nis nPk=n!/(n−k)!. Now, since we are no longer concerned with the order of the chosen objects, which can be internally arranged in k! diﬀerent ways, the number of combinations of k objects from nis n! (n−k)!k!≡nCk≡parenleftbiggn kparenrightbigg for 0≤k≤n, (26.33) where, as noted in chapter 1,nCkis called the binomial coeﬃcient since it also appears in the binomial expansion for positive integer n,n a m e l y (a+b)n=nsummationdisplay k=0nCkakbn−k. (26.34)IA hand of 13 playing cards is dealt from a well-shuﬄed deck of 52. What is the probability that the hand contains two aces? Since the order of the cards in the hand is immaterial, the total number of distinct handsis simply equal to the number of combinations of 13 objects drawn from 52, i.e. 52C13. However, the number of hands containing two aces is equal to the number of ways,4C2, in which the two aces can be drawn from the four available, multiplied by the number ofways, 48C11, in which the remaining 11 cards in the hand can be drawn from the 48 cards that are not aces. Thus the required probability is given by 4C248C11 52C13=4! 2!2!48! 11!37!13!39! 52! =(3)(4) 2(12)(13)(38)(39) (49)(50)(51)(52)=0.213 J 977 PROBABILITY Another useful result that may be derived using the binomial coeﬃcients is the number of ways in which ndistinguishable objects can be divided into mpiles, with niobjects in the ith pile, i=1,2,...,m (the ordering of objects within each pile being unimportant). This may be straightforwardly calculated as follows. We may choose the n1objects in the ﬁrst pile from the original nobjects innCn1ways. Then2objects in the second pile can then be chosen from the n−n1remaining objects inn−n1Cn2ways, etc. We may continue in this fashion until we reach the (m−1)th pile, which may be formed inn−n1−···−nm−2Cnm−1ways. The remaining objects then form the mth pile and so can only be ‘chosen’ in one way. Thus the total number of ways of dividing the original nobjects into mpiles is given by the product N=nCn1n−n1Cn2···n−n1−···−nm−2Cnm−1 =n! n1!(n−n1)!(n−n1)! n2!(n−n1−n2)!···(n−n1−n2−···−nm−2)! nm−1!(n−n1−n2−···−nm−2−nm−1)! =n! n1!(n−n1)!(n−n1)! n2!(n−n1−n2)!···(n−n1−n2−···−nm−2)! nm−1!nm! =n! n1!n2!···nm!. (26.35) These numbers are called multinomial coeﬃcients since (26.35) is the coeﬃcient of xn1 1xn2 2···xnmmin the multinomial expansion of ( x1+x2+···+xm)n, i.e. for positive integer n (x1+x2+···+xm)n=summationdisplay n1,n2,... ,nm n1+n2+···+nm=nn! n1!n2!···nm!xn1 1xn2 2···xnmm. For the case m=2 ,n1=k,n2=n−k, (26.35) reduces to the binomial coeﬃcient nCk. Furthermore, we note that the multinomial coeﬃcient (26.35) is identical to the expression (26.32) for the number of distinguishable permutations of nobjects, niof which are identical and of type i(fori=1,2,...,m andn1+n2+···+nm=n). A few moments’ thought should convince the reader that the two expressions(26.35) and (26.32) must be identical.IIn the card game of bridge, each of four players is dealt 13 cards from a full pack of 52. What is the probability that each player is dealt an ace? From (26.35), the total number of distinct bridge dealings is 52! /(13!13!13!13!). However, the number of ways in which the four aces can be distributed with one in each hand is 4!/(1!1!1!1!) = 4!; the remaining 48 cards can then be dealt out in 48! /(12!12!12!12!) ways. Thus the probability that each player receives an ace is 4!48! (12!)4(13!)4 52!=24(13)4 (49)(50)(51)(52)=0.105. J As in the case of permutations we might ask how many combinations of k objects can be chosen from nwith replacement (repetition). To calculate this, we 978 26.3 PERMUTATIONS AND COMBINATIONS may imagine the n(distinguishable) objects set out on a table. Each combination ofkobjects can then be made by pointing to kof the no b j e c t si nt u r n( w i t h repetitions allowed). These kequivalent selections distributed amongst ndiﬀerent but re-choosable objects are strictly analogous to the placing of kindistinguishable ‘balls’ in ndiﬀerent boxes with no restriction on the number of balls in each box. A particular selection in the case k=7 , n= 5 may be symbolised as xxx||x|xx|x. This denotes three balls in the ﬁrst box, none in the second, one in the third, two in the fourth and one in the ﬁfth. We therefore need only to consider the number of (distinguishable) ways in which kcrosses and n−1 vertical lines can be arranged, i.e. the number of permutations of k+n−1 objects of which kare identical crosses and n−1 are identical lines. This is given by (26.33) as (k+n−1)! k!(n−1)!=n+k−1Ck. (26.36) We note that this expression also occurs in the binomial expansion for negative integer powers. If nis a positive integer, it is straightforward to show that (see chapter 1) (a+b)−n=∞summationdisplay k=0(−1)kn+k−1Cka−n−kbk, where ai st a k e nt ob el a r g e rt h a n b.IA system contains a number Nof (non-interacting) particles, each of which can be in any of the quantum states of the system. The structure of the set of quantum states is suchthat there exist Renergy levels with corresponding energies E iand degeneracies gi(i.e. the ith energy level contains giquantum states). Find the numbers of distinct ways in which the particles can be distributed among the quantum states of the system such that the ith energy level contains niparticles, for i=1,2,...,R , in the cases where the particles are (i) distinguishable with no restriction on the number in each state; (ii) indistinguishable with no restriction on the number in each state; (iii) indistinguishable with a maxim um of one particle in each state; (iv) distinguishable with a maximum of one particle in each state. It is easiest to solve this problem in two stages. Let us ﬁrst consider distributing the N particles among the Renergy levels, without regard for the individual degenerate quantum states that comprise each level. If the particles are distinguishable then the number of distinct arrangements with niparticles in the ith level, i=1,2,...,R , is given by (26.35) as N! n1!n2!···nR!. If, however, the particles are indistinguishable then clearly there exists only one distinct arrangement having niparticles in the ith level, i=1,2,...,R . Now let us suppose there exist wiways in which the niparticles in the ith energy level can be distributed among thegidegenerate states. Thus it follows that the number of distinct ways in which the N 979 PROBABILITY particles can be distributed among all Rquantum states of the system, with niparticles in theith level, is given by W{ni}= /8/>/> /> />/</>/>/> />/: N! n1!n2!···nR!RY i=1wifor distinguishable particles , RY i=1wi for indistinguishable particles .(26.37) It therefore remains only for us to ﬁnd the appropriate expression for wiin each of the cases (i)–(iv) above. Case (i). If there is no restriction on the number of particles in each quantum state, then in the ith energy level each particle can reside in any of the gidegenerate quantum states. Thus, if the particles are distinguishable then the number of distinct arrangementsis simply w i=gni i. Thus, from (26.37), W{ni}=N! n1!n2!···nR!RY i=1gni i=N!RY i=1gni i ni!. Such a system of particles (for example atoms or molecules in a classical gas) is said to obey Maxwell–Boltzmann statistics. Case (ii). If the particles are indistinguishable and there is no restriction on the number in each state then, from (26.36), the number of distinct arrangements of the niparticles among the gistates in the ith energy level is wi=(ni+gi−1)! ni!(gi−1)!. Substituting this expression in (26.37), we obtain W{ni}=RY i=1(ni+gi−1)! ni!(gi−1)!. Such a system of particles (for example a gas of photons) is said to obey Bose–Einstein statistics. Case (iii). If a maximum of one particle can reside in each of the gidegenerate quantum states in the ith energy level then the number of particles in each state is either 0 or 1. Since the particles are indistinguishable, wiis equal to the number of distinct arrangements in which nistates are occupied and gi−nistates are unoccupied; this is given by wi=giCni=gi! ni!(gi−ni)!. Thus, from (26.37), we have W{ni}=RY i=1gi! ni!(gi−ni)!. Such a system is said to obey Fermi–Dirac statistics, and an example is provided by an electron gas. Case (iv). Again, the number of particles in each state is either 0 or 1. If the particles are distinguishable, however, each arrangement identiﬁed in case (iii) can be reordered inn i! diﬀerent ways, so that wi=giPni=gi! (gi−ni)!. 980 26.4 RANDOM VARIABLES AND DISTRIBUTIONS Substituting this expression into (26.37) gives W{ni}=N!RY i=1gi! ni!(gi−ni)!. Such a system of particles has the names of no famous scientists attached to it, since it appears that it never occurs in nature. J 26.4 Random variables and distributions Suppose an experiment has an outcome sample space S. A real variable Xthat is deﬁned for all possible outcomes in S(so that a real number – not necessarily unique – is assigned to each possible outcome) is called a random variable (RV). The outcome of the experiment may already be a real number and hence a randomvariable, e.g. the number of heads obtained in 10 throws of a coin, or the sum of the values if two dice are thrown. However, more arbitrary assignments are possi- ble, e.g. the assignment of a ‘quality’ rating to each successive item produced by amanufacturing process. Furthermore, assuming that a probability can be assignedto all possible outcomes in a sample space S, it is possible to assign a probability distribution to any random variable. Random variables may be divided into two classes, discrete and continuous, and we now examine each of these in turn. 26.4.1 Discrete random variables A random variable Xthat takes only discrete values x 1,x2,...,x n, with proba- bilities p1,p2,...,p n, is called a discrete random variable. The number of values nfor which Xhas a non-zero probability is ﬁnite or at most countably inﬁnite. As mentioned above, an example of a discrete random variable is the number ofheads obtained in 10 throws of a coin. If Xis a discrete random variable, we can deﬁne a probability function (PF) f(x) that assigns probabilities to all the distinct values that Xcan take, such that f(x)=P r ( X=x)=braceleftBigg p iifx=xi, 0o t h e r w i s e .(26.38) A typical PF (see ﬁgure 26.6) thus consists of spikes, at valid values ofX, whose height at xcorresponds to the probability that X=x. Since the probabilities must sum to unity, we require nsummationdisplay i=1f(xi)=1 . (26.39) We may also deﬁne the cumulative probability function (CPF) of X,F(x), whose value gives the probability that X≤x,s ot h a t F(x)=P r ( X≤x)=summationdisplay xi≤xf(xi). (26.40) 981 PROBABILITY xf(x) F(x) 2p p 1 2p 11 12 2 3 3 4 4 5 5 6 6 (a) (b) Figure 26.6 ( a) A typical probability function for a discrete distribution, that for the biased die discussed earlier. Since the probabilities must sum to unitywe require p=2/13. (b) The cumulative probability function for the same discrete distribution. (Note that a diﬀerent scale has been used for ( b).) Hence F(x) is a step function that has upward jumps of piatx=xi,i= 1,2,...,n, and is constant between possible values of X. We may also calculate the probability that Xlies between two limits, l1andl2(l1<l2); this is given by Pr(l1<X≤l2)=summationdisplay l1<xi≤l2f(xi)=F(l2)−F(l1), (26.41) i.e. it is the sum of all the probabilities for which xilies within the relevant interval.IA bag contains seven red balls and three white balls. Three balls are drawn at random and not replaced. Find the probability function for the number of red balls drawn. LetXbe the number of red balls drawn. Then Pr(X=0 )= f(0) =3 10×2 9×1 8=1 120, Pr(X=1 )= f(1) =3 10×2 9×7 8×3=7 40, Pr(X=2 )= f(2) =3 10×7 9×6 8×3=21 40, Pr(X=3 )= f(3) =7 10×6 9×5 8=7 24. It should be noted that P3 i=0f(i) = 1, as expected. J 26.4.2 Continuous random variables A random variable Xis said to have a continuous distribution if Xis deﬁned for a continuous range of values between given limits (often −∞to∞). An example of a continuous random variable is the height of a person drawn from a population,which can take anyvalue (within limits!). We can deﬁne the probability density function (PDF) f(x) of a continuous random variable Xsuch that Pr(x<X≤x+dx)=f(x)dx, 982 26.4 RANDOM VARIABLES AND DISTRIBUTIONS l1 l2 abxf(x) Figure 26.7 The probability density function for a continuous random vari- ableXthat can take values only between the limits l1andl2. The shaded area under the curve gives Pr( a<X≤b), whereas the total area under the curve, between the limits l1andl2, is equal to unity. i.e.f(x)dxis the probability that Xlies in the interval x<X≤x+dx. Clearly f(x) must be a real function that is everywhere ≥0. If Xcan only take values between the limits l1andl2then in order for the sum of the probabilities of all possible outcomes to be equal to unity, we require integraldisplayl2 l1f(x)dx=1. Often Xcan take any value between −∞and∞and so integraldisplay∞ −∞f(x)dx=1. The probability that Xlies in the interval a<X≤bis then given by Pr(a<X≤b)=integraldisplayb af(x)dx, (26.42) i.e. Pr( a<X≤b) is equal to the area under the curve of f(x) between these limits (see ﬁgure 26.7). We may also deﬁne the cumulative probability function F(x) for a continuous random variable by F(x)=P r ( X≤x)=integraldisplayx l1f(u)du, (26.43) where uis a (dummy) integration variable. We can then write Pr(a<X≤b)=F(b)−F(a). From (26.43) it is clear that f(x)=dF(x)/dx. 983 PROBABILITYIA random variable Xhas a PDF f(x)given by Ae−xin the interval 0<x<∞and zero elsewhere. Find the value of the constant Aand hence calculate the probability that Xlies in the interval 1<X≤2. We require the integral of f(x) between 0 and ∞to equal unity. Evaluating this integral, we ﬁndZ∞ 0Ae−xdx= / −Ae−x /∞ 0=A, and hence A= 1. From (26.42), we then obtain Pr(1<X≤2) = Z2 1f(x)dx= Z2 1e−xdx=−e−2−(−e−1)=0 .23. J It is worth mentioning here that a discrete RV can in fact be treated as continuous and assigned a corresponding probability density function. If Xis a discrete RV that takes only the values x1,x2,...,x nwith probabilities p1,p2,...,p n then we may describe Xas a continuous RV with PDF f(x)=nsummationdisplay i=1piδ(x−xi), (26.44) where δ(x) is the Dirac delta function discussed in subsection 13.1.3. From (26.42) and the fundamental property of the delta function (13.12), we see that Pr(a<X≤b)=integraldisplayb af(x)dx, =nsummationdisplay i=1piintegraldisplayb aδ(x−xi)dx=summationdisplay ipi, where the ﬁnal sum extends over those values of ifor which a<x i≤b. 26.4.3 Sets of random variables It is common in practice to consider two or more random variables simultane- ously. For example, one might be interested in both the height and weight ofa person drawn at random from a population. In the general case, these vari-ables may depend on one another and are described by joint probability density functions . These are discussed fully in section 26.11, and we simply note here that, if we have (say) two random variables XandY, then by analogy with the single-variable case we deﬁne their joint probability density function f(x, y)i n such a way that, if XandYare discrete RVs, Pr(X=x i,Y=yj)=f(xi,yj), or, if XandYare continuous RVs, Pr(x<X≤x+dx, y < Y≤y+dy)=f(x, y)dx dy. 984 26.5 PROPERTIES OF DISTRIBUTIONS In many circumstances, however, random variables do not depend on one another, i.e. they are independent . As an example, for a person drawn at random from a population, we might expect height and IQ to be independent randomvariables. Let us suppose that XandYare two random variables with probability density functions g(x)a n d h(y) respectively. In mathematical terms, XandYare independent RVs if their joint probability density function is given by f(x, y)= g(x)h(y). Thus, for independent RVs, if XandYare both discrete then Pr(X=x i,Y=yj)=g(xi)h(yj) or, if XandYare both continuous, then Pr(x<X≤x+dx, y < Y≤y+dy)=g(x)h(y)dx dy. The important point in each case is that the RHS is simply the product of the individual probability density functions (compare with the expression for Pr( A∪B) in (26.22) for statistically independent events AandB). By a simple extension, one may also consider the case where one of the random variables is discrete and the other continuous. The above discussion may also be trivially extended to any number of independent RVs Xi,i=1,2,...,N .IThe independent random variables XandYhave the PDFs g(x)=e−xandh(y)=2 e−2y respectively. Calculate the probability that Xlies in the interval 1<X≤2andYlies in the interval 0<Y≤1. Since XandYare independent RVs, the required probability is given by Pr(1<X≤2,0<Y≤1) = Z2 1g(x)dx Z1 0h(y)dy = Z2 1e−xdx Z1 02e−2ydy = / −e−x /2 1× / −e−2y /1 0=0.23×0.86 = 0 .20. J 26.5 Properties of distributions For a single random variable X, the probability density function f(x) contains all possible information about how the variable is distributed. However, for the purposes of comparison, it is conventional and useful to characterise f(x)b y certain of its properties. Most of these standard properties are deﬁned in termsofaverages orexpectation values . In the most general case, the expectation value E[g(X)] of any function g(X) of the random variable Xis deﬁned as E[g(X)] =braceleftBiggsummationtext ig(xi)f(xi) for a discrete distribution,integraltext g(x)f(x)dxfor a continuous distribution,(26.45) 985 PROBABILITY where the sum or integral is over all allowed values of X. It is assumed that the series is absolutely convergent or that the integral exists, as the case may be.From its deﬁnition it is straightforward to show that the expectation value hasthe following properties: (i) if ais a constant then E[a]=a; (ii) if ais a constant then E[ag(X)] =aE[g(X)]; (iii) if g(X)=s(X)+t(X)t h e n E[g(X)] =E[s(X)] +E[t(X)]. It should be noted that the expectation value is not a function of Xbut is instead a number that depends on the form of the probability density functionf(x) and the function g(x). Most of the standard quantities used to characterise f(x) are simply the expectation values of various functions of the random variable X. We now consider these standard quantities. 26.5.1 Mean The property most commonly used to characterise a probability distribution is itsmean, which is deﬁned simply as the expectation value E[X] of the variable X itself. Thus, the mean is given by E[X]=braceleftBiggsummationtext ixif(xi) for a discrete distribution,integraltext xf(x)dxfor a continuous distribution.(26.46) The alternative notations µand/angbracketleftx/angbracketrightare also commonly used to denote the mean. If in (26.46) the series is not absolutely convergent, or the integral does not exist, we say that the distribution does not have a mean, but this is very rare in physicalapplications.IThe probability of ﬁnding a 1selectron in a hydrogen atom in a given inﬁnitesimal volume dVisψ∗ψd V, where the quantum mechanical wavefunction ψis given by ψ=Ae−r/a0. Find the value of the real constant Aand thereby deduce the mean distance of the electron from the origin. Let us consider the random variable R= ‘distance of the electron from the origin’. Since the 1s orbital has no θ-o rφ-dependence (it is spherically symmetric), we may consider the inﬁnitesimal volume element dVas the spherical shell with inner radius rand outer radius r+dr. Thus, dV=4πr2drand the PDF of Ris simply Pr(r<R≤r+dr)≡f(r)dr=4πr2A2e−2r/a0dr. The value of Ais found by requiring the total probability (i.e. the probability that the electron is somewhere ) to be unity. Since Rmust lie between zero and inﬁnity, we require that A2 Z∞ 0e−2r/a04πr2dr=1. 986 26.5 PROPERTIES OF DISTRIBUTIONS Integrating by parts we ﬁnd A=1/(πa3 0)1/2. Now, using the deﬁnition of the mean (26.46), we ﬁnd E[R]= Z∞ 0rf(r)dr=4 a3 0 Z∞ 0r3e−2r/a0dr. The expression on the RHS may be integrated by parts and takes the value 3 a4 0/8; consequently we ﬁnd that E[R]=3 a0/2. J 26.5.2 Mode and median Although the mean discussed in the last section is the most common measure of the ‘average’ of a distribution, two other measures, which do not rely on theconcept of expectation values, are frequently encountered. Themodeof a distribution is the value of the random variable Xat which the probability (density) function f(x) has its greatest value. If there is more than one value of Xfor which this is true then each value may equally be called the mode of the distribution. Themedian Mof a distribution is the value of the random variable Xat which the cumulative probability function F(x) takes the value 1 2,i . e .F(M)=1 2. Related to the median are the lower and upper quartiles QlandQuof the PDF, which are deﬁned such that F(Ql)=1 4,F (Qu)=3 4. Thus the median and lower and upper quartiles divide the PDF into four regions each containing one quarter of the probability. Smaller subdivisions are alsopossible, e.g. the nth percentile, P n, of a PDF is deﬁned by F(Pn)=n/100.IFind the mode of the PDF for the distance from the origin of the electron whose wave- function was given in the previous example. We found in the previous example that the PDF for the electron’s distance from the originwas given by f(r)=4r 2 a3 0e−2r/a0. (26.47) Diﬀerentiating f(r) with respect to r,w eo b t a i n df dr=8r a3 0 / 1−r a0 / e−2r/a0. Thus f(r) has turning points at r=0a n d r=a0,w h e r e df/dr = 0. It is straightforward to show that r= 0 is a minimum and r=a0is a maximum. Moreover, it is also clear that r=a0is a global maximum (as opposed to just a local one). Thus the mode of f(r) occurs atr=a0. J 987 PROBABILITY 26.5.3 Variance and standard deviation Thevariance of a distribution, V[X], also written σ2, is deﬁned by V[X]=Ebracketleftbig (X−µ)2bracketrightbig =braceleftBiggsummationtext j(xj−µ)2f(xj) for a discrete distribution,integraltext (x−µ)2f(x)dxfor a continuous distribution. (26.48) Here µhas been written for the expectation value E[X]o fX. As in the case of the mean, unless the series and the integral in (26.48) converge the distributiondoes not have a variance. From the deﬁnition (26.48) we may easily derive thefollowing useful properties of V[X]. Ifaandbare constants then (i)V[a]=0 , (ii)V[aX+b]=a 2V[X]. The variance of a distribution is always positive; its positive square root is known as the standard deviation of the distribution and is often denoted by σ. Roughly speaking, σmeasures the spread (about x=µ) of the values that Xcan assume.IFind the standard deviation of the PDF for the distance from the origin of the electron whose wavefunction was discussed in the previous two examples. Inserting the expression (26.47) for the PDF f(r) into (26.48), the variance of the random variable Ris given by V[R]= Z∞ 0(r−µ)24r2 a3 0e−2r/a0dr=4 a3 0 Z∞ 0(r4−2r3µ+r2µ2)e−2r/a0dr, where the mean µ=E[R]=3 a0/2. Integrating each term in the integrand by parts we obtain V[R]=3 a2 0−3µa0+µ2=3a2 0 4. Thus the standard deviation of the distribution is σ=√ 3a0/2. J We may also use the deﬁnition (26.48) to derive the Bienaym ´e–Chebyshev inequality , which provides a useful upper limit on the probability that random variable Xtakes values outside a given range centred on the mean. Let us consider the case of a continuous random variable, for which Pr(|X−µ|≥c)=integraldisplay |x−µ|≥cf(x)dx, where the integral on the RHS extends over all values of xsatisfying the inequality 988 26.5 PROPERTIES OF DISTRIBUTIONS |x−µ|≥c. From (26.48), we ﬁnd that σ2≥integraldisplay |x−µ|≥c(x−µ)2f(x)dx≥c2integraldisplay |x−µ|≥cf(x)dx. (26.49) The ﬁrst inequality holds because both ( x−µ)2andf(x) are non-negative for allx, and the second inequality holds because ( x−µ)2≥c2over the range of integration. However, the RHS of (26.49) is simply equal to c2Pr(|X−µ|≥c), and thus we obtain the required inequality Pr(|X−µ|≥c)≤σ2 c2. A similar derivation may be carried through for the case of a discrete random variable. Thus, for anydistribution f(x) that possesses a variance we have, for example, Pr(|X−µ|≥2σ)≤1 4and Pr(|X−µ|≥3σ)≤1 9. 26.5.4 Moments The mean (or expectation) of Xis sometimes called the ﬁrst moment ofX,s i n c e it is deﬁned as the sum or integral of the probability density function multiplied by the ﬁrst power of x. By a simple extension the kth moment of a distribution is deﬁned by µk≡E[Xk]=braceleftBiggsummationtext jxk jf(xj) for a discrete distribution,integraltext xkf(x)dxfor a continuous distribution.(26.50) For notational convenience, we have introduced the symbol µkto denote E[Xk], thekth moment of the distribution. Clearly, the mean of the distribution is then denoted by µ1, often abbreviated simply to µ, as in the previous subsection, as this rarely causes confusion. A useful result that relates the second moment, the mean and the variance of a distribution is proved using the properties of the expectation operator: V[X]=Ebracketleftbig (X−µ)2bracketrightbig =Ebracketleftbig X2−2µX+µ2bracketrightbig =Ebracketleftbig X2bracketrightbig −2µE[X]+µ2 =Ebracketleftbig X2bracketrightbig −2µ2+µ2 =Ebracketleftbig X2bracketrightbig −µ2. (26.51) In alternative notations, this result can be written /angbracketleft(x−µ)2/angbracketright=/angbracketleftx2/angbracketright−/angbracketleftx/angbracketright2or σ2=µ2−µ2 1. 989 PROBABILITYIA biased die has probabilities p/2,p,p,p,p,2pof showing 1, 2, 3, 4, 5, 6 respectively. Find (i) the mean, (ii) the second moment and (iii) th e variance of this probability distribution. By demanding that the sum of the probabilities equals unity we require p=2/13. Now, using the deﬁnition of the mean (26.46) for a discrete distribution, E[X]= X jxjf(xj)=1×1 2p+2×p+3×p+4×p+5×p+6×2p =53 2p=53 2×2 13=53 13. Similarly, using the deﬁnition of the second moment (26.50), E[X2]= X jx2 jf(xj)=12×1 2p+22p+32p+42p+52p+62×2p =253 2p=253 13. Finally, using the deﬁnition of the variance (26.48), with µ=5 3/13, we obtain V[X]= X j(xj−µ)2f(xj) =( 1−µ)21 2p+( 2−µ)2p+( 3−µ)2p+( 4−µ)2p+( 5−µ)2p+( 6−µ)22p = /3120 169 / p=480 169. It is easy to verify that V[X]=E / X2 / −(E[X])2. J In practice, to calculate the moments of a distribution it is often simpler to use the moment generating function discussed in subsection 26.7.2. This is particularlytrue for higher-order moments, where direct evaluation of the sum or integral in (26.50) can be somewhat laborious. 26.5.5 Central moments The variance V[X] is sometimes called the second central moment of the distribu- tion, since it is deﬁned as the sum or integral of the probability density functionmultiplied by the second power of x−µ. The origin of the term ‘central’ is that by subtracting µfrom xbefore squaring we are considering the moment about the mean of the distribution, rather than about x= 0. Thus the kthcentral moment of a distribution is deﬁned as ν k≡Ebracketleftbig (X−µ)kbracketrightbig =braceleftBiggsummationtext j(xj−µ)kf(xj) for a discrete distribution,integraltext (x−µ)kf(x)dxfor a continuous distribution.(26.52) It is convenient to introduce the notation νkfor the kth central moment. Thus V[X]≡ν2and we may write (26.51) as ν2=µ2−µ2 1. Clearly, the ﬁrst central 990 26.5 PROPERTIES OF DISTRIBUTIONS moment of a distribution is always zero since, for example in the continuous case, ν1=integraldisplay (x−µ)f(x)dx=integraldisplay xf(x)dx−µintegraldisplay f(x)dx=µ−(µ×1) = 0 . We note that the notation µkand νkfor the moments and central moments respectively is not universal. Indeed, in some books their meanings are reversed. We can write the kth central moment of a distribution in terms of its kth and lower-order moments by expanding ( X−µ)kin powers of X. We have already noted that ν2=µ2−µ2 1, and similar expressions may be obtained for higher-order central moments. For example, ν3=Ebracketleftbig (X−µ1)3bracketrightbig =Ebracketleftbig X3−3µ1X2+3µ2 1X−µ3 1bracketrightbig =µ3−3µ1µ2+3µ2 1µ1−µ3 1 =µ3−3µ1µ2+2µ3 1. (26.53) In general, it is straightforward to show that νk=µk−kC1µk−1µ1+···+(−1)rkCrµk−rµr 1+···+(−1)k−1(kCk−1−1)µk 1. (26.54) Once again, direct evaluation of the sum or integral in (26.52) can be rather tedious for higher moments, and it is usually quicker to use the moment generatingfunction (see subsection 26.7.2), from which the central moments can be easilyevaluated as well.IThe PDF for a Gaussian distribution (see subsection 26.9.1) with mean µand variance σ2is given by f(x)=1 σ√ 2πexp / −(x−µ)2 2σ2 / . Obtain an expression for the kth central moment of this distribution. As an illustration, we will perform this calculation by evaluating the integral in (26.52) directly. Thus, the kth central moment of f(x)i sg i v e nb y νk= Z∞ −∞(x−µ)kf(x)dx =1 σ√ 2π Z∞ −∞(x−µ)kexp / −(x−µ)2 2σ2 / dx =1 σ√ 2π Z∞ −∞ykexp / −y2 2σ2 / dy, (26.55) where in the last line we have made the substitution y=x−µ. It is clear that if kis odd then the integrand is an odd function of yand hence the integral equals zero. Thus, νk=0i f kis odd. When kis even, we could calculate νkby integrating by parts to obtain a reduction formula, but it is more elegant to consider instead the standard integral (seesubsection 6.4.2) I= Z∞ −∞exp(−αy2)dy=π1/2α−1/2, 991 PROBABILITY and diﬀerentiate it repeatedly with respect to α(see section 5.12). Thus, we obtain dI dα=− Z∞ −∞y2exp(−αy2)dy=−1 2π1/2α−3/2 d2I dα2= Z∞ −∞y4exp(−αy2)dy=(1 2)(3 2)π1/2α−5/2 ... dnI dαn=(−1)n Z∞ −∞y2nexp(−αy2)dy=(−1)n(1 2)(3 2)···(1 2(2n−1))π1/2α−(2n+1)/2. Setting α=1/(2σ2) and substituting the above result into (26.55), we ﬁnd (for keven) νk=(1 2)(3 2)···(1 2(k−1))(2σ2)k/2= (1)(3) ···(k−1)σk. J One may also characterise a probability distribution f(x) using the closely related normalised and dimensionless central moments γk≡νk νk/2 2=νk σk. From this set, γ3andγ4are more commonly called, respectively, the skewness andkurtosis of the distribution. The skewness γ3of a distribution is zero if it is symmetrical about its mean. If the distribution is skewed to values of xsmaller than the mean then γ3<0. Similarly γ3>0 if the distribution is skewed to higher values of x. From the above example, we see that the kurtosis of the Gaussian distribution (subsection 26.9.1) is given by γ4=ν4 ν2 2=3σ4 σ4=3. It is therefore common practice to deﬁne the excess kurtosis of a distribution asγ4−3. A positive value of the excess kurtosis implies a relatively narrower peak and wider wings than the Gaussian distribution with the same mean andvariance. A negative excess kurtosis implies a wider peak and shorter wings. Finally, we note here that one can also describe a probability density function f(x) in terms of its cumulants , which are again related to the central moments. However, we defer the discussion of cumulants until subsection 26.7.4, since their deﬁnition is most easily understood in terms of generating functions. 26.6 Functions of random variables Suppose Xis some random variable for which the probability density function f(x) is known. In many cases, we may be more interested in the related random variable Y=Y(X), where Y(X) is some function of X. What is the probability 992 26.6 FUNCTIONS OF RANDOM VARIABLES density function g(y) for the new random variable Y? We now discuss how to obtain this function. 26.6.1 Discrete random variables IfXis a discrete RV that takes only the values xi,i=1,2,...,n,t h e n Ymust also be discrete and takes the values yi=Y(xi), although some of these values may be identical. The probability function for Yis given by g(y)=braceleftBiggsummationtext jf(xj)i f y=yi, 0o t h e r w i s e ,(26.56) where the sum extends over those values of jfor which yi=Y(xj). The simplest case arises when the function Y(X) possesses a single-valued inverse X(Y). In this case, only one x-value corresponds to each y-value, and we obtain a closed-form expression for g(y)g i v e nb y g(y)=braceleftBigg f(x(y)) if y=yi, 0o t h e r w i s e . IfY(X) does not possess a single-valued inverse then the situation is more complicated and it may not be possible to obtain a closed-form expression forg(y). Nevertheless, whatever the form of Y(X), one can always use (26.56) to obtain the numerical values of the probability function g(y)a ty=y i. 26.6.2 Continuous random variables IfXis a continuous RV, then so too is the new random variable Y=Y(X). The probability that Ylies in the range ytoy+dyis given by g(y)dy=integraldisplay dSf(x)dx, (26.57) where dScorresponds to all values of xfor which Ylies in the range ytoy+dy. Once again the simplest case occurs when Y(X) possesses a single-valued inverse X(Y). In this case, we may write g(y)dy=vextendsinglevextendsinglevextendsinglevextendsingleintegraldisplayx(y+dy) x(y)f(x/prime)dx/primevextendsinglevextendsinglevextendsinglevextendsingle=integraldisplayx(y)+|dx dy|dy x(y)f(x/prime)dx/prime, from which we obtain g(y)=f(x(y))vextendsinglevextendsinglevextendsinglevextendsingledx dyvextendsinglevextendsinglevextendsinglevextendsingle. (26.58) 993 PROBABILITY lighthouse beam L 0θ coastliney Figure 26.8 The illumination of a coastline by the beam from a lighthouse.IA lighthouse is situated at a distance Lfrom a straight coastline, opposite a point O, and sends out a narrow continuous beam of light simultaneously in opposite directions. The beamrotates with constant angular velocity. If the random variable Yis the distance along the coastline, measured from O, of the spot that the light beam illuminates, ﬁnd its probability density function. The situation is illustrated in ﬁgure 26.8. Since the light beam rotates at a constant angularvelocity, θis distributed uniformly between −π/2a n d π/2, and so f(θ)=1 /π.N o w y=Ltanθ, which possesses the single-valued inverse θ=t a n −1(y/L), provided that θlies between−π/2a n d π/2. Since dy/dθ =Lsec2θ=L(1 + tan2θ)=L[1 + ( x/L)2], from (26.58) we ﬁnd g(y)=1 π ////dθ dy ////=1 πL[1 + ( y/L)2]for−∞<y<∞. A distribution of this form is called a Cauchy distribution and is discussed in subsec- tion 26.9.5. J IfY(X) does not possess a single-valued inverse then we encounter complica- tions, since there exist several intervals in the X-domain for which Ylies between yandy+dy. This is illustrated in ﬁgure 26.9, which shows a function Y(X) such that X(Y) is a double-valued function of Y. Thus the range ytoy+dy corresponds to X’s being either in the range x1tox1+dx1or in the range x2to x2+dx2. In general, it may not be possible to obtain an expression for g(y)i n closed form, although the distribution may always be obtained numerically using(26.57). However, a closed-form expression may be obtained in the case wherethere exist single-valued functions x 1(y)a n d x2(y) giving the two values of xthat correspond to any given value of y.I nt h i sc a s e , g(y)dy=vextendsinglevextendsinglevextendsinglevextendsingleintegraldisplay x1(y+dy) x1(y)f(x)dxvextendsinglevextendsinglevextendsinglevextendsingle+vextendsinglevextendsinglevextendsinglevextendsingleintegraldisplay x2(y+dy) x2(y)f(x)dxvextendsinglevextendsinglevextendsinglevextendsingle, from which we obtain g(y)=f(x 1(y))vextendsinglevextendsinglevextendsinglevextendsingledx1 dyvextendsinglevextendsinglevextendsinglevextendsingle+f(x2(y))vextendsinglevextendsinglevextendsinglevextendsingledx2 dyvextendsinglevextendsinglevextendsinglevextendsingle. (26.59) 994 26.6 FUNCTIONS OF RANDOM VARIABLES yy+dy dx1 dx2XY Figure 26.9 Illustration of a function Y(X) such that its inverse X(Y)i sa double-valued function of Y. The range ytoy+dycorresponds to Xbeing either in the range x1tox1+dx1or in the range x2tox2+dx2. This result may be generalised straightforwardly to the case where the range yto y+dycorresponds to more than two x-intervals.IThe random variable Xis Gaussian distributed (see subsection 26.9.1) with mean µand variance σ2. Find the PDF of the new variable Y=(X−µ)2/σ2. It is clear that X(Y) is a double-valued function of Y. However, in this case, it is straightforward to obtain single-valued functions giving the two values of xthat correspond to a given value of y;t h e s ea r e x1=µ−σ√yandx2=µ+σ√y,w h e r e√yis taken to mean the positive square root. The PDF of Xis given by f(x)=1 σ√ 2πexp / −(x−µ)2 2σ2 / . Since dx1/dy=−σ/(2√y)a n d dx2/dy=σ/(2√y), from (26.59) we obtain g(y)=1 σ√ 2πexp(−1 2y) ////−σ 2√y ////+1 σ√ 2πexp(−1 2y) ////σ 2√y //// =1 2√π(1 2y)−1/2exp(−1 2y). As we shall see in subsection 26.9.3, this is the gamma distribution γ(1 2,1 2). J 26.6.3 Functions of several random variables We may extend our discussion further, to the case in which the new random variable is a function of several other random variables. For deﬁniteness, let us consider the random variable Z=Z(X,Y), which is a function of two other RVs XandY. Given that these variables are described by the joint probability density function f(x, y), we wish to ﬁnd the probability density function p(z)o f the variable Z. 995 PROBABILITY IfXandYare both discrete RVs then p(z)=summationdisplay i,jf(xi,yj), (26.60) where the sum extends over all values of iandjfor which Z(xi,yj)=z. Similarly, ifXandYare both continuous RVs then p(z) is found by requiring that p(z)dz=integraldisplayintegraldisplay dSf(x, y)dx dy, (26.61) where dSis the inﬁnitesimal area in the xy-plane lying between the curves Z(x, y)=zandZ(x, y)=z+dz.ISuppose XandYare independent continuous random variables in the range −∞to∞, with PDFs g(x)andh(y)respectively. Obtain expressions for the PDFs of Z=X+Yand W=XY. Since XandYare independent RVs, their joint PDF is simply f(x, y)=g(x)h(y). Thus, from (26.61), the PDF of the sum Z=X+Yis given by p(z)dz= Z∞ −∞dx g(x) Zz+dz−x z−xdy h(y) = /Z∞ −∞g(x)h(z−x)dx / dz. Thus p(z)i st h e convolution of the PDFs of gandh(i.e.p=g∗h, see subsection 13.1.7). In a similar way, the PDF of the product W=XYis given by q(w)dw= Z∞ −∞dx g(x) Z(w+dw)/|x| w/|x|dy h(y) = /Z∞ −∞g(x)h(w/x)dx |x| / dw J The prescription (26.61) is readily generalised to functions of nrandom variables Z=Z(X1,X2,...,X n), in which case the inﬁnitesimal ‘volume’ element dSis the region in x1x2···xn-space between the (hyper)surfaces Z(x1,x2,...,x n)=zand Z(x1,x2,...,x n)=z+dz. In practice, however, the integral is diﬃcult to evaluate, since one is faced with the complicated geometrical problem of determining thelimits of integration. Fortunately, an alternative (and powerful) technique existsfor evaluating integrals of this kind. One eliminates the geometrical problem byintegrating over allvalues of the variables x iwithout restriction, while shifting the constraint on the variables to the integrand. This is readily achieved bymultiplying the integrand by a function that equals unity in the inﬁnitesimal region dSand zero elsewhere. From the discussion of the Dirac delta function in subsection 13.1.3, we see that δ(Z(x 1,x2,...,x n)−z)dzsatisﬁes these requirements, and so in the most general case we have p(z)=integraldisplayintegraldisplay ···integraldisplay f(x1,x2,...,x n)δ(Z(x1,x2,...,x n)−z)dx1dx2...d x n, (26.62) 996 26.6 FUNCTIONS OF RANDOM VARIABLES where the range of integration is over all possible values of the variables xi.T h i s integral is most readily evaluated by substituting in (26.62) the Fourier integralrepresentation of the Dirac delta function discussed in subsection 13.1.4, namely δ(Z(x 1,x2,...,x n)−z)=1 2πintegraldisplay∞ −∞eik(Z(x1,x2,...,x n)−z)dk. (26.63) This is best illustrated by considering a speciﬁc example.IA general one-dimensional random walk consists of nindependent steps, each of which can be of a diﬀerent length and in either direction along the x-axis. If g(x)is the PDF for the (positive or negative) displacement Xalong the x-axis achieved in a single step, obtain an expression for the PDF of the total displacement Safter nsteps. The total displacement Sis simply the algebraic sum of the displacements Xiachieved in each of the nsteps, so that S=X1+X2+···+Xn. Since the random variables Xiare independent and have the same PDF g(x), their joint PDF is simply g(x1)g(x2)···g(xn). Substituting this into (26.62), together with (26.63), we obtain p(s)= Z∞ −∞ Z∞ −∞··· Z∞ −∞g(x1)g(x2)···g(xn)1 2π Z∞ −∞eik[(x1+x2+···+xn)−s]dk dx 1dx2···dxn =1 2π Z∞ −∞dk e−iks /Z∞ −∞g(x)eikxdx /n . (26.64) It is convenient to deﬁne the characteristic function C(k) of the variable Xas C(k)= Z∞ −∞g(x)eikxdx, which is simply related to the Fourier transform of g(x). Then (26.64) may be written as p(s)=1 2π Z∞ −∞e−iks[C(k)]ndk. Thus p(s) can be found by evaluating two Fourier in tegrals. Characteristic functions will be discussed in more detail in subsection 26.7.3. J 26.6.4 Expectation values and variances In some cases, one is interested only in the expectation value or the variance of the new variable Zrather than in its full probability density function. For deﬁniteness, let us consider the random variable Z=Z(X,Y), which is a function of two RVs XandYwith a known joint distribution f(x, y); the results we will obtain are readily generalised to more (or fewer) variables. It is clear that E[Z]a n d V[Z] can be obtained, in principle, by ﬁrst using the methods discussed above to obtain p(z) and then evaluating the appropriate sums or integrals. The intermediate step of calculating p(z) is not necessary, however, since it is straightforward to obtain expressions for E[Z]a n d V[Z] in terms of 997 PROBABILITY the variables XandY. For example, if XandYare continuous RVs then the expectation value of Zis given by E[Z]=integraldisplay zp(z)dz=integraldisplayintegraldisplay Z(x, y)f(x, y)dx dy. (26.65) An analogous result exists for discrete random variables. Integrals of the form (26.65) are often diﬃcult to evaluate. Nevertheless, we may use (26.65) to derive an important general result concerning expectationvalues. If XandYareanytwo random variables and aandbare arbitrary constants then by letting Z=aX+bYwe ﬁnd E[aX+bY]=aE[X]+bE[Y]. Furthermore, we may use this result to obtain an approximate expression for the expectation value E[Z(X,Y)] of any arbitrary function of XandY. Letting µ X= E[X]a n d µY=E[Y], then, provided Z(X,Y) can be reasonably approximated by the linear terms of its Taylor expansion about the point ( µX,µY), we have Z(X,Y)≈Z(µX,µY)+parenleftbigg∂Z ∂Xparenrightbigg (X−µX)+parenleftbigg∂Z ∂Yparenrightbigg (Y−µY), (26.66) where the partial derivatives are evaluated at X=µXandY=µY.T a k i n gt h e expectation value of both sides, we ﬁnd E[Z(X,Y)]≈Z(µX,µY)+parenleftbigg∂Z ∂Xparenrightbigg (E[X]−µX)+parenleftbigg∂Z ∂Yparenrightbigg (E[Y]−µY)=Z(µX,µY), which gives the approximate result E[Z(X,Y)]≈Z(µX,µY). By analogy with (26.65), the variance of Z=Z(X,Y)i sg i v e nb y V[Z]=integraldisplay (z−µZ)2p(z)dz=integraldisplayintegraldisplay [Z(x, y)−µZ]2f(x, y)dx dy, (26.67) where µZ=E[Z]. We may use this expression to derive a useful general result. If XandYare two independent random variables, so that f(x, y)=g(x)h(y), and a,bandcare constants then by setting Z=aX+bY+cin (26.67) we obtain V[aX+bY+c]=a2V[X]+b2V[Y]. (26.68) From (26.68) we also obtain the important special case V[X+Y]=V[X−Y]=V[X]+V[Y]. Provided XandYare indeed independent random variables, we may obtain an approximate expression for V[Z(X,Y)], for any arbitrary function Z(X,Y), in a similar manner to that used in approximating E[Z(X,Y)] above. Taking the 998 26.7 GENERATING FUNCTIONS variance of both sides of (26.66), and using (26.68), we ﬁnd V[Z(X,Y)]≈parenleftbigg∂Z ∂Xparenrightbigg2 V[X]+parenleftbigg∂Z ∂Yparenrightbigg2 V[Y], (26.69) the partial derivatives being evaluated at X=µXandY=µY. 26.7 Generating functions As we saw in chapter 16, when dealing with particular sets of functions fn, each member of the set being characterised by a diﬀerent non-negative integern, it is sometimes possible to summarise the whole set by a single function of a dummy variable (say t), called a generating function. The relationship between the generating function and the nth member f nof the set is that if the generating function is expanded as a power series in tthen fnis the coeﬃcient of tn.F o r example, in the expansion of the generating function G(z,t)=( 1−2zt+t2)−1/2, the coeﬃcient of tnis the nth Legendre polynomial Pn(z), i.e. G(z,t)=( 1−2zt+t2)−1/2=∞summationdisplay n=0Pn(z)tn. We found that many useful properties of, and relationships between, the members of a set of functions could be established using the generating function and otherfunctions obtained from it, e.g. its derivatives. Similar ideas can be used in the area of probability theory, and two types of generating function can be usefully deﬁned, one more generally applicable than the other. The more restricted of the two, applicable only to discrete integral distributions, is called a probability generating function; this is discussed in thenext section. The second type, a moment generating function, can be used withboth discrete and continuous distributio ns and is considered in subsection 26.7.2. From the moment generating function, we may also construct the closely re-lated characteristic and cumulant generating functions; these are discussed insubsections 26.7.3 and 26.7.4 respectively. 26.7.1 Probability generating functions As already indicated, probability generating functions are restricted in applicabil- ity to integer distributions, of which the most common (the binomial, the Poissonand the geometric) are considered in this and later subsections. In such distribu-tions a random variable may take only non-negative integer values. The actualpossible values may be ﬁnite or inﬁnite in number, but, for formal purposes, all integers, 0 ,1,2,...are considered possible. If only a ﬁnite number of integer values can occur in any particular case then those that cannot occur are includedbut are assigned zero probability. 999 PROBABILITY If, as previously, the probability that the random variable Xtakes the value xn isf(xn), then summationdisplay nf(xn)=1 . In the present case, however, only non-negative integer values of xnare possible, and we can, without ambiguity, write the probability that Xtakes the value nas fn, with ∞summationdisplay n=0fn=1. (26.70) We may now deﬁne the probability generating function ΦX(t)b y ΦX(t)≡∞summationdisplay n=0fntn. (26.71) It is immediately apparent that Φ X(t)=E[tX] and that, by virtue of (26.70), ΦX(1) = 1. Probably the simplest example of a probability generating function (PGF) is provided by the random variable Xdeﬁned by X=braceleftBigg 1 if the outcome of a single trial is a ‘success’, 0 if the trial ends in ‘failure’. If the probability of success is pand that of failure q(= 1−p)t h e n ΦX(t)=qt0+pt1+0+0+ ···=q+pt. (26.72) This type of random variable is discussed much more fully in subsection 26.8.1. In a similar but slightly more complicated way, a Poisson-distributed integervariable with mean λ(see subsection 26.8.4) has a PGF Φ X(t)=∞summationdisplay n=0e−λλn n!tn=e−λeλt. (26.73) We note that, as required, Φ X(1) = 1 in both cases. Useful results will be obtained from this kind of approach only if the summation (26.71) can be carried out explicitly in particular cases and the functions derivedfrom Φ X(t) can be shown to be related to meaningful parameters. Two such relationships can be obtained by diﬀerentiating (26.71) with respect to t. Taking the ﬁrst derivative we ﬁnd dΦX(t) dt=∞summationdisplay n=0nfntn−1⇒Φ/prime X(1) =∞summationdisplay n=0nfn=E[X], (26.74) 1000 26.7 GENERATING FUNCTIONS and diﬀerentiating once more we obtain d2ΦX(t) dt2=∞summationdisplay n=0n(n−1)fntn−2⇒Φ/prime/prime X(1) =∞summationdisplay n=0n(n−1)fn=E[X(X−1)]. (26.75) Equation (26.74) shows that Φ/prime X(1) gives the mean of X. Using both (26.75) and (26.51) allows us to write Φ/prime/prime X(1) + Φ/primeX(1)−bracketleftbig Φ/prime X(1)bracketrightbig2=E[X(X−1)] + E[X]−(E[X])2 =Ebracketleftbig X2bracketrightbig −E[X]+E[X]−(E[X])2 =Ebracketleftbig X2bracketrightbig −(E[X])2 =V[X], (26.76) and so express the variance of Xin terms of the derivatives of its probability generating function.IA random variable Xis given by the number of trials needed to obtain a ﬁrst success when the chance of success at each trial is constant and equal to p. Find the probability generating function for Xand use it to determine the mean and variance of X. Clearly, at least one trial is needed, and so f0=0 .I f n(≥1) trials are needed for the ﬁrst success, the ﬁrst n−1 trials must have resulted in failure. Thus Pr(X=n)=qn−1p, n≥1, (26.77) where q=1−pis the probability of failure in each individual trial. The corresponding probability generating function is thus ΦX(t)=∞X n=0fntn=∞X n=1(qn−1p)tn =p q∞X n=1(qt)n=p q×qt 1−qt=pt 1−qt, (26.78) where we have used the result for the sum of a geometric series, given in chapter 4, to obtain a closed-form expression for Φ X(t). Again, as must be the case, Φ X(1) = 1. To ﬁnd the mean and variance of Xwe need to evaluate Φ/prime X(1) and Φ/prime/primeX(1). Diﬀerentiating (26.78) gives Φ/prime X(t)=p (1−qt)2⇒Φ/prime X(1) =p p2=1 p, Φ/prime/prime X(t)=2pq (1−qt)3⇒Φ/prime/prime X(1) =2pq p3=2q p2. Thus, using (26.74) and (26.76), E[X]=Φ/prime X(1) =1 p, V[X]=Φ/prime/prime X(1) + Φ/primeX(1)−[Φ/prime X(1)]2 =2q p2+1 p−1 p2=q p2. A distribution with probabilities of the general form (26.77) is known as a geometric distribution and is discussed in subsection 26.8.2. This form of distribution is common in ‘waiting time’ problems (subsection 26.9.3). J 1001 PROBABILITY n r=n r Figure 26.10 The pairs of values of nandrused in the evaluation of Φ X+Y(t). Sums of random variables We now turn to considering the sum of two or more independent random variables, say XandY, and denote by S2the random variable S2=X+Y. If Φ S2(t)i st h eP G Ff o r S2, the coeﬃcient of tnin its expansion is given by the probability that X+Y=nand is thus equal to the sum of the probabilities that X=randY=n−rfor all values of rin 0≤r≤n. Since such outcomes for diﬀerent values of rare mutually exclusive, we have Pr(X+Y=n)=∞summationdisplay r=0Pr(X=r)P r (Y=n−r). (26.79) Multiplying both sides of (26.79) by tnand summing over all values of nenables us to express this relationship in terms of probability generating functions asfollows: Φ X+Y(t)=∞summationdisplay n=0Pr(X+Y=n)tn=∞summationdisplay n=0nsummationdisplay r=0Pr(X=r)trPr(Y=n−r)tn−r =∞summationdisplay r=0∞summationdisplay n=rPr(X=r)trPr(Y=n−r)tn−r. The change in summation order is justiﬁed by reference to ﬁgure 26.10, which illustrates that the summations are over exactly the same pairs of values of nand r, but with the ﬁrst (inner) summation over the points in a column rather than over the points in a row. Now, setting n=r+sgives the ﬁnal result, ΦX+Y(t)=∞summationdisplay r=0Pr(X=r)tr∞summationdisplay s=0Pr(Y=s)ts =Φ X(t)ΦY(t), (26.80) 1002 26.7 GENERATING FUNCTIONS i.e. the PGF of the sum of two independent random variables is equal to the product of their individual PGFs. The same result can be deduced in a less formalway by noting that if XandYare independent then Ebracketleftbig t X+Ybracketrightbig =Ebracketleftbig tXbracketrightbig Ebracketleftbig tYbracketrightbig . Clearly result (26.80) can be extended to more than two random variables by writing S3=S2+Zetc., to give Φ( Pn i=1Xi)(t)=nproductdisplay i=1ΦXi(t), (26.81) and, further, if all the Xihave the same probability distribution, Φ( Pn i=1Xi)(t)=[ΦX(t)]n. (26.82) This latter result has immediate application in the deduction of the PGF for the binomial distribution from that for a single trial, equation (26.72). Variable-length sums of random variables As a ﬁnal result in the theory of probability generating functions we show how to calculate the PGF for a sum of Nrandom variables, all with the same probability distribution, when the value of Nis itself a random variable but one with a known probability distribution. In symbols, we wish to ﬁnd the distribution of SN=X1+X2+···+XN, (26.83) where Nis a random variable with Pr( N=n)= hnand PGF χN(t)=summationtexthntn. The probability ξkthatSN=kis given by a sum of conditional probabilities, namely† ξk=∞summationdisplay n=0Pr(N=n)P r(X0+X1+X2+···+Xn=k) =∞summationdisplay n=0hn×coeﬃcient of tkin [Φ X(t)]n. Multiplying both sides of this equation by tkand summing over all k,w eo b t a i n †Formally X0= 0 has to be included, since Pr( N= 0) may be non-zero. 1003 PROBABILITY an expression for the PGF Ξ S(t)o fSN: ΞS(t)=∞summationdisplay k=0ξktk=∞summationdisplay k=0tk∞summationdisplay n=0hn×coeﬃcient of tkin [Φ X(t)]n =∞summationdisplay n=0hn∞summationdisplay k=0tk×coeﬃcient of tkin [Φ X(t)]n =∞summationdisplay n=0hn[ΦX(t)]n =χN(ΦX(t)). (26.84) In words, the PGF of the sum SNis given by the compound function χN(ΦX(t)) obtained by substituting Φ X(t)f o r tin the PGF for the number of terms Nin the sum. We illustrate this with the following example.IThe probability distribution for the number of eggs in a clutch is Poisson distributed with mean λ, and the probability that each egg will hatch is p(and is independent of the size of the clutch). Use the results stated in (26.72) and (26.73) to show that the PGF (and hencethe probability distribution) for the numbe r of chicks that hatch corresponds to a Poisson distribution having mean λp. The number of chicks that hatch is given by a sum of the form (26.83) in which Xi=1i f theith chick hatches and Xi= 0 if it does not. As given by (26.72), Φ X(t) is thus (1−p)+pt. The value of Nis given by a Poisson distribution with mean λ; thus, from (26.73), in the terminology of our previous discussion, χN(t)=e−λeλt. We now substitute these forms into (26.84) to obtain ΞS(t)=e x p (−λ)exp[ λΦX(t)] =e x p (−λ)exp{λ[(1−p)+pt]} =e x p (−λp)exp( λpt). But this is exactly the PGF of a Poisson distribution with mean λp. That this implies that the probability is Poisson distributed is intuitively obvious since, in the expansion of the PGF as a power series in t, every coeﬃcient will be precisely that implied by such a distribution. A solution of the same problem by direct calculation appears in the answer to exercise 26.29. J 26.7.2 Moment generating functions As we saw in section 26.5 a probability function is often expressed in terms of its moments. This leads naturally to the second type of generating function, amoment generating function . For a random variable X, and a real number t,t h e moment generating function (MGF) is deﬁned by M X(t)=Ebracketleftbig etXbracketrightbig =braceleftBiggsummationtext ietxif(xi) for a discrete distribution,integraltext etxf(x)dxfor a continuous distribution.(26.85) 1004 26.7 GENERATING FUNCTIONS The MGF will exist for all values of tprovided that Xis bounded and always exists at the point t=0w h e r e M(0) = E(1) = 1. It will be apparent that the PGF and the MGF for a random variable X are closely related. The former is the expectation of tXwhilst the latter is the expectation of etX: ΦX(t)=Ebracketleftbig tXbracketrightbig ,M X(t)=Ebracketleftbig etXbracketrightbig . The MGF can thus be obtained from the PGF by replacing tbyet,a n dv i c e versa. The MGF has more general applicability, however, since it can be used with both continuous and discrete distributions whilst the PGF is restricted to non-negative integer distributions. As its name suggests, the MGF is particularly useful for obtaining the moments of a distribution, as is easily seen by noting that Ebracketleftbig etXbracketrightbig =Ebracketleftbigg 1+tX+t2X2 2!+···bracketrightbigg =1+ E[X]t+Ebracketleftbig X2bracketrightbigt2 2!+···. Assuming that the MGF exists for all taround the point t= 0, we can deduce that the moments of a distribution are given in terms of its MGF by E[Xn]=dnMX(t) dtnvextendsinglevextendsinglevextendsinglevextendsingle t=0. (26.86) Similarly, by substitution in (26.51), the variance of the distribution is given by V[X]=M/prime/prime X(0)−bracketleftbig M/prime X(0)bracketrightbig2, (26.87) where the prime denotes diﬀerentiation with respect to t.IThe MGF for the Gaussian distribution (see the end of subsection 26.9.1) is given by MX(t)=e x p /; µt+1 2σ2t2 / . Find the expectation and variance of this distribution. Using (26.86), M/prime X(t)= /; µ+σ2t / exp /; µt+1 2σ2t2 / ⇒ E[X]=M/prime X(0) = µ, M/prime/prime X(t)= / σ2+(µ+σ2t)2 / exp /; µt+1 2σ2t2 / ⇒ M/prime/prime X(0) = σ2+µ2. Thus, using (26.87), V[X]=σ2+µ2−µ2=σ2. That the mean is found to be µand the variance σ2justiﬁes the use of these symbols in the Gaussian distribution. J The moment generating function has several useful properties that follow from its deﬁnition and can be employed in simplifying calculations. 1005 PROBABILITY Scaling and shifting IfY=aX+b,w h e r e aandbare arbitrary constants, then MY(t)=Ebracketleftbig etYbracketrightbig =Ebracketleftbig et(aX+b)bracketrightbig =ebtEbracketleftbig eatXbracketrightbig =ebtMX(at). (26.88) This result is often useful for obtaining the central moments of a distribution. If the MFG of XisMX(t) then the variable Y=X−µhas the MGF MY(t)=e−µtMX(t), which clearly generates the central moments of X,i . e . E[(X−µ)n]=E[Yn]=M(n) Y(0) =parenleftbiggdn dtn[e−µtMX(t)]parenrightbigg t=0. Sums of random variables IfX1,X2,...,X Nare independent random variables and SN=X1+X2+···+XN then MSN(t)=Ebracketleftbig etSNbracketrightbig =Ebracketleftbig et(X1+X2+···+XN)bracketrightbig =EbracketleftBiggNproductdisplay i=1etXibracketrightBigg . Since the Xiareindependent , MSN(t)=Nproductdisplay i=1Ebracketleftbig etXibracketrightbig =Nproductdisplay i=1MXi(t). (26.89) In words, the MGF of the sum of Nindependent random variables is the product of their individual MGFs. By combining (26.89) with (26.88), we obtain the moregeneral result that the MGF of S N=c1X1+c2X2+···+cNXN(where the ciare constants) is given by MSN(t)=Nproductdisplay i=1MXi(cit). (26.90) Variable-length sums of random variables Let us consider the sum of Nindependent random variables Xi(i=1,2,...,N ), all with the same probability distribution, and let us suppose that Nis itself a random variable with a known distribution. Following the notation of section 26.7.1, SN=X1+X2+···+XN, where Nis a random variable with Pr( N=n)=hnand probability generating function χN(t)=summationtexthntn. For deﬁniteness, let us assume that the Xiare continuous RVs (an analogous discussion can be given in the discrete case). Thus, the 1006 26.7 GENERATING FUNCTIONS probability that value of SNlies in the interval stos+dsis given by† Pr(s<S N≤s+ds)=∞summationdisplay n=0Pr(N=n)P r (s<X 0+X1+X2···+Xn≤s+ds). Let us denote Pr( s<S N≤s+ds)b yfN(s)dsand Pr( s<X 0+X1+X2···+Xn≤ s+ds)b yfn(s)ds. Thus, the kth moment of the PDF fN(s)i sg i v e nb y µk=integraldisplay skfN(s)ds=integraldisplay sk∞summationdisplay n=0Pr(N=n)fn(s)ds =∞summationdisplay n=0Pr(N=n)integraldisplay skfn(s)ds =∞summationdisplay n=0hn×(k!×coeﬃcient of tkin [MX(t)]n) Thus the MGF of SNis given by MSN(t)=∞summationdisplay k=0µk k!tk=∞summationdisplay n=0hn∞summationdisplay k=0tk×coeﬃcient of tkin [MX(t)]n =∞summationdisplay n=0hn[MX(t)]n =χN(MX(t)). In words, the MGF of the sum SNis given by the compound function χN(MX(t)) obtained by substituting MX(t)f o r tin the PGF for the number of terms Nin the sum. Uniqueness If the MGF of the random variable X1is identical to that for X2then the probability distributions of X1andX2are identical. This is intuitively reasonable although a rigorous proof is complicated, ‡and beyond the scope of this book. 26.7.3 Characteristic function Thecharacteristic function (CF) of a random variable Xis deﬁned as CX(t)=Ebracketleftbig eitXbracketrightbig =braceleftBiggsummationtext jeitxjf(xj) for a discrete distribution,integraltext eitxf(x)dxfor a continuous distribution(26.91) so that CX(t)=MX(it), where MX(t)i st h eM G Fo f X. Clearly, the characteristic †As in the previous section, X0has to be formally included, since Pr( N= 0) may be non-zero. ‡See, for example, Moran, An Introduction to Probability Theory (Oxford Science Publications). 1007 PROBABILITY function and the MGF are very closely related and can be used interchangeably. Because of the formal similarity between the deﬁnitions of CX(t)a n d MX(t), the characteristic function possesses analogous properties to those listed in the previ-ous section for the MGF, with only minor modiﬁcations. Indeed, by substituting it fortin any of the relations obeyed by the MGF and noting that C X(t)=MX(it), we obtain the corresponding relationship for the characteristic function. Thus, forexample, the moments of Xare given in terms of the derivatives of C X(t)b y E[Xn]=(−i)nC(n) X(0). Similarly, if Y=aX+bthen CY(t)=eibtCX(t). Whether to describe a random variable by its characteristic function or by its MGF is partly a matter of personal preference. However, the use of the CF doeshave some advantages. Most importantly, the replacement of the exponential e tX in the deﬁnition of the MGF by the complex oscillatory function eitXin the CF means that in the latter we avoid any diﬃculties associated with convergence of the relevant sum or integral. Furthermore, when Xis a continous RV, we see from (26.91) that CX(t) is related to the Fourier transform of the PDF f(x). As a consequence of Fourier’s inversion theorem, we may obtain f(x)f r o m CX(t)b y performing the inverse transform f(x)=1 2πintegraldisplay∞ −∞CX(t)e−itxdt. 26.7.4 Cumulant generating function As mentioned at the end of subsection 26.5.5, we may also describe a probability density function f(x) in terms of its cumulants . These quantities may be expressed in terms of the moments of the distribution and are important in sampling theory,which we discuss in the next chapter. The cumulants of a distribution are bestdeﬁned in terms of its cumulant generating function (CGF), given by K X(t)= lnMX(t)w h e r e MX(t) is the MGF of the distribution. If KX(t)i se x p a n d e da sa power series in tthen the kth cumulant κkoff(x) is the coeﬃcient of tk/k!: KX(t)=l n MX(t)≡κ1t+κ2t2 2!+κ3t3 3!+···. (26.92) Since MX(0) = 1, KX(t) contains no constant term.IFind all the cumulants of the Gaussian distribution discussed in the previous example. The moment generating function for the Gaussian distribution is MX(t)=e x p /; µt+1 2σ2t2 / . Thus, the cumulant generating function has the simple form KX(t)=l n MX(t)=µt+1 2σ2t2. Comparing this expression with (26.92), we ﬁnd that κ1=µ,κ2=σ2and all other cumulants are equal to zero. J 1008 26.8 IMPORTANT DISCRETE DISTRIBUTIONS We may obtain expressions for the cumulants of a distribution in terms of its moments by diﬀerentiating (26.92) with respect to tto give dKX dt=1 MXdMX dt. Expanding each term as power series in tand cross-multiplying, we obtain parenleftbigg κ1+κ2t+κ3t2 2!+···parenrightbiggparenleftbigg 1+µ1t+µ2t2 2!+···parenrightbigg =parenleftbigg µ1+µ2t+µ3t2 2!+···parenrightbigg , and, on equating coeﬃcients of like powers of ton each side, we ﬁnd µ1=κ1, µ2=κ2+κ1µ1, µ3=κ3+2κ2µ1+κ1µ2, µ4=κ4+3κ3µ1+3κ2µ2+κ1µ3, ... µk=κk+k−1C1κk−1µ1+···+k−1Crκk−rµr+···+κ1µk−1. Solving these equations for the κk, we obtain (for the ﬁrst four cumulants) κ1=µ1, κ2=µ2−µ2 1=ν2, κ3=µ3−3µ2µ1+2µ3 1=ν3, κ4=µ4−4µ3µ1+1 2µ2µ2 1−3µ2 2−6µ4 1=ν4−3ν2 2. (26.93) Higher-order cumulants may be calculated in the same way but become increas- ingly lengthy to write out in full. The principal property of cumulants is their additivity, which may be proved by combining (26.92) with (26.90). If X1,X2,...,XNare independent random variables and KXi(t)f o r i=1,2,...,N is the CGF for Xithen the CGF of SN=c1X1+c2X2+···+cNXN(where the ciare constants) is given by KSN(t)=Nsummationdisplay i=1KXi(cit). Cumulants also have the useful property that, under a change of origin X→ X+athe ﬁrst cumulant undergoes the change κ1→κ1+abut all higher-order cumulants remain unchanged. Under a change of scale X→bX, cumulant κr undergoes the change κr→brκr. 26.8 Important discrete distributions Having discussed the some general properties of distributions, we now consider the more important discrete distributions encountered in physical applications. 1009 PROBABILITY Distribution Probability law f(x)M G F E[X] V[X] binomialnCxpxqn−x(pet+q)nnp npq negative binomialr+x−1Cxprqx /p 1−qet /rrq prq p2 geometric qx−1ppet 1−qet1 pq p2 hypergeometric(Np)!(Nq)!n!(N−n)! x!(Np−x)!(n−x)!(Nq−n+x)!N!npN−n N−1npq Poissonλx x!e−λeλ(et−1)λλ Table 26.1 Some important discrete probability distributions. These are discussed in detail below, and summarised for convenience in table 26.1; we refer the reader to the relevant section below for an explanation of the symbolsused. 26.8.1 The binomial distribution Perhaps the most important discrete probability distribution is the binomial dis- tribution . This distribution describes processes that consist of a number of inde- pendent identical trials with two possible outcomes, AandB=¯A.W em a yc a l l these outcomes ‘success’ and ‘failure’ respectively. If the probability of a success is Pr( A)=p, then the probability of a failure is Pr( B)=q=1−p.I fw ep e r f o r m ntrials then the discrete random variable X= number of times Aoccurs can take the values 0 ,1,2,...,n; its distribution amongst these values is described by the binomial distribution . We now calculate the probability that in ntrials we obtain xsuccesses (and so n−xfailures). One way of obtaining such a result is to have xsuccesses followed byn−xfailures. Since the trials are assumed independent, the probability of this is pp···pbracehtipupleft bracehtipdownrightbracehtipdownleftbracehtipupright xtimes×qq···qbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright n−xtimes=pxqn−x. This is, however, just one permutation of xsuccesses and n−xfailures. The total number of permutations of nobjects, of which xare identical and of type 1 and n−xare identical and of type 2, is given by (26.33) as n! x!(n−x)!≡nCx. 1010 26.8 IMPORTANT DISCRETE DISTRIBUTIONS 1 11 1 2 22 2 33 33 444 4 55 5 5 6 6 7 78 8 9 910 100 0 00000 00.1 0.1 0.1 0.10.2 0.2 0.2 0.20.3 0.3 0.3 0.30.4 0.4 0.4 0.4 x xx x f(x) f(x)f(x) f(x) n=5 , p=0.6 n=5 , p=0.167 n= 10, p=0.6 n= 10, p=0.167 Figure 26.11 Some typical binomial distributions with various combinations of parameters nandp. Therefore, the total probability of obtaining xsuccesses from ntrials is f(x)=P r ( X=x)=nCxpxqn−x=nCxpx(1−p)n−x, (26.94) which is the binomial probability distribution formula . When a random variable Xfollows the binomial distribution for ntrials, with a probability of success p, we write X∼Bin(n, p). Then the random variable Xi so f t e nr e f e r r e dt oa sa binomial variate . Some typical binomial distributions are shown in ﬁgure 26.11.IIf a single six-sided die is rolled ﬁve times, what is the probability that a six is thrown exactly three times? Here the number of ‘trials’ n= 5, and we are interested in the random variable X= number of sixes thrown. Since the probability of a ‘success’ is p=1 6, the probability of obtaining exactly three sixes in ﬁve throws is given by (26.94) as Pr(X=3 )=5! 3!(5−3)! /1 6 /3 /5 6 /(5−3) =0.032. J In evaluating binomial probabilities a useful result is the binomial recurrence formula Pr(X=x+1 )=p qparenleftbiggn−x x+1parenrightbigg Pr(X=x), (26.95) 1011 PROBABILITY which enables successive probabilities Pr( X=x+k),k=1,2,..., to be calculated once Pr( X=x) is known; it is often quicker to use than (26.94).IThe random variable Xis distributed as X∼Bin(3 ,1 2). Evaluate the probability function f(x)using the binomial recurrence formula. The probability Pr( X= 0) may be calculated using (26.94) and is Pr(X=0 )=3C0 /;1 2 /0 /;1 2 /3=1 8. The ratio p/q=1 2/1 2= 1 in this case and so, using the binomial recurrence formula (26.95), we ﬁnd Pr(X=1 )=1×3−0 0+1×1 8=3 8, Pr(X=2 )=1×3−1 1+1×3 8=3 8, Pr(X=3 )=1×3−2 2+1×3 8=1 8, results which may be veriﬁed by direct application of (26.94). J We note that, as required, the binomial distribution satiﬁes nsummationdisplay x=0f(x)=nsummationdisplay x=0nCxpxqn−x=(p+q)n=1. Furthermore, from the deﬁnitions of E[X]a n d V[X] for a discrete distribution, we may show that for the binomial distribution E[X]=npandV[X]=npq.T h e direct summations involved are, however, rather cumbersome and these resultsare obtained much more simply using the moment generating function. The moment generating function for the binomial distribution To ﬁnd the MGF for the binomial distribution we consider the binomial random variable Xto be the sum of the random variables X i,i=1,2,...,n,w h i c ha r e deﬁned by Xi=braceleftBigg 1 if a ‘success’ occurs on the ith trial, 0 if a ‘failure’ occurs on the ith trial. Thus Mi(t)=Ebracketleftbig etXibracketrightbig =e0t×Pr(Xi=0 )+ e1t×Pr(Xi=1 ) =1×q+et×p =pet+q. From (26.89), it follows that the MGF for the binomial distribution is given by M(t)=nproductdisplay i=1Mi(t)=(pet+q)n. (26.96) 1012 26.8 IMPORTANT DISCRETE DISTRIBUTIONS We can now use the moment generating function to derive the mean and variance of the binomial distribution. From (26.96) M/prime(t)=npet(pet+q)n−1, and from (26.86) E[X]=M/prime(0) = np(p+q)n−1=np, where the last equality follows from p+q=1 . Diﬀerentiating with respect to tonce more gives M/prime/prime(t)=et(n−1)np2(pet+q)n−2+etnp(pet+q)n−1, and from (26.86) E[X2]=M/prime/prime(0) = n2p2−np2+np. Thus, using (26.87) V[X]=M/prime/prime(0)−bracketleftbig M/prime(0)bracketrightbig2=n2p2−np2+np−n2p2=np(1−p)=npq. Multiple binomial distributions Suppose Xand Yare two independent random variables, both of which are described by binomial distributions with a common probability of success p, but with (in general) diﬀerent numbers of trials n1andn2,s ot h a t X∼Bin(n1,p) andY∼Bin(n2,p). Now consider the random variable Z=X+Y.W ec o u l d calculate the probability distribution of Zdirectly using (26.60), but it is much easier to use the MGF (26.96). Since XandYare independent random variables, the MGF MZ(t) of the new variable Z=X+Yis given simply by the product of the individual MGFs MX(t)a n d MY(t). Thus, we obtain MZ(t)=MX(t)MY(t)=(pet+q)n1(pet+q)n1=(pet+q)n1+n2, which we recognise as the MGF of Z∼Bin(n1+n2,p). Hence Zis also described by a binomial distribution. This result may be extended to any number of binomial distributions. If Xi, i=1,2,...,N , is distributed as Xi∼Bin(ni,p)t h e n Z=X1+X2+···+XNis distributed as Z∼Bin(n1+n2+···+nN,p), as would be expected since the result ofsummationtext initrials cannot depend on how they are split up. A similar proof is also possible using either the probability or cumulant generating functions. Unfortunately, no equivalent simple result exists for the probability distribution of the diﬀerence Z=X−Yof two binomially distributed variables. 1013 PROBABILITY 26.8.2 The geometric and negative binomial distributions A special case of the binomial distribution occurs when instead of the number of successes we consider the discrete random variable X= number of trials required to obtain the ﬁrst success . The probability that xtrials are required in order to obtain the ﬁrst success, is simply the probability of obtaining x−1 failures followed by one success. If the probability of a success on each trial is p,t h e nf o r x>0 f(x)=P r ( X=x)=( 1−p)x−1p=qx−1p, where q=1−p. This distribution is sometimes called the geometric distribution . The probability generating function for this distribution is given in (26.78). Byreplacing tbye tin (26.78) we immediately obtain the MGF of the geometric distribution M(t)=pet 1−qet, from which its mean and variance are found to be E[X]=1 p,V [X]=q p2. Another distribution closely related to the binomial is the negative binomial distribution. This describes the probability distribution of the random variable X= number of failures before the rth success . One way of obtaining xfailures before the rth success is to have r−1s u c c e s s e s followed by xfailures followed by the rth success, for which the probability is pp···pbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright r−1t i m e s×qq···qbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright xtimes×p=prqx. However, the ﬁrst r+x−1 factors constitute just one permutation of r−1 successes and xfailures. The total number of permutations of these r+x−1 objects, of which r−1 are identical and of type 1 and xare identical and of type 2, isr+x−1Cx. Therefore, the total probability of obtaining xfailures before the rth success is f(x)=P r ( X=x)=r+x−1Cxprqx, which is called the negative binomial distribution (see the related discussion on p. 979). It is straightforward to show that the MGF of this distribution is M(t)=parenleftbiggp 1−qetparenrightbiggr , 1014 26.8 IMPORTANT DISCRETE DISTRIBUTIONS and that its mean and variance are given by E[X]=rq pand V[X]=rq p2. 26.8.3 The hypergeometric distribution In subsection 26.8.1 we saw that the probability of obtaining xsuccesses in n independent trials was given by the binomial distribution. Suppose that these n ‘trials’ actually consist of drawing at random nballs, from a set of Nsuch balls of which Mare red and the rest white. Let us consider the random variable X= number of red balls drawn. On the one hand, if the balls are drawn with replacement then the trials are independent and the probability of drawing a red ball is p=M/N each time. Therefore, the probability of drawing xred balls in ntrials is given by the binomial distribution as Pr(X=x)=n! x!(n−x)!px(1−p)n−x. On the other hand, if the balls are drawn without replacement the trials are not independent and the probability of drawing a red ball depends on how many redballs have already been drawn. We can, however, still derive a general formulafor the probability of drawing xred balls in ntrials, as follows. The number of ways of drawing xred balls from Mis MCx, and the number of ways of drawing n−xwhite balls from N−MisN−MCn−x. Therefore, the total number of ways to obtain xred balls in ntrials isMCxN−MCn−x. However, the total number of ways of drawing nobjects from Nis simplyNCn. Hence the probability of obtaining xred balls in ntrials is Pr(X=x)=MCxN−MCn−x NCn =M! x!(M−x)!(N−M)! (n−x)!(N−M−n+x)!n!(N−n)! N!,(26.97) =(Np)!(Nq)!n!(N−n)! x!(Np−x)!(n−x)!(Nq−n+x)!N!, (26.98) where in the last line p=M/N andq=1−p.T h i si sc a l l e dt h e hypergeometric distribution . By performing the relevant summations directly, it may be shown that the hypergeometric distribution has mean E[X]=nM N=np and variance V[X]=nM(N−M)(N−n) N2(N−1)=N−n N−1npq. 1015 PROBABILITYIIn the UK National Lottery each participant chooses six diﬀerent numbers between 1 and49. In each weekly draw six numbered winning balls are subsequently drawn. Find the probabilities that a participant chooses 0, 1,2,3, 4,5,6 winning numbers correctly. The probabilities are given by a hypergeometric distribution with N(the total number of balls) = 49, M(the number of winning balls drawn) = 6, and n(the number of numbers chosen by each participant) = 6. Thus, substituting in (26.97), we ﬁnd Pr(0) =6C043C6 49C6=1 2.29,Pr(1) =6C143C5 49C6=1 2.42, Pr(2) =6C243C4 49C6=1 7.55,Pr(3) =6C343C3 49C6=1 56.6, Pr(4) =6C443C2 49C6=1 1032,Pr(5) =6C543C1 49C6=1 54200, Pr(6) =6C643C0 49C6=1 13.98×106. It can easily be seen that 6X i=0Pr(i)=0 .44 + 0 .41 + 0 .13 + 0 .02 + O(10−3)=1 , as expected. J Note that if the number of trials (balls drawn) is small compared with N,M andN−Mthen not replacing the balls is of little consequence, and we may approximate the hypergeometric distribution by the binomial distribution (withp=M/N); this is much easier to evaluate. 26.8.4 The Poisson distribution We have seen that the binomial distribution describes the number of successful outcomes in a certain number of trials n. The Poisson distribution also describes the probability of obtaining a given number of successes but for situationsin which the number of ‘trials’ cannot be enumerated; rather it describes thesituation in which discrete events occur in a continuum. Typical examples ofdiscrete random variables Xdescribed by a Poisson distribution are the number of telephone calls received by a switchboard in a given interval, or the numberof stars above a certain brightness in a particular area of the sky. Given a mean rate of occurrence λof these events in the relevant interval or area, the Poisson distribution gives the probability Pr( X=x) that exactly xevents will occur. We may derive the form of the Poisson distribution as the limit of the binomial distribution when the number of trials n→∞ and the probability of ‘success’ p→0, in such a way that np=λremains ﬁnite. Thus, in our example of a telephone switchboard, suppose we wish to ﬁnd the probability that exactly x calls are received during some time interval, given that the mean number of calls 1016 26.8 IMPORTANT DISCRETE DISTRIBUTIONS in such an interval is λ. Let us begin by dividing the time interval into a large number, n, of equal shorter intervals, in each of which the probability of receiving ac a l li s p.A sw el e t n→∞ then p→0, but since we require the mean number of calls in the interval to equal λ, we must have np=λ. The probability of x successes in ntrials is given by the binomial formula as Pr(X=x)=n! x!(n−x)!px(1−p)n−x. (26.99) Now as n→∞, with xﬁnite, the ratio of the n-dependent factorials in (26.99) behaves asymptotically as a power of n,i . e . lim n→∞n! (n−x)!= lim n→∞n(n−1)(n−2)···(n−x+1 )∼nx. Also lim n→∞lim p→0(1−p)n−x= lim p→0(1−p)λ/p (1−p)x=e−λ 1, Thus, using λ=np, (26.99) tends to the Poisson distribution f(x)=P r ( X=x)=e−λλx x!, (26.100) which gives the probability of obtaining exactly xcalls in the given time interval. As we shall show below, λis the mean of the distribution. Events following a Poisson distribution are usually said to occur randomly in time. Alternatively we may derive the Poisson distribution directly, without consid- ering a limit of the binomial distribution. Let us again consider our exampleof a telephone switchboard. Suppose that the probability that xcalls have been received in a time interval tisP x(t). If the average number of calls received in a unit time is λthen in a further small time interval ∆ tthe probability of receiving ac a l li s λ∆t,p r o v i d e d∆ tis short enough that the probability of receiving two or more calls in this small interval is negligible. Similarly the probability of receivingno call during the same small interval is simply 1 −λ∆t. Thus, for x>0, the probability of receiving exactly xcalls in the total interval t+∆tis given by P x(t+∆t)=Px(t)(1−λ∆t)+Px−1(t)λ∆t. Rearranging the equation, dividing through by ∆ tand letting ∆ t→0, we obtain the diﬀerential recurrence equation dPx(t) dt=λPx−1(t)−λPx(t). (26.101) Forx= 0 (i.e. no calls received), however, (26.101) simpliﬁes to dP0(t) dt=−λP0(t), 1017 PROBABILITY which may be integrated to give P0(t)=P0(0)e−λt. But since the probability P0(0) of receiving no calls in a zero time interval must equal unity, we have P0(t)=e−λt. This expression for P0(t) may then be substituted back into (26.101) with x=1 to obtain a diﬀerential equation for P1(t) that has the solution P1(t)=λte−λt. We may repeat this process to obtain expressions for P2(t),P3(t),...,P x(t), and we ﬁnd Px(t)=(λt)x x!e−λt. (26.102) By setting t= 1 in (26.102), we again obtain the Poisson distribution (26.100) for obtaining exactly xcalls in a unit time interval. If a discrete random variable is described by a Poisson distribution of mean λ then we write X∼Po(λ). As it must be, the sum of the probabilities is unity: ∞summationdisplay x=0Pr(X=x)=e−λ∞summationdisplay x=0λx x!=e−λeλ=1. From (26.100) we may also derive the Poisson recurrence formula , Pr(X=x+1 )=λ x+1Pr(X=x)f o r x=0,1,2,..., (26.103) which enables successive probabilities to be calculated easily once one is known.IA person receives on average one e-mail message per half-hour interval. Assuming that the e-mails are received randomly in time, ﬁnd the probabilities that in any particular hour 0,1,2,3,4,5messages are received. LetX= number of e-mails received per hour. Clearly the mean number of e-mails per hour is two, and so Xfollows a Poisson distribution with λ=2 ,i . e . Pr(X=x)=2x x!e−2. Thus Pr( X=0 )= e−2=0.135, Pr( X=1 )=2 e−2=0.271, Pr( X=2 )=22e−2/2! = 0 .271, Pr(X=3 )=23e−2/3! = 0 .180, Pr( X=4 )=24e−2/4! = 0 .090, Pr( X=5 )=25e−2/5! = 0.036. These results may also be calculated using the recurrence formula (26.103). J The above example illustrates the point that a Poisson distribution typically rises and then falls. It either has a maximum when xis equal to the integer part ofλor, if λhappens to be an integer, has equal maximal values at x=λ−1a n d x=λ. The Poisson distribution always has a long ‘tail’ towards higher values of X but the higher the value of the mean the more symmetric the distribution becomes.Typical Poisson distributions are shown in ﬁgure 26.12. Using the deﬁnitions ofmean and variance, we may show that, for the Poisson distribution, E[X]=λand V[X]=λ. Nevertheless, as in the case of the binomial distribution, performing the relevant summations directly is rather tiresome, and these results are muchmore easily proved using the MGF. 1018 26.8 IMPORTANT DISCRETE DISTRIBUTIONS 1 112 22 3 33 4 44 5 55 6 67 78910110 000 000.1 0.1 0.10.20.2 0.2 0.30.3 0.3 x xxf(x) f(x)f(x) λ=1 λ=2 λ=5 Figure 26.12 Three Poisson distributions for diﬀerent values of the parame- terλ. The moment generating function for the Poisson distribution The MGF of the Poisson distribution is given by MX(t)=Ebracketleftbig etXbracketrightbig =∞summationdisplay x=0etxe−λλx x!=e−λ∞summationdisplay x=0(λet)x x!=e−λeλet=eλ(et−1) (26.104) from which we obtain M/prime X(t)=λeteλ(et−1), M/prime/prime X(t)=(λ2e2t+λet)eλ(et−1). Thus, the mean and variance of the Poisson distribution are given by E[X]=M/prime X(0) = λ and V[X]=M/prime/prime X(0)−[M/prime X(0)]2=λ. The Poisson approximation to the binomial distribution Earlier we derived the Poisson distribution as the limit of the binomial distribution when n→∞andp→0i ns u c haw a yt h a t np=λremains ﬁnite, where λis the 1019 PROBABILITY mean of the Poisson distribution. It is not surprising, therefore, that the Poisson distribution is a very good approximation to the binomial distribution for largen(≥50, say) and small p(≤0.1, say). Moreover, it is easier to calculate as it involves fewer factorials.IIn a large batch of light bulbs, the probability that a bulb is defective is 0.5%.F o ra sample of 200bulbs taken at random, ﬁnd the approximate probabilities that 0,1and2of the bulbs respectively are defective. Let the random variable X= number of defective bulbs in a sample. This is distributed asX∼Bin(200, 0.005), implying that λ=np=1.0. Since nis large and psmall, we may approximate the distribution as X∼Po(1), giving Pr(X=x)≈e−11x x!, from which we ﬁnd Pr( X=0 )≈0.37, Pr( X=1 )≈0.37, Pr( X=2 )≈0.18. For comparison, it may be noted that the exact values calculated from the binomial distribution are identical to those found here to two decimal places. J Multiple Poisson distributions Mirroring our discussion of multiple binomial distributions in subsection 26.8.1, let us suppose XandYare two independent random variables, both of which are described by Poisson distributions with (in general) diﬀerent means, so thatX∼Po(λ 1)a n d Y∼Po(λ2). Now consider the random variable Z=X+Y.W e may calculate the probability distribution of Zdirectly using (26.60), but we may derive the result much more easily by using the moment generating function (or indeed the probability or cumulant generating functions). Since XandYare independent RVs, the MGF for Zis simply the product of the individual MGFs for XandY. Thus, from (26.104), MZ(t)=MX(t)MY(t)=eλ1(et−1)eλ2(et−1)=e(λ1+λ2)(et−1), which we recognise as the MGF of Z∼Po(λ1+λ2). Hence Zis also Poisson distributed and has mean λ1+λ2. Unfortunately, no such simple result holds for thediﬀerence Z=X−Yof two independent Poisson variates. A closed-form expression for the PDF of this Zdoes exist, but it is a rather complicated combination of exponentials and a modiﬁed Bessel function. †ITwo types of e-mail arrive independently and at random: external e-mails at a mean rate of one every ﬁve minutes and internal e-mails at a rate of two every ﬁve minutes. Calculatethe probability of receiving two or more e-mails in any two-minute interval. Let X= number of external e-mails per two-minute interval, Y= number of internal e-mails per two-minute interval. †For a derivation see, for example, Hobson & Lasenby, Monthly Notices of the Royal Astronomical Society ,298, 905 (1998). 1020 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS Distribution Probability law f(x)M G F E[X] V[X] Gaussian1 σ√ 2πexp / −(x−µ)2 2σ2 / exp(µt+1 2σ2t2) µσ2 exponential λe−λx /λ λ−t /1 λ1 λ2 gammaλ Γ(r)(λx)r−1e−λx /λ λ−t /rr λr λ2 chi-squared1 2n/2Γ(n/2)x(n/2)−1e−x/2 /1 1−2t /n/2 n 2n uniform1 b−aebt−eat (b−a)ta+b 2(b−a)2 12 Table 26.2 Some important continuous probability distributions. Since we expect on average one external e-mail and two internal e-mails every ﬁve minutes we have X∼Po(0.4) and Y∼Po(0.8). Letting Z=X+Ywe have Z∼Po(0.4+0 .8) = Po(1.2). Now Pr(Z≥2) = 1−Pr(Z<2) = 1−Pr(Z=0 )−Pr(Z=1 ) and Pr(Z=0 )= e−1.2=0.301, Pr(Z=1 )= e−1.21.2 1=0.361. Hence Pr( Z≥2) = 1−0.301−0.361 = 0 .338. J The above result can be extended, of course, to any number of Poisson processes, so that if Xi=P o ( λi),i=1,2,...,n then the random variable Z=X1+X2+ ···+Xnis distributed as Z∼Po(λ1+λ2+···+λn). 26.9 Important continuous distributions Having discussed the most commonly encountered discrete probability distri- butions, we now consider some of the more important continuous probability distributions. These are summarised for convenience in table 26.2; we refer thereader to the relevant subsection below for an explanation of the symbols used. 26.9.1 The Gaussian distribution By far the most important continuous probability distribution is the Gaussian ornormal distribution. The reason for its importance is that a great many random variables of interest, in all areas of the physical sciences and beyond, are described either exactly or approximately by a Gaussian distribution. Moreover, the Gaussian distribution can be used to approximate other, more complicated,probability distributions. 1021 PROBABILITY −6−4−2 2346810 120.10.20.30.4 σ=1 σ=2 σ=3µ=3 Figure 26.13 The Gaussian or normal distribution for mean µ=3a n d various values of the standard deviation σ. The probability density function for a Gaussian distribution of a random variable X, with mean E[X]=µand variance V[X]=σ2,t a k e st h ef o r m f(x)=1 σ√ 2πexpbracketleftbigg −1 2parenleftBigx−µ σparenrightBig2bracketrightbigg . (26.105) The factor 1 /√ 2πarises from the normalisation of the distribution, integraldisplay∞ −∞f(x)dx=1 ; the evaluation of this integral is discussed in subsection 6.4.2. The Gaussian distribution is symmetric about the point x=µand has the characteristic ‘bell’ shape shown in ﬁgure 26.13. The width of the curve is described by the standarddeviation σ:i fσis large then the curve is broad, and if σis small then the curve is narrow (see the ﬁgure). At x=µ±σ,f(x) falls to e −1/2≈0.61 of its peak value; these points are points of inﬂection, where d2f/dx2= 0. When a random variable Xfollows a Gaussian distribution with mean µand variance σ2,w ew r i t e X∼N(µ, σ2). The eﬀects of changing µandσare only to shift the curve along the x-axis or to broaden or narrow it, respectively. Thus all Gaussians are equivalent in thata change of origin and scale can reduce them to a standard form. We thereforeconsider the random variable Z=(X−µ)/σ, for which the PDF takes the form φ(z)=1 √ 2πexpparenleftbigg −z2 2parenrightbigg , (26.106) which is called the standard Gaussian distribution and has mean µ=0a n d variance σ2= 1. The random variable Zis called the standard variable . 1022 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS y −4 −12−2 −2 0 a1 2 4 az zφ(z) Φ(z) Φ(a)Φ(a) 0.20.40.60.8 0.10.20.30.4 Figure 26.14 On the left, the standard Gaussian distribution φ(z); the shaded area gives Pr( Z<a )=Φ ( a). On the right, the cumulative probability function Φ(z) for a standard Gaussian distribution φ(z). From (26.105) we can deﬁne the cumulative probability function for a Gaussian distribution as F(x)=P r ( X<x )=1 σ√ 2πintegraldisplayx −∞expbracketleftbigg −1 2parenleftBigu−µ σparenrightBig2bracketrightbigg du, (26.107) where uis a (dummy) integration variable. Unfortunately, this (indeﬁnite) integral cannot be evaluated analytically. It is therefore standard practice to tabulate val-ues of the cumulative probability function for the standard Gaussian distribution(see ﬁgure 26.14), i.e. Φ(z)=P r ( Z<z )=1 √ 2πintegraldisplayz −∞expparenleftbigg −u2 2parenrightbigg du. (26.108) It is usual only to tabulate Φ( z)f o r z>0, since it can be seen easily, from ﬁgure 26.14 and the symmetry of the Gaussian distribution, that Φ( −z)=1−Φ(z); see table 26.3. Using such a table it is then straightforward to evaluate the probability that Zlies in a given range of z-values. For example, for aandb constant, Pr(Z<a )=Φ ( a), Pr(Z>a )=1−Φ(a), Pr(a<Z≤b)=Φ ( b)−Φ(a). Remembering that Z=(X−µ)/σand comparing (26.107) and (26.108), we see that F(x)=ΦparenleftBigx−µ σparenrightBig , and so we may also calculate the probability that the original random variable 1023 PROBABILITY Φ(z).00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998 Table 26.3 The cumulative probability function Φ( z) for the standard Gaus- sian distribution, as given by (26.108). The units and the ﬁrst decimal place ofzare speciﬁed in the column under Φ( z) and the second decimal place is speciﬁed by the column headings. Thus, for example, Φ(1 .23) = 0 .8907. 1024 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS Xlies in a given x-range. For example, Pr(a<X≤b)=1 σ√ 2πintegraldisplayb aexpbracketleftbigg −1 2parenleftBigu−µ σparenrightBig2bracketrightbigg du (26.109) =F(b)−F(a) (26.110) =Φparenleftbiggb−µ σparenrightbigg −ΦparenleftBiga−µ σparenrightBig . (26.111)IIfXis described by a Gaussian distribution of mean µand variance σ2,c a l c u l a t et h e probabilities that Xlies within 1σ,2σand3σof the mean. From (26.111) Pr(µ−nσ < X≤µ+nσ)=Φ ( n)−Φ(−n)=Φ ( n)−[1−Φ(n)], and so from table 26.3 Pr(µ−σ<X≤µ+σ)=2 Φ ( 1 )−1=0 .6826≈68.3%, Pr(µ−2σ<X≤µ+2σ)=2 Φ ( 2 )−1=0 .9544≈95.4%, Pr(µ−3σ<X≤µ+3σ)=2 Φ ( 3 )−1=0 .9974≈99.7%. Thus we expect Xto be distributed in such a way that about two thirds of the values will lie between µ−σandµ+σ, 95% will lie within 2 σof the mean and 99 .7% will lie within 3σof the mean. These limits are called the one-, t wo- and three-sigma limits respectively; it is particularly important to note that they are independent of the actual values of the mean and variance. J There are many other ways in which the Gaussian distribution may be used. We now illustrate some of the uses in more complicated examples.ISawmill Aproduces boards whose lengths are Gaussian distributed with mean 209.4 cm and standard deviation 5.0 cm . A board is accepted if it is longer than 200 cm but is rejected otherwise. Show that 3%of boards are rejected. Sawmill Bproduces boards of the same standard deviation but of mean length 210.1 cm . Find the proportion of boards rejected if they are drawn at random from the outputs of A andBin the ratio 3:1. LetX= length of boards from A,s ot h a t X∼N(209.4,(5.0)2)a n d Pr(X<200) = Φ /200−µ σ / =Φ /200−209.4 5.0 / =Φ (−1.88). But, since Φ( −z)=1−Φ(z) we have, using table 26.3, Pr(X<200) = 1−Φ(1.88) = 1−0.9699 = 0 .0301, i.e. 3.0% of boards are rejected. Now let Y= length of boards from B,s ot h a t Y∼N(210.1,(5.0)2)a n d Pr(Y<200) = Φ /200−210.1 5.0 / =Φ (−2.02) =1−Φ(2.02) =1−0.9783 = 0 .0217. 1025 PROBABILITY Therefore, when taken alone, only 2 .2% of boards from Bare rejected. If, however, boards are drawn at random from AandBin the ratio 3 : 1 then the proportion rejected is 1 4(3×0.030 + 1×0.022) = 0 .028 = 2 .8%. J We may sometimes work backwards to derive the mean and standard deviation of a population that is known to be Gaussian distributed.IThe time taken for a computer ‘packet’ to travel from Cambridge UK to Cambridge MA is Gaussian distributed. 6.8% of the packets take over 200 ms to make the journey, and 3.0% take under 140 ms . Find the mean and standard deviation of the distribution. LetX= journey time in ms; we are told that X∼N(µ, σ2)w h e r e µandσare unknown. Since 6.8% of journey times are longer than 200 ms, Pr(X>200) = 1−Φ /200−µ σ / =0.068, from which we ﬁnd Φ /200−µ σ / =1−0.068 = 0 .932. Using table 26.3, we have therefore 200−µ σ=1.49. (26.112) Also, 3 .0% of journey times are under 140 ms, so Pr(X<140) = Φ /140−µ σ / =0.030. Now using Φ( −z)=1−Φ(z)g i v e s Φ /µ−140 σ / =1−0.030 = 0 .970. Using table 26.3 again, we ﬁnd µ−140 σ=1.88. (26.113) Solving the simultaneous equations (26.112) and (26.113) gives µ= 173 .5,σ=1 7.8. J The moment generating function for the Gaussian distribution Using the deﬁnition of the MGF (26.85), MX(t)=Ebracketleftbig etXbracketrightbig =integraldisplay∞ −∞1 σ√ 2πexpbracketleftbigg tx−(x−µ)2 2σ2bracketrightbigg dx =cexpparenleftbig µt+1 2σ2t2parenrightbig , where the ﬁnal equality is established by completing the square in the argument of the exponential and writing c=integraldisplay∞ −∞1 σ√ 2πexpbraceleftbigg −[x−(µ+σ2t)]2 2σ2bracerightbigg dx. 1026 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS However, the ﬁnal integral is simply the normalisation integral for the Gaussian distribution, and so c=1a n dt h eM G Fi sg i v e nb y MX(t)=e x pparenleftbig µt+1 2σ2t2parenrightbig . (26.114) We showed in subsection 26.7.2 that this MGF leads to E[X]=µandV[X]=σ2, as required. Gaussian approximation to the binomial distribution We may consider the Gaussian distribution as the limit of the binomial distribu- tion when the number of trials n→∞but the probability of a success premains ﬁnite, so that np→∞ also. (This contrasts with the Poisson distribution, which corresponds to the limit n→∞ andp→0 with np=λremaining ﬁnite.) In other words, a Gaussian distribution results when an experiment with a ﬁnite probability of success is repeated a large number of times. We now show howthis Gaussian limit arises. The binomial probability function gives the probability of xsuccesses in ntrials as f(x)=n! x!(n−x)!px(1−p)n−x. Taking the limit as n→∞ (and x→∞) we may approximate the factorials by Stirling’s approximation n!∼√ 2πnparenleftBign eparenrightBign to obtain f(x)≈1√ 2πnparenleftBigx nparenrightBig−x−1/2parenleftBign−x nparenrightBig−n+x−1/2 px(1−p)n−x =1√ 2πnexpbracketleftBig −parenleftbig x+1 2parenrightbig lnx n−parenleftbig n−x+1 2parenrightbig lnn−x n +xlnp+(n−x)l n ( 1−p)bracketrightBig . By expanding the argument of the exponential in terms of y=x−np,w h e r e 1/lessmuchy/lessmuchnpand keeping only the dominant terms, it can be shown that f(x)≈1√ 2πn1√p(1−p)expbracketleftbigg −1 2(x−np)2 np(1−p)bracketrightbigg , which is of Gaussian form with µ=npandσ=√np(1−p). Thus we see that the value of the Gaussian probability density function f(x)i s a good approximation to the probability of obtaining xsuccesses in ntrials. This approximation is actually very good even for relatively small n. For example, if n=1 0a n d p=0.6 then the Gaussian approximation to the binomial distribution is (26.105) with µ=1 0×0.6=6a n d σ=√10×0.6(1−0.6) = 1 .549. The 1027 PROBABILITY xf (x) (binomial) f(x) (Gaussian) 0 0.0001 0.0001 1 0.0016 0.00142 0.0106 0.0092 3 0.0425 0.0395 4 0.1115 0.11195 0.2007 0.2091 6 0.2508 0.2575 7 0.2150 0.20918 0.1209 0.1119 9 0.0403 0.0395 10 0.0060 0.0092 Table 26.4 Comparison of the binomial distribution for n=1 0a n d p=0.6 with its Gaussian approximation. probability functions f(x) for the binomial and associated Gaussian distributions for these parameters are given in table 26.4, and it can be seen that the Gaussianapproximation is a good one. Strictly speaking, however, since the Gaussian distribution is continuous and the binomial distribution is discrete, we should use the integral of f(x)f o rt h e Gaussian distribution in the calculation of approximate binomial probabilities.More speciﬁcally, we should apply a continuity correction so that the discrete integer xin the binomial distribution becomes the interval [ x−0.5,x+0.5] in the Gaussian distribution. Explicitly, Pr(X=x)≈1 σ√ 2πintegraldisplayx+0.5 x−0.5expbracketleftbigg −1 2parenleftBigu−µ σparenrightBig2bracketrightbigg du. The Gaussian approximation is particularly useful for estimating the binomial probability that Xlies between the (integer) values x1andx2, Pr(x1<X≤x2)≈1 σ√ 2πintegraldisplayx2+0.5 x1−0.5expbracketleftbigg −1 2parenleftBigu−µ σparenrightBig2bracketrightbigg du.IA manufacturer makes computer chips of which 10%are defective. For a random sample of200chips, ﬁnd the approximate probability that more than 15are defective. We ﬁrst deﬁne the random variable X= number of defective chips in the sample , which has a binomial distribution X∼Bin(200,0.1). Therefore, t he mean and variance of this distribution are E[X] = 200×0.1 = 20 and V[X] = 200×0.1×(1−0.1) = 18 , and we may approximate the binomial distribution with a Gaussian distribution such that 1028 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS X∼N(20,18). The standard variable is Z=X−20√ 18, and so, using X=1 5.5 to allow for the continuity correction, Pr(X>15.5) = Pr / Z>15.5−20√ 18 / =P r ( Z>−1.06) =P r ( Z<1.06) = 0 .86. J Gaussian approximation to the Poisson distribution We ﬁrst met the Poisson distribution as the limit of the binomial distribution for n→∞ andp→0 ,t a k e ni ns u c haw a yt h a t np=λremains ﬁnite. Further, in the previous subsection, we considered the Gaussian distribution as the limit ofthe binomial distribution when n→∞butpremains ﬁnite, so that np→∞also. It should come as no surprise, therefore, that the Gaussian distribution can also be used to approximate the Poisson distribution when the mean λbecomes large. The probability function for the Poisson distribution is f(x)=e −λλx x!, which, on taking the logarithm of both sides, gives lnf(x)=−λ+xlnλ−lnx!. (26.115) Stirling’s approximation for large xgives x!≈√ 2πxparenleftBigx eparenrightBigx implying that lnx!≈ln√ 2πx+xlnx−x, which, on substituting into (26.115), yields lnf(x)≈−λ+xlnλ−(xlnx−x)−ln√ 2πx. Since we expect the Poisson distribution to peak around x=λ, we substitute /epsilon1=x−λto obtain lnf(x)≈−λ+(λ+/epsilon1)braceleftBig lnλ−lnbracketleftBig λparenleftBig 1+/epsilon1 λparenrightBigbracketrightBigbracerightBig +(λ+/epsilon1)−lnradicalbig 2π(λ+/epsilon1). Using the expansion ln(1 + z)=z−z2/2+···, we ﬁnd lnf(x)≈/epsilon1−(λ+/epsilon1)parenleftbigg/epsilon1 λ−/epsilon12 2λ2parenrightbigg −ln√ 2πλ−parenleftbigg/epsilon1 λ−/epsilon12 2λ2parenrightbigg ≈−/epsilon12 2λ−ln√ 2πλ, 1029 PROBABILITY when only the dominant terms are retained, after using the fact that /epsilon1is of the order of the standard deviation of x,i . e .o fo r d e r λ1/2. On exponentiating this result we obtain f(x)≈1√ 2πλexpbracketleftbigg −(x−λ)2 2λbracketrightbigg , which is the Gaussian distribution with µ=λandσ2=λ. The larger the value of λ, the better is the Gaussian approximation to the Poisson distribution; the approximation is reasonable even for λ= 5, but λ≥10 is safer. As in the case of the Gaussian approximation to the binomial distribution,a continuity correction is necessary since the Poisson distribution is discrete.IE-mail messages are received by an author at an average rate of one per hour. Find the probability that in a day the author receives 24messages or more. We ﬁrst deﬁne the random variable X= number of messages received in a day . Thus E[X]=1×24 = 24, and so X∼Po(24). Since λ>10 we may approximate the Poisson distribution by X∼N(24,24). Now the standard variable is Z=X−24√ 24, and, using the continuity correction, we ﬁnd Pr(X>23.5) = P / Z>23.5−24√ 24 / =P r ( Z>−0.102) = Pr( Z<0.102) = 0 .54. J In fact, almost all probability distributions tend towards a Gaussian when the numbers involved become large – that this should happen is required by thecentral limit theorem, which we discuss in section 26.10. Multiple Gaussian distributions Suppose Xand Yareindependent Gaussian-distributed random variables, so thatX∼N(µ 1,σ2 1)a n d Y∼N(µ2,σ2 2). Let us now consider the random variable Z=X+Y. The PDF for this random variable may be found directly using (26.61), but it is easier to use the MGF. From (26.114), the MGFs of XandY are MX(t)=e x pparenleftbig µ1t+1 2σ2 1t2parenrightbig ,M Y(t)=e x pparenleftbig µ2t+1 2σ2 2t2parenrightbig . Using (26.89), since XandYare independent RVs, the MGF of Z=X+Yis simply the product of MX(t)a n d MY(t). Thus, we have MZ(t)=MX(t)MY(t)=e x pparenleftbig µ1t+1 2σ2 1t2parenrightbig expparenleftbig µ2t+1 2σ2 2t2parenrightbig =e x pbracketleftbig (µ1+µ2)t+1 2(σ2 1+σ2 2)t2bracketrightbig , 1030 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS which we recognise as the MGF for a Gaussian with mean µ1+µ2and variance σ2 1+σ2 2. Thus, Zis also Gaussian distributed: Z∼N(µ1+µ2,σ2 1+σ2 2). A similar calculation may be performed to calculate the PDF of the random variable W=X−Y. If we introduce the variable ˜Y=−Ythen W=X+˜Y, where ˜Y∼N(−µ1,σ2 1). Thus, using the result above, we ﬁnd W∼N(µ1− µ2,σ2 1+σ2 2).IAn executive travels home from her oﬃce every evening. Her journey consists of a train ride, followed by a bicycle ride. The time spent on the train is Gaussian distributed withmean 52minutes and standard deviation 1.8minutes, while the time for the bicycle journey is Gaussian distributed with mean 8minutes and standard deviation 2.6minutes. Assuming these two factors are independent, estimate the percentage of occasions on which the whole journey takes more than 65minutes. We ﬁrst deﬁne the random variables X=t i m es p e n to nt r a i n ,Y = time spent on bicycle, so that X∼N(52,(1 .8)2)a n d Y∼N(8,(2.6)2). Since XandYare independent, the total journey time T=X+Yis distributed as T∼N(52 + 8 ,(1.8)2+( 2.6)2)=N(60,(3.16)2). The standard variable is thus Z=T−60 3.16, and the required probability is given by Pr(T>65) = Pr / Z>65−60 3.16 / =P r ( Z>1.58) = 1−0.943 = 0 .057. Thus the total journey time exceeds 65 minutes on 5 .7% of occasions. J The above results may be extended. For example, if the random variables Xi,i=1,2,...,n, are distributed as Xi∼N(µi,σ2 i) then the random variable Z=summationtext iciXi(where the ciare constants) is distributed as Z∼N(summationtext iciµi,summationtext ic2 iσ2 i). 26.9.2 The log-normal distribution If the random variable Xfollows a Gaussian distribution then the variable Y=eXis described by a log-normal distribution. Clearly, if Xcan take values in the range −∞to∞,t h e n Ywill lie between 0 and ∞. The probability density function for Yis found using the result (26.58). It is g(y)=f(x(y))vextendsinglevextendsinglevextendsinglevextendsingledx dyvextendsinglevextendsinglevextendsinglevextendsingle=1 σ√ 2π1 yexpbracketleftbigg −(lny−µ)2 2σ2bracketrightbigg . We note that µand σ2are not the mean and variance of the log-normal distribution, but rather the parameters of t he corresponding Gaussian distribution forX. The mean and variance of Y, however, can be found straightforwardly 1031 PROBABILITY 000.20.40.60.8 11 2 3 4yg(y) µ=0 , σ=0 µ=0 , σ=0.5 µ=0 , σ=1.5 µ=1 , σ=1 Figure 26.15 The PDF g(y) for the log-normal distribution for various values of the parameters µandσ. using the MGF of X,w h i c hr e a d s MX(t)=E[etX]=e x p ( µt+1 2σ2t2). Thus, the mean of Yis given by E[Y]=E[eX]=MX(1) = exp( µ+1 2σ2), and the variance of Yreads V[Y]=E[Y2]−(E[Y])2=E[e2X]−(E[eX])2 =MX(2)−[MX(1)]2=e x p ( 2 µ+σ2)[exp( σ2)−1]. In ﬁgure 26.15, we plot some examples of the log-normal distribution for various values of the parameters µandσ2. 26.9.3 The exponential and gamma distributions The exponential distribution with positive parameter λis given by f(x)=braceleftBigg λe−λxforx>0, 0f o r x≤0(26.116) and satisﬁesintegraltext∞ −∞f(x)dx= 1 as required. The exponential distribution occurs nat- urally if we consider the distribution of the length of intervals between successive events in a Poisson process or, equivalently, the distribution of the interval (i.e.the waiting time) before the ﬁrst event. If the average number of events per unitinterval is λthen on average there are λxevents in interval x, so that from the Poisson distribution the probability that there will be no events in this interval isgiven by Pr(no events in interval x)=e −λx. 1032 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS The probability that an event occurs in the next inﬁnitestimal interval [ x, x+dx] is given by λd x,s ot h a t Pr(the ﬁrst event occurs in interval [ x, x+dx]) =e−λxλd x . Hence the required probability density function is given by f(x)=λe−λx. The expectation and variance of the exponential distribution can be evaluated as 1/λand (1 /λ)2respectively. The MGF is given by M(t)=λ λ−t. (26.117) We may generalise the above discussion to obtain the PDF for the interval between every rth event in a Poisson process or, equivalently, the interval (waiting time) before the rth event. We begin by using the Poisson distribution to give Pr(r−1 events occur in interval x)=e−λx(λx)r−1 (r−1)!, from which we obtain Pr(rth event occurs in the interval [ x, x+dx]) =e−λx(λx)r−1 (r−1)!λd x . Thus the required PDF is f(x)=λ (r−1)!(λx)r−1e−λx, (26.118) which is known as the gamma distribution of order rwith parameter λ. Although our derivation applies only when ris a positive integer, the gamma distribution is deﬁned for all positive rby replacing ( r−1)! by Γ( r) in (26.118); see the appendix for a discussion of the gamma function Γ( x). If a random variable Xis described by a gamma distribution of order rwith parameter λ,w ew r i t e X∼γ(λ, r); we note that the exponential distribution is the special case γ(λ,1). The gamma distribution γ(λ, r) is plotted in ﬁgure 26.16 for λ=1a n d r=1,2,5,10. For large r, the gamma distribution tends to the Gaussian distribution whose mean and variance are speciﬁed by (26.120) below. The MGF for the gamma distribution is obtained from that for the exponential distribution, by noting that we may consider the interval between every rth event in a Poisson process as the sum of rintervals between successive events. Thus the rth-order gamma variate is the sum of rindependent exponentially distributed random variables. From (26.117) and (26.90), the MGF of the gamma distribution is therefore given by M(t)=parenleftbiggλ λ−tparenrightbiggr , (26.119) 1033 PROBABILITY 0 024 68 1 0 12 14 16 18 200.20.40.60.81 r=1 r=2 r=5 r=1 0 xf(x) Figure 26.16 The PDF f(x) for the gamma distributions γ(λ, r)w i t h λ=1 andr=1,2,5,10. from which the mean and variance are found to be E[X]=r λ,V [X]=r λ2. (26.120) We may also use the above MGF to prove another useful theorem regarding multiple gamma distributions. If Xi∼γ(λ, ri),i=1,2,...,n, are independent gamma variates then the random variable Y=X1+X2+···+Xnhas MGF M(t)=nproductdisplay i=1parenleftbiggλ λ−tparenrightbiggri =parenleftbiggλ λ−tparenrightbiggr1+r2+···+rn . (26.121) Thus Yis also a gamma variate, distributed as Y∼γ(λ, r1+r2+···+rn). 26.9.4 The chi-squared distribution In subsection 26.6.2, we showed that if Xis Gaussian distributed with mean µand variance σ2, such that X∼N(µ, σ2), then the random variable Y=(x−µ)2/σ2 is distributed as the gamma distribution Y∼γ(1 2,1 2). Let us now consider n independent Gaussian random variables Xi∼N(µi,σ2 i),i=1,2,...,n, and deﬁne the new variable χ2 n=nsummationdisplay i=1(Xi−µi)2 σ2 i. (26.122) 1034 26.9 IMPORTANT CONTINUOUS DISTRIBUTIONS Using the result (26.121) for multiple gamma distributions, χ2 nmust be distributed as the gamma variate χ2 n∼γ(1 2,1 2n), which from (26.118) has the PDF f(χ2 n)=1 2 Γ(1 2n)(1 2χ2 n)(n/2)−1exp(−1 2χ2 n) =1 2n/2Γ(1 2n)(χ2 n)(n/2)−1exp(−1 2χ2 n). (26.123) This is known as the chi-squared distribution of order nand has numerous applications in statistics (see chapter 27). Setting λ=1 2andr=1 2nin (26.120), we ﬁnd that E[χ2 n]=n, V [χ2 n]=2 n. An important generalisation occurs when the nGaussian variables Xiarenot linearly independent but are instead required to satisfy a linear constraint of theform c 1X1+c2X2+···+cnXn=0, (26.124) in which the constants ciare not all zero. In this case, it may be shown (see exercise 26.40) that the variable χ2 ndeﬁned in (26.122) is still described by a chi- squared distribution, but one of order n−1. Indeed, this result may be trivially extended to show that if the nGaussian variables Xisatisfy mlinear constraints of the form (26.124) then the variable χ2 ndeﬁned in (26.122) is described by a chi-squared distribution of order n−m. 26.9.5 The Cauchy and Breit–Wigner distributions A random variable X(in the range −∞to∞)t h a to b e y st h e Cauchy distribution is described by the PDF f(x)=1 π1 1+x2. This is a special case of the Breit–Wigner distribution f(x)=1 π1 2Γ 1 4Γ2+(x−x0)2, which is encountered in the study of nuclear and particle physics. In ﬁgure 26.17, we plot some examples of the Breit–Wigner distribution for several values of theparameters x 0and Γ. We see from the ﬁgure that the peak (or mode) of the distribution occurs atx=x0. It is also straightforward to show that the parameter Γ is equal to the width of the peak at half the maximum height. Although the Breit–Wignerdistribution is symmetric about its peak, it does not formally possess a mean since 1035 PROBABILITY 000.20.40.60.8 −4−22 4xf(x) x0=0 , x0=0 ,x0=2 , Γ=1 Γ=1 Γ=3 Figure 26.17 The PDF f(x) for the Breit–Wigner distribution for diﬀerent values of the parameters x0and Γ. the integralsintegraltext0 −∞xf(x)dxandintegraltext∞ 0xf(x)dxboth diverge. Similar divergences occur for all higher moments of the distribution. 26.9.6 The uniform distribution Finally we mention the very simple, but common, uniform distribution ,w h i c h describes a continuous random variable that has a constant PDF over its allowedrange of values. If the limits on Xareaandbthen f(x)=braceleftBigg 1/(b−a)f o r a≤x≤b, 0o t h e r w i s e . The MGF of the uniform distribution is found to be M(t)=e bt−eat (b−a)t, and its mean and variance are given by E[X]=a+b 2,V [X]=(b−a)2 12. 26.10 The central limit theorem In subsection 26.9.1 we discussed approximating the binomial and Poisson distri- butions by the Gaussian distribution when the number of trials is large. We now discuss why the Gaussian distribution is so common and therefore so important.Thecentral limit theorem may be stated as follows. 1036 26.10 THE CENTRAL LIMIT THEOREM Central limit theorem. Suppose that Xi,i=1,2,...,n,a r eindependent random variables, each of which is described by a probability density function fi(x)(these may all be diﬀerent) with a mean µiand a variance σ2 i. The random variable Z=parenleftbigsummationtext iXiparenrightbig /n, i.e. the ‘mean’ of the Xi, has the following properties: (i)its expectation value is given by E[Z]=parenleftbigsummationtext iµiparenrightbig /n; (ii)its variance is given by V[Z]=parenleftbigsummationtext iσ2 iparenrightbig /n2; (iii)asn→∞ the probability function of Ztends to a Gaussian with corre- sponding mean and variance. We note that for the theorem to hold, the probability density functions fi(x) must possess formal means and variances. Thus, for example, if each Xiwere described by a Cauchy distribution then the theorem would not apply. Properties (i) and (ii) of the theorem are easily proved, as follows. Firstly E[Z]=1 n(E[X1]+E[X2]+···+E[Xn]) =1 n(µ1+µ2+···+µn)=summationtext iµi n, a result which does notrequire that the Xiareindependent random variables. If µi=µfor all ithen this becomes E[Z]=nµ n=µ. Secondly, if the Xiareindependent, it follows from an obvious extension of (26.68) that V[Z]=Vbracketleftbigg1 n(X1+X2+···+Xn)bracketrightbigg =1 n2(V[X1]+V[X2]+···+V[Xn])=summationtext iσ2 i n2. Let us now consider property (iii), which is the reason for the ubiquity of the Gaussian distribution and is most easily proved by considering the momentgenerating function M Z(t)o fZ. From (26.90), this MGF is given by MZ(t)=nproductdisplay i=1MXiparenleftbiggt nparenrightbigg , where MXi(t)i st h eM G Fo f fi(x). Now MXiparenleftbiggt nparenrightbigg =1+t nE[Xi]+1 2t2 n2E[X2 i]+··· =1+ µit n+1 2(σ2 i+µ2 i)t2 n2+···, and as nbecomes large MXiparenleftbiggt nparenrightbigg ≈expparenleftbiggµit n+1 2σ2 it2 n2parenrightbigg , 1037 PROBABILITY as may be veriﬁed by expanding the exponential up to terms including ( t/n)2. Therefore MZ(t)≈nproductdisplay i=1expparenleftbiggµit n+1 2σ2 it2 n2parenrightbigg =e x pparenleftbiggsummationtext iµi nt+1 2summationtext iσ2 i n2t2parenrightbigg . Comparing this with the form of the MGF for a Gaussian distribution, (26.114), we can see that the probability density function g(z)o fZtends to a Gaussian dis- tribution with meansummationtext iµi/nand variancesummationtext iσ2 i/n2. In particular, if we consider Zto be the mean of nindependent measurements of the samerandom variable X (so that Xi=Xfori=1,2,...,n) then, as n→∞,Zhas a Gaussian distribution with mean µand variance σ2/n. We may use the central limit theorem to derive an analogous result to (iii) above for the product W=X1X2···Xnof the nindependent random variables Xi. Provided the Xionly take values between zero and inﬁnity, we may write lnW=l nX1+l nX2+···+l nXn, which is simply the sum of nnew random variables ln Xi. Thus, provided these new variables each possess a formal mean and variance, the PDF of ln Wwill tend to a Gaussian in the limit n→∞, and so the product Wwill be described by a log-normal distribution (see subsection 26.9.2). 26.11 Joint distributions As mentioned brieﬂy in subsection 26.4.3, it is common in the physical sciences to consider simultaneously two or more random variables that are not independent, in general, and are thus described by joint probability density functions . We will return to the subject of the interdependence of random variables after ﬁrstpresenting some of the general ways of characterising joint distributions. Wewill concentrate mainly on bivariate distributions, i.e. distributions of only two random variables, though the results may be extended readily to multivariatedistributions. The subject of multivariate distributions is large and a detailed study is beyond the scope of this book; the interested reader should therefore consult one of the many specialised texts. However, we do discuss the multinomialand multivariate Gaussian distributions, in section 26.15. The ﬁrst thing to note when dealing with bivariate distributions is that the distinction between discrete and continuous distributions may not be as clear asfor the single variable case; the random variables can both be discrete, or bothcontinuous, or one discrete and the other continuous. In general, for the randomvariables XandY, the joint distribution will take an inﬁnite number of values unless both XandYhave only a ﬁnite number of values. In this chapter we will consider only the cases where XandYare either both discrete or both continuous random variables. 1038 26.11 JOINT DISTRIBUTIONS 26.11.1 Discrete bivariate distributions In direct analogy with the one-variable (univariate) case, if Xis a discrete random variable that takes the values {xi}andYone that takes the values {yj}then the probability function of the joint distribution is deﬁned as f(x, y)=braceleftBigg Pr(X=xi,Y=yj)f o r x=xi,y=yj, 0o t h e r w i s e . We may therefore think of f(x, y) as a set of spikes at valid points in the xy-plane, whose heights represent the probability of obtaining X=xiandY=yj.T h e normalisation of f(x, y) implies summationdisplay isummationdisplay jf(xi,yj)=1 , (26.125) where the sums over iandjtake all valid pairs of values. We can also deﬁne the cumulative probability function F(x, y)=summationdisplay xi≤xsummationdisplay yj≤yf(xi,yj), (26.126) from which it follows that the probability that Xlies in the range [ a1,a2]a n d Y lies in the range [ b1,b2]i sg i v e nb y Pr(a1<X≤a2,b1<Y≤b2)=F(a2,b2)−F(a1,b2)−F(a2,b1)+F(a1,b1). Finally, we deﬁne XandYto beindependent if we can write their joint distribution in the form f(x, y)=fX(x)fY(y), (26.127) i.e. as the product of two univariate distributions. 26.11.2 Continuous bivariate distributions In the case where both XandYare continuous random variables, the PDF of the joint distribution is deﬁned by f(x, y)dx dy =P r ( x<X≤x+dx, y < Y≤y+dy), (26.128) sof(x, y)dx dy is the probability that xlies in the range [ x, x+dx]a n d ylies in the range [ y,y+dy]. It is clear that the two-dimensional function f(x, y) must be everywhere non-negative and that normalisation requires integraldisplay∞ −∞integraldisplay∞ −∞f(x, y)dx dy =1. 1039 PROBABILITY It follows further that Pr(a1<X≤a2,b1<Y≤b2)=integraldisplayb2 b1integraldisplaya2 a1f(x, y)dx dy. (26.129) We can also deﬁne the cumulative probability function by F(x, y)=P r ( X≤x, Y≤y)=integraldisplayx −∞integraldisplayy −∞f(u, v)du dv, from which we see that (as for the discrete case), Pr(a1<X≤a2,b1<Y≤b2)=F(a2,b2)−F(a1,b2)−F(a2,b1)+F(a1,b1). Finally we note that the deﬁnition of independence (26.127) for discrete bivariate distributions also applies to continuous bivariate distributions.IA ﬂat table is ruled with parallel straight lines a distance Dapart, and a thin needle of length l<D is tossed onto the table at random. What is the probability that the needle will cross a line? Letθbe the angle that the needle makes with the lines, and let xbe the distance from the centre of the needle to the nearest line. Since the needle is tossed ‘at random’ ontothe table, the angle θis uniformly distributed in the interval [0 ,π], and the distance x is uniformly distributed in the interval [0 ,D/2]. Assuming that θandxare independent, their joint distribution is just the product of their individual distributions, and is given by f(θ,x)=1 π1 D/2=2 πD. The needle will cross a line if the distance xof its centre from that line is less than1 2lsinθ. Thus the required probability is 2 πD Zπ 0 Z1 2lsinθ 0dx dθ =2 πDl 2 Zπ 0sinθd θ=2l πD. This gives an experimental (but cumbersome) method of determining π. J 26.11.3 Marginal and conditional distributions Given a bivariate distribution f(x, y), we may only be interested in the proba- bility function for Xirrespective of the value of Y(or vice versa). This marginal distribution of Xis obtained by summing or integrating, as appropriate, the joint probability distribution over all allowed values of Y. Thus, the marginal distribution of X(for example) is given by fX(x)=braceleftBiggsummationtext jf(x, yj) for a discrete distribution,integraltext f(x, y)dyfor a continuous distribution.(26.130) It is clear that an analogous deﬁnition exists for the marginal distribution of Y. Alternatively, one might be interested in the probability function of Xgiven 1040 26.12 PROPERTIES OF JOINT DISTRIBUTIONS thatYtakes some speciﬁc value of Y=y0,i . e .P r ( X=x|Y=y0). This conditional distribution of Xis given by g(x)=f(x, y0) fY(y0), where fY(y) is the marginal distribution of Y. The division by fY(y0) is necessary in order that g(x) is properly normalised. 26.12 Properties of joint distributions The probability density function f(x, y) contains all the information on the joint probability distribution of two random variables XandY. In a similar manner to that presented for univariate distributions, however, it is conventional tocharacterise f(x, y) by certain of its properties, which we now discuss. Once again, most of these properties are based on the concept of expectation values,which are deﬁned for joint distributions in an analogous way to those for single- variable distributions (26.46). Thus, the expectation value of any function g(X,Y) of the random variables XandYis given by E[g(X,Y)] =braceleftBiggsummationtext isummationtext jg(xi,yj)f(xi,yj) for the discrete case,integraltext∞ −∞integraltext∞ −∞g(x, y)f(x, y)dx dy for the continuous case. 26.12.1 Means The means of XandYare deﬁned respectively as the expectation values of the variables XandY. Thus, the mean of Xis given by E[X]=µX=braceleftBiggsummationtext isummationtext jxif(xi,yj) for the discrete case,integraltext∞ −∞integraltext∞ −∞xf(x, y)dx dy for the continuous case.(26.131) E[Y] is obtained in a similar manner.IShow that if XandYare independent random variables then E[XY]=E[X]E[Y]. Let us consider the case where XandYare continuous random variables. Since Xand Yare independent f(x, y)=fX(x)fY(y), so that E[XY]= Z∞ −∞ Z∞ −∞xyf X(x)fY(y)dx dy= Z∞ −∞xfX(x)dx Z∞ −∞yfY(y)dy=E[X]E[Y]. An analogous proof exists for the discrete case. J 1041 PROBABILITY 26.12.2 Variances The deﬁnitions of the variances of Xand Yare analogous to those for the single-variable case (26.48), i.e. the variance of Xis given by V[X]=σ2 X=braceleftBiggsummationtext isummationtext j(xi−µX)2f(xi,yj) for the discrete case,integraltext∞ −∞integraltext∞ −∞(x−µX)2f(x, y)dx dy for the continuous case.(26.132) Equivalent deﬁnitions exist for the variance of Y. 26.12.3 Covariance and correlation Means and variances of joint distributions provide useful information about their marginal distributions, but we have not yet given any indication of how to measure the relationship between the two random variables. Of course, it may be that the two random variables are independent, but often this is not so. Forexample, if we measure the heights and weights of a sample of people we wouldnot be surprised to ﬁnd a tendency for tall people to be heavier than short peopleand vice versa. We will show in this section that two functions, the covariance and the correlation , can be deﬁned for a bivariate distribution and that these are useful in characterising the relationship between the two random variables. Thecovariance of two random variables XandYis deﬁned by Cov[X,Y]=E[(X−µ X)(Y−µY)], (26.133) where µXandµYare the expectation values of XandYrespectively. Clearly related to the covariance is the correlation of the two random variables, deﬁned by Corr[ X,Y]=Cov[X,Y] σXσY, (26.134) where σXandσYare the standard deviations of XandYrespectively. It can be shown that the correlation function lies between −1 and +1. If the value assumed is negative, XandYare said to be negatively correlated , if it is positive they are said to be positively correlated a n di fi ti sz e r ot h e ya r es a i dt ob e uncorrelated . We will now justify the use of these terms. One particularly useful consequence of its deﬁnition is that the covariance of two independent variables, Xand Y, is zero. It immediately follows from (26.134) that their correlation is also zero, and this justiﬁes the use of the term‘uncorrelated’ for two such variables. To show this extremely important property 1042 26.12 PROPERTIES OF JOINT DISTRIBUTIONS we ﬁrst note that Cov[X,Y]=E[(X−µX)(Y−µY)] =E[XY−µXY−µYX+µXµY] =E[XY]−µXE[Y]−µYE[X]+µXµY =E[XY]−µXµY. (26.135) Now, if XandYare independent then E[XY]=E[X]E[Y]=µXµYand so Cov[X,Y] = 0. It is important to note that the converse of this result is not necessarily true; two variables dependent on each other can still be uncorrelated. In other words, it is possible (and not uncommon) for two variables XandY to be described by a joint distribution f(x, y)t h a t cannot be factorised into a product of the form g(x)h(y), but for which Corr[ X,Y] = 0. Indeed, from the deﬁnition (26.133), we see that for any joint distribution f(x, y) that is symmetric inxabout µX(or similarly in y) we have Corr[ X,Y]=0 . We have already asserted that if the correlation of two random variables is positive (negative) they are said to be positively (negatively) correlated. We havealso stated that the correlation lies between −1 and +1. The terminology suggests that if the two RVs are identical (i.e. X=Y) then they are completely correlated and that their correlation should be +1. Likewise, if X=−Ythen the functions are completely anticorrelated and their correlation should be −1. Values of the correlation function between these extremes show the existence of some degreeof correlation. In fact it is not necessary that X=Yfor Corr[ X,Y] = 1; it is suﬃcient that Yis a linear function of X,i . e .Y=aX+b(with apositive). If a is negative then Corr[ X,Y]=−1. To show this we ﬁrst note that µ Y=aµX+b. Now Y=aX+b=aX+µY−aµX⇒ Y−µY=a(X−µX), and so using the deﬁnition of the covariance (26.133) Cov[X,Y]=aE[(X−µX)2]=aσ2 X. It follows from the properties of the variance (subsection 26.5.3) that σY=|a|σX and so, using the deﬁnition (26.134) of the correlation, Corr[ X,Y]=aσ2 X |a|σ2 X=a |a|, which is the stated result. It should be noted that, even if the possibilities of XandYbeing non-zero are mutually exclusive, Corr[ X,Y] need not have value ±1. 1043 PROBABILITYIA biased die gives probabilities1 2p,p,p,p,p,2pof throwing 1, 2, 3, 4, 5, 6 respectively. If the random variable Xis the number shown on the die and the random variable Yis deﬁned as X2, calculate the covariance and correlation of XandY. We have already calculated in subsections 26.2.1 and 26.5.4 that p=2 13,E[X]=53 13, E / X2 / =253 13andV[X]=480 169. Using (26.135) Cov[X,Y]=C o v [ X,X2]=E[X3]−E[X]E[X2]. Now E[X3]i sg i v e nb y E[X3]=13×1 2p+( 23+33+43+53)p+63×2p =1313 2p= 101 , and the covariance of XandYis given by Cov[X,Y] = 101−53 13×253 13=3660 169. The correlation is deﬁned by Corr[ X,Y]=C o v [ X,Y]/σXσY. The standard deviation of Ymay be calculated from the deﬁnition of the variance. Letting µY=E[X2]=253 13gives σ2 Y=p 2 /; 12−µY /2+p /; 22−µY /2+p /; 32−µY /2+p /; 42−µY /2 +p /; 52−µY /2+2p /; 62−µY /2 =187356 169p=28824 169. We deduce that Corr[ X,Y]=3660 169 r 169 28824 r 169 480≈0.984. Thus the random variables XandYdisplay a strong degree of positive correlation, as we would expect. J We note that the covariance of XandYoccurs in various expressions. For example, if XandYarenotindependent then V[X+Y]=Ebracketleftbig (X+Y)2bracketrightbig −(E[X+Y])2 =Ebracketleftbig X2bracketrightbig +2E[XY]+Ebracketleftbig Y2bracketrightbig −{(E[X])2+2E[X]E[Y]+(E[Y])2} =V[X]+V[Y]+2 ( E[XY]−E[X]E[Y]) =V[X]+V[Y]+2C o v [ X,Y]. More generally, we ﬁnd (for a,bandcconstant) V[aX+bY+c]=a2V[X]+b2V[Y]+2abCov[X,Y]. (26.136) 1044 26.12 PROPERTIES OF JOINT DISTRIBUTIONS Note that if XandYare in fact independent then Cov[ X,Y]=0a n dw er e c o v e r the expression (26.68) in subsection 26.6.4. We may use (26.136) to obtain an approximate expression for V[f(X,Y)] for any arbitrary function f, even when the random variables Xand Yare correlated. Approximating f(X,Y) by the linear terms of its Taylor expansion about the point ( µX,µY), we have f(X,Y)≈f(µX,µY)+parenleftbigg∂f ∂Xparenrightbigg (X−µX)+parenleftbigg∂f ∂Yparenrightbigg (Y−µY), (26.137) where the partial derivatives are evaluated at X=µXandY=µY.T a k i n gt h e variance of both sides, and using (26.136), we ﬁnd V[f(X,Y)]≈parenleftbigg∂f ∂Xparenrightbigg2 V[X]+parenleftbigg∂f ∂Yparenrightbigg2 V[Y]+2parenleftbigg∂f ∂Xparenrightbiggparenleftbigg∂f ∂Yparenrightbigg Cov[X,Y]. (26.138) Clearly, if Cov[ X,Y] = 0, we recover the result (26.69) derived in subsection 26.6.4. We note that (26.138) is exact if f(X,Y) is linear in XandY. For several variables Xi,i=1,2,...,n, we can deﬁne the symmetric (positive deﬁnite) covariance matrix whose elements are Vij=C o v [ Xi,Xj], (26.139) and the symmetric (positive deﬁnite) correlation matrix ρij=C o r r [ Xi,Xj]. The diagonal elements of the covariance matrix are the variances of the variables, whilst those of the correlation matrix are unity. For several variables, (26.138)generalises to V[f(X 1,X2,...,X n)]≈summationdisplay iparenleftbigg∂f ∂Xiparenrightbigg2 V[Xi]+summationdisplay isummationdisplay j/negationslash=iparenleftbigg∂f ∂Xiparenrightbiggparenleftbigg∂f ∂Xjparenrightbigg Cov[Xi,Xj], where the partial derivatives are evaluated at Xi=µXi. 1045 PROBABILITYIA card is drawn at random from a normal 52-card pack and its identity noted. The card is replaced, the pack shuﬄed and the process repeated. Random variables W, X, Y, Z are deﬁned as follows: W=2 if the drawn card is a heart; W=0otherwise. X=4 if the drawn card is an ace, king, or queen; X=2if the card is aj a c ko rt e n ; X=0otherwise. Y=1 if the drawn card is red; Y=0otherwise. Z=2 if the drawn card is black and an ace, king or queen; Z=0 otherwise. Establish the correlation matrix for W, X, Y, Z . The means of the variables are given by µW=2×1 4=1 2,µ X= /; 4×3 13 / + /; 2×2 13 / =16 13, µY=1×1 2=1 2,µ Z=2×6 52=3 13. The variances, calculated from σ2 U=V[U]=E / U2 / −(E[U])2,w h e r e U=W,X,Yor Z,a r e σ2 W= /; 4×1 4 / − /;1 2 /2=3 4,σ2 X= /; 16×3 13 / + /; 4×2 13 / − /;16 13 /2=472 169, σ2 Y= /; 1×1 2 / − /;1 2 /2=1 4,σ2 Z= /; 4×6 52 / − /;3 13 /2=69 169. The covariances are found by ﬁrst calculating E[WX] etc. and then forming E[WX]−µWµX etc. E[WX]=2(4) /;3 52 / +2(2) /;2 52 / =8 13,Cov[W,X]=8 13−1 2 /;16 13 / =0, E[WY] = 2(1) /;1 4 / =1 2, Cov[W,Y]=1 2−1 2 /;1 2 / =1 4, E[WZ]=0 , Cov[W,Z]=0−1 2 /;3 13 / =−3 26, E[XY] = 4(1) /;6 52 / +2 ( 1 ) /;4 52 / =8 13,Cov[X,Y]=8 13−16 13 /;1 2 / =0, E[XZ] = 4(2) /;6 52 / =12 13, Cov[X,Z]=12 13−16 13 /;3 13 / =108 169, E[YZ]=0 , Cov[Y,Z]=0−1 2 /;3 13 / =−3 26. The correlations Corr[ W,X] and Corr[ X,Y] are clearly zero; the remainder are given by Corr[ W,Y]=1 4 /;3 4×1 4 /−1/2=0.577, Corr[ W,Z]=−3 26 /;3 4×69 169 /−1/2=−0.209, Corr[ X,Z]=108 169 /;472 169×69 169 /−1/2=0.598, Corr[ Y,Z]=−3 26 /;1 4×69 169 /−1/2=−0.361. Finally, then, we can write down the correlation matrix: ρ= /0B/@10 0 .58−0.21 010 0 .60 0.58 0 1 −0.36 −0.21 0 .60−0.36 1 /1CA. 1046 26.13 GENERATING FUNCTIONS FOR JOINT DISTRIBUTIONS As would be expected, Xis uncorrelated with either WorY, colour and face-value being two independent characteristics. Positive correlations are to be expected between Wand Yand between XandZ; both correlations are fairly strong. Moderate anticorrelations exist between Zand both WandY, reﬂecting the fact that it is impossible for WandY to be positive if Zis positive. J Finally, let us suppose that the random variables Xi,i=1,2,...,n, are related to a second set of random variables Yk=Yk(X1,X2,...,X n),k=1,2,...,m .B y expanding each Ykas a Taylor series as in (26.137) and inserting the resulting expressions into the deﬁnition of the covariance (26.133), we ﬁnd that the elements of the covariance matrix for the Ykvariables are given by Cov[Yk,Yl]≈summationdisplay isummationdisplay jparenleftbigg∂Yk ∂Xiparenrightbiggparenleftbigg∂Yl ∂Xjparenrightbigg Cov[Xi,Xj]. (26.140) It is straightforward to show that this relation is exact if the Ykare linear combinations of the Xi. Equation (26.140) can then be written in matrix form as VY=SVXST, (26.141) where VYand VXare the covariance matrices of the YkandXivariables re- spectively and Sis the rectangular m×nmatrix with elements Ski=∂Yk/∂X i. 26.13 Generating functions for joint distributions It is straightforward to generalise the discussion of generating function in section 26.7 to joint distributions. For a multivariate distribution f(X1,X2,...,X n)o f non-negative integer random variables Xi,i=1,2,...,n, we deﬁne the probability generating function to be Φ(t1,t2,...,t n)=E[tX1 1tX2 2···tXnn]. As in the single-variable case, we may also deﬁne the closely related moment generating function, which has wider applicability since it is not restricted tonon-negative integer random variables but can be used with any set of discreteor continuous random variables X i(i=1,2,...,n). The MGF of the multivariate distribution f(X1,X2,...,X n) is deﬁned as M(t1,t2,...,t n)=E[et1X1et2X2···etnXn]=E[et1X1+t2X2+···+tnXn] (26.142) and may be used to evaluate (joint) moments of f(X1,X2,...,X n). By performing a derivation analogous to that presented for the single-variable case in subsection26.7.2, it can be shown that E[X m1 1Xm2 2···Xmn n]=∂m1+m2+···+mnM(0,0,...,0) ∂tm1 1∂tm2 2···∂tmnn. (26.143) 1047 PROBABILITY Finally we note that, by analogy with the single-variable case, the characteristic function and the cumulant generating function of a multivariate distribution aredeﬁned respectively as C(t 1,t2,...,t n)=M(it1,i t2,...,i t n)a n d K(t1,t2,...,t n)=l n M(t1,t2,...,t n).ISuppose that the random variables Xi,i=1,2,...,n,a r ed e s c r i b e db yt h eP D F f(x)=f(x1,x2,...,x n)=Nexp(−1 2xTAx), where the column vector x=(x1x2··· xn)T,Ais an n×nsymmetric matrix and N is a normalisation constant such thatZ ∞f(x)dnx≡ Z∞ −∞ Z∞ −∞··· Z∞ −∞f(x1,x2,...,x n)dx1dx2···dxn=1. Find the MGF of f(x). From (26.142), the MGF is given by M(t1,t2,...,t n)=N Z ∞exp(−1 2xTAx+tTx)dnx, (26.144) where the column vector t=(t1t2··· tn)T. In order to evaluate this multiple integral, we begin by noting that xTAx−2tTx=(x−A−1t)TA(x−A−1t)−tTA−1t, which is the matrix equivalent of ‘completing the square’. Using this expression in (26.144) and making the substitution y=x−A−1t,w eo b t a i n M(t1,t2,...,t n)=cexp(1 2tTA−1t), (26.145) where the constant cis given by c=N Z ∞exp(−1 2yTAy)dny. From the normalisation condition for N,w es e et h a t c= 1, as indeed it must be in order thatM(0,0,...,0) = 1. J 26.14 Transformation of variables in joint distributions Suppose the random variables Xi,i=1,2,...,n, are described by the multivariate PDF f(x1,x2...,x n). If we wish to consider random variables Yj,j=1,2,...,m , related to the XibyYj=Yj(X1,X2,...,X m) then we may calculate g(y1,y2,...,y m), the PDF for the Yj, in a similar way to that in the univariate case by demanding that |f(x1,x2...,x n)dx1dx2···dxn|=|g(y1,y2,...,y m)dy1dy2···dym|. From the discussion of changing the variables in multiple integrals given in chapter 6 it follows that, in the special case where n=m, g(y1,y2,...,y m)=f(x1,x2...,x n)|J|, 1048 26.15 IMPORTANT JOINT DISTRIBUTIONS where J≡∂(x1,x2...,x n) ∂(y1,y2,...,y n)=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle∂x 1 ∂y1...∂xn ∂y1......... ∂x1 ∂yn...∂xn ∂ynvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle, is the Jacobian of the x iwith respect to the yj.ISuppose that the random variables Xi,i=1,2,...,n, are independent and Gaussian dis- tributed with means µiand variances σ2 irespectively. Find the PDF for the new variables Zi=(Xi−µi)/σi,i=1,2,...,n. By considering an elemental spherical shell in Z-space, ﬁnd the PDF of the chi-squared random variable χ2 n= Pn i=1Z2 i. Since the Xiare independent random variables, f(x1,x2,...,x n)=f(x1)f(x2)···f(xn)=1 (2π)n/2σ1σ2···σnexp /" −nX i=1(xi−µi)2 2σ2 i /# . To derive the PDF for the variables Zi,w er e q u i r e |f(x1,x2,...,x n)dx1dx2···dxn|=|g(z1,z2,...,z n)dz1dz2···dzn|, and, noting that dzi=dxi/σi,w eo b t a i n g(z1,z2,...,z n)=1 (2π)n/2exp / −1 2nX i=1z2 i /! . Let us now consider the random variable χ2 n= Pn i=1Z2 i, which we may regard as the square of the distance from the origin in the n-dimensional Z-space. We now require that g(z1,z2,...,z n)dz1dz2···dzn=h(χ2 n)dχ2 n. If we consider the inﬁnitesimal volume dV=dz1dz2···dznto be that enclosed by the n-dimensional spherical shell of radius χnand thickness dχnthen we may write dV= Aχn−1 ndχn, for some constant A. We thus obtain h(χ2 n)dχ2 n∝exp(−1 2χ2 n)χn−1 ndχn∝exp(−1 2χ2 n)χn−2 ndχ2n, where we have used the fact that dχ2 n=2χndχn. Thus we see that the PDF for χ2 nis given by h(χ2 n)=Bexp(−1 2χ2 n)χn−2 n, for some constant B. This constant may be determined from the normalisation conditionZ∞ 0h(χ2 n)dχ2 n=1 and is found to be B=[ 2n/2Γ(1 2n)]−1. This is the nth-order chi-squared distribution discussed in subsection 26.9.4. J 26.15 Important joint distributions In this section we will examine two important multivariate distributions, the multinomial distribution , which is an extension of the binomial distribution, and themultivariate Gaussian distribution . 1049 PROBABILITY 26.15.1 The multinomial distribution The binomial distribution describes the probability of obtaining x‘successes’ from nindependent trials, where each trial has only two possible outcomes. This may be generalised to the case where each trial has kpossible outcomes with respective probabilities p1,p2,...,pk. If we consider the random variables Xi,i=1,2,...,n, to be the number of outcomes of type iinntrials then we may calculate their joint probability function f(x1,x2,...,x k)=P r ( X1=x1,X2=x2, ..., X k=xk), w h e r ew em u s th a v esummationtextk i=1xi=n.I n ntrials the probability of obtaining x1 outcomes of type 1, followed by x2outcomes of type 2 etc. is given by px1 1px2 2···pxk k. However, the number of distinguishable permutations of this result is n! x1!x2!···xk!, and thus f(x1,x2,...,x k)=n! x1!x2!···xk!px1 1px2 2···pxk k. (26.146) This is the multinomial probability distribution . Ifk= 2 then the multinomial distribution reduces to the familiar binomial distribution. Although in this form the binomial distribution appears to be afunction of two random variables, it must be remembered that, in fact, sincep 2=1−p1andx2=n−x1, the distribution of X1is entirely determined by the parameters pandn.T h a t X1has abinomial distribution is shown by remembering that it represents the number of objects of a particular type obtained fromsampling with replacement, which led to the original deﬁnition of the binomialdistribution. In fact, any of the random variables X ihas a binomial distribution, i.e. the marginal distribution of each Xiis binomial with parameters nandpi.I t immediately follows that E[Xi]=npiand V[Xi]2=npi(1−pi). (26.147)IAt a village f ˆete patrons were invited, for a 10pentry fee, to pick without looking six tickets from a drum containing equal large numbers of red, blue and green tickets. If ﬁveor more of the tickets were of the same colour a prize of 100 p was awarded. A consolation award of 40pwas made if two tickets of each colour were picked. Was a good time had by all? In this case, all types of outcome (red, blue and green) have the same probabilities. Theprobability of obtaining any given combination of tickets is given by the multinomialdistribution with n=6 , k=3a n d p i=1 3,i=1,2,3. 1050 26.15 IMPORTANT JOINT DISTRIBUTIONS (i) The probability of picking six tickets of the same colour is given by Pr (six of the same colour) = 3 ×6! 6!0!0! /1 3 /6 /1 3 /0 /1 3 /0 =1 243. The factor of 3 is present because there are three diﬀerent colours. (ii) The probability of picking ﬁve tickets of one colour and one ticket of another colour is Pr(ﬁve of one colour; one of another) = 3 ×2×6! 5!1!0! /1 3 /5 /1 3 /1 /1 3 /0 =4 81. The factors of 3 and 2 are included because there are three ways to choose the colour of the ﬁve matching tickets, and then two ways to choose the colour of theremaining ticket. (iii) Finally, the probability of picking two tickets of each colour is Pr (two of each colour) =6! 2!2!2! /1 3 /2 /1 3 /2 /1 3 /2 =10 81. Thus the expected return to any patron was, in pence, 100 /1 243+4 81 / + / 40×10 81 / =1 0.29. A good time was had by all but the stallholder! J 26.15.2 The multivariate Gaussian distribution A particularly interesting multivariate distribution is provided by the generalisa- tion of the Gaussian distribution to multiple random variables Xi,i=1,2,...,n. If the expectation value of XiisE(Xi)=µithen the general form of the PDF is given by f(x1,x2,...,x n)=Nexpbracketleftbigg −1 2summationdisplay isummationdisplay jaij(xi−µi)(xj−µj)bracketrightbigg , where aij=ajiandNis a normalisation constant that we give below. If we write the column vectors x=(x1x2··· xn)Tandµ=(µ1µ2··· µn)T,a n d denote the matrix with elements aijbyAthen f(x)=f(x1,x2,...,x n)=Nexpbracketleftbig −1 2(x−µ)TA(x−µ)bracketrightbig , where Ais symmetric. Using the same method as that used to derive (26.145) it is straightforward to show that the MGF of f(x)i sg i v e nb y M(t1,t2,...,t n)=e x pparenleftbig µTt+1 2tTA−1tparenrightbig , where the column matrix t=(t1t2··· tn)T. From the MGF, we ﬁnd that E[XiXj]=∂2M(0,0,...,0) ∂ti∂tj=µiµj+(A−1)ij, 1051 PROBABILITY and thus, using (26.135), we obtain Cov[Xi,Xj]=E[(Xi−µi)(Xj−µj)] = ( A−1)ij. Hence Ais equal to the inverse of the covariance matrix Vof the Xi, see (26.139). Thus, with the correct normalisation, f(x)i sg i v e nb y f(x)=1 (2π)n/2(det V)1/2expbracketleftbig −1 2(x−µ)TV−1(x−µ)bracketrightbig . (26.148)IEvaluate the integral I= Z ∞exp / −1 2(x−µ)TV−1(x−µ) / dnx, where Vis a symmetric matrix, and hence verify the normalisation in (26.148). We begin by making the substitution y=x−µto obtain I= Z ∞exp(−1 2yTV−1y)dny. Since Vis a symmetric matrix, it may be diagonalised by an orthogonal transformation to the new set of variables y/prime=STy,w h e r e Sis the orthogonal matrix with the normalised eigenvectors of Vas its columns (see section 8.16). In this new basis, the matrix Vbecomes V/prime=STVS=d i a g ( λ1,λ2,...,λ n), where the λiare the eigenvalues of V. Also, since Sis orthogonal, det S=±1, and so dny=|detS|dny/prime=dny/prime. Thus we can write Ias I= Z∞ −∞ Z∞ −∞··· Z∞ −∞exp / −nX i=1y/prime i2 2λi /! dy/prime 1dy/prime 2···dy/prime n =nY i=1 Z∞ −∞exp / −y/prime i2 2λi /! dy/prime i=( 2π)n/2(λ1λ2···λn)1/2, (26.149) where we have used the standard integral R∞ −∞exp(−αy2)dy=(π/α)1/2(see subsection 6.4.2). From section 8.16, however, we note that the product of eigenvalues in (26.149) ise q u a lt od e t V. Thus we ﬁnally obtain I=( 2π) n/2(det V)1/2, and hence the normalisation in (26.148) ensures that f(x) integrates to unity. J The above example illustrates some importants points concerning the multi- variate Gaussian distribution. In particular, we note that the Y/prime iareindependent Gaussian variables with mean zero and variance λi. Thus, given a general set of nGaussian variables xwith means µand covariance matrix V, one can always perform the above transformation to obtain a new set of variables y/prime,w h i c ha r e linear combinations of the old ones and are distributed as independent Gaussians with zero mean and variances λi. This result is extremely useful in proving many of the properties of the mul- 1052 26.16 EXERCISES tivariate Gaussian. For example, let us consider the quadratic form (multiplied by 2) appearing in the exponent of (26.148) and write it as χ2 n,i . e . χ2 n=(x−µ)TV−1(x−µ). (26.150) From (26.149), we see that we may also write it as χ2 n=nsummationdisplay i=1y/prime i2 λi, which is the sum of nindependent Gaussian variables with mean zero and unit variance. Thus, as our notation implies, the quantity χ2 nis distributed as a chi- squared variable of order n. As illustrated in exercise 26.40, if the variables Xiare required to satisfy mlinear constraints of the formsummationtextn i=1ciXi=0t h e n χ2 ndeﬁned in (26.150) is distributed as a chi-squared variable of order n−m. 26.16 Exercises 26.1 By shading Venn diagrams, determine which of the following are valid rela- tionships between events. For those that are, prove them using de Morgan’slaws. (a) (¯X∪Y)=X∩¯Y. (b)¯X∪¯Y=(X∪Y). (c) ( X∪Y)∩Z=(X∪Z)∩Y. (d)X∪(Y∩Z)=(X∪¯Y)∩¯Z. (e)X∪(Y∩Z)=(X∪¯Y)∪¯Z. 26.2 Given that events X,YandZsatisfy (X∩Y)∪(Z∩X)∪(¯X∪¯Y)=(Z∪¯Y)∪{[(¯Z∪¯X)∪(¯X∩Z)]∩Y}, prove that X⊇Yand either Y∩Z=∅orY⊇Z. 26.3 AandBeach have two unbiased four-faced dice, the four faces being numbered 1, 2, 3, 4. Without looking, Btries to guess the sum xof the numbers on the bottom faces of A’s two dice after they have been thrown onto a table. If the guess is correct Breceives x2euros, but if not he loses xeuros. Determine B’s expected gain per throw of A’s dice when he adopts each of the following strategies: (a) he selects xat random in the range 2 ≤x≤8; (b) he throws his own two dice and guesses xto be whatever they indicate; (c) he takes your advice and always chooses the same value for x. Which number would you advise? 26.4 Use the method of induction to prove e quation (26.16), the probability addition law for the union of ngeneral events. 26.5 Two duellists, AandB, take alternate shots at each other, and the duel is over when a shot (fatal or otherwise!) hits its target. Each shot ﬁred by Ahas a probability αof hitting B, and each shot ﬁred by Bhas a probability βof hitting A. Calculate the probabilities P1andP2, deﬁned as follows, that Awill win such a duel: P1,Aﬁres the ﬁrst shot; P2,Bﬁres the ﬁrst shot. If they agree to ﬁre simultaneously, rath er than alternately, what is the proba- bility P3thatAwill win? Verify that your results satisfy the intuitive inequality P1≥P3≥P2. 1053 PROBABILITY 26.6 X1,X2,...,X nare independent identically distributed random variables drawn from a uniform distribution on [0 ,1]. The random variables AandBare deﬁned by A=m i n ( X1,X2,...,X n),B = max( X1,X2,...,X n). For any ﬁxed ksuch that 0 ≤k≤1 2, ﬁnd the probability pnthat both A≤k and B≥1−k. Check your general formula by considering directly the cases (a) k=0 ,( b ) k=1 2, (c)n=1a n d( d ) n=2 . 26.7 A tennis tournament is arranged on a straight knockout basis for 2nplayers and for each round, except the ﬁnal, opponents for those still in the competition aredrawn at random. The quality of the ﬁeld is so even that in any match it isequally likely that either player will win. Two of the players have surnames thatbegin with ‘ Q’. Find the probabilities that they play each other (a) in the ﬁnal, (b) at some stage in the tournament. 26.8 (a) Gamblers AandBeach roll a fair six-faced die, and Bwins if his score is strictly greater than A’s. Show that the odds are 7 to 5 in A’s favour. (b) Calculate the probabilities of scoring a total Tfrom two rolls of a fair die forT=2,3,...,12. Gamblers CandDeach roll a fair die twice and score respective totals T CandTD,Dwinning if TD>T C. Realising that the odds are not equal, Dinsists that Cshould increase her stake for each game. C agrees to stake £1.10 per game, as compared to D’s£1.00 stake. Who will show a proﬁt? 26.9 An electronics assembly ﬁrm buys its microchips from three diﬀerent suppliers; half of them are bought from ﬁrm X, whilst ﬁrms YandZsupply 30% and 20% respectively. The suppliers use diﬀerent quality-control procedures and thepercentages of defective chips are 2%, 4% and 4% for X,YandZrespectively. The probabilities that a defective chip will fail two or more assembly-line testsare 40%, 60% and 80% respectively, whilst all defective chips have a 10% chanceof escaping detection. An assembler ﬁnds a chip that fails only one test. What isthe probability that it came from supplier X? 26.10 As every student of probability theory will know, Bayesylvania is awash with natives, not all of whom can be trusted to tell the truth, and lost and apparentlysomewhat deaf travellers who ask the sam e question several times in an attempt to get directions to the nearest village. One such traveller ﬁnds himself at a T-junction in an area populated by the Asciis and Bisciis in the ratio 11 to 5. As is well known, the Biscii always lie butthe Ascii tell the truth three quarters o f the time, giving independent answers to all questions, even to immediately repeated ones. (a) The traveller asks one particular native twice whether he should go to the left or to the right to reach the local village. Each time he is told ‘left’. Shouldhe take this advice, and, if he does, what are his chances of reaching thevillage? (b) The traveller then asks the same native the same question a third time and for a third time receives the answer ‘left’. What should the traveller do now?Have his chances of ﬁnding the village been altered by asking the thirdquestion? 26.11 A boy is selected at random from amongs t the children belonging to families with nchildren. It is known that he has at least two sisters. Show that the probability 1054 26.16 EXERCISES that he has k−1b r o t h e r si s (n−1)! (2n−1−n)(k−1)!(n−k)!, for 1≤k≤n−2 and zero for other values of k. 26.12 Villages A,B,CandDare connected by overhead telephone lines joining AB, AC,BC,BDandCD. As a result of severe gales, there is a probability p(the same for each link) that any particular link is broken. (a) Show that the probability that a call can be made from AtoBis 1−2p2+p3. (b) Show that the probability that a call can be made from DtoAis 1−2p2−2p3+5p4−2p5. 26.13 A set of 2 N+ 1 rods consists of one of each integer length 1 ,2,... ,2N,2N+1 . Three, of lengths a,bandc, are selected, of which ais the longest. By considering the possible values of bandc, determine the number of ways in which a non- degenerate triangle (i.e. one of non-zero area) can be formed (i) if ais even, and (ii) if ais odd. Combine these results appropriately to determine the total number of non-degenerate triangles that can be formed with the 2 N+ 1 rods, and hence show that the probability that such a triangle can be formed from arandom selection (without replacement) of three rods is (N−1)(4N+1 ) 2(4N2−1). 26.14 A certain marksman never misses his target, which consists of a disc of unit radius with centre O. The probability that any given shot will hit the target within a distance tofOist2for 0≤t≤1. The marksman ﬁres nindependendent shots at the target, and the random variable Yis the radius of the smallest circle with centre Othat encloses all the shots. Determine the PDF for Yand hence ﬁnd the expected area of the circle. The shot that is furthest from Ois now rejected and the corresponding circle determined for the remaining n−1 shots. Show that its expected area is n−1 n+1π. 26.15 The duration of a telephone call made from a public call-box is a random variable T. The probability density function of Tis f(t)= /8/>/</>/:0 t<0, 1 20≤t<1, ke−2tt≥1, where kis a constant. To pay for the call, 20 pence has to be inserted at the beginning, and a further 20 pence after each subsequent half-minute. Determineby how much the average cost of a call exceeds the cost of a call of averagelength charged at 40 pence per minute. 26.16 Kittens from diﬀerent litters do not get on with each other and ﬁghting breaks out whenever two kittens from diﬀerent litters are present together. A cage initiallycontains xkittens from one litter and yfrom another. To quell the ﬁghting, kittens are removed at random, one at a time, until peace is restored. Show, byinduction, that the expected number of kittens ﬁnally remaining is N(x, y)=x y+1+y x+1. 1055 PROBABILITY 26.17 ( A more diﬃcult question. ) If the scores in a cup football match are equal at the end of the normal period of play, a ‘penalty shoot-out’ is held in which each side takes up to ﬁveshots (from the penalty spot) alternately, the shoot-out being stopped if oneside acquires an unassailable lead (i.e. has a lead greater than its opponentshave shots remaining). If the scores are still level after the shoot-out a ‘suddendeath’ competition takes place. In sudden death each side takes one shot and thecompetition is over if one side scores and the other does not; if both score, orboth fail to score, a further shot is taken by each side, and so on. Team 1, whichtakes the ﬁrst penalty, has a probability p 1, which is independent of the player involved, of scoring and a probability q1(= 1−p1) of missing; p2andq2are deﬁned likewise. Deﬁne Pr( i:x, y) as the probability that team ihas scored xgoals after y attempts, and let f(M) be the probability that the shoot-out terminates after a totalofMshots. (a) Prove that the probability tha t ‘sudden death’ will be needed is f(11+) =5X r=0(5Cr)2(p1p2)r(q1q2)5−r. (b) Give reasoned arguments (preferably without ﬁrst looking at the expressions involved) which show that f(M=2N)=2N−6X r=0 / p2Pr(1: r,N)P r(2:5−N+r,N−1) +q2Pr(1:6−N+r,N)P r(2: r,N−1) / forN=3,4,5a n d f(M=2N+1 )=2N−5X r=0 / p1Pr(1:5−N+r,N)P r(2: r,N) +q1Pr(1: r,N)P r( 2:5−N+r,N) / forN=3,4. (c) Give an explicit expression for Pr( i:x, y) and hence show that if the teams are so well matched that p1=p2=1/2t h e n f(2N)=2N−6X r=0 /1 22N /N!(N−1)!6 r!(N−r)!(6−N+r)!(2N−6−r)!, f(2N+1 )=2N−5X r=0 /1 22N /(N!)2 r!(N−r)!(5−N+r)!(2N−5−r)!. (d) Evaluate these expressions to show that, expressing f(M) in units of 2−8,w e have M 6 7 8 9 10 11+ f(M) 8 24 42 56 63 63 Give a simple explanation of why f(10) = f(11+). 26.18 A particle is conﬁned to the one-dimensional space 0 ≤x≤aand classically it can be in any small interval dxwith equal probability. However, quantum mechanics gives the result that the probability distribution is proportional tosin 2(nπx/a ), where nis an integer. Find the variance in the particle’s position in both the classical and quantum mechanical pictures and show that, althoughthey diﬀer, the latter tends to the former in the limit of large n, in agreement with the correspondence principle of physics. 1056 26.16 EXERCISES 26.19 A continuous random variable Xhas a probability density function f(x); the corresponding cumulative probability function is F(x). Show that the random variable Y=F(X) is uniformly distributed between 0 and 1. 26.20 For a non-negative integer random variable X, in addition to the probability generating function Φ X(t) deﬁned in equation (26.71) it is possible to deﬁne the probability generating function ΨX(t)=∞X n=0gntn, where gnis the probability that X>n . (a) Prove that Φ Xand Ψ Xare related by ΨX(t)=1−ΦX(t) 1−t. (b) Show that E[X]i sg i v e nb yΨ X(1) and that the variance of Xcan be expressed as 2Ψ/prime X(1) + Ψ X(1)−[ΨX(1)]2. (c) For a particular random variable X, the probability that X>n is equal to αn+1with 0 <α< 1. Use the results in (b) to show that V[X]=α(1−α)−2. 26.21 (a) In two sets of binomial trials Tandtthe probabilities that a trial has a successful outcome are Pandprespectively, with corresponding probabilites of failure of Q=1−Pandq=1−p. One ‘game’ consists of a trial T followed, if Tis successful, by a trial tand then a further trial T.T h et w o trials continue to alternate until one of the Ttrials fails, at which point the game ends. The score Sfor the game is the total number of successes in the t-trials. Find the PGF for Sand use it to show that E[S]=Pp Q,V [S]=Pp(1−Pq) Q2. (b) Two normal unbiased six-faced dice AandBare rolled alternately starting with A;i fAshows a 6 the experiment ends. If Bshows an odd number no points are scored, if it shows a 2 or a 4 then one point is scored, whilst ifit records a 6 then two points are awarded. Find the average and standarddeviation of the score for the experiment and show that the latter is thegreater. 26.22 Use the formula obtained in subsection 26.8.2 for the moment generating function of the negative binomial distribution to determine the CGF K n(t) for the number of trials needed to record nsuccesses. Evaluate the ﬁrst four cumulants and use them to conﬁrm the stated results for the mean and variance and to show thatthe distribution has skewness and kurtosis given respectively by 2−p √n(1−p)and 3 +6−6p+p2 √n(1−p). 26.23 A point Pis chosen at random on the circle x2+y2= 1. The random variable Xdenotes the distance of Pfrom (1 ,0). Find the mean and variance of Xand the probability that Xis greater than its mean. 26.24 As assistant to a celebrated and imperious newspaper proprietor, you are given the job of running a lottery in which each of his ﬁve million readers will havean equal independent chance pof winning a million pounds ; you have the job of choosing p. However, if nobody wins it will be bad for publicity whilst if more than two readers do so, the prize cost will more than oﬀset the proﬁt from extracirculation – in either case you will be sacked! Show that, however you choosep, there is more than a 40% chance you will soon be clearing your desk. 1057 PROBABILITY 26.25 The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean µ. The cost of the ﬁrst correction on any page is αand that of each subsequent correction on the same page is β. Prove that the average cost of correcting a page is α+β(µ−1)−(α−β)e−µ. 26.26 In the game of Blackball, at each turn Muggins draws a ball at random from a bag containing ﬁve white balls, three red balls and two black balls; after being recorded, the ball is replaced in the bag. A white ball earns him $1 whilst a redball gets him $2; in either case he also has the option of leaving with his currentwinnings or of taking a further turn on the same basis. If he draws a black ballthe game ends and he loses all he may have gained previously. Find an expressionfor Muggins’ expected return if he adopts the strategy to drawing up to nballs if he has not been eliminated by then. Show that, as the entry fee to play is $3, Muggins should be dissuaded from playing Blackball, but if that cannot be done what value of nwould you advise him to adopt? 26.27 Show that for large rthe value at the maximum of the PDF for the gamma distribution of order rwith parameter λis approximately λ/√ 2π(r−1). 26.28 A husband and wife decide that their family will be complete when it includes two boys and two girls – but that this would then be enough! The probabilitythat a new baby will be a girl is p. Ignoring the possibility of identical twins, show that the expected size of their family is 2 /1 pq−1−pq / , where q=1−p. 26.29 The probability distribution for the number of eggs in a clutch is Po( λ), and the probability that each egg will hatch is p(independently of the size of the clutch). Show by direct calculation that the probability distribution for the number ofchicks that hatch is Po( λp) and so justify the assumptions made in the worked example at the end of subsection 26.7.1. 26.30 A shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the ﬁnal digit (the number of pence) in the price is uniformly and randomly distributed from 0 to 9. Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probabilityif the price ends in a ‘5’. Should she suspect a mistake if the cashier asks her for23 pence more than she estimated? 26.31 Under EU legislation on harmonisation, all kippers are to weigh 0.2000 kg and vendors who sell underweight kippers must be ﬁned by their government. Theweight of a kipper is normally distributed with a mean of 0.2000 kg and astandard deviation of 0.0100 kg. They are packed in cartons of 100 and largequantities of them are sold. Every day a carton is to be selected at random from each vendor and tested according to one of the following schemes, which have been approved for thepurpose. (a) The entire carton is weighed and the vendor is ﬁned 2500 euros if the average weight of a kipper is less than 0.1975 kg. (b) Twenty-ﬁve kippers are selected at random from the carton; the vendor is ﬁned 100 euros if the average weight of a kipper is less than 0.1980 kg. (c) Kippers are removed one at a time, at random, until one has been found that weighs morethan 0.2000 kg; the vendor is ﬁned n(n−1) euros, where n is the number of kippers removed. 1058 26.16 EXERCISES Which scheme should the Chancellor of the Exchequer be urging his government to adopt? 26.32 In a certain parliament the government consists of 75 New Socialites and the opposition consists of 25 Preservatives. Preservatives never change their mind, al-ways voting against government policy without a second thought; New Socialitesvote randomly, but with probability pthat they will vote for their party leader’s policies. Following a decision by the New Socialites’ leader to drop certain manifesto commitments, Nof his party decide to vote consistently with the opposition. The leader’s advisors reluctantly admit that an election must be called if Nis such that, at any vote on government policy, the chance of a simple majority in favour would be less than 80%. Given that p=0.8, estimate the lowest value of Nthat would precipitate an election. 26.33 A practical-class demonstrator sends his 12 students to the storeroom to collect apparatus for an experiment, but forgets to tell each which type of componentto bring. There are three types, A,BandC,h e l di nt h es t o r e s( i nl a r g en u m b e r s ) in the proportions 20%, 30% and 50% respectively, and each student picks acomponent at random. In order to set up one experiment, one unit each of Aand Band two units of Care needed. Find an expression for the probability Pr( N) that at least Nexperiments can be set up. (a) Evaluate Pr(3). (b) Show that Pr(2) can be written in the form Pr(2) = (0 .5) 126X i=212Ci(0.4)i8−iX j=212−iCj(0.6)j. (c) By considering the conditions under which no experiments can be set up, show that Pr(1) = 0 .9145. 26.34 The random variables XandYtake integer values ≥1 such that 2 x+y≤2a, where ais an integer greater than 1. The joint probability within this region is given by Pr(X=x, Y=y)=c(2x+y), where cis a constant, and it is zero elsewhere. Show that the marginal probability Pr( X=x)i s Pr(X=x)=6(a−x)(2x+2a+1 ) a(a−1)(8a+5 ), and obtain expressions for Pr( Y=y), (a) when yis even and (b) when yis odd. Show further that E[Y]=6a2+4a+1 8a+5. (You will need the results about series involving the natural numbers given in subsection 4.2.5.) 26.35 The continuous random variables XandYhave a joint PDF proportional to xy(x−y)2with 0≤x≤1a n d0≤y≤1. Find the marginal distributions forXand Yand show that they are negatively correlated with correlation coeﬃcient−2 3. 1059 PROBABILITY 26.36 A discrete random variable Xtakes integer values n=0,1,... ,N with probabil- itiespn. A second random variable Yis deﬁned as Y=(X−µ)2,w h e r e µis the expectation value of X. Prove that the covariance of XandYis given by Cov[X,Y]=NX n=0n3pn−3µNX n=0n2pn+2µ3. Now suppose that Xtakes all its possible values with equal probability and hence demonstrate that two random variables can be uncorrelated even though one isdeﬁned in terms of the other. 26.37 Two continuous random variables XandYhave a joint probability distribution f(x, y)=A(x 2+y2), where Ais a constant and 0 ≤x≤a,0≤y≤a. Show that XandYare negatively correlated with correlation coeﬃcient −15/73. By sketching a rough contour map of f(x, y) and marking oﬀ the regions of positive and negative correlation, convince yourself that this (perhaps counter-intuitive) result is plausible. 26.38 A continuous random variable Xis uniformly distributed over the interval [ −c, c]. As a m p l eo f2 n+ 1 values of Xis selected at random and the random variable Zis deﬁned as the median of that sample. Show that Zis distributed over [ −c, c] with probability density function fn(z)=(2n+1 ) ! (n!)2(2c)2n+1(c2−z2)n. Find the variance of Z. 26.39 Show that, as the number of trials nbecomes large but npi=λi,i=1,2,...,k−1, remains ﬁnite, the multinomial p robability distribution (26.146), Mn(x1,x2,...,x k)=n! x1!x2!···xk!px1 1px2 2···pxk k, can be approximated by a multiple Poisson distribution (with k−1 factors) M/prime n(x1,x2,...,x k−1)=k−1Y i=1e−λiλxi i xi!. (Write Pk−1 ipi=δand express all terms involving subscript kin terms of nand δ, either exactly or approximately. You will need to use n!≈n/epsilon1[(n−/epsilon1)!] and (1−a/n)n≈e−afor large n.) (a) Verify that the terms of M/prime nwhen summed over all values of x1,x2,...,x k−1 a d du pt ou n i t y . (b) If k=7a n d λi=9f o ra l l i=1,2,...,6, estimate, using the appropriate Gaussian approximation, the chance that at least three of x1,x2,...,x 6will be 15 or greater. 26.40 The variables Xi,i=1,2,...,n, are distributed as a multivariate Gaussian, with means µiand a covariance matrix V.I ft h e Xiare required to satisfy the linear constraint Pn i=1ciXi=0 ,w h e r et h e ciare constants (and not all equal to zero), show that the variable χ2 n=(x−µ)TV−1(x−µ) follows a chi-squared distribution of order n−1. 1060 26.17 HINTS AND ANSWERS 26.17 Hints and answers 26.1 (a) Yes, (b) no, (c) no, (d) no, (e) yes. 26.2 Reduce the equality to X∩(Y∪Z)=Y. 26.3 Show that if px/16 is the probability that the total will be xthen the corrsponding gain is [ px(x2+x)−16x]/16. (a) A loss of 2.5 euros; (b) a gain of27 64euros; (c) a gain of 2.5 euros, provided he takes your advice and guesses ‘5’ each time. 26.4 Let Bbe the union of events A1,A2,... ,A nand apply (26.9) with AasAn+1. Evaluate Pr( B∩An+1) by applying the assumed result to the set of nevents Ci= Ai∩An+1fori=1,2,...,n and noting that Ci∩Cj∩···∩Cm=Ai∩Aj∩···∩Am∩An+1. 26.5 P1=α(α+β−αβ)−1;P2=α(1−β)(α+β−αβ)−1;P3=α(α+β)−1. 26.6 Find simple expressions for the separate probabilities that A≥kandB≤1−k, and also for the two conditions at the same time. Then applying (26.11) andidentities typiﬁed by Pr( C)=P r ( CandD)+Pr( Cand¯D), show that p n=1−2(1− k)n+( 1−2k)n. (a) 0, (b) 1 −2−(n−1),( c )0 ,( d )2 k2. 26.7 If pris the probability that before the rth round both players are still in the tournament (and have not met each other), show that pr+1=1 42n+1−r−2 2n+1−r−1prand hence that pr= /1 2 /r−12n+1−r−1 2n−1. (a) The probability that they meet in the ﬁnal is pn=2−(n−1)(2n−1)−1. (b) The probability that they meet at some stage in the tournament is given by the sum Pn r=1pr(2n+1−r−1)−1=2−(n−1). 26.8 (b) Pr( TD>T C)=0 .5{1−[146/(36)2]}=0.4437; C’s expected return is equal to £2.10(1−0.4437)≈£1.17 for a £1.10 stake. 26.9 The relative probabilities are X:Y:Z= 50 : 36 : 8 (in units of 10−4);25 47. 26.10 (a) Show that the probability that an Ascii gives the same answer twice in succession to the same question is 5 /8 and that if he gives the same answer twice the probability that he is telling the truth is 9 /10. Conclude that the probability that the native questioned is an Ascii is 55 /95 and that the probability that the traveller is being correctly directed is 99 /190. As this is more than1 2, he should go left. (b) For the same answer given three times the corresponding fractions are 28 /64, 27/28 and 308 /628. The chance that the traveller is being told the truth has d r o p p e dt o2 9 7 /628, and, as this is less than one half, he should go ‘ right’ with a 331 /628 chance of success. This is a (very) slight improvement on his previous situation. 26.11 Take Ajas the event that a family consists of jboys and n−jgirls, and Bas the event that the boy has at least two sisters. Apply Bayes’ theorem. 26.12 If q=1−p, the probability is q3+3pq2+p2q, the separate terms corresponding to zero, one and a particular set of double breaks; (b) similarly, the probabilityisq 5+5q4p+( 1 0−2)p2q3+2p3q2. 26.13 (i) For aeven, the number of ways is 1 + 3 + 5 + ···+(a−3), and (ii) for aodd i ti s2+4+6+ ···+(a−3). Combine the results for a=2manda=2m+1 , with mrunning from 2 to N, to show that the total number of non-degenerate triangles is given by N(4N+1 ) ( N−1)/6. The number of possible selections of a set of three rods is (2 N+ 1)(2 N)(2N−1)/6. 26.14 The CPF for Yisy2nand the PDF is the derivative of this, namely 2 ny2n−1. This leads to an expected area equal to nπ/(n+ 1). The same PDF gives the distribution of the rejected shot and, for a given y, the remaining n−1 shots, all lying within yofO, have a CPF of ( z2/y2)n−1. Show, from the corresponding PDF, that the expected area is then ( n−1)πy2/nand that when this is averaged over ythe stated result is obtained. 1061 PROBABILITY 26.15 Show that k=e2and that the average duration of a call is 1 minute. Let pn be the probability that the call ends during the interval 0 .5(n−1)≤t<0.5n and cn=2 0 nbe the corresponding cost. Prove that p1=p2=1 4and that pn=1 2e2(e−1)e−nforn≥3. It follows that the average cost is E[C]=30 2+2 0e2(e−1) 2∞X n=3ne−n. The arithmetico-geome tric series has sum (3 e−1−2e−2)/(e−1)2and the total charge is 5( e+1 )/(e−1) = 10 .82 pence more than the 40 pence a uniform rate would cost. 26.16 Establish that N(x, y+1 )=[ ( y+1 )N(x, y)+xN(x−1,y+1 ) ] /(x+y+1 ) . 26.17 (a) The scores must be equal, at reach, after ﬁve attempts each. (b)Mcan only be even if team 2 gets too far ahead (or drops too far behind) to be caught (or catch up), with conditional probability p2(orq2). Conversely M can only be odd as a result of a ﬁnal action by team 1.(c) Pr( i:x, y)= yCxpx iqy−x i. (d) if the match is still alive at the tenth kick, team 2 is just as likely to lose it asto take it into sudden death. 26.18 a 2/12;a2/12−a2/(2π2n2). 26.19 Show that dY /dX =fand use g(y)=f(x)|dx/dy|. 26.20 (a) Note that gn−gn−1=−fnand that g0=1−f0. (b) Show that Φ/prime X(1) = Ψ X(1) and relate Φ/prime/primeX(1) to Ψ/primeX(1). (c) Ψ X(t)=α/(1−αt). 26.21 (a) Use result (26.84) to show that the PGF for SisQ/(1−Pq−Ppt). Then use equations (26.74) and (26.76). (b) The PGF for the score is 6 /(21−10t−5t2) and the average score is 10 /3. The variance is 145 /9 and the standard deviation is 4 .01. 26.22 Kn(t)=nlnp+nt+n P∞ r=1r−1(1−p)retr. This gives the ﬁrst four cumulants as n/p,n(1−p)/p2,n(1−p)(2−p)/p3andn(1−p)(6−6p+p2)/p4. 26.23 Mean = 4 /π.V a r i a n c e=2 −(16/π2). Probability that Xexceeds its mean =1−(2/π)sin−1(2/π)=0 .561. 26.24 Write x=5×106p. Show that ( x+1 2x2)e−xhas a maximum value of 0 .587 whatever the value of x, and hence of p. 26.25 Consider 0, 1 and ≥2 errors on a page separately. 26.26 Show that the expected return is/4 5 /nnX r=0nCr /3 8 /r /5 8 /n−r =11 8n /4 5 /n . This is maximal when n=( l n 5 /4)−1=4.48;n=4a n d n= 5 both give an expected return of $2.2528, i.e. less than the entry fee, but are the best that canbe advised. 26.27 Show that the maximum occurs at x=(r−1)/λand then use Stirling’s approxi- mation to ﬁnd the maximum value. 26.28 Show that the probability that the ‘trials’ end with the nth child, ( n≥4), is given by ( n−1C1pqn−2)p+(n−1C1qpn−2)q. The expectation value for nis then given by the sum P∞ n=4n(n−1)(p2qn−2+q2pn−2). By twice diﬀerentiating the result for the sum of a geometric series, prove that P∞ n=2n(n−1)rn−2=2/(1−r)3.U s et h i s result, after explicitly removing the ﬁrst two terms, to show that E[n]i sa sg i v e n . 26.29 Pr( kchicks hatching) = P∞ n=kPo(n, λ)B i n ( n, p). 26.30 Show that the variance of the distribution that has probabilities of 1 /20 for i=−5 andi= 5, and probabilities of 1 /10 for i=−4,−3,...,4i s1 7 /2. Conclude that 1062 26.17 HINTS AND ANSWERS 23 pence is only 1 .3×the standard deviation expected for the total bill and that a bigger discrepancy would occur about 20% of the time. 26.31 There is not much to choose between the schemes. In (a) the critical value of the standard variable is −2.5 and the average ﬁne would be 15.5 euros. For (b) the corresponding ﬁgures are −1.0 and 15.9 euros. Scheme (c) is governed by a geometric distribution with p=q=1 2, and leads to an expected ﬁne of P∞ n=1n(n−1)(1 2)n. The sum can be evaluated by diﬀerentiating the resultP∞ n=1pn=p/(1−p) with respect to p, and gives the expected ﬁne as 16 euros. 26.32 By making a Gaussian approximation to the binomial distribution, establish that Nmust be such that 25−75q−Np=0.841 p (75−N)pq. With p=0.8a n d q=0.2, this has solution N=9.1. 26.33 (a) [12!(0 .5)6(0.3)3(0.2)3]/(6!3!3!) = 0 .063. 26.34 Show that Pr( X=x)=c(a−x)(2x+2a+ 1) and use the fact that Pa−1 x=1Pr(X= x) = 1 to prove that c=6/[a(a−1)(8a+5)]. When evaluating Pr( Y=y)c o n s i d e r carefully the value of the upper limit in the summation over x. (a) Pr( Y=y)=3(2a−y)(2a+y+2 ) 2a(a−1)(8a+5 ), (b) Pr( Y=y)=3(2a−y−1)(2a+y+1 ) 2a(a−1)(8a+5 ). Express the expectation value as a summation over mfrom 1 to a−1, combining the terms involving y=2m−1a n d y=2m. 26.35 You will need to establish the normalisation constant for the distribution (36), the common mean value (3 /5), and the common standard deviation (3 /10). The marginal distributions are f(x)=3 x(6x2−8x+ 3) and the same function of y. The covariance has the value −3/50, yielding a correlation of −2/3. 26.36 E[XY]= PN n=0n3pn−2µ PN n=0n2pn+µ3. Setpn=1/(N+1 )f o ra l l nand use the results for series involving the natural numbers given in subsection 4.2.5 to show that Cov[ X,Y]=0 . 26.37 A=3/(24a4);µX=µY=5a/8;σ2 X=σ2 Y=7 3 a2/960; E[XY]=3 a2/8; Cov[X,Y]=−a2/64. 26.38 This is the multinomial distribution for nRVs in each of the intervals [ −c, z], [z+dz, c] and one RV in the interval [ z,z+dz]. The corresponding basic probabilities are [(c±z)/(2c)]nanddz/(2c). Use the fact that fnandfn+1are normalised to deduce the value of R z2fn(z)dz. The variance is c2/(2n+3 ) . 26.39 (b) With the continuity correction Pr( xi≥15) = 0 .0334. The probability that at least three are 15 or greater is 7 .5×10−4. 26.40 Perform successive transformations of variables y/prime=ST(x−µ)a n d zi=y/prime i/√λi, where the columns of Sare eigenvectors of Vand the λiare eigenvalues of V. Then bothχ 2 n= Pn i=1z2 iandziare independent Gaussian variables with mean zero and unit variance, which are required to satisfy the linear constraint Pn i=1c/prime izi=0f o r some constants c/prime i. Now require that f(z1,z2,...,z n)dz1dz2···dzn=h(χ2 n)dχ2 n, where dz1dz2···dznis the inﬁnitesimal volume enclosed by the intersection of then-dimensional spherical shell of radius χ2 nand thickness dχ2 nwith the ( n−1)- dimensional hyperplane Pn i=1c/prime izi=0 . 1063 27 Statistics In this chapter, we turn to the study of statistics, which is concerned with the analysis of experimental data. In a book of this nature we cannot hopeto do justice to such a large subject; indeed, many would argue that statisticsbelongs to the realm of experimental science rather than in a mathematicstextbook. Nevertheless, physical scientists and engineers are regularly called uponto perform a statistical analysis of their data and to present their results in astatistical context. Therefore, we will concentrate on this aspect of a much more extensive subject. † 27.1 Experiments, samples and populations We may regard the product of any experiment as a set of Nmeasurements of some quantity xor set of quantities x ,y,...,z . This set of measurements constitutes the data. Each measurement (or data item ) consists accordingly of a single number x i or a set of numbers ( xi,yi,...,,z i), where i=1,...,,N . For the moment, we will assume that each data item is a single number, although our discussion can beextended to the more general case. As a result of inaccuracies in the measurement process, or because of intrinsic variability in the quantity xbeing measured, one would expect the Nmeasured values x 1,x2,...,x Nto be diﬀerent each time the experiment is performed. We may therefore consider the xias a set of Nrandom variables. In the most general case, †There are, in fact, two separate schools of thought concerning statistics: the frequentist approach and the Bayesian approach. Indeed, which of these approaches is the more fundamental is still a matter of heated debate. Here we shall concentrate primarily on the more traditional frequentistapproach (despite the preference of some of the authors for the Bayesian viewpoint!). For a fuller discussion of the frequentist approach one could refer to, for example, Stuart & Ord, Kendall’s Advanced Theory of Statistics Vol. I (Edward Arnold) or Kenney & Keeping, Mathematics of Statistics (Van Nostrand). For a discussion of the Bayesian approach one might consult, for example, Sivia, Data Analysis: A Bayesian Tutorial (OUP). 1064 27.2 SAMPLE STATISTICS these random variables will be described by some N-dimensional joint probability density function P(x1,x2,...,x N).†In other words, an experiment consisting of N measurements is considered as a single random sample from the joint distribution (orpopulation )P(x), where xdenotes a point in the N-dimensional data space having coordinates ( x1,x2,...,x N). The situation is simpliﬁed considerably if the sample values xiareindependent . In this case, the N-dimensional joint distribution P(x) factorises into the product ofNone-dimensional distributions, P(x)=P(x1)P(x2)···P(xN). (27.1) In the general case, each of the one-dimensional distributions P(xi)m a yb e diﬀerent. A typical example of this occurs when Nindependent measurements are made of some quantity xbut the accuracy of the measuring procedure varies between measurements. It is often the case, however, that each sample value xiis drawn independently from the samepopulation. In this case, P(x) is of the form (27.1), but, in addition, P(xi) has the same form for each value of i. The measurements x1,x2,...,x N a r et h e ns a i dt of o r ma random sample of size Nfrom the one-dimensional population P(x). This is the most common situation met in practice and, unless stated otherwise, we will assume from now on that this is the case. 27.2 Sample statistics Suppose we have a set of Nmeasurements x1,x2,...,x N. Any function of these measurements (that contains no unknown parameters) is called a sample statistic , or often simply a statistic . Sample statistics provide a means of characterising the data. Although the resulting characteris ation is inevitably incomplete, it is useful to be able to describe a set of data in terms of a few pertinent numbers. We nowdiscuss the most commonly used sample statistics. 27.2.1 Averages The simplest number used to characterise a sample is the mean,w h i c hf o r N values x i,i=1,2,...,N , is deﬁned by ¯x=1 NNsummationdisplay i=1xi. (27.2) †In this chapter, we will adopt the common convention that P(x) denotes the particular probability density function that applies to its argument, x. This obviates the need to use a diﬀerent letter for the PDF of each new variable. For example, if XandYare random variables with diﬀerent PDFs, then properly one should denote these distributions by f(x)a n d g(y), say. In our shorthand notation, these PDFs are denoted by P(x)a n d P(y), where it is understood that the functional form of the PDF may be diﬀerent in each case. 1065 STATISTICS 188.7 204.7 193.2 169.0 168.1 189.8 166.3 200.0 Table 27.1 Experimental data giving eight measurements of the round trip time in milliseconds for a computer ‘packet’ to travel from Cambridge UK to Cambridge MA. In words, the sample mean is the sum of the sample values divided by the number of values in the sample.ITable 27.1 gives eight values for the round trip time in milliseconds for a computer ‘packet’ to travel from Cambridge UK to Cambridge MA. Find the sample mean. Using (27.2) the sample mean in milliseconds is given by ¯x=1 8(188.7 + 204 .7 + 193 .2 + 169 .0 + 168 .1 + 189 .8 + 166 .3 + 200 .0) =1479.8 8= 184 .975. Since the sample values in table 27.1 are quoted to an accuracy of one decimal place, it is usual to quote the mean to the same accuracy, i.e. as ¯x= 185 .0. J Strictly speaking the mean given by (27.2) is the arithmetic mean and this is by far the most common deﬁnition used for a mean. Other deﬁnitions of the meanare possible, though less common, and include (i) the geometric mean , ¯x g=parenleftBiggNproductdisplay i=1xiparenrightBigg1/N , (27.3) (ii) the harmonic mean , ¯xh=NsummationtextN i=11/xi, (27.4) (iii) the root mean square , ¯xrms=parenleftBiggsummationtextN i=1x2 i NparenrightBigg1/2 . (27.5) It should be noted that, ¯x,¯xhand¯xrmswould remain well deﬁned even if some sample values were negative, but the value of ¯xgcould then become complex. The geometric mean should not be used in such cases. 1066 27.2 SAMPLE STATISTICSICalculate ¯xg,¯xhand¯xrmsfor the sample given in table 27.1. The geometric mean is given by (27.3) to be ¯xg= (188 .7×204.7×···×200.0)1/8= 184 .4. The harmonic mean is given by (27.4) to be ¯xh=8 (1/188.7) + (1 /204.7) +···+( 1/200.0)= 183 .9. Finally, the root mean square is given by (27.5) to be ¯xrms= /1 8(188.72+ 204 .72+···+ 200 .02) /1/2= 185 .5. J Two other measures of the ‘average’ of a sample are its modeandmedian .T h e mode is simply the most commonly occurring value in the sample. A sample maypossess several modes, however, and it can thus be misleading in such cases touse the mode as a measure of the average of the sample. The median of a sampleis the halfway point when the sample values x i(i=1,2,...,N ) are arranged in ascending (or descending) order. Clearly, this depends on whether the size of the sample, N, is odd or even. If Nis odd then the median is simply equal to x(N+1)/2,w h e r e a si f Nis even the median of the sample is usually taken to be 1 2(xN/2+x(N/2)+1).IFind the mode and median of the sample given in table 27.1. From the table we see that each sample value occurs exactly once, and so any value may be called the mode of the sample. To ﬁnd the sample median, we ﬁrst arrange the sample values in ascending order and obtain 166.3, 168.1, 169.0, 188.7, 189.8, 193.2, 200.0, 204.7 . Since the number of sample values N= 8, which is even, the median of the sample is 1 2(x4+x5)=1 2(188.7 + 189 .8) = 189 .25. J 27.2.2 Variance and standard deviation The variance and standard deviation both give a measure of the spread of values in a sample about the sample mean ¯x.T h e sample variance is deﬁned by s2=1 NNsummationdisplay i=1(xi−¯x)2, (27.6) and the sample standard deviation is the positive square root of the sample variance, i.e. s=radicaltpradicalvertexradicalvertexradicalbt1 NNsummationdisplay i=1(xi−¯x)2. (27.7) 1067 STATISTICSIFind the sample variance and sample standard deviation of the data given in table 27.1. We have already found that the sample mean is 185.0 to one decimal place. However, when the mean is to be used in the subsequent calculation of the sample variance it isbetter to use the most accurate value available. In this case the exact value is 184.975, andso using (27.6), s 2=1 8 / (188.7−184.975)2+···+ (200 .0−184.975)2 / =1608.36 8= 201 .0, where once again we have quoted the result to one decimal place. The sample standard deviation is then given by s=√ 201.0=1 4 .2. As it happens, in this case the diﬀerence between the true mean and the rounded value is very small compared to the variation of the individual readings about the mean and using the rounded value makes negligiblediﬀerence; however, this would not be so if the diﬀerence were comparable to the samplestandard deviation.J Using the deﬁnition (27.7), it is clear that in order to calculate the standard deviation of a sample we must ﬁrst calculate the sample mean. This requirementcan be avoided, however, by using an alternative form for s 2. From (27.6), we see that s2=1 NNsummationdisplay i=1(xi−¯x)2 =1 NNsummationdisplay i=1x2 i−1 NNsummationdisplay i=12xi¯x+1 NNsummationdisplay i=1¯x2 =x2−2¯x2+¯x2=x2−¯x2 We may therefore write the sample variance s2as s2=x2−¯x2=1 NNsummationdisplay i=1x2 i−parenleftBigg 1 NNsummationdisplay i=1xiparenrightBigg2 , (27.8) from which the sample standard deviation is found by taking the positive square root. Thus, by evaluating the quantitiessummationtextN i=1xiandsummationtextN i=1x2 ifor our sample, we can calculate the sample mean and sample standard deviation at the same time.ICalculate PN i=1xiand PN i=1x2 ifor the data given in table 27.1 and hence ﬁnd the mean and standard deviation of the sample. From table 27.1, we obtain NX i=1xi= 188 .7 + 204 .7+···+ 200 .0 = 1479 .8, NX i=1x2 i= (188 .7)2+ (204 .7)2+···+ (200 .0)2= 275334 .36. 1068 27.2 SAMPLE STATISTICS Since N= 8, we ﬁnd as before (quoting the ﬁnal results to one decimal place) ¯x=1479.8 8= 185 .0,s = s 275334 .36 8− /1479.8 8 /2 =1 4.2. J 27.2.3 Moments and central moments By analogy with our discussion of probability distributions in section 26.5, the sample mean and variance may also be described respectively as the ﬁrst momentand second central moment of the sample. In general, for a sample x i,i= 1,2,...,N , we deﬁne the rth moment mrandrth central moment nras mr=1 NNsummationdisplay i=1xr i, (27.9) nr=1 NNsummationdisplay i=1(xi−m1)r. (27.10) Thus the sample mean ¯xand variance s2m a ya l s ob ew r i t t e na s m1andn2 respectively. As is common practice, we have introduced a notation in which a sample statistic is denoted by the Roman letter corresponding to whichever Greekletter is used to describe the corresponding population statistic. Thus, we use m r andnrto denote the moment and central moment of a sample, since in section 26.5, we denoted the rth moment and central moment of a population by µrand νrrespectively. This notation is particularly useful, since the rth central moment of a sample, mr, may be expressed in terms of the rth- and lower-order sample moments nrin a way exactly analogous to that derived in subsection 26.5.5 for the correspondingpopulation statistics. For example, as discussed in the previous section, the samplevariance is given by s 2=x2−¯x2but this may also be written as n2=m2−m2 1, which is to be compared with the corresponding relation ν2=µ2−µ2 1derived in subsection 26.5.3 for population statistics. This correspondence also holds for higher-order central moments of the sample. For example, n3=1 NNsummationdisplay i=1(xi−m1)3 =1 NNsummationdisplay i=1(x3 i−3m1x2 i+3m2 1xi−m3 1) =m3−3m1m2+3m2 1m1−m3 1 =m3−3m1m2+2m3 1, (27.11) which may be compared with equation (26.53) in the previous chapter. 1069 STATISTICS Mirroring our discussion of the normalised central moments γrof a population in subsection 26.5.5, we may also describe a sample in terms of the dimensionlessquantities g k=nk nk/2 2=nk sk; g3andg4are called the sample skewness and kurtosis. Likwise, it is common to deﬁne the excess kurtosis of a sample by g4−3. 27.2.4 Covariance and correlation So far we have assumed that each data item of the sample consists of a single number. Now let us suppose that each item of data consists of a pair of numbers,so that the sample is given by ( x i,yi)(i=1,2,...,N ). We may calculate the sample means, ¯xand¯y, and sample variances, s2 xand s2 y,o ft h e xiandyivalues individually but these statistics do not provide any measure of the relationship between the xiandyi. By analogy with our discussion in subsection 26.12.3 we measure any interdependence between the xiandyiin terms of the sample covariance , which is given by Vxy=1 NNsummationdisplay i=1(xi−¯x)(yi−¯y) =(x−¯x)(y−¯y) =xy−¯x¯y. (27.12) Writing out the last expression in full, we obtain the form most useful for calculations, which reads Vxy=1 NparenleftBiggNsummationdisplay i=1xiyiparenrightBigg −1 N2parenleftBiggNsummationdisplay i=1xiparenrightBiggparenleftBiggNsummationdisplay i=1yiparenrightBigg . We may also deﬁne the closely related sample correlation by rxy=Vxy sxsy, which can take values between −1 and +1. If the xiandyiare independent then Vxy=0= rxy, and from (27.12) we see that xy=¯x¯y. It should also be noted that the value of rxyis not altered by shifts in the origin or by changes in the scale of the xioryi. In other words, if x/prime=ax+bandy/prime=cy+d,w h e r e a, b,c,dare constants, then rx/primey/prime=rxy. Figure 27.1 shows scatter plots for several two-dimensional random samples xi,yiof size N= 1000, each with a diﬀerent value of rxy. 1070 27.2 SAMPLE STATISTICS rxy=0.0 rxy=0.1 rxy=0.5 rxy=−0.7 rxy=−0.9 rxy=0.99xy Figure 27.1 Scatter plots for two-dimensional data samples of size N= 1000, with various values of the correlation r. No scales are plotted, since the value ofris unaﬀected by shifts of origin or changes of scale in xandy.ITen UK citizens are selected at random and their heights and weights are found to be as follows (to the nearest cmorkgrespectively): Person ABCDEFGH IJ Height (cm) 194 168 177 180 171 190 151 169 175 182Weight (kg) 75 53 72 80 75 75 57 67 46 68 Calculate the sample correlation between the heights and weights. In order to ﬁnd the sample correlation, we begin by calculating the following sums (where xiare the heights and yiare the weights)X ixi= 1757 , X iyi= 668 ,X ix2 i= 310041 , X iy2 i= 45746 , X ixiyi= 118029 . T h es a m p l ec o n s i s t so f N= 10 pairs of numbers, so the means of the xiand of the yiare given by ¯x= 175 .7a n d ¯y=6 6 .8. Also, xy= 11802 .9. Similarly, the standard deviations of the xiandyiare calculated, using (27.8), as sx= s 310041 10− /1757 10 /2 =1 1.6, sy= s 45746 10− /668 10 /2 =1 0.6. 1071 STATISTICS Thus the sample correlation is given by rxy=xy−¯x¯y sxsy=11 802 .9−(175.7)(66 .8) (11.6)(10 .6)=0.54. Thus there is a moderate positive correlation between the heights and weights of the people measured. J It is straightforward to generalise the above discussion to data samples of arbitrary dimension, the only complication being one of notation. We chooseto denote the ith data item from an n-dimensional sample as ( x (1) i,x(2)i,...,x(n) i), where the bracketted superscript runs from 1 to nand labels the elements within a given data item whereas the subscript iruns from 1 to Nand labels the data items within the sample. In this n-dimensional case, we can deﬁne the sample covariance matrix whose elements are Vkl=x(k)x(l)−x(k)x(l) and the sample correlation matrix with elements rkl=Vkl sksl. Both these matrices are clearly symmetric but are notnecessarily positive deﬁnite. 27.3 Estimators and sampling distributions In general, the population P(x) from which a sample x1,x2,...,x Nis drawn isunknown .T h e central aim of statistics is to use the sample values xito infer certain properties of the unknown population P(x), such as its mean, variance and higher moments. To keep our discussion in general terms, let us denote the variousproperties of the population by a 1,a2,..., or collectively by a. Moreover, we make the dependence of the population on the values of these quantities explicit by writing the population as P(x|a). For the moment, we are assuming that the sample values xiare independent and drawn from the same (one-dimensional) population P(x|a), in which case P(x|a)=P(x1|a)P(x2|a)···P(xN|a). Suppose, we wish to estimate the value of one of the quantities a1,a2,...,w h i c h we will denote simply by a. Since the sample values xiare our only source of information, any estimate of amust be some function of the xi, i.e. some sample statistic. Such a statistic is called an estimator ofaand is usually denoted by ˆa(x), where xdenotes the sample elements x1,x2,...,x N. Since an estimator ˆais a function of the sample values of the random variables x1,x2,...,x N, it too must be a random variable. In other words, if a number of random samples, each of the same size N, are taken from the (one-dimensional) 1072 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS population P(x|a) then the value of the estimator ˆawill vary from one sample to the next and in general will not be equal to the true value a. This variation of the estimator is described by its sampling distribution P(ˆa|a). From section 26.14, this is given by P(ˆa|a)dˆa=P(x|a)dNx, where dNxis the inﬁnitesimal ‘volume’ in x-space lying between the ‘surfaces’ ˆa(x)=ˆaandˆa(x)=ˆa+dˆa. The form of the sampling distribution generally depends upon the estimator under consideration and upon the form of thepopulation from which the sample was drawn, including, as indicated, the truevalues of the quantities a. It is also usually dependent on the sample size N.IThe sample values x1,x2,...,x Nare drawn independently from a Gaussian distribution with mean µand variance σ. Suppose that we choose the sample mean ¯xas our estimator ˆµof the population mean. Find the sampling distributions of this estimator. The sample mean ¯xis given by ¯x=1 N(x1+x2+···+xN), where the xiare independent random variables distributed as xi∼N(µ, σ2). From our discussion of multiple Gaussian distributions on page 1030, we see immediately that ¯xwill also be Gaussian distributed as N(µ, σ2/N). In other words, the sampling distribution of ¯xis given by P(¯x|µ, σ)=1p 2πσ2/Nexp / −(¯x−µ)2 2σ2/N / . (27.13) Note that the variance of this distribution is σ2/N. J 27.3.1 Consistency, bias and eﬃciency of estimators For any particular quantity a, we may in fact deﬁne any number of diﬀerent estimators, each of which will have its own sampling distribution. The qualityof a given estimator ˆamay be assessed by investigating certain properties of its sampling distribution P(ˆa|a). In particular, an estimator ˆais usually judged on the three criteria of consistency ,biasandeﬃciency , each of which we now discuss. Consistency An estimator ˆaisconsistent if its value tends to the true value ain the large-sample limit, i.e. lim N→∞ˆa=a. Consistency is usually a minimum requirement for a useful estimator. An equiv- alent statement of consistency is that in the limit of large Nthe sampling 1073 STATISTICS distribution P(ˆa|a) of the estimator must satisfy lim N→∞P(ˆa|a)→δ(ˆa−a). Bias The expectation value of an estimator ˆais given by E[ˆa]=integraldisplay ˆaP(ˆa|a)dˆa=integraldisplay ˆa(x)P(x|a)dNx, (27.14) where the second integral extends over all possible values that can be taken by the sample elements x1,x2,...,x N. This expression gives the expected mean value ofˆafrom an inﬁnite number of samples, each of size N.T h ebiasof an estimator ˆais then deﬁned as b(a)=E[ˆa]−a. (27.15) We note that the bias bdoes not depend on the measured sample values x1,x2,...,x N. In general, though, it will depend on the sample size N, the func- tional form of the estimator ˆaand, as indicated, on the true properties aof the population, including the true value of aitself. If b=0t h e n ˆais called an unbiased estimator of a.IAn estimator ˆais biased in such a way that E[ˆa]=a+b(a),w h e r et h eb i a s b(a)is given by(b1−1)a+b2andb1andb2are known constants. Construct an unbiased estimator of a. Let us ﬁrst write E[ˆa] is the clearer form E[ˆa]=a+(b1−1)a+b2=b1a+b2. The task of constructing an unbiased estimator is now trivial, and an appropriate choice isˆa/prime=(ˆa−b2)/b1, which (as required) has the expectation value E[ˆa/prime]=E[ˆa]−b2 b1=a. J Eﬃciency The variance of an estimator is given by V[ˆa]=integraldisplay (ˆa−E[ˆa])2P(ˆa|a)dˆa=integraldisplay (ˆa(x)−E[ˆa])2P(x|a)dNx (27.16) and describes the spread of values ˆaabout E[ˆa] that would result from a large number of samples, each of size N. An estimator with a smaller variance is said to be more eﬃcient than one with a larger variance. As we show in the next section, for any given quantity aof the population there exists a theoretical lower 1074 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS limiton the variance of anyestimator ˆa. This result is known as Fisher’s inequality (or the Cram´er–Rao inequality )a n dr e a d s V[ˆa]≥parenleftbigg 1+∂b ∂aparenrightbigg2slashbigg Ebracketleftbigg −∂2lnP ∂a2bracketrightbigg , (27.17) where Pstands for the population P(x|a)a n d bis the bias of the estimator. Denoting the quantity on the RHS of (27.17) by Vmin,t h e eﬃciency eof an estimator is deﬁned as e=Vmin/V[ˆa]. An estimator for which e= 1 is called a minimum-variance oreﬃcient estimator. Otherwise, if e<1,ˆais called an ineﬃcient estimator. It should be noted that, in general, there is no unique ‘optimal’ estimator ˆafor a particular property a. To some extent, there is always a trade-oﬀ between bias and eﬃciency. One must often weigh the relative merits of an unbiased, ineﬃcientestimator against another that is more eﬃcient but slightly biased. Nevertheless, a common choice is the best unbiased estimator (BUE), which is simply the unbiased estimator ˆahaving the smallest variance V[ˆa]. Finally, we note that some qualities of estimators are related. For example, suppose ˆais an unbiased estimator, so that E[ˆa]=aandV[ˆa]→0a s N→ ∞. Using the Bienaym ´e–Chebyshev inequality discussed in subsection 26.5.3, it follows immediately that ˆais also a consistent estimator. Nevertheless, it does not follow that a consistent estimator is unbiased.IThe sample values x1,x2,...,x Nare drawn independently from a Gaussian distribution with mean µand variance σ. Show that the sample mean ¯xis a consistent, unbiased, minimum-variance estimator of µ. We found earlier that the sampling distribution of ¯xis given by P(¯x|µ, σ)=1p 2πσ2/Nexp / −(¯x−µ)2 2σ2/N / , from which we see immediately that E[¯x]=µandV[¯x]=σ2/N. Thus ¯xis an unbiased estimator of µ. Moreover, since it is also true that V[¯x]→0a sN→∞,¯xis a consistent estimator of µ. In order to determine whether ¯xis a minimum-variance estimator of µ,w em u s tu s e Fisher’s inequality (27.17). Since the sample values xiare independent and drawn from a Gaussian of mean µand standard deviation σ, we have lnP(x|µ, σ)=−1 2NX i=1 / ln(2πσ2)+(xi−µ)2 σ2 / , and, on diﬀerentiating twice with respect to µ, we ﬁnd ∂2lnP ∂µ2=−N σ2. This is independent of the xiand so its expectation value is also equal to −N/σ2.W i t h b 1075 STATISTICS set equal to zero in (27.17), Fisher’s inequality thus states that, for anyunbiased estimator ˆµof the population mean, V[ˆµ]≥σ2 N. Since V[¯x]=σ2/N, the sample mean ¯xis a minimum-variance estimator of µ. J 27.3.2 Fisher’s inequality As mentioned above, Fisher’s inequality provides a lower limit on the variance of anyestimator ˆaof the quantity a;i tr e a d s V[ˆa]≥parenleftbigg 1+∂b ∂aparenrightbigg2slashbigg Ebracketleftbigg −∂2lnP ∂a2bracketrightbigg , (27.18) where Pstands for the population P(x|a)a n d bis the bias of the estimator. We now present a proof of this inequality. Since the derivation is somewhatcomplicated, and many of the details are unimportant, this section can be omittedon a ﬁrst reading. Nevertheless, some aspects of the proof will be useful whenthe eﬃciency of maximum-likelihood estimators is discussed in section 27.5.IProve Fisher’s inequality (27.18). The normalisation of P(x|a)i sg i v e nb yZ P(x|a)dNx=1, (27.19) where dNx=dx1dx2···dxNand the integral extends over all the allowed values of the sample items xi. Diﬀerentiating (27.19) with respect to the parameter a,w eo b t a i nZ∂P ∂adNx= Z∂lnP ∂aPdNx=0. (27.20) We note that the second integral is simply the expectation value of ∂lnP/∂a,w h e r et h e average is taken over all possible samples xi,i=1,2,...,N .F u r t h e r ,b ye q u a t i n gt w o expressions for ∂E[ˆa]/∂a, obtained by diﬀerentiating (27.15) and (27.14) with respect to a we obtain, dropping the functional dependencies, a second relationship, 1+∂b ∂a= Z ˆa∂P ∂adNx= Z ˆa∂lnP ∂aPdNx. (27.21) Now, multiplying (27.20) by α(a), where α(a)i sanyfunction of a, and subtracting the result from (27.21), we obtainZ [ˆa−α(a)]∂lnP ∂aPdNx=1+∂b ∂a. At this point we must invoke the Schwarz inequality proved in subsection 8.1.3. The proof is trivially extended to multiple integrals and shows that for two real functions, g(x)a n d h(x),/Z g2(x)dNx //Z h2(x)dNx / ≥ /Z g(x)h(x)dNx /2 . (27.22) 1076 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS If we now let g=[ˆa−α(a)]√ Pandh=(∂lnP/∂a)√ P, we ﬁnd/Z [ˆa−α(a)]2PdNx / /"Z /∂lnP ∂a /2 PdNx /# ≥ / 1+∂b ∂a /2 . On the LHS, the factor in braces represents the expected spread of ˆa-values around the point α(a). The minimum value that this integral may take occurs when α(a)=E[ˆa]. Making this substitution, we recognise the integral as the variance V[ˆa], and so obtain the result V[ˆa]≥ / 1+∂b ∂a /2 /"Z /∂lnP ∂a /2 PdNx /#−1 . (27.23) We note that the factor in brackets is the expectation value of ( ∂lnP/∂a)2. Fisher’s inequality is, in fact, often quoted in the form (27.23). We may recover the form (27.18) by noting that on diﬀerentiating (27.20) with respect to awe obtainZ /∂2lnP ∂a2P+∂lnP ∂a∂P ∂a / dNx=0. Writing ∂P/∂a as (∂lnP/∂a)Pand rearranging we ﬁnd thatZ /∂lnP ∂a /2 PdNx=− Z∂2lnP ∂a2PdNx. Substituting this result in (27.23) gives V[ˆa]≥− / 1+∂b ∂a /2 /Z∂2lnP ∂a2PdNx /−1 . Since the factor in brackets is the expectation value of ∂2lnP/∂a2, we have recovered result (27.18). J 27.3.3 Standard errors on estimators For a given sample x1,x2,...,x N, we may calculate the value of an estimator ˆa(x) for the quantity a. It is also necessary, however, to give some measure of the statistical uncertainty in this estimate. One way of characterising this uncertainty is with the standard deviation of the sampling distribution P(ˆa|a), which is given simply by σˆa=(V[ˆa])1/2. (27.24) If the estimator ˆa(x) were calculated for a large number of samples, each of size N, then the standard deviation of the resulting ˆavalues would be given by (27.24). Consequently, σˆais called the standard error on our estimate. In general, however, the standard error σˆadepends on the true values of some or all of the quantities aand they may be unknown. When this occurs, one must substitute estimated values of any unknown quantities into the expression for σˆa in order to obtain an estimated standard error ˆσˆa. One then quotes the result as a=ˆa±ˆσˆa. 1077 STATISTICSITen independent sample values xi,i=1,2,...,10, are drawn at random from a Gaussian distribution with standard deviation σ=1. The sample values are as follows (to two decimal places): 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Estimate the population mean µ, quoting the standard error on your result. We have shown in the ﬁnal worked example of subsection 27.3.1 that, in this case, ¯xis a consistent, unbiased, minimum-variance estimator of µand has variance V[¯x]=σ2/N. Thus, our estimate of the population mean with its associated standard error is ˆµ=¯x±σ√ N=1.11±0.32. If the true value of σhad not been known, we would have needed to use an estimated value ˆσin the expression for the standard error. Useful basic estimators of σare discussed in subsection 27.4.2. J It should be noted that the above approach is most meaningful for unbiased estimators. In this case, E[ˆa]=aand so σˆadescribes the spread of ˆa-values about the true value a. For a biased estimator, however, the spread about the true value ais given by the root mean square error /epsilon1ˆa, which is deﬁned by /epsilon12 ˆa=E[(ˆa−a)2] =E[(ˆa−E[ˆa])2]+(E[ˆa]−a)2 =V[ˆa]+b(a)2. We see that /epsilon12 ˆais the sum of the variance of ˆaand the square of the bias and so can be interpreted as the sum of squares of statistical and systematic errors. For a biased estimator, it is often more appropriate to quote the result as a=ˆa±/epsilon1ˆa. As above, it may be necessary to use estimated values ˆain the expression for the root mean square error and thus to quote only an estimate ˆ/epsilon1ˆaoftheerror . 27.3.4 Conﬁdence limits on estimators An alternative (and often equivalent) way of quoting a statistical error is with a conﬁdence interval . Let us assume that, other than the quantity of interest a,t h e quantities ahave known ﬁxed values. Thus we denote the sampling distribution ofˆabyP(ˆa|a). For any particular value of a, one can determine the two values ˆaα(a)a n d ˆaβ(a) such that Pr(ˆa<ˆaα(a)) =integraldisplayˆaα(a) −∞P(ˆa|a)dˆa=α, (27.25) Pr(ˆa>ˆaβ(a)) =integraldisplay∞ ˆaβ(a)P(ˆa|a)dˆa=β. (27.26) 1078 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS ˆaP(ˆa|a) ˆaα(a) ˆaβ(a)α β Figure 27.2 The sampling distribution P(ˆa|a)o fs o m ee s t i m a t o r ˆafor a given value of a. The shaded regions indicate the two probabilities Pr( ˆa<ˆaα(a)) =α and Pr( ˆa>ˆaβ(a)) =β. This is illustrated in ﬁgure 27.2. Thus, for any particular value of a, the probability that the estimator ˆalies within the limits ˆaα(a)a n d ˆaβ(a)i sg i v e nb y Pr(ˆaα(a)<ˆa<ˆaβ(a)) =integraldisplayˆaβ(a) ˆaα(a)P(ˆa|a)dˆa=1−α−β. Now, let us suppose that from our sample x1,x2,...,x N, we actually obtain the value ˆaobsfor our estimator. If ˆais a good estimator of athen we would expect ˆaα(a)a n d ˆaβ(a) to be monotonically increasing functions of a(i.e.ˆaαandˆaβboth change in the samesense as awhen the latter is varied). Assuming this to be the case, we can uniquely deﬁne the two numbers a−anda+by the relationships ˆaα(a+)=ˆaobs and ˆaβ(a−)=ˆaobs. From (27.25) and (27.26) it follows that Pr(a+<a)=α and Pr( a−>a)=β, which when taken together imply Pr(a−<a<a +)=1−α−β. (27.27) Thus, from our estimate ˆaobs, we have determined two values a−anda+such that this interval contains the true value of awith probability 1 −α−β. It should be emphasised that a−anda+are random variables. If a large number of samples, each of size N, were analysed then the interval [ a−,a+] would contain the true value aon a fraction 1 −α−βof occasions. The interval [ a−,a+] is called a conﬁdence interval onaat the conﬁdence level1−α−β. The values a−and a+themselves are called respectively the lower conﬁdence limit and the upper conﬁdence limit at this conﬁdence level. In practice, the conﬁdence level is often quoted as a percentage. A convenient way 1079 STATISTICS ˆaP(ˆa|a−) P(ˆa|a+) ˆaobsα β Figure 27.3 An illustration of how the observed value of the estimator, ˆaobs, and the given values αandβdetermine the two conﬁdence limits a−anda+, which are such that ˆaα(a+)=ˆaobs=ˆaβ(a−). of presenting our results isintegraldisplayˆaobs −∞P(ˆa|a+)dˆa=α, (27.28) integraldisplay∞ ˆaobsP(ˆa|a−)dˆa=β. (27.29) The conﬁdence limits may then be found by solving these equations for a−and a+either analytically or numerically. Occasionally one might not combine the results (27.28) and (27.29) but use either one or the other to provide a one-sided conﬁdence interval on a. Whenever the results are combined to provide a two-sided conﬁdence interval, the interval isnotspeciﬁed uniquely by the conﬁdence level 1 −α−β. In other words, there are generally an inﬁnite number of intervals [ a−,a+] for which (27.27) holds. To specify a unique interval, one often chooses α=β, resulting in the central conﬁdence interval ona. All cases can be covered by calculating the quantities c=ˆa−a−andd=a+−ˆaand quoting the result of an estimate as a=ˆa+d −c. We have so far assumed that the quantities aother than the quantity of interest aare known in advance. If this is not the case then, in principle, the construction of conﬁdence limits is considerably more complicated. This is discussed brieﬂy insubsection 27.3.6. 27.3.5 Conﬁdence limits for a Gaussian sampling distribution An important special case occurs when the sampling distribution is Gaussian; if the mean is aand the standard deviation is σ ˆathen P(ˆa|a, σˆa)=1radicalBig 2πσ2 ˆaexpbracketleftbigg −(ˆa−a)2 2σ2 ˆabracketrightbigg . (27.30) 1080 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS For almost any (consistent) estimator ˆa, the sampling distribution will tend to this form in the large-sample limit N→∞, as a consequence of the central limit theorem. For a sampling distribution of the form (27.30), the above procedurefor determining conﬁdence intervals becomes straightforward. Suppose, from our sample, we obtain the value ˆa obsfor our estimator. In this case, equations (27.28) and (27.29) become Φparenleftbiggˆaobs−a+ σˆaparenrightbigg =α, 1−Φparenleftbiggˆaobs−a− σˆaparenrightbigg =β, where Φ( z) is the cumulative probability function for the standard Gaussian distri- bution, discussed in subsection 26.9.1. Solving these equations for a−anda+gives a−=ˆaobs−σˆaΦ−1(1−β), (27.31) a+=ˆaobs+σˆaΦ−1(1−α); (27.32) we have used the fact that Φ−1(α)=−Φ−1(1−α) to make the equations symmetric. The value of the inverse function Φ−1(z) can be read oﬀ directly from table 26.3, given in subsection 26.9.1. For the normally-used central conﬁdence interval onehasα=β. In this case, we see that quoting a result using the standard error, as a=ˆa±σ ˆa, (27.33) is equivalent to taking Φ−1(1−α) = 1. From table 26.3, we ﬁnd α=1−0.8413 = 0.1587, and so this corresponds to a conﬁdence level of 1 −2(0.1587)≈0.683. Thus, the standard error limits give the 68.3% central conﬁdence interval.ITen independent sample values xi(i=1,2,...,10)are drawn at random from a Gaussian distribution with standard deviation σ=1. The sample values are as follows (to two decimal places): 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Find the 90% central conﬁdence interval on the population mean µ. Our estimator ˆµis the sample mean ¯x. As shown towards the end of section 27.3, the sampling distribution of ¯xis Gaussian with mean E[¯x] and variance V[¯x]=σ2/N.S i n c e σ= 1 in this case, the standard error is given by σˆx=σ/√ N=0.32. Moreover, in subsection 27.3.3, we found the mean of the above sample to be ¯x=1.11. For the 90% central conﬁdence interval, we require α=β=0.05. From table 26.3, we ﬁnd Φ−1(1−α)=Φ−1(0.95) = 1 .65, and using (27.31) and (27.32) we obtain a−=¯x−1.65σ¯x=1.11−(1.65)(0 .32) = 0 .58, a+=¯x+1.65σ¯x=1.11 + (1 .65)(0 .32) = 1 .64. Thus, the 90% central conﬁdence interval on µis [0.58,1.64]. For comparison, the true value used to create the sample was µ=1 . J 1081 STATISTICS In the case where the standard error σˆain (27.33) is not known in advance, one must use a value ˆσˆaestimated from the sample. In principle, this complicates somewhat the construction of conﬁdence intervals, since properly one shouldconsider the two-dimensional joint sampling distribution P(ˆa,ˆσ ˆa|a). Nevertheless, in practice, provided ˆσˆais a fairly good estimate of σˆathe above procedure may be applied with reasonable accuracy. In the special case where the sample valuesx iare drawn from a Gaussian distribution with unknown µandσ,i ti si nf a c t possible to obtain exact conﬁdence intervals on the mean µ, for a sample of any sizeN, using Student’s t-distribution. This is discussed in section 27.7.5. 27.3.6 Estimation of several quantities simultaneously Suppose one uses a sample x1,x2,...,x Nto calculate the values of several es- timators ˆa1,ˆa2,...,ˆaM(collectively denoted by ˆa)o ft h eq u a n t i t i e s a1,a2,...,a M (collectively denoted by a) that describe the population from which the sample was drawn. The joint sampling distribution of these estimators is an M-dimensional PDF P(ˆa|a)g i v e nb y P(ˆa|a)dMˆa=P(x|a)dNx.ISample values x1,x2,...,x Nare drawn independently from a Gaussian distribution with mean µand standard deviation σ. Suppose we choose the sample mean ¯xand sample stan- dard deviation srespectively as estimators ˆµandˆσ. Find the joint sampling distribution of these estimators. Since each data value xiin the sample is assumed to be independent of the others, the joint probability distribution of sample values is given by P(x|µ, σ)=( 2 πσ2)−N/2exp / − P i(xi−µ)2 2σ2 / . We may rewrite the sum in the exponent as follows:X i(xi−µ)2= X i(xi−¯x+¯x−µ)2 = X i(xi−¯x)2+2 (¯x−µ) X i(xi−¯x)+ X i(¯x−µ)2 =Ns2+N(¯x−µ)2, where in the last line we have used the fact that P i(xi−¯x) = 0. Hence, for given values ofµandσ, the sampling distribution is in fact a function only of the sample mean ¯xand the standard deviation s. Thus the sampling distribution of ¯xandsmust satisfy P(¯x, s|µ, σ)d¯xd s=( 2πσ2)−N/2exp / −N[(¯x−µ)2+s2] 2σ2 / dV, (27.34) where dV=dx1dx2···dxNis an element of volume in the sample space which yields simultaneously values of ¯xandsthat lie within the region bounded by [ ¯x,¯x+d¯x]a n d [s, s+ds]. Thus our only remaining task is to express dVin terms of ¯xandsand their diﬀerentials. 1082 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS LetSbe the point in sample space representing the sample ( x1,x2,...,x N). For given values of ¯xands, we require the sample values to satisfy both the conditionX ixi=N¯x, which deﬁnes an ( N−1)-dimensional hyperplane in the sample space, and the conditionX i(xi−¯x)2=Ns2, which deﬁnes an ( N−1)-dimensional hypersphere. Thus Sis constrained to lie in the intersection of these two hypersurfaces, which is itself an ( N−2)-dimensional hypersphere. Now, the volume of an ( N−2)-dimensional hypersphere is proportional to sN−1. It follows from this that the volume dVbetween two concentric ( N−2)-dimensional hyperspheres of radius√ Nsand√ N(s+ds)a n dt w o( N−1)-dimensional hyperplanes corresponding to¯xand¯x+d¯xis dV=AsN−2ds d¯x, where Ais some constant. Thus, substituting this expression for dVinto (27.34), we ﬁnd P(¯x, s|µ, σ)=C1exp / −N(¯x−µ)2 2σ2 / C2sN−2exp / −Ns2 2σ2 / =P(¯x|µ, σ)P(s|σ), (27.35) where C1andC2are constants. We have written P(¯x, s|µ, σ) in this form to illustrate that it separates naturally into two parts, one depending only on ¯xand the other only on s. Thus, ¯xandsareindependent variables. Separate normalisations of the two factors in (27.35) require C1= /N 2πσ2 /1/2 and C2=2 /N 2σ2 /(N−1)/21 Γ /;1 2(N−1) /, where the calculation of C2requires the use of the gamma function discussed in the Appendix. J Themarginal sampling distribution of any one of the estimators ˆaiis given simply by P(ˆai|a)=integraldisplay ···integraldisplay P(ˆa|a)dˆa1···dˆai−1dˆai+1···dˆaM, and the expectation value E[ˆai] and variance V[ˆai]o fˆaiare again given by (27.14) and (27.16) respectively. By analogy with the one-dimensional case, the standarderror σ ˆaion the estimator ˆaiis given by the positive square root of V[ˆai]. With several estimators, however, it is usual to quote their full covariance matrix. This M×Mmatrix has elements Vij=C o v [ ˆai,ˆaj]=integraldisplay (ˆai−E[ˆai])(ˆaj−E[ˆaj])P(ˆa|a)dMˆa =integraldisplay (ˆai−E[ˆai])(ˆaj−E[ˆaj])P(x|a)dNx. Fisher’s inequality can be generalised to the multi-dimensional case. Adapting the proof given in subsection 27.3.2, one may show that, in the case where the 1083 STATISTICS estimators are eﬃcient and have zero bias, the elements of the inverse of the covariance matrix are given by (V−1)ij=Ebracketleftbigg −∂2lnP ∂ai∂ajbracketrightbigg , (27.36) where Pdenotes the population P(x|a) from which the sample is drawn. The quantity on the RHS of (27.36) is the element Fijof the so-called Fisher matrix Fof the estimators.ICalculate the covariance matrix of the estimators ¯xandsin the previous example. As shown in (27.35), the joint sampling distribution P(¯x, s|µ, σ) factorises, and so the estimators ¯xandsare independent. Thus, we conclude immediately that Cov[¯x, s]=0 . Since we have already shown in the worked example at the end of subsection (27.3.1) that V[¯x]=σ2/N, it only remains to calculate V[s]. From (27.35), we ﬁnd E[sr]=C2 Z∞ 0sN−2+rexp / −Ns2 2σ2 / ds= /2 N /r/2Γ /;1 2(N−1+r) / Γ /;1 2(N−1) /σr, where we have evaluated the integral using the deﬁnition of the gamma function given in the Appendix. Thus, the expectation value of the sample standard deviation is E[s]= /2 N /1/2Γ /;1 2N / Γ /;1 2(N−1) /σ, (27.37) and its variance is given by V[s]=E[s2]−(E[s])2=σ2 N /8/</:N−1−2 /" Γ /;1 2N / Γ /;1 2(N−1) / /#2 /9/=/;; We note, in passing, that (27.37) shows that sis abiased estimator of σ. J The idea of a conﬁdence interval can also be extended to the case where several quantities are estimated simultaneously but then the practical construction of an interval is considerably more complicated. The general approach is to constructanM-dimensional conﬁdence region Rina-space. By analogy with the one- dimensional case, for a given conﬁdence level of (say) 1 −α, one ﬁrst constructs a region ˆRinˆa-space, such that integraldisplayintegraldisplay ˆRP(ˆa|a)dMˆa=1−α. A common choice for such a region is that bounded by the ‘surface’ P(ˆa|a)= constant. By considering all possible values aand the values of ˆalying within the region ˆR, one can construct a 2 M-dimensional region in the combined space (ˆa,a). Suppose now that, from our sample x, the values of the estimators are ˆai,obs,i=1,2,...,M . The intersection of the M‘hyperplanes’ ˆai=ˆai,obswith the 2 M-dimensional region will determine an M-dimensional region which, when 1084 27.3 ESTIMATORS AND SAMPLING DISTRIBUTIONS a1a2 ˆa1ˆa2 atrue atrue ˆaobs ˆaobs(a) (b) Figure 27.4 (a) The ellipse Q(ˆa,a)=cinˆa-space. (b) The ellipse Q(a,ˆaobs)=c ina-space that corresponds to a conﬁdence region Rat the level 1 −α,w h e n csatisﬁes (27.39). projected onto a-space, will determine a conﬁdence limit Rat the conﬁdence level 1−α. It is usually the case that this conﬁdence region has to be evaluated numerically. The above procedure is clearly rather complicated in general and a simpler approximate method that uses the likelihood function is discussed in subsec-tion 27.5.5. As a consequence of the central limit theorem, however, in thelarge-sample limit, N→∞, the joint sampling distribution P(ˆa|a) will tend, in general, towards the multivariate Gaussian P(ˆa|a)=1 (2π)M/2|V|1/2expbracketleftbig −1 2Q(ˆa,a)bracketrightbig , (27.38) where Vis the covariance matrix of the estimators and the quadratic form Qis given by Q(ˆa,a)=(ˆa−a)TV−1(ˆa−a). Moreover, in the limit of large N, the inverse covariance matrix tends to the Fisher matrix Fgiven in (27.36), i.e. V−1→F. For the Gaussian sampling distribution (27.38), the process of obtaining conﬁ- dence intervals is greatly simpliﬁed. The surfaces of constant P(ˆa|a) correspond to surfaces of constant Q(ˆa,a), which have the shape of M-dimensional ellipsoids inˆa-space, centred on the true values a. In particular, let us suppose that the ellipsoid Q(ˆa,a)=c(where cis some constant) contains a fraction 1 −αsay of the total probability. Now suppose that, from our sample x, we obtain the values ˆaobsfor our estimators. Because of the obvious symmetry of the quadratic form Qwith respect to aandˆa, it is clear that the ellipsoid Q(a,ˆaobs)=cina-space that is centred on ˆaobsshould contain the true values awith probability 1 −α. Thus Q(a,ˆaobs)=cdeﬁnes our required conﬁdence region Rat this conﬁdence level. This is illustrated in ﬁgure 27.4 for the two-dimensional case. 1085 STATISTICS It remains only to determine the constant ccorresponding to the conﬁdence level 1−α. As discussed in subsection 26.15.2, the quantity Q(ˆa,a) is distributed as a χ2variable of order M. Thus, the conﬁdence region corresponding to the conﬁdence level 1 −αis given by Q(a,ˆaobs)=c, where the constant csatisﬁes integraldisplayc 0P(χ2 M)d(χ2 M)=1−α, (27.39) andP(χ2 M) is the chi-squared PDF of order M, discussed in subsection 26.9.4. This integral may be evaluated numerically to determine the constant c. Alternatively, some reference books tabulate the values of ccorresponding to given conﬁdence levels and various values of M. 27.4 Some basic estimators In many cases, one does not know the functional form of the population from which a sample is drawn. Nevertheless, in a case where the sample valuesx 1,x2,...,x Nare each drawn independently from a one-dimensional population P(x), it is possible to construct some basic estimators for the moments and central moments of P(x). In this section, we investigate the estimating properties of the common sample statistics presented in section 27.2. In fact, expectation values and variances of these sample statistics can be calculated without prior knowledge of the functional form of the population; they depend only on the sample size N and certain moments and central moments of P(x). 27.4.1 Population mean µ Let us suppose that the parent population P(x) has mean µand variance σ2.A n obvious estimator ˆµof the population mean is the sample mean ¯x.P r o v i d e d µ andσ2are both ﬁnite, we may apply the central limit theorem directly to obtain exact expressions, valid for samples of any size N, for the expectation value and variance of ¯x. From parts (i) and (ii) of the central limit theorem, discussed in section 26.10, we immediately obtain E[¯x]=µ, V [¯x]=σ2 N. (27.40) Thus we see that ¯xis an unbiased estimator of µ. Moreover, we note that the standard error in ¯xisσ/√ N, and so the sampling distribution of ¯xbecomes more tightly centred around µas the sample size Nincreases. Indeed, since V[¯x]→0 asN→∞,¯xis also a consistent estimator of µ. In the limit of large N, we may in fact obtain an approximate form for the full sampling distribution of ¯x. Part (iii) of the central limit theorem (see section 26.10) tells us immediately that, for large N, the sampling distribution of ¯xis 1086 27.4 SOME BASIC ESTIMATORS given approximately by the Gaussian form P(¯x|µ, σ)≈1radicalbig 2πσ2/Nexpbracketleftbigg −(¯x−µ)2 2σ2/Nbracketrightbigg . Note that this does notdepend on the form of the original parent population. If, however, the parent population is in fact Gaussian then this result is exact for samples of anysizeN(as is immediately apparent from our discussion of multiple Gaussian distributions in subsection 26.9.1). 27.4.2 Population variance σ2 An estimator for the population variance σ2is not so straightforward to deﬁne as one for the mean. Complications arise because, in many cases, the true mean of the population µis not known. Nevertheless, let us begin by considering the case where in fact µis known. In this event, a useful estimator is hatwideσ2=1 NNsummationdisplay i=1(xi−µ)2=parenleftBigg 1 NNsummationdisplay i=1x2 iparenrightBigg −µ2. (27.41)IShow that bσ2is an unbiased and consistent estimator of the population variance σ2. The expectation value of bσ2is given by E[ bσ2]=1 NE /"NX i=1x2 i /# −µ2=E[x2 i]−µ2=µ2−µ2=σ2, from which we see that the estimator is unbiased. The variance of the estimator is V[ bσ2]=1 N2V /"NX i=1x2 i /# +V[µ2]=1 NV[x2 i]=1 N(µ4−µ2 2), in which we have used that fact that V[µ2]=0a n d V[x2 i]=E[x4 i]−(E[x2 i])2=µ4−µ2 2, where µris the rth population moment. Since bσ2is unbiased and V[ bσ2]→0a s N→∞, showing that it is also a consistent estimator of σ2, the result is established. J If the true mean of the population is unknown, however, a natural alternative is to replace µby¯xin (27.41), so that our estimator is simply the sample variance s2given by s2=1 NNsummationdisplay i=1x2 i−parenleftBigg 1 NNsummationdisplay i=1xiparenrightBigg2 . In order to determine the properties of this estimator, we must calculate E[s2] andV[s2]. This task is straightforward but lengthy. However, for the investigation of the properties of a central moment of the sample, there exists a useful trick that simpliﬁes the calculation. We can assume, with no loss of generality, that 1087 STATISTICS the mean µ1of the population from which the sample is drawn is equal to zero. With this assumption, the population central moments, νr,a r ei d e n t i c a lt ot h e corresponding moments µr, and we may perform our calculation in terms of the latter. At the end, however, we replace µrbyνrin the ﬁnal result and so obtain a general expression that is valid even in cases where µ1/negationslash=0 .ICalculate E[s2]andV[s2]for a sample of size N. The expectation value of the sample variance s2for a sample of size Nis given by E[s2]=1 NE /"X ix2 i /# −1 N2E /2/4 / X ixi /!2 /3/5 =1 NNE[x2 i]−1 N2E /2/6/4 X ix2 i+ X i,j j/negationslash=ixixj /3/7/5. (27.42) The number of terms in the double summation in (27.42) is N(N−1), so we ﬁnd E[s2]=E[x2 i]−1 N2(NE[x2 i]+N(N−1)E[xixj]). Now, since the sample elements xiandxjare independent, E[xixj]=E[xi]E[xj]=0 , assuming the mean µ1of the parent population to be zero. Denoting the rth moment of the population by µr, we thus obtain E[s2]=µ2−µ2 N=N−1 Nµ2=N−1 Nσ2, (27.43) where in the last line we have used the fact that the population mean is zero, and so µ2=ν2=σ2. However, the ﬁnal result is also valid in the case where µ1/negationslash=0 . Using the above method, we can also ﬁnd the variance of s2, although the algebra is rather heavy going. The variance of s2is given by V[s2]=E[s4]−(E[s2])2, (27.44) where E[s2] is given by (27.43). We therefore need only consider how to calculate E[s4], where s4is given by s4= /"P ix2 i N− / P ixi N /2 /#2 =( P ix2 i)2 N2−2( P ix2 i)( P ixi)2 N3+( P ixi)4 N4. (27.45) We will consider in turn each of the three terms on the RHS. In the ﬁrst term, the sum ( P ix2 i)2can be written as/ X ix2 i /!2 = X ix4i+ X i,j j/negationslash=ix2 ix2j, where the ﬁrst sum contains Nterms and the second contains N(N−1) terms. Since the sample elements xiandxjare assumed independent, we have E[x2 ix2j]=E[x2 i]E[x2 j]=µ2 2, 1088 27.4 SOME BASIC ESTIMATORS and so E /2/4 / X ix2 i /!2 /3/5=Nµ4+N(N−1)µ2 2. Turning to the second term on the RHS of (27.45),/ X ix2 i /!/ X ixi /!2 = X ix4 i+ X i,j j/negationslash=ix3 ixj+ X i,j j/negationslash=ix2 ix2j+ X i,j,k k/negationslash=j/negationslash=ix2 ixjxk. Since the mean of the population has been assumed to equal zero, the expectation values of the second and fourth sums on the RHS vanish. The ﬁrst and third sums contain N andN(N−1) terms respectively, and so E /2/4 / X ix2 i /!/ X ixi /!2 /3/5=Nµ4+N(N−1)µ2 2. Finally, we consider the third term on the RHS of (27.45), and write/ X ixi /!4 = X ix4 i+ X i,j j/negationslash=ix3 ixj+ X i,j j/negationslash=ix2 ix2j+ X i,j,k k/negationslash=j/negationslash=ix2 ixjxk+ X i,j,k,l l/negationslash=k/negationslash=j/negationslash=ixixjxkxl. The expectation values of the second, fourth and ﬁfth sums are zero, and the ﬁrst and third sums contain Nand 3 N(N−1) terms respectively (for the third sum, there are N(N−1)/2 ways of choosing iandj, and the multinomial coeﬃcient of x2 ix2jis 4!/(2!2!) = 6). Thus E /2/4 / X ixi /!4 /3/5=Nµ4+3N(N−1)µ2 2. Collecting together terms, we therefore obtain E[s4]=(N−1)2 N3µ4+(N−1)(N2−2N+3 ) N3µ2 2, (27.46) which, together with the result (27.43), may be substituted into (27.44) to obtain ﬁnally V[s2]=(N−1)2 N3µ4−(N−1)(N−3) N3µ2 2 =N−1 N3[(N−1)ν4−(N−3)ν2 2], (27.47) where in the last line we have used again the fact that, since the population mean is zero, µr=νr. However, result (27.47) holds even when the population mean is not zero. J From (27.43), we see that s2is abiased estimator of σ2, although the bias becomes negligible for large N. However, it immediately follows that an unbiased estimator of σ2is given simply by hatwideσ2=N N−1s2, (27.48) where the multiplicative factor N/(N−1) is often called Bessel’s correction . Thus 1089 STATISTICS in terms of the sample values xi,i=1,2,...,N , an unbiased estimator of the population variance σ2is given by hatwideσ2=1 N−1Nsummationdisplay i=1(xi−¯x)2. (27.49) Using (27.47), we ﬁnd that the variance of the estimatorhatwideσ2is V[hatwideσ2]=parenleftbiggN N−1parenrightbigg2 V[s2]=1 Nparenleftbigg ν4−N−3 N−1ν2 2parenrightbigg , where νris the rth central moment of the parent population. We note that, since E[hatwideσ2]=σ2andV[hatwideσ2]→0a s N→∞, the statistichatwideσ2is also a consistent estimator of the population variance. 27.4.3 Population standard deviation σ The standard deviation σof a population is deﬁned as the positive square root of the population variance σ2(as, indeed, our notation suggests). Thus, it is common practice to take the positive square root of the variance estimator as our estimatorforσ.T h u s ,w et a k e ˆσ=parenleftBighatwideσ 2parenrightBig1/2 , (27.50) wherehatwideσ2is given by either (27.41) or (27.48), depending on whether the population mean µis known or unknown. Because of the square root in the deﬁnition of ˆσ, it is not possible in either case to obtain an exact expression for E[ˆσ]a n d V[ˆσ]. Indeed, although in each case the estimator is the positive square root of an unbiased estimator of σ2,i ti snotitself an unbiased estimator of σ. However, the bias does becomes negligible for large N.IObtain approximate expressions for E[ˆσ]andV[ˆσ]for a sample of size Nin the case where the population mean µis unknown. As the population mean is unknown, from (27.50) and (27.48) our estimator is given by ˆσ= /N N−1 /1/2 s, where sis the sample standard deviation. The expectation value of this estimator is given by E[ˆσ]= /N N−1 /1/2 E[(s2)1/2]≈ /N N−1 /1/2 (E[s2])1/2=σ. An approximate expression for the variance of ˆσmay be found using (27.47) and is given 1090 27.4 SOME BASIC ESTIMATORS by V[ˆσ]=N N−1V[(s2)1/2]≈N N−1 /d d(s2)(s2)1/2 /2 s2=E[s2]V[s2] ≈N N−1 /1 4s2 / s2=E[s2]V[s2]. Using the expressions (27.43) and (27.47) for E[s2]a n d V[s2] respectively, we obtain V[ˆσ]≈1 4Nν2 / ν4−N−3 N−1ν2 2 / . J 27.4.4 Population moments µr We may straightforwardly generalise our discussion of estimation of the popu- lation mean µ(=µ1) in section 27.4.1 to the estimation of the rth population moment µr. An obvious choice of estimator is the rth sample moment mr.T h e expectation value of mris given by E[mr]=1 NNsummationdisplay i=1E[xr i]=Nµr N=µr, and so it is an unbiased estimator of µr. The variance of mrmay be found in a similar manner, although the calculation is a little more complicated. We ﬁnd that V[mr]=E[(mr−µr)2] =1 N2E parenleftBiggsummationdisplay ixr i−NµrparenrightBigg2  =1 N2E summationdisplay ix2r i+summationdisplay isummationdisplay j/negationslash=ixr ixrj−2Nµrsummationdisplay ixr i+N2µ2 r  =1 Nµ2r−µ2 r+1 N2summationdisplay isummationdisplay j/negationslash=iE[xr ixrj]. (27.51) However, since the sample values xiare assumed to be independent, we have E[xr ixrj]=E[xr i]E[xr j]=µ2 r. (27.52) The number of terms in the sum on the RHS of (27.51) is N(N−1), and so we ﬁnd V[mr]=1 Nµ2r−µ2 r+N−1 Nµ2 r=µ2r−µ2 r N. (27.53) Since E[mr]=µrandV[mr]→0a sN→∞,t h e rth sample moment mris also a consistent estimator of µr. 1091 STATISTICSIFind the covariance of the sample moments mrandmsfor a sample of size N. We obtain the covariance of the sample moments mrandmsin a similar manner to that used above to obtain the variance of mr. From the deﬁnition of covariance, we have Cov[mr,ms]=E[(mr−µr)(ms−µs)] =1 N2E /" / X ixr i−Nµr /!/ X jxs j−Nµs /!/# =1 N2E /2/4 X ixr+s i+ X i X j/negationslash=ixr ixsj−Nµr X jxs j−Nµs X ixr i+N2µrµs /3/5 Assuming the xito be independent, we may again use result (27.52) to obtain Cov[mr,ms]=1 N2[Nµr+s+N(N−1)µrµs−N2µrµs−N2µsµr+N2µrµs] =1 Nµr+s+N−1 Nµrµs−µrµs =µr+s−µrµs N. We note that by setting r=s, we recover the expression (27.53) for V[mr]. J 27.4.5 Population central moments νr We may generalise the discussion of estimators for the second central moment ν2 (or equivalently σ2) given in subsection 27.4.2 to the estimation of the rth central moment νr. In particular, we saw in that subsection that our choice of estimator forν2depended on whether the population mean µ1is known; the same is true for the estimation of νr. Let us ﬁrst consider the case in which µ1is known. From (26.54), we may write νras νr=µr−rC1µr−1µ1+···+(−1)krCkµr−kµk 1+···+(−1)r−1(rCr−1−1)µr 1. Ifµ1is known, a suitable estimator is obviously ˆνr=mr−rC1mr−1µ1+···+(−1)krCkmr−kµk 1+···+(−1)r−1(rCr−1−1)µr 1, where mris the rth sample moment. Since µ1and the binomial coeﬃcients are (known) constants, it is immediately clear that E[ˆνr]=νr,a n ds o ˆνris an unbiased estimator of νr.I ti sa l s op o s s i b l et oo b t a i na ne x p r e s s i o nf o r V[ˆνr], though the calculation is somewhat lengthy. In the case where the population mean µ1isnotknown, the situation is more complicated. We saw in subsection 27.4.2 that the second sample moment n2(or s2)i snotan unbiased estimator of ν2(orσ2). Similarly, the rthcentral moment of a sample, nr, is not an unbiased estimator of the rth population central moment νr. However, in all cases the bias becomes negligible in the limit of large N. 1092 27.4 SOME BASIC ESTIMATORS As we also found in the same subsection, it is rather complicated to calculate the expectation and variance of n2; this complication increases considerably for general r. Nevertheless, we have derived already in this chapter exact expressions for the expectation value of the ﬁrst few sample central moments, which are valid for samples of any size N. From (27.40), (27.43) and (27.46), we ﬁnd E[n1]=0 , E[n2]=N−1 Nν2, (27.54) E[n2 2]=N−1 N3[(N−1)ν4+(N2−2N+3 )ν2 2]. By similar arguments it can be shown that E[n3]=(N−1)(N−2) N2ν3, (27.55) E[n4]=N−1 N3[(N2−3N+3 )ν4+3 ( 2 N−3)ν2 2]. (27.56) From (27.54) and (27.55), we see that unbiased estimators of ν2andν3are ˆν2=N N−1n2, (27.57) ˆν3=N2 (N−1)(N−2)n3, (27.58) where (27.57) simply re-establishes our earlier result thathatwideσ2=Ns2/(N−1) is an unbiased estimator of σ2. Unfortunately, the pattern that appears to be emerging in (27.57) and (27.58) isnotcontinued for higher r, as is seen immediately from (27.56). Nevertheless, in the limit of large N, the bias becomes negligible, and often one simply takes ˆνr=nr. For large N,i tm a yb es h o w nt h a t E[nr]≈νr V[nr]≈1 N(ν2r−ν2 r+r2ν2ν2 r−1−2rνr−1νr+1) Cov[nr,ns]≈1 N(νr+s−νrνs+rsν2νr−1νs−1−rνr−1νs+1−sνs−1νr+1) 27.4.6 Population covariance Cov [x, y]and correlation Corr [x, y] So far we have assumed that each of our Nindependent samples consists of a single number xi. Let us now extend our discussion to a situation in which each sample consists of two numbers xi,yi, which we may consider as being drawn randomly from a two-dimensional population P(x, y). In particular, we now consider estimators for the population covariance Cov[ x, y] and for the correlation Corr[ x, y]. 1093 STATISTICS When µxandµyareknown , an appropriate estimator of the population covari- ance is hatwidestCov[x, y]=xy−µxµy=parenleftBigg 1 NNsummationdisplay i=1xiyiparenrightBigg −µxµy. (27.59) This estimator is unbiased since EbracketleftBig hatwidestCov[x, y]bracketrightBig =1 NEbracketleftBiggNsummationdisplay i=1xiyibracketrightBigg −µxµy=E[xiyi]−µxµy=C o v [ x, y]. Alternatively, if µxandµyareunknown , it is natural to replace µxandµyin (27.59) by the sample means ¯xand¯yrespectively, in which case we recover the sample covariance Vxy=xy−¯x¯ydiscussed in subsection 27.2.4. This estimator is biased but an unbiased estimator of the population covariance is obtained byforming hatwidestCov[x, y]=N N−1Vxy. (27.60)ICalculate the expectation value of the sample covariance Vxyfor a sample of size N. The sample covariance is given by Vxy= / 1 N X ixiyi /! − / 1 N X ixi /!/ 1 N X jyj /! . Thus its expectation value is given by E[Vxy]=1 NE /"X ixiyi /# −1 N2E /"/ X ixi /!/ X jxj /! /# =E[xiyi]−1 N2E /2/6/4 X ixiyi+ X i,j j/negationslash=ixiyj /3/7/5 Since the number of terms in the double sum on the RHS is N(N−1), we have E[Vxy]=E[xiyi]−1 N2(NE[xiyi]+N(N−1)E[xiyj]) =E[xiyi]−1 N2(NE[xiyi]+N(N−1)E[xi]E[yj]) =E[xiyi]−1 N /; E[xiyi]+(N−1)µxµy / =N−1 NCov[x, y], where we have used the fact that, since the samples are independent, E[xiyj]=E[xi]E[yj]. J 1094 27.4 SOME BASIC ESTIMATORS It is possible to obtain expressions for the variances of the estimators (27.59) and (27.60) but these quantities depend upon higher moments of the populationP(x, y) and are extremely lengthy to calculate. Whether the means µ xandµyare known or unknown, an estimator of the population correlation Corr[ x, y]i sg i v e nb y /hatwideCorr[ x, y]=hatwidestCov[x, y] ˆσxˆσy, (27.61) wherehatwidestCov[x, y],ˆσxandˆσyare the appropriate estimators of the population co- variance and standard deviations. Although this estimator is only asymptoticallyunbiased, i.e. for large N, it is widely used because of its simplicity. Once again the variance of the estimator depends on the higher moments of P(x, y)a n di s diﬃcult to calculate. In the case in which the means µ xandµyare unknown, a suitable (but biased) estimator is /hatwideCorr[ x, y]=N N−1Vxy sxsy=N N−1rxy, (27.62) where sxandsyare the sample standard deviations of the xiandyirespectively andrxyis the sample correlation. In the special case when the parent population P(x, y) is Gaussian, it may be shown that, if ρ= Corr[ x, y], E[rxy]=ρ−ρ(1−ρ2) 2N+O ( N−2), (27.63) V[rxy]=1 N(1−ρ2)2+O ( N−2), (27.64) from which the expectation value and variance of the estimator /hatwideCorr[ x, y]m a y be found immediately. We note ﬁnally that our discussion may be extended, without signiﬁcant al- teration, to the general case in which each data item consists of nnumbers xi,yi,...,z i. 27.4.7 A worked example We conclude our discussion of basic estimators by reconsidering the set of experimental data given in subsection 27.2.4. 1095 STATISTICSITen UK citizens are selected at random and their heights and weights are found to be as follows (to the nearest cmorkgrespectively): Person ABCDEFGH IJ Height (cm) 194 168 177 180 171 190 151 169 175 182Weight (kg) 75 53 72 80 75 75 57 67 46 68 Estimate the means, µ xandµy, and standard deviations, σxandσy, of the two-dimensional joint population from which the sample was drawn, quoting the standard error on the esti-mate in each case. Estimate also the correlation Corr[ x, y]of the population, and quote the standard error on the estimate under the assumption that the population is a multivariate Gaussian. In subsection 27.2.4, we calculated various sample statistics for these data. In particular,we found that for our sample of size N= 10, ¯x= 175 .7,¯y=6 6.8, s x=1 1.6,s y=1 0.6,r xy=0.54. Let us begin by estimating the means µxandµy. As discussed in subsection 27.4.1, the sample mean is an unbiased, consistent estimator of the population mean. Moreover, the standard error on ¯x(say) is σx/√ N. In this case, however, we do not know the true value ofσxand we must estimate it using bσx= p N/(N−1)sx. Thus, our estimates of µxand µy, with associated standard errors, are ˆµx=¯x±sx√ N−1= 175 .7±3.9, ˆµy=¯y±sy√ N−1=6 6.8±3.5. We now turn to estimating σxandσy. As just mentioned, our estimate of σx(say) is bσx= p N/(N−1)sx. Its variance (see the ﬁnal line of subsection 27.4.3) is given approximately by V[ˆσ]≈1 4Nν2 / ν4−N−3 N−1ν2 2 / . Since we do not know the true values of the population central moments ν2andν4,w e must use their estimated values in this expression. We may take ˆν2= bσ2 x=(ˆσ)2,w h i c hw e have already calculated. It still remains, however, to estimate ν4.A si m p l i e dn e a rt h ee n d of subsection 27.4.5, it is acceptable to take ˆν4=n4. Thus for the xiandyivalues, we have (ˆν4)x=1 NNX i=1(xi−¯x)4= 53411 .6 (ˆν4)y=1 NNX i=1(yi−¯y)4= 27732 .5 Substituting these values into (27.50), we obtain ˆσx= /N N−1 /1/2 sx±(ˆV[ˆσx])1/2=1 2.2±6.7, (27.65) ˆσy= /N N−1 /1/2 sy±(ˆV[ˆσy])1/2=1 1.2±3.6. (27.66) 1096 27.5 MAXIMUM-LIKELIHOOD METHOD Finally, we estimate the population correlation Corr[ x, y], which we shall denote by ρ. From (27.62), we have ˆρ=N N−1rxy=0.60. Under the assumption that the sample was drawn from a two-dimensional Gaussian population P(x, y), the variance of our estimator is given by (27.64). Since we do not know the true value of ρ, we must use our estimate ˆρ. Thus, we ﬁnd that the standard error ∆ ρ in our estimate is given approximately by ∆ρ≈10 9 /1 10 / [1−(0.60)2]2=0.05. J 27.5 Maximum-likelihood method The population from which the sample x1,x2,...,x Nis drawn is, in general, unknown . In the previous section, we assumed that the sample values were inde- pendent and drawn from a one-dimensional population P(x), and we considered basic estimators of the moments and central moments of P(x). We did not,h o w - ever, assume a particular functional form for P(x). We now discuss the process ofdata modelling , in which a speciﬁc form is assumed for the population. In the most general case, it will not be known whether the sample values are independent, and so let us consider the full joint population P(x), where xis the point in the N-dimensional data space with coordinates x1,x2,...,x N.W et h e n adopt the hypothesis Hthat the probability distribution of the sample values has some particular functional form L(x;a), dependent on the values of some set of parameters ai,i=1,2,...,m . Thus, we have P(x|a,H)=L(x;a), where we make explicit the conditioning on both the assumed functional form and on the parameter values; L(x;a) is called the likelihood function . Hypotheses of this type form the basis of data modelling andparameter estimation . One proposes a particular model for the underlying population and then attempts to estimate fromthe sample values x 1,x2,...,x Nthe values of the parameters adeﬁning this model.IA company measures the duration (in minutes) of the Nintervals xi,i=1,2,...,N between successive telephone calls received by its switchboard. Suppose that the samplevalues x iare drawn independently from the distribution P(x|τ)=( 1 /τ)e x p (−x/τ),w h e r e τ is the mean interval between calls. C alculate the likelihood function L(x;τ). Since the sample values are independent and drawn from the stated distribution, the likelihood is given by L(x;τ)=P(xi|τ)P(x2|τ)···P(xN|τ) =1 τexp / −x1 τ /1 τexp / −x2 τ / ···1 τexp / −xN τ / =1 τNexp / −1 τ(x1+x2+···+xN) / . (27.67) which is to be considered as a function of τ, given that the sample values xiare ﬁxed. J 1097 STATISTICS 0024 681 0 12 14 16 18 200.51 N=5L(x;τ) τ0024 681 0 12 14 16 18 200.51 N=1 0L(x;τ) τ 0024 681 0 12 14 16 18 200.51 N=2 0L(x;τ) τ0024 681 0 12 14 16 18 200.51 N=5 0L(x;τ) τ Figure 27.5 Examples of the likelihood function (27.67) for samples of dif- ferent size N. In each case, the true value of the parameter is τ=4a n dt h e sample values xiare indicated by the short vertical lines. For the purposes of illustration, in each case the likelihood function is normalised so that its maximum value is unity. The likelihood function (27.67) depends on just a single parameter τ.P l o t so f the likelihood function, considered as a function of τ, are shown in ﬁgure 27.5 for samples of diﬀerent size N. The true value of the parameter τused to generate the sample values was 4. In each case, the sample values xiare indicated by the short vertical lines. For the purposes of illustration, the likelihood function in each case has been scaled so that its maximum value is unity (this is, in fact, commonpractice). We see that when the sample size is small, the likelihood function is verybroad. As Nincreases, however, the likelihood becomes narrower (it is inversely proportional to√ N) and tends to a Gaussian-like shape, with its peak centred on 4, the true value of τ. We discuss these properties of the likelihood function in more detail in subsection 27.5.6. 27.5.1 The maximum-likelihood estimator Since the likelihood function L(x;a) gives the probability density associated with any particular set of values of the parameters a, our best estimate ˆaof these parameters is given by the values of afor which L(x;a) is a maximum. This is called the maximum-likelihood estimator (or ML estimator). In general, the likelihood function can have a complicated shape when con- 1098 27.5 MAXIMUM-LIKELIHOOD METHOD a aa a L(x;a) L(x;a)L(x;a) L(x;a) ˆa ˆaˆa ˆa(a)( b) (c)( d) Figure 27.6 Typical shapes of one-dimensional likelihood functions L(x;a) encountered in practice, where, for illustration purposes, it is assumed that the parameter ais restricted to the range zero to inﬁnity. The ML estimator in the various cases occurs at: ( a) the only stationary point; ( b) one of several stationary points; ( c) an end-point of the allowed parameter range that is not a stationary point (although stationary points do exist); ( d) an end-point of the allowed parameter range in whic h no stationary point exists. sidered as a function of a, particularly when the dimensionality of the space of parameters a1,a2,...,a Mis large. It may be that the values of some parameters are either known or assumed in advance, in which case the eﬀective dimension-ality of the likelihood function is reduced accordingly. However, even when thelikelihood depends on just a single parameter a(either intrinsically or as the result of assuming particular values for the remaining parameters), its form maybe complicated when the sample size Nis small. Frequently occurring shapes of one-dimensional likelihood functions are illustrated in ﬁgure 27.6, where we have assumed, for deﬁniteness, that the allowed range of the parameter ais zero to inﬁnity. In each case, the ML estimate ˆais also indicated. Of course, the ‘shape’ of higher-dimensional likelihood functions may be considerably more complicated. In many simple cases, however, the likelihood function L(x;a) has a single maximum that occurs at a stationary point (the likelihood function is then termedunimodal ). In this case, the ML estimators of the parameters a i,i=1,2,...,M , may be found without evaluating the full likelihood function L(x;a). Instead, one simply solves the Msimultaneous equations ∂L ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=ˆa=0 f o r i=1,2,...,M. (27.68) 1099 STATISTICS Since ln zis a monotonically increasing function of z(and therefore has the same stationary points), it is often more convenient, in fact, to maximise thelog-likelihood function ,l nL(x;a), with respect to the a i. Thus, one may, as an alternative, solve the equations ∂lnL ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=ˆa=0 f o r i=1,2,...,M. (27.69) Clearly, (27.68) and (27.69) will lead to the same ML estimates ˆaof the parameters. In either case, it is, of course, prudent to check that the point a=ˆais a local maximum.IFind the ML estimate of the parameter τin the previous example, in terms of the measured values xi,i=1,2,...,N . From (27.67), the log-likelihood function in this case is given by lnL(x;τ)=NX i=1ln /1 τe−xi/τ / =−NX i=1 / lnτ+xi τ / . (27.70) Diﬀerentiating with respect to the parameter τand setting the result equal to zero, we ﬁnd ∂lnL ∂τ=−NX i=1 /1 τ−xi τ2 / =0. Thus the ML estimate of the parameter τis given by ˆτ=1 NNX i=1xi, (27.71) which is simply the sample mean of the Nmeasured intervals. J In the previous example we assumed that the sample values xiwere drawn independently from the sameparent distribution. The ML method is more ﬂexible than this restriction might seem to imply and it can equally well be applied to thecommon case in which the samples x iare independent but each is drawn from a diﬀerent distribution.IIn an experiment, Nindependent measurements xiof some quantity are made. Suppose that the random measurement error on the ith sample value is Gaussian distributed with mean zero and known standard deviation σi. Calculate the ML estimate of the true value µof the quantity being measured. As the measurements are independent, the likelihood factorises: L(x;µ,{σk})=NY i=1P(xi|µ, σ i), where{σk}denotes collectively the set of known standard deviations σ1,σ2,...,σ N.T h e individual distributions are given by P(xi|µ, σ i)=1p 2πσ2 iexp / −(xi−µ)2 2σ2 i / . 1100 27.5 MAXIMUM-LIKELIHOOD METHOD and so the full log-likelihood function is given by lnL(x;µ,{σk})=−1 2NX i=1 / ln(2πσ2 i)+(xi−µ)2 σ2 i / . Diﬀerentiating this expression with respect to µand setting the result equal to zero, we ﬁnd ∂lnL ∂µ=NX i=1xi−µ σ2 i=0, from which we obtain the ML estimator ˆµ= PN i=1(xi/σ2 i)PN i=1(1/σ2 i). (27.72) This estimator is commonly used when averaging data with diﬀerent statistical weights wi=1/σ2 i. We note that when all the variances σ2 ihave the same value the estimator reduces to the sample mean of the data xi. J There is, in fact, no requirement in the ML method that the sample values be independent. As an illustration, we shall generalise the above example to a case in which the measurements xiare not all independent. This would occur, for example, if these measurements were based at least in part on the same data.IIn an experiment Nmeasurements xiof some quantity are made. Suppose that the random measurement errors on the samples are drawn from a joint Gaussian distribution with meanzero and known covariance matrix V. Calculate the ML estimate of the true value µof the quantity being measured. From (26.148), the likelihood in this case is given by L(x;µ,V)=1 (2π)N/2|V|1/2exp / −1 2(x−µ1)TV−1(x−µ1) / , where xis the column vector with components x1,x2,...,x Nand 1is the column vector with all components equal to unity. Thus, the log-likelihood function is given by lnL(x;µ,V)=−1 2 / Nln(2π)+l n|V|+(x−µ1)TV−1(x−µ1) / . Diﬀerentiating with respect to µand setting the result equal to zero gives ∂lnL ∂µ=1TV−1(x−µ1)=0 . Thus, the ML estimator is given by ˆµ=1TV−1x 1TV−11= P i,j(V−1)ijxjP i,j(V−1)ij. In the case of uncorrelated errors in measurement, ( V−1)ij=δij/σ2 iand our estimator reduces to that given in (27.72). J In all the examples considered so far, the likelihood function has been eﬀectively one-dimensional, either instrinsically or under the assumption that the values ofall but one of the parameters are known in advance. As the following example 1101 STATISTICS involving two parameters shows, the application of the ML method to the estimation of several parameters simultaneously is straightforward.IIn an experiment Nmeasurements xiof some quantity are made. Suppose the random error on each sample value is drawn independently from a Gaussian distribution of mean zero butunknown standard deviation σ(which is the same for each measurement). Calculate the ML estimates of the true value µof the quantity being measured and the standard deviation σ of the random errors. In this case the log-likelihood function is given by lnL(x;µ, σ)=−1 2NX i=1 / ln(2πσ2)+(xi−µ)2 σ2 / . Taking partial derivatives of ln Lwith respect to µandσand setting the results equal to zero at the joint estimate ˆµ,ˆσ,w eo b t a i n NX i=1xi−ˆµ ˆσ2=0, (27.73) NX i=1(xi−ˆµ)2 ˆσ3−NX i=11 ˆσ=0. (27.74) In principle, one should solve these two equations simultaneously for ˆµandˆσ, but in this case we notice that the ﬁrst is solved immediately by ˆµ=1 NNX i=1xi=¯x, where ¯xis the sample mean. Substituting this result into the second equation, we ﬁnd ˆσ= vuut1 NNX i=1(xi−¯x)2=s, where sis the sample standard deviation. As shown in subsection 27.4.3, sis a biased estimator of σ. The reason why the ML method may produce a biased estimator is discussed in the next subsection. J 27.5.2 Transformation invariance and bias of ML estimators An extremely useful property of ML estimators is that they are invariant to parameter transformations. Suppose that, instead of estimating some parameter aof the assumed population, we wish to estimate some function α(a)o ft h e parameter. The ML estimator ˆα(a) is given by the value assumed by the function α(a) at the maximum point of the likelihood, which is simply equal to α(ˆa). Thus, we have the very convenient property ˆα(a)=α(ˆa). We do not have to worry about the distinction between the two cases, estimating aand estimating a function of a.T h i si s nottrue, in general, for other estimation procedures. 1102 27.5 MAXIMUM-LIKELIHOOD METHODIA company measures the duration (in minutes) of the Nintervals xi,i=1,2,...,N , between successive telephone calls received by its switchboard. Suppose that the samplevalues x iare drawn independently from the distribution P(x|τ)=( 1 /τ)exp(−x/τ).F i n dt h e ML estimate of the parameter λ=1/τ. This is the same problem as that considered at the start of section 27.5.1. In terms of the new parameter λ, the log-likelihood function is given by lnL(x;λ)=NX i=1ln(λe−λxi)=NX i=1(lnλ−λxi). Diﬀerentiating with respect to λand setting the result equal to zero, we have ∂lnL ∂λ=NX i=1 /1 λ−xi / =0. Thus, the ML estimator of the parameter λis given by ˆλ= / 1 NNX i=1xi /!−1 =¯x−1. (27.75) Referring back to (27.71), we see that, as expected, the ML estimators of λandτare related by ˆλ=1/ˆτ. J Although this invariance property is useful it also means that, in general, ML estimators may be biased . In particular, one must be aware of the fact that even ifˆais an unbiased ML estimator of ait does notfollow that the estimator ˆα(a)i s also unbiased. In the limit of large N, however, the bias of ML estimators always tends to zero. As an illustration, it is straightforward to show (see exercise 27.8)that the ML estimators ˆτandˆλin the above example have expectation values E[ˆτ]=τand E[ˆλ]=N N−1λ. (27.76) In fact, since ˆτ=¯xand the sample values are independent, the ﬁrst result follows immediately from (27.40). Thus, ˆτis unbiased, but ˆλ=1/ˆτis biased, albeit that the bias tends to zero for large N. 27.5.3 Eﬃciency of ML estimators We showed in subsection 27.3.2 that Fisher’s inequality puts a lower limit on the variance V[ˆa] of any estimator of the parameter a. Under our hypothesis Hon p. 1097, the functional form of the population is given by the likelihood function, i.e.P(x|a,H)=L(x;a). Thus, if this hypothesis is correct, we may replace Pby Lin Fisher’s inequality (27.18), which then reads V[ˆa]≥parenleftbigg 1+∂b ∂aparenrightbigg2slashbigg Ebracketleftbigg −∂2lnL ∂a2bracketrightbigg , where bis the bias in the estimator ˆa. We usually denote the RHS by Vmin. 1103 STATISTICS An important property of ML estimators is that ifthere exists an eﬃcient estimator ˆaeﬀ,i . e .o n ef o rw h i c h V[ˆaeﬀ]=Vmin,t h e ni t mustbe the ML estimator or some function thereof. This is easily shown by replacing PbyLin the proof of Fisher’s inequality given in subsection 27.3.2. In particular, we note that the equality in (27.22) holds only if h(x)=cg(x), where cis a constant. Thus, if an eﬃcient estimator ˆaeﬀexists, this is equivalent to demanding that ∂lnL ∂a=c[ˆaeﬀ−α(a)]. Now, the ML estimator ˆaMLis given by ∂lnL ∂avextendsinglevextendsinglevextendsinglevextendsingle a=ˆaML=0⇒ c[ˆaeﬀ−α(ˆaML)] = 0 , which, in turn, implies that ˆaeﬀmust be some function of ˆaML.IShow that the ML estimator ˆτgiven in (27.71) is an eﬃcient estimator of the parameter τ. As shown in (27.70), the log-likelihood function in this case is lnL(x;τ)=−NX i=1 / lnτ+xi τ / . Diﬀerentiating twice with respect to τ, we ﬁnd ∂2lnL ∂τ2=NX i=1 /1 τ2−2xi τ3 / =N τ2 / 1−2 τNNX i=1xi /! , (27.77) and so the expectation value of this expression is E /∂2lnL ∂τ2 / =N τ2 / 1−2 τE[xi] / =−N τ2, where we have used the fact that E[x]=τ. Setting b= 0 in (27.18), we thus ﬁnd that for anyunbiased estimator of τ, V[ˆτ]≥τ2 N. From (27.76), we see that the ML estimator ˆτ= P ixi/Nis unbiased. Moreover, using the fact that V[x]=τ2, it follows immediately from (27.40) that V[ˆτ]=τ2/N. Thus ˆτis a minimum-variance estimator of τ. J 27.5.4 Standard errors and conﬁdence limits on ML estimators The ML method provides a procedure for obtaining a particular set of estimators ˆaMLfor the parameters aof the assumed population P(x|a). As for any other set of estimators, the associated standard errors, covariances and conﬁdence intervalscan be found as described in subsections 27.3.3 and 27.3.4. 1104 27.5 MAXIMUM-LIKELIHOOD METHOD 0 00.10.20.30.4 24 681 0 12 14P(ˆτ|τ) ˆτ Figure 27.7 The sampling distribution P(ˆτ|τ) for the estimator ˆτfor the case τ=4a n d N= 10.IA company measures the duration (in minutes) of the 10intervals xi,i=1,2,...,10, between successive telephone calls made to its switchboard to be as follows: 0.43 0 .24 3 .03 1 .93 1 .16 8 .65 5 .33 6 .06 5 .62 5 .22. Supposing that the sample value s are drawn independently from the probability distribution P(x|τ)=( 1 /τ)e x p (−x/τ), ﬁnd the ML estimate of the mean τand quote an estimate of the standard error on your result. As shown in (27.71) the (unbiased) ML estimator ˆτin this case is simply the sample mean ¯x=3.77. Also, as shown in subsection 27.5.3, ˆτis a minimum-variance estimator with V[ˆτ]=τ2/N. Thus, the standard error in ˆτis simply σˆτ=τ√ N. (27.78) Since we do not know the true value of τ, however, we must instead quote an estimate ˆσˆτ of the standard error, obtained by substituting our estimate ˆτforτin (27.78). Thus, we quote our ﬁnal result as τ=ˆτ±ˆτ√ N=3.77±1.19. (27.79) For comparison, the true value used to create the sample was τ=4 . J For the particular problem considered in the above example, it is in fact possible to derive the full sampling distribution of the ML estimator ˆτusing characteristic functions, and it is given by P(ˆτ|τ)=NN (N−1)!ˆτN−1 τNexpparenleftbigg −Nˆτ τparenrightbigg , (27.80) where Nis the size of the sample. This function is plotted in ﬁgure 27.7 for the case τ=4a n d N= 10, which pertains to the above example. Knowledge of the analytic form of the sampling distribution allows one to place conﬁdence limits on the estimate ˆτobtained, as discussed in subsection 27.3.4. 1105 STATISTICSIUsing the sample values in the above example, obtain the 68% central conﬁdence interval on the value of τ. For the sample values given, our observed value of the ML estimator is ˆτobs=3.77. Thus, from (27.28) and (27.29), the 68% central conﬁdence interval [ τ−,τ+] on the value of τis found by solving the equationsZˆτobs −∞P(ˆτ|τ+)dˆτ=0.16,Z∞ ˆτobsP(ˆτ|τ−)dˆτ=0.16, where P(ˆτ|τ) is given by (27.80) with N= 10. The above integrals can be evaluated analytically but the calculations are rather cumbersome. It is much simpler to evaluatethem by numerical integration, from which we ﬁnd [ τ −,τ+]=[ 2 .86,5.46]. Alternatively, we could quote the estimate and its 68% conﬁdence interval as τ=3.77+1.69 −0.91. Thus we see that the 68% central conﬁdence interval is not symmetric about the estimated value, and diﬀers from the standard error calculated above. This is a result of the (non-Gaussian) shape of the sampling distribution P(ˆτ|τ), apparent in ﬁgure 27.7.J In many problems, however, it is not possible to derive the full sampling distribution of an ML estimator ˆain order to obtain its conﬁdence intervals. Indeed, one may not even be able to obtain an analytic formula for its standarderror σ ˆa. This is particularly true when one is estimating several parameter ˆa simultaneously, since the joint sampling distribution will be, in general, very complicated. Nevertheless, as we discuss below, the likelihood function L(x;a) itselfcan be used very simply to obtain standard errors and conﬁdence intervals. The justiﬁcation for this has its roots in the Bayesian approach to statistics, as opposed to the more traditional frequentist approach we have adopted here. We now give a brief discussion of the Bayesian viewpoint on parameter estimation. 27.5.5 The Bayesian interpretation of the likelihood function As stated at the beginning of section 27.5, the likelihood function L(x;a)i s deﬁned by P(x|a,H)=L(x;a), where Hdenotes our hypothesis of an assumed functional form. Now, using Bayes’ theorem (see subsection 26.2.3), we may write P(a|x,H)=P(x|a,H)P(a|H) P(x|H), (27.81) which provides us with an expression for the probability distribution P(a|x,H) of the parameters a, given the (ﬁxed) data xand our hypothesis H, in terms of 1106 27.5 MAXIMUM-LIKELIHOOD METHOD other quantities that we may assign. The various terms in (27.81) have special formal names, as follows. •The quantity P(a|H)o nt h eR H Si st h e priorprobability, which represents our state of knowledge of the parameter values (given the hypothesis H)before we have analysed the data. •This probability is modiﬁed by the experimental data xthrough the likelihood P(x|a,H). •When appropriately normalised by the evidence P(x|H), this yields the posterior probability P(a|x,H), which is the quantity of interest. •The posterior encodes allour inferences about the values of the parameters a. Strictly speaking, from a Bayesian viewpoint, this entire function ,P(a|x,H), is the ‘answer’ to a parameter estimation problem. Given a particular hypothesis, the (normalising) evidence factor P(x|H)i s unimportant, since it does not depend explicitly upon the parameter values a. Thus, it is often omitted and one considers only the proportionality relation P(a|x,H)∝P(x|a,H)P(a|H). (27.82) If necessary, the posterior distribution can be normalised empirically, by requiring that it integrates to unity, i.e.integraltext P(a|x,H)dma= 1, where the integral extends over all values of the parameters a1,a2,...,a m. The prior P(a|H) in (27.82) should reﬂect our entire knowledge concerning the values of the parameters a,before the analysis of the current data x. For example, there may be some physical reason to require some or all of the parameters to lie in a given range. If we are largely ignorant of the values of the parameters,we often indicate this by choosing a uniform (or very broad) prior, P(a|H) = constant , in which case the posterior distribution is simply proportional to the likelihood. In this case, we thus have P(a|x,H)∝L(x;a). (27.83) In other words, if we assume a uniform prior then we can identify the posterior distribution (up to a normalising factor) with L(x;a), considered as a function of the parameters a. Thus, a Bayesian statistician considers the ML estimates ˆa MLof the parameters to be the values that maximise the posterior P(a|x,H) under the assumption of a uniform prior. More importantly, however, a Bayesian would notcalculate the standard error or conﬁdence interval on this estimate using the (classical) methodemployed in subsection 27.3.4. Instead, a far more straightforward approach is 1107 STATISTICS adopted. Let us assume, for the moment, that one is estimating just a single parameter a. Using (27.83), we may determine the values a−anda+such that Pr(a<a−|x,H)=integraldisplaya− ∞L(x;a)da=α, Pr(a>a +|x,H)=integraldisplay∞ a+L(x;a)da=β. where it is assumed that the likelihood has been normalised in such a way thatintegraltext L(x;a)da= 1. Combining these equations gives Pr(a−≤a<a +|x,H)=integraldisplaya+ a−L(x;a)da=1−α−β, (27.84) and [ a−,a+]i st h e Bayesian conﬁdence interval on the value of aat the conﬁdence level 1−α−β. As in the case of classical conﬁdence intervals, one often quotes the central conﬁdence interval, for which α=β. Another common choice (where possible) is to use the two values a−anda+satisfying (27.84), for which L(x;a−)= L(x;a+). It should be understood that a frequentist would consider the Bayesian conﬁ- dence interval as an approximation to the (classical) conﬁdence interval discussed in subsection 27.3.4. Conversely, a Bayesian would consider the conﬁdence inter-val deﬁned in (27.84) to be the more meaningful. In fact, the diﬀerence betweenthe Bayesian and classical conﬁdence intervals is rather subtle. The classical con-ﬁdence interval is deﬁned in such a way that if one took a large number of samples each of size Nand constructed the conﬁdence interval in each case then the proportion of cases in which the true value of awould be contained within the interval is 1 −α−β. For the Bayesian conﬁdence interval, one does not rely on the frequentist concept of a large number of repeated samples. Instead, its meaning isthat, given the single sample x(and our hypothesis Hfor the functional form of the population), the probability that alies within the interval [ a −,a+]i s1−α−β. By adopting the Bayesian viewpoint, the likelihood function L(x;a) may also be used to obtain an approximation ˆσˆato the standard error in the ML estimator; the approximation is given by ˆσˆa=parenleftbigg −∂2lnL ∂a2vextendsinglevextendsinglevextendsinglevextendsingle a=ˆaparenrightbigg−1/2 . (27.85) Clearly, if L(x;a) were a Gaussian centred on a=ˆathenˆσˆawould be its standard deviation. Indeed, in this case, the resulting ‘one-sigma’ limits would constitute a 68.3% Bayesian central conﬁdence interval. Even when L(x;a) is not Gaussian, however, (27.85) is often used as a measure of the standard error. 1108 27.5 MAXIMUM-LIKELIHOOD METHOD 0 00.10.20.30.4 24 681 0 12 14L(x;τ) τ Figure 27.8 The likelihood function L(x;τ) (normalised to unit area) for the sample values given in the worked example in subsection 27.5.4, and indicated here by short vertical lines.IFor the sample data given in section 27.5.4, use the likelihood function to estimate the standard error ˆσˆτin the ML estimator ˆτand obtain the Bayesian 68% central conﬁdence interval on τ. We showed in (27.67) that the likelihood function in this case is given by L(x;τ)=1 τNexp[−1 τ(x1+x2+···+xN)]. where xi,i=1,2,...,N , denotes the sample value and N= 10. This likelihood function is plotted in ﬁgure 27.8, after normalising (numerically) to unit area. The short vertical linesin the ﬁgure indicate the sample values. We see that the likelihood function peaks at theML estimate ˆτ=3.77 that we found in subsection 27.5.4. Also, from (27.77), we have ∂ 2lnL ∂τ2=N τ2 / 1−2 τNNX i=1xi /! , Remembering that ˆτ= P ixi/N, our estimate of the standard error in ˆτis ˆσˆτ= / −∂2lnL ∂τ2 //// τ=ˆτ /−1/2 =ˆτ√ N=1.19, which is precisely the estimate of the standard error we obtained in subsection 27.5.4. It should be noted, however, that in general we would not expect the two estimates ofstandard error made by the diﬀerent methods to be identical. In order to calculate the Bayesian 68% central conﬁdence interval, we must determine the values a −anda+that satisfy (27.84) with α=β=0.16. In this case, the calculation can be performed analytically but is somewhat tedious. It is trivial, however, to determine a−anda+numerically and we ﬁnd the conﬁdence interval to be [3 .16,6.20]. Thus we can quote our result with 68% central conﬁdence limits as τ=3.77+2.43 −0.61. By comparing this result with that given towards the end of subsection 27.5.4, we see that, as we might expect, the Bayesian and classical conﬁdence intervals diﬀer somewhat. J 1109 STATISTICS The above discussion is generalised straightforwardly to the estimation of several parameters a1,a2,...,a Msimultaneously. The elements of the covariance matrix of the ML estimators can be approximated by ˆVij=hatwidestCov[ˆai,ˆaj]=parenleftbigg −∂2lnL ∂ai∂ajvextendsinglevextendsinglevextendsinglevextendsingle a=ˆaparenrightbigg−1 . (27.86) From (27.36), we see that (at least for unbiased estimators) the expectation value of (27.86) is equal to the element Fijof the Fisher matrix. The construction of a multi-dimensional Bayesian conﬁdence region is also straightforward. For a given conﬁdence level 1 −α(say), it is most common to construct the conﬁdence region as the M-dimensional region Rina-space, bounded by the ‘surface’ L(x;a) = constant, for which integraldisplay RL(x;a)dMa=1−α, where it is assumed that L(x;a) is normalised to unit volume. Moreover, we see from (27.83) that (assuming a uniform prior probability) we may obtain themarginal posterior distribution for any parameter a isimply by integrating the likelihood function L(x;a) over the other parameters: P(ai|x,H)=integraldisplay ···integraldisplay L(x;a)da1···dai−1dai+1···daM. Here the integral extends over all possible values of the parameters, and again is it assumed that the likelihood function is normalised in such a way thatintegraltext L(x;a)dMa= 1. This marginal distribution can then be used as above to determine Bayesian conﬁdence intervals on each aiseparately.ITen independent sample values xi,i=1,2,...,10, are drawn at random from a Gaussian distribution with unknown mean µand standard deviation σ. The samples values are as follows (to two decimal places): 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Find the Bayesian 95% central conﬁdence intervals on µandσseparately. The likelihood function in this case is L(x;µ, σ)=( 2 πσ2)−N/2exp /" −1 2σ2NX i=1(xi−µ)2 /# . (27.87) Assuming uniform priors on µandσ(over their natural ranges of −∞ → ∞ and 0→∞ respectively), we may identify this likelihood function with the posterior probability, as in(27.83). Thus, the marginal posterior distribution on µis given by P(µ|x,H)∝ Z∞ 01 σNexp /" −1 2σ2NX i=1(xi−µ)2 /# dσ. 1110 27.5 MAXIMUM-LIKELIHOOD METHOD By substituting σ=1/u(so that dσ=−du/u2) and integrating by parts either ( N−2)/2 or (N−3)/2 times, we ﬁnd P(µ|x,H)∝ / N(¯x−µ)2+Ns2 /−(N−1)/2, where we have used the fact that P i(xi−µ)2=N(¯x−µ)2+Ns2,¯xbeing the sample mean and s2the sample variance. We may now ﬁnd the 95% central conﬁdence interval by ﬁnding the values µ−andµ+for whichZµ− −∞P(µ|x,H)dµ=0.025 and Z∞ µ+P(µ|x,H)dµ=0.025. The normalisation of the posterior distribution and the values µ−and µ+are easily obtained by numerical integration. Substituting in the appropriate values N= 10, ¯x=1.11 ands=1.01, we ﬁnd the required conﬁdence interval to be [0 .29,1.97]. To obtain a conﬁdence interval on σ, we must ﬁrst obtain the corresponding marginal posterior distribution. From (27.87), again using the fact that P i(xi−µ)2=N(¯x−µ)2+Ns2, this is given by P(σ|x,H)∝1 σNexp / −Ns2 2σ2 /Z∞ −∞exp / −N(¯x−µ)2 2σ2 / dµ. Noting that the integral of a one-dimensional Gaussian is proportional to σ, we conclude that P(σ|x,H)∝1 σN−1exp / −Ns2 2σ2 / . The 95% central conﬁdence interval on σcan then be found in an analogous manner to that on µ, by solving numerically the equationsZσ− 0P(σ|x,H)dσ=0.025 and Z∞ σ+P(σ|x,H)dσ=0.025. We ﬁnd the required interval to be [0 .76,2.16]. J 27.5.6 Behaviour of ML estimators for large N As mentioned in subsection 27.3.6, in the large-sample limit N→∞, the sampling distribution of a set of (consistent) estimators ˆa, whether ML or not, will tend, in general, to a multivariate Gaussian centred on the true values a.T h i si sa direct consequence of the central limit theorem. Similarly, in the limit N→∞the likelihood function L(x;a)alsotends towards a multivariate Gaussian but one centred on the ML estimate(s) ˆa. Thus ML estimators are always asymptotically consistent . This limiting process was illustrated for the one-dimensional case by ﬁgure 27.5. Thus, as Nbecomes large, the likelihood function tends to the form L(x;a)=Lmaxexpbracketleftbig −1 2Q(a,ˆa)bracketrightbig , where Qdenotes the quadratic form Q(a,ˆa)=(a−ˆa)TV−1(a−ˆa) 1111 STATISTICS and the matrix V−1is given by parenleftbig V−1parenrightbig ij=−∂2lnL ∂ai∂ajvextendsinglevextendsinglevextendsinglevextendsingle a=ˆa. Moreover, in the limit of large N, this matrix tends to the Fisher matrix given in (27.36), i.e. V−1→F. Hence ML estimators are asymptotically minimum-variance . Comparison of the above results with those in subsection 27.3.6 shows that the large-sample limit of the likelihood function L(x;a) has the same form as the large-sample limit of the joint estimator sampling distribution P(ˆa|a). The only diﬀerence is that P(ˆa|a) is centred in ˆa-space on the true values ˆa=awhereas L(x;a) is centred in a-space on the ML estimates a=ˆa. From ﬁgure 27.4 and its accompanying discussion, we therefore conclude that, in the large-sample limit, the Bayesian and classical conﬁdence limits on the parameters coincide . 27.5.7 Extended maximum-likelihood method It is sometimes the case that the number of data items Nin our sample is itself a random variable. Such experiments are typically those in which data are collectedfor a certain period of time during which events occur at random in some way,as opposed to those in which a prearranged number of data items are collected.In particular, let us consider the case where the sample values x 1,x2,...,x Nare drawn independently from some distribution P(x|a) and the sample size Nis a random variable described by a Poisson distribution with mean λ,i . e .N∼Po(λ). The likelihood function in this case is given by L(x;λ,a)=λN N!e−λNproductdisplay i=1P(xi|a), (27.88) a n di so f t e nc a l l e dt h e extended likelihood function . The function L(x;λ,a)c a n be used as before to estimate parameter values or obtain conﬁdence intervals.Two distinct cases arise in the use of the extended likelihood function, dependingon whether the Poisson parameter λis a function of the parameters aor is an independent parameter. Let us ﬁrst consider the case in which λis a function of the parameters a.F r o m (27.88), we can write the extended log-likelihood function as lnL=Nlnλ(a)−λ(a)+ Nsummationdisplay i=1lnP(xi|a)=−λ(a)+Nsummationdisplay i=1ln[λ(a)P(xi|a)]. where we have ignored terms not depending on a. The ML estimates ˆaof the parameters can then be found in the usual way, and the ML estimate of the Poisson parameter is simply ˆλ=λ(ˆa). The errors on our estimators ˆawill be, in general, smaller than those obtained in the usual likelihood approach, since ourestimate includes information from the value of Nas well as the sample values x i. 1112 27.6 THE METHOD OF LEAST SQUARES The other possibility is that λis an independent parameter and not a function of the parameters a. In this case, the extended log-likelihood function is lnL=Nlnλ−λ+Nsummationdisplay i=1lnP(xi|a), (27.89) where we have omitted terms not depending on λora. Diﬀerentiating with respect to λand setting the result equal to zero, we ﬁnd that the ML estimate of λis simply ˆλ=N. By diﬀerentiating (27.89) with respect to the parameters aiand setting the results equal to zero, we obtain the usual ML estimates ˆaiof their values. In this case, however, the errors in our estimates will be larger, in general, than those in thestandard likelihood approach, since they must include the eﬀect of statisticaluncertainty in the parameter λ. 27.6 The method of least squares The method of least squares is, in fact, just a special case of the method of maximum-likelihood. Nevertheless, it is so widely used as a method of parameterestimation that it has acquired a special name of its own. At the outset, let ussuppose that a data sample consists of a set of pairs ( x i,yi),i=1,2,...,N .F o r example, these data might correspond to the temperature yimeasured at various points xialong some metal rod. For the moment, we will suppose that the xiare known exactly, whereas there exists a measurement error (or noise)nion each of the values yi. Moreover, let us assume that the true value of yat any position xis given by some function y=f(x;a) that depends on the Munknown parameters a.T h e n yi=f(xi;a)+ni. Our aim is to estimate the values of the parameters afrom the data sample. Bearing in mind the central limit theorem, let us suppose that the niare drawn from a Gaussian distribution with zero mean and no systematic bias. In the most general case the measurement errors nimight notbe independent but described by an N-dimensional multivariate Gaussian with non-trivial covariance matrix N, whose elements Nij=C o v [ ni,nj] we assume to be known. Under these assumptions it follows from (26.148), that the likelihood function is L(x,y;a)=1 (2π)N/2|N|1/2expbracketleftbig −1 2χ2(a)bracketrightbig , 1113 STATISTICS where the quantity denoted by χ2is given by the quadratic form χ2(a)=Nsummationdisplay i,j=1[yi−f(xi;a)](N−1)ij[yj−f(xj;a)] = ( y−f)TN−1(y−f). (27.90) In the last equality, we have rewritten the expression in matrix notation by deﬁning the column vector fwith elements fi=f(xi;a). We note that in the (common) special case in which the measurement errors niareindependent ,t h e i r covariance matrix takes the diagonal form N=d i a g ( σ2 1,σ2 2,...,σ2 N), where σiis the standard deviation of the measurement error ni. In this case, the expression (27.90) for χ2reduces to χ2(a)=Nsummationdisplay i=1bracketleftbiggyi−f(xi;a) σibracketrightbigg2 . The least squares (LS) estimators ˆaLSof the parameter values are deﬁned as those that minimise the value of χ2(a); they are usually determined by solving theMequations ∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=ˆaLS=0 f o r i=1,2,...,M. (27.91) Clearly, if the measurement errors niare indeed Gaussian distributed, as assumed above, then the LS and ML estimators of the parameters acoincide. Because of its relative simplicity, the method of least squares is often applied to cases in which the niare not Gaussian distributed. The resulting estimators ˆaLSarenotthe ML estimators, and the best that can be said in justiﬁcation is that the method isan obviously sensible procedure for parameter estimation that has stood the testof time. Finally, we note that the method of least squares is easily extended to the case in which each measurement y idepends on several variables, which we denote byxi. For example, yimight represent the temperature measured at the (three- dimensional) position xiin a room. In this case, the data is modelled by a function y=f(xi;a), and the remainder of the above discussion carries through unchanged. 27.6.1 Linear least squares We have so far made no restriction on the form of the function f(x;a). It so happens, however, that, for a model in which f(x;a)i sa linear function of the parameters a1,a2,...,a M, one can always obtain analytic expressions for the LS estimators ˆaLSand their variances. The general form of this kind of model is f(x;a)=Msummationdisplay i=1aihi(x), (27.92) 1114 27.6 THE METHOD OF LEAST SQUARES where h1(x),h2(x),...,h M(x) are some set of linearly independent ﬁxed functions ofx,o f t e nc a l l e dt h e basis functions . Note that the functions hi(x) themselves may be highly non-linear functions of x. The ‘linear’ nature of the model (27.92) refers only to its dependence on the parameters ai. Furthermore, in this case, it may be shown that the LS estimators ˆaihave zero bias and are minimum-variance, irrespective of the probability density function from which the measurement errorsn iare drawn. In order to obtain analytic expressions for the LS estimators ˆaLS,i ti sc o n v e n i e n t to write (27.92) in the form f(x;a)=Msummationdisplay j=1Rijaj, (27.93) where Rij=hj(xi) is an element of the response matrix Rof the experiment. The expression for χ2given in (27.90) can then be written, in matrix notation, as χ2(a)=( y−Ra)TN−1(y−Ra). (27.94) The LS estimates of the parameters aare now found, as shown in (27.91), by diﬀerentiating (27.94) with respect to the aiand setting the resulting expressions equal to zero. Denoting by ∇χ2the vector with elements ∂χ2/∂a i, we ﬁnd ∇χ2=−2RTN−1(y−Ra). (27.95) This can be veriﬁed by writing out the expression (27.94) in component form and diﬀerentiating directly.IVerify result (27.95) by formulating the calculation in component form. To make the derivation less cumbersome, let us adopt the summation convention discussed in section 21.1, in which it is understood that any subscript that appears exactly twice in any term of an expression is to be summed over all the values that a subscript in thatposition can take. Thus, writing (27.94) in component form, we have χ 2(a)=(yi−Rikak)(N−1)ij(yj−Rjlal). Diﬀerentiating with respect to apgives ∂χ2 ∂ap=−Rikδkp(N−1)ij(yj−Rjlal)+(yi−Rikak)(N−1)ij(−Rjlδlp) =−Rip(N−1)ij(yj−Rjlal)−(yi−Rikak)(N−1)ijRjp, (27.96) where δijis the Kronecker delta symbol discussed in section 21.1. By swapping the indices iandjin the second term on the RHS of (27.96) and using the fact that the matrix N−1 is symmetric, we obtain ∂χ2 ∂ap=−2Rip(N−1)ij(yj−Rjkak) =−2(RT)pi(N−1)ij(yj−Rjkak). (27.97) If we denote the vector with components ∂χ2/∂a p,p=1,2,...,M ,b y∇χ2and write the RHS of (27.97) in matrix notation, we recover the result (27.95). J 1115 STATISTICS Setting the expression (27.95) equal to zero at a=ˆa, we ﬁnd −2RTN−1y+2RTN−1Rˆa=0. Provided the matrix RTN−1Ris not singular, we may solve this equation for ˆato obtain ˆa=(RTN−1R)−1RTN−1y≡Sy, (27.98) thus deﬁning the M×Nmatrix S. It follows that the LS estimates ˆai,i=1,2,...,M , are linear functions of the original measurements yj,j=1,2,...,N .M o r e o v e r , using the error propagation formula (26.141) derived in subsection 26.12.3, weﬁnd that the covariance matrix of the estimators ˆa iis given by V≡Cov[ˆai,ˆaj]=SNST=(RTN−1R)−1. (27.99) The two equations (27.98) and (27.99) contain the complete method of least squares. In particular, we note that, if one calculates the LS estimates using(27.98) then one has already obtained their covariance matrix (27.99).IProve result (27.99). Using the deﬁnition of Sgiven in (27.98), the covariance matrix (27.99) becomes V=SNST =[ ( RTN−1R)−1RTN−1]N[(RTN−1R)−1RTN−1]T. Using the result ( AB···C)T=CT···BTATfor the transpose of a product of matrices and noting that, for any non-singular matrix, ( A−1)T=(AT)−1we ﬁnd V=(RTN−1R)−1RTN−1N(NT)−1R[(RTN−1R)T]−1 =(RTN−1R)−1RTN−1R(RTN−1R)−1 =(RTN−1R)−1, where we have also used the fact that Nis symmetric and so NT=N. J It is worth noting that one may also write the elements of the (inverse) covariance matrix as (V−1)ij=1 2parenleftbigg∂2χ2 ∂ai∂ajparenrightbigg a=ˆa, which is the same as the Fisher matrix (27.36) in cases where the measurement errors are Gaussian distributed (and so the log-likelihood is ln L=−χ2/2). This p r o v e s ,a tl e a s tf o rt h i sc a s e ,o u re a r l i e rs t a t e m e n tt h a tt h eL Se s t i m a t o r sa r eminimum-variance. In fact, since f(x;a) is linear in the parameters a,o n ec a n write χ 2exactly as χ2(a)=χ2(ˆa)+1 2Msummationdisplay i,j=1parenleftbigg∂2χ2 ∂ai∂ajparenrightbigg a=ˆa(ai−ˆai)(aj−ˆaj), which is quadratic in the parameters ai. Hence the likelihood function L∝ 1116 27.6 THE METHOD OF LEAST SQUARES 0 01 12 23 34 4 5567y x Figure 27.9 A set of data points with error bars indicating the uncertainty σ=0.5o nt h e y-values. The straight line is y=ˆmx+ˆc,w h e r e ˆmandˆcare the least squares estimates of the slope and intercept. exp(−χ2/2) is Gaussian. From the discussions of sections 27.3.6 and 27.5.6, it follows that the ‘surfaces’ χ2(a)=c,w h e r e cis a constant, bound ellipsoidal conﬁdence regions for the parameters ai. The relationship between the value of the constant cand the conﬁdence level is given by (27.39).IAn experiment produces the following data sample pairs (xi,yi): xi:1.85 2 .72 2 .81 3 .06 3 .42 3 .76 4 .31 4 .47 4 .64 4 .99 yi:2.26 3 .10 3 .80 4 .11 4 .74 4 .31 5 .24 4 .03 5 .69 6 .57 where the xi-values are known exactly but each yi-value is measured only to an accuracy ofσ=0.5. Assuming the underlying model for the data to be a straight line y=mx+c, ﬁnd the LS estimates of the slope mand intercept cand quote the standard error on each estimate. The data are plotted in ﬁgure 27.9, together wi th error bars indicating the uncertainty in theyi-values. Our model of the data is a straight line, and so we have f(x;c, m)=c+mx. In the language of (27.92), our basis functions are h1(x)=1a n d h2(x)=xand our model parameters are a1=canda2=m. From (27.93) the elements of the response matrix are Rij=hj(xi), so that R= /0BBB/@1x1 1x2 ...... 1xN /1CCCA, (27.100) where xiare the data values and N= 10 in our case. Further, since the standard deviation on each measurement error is σ, we have N=σ2I,w h e r e Iis the N×Nidentity matrix. Because of this simple form for N, the expression (27.98) for the LS estimates reduces to ˆa=σ2(RTR)−11 σ2RTy=(RTR)−1RTy. (27.101) Note that we cannot expand the inverse in the last line, since Ritself is not square and 1117 STATISTICS hence does not possess an inverse. Inserting the form for Rin (27.100) into the expression (27.101), we ﬁnd/ ˆc ˆm / = /P i1 P ixiP ixi P ix2 i /−1 /P iyiP ixiyi / =1 N(x2−¯x2) / x2−¯x −¯x1 // N¯y Nxy / . We thus obtain the LS estimates ˆm=xy−¯x¯y x2−¯x2and ˆc=x2¯y−¯xxy x2−¯x2=¯y−ˆm¯x, (27.102) where the last expression for ˆcshows that the best-ﬁt line passes through the ‘centre of mass’ ( ¯x,¯y) of the data sample. To ﬁnd the standard errors on our results, we must calculate the covariance matrix of the estimators. This is given by (27.99), which in ourcase reduces to V=σ 2(RTR)−1=σ2 N(x2−¯x2) / x2−¯x −¯x1 / . (27.103) The standard error on each estimator is simply the positive square root of the corresponding diagonal element, i.e. σˆc=√V11andσˆm=√V22, and the covariance of the estimators ˆm andˆcis given by Cov[ ˆc,ˆm]=V12=V21. Inserting the data sample averages and moments into (27.102) and (27.103), we ﬁnd c=ˆc±σˆc=0.40±0.62 and m=ˆm±σˆm=1.11±0.17. The ‘best-ﬁt’ straight line y=ˆmx+ˆcis plotted in ﬁgure 27.9. For comparison, the true values used to create the data were m=1a n d c=1 . J The extension to the ﬁtting the data to a higher-order polynomial, such as f(x;a)=a1+a2x+a3x2, is obvious. Nevertheless, as the order of the polynomial increases the matrix inversions become rather complicated. Indeed, even when the matrices are inverted numerically, the inversion is prone to numerical instabilities. A better approach is to replace the basis functions hm(x)=xm,m=1,2,...,M , with a set of polynomials that are ‘orthogonal over the data’, i.e. such that Nsummationdisplay i=1hl(xi)hm(xi)=0 f o r l/negationslash=m. Such a set of polynomial basis functions can always be found by using the Gram– Schmidt orthogonalisation procedure presented in section 17.1. The details of thisapproach are beyond the scope of our discussion but we note that, in this case, the matrix R TRis diagonal and may be inverted easily. 27.6.2 Non-linear least squares If the function f(x;a)i snotlinear in the parameters athen, in general, it is not possible to obtain an explicit expression for the LS estimates ˆa.I n s t e a d ,o n e must use an iterative (numerical) procedure, which we now outline. In practice, 1118 27.7 HYPOTHESIS TESTING however, such problems are best solved using one of the many commercially available software packages. One begins by making a ﬁrst guess a0for the values of the parameters. At this point in parameter space, the components of the gradient ∇χ2will, not be equal to zero, in general (unless one makes a very lucky guess!). Thus, for at least some values of i, we have ∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=a0/negationslash=0. Our aim is to ﬁnd a small increment δain the values of the parameters, such that ∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=a0+δa=0 f o ra l l i. (27.104) If our ﬁrst guess a0were suﬃciently close to the true (local) minimum of χ2, we could ﬁnd the required increment δaby expanding the LHS of (27.104) as a Taylor series about a=a0, keeping only the zeroth-order and ﬁrst-order terms: ∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=a0+δa≈∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=a0+Msummationdisplay j=1∂2χ2 ∂ai∂ajvextendsinglevextendsinglevextendsinglevextendsingle a=a0δaj. (27.105) Setting this expression to zero, we ﬁnd that the increments δajmay be found by solving the set of Mlinear equations Msummationdisplay j=1∂2χ2 ∂ai∂ajvextendsinglevextendsinglevextendsinglevextendsingle a=a0δaj=−∂χ2 ∂aivextendsinglevextendsinglevextendsinglevextendsingle a=a0. It most cases, however, our ﬁrst guess a0will not be suﬃciently close to the true minimum for (27.105) to be an accurate approximation, and consequently (27.104)will not be satisﬁed. In this case, a 1=a0+δais (hopefully) an improved guess at the parameter values; the whole process is then repeated until convergence isachieved. It is worth noting that, when one is estimating several parameters a,t h e function χ 2(a)m a yb e verycomplicated. In particular, it may possess numerous local extrema. The procedure outlined above will converge to the local extremum ‘nearest’ to the ﬁrst guess a0. Since, in fact,we are interested only in the local minimum that has the absolute lowest value of χ2(a), it is clear that a large part of solving the problem is to make a ‘good’ ﬁrst guess. 27.7 Hypothesis testing So far we have concentrated on using a data sample to obtain a number or a set of numbers. These numbers may be estimated values for the moments or central moments of the population from which the sample was drawn or, more generally,the values of some parameters ain an assumed model for the data. Sometimes, 1119 STATISTICS however, one wishes to use the data to give a ‘yes’ or ‘no’ answer to a particular question. For example, one might wish to know whether some assumed modeldoes, in fact, provide a good ﬁt to the data, or whether two parameters have thesame value. 27.7.1 Simple and composite hypotheses In order to use data to answer questions of this sort, the question must be posed precisely. This is done by ﬁrst asserting that some hypothesis is true. The hypothesis under consideration is traditionally called the null hypothesis and is denoted by H 0. In particular, this usually speciﬁes some form P(x|H0) for the probability density function from which the data xare drawn. If the hypothesis determines the PDF uniquely, then it is said to be a simple hypothesis . If, however, the hypothesis determines the functional form of the PDF but not thevalues of certain parameters aon which it depends then it is called a composite hypothesis . One decides whether to accept orreject the null hypothesis H 0by performing some statistical test , as described below in subsection 27.7.2. In fact, formally one uses a statistical test to decide between the null hypothesis H0and the alternative hypothesis H1. We deﬁne the latter to be the complement H0of the null hypothesis within some restricted hypothesis space known (or assumed) in advance . Hence, rejection of H0implies acceptance of H1, and vice versa. As an example, let us consider the case in which a sample xis drawn from a Gaussian distribution with a known variance σ2but with an unknown mean µ. If one adopts the null hypothesis H0that µ=0w h i c hw ew r i t ea s H0:µ=0 , then the corresponding alternative hypothesis must be H1:µ/negationslash= 0. Note that, in this case, H0is a simple hypothesis whereas H1is a composite hypothesis. If, however, one adopted the null hypothesis H0:µ<0 then the alternative hypothesis would be H1:µ≥0, so that both H0andH1would be composite hypotheses. Very occasionally both H0andH1will be simple hypotheses. In our illustration, this would occur, for example, if one knew in advance that the meanµof the Gaussian distribution were equal to either zero or unity. In this case, if one adopted the null hypothesis H 0:µ= 0 then the alternative hypothesis would beH1:µ=1 . 27.7.2 Statistical tests In our discussion of hypothesis testing we will restrict our attention to cases in which the null hypothesis H0issimple (see above). We begin by constructing a test statistic t(x) from the data sample. Although, in general, the test statistic need not be just a (scalar) number, and could be a multi-dimensional (vector) quantity,we will restrict our attention to the former case. Like any statistic, t(x) will be a 1120 27.7 HYPOTHESIS TESTING tP(t|H0) tcritα t tcritP(t|H1) β Figure 27.10 The sampling distributions P(t|H0)a n d P(t|H1) of a test statistic t. The shaded areas indicate the (one-tailed) regions for which Pr( t>t crit|H0)= αand Pr( t<t crit|H1)=βrespectively. random variable. Moreover, given the simple null hypothesis H0concerning the PDF from which the sample was drawn, we may determine (in principle) thesampling distribution P(t|H 0) of the test statistic. A typical example of such a sampling distribution is shown in ﬁgure 27.10. One deﬁnes for tarejection region containing some fraction αof the total probability. For example, the (one-tailed) rejection region could consist of values of tgreater than some value tcrit,f o r which Pr(t>t crit|H0)=integraldisplay∞ tcritP(t|H0)dt=α; (27.106) this is indicated by the shaded region in the upper half of ﬁgure 27.10. Equally, a (one-tailed) rejection region could consist of values of tless than some value tcrit. Alternatively, one could deﬁne a (two-tailed) rejection region by two values t1andt2such that Pr( t1<t<t 2|H0)=α. In all cases, if the observed value of t lies in the rejection region then H0isrejected atsigniﬁcance level α;o t h e r w i s e H0 isaccepted at this same level. It is clear that there is a probability αof rejecting the null hypothesis H0 even if it is true. This is called an error of the ﬁrst kind .C o n v e r s e l y ,a n error of the second kind occurs when the hypothesis H0i sa c c e p t e de v e nt h o u g hi ti s 1121 STATISTICS false (in which case H1is true). The probability β( s a y )t h a ts u c ha ne r r o rw i l l occur is, in general, diﬃcult to calculate, since the alternative hypothesis H1is often composite. Nevertheless, in the case where H1is a simple hypothesis, it is straightforward (in principle) to calculate β. Denoting the corresponding sampling distribution of tbyP(t|H1), the probability βis the integral of P(t|H1)o v e rt h e complement of the rejection region, called the acceptance region . For example, in the case corresponding to (27.106) this probability is given by β=P r ( t<t crit|H1)=integraldisplaytcrit −∞P(t|H1)dt. This is illustrated in ﬁgure 27.10. The quantity 1 −βis called the power of the statistical test to reject the wrong hypothesis. 27.7.3 The Neyman–Pearson test In the case where H0andH1are both simple hypotheses, the Neyman–Pearson lemma (which we shall not prove) allows one to determine the ‘best’ rejection region and test statistic to use. We consider ﬁrst the choice of rejection region. Even in the general case, in which the test statistic tis a multi-dimensional (vector) quantity, the Neyman– Pearson lemma states that, for a given signiﬁcance level α, the rejection region for H0giving the highest power for the test is the region of t-space for which P(t|H0) P(t|H1)>c , (27.107) where cis some constant determined by the required signiﬁcance level. In the case where the test statistic tis a simple scalar quantity, the Neyman– Pearson lemma is also useful in deciding which such statistic is the ‘best’ inthe sense of having the maximum power for a given signiﬁcance level α.F r o m (27.107), we can see that the best statistic is given by the likelihood ratio t(x)=P(x|H 0) P(x|H1). (27.108) and that the corresponding rejection region for H0is given by t<t crit.I nf a c t , it is clear that any statistic u=f(t) will be equally good, provided that f(t)i sa monotonically increasing function of t. The rejection region is then u<f (tcrit). Alternatively, one may use any test statistic v=g(t)w h e r e g(t) is a monotonically decreasing function of t; in this case the rejection region becomes v>g(tcrit). To construct such statistics, however, one must know P(x|H0)a n d P(x|H1) explicitly, and such cases are rare. 1122 27.7 HYPOTHESIS TESTINGITen independent sample values xi,i=1,2,...,10, are drawn at random from a Gaussian distribution with standard deviation σ=1. The mean µof the distribution is known to equal either zero or unity. The sample values are as follows: 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Test the null hypothesis H0:µ=0at the 10% signiﬁcance level. The restricted nature of the hypothesis space means that our null and alternative hypotheses areH0:µ=0a n d H1:µ= 1 respectively. Since H0andH1are both simple hypotheses, the best test statistic is given by the likelihood ratio (27.108). Thus, denoting the meansbyµ 0andµ1, we have t(x)=exp / −1 2 P i(xi−µ0)2 / exp / −1 2 P i(xi−µ1)2 /=exp / −1 2 P i(x2 i−2µ0xi+µ2 0) / exp / −1 2 P i(x2 i−2µ1xi+µ2 1) / =e x p / (µ0−µ1) P ixi−1 2N(µ2 0−µ2 1) / . Inserting the values, µ0=0 ,a n d µ1= 1, yields t=e x p (−N¯x+1 2N), where ¯xis the sample mean. Since −lntis a monotonically decreasing function of t, however, we may equivalently use as our test statistic v=−1 Nlnt+1 2=¯x, where we have divided by the sample size Nand added1 2for convenience. Thus we may take the sample mean as our test statistic. From (27.13), we know that the samplingdistribution of the sample mean under our null hypothesis H 0is the Gaussian distribution N(µ0,σ2/N), where µ0=0 , σ2=1a n d N= 10. Thus ¯x∼N(0,0.1). Since ¯xis a monotonically decreasing function of t, our best rejection region for a given signiﬁcance αis¯x>¯xcrit,w h e r e ¯xcritdepends on α. Thus, in our case, ¯xcritis given by α=1−Φ /¯xcrit−µ0 σ / =1−Φ(10¯xcrit), where Φ( z) is the cumulative distribution function for the standard Gaussian. For a 10% signiﬁcance level we have α=0.1 and, from table 26.3 in subsection 26.9.1, we ﬁnd ¯xcrit=0.128. Thus the rejection region on ¯xis ¯x>0.128. From the sample, we deduce that ¯x=1.11, and so we can clearly reject the null hypothesis H0:µ= 0 at the 10% signiﬁcance level It can, in fact, be rejected at a much higher signiﬁcance level. As revealed on p. 1081, the data was generated using µ=1 . J 27.7.4 The generalised likelihood-ratio test If the null hypothesis H0or the alternative hypothesis H1(or both) is composite then the corresponding distributions P(x|H0)a n d P(x|H1) are not uniquely de- termined, in general, and so we cannot use the Neyman–Pearson lemma to obtain the ‘best’ test statistic t. Nevertheless, in many cases, there still exists a general procedure for constructing a test statistic twhich has useful properties and which 1123 STATISTICS reduces to the Neyman–Pearson statistic (27.108) in the special case where H0 andH1are both simple hypotheses. Consider the quite general, and commonly occurring, case in which the data sample xis drawn from a population P(x|a) with a known (or as- sumed) functional form but depends on the unknown values of some parametersa 1,a2,...,a M. Moreover, suppose we wish to test the null hypothesis H0that the parameter values alie in some subspace Sof the full parameter space A. In other words, on the basis of the sample xit is desired to test the null hypothesis H0:(a1,a2,...,a Mlies in S) against the alternative hypothesis H1:(a1,a2,...,a Mlies in S), where SisA−S. Since the functional form of the population is known, we may write down the likelihood function L(x;a) for the sample. Ordinarily, the likelihood will have a maximum as the parameters aare varied over the entire parameter space A. This is the usual maximum-likelihood estimate of the parameter values, which we denote by ˆa. If, however, the parameter values are allowed to vary only over the subspace Sthen the likelihood function will be maximised at the point ˆaS, which may or may not coincide with the global maximum ˆa. Now, let us take as our test statistic the generalised likelihood ratio t(x)=L(x;ˆaS) L(x;ˆa), (27.109) where L(x;ˆaS) is the maximum value of the likelihood function in the subspace SandL(x;ˆa) is its maximum value in the entire parameter space A.I ti sc l e a r thattis a function of the sample values only and must lie between 0 and 1. We will concentrate on the special case where H0is the simple hypothesis H0:a=a0. The subspace Sthen consists of only the single point a0. Thus (27.109) becomes t(x)=L(x;ˆa0) L(x;ˆa), (27.110) and the sampling distribution P(t|H0) can be determined (in principle). As in the previous subsection, the best rejection region for a given signiﬁcance αis simply t<t crit, where the value tcritdepends on α. Moreover, as before, an equivalent procedure is to use as a test statistic u=f(t), where f(t) is any monotonically increasing function of t; the corresponding rejection region is then u<f (tcrit). Similarly, one may use a test statistic v=g(t), where g(t) is any monotonically decreasing function of t; the rejection region then becomes v>g(tcrit). Finally, we note that if H1is also a simple hypothesis H1:a=a1, then (27.110) reduces to the Neyman–Pearson test statistic (27.108). 1124 27.7 HYPOTHESIS TESTINGITen independent sample values xi(i=1,2,...,10)are drawn at random from a Gaussian distribution with standard deviation σ=1. The sample values are as follows: 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Test the null hypothesis H0:µ=0at the 10% signiﬁcance level. We must test the (simple) null hypothesis H0:µ= 0 against the (composite) alternative hypothesis H1:µ/negationslash= 0. Thus, the subspace Sis the single point µ=0 ,w h e r e a s Ais the entire µ-axis. The likelihood function is L(x;µ)=1 (2π)N/2exp / −1 2 P i(xi−µ)2 / , which has its global maximum at µ=¯x. The test statistic tis then given by t(x)=L(x;0) L(x;¯x)=exp / −1 2 P ix2 i / exp / −1 2 P i(xi−¯x)2 /=e x p /; −1 2N¯x2 / . It is in fact more convenient to consider the test statistic v=−2lnt=N¯x2. Since−2l ntis a monotonically decreasing function of t, the rejection region now becomes v>v crit,w h e r eZ∞ vcritP(v|H0)dv=α, (27.111) αbeing the signiﬁcance level of the test. Thus it only remains to determine the sampling distribution P(v|H0). Under the null hypothesis H0, we expect ¯xto be Gaussian distributed, with mean zero and variance 1 /N. Thus, from subsection 26.9.4, vwill follow a chi-squared distribution of order 1. Substituting the appropriate form for P(v|H0) in (27.111) and setting α=0.1, we ﬁnd by numerical integration (or from tables of the cumulative chi- squared distribution) that vcrit=N¯x2 crit=2.71. Since N= 10, the rejection region on ¯xat the 10% signiﬁcance level is thus ¯x<−0.52 and ¯x>0.52. As noted before, for this sample ¯x=1.11, and so we may reject the null hypothesis H0:µ= 0 at the 10% signiﬁcance level. J The above example illustrates the general situation that if the maximum- likelihood estimates ˆaof the parameters fall in or near the subspace Sthen the sample will be considered consistent with H0and the value of twill be near unity. If ˆais distant from Sthen the sample will not be in accord with H0and ordinarily twill have a small (positive) value. It is clear that in order to prescribe the rejection region for t, or for a related statistic uorv, it is necessary to know the sampling distribution P(t|H0). IfH0 is simple then one can in principle determine P(t|H0), although this may prove diﬃcult in practice. Moreover, if H0is composite, then it may not be possible to obtain P(t|H0), even in principle. Nevertheless, a useful approximate form for P(t|H0) exists in the large-sample limit. Consider the null hypothesis H0:(a1=a0 1,a2=a0 2,...,a R=a0 R),where R≤M 1125 STATISTICS and the a0 iare ﬁxed numbers. (In fact, we may ﬁx the values of any subset containing Rof the Mparameters.) If H0is true then it follows from our discussion in subsection 27.5.6 (although we shall not prove it) that, when thesample size Nis large, the quantity −2l ntfollows approximately a chi-squared distribution of order R. 27.7.5 Student’s t-test Student’s t-test is just a special case of the generalised likelihood ratio test applied to a sample x 1,x2,...,x Ndrawn independently from a Gaussian distribution for which boththe mean µand variance σ2are unknown, and for which one wishes to distinguish between the hypotheses H0:µ=µ0,0<σ2<∞,and H1:µ/negationslash=µ0,0<σ2<∞, where µ0is a given number. Here, the parameter space Ais the half-plane −∞<µ<∞,0<σ2<∞, whereas the subspace Scharacterised by the null hypothesis H0is the line µ=µ0,0<σ2<∞. The likelihood function for this situation is given by L(x;µ, σ2)=1 (2πσ2)N/2expbracketleftbigg −summationtext i(xi−µ)2 2σ2bracketrightbigg . On the one hand, as shown in subsection 27.5.1, the values of µandσ2that maximise LinAareµ=¯xandσ2=s2,w h e r e ¯xis the sample mean and s2is the sample variance. On the other hand, to maximise Lin the subspace Swe set µ=µ0, and the only remaining parameter is σ2; the value of σ2that maximises Lis then easily found to be hatwideσ2=1 NNsummationdisplay i=1(xi−µ0)2. To retain, in due course, the standard notation for Student’s t-test, in this section we will denote the generalised likelihood ratio by λ(rather than t); it is thus given by λ(x)=L(x;µ0,hatwideσ2) L(x;¯x, s2) =[(2π/N)summationtext i(xi−µ0)2]−N/2exp(−N/2) [(2π/N)summationtext i(xi−¯x)2]−N/2exp(−N/2)=bracketleftbiggsummationtext i(xi−¯x)2 summationtext i(xi−µ0)2bracketrightbiggN/2 .(27.112) Normally, our next step would be to ﬁnd the sampling distribution of λunder the assumption that H0were true. It is more conventional, however, to work in terms of a related test statistic t, which was ﬁrst devised by William Gossett, who wrote under the pen name of ‘Student’. 1126 27.7 HYPOTHESIS TESTING The sum of squares in the denominator of (27.112) may be put into the form summationtext i(xi−µ0)2=N(¯x−µ0)2+summationtext i(xi−¯x)2. Thus, on dividing the numerator and denominator in (27.112) bysummationtext i(xi−¯x)2and rearranging, the generalised likelihood ratio λcan be written λ=parenleftbigg 1+t2 N−1parenrightbigg−N/2 , where we have deﬁned the new variable t=¯x−µ0 s/√ N−1. (27.113) Since t2is a monotonically decreasing function of λ, the corresponding rejection region is t2>c,w h e r e cis a positive constant depending on the required signiﬁcance level α. It is conventional, however, to use titself as our test statistic, in which case our rejection region becomes two-tailed and is given by t<−tcrit and t>t crit, (27.114) where tcritis the positive square root of the constant c. The deﬁnition (27.113) and the rejection region (27.114) form the basis of Student’s t-test. It only remains to determine the sampling distribution P(t|H0). At the outset, it is worth noting that if we write the expression (27.113) for t in terms of the standard estimator ˆσ=radicalbig Ns2/(N−1) of the standard deviation then we obtain t=¯x−µ0 ˆσ/√ N. (27.115) If, in fact, we knew the true value of σand used it in this expression for tthen it is clear from our discussion in section 27.3 that twould follow a Gaussian distribution with mean 0 and variance 1, i.e. t∼N(0,1). When σis not known, however, we have to use our estimate ˆσin (27.115), with the result that tis no longer distributed as the standard Gaussian. As one might expect fromthe central limit theorem, however, the distribution of tdoes tend towards the standard Gaussian for large values of N. As noted earlier, the exact distribution of t, valid for any value of N, was ﬁrst discovered by William Gossett. From (27.35), if the hypothesis H 0is true then the joint sampling distribution of ¯xandsis given by P(¯x, s|H0)=CsN−2expparenleftbigg −Ns2 2σ2parenrightbigg expbracketleftbigg −N(¯x−µ)2 2σ2bracketrightbigg , (27.116) where Cis a normalisation constant. We can use this result to obtain the joint sampling distribution of sandtby demanding that P(¯x, s|H0)d¯xd s=P(t, s|H0)dt ds. 1127 STATISTICS Using (27.113) to substitute for ¯x−µin (27.116), and noting that d¯x= (s/√ N−1)dt, we ﬁnd P(¯x, s|H0)d¯xd s=AsN−1expbracketleftbigg −Ns2 2σ2parenleftbigg 1+t2 N−1parenrightbiggbracketrightbigg dt ds, where Ais another normalisation constant. In order to obtain the sampling distribution of talone, we must integrate P(t, s|H0)w i t hr e s p e c tt o sover its a l l o w e dr a n g e ,f r o m0t o ∞. Thus, the required distribution of talone is given by P(t|H0)=integraldisplay∞ 0P(t, s|H0)ds=Aintegraldisplay∞ 0sN−1expbracketleftbigg −Ns2 2σ2parenleftbigg 1+t2 N−1parenrightbiggbracketrightbigg ds. (27.117) To perform this integration, we make the change of variable y=s{1+[t2/(N− 1)]}1/2, which on substitution into (27.117) yields P(t|H0)=Aparenleftbigg 1+t2 N−1parenrightbigg−N/2integraldisplay∞ 0yN−1expparenleftbigg −Ny2 2σ2parenrightbigg dy. Since the integral over ydoes not depend on t, it is simply a constant. We thus ﬁnd that that the sampling distribution of the variable tis P(t|H0)=1√(N−1)πΓparenleftbig1 2Nparenrightbig Γparenleftbig1 2(N−1)parenrightbigparenleftbigg 1+t2 N−1parenrightbigg−N/2 , (27.118) w h e r ew eh a v eu s e dt h ec o n d i t i o nintegraltext∞ −∞P(t|H0)dt= 1 to determine the normali- sation constant (see exercise 27.18). The distribution (27.118) is called Student’s t-distribution with N−1degrees of freedom . A plot of Student’s t-distribution is shown in ﬁgure 27.11 for various values of N. For comparison, we also plot the standard Gaussian distribution, to which the t-distribution tends for large N. As is clear from the ﬁgure, the t-distribution is symmetric about t= 0. In table 27.2 we list some critical points of the cumulative probability function Cn(t)o ft h e t-distribution, which is deﬁned by Cn(t)=integraldisplayt −∞P(t/prime|H0)dt/prime, where n=N−1 is the number of degrees of freedom. Clearly, Cn(t) is analogous to the cumulative probability function Φ( z) of the Gaussian distribution, discussed in subsection 26.9.1. For comparison purposes, we also list the critical points ofΦ(z), which corresponds to the t-distribution for N=∞. 1128 27.7 HYPOTHESIS TESTING 0 00.10.20.30.40.5 −4−3−2−11 2 3 4tP(t|H0) N=2N=3N=5N=1 0 Figure 27.11 Student’s t-distribution for various values of N. The broken curve shows the standard Gaussian distribution for comparison.ITen independent sample values xi,i=1,2,...,10, are drawn at random from a Gaussian distribution with unknown mean µand unknown standard deviation σ. The sample values are as follows: 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Test the null hypothesis, H0:µ=0at the 10% signiﬁcance level. For our null hypothesis µ0=0 .S i n c ef o rt h i ss a m p l e ¯x=1.11,s=1.01 and N= 10, it follows from (27.113) that t=¯x s/√ N−1=3.33. The rejection region for tis given by (27.114) where tcritis such that CN−1(tcrit)=1−α/2, andαis the required signiﬁcance of the test. In our case α=0.1a n d N= 10, and from table 27.2 we ﬁnd tcrit=1.83. Thus our rejection region for H0at the 10% signiﬁcance level is t<−1.83 and t>1.83. For our sample t=3.30 and so we can clearly reject the null hypothesis H0:µ=0a tt h i s level. J It is worth noting the connection between the t-test and the classical conﬁdence interval on the mean µ. The central conﬁdence interval on µat the conﬁdence level 1−α, is the set of values for which −tcrit<¯x−µ s/√ N−1<tcrit, 1129 STATISTICS Cn(t)0.5 0.6 0.7 0.8 0.9 0.950 0.975 0.990 0.995 0.999 n=1 0.00 0.33 0.73 1.38 3.08 6.31 12.7 31.8 63.7 318.3 2 0.00 0.29 0.62 1.06 1.89 2.92 4.30 6.97 9.93 22.3 3 0.00 0.28 0.58 0.98 1.64 2.35 3.18 4.54 5.84 10.2 4 0.00 0.27 0.57 0.94 1.53 2.13 2.78 3.75 4.60 7.17 5 0.00 0.27 0.56 0.92 1.48 2.02 2.57 3.37 4.03 5.89 6 0.00 0.27 0.55 0.91 1.44 1.94 2.45 3.14 3.71 5.21 7 0.00 0.26 0.55 0.90 1.42 1.90 2.37 3.00 3.50 4.79 8 0.00 0.26 0.55 0.89 1.40 1.86 2.31 2.90 3.36 4.50 9 0.00 0.26 0.54 0.88 1.38 1.83 2.26 2.82 3.25 4.30 10 0.00 0.26 0.54 0.88 1.37 1.81 2.23 2.76 3.17 4.14 11 0.00 0.26 0.54 0.88 1.36 1.80 2.20 2.72 3.11 4.03 12 0.00 0.26 0.54 0.87 1.36 1.78 2.18 2.68 3.06 3.93 13 0.00 0.26 0.54 0.87 1.35 1.77 2.16 2.65 3.01 3.85 14 0.00 0.26 0.54 0.87 1.35 1.76 2.15 2.62 2.98 3.79 15 0.00 0.26 0.54 0.87 1.34 1.75 2.13 2.60 2.95 3.73 16 0.00 0.26 0.54 0.87 1.34 1.75 2.12 2.58 2.92 3.69 17 0.00 0.26 0.53 0.86 1.33 1.74 2.11 2.57 2.90 3.65 18 0.00 0.26 0.53 0.86 1.33 1.73 2.10 2.55 2.88 3.61 19 0.00 0.26 0.53 0.86 1.33 1.73 2.09 2.54 2.86 3.58 20 0.00 0.26 0.53 0.86 1.33 1.73 2.09 2.53 2.85 3.55 25 0.00 0.26 0.53 0.86 1.32 1.71 2.06 2.49 2.79 3.46 30 0.00 0.26 0.53 0.85 1.31 1.70 2.04 2.46 2.75 3.39 40 0.00 0.26 0.53 0.85 1.30 1.68 2.02 2.42 2.70 3.31 50 0.00 0.26 0.53 0.85 1.30 1.68 2.01 2.40 2.68 3.26 100 0.00 0.25 0.53 0.85 1.29 1.66 1.98 2.37 2.63 3.17 200 0.00 0.25 0.53 0.84 1.29 1.65 1.97 2.35 2.60 3.13 ∞ 0.00 0.25 0.52 0.84 1.28 1.65 1.96 2.33 2.58 3.09 Table 27.2 The conﬁdence limits tof the cumulative probability function Cn(t) for Student’s t-distribution with ndegrees of freedom. For example, C5(0.92) = 0 .8. The n=∞row is also the corresponding result for the standard Gaussian distribution. where tcritsatisﬁes CN−1(tcrit)=α/2. Thus the required conﬁdence interval is ¯x−tcrits√ N−1<µ< ¯x+tcrits√ N−1. Hence, in the above example, the 90% classical central conﬁdence interval on µ is 0.49<µ< 1.73. Thet-distribution may also be used to compare diﬀerent samples from Gaussian distributions. In particular, let us consider the case where we have two independent 1130 27.7 HYPOTHESIS TESTING samples of sizes N1andN2, drawn respectively from Gaussian distributions with a common variance σ2but with possibly diﬀerent means µ1andµ2.O n et h eb a s i s of the samples, one wishes to distinguish between the hypotheses H0:µ1=µ2,0<σ2<∞ and H1:µ1/negationslash=µ2,0<σ2<∞. In other words, we wish to test the null hypothesis that the samples are drawn from populations having the same mean. Suppose that the measured samplemeans and standard deviations are ¯x 1,¯x2ands1,s2respectively. In an analogous way to that presented above, one may show that the generalised likelihood ratiocan be written as λ=parenleftbigg 1+t 2 N1+N2−2parenrightbigg−(N1+N2)/2 . In this case, the variable tis given by t=¯w−ω ˆσparenleftbiggN1N2 N1+N2parenrightbigg1/2 , (27.119) where ¯w=¯x1−¯x2,ω=µ1−µ2and ˆσ=bracketleftbiggN1s2 1+N2s2 2 N1+N2−2bracketrightbigg1/2 . It is straightforward (albeit with complicated algebra) to show that the variable t in (27.119) follows Student’s t-distribution with N1+N2−2 degrees of freedom, and so we may use an appropriate form of Student’s t-test to investigate the null hypothesis H0:µ1=µ2(or equivalently H0:ω= 0). As above, the t-test can be used to place a conﬁdence interval on ω=µ1−µ2.ISuppose that two classes of students take the same mathematics examination and the following percentage marks are obtained: C l a s s 1 :6 66 23 45 57 78 05 56 06 94 75 0 C l a s s 2 :6 49 07 65 68 17 27 0 Assuming that the two sets of examinations marks are drawn from Gaussian distributions with a common variance, test the hypothesis H0:µ1=µ2at the 5% signiﬁcance level. Use your result to obtain the 95% classical central conﬁdence interval on ω=µ1−µ2. We begin by calculating the mean and standard deviation of each sample. The number of values in each sample is N1=1 1a n d N2= 7 respectively, and we ﬁnd ¯x1=5 9.5,s1=1 2.8a n d ¯x2=7 2.7,s2=1 0.3, leading to ¯w=¯x1−¯x2=−13.2a n d ˆσ=1 2 .6. Setting ω= 0 in (27.119), we thus ﬁnd t=−2.17. The rejection region for H0is given by (27.114), where tcritsatisﬁes CN1+N2−2(tcrit)=1−α/2, (27.120) 1131 STATISTICS where αis the required signiﬁcance level of the test. In our case we set α=0.05, and from table 27.2 with n= 16 we ﬁnd that tcrit=2.12. The rejection region is therefore t<−2.12 and t>2.12. Since t=−2.17 for our samples, we can reject the null hypothesis H0:µ1=µ2, although only by a small margin. (Indeed, it is easily shown that one cannot reject H0at the 2% signiﬁcance level). The 95% central conﬁdence interval on ω=µ1−µ2is given by ¯w−ˆσtcrit /N1+N2 N1N2 /1/2 <ω< ¯w+ˆσtcrit /N1+N2 N1N2 /1/2 , where tcritis given by (27.120). Thus, we ﬁnd −26.1<ω<−0.28, which, as expected, does not (quite) contain ω=0 . J In order to apply Student’s t-test in the above example, we had to make the assumption that the samples were drawn from Gaussian distributions possessing acommon variance, which is clearly unjustiﬁed ap r i o r i . We can, however, perform another test on the data to investigate whether the additional hypothesis σ 2 1=σ2 2 is reasonable; this test is discussed in the next subsection. If this additional test shows that the hypothesis σ2 1=σ2 2may be accepted (at some suitable signiﬁcance level), then we may indeed use the analysis in the above example to infer that the null hypothesis H0:µ1=µ2may be rejected at the 5% signiﬁcance level. If, however, we ﬁnd that the additional hypothesis σ2 1=σ2 2must be rejected, then we can only infer from the above example that the hypothesis that the twosamples were drawn from the same Gaussian distribution may be rejected at the5% signiﬁcance level. Throughout the above discussion, we have assumed that samples are drawn from a Gaussian distribution. Although this is true for many random variables, in practice it is usually impossible to know ap r i o r i whether this is case. It can be shown, however, that Student’s t-test remains reasonably accurate even if the sampled distribution(s) diﬀer considerably from a Gaussian. Indeed, for sampleddistributions that diﬀer only slightly from a Gaussian form, the accuracy ofthe test is remarkably good. Nevertheless, when applying the t-test, it is always important to remember that the assumption of a Gaussian parent population is central to the method. 27.7.6 Fisher’s F-test Having concentrated on tests for the mean µof a Gaussian distribution, we now consider tests for its standard deviation σ. Before discussing Fisher’s F-test for comparing the standard deviations of two samples, we begin by considering the case when an independent sample x 1,x2,...,x Nis drawn from a Gaussian distribution with unknown µandσ, and we wish to distinguish between the two 1132 27.7 HYPOTHESIS TESTING 0 00.050.10 10 20 30 40λ(u) uλcrit ab Figure 27.12 The sampling distribution P(u|H0)f o r N= 10; this is a chi- squared distribution for N−1 degrees of freedom. hypotheses H0:σ2=σ2 0,−∞<µ<∞ and H1:σ2/negationslash=σ2 0,−∞<µ<∞, where σ2 0is a given number. Here, the parameter space Ais the half-plane −∞<µ<∞,0<σ2<∞, whereas the subspace Scharacterised by the null hypothesis H0is the line σ2=σ2 0,−∞<µ<∞. The likelihood function for this situation is given by L(x;µ, σ2)=1 (2πσ2)N/2expbracketleftbigg −summationtext i(xi−µ)2 2σ2bracketrightbigg . The maximum of LinAoccurs at µ=¯xandσ2=s2, whereas the maximum of LinSis at µ=¯xandσ2=σ2 0. Thus, the generalised likelihood ratio is given by λ(x)=L(x;¯x, σ2 0) L(x;¯x, s2)=parenleftBigu NparenrightBigN/2 expbracketleftbig −1 2(u−N)bracketrightbig , where we have introduced the variable u=Ns2 σ2 0=summationtext i(xi−¯x)2 σ2 0. (27.121) An example of this distribution is plotted in ﬁgure 27.12 for N= 10. From the ﬁgure, we see that the rejection region λ<λ critcorresponds to a two-tailed rejection region on ugiven by 0<u<a and b<u<∞, where aandbare such that λcrit(a)=λcrit(b), as shown in ﬁgure 27.12. In practice, 1133 STATISTICS however, it is diﬃcult to determine aandbfor a given signiﬁcance level α,s oa slightly diﬀerent rejection region, which we now describe, is usually adopted. The sampling distribution P(u|H0) may be found straightforwardly from the sampling distribution of sgiven in (27.35). Let us ﬁrst determine P(s2|H0)b y demanding that P(s|H0)ds=P(s2|H0)d(s2), from which we ﬁnd P(s2|H0)=P(s|H0) 2s=(N/2σ2 0)(N−1)/2 Γparenleftbig1 2(N−1)parenrightbig(s2)(N−3)/2expparenleftbigg −Ns2 2σ2 0parenrightbigg . (27.122) Thus, the sampling distribution of u=Ns2/σ2 0is given by P(u|H0)=1 2(N−1)/2Γparenleftbig1 2(N−1)parenrightbigu(N−3)/2expparenleftbig −1 2uparenrightbig . We note, in passing, that the distribution of uis precisely that of an ( N−1) th- order chi-squared variable (see subsection 26.9.4), i.e. u∼χ2 N−1. Although it does not give quite the best test, one then takes the rejection region to be 0<u<a and b<u<∞, with aandbchosen such that the two tails have equal areas ; the advantage of this choice is that tabulations of the chi-squared distribution make the size of this region relatively easy to estimate. Thus, for a given signiﬁcance level α, we have integraldisplaya 0P(u|H0)du=α/2a n dintegraldisplay∞ bP(u|H0)du=α/2.ITen independent sample values xi,i=1,2,...,10, are drawn at random from a Gaussian distribution with unknown mean µand standard deviation σ. The sample values are as follows: 2.22 2 .56 1 .07 0 .24 0 .18 0 .95 0 .73−0.79 2 .09 1 .81 Test the null hypothesis H0:σ2=2at the 10% signiﬁcance level. For our null hypothesis σ2 0=2 .S i n c ef o rt h i ss a m p l e s=1.01 and N= 10, from (27.121) we have u=5.10. For α=0.1 we ﬁnd, either numerically or using tables, that a=3.30 andb=1 6.92. Thus, our rejection region is 0<u< 3.33 and 16 .92<u<∞. The value u=5.10 from our sample does not lie in the rejection region, and so we cannot reject the null hypothesis H0:σ2=2 . J 1134 27.7 HYPOTHESIS TESTING We now turn to Fisher’s F-test. Let us suppose that two independent samples of sizes N1and N2are drawn from Gaussian distributions with means and variances µ1,σ2 1andµ2,σ2 2respectively, and we wish to distinguish between the two hypotheses H0:σ2 1=σ2 2and H1:σ2 1/negationslash=σ2 2. In this case, the generalised likelihood ratio is found to be λ=(N1+N2)(N1+N2)/2 NN1/2 1NN2/2 2bracketleftbig F(N1−1)/(N2−1)bracketrightbigN1/2 bracketleftbig 1+F(N1−1)/(N2−1)bracketrightbig(N1+N2)/2, where Fis given by the variance ratio F=N1s2 1/(N1−1) N2s2 2/(N2−1)≡u2 v2(27.123) ands1ands2are the standard deviations of the two samples. On plotting λas a function of F, it is apparent that the rejection region λ<λ critcorresponds to a two-tailed test on F. Nevertheless, as will shall see below, by deﬁning the fraction (27.123) appropriately, it is customary to make a one-tailed test on F. The distribution of Fmay be obtained in a reasonably straightforward manner by making use of the distribution of the sample variance s2given in (27.122). Under our null hypothesis H0, the two Gaussian distributions share a common variance, which we denote by σ2. Changing the variable in (27.122) from s2tou2 we ﬁnd that u2has the sampling distribution P(u2|H0)=parenleftbiggN−1 2σ2parenrightbigg(N−1)/21 Γparenleftbig1 2(N−1)parenrightbig(u2)(N−3)/2expbracketleftbigg −(N−1)u2 2σ2bracketrightbigg . Since u2andv2are independent, their joint distribution is simply the product of their individual distributions and is given by P(u2|H0)P(v2|H0)=A(u2)(N1−3)/2(v2)(N2−3)/2expbracketleftbigg −(N1−1)u2+(N2−1)v2 2σ2bracketrightbigg , where the constant Ais given by A=(N1−1)(N1−1)/2(N2−1)(N2−1)/2 2(N1+N2−2)/2σ(N1+N2−2)Γparenleftbig1 2(N1−1)parenrightbig Γparenleftbig1 2(N2−1)parenrightbig. (27.124) Now, for ﬁxed vwe have u2=Fv2andd(u2)=v2dF. Thus, the joint sampling 1135 STATISTICS distribution P(v2,F|H0) is obtained by requiring that P(v2,F|H0)d(v2)dF=P(u2|H0)P(v2|H0)d(u2)d(v2). (27.125) In order to ﬁnd the distribution of Falone, we now integrate P(v2,F|H0) with respect to v2from 0 to∞,f r o mw h i c hw eo b t a i n P(F|H0)= parenleftbiggN1−1 N2−1parenrightbigg(N1−1)/21 Bparenleftbig1 2(N1−1),1 2(N2−1)parenrightbigF(N1−3)/2parenleftbigg 1+N1−1 N2−1Fparenrightbigg−(N1+N2−2)/2 , (27.126) where Bparenleftbig1 2(N1−1),1 2(N2−1)parenrightbig is the beta function deﬁned in the Appendix. P(F|H0) is called the F-distribution (or occasionally the Fisher distribution ) with (N1−1,N2−1) degrees of freedom..IEvaluate the integral R∞ 0P(v2,F|H0)d(v2)to obtain result (27.126). From (27.125), we have P(F|H0)=AF(N1−3)/2 Z∞ 0(v2)(N1+N2−4)/2exp / −[(N1−1)F+(N2−1)]v2 2σ2 / d(v2). Making the substitution x=[ (N1−1)F+(N2−1)]v2/(2σ2), we obtain P(F|H0)=A /2σ2 (N1−1)F+(N2−1) /(N1+N2−2)/2 F(N1−3)/2 Z∞ 0x(N1+N2−4)/2e−xdx =A /2σ2 (N1−1)F+(N2−1) /(N1+N2−2)/2 F(N1−3)/2Γ /;1 2(N1+N2−2) / , where in the last line we have used the deﬁnition of the gamma function given in the Appendix. Using the further result (A11), which expresses the beta function in terms ofthe gamma function, and the expression for Agiven in (27.124), we see that P(F|H 0)i s indeed given by (27.126). J As it does not matter whether the ratio Fgiven in (27.123) is deﬁned as u2/v2 or as v2/u2, it is conventional to put the larger sample variance on the top, so thatFis always greater than or equal to unity. A large value of Findicates that the sample variances u2andv2are very diﬀerent whereas a value of Fclose to unity means that they are very similar. Therefore, for a given signiﬁcance α,i ti s 1136 27.7 HYPOTHESIS TESTING Cn1,n2(F)n1= 1 2345678 n2=1 161 200 216 225 230 234 237 239 2 18.5 19.0 19.2 19.2 19.3 19.3 19.4 19.4 3 10.1 9.55 9.28 9.12 9.01 8.94 8.89 8.85 4 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 50 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 100 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 ∞ 3.84 3.00 2.60 2.37 2.21 2.10 2.01 1.94 n1= 9 10 20 30 40 50 100 ∞ n2=1 241 242 248 250 251 252 253 254 2 19.4 19.4 19.4 19.5 19.5 19.5 19.5 19.5 3 8.81 8.79 8.66 8.62 8.59 8.58 8.55 8.53 4 6.00 5.96 5.80 5.75 5.72 5.70 5.66 5.63 5 4.77 4.74 4.56 4.50 4.46 4.44 4.41 4.37 6 4.10 4.06 3.87 3.81 3.77 3.75 3.71 3.67 7 3.68 3.64 3.44 3.38 3.34 3.32 3.27 3.23 8 3.39 3.35 3.15 3.08 3.04 3.02 2.97 2.93 9 3.18 3.14 2.94 2.86 2.83 2.80 2.76 2.71 10 3.02 2.98 2.77 2.70 2.66 2.64 2.59 2.54 20 2.39 2.35 2.12 2.04 1.99 1.97 1.91 1.84 30 2.21 2.16 1.93 2.69 1.79 1.76 1.70 1.62 40 2.12 2.08 1.84 1.74 1.69 1.66 1.59 1.51 50 2.07 2.03 1.78 1.69 1.63 1.60 1.52 1.44 100 1.97 1.93 1.68 1.57 1.52 1.48 1.39 1.28 ∞ 1.88 1.83 1.57 1.46 1.39 1.35 1.24 1.00 Table 27.3 Values of Ffor which the cumulative probability function Cn1,n2(F) of the F-distribution with ( n1,n2) degrees of freedom has the value 0.95. For example, for n1=1 0a n d n2=6 , Cn1,n2(4.06) = 0 .95. customary to deﬁne the rejection region on FasF>F crit,w h e r e Cn1,n2(Fcrit)=integraldisplayFcrit 1P(F|H0)dF=α, and n1=N1−1a n d n2=N2−1 are the numbers of degrees of freedom. Table 27.3 lists values of Fcritcorresponding to the 5% signiﬁcance level (i.e. α=0.05) for various values of n1andn2. 1137 STATISTICSISuppose that two classes of students take the same mathematics examination and the following percentage marks are obtained: C l a s s 1 :6 66 23 45 57 78 05 56 06 94 75 0 C l a s s 2 :6 49 07 65 68 17 27 0 Assuming that the two sets of examinations marks are drawn from Gaussian distributions, test the hypothesis H0:σ2 1=σ2 2at the 5% signiﬁcance level. The variances of the two samples are s2 1=( 1 2 .8)2ands2 2=( 1 0 .3)2and the sample sizes areN1=1 1a n d N2= 7. Thus, we have u2=N1s2 1 N1−1= 180 .2a n d v2=N2s2 2 N2−1= 123 .8, where we have taken u2to be the larger value. Thus, F=u2/v2=1.46 to two decimal places. Since the ﬁrst sample contains eleven values and the second contains seven values, we take n1=1 0a n d n2= 6. Consulting table 27.3, we see that, at the 5% signiﬁcance level, Fcrit=4.06. Since our value lies comfortably below this, we conclude that there is no statistical evidence for rejecting the hypothesis that the two samples were drawn fromGaussian distributions with a common variance.J It is also common to deﬁne the variable z=1 2lnF, the distribution of which can be found straightfowardly from (27.126). This is a useful change of variablesince it can be shown that, for large values of n 1and n2, the variable zis distributed approximately as a Gaussian with mean1 2(n−1 2−n−1 1) and variance 1 2(n−1 2+n−1 1). 27.7.7 Goodness of ﬁt in least squares problems We conclude our discussion of hypothesis testing with an example of a goodness of ﬁt test. In section 27.6, we discussed the use of the method of least squares inestimating the best-ﬁt values of a set of parameters ain a given model y=f(x;a) for a data set( x i,yi),i=1,2,...,N . We have not addressed, however, the question of whether the best-ﬁt model y=f(x;ˆa) does, in fact, provide a good ﬁt of the data. In other words, we have not considered thus far how to verify that thefunctional form fof our assumed model is indeed correct. In the language of hypothesis testing, we wish to distinguish between the two hypotheses H 0: model is correct and H1: model is incorrect . Given the vague nature of the alternative hypothesis H1, we clearly cannot use the generalised likelihood-ratio test. Nevertheless, it is still possible to test the null hypothesis H0at a given signiﬁcance level α. The least squares estimates of the parameters ˆa1,ˆa2,...,ˆaM, as discussed in section 27.6, are those values that minimise the quantity χ2(a)=Nsummationdisplay i,j=1[yi−f(xi;a)](N−1)ij[yj−f(xj;a)] = ( y−f)TN−1(y−f). 1138 27.7 HYPOTHESIS TESTING In the last equality, we rewrote the expression in matrix notation by deﬁning the column vector fwith elements fi=f(xi;a). The value χ2(ˆa) at this minimum can be used as a statistic to test the null hypothesis H0, as follows. The Nquantities yi−f(xi;a) are Gaussian distributed. However, provided the function f(xj;a)i s linear in the parameters a, the equations (27.98) that determine the least squares estimate ˆaconstitute a set of Mlinear constraints on these Nquantities. Thus, as discussed in subsection 26.15.2, the sampling distribution of the quantity χ2(ˆa) will be a chi-squared distribution with N−Mdegrees of freedom (d.o.f), which has the expectation value and variance E[χ2(ˆa)] =N−M and V[χ2(ˆa)] = 2( N−M). Thus we would expect the value of χ2(ˆa) to lie typically in the range ( N−M)±√2(N−M). A value lying outside this range may suggest that the assumed model for the data is incorrect. A very small value of χ2(ˆa) is usually an indication that the model has too many free parameters and has ‘over-ﬁtted’ the data. Morecommonly, the assumed model is simply incorrect, and this usually results in avalue of χ 2(ˆa) that is larger than expected. One can choose to perform either a one-tailed or a two-tailed test on the value of χ2(ˆa). It is usual, for a given signiﬁcance level α, to deﬁne the one-tailed rejection region to be χ2(ˆa)>c,w h e r et h ec o n s t a n t csatisﬁes integraldisplay∞ cP(χ2 n)dχ2 n=α (27.127) andP(χ2 n) is the PDF of the chi-squared distributiom with n=N−Mdegrees of freedom (see subsection 26.9.4).IAn experiment produces the following data sample pairs (xi,yi): xi:1.85 2 .72 2 .81 3 .06 3 .42 3 .76 4 .31 4 .47 4 .64 4 .99 yi:2.26 3 .10 3 .80 4 .11 4 .74 4 .31 5 .24 4 .03 5 .69 6 .57 where the xi-values are known exactly but each yi-value is measured only to an accuracy ofσ=0.5. At the one-tailed 5% signiﬁcance level, tes t the null hypothesis H0that the underlying model for the data is a straight line y=mx+c. These data are the same as those investigated in section 27.6 and plotted in ﬁgure 27.9. As shown previously, the least squares estimates of the slope mand intercept care given by ˆm=1.11 and ˆc=0.4. (27.128) Since the error on each yi-value is drawn independently from a Gaussian distribution with standard deviation σ, we have χ2(a)=NX i=1 /yi−f(xi;a) σ /2 =NX i=1 hyi−mxi−c σ i2 . (27.129) Inserting the values (27.128) into (27.129), we obtain χ2(ˆm,ˆc)=1 1 .5. In our case, the number of data points is N= 10 and the number of ﬁtted parameters is M= 2. Thus, the 1139 STATISTICS number of degrees of freedom is n=N−M= 8. Setting n=8a n d α=0.05 in (27.127) we ﬁnd, either numerically or from tables, that c=1 5.51. Hence our rejection region is χ2(ˆm,ˆc)>15.51. Since we found that χ2(ˆm,ˆc)=1 1 .5, we cannot reject the null hypothesis that the underlying model for the data is a straight line y=mx+c. J As mentioned above, our analysis is only valid if the function f(x;a) is linear in the parameters a. Nevertheless, it is so convenient that it is sometimes applied in non-linear cases, provided the non-linearity is not too severe. 27.8 Exercises 27.1 A group of students uses a pendulum experiment to measure g, the acceleration of free fall, and obtain the following values (in m s−2): 9.80, 9.84, 9.72, 9.74, 9.87, 9.77, 9.28, 9.86, 9.81, 9.79, 9.82. What would you give as the best value andstandard error for gas measured by the group? 27.2 Measurements of a certain quantity gave the following values: 296, 316, 307, 278, 312, 317, 314, 307, 313, 306, 320, 309. Within what limits would you say there isa 50% chance that the correct value lies? 27.3 The following are the values obtained by a class of 14 students when measuring a physical quantity x: 53.8, 53.1, 56.9, 54.7, 58.2, 54.1, 56.4, 54.8, 57.3, 51.0, 55.1, 55.0, 54.2, 56.6. (a) Display these results as a histogram and state what you would give as the best value for x. (b) Without calculation estimate how much reliance could be placed upon your answer to (a). (c) Databooks give the value of xas 53.6 with negligible error. Are the data obtained by the students in conﬂict with this? 27.4 Two physical quantities xandyare connected by the equation y 1/2=x ax1/2+b, and measured pairs of values for xandyare as follows: x:1 01 21 6 2 0 y: 409 196 114 94. Determine the best values for aandbby graphical means and (either by hand or by using a built-in calculator routine) by a least squares ﬁt to an appropriatestraight line, 27.5 Measured quantities xandyare known to be connected by the formula y=ax x2+b, where aandbare constants. Pairs of values obtained experimentally are x: 2.0 3.0 4.0 5.0 6.0 y: 0.32 0.29 0.25 0.21 0.18. Use these data to make best estimates of the values of ythat would be obtained for (a) x=7.0, and (b) x=−3.5. As measured by fractional error, which estimate is likely to be the more accurate? 1140 27.8 EXERCISES 27.6 Prove that the sample mean is the best linear unbiased estimator of the population mean µas follows. (a) If the real numbers a1,a2,...,a nsatisfy the constraint Pn i=1ai=C,w h e r e C is a given constant, show that Pn i=1a2 iis minimised by ai=C/nfor all i. (b) Consider the linear estimator ˆµ= Pn i=1aixi. Impose the conditions (i) that it isunbiased , and (ii) that it is as eﬃcient as possible. 27.7 A population contains individuals of ktypes in equal proportions. A quantity X has mean µiamongst individuals of type i, and variance σ2which has the same value for all types. In order to estimate the mean of Xover the whole population, two schemes are considered; each involves a total sample size of nk. In the ﬁrst the sample is drawn randomly from the whole population, whilst in the second(stratiﬁed sampling )nindividuals are randomly selected from each of the ktypes. Show that in both cases the estimate has expectation µ=1 kkX i=1µi, but that the variance of the ﬁrst scheme exceeds that of the second by an amount 1 k2nkX i=1(µi−µ)2. 27.8 Carry through the following proofs of statements made in subsections 27.5.2 and 27.5.3 about the ML estimators ˆτandˆλ. (a) Find the expectation values of the ML estimators ˆτandˆλgiven respectively in (27.71) and (27.75). Hence verify equations (27.76) which show that, eventhough an ML estimator is unbiased, it does not follow that functions of itare also unbiased. (b) Show that E[ˆτ 2]=(N+1)τ2/Nand hence prove that ˆτis a minimum-variance estimator of τ. 27.9 An experiment consists of a large, but unknown, number n(/greatermuch1) of trials in each of which the probability of success pis the same, but also unkown. In the ith trial, i=1,2,...,N , the total number of successes is xi(/greatermuch1). Determine the log-likelihood function.Using Stirling’s approximation to ln( n−x), show that dln(n−x) dn≈1 2(n−x)+l n ( n−x), and hence evaluate ∂(nCx)/∂n. By ﬁnding the (coupled) equations determining the ML estimators ˆpandˆn,s h o w that, to order N−1, they must satisfy the simultaneous ‘arithmetic’ and ‘geometric’ mean constraints ˆnˆp=1 NNX i=1xiand (1−ˆp)N=NY i=1 / 1−xi ˆn / . 1141 STATISTICS 27.10 This exercise is intended to illustrate the dangers of applying formalised estimator techniques to distributions that are not well behaved in a statistical sense. The following are ﬁve sets of 10 values, all drawn from the same Cauchy distribution with parameter a. (i) 4 .81−1.24 1 .30−0.23 2 .98 −1.13−8.32 2 .62−0.79−2.85 (ii) 0 .07 1 .54 0 .38−2.76−8.82 1.86−4.75 4 .81 1 .14−0.66 (iii) 0 .72 4 .57 0 .86−3.86 0 .30 −2.00 2 .65−17.44−2.26−8.83 (iv)−0.15 202 .76−0.21−0.58−0.14 0.36 0 .44 3 .36−2.96 5 .51 (v) 0 .24−3.33−1.30 3 .05 3 .99 1.59−7.76 0 .91 2 .80−6.46 Ignoring the fact that the Cauchy distribution does not have a ﬁnite variance (or even a formal mean), show that ˆa,t h eM Le s t i m a t o ro f a, has to satisfy s(ˆa)=10X i=11 1+x2 i/ˆa2=5.(∗) Using a programmable calculator, spreadsheet or computer, ﬁnd the value of ˆathat satisﬁes (*) for each of the data sets and compare it with the value a=1.6 used to generate the data. Form an opinion regarding the variance of the estimator. Show further that if it is assumed that (E[ˆa])2=E[ˆa2]t h e n E[ˆa]=ν1/2 2,w h e r e ν2is the second (central) moment of the distribution, which for the Cauchy distribution is inﬁnite! 27.11 According to a particular theory, two dimensionless quantities XandYhave equal values. Nine measurements of Xgave values of 22, 11, 19, 19, 14, 27, 8, 24 and 18, whilst seven measured values of Ywere 11, 14, 17, 14, 19, 16 and 14. Assuming that the measurements of both quantities are Gaussian distributedwith a common variance, are they consistent with the theory? An alternativetheory predicts that Y 2=π2X; is the data consistent with this proposal? 27.12 On a certain (testing) steeplechase course there are 12 fences to be jumped and any horse that falls is not allowed to continue in the race. In a season of racing a total of 500 horses started the course and the following numbers fell at eachfence: F e n c e :123456789 1 0 1 1 1 2 F a l l s : 6 27 54 92 93 32 53 01 71 91 11 51 2 Use this data to determine the overall probability of a horse falling at a fence, and test the hypothesis that it is the same for all horses and fences as follows. (a) draw up a table of the expected number of falls at each fence on the basis of the hypothesis; (b) consider for each fence ithe standardised variable z i=estimated falls −actual falls standard deviation of estimated falls and use it in an appropriate χ2test; (c) show that the data indicates that the odds against all fences being equally testing are about 40 to 1. Identify the fences that are signiﬁcantly easier orharder than the average. 1142 27.8 EXERCISES 27.13 A similar technique to that employed in exercise 27.12 can be used to test correlations between characteristics of sampled data. To illustrate this considerthe following problem. During an investigation into possible links between mathematics and classical music, pupils at a school were asked whether they had preferences (a) betweenmathematics and english, and (b) between classical and pop music. The resultsare given below. Classical None Pop Mathematics 23 13 14None 17 17 36English 30 10 40 By computing tables of expected numbers, based on the assumption that no correlations exist, and calculating the relevant values of χ 2, determine whether there is any evidence for (a) a link between academic and musical tastes, and (b) a claim that pupils either had preferences in both areas or had no preference. You will need to consider the appropriate value for the number of degrees of freedom to use when applying the χ2test. 27.14 Three candidates X,YandZwere standing for election to a vacant seat on their college’s Student Committee. The members of the electorate (current ﬁrst-yearstudents, consisting of 150 men and 105 women) were each allowed to crossout the name of the candidate they least wished to be elected, the other twocandidates then being credited with one vote each. the following data are known. (a)Xreceived 100 votes from men, whilst Yreceived 65 votes from women. (b)Zreceived ﬁve more votes from men than Xreceived from women. (c) The total votes cast for XandYwere equal. Analyse this data in such a way that a χ 2test can be used to determine whether voting was other than random (i) amongst men, and (ii) amongst women. 27.15 A particle detector consisting of a shielded scintillator is being tested by placing it near a particle source of controlled intensity (by the use of absorbers). It mightregister counts even in the absence of particles from the source because of thecosmic ray background. The number of counts nregistered in a ﬁxed time interval as a function of the source strength sis given in the following table: Source strength: s0123456 Counts n: 6 11 20 42 44 62 61 At any given source strength the number of counts is expected to be Poisson distributed with mean n=a+bs, where aandbare constants. Analyse the data for a ﬁt to this relationship and obtain the best values for aandbtogether with their standard errors. (a) How well is the cosmic ray background determined? (b) What is the value of the correlation coeﬃcient between aandb?I st h i s consistent with what would happen if the cosmic ray background were imagined to be negligible? (c) Does the data ﬁt the expected relationship well? Is there any evidence that t h er e p o r t e dd a t a‘ i st o og o o daﬁ t ’ ? 1143 STATISTICS 27.16 The function y(x) is known to be a quadratic function of x. The following table gives the measured values and uncorrelated standard errors of ymeasured at various values of x(in which there is negligible error): x 1234 5 y(x)3 .5±0.52 .0±0.53 .0±0.56 .5±1.01 0 .5±1.0 Construct the response matrix Rusing as basis functions 1 ,x ,x2. Calculate the matrix RTN−1Rand show that its inverse, the covariance matrix V,h a st h ef o r m V=1 9184 /0/@12592−9708 1580 −9708 8413 −1461 1580−1461 269 /1A. Use this matrix to ﬁnd the best values, and their uncertainties, for the coeﬃcients of the quadratic form for y(x). 27.17 The following are the values and standard errors of a physical quantity f(θ) measured at various values of θ(in which there is negligible error): θ 0 π/6 π/4 π/3 f(θ)3 .72±0.21 .98±0.1−0.06±0.1−2.05±0.1 θπ / 22 π/33 π/4 π f(θ)−2.83±0.21 .15±0.13 .99±0.29 .71±0.4 Theory suggests that fshould be of the form a1+a2cosθ+a3cos 2θ. Show that the normal equations for the coeﬃcients aiare 481.3a1+ 158 .4a2−43.8a3= 284 .7, 158.4a1+ 218 .8a2+6 2.1a3=−31.1, −43.8a1+6 2.1a2+ 131 .3a3= 368 .4. (a) If you have matrix inversion routines available on a computer, determine the best values and variances for the coeﬃcients aiand the correlation between the coeﬃcients a1anda2. (b) If you have only a calculator available, solve for the values using Gauss– Seidel iteration starting from the approximate solution a1=2,a2=−2,a3= 4. 27.18 Prove that the expression given for the Student’s t-distribution in equation (27.118) is correctly normalised. 27.19 Verify that the F-distribution P(F) given explicitly in equation (27.126) is symme- tric between the two data samples, i.e. that it retains the same form but with N1 andN2interchanged, if Fis replaced by F/prime=F−1. Symbolically, if P/prime(F/prime)i st h e distribution of F/primeandP(F)=η(F,N 1,N2), then P/prime(F/prime)=η(F/prime,N2,N1). 27.20 It is claimed that the two following sets of values were obtained (a) by ran- domly drawing from a normal distribution that is N(0,1) and then (b) randomly assigning each reading to one of the two sets A and B. Set A−0.314 0 .603−0.551−0.537−0.160−1.635 0 .719 0.610 0 .482−1.757 0 .058 Set B−0.691 1 .515−1.642−1.736 1 .224 1 .423 1 .165 Make tests, including t-a n d F-tests, to establish whether there is any evidence that either claims is, or both claims are, false. 1144 27.9 HINTS AND ANSWERS 27.9 Hints and answers 27.1 Note that the reading of 9.28 m s−2is clearly in error and should not be used in the calculation; 9 .80±0.02 m s−2. 27.2 The reading of 278 should probably be rejected. The other readings do not look as though they are Gaussian distributed and the best estimate is probably givenby the inter-quartile range of the remaining 11 readings, i.e. 307 – 316. 27.3 (a) 55.1. (b) Note that two thirds of the readings lie within ±2 of the mean and that 14 readings are being used. This gives a standard error in the mean ≈0.6. (c) Student’s thas a value of about 2.5 for 13 d.o.f. (degrees of freedom), and therefore it is likely at the 3% signiﬁcance level that the data are in conﬂict withthe accepted value. 27.4 Plot either xy −1/2versus x1/2or (x/y)1/2versus x−1/2;a=1.20,b=−3.29. 27.5 Plot or calculate a least squares ﬁt of either x2versus x/yorxyversus y/x to obtain a≈1.19 and b≈3.4. (a) 0.16; (b) −0.27. Estimate (b) is the more accurate because, using the fact that y(−x)=−y(x), it is eﬀectively obtained by interpolation rather than extrapolation. 27.6 (a) Use Lagrange multipliers. (b) Write xiasµ+zi,w h e r e/angbracketleftz2 i/angbracketright=σ2for all i,a n d use the result of (a) to evaluate E[(ˆµ−µ)2]. 27.7 Recall that, because of the equal proportions of each type, the expected numbers of each type in the ﬁrst scheme is n. Show that the variance of the estimator for the second scheme is σ2/(kn). When calculating that for the ﬁrst scheme, recall that¯x2 i=µ2 i+σ2and note that µ2 ican be written as ( µi−µ+µ)2. 27.8 (a) Note that R P(x|τ) Q j/negationslash=idxj=τ−1exp(−xi/τ). With E[ˆλ]= R N( Pxi)−1λNexp(−λ Pxi)dx, ﬁnd the ﬁrst-order diﬀerential equation involving dE[ˆλ]/dλ. The relevant integrating factor is λ−N. (b) Denoting τ−1 R xrexp(−x/τ)dxbyJr, show that E[ˆτ2]=N−1[J2+(N−1)J2 1]. 27.9 The log-likelihood function is lnL=NX i=1lnnCxi+NX i=1xilnp+ / Nn−NX i=1xi /! ln(1−p); ∂(nCx) ∂n≈ln /n n−x / −x 2n(n−x). Ignore the second term on the RHS of the above to obtain NX i=1ln /n n−xi / +Nln(1−p)=0 . 27.10 Remember that aappears in the normalisation constant of the Cauchy distribu- tion.(i) 1.85; (ii) 1.66; (iii) 2.46; (iv) 0.68; (v) 2.44. Although the estimates have thecorrect order of magnitude, there is clearly a very large (perhaps inﬁnite) sam-pling variance. Even when all 50 samples are combined the estimated value of1.84 is still 0.24 diﬀerent from that used to generate the data. 27.11 ¯X=1 8 .0±2.2,¯Y=1 5 .0±1.1.ˆσ=4.92 giving t=1.21 for 14 d.o.f., and is signiﬁcant only at the 75% level. Thus there is no signiﬁcant disagreementbetween the data and the theory. For the second theory, only the mean values canbe tested as Y 2will not be Gaussian distributed. ¯Y2−π2¯X=4 7±38 and is not signiﬁcantly diﬀerent from zero. Again the data is consistent with the proposedtheory. 1145 STATISTICS 27.12 Whilst the distribution of falls at each fence is formally binomial, it can be approximated for the purposes of the question by a Poisson distribution. Thetotal number of falls is 377. The total number of attempted jumps = 3202.Overall probability of a fall is 0.1177.(a) 58.9, 51.6, 42.7, 37.0, 33.6, 29.7, 26.7, 23.2, 21.2, 19.0, 17.7, 15.9. (b) χ 2=2 1.2 for 11 d.o.f. (c) The χ2-value is close to the 97.5% conﬁdence limit. Fence 2 is much harder than the rest, fence 10 is easier, and fences 4 and 8 are somewhateasier than the average. 27.13 Consider how many entries may be chosen freely in the table if all row and column totals are to match the observed values. It should be clear that for anm×ntable the number of degrees of freedom is ( m−1)(n−1). (a) In order to make the fractions expre ssing each preference or lack of prefer- ence correct, the expected distribution, if there were no correlation, is Classical None Pop Mathematics 17.5 10 22.5 None 24.5 14 31.5 English 28 16 36 This gives a χ 2of 12.3 for 4 d.o.f., making it less than 2% likely, that no correlation exists. (b) The expected distribution, if there were no correlation, is Music preference No music preference Academic preference 104 26No academic preference 56 14 This gives a χ 2of 1.2 for one d.o.f and no evidence for the claim. 27.14 The votes are not statistically independent and, once established, must be con- verted to a table of deletions before any χ2test is applied. This table is NotXNotYNotZ Men 50 35 65Women 25 40 40 The relevant values of χ 2are (i) 9.0 and (ii) 4.3, both for 2 d.o.f., suggesting that the voting by men was almost certainly not random but that by women mayhave been. 27.15 As the distribution at each value of sis Poisson, the best estimate of the measurement error is the square root of the number of counts, i.e.√ n(s). Linear regression gives a=4.3±2.1a n d b=1 0.06±0.94. (a) The cosmic ray background must be present, since n(0)/negationslash= 0 but its value of about 4 is uncertain to within a factor of 2. (b) The correlation coeﬃcient between aandbis−0.63. Yes; if awere reduced towards zero then bwould have to be increased to compensate. (c) Yes, χ2=4.9 for 5 d.o.f., which is almost exactly the ‘expected’ value, neither t o og o o dn o rt o ob a d . 27.16 The matrix RTN−1Rhas entries 14, 33, 97; 33, 97, 333; 97, 333, 1273, whilst RTN−1b,w h e r e bis the data vector, has entries 51, 144.5, 520.5. y(x)=( 6 .73±1.17)−(4.34±0.96)x+( 1.03±0.17)x2. 27.17 a1=2.02±0.06,a2=−2.99±0.09,a3=4.90±0.10;r12=−0.60. 1146 27.9 HINTS AND ANSWERS 27.18 Make the substitution t=√ N−1t a n θto reduce the integral of the t-dependent part of (27.118) to 2√ N−1 Rπ/2 0cosN−2θd θ. Relate this to the beta function B /;1 2,1 2(N−1) / , and hence to the gamma functions, using the relationships given in the Appendix. Note that Γ /;1 2 / =√π. 27.19 Note that |dF|=|dF/prime/F/prime2|and write 1+N1−1 (N2−1)F/primeas /N1−1 (N2−1)F/prime // 1+(N2−1)F/prime N1−1 / . 27.20 (a) The mean and variance of the whole sample are −0.068 and 1 .180, which are obviously compatible with N(0,1) without the need for statistical tests. (b) The means and variances of the two sets are: A, −0.226 and 0 .815; B, 0 .180 and 2 .554. The value of tfor the diﬀerence between the two means is 0 .692; for 16 degrees of freedom, this or a greater value of tcan be expected in marginally more than half all cases. The value of Fis 3.13. For n1=6a n d n2= 10, this value is very close to the 95% conﬁdence limit of 3 .22. Thus it is rather unlikely that the allocation between the two groups was made at random – set B hassigniﬁcantly more readings that are more than one standard deviation from themean for a N(0,1) distribution than it should. 1147 28 Numerical methods It happens frequently that the end product of a calculation or piece of analysis is one or more algebraic or diﬀerential equations, or an integral that cannot beevaluated in closed form or in terms of tabulated or pre-programmed functions.From the point of view of the physical scientist or engineer, who needs numericalvalues for prediction or comparison with experiment, the calculation or analysisis thus incomplete. With the ready availability of standard packages on powerful computers for the numerical solution of equations, both algebraic and diﬀerential, and for theevaluation of integrals, in principle there is no need for the investigator to doother than turn to them. However, it should be a part of every engineer’s orscientist’s competence to have some understanding of the kinds of procedure thatare being put into practice within those packages. The present chapter indicates (at a simple level) some of the ways in which analytically intractable problems can be tackled using numerical methods. In the restricted space available in a book of this nature it is clearly not possible to give anything like a full discussion, even of the elementary points thatwill be made in this chapter. The limited objective adopted is that of explainingand illustrating by simple examples some of the basic principles involved. Inmany cases, the examples used can be solved in closed form anyway, but this ‘obviousness’ of the answers should not detract from their illustrative usefulness, and it is hoped that their transparency will help the reader to appreciate some ofthe inner workings of the methods described. The student who proposes to study complicated sets of equations or make repeated use of the same procedures by, for example, writing computer programsto carry out the computations, will ﬁnd it essential to acquire a good under- standing of topics hardly mentioned here. Amongst these are the sensitivity of the adopted procedures to errors introduced by the limited accuracy with whicha numerical value can be stored in a computer (rounding errors) and to the 1148 28.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS errors introduced as a result of approximations made in setting up the numerical procedures (truncation errors). For this scale of application, books speciﬁcallydevoted to numerical analysis, data analysis and computer programming shouldbe consulted. So far as is possible, the method of presentation here is that of indicating and discussing in a qualitative way the main steps in the procedure, and thenof following this with an elementary worked example. The examples have beenrestricted in complexity to a level at which they can be carried out with a pocketcalculator. Naturally it will not be possible for the student to check all thenumerical values presented unless he or she has a programmable calculator orcomputer readily available, and even then it might be tedious to do so. However, it is advisable to check the initial step and at least one step in the middle of each repetitive calculation given in the text, so that how the symbolic equationsare used with actual numbers is understood. Clearly the intermediate step shouldbe chosen to be at a point in the calculation at which the changes are stillsuﬃciently large that they can be detected by whatever calculating device isused. Where alternative methods for solving the same type of problem are discussed, for example in ﬁnding the roots of a polynomial equation, we have usually taken the same example to illustrate each method. This could give the mistakenimpression that the methods are very restricted in applicability, but it is felt bythe authors that using the same examples repeatedly has suﬃcient advantages, interms of illustrating the relative characteristics of competing methods, to justify doing so. Once the principles are clear, little is to be gained by using new exampleseach time and, in fact, having some prior knowledge of the ‘correct answer’ should allow the reader to judge the eﬃciency and dangers of particular methods as the successive steps are followed through. One other point remains to be mentioned. Here, in contrast with every other chapter of this book, the value of a large selection of exercises is not clear cut.The reader with suﬃcient computing resources to tackle them can easily devisealgebraic or diﬀerential equations to be solved, or functions to be integrated (which perhaps have arisen in other contexts). Further, the solutions of these problems will be self-checking, for the most part. Consequently, although anumber of exercises are included, no attempt has been made to test the full rangeof ideas treated in this chapter. 28.1 Algebraic and transcendental equations The problem of ﬁnding the real roots of an equation of the form f(x)=0 ,w h e r e f(x) is an algebraic or transcendental function of x, is one that can sometimes be treated numerically even if explicit solutions in closed form are not feasible. 1149 NUMERICAL METHODS 0.20.40.60.81.01.21.41.61.8 −4−202468101214 xf(x) f(x)=x5−2x2−3 Figure 28.1 A graph of the function f(x)=x5−2x2−3f o r xin the range 0≤x≤1.9. Examples of the types of equation mentioned are the quartic equation ax4+bx+c=0, and the transcendental equation x−3t a nh x=0. The latter type is characterised by the fact that it contains in eﬀect a polynomial of inﬁnite order on the left-hand side. We will discuss four methods that, in various circumstances, can be used to obtain the real roots of equations of the above types. In all cases we will take as the speciﬁc equation to be solved the ﬁfth-order polynomial equation f(x)≡x5−2x2−3=0 . (28.1) The reasons for using the same equation each time were discussed in the intro- duction to this chapter. For future reference and so that the reader may follow some of the calculations leading to the evaluation of the real root of (28.1), a graph of f(x) in the range 0≤x≤1.9 is shown in ﬁgure 28.1. Equation (28.1) is one for which no solution can be found in closed form, that is in the form x=awhere adoes not explicitly contain x. The general scheme to be employed will be an iterative one in which successive approximations to a real root of (28.1) will be obtained, each approximation, it is to be hoped, being better than the preceding one; certainly, we require that the approximations convergeand that they have as their limit the sought-for root. Let us denote the required 1150 28.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS root by ξand the values of successive approximations by x1,x2,...,xn,....T h e n for any particular method to be successful, lim n→∞xn=ξwhere f(ξ)=0 . (28.2) However, success as deﬁned here is not the only criterion. Since, in practice, only a ﬁnite number of iterations will be possible, it is important that the values ofxnbe close to that of ξfor all n>N ,w h e r e Nis a relatively low number; exactly how low it is naturally depends on the computing resources available andthe accuracy required in the ﬁnal answer. So that the reader may assess the progress of the calculations that follow, we record that to nine signiﬁcant ﬁgures the real root of (28.1) has the value ξ=1.495 106 40 . (28.3) We now consider in turn four methods for determining the value of this root. 28.1.1 Rearrangement of the equation If equation (28.1), f(x) = 0, can be recast into the form x=φ(x) (28.4) where φ(x)i saslowly varying function of xthen an iteration scheme x n+1=φ(xn) (28.5) will often produce a fair approximation to the root ξafter a few iterations, as follows. Clearly ξ=φ(ξ)s i n c e f(ξ) = 0; thus when xnis close to ξthe next approximation, xn+1, will diﬀer little from xn, the actual size of the diﬀerence giving an order-of-magnitude indication of the inaccuracy in xn+1(when compared with ξ). In the present case the equation can be written x=( 2x2+3 )1/5. (28.6) Because of the presence of the one-ﬁfth power, the RHS is rather insensitive to the value of xused to compute it, and so the form (28.6) ﬁts the general requirements for the method to work satisfactorily. It remains only to choose astarting approximation. It is easy to see from ﬁgure 28.1 that the value x=1.5 would be a good starting point but, so that the behaviour of the procedure at values some way from the actual root can be studied, we will make a poorerchoice, x 1=1.7. With this starting value and the general recurrence relationship xn+1=( 2x2 n+3 )1/5, (28.7) 1151 NUMERICAL METHODS nx n f(xn) 1 1.7 5.42 2 1.544 18 1.013 1.506 86 2 .28×10 −1 4 1.497 92 5 .37×10−2 5 1.495 78 1 .28×10−2 6 1.495 27 3 .11×10−3 7 1.495 14 7 .34×10−4 8 1.495 12 1 .76×10−4 Table 28.1 Successive approximations to the root of (28.1) using the rear- rangement method. nA n f(An) Bnf(Bn) xn f(xn) 11 . 0 −4.0000 1.7 5.4186 1.2973 −2.6916 2 1.2973 −2.6916 1.7 5.4186 1.4310 −1.0957 3 1.4310 −1.0957 1.7 5.4186 1.4762 −0.3482 4 1.4762 −0.3482 1.7 5.4186 1.4897 −0.1016 5 1.4897 −0.1016 1.7 5.4186 1.4936 −0.0289 6 1.4936 −0.0289 1.7 5.4186 1.4947 −0.0082 Table 28.2 Successive approximations to the root of (28.1) using linear interpolation. successive values can be found. These are recorded in table 28.1. Although not strictly necessary, the value of f(xn)≡x5 n−2x2 n−3 is also shown at each stage. It will be seen that x7and all later xnagree with the precise answer (28.3) to within one part in 104. However, f(xn)a n d xn−ξare both reduced by a factor of only about 4 for each iteration; thus a large number of iterations wouldbe needed to produce a very accurate answer. The factor 4 is of course speciﬁcto this particular problem and would be diﬀerent for a diﬀerent equation. The successive values of x nare shown in graph ( a) of ﬁgure 28.2. 28.1.2 Linear interpolation In this approach two values A1and B1ofxare chosen with A1<B 1and such that f(A1)a n d f(B1) have opposite signs. The chord joining the two points (A1,f(A1)) and ( B1,f(B1)) is then notionally constructed, as illustrated in graph (b) of ﬁgure 28.2, and the value x1at which the chord cuts the x-axis is determined by the interpolation formula xn=Anf(Bn)−Bnf(An) f(Bn)−f(An), (28.8) 1152 28.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS 1.01.0 1.0 1.01 .21.2 1.2 1.21 .41.4 1.4 1.4 1.61.6 1.6 1.6 −4−4 −4 −4−2−2 −2 −222 2 244 4 466 6 6x1x1 x2x2x3 x3 x4 x1 x1x2 x2x3 x3ξ ξ ξξ(a) (b) (c) (d) Figure 28.2 Graphical illustrations o f the iteration methods discussed in the text: ( a) rearrangement; ( b) linear interpolation; ( c) binary chopping; (d) Newton–Raphson. with n=1 .N e x t f(x1) is evaluated and the process repeated after replacing by x1 either A1orB1, according to whether f(x1) has the same sign as f(A1)o rf(B1) respectively. In ﬁgure 28.2( b),A1is the one replaced. As can be seen in the particular example that we are considering, with this method there is a tendency, if the curvature of f(x)i so fc o n s t a n ts i g nn e a r the root, for one of the two ends of the successive chords to remain un-changed. Starting with the initial values A 1=1a n d B1=1.7, the results of the ﬁrst ﬁve iterations using (28.8) are given in table 28.2 and indicated in graph ( b)o f ﬁgure 28.2. As with the rearrangement method, the improvement in accuracy,as measured by f(x n)a n d xn−ξ, is a fairly constant factor at each iteration (approximately 3 in this case), and for our particular example there is little tochoose between the two. Both tend to their limiting value of ξmonotonically, from either higher or lower values, and this makes it diﬃcult to estimate limits within which ξcan safely be presumed to lie. The next method to be described gives at any stage a range of values within which ξisknown to lie. 1153 NUMERICAL METHODS nA n f(An) Bn f(Bn) xn f(xn) 1 1.0000 −4.0000 1.7000 5.4186 1.3500 −2.1610 2 1.3500 −2.1610 1.7000 5.4186 1.5250 0 .5968 3 1.3500 −2.1610 1.5250 0.5968 1.4375 −0.9946 4 1.4375 −0.9946 1.5250 0.5968 1.4813 −0.2573 5 1.4813 −0.2573 1.5250 0.5968 1.5031 0 .1544 6 1.4813 −0.2573 1.5031 0.1544 1.4922 −0.0552 7 1.4922 −0.0552 1.5031 0.1544 1.4977 0 .0487 8 1.4922 −0.0552 1.4977 0.0487 1.4949 −0.0085 Table 28.3 Successive approximations to the root of (28.1) using binary chopping. 28.1.3 Binary chopping Again two values of x,A1andB1, that straddle the root are chosen, such that A1<B 1andf(A1)a n d f(B1) have opposite signs. The interval between them is then halved by forming xn=1 2(An+Bn), (28.9) with n=1 ,a n d f(x1) is evaluated. It should be noted that x1is determined solely by A1andB1, and not by the values of f(A1)a n d f(B1) as in the linear interpolation method. Now x1is used to replace either A1orB1, depending on which of f(A1)o rf(B1) has the same sign as f(x1), i.e. if f(A1)a n d f(x1) have the same sign then x1replaces A1. The process is then repeated to obtain x2,x3etc. This has been carried through in table 28.3 for our standard equation (28.1) and is illustrated in ﬁgure 28.2( c). The entries have been rounded to four places of decimals. It is suggested that the reader follows through the sequential replace-ments of the A nandBnin the table and correlates the ﬁrst few of these with graph ( c) of ﬁgure 28.2. Clearly the accuracy with which ξis known in this approach increases by only a factor of 2 at each step, but this accuracy is predictable at the outset of the calculation and (unless f(x) has very violent behaviour near x=ξ)ar a n g eo f x in which ξlies can be safely stated at any stage. At the stage reached in the last line of table 28.3 it may be stated that 1 .4949 <ξ< 1.4977. Thus binary chopping gives a simple approximation method (it involves less multiplication than linearinterpolation, for example) that is predictable and relatively safe, although itsconvergence is slow. 28.1.4 Newton–Raphson method The Newton–Raphson (NR) procedure is somewhat similar to the interpolation method but, as will be seen, has one distinct advantage over the latter. Instead 1154 28.1 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS nx n f(xn) 1 1.7 5.42 2 1.545 01 1.033 1.498 87 7 .20×10 −2 4 1.495 13 4 .49×10−4 5 1.495 106 40 2 .6×10−8 6 1.495 106 40 – Table 28.4 Successive approximations to the root of (28.1) using the Newton– Raphson method. of (notionally) constructing the chord between two points on the curve of f(x) against x, the tangent to the curve is notionally constructed at each successive value of xnand the next value xn+1taken as the point at which the tangent cuts the axis f(x) = 0. This is illustrated in graph ( d) of ﬁgure 28.2. If the nth value is xn, the tangent to the curve of f(x) at that point has slope f/prime(xn) and passes through the point x=xn,y=f(xn). Its equation is thus y(x)=(x−xn)f/prime(xn)+f(xn). (28.10) The value of xat which y= 0 is then taken as xn+1; thus the condition y(xn+1)=0 yields from (28.10) the iteration scheme xn+1=xn−f(xn) f/prime(xn). (28.11) This is the Newton–Raphson iteration formula . Clearly,if xnis close to ξthen xn+1 is close to xn, as it should be. It is also apparent that if any of the xncomes close to a stationary point of f,s ot h a t f/prime(xn) is close to zero, the scheme is not going to work well. For our standard example, (28.11) becomes xn+1=xn−x5 n−2x2 n−3 5x4n−4xn=4x5 n−2x2 n+3 5x4n−4xn. (28.12) Again taking a starting value of x1=1.7 we obtain in succession the entries in table 28.4. The diﬀerent values are given to an increasing number of decimal places as the calculation proceeds; f(xn) is also recorded. It is apparent that this method is unlike the previous ones in that the increase in accuracy of the answer is not constant throughout the iterations but improvesdramatically as the required root is approached. Away from the root the behaviour of the series is less satisfactory and from its geometrical interpretation it can be seen that if, for example, there were a maximum or minimum near the root thenthe series could oscillate between values on either side of it (instead of ‘homing 1155 NUMERICAL METHODS in’ on the root). The reason for the good convergence near the root is discussed in the next section. Of the four methods mentioned, no single one is ideal and, in practice, some mixture of them is usually to be preferred. The particular combination of methodsselected will depend a great deal on how easily the progress of the calculationmay be monitored, but some combination of the ﬁrst three methods mentioned,followed by the NR scheme if great accuracy were required, would be suitable for most situations. 28.2 Convergence of iteration schemes For iteration schemes in which x n+1can be expressed as a diﬀerentiable function ofxn, e.g. the rearrangement or NR methods of the previous section, a partial analysis of the conditions necessary for a successful scheme can be made as follows. Suppose the general iteration formula is expressed as xn+1=F(xn) (28.13) ((28.7) and (28.12) are examples). Then the sequence of values x1,x2,...,x n,...is required to converge to the value ξthat satisﬁes both f(ξ)=0 a n d ξ=F(ξ). (28.14) If the error in the solution at the nth stage is /epsilon1n,i . e .xn=ξ+/epsilon1n,t h e n ξ+/epsilon1n+1=xn+1=F(xn)=F(ξ+/epsilon1n). (28.15) For the iteration process to converge, a decreasing error is required, i.e. |/epsilon1n+1|< |/epsilon1n|. To see what this implies about F, we expand the right-hand term of (28.15) by means of a Taylor series and use (28.14) to replace (28.15) by ξ+/epsilon1n+1=ξ+/epsilon1nF/prime(ξ)+1 2/epsilon12 nF/prime/prime(ξ)+···. (28.16) This shows that, for small /epsilon1n, /epsilon1n+1≈F/prime(ξ)/epsilon1n and that a necessary (but not suﬃcient) condition for convergence is that |F/prime(ξ)|<1. (28.17) It should be noticed that this is a condition on F/prime(ξ) and not on f/prime(ξ), which may have any ﬁnite value. Figure 28.3 illustrates in a graphical way how theconvergence proceeds for the case 0 <F /prime(ξ)<1. 1156 28.2 CONVERGENCE OF ITERATION SCHEMES xn xn+1xn+2y=x y=F(x) ξ xy Figure 28.3 Illustration of the convergence of the iteration scheme xn+1= F(xn)w h e n0 <F/prime(ξ)<1, where ξ=F(ξ). The line y=xmakes an angle π/4 with the axes. The broken line makes an angle tan−1F/prime(ξ)w i t ht h e x-axis. Equation (28.16) suggests that if F(x) can be chosen so that F/prime(ξ) = 0 then the ratio|/epsilon1n+1//epsilon1n|could be made very small, of order /epsilon1ni nf a c t .T og oe v e nf u r t h e r , if it can be arranged that the ﬁrst few derivatives of Fvanish at x=ξthen the convergence, once xnhas become close to ξ, could be very rapid indeed. If the ﬁrstN−1 derivatives of Fvanish at x=ξ,i . e . F/prime(ξ)=F/prime/prime(ξ)=···=F(N−1)(ξ) = 0 (28.18) and consequently /epsilon1n+1=O ( /epsilon1N n), (28.19) then the scheme is said to have Nth-order convergence . This is the explanation of the signiﬁcant diﬀerence in convergence between the NR scheme and the others discussed (judged by reference to (28.19), so that thediﬀerentiability of the function Fis not a prerequisite). The NR procedure has second-order convergence, as is shown by the following analysis. Since F(x)=x−f(x) f/prime(x), F/prime(x)=1−f/prime(x) f/prime(x)+f(x)f/prime/prime(x) [f/prime(x)]2=f(x)f/prime/prime(x) [f/prime(x)]2. Now, provided f/prime(ξ)/negationslash= 0, it follows that F/prime(ξ)=0b e c a u s e f(x)=0a t x=ξ. 1157 NUMERICAL METHODS nx n+1 /epsilon1n 18 . 5 4 . 5 2 5.191 1.193 4.137 1 .4×10 −1 4 4.002257 2 .3×10−3 5 4.000000637 6 .4×10−7 64 — Table 28.5 Successive approximations to√ 16 using the iteration scheme (28.20).IThe following is an iteration scheme for ﬁnding the square root of X: xn+1=1 2 / xn+X xn / . (28.20) Show that it has second-order convergence and illustrate its eﬃciency by ﬁnding, say,√ 16 starting with a very poor guess√ 16 = 1 . If this scheme does converge to ξthen ξwill satisfy ξ=1 2 / ξ+X ξ / ⇒ ξ2=X, as required. The iteration function Fis given by F(x)=1 2 / x+X x / , and so, since ξ2=X, F/prime(ξ)=1 2 / 1−X x2 / x=ξ=0, whilst F/prime/prime(ξ)= /X x3 / x=ξ=1 ξ/negationslash=0. Thus the procedure has second-order, but not third-order, convergence. We now show the procedure in action. Table 28.5 gives successive values of xnand of /epsilon1n, the diﬀerence between xnand the true value, 4. As we can see the scheme is crude initially, but once xngets close to ξ, it homes in on the true value extremely rapidly. J 28.3 Simultaneous linear equations As we saw in chapter 8, many situations in physical science can be described approximately or exactly by a set of Nsimultaneous linear equations in N 1158 28.3 SIMULTANEOUS LINEAR EQUATIONS variables (unknowns), xi,i=1,2,...,N . The equations take the general form A11x1+A12x2+···+A1NxN=b1, A21x1+A22x2+···+A2NxN=b2, (28.21) ... AN1x1+AN2x2+···+ANNxN=bN, where the Aijare constants and form the elements of a square matrix A.T h e bi are given and form a column matrix b.I fAis non-singular then (28.21) can be solved for the xiusing the inverse of A, according to the formula x=A−1b. This approach was discussed at length in chapter 8 and will not be considered further here. 28.3.1 Gaussian elimination We follow instead a continuation of one of the earliest techniques acquired by a student of algebra, namely the solving of simultaneous equations (initially only two in number) by the successive elimination of all the variables but one. This (known as Gaussian elimination ) is achieved by using, at each stage, one of the equations to obtain an explicit expression for one of the remaining xiin terms of the others and then substituting for that xiin all other remaining equations. Eventually a single linear equation in just one of the unknowns is obtained. Thisis then solved and the result re-substituted in previously derived equations (inreverse order) to establish values for all the x i. This method is probably very familiar to the reader and so a speciﬁc example to illustrate this alone seems unnecessary. Instead, we will show how a calculationalong such lines might be arranged so that the errors due to the inherent lack ofprecision in any calculating equipment do not become excessive. This can happen if the value of Nis large and particularly (and we will merely state this) if the elements A 11,A22,...,A NNon the leading diagonal of the matrix in (28.21) are small compared with the oﬀ-diagonal elements. The process to be described is known as Gaussian elimination with interchange . The only, but essential, diﬀerence from straightforward elimination is that before each variable xiis eliminated, the equations are reordered to put the largest (in modulus) remaining coeﬃcient of xion the leading diagonal. We will take as an illustration a straightforward three-variable example, which can in fact be solved perfectly well without any interchange since, with simple numbers and only two eliminations to perform, rounding errors do not havea chance to build up. However, the important thing is that the reader should 1159 NUMERICAL METHODS appreciate how this would apply in (say) a computer program for a 1000-variable case, perhaps with unforseeable zeroes or very small numbers appearing on theleading diagonal.ISolve the simultaneous equations (a) x1+6x2−4x3=8, (b) 3 x1−20x2+x3=1 2 , (c)−x1+3x2+5x3=3.(28.22) Firstly, we interchange rows (a) and (b) to bring the term 3 x1onto the leading diagonal. In the following, we label the important equations (I), (II), (III), and the others alphabetically. (I) 3 x1−20x2+x3=1 2 , (d) x1+6x2−4x3=8, (e)−x1+3x2+5x3=3. For ( j) = (d) and (e), replace row (j) by row ( j)−aj1 3×row (I) , where aj1is the coeﬃcient of x1in row ( j), to give the two equations (II) /; 6+20 3 / x2+ /; −4−1 3 / x3=8−12 3, (f) /; 3−20 3 / x2+ /; 5+1 3 / x3=3 +12 3. Now|6+20 3|>|3−20 3|and so no interchange is needed before the next elimination. To eliminate x2, replace row (f) by row (f)− /; −11 3 / 38 3×row (II) . This gives (III) /16 3+11 38×(−13) 3 / x3=7+11 38×4. Collecting together and tidying up the ﬁnal equations, we have (I) 3 x1−20x2+x3=1 2 , (II) 38 x2−13x3=1 2 , (III) x3=2. Starting with (III) and working backwards it is now a simple matter to obtain x1=1 0,x 2=1,x 3=2. J 28.3.2 Gauss–Seidel iteration In the example considered in the previous subsection an explicit way of solving a set of simultaneous equations was given, the accuracy obtainable being limited only by the rounding errors in the calculating facilities available, and the calcula- tion was planned to minimise these. However, in some situations it may be thatonly an approximate solution is needed. If, for a large number of variables, this is 1160 28.3 SIMULTANEOUS LINEAR EQUATIONS the case then an iterative method may produce a satisfactory degree of precision with less calculation. Such a method, known as Gauss–Seidel iteration ,i sb a s e d upon the following analysis. The problem is again that of ﬁnding the components of the column matrix x that satisﬁes Ax=b (28.23) when Aand bare a given matrix and column matrix respectively. The steps of the Gauss–Seidel scheme are as follows. (i) Rearrange the equations (usually by simple division on both sides of each equation) so that all diagonal elements of the new matrix Care unity, i.e. (28.23) becomes Cx=d, (28.24) where C=I−F,a n d Fhas zeroes as its diagonal elements. (ii) Step (i) produces Fx+d=Ix=x, (28.25) and this forms the basis of an iteration scheme xn+1=Fxn+d, (28.26) where xnis the nth approximation to the required solution vector ξ. (iii) To improve the convergence, the matrix F, which has zeroes on its leading diagonal, can be written as the sum of two matrices Land Uthat have non-zero elements only below and above the leading diagonal respectively: Lij=braceleftBigg Fijifi>j , 0o t h e r w i s e , (28.27) Uij=braceleftBigg Fijifi<j , 0o t h e r w i s e . This allows the latest values of the components of xto be used at each stage and an improved form of (28.26) to be obtained, xn+1=Lxn+1+Uxn+d. (28.28) To see why this is possible we note, for example, that when calculating, say, the fourth component of xn+1, its ﬁrst three components are already known, and, because of the structure of L, these are the only ones needed to evaluate the fourth component of Lxn+1. 1161 NUMERICAL METHODS nx 1 x2 x3 12 2 2 24 0 . 1 1 . 3 43 12.76 1.381 2.323 4 9.008 0.867 1.881 5 10.321 1.042 2.0396 9.902 0.987 1.988 7 10.029 1.004 2.004 Table 28.6 Successive approximations to the solution of simultaneous equa-tions (28.29) using the Gauss–Seidel iteration method.IObtain an approximate solution to the simultaneous equations x1+6x2−4x3=8, 3x1−20x2+x3=1 2 , −x1+3x2+5x3=3.(28.29) These are the same equations as were solved in subsection 28.3.1. Divide the equations by 1, −20 and 5 respectively to give x1+6x2−4x3=8, −0.15x1+x2−0.05x3=−0.6, −0.2x1+0.6x2+x3=0.6. Thus, set out in matrix form, (28.28) is in this case/0/@x1 x2 x3 /1 A n+1= /0/@00 0 0.1 500 0.2−0.60 /1A /0/@x1 x2 x3 /1 A n+1 + /0/@0−64 000 .05 00 0 /1A /0/@x1 x2 x3 /1 A n+ /0 /@ 8 −0.6 0.6 /1A. Suppose initially ( n= 1) we guess each component to have the value 2. Then the successive sets of values of the three quantities generated by this scheme are as shown in table 28.6.Even with the rather poor initial guess, a close approximation to the exact result x 1= 10, x2=1 , x3= 2 is obtained in only a few iterations. J 28.3.3 Tridiagonal matrices Although for the solution of most matrix equations Ax=bthe number of operations needed increases rapidly with the size N×Nof the matrix (roughly as N3), for one particularly simple kind of matrix the computing required increases only linearly with N. This type often occurs in physical situations in which objects in an ordered set interact only with their nearest neighbours and is one in whichonly the leading diagonal and the diagonals immediately above and below it 1162 28.3 SIMULTANEOUS LINEAR EQUATIONS contain non-zero entries. Such matrices are known as tridiagonal matrices. They may also be used in numerical approximations to the solutions of certain typesof diﬀerential equation. A typical matrix equation involving a tridiagonal matrix is thus 00b1c1 b2c2 b3c3 bNa2 a3 aN–1 aN. . .x1 x2 x3 . . . xN–1 xNy1 y2 y3 . . . yN–1 yN= bN–1cN–1. . . . . .(28.30) So as to keep the entries in the matrix as free from subscripts as possible, we have used a,bandcto indicate subdiagonal, leading diagonal and superdiagonal elements respectively. As a consequence we have had to change the notation forthe column matrix on the right-hand side from bto (say) y. In such an equation the ﬁrst and last rows involve x 1andxNrespectively, and so the solution could be found by letting x1be unknown and then solving in turn each row of the equation in terms of x1, and ﬁnally determining x1by requiring the next-to-last line to generate for xNan equation compatible with that given by the last line. However, if the matrix is large then this becomes a very cumbersome operation, and a simpler method is to assume a form of solution xi−1=θi−1xi+φi−1. (28.31) Since the ith line of the matrix equation is aixi−1+bixi+cixi+1=yi, we must have, by substituting for xi−1,t h a t (aiθi−1+bi)xi+cixi+1=yi−aiφi−1. This is also in the form of (28.31), but with ireplaced by i+1. Thus the recurrence formulae for θiandφiare θi=−ci aiθi−1+bi,φ i=yi−aiφi−1 aiθi−1+bi, (28.32) provided the denominator does not vanish for any i. From the ﬁrst of the matrix equations it follows that θ1=−c1/b1andφ1=y1/b1. The equations may now be solved for the xiin two stages without carrying through an unknown quantity. First, all the θiandφiare generated using (28.32) and the values of θ1andφ1 and then, as a second stage, (28.31) is used to evaluate the xi, starting with xN (=φN) and working backwards. 1163 NUMERICAL METHODSISolve the following tridiagonal matrix equation, in which only non-zero elements are shown./0BBBBB/@12 −12 1 2−12 311 342 −22 /1CCCCCA /0BBBBB/@x1 x2 x3 x4 x5 x6 /1CCCCCA= /0BBBBB/@4 3 −3 10 7 −2 /1CCCCCA. (28.33) The solution is set out in table 28.7, in which the arrows indicate the general ﬂow of the calculation. First, the columns of ai,bi,ciandyiare ﬁlled in from the original equation (28.33) and then the recurrence relations ( 28.32) are used to ﬁll in the successive rows starting from the top; on each row we work from left to right as far as and including the φi column. Finally, the bottom entry in the the xicolumn is set equal to the bottom entry in the completed φicolumn and the rest of the xicolumn completed by using (28.31) and working up from the bottom. Thus the solution is x1=2 ;x2=1 ;x3=3 ;x4=−1;x5=2 ;x6=1 . J ai bici aiθi−1+bi θi yi aiφi−1 φi xi ↓01 2 → 1−2 40 4 2↑ ↓− 121 → 4−1/4 3−4 7/4 1↑ ↓2−12→− 3/2 4/3 −3 7/2 13/3 3↑ ↓31 1 → 5−1/510 13 −3/5−1↑ ↓34 2 → 17/5−10/17 7−9/5 44/17 2↑ ↓− 220 → 54/17 0 −2−88/17 1→ 1↑ Table 28.7 The solution of tridiagonal matrix equation (28.33). The arrows indicate the general ﬂow of the calculation, as described in the text. 28.4 Numerical integration As noted at the start of this chapter, with modern computers and computer packages – some of which will present solutions in algebraic form, where thatis possible – the inability to ﬁnd a closed-form expression for an integral no longer presents a problem. But, just as for the solution of algebraic equations, it is extremely important that scientists and engineers should have some idea of theprocedures on which such packages are based. In this section we discuss some ofthe more elementary methods used to evaluate integrals numerically and at thesame time indicate the basis of more sophisticated procedures. The standard integral evaluation has the form I=integraldisplay b af(x)dx, (28.34) where the integrand f(x) may be given in analytic or tabulated form, but for the cases under consideration no closed-form expression for Ican be obtained. All 1164 28.4 NUMERICAL INTEGRATION xi xi xi xi+1/2 xi+1 xi+1 xi+1 xi−1hh h hf(x) f(x)(a) (b) (c) fififi+1 fi+1 fi+1 fi−1 Figure 28.4 ( a) Deﬁnition of nomenclature. ( b) The approximation in using the trapezium rule; f(x) is indicated by the broken curve. ( c) Simpson’s rule approximation; f(x) is indicated by the broken curve. The solid curve is part of the approximating parabola. numerical evaluations of Iare based on regarding Ias the area under the curve off(x) between the limits x=aandx=band attempting to estimate that area. The simplest methods of doing this involve dividing up the interval a≤x≤b into Nequal sections, each of length h=(b−a)/N. The dividing points are labelled xiwith x0=a,xN=b,irunning from 0 to N. The point xiis a distance ihfrom a. The central value of xin a strip ( x=xi+h/2) is denoted for brevity byxi+1/2, and for the same reason f(xi) is written as fi. This nomenclature is indicated graphically in ﬁgure 28.4( a). So that we may compare later estimates of the area under the curve with the true value, we next obtain an exact expression for I, even though we cannot evaluate it. To do this we need to consider only one strip, say that between xi andxi+1. For this strip the area is, using Taylor’s expansion, integraldisplayh/2 −h/2f(xi+1/2+y)dy=integraldisplayh/2 −h/2∞summationdisplay n=0f(n)(xi+1/2)yn n!dy =∞summationdisplay n=0f(n) i+1/2integraldisplayh/2 −h/2yn n!dy =∞summationdisplay nevenf(n) i+1/22 (n+1 ) !parenleftbiggh 2parenrightbiggn+1 . (28.35) It should be noticed that, in this exact expression, only the even derivatives offsurvive the integration and all derivatives are evaluated at xi+1/2. Clearly 1165 NUMERICAL METHODS other exact expressions are possible, e.g. the integral of f(xi+y) over the range 0≤y≤h, but we will ﬁnd (28.35) the most useful for our purposes. We now turn to practical ways of approximating I, given the values of fi,o ra means to calculate them, for i=0,1,...,N . 28.4.1 Trapezium rule In this simple case the area shown in ﬁgure 28.4( a) is approximated as shown in ﬁgure 28.4( b), i.e. by a trapezium. The area Aiof the trapezium is Ai=1 2(fi+fi+1)h, (28.36) and if such contributions from all strips are added together then the estimate of the total, and hence of I,i s I(estim.) =N−1summationdisplay i=0Ai=h 2(f0+2f1+2f2+···+2fN−1+fN). (28.37) This provides a very simple expression for estimating integral (28.34); its accuracy is limited only by the extent to which hcan be made very small (and hence N very large) without making the calculation excessively long. Clearly the estimateprovided is only exact if f(x) is a linear function of x. The error made in calculating the area of the strip when the trapezium rule is used may be estimated as follows. The values used are f iandfi+1, as in (28.36). These can be expressed accurately in terms of fi+1/2and its derivatives by the Taylor series fi+1/2±1/2=fi+1/2±h 2f/prime i+1/2+1 2!parenleftbiggh 2parenrightbigg2 f/prime/prime i+1/2±1 3!parenleftbiggh 2parenrightbigg3 f(3) i+1/2+···. Thus Ai(estim.) =1 2h(fi+fi+1), =hbracketleftBigg fi+1/2+1 2!parenleftbiggh 2parenrightbigg2 f/prime/prime i+1/2+O ( h4)bracketrightBigg , whilst, from the ﬁrst few terms of the exact result (28.35), Ai(exact) = hfi+1/2+2 3!parenleftbiggh 2parenrightbigg3 f/prime/prime i+1/2+O ( h5). Thus the error ∆ Ai=Ai(estim.)−Ai(exact) is given by ∆Ai=parenleftbig1 8−1 24parenrightbig h3f/prime/prime i+1/2+O ( h5) ≈1 12h3f/prime/prime i+1/2. 1166 28.4 NUMERICAL INTEGRATION The total error in I(estim.) is thus given approximately by ∆I(estim.)≈1 12nh3/angbracketleftf/prime/prime/angbracketright=1 12(b−a)h2/angbracketleftf/prime/prime/angbracketright, (28.38) where/angbracketleftf/prime/prime/angbracketrightrepresents an average value for the second derivative of fover the interval atob.IUse the trapezium rule with h=0.5to evaluate I= Z2 0(x2−3x+4 )dx, and, by evaluating the integral exactly, examine how well (28.38) estimates the error. With h=0.5, we will need ﬁve values of f(x)=x2−3x+ 4 for use in formula (28.37). They are f(0) = 4, f(0.5) = 2 .75,f(1) = 2, f(1.5) = 1 .75 and f(2) = 2. Putting these into (28.37) gives I(estim.) =0.5 2(4 + 2×2.75 + 2×2+2×1.75 + 2) = 4 .75. The exact value is I(exact) = /x3 3−3x2 2+4x /2 0=42 3. The diﬀerence between the estimate of the integral and the exact answer is 1 /12. Equation (28.38) estimates this error as 2 ×0.25×/angbracketleftf/prime/prime/angbracketright/12. Our (deliberately chosen!) integrand is one for which /angbracketleftf/prime/prime/angbracketrightcan be evaluated trivially. Because f(x) is a quadratic function of x, its second derivative is constant, and equal to 2 in this case. Thus /angbracketleftf/prime/prime/angbracketrighthas value 2 and (28.38) estimates the error as 1 /12; that the estimate is exactly right should be no surprise since the Taylor expansion for a quadratic polynomial about any point always terminatesafter three terms and so no higher-order terms in hhave been ignored in (28.38).J 28.4.2 Simpson’s rule Whereas the trapezium rule makes a linear interpolation of f, Simpson’s rule eﬀectively mimics the local variation of f(x) using parabolas. The strips are treated two at a time (ﬁgure 28.4( c)) and therefore their number, N, should be made even. In the neighbourhood of xi,f o riodd, it is supposed that f(x) can be adequately represented by a quadratic form, f(xi+y)=fi+ay+by2. (28.39) In particular, applying this to y=±hyields two expressions involving b, fi+1=f(xi+h)=fi+ah+bh2, fi−1=f(xi−h)=fi−ah+bh2; thus bh2=1 2(fi+1+fi−1−2fi). 1167 NUMERICAL METHODS Now, in the representation (28.39), the area of the double strip from xi−1to xi+1is given by Ai(estim.) =integraldisplayh −h(fi+ay+by2)dy=2hfi+2 3bh3. Substituting for bh2then yields for the estimated area Ai(estim.) = 2 hfi+2 3h×1 2(fi+1+fi−1−2fi) =1 3h(4fi+fi+1+fi−1), an expression involving only given quantities. It should be noted that the values of neither bnoraneed be calculated. For the full integral I(estim.) =1 3h(f0+fN+4summationdisplay moddfm+2summationdisplay mevenfm). (28.40) It can be shown, by following the same procedure as in the trapezium rule case, that the error in the estimated area is approximately ∆I(estim.)≈(b−a) 180h4/angbracketleftf(4)/angbracketright. 28.4.3 Gaussian integration In the cases considered in the previous two subsections, the function fwas mimicked by linear and quadratic functions. These yield exact answers if f itself is a linear or quadratic function (respectively) of x. This process could be continued by increasing the order of the polynomial mimicking-function soas to increase the accuracy with which more complicated functions fcould be numerically integrated; but the same eﬀect can be achieved with less eﬀort by not insisting upon equally spaced points x i. The detailed analysis of such methods of numerical integration, in which the integration points are not equally spaced and the weightings given to the values at each point do not fall into a few simple groups, is too long to be given here.The reader is referred to books devoted speciﬁcally to the theory of numericalanalysis, where details of the integration points and weights for many schemeswill be found. † We will content ourselves here with describing Gaussian integration, which is based upon the orthogonality properties, in the interval −1≤x≤1, of the Legendre polynomials P /lscript(x), discussed in subsection 16.6.2. In order to use these properties, the integral between limits aandbin (28.34) has to be changed to †The points and weights may be found in, e.g. Abramowitz and Stegun, Handbook of Mathematical Functions (Dover, 1965). 1168 28.4 NUMERICAL INTEGRATION one between the limits −1 and +1. This is easily done with a change of variable from xtozgiven by z=2x−b−a b−a, so that Ibecomes I=b−a 2integraldisplay1 −1g(z)dz, (28.41) in which g(z)≡f(x). Thenintegration points xifor an n-point Gaussian integration are given by the zeroes of Pn(x), i.e. the xiare such that Pn(xi) = 0. The integrand g(x)i s mimicked by the ( n−1)th-degree polynomial G(x)=nsummationdisplay i=1Pn(x) (x−xi)P/primen(xi)g(xi), which coincides with g(x) at each of the points xi,i=1,2,...,n. To see this it should be noted that lim x→xkPn(x) (x−xi)P/primen(xi)=δik. It then follows, to the extent that g(x) is well reproduced by G(x), that integraldisplay1 −1g(x)dx≈nsummationdisplay i=1g(xi) P/primen(xi)integraldisplay1 −1Pn(x) x−xidx. (28.42) The expression w(xi)≡1 P/primen(xi)integraldisplay1 −1Pn(x) x−xidx can be shown, using the properties of Legendre polynomials, to be equal to wi=2 (1−x2 i)|P/primen(xi)|2, and is thus the weighting to be attached to the factor g(xi) in the sum (28.42), which becomes integraldisplay1 −1g(x)dx≈nsummationdisplay i=1wig(xi). (28.43) In fact, because of the particular properties of Legendre polynomials, it can be shown that (28.43) integrates exactly any polynomial of degree up to 2 n−1. The error in the approximate equality is of the order of the 2 nth derivative of gand so, provided g(x) is a reasonably smooth function, the approximation is a good one. 1169 NUMERICAL METHODS As an example, for a three-point integration, the three xiare the zeroes of P3(x)=1 2(5x3−3x), namely 0 and ±0.774 60, and the corresponding weights are 2 1×parenleftbig −3 2parenrightbig2=8 9and2 (1−0.6)×parenleftbig6 2parenrightbig2=5 9. For other forms of integrand, formulae based on other sets of orthogonal functions give better results. For example, integrals over ﬁnite ranges involvingfactors of the form (1 −x 2)±1/2in the integrand are best treated using formulae based on Chebyshev polynomials, whilst inﬁnite integrals containing e−x(0≤ x<∞)o re−x2(−∞<x<∞) are best handled using schemes based on Laguerre or Hermite polynomials respectively.IUsing a three-point formula in each case, evaluate the integral I= Z1 01 1+x2dx, (i) using the trapezium rule, (ii) using Simpson’s rule, (iii) using Gaussian integration. Also evaluate the integral analytically and compare the results. (i) Using the trapezium rule, we obtain I=1 2×1 2 / f(0) + 2 f /;1 2 / +f(1) / =1 4 / 1+8 5+1 2 / =0.7750. (ii) Using Simpson’s rule, we obtain I=1 3×1 2 / f(0) + 4 f /;1 2 / +f(1) / =1 6 / 1+16 5+1 2 / =0.7833. (iii) Using Gaussian integration, we obtain I=1−0 2 Z1 −1dz 1+1 4(z+1 )2 =1 2 n 0.55556 [f(−0.77460) + f(0.77460) ]+0.88889 f(0) o =1 2 n 0.55556 [0.987458 + 0 .559503 ]+0.88889×0.8 o =0.78527 . (iv) Exact evaluation gives I= Z1 0dx 1+x2= / tan−1x /1 0=π 4=0.78540 . In practice, a compromise has to be struck between the accuracy of the result achieved and the calculational labour that goes into obtaining it. J 28.4.4 Monte Carlo methods Surprising as it may at ﬁrst seem, random numbers may be used to carry out numerical integration. The random element comes in principally when selecting 1170 28.4 NUMERICAL INTEGRATION the points at which the integrand is evaluated, and naturally does not extend to the actual values of the integrand! For the most part we will continue to use as our model one-dimensional integrals between ﬁnite limits, as typiﬁed by equation (28.34). Extensions to cover inﬁnite or multidimensional integrals will be indicated brieﬂy at the end of thesection. It should be noted here, however, that Monte Carlo methods – the namehas become attached to methods based on randomly generated numbers – inmany ways come into their own when used on multidimensional integrals overregions with complicated boundaries. It goes without saying that in order to use random numbers for calculational purposes a supply of them must be available. There was a time when theywere provided in book form as a two-dimensional array of random digits inthe range 0 to 9, and the user could generate a random number of any desiredlength by selecting the positions in the table of its successive digits in anypredetermined and systematic way. Nowadays all computers and nearly all pocket calculators oﬀer a function which supplies a sequence of decimal numbers ξthat, for all practical purposes, are randomly and uniformly chosen in the range0≤ξ<1. The maximum number of signiﬁcant ﬁgures available in each random number depends on the precision of the generating device. We will defer thedetails of how these numbers are produced to a later subsection, where it willalso be shown how random numbers distributed in a prescribed way can begenerated. All integrals of the general form shown in equation (28.34) can, by a suitable change of variable, be brought to the form θ=integraldisplay 1 0f(x)dx, (28.44) and we will use this as our standard model. All approaches to integral evaluation based on random numbers proceed by estimating a quantity whose expectation value is equal to the sought-for value θ. The estimator tmust be unbiased, i.e. we must have E[t]=θ, and the method must provide some measure of the likely error in the result. The latter will appeargenerally as the variance of the estimate, with its usual statistical interpretation,and not as a band in which the true answer is known to lie with certainty. The various approaches really diﬀer from each other only in the degree of sophistication employed to keep the variance of the estimate of θsmall. The overall eﬃciency of any particular method has to take into account not only thevariance of the estimate but also the computing and book-keeping eﬀort requiredto achieve it. We do not have the space to describe even the more elementary methods in full detail, but the main thrust of each approach should be apparent to the readerfrom the brief descriptions that follow. 1171 NUMERICAL METHODS Crude Monte Carlo The most straightforward application is one in which the random numbers areused to pick sample points at which f(x) is evaluated. These values are then averaged: t=1 nnsummationdisplay i=1f(ξi). (28.45) Stratiﬁed sampling Here the range of xis broken up into ksubranges, 0=α0<α1<···<α k=1, and crude Monte Carlo evaluation is carried out in each subrange. The estimate E[t] is then calculated as E[t]=ksummationdisplay j=1njsummationdisplay i=1αj−αj−1 njfparenleftbig αj−1+ξij(αj−αj−1)parenrightbig . (28.46) This is an unbiased estimator of θwith variance σ2 t=ksummationdisplay j=1αj−αj−1 njintegraldisplayαj αj−1[f(x)]2dx−ksummationdisplay j=11 njbracketleftBiggintegraldisplayαj αj−1f(x)dxbracketrightBigg2 . This variance can be made less than that for crude Monte Carlo, whilst using the same total number of random numbers, n=summationtextnj, if the diﬀerences between the average values of f(x) in the various subranges are signiﬁcantly greater than the variations in fwithin each subrange. It is easier administratively to make all subranges equal in length but better, if it can be managed, to make them suchthat the variations in fare approximately equal in all the individual subranges. Importance sampling Although we cannot integrate f(x) analytically – we would not be using Monte Carlo methods if we could – if we can ﬁnd another function g(x)t h a t canbe integrated analytically and mimics the shape of fthen the variance in the estimate ofθcan be reduced signiﬁcantly compared with that resulting from the use of crude Monte Carlo evaluation. Firstly, if necessary the function gmust be renormalised, so that G(x)=integraltext x 0g(y)dyhas the property G(1) = 1. Clearly, it also has the property G(0) = 0. Then, since θ=integraldisplay1 0f(x) g(x)dG(x), it follows that ﬁnding the expectation value of f(η)/g(η) using a random number η, distributed in such a way that ξ=G(η) is uniformly distributed on (0 ,1), is equivalent to estimating θ. This involves being able to ﬁnd the inverse function 1172 28.4 NUMERICAL INTEGRATION ofG; a discussion of how to do this is given in a later subsection. If g(η) mimics f(η) well, f(η)/g(η) will be nearly constant and the estimation will have a very small variance. Further, any error in inverting the relationship between ηandξ will not be important since f(η)/g(η) will be largely independent of the value ofη. As an example, consider the function f(x)=[ t a n−1(x)]1/2, which is not analyti- cally integrable over the range (0 ,1) but is well mimicked by the easily-integrated function g(x)=x1/2(1−x2/6). The ratio of the two varies from 1 .00 to 1 .06 as x varies from 0 to 1. The integral of gover this range is 0 .619 048, and so it has to be renormalised by the factor 1 .615 38. The value of the integral of f(x)f r o m0 to 1 can then be estimated by averaging the value of [tan−1(η)]1/2 1.615 38 η1/2(1−1 6η2) for random variables ηwhich are such that G(η) is uniformly distributed on (0,1). Using batches of as few as 10 random numbers gave a value 0 .630 for θ, with standard deviation 0 .003. The corresponding result for crude Monte Carlo, using the same random numbers, was 0 .634±0.065. The increase in precision is obvious, though the additional labour involved would not be justiﬁed for a single application. Control variates The control-variate method is similar to, but not the same as, importance sam-pling. Again, an analytically integrable function that mimics f(x) in shape has to be found. The function, known as the control variate, is ﬁrst scaled so as tomatch fas closely as possible in magnitude and then its integral is found in closed form. If we denote the scaled control variate by h(x) then the estimate of θis computed as t=integraldisplay 1 0[f(x)−h(x)]dx+integraldisplay1 0h(x)dx. (28.47) The ﬁrst integral in (28.47) is evaluated using (crude) Monte Carlo, whilst the second is known analytically. Although the ﬁrst integral should have been ren-dered small by the choice of h(x), it is its variance that matters. The method relies on the result (see equation (26.136)) V[t−t /prime]=V[t]+V[t/prime]−2C o v [ t, t/prime] a n do nt h ef a c tt h a ti f testimates θwhilst t/primeestimates θ/primeusing the same random numbers then the covariance of tandt/primecan be larger than the variance of t/prime,a n d indeed will be so if the integrands producing θandθ/primeare highly correlated. To evaluate the same integral as was estimated previously using importance sampling, we take as h(x) the function g(x) used there, before it was renormalised. Again using batches of 10 random numbers, the estimated value for θwas found 1173 NUMERICAL METHODS to be 0 .629±0.004, a result almost identical to that obtained using importance sampling, in both value and precision. Since we knew already that f(x)a n d g(x) diverge monotonically by about 6% as xvaries over the range (0 ,1), we could have made a small improvement to our control variate by scaling it by 1 .03 before using it in equation (28.47). Antithetic variates As a ﬁnal example of a method that improves on crude Monte Carlo, and one thatis particularly useful when monotonic functions are to be integrated, we mentionthe use of antithetic variates. This method relies on ﬁnding two estimates tand t /primeofθthat are strongly anticorrelated (i.e. Cov[ t, t/prime] is large and negative) and using the result V[1 2(t+t/prime)] =1 4V[t]+1 4V[t/prime]+1 2Cov[t, t/prime]. For example, the use of1 2[f(ξ)+f(1−ξ)] instead of f(ξ) involves only twice as many evaluations of f, and no more random variables, but generally gives an improvement in precision signiﬁcantly greater than this. For the integral of f(x)=[ t a n−1(x)]1/2, using as previously a batch of 10 random variables, an estimate of 0 .623±0.018 was found. This to be compared with the crude Monte Carlo result, 0 .634±0.065, obtained using the same number of random variables. For a fuller discussion of these methods, and of theoretical estimates of their eﬃciencies, the reader is referred to more specialist treatments. For practicalimplementation schemes, a book dedicated to scientiﬁc computing should beconsulted.† Hit or miss method We now come to the approach that, in spirit, is closest to the activities that gave Monte Carlo methods their name. In this approach, one or more straightforwardyes/no decisions are made on the basis of numbers drawn at random – the endresult of each trial is either a hit or a miss! In this section we are concernedwith numerical integration, but the general Monte Carlo approach, in whichone estimates a physical quantity that is hard or impossible to calculate directlyby simulating the physical processes that determine it, is widespread in modern science. For example, the calculation of the eﬃciencies of detector arrays in experiments to study elementary particle interactions are nearly always carriedout in this way. Indeed, in a normal experiment, far more simulated interactionsare generated in computers than ever actually occur when the experiment istaking real data. As was noted in chapter 2, the process of evaluating a one-dimensional integralintegraltext b af(x)dxcan be regarded as that of ﬁnding the area between the curve y=f(x) †e.g.,Numerical Recipes ,W .H .P r e s s et al. (Cambridge University Press). 1174 28.4 NUMERICAL INTEGRATION xy=f(x) y=c x=ax =b Figure 28.5 A simple rectangular ﬁgure enclosing the area (shown shaded) which is equal to Rb af(x)dx. and the x-axis in the range a≤x≤b. It may not be possible to do this analytically but if, as shown in ﬁgure 28.5, we can enclose the curve in a simpleﬁgure whose area can be found trivially then the ratio of the required area (shown shaded) to that of the bounding ﬁgure, c(b−a), is the same as the probability that a randomly selected point inside the boundary will lie below the line. In order to accommodate cases in which f(x) can be negative in part of the x-range, we treat a slightly more general case. Suppose that, for a≤x≤b,f(x) is bounded and known to lie in the range A≤f(x)≤B, then the transformation z=x−a b−a will reduce the integralintegraltextb af(x)dxto the form A(b−a)+(B−A)(b−a)integraldisplay1 0h(z)dz, (28.48) where h(z)=1 B−A[f((b−a)z+a)−A]. In this form zlies in the range 0 ≤z≤1a n d h(z) lies in the range 0 ≤h(z)≤1, i.e. both are suitable for simulation using the standard random-number generator.It should be noted that for an eﬃcient estimation the bounds AandBshould be drawn as tightly as possible – preferably, but not necessarily, they should be equal to the minimum and maximum values of fin the range. The reason for this is that random numbers corresponding to values which f(x) cannot reach add nothing to the estimation but do increase its variance. It only remains to estimate the ﬁnal integral on the RHS of equation (28.48). This we do by selecting pairs of random numbers ξ 1andξ2and testing whether 1175 NUMERICAL METHODS h(ξ1)>ξ2. The fraction of times that this inequality is satisﬁed estimates the value of the integral (without the scaling factors ( B−A)(b−a)) since the expectation value of this fraction is the ratio of the area below the curve y=h(z)t ot h ea r e a of a unit square. To illustrate the evaluation of multiple integrals using Monte Carlo techniques, consider the relatively elementary problem of ﬁnding the volume of an irregularsolid bounded by planes, say an octahedron. In order to keep the descriptionbrief, but at the same time illustrate the general principles involved, let us supposethat the octahedron has two vertices on each of the three Cartesian axes, one on either side of the origin for each axis. Denote those on the x-axis by x 1(<0) and x2(>0), and similarly for the y-a n d z-axes. Then the whole of the octahedron can be enclosed by the rectangular parallelepiped x1≤x≤x2,y 1≤y≤y2,z 1≤z≤z2. Any point in the octahedron lies inside or on the parallelepiped, but any point in the parallelepiped may or may not lie inside the octahedron. The equation of the plane containing the three vertex points ( xi,0,0),(0,yj,0) and (0 ,0,zk)i s x xi+y yj+z zk=1 f o r i, j, k=1,2, (28.49) and the condition that any general point ( x, y, z) lies on the same side of the plane as the origin is that x xi+y yj+z zk−1≤0. (28.50) For the point to be inside or on the octahedron, equation (28.50) must therefore be satisﬁed for all eight of the sets of i, jandkgiven in (28.49). Thus an estimate of the volume of the octahedron can be made by generating random numbers ξfrom the usual uniform distribution and then using them in sets of three, according to the following scheme. With integer mlabelling the mth set of three random numbers, calculate x=x1+ξ3m−2(x2−x1), y=y1+ξ3m−1(y2−y1), z=z1+ξ3m(z2−z1). Deﬁne a variable nmas 1 if (28.50) is satisﬁed for all eight combinations of i, j, k values and as 0 otherwise. The volume Vcan then be estimated using 3 Mrandom numbers from the formula V (x2−x1)(y2−y1)(z2−z1)=1 MMsummationdisplay m=1nm. 1176 28.4 NUMERICAL INTEGRATION It will be seen that, by replacing each nmin the summation by f(x, y, z)nm,t h i s procedure could be extended to estimating the integral of the function fover the volume of the solid. The method has special value if fis too complicated to have analytic integrals with respect to x, yandzor if the limits of any of these integrals are determined by anything other than the simplest combinations of the other variables. If large values of fare known to be concentrated in particular regions of the integration volume then some form of stratiﬁed sampling shouldbe used. It will be apparent that this general method can be extended to integrals of general functions, bounded but not necessarily continuous, over volumes withcomplicated bounding surfaces and, if appropriate, in more than three dimensions. Random number generation Earlier in this subsection we showed how to evaluate integrals using sequences of numbers that we took to be distributed uniformly on the interval 0 ≤ξ<1. In reality the sequence of numbers is not truly random, since each is generated in amechanistic way from its predecessor and eventually the sequence repeats itself.However, the cycle is so long that in practice this is unlikely to be a problem,and the reproducibility of the sequence can even be turned to advantage whenchecking the accuracy of the rest of a calculational program. Much research has gone into the best ways to produce such ‘pseudo-random’ sequences of numbers. We do not have space to pursue them here and will limit ourselves to one recipethat works well in practice. Given any particular starting (integer) value x 0, the following algorithm will generate a full cycle of mvalues for ξi, uniformly distributed on 0 ≤ξi<1, before repeats appear: xi=axi−1+c(mod m); ξi=xi m. Here cis an odd integer and ahas the form a=4k+1w i t h kan integer. For practical reasons, in computers and calculators mis taken as a (fairly high) power of 2, typically 32. The uniform distribution can be used to generate random numbers ydistributed according to a more general probability distribution f(y) on the range a≤y≤b if the inverse of the indeﬁnite integral of fcan be found, either analytically or by means of a look-up table. In other words, if F(y)=integraldisplayy af(t)dt, for which F(a)=0a n d F(b)=1t h e n F(y) is uniformly distributed on (0 ,1). This approach is not limited to ﬁnite aandb;acould be−∞andbcould be∞. The procedure is thus to select a random number ξfrom a uniform distribution 1177 NUMERICAL METHODS on (0 ,1) and then take as the random number ythe value of F−1(ξ). We now illustrate this with a worked example.IFind an explicit formula that will generate a random number ydistributed on (−∞,∞) according to the Cauchy distribution f(y)dy= /a π /dy a2+y2, given a random number ξuniformly distributed on (0,1). The ﬁrst task is to determine the indeﬁnite integral F(y)= Zy −∞ /a π /dt a2+t2=1 πtan−1y a+1 2. Now, if yis distributed as we wish then F(y) is uniformly distributed on (0 ,1). This follows from the fact that the derivative of F(y)i sf(y). We therefore set F(y)e q u a lt o ξand obtain ξ=1 πtan−1y a+1 2, yielding y=atan[π(ξ−1 2)]. This explicit formula shows how to change a random number ξdrawn from a population uniformly distributed on (0 ,1) into a random number ydistributed according to the Cauchy distribution. J Look-up tables operate as described below for cumulative distributions F(y) that are non-invertible, i.e. F−1(y) cannot be expressed in closed form. They are especially useful if many random numbers are needed but great samplingaccuracy is not essential. The method for an N-entry table can be summarised as follows. Deﬁne w mbyF(wm)=m/Nform=1,2,...,N , and store a table of y(m)=1 2(wm+wm−1). As each random number yis needed, calculate kas the integral part of Nξand take yas given by y(k). Normally, such a look-up table would have to be used for generating random numbers with a Gaussian distribution, as the cumulative integral of a Gaussian isnon-invertible. It would be in essence table 26.3, with the roles of argument andvalue interchanged. In this particular case an alternative, based on the centrallimit theorem, can be considered. With ξ igenerated in the usual way, i.e. uniformly distibuted on the interval 0≤ξ<1, the random variable y=nsummationdisplay i=1ξi−1 2n (28.51) is normally distributed with mean 0 and variance n/12 when nis large. This 1178 28.5 FINITE DIFFERENCES approach does produce a continuous spectrum of possible values for y, but needs many values of ξifor each value of yand is a very poor approximation if the wings of the Gaussian distribution have to be sampled accurately. For nearly allpractical purposes a Gaussian look-up table is to be preferred. 28.5 Finite diﬀerences It will have been noticed that earlier sections included several equations linking sequential values of f iand the derivatives of fevaluated at one of the xi.I n this section, by way of preparation for the numerical treatment of diﬀerentialequations, we establish these relationships in a more systematic way. Again we consider a set of values f iof a function f(x)e v a l u a t e da te q u a l l y spaced points xi, their separation being h. As before, the basis for our discussion will be a Taylor series expansion, but on this occasion about the point xi: fi±1=fi±hf/prime i+h2 2!f/prime/prime i±h3 3!f(3) i+···. (28.52) In this section, and subsequently, we denote the nth derivative evaluated at xi byf(n) i. From (28.52), three diﬀerent expressions that approximate f(1) ican be derived. The ﬁrst of these, obtained by subtracting the ±equations, is f(1) i≡parenleftbiggdf dxparenrightbigg xi=fi+1−fi−1 2h−h2 3!f(3) i−···. (28.53) The quantity ( fi+1−fi−1)/(2h) is known as the central diﬀerence approximation tof(1) iand can be seen from (28.53) to be in error by approximately ( h2/6)f(3) i. An alternative approximation, obtained from (28.52+) alone, is given by f(1) i≡parenleftbiggdf dxparenrightbigg xi=fi+1−fi h−h 2!f(2) i−···. (28.54) Theforward diﬀerence approximation, ( fi+1−fi)/h, is clearly a poorer approxima- tion, since it is in error by approximately ( h/2)f(2) i, as compared with ( h2/6)f(3) i. Similarly, the backward diﬀerence ( fi−fi−1)/hobtained from (28.52 −) is not as good as the central diﬀerence; the sign of the error is reversed in this case. This type of diﬀerencing approximation can be continued to the higher deriva- tives of fin an obvious manner. By adding the two equations (28.52 ±), a central diﬀerence approximation to f(2) ican be obtained: f(2) i≡parenleftbiggd2f dx2parenrightbigg ≈fi+1−2fi+fi−1 h2. (28.55) The error in this approximation (also known as the second diﬀerence of f)i s e a s i l ys h o w nt ob ea b o u t( h2/12)f(4) i. Of course, if the function f(x) is a suﬃciently simple polynomial in x,a l l 1179 NUMERICAL METHODS derivatives beyond a particular one will vanish and there is no error in taking the diﬀerences to obtain the derivatives.IThe following is copied from the tabulation of a second-degree polynomial f(x)at values ofxfrom1to12inclusive, 2,2,?,8,14,22,32,46,?,74,92,112. The entries marked ?were illegible and in addition one er ror was made in transcription. Complete and correct the table. Would your procedure have worked if the copying errorhad been in f(6)? Write out the entries again in row (a) below, and where possible calculate ﬁrst diﬀerences in row (b) and second diﬀerences in row (c). Denote the jth entry in row ( n)b y( n)j. (a) 2 2 ? 8 14 22 32 46 ? 74 92 112 (b) 0 ? ? 6 8 10 14 ? ? 18 20(c) ? ? ? 2 2 4 ? ? ? 2 Because the polynomial is second-degree the second diﬀerences (c) j, which are proportional tod2f/dx2, should be constant, and clearly the constant should be 2. That is, (c) 6should equal 2 and (b) 7should equal 12 (not 14). Since all the (c) j= 2, we can conclude that (b)2=2 ,( b ) 3=4 ,( b ) 8= 14, and (b) 9= 16. Working these changes back to row (a) shows that (a) 3=4 ,( a ) 8= 44 (not 46), and (a) 9= 58. The entries therefore should read (a) 2,2,4,8,14,22,32,44,58,74,92,112, where the amended entries are shown in bold type. It is easily veriﬁed that if the error were in f(6) no two computable entries in row (c) would be equal, and it would not be clear what the correct common entry should be.Nevertheless, trial and error might arrive at a self-consistent scheme.J 28.6 Diﬀerential equations For the remaining sections of this chapter our attention will be on the solution of diﬀerential equations by numerical methods. Some of the general diﬃcultiesof applying numerical methods to diﬀerentia l equations will be all too apparent. Initially we consider only the simplest kind of equation – one of ﬁrst order, typically represented by dy dx=f(x, y), (28.56) where yis taken as the dependent variable and xthe independent one. If this equation can be solved analytically then that is the best course to adopt. Butsometimes it is not possible to do so and a numerical approach becomes theonly one available. In fact, most of the examples that we will use can be solved easily by an explicit integration, but, for the purposes of illustration, this is an advantage rather than the reverse since useful comparisons can then be madebetween the numerically derived solution and the exact one. 1180 28.6 DIFFERENTIAL EQUATIONS x h y(exact) 0.01 0.1 0.5 1.0 1.5 2 3 0 (1) (1) (1) (1) (1) (1) (1) (1) 0.5 0.605 0.590 0.500 0 −0.500−1−20.607 1.0 0.366 0.349 0.250 0 0 .250 1 4 0.368 1.5 0.221 0.206 0.125 0 −0.125−1−80.223 2.0 0.134 0.122 0.063 0 0 .063 1 16 0.135 2.5 0.081 0.072 0.032 0 −0.032−1−32 0.082 3.0 0.049 0.042 0.016 0 0 .016 1 64 0.050 Table 28.8 The solution yof diﬀerential equation (28.57) using the Euler forward diﬀerence method for various values of h. The exact solution is also shown. 28.6.1 Diﬀerence equations Consider the diﬀerential equation dy dx=−y, y (0) = 1 , (28.57) and the possibility of solving it numerically by approximating dy/dx by a ﬁnite diﬀerence along the lines indicated in section 28.5. We start with the forwarddiﬀerence parenleftbiggdy dxparenrightbigg xi≈yi+1−yi h, (28.58) where we use the notation of section 28.5 but with freplaced by y.I nt h i s particular case, it leads to the recurrence relation yi+1=yi+hparenleftbiggdy dxparenrightbigg i=yi−hyi=( 1−h)yi. (28.59) Thus, since y0=y(0) = 1 is given, y1=y(0 +h)=y(h) can be calculated, and so on (this is the Euler method). Table 28.8 shows the values of y(x) obtained if this is done using various values of hand for selected values of x. The exact solution, y(x)=e x p (−x), is also shown. It is clear that to maintain anything like a reasonable accuracy only very small steps hcan be used. Indeed, if his taken to be too large, not only is the accuracy bad but, as can be seen, for h>1 the calculated solution oscillates (when it should be monotonic) and for h>2 it diverges. Equation (28.59) is of the form yi+1=λyi, and a necessary condition for non-divergence is |λ|<1, i.e. 0 <h< 2, though in no way does this ensure accuracy. Part of this diﬃculty arises from the poor approximation (28.58); its right- hand side is a closer approximation to dy/dx evaluated at x=xi+h/2t h a nt o dy/dx atx=xi. This is the result of using a forward diﬀerence rather than the 1181 NUMERICAL METHODS xy (estim.) y(exact) −0.5 (1.648) – 0 (1.000) (1.000) 0.5 0.648 0.607 1.0 0.352 0.368 1.5 0.296 0.223 2.0 0.056 0.135 2.5 0.240 0.082 3.0−0.184 0.050 Table 28.9 The solution of diﬀerential equation (28.57) using the Milne central diﬀerence method with h=0.5 and accurate starting values. more accurate, but of course still approximate, central diﬀerence. A more accurate method based on central diﬀerences ( Milne’s method ) gives the recurrence relation yi+1=yi−1+2hparenleftbiggdy dxparenrightbigg i(28.60) in general and, in this particular case, yi+1=yi−1−2hyi. (28.61) An additional diﬃculty now arises, since two initial values of yare needed. The second must be estimated by other means (e.g. by using a Taylor series,as discussed later) but for illustration purposes we will take the accurate value,y(−h)=e x p h, as the value of y −1.I fhis taken as, say, 0.5 and (28.61) applied repeatedly then the results shown in table 28.9 are obtained. Although some improvement in the early values of the calculated y(x)i s noticeable, as compared with the corresponding ( h=0.5) column of table 28.8, this scheme soon runs into diﬃculties, as is obvious from the last two rows of thetable. Some part of this poor performance is not really attributable to the approxi- mations made in estimating dy/dx but to the form of the equation itself and hence of its solution. Anyrounding error occurring in the evaluation eﬀectively introduces into ysome contamination by the solution of dy dx=+y. This equation has the solution y(x)=e x p xand so grows without limit; ultimately it will dominate the sought-for solution and thus render the calculations totally inaccurate. We have only illustrated, rather than analysed, some of the diﬃculties associated with simple ﬁnite-diﬀerence iteration schemes for ﬁrst-order diﬀerential equations, 1182 28.6 DIFFERENTIAL EQUATIONS but they may be summarised as (i) insuﬃciently precise approximations to the derivatives and (ii) inherent instability due to rounding errors. 28.6.2 Taylor series solutions Since a Taylor series expansion is exact if all its terms are included, and the limits of convergence are not exceeded, we may seek to use one to evaluate y1,y2etc. for an equation dy dx=f(x, y), (28.62) when the initial value y(x0)=y0is given. The Taylor series is y(x+h)=y(x)+hy/prime(x)+h2 2!y/prime/prime(x)+h3 3!y(3)(x)+···. (28.63) In the present notation, at the point x=xithis is written yi+1=yi+hy(1) i+h2 2!y(2) i+h3 3!y(3) i+···. (28.64) But, for the required solution y(x), we know that y(1) i≡parenleftbiggdy dxparenrightbigg xi=f(xi,yi) (28.65) and the value of the second derivative at x=xi,y=yican be obtained from it: y(2) i=∂f ∂x+∂f ∂ydy dx=∂f ∂x+f∂f ∂y. (28.66) This process can be continued for the third and higher derivatives, all of which are to be evaluated at ( xi,yi). Having obtained expressions for the derivatives y(n) iin (28.63), two alternative ways of proceeding are open: (i) equation (28.64) is used to evaluate yi+1and then the whole process is repeated to obtain yi+2a n ds oo n ; (ii) equation (28.64) is applied several times but using a diﬀerent value of h each time, and so the corresponding values of y(x+h) are obtained. It is clear that, on the one hand, approach (i) does not require so many terms of (28.63) to be kept but, on the other hand, the yi(n) have to be recalculated at each step. With approach (ii), fairly accurate results for ymay be obtained for values ofxclose to the given starting value, but for large values of ha large number of terms of (28.63) must be kept. As an example of approach (ii) we solve thefollowing problem. 1183 NUMERICAL METHODS xy (estim.) y(exact) 0 1.0000 1.0000 0.1 1.2346 1.23460.2 1.5619 1.5625 0.3 2.0331 2.0408 0.4 2.7254 2.77780.5 3.7500 4.0000 Table 28.10 The solution of diﬀerential equation (28.67) using a Taylor series.IFind the numerical solution of the equation dy dx=2y3/2,y (0) = 1 , (28.67) forx=0.1to0.5in steps of 0.1. Compare it with the exact solution obtained analytically. Since the right-hand side of the equation does not contain xexplicitly, (28.66) is greatly simpliﬁed and the calculation becomes a repeated application of y(n+1) i=∂y(n) ∂ydy dx=f∂y(n) ∂y. The necessary derivatives and their values at x=0 ,w h e r e y= 1, are given below. y(0) = 1 1 y/prime=2y3/22 y/prime/prime=( 3/2)(2y1/2)(2y3/2)=6 y26 y(3)=( 1 2 y)2y3/2=2 4y5/224 y(4)=( 6 0 y3/2)2y3/2= 120 y3120 y(5)= (360 y2)2y3/2= 720 y7/2720 Thus the Taylor expansion of the solution about the origin (in fact a Maclaurin series) is y(x)=1+2 x+6 2!x2+24 3!x3+120 4!x4+720 5!x5+···. Hence, y(estim.) = 1 + 2 x+3x2+4x3+5x4+6x5. Values calculated from this are given in table 28.10. Comparison with the exact values shows that using the ﬁrst six terms givesa value that is correct to one part in 100, up to x=0.3.J 28.6.3 Prediction and correction An improvement in the accuracy obtainable using diﬀerence methods is possible if steps are taken, sometimes retrospectively, to allow for inaccuracies in approx- imating derivatives by diﬀerences. We will describe only the simplest schemes ofthis kind and begin with a prediction method, usually called the Adams method . 1184 28.6 DIFFERENTIAL EQUATIONS The forward diﬀerence estimate of yi+1,n a m e l y yi+1=yi+hparenleftbiggdy dxparenrightbigg i=yi+hf(xi,yi), (28.68) would give exact results if ywere a linear function of xin the range xi≤x≤xi+h. The idea behind the Adams method is to allow some relaxation of this andsuppose that ycan be adequately approximated by a parabola over the interval x i−1≤x≤xi+1. In the same interval dy/dx can then be approximated by a linear function: f(x, y)=dy dx≈a+b(x−xi)f o r xi−h≤x≤xi+h. The values of aandbare ﬁxed by the calculated values of fatxi−1andxi,w h i c h we may denote by fi−1andfi: a=fi,b =fi−fi−1 h. Thus yi+1−yi≈integraldisplayxi+h xibracketleftbigg fi+(fi−fi−1) h(x−xi)bracketrightbigg dx, which yields yi+1=yi+hfi+1 2h(fi−fi−1). (28.69) The last term of this expression is seen to be a correction to result (28.68). That it is, in some sense, the second-order correction 1 2h2y(2) i−1/2 to a ﬁrst-order formula is apparent. Such a procedure requires, in addition to a value for y0, a value for either y1or y−1,s ot h a t f1orf−1can be used to initiate the iteration. This has to be obtained by other methods, e.g. a Taylor series expansion. Improvements to simple diﬀerence formulae can also be obtained by using correction methods. Here a rough prediction of the value yi+1is ﬁrst made and then this is used in a better formula, not originally usable since it in turn requires a value of yi+1for its evaluation. The value of yi+1is then recalculated using this better formula. Such a scheme based on the forward diﬀerence formula might be as follows: (i) predict yi+1using yi+1=yi+hfi; (ii) calculate fi+1using this value; (iii) recalculate yi+1using yi+1=yi+h(fi+fi+1)/2. Here ( fi+fi+1)/2h a s replaced the fiused in (i), since it better represents the average value of dy/dx in the interval xi≤x≤xi+h. 1185 NUMERICAL METHODS Steps (ii) and (iii) can be iterated to improve further the approximation to the average value of dy/dx , but this will not compensate for the omission of higher- order derivatives in the forward diﬀerence formula. Many more complex schemes of prediction and correction, in most cases combining the two in the same process, have been devised, but the reader isreferred to more specialist texts for discussions of them. However, because itoﬀers some clear advantages, one group of methods will be set out explicitly inthe next subsection. This is the general class of schemes known as Runge–Kuttamethods. 28.6.4 Runge–Kutta methods The Runge–Kutta method of integrating dy dx=f(x, y) (28.70) is a step-by-step process of obtaining an approximation for yi+1by starting from the value of yi. Among its advantages are that no functions other than fare used, no subsidiary diﬀerentiation is needed and no additional starting values need be calculated. To be set against these advantages is the fact that fis evaluated using somewhat complicated arguments and that this has to be done several times for each increase in the value of i. However, once a procedure has been established, for example on a computer, the method usually gives good results. The basis of the method is to simulate the (accurate) Taylor series for y(xi+h), not by calculating all the higher derivatives of yat the point xibut by taking a particular combination of the values of the ﬁrst derivative of yevaluated at a number of carefully chosen points. Equation (28.70) is used to evaluate thesed e r i v a t i v e s .T h ea c c u r a c yc a nb em a d et ob eu pt ow h a t e v e rp o w e ro f his desired but, naturally, the greater the accuracy the more complex the calculation and, inany case, rounding errors cannot ultimately be avoided. The setting up of the calculational scheme may be illustrated by considering the particular case in which second-order accuracy in his required. To second order, the Taylor expansion is y i+1=yi+hfi+h2 2parenleftbiggdf dxparenrightbigg xi, (28.71) whereparenleftbiggdf dxparenrightbigg xi=parenleftbigg∂f ∂x+f∂f ∂yparenrightbigg xi≡∂fi ∂x+fi∂fi ∂y, the last step being merely the deﬁnition of an abbreviated notation. 1186 28.6 DIFFERENTIAL EQUATIONS We assume that this can be simulated by a form yi+1=yi+α1hfi+α2hf(xi+β1h, y i+β2hfi), (28.72) which in eﬀect uses a weighted mean of the value of dy/dx atxiand its value at some point yet to be determined. The object is to choose values of α1,α2,β1and β2such that (28.72) coincides with (28.71) up to the coeﬃcient of h2. Expanding the function fin the last term of (28.72) in a Taylor series of its own we obtain f(xi+β1h, y i+β2hfi)=f(xi,yi)+β1h∂fi ∂x+β2hfi∂fi ∂y+O ( h2). Putting this result into (28.72) and rearranging in powers of hwe obtain yi+1=yi+(α1+α2)hfi+α2h2parenleftbigg β1∂fi ∂x+β2fi∂fi ∂yparenrightbigg . (28.73) Comparing this with (28.71) shows that there is in fact some freedom remaining in the choice of the α’s and β’s. In terms of an arbitrary α1(/negationslash=1 ) α2=1−α1,β 1=β2=1 2(1−α1). One possible choice is α1=0.5, giving α2=0.5,β1=β2= 1. In this case the procedure (equation (28.72)) can be summarised by yi+1=yi+1 2(a1+a2), (28.74) where a1=hf(xi,yi), a2=hf(xi+h, y i+a1). Similar schemes giving higher-order accuracy in hcan be devised. Two such schemes, given without derivation, are (i) to order h3, yi+1=yi+1 6(b1+4b2+b3), (28.75) where b1=hf(xi,yi), b2=hf(xi+1 2h, y i+1 2b1), b3=hf(xi+h, y i+2b2−b1), 1187 NUMERICAL METHODS (ii) to order h4, yi+1=yi+1 6(c1+2c2+2c3+c4), (28.76) where c1=hf(xi,yi), c2=hf(xi+1 2h, y i+1 2c1), c3=hf(xi+1 2h, y i+1 2c2), c4=hf(xi+h, y i+c3). 28.6.5 Isoclines The ﬁnal method to be described for ﬁrst-order diﬀerential equations is not so much numerical as graphical, but since it is sometimes useful it is included here.The method, known as that of isoclines , involves sketching for a number of values of a parameter cthose curves (the isoclines) in the xy-plane along which f(x, y)=c, i.e. those curves along which dy/dx is a constant of known value. It should be noted that they are not generally straight lines. Since a straight lineof slope dy/dx at and through any particular point is a tangent to the curve y=y(x) at that point, small elements of straight lines, with slopes appropriate to the isoclines they cut, eﬀectively form the curve y=y(x). Figure 28.6 illustrates in outline the method as applied to the solution of dy dx=−2xy. (28.77) The thinner curves (rectangular hyperbolae) are a selection of the isoclines along which−2xyis constant and equal to the corresponding value of c.T h es m a l l cross lines on each curve show the slopes (= c) that solutions of (28.77) must have if they cross the curve. The thick line is the solution for which y=1a t x= 0; it takes the slope dictated by the value of con each isocline it crosses. The analytic solution with these properties is y(x)=e x p (−x2). 28.7 Higher-order equations So far the discussion of numerical solutions of diﬀerential equations has been in terms of one dependent and one independent variable related by a ﬁrst-orderequation. It is straightforward to carry out an extension to the case of severaldependent variables y [r]governed by Rﬁrst-order equations dy[r] dx=f[r](x, y[1],y[2],...,y [R]),r =1,2,...,R. We have enclosed the label rin brackets so that there is no confusion between, say, the second dependent variable y[2]and the value y2of a variable yat the 1188 28.7 HIGHER-ORDER EQUATIONS 0.20.2 0.40.4 0.60.6 0.80.8 1.01.0 cy y x−1.0 −0.8 −0.6 −0.4 −0.2 −0.1 Figure 28.6 The isocline method. The cross lines on each isocline show the slopes that solutions of dy/dx =−2xymust have at the points where they cross the isoclines. The heavy line is the solution with y(0) = 1, namely exp(−x2). second calculational point x2. The integration of these equations by the methods discussed in the previous section presents no particular diﬃculty, provided that all the equations are advanced through each particular step before any of them is taken through the following step. Higher-order equations in one dependent and one independent variable can be reduced to a set of simultaneous equations, provided that they can be written inthe form d Ry dxR=f(x, y, y/prime,...,y(R−1)), (28.78) where Ris the order of the equation. To do this, a new set of variables p[r]is deﬁned by p[r]=dry dxr,r =1,2,...,R−1. (28.79) Equation (28.78) is then equivalent to the set of simultaneous ﬁrst-order equations dy dx=p[1], dp[r] dx=p[r+1],r =1,2,...,R−2, (28.80) dp[R−1] dx=f(x, y, p [1],...,p [R−1]). These can then be treated in the way indicated in the previous paragraph. The extension to more than one dependent variable is straightforward. 1189 NUMERICAL METHODS In practical problems it often happens that boundary conditions applicable to a higher-order equation consist not of the values of the function and all itsderivatives at one particular point but of, say, the values of the function at twoseparate end-points. In these cases a solution cannot be found using an explicit step-by-step ‘marching’ scheme, in which the solutions at successive values of the independent variable are calculated using solution values previously found. Othermethods have to be tried. One obvious method is to treat the problem as a ‘marching one’, but to use a number of (intelligently guessed) initial values for the derivatives at the startingpoint. The aim is then to ﬁnd, by interpolation or some other form of iteration,those starting values for the derivatives that will produce the given value of the function at the ﬁnishing point. In some cases the problem can be reduced by a diﬀerencing scheme to a matrix equation. Such a case is that of a second-order equation for y(x) with constant coeﬃcients and given values of yat the two end-points. Consider the second-order equation y /prime/prime+2ky/prime+µy=f(x), (28.81) with the boundary conditions y(0) = A, y (1) = B. If (28.81) is replaced by a central diﬀerence equation, yi+1−2yi+yi−1 h2+2kyi+1−yi−1 2h+µyi=f(xi), we obtain from it the recurrence relation (1 +kh)yi+1+(µh2−2)yi+( 1−kh)yi−1=h2f(xi). Forh=1/(N−1) this is in exactly the form of the N×Ntridiagonal matrix equation (28.30), with b1=bN=1,c 1=aN=0, ai=1−kh, b i=µh2−2,c i=1+ kh, i =2,3,...,N−1, andy1replaced by A,yNbyBandyibyh2f(xi)f o r i=2,3,...,N−1. The solutions can be obtained as in (28.31) and (28.32). 28.8 Partial diﬀerential equations The extension of previous methods to partial diﬀerential equations, thus involving two or more independent variables, proceeds in a more or less obvious way. Ratherthan an interval divided into equal steps by the points at which solutions to the 1190 28.8 PARTIAL DIFFERENTIAL EQUATIONS equations are to be found, a mesh of points in two or more dimensions has to be set up and all the variables given an increased number of subscripts. Considerations of the stability, accuracy and feasibility of particular calcula- tional schemes in principle are the same as for the one-dimensional case, but in practice are too complicated to be discussed here. Rather than note generalities that we are unable to pursue in any quantitative way, we will conclude this chapter by indicating in outline how two familiarpartial diﬀerential equations of physical science can be set up for numericalsolution. The ﬁrst of these is Laplace’s equation in two dimensions, ∂ 2φ ∂x2+∂2φ ∂y2= 0 (28.82) the value of φbeing given on the perimeter of a closed domain. A grid with spacings ∆ xand ∆ yin the two directions is ﬁrst chosen, so that, for example, xistands for the point x0+i∆xandφi,jfor the value φ(xi,yj). Next, using a second central diﬀerence formula, (28.82) is turned into φi+1,j−2φi,j+φi−1,j (∆x)2+φi,j+1−2φi,j+φi,j−1 (∆y)2=0, (28.83) fori=0,1,...,N andj=0,1,...,M .I f( ∆ x)2=λ(∆y)2then this becomes the recurrence relationship φi+1,j+φi−1,j+λ(φi,j+1+φi,j−1)=2 ( 1+ λ)φi,j. (28.84) The boundary conditions in their simplest form (i.e. for a rectangular domain) mean that φ0,j,φ N,j,φ i,0,φ i,M (28.85) have predetermined values. Non-rectangular boundaries can be accommodated, either by more complex boundary-value prescriptions or by using non-Cartesiancoordinates. To ﬁnd a set of values satisfying (28.84), an initial guess at a complete set of values for the φ i,jis made, subject to the requirement that the quantities listed in (28.85) have the given ﬁxed values; those values that are not on the boundaryare then adjusted iteratively in order to try to bring about condition (28.84)everywhere. Clearly one scheme is to set λ= 1 and recalculate each φ i,jas the mean of the four current values at neighbouring grid-points, using (28.84) directly,and then to iterate this recalculation until no value of φchanges signiﬁcantly after a complete cycle through all values of iandj. This procedure is the simplest of such ‘relaxation’ methods; for a slightly more sophisticated scheme see exercise 28.22 at the end of the chapter. The reader is referred to specialist books forfuller accounts of how this approach can be made faster and more accurate. 1191 NUMERICAL METHODS Our ﬁnal example is based upon the one-dimensional diﬀusion equation for the temperature φof a system, ∂φ ∂t=κ∂2φ ∂x2. (28.86) Ifφi,jstands for φ(x0+i∆x, t0+j∆t) then a forward diﬀerence representation of the time derivative and a central diﬀerence representation for the spatialderivative lead to the following relationship: φ i,j+1−φi,j ∆t=κφi+1,j−2φi,j+φi−1,j (∆x)2. (28.87) This allows the construction of an explicit scheme for generating the temperature distribution at later times, given that it is known at some earlier time: φi,j+1=α(φi+1,j+φi−1,j)+( 1−2α)φi,j, (28.88) where α=κ∆t/(∆x)2. Although this scheme is explicit it is not a good one, because of the asymmetric way in which the diﬀerences are formed. However, the eﬀect of this can be minimised if we study and correct for the errors introduced, in the following way. Taylor’s series for the time variable gives φi,j+1=φi,j+∆t∂φi,j ∂t+(∆t)2 2!∂2φi,j ∂t2+···, (28.89) using the same notation as previously. Thus the ﬁrst correction term to the left-hand side of (28.87) is −∆t 2∂2φi,j ∂t2. (28.90) The ﬁrst term omitted on the right-hand side of the same equation is, by a similar argument, −κ2(∆x)2 4!∂4φi,j ∂x4. (28.91) But, using the fact that φsatisﬁes (28.86) we obtain ∂2φ ∂t2=∂ ∂tparenleftbigg κ∂2φ ∂x2parenrightbigg =κ∂2 ∂x2parenleftbigg∂φ ∂tparenrightbigg =κ2∂4φ ∂x4, (28.92) and so, to this accuracy, the two errors (28.90) and (28.91) can be made to cancel ifαis chosen such that −κ2∆t 2=−2κ(∆x)2 4!,i.e.α=1 6. 1192 28.9 EXERCISES 28.9 Exercises 28.1 Use an iteration procedure to ﬁnd to four signiﬁcant ﬁgures the root of the equation 40 x=e x p x. 28.2 Using the Newton–Raphson procedure ﬁnd, correct to three decimal places, the root nearest to 7 of the equation 4 x3+2x2−200x−50 = 0. 28.3 (a) Show that if a polynomial equation g(x)≡xm−f(x)=0 ,where f(x)i sa polynomial of degree less than mand for which f(0)/negationslash= 0, is solved using a rearrangement iteration scheme xn+1=[f(xn)]1/m, then, in general, the scheme will have only ﬁrst-order convergence. (b) By considering the cubic equation x3−ax2+2abx−(b3+ab2)=0 for arbitrary non-zero values of aandb, demonstrate that, in special cases, a rearrangement scheme can give second- (or higher-) order convergence. 28.4 The square root of a number Nis to be determined by means of the iteration scheme xn+1=xn / 1− /; N−x2 n / f(N) / . Determine how to choose f(N) so that the process has second-order convergence. Given that√ 7≈2.65, calculate√ 7 as accurately as a single application of the formula will allow. 28.5 Solve the following set of simultaneous equations using Gaussian elimination (including interchange where it is formally desirable), x1+3x2+4x3+2x4=0, 2x1+1 0x2−5x3+x4=6, 4x2+3x3+3x4=2 0, −3x1+6x2+1 2x3−4x4=1 6. 28.6 The following table of values of a polynomial p(x) of low degree contains an error. Identify and correct the erroneous value and extend the table up to x=1.2. xp (x) xp (x) 0.0 0.000 0.5 0.165 0.1 0.011 0.6 0.2160.2 0.040 0.7 0.2450.3 0.081 0.8 0.2560.4 0.128 0.9 0.243 28.7 Simultaneous linear equations that result in tridiagonal matrices can be treated sometimes as three-term recurrence relations and their solution found in a similar manner to that described in chapter 15. Consider the tridiagonal simultaneousequations x i−1+4xi+xi+1=3 (δi+1,0−δi−1,0),i=0,±1,±2,... . Prove that for i>0 the equations have a general solution of the form xi= αpi+βqi,w h e r e pandqare the roots of a certain quadratic equation. Show that a similar result holds for i<0. In each case express x0in terms of the arbitrary constants α, β, . . . . Now impose the condition that xiis bounded as i→±∞ and obtain a unique solution. 1193 NUMERICAL METHODS 28.8 A possible rule for obtaining an approximation to an integral is the mid-point rule,g i v e nb yZx0+∆x x0f(x)dx=∆xf(x0+1 2∆x)+O ( ∆ x3). Writing hfor ∆ x, and evaluating all derivates at the mid-point of the interval (x, x+∆x), use a Taylor series expansion to ﬁnd, up to O( h5), the coeﬃcients of the higher-order errors in both the trapezium and midpoint rules. Hence ﬁnd alinear combination of these two rules that gives O( h 5)a c c u r a c yf o re a c hs t e p∆ x. 28.9 Given a random number ηuniformly distributed on (0 ,1), determine the function ξ=ξ(η) that would generate a random number ξdistributed as (a) 2 ξon 0≤ξ<1. (b)3 2√ξon 0≤ξ<1. (c)π 4acosπξ 2aon−a≤ξ<a. (d)1 2exp(−|ξ|)o n−∞ <ξ<∞. 28.10 A, Band Care three circles of unit radius with centres in the xy-plane at (1,2),(2.5,1.5) and (2 ,3) respectively. Devise a hit or miss Monte Carlo calculation to determine the size of the area that lies outside Cbut inside AandB, as well as inside the square centred on (2 ,2.5) that has sides of length 2 parallel to the coordinate axes. You should choose your sampling region so as to make the estimation as eﬃcient as possible. Take the random number distribution to be uniform on (0 ,1) and determine the inequalities that have to be tested using the random numbers chosen. 28.11 Use a Taylor series to solve the equation dy dx+xy=0,y (0) = 1 , evaluating y(x)f o r x=0.0 to 0.5 in steps of 0.1. 28.12 Consider the application of the predictor–corrector method described near the end of subsection 28.6.3 to the equation dy dx=x+y. Show, by comparison with a Taylor series expansion, that the expression obtained foryi+1in terms of xiandyiby applying the three steps indicated (without any repeat of the last two) is correct to O( h2). Using steps of h=0.1 compute the value of y(0.3) and compare it with the value obtained by solving the equation analytically. 28.13 A more reﬁned form of the Adams predictor–corrector method for solving the ﬁrst-order diﬀerential equation dy dx=f(x, y) is known as the Adams–Moulton–Bashforth scheme. At any stage (say the nth) in an Nth-order scheme the values of xandyat the previous Nsolution points are ﬁrst used to predict the value of yn+1. This approximate value of yat the next solution point xn+1, denoted by ¯yn+1, is then used together with those at the previous N−1 solution points to make a more reﬁned ( corrected ) estimation of y(xn+1). The calculational procedure for a third-order scheme is summarised by the two equations ¯yn+1=yn+h(a1fn+a2fn−1+a3fn−2)( p r e d i c t o r ) , yn+1=yn+h(b1f(xn+1,¯yn+1)+b2fn+b3fn−1) (corrector) . 1194 28.9 EXERCISES (a) Find Taylor series expansions for fn−1andfn−2in terms of the function fn=f(xn,yn) and its derivatives at xn. (b) Substitute them into the predictor equation and, by making that expression for¯yn+1coincide with the true Taylor series for yn+1up to order h3, establish simultaneous equations that determine the values of a1,a2anda3. (c) Find the Taylor series for fn+1and substitute it and that for fn−1into the corrector equation. Make the corrected prediction for yn+1coincide with the true Taylor series by choosing the weights b1,b2andb3appropriately. (d) The values of the numerical solution of the diﬀerential equation dy dx=2(1 + x)y+x3/2 2x(1 +x) at three values of xare given in the following table: x0.1 0.2 0.3 y(x) 0.030628 0.084107 0.150328. Use the above predictor–corrector scheme to ﬁnd the value of y(0.4) and compare your answer with the accurate value, 0.225577. 28.14 If dy/dx =f(x, y) then show that d2f dx2=∂2f ∂x2+2f∂2f ∂x∂y+f2∂2f ∂y2+∂f ∂x∂f ∂y+f /df dy /2 . Hence verify, by substitution and the subsequent expansion of arguments in Taylor series of their own, that the scheme given in (28.75) coincides with theTaylor expansion (28.64), i.e. y i+1=yi+hy(1) i+h2 2!y(2) i+h3 3!y(3) i+···. up to terms in h3. 28.15 To solve the ordinary diﬀerential equation du dt=f(u, t) forf=f(t), the explicit two-step ﬁnite diﬀerence scheme un+1=αun+βun−1+h(µfn+νfn−1) may be used. Here, in the usual notation, his the time step, tn=nh,un=u(tn) andfn=f(un,tn);α,β,µ,a n d νare constants. (a) A particular scheme has α=1 ,β=0,µ=3/2a n d ν=−1/2. By considering Taylor expansions about t=tnfor both un+jandfn+j, show that this scheme gives errors of order h3. (b) Find the values of α,β,µ,a n d νthat will give the greatest accuracy. 28.16 Set up a ﬁnite diﬀerence scheme to solve the ordinary diﬀerential equation xd2φ dx2+dφ dx=0 in the range 1 ≤x≤4 and subject to the boundary conditions φ(1) = 2 and dφ/dx =2a t x=4 .U s i n g Nequal increments, ∆ x,i nx, obtain the general diﬀerence equation and state how the boundary conditions are incorporatedinto the scheme. Setting ∆ xequal to the (crude) value 1, obtain the relevant simultaneous equations and so obtain rough estimates for φ(2),φ(3) and φ(4). Finally, solve the original equation analytically and compare your numerical estimates with the accurate values. 1195 NUMERICAL METHODS 28.17 Write a computer program that would solve, for a range of values of λ,t h e diﬀerential equation dy dx=1p x2+λy2,y (0) = 1 , using a third-order Runge–Kutta scheme. Consider the diﬃculties that might arise when λ<0. 28.18 Use the isocline approach to sketch the family of curves that satisﬁes the non- linear ﬁrst-order diﬀerential equation dy dx=ap x2+y2. 28.19 For some problems, numerical or algebraic experimentation may suggest the form of the complete solution. Consider the problem of numerically integratingthe ﬁrst-order wave equation ∂u ∂t+A∂u ∂x=0, in which Ais a positive constant. A ﬁnite diﬀerence scheme for this partial diﬀerential equation is u(p, n+1 )−u(p, n) ∆t+Au(p, n)−u(p−1,n) ∆x=0, where x=p∆xandt=n∆t,w i t h pany integer and na non-negative integer. The initial values are u(0,0) = 1 and u(p,0) = 0 for p/negationslash=0 . (a) Carry the diﬀerence equation forward in time for two or three steps and attempt to identify the pattern of solution. Establish the criterion for themethod to be numerically stable. (b) Suggest a general form for u(p, n), expressing it in generator function form, i.e. ‘as u(p, n) is the coeﬃcient of s pin the expansion of G(n, s)’. (c) Using your form of solution (or that given in the answers!), obtain an explicit general expression for u(p, n) and verify it by direct substitution into the diﬀerence equation. (d) An analytic solution of the original PDE indicates that an initial distur- bance propagates undistorted. Under what circumstances would the diﬀer- ence scheme reproduce that behaviour? 28.20 In the previous question the diﬀerence scheme for solving ∂u ∂t+∂u ∂x=0, in which Ahas been set equal to unity, was one-sided in both space ( x)a n d time ( t). A more accurate procedure (known as the Lax–Wendroﬀ scheme) is u(p, n+1 )−u(p, n) ∆t+u(p+1,n)−u(p−1,n) 2∆x =∆t 2 /u(p+1,n)−2u(p, n)+u(p−1,n) (∆x)2 / . (a) Establish the orders of accuracy of the two ﬁnite diﬀerence approximations on the LHS of the equation. (b) Establish the accuracy with which the expression in the brackets approxi- mates ∂2u/∂x2. (c) Show that the RHS of the equation is such as to make the whole diﬀerence scheme accurate to second order in both space and time. 1196 28.9 EXERCISES 28.21 Laplace’s equation ∂2V ∂x2+∂2V ∂y2=0, is to be solved for the region and boundary conditions shown in ﬁgure 28.7. 40 40 40 40 40 40 40 20 20 20V=8 0 V=0∞ −∞ Figure 28.7 Region, boundary values and initial guessed solution for exer- cise 28.21. Starting from the given initial guess for the potential values Vand using the simplest possible form of relaxation, obtain a better approximation to the actualsolution. Do not aim to be more accurate than ±0.5 units and so terminate the process when subsequent changes would be no greater than this. 28.22 Consider the solution φ(x, y) of Laplace’s equation in two dimensions using a relaxation method on a square grid with common spacing h.A si nt h em a i nt e x t , denote φ(x 0+ih, y 0+jh)b yφi,j. Further, deﬁne φm,n i,jby φm,n i,j≡∂m+nφ ∂xm∂yn evaluated at ( x0+ih, y 0+jh). (a) Show that φ4,0 i,j+2φ2,2 i,j+φ0,4 i,j=0. (b) Working up to terms of order h5, ﬁnd Taylor series expansions, expressed in terms of the φm,n i,j,f o r S±,0=φi+1,j+φi−1,j S0,±=φi,j+1+φi,j−1. (c) Find a corresponding expansion, to the same order of accuracy, for φi±1,j+1+ φi±1,j−1and hence show that S±,±=φi+1,j+1+φi+1,j−1+φi−1,j+1+φi−1,j−1 has the form 4φ0,0 i,j+2h2(φ2,0 i,j+φ0,2 i,j)+h4 6(φ4,0 i,j+6φ2,2 i,j+φ0,4 i,j). 1197 NUMERICAL METHODS (d) Evaluate the expression 4( S±,0+S0,±)+S±,±and hence deduce that a possible relaxation scheme, good to the ﬁfth order in h, is to recalculate each φi,jas the weighted mean of the current values of its four nearest neighbours (eachwith weight 1 5) and its four next-nearest neighbours (each with weight1 20). 28.23 The Schr ¨odinger equation for a quantum mechanical particle of mass mmoving in a one-dimensional harmonic oscillator potential V(x)=kx2/2i s − /~2 2md2ψ dx2+kx2ψ 2=Eψ. For physically acceptable solutions the wavefunction ψ(x) must be ﬁnite at x=0 , tend to zero as x→± ∞ and be normalised so that R |ψ|2dx=1 .I np r a c t i c e these constraints mean that only certain (quantised) values of E, the energy of the particle, are allowed. The allowed values fall into two groups, those for whichthe corresponding y(0) = 0 and those for which the corresponding y(0)/negationslash=0 . Show that if the unit of length is taken as [/~2/(mk)]1/4and the unit of energy as /~(k/m)1/2then the Schr ¨odinger equation takes the form d2ψ dy2+( 2E/prime−y2)ψ=0. Devise an outline computerised scheme, using Runge–Kutta integration, that will enable you to: (a) determine the three lowest allowed values of E; (b) tabulate the normalised wavefunction corresponding to the lowest allowed energy. You should consider explicitly: (i) the variables to use in the numerical integration; (ii) how starting values near y= 0 are to be chosen; (iii) how the condition on ψasy→±∞ is to be implemented; (iv) how the required values of Eare to be extracted from the results of the integration; (v) how the normalisation is to be carried out. 28.10 Hints and answers 28.1 5.370. 28.2 6.951 after two iterations.28.3 (a) ξ/negationslash=0a n d f /prime(ξ)/negationslash= 0 in general; (b) ξ=b, but f/prime(b) = 0 whilst f(b)/negationslash=0 . 28.4 f(N)=(N−3x2)−1=−(2N)−1; 2.6457411, accurate value 2.6457513. 28.5 Interchange is formally needed for the ﬁrst two steps, though in this case no error will result if it is not carried out; x1=−12,x2=2,x3=−1,x4=5. 28.6 p(0.5) = 0 .175,p(1.0) = 0 .200,p(1.1) = 0 .121,p(1.2) = 0 .000. 28.7 The quadratic equation is z2+4z+1=0 ; α+β−3=x0=α/prime+β/prime+3 . With p=−2+√3a n d q=−2−√3,βmust be zero for i>0a n d α/primemust be zero for i<0;xi=3 (−2+√3)ifori>0,xi=0f o r i=0,xi=−3(−2−√3)i fori<0. 28.8 Iexact=hf+h3f/24;IT=hf+h3f/8;IM=hf. Thus Iexact=1 3IT+2 3IM+O (h5). 28.9 Listed below are the relevant indeﬁnite integrals F(y) of the distributions together with the functions ξ=ξ(η): (a)y2;ξ=√η. (b)y3/2;ξ=η2/3. 1198 28.10 HINTS AND ANSWERS aa 2a2a −a −a−2a −2axy Figure 28.8 Typical solutions y=y(x), shown by solid lines, of dy/dx = a(x2+y2)−1/2. The short arrows give the direction that the tangent to any solution must have at that point. (c)1 2[sin(πy/2a)+1 ] ; ξ=( 2a/π)sin−1(2η−1). (d)1 2exp(y)f o r y≤0;1 2(1−exp(−y)) for y>0.ξ=l n ( 2 η)f o r0 <η≤1 2; ξ=−ln[2(1−η)] for frac12<η< 1. 28.10 Show that the corners of the required area (with three curved sides and one straight one) are at (1 .5,1.5),(1.866,1.5),(2,2) and (1 .669,2.056). For a pair of random numbers ( ξ1,ξ2), take x=1.5+αξ1andy=1.5+βξ2. The best values forαandβare 0.5 and 0.556 respectively. Test the inequalities (αξ1+0.5)2+(βξ2−0.5)2≤1, (αξ1−1)2+(βξ2)2≤1, (αξ1−0.5)2+(βξ2−1.5)2≥1. If all three conditions are satisﬁed for nout of Npairs, the area can be estimated asnαβ/N . 28.11 1−x2/2+x4/8−x6/48; 1.0000, 0.9950, 0.9802, 0.9560, 0.9231, 0.8825; exact solution y=e x p (−x2/2). 28.12 yi+1=yi+h(xi+yi)+1 2h2(1 + xi+yi). The numerical estimate is 0.04923; the analytic solution y(x)=ex−1−xgives 0.04986. 28.13 (b) a1=2 3/12,a2=−4/3,a3=5/12. (c)b1=5/12,b2=2/3,b3=−1/12. (d)¯y(0.4) = 0 .224582 ,y(0.4) = 0 .225527 after correction. 28.15 (a) The error is5 12h3un+O ( h4). (b)α=−4,β=5,µ=4 ,a n d ν=2 1199 NUMERICAL METHODS 40 40 2041.5 41.5 46.5 46.5 48 16.5 16.5V=8 0 V=0∞ −∞ Figure 28.9 The solution to exercise 28.21. 28.16 With xj=1+ j∆x,N∆x=3a n d φj=φ(xj), [2 + (2 j+1 ) ∆ x]φj+1−(4 + 4 j∆x)φj+[ 2+( 2 j−1)∆x]φj−1=0 forj=1,2,...,N−2. In addition φ0=2a n d φN=φN−1+2 ∆ x. Analytically φ(x)=2+8l n x. Estimated (accurate) values are 6.67 (7.55), 9.47 (10.79), 11.47 (13.09). 28.18 See ﬁgure 28.8.28.19 (a) Setting A∆t=c∆xgives, for example, u(0,2) = (1−c) 2,u(1,2) = 2 c(1−c), u(2,2) = c2. For stability 0 <c< 1. (b)G(n, s)=[ ( 1−c)+cs]nfor 0≤p≤n. (c) [ n!(1−c)n−pcp]/[p!(n−p)!]. (d) When c= 1 and the diﬀerence equation becomes u(p, n+1 )= u(p−1,n). 28.20 (a) First order in time and second order in space; (b) O(∆ x)2;( c )s h o wt h a t ∂2u/∂t2=∂2u/∂x2and write the second-order correction to ∂u/∂t in the form 1 2(∆t)2∂2u/∂x2. 28.21 See ﬁgure 28.9. 28.22 (a) Write Laplace’s equation as φ2,0 i,j+φ0,2 i,j= 0 and diﬀerentiate twice with respect toxandyseparately. Then add the resulting equations. (b)S±,0=2φ0,0 i,j+h2φ2,0 i,j+1 12h4φ4,0 i,jand corresponding results for S0,±. (c) Note that φ0,0 i±1,j,φ0,2 i±1,jandφ0,4 i±1,jhave themselves to be expanded as Taylor series in x, the number of terms to be retained being determined by the power of ∆y=hthat multiplies them. The roles of xandycould be reversed. (d) Use Laplace’s equation and result (a) to show that the given expression has the value 20 φi,j. 1200 Appendix Gamma, beta and error functions In several places in this book we have made mention of the gamma, beta and error functions. These convenient functions appear in a number of contexts and herewe gather together some of their properties. This appendix should be regardedmerely as a reference containing some useful relations with a minimum of formalproofs. A1.1 The gamma function Thegamma function Γ(n) is deﬁned by Γ(n)=integraldisplay ∞ 0xn−1e−xdx, (A1) which converges for n>0. Replacing nbyn+1 in (A1) and integrating the RHS by parts, we ﬁnd Γ(n+1 )=integraldisplay∞ 0xne−xdx =bracketleftbig −xne−xbracketrightbig∞ 0+integraldisplay∞ 0nxn−1e−xdx =nintegraldisplay∞ 0xn−1e−xdx, from which we obtain the important result Γ(n+1 )= nΓ(n). (A2) From (A1), we see that Γ(1) = 1, and so, if nis a positive integer, Γ(n+1 )= n!. (A3) 1201 GAMMA, BETA AND ERROR FUNCTIONS Γ( ) 1 22 344 −1−2 −2−3−4 −4 −66 nn Figure A1.1 The gamma function Γ( n). In fact, equation (A3) serves as a deﬁnition of the factorial function even for non-integer n. For negative nthe factorial function is deﬁned by n!=(n+m)! (n+m)(n+m−1)···(n+1 ), (A4) where mis any positive integer that makes n+m>0. Diﬀerent choices of m (>−n) do not lead to diﬀerent values for n!. A plot of the gamma function is given in ﬁgure A1.1, where it can be seen that the function is inﬁnite for negativeinteger values of n, in accordance with (A4). By letting x=y 2in (A1), we immediately obtain another useful representation of the gamma function given by Γ(n)=2integraldisplay∞ 0y2n−1e−y2dy. (A5) Setting n=1 2we ﬁnd the result Γparenleftbig1 2parenrightbig =2integraldisplay∞ 0e−y2dy=integraldisplay∞ −∞e−y2dy=√π, where have used the standard integral discussed in section 6.4.2. From this result, Γ(n) for half-integral ncan be found using (A2). Some immediately derivable factorial values of half integers are parenleftbig −3 2parenrightbig !=−2√π,parenleftbig −1 2parenrightbig !=√π,parenleftbig1 2parenrightbig !=1 2√π,parenleftbig3 2parenrightbig !=3 4√π. 1202 A1.2 THE BETA FUNCTION It can also be shown that the gamma function is given by Γ(n+1 )=√ 2πn nne−nparenleftbigg 1+1 12n+1 288n2−139 51 840 n3+...parenrightbigg =n!,(A6) which is known as Stirling’s asymptotic series . For large nthe ﬁrst term dominates and so n!≈√ 2πn nne−n; (A7) this is known as Stirling’s approximation . This approximation is particularly useful in statistical thermodynamics, when arrangements of a large number of particlesa r et ob ec o n s i d e r e d .IProve Stirling’s approximation n!≈√ 2πn nne−nfor large n. From (A1), the extended deﬁnition of the factorial function (which is valid for n>−1) is given by n!= Z∞ 0xne−xdx= Z∞ 0enlnx−xdx. (A8) If we let x=n+y,t h e n lnx=l nn+l n / 1+y n / =l nn+y n−y2 2n2+y3 3n3−···. Substituting this result into (A8), we obtain n!= Z∞ −nexp / n / lnn+y n−y2 2n2+··· / −n−y / dy. Thus, when nis suﬃciently large, we may approximate n!b y n!≈enlnn−n Z∞ −∞e−y2/(2n)dy=enlnn−n√ 2πn=√ 2πn nne−n, which is Stirling’s approximation (A7). J A1.2 The beta function Thebeta function is deﬁned by B(m, n)=integraldisplay1 0xm−1(1−x)n−1dx, (A9) which converges for m>0,n>0. By letting x=1−yin (A9) it is easy to show thatB(m, n)=B(n, m). Other useful representations of the beta function may be obtained by suitable changes of variable. For example, putting x=( 1+ y)−1in (A9), we ﬁnd that B(m, n)=integraldisplay∞ 0yn−1dy (1 +y)m+n. 1203 GAMMA, BETA AND ERROR FUNCTIONS Alternatively, if we let x=s i n2θin (A9), we obtain immediately B(m, n)=2integraldisplayπ/2 0sin2m−1θcos2n−1θd θ . (A10) The beta function may also be written in terms of the gamma function as B(m, n)=Γ(m)Γ(n) Γ(m+n). (A11)IProve the result (A11). Using (A5), we have Γ(n)Γ(m)=4 Z∞ 0x2n−1e−x2dx Z∞ 0y2m−1e−y2dy =4 Z∞ 0 Z∞ 0x2n−1y2m−1e−(x2+y2)dx dy. Changing variables to plane polar coordinates ( ρ, φ)g i v e nb y x=ρcosφ,y=ρsinφ,w e obtain Γ(n)Γ(m)=4 Zπ/2 0 Z∞ 0ρ2(m+n−1)e−ρ2sin2m−1θcos2n−1θ ρ dρ dθ =4 Zπ/2 0sin2m−1θcos2n−1θd θ Z∞ 0ρ2(m+n)−1e−ρ2dρ =B(m, n)Γ(m+n), where in the last line we have used the results (A5) and (A10). J A1.3 The error function Finally we mention the error function , which is encountered in probability theory and in the solutions of some partial diﬀerential equations, and which is deﬁnedby erf(x)=2 √πintegraldisplayx 0e−u2du=1−2√πintegraldisplay∞ xe−u2du. (A12) From this deﬁnition we can easily see that erf(0) = 0 ,erf(∞)=1 ,erf(−x)=−erf(x). By making the substitution y=√ 2uin (A12), we ﬁnd erf(x)=1√ 2πintegraldisplay√ 2x 0e−y2/2dy. The cumulative probability function Φ( x) for the standard Gaussian distribution (discussed in section 26.9.1) may be written in terms of the error function as 1204 A1.3 THE ERROR FUNCTION follows Φ(x)=1√ 2πintegraldisplayx −∞e−y2/2dy =1 2+1√ 2πintegraldisplayx 0e−y2/2dy =1 2+e r fparenleftbiggx√ 2parenrightbigg . It is also sometimes useful to deﬁne the complementary error function erfc(x)=1−erf(x)=2√πintegraldisplay∞ xe−u2du. (A13) 1205 Index F-distribution (Fisher), 1132–1138 critical points table, 1137logarithmic form, 1138 t-test,seeStudent’s t-test correlation, chi-squared test , 1143 Cram ´er-Rao (Fisher’s) inequality , 1075, 1076 Fisher’s inequality , 1075, 1076maximum-likelihood, method of extended, 1112 A, B, one-dimensional irreps, 932, 944, 951 Abelian groups, 886absolute convergence of series, 127, 717 absolute derivative, 824–826 acceleration vector, 341Adams method, 1184Adams–Moulton–Bashforth, predictor-corrector scheme, 1194 addition rule for probabilities, 967, 972 adjoint, seeHermitian conjugate adjoint operators, 587–591adjustment of parameters, 854–855algebra of complex numbers, 88–89 functions in a vector space, 583 matrices, 256–257power series, 137series, 134tensors, 787–790vectors, 217–218 in a vector space, 247 in component form, 222 algebraic equations, numerical methods for, see numerical methods for algebraic equations alternating group, 958 alternating series test, 133 ammonia molecule, symmetries of, 884Amp`ere’s rule (law), 387, 414 amplitude modulation of radio waves, 450analytic (regular) functions, 712angle between two vectors, 225 angular frequency, 626n in Fourier series, 425 angular momentum, 782, 798 and irreps, 935of particle system, 799–801of particles, 344of solid body, 402, 800 vector representation, 241 angular velocity, vector representation, 227, 241, 359 anti-Hermitian matrices, 276 eigenvalues, 281–283 imaginary nature, 282–283 eigenvectors, 281–283 orthogonality, 282–283 anticommutativity of vector/cross product, 226 antisymmetric functions, 422 and Fourier series, 425–426and Fourier transforms, 451 antisymmetric matrices, 275 general properties, seeanti-Hermitian matrices antisymmetric tensors, 787, 790antithetic variates, in Monte Carlo methods, 1174 aperture function, 443approximately equal ≈, deﬁnition, 135 arbitrary parameters for ODE, 475 arc length of plane curves, 74–75space curves, 347 arccosech, arccosh, arccoth, arcsech, arcsinh, arctanh, seehyperbolic functions, inverses Archimedean upthrust, 402, 416area element in Cartesian coordinates, 191plane polars, 205 area of circle, 72 ellipse, 72, 210 1206 INDEX parallelogram, 227 region, using multiple integrals, 194–196surfaces, 352 as vector, 399–401, 414 area, maximal enclosure, 838arg, argument of a complex number, 90Argand diagram, 87, 711 argument, principle of the, 755 arithmetic series, 120arithmetico-geometric series, 121arrays, seematrices associated Legendre equation, 594–595, 666, 670, 703 associated Legendre functions P m /lscript(x), 666 generating function, 594 orthogonality, 594Rodrigues’ formula, 594 associated Legendre functions P m /lscript(x), 703 associative law for addition in a vector space of ﬁnite dimensionality, 247 in a vector space of inﬁnite dimensionality,583of complex numbers, 89of matrices, 256of vectors, 217 convolution, 453, 464 group operations, 885 linear operators, 254multiplication of a matrix by a scalar, 256o fav e c t o rb yas c a l a r ,2 1 8of complex numbers, 91of matrices, 258 multiplication by a scalar in a vector space of ﬁnite dimensionality, 247 in a vector space of inﬁnite dimensionality,583 atomic orbitals, 957 d-states, 948, 950, 956 p-states, 948 s-states, 986 auto-correlation functions, 456automorphism, 903auxiliary equation, 499 repeated roots, 499 average value, seemean value axial vectors, 798 backward diﬀerences, 1179 basis functions for linear least squares estimation, 1115in a vector space of inﬁnite dimensionality, 583–584 of a representation, 920 change in, 926–927, 929, 934 basis vectors, 221–222, 248–249, 778, 920 derivatives, 814–816Christoﬀel symbol Γ k ij, 814 for particular irrep, 948–950, 958 linear dependence and independence, 221 non-orthogonal, 250 orthonormal, 249–250 required properties, 221 Bayes’ theorem, 974–975Bernoulli equation, 483 Bessel correction to variance estimate, 1090 Bessel equation, 541, 564–568, 595, 674Bessel functions J ν(z) zeroes of, 662, 673 Bessel functions Jν(z), 662, 672 generating function, 573, 595 integral relationships, 572 integral representation, 574–575orthogonality, 570–573 recurrence relations, 569–570 second kind, Y ν(z), 568 series, 565, 566, 595 ν= 0, 567 ν=±1/2, 566 spherical, 675 Bessel inequality, 251, 586best unbiased estimator, 1075 beta function, 1203 bias, of estimator, 1074bilinear transformation, general, 113 binary chopping, 1154 binomial coeﬃcient nCk, 27–30 elementary properties, 26 identities, 27 negative n,2 9 non-integral n,2 9 binomial coeﬃcientnCk, 977–979 in Leibniz’ theorem, 50 binomial distribution Bin( n, p), 1010–1013 and Gaussian distribution, 1027and Poisson distribution, 1016, 1019 mean and variance, 1013 MGF, 1012recurrence formula, 1011 binomial expansion, 143 binormal to space curves, 348birthdays, diﬀerent, 976 bivariate distributions, 1038–1049 conditional, 1040 continuous, 1039 correlation, 1042–1049 and independence, 1042 matrix, 1045–1049 positive/negative, 1042 covariance, 1042–1049 matrix, 1045 expectation (mean), 1041independent (uncorrelated), 1039 marginal, 1040 variance, 1042 Boltzmann distribution from constraints on total energy, 174–176 1207 INDEX bonding in molecules, 945, 947–950 Born approximation, 152, 606Bose–Einstein statistics, 980boundary conditions and characteristics, 633and Laplace equation, 699, 701 for Green’s functions, 518, 520–522 inhomogeneous, 521 for ODE, 474, 476, 507for PDE, 614, 618–620for Sturm–Liouville equations, 592homogeneous and inhomogeneous, 618, 656, 686, 688 superposition solutions, 651–657types, 635–638 brachistochrone problem, 843Bragg formula, 241branch cut, 722branch points, 721Bromwich integral, 765bulk modulus, 829 calculus of residues, seezeroes of a function of a complex variable andcontour integration calculus of variations constrained variation, 844–846estimation of ODE eigenvalues, 849 Euler–Lagrange equation, 835–836 Fermat’s principle, 846Hamilton’s principle, 847higher-order derivatives, 841several dependent variables, 841several independent variables, 841soap ﬁlms, 839–840variable end-points, 841–844 calculus, elementary, 42–77cancellation law in a group, 888 canonical form, for second-order ODE, 522 card drawing, seeprobability carrier frequency of radio waves, 451Cartesian coordinates, 221–222Cartesian tensors, 779–804 algebra, 787–790contraction, 788deﬁnition, 784ﬁrst-order, 781–784 from scalar, 783 general order, 784–803 integral theorems, 803–804isotropic, 793–795physical applications, 783, 788–790, 799–803second-order, 784–803, 817symmetry and antisymmetry, 787tensor ﬁelds, 803zero-order, 781–784 from vector, 784 Cartesian tensors, particular conductivity, 801inertia, 800strain, 802stress, 802 susceptibility, 801 catenary, 840, 846Cauchy boundary conditions, 635distribution, 994inequality, 747integrals, 745–747product, 134root test, 132, 717theorem, 742 Cauchy–Riemann relations, 713–716, 727, 729, 743 in terms of zandz ∗, 715 central diﬀerences, 1179central limit theorem, 1036–1038, 1178central moments, seemoments, central centre of a group, 911centre of mass, 198 of hemisphere, 198of semicircular lamina, 200 centroid, 198 of plane area, 198of plane curve, 200of triangle, 220–221 CF,seecomplementary function chain rule for functions of one real variable, 47–48several real variables, 160–161 change of basis, seesimilarity transformations change of variables and coordinate systems, 161–163in multiple integrals, 202–210 evaluation of Gaussian integral, 205–207 general properties, 209–210 in RVD, 992–999 character tables, 935 4mmorC 4v, 944, 948, 950 S4or 432 or O, 956, 957 ¯43morTd, 957 3mor 32 or C3vorS3, 935, 939, 950, 952, 958, 959 construction of, 942–944 characteristic equation, 285 normal mode form, 325of recurrence relation, 505 characteristic function, seemoment generating functions (MGF) characteristics and boundary curves, 633 multiple intersections, 633, 638 and the existence of solutions, 632–638ﬁrst-order equations, 632–633second-order equations, 636 and equation type, 636 characters, 934–938, 942–944 and conjugacy classes, 934, 937character tables 4mmorC 4vorD4, 955 A4, 958 1208 INDEX D5, 958 quaternion, 955 counting irreps, 937–938 deﬁnition, 934of product representation, 946–947 orthogonality properties, 936, 944 summation rules, 939 charge (point), Dirac δ-function respresentation, 447 charged particle in electromagnetic ﬁelds, 376 Chebyshev equation, 541, 597 polynomial solutions, 578 Chebyshev polynomials T n(x) generating function, 597 orthogonality, 597Rodrigues’ formula, 597 chi-squared ( χ 2) distribution and likelihood-ratio test, 1125 chi-squared ( χ2) distribution, 1034 and goodness of ﬁt, 1139 and likelihood-ratio test, 1134and multiple estimators, 1086 test for correlation, 1143 Cholesky separation, 318Christoﬀel symbol Γ k ij, 814–816 from metric tensor, 815, 822 circle of convergence, 717–718circle, area of, 72 circuits, electrical transients, 491 Clairaut equation, 489 classes and equivalence relations, 906 closure of a group, 885closure property of eigenfunctions of an Hermitian operator, 601 cofactor of a matrix element, 264 column matrix, 255 column vector, 255combinations (probability), 975–981 common ratio in geometric series, 120 commutation law for group elements, 886commutative law for addition in a vector space of ﬁnite dimensionality,247 in a vector space of inﬁnite dimensionality, 583 of complex numbers, 89 of matrices, 256of vectors, 217 complex scalar/dot product, 226 convolution, 453, 464inner product, 249 multiplication o fav e c t o rb yas c a l a r ,2 1 8of complex numbers, 91 scalar/dot product, 224 commutator, of two matrices, 314comparison test, 128 complement, 963probability for, 967 complementary equation, 496complementary error function, 1205complementary function (CF), 497 for ODE, 498partially known, 512repeated roots of auxiliary equation, 499 completeness of basis vectors, 248eigenfunctions of an Hermitian operator, 588, 601 eigenvectors of a normal matrix, 280spherical harmonics Y m /lscript(θ, φ), 670 completing the square as a means of integration, 67–68for quadratic equations, 35to evaluate Gaussian integral, 442, 684 complex conjugate z ∗, of complex number, 92–94, 715 of a matrix, 261–263of scalar/dot product, 226properties of, 93 complex exponential function, 95, 719complex Fourier series, 430–431complex integrals, 738–742, see also zeroes of a function of a complex variable andcontour integration Cauchy’s, 745–747Cauchy’s theorem, 742deﬁnition, 739 Jordan’s lemma, 761–762 Morera’s theorem, 744ofz −1, 740 principal value, 760residue theorem, 752–754 complex logarithms, 102–103, 720 principal value of, 103, 720 complex numbers, 86–117 addition and subtraction of, 88–89applications to diﬀerentiation and integration, 104 argument of, 90associativity of addition and subtraction, 89 multiplication, 91 commutativity of addition and subtraction, 89multiplication, 91 complex conjugate of, seecomplex conjugate components of, 87de Moivre’s theorem, seede Moivre’s theorem division of, 94–95, 97–98 properties, 95 from roots of polynomial equations, 86–87imaginary part of, 86–87modulus of, 90multiplication of, 91–92, 97–98 as rotation in the Argand diagram, 91–92 notation, 87polar representation of, 95–98 1209 INDEX real part of, 86–87 trigonometric representation of, 96 complex potentials, 725–730 and ﬂuid ﬂow, 727–728equipotentials and ﬁeld lines, 726for circular and elliptic cylinders, 728, 730for parallel cylinders, 770for plates, 736–738, 770for strip, 770for wedges, 737under conformal transformations, 730–738 complex power series, 136 complex powers, 102–103complex variables, seefunctions of a complex variable andpower series in a complex variable andcomplex integrals components of a complex number, 87of a vector, 221–222 in a non-orthogonal basis, 238uniqueness, 248 conditional (constrained) variation, 844–846conditional convergence, 127conditional distributions, 1040conditional probability, seeprobability, conditional cone surface area of, 75–76volume of, 76 conﬁdence interval, 1079 conﬁdence region, 1084 conformal transformations (mappings), 730–738 applications, 735–738examples, 732–735properties, 730–732Schwarz–Christoﬀel transformation, 733–735 congruence, 907conic sections, 15 eccentricity, 16parametric forms, 17standard forms, 16 conjugacy classes, 910–912 in a class by itself, 910 conjugate roots of polynomial equations, 102connectivity of regions, 389conservative ﬁelds, 393–395 necessary and suﬃcient conditions, 393–395potential (function), 395 consistency, of estimator, 1073constant coeﬃcients in ODE, 498–509 auxiliary equation, 499 constants of integration, 63, 474constrained variation, 844–846 constraints, stationary values under, see Lagrange undetermined multiplers continuity correction for discrete RV, 1028continuity equation, 410contour integration, 758–768 inﬁnite integrals, 759–764inverse Laplace transforms, 765–768residue theorem, 752–754, 758–768 sinusoidal functions, 758–759summing series, 764–765 contraction of tensors, 788contradiction, proof by, 32–34contravariant basis vectors, 810 derivative, 814 components of tensor, 805–806 deﬁnition, 810 control variates, in Monte Carlo methods, 1173convergence of inﬁnite series, 717–718 absolute, 127, 717complex power series, 136conditional, 127necessary condition, 128power series, 135 under various manipulations, seepower series, manipulation ratio test, 718rearrangement of terms, 127tests for convergence, 128–134 alternating series test, 133comparison test, 128grouping terms, 132integral test, 131quotient test, 130ratio comparison test, 130ratio test (D’Alembert), 129, 135root test (Cauchy), 132, 717 convergence of numerical iteration schemes, 1156–1158 convolution Fourier tranforms, seeFourier transforms, convolution Laplace tranforms, seeLaplace transforms, convolution convolution theorem Fourier transforms, 454Laplace transforms, 463 Coordinate geometry, 15–18 straight line, 15 coordinate systems, seeCartesian, curvilinear, cylindrical polar, plane polar andspherical polar coordinates coordinate transformations and integrals, seechange of variables and matrices, seesimilarity transformations general, 809–814 relative tensors, 812 tensor transformations, 811weight, 813 orthogonal, 781 coplanar vectors, 229correlation functions, 455–457 auto-correlation, 456cross-correlation, 455energy spectrum, 456Parseval’s theorem, 457Wiener–Kinchin theorem, 456 1210 INDEX correlation matrix, of sample, 1072 correlation of bivariate distributions, 1042–1049correlation of sample data, 1072correspondence principle in quantum mechanics, 1056 cosets and congruence, 907cosh, hyperbolic cosine, 105, 720, see also hyperbolic functions cosine in terms of exponential functions, 105Maclaurin series for, 143orthogonality relations, 423 counting irreps, seecharacters, counting irreps coupled pendulums, 335, 337covariance matrix, of linear least squares estimators, 1116 covariance matrix, of sample, 1072covariance of bivariate distributions, 1042–1049covariance of sample data, 1072covariant basis vector, 810 derivative, 814 components of tensor, 805–806 deﬁnition, 810 derivative, 817 of scalar, 820semi-colon notation, 818 diﬀerentiation, 817–820 CPF,seeprobability functions, cumulative Cramer determinant, 304–305Cramer’s rule, 304–305cross product, seevector product cross-correlation functions, 455crystal lattice, 151crystal point groups, 924cube roots of unity, 101cube, rotational symmetries of, 956curl of a vector ﬁeld, 359 as a determinant, 359 as integral, 404, 406 curl curl, 362in curvilinear coordinates, 374in cylindrical polars, 366in spherical polars, 368Stoke’s theorem, 412–415tensor form, 823 current-carrying wire, magnetic potential, 662Curvature, 53–56 circle of, 54of a function, 53radius of, 54 curvature of space curves, 348curves, seeplane curves andspace curves curvilinear coordinates, 370–375 basis vectors, 370length and volume elements, 371scale factors, 370surfaces and curves, 370tensors, 804–826vector operators, 373–375cut plane, 762 cycle notation for permutations, 899cyclic groups, 903, 940cyclic relation for partial derivatives, 160cycloid, 376, 844cylinders, conducting, seecomplex potentials, for circular and elliptic cylinders cylindrical polar coordinates, 363–367 area element, 366basis vectors, 364Laplace equation, 661–664length element, 366vector operators, 363–367volume element, 366 δ-function (Dirac), seeDirac δ-function δ ij(δj i), Kronecker delta, tensor, seeKronecker delta, δij(δj i), tensor D’Alembert’s ratio test, 129, 718 in convergence of power series, 135 D’Alembert’s solution to wave equation, 627damped harmonic oscillators, 243 and Parseval’s theorem, 457 data modelling, maximum-likelihood, 1097de Broglie relation, 442, 642, 703de Moivre’s theorem, 98, 758 applications, 98–102 ﬁnding the nth roots of unity, 100–101 solving polynomial equations, 101–102trigonometric identities, 98–100 deconvolution, seeFourier transforms, deconvolution defective matrices, 283, 316degeneracy breaking of, 953–955of normal modes, 952 degenerate eigenvalues, 280, 287degenerate kernel, seekernel of integral equations, separable degree of polynomial equation, 2 degree of ODE, 474del∇,seegradient operator (grad) del squared ∇ 2(Laplacian), 358, 609 as integral, 406in curvilinear coordinates, 374in cylindrical polar coordinates, 366in polar coordinates, 658 in spherical polar coordinates, 368, 675 tensor form, 822 delta function (Dirac), seeDirac δ-function dependent random variables, 1038–1047derivative, see also diﬀerentiation absolute, 824–826covariant, 817Fourier transform of, 450Laplace transform of, 461normal, 356of a function of a complex variable, 711 1211 INDEX of a vector, 340 of basis vectors, 342–343of composite vector expressions, 343–344of function of a function, 47–48of hyperbolic functions, 109–112of products, 45–47, 49–51of quotients, 48of simple functions, 45ordinary, ﬁrst, second and nth, 43–44 partial, seepartial diﬀerentiation total, 157 derivative method for second series solution of ODE, 551–554 determinant form and/epsilon1 ijk, 791 for curl, 359 determinants, 264–268 adding rows or columns, 267and singular matrices, 268as product of eigenvalues, 292evaluation using /epsilon1 ijk, 791 using Laplace expansion, 264–265 identical rows or columns, 267in terms of cofactors, 264–265 interchanging two rows or two columns, 267 Jacobian representation, 204, 208, 209notation, 264of Hermitian conjugate matrices, 267of order three, in components, 265of transpose matrices, 266product rule, 267properties, 266–268, 827relationship with rank, 272–273removing factors, 267secular, 285 diagonal matrices, 273diagonalisation of matrices, 290–293 normal matrices, 291–292properties of eigenvalues, 292–293simultaneous, 337–338 diamond, unit cell, 239die throwing, seeprobability diﬀerence method for summation of series, 122–123 diﬀerence schemes for diﬀerential equations, 1180–1183, 1190–1192 diﬀerence, ﬁnite, seeﬁnite diﬀerences diﬀerentiable function of a complex variable, 711–713function of a real variable, 43 diﬀerential deﬁnition, 44 exact and inexact, 158–159 of vectors, 344, 350total, 157 diﬀerential equations, seeordinary diﬀerential equations andpartial diﬀerential equations diﬀerential equations, particular Bernoulli, 483Bessel, 541, 564–568 Chebyshev, 541Clairaut, 489diﬀusion, 611, 628–631, 649, 656–657, 1192Euler, 510Euler–Lagrange, 835–836Helmholtz, 671–676Hermite, 541Lagrange, 848Laguerre, 541Laplace, 612, 623, 650, 651, 1191Legendre, 540, 541, 555–564Legendre linear, 509–511Poisson, 612, 678–681Schr¨odinger, 675, 703 Schr¨odinger, 612, 854 simple harmonic oscillator, 541Sturm–Liouville, 849wave, 609–610, 622, 626–628, 647, 671, 849 diﬀerential operators, seelinear diﬀerential operator diﬀerentiation, see also derivative as gradient, 43as rate of change, 42chain rule, 47–48covariant, 817–820from ﬁrst principles, 42–45implicit, 48–49logarithmic, 49notation, 44of Fourier series, 430of integrals, 181–182of power series, 138partial, seepartial diﬀerentiation product rule, 45–47, 49–51quotient rule, 48theorems, 56–58using complex numbers, 104 diﬀraction, seeFraunhofer diﬀraction diﬀusion equation, 611, 621, 628–631 combination of variables, 629–631integral transforms, 681–683numerical methods, 1192separation of variables, 649simple solution, 629superposition, 656–657 diﬀusion of solute, 611, 629, 681–683dihedral group, 955, 958dimension of irrep, 930dimensionality of vector space, 248dipole matrix elements, 211, 950, 957dipole moments of molecules, 919–920Dirac δ-function, 361, 411, 445–449 and convolution, 453and Green’s functions, 517, 518as limit of various distributions, 449as sum of harmonic waves, 448deﬁnition, 445Fourier transform of, 449impulses, 447 1212 INDEX point charges, 447 properties, 445reality of, 449relation to Fourier transforms, 448–449relation to Heaviside (unit step) function, 447three-dimensional, 447, 458 direct product, of groups, 914direct sum⊕, 928 direction cosines, 225Dirichlet boundary conditions, 635, 745n Green’s functions, 688, 690–699 method of images, 693–699 Dirichlet conditions, for Fourier series, 421–422disc, moment of inertia, 211discontinuous functions and Fourier series, 426–428 discrete Fourier transforms, 468disjoint events, seemutually exclusive events displacement kernel, seekernel of integral equations, displacement distance from a line to a line, 235–236line to a plane, 236–237point to a line, 233–234point to a plane, 234–235 distributive law for addition of matrix products, 259convolution, 453, 464inner product, 249linear operators, 254multiplication of a matrix by a scalar, 256of a vector by a complex scalar, 226o fav e c t o rb yas c a l a r ,2 1 8 multiplication by a scalar in a vector space of ﬁnite dimensionality,247in a vector space of inﬁnite dimensionality, 583 scalar/dot product, 224vector/cross product, 226 div,seedivergence of vector ﬁelds divergence of vector ﬁelds, 358 as integral, 404–405 in curvilinear coordinates, 373 in cylindrical polars, 366in spherical polars, 368tensor form, 821–822 divergence theorem for tensors, 803for vectors, 407–408physical applications, 410–411related theorems, 409 division axiom in a group, 888division of complex numbers, 94–95dot product, seescalar product double integrals, seemultiple integrals drumskin, seemembrane dual tensors, 798–799dummy variable, 62/epsilon1 ijk, Levi-Civita symbol, tensor, 790–795 and determinant, 791identities, 792–793isotropic, 794vector products, 791weight, 813 e x,seeexponential function E, two-dimensional irrep, 932, 944, 950eccentricity, of conic sections, 16 eﬃciency, of estimator, 1074 eigenequation for diﬀerential operators, 581 more general form, 582–583, 601–602 eigenfrequencies, 325 estimation using Rayleigh–Ritz method, 333–335 eigenfunctions completeness for an Hermitian operator, 588construction of a real set for an Hermitian operator, 590–591 deﬁnition, 582of integral equations, 876of Legendre equation, 582of simple harmonic oscillators, 582orthogonality for Hermitian operators, 589–590 eigenvalues, 277–287 characteristic equation, 285deﬁnition, 277degenerate, 287determination, 285–287estimation for ODE, 849estimation using Rayleigh–Ritz method, 333–335 notation, 278of a general square matrix, 283of a representative matrix, 942 of a unitary matrix, 283 of an Hermitian operator reality, 588–589 of anti-Hermitian matrices, see anti-Hermitian matrices of Fredholm equations, 867of Hermitian matrices, seeHermitian matrices of integral equations, 867, 875of linear diﬀerential operators adjustment of parameters, 854–855deﬁnition, 582 error in estimate of, 852 estimation, 849–855higher eigenvalues, 852, 859Legendre equation, 582simple harmonic oscillator, 582 of linear operators, 277of normal matrices, 278–281under similarity transformation, 292–293 eigenvectors, 277–287 characteristic equation, 285deﬁnition, 277determination, 285–287normalisation condition, 278 1213 INDEX notation, 278 of a general square matrix, 283of a unitary matrix, 283of anti-Hermitian matrices, see anti-Hermitian matrices of commuting matrices, 283of Hermitian matrices, seeHermitian matrices of linear operators, 277of normal matrices, 278–281stationary properties for quadratic/Hermitian forms, 295–296 Einstein relation, 442, 642, 703elastic deformations, 802–803 electromagnetic ﬁelds ﬂux, 401Maxwell’s equations, 379, 414, 828 electrostatic ﬁelds and potentials charged split sphere, 668conducting cylinder in uniform ﬁeld, 730conducting sphere in uniform ﬁeld, 667from charge density, 680, 692from complex potential, 727inﬁnite charged plate, 694, 736inﬁnite wedge with line charge, 738ininite charged wedge, 736of line charges, 696, 726semi-inﬁnite charged plate, 736sphere with point charge, 698 ellipse area of, 72, 210, 391as section of quadratic surface, 297 ellipsoid, volume of, 210 elliptic PDE, 620, 623 empty event ∅, 963 end-points for variations contributions from, 841ﬁxed, 836variable, 841–844 energy levels of particle in a box, 703simple harmonic oscillator, 604 energy spectrum and Fourier transforms, 456, 457 envelopes, 176–178 equations of, 177to a family of curves, 176 epimorphism, 903equilateral triangle, symmetries of, 889, 894, 923–924, 952 equivalence relations, 906–908, 910 and classes, 906congruence, 907–909examples, 912 equivalence transformations, seesimilarity transformations equivalent representations, 926–928, 941error function, erf, 630, 682, 1204error terms in Fourier series, 436–437in Taylor series, 142–143errors, ﬁrst and second kind, 1122 essential singularity, 724, 750estimation of eigenvalues linear diﬀerential operator, 851–854Rayleigh–Ritz method, 333–335 estimator, maximum-likelihood, 1098 estimators (statistics), 1072 best unbiased, 1075bias, 1074central conﬁdence interval, 1080conﬁdence interval, 1079conﬁdence limits, 1079conﬁdence region, 1084consistency, 1073eﬃciency, 1074 minimum-variance, 1075 standard error, 1077 Euler equation diﬀerential, 510, 528trigonometric, 96 Euler method, numerical, 1181Euler–Lagrange equation, 835–836 special cases, 836–840 even functions, seesymmetric functions events, 962 complement of, 963 empty∅, 963 intersection of ∩, 962 mutually exclusive, 971statistically independent, 971union of∪, 963 exact diﬀerentials, 158–159exact equations, 478, 511–512 condition for, 478non-linear, 525 expectation values, seeprobability distributions, mean exponential distribution, 1032–1033 from Poisson, 1032MGF, 1033 exponential function Maclaurin series for, 143of a complex variable, 95, 719relation with hyperbolic functions, 105 Fabry–P ´erot interferometer, 149 factorial function, general, 1202factorisation, of a polynomial equation, 7faithful representation, 925, 940Fermat’s principle, 846, 857Fermi–Dirac statistics, 980 Fibonacci series, 531 ﬁeld lines and complex potentials, 726ﬁelds conservative, 393–395scalar, 353tensor, 803vector, 353 ﬁelds, electrostatic, seeelectrostatic ﬁelds and potentials 1214 INDEX ﬁelds, gravitational, seegravitational ﬁelds and potentials ﬁnite diﬀerences, 1179–1180 central, 1179for diﬀerential equations, 1180–1183forward and backward, 1179from Taylor series, 1179, 1186schemes for diﬀerential equations, 1190–1192 ﬁnite groups, 885ﬁrst law of thermodynamics, 179ﬁrst-order diﬀerential equations, seeordinary diﬀerential equations Fisher distribution, seeF-distribution (Fisher) ﬂuids Archimedean upthrust, 402, 416complex velocity potential, 727continuity equation, 410cylinder in uniform ﬂow, 728ﬂow, 727–728ﬂux, 401, 729irrotational ﬂow, 359 sources and sinks, 410–411, 727 stagnation points, 727velocity potential, 415, 612vortex ﬂow, 414, 728 forward diﬀerences, 1179Fourier cosine transforms, 452Fourier series, 421–438 and separation of variables, 652–655, 657coeﬃcients, 423–425, 431complex, 430–431diﬀerentiation, 430Dirichlet conditions, 421–422discontinuous functions, 426–428error term, 436–437examples square-wave, 424–425x, 430, 431 x 2, 428–429 x3, 430 integration, 430non-periodic functions, 428–430orthogonality of terms, 423 complex case, 431 Parseval’s theorem, 432–433raison d’ ˆetre, 421 standard form, 423summation of series, 433symmetry considerations, 425–426 Fourier sine transforms, 451Fourier transforms, 439–459 as generalisation of Fourier series, 439–441 convolution, 452–455 and the Dirac δ-function, 453 associativity, commutativity, distributivity, 453 deﬁnition, 453resolution function, 452 convolution theorem, 454correlation functions, 455–457cosine transforms, 452 deconvolution, 455deﬁnition, 441discrete, 468evaluation using convolution theorem, 454for integral equations, 868–871for PDE, 683–686Fourier-related (conjugate) variables, 442in higher dimensions, 457–459inverse, deﬁnition, 441odd and even functions, 451Parseval’s theorem, 456–457properties: diﬀerentiation, exponential multiplication, integration, scaling, translation, 450 relation to Dirac δ-function, 448–449 sine transforms, 451 Fourier transforms, examples convolution, 454damped harmonic oscillator, 457Dirac δ-function, 449 exponential decay function, 441 Gaussian (normal) distribution, 441rectangular distribution, 448spherically symmetric functions, 458two narrow slits, 454two wide slits, 444, 454 Fourier’s inversion theorem, 441Fraunhofer diﬀraction, 443–445 diﬀraction grating, 467two narrow slits, 454two wide slits, 444, 454 Fredholm integral equations, 864 eigenvalues, 867operator form, 865with separable kernel, 866–867 Fredholm theory, 874–875Frenet–Serret formulae, 349Frobenius series, 545Fuch’s theorem, 544function of a matrix, 260functional, 835functions of a complex variable, 711–725, 747–752 analyticity, 712behaviour at inﬁnity, 725branch points, 721Cauchy’s integrals, 745–747Cauchy–Riemann relations, 713–716conformal transformations, 730–738derivative, 711diﬀerentiation, 711–716identity theorem, 748Laplace equation, 715, 725Laurent expansion, 749–752multivalued and branch cuts, 721–723, 766particular functions, 718–721poles, 724power series, 716–718real and imaginary parts, 711, 716 1215 INDEX singularities, 712, 723–725 Taylor expansion, 747–748zeroes, 725, 754–758 functions of one real variable decomposition into even and odd functions, 422 diﬀerentiation of, 42–51Fourier series, seeFourier series integration of, 60–73limits, seelimits maxima and minima of, 51–53stationary values of, 51–53 Taylor series, seeTaylor series functions of several real variables chain rule, 160–161diﬀerentiation of, 154–182integration of, seemultiple integrals, evaluation maxima and minima, 165–170 points of inﬂection, 165–170rates of change, 156–158saddle points, 165–170stationary values, 165–170Taylor series, 163–165 fundamental solution, 691–693fundamental theorem of algebra, 86, 88, 770calculus, 62–63complex numbers, seede Moivre’s theorem gamma function as general factorial function, 1202deﬁnition and properties, 1201 Gauss’s theorem, 700 Gauss–Seidel iteration, 1160–1162Gaussian (normal) distribution N(µ, σ 2), 1021–1031 and Binomial distribution, 1027and central limit theorem, 1037 and Poisson distribution, 1029–1030 continuity correction, 1028CPF, 1023 tabulation, 1024 Fourier transform, 441integration with inﬁnite limits, 205integration with inﬁnite limits, 207mean and variance, 1022–1026MGF, 1027, 1030multiple, 1030–1031multivariate, 1051 sigma limits, 1025 standard variable, 1022 Gaussian (normal) distribution N(µ, σ 2), cumulative probability function, 1178random number generation, 1178 Gaussian elimination with interchange, 1159–1160 Gaussian integration, 1168–1170general tensors algebra, 787–790contraction, 788 contravariant, 810covariant, 810dual, 798–799metric, 806–809physical applications, 806–809, 825–826pseudotensors, 813tensor densities, 813 generalised likelihood ratio, 1124generating functions associated Legendre polynomials, 594Bessel functions, 573, 595Chebyshev polynomials, 597Hermite polynomials, 578, 596Laguerre polynomials, 597 Legendre polynomials, 562–564, 594 generating functions, probability, 999–1009, see alsomoment generating functions and probability generating functions geodesics, 825–826, 831, 856geometric distribution, 1001 geometric series, 120 Gibbs’ free energy, 181Gibbs’ phenonmenon, 427gradient of a function of one variable, 43several real variables, 156–158 gradient of scalar, 354–358 tensor form, 821 gradient of vector, 785, 818gradient operator (grad), 354 as integral, 404in curvilinear coordinates, 373in cylindrical polars, 366in spherical polars, 368tensor form, 821 Gram–Schmidt orthogonalisation of eigenfunctions of Hermitian operators, 589–590 eigenvectors of Hermitian matrices, 282normal matrices, 280 functions in a Hilbert space, 584–586 gravitational ﬁelds and potentials Laplace equation, 612Newton’s law, 345Poisson equation, 612, 678uniform disc, 706uniform ring, 676 Green’s functions, 597–601, 686–702 and boundary conditions, 518, 520and Dirac δ-function, 517 and partial diﬀerential operators, 687and Wronskian, 533diﬀusion equation, 684Dirichlet problems, 690–699for ODE, 188, 517–522Neumann problems, 700–702particular integrals from, 520Poisson’s equation, 689 1216 INDEX Green’s theorems applications, 639, 689, 743in a plane, 390–393, 413in three dimensions, 408 ground-state energy harmonic oscillator, 855 hydrogen atom, 860 group multiplication tables, 892 order ﬁve, 904order four, 892, 894, 903order six, 897, 903order three, 904 grouping terms as a test for convergence, 132groups Abelian, 886 associative law, 885cancellation law, 888centre, 911closure, 885cyclic, 903deﬁnition, 885–888direct product, 914division axiom, 888elements, 885 order, 889 ﬁnite, 885identity element, 885–888inverse, 885, 888isomorphic, 893mappings between, 901–903 homomorphic, 901–903image, 901isomorphic, 901 nomenclature, 944–945 non-Abelian, 894–898order, 885, 923, 924, 936, 939, 942permutation law, 889subgroups, seesubgroups groups, examples ±1 under multiplication, 885 alternating, 958complex numbers e iθ, 890 functions, 897 general linear, 914integers under addition, 885integers under multiplication (mod N), 891–893 matrices, 896 permutations, 898–900quaternion, 915rotation matrices, 890symmetries of an equilateral triangle, 889 H n(x),seeHermite polynomials Hamilton’s principle, 847Hamiltonian, 855 Hankel transforms, 465 harmonic oscillators damped, 243, 457ground-state energy, 855Schr¨odinger equation, 855 simple, seesimple harmonic oscillator heat ﬂow diﬀusion equation, 611, 629, 656in bar, 656–657, 683, 705in thin sheet, 631 Heaviside function, 447 relation to Dirac δ-function, 447 Heisenberg’s uncertainty principle, 441–443 Helmholtz equation, 671–676 cylindrical polars, 673plane polars, 672–673spherical polars, 674–676 Helmholtz potential, 180hemisphere, centre of mass and centroid, 198Hermite equation, 541, 593, 596Hermite polynomials H n(x) generating function, 578, 596orthogonality, 596Rodrigues’ formula, 596 Hermitian conjugate, 261–263 and inner product, 263product rule, 262 Hermitian forms, 293–297 positive deﬁnite and semi-deﬁnite, 295stationary properties of eigenvectors, 295–296 Hermitian kernel, seekernel of integral equations, Hermitian Hermitian matrices, 276 eigenvalues, 281–283 reality, 281–282 eigenvectors, 281–283 orthogonality, 282 Hermitian operators, 587–591 boundary condition for simple harmonic oscillators, 587–588 eigenfunctions completeness, 588orthogonality, 589–590 eigenvalues reality, 588–589 Green’s functions, 597–601importance of, 583, 588in Sturm–Liouville equations, 591–592properties, 588–591superposition methods, 597–601 higher-order diﬀerential equations, seeordinary diﬀerential equations Hilbert spaces, 584–586hit or miss, in Monte Carlo methods, 1174homogeneous boundary conditions, seeboundary conditions, homogeneous and inhomogeneous diﬀerential equations, 496dimensionally, 481, 527–528simultaneous linear equations, 298 homomorphism, 901–903 kernel of, 902representation as, 925 1217 INDEX Hooke’s law, 802 hydrogen atom, 604 s-states, 986 electron wavefunction, 211ground-state energy, 860 hydrogen molecule, symmetries of, 883 hyperbola, as section of quadratic surface, 297 hyperbolic functions, 105–112, 720 calculus of, 109–112deﬁnitions, 105, 720graphs, 105identities, 107in equations, 108inverses, 108–109 graphs, 109 trigonometric analogies, 105–107 hyperbolic PDE, 620, 623 hypergeometric distribution, 1015–1016 mean and variance, 1015 hypergeometric equation, 603hypothesis testing, 1119–1140 errors, ﬁrst and second kind, 1122generalised likelihood ratio, 1124generalised likelihood ratio test, 1123goodness of ﬁt, 1138Neyman–Pearson test, 1122null, 1120 power, 1122 rejection region, 1121simple or composite, 1120statistical tests, 1120test statistic, 1120 i,j,k(unit vectors), 223 i,s q u a r er o o to f −1, 87 identity element of a group, 885–888 uniqueness, 885, 887 identity matrices, 259, 260identity operator, 254images, method of, seemethod of images imaginary part/term of a complex number, 86–87 importance sampling, in Monte Carlo methods, 1172 improper integrals, 71rotations, 795–797 impulses, δ-function respresentation, 447 independent random variables, 998, 1042index of a subgroup, 908indices, of regular singular points, 546 indicial equation, 545 distinct roots with non-integral diﬀerence, 546–547 repeated roots, 547, 551, 553roots diﬀer by integer, 548–549, 552 induction, proof by, 31–32 inequalities amongst integrals, 73Bessel, 251, 586Schwarz, 251, 586 triangle, 251, 586 inertia, see also moments of inertia moments and products, 800tensor, 800 inexact diﬀerentials, 158–159inexact equation, 479inﬁnite integrals, 71 contour integration, 759–764 inﬁnite series, seeseries inﬂection general points of, 53 stationary points of, 51–53 inhomogeneous boundary conditions, seeboundary conditions, homogeneous and inhomogeneous diﬀerential equations, 496simultaneous linear equations, 298 inner product in a vector space, see also scalar product of ﬁnite dimensionality, 249–250 and Hermitian conjugate, 263commutativity, 249distributivity over addition, 249 of inﬁnite dimensionality, 584 integral equations eigenfunctions, 876eigenvalues, 867, 875Fredholm, 864 from diﬀerential equations, 862–863 homogeneous, 864linear, 863 ﬁrst kind, 864second kind, 864 nomenclature, 863singular, 864Volterra, 864 integral equations, methods for diﬀerentiation, 871Fredholm theory, 874–875integral transforms, 868–871 Fourier, 868–871Laplace, 869 Neumann series, 872–874Schmidt–Hilbert theory, 875–878separable (degenerate) kernels, 866–867 integral test for convergence of series, 131integral transforms, see also Fourier transforms andLaplace transforms general form, 465Hankel transforms, 465Mellin transforms, 465 integrals, see also integration complex, seecomplex integrals deﬁnite, 60double, seemultiple integrals Fourier transform of, 450improper, 71indeﬁnite, 63 1218 INDEX inequalities, 73, 586 inﬁnite, 71Laplace transform of, 462limits containing variables, 191ﬁxed, 60variable, 62 line,seeline integrals multiple, seemultiple integrals non-zero, 946–947properties, 61triple, seemultiple integrals undeﬁned, 60 integrals of vectors, seevectors, calculus of, integration integrand, 60integrating factor (IF), 512 ﬁrst-order ODE, 479–481 integration, see also integrals applications, 73–77 ﬁnding the length of a curve, 74–75mean value of a function, 73–74surfaces of revolution, 75–76volumes of revolution, 76–77 as area under a curve, 60as the inverse of diﬀerentiation, 62–63formal deﬁnition, 60from ﬁrst principles, 60–61in plane polar coordinates, 71–72logarithmic, 65multiple, seemultiple integrals multivalued functions, 762–764, 766of Fourier series, 430 of functions of several real variables, see multiple integrals of hyperbolic functions, 109–112of power series, 138of simple functions, 63–64of singular functions, 71of sinusoidal functions, 64–65 integration constant, 63integration, methods for by inspection, 63–64by parts, 68–70by substitution, 66–68 tsubstitution, 66–67 completing the square, 68ing the square, 67 change of variables, seechange of variables contour, seecontour integration Gaussian, 1168–1170 numerical, 1164–1170 partial fractions, 65–66reduction formulae, 70trigonometrical expansions, 64–65using complex numbers, 104 intersection ∩, probability for, seeprobability, for intersection intrinsic derivative, seeabsolute derivative invariant tensors, seeisotropic tensorsinverse hyperbolic functions, 108–109 inverse integral transforms Fourier, 441Laplace, 460, 765–768 uniqueness, 460 inverse matrices, 268–271 elements, 269in solution of simultaneous linear equations, 300–301 product rule, 271 properties, 270–271 inverse of a linear operator, 254inverse of a product in a group, 888inverse of element in a group uniqueness, 885, 888 inversion theorem, Fourier’s, 441inversions as improper rotations, 795symmetry operations, 883–884 irregular singular points, 540 irreps, 929 n-dimensional, 931, 944 counting, 937–938dimension n λ, 939 direct sum⊕, 928 identity A 1, 942, 946 n-dimensional, 930 number in a representation, 929, 937one-dimensional, 930, 931, 935, 941, 944 orthogonality theorem, 932–934 projection operators for, 949reduction to, 938summation rules for, 939–941 irrotational vectors, 359isobaric ODE, 482 non-linear, 527–528 isoclines, method of, 1188, 1196isomorphic groups, 893–898, 900, 901 isomorphism (mapping), 902 isotope decay, 490, 531isotropic (invariant) tensors, 793–795, 802iteration schemes convergence of, 1156–1158for algebraic equations, 1150–1158for diﬀerential equations, 1185for integral equations, 872–875Gauss–Seidel, 1160–1162 order of convergence, 1157 J ν(z),seeBessel functions j,s q u a r er o o to f −1, 87 j/lscript(z),seespherical Bessel functions Jacobians analogy with derivatives, 210and change of variables, 209–210deﬁnition in three dimensions, 208 two dimensions, 204 general properties, 209–210in terms of a determinant, 204, 208, 209 1219 INDEX joint distributions, seebivariate distributions andmultivariate distributions Jordan’s lemma, 761–762 kernel of a homomorphism, 902, 905 kernel of an integral transform, 465 kernel of integral equations displacement, 868Hermitian, 875 of form exp( −ixz), 869–871 of linear integral equations, 863 resolvent, 873, 874 separable (degenerate), 866–867 kinetic energy of oscillating system, 322 Klein-Gordon equation, 643 Kronecker delta δ ijand orthogonality, 249 Kronecker delta, δij(δj i), tensor, 777, 790–795, 805, 811 identities, 792–793 isotropic, 794 vector products, 791 L’Hˆopital’s rule, 145–147 Lagrange equations, 848 and energy conservation, 856 Lagrange undetermined multipliers, 170–176 and ODE eigenvalue estimation, 851 application to stationary properties of the eigenvectors of quadratic/Hermitian forms, 295 for functions of more than two variables, 172–176 in deriving the Boltzmann distribution, 174–176 integral constraints, 844 with several constraints, 172–176 Lagrange’s identity, 230 Lagrange’s theorem, 907–908 and the order of a subgroup, 904and the order of an element, 904 Lagrangian, 848, 856 Laguerre equation, 541, 596–597 Laguerre polynomials L n(x) generating function, 597 orthogonality, 596 Rodrigues’ formula, 577, 596 Lam´e constants, 802 lamina: mass, centre of mass and centroid, 196–198 Laplace equation, 612 expansion methods, 676–678 in three dimensions cylindrical polars, 661–664 spherical polars, 664–671 in two dimensions, 621, 623, 650, 651 and analytic functions, 715 and conformal transformations, 735–738 numerical method for, 1191, 1197 plane polars, 658–660 separated variables, 650uniqueness of solution, 676 with speciﬁed boundary values, 699, 701 Laplace expansion, 264–265Laplace transforms, 459–465, 765 convolution associativity, commutativity, distibutivity,464deﬁnition, 463 convolution theorem, 463deﬁnition, 459for ODE with constant coeﬃcients, 507–509for PDE, 681–683inverse, 460, 765–768 uniqueness, 460 properties: translation, exponential multiplication, etc., 462 table for common functions, 461 Laplace transforms, examples constant, 459derivatives, 461exponential function, 459integrals, 462polynomial, 459 Laplacian, seedel squared ∇ 2(Laplacian) Laurent expansion, 749–752 analytic and principal parts, 749region of convergence, 749 least squares, method of, 1113–1119 basis functions, 1115linear, 1114non-linear, 1118response matrix, 1115 Legendre equation, 540, 541, 555–564, 582, 593–594 as an example of a Sturm–Liouville equation, 591 associated, seeassociated Legendre equation general series solution, 556 Legendre functions, 556 of second kind, 557 Legendre functions P /lscript(x) associated Legendre functions, 703 Legendre linear equation, 509Legendre polynomials P /lscript(x) orthogonality, 668 Legendre polynomials P/lscript(x), 557 associated Legendre functions, 666generating function, 562–564, 594graph of, 557in Gaussian integration, 1168normalisation, 557, 559orthogonality, 560, 594recurrence relation, 562Rodrigues’ formula, 559, 594 Leibnitz’ rule for diﬀerentiation of integrals, 181Leibniz’ theorem, 49–51length of a vector, 222–223plane curves, 74–75, 347space curves, 347 1220 INDEX tensor form, 831 Levi-Civita symbol, see/epsilon1ijk, Levi-Civita symbol, tensor likelihood function, 1097limits, 144–147 deﬁnition, 144L’Hˆopital’s rule, 145–147 of functions containing exponents, 145 of integrals, 60 containing variables, 191 of products, 144of quotients, 144–147of sums, 144 line charge, electrostatic potential, 726, 738 line integrals and Cauchy integrals, 745–747and Stokes’ theorem, 412–415of scalars, 383–393 of vectors, 383–395 physical examples, 387round closed loop, 392 line, vector equation of, 230–231linear dependence and independence deﬁnition in a vector space, 247 of basis vectors, 221 relationship with rank, 272 linear diﬀerential operator L, 517, 551, 581 adjoint L †, 587 eigenfunctions, seeeigenfunctions eigenvalues, seeeigenvalues, of linear diﬀerential operators for Sturm-Liouville equation, 591–593Hermitian (self-adjoint), 583, 587–591 linear equations, diﬀerential ﬁrst-order ODE, 480general ODE, 496–523ODE with constant coeﬃcients, 498–509ODE with variable coeﬃcients, 509–523 linear equations, simultaneous, seesimultaneous linear equations linear independence of functions, 497 Wronskian test, 497, 538 linear integral operator K, 864 and Schmidt–Hilbert theory, 875–877Hermitian conjugate, 864inverse, 865 linear interpolation for algebraic equations, 1152–1153 linear least squares, method of, 1114linear molecules normal modes of, 326–328 symmetries of, 919 linear operators, 252–254 associativity, 254distributivity over addition, 254eigenvalues and eigenvectors, 277in a particular basis, 253 inverse, 254 non-commutativity, 254particular: identity, null/zero, singular/non-singular, 254 properties, 254 linear vector spaces, seevector spaces Liouville’s theorem, 747Ln of a complex number, 102–103, 720ln Maclaurin series for, 143of a complex number, 102–103, 720 log-likelihood function, 1100longitudinal vibrations in a rod, 610lottery (UK), and hypergeometric distribution, 1016 lower triangular matrices, 274 Maclaurin series, 141 standard expressions, 143 Madelung constant, 151 magnetic dipole, 224 magnitude of a vector, 222–223 in terms of scalar/dot product, 225 mappings between groups, seegroups, mappings between marginal distributions, 1040mass of non-uniform bodies, 196matrices, 246–312 as a vector space, 257 as arrays of numbers, 254as representation of a linear operator, 254column, 255elements, 254 minors and cofactors, 264 identity/unit, 259row, 255zero/null, 259 matrices, algebra of, 255 Cholesky separation , 318addition, 256–257and normal modes, seenormal modes change of basis, 288–290 diagonalisation, seediagonalisation of matrices multiplication, 257–259 and common eigenvalues, 283commutator, 314non-commutativity, 259 multiplication by a scalar, 256–257 numerical methods, seenumerical methods for simultaneous linear equations similarity transformations, seesimilarity transformations simultaneous linear equations, see simultaneous linear equations subtraction, 256 matrices, derived adjoint, 261–263 complex conjugate, 261–263Hermitian conjugate, 261–263inverse, seeinverse matrices transpose, 255 1221 INDEX matrices, properties of anti-Hermitian, seeanti-Hermitian matrices antisymmetric/skew-symmetric, 275determinant, seedeterminants diagonal, 273eigenvalues, seeeigenvalues eigenvectors, seeeigenvectors Hermitian, seeHermitian matrices normal, seenormal matrices nullity, 298order, 254orthogonal, 275–276rank, 272square, 254symmetric, 275trace/spur, 263triangular, 274tridiagonal, 1162–1164, 1190unitary, seeunitary matrices matrix elements in quantum mechanics as integrals, 945dipole, 950–951, 957 maxima and minima (local) of a function of constrained variables, seeLagrange undetermined multipliers one real variable, 51–53 suﬃcient conditions, 52 several real variables, 165–170 suﬃcient conditions, 167, 170 maximum modulus theorem, 756maximum-likelihood, method of, 1097–1113 bias, 1102data modelling, 1097estimator, 1098log-likelihood function, 1100parameter estimation, 1097transformation invariance, 1102 Maxwell’s electromagnetic equations, 379, 414, 828thermodynamic relations, 179–181 Maxwell–Boltzmann statistics, 980mean µ from MGF, 1005from PGF, 1000of RVD, 986–987of sample, 1066of sample: geometric, harmonic, root mean square, 1066 mean value of a function of one variable, 73–74several variables, 202 mean value theorem, 57–58median of RVD, 987membrane deformed rim, 658–660normal modes, 673, 954transverse vibrations, 610, 673, 702, 954 method of images, 639, 693–699, 738 disc (section of cylinder), 699, 701inﬁnite plate, 694intersecting plates in two dimensions, 696 sphere, 697–698, 706 metric tensor, 806–809, 812 and Christoﬀel symbols, 815and scale factors, 806, 821covariant derivative of, 831determinant, 806, 813 derivative of, 822 length element, 806raising/lowering index, 808, 812scalar product, 807volume element, 806, 830 MGF, seemoment generating functions Milne’s method, 1182minimum-variance estimator, 1075minor of a matrix element, 264mixed, components of tensor, 806, 811, 818ML estimator, 1098ML estimators, 1098 bias, 1102conﬁdence limits, 1104eﬃciency, 1103transformation invariance, 1102 mod N, multiplication, 891 mode of RVD, 987modulo, seemod N, multiplication modulus of a complex number, 90of a vector, seemagnitude of a vector molecules bonding in, 945, 947–950dipole moments of, 919–920symmetries of, 919 moment generating functions (MGF), 1004–1009 and central limit theorem, 1037–1038and PGF, 1005mean and variance, 1005particular distributions binomial, 1012exponential, 1033Gaussian, 1005, 1027Poisson, 1019 properties, 1005 moments central, 990of RVD, 989 moments of inertia and inertia tensor, 800deﬁnition, 201of disc, 211of rectangular lamina, 201of right circular cylinder, 212of sphere, 208perpendicular axes theorem, 212 moments, vector representation of, 227momentum as ﬁrst-order tensor, 782monomorphism, 903Monte Carlo methods antithetic variates, 1174control variates, 1173 1222 INDEX crude, 1172 hit or miss, 1174importance sampling, 1172multiple integrals, 1176random number generation, 1177stratiﬁed sampling, 1172 Monte Carlo methods, of integration, 1170–1177 Morera’s theorem, 744multinomial distribution, 1050–1051 and multiple Poisson distribution, 1060 multiple angles, trigonometric formulae, 10multiple integrals application in ﬁnding area and volume, 194–196mass, centre of mass and centroid, 196–198 mean value of a function of several variables, 202moments of inertia, 201 change of variables double integrals, 203–207general properties, 209–210 triple integrals, 207–209 deﬁnitions of double integrals, 190–191triple integrals, 193 evaluation, 191–193notation, 191, 192, 194order of integration, 191–192, 194 caveats, 193 multiplication tables for groups, seegroup multiplication tables multiplication theorem, seeParseval’s theorem multivalued functions, 721–723 integration of, 762–764 multivariate distributions, 1038, 1049–1053 change of variables, 1048–1049 Gaussian, 1051multinomial, 1050–1051 mutually exclusive events, 962, 971 n /lscript(z),seespherical Bessel functions nabla∇,seegradient operator (grad) natural logarithm, seelnandLn natural numbers, in series, 31, 124–125natural representations, 923, 952Necessary and suﬃcient conditions, 34–35negative function, 583 vector, 247 Neumann boundary conditions, 635 Green’s functions, 688, 700–702method of images, 700–702self-consistency, 700 Neumann series, 872–874Newton–Raphson (NR) method, 1154–1156 order of convergence, 1157 Neyman–Pearson test, 1122 nodes of oscillation, 626non-Abelian groups, 894–898 functions, 897matrices, 896 permutations, 898–900rotations–reﬂections, 894 non-Cartesian coordinates, seecurvilinear, cylindrical polar, plane polar andspherical polar coordinates non-linear diﬀerential equations, seeordinary diﬀerential equations, non-linear non-linear least squares, method of, 1118norm of function, 584vector, 249 normal to a plane, 232to coordinate surface, 372to surface, 352, 356, 396 normal derivative, 356normal distribution, seeGaussian (normal) distribution normal matrices, 277 eigenvectors completeness, 280orthogonality, 280–281 eigenvectors and eigenvalues, 278–281 normal modes, 322–335 characteristic equation, 325coupled pendulums, 335, 337deﬁnition, 326degeneracy, 952–955frequencies of, 325linear molecular system, 326–328membrane, 673, 954normal coordinates, 326normal equations, 326rod–string system, 323–326symmetries of, 328 normalisation of eigenfunctions, 589eigenvectors, 278functions, 585vectors, 223 null (zero) matrix, 259, 260operator, 254space, of a matrix, 298vector, 218, 247, 583 null operation, as identity element of group, 886nullity, of a matrix, 298numerical methods for algebraic equations, 1149–1156 binary chopping, 1154convergence of iteration schemes, 1156–1158linear interpolation, 1152–1153Newton–Raphson, 1154–1156rearrangement methods, 1151–1152 numerical methods for integration, 1164–1170 Monte Carlo, 1170Gaussian integration, 1168–1170midpoint rule, 1194nomenclature, 1165 1223 INDEX Simpson’s rule, 1167 trapezium rule, 1166–1167 numerical methods for ordinary diﬀerential equations, 1180–1190 accuracy and convergence, 1181Adams method, 1184diﬀerence schemes, 1181–1183Euler method, 1181ﬁrst-order equations, 1181–1188higher-order equations, 1188–1190isoclines, 1188Milne’s method, 1182prediction and correction, 1184–1186reduction to matrix form, 1190Runge–Kutta methods, 1186–1188 Taylor series methods, 1183–1184 numerical methods for partial diﬀerential equations, 1190–1192 diﬀusion equation, 1192Laplace’s equation, 1191minimising error, 1192 numerical methods for simultaneous linear equations, 1158–1164 Gauss–Seidel iteration, 1160–1162Gaussian elimination with interchange, 1159–1160 matrix form, 1158–1164tridiagonal matrices, 1162–1164 O(x), order of, 135 observables in quantum mechanics, 282, 588odd functions, seeantisymmetric functions ODE, seeordinary diﬀerential equations operators Hermitian, seeHermitian operators linear, seelinear operators andlinear diﬀerential operator andlinear integral operator order of approximation in Taylor series, 140nconvergence of iteration schemes, 1157group, 885group element, 889ODE, 474permutation, 900subgroup, 903 and Lagrange’s theorem, 907 tensor, 779 ordinary diﬀerential equations (ODE), see also diﬀerential equations, particular boundary conditions, 474, 476, 507complementary function, 497degree, 474dimensionally homogeneous, 481exact, 478, 511–512 ﬁrst-order, 474–490 ﬁrst-order higher-degree, 486–490 soluble for p, 486 soluble for x, 487 soluble for y, 488general form of solution, 474–476 higher-order, 496–529homogeneous, 496inexact, 479isobaric, 482, 527–528linear, 480, 496–523non-linear, 524–529 xabsent, 524 yabsent, 524 exact, 525isobaric (homogeneous), 527–528 order, 474 ordinary point, seeordinary points of ODE particular integral (solution), 475, 498, 500–501 singular point, seesingular points of ODE singular solution, 475, 487, 488, 490 ordinary diﬀerential equations, methods for canonical form for second-order equations, 522 eigenfunctions, 581–602equations containing linear forms, 484–486equations with constant coeﬃcients, 498–509Green’s functions, 517–522integrating factors, 479–481Laplace transforms, 507–509numerical, 1180–1190partially known CF, 512separable variables, 477series solutions, 537–558, 564–568undetermined coeﬃcients, 500variation of parameters, 514–516 ordinary points of ODE, 539, 541–544 indicial equation, 549 orthogonal lines, condition for, 12orthogonal matrices, 275–276, 778, 779 general properties, seeunitary matrices orthogonal systems of coordinates, 370orthogonal transformations, 781 orthogonalisation (Gram–Schmidt) of eigenfunctions of an Hermitian operator, 589–590 eigenvectors of a normal matrix, 280functions in a Hilbert space, 584–586 orthogonality of eigenfunctions of an Hermitian operator, 589–590 eigenvectors of a normal matrix, 280–281eigenvectors of an Hermitian matrix, 282functions, 584terms in Fourier series, 423, 431vectors, 223, 249 orthogonality properties of characters, 936, 944orthogonality theorem for irreps, 932–934orthonormal basis functions, 584basis vectors, 249–250 under unitary transformation, 290 oscillations, seenormal modes outcome, of trial, 961 1224 INDEX outer product of two vectors, 785 P/lscript(x),seeLegendre polynomials Pm /lscript(x),seeassociated Legendre functions Pappus’ theorems, 198–200parabolic PDE, 620, 623parallel axis theorem, 242parallel vectors, 227parallelepiped, volume of, 229–230parallelogram equality, 252parallelogram, area of, 227, 228parameter estimation (statistics), 1072–1097, 1140 Bessel correction, 1090error in mean, 1140mean, 1086variance, 1087–1090 parameter estimation, maximum-likelihood, 1097parameters, variation of, 514–516parametric equations of cycloid, 376, 844of space curves, 346 of surfaces, 351 parity inversion, 944Parseval’s theorem conservation of energy, 457for Fourier series, 432–433for Fourier transforms, 456–457 partial derivative, seepartial diﬀerentiation partial diﬀerential equations (PDE), 608–640, 646–702, see also diﬀerential equations, particular arbitrary functions, 613–618boundary conditions, 614, 632–640, 656characteristics, 632–638 and equation type, 636 equation types, 620, 643ﬁrst-order, 614–620general solution, 614–625homogeneous, 618inhomogeneous equation and problem, 618–620, 678–681, 686–702 particular solutions (integrals), 618–625 second-order, 620–631 partial diﬀerential equations (PDE), methods for change of variables, 624–625, 629–631constant coeﬃcients, 620 general solution, 622 integral transform methods, 681–686method of images, seemethod of images numerical, 1190–1192separation of variables, seeseparation of variables superposition methods, 650–657with no undiﬀerentiated term, 617–618 partial diﬀerentiation, 154–182 as gradient of a function of several real variables, 154–155 chain rule, 160–161change of variables, 161–163deﬁnitions, 154–156 properties, 160 cyclic relation, 160reciprocity relation, 160 Partial fractions, 18–25 and degree of numerator, 21complex roots, 22repeated roots, 23 partial fractions as a means of integration, 65–66in inverse Laplace transforms, 460, 508 partial sum, 118 particular integrals (PI), 475, see also ordinary diﬀerential equation, methods for and partial diﬀerential equations, methods for partition of a group, 906 set, 907 parts, integration by, 68–70path integrals, seeline integrals PDE, seepartial diﬀerential equations PDF, seeprobability functions, density functions penalty shoot-out, 1056pendulums, coupled, 335, 337periodic function representation, seeFourier series permutation groups S n, 898–900 cycle notation, 899 permutation law in a group, 889permutations, 975–981 degree, 898 distinguishable, 977order of, 900symbol nPk, 975 perpendicular axes theorem, 212perpendicular vectors, 223, 249PF,seeprobability functions PGF, seeprobability generating functions PI,seeparticular integrals plane curves, length of, 74–75 in Cartesian coordinates, 74in plane polar coordinates, 75 plane polar coordinates, 71, 342 arc length, 75, 367area element, 205, 367basis vectors, 342velocity and acceleration, 343 plane waves, 628, 649planes and simultaneous linear equations, 305–306vector equation of, 231–232 plates, conducting, see also complex potentials, for plates l i n ec h a r g en e a r ,6 9 6point charge near, 694 point charges, δ-function respresentation, 447 point groups, 924points of inﬂection of a function of one real variable, 51–53several real variables, 165–170 1225 INDEX Poisson distribution Po( λ), 1016–1021 and Gaussian distribution, 1029–1030as limit of Binomial distribution, 1016, 1019mean and variance, 1018MGF, 1019multiple, 1020–1021 recurrence formula, 1018 Poisson equation, 606, 612, 678–681 fundamental solution, 691–693Green’s functions, 688–702uniqueness, 638–640 Poisson summation formula, 467Poisson’s ratio, 802polar coordinates, seeplane polar andcylindrical polar andspherical polar coordinates polar representation of complex numbers, 95–98polar vectors, 798pole, of a function of a complex variable contours containing, 758–768order, 724, 750residue, 750–752 polynomial equations, 1–10 conjugate roots, 102factorisation, 7 multiplicities of roots, 4 number of roots, 86, 88, 770properties of roots, 9real roots, 1solution of using de Moivre’s theorem, 101–102 polynomial solutions of ODE, 544, 554–555 populations, sampling of, 1065 positive deﬁnite and semi-deﬁnite quadratic/Hermitian forms, 295 positive semi-deﬁnite norm, 249potential energy of ion in a crystal lattice, 151magnetic dipoles vector representation, 224 oscillating system, 323 potential function and conservative ﬁelds, 395complex, 725–730electrostatic, seeelectrostatic ﬁelds and potentials gravitational, seegravitational ﬁelds and potentials vector, 395 power series and diﬀerential equations, seeseries solutions of diﬀerential equations interval of convergence, 135Maclaurin, seeMaclaurin series manipulation: diﬀerence, diﬀerentiation, integration, product, substitution, sum, 137–138 Taylor, seeTaylor series power series in a complex variable, 136, 716–718 analyticity, 718circle and radius of convergence, 136, 717–718convergence tests, 717, 718 form, 716 power, in hypothesis testing, 1122powers, complex, 102–103, 719prediction and correction methods, 1184–1186, 1194 prime, non-existence of largest, 34principal axes of Cartesian tensors, 800–802conductivity tensors, 801inertia tensors, 800quadratic surfaces, 297rotation symmetry, 944 principal normals of space curves, 348principal value of complex integrals, 760complex logarithms, 103, 720 principle of the argument, 755probability, 966–1053 axioms, 967 conditional, 970–975 Bayes’ theorem, 974–975combining, 972 deﬁnition, 967for intersection ∩, 962 for union∪, 963, 967–970 probability distributions, 981, see also individual distributions bivariate, seebivariate distributions change of variables, 992–999generating functions, seemoment generating functions andprobability generating functions mean µ, 986–987 mean of functions, 987mode, median and quartiles, 987 moments, 989–992 multivariate, seemultivariate distributions standard deviation σ, 988 variance σ 2, 988 probability functions (PF), 981 cumulative (CPF), 981, 983density functions (PDF), 982 probability generating functions (PGF), 999–1004 and MGF, 1005binomial, 1003 deﬁnition, 1000 geometric, 1001mean and variance, 1000–1001Poisson, 1000sums of RV, 1003trials, 1000variable sums of RV, 1003–1004 product rule for diﬀerentiation, 45–47, 49–51products of inertia, 800projection operators for irreps, 949, 958projection tensors, 828proper rotations, 795proper subgroups, 903 1226 INDEX pseudoscalars, 796, 799 pseudotensors, 795–799, 813pseudovectors, 795–799 quadratic equations properties of roots, 10roots of, 2 quadratic equations, complex roots of, 86–87 quadratic forms, 293–297 positive deﬁnite and semi-deﬁnite, 295quadratic surfaces, 297removing cross terms, 294stationary properties of eigenvectors, 295–296 quartiles, of RVD, 987 quaternion group, 915, 955 quotient law for tensors, 788–790quotient rule, for diﬀerentiation, 48quotient test for series, 130 radius of convergence, 136, 717 radius of curvature of space curves, 348radius of torsion of space curves, 349random number generation, 1177random numbers non-uniform distribution, 1194 random variable distributions, seeprobability distributions random variables (RV), 961, 981–985 continuous, 982–985dependent, 1038–1047discrete, 981–982independent, 998, 1042sums of, 1002–1004 range, of a matrix, 298 rank of matrices, 272 and determinants, 272–273and linear dependence, 272 rank of tensors, seeorder of, tensors rate of change of a function of one real variable, 42several real variables, 156–158 ratio comparison test, 130ratio test (D’Alembert), 129, 718 in convergence of power series, 135 ratio theorem, 219 and centroid of a triangle, 220–221 Rayleigh–Ritz method, 333–335, 859real part/term of a complex number, 86–87real roots, of a polynomial equation, 1 rearrangement methods for algebraic equations, 1151–1152 reciprocal vectors, 237–238, 372, 804, 808 reciprocity relation for partial derivatives, 160 rectangular distribution, 1036 Fourier transform of, 448 recurrence relations, 502–507 characteristic equation, 505coeﬃcients, 542, 543, 1163 ﬁrst-order, 503 functions, 562, 569–570higher-order, 507 second-order, 505 reducible representations, 926, 928reduction formulae for integrals, 70reﬂections and improper rotations, 795as symmetry operations, 883–884 reﬂexivity, and equivalence relations, 906regular functions, seeanalytic functions regular representations, 939, 952regular singular points, 540, 544–546relative velocities, 222remainder term in Taylor series, 141repeated roots of auxiliary equation, 499representations, 918 deﬁnition, 924dimension of, 920, 924equivalent, 926–928faithful, 925, 940generation of, 920–926, 954irreducible, seeirreps natural, 923, 952 product, 945–947 reducible, 926, 928regular, 939, 952 counting irreps, 940 unitary, 928 representative matrices, 921 block-diagonal, 928eigenvalues, 942inverse, 925number needed, and order of group, 924of identity, 924 residue at a pole, 750–752theorem, 752–754 resolution function, 452resolvent kernel, 873, 874response matrix, for linear least squares, 1115rhomboid, volume of, 241Riemann tensor, 830Riemann theorem for conditional convergence, 127 Riemann zeta series, 131, 132right hand screw rule, 226Rodrigues’ formula for associated Legendre functions, 594Chebyshev polynomials, 597 Hermite polynomials, 596 Laguerre polynomials, 577, 596Legendre polynomials, 559, 594 Rolle’s theorem, 56root test (Cauchy), 132, 717roots of a polynomial equation, properties, 9 roots of unity, 100–101roots, of a polynomial equation, 2rope, suspended at its ends, 845rotation groups (continuous), invariant subspaces, 930 1227 INDEX rotation matrices as a group, 890 rotation of a vector, seecurl rotations as symmetry operations, 883–884axes and orthogonal matrices, 779, 780, 810improper, 795–797 invariance under, 783 product of, 780proper, 795 Rouch ´e’s theorem, 755–757 row matrix, 255Runge–Kutta methods, 1186–1188RV,seerandom variables RVD (random variable distributions), see probability distributions saddle points, 165 suﬃcient conditions, 167, 170 sampling correlation, 1070covariance, 1070space, 961statistics, 1065–1072with/without replacement, 971 scalar ﬁelds, 353 derivative along a space curve, 355gradient, 354–358line integrals, 383–393rate of change, 355 scalar product, 223–226 and inner product, 249and metric tensor, 807and perpendicular vectors, 223, 249for vectors with complex components, 225in Cartesian coordinates, 225 invariance, 779, 788 scalar triple product, 228–230 cyclic permutation of, 229in Cartesian coordinates, 229 determinant form, 229 interchange of dot and cross, 229 scalars, 216–217 invariance, 779zero-order tensors, 782 scale factors, 365, 368, 370 and metric tensor, 806, 821 scattering in quantum mechanics, 469Schmidt–Hilbert theory, 875–878Schr¨odinger equation, 612 constant potential, 703hydrogen atom, 675numerical solution, 1198variational approach, 854 Schwarz inequality, 251, 586Schwarz–Christoﬀel transformation, 733–735 second diﬀerences, 1179 second-order diﬀerential equations, seeordinary diﬀerential equations andpartial diﬀerential equations secular determinant, 285self-adjoint operators, seeHermitian operators semicircle, angle in, 18semicircular lamina, centre of mass, 200separable kernel in integral equations, 866–867variables in ODE, 477 separation constants, 648, 650separation of variables, for PDE, 646–681 diﬀusion equation, 649, 655–657, 671, 685expansion methods, 676–678general method, 646–650Helmholtz equation, 671–675inhomogeneous boundary conditions, 655–657inhomogeneous equations, 678–681Laplace equation, 650–655, 658–671, 676 polar coordinates, 658–681 separation constants, 648, 650superposition methods, 650–657wave equation, 647–649, 671, 673 series, 118–144 convergence of, seeconvergence of inﬁnite series diﬀerentiation of, 134ﬁnite and inﬁnite, 119integration of, 134multiplication by a scalar, 134multiplication of (Cauchy product), 134notation, 119operations, 134summation, seesummation of series series, particular arithmetic, 120arithmetico-geometric, 121 Fourier, seeFourier series geometric, 120Maclaurin, 141, 143power, seepower series powers of natural numbers, 124–125Riemann zeta, 131, 132Taylor, seeTaylor series series solutions of diﬀerential equations, 537–558, 564–568 about ordinary points, 541–544about regular singular points, 544–546 Frobenius series, 545 convergence, 541indicial equation, 545linear independence, 546polynomial solutions, 544, 554–555recurrence relation, 542, 543second solution, 542, 549–554 derivative method, 551–554 Wronskian method, 550, 558 shortest path, 837 and geodesics, 825, 831 similarity transformations, 288–290, 778–779, 934 properties of matrix under, 289–290unitary transformations, 290 simple harmonic oscillator, 582, 595 1228 INDEX energy levels of, 604 equation, 541 simple poles, 724Simpson’s rule, 1167simultaneous linear equations, 297–312 and intersection of planes, 305–306homogeneous and inhomogeneous, 298singular value decomposition, 306–312solution using Cramer’s rule, 304–305inverse matrix, 300–301numerical methods, seenumerical methods for simultaneous linear equations sine in terms of exponential functions, 105Maclaurin series for, 143orthogonality relations, 423 singular and non-singular integral equations, 864linear operators, 254matrices, 268 singular integrals, seeimproper, integrals singular points (singularities), 712, 723–725 essential, 724, 750removable, 725 singular points of ODE, 539 irregular, 540particular equations, 541regular, 540 singular solution of ODE, 475, 487, 488, 490singular value decomposition and simultaneous linear equations, 306–312singular values, 307 sinh, hyperbolic sine, 105, 720, see also hyperbolic functions skew-symmetric matrices, 275Snell’s law, 847soap ﬁlms, 839–840solenoidal vectors, 358, 395solid angle as surface integral, 401subtended by rectangle, 417 solid: mass, centre of mass and centroid, 196–198 source density, 612space curves, 346–350 arc length, 347binormal, 348curvature, 348Frenet–Serret formulae, 349parametric equations, 346principal normal, 348radius of curvature, 348radius of torsion, 349tangent vector, 348torsion, 348 spaces, seevector spaces span of a set of vectors, 247sphere, vector equation of, 232spherical Bessel functions j /lscript(z), 675spherical harmonics Ym /lscript(θ, φ), 670–671 spherical polar coordinates, 367–369 area element, 368basis vectors, 368length element, 368vector operators, 367–369volume element, 208, 368 spur of a matrix, 263–264spur, of a matrix, seetrace, of a matrix square matrices, 254square, symmetries of, 942square-wave, Fourier series for, 424–425stagnation points of ﬂuid ﬂow, 727standard deviation σ, 988 of sample, 1067 standing waves, 626stationary values of functions of one real variable, 51–53several real variables, 165–170 of integrals, 835under constraints, seeLagrange undetermined multipliers statistical tests, and hypothesis testing , 1120statistics, 961, 1064–1140 describing data, 1065–1072estimating parameters, 1072–1097, 1140 Stirling’s approximation, 1027, 1203asymptotic series, 1203 Stokes’ equation, 858Stokes’ theorem, 394, 412–415 for tensors, 804physical applications, 414related theorems, 413 strain tensor, 802stratiﬁed sampling, in Monte Carlo methods, 1172 streamlines and complex potentials, 727stress tensor, 802stress waves, 829string loaded, 857plucked, 705transverse vibrations of, 609, 848 Student’s t-distribution comparison of means, 1131critical points table, 1130normalisation, 1128plots, 1129 Student’s t-test, 1126–1132 Student’s t-distribution one/two-tailed conﬁdence limits, 1130 Sturm–Liouville equations, 591–597 boundary conditions, 592examples, 593–597 associated Legendre equation, 594–595Bessel equation, 595Chebyshev equation, 597Hermite equation, 596 1229 INDEX hypergeometric equation, 603 Laguerre equation, 596–597Legendre equation, 593–594simple harmonic oscillator, 595 manipulation to self-adjoint form, 592–593two independent variables, 860variational approach, 849–854weight function, 849 Sturm-Liouville equations zeroes of eigenfunctions, 603 subgroups, 903–905 index, 908normal, 905order, 903 Lagrange’s theorem, 907 proper, 903trivial, 903 submatrices, 272–273subscripts and superscripts, 777 contra- and covariant, 805covariant derivative, 818dummy, 777free, 777partial derivative, 818summation convention, 777, 804 substitution, integration by, 66–68summation convention, 777, 804summation of series, 119–127 arithmetic, 120arithmetico-geometric, 121contour integration method, 764–765diﬀerence method, 122–123Fourier series method, 433geometric, 120powers of natural numbers, 124–125transformation methods, 125–127 diﬀerentiation, 125integration, 125substitution, 126 superposition methods for ODE, 581, 597–601for PDE, 650–657 surface integrals and divergence theorem, 407Archimedean upthrust, 402, 416of scalars, vectors, 395–402physical examples, 401 surfaces, 351–353 area of, 352 cone, 75–76solid, and Pappus’ theorem, 198–200sphere, 352 coordinate curves, 352normal to, 352, 356of revolution, 75–76parametric equations, 351quadratic, 297tangent plane, 352 symmetric functions, 422 and Fourier series, 425–426and Fourier transforms, 451 symmetric matrices, 275 general properties, seeHermitian matrices symmetric tensors, 787symmetry operations on molecules, 883 order of application, 886 symmetry, and equivalence relations, 906 tsubstitution, 66–67 tan −1x, Maclaurin series for, 143 tangent planes to surfaces, 352tangent vectors to space curves, 348tanh, hyperbolic tangent, seehyperbolic functions Taylor series, 139–144 and ﬁnite diﬀerences, 1179, 1186 and Taylor’s theorem, 139–142, 747approximation errors, 142–143 in numerical methods, 1156, 1166 as solution of ODE, 1183–1184for functions of a complex variable, 747–748for functions of several real variables, 163–165remainder term, 141required properties, 139standard forms, 139 tensors, seeCartesian tensors andCartesian tensors, particular andgeneral tensors test statistic, 1120tetrahedral group, 957tetrahedron mass of, 197volume of, 195 thermodynamics ﬁrst law of, 179Maxwell’s relations, 179–181 top-hat function, seerectangular distribution torque, vector representation of, 227torsion of space curves, 348total derivative, 157total diﬀerential, 157trace of a matrix, 263–264 and second-order tensors, 788as sum of eigenvalues, 285, 292 invariance under similarity transformations, 289, 934 trace formula, 292 transcendental equations, 1150transformation matrix, 288, 294transformations active and passive, 797 conformal, 730–738coordinate, seecoordinate transformations similarity, seesimilarity transformations transforms, integral, seeintegral transforms and Fourier transforms andLaplace transforms transients in diﬀusion equation, 656in electric circuits, 491 transitivity, and equivalence relations, 906 1230 INDEX transpose of a matrix, 255, 260–261 product rule, 261 transverse vibrations membrane, 610, 673, 702 rod, 704 string, 609 trapezium rule, 1166–1167trial functions for eigenvalue estimation, 852 for particular integrals of ODE, 500 trials, 961triangle inequality, 251, 586triangle, centroid of, 220–221triangular matrices, 274tridiagonal matrices, 1162–1164, 1190, 1193 trignometric identities, 15 trigonometric identities, 10triple integrals, seemultiple integrals triple scalar product, seescalar triple product triple vector product, seevector triple product uncertainty principle (Heisenberg), 441–443 undetermined coeﬃcients, method of, 500undetermined multipliers, seeLagrange undetermined multipliers uniform distribution, 1036union∪, probability for, seeprobability for union uniqueness theorem Laplace equation, 676Poisson equation, 638–640 unit step function, seeHeaviside function unit vectors, 223 unitary matrices, 276 eigenvalues and eigenvectors, 283 representations, 928transformations, 290 upper triangular matrices, 274 variable end-points, seeend-points for variations, variable variable, dummy, 62 variables, separation of, seeseparation of variables variance σ 2, 988 from MGF, 1005 from PGF, 1001of dependent RV, 1044of sample, 1067 variation of parameters, 514–516variation, constrained, 844–846 variational principles, physical, 846–849 Fermat, 846Hamilton, 847 variations, calculus of, seecalculus of variations vector operators, 353–375 acting on sums and products, 360–361 combinations of, 361–363 curl, 359, 374del∇, 354 del squared ∇ 2, 358 divergence (div), 358geometrical deﬁnitions, 404–406gradient operator (grad), 354–358, 373identities, 362, 827Laplacian, 358, 374non-Cartesian, 363–375tensor forms, 820–824 curl, 823divergence, 821–822gradient, 821Laplacian, 822 vector product, 226–228 anticommutativity, 226deﬁnition, 226determinant form, 228in Cartesian coordinates, 228non-associativity, 226 vector spaces, 247–252, 955 action of group on, 930associativity of addition, 247basis vectors, 248–249commutativity of addition, 247complex, 247deﬁning properties, 247dimensionality, 248inequalities: Bessel, Schwarz, triangle, 251–252invariant, 930, 955matrices as an example, 257of inﬁnite dimensionality, 583–586 associativity of addition, 583basis functions, 583–584commutativity of addition, 583deﬁning properties, 583Hilbert spaces, 584–586inequalities: Bessel, Schwarz, triangle, 586 parallelogram equality, 252real, 247span of a set of vectors in, 247 vector triple product, 230 identities, 230non-associativity, 230 vectors as ﬁrst-order tensors, 781as geometrical objects, 246base, 342column, 255compared with scalars, 216–217component form, 221–222examples of, 216graphical representation of, 216–217irrotational, 359magnitude of, 222–223non-Cartesian, 342, 364, 368notation, 216polar and axial, 798solenoidal, 358, 395span of, 247 vectors, algebra of, 216–238 1231 INDEX addition and subtraction, 217–218 in component form, 222 angle between, 225associativity of addition and subtraction, 217commutativity of addition and subtraction, 217 multiplication by a complex scalar, 226multiplication by a scalar, 218multiplication of, seescalar product and vector product outer product, 785 vectors, applications centroid of a triangle, 220–221equation of a line, 230–231equation of a plane, 231–232equation of a sphere, 232ﬁnding distance from a line to a line, 235–236line to a plane, 236–237point to a line, 233–234point to a plane, 234–235 intersection of two planes, 232 vectors, calculus of, 340–375 diﬀerentiation, 340–345, 350integration, 345–346line integrals, 383–395surface integrals, 395–402volume integrals, 402–403 vectors, derived quantities curl, 359 derivative, 340diﬀerential, 344, 350divergence (div), 358reciprocal, 237–238, 372, 804, 808vector ﬁelds, 353 curl, 412divergence, 358ﬂux, 401rate of change, 356 vectors, physical acceleration, 341angular momentum, 241angular velocity, 227, 241, 359area, 399–401, 414area of parallelogram, 227, 228force, 216, 217, 224moment/torque of a force, 227velocity, 341 velocity vectors, 341Venn diagrams, 961–966vibrations internal, seenormal modes longitudinal, in a rod, 610transverse membrane, 610, 673, 702, 858, 860rod, 704string, 609, 848 Volterra integral equation, 863, 864 diﬀerentiation methods, 871Laplace transform methods, 869volume elements curvilinear coordinates, 371 cylindrical polars, 366 spherical polars, 208, 368 volume integrals, 402–403 and divergence theorem, 407 volume of cone, 76ellipsoid, 210 parallelepiped, 229 rhomboid, 241tetrahedron, 195 volumes as surface integrals, 403, 407 in many dimensions, 213of regions, using multiple integrals, 194–196 volumes of revolution, 76–77 and surface area & centroid, 198–200 wave equation, 609–610, 621, 849 boundary conditions, 626–628 characteristics, 637 from Maxwell’s equations, 379in one dimension, 622, 626–628 in three dimensions, 628, 647, 671 standing waves, 626 wave number, 443, 626nwave packet, 442 wave vector, k, 443 wavefunction of electron in hydrogen atom, 211wedge product, seevector product weight of relative tensor, 813of variable, 483 weight function, 582–583, 849 Wiener–Kinchin theorem, 456 work done by force, 387 vector representation, 224 Wronskian and Green’s functions, 533for second solution of ODE, 550, 558 from ODE, 538 test for linear independence, 497, 538 X-ray scattering, 241 Y m /lscript(θ, φ),seespherical harmonics Yν(z), Bessel functions of second kind, 568 Young’s modulus, 610, 802 z, as a complex number, 87 z∗, as complex conjugate, 92–94 zero (null) matrix, 259, 260 operator, 254 vector, 218, 247, 583 zero-order tensors, 781–784 zeroes of a function of a complex variable, 725 location of, 754–758, 771 1232 INDEX order, 725, 750 principle of the argument, 755 Rouch ´e’s theorem, 755, 757 zeroes of Sturm-Liouville eigenfunctions, 603 zeroes, of a polynomial, 2 zeta series (Riemann), 131, 132 z-plane, seeArgand diagram 1233