Lebedev-N-Et-Al-Worked-Problems-in-Applied-MathematicsNo-Pp

WORKED PROBLEMS IN APPLIED MATHEMATICS N. N. LEBEDEV I. P. SKALSKAYA Y. S. UFLYAND A. F. Ioffe Physico-Technical Institute Academy of Scienoes, U. S. S. R. Revised, Enlurged and Corrected English Edition Translated and Edited by Richard A. Silverman With a Supplement hy Edward L. Reiss Courant Institute of Mathematical Sciences New York University DOVER PUBLICATIONS, INC. NEW YORK Copyright 1965 by Richard A. Silverman. All rights reserved under Pan American and International Copyright Conventions. Published in Canada by General Publishing Com- pany, Ltd., 30 Lesrnill Road, Don Mills, Toronto, Ontario. Published in the United Kingdom by Constable and Con~pany, Ltd., This Dover edition, first published in 1979, is an unabridged republication of the work originally published by Prentice-Hall, Inc., in 1965 under the title Problems of Mathematicul Physics. International Standard Booh Number: 0-486-63730-1 Library of Congress Catalog Card Number: 78-67857 Manufactured in the United States of America Dover Publications, Inc. 180 Varick Street New York, N.Y. 10014 AUTHORS' PREFACE The aim of the present book is to'help the reader ac- quire the proficiency needed to successfully apply the methods of mathematical physics to a variety of prob- lems drawn from mechanics, the theory of heat conduc- tion, and the theory of electric and magnetic phenomena. A wide range of topics is covered, including not only problems of the simpler sort, but also problems of a more complicated nature involving such things as curvilinear coordinates, integral transforms, certain kinds of integral equations, etc. The book is intended both for students concomitantly studying the cor- responding topics in courses of mathematical physics, and for research scientists who in their work find it necessary to carry out calculations using the methods described here. We also think that quite apart from its value as a tool for acquiring technique, the book can also serve as a handbook, especially in view of the fact that answers to the problems are included. A rather solid background in applied mathematics is needed to profit from the book in its entirety. However, most of the problems appearing in Chapters 2 to 5 will be accessible to those who have taken only the usual first course in methods of mathematical physics. Chap- ters 6 to 8 are more specialized, and presuppose some familiarity with special functions, integral transforms, integral equations, and so on. To make the book easier to use, each section begins with a brief introduction describing its contents and presenting a certain amount of relevant background information. However, it is not claimed that this in- formation is complete in any sense, and the reader v desiring further details must consult the literature, e.g., the books and monographs cited at the end of each chapter. The majority of problems in this collection are ac- companied by hints, facilitating the choice of meaning- ful methods of solution. In addition, certain problems, whose numbers are equipped with asterisks (e.g., *52, *148, etc.), are solved in detail in a special section at the end of the book. The problems singled out in this way have been selected either because they illustrate the application of certain specific methods, or because of their special difficulty or particular importance in the applications. Because of the applied character of the book, we restrict ourselves to formal solutions, whose rigorous justification can be supplied by the interested reader. In compiling the collection, we have consulted not only the classic works on mathematical physics, but also a number of journal articles. Material accumulated during years of teaching and research in the Department of Mathematical Physics at the Leningrad Polytechnic Institute, as well as work done in connection with in- dustrial projects, plays a role in the material presented here. It would be impractical, and in many cases impossible, to cite the original source where a given problem was solved for the first time. Thus references to the literature have been confined to cases we find particularly relevant. We would like to take this opportunity to thank Prof. G. A. Grinberg for many valuable suggestions made in the course of writing the book. TRANSLATOR'S PREFACE The present edition differs from the Russian original in various respects, of which three merit particular mention: 1. The Bibliography has been expanded and up- dated. For example, the original sources of works translated into Russian have been tracked down, all references have been equipped with titles, further references (especially, later editions and English translations) have been added, and so on. As in other volumes of this series, the system of references is in "letter-number form." Thus L10 refers to the tenth paper (or book) whose (first) author's surname begins with the letter L, where the entire Bibliog- raphy is arranged in lexicographic order, and chronological order as well, whenever there are several papers by the same author. 2. Working from an extensive list of errata sent me by the authors, I have corrected numerous misprints and mistakes present in the Russian edi- tion. I am particularly grateful for their help, since the task of eliminating errors from a book of this type (consisting primarily of problems and answers) is both imperative and one which only the authors themselves can perform in finite time! The authors have also been kind enough to answer a number of specific questions that arose in the course of the translation. 3. It was felt that the English-language edition would benefit greatly by the addition of material on the approximate solution of problems of mathe- matical physics, since the emphasis of the Russian vii edition is on exact solutions. This led to the writ- ing of a Supplement on variational and related methods by Professor Edward L. Reiss of the Courant Institute of Mathematical Sciences of New York University. The Supplement is independent of the rest of the book, even to the extent of having its own references. R. A. S. CONTENTS PART 1 PROBLEMS, Page 1. DERIVATION OF EQUATIONS AND FORMU- 1 LATION OF PROBLEMS, Page 3. 1. Mechanics, 3. 2. Heat Conduction, 9. 3. Electricity and Magnetism, 11. 2 SOME SPECIAL METHODS FOR SOLVING HYPERBOLIC AND ELLIPTIC EQUATIONS, Page 19. 1. Hyperbolic Equations, 19. 2. Elliptic Equations: The Green's Function Method, 27. 3. Elliptic Equations: The Method of Conformal Mapping, 33. 3 STEADY-STATE HARMONIC OSCILLATIONS, Page 42. 1. Elastic Bodies: Free Oscillations, 43. 2. Elastic Bodies: Forced Oscillations, 46. 3. Electromagnetic Oscillations, 49. 4 THE FOURIER METHOD, Page 55. 1. Mechanics: Vibrating Systems, Acoustics, 60. 2. Mechanics: Statics of Deformable Media, Fluid Dynamics, 73. ix X CONTENTS 4 THE FOUKIER METHOD-Continued 3. Heat Conduction: Nonstationary Problems, 77. 4. Heat Conduction: Stationary Problems, 83. 5. Electricity and Magnetism, 91. 5 THE EIGENFUNCTION METHOD FOR SOLV- ING INHOMOGENEOUS PROBLEMS, Page 103. 1. Mechanics: Vibrating Systems, 107. 2. Mechanics: Statics of Deformable Media, 114. 3. Heat Conduction: Nonstationary Problems, 119. 4. Heat Conduction: Stationary Problems, 124. 5. Electricity and Magnetism, 13 1. 6 INTEGRAL TRANSFORMS, Page 143. 1. The Fourier Transform, 146. 2. The Hankel Transform, 160. 3. The Laplace Transform, 169. 4. The Mellin Transform, 189. 5. Integral Transforms Involving Cylinder Functions of Imaginary Order, 194. 7 CURVILINEAR COORDINATES, Page 203. 1. Elliptic Coordinates, 204. 2. Parabolic Coordinates, 210. 3. Two-Dimensional Bipolar Coordinates, 212. 4. Spheroidal Coordinates, 219. 5. Paraboloidal Coordinates, 23 1. 6. Toroidal Coordinates, 233. 7. Three-Dimensional Bipolar Coordinates, 242. 8. Some General Problems on Separation of Var- iables, 247. 8 INTEGRAL EQUATIONS, Page 253. 1. Diffraction Theory, 254. 2. Electrostatics, 259. CONTENTS Xi PART 2 SOLUTIONS, Page 273. MATHEMATICAL APPENDIX, Page 38 1. 1. Special Functions Appearing in the Text, 381. 2. Expansions in Series of Orthogonal Functions, 384. 3. Some Definite Integrals Frequently Encountered in the Applications, 386. 4. Expansion of Some Differential Operators in Orthogonal Curvilinear Coordinates, 388. Supplement. VARIATIONAL AND RELATED METHODS, Page 391. 1. Variational Methods, 392. 1.1. Formulation of Variational Problems, 392. 1.2. The Ritz Method, 396. 1.3. Kantorovich's Method, 401. 2. Related Methods, 404. 2.1. Galerkin's Method, 404. 2.2. Collocation, 407. 2.3. Least Squares, 411. 3. References, 412. BIBLIOGRAPHY, Page 415, NAME INDEX, Page 423. SUBJECT INDEX, Page 427. SOLLVWAHLVW CalIddV NI SWAIKOUd CHAWOM Part1 PROBLEMS DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS Chapter 1 is devoted to problem material on the derivation of the equations of mathematical physics and the formulation of appropriate initial ahd boundary conditions. It also serves as a convenient place to list the basic equations appearing later in the book. Throughout, we assume that the reader is familiar with the physical laws underlying the mathematical formulation of the problems which arise in various branches of physics. The chapter consists of three sections devoted in turn to problems of mechanics, heat conduction and the theory of electric and magnetic phe- nomena. Each section starts with the basic equations governing the corre- sponding set of problems, with appropriate references to sources where the derivations can be found. Special attention is devoted to the formulation of problems of electrodynamics, since this subject is inadequately covered in the available 1iterature.l I. Mechanics This section contains problems on the derivation of equations of motion and formulation of initial and boundary conditions for vibrating strings, membranes, rods and plates, as well as some examples pertaining to the statics of deformable media. It will be assumed that the reader has already ' Those particularly interested in mathematical aspects of the formulation of physical problems can find relevant material in C5, GI, L1, P2, S1 and S13. (The reference scheme is explained in the Translator's Preface.) 4 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS encountered the basic equations in a first course on mathematical physic^.^ Thus we shall merely list the equations concisely, at the same time explaining the notation to be used in the book. 1. The equation of a vibrating string is where u(x, t) is the displacement of the point of the string with abscissa x at the time t, q(x, t) is the external load per unit length, Tis the tension, and p is the linear density. 2. The equation for longitudinal oscillations of a rod of conztant cross section is where u(x, t) is the displacement of the cross section of the rod with abscissa x at the time t, E is Young's modulus, and p is the density. 3. The equation for transverse oscillations of a rod (beam) is where u(x, t) is the displacement of the points along the midline of the rod, q(x, t) is the external load per unit length, E is Young's modulus, J is the moment of inertia of a transverse cross section, p is the density, and S is the cross-sectional area. 4. The equation of a vibrating membrane is where u(x, y, t) is the displacement of the point (x, y) of the membrane at the time t, q(x, y, t) is the external load per unit area, Tis the tension per unit length of the boundary of the membrane, and p is the surface density. 5. The equation for transverse oscillations of a thin elastic plate is a See S6 (Vol. 11), S14, T1 and T2. Concerning the derivation of the equations of vibrating plates, see T4. FROB. 1 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 5 where u(x, t) is the displacement of the point (x, y) of the midplane of the plate at the time t, q(x, y, t) is the density of the external load, D is the flexural rigidity, h is the thickness, p is the density, and is the two-dimensional Laplacian operator. The above equations lead to corresponding equations for static deflections, if we regard the external load q and the unknown displace- ment u as independent of the time t. For example, the equilibrium equation for the membrane is :the static deflection of the plate satisfies the equation and so on. Among the other equations governing the statics of elastic bodies which will figure in this book, we cite the familiar equation for twisting of a prismatic rod, where u(x, y) is the torsion function. We now give some problems on the formulation of initial and boundary conditions for these equations, and also some problems on the derivation of other differential equations. 1. Describe the initial and boundary conditions for a vibrating string with fixed ends (0 g x < I), which is stretched at the point x = c and time t = 0 to a height h, and then released without initial velocity. Ans. 6 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS PROB. 2 2. A concentrated load of mass mo is fastened at the point x = c of a string 0 < x < 1 of length I. Find the equations describing vibrations of the string with arbitrary initial conditions, assuming that the ends of the string are fastened. Ans. with initial conditions and boundary conditions 3. Formulate initial and boundary conditions for the problem of longi- tudinal oscillations of a rod in the following special cases: a) A rod of length 1 is clamped at the end x = 0 and stretched by a force F applied to the other end; at the time t = 0 the force is suddenly discontinued; b) A tensile force F(t) is applied at the time t = 0 to the end x = 1 of a cantilever in equilibrium; c) A cantilever clamped at the point x = 0, with a load of mass Mo at the free end x = 1, undergoes longitudinal oscillations subject to arbitrary initial conditions. Ans. 4. Derive the differential equation for longitudinal oscillations of a thin rod of variable cross section S = S(x). As an example, derive the equation for oscillations of a conical rod. /ins. PROB. 8 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 7 5. Derive the equation for torsional oscillations of a shaft of circular cross section. Ans. where 8(x, t) is the angular displacement of the cross section x relative to the equilibrium position, u = JG/p, p is the density, and G is the shear modulus. Hint. The torque at the cross section x is given by the expression where J is the polar moment of inertia of a cross section of the shaft. 6. Formulate initial and boundary conditions for the problem of torsional oscillations of a shaft of circular cross section and length 1, where the end x = 0 isclamped and a disk-shaped mass with moment of inertia J, is attached to the other end. At the time t = 0, the disk is rotated through a given angle u and then released without initial velocity. Ans. 7. A cantilever of length 1 is clamped at one end x = 0 and loaded by a force Fat the other end. At the time t = 0, the action of the force is suddenly discontinued. Formulate initial and boundary conditions for the corre- sponding oscillations. Ans. Initial conditions and boundary conditions 8. Describe initial and boundary conditions for the problem of free oscillations of a disk-shaped plate with clamped edge, whose initial deforma- tion is due to a concentrated force F applied at the center of the disk. Ans. 8 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS PROB. 9 Hint. To determine the static deflection due to the concentrated force, consider the force as the limiting case of a load of density F/T&~ uniformly distributed over a small disk of radius E. 9. Show that the problem of the deflection of a plate with a simply supported polygonal boundary reduces to the solution of Poisson's equation Aw =f (x, y>, with boundary condition wl, = 0 (fis a known function). Hint. Note that in the present case, the boundary conditions on the supported edge can be written in the form ul, = 0, Aul, = 0. 10. Show that the velocity potential for the three-dimensional flow of an ideal incompressible fluid containing no sources is described by Laplace's equation Au = 0. Hint. Use the condition (v is the vector describing the velocity of fluid particles at a given point, S is an arbitrary closed surface inside the flow, and n is the exterior normal to the surface S) and the condition v = -grad u for potential flow. 11. Formulate mathematically the problem of the flow of an ideal fluid past an object bounded by a surface S, where fluid emanates from a point source of strength m located at a point M, in the region exterior to S. Ans. The problem reduces to finding a solution of the equation which is regular (i.e., has no singularities) in the region exterior to S, except at the point M,. In a neighborhood of M,, m u = + a regular function 4v IMMOI where M is a point near M, and p is the density of the fluid (IMM,! denotes the distance between M and M,). The desired function u must satisfy the boundary condition PRVB. 12 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 9 and the condition u = O(R-I), R -t m at infinity. 2. Heat Conduction As proved in courses on mathematical physics (see S1, Tl), the flow of heat in a body of thermal conductivity k, specific heat c and density p is governed by Fourier's equation where T(M, t) is the temperature at the point M, and Q is the density of heat sources within the body.3 The boundary conditions to be satisfied on the surface of the body (or its parts) depend on the particular problem under consideration. Most often it is assumed that the surface of the body has a given temperature ~1~ = f(P, t), where P is a point of the surface S, or that the body radiates heat into the surrounding medium according to Newton's law, which states that the amount of heat radiated by a unit area of the surface per unit time is proportional to the difference between the temperature of the surface and that of the surrounding medium. In the latter case, the boundary condition takes the form where a/an indicates differentiation with respect to the exterior normal to S, Tmed is the temperature of the surrounding medium, and h is the heat exchange coefficient or emissivity. Without loss of generality, we can assume that Tmed = 0; this assumption is made in all the problems involving heat conduction except Prob. 1 5K4 We now give a few problems on the formulation of initial and boundary conditions for the equation of heat conduction (and for the related diffusion equation). 12. Let the temperature of a conductor in the form of an infinite cylinder of radius a be initially the same as that of the surrounding medium. Suppose that starting from the time t = 0, the conductor is heated by a constant The density of heat current (i.e., the heat flux) is described by the vector q = -k grad T. Examples of other boundary conditions encountered in the applications are given in Probs. 365, 367 and 370. 10 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS PROB. 13 electric current releasing an amount of heat Q per unit volume of the con- ductor. Give a mathematical formulation of the corresponding problem of heat conduction, assuming that the heat exchange at the surface of the con- ductor obeys Newton's law.5 Ans. The temperature T(r, t) satisfies the equation with initial condition TI,,, = 0 and boundary condition 13. A homogeneous sphere of radius a is heated for a long time by heat sources uniformly distributed throughout its volume with density Q. Write the equations which describe the cooling of the sphere after the sources are turned off, assuming that the heat exchange between the surface of the sphere and the surrounding medium, during both the heating and cooling, obeys Newton's law. Ans. 14. Two slabs of thicknesses a, and a,, made from different materials and heated to temperatures T,O and T,O, are put into contact with each other at the time t = 0. Write the equations governing the resulting process of tempera- ture equalization, assuming that the free surfaces are thermally insulated from the surrounding medium. Ans. It is recommended that the problem be solved directly from underlying physical principles, without regarding Fourier's equation as known. PROB. 16 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS I 1 and boundary conditions 15. A nonuniformly heated body in the form of a circular ring of radius a with a small cross section cools by giving off heat from its lateral surface. Write the equations describing the corresponding process of temperature equalization, assuming that the temperature drop inside the ring can be neglected and that the surface cooling obeys Newton's law. Ans. where p is the perimeter, S the cross-sectional area and h the heat exchange coefficient. The temperature, which must be a periodic function of the angular coordinate cp, satisfies the initial condition TI,=, = f (Y), where f is a given function. 16. Show that the concentration C(x, y, z, t) of a substance diffusing in a gas or liquid obeys the differential equation where Q is the source density of the diffusing substance and D is the diffusion coefficient. Hint. Starting from Nernst's law q = -grad C (where the vector q is the density of flow of the diffusing substance), write a conservation equation for an arbitrary volume element. 3. Electricity and Magnetism An important class of problems of mathematical physics involves integra- tion of the differential equations arising in various branches of electromagnetic theory. Assuming that the reader has previously encountered this subject (see G5, 56, Pl), we shall regard the following basic equations as known: 1. The equations of electrostatics 4T Au=-- , E=-gradu, E where u is the potential of the electrostatic field E, p = p(M) is the volume density of charge at the point M, E is the dielectric constant of the medium, and A is the Laplacian operator. 12 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 2. The equations for the distribution of d-c current density inside a homogeneous conductor, where u is the potential of the current field, j is the current density vector, Q = Q(M~ is the volume density of current sources (in particular, Q may vanish), and o is the conductivity. The equations for the magnetic field due to d-c currents, where A is the vector poten- tial of the magnetic field H, the vector j(') is the density of the (external) currents producing the magnetic field, p is the magnetic permeability of the medium, c is the velocity of light in vacuum, and A is the Laplacian operator." Maxwell's equations E aE 4x0 477 curl H = - - + - E + - j(')), at C 477~ div E = - , E div H = 0 for the electromagnetic field in a homogeneous isotropic medium, where E and H are the electric and magnetic field vectors, E, p and a are the dielectric constant, the magnetic permeability and8 the conduc- tivity of the medium, c is the velocity of light, and p and j(') are the charge and current densities producing the field.7 The components of the vector AA in a Cartesian coordinate system are AA,, AA, and AA,. To calculate the components of the vector. AA in other coordinate systems, one should use the relation AA = grad div A - curl curl A. Expressions for the components of AA in cylindrical and spherical coordinates are given on p. 389-390. It should be noted that if is given, then p cannot be chosen arbitrarily, but must satisfy the differential equation implied by the first and third Maxwell equations. PROB. 18 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 13 If we use the relations 1 1 aA H = - curl A, E = -grad u - - - P at to introduce the vector and scalar potentials A and u,~ the problem of deter- mining the electromagnetic field reduces to integrating the system of equations We now consider the mathematical formulation of various problems involving electric and magnetic fields (both static and variable), as well as some problems on transformations of the differential equations of electro- dynamics which are useful in special cases. 17. Formulate mathematically the problem of finding the three-dimen- sional electrostatic field between N conductors of arbitrary shape at given potentials Vi (i = 1, . . . , N). Ans. In the region D bounded by the surfaces Si (i = 1, . . . , N) of the conductors, the potential u satisfies Laplace's equation The boundary conditions have the form ulSi = 6, i = 1,. . . , N, where, in the case where the point at infinity belongs to D, these conditions must be supplemented by the requirement that at infinity the potential u approach zero uniformly in all directions. Comment. If none of the surfaces Si extends to infinity, then the products Ru and R2 grad u (where R2 = x2 + y2 + z2) remain uniformly bounded as R -+ co. However, these conditions need not be included in the formulation of the problem, since the uniqueness of the solution is guaranteed by the above requirement that the potential u approach zero uniformly as R +a. 18. A charge Q is placed at the point M, = (x,, yo, z,) near a conductor at potential V, bounded by a surface S. Formulate the corresponding problem of electrostatics. The quantities A and u are not independent, but are connected by the relation 14 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS PROB. 19 Ans. The potential u satisfies Laplace's equation at every point of the region surrounding the conductor, except at the point M,, near which U=- + a regular function, R (R = [MOM( is the distance between the points M and M,). Moreover, the potential satisfies the boundary condition ul, = V and the condition that ulW 3 0 uniformly in all directions. 19. A thin charged wire of charge q per unit length is placed inside a grounded cylindrical shell whose generators are parallel to the wire (see Figure 1). Formulate the corresponding two- dimensional electrostatic problem. Ans. The potential u satisfies the two- dimensional Laplace equation in the whole region D except at the point M,, where the potential has a logarithmic singularity u = -29 In R + a regular function. The boundary condition is ul, = 0. 20. Reformulate the preceding problem for the case where the charged wire is placed outside the conductor, and the total charge per unit length of the conductor is specified instead of its potential. Ans. The boundary condition is now (alas denotes differentiation along the tangent to the contour I'), while the condition at infinity becomes ul, = -2(Q + q) In R + a bounded function, where Q is the charge per unit length of the conductor. 21. Show that the problem of the current distribution in a thin conducting shell (see Figure 2) reduces to integration of the equation PROB. 22 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS 15 where u is the potential of the current distribution in the shell (j = -5 grad u is the current density vector), o is the conductivity and h the thickness of the shell, j?' is the normal component of the density of current applied to the shell by using suitable electrodes, and A is the appropriate two-dimensional Laplacian, i.e., (ds is the element of arc length on the surface of the shell). Hint. Average equation (I), p. 12 (giving the volume distribution of current) over the thickness of the shell. 22. Suppose an object of arbitrary shape, made of magnetic material of permeability p, is magnetized by being introduced into a homogeneous magnetic field H, (see Figure 3). Formulate mathematically the correspond- ing problem of magnetostatics. Ans. If u is the potential of the magnetic field H (i.e., H = -grad u), the problem reduces to integration of the equation with the boundary conditions and the following conditions at infinity -grad u,(, = H,, ull, is bounded. Hint. It helps to keep in mind that in the source-free part of space, the magnetic field H satisfies the equations curl H = 0, div H = 0, which imply H = --grad u, Au = 0. 16 DERIVATION OF EQUATIONS AND FORMULATION OF PROBLEMS PROB. 23 23. The differential equations for wave propagation down a long trans- mission line, with self-inductance L, capacitance C, resistance R and leakage conductance G per unit length, take the form X=O ar au-~-+~~, --= c + GU, ax at where u(x, t) and I(x, t) are the values of the I voltage and current at the point x at the time FIGURE 4 t (see T1, p. 18). Formulate initial and boundary conditions for wave propagation along such a line, if at one end a constant voltage E is switched on in series with a lumped resistance R,, while the other end is terminated by a coil of self-inductance L (see Figure 4). 24. Show that if j(" = 0 and = 0, then the differential equations for the electromagnetic potentials A and u can be satisfied by setting .where II is the Hertz vector satisfying the equation Ep a2n 4~~~ an An------- - - 0. cz atz cz at Derive expressions for the vectors E and H in terms of the vector II. Ans. , E = curl curl n. C Hint. According to footnote 7, p. 12, it follows from fe) = 0 and ~l,,, = 0 that Q = 0 for arbitrary t. 25. Verify that if j(" = 0 and = 0, then the vectors E and H satisfy the same differential equation as the Hertz vector, i.e.,9 In some problems it is more convenient to start from these equations than from the equations for the electromagnetic potentials or for the Hertz vector. 26. Show that for steady-state harmonic oscillations of frequency w, in which the time dependence of the quantities defining the electromagnetic field (i.e., the vectors E and H, the charge and current densities which are the sources of the field, etc.) is characterized by a factor eiUt, Maxwell's equations (see p. 12) take the form AA* + k2~* = - where f * denotes the complex amplitude of a scalar or vector functionf, andlo k = , Im lc < 0. Comment. The importance of this problem consists in showing that only one unknown function (rather than two) is needed to calculate the electro- magnetic field in the case of harmonic time dependence. 27. Starting from Maxwell's equations for the case of steady-state harmonic oscillations, deduce the corresponding differential equations for the two-dimensional (planar) electromagnetic problem, where (e)* - .(e)* = 0 j?)* = j(X, y), j, - J, E,* = E,* = 0, ~f = E(x, y), H,* = H,(x, y), H,* = H,(x, y), Hf = 0. Ans. c aE H =--- c aE , H,=--. piw ay piw ax Hint. For harmonic time dependence, the connection between j("'%nd p* is given by which in the present case implies p* = 0. lo For example, if j'O' = jAel sin ot, where j:" is real, then the actual values of E and H are given by the imaginary parts of ,the expressions ~*e"' and ~*e~('". 28. Derive the equations for steady-state harmonic electromagnetic oscillations for the case of axial symmetry, where 4e)* - .(e)* - Jr - Jq - 0, j!')* = j(r, z) E: = E,(r, z), E: = 0, E: = E,(r, z), H,* = HZ = 0, H,* = H(r, z). Ans. piw aH E, = - ck2 ' Hint. Note that References Courant and Hilbert (CS), Frank and von Mises (F6), Garabedian (Gl), Grinberg (G5), Morse and Feshbach (M9), Petrovski (P2), Smirnov (S6, Vol. 11), Sornrnerfeld (S14), Tikhonov and Samarski (TI), Timoshenko (T2), Webster (W5). SOME SPECIAL METHODS FOR SOLVING HYPERBOLIC AND ELLIPTIC EQUATIONS This chapter deals with some special methods which, unlike those considered later, can only be used to solve problems pertaining to partial differential equations of a particular type, e.g., of the hyperbolic or elliptic type.' Among such methods, we mention Riemann's method for solving the Cauchy problem for hyperbolic equations, the Green's function method for solving boundary value problems involving elliptic equations, complex variable methods for solving the two-dimensional problems of potential theory and so on. There are a great many such special methods, which in some cases belong to the more difficult problems of the theory of partial differential equations. Thus it will be impossible to go into very much detail here. Instead we confine ourselves to a few simple problems illustrating the methods most frequently encountered in practice. I. Hyperbolic Equations It will be recalled that problems of mathematical physics involving the propagation of various kinds of waves (elastic, electromagnetic, etc.) in one, two or three dimensions lead to the consideration of partial differential equations of the hyperbolic type, subject to extra conditions. Depending on the character of these conditions, the problem is classified as a Cauchy problem or as a mixed problem. By the mixedproblem for an equation of the Concerning the classification of partial differential equations, see C5, GI, TI, etc. 19 22 SOME SPECIAL METHODS PROB. 32 Before the time t = xlv, the point x is at rest. During the interval i.e., as the source approaches, the point x undergoes harmonic oscillations of frequency For t > x/uo, i.e., as the source recedes, the frequency of the observed oscilla- tions is WV W2 = - < w 0 + 00 (the Doppler effect). 32. A semi-infinite rod, clamped at the end x = co and free from forces at the end x = 0, undergoes longitudinal oscillations. Investigate the nature of these oscillations, assuming that the initial conditions are of the form and that f(x) + 0 as x + co. Ans. 1 a[f(~t - XI +f(vt + XI], 0 < x < vt, U(X, t) = 2J[ f (x - vt) + f (x + vt)], vt < x < co. Hint. Make the even extension of the function f (x) to the negative x-axis, and use the solution of the Cauchy problem for an infinite string. 33. Find the distribution of voltage and current along an infinite trans- mission line with parameters L, C, R and G, given the initial conditions assuming that the parameters of the line are connected by the relation (a distortionless line). 1 [cp(x - vt) + cp(x + vt)] + - [$(x - vt) - $(x + vt)] 2Cu 6 1 0 = cut - [$(x - vt) + $(x + vt)] + - [cp(x - vt) - cp(x + vt)] , 2Lv where a = RIL, v = l/JE. I PROB. 35 SOME SPECIAL METHODS 23 Hint. Use Prob. 23 to write the differential equation for u, introduce a new unknown function w by writing u = e-~tw, and choose y such that the coefficient of awlat vanishes. 34. Show that the solution of the wave equation with radially symmetric initial conditions is given by the formula u(r, t) = (r - vt)cp(r - vt) + (r + vt)cp(r + vt) 1 '+Ot 2r + P$(P) dP9 where the values of the functions cp and $ for negative arguments are given by the relations cp( - r) = cp(r), $( - r) = $W. Hint. Transform the equation by setting ru = w, where w is a new unknown function. Then bear in mind that u remains bounded as r -t 0. 35. Study the oscillations occurring in a gas initially at rest when a local condensation so is formed inside a sphere of radius a contained in the gas. Am. The condensation of the gas at an arbitrary point r is where v is the velocity of wave propagation in the gas. Hint. By the condensation is meant the quantity P - Po SF- Po (i.e., the relative change in density of the oscillating gas), which satisfies the differential equation where v = Jcppo/c,p0, c, and c, are the specific heats at constant pressure and volume, and p, and po are the initial values of the pressure and density. This problem is a special case of the preceding problem, corresponding to initial conditions 36. In a gas initially at rest, a condensation s = so localized in the volume bounded by a surface a is created at the time t = 0. Show that the condensa- tion at the point M = (x, y, z) at an arbitrary time t is given by the expression where S,, is the sphere of radius vt with center at the point M, and a, is the part cut out of S,, by the surface a. Hint. Use the general solution of the homogeneous wave equation for arbitrary initial conditions (see G1, p. 197). 37. The solution of the Cauchy problem for the three-dimensional in- homogeneous wave equation 1 aZu Au - - - = -4np(x, y, z, t), v2 at2 with initial conditions is of the form where r. = J(x - <)2 + (y - q)' + (z PC)~, h-. ,-------- - region G,, bounded by this sphere cp,, and J,, are the average values of the functions cp and + over a sphere of radius vt with its center at the point M = 7 (x, y, z), and the integration is over the vf (x;Y)' (see K4, p. 101). Starting from this fact, E solve the corresponding problem for FIGURE 6 the two-dimensional inhornogeneous PROB. 39 equation with initial conditions Ans. S is the disk of radius vt with center at the point (x, y), and D is the right circular cone shown in Figure 6. 38. Show that one solution of the two-dimensional wave equation is the function u = Re f(8), where f is an arbitrary analytic function of the argument 8, related to the variables x, y, and t by the relation Comment. This class of solutions of the wave equation is widely used in diffraction theory and other applications (see K4, p. 114 and S6, Vol. 111, Pt. 2, p. 176). 39. Applying Riemann's method (see Ti, p. 116), solve the hyperbolic equation (where v and c are given constants), with arbitrary initial conditions PROB. 40 Ans. where 40. Find the distribution of current along an infinite transmission line with parameters L, C, R and G, assuming that at the time f = 0 the current vanishes while the voltage is nonzero and equal to a constant V on the section of the line 1x1 < a. Ans. where and Io(x) is the Bessel function of imaginary argument. Hint. Use the result of the preceding problem. 41. A semi-infinite rod of variable cross section S(x) = S(0)e-ax, where the end x = 0 is clamped, undergoes longitudinal oscillations with initial conditions Find the displacement of an arbitrary cross section of the rod at an arbitrary time t. Ans. eax/z~*v>l (+) (oe-a~~ -- dE if vt < x < m, 4 x-vt Rl where RlSz = J(vt)' - (x 7 0'. Hint. By introducing a new unknown function w = JS(x)u, reduce this problem to the integration of the equation in Prob. 39. 2. Elliptic Equations: The Green's Function Method A typical problem of the kind to be considered in this section is to find a solution of a partial differential equation of the elliptic type which is well- behaved in a given spatial region D and satisfies certain conditions on the boundary S of D. The simplest such problem is to find a function u which is harmonic in a region D with boundary S,3 and satisfies one of the following boundary conditions uls = f (PI, (24 where f(P) is a given function of a variable point P of S, ni is the interior normal to S at P, and h is a positive constant. The problem is called theJirst boundary value problem (of potential theory) or the Dirichlet problem if the boundary condition is of the form (2a), the second boundary value problem or the Neumann problem if the boundary condition is of the form (2b), and the third boundary value problem or the Robin problem if the boundary condition A function u is said to be harmonic in a (two or three-dimensional) region D if u and its first and second partial derivatives are continuous and satisfy Laplace's equation Au = 0 in D. If D is unbounded, certain extra requirements must be imposed on the behavior of u at infinity (see T1, p. 265). is of the form (2c). Similar problems can also be formulated for Helmholtz's equation Au + k2u = 0 and other equations of elliptic type encountered in mathematical physics. One of the special methods for solving boundary value problems of this kind is based on the use of the Green's function (see S6, Vol. IV and TI, Chap. 4). The key result of this theory is that the solution of the boundary value problem for Poisson's equation subject to any of the boundary conditions (2a)-(2c), can be written in quadra- ture~, once we know the Green's function. The Green's function does not depend on the form of the functions f(P) and F(M), and can be found by considering a special boundary value problem (see below). Thus, for example, the solution of the first boundary value problem for the equation (3) can be written in the form u(Mo) = f (P) - do + f (M)G d~, aG SD Ss aui where Ad is a variable point and Mo a fixed point of the region D (do is the element of surface area and d~ the element of volume). Here the Green's function G(M, Mo) is the function such that 1. G is harmonic in D except at the point Mo, near which G is of the form where the function u is regular (i.e., has no singularities) in D; 2. G satisfies the boundary condition It follows that u is harmonic and satisfies the boundary condition i.e., v is the solution of a special case of the Dirichlet problem. The same formula (4) gives the solution of the first boundary value prob- lem in two dimensions, if by the Green's function we now mean a function such that PROB. 43 SOME SPECIAL METHODS 29 1. G is harmonic in a planar region D except at the point M,, near which G is of the form 1 G(M, M,) = - In - + a regular function ; 2~ IMMOI 2. G satisfies the boundary condition GI, = 0 on the contour bounding D. Formulas of a similar kind can be found giving solutions of other bound- ary value problems, involving Laplace's equation, Poisson's equation, Helmholtz's equation, etc. Green's functions for regions of various shapes can be found by using the methods considered in Chaps. 4-7, and also by using certain special techniques, like the method of images and the method of inver~ion.~ The method of images allows us to construct the Green's function for a half-space and for a sphere (or, in two dimensions, for a half-plane and a circle) and for certain regions of a more complicated shape, e.g., the layer bounded by two parallel planes or the interior of an angle of xltz radians (n = 1, 2, . . . ). Starting from the Green's function for a region D and using the method of inversion, we can find the Green's function for the region D* obtained by inverting D in a sphere lying outside D. Thus, for example, we can find the Green's function for a sphere from a knowledge of the Green's function for a half-space, the Green's function for the region bounded by two intersecting spheres from the Green's function for the region bounded by two intersecting planes, and so on. We now give some problems illustrating these methods of constructing Green's functions, and also a few problems of a more theoretical nature. 42. Construct the Green's function for the two-dimensional Dirichlet problem in the case where the region D is the first quadrant x > 0, y > 0. Ans. where M = (x, y, z), Mo = (x,, yo, zd, M, = (-x,, yo, zo), M3 = (-x,, -yo, zo) and M, = (xo, -yo, zo). Hint. Use the method of images. 43. Using the method of images, construct the Green's function for the Dirichlet problem in the case where the region D is the part of space lying between two parallel planes z = f 112. "See TI, and in particular G5, which contains a number of interesting applications of the method of inversion to problems of electrostatics. PROB. 44 Ans. where 44. Use the method of inversion to deduce the Green's function for the Dirichlet problem in the case where D is a sphere of radius a with its center at the origin 0, assuming that the expression for the Green's function of a half- space is known. Ans. where MI is the image of the point Mo in the sphere. 45. Find the Green's function for a hemisphere of radius a. where M, is the image of Mo in the corresponding full sphere, M, is the image of Mo in the diametral plane of the hemisphere, and M, is the image of M, in the full sphere. 46. The Green's function G = G(M, Mo) for the Neumann problem6 au Au = -F(M), - 1 = f(P) ani s is defined by the conditions 1. G is harmonic in D except at the fixed point Mo, near which G is of the form G = + a regular function; 4x IMMOI Here M is a point of the three-dimensional region D, P is a point of the surface S bounding D, and the functions f and Fsatisfy the condition lsfdo =IDFd~ for the solvability of the Neumann problem. If the Green's function is known, the solution is given by the formula u(M,) = - f (P)G do + is SOME SPECIAL METHODS 3 1 where So is the area of the surface S; Verify that in the special case where D is a sphere of radius a with center at the origin 0, the Green's function is where M, is the image of the point M, in the sphere, and Q is the foot of the perpendicular dropped from the point M onto the line OM,. 47. A conductor bounded by a closed surface S and held at a given potential V is introduced into an arbitrary external field with potential uo. Suppose we know the charge density p(P, Mo) at the point P of the surface S in the case where the surface is grounded and the external field is due to a unit charge at an arbitrary point M, outside the conductor. Show that the potential distribution in the general case is given by the formula represents the solution of the boundary value problem hw = 0 outside S, Wl,=f(P), WI, =O in terms of the Green's function. Apply this formula to the function w = u - uo, bearing in mind the electrostatic interpretation of the Green's function. 48. Find the Green's function for the two-dimensional Dirichlet problem, assuming that we know the function < = <(z) mapping a given region in the z-plane conformally onto the upper half of the <-plane (Im < > 0). Use the result to construct the Green's function for the half-strip x 0, 0 g y g n. PROB. 49 Ans. where [ and [, are the points of the half-plane corresponding to M = (x, y) and Mo = (x,, yo). In the case of the half-strip, 1 In [COS~ (X + xO) - cos (Y + yO)I[co~h (X - XJ - cos (Y - yo)] --- -. 47t [cosh (x + x,) - cos (y - yo)][cosh (x - x,) - cos (y + yo)] ' Hint. The conformal mapping of the half-strip onto the half-plane is accomplished by the function 1: = cosh z. 49. The boundary value problem (A is the two-dimensional Laplacian, and M is a point of a planar region D bounded by a contour S) is encountered in the theory of bending of thin elastic plates. The solution of this problem can be written in the form (dr is an element of area), where the Green's function G = G(M, M,) is defined by the conditions 1. G is the solution of the biharmonic equation A2u = 0 which is regular (i.e., free of singularities) in D, except at the fixed point M,, near which G is of the form 1 G = - IMM01' In IMM,I + a regular function; 87t Verify that in the special case where D is a disk of radius n with its center at the origin 0, the Green's function is given by where M, is the image of the point M, in the circle bounding D. SOME SPECIAL METHODS 33 3. Elliptic Equations: The Method of Conformal Mapping In mathematical physics one often encounters the problem of finding a function which is harmonic in a two-dimensional region D and satisfies the boundary condition 4s =f (53) where f is a given function and n is the normal to the contour S bounding D. For example, such problems arise in studying electrostatics, magnetostatics, heat conduction, flow of ideal fluids, filtration phenomena, and so on. An effective method of solving problen~s of this kind is to construct a function of a complex variable c = F(z) such that F(z) is analytic in D and maps D conformally onto a region D* (with boundary S*) of a special form for which the solution of the given problem is either known or can be found more simply than for the original region D. Here it is assumed that Ff(z) is non- zero in the region D, a condition which guarantees that the mapping is one-to-one. In asserting that this method leads to a solution of the boundary value problem, we rely on the fact that the Laplacian and the boundary conditions (5a) and (5b) preserve their form6 under the transformation from the variables x and y to the new variables [ and q defined by the relation The method of conformal mapping can also be used to deal with more complicated boundary value problems, e.g., problems where the value of the unknown function u is specified on parts of the contour S while the value of aupn is specified on the rest of S, problems of jet flow of an ideal flow where the form of S is not known in advance but is determined in the course of solving the problem, and so on. In many cases, the construction of the function < = F(z) mapping the region D onto the region D* can be accomplished by consecutive application of several mappings which involve elementary functions. Of particular importance in applied work is the case where D is a polygon and D* is the upper half-plane. Then the function effecting the mapping can be found by using the familiar Schwarz-Christoffel transformation (see Wl). The use of conformal mapping to solve problems of mathematical physics, involving In the case of the boundary condition (5b), the value taken by the normal derivative on the contour S* is 1 34 SOME SPECIAL METHODS PROB. 50 the biharmonic equation as well as Laplace's equation, is amply discussed in books on complex variable theory and in special monographs (see B3, F10, M10, S7, etc.). Hence we confine ourselves here to a few typical problems which illustrate the technique of the method, assuming that the reader is already familiar with the elementary theory of conformal mapping. In most of the problems, the required conformal mapping can be found by using the Schwarz-Christoffel transformation. Problems 51, 57 and 59 require knowledge of the properties of elliptic integrals and Jacobian elliptic functions. In connection with Probs. 50-54, the following remarks will be found helpful: If cp is the potential of a stationary plane-parallel flow of an ideal fluid, described by the velocity field v = -grad cp, then by the complex potential w = w(z) is meant a function of the complex variable z = x + iy whose real part equals cp. In other words, w = cp + i$, where $ is related to cp by the Cauchy-Riemann equations The lines of flow or streamlines are described by the family of curves $ = const, and hence the function x is called the stream function. The amount of fluid Q flowing per unit time between two streamlines + = +, and $ = +, (in a slab of unit thickness parallel to the xy-plane) is given by The components of the velocity vector v = v, + io, are related to the deriva- tive of the complex potential by the formula The complex potential is a valuable tool for studying plane-parallel flows.' 50. An ideal fluid, whose velocity at infinity equals v, = v,, v, = 0, flows past an obstacle in the shape of an elliptical cylinder Use the method of conformal mapping to find the complex potential of the flow. Similarly, in the theory of stationary heat flow and in electrostatics, one can introduce coniplex potentials, with the role of v and cp being played by qlk (the ratio of the heat flow density to the thermal conductivity) and the temperature Tin the first case, and by the electric vector E and the electrostatic potential cp in the second case. PROB. 52 SOME SPECIAL METHODS 35 Ans. The relation between the complex potential w and the variable z is given in parametric form by the equations where t belongs to the region It1 > a, 0 < arg t < n. Hint. First make a conformal mapping of the part of the region occupied by the flow and lying above the axis of symmetry onto the half-plane with a semi-circular cut of radius a, and then map this region conformally onto the upper half of the <-plane in such a way that the semi-circular arc of radius a goes into the interval (-a, a) of the real axis. 51. Solve the preceding problem for the case where the obstacle is a cylinder -a < x < a, -b < y < b of rectangular cross section. Ans. The complex potential has the parametric representation where and the modulus of the elliptic integrals is determined from the condition Hint. Use the Schwarz-Christoffel transformation to map the region occupied by the flow and lying above the axis of symmetry y = 0 onto the half-plane in such a way that the vertices fa, *a & ib go into the points &Ilk, 51. "52. Study the two-dimensional motion of an ideal fluid in the channel of variable cross section shown in Figure 7, assuming that at infinity the direction of the flow coincides with - 20 - x the x-axis and has the values (av, = bu,). FIGURE 7 36 SOME SPECIAL METHODS PROB. 53 Find the distribution of velocity along the axis of symmetry of the Ans. The velocity distribution in parametric form is given by the equations Hint. Using the Schwarz-Christoffel transformation, map the domain FIGURE 8 Ans. . s ABCDE onto the upper half-plane of the complex variable T: = t + iq, requir- ing the points B, C and D to go into the points T:= -1, <=--A and C=0, where h is a number between 0 and 1 which subsequent calculations show to be equal to the ratio b2/a2. 53. Solve the preceding problem for the case where the channel has the form shown in Figure 8, assun~ing that vale+-m = 11,. Hint. Transform the region bounded by the wall of the channel and the axis of symmetry of the flow into the upper half-plane of the complex vari- able T:, making the vertices B and C go into the points - 1 and 0. 4 ,\ i j/ -.. b-204 / "54. Investigate the jet flow of a 2 * X liquid through a slot of width 20 in a plane wall (see Figure 9), assuming that the amount of fluid flowing through the slot per unit time (in a slab of unit thickness parallel to the xy-plane) equals Q. Find the form of 1 the jet. ,2*, Ill 11i Ans. In parametric form, the equa- '9 tion of the curve bounding the jet is FIGURE 9 In Probs. 52-54, where the flow is symmetric with respect to an axis, it is convenient to assume that + = 0 along this axis. The value of + along any other streamline can be found by using the formula Q = I+, - +,I. SOME SPECIAL METHODS 37 given by x Ji + (~12) Y - = - - - a 1 + (~12) ' a 1 + (42) 2 1-41-1 (0 < t g 1). The width of the jet at a great distance from the slot is Hint. Use Kirkhhoff's method (see K2, p. 332 ff.). 55. A pipe of radius a lies below the ground at depth h (see Figure 10). Find the stationary temperature distribution in the region surrounding the pipe, assuming that the temperature of the earth's surface is zero, while the temperature of the pipe is To. Ans. T:n n where c = Jh2 - a2. Hint. Use a fractional linear transformation I FIGURE 10 to map the given region into a circular ring. 56. Find the stationary temperature distribution in a wall of thickness a near the corner of a building (see Figure I I), assuming that the temperature of the inside surface of the wall is To, while the temperature of the outside surface is' zero. Ans. - ~[~in~arctant a x2 1-t + 1-ki, where I X and In and arc tan denote the branches which FIGURE 11 go to zero as t + 0. 38 SOME SPECIAL METHODS PROB. 57 Hint. Use the Schwarz-Christoffel transformation to map the figure ABCDA onto the upper half-plane of the complex variable <, making the points B, C and D go into the points -1, 0, A, where A is to be determined (a calculation shows that A = 1). 57. Solve the problem of the stationary temperature distribution in a homogeneous slab --a < x < co, -b < y < b of thickness 2b, inside which there is another thin slab of thickness 2a (a < b) sharing the same midplane and held at temperature To. It is assumed that the temperature of the outside surfaces of the slab equals zero. Calculate the flow of heat Q given off by the source per unit time. Ans. where the relation between the complex variables < and z is given by the equation 1 xz sn 7; = - tanh - , k 2b and the modulus of the elliptic function is xa k = tanh - . 2b Moreover, where x is the thermalconductivity of the slab, while K and K'are the complete elliptic integrals with moduli k and k' = J1 - k2. Hint. using the transformation -----X where k has the value indicated above, map the strip - co < x < co, 0 < y < b onto the upper half-plane of the variable t. Then use the Schwarz-Christoffel transformation to transform this strip into a rectangle with vertices at the points &K, iiK' in the <-plane. i I 58. A wire with charge q per unit length is located near the rectangular edge of a grounded conductor FIGURE 12 (see Figure 12). Find the distribution of the electric PROB. 59 SOME SPECIAL METHODS 39 field in the symmetry plane of the region between the wire and the conductor. Ans. In parametric form, the field is given by the formulas where q, is the value of the parameter q corresponding toy = h. Hint. Map the part of the z-plane lying outside the conductor onto the upper half-plane of the variable < = + iq, making the corners go into the points < = f I. *59. On the axis of a box -a G x < a, 0 < y < b of rectangular cross section with grounded walls, there is a thin wire with charge q per unit length. Write an expression for an appropriate complex potential, and calculate the distribution of charge density on the walls of the box. Ans. where sn z is a Jacobian elliptic function with modulus k. The modulus k is determined from the equation b K' --- - a K' where K = K(k) is the complete elliptic integral of the first kind and K' = K(JI - k2). The distribution of charge density on the wall -a < x < a, y = 0 is given by the formulas where cn z and dn z are Jacobian elliptic functions. Hint. Use the Schwarz-Christoffel transformation to map the interior of the rectangle onto the upper half-plane, making the vertices fa, *a + ib of the rectangle go into the points 4 1, f llk. During the calculations, bear in mind that iK' i sn=- 2 Jk' 60. Find the electrostatic field on the axis of an electronic lens made two pairs of plates at potentials + V and - V, separated by a space (see Figure 13). 't Ans. - JZ E&=O Eo 1 - A2E2 ' - 1 + E A2F E-- - - 2b 2b 2 1 1-A2 (-1 < F < I), where and A is a number between 0 and 1 determined from the equation Hint. Map the upper half-plane of the variable z = x + iy cut along the line segments (--a + ib, -a + ib) and (a + ib, m + ib) onto the upper half-plane of the variable c, in such a way that the corners go into the points 5 1, f l/A. Then transform the half-plane onto the half-strip 61. Find the field on the axis of the electronic lens shown in Figure 14. Ans. (-1 < E < I), where A is determined from the equation and we introduce the abbreviations a-b V, - V, 1 - (l/h2) - =Y, Eo=--- a + b a+b 1+yh Hint. Map the domain ABCDEA onto the upper half-plane of the variable c = + iq, making the points B, C, D and Ego into the points --A, -1, 1 and p. After determining the function z = z(<), carry out the transformation c = sin t. 62. Find the magnetic field in the midplane of the magnet whose poles have the rectangular shape shown in Figure 15, assuming that the magnet is made of iron with infinite magnetic permeability (p = co). Ans. xx 1 1 1-t H,lr-o= ,, - =-+ -In-- Ha 2h t 2 l+t X I (0 < t ,< I), where H, is the homogeneous field in the mid- plane of the magnet at a great distance from the edge. FIGURE 15 Hint. Map the region ABCD onto the upper half-plane, making the points B and C go into the points -1 and 0. 63. The region x 2 0, y < 0 is filled with iron of magnetic permeability p = co. Find the magnetic field due to a linear current source J passing through the point (-a, 0). Ans. The components of the field are determined by the relation where c is the velocity of light. Hint. Bear in mind that near the current source, the complex potential of the magnetic field has a logarithmic singularity: 2iJ w = - In (z - zo) $ a regular function. C References Betz (B2), Courant and Hilbert (CS), Kupradze (KS), Morse and Feshbach (M9), Petrovski (P2), Smythe (S7), Sneddon (S1 l), Sternberg and Smith (S16), Tikhonov and Samarski (Tl), Walker (Wl). STEADY-STATE HARMONIC OSCILLATIONS A solution of a partial differential equation is said to be a steady-state harmonic oscillation if its time dependence is described by the factor eiUt, where w is the frequency.' Problems involving steady-state harmonic oscillations are among the simplest and the most important encountered in mathematical physics. Because of the particularly simple form of the time dependence, we can eliminate the variable t from the original equation, thereby reducing the problem to the determination of complex amplitudes depending only on the spatial coordinates. In the special case where the solution depends only on a single spatial coordinate, the equation for the complex amplitude reduces to an ordinary differential equation. This category, to which most of the problems in the present chapter belong, is of considerable interest because of its numerous applications to concrete problems of mechanics, electromagnetic theory, etc. Moreover, such prob- lems are very important from a methodological standpoint, since they serve as the best introduction to the technique of particular solutions to be con- sidered in Chapter 4. Thus, for example, the problem of determining natural frequencies anticipates the problem of determining eigenvalues, and the problem of forced oscillations gives insight into ways of overcoming difficulties associated with the application of the Fourier method to inhomogeneous equations. In using complex quantities in intermediate steps of our calculations, we rely on the fact that the equations of mathematical physics (at least, those considered here) are linear. Thus, to obtain the final answer, we need only take the real or imaginary part of some expression (depending on the conditions of the particular problem). 42 p~oe. 66 STEADY-STATE HARMONIC OSCILLATIONS 43 This chapter contains three sections. The first is devoted to problems on the determination of natural frequencies of vibrating systems (strings, rods, membranes and plates), while the second deals with forced oscillations of such system^.^ The third section is concerned with problems on steady-state electromagnetic oscillations in transmission lines and cavity resonators, certain related problems on the propagation of electromagnetic waves in waveguides of given cross section, etc. I. Elastic Bodies: Free Oscillations 64. Find the natural frequencies for longitudinal oscillations of a canti- lever beam of length I. where v = Jz, E is Young's modulus, and p is the density. 65. Solve the preceding problem, assuming that the free end of the beam is loaded by a mass Mo. Ans. V w,=-y,, n=l,2 ,..., 1 where the y, are consecutive positive roots of the equation M y tan y .= - Mo (i.e., 0 < y, < . . . < y, < . . .), and M is the mass of the beam. 66. Determine the natural frequencies for torsional oscillations of a rod of length I, one end of which is clamped, while the other end is attached to a disk whose moment of inertia with respect to the axis of rotation is Jo. Ans. v W =-y 1 "' n=l,2, ..., where v = JS, G is the shear modulus, p is the density, the y, are con- secutive positive roots of the equation J ytany =-, Jo and J is the moment of inertia of the rod. The forced oscillations studied in this chapter will always have the same frequency as the perturbing force itself. 44 STEADY-STATE HARMONIC OSCILLATIONS PROB. 67 67. Find the natural frequencies for transverse oscillations of a beam of length I with simply supported ends. Ans. where n2 = JEJI~S, E is Young's modulus, J is the moment of inertia and S the area of a cross section, and p is the density. 68. Find the natural frequencies for transverse oscillations of a beam of length I with clamped ends. Ans. where the constant a is the same as in the preceding problem and the y, are consecutive positivt roots of the equation cosh y cos y = 1. *69. Solve the preceding problem, assuming that one end of the beam (of mass M) is clamped, while the other is loaded by a mass Ma. Using the method of successive approximations, calculate the values of the first three frequencies, given that Ans. where the y,, are consecutive positive roots of the equation M 1 + cosh y cos y = 2 y(sin y cosh y - cos y sinh y). M 70. Find the natural frequencies for radial oscillations of a circular membrane of radius a. Am. where o = JK, T is the tension per unit length of the boundary, p is the surface density, and the y, are consecutive positive roots of the equation Jo(y) = 0 involving the Bessel function of order zero. 71. Find the natural frequencies for oscillations of a rectangular mern- brane with sides a and 6. pROB. 74 STEADY-STATE HARMONIC OSCILLATIONS 45 Ans. where v is the same as in the preceding problem. 72. Find the natural frequencies for transverse radial oscillations of a circular plate of radius a whose edge is clamped. Calculate the first three roots of the transcendental equation determining the frequencies. Ans. where b" JJDlpk, D is the flexural rigidity, p the density and h the thickness of the plate, and the. y,, are consecutive positive roots of the equation (the notation is the same as in the theory of cylinder functions). Numerical calculations show that the first three roots are y, = 3.20, y, = 6.30, y3 = 9.44. 73. Find the maximum wavelength Amax of a nonplanar sound wave3 which can propagate inside a hollow cylinder tube with perfectly reflecting walls, whose cross section is a) a rectangle with sides a and b; b) a circle of radius a. Ans. where y, = 3.832 is the smallest positive root of the equation J,(y) = 0 (for waves which are symmetric with respect to the diarnetral plane). 74. Find the natural frequencies for acoustic oscillations in an enclosure shaped like a rectangular parallelepiped with sides a, b and c. Ans. where u is the velocity of wave propagation (m, n, p cannot all vanish simulta- neously). section. 46 STEADY-STATE HARMONIC OSCILLATIONS PROB. 75 75. Find the natural frequencies of an acoustic re~onator,~ where the oscillations have axial symmetry and the resonator is a) a sphere of radius a; b) a circular cylinder of radius a and height I. Ans. where the y,,, are consecutive positive roots of the equation 2ymJL+ ~m (~m) = Jm+ (yrn), Jm+ %(x) is the Bessel function of order m + 4, and u is the velocity of wave propagation ; b) m=0,1,2 ,..., where the y, are consecutive positive roots of the equation J,(y) = 0. 2. Elastic Bodies: Forced Oscillations 76. A string of length I with ends fastened at the points x = 0 and x = I undergoes oscillations under the action of a concentrated force A sin (at + cp) applied at some point x = c of the string. Find the form of the forced oscillations. Ans. ox . o(1- c) x i sin - s~n --- , O<x<c, Av sin (at + cp) v v U(X, t) = - oT sin (ollu) oc o(l - x) sin - sin --- , C<X<~, u v where u = t/~/p, T is the tension and p the linear density of the string. 77. Solve the preceding problem for the case where the external force is uniformly distributed over the whole length of the string. Ans. ox . w(1- x) sin - sln --- 2qu2 2v 2 v u(x, t) = - sin (a1 + cp), W~T o 1 COS - 2 v where q is the amplitude of the load per unit length of the string. An acoustic resonator is a device used to amplify acoustic oscillations, and consists of an enclosure whose walls reflect sound. PROB 81 STEADY-STATE HARMONIC OSCILLATIONS 47 78. Find the forced longitudinal oscillations of a rod of length I, if the end x 0 is clamped while the end x = 1 is acted upon by a force A sin (ot + cp). Ans. ox A v sin - D U(X, t) = - - sin (at + cp), ESo wl COS - where v = Jz, E is Young's modulus, p is the density and S the cross- sectional area of the rod. 79. Find the forced oscillations of a beam simply supported at the ends x = 0 and x = I, under the action of a uniformly distributed pulsating load q sin at. Ans. a JO 1 2 cos - - - 1 sin of, I where a2 = JEJI~S, E is Young's modulus, J is the moment of inertia, p the density and S the cross-sectional area of the beam. 80. Solve the preceding problem under the assumption that the oscillations are due to a concentrated force A sin of applied to the point x = c. Ans. ~a~ sin ot U(X, t) = JZl ZEJ~J; sin @ sinh - a a JWl . JO(Z - c) JWx JOl JG(1- c) JG ( sinh sin - - sin - sinh sinh - , a a a a a O . (1 - x JOc A \ sinh m JOi JO(1- X) JWC sin - - sin - sinh sinh - , a 2 a a a 81. A cantilever is clamped at one end x = 0 and loaded at the other end x = I by a force A sin of. Find the resulting forced oscillations. Ans. With the notation of the preceding problem, Aa3 sin wt 2~~wJw JZX) ( J;l &A) [(cash @ - cos - sinh - + sin - a a I Jwl Jwl 1 + cos - cosh - Jwx (sinh T - sin @) (cosh - dm1 + cos - - a a a ,- ,- ,lwl ,lwl 1 + cos - cosh - 82. Find the forced oscillations of a circular membrane of radius a due to a pulsating load q sin (of + cp) uniformly distributed over the mem- brane. Ans. u(r, t) = - - 1 - A sin (at + cp), '[ Jo(oa/v) Pa where v = JTI~, T is the tension per unit length of the contour, p is the surface density of the membrane, and Jo(x) is the Bessel function of order zero. *83. Solve the preceding problem, assuming that the load is uniformly distributed over a disk of radius b < a. Ans. nbqv u(r, t) = - - sin (at + cp) 2wT 2u JOwr'v [Y (7) J () - J (7) Y (I, o < r < b where J,(x) and Y,(x) are cylinder functions. 84. Study the forced oscillations of a circular plate of radius a with a clamped edge under the action of a uniformly distributed pulsating load q sin (at + cp). PROB. 86 STEADY-STATE HARMONIC OSCILLATIONS 49 Ans. With the usual notation from the theory of cylinder functions, where b2 = JDlph, D is the flexural rigidity, 11 the thickness and p the density of the plate. 85. Find the steady-state harmonic oscillations of frequency w inside a spherical resonator due to a point source of sound located at the center of the sphere, bearing in mind that the potential of a point source of frequency w in free space is given by sin (wt - kR) u, = A R 9 where k = w/u is the wave number and R the distance from the source. Ans. The velocity potential is sin (wt - kr) ka cos (wt - ka) + sin (wt - ku) sin Itr u(r, t) = A +A , r ka cos ka - sin ka r where a is the radius of the sphere. 3. Electromagnetic Oscillations 86. Find the steady-state harmonic oscillations of voltage in a long transmission line with parameters L, C and R, if the end x = 0 is attached to a source of variable voltage E sin (wt + cp), while the end x = I is terminated by a resistance R,. Ans. w(1 - X) R w(l - x) sin --- + P cos --- U* rZ* U* rr(x, t) = Im wl R sin - + 2 cos - u* iZ* u* where 1 u* = - 1 JE Jz' wL are the complex propagation velocity and wave resistance of the line. 87. Solve the preceding problem, assuming that the load terminating the line is a concentrated inductance Lo, instead of a resistance R,. Am. w(l - x) oL w(l - x) sin ---- -f- A cos --- v* z* wl wL wl sin - + cos - v* z* v* 88. Find the components of the electromagnetic field in a transverse magnetic wave propagating in a waveguide whose cross section is a rectangle with sides a and b.5 Calculate the corresponding cutoff wavelength Amax (i.e., the maximum wavelength passed by the waveguide). Ans. nx mxx nny -i(vz-ol) H, = ik - sin - cos - e b b 2 a where k = wlc is the wave number. An arbitrary constant factor has been omitted in all the expressions for the components of the electromagnetic field. 89. Solve the preceding problem for a transverse electric wave. By a transverse magnetic wave (TM-wave) is meant a wave in which the magnetic field vector H is perpendicular to the direction of wave propagation. Similarly, a transverse electric wave (TE-wave) is a wave in which the electric field vector E is perpendicular to the direction of propagation, and a transverse electrortragnetic wave (TEM-wave) is a wave in which both vectors E and H are perpendicular to the direction of propagation (see S3, p. 154). PROB 90 STEADY-STATE HARMONIC OSCILLATIONS 5 1 Ans. Amax = 2a if a > b (m and n cannot vanish simultaneously). 90. Find the components of the electromagnetic field in a transverse magnetic wave propagating in a waveguide whose cross section is a circle of radius a, and determine the corresponding cut-off frequency Amax. Y mn ( f) -i(vz-wt), H, = -ik - cos mcp Jk y,,,, e a H, = 0, Ymn E, = - iv - cos mcp Jk a m E, = iv - sin m cp J, r where k = w/c is the wave number, and they,,,, are consecutive positive roots of the equation Jm(y) = 0 (m = 0, 1, 2, . . .) involving the Bessel function of order m. "91. Calculate the cutoff wavelength Amax for a TM-wave propagating in a waveguide whose cross section is a circular sector of radius a and central angle a. Ans. where yo is the smallest positive root of the equations involving the Bessel function J,(x). 92. Describe the free harmonic oscillations in an electromagnetic resonator in the form of a rectangular parallelepiped with sides a, b, c and perfectly conducting walls. Ans. mxx nxy . pxz mxx nxy pxz Ex = A cos - sin - sln - , H, = M sin - cos - cos - , a b c a b c mrrx nxy pnz mnx . nny pnz Ey = B sin - cos - sin - , H, = N cos - sln - cos - , a b c a b c mxx nxy pnz mnx nxy . pxz E, = C sin - sin - cos - , Hz = P cos - cos - sln - , a b c a b c where m, n and p are integers, and the constants A, B, C, M, N and P are connected by the relations (k is the wave number). 93. Solve the preceding problem for a resonator in the form of a circular cylinder of radius a and length I. pROB. 94 Ans. STEADY-STATE HARMONIC OSCILLATIONS 53 r cos rncp nxz 4 = AJL (Ymn Jsin sin T 9 1 r sin mcp nxz E = B - ( -) sin - , r acosmcp 1 r cos rncp nxz Ez = CJ, (y,,,, -) cos - , a sin mcp 1 1 r sin mcp nnz H, = M - J, (y,. -) cos - , r acosmcp 1 r cos my nxz H, = NJ; jymn -) cos - , asinmcp 1 nxz H,= P-JA r ( y,,- Jr:szsin-, 1 where m and n are integers, the constants A, B, C, M, Nand Pare connected by the relations nx Akn+C-=0, a 1 and the y,, are consecutive roots of the equation J,(y) = 0. 94. A high-frequency current I sin wt flows along a cylindrical conductor of radius a, made of material of conductivity o and magnetic permeability p. Find the distribution of current density along the cross section of the wire, and calculate the active resistance of the conductor at the frequency w (the skin efect problem). Ans. The complex amplitude of the current density is given by the formula JdW j(a> - Jdka) ' 54 STEADY-STATE HARMONLC OSCILLATIONS PROB. 95 where The resistance per unit length of the conductor is (the overbar denotes the complex conjugate), where Taking account of the asymptotic behavior of the Bessel functions for large values of the argument, we find that R, a R3 - ' R, 2s' where C 6 = - J27rawt* and 1 R, = --, rca a is the d-c resistance. 95. Solve the skin effect problem for a conductor whose cross section is a strip of width 2a. Find the corresponding current distribution and resistance. Ans. I k J~, j(a)--~~tka, j(a) cos ka 2 R, = Ik2 (k cot La - k cot ka), 20(k2 - k2) where I is the amplitude of the total current. For high frequencies, where 6 is the same as in the preceding problem. References Marcuvitz (M4), Morse (M8), Rayleigh (RI), Schelkunoff (S3), Stratton (S17), Timoshenko (T2). THE FOURIER METHOD The Fourier method is one of the most general techniques of mathematical physics, and is effective in solving a very wide class of problems. Its use, which is not restricted to equations of any particular type (e.g., hyperbolic or elliptic), relies on the fact that linear problems obey the superposition principle, i.e., any linear combination of solutions of a homogeneous linear partial differential equation is itself a solution of the equation. Thus, if a linear equation Lu = 0 has a certain set of particular solutions u=u,, n=1,2 ,..., the sum of the series is also a solution, provided the convergence of the series permits interchanging the operations L and C. Similarly, if Lu = 0 has a set of particular solutions which depend continuously on the parameter h in the interval (p, v), then the integral is also a solution, provided the operations L and S can be interchanged. Given a problem of mathematical physics involving the integration of a differential equation Lu = 0 subject to certain initial and boundary con- ditions, the basic idea of the Fourier method is to construct a solution by 56 THE FOURIER METHOD superposition of particular solutions. If the operator Lu = 0 has an appro- priate structure, we can "separate variables," i.e., the particular solutions can be written as products of factors, each involving only one independent variable and satisfying an ordinary differential equation. By suitably choosing some of the parameters figuring in this relatively simple problem. it is usually possible to satisfy all the homogeneous boundary conditions, thereby singling out a countable or uncountable set of particular solutions of the required type. Then, after making a superposition of these solutions, we choose the remaining parameters in such a way as to satisfy the inhomo- geneous boundary conditions. Having made these general remarks, we now confine ourselves in this chapter to problems of mathematical physics which lead to integration of the differential equation + M,u = 0 (a < x < b, c < y < d), (1) where M, is a differential operator of the form A, B and C are given constants, and p(x), q(x) and r(x) are given continuous functions such that p(x) and q(x) are positive, and p(x) is continuously differentiab1e.l [In the next chapter, we shall consider the inhomogeneous case, where the right-hand side of (1) is a given function F(x, y).] For the time being, we assume that the interval (a, b) is finite and that the behavior of the functions p, q and r at the end points a and b is such that the ratios all approach finite limits as x + a and x + b. Moreover, we require the solution of (1) to satisfy homogeneous boundary conditions at the end points of (a, b), of the form ' Equation (1) is not the most general second-order equation with two independent variables which permits separation of variables, but it includes as special cases most of the commonly encountered equations of mathematical physics. THE FOURIER METHOD 57 where a,, ab, Pa and Pb are given constants, some of which may equal zero,2 and inhomogeneous boundary conditions at the end points of (c, d), whose form depends on whether the differential equation is of hyperbolic, parabolic or elliptic type (cf. footnote I, p. 20). If the equation is of elliptic type, it will be sufficiently general for our purposes to assume that these conditions are of the form where y,, y,, 8, and 8, are given constants, while g,(x) and g,(x) are given functions defined in the interval (c, d). On the other hand, if the equation is of the hyperbolic or parabolic type, which corresponds to problems of mathematical physics where the variable y plays the role of a time varying over an injinite interval (c, a), then the inhomogeneous boundary conditions take the form of initial conditions, i.e., in the hyperbolic case, and uly=c =f(4 in the parabolic case. We now look for a function u = u(x, y) satisfying the differential equation (1) and the boundary conditions (3) and (4) [or (4'), (4")l. Following the basic procedure already mentioned, we consider particular solutions of equation (1) of the form u = X(x) Y(y). (5) After substituting (5) into (I), the variables separate, and the result is a pair of ordinary differential equations a In particular, we obtain boundary conditions of the first kind uIZ=, = u],,~ = 0 if a, = a, =. 0, p, = pb = 1, boundary conditions of the second kind if a, = a, = 1, pa = P, = 0, and so on. In the applications, one also encounters boundary conditions of the form which are not comprised in the formulas (3). determining the factors X and Y, where A is an arbitrary parameter. The requirement that the particular solutions (5) satisfy the homogeneous boundary conditions (3) leads to corresponding homogeneous boundary conditions for the function X: The problem of solving equation (6) subject to the boundary conditions (8) is called the Sturm-Liouuille problem. For arbitrary A, this problem will in general have no solution other than the trivial solution X -- 0. However, for certain values of A, called eigenualues, there are nontrivial solutions, called eigenfunctions. In the theory of the Sturm-Liouville problem, it is shown that wiih our assumptions concerning the interval (a, b) and the functions p, q and r, the spectrum (i.e., the set of all eigenvalues) is discrete, consisting of countably many real eigenvalues A = h, (n = 1, 2, . . .), each associated with a single eigenfunction X = Xn(x) which is uniquely defined (except for a constant factor). The eigenfunctions Xn(x) are found to be orthogonal on the interval (a, 6) with weight r(x), i.e., Moreover, under certain condition^,^ a function f (x) defined in (a, b) can be expanded as a series of the form n=l with coefficients The calculation of the eigenvalues and the corresponding eigenfunctions is easily carried out in the case where the linearly independent solutions and hence the general solution of (6) are known for arbitary h. In fact, substitution of the general solution into the boundary conditions (8) then gives a homo- geneous linear system for the arbitrary constants, and the condition that the determinant of this system vanish leads at once to a transcendental equation for the permissible values of A. After the eigenvalues and eigenfunctions have been determined, we find the second factor Y(y) in (5) by solving (7), with h = A,. If the original equation is of hyperbolic or elliptic type, the general solution of (7) can be written in the form For example, if f(x) is piecewise smooth in (a, b). THE FOURIER METHOD 59 where Yt' and YF) are linearly independent solutions of (7) and c:), c:) are arbitrary constank4 In this way, we arrive at a set of particular solutions and the solution of the problem is then constructed in the form of a series where the coefficients cf' are found by substituting this series into the boundary conditions (4).5 If the function u satisfies boundary conditions of the type (3') instead of (3, the above method carries over virtually without change, except that now two linearly independent eigenfunctions may correspond to the same eigen- value. Things become more complicated if the interval (a, b) is infinite, or if one (or both) of the end points of (a, b) is singular, i.e., if one of the ratios (2) becomes infinite as we approach the given end point. In such cases, which are among the most interesting encountered in practice, the boundary condition involving the singular end point or the point x = b = cc cannot be prescribed arbitrarily, but rather is replaced by a condition whose formulation in concrete situations usually presents no special difficulties (most often, the condition consists in the requirement that the solution remain bounded as the singular point is approached). In the case where the interval (a, b) is finite and only one end point is singular, the eigenfunctions are found as the nontrivial solutions of equation (6) satisfying some condition of the type just mentioned at the singular point and a condition like (3) at the other end point. The same approach can be used to find the eigenfunctions for a finite interval with two singular end points, for an infinite interval, and so on. The essential difference between these cases and the case analyzed above is that the spectrum may now be either discrete or not, depending on the structure of the differential equation and the nature of the boundary con- ditions. If the spectrum is still discrete, despite the presence of an infinite interval or of a singular end point, the Fourier method can be applied with no essential changes. On the other hand, if the spectrum is no longer discrete, the character of the solution changes. In the case of a continuous spectrum," * If the equation is of parabolic type, the general solution is of the form For rigorous justification of the application of the Fourier method to problems of mathematical physics of this or more complicated types, see L1, TI, T7, etc. In many cases, however, it is an easy matter to verify directly the validity of the solution found fornlally by the procedure just described. Chapters 4-5 are devoted exclusively to problems with discrete spectra. Problems with continuous spectra will be considered in Chapter 6. 60 THE FOURIER METHOD PROB. 96 the solution is constructed from particular solutions by integrating instead of summing with respect to the parameter A, and the unknown functions appearing in the integrand are determined by using the theory of integral transforms instead of the theory of expansions in series of eigenfi~nctions.~ The problems in the present chapter can all be solved by the Fourier method (in most cases by the method just described), and are grouped into five sections, two devoted to mechanics, two to heat conduction (including a few problems on diffusion), and one to electricity and magnetism. We also include a few problems involving inhomogeneous equations and inhomo- geneous boundary conditions, which can be solved by the Fourier method after being reduced to homogeneous problems by the use of appropriate tricks. However, inhomogeneous problems will for the most part be con- sidered in Chap. 5, where they are studied systematically. Since an entire chapter (Chap. 7) will be devoted to the less familiar special coordinate systems, we confine ourselves here to rectangular and polar coordinates (both cylindrical and spherical). To illustrate further extensions of the Fourier method, we include a few problems of a more complicated type, e.g., problems involving three variables, elasticity theory, fourth-order differ- ential equations, etc. I. Mechanics: Vibrating Systems, Acoustics *96. At the time t = 0, a string with ends fastened at the points x = 0 and x = I is plucked at the point x = c, and then released without initial velocity. Find the displacement u(x, t) of an arbitrary point of the string if U(C, 0) = h. Ans. 2h 1' Z sin (nncll) . nxx nxvt U(X, t) = -- --- sln - cos - , x2 c(1 - c) .=, n2 1 1 where u = 45, T is the tension and p is the linear density. Ut 97. Find the vibrations of a FIGURE 16 string. For information concerning such integral expansions and Sturm-Liouville theory in general (especially the singular case), see Al, L13, S6, Vol. V and T6. Ans. 8h1 2 cos [(2n + 1)7cn/211 (2n + l)xx cos (2n + lbvt U(X, 0 = COS (1 - a)n2 .=, (2n + 1)' 21 21 98. Solve the preceding problem, assuming that the initial form of the string is a parabola symmetric with respect to the center of the string and that the maximum initial displacement from equilibrium is h. Ans. 99. At the time t = 0, the center of a string of length 21 fastened at the points x = -1 and x = 1 receives an impulse P. Find the subsequent vibrations of the string 2P cos [(2n $ l)rcx/21] sin (211 + 1)xvt u(x, t) = - 2 xvp .=, 2n + 1 2 1 Hint. Consider the vibrations of the string subject to the initial conditions and then take the limit as E --, 0. 100. Study the vibrations of a string fastened at the points x = 0 and x = 1 due to a suddenly applied load distributed along the string with constant density q which subsequently remains constant. The string is assumed to be at rest at the time t = 0. Ans. Hint. Before applymg the Fourier method, make the problem homo- geneous by subtracting out the static deflection of the string under the uniform load. 101. Find the vibrations of a string -I < x < I of mass m loaded at the point x = 0 by a concentrated mass m,. In solving the problem, assume that the load is initially displaced by a small amount h, and that the initial velocity of the string is zero. 62 THE FOURIER METHOD Ans. PROB. 102 where a = m/mo and the yn are consecutive positive roots of the equation tan y = a/y. 102. A rod of length 1, density p and cross-sectional area S is clamped at the end x = 0 and stretched by a force F applied at the other end x = 1. Study the longitudinal oscillations of the rod if the force is suddenly dis- continued at the time t = 0. Ans. 8F1 U(X, t) = - (- 1)" sin (2n + 1)xx (2n + 1)nvt n2ES zo (2n + 1)' COS 21 2 1 9 where v = JE/(, and E is Young's modulus. 103. Find the general solution of the problem of longitudinal oscillations of a rod of length I with arbitrary initial conditions if the end x = .O is clamped and the end x = 1 is free. Ans. x sin (2n + l)rr~t/'~(~) sin (2n + 11x4 &I sin (2" + lhx 21 0 21 2 1 104. Investigate the longitudinal oscillations of a cantilever of length 1 and mass M if the end x = 0 is clamped while the end x = I is loaded by a concentrated mass Mo, which at the time t = 0 experiences a displacement 6 without acquiring any initial velocity. Ans. Y x sin M cos y, u(x, t) = 26 - 2 7 1 Y v* COS , Mo n=~ Yn sin 2y, 1 I+- 2~n where the y, are consecutive positive roots of the equation 105. Find the longitudinal oscillations of a rod of length I if the end x = 0 is clamped while the end x = I receives an impulse P at the time t = 0. The rod is assumed to be at rest before the impulse acts. Ans. In the notation of the preceding problem, Hint. Solve the problem of oscillations with the initial conditions where S is the cross-sectional area of the rod, and then take the limit as E + 0. 106. Find the displacement of the points of a rod of length I clamped at the end x = 0, which undergoes longitudinal oscillations under the action of a pulsating force A sin ot applied to the free end x = I. The rod is assumed to be at rest before the force begins to act. Ans. sin wt COS - U sin (2n + 1)nx sin (2n + 1)xut 21 Hint. To make the problem homogeneous, represent the displacement as a sum of free and forced oscillations (see Prob. 78). Another method of solution is given in Chap. 5 (see Prob. 21 1). 107. A conical cantilever with the dimensions shown in Figure 17 is stretched by a force F applied at the end x = I. Study the longitudinal oscillations which result when the force is suddenly discontinued. Ans. 2F cot a U(X, t) = 2 COS Y n Yx YUt sin " cos -E- , xE(a - x tan a) y,[(sin 2yn/2yn) - 11 1 1 PROB. 108 where the y, are consecutive positive roots of the equation tan y = 1 - -cot a y. (r *108. Solve the problem of the longitudinal oscillations of the pyramid- shaped cantilever of rectangular cross section and constant thickness shown in Figure 18, subject to a given initial deformation ul,=, = f(x). FIGURE 18 Ans. 2 tana Xy,(a - x tan a) vty, tan a u(x;o=-Z ,,, COS b2 (4a2/n ynb ) - ~;,(b) a where Jo(x) and Yo(x) are cylinder functions of order zero), and the y, are con- secutive positive roots of the equation X;(b) = 0. -1 *109. Find the general solution of the problem of longitudinal oscil- lations of a rod consisting of two FIGURE 19 rigidly fastened sections withdifferent dimensions and elastic properties (see Figure 19). It is assumed that the ends of the rods are clamped and that the initial state of the rod is characterized by the conditions Ans the y, are consecutive positive roots of the equation the two sections have Young's moduli Ei, cross-sectional areas Si and densities pi (i = 1, 2), and vi = 4%. 110. A pointer is fastened to the free end of a rod of length I clamped at the end x = 0. Study the torsional oscillations which result if at the time t = 0 the pointer is twisted through an angle a and then released without initial velocity, given that the moment of inertia of the pointer with respect to the axis of rotation is Jo. Ans. YnX sin - J cosy, 0(x, t) = 2cc - 2 -7y- I Y vt cos -?-- , Jo ,=I Yn sin2yn 1 I+- 2yn where the y, are consecutive positive roots of the equation J is the moment of inertia, G the shear modulus and p the density of the rod, and u = 4%. 111. Solve the preceding problem with arbitrary initial conditions Ans. with the previous notation. "112. A disk with moment of inertia Ju is fastened to the point x = c of a cylindrical shaft with clamped ends x = 0 and x = I. Find the torsional oscillations of the shaft if the disk is twisted through the angle a at the time t = 0 and then released without initial velocity. Ans. 0(x, 1) = -- Y nut 201 sin y,(~ - :) sin ~,,(x) cos - a - f) n-1 1 1 D n where a ~in'2~,(1 - (all )) + I--+ [ 1 2~n lsinz?), the y, are consecutive positive roots of the equation JOY . YO sin y = - sin - sin y J 1 and J is the moment of inertia of the shaft. 113. A disk with moment of inertia Jo is fastened to one end x = 0 of a circular shaft, and another disk with moment of inertia J, is fastened to the other end x = I. Find the torsional oscillations 0.f the shaft if the relative angle of rotation of the disks equals a. Ans. J J - (1 - cos y,) + y, sin y, e(x, t) = 2a - 2 Jo sin 2y, J Y.,[($ + y:) - 6 - y:) - + - (I - cos 2y.) 2~n Jo I where the y,, are consecutive positive roots of the equation tan y = and J is the moment of inertia of the shaft. *114. Find the general solution of the problem of transverse oscillations of a beam of length I, simply supported at its ends x = 0 and x = I, with arbitrary initial conditions Ans. n4 u(x, t) = 22 [cos -lf (4) sin - df ln=l 1 where a2 = JEJI~S, E is Young's modulus, J the moment of inertia of a cross section, p the density and S the cross-sectional area of the beam. 115. Investigate the transverse oscillations of a beam of length I, simply supported at its ends x = 0 and x = I, under the action of an impulse P applied to the point x = c at the time t = 0. Ans. 21P sin (nxcll) nxx n2n2a2t U(X, t) = -3 sin - sin - . n2 JEJ~S ,=, n2 I l2 Hint. Solve the problem of the oscillations of the beam with the initial conditions P all vo = - , C-E<X<C+E, u,t=o = 0, [ 2.s~ 0, otherwise, and then take the limit as E + 0. 116. A beam of length 21, clamped at its ends x = il, undergoes transverse oscillations with initial conditions Find the oscillations of the beam, assuming that the initial deflection of the beam is symmetric with respect to the center of the beam and that there is no initial velocity. Ans. 1 " cos (y:a2t/12) ( U(X, t) = -2 COS~ y, cos - Ynx - cos y, COS~ - 1 n=l cos2 yn cosh2 yn 1 YE cosh y, cos -2t- - cos y, cosh I where the y, are consecutive positive roots of theequation tan y + tanh y = 0. 117. A beam simply supported at the points x = 0 and x = 1 is in equilibrium under the action of a concentrated force F applied at the point x = c. Find the transverse oscillations which result if the force is suddenly removed. Ans. 2F13 " sin (nxcll) nxx n2x2a2t U(X, t) = - C sin - cos - . x4EJ n4 1 l2 *118. Find the transverse oscillations of a cantilever of length I if the initial deflection is due to a concentrated force F applied to the free end x = I and is suddenly removed at the time t = 0. Ans. 2~1~ X,(x) cos (y:a2t/12) U(X, t) = - C EJ ,=, (cos y, sinh y, - sin y, cosh y,)yj ' where X,(x) = (sin y, + sinh y,) and the y, are consecutive positive roots of the equation cos y cosh y + 1 = 0. 119. A beam of length I is simply supported at the end x = 0 and clamped at the end x = 1. Find the transverse oscillations of the beam under the action of a suddenly applied uniformly distributed load q. Ans. q14 rn sinh y, - 2 cosh y, sin y, + sin y, Y:a2t +-C X,(x) cos - EJ .=I y: sinh2 y, sin", l2 ' Y X YnX X,(x) = sinh y, sin fl - sin y, sinh - , 1 1 where the y, are consecutive positive roots of the equation tan y = tanh y. Hint. Make the problem homogeneous by subtracting out the static deflection of the beam. "120. Study the axially symmetric vibrations of a circular membrane of radius a due to an impulse P applied at the time t = 0 and distributed over a disk of radius E. Ans. where Jo(x) and Jl(x) are Bessel functions, the y, are consecutive positive roots of the equation Jo(y) = 0, T is the tension per unit length of the boundary, p is the surface density of the membrane, and v = JT/p, Hint. The initial conditions have the form 121. Find the general solution of the problem of vibrations of a ring- shaped membrane fastened to the circles r = a and r = 6, and subject to arbitrary initial conditions Ans. 70 THE FOURIER METHOD PROB. 122 is a linear combination of Bessel functions of the first and second kinds, and the y, are consecutive positive roots of the equation Ry(a) = 0. 122. Determine the axially symmetric vibrations of a circular membrane of radius a due to a pulsating load q sin at which is uniformly distributed over the whole membrane and begins to act at the time t = 0. Ans. 2uaq Jo(ynr/a) sin (ynvt/a) u(r, t) = - - I - (or/u)] sin ot + - 2 -;- "1 Jo(oa/v) PW wT .=I ynJdy,) 1 - (~y,loa)~ ' where the y, are consecutive positive roots of the equation Jo(y) = 0. Hint. Make the problem homogeneous by subtracting out the forced oscillations (see Prob. 82). 123. Find the vibrations of a rectangular membrane -a < x < a, -b < y < b with initial conditions ult=o = f (x, Y), where f is a given function which is even Ans. in each of the variables. where = 4 /'So f (x, y) cos (2m + 1)xx cos (2n + ~)XY dx dy. Am, ab o o 2a 2b *124. Study the transverse oscillations of a circular plate of radius a with a clamped edge, for arbitrary initial conditions Ans. u(r, t) = - 2 Ryn(r) [ COS Yy~Pf(F%o) - dp a2 .=I I:(Y,)J:(Y,) where ROB. 127 THE FOURIER METHOD 7 1 is a linear combination of cylinder functions, the y, are consecutive roots of the equation Rl(a) = 0, b2 = JDlph, and D is the flexural rigidity, h the thickness and p the density of the plate. 125. Solve the preceding problem for the casc wherp the oscillations are due to an impulse P applied at the center of the plate at the time t = 0. Ans. Pb2 " [Idyn) - Jo(yn)lRy,(r) y;b2t ~(r, t) = - 2 sin - 2xD n=1 y;Ii(yn)J:(~n) a Hint. Solve the problem with the initial conditions and then take the limit as E + 0. 126. Investigate the transverse oscillations of a circular plate of radius a with a clamped edge under the action of a concentrated force F applied to the center of the plate. The plate is assumed to be at rest at the time t = 0. Ans. with the notation of Prob. 124. Hint. Make the problem homogeneous by subtracting out the static deflection of the plate. 127. Study the radial oscillations of a gas confined in a spherical reso- nat~r,~ assuming that the initial values of the velocity potential and its time derivative are u (r), % / = 0. at t=o The velocity potential of an oscillating gas satisfies the wave equation (see T1, p. 25). In Probs. 127-130 it is assumed that the walls are perfectly reflecting, i.e. that Ans. 2 sin (y,r/a) cos (y,vt/a) u(r, t) = - 2 a ,=I r sin2 y, a where the y, are consecutive positive roots of the equation tan y = y, a is the radius of the sphere and v is the velocity of wave propagation in the gas. 128. Investigate the steady-state acoustic oscillations in a semi-infinite cylindrical pipe of radius a, assuming that the distribution of the normal component of the velocity of the air particles in the plane z = 0 is a given function (vr)zlz=o = f (r) sin wt. Consider the special cases where y is the smallest positive root of the equation J,(y) = 0. Ans. The velocity potential is given by the formula where k = wlv, the yn are consecutive nonnegative roots of the equation J,(y) = 0 (yo = O), and Jo(x), Jl(x) are Bessel functions. In the special cases, D u(z, t) = - --O cos (at - kz), k 129. Find the steady-state harmonic oscillations of sound inside a conical horn a < r < co, 0 ,< 0 < a, assuming that the velocity distribution along the base of the horn is given by (~r>~l~=~ = f (0) sin wt. Consider the special case f (0) = v,. Ans. x Pvn(cOS Soa (B)P,~~~~ 0) sin 0 ~P;,,,(cos 1 avm Ivm=vn eiol - 1; (0) sin 0 do], k(l - cos a) H@&ka) p~oe. 131 THE FOURIER METHOD 73 where the v, are consecutive roots of the equation Pin(cos a) = 0, PY(x) is the Legendre function and HJ:'% (x) the second Hankel function. In the special case, a sin [at - k(r - a)]- 2ka cos [wt - k(r - a)] u(r, t) = 2v0a - Jr 1 + 4k2a2 130. Solve the problem of diffraction of a plane sound wave uoei("t-kz) by a spherical obstacle of radius a. Ans. 9- where Jn,.;$(x) is the Bessel function of the first kind, H(x):l% the second Hankel function, and Pn(x) the Legendre polynomial. Hint. If the velocity potential is written as a sum then solving the problem reduces to integrating Helmholtz's equation Aul + k2u, = O with the boundary condition where u, must satisfy the radiation condition at infinity. 2. Mechanics: Statics of Deformable Media, Fluid Dynamics 131. Find the equilibrium shape of a rectangular membrane with sides 2a and 2b under the action of a uniformly distributed load q, choosing the origin at the center of the membrane. Calculate the deflection of the center of the membrane, assuming that the ratio bla takes the values 1, 2 and 3. Ans. qa2 1 16 (- l)n+l cosh [(2n + l)xy/2a] cos 1-- +-2--- = -( T2 ) x3 n=o (2n + 1)' corh [(2n + l)rrb/2a] 2a 74 THE FOURIER METHOD PROB. 132 where T is the tension per unit length of the boundary. Numerical calcula- tions show that u(0,O) = kQ/T, where Q = qab is the total load, and Hint. Make the problem homogeneous by subtracting out the particular solution of the equation for equilibrium of the membrane which depends only on the coordinate x and satisfies the boundary conditions on the sides x = +a. *132. Find the equilibrium shape of a semicircular membrane of radius a (see Figure 20) under a uniformly distributed load q. Ans. 2ar sin cp -?(if +$) cos2cp - 2 2 a2 a2 - r2 +- ( --- ::) sin 2cp ~n a2 + r2 -- 2ar cos cp a2 + r2 + 2ar cos cp where T is the tension per unit length of the boundary. Hint. To apply the Fourier method, sub- tract out the particular solution qr2 . u, = - - sln2 cp 2 T 0 of the equilibrium equation. To write the solution in closed form, it is necessary to sum a FIGURE 20 series (this has been done in the answer). 133. Study the twisting of a rod whose cross section is a rectangle with sides a and b. Find the torsion function and the torsional rigidity. Ans. The torsion function is 8a2 " sin [(2n + l)nx/a] cosh[(2n + l)($b - y)x/a] U(X, y) = x(a - x) - - 2 z3 ,,=,, (2n + lI3 cosh [(2n -i- I)xb/2a] ' (OG x~ a,OG y < b). and the torsional rigidity is where G is the shear modulus. PROB. 136 THE FOURIER METHOD 75 Hint. Make the problem homogeneous by subtracting out the particular solution of the differential equation for the torsion function which depends only on the coordinate x and satisfies the boundary conditions for x = 0 and x = a. 134. A rectangular plate with sides 2a and 2b, simply supported on its edges, is acted upon by a uniformly distributed load q. Find the deformation of the plate, choosing the origin at the center of the plate. Derive an ex- pression for the deflection of the plate. Ans. r 7 2 + (2n + 1)xb tanh (2n 2a 2 cosh (2n + 1)xb 2a where D is the flexural rigidity of the plate. Hint. Subtract out the particular solution of the deflection equation which depends only on the coordinate x and satisfies the boundary conditions for x = fa. 135. Solve the preceding problem, assuming that the boundaries x = *a are simply supported, while the boundaries y = f b are free. Calculate the deflection at the center of the plate. Ans. (3 + v) sinh (211 + l)xb cosll (2n + l)xb - (1 - v) (2n + 1)xb 2a 2a 2a where v is Poisson's ratio. "136. A semicircular plate of radius a is clamped along the semicircular arc and simply supported along its rectilinear edge. Find the deflection of the plate under a uniform load. Write a formula for the deflection of the axis of symmetry of the plate, and represent the result in the form of a graph. Ans. 1 r3 r a 5--11-$3-+3- G a3 a r r3 1 (see Figure 21). Hint. Make the problem homogeneous by subtracting out the particular solution 4 - r4 sin4 cp, 0 < cp ,< rr 24 D of the deflection equation satisfying the boundary conditions on the rectilinear edge. 137. An infinite cylinder of radius a is placed in a plane-parallel flow of an ideal fliid. Find the velocity potential, YI choosing the origin at the center of the cylinder and the direction of the x-axis opposite to the direction of flow (see Fig- Ans. V- a(r, y) = u, (r 4- :) cos y + const, FIGURE 22 where v, is the value of the flow velocity far from the cylinder. 138. Find the velocity potential for flow of an ideal fluid emanating from a source of strength rn and flowing past an infinite cylinder of radius a, where the configuration of the cylinder and the source is shown in Figure 23. Ans. m rb u(r, cp) = - In 1 + const, 2~ PP where b6 = n2 and the meaning of the ious symbols is indicated in the figure. Hint. Subtract the source potential from the solution. var- - 139. Solve the problem of plane-parallel flow of an ideal fluid past a sphere of radius a, choosing the origin of a system of spherical coordinates ROB 143 THE FOURIER METHOD 77 r, 0, cp at the center of the sphere, with the direction of the z-axis opposite to that of the flow. Ans. u(r, 8) = urn (r + $) cos 8 + const, where v, is the value of the flow velocity far from the sphere. *140. Solve the problem of flow past a sphere of radius a due to a source of strength rn at a distance b from the center. Ans. with the same notation as in Figure 23, except that the x-axis now becomes the z-axis. 3. Heat Conduction: Nonstationary Problems 141. A slab of thickness 2a, thermal conductivity k, specific heat c and density p is heated to temperature To, and its faces are then held at tem- perature To, starting from the time t = 0 (see Figure 24). Find the temperature distribution T(x, t) in the slab. Ans. . - e-(2n+~)2n2r/4a2 COS (2n + l)nx 20 9 FIGURE 24 where T = ktlcp. 142. Describe the equalization of a given initial temperature distribution T(x, 0) = f(x) in a slab whose faces x = 0 and x = a do not transmit heat. Ans. 2" nnt f (4) d4 + - 2 e~n22Tiaacos If@,) cos - d4. a .=l a a 143. Starting from the time t = 0, a slab -a < x < c of thickness 2a with a given initial temperature distribution T(x, 0) = f(x) radiates heat into the surrounding medium, whose temperature is taken to be zero. Assuming that the radiation obeys Newton's law, find the temperature distribution in the slab for arbitrary time r. where T = ktlcp, the y:) are consecutive positive roots of the equation h is the heat exchange coefficient, and the yp) are the corresponding roots of the equation 144. Starting from the time t = 0, heat is produced with constant density Q in a slab -a < x < a of thickness 2a. Find the temperature distribution in the slab, assuming that its faces are held at temperature zero and that the initial temperature is also zero. Ans. Hint. Make the problem homogeneous by subtracting out the solution of the corresponding stationary problem. "145. An inhomogeneous slab consisting of two layers with different thermal properties is heated to a certain temperature To, and then cooled by having its faces held at temperature zero starting from the time t = 0. Assuming that the faces of the slab are at x = 0 and x = a, + a, (where a, and a, are the thicknesses of the two layers), find the temperature distribution in the slab. ROB. 147 THE FOURIER METHOD 79 Ans. where the y, are consecutive positive roots of the equatlon a2Jbzy Jb, k2 tan y + Jb, k, tan - - 0, a,J& - the two layers have specific heats ci, densities pi and thermal conductivities ki (i = 1, 2), and bi = cipi/ki. 146. The ends of a thin rod of length I are held at different temperatures, while the lateral surface of the rod gives off heat into the surrounding medium according to Newton's law. Find the temperature distribution along the rod, assuming that the ends of the rod x = 0 and x = 1 have temperatures zero and To, respectively, and that the initial temperature equals zero. Ans. sinh Jix + - 2 e-Y,z (-1)"n sin e-nhziji2 T(x, t) = T, --- rsinh &I i~ ..I n2 + (p12/n? I where p and S are the perimeter and cross-sectional area of the rod, k is the heat exchange coefficient figuring in Newton's law, and p = ph/S. Hint. The problem reduces to integration of the differential equation (see C3, p. 134). 147. A cylinder of radius a is heated to temperature To and then cooled by having its surface held at temperature zero starting from the time t = 0. Find the subsequent temperature distribution in the cylinder, assuming that all cross sections have the same temperature distributi~n.~ Ans. This corresponds to a long cylinder (theoretically, infinitely long). where r is the distance from the axis of the cylinder, Jo(x) and Jl(x) are Bessel functions, the yn are consecutive roots of the equation Jo(y) = 0, T = ktlcp where k is the thermal conductivity, c the specific heat and p the density of the cylinder. *148. Describe the equalization of a given axially symmetric initial temperature distribution T(r, 0) = f(r) in an infinite cylinder of radius a, whose lateral surface does not transmit heat. Ans. where the yn are consecutive positive roots of the equation Jl(y) = 0. 149. An infinite cylinder of radius a, initially heated to the temperature To, subsequently cools off by radiating heat into the surrounding medium according to Newton's law. Describe the cooling process. Ans. -y,BT/a2 Jl(~n)Jo(~nr/a) e T(r, t) = 2~,2 n=1 JXYn) + J;(Y~) Yn ' where the yn are consecutive positive roots of the equation YJ~(Y) = ahJo(y). 150. Starting from the time t = 0, Joule heat is produced with density Q in a cylindrical conductor of radius a. Find the temperature distribution over a cross section, assuming that both the initial temperature and the surface temperature equal zero. Ans. where the yn are consecutive positive roots of the equation Jo(y) = 0. Hint. Make the problem homogeneous by subtracting out the particular solution corresponding to the stationary distribution of temperature in the cylinder. 151. A cylindrical conductor of radius a is heated for a long time by an electric current producing heat in the conductor with density Q. Study the process of cooling that ensues after the current is turned off, assuming that the cooling from the surface always obeys Newton's law and that the tem- perature of the surrounding medium equals zero. Am. where the y, are consecutive positive roots of the equation and h is the heat exchange coefficient. Hint. To determine the initial condition for the cooling problem, find the stationary distribution of temperature during the period of heating. 152. Find the temperature distribution in a cylindrical pipe a < r < b if there is a constant heat current of density q through the inner surface r = a, while the outer surface r = b is held at temperature zero. The initial tem- perature of the pipe is assumed to be zero. Ans. where where Jo(x) and Yo(x) are Bessel functions, and the y, are consecutive positive foots of the equation R;(a) = 0. Hint. Subtract out the particular solution corresponding to the stationary distribution of temperature in the pipe. "153. Find the general solution of the problem of the cooling of a sphere of radius a, given that the initial temperature distribution of the sphere is T(r, 0) = f (r), while the surface temperature equals zero. Ans. 2 " 12 , = - - n a nYr a nn P sin -1 f (p) sin - p dp, ar .=I a o a where k is the thermal conductivity, c the specific heat and p the density of the sphere, and r = ktlcp. 154. Find the temperature distribution in a sphere of radius a whose surface radiates heat starting from the time t = 0 according to Newton's law, if the initial temperature is To. Ans. 2T0a ah Zcos y,, sin (ynr/a) e-~%~a2 T(r, t) = - - r 1 - ah ,,, y, 1 - (sin 2yn/2yn) > where h is the heat exchange coefficient, and the y, are consecutive positive roots of the equation Y tan y = - 1 - ah 82 THE FOURIER METHOD PROB. 155 155. A spherical object of radius a is heated for a long time by a source producing heat with volume density Q. Study the process of cooling that ensues after the heating is stopped, assuming that the cooling is due to radiation from the surface and that the temperature of the air in the chamber where the heating occurred is To. Ans. 2Qa4h 2 COS y, -pi~/a~ Y,Y T(r, f) = To + e sin - , (1 - ah)kr ,=, y:[1 - (sin 2~,/2~,)l a where the y, are consecutive positive roots of the equation Y tan y = - 1 -ah' Hint. To determine the initial temperature distribution in the sphere, solve the corresponding stationary problem. 156. The region between two parallel planes x = 0 and x = a is occupied by a solution with a given initial concentration C(x, 0) = f(x). Describe the subsequent equalization of concentration, assuming that the walls are impermeable. Examine the special case Ans. In the special case, e-nzn"ULg nnx cos -1, a where D is the diffusion coefficient. 157. Find the concentration in a solution inside a cylindrical pipe a < r < b with impermeable walls, if the initial concentration distribution is Ans. where where J,(x) and Y,(x) are Bessel functions, and the y, are consecutive positive roots of the equation Ri(b) = 0. 158. Find the concentration of a gas inside a cylindrical metal object of a, assuming that the initial concentration of the gas is C(t, 0) = f(r) and that the object is surrounded by a medium in which the gas is maintained at constant concentration C,. Consider the special caseJ(r) = Co. where the y, are consecutive positive roots of the equation aa YJ~Y) = 5 JO(l). In the special case, -y;gtlaa a 2a2a2[Co - (al/a)C1] e C(r, t) = C, + Jo(ynr/a) a D n=1 ~nJl(yn)[y", (a2a2/D2)I Hint. The problem reduces to solving the differential equation with initial condition ~It=o =fW and boundary condition where a and a, are the coefficients characterizing the emission and re- absorption of the gas by the surface of the metal (see G3). 4. Heat Conduction: Stationary Problems 159. Find the stationary temperature distribution T(x, y) in an infinite bar of rectangular cross sectioil (see Figure 25) if three faces are held at temperature zero, while a given temperature distribution T(x, b) = f(x) is maintained on the fourth side. Apply the resulting general formulas to the special case f(x) = To. Ans. 2 " sinh (nnyla) nnx T(x, Y) = - 2 sin - a .=, sinh (nnbla) a X nx< . x J"/Q sin - d~. a FIGURE 25 PROB. 160 In the special case, 4T0 " sinh [(2n + l)ny/a] sin [(2n + l)nx/a] T(x, Y) = - 2 n ,=, sinh [(2n + l)nb/a] 2n + 1 160. Find the distribution of temperature in a bar of rectangular cross section if the two opposite faces y = 0 and y = b are held at temperatures zero and To, respectively, while the other two faces x = *a radiate heat into the surrounding medium (assumed to have temperature zero) according to Newton's law. Ans. where the yn are consecutive positive roots of the equation ah tan y = - Y and h is the heat exchange coefficient. 161. Find the stationary temperature distribution in a conductor of rectangular cross section -a < x < a, -b < y < b, heated by an electric current producing Joule heat Q per unit volume, if the faces of the conductor are held at temperature zero. Ans. k cosh [(2n + l)ny/2al cos cosh [(2n + l)nb/2a] where k is the thermal conductivity of the conductor. Hint. Subtract out the particular solution of the inhomogeneous heat conduction equation which depends only on the coordinate x and satisfies the boundary conditions for x = ha. 162. Solve the preceding problem, assuming that all faces of the conductor radiate heat into the surrounding medium (assumed to have temperature zero) according to Newton's law. Ans. T(x, Y) = - where the 2ah 2 sin Y, cos (y,xla) cash (Y,Y/~) y:[1 + (sin 2y,/2yn)l[yn sinh (y,b/a) + ah cosh (y,b/a)l y, are consecutive positive roots of the equation ah tan y = - Y 163. A bar of rectangular cross section 0 < x < a, 0 < y < b is heated by a constant thermal current of density q incident on one face y = b of the bar. Find the stationary temperature distribution over a cross section of the bar, assuming that heat is lost by radiation into the surrounding medium according to Newton's law. Ans. q "- y, sin y, + ah(1 - cos y,) T(x9 ') = 6 2ah + yi + (ah)' Y Y ah . YnY cosh ^ + - smh - Yn a X ynb Y: + (ah)'sinh ynb cosh - + a 2ahyn a where the y, are consecutive positive roots of the equation 2ahy tan y = Y2 - (ah)' 164. A rectangular bar consists of two sections with different thermal conductiv- ities k, and k,, respectively (see Figure 26). Find the temperature distribution in the bar, assuming that two opposite faces y = f b are at temperature To, while the other two sides are at temperature zero. PROB. 165 where the y, are consecutive positive roots of the equation k YO2 tan y + --' tan - = 0. k2 al 165. Determine the stationary temperature cpzo distribution in a bar whose cross section is a Wp I =O 46 curvilinear rectangle," with two faces consisting \ \ 1 / of arcs of concentric circles and the other two \ /' \ / faces of segments of radii of the larger circle '$ (see Figure 27). It is assumed that one of the curved faces r = 6 has temperature To, while FIGURE 27 the other faces are held at temperature zero. Ans. 166. Solve the preceding problem, assuming that one of the plane faces cp = a is held at temperature To, while the other faces are held at temperature zero. Ans. 167. Find the stationary temperature distribution in a cylinder of radius a and length I (see Figure 28) with ends held at temperature zero and lateral surface held at temperature To. Calculate the temperature distribution along the axis of the cylinder, assuming that the ratio all equals 0.5, 1, 2. Ans. 4T0 " Io[(2n + l)nr/l] sin [(2n {- l)nz/l] T(r, z) = -2 n n=o Io[(2n + l)na/l] 2n -I- 1 where Io(x) is the Bessel function of imaginary argument. The results of nun~erical calculations of the quantity FIGURE 28 are given in the following table 168. Solve the preceding problem, assuming that the ends of the cylinder do not transmit heat, while a given temperature distribution TI,=, = f (2) is maintained along the lateral surface of the cylinder. Ans. *169. Solve Prob. 167, assuming that the ends of the cylinder cool off according to Newton's law and choosing the origin at the center of the cylinder. Ans. where the yn are consecutive positive roots of the equation hl tan y = - 2~ and h is the heat exchange coefficient. 170. The walls of a cylindrical hole drilled in an infinite slab of thickness h (see Figure 29) are held at a given temperature To. Find the stationary tern- perature distribution in the slab, if its plane faces have temperature zero. PROB. 171 Ans. 4T0 sin [(2n + 1)7iz/h] T(r, z) = -2 7~ n=~ 2n + 1 X K0[(2n + 1)xrlhI Ko[(2n + l)xa/h] ' where Ko(x) is Macdonald's function. 171. Find the stationary temperature distribution in a cylinder 0 < r < a, 0 < z g I if the upper end is at tem- perature To while the rest of the surface is at temperature zero (cf. Prob. 167). Ans. J ( rla) sinh (ynz/a) T(r, z) = 2~~2 OY'- n=1 ~nJl(yn) sinh (ynlla) ' where the y, are consecutive positive roots of the equation Jo(y) = 0. 172. Heat is produced with constant density Q in a cylinder of radius a, length I and thermal conductivity k. Find the stationary temperature distri- bution if heat leaves the cylinder through the part of the upper end bounded by the circler = b < a, but not through the rest of the surface of the cylinder. It is assumed that the flow of heat out of the cylinder is uniformly distributed over the disk r g b. Ans. J~(~nbla)Jo(~nrla) cash (~nzla) I + const, y:~;(y,) sinh (ynI/a) where the yn are consecutive positive roots of the equation Jl(y) = 0. Hint. Subtract out a particular solution of the inhomogeneous heat conduction problem which depends only on the coordinate z. 173. A cylinder standing on a thermally in- sulating slab is heated from above by a uniformly distributed thermal current (see Figure 30), and radiates heat from its lateral surface into the sur- rounding medium (assumed to be at temperature zero) according to Newton's law. Find the station- ary distribution of heat in the cylinder. Ans. where the y, are consecutive positive roots of the equation YJdY) = ahJo(y1, h is the heat exchange coefficient and q is the density of the incident heat current. 174. A semi-infinite cylindrical pipe a ,< r g b, 0 g z < m is heated at the end z = 0 held at temperature To, and cooled at its lateral surfaces r = a and r = b held at temperature zero. Find the stationary temperature distri- bution in the pipe. Ans. where is a linear combination of Bessel functions, and the y, are consecutive positive roots of the equation *175. An inhomogeneous cylinder formed of two sections with different thermal conductivities k, and k, (see Figure 31) is heated at its lateral surface held at temperature To and cooled at its ends held at temperature zero. Find the stationary temperature distribution T(r, z) in the cylinder. Ans. a tan y,(cos y, - cos - z Ynh2 h "=' yn(? sin - + 3 sin2 y, hl h, where PROA. 176 and the y, are consecutive positive roots of the equation 1~1 yh'2 tan y + - tan - = 0. k' hl *176. Find the stationary temperature distributioli in a, if one part of its surface S, is held at constant temperature To, while the remaining part S, is held at temperature zero (see Figure 32). Ans. a sphere of radius in terms of the Legendre polynomials P,(x). FIGURE 32 177. Solve the preceding problem, assuming that heat is produced in the sphere with volume density Q, and that heat leaves the sphere through the surface S, flowing in the normal direction with constant density (the surface S, does not transmit heat). Ans. where k is the thermal conductivity of the sphere. Hint. Subtract out a particular solution of the inhomogeneous heat conduction equation which depends only on the variable r. 178. A sphere of radius a is heated by a plane-parallel thermal current of density q incident on its surface, and gives off heat into the surrounding medium according to Newton's law. Find the stationary temperature dis- tribution in the sphere. Ans. pROB. 181 THE FOURIER METHOD 9 1 in terms of the Legendre polynomials P,(x). Note that Hint. Here the boundary condition takes the form 5. Electricity and Magnetism 179. Find the electrostatic potential u(x, y) inside an elongated box of rec- Yt tangular cross section (see Figure 33), if two opposite sides are at potential V and the u=o '7 other two sides are grounded. Ans. 4V " cosh [(2n + l)xx/b] u(x, y) = - 2 (- 1ln - x ,=, cosh [(2n + l)xa/b] FIGURE 33 180. Find the electrostatic potential u(x, y) inside a semi-infinite rectangular box (see Figure 34), if the vertical wall is held at potential V and the horizon- tal walls are held at potential zero. Ans. 2 V sin (xylb) u(x, y) = - arc tan x sinh (xxlb) ' Hint. To represent the solution in closed form, use the expansion 3'"""'" 1 sin (2n + I) y = - arc tan - .=o 2n + 1 2 smh x ' x> 0. 181. Find the electrostatic potential u(x, y) X between two infinite parallel sheets if one 0 sheet y = 0 is at potential zero, while a given FIGURE 34 periodic potential &b = f (-4 is maintained on the other sheet (where f is a function with a given period 2a). Ans. 1 " sinh (nnyla) [ n;xr u(x, Y) = - 2 nnS cos - f (S) cos - dt a .=, sinh (nnbla) a (s) sin dt] + 2 SP/(F) d~. a a 2ab o 182. A thin charged wire with linear charge density q is placed inside and parallel to a conducting cylinder of radius a held at potential zero. Use the familiar method of images to solve the cor- responding electrostatic problem, assuming that the wire is a distance b from the axis of the cylinder. Ans. where R and a are the distances shown in Figure 35, and a2 = b6. 183. Solve the preceding problem if the wire is placed outside the cylinder, and if the cylinder has total charge Q per unit length. Ans. where R and a are the distances shown in Figure 36, and a2 = b6. PROB. 187 THE FOUR~ER METHOD 93 184. Find the electrostatic potential u(r, cp) in the space between two conducting infinite half-cylinders, one of which is held at potential V and the other at potential zero (see Figure 37). It is assumed that the half-cylinders are separated by thin layers of insulating material along the lines where they meet. Ans. 2ar cos cp 2 a2 - r I Hint. To solve the problem in closed form, use the expansion 1 2ar cos cp cos (2n + 1)cp = - arc tan r < a. 2 a2 - r2 ' 185. A cylinder of radius a made from material with dielectric constant E is introduced into a plane-parallel electric field with components E, = -Eo, E, = E, = 0. Find the resulting potential distribution, and show that the field inside the cylinder is homogeneous. Ans. The potential distribution is u=Ezx [ I-- : + : (:I] + const outside the cylinder, u = - Eox + const inside the cylinder. E+1 The field inside the cylinder is 186. Find the electrostatic potential u(r, z) inside a closed cylindrical' surface of length 1 and radius a, if the base and lateral surface are held at potential V, while the upper surface is held at potential zero. Ans. where the yn are consecutive positive roots of the equation Jo(y) = 0. Hint. 187. Two metallic hemispheres of radius a, separated by a thin insulating washer, are held at potential V and zero, respectively, corresponding to the 94 THE FOURIER METHOD boundary condition PROB. 188 Find the electrostatic potential u(r, 0) in the space between the hemispheres. Ans. in terms of the Legendre polynomials P,(x), where 188. Find the electrostatic field of a point charge q placed at distance b from the center of a conducting sphere of radius a (a < b) held at potential zero.1° Ans. The electrostatic potential is where R = Jb2 + r2 - 2br cos 8, R = Jh2 + r2 - 26r cos 0, 189. Solve the preceding problem, assuming that the sphere is made from material of dielectric constant E. Ans. The potential is inside the sphere and outside the sphere, in terms of the Legendre polynomials P,(x). lo This problem can either be solved by the method of images or by the method of inversion (starting from the familiar solution of the problem of a point charge placed over a conducting plane). ROB. 193 THE FOURIER METHOD 95 "190. Find the distribution of d-c current in a thin rectangular sheet, if the current is applied by electrodes at the points x = -a, y = 0 and x = a, y = 0 (see Figure 38). Ans. The potential of the current distribution in the sheet is J [x 2 " sinh (nmlb) U(X, y) = - - - + - C 2ah b n .=, n cosh (malb) where o is the conductivity and h the thickness of the sheet, and J is the total current flowing through the sheet. Hint. The differential equation for the potential of the current distribution in a thin JS74: -x conducting shell is given in Prob. 21. 191. Find the distribution of d-c current FIGURE 39 in a thin disk of radius a, if the current is applied by electrodes at ths points r = a, cp = 0 and r = a, cp = rr (see Figure 39). Ans. In a ' az u(r, cp) = - + const. 2noh 2 r r2 l+-coscp+~ "192. Find the distribution of d-c current in a cylindrical shell of radius a, height 21 and thickness h, if the current is applied by electrodes at the points r. = a, cp = 0, z = h1. Ans. J z sinh (nzla) cos ncp u(cp, z) = -[- + 22 -1 + const. 2noh a ,,=, cosh (nlla) n 193. Find the distribution of d-c current in a hemispherical cap of radius a, if the current is applied by electrodes at the points r = a, 8 = "12, cp = 0 and r = a, 8 = "12, cp = rr (see Figure 40). Ans. 0 0 1 + 2 tan - cos cp + tan2 - - QJ u(e, cp) = -i- ln 2 e e J + const. 2noh I-2tan-coscp+tan2- FIGURE 40 2 2 194. A d-c current J enters one end of a cylindrical conductor of radius a made from material of conductivity o and leaves the other end, via electrodes in the shape of disks of radius r < a (see Figure 41). Find the current distribution inside the conductor, assuming that the current is uniformly distributed over the electrodes. It , Ans. Jz 25 " sinh (y,z/a) u(r, z) = - + - 2 xa20 xbo ,=, cosh (y,h/a) X Jl(~nb/a)Jd~nr/a) + const, Y:J:(Y~) where the y, are consecutive positive roots of x ,----T--- the equation J,(y) = 0. Hint. 195. Find the current distribution in a homogeneous conductor in the form of a rectangular parallelepiped -a < x < a, -b < y < b, -c < z < c, p~oe. 196 THE FOURIER METHOD 97 assuming that current enters and leaves via rectangular electrodes of dimen- sions 26 x 2~ applied at the boundaries z = f c. The current distribution is assumed to be uniform over the area of the.electrodes. A ns. mx8 mxz sin - sinh - JZ +Ls- a a mnx u(x, Y, z) = - COS - 4aab 2x2a8c ,,, m2 mxc a cosh - nxc nxz sin -- sinh - b b nny +"Z--- cos - 2n2aca n2 nxc b cosh - mn8 . nxc sin - sln - a b ++22 -2 x ,,,=I ,,=I mn /~ll sinh + $ nz mxx nxy X .- cos - cos - + const, where J is the current and o the conductivity. *196. A cylindrical pipe a < r < b made from material of magnetic permeability p is placed in a homogeneous magnetic field H,. Find the resulting distribution of magnetic potential. Plot the lines of force for the values p = 5 and bla = 1.5. A ns. (p2 - l)(b2 - a2) u, = H0x 1 - + const, b < r i m, [ byii + 1)' - a2(p - 1)' r The lines of force are plotted in Figure 42. 98 THE FOURIER METHOD PROB. 197 197. Find the magnetic field due to a current J flowing in a wire placed inside a cylindrical hole of radius a drilled in iron of magnetic permeability p, if the wire is at distance b from the axis of the hole. Plot the lines of force for the values p = 3, b/n = 0.5. Ans. where A, and A2 are the values of the z-component of the vector potential of the magnetic field in the air and in the iron, and R = Jr2 + be - 2br cos 9, K = dr2 + 6' - 2hr cos cp, b6 = a'. The lines of force are shown in Figure 43. 198. Solve the preceding problem for the limiting case p = co. Find the equation of the lines of force in the air and in the iron. Ans. PROB. 201 THE FOURIER METHOD 99 The lines of force are ovals of Cassini RR = const in the air, and circles r = const in the iron. 199. A sphere of radius a made from material of magnetic permeability p is introduced into a honiogeneous magnetic field with components Hz = H, = 0, Hz = -Ho. Show that the field inside the sphere is homo- geneous, and find its value. Ans. 200. A hollow sphere a < r G b of magnetic permeability p is placed in a hon~ogeneous magnetic field Hz = H, = 0, Hz = -H,. Solve the corre- sponding problem of magnetostatics. Ans. b3(a3 -- b3)(p - 1)(2p + l)Hoz + const, b < r. < a. HOz + r31b3(p + 2)(2p + 1) - 2a3([r - I)'] *201. Find the magnetic field due to a d-c current J flowing in a circular loop of radius ro inside a hollow spherical shield made from material of magnetic pelmeability p (see Figure 44). Ans. The components of the vector potential of the magnetic field are A, = A" = 0 2nJp " A, = A(r, 8) = - c .=, (212 + 1)(2n + 2) FIGURE 44 100 THE FOURIER METHOD PROB. 202 in terms of the associated Legendre functions P,',+,(x). Note that 202. A lossless open-ended transmission line of length 1 with parameters L and C is charged to a constant potential E (cf. Prob. 23). Determine the current distribution along the line, assuming that a coil of self-inductance Lo is connected across the end x = I at the time t = 0. Ans. 2aE * sin (ynvt/l) sin (ynx/l) I(x, t) = -2 z ,=I cos yn[yE + 41 + all ' where the y, are consecutive positive roots of the equation a tan y = - , Y a = LIIL,, v = 11 JLC is the velocity of wave propagation along the line, and Z = Lv is the wave resistance of the line. 203. A transmission line with parameters L, C and R is short-circuited at one end x = 1 and connected at the other end x = 0 to a source of constant e.m.f. E. Find the voltage distribution along the line, for the case of zero initial conditions. Ans. nxu*t Rl nxv*t sin (nxxll) X cos - +- ( 1 2nx,* sin - 1 1 n where Hint. Make the boundary conditions homogeneous by subtracting out a particular solution of the differential equation depending only on the coordinate x. 204. A plane electromagnetic wave with electric field components E, = Ey = 0, E, = Eoef("-k") (where k = wlc is the wave number) is incident on an infinite perfectly conducting cylinder of radius a. Find the resulting diffracted electric field. THE FOURIER METHOD 101 Ans. where Jn(x) and HF'(x) are Bessel functions of the first and third kinds. 205. Solve the preceding problem, assuming that the cylinder is made from material of conductivity o and dielectric constant E. Ans. + 2 2 e-inn/2 klJn(/W)JXkla) - k~Jn(k1a)JXk~a) ~:'(k,r) cos n cp] e'Ut, n=l k2H(,2)(kla)JQ(k2a) - klJn(k2a)H(,2)'(kla) m Jn(k2r) cos ncp ei~t + 2Ce-inx12 n-1 kl~n(k2a)H~"(kla) - k2H~)(kla)J',(k2a) 1 Ogrga, "206. Find the electromagnetic oscillations in a spherical resonator of radius a excited by a dipole of moment P located at the center of the sphere, assuming that the direction of the dipole coincides with the direction of the z-axis. Ans. The complex amplitudes of the field components are + kecika 1 + ika - k2a2 sin kr cos kr) ] sin 8, ka cos ka + (k2a2 - 1) sin ka i a E - - ----- (H, sin 8), kr sin 8 a€l 102 THE FOURIER METHOD PROB. 206 in terms of the spherical coordinates r., 8 and cp, where w is the frequency of the oscillations and k = wlc is the wave number. References Bateman (B2), Frank and von Mises (F6), Franklin (F7), Gray and Mathews (G2), Grinberg (G5), Jackson (Jl), Jeffreys and Jeffreys (J4), Lebedev (L9, Chaps. 6 and 8), McLachlan (M5), Morse and Feshbach (M9), Tikhonov and Samarski (Tl), Tolstov (T7), Webster (W5). For further problems, see Budak, Samarski and Tikhonov (B6), Gyunter and Kuzmin (G7, Chap. 15), Smirnov (S5). - THE EIGENFUNCTION METHOD FOR SOLVING INHOMOGENEOUS PROBLEMS In this chapter we study various inhomogeneous problems of mathematical physics leading to integration of the equation (1) which is the same as equation (1) of Chap. 4, except for the presence of the given function F(x, y) in the right-hand side.' This time we require that the solution of (I) satisfy inllomogeneous boundary conditions where a,, a,, pa, P, are constants and fa(y), f,(y) are given functions. In the elliptic case, where again y,, ydr Sc, Sd are constants and gc(x), gd(x) are given functions. In the hyperbolic and parabolic cases, ;he boundary conditions (3) are replaced by the conditions (4') and (4"), p. 57. It is sometimes possible to find a particular solution u* of equation (1) In particular, the functions p(x), q(x), i(x) and the differential operator M, have the same meaning as on p. 56. 104 THE EIGENFUNCTION METHOD satisfying the conditions (2), and then the substitution u = u* + v reduces the present problem to the homogeneous problem which can be solved by the Fourier method. The problem can also be solved easily in the case where only the differential equation (1) is inhomogeneous, but not the boundary conditions (2), so that fa =f, = 0. Then we can look for a solution in the form of an expansion with respect to the eigenfunctions Xn(x) of the homogeneous problem, i.e., the nontrivial solutions of the equation @X')+ (hr -q)X = 0 (5) satisfying the homogeneous boundary conditions Suppose the right-hand side of (1) can be expanded in a series with respect to the functions Xn(x), so that Then, after substituting (4) into (I), the problem reduces to the integration of the ordinary differential equation where the A, are the eigenvalues of the homogeneous problem. To determine the resulting constants of integration, we substitute (4) into (3) [or into equations (4'), (4"), p. 571, expand the functions on the right in terms of the eigenfunctions Xn(x), and then equate corresponding coefficients of the functions Xn(x). The general case of inhomogeneous boundary conditions can be reduced to the problem just considered (an inhomogeneous differential equation and homogeneous boundary conditions) by looking for a solution of the form u = u* + v, where u* is a sufficiently smooth function which satisfies the boundary conditions (2) but, unlike the case mentioned above, is not necessarily a solution of the differential equation. For example, if the boundary conditions are of the first kind, i.e., we can choose u* to be the following linear function of x: x-a b-x u* = -fb(Y) + -fa(Y). b-a b-a Similarly, if the boundary conditions are of the second kind, i.e., we can choose - 1 (x - a)' U* - - ---- 1 (b - x)' Y - 2 b--o 2 b-a fa(^), and so on. However, it should be noted that this method, involving as it does a function u* which is to a large extent arbitrary, is not always successful (for example, in cases where the boundary conditions are discontinuous). In fact, improper choice of u* [even such simple functions as (7) and (8)] can lead to great complication in later stages of the calculations. A more adequate method of solving inhomogeneous problems has been proposed by Grinberg (G4),2 and is free from the need to choose the function u* in each particular case (which sometimes requires great ingenuity). In Grinberg's method, we try to solve the inhomogeneous problem by again representing the solution as a series of the form (4), whose coefficients are given by the formula jabru x,,(x) dx - - cn , %(Y) = jabrx:(x) dx jabrx:(x) dx (9) in keeping with the general theory of expansion in series of orthogonal functions. Thus, to obtain a formal solution of the problem, we need only find the value of the integral 6,. This can be done by the following device: First we multiply equation (1) by Xn(x) and integrate the result from a to b. Then we integrate by parts twice, obtaining In cases where the boundary conditions are homogeneous and only the differential equation is inhomogeneous, Grinberg's method gives the same result as the classical method of solution. Taking account of equation (5) and the boundary conditions (2) and (6), we can write (10) in the form3 '46) p(a) Mviin - = Fn - - Xn(b)fD(~) f - Xn(a)fa(~) (I1) '% aa in terms of the eigenvalues A,, where F, ==~%Fx, a dx. Equation (1 1) serves to determine d,, since its right-hand side involves only known functions. The resulting constants of integration are found from the equations which result when the same method [i.e., multiplication by rX,(x), followed by integration from a to b] is applied to equation (3) [or to equations (47, (4'7, p. 571. The method just described can also be applied to problems of mathe- matical physics involving the Sturm-Liouville problem with singular end points (see p. 59), provided that the eigenvalue spectrum is discrete. More- over, the method can be extended to certain problems involving higher-order equations (see Probs. 236-241), or to problems where the solution depends on a larger number of variables. It should be pointed out that for inhomogeneous boundary conditions of the first kind, the series representing the solution will not be uniformly convergent near the end points of the interval (a, b).4 To improve the con- vergence, we can apply the methods ordinarily used in such cases.5 In the simplest problems, we can improve the convergence by separating out the slowly converging part of the series and summing it by using the tables given in Sec. 2 of the Mathematical Appendix (see p. 381). The problems in this chapter, as in the preceding one, are grouped into five sections, two on mechanics, two on heat conduction (including a problem on diffusion), and one on electricity and magnetism. Problems involving coordinate systems more complicated than rectangular or polar coordi- nates (both cylindrical and spherical) will be deferred until Chap. 7. Problems with concentrated sources are usually regarded as limiting cases In the case of boundary conditions of the first kind (a, = a, = O), the right-hand side of (I I) should be replaced by * If the boundary conditions are inhomogeneous only at one end point x = a, this statenlent appties only at x = a. In the case of boundary conditions of the second kind, the series representing the derivative a~i/ax exhibits similar behavior. See K1, Chap. 1, Sec. 5. Another method, of a completely general character, is given by Grinberg ((35, Chap. 12). of the corresponding problems with distributed sources; this greatly simplifies the calculations, allowing us to write the solutions in compact and symmetric form. For example, the field due to a linear oscillator inside a cylindrical resonator can easily be solved in this way (see Prob. 256), whereas the usual method of solution (which involves subtracting out the singularity) leads to very complicated calculations. In the case of problems with inhomogeneous boundary conditions, the choice of a method of solution is left to the reader, although we are of the opinion that in such cases, Grinberg's method has indisputable methodolo- gical advantages. Of course, by proper choice of u*, certain problems can be solved quite easily, without recourse to this method. As a r~~le, the answers are given in the form of series, obtained after improving convergence, or in closed form. In some cases, the solution is given in two forms, corresponding to expansions in functions of each of the two independent variables. I. Mechanics: Vibrating Systems 207. A string of length I with fastened ends vibrates under the action of a uniformly distributed pulsating load q sin wt. Describe the vibrations, assum- ing that the string is at rest at the time t = 0. Ans. sin (2n + 1)mt - (2n + l)xvsin wt sin (2n + 1)rx 4qul 1 wl U(X, t) = -C 1 (2n + 1)' ' X%T,=~ O<x< where u = JTI~, T is the tension and p is the linear density of the string. 208. Solve the preceding problem, assuming that the pulsating load acts only on the section a < x < b of the string. Ans. 209. Study the vibrations of a string due to a concentrated pulsating load A sin wt applied at the time t = 0 to an arbitrary point x = c of the string. PROB. 210 Ans. nnvt nnv nxc nxx sin - - - sin wl sin - sin -- 1 w 1 1 I u(x, t) = - n Hint. Pass to the limit in the solution of Prob. 208. *210. Find the general solution of the problem of a vibrating string under the action of an external load q(x, t), assuming that the string is at rest at the time t = 0. Ans. 20 1 . nnx nnt U(X, t) = - 2 - sm -[sin nxv(t - diJLq(t, i) sin - dt, XT n=l n 1 1 o 1 211. Solve Prob. 106 on the longitudinal .solved by another method in Chap. 4. O<x<l. oscillations of a rod, which was 201 (2n + 1)nvt 2A1 " sin at - U(X, t) = - 2 (- l)n (2n + 1)nv Sin (2n + 1)nx ES n-0 (2n + 1)~ 2 1 212. Investigate the vertical longitudinal oscillations of a rod of length 1 suspended from the end x = 0 under the action of its own weight, subject to zero initial conditions. Ans. 16g/Z " 1 - cos [(2n + l)nvt/2l] sin (2n + 1)xx U(X, t) = ----I n3v2 n=O (2n + 2 1 where g is the acceleration of gravity, E is Young's modulus, p is the density, and v = 4%. 213. Investigate the longitudinal oscillations of the pyramid-shaped cantilever of square cross section shown in Figure 45, due to a force A sin at applied at the time t = 0 to its free end. To verify that the two forms of the solution given in the answers to Probs. 106 and 21 1 coincide, use the expansion sin wt -x YnVt YnV sin - - - sin wt YnX sin - A v 2 1 wl sin yn 1 U(X, 1) = 2awE(b - x tan a) ,=I Yn sin 2yn1 I-- 2~n where the yn are consecutive positive roots of the equation 214. An inhomogeneous rod con- f---Wj p = sisting of two sections made from different materials is clamped at one a end and is initially at rest. Find the FIGURE 46 longitudinal oscillations which result if a constant force P is applied to the free end of the rod (see Figure 46). a: " sin y, u(x, 1) = 2P, 2,- (1 - cos y,t)X,(x) v1 Y, plS1al cos2 (yna2vlla1v,) + p2Szaz sin2 y, ' where the y, are consecutive positive roots of the equation yaz4 v1EzS2 tan y = vzE1Sl cot -, "1'4 the two sections have Young's moduli Ei, cross-sectional areas Si and -- densities pi (i = 1, 2), and vi = 1 10 THE EIGENFUNCTION METHOD PROB. 215 215. A beam of length I, simply supported at its ends, is originally in a state of equilibrium. Investigate the transverse oscillations of the beam after applying an arbitrary load, uniformly distributed over the section XI < x < x,. A ns. nxx2 sin (nnxll) U(X, t) = - 1 n3 x lq(3 sin n2x2a2(t - T) l2 dz, where a2 = JEJ/~s, E is Young's modulus, J the moment of inertia of a cross section, p the density and S the cross-sectional area af the beam. 216. Solve the preceding problem, assuming that a) the load is uniformly distributed over the whole length of the beam and is a periodic function of time q(t) = q sin wt; b) a concentrated pulsating force A sin wt is applied to the point x = c of the beam. Ans. a) (2n + 1)2~2a2t - (2, + 1)2x2a2 sin wt sin (2n + 1)xx sin 412a2q f l2 w12 u(x, 1) = - 1. X'~EJ ,L=O I - [(Zn + l)2x2a2]2 (2n + 1)3 ' w12 *217. Find the transverse oscillations of a beam -1 < x < 1 with clamped ends under the action of a pulsating' force q sin at, uniformly distributed over the whole length of the beam, assuming that the beam is at rest before the load is applied. Ans. a sin wt pROB. 220 THE EIGENFUNCTION METHOD I I I 1- where a2 = v EJ/pS, in the notation of Prob. 215, and the y, are consecutive positive roots of the equation tan y + tanh y = 0. 218. Solve the preceding problem for the case where the external load is a concentrated force A sin wt applied to the center of the beam. Ans. y2aZt sin "- - - sin wt Ala2 " cosh y, - cosy, l2 w12 u(x, t) = - t: , 2wEJ .=I y, cos y, cosh2 y, Hint. First replace the concentrated load by a load uniformly distributed over the section -E < x < E of the beam, and then take the limit as E + 0. 219. Solve Prob. 217 for a beam 0 < x < 1 if the end x = 0 is simply supported, while the end x = 1 is clamped. Ans. q12a2 * sinh y, - 2 cosh y, sin y, + sm y, U(X, t) = - 2 wEJ ,=I Yz sinh2 y, sin2 y, y2a2t sin "- - - sin wt X l2 wL2 Xn(x), where YnX X,(x) = sinh y, sin - - 1 YnX sin y, sinh - , 1 and the y, are consecutive positive roots of the equation tan y = tanh y. 220. A concentrated force P is applied to the free end of a cantilever initially in equilibrium (see Figure 47). Investigate the resulting transverse oscillations, assuming that the force does not change subsequently. Ans. x 2p13 " 1 - cos (yta2t/12) U(X, t) = -2 Xn(x), EJ .=, yk(sinl1 y, + sin y,) FIGURE 47 1 12 THE EIGENFUNCTION METHOD PROB. 221 where X,(x) = (cosh y, + cos y,) - (sinh y, + sin y,) cosh - - ( ';" Y ,*) cos - , I and the y, are consecutive positive roots of the equation cosh y cos y = - 1. 221. Solve the preceding problem for the case where the force is a periodic function of time P = A sin wt. Ans. *222. Solve Prob. 220 for the case where the force P = P(t) is arbitrary. 21a2 " U(X, t) = -2 Xn(x> /'P(~) sin yPa2(t - dr. EJ ,=I yP(sinh y, + sin y,) o l2 *223. Investigate the transverse oscillations of a beam of mass M clamped at the points x = 0 and x = 1, due to a concentrated pulsating load A sin at moving along the beam with constant velocity v. Assume that at the time t = 0, the beam is at rest and the moving load is at the point x = 0.7 Ans. n2n2a2t cos - - cos 1 + - at l2 3 n2n2a2t nnx - sin - . I 224. Investigate the vibrations of a circular membrane of radius a due to a load applied at the time t = 0, if the load is uniformly distributed with density q(t) over the circular ring r, < r < r,. Consider the special case q(t) = q sin wt. ' This is the problem of a locomotive moving along a railway bridge (see T2, Sec. 59). THE EIGENFUNCTION METHOD 1 13 Ans. where the y, are consecutive positive roots of the equation Jo(y) = 0, p is the surface density, T is the tension per unit length of the boundary of the membrane, and v = JG. In the special case, sin @ - sin at r2Jlr$) - rlJl(y) a w a u(r, t) = - YZJXY~) JO (y ) *225. Investigate the vibrations of a circular membrane of radius a due to a pulsating loadp sin wt applied at the time t = 0 along the circumference of a circle of radius b < a. Ans. Hint. Replace the load by a load distributed with constant density over the area of the ring b - E < r < b + E, and then take the limit as E + 0. 226. A circular elastic plate of radius a, clamped along its boundary, begins to oscillate under the action of a suddenly applied pulsating load q sin wt, uniformly distributed over the area of the plate. Fird the resulting transverse oscillations. Ans. is a linear combination of cylinder functions, the y, are consecutive positive 1 14 THE BIGENFUNCTION METHOD PROB. 227 roots of the equation Ri(a) = 0, D is the flexural rigidity, h the thickness and p the density of the plate, and b2 = JDlph. *227. Solve the preceding problem, assuming that the oscillations are due to a concentrated pulsating force A sin wt applied at the center of the plate (oscillations of the diaphragm of a loudspeaker). Ans. Hint. Replace the concentrated load by a load distributed over a disk of small radius E, and then take the limit E -+ 0. 2. Mechanics: Statics of Deformable Media 228. Find the deflection of a rectangular membrane -a < x g a, -b g y g b due to a load uniformly distributed with density q over the rectangle -c < x < c, -d < y < d forming part of the membrane. Ans. 16qa2 sin [(2n + l)nc/2a] U(X, ~)ll~l<~ = -4 n T ,=, (2n + cosh (2n + l)ny (2n + lh(b - d) COS (2n + 1)nx cosh 2k 9 cosh (2n + 1)nb 2a 2a 16qa2 sin [(2n + l)nc/2a] sinh [(2n + l)nd/2a] =-z n3T n=O (2n + 113 sinh (2n f l)n(b - IYI) 2a (2n + 1)nx X COS (2n + l)xb 2a 9 cosh 2a where T is the tension per unit length of the boundary of the membrane. pROB. 232 THE EIGENFUNCTION METHOD 1 15 229. Find the deflection of a uniformly loaded rectangular membrane (this is a special case of the preceding problem), and compare the answer with that found earlier in Prob. 131.8 Ans. 16qa2 " cosh [(2n + l)xy/2a] u(x3 Y) = Yy- 2 x T .., (:~:);y (l - cosh [(2n + l)rrb/2a] *230. Find the static deflection of a rectangular membrane under the action of a line load p uniformly distributed along an axis of symmetry (see Figure 48). Ans. 4pa " cosh [(2n+l)xy/2a] u(x,~)=T ~(1- x T .=, cosh [(2n+ l)xb/2a] X cos [(2n + l)xx/2a] (2n + (12) Another form of the answer is 231. Find the deflection of a circular membrane of radius a due to the action of a line load p uniformly distributed along a diameter. Ans. where the series can be summed easily. Hint. To solve the problem, replace the line load by a load uniformly distributed over the sector -E < cp < E, x - E < cp < x + E, and then take the limit as E + 0. 232. Investigate the twisting of a rod whose cross section is a semicircle of radius a. Calculate the tangential stresses T on the surface of the rod. To compare the two answers, use formula 16, p. 385. 1 16 THE EIGENFUNCTION METHOD Ans. PROB. 233 z---- 1 - 2 + TI+O = ~1~-n = -1 T r2 a-r where 0 is the angle of twist per unit length and G is the shear modulus. Hint. The sum of the series needed to represent the solution in closed form is found in the solution to Prob. 132. 233. Find the torsion function u(r, cp) for the twisting of a circular shaft of radius a weakened by a radial crack going from the surface of the shaft to its axis. Calculate the torsional rigidity C of the shaft. 234. Investigate the twisting of a rod whose cross section is a circular sector of radius a and vertex angle a. Ans. The torsion function is 235. Solve the preceding problem for a rod whose cross section is a "curvilinear rectangle" a < r g b, 0 < cp < a. FROB. 238 THE EIGENFUNCTION METHOD I 17 A ns. sin (2n + 1)ny 8b2 M u(r, 9) = - 2 236. A rectangular elastic plate 0 ,< x ,< a, -612 g y < 612 is simply supported along its boundary and loaded by a concentrated force P applied at the center of the plate. Find the deflection along the midline y = 0. Ans. sinh (2n + 1)xb (2n + 1)nb - pa (- 1)" a a sin (2n + 1)xx UJy=O = 4rre~ zo (2n + 1)' (2n + 1)nb a 7 cosh2 2a where D is the flexural rigidity of the plate. Hint. Replace the concentrated load by a load uniformly distributed over the rectangle and then take the limit as 6, E -+ 0. 237. Solve Prob. 134, using the method of this chapter. (2n + 1)zb tanh (2n + 1)nb 2a 2 cosh (2n + 1)nb j. 2a 238. A rectangular elastic plate with sides a and b is simply supported along the edges x = 0 and x = a and clamped along the edges y = f b/2. To reduce the solution to the form given in Prob. 134, use the formula Find the deflection of the plate under the action of a loadp applied along the midline x = 42. Ans. 2pa3 sin [(2n + l)xx/a] U(X, Y) = 7 C(-lIn n D n=O (2n + 114 (2n + 1)xb + (2n + l)~b cash (2n + l)nb]cosh (2n + ~)ZY sinh 2a 2a a sinh (2n + 1)nb + (2n + l)xb 239. A rectangular elastic plate, simply supported along its boundary, is acted upon by bending moments m uniformly distributed along two opposite edges (see Figure 49). Find the deflection of an arbitrary point of the X a - (1 - :) FIGURE 49 ROB. 242 THE EIGENFUNCTION METHOD I I9 *240. Solve the preceding problem, assuming that the edges y = &b/2 are clamped. (2n + l)nb + (2n + lhb cash (2n + cash (2n + l)ny - 22 2a 2a a n=O sinh (2n + 1)xb (2n + 1)nb a + a ](2n + 1)' sinh (2n + 1)nb + (2n + l)nb] (2n + 1)3 ) a a sin *241. Find the deflection of the center of a circular plate of radius a with a clamped boundary under the action of a line load p uniformly dis- tributed along one of its radii. Ans. 3. Heat Conduction: Nonstationary Problems *242. A slab is heated by a thermal current of constant density q flowing through the face x = 0 starting from the time t = 0, while the face x = a is held at temperature To. Find the subsequent temperature distribution in the slab, assuming that the initial temperature of the slab is zero. Ans. where k is the thermal conductivity, c the specific heat and p the density of the slab, and T = ktlcp. 120 THE EIGENFUNCTION METHOD PROB. 243 243. Solve the preceding problem, assuming that the face x = 0 is held at temperature T = f(~), while the other face x = a is held at temperature zero. Consider the special case f (7) = AT. Ans. 2x " z a T(x, t) = -2 n sin s. a2 n,l a In the special case. sin "T*] . a 244. Find the temperature distribution in a slab if the face x = 0 radiates heat into the surrounding medium according to Newton's law, while the other face x = a is held at the temperature T, equal to the initial temperature of the slab. A ns. where h is the heat exchange coefficient figuring in Newton's law, and the y, are consecutive positive roots of the equation Y tan y = - -. ah X 245. Find the temperature distribution in a con- ductor with the cross section shown in Figure 50, heated from the time t = 0 by a d-c current producing Joule heat with density Q. It is assumed that the initial temperature is zero, and that the loss of heat FIGURE 50 into the surrounding medium is described by Newton's law. A ns. 1 a, sin e-~%/a2 T(x,t)=-- 1 -- +- -x cos P), 2Qa11 k 4 ( ) Zah y"l + (in 2y,/2yn)] a where the y, are consecutive positive roots of the equation ah tan y = - . Y Hint. Unless a particular solution of the inhomogeneous equation is subtracted out first, the expansion 1 a sin yn cos YnX - , -a < x <a (sin 2yn/2y,)l a must be used to reduce the solution to the form given in the answer. p~oe. 249 THE EIGENFUNCTION METHOD 12 1 246. Find the temperature distribution T(r, t) in a cylinder of radius a whose surface temperature varies according to the law TI,=, = f (T), where T = ktlcp, assuming that the initial temperature of the cylinder is zero. Consider the special cases a) f (T) = AT; b) f (T) = A sin 07. Ans. where the y, are consecutive positive roots of the equation Jo(y) = 0. In the suecia1 cases, I m b) T(r, t) = A sin WT + 2aa2C Yn n=l (7: + a4u2)Jl(yn) Hint. Use formula 17, p. 385. *247. Find the temperature distribution in a cylindrical conductor of radius a heated from the time t = 0 by a d-c current producing Joule heat with density Q. It is assumed that the initial temperature distribution is zero and that the loss of heat from the surface of the cylinder is described by Newton's law. Ans. where the y, are consecutive positive roots of the equation YJl(Y) = allJo(y). 248. Solve Prob. 150, using the method of this chapter. Ans.1° 249. Find the temperature distribution in a cylinder of radius a in which heat is produced with volume density Q, assuming that the initial temperature lo To reduce the answer to the form given in Prob. 150, use formula 18, p. 385. 122 THE EIGENFUNCTION METHOD PR~B. 250 of the cylinder is zero and that heat flows out of the cylinder with surface density q. Ans. Q 29 2qa m e-~?,~/aa T(r,t)= - -- T+- 1 - 2- +-2 -,Io(?), (k ka) :;( 1') k y:~o(yn) where the y, are consecutive positive roots of the equation Jl(y) = 0. Hint. Use formula 19, p. 385. 250. The outer surface of a cylindrical pipe a < r < b is held at tem- perature TI,,, = f(.r), while the inner surface is held at temperature zero. Find the temperature distribution, assuming that the initial temperature is zero. Ans. where is a linear combination of Bessel functions of the first and second kinds, and the y, are consecutive positive roots of the equation R,(b) = 0. 251. Find the temperature distribution in a cylinder 0 < r < a, 0 < z g I, assuming that the initial temperature is zero, and that starting from the time t = 0, the face z = 1 of the cylinder is held at temperature To, while the rest of the surface is held at temperature zero. Ans. 2 * (-1 I(mr/) mxz T(r, 1, 1) = T,[- + ; 1 - sin - m IdmxalO 1 where Jo(x), J,(x) and I&) are Bessel functions, and the yn are consecutive positive roots of the equation J,(y) = 0. Hint. Make the boundary conditions homogeneous by setting PROB. 255 THE EIGENFUNCTION METHOD I23 252. Solve the preceding problem, assuming that the surface of the cylinder is held at temperature zero and that heat is produced inside the cylinder with density Q. Ans. T(r, r, t) = 8Qa2 f f Jo(ynr/a) 1 - e-[(ynla)2f(m"1L)21T sin (mnzll) nk ,=1,3,5 ,... .=I ynJdyn) Y: + (mxall)2 9 m where the yn are consecutive positive roots of the equation J,(y) = 0. 253. Find the temperature distribution in a sphere of radius a inside which heat is produced with density Q, starting from the time t = 0. It is assumed that the sphere is initially at temperature zero and that its surface is held at constant temperature zero. Ans. 254. Solve the preceding problem if a) heat flows out of the sphere with surface density q; b) heat is radiated into the surrounding medium accord- ing to Newton's law. Ans. Ynr sin - , a where the yn are consecutive positive roots of the equation tan y = y; where h is the heat exchange coefficient and the yn are consecutive positive roots of the equation Y tan y = - . 1 - ah 255. A diffusing substance enters a thin tube of length 1 with impermeable walls. Find the concentration distribution in the tube if the density with which the substance flows into the end x = 0 is a given function of time q(t). It is assumed that the initial concentration in the tube is zero and that the other end of the tube is joined to a vessel in which a given concentration C, is maintained. PROB. 256 Ans. - !. [e-~(~n+l)2na(i-~,1412 q(~) d~] cos (2n + 1)xx 1 21 ' where D is the diffusion coefficient. 4. Heat Conduction: Stationary Problems X 256. Find the stationary tempera- ture distribution in a bar of rectangular FIGURE 51 Ans. cross section, given the temperature distribution on its faces (see Figure 51). nxS T(x, y) = 2 3 (sinh "~YJ:~~(S) sin - dS a .=I a a where (sinh sinh ndb - Y) , Gn(q' = 1 nxy nn(b - q) sinh - sinh 257. Study the special case of the preceding problem corresponding to the boundary conditions "EY sinh - + (-1)" sinh - 2" a a nnx sin - nxb a n sinh - a Hint. Use formula 2, p. 384. 258. A heat current Q flows into a bar of rectangular cross section through two opposite faces and leaves the bar through the other two faces (see Figure 52). Find the stationary temperature distribution in the bar, assuming that both the incoming and the out- going currents are uniformly distributed over the faces. Ans. Q T(x, Y) = - [~(b - Y) - x(a - x)l, 2abk where k is the thermal conductivity. Hint. To obtain the solution in closed form, FIGURE 52 use formula 9, p. 385. 259. Solve the preceding problem for an arbitrary distribution of current density on the face, i.e., where the functions on the right satisfy the condition for the solvability of the Neumann problem. Ans. 2" + - 2 (COS~ nx(b - ~)l(~~(t) cos - nxt dt xk .-I a a -l[(-l)nja(q) - ~O(T)IG~(?, Y) dq] cos (nxxla) + const, n sinh (nxbla) where nxr) nx(b - y) cosh - cosh 9 ?<Y, a a Gn(% Y) = nny nx(b - r)) cosh - cosh , ri>y. a a 260. Two faces of a rectangular bar are thermally insulated, and the other two are held at temperature zero (see Figure 53). Find the stationary tem- perature distribution, assuming that heat is produced with density Q inside the bar. Ans. 16Qa2 " cosh [(2n + l)n~/2al T(x, Y) = - 2 (:n-:;)3(1 - cosh [(2n + i)nb/2a] n3k *261. Heat is produced with density Q in the bar shown in Figure 54. Find the stationary temperature distribution, assuming that a heat current of constant density q leaves the bar through the section 1x1 < c of the upper face, while the rest of the surface of the bar is thermally insulated. Ans. sin (nxcja) cosh (nnyla) cos F] n2 sinh (nnbla) a + const, (14) where k is the thermal conductivity. Another form of the solution, suitable for alb > 1, is a X2 - - y2 2 2ab2 " (- 1)" sinh [nx(a - c)jb] + n2 sinh (nxajb) 1x1 < C, 2ab22(- I)" sinh (nncjb) cash -- n2c ,,,, n2 sinh (nxalb) b b 1x1 > c. 262. Find the stationary temperature distribution in a conductor of rectangular cross section heated by a d-c current producing heat with density Q, if the surface of the conductor gives off heat according to Newton's law. THE EIGENFUNCTION METHOD 127 Ans. where h is the heat exchange coefficient, and the yn are consecutive positive roots of the equation ah tan y = - . Y 263. Find the stationary temperature distribution in a rectangular parallelepiped 0 G x g a, 0 G y G b, 0 G z G c, if the faces x = 0, y = 0, z = 0 are held at temperature zero, while the other faces have the temperature distribution 4"" mxt nxq ~(x, y, z) = - 2 2 jSinh ymnz jajbfc(t, 1) sin - sin - dc ab sinh ymnc o o a b - -.!- Lcmn(:, z) [(- 1)m m'rJb/,(q, <) sin dq Ymn a o b where + (- 1). %Ja/&, 0 sin o a sinh ymnC sinh y,,,(c - z), C G z, Gmn(C, z) = sinh ymnz sinh ym,(c - C), < z. 264. A heat current Q enters a bar of semicircular cross section through its plane face and leaves through the curved face (see Figure 55). Find the stationary temperature distribution in the bar, assuming that the incoming and outgoing currents have constant density. Ans. At/ 1 - (1 /2n)(~/a)~~-' 1 x cos 2ncp + const, An2 - 1 where k is the thermal conductivity. FIGURE 55 128 THE EIGENFUNCTION METHOD PROB. 265 265. Find the temperature distribution in a bar whose cross section is the "curvilinear rectangle" a < r < b, 0 < cp < a, given the following temperature distribution on the faces of the bar: nncp (-1). sinhnn(. - + sinh - x2 In (b/a) In @la) sin nnu n sinh .--- ln (0) Another form of the solution is [I - ($"'"I (3"'"- - (- I). ($)"'"I (y sin nv - n=I [(:)"/a- ($n'a] Cc 266. Find the stationary temperature distribution T(r, z) in a cylinder 0 < r < a, 0 g z G 1 with an arbitrary axially symmetric temperature distribution along its surface: T(r, z) = -2 --- I (lo (7) J'02cp(<) sin @ d< 1 ,=, Io(nnall) 1 PROB. 268 THE LIGENFUNCTION METHOD 129 where Io(x) and Ko(x) are cylinder functions of imaginary argument. Another form of the solution is (sinh sinh Y"(' - g z, a a where the yn are consecutive positive roots of the equation Jo(y) = 0. 267. A heat current Q enters a cylinder 0 ,< r < a, 0 ,< z < 1 through its ends and leaves through the lateral surface. Find the temperature distribution in the cylinder, assuming that the incoming and outgoing currents have constant density. Ans. where k is the thermal conductivity. 268. Find the temperature distribution in a cylinder 0 ,( r g a, - I ,< z < 1 inside which heat is produced with density Q, if the surface radiates heat into the surrounding medium according to Newton's law. Ans. 2~1~ sin y, T(r, z) = -2 k ,=I y;[1 + (sin ~Y,/~Y~)I 130 THE EIGENFUNCTION METHOD PROB. 269 where h is the heat exchange coefficient, Zo(x) and Il(x) are Bessel functions of imaginary argument, and the yn are consecutive positive roots of the equation hl tany=-. Y Another form of the solution is where the yn are consecutive positive roots of the equation *269. A thin wire heated by a d-c current producing Joule heat Q per unit length is placed inside a cylindrical object (see Figure 56). Find the temperature distribution in the object, assuming that the lateral surface of the cylinder is held at temperature zero, while the ends radiate heat into the surrounding medium according to Newton's law. Ans. x [I - ah cosh (y,z/a) yn sinh (ynl/a) + ah cosh (ynl/a) 56 where the y,, are consecutive positive roots of the equation JO(Y) = 0. Hint. Replace the line source by a source distributed over a cylinder of small radius E, and then take the limit as E --t 0. 270. Find the stationary temperature distribution T(r, 8) in a sphere of radius a, assuming that heat is produced with density Q inside the sphere, while the boundary condition involving a given function f (O), is satisfied on the surface of the sphere. THE EIGENFUNCTION METHOD 13 1 A ns. "2n+1 r + f 2 - (-J~,(cos e)J)(o)Pn(cos 8) sin 0 do. ah + n a 5. Electricity and Magnetism 271. Calculate the two-dimensional electrostatic field due to the elec- trodes shown in Figures 57(a) and 57(b). 4v " - - x ((-1)" sinh nx[a - J2(y - x)] 7~ n=l 2a 2a 2a n sinh nx 2 " cosh (nxyla) sin (nxxla) b) u(x, y) = V - + - 2 k x cosh (nnbla) n where u(x, y) is the electrostatic potential. *272. Find the electrostatic field in the electron-optical device shown in Figure 58." What is the distribution of potential along the axis of symmetry? By an electron-optical device (for example, a lens), we mean a system of conductors at given potentiais producing an electrostatic field used to govern the trajectories of charged particles. 132 THE EIGENFUNCT~ON METHOD PROB. 273 Ans. FIGURE 58 x cosh (2n + lIna e-(2n+l)nz/2b 26 273. Find the electrostatic potential u(x, y) inside a box of rectangular cross section 0 < x s a, 0 < y < b with grounded walls, due to a charged wire along the line x = x,, y =yo. Ans. nx(a - x,) nxy, m sinh sin - 4% Y) = 8q2 b b nxx nxy sinh - sin - , x < x,, n=l nna b b n sinh - b where q is the charge per unit length of the wire. Hint. Solve Poisson's equation, regarding the charge as uniformly dis- tributed over the rectangle xo - 6 < x < xo + 6, yo - E < y < yo + E, and then take the limit as 8,~ + 0. 274. Find the electrostatic field u(x, y, z) due to a charge at the point xo, yo, z0 inside a rectangular parallelepiped 0 s x s a, 0 s y G b, 0 G z g c with grounded walls. Ans. PROB. 276 THE EIGENFUNCTION METHOD 133 Hint. First assume that the charge is uniformly distributed over a small volume, and then pass to the limit. 275. A charged wire, with charge q per unit length, is placed inside a grounded metal box whose cross section is a "curvilinear rectangle" a < r < b, 0 < cp g a. Find the elxtrostatic potential u(r, cp) inside the box. Ans. r > ro, where r,, yo are the polar coordinates of the wire. 276. Examine the following special cases of the preceding problem: a) a = 0, b = m (charged wire inside a wedge); b) a = 0, b = m, a = ~TF (charged wire near the edge of a conducting half-plane) ; c) b = GO, a = TF, cpO = x/2 (charged wire over a plane with a semi- cylindrical boss). Ans. 134 THE EIGENFUNCTION METHOD PROB. 277 *277. Find the electrostatic field inside a grounded cylindrical shell 0 < r g a, 0 G z G 1 due to a charge q at the point r = 0, z = c. Ans. The electrostatic potential is 4q * sinh [yn(l - c)/al ynz Jdynrla) u(r, z) = - 2 sinh - --- , Z<C, a sinh (ynl/a) a Y~J:(Y~) 4q * sinh (~nc/a) yn(l - Z) Jo(Y~~/Q) , , c, u(r, z) = -2 sinh a sinh (ynl/a) a ~nJXyn) where the yn are consecutive positive roots of the equation Jo(y) = 0. Another form of the solution is 4q * Io(nna/l)Ko(nnr/l) - Ko(nna/l) Io(nxr/l) nnc nnz u(r, z) = -2 sin - sin - . 1 n=l Io(nxa/l) I 1 278. Find the electrostatic field inside a cylindrical shell 0 G r < a, 0 < z f 1 whose ends and lateral surface are at the potentials Vo, V, and V, respectively. Ans. The electrostatic potential is [I - (-1)"]V + (-l)nVL - V I (nnr/l) nnz 00 sin - . n=l n Io(nxa/l) 1 Another form of the solution is where Jo(x), Jl(x) and Io(x) are cylinder functions, and the y, are consecutive positive roots of the equation Jo(y) = 0. 279. Find the electrostatic potential along the axis of a cylmdrical shell 0 < r < a, 0 < z < 1 if the lateral surface is held at a given potential while the ends are held at potentials Vo = 0 and V, = V. Ans. Vz 2" Vl(-l)n nnc sin (nnzll) 4r=0 = - + [; - +JhJ sin d~] 1 .=I n IO(n44 ' where Io(x) is the Bessel function of imaginary argument. PROB. 282 THE EIGENFUNCTION METHOD 135 280. Examine the special cases of the preceding problem which correspond to the following potential distributions on the lateral surface of the cylinder: Ans. Vz 2 " sin (n~zll) . a> ulr=o = - + - 2 [(-~Y(v - Va) + Val 1 x nZO(ma/l) ' n~ (2k - 1)nx sin (nxzll) . (-1)nV+2sin-x~ksin n=l 2N ,=I 2N 1 nI,(nxa/l) ' cos (nrcc/l) sin (nnzll) X n=l n I,(nna/l) ' Comment. Case b corresponds to a piecewise constant potential, pro- duced in electronic practice by the use of a voltage divider. Case c is the problem of the distribution of electrostatic potential between two conducting cylindrical caps separated by a negligibly small space. 281. What potential distribution must be maintained along the lateral surface of the cylinder of Prob. 279 in order to obtain the distribution N ulr=o = v [T + 2 an sin F] n=l 1 along the axis, where the an are any given numbers? Ans. *282. Determine the electric field on the axis of the electron-optical lens shown in Figure 59, consisting of two cylinders at potentials V, if the potential distribution in the space between the cylinders is given approximately by the formula XZ ~1,~,-6<~<6 = Vsin - . 28 PROB. 284 Ans. El,=, = - V[l + 22 cos (nx811) cos (nnzll) 1 n=l 1 - (2nS/1)2 IO(n.rca/l) 283. Find the potential distribu- Ol--_u:V-- -1- 20 --u:O-- 4-m. to in the electron-optical device shown in Figure 60. FIGURE 60 Ans. where the yn are consecutive positive roots of the equation J&y) = 0. Hint. Use formula 17, p. 385. 284. Find the potential distribution in the electron-optical device shown in Figure 61, con- sisting of two semicylinders (with closed ends) at potentials u = 0 and u = V, separated by a negligibly small space. Ans. X TL m=l n=l m FIGURE 61 where Jm(x) is the Bessel function of order m, a is the radius and 1 the length of the semicylinders, and the y,, are consecutive positive roots of the equation Jm(y) = 0. ROB. 287 THE EIGENFUNCTION METHOD 137 285. Find the distribution of d-c current in a thin conducting sheet, if a current J enters and leaves via point electrodes applied at the points (&c, 0) [see Figure 62].12 fY Ans. The potential of the current distribution is 25 " sin [(2n + l)xc/2a] 4% Y) = - - C noh n=o (2n + 1) sinh [(2n + l)xb/2a] x cosh (2n + l)x(b - lul) sin (2n + 11~~ + const, 2a 2a where h is the thickness and a the conductivity of the sheet. Hint. Regard the current as distributed over two small rectangles, and then pass to the limit. 286. Find the distribution of d-c current in a thin conducting disk of radius a, if a current J enters and leaves via point electrodes applied at the points r = b, cp = 0 and r = b, cp = x (b < a). Ans. Hint. To represent the solution in closed form, use the expansion 287. Find the distribution of d-c current in a thin cylindrical shell of radius a, if a current J enters and leaves via point electrodes applied at the points (a, -x/2,0) and (a, x/2,O) [see Figure 631. la The differential equation for the potential of the current distribution in a thin conducting shell is given in Prob. 21. 138 THE EIGENFUNCTION METHOD Ans. where h is the thickness and a the conductivity of the shell. 288. Solve Prob. 287 for the limiting case of a cylinder of infinite length. Ans. ---+--- J In cosh (zla) - sin cp u(cp, 2) = - 2xah cash (zla) + sin cp *289. A thin conducting shell of hemispher- ical shape lies on a plane base, made of a good conductor (see Figure 64). Find the distribu- tion of d-c current in the shell, assuming that a FIGURE 63 current J enters the shell by an electrode applied to the hemisphere at the point r = a, 8 = O,, cp = 0, while the current leaves through the rim of the hemisphere (in contact with the plane). Ans. 0 0 0 1 - 2 tan 2 tan - cos cp + tan2 tan2 - ~(e, cp) = A- ln 2 2 2 2 4xoh e 0 0 2 00 tan2- - 2 tan -2 tan - cos cp + tan - 2 2 2 2 Hint. Introduce tan (812) as a new independent variable. 290. Suppose an infinite slab of conductivity a contains a line current source (see Figure 65), from which a current J per unit length flows into the slab. Find the distribution of current in the slab, assuming that the slab is surrounded by a nonconducting medium. ROB 293 THE EIGENFUNCTION METHOD 139 Ans. The potential of the current field is 291. Find the voltage distribution in a lossless transmission line of length 1, if the end x = 0 is connected at the time t = 0 to a source of variable e.4.f. Eecat and the end x = I is kept open. It is assumed that the current and voltage in the line are initially zero. Ans. sin (2n + l)mt (2n + 1)m 2Eu " 2 1 21a 21 + U(X, t) = -2 a1 ,,=, (2n + 1)xx x sin 2 1 9 where L and C are the self-inductance and capacitance of the line per unit length, and u = l/JZ is the velocity of wave propagation along the line. 292. One end x = 0 of a transmission line of length I with parameters L, C and R is connected to a source of constant e.m.f. E, while the other end x = I is connected to a resistance R,. Find the voltage in the line if the load Ro is suddenly disconnected. Ans. (2n + 1)xx (- l), sin 21 X (2n + 1)' , where 293. Find the steady-state electromagnetic oscillations in conducting waveguide whose cross section is a rectangle 21 J a perfectly O<x<a, 140 THE EIGENFUNCTION METHOD PROB. 294 0 G y < b, assuming that the oscillations are excited by an infinite line current source J = Jo sin wt passing through the point (x,, yo). Ans. The complex amplitude of the vector potential of the electro- magnetic field is 8nJ0 " sin sinh (b - yo) a A = A,(x, y) = - 2 J(!!?J- k2 Sinh J (4)"- k2 b a nnx xsinh J(7)P-/czysin-, a ~<y<y,,, where 0 G x < a and k = o/c is the wave nunlber. Hint. Integrate the inhomogeneous wave equation for A, assuming that the current J is uniformly distributed over a rectangle whose dimensions are then made to approach zero. 294. Find the electromagnetic field due to an infinite linear current source J = Jo sin wt placed between two parallel perfectly conducting planes (see Figure 66). Ans. The complex amplitude of the vector po- tential of the electromagnetic field is 295. Find the steady-state electromagnetic oscillations in a perfectly conducting waveguide whose cross section is a circular sector 0 < r < a, 0 < cp < u, assuming that the oscillations are excited by an infinite line current source J = Jo sin wt passing through the point r = r,, cp = yo. Ans. The vector potential of the electromagnetic fieId is A = Im{Aoe'at), where J (kr) nwo sin - , x LL- sin - r < To, J&a) a a no nw% sin 9 , x - sin - r > ro, Jn(ka) a a k = w/c is the wave number, and Jn(x), Ht)(x), H?)(X) are cylinder functions. *296. Find the electromagnetic oscillations in a cylindrical resonator 0 < r < a, -I < z < 1 excited by a dipole of moment P located at the origin of coordinates and directed along the z-axis. Ans. The complex amplitude of the z-component of the vector potential is given by the Fourier expansion where the coefficients an have the values k is the wave number, and Io(x), Ko(x) are Bessel functions of imaginary argument. 297. Find the electromagnetic field in an infinite cylindrical waveguide with perfectly conducting walls, assuming that the source of the oscillations is a current Jsin wt in a coil of given dimensions, with a single uniformly wound layer (see Figure 67). Ans. The complex amplitudes of the compo- nents of the electromagnetic field are E, = E, = 0, a h -a,, sinh e , "i 2 z > h/2, c aE, la = -- Hz -- (r), H,= 0, iw az iw r ar 142 THE EIGENFUNCTION METHOD PROB. 297 where J&x) and Jl(x) are Bessel functions, k is the wave number, c the velocity of light, N the number of turns in the coil, and the y, are consecutive positive roots of the equation Jl(y) = 0. References Grinberg (G4, G5, G6), Morse and Feshbach (M9), Tikhonov and Samarski (Tl). INTEGRAL TRANSFORMS If application of the Fourier method to a given problem leads to a set of particular solutions depending continuously on some real or complex parameter, we say that the problem has a continuous spectrum.l Character- istically, the solution of a problem of this kind is constructed from appropriate particular solutions by integrating with respect to the parameter, i.e., the solution takes the form of an integral expansion involving the eigenfunctions (the continuous analogue of the series expansions considered in Chaps. 4 and 5).2 Problems with continuous spectra are encountered in all branches of mathematical physics, and can often be solved by the method of integral transforms, to which the present chapter is devoted. We begin by reminding the reader of the necessary background information. By an integral transform of a function f (x), defined in an interval (a, m), we mean an expression of the form where a and c are real numbers (the value - m is allowed), and K is a function called the kernel of the transform. More generally, we allow K to depend on a complex parameter p = a + ir varying over some region D of the complex As a rule, such problems involve unbounded domains. a The theory of integral expansions has undergone considerable development in recent years (see e.g., Al, L13, L14, L15, T6, T7 and S6, Vol. V. We also mention the classic paper by Weyl (W7). 143 144 INTEGRAL TRANSFORMS plane. Then (1) is replaced by /(PI =jam f (x)~(x, P) dx, P E D. Examples of transformations of type (1): 1. The Fourier transform 2. The Fourier cosine transform 3. The Fourier sine transform K(x, 7) = - sin TX, J: 4. The Hankel transform K(x, T) = xJ~(Tx), a = 0, c = 0, where JY(x) is the Bessel function of the first kind of order v > -8. 5. The transform where K,(x) is Macdonald's cylinder f~nction.~ 6. The Mehler-Fock transform K(x, .c) = P-M+i+(~), a = 1, c = 0, where Pv(x) is the Legendre function of the first kind. Examples of transformations of type (2): 7. The Laplace transform K(x, p) = ecPX, a = 0, where D is the half-plane lying to the right of some line a = a, par- allel to the imaginary axis. 8. The Mellin transform K(x, p) = xB-l, a = 0, where D is the strip between the parallel lines a = el and a = a,. The second expression for K(x,T) leads to a more symmetric inversion formula [see formula (21), p. 1951. INTEGRAL TRANSFORMS 145 Provided that the function f (x) belongs to an appropriate class (depending on the integral transform in question), we can express f(x) in terms of its integral transform by using a suitable inversion formula, which for transforms of type (1) takes the form Here M(x, r) is a suitable function defined in the region a < x < CO, c < r < co and called the kernel of the inverse tranqorm. In the case of the transforms 1-6 just enumerated, we have 2r sinh XT Ki,(x) 27 sinh x~ Ki,(x) 5. M(x, r) = or 2 4. ' x2 x 6. M(x, r) = r tanh n~P-~+~,(x), In the case of transforms of type (2), the inversion formula takes the form where M(x, p) is the kernel of the inverse transform, defined for all x in the interval (a, co) and p in the region D, while F is a suitable path of integration contained in D. For example, for the Laplace transform, while for the Mellin transform. In both cases, F is a straight line parallel to the imaginary axis and lying in the region D. We now turn to the integral transform method for solving partial differ- ential equations. The basic idea is to look for some integral transform zi of the solution, rather than for u itself, deferring the calculation of u until the end of the problem. In many cases, we can choose the kernel K in such a way that the original equation for u is transformed int0.a simpler equation for ti, with one less independent variable. Of course, the extra conditions on the function u are transformed into corresponding conditions on its integral transform, but the conditions involving the behavior of u as x + a, x -t w are automatically taken into account when transforming the original equation for u. The integral transform method has many advantages, e.g., it is applicable to both homogeneous and inhomogeneous problems, it simplifies calculations and singles out the purely computational part of the solution, it allows us to construct an operational calculus for a given kernel by using tables of direct and inverse transforms of the functions most commonly encountered in the applications, and so on.4 The present chapter is devoted to the solution of problems with con- tinuous spectra by writing the solutions as integral expansions involving suitable functions or by using the method of integral transform^.^ The problems are not classified by physical content, but rather by the particular transform used. There are five sections, the first on the Fourier integral and the Fourier transform, the second on Hankel's expansion and the related transform, the third and fourth on the Laplace and Mellin transforms, and the fifth on expansions with respect to cylinder functions of imaginary arg~ment.~ Many of the more difficult problems are equipped with solutions. I. The Fourier Transform Given a real function f(x), defined in the interval (- co, a), suppose that 1. f(x) is piecewise continuous and of bounded variation in every finite subinterval [a, b], where -w < a < b < co;' 2. The integral is finite. Although the literature on the application of integral transforms to physical problems emphasizes the Laplace and Fourier transforms, a number of works have appeared in recent years on the application of various other integral transforms (see e.g., G5, H3, K3, L8, L10, S8, S9, S10, T8). AS already noted, every integral expansion of the form (1) or (2) is accompanied by an inversion formula of the form (3) or (4), and conversely, and hence the distinction between the method of integral expansions and that of integral transforms is purely formal. Thus problems on the Fourier integral will be grouped with those on Fourier transforms, problems on Hankel's integral formula with those on Hankel transforms, and so on. Other integral expansions and transforms will be found in Chap. 7, which is concerned with the method of curvilinear coordinates. ' In particular, this condition is satisfied if f(x) is piecewise smooth in [a, b], or if f(x) satisfies so-called Dirichlef condifions in [a, b] (see W8, p. 161). INTEGRAL TRANSFORMS 147 Then f (x) satisfies the Fourier integral theorem where, if f(x) has a jump discontinuity at the point x = c, the left-hand side should be replaced by the sum (see T5, p. 13). Formula (5) is valid under other conditions (see T5, Chap. l), and can be written in the alternative form j(x) = A Jm [cos hxJ_",j(~) cos AS d~ + sin hxJ:f(t) sin AS dt] dl, X 0 The Fourier transform of a function satisfying the above conditions is defined as Then, according to formula (9, the inverse of (6) is given by Formulas (5)-(7) play an important role in solving a wide variety of physical problems, in particular, boundary value problems for the Laplace and Helmholtz equations involving infinite strips, infinite cylinders, etc. In general, the application of these formulas is called for in problems leading to integration of the equation where L is a linear differential operator which does not contain x, and f (x, . . .) is a given function. Besides the formulas already written, many problems of mathematical physics involve the application of the Fourier sine and cosine integrals j(x) = 2 Jmcos Ax dh/owf (s) cos AS dS, 0 < x < m, X 0 (9) 148 INTEGRAL TRANSFORMS PROB. 298 valid for functions obeying the obvious analogues of the above conditions, i.e., such that 1. f (x) is piecewise continuous and of bounded variation in every finite subinterval [a, b], where 0 < a < b < co; 2. The integral is finite. The analogues of (6) and (7) are then - L(A) =J? J Wf (x) sin hr dx, f(x) =Jz [%(A) sin hr dl, (10) X 0 7C 0 Formulas (8)-(11) are encountered in solving boundary value problems for the Laplace and Helmholtz equations involving half-strips, semi-infinite cylinders, etc. The problems which follow are taken from various branches of physics, and are all susceptible to solution by using expansions or transforms like formulas (5)-(11). 298. Solve the problem of the temperature distribution in an infinite rod, with the following special initial temperature distributions TI,=, = f (x): 1 To, I4 < xo, f'(x> = 0, 1x1 > xo; b) f (x) = ~~e-~~~~. where O(x) is the probability integral; Here k is the thermal conductivity, c the specific heat and p the density of the rod, and T = ktlcp' 299..A semi-infinite body bounded by the plane x = 0 has a given initial temperature distribution ROB. 302 INTEGRAL TRANSFORMS 149 Find the subsequent temperature distribution in the body, assuming that its boundary is held at temperature zero starting from the time t = 0. Apply the general result to the special case f (x) = To. where in the special case. v 300. Find the stationary temperature distribution T(x, y) in a semi- infinite body bounded by the plane y = 0, if the part 1x1 < a is held at tem- perature To, while the other part 1x1 > a is held at temperature zero (see Figure 68). Ans. -- 'I T(x, y) = I_o $, x where $ is the angle subtended by the segment -a < x < a, y = 0 at the X point P = (x, y). 301. Find the stationary tempera- FIGURE 68 ture distribution T(x, y) in a semi- infinite slab 0 < x < a, 0 < y < b if the face y = b is held at tempera- ture To, while the other two faces are held at temperature zero. Ans. Hint. Use the formula sin Ax I=?/ -dl, x>O. xo A 302. A heat current Q enters a semi-infinite body through the section 1x1 < a of its plane boundary (see Figure 69). Find the stationary tem- perature distribution in the body, assuming that the current is uniformly distributed and that the surface of the body radiates heat into the surrounding medium according to Newton's law. Ans. e-XU cos Ax dA, 150 INTEGRAL TRANSFORMS PROB. 303 where k is the thermal conductivity of the body and h is its heat exchange coefficient. In particular, the temperature of the part of the surface 1x1 < a has the representation in closed form 1 TI,=^,,^,.^ = q[l - cos ah cos xh + - {Si [(a + x)h] cos (a + x)h 2kah x + Si [(a - x)h] cos (a - x)h 7 - Ci [(a + x)h] sin (a + x)h - Ci [(a - x)h] sin (a - x)h) where Si (x) and Ci (x) are the sine and cosine integrals. *303. Solve the two-dimensional stationary heat conduction problem for a quadrant of thermal conductivity k (see Figure 70), if the face y = 0 is held at temperature zero, while the other face is covered by a thermal insulator except for the section 0 < y < b through which heat flows with constant density q. Find the distribution of heat current through the face y = 0. Ans. T(x, Y) = - sin Ay dl, q(x, 0) = X 304. Find the stationary temperature distribution in the quadrant x 2 0, y > 0 if the face y = 0 is held at temperature To, while the face x = 0 radiates heat into the surrounding medium according to Newton's law. Find the temperature distribution along the radiating face. Ans. sin Ay - Si (yh) cos yh + Ci (yh) sin yh , TI.-0 = ,[(, I where Si (x) and Ci (x) are the sine and cosine integrals. PROB. 307 INTEGRAL TRANSFORMS 15 1 Hint. Take the Fourier sine transform of the required function T(x, y), i.e., multiply the relevant differential equation by sin Ay and integrate with respect toy from 0 to cc. 305. The end of a semi-infinite cylinder 0 < r < a, 0 < z g cc is held at constant temperature To, while the lateral surface is held at temperature zero. Find the stationary temperature distribution in the cylinder, by expanding the required function in a Fourier sine integral with respect to z. Ans. I (Ar) sin Az T= To[i-IS n o e-dl], Io(Aa) A where Io(x) is the Bessel function of imaginary argument. Hint. Introduce a new unknown function u = T - To, and use the integral 306. Solve the preceding problem, assuming that a given temperature distribution TI,=, = f(r) is maintained on the end of the cylinder, while the lateral surface radiates heat into the surrounding medium according to Newton's law. where D(A) = AI,(Aa) + hIo(Aa), IJz) and K,(z) are Bessel functions of imaginary argument, and h is the heat exchange coefficient. 307. Find the stationary temperature distribution in a semi-infinite cylinder 0 < r < a, 0 < z < co if the lateral surface is maintained at the temperature TI,,, = f (z), while the end radiates heat into the surrounding medium. Ans. yA(z) = A cos AZ + h sin Az. 152 INTEGRAL TRANSFORMS PROB. 308 Hint. To solve the problem, use the following generalization of the Fourier integral theorem (see L13, p. 79): 308. Find the two-dimensional electrostatic potential in the half-space - oo < x < co, y z 0, if the potential distribution UI,=~ =f (x) is maintained on the plane y = 0. Ans. Hint. To reduce the solution to final form, use the integral a S:r-" cos bx dx = - a > 0. a2 + ba ' 309. Examine the special case of the preceding problem corresponding to the piecewise constant potential distribution in the plane y = 0 shown in Figure 7 1. Ans. where Vk is the value of the potential in the interval (x,-,, x,) and $, is the angle subtended by (x,,, xk) at the point P = (x, y). 310. Find the distribution of electrostatic potential in the planar electron- optical lens shown in Figure 72 (cf. Prob. 282).8 Ans. V2 + Vl V2 - Vl " cosh hy sin Ax u(x, Y) = - --S -- dh. 2 x o cosh hh A Note that the integrals representing the solutions of Probs. 310-31 1 can be expressed in terms of elementary functions. PROB. 3 12 INTEGRAL TRANSFORMS 153 Hint. Subtract out the particular solution +(V2 + Vl) of Laplace's equation, and then use the expansion 311. Find the distribution of electrostatic potential on the axis of the planar electron-optical lens shown in Figure 73. Ans. " sin Aa cosh Ay U(X, y) = V2 + 2- - - cos Ax dh. x - ''1 A cosh Ah 312. A thin charged wire of charge q per unit length is placed between two parallel conducting planes (see Fig- ure 74). Find the resulting distribution of electrostatic potential, and also the density of charge on the planes y = 0 and y = h. Ans. The potential distribution is " sinh A(h - a) 4x9 Y) = 44J A Sinh Ah x sinh Ay cos Ax dh, where the corresponding formula for y > a is obtained by permuting y and a. The charge density on the planes is 1 y=o, cosh (xxlh) - cos (nalh) - 1 y=h. cash (xxlh) + cos (nalh) Hint. First assume that the charge is uniformly distributed over the rectangle -8 < x < 8, a - c < y < a + E, and then take the limit as 8, E -+ 0. To solve the corresponding Poisson equation, take the Fourier cosine 154 INTEGRAL TRANSFORMS PROB. 313 transform of the unknown function, by multiplying the equation by cos Ax and integrating with respect to x from 0 to co. *313. Find the electrostatic field of a thin charged wire of charge q per unit length located near the plane interface between two dielectric slabs (see Figure 75). Ans. I where FIGURE 75 R;,, = x2 + (Y F a)2. Hint. To avoid any difficulties associated with the behavior of the logarithmic potential at infinity, set up a system of equations for the com- ponents of the electrostatic field. 314. Find the potential distribution in the electron-optical lens shown in Figure 76. Ans. where Io(x) is the Bessel function of imaginary argument. Hint. Reduce the problem to one with boundary conditions which are odd in the variable z, and then make a sine expansion, using the formula " sin Az I=?! -dl, r>O, KO A 315. Find the potential of the electrostatic field due to a point charge q placed on the axis of an infinite conducting cylinder of radius a. INTEGRAL TRANSFORMS 155 Ans. 4 w , Z) = -- - 241 lo(lir) cos Az dl, Jr2 + z2 x u I,(Aa) where Io(x) and Ko(x) are Bessel functions of imaginary argument. Hint. In the course of the calculations, use the following integral rep- resentation of Macdonald's function: cos Az , dz. ~0~1 =Jow jG 316. Find the distribution of electrostatic potential inside a conducting cone 0 < r < co, 0 < 0 < 0, due to a point charge q on its axis (see Figure 77). Ans. 4 u(r, 0) = Ja2 - 2ar cos e + r2 m J P-%+iT(-co~ 00) P-%+,(cos 0) cos r In - - Jra 0 p-%+,(cos 8,) ( koLy where P,(x) is the Legendre function of the first kind. Hint. Introduce new variables ' t I r I x = In - , u = r-'h. a To expand the source, use the following integral representation of the Legendre function: COS 7X P-x+iT(cos a) = x - cosh nr ~J2coshx-2cosa dx. I I 317. A point current source is placed on the axis of a cylindrical tube filled with a medium of conductivity a, and v 0 surrounded by a medium of conductivity o,. Find the potential FIGURE 77 of the current field in each medi~rn.~ Ans. Jw Ko(Ar) cos Az uZ(r, z) = - dh - 2x2a o olKo(Aa)Il(Aa) + oJ,(ha)K,(Aa) h ' This is the problem of "electrical coring" (see Fock's paper F2). 156 INTEGRAL TRANSFORMS PROB. 318 where In(x) and KJx) are Bessel functions of imaginary argument, J is the current emanating from the electrode, and a is the radius of the tube. 318. A line current J is placed between the boundary planes of two massive bodies made from iron . - of magnetic permeability p (see Figure 78). Find the magnetic field in the air space. Ans. (p - 1) e-" sinh 'Y cos Ax dA] x ~omcosh hb + p sinh Ab Y e-Ab cash 'y sin Ax d),] x fcOsh Ab + p sinh Ab Hint. Take the Fourier transform of the equations for.the.components of the magnetic field. 319. Solve the preceding problem, assuming that the iron )has infinite magnetic permeability. Ans. me-~b cash AY sin Ax A]. sinh Ab 320. A current J flows in a circular loop placed on a cylindrical core made from material of magnetic permeability p (see Figure 79). Find the distribution of magnetic field on the axis of the core (see Lebedev's paper L3). Ans. K,(Aa) cos Az dl, HI,=, = - - 4!JrIo(Aa)K,(Aa) + p ldAa) Ko(Aa) where f,,(x) and Kn(x) are Bessel functions of imaginary argu- ment, and c is the velocity of light. *321. Find the electromagnetic field radiated by a line cur- rent Joeiwt placed inside an ideally conducting shield of rectangu- lar cross section (see Figure 80). Investigate the limiting case b-t co. FIGURE 79 PROB. 323 INTEGRAL TRANSFORMS 157 Ans. 4ikJ"S a sin Aa sinh [Jk2 - k2 (b - Iyl)] sin Ax dA, E,(x, y) = - - - c 0 Jh2 - k2 cash JA2 - k2 b where k = w/c is the wave number and E, is the complex amplitude. For b 3 a, we have C - H%~J(x - + y2)l in terms of the Hankel function Hi2'(x), 322. Find the electromagnetic field produced in a cylindrical waveguide by a dipole of moment P placed at the origin and directed along the axis of the cylinder. Find an expression for the longitudinal component of the electric field. Jeiu+ A - .. + X which gives the familiar law for reflection by Ans. The complex amplitude of the z-component of the electric field is u where a is the radius of the cylinder, and Io(x), Ko(x) are Bessel functions of imaginary argument. a conducting plane of the radiation due to a line source. FIGURE 80 323. Find the steady-state oscillations produced by a point source of sound of frequency w placed on the axis of an infinite cylindrical tube with ideally reflecting walls. Ans. The velocity potential is sin (at - kJr2 + z2) ~(r, Z, t) = A Jm 2A iot K~(JA~ - k2 a) ,- +~m[;eS, - ~,(dh~ - k2 r) cos Az d~ , I~(JA' - k2 a) I where I,(x), Il(x) and K,(x) are Bessel functions of imaginary argument. Hint. Concerning the character of the singularity at the source point, see Prob. 85. 158 INTEGRAL TRANSFORMS PROB. 324 *324. Study the stress distribution in an elastic half-plane due to arbitrary stresses ovl,=O = f (x)r T~~(~=O = g(x) applied to its boundary. Ans. Hint. Take the Fourier transform of the system of equations from two-dimensional elasticity theory. 325. Examine the special case of the preceding problem obtained when a concentrated force P with components P, = 0 and P, = P is applied at the origin.1° Ans. 0 =- 2Px2y 5, = - 2pY3 x(x2 + y2)2 ) x(x2 + y2)2 ' T,, = - 2Pxy3 n(x2 + y2)' ' 326. Study the stress distribution in an elastic half-plane y > 0 due to a concentrated force P applied at the point x = 0, y = a and directed along the y-axis." Find an expression for the shear stress T,,. Ans. lo This is Flamant's problem, solved by inspection (without recourse to Fourier trans- forms) in courses on elasticity. l1 Another way of solving this problem is given in M6. PROB. 328 INTEGRAL TRANSFORMS 159 where R,,, = Jx2 + (Y f a)' and v is Poisson's ratio. Hint. Regard the force P as a distributed body force with components X and Y, and use the equations a2 a2 a27x, -2 (ox - va,) + -2 (a, - vo,) = 2(1 + v) - ay ax ax ay from two-dimensional elasticity theory. 327. Study the two-dimensional stress distribution in an elastic strip compressed by two concentrated forces P applied at the points x = 0, y = f b (see Figure 81). Find the normal stress erg along the axis of symmetry. Ans. a,I,=o = - "J " sinh Ab + Ab cosh Ab cos Ax dA. x o 2Ab + sinh 2Ab Hint. See Prob. 324. *328. A semi-infinite thin elastic plate, clamped along the edge y = 0, is loaded by a concentrated force P at the point (0, b). Find the bending moment M and the shear force N along the clamped edge (see Figure 82). Ans. Hint. Replace the concentrated force by a force uniformly distributed over the rectangle -6 < x < 8, b - E < y < b + E, and take the Fourier cosine transform with respect to the variable x of the differential equation for deflection of the plate. Then pass to the limit 6, E -+ 0. 160 INTEGRAL TRANSFORMS PROB. 329 329. A thin elastic plate, in the form of an infinite strip - w < x < m, 0 < y < b of width b, is clamped along its edges and loaded by a con- centrated force P at the point (0, a). Find the bending moment along the edge y = 0. Ans. -Ml,=o m cos Ax dh = 1 [a sinh Ab sinh h(b - a) - (b - a)hb sinh ha] x o sinh2 Ab - h2b2 330. Solve the preceding problem, assuming that the edges of the strip are simply supported and that the force is applied at the point (0, b/2). Find the deflection of the center of the strip due to the force. Ans. where D is the flexural rigidity of the plate. 2. The Hankel Transform Given a real function f (r), defined in the interval (0, a), suppose that 1. f (r) is piecewise continuous and of bounded variation in every finite subinterval [a, b], where 0 < a < b < a; 2. The integral jom J; I ~WI dr is finite. Then f (r) satisfies Hankel's integral theorem12 where JV(x) is the Bessel function of the first kind of order v > -4. If f(x) has a jump discontinuity at the point r = c, the left-hand side should be replaced by the sum Hf (c - 0) + f (c + 0)1 (see W4, p. 456 ff.). Formula (12) is one of the most important integral expansions encountered in mathematical physics. The Hankel transform of a function satisfying the above conditions is defined as ()=f(r)~(r)rdr, O<h<a. (13) la Sometimes called the Fourier-Bessel integral. Then, according to formula (12), the inverse of (13) is given by There is a generalization of formula (12), known as Weber's integral (see T6, p. 75) f(r) =Iw dh rf(p)f(p)pdp)p dp, a < r < m, (15) o J;(ha) + ha) involving the linear combination of Bessel functions of the first and second kinds (v > -+). A sufficient condition for validity of (15) is that f(r) be piecewise continuous and of bounded variation in every finite subinterval [a, PI, where a < a < P < w, and that the integral be finite. It should be noted that Weber's integral reduces to Hankel's integral'as a -> 0. Hankel's integral expansion and the Hankel transform can be used to solve a number of problems of mathematical physics, e.g., boundary value problems for the Laplace and Helmholtz equations involving half-spaces and regions bounded by parallel planes, certain problems of elasticity theory, etc.13 The problems that follow can be solved quite readily, as soon as one has acquired the necessary experience in handling Bessel functions. 331. Find the stationary temperature distribution in the half-space z > 0, if a given temperature distribution TI,=, = f (r) is maintained on the boundary z = 0. Examine the special case l3 In general, application of these formulas is called for in problems leading to in- tegration of the equation where L is a linear operator which does not contain r, and f(r, . . .) is a given function. Weber's expansion plays the same role for the interval a < r < co (see Probs. 335-337). In the special case, where Jo(x) and Jl(x) are Bessel functions. Hint. To evaluate the integral use the differential equation for the Bessel function. 332. Solve the preceding problem, assuming that the half-space is heated by a thermal current of constant density q, incident on the disk of radius a with center at the origin, while the rest of the boundary exchanges heat with the surrounding medium according to Newton's law. Ans. m e-A~ T(r, z) = - J,(ha)Jo(Ar) dh, k 0 h+h where h is the heat exchange coefficient. 333. A cylindrical rod of radius a, heated to temperature To, is intro- duced into an unbounded medium whose initial temperature is zero. Find the temperature distribution T(r, t), assuming that the medium and the rod have the same thermal conductivity k, specific heat c and density p. Ans. m T = Toa jo e-"~,(ha)~,(hr) dh, where 7. = ktlcp. *334. Examine the process of temperature equalization (in unbounded space) of an arbitrary axially symmetric initial temperature distribution T = fr, 0 < r < a. Ans. where Io(x) is the Bessel function of imaginary argument. Hint. To calculate the ooefficient in the Fourier-Bessel integral, use the formula (see W4, p. 395). *335. A cylindrical hole of radius a is drilled in an infinite body, and the walls of the hole are maintained at temperature To starting from the time t =O. Examine the evolution of the temperature distribution in the body, assuming that its initial temperature is zero. Ans. where PA(^) = Jo(Aa) YO(W - Yo(Aa)Jo(A& and Jo(x) and Yo(x) are Bessel functions of the first and second kinds. Hint. Set v = 0 in formula (15). 336. A cylindrical conductor of radius a heated by a d-c current passes through an infinite slab of width 2h (see Figure 83). Find +he stationary temperature distribution in the slab, assuming that the surface temperature of the conductor is To, while the faces of the slab have temperature zero. Ans. [ 2 1' ydr) cosh hz dA] T(r, z) = To 1 - - x o J;(Aa) + Y;(AU) cosh Ah A ' where cp,(r) has the same meaning as in the preceding problem. 337. The walls of a cylindrical hole terminating at the plane surface of an infinite body (see Figure 84) are held at a given tempevature To. Find the stationary temperature distribution in the body, assuming that it radiates 164 INTEGRAL TRANSFORMS PROB. 338 heat from its surface into the surrounding medium according to Newton's law. Ans. T(r,z)=To[l-2hJa %(r)e-"dA n o A(i + h)[~;(Aa) + Y;(A~)] 1 ' 338. Find the distribution of electrostatic potential in the space between two grounded plane electrodes z = a, due to a point charge q at the point r=O,z=O. Ans. e-,, cosh Az u(r, z) = - cosh Aa Jo(W dh, Jr2 -I- z2 in terms of the Bessel function Jo(x). Hint. Use the formula 339. Find the electrostatic field due to a point charge q located near the plane interface between two media with different dielectric constants (see Figure 85). Ans. Hint. To represent the solution in closed form, use the hint to Prob. 338. PROB. 343 INTEGRAL TRANSFORMS 165 340. Find the electrostatic field produced by two point charges +q and -9, between which there is a slab of material of dielectric constant E (see Figure 86). Calculate the field on the line joining the charges. Ans. - 20 [" ~e-""~) cosh Az dl lzl < h. -'Jo sinh Ab + E cash Ab 0-1 - -7 . I 8, wr,, JO sinh Ab + 6 cosh Ab * 341. A.d-c current J enters the ground through an electrode making contact with the earth's surface (z = 0) over the area of a disk of radius a. Find the current distribution in the earth, and examine the limiting case of a point contact. Ans. The potential of the current field is d A e-hz~l(~a)~o(~r) - , z > 0, h where Jo(x) and J,(x) are Bessel functions, and o is the conductivity of the earth. In the limiting case, J u(r,.z) = - 2x0 Jr2 + z2 ' 342. A point electrode carrying current J is placed on terrain consisting of two layers of different conductivities (see Figure 87). Calculate the potential of the current field on the earth's surface. Ans. J J a e-ha J~(A~) uIz,o = - + - (" - 02)l oi sinh Aa + o2 co~h Aa dh. 2x0,~- 2xol 343. Determine the electromagnetic field of a vertical radiator (antenna) placed at height h over the plane surface of the earth, assumed to be perfectly conducting (see Figure 88). 166 INTEGRAL TRANSFORMS PROB. 344 Ans. The z-component of the vector potential of the electromagnetic field is - (all other components vanish), where R = J(z - h)2 + r2, R = =(z + h)2 i- r2, w is the frequency of the oscillations, c is the velocity of light, P is the moment of the radiating dipole, and k = w/c. Hint. Use the expansion e-ikdm JW 344. Solve the preceding conductivity. problem, assuming that the earth has finite Ans. The vector potential of the electromagnetic field is where w J(w - 4noi)w k,=-, k2= C C in terms of the earth's dielectric constant E and conductivity o.14 345. Using the solution of Prob. 344, find an expression for the normal component of the electric field on the earth's surface for the case where the dipole is placed directly on the surface itself (h -, 0). Ans. l4 Details on the transformation of theseexpressions into a form suitable for calculation as well as an analysis of the corresponding physical picture of wave propagation, can be found in the specialized literature (see e.g., F6, Chap. 23, Sec. 1 and S14, Secs. 31-32). Hint. Use the Van der Pol substitution 346, Determine the electromagnetic field of a horizontal radiator located at height z A h above the plane surface of the earth, assumed to be a perfect conductor (see ( I 1 Figure 89). Ans. The vector potential of the electro- -Y magnetic field has the components - A, = A(r, z) = - FIGURE 89 A, = A, = 0. 347. Solve the preceding problem, assuming that the earth has finite conductivity. 2P A;) = - (k; - ki) cos cp C - - h2e-d~a-klah+ d~'-ka'z J1(hr) - - - dh. x r(k:Ji2 - k; + E: JAL kk:)(Jh2 - k: + Jh2 - k:) For further details, see F6, Chap. 23, Sec. 2 and S14, Sec. 33. 348. Find the magnetic field of a horizontal radiator lying on the plane surface of the earth, assumed to have dielectric constant E and conductivity 0. 168 INTEGRAL TRANSFORMS PROB. 349 Show that the magnetic field on the earth's surface can be expressed in terms of elementary functions. Ans. 349. Find the steady-state acoustic vibrations due to a disk-shaped piston of radius a inserted in an infinite screen and vibrating with velocity vo sin wt. Ans. The velocity potential is where k = w/c and Jo(x), Jl(x) are Bessel functions. 350. A concentrated normal force P is applied to the plane surface of a semi-infinite elastic body z > 0. Study the resulting stress distribution in the body, and find expressions for o, and r,,. Ans. where the force P is assumed to be applied at the point r = z = 0. Hint. Use the formulas expressing the stresses o, and r,, in terms of the biharmonic stress function (v is Poisson's ratio). Then expand the quantities Au and a2u/az2 in Hankel integrals (see also the solution of Prob. 351). In the boundary conditions, first replace the concentrated force P by a force uniformly distributed over a small disk of radius E, and then take the limit as E -+ 0. *351. Generalize the preceding problem to the case of a concentrated force P with components P, = P, = 0, P, = P, applied at an arbitrary interior point of the body (with coordinates r = 0, z = a). Find an expression for the stress (T,. Ans. 0 =- 3(z - a)3 (1 - 2v)(z - a) - - - + 8x(1 - v) id: dl) R: where R,,, = dr2 + (Z 7 Hint. Use the stress function corresponding to a concentrated force P (with components P, = P, = 0, F, = P) applied to the point r = z = 0 of an infinite elastic body (see T4, p. 355). 352. Study the transverse oscillations of an infinite elastic plate due to a concentrated force P(t) applied at the point r = z = 0 starting from the time t= 0. Ans. where D is the flexural rigidity and p the density of the plate, and Si(x) is the sine integral. 3. The Laplace Transform The Laplace transform is acknowledged to be the most effective tool for dealing with the nonstationary problems of mathematical physics. Since the whole subject has been thoroughly treated in the literature,16 we shall confine ourselves to a few brief remarks, mainly for reference purposes. Let f (t) be a real function defined in the interval (0, co) such that 1. f(t) is piecewise continuous in every finite subinterval [a, TI, where O<a<T<cO; 2. The product f(t)e-st is absolutely integrable on (0, w) for some suitable Dl > 0. Then the Laplace tranqorm of f(t) is defined by the formula where p = a + i~ is any complex number in the half-plane Rep > al.ls If it is also assumed that f (t) is of bounded variation in every finite subinterval l6 See the relevant books cited at the end of this chapter (p. 202). 'Taplace transforms can also be defined for functions satisfying weaker conditions. Note that the function f is an analytic function ofp in the domain Rep > a,. The values off in the rest of the complex plane can be determined by analytic continuation. [a, TI," then formula (16) can be inverted by using the Fourier-Mellin theorem where l? is a straight line parallel to the imaginary axis lying to the right of the line Rep = o, (see Figure 90). Conversely, (17) implies (16) if J'@) satisfies appropriate conditions. The application of the Laplace transform method is called for in nonstationary problems leading to integration of the equation where L is a linear differential operator which does not contain t, a and b are given constants, and f(t, . . .) is a given function. Its use allows FIGURE 90 us to eliminate the time t, thereby reducing the problem to the determination of a function fi satisfying a simpler equation. In particular, if the unknown function u depends only on one spatial variable (in addition to the time), the equation for t7 will be an ordinary differential equation. After finding t7, the problem can be solved by using the inversion formula (17), where the path of integration l? must be chosen in such a way that all the singular points of ii lie to the left of r. The actual calculation of the complex integral (17) can be carried out by various methods, the most important of which involve the use of Cauchy's theorem and residue theory, expansion in series, application of the convolution theorem, use of appropriate tables,18 etc. The variety of available methods makes it possible to obtain the solution of the problem quickly, in the form most suitable for understand- ing the physics of the situation and making subsequent numerical calculations. This constitutes the great advantage of the Laplace transform method, which is particularly suitable for studying wave propagation along transmission lines, physical problems with boundary conditions involving time derivatives (see Probs. 365, 367, 370), and so on. This section contains a variety of nonstationary problems, dealing first with heat conduction, then with electricity and magnetism, and finally with mechanics. Because of the abundance of specialized literature on Lapiace transforms, we have omitted the simplest problems belonging to these categories. At the end of the section, we give a few problems of a more l7 In particular, this condition is satisfied if f(t) is piecewise smooth in [a, TI, or if f(t) satisfies Dirichlet conditions in [a, TI. l8 The tables in E3 are particularly complete. PROB. 355 INTEGRAL TRANSFORMS 17 1 complicated nature (e.g., Probs. 391, 405, 406), to be solved by combining the Laplace transform with some other integral transform (e.g., the Fourier transform or the Hankel transform). 353. Starting from the time t = 0, the plane boundary of a semi-infinite body of thermal conductivity k, specific heat c and density p is maintained at the temperatureT I,=, = f (T), where T = ktlcp. Find the subsequent tem- perature distribution in the body, assuming that the initial temperature is zero. Ans. 354. Consider the following special cases of the preceding problem: a) f (7) = To; b) f(~) = AT; To, 0 < 7 < 70, d) f(r) = To sin WT. Ans. where @(x) is the probability integral. *355. Solve Prob. 353 for a given density q of heat current incident on the boundary (instead of a given surface temperature distribution). Examine the special caseslg a) q = go; b) q = go sin WT. Ans. IB In Case b, consider only the surface temperature. 172 INTEGRAL TRANSFORMS PROB. 356 where C(x) and S(x) are the Fresnel integrals, and k is the thermal con- ductivi ty. 356. Find the evolution in time of the temperature on the plane boundary of a semi-infinite body, if the density of heat current incident on the body is a given function of time q = q(~). Consider the special case q = const. Ans. In the special case, v Y "357, A semi-infinite body, heated to the initial temperature To, radiates heat from its plane boundary x = 0. Find the distribution of temperature in the body, assuming that the radiation obeys Newton's law and that the temperature of the surrounding medium is zero. Ans. where h is the heat exchange coefficient. Hint. To simplify the calculations, substitute the integral representation into the inversion formula, and then reverse the order of integration. 358. Starting from the time t = 0, a train of heat current pulses q = f(~) such that flows through the plane boundary of a semi-infinite body. Find the tem- perature distribution in the body after a large number of cycles, assuming that the initial temperature is zero and neglecting heat exchange between the surface of the body and the surrounding medium. Ans. For finite x, 359. Two semi-infinite bodies made from different materials, one heated to temperature To and the other held at temperature zero, are put into ROB. 362 INTEGRAL TRANSFORMS 173 contact starting from the time t = 0 (see Figure 91). Describe the subsequent equalization of temperature. Ans. T(x,t)=~[l+:B(&)], l+a x>O, 24 X where @(x) is the probability integral, and 360. The temperature distribution is maintained on the plane boundary of the half-space 0 < x < oo, --a < y < co, starting from the time t = 0. Solve the corresponding problem of heat conduction, assuming that the initial temperature equals zero. Ans. Hint. Take Laplace and Fourier transforms in succession. 361. Find the temperature distribution inside a body shaped like a quadrant (x > 0, y > 0), whose surface is held at temperature To starting from the time t = 0 (the initial temperature is assumed to be zero). Plot the corresponding isotherms. Ans. The result of the calculations is shown in Figure 92. 1 Hint. Look for a solution of the form FIGURE 92 T = T,[1 + u(x, t)v(j, t)], and then reduce the problem to Prob. 354, Case a. 362. Find the temperature distribution inside a body shaped like an octant (x > 0, y > 0, z > 0), whose surface is held at temperature To 174 INTEGRAL TRANSFORMS PROB. A,, starting from the time t = 0 (the initial temperature is assumed to be zero). 363. Find the temperature T(x, t) in a slab of finite thickness, if one face x = 0 is held at temperature To starting from the time t = 0, while the other x = a is held at temperature zero. It is assumed that the whole slab is initially at temperature zero. Give two forms of the solution, one suitable for large t, the other for small t. Ans. Hint. To obtain the second form of the solution, expand the Laplace transform of the desired function in ascending powers of the quantity ecx2/;. 364. Solve the preceding problem, assuming that a thermal current of constant density q is incident on the face x = a, while the face x = 0 radiates heat according to Newton's law. Ans. where the y, are consecutive positive roots of the equation Y cot y = - , ah and h is the heat exchange coefficient. 365. Solve Prob. 363 assuming that the face x = 0 is held at constant temperature To, while the face x = a is connected to a thermal capa~itance.~~ Derive expressions for the density of heat current on the faces of the slab. 2akTo 41.-a = -2 Yn e-~n2r/a2 a (1 + a + a2y2,) sin yl, , 20 By a "thermal capacitance" we mean a body in which any temperature drop can be neglected. In Probs. 365, 367, etc., C, denotes the amount of heat needed to raise the temperature of the body by 1 degree, referred to unit area, unit length, etc. FROB. 367 INTEGRAL TRANSFORMS 175 where the y, are consecutive positive roots of the equation co Cot y = uy, a = - cw Hint. The boundary condition at x = a has the form where Co is the thermal capacitance per unit area and k is the thermal conductivity. 366. Find the temperature distribu- O(T, tion in a slab -a < x < a in which, starting from the time t = 0, there is a process periodically producing heat according to the law shown in Figure 93. slab and the initial temperature are assumed to be zero. The temperature of the faces of the To 2To 3To 4To FIGURE 93 Ans. 16a2 * (- 1)" cos (~,A;x/a) 2 42 7z0 (zn + + e~os~n4/1al la ,[(I-4 cosh A,(x +a) cos An(x - a) + cosh A,(x -a) cos An(x +a) COS 2~2,~ cosh 2Ana + cos 2A,a + 4 sinh A,(x - a) sin An(x + a) + sinh An(x + a) sin An(x - a) cosh 2Ana + cos 2Ana 367. A thin cylindrical rod (probe), heated to temperature To, is inserted into the ground, in order to measure the ground's thermal properties. Describe how the temperature of the probe varies with time, assuming that the temperature drop inside the rod can be neglected (see T10). Ans. 176 INTEGRAL TRANSFORMS PROB. 368 where J,(x) and Y,(x) are Bessel functions of the first and second kinds, a = Co/2xa2cp, Co is the thermal capacitance of the probe per unit length, a is the radius of the probe, k is the thermal conductivity, c the specific heat and p the density of the ground, and T = ktlcp. Hint. Solve the heat conduction problem for the domain r > a with the boundary condition 368. Use the Laplace transform to solve Prob. 335, and then show that the two answers are equivalent. Ans. The equivalence of this result and the answer to ?rob. 335 follows from the expansion 369. Investigate the heating of a cylindrical cable if, starting from the time t = 0, heat is produced with density Q in the core of the cable, while its outer surface is held at temperature To (see Figure 94). Find the tem- perature of the core, neglecting any temperature drop inside the core. Ans. where is a linear combination of Bessel functions, and the y, are consecutive positive roots of the equation Rl(a) = 0. 370. Starting from the time t = 0, heat is produced with density Q in a cylindrical conductor of radius a. Find the temperature along the axis of the PROB. 372 INTEGRAL TRANSFORMS 177 conductor, assuming that heat leaves its surface by way of a thermal capaci- tance and the initial temperature is zero. Ans. where the yn are consecutive positive roots of the equation Co is the thermal capacitance per unit length, and a = C0/2rra2cp. *371. At the time t = 0, a cold cylinder of radius a is encased in a thin heated cylindrical sleeve covered on the outside by a thermally insulating layer (sze Figure 95). Find the temperature distri- bution in the cylinder, assuming that the initial temperatures of the cylinder and the sleeve are 0 and To, respectively, and neglecting any tem- perature drop inside the sleeve. Ans. T(r, 0 I 1 =To -+z 1 1 + - J,(Y,) (I + - + - FIGURE 95 2u 2a 2 where the yn are consecutive positive roots of the equation Co is the thermal capacitance of the sleeve per unit length, and u = Co/2xa2cp. 372. A diffusing substance is distributed in the half-space x > 0 with a given initial concentration Find the density of the substance through the boundary x = 0, assuming that the concentration on the boundary is maintained at zero starting from the time t = 0. Ans. where D is the diffusion coefficient. 178 INTEGRAL TRANSFORMS PROB. 373 373. Find the concentration distribution of a diffusing substance in the half-space x > 0 bounded by an impermeable wall, assuming that the initial concentration equals zero and that the substance is released with constant density Q in the layer 0 < x < a during a finite time interval T. Derive an expression for the concentration of the substance on the wall x = 0. Ans. 374. A substance diffuses outward through the lateral surface of an infinite cylinder of radius a into the surrounding medium, where the con- centration of the substance equals zero at the time t = 0. Find the subsequent concentration distribution, assuming that the substance flows out of the cylinder with constant density q. Derive a formula for the concentration of the substance on the surface of the cylinder. Ans. m 1 (1 - e-x2~i/h dx 179 r in terms of the Bessel functions J,(x) and Y,(x). "375. A diffusing substance emanates from a thin cylindrical tube of length I FIGURE 96 closed at one end, and enters the half- space z > 0 through an opening in the impermeable wall z = 0 (see Figure 96). Find the amount of substance inside the tube as a function of time, assuming that the flow of current is constant over a cross section of the tube and that the initial values of the concentration of the substance in the tube and in the half-space equal Co (per unit length) and 0, respectively. Aqs. ax -,t2/12 (I - cos -i)e X X 1 + sin2 x - 2 sin x sin pROB. 379 INTEGRAL TRANSFORMS 179 where a is the radius of the tube and M, is the initial amount of substance inside the tube (M, = C,I). 376. The end x = 0 of an infinite transmission line, with self-inductance L and capacitance C per unit length, is joined at the time t = 0 to a source of e.m.f. E = f(t). Find the voltage u(x, t) at every point of the line. Ans. where u = IIJZ is the velocity of wave propagation along the line. 377. Solve the preceding problem for the case of self-inductance L, capacitance C, resistance R and leakage conductance G per unit length, chosen to satisfy the relation RC = LC (a distortionless line). Am. 378. A condensor of capacitance C,, charged to the potential V, is discharged at the time t = 0 into an infinite line with parameters L and C. Find the distribution of current I(x, t) in the line. Ans. where Z = J~c is the wave resistance of the line, and u = 1ICJ. 379. The end x = 0 of an infinite line with self-inductance L, capacitance C and resistance R per unit length is connected at the time t = 0 to a source of constant e.m.f. Study the resulting process of propagation of a voltage wave along the line (see C2, p. 202). Ans. 180 INTEGRAL TRANSFORMS PROB. 380 where E is the size of the applied e.m.f., u == x = ! R/2L and I,(x) is the Bessel function of imag- $y{ inary 380. argument. A line of length 1 with parameters L and C is terminated at the end x = I by a resist- FIGURE 97 ance Ro (see Figure 97). Find the subsequent voltage in the load R,, assuming that the end x = 0 is suddenly connected at the time t = 0 to a source of constant e.m.f. E. Under what conditions is there no reflection of waves from the end of the line? Ans. O<t<T, where Z = J~c is the wave resistance of the line, and T = I/v is the time it takes a wave to go from one end of the line to the other. There is no wave reflection if the resistance Ro equals the wave resistance Z. 381. Solve the preceding problem, assuming that the line is terminated by a lumped capacitance Co rather than by a resistance R,. Derive an expression for the voltage across the capacitance in two forms: a) as a trigono- metric series; b) in closed form for the first few reflections. Ans. m a) sin Y n ,=, 2y, + sin 2y, T where the yn are consecutive positive roots of the equation and so on. 382. Solve Prob. 381 for the case where the line is terminated by a lumped inductance Lo. Ans. 1 -4 sin yn cos GI, a) ulx=c = EL + a ~y,, - sin 2yn T where the y, are consecutive positive roots of the equation Y tan y = - - 1L , g=--' u Lo ' and so on. 383. Write the general expression for the reflected waves in Prob. 382.21 Ans. where (2N - l)T < t < (2N + 1)T, N = 1, 2, 3, . . . , and L,(x) is the Laguerre polynomial, defined by Hint. Note that the Laplace transform of the Laguerre polynomial is 384. Using the residue theorem, give the solution of Prob. 380 in the form of a Fourier series. Ans. I R oe-"ul/l m - nn sin (nxutll) - ucos(nnut/l) u/*=~ = 1 - . E UJZ~ - R: u2 + n2n2 (n - ;)n cos [(n - i)nut/l] Psin [(n - -I- n=l " The details are given in L9, Sec. 4.25. PROB. 385 where I Ro+Z 1 '+Ro, p=-ln- u = - 1 z = LO. cr =-In- 2 2-Ro 2 R,-Z' JE' 385. Study the propagation of waves along an inhomogeneous transmission line consisting of a finite section of length I with wave resistance Z,, fol- lowed by an infinite section with wave FIGURE 98 resistance Z, (see Figure 98). Find the reflected and refracted waves appearing at the junction, assuming that at the time t = 0 an arbitrary e.m.f. E = f(t) is applied at the end x = 0. Ans. where o1 = 1/JL,C,and v2 = l/JL,C,are the velocities of wave propagation along the two parts of the line, and T = I/vl. *386. A voltage wave E = Eoe-at produced by a lightning discharge at the end x = 0 of a transmission line activates a lightning rod at the point x = I. Find the voltage in the section of the line after the lightning rod, assuming that the rod behaves like an ohmic resistance R, during its time of operation. PROB. 389 Ans. where v is the velocity of wave propagation along the line, and Z is the wave resistance. 387. A constant e.m.f. E is applied at the time t = 0 to the end x = 0 of a semi-infinite cable (a line with parameters R and C). Find the voltage at every point of the cable. Ans. x JRC 4x2 = ~[l - @ (-)I, where @(x) is the probability integral. 388. Find the voltage in a cable of length I if a source of constant e.m.f. E is applied to the end x = 0, while the end x = I is terminated by an ohmic load Ro. Ans. sin (ynx/l) e-yn2t/~c'2 U(X, t) = E l+a [1 + a + (y,%)I sin yn cos Y,, where the yn are consecutive positive roots of the equation tany = -1, CL C and R are the capacitance and resistance of the cable per unit length, and a = Rl/Ro. 389. Solve Prob. 387 for a cable with leakage conductance G per unit length. Ans. 184 INTEGRAL TRANSFORMS PROB. 390 390. Study the propagation of voltage waves in the compound line with equivalent circuit shown in Figure 99, caused by switching on a constant e.m.f. E at the end x = 0. sin (ax tan cp) cos (pt sin cp) cot cp dcp , I where L, C and K are the self-inductance and capacitances per unit length of the line, cc = JFK and P = l/Jz. Hint. The equations governing the current and voltage in the line are in terms of the voltage u(x, t), the total current I(x, t), and the currents IL(x, t) and IK(x, t) flowing through the self-inductance and capacitances L and K. 391. Near the plane interface between two slabs of material with dielectric constants and E,, there is a source of electromagnetic oscillations radiating a spherical wave whose Hertz vector has components where f (t) = 0 if < 0, R is the distance from the source to the observation - point, and v1 = c/JE~ is the velocity of propagation of electromagnetic waves. Find the Hertz vector on the interface for the limiting case where the source is located on the interface itself. Ans. nl,=, = n1 - n,, INTEGRAL TRANSFORMS 185 - and vi = c/Jci (i = I, 2) are the velocities of propagation of electromagnetic waves in the two media. Hint. Take Laplace and Hankel transforms in succession. 392. Consider the special case of the preceding problem corresponding to a wave with the steep front described by the function Ans. 393. A force F(t) is applied at the time t = 0 to the end x = 0 of a semi- infinite rod. Study the resulting propagation of elastic waves in the rod. Ans. The displacement of an arbitrary point of the rod is where E is Young's modulus, p the density and S the cross-sectional area of the rod, and v = 4%. 186 INTEGRAL TRANSFORMS PROB. 394 394. Suppose one end x = 0 of a F(f'b///AA rod of length I is clamped, while a compressive force F(t) with the saw- tooth wave form shown in Figure 100 IIIII + is applied to the other end x = 1, 0 T 2~ 3~ 4~ 5T starting from the time t = 0. Investi- FIGURE 100 gate the resulting longitudinal oscil- lations, and find the reaction at the fastened end, assuming that the period T equals the time T = I/v it takes an elastic wave to traverse the rod (u is the velocity of wave propagation). Ans. 0, O<t<T, (2n+l)T<t<(2n+3)T, n=1,3,5 ,..., 2A n, nT<t<(n+l)T, n=1,2,5,6,9,10 ,.... 395. A cantilever clamped at the end x = 0 begins to oscillate under the action of an impulse delivered to a concentrated mass M,, fastened to the free end x = I. Find the dynamic reaction at the clamped end, assuming that a velocity v,, is imparted to the mass M, by the impulse. Ans. RI,=, = 3 ESf(t), 2, where 0, O<t<T, in terms of the velocity of wave propagation u and the constants a = ES/M,,v, T = 110. Hint. The boundary conditions for the displacement u(x, t) at the end x = 1 are 396. Use the Laplace transform to solve Prob. 106. u(x, t) = - sin wt + " (- 1)" sin [(2n + l)nx/21] sin [(2n + l)xvt/21] n=~ 2n + 1 1 - [(2n + l)xv/21wI2 397. A constant force Q is applied to the end x = 0 of a semi-infinite beam, starting from the time t = 0. Find the deflection at any point of the beam, assuming that the beam is initially at rest.22 Ans. where 2 sin x2 cos x2 +(I -~3- X X 19 E is Young's modulus, J the moment'of inertia of a cross section, p the density and S the cross-sectional area, aa = J~s, and C(s), S(x) are the Fresnel integrals. 398. Solve the preceding problem, assuming that a constant bending moment (rather than a constant force) is applied to the end x = 0. Find the bending moment along the beam at any time t. Ans. 399. Find the displacement of the end x = 0 of a semi-infinite beam struck at the time t = 0 by a mass M, moving with velocity uo. Ans. where cr = 2 J2 paS/M and @(x) is the probability integral. 400. Find the transverse oscillations of a beam -1 9 x 9 1, simply supported at both ends, due to an impulse P acting at the center of the beam. Write an expression for the deflection of the center of the beam. Ans. 401. Find the deflection of an infinite elastic plate, if at the time t = 0 a constant force Q is applied to the point x = y = 0 (see L16, p. 424). Ans. B1 Problems 397-399 are treated in Lurye's book L16. 188 INTEGRAL TRANSFORMS PROB. 402 where D is the flexural rigidity and p the surface density of the plate, T = tdz and Si(x), Ci(x) are the sine and cosine integrals. Hint. At the point where the force is applied, UI,=~ must be bounded, and moreover *402. Solve the preceding problem, assuming that an impulse P (rather than a force Q) is applied to the point x = y = 0. Ans. 403. At the time t = 0 an impulse with components P, = 0, P, = P is applied to the point x = y = 0 of an infinite elastic plate. Describe the resulting process of wave propagation. Ans. The elastic potentials are given by the formulas where are the velocities of propagation ofthe longitudinaland transverse oscillations, h and p are Lamt's constants, and p is the density. 404. Solve the preceding problem, assuming that the source of the oscillations is a concentrated force with components Q, = 0, Q, = Q. A ns. where the elastic potentials are zero for smaller values of the time. 405. Show that the solution of the two-dimensional wave equation in the domain x > 0, subject to zero initial conditions and the boundary condition uJ,=, = fCv>, can be written in the form Hint. Take Laplace and Fourier transforms in succession. *406. Show that the solution of the wave equation for a medium with attenuation, subject to zero initial conditions and the boundary condition u],=, = f(y), can be written in the form where Deduce the solutions of Probs. 308, 360 and 405 as special cases. 4. The Mellin Transform Let f(r) be a real function defined in the interval (0, co) such that 1. f(r) is piecewise continuous and of bounded variation in every finite subinterval [a, b], where 0 < a < b < a; 2. Both integrals are finite for suitably chosen real numbers a, and a,. 190 INTEGRAL TRANSFORMS PROB. 407 Then the MeNin transform of f(r) is defined by the formula !(PI = JOmf(r~rw-l dr, (19) where p = a + i~ is any complex number in the strip -a a, < Rep < a, (see Figure 101). The inversion of (18) is given by the formula FIGURE 101 where F is a straight line parallel to the imaginary axis lying inside the strip.23 The Mellin transform is related to the Laplace and Fourier transforms, and is the appropriate tool to use for solving problems of two-dimensional elasticity theory and potential theory involving angular regions. The required technique can easily be acquired by working through the following small set of problems.24 *407. Find the stationary temperature distribution inside the dihedral angle 0 < r < a, 0 < cp < a < x, if one boundary is held at temperature zero, while the temperature distribution is maintained on the other boundary. Ans. (:ysin 3 To u T(r, cp) = - arc tan x 408. Solve the preceding problem, assuming that a given distribution of heat current q,, a-~<r<a+~, qvlv=a = (0 otherwise la See e.g., T5, Secs. 1.5 and 1.29. The conditions imposed on f(r) can be weakened. 24 A few remarks are in order concerning the choice of the path of integration r in the inversion formula (20). If the behavior of the function f as r + 0 and r + rn is known in advance (e.g., from physical considerations), then the boundaries of the strip (o,, o,) can be found from the requirement that both integrals (18) be finite. If the behavior of the function f is known oniy at one end point of the interval (0, a), say as r + 0, we can first determine the left-hand boundary o,, and the line I' must then lie to rhe right of r and to the left of the nearest singular point of the function f figuring in the integral (20). ROB. 4 1 1 INTEGRAL TRANSFORMS 1 9 1 is maintained on the boundary cp = a. Consider the case of a concentrated current Q entering the boundary along the line r = a, cp = a. Ans. x ln (ria) 2nk cosh --- 2a x In (rla) cosh --- = cp + sin - 2a 2a x ln (ria) X'P ' cosh - - sin - 2a 2a where k is the thermal conductivity. 409. Use the Mellin transform to solve Prob. 303. Ans. cos [T In (rlb)] + T sin [T In (rlb)] sinh TV) T(r, 9) = - dr. ~(1 + T2) cosh (~7~12) 410. A thin charged wire, with charge q per unit length, is placed along the line r = r,, cp = cp, inside the dihedral angle 0 < r < m, 0 < cp < a, whose boundaries are held at potential zero. Find the potential of the resulting electrostatic field. Ans. 3Joiim sin p(a - rpo) sin pp z U-im p sin pa = 3Joiiw sin pep, sin p(a - p) ~(r, cp)Iw>'Po . 1 O-im p sin pa where 101 < x/a. In particular, the imaginary axis can be chosen as the path of integration. 411. Calculate the following special cases of the preceding problem: a) a = 2x, ro = a, cpo = x (line charge opposite the edge of a conducting half plane); b) a = 3x12, ro = a, cp, = n (line charge near a conducting right-angular corner) ; c) a = n/2 (line charge inside a dihedral angle). Ans. 192 INTEGRAL TRANSFORMS PROB. 412 1 - 2 - cos 2(cp + yo) + (kr c) u(r, 9) = q ln (;J 1 - 2 - cos 2(9 - cp,) + - (2 (:J . 412. The common boundary of two media of dielectric constants E, and E~ consists of two planes intersecting at the angle 2a (see Figure 102). Find the electrostatic field due to a charged wire lying in the plane of symmetry. Am. l+im sin 2a(p - 1) cos (x - cp)(p - 1) x sin x(p - I)[sinx(p - I) - p sin (x - 2a)(p - I)] p = - '" COS CP(P - 1) (!T-'dp, c2 + I-im sin x(p - 1) - P sin (x - 2a)(p - 1) r I+im sin 2a(p - 1) sin (x - cp)(p - 1) - sin (p - I)[sin n(p - I) - p sin (x - 23(p - I)] ~lp2) = - '" sin cp(p - 1) E~ + I-im sin x(p - 1) - P sin (x - 2cr)(p - 1) where and R is the distance from the charge to the observation point (see G5, Chap. 14). PROB. 4 1 5 INTEGRAL TRANSFORMS 193 413. Investigate the bending of a thin wedge-shaped elastic plate with simply supported edges, loaded by a concentrated force P applied at an arbitrary point of the axis of symmetry (see Figure 103).25 Ans. The deflection of an arbitrary point of the plate is given by the formula u(r, cp) = - {cos cp sinh (a - cp)r - cos (a - cp) sinh cpr Sm 4xD o + sin cp cosh (a - cp)r - sin (a - cp) cosh cp'r]) cos ['r In (rlr,)] d~ cosh ar + cos a r(r2 + 1) ' where D is the flexural rigidity of the plate. 414. Let a = x/2 in the preceding problem. Show that the deflection of the points on the axis of symmetry of the plate is given by the formula *415. A thin elastic plate 0 g r < co, 0 < cp < a is clamped along its edges and loaded at the point (r,, yo) by a concentrated force P. Find the bending moment and shear force along the edge cp = OeZ6 Ans. MI,=o = Pr, rm [sin yo sinh ar sinh (a - po)r xr. o sinh2 a7 - r2 sin2 a - sin a sin (a - yo) . r sinh cp,r ] cos (, ln :)dr, sinh2 a'r - 'r2 sin2 a m - % {cos yo sinh ar sinh (a - y,)r qv=0 - xr2 + r[sin a cos (a - cp,) cosh ar sinh (a - cp,)r - cos a sin (a - cp,) sinh ar cos [r In (r,/r)] x cosh (a - cpo)r] - r2 sin a sin (a - yo) cosh cpor} d'r. smh2 a'r - 'r2 sin2 a 25 Problems 413, 414 and 417 are treated in Uflyand's paper U3. ="nother way of solving this problem is due to Sakharov (S2). 194 INTEGRAL TRANSFORMS PROB. 416 416. Use the Mellin transform to solve Prob. 328 (a special case of the preceding problem). Ans. P r: sin2 9, 2P r, sin yo 3 3 - - p=o - , N/,+?=,=- n r2 + r: - 2r0r cos yo x (r2 + ri - 2ror cos Hint. Use the formula 1 sin cos T+ d~ = - 2 cosh + + cos 417. Solve Prob. 415 assuming that one of the edges of the plate (9 = a) is supported. Consider the special cases a) u = n/2 and b) u = n (the quadrant and the half-plane). Ans. m MI,=, = %/ [sin p0 sinh (2. - 9,)r - sin (2~ - po) sinh par] nr o cos [T In (rlr,)] x d~. sinh ~UT - T sin 2u In the special cases, we have 2Pr0 sin po sin 29, . a) MI,=, = - - xr r2 r2 --, + --O - 2cos 29, ro r2 5. Integral Transforms Involving Cylinder Functions of Imaginary Order Let f(x) be a real function defined in the interval (0, co) such that 1. f(x) is piecewise continuous and of bounded variation in every finite subinterval [a, b], where 0 < a < b < co; 2. Both integrals are finite. INTEGRAL TRANSFORMS 195 Then f(x) satisfies the formula where K,,(x) is Macdonald's function. If we write it follows from (21) that Besides the more familiar transforms considered so far, formula (21), proved by one of the authors of this book (see L6, L8), plays a role in certain physical problems. If we use the formulas x = Ar, E = Ap, f(x)& = g(r) (A > 0) to introduce new variables, (21) takes the form 2 " m g(r) = 2 l K,,(Ar)r sinh nr drl g(p) dp, 0 < r < m, (24) P which, although less symmetric than (21), is more suitable for solving the problems encountered in mathematical physics. Formula (24) holds provided the integrals are finite. The following expansion of this type is useful in the application^:^^ In addition to the above formulas involving Macdonald's function, there is an analogous expansion in Hankel functions and a corresponding inversion formula, which play a role in certain applications. These relations can be deduced formally from (24) by setting A equal to a pure imaginary (A = ik), However, note that the first of the integrals (25) is not finite forg(r) = e-hr. and then using the relation between the functions KV(z) and Hp'(z). In this way, we find the formulasz8 cp(r) = - /"xr)enT12 HiB'(ir) r sinh m dr, 0 < r < m. (28) 2 0 The integral expansion (24) can be used to solve the Dirichlet problem and other problems of potential theory for regions bounded by two inter- secting planes (the three-dimensional problem), for wedge-shaped regions bounded by two parallel planes and two intersecting planes (perpendicular to the parallel planes), and so on. Formulas of the type (27) and (28) are encountered in solving problems involving the diffraction of acoustic and electromagnetic waves by an obstacle in the shape of a dihedral angle or a cone. The following problems illustrate z various physical applications of the above ex- pansions. *418. Find the stationary temperature distri- // bution in a wedge-shaped body of thickness I (see Figure 104), if the temperature distribution 0 is maintained on the boundary cp = a, while the other boundaries are held at temperature zero. Ans. nxp dp sinh cpr x [[f(p) - e-""Y(O)lKi. - KiT (7) d~, p smh ar where KV(z) is Macdonald's function. 419. Solve the preceding problem for an arbitrary temperature dis- tribution TI,=, = f (r, z) on the face cp = a. Is For conditions under which (28) implies (27), see the paper K3. INTEGRAL TRANSFORMS 197 PROB. 422 A ns. sinh cpr X - &(?)dr, sinh ar nRZ . f(r, Z) sin - dz. 1 Hint. Expand the function f(r, z) in a Fourier series with respect to sin (nnzll), and then use the result of Prob. 418. 420. Find the stationary temperature distribution in the "quadrant- shaped" slab 0 < x < co, 0 < y < co, 0 < z r I, if the boundary x = 0 is held at constant temperature To, while the other boundaries are held at temperature zero. Ans. 8T0 " sin [(2n + l)~z/l] mcosh sinh (PT T=-2 J^o 2 sinh (rrr/2) K, [(2n + l)nr/l] dr. n2 ,,, 2n + 1 By using the representation KiT(x) = rcos (x sinh t) cos rt dr cosh (xr/2) of Macdonald's function, this result can be brought into simpler form: " sinh [(2n + l)~z/l] cos [(2n + l)m sinh t/l] T = 9sin2cp2 dt. X ,=, 2n + 1 So COS~ 21 + cos 29 421. Find the distribution of the electric charge density induced by a point charge q placed near the edge of a thin conducting half-plane (see Fig- z ure 105). Ans. o=- 44; 2~"Jx[(x f a)' + z2] ' *422. Solve the preceding problem, as- suming that the charge q is located at an arbitrary point r = r,, cp = cp,, z = 0. I Find the distribution of electrostatic t Y potential. FIGURE 105 PROB. 423 where R,2,, = r2 + z2 + r: - 2r0r cos ((P F (PO), cash 8~ + cos H(P F (PO) r2 + ri + z2 , cosh u = +1~2 dcosh $a - cos $((P + yo) 2ror 423. A point charge q is placed near the edge of a conductor of rectangular shape held at potential u = 0 (see Figure 106). Find the distribution of charge density on the boundaries of the conduct~r.~~ Ans. where cosh A = r2 + z2 + a2 2a r 424. Find the current distribution produced in the ground by a point electrode located near a wedge-shaped layer, assuming that the layer has conductivity o, while the rest of the ground has conductivity 0, (see Figure 107). Write an expression for the potential distribution on the earth's surface. Ans. K,,(Aa)K,,(Ar) sinh 2(x - U)T cos Az dA UJ'=O = - 2 fo sinh rrr + (3 sinh (rr - 2u)r d7, 4J f Qc~s Az dAf " K,,(Aa)K,,(Ar) sinh xr = o sinh XT + (3 sinh (x - 2u)~ d~, x3(01 + 02) o " This problem was first solved by Macdonald (Ml). ROB. 425 where INTEGRAL TRANSFORMS 199 J is the current and KY(x) is Macdonald's function (see S4). 425. Show that the solution of the preceding problem can be reduced to the form P 6 = arc cos - 2 if u = n/4, and to the form 1+P 8 = arc cos - 2 200 INTEGRAL TRANSFORMS PROB. 426 426. Solve the problem of diffraction of a / plane electromagnetic wave Y - (where k = wlc is the wave number) incident on - a thin perfectly conducting sheet (see Figure ~GURE 108 108) making an angle a with the direction of wave propagation (Sommerfeld's problem). Ans. The complex amplitude of the z-component of the total field is 427. Using the result of the preceding problem, find the current dis- tribution on each side of the sheet. Consider the special cases where a) a = 0; b) a = n/2. Ans. The required densities are determined by the system of linear eauations E~~J~ [c a 2ei44 LV'ZT sin %a j,+j,=j=- - COS - + - sin ae-ikr cosa epis2 ds] , 2x xJikr 2 Jx where j, and j, are the densities on the upper and respectively. In the special cases, ? lower sides of the sheet, 428. A line source of a-c current J = Joeiwt is placed parallel to the edge of a thin conducting sheet 0 < x < a, -a < y < a. Find the distribution of induced currents if the source lies in the plane of the sheet at a distance a from its edge. Ans. The complex amplitude of the current density is PROB. 431 INTEGRAL TRANSFORMS 20 1 429. Find the electromagnetic field of a dipole of moment P located on the axis of a perfectly conducting conical reflector of vertex angle 2a, if the dipole lies at a distance a from the vertex of the cone (see L10). Ans. If r < a, the complex amplitude of the magnetic field is given by the series ~(2) v,+dka) Jvn+%(kr) Pvn(-cos a)Ptn(cos 0, , A 1 J' sin rrv. raPv(cos a) where the v, are consecutive positive roots of the equation PJCOS a) = 0 involving the Legendre function PV(x), and JV(x), H:2) are cylinder functions. The corresponding formula for r > a is obtained by permuting the symbols r and a in the general term of the series. 430. A plane acoustic wave is incident on a screen in the form of a half-plane r > 0, y, = a. Find the wave reflected from the screen. Ans. The complex amplitude of the velocity potential at an arbitrary point is eix/4 dG sin %rp , -ikr cos rp [i -+- e-">S] JX o 431. A point source of sound, radiating a spherical wave sin (at - kR) u = Uo R 9 is placed on the axis of a conical resonator 0 < 0 < a with perfectly reflecting walls. Find the velocity potential inside the cone. Ans. The complex amplitude of the velocity potential is where the v, are consecutive positive roots of the equation P:(cos a) = 0, and a is the distance from the source to the vertex of the cone. The corre- sponding formula for r > a is obtained by permuting the symbols r and a. 202 INTEGRAL TRANSFORMS References Books: Campbell and Foster (Cl), Carslaw and Jaeger (C2), Churchill (C4), Doetsch (Dl), Fuchs and Levin (F9), Gray and Mathews (G2), Grinberg (G5), Lebedev (L8, L9), Levitan (L13), Lurye (L16), Morse and Feshbach (M9), Sneddon (SIO), Titchmarsh (T5, T6), Tranter (T8), Van der Pol and Bremmer (Vl), Watson (W4), Widder (W9). Papers: Fock (F4), Kontorovich and Lebedev (K3), Lebedev (L6), Lebedev and Kontorovich (LlO), Sneddon (S8, S9), Weyl (W7). CURVILINEAR COORDINATES A physical problem can often be greatly simplified by the introduction of a suitable system of orthogonal curvilinear coordinates, facilitating the formulation of the boundary conditions and making it possible to solve the problem by using the techniques of Chaps. 4-6. These earlier chapters contain an abundance of examples illustrating the simplest systems of curvi- linear coordinates, i.e., polar, cylindrical and spherical coordinates. We'now turn to more complicated coordinate systems, whose effective use will allow the reader to solve a much larger class of problems. Perhaps the most important use of curvilinear coordinates is to solve boundary value problems for the Laplace and Helmholtz equations. How- ever, neither the three-dimensional Laplace equation nor the Helmholtz equation permits separation of variables when written in arbitrary orthogonal curvilinear coordinates, a fact which prevents us from applying the methods developed in the preceding three chapters. Therefore n problem of great theoretical and practical interest is to find all coordinate systems which actually lead to separation of variables in these equations. Some special results pertaining to this problem, which has not yet been solved completely, will be found in concise form in Sec. 8, p. 247.' The material given here is organized as follows: All problems involving a given coordinate system are grouped together, regardless of their physical content or spectral character (the latter determines whether the solution takes the form of a series or an integral). In the case of two-dimensional systems, considered in Secs. 1-3, all the necessary preliminary material is presented in problem form. However, in the case of three-dimensional See also the papers cited at the end of the chapter (p. 252). 204 CURVILINEAR COORDINATES PROB. 432 systems, considered in Secs. 4-7, more background information on differ- ential equations, special functions, etc. is needed, and this material is summarized at the beginning of each section. Besides problems of the simpler kind, this chapter contains some relatively difficult problems, whose solution requires the use of various integral trans- forms, knowledge of the properties of certain special functions, and so on (see e.g., Probs. 483, 497, 502-504). These problems are intended for the adequately prepared reader, and can serve as practice material for those trying to deepen their understanding of the methods of mathematical physics. Finally, it should be kept in mind that some of the problems can be solved more simply by using other methods (e.g., conformal mapping or inversion). I. Elliptic Coordinates 432. Study the system of elliptic coordinates a, (3 related to the rectangular coordinates x, y by the formula x + iy = c cosh (a + ip) (0 < a < 03, -n < p < n). (1) Show that the curves a = const, P = const form an orthogonal system of confocal ellipses and hyperbolas (see Figure 109). What is the appropriate expression for ds2, the square of the element of arc length? Write Laplace's equation in elliptic coordinates. Ans. ds2 = c2(cosh2 a - cos2 P)(da2 + dp2), 433. A conducting elliptic cylinder with semiaxes a and b is placed in a homogeneous electric field (see Figure 110). Find the distribution of electric charge density on the surface of the cylinder. pROB. 435 CURVILINEAR COORDINATES 205 Ans. E ="(a + b) cos (P - Y> 4x JaZ sin2 p + b2 cos2 P ' u - do where E, is the external field. - Hint. Introduce elliptic coordinates a, P (see x the preceding problem), where the parameter d \ c equals the eccentricity of the given ellipse. & 2C \ \p.-p *434. A wire with charge q per unit length is placed inside a hollow conducting elliptic FIGURE 11 1 cylinder with semiaxes a and b. Find the poten- tial distribution inside the cylinder, assuming that the wire is parallel to the axis of the cylinder (see Figure 11 1). Ans. " sinh n(a, - a) cos np* cos np , n=l n cosh na, I in terms of the elliptic coordinates a and P, where a, and P* are the parameters defined by the relations b d tanh a, = - , cos p* = - (d < c). a c Hint. Regard the charge q as uniformly distributed over the "curvilinear rectangle" O<a<8, p*-><PI <P*tf, 2 2 and then take the limit as 8, E + 0. 435. Solve the preceding problem, assuming that the charged wire is placed outside the cylinder at the point x = d (d > a), y = 0. Ans. cosh (a* + a - 2a0) - cos p 4% P) = 4 1n d , cosh a* = - , c = Ja2 - b2. cash (a* - a) - cos P c Hint. To sum the series, use the formula io e-nv cos nx In (2 cosh y - 2 cos x) = y - 22 , Y>O. n=l n 206 CURVILINEAR COORDINATES PROB. 436 436. Find the distribution of induced charge on an infinitely long con- ducting strip, near which there is a wire with charge q per unit length, as shown in Figure 112. Ans. *437. An elliptic cylinder of given dimensions, made from material of magnetic permeability p, is introduced into a homogeneous magnetic field making angle y with the major axis (see Figure 110). Find the magnetic potential, and show that the field outside the cylinder is homogeneous. Ans. u = Ho(x cos y + y sin y) a + b cos y cos p sin y sin p + H,(1 - dab + dmu( a + pb b + pa ) + const outside the cylinder, cos y sin y u = H,(a + b) - y + const inside the cylinder, (a+pbx+G ) where Ho is the external field, and the ellipse has semiaxes a and b. Hint. The choice of the particular solutions for the region outside the cylinder is dictated by the requirement that grad u be bounded. 438. A hollow elliptic cylinder, made from material of magnetic per- meability p, has a cross section bounded by the confocal ellipses Suppose the cylinder is introduced into a homogeneous magnetic field with components Hz=-H,, H,=H,=O. Find the distribution of potential in the body of the cylinder. Ans. sinh a, sin11 (a - a,) + p cosh a, cosh (a - a,) = H,c eaz cos p, sinh (a2 - aJsinh a, + p2 cash al) + peal cosh (a2 - al) where b. tanh ai = (i = 1, 2). a i 439. A cavity in the shape of an elliptic cylinder with semiaxes a and b is hollowed out of iron of magnetic permeability p, and contains a line current J whose direction is parallel to the axis of the cylinder. Find the vector potential of the magnetic field, assuming that the current passes through the point x, < Jaz - b2, yo = 0 of the semimajor axis. Ans. xz ePao cos nPo cosh na cos np + const, 0 < a < a,, ,=, n(cosh na, + p sinh nu,) x2 sinh na, cos np, ePna cos np + const, a > a,, ,=, n(cosh na, + p sinh na,) where A, and A, are the values of the z-component of the vector potential in the air and in the iron, respectively, R is the distance from the point (x,, 0) to the point (x, y), a and p are elliptic coordinates, c is the velocity of light, and b x 0 tanha, = - , cos p, = -. a Ja2 - b2 440. Solve the preceding problem for the limiting case p = a. Find the tangential component of the magnetic field on the interface between the air and the iron. Ans. 25 4 J *. e-nao cos np, A,=--1nRf-2 cosh na cos np + const, c c n=l n sinh na, 208 CURVILINEAR COORDINATES PROB. 441 The tangential component of the magnetic field on the interface is where p is the length of the perpendicular dropped from the origin of co- ordinates onto the tangent to the ellipse at the point M = (a,, p). 441. A d-c current flows in a conductor whose cross section is an ellipse with semiaxes a and b, producing heat with volume density Q. Find the temperature distribution inside the conductor, assuming that its surface is held at temperature zero, Ans. Q cosh 2a T(a, (3) = - (a2 - b2) (cosh 2ao - cos 2p), 8k cosh 2u0 where k is the thermal conductivity and b tanh a, = - . a Hint. Subtract out a particular solution u = P(x, y) of the inhomogeneous heat conduction equation, where P(x, y) is a polynomial in x and y. 442. A thin sheet of width 2a is placed in a plane-parallel flow of an ideal fluid. / Find the velocity potential, assuming that the direction of the flow makes angle y with the plane of the sheet (see Figure -a 2i1 +a$ix 113). Ans. I u = u,(x cos y + y sin y + a sin ye-" sin p), FIGURE 113 where v, is the velocity of the flow far from the sheet. *443. Solve the problem of the twisting of a rod of elliptical cross section with two cuts extending to its foci, as shown in Figure 114. Calculate the torsional rigidity C numerically for the cases where the ratio of the semiaxes is a, and #. Ans. The torsion function is 8b2 " cosh (2n + l)a sin (2n + 1)P u(a, p) = -c2 sinh2 a sin2 p + - 2 x .=, cosh (2n + l)a, (1 - 4n2)(2n + 3) ' CURVILINEAR COORDINATES 209 The torsional rigidity is (2n - 1) sinh (2n + 3)a0 + (2n + 3) sinh (2n - l)ao n=O 2(2n + 1) where tanh a, = b/a, and C, is the tor- sional rigidity of the ellipse without the Y4 cut. The result of the numerical calcula- tions are I = 0.997, 1 = 0.970, Co b/a=1/4 C, b/a=1/2 L I = 0.826. Co b/a=3/4 d ' -a, Hint. Subtract out the particular so- lution -y2. 444. Find the torsion function of a rod of semielliptic cross section. Ans. 8b2 " sinh (2n + 1)a sin (2n + 1)P u(a, p) = -c2 sinh2 a sin2 p - - 2 TC n=O sinh (2n + l)a, (4n2 - 1)(2n + 3) ' where a and b are the semiaxes of the ellipse, and b tanh a, = - , c2 = a2 - b2. a 445. Find the stationary temperature distribution in a body whose surface is the hyperbolic cylinder given the temperature distribution on the surface. Ans. cosh AP Sinh " sin la d~, cos Aa + f, - sinh AP, I 2 10 CURVILINEAR COORDINATES where cn f, =jymf(a) cos ha da, fs =Ipcn f (a) sin ha da are the Fourier cosine and sine transforms of the function f (u) figuring in the boundary condition Hint. In Probs. 445-447 use elliptic coordinates defined by x + iy = c cosh (a f ip) (-co<a<co,O<p<n), FIGURE 1 15 instead of by formula (I), p. 204. 446. Find the density of electric charge on two perpendicular grounded planes between which there is a charged wire, as shown in Figure 115. 447. A charged wire with charge q per unit length is placed on the axis of symmetry of a slot of width 2a cut in a grounded conducting metal plane. Find the resulting electrostatic potential u. What is the charge density on the two parts of the plane? Ans. cosh a + sin p u(., P) = 4 In cosh a - sin B The charge density is 4 0 = - a x > a. 2xx JX7 ' Hint. In elliptic coordinates, the two parts of the plane have equations p = 0 and p = x. 2. Parabolic Coordinates 448. Study the system of parabolic coordinates a, P, related to the rectangular coordinates x, y by the formula PROB. 450 CURVILINEAR COORDINATES 2 1 1 Show that the curves a = const, P = const form two orthogonal families of parabolas d. ,,,, (see Figure 116). What is the appropriate expression for the square of the element of arc length? Write Laplace's equation in a >O parabolic coordinates. Ans. X ds2 = c2(a2 + P2)(da2 + dP2), AU = c?a+ p2) (?2+$)=o. aa2 *449. A charged wire with charge q per unit length is placed at the focus of a FIGURE 116 conducting screen in the form of a para- bolic cylinder. Find the resulting electrostatic field. Ans. The electrostatic potential is " sinh h(po - (3) cos ?a dh. in terms of Po, the value of the coordinate (3 on the surface of the cylinder, given by Po = JpT, where p is the focal distance of the parabola and c is the scale factor figuring in formula (2). Using formula 13, p. 385, we can write the solution in closed form : XU XP cosh - + cos - ~(a, (3) = 24 In 2Po 2Po 7Ca ~(3 ' (3) cosh - - cos - 2Po 2Bo 450. Write a solution of the preceding problem in the form of a series of functions depending on the variable P. Ans. Using the formula 2 pZn+' cos (2n + 1)x = ! I + 2p cos x + p2 $ < I n=o 2n + 1 4 1 - 2p cos x + p2 ' to sum the series, we arrive at formula (3). 2 12 CURVILINEAR COORDINATES PROB. 451 451. A charged wire with charge q per unit length is placed parallel to the edge of a thin conducting half-plane (see Figure 117). Find the resulting charge distribution on the half-plane. Ans. 0 = Hint. The equation of the half-plane in parabolic coordinates is P = 0. In solving the problem, regard the charge as uniformly distributed over the area of a curvilinear rectangle bounded by appropriate curves a = const, (3 = const, and then make the dimensions of the rectangle go to zero. 3. Two-Dimensional Bipolar Coordinates 452. Study the system of two-dimensional bipolar coordinates a, P, related to the rectangular coordinates x, y by the formula '+ iP (-a <a< m, rr< <XI. (4) x + iy = c tanh - 2 Show that the curves P = const are circles c2 x2 + (y - c cot f3)2 = - sin" PROB. 454 CURVILINEAR COORDINATES 2 13 going through the points x = f c, while the curves a = const are the orthogonal circles (see Figure 118). What is the appropriate expression for the square of the element of arc length? Write Laplace's equation in two-dimensional bipolar coordinates. Ans. ds2 = c2 (cosh a + cos (3)2 (da2 + dp2), 453. Find the electrostatic potential in the region between two parallel cylinders of radius a, held at potentials & V, respectively, if the axes of the cylinders are a distance 21 apart (see Figure 119). Calculate the capacitance per unit length between the pair of cylinders. Am. In terms of bipolar coordinates cr, P with parameter c = dl2 - cr2, (the two cylinders have equations a = * ~0). 454. A cylindrical pipe of radius a is buried in the ground at depth b FIGURE 119 (see Figure 120). Find the stationary temperature distribution in the region surrounding the pipe, if the tempera- ture of the ground is zero while a heat current Q, uniformly distributed with respect to angle, leaves the pipe's surface. Ans. Q T(a, PI = sinh nu cos np , I where cosh a, = b/a and k is the thermal conductivity. 2 14 CURVILINEAR COORDINATES PROB. 455 455. Two parallel cylinders of radius a with axes a distance 21 apart are placed in a plane-parallel flow of an ideal fluid, making angle a with the line joining the centers of the cylinders. Find the resulting velocity potential. Ans. u(a, = vmJ12 - a2 sinh a e-nao + 2e(-l)n --- sinh na cos np cosh a + cos p cosh na, I sin p e-n.ro f sin y + 23(-1)" --- cosh na sin np , cosh a + cos p sinh na, 1) where cosh a, = //a and v, is the velocity of the flow far from the cylinders. 456. Solve the problem of the twisting of a circular shaft weakened by an eccentrically drilled hole, as shown in Figure 121 (see W6). Ans. The torsion function is YI u(a, p) = o; sinh2 a, coth a, + ~0th az [ - cash a cosh a + cos p t2f (-I)?' sinh n(a, - a,) I x [e+%' coth a, sinh n(a2 - a) + ePZa coth a2 sinh n(a - where a, and a, are determined from the relations a; - + d2 a; - a; - d2 cosh a, = , COS~ a2 = 2dal 2da2 PROB. 457 CURVILINEAR COORDINATES 2 15 *457. A narrow slot is cut in a circular shaft subject to twisting (see Figure 122). Find the torsion function and calculate the torsional rigidity. Calculate the rigidity numerically for the case hla = 4. Ans. The torsion function is I sin2 (3 u(a, (3) = a2 sinh2 a, - (cosh a + cos (3)2 where 2n sin2 (3 sin (n + an =S, l)P dp, (cosh a, + cos (3)' h(2a - h) sinh a, = -- (h < a). 2a(a - h) The rigidity is m 1 " C = ~o~ sinh"ao[2 sinh a, 2 o,b, - - 2 (2n + 1)a: - iT n=O 2~ .=o 2 sinh4 a where '" sin2 p sin (n + $)(3 (cosh a + cos (3)3 4 In the case h/a = 4, it is found that C = 1.28Ga4 (compare with the result of Prob. 233). Hint. Subtract out the particular solution -y2. To calculate the rigidity, use the formula In the numerical calculation of the coefficients a, and b,, use the relations m A, = - [L + (2n + IJX (-l)me-nL1O sinh a, 2n + 1 m=l (n + :)'- m m [A, cosh a. + (2n + 1) m(- l)me-mlo B, = - sinh2 a, m=l (n + 4)'- m2 2 16 CURVILINEAR COORDINATES PROB. 458 458. An eccentrically drilled tube, with the cross section shown in Figure 121, is subjected to a pressure uniformly distributed over its interior surface. Find the resulting (two-dimensional) deformation of the tube, if no forces act on its outer surface. Calculate the normal stresses on the inner surface of the tube (see J3). Ans. where p is the pressure. 459. A charged wire with charge q per unit length is placed at height h inside a long tunnel of semicircular profile. Find the electrostatic field in the plane of symmetry, assuming that the walls of the tunnel constitute an equipotential surface. Ans. 4q sin 2P*(1 + cos P) ~l,,o = - a cos 2P - cos 2P* ' where a is the radius of the semicircle, and p* is determined from the formula Hint. In a system of bipolar coordinates, the region in question is bounded by the coordinate surfaces P = 0 and P = 742. Expand the solution in a Fourier integral with respect to the variable a. 460. A conducting plane has a semicylindrical boss of radius a, as shown in Figure 123. Find the distri- x bution of electric charge induced on the surface of the conductor by a wire carrying charge q per unit length - -- FIGURE 123 placed Tn the plane of symmetry (see Figure 123). Calculate the maximum value of the electric field on the surface. Ans. The charge density is q sin 2P*(cosh a - 1) a 4x1 = - - , x = a coth - 2xa sinh2 cr + sin2 p* 2 CURVILINEAR COORDINATES 2 17 on the plane, and q sin 2P* cosh u ~($3) = - - 1 , sln cp = - 2na cosh2 u - sin2 p* cosh a on the boss, where p* is determined from the formula P* tan - = - . 2 a The maximum field is 8qh Emax = /I2 - aB Hint. Use formula 15, p. 385. 461. Solve the preceding problem, assuming that there is a semicylindrical groove in the plane (rather than a boss). Ans. The charge density is sin /ns) (cash a - on the plane, and sin 2(n - P*) COSh ~($3) = - 4 3 1 sin cp = - 3na 2a 2(x - p*) ' cosh a cosh - + cos 3 3 on the surface of the groove. 462. A cylindrical body with cross section in the shape of a symmetrical circular lune is placed in a homogeneous plane-parallel flow of an ideal fluid, with velocity components u, = -v,, u, = u, = 0 (see Figure 124). Find the resulting velocity potential. Ans. 4% P) X sinh a = UmJaz- b2[cosh u + cos p /3z2~-flo I sinh Ap, cosh A(n - P) + 2l sinh TCA sinh (n - Po)A where PROB. 463 Po b tan - = - 2 a and v, is the velocity of the flow far from the body. Hht. Use the system of bipolar co- Yt ordinates a -k ip x + iy = c tanh - 2 x (-a <o: < 03, Po < p <2~- Po), instead of the system given by formula (4), p. 212. 463. Find the torsion function for a FIGURE 125 cylinder whose cross section is a circular lune bounded by the curves P = P, and p = P2 in bipolar coordinates, as shown in Figure 125 (see UI). A ns. S sinh Afi2 sinh A(P - pl) u(a, p> = c2 (cash ?lpcos p - 2 cot p2 cos Aa dh o sinh Ax sinh h(P2 - pl) " sinh A$, sinh A(P2 - cos ha dA), sinh Ax sinh A(P2 - P1) where the parameters p, and p, are determined from the relations Hint. To make the problem homogeneous, subtract out the particular solution &(c2 - x2 - y2) from the equation for the torsion function, where c is the scale factor of the system of bipolar coordinates. 464. A semicircular elastic plate of radius a is clamped along its edges and loaded by a concentrated force P applied at an arbitrary point of its axis of symmetry. Find the distribution of bending moments along the edges of the plate.2 Problcnls 464-466 are treated in Uflyand's book U2. PROB. 466 CURV~L~NEAR COORDINATES 2 19 Ans. P m M/~=~ = - (cosh a + I)/ [sinh sinh A(: - P*) 2x 2 - R cot p* sinh AP* cos ha dh 1 sinh2 (Ax/2) - h2 ' P m hx M]~=,,, = - - cosh a1 [sinh - sinh AP* cot p* 2x 2 - h sinh h (T - P*)] cos ha dA sinh2 (hx/2) - h2 ' where p* is determined from the relation P* b tan - = - , 2 a b is the distance from the point of application of the force to the rectilinear edge of the plate, and a, P is a system of bipolar coordinates. 465. Solve the preceding problem for the case of a uniformly distributed external load q. Write an. expression for the deflection along the axis of symmetry. Ans. Ax u~,=o = (r[sinh Apcos P-hsin p cosh AP-coth- sinh AP 32 D cos2(P/2) 2 where D is the flexural rigidity of the plate. 466. Find the distribution of bending moments along the edges of an elastic plate in the form of a symmetric circular lune -Po < P g Po, due to a concentrated load P applied at the center of the plate. A ns. P sin (3, f sinh AP, cos ha Ml~=i~a = - (COS~ u f cos Po) dh. o, sinh 2AP0 + A sin 2P0 4. Spheroidal Coordinates Turning to three-dimensional coordinate systems, we first consider the case where the region of interest is an ellipsoid. If all three semiaxes of the ellipsoid, are different, it is necessary to deal with Lam6 functions, whose 220 CURVlLrNEAR COORDINATES theory lies beyond the scope of this book.= However, in most cases of practical interest, two of the semiaxes of the ellipsoid are equal. Then the ellipsoid reduces to a spheroid, i.e., an ellipsoid of revolution, the corre- sponding coordinate systems are called spheroidal coordinates, and the appropriate particular solutions of Laplace's equation can be written in terms of elementary functions and spherical harmonics, whose theory, unlike that of Lam6 functions, has been fully developed. Moreover, these particular solutions can be used to solve boundary value problems for the region bounded by a hyperboloid of revolution (see Probs. 483-485).4 By prolate spheroidal coordinates we mean coordinates a, p, cp related to the rectangular coordinates x, y, z by the formulas x = c sinh a sin p cos cp, y = c sinh a sin p sin cp, z = c cosh a cos p, where O<a<co, O<P<x, -x<cp<x, and c > 0 is a scale fa~tor.~ Then every point of space is characterized by a unique triple of numbers a: P, cp. The corresponding triply orthogonal system of surfaces consists of the prolate spheroids a = const with foci at the points (0, 0, kc), the double-sheeted hyperboloids of revolution P = const, which are confocal with the spheroids, and the planes cp = const passing through the z-axis (see Figure 126). The square of the element of arc length and Laplace's equation take the form All = 1 [Lz(sinh a$) + '(sinp6) c2(sinh2 a + sin2 p) sinh a aa If there is no dependence on the angle cp, the appropriate particular solutions For the general theory of ellipsoidal coordinates and Lame functions, see e.g., H4 and W4. Some problen~s involving ellipsoidal coordinates, but not requiring knowledge of Lam6 functions, are given at the end of this section (see Probs. 486-489). Spheroidal coordinates can also be used to solve boundary value problems for Helmholtz's equation, but then the particular solutions involve more complicated functions, called spl~eroidal wave futrctiot~s (see S18, S19). "f a point has cylindrical coordinates r, y, z, then z + ir = c cosh (a+ iP). CURVILINEAR COORDINATES 22 1 of Laplace's equation for dealing with boundary conditions specified on the surface of a prolate spheroid (a = a,) are given by in the case of the interior problem (0 < a < a,), and by in the case of the exterior problem (a, < a < Here P,(z) is the Legendre polynomial of degree n, Q,(z) is the Legendre function of the second kind (of degree n), and M,, N, are arbitrary constants.' Similarly, if there is no dependence on the angle cp, the use of the super- position method to solve boundary value problems for the region bounded by the hyperboloid of revolution P = Po starts from the following particular sol- utions of Laplace's equation, which depend continuously on the parameter r: u = u, = M,P-lm+,,(cosh a)P-u,+iT(f cos P), 7 0. (5) Here PY(z) is the Legendre function of the first kind, and the plus sign pertains to the interior region 0 < p < Po and the minus sign to the exterior region Po < P < X. The general solution is now constructed by integrating (5) with respect to 7. To determine M,, we use the Mehler-Fock the~rern,~ instead of the theory of expansions in series of spherical harmonics. Next we consider oblate spheroidal coordinates a, P, cp related to the rectangular coordinates x, y, z by the formulas x = c cosh a sin p cos cp, y = c cosh a sin p sin cp, z = c sinh u cos p, where O<a<co, O<p<n, -~<cp<x, and c > 0 is a scale fact~r.~ In this case, the triply orthogonal system of For particular solutions in the more general case of dependence on cp, see e.g., L9, p. 218. The functions Q,(z) can be expressed in terms of elementary functions by using the recurrence relation together with the formulas See L9, Sec. 8.9 and also Probs. 483-485. If a point has cylindrical coordinates r, cp, z, we now have z + ir = c sinh (a + iP). 222 CURVILINEAR COORDINATES PROB. 467 surfaces consists of the oblate spheroids a = const, the single-sheeted hyperboloids of revolution P = const and the planes cp = const (see Figure 127). The square of the element of arc length and Laplace's equation now take the form a(sinp$) x - cosha- + -- " 3 sm, a, a, If there is no dependence on the angle cp, the appropriate particular solutions of Laplace's equation for dealing with boundary conditions specified on the surface of an oblate spheroid (a = a,) are given by for the interior problem (0 < a < a,), and by I( = u, = N,Q,,(i sinh a)P,(cos p) for the exterior problem (a, < a < a). Here the boundedness of grad u plays a role (see L9, p. 217). Having made these preliminary remarks, we now give a number of physical problems whose solution involves the use of spheroidal coordinates. 467. Find the charge density on the surface of a conductor in the form of a prolate spheroid with semiaxes a and b, carrying total charge Q. What is the capacitance of the spheroid? Ans. 0 = Q 1 - Q -- 1 4nc2 sinh a, Jcosh2 a,, - cos2 P 4xab2 where b c = Ja2 - b2, tanh a, = - a Hint. Introduce a system of prolate spheroidal coordinates such that the surface of the ellipsoid has equation a = a,. 468. A point charge q is placed at the center of a hollow conducting shield in the form of a prolate ellipsoid with semiaxes a and b. Find the potential distribution inside the shield, assuming that its surface is at zero potential. Ans. where Pn(x) and Qn(x) are the Legendre functions of the first and second kind, tanh a, = bla, and Hint. Subtract the potential of the point charge from the solution. To express the solution in series, use the integral S P2n(x) 'fx = 2P2,(0)Q2,(cosh a). -1 Jsinh2 a + x2 469. Solve Prob. 467 for the case of an oblate spheroid. Ans. 0 = Q 1 - e 1 - 4xc2 cosh a, Jcosh2 a, - sin2 p 4xa2b C c=-, C arc sin - a where b c = Ja2 - b2, tanh ctO = - . a 470. Find the charge density on the surface of a conducting disk of radius a, carrying total charge Q. What is the capacitance of the disk? Ans. where r is the distance from the center of the disk. 224 CURVILINEAR COORDINATES PROB. 471 *471. Find the surface charge density induced on a disk by a point charge q located at an arbitrary point of its axis of symmetry (see Figure 128).1° Ans. 2 i 4xa2 d2 +- JX cos p in terms of the Legendre functions of the first and second kind Pn(z) and Q,(z), where (3 = arc sin (rla), and r is the distance from the center of the disk to an arbitrary point on its surface. Hint. Use the expansion 1 00 = ix(4n + 1)P2,(0)Q2,(i sinh a)PZn(x). Jcosh2 a - x2 n=O 472. A grounded plane screen with a circular aperture of radius a is placed in an electric field which is homogeneous at a great distance from the screen. Suppose the field has the value El to the left of the screen and the value E2 to the right of the screen (see Figure 129). Find the potential in the surrounding space, and calculate the field along the axis of symmetry (a problem of interest in electron optics). Ans. The potential is 't a u],,o = - (El - E,) X 4 - 1 - sinh a arc tan - cos p - E,z, - smh a - I - - - oi - I 1 - sinh a arc tan - cos p - E,z, slnh a FIGURE 129 while .1 sinh a E~~=o,~>o = Ez + Ed (arc tan - - - x sinh a cosh2 a 1 sinh a Ez - tan - - - EIT=o,z<o = El + - x s~nh a cosh2 a is the field along the axis. ? lo This problem can be solved more easily by using integral equations (see Prob. 551b). PROB. 475 CURVILINEAR COORDINATES 225 Hint. Introduce spheroidal coordinates with parameter c equal to the radius of the aperture, and look for a solution of the form ul,,, = A2(a) cos fi - E2z. 473. An oblate dielectric spheroid, with semiaxes a and b and dielectric constant E, is placed in a homogeneous electric field E, directed along its axis of symmetry (in the negative z-direction). Solve the resulting problem of electrostatics. Ans. The potential is U = Eoz + const E cosh2 a, - sinh2 a, - (E - 1) sinh a, cosh2 a, arc cot sinh a, in the dielectric, and E,c(E - 1) sinh a, cosh2 a,(l - sinh a arc cot sinh u) cos P u = E,z - E cosh2 a, - sinh2 a, - (E - 1) sinh a, cosh2 a, arc cot sinh a, + const in the air, where b tanh a, = - a 474. Find the resistance of a grounding rod inserted in ground of con- ductivity a (see Figure 130), assuming that the rod is shaped like half of a prolate spheroid with semiaxes a and b, where a > b (see 01). Ans. 475. A constant current J enters the ground through a point contact placed on the earth's surface over a hole filled with material of conductivity a,, different from the conductivity o, of the rest of the ground (see Figure 131). 226 CURVILINEAR COORDINATES PROB. 476 Find the current distribution in the ground, assuming that the boundary between the two media is the prolate spheroid with equation Ans. The potentials of the current field in the two media are given by X Qzn(c0sh a)P,n(cos p) olQzn(cosh a,)P~,(cosh a,) - 02P2,(cosh a,)Qh,(cosh a,) ' where R is the distance from the source to the field point, tan11 a, = bla, Pn(x) and Qn(x) are Legendre functions, and 476. A d-c current enters ground of conductivity s through a grounding plate in the form of a disk of radius a (see Figure ! 32). Find the distribution of current under the plate, and calculate the resistance of the plate. r Ans. The potential of the current field is 2v u = - arc cot sinh a, 7C where V is the potential of the plate. The resistance is Hint. Introduce a system of spheroidal coordinates (0 g a < m, 0 < p < ~12). 477. A prolate spheroid made from material of magnetic permeability p is introduced into a homogeneous magnetic field H, directed along its axis of symmetry (in the negative z-direction). Solve the resulting problem of magnetostatics, and show that the field inside the spheroid is homogeneous. Ans. The field inside the spheroid is cosh2 a, - p sinh2 a, + (p - 1) sinh2 a, cosh a, In coth 3 2 where b tanh a, = - . a Substitution for a, leads to the expression 478. Find the stationary temperature distribution in a prolate spheroid u = a,, if a given axially symmetric temperature distribution T(a9 P)I,=a, = f (P) is maintained on its surface. Consider the special case where one half of the surface of the spheroid (z < 0) is held at temperature zero, while the other half (z > 0) is held at temperature To. Ans. in terms of the Legendre polynolnials P,(x). In the special case, " 4ni3 T(a, P) = - 1 + 2 --- P2n(0) PZn+,(cos P)P2n+l(cosh a)]. :[ .., 2(n + 1) P2,+,(cosh a,) Hint. To calculate the integral lo1 Pn(x) dx, use the recurrence relation (2n + l)Pn(x) = Pn+,(x) - Pa-,(XI. 479. Find the stationary temperature distribution in a prolate spheroid with semiaxes a and b, whose surface is held at temperature zero, if heat is produced inside the spheroid with constant density Q. Ans. in terms of the spheroidal coordinates a and P, where c = Ja2 - b2 and k is the thermal conductivity. 228 CURVILINEAR COORDINATES PROB. 480 Hint. Make the problem homogeneous by subtracting out the particular solution - Qr2/4k of the inhomogeneous heat conduction equation. 480. A body in the shape of a prolate spheroid with semiaxes a and b is placed in a homogeneous flow of an ideal fluid, directed along its axis of symmetry (in the negative z-direction). Find the resulting velocity potential. Ans. a cosh a In coth - - 1 cos p + const, 2 a-c where c = Ja2 - b2 and v, is the velocity far from the body. *481. Calculate the gravitational potential due to a homogeneous prolate spheroid with semiaxes a and b, and find an asymptotic expression for the potential in the case of small eccentricity c. Ans. The potential outside the spheroid is + [2(sin2 p - sinh2 a) + 3 sin2 p sinh2 a] In coth where p is the density, and the gravitational constant is taken to be unity. For small c, where is the mass of the ellipsoid, and r 3x2 - 1 R = Jr2 + z2, 0 = arc tan - , p2[.x) = 7 . z Hint. Inside the spheroid, subtract out the particular solution -xpr2 of the inhomogeneous equation. 482. Solve the preceding problem for the case of an oblate spheroid. Ans. Outside the spheroid the gravitational potential is - 3 cosh2 a sin2 P] arc cot sinh a - sinh a(3 cos2 P - 1)). For small c, where M = Qxpa2b. Hint. Inside the spheroid subtract out the particular solution of the inhomogeneous equation. *483. A point charge q is placed at FIGURE 133 the focus of a grounded conducting screen shaped like a hyperboloid of revolution (see Figure 133). Solve the resulting problem of electrostatics. Ans. The electrostatic potential is T tanh xr P-w+,,(-cos Po) u(a, P) = - - - P_~+iT(cosh a) cash XT P-x+,,(cos Po) x P-w+,,(cos P) d~, 0 < P < Po, where PY(x) is the Legendre function of the first kind. Hint. Introduce prolate spheroidal coordinates a, P, cp such that the hyperboloid has equation P = Po, and make use of the Mehler-Fock theorem (see L9, p. 221). 484. A point charge q is placed near the vertex of an electrode shaped like a hyperboloid of revolution. Find the potential in the surrounding space, assuming that the charge lies on the axis of the hyperboloid (see Figure 134). Ans. 4 U=--- -SiD,tanhnTr(t + F) R 2cJn o z i~ P-I*,+,,(-cos P) r(i - -) 2 P-1,4+,,(-cos Po) x P-w+,,(cos P,)P-~+,,(cosh a) d~, 485. A d-c current J flows into ground of conductivity o through an 230 CURVILINE:.R COORDINATES PROB. 486 electrode placed at the bottom of a hollow shaped like a hyperboloid of revolution, with equation p = So in spheroidal coordinates (see Figure 135). Find the current distribution in the ground. Ans. The potential of the current field is J J ,J=--- 2noR 2xac r r sinh x.c PLx+,,(cos Po) where R is the distance from the source to the field point, c is the eccentricity of the hyperbola P = Po, and P,(x) is Legendre's function. 486. Find the charge density on the surface of an ellipsoidal conductor with semiaxes a, b and c, carrying total charge Q. What is the capacitance of the ellipsoid ? Ans. I\ I SL 1 0 = - 4xa bc 2 ds Hint. Introduce ellipsoidal coordinates u, P, y, defined as the roots of the cubic equation x2 y2 +-=I. z2 +- a2+A b2+A c2+A Then look for a solution depending only on~cr. 487. Find the charge density on a thin.elliptic plate with semiaxes a and b, carrying total charge Q. What is the capacitance of the plate? a C= ~(~7) ' where K(k) is the complete elliptic integral of the first kind. Hint. Take the limit c -+ 0 in the solution of the preceding problem. 488. An ellipsoid with semiaxes a, b and c, made from material of magnetic permeability [A, is placed in a homogeneous magnetic field Ho directed along its major axis. Find the resulting magnetic field inside the ellipsoid. Ans. The direction of the field coincides with that of the external field. The magnitude of the field equals 489. Calculate the gravitational potential of a homogeneous ellipsoid of density p (see S16, p. 161). Ans. where A is the positive root of the equation and the gravitational constant is taken to be unity. 5. Paraboloidal Coordinates Physical problems involving a region bounded by a paraboloid of revolu- tion can be solved by introducing paraboloidal coordinates a, P, cp related to the rectangular coordinates x, y, z by the formulas C x = cap cos cp, y = cap sin cp, z = - (a2 - P2), 2 where O<a<co, O<p<c~, -n<cp<x, and c > 0 is a scale factor.ll In this case, the triply orthogonal system of coordinate surfaces consists of the two families of paraboloids of revolution l1 If a point has cylindrical coordinates r, cp, z, then 232 CURVlLlNEAR COORDINATES PROB. 490 u = const and p = const, together with the ,,,,+ planes cp = const (see Figure 136). The square of the element of arc length and Laplacels equation take the form ds2 = c2(u2 + P2)(du2 + dp2) + c2a2p2 dcp2, z Au = I a ( a,) (; I) a%] +-- p- + -+- -. pap aP P2 a(f2 If there is no dependence on the angle cp,l2 the use of the superposition method to FIGURE 136 solve boundary value problems for the region bounded by a paraboloid of revolution p = Po starts from the following particular solutions of Laplace's equation, which depend continuously on the parameter A:'= Here Io(x), Jo(x) and Ko(x) are cylinder functions, the upper row pertains to the interior region (0 < p < Po) and the lower row to the exterior region (Po < P < 03). The general solution is now constructed by integrating (6) with respect to A, where, to determine MA, we use Hankel's integral theorem [see formula (12), p. 1601. Paraboloidal coordinates can also be used to solve boundary value problems for Helmholtz's equation, but then the particular solutions involve confluent hypergeometric functions (see E2, Vol. 2, Secs. 8.7-8.8). 490. Solve Prob. 483, assuming that the conducting screen is shaped like a paraboloid of revolution, with equation P = Po in paraboloidal coordinates. Ans. u(u, P) = 2q - 2/" %@!d Io(h~)Jo(ha) h dh, 4." P2) C 0 IO(AP0) in terms of the Bessel function of the first kind Jo(x) and the Bessel functions of imaginary argument I,(x) and Ko(x). Note that l2 See Prob. 492 for the case where dependence on g, is present. l3 Formula (6) is an abbreviated way of writing two formulas, one involving the function Io(AP), the other Ko(AP). PROB. 492 CURVILINEAR COORDINATES 233 wherep is the focal distance and c is the scale factor figuring in the definition of the paraboloidal coordinates. Hint. Use the integral 491. Find the stationary temperature distribution in a body shaped like a paraboloid of revolution P = Po, if a given axially symmetric temperature distribution T(% P)lp=p, = f (a) is maintained on its surface. Ans. 492. Solve the Dirichlet problem for the domain bounded by the parab- oloid of revolution P = Po, assuming that the boundary condition is of the form cos ncp ~lp=po = fn(~) , n=0,1,2 ,..., sin ncp where f,(a) is a given function. Use the result to construct solutions for arbitrary boundary conditions depending on cp. Ans. Inside the paraboloid, m cos ncp .(a, P, cp) =I fn(A) Jn(Aa)h dh 0 I,@Po) sin ncp ' wherefn(h) is the Hankel transform of fn(a):14 6. Toroidal Coordinates Besides spherical and spheroidal coordinates, there are other coordinate systems whose use is intimately connected with Legendre functions. First we consider toroidal coordinates u, P, cp related to the rectangular coordinates x, y, z by the formulas c sinh u cos cp c sinh a sin cp c sin p X = 9 Y= , z= cosh a - cos p cosh u - cos p cosh u - cos p 9 (7) l4 Cf. formula (13), p. 160. where Ogu<co, -n<p<n, --n<cpgn, and c > 0 is a scale fact~r.~~.~~ The corresponding triply orthogonal system of surfaces consists of the toroidal surfaces u = const, satisfying the equation (r - c coth + z2 = - isis .)" where r = Jx2 f y2, the spheres fi = const, satisfying the equation and the planes intersect in the (r - c cot p) + r2 = - (si: J' cp = const (see Figure 137). Note that all the spheres (8) circle r = c, z = 0. It is clear from (7) that x, y and z are l6 In the next section, we shall consider a closely related coordinate system, i.e., three- dimensional bipolar coordinates. la If a point has cylindrical coordinates r, rp, z, then c sinh a c sin p r = z = cosh a - cos p ' cosh a - cos p ' or more concisely, a + ip z $ ir = iccoth - 2 ' periodic in p and cp, with period 2x. Therefore we can choose Pl < P < + 2n, y1 < cp g y1 + 2~ instead of -x < P < x, -7~ < cp g 7~ (which corresponds to the particular choice P, = cp,.= -x), and it is sometimes convenient to do so. In toroidal coordinates, the square of the element of arc length is ds2 = c2 (da2 + dp2 + sinh2 a dY2), (cosh a - cos P)' and Laplace's equation takes the form sinh a &) +$( sinh a COS~ a - cos p aa COS~ a - cos p ap + 1 aZu - = 0. (9) sinh a(cosh a - cos p) ay2 Unlike the cases considered so far, equation (9) does not permit separation of variables directly. However, if we first introduce a new function v by making the substitution u = J2 cosh a - 2 cos p v, (9) goes into a new equation belonging to the class which permits separation of variables (see L9, p. 223). If there is no dependence on the angle cp, it turns out that Laplace's equation (9) has particular solutions of the form u = u, = J2 cosh a - 2 cos p [A,PV-~(co~h a) + B,Q,-%(cash a)] x [C, cos vp + D, sin $1, in terms of the Legendre functions of the first and second kinds, where v is a parameter and A,, . . . , Dv are arbitrary constants. In boundary value problems involving the region bounded by a torus, the parameter v is deter- mined by the requirement that the solution be periodic in P. This leads to the particular solutions Q,-x(cosh a) u = u, = J2 cosh a - 2 cos p [M, cos np + N, sin np1 Pn-x(cosh a) ' where the upper row pertains to the interior problem (a, < a < co) and the lower row to the exterior problem (0 < a < a,). In problems involving the region bounded by two intersecting spheres /3 = P, and P = P,, the appro- priate particular solutions are obtained by choosing v = ir (r >, O), and are of the form 236 CURVILINEAR COORDINATES PROB. 493 where (3, < P < P, for the interior problem and P, < P < 2x + P1 for the exterior problem. Then the solution of the problem is constructed by integrating (10) with respect to T, where the factors M, and N, are determined by using the Mehler-Fock theorem (see L9, Sec. 8.12). This section contains a number of physical problems which can be solved by using toroidal coordinates. Most of the problems are rather difficult, and are intended for those with the necessary background in the theory of special functions.17 493. Find the electrostatic potential due to a charged toroidal conductor at potential V, with the dimensions shown in Figure 138. Calculate the capacitance of the conductor. Ans. The potential is v u(a, p) = - J2 cosh cr - 2 cos p X z x [e-112(C0Sh ~-~~~(cosh a) P-1,2(cosh a,) + 22 Qn-x(cOsh ~.-%(cosh a) cos np , .=I Pn-w(cosh uo) and the capacitance is. I where Pv(x) and Qv(x) are the Legendre functions of the first and second kind, and Hint. Introduce toroidal coordinates a, P, cp with parameter c, such that the surface of the conductor has equation u = a,. In the course of the solution, use the integral S " cosnpdp = Qn-%(cash a,). o J2 cosh a, - 2 cos p *494. Find the distribution of electrostatic potential on the axis of a grounded conducting torus introduced into a homogeneous electric field E,, directed along its axis of symmetry (in the negative z-direction). l7 Some of the problems can be solved more easily by using other methods (by inversion, say). PROB. 497 CURVILINEAR COORDINATES 237 Ans. 8 - P " nQn-dcosh a01 sin np, ul,=, = E,z - - E,JP - a2 sin - x x 2 Pn-%(co~h a,) where 1 COS~ a, = - , a and the dimensions I and a are the same as in Figure 138. 495. Solve the preceding problem, assuming that the external field is due to a point charge q at the center of the torus. Ans. 9 29 sin iP Q-i/z(cosh ao) + 2 (- Qn-%(cosh a,) ul,,, = - - cos np . r ,dm [P-lldcosh a,) ..I P,-%(cash a,) 1 496. A current Jflows in a ring-shaped conductor of circular cross section (see Figure 138). Find the resulting magnetic field along the z-axis, assuming that the current J is uniformly distributed over the cross section of the ring. Ans. CO + 22 [~;-%(cosh u~)Q~-~(cos~ a,) - Q?-%(cash ao)~n'-%(cash a,)] cos np , n=l where 1 1 COS~ a, = - , a c is the velocity of light, and Qi(x), Qt(x) are associated Legendre functions of the second kind. 497. Find the distribution of a-c current along the surface of a perfect conductor shaped like a ring with circular cross section. Calculate the self- inductance L of the ring.ls Ans. cosh a, 1 Qk-%(cash a,) - 22 - 4n2 - 1 Pi-%(cash a,) l8 This is the skin effect problem (see Fl). 238 CURVILINEAR COORDINATES PROB. 498 the dimensions I and a are the same as in Figure 138, and Pt(x), Q:(x) are associated Legendre functions of the first and second kind. The distribution of current density along the periphery of the ring is LJ [2(cosh a, - cos 8)13/' +22 cos np j(P) = - -3 32x sinh a0(l2 - a2) P-l12(cosh a,) ,=I P;-%(cosh ao) where J is the total current. *498. Suppose a d-c,current flows in a ring-shaped conductor with the dimensions shown in Figure 138, producing heat with density Q. Find the temperature distribution inside the conductor, assuming that its surface is held at temperature zero. Ans. Q(1" a2) sinh2 a T(a, B) = - - 42 cosh a - 2 cos IJ k ((2 cosh a - 2 cos p)' 2 sinh2 a,? Q;-%(cosh a,) + L Q,-%(cos~ a) cos nB , 3x .=I Qn-%(cosh a,) I I in terms of the Legendre function of the second kind Q,(z), where 1 cosh a, = - , a and k is the thermal conductivity. Hint. Subtract out the particular solution - Qr2/4k of the inhomogeneous heat conduction equation. Use the integral S " cosnpdp 1 = - Qi-l,t;(cosh a). o (2 cosh a - 2 cos p)5/2 3 499. Calculate the gravitational potential of a homogeneous torus of density p, with the dimensions shown in Figure 138, assuming that the gravitational constant equals unity. Ans. 4pc2 u(a, B) = - - sinh2 a, J2 cosh a. - 2 cos p [Q-l12(c~sh ao)~?l~z(cosh a,) 3 PROB. 501 CURVILINEAR COORDINATES 239 where Pv(x) and Qv(x) are Legendre functions, Qt(x) is the associated Legendre function of the second kind, and 500. A torus with the dimensions shown in Figure 138 is introduced into a homogeneous flow of an ideal fluid, whose direction coincides with the axis of symmetry of the torus. Solve the resulting hydrodynamical problem, and find the velocity distribution along the axis. Ans. The stream function is vwrS v = ---- + sinh a 2 J2 cosh a - 2 cos p V) x [: P?,,,(cosh a) + 2 c$'h-~(cosh a) cos na , n=l where P~(x) and Qt(x) are associated Legendre functions, I + 2A [sinh aoQ',-x(cosh a,) - 11 (4n2 - 1) sinh a, cosh a, = //a, and uw is the velocity of the flow far from the torus. The constant A is determined from the condition 501. Find the surface density of free charge on a thin charged conductor shaped like a spherical bowl of radius a (see Figure 139). Calculate the capacitance of the bowl (see 52, p. 250). Ans. The charged density is ' t v J2 cosh o: - 2 cos p, 0. = - 4x2a [ 2 cos 42 cosh a - 2 cos po - arc tan 2 cos +Po on the inner surface of the bowl, and I PROB. 502 on the outer surface, where V is the potential of the bowl and C sin Po = - . a Using the formula 4a2 - b2 cash u - cos Po -- - b2 - p2 2 cos2 iP0 ' where the distances b and p are shown in Figure 139, we find that The capacitance of the bowl is b ------- 2a b c = -. J4a2 - b2 + - arc tan - 2~a x J4a2 - b2 ' Hint. To calculate the density, use the integral x [I+ 2 cos BBo 2 cos 4po arc tan - J2 cosh - 2 cos po J2 cosh a - 2 cos p0 I. *502. Find the surface density of induced charge on a thin conductor shaped like a spherical bowl of radius a, due to a point charge q located at the point r = z = 0 (see Figure 140). Ans. a. = qb2J4a2 - b2 8n3a2R3 b2 - 2a2 z where a, and ai are the bowl. the charge densities on the outer and inner surfaces of Hint. Subtract out the potential of the point charge. To expand this potential in a Mehler-Fock integral, use the relation 1 ~-~+,(cosh u) d~. J2 cosh a + 2 cos po 503. Find the potential distribution in the space surrounding a charged conductor shaped like the "spherical zone" shown in Figure 141. Ans. u(a, p) = vJ2 cosh a - 2 cos P 8: r x r[sinh (Ix + po - P)r - cosh (x - p0)r P-%+Jcosh a) x sinh (n - P)r] dry sinh (IG + PO)r cash nr where PY(x) is the Legendre function of the 2@-z first kind, V is the potential of the conductor FIGURE 141 and sin po = c/a. 504. Use the result of the preceding problem to calculate the capacitance of a hemisphere of radius a. Ans. 505. A lens-shaped conductor at zero potential is introduced into a homogeneous electric field Eo directed along its axis of symmetry (in the negative z-direction), as shown in Figure 142. Find the resulting potential distributibn. Ans. = Eoz - 2~,cJ2 cosh a - 2 cos P r sinh (x - Po)i cosh xr sinh p~ x --- P-%+,(cosh a) dr. sinh P,r I 506. Find the gravitational po- FIGURE 142 tential of a homogeneous hemisphere of density p and radius a. Ans. The potential outside the hemisphere is 2npa2 ~(a, p) = - J2 cosh a - 2 cos p 3 cosh XT - sinh (p - n)r + (I + 27') sinh ($ - p)r] cosh (xr/2) P-~+,(cosh a) x dr sinh (3x712) cosh xr Hint. Inside the hemisphere, subtract out the particular solution -2xpz2 of the inhomogeneous equation. Use the integral 3 sin2 p P-%+,(cash a) d~ (2 cosh a - 2 cos p)6/2 cosh XT +Soa T' cosh (X - @T P-lm+,(~osh a) d~. cosh XT 7. Three-Dimensional Bipolar Coordinates By three-dimensional bipolar coordinates, we mean coordinates a, p, rg related to the rectangular coordinates x, y, z by the formulas c sin u cos cp c sin a sin cp X = c sinh p 9 y= , z= cosh p - cos a cosh p - cos a cosh p - cos u 9 (11) where O<a<x, -co<p<co, -x<cp<x, and c > 0 is a scale factor.19 The close resemblance between (11) and the formulas defining toroidal coordinates should be noted (see p. 233). The corresponding triply orthogonal system of surfaces consists of the spindle- shaped surfaces of revolution a = const, satisfying the equation (r - c cot a)' + z2 = (k', the spheres P = const satisfying the equation (z - c coth P)' + r2 = and the planes cp = const (see Figure 143). The square of the element of arc length is ds2 = c2 (da2 + dp2 + sin2 a dcp2), (cosh p - cos a)' l9 If a point has cylindrical coordinates r, 9, z, then c sin a c sinh p r = z = cosh p - cos a ' cosh p - cos a ' or more concisely u + ip z + ir = iccot - 2 ' and hence Laplace's equation takes the form + 1 aZu - = 0. (12) sin a(cosh p - cos a) aY2 To separate variables in (12), we first introduce a new function v by making the substitution u = 42 cosh p - 2 cos a r, as in the case of toroidal coordinates. If there is no dependence on the angle 9, it turns out that Laplace's equation (12) has particular solutions of the form u = U, = J2 cosh p - 2 cos a [A,P,(cos a) + B,(cos a)] x [C, cosh (v + 8)p + D, sinh (v + @I, in terms of the Legendre functions of the first and second kinds, where v is a parameter and A,, . . . , Dy are arbitrary constants (see L9, p. 232). In boundary value problems involving a region (3, < P < P, bounded by two nonintersecting spheres P = PI and (3 = P,, it is easy to see that the appro- priate particular solutions are u = u, = J2 cosh p - 2 cos a [M, cosh (n + t)p + N, sinh (n + +)P]P,(cos a), n = 0, 1, 2, . . . , in terms of the Legendre polynomials P,(x), and the general solution is constructed by summing these solutions. In problems involving the region 244 CURVILINEAR COORDINATES PROB. 507 bounded by the spindle-shaped surface u = a,, the appropriate particular solutions are obtained by choosing v = -a + i~ (T >, O), and are of the form u = u, = 42 cosh p - 2 cos u [M, cos ~p + N, sin T~]P-~+~~(~cos a), 7 > 0, (13) where the plus sign corresponds to the exterior problem (0 < a < a,) and the minus sign to the interior problem (ao < u < TC). In this case, the general solution is obtained by integrating (13) with respect to T, and the factors M, and N, are determined by taking Fourier cosine and sine trans- forms with respect to p. This section contains problems from various branches of mathematical physics which can be solved by using three-dimensional bipolar coordinates. The last three problems (Probs. 512- 514) involve limiting cases of bipolar and toroidal coordinates, and lead to elegant formulas for the capacitance Z of such objects as a pair of spheres in contact or the surface obtained by " rotating a circle about a tangent line. I 507. Find the electrostatic field in FIGURE 144 a spark gap consisting of two con- ducting spheres of radius a, with centers a distance 21 apart, if the spheres are at potentials V, and V, respectively (see Figure 144). Ans. The electrostatic potential is in terms of the Legendre polynomials Pn(x), where 1 cosh Po = - a Hint. Use the expansion 1 a) - - 2 e-(n+lm)~~n(cos a). J2 cash p - 2 cos a ,=, *508. Find the capacitances C,,, C,, and C,, of a system of conductors consisting of two spheres of radii a, and a,, with centers a distance 21 aparLZ0 a0 Concerning the meaning of C,,, C,, and C,,, see the solution, p. 370. PROB. 509 CURVILINEAR COORDINATES 245 Assuming that the radii are equal (a, = a, = a), tabulate C12 as a function of the ratio //a. Ans. w ( + 2 [e-(n+")@a cosh (n + +)(PI + P2) - e-("+")Dl] C2, = c - 2 sinh p2 ,=, where p,, P, and c are determined from the relations 2 2 412 + '1 - '2 , cash (3, = 41, - a: + a: cosh = 410, 4Ia, , c = a, sinh p, = a, sinh p,. Hint. In three-dimensional bipolar coordinates a, P, cp, the surfaces of the conductors have equations p = -pl and p = P,. 509. A conducting sphere of radius a is buried to a given depth in a liquid of dielectric constant E. Find the potential distribution outside the sphere, assuming that the sphere is at potential V (see Figure 145). Calculate the capacitance of the sphere. Ans. 00 e(n+x)(~-~o)p (COS CC) u, = VEJ~COS~ P - 2cosaC .=o sinh (n + $)Po + E cosh (n + $)Po ' PROB. 510 in terms of the Legendre polynomials P,(x), where 1 cosh Po = - a 510. Find the potential distribution outside a charged spindle-shaped conductor at potential V (see Figure 146). Ans. ~(a, p) = V J2 cosh P - 2 cos a cos $3 P-%+iT(-~o~ a,) x jm- P-lm+,(cos a) dr, o cosh rn P-%+,(cos a,) in terms of the Legendre function PY(x), where C sln a, = - . a Hint. In bipolar coordinates a, p, 9, the surface of the conductor has equation a = a,. In the course of the solution, use the integral representation cos rP dp P-lm+,(-cos a,) = 511. Solve the preceding problem, assuming that the conductor is placed in a homogeneous electric field E, directed along the axis of rotation (in the negative z-direction). Ans. u = Eoz - 2~,c J2 cosh P - 2 cos a - rcos; rn P-lm+,(-cos a,) x P-N+iT(~~~ a) sin Pr dr. P-!A+,(cos a,) PROB. 515 CURVILINEAR COORDINATES 247 *512. Calculate the capacitance of a conductor consisting of two touching spheres of equal radius (see Figure 147). Ans. C = 2a In 2. 't Hint. Introduce degenerate bipolar I u = const coordinates, defined by the formula C z+ir=- a+ ip' U=O -2 which can be obtained from the formula a + iP z + ir = ic cot - 2 (cf. footnote 19, p. 242) by replacing FIGURE 147 a, p, c by UE, PE, ~CE and taking the limit as E + 0. Then the surfaces of the spheres have equations p = &Po. 513. Calculate the capacitance of a conducting sphere of radius a lying on a plane with dielectric constant E (see Figure 148). Ans. E+ll,,E+l C=a- E-1 2 514. Calculate the capacitance of a conductor in the shape of the surface r obtained by rotating a circle of radius a about one of its tangents (a "doughnut without a hole"). Ans. FIGURE 148 c = %Im dx, x 0 Idx) where Io(x) and Ko(x) are Bessel functions of imaginary argument. Hint. The surface of the conductor has the equation a = U, in degenerate bipolar coordinates (see the hint to Prob. 512). 8. Some General Problems on Separation of Variables 515. Show that a necessary and sufficient condition for being able to separate variables in Helmholtz's equation Au + k2u = 0 (where A is the two-dimensional Laplace operator) in a system of curvilinear coordinates a, P defined by the formula x+iy=f(a+ip) (14) 248 CURVILINEAR COORDINATES PROB. 5 16 (where f is analytic) is that f be the solution of the third-order linear differ- ential equation f '"(t) - Af '(Z) = 0 (A is an arbitrary constant). 516. Using the result of the preceding problem, show that apart from linear transformations (corresponding to translation and rotation of the coordinate axes or change of scale in the xy and @-planes) the only trans- formations of the form (14) leading to separation of variables in Helmholtz's equation are the following: x + iy = ea+@ (polar coordinates), x + iy = cosh (a + ip) (elliptic coordinates), x + iy = (a + ip)2 (parabolic coordinates). 517. Show that Laplace's equation has infinitely many particular solutions of the form where a, p, cp are a system of orthogonal curvilinear coordinates defined by the formula z+ ir = f(a+ ip), and f (Z) is a solution of the differential equation where the A, are arbitrary real constants (see L2). 518. Show that all the three-dimensional coordinate systems considered in this chapter (as well as .cylindrical and spherical coordinates) can be obtained as special cases of the coordinate system of the preceding problem. Ans. Cylindrical coordinates: f(<)=Z, Ao=l, A,=A2=A,=A4=0, u = [AJ,(vr) + BY,(vr)][C cosh vz f D sinh vz] 'OS sin pcp ' where J&) and Y,(x) are Bessel functions of the first and second kind. PROB. 518 Spherical coordinates: f(<)=es, A,= 1, Ao=A,=A3=A,=0, a=Inr, p=0, where Pf(x) and Q;(x) are associated Legendre functions of the first and second kind for the interval (-1, Prolate spheroidal coordinates: ~(<)=ccos~<, A,= -cZ, A2= 1, Al=h3= A4=0, u = [APf(cosh a) + BQXcosh a)][CPr(cos P) + DQ:(cos a)] C:: L:. Oblate spheroidal coordinates : f(<)=csinh<, A0=c2, A2=l, Al=A3=A4=0, u = [APr(i sinh a) + BQ:(i sinh a)][C,Pr(cos P) + DQr(cos P)] 'OS sin py Paraboloidal coordinates : cos tLY u = [AJ,(va) + By,(va)l[clJvP) + DK,(vP)I sin prp ' where I,(x) and K,(x) are Bessel functions of imaginary argument. Toroidal coordinates : x [C cos vp + D sin vpl 'OS pY sin py ' Three-dimensional bipolar coordinates: c2 1 A --- 1 Al=hB=O, < --- --, 4- f(5) = ci cot - 2 ' 0 - 4' z-2 4cZ ' u = J2 cosh p - 2 cos a[AP:(cos a) + BQ:(cos a)] x [C cash (v + 1)P + D sinh (v + h)P] :: L: . See e.g., L9, p. 193. 250 CURVILINEAR COORDINATES PROB. 519 519. Prove that besides the coordinate systems listed in Prob. 518, separation of variables in Laplace's equation is also possible in coordinates defined by the formula z+ ir= f(a+ ip), where f (C) is one of the Jacobian elliptic functions sn C, cn C, dn C.22 Con- struct particular solutions of the form for each of these three functions. where A(a) and B(p) are solutions of the differential equations and p, v are arbitrary parameters. 22 See L2, W2, W3, and also the paper L4, where a system of solutions of Laplace's equation suitable for solving boundary value problems for a ring of oval cross section is constructed. PROB. 521 CURVILINEAR COORDINATES 25 1 520. Verify that the biharmonic equation A2u = 0 (where A is the two- dimensional Laplace operator) has infinitely many particular solutions of the form cos AP u = If '(a + iP)l Nu) sin Ap , where a, p is a system of two-dimensional curvilinear coordinates defined by the formula x + (Y = f(a + iP), where dl: fco =Iz). and F(l:) is the solution of the differential equation (A and p are arbitrary parameters). 521. Using the result of the preceding problem, show that the two- dimensional biharmonic equation permits separation of variables in rec- tangular, polar, two-dimensional bipolar and degenerate bipolar coordinates, and construct the corresponding particular solutions. Ans. The general transformation called for here is of the form where a, b and d are arbitrary constants. 1. Rectangular coordinates: f(l:)=l:, p-0, a=l, b=d=O, u=x, p=y, u = (A cosh Ax + B sinh Ax + Cx cosh Ax + Dx sin Ax) cos Ay sin Ay ' 2. Polar coordinates : a = In r, P = cp, cos Acp u = (~r" Br-A + crG2 - Dr-") sin 252 CURVILINEAR COORDINATES PROB. 521 3. Bipolar coordinates: C u = [A cosh (A + 1)a + B sinh (A + l)a cosh a + cos @ cos A@ + C cosh (A - 1)a + D sinh (A - 1)aJ sin Ap . 4. Degenerate bipolar coordinates : C [A cosh Aa + B sinh Aa + Ca cosh Aa + Da sinh Aa] cos A@ u = (a" P2I2 sin A@ ' References Books: Bateman (B2), Hobson (H4), Lebedev (L9), Lense (L1 l), Magnus and Oberhettinger (M3), Morse and Feshbach (M9), Smythe (S7), Snow (S12), Stratton et al. (see p. 358) (S18), Strutt (S19). Papers: Bacher (B5), Eisenhart (El), Haentzschel (Hl, H2), Lagrange (L2), Stepanov (S15), Wangerin (W2, W3). INTEGRAL EQUATIONS The use of integral equations to prove existence theorems for problems of mathematical physics, or to find approximate solutions, is a classical subject, which lies outside the scope of this book but is treated in con- siderable detail in the available literature. The purpose of this chapter is simply to show how integral equations can be used to find exact solutions of certain physical problems. The methods we have in mind are admittedly quite special, but very effective in the cases to which they apply, and their full possibilities do not yet seem to have been exploited. As an example of the successful application of integral equations to physical problems, we cite the work of Grinberg, summarized in his book G5, devoted to the solution of a number of interesting problems from the theory of electricity and magnetism. This chapter consists of two sections. The first is devoted to some nonstationary problems of diffraction theory which can be reduced to the solution of familiar integral equations, e.g., Abel's equation, Volterra's equation with a difference kernel, etc. The second section, stemming from Grinberg's work, is primarily concerned with stationary problems stated in terms of electrostatics, but with obvious analogues involving magneto- statics, heat conduction or d-c current flow. Because of their relatively greater difficulty, we omit problems whose solution requires the use of the Wiener-Hopf method, or problems which involve singular integral equations containing integrals of the Cauchy type. Concerning these topics, the reader should consult the relevant references cited at the end of the chapter (see p. 271). PROB. 52'2 I. Diffraction Theory *522. A plane electromagnetic wave with electric field components is incident on a perfectly conducting half-plane (screen) x > 0, z = 0. Denoting the components of the resulting electric field (the sum of the incident and reflected waves) by 0, 0, E and setting show that the reflected wave u can be represented in the form where the function y(s) satisfies Abel's integral equation Hint. Look for a solution of the wave equation depending only on E and 1.. *523. Solve Prob. 522, assuming t I that the incident wave encounters the Quiescent zone -1 screen at the time t = 0, i.e., Describe the diffraction process graph- ically. Ans. /J;* a ds, u = ui > 0, JE - s where The diffraction process is illustrated in Figure 149. 524. Solve Prob. 523 for the special case where a) g(6) = 1 (a wave with a rectilinear front); b) g(E) = sin WE. Ans. In the notation of Prob. 523, the reflected wave u has the following representation in the excited zone : 525. By passing to the limit t + oo in the formulas of Prob. 524, solve the well-known Sommerfeld problem on the steady-state sinusoidal electro- magnetic oscillations due to a plane wave incident on the edge of a con- ducting screen (see Prob. 426). Ans. where 2. m = ,; ,m/4 e -ikx e-is2 ds, k=-. w Ids) v 526. A plane electromagnetic wave with components is incident on a perfectly conducting screen shaped like a parabolic cylinder r = x + 2a. Setting where E is the z-component of the electric field, show that the reflected wave can be represented in the form where cp(s) is the solution of the integral equation Hint. Look for a solution of the waveequation dependingonly on E and q. 256 INTEGRAL EQUATIONS PROB. 527 527. Solve the preceding problzm of diffraction theory, assuming that the wave makes contact with the screen at the time t = 0 and is continuous along its front,' i.e., Describe the diffraction process graphically. Ans. where lo, Here and Rare the Laplace transforms of the functions g(E) and so that where @(x) is the probability integral and the path of integration r is a Quiescent rX straight line to the imaginary axis lying to the right of the singular points of the integrand. The diffraction process is indicated in Figure 150. The boundary of the excited zone is the envelope of the secondary waves reflected from points of the screen, in keep- ing with Huygens' principle. 528. Suppose the incident wave in Frob. 547 has the equation Show that the reflected wave u can then be represented in the form 21 = - arc tan , q>O. X The case of a discontinuity on the wave front can be treated by passing to the limit. 529. Consider the problem of diffraction of a plane sound wave by an obstacle shaped like a parabolic cylinder. Show that the reflected wave has the representation2 where ~(s) is the solution of the Volterra integral equation - 1 2a .p(E) + - Jq ~('1 2 v -a [E - s + (2a/v)1312 21 (see F8.) 530. Solve the preceding problem, assuming that the wave encounters the obstacle at the time t = 0: Ans. " cpw tl = [! 0 ,/E - s + (201~) ds, ri > 0, -q < 0. where, in the notation of Prob. 527, "531. Consider the problem of diffraction of a plane sound wave by an obstacle shaped like a paraboloid of revolution r = z + 2a. Applying the technique of the preceding problems, show that the reflected wave has the representation Vn problems on diffraction of acoustic waves (unlike the case of electromagnetic waves), we write the total solution in the form f(c) + rr. PROB. 532 where y(s) is the solution of the Volterra integral equation (see F8). *532. Solve the preceding problem, assuming that the incident wave has the equation Ans. In the excited zone (q > O), where In the last formula, g and R are the Laplace transforms of g(t) and the kernel and the path of integration f is a straight line parallel to the imaginary axis lying to the right of the singular points of the integrand. Note that in terms of the exponential integral Eitx). 533. Consider the problem of diffraction of a plane wave by a paraboloid of revolution r = z f 2a with homogeneous boundary conditions of the first kind. Show that. the reflected wave has the representation where y(s) is the solution of the integral equation PROB. 536 INTEGRAL EQUATIONS 259 534. Using the Laplace transform, solve the integral equation of Prob. 533 for the case of a wave of the form Ans. In the notation of Prob. 532, 2. Electrostatics 535. A conductor of arbitrary shape, bounded by a surface 2, is intro- duced into a given external field E0 (see Figure 151). Show that the density of charge induced on the conductor satisfies the integral equation n t o(N) = - , cos (r ,I,,, 11) dS (1) where M and N are two arbitrary points of the surface C, dS is the element of area, r,,,,, is the vector joining M to N, n is the unit exterior normal to C at the point N, and Ei = E0 n is the projection of E0 onto n. FIGURE 151 536. Show that in the special case where the surface of the conductor is an infinite plane, the solution of the integral equation (1) is given by3 Use this result to find the charge density induced on a conducting plane by a point charge q placed at height h above the plane. Ans. where R is the distance from the charge to the point N of the plane. Naturally, this result can be found in other ways. The present method is of interest mainly because the final result is obtained practically without calculations. 260 INTEGRAL EQUATIONS PROB. 537 537. A metallic sphere of radius a at potential V is introduced into an external electric field EO. Starting from the integral equation (I), show that the density of charge induced on the surface of the sphere is given by where u0 is the potential of the external field. Examine the special case where the source of the field E0 is a point charge q at distance b (b > a) from the center of the sphere. Am. V q b2 - a' o(N) = - - - - 4xa 4x aR3 ' where R is the distance from the charge to the given point N of the surface of the sphere. 538. Solve the preceding problem, given the total charge Q of the sphere (rather than its potential). Use the formula so obtained to solve the problem of the charge distribution on the surface of an initially uncharged insulated sphere introduced into a homogeneous external field EO. A ns. where ziO is the average over the sphere of the potential of the external field: 1 a0 = - uO(N) d~. 4xa2 In the special case where 8 is the angle between the direction of the external field E0 and the radius vector drawn from the center of the sphere to the point N. 539. A cylindrical conductor with cross section bounded by an arbitrary contour I' (see Figure 152) is introduced into a given plane-parallel field EO. Show that the density of charge satisfies the integral n equation o(N) = Ez) + cos (rMN, n) d~, (2) 27-c x ~I~MNI where M and N are two arbitrary points of the contour 1', ds is the element of arc length, ~MN is the vector joining M to N, n is the unit exterior normal to I' at the FIGURE 152 point N, and E: = EO . n is the projection of E0 onto n. 540. Suppose a conductor shaped like an infinite circular cylinder of radius a, carrying charge Q per unit length, is introduced into an external plane-parallel field EO. Show that the density of induced charge on the surface of the conductor is given by Consider the special case where a) The external field is homogeneous; b) The source of the external field is a line charge with charge q per unit length, placed outside the cylinder at the distance b from its axis. Ans. where 0 is the angle between the direction of the homogeneous field (of strength E) and the vector drawn from the center of the cylinder to the given point N on the surface of the conductor; where R is the distance from the line charge to N. 541. Find the distribution of charge density on the inner surface of a grounded cylindrical shell of radius a, assuming that the external field is produced by line charges parallel to the axis of the cylinder passing through the points Mk = (a,, cp,), k = 1, 2, . . . , n. Ans. where qk is the charge per unit length of the line charge passing through the point M,, and R, is the distance between the points Mk and N. *542. The electrostatic field in the region 0 < y < h between two grounded parallel planes is due to line sources whose free-space field is EO. Show that the densities oo(x) and oh(x) of induced charge on the platizs y = 0 and y = h satisfy the system of integral equations and then solve this system. 262 INTEGRAL EQUATIONS PROB. 543 Ans. where f is the Fourier transform off (x), i.e.,4 m f =I /(x)eih dx, -00 and fo(~> = Eilg=o, fh(~) = E:lv=h. Hint. Take the Fourier transform of each of the equations (3). 543. Solve the preceding problem for the special case where the field EO is due to a line source with charge q per unit length, passing through the point Mo = (0, b). Ans. nb xb sin - sin - 4 GO(X) = - - h 4 G/,(x) = - - h 2h xx 7cb ' cosh - - cos - xb ' 2h cosh + cos - h h h h Hint. To obtain the solution in closed form, use formula 15, p. 385. 544. Suppose a system of line sources, whose free-space field E0 has components E,O, E!, 0 in cylindrical coordinates, is placed inside a dihedral angle 0 < cp < a with grounded conducting walls. Show that the charge densities oo(r) and o,(r) on the walls satisfy the system of integral equations 1 sin a " odr) = - E;l,=, p~a(P) 27c - xJ0 p2 + rz - 2rp cos a dp, 1 sin a " oa(r) = - - EOI P~O(P) 2x - xS0 p2 + r2 - 2rp cos a df? and solve this system, using the Mellin transform. Ans. sin x(p - 1) +fa sin (x - a)(p - 1) sin (2x - a)(p - 1) . sin a(p - 1) X sin x(p - 1)r-D dp, sin x(p - 1) + ,f', sin (x - a)(p - 1) rSa(r) = - - 4x2i 1-im sin (27c - a)(p - 1) . sin a(p - 1) x sin n(p - 1)r-' dp, * This definition of the Fourier transform differs from the customary one by a numerical factor. where f is the Mellin transform off(r), i.e., m f =S f (r)rv-' dr, and fo(r) = E:l,=o, fa(r) = ~Jq=a. Hint. t "dt - x sin (x - a)s -- -1 <Res< 1. t2 - 2t cos a + 1 sin a sin xs ' 545. Solve the preceding problem, assuming that the field E0 is due to a line source with charge q per unit length, passing through the point Mo = (r,, yo). Use the formula so obtained to find the electrostatic field due to a charged line placed at distance u from the edge of a conducting half-plane (a = 2x, r, = a, yo = x) or near a right-angular corner (a = 3x12, ro = a, 'Po = n). A ns. sin =(Po ='Po sin - 546. A conductor shaped like an open surface of arbitrary form (see Figure 153) is placed in an external field EO. Show that the sum of the charge densities on opposite sides of the surface satisfies the integral equation of the first kind 1 "0 d~ = Y - UO(N), (4) n r ~MNI where o(N) = o,(N) + o,(N), u, is the potential of the external field and V is the potential of the conductor, while the difference between the charge densities is given by the formula E'(N) + S + cos (rMN, n) d~. o,(N) - o,(N) = - 2x 2~ lrMNl FIGURE 153 (5) Thus, to solve the electrostatic problem completely, it is sufficient to know the solution of the integral equation (4) [see G5, Chap. 201. PROB. 547 547. Show that equation (5) takes the form for a plane surface, and the form for a spherical surface of radius R, regardless of the form of the boundary curve. 548. Write the integral equation (4) for the case where the surface of the conductor is a disk of radius R or a thin spheri- e, a cal bowl r = R, 0 < 8 < u (see Figure 154), assuming that the external field has rotational symmetry with respect to the FIGURE 154 z-axis.6 Ans. 4/R&)K(*)dp=V--u0(r), 0 pf r pi-r O<r<R for the disk, and 2R la o(6) sin 6 (Jsin o sin 0 d6 = V - uO(0), 0 g 8 < a o sin 4(6 + 0) sin i(6 + 0) for the bowl, where K(k) is the complete elliptic integral of the first kind. *549. Show that the integral equations of the preceding problem can be reduced to the integral equation and solve this equation. Ans. 550. Using the results of Probs. 548 and 549, find the distribution of charge on a disk of radius R at potential V introduced into an arbitrary axially symmetric external field. Problems 548-555 are considered in Lebedev's paper L5. paoe. 552 INTEGRAL EQUATIONS 265 where E,O(r) is the normal component of the external field on the surface of the disk. 551. Find the charge density on a thin conducting disk of radius R for the following cases : a) The disk is charged to potential V, and there is no external field (freely charged disk); b) A point charge q is placed on the axis of symmetry of the disk, at distance h from the disk. Am. b) 01,dr) = - 2n2(h2 qh + r2)3'2 [arc tan Js + ,Jx R -r * where a, is the charge density on the side facing the charge. 552. Find the charge distribution on a thin spherical bowl r = R, 0 < 8 g a at potential V placed in an arbitrary axially symmetric external field EO. Am. xq "V - uO(t)] tan it dt ds 0 Jtan2 4s - tan2 4t cos it ' where uO(0) and E,O(B) are the values of the potential and the normal com- ponent of the electric field on the surface of the bowl, while a, and a, are the charge densities on the convex and concave sides of the surface. 266 INTEGRAL EQUATIONS PROB. 553 553. Solve the preceding problem for the following specialcases: a) There is no external field (free charge distribution); b) The external field is homogeneous, and the potential V is zero (a thin conducting spherical shield with a circular hole, placed in a homo- geneous field EO). Ans. v Jsin2 $a - sin2 ge + cos 4'2. a,,@) = -[arc tan 4n2R cos ga Jsin2 ha - sin2 $0 A", 3E0 cos 8 Jsin2 &a - sin2 ae ai,z(f9 = - [f + arc tan 42 cos ha I 554. Suppose a conducting plane at potential V, with a circular hole of radius R, is placed in an arbitrary axially symmetric external field (see Figure 155). Show that the problem of determining the charge distribution on the plane reduces to solving the integral ' t equation (4), and find the distribution. Ans. where 0, is the charge density on the upper surface of the plane, while uO(r) is the potential and E,O(r) the normal component of the external field at the point r. 555. Solve the preceding problem, assuming that the external field is due to a point charge q on the axis of symmetry of the hole at distance h from the plane. Ans. r2 - R' R r2 + hZ o1,2(r) = - 2n2(h2 qh + r2)3/2 [arctaniJ-+- r2 + h2 h J-A;]. r2 - R' 556. Show that the problem of the charge density on a grounded thin conducting half-plane, introduced into a given plane-parallel external field EO, reduces to the solution of the integral equation of the first kind where o(x) is the total charge per unit length, o(x) = o,(x) + ~,(x),~ and f(x) = iuO(x), in terms of the potential uO(x) of the external field at the point x. Solve this integral equation. where Hint. Set x = 0 in (6) and subtract the result from the original equation. Then take the Mellin transform of the equation so obtained. 557. Use the result of the preceding problem to find the charge dis- tribution on the surface of a thin conducting sheet x > 0, if there is a line source with charge q per unit length near the edge of the sheet. Ans. 'Po sin - 4 = - - 2 - 2dGL +il+2J2c0s3' '-0 r0 2 where r,, yo are the polar coordinates of the point M, and the upper sign pertains to the den- sity on the side of the sheet facing the charge. 558. Two media with dielectric constants E, I and E~ are separated by a surface C (see Figure FIGURE 156 156). Consider the electrostatic field in the resulting inhomogeneous medium due to sources whose free-space field is E,. Show that the density of polarized charge on the surface C, determining the secondary field,7 satisfies the integral equation where M and N are two arbitrary points of the surface C, dS is the element of area, r,, is the vector joining M to N, n is the unit exterior normal to The difference between the densities is given by the previous formula 1 o,(x) - oz(x) = - E,O(x). 2n 7 The potential of the secondary field in each medium is given by 268 INTEGRAL EQUATIONS PROB. 559 C at the point N, pointing from the medium with dielectric constant E, to the medium with dielectric constant E,, and EE = EO. n is the projection of the external field onto n (see G5, Chap. 14). 559. Using the integral equation (7), find the distribution of polarized charge for the case where the surface C is an infinite plane (this generalizes Prob. 536). Ans. 560. Derive the two-dimensional analogue of equation (7), corresponding to the plane-parallel electrostatic problem of an inhomogeneous medium made up of two homogeneous media with dielectric constants E, and E,. where n is the unit normal to the contour I? representing the interface between the dielectrics. *561. Consider a dihedral angle whose interior 0 < cp < a is filled with a medium of dielectric constant E,, and whose exterior a < cp < x is filled with a medium of dielectric constant E, (see Figure 157). Show that the corre- sponding two-dimensional electrostatic problem reduces to solving the system of integral equations x [-EO,~,=, + 2 sin a - 2rp cos a where a, and a, are the densities of polarized charge on the faces cp = 0 and cp = a. Solve this system by using the Mellin transform. Ans. fa$ sin (x - a)(p - 1) - ji sin x(p - 1) ao(r) = - JiiW - 4n2i I-iw sin2 n(p - 1) - P2 sin2 (x - a)(p - 1) x r-' sin n(p - 1) dp, sin n(p - 1) -jib sin (x - u)(p - 1) aa(r) = - n(p - 1) - p2 sin2 (x - a)(p - 1) x r-" sin n(p - 1) dp, where f; is the Mellin transform 562. Solve the preceding problem for the special case where a = x/2 and the external field E0 is due to a line source with charge q per unit length, located in the medium with dielectric constant E, at the point r = a, cp = x. Ans. l+im sin in(p - 1) sin x(p - 1) %/2(r) = - - (5). dp. a&, 2xi I-im sin2 x(p - 1) - P2 sin2 $n(p - 1) r 563. A slab of dielectric constant el, bounded by the parallel planes y = 0 and y = h and surrounded by a medium of dielectric constant E, (see Figure 158), is introduced into an arbitrary plane-parallel field EO. Show that the resulting electrostatic problem reduces to solving the following system of integral equations for the polarized charge densities oo(x) and oh(x): where f is the Fourier transform of f(x), i.e., m /=I f (x)ei" dx, -a 564. Solve the preceding problem for the special case where the external field E0 is due to a line source with charge q per unit length passing through the point x = 0, y = h/2. Ans. oo(x) = oh(x) = - cos Ax dA. *565. A perfectly conducting half-plane x 2 0, y = 0 is introduced into an external electromagnetic field with components Show that the sum j = J, + j, of the current densities flowing in the upper and lower sides of the half-plane satisfy the integral equation E(x) = EO(x, 0), k = c2 and E, p and o are the dielectric constant, the magnetic permeability and the conductivity of the medium, while the difference between the current densities is Solve the integral equation by using the transform (27), p. 196. Ans. where [it is assumed that f(x) approaches zero as x -t 0 in such a way that the integral converges at its lower limit]. *566. A plane electromagnetic wave with components PROB. 566 INTEGRAL EQUATIONS 27 1 is incident on a perfectly conducting half-plane r 0, cp = a. Using the result of the preceding problem, find the distribution of current density on the half-plane. Ans. References Books: Grinberg (G5), Kupradze (K5), Mikhlin (M7), Morse and Feshbach (M9), Muskhelishvili (M10, Mll), Noble (Nl), Smirnov (S6, Vol. IV), Titchmarsh (T5), Tricomi (T9). Papers: Fock (F3, F5), Lebedev (L5), Wiener and Hopf (W10). SOLUTIONS SOLUTIONS 52. The solution of the problem reduces to the determination of the complex flow potential w = cp + i$, whose imaginary part is a harmonic function which equals zero on the axis of symmetry and takes thevalue $=- v,a on the walls of the channel. To determine w, we need only find a conformal mapping of the region ABCDE onto the upper half-plane of the variable < = + ir]. Suppose that in applying the Schwarz-Christoffel trans- formation, we make the points of the z and <-planes correspond in the way suggested in the hint to the problem. Then the relation between z and < is obtained by integrating the equation where M is a constant to be determined later. Bearing in mind that z = ib if < = -1, we find that where the integration is along any path joining the point = -1 to a given point < in the upper half-plane. It follows from the condition lim [ZI~=-~ - ~l~=~] = ia E-0 that M = adX/x, and hence it only remains to determine the value of the parameter A. This is done by using the correspondence between the points z = ia and < = -A. Since in evaluating the integral with < = -A as its upper limit, we can integrate along the line segment joining the points < = -1 and < = --A, on which < + 1 = 1 - s, < + A = ei"(s - A), < = -s (A < s < I), 275 PROB. 54 the last requirement leads to the formula which, after carrying out the integration, implies A = (b/~)~ and hence M = b/z. AS is easily verified, the complex potential in the <-plane is which, together with the transformation z = z(c) just derived, gives a para- metric solution of the problem. To calculate the velocity along the axis of symmetry (5 > 0, = O), we use the formula which implies where the relation between x and c must be established by using the trans- formation z = z(<). Choosing the path of integration to be a curve con- sisting of the segment (- 1, -R) of the real axis, an arc of a circle of radius R and the segment (R, c), and then taking the limit as R -+ a, we find that After some simple calculations, this leads to The final formulas, given in the answer to the problem, are obtained by introducing the new parameter 54. This problem belongs to a category which is both of considerable mathematical interest and of great importance in the applications, i.e., problems .involving the formation of a jet at the boundary of an obstacle placed in a stationary plane-parallel flow of an ideal liquid. In such problems, the form of the jet is not known in advance, but must be determined from the condition that the velocity vector have a constant value on the free surface PROB. 54 SOLUTIONS 277 of the jet. For the case where the walls of the obstacle impeding the flow consist of line segments, an effective method of solving such problems is based on the possibility of establishing a connection between the complex potential w and the derivative dwldz, starting from examination of the kinematic picture of the fluid motion. Thus consider the part of the region occupied by the flow which is bounded by the axis of symmetry AB, the free boundary of the flow BC and the wall CD. The behavior of the velocity components u, and v, along the boundary of this region is determined by the following relations (where v, = Q/2b and 2b is the width of the jet at a great distance from the slot AB): Introducing the auxiliary complex variable r = dwldz and taking account of the formula we find on the basis of the above picture of the flow that the region ABCD is mapped conformally onto the interior of the circular sector X 1 < , - - < arg r < 0 2 in the [-plane, with the boundary of,the jet going into the arc of the circle Under the transformation1 r = tlb(JY - Jtl), this sector is mapped into the upper half-plane of the complex variable t, with the curves AB, BC and CD in the original plane going into the negative real axis, the segment (0, 1) and the segment (1, w). In the t-plane, the determination of the complex potential reduces to constructing a function analytic in the upper half-plane whose imaginary part takes the value zero on the negative real axis and the constant value -Q/2 on the positive real axis. It is easy to see that the solution of this problem is To obtain this expression, it is convenient to first transform the sector into a half-strip by using the transformation with the Schwarz-Christoffel transformation being applied afterwards. PROB. 59 which, together with the results found earlier, gives Integrating (1) and bearing in mind that the point z = a must correspond to the point t = 1, we obtain where we choose the branch of the arc tangent which vanishes as t -t 1. The functions w = w(z) and z = z(t) establish the required connection between the complex variable w and the variable z in parametric form. To determine the form of the boundary of the jet, we need only separate real and imaginary parts in (2), assuming that the variable t belongs to the interval (0, 1). This gives the following parametric representation of the curve bounding the jet: The width 26 of the jet at a great distance from the slot, and the corre- sponding value of the velocity v, = Q/2b are found from the condition that x = b for t = 0, which implies and immediately leads to the formulas given in the answer. 59. Guided by the hint to the problem, we construct the function z = z([) mapping the interior of the rectangle onto the upper half-plane. Using the Schwarz-Christoffel transformation, we find that which implies since the symmetry requires that the point z = 0 correspond to the point < = 0. The values of the constants M and k are determined from the con- dition that the points z = a and z = a + ib correspond to the points [ = 1 and l: = Ilk. This leads to the formulas = MJ1 dl: = MK(k), 0 J(1 - l:"(l - ky2) l/k dl: = MK(kf) (kf = Jm), J(1 - CZ)(1 - k2CZ) where K(k) is the complete integral of the first kind.2 Eliminating M, we obtain the relations the first of which is an equation for determining the modulus k, while the second solves the given problem of conformal mapping. According to the theory of elliptic functions, the inversion of the integral in the last expression is given by the formula Kz l:=sn-, a where sn z is Jacobi's elliptic function. Under the conformal mapping, the - point z = z, = Bib goes into the point 1: = c, = sn (iKb/2a) = i/J/c. The expression for the complex potential in the c-plane is To calculate the distribution of charge density on the walls of the box, we use the relations where En denotes the field normal to the surface of the conductor at the point where the value of the density o is being determined. Applying these formulas, we find that the charge distribution on the wall -a < x < a, y = 0 is the expression given in the answer to the problem. 69. The displacement u(x, t) of any point of the midline of the beam satisfies the differential equation To reduce the second integral to canonical forrn,use the substitution 41 - k2Cz = k't. 280 SOLUTIONS PROB. 69 and the boundary conditions To find the natural frequencies for transverse oscillations, we write U(X, t) = U(X) sin (at + cp). Then, after substituting this expression into the differential equation and the boundary conditions we find the following conditions determining the amplitude v(x) : The general solution of this equation is JWx v(x) = A, cos - JW x '' + A, cosh - JW x + B, sin - + B, sinh - . a a a a The fact that the end x = 0 is clamped allows us the determine two of these constants, and leads to the expression Then, imposing the remaining conditions at x = I, we obtain the following homogeneous system of equations for the quantities A and B (for brevity, we set y = Jw //a) : A(cos y + cosh y) + B(sin y + sinh y) = 0, (sin y - sinh y) - (cos y + cosh y) = 0. I The equation determining the natural frequencies is obtained by setting the determinant of this system equal to zero. The result is M 1 + cos y cosh y = -O y(sin y cosh y - cos y sinh y). M If the roots of this equation are denoted by y, (n = 1, 2, . . .), the natural frequencies are a' .. n=l,2 ,... 83. The problem of finding the forced oscillations of the membrane under the action of a load q sin (at + cp) distributed over a disk of radius b < a can be posed as follows: Find the solution of the differential equation governing the oscillations of the membrane, where which satisfies the boundary condition and has the same frequency w as the perturbing force. Writing ~(r, t) = w(r) sin (wt + cp), and substituting this expression into the differential equation and boundary condition, we find that The solution of this inhomogeneous equation, obtained by variation of con- stants, has the form The constant B equals zero because of the requirement that the solution be bounded at the point r = 0. The constant A is determined from the condition w(a) = 0, which gives After some manipulation, the desired expression for the amplitude takes the form = /&)G(P? I)P dp, (4) 282 SOLUTIONS PROB. 91 where Substituting (3) and (5) into (4), and using the formulas (5) and the familiar expression 2 Jo(x)Y;(x) - Yo(x)J6(x) = - nx for the Wronskian of the Bessel functions, we finally obtain the answer on p. 48. 91. If the z-axis is parallel to the generators of the wave guide, then the only component of the electric field of the TM-wave is (a is the frequency of the oscillations and v is the propagation constant), whose amplitude satisfies Helmholtz's equation (k = olc = 24A, where A is the wavelength) and the homogeneous boundary conditions El,=, = 0, El,=, = E,,, = 0. These equations have infinitely many nonzero solutions of the form where the ymn are the roots of the equation and the value of the propagation constant corresponding to ymn is A wave with an imaginary value of vmn falls off exponentially in the z-direc- tion and is essentially unable to propagate in the wave guide, i.e., a wave can propagate in the guide only if v,, is real. This leads to the inequality 2xa A<-. Ymn The maximum wavelength which can propagate in the guide is given by the formula 2xa Amax = - , Yo where yo is the sn~allest positive root of the equations J,,,,Jy) = 0, m = 1, 2, . . . 96. The problem reduces to integration of the equation aZll 1 aZu ---- axz 2 a,. = O' with initial conditions and boundary conditions ule=O = u],=~ = 0. Setting u(x, t) = X(x)T(t) and separating variables, we arrive at the equations XN+AX=O, TN+Av2T=0. Solving the first of these equations with the boundary conditions X(0) = X(1) = 0, we find the corresponding eigenvalues and eigenfunctions nxx X = Xn(x) = sin - 1 The solution of the second equation satisfying the conditions T'(0) = 0 is given by nxvt T = TJt) = cn cos - . 1 Therefore the set of particular solutions of the equation of the vibrating string satisfying all the homogeneous conditions is . nxx nxvt u=u,=c,sm-cos-, n=1,2 ,... 1 1 284 SOLUTIONS PROB. 108 According to the basic idea of the Fourier method, we now look for a solution of the given problem in the form of a series CO nxx nxot U(X, t) =Ccn sin - cos - , n=l 1 I where the coefficients c, are determined from the condition utl=o = f (x), i.e., coincide with the coefficients of the expansion of the function f(x) in a Fourier series 02 . nxx f(x) =Xcnsm-, 0 < x < I. n=l 1 As is well known (see T7, p. 35), nxx cn = Tlf(x) sin - I dx, and hence in the present case 2h12 nxc C, = sin - , n2x2c(l - c) 1 which leads to the answer on p. 60. It can be shown that this series represents a piecewise smooth function of the variables x and t, satisfying the equation of the vibrating string and all the initial and boundary conditions. 108. In the present case, the differential equation for longitudinal oscillations of a beam of variable cross section takes 'the form 1 a where y(x) = u - x tan u is the variable height of the cross section at x measured from the axis of symmetry of the beam. Setting u(x, t) = X(x)T(t) and separating variables, we find that the factors X and T satisfy the equations tan u xu - - X' + AX = 0, T" + AV~T = 0. Y (6) The first of these equations reduces to Bessel's equation in the variable y d2x 1 dx (Ji JX = 0, -+--+ - dy2 y dy tan a with general solution X = A"($) + BYo(s). Using the boundary conditions which imply the conditions we obtain the eigenvalues y, tan a with corresponding eigenfunctions X(x) = X,,"(x), where and the y, are consecutive positive roots of the equation Xyf(b) = 0. Inte- grating the second of the equations (6) and taking account of the condition Tf(0) = 0, we find that uty, tan a T = c, cos a It follows that the set of particular solutions satisfying the homogeneous conditions is vty, tan a u = un = cnXvn(y) cos , n=l,2, ... a The solution of our problem is then constructed in the form of a series where the coefficients c, are determined from the condition m u(t=o = f (XI = 2 c,x,,(Y), b < Y < a. n=l Using the formulas 286 SOLUTIONS PROB. 109 where the relation .-, has also been used. In this way we finally arrive at the answer on p. 64. 109. The problem reduces to integration of the system of equations a2i, 1 aZu, - 0, 0 < x < a,, ax2 V: at2 with initial conditions and boundary conditions Separation of variables leads to the expression for the displacement m u(x, t) = 2 cnxn(x) cos , n=l a1 satisfying all the conditions of the problem except the first initial condition. Here where the y, are consecutive positive roots of the equation S, JE,p2 tan y + S1 JE,pl tan = 0. a 1% It can be shown that the eigenfunctions Xn(x) of the problem are orthogonal on the interval -al < x g a, with weight r(x) = Slpl, -a1 g x < 0, 0 < x g a,. SOLUTIONS 287 Therefore the initial condition ul,=, = f(x) implies Substituting the eigenfunctions (7) into (8), we find that the denominator becomes Thus the solution finally takes the form given in the answer on p. 65. 312. To solve the problem, we have to integrate the differential equation for torsional oscillation of the shaft, subject to the following initial and boundary conditions (6, denotes the polar moment of inertia per unit length of the shaft) Separating variables, and taking account of the fact that the ends are clamped and there is no initial velocity, we find the following particular solutions: dl) sin JA x cos JX vt, O~xga, d2) sin JA 1 - - cos Ji\ vt, a G x g I. - 1) Using the fact that the two sections of the shaft are joined at the point x = a, we obtain the eigenvalues ),=A n =Y: 12 ' and the corresponding eigenfunctions (n = 1,2, . . .), where the yn are consecutive positive roots of the equation JOY Ya sin y = - sin - sin y J 1 If the solution of the problem is written as a series then the coefficients cn must satisfy the relation m f (x) = 2 cnO,(x), 0 < x < 1. n=1 In the present case, the functions $,(x) = 0',(x) are orthogonal, i.e., which leads to the answer on p. 66. 114. The problem reduces to integrating the differential equation for transverse oscillations of the beam, with initial conditions PROB. 1 18 and boundary conditions Writing u(x, t) = X(x)T(t) and separating variables, we find the differential equations X(i") - AX = 0, T" + a4AT = 0 for the separate factors, with general solutions X = A cos $'xx + B sin $'Ax + C cosh $'hx + D sinh $'%, T = M cos JAa2t + N sin Jrazt. Using the boundary conditions X(0) = we arrive at the eigenvalues and eigenfunctions Determination of the constants M,, and Nn in the expansion m n%'a2 t U(X, t) = 2 (Mn cos - nz~a2t) sin 7 la + Nn sin - n=l l2 reduces to evaluation of the Fourier coefficients of the functions m n7rx 7r2a2 n7rx f (x) =x M, sin - , g(x) = --p-zn'~~ sin - (0 < x < I). n=1 1 n=l 1 118. We want the solution of the equation satisfying the initial conditions and boundary conditions 290 SOLUTIONS PROB. 1 18 where the initial deflection f (x) is the solution of the following static problem:3 Writing u(x, t) = X(x)T(t) and separating variables, we find that X= ~cos-$'hx+~sin-$'hx+ ~coshYS;x+ ~sinhqhx, T = M cos JA a2t + N sin JAa2t. Using the boundary conditions X(0) = X'(0) = X"(1) = X"'(1) = 0, we obtain the eigenvalues and eigenfunctions ,X,(x) = (sin y, + sinh y,) - (cos y, + cosh y,) where the y, are consecutive positive roots of the transcendental equation cos y cosh y + 1 = 0. Next we show that the functions X,(x) are orthogonal on the interval (0, I). Multiplying the first of the equations by X,, and the second by X,, we subtract the results from each other and integrate with respect to x from 0 to I. Taking account of the boundary conditions, we obtain (A, - A,)[' XnXm dx = (X,X, - XEX, + X,X& - X;X,) = 0 I: after integration by parts. This immediately implies the required ortho- gonality of the functions X,(x). Using the general theory of expansions in series of orthogonal functions, we can represent the solution of the problem in the form An explicit expression for f(x) is given in Prob. 7, but will not be used in our method of solution. PROB. 120 SOLUTIONS 29 1 The integral in the numerator is easily evaluated by replacing X, by Xp)/l, and integrating by parts, which gives To evaluate the integral in the denominator, we use the formula 1 S,'X:(S) dS = - [x:(I) + XAZ(l) - 2X:(l)X"l)], 4 (see T2, p. 336), which in the present case takes the form 1 S,'xm dS = - 4 X%). Substituting these integrals into (9), we find that The form of the solution given in the answer on p. 68 is obtained after making the substitution X,(I) = 2(cos y, sinh y, - sin y, cosh y,). 120. The problem reduces to integration of the equation for a vibrating membrane, with initial conditions , O<~<E, uJ,=,,=o, &/ =f(r)= at t=o ~<rga, and boundary condition Writing u(r, t) = R(r)T(t) and separating variables, we arrive at the equations The permissible values of the parameter A are obtained from the requirement that the first of these equations have solutions which are bounded in the region 0 < r < a and satisfy the boundary condition R(a) = 0. This leads PROB. 124 to the eigenvalues and eigenfunctions where the y, are consecutive positive roots of the equation Jo(y) = 0. The solution of the equation for a vibrating membrane satisfying all the homogeneous conditions is The constants cn are determined from the condition which, after substitution of the series for u(r, t), takes the form ,According to the well-known forn~ula for the coefficients of expansions in Fourier-Bessel series (see T7, p. 221), we have which implies the answer on p. 69. 124. We want the solution of the equation for transverse oscillations of a plate which satisfies the initial conditions and boundary conditions Separating variables, we obtain u(r, t) = R(r)T(t), T" + b4AT = 0. The functions R(r) remaining finite at the center of the plate are of the form R(r) = AJ,(VK~) + BI,(VA r). SOLUTIONS 293 It follows from the boundary conditions R(a) = R1(a) = 0 that where and the y, are consecutive positive roots of the equation Rt(a) = 0. The eigenfunctions R,,(r) are orthogonal on the interval (0, a) with weight r, since where we introduce the abbreviation RYk = R, and use the boundary con- ditions for the function R,(r). The solution of the problem is given by the formula dp a + - sin W[g(p)~,(p)p d bZy% a2 The value of the integral j, R%P dP can be found from the relation which takes the form after some simple calculations. 132. The problem reduces to integration of the equation 294 SOLUTIONS PROB. 132 with zero boundary conditions u(,=, = ulp=O = uIp=,?= 0. If we write u = u, + u,, where the function u, must satisfy Laplace's equation with the boundary conditions up = u = 0, uZ(,.. sin2 cp. 2T The substitution u,(r, A) = R(r)cD(cp) leads to the equations with general solutions R = ~rdh + ~dh, (D = C cos JA cp + D sin JX cp. By satisfying the homogeneous boundary conditions @(O) = cD(x) = 0, we obtain the eigenvalues A, = n2 and the eigenfunctions cDn(cp) = sin ncp, n = 1, 2, . . . Because of the finiteness of the sclution for r = 0, the constant B must be set equal to zero. Thus the function u, can be represented as the sum of the series It follows from the boundary condition for r = a that m ~sin2cp=~A,sinncp, O<cp<r, 2 T n=l and hence, by the theory of Fourier series, PROB. 136 SOLUTIONS 295 Substituting the values of the coefficients A, into (lo), we obtain qr2 sin2 cp 4qa2 2 (;Tn+l sin (2n + 1)cp u(r, cp) = - +- 2 T XT .=, (2n + 1)[4 - (2n + 1)'l This result can be written in closed form by using the expansions 1 2p sin cp 2 J?E- sin (2n + 1)cp = - arc tan -- .=, 2n + 1 2 1-P2' 1 1+2pcoscp+p2 2 J?E cos (2n + 1)cp = - In -- .no 2n + 1 4 1-2pc0scp+~~' where lpl < 1, lcpl < n. After some manipulation, we obtain 2 $"+'sin (2n + 1)cp = :[I - :(p2 + :) cos ~cp] arc tan 3 2p sin p (2n + 1)[4 - (2n + 1)21 8 1+2pcoscp+ p2-- ( -- P) sin p, 1-2pcoscpf p2 8 p which immediately implies the form of the solution given in the answer on p. 74. 136. We want the solution of the equilibrium equation for a semicircular plate which satisfies the boundary conditions Setting we find that the function v(r, 9) satisfies the equation A2v = 0 with the boundary conditions homogeneous biharmonic vI,=, = sin4 cp, We can separate variables in the biharmonic equation by looking for par- ticular solutions of the iorm v = v,(r, cp) = (A cos pcp + B sin pcp)R(r). 296 SOLUTIONS PROB. 140 It follows from the boundary conditions for cp = 0 and cp = a that Moreover, R(r) = Cnrn + Dnrfl+2, since the deflection at the center of the plate is bounded. Therefore, summing particular solutions, we find that V(I, cp) = 2 [M, (I)'+ N, (:rE] sin n p. R-1 a The values of the constants M, and Nn are determined from the boundary conditions on the arc r = a: m sin4 p = C(M, + N,) sin ncp, n=l nM 4 sin4 y = 1 [-;? + (" + 2)Nn] sin np, o < p < a. a n-1 a This gives the system M, + N,, = 2 Ssin4 p sin np dy, a 0 After some simple calculations, we find that J, =I:sin4 p sin ny dy - - 8 which implies n - 2 4-n M, = - J,, N, = - X Jn. 7C Using the expansion we can sum the series (1 1) for y = x/2, thereby expressing the deflection of the axis of symmetry of the plate in closed form in terms of elementary functions. 140. To reduce the problem to a special case of the Neumann problem, we subtract out the velocity potential of the source, by setting u = u, + 241, where m uo = - + const 4xP (p is the distance from the source to an arbitrary point of the flow). Then the function u1 must be a solution of the equation which is regular outside the sphere, and satisfies the boundary condition and the condition ull,+, -f. 0 at infinity. Setting ul(r, 0) = R(r)O(0) and separating variables, we arrive at the equations 1 (r2R')' - AR = 0, - (sin 0 . 0')' + A@ = 0. sin 0 This equation has finite solutions for 0 = 0 and 0 = x if and only if which determines the eigenvalues of the problem. The corresponding eigen- functions are @,(0) = Pn(cos 0), where P,(x) is the Legendre polynomial of degree n. Similarly, for Rn(r) we obtain Rn(r) = Anrn + Bnrpn-l, where A, = 0, because of the condition at infinity. Thus we find that w ul(r, 0) = 2 Bnr-n-l~n(cos 0), n=O (12) and to determine the constants B,, we need only satisfy the boundary condi- tion In the present case, we can calculate the coefficients B, in (12) by differentiat- ing the expansion of the generating function thereby obtaining PROB. 145 Substitution of (13) into (12) gives This series can be summed by integrating the expansion r -= r a2 - = 2 (~);)n(cos o), 6 = - < a P Jr2 - 2br cos o + b2 ,,=, b with respect to the parameter 6 from 0 to 6, which leads to the relation --= S" db r(1 + cos 0) = In n=O n + 1 rn+' o Jr2-26rCoS~+62 ij+ ~COSO-6' Writing (14) in the form (15) and using (15), we arrive at the expression given in the answer on p. 77. 145. The problem reduces to solving the system of equations with initial condition TI,=, = T, and boundary conditions Application of the Fourier method leads to the expression w ~(x, 1) = 2 n=l for the required temperature distribution, satisfying the homogeneous boundary conditions for arbitrary values of the coefficients C,, where PROB. 148 SOLUTIONS and the y, are consecutive positive roots of the transcendental equation By the usual procedure, it can be shown that the eigenfunctions X,(x) are orthogonal on the interval 0 < x < a, + a, with weight Therefore the coefficients C, can be calculated by using the formula Substituting from (16), we find that the denominator of (17) equals 2 a26 Y ~[alclpl sin - 2 + azc,p2 sin2 y, a1,K I Then setting f(E) = To in (17) and making some simple calculations, we arrive at the answer on p. 79. 148. We want the solution of the equation satisfying the initial condition TI,=, = f (r) and the boundary condition Setting T(r, r) = R(r)O(.r) and separating variables, we obtain Then R = J,(JA r) is the solution of the first equation which is finite on the axis of the cylinder. From the boundary condition R1(a) = 0, we find the eigenvalues 300 SOLUTIONS and corresponding eigenfunctions where yo = 0, y,, . . . , y,, . . . are consecutive nonnegative roots of the equation J,(y) = 0. The general solution of the second equation is @ = e-~na~~a2 n n 9 and the expression m T(r, 7) = 2 cne-yn2T1a2Jo (y) , n=O obtained by summation of particular solutions, satisfies all the conditions of the problem, except the initial condition. Since the eigenfunctions are orthogonal with weight r on the interval (0, a), it follows from the initial condition that the coefficients cn are given by 153. The problem with initial condition reduces to integration of the differential equation and boundary condition TI,,, = 0. Setting T(r, 7) = R(r)@(r) and separating variables, we obtain the equations whose general solutions are sin JXr cos JAr R=A- + B ----- , @ = ce-". From the condition that R be finite at the center of the sphere, we find that B = 0, while the boundary condition R(a) = 0 leads to the eigenvalues and corresponding eigenfunctions 1 nrr R= Rn=-sin-. r a Summing particular solutions, we obtain 1" T(r, T) = - 2 cne-nana~laP sin - . r n=l a The coefficients c, must be determined from the initial condition I" nxr TJ,,~ = f(r) = - 2 cn sin - , o < r < a, n=l r which, by the theory of Fourier series, implies c. = -2 la (p) sin % p dp. a o a This leads at once to the answer on p. 81. 169. This temperature distribution problem leads to integration of Laplace's equation Writing T(r, z) = R(r)Z(z) and separating variables, we obtain the ordinary differential equations with general solutions R = ~1,(Jhr1 + BK,(J~~), z = c cos Jhz + D sin :h z. The constant B equals zero because of the requirements that the temperature be finite on the axis of the cylinder. The boundary conditions (2' + hZ)I,=*l,, = 0 lead to the eigenvalues 302 SOLUTIONS PROB. 175 and eigenfunctions 2~ nz Z = Zn = COS - 1 ' where the yn are consecutive positive roots of the equation h 1 tan y = - 2v The expression satisfies Laplace's equation and the boundary conditions on the ends of the cylinder. To determine the constants c,, we use the boundary condition on the lateral surface Because of the orthogonality of the eigenfunctions, this gives Substituting (19) into (18), we obtain the answer on p. 87. 175. We have to integrate the system of differential equations Application of the Fourier method leads to the expression PROB. 175 SOLUTIONS 303 for the required temperature distribution, satisfying all the conditions of the problem except the boundary condition on the lateral surface, where ynh~ yn(z + hl) -hl o, Z:'(Z) = sin - sin hl hl 9 Zn(4 = Z?)(Z) = sin yn sin ~n(h, - Z) , 0 G z < h,, hl and the yn are consecutive positive roots of the equation kl yh, tan y + - tan - = 0. k, hl The eigenfunctions Zn(z) are orthogonal on the interval (-k,, h,) with weight -hl G z < 0, r(z) = : 0 < z < h,. To see this, we multiply the equations by r(z)Z,(z) and r(z)Zn(z), respectively, subtract the results from each other, and then integrate with respect to z from -hl to h,. This gives where we have used the boundary conditions (1) 2, (- hl) = ZZ'(h,) = 0, z~'(o) = z~'(o), k1zS"(0) = ~,z:"(o). The orthogonality of the functions Z;(z), together with the condition TI,=, = To, implies Evaluating the integrals in (21), we obtain cos yn - cos - Cn = (Y) t~nhz hn 2 )' ynIo " - sin - + - sin yn hl hl Substitution of these coefficients into (20) gives the answer on p. 89. 304 SOLUTIONS PROB. 176 176. The problem reduces to integration of Laplace's equation with boundary condition To, 0 < 0 < a, TI,.. = f (0) = ( 0, a<e<n. The required harmonic function is constructed as a series where P,(x) is the Legendre polynomial of degree n. Because of the boundary condition, the coefficients cn must coincide with the expansion coefficients of the function f (0) with respect to the Legendre polynomials, i.e., m j(o) = 2 cn~n(~os e), o G e < x, n--1 which implies 2n + 1 2n + 1 en = -[j(~)Pn(cos 0) sin e dB = - 2 2 T~J~:~ :n(X) dx. For n = 0 we immediately find 1 Ico8a~o(x) dx = 1 - cos a. For arbitrary n, we use the recurrence formula 1 1 JCo8 a Pn(x) dx = - [Pn-l(~~~ a) - Pn+l(~~~ a)], n = 1,2, . . . . 2n + 1 Substituting the values of cn obtained in this way into (22), we find the answer on p. 90. 190. To solve the problem, we find it convenient to assume that the current J is uniformly distributed with density J/2&h over a small section lyl < E of the sheet (where h is the thickness of the sheet), afterwards taking the limit as E --+ 0. Then the problem reduces to integration of the two- dimensional Laplace equation PROB. 192 SOLUTIONS 305 with boundary conditions Application of the Fourier method leads to an expansion of the form m nxx nny u = cox + 2 cn sinh - cos - + const, n=l b b whose coefficients are calculated from the formulas 2 co = lf (u) dy, en = b nx cosh (nxalb) Substituting for f(y), evaluating the integrals and taking the limit as E -+ 0, we find the answer on p. 95. 192. The potential of the surface current must satisfy Laplace's equation (cf. Prob. 21). To formulate boundary conditions for the problem, we first assume that the current J is distributed with constant density over the section IcpI < E, z = $1, so that and separating variables, we obtain Qn(cp) = A, cos ncp + B, sin ncp, n z nz Zn(z) = Cn cosh - + D, sinh - , Z, = Co + Doz, a a where we use the required periodicity of the solution in the angular variable cp. Because of the symmetry with respect to the plane cp = 0, the coefficients Bn vanish, and hence the solution of the problem can be written as a series m u(y, 1) = 2 (M, cosh "1 + Nn sinh cos ncp + iV,z + const. n=l a a PROB. 196 The boundary conditions give nl Mn = 0, f (cp) = 2 - N, cosh - cos ncp + No, n=l a a l.e., n Nn - 2 J sin ne cosh - n1 = - Snj(cp) cos ncp dg = - - -- . a a TCO x 2aeho n Taking the limit as E -+ 0, we obtain Nn = J xnho cosh (nlla)' which leads to the answer on p. 95. 196. The problem reduces to integration of the system of differential equations for the potential of the magnetic field, with boundary conditions Bearing in mind that the required potentials are even periodic functions of the variable cp, we represent the functions ui as series m u, = Hor cos cp + 2 ~,r-" cos ncp, n=l m = CO In r + 2(Enrn + Cnr-n) cos ncp, n=l do ua = 2 Dnrn cos ncp (where arbitrary additive constants have been omitted). Because of the boundary conditions, the constants Co and C,, (n 2 2) vanish. To determine pROB. 201 SOLUTIONS 307 the remaining constants, we use the system of equations pa2Bl - pC1 - a2D, = 0. It follows that where A = byp + - a2(p - 1)'. Substituting these coefficients into the series for the ui, we obtain the answer on p. 97. 201. In this problem, it is convenient to characterize the magnetic field by a vector potential which in each of the media (air, magnetic material, air) has a single component A=Ai(r,), i=l,2,3. Setting A(') = A, + A,, A(2) = A,, A(3) = A 39 where cos cp drp A,(r, 8) = - 2:"I:Jr2 - 2rr0 sin e cos rp + ri is the vector potential of the loop, we reduce the problem to determination of the functions A, satisfying the differential equations4 1 a A l "r.2) + - -(sine 2) - - - ar sin 8 a0 sin2 8 -0, O<r<a, 1 a A , "r.2) +--(Sin 0%) -- - -0, b<r<m ar sin 8 38 sin2 0 Note that PROB. 201 and the boundary conditions Looking for solutions of these equations of the form A = R(r)@(0), we obtain the equations 1 (r2R')' - AR = 0, - sin 0 The permissible values of the parameter A are determined from the condition that the second of the equations (24) have solutions which are regular in the closed interval 0 < 0 < 7c. This requirement leads to the eigenvalues and corresponding eigenfunctions where the Pi(x) are associated Legendre functions of the first kind. The general solution of the first of the equations (24) is Taking account of the behavior of the functions Ai near r = 0 and r = coy we find that they can be represented as series of the form The vector potential of the source can also be represented as an expansion in terms of Legendre functions, by starting from the formula I 1 " = - + 2 Pn(sin 0 cos cp) , r > ro. (25) Jr2 - 2rr, sin 0 cos cp + r: r .=I Using the addition formula for spherical harmonics " P(n - m + 1) Pn(sin 0 cos cp) = Pn(0)Pn(cos 0) + 2 2 m=l p(n + m + 1) x P~(O)P~(cos 0) cos mcp, PROB. 206 SOLUTIONS 309 and substituting (25) into the integral (23) for A,, we find after some simple calculations that Then the boundary conditions lead to the following system of equations for determining the coefficients A,, B,, C, and D,: Solving this system we obtain which leads to the solution in the region outside the shield given on p. 99, if we bear in mind that Pik(O) = 0. 206. The magnetic field in the spherical resonator has only a cp-component with complex amplitude H, = H(r, 8). Writing H -- H, + HI, where6 P sin 8 H, = - (1 + ikr)eWikT crZ is the magnetic field of the source, we find that HI satisfies the equation 1 AH, + (k2 - -) H, = 0. r2 sin2 8 Next we introduce a new unknown function u = u(r, 8) such that ' This expression can be obtained from the relations p e-ik~ p ,-ikr H, = (curl A'O'),, A!.') = - - cos 0, A(,) = - - - e sin 0. c r c r PROB. 210 Then u is the solution of Helmholtz's equation which is regular inside the sphere. Since the tangential component of the electric field must vanish on the surface of the sphere, it follows that Using the Fourier method to solve the differential equation for u, we find that 1 " u(r, 0) = -- 2 ~,J,+~(kr)P~(cos F)), Jr n=O (26) in terms of the Legendre polynomials P,(x) and the Bessel functions of half- integral order J,+&). Using the boundary condition and the familiar relation - 2 sin z 2 = ( Z - COS .), we find that the coefficients c, equal xk Pk -ika 1 + ika - k2a2 O' = A T (1 - kga') sin ka - ka cos ka' Substituting these values of c, into (26) and differentiating with respect to 0, we arrive at the expression for H, = H(r, 8) given on p. 101. 210. The problem reduces to solving the equation of the vibrating string, with zero initial conditions and homogeneous boundary conditions u(,=" = uI,,t = 0 pROB. 210 SOLUTIONS 3 1 1 of the first kind. We look for a solution in the form of an expansion with respect to the eigenfunctions X,(x) of the corresponding homogeneous problem, where the weight r equals 1 and C, =/;UX,(X) dx. The functions X,(x) are the nontrivial solutions of the equation satisfying the homogeneous boundary conditions X(0) = X(1) = 0. Such solutions exist for and are of the form nxx X = X,(x) = sin - . 1 To determine the coefficients C,, we multiply (27) by X,(x) and integrate with respect to x from 0 to Integrating by parts twice and taking account of the boundary conditions, we obtain nm e; + (?Isn = Jb(x, t) sin - dx. To 1 The solution of this equation can be found by variation of constants: nxvt nmt ul Stsin nm(; - 1) fin= A,cos-+ Bnsin-+- 1 I nxT o To calculate the constants A, and B,, we use the initial conditions for the function C,, which are obtained by multiplying the original initial conditions by Xn(x) and integrating with respect to x from 0 to I. The result is which implies A, = B, = 0. In this way, we arrive at the answer on p. 108. In the interest of using a unified approach, we follow the general scheme on p. 105. For problems of the type under consideration, this method is entirely equivalent to that described on p. 104. 3 12 SOLUTIONS PROB. 217 217. We have to solve the inhomogeneous equation a4u 1 8% q sin at - +--=- ax4 a4 at2 EJ for transverse oscillations of the beam, with zero initial conditions and homogeneous boundary conditions A feature of this problem is that it involves an expansion in terms of eigen- functions of a fourth-order differential operator. Following the usual method, we represent the solution as an expansion with respect to the eigenfunctions X,(x) of the homogeneous problem, where 8, =/lCruxn(x) dx. In the present case, the weight r = 1, and the functions X,(x) are the solutions of the equation X(iv) - AX = 0 satisfying the boundary conditions Simple calculations show that7 YnX X,(x) = cosh y, cos - - 1 YnX cos y, cosh - , Z where the y, are consecutive positive roots of the equation tan y + tanh y = 0. To determine the functions d,, we multiply (28) by X,(x) and integrate with respect to x from -1 to I. Integrating by parts four times and taking account of the boundary conditions we obtain ' Concerning the orthogonality of the functions X,,(x), see the solution of Prob. 118. PROB. 222 SOLUTIONS 3 13 The solution of this equation satisfying the zero initial conditions a,(o) = n;(o) = o obtained by multiplying the original initial conditions by Xn(x) and integrat- ing with respect to x from -1 to I, is given by aZy2 t sin ot w sin 2 - - - qa2P un = - l2 l2 EJYI The final form of the solution, as given in the answer on p. 110, is found by taking account of the easily verified formulas 41 sin yn cosh y, 9 Yn I S_il X:(x) dx = - Xk2(l) = 21 cash", cos2 y,. 2 222. The problem reduces to integrating the equation for the oscillating beam, with zero initial conditions and inhomogeneous boundary conditions Applying Grinberg's method, we look for a solution in the form of an expan- sion with respect to the eigenfunctions X,(x) of the homogeneous problem, where Explicit expressions equation - flruxn(x) dx. un for the functions Xn(x) are obtained by solving the x?) - AX = 0, 3 14 SOLUTIONS PROB. 223 with homogeneous boundary conditions X(0) = X'(0) = X"(1) = X"'(1) = 0. This gives X,(x) = (sin y, + sinh y,) - (COS y, + cosh y,) where they, are consecutive positive roots of the equation cos y cosh y + 1 = 0. The functions X,(x) are orthogonal on the interval (0, I) with weight r = 1 (see the solution to Prob. 118), and the integral of X:(x) is 1 ~oL~~x) dx = - ~:(l) = l(sinh y, + sin yJ2. 4 To determine the coefficients &,, we multiply (29) by X,(x) and integrate with respect to x from 0 to I. After a bit of n~anipulation, we arrive at the equation The solution of this equation satisfying the initial conditions a,(o) = n;(o) = o, obtained by multiplying the original initial conditions integrating with respect to x from 0 to I, is given by by X,(x) and - a212 U, = - sin y:a2(t - d~, EJY: l2 and immediately leads to the answer on p. 112. 223. Clearly we can express the dependence of the external load on the coordinate x and the time t in the form \O otherwise, where E > 0 is arbitrarily small. Let 2" nxx U(X, t) = - 2 u -,, sin - , 1 ,=I 1 where SOLUTIONS 3 15 ' nxx ii, =[ u sin 7- dx. Multiplying the equation for the oscillations by sin (nxxll), integrating with respect to x from 0 to I, and taking account of the boundary conditions, we obtain nxx ii? (~rti, = $ lq(x, t) sin - dx, 1 or nxvt 6, = sin ot sin - EJ 1 after passing to the limit E -+ 0. The general solution of this equation is a2n2x2t a2n2x2t ii, = A, cos - + B, sin - l2 l2 where we introduce the abbreviation Using the initial conditions we find that iin(t=.O = ii,$=O = 0, and hence Substituting these values of the coefficients into (30), and letting M denote the mass of the beam, we obtain the answer on p. 112. 225. To solve the problem, we assume that the external load is distributed over the membrane with density p sin ot forb-~<r<b+~, (0 otherwise, 3 16 SOLUTIONS PROB. 225 where E > 0 is arbitrarily small. The deflection u(r, t) of an arbitrary point of the membrane is then the solution of the inhomogeneous equation with homogeneous initial and boundary conditions The desired solution is constructed as an expansion with respect to the eigenfunctions R,(r) of the homogeneous problem. The latter are the solutions of the equation (rR')' + ArR = 0 which are unbounded in the closed interval [0, a] and satisfy the boundary condition R(a) = 0. As usual, ii, denotes the integral It is easily verified that the eigenvalues and eigenfunctions are given by where J,,(x) is the Bessel function and the y, are consecutive positive roots of the equation J,(y) = 0. Applying the usual method for determining the coeffi~ients ii,, we obtain sin at *+ Jb-z J'(?) dr' 6; + Y"U c, = --- pbv"'f) J, - sin at (,I 7- after passing to the limit E -+ 0. Integrating this equation with zero initial conditions ii,(O) = iiA(0) = 0, PROB. 227 SOLUTIONS 3 17 we find that Ynvt Yn" sin- - - sin wt J, 2abpv a wa - un = - wT Yn which leads to the answer on p. 113. 227. To solve the problem, we regard the concentrated load as the limiting case of a load distributed over a disk of small radius c. Then, to determine the transverse oscillations of the plate due to this load, we integrate the equation 1 3% q(r, t) where A sin wt , Ogr<c, subject to zero initial conditions and the boundary conditions Let where the Rn(r) are the eigenfunctions of the homogeneous problem and 8% = 1: ruRn(r) dr. The functions R,(r) are the solutions of the differential equation which are bounded for r = 0 and satisfy the conditions R(a) = R1(a) = 0. Therefore where the yn are consecutive positive roots of the equation R;(a) = 0. The corresponding eigenvalues are 3 18 SOLUTIONS PROB. 230 and the functions Ry,(r) are orthogonal on the interval (0, a) with weight r (see the solution to Prob. 124). Multiplying the original differential equation by rRy,(r), integrating with respect to r from 0 to a, and then integrating by parts four times, we obtain Ab4 - R,,(O) sin wt. 2x D after taking the limit as 0 -+ 0. The solution of this equation satisfying the boundary conditions n,(o) = n;(o) = o is Substituting these values of ii, into the series for u(r, t) and using the formula we finally arrive at the form of the solution given in the answer on p. 114. 230. To solve the problem, we replace the line loadp by a load uniformly distributed over the strip -0 < x < 0, -b < y < b of width 20, i.e., we reduce the problem to integration of Poisson's equation with homogeneous boundary conditions of the first kind : Two forms of the solution can be found. To obtain the first, we represent the displacement as a series with respect to the eigenfunctions of the correspond- ing homogeneous problem which depend on the variable x: 2" u = - 2 ii, cos (2n + 1)xx " (2n + 1)m dx. 9 fin=l u COS a .=o 2a 2a PROB. 230 SOLUTIONS 3 19 Multiplying (31) by cos [(2n + l)xx/2a], integrating with respect to x from 0 to a, and then integrating by parts twice, we obtain8 after taking the limit as E 40. The solution of this equation satisfying the conditions iinly= * = 0 is cosh 2a cosh (2n + 1)rb I' which immediately implies formula (1 2), p. 11 5. The second form of the solution is obtained by expanding u in a series with respect to eigenfunctions which depend on the variable y: 2" U = - 2 ii, COS (2n + 1)xy u COS 2b 2b dy. b n=o This time the coefficients ii, are functions of the variable x (rather than of y), and are determined by the equation The solution of this equation satisfying the conditions ii,l,=,, = 0 is cosh (2n + 1)xx cosh (2n + 1)xa 2b sinh Substituting for f(E) and taking the limit as E -+ 0, we obtain We also take account of the boundary condition ul,=, = 0 and the relation (&/ax) I,=, = 0 implied by the symmetry of the problem. 320 SOLUTIONS PROB. 240 which implies formula (13), p. 115. Which form of the solution to use in making calculations depends on which series converges more rapidly (this depends primarily on the ratio alb of the sides of the rectangle). 240. The problem reduces to solving the biharmonic equation with boundary conditions m uIxdi = uI,=, = 0, x=a D' It is easy to construct a function satisfying both the differential equation and the boundary conditions at x = 0 and x = a. If we set u=u* f v, then the new unknown function v must be a solution of the homogeneous biharmonic equation satisfying homogeneous boundary conditions in x: This enables us to use the Fourier method, where the boundary conditions in the variables y take the form Taking account of the boundary conditions in the variable x, we look for particular solutions of the biharmonic equation A2v = 0 of the form nxx v = vn(y) sin - , n = 1, 2, . . . a The amplitude u, must then be a solution of the differential equation "y - (!!2)'.; + (!gun = 0 which is even in y, and hence "xY n'=Y o,, = A, cosh - + B,y sinh - . a a Writing u as a series we determine the coefficients A, and B, from the remaining conditions This requires expanding the known function u*(x) in a Fourier series with respect to sin (nxxla). In this way, we eventually arrive at the answer on p. 119. 241. Suppose the line load p is replaced by a load uniformly distributed over the sector -E < cp < E, 0 < r < a with central angle 20, where E > 0 is arbitrarily small. Then the problem reduces to solving the inhomogeneous biharmonic equation with homogeneous boundary conditions With our way of measuring angles, u is an even function of p and hence can be written as a cosine series 1 2" u(r, cp) = - a,, + - 2 17, cos ncp, x TE n-1 where a, =cu cos np dp. To find a,, we multiply the equation for u by cos np and integrate with respect to cp from 0 to n. Then, integrating by parts four times, we find that p sin nE r dr r2 caDn where the right-hand side can be replaced by p/aD after taking the limit as E + 0. We are interested in the solution of this equation which is regular for r = 0, i.e., ii, = A,rn + ~,r"+' + a,*, where PROB. 242 except for the cases n = 2 and n = 4: The constants An and B, are determined from the conditions an(a) = 17L(a) = 0. 242. The problem reduces to integration of the heat conduction equation a2~ aT --- - ax2 a7. ' with the zero initial condition Tl,=o = 0 and inhomogeneous boundary conditions It is easy to see that the linear function 4 T* = To + -(a - x) k is a solution of (32) satisfying both inhomogeneous boundary conditions in the variable x. Therefore, writing T= T*-U, we find that u satisfies the differential equation with initial condition u(T-0 4 = T*(x) = To + - (a - x) k and homogeneous boundary conditions Application of the Fourier method gives where the cn are the coefficients of the Fourier expassion of T*(x) with respect to the functions cos [(2n + l)nx/2a]. 247. We have to solve the inhomogeneous equation with the zero initial condition T(,,, = 0 and homogeneous boundary condition of the third kind: Suppose the solution is of the form where T,, = j;T~,(r)r dr, in terms of the eigenfunctions Rn(r) of the homogeneous problem. The latter must satisfy the equation the boundary condition R'(a) + hR(a) =. 0 and the requirement that R(0) be bounded. Solutions of the required type exist if where the y, are consecutive positive roots of the equation The corresponding solutions of (34) are These functions are orthogonal with weight r on the interval (0, a), and more- over 324 SOLUTIONS PROB. 261 To find the functions Tn, we multiply (33) by Rn(r) and integrate from 0 to a. Then, integrating by parts twice and taking account of the boundary conditions, we find that The solution of (35) satisfying the condition Tn = 0 is Therefore the desired temperature distribution can be represented as a series This form of the solution is suitable only for small values of T, i.e., during the initial stages of the heating. For large values of T, it is convenient to subtract out the terms of the series which are independent of time, by using the formula Then T(r, T) takes the form given in the answer on p. 121. 261. The problem reduces to finding the solution of the equation which satisfies the boundary conditions of the second kindg The solution can be obtained in two different forms, either as a series with respect to the eigenfunctions Xn(x) satisfying homogeneous boundary con- ditions in the variable x, or as a series with respect to the eigenfunctions YJy) satisfying homogeneous boundary conditions at the end points of the interval 0 < y < b. The first form of the solution is 1 2" nxx T(x,y)=-~~+-~T~~os-, a a .=I a The density q of the heat current through the section I;rl < c, y = b can be expressed in terms of the density Q of heat produced inside the bar by using the condition qc = Qab for solvability of the problem. where nxx To determine Tn, we multiply the original inhoniogeneous equation by cos (nxxla) and integrate with respect to x from 0 to a. This gives which implies T o - Quy2 + A. + Boy, 2k nxY n=Y Tn = A, cosh - -/- B, sinh - , n > 1. a a Using the boundary conditions in y, we find that dT, nnx Qa2b nxc J,=b=cj(x) coS - dx = - - sin - , a nxkc a which leads to the following values of the constants:I0 Qa3b sin (nxcla) A,=-- B, = 0. n2x2kc sinh (nxbla) ' Substituting A, and Bn into (36), we obtain formula (14), p. 126. To obtain the other form of the solution, we set 1 2" n=Y T(x, Y) = - To + - Tn cos - , b b .=I b where nxy T = T cos- dy. @b Then, by the same procedure as before, we obtain the differential equation determining the coefficients Tn. The solution of (37) satisfying the conditionsl1 lo The constant A, remains indeterminate. l1 The desired solution T(x, y) is an even function of x, and hence, from now on, we need only consider the region 0 < x < a. PROB. 261 can be found by variation of constants, and turns out to be r Similar calculations for the case n = 0 lead to the following expression: Qab3 (-l)n+l Tn=--- c n2x2k After some manipulation, we find that nxx cosh - (sinh 7 - sinh nx(a - c) nxa sinh - b b nxy (- 1)" cos - 2Qab2 b +YC x kc nxa n2 sinh - b nxx 1 - cosh - , b cash nx(x - c) b nx(a - c) nxx nxa sinh b cosh--sinh-, Ixl<c, b b " [ nr n4a - 1x1) - sinh - cosh b' The form of the solution given in the answer on p. 126 is obtained if we improve the convergence by using the formula to carry out partial summation of the series. 269. To solve the problem, we assume that heat is produced with uniform density Q/m2 inside a cylinder of arbitrarily small radius E. Then the problem reduces to integrating Poisson's equation with boundary conditions Expanding the solution in a series of eigenfunctions of the corresponding homogeneous problem depending on the variable I., we find that 2 " Tn T, z) = - - J), Tn =&TJo(y) r dr, a2 n=1 JXyn) where the y, are consecutive positive roots of the equation Jo(y) = 0. Multiplying the original equation by rJo(ynr/a) and integrating with respect to r from 0 to a, we obtain after taking the limit as E -+ 0. The solution of this equation satisfying the boundary conditions ah cosh (y,z/a) Tn = - 2:;:; - yn sinh (ynl/a) + ah cosh (ynl/a) which leads to the answer given on p. 130. 272. This problem of electrostatics reduces to finding a solution of Laplace's equation a2~ aZu -+,=o ax2 ay satisfying the following inhomogeneous boundary conditions of the first kind : 328 SOLUTIONS PROB. 272 Following Grinberg's method, we look for a solution in the form of an expansion with respect to the eigenfunctions of the corresponding homo- where 2" u(x, y) = - 2 17, cos (2n + 1)xy 2b b n=o 3 To determine the unknown quantities fin, we multiply Laplace's equation by cos [(2n + I)xy/2b] and integrate with respect to y from 0 to b. Taking account of the boundary conditions, we obtain We want the solution of (38) which is bounded at infinity and satisfies the condition It is easy to see that this solution can be written in the form 172) = B, sinh (2n + 1)xx 2bV(- l)n 26 + 9 X<a, U, = (2n + l)x x > a, where the values of the constants B, and C, are determined from the "contact conditions" B, = 2b V(- 'In+' e-(2n+l)naj2b 3 Cn = COS~ 2bV(- 1)" (2n + 1)xa (2n + l)x (2n + l)x 26 Substitution of these values of the coefficients into 17, leads to the following series solution of the problem: l2 Choosing the other form of the solution leads to an expansion in a Fourier sine integral over the integral (0, m). To obtain the final form of the solution, we improve the convergence by using the formula X X 2 cos (2n + 1)x = - 1x1 < 5 n=o 2n + 1 4 ' to sum the slowly convergent part of the first series. It would be noted that the solution can also be written in closed form. 277. To solve the problem, we first assume that the charge q is uniformly distributed with density p over an arbitrarily small cylinder 0 < r < 8, c - 3~ < z < c + 3~, i.e., we reduce the problem to integration of Poisson's equation where b otherwise, subject to the boundary conditions One of the two possible forms of the solution is an expansion with respect to the functions Jo(ynr/a), which are the eigenfunctions of the corresponding homogeneous problem, i.e., where the y, are consecutive positive roots of the equation Jo(y) = 0. Multiplying the original equation by rJ,(y,r/a) and integrating with respect to r from 0 to a, we arrive at the equation which is to be solved with zero boundary conditions The general solution of (39) satisfying the first of these conditions is sinh (ynz/a) a, = A, sinh (ynlla) Y, a 330 SOLUTIONS PROB. 282 Using the other boundary condition to calculate A,, and then passing to the limit 6, E -+ 0, we obtain 2aq yn(l - c) A, = - sinh Yn a Thus the coefficients 0, are equal to sinh yn(l - C) sinh , O<z<c, 2aq ( a a Un = ~nl YC Y, cCz<l, Yn - sinh 2 sinh a a a which immediately leads to the answer on p. 134. The other form of the solution can be obtained by expanding u(r, z) in a series with respect to the eigenfunctions in the variable z, i.e., 2 - . nxz ' nxz u(r, z) = -2 U, sm - , an = u sin - dz. 1 ,=I 1 1 282. Since the potential distribution must be an odd function of the coordinate z, the problem reduces to solving Laplace's equation with boundary conditions uIz=o = 0, uI,=t = v, = f(4, where To obtain homogeneous boundary conditions in the variable z, we set Then the function v(r, z) will be the solution of Laplace's equation satisfying the homogeneous conditions vl,,o = UI,,~ = 0 and the following boundary condition on the lateral surface PROB. 289 SOLUTIONS 33 1 To find the function v, we can now use the Fourier method, which, after separation of variables and determination of eigenvalues and eigenfunctions, leads to the expansion where the coefficients cn are found from the condition After determining the potential u(r, z), the electric field on the axis of the lens can be calculated from the formula .Ezlp0 = - a,/ . az T=~ 289. Suppose the current is distributed with uniform density over the arbitrarily small area Then the problem reduces to integration of the equation I a 1 aZu a2 sin 8 88 7t Og 8g-, -x<cp<z (40) 2 (see Prob. 21, p. 14), where J 6 6 E for 8,- <e<eo+- <I, 2 2 ' otherwise, subject to the boundary condition UIO=~,Z = 0. If we introduce a new variable by writing then (40) takes the simpler form PROB. 289 whose solution can be constructed as a Fourier series 1 2 u = - tiO + - CG,COS ncp, X 7C n=l where To determine the coefficients g,, we follow the usual approach, obtaining the differential equation whose general solution is The constants A, and B, are determined from the boundary condition and the condition that ii, be bounded for $ = 0. Passing to the limit 6, E .+ 0, we find that13 J 8 An = - - ( - ) $o = tan 2, 4ohn 2 Bn=O, n= 1,2 ,..., which implies Therefore the desired solution has the following series representation: J m cos ncp 2xah n=l n J cos ncp uI+o~+~l = - 2xoh n=l n la The coefficients B, vanish for arbitrary values of 6 and E. PROB. 296 SOLUTIONS 33 3 Using the formula to sum the series, we arrive at the answer given on p. 138. 296. In this problem it is convenient to characterize the electromagnetic field by the vector potential AeiUt, whose complex amplitude has components A, = A, = 0, A, = A(r, z). Suppose the current in the dipole is replaced by a current distributed over the volume of an arbitrarily small cylinder Then A(r, z) is determined by the differential equation where to otherwise. The tangential component of the electric field must vanish on the surface of the resonator, and hence We look for a solution of the problem in the form of a Fourier cosine series 1 2" nnz A(r,z) = -KO 1 + -~A;,cos-, 1 n-1 1 where = A cos - dz. S,' "7 The usual argument implies where the last condition is equivalent to 334 SOLUTIONS PROB. 303 because of the differential equation for A,. Using the method of variation of constants, we find that which immediately implies the answer on p. 141. 303. We want the solution of Laplace's equation a2T a2T -+-=o (O<x< m,o< y< co) ax2 ay2 satisfying the boundary conditions Application of the Fourier method leads to the particular solutions T = T, = ~,e-'" sin Ay, A 0, which are bounded in the quadrant 0 < x < co, 0 < y < c~ and vanish for y = 0. Integrating with respect to the parameter A, we obtain T(x, y) = JOm ~,e-" sin hy dl, where the coefficient B, is determined from the boundary condition Because of the theorem on expansion in a Fourier sine integral, we have B,=-- 2q 1 - cos Ab f(y) sin Ay dy = - Sm nh o k A2 ' which is the same as the expression for T(x, y) given on p. 150. PROB. 313 SOLUTIONS 335 313. To avoid the difficulties associated with the fact that the logarithmic potential does not go to zero at infinity, we look for the components of the electric field in the two media: Exl, Evl, Ez2, Em Setting (0) (1) (2) E,, = Ex + E, , E,, = E:' + E:', Ex, = E:', E,, SE E, , where EcO) is the field due to the charged wire in an unbounded medium of dielectric constant E,, with components EL0) = qx q) = dy - a) E~[x~ + (Y - a)'] ' EI[X~ + (y - ' we obtain the system of differential equations which, together with the boundary conditions (0) (1) (2) (0) (1) (2) [Ex + Ex I,=, = Ex I,=,, %IE, + Ey 1,=0 = EZEW ly=O, (1) (1) (2) (2) Ex ,E, /,*+m+o, Ex, E, 1 y+-m +o, E!), E(~)[ (42) y 2+*m -+O, determine the functions E:', E:) (i = 1,2). A convenient way of solving (41) is to use the method of integral transforms, by taking the sine transform of E:) and the cosine transform of E:' (i = 1, 2).14 Thus we multiply the first of the equations (41) and the second of each pair of boundary conditions (42) by cos Ax, and the second of the equations (41) and the first of each pair of boundary conditions by sin Ax. Then, integrating from 0 to co, we find that Ep, E:)lw++m + 0, p, E:)lw+-m + 0, where Et) = 1" E") sin dx, Ep = Jom ~:)cos AX dx. The solution of the system (43) is l4 Note that the cosine transform of Eb" and the sine transform of EA" vanish, because of the symmetry of the problem. 3 3 6 SOLUTIONS Using the inversion formulas and making a few simple calculations, we arrive at the expressions for the components of the electric field given in the answer on p. 154. 321. The electric field has only a z-component, whose complex amplitude we denote by E(x, y). If we regard the current as distributed over an arbi- trarily small rectangle a - 6 < x < a + 6, lyl < E, then the solution of the problem reduces to integration of the inhomogeneous Helmholtz equation 4xiw AE + k2~ = - j(x, y), c2 where for a-~<x<a+~, IYI<E, j(x, y) = \- 46~ (0 otherwise, with boundary conditions To solve the problem, we first make a Fourier sine transform, carrying (44) into the ordinary differential equation k2)E = 4*S0wj(<, y) sin A< d< c2 for the quantity w E = lo E sin AX dx. The solution of (45) satisfying the boundary conditions can be obtained by variation of constants. Then, taking the limit as 6, E -+ 0, we find after some simple calculations that = - 2nikJ sinh Jh2 - kyb - lyl) sin ha. cJh2 - k2 cosh Jh2 - k2 b This immediately leads to the answer on p. 157, if we use the inversion formula w E =2 [ Esin Axdh. PROB. 324 SOLUTIONS 337 324. We want the stresses ox, r,, and o, satisfying the system of equations sox arXv arxv a. - + - = 0, - + = 0, (equilibrium equations) ax ay ax ay (46) aZox a2~xv -- aao 2- +-,=o (compatibility equation) ay2 ax ay ax2 and the boundary conditions o,l,-o =fW, ~xVlyPO = gw. Introducing Fourier transforms of the unknown functions, we multiply each of the equations (46) by eiAx and integrate with respect to x from -cc to cc, taking account of the behavior of the stresses as x + f w.I6 This gives the system of ordinary differential equations -ihEx + Tk, = 0, -iATxv + 5; = 0, a; + 2iASh, - A%, = 0, (47) which must be solved with the boundary conditions - %lv=o =A ?xY(Y=o = g and the conditions at infinity. - ox, T~, E,+O as y+ cc . The solution of the system (47) satisfying all the conditions of the problem is where the constants A and B have the form A = g, I4 - B=f--g. A l6 We assume that the stresses and their first derivatives approach zero at infinity. It should be noted that the problem cannot be solved in this way for the Airy stress function, since the latter cannot be expanded as a Fourier integral. The final form of the solution given on p. 158, involving various integrals, is found by using the inversion formula to go back from the quantities T,,, Ex, Ey to the stresses T,,, ox, oU themselves. 328. Replacing the concentrated force P by a load uniformly distributed over the arbitrarily small rectangle we reduce the problem to integration of the inhomogeneous biharmonic equation where 6 6 E for - - <x<- 6--<y<b+" 4(x, Y) = 2 2 ' 2 2 ' [r otherwise, subject to the boundary conditions Taking the Fourier transform of (48), where li = IOw u cos Ax dx, we obtain the following equation for li:16 The general solution of (49) can be obtained by variation of constants, and has the form l6 It is assumed that u and its first three derivatives with respect to x go to zero as X'cO. PROB. 334 SOLUTIONS 339 At this stage, it is convenient to simplify the calculations by taking the limit as 6, E + 0. The result is The constants C and E are determined from the condition q,+, + 0, which gives The other two constants are found from the boundary conditions -I C(~=O = u l,=O = 0, which implies A = -C, B = -2C - E. The value of the deflection u(x, y) is obtained by using the inversion formula. To find the bending moment and the shear force on the clamped edge, we differentiate the expression pePhb qy<b = - [(I + Ab + A2by) sinh Ay - Ay(1 + Ab) cosh Ay], 4 Dl3 obtaining (1 + Ab)~e-'~ 2 D The values of M and N are then found by substituting the corresponding values of R and into the appropriate inversion formulas. 334. The problem reduces to integration of the heat conduction equation with the initial condition TI,=, = f (r). Writing T = R(r)@(r) and separating variables, we obtain 340 SOLUTIONS PROB. 335 Integrating these equations, and taking account of the boundedness of T as r + 0, we find that T = TA = C,~-~~J,( JA r). It follows from the boundedness of T as r + rn that the parameter A can only take positive values A = p2. This leads to the following set of particular solutions depending continuously on p: T = T, = c,e-~2'~o(yr), 0 < y < rn. The general solution is then constructed as an integral of the form T(r, r) = jOm ~,e-~7,(~r) dp. (50) The coefficients c, are determined from the initial condition, which gives if we take account of Hankel's integral theorem. Substituting (51) into (50), reversing the order of integration and then integrating with respect to y, we find the form of the solution given in the answer on p. 162. 335. We want the solution of the equation satisfying the initial condition TI,=, = 0 and the boundary conditions17 we carry out a "Weber transform" by multiplying (52) by ryh(r) and inte- grating from a to rn. Taking account of the behavior of the various functions as r + rn and the relations 2 = 0, cp;(a) = - 3 xa we find that where l7 It is assumed that 4; T and d;(aTlar) approach zero as r + oo, and that the integral converges. SOLUTIONS 34 1 The solution of (53) satisfying the condition TI,=, is To determine the solution T(r, 7) from its Weber transform, we use the inversion formula 351. As is well known (see T4, p. 343), in the case of axially symmetric problems of elasticity theory, the stresses can be expressed in terms of a solution u(r, z) of the biharmonic equation (it is assumed that there are no body forces). To subtract out the singularity at the point of application of the force, we write u = uo + u1, where is the stress function corresponding to a concentrated force P applied to an infinite elastic body, and u, is a biharmonic function regular in the region z > 0. Since the unknown stress 5, is related to the function u by the formula to solve the problem we need only find the quantities Au, and a2u,/az2. The first quantity is harmonic in the region z > 0 and can be written as an integral w Au, = fo ~,e-"~~(hr)h dh, (54) while the second quantity is biharmonic in the region z > 0 and can be written in the form (note that the integrand is biharmonic). Comparing the result of differ- entiating (54) twice with respect to z with the result of applying the operator 342 SOLUTIONS PROB. 355 to (55), we find that CA = -&AAA. To determine the remaining constants, we have to use the boundary conditions which can be written as conditions on the function u,:~~ Performing the differentiations on the right, expanding the results in Hankel integrals and substituting from (54) and (55), we obtain a system of linear equations determining the constants Ah and BA. The formula given in the answer on p. 168 is obtained after evaluating certain integrals of a familiar tY Pea 355. The problem reduces to integration of the one-dimensional heat conduction eiuation a2~ aT -=- ax2 aT ' with the initial condition TI,,, = 0 the boundary condition and the condition at infinity TI,,, -+ 0. Introducing the Laplace transform we multiply the differential equation and boundary conditions by e-9T and integrate with respect to T from 0 to co. If we take account of the initial condition, this gives l8 The second of these equations follows from the formula PROB. 355 SOLUTIONS 343 The problem is now solved by using the Fourier-Mellin inversion theorem where T' is a straight line parallel to the imaginary axis lying to the right of all the singular points of the integrand. In Case a, where q = go = const, we have T = 9 -L eur-d;xdp (56) 1 PJI. T A k 2xi r As the next step, we calculate the deriva- p =g +i~ tive -=--- ax P Applying Cauchy's integral theorem to the - (r contour shown in Figure 159, and then tak- ing the limit as E + 0, R -t a~, we obtainlB -- aT - - ~m[~ - 2 /we-TT sin J; dy] ax k x o r FIGURE 159 where O(x) is the probability integral. It follows that and the final form of the solution given in the answer on p. 171 is obtained from this formula by integrating by parts. In Case b, The temperature of the surface of the body can be found by using the con- volution theorem, which gives and leads at once to the answer on p. 171. Direct application of the method of contour integration to the integral (56) itself is impossible, since the corresponding integral along the circle of radius E becomes infinite as E -+ 0. 344 SOLUTIONS PROB. 357 357. Using the Laplace transform, we write the solution in the form of a contour integral where I? is a straight line parallel to and on the right of the imaginary axis, and Jj denotes the branch of the square root whose real part is positive.20 A simple way of calculating the integral is to make the substitution and then reverse the order of integration. Together with the result obtained in the solution of Prob. 355, this gives Integrating by parts, we find that which leads at once to the answer on p. 172. 371. The problem reduces to finding a solution of the equation satisfying the initial conditions and the boundary condition Taking Laplace transforms and using the initial condition, we obtain the equation For this branch, di # -h, and hencep = 0 is the only singular point of the inte- grand. PROB. 375 SOLUTIONS 345 and the condition which together imply The solution of the problem is given by the inversion formula The contour integral can be evaluated by residues, since the integrand is single-valued. The singular points of the integrand consist of poles at the points p = 0 and p = p, = -y,2/a2, where the y, are consecutive positive roots of the equation Jdy> + aJo(y) = 0. Calculating the residues at these points, we immediately find the answer on p. 177.21 AS in other problems with boundary conditions involving time derivatives, the solution of this problem is greatly simplified by the use of Laplace transforms. 375. In the first region 0 < r < co, 0 < z < a, the concentration Cl(r, z, t) satisfies the equation the the i a ac azc 1 ac, ;-a;(rg) ++=-- D at ' initial condition ClI t=o> boundary condition [where f(t) is a function to be determined later], and the conditions at infinity CIIr-m + 0, Cllz-m '0. In the second region (the tube), the concentration C2(z, t) satisfies the one- dimensional equation a1 It is easy to see that the integral along the large circle of radius R completing the contour of integration goes to zero as R 4 a. PROB. 375 the initial condition qt=o = co, and the boundary conditions Taking first the Laplace transform and then the Hankel transform of (57) and (58), and using the initial condition and the condition c,[,,, 4 0, we obtain where a single overbar denotes the Laplace transform and a double overbar the Laplace transform followed by the Hankel transform. Integrating the - equation for C,, we obtain after inverting the Hankel transform. Similarly, taking the Laplace transform of (59) and (60), we find that C, fcosh Ja(z + 1) e--+ 2 - - -. Y Jpl~ sin11 ,/p/~ 1 In the present approximation, we can find the unknown quantity,f by using the relation ellr=z=O = C2Iz=03 which implies The amount of substance M in the tube can now be calculated from the formula M = C2(z, t) dz = Mo + C epL dp -9 - 1 - coth Jp/~l] PROB. 386 SOLUTIONS 347 where Mo = CoI is the initial amount of substance inside the tube. Integrating along the contour shown in Figure 159, we obtain the answer on p. 178. 386. We want the solution of the system arl aul aul ar, L-+-=o, c-+-=o at ax at ax (0 < x < I), ar, au, a~, ar, L-+-=o, C-+-=o at ax at ax (1 < X < 03) satisfying zero initial conditions and the boundary conditions Eliminating the variable t by taking Laplace transforms, we obtain These equations can be solved for ii,, G2, P; and I,. In particular, for ii, we obtain the expression Eo e-~(x-t)/v fjz = - p + a cosh pT + [l + (Z/Ro)] sinh pT ' - where u = l/ JLC is the propagation velocity, T = I/v is the time it takes a wave to traverse the part ofthe line going from x = 0 to x = 1, and Z = .- JL/C is the wave resistance. Then the Fourier-Mellin inversion formula leads to the following representation of u, as a contour integral: " S eCt-(x/v)ST1 u&, t) = - - dp , x > 1. 2ni r cosh pT+ [I + (ZIR,)] sinh pT p + cx The most interesting form of the solution can be obtained by using the expansion 1 - 2R eVpT " - L 2 (+re-2naT cash pT + [I + (Z/Ro)] sinh pT 2Ro + Z .,, 2R0 + PROB. 402 and then integrating term by term. This gives According to the formula all the terms of this series vanish for fixed x and t, starting from some value of n. In particular, we have and so on. The general result given in the answer on p. 183 can easily be obtained by induction, with the help of the formula for summing a finite geometric series. The jumps in the voltage can be interpreted as the arrival at the point x of successive refracted waves. 402. The problem involves integration of the equation subject to the conditions a !k! 1 = 0, ( - (A)) 1 = /(T) = ar ~=o ar r =O + 0) at the point of application of the force, and to the condition at infinity Going over to the Laplace transform 2, we find that ii satisfies the differential equation Azii + p2ii = 0, (61) the boundary conditions dii P , ~~~-+m+o r=o 2x0 SOLUTIONS 349 and the condition at infinity -+ 0. The solution of (61) vanishing at infinity is 17 = A ~,(rfi eixt4) +- B ~~(r ,/p e-'"I4), where di denotes the principal branch of the square root (largpl < n). Taking account of the behavior of the Macdonald function near the point r = 0 and using the boundary conditions, we find that B=--A= P hip JDp ' Application of the convolution theorem gives u(r, r) = - jiHT) dr, 4zin o where Transforming this expression by integrating along the contour shown in Figure 159, we eventually obtainzz to evaluate the integral, we find that which immediately leads to the answer on p. 188, after integrating with respect to T. 22 Note that the function K&) is analytic in the z-plane cut along the line (-GO, 0), and use the formula Ko(eiXy) = Ko(y) - inI,(y). 350 SOLUTIONS PROB. 406 406. Taking first the Laplace transform of the original differential equation and boundary conditions, and then the Fourier transform with respect to the variable y, we find that 1 0 = - -+ 0, P where ;=I-: w e'"dyIomue-"dl, F=Im -m f($eihdq. It follows that = ! Fe-d(P~v)a+b?J+~aP: P where the radical denotes the branch of the square.root which has positive real part. Using the inversion formula and reversing the order of integration, after substituting for and F, we obtain The inner integral can be evaluated,by using the formulaz3 involving Macdonald's function Kl(x). Then the solution can be represented as the following double integral: TO deduce (62), substitute p = 1, Y = -) into formula (5.15.6) on p. 134 of L9, We now use Cauchy's theorem to evaluate the contour integral bypassing the branch pointsp = -c andp = 0 both on the upper and lower branches of the cut. The result is where in the course of the calculations, we use the formula /:2~o(zl sin x) cosh (z, cos x) sin x dx = sinh Jzi - z: J2 z2 - z1 z2> " [easily deduced from formula (4.455) of R2, p. 240 by setting p = 0, q = -4. It follows from (63) and (64) that The answer on p. 189 is easily obtained by evaluating the inner integral. 407. This problem can easily be solved by using the Mellin transform. Suppose the function T being sought is such that where s is some positive number.24 Multiplying Laplace's equation AT = 0 by rYfl, wherep is a complex number such that 0 < Rep < s, and integrating the result from 0 to a, we obtain 24 The existence of a solution with these properties can be anticipated from physical considerations. After the solution has been obtained, we can easily verify that it actually satisfies all the conditions of the problem. 352 SOLUTIONS PROB. 407 where is the Mellin transform of the function T. It follows from (65) that the term vanishes if 0 < Rep < s,z6 thereby reducing (66) to together with the boundary conditions a" T )$,=o = 0, T = TO - P (implied by those obeyed by the function T). Thus we see at once that a" sin p T=T,-- p sin pa ' which, in particular, implies that s = xlu. The temperature distribution T is now determined by using the inversion formula where 0 < o < xlu. The line integral can be evaluated by using residue theory, after completing the contour of integration by the arc of a circle of sufficiently large radius, lying to the left of the line Rep = o if r < a and to the right of this line if r > a. After some simple calculations, we obtain . nncp sln-, O<r<a, a Using the formula p sin x pn sin nx = arc tan , p2<L n=l 1 + p COS X to sum the series, we arrive at the single analytical expression for the function T(r, cp) given in the answer on p. 190. 25 Note that it also follows from (65) that the integral is analytic in the strip 0 < Rep < s, being uniformly convergent in every closed subset of the strip. PROB. 415 SOLUTIONS 353 415. If we replace the concentrated load by a uniformly distributed load with density 6 6. for r--rr0+- po-I< p<~~+i, q(r, 9) = ro 6~ [ otherwise, 2 2 where 6 and E are arbitrarily small positive numbers, then the problem reduces to solving the inhomogeneous biharmonic equation subject to the boundary conditions Multiplying (68) by rDf2 (wherep is a suitably chosen complex number), and integrating from 0 to co, we find that 2e m = (r, )r2dr, (69) D 0 where ii = jom~rv-z dr. (70) Suppose the function u is such that the quantities r-lu, aular, r Au and r2(aAu/ar) are all O(rQ) as r + 0 and all O(rS2) as r 4 oo, where s, > 0, sz > 0. Then the integrated term I{. . .)I," in (69) vanishes if -sl < Rep < s2, thereby reducing (69) to the ordinary differential equationz7 26 In problems of elasticity theory involving integration of the biharmonic equation, it is best to use a modification of the Mellin transform, in which the exponentp is replaced byp - 1. By the same token, the integral (70) is analytic in the strip -s, < Rep < s,, being uniformly convergent in every closed subset of the strip. 354 SOLUTIONS Using the method of variation of constants, we find that fi = A cos (p - 1)cp + B sin (p - 1)cp + C cos (p + l)cp 4- E sin (P + l)cp sin (P - l)(p - 1) sin (P + 1Xcp - 1) q(p, i)pp+~ dp, - - p-1 4Dp p+l I I" where the boundary conditions serve to determine the coefficients A, B, C and E. Passing to the limit 8, E + 0 and solving for these coefficients, we find that fi is a meromorphic function with poles at the points where the expression p2 sin2 u - sin2pu vanishes, and moreover that the number s, = s2 is the smallest root of the equation p2 sin2 a - sin2yu = 0. The bending moment M and the shear stress N along the edge cp = 0 can be determined from the relations Using the inversion formula for the Mellin transform, and choosing the imaginary axis as the path of integration, we find, after a certain amount of calculation, that M and N are the same as in the answer on p. 193. 418. Following the Fourier method, we look for particular solutions of Laplace's equation of the form nrrz T = R(r)cD(cp) sin - 1 Separating variables and integrating the resulting equations, we find that where Iv(x) and K,(x) are cylinder functions of imaginary argument. Because of the behavior of Iv(x) and K,(x) as r -, 0 and r -t co, the boundedness of the solutions T requires that A = 0 and A > 0. Thus the particular solutions needed to solve :he boundary value problem, which has a continuous spec- trum (0, co), are of the form . nnz T = T, = (M, cosh T(P + N, sinh rcp)KiT ("9 - sm T , PROB. 422 SOLUTIONS 355 The general so1ution.i~ constructed by.integrating with respect to the parameter T. Noting that TI,,, = 0, .we have nxz T= sin -S:N, sinh rrp Ki,(y) d~, 1 The coefficient NT is determined from .the boundary condition nxz TI,,, = f(r) sin - , I which gives For a certain class of functions f(r), we can invert (71), obtainingz8 2 Ki,(nxr/l) N, sinh ~a = - r sinh xrl j(r) dr. x2 r The conditions for using this formula are usually satisfied, except that f(r) may not go to zero sufficiently rapidly as r --+ 0. If f(0) # 0, then in most cases encountered in practice we can use the formula29 2 2 ww,~ KiT(nnr/l) N, sinhra = - f(0) + -7 sinh xr [f(r) - f(0)e ] dr (72) x xz r to determine NT (see L9, pp. 150-153). Assuming that the conditions imposed on f(r) are sufficient to guarantee the applicability of (72), we arrive at the result given in the answer on p. 196. 422. We subtract out the source potential, by writing where R is the distance from the charge q to an arbitrary point of space. Then the problem reduces to integration of Laplace's equation (0< r < co, O< cp <2x,-co <z < co), with boundary conditions This follows from formula (24), p. 195. aa Implied by formulas (24) and (26), p. 195. 3 56 SOLUTlONS PROB. 422 Expanding the function v in a Fourier cosine integral, i.e., setting we find that 0 satisfies the equation and the boundary conditions which, after evaluating the integrals take the form filrp=O = filw=2n = q~O(oJr2 + ri - 2rr0 cos (pO), in terms of Macdonald's function Ko(x). Using the Fourier method to integrate (73), we represent 6 as an integral " cosh (x - cp)r Ki,(or) dr cosh xr (see the solution to Prob. 418), where the coefficient Ma,, is determined by the condition K0(o Jr2 + r: - 2rr0 cos h) = ~om~a,,~iT(or) dr, 0 < r < m. (74) To avoid the difficulties associated with direct application of the inversion theorem, we write the left-hand side of (74) in the form ~,(oJr' + r: - 2rro cos yo) = [~,(a Jr2 + r: - 2rr0 cos yo) - Ko(or0)] + Ko(~ro)[l - e-"1 + Ko(ar0)eC"' and use the formula Then the inversion formula implies 2 2 1 - e-QT Ma,, = - Ko(oro) + Ko(oro)r sinh xr ---- S." r K,,(or) dr X X 2 m dr + ;;, r sinh ~rl [KJO Jr2 + r: - 2rr0 cos pa) - Ko(oro)] KJor) - . r PROB. 422 SOLUTIONS 357 The integrals appearing in (75) can be calculated by using the formulas cash (n - y)r 1 Ko(y) sinh xr dx x tanh 4x7 r(1- cX)KtT(x) - = - ---- x 27 cosh 4x7. ' which lead to the result 2 M,,, = - cosh (n - yO)~ KiT(orO). n Thus the solution of the problem takes the form v = - cosh (n - yo)r Ki,(crro)Ki,(or) d~. x2 cosh XT Substituting the known integral representation (76) into (76), reversing the order of integration and evaluating inner integrals,30 we find that 1 + 1 cosh ?js - cos $(cp + yo) cosh ?js + cos &(y - yo) 1 sinh 4s ds 9 (77) JCOS~ s - COS~ h where cosh A = z2 + r2 + r: 2rr0 To obtain the final form of the solution, given in the answer on p. 198, we evaluate the integral in (77) by making the substitution s A cosh - = cosh - cosh t. 2 2 30 The formulas m cos 7s ds - 1 -- sin TS ds Sa v'cosh h + cosh s SinhnT Jh icosh s - cosh h ' are used in the course of the calculation. PROB. 434 434. Introducing elliptic coordinates u and P, where x = c cosh u cos p, y = c sinh u sin p and c is the eccentricity of the given ellipse, we assume that the charge q is uniformly distributed over the curvilinear rectangle O<u<6, p*-!<[PI <P*f4. 2 2 The problem then reduces to integration of Poisson's equation where is the charge density inside the elliptic cylinder, and is the metric coefficient. Since u must be even in the variable P, we look for a solution of the form 1 m u = - $ + ? 2 C, cos np, n n n=l where ti, =/au cos np dp. Multiplying (78) by cos np and integrating from 0 to n,. we obtain the equation ii: - nzCn = -4njoxphz cos np dp, whose solution is easily found by variation of constants: 11, = A, cosh nu + B, sinh nu - ph2 sinh n(u - t) dt. The condition that the components of the electric field be bounded at the foci of the ellipse implies B, = 0. The value of the second constant A, is determined from the boundary condition finla=ao = 0, where uo is the vaiue of the coordinate u on the surface of the. cylinder. PROB. 437 SOLUTIONS 359 Taking the limit as 8, E -+ 0, we find that 2xq sinh n(a, -- a) 17, = - cos np* , n=0,1,2 ,..., n cosh na, which immediately implies the answer on p. 205. 437. Since the regular solutions of the two-dimensional Laplace equation are of the form u = u, = A, cosh na cos np + B, sinh na sin np, n = 0,1,2, . inside the ellipse a = and of the form 11 = u,, = e-'"(C, cos np -1 D, sin np), n = 0, 1, 2,. . outside the ellipse, we look for a magnetic potential of the form m dL) = H0(x cos y + y sin y) + ~e-""(~, cos np + D, sin np) n=l in the air, and m d2) = 2 (A, cosh na cos np + B,, sinh na sin np) in the magnetic medium (arbitrary additive constants are omitted). The values of the coefficients A,, . . . , D, are determined from the condition that the tangential component of the magnetic field and the normal com- ponent of the magnetic induction be continuous on the boundary surface, l.e., This gives euo cosh a, sinh a, cos y C, = (1 - p)H0c cosh a, + p sinh a, ' eao cosh a, sinh a, sin y Dl = (1 - p)H0c sinh a, + [L cash Hoceao cos y Hoceao sin y A, = , B1 = cosh a, + p sinh a, sinh a, + p cosh a, ' where all the other coefficients vanish. The final expressions for u(') and u(~) given on p. 206 are obtained by using the relations a b cosh a, = - , sinh a, = - . C C The other combinations of products of hyperbolic and trigonometric functions lead to infinite values of grad u at the foci of the ellipse. 360 SOLUTIONS PROB. 443 443. To make the problem homogeneous, we write the torsion function as a sum u = -y2 $0. Then v is a solution of Laplace's equation regular inside the cut ellipse (i.e., in the region la1 < a,, 0 < P < x) and satisfying the boundary con- ditions vIP=, = vplVn = 0, ~l~=+~~ = (C sinh a, sin p)'. Applying the Fourier method and using the evenness of v in the variable a, we construct a solution of the form The constants A, are determined from the boundary condition for or = a,, which gives Thus the torsion function is given by the series 8b2 " cosh (2n + 1)a sin (2n + 1)P u=-y2+-C x ,,, cosh (2n + l)ao (2n + 3)(1 - 4n2) ' (79) while the torsional rigidity can be calculated from the formula where C = uh~a ddp. is the metric coefficient. Substituting (79) into (80) and evaluating the double integral, we arrive at the expression given in the answer on p. 209. 449. Choosing a system of parabolic coordinates a, P such that the surface of the cylinder has equation dp = Po, and regarding the charge q as uniformly distributed over the small area bounded by the curves la1 = 8, p = E, we reduce the problem to integration of Poisson's equation (0 otherwise, PROB. 457 SOLUTIONS 36 1 is the metric coefficient. The problem is solved by taking the Fourier cosine transform. Writing ii =JOmu cos Aa da, we multiply Poisson's equation by cos Aa and integrate from 0 to c~. This gives the equation m P" - h2ii = -4nj0 ph"os ha da, whose solution is easily found by variation of constants: ii = A cosh AP + B sinh AP - ph2 sinh A(P - q) dq. A The requirement that grad u be bounded at the focus of the parabola implies B = 0. The value of the constant A is determined from the boundary condition t'lp=p0 = 0, which, in the limit 6, E + 0, gives 2x9 A = - tanh ?,Po. A The corresponding value of ii is - 2xqsinhA(P,-(3) u=- A cosh AP, ' and the final answer (see p. 21 1) is obtained by using the inversion formula m u = I cos ha dl. X 0 457. If we introduce bipolar coordinates a, P as shown in Figure 122, p. 215, and represent the torsion function u as a sum u= - 2 y +v, the problem reduces to determining the function v which is harmonic in the domain a, < a < a, 0 < p < 2x and satisfies the conditions c2 sin2 p ulp=o = ulp=zX = 0, uIa=a, = y2(a=a0 = (cosh uo + cos P)' ' vl,,, -+ 0. The solution is constructed as a series whose coefficients, according to the theory of Fourier series, are given by '" sin2 p sin 4np (cosh uo + cos p)' dP. To calculate the torsional rigidity, we use the relation implied by the formula given in the hint to the problem (see p. 215) after setting II, = -y2. 471. Setting 4 u = - + u,, R where R is the distance from the source to an arbitrary point of space, we reduce the problem to integration of Laplace's equation AM, = 0, with the boundary condition Ulla=O = - - 4 d , sinh uo = - cJsinh2 cro + sin2 P a and the condition at infinity ~lla-trn +O. In keeping with the discussion on p. 222, we look for a secondary potential in the form of a series m u, = 2 A,Q,(i sinh u)P,(cos P), 7,=0 where the coefficients A, are determined from the boundary condition. Using the theorem on expansion of an arbitrary function in a series of Legendre polynomials, we find that for even n, while A, = 0 for odd n. To evaluate (81) for even n, we use the integral PROB. 47 1 SOLUTIONS 363 which can be evaluated by expanding (b2 - x~)-~/~ in a power series and then integrating term by term. Using well-known formulas, we find that where The result can be expressed in terms of the hypergeometric function Using the familiar formula we find that J =- 1 r"n+') F n+b,n+1;2n+g;-~), ,/x I'(2n + +) sinh ""a, smh a, or J, = 2iPzn(0)Qzn(i sinh a,), because of the definition of the Legendre function of the second kind. Thus the required values of the coefficients A, are A,, = 2 (4n + 1)Q2,(i sinh a,), XC which leads to the potential distribution 4 29" u = - + - - 2 (4n + 1)QZn(i sinh a,)Q,,(i sinh a)Pzn(cos p), R x cn=0 if we note that 364 SOLUTIONS PROB. 481 The distribution of charge on the surface of the disk is now found by differentiation, according to the formula 481. The problem reduces to solving the system of equations AM(') = -4np (0 < a < a,), = 0 (a, < a < a) for the gravitational potentials dl) and u(,), with boundary conditions Setting = u, $ u,, u',' = u2, where u, = -npr2 = -npc2 sinh2 a sin2 p, and noting that u, is harmonic inside the spheroid (a < a,), while u2 is ,harmonic outside the spheroid (a > a,), we have m m u1 = 2 A~P~(COS~ ~)P,,(cos P), u2 = 2 B,Q,,(COS~ a)P,(cos P). n=O n=O Using the boundary conditions, we obtain the formulas m -&pc2 COS~ aO[l - P~(COS P)] + C A,P~(COS~ ao)Pn(cos p) n=O m = 1 ~~~~(cosha,)~,(cos P) n=O determining the coefficient^,^^ which imply that Thus A,, B,, A,, B2 satisfy the system of equations A,P,(cosh a,) - B,Q,(cosh a,) = $npc2 sinh2 a,, A,P;(cosh a,) - B,Q;(cosh a,) = 4rcpc2 cosh a,, A2P,(cosh a,) - B2Q,(cosh a,) = -gnpc2 sinhZ a,, A2P;(cosh a,) - B2Q;(cosh a,) = --&pc2 cosh a,, 32 Note that sina P = $[1 - P2(cos P)]. SOLUTIONS 365 whose solution is33 A, = #xpc2 sinh2 a, 1 + 2 cosh a, In coth A, = +xpc2 sinh2 a, 1 - cosh a, In coth B, = +xpc2 cosh a,, sinh2 a,, B2 = -+xpc2 cosh a, sinh2 a,. Substituting for B, and B2 in the formula for u,, we find that the gravitational potential outside the spheroid is [2(sin2 p - sinh2 a) + 3 sinh2 a sin2 P] In coth C 2 + cosh a(3 cos2 p - 1) . I To obtain an asymptotic representation of the gravitational potential for small eccentricity c, we introduce spherical coordinates R and 8, and use the formulas C C z = - (e" + e-") cos p = R cos 0, r = - (ea - e-") sin p = R sin 8. 2 2 Solving for a and P, we find that It follows that as c + 0, c2 cosh a = - 1 -1 - sin2 C "[ 2R2 cos p = cos 0 + 0 - . (3 33 Here we use the expression for the Wronskian of the Legendre functions, as well as the formulas 366 SOLUTIONS PROB. 483 Using these formulas and the exact solution found previously, we obtain where M is the mass of the spheroid (cf. the answer on p. 228). 483. If we write U = U, + u1, where u, = q/R is the source potential (R is the distance from the charge to an arbitrary point of space), and introduce spheroidal coordinates a, P, such that the hyperbola has the equation P = Po, then the problem reduces to finding the function u, which is harmonic in the region 0 < P < Po and satisfies the boundary condition ~1[~=@, = - - 4 c(cosh a - cos Po) ' In prolate spheroidal coordinates, Laplace's equation takes the form J- "iSinh a 2) + sinh a aa if we assume that u, is independent of 9. Setting Ul = A(coB(P), we obtain the equations 1 (sinh a . A')' + hA = 0, - (sin p . B1)' - hB = 0 sinh a sin p for the separate factors. Therefore where P,(z) and Q,(z) are Legendre functions of the first and second kind, and v is an auxiliary parameter related to A by the formula Taking account of the behavior of the Legendre functions near the points z = 1 and z = co, we see that in order for the solutions (82) to represent bounded real functions in the region 0 < a < a, 0 < P < Po, the parameter v must be chosen equal to -3 + i~ (T > O), and the constants N and D must be set equal to zero.34 Thus we arrive at the particular solutions To construct the general solution, we integrate over the parameter T, obtaining where the coefficients C, must satisfy the boundary condition =j;~,~-~+~,(cos ~,)~-~+~,(cosh o) dr, a > 0. Using the inversion formula implied by the Mehler-Fock theorem, we find that CT = - 4T tanh rr S sinh a P-x+i,(cosh a) da cP-~+~~(cos PO) o cash a - cos Po - q T tanhm --- dF. Evaluating the integral, we arrive at the formula for the electrostatic potential given in the answer on p. 229.35 34 In particular, we use the formula which shows that a bounded solution in the interval (0, co) exists only if -1 < Re v < 0. The extra requirement that 11, be real compels us to set v = -4 + i~. 35 TO prove the formula (see L7), use the integral representation cos TS ds P-!4 +iT(cosh a) = - x Sa o 62 COS~ a - 2 cosh s ' and then reverse the order of integration with respect to a and s. After evaluating the inner integral, this gives where we have used another integral representation of P-x+,,(x). 368 SOLUTIONS PROB, 494 494. Bearing in mind that the potential u can be represented in the forIll u = Eoz + u,, where u, is harmonic outside the torus and goes to zero at infinity, we look for a solution of the form m Pn-~(cosh a) u = E,z + 42 cosh a - 2 cos (3 2 An sin np. n=l P,-x(cos~ a,) The coefficients An are found from the boundary condition and coincide with the coefficients of the function -2E,c sin P(2 cosh a, - 2 cos p)Pf2 when expanded in asFourier sine series in the interval (0, x). Thus we find that A = rr sin p sin np dp n 4E0cS x o (2 cash a, - 2 cos p)3/2 ' Integrating by parts and using the formula given in the hint to Prob. 493, we arrive at the answer given on p. 237. 498. Setting we reduce the problem to finding the function u. This function is harmonic outside the torus (0 < a < a,) and goes to zero at infinity (i.e., as a -+ 0, p -t 0). We look for a solution of the form where the coefficients An are determined from the boundary condition It follows from the theory of Fourier series that A = dP Qc2 F2 '0/:(2 COS~ a, - 2 cos fi)512 ' An = 2Qc2 sinh2 aoJ" cos np dp n=l,2, ... kx o (2 cosh a, - 2 cos p)6f' ' Evaluating these integrals, we eventually arrive at the answer on p. 238. 502. If we subtract out the singularity at the point r = z = 0 by setting the potential u, of the secondary field is harmonic in the region 0 < a < a, p, < P < 2x + p, outside the conductor, and vanishes as a + 0, p -t 2x. The function u, can be represented in the form of an integral u1 = - 4_ J2 cosh a - 2 co OM, COSh (x + PO - p) P-%+~,(cos~ a) d~. c cosh x.r It follows from the boundary condition 43=po = ulp=2,+po = 0 that M, coincides with the coefficients of the expansion of the function (2 cosh a + 2 cos PO)-ll2 in a Mehler-Fock integral with respect to the functions P-x+ir(~~~h a), i.e., (2 cosh a + 2 cos Po)-112 =IoW~,~-x +,,(cash a) dr, a > 0. In the present case, we cannot determine M, directly by using the inversion formula implied by the Mehler-Fock theorem, since the function being expanded does not belong to the class for which the theorem holds (see L9, p. 228). However, it can be shown without recourse to the Mehler-Fock theorem (ibid., p. 229) that (2 cosh a + 2 cos Po)-'I2 - P-W+i,(cosh a) d~, and hence cosh p0r M, = ----- . cosh n-r Therefore the solution of the problem is u, = -9 J2cosha - 2 cos p C cosh p,r cosh (x + p, - p)-r P-x+i,(cosha) dr, cosh2 xr The charge density on the inner and outer surfaces of the spherical bowl are given by COS~ a - cos p, au COS~ a - cos p, au 5. = - , 0, = ~XC G L=p0 ~XC -1 ap p=2x+po Performing the differentiation with respect to P and evaluating the resulting integrals by replacing the Legendre function by its integral representation cos .r+ dJi P-!4+iT(cosh a) = - cash xr J2 cosh Ji + 2 cosh a ' x we eventually arrive at the closed-form expressions for 5, and oi given in the answer on p. 240. 508. To calculate the capacitances, we must first solve the electrostatic problem, assuming that the spheres have arbitrary given potentials Vl and V2. Introducing a system of bipolar coordinates a, P, cp in which the spheres under consideration have equations p = -p, and p = p,, we reduce the problem to determining a function u which is harmonic in the region -p1 < P < Pz and goes to zero as a -+ 0, p -t 0. The desired solution can be constructed in the form of a series m u = J2 cosh (3 - 2 cos a 2 [A, cosh (n + &)p + B, sinh (n + $)P]P,(cosor), n=O where the coefficients A, and B, are found from the boundary conditions which immediately lead to a system of linear equations for A, and B, if we use the familiar expansion 1 00 - 2 &n+%)P - P,(cos a), p > 0. J2 cash p - 2 cos a ,,=o After determining u, the charges on each conductor can be calculated from the formulas To find the capacitances C,,, C,, and C,,, we use the relations - Q2(V,=-1,v2=o, C22 = Qzlv,=v,=~- C11 = QllVI=Vz=lr CIZ = Q~~v,=o,v,=-1 - 512. In the new coordinate system, the problem reduces to solving the equation with the boundary conditions ulp=*Po = v. Variables can be separated by setting Integrating the resulting equations for A(a) and B(P), and noting that u must be even in p and bounded at a = 0, we arrive at the following particular solutions : SOLUTIONS 37 1 The solution is then constructed in the form w cosh pP u = V~U" pz/ N, -- o cosh pPO JO(l-4. Using the well-known formula and taking account of the boundary condition ulp=po = V, we find that Nu = e-l*Po, and hence m cosh pP U = v Ja2 + p2 e-r*Po - cosh pPo Jo(p4 dp- To calculate the total charge Q on the conductor, we start from the relation After substituting (83) into (84) and reversing the order of integration, we obtain the expressi~n~~ m e-r*Po e = .V/ -- dp = 2Va In 2. o cosh pPo The capacitance C is now determined from the formula Q = VC. 522-523. If we set the problem reduces to finding a solution of the wave equation satisfying zero initial conditions, the boundary condition and the condition that u vanish at infinity. Introducing new variables 36 Note that the parameter c is related to the radius a by the formula c = 2ap0. 372 SOLUTIONS PROB. 523 and looking for a solution which is a function only oft and 3, we obtain the equation aZu - i au $ ------ - = 0, at a, 2(5 - "I) a, whose solution is where cp and $ are arbitrary functions. Moreover +(t) = 0, since On the screen 3 = 5, u = f (51, and hence y(s) must satisfy Abel's integral equation with solution37 If s < 0, then the integrand vanishes identically by hypothesis, which implies y(s) = 0. It follows from (85) that uIV<0 = 0, i.e., the excited zone is bounded by the circle -q = 0 (see Figure 149, p. 254). Outside the excited zone, and in particular, E = 0 for x > vt. For q > 0, i.e., in the excited zone, where p(s) = .! d Js& dE. nds o Js- 5 37 See e.g., S6, Vol. 11, p. 220. PROB. 532 SOLUTIONS 373 531-532. In the present problem, considerations like those given in the solution of Probs. 522-523 show that the reflected wave can be represented in the form '(') ds, 5 - s + (2a/u) where q(s) satisfies the Volterra integral equation and the variables 5 and q are defined by r-a z+a, t=t-- (r = Jx2 + y2 + 2). U U we find that y(s) = 0 if s < 0, and hence u = 0 if q < 0. Thus the boundary of the excited zone is determined by the equation q = 0. The value of u inside the excited zone (q > 0) is given by (86) and (87) with f(t) replaced by g(E) and the intervals of integration (-a, E), (- co, q) replaced by (0, E,), (0, q). The integral equation belongs to the class which can be solved readily by the use of the Laplace transf~rm.~~ Writing multiplying (88) by ecpC and integrating with respect to E from 0 to a, we find that u' where 38 In the applications, it is sometimes more convenient to construct the resolvent of the given equation, without recourse to the Laplace transform (see e.g., F8). 374 SOLUTIONS PROB. 542 and Ei (2) is the exponential integral. The final answer can be obtained by using the inversion formula where the line I? lies to the right of the singular points of the function T. 542. To find integral equations for the charge densities, first let M = x be a fixed point in the plane y = 0. Then cos (r,,,, n) = 0 if the variable point N = 5 also belongs to the plane y = 0, while if N = 5 belongs to the plane y = h. Therefore the integral equation (2) on p. 260 takes the form In just the same way, choosing M = x in the plane y = h, we obtain the integral equation This system of integral equations can be solved by using Fourier transforms. Multiplying each equation by edx and integrating with respect to x from - co to w, we obtain where (for brevity). Reversing the order of integration and using the well-known formula COS AT X So ;"Ti;. dq = - e-'*Ih, 2h PROB. 549 SOLUTIONS 375 we obtain the following system of linear algebraic equations for 5, and 5,: It follows that The answer on p. 262 is now an immediate consequence of the inversion formula 549. The integral equation can be solved as follows: Writing (89) in the form we make a Landen transformation obtaining Because of the formulas 1 - x K(f) =SU ds 0 J(x" s2)(y" sZ) ' (90) becomes 376 SOLUTIONS PROB. 561 The expression on the left can be represented as a double integral over the trapezoid bounded by the lines s = 0, s = x, s = y and x = a. Changing the order of integration in this integral, we obtain If we write (91) goes into Schlomilch's integral equation . . with solution (see W8, p. 229). To deduce f(x) from a knowledge of Y(s), we use the formula (see Bl). Substituting for Y(s), we arrive at the answer on p. 264. 561. To find integral equations for the virtual charge densities on the planes cp = 0 and cp = u, we note that if M = r is a fixed point in the first of these planes and if N = p is an arbitrary point of the interface between the two dielectrics, then p sin u if N belongs to the plane cp = u, + p2 - 2rp cos u if N belongs to the plane cp = 0. Applying formula (7), p. 267, we find that P 0 P m oo(r) = - - E&,, + - sin u .U(P)P 2x x So r2 + p2 - 2rp coso dp, where El - E2 p=-. El + E2 By a similar argument, if we choose the fixed point M in the plane cp = a, then P P m oa(r) = - ~il~=~ + - sin u GO( P) P 2x x So r2 + p2-- 2rp cosu dp. PROB. 565 SOLUTIONS 377 This system of integral equations can be solved by using Mellin transforms. Multiplying each equation by rp-I and integrating with respect to r from 0 to co, we eventually obtain the system of linear algebraic equations - sin (x - a)(p - 1) - 00 - P P - 0a = - -fO, sin x(p - 1) 2x \, sin (x - a)(p - 1) - -P P - 00 + a = -fa sin x(p - 1) ~TC for 5, and iJa, where (for brevity). To guarantee the convergence of the integrals appearing in (92), we choose p to be a complex number of the form p = 1 + i~ (- CCI < T < a).39 Solving the system (92) for Eo and Z,, we find the values of the charge densities by using the inversion formulas oo = - I"'" OOr-Bdp, 0, = - 1"'" Gr-" dp. 2xi I-im 2xi I-im 565. The requirement that the tangential component of the electric field be zero on the surface of the conductor leads to the integral equation of the first kind where j(c) is the total density of current flowing on both sides of the half- plane, and E(x) is the tangential component of the external field at the point x.~O This integral equation can be solved by using the integral transform (27), p. 196. To reduce (93) to a form suitable for application of this method, we multiply the equation c r~b~)ikF)j(~) 4 = XU - ~(0) 39 Each of the densities a, and 0, is 0(rSi) as r + 0 and O(rS2) as r + w, where s, < 1 and s, > 1. The functions f, and f, are assumed to be O(1) as r -0 and O(rS) as r -. a, where s > 1. 40 Here we have . . ,j(Q = j.~Y=+o + jllY=-O = jl + j2, E(x) = E.OlyEO. The difference between the current densities is given by formula (9), p. 270, implied by the conservation law for the circulation of the magnetic field around a closed contour. by e-ikx and subtract it from (93). The result is Multiplying both sides of (94) by and integrating with respect to x from 0 to co, we obtain c2ex~jB l" E(x) - CikXE(O) - Hj:'(ltx) dx, rrW X assuming that it is legitimate to reverse the order of integration. The inner integral in the left-hand side can be evaluated after making the preliminary transformation It then follows from the addition theorem for Hankel functions, [see L9, formula (5.12.1 I), p. 1261 that PROB. 566 SOLUTIONS 379 The integrals on the right can be evaluated in closed form, eventually leading to the expressions 1 CO 11 = , 2 ,,I2 (2) IT sinh &KT HSY~F) - xir2 - e H,, (lc5)[l + 2 m=l 2 -1 m - - 1 2 HP'(k[) - - enTl2 coth xr Hj:'(kE), ir sinh 4x7 ir 1, = Hp(k~[ 2 - ir sinh i~ sinh &xr Thus we finally have I - - [Hf)(k[) - PI2 coth rrr H!B'(ke)]. i~ sinh xr If we now introduce the integral transform (95) takes the form which implies The final form of the solution given in the answer on p. 270 is obtained using the formula4' $(~)e~"~r sinh nr Hj:)(kx) dr, 566. The problem reduces to solving the integral We assume that k is of the form k = Jkl e-'y(O < y o<x<m. equation O<r<co. < x/2) as in Prob. 565, " The applicability of this formula is guaranteed by the requirement that k be a complex number of the form k = lkl e-iu (0 < y < 7c/2), and that the external field be due to line sources located in the finite part of the xy-plane. 380 SOLUTIONS PROB. 566 and also that the angle between the half-plane and the direction of prop- agation of the incident wave is less than y.42 Then the problem can be solved by using the general formulas obtained in Prob. 565, with x = r, = p, E(x) = EOe-"" cos a. Bearing in mind that - 2iE0 -- (cosh ra - I), ~sinh KT we have c2E0 rn P j(~) = - - 1 r tanh xr en"%osh ra H::'(~P) dr. 2x0, 0 The last integral can be written in the form where 1 " cash Pr (2, $(PI = -- - --- 2i o cosh xr Hi, (kp) d7, IPI < x + Y. In the paper K3, it is shown that Using this formula and performing the differentiation with respect to a, we obtain the expression for j = j, + j, given in the answer on p.27 1. " These restrictions are needed to guarantee the convergence of the integrals and to justify using the inversion formula, but can be dropped in the final results. In particular, the expressions for the current densities,/, and j, found here are also valid for real k, in which case they coincide with the corresponding fornlulas for the Sommerfeld problem (see Prob. 427). MATHEMATICAL APPENDIX I. Special Functions Appearing in the Text Certain basic functions The gamma function The probability integral 2 Q(Z) = - J;~-L' dt. ,lX The Fresnel integrals The exponential integral dt, 0 < arg z < 2x. The sine integral The cosine integral COS t ~i (2) =J, dt, larg z/ < n. Orthogonal polynomials The Legendre polynomials The Hermite polynomials H.(x) = (- 1 )"ex' epx2 , n=0,1,2 ,... dxn The Laguerre polynomials L,"(x) = Ln(x). Cylinder functions The Bessel function of the first kind (- l)k(z/2)v+2k JV(z) = 2 larg zl < K. k-o r(k + l)r(k + v + 1) ' The Bessel function of the second kind Jv(z) cos vx - J-,(z) Yv(z) = , larg zl < n. sin VK The Bessel function of the second kind of integral order (n = 0, 1,2, . . .) 2 z 1 n-l (n - k - 1) ! (ir-n Yn(z) = lim Yv(z) = - Jn(z) In - - - 2 v+n x 2 n k-0 k! - 2 (- 1)k(z/2)n+2k r,=, k! (n + k)! [Nk + 1) + Nk + n + I)], larg zl < n, where #(z) is the logarithmic derivative of the gamma function (the first sum is omitted if n = 0). The first and second Hankel functions H!"(z) = Jv(z) + i Yv(z), H?'(z) = Jv(z) - i YV(z), larg zl < n. The Bessel function of imaginary argument m (~/2)~+~~ Iv(z) = 2 larg zl < n. ,=o qk + l)F(k + v + 1) ' The Macdonald function n v - v , larg zl < n. Kv(z) = - 2 sin vx MATHEMATiCAL APPENDIX 383 The Macdonald function of integral order (n = 0, 1, 2, . . .) k=O k! (k + n)! [ICl(k + 1) + $(k + n 4- I)], where +(z) is the logarithmic derivative of the gamma function (the first sum is omitted if n = 0). Spherical harmonics The Legendre functions of the first and second kinds where is the hypergeometric series.' For real x in the interval (-1, l), the Legendre function of the second kind is defined by the formula QV@) = &[Qv<x + io, + Qv<x - i0)l. Analytic expressions for the spherical harmonics appearing in this book can be found in H4, L9 and M2. The associated Legendre functions The functions Pv(z) and Qv(z) are defined outside the indicated regions by using analytic continuation (see e.g., L9, Sec. 7.3). The associated Legendre functions for the interval (- 1, 1) drnP,(x) P, (x) = (- l)m(l - x~)~"'~ - , dxm Elliytic integrals and functions The elliptic integrals of the first and second kinds The complete elliptic integrals of the first and second kinds The Jacobian elliptic functions sn z = sin cp, cn z = cos cp, dn z = 41 - k2 sin2 cp, where cp is the inversion of the elliptic integral of the first kind, i.e., Further information on special functions can be found in such books as ErdClyi et al. (E2), Gray and Mathews (G2), Hobson (H4), Jackson (Jl), Lebedev (L9), Lense (L11, L12), MacRobert (M2), Magnusand Oberhettinger (M3), McLachlan (M5), Ryshik and Gradstein (R2), Smirnov (S6, Vol. 111, Pt. 2), Snow (S12), Watson (W4), and Whittaker and Watson (W8). 2. Some Expansions in Series of Orthogonal Functions 2. 2 cos (nxx/a) n=l n m 3. C(-l)"-' sin (nnxla) xx - - - , O<x<a. n=l n 2a m 4. C(-l)"-' cos (nnx/a) n=l n MATHEMATICAL APPENDIX 385 5. 2 sin [(2n + l)nx/2a] n - - - 4 ' O<x<a. n=O 2n + 1 2 cos [(2n + l)nx/2a] 1 nx =-Incot-, O<x<a. n=o 2n f 1 2 2a m 7. C(-l)" sin [(2n + l)nx/2a] 1 nx =-Intan--, O<x<a. n=O 2n + 1 2 2a m 8. 2 (-1)" cos [(2n + l)nx/2a] - 2 - 4 ' O~x<a. n=O 2n f 1 g, 2 cos (n~xia) - -2- 1 -- x A), 0 G x c 0. n=l n2 6 2a 4a2 m 10, 2 (I)" COS XI^) n=l tl " cos [(2n $ 1)nx/2a] 11. 2 1) 0.x.. . n=o (2n + 8 m 12. C(-ly sin [(2n + l)nx/2a] - x> - , Ogx~a. n=O (2n + 1)' 8a 2 sin (nnx/a) x x2 0.x. a. n=l n3 14. Z(-l)%-l sin (nxxla) (1) O<x<a. 71=1 n 12a " sin [(2n + l)nx/2a] 15. 2 =&(I -%), O< x< a. n=o (2n f 16a 16. Z(-l)" cos [(2n + l)nx/2al ?( -xx I---, Osxga. n=O (2n + I)3 32 " Jdynrla) - 1 17. 2- - - O<r<a, n=1 ynJl(~n) 2 ' where the yn are the positive roots of the equation Jo(y) = 0. where the y, are the positive roots of the equation Jo(y) = 0. where the y, are the positive roots of the equation Jl(y) = 0. For various other expansions in orthogonal functions (and series of a different kind), we refer to the handbooks by Jolley (J5) and Ryshik and Gradstein (R2). 3. Some Definite Integrals Frequently Encountered in the Applications n dx = - , O<Rev<I. sin TCV 2. Sm xY dx n sin (TC - cp)v -- - -1 < Rev < 1, o 1 - 2x cos cp + xbin TCV sin cp 4. lm cos ax - cos bx b dx=ln-, a>O, b>0. x a b 8 re-" sin bx dx = - a > 0. a2 + b2 ' 4; e-b2,4": a > 0, eCa ' cos bx dx = - 2a Jam Ji 12. rsin x2 dx = cos x2 dx = - . 2 2 sinh -sin ' " sinh px x 2a 2a sin rx dx = - , o<p<q. xr P cosh - + cos - nr sinh - " cosh px n sin rx dx = - 4 , O<p<q. XP 2q cosh + cos - 4 4 sin " sinh px n cos rx dx = - 4 , o<p<q. XP 2q cosh + cos - cosh - cos " cosh px x cos rx dx = - 2q 2q , O<p<q. x t' " P cosh - + cos - 4 4 22 bV e-b2/4a2 e-a " JV(bx)xv+' dx = - , b > 0, Rev > -1. (2~7~)~~' Among the handbooks on definite integrals, we cite those by Dwight (D2) and Ryshik and Gradstein (R2), as well as the celebrated compendium of Bierens de Haan (B4). 4. Expansion of Some Differential Operators in Orthogonal Curvilinear Coordinates General formulas Let (ql, q,, q,) be a system of orthogonal curvilinear coordinates related to Cartesian coordinates (x, y, z) by the formula Suppose the square of the element of arc length in the given system is where the hi are the metric coefficients Then the differential operators (grad u),~, div A&, (curl A),i [where u and A are given functions of the coordinates, and the index qi denotes the corresponding vector component] take the form 1 au (grad u),, = - - , hi a,, The coordinates of the vector AA = grad div A - curl curl A are: 1 2 aA (AA), = AA, - - A, - - , r2 r2 acp 1 2 aA (AA), = AA, - - A, + -', r2 r 89 (AA), = AA,. Spherical coordinates x = r sin 0 cos cp, y = r sin 0 sin cp, z = r cos 0, (O<r<co, O<0<n, -n<cp<x), ds2 = dr2 + r2 do2 + r2 sin2 0 dcp2, h, = 1, ho = r, h, = r sin 0, au 1 au 1 au (grad u), = - , (grad u)e = - - , (grad u), = - - ar r 80 r sin 0 acp ' I a 1 a div A = - - (r2A,) + - - 1 aA, (Ao sin 0) + - - , r2 ar r sin 0 80 r sin 0 acp I a 1 a. a 1 aZu ~u=--(r'$) +-- s1n0- +-- r2 ar r2 sin 8 ae ( a:) r2sin20 ay2 ' 1 a 1 aAo (curl A), = - - (A, sin 0) - - - , r sln 0 80 r sin 0 acp 1 aA, i a (curl A)e = - - - - - r sln 0 acp r ar (rA,L i a (curl A), = - - (rA 1 aA, 0)---, r ar r a0 2 2 a 2 aA, (AA), = AA, - - A, - - - (Ao sin 0) - - - , r r2 sln 0 30 r2 sln 0 acp 1 2 aA, 2 cos 0 aA, (AA), = AAo - - A~+----- r2 sin2 0 r2 a0 r2 sin2 0 acp ' 1 2 a~, 2~0s 0 a~~ (AA), = AA, - A, + - - + ------ - . r sin 0 r2 sln 0 ay, r2 sin2 0 acp Expressions for the above differential operators in other special orthog- onal curvilinear coordinate systems can be found in Chapter 7 of this book, and in the handbook by Magnus and Oberhettinger (M3). Supplement VARIATIONAL AND RELATED METHODS' Many, and perhaps most, mathematical problems encountered in science and engineering are difficult or impossible to solve by analytical methods. It is also found that explicitly obtained exact solutions are often too cumber- some for interpretation and numerical evaluation. Therefore, in these in- stances, it is either necessary or convenient to employ approximate methods which yield accurate numerical estimates of the solution. The recent develop- ment of high speed electronic digital computers has made practical the success- ful application of many of these methods to complex problems. This supplement contains a collection of typical problems that illustrate a special class of approximate methods. They are related either directly or indirectly to the variational formulation of physical problems. Almost all of the examples are concerned with boundary value problems for ordinary or partial differential equations. However, with suitable and sometimes trivial modifications, the methods presented can often be applied to other situations, e.g., eigenvalue problems or problems involving integral equations or integro- differential equations. The selection of problems was, in large measure, influenced by the amount of computational work necessary to obtain a solution. Hence, by necessity, they are essentially "simple." However, the methods employed can usually be applied directly to more complicated problems, the only additional difficulty being that the calculations are more involved. The supplement is independent of the main body of the book in the sense This supplement was written by Edward L. Reiss, Courant Institute of Mathematical Sciences, New York University. 39 1 that all equation and problem numbers refer only to those in the supplement. Literature references, indicated in brackets, are to items in the references section on p. 412. I. Variational Methods I. I. FORMULATION OF VARIATIONAL PROBLEMS2 Physical problerhs can frequently be formulated mathematically as minimum problems, as well as in terms of differential or integral equations. The solution is then a function, selected from a certain class called admissible functions, which minimizes a specified functional3 with respect to all admis- sible functions. For example, Hamilton's principle is an alternative to New- ton's equations of motion as a formulation of the laws of mechanics. It states that if u(x, t) is a vector describing the motion of a mechanical system: then between any two times to and t,, the actual (stable) motion is an admis- sible vector which coincides with the actual motion at t = to and t = t, and makes the functional a minimum. Here T and U are the kinetic and potential energy functionals of the system. The admissibility conditions usually take the form of boundary conditions and of continuity requirements on u and its derivatives. If the mechanical system is in equilibrium, so that T= 0 and u is independent oft, then Hamilton's principle becomes the principle of minimumpotentia/ energy: the actual (stable) displacement of the system is an admissible vector that minimizes the potential energy functional. It is usually not difficult to show that the admissible function (or vector) that minimizes the functional is the solution of a system of differential equations (or sometimes integro-differential equations, or integral equations) called the Euler equations for the functional. Thus in mechanics we obtain For a fuller discussion of the calculus of variations and its applications, see [3, 4, 81. Here we use the general term /Itnctiotral to denote any mapping of a set of functions (e.g., admissible functions) into real numbers. Thus, for example, is a functional, where f(x) is any piecewise continuous function on the unit interval. In our applications, the domain of the functional is the set of admissible functions. ' We use bold face to indicate a vector, and x = (x,, x,, . . . , x,) is the vector of p independent variables. The function u is a vector-valued function of p + 1 variables. SUPPLEMENT 393 the equations of motion as the Euler equations for Hamilton's principle and the equilibrium equations as the Euler equations for the principle of minimum potential energy. To illustrate these remarks, consider the functional for the scalar function u(x) of the single variable x. Here a prime is used to denote differentiation. The admissibility conditions are the following: u(x), ul(x) and u"(x) are continuous functions in the closed interval [a, b] which satisfy the boundary conditions u(a) = u,, u(b) = u,, (2) where u, and u, are prescribed numbers. The prescribed functionsp(x),pf(x), q(x) and f(x) are continuous in [a, b]. We shall now show that if the admis- sible function u(x) minimizes I, i.e., I[u] < I[v] for all admissible functions v, then u is a solution of the Euler equation To see this, we introduce ii(x), the variation of u, namely a function defined in the interval [a, b], which has a continuous second derivative and satisfies the homogeneous boundary conditions (2) [i.e., vanishes at the end points a and b], but is otherwise arbitrary. Consider the admissible function v = u + ~ii, where the real number E is a parameter. Then is a quadratic function of E, given by where It is easy to show, by using integration by parts and the conditions Q(a) = ii(b) = 0, that Since I[v] is minimized when E = 0 and hence J(E) is a minimum at E = 0, it follows from (4) that I,[u, ii] = 0 for all variations ii(x). Thus we conclude from (5) that u satisfies the Euler equation (3). 394 SUPPLEMENT PROB. 1.1.1 1.1.1. Determine the Euler equation corresponding to the functional (1) if u(a) = u, and no conditions are specified at x = b. Determine the bound- ary conditions that u must satisfy at x = b in order to minimize I (they are called natural boundary conditions.) Ans. Lu =f, ul(b) = 0. Hint. Let i(a) = 0 and u(b) be arbitrar~.~ 1.1.2. Let F(x, u, u') be a specified twice continuously differentiable function of its arguments x, u and u'. Determine the Euler equation of the functional I[u] =C(x, u, u') dx, assuming as admissibility conditions that u, u' and u" are continuous in [a, b] and satisfy (2). Ans. 1.1.3. Let F(x, u, u', . . . , dn)) be a specified twice continuously differen- tiable function of its arguments, where u(") -- dnu/dxn . Determine the Euler equation of the functional I[.] =jabp(x, u, ul, . . . , dn)) dx, assuming as admissibility conditions that u, u', . . . , u'") are continuous in [a, b] and have prescribed values at x = a, b. Ans. 1.1.4. Determine the Euler equation of the functional I[ul =//[ut + ui +.2f(x, y)ul dx dy D for the functions u(x, y) defined on the domain D in the xy-plane bounded by the contour C (the subscripts denote the corresponding partial derivatives, e.g., u, = &/ax). The admissibility conditions are that u and its first and second partial derivatives be continuous and that u satisfy the boundary condition u = cp(s) on C, (6) where s is arc length along C and cp is a specified function on C. The function f is prescribed and continuous on C. Ans. Au = u,, + u,, = f (x, y). For a discussion of more general boundary conditions, see [I], pp. 203-207. PROB. 1.1.8 SUPPLEMENT 395 1.1.5. In Prob. 1.1.4, alter the admissibility conditions so that (6) is satisfied on a subarc C, of C. On C, - C - C,, there are no specified boundary conditions. Determine the Euler equation and the natural bound- ary condition that must be satisfied on C, in order that u minimize I. Ans. Au = f (x, y), U, = 0 on C,, where the subscript n denotes differentiation with respect to the unit outward normal n to D. 1.1.6. Determine the Euler equation of the functional using the admissibility conditions of Prob. 1.1.4. Here a, b, c and f are pre- scribed continuous functions on D, and a and b have continuous first partial derivatives. Ans. (au,), + (bug), - cu = f. 1.1.7. Determine the Euler equation and natural boundary condition for the functional where C is the contour bounding D, the functions a, c and f are prescribed and continuous on D, and a has continuous first partial derivatives. The prescribed functions A(s) and cp(s) are continuous on C. Ans. (au,), + (q), - cu =f, u, + Au = cp. 1.1.8. Determine the Euler equation of the functional where f is a prescribed continuous function on D. The admissible functions u(x, y) have continuous partial derivatives up to and including the fourth order, and satisfy the boundary conditions u = cp(s), u, = +(s) on C. (7) Ans. A2u -- u,, + 2u,,,y + UY,YY =f (x, y). Hint. Let u = ii, = 0 for x, y on C. PROB. 1.1.9 1.1.9. Determine the Euler equation of the functional " I[U] =i J [(Au)' - 2(1 - v)(u,,u,, - u:,) - 2ful dx dy, D where the admissible functions have the same continuity properties as in Prob. 1.1.8 and the condition (7) is satisfied on the subarc C, of C. The remaining part of the boundary C, - C - C, is "free," i.e., no conditions are specified on C,. Determine the natural boundary conditions on C,. When C, = 0, compare the results with those obtained in the previous problem. The constant v is a specified number in the range 0 < v < 8. Ans. A2u = f, v Au + (1 - v)(u,,nl + 2u,,nln, + u,,ni) = 0, (Au), + (1 - v)[(L~,, - u,x)n,n, + ux,(4 - n3l,, where n, and n, are the x and y-components of the outward unit normal to D, and the subscripts denotes differentiation with respect to arc length s along C. Hint. On C,, 6 and fin are arbitrary. 1.1.10. Determine admissibility conditions and a functional whose Euler equation and natural boundary condition .yield the following boundary value problem for the region D in the xu-plane with contour C: A2u=f(x,y) forx,yinD, ti = 0, v AU + (1 - v)(u,,n? + 2u,,nln, + u,,,ni) = 0 for x, y on C. Ans. The functional is given in the preceding problem. The admissible functions have the same continuity properties as in Prob. 1.1.8, and in addi- tion, u = 0 on C. 1.2. THE RlTZ METHOD [I41 The minimum property of the solutions of boundary value problems suggests a method for their approximate determination. Suppose that a sequence of admissible functions is constructed whose limit minimizes an appropriate functional. Then the function obtained by truncating the se- quence after a finite number of terms may provide an approximation to the minimizing function. The approximation is presumably more accurate when more terms in the sequence are retained. Specifically, we select a family of admissible functions u = U(x; c) (8) depending on n (unknown) parameters c = (c,, c,, . . . , c,). Inserting these functions into the functional and performing the necessary integrations, we obtain I[U(x; c)l = 'Wc), (9) SUPPLEMENT 397 where Q is a function of then parameters c. Necessary conditions for @ to be The m solutions c = cj, j = 1, 2, . . . , m of the algebraic equations (10) give the stationary points of @. Let c = c0 be a stationary point which also fur- nishes a minimum of @. Then we expect that the function u = U(x; cO), which is called a Ritz approximation and which minimizes I with respect to all admissible functions of the form (8), is an approximation to an admissible function that minimizes I.6 In practice, the family of admissible functions is usually formed by taking a linear combination n n(x; C) = uO(X) + C c$(x), j=1 (11) where u0 is an admissible function and the uj (j = 1, 2, . . . , n) are variations. We shall refer to U in the form (1 1) as a trial solution. For linear problems, the functional I is quadratic in u and its derivatives. Then substitution of (I 1) into the equations (10) leads to a system of linear algebraic equations for c. Naturally, we should try to choose the functions u0 and lii so that they approximate the solution as closely as possible. However, there are several practical considerations governing their selection. First of all, they should be chosen so that the integrals necessary to obtain 0 are "easy" to evaluate. Furthermore, the iii must be sufficiently different. If, for example, two of the functions are identical, then the resulting system of linear algebraic equations for c will have a zero determinant. If two or more of the functions Ci differ only slightly, then the determinant may be small and it will be difficult to solve the algebraic equations accurately. If natural boundary conditions are to be satisfied on some portion of the boundary, then, as we have seen in Sec. 1.1, it is not necessary to impose them as part of the admissibility con- ditions, since the solution of the minimum problem automatically satisfies them. However, if it is easy to select uo and ui which satisfy the natural boundary conditions, then it is advantageous to do so in the Ritz method. To illustrate the application of the Ritz method, consider the boundary value problem consisting of the differential equation (3) and the boundary conditions (2), where the associated functional is (I). For simplicitly, we take uo = u, = 0, so that u0 = 0.' Then substituting (1 1) into (1) and Convergence properties and the sense of approximation afforded by the Ritz method have been established in special cases (see [5, 111). For the variations we may take, for example, irj = (6 - x)(a - x)xj, j = I, 2, . . . , 11, or 398 SUPPLEMENT PROB. 1.2.1 performing the necessary differentiations, we obtain I[u] = @(c). Applying the stationary conditions (lo), we find that the parameters cj satisfy the system of algebraic equations f A,,c,+B,=o, i=1,2 ,..., n, j=l (12) where 1.2.1. Prove that if the zjj are linearly independent functions and the coefficientsp(x) and q(x) satisfy the conditions p(x) > 0, q(x) > 0 for all x in [a, b], then the system (12) has a unique solution. Hint. Show by contradiction that the homogeneous form of the system (12), i.e., with Bi = 0 (i = 1, 2, . . . , n), has only the solution c = 0. 1.2.2. Use the Ritz method to obtain an approximate solution of the boundary value problem u" + u + x = 0, u(0) = u(1) = 0 for each of the following trial solutions: a) U = cx(1 - x); b) U = c,x(l - x) + c,x2(1 - x); c) U = clx(l - x) + c2(l - x2). Why are these legitimate trial solutions? Compare the approximations so obtained for u and u' with the exact solution. Ans. (see [ll], p. 269 and [I], p. 220). 1.2.3. Use the Ritz method to obtain an approximate solution of Bessel's equation x2u" + XU' + (x2 - 1)u = 0 in the interval 1 < x < 2, where u(1) = 1, 42) = 2. Compare the result with the exact solution. Hint. First write Bessel's equation in the form (3). 1.2.4. Use the Ritz method to obtain an approximate solution of the boundary value problem (XU')' + u = X, u(0) = 0, u(1) = 1, of the form U = x + x(l - x)(cl + c,x). Ans. 85 C1 = - 3 5 26, c2= -- see [91).) 13 PROB. 1.2.8 SUPPLEMENT 399 1.2.5. Use the Ritz method to obtain an approximate solution of the boundary value problem u" + (I + x2)u + 1 = 0, u(-1) = u(1) = 0 of the form a) U = c,(l - x2) + cZ(l - x4); b) U = cl(l - x2) + c2(1 - x4) + c3(l - x6). Ans. (see [l], p. 209). 1.2.6. Obtain a Ritz approximation to the solution of the boundary value problem [(2 - x2)u"]" + 40~ = 2 - x2, y"(& 1) = ym(& 1) = 0, of the form U = cl + c2x2 + c3x4. Ans. C1 = 143363 953 , c2=-- 189 C3 = - 40.79301 79301 ' 79301 (see [I], p. 219). Hint. Use Prob. 1.1.3 to formulate the functional. Note that the boundary conditions are natural boundary coi~ditions. Determine how accurately the boundary conditions are satisfied by the approximate solution. 1.2.7. Use the Ritz method to obtain an approximate solution of the Poisson equation AU = u,, + u,, = -2, subject to the condition u = 0 on the boundary of the rectangle 1x1 < a, IyI < b, where the trial solution is of the form a) U = c(x2 - a2)(y2 - b2); b) U = (x2 - a2)(y2 - a2)[clx + c2(x2 + y2)] (for the square b = a). Ans. (see [l 11, p. 281). Hint. Use Prob. 1.1.4. 1.2.8. Solve Prob. 1.2.7, using the Ritz method with mrrx nny U = -f c,, cos -FOE - . m=1.3,6,. . . n=1,3,6.. . . 2a 2b 400 SUPPLEMENT PROB. 1.2.9 Show that the constants c,, so obtained coincide with those found by the method of separation of variables. Ans. c,, = 128~-~a~b~(- l)x(m+n)-1[mn(b2m2 + a2n2)]-I (see [ll]. p. 282). 1.2.9. Apply the Ritz method to construct a solution of Au = - 1 satis- fying the boundary condition u, + u = 0 on the sides of the square 1x1 < 1, IyI < 1 (see Prob. 1.1.7), where the trial solution is of the form a) U = c1 + c2(x2 + y2); b) U = cl + c2(x2 + y2) + c3x2y2. Note that the trial solutions need not satisfy the boundary conditions, since they are natural conditions. Ans. (see [l], p. 429). 1.2.10. Solve Prob. 1.2.9 by the Ritz method, selecting trial solutions that satisfy the natural boundary conditions. Make use of the symmetry of the solutions in x and y. Compare with the answer to Prob. 1.2.9. 1.2.11. Find Ritz approximations to the solution of on the rectangle 1x1 ,< 4, lyl G 1, where u = 0 on the edges of the rectangle. As trial solutions, use Ans. 7 a) c = - . b) 104c1 ,-- 10.185, 104c2 m 4.84 7264 ' (see [I], p. 459). Hint. Use Prob. 1.1.6. 1.2.12. Obtain a Ritz approximation to the solution of the biharrnonic equation A2u = 0, satisfying the following boundary conditions on the edges of the square 1x1 < 1, lyl < 1: u,, = 0 for x = 51, y = f 1, u,,=l-y2 for x&1, As trial solutions, use a) U= - 1 ey 2 ) + cdx2 - 1Y(y2 - 1); b) U = +y2(1 - +y2) + (x2 - 1)2(y2 - 1)2(~1 + c2x2 + c3y2). Ans. a)c,w0.0425; b)clw0.0404, c2=c3~0.0117 (see [17], p. 167). Hint. Transform the boundary conditions into the form (7), and then use Prob. 1.1.8. 1.2.13. Determine a Ritz approximation to the solution of h2u = f (x, y) in the rectangle 0 < x < a, 0 < y < b, satisfying the boundary conditions of Prob. 1.1.10 on the edges of the rectangle. Use a trial solution of the form mxx nxy o = 3 2 c,. sin - sin - . m=l n=l a b Ans. 2-2u b mxx nTcy cmn = & [(:P+ (:)I S, SJ(x, Y) sin - a sin - b dx dy (see [16], p. 345). 1.2.14. Use the Ritz method to obtain an approximate solution of the clamped rectangular plate problem A2u = f where f is a constant (see Prob. 1.1.8), subject to the conditions u = un = 0 on the boundaries of the rectangle 0 < x < a, 0 < y < b. Use a trial solution of the form Ans. (see [18], p. 288). 1.3. KANTOROVICH'S METHOD8 Kantorovich's method, which is sometimes called the mixed Ritz method or the method of reduction to ordinary differential equations, is essentially a generalization of the Ritz method. More "freedom" is permitted in the selec- tion of the trial solutions (8) and (1 1) by allowing the parameters c to be functions of one of the independent variables x, say x. The functional I then reduces to a functional I[U(x; +))I = Y[Wl (13) of n functions cj(x), which are determined so as to furnish a minimum of Y. For a general description and analysis of this method, see [ll]. Thus the cj(x) are solutions of a system of n ordinary differential equations which are the Euler equations of Y. The solutions of these equations subject to appropriate boundary conditions yield the approximation U(x; c(x)). For simplicity, we shall consider Kantorovich's method only for a rectangular region no < x < a,, 6, < y < b, in the xy-plane. However, the method can be applied to regions of more general shape (see [ll]). We shall employ trial solutions of the form n qx, y; 4~)) = uO(x, Y) + C c,(x)iii(x, Y), j=1 (14) where u0 satisfies inhomogeneous boundary conditions and the zij homo- geneous boundary conditions on y = b,, b,. The boundary conditions on x = a,, a, yield the values of cj(ao) and cj(al), j = 1, 2, . . . , n. As an example (see [ll], p. 304), consider the problem of solving the equation hu = -1 for x, y in the square 1x1 < 1, lyl < 1, subject to the boundary c0ndition.u = 0 on the edges of the square. As a trial solution, we take U = (1 - y2)c(x), which satisfies the boundary conditions on y = f 1. To make the trial solution satisfy the conditions on x = & 1, we require that c(- 1) = c(1) = 0. Then the associated functional (see Prob. 1.1.4) reduces to 8 I[U]= ~[c(x)l= -J (2 cJ2 + c2 - c 3 -1 5 The Euler equation of 'P is obtained by using (1) and (3), and is given by 5 5 Cn--C= -- 2 4 ' Solvingthisequation and applying the boundary conditions c(- 1) = c(1) = 0, we obtain 'i cosh bx), J: c(x) = - 1 - --- I< = , 2 cosh k cosh kx U = -(1 - y2) 1 - --- 2 ( cosh k 1.3.1. Solve the above boundary value problem by Kantorovich's method, using the trial solution u = (1 - y2)[c1(x) + c2(x)y21. Compare with the result of the Ritz approximation obtained in Prob. 1.2.7. Ans. 1 cosh a-x cosh a+x c~(x) w -- + 0.516 ------ - 0.0156 ------ , 2 cosh a- cosh a+ where a, = (14 f J%)l/2 are the roots of the characteristic equation E4 - 28E2 + 63 = 0 (see [I I], p. 317). 1.3.2. Solve the above boundary value problem by Kantorovich's method, using the trial solution m U = 2 cj(x) cos (j + +)TFY. j=O Verify that this yields the infinite series representation of the exact solution. Ans. cj(x) = (- 1 )j-12x-3(j - cosh (j - $)nx I , j = 1,2, . . . cosh (j - $)TC (see [l 11, p. 320). 1.3.3. Use Kantorovich's method to solve the clamped rectangular plate problem, i.e., A2u = 1 in the rectangle 1x1 < a, lyl < b, with boundary con- ditions u = u, = 0 on the edges of the rectangle. Use U = (y2 - b2)%(x) as a trial solution. Ans. 24c(x) = A cosh a< cos P< + B sinh E, sin P< + 1, where E, = xlb, A = dl/do, B = d2/do, do = p sinh ar cosh ar 4- a sin pr cos pr, -dl = a cosh ar sin pr + P sinh ar cos pr, d2 = a sinh ar cos pr - P cosh ar sin pr, r = alb, a w 2.075 and P w 1.143 (see [I I], p. 322). 1.3.4. Use Kantorovich's method to obtain an approximate solution of A2u = 0 on the semi-infinite strip 0 ,< x < co, lyl g 1, subject to the follow- ing boundary conditions: lim u,,(x, y) = lirn u,,(x, y) = 0 uniformly in y. x+ m x+m Use the trial solution (note that U satisfies the boundary conditions on y = 5 1). Ans. where y = a + pi w 2.075 + 1.143i is a root of y4 - 6y2 + a$ = =(see [lo]). 2. Related Methods The application of the Ritz method to the solution of boundary value problems requires a variational principle. However, in some problems there is no such principle, while in others, it is difficult to determine the proper functional or cumbersome to evaluate the integrals needed in the Ritz method. Thus, in this section, we shall discuss three procedures for obtaining approxi- mate solutions which do not require a variational functional, although they l,ead to approximations related to those obtained by the Ritz method. For simplicity, consider the following boundary value problem involving a single function u(x): Lu = f for x in D, Bu = g for x on C. (1 5) Here L is a differential operator defined in a domain D, B is a boundary operator defined only on the boundary C of D, and f and g are prescribed functions. Thus Bu = g is the boundary condition for the single differential equation Lu =J As in the Ritz method, we seek an approximate solution of (15) of the form u = U(x; c), depending on n parameters c = (c,, c2, . . . , c,). We shall assume, unless it is otherwise specified, that c is independent of x. In general, the approximate solution U does not satisfy the differential equation and the boundary con- dition, and in fact LU - f = e(x; c) for x in D, BU - g = E(x; c) for x on C, (16) where e and E, called the interior error and the boundary error, are algebraic functions of x and c. If c is a function of one independent variable, then e will be an ordinary differential operator acting on c, and E will contain initial or boundary conditions for c. If the function U is selected so that E = 0 for all x on C, the procedure used to determine c is called an interior method, while if e - 0 for all x in D, the procedure is called a boundary method. We wish to determine c so that the errors are, in some sense, as small as possible. Essentially, each of the methods described below amounts to ascribing a definite meaning to the term "small." 2.1. GALERKIN'S METHOD [7] In Galerkin's method, the n parameters are chosen to make the errors orthogonal to a set of n independent functions wl(x), w2(x), . . . , wn(x), PROB. 2.1.2 SUPPLEMENT 405 usually taken to be orthogonal. This gives n conditions of the form where ds is an element of area on C. These are n algebraic equations for determining then parameters c. In fact, the equations are linear if L and Bare linear operators and U is chosen in the form as is customary in practice. The interior Galerkin methodB corresponds to choosing u0 and iii, j = 1,2, . . . , n to satisfy the inhomogeneous and homo- geneous boundary conditions, respectively. In the applications, it is custom- ary (but not essential) to set wi = z.2, j = 1, 2, . . . , n, and we shall do so in all the problems that follow. If, as n + co, the wi form a complete set of functions, then e -t 0 as n 4 co (being orthogonal to every function of a com- plete set). Some convergence properties of Galerkin's method are discussed in [12]. Practical selection of the functions iij and wi is governed by the same considerations as in the Ritz method, i.e., they should make evaluation of the integrals in (17) easy and they should be sufficiently dissimilar (say orthogonal) to lead to a "well-conditioned" system of algebraic equations. If the boundary value problem (15) can be derived from a variational principle, then, in many cases, it can be shown that Ritz's method coincides with Galerkin's. If the parameters cj in (18) are permitted to be functions of one variable, we obtain the Galerkin-Kantorovich method. The conditions (17) then give ordinary differential equations and boundary conditions for determining c. 2.1.1. Given the differential equation (3) and the boundary conditions (2), with uo = u, = 0, show that the Ritz and Galerkin methods lead to the same system of algebraic equations (12) for determining the coefficients c. Hint. Use integration by parts. 2.1.2. Given the differential equation (see the answer to Prob. 1.1.6) and the boundary condition u = 0 on C, show that the Ritz and Galerkin methods lead to the same system of algebraic equations for the coefficients c. The expression Galerkitr's method conventionally denotes the interior Galerkin method. 2.1.3. Solve Prob. 1.2.3 by Galerkin's method without transforming Bessel's equation into the form (3). Compare with the exact solution, and also with the Ritz approximation using the same number of parameters. 2.1.4. Use Galerkin's method to obtain an approximate solution of the boundary value problem u" + xu1+ u = 2x, u(0) = 1, u(1) = 0, choosing a trial solution of the form U = (1 - x)(l + clx + c2x2 + c3x3). Ans. c, w -0.209, c2 w -0.789, c3 m 0.209 (see [13], p. 115). 2.1.5. Solve Prob. 1.2.4 by Galerkin's method, using the same trial solution. Verify that c, and c, satisfy the same algebraic equations as in the Ritz method. 2.1.6. Use Galerkin's method to solve the boundary value problem choosing a trial solution of the form U = c, sin TCX + c2 sin 3nx. Ans. cl = 4n-'(x4 + I)-', c2 = 4[3x(81x4 + I)]-' (see [6], p. 233). 2.1.7. Solve Probs. 1.2.7 and 1.2.12 by Galerkin's method, using the same trial solutions. Verify that the coefficients cj satisfy the same algebraic equations as in the Ritz method. 2.1.8. Use Galerkin's method to solve Prob. 1.2.9, choosing the following trial solutions which satisfy the (natural) boundary conditions: a) U = c[9 - 3(x2 + y2) + x2y2]; b) U = ~1[9 - 3(x2 + y2) + x2y2] + c2[30 - 5(x2 + y2) - 3(x4 + y4) + x2y2(x2 + y2)]. Ans. 5 a) c = - . b) 103c1 w 73.3, 103c2 m 5.38 54 ' (see [I], p. 413). 2.1.9. Use the Galerkin-Kantorovich method to obtain an approximate solution of the heat equation u,, = u, in the semi-infinite strip 0 < x < 1, t > 0. The boundary and initial conditions are PROB. 2.1.10 SUPPLEMENT 407 and u must remain bounded as t + a. Use a trial solution of the form which satisfies the boundary conditions but not the intial conditions. Ans. (see [6], p. 372). Hint. In applying (17), set the area integral over thestrip and the boundary integral over the initial line separately equal to zero. 2.1.10. Use the Galerkin-Kantorovich method to obtain an approximate solution of the wave equation u,, = u,, in the semi-infinite strip 0 < x < 1, t > 0, where the boundary and initial conditions are u(O,t)=u(l,t)=O, t>0, u(x, 0) = x(l - x), u,(x, 0) = 0, 0 < x g 1. Use a trial solution of the form u = x(l - x)[c,(t) + c,(t)x(l - x)]. Ans. c1 w 0.804 cos at + 0.197 cos Pt, c, w 0.91 l(cos at - cos Pt), where a = x, p w 10.1 1 (see [6], p. 375). COLLOCATION Of all the approximation procedures under consideration, the collocation method is perhaps the simplest to apply. In this method, the n parameters are determined by requiring the errors in (16) to vanish at iz points x,, x,, . . . , x, in D + C called the collocationpoints. Of course, these points must be chosen so that the resulting system of equations has a solution, say cO(xj). The ideal collocation points are those for which cO(xj) minimizes the maximum error for all x in D + C. For example, if we define &(xj) = max leix; cO(xj)) 1 + max I E(x; co(xj))l, xin D xon C (19) then as the collocation points we should take the values xj, j = 1,2, . . . , n for which & is a minimum. However, no general procedures are presently available for ayiiori selection of points satisfying this criterion; in fact, they 408 SUPPLEMENT PROB. 2.2.1 are usually determined by intuition or by practical considerations such as computational simplicity. Only interior or boundary points need be con- sidered as collocation points, depending on whether interior or boundary collocation is employed. A disadvantage of collocation is that the approximate solution may vary considerably with the position of the collocation points. One way to minimize this is to take a sufficient number of points and distribute them over the domain and the boundary. An obvious generalization of the collocation method is to allow the parameters to be functions of one variable, say x. Then the errors will depend on c(x) and its derivatives, and collocation may yield a system of differential equations and boundary conditions for determining the param- eters. 2.2.1. Solving Prob. 1.2.2 by interior collocation, using U = cx(1 - X) as a trial solution and the following collocation points: Compare with the Ritz approximation and the exact solution. In each ease, evaluate & -- max [el 0<&1 [cf. (19)l. Does the approximation with smallest & have the smallest deviation from the exact solution? Ans. - 2.2.2. Solve Prob. 1.2.5 by interior collocation, using as a trial solution and the following collocation points: a) x =a,% (set c, = 0); b) x = 6, 2, ". o Ans. (see [I], p. 182). 2.2.3. Solve Prob. 2.1.6 by interior collocation, using the same trial solution and x = a, + as coilocation points. Compare with the approxima- tion obtained by Galerkin's method. Ans. (see [6], p. 233). 2.2.4. Solve Prob. 1.2.7 for the square b = a by interior collocation, using the same trial solutions. For the trial solution a, use x = y = 0 as the collocation point, and for the trial solution b, use the points x = y = 0 and x = y = a/2. Compare with the Ritz approximation and the infinite series solution. Ans. (see [15], p. 437). 2.2.5. Use boundary collocation to solve Prob. 1.2.7 for the square b = a. To select trial solutions, it is convenient to introduce polar coordinates 1 Y r2 = x2 + y2, 0 = tan- - . X Then the function is a solution of the differential equation. Determine the parameters c,, c, and c,, using the collocation points Compare with the solutions obtained by interior collocation (Prob. 2.2.4) and by the Ritz method (Prob. 1.2.7). Also compare with the infinite series solution (Prob. 1.2.8). Ans. el w 0.590a2, a2c, w -0.0924, a6c3 w 0.00254 (see PI). 2.2.6. Use boundary collocation to determine an approximate solution of Au = -2 where u = 0 on the boundary of a regular hexagon with sides of length 2a/ J3 whose vertical sides lie on x = fa. As a trial solution, use the function la2 U = - - + c, + c2rs cos 60 + c3r12 cos 120, 2 which solves the differential equation. Choose polar coordinates with respect to the center of the hexagon, and use the collocation points 2.2.7. Solve Prob. 1.2.9 by interior collocation, using the trial solutions of Prob. 2.1.8. For the one-parameter approximation, use the collocation points a)x=y=O; b)x=y=Q; c)x=y=Z 3. For the two-parameter approximation, use the collocation points Compare these approximate solutions with those obtained by the Ritz and Galerkin methods. Am. (see [I], p. 41 1). 2.2.8. Solve Prob. 1.2.9 by boundary collocation, using the following trial solutions and collocation points: (Both trial functions are solutions of the differential equation.) Compare with the results of Prob. 2.2.7. A ns. a) cw0.813; b) ~~w0.821, c,w -0.0144 (see [I], p. 413). 2.2.9. Use boundary collocation to solve Prob. 1.2.14 for the square b = a. Let r and O be polar coordinates with respect to the center of the square, and use the trial solution and the collocation points Verify that U is a solution of the differential equation. Compare with the approximate solutiori obtained by the Ritz method in Prob. 1.2.14. Ans. c, m 1.296F, a2c, w -2.256F, a4c, m -0.3603F, where F = fa4/64 (see [2]). 2.3. LEAST SQUARES In the method of least squares we seek an approximate solution in the form u = U(x; c), as before, but the parameters c are determined to minimize the "mean square error" of the errors e and E in (16), i.e., ID o(x)e2(x; c) dx + IC Q(x)E~(x; e) ds = minimum, (20) where the weighting functions w(x) > 0 for x in D and Q(x) > 0 for x on C are at our disposal. Usually it is convenient to take o = i2 - 1, and we shall do so in the problems below. Necessary conditions for the mean square error (20) to be a minimum are obtained by differentiating (20) with respect to each cj: This gives n algebraic equations for determining the n parameters cj by the method of least squares. The method of least squares is usually less convenient than collocation, since the additional integrals in (21) may be difficult to evaluate. On the other hand, the method of least squares is more systematic than collocation, since there is no arbitrariness corresponding to the selection of collocation points. 2.3.1. Solve Prob. 1.2.2 by interior least squares. As the trial solution, use the function U = c,x(l - x) + c2x(l - x2), which satisfies the boundary conditions. Compare the results with the Ritz and collocation approximations (see Prob. 2.2.1). Ans. C1 = 4448 413 , cg=- 101 -2437 2437 (see [I], p. 220). 4 12 SUPPLEMENT PROB. 2.3.2 2.3.2. Use interior least squares and the trial solution to solve Prob. 1.2.7 for the square b = a. Compare the resulting approxima- tion with those obtained by the Ritz and collocation methods (Probs. 1.2.7 and 2.2.4). Also compare with the infinite series representation of the solu- tion obtained by separation of variables. Ans. 2.3.3. Solve Prob. 1.2.9 by interior least squares, using the trial solution U = c[9 - 3(x2 + y2) + x2y2] which satisfies the boundary conditions. Compare with the approximations obtained by the Ritz, Galerkin and collocation methods (Probs. 1.2.9, 2.1.8 and 2.2.7). Ans. 15 c=- 161 (see [I], p. 414). 2.3.4. Solve the equation Au = x2 - 1 in the rectangle 1x1 < 1, lyl < t by the boundary least squares method, where u = 0 on the edges of the rec- tangle. Use the trial solution which is a solution of the differential equation. Ans. (see [I], p. 417). 1 Collatz, L., The Numerical Treatment of Dtffferential Eyuatiotrs, third edition, Springer- Verlag, Berlin (1960). 2 Conway, H. D., The approximate analysis of certain boundary value problems, J. Appl. Mech., 27, 275 (1960). 3 Courant, R., Variatiotralmethods for the solution ofprobletns of equilibrium atrdvibratiotrs, Bull. Amer. Math. Soc., 49, 1 (1943). 4 Courant, R. and D. 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Lebedev, On a method of solving some problems of drffraction theory and related problems, Zh. ~ks~er. ~eor. ~iz., 8, 1192 (1938). K4 Koshlyakov, N. S., M. M. Smirnov and E. B. Gliner, Differential Equations of Mathematical Physics (translation edited by H. J. Eagle), John Wiley and Sons, Inc., New York (1964). K5 Kupradze, V. D., Randwertalrfgaben der Schwingungstheorie rtnrl Integralglei- chungen, VEB Deutscher Verlag der Wissenschaften, Berlin (1956). L1 Ladyzhenskaya, 0. A., The Mixed Problem for a Hyperbolic Equation (in Russian), Gos. Izd. Tekh.-Teor. Lit., Moscow (1953). L2 Lagrange, R., Les familles de surfaces de re'uolution qui possedent des harmoniques, Acta Math., 71, 283 (1939). L3 Lebedev, N. N., The coeficient oJ'mutua1 induction between coils wound on a cylindrical core with magnetic permeability y, Zh. Tekh. Fiz., 6, 530 (1936). L4 Lebedev, N. N., The jitnctiotis associated with a ring of oval cross-section, Tech. Phys. USSR, 4, 3 (1937). L5 Lebedev, N. 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W1 Walker, M., The Schwarz-Christoffel Transformation and its Applications- A Simple Exposition, Dover Publications, Inc., New York (1964). W2 Wangerin, A., Reduction der Potentialgleichung fur gewisse Rotationskorper auf eine gewohnliche Dlferentialgleichllng, Preisschr. der Jabl. Ges., Leipzig (1875). W3 Wangerin, A., #ber ein dreifach orthogonales Fliichensystem, gebildet nus gewissen Flachen vierter Ordnung, Borchardt J., 82, 145 (1876); Notiz zu dem Aufsatz uber ein dreifnch orthogonales Fliichensystem etc., ibid., 82,348 (1 876). W4 Watson, G. N., A Treatise on the Theory of Bessel Functions, second edition, Cambridge University Press, London (1962). W5 Webster, A. G., Partial Differential Eyuafions of Mathematicnl Physics, second edition (edited by S. J. Plimpton), Dover Publications, Inc., New York (1955). W6 Weinel, E., Das Torsionsproblem fur den exzentrischen Kreisring, 1ng.-Arch., 3, 67 (1932). W7 Weyl, H., Uber gewohnliche Dt~erentialgleichungen mif Singularitaten rrnd die zugehijvigen Entwicklutlgen willkurlicher Funktionen, Math. Ann., 68, 220 (1910). W8 Whittaker, E. T. and G. N. Watson, A Course of Modern Analysis, fourth edition, Cambridge University Press, London (1963). W9 Widder, D. V., The Laplace Transform, Princeton University Press, Princeton, N.J. (1941). W10 Wiener, N. and E. Hopf, Uber eine Klasse singularer Integralgleichungen, S. B. Preuss. Akad. Wiss., 696 (1931). NAME INDEX Akhiezer, N. I., 60, 143, 415 Armstrong, A. H., 419 Bateman, H., 102, 252, 376, 415, 416 Benster, C. D., 413, 417 Berry, J., 416 Betz, A., 34, 41, 415 Bierens de Haan, D., 388, 415 Bacher, M., 252, 415 Boyanovitch, D., 417 Bremmer, H., 202,421 Brink, D. M., 415 Browder, F. E., 420 Budak, B. M., 102,415 Campbell, G. A., 202, 415 Carslaw, H. S., 79, 179, 202, 415 Chu, L. J., 220, 252, 420 Churchill, R. V., 202, 4 15 Collatz, L., 394, 398, 399, 400, 406, 410, 411, 412 Conway, H. D., 409, 411,412 Corbat6, F., 220, 252, 420 Courant, R., 3, 18, 19, 41, 392, 397, 413,416 Crandall, S., 406, 407, 408, 413 Doetsch, G., 202, 416 Dwight, H. B., 388, 416 Eagle, H. J., 418 Eisenhart, L. P., 252, 416 Erdilyi, A., 170, 232, 384, 416 Feshbach, H., 18, 41, 102, 142, 202, 252, 271,419 Fock, V. A., 155, 202,237, 271,416 Fomin, S. V., 392, 413 Foster, R. M., 202, 415 Frank, P., 18, 102, 166, 167, 416 Franklin, P., 102, 416 Freeman, I., 417 Friedlander, F. G., 257, 258, 373, 416 Fuchs, B. A,, 34, 202,416 Galerkin, B. G., 404, 413 Garabedian, P. R., 3, 18, 19, 24, 41'6 Gelfand, I. M., 392, 413 Glazman, I. M., 60, 143, 415 Gliner, E. B., 24, 25,418 Goodier, J. N., 4, 169, 341, 401, 413, 421 Gradstein, I. S., 351, 384, 386, 388, 419 Gray, A,, 102, 202, 384, 417 Grinberg, G. A., 11, 18, 29, 83, 102, 105, 106, 142, 146, 192, 202, 253, 263, 268, 271, 417 Gyrmter, N. M., 102, 417 Haentzschel, E., 252, 417 Harding, I. W., 146, 417 423 Hilbert, D., 3, 18, 19, 41, 392, 397, 412, 413, 416 Hildebrand, F. B., 398, 413 Hobson, E. W., 220, 252, 384, 417 Hopf, E., 27 1,422 Horvay, G., 403, 413 Jackson, D., 102, 384, 417 Jaeger, J. C., 79, 179, 202,415 Jeans, J., 239, 417 Jeffery, G. B., 216, 417 Jeffreys, B. S., 102, 417 Jeffreys, H., 102, 417 Jolley, L. B. W., 386, 417 Joos, G., 11, 417 Kantorovich, L. V., 106, 397, 398, 399, 400, 401, 402, 403,413, 417 Kibel, I. A., 37, 417 Kochin, N. E., 37, 417 Kontorovich, M. I., 146, 196, 200, 201, 202, 418 Koshlyakov, N. S., 24, 25, 418 Krylov, V. I., 106, 397, 398, 399, 400, 401,402, 403, 413,417 Kupradze, V. D., 41, 271,418 Kuzrnin, R. O., 102, 417 Ladyzhenskaya, 0. A., 3, 59, 418 Lagrange, R., 248, 250, 252, 418 Lebedev, N. N., 102, 146, 156, 181, 195, 196, 200, 201, 202, 221, 222, 229, 235, 236, 243, 249, 250, 252, 264, 271, 350, 355, 367, 369, 378, 383, 384,418 Lense, J., 252, 384, 418 Levin, V. I., 202, 416 Levitan, B. M., 60, 143, 152, 202, 418 Little, J. D. C., 220, 252, 420 Lurye, A. I., 187, 202,419 Macdonald, H. M., 198, 419 MacRobert, T. M., 384, 417, 419 Magnus, W., 170, 232,252, 384, 390,416, 419 Marcuvitz, N., 54, 419 Mathews, G. B., 102, 202, 384, 417 McLachlan, N. W., 102, 384, 419 Melan, E., 158, 419 Mikhlin, S. G., 271, 405, 413, 419 Milne, W. E., 406, 413 Morse, P. M., 18, 41, 54, 102, 142, 202, 220,252,271,419, 420 Muskhelishvili, N. I., 34, 271, 419 Nestell, M., 415 Noble, B., 271, 419 Oberhettinger, F., 170, 232, 252, 384, 390,416, 419 Ollendorff, F., 225, 419 Panovsky, W. K. H., 11, 419 Petrovski, I. G., 3, 18, 41, 419 Phillips, M., 11, 419 Plimpton, S. I., 422 Radding, S., 421 Radok, J. R. M., 417, 419 Rayleigh, Lord, 54, 419 Reed, J. W., 416 Reiss, E. L., 391 Ritz, W., 396, 41 3 Robson, A. R. M., 415 Roze, N. V., 37, 417 Ryshik, I. M., 351, 384, 386, 388, 419 Sagan, H., 3, 9, 420 Sakharov, I. Y., 193, 420 Samarski, A. A., 4, 9, 16, 18, 19, 20, 25, 27, 28, 29, 41, 59, 71, 102, 142, 415, 42 1 Schelkunoff, S. A., 50, 54, 420 Shabat, B. V., 34,416 Shenitzer, A,, 419 Silverman, R. A., 413, 418, 421 Skalskaya, I. P., 199, 420 Smirnov, M. M., 24, 25, 102,418, 420 Smirnov, V. I., 4, 18, 25, 28, 60, 143, 271, 372, 384,420 Smith, T. L., 41, 420 Smythe, W. R., 34, 41, 252, 420 Sneddon, I. N., 41, 146, 202,417, 420 Snow, C., 252, 384,420 Sobolev, S. L., 3, 420 Sokolnikoff, I. S., 409, 412, 413 Sommerfeld, A., 4, 18, 166, 167, 420 Stepanov, V. V., 252,420 Sternberg, W. J., 41, 420 Stratton, J. A., 54, 220, 252, 420 Straus, E. G., 420 Strutt, J. W. (see Lord Rayleigh) Strutt, M. J. O., 220, 252, 420 Tikhonov, A. N., 4, 9, 16, 18, 19, 20, 25, 27, 28, 29, 41, 59, 71, 102, 142, 415, 42 1 Timoshenko, S., 4, 18, 54, 1 12, 169, 29 1, 341, 401, 413, 421 Titchmarsh, E. C., 60, 143, 147, 161, 190, 202, 27 1, 421 Tolstov, G. P., 59, 102, 143, 284, 292, 42 1 Tranter, C. J., 146, 202, 421 Tricomi, F. G., 170, 232, 271, 384, 416, 42 1 Tsukkerman, I. I., 175, 421 Uflyand, Y. S., 193,218,421 Van der Pol, B., 202, 421 Von Mises, R., 18, 102, 166, 167, 416 Walker, M., 33, 41, 421 Wang, C. T., 401, 413 Wangerin, A,, 250, 252, 421 Watson, G. N., 146, 160, 162, 202, 220, 376, 384, 421,422 Webster, A. G., 18, 102, 422 Weinel, E., 214, 422 Wermer, J., 419 Weyl, H., 143, 202, 422 Whittaker, E. T., 146, 376, 384, 422 Widder, D. V., 202, 422 Wiener, N., 27 1, 422 SUBJECT INDEX Abel's equation, 253 Acoustic resonator, 46 Addition theorem for Hankel functions, 378 Admissible functions, 392 Boundary error, 404 Boundary method, 404 Boundary value problems of potential theory, 27 Cauchy problem, 20 Collocation, 407 Collocation points, 407 Complex potential, 34 Conformal mapping, method of, 33 Continuous spectrum, 59, 143 Curvilinear coordinates, 203 Cutoff wavelength, 50 Cylinder functions of ima~inary order, integral transforms involving, 194 Definite integrals encountered in applica- tions, 386 Degenerate bipolar coordinates, 247 Diffraction theory, 254 Diffusion equation, 11 Dirichlet conditions, 146 Dirichlet problem, 27 Discrete spectrum, 58 Distribution of d-c current, equations for, 12 Eigenfunction method, 103 Eigenfunctions, 58 orthogonality of, with weight r(x), 58 Eigenvalues, 58 Electromagnetic oscillations, 49 Electron-optical device, 13 1 Electrostatic problems, solution by in- tegral equations, 259 Electrostatics, equations of, 11 Elliptic coordinates, 204 Elliptic equations, 27 Euler equations, 392 Expansions in series of orthogonal func- tions, 384 Expansion of differential operators in curvilinear coordinates, 388 Flamant's problem, 158 Forced oscillations of elastic bodies, 46 Fourier method, 55 Fourier-Bessel integral, 160 Fourier-Bessel series, 292 Fourier-Mellin theorem, 170 1 Because of the contents of this book (problems and their solutions), the subject index is necessarily eclectic, consisting mainly of first occurrences of key terms. 427 Fourier cosine transform, 144 Laplace's equation, 8 Fourier integral theorem, 147 Laplacian operator, 5 Fourier sine transform, 144 Least squares, method of, 41 1 Fourier transform, 144, 146 Longitudinal oscillations of a rod, equa- Fourier's equation, 9 tion for, 4 Free oscillations of elastic bodies, 43 Functional, 392 M G Maxwell's equations, 12 Mean square error, 41 1 Galerkin-Kantorovich method, 405 Mehler-Fock theorem, 221 Galerkin's method, 404 Mehler-Fock transform, 144 Green's function, 28 Mellin transform, 144, 189 Grinberg's method, 105 Minimum potential energy, principle of, 392 Mixed problem, 19 Hamilton's principle, 392 Hankel transform, 144, 160 N Hankel's integral theorem, 160 Harmonic function, 27 Natural boundary conditions, 394 Helmholtz's equation, 28 Nernst's law, 11 Hertz vector, 16 Neumann problem, 27 Hyperbolic equations, 19 Newton's law, 9 Images, method of, 29 Parabolic coordinates, 210 Inhomogeneous boundary conditions, 103 Paraboloidal coordinates, 23 1 Integral equations, 253 Poisson's equation, 8, 28 Integral transform, 143 kernel of, 143 Interior error, 404 R Interior method, 404 Inverse transform, kernel of, 145 Riemann's method, 19 Inversion formula, 145 Ritz approximation, 397 Inversion, method of, 29 Ritz method, 396 Robin problem, 27 Jet flow, 36 Scalar potential, 13 Schlomilch's integral equation, 376 Schwarz-Christoffel transformation, 34 Kantorovich's method, 401 Kirkhhoff's method, 37 Lam6 functions, 219 Laplace transform, 144, 169 Separation of variables, general theory of, 247 Skin effect, 53, 237 Singular end points, 59 Sommerfeld problem, 255 Special functions, glossary of, 381 Spectrum, 58 Spheroidal coordinates, 219 oblate, 221 prolate, 220 Spheroidal wave functions, 220 Steady-state harmonic oscillations, 42 Stream function, 34 Streamlines, 34 Sturm-Liouville problem, 58 Superposition principle, 50 Thermal capacitance, 174 Three-dimensional bipolar coordinates, 242 Toroidal coordinates, 233 Transmission line equations, 16 Transverse electric wave, 50 Transverse electromagnetic wave, 50 Transverse magnetic wave, 50 Transverse oscillations: of a plate, equation for, 4 of a rod, equation for, 4 Trial solution, 397 Twisting of a prismatic rod, equation for, 5 Two-dimensional bipolar coordinates, 212 Variation, 393 Variational and related methods, 391 Variational problems, formulation of, 392 Vector potential, 12 Vibrating membrane, equation for, 4 Vibrating string, equation for, 4 Volterra's equation, 253 Waveguide problems, 50 Weber transform, 340 Weber's integral, 16 1 Wiener-Hopf method, 253 DOVER BOOKS ON MATHEMATICS ARITHMETIC REFRESHER, A. 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