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Tang K.T. Mathematical Methods for Engineers and Scientists. P. 2 (Springer, 2007)(ISBN 3540302689)(349s)_M_

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Second volume of Kwong-Tin Tang's Mathematical Methods for Engineers and Scientists, published by Springer in 2007. The contents list covers vector analysis (vector operations, vector calculus, divergence and Stokes' theorems, Helmholtz's theorem, curved coordinates), followed by ordinary differential equations and Laplace transforms. It is a downloaded reference book by another author, kept in Phil's collection of math methods books.

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K.T. T ang V ector Analysis, Ordinary Differential Equations and Laplace Transforms 123 With 73 Figures and 4 TablesMathematical Methods 2for Engineers and Scientists Paciﬁc Lutheran University Department of Physics T acoma, W A 98447, USA E-mail: [email protected] ISBN-10 3-540-30268-9 Springer Berlin Heidelberg New Y ork ISBN-13 978-3-540-30268-1 Springer Berlin Heidelberg New Y ork This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broad- casting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media. springer.com © Springer-V erlag Berlin Heidelberg 2007 The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant pro- tective laws and regulations and therefore free for general use. Printed on acid-free paper SPIN 11580966 543210L i b r a r yo fC o n g r e s sC o n t r o lN u m b e r : 2006932619 Cover design: eStudio Calamar Steinen 57/3100/SPi Professor Dr. Kwong-Tin Tang AE Typesetting by the author and SPi using a Springer L TX macro package Preface For some thirty years, I have taught two “Mathematical Physics” courses. One of them was previously named “Engineering Analysis”. There are several textbooks of unquestionable merit for such courses, but I could not ﬁnd one that ﬁtted our needs. It seemed to me that students might have an easier time if some changes were made in these books. I ended up using class notes. Actually I felt the same about my own notes, so they got changed again and again. Throughout the years, many students and colleagues have urged me to publish them. I resisted until now, because the topics were not new and I was not sure that my way of presenting them was really that much better than others. In recent years, some former students came back to tell me that they still found my notes useful and looked at them from time to time. The fact that they always singled out these courses, among many others I have taught, made me think that besides being kind, they might even mean it. Perhaps it is worthwhile to share these notes with a wider audience. It took far more work than expected to transcribe the lecture notes into printed pages. The notes were written in an abbreviated way without much explanation between any two equations, because I was supposed to supply the missing links in person. How much detail I would go into depended on the reaction of the students. Now without them in front of me, I had to decide the appropriate amount of derivation to be included. I chose to err on the side of too much detail rather than too little. As a result, the deriva- tion does not look very elegant, but I also hope it does not leave any gap in students’ comprehension. Precisely stated and elegantly proved theorems looked great to me when I was a young faculty member. But in later years, I found that elegance in the eyes of the teacher might be stumbling blocks for students. Now I am convinced that before the student can use a mathematical theorem with con- ﬁdence, he must ﬁrst develop an intuitive feeling. The most eﬀective way to do that is to follow a suﬃcient number of examples. This book is written for students who want to learn but need a ﬁrm hand- holding. I hope they will ﬁnd the book readable and easy to learn from. VI Preface Learning, as always, has to be done by the student herself or himself. No one can acquire mathematical skill without doing problems, the more the better. However, realistically students have a ﬁnite amount of time. They will be overwhelmed if problems are too numerous, and frustrated if problems are too diﬃcult. A common practice in textbooks is to list a large number of problems and let the instructor to choose a few for assignments. It seems to me that is not a conﬁdence building strategy. A self-learning person would not know what to choose. Therefore a moderate number of not overly diﬃcult problems, with answers, are selected at the end of each chapter. Hopefully after the student has successfully solved all of them, he will be encouraged to seek more challenging ones. There are plenty of problems in other books. Of course, an instructor can always assign more problems at levels suitable to the class. Professor I.I. Rabi used to say “All textbooks are written with the principle of least astonishment”. Well, there is a good reason for that. After all, text- books are supposed to explain the mysteries and make the profound obvious. This book is no exception. Nevertheless, I still hope the reader will ﬁnd some- thing in this book exciting. On certain topics, I went farther than most other similar books. For example, most textbooks of mathematical physics discuss viscous damping of an oscillator, in which the friction force is proportional to velocity. Yet every student in freshman physics learnt that the friction force is propor- tional to the normal force between the planes of contact. This is known as Coulomb damping. Usually Coulomb damping is not even mentioned. In this book, Coulomb damping and viscous damping are discussed side by side. Volume I consists of complex analysis and matrix theory. In this volume, we discuss vector and tensor analysis, ordinary diﬀerential equations and Laplace transforms. Fourier analysis and partial diﬀerential equations will be discussed in volume III. Students are supposed to have already completed two or three semesters of calculus and a year of college physics. This book is dedicated to my students. I want to thank my A and B students, their diligence and enthusiasm have made teaching enjoyable and worthwhile. I want to thank my C and D students, their diﬃculties and mis- takes made me search for better explanations. I want to thank Brad Oraw for drawing many ﬁgures in this book, and Mathew Hacker for helping me to typeset the manuscript. I want to express my deepest gratitude to Professor S.H. Patil, Indian Institute of Technology, Bombay. He has read the entire manuscript and provided many excellent suggestions. He has also checked the equations and the problems and corrected numerous errors. The responsibility for remaining errors is, of course, entirely mine. I will greatly appreciate if they are brought to my attention. Tacoma, Washington K.T. Tang December 2005 Contents Part I Vector Analysis 1V e c t o r s .................................................... 3 1 . 1 B o u n da n dF r e eV e c t o r s.................................. 4 1.2 Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2.1 Multiplication by a Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2.2 Unit Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2.3 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2.4 Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2.5 Vector Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.6 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.2.7 Triple Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1 . 3 L i n e sa n dP l a n e s ........................................ 2 3 1.3.1 Straight Lines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.3.2 Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 E x e r c i s e s ................................................... 3 1 2 Vector Calculus ............................................ 3 5 2.1 The Time Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 2.1.1 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . 36 2.1.2 Angular Velocity Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.2 Diﬀerentiation in Noninertial Reference Systems . . . . . . . . . . . . . 42 2 . 3 T h e o r yo fS p a c eC u r v e................................... 4 7 2.4 The Gradient Operator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.4.1 The Gradient of a Scalar Function . . . . . . . . . . . . . . . . . . . 51 2.4.2 Geometrical Interpretation of Gradient . . . . . . . . . . . . . . . 53 2.4.3 Line Integral of a Gradient Vector . . . . . . . . . . . . . . . . . . . 56 2.5 The Divergence of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 2.5.1 The Flux of a Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . 62 2.5.2 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 2.5.3 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 VIII Contents 2 . 6 T h eC u r lo faV e c t o r..................................... 7 0 2.6.1 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 2.7 Further Vector Diﬀerential Operations . . . . . . . . . . . . . . . . . . . . . 78 2.7.1 Product Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 2.7.2 Second Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 2 . 8 F u r t h e rI n t e g r a lT h e o r e m s................................ 8 5 2.8.1 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 2.8.2 Other Related Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 2.9 Classiﬁcation of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 2.9.1 Irrotational Field and Scalar Potential . . . . . . . . . . . . . . . 89 2.9.2 Solenoidal Field and Vector Potential . . . . . . . . . . . . . . . . 92 2.10 Theory of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 2.10.1 Functions of Relative Coordinates . . . . . . . . . . . . . . . . . . . 95 2.10.2 Divergence of /hatwideR/|R|2as a Delta Function . . . . . . . . . . . . 98 2.10.3 Helmholtz’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2.10.4 Poisson’s and Laplace’s Equations . . . . . . . . . . . . . . . . . . . 104 2.10.5 Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 E x e r c i s e s ...................................................1 0 6 3 Curved Coordinates .......................................1 1 3 3 . 1 C y l i n d r i c a lC o o r d i n a t e s ..................................1 1 3 3.1.1 Diﬀerential Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 3.1.2 Inﬁnitesimal Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 3.2 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 3.2.1 Diﬀerential Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 3.2.2 Inﬁnitesimal Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 3.3 General Curvilinear Coordinate System . . . . . . . . . . . . . . . . . . . . 130 3.3.1 Coordinate Surfaces and Coordinate Curves . . . . . . . . . . 130 3.3.2 Diﬀerential Operations in Curvilinear Coordinate S y s t e m s..........................................1 3 3 3.4 Elliptical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 3.4.1 Coordinate Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 3.4.2 Relations with Rectangular Coordinates . . . . . . . . . . . . . . 141 3.4.3 Prolate Spheroidal Coordinates . . . . . . . . . . . . . . . . . . . . . 144 3.5 Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 3.5.1 Jacobian for Double Integral . . . . . . . . . . . . . . . . . . . . . . . . 145 3.5.2 Jacobians for Multiple Integrals . . . . . . . . . . . . . . . . . . . . . 147 E x e r c i s e s ...................................................1 5 0 4 Vector Transformation and Cartesian Tensors .............1 5 5 4.1 Transformation Properties of Vectors . . . . . . . . . . . . . . . . . . . . . . 156 4.1.1 Transformation of Position Vector . . . . . . . . . . . . . . . . . . . 156 4.1.2 Vector Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 4.1.3 Euler Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.1.4 Properties of Rotation Matrices . . . . . . . . . . . . . . . . . . . . . 162 Contents IX 4.1.5 Deﬁnition of a Scalar and a Vector in Terms of Transformation Properties . . . . . . . . . . . . . . . 165 4 . 2 C a r t e s i a nT e n s o r s .......................................1 6 9 4.2.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 4.2.2 Kronecker and Levi-Civita Tensors . . . . . . . . . . . . . . . . . . 171 4.2.3 Outer Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 4.2.4 Contraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 4.2.5 Summation Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 4.2.6 Tensor Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 4.2.7 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 4.2.8 Symmetry Properties of Tensors . . . . . . . . . . . . . . . . . . . . . 183 4.2.9 Pseudotensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 4 . 3 S o m eP h y s i c a lE x a m p l e s .................................1 8 9 4.3.1 Moment of Inertia Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 189 4.3.2 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 4.3.3 Strain Tensor and Hooke’s Law . . . . . . . . . . . . . . . . . . . . . 193 E x e r c i s e s ...................................................1 9 5 Part II Diﬀerential Equations and Laplace Transforms 5 Ordinary Diﬀerential Equations ............................2 0 1 5.1 First-Order Diﬀerential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 201 5.1.1 Equations with Separable Variables . . . . . . . . . . . . . . . . . . 202 5.1.2 Equations Reducible to Separable Type . . . . . . . . . . . . . . 204 5.1.3 Exact Diﬀerential Equations . . . . . . . . . . . . . . . . . . . . . . . . 205 5.1.4 Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 5.2 First-Order Linear Diﬀerential Equations . . . . . . . . . . . . . . . . . . . 210 5.2.1 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 5.3 Linear Diﬀerential Equations of Higher Order . . . . . . . . . . . . . . . 214 5.4 Homogeneous Linear Diﬀerential Equations with Constant Coeﬃcients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 5.4.1 Characteristic Equation with Distinct Roots . . . . . . . . . 217 5.4.2 Characteristic Equation with Equal Roots . . . . . . . . . . . 218 5.4.3 Characteristic Equation with Complex Roots . . . . . . . . . 218 5.5 Nonhomogeneous Linear Diﬀerential Equations with Constant Coeﬃcients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 5.5.1 Method of Undetermined Coeﬃcients . . . . . . . . . . . . . . . . 222 5.5.2 Use of Complex Exponentials . . . . . . . . . . . . . . . . . . . . . . . 229 5.5.3 Euler–Cauchy Diﬀerential Equations . . . . . . . . . . . . . . . . . 230 5.5.4 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 5.6 Mechanical Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 5.6.1 Free Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 5.6.2 Free Vibration with Viscous Damping . . . . . . . . . . . . . . . . 238 5.6.3 Free Vibration with Coulomb Damping . . . . . . . . . . . . . . 241 X Contents 5.6.4 Forced Vibration without Damping . . . . . . . . . . . . . . . . . . 244 5.6.5 Forced Vibration with Viscous Damping. . . . . . . . . . . . . . 247 5 . 7 E l e c t r i cC i r c u i t s.........................................2 4 9 5.7.1 Analog Computation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 5.7.2 Complex Solution and Impedance . . . . . . . . . . . . . . . . . . . 252 5.8 Systems of Simultaneous Linear Diﬀerential Equations . . . . . . . 254 5.8.1 The Reduction of a System to a Single Equation . . . . . . 254 5.8.2 Cramer’s Rule for Simultaneous Diﬀerential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 5.8.3 Simultaneous Equations as an Eigenvalue Problem. . . . . 257 5.8.4 Transformation of an nth Order Equation into a System of nFirst-Order Equations . . . . . . . . . . . . . 259 5.8.5 Coupled Oscillators and Normal Modes . . . . . . . . . . . . . . 261 5.9 Other Methods and Resources for Diﬀerential Equations. . . . . . 264 E x e r c i s e s ...................................................2 6 5 6 Laplace Transforms ........................................2 7 1 6.1 Deﬁnition and Properties of Laplace Transforms . . . . . . . . . . . . . 271 6.1.1 Laplace Transform – A Linear Operator . . . . . . . . . . . . . . 271 6.1.2 Laplace Transforms of Derivatives . . . . . . . . . . . . . . . . . . . 274 6.1.3 Substitution: s-Shifting. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 6.1.4 Derivative of a Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 276 6.1.5 A Short Table of Laplace Transforms . . . . . . . . . . . . . . . . 276 6.2 Solving Diﬀerential Equation with Laplace Transform . . . . . . . . 278 6.2.1 Inverse Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 278 6.2.2 Solving Diﬀerential Equations . . . . . . . . . . . . . . . . . . . . . . . 288 6.3 Laplace Transform of Impulse and Step Functions . . . . . . . . . . . 291 6.3.1 The Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . 291 6.3.2 The Heaviside Unit Step Function . . . . . . . . . . . . . . . . . . . 294 6.4 Diﬀerential Equations with Discontinuous Forcing Functions . . 297 6.5 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 6.5.1 The Duhamel Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 6.5.2 The Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 304 6 . 6 F u r t h e rP r o p e r t i e so fL a p l a c eT r a n s f o r m s...................3 0 7 6.6.1 Transforms of Integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 6.6.2 Integration of Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . 307 6.6.3 Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 6.6.4 Laplace Transforms of Periodic Functions . . . . . . . . . . . . 309 6.6.5 Inverse Laplace Transforms Involving Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . 311 6.6.6 Laplace Transforms and Gamma Functions . . . . . . . . . . . 312 6.7 Summary of Operations of Laplace Transforms . . . . . . . . . . . . . . 313 6.8 Additional Applications of Laplace Transforms . . . . . . . . . . . . . . 316 6.8.1 Evaluating Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Contents XI 6.8.2 Diﬀerential Equation with Variable Coeﬃcients . . . . . . . 319 6.8.3 Integral and Integrodiﬀerential Equations. . . . . . . . . . . . . 321 6.9 Inversion by Contour Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 323 6.10 Computer Algebraic Systems for Laplace Transforms . . . . . . . . . 326 E x e r c i s e s ...................................................3 2 8 References .....................................................3 3 3 Index ..........................................................3 3 5 Part I Vector Analysis 1 Vectors Vectors are used when both the magnitude and the direction of some physical quantity are required. Examples of such quantities are velocity, acceleration, force, electric and magnetic ﬁelds. A quantity that is completely character- ized by its magnitude is known as a scalar. Mass and temperature are scalar quantities. A vector is characterized by both magnitude and direction, but not all quantities that have magnitude and direction are vectors. For example, in the study of strength of materials, stress has both magnitude and direction. But stress is a second rank tensor, which we will study in a later chapter. Vectors can be analyzed either with geometry or with algebra. The algebraic approach centers on the transformation properties of vectors. It is capable of generalization and leads to tensor analysis. Therefore it is funda- mentally important in many problems of mathematical physics. However, for pedagogical reasons we will begin with geometrical vectors, since they are easier to visualize. Besides, most readers probably already have some knowledge of the graphical approach of vector analysis. A vector is usually indicated by a boldfaced letter, such as V,or an arrow over a letter− →V. While there are other ways to express a vector, whatever convention you choose, it is very important that vector and scalar quantities are represented by diﬀerent types of symbols. A vector is graphically repre- sented by a directed line segment. The length of the segment is proportional to the magnitude of the vector quantity with a suitable scale. The direction of the vector is indicated by an arrowhead at one end of the segment, which is known as the tip of the vector. The other end is called the tail. The magnitude of the vector is called the norm of the vector. In what follows, the letter Vis used to mean the norm of V.Sometimes, the norm of Vis also represented by |V|or/bardblV/bardbl. 4 1 Vectors 1.1 Bound and Free Vectors There are two kinds of vectors; bound vector andfree vector . Bound vectors are ﬁxed in position. For example, in dealing with forces whose points of application or lines of action cannot be shifted, it is necessary to think of them as bound vectors. Consider the cases shown in Fig. 1.1. Two forces of the same magnitude and direction act at two diﬀerent points along a beam. Clearly the torques produced at the supporting ends and the displacements at the free ends are totally diﬀerent in these two cases. Therefore these forces are bound vectors. Usually in statics, structures, and strength of materials, forces are bound vectors; attention must be paid to their magnitude, direction, and the point of application. Afree vector is completely characterized by its magnitude and direction. These vectors are the ones discussed in mathematical analysis. In what follows, vectors are understood to be free vectors unless otherwise speciﬁed. Two free vectors whose magnitudes, or lengths, are equal and whose directions are the same are said to be equal, regardless of the points in space from which they may be drawn. In other words, a vector quantity can be represented equally well by any of the inﬁnite many line segments, all having the same length and the same direction. It is, therefore, customary to say that a vector can be moved parallel to itself without change. 1.2 Vector Operations Mathematical operations deﬁned for scalars, such as addition and multipli- cation, are not applicable to vectors, since vectors not only have magnitude but also direction. Therefore a set of vector operations must be introduced. These operations are the rules of combining a vector with another vector or a vector with a scalar. There are various ways of combining them. Some useful combinations are deﬁned in this section. F F Fig. 1.1. Bound vectors representing the forces acting on the beam cannot be moved parallel to themselves 1.2 Vector Operations 5 1.2.1 Multiplication by a Scalar Ifcis a positive number, the equation A=cB (1.1) means that the direction of the vector Ais the same as that of B,and the magnitude of Aisctimes that of B.Ifcis negative, the equation means that the direction of Ais opposite to that of Band the magnitude of Aisctimes that of B. 1.2.2 Unit Vector Aunit vector is a vector having a magnitude of one unit. If we divide a vector Vby its magnitude V, we obtain a unit vector in the direction of V.T h u s , the unit vector nin the direction of Vis given by n=1 VV. (1.2) Very often a hat is put on the vector symbol ( /hatwiden) to indicate that it is a unit vector. Thus A=A/hatwideAand the statement “ nis an unit vector in the direction ofA” can be expressed as n=/hatwideA. 1.2.3 Addition and Subtraction Two vectors AandBare added by placing the tip of one at the tail of the other, as shown in Fig. 1.2. The sum A+Bis the vector obtained by con- necting the tail of the ﬁrst vector to the tip of the second vector. In Fig. 1.2a, Bis moved parallel to itself, in Fig. 1.2b Ais moved parallel to itself. Clearly A+B=B+A. (1.3) B B C = A + B(a) (b) (c) C = B + A C = A + B = B + AB BB A AA AA CC C Fig. 1.2. Addition of two vectors: ( a) connecting the tail of Bto the tip of A; (b) connecting the tail of Ato the tip of B;(c) parallelogram law which is valid for both free and bound vectors 6 1 Vectors −B −BB B(a) (b) (c) D = A + ( −B) = A − BD + B = A D = A − BB + D = A D = A − BBBAA ADD D Fig. 1.3. Subtraction of two vectors: ( a) as addition of a negative vector; ( b)a sa n inverse of addition; ( c) as the tip-to-tip vector which is the most useful interpretation of vector subtraction If the two vectors to be added are considered to be the sides of a parallelogram, the sum is seen to be the diagonal as shown in Fig. 1.2c. This parallelogram ruleis valid for both free vectors and bound vectors, and is often used to deﬁne the sum of two vectors. It is also the basis for decomposing a vector into its components. Subtraction of vectors is illustrated in Fig. 1.3. In Fig. 1.3a subtraction is taken as a special case of addition A−B=A+(−B). (1.4) In Fig. 1.3b, subtraction is taken as an inverse operation of addition. Clearly, they are equivalent. The most often and the most useful deﬁnition of vector subtraction is illustrated in Fig. 1.3c, namely A−Bis the tip-to-tip vector D,starting from the tip of Bdirected towards the tip of A. Graphically it can also be easily shown that vector addition is associative A+(B+C)=(B+A)+C. (1.5) IfA,B,a n dCare the three sides of a parallelepied, then A+B+Cis the vector along the longest diagonal. 1.2.4 Dot Product The dot product (also known as the scalar product ) of two vectors is deﬁned to be A·B=ABcosθ, (1.6) where θis the angle that AandBform when placed tail-to-tail. Since it is a scalar, clearly the product is commutative A·B=B·A. (1.7) Geometrically, A·B=ABAwhere BAis the projection of BonA,a s shown in Fig. 1.4. It is also equal to BAB, where ABis the projection of AonB. 1.2 Vector Operations 7 AB AB = A cos q BA = B cos qq Fig. 1.4. Dot product of two vectors. A·B=ABA=BAB=ABcosθ If the two vectors are parallel, then θ=0a n d A·B=AB.In particular, A·A=A2, (1.8) which says that the square of the magnitude of any vector is equal to its dot product with itself. IfAandBare perpendicular, then θ=9 0◦andA·B=0.Conversely, if we can show A·B=0,then we have proved that Ais perpendicular to B. It is clear from Fig. 1.5 that A(B+C)A=ABA+ACA. This shows that the distributive law holds for the dot product A·(B+C)=A·B+A·C. (1.9) With vector notations, many geometrical facts can be readily demon- strated. C B (B + C)AA BA CAB + C Fig. 1.5. Distributive law of dot product of two vectors. A·(B+C)=A·B+A·C 8 1 Vectors Example 1.2.1. Law of cosines . If A, B, C are the three sides of a triangle, andθis the interior angle between AandB, show that C2=A2+B2−2ABcosθ. q AC Bα β Fig. 1.6. The law of cosine can be readily shown with dot product of vectors, and the law of sine, with cross product Solution 1.2.1. Let the triangle be formed by the three vectors A,B,a n d Cas shown in Fig. 1.6. Since C=A−B, C·C=(A−B)·(A−B)=A·A−A·B−B·A+B·B. It follows C2=A2+B2−2ABcosθ. Example 1.2.2. Prove that the diagonals of a parallelogram bisect each other. Solution 1.2.2. Let the two adjacent sides of the parallelogram be repre- sented by vectors AandBas shown in Fig. 1.7. The two diagonals are A−B andA+B.The vector from the bottom left corner to the mid-point of the diagonal A−Bis B+1 2(A−B)=1 2(A+B), which is also the half of the other diagonal A+B.Therefore they bisect each other. B AA − BA + B Fig. 1.7. Diagonals of a parallelogram bisect each other; diagonals of a rhombus (A=B) are perpendicular to each other 1.2 Vector Operations 9 Example 1.2.3. Prove that the diagonals of a rhombus (a parallelogram with equal sides) are orthogonal (perpendicular to each other). Solution 1.2.3. Again let the two adjacent sides be AandB.The dot product of the two diagonals (Fig. 1.7) is (A+B)·(A−B)=A·A+B·A−A·B−B·B=A2−B2. For a rhombus, A=B.Therefore the dot product of the diagonals is equal to zero. Hence they are perpendicular to each other. Example 1.2.4. Show that in a parallelogram, the two lines from one corner to the midpoints of the two opposite sides trisect the diagonal they cross. A BP M N OR S T Fig. 1.8. Two lines from one corner of a parallelogram to the midpoints of the two opposite sides trisect the diagonal they cross Solution 1.2.4. With the parallelogram shown in Fig. 1.8, it is clear that the line from O to the midpoint of RS is represented by the vector A+1 2B. A vector drawn from O to any point on this line can be written as r(λ)=λ/parenleftbigg A+1 2B/parenrightbigg , where λis a real number which adjusts the length of OP. The diagonal RT is represented by the vector B−A.A vector drawn from O to any point on this diagonal is r(µ)=A+µ(B−A), where the parameter µadjusts the length of the diagonal. The two lines meet when λ/parenleftbigg A+1 2B/parenrightbigg =A+µ(B−A), which can be written as (λ−1+µ)A+/parenleftbigg1 2λ−u/parenrightbigg B=0. 10 1 Vectors This gives µ=1 3andλ=2 3,so the length of RM is one-third of RT. Similarly, we can show the length NT is one-third of RT. Example 1.2.5. Show that an angle inscribed in a semicircle is a right angle. 0P AB −AB − AB + A Fig. 1.9. The circum-angle of a semicircle is a right angle Solution 1.2.5. With the semicircle shown in Fig. 1.9, it is clear the magni- tude of Ais the same as the magnitude of B,since they both equal to the radius of the circle A=B.Thus ( B−A)·(B+A)=B2−A2=0.Therefore (B−A) is perpendicular to ( B+A). 1.2.5 Vector Components For algebraic description of vectors, we introduce a coordinate system for the reference frame, although it is important to keep in mind that the magnitude and direction of a vector is independent of the reference frame. We will ﬁrst use the rectangular Cartesian coordinates to express vectors in terms of their components. Let ibe a unit vector in the positive xdirection, and jandkbe unit vectors in the positive yandzdirections. An arbitrary vector Acan be expanded in terms of these basis vectors as shown in Fig. 1.10: A=Axi+Ayj+Azk, (1.10) where Ax,Ay,andAzare the projections of Aalong the three coordinate axes, they are called components of A. Sincei,j,andkare mutually perpendicular unit vectors, by the deﬁnition of dot product i·i=j·j=k·k=1, (1.11) i·j=j·k=k·i=0. (1.12) Because the dot product is distributive, it follows that A·i=(Axi+Ayj+Azk)·i =Axi·i+Ayj·i+Azk·i=Ax, 1.2 Vector Operations 11 z A Ax i Ay jAz k Axk xyji Fig. 1.10. Vector components. i,j,kare three unit vectors pointing in the direction of positive x-,y-a n d z-axis, respectively. Ax,Ay,Azare the projections of Aon these axes. They are components of AandA=Axi+Ayj+Azk A·j=Ay,A·k=Az, the dot product of Awith any unit vector is the projection of A along the direction of that unit vector (or the component of Aalong that direction). Thus, (1.10) can be written as A=(A·i)i+(A·j)j+(A·k)k. (1.13) Furthermore, using the distributive law of dot product and (1.11) and (1.12), we have A·B=(Axi+Ayj+Azk)·(Bxi+Byj+Bzk) =AxBx+AyBy+AzBz, (1.14) and A·A=A2 x+A2 y+A2 z=A2. (1.15) SinceA·B=ABcosθ,the angle between AandBis given by θ=c o s−1A·B AB=c o s−1/parenleftbiggAxBx+AyBy+AzBz AB/parenrightbigg . (1.16) Example 1.2.6. Find the angle between A=3i+6j+9kandB=−2i+3j+k. Solution 1.2.6. A=( 32+62+92)1/2=3√ 14;B=/parenleftbig (−2)2+32+12/parenrightbig1/2=√ 14, A·B=3×(−2) + 6 ×3+9×1=2 1 cosθ=A·B AB=21 3√ 14√ 14=7 14=1 2, θ=c o s−1/parenleftbigg1 2/parenrightbigg =6 0◦. 12 1 Vectors Example 1.2.7. Find the angle θbetween the face diagonals AandBof a cube shown in Fig. 1.11. BA - B A ya a aq xz Fig. 1.11. The angle between the two face diagonals of a cube is 60◦ Solution 1.2.7. The answer can be easily found from geometry. The triangle formed by A,BandA−Bis clearly an equilateral triangle, therefore θ=6 0◦. Now with dot product approach, we have A=aj+ak;B=ai+ak; A=√ 2a=B, A·B=a·0+0·a+a·a=a2=ABcosθ=2a2cosθ. Therefore cosθ=a2 2a2=1 2,θ =6 0◦. Example 1.2.8. IfA=3i+6j+9kandB=−2i+3j+k,ﬁnd the projection ofAonB. Solution 1.2.8. The unit vector along Bis n=B B=−2i+3j+k√ 14. The projection of AonBis then A·n=1 BA·B=1√ 14(3i+6j+9k)·(−2i+3j+k)=21√ 14. Example 1.2.9. The angles between the vector Aand the three basis vectors i,j,andkare, respectively, α, β, andγ.Show that cos2α+cos2β+cos2γ=1. 1.2 Vector Operations 13 Solution 1.2.9. The projections of Aoni,j,kare, respectively, Ax=A·i=Acosα;Ay=A·j=Acosβ;Az=A·k=Acosγ. Thus A2 x+A2 y+A2 z=A2cos2α+A2cos2β+A2cos2γ=A2/parenleftbig cos2α+c o s2β+c o s2γ/parenrightbig . Since A2 x+A2 y+A2 z=A2,therefore cos2α+c o s2β+c o s2γ=1. The quantities cos α,cosβ,and cos γare often denoted l, m, andn,respec- tively, and they are called the direction cosine ofA. 1.2.6 Cross Product The vector cross product written as C=A×B (1.17) is another particular combination of the two vectors AandB,which is also very useful. It is deﬁned as a vector (therefore the alternative name: vector product ) with a magnitude C=ABsinθ, (1.18) where θis the angle between AandB,and a direction perpendicular to the plane of AandBin the sense of the advance of a right-hand screw as it is turned from AtoB.In other words, if the ﬁngers of your right hand point in the direction of the ﬁrst vector Aand curl around toward the second vector B,then your thumb will indicate the positive direction of Cas shown in Fig. 1.12 . ABC = A × B A B C = A × B Fig. 1.12. Right-hand rule of cross product A×B=C.If the ﬁngers of your right hand point in the direction of the ﬁrst vector Aand curl around toward the second vector B,then your thumb will indicate the positive direction of C 14 1 Vectors With this choice of direction, we see that cross product is anticommutative A×B=−B×A. (1.19) It is also clear that if AandBare parallel, then A×B=0,sinceθis equal to zero. From this deﬁnition, the cross products of the basis vectors ( i,j,k)c a n be easily obtained i×i=j×j=k×k=0, (1.20) i×j=−j×i=k, j×k=−k×j=i, k×i=−i×k=j. (1.21) The following example illustrates the cross product of two nonorthogonal vectors. If Vis a vector in the xz-plane and the angle between Vandk,the unit vector along the z-axis, is θas shown in Fig. 1.13, then k×V=Vsinθj. Since |k×V|=|k||V|sinθ=Vsinθis equal to the projection of Von the xy-plane, the vector k×Vis the result of rotating this projection 90◦around thezaxis. With this understanding, we can readily demonstrate the distributive law of the cross product A×(B+C)=A×B+A×C. (1.22) k × Vk V V sin qq yz x Fig. 1.13. The cross product of k,the unit vector along the z-axis, and V,av e c t o r in the xz-plane 1.2 Vector Operations 15 C MB A^ A^ × (B + C)A^ × B A^ × CB + CQ Q9 Q0 P0P9P O Fig. 1.14. Distributive law of cross product A×(B+C)=A×B+A×C Let the triangle formed by the vectors B,C,andB+Cbe arbitrarily oriented with respect to the vector Aas shown in Fig. 1.14. Its projection on the plane Mperpendicular to Ais the triangle OP/primeQ/prime. Turn this triangle 90◦around A, we obtain another triangle OP/prime/primeQ/prime/prime. The three sides of the triangle OP/prime/primeQ/prime/primeare /hatwideA×B,/hatwideA×C,a n d/hatwideA×(B+C),where/hatwideAis the unit vector along the direction ofA. It follows from the rule of vector addition that /hatwideA×(B+C)=/hatwideA×B+/hatwideA×C. Multiplying both sides by the magnitude A,we obtain (1.22). With the distributive law and (1.20) and (1.21), we can easily express the cross product A×Bin terms of the components of AandB: A×B=(Axi+Ayj+Azk)×(Bxi+Byj+Bzk) =AxBxi×i+AxByi×j+AxBzi×k +AyBxj×i+AyByj×j+AyBzj×k +AzBxk×i+AzByk×j+AzBzk×k =(AyBz−AzBy)i+(AzBx−AxBz)j+(AxBy−AyBx)k.(1.23) This cumbersome equation can be more neatly expressed as the determinant A×B=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk AxAyAz BxByBz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, (1.24) with the understanding that it is to be expanded about its ﬁrst row. The determinant form is not only easier to remember but also more convenient to use. The cross product has a useful geometrical interpretation. Figure 1.15 shows a parallelogram having AandBas co-terminal edges. The area of this parallelogram is equal to the base Atimes the height h.B u t h=Bsinθ, so 16 1 Vectors B Ah q Fig. 1.15. The area of the parallelogram formed by AandBis equal to the magnitude of A×B Parallelogram Area = Ah=ABsinθ=|A×B|. (1.25) Thus the magnitude of A×Bis equal to the area of the parallelogram formed byAandB,its direction is normal to the plane of this parallelogram. This suggests that area may be treated as a vector quantity. Since the area of the triangle formed by AandBas co-terminal edges is clearly half of the area of the parallelogram, so we also have Triangle Area =1 2|A×B|. (1.26) Example 1.2.10. The law of sine. With the triangle in Fig. 1.6, show that sinθ C=sinα A=sinβ B. Solution 1.2.10. The area of the triangle is equal to1 2|A×B|=1 2ABsinθ. The same area is also given by1 2|A×C|=1 2ACsinβ.Therefore, ABsinθ=ACsinβ. It followssinθ C=sinβ B.Similarly,sinθ C=sinα A.Hence sinθ C=sinα A=sinβ B. Lagrange Identity The magnitude of |A×B|c a nb ee x p r e s s e di nt e r m so f A,B,andA·B through the equation |A×B|2=A2B2−(A·B)2, (1.27) known as the Lagrange identity. This relation follows from the fact 1.2 Vector Operations 17 |A×B|2=(ABsinθ)2=A2B2(1−cos2θ) =A2B2−A2B2cos2θ=A2B2−(A·B)2. This relation can also be shown by the components of the vectors. It follows from (1 .24) that |A×B|2=(AyBz−AzBy)2+(AzBx−AxBz)2+(AxBy−AyBx)2(1.28) and A2B2−(A·B)2=(A2 x+A2 y+A2 z)(B2 x+B2 y+B2 z)−(AxBx+AyBy+AzBz)2. (1.29) Multiplying out the right-hand sides of these two equations, we see that they are identical term by term. 1.2.7 Triple Products Scalar Triple Product The combination ( A×B)·Cis known as the triple scalar product. A×Bis a vector. The dot product of this vector with the vector Cgives a scalar. The triple scalar product has a direct geometrical interpretation. The three vectors can be used to deﬁne a parallelopiped as shown in Fig. 1.16. The magnitude ofA×Bis the area of the parallelogram base and its direction is normal (perpendicular) to the base. The projection of Conto the unit normal of the base is the height hof the parallelopiped. Therefore, ( A×B)·Cis equal to the area of the base times the height which is the volume of the parallelopiped: Parallelopiped Volume = Area×h=|A×B|h=(A×B)·C. The volume of a tetrahedron is equal to one-third of the height times the area of the triangular base. Thus the volume of the tetrahedron formed by the A 3 B C ABh Fig. 1.16. The volume of the parallelopiped is equal to the triple scalar product of its edges as vectors 18 1 Vectors vectors A,B,andCas concurrent edges is equal to one-sixth of the scalar triple product of these three vectors: Tetrahedron Volume =1 3h×1 2|A×B|=1 6(A×B)·C. In calculating the volume of the parallelopiped we can consider just as wellB×CorC×Aas the base. Since the volume is the same regardless of which side we choose as the base, we see that (A×B)·C=A·(B×C)=(C×A)·B. (1.30) The parentheses in this equation are often omitted, since the cross product must be performed ﬁrst. If the dot product were performed ﬁrst, the expres- sion would become a scalar crossed into a vector, which is an undeﬁned and meaningless operation. Without the parentheses, A×B·C=A·B×C,w e see that in any scalar triple product, the dot and the cross can be interchanged without altering the value of the product. This is an easy way to remember this relation. It is clear that if his reduced to zero, the volume will become zero also. Therefore if Cis in the same plane as AandB,the scalar triple product (A×B)·Cvanishes. In particular (A×B)·A=(A×B)·B=0. (1.31) A convenient expression in terms of components for the triple scalar prod- uct is provided by the determinant A·(B×C)=(Axi+Ayj+Azk)·/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk BxByBz CxCyCz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleAxAyAz BxByBz CxCyCz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle.(1.32) The rules for interchanging rows of a determinant provide another veriﬁcation of (1.30) (A×B)·C=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleCxCyCz AxAyAz BxByBz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleAxAyAz BxByBz CxCyCz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleBxByBz CxCyCz AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Vector Triple Product The triple product A×(B×C) is a meaningful operation, because B×C is a vector, and can form cross product with Ato give another vector (hence the name vector triple product). In this case, the parentheses are necessary, because A×(B×C) and ( A×B)×Care two diﬀerent vectors. For example, i×(i×j)=i×k=−jand ( i×i)×j=0×j=0. 1.2 Vector Operations 19 The relation A×(B×C)=(A·C)B−(A·B)C (1.33) is a very important identity. Because of its frequent use in a variety of prob- lems, this relation should be memorized. This relation (sometimes known as ACB–ABC rule) can be veriﬁed by the direct but tedious method of expand- ing both sides into their cartesian components. A vector equation is, of course, independent of any particular coordinate system. Therefore, it might be more instructive to prove (1 .33) without coordinate components. Let (B×C)=D,henceDis perpendicular to the plane of BandC.Now the vector A×(B×C)=A×Dis perpendicular to D,therefore it is in the plane of BandC.Thus we can write A×(B×C)=αB+βC, (1.34) where αandβare scalar constants. Furthermore, A×(B×C)i sa l s op e r - pendicular to A.So, the dot product of Awith this vector must be zero: A·[A×(B×C)] =αA·B+βA·C=0. It follows that β=−αA·B A·C and (1.34) becomes A×(B×C)=α A·C[(A·C)B−(A·B)C]. (1.35) This equation is valid for any set of vectors. For the special case B=A,this equation reduces to A×(A×C)=α A·C[(A·C)A−(A·A)C]. (1.36) Take the dot product with C,we have C·[A×(A×C)] =α A·C[(A·C)2−A2C2]. (1.37) Recall the property of the scalar triple product C·(A×D)=(C×A)·D, withD=(A×C) the left-hand side of the last equation becomes C·[A×(A×C)] = (C×A)·(A×C)=−|A×C|2. Using the Lagrange identity (1 .27) to express |A×C|2,we have C·[A×(A×C)] =−[(A2C2−(A·C)2]. (1.38) Comparing (1 .37) and (1 .38) we see that α A·C=1, and (1 .35) reduces to the ACB–ABC rule of (1 .33). All higher vector products can be simpliﬁed by repeated application of scalar and vector triple products. 20 1 Vectors Example 1.2.11. Use the scalar triple product to prove the distributive law of cross product: A×(B+C)=A×B+A×C. Solution 1.2.11. First take a dot product D·A×(B+C) with an arbitrary vector D,then regard ( B+C) as one vector: D·A×(B+C)=D×A·(B+C)=D×A·B+D×A·C =D·A×B+D·A×C=D·[A×B+A×C]. (The ﬁrst step is evident because dot and cross can be interchanged in the scalar triple product; in the second step we regard D×Aas one vector and use the distributive law of the dot product; in the third step we interchange dot and cross again; in the last step we use again the distributive law of the dot product to factor out D.) Since Dcan be any vector, it follows that A×(B+C)=A×B+A×C. Example 1.2.12. Prove the general form of the Lagrange identity: (A×B)·(C×D)=(A·C)(B·D)−(A·D)(B·C). Solution 1.2.12. First regard ( C×D) as one vector and interchange the cross and the dot in the scalar triple product ( A×B)·(C×D): (A×B)·(C×D)=A×B·(C×D)=A·B×(C×D). Then we expand the vector triple product B×(C×D)i n A·B×(C×D)=A·[(B·D)C−(B·C)D]. Since ( B·D) and ( B·C) are scalars, the distributive law of dot product gives A·[(B·D)C−(B·C)D]=(A·C)(B·D)−(A·D)(B·C). Example 1.2.13. The dot and cross products of uwithAare given by A·u=C; A×u=B. Express uin terms of A,B,andC. Solution 1.2.13. A×(A×u)=(A·u)A−(A·A)u=CA−A2u, A×(A×u)=A×B. CA−A2u=A×B u=1 A2[CA−A×B]. 1.2 Vector Operations 21 Example 1.2.14. The force Fexperienced by the charge qmoving with velocity Vin the magnetic ﬁeld Bis given by the Lorentz force equation F=q(V×B). In three separate experiments, it was found V=i,F/q=2k−4j, V=j,F/q=4i−k, V=k,F/q=j−2i. From these results determine the magnetic ﬁeld B. Solution 1.2.14. These results can be expressed as i×B=2k−4j(1) ;j×B=4i−k(2) ;k×B=j−2i(3). From (1) i×(i×B)=i×(2k−4j)=−2j−4k, i×(i×B)=(i·B)i−(i·i)B=Bxi−B; therefore, Bxi−B=−2j−4korB=Bxi+2j+4k. From (2) k·(j×B)=k·(4i−k)=−1, k·(j×B)=(k×j)·B=−i·B. Thus, −i·B=−1,orBx=1. The ﬁnal result is obtained just from these two conditions B=i+2j+4k. We can use the third condition as a consistency check k×B=k×(i+2j+4k)=j−2i, which is in agreement with (3) . 22 1 Vectors Example 1.2.15. Reciprocal vectors. Ifa,b,care three noncoplanar vectors, a/prime=b×c a·b×c,b/prime=c×a a·b×c,c/prime=a×b a·b×c are known as the reciprocal vectors. Show that any vector rcan be expressed as r=(r·a/prime)a+/parenleftbig r·b/prime/parenrightbig b+(r·c/prime)c. Solution 1.2.15. Method I. Consider the vector product ( r×a)×(b×c). First, regard ( r×a) as one vector and expand (r×a)×(b×c)=[ (r×a)·c]b−[(r×a)·b]c. Then, regard ( b×c) as one vector and expand (r×a)×(b×c)=[ (b×c)·r]a−[(b×c)·a]r. Therefore, [(r×a)·c]b−[(r×a)·b]c=[(b×c)·r]a−[(b×c)·a]r or [(b×c)·a]r=[ (b×c)·r]a−[(r×a)·c]b+[ (r×a)·b]c. Since −(r×a)·c=−r·(a×c)=r·(c×a), (r×a)·b=r·(a×b), it follows that r=r·(b×c) (b×c)·aa+r·(c×a) (b×c)·ab+r·(a×b) (b×c)·ac =(r·a/prime)a+/parenleftbig r·b/prime/parenrightbig b+(r·c/prime)c. Method II. Let r=q1a+q2b+q3c. r·(b×c)=q1a·(b×c)+q2b·(b×c)+q3c·(b×c). Since ( b×c) is perpendicular to band perpendicular to c,therefore b·(b×c)=0,c·(b×c)=0. Thus q1=r·(b×c) a·(b×c)=r·a/prime. 1.3 Lines and Planes 23 Similarly, q2=r·(c×a) b·(c×a)=r·(c×a) a·(b×c)=r·b/prime, q3=r·(a×b) c·(a×b)=r·c/prime. It follows that r=(r·a/prime)a+/parenleftbig r·b/prime/parenrightbig b+(r·c/prime)c. 1.3 Lines and Planes Much of analytic geometry can be simpliﬁed by the use of vectors. In analytic geometry, a point is a set of three coordinates ( x,y,z).All points in space can be deﬁned by the position vector r(x,y,z) (or just r) r=xi+yj+zk, (1.39) drawn from the origin to the point ( x,y,z).To specify a particular point (x0,y0,z0),we use the notation r0(x0,y0,z0) r0=x0i+y0j+z0k. (1.40) With these notations, we can deﬁne lines and planes in space. 1.3.1 Straight Lines There are several ways to specify a straight line in space. Let us ﬁrst consider a line through a given point ( x0,y0,z0) in the direction of a known vector v=ai+bj+ck.Ifr(x,y,z) is any other point on the line, the vector r−r0 is parallel to v.Thus, we can write the equation of a straight line as r−r0=tv, (1.41) where tis any real number. This equation is called the parametric form of a straight line. It is inﬁnitely long and ﬁxed in space, as shown in Fig. 1.17. It cannot be moved parallel to itself as a free vector. This equation in the form of its components (x−x0)i+(y−y0)j+(z−z0)k=tai+tbj+tck (1.42) represents three equations (x−x0)=ta, (y−y0)=tb, (z−z0)=tc. (1.43) 24 1 Vectors r0 xyz r Vr = r0 + t Vr − r0 (x, y, z)(x0, y0, z0) Fig. 1.17. A straight line in the parametric form Now if a,b,c are not zero, we can solve for tin each of the three equations. The solutions must be equal to each other, since they are all equal to the same t.x−x0 a=y−y0 b=z−z0 c. (1.44) This is called the symmetric form of the equation of a line. If vis a normalized unit vector, then a,b,c are the direction cosines of the line. Ifchappens to be zero, then (1.43) should be written as x−x0 a=y−y0 b;z=z0. (1.45) The equation z=z0means that the line lies in the plane perpendicular to thez-axis, and the slope of the line isb a.If both bandcare zero, then clearly the line is the intersection of the planes y=y0andz=z0. The parametric equation (1.41) has a useful interpretation when the para- meter tmeans time. Consider a particle moving along this straight line. The equation r=r0+tvindicates when t=0,the particle is at r0. As time goes on, the particle is moving with a constant velocity v,ordr dt=v. Perpendicular Distance Between Two Skew Lines Two lines which are not parallel and which do not meet are said to be skew lines. To ﬁnd the perpendicular distance between them is a diﬃcult problem in analytical geometry. With vectors, it is relatively easy. Let the equations of two such lines be r=r1+tv1, (1.46) r=r2+t/primev2. (1.47) 1.3 Lines and Planes 25 Letaon line 1 and bon line 2 be the end points of the common perpendicular on these two lines. We shall suppose that the position vector rafrom origin toais given by (1 .46) with t=t1,and the position vector rb,by (1.47) with t/prime=t2.Accordingly ra=r1+t1v1, (1.48) rb=r2+t2v2. (1.49) Sincerb−rais perpendicular to both v1andv2,it must be in the direction of (v1×v2).Ifdis the length of rb−ra,then rb−ra=v1×v2 |v1×v2|d. Sincerb−ra=r2−r1+t2v2−t1v1, r2−r1+t2v2−t1v1=v1×v2 |v1×v2|d. (1.50) Then take the dot product with v1×v2on both sides of this equation .Since v1×v2·v1=v1×v2·v2=0,the equation becomes (r2−r1)·(v1×v2)=|v1×v2|d; therefore, d=(r2−r1)·(v1×v2) |v1×v2|. (1.51) This must be the perpendicular distance between the two lines. Clearly, if d=0,the two lines meet. Therefore the condition for the two lines to meet is (r2−r1)·(v1×v2)=0. (1.52) To determine the coordinates of aandb, take the dot product of (1 .50) ﬁrst with v1,then with v2 (r2−r1)·v1+t2v2·v1−t1v1·v1=0, (1.53) (r2−r1)·v2+t2v2·v2−t1v1·v2=0. (1.54) These two equations can be solved for t1andt2.With them, raandrbcan be found from (1 .48) and (1 .49). Example 1.3.1. Find the coordinates of the end points aandbof the common perpendicular to the following two lines r=9j+2k+t(3i−j+k), r=−6i−5j+1 0k+t/prime(−3i+2j+4k). 26 1 Vectors Solution 1.3.1. The ﬁrst line passes through the point r1(0,9,2) in the direction of v1=3i−j+k.The second line passes the point r2(−6,−5,10) in the direction of v2=−3i+2j+4k.From (1 .53) and (1 .54), (r2−r1)·v1+t2v2·v1−t1v1·v1=4−7t2−11t1=0, (r2−r1)·v2+t2v2·v2−t1v1·v2=2 2+2 9 t2+7t1=0. The solution of these two equations is t1=1,t2=−1. Therefore, by (1 .48) and (1 .49), ra=r1+t1v1=3i+8j+3k, rb=r2+t2v2=−3i−7j+6k. Example 1.3.2. Find the perpendicular distance between the two lines of the previous example, and an equation for the perpendicular line. Solution 1.3.2. The perpendicular distance dis simply d=|ra−rb|=|6i+1 5j−3k|=3√ 30. It can be readily veriﬁed that this is the same as given by (1 .51).The perpen- dicular line can be represented by the equation r=ra+t(ra−rb)=( 3+6 t)i+( 8+1 5 t)j+( 3−3t)k, or equivalently x−3 6=y−8 15=z−3 −3. This line can also be represented by r=rb+s(ra−rb)=(−3+6s)i+(−7+1 5 s)j+( 6−3s)k, orx+3 6=y+7 15=z−6 −3. Example 1.3.3. Find (a) the perpendicular distance of the point (5 ,4,2) from the linex+1 2=y−3 3=z−1 −1, and (b) also the coordinates of the point where the perpendicular meets the line, and (c) an equation for the line of the perpendicular. 1.3 Lines and Planes 27 Solution 1.3.3. The line passes the point r0=−i+3j+ka n di si nt h e direction of v=2i+3j−k.The parametric form of the line is r=r0+tv=−i+3j+k+t(2i+3j−k). Let the position vector to (5 ,4,2) be r1=5i+4j+2k. The distance dfrom the point r1(5,4,2) to the line is the cross product of (r1−r0) with the unit vector in the vdirection d=/vextendsingle/vextendsingle/vextendsingle(r1−r0)×v v/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle(6i+j+k)×2i+3j−k√4+9+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2√ 6. Letpbe the point where the perpendicular meets the line .Since pis on the given line, the position vector to pmust satisfy the equation of the given line with a speciﬁc t.Let that speciﬁc tbet1, rp=−i+3j+k+t1(2i+3j−k). Since ( rp−r0) is perpendicular to v,their dot product must be zero (r1−rp)·v=[ 6i+j+k−t1(2i+3j−k)]·(2i+3j−k)=0. This gives t1=1.It follows that rp=−i+3j+k+1( 2i+3j−k)=i+6j. In other words, the coordinates of the foot of the perpendicular is (1 ,6,0). The equation of the perpendicular can be obtained from the fact that it passes r1(orrp) and is in the direction of the vector from rptor1.So the equation can be written as r=r1+t(r1−rp)=5i+4j+2k+t(4i−2j+2k) orx−5 4=y−4 −2=z−2 2. 1.3.2 Planes in Space A set of parallel planes in space can be determined by a vector normal (per- pendicular) to these planes. A particular plane can be speciﬁed by an addi- tional condition, such as the perpendicular distance between the origin and the plane, or a given point that lies on the plane. 28 1 Vectors z (x,y,z) yr n n D x Fig. 1.18. A plane in space. The position vector rfrom the origin to any point on the plane must satisfy the equation r·n=Dwhere nis the unit normal to the plane and Dis the perpendicular distance between the origin and the plane Suppose the unit normal to the plane is known to be n=Ai+Bj+Ck (1.55) and the distance between this plane and the origin is Das shown in Fig. 1.18. If (x,y,z) is a point (any point) on the plane, then it is clear from the ﬁgure that the projection of r(x,y,z)o nnmust satisfy the equation r·n=D. (1.56) Multiplying out its components, we have the familiar equation of a plane Ax+By+Cz=D. (1.57) This equation will not be changed if both sides are multiplied by the same constant. The result represents, of course, the same plane. However, it is to be emphasized that if the right-hand side Dis interpreted as the distance between the plane and the origin, then the coeﬃcients A,B,C on the left- hand side must satisfy the condition A2+B2+C2=1,since they should be the direction cosine of the unit normal. The plane can also be uniquely speciﬁed if in addition to the unit normal n,a point ( x0,y0,z0) that lies on the plane is known. In this case, the vector r−r0must be perpendicular to n. That means the dot product with nmust vanish, (r−r0)·n=0. (1.58) Multiplying out the components, we can write this equation as Ax+By+Cz=Ax0+By0+Cz0. (1.59) 1.3 Lines and Planes 29 This is in the same form of (1 .57).Clearly the distance between this plane and the origin is D=Ax0+By0+Cz0. (1.60) In general, to ﬁnd distances between points and lines or planes, it is far simpler to use vectors as compared with calculations in analytic geometry without vectors. Example 1.3.4. Find the perpendicular distance from the point (1 ,2,3) to the plane described by the equation 3 x−2y+5z=1 0. Solution 1.3.4. The unit normal to the plane is n=3√9+4+2 5i−2√9+4+2 5j+5√9+4+2 5k. The distance from the origin to the plane is D=10√9+4+2 5=10√ 38. The length of the projection of r1=i+2j+3konnis /lscript=r1·n=3√ 38−4√ 38+15√ 38=14√ 38. The distance from (1 ,2,3) to the plane is therefore d=|/lscript−D|=14√ 38−10√ 38=4√ 38. Another way to ﬁnd the solution is to note that the required distance is equal to the projection on nof any vector joining the given point with a point on the plane. Note that r0=Dn=30 38i−20 38j+50 38k is the position vector of the foot of the perpendicular from the origin to the plane. Therefore d=|(r1−r0)·n|=4√ 38. Example 1.3.5. Find the coordinates of the foot of the perpendicular from the point (1 ,2,3) to the plane of the last example. 30 1 Vectors Solution 1.3.5. Let the position vector from the origin to the foot of the perpendicular be rp.The vector r1−rpis perpendicular to the plane, therefore it is parallel to the unit normal vector nof the plane, r1−rp=kn. It follows that |r1−rp|=k.Since|r1−rp|=d,sok=d.Thus, rp=r1−dn=i+2j+3k−4√ 38/parenleftbigg3√ 38i−2√ 38j+5√9+4+2 5k/parenrightbigg . Hence, the coordinates of the foot of the perpendicular are/parenleftbig26 38,84 38,94 38/parenrightbig . Example 1.3.6. A plane intersects the x,y,a n dzaxes, respectively, at ( a,0,0), (0,b,0),and (0 ,0,c) (Fig. 1.19). Find ( a) a unit normal to this plane, ( b)t h e perpendicular distance between the origin and this plane, ( c) the equation for this plane. xyz n (a,0,0)(0,b,0)(0,0,c) Fig. 1.19. The plane bcx+acy+abz=abccuts the three axes at ( a,0,0), (0,a ,0),(0,0,a),respectively Solution 1.3.6. Letr1=ai,r2=bj,r3=ck.The vector from ( a,0,0) to (0,b,0) isr2−r1=bj−ai,and the vector from ( a,0,0) to (0 ,0,c)i s r3−r1=ck−ai.The unit normal to this plane must be in the same direction as the cross product of these two vectors: n=(bj−ai)×(ck−ai) |(bj−ai)×(ck−ai)|, (bj−ai)×(ck−ai)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k −ab0 −a0c/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=bci+acj+abk, 1.3 Lines and Planes 31 n=bci+acj+abk /parenleftBig (bc)2+(ac)2+(ab)2/parenrightBig1/2. If the perpendicular distance from the origin to the plane is D,then D=r1·n=r2·n=r3·n, D=abc /parenleftBig (bc)2+(ac)2+(ab)2/parenrightBig1/2. In general, the position vector r=xi+yj+zkto any point ( x,y,z)o nt h e plane must satisfy the equation r·n=D, xbc+yac+zab /parenleftBig (bc)2+(ac)2+(ab)2/parenrightBig1/2=abc /parenleftBig (bc)2+(ac)2+(ab)2/parenrightBig1/2. Therefore the equation of this plane can be written as bcx+acy+abz=abc or asx a+y b+z c=1. Another way to ﬁnd an equation for the plane is to note that the scalar triple product of three coplanar vectors is equal to zero. If the position vector from the origin to any point ( x,y,z) on the plane is r=xi+yj+zk,then the three vectors r−ai,bj−aiandck−aiare in the same plane. Therefore, (r−ai)·(bj−ai)×(ck−ai)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglex−ayz −ab0 −a0c/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =bc(x−a)+acy+abz=0 or bcx+acy+abz=abc. Exercises 1. If the vectors A=2i+3kandB=i−k,ﬁnd|A|,|B|,A+B,A−B, andA·B.What is the angle between the vectors AandB? Ans.√ 13,√ 2,3i+2k,i+4k,−1,101◦. 32 1 Vectors 2. For what value of care the vectors ci+j+kand−i+2kperpendicular? Ans. 2. 3. IfA=i+2j+2kandB=−6i+2j+3k,ﬁnd the projection of AonB, and the projection of BonA. Ans. 4/7, 4/3. 4. Show that the vectors A=3i−2j+k,B=i−3j+5k,a n dC=2i+j−4k form a right triangle. 5. Use vectors to prove that the line joining the midpoints of two sides of any triangle is parallel to the third side and half its length. 6. Use vectors to show that for any triangle, the medians (the three lines drawn from each vertex to the midpoint of the opposite side) all pass the same point. The point is at two-thirds of the way of the median from the vertex. 7. IfA=2i−3j−kandB=i+4j−2k,ﬁndA×BandB×A. Ans. 10 i+3j+1 1k,−10i−3j−11k. 8. Find the area of a parallelogram having diagonals A=3i+j−2kand B=i−3j+4k. Ans. 5√ 3. 9. Evaluate (2 i−3j)·[(i+j−k)×(3j−k)]. Ans. 4. 10. Find the volume of the parallelepied whose edges are represented by A= 2i−3j+4k,B=i+2j−k,a n dC=3i−j+2k. Ans. 7. 11. Find the constant asuch that the vectors 2 i−j+k,i+2j−3kand 3i+aj+5kare coplanar. Ans.a=−4. 12. Show that (a) ( b×c)×(c×a)=c(a·b×c); (b) (a×b)·(b×c)×(c×a)=(a·b×c)2. Hint: to prove (a) ﬁrst regard ( b×c) as one vector, then note b×c·c=0 1.3 Lines and Planes 33 13. Ifa,b,care non-coplanar (so a·b×c/negationslash=0),and a/prime=b×c a·b×c,b/prime=c×a a·b×c,c/prime=a×b a·b×c, show that (a)a/prime·a=b/prime·b=c/prime·c=1, (b)a/prime·b=a/prime·c=0,b/prime·a=b/prime·c=0,c/prime·a=c/prime·b=0, (c) ifa·b×c=Vthena/prime·b/prime×c/prime=1/V. 14. Find the perpendicular distance from the point ( −1,0,1) to the line r= 3i+2j+3k+t(i+2j+3k). Ans.√ 10. 15. Find the coordinates of the foot of the perpendicular from the point (1,2,1) to the line joining the origin to the point (2 ,2,5). Ans. (2 /3,2/3,5/3). 16. Find the length and equation of the line which is the common perpendic- ular to the two lines x−4 2=y+2 1=z−3 −1,x+7 3=y+2 2=z−1 1. Ans.√ 35,x−2 3=y+3 −5=z−4 1. 17. Find the distance from ( −2,4,5) to the plane 2 x+6y−3z=1 0. Ans. 5/7 18. Find the equation of the plane that is perpendicular to the vector i+j−k and passes through the point (1 ,2,1). Ans.x+y−z=2. 19. Find an equation for the plane determined by the points (2 ,−1,1), (3,2,−1), and ( −1,3,2). Ans. 11 x+5y+1 3z=3 0. 2 Vector Calculus So far we have been discussing constant vectors, but the most interesting applications of vectors involve vector functions. The simplest example is a position vector that depends on time. Such a vector can be diﬀerentiated with respect to time. The ﬁrst and second derivatives are simply the velocity and acceleration of the particle whose position is given by the position vector. In this case, the coordinates of the tip of the position vector are functions of time. Even more interesting are quantities which depend on the position in space. Such quantities are said to form ﬁelds. The word “ﬁeld” has the connotation that the space has some physical properties. For example, the electrical ﬁeld created by a static charge is that space surrounding the charge which has now been given a certain property, known as the electrical ﬁeld. Every point in this ﬁeld is associated with an electric ﬁeld vector whose magnitude and direction depend on the location of the point. This electric ﬁeld vector will manifest itself when another charge is brought to that point. Mathematically a vector ﬁeld is simply a vector function, each of its three components depends on the coordinates of the point. The ﬁeld may also dependent on time, such as the electric ﬁeld in the electromagnetic wave. There are scalar ﬁelds, by which we mean that the ﬁeld is characterized at each point by a single number. Of course the number may change in time, but usually we are talking about the ﬁeld at a given instant. For example, temperature at diﬀerent point in space is diﬀerent, so the temperature is a function of the position. Thus temperature is a scalar ﬁeld. It is possible to derive one kind of ﬁeld from another. For example, the directional derivatives of a scalar ﬁeld lead to a vector ﬁeld, known as the gradient. With vector diﬀerential calculus, we develop a set of precise terms, such as gradient, divergence, and curl, to describe the rate of change of vector func- tions with respect to the spatial coordinates. With vector integral calculus, we establish relationships between line, surface, and volume integrals through the theorems of Gauss and Stokes. These are important attributes of vector ﬁelds, in terms of which many fundamental laws of physics are expressed. 36 2 Vector Calculus In this chapter, we shall assume that the vector functions are continu- ous and diﬀerentiable, and the region of interests is simply connected unless otherwise speciﬁed. However, this does not mean that singularities and multiple connected regions are not of our concern. They have important impli- cations in physical problems. We will more carefully deﬁne and discuss these terms at appropriate places. 2.1 The Time Derivative Diﬀerentiating a vector function is a simple extension of diﬀerentiating scalar quantities. If the vector Adepends on time tonly,then the derivative of A with respect to tis deﬁned as dA dt= lim ∆t→0A(t+∆t)−A(t) ∆t= lim ∆t→0∆A ∆t. (2.1) From this deﬁnition it follows that the sums and products involving vector quantities can be diﬀerentiated as in ordinary calculus; that is d dt(A+B)=dA dt+dB dt, (2.2) d dt(A·B)=AdB dt+dA dt·B, (2.3) d dt(A×B)=A×dB dt+dA dt×B. (2.4) Since ∆ Ahas components ∆ Ax,∆Ay, and ∆ Az, dA dt= lim ∆t→0∆Axi+∆Ayj+∆Azk ∆t=dAx dti+dAy dtj+dAz dtk. (2.5) The time derivatives of a vector is thus equal to the vector sum of the time derivative of its components. 2.1.1 Velocity and Acceleration Of particular importance is the case where Ais the position vector r, r(t)=x(t)i+y(t)j+z(t)k. (2.6) Iftchanges, the tip of rtraces out a space curve as shown in Fig. 2.1. If a particle is moving along this space curve, then d r/dtis clearly the velocity v of the particle along this trajectory v=dr dt= lim ∆t→0∆r ∆t= lim ∆t→0∆xi+∆yj+∆zk ∆t=dx dti+dy dtj+dz dtk=vxi+vyj+vzk. (2.7) 2.1 The Time Derivative 37 xyz r(t ) r(t + ∆t )∆r Fig. 2.1. The tip of rtraces out the trajectory of a particle moving in space, ∆ ris independent of the origin It is important to note that the direction of ∆ ris unrelated to the direction ofr.In other words the velocity is independent of the origin chosen. Similarly, the acceleration is deﬁned as the rate of change of velocity a=dv dt=dvx dti+dvy dtj+dvz dtk=d2x dt2i+d2y dt2j+d2z dt2k=d2r dt2.(2.8) The acceleration is also independent of the origin. Notation of diﬀerentiation with respect to time. A convenient and widely used notation (Newton’s notation) is that a single dot above a symbol denotes the ﬁrst time derivative and two dots denote the second time derivative, and so on. Thus v=dr dt=/squaresmallsolidr=/squaresmallsolidxi+/squaresmallsolidyj+/squaresmallsolidzk, (2.9) a=/squaresmallsolidv=/squaresmallsolid/squaresmallsolidr=/squaresmallsolid/squaresmallsolidxi+/squaresmallsolid/squaresmallsolidyj+/squaresmallsolid/squaresmallsolidzk. (2.10) 2.1.2 Angular Velocity Vector For a particle moving around a circle, shown in Fig. 2.2, the rate of change of the angular position is called angular speed ω: ω= lim ∆t→0∆θ ∆t=dθ dt=/squaresmallsolid θ. (2.11) The velocity vof the particle is, by deﬁnition, v=dr dt=/squaresmallsolidr, (2.12) 38 2 Vector Calculus w nar ∆s∆q Or(t) r(t + ∆t) Fig. 2.2. Angular velocity vector ω.The velocity vof a particle moving around a circle is given by v=ω×r where ris the position vector drawn from the origin to the position of the particle. The magnitude of the velocity is given by v=|v|= lim ∆t→0∆s ∆t= lim ∆t→0/rho1∆θ ∆t=/rho1ω, (2.13) where /rho1is the radius of the circle. The direction of the velocity is, of course, tangent to the circle. Now, let nbe the unit vector drawn from the origin to the center of the circle, pointing in the positive direction of advance of a right-hand screw when turned in the same sense as the rotation of the particle. Since |n×r|=rsinα=/rho1,as shown in Fig. 2.2, the magnitude of the velocity can be written as v=/rho1ω=|n×r|ω. (2.14) If we deﬁne the angular velocity vector ωas ω=ωn, (2.15) then we can write the velocity vas v=/squaresmallsolidr=ω×r. (2.16) Recalling the deﬁnition of cross product of two vectors, one can easily see that both direction and magnitude of the velocity are given by this equation. A particle moving in space, even though not in a circle, may always be considered at a given instant to be moving in a circular path. Even a straight line can be considered as a circle with inﬁnite radius. The path, which the particle describes during an inﬁnitesimal time interval δt,may be represented as an inﬁnitesimal arc of a circle. Therefore, at any moment, an instantaneous angular velocity vector can be deﬁned to describe the general motion. The instantaneous velocity is then given by (2.16). 2.1 The Time Derivative 39 Example 2.1.1. Show that the linear momentum, deﬁned as p=m/squaresmallsolidr, always lies in a ﬁxed plane in a central force ﬁeld. (A central force ﬁeld means that the force Fis in the radial direction, such as gravitational and electrostatic forces, in other words Fis parallel to r.) Solution 2.1.1. Let us form the angular momentum L L=r×p=r×m/squaresmallsolidr. Diﬀerentiating with respect to time, we have /squaresmallsolid L=/squaresmallsolidr×p+r×/squaresmallsolidp. Now/squaresmallsolidr×p=/squaresmallsolidr×m/squaresmallsolidr=0andr×/squaresmallsolidp=r×m/squaresmallsolid/squaresmallsolidr.According to Newton’s second lawm/squaresmallsolid/squaresmallsolidr=FandFis parallel to r,therefore r×F=r×m/squaresmallsolid/squaresmallsolidr=0.Thus,/squaresmallsolid L=0. In other words, Lis a constant vector. Furthermore, Lis perpendicular to p, sincer×pis perpendicular to p. Therefore pmust always lie in the plane perpendicular to the constant vector L. Example 2.1.2. Suppose a particle is rotating around the z-axis with a con- stant angular velocity ωas shown in Fig. 2.3. Find the velocity and acceleration of the particle. Y XrV(x, y) P Oy xq = w t Fig. 2.3. Particle rotating around z-axis with a constant angular velocity ω Solution 2.1.2. Method I . Since the particle is moving in a circular path and zis not changing in this motion, we will consider only the xandycomponents. The angular velocity vector is in the k(unit vector along the z axis) direction, ω=ωk,and the position vector rdrawn from the origin to the point ( x,y) is perpendicular to k.Therefore v=ω×r. 40 2 Vector Calculus The direction of vis perpendicular to kand perpendicular to r,t h a ti s ,i n the tangential direction of the circle. The magnitude of the velocity is v=ωrsin(π/2) =ωr. Explicitly v=ω×r=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk 00ω xy0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−ωyi+ωxj. The acceleration is given by a=dv dt=d dt(ω×r)=/squaresmallsolidω×r+ω×/squaresmallsolidr. Since ωis a constant,/squaresmallsolidω=0.Moreover,/squaresmallsolidr=v=ω×r.Thus a=ω×(ω×r)=ω2k×(k×r). Hence, ais in the −rdirection and its magnitude is equal to ω2r. Method II . The position vector can be explicitly written as r=x(t)i+y(t)j=rcosωti+rsinωtj. (2.17) The velocity and acceleration are, respectively, v=/squaresmallsolidr=/squaresmallsolidxi+/squaresmallsolidyj=−ωrsinωti+ωrcosωtj=ω(−yi+xj), (2.18) v=(v·v)1/2=/parenleftbig ω2r2sin2ωti+ω2r2cos2ωt/parenrightbig1/2=ωr (2.19) a=/squaresmallsolidv=−ω2cosωti−ω2sinωtj=−ω2(xi+yj)=−ω2r=−v2 r2r.(2.20) We see immediately that the acceleration is toward the center with a magnitude of ω2r.This is the familiar centripetal acceleration . Also the velocity is perpendicular to the position vector since v·r=ω(−yi+xj)·(xi+yj)=0. In this example, the magnitude of ris a constant, we have explicitly shown that the velocity is perpendicular to r.This fact is also a consequence of the following general theorem. If the magnitude of a vector is not changing, the vector is always ortho- gonal (perpendicular) to its derivative . 2.1 The Time Derivative 41 This follows from the fact that if r·r=r2 0andr0is a constant, then d dt(r·r)=d dtr2 0=0, d dt(r·r)=dr dt·r+r·dr dt=2dr dt·r=0. When the dot product of two vectors is zero, the two vectors are perpendicular to each other. This can also be seen from geometry. In Fig. 2.1, if r(t)a n dr(t+∆t)h a v e the same length, then ∆ ris the base of an isosceles triangle. When ∆ t→0, the angle between r(t)a n dr(t+∆t) also goes to zero. In that case, the two base angles approach 90◦, which means ∆ ris perpendicular to r.Since ∆ t is a scalar, the direction of ∆ r/∆tis determined by ∆ r.Therefore d r/dtis perpendicular to r. This theorem is limited neither to the position vector, nor to the time derivative. For example, if vector Ais a function of the arc distance smeasured from some ﬁxed point, as long as the magnitude of Ais a constant, it can be shown in the same way that d A/dsis always perpendicular to A.This theorem is of considerable importance and should always be kept in mind. Velocity Vector Field. Sometimes the name velocity (or acceleration) vector ﬁeld is used to mean that at every point ( x,y,z) there is a velocity vector whose magnitude and direction depend on where the point is. In other words, the velocity is a vector function which has three components. Each component can be a function of ( x,y,z). For example, consider a rotating body. The velocity of the material of the body at any point is a vector which is a function of position. In general, a vector function may also explicitly dependent on timet. For example, in a continuum, such as a ﬂuid, the velocity of the particles in the continuum is a vector ﬁeld which is not only a function of position but may also of time. To ﬁnd the acceleration, we can use the chain rule: a=dv dt=∂v ∂xdx dt+∂v ∂ydy dt+∂v ∂zdz dt+∂v ∂t. (2.21) Sincedx dt=vx,dy dt=vy,dz dt=vz, It follows that a=vx∂v ∂x+vy∂v ∂y+vz∂v ∂z+∂v ∂t. (2.22) Therefore the acceleration may also be a vector ﬁeld. Example 2.1.3. A body is rotating around the z-axis with an angular velocity ω,ﬁnd the velocity of the particles in the body as a function of the position, and use (2.22) to ﬁnd the acceleration of these particles. 42 2 Vector Calculus Solution 2.1.3. The angular velocity vector is ω=ωk,and the velocity of any point in the body is v=ω×r=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk 00ω xyz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−ωyi+ωxj. Thus the components of the velocity vector are vx(x,y,z)=−ωy, v y(x,y,z)=ωx, v z(x,y,z)=0, and∂v ∂x=ωj,∂v ∂y=−ωi,∂v ∂z=0,∂v ∂t=0. Hence according to (2.22) a=(−ωy)ωj+(ωx)(−ωi)=−ω2(xi+yj). If we deﬁne /rho1=xi+yj,then the magnitude of /rho1is the perpendicular distance between the particle and the rotating axis. Thus a=−ω2/rho1, which shows that every particle has a centripetal acceleration ω2ρ,as expected. 2.2 Diﬀerentiation in Noninertial Reference Systems The acceleration ain Newton’s equation F=mais to be measured in an inertial reference system. An inertial reference system is either a coordinate system ﬁxed in space, or a system moving with a constant velocity relative to the ﬁxed system. A coordinate system ﬁxed on the earth is not an inertial system because the earth is rotating. The derivatives of a vector in a noninertial system are, of course, diﬀerent from those in a ﬁxed system. To ﬁnd the relationships between them, let us ﬁrst consider a moving coordinate system that has the same origin as a ﬁxed system. Intuition tells us that, in this case, the only possible relative motion between the coordinate systems is a rotation. To transform the derivatives from one system to the other, we need to take this rotation into account. Let us denote the quantities associated with the moving system by a prime. The position vector of a particle expressed in terms of the basis vector of the ﬁxed system is r=xi+yj+zk. (2.23) 2.2 Diﬀerentiation in Noninertial Reference Systems 43 The same position vector expressed in the moving coordinate system whose origin coincides with that of the ﬁxed system becomes r=x/primei/prime+y/primej/prime+z/primek/prime, (2.24) where i/prime,j/prime,k/primeare the basis vectors of the moving system. The velocity vis by deﬁnition the time derivative of the position vector in the ﬁxed system v=dr dt=dx dti+dy dtj+dz dtk. (2.25) If we express the time derivative of rin the moving system, then with (2.24) we have v=dr dt=dx/prime dti/prime+dy/prime dtj/prime+dz/prime dtk/prime+x/primedi/prime dt+y/primedj/prime dt+z/primedk/prime dt, (2.26) sincei/prime,j/prime,k/primeare ﬁxed in the moving system, so they are not constant in time. Clearly, the velocity seen in the moving system is v/prime=dx/prime dti/prime+dy/prime dtj/prime+dz/prime dtk/prime=Dr Dt. (2.27) This equation also deﬁnes the operation D /Dt,which simply means the time derivative in the moving system. The notation Dr Dt=/squaresmallsolidr (2.28) is also often used. As mentioned earlier, a dot on top of a symbol means the time derivative. In addition, it usually means the time derivative in the moving system. Note that while the position vector has the same appearance in both the ﬁxed and the moving system as seen in (2.23) and (2.24), the velocity vector, or any other derivative, has more terms in the moving system than in the ﬁxed system as seen in (2.26) and (2.25) . The three derivatives (dx/prime/dt,dy/prime/dt,dz/prime/dt) are not the components of the velocity vector vin the moving system, they only appear to be the velocity components to a stationary observer in the moving system. The velocity vector vexpressed in the moving system is given by (2.26), which can be written as v=Dr Dt+x/primedi/prime dt+y/primedj/prime dt+z/primedk/prime dt. (2.29) Sincei/prime,j/prime,k/primeare unit vectors, their magnitudes are not changing. Therefore their derivatives must be perpendicular to themselves. For example, d i/prime/dtis perpendicular to i/prime, and lies in the plane of j/primeandk/prime.Thus we can write di/prime dt=cj/prime−bk/prime, (2.30) 44 2 Vector Calculus where cand−bare two constants. (The reason for choosing these particular symbols for the coeﬃcients of the linear combination is for convenience, as will be clear in a moment). Similarly dj/prime dt=ak/prime−fi/prime, (2.31) dk/prime dt=ei/prime−dj/prime. (2.32) Buti/prime=j/prime×k/prime,s o di/prime dt=dj/prime dt×k/prime+j/prime×dk/prime dt=(ak/prime−fi/prime)×k/prime+j/prime×(ei/prime−dj/prime)=fj/prime−ek/prime.(2.33) Comparing (2.30) and (2.33), we see that f=cande=b.Similarly, from j/prime=k/prime×i/primeone can show that dj/prime dt=dk/prime−ci/prime. (2.34) It is clear from (2.31) and (2.34) that d=aandc=f. It follows that x/primedi/prime dt+y/primedj/prime dt+z/primedk/prime dt=x/prime(cj/prime−bk/prime)+y/prime(ak/prime−ci/prime)+z/prime(bi/prime−aj/prime) =i/prime(bz/prime−cy/prime)+j/prime(cx/prime−az/prime)+k/prime(ay/prime−bx/prime) =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglei/primej/primek/prime abc x/primey/primez/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (2.35) If we deﬁne ω=ai/prime+bj/prime+ck/prime, (2.36) withrgiven by (2.24), we can write (2.35) as x/primedi/prime dt+y/primedj/prime dt+z/primedk/prime dt=ω×r. (2.37) The meaning of ω×ris exactly the same as in (2.16). We have thus demon- strated explicitly that the most general relative motion of two coordinate systems having a common origin is a rotation with an instantaneous angular velocity ω.Furthermore, (2.29) becomes v=v/prime+ω×r. (2.38) Often this equation is written in the form dr dt=/parenleftbiggD Dt+ω×/parenrightbigg r=/squaresmallsolidr+ω×r, (2.39) 2.2 Diﬀerentiation in Noninertial Reference Systems 45 with the understanding that the time derivative on the left-hand side is in the ﬁxed system, and on the right-hand side, all quantities are to be expressed in the rotating system. This analysis is not limited to the position vector. For any vector A,w e can follow exactly the same procedure and show that dA dt=D DtA+ω×A, (2.40) where D DtA=/squaresmallsolid A=/squaresmallsolid A/prime xi/prime+/squaresmallsolid A/prime yj/prime+/squaresmallsolid A/prime zk/prime, ω×A=A/prime xdi/prime dt+A/prime ydj/prime dt+A/prime zdk/prime dt. Example 2.2.1. Show that the time derivative of the angular velocity vector is the same in either the ﬁxed or the rotating system. Solution 2.2.1. Since dω dt=D Dtω+ω×ω, butω×ω=0,therefore the time derivative in the rotating system/squaresmallsolidωis the same time derivative in the ﬁxed system. Example 2.2.2. If the rotating system and the ﬁxed system have the same origin, express the acceleration ain the ﬁxed system in terms of a/prime,v/prime,ω,/squaresmallsolidω of the rotating system. Solution 2.2.2. By deﬁnition a=dv/dt. So by (2.40) a=dv dt=D Dtv+ω×v. Since v=dr dt=D Dtr+ω×r=/squaresmallsolidr+ω×r, D Dtv=D Dt(/squaresmallsolidr+ω×r)=/squaresmallsolid/squaresmallsolidr+/squaresmallsolidω×r+ω×/squaresmallsolidr, ω×v=ω×(/squaresmallsolidr+ω×r)=ω×/squaresmallsolidr+ω×(ω×r). 46 2 Vector Calculus Therefore, a=D Dtv+ω×v=/squaresmallsolid/squaresmallsolidr+/squaresmallsolidω×r+ω×/squaresmallsolidr+ω×/squaresmallsolidr+ω×(ω×r) =/squaresmallsolid/squaresmallsolidr+/squaresmallsolidω×r+2ω×/squaresmallsolidr+ω×(ω×r). Since/squaresmallsolid/squaresmallsolidr=a/primeand/squaresmallsolidr=v/prime, a=a/prime+/squaresmallsolidω×r+2ω×v/prime+ω×(ω×r). In general, the primed system may have both translational and rotational motion. This can be thought as a translation followed by a rotation. It is clear from Fig. 2.4 that r=r/prime+r0,so v=dr dt=dr/prime dt+dr0 dt =/parenleftbiggDr/prime Dt+ω×r/prime/parenrightbigg +dr0 dt. (2.41) The translational velocity of the coordinates is simply v0=dr0/dt. Similarly the linear acceleration of the coordinates is a0=d2r0/dt2.There- fore, the acceleration of the particle is given by a=dv dt=d dt/parenleftbiggDr/prime Dt+ω×r/prime/parenrightbigg +d2r0 dt2 =D Dt/parenleftbiggDr/prime Dt+ω×r/prime/parenrightbigg +ω×/parenleftbiggDr/prime Dt+ω×r/prime/parenrightbigg +a0. (2.42) z xyy 9 x 9z 9 k ijj 9 i 9k 9r r0r 9P OO9 Fig. 2.4. Geometry of the coordinate systems. The primed system is a noninertial reference frame which has both translational and rotational motion relative to the ﬁxed system 2.3 Theory of Space Curve 47 Therefore the general equations for the transformation from a ﬁxed system to a moving system are v=v/prime+ω×r/prime+v0, (2.43) a=a/prime+/squaresmallsolidω×r/prime+2ω×v/prime+ω×(ω×r/prime)+a0. (2.44) The term/squaresmallsolidω×r/primeis known as the transverse acceleration and the term 2 ω×v/prime is called the coriolis acceleration .T h e centripetal acceleration ω×(ω×r)i s always directed toward the center. 2.3 Theory of Space Curve Suppose we have a particle moving on a space curve as shown in Fig. 2.5. At certain time, the particle is at some point P. In a time interval ∆ tthe particle moves to another point P/primealong the path. The arc distance between Pand P/primeis ∆s.Lettbe the unit vector in the direction of the tangent of the curve atP. The velocity of the particle is of course in the direction of t, v=dr dt=vt, (2.45) where vis the magnitude of the velocity, which is given by v= lim ∆t→0∆s ∆t=ds dt. (2.46) The point Pcan also be speciﬁed by s,the distance along the curve measured from a ﬁxed point to P. Then by chain rule, v=dr dt=ds dtdr ds=vdr ds. (2.47) z y x∆q∆s rr9 nn9 tt9 PP9 r O Fig. 2.5. Space curve. The tangent vector tand the normal vector ndetermine the osculating plane which may turn in space 48 2 Vector Calculus Comparing (2.45) and (2.47), we have dr ds=t, (2.48) hardly a surprising result. Clearly, as ∆ s→0,/vextendsingle/vextendsingle/vextendsingle/vextendsingle∆r ∆s/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1. Now,tis a unit vector, which means its magnitude is not changing, there- fore its derivative must be perpendicular to itself. Let nbe a unit vector perpendicular to t,so we can write dt ds=κn, (2.49) where κis the magnitude of d t/dsand is called the curvature. The vector n is known as the normal vector and is perpendicular to t. The reciprocal of the curvature /rho1=1/κis known as the radius of the curvature. Equation (2.49) deﬁnes both κandn, and tells us how fast the unit tangent vector changes direction as we move along the curve. The acceleration of the particle is a=dv dt=d dt(vt)=/squaresmallsolidvt+v/squaresmallsolid t, where/squaresmallsolid t=dt dt=ds dtdt ds=vdt ds=vκn. (2.50) Therefore the acceleration can be written as a=/squaresmallsolidvt+v2κn=/squaresmallsolidvt+v2 /rho1n. (2.51) The tangential component of acorresponds to the change in the magnitude ofv,and the normal component of acorresponds to the change in direction ofv.The normal component is the familiar centripetal acceleration. For the circular motion in example 2.1.2, r=rcosωti+rsinωtjand v=ωr, t=1 vv=1 ωr(−rωsinωti+rωcosωtj)=−sinωti+ωcosωtj. Since by (2.50), dt ds=1 vdt dt=1 ωr(−ωcosωti−ωsinωtj)=−1 r2(rcosωti+rsinωtj)=−1 r2r. But by deﬁnition, d t/ds=κn.So, in this case, n=−r/r,κ=1/r,and /rho1=1/κ=r.In other words, the radius of curvature in a circular motion is equal to the radius of the circle. In a small region, we can approximate ∆ sby the arc of a circle. The radius of this circle is the radius of curvature of the curve as shown in Fig. 2.5. 2.3 Theory of Space Curve 49 The motion may not be conﬁned in a plane, although both the velocity and acceleration lie in the plane of tandn,known as the osculating plane . In general, there is another degree of freedom for the motion, namely the arc as a whole may turn. In other words, the osculating plane is not necessarily ﬁxed in space. We need another factor to compute the derivatives of the acceleration. Let us deﬁne a third vector, known as binormal vector , b=t×n. (2.52) Since both tandnare unit vectors and they are perpendicular to each other, therefore bis also a unit vector and is perpendicular to both tandn.It follows from the deﬁnition that, in a right-hand system, b×t=n,b×n=−t,n×t=−b. (2.53) All vectors associated with the curve at the point Pcan be written as a linear combination of t,n,andbwhich form a basis at P. Now we evaluate d b/ds and dn/ds. Sincebis perpendicular to t,s ob·t= 0. Afer diﬀerentiating we have d ds(b·t)=db ds·t+b·dt ds=db ds·t+b·κn=0. (2.54) Hence, d b/ds·t=−κb·n.Sincebis perpendicular to n,b·n=0 .T h u s , db/ds·t= 0, which means d b/dsis perpendicular to t. On the other hand, sincebis a unit vector, so d b/dsis perpendicular b. Therefore d b/dsmust be in the direction of n.Let db/ds=γn, (2.55) where γ, by deﬁnition, is the magnitude of d b/dsand is called the torsion of the curve . To obtain d n/ds,we use (2.53), dn ds=d ds(b×t)=db ds×t+b×dt ds=γn×t+b×κn=−γb−κt.(2.56) The set of equations dt ds=κn,dn ds=−(γb+κt),db ds=γn (2.57) are the famous Frenet–Serret formulas . They are fundamental equations in diﬀerential geometry. 50 2 Vector Calculus Example 2.3.1. Find the arc length sof the curve r(t)=acosti+asintj between t=0a n d t=T.Express ras a function of s. Solution 2.3.1. Since d s/dt=vandv=(v·v)1/2=(/squaresmallsolidr·/squaresmallsolidr)1/2, ds=vdt=(/squaresmallsolidr·/squaresmallsolidr)1/2dt. This seemingly trivial formula is actually very useful in a variety of problems. In the present case ds=[ (−asinti+acostj)·(−asinti+acostj)]1/2dt=adt, s=/integraldisplayT 0adt=aT. In general s=atandt=s/a,thus r(s)=acoss ai+asins aj. Example 2.3.2. A circular helix is given by r=acosti+asintj+btk,calcu- latet,n,bandκ,ρ,γ for this curve. Solution 2.3.2. v=/squaresmallsolidr=−asinti+acostj+bk, v=(v·v)1/2=[/parenleftbig a2sin2t+a2cos2t/parenrightbig +b2]1/2=(a2+b2)1/2, t=1 vv=1 (a2+b2)1/2(−asinti+acostj+bk). Since dt dt=ds dtdt ds=vdt ds, dt ds=1 vdt dt=1 (a2+b2)(−acosti−asintj)=κn, κ2=(κn·κn)=1 (a2+b2)2(a2cos2t+a2sin2t)=a2 (a2+b2)2, κ=a (a2+b2),ρ =1 κ=a2+b2 a, n=1 κdt ds=−costi−sintj. 2.4 The Gradient Operator 51 b=t×n=1 (a2+b2)1/2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k −asintacostb −cost−sint0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =1 (a2+b2)1/2[bsinti−bcostj+(asin2t+acos2t)k] =1 (a2+b2)1/2[bsinti−bcostj+ak]. Use chain rule again, db dt=ds dtdb ds=vdb ds, so db ds=1 vdb dt=b (a2+b2)[cotti+s i ntj]=γn, γ=−b (a2+b2). 2.4 The Gradient Operator The application of vector methods to physical problems most frequently takes the form of diﬀerential operations. We have discussed the rate of change with respect to time which allows us to deﬁne velocity and acceleration vectors for the motion of a particle. Now we begin a more systematic study of the rate of change with respect to the spatial coordinates. The most important diﬀerential operator is the gradient. 2.4.1 The Gradient of a Scalar Function Before we discuss the gradient, let us brieﬂy review the notation of derivative and partial derivative in calculus df(x) dx= lim ∆x→0f(x+∆x)−f(x) ∆x= lim ∆x→0∆f ∆x. If ∆x→0 is implicitly understood, we can multiply both sides by ∆ xand write ∆f=f(x+∆x)−f(x)=df dx∆x. (2.58) This equation can also be derived from the Taylor expansion of f(x+∆x) around x: f(x+∆x)=f(x)+df dx∆x+1 2d2f dx2(∆x)2+···· Equation (2.58) is obtained if one drops terms of (∆ x)nwithn/greaterdblequal2a s∆ x→0. 52 2 Vector Calculus Similarly in terms of partial derivatives f(x+∆x,y)−f(x,y)=∂f ∂x∆x, or f(x,y+∆y,z+∆z)−f(x,y,z +∆z)=∂f ∂y∆y. Let the diﬀerence of the scalar function ϕbetween two nearby points (x+∆x,y+∆y,z+∆z) and ( x,y,z)b e∆ ϕ: ∆ϕ=ϕ(x+∆x,y+∆y,z+∆z)−ϕ(x,y,z). This equation can be written as ∆ϕ=ϕ(x+∆x,y+∆y,z+∆z) −[ϕ(x,y+∆y,z+∆z)−ϕ(x,y+∆y,z+∆z)] −[ϕ(x,y,z +∆z)−ϕ(x,y,z +∆z)]−ϕ(x,y,z), since the quantities in the two brackets cancel out. Removing the brackets, we have ∆ϕ=ϕ(x+∆x,y+∆y,z+∆z)−ϕ(x,y+∆y,z+∆z) +ϕ(x,y+∆y,z+∆z)−ϕ(x,y,z +∆z) +ϕ(x,y,z +∆z)−ϕ(x,y,z). With the deﬁnition of partial derivative, the above equation can be written as ∆ϕ=∂ϕ ∂x∆x+∂ϕ ∂y∆y+∂ϕ ∂z∆z. (2.59) The displacement vector from ( x,y,z)t o( x+∆x,y+∆y,z+∆z)i s ,o f course, ∆r=∆xi+∆yj+∆zk. One can readily verify that /parenleftbigg i∂ϕ ∂x+j∂ϕ ∂y+k∂ϕ ∂z/parenrightbigg ·(∆xi+∆yj+∆zk)=∂ϕ ∂x∆x+∂ϕ ∂y∆y+∂ϕ ∂z∆z. Thus, ∆ϕ=/parenleftbigg i∂ϕ ∂x+j∂ϕ ∂y+k∂ϕ ∂z/parenrightbigg ·∆r. (2.60) The vector in the parenthesis is called the gradient ofϕ,and is usually written as grad ϕor∇ϕ, ∇ϕ=i∂ϕ ∂x+j∂ϕ ∂y+k∂ϕ ∂z. (2.61) Since ϕis an arbitrary scalar function, it is convenient to deﬁne the diﬀerential operation in terms of the gradient operator ∇(sometimes known as del or 2.4 The Gradient Operator 53 del operator) ∇=i∂ ∂x+j∂ ∂y+k∂ ∂z. (2.62) This is a vector operator and obeys the same convention as the derivative notation. If a function is placed on the left-hand side of it, ϕ∇is still an opera- tor and by itself means nothing. What is to be diﬀerentiated must be placed on the right of ∇.When it operates on a scalar function, it turns ∇ϕinto a vector with deﬁnite magnitude and direction. It also has a deﬁnite physical meaning. Example 2.4.1. Show that ∇r=/hatwiderand∇f(r)=/hatwiderdf/dr, where/hatwideris a unit vector along the position vector r=xi+yj+zkandris the magnitude of r. Solution 2.4.1. ∇r=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg r, i∂r ∂x=i∂ ∂x/parenleftbig x2+y2+z2/parenrightbig1/2=ix (x2+y2+z2)1/2=ix r,etc. ∇r=ix r+jy r+kz r=xi+yj+zk r=r r=/hatwider. ∇f(r)=i∂f ∂x+j∂f ∂y+k∂f ∂z, i∂f ∂x=idf dr∂r ∂x=idf drx r,etc. ∇f(r)=idf drx r+jdf dry r+kdf drz r=xi+yj+zk rdf dr=/hatwiderdf dr. Example 2.4.2. Show that ( A·∇)r=A. Solution 2.4.2. (A·∇)r=/bracketleftbigg (Axi+Ayj+Azk)·/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg/bracketrightbigg r =/parenleftbigg Ax∂ ∂x+Ay∂ ∂y+Az∂ ∂z/parenrightbigg (xi+yj+zk) =Axi+Ayj+Azk=A. 2.4.2 Geometrical Interpretation of Gradient To see the physical meaning of ∇ϕ,let us substitute (2.61) into (2.60) ∆ϕ=∇ϕ·∆r. 54 2 Vector Calculus Taking the limit as ∆ rapproaches zero yields the diﬀerential form of this equation: dϕ=∇ϕ·dr=∂ϕ ∂xdx+∂ϕ ∂ydy+∂ϕ ∂zdz. (2.63) Now,ϕ(x,y,z)=Crepresents a surface in space. For example, ϕ(x,y,z)= x+y+zandϕ=Crepresents a family of parallel planes, as discussed in the previous chapter. Diﬀerent values of Csimply mean the diﬀerent per- pendicular distances between the plane and the origin. Another example, ϕ(x,y,z)=x2+y2+z2= 4 is the surface of a sphere of radius 2 .Changing 4 to 9 simply means another sphere of radius 3 . If the two near-by points lie on the same surface ϕ=C,then clearly dϕ=0,sinceϕ(x+dx,y+dy,z+dz)=ϕ(x,y,z)=C.In this case d ris, of course, a vector on this surface and dϕ=∇ϕ·dr=0. (2.64) Since the dot product of ∇ϕand dris equal to zero, ∇ϕmust be perpendic- ular to d r.Therefore ∇ϕis normal (perpencicular) to the surface ϕ=Cas shown in Fig. 2.6. We can look at it in another way. Let the unit vector in the direction d r bedand the magnitude of d rbe dr,then the dot product of (2.63) can be written as dϕ=∇ϕ·ddr, ordϕ dr=∇ϕ·d. (2.65) This means that the rate of change of ϕin the direction of dis equal to ∇ϕ·d. Furthermore, since ∇ϕ·d=|∇ϕ|cosθ, xyz r + drrdr∇j j = Constant Fig. 2.6. Gradient of a scalar function. ∇ϕis a vector normal to the surface of ϕ= constant 2.4 The Gradient Operator 55 where θis the angle between d rand∇ϕ,the maximum rate of change occurs atθ=0.This means that if d rand∇ϕare in the same direction, the change ofϕis the largest. Therefore the meaning of ∇ϕcan be summarized as follows: The vector ∇ϕis in the direction of the steepest increase in ϕand the magnitude of the vector ∇ϕis equal to the rate of increase in that direction. Example 2.4.3. Find the unit normal to the surface described by ϕ(x,y,z)= 2x2+4yz−5z2=−10 at (3 ,−1,2). Solution 2.4.3. First check the point (3 ,−1,2) is indeed on the surface: 2(3)2+4 (−1)2−5(2)2=−10.Recall the unit normal to the surface at any point is n=∇ϕ/|∇ϕ|. ∇ϕ=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg/parenleftbig 2x2+4yz−5z2/parenrightbig =4xi+4zj+( 4y−10z)k. n=/bracketleftbigg∇ϕ |∇ϕ|/bracketrightbigg 3,−1,2=12i+8j−24k (122+82+2 42)1/2=3i+2j−6k√ 46. Example 2.4.4. Find the maximum rate of increase for the surface ϕ(x,y,z)= 100 + xyzat the point (1 ,3,2). In which direction is the maximum rate of increase? Solution 2.4.4. The maximum rate of increase is |∇ϕ|1,3,2. ∇ϕ=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg (100 + xyz)=yzi+xzj+xyk, |∇ϕ|1,3,2=|6i+2j+3k|=( 3 6+4+9 )1/2=9. The direction of the maximum increase is given by ∇ϕ|1,3,2=6i+2j+3k. Example 2.4.5. Find the rate of increase for the surface ϕ(x,y,z)=xy2+yz3 at the point (2 ,−1,1) in the direction of i+2j+2k. Solution 2.4.5. ∇ϕ=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg/parenleftbig xy2+yz3/parenrightbig =y2i+/parenleftbig 2xy+z3/parenrightbig j+3yz2k, ∇ϕ2,−1,1=i−3j−3k. 56 2 Vector Calculus The unit vector along i+2j+2kis n=i+2j+2k√1+4+4=1 3(i+2j+2k). The rate of increase is dϕ dr=∇ϕ·n=(i−3j−3k)·1 3(i+2j+2k)=−11 3. Example 2.4.6. Find the equation of the tangent plane to the surface described byϕ(x,y,z)=2xz2−3xy−4x= 7 at the point (1 ,−1,2). Solution 2.4.6. Ifr0is a vector from the origin to the point (1 ,−1,2) and r is a vector to any point in the tangent plane, then r−r0lies in the tangent plane. The tangent plane at (1 ,−1,2) is normal to the gradient at that point, so we have ∇ϕ|1,−1,2·(r−r0)=0. ∇ϕ|1,−1,2=/bracketleftbig/parenleftbig 2z2−3y−4/parenrightbig i−3xj−4xzk/bracketrightbig 1,−1,2=7i−3j+8k. Therefore the tangent plane is given by the equation (7i−3j+8k)·[(x−1)i+(y+1 )j+(z−2)k]=0, 7(x−1)−3(y+1 )+8 ( z−2) = 0 , 7x−3y+8z=2 6. 2.4.3 Line Integral of a Gradient Vector Line integrals occur frequently in physical sciences. The most familiar is prob- ably the work done by a force Fbetween AandBalong some path Γ: Work( A→B)=/integraldisplayB A,ΓF·dr, where dr=idx+jdy+kdz is the diﬀerential displacement vector from ( x,y,z)t o( x+dx,y+dy,z+ dz).Sometimes d lis used in place of d rto emphasize that the diﬀerential displacement vector is along a certain path for the line integral. We will not use this convention here. For any vector ﬁeld A(x,y,z), the line integral /integraldisplayB A,ΓA·dr=/integraldisplayB A,Γ(Axdx+Aydy+Azdz) (2.66) 2.4 The Gradient Operator 57 drB A Fig. 2.7. Path of a line integral. The diﬀerential displacement d ris along a speciﬁed curve in space between AandB B A(a) Γ2Γ1 B A(b) Γ2Γ1 B,A(c) Γ1 Fig. 2.8. Path independence of the line integral of ∇ϕ·dr.(a) The value of the integral from AtoBalong Γ1i st h es a m ea sa l o n g Γ2.(b) They are still the same even though Γ2is much shorter. ( c)A sΓ2shrinks to zero, the integral along Γ2 vanishes. The integral along Γ1,which becomes a loop integral, must also be zero is the sum of contributions A·drfor each diﬀerential displacement d ralong the curve Γin space from AtoBas shown in Fig. 2.7. The line integral is also called path integral because it is carried out along a speciﬁc path Γ. In general, the result depends on the path taken between AandB. However, if A=∇ϕfor some scalar function ϕ,the integral is independent of the path. /integraldisplayB A,ΓA·dr=/integraldisplayB A,Γ∇ϕ·dr=/integraldisplayB Adϕ=ϕ(B)−ϕ(A), (2.67) where we have used (2.63) to convert ∇ϕ·drto the total diﬀerential d ϕ.Since the result depends only on the position of the two end points, the integral is independent of path. In this case, the integral from AtoBin Fig. 2.8 gives the same value whether it is carried out along Γ1or along Γ2.This remains to be true as we bring the two points AandBcloser, no matter how short Γ2becomes. When Bis brought to the same place as A, the line integral over Γ2obviously vanishes because the length of Γ2is equal to zero. So the line integral over Γ1must also be zero. The line integral over Γ1is an integral around a closed loop: /contintegraldisplay ∇ϕ·dr=0. (2.68) 58 2 Vector Calculus The symbol/contintegraldisplay means the integration is over a closed loop. The line integral around the closed loop is called the circulation of the vector ﬁeld Aaround the closed loop Γ.Thus we have the following result: When a vector ﬁeld Ais the gradient of a scalar function ϕ,the circulation of Aaround any closed curve is zero. Sometimes this is called the fundamental theorem of gradient . The argu- ment can be reversed. If/contintegraldisplay A·dr=0,then/integraldisplayB AA·dris independent of path. In that case, Ais the gradient of some scalar function ϕ. Example 2.4.7. Find the work done by the force F=6xyi+/parenleftbig 3x2−3y2/parenrightbig jin a plane along the curve C:y=x2−xfrom (0 ,0) to (2 ,2). Solution 2.4.7. The work done is by deﬁnition the line integral W=/integraldisplay CF·dr=/integraldisplay C/bracketleftbig 6xyi+/parenleftbig 3x2−3y2/parenrightbig j/bracketrightbig ·(idx+jdy+kdz) =/integraldisplay C/bracketleftbig 6xydx+( 3x2−3y2)dy/bracketrightbig . There are more than one way to carry out this integration along curve C. Method I . Change all variables into x. y=x2−x,dy=( 2x−1)dx, W=/integraldisplay C/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =/integraldisplay2 0{6x/parenleftbig x2−x/parenrightbig dx+[ 3x2−3(x2−x)2](2x−1)}dx =/integraldisplay2 0{−6x5+1 5x4−6x2}dx=/bracketleftbig −x6+3x5−2x3/bracketrightbig2 0=1 6. Method II . The curve Ccan be considered as the trajectory described by the tip of the position vector r(t)=x(t)i+y(t)jwithtas a parameter. It can be readily veriﬁed that with x=tandy=t2−t, the curve y=x2−xis traced out. Therefore the curve Cis given by r(t)=x(t)i+y(t)j=ti+/parenleftbig t2−t/parenrightbig j. The point (0 ,0) corresponds to t=0,and (2 ,2) corresponds to t=2.Now we can change all variables into t. 2.4 The Gradient Operator 59 F(x,y,z)·dr=F(x(t),y(t),z(t))·dr dtdt, dr dt=i+( 2t−1)j, F=6xyi+/parenleftbig 3x2−3y2/parenrightbig j=6t/parenleftbig t2−t/parenrightbig i+[ 3t2−3/parenleftbig x2−x/parenrightbig2]j, F·dr={6t/parenleftbig t2−t/parenrightbig +[ 3t2−3/parenleftbig x2−x/parenrightbig2](2t−1)}dt, W=/integraldisplay CF·dr=/integraldisplay2 0{6t/parenleftbig t2−t/parenrightbig +[ 3t2−3/parenleftbig x2−x/parenrightbig2](2t−1)}dt=1 6. Example 2.4.8. Calculate the line integral of the last example from the point (0,0) to the point ( x1,y1) along the path which runs straight from (0 ,0) to (x1,0) and thence to ( x1,y1). Make a similar calculation for the path which runs along the other two sides of the rectangle, via the point (0 ,y1). If (x1,y1)=( 2 ,2), what is the value of the integral? Solution 2.4.8. I1(x1,y1)=/integraldisplay C1F·dr=/integraldisplay C1/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig , C1:( 0,0)→(x1,0)→(x1,y1). From (0 ,0)→(x1,0) :y=0,dy=0, /integraldisplayx1,0 0,0/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =0. From ( x1,0)→(x1,y1):x=x1,dx=0, /integraldisplayx1,y1 x1,0/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =/integraldisplayy1 0(3x2 1−3y2)dy =/bracketleftbig 3x2 1y−y3/bracketrightbigy1 0=3x2 1y1−y3 1, I1(x1,y1)=/bracketleftbigg/integraldisplayx1,0 0,0+/integraldisplayx1,y1 x1,0/bracketrightbigg/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =3x2 1y1−y3 1, I2(x1,y1)=/integraldisplay C2/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig , C2:( 0,0)→(0,y1)→(x1,y1). From (0 ,0)→(0,y1):x=0,dx=0, /integraldisplay0,y1 0,0/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =/integraldisplayy1 0(−3y2)dy=[−y3]y1 0=−y3 1. 60 2 Vector Calculus From (0 ,y1)→(x1,y1):y=y1,dy=0, /integraldisplayx1,y1 0,y1/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =/integraldisplayx1 06xy1dx=3x2 1y1, I2(x1,y1)=/bracketleftbigg/integraldisplay0,y1 0,0+/integraldisplayx1,y1 0,y1/bracketrightbigg/bracketleftbig 6xydx+/parenleftbig 3x2−3y2/parenrightbig dy/bracketrightbig =−y3 1+3x2 1y1. Clearly I1(x1,y1)=I2(x1,y1),andI1(2,2) = 3(2)22−(2)3=1 6. Example 2.4.9. From the last two examples, it is clear that the line integral /integraldisplay CF·drwithF=6xyi+/parenleftbig 3x2−3y2/parenrightbig jdepends only on the end points and is independent of the path of integration, therefore F=∇ϕ.Findϕ(x,y)a n d show that/integraltext2,2 0,0F·dr=ϕ(2,2)−ϕ(0,0). Solution 2.4.9. ∇ϕ=i∂ϕ ∂x+j∂ϕ ∂y=6xyi+/parenleftbig 3x2−3y2/parenrightbig j=F, ∂ϕ ∂x=6xy=⇒ϕ=3x2y+f(y), ∂ϕ ∂y=3x2−3y2=3x2+df(y) dy, df(y) dy=−3y2=⇒f(y)=−y3+k(kis a constant) . Thus, ϕ(x,y)=3x2y−y3+k. /integraldisplay2,2 0,0F·dr=/integraldisplay2,2 0,0∇ϕ·dr=ϕ(2,2)−ϕ(0,0) = 16 + k−k=1 6. Note that/integraldisplayx1,y1 0,0F·dr=ϕ(x1,y1)−ϕ(0,0) = 3 x2 1y1−y3 1, in agreement with the result of the last example. 2.5 The Divergence of a Vector 61 Example 2.4.10. Find the line integral/integraldisplay2,1 0,0F·drwithF=xyi−y2jalong the path ( a)y=( 1/2)x,(b)y=( 1/4)x2,(c)f r o m( 0 ,0) straight up to (0 ,1) and then along a horizontal line to (2 ,1). Solution 2.4.10./integraltext2,1 0,0F·dr=/integraltext2,1 0,0/parenleftbig xydx−y2dy/parenrightbig along (a)y=1 2x,so dy=1 2dx, /integraldisplay2,1 0,0/parenleftbig xydx−y2dy/parenrightbig =/integraldisplay2 0/parenleftbigg1 2x2dx−1 8x2dx/parenrightbigg =/bracketleftbigg3 8·1 3x3/bracketrightbigg2 0=1. (b)y=1 4x2,so d y=1 2xdx, /integraldisplay2,1 0,0/parenleftbig xydx−y2dy/parenrightbig =/integraldisplay2 0/parenleftBig1 4x3dx−1 32x5dx/parenrightBig =/bracketleftBig1 16x4−1 32·6x6/bracketrightBig2 0=2 3. (c)F r o m( 0 ,0) straight up to (0 ,1) :x=0sodx=0 ; then from (0 ,1) along a horizontal line to (2 ,1) :y=1a n dd y=0, /integraldisplay2,1 0,0/parenleftbig xydx−y2dy/parenrightbig =/integraldisplay0,1 0,0/parenleftbig xydx−y2dy/parenrightbig +/integraldisplay2,1 0,1/parenleftbig xydx−y2dy/parenrightbig =−/integraldisplay1 0y2dy+/integraldisplay2 0xdx=−1 3+2=5 3. In general the line integral/integraldisplay CF·drdepends on the path of integration as shown in the last example. However, if F=∇ϕ,the line integral is indepen- dent of the path of integration. We are going to discuss the condition under whichFcan be expressed as the gradient of a scalar function. 2.5 The Divergence of a Vector Just as we can operate with ∇on a scalar ﬁeld, we can also operate with ∇ on a vector ﬁeld Aby taking the dot product. With their components, this operation gives ∇·A=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·(iAx+jAy+kAz) =∂Ax ∂x+∂Ay ∂y+∂Az ∂z. (2.69) 62 2 Vector Calculus Just as the dot product of two vectors is a scalar, ∇·Ais also a scalar. This sum, called the divergence ofA(or div A), is a special combination of derivatives. Example 2.5.1. Show that ∇·r=3 a n d ∇·rf(r)=3f(r)+r(df/dr). Solution 2.5.1. ∇·r=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·(ix+jy+kz) =∂x ∂x+∂y ∂y+∂z ∂z=3. ∇·rf(r)=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·(ixf(r)+jyf(r)+kzf(r)) =∂ ∂x[xf(r)] +∂ ∂y[yf(r)] +∂ ∂z[zf(r)] =f(r)+x∂f ∂x+f(r)+y∂f ∂y+f(r)+z∂f ∂z =3f(r)+xdf dr∂r ∂x+ydf dr∂r ∂y+zdf dr∂r ∂z. ∂r ∂x=∂ ∂x/parenleftbig x2+y2+z2/parenrightbig1/2=1 22x (x2+y2+z2)1/2=x r; ∂r ∂y=y r;∂r ∂z=z r. ∇·rf(r)=3f(r)+x2 rdf dr+y2 rdf dr+z2 rdf dr =3f(r)+x2+y2+z2 rdf dr=3f(r)+rdf dr. 2.5.1 The Flux of a Vector Field To gain some physical feeling for the divergence of a vector ﬁeld, it is helpful to introduce the concept of ﬂux(Latin for “ﬂow”). Consider a ﬂuid of density /rho1moving with velocity v.We ask for the total mass of ﬂuid which crosses an area ∆ aperpendicular to the direction of ﬂow in a time ∆ t.As shown in Fig. 2.9a, all the ﬂuid in the rectangular pipe of length v∆twith the patch ∆ a as its base will cross ∆ ain the time interval ∆ t.The volume of this pipe is 2.5 The Divergence of a Vector 63 v∆t v∆tv∆a(a) v ∆a(b) n q q Fig. 2.9. Flux through the base. ( a) Flux through ∆ a=ρv∆a.(b) Flux through the tilted ∆ a=ρv·n∆a v∆t∆a,and it contains a total mass /rho1v∆t∆a.Dividing ∆ twill give the mass across ∆ aper unit time, which by deﬁnition is the rate of the ﬂow Rate of ﬂow through ∆ a=/rho1v∆a. Now let us consider the case shown in Fig. 2.9b. In this case the area ∆ a is not perpendicular to the direction of the ﬂow. The total mass which will ﬂow through this tilted ∆ ain time ∆ tis just the density times the volume of this pipe with the slanted bases. That volume is v∆t∆acosθ,where θis the angle between the velocity vector vandn, the unit normal to ∆ a.But vcosθ=v·n.So, multiplying by /rho1and dividing ∆ t,we have Rate of ﬂow through tilted ∆ a=/rho1v·n∆a. To get the total ﬂow through any surface S, ﬁrst we divide the whole surface into little patches which are so small that over any one patch the surface is practically ﬂat. Then we sum up the contributions from all the patches. As the patches become smaller and more numerous without limit, the sum becomes a surface integral. Accordingly, Total ﬂow through S=/integraldisplay/integraldisplay S/rho1v·nda. (2.70) If we deﬁne J=/rho1v,(2.70) is known as the ﬂux of Jthrough the surface S Flux of Jthrough S=/integraldisplay/integraldisplay SJ·nda. (2.71) Originally it means the rate of the ﬂow, the word ﬂux is now generalized to mean the surface integral of the normal component of a vector. For example, the vector might be the electric ﬁeld E.Although electric ﬁeld is not ﬂowing in the sense in which ﬂuid ﬂows, we still say things like “the ﬂux of Ethrough a closed surface is equal to the total charge inside” to help us to visualize the electric ﬁeld lines “ﬂowing” out of the electric charges. 64 2 Vector Calculus Example 2.5.2. LetJ=/rho1v0kwhere /rho1is the density of the ﬂuid and v0kis its velocity ( kis the unit vector in the zdirection). Calculate the ﬂux of J(the ﬂow rate of the ﬂuid) through a hemispherical surface of radius b. z xy dydxdank (0, b, 0)(0, 0, b) (b, 0, 0) Fig. 2.10. Surface element on a hemisphere. The projected area on the xyplane is dxdy=c o s θdawhere θis the angle between the tangent plane at d aand the xy plane Solution 2.5.2. The equation of the spherical surface is ϕ(x,y,z)=x2+ y2+z2=b2.Therefore the unit normal to the surface is n=∇ϕ |∇ϕ|=2xi+2yj+2zk (4x2+4y2+4z2)1/2=xi+yj+zk b. Let the ﬂux of Jthrough the hemisphere be Φ, which is given by Φ=/integraldisplay/integraldisplay SJ·nda=/integraldisplay/integraldisplay S/rho1v0k·nda, where d ais an element of the surface area on the hemisphere as shown in Fig. 2.10. This surface area projects onto d xdyin the xyplane. Let θbe the acute angle between d a(actually the tangent plane at d a) and the xyplane. Then we have d xdy=c o s θda.The integral becomes Φ=/integraldisplay/integraldisplay S/rho1v0k·nda=/integraldisplay/integraldisplay S/rho1v0k·n1 cosθdxdy, where the limit on xandymust be such that we integrate over the projected area in the xyplane which is inside a circle of radius b.The angle between two planes is the same as the angle between the normals to the planes. Since nis the unit normal to d aandkis the unit normal to xyplane, the angle θ is between nandk.Thus, cos θ=n·k.Therefore, 2.5 The Divergence of a Vector 65 Φ=/integraldisplay/integraldisplay S/rho1v0k·n1 cosθdxdy=/integraldisplay/integraldisplay S/rho1v0dxdy=/rho1v0πb2. Note that this result is the same as the ﬂux through the circular ﬂat area in thexyplane. In fact, it is exactly the same as the ﬂux through any surface whose boundary is the circle of radius bin the xyplane, since we obtained the result without using the explicit expression of n. 2.5.2 Divergence Theorem Thedivergence theorem is also known as Gauss’ theorem . It relates the ﬂux of a vector ﬁeld through a closed surface Sto the divergence of the vector ﬁeld in the enclosed volume /integraldisplay/integraldisplay closed surface SA·nda=/integraldisplay/integraldisplay/integraldisplay vol.enclosed in S∇·AdV. (2.72) The surface integral is over a closed surface as shown in Fig. 2.11. The unit normal vector nis pointing outward from the enclosed volume. The right- hand side of this equation is the integration of the divergence over the volume that is enclosed in the surface. To prove this theorem, we cut the volume Vup into a very large number of tiny (diﬀerential) cubes. The volume integral is the sum of the integrals over all the cubes. Imagine we have a parallelepiped with six surfaces enclosing a volume V. We separate the volume into two cubes by a cut as in Fig. 2.12. Note that the sum of the ﬂux through the six surfaces of the cube on the left and the ﬂux through the six surfaces of the cube on the right is equal to the ﬂux through the six surfaces of the original parallelepiped before it was cut. This is because n da dv Fig. 2.11. The divergence theorem. The volume is enclosed by the surface. The integral of the divergence over the volume inside is equal to the ﬂux through the outside surface 66 2 Vector Calculus ADC BC D ABC B AD n n(a) (b) Fig. 2.12. The ﬂux out of the touching sides of the two neighboring cubes cancel each other the unit normal vectors on the touching sides of the two neighboring cubes are equal and opposite to each other. So the contributions to the ﬂux for the two cubes from these two sides exactly cancel. Thus it must be generally true that the sum of the ﬂux through the surfaces of all the cubes is equal to the integral over those surfaces that have no touching neighbors, i.e., over the original outside surface. So if we can prove the result for a small cube, we can prove it for any arbitrary volume. Consider the ﬂux of Athrough the surfaces of the small cube of volume ∆V=∆x∆y∆zshown in Fig. 2.13. The unit vector perpendicular to the sur- face ABCD is clearly −j(n=−j). The ﬂux through this surface is therefore given by A(x,y,z)·(−j)∆a=−Ay(x,y,z)∆x∆z. The ﬂux is deﬁned as the outgoing “ﬂow.” The minus sign simply means the ﬂux is ﬂowing into the volume. Similarly, the unit normal to the surface EFGH isj,and the ﬂux through this surface is A(x,y+∆y,z)·j∆a=Ay(x,y+∆y,z)∆x∆z. xyz ∆z ∆y∆xAy ∆x ∆z A( y + ∆y ) ∆x ∆z BC D A EFG H ∂y∂Ay Fig. 2.13. The ﬂux through the left and right face of a inﬁnitesimal cube 2.5 The Divergence of a Vector 67 Note that for every point ( x,y,z) on ABCD, the corresponding point on EFGH is (x,y+∆y,z).The net ﬂux through these two surfaces is simply the sum of the two: [Ay(x,y+∆y,z)−Ay(x,y,z)] ∆x∆z=∂Ay ∂y∆y∆x∆z=∂Ay ∂y∆V.(2.73) By applying similar reasoning to the ﬂux components in the two other direc- tions, we ﬁnd the total ﬂux through all the surfaces of the cube is /summationdisplay cubeA·nda=/parenleftbigg∂Ax ∂x+∂Ay ∂y+∂Az ∂z/parenrightbigg ∆V=(∇·A)∆V. (2.74) This shows that the outward ﬂux from the surface of an inﬁnitesimal cube is equal to the divergence of the vector multiplied by the volume of the cube. Thus the divergence has the following physical meaning: The divergence of a vector Aat a point is the total outward ﬂux of A per unit volume in the neighborhood of that point. For any ﬁnite volume, the total ﬂux of Athrough the outside surface enclosing the volume is equal to the sum of the ﬂuxes out of all the inﬁnitesimal interior cubes, and the ﬂux out of each cube is equal to the divergence of A times the volume of the cube. Therefore the integral of the normal component of a vector over any closed surface is equal to the integral of the divergence of the vector over the volume enclosed by the surface, /integraldisplay/integraldisplay /circlecopyrt SA·nda=/integraldisplay/integraldisplay/integraldisplay V∇·AdV. (2.75) The small circle on the double integral sign means the surface integral is over a closed surface. The volume integral is understood to be over the entire region inside the closed surface. This is the divergent theorem of (2.72), which is sometimes called the fundamental theorem for divergence. A ﬂow ﬁeld Ais said to be solenoidal if everywhere the divergence of A is equal to zero ( ∇·A=0 ).An incompressible ﬂuid must ﬂow out of a given volume as rapidly as it ﬂows in. The divergence of such a ﬂow ﬁeld must be zero, therefore the ﬁeld is solenoidal. On the other hand if Ais such a ﬁeld that at certain point ∇·A/negationslash=0,then there is a net outward ﬂow from a small volume surrounding that point. Fluid must be “created” or “put in” at that point. If ∇·Ais negative, ﬂuid must be “taken out” at that point. Therefore we come to the following conclusion. The divergence of ﬂow ﬁeld at a point is a measure of the strength of the source (or sink) of the ﬂow at that point. 68 2 Vector Calculus Example 2.5.3. Verify the divergence theorem by evaluating both sides of (2.72) with A=xi+yj+zkover a cylinder described by x2+y2=4 and 0 ≤z≤4. Solution 2.5.3. Since ∇·A=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·(xi+yj+zk)=3,the volume integral is /integraldisplay/integraldisplay/integraldisplay V∇·AdV=3/integraldisplay/integraldisplay/integraldisplay VdV=3 (π22)4 = 48 π, which is simply three times the volume of the cylinder. The surface of the cylinder consists of the top, bottom, and curved side surfaces. Therefore the surface integral can be divided into three parts /integraldisplay/integraldisplay /circlecopyrt SA·nda=/integraldisplay/integraldisplay topA·nda+/integraldisplay/integraldisplay bottomA·nda+/integraldisplay/integraldisplay curvedA·nda. For the top surface /integraldisplay/integraldisplay topA·nda=/integraldisplay/integraldisplay top(xi+yj+4k)·kda=4/integraldisplay/integraldisplay topda=4π22=1 6π. For the bottom surface /integraldisplay/integraldisplay bottomA·nda=/integraldisplay/integraldisplay bottom(xi+yj+0k)·(−k)da=0. For the curved side surface, we must ﬁrst ﬁnd the unit normal n.Since the surface is described by ϕ(x,y)=x2+y2=4, n=∇ϕ |∇ϕ|=2xi+2yj (4x2+4y2)1/2=xi+yj 2, A·n=(xi+yj+zk)·xi+yj 2=1 2/parenleftbig x2+y2/parenrightbig =2, /integraldisplay/integraldisplay curvedA·nda=2/integraldisplay/integraldisplay curvedda = 2(2 π·2) 4 = 32 π. Therefore /integraldisplay/integraldisplay /circlecopyrt SA·nda=1 6π+0+3 2 π=4 8π, which is the same as the volume integral. 2.5 The Divergence of a Vector 69 2.5.3 Continuity Equation One of the most important applications of the divergence theorem is using it to express the conservation laws in diﬀerential forms. As an example, consider a ﬂuid of density /rho1moving with velocity v.According to (2.70), the rate at which the ﬂuid ﬂows out of a closed surface is Rate of outward ﬂow through a closed surface =/integraldisplay/integraldisplay /circlecopyrt S/rho1v·nda. (2.76) Now because of the conservation of mass, this rate of out ﬂow must be equal to the rate of decrease of the ﬂuid inside the volume that is enclosed by the surface. Therefore /integraldisplay/integraldisplay /circlecopyrt S/rho1v·nda=−/integraldisplay/integraldisplay/integraldisplay V∂/rho1 ∂tdV. (2.77) The negative sign accounts for the fact that the ﬂuid inside is decreasing if the ﬂow is outward. Using the divergence theorem /integraldisplay/integraldisplay /circlecopyrt S/rho1v·nda=/integraldisplay/integraldisplay/integraldisplay V∇·(/rho1v)dV, (2.78) we have /integraldisplay/integraldisplay/integraldisplay V/bracketleftbigg ∇·(/rho1v)+∂/rho1 ∂t/bracketrightbigg dV=0. (2.79) Since the volume Vin this equation, the integrand must equal to zero, or ∇·(/rho1v)+∂/rho1 ∂t=0 . (2.80) This important equation, known as the continuity equation, relates the den- sity and the velocity at the same point in a diﬀerential form. Many other conservation laws can be similarly expressed. For an incompressible ﬂuid, /rho1is not changing in time. In that case, the divergence of the velocity must be zero, ∇·v=0. (2.81) Singularities in the Field. In deriving these integral theorems, we require the scalar and vector ﬁelds to be continuous and ﬁnite at every point. Often there are points, lines, or surfaces in space at which ﬁelds become discon- tinuous or even inﬁnite. Examples are the electric ﬁelds produced by point, line, or surface charges. One way of dealing with this situation is to eliminate these volume elements, by appropriate surfaces, from the domain to which the theorems are to be applied. Another scheme is to “smear out” the discontinu- ous quantities, such as using charge densities, so that the ﬁelds are again well behaved. Still another powerful way is to use Dirac delta function. Sect. 2.10.2 is a speciﬁc example of these procedures. 70 2 Vector Calculus 2.6 The Curl of a Vector The cross product of the gradient operator ∇with vector Agives us another special combination of the derivatives of the components of A ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =i/parenleftbigg∂Az ∂y−∂Ay ∂z/parenrightbigg +j/parenleftbigg∂Ax ∂z−∂Az ∂x/parenrightbigg +k/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg . (2.82) It is a vector known as the curlofA.The name curl suggests that it has something to do with rotation. In fact, in European texts the word rotation (or rot) is used in place of curl. In Example 2.1.3, we have considered the motion of a body rotating around the z-axis with angular velocity ω.The velocity of the particles in the body is v=−ωyi+ωxj.The circular characteristic of this velocity ﬁeld is manifested in the curl of the velocity ∇×v=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z −ωy ωx 0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2ωk, (2.83) which shows that the curl of vis twice the angular velocity of the rotating body. If this velocity ﬁeld describes a ﬂuid ﬂow, curl vis called the vorticity vector of the ﬂuid. It points in the direction around which a vortex motion takes place and is a measure of the the angular velocity of the ﬂow. A small paddle wheel placed in the ﬁeld will tend to rotate in regions where ∇×v/negationslash=0. The paddle wheel will remain stationary in those regions where ∇×v=0. If the curl of a vector ﬁeld is equal to zero everywhere, the ﬁeld is called irrotational . Example 2.6.1. Show that ( a)∇×r=0;(b)∇×rf(r)=0where ris the position vector. Solution 2.6.1. (a) Since r=xi+yj+zk,so ∇×r=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z xyz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. 2.6 The Curl of a Vector 71 (b) ∇×rf(r)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z xf(r)yf(r)zf(r)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/braceleftbigg∂ ∂y[zf(r)]−∂ ∂z[yf(r)]/bracerightbigg i +/braceleftbigg∂ ∂z[xf(r)]−∂ ∂x[zf(r)]/bracerightbigg j+/braceleftbigg∂ ∂x[yf(r)]−∂ ∂y[xf(r)]/bracerightbigg k =/braceleftbigg z∂ ∂yf(r)−y∂ ∂zf(r)/bracerightbigg i+/braceleftbigg x∂ ∂zf(r)−z∂ ∂xf(r)/bracerightbigg j +/braceleftbigg y∂ ∂xf(r)−x∂ ∂yf(r)/bracerightbigg k. Since ∂ ∂yf(r)=df dr∂r ∂yandr=/parenleftbig x2+y2+z2/parenrightbig1/2, ∂ ∂yf(r)=df dr/parenleftBigg −y (x2+y2+z2)1/2/parenrightBigg =−df dry r, ∂ ∂xf(r)=−df drx r;∂ ∂zf(r)=−df drz r. Therefore ∇×rf(r)=df dr/braceleftBig −zy r+yz r/bracerightBig i+df dr/braceleftBig −xz r+zx r/bracerightBig j+df dr/braceleftBig −yx r+xy r/bracerightBig k=0. 2.6.1 Stokes’ Theorem Stokes’ theorem relates the line integral of a vector function around a closed loopCto a surface integral of the curl of that vector over a surface Sthat spans the loop. The theorem states that /integraldisplay closed loop CA·dr=/integraldisplay/integraldisplay area bounded by C(∇×A)·nda, (2.84) where d ris a directed line element along a closed curve CandSis any surface bounded by C. At any point on the surface, the unit normal vector n is perpendicular to the surface element d aat that point as shown in Fig. 2.14. The sign of nis determined by the convention of the “right-hand rule.” Curl the ﬁngers of your right hand in the direction d r,then your thumb points in the positive direction of n.If the curve Clies in a plane, the simplest surface spans Cis a ﬂat surface. Now imagine the ﬂat surface is a ﬂexible membrane 72 2 Vector Calculus dr n da Fig. 2.14. Stokes’ theorem. The integral of the curl over the surface is equal to the line integral around the closed boundary curve which can continuously expand but remains attached to curve C. A sequence of curved surfaces is generated. Stokes theorem applies to all such surfaces. The positive direction of nfor the ﬂat surface is clear from the right-hand rule. As the surface expands, nmoves along with it. For example, with the direction of d rshown in Fig. 2.14, the normal vector nof the nearly ﬂat surface bounded by Cis pointing “downward” according to the right-hand rule. When the surface is expanded into the ﬁnal shape, nis turned to “outward” direction as shown in the ﬁgure. The surface in Stokes’ theorem must be two sided. A one-sided surface can be constructed by taking a long strip of paper, giving it a half twist, and joining the ends. If we tried to color one side of the surface we would ﬁnd the whole thing colored. A belt of this shape is called a Moebius surface. Such a surface is not orientable since we cannot deﬁne the sense of the normal vector n.Stokes’ theorem applies only to the orientable surface , furthermore, the boundary curve of the surface must not cross itself. To prove Stokes’ theorem, we divide the surface into a large number of small rectangles. The surface integral on the right-hand side of (2.84) is of course just the sum of the surface integrals over all the small rectangles. If the line integrals around all the small rectangles are traced in the same direction, each interior line will be traced twice – once in each direction. Thus the line integrals of A·drfrom all the interior lines will sum up to zero, since each term will appear twice with opposite sign. Therefore the sum of the line integrals around all the small rectangles will equal to the line integral around the boundary curve C, as shown in Fig. 2.15. So if we prove the result for a small rectangle, we will have proved it for any closed curve C. Since the surface is to be composed of an inﬁnitely large number of these inﬁnitesimal rectangles, we may consider each to be a plane rectangle. Let us orient the coordinate axes so that one of these rectangles lies in the xy- plane, the sides will be of length ∆ xand ∆ yas shown in Fig. 2.16. Let the coordinates of the center of the loop be ( x,y,z).We designate the corners of this rectangle by A, B, C, D. So, the line integral around this rectangle is /contintegraldisplay ABCDA·dr=/integraldisplay ABA·(idx)+/integraldisplay BCA·(jdy)+/integraldisplay CDA·(−idx)+/integraldisplay DAA·(−jdy). (2.85) 2.6 The Curl of a Vector 73 = Fig. 2.15. Proof of Stokes’ theorem. The surface is divided into diﬀerential surface elements. Circulation along interior lines cancel and the result is a circulation around the perimeter of the original surface ADC By x∆y ∆x(x, y) Fig. 2.16. T h el i n ei n t e g r a lo f A·draround the four sides of the inﬁnitesimal square is equal to the surface integral of ∇×Ao v e rt h ea r e ao ft h i ss q u a r e We use the symbol/contintegraldisplay to mean the line integral is over a closed loop. Now we may approximate the integral by the average value of the integrand multiplied by the length of the integration interval. The average value of A on the line AB may be taken to be the value of Aat the midpoint of AB. The coordinates at the midpoint of AB is/parenleftbig x,y−1 2∆y,z/parenrightbig .Thus /integraldisplay ABA·(idx)=/integraldisplay ABAxdx=Ax/parenleftbigg x,y−1 2∆y,z/parenrightbigg ∆x. Similarly, /integraldisplay BCA·(jdy)=/integraldisplay BCAydy=Ay/parenleftbigg x+1 2∆x,y,z/parenrightbigg ∆y, /integraldisplay CDA·(−idx)=−/integraldisplay CDAxdx=−Ax/parenleftbigg x,y+1 2∆y,z/parenrightbigg ∆x, /integraldisplay DAA·(−jdy)=−/integraldisplay DAAydy=−Ay/parenleftbigg x−1 2∆x,y,z/parenrightbigg ∆y. 74 2 Vector Calculus Summing all these contributions, we obtain /contintegraldisplay ABCDA·dr=/parenleftbigg Ax/parenleftbigg x,y−1 2∆y,z/parenrightbigg −Ax/parenleftbigg x,y+1 2∆y,z/parenrightbigg/parenrightbigg ∆x +/parenleftbigg Ay/parenleftbigg x+1 2∆x,y,z/parenrightbigg −Ay/parenleftbigg x−1 2∆x,y,z/parenrightbigg/parenrightbigg ∆y. Since Ay/parenleftbigg x+1 2∆x,y,z/parenrightbigg −Ay/parenleftbigg x−1 2∆x,y,z/parenrightbigg =∂Ay ∂x∆x, Ax/parenleftbigg x,y−1 2∆y,z/parenrightbigg −Ax/parenleftbigg x,y+1 2∆y,z/parenrightbigg =−∂Ax ∂y∆y, so we have /contintegraldisplay ABCDA·dr=/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg ∆x∆y. (2.86) Next consider the surface integral of ∇×Aover ABCD. In this case the unit normal nis just k.Again we take the integral to be equal to the average value of the integrand over the area multiplied by the area of the integration. The average value of Ais simply the value of Aevaluated at the center. Therefore /integraldisplay/integraldisplay ABCD(∇×A)·nda=(∇×A)·k∆x∆y=/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg ∆x∆y, (2.87) which is the same as (2.86). This result can be interpreted as follows: The component of ∇×Ain the direction of nat a point Pis the circulation of Aper unit area around Pin the plane normal to n. The circulation of a vector ﬁeld around any closed loop can now be easily related to the curl of that ﬁeld. We ﬁll the loop with a surface Sand add the circulations around a set of inﬁnitesimal squares covering this surface. Thus we have/summationdisplay/contintegraldisplay ABCDA·dr=/summationdisplay/integraldisplay/integraldisplay ABCD(∇×A)·nda, (2.88) which can be written as /contintegraldisplay CA·dr==/integraldisplay/integraldisplay S(∇×A)·nda. (2.89) This is Stokes’ theorem. Sometimes this theorem is referred to as the funda- mental theorem for curls . Note that this theorem holds for any surface Sas long as the boundary of the surface is the closed loop C. 2.6 The Curl of a Vector 75 Example 2.6.2. Verify Stokes’ theorem by ﬁnding the circulation of the vector ﬁeldA=4yi+xj+2zkaround a square of radius 2 ain the xyplane, centered at the origin and the surface integral/integraldisplay/integraldisplay (∇×A)·ndaover the surface of the square. Solution 2.6.2. We compute the circulation by calculating the line inte- gral around each of the four sides of the square. From ( a,−a,0) to ( a,a,0), x=a,dx=0,z=0 : I1=/integraldisplaya,a,0 a,−a,0A·dr=/integraldisplaya,a,0 a,−a,0(4yi+xj+2zk)·(idx+jdy+kdz) =/integraldisplaya,a,0 a,−a,0(4ydx+xdy+2zdz)=/integraldisplaya −aady=2a2. From ( a,a,0) to ( −a,a,0),y=a,dy=0,z=0 : I2=/integraldisplay−a,a,0 a,a,0A·dr=/integraldisplay−a a4adx=−8a2. From ( −a,a,0) to ( −a,−a,0),x=−a,dx=0,z=0 : I3=/integraldisplay−a,−a,0 −a,a,0A·dr=/integraldisplay−a a(−a)dy=2a2. From ( −a,−a,0) to ( a,−a,0),y=−a,dy=0,z=0 : I4=/integraldisplaya,−a,0 −a,−a,0A·dr=/integraldisplaya −a4(−a)dx=−8a2. Therefore the circulation is/contintegraldisplay CA·dr=I1+I2+I3+I4=−12a2. Now we compute the surface integral. First nda=kdxdyover the square and the curl of Ais ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z 4yx 2z/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−3k. Thus/integraldisplay/integraldisplay S∇×A·nda=−3/integraldisplay/integraldisplay Sdxdy=−3(2a)2=−12a2, which is the same as the circulation, satisfying Stokes’ theorem. 76 2 Vector Calculus Example 2.6.3. Verify Stokes’ theorem by evaluating both sides of (2.84) with A=4yi+xj+2zk.This time the surface is over the hemisphere described by ϕ(x,y,z)=x2+y2+z2= 4 and the loop Cis given by the circle x2+y2=4. Solution 2.6.3. Since ∇×A=−3k, /integraldisplay/integraldisplay S∇×A·nda=−3/integraldisplay/integraldisplay Sk·nda. The surface is over a hemisphere. The geometry is shown in Fig. 2.10. The surface integral can be evaluated over the projection of the hemisphere on the xyplane using the relation da=1 cosθdxdy=1 k·ndxdy. The integration is simply over the disk of radius 2: /integraldisplay/integraldisplay S∇×A·nda=−3/integraldisplay/integraldisplay k·n1 k·ndxdy=−3/integraldisplay/integraldisplay dxdy=−3(4π)=−12π. To evaluate the line integral around the circle, it is convenient to write the circle in the parametric form r=xi+yj,x= 2cos θ, y =2s i n θ,0≤θ≤2π. dr dθ=−2sinθi+ 2cos θj,A=4yi+xj+2zk=8s i n θi+ 2cos θj+2zk. /contintegraldisplay CA·dr=/contintegraldisplay CA·dr dθdθ=/integraldisplay2π 0(−16sin2θ+ 4cos2θ)dθ=−12π. Thus, Stokes’ theorem is veriﬁed, /contintegraldisplay CA·dr=/integraldisplay/integraldisplay S∇×A·nda. Example 2.6.4. Use Stokes’ theorem to evaluate the line integral/contintegraldisplay CA·dr withA=2yzi+xj+z2kalong the circle described by x2+y2=1. Solution 2.6.4. The curl of Ais ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z 2yz x z2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=2yj+( 1−2z)k, 2.6 The Curl of a Vector 77 and according to Stokes’ theorem /contintegraldisplay CA·dr=/integraldisplay/integraldisplay S∇×A·nda=/integraldisplay/integraldisplay S[2yj+( 1−2z)k]·nda. Since Scan be any surface as long as it is bounded by the circle, the simplest way to do this problem is to use the ﬂat surface inside the circle. In that case z=0a n d n=k.Hence, /contintegraldisplay A·dr=/integraldisplay/integraldisplay Sda=π. Connectivity of Space. Stokes’ theorem is valid in a simply connected region . A region is simply connected if any closed loop in the region can be shrunk to a point without encountering any points not in the region. In a simply connected region, any two curves between two points can be distorted into each other within the region. The space inside a torus (doughnut) is multiply connected since a closed curve surrounds the hole cannot be shrunk to a point within the region. The space between two inﬁnitely long concen- tric cylinders is also not simply connected. However, the region between two concentric spheres is simply connected. If∇×F=0in a simply connected region, we can use Stokes’ theorem /contintegraldisplay CF·dr=/integraldisplay/integraldisplay S∇×F·nda=0 to conclude that the line integral/integraltextB AF·dris independent of the path. If∇×F=0in a multiply connected region, then/integraltextB AF·dris not unique. In such a case, we often “cut” the region so as to make it simply connected. Then/integraltextB AF·dris independent of the path inside the simply connected region, but/integraltextB AF·dracross the cut line may give a ﬁnite jump. For example, consider the loop integral/contintegraltext F·drwith F=−y x2+y2i+x x2+y2j around a unit circle centered at the origin. Since ∇×F=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z −y x2+y2x x2+y20/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0, one might conclude that /contintegraldisplay F·dr=/integraldisplay/integraldisplay S∇×F·nda=0. 78 2 Vector Calculus y x Fig. 2.17. If the function is singular on the z-axis, the region is multiply connected. A “cut” can be made to change it into a simply connected region. However, a lined integral across the cut line may give a sudden jump This is incorrect, as one can readily see that /contintegraldisplay F·dr=/contintegraldisplay/parenleftbigg −y x2+y2dx+x x2+y2dy/parenrightbigg . Withx=c o s θ, y=s i nθ,(so dx=−sinθdθ,dy=c o s θdθ,andx2+y2=1 ), this integral is seen to be /contintegraldisplay F·dr=/contintegraldisplay/parenleftbig sin2θ+c o s2θ/parenrightbig dθ=/contintegraldisplay dθ=2π, which is certainly not zero. The source of the problem is that at x=0a n d y= 0 the function blows up. Thus we can only say that the curl of the function is zero except along the z-axis. If we try to exclude the z-axis from the region of integration, the region becomes multiply connected. In a multiply connected region, Stokes’ theorem does not apply. To make it simply connected, we can cut the region, such as along the y= 0 plane shown in Fig. 2.17 (or along any other direction). Within the simply connected region/integraltextB AF·dr=θB−θA.It will be equal to zero if A and B are inﬁnitesimally close. However, if the integral is across the cut line, as long as A and B are on the diﬀerent side of the cut, no matter how close are A and B, there is a sudden jump of 2 π. 2.7 Further Vector Diﬀerential Operations There are several combinations of vector operations involving the del ∇oper- ator which appear frequently in applications. They all follow the general rules of ordinary derivatives. The distributive rules are straightforward. With the deﬁnition the del operator ∇, one can readily verify ∇(ϕ1+ϕ2)=∇ϕ1+∇ϕ2; (2.90) ∇·(A+B)=∇·A+∇·B; (2.91) ∇×(A+B)=∇×A+∇×B. (2.92) However, the product rules are not so simple because there are more than one way to form a vector product. 2.7 Further Vector Diﬀerential Operations 79 2.7.1 Product Rules The following is a list of useful product rules: ∇(ϕψ)=ϕ∇ψ+ψ∇ϕ, (2.93) ∇·(ϕA)=∇ϕ·A+ϕ∇·A, (2.94) ∇×(ϕA)=∇ϕ×A+ϕ∇×A, (2.95) ∇·(A×B)=(∇×A)·B−(∇×B)·A, (2.96) ∇×(A×B)=(∇·B)A−(∇·A)B+(B·∇)A−(A·∇)B,(2.97) ∇(A·B)=(A·∇)B+(B·∇)A+A×(∇×B)+B×(∇×A).(2.98) They can be veriﬁed by expanding both sides in terms of their Cartesian components. For example, ∇(ϕψ)=i∂ ∂x(ϕψ)+j∂ ∂y(ϕψ)+k∂ ∂z(ϕψ) =iϕ∂ ∂xψ+jϕ∂ ∂yψ+kϕ∂ ∂zψ +iψ∂ ∂xϕ+jψ∂ ∂yϕ+kψ∂ ∂zϕ =ϕ∇ψ+ψ∇ϕ. (2.99) Similarly, ∇·(ϕA)=∂ ∂x(ϕAx)+∂ ∂y(ϕAy)+∂ ∂z(ϕAz) =/parenleftbigg∂ϕ ∂xAx+ϕ∂Ax ∂x/parenrightbigg +/parenleftbigg∂ϕ ∂yAy+ϕ∂Ay ∂y/parenrightbigg +/parenleftbigg∂ϕ ∂zAz+ϕ∂Az ∂z/parenrightbigg =/parenleftbigg∂ϕ ∂xAx+∂ϕ ∂yAy+∂ϕ ∂zAz/parenrightbigg +ϕ/parenleftbigg∂Ax ∂x+∂Ay ∂y+∂Az ∂z/parenrightbigg =∇ϕ·A+ϕ∇·A. (2.100) Clearly it will be very tedious to explicitly prove the rest of the product rules in this way. More “elegant” proofs will be given in the chapter of tensor analysis. Here, we will use the following formal procedure to establish these relations. The procedure consists of (1) ﬁrst using ∇as a diﬀerential operator and (2) then treating ∇as if it were a regular vector. This procedure is a mnemonic device to give correct results. Since ∇is a linear combination of diﬀerential operators, we require it to obey the product rule of diﬀerentiation. When ∇operates on a product, the result is the sum of two derivatives obtained by holding one of the factors constant and allowing the other to be operated on by ∇.As a matter of notation, we attach to ∇a subscript indicating the one factor upon which 80 2 Vector Calculus it is currently allowed to operate, and the other factor is kept constant. For instance, ∇×(ϕA)=∇ϕ×(ϕA)+∇A×(ϕA). Since ∇A×(ϕA) means that ϕis a constant, it is then clear ∇A×(ϕA)=ϕ∇A×A=ϕ∇×A, where the subscript Ais omitted from the right-hand side, since it is clear what ∇operates on when it is followed by just one factor. Similarly, ∇ϕ×(ϕA) means Ais constant. In this case it is easy to show ∇ϕ×(ϕA)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z ϕAxϕAyϕAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ϕ ∂x∂ϕ ∂y∂ϕ ∂z AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=∇ϕ×A. Thus, ∇×(ϕA)=∇ϕ×(ϕA)+∇A×(ϕA)=∇ϕ×A+ϕ∇×A. (2.101) For the divergence of a cross product, we start with ∇·(A×B)=∇A·(A×B)+∇B·(A×B) =∇A·(A×B)−∇B·(B×A). Recall the scalar triple product a·(b×c),the dot ( ·) and the cross ( ×)c a n be interchanged a·(b×c)=(a×b)·c.Treating ∇Aas a vector, we have ∇A·(A×B)=(∇A×A)·B=(∇×A)·B, where the subscript Ais dropped in the last step because the meaning is clear without it. Similarly, ∇B·(B×A)=(∇×B)·A. Therefore, ∇·(A×B)=(∇×A)·B−(∇×B)·A. (2.102) For the curl of a cross product, we will use the analogy of the vector triple product a×(b×c)=(a·c)b−(a·b)c. ∇×(A×B)=∇A×(A×B)+∇B×(A×B), ∇A×(A×B)=(∇A·B)A−(∇A·A)B=(B·∇A)A−(∇A·A)B. In the last step, we have used the relation ( ∇A·B)A=(B·∇A)A,sinceB is regarded as a constant. Similarly, ∇B×(A×B)=(∇B·B)A−(∇B·A)B=(∇B·B)A−(A·∇B)B. 2.7 Further Vector Diﬀerential Operations 81 Therefore ∇×(A×B)=(B·∇)A−(∇·A)B+(∇·B)A−(A·∇)B,(2.103) where we have dropped the subscripts because the meaning is clear without them. The product rule of the gradient of a dot product vis more cumbersome, ∇(A·B)=∇A(A·B)+∇B(A·B). To work out ∇A(A·B),we use the property of the vector triple product a×(b×c)=b(a·c)−(a·b)c, A×(∇B×B)=∇B(A·B)−(A·∇B)B. Hence, ∇B(A·B)=(A·∇B)B+A×(∇B×B). Similarly, ∇A(A·B)=∇A(B·A)=(B·∇A)A+B×(∇A×A). Dropping the subscripts when they are not necessary, we have ∇(A·B)=(B·∇)A+B×(∇×A)+(A·∇)B+A×(∇×B).(2.104) 2.7.2 Second Derivatives Several second derivatives can be constructed by applying ∇twice. The fol- lowing four identities of second derivatives are of great interests: ∇×∇ϕ=0, (2.105) ∇·∇×A=0, (2.106) ∇×(∇×A)=∇(∇·A)−∇2A, (2.107) ∇·(∇ϕ×∇ψ)=0. (2.108) The ﬁrst identity states that the curl of the gradient of a scalar function is identically equal to zero. This can be shown by direct expansion. ∇×∇ϕ=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z ∂ϕ ∂x∂ϕ ∂y∂ϕ ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=i/parenleftbigg∂2ϕ ∂y∂z−∂2ϕ ∂z∂y/parenrightbigg +j/parenleftbigg∂2ϕ ∂z∂x−∂2ϕ ∂x∂z/parenrightbigg +k/parenleftbigg∂2ϕ ∂x∂y−∂2ϕ ∂y∂x/parenrightbigg =0, (2.109) 82 2 Vector Calculus provided the second cross partial derivatives of ϕare continuous (which are generally satisﬁed by functions of interests). In such a case, the order of dif- ferentiation is immaterial. The second identity states that the divergence of curl of a vector function is identically equal to zero. This can also be shown by direct calculation. ∇·∇×A=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂ ∂x∂ ∂y∂ ∂z ∂ ∂x∂ ∂y∂ ∂z AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. (2.110) It is understood that the determinant is to be expanded along the ﬁrst row. Again if the partial derivatives are continuous, this determinant with two identical rows is equal to zero. The curl curl identity is equally important and is worthwhile to commit to memory. For the mnemonic purpose, we can use the analogy of the vec- tor triple product a×(b×c)=b(a·c)−(a·b)c,with∇,∇,Aasa,b,c, respectively. Thus, the vector triple product suggests ∇×(∇×A)=∇(∇·A)−(∇·∇)A. (2.111) The∇·∇is a scalar operator. Because it appears often in physics, it has given a special name – the Laplacian , or just ∇2 ∇·∇=/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg ·/parenleftbigg i∂ ∂x+j∂ ∂y+k∂ ∂z/parenrightbigg =∂2 ∂x2+∂2 ∂y2+∂2 ∂z2=∇2. (2.112) Therefore, (2.111) can be written as ∇×(∇×A)=∇(∇·A)−∇2A. (2.113) Expanding both sides of this equation in rectangular coordinates, one can readily verify that this is indeed an identity. Since∇2is a scalar operator, when it operates on a vector, it means the same operation on each component of the vector ∇2A=i∇2Ax+j∇2Ay+k∇2Az. (2.114) 2.7 Further Vector Diﬀerential Operations 83 The identity (2.108) follows from ∇·(A×B)=∇×A·B−∇×B·A. Since ∇ϕand∇ψare two diﬀerent vectors, ∇·(∇ϕ×∇ψ)=∇×∇ϕ·∇ψ−∇×∇ψ·∇ϕ. Now∇×∇ϕ=∇×∇ψ=0,therefore ∇·(∇ϕ×∇ψ)=0. (2.115) Example 2.7.1. Show that ∇×A=B,ifA=1 2B×randBis a constant vector, ﬁrst by direct expansion, then by the formula of the curl of a cross product. Solution 2.7.1. Method I ∇×A=1 2∇×(B×r)=1 2∇×/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk BxByBz xyz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =1 2∇×[i(Byz−Bzy)+j(Bzx−Bxz)+k(Bxy−Byx)] =1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z (Byz−Bzy)(Bzx−Bxz)(Bxy−Byx)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =1 2[i2Bx+j2By+k2Bz]=B. Method II 1 2∇×(B×r)=1 2[(∇·r)B−(∇·B)r+(r·∇)B−(B·∇)r] =1 2[(∇·r)B−(B·∇)r] (since Bis a constant) , (∇·r)B=3B(see Example 2.5.1) , (B·∇)r=B(see Example 2.4.2) , 1 2∇×(B×r)=1 2[3B−B]=B. 84 2 Vector Calculus Example 2.7.2. Show that ∇×/parenleftbig ∇2A/parenrightbig =∇2(∇×A). Solution 2.7.2. Since ∇×(∇×A)=∇(∇·A)−∇2A, ∇2A=∇(∇·A)−∇×(∇×A), ∇×/parenleftbig ∇2A/parenrightbig =∇×[∇(∇·A)−∇×(∇×A)]. Since curl gradient is equal to zero, ∇×∇(∇·A)=0.Using curl curl formula again, we have ∇×/parenleftbig ∇2A/parenrightbig =−∇×∇×(∇×A)=−/braceleftbig ∇(∇·(∇×A)−∇2(∇×A)/bracerightbig . Since divergence of a curl is equal to zero, ∇·(∇×A)=0,therefore ∇×/parenleftbig ∇2A/parenrightbig =∇2(∇×A). Example 2.7.3. If ∇·E=0,∇×E=−∂ ∂tH, ∇·H=0,∇×H=∂ ∂tE, show that ∇2E=∂2 ∂t2E;∇2H=∂2 ∂t2H. Solution 2.7.3. ∇×(∇×E)=∇×/parenleftbigg −∂ ∂tH/parenrightbigg =−∂ ∂t(∇×H)=−∂ ∂t/parenleftbigg∂ ∂tE/parenrightbigg =−∂2 ∂t2E, ∇×(∇×E)=∇(∇·E)−∇2E=−∇2E(since ∇·E=0 ). Therefore ∇2E=∂2 ∂t2E. Similarly, ∇×(∇×H)=∇×/parenleftbigg∂ ∂tE/parenrightbigg =∂ ∂t(∇×E)=∂ ∂t/parenleftbigg −∂ ∂tH/parenrightbigg =−∂2 ∂t2H, ∇×(∇×H)=∇(∇·H)−∇2H=−∇2H(since ∇·H=0 ). It follows that ∇2H=∂2 ∂t2H. 2.8 Further Integral Theorems 85 2.8 Further Integral Theorems There are many other integral identities that are useful in physical applica- tions. They can be derived in a variety of ways. Here we discuss some of the most useful ones and show that they all follow from the fundamental theorems of gradient, divergence, and curl. 2.8.1 Green’s Theorem The following integral identities are all named after George Green (1793– 1841). To distinguish them, we adopt the following terminology. Green’s Lemma: /contintegraldisplay C[f(x,y)dx+g(x,y)dy]=/integraldisplay/integraldisplay S/parenleftbigg∂g ∂x−∂f ∂y/parenrightbigg dxdy, (2.116) Green’s Theorem: /integraldisplay/integraldisplay /circlecopyrt Sϕ∇ψ·nda=/integraldisplay/integraldisplay/integraldisplay V(∇ϕ·∇ψ+ϕ∇2ψ)dV, (2.117) Symmetrical form of Green’s Theorem: /integraldisplay/integraldisplay /circlecopyrt S(ϕ∇ψ−ψ∇ϕ)·nda=/integraldisplay/integraldisplay/integraldisplay V(ϕ∇2ψ−ψ∇2ϕ)dV. (2.118) To prove Green’s Lemma, we start with Stokes’ theorem /contintegraldisplay CA·dr=/integraldisplay S(∇×A)·nda. With the curve C lying entirely on the xyplane, A·dr=(iAx+jAy+kAz)·(idx+jdy)=Axdx+Aydy, andnis equal to k,the unit vector in the zdirection, (∇×A)·nda=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z AxAyAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle·kdxdy=/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg dxdy. Thus we have /contintegraldisplay C(Axdx+Aydy)=/integraldisplay/integraldisplay S/parenleftbigg∂Ay ∂x−∂Ax ∂y/parenrightbigg dxdy. 86 2 Vector Calculus SinceAin Stokes’ theorem can be any vector function, Green’s Lemma follows withAx=f(x,y),andAy=g(x,y). To prove Green’s theorem, we start with the divergence theorem /integraldisplay/integraldisplay/integraldisplay V∇·(ϕ∇ψ)dV=/integraldisplay/integraldisplay /circlecopyrt Sϕ∇ψ·nda. Using the identity ∇·(ϕ∇ψ)=∇ϕ·∇ψ+ϕ∇2ψ, we have /integraldisplay/integraldisplay/integraldisplay V(∇ϕ·∇ψ+ϕ∇2ψ)dV=/integraldisplay/integraldisplay /circlecopyrt Sϕ∇ψ·nda, (2.119) which is Green’s theorem (2.117). Clearly (2.119) is equally valid when ϕandψare interchanged /integraldisplay/integraldisplay/integraldisplay V(∇ψ·∇ϕ+ψ∇2ϕ)dV=/integraldisplay/integraldisplay /circlecopyrt Sψ∇ϕ·nda. (2.120) Taking the diﬀerence of the last two equations, we obtain the symmetric form of the Green’s theorem /integraldisplay/integraldisplay/integraldisplay V(ϕ∇2ψ−ψ∇2ϕ)dV=/integraldisplay/integraldisplay /circlecopyrt S(ϕ∇ψ−ψ∇ϕ)·nda. 2.8.2 Other Related Integrals The divergence theorem can take some other alternative forms. Let ϕbe a scalar function and Cbe an arbitrary constant vector. Then, /integraldisplay/integraldisplay/integraldisplay V∇·(ϕC)dV=/integraldisplay/integraldisplay /circlecopyrt SϕC·nda, ∇·(ϕC)=∇ϕ·C+ϕ∇·C=∇ϕ·C, sinceCis constant and ∇·C=0. /integraldisplay/integraldisplay/integraldisplay V∇·(ϕC)dV=/integraldisplay/integraldisplay/integraldisplay V∇ϕ·CdV=C·/integraldisplay/integraldisplay/integraldisplay V∇ϕdV. /integraldisplay/integraldisplay /circlecopyrt SϕC·nda=C·/integraldisplay/integraldisplay /circlecopyrt Sϕnda. Therefore the divergence theorem can be written as C·/bracketleftbigg/integraldisplay/integraldisplay/integraldisplay V∇ϕdV−/integraldisplay/integraldisplay /circlecopyrt Sϕnda/bracketrightbigg =0. 2.8 Further Integral Theorems 87 SinceCis arbitrary, the terms in the brackets must be zero. Thus we have another interesting relation between volume integral and surface integral /integraldisplay/integraldisplay/integraldisplay V∇ϕdV=/integraldisplay/integraldisplay /circlecopyrt Sϕnda. (2.121) Similarly, let Abe a vector function and C, an arbitrary constant vector. A×Cis another vector function. The divergence theorem can be written as /integraldisplay/integraldisplay/integraldisplay V∇·(A×C)dV=/integraldisplay/integraldisplay /circlecopyrt S(A×C)·nda. Since ∇·(A×C)=(∇×A)·C−(∇×C)·A=(∇×A)·C, (A×C)·n=−(C×A)·n=−C·A×n, therefore /integraldisplay/integraldisplay/integraldisplay V∇·(A×C)dV=C·/integraldisplay/integraldisplay/integraldisplay V∇×AdV, /integraldisplay/integraldisplay /circlecopyrt S(A×C)·nda=−C·/integraldisplay/integraldisplay /circlecopyrt SA×nda. Thus we have another form of the divergence theorem /integraldisplay/integraldisplay/integraldisplay V∇×AdV=−/integraldisplay/integraldisplay /circlecopyrt SA×nda. (2.122) This exploitation of the arbitrary nature of a part of a problem is a very useful technique. In the following examples some alternative forms of Stokes’ theorem will be derived using this technique. Example 2.8.1. Show that/contintegraldisplay Cϕdr=−/integraldisplay/integraldisplay S∇ϕ×nda. Solution 2.8.1. LetCbe an arbitrary constant vector. By Stokes’ theorem we have /contintegraldisplay CϕC·dr=/integraldisplay/integraldisplay S∇×(ϕC)·nda. SinceCis a constant and ∇×C=0, ∇×ϕC=∇ϕ×C+ϕ∇×C=∇ϕ×C, /integraldisplay/integraldisplay S∇×(ϕC)·nda=/integraldisplay/integraldisplay S∇ϕ×C·nda. 88 2 Vector Calculus Furthermore, ∇ϕ×C·n=−C×∇ϕ·n=−C·(∇ϕ×n). Therefore /integraldisplay/integraldisplay S∇×(ϕC)·nda=−C·/integraldisplay/integraldisplay S∇ϕ×nda. With /contintegraldisplay CϕC·dr=C·/contintegraldisplay Cϕdr, we can write Stokes’ theorem as C·/contintegraldisplay Cϕdr=−C·/integraldisplay S∇ϕ×nda. Again since Cis an arbitrary constant vector, it follows that /contintegraldisplay Cϕdr=−/integraldisplay/integraldisplay S∇ϕ×nda. (2.123) Example 2.8.2. Show that/contintegraldisplay Cr×dr=2/integraldisplay Sndawhereris the position vector from an origin that can be chosen at any point in space. Solution 2.8.2. To prove this, we use an arbitrary constant vector Cand start with Stokes’ theorem, /contintegraldisplay C(C×r)·dr=/integraldisplay/integraldisplay S∇×(C×r)·nda. Since /contintegraldisplay C(C×r)·dr=/contintegraldisplay CC·r×dr=C·/contintegraldisplay Cr×dr, and ∇×(C×r)=2C(see example 2.7.1) , /integraldisplay/integraldisplay S∇×(C×r)·nda=/integraldisplay/integraldisplay S2C·nda=2C·/integraldisplay/integraldisplay Snda, it follows C·/contintegraldisplay Cr×dr=C·2/integraldisplay/integraldisplay Snda. SinceCis an arbitrary constant vector, the integral identity /contintegraldisplay Cr×dr=2/integraldisplay/integraldisplay Snda (2.124) 2.9 Classiﬁcation of Vector Fields 89 must hold. This integral identity is of some importance in electrodynamics. This integral also shows that the area Aof a ﬂat surface Senclosed by a curve C is given by A=/integraldisplay/integraldisplay Sda=1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/contintegraldisplay Cr×dr/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (2.125) 2.9 Classiﬁcation of Vector Fields 2.9.1 Irrotational Field and Scalar Potential If∇×F=0in a simply connected region, we say Fis anirrotational vector ﬁeld. An irrotational ﬁeld is also known as a conservative vector ﬁeld .W eh a v e seen that if ∇×F=0,the line integral/integraltextB AF·dris independent of path. This means, as shown in Sect. 2.4.3, that Fcan be expressed as the gradient of a scalar function ϕ,known as the scalar potential. Because of Stokes’ theorem /contintegraldisplay CF·dr=/integraldisplay/integraldisplay S∇×F·nda, an irrotational ﬁeld Fis characterized by any of the following equivalent conditions: (a)∇×F=0, (b)/contintegraldisplay F·dr= 0 for any closed loop, (c)/integraltextB AF·dris independent of path, (d)F=−∇ϕ. The sign in (d) is arbitrary, since ϕis yet to be speciﬁed. In hydrodynamics, often a plus sign (+) is chosen for the velocity potential. Here we have followed the convention in choosing a minus sign ( −) for the convenience of establishing the principle of conservation of energy. Conservative Force Field. To see why an irrotational ﬁeld is also called a conservative vector ﬁeld, consider F(x,y,z) as the force in Newton’s equation of motion F(x,y,z)=ma=mdv dt. (2.126) SinceFis irrotational, so F(x,y,z)=−∇ϕ(x,y,z). (2.127) 90 2 Vector Calculus Therefore mdv dt=−∇ϕ. (2.128) Take dot product of both sides with d rand integrate. The left-hand side becomes /integraldisplay mdv dt·dr=/integraldisplay mdv dt·dr dtdt=/integraldisplay mdv dt·vdt =/integraldisplayd dt/parenleftbigg1 2mv·v/parenrightbigg dt=/integraldisplay d/parenleftbigg1 2mv2/parenrightbigg =1 2mv2+ constant. (2.129) The right-hand side becomes /integraldisplay (−∇ϕ)·dr=−/integraldisplay dϕ=−ϕ+ constant. (2.130) Equating the results of the two sides of (2.128) gives 1 2mv2+ϕ= constant. (2.131) The expression1 2mv2is deﬁned as the kinetic energy and ϕ(x,y,z)i s t h e potential energy in classical mechanics. The sum of the two is the total energy. The last equation says that no matter where and when the total energy is evaluated, it must be equal to the same constant. This is the principle of conservation of energy. Although we have used classical mechanics to introduce the idea of con- servative ﬁeld, the idea can be generalized. Any vector ﬁeld v(x,y,z)w h i c h can be expressed as the gradient of a scalar ﬁeld ϕ(x,y,z) is called a conser- vative ﬁeld and the scalar function ϕis called the scalar potential. Since ∇ϕ=∇(ϕ+ constant), the scalar potential is deﬁned up to an additive constant. Example 2.9.1. Determine which of the following is an irrotational (or conser- vative) ﬁeld: (a) F1=6xyi+( 3x2−3y2)j,(b)F2=xyi−yj Solution 2.9.1. ∇×F1=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z 6xy3x2−3y20/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−i∂ ∂z/parenleftbig 3x2−3y2/parenrightbig +j∂ ∂z(6xy) +k/parenleftbigg∂ ∂x/parenleftbig 3x2−3y2/parenrightbig −∂ ∂y(6xy)/parenrightbigg =k(6x−6x)=0. 2.9 Classiﬁcation of Vector Fields 91 ∇×F2=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z xy y 0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−i∂ ∂zy+j∂ ∂z(xy) +k/parenleftbigg∂ ∂xy−∂ ∂y(xy)/parenrightbigg =−xk/negationslash=0. Therefore F1is an irrotational ﬁeld and F2is not an irrotational ﬁeld. We have shown explicitly, in the examples of Sect. 2.4.3, that the line integral/integraltextB AF1·dris independent of path and/integraltextB AF2·dris dependent on the path. Example 2.9.2. Show that the force ﬁeld F=−(2ax+by)i−bxj−ckis conservative, and ﬁnd ϕsuch that −∇ϕ=F. Solution 2.9.2. ∇×F=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z −(2ax+by)−bx−c/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. Therefore, Fis conservative, there must exist a ϕsuch that −∇ϕ=F. −∂ϕ ∂x=Fx=−(2ax+by)=⇒ϕ=ax2+bxy+f(y,z). −∂ϕ ∂y=Fy=−bx,but−∂ϕ ∂y=−bx−∂ ∂yf(y,z) ∂ ∂yf(y,z)= 0= ⇒f(y,z)=g(z) −∂ϕ ∂z=Fz=−c,but−∂ϕ ∂z=−∂ ∂zg(z) ∂ ∂zg(z)=c=⇒g(z)=cz+k. ϕ=ax2+bxy+cz+k. Example 2.9.3. Suppose a particle of mass mis moving in the force ﬁeld of the last example, and at t= 0 the particle passes through the origin with speed v0.What will the speed of the particle be if and when it passes through the pointr=i+2j+k? Solution 2.9.3. The conservation of energy requires 1 2mv2+ϕ(r)=1 2mv2 0+ϕ(0). 92 2 Vector Calculus v2=v2 0+2 m/bracketleftbig k−/parenleftbig ax2+bxy+cz+k/parenrightbig/bracketrightbig . Atx=1,y=2,z=1 : v2=v2 0+2 m(a+2b+c). 2.9.2 Solenoidal Field and Vector Potential If the ﬁeld Fis divergence-less (that is ∇·F= 0) everywhere in a simply con- nected region ,the ﬁeld is called solenoidal . For a solenoidal ﬁeld, the surface integral of F·ndaover any closed surface is zero, since by the divergence theorem /integraldisplay/integraldisplay /circlecopyrt SF·nda=/integraldisplay/integraldisplay/integraldisplay ∇·FdV=0. Furthermore, Fcan be expressed as the curl of another vector function A, F=∇×A. The vector function Ais known as the vector potential of the ﬁeld F. The existence of vector potentials for solenoidal ﬁelds can be shown in the following way. For any given solenoidal ﬁeld F(that is, Fx(x,y,z),Fy(x,y,z), andFz(x,y,z) are known), we shall ﬁrst show that it is possible to ﬁnd a vector function Awith one zero component to satisfy F=∇×A.Then a general formula for all possible vector potentials can be found. Let us take Az=0,and try to ﬁnd AxandAyinA=Axi+Ayjso that ∇×A=F: ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleijk ∂ ∂x∂ ∂y∂ ∂z AxAy0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−i∂ ∂zAy+j∂ ∂zAx+k/parenleftbigg∂ ∂xAy−∂ ∂yAx/parenrightbigg =iFx+jFy+kFz. F o rt h i st oh o l d ,w em u s th a v e ∂ ∂zAy=−Fx,∂ ∂zAx=Fy,∂ ∂xAy−∂ ∂yAx=Fz. (2.132) From the ﬁrst two equations we have Ay=−/integraldisplay Fx(x,y,z)dz+f(x,y), (2.133) Ax=/integraldisplay Fy(x,y,z)dz+g(x,y). (2.134) 2.9 Classiﬁcation of Vector Fields 93 With AyandAxso obtained, if we can show∂ ∂xAy−∂ ∂yAx=Fz,then we would have proved that ∇×A=F. Using (2.133) and (2.134), we have ∂ ∂xAy−∂ ∂yAx=−/integraldisplay/parenleftbigg∂ ∂xFx+∂ ∂yFy/parenrightbigg dz+h(x,y). SinceFis solenoidal, ∇·F= 0 which can be written as ∂ ∂xFx+∂ ∂yFy=−∂ ∂zFz. Thus, ∂ ∂xAy−∂ ∂yAx=/integraldisplay∂ ∂zFzdz+h(x,y). With proper choice of h(x,y),we can certainly make /integraldisplay∂ ∂zFzdz+h(x,y)=Fz. This proof clearly indicates that Ais not unique. If A/primeis another vector potential, then both ∇×Aand∇×A/primeare equal to F.Therefore ∇×(A/prime− A)=0.Since ( A/prime−A) is irrotational, it follows that A/prime−A=∇ψ.Thus we conclude that with one Aobtained from the above procedure, all other vector potentials are of the form A+∇ψwhere ψis any scalar function. It is also possible for us to require the vector potential to be solenoidal. If we ﬁnd a vector potential Awhich is not solenoidal (that is, ∇×A=Fand ∇·A/negationslash= 0), we can construct another vector potential A/primewhich is solenoidal (∇·A/prime= 0). Let A/prime=A+∇ψ, ∇×A/prime=∇×A+∇×∇ψ=∇×A, ∇·A/prime=∇·A+∇2ψ. If we choose ψsuch that ∇2ψ+∇·A=0,then we will have ∇·A/prime=0.The following example will make this clear. Example 2.9.4. Show that F=x2i+3xz2j−2xzkis solenoidal, and ﬁnd a vector potential Asuch that ∇×A=Fand∇·A=0. Solution 2.9.4. Since ∇·F=∂ ∂xx2+∂ ∂y/parenleftbig 3xz2/parenrightbig +∂ ∂z(−2xz)=0, this shows that Fis solenoidal. Let A1=Axi+Ayjand∇×A1=F.B y (2.132) 94 2 Vector Calculus ∂ ∂zAy=−Fx=−x2,=⇒Ay=−x2z+f(x,y), ∂ ∂zAx=Fy=3xz2,=⇒Ax=xz3+g(x,y), ∂ ∂xAy−∂ ∂yAx=Fz=−2xz, =⇒−2xy+∂f ∂x+∂g ∂y=−2xy. Since fandgare arbitrary, the simplest choice is to make f=g=0.Thus, A1=xz3i+−x2zj,but∇·A1=z3/negationslash=0.Let A=A1+∇ψ,∇·A=∇·A1+∇2ψ=z3+∇2ψ. If∇·A=0,then∇2ψ=−z3.A simple solution of this equation is ψ=−1 20z5. Since ∇ψ=∇/parenleftbigg −1 20z5/parenrightbigg =−1 4z4k, A=A1+∇ψ=xz3i−x2zj−1 4z4k. It can be readily veriﬁed that ∇·A=∂ ∂x/parenleftbig xz3/parenrightbig +∂ ∂y/parenleftbig −x2z/parenrightbig +∂ ∂z(−1 4z4)=0, ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z xz3−x2z−1 4z4/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=x2i+3xz2j−2xzk=F. This vector potential is still not unique. For example, we can assume A2=Ayj+ Azkand∇×A2=F. Following the same procedure, we obtain A2=−x2zj−3 2x2z2k. Now, ∇·A2=−3x2z/negationslash=0.We can ﬁnd A/primesuch that A/prime=A2+∇ψand ∇·A/prime=0.It follows that ∇2ψ=−∇·A2=3x2z.A simple solution is ψ=1 4x4z.Therefore ∇ψ=x3zi+1 4x4k,and A/prime=A2+∇ψ=x3zi−x2zj+/parenleftbigg1 4x4−3 2x2z2/parenrightbigg k. Again, it can be readily veriﬁed that ∇×A/prime=Fand∇·A/prime=0. 2.10 Theory of Vector Fields 95 Clearly, AandA/primeare not identical. They must diﬀer by an additive gradient A/prime=A+∇χ. (2.135) Now∇·A/prime=∇·A+∇2χand∇·A/prime=∇·A=0.Therefore ∇2χ=0. (2.136) In this particular case, ∇χ=A/prime−A=/parenleftbig x3z−xz3/parenrightbig i+/parenleftbigg1 4x4−3 2x2z2+1 4z4/parenrightbigg k, χ=1 4x4z−1 2x2z3+1 20z5, ∇2χ=/parenleftbigg∂2 ∂x2+∂2 ∂y2+∂2 ∂z2/parenrightbigg/parenleftbigg1 4x4z−1 2x2z3+1 20z5/parenrightbigg =0. Equation (2.135) is an example of what is known as a gauge transfor- mation . The requirement (2.136) leads to the so-called Coulomb gauge .T h e vector potential is not as useful as the scalar potential in computation. It is in the conceptual development of time-dependent problems, especially in electrodynamics, that the vector potential is essential. 2.10 Theory of Vector Fields 2.10.1 Functions of Relative Coordinates Very often we deal with functions that depend only on the diﬀerence of the coordinates. For example, the electric ﬁeld at the point ( x,y,z) due to the ap o i n tc h a r g ea t( x/prime,y/prime,z/prime) is a function solely of ( x−x/prime),(y−y/prime),(z−z/prime). The point ( x,y,z) is called the ﬁeld point and the point ( x/prime,y/prime,z/prime) is called source point. The relative position vector Rshown in Fig. 2.18 can be written as R=r−r/prime=(x−x/prime)i+(y−y/prime)j+(z−z/prime)k. (2.137) The distance between these two points is R=|r−r/prime|=[ (x−x/prime)2+(y−y/prime)2+(z−z/prime)2]1/2. (2.138) Letf(R) be a function of the relative position vector. This function could be a scalar or a component of a vector. Functions of this type have some important properties. Let us deﬁne X=(x−x/prime),Y=(y−y/prime),Z=(z−z/prime). Using the chain rule of diﬀerentiation, we ﬁnd ∂f ∂x=∂f ∂X∂X ∂x=∂f ∂X;∂f ∂x/prime=∂f ∂X∂X ∂x/prime=−∂f ∂X. 96 2 Vector Calculus xyz rR r9(x, y, z) (x 9, y 9, z 9) Fig. 2.18. Relative coordinates R=r−r/prime Similar expressions can be found for the yandzderivatives. It follows ∂f ∂x=−∂f ∂x/prime,∂f ∂y=−∂f ∂y/prime,∂f ∂z=−∂f ∂z/prime. (2.139) Corresponding to the gradient ∇with respect to the ﬁeld point ∇f=i∂f ∂x+j∂f ∂y+k∂f ∂z, we deﬁne the gradient ∇/primewith respect to the source point ∇/primef=i∂f ∂x/prime+j∂f ∂y/prime+k∂f ∂z/prime. It follows from (2.139) that ∇f=−∇/primef. (2.140) This shows that when we deal with functions of the relative coordinates the ∇and∇/primeoperator can be interchanged provided the sign is also changed. Similar calculations can be used to show ∇·A(R)=−∇/prime·A(R), (2.141) ∇×A(R)=−∇/prime×A(R), (2.142) and ∇2f(R)=∇/prime2f(R). (2.143) 2.10 Theory of Vector Fields 97 Example 2.10.1. Show that (a) ∇·R=3,(b)∇×R=0,(c)∇×f(R) R=0,and (d) ∇·f(R)R=d(R) dRR+3f(R). Solution 2.10.1. ∇·R=∂ ∂x(x−x/prime)+∂ ∂y(y−y/prime)+∂ ∂z(z−z/prime)=3, ∇×R=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂ ∂x∂ ∂y∂ ∂z (x−x/prime)(y−y/prime)(z−z/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0, ∇×f(R)R=∇f(R)×R+f(R)∇×R =df(R) dR/hatwideR×R=0, ∇·f(R)R=∇f(R)·R+f(R)∇·R =df(R) dR/hatwideR·R+3f(R)=df(R) dRR+3f(R). For functions that depend only on the distance between the two points, the gradient takes a simple form: ∇f(R)=i∂f(R) ∂x+j∂f(R) ∂y+k∂f(R) ∂z. By the chain rule ∂f(R) ∂x=df(R) dR∂R ∂x, ∂R ∂x=∂ ∂x/radicalBig (x−x/prime)2+(y−y/prime)2+(z−z/prime)2=x−x/prime R. With similar expressions for yandzderivatives, we have ∇f(R)=df(R) dR/parenleftbigg ix−x/prime R+jy−y/prime R+kz−z/prime R/parenrightbigg =df(R) dRR R=df(R) dR/hatwiderR, (2.144) 98 2 Vector Calculus where/hatwideRis the unit vector in the direction of R.In particular ∇R=/hatwiderR, (2.145) ∇Rn=nRn−1/hatwideR. (2.146) Forn=−1 ∇1 R=−1 R2/hatwideR. (2.147) This last expression is an especially important case because −/hatwideR/R2is the “radial inverse-square-law” ﬁeld. This vector ﬁeld (with appropriate multi- plicative constants) describes two of the most important fundamental forces in nature, namely the gravitational force ﬁeld and the Coulomb force ﬁeld of a static electric charge. The divergence of this ﬁeld requires our special attention. 2.10.2 Divergence of /hatwideR/|R|2as a Delta Function The divergence of /hatwideR/R2has some peculiar and important properties. Calculated directly ∇·/hatwideR R2=∇·1 R3R=/parenleftbigg ∇1 R3/parenrightbigg ·R+1 R3∇·R =−31 R4/hatwideR·R+31 R3=0, we get zero. On the other hand, as we discussed earlier, the divergence is a measure of the strength of the source of the vector ﬁeld. If it were zero every- where, how could there be any gravitational and electric ﬁelds? Furthermore, if we apply the divergence theorem (2.72) to this function over a sphere of radius Raround the point ( x/prime,y/prime,z/prime), we will get a nonzero result, /integraldisplay/integraldisplay/integraldisplay V∇·/hatwideR R2dV=/integraldisplay/integraldisplay /circlecopyrt S/hatwideR R2·nda=1 R2/integraldisplay/integraldisplay /circlecopyrt S/hatwideR·/hatwideRda =1 R2/integraldisplay/integraldisplay /circlecopyrt Sda=1 R24πR2=4π. (2.148) In the integral, we have used the facts that on the surface of a sphere, the unit normal nis equal to /hatwideRandRis a constant. This integral would be zero if∇·(/hatwideR/R2) were equal to zero everywhere. The source of the problem is at the point R= 0 where /hatwideR/R2blows up and the derivative in the usual sense does not exist. Thus we can only say that the divergence is zero everywhere except at R=0.To ﬁnd out the divergence at R=0,we note that the volume integral (2.148) of the divergence over a sphere is equal to 4 πno matter how small Ris. Evidently the entire contribution must 2.10 Theory of Vector Fields 99 be coming from the point R=0.A useful way to describe this behavior is through the Dirac delta function δ3(r−r/prime). A more detailed description of the delta function is given a later chapter. Here it suﬃces to know that the delta function δ3(r−r/prime) is a sharply peaked function at r=r/primewith the properties δ3(r−r/prime)=/braceleftbigg 0r/negationslash=r/prime ∞r=r/prime (2.149) and /integraldisplay/integraldisplay/integraldisplay all spaceδ3(r−r/prime)d3r=1, (2.150) where d3ris a commonly used symbol for the volume element around the ﬁeld point d3r=dV=dxdydz.It follows that the delta function is characterized by the shifting property /integraldisplay/integraldisplay/integraldisplay all spacef(r)δ3(r−r/prime)d3r=f(r/prime), (2.151) because /integraldisplay/integraldisplay/integraldisplay all spacef(r)δ3(r−r/prime)d3r=/integraldisplay/integraldisplay/integraldisplay all spacef(r/prime)δ3(r−r/prime)d3r, since the value of f(r) is immaterial for r/negationslash=r/primeas the integrand is going to be zero anyway. Furthermore, /integraldisplay/integraldisplay/integraldisplay all spacef(r/prime)δ3(r−r/prime)d3r=f(r/prime)/integraldisplay/integraldisplay/integraldisplay all spaceδ3(r−r/prime)d3r=f(r/prime), since the integration is over d3r.This property can also be written as /integraldisplay/integraldisplay/integraldisplay all spacef(r/prime)δ3(r−r/prime)d3r/prime=f(r), (2.152) where d3r/prime=dx/primedy/primedz/prime. With the delta function, the divergence of /hatwideR/R2can be precisely expressed as ∇·/hatwideR R2=∇·r−r/prime |r−r/prime|3=4πδ3(r−r/prime). (2.153) With this understanding, we see that /integraldisplay/integraldisplay/integraldisplay V∇·/hatwideR R2dV=/integraldisplay/integraldisplay/integraldisplay V∇·r−r/prime |r−r/prime|3d3r=4π/integraldisplay/integraldisplay/integraldisplay Vδ3(r−r/prime)d3r =/braceleftbigg 4πif the volume includes r/prime 0i fr/primeis outside the body.(2.154) 100 2 Vector Calculus Since ∇1 R=−/hatwideR R2, it follows that the Laplacian of/parenleftbig1 R/parenrightbig is given by ∇21 R=∇·∇1 R=−∇·/hatwideR R2=−4πδ3(r−r/prime). (2.155) Example 2.10.2. Evaluate the integral I=/integraldisplay/integraldisplay/integraldisplay V(r3+1 )∇·/hatwider r2dV, where Vis a sphere of radius bcentered at the origin. Solution 2.10.2. Method I . Use the delta function. Since ∇·/hatwider r2=4πδ3(r), I=/integraldisplay/integraldisplay/integraldisplay V(r3+ 1)4πδ3(r)dV=4π(0 + 1) = 4 π. Method II . Use integration by parts. Since f∇·A=∇·(fA)−∇f·A, I=/integraldisplay/integraldisplay/integraldisplay V(r3+1)∇·/hatwider r2dV=/integraldisplay/integraldisplay/integraldisplay V∇·/bracketleftbigg (r3+1 )/hatwider r2/bracketrightbigg dV−/integraldisplay/integraldisplay/integraldisplay V∇(r3+1)·/hatwider r2dV. By the divergence theorem /integraldisplay/integraldisplay/integraldisplay V∇·/bracketleftbigg (r3+1 )/hatwider r2/bracketrightbigg dV=/integraldisplay/integraldisplay /circlecopyrt S(r3+1 )/hatwider r2·/hatwiderda=/integraldisplay/integraldisplay /circlecopyrt S(r+1 r2)da, where Sis the surface of the sphere of radius b.Since on this surface r=b everywhere, therefore /integraldisplay/integraldisplay/integraldisplay V∇·/bracketleftbigg (r3+1 )/hatwider r2/bracketrightbigg dV=/parenleftbigg b+1 b2/parenrightbigg/integraldisplay/integraldisplay /circlecopyrt Sda=/parenleftbigg b+1 b2/parenrightbigg 4πb2=4πb3+4π. Since ∇(r3+1 )=3 r2/hatwider, /integraldisplay/integraldisplay/integraldisplay V∇(r3+1 )·/hatwider r2dV=/integraldisplay/integraldisplay/integraldisplay V3r2/hatwider·/hatwider r2dV=3/integraldisplay/integraldisplay/integraldisplay VdV=34 3πb3=4πb3. Thus we have I=4πb3+4π−4πb3=4π, which is the same as the result of delta function method. This example illus- trates the validity and power of the delta function method. If the volume is not a sphere, as long as it includes the origin, the delta function result is still valid, but the direct integration will be much more diﬃcult to do. 2.10 Theory of Vector Fields 101 2.10.3 Helmholtz’s Theorem The Helmholtz theorem deals with the question of what information we need to determine a vector ﬁeld. Basically, the answer is that if the divergence and the curl of a vector ﬁeld are known, with some boundary conditions the vector ﬁeld can be found uniquely. The Helmholtz theorem states that any vector ﬁeld Fmay be decomposed into the sum of two vectors, one is the gradient of a scalar potential ϕand the other the curl of a vector potential A, F=−∇ϕ+∇×A. (2.156) Furthermore, if F→0on the surface at inﬁnity faster than 1 /Rand∇·F and∇×Fare known everywhere, then ϕ(r)=1 4π/integraldisplay/integraldisplay/integraldisplay∇/prime·F(r/prime) |r−r/prime|d3r/prime, (2.157) A(r)=1 4π/integraldisplay/integraldisplay/integraldisplay∇/prime×F(r/prime) |r−r/prime|d3r/prime. (2.158) To prove this theorem, we ﬁrst construct a vector function G G(r)=/integraldisplay/integraldisplay/integraldisplayF(r/prime) |r−r/prime|d3r/prime. (2.159) Let us apply the Laplacian ∇2to both sides of this equation. Because ∇2 operates only on rand only |r−r/prime|−1contains r,we have ∇2G(r)=/integraldisplay/integraldisplay/integraldisplay/parenleftbigg ∇21 |r−r/prime|/parenrightbigg F(r/prime)d3r/prime. (2.160) Since by (2.155) ∇21 |r−r/prime|=−4πδ3(r−r/prime), it follows from the deﬁnition of the delta function that ∇2G(r)=/integraldisplay/integraldisplay/integraldisplay/parenleftbig −4πδ3(r−r/prime)/parenrightbig F(r/prime)d3r/prime=−4πF(r). (2.161) Therefore F(r)=−1 4π∇2G(r). (2.162) Using the vector identity ∇×(∇×G)=∇(∇·G)−∇2G,we have ∇2G=∇(∇·G)−∇×(∇×G). 102 2 Vector Calculus Thus with ϕ=1 4π(∇·G),A=1 4π(∇×G), the ﬁrst part of the theorem follows from (2.162) F(r)=−1 4π∇2G(r)=−∇ϕ+∇×A. To ﬁnd the explicit expression for ϕ,we start with ϕ(r)=1 4π(∇·G)=1 4π∇·/integraldisplay/integraldisplay/integraldisplayF(r/prime) |r−r/prime|d3r/prime. Since ∇operates only on r,and only |r−r/prime|contains r, ∇·/integraldisplay/integraldisplay/integraldisplayF(r/prime) |r−r/prime|d3r/prime=/integraldisplay/integraldisplay/integraldisplay ∇·F(r/prime) |r−r/prime|d3r/prime=/integraldisplay/integraldisplay/integraldisplay/parenleftbigg ∇1 |r−r/prime|/parenrightbigg ·F(r/prime)d3r/prime. Now ∇1 |r−r/prime|=−∇/prime1 |r−r/prime| and /parenleftbigg ∇/prime1 |r−r/prime|/parenrightbigg ·F(r/prime)=∇/prime·F(r/prime) |r−r/prime|−1 |r−r/prime|∇/prime·F(r/prime), so ϕ(r)=−1 4π/integraldisplay/integraldisplay/integraldisplay ∇/prime·F(r/prime) |r−r/prime|d3r/prime+1 4π/integraldisplay/integraldisplay/integraldisplay1 |r−r/prime|∇/prime·F(r/prime)d3r/prime.(2.163) The ﬁrst integral on the right-hand side can be changed to a surface integral at inﬁnity by the divergence theorem /integraldisplay/integraldisplay/integraldisplay all space∇/prime·F(r/prime) |r−r/prime|d3r/prime=/integraldisplay/integraldisplay S→∞1 |r−r/prime|F(r/prime)·nda/prime. Asr/prime→∞,F(r/prime) goes to zero faster than 1 /r/prime.Hence the surface integral is equal to zero. This follows from the fact that the surface is only proportional tor/prime2,andF(r/prime)/|r−r/prime|goes to zero faster than 1 /r/prime2.Thus only the second integral on the right-hand side of (2.163) remains ϕ(r)=1 4π/integraldisplay/integraldisplay/integraldisplay1 |r−r/prime|∇/prime·F(r/prime)d3r/prime. Similarly, for the vector potential we start with A(r)=1 4π(∇×G)=1 4π/integraldisplay/integraldisplay/integraldisplay ∇×1 |r−r/prime|F(r/prime)d3r/prime =1 4π/integraldisplay/integraldisplay/integraldisplay ∇1 |r−r/prime|×F(r/prime)d3r/prime. 2.10 Theory of Vector Fields 103 Using the identities ∇1 |r−r/prime|×F(r/prime)=−∇/prime1 |r−r/prime|×F(r/prime), ∇/prime1 |r−r/prime|×F(r/prime)=∇/prime×1 |r−r/prime|F(r/prime)−1 |r−r/prime|∇/prime×F(r/prime), we have A(r)=−1 4π/integraldisplay/integraldisplay/integraldisplay ∇/prime×1 |r−r/prime|F(r/prime)d3r/prime+1 4π/integraldisplay/integraldisplay/integraldisplay1 |r−r/prime|∇/prime×F(r/prime)d3r/prime. (2.164) By the integral theorem (2.122) /integraldisplay/integraldisplay/integraldisplay V∇×Pd3r/prime=−/integraldisplay/integraldisplay SP×nda, the ﬁrst integral on the right-hand side of (2.164) can be transformed into a surface integral −1 4π/integraldisplay/integraldisplay/integraldisplay all space∇/prime×1 |r−r/prime|F(r/prime)d3r/prime=1 4π/integraldisplay/integraldisplay S→∞1 |r−r/prime|F(r/prime)×nda/prime, which is zero because F(r/prime)→0 on the surface at inﬁnity faster than 1 /r/prime. Thus (2.164) becomes A(r)=1 4π/integraldisplay/integraldisplay/integraldisplay1 |r−r/prime|∇/prime×F(r/prime)d3r/prime. (2.165) This completes the proof. The divergence and curl of Fare often called the sources of the ﬁeld, since Fcan be found from the knowledge of them. The point rwhere we evaluate Fis called the ﬁeld point. The point r/primewhere the sources are evaluated for the purpose of integration is called the source point. The volume element d3r/primeis at the source point. The function ϕand Aare called scalar and vector potentials, respectively, because Fis obtained from them by diﬀerentiation. It should be noted that while the ﬁeld F(r) so determined is unique, the potentials ϕandAare not. Any constant can be added to ϕ, since ∇(ϕ+C)=∇ϕ.The gradient of any scalar function can be added to A, since∇×(A+∇ψ)=∇×A. Example 2.10.3. IfA(r)=1 4π/integraldisplay/integraldisplay/integraldisplay1 |r−r/prime|∇/prime×F(r/prime)d3r/primeandF(r/prime)g o e st o zero on the surface at inﬁnity faster than 1 /r/prime,show that ∇·A(r)=0. 104 2 Vector Calculus Solution 2.10.3. Since ∇operates only on r, ∇·A(r)=1 4π/integraldisplay/integraldisplay/integraldisplay ∇·1 |r−r/prime|∇/prime×F(r/prime)d3r/prime =1 4π/integraldisplay/integraldisplay/integraldisplay ∇1 |r−r/prime|·∇/prime×F(r/prime)d3r/prime. Now ∇1 |r−r/prime|·∇/prime×F(r/prime)=−∇/prime1 |r−r/prime|·∇/prime×F(r/prime) and ∇/prime·/bracketleftbigg1 |r−r/prime|∇/prime×F(r/prime)/bracketrightbigg =∇/prime1 |r−r/prime|·∇/prime×F(r/prime)+1 |r−r/prime|∇/prime·∇/prime×F(r/prime) =∇/prime1 |r−r/prime|·∇/prime×F(r/prime), because the divergence of a curl is equal to zero. Therefore we have ∇·A(r)=−1 4π/integraldisplay/integraldisplay/integraldisplay/bracketleftbigg ∇/prime·1 |r−r/prime|∇/prime×F(r/prime)/bracketrightbigg d3r/prime =−1 4π/integraldisplay/integraldisplay /circlecopyrt S1 |r−r/prime|∇/prime×F(r/prime)·nda. AsS→∞,∇·A(r)=0. 2.10.4 Poisson’s and Laplace’s Equations The Helmholtz’s theorem shows that the vector ﬁeld is uniquely determined by its divergence and curl. To derive the expressions for the divergence and curl from experimental observations is therefore of great importance. One of the most important vector ﬁelds is the radial inverse square law ﬁeld, which is the mathematical statement of the gravitational law and the Coulomb’s law, the two fundamental laws in nature. For example, together with the principle of superposition, the electric ﬁeld E(r) produced by static charges can be written as E(r)=1 4π/integraldisplay/integraldisplay/integraldisplay /rho1(r/prime)/hatwideR R2d3r/prime=1 4π/integraldisplay/integraldisplay/integraldisplay /rho1(r/prime)r−r/prime (r−r/prime)3d3r/prime, (2.166) where /rho1(r/prime) is the charge density (electric charge per unit volume) in the neighborhood of r/prime.The constant 1 /4πis a matter of units and need not concern us here. The divergence of E(r)i s 2.10 Theory of Vector Fields 105 ∇·E(r)=1 4π/integraldisplay/integraldisplay/integraldisplay /rho1(r/prime)/parenleftBigg ∇·/hatwideR R2/parenrightBigg d3r/prime, since∇operates only on r.But, ∇·/hatwideR R2=4πδ3(r−r/prime) as shown in (2.153). Thus, ∇·E(r)=1 4π/integraldisplay/integraldisplay/integraldisplay /rho1(r/prime)4πδ3(r−r/prime)d3r/prime=/rho1(r). (2.167) The fact that we can relate the divergence of Eatrto the charge density at same point ris remarkable. Coulomb’s law of (2.166) is the experimental result, which says that the electric ﬁeld Eatris due to all other charges at diﬀerent places r/prime.Yet through vector analysis, we ﬁnd ∇·Eatris equal to the charge density /rho1(r) at the same place where Eis to be evalued. This type of equation is called ﬁeld equation which describes the property of the ﬁeld at each point in space. Since curl of ( /hatwideR/R2) is equal to zero, Ecan be expressed as the gradient of scalar potential E=−∇ϕ.Thus, ∇·E=−∇·∇ϕ=/rho1. Therefore, ∇2ϕ=−/rho1. (2.168) This result is known as Poisson’s equation which speciﬁes the relationship between the source density and the scalar potential for an irrotational ﬁeld. In that part of the space where there is no charge ( /rho1= 0), the equation reduces to ∇2ϕ=0, (2.169) which is known as Laplace’s equation . The equations of Poisson and Laplace are two of the most important equa- tions in mathematical physics. They are encountered repeatedly in a variety of problems. 2.10.5 Uniqueness Theorem In the following chapters, we shall describe various methods of solving Laplace’s equation. It does not matter which method we use, as long as we can ﬁnd a scalar function ϕthat satisﬁes the equation and the boundary con- ditions, the vector ﬁeld derived from it is uniquely determined. This is known asuniqueness theorem . Let the region of interests be surrounded by surface S, (if the boundary consists of many surfaces including the surface at inﬁnity, then Srepresents 106 2 Vector Calculus all of them). There are two kinds of boundary conditions (1) the values of ϕare speciﬁed on S, known as Dirichlet boundary condition and (2) the normal derivatives ∂ϕ/∂n overSare speciﬁed, known as Neumann boundary condition. The theorem says: Two solutions ϕ1andϕ2of the Laplace equation which satisfy the ﬁrst kind of boundary conditions must be idential. Two solutions ϕ1and ϕ2of the Laplace equation which satisfy the second kind of boundary conditions can diﬀer at most by an additive constant. To prove this theorem, we deﬁne a new function Φ=ϕ1−ϕ2. Obviously, ∇2Φ=∇2ϕ1−∇2ϕ2=0.Furthermore, either Φor∂Φ/∂n =∇Φ·nvanishes onS. Applying the divergence theorem to Φ∇Φ,we have /integraldisplay/integraldisplay/integraldisplay ∇·(Φ∇Φ)dV=/integraldisplay/integraldisplay SΦ∇Φ·nda=0, since the integral on the right-hand side vanishes. But ∇·(Φ∇Φ)=∇Φ·∇Φ+Φ∇2Φ and∇2Φ= 0 at all points, so the divergence theorem in this case becomes /integraldisplay/integraldisplay/integraldisplay ∇Φ·∇ΦdV=0. Now∇Φ·∇Φ=(∇Φ)2must be positive or zero, and since the integral is zero, it follows that the only possibility is ∇Φ= 0 everywhere inside the volume .A function whose gradient is zero at all points cannot change, hence Φhas the same value that it has on the boundary S. For the ﬁrst kind of boundary condition, Φ=0o n S,a n d Φm u s te q u a lt oz e r oa te v e r yp o i n ti n the region. Therefore ϕ1=ϕ2.For the second kind of boundary conditions, ∇Φequal to zero at all points in the region and ∇Φ·n=0 on S, the only possible solution is Φequal to a constant. Thus ϕ1andϕ2can diﬀer at most by a constant. In either case, the vector ﬁeld ∇ϕis uniquely deﬁned. Exercises 1. Finddr dt,d2r dt2,/vextendsingle/vextendsingle/vextendsingle/vextendsingledr dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle,/vextendsingle/vextendsingle/vextendsingle/vextendsingled2r dt2/vextendsingle/vextendsingle/vextendsingle/vextendsingle,ifr(t)=s i n ti+c o s tj+tk. Ans. cos ti−sintj+k,−sinti−costj,√ 2,1. 2. Show that A·dA dt=AdA dt. 2.10 Theory of Vector Fields 107 3. A particle moves along the curve r(t)=2 t2i+/parenleftbig t2−4t/parenrightbig j+( 3t−5)k, where tis time. Find its velocity and acceleration at t=1. Ans. 4i−2j+3k,4i+2j. 4. A particle moves along the curve r(t)=/parenleftbig t3−4t/parenrightbig i+/parenleftbig t2+4t/parenrightbig j+/parenleftbig 8t2−3t3/parenrightbig k,where tis time. Find the magnitudes of the tangential and normal components of its acceleration at t=2. Ans. 16 ,2√ 73. 5. A velocity ﬁeld is given by v=x2i−2xyj+4tk.Determine the acceler- ation at the point (2 ,1,−4). Ans. 16 i+8j+4k. Hint:a=∂v ∂xdx dt+∂v ∂ydy dt+∂v ∂zdz dt+∂v ∂t 6. A wheel of radius brolls along the ground with a constant forward speed v0.Find the acceleration of any point on the rim of the wheel. Ans.v2 0/btoward the center of the wheel. Hint: Let the moving origin be at the center of the wheel with x/primeaxis passing through the point in question, thus r/prime=bi,v/prime=0,a/prime=0.The angular velocity vector is ω=(v0/b)k/prime.Then use (2.44) 7. Find the arc length of r(t)=acosti+asintj+btkfromt=0t o t=2π. Ans.s=2π√ a2+b2. Hint: d s=vdt=(/squaresmallsolidr·/squaresmallsolidr)1/2)dt 8. Find the arc length of r(t) = (cos t+tsint)i+(sin t−tcost)jfromt=0 tot=π. Ans.s=π2/2. 9. Given the space curve r=ti+t2j+2 3t3k,ﬁnd (a) the curvature κand (b) the torsion γ. Ans.2 (1+2t2)2,2 (1+2t2)2. 10. Show that/squaresmallsolid/squaresmallsolidr=/squaresmallsolidvt+v2κn. 11. Show that the curvature κof a space curve r=r(t) is given by κ=/vextendsingle/vextendsingle/vextendsingle/squaresmallsolidr×/squaresmallsolid/squaresmallsolidr/vextendsingle/vextendsingle/vextendsingle//vextendsingle/vextendsingle/vextendsingle/squaresmallsolidr/vextendsingle/vextendsingle/vextendsingle3 , where dots denote diﬀerentiation with respect to time t. Hint: ﬁrst show that/squaresmallsolidr×/squaresmallsolid/squaresmallsolidr=vt×(/squaresmallsolidvt+v2κn) 108 2 Vector Calculus 12. Show that the torsion γof a space curve is given numerically by γ=/vextendsingle/vextendsingle/vextendsingle/squaresmallsolidr·/squaresmallsolid/squaresmallsolidr×/squaresmallsolid/squaresmallsolid/squaresmallsolidr/vextendsingle/vextendsingle/vextendsingle//vextendsingle/vextendsingle/vextendsingle/squaresmallsolidr×/squaresmallsolid/squaresmallsolidr/vextendsingle/vextendsingle/vextendsingle2 . Hint: ﬁrst show that/squaresmallsolidr·/squaresmallsolid/squaresmallsolidr×/squaresmallsolid/squaresmallsolid/squaresmallsolidr=−v6κ2γ,then use the result of the previous problem 13. Find the gradient of the scalar ﬁeld φ=xyz,and evaluate it at the point (1,2,3), ﬁnd the derivative of φin the direction of i+j. Ans.yzi+xzj+xyk,6i+3j+2k,9/√ 2. 14. Find the unit normal to each of the following surfaces at the point indicated: (a) x2+y2−z= 0 at (1,1,2), (b) x2+y2=5a t (2,1,0),and (c) y=x2+z3at (1,2,1). Ans. (2 i+2j−k)/3,(3i+4j)/5,(−i−j−3k)/√ 11. 15. The temperature Tis given by T=x2+xy+yz. What is the unit vector that points in the direction of maximum change of temperature at (2,1,4)? What is the value of the derivative of the temperature in the x direction at that point? Ans. (5 i+6j+k)/√ 62,5. 16. Determine the equation of the plane tangent to the given surface at the point indicated: (a) x2+y2+z2=2 5( 3 ,4,0),and (b) x2−2xy=0 (2, 2, 1). Ans. 3 x+4y=2 5,y=2. 17. Find the divergence of each of the following vector ﬁelds at the point (2,1,−1).(a)F=x2i+yzj+y2k,(b)F=xi+yj+yk,and (c) F= r/r=(xi+yj+yk)//radicalbig x2+y2+z2. Ans. 3, 3,√ 6/3. 18. Verify the divergence theorem by calculating both the volume integral and the surface integral for the vector ﬁeld F=yi+xj+(z−x)kand the volume of the unit cube 0 ≤x,y,z ≤1. 19. By using the divergence theorem, evaluate (a)/integraldisplay/integraldisplay /circlecopyrt S(xi+yj+zk)·nda, where Sis the surface of the sphere x2+y2+z2=9 ; (b)/integraldisplay/integraldisplay /circlecopyrt S/parenleftbig xi+xj+z2k/parenrightbig ·nda, 2.10 Theory of Vector Fields 109 where Sis the surface of the cylinder x2+y2=4,0≤z≤8; (c)/integraldisplay/integraldisplay /circlecopyrt S/parenleftbig xsinyi+c o s2xj−zsinyk/parenrightbig ·nda, where Sis the surface of the sphere x2+y2+(z−2)2=1. Ans. 108 π,288π,0. 20. Show that /integraldisplay/integraldisplay /circlecopyrt Sr·nda=3V, where Vis the volume bounded by the closed surface S. 21. Recognizing that i·nda=dydz;j·nda=dxdz;k·nda=dxdy (see Example 2.5.2), evaluating the following integral using the divergence theorem /integraldisplay/integraldisplay /circlecopyrt S(xdydz+ydxdz+zdxdy), where Sis the surface of the cylindr x2+y2=9,0≤z≤3. Ans. 81 π. H i n t :ﬁ r s ts h o wt h a t( xdydz+ydxdz+zdxdy)=(xi+yj+zk)·nda (see Example 2.5.2) 22. Evaluating the following integral using the divergence theorem /integraldisplay/integraldisplay /circlecopyrt S/parenleftbig xdydz+2ydxdz+y2dxdy/parenrightbig , where Sis the surface of the sphere x2+y2+z2=4. Ans. 32 π. 23. Use the divergence theorem to evaluate the surface integral /integraldisplay/integraldisplay S[(x+y)i+z2j+x2k]·nda, where Sis the surface of the hemisphere x2+y2+z2= 1 with z>0a n d nis the outward unit normal. Note that the surface is not closed. Ans.11 12π. Hint: the integral is equal to the closed surface integral over the hemisphere subtract the integral over the base. 24. Find the curl of each of the following vector ﬁelds at the point ( −2,4,1). (a)F=x2i+y2j+z2kand (b)F=xyi+y2j+xzk. Ans. 0 ,−j+2k. 110 2 Vector Calculus 25. Verify Stokes’ theorem by evaluating both the line and surface integral for the vector ﬁeld A=( 2x−y)i−y2j+y2zkand the surface Sgiven by the disc z=0,x2+y2≤1. 26. Ampere’s law states that the total ﬂux of electric current ﬂowing through a loop is proportional to the line integral of the magnetic ﬁeld around the loop, that is/contintegraltext CB·dr=µ0/integraltext/integraltext SJ·ndawhere B is the magnetic ﬁeld, Jis the current density and µ0is a proportional constant. If this is true for any loop C, show that ∇×B=µ0J. 27. Show that/contintegraltext Cr·dr= 0 for any closed curve C. 28. Calculate the circulation of the vector F=y2i+xyj+z2k/parenleftbig/contintegraltext F·dr/parenrightbig around a triangle with vertices at the origin, (2 ,2,0), and (0 ,2,0) by (a) direct integration, and (b) using Stokes’ theorem. Ans. 8/3. 29. Calculate the circulation of F=yi−xj+zkaround a unit circle in the xyplane with center at the origin by (a) direct integration and (b) using Stokes’ theorem. Ans.−2π. 30. Evaluate the circulation of the following vector ﬁelds around the curves speciﬁed. Use either direct integration or Stokes’ theorem. (a) F=2zi+ yj+xkaround a triangle with vertices at the origin, (1 ,0,0) and (0 ,0,4). (b)F=x2i+y2j+z2karound a unit circle in the xyplane with center at the origin. Ans. 2 ,0. 31. Check the product rule ∇·(A×B)=(∇×A)·B−(∇×B)·A by calculating each term separately for the functions A=y2i+2xyj+ z2k,B=s i nyi+s i nxj+z3k. 32. Check the relation ∇×(∇×A)=∇(∇·A)−∇2A by calculating each term separately for the function A=y2i+2xyj+z2k. 33. Show that ∇×(ϕ∇ϕ)=0. 34. Show that 2.10 Theory of Vector Fields 111 /integraldisplay/integraldisplay/integraldisplay V(∇×A)·BdV=/integraldisplay/integraldisplay/integraldisplay V(∇×B)·AdV+/integraldisplay/integraldisplay /circlecopyrt S(A×B)·nda, where Sis the surface bounding the volume V. 35. Show that for any closed surface S /integraldisplay/integraldisplay /circlecopyrt S(∇×B)·nda=0. 36. For what values, if any, of the constants aandbis the following vector ﬁeld irrotational? F=(ycosx+axz)i+(bsinx+z)j+/parenleftbig x2+y/parenrightbig k. Ans.a=2,b=1. 37. (a) Show that F=( 2xy+3 )i+/parenleftbig x2−4z/parenrightbig j−4ykis a conservative ﬁeld. (b) Find a scalar potential ϕsuch that ∇ϕ=−F.(c) Evaluate the integral/integraltext2,1,−1 3,−1,2F·dr. Ans.∇×F=0,ϕ=−x2y−3x+4yz,6. 38. (a) Show that F=y2zi−(z2siny−2xyz)j+( 2zcosy+y2x)kis irrotational. (b) ﬁnd a function ϕsuch that ∇ϕ=F. (c) Evaluate the integral/integraltext ΓF·drwhere Γis along the curve x= sin(πt/2),y=t2−t, z=t4,0≤t≤1. Ans.∇×F=0,ϕ=z2cosy+xy2z,1. 39. IfAis irrotational, show that A×ris solenoidal. 40. Vector Bis formed by the product of two gradients B=(∇u)×(∇v), where uandvare scalar functions. (a) Show that Bis solenoidal. (b) Show that A=1 2(u∇v−v∇u) is a vector potential for Bin that B=∇×A. 41. Show that if ∇2ϕ= 0 in the volume V, then /integraldisplay/integraldisplay /circlecopyrt S∇ϕ·nda=0, where Sis the surface bounding the volume. 112 2 Vector Calculus 42. Two ﬁelds fandgare related by Poisson’s equation, ∇2f=g.Show that /integraldisplay/integraldisplay/integraldisplay VgdV=/integraldisplay/integraldisplay /circlecopyrt S∇f·nda, where Sis the bounding surface of V. 43. Use Stokes’ theorem to show that /contintegraldisplay Cf∇g·dr=−/contintegraldisplay Cg∇f·dr for any closed curve C and diﬀerentiable ﬁelds fandg. Hint: ﬁrst show/contintegraltext Cf∇g·dr=/integraltext/integraltext S∇f×∇g·nda 3 Curved Coordinates Up to now we have used only Cartesian (rectangular) coordinates with their constant unit vectors. Frequently, because of the geometry of the problems, other coordinate systems are much more convenient. There are many coordi- nate systems, each of them can be regarded as a particular case of the general curvilinear coordinate system. It would be most eﬃcient if we ﬁrst develop a theory of curvilinear coordinates and then introduce each coordinate system as a special example. However, for pedagogic reasons, we will do that after we ﬁrst directly transform the vector expressions of the rectangular coordi- nates into the corresponding ones in the two most commonly used systems, namely cylindrical and spherical coordinates. This procedure has the advan- tage of emphasizing that the physical meaning of gradient, divergence, curl, and Laplacian operations remain the same in diﬀerent coordinate systems. Their appearances are diﬀerent only because they are expressed with diﬀerent notations. Furthermore, expressions in cylindrical and spherical coordinates will serve as familiar examples to clarify the terms in the general curvilinear system. As a further example, the elliptical coordinate system is discussed in some detail because of its importance in dealing with two center problems. Within the framework of curvilinear coordinates, we introduce the Jacobian determinant for multiple integrals in Sect. 3.5. 3.1 Cylindrical Coordinates The cylindrical coordinate system is formally known as the circular cylindrical or cylindrical polar coordinate system. In this system, the position of a point is speciﬁed by ( ρ,ϕ,z ) as shown in Fig. 3.1a: ρis the perpendicular distance from the z-axis, ϕis the angle between the x-axis and the projection of ρon the xy-plane, and zis the same as in the rectangular coordinates. The three unit vectors, eρ,eϕ,ez,point in the direction of increase of the corresponding coordinates. The relation to Cartesian coordinates can be easily seen from Fig. 3.1b where we have moved eρ,eϕto the origin: 114 3 Curved Coordinates xyz jerejez zry x xyj i(b) (a) ejj jerejer rr Fig. 3.1. Cylindrical coordinates. ( a) A point is speciﬁed by ( ρ, ϕ, z ),the unit vectors eρ,eϕ,ezare pointing in the direction of increase of the corresponding coordinates, ( b)eρ,eϕare moved to the origin to ﬁnd the relationships with i,j of the rectangular coordinate system x=ρcosϕ, y =ρsinϕ, ρ =(x2+y2)1/2,ϕ= tan−1y x. (3.1) eρ=c o s ϕi+si nϕj=x (x2+y2)1/2i+y (x2+y2)1/2j, (3.2) eϕ=−sinϕi+co sϕj=−y (x2+y2)1/2i+x (x2+y2)1/2j. (3.3) i=c o s ϕeρ−sinϕeϕ, (3.4) j=s i nϕeρ+c o s ϕeϕ. (3.5) It follows ∂x ∂ρ=c o s ϕ,∂x ∂ϕ=−ρsinϕ,∂y ∂ρ=s i nϕ,∂y ∂ϕ=ρcosϕ, (3.6) ∂ρ ∂x=∂ ∂x(x2+y2)1/2=x (x2+y2)1/2=ρcosϕ ρ=c o s ϕ, (3.7) ∂ρ ∂y=∂ ∂y(x2+y2)1/2=y (x2+y2)1/2=ρsinϕ ρ=s i nϕ, (3.8) ∂ϕ ∂x=∂ ∂xtan−1/parenleftBigy x/parenrightBig =−y x2+y2=−sinϕ ρ, (3.9) ∂ϕ ∂y=∂ ∂ytan−1/parenleftBigy x/parenrightBig =x x2+y2=cosϕ ρ. (3.10) 3.1 Cylindrical Coordinates 115 The relationships between eρ,eϕ,ezcan be easily worked out, for example, eρ·eρ= (cos ϕi+si nϕj)·(cosϕi+si nϕj) = cos2ϕ+s i n2ϕ=1, eρ·eϕ= (cos ϕi+si nϕj)·(−sinϕi+co sϕj)=0, eρ×eϕ= (cos ϕi+si nϕj)×(−sinϕi+co sϕj) = cos2ϕk+s i n2ϕk=k=ez. Taken together, they form an orthonormal basis set eρ·eρ=eϕ·eϕ=ez·ez=1, eρ·eϕ=eϕ·ez=ez·eρ=0, (3.11) eρ×eϕ=ez,eϕ×ez=eρ,ez×eρ=eϕ. The position vector r,from the origin to any point in space, is clearly seen in Fig. 3.1 to be r=ρeρ+zez. (3.12) This expression can also be obtained from directly transforming r=xi+yj+zk into the cylindrical coordinates. Any vector can be expressed in terms of them. If the vector is a function of the position, then A(ρ,ϕ,z )=Aρ(ρ,ϕ,z )eρ+Aϕ(ρ,ϕ,z )eϕ+Az(ρ,ϕ,z )ez. (3.13) In general, each component is a function of ρ,ϕ,z. Unlike the constant unit vector i,j,kin the rectangular coordinate system, only ez=kis ﬁxed in space, the directions of eρ,eϕchange as the point is moved around. Note that both eρandeϕdepend on ϕ.In particular, ∂ ∂ϕeρ=∂ ∂ϕ(cosϕi+si nϕj)=−sinϕi+co sϕj=eϕ, ∂ ∂ϕeϕ=∂ ∂ϕ(−sinϕi+co sϕj)=−(cosϕi+si nϕj)=−eρ,(3.14) ∂ ∂ρeρ=∂ ∂ρeϕ=0. Example 3.1.1. Show that the acceleration of a particle expressed in cylindri- cal coordinates is given by a=/parenleftbigg /squaresmallsolid/squaresmallsolidρ−ρ/squaresmallsolidϕ2/parenrightbigg eρ+/parenleftBig ρ/squaresmallsolid/squaresmallsolidϕ+2/squaresmallsolidρ/squaresmallsolidϕ/parenrightBig eϕ+/squaresmallsolid/squaresmallsolidzez, where dots denote diﬀerentiation with respect to time t. Solution 3.1.1. Since the position vector is given by r=ρeρ+zez,the velocity is v=/squaresmallsolidr, 116 3 Curved Coordinates /squaresmallsolidr=/squaresmallsolidρeρ+ρ/squaresmallsolideρ+/squaresmallsolidzez, where ezis a constant unit vector and eρdepends on ϕ.Since by (3.14) /squaresmallsolideρ=deρ dt=dϕ dtdeρ dϕ=/squaresmallsolidϕeϕ, /squaresmallsolidr=/squaresmallsolidρeρ+ρ/squaresmallsolidϕeϕ+/squaresmallsolidzez. The acceleration is the rate of change of velocity, therefore a=/squaresmallsolidv=/squaresmallsolid/squaresmallsolidr, /squaresmallsolid/squaresmallsolidr=/squaresmallsolid/squaresmallsolidρeρ+/squaresmallsolidρ/squaresmallsolideρ+/squaresmallsolidρ/squaresmallsolidϕeϕ+ρ/squaresmallsolid/squaresmallsolidϕeϕ+ρ/squaresmallsolidϕ/squaresmallsolideϕ+/squaresmallsolid/squaresmallsolidzez. Again by (3.14), /squaresmallsolideϕ=deϕ dt=dϕ dtdeϕ dϕ=/squaresmallsolidϕ(−eρ), /squaresmallsolid/squaresmallsolidr=/squaresmallsolid/squaresmallsolidρeρ+/squaresmallsolidρ/squaresmallsolidϕeϕ+/squaresmallsolidρ/squaresmallsolidϕeϕ+ρ/squaresmallsolid/squaresmallsolidϕeϕ−ρ/squaresmallsolid2ϕeρ+/squaresmallsolid/squaresmallsolidzez. Therefore a=/squaresmallsolid/squaresmallsolidr=/parenleftBig/squaresmallsolid/squaresmallsolidρ−ρ/squaresmallsolidϕ2/parenrightBig eρ+( 2/squaresmallsolidρ/squaresmallsolidϕ+ρ/squaresmallsolid/squaresmallsolidϕ)eϕ+/squaresmallsolid/squaresmallsolidzez. 3.1.1 Diﬀerential Operations Gradient. Starting from the deﬁnition of gradient in the Cartesian coordi- nates, we can use the coordinate transformation to express it in terms of (ρ,ϕ,z ).Using (3.4) and (3.5), ∇Φ=i∂Φ ∂x+j∂Φ ∂y+k∂Φ ∂z = (cos ϕeρ−sinϕeϕ)∂Φ ∂x+( s i n ϕeρ+c o s ϕeϕ)∂Φ ∂y+ez∂Φ ∂z =/parenleftbigg cosϕ∂Φ ∂x+s i nϕ∂Φ ∂y/parenrightbigg eρ+/parenleftbigg −sinϕ∂Φ ∂x+c o s ϕ∂Φ ∂y/parenrightbigg eϕ+∂Φ ∂zez.(3.15) By chain rule and (3.6) ∂Φ ∂ρ=∂x ∂ρ∂Φ ∂x+∂y ∂ρ∂Φ ∂y=c o s ϕ∂Φ ∂x+s i nϕ∂Φ ∂y, (3.16) ∂Φ ∂ϕ=∂x ∂ϕ∂Φ ∂x+∂y ∂ϕ∂Φ ∂y=−ρsinϕ∂Φ ∂x+ρcosϕ∂Φ ∂y. (3.17) With these expressions, (3.15) becomes ∇Φ=∂Φ ∂ρeρ+1 ρ∂Φ ∂ϕeϕ+∂Φ ∂zez. (3.18) 3.1 Cylindrical Coordinates 117 Thus, the gradient operator in the cylindrical coordinates can be written as ∇=eρ∂ ∂ρ+eϕ1 ρ∂ ∂ϕ+ez∂ ∂z. (3.19) An immediate consequence is ∇ρ=eρ,∇ϕ=1 ρeϕ,∇z=ez. (3.20) This is not a surprising result. After all, ∇uis a vector perpendicular to the surface u= constant. Divergence. The divergence of a vector ∇·V=∇·(Vρeρ+Vϕeϕ+Vzez) can be expanded ﬁrst by the distributive law of dot product. Now, ∇·Vρeρ=∇·Vρ(eϕ×ez)=∇·Vρ(ρ∇ϕ×∇z) =∇(ρVρ)·(∇ϕ×∇z)+ρVρ∇·(∇ϕ×∇z). But∇·(∇ϕ×∇z)=∇×∇ϕ·∇z−∇×∇z·∇ϕ=0,so ∇·Vρeρ=∇(ρVρ)·(∇ϕ×∇z)=∇(ρVρ)·/parenleftbigg1 ρeϕ×ez/parenrightbigg =1 ρ∇(ρVρ)·eρ =1 ρ/parenleftbigg eρ∂ρVρ ∂ρ+eϕ1 ρ∂ρVρ ∂ϕ+ez∂ρVρ ∂z/parenrightbigg ·eρ=1 ρ∂ ∂ρ(ρVρ).(3.21) ∇·Vϕeϕ=∇·Vϕ(ez×eρ)=∇·Vϕ(∇z×∇ρ) =∇Vϕ·(∇z×∇ρ)+Vϕ∇·(∇z×∇ρ) =∇Vϕ·(∇z×∇ρ)=∇Vϕ·(ez×eρ)=∇Vϕ·eϕ =/parenleftbigg eρ∂Vϕ ∂ρ+eϕ1 ρ∂Vϕ ∂ϕ+ez∂Vϕ ∂z/parenrightbigg ·eϕ=1 ρ∂Vϕ ∂ϕ. (3.22) Therefore, ∇·V=1 ρ∂ ∂ρ(ρVρ)+1 ρ∂Vϕ ∂ϕ+∂Vz ∂z. (3.23) Laplacian. By deﬁnition the Laplacian of Φis given by ∇2Φ=∇·∇Φ=∇·/parenleftbigg∂Φ ∂ρeρ+1 ρ∂Φ ∂ϕeϕ+∂Φ ∂zez/parenrightbigg . (3.24) Using the expression of the divergence, we have ∇·∇Φ=1 ρ∂ ∂ρ/parenleftbigg ρ∂Φ ∂ρ/parenrightbigg +1 ρ∂ ∂ϕ/parenleftbigg1 ρ∂Φ ∂ϕ/parenrightbigg +∂ ∂z/parenleftbigg∂Φ ∂z/parenrightbigg . 118 3 Curved Coordinates Therefore ∇2Φ=∂2Φ ∂ρ2+1 ρ∂Φ ∂ρ+1 ρ2∂2Φ ∂ϕ2+∂2Φ ∂z2. (3.25) Since the Laplacian is a scalar operator, it is instructive to convert it directly from its deﬁnition in the rectangular coordinates ∇2Φ=∂2Φ ∂x2+∂2Φ ∂y2+∂2Φ ∂z2. Now with chain rule and (3.7) and (3.9), we have ∂Φ ∂x=∂ρ ∂x∂Φ ∂ρ+∂ϕ ∂x∂Φ ∂ϕ=c o s ϕ∂Φ ∂ρ−sinϕ ρ∂Φ ∂ϕ, ∂2Φ ∂x2=∂ ∂x/bracketleftbigg∂Φ ∂x/bracketrightbigg =∂ρ ∂x∂ ∂ρ/bracketleftbigg∂Φ ∂x/bracketrightbigg +∂ϕ ∂x∂ ∂ϕ/bracketleftbigg∂Φ ∂x/bracketrightbigg =c o s ϕ∂ ∂ρ/bracketleftbigg cosϕ∂Φ ∂ρ−sinϕ ρ∂Φ ∂ϕ/bracketrightbigg −sinϕ ρ∂ ∂ϕ/bracketleftbigg cosϕ∂Φ ∂ρ−sinϕ ρ∂Φ ∂ϕ/bracketrightbigg =c o s2ϕ∂2Φ ∂ρ2+cosϕsinϕ ρ2∂Φ ∂ϕ−cosϕsinϕ ρ∂2Φ ∂ρ∂ϕ +sin2ϕ ρ∂Φ ∂ρ−sinϕcosϕ ρ∂2Φ ∂ϕ∂ρ+sinϕcosϕ ρ2∂Φ ∂ϕ+sin2ϕ ρ2∂2Φ ∂ϕ2. Similarly, ∂Φ ∂y=∂ρ ∂y∂Φ ∂ρ+∂ϕ ∂y∂Φ ∂ϕ=s i nϕ∂Φ ∂ρ+cosϕ ρ∂Φ ∂ϕ, ∂2Φ ∂y2=∂ ∂y/bracketleftbigg∂Φ ∂y/bracketrightbigg =∂ρ ∂y∂ ∂ρ/bracketleftbigg∂Φ ∂y/bracketrightbigg +∂ϕ ∂y∂ ∂ϕ/bracketleftbigg∂Φ ∂y/bracketrightbigg =s i nϕ∂ ∂ρ/bracketleftbigg sinϕ∂Φ ∂ρ+cosϕ ρ∂Φ ∂ϕ/bracketrightbigg +cosϕ ρ∂ ∂ϕ/bracketleftbigg sinϕ∂Φ ∂ρ+cosϕ ρ∂Φ ∂ϕ/bracketrightbigg =s i n2ϕ∂2Φ ∂ρ2−sinϕcosϕ ρ2∂Φ ∂ϕ+sinϕcosϕ ρ∂2Φ ∂ρ∂ϕ +cos2ϕ ρ∂Φ ∂ρ+cosϕsinϕ ρ∂2Φ ∂ϕ∂ρ−cosϕsinϕ ρ2∂Φ ∂ϕ+cos2ϕ ρ2∂2Φ ∂ϕ2. Thus, ∂2Φ ∂x2+∂2Φ ∂y2= (cos2ϕ+s i n2ϕ)∂2Φ ∂ρ2+sin2ϕ+c o s2ϕ ρ∂Φ ∂ρ +sin2ϕ+c o s2ϕ ρ2∂2Φ ∂ρ2=∂2Φ ∂ρ2+1 ρ∂Φ ∂ρ+1 ρ2∂2Φ ∂ϕ2. Clearly the Laplacian obtained this way is identical to (3.25). 3.1 Cylindrical Coordinates 119 Curl. The curl of a vector can be written as ∇×V=∇×(Vρeρ+Vϕeϕ+Vzez). (3.26) Now ∇×Vρeρ=∇×Vρ∇ρ=∇Vρ×∇ρ+Vρ∇×∇ρ. Since ∇×∇ρ=0, ∇×Vρeρ=∇Vρ×∇ρ=∇Vρ×eρ =/parenleftbigg eρ∂Vρ ∂ρ+eϕ1 ρ∂Vρ ∂ϕ+ez∂Vρ ∂z/parenrightbigg ×eρ =−1 ρ∂Vρ ∂ϕez+∂Vρ ∂zeϕ, (3.27) ∇×Vϕeϕ=∇×Vϕ(ρ∇ϕ)=∇(ρVϕ)×∇ϕ+ρVϕ∇×∇ϕ =∇(ρVϕ)×∇ϕ=∇(ρVϕ)×1 ρeϕ =1 ρ/parenleftbigg eρ∂ρVϕ ∂ρ+eϕ1 ρ∂ρVϕ ∂ϕ+ez∂ρVϕ ∂z/parenrightbigg ×eϕ =1 ρ∂ρVϕ ∂ρez−1 ρ∂ρVϕ ∂zeρ, (3.28) ∇×Vzez=∇Vz×ez=/parenleftbigg eρ∂Vz ∂ρ+eϕ1 ρ∂Vz ∂ϕ+ez∂Vz ∂z/parenrightbigg ×ez =−∂Vz ∂ρeϕ+1 ρ∂Vz ∂ϕeρ. Thus, ∇×V=/parenleftbigg1 ρ∂Vz ∂ϕ−1 ρ∂ρVϕ ∂z/parenrightbigg eρ+/parenleftbigg∂Vρ ∂z−∂Vz ∂ρ/parenrightbigg eϕ +/parenleftbigg1 ρ∂ ∂ρ(ρVϕ)−1 ρ∂Vρ ∂ϕ/parenrightbigg ez. (3.29) Example 3.1.2. (a) Show that the vector ﬁeld F=/parenleftbigg A−B ρ2/parenrightbigg cosϕeρ−/parenleftbigg A+B ρ2/parenrightbigg sinϕeϕ is irrotational ( ∇×F=0).(b) Find a scalar potential Φsuch that ∇Φ=F. (c) Show that Φsatisﬁes the Laplace’s equation ( ∇2Φ=0 ). 120 3 Curved Coordinates Solution 3.1.2. (a) All derivatives with respect to zare equal to zero, since there is no zdependence. Furthermore, Vz=0.Therefore ∇×F=/parenleftbigg1 ρ∂ ∂ρ(ρFϕ)−1 ρ∂Fρ ∂ϕ/parenrightbigg ez =1 ρ∂ ∂ρ/bracketleftbigg ρ/parenleftbigg −A−B ρ2/parenrightbigg sinϕ/bracketrightbigg ez−1 ρ∂ ∂ϕ/bracketleftbigg/parenleftbigg A−B ρ2/parenrightbigg cosϕ/bracketrightbigg ez =1 ρ/bracketleftbigg/parenleftbigg −A+B ρ2/parenrightbigg sinϕ+/parenleftbigg A−B ρ2/parenrightbigg sinϕ/bracketrightbigg ez=0. (b) ∇Φ=eρ∂Φ ∂ρ+eϕ1 ρ∂Φ ∂ϕ+ez∂Φ ∂z =/parenleftbigg A−B ρ2/parenrightbigg cosϕeρ−/parenleftbigg A+B ρ2/parenrightbigg sinϕeϕ, ∂Φ ∂ρ=/parenleftbigg A−B ρ2/parenrightbigg cosϕ;1 ρ∂Φ ∂ϕ=−/parenleftbigg A+B ρ2/parenrightbigg sinϕ. It is clear, up to an additive constant, Φ=/parenleftbigg Aρ+B ρ/parenrightbigg cosϕ. (c) ∇2Φ=1 ρ∂ ∂ρ/parenleftbigg ρ∂Φ ∂ρ/parenrightbigg +1 ρ2∂2Φ ∂ϕ2+∂2Φ ∂z2 =1 ρ∂ ∂ρ/bracketleftbigg ρ∂ ∂ρ/parenleftbigg Aρ+B ρ/parenrightbigg cosϕ/bracketrightbigg +1 ρ2∂2 ∂ϕ2/parenleftbigg Aρ+B ρ/parenrightbigg cosϕ =1 ρ/parenleftbigg A+B ρ2/parenrightbigg cosϕ−1 ρ/parenleftbigg A+B ρ2/parenrightbigg cosϕ=0. 3.1.2 Inﬁnitesimal Elements W h e nap o i n ta t( x,y,z) is moved to ( x+dx,y+dy,z+dz), the inﬁnitesimal displacement vector is d r=idx+jdy+kdz.Similarly, when the point at (ρ,ϕ,z ) in the cylindrical coordinates is moved to ( ρ+dρ,ϕ+dϕ,z+dz), the inﬁnitesimal displacement vector is dr=eρdρ+eϕρdϕ+ezdz. (3.30) Notice the distance in eϕdirection is ρdϕas shown in Fig. 3.2. The inﬁnites- imal length element is 3.1 Cylindrical Coordinates 121 xyz dzdr drr dj rjdj Fig. 3.2. Diﬀerential elements in cylindrical coordinates. Note that the diﬀerential length in the direction of increasing ϕisρdϕ.The diﬀerential volume element is ρdϕdρdz. ds=( dr·dr)1/2=/bracketleftBig (dρ)2+(ρdϕ)2+( dz)2/bracketrightBig1/2 . (3.31) The gradient is deﬁned as a vector of derivatives with respect to the distances in three perpendicular directions. Thus the gradient in cylindrical coordinates should be ∇=eρ∂ ∂ρ+eϕ1 ρ∂ ∂ϕ+ez∂ ∂z, which is, of course, identical to (3.19) obtained from direct transformation. The inﬁnitesimal volume element d Vis the product of the perpendicular inﬁnitesimal displacements dV=( dρ)(ρdϕ)(dz)=ρdρdϕdz. (3.32) The possible range of ρis 0 to ∞,ϕgoes from 0 to 2 π,a n dzfrom−∞to∞. The inﬁnitesimal surface element depends on the orientation of the surface. For example, on the side surface of a cylinder parallel to z-axis and of constant radius ρ,the surface element directed outward is nda=ρdϕdzeρ.The surface element on the xy-plane directed upward is nda=ρdϕdρez. Example 3.1.3. Verify the divergence theorem /integraldisplay/integraldisplay/integraldisplay V∇·FdV=/integraldisplay/integraldisplay /circlecopyrt SF·nda with a vector ﬁeld F=ρ/parenleftbig 2+s i n2ϕ/parenrightbig eρ+ρsinϕcosϕeϕ+3z2ez over a cylinder with base of radius 2 and height 5 . 122 3 Curved Coordinates Solution 3.1.3. ∇·F=1 ρ∂ ∂ρ(ρFρ)+1 ρ∂Fϕ ∂ϕ+∂Fz ∂z =1 ρ2ρ/parenleftbig 2+s i n2ϕ/parenrightbig +1 ρρ/parenleftbig cos2ϕ−sin2ϕ/parenrightbig +6z =4+s i n2ϕ+c o s2ϕ+6z=5+6 z. /integraldisplay/integraldisplay/integraldisplay V∇·FdV=/integraldisplay/integraldisplay/integraldisplay V(5 + 6 z)ρdϕdρdz =/integraldisplay2π 0dϕ/integraldisplay2 0ρdρ/integraldisplay5 0(5 + 6 z)dz= 400 π. /integraldisplay/integraldisplay /circlecopyrt SF·nda=/integraldisplay/integraldisplay S1F·nda+/integraldisplay/integraldisplay S2F·nda+/integraldisplay/integraldisplay S3F·nda, where S1is the side surface of the cylinder, S2andS3are, respective, the bottom and top surfaces of the cylinder. /integraldisplay/integraldisplay S1F·nda=/integraldisplay/integraldisplay S1F·eρda=/integraldisplay2π 0/integraldisplay5 0/bracketleftbig ρ/parenleftbig 2+s i n2ϕ/parenrightbig ρ/bracketrightbig ρ=2dϕdz= 100 π. /integraldisplay/integraldisplay S2F·nda=/integraldisplay/integraldisplay S2F·(−ez)da=/integraldisplay/integraldisplay S2/bracketleftbig −3z2/bracketrightbig z=0da=0, /integraldisplay/integraldisplay S3F·nda=/integraldisplay/integraldisplay S3F·(ez)da=/integraldisplay2π 0/integraldisplay2 0/bracketleftbig 3z2/bracketrightbig z=5ρdϕdρ= 300 π. Therefore, /integraldisplay/integraldisplay /circlecopyrt SF·nda= 100 π+ 300 π= 400 π. Clearly, /integraldisplay/integraldisplay/integraldisplay V∇·FdV=/integraldisplay/integraldisplay /circlecopyrt SF·nda. 3.2 Spherical Coordinates The spherical polar coordinate system is commonly known just as the spherical coordinates. The location of a point is speciﬁed by ( r, θ, ϕ) as shown in Fig. 3.3, where ris the distance from the origin, θis the angle made by the position vector rwith the positive z-axis which is often called polar angle, and ϕis the angle made with the positive x-axis by the projection of ron the xy-plane, this angle is known as azimuthal angle. The relations between the rectangular and spherical coordinates are seen from Fig. 3.3b and c. 3.2 Spherical Coordinates 123 xyz q jeq ejejerz eqer eqerqkr er BOA BOA j ej BO ererjej i xy(b) (a) (c) r Fig. 3.3. Spherical coordinates. (a)A point is speciﬁed by ( r, θ, ϕ)w h e r e ris the distance from the origin, θis the angle made by rwith the positive z-axis, and ϕis the angle made with the positive x-axis by the projection of ron the xy-plane. The three unit vectors er,eθ,eϕare in the direction of increasing r, θ, ϕ, respectively. The auxiliary unit vector e/rho1is in the direction of the projection of ron the xy-plane. (b)The unit vectors erandeθare moved to the origin in the AOB plane to ﬁnd their relationships with kande/rho1.(c)In the xy-plane, eϕis moved to the origin to ﬁnd the relationships between eϕ,e/rho1andi,j x=rsinθcosϕ, y =rsinθsinϕ, z =rcosθ (3.33) and r=/parenleftbig x2+y2+z2/parenrightbig1/2,tanθ=/parenleftbig x2+y2/parenrightbig1/2 z,tanϕ=y x, (3.34) sinθ=/parenleftbig x2+y2/parenrightbig1/2 (x2+y2+z2)1/2,cosθ=z (x2+y2+z2)1/2, (3.35) sinϕ=y (x2+y2)1/2, cosϕ=x (x2+y2)1/2. (3.36) Figure 3.3 also shows a set of mutually perpendicular unit vectors er,eθ,eϕ in the sense of increasing r, θ, ϕ, respectively. In this system, the position vector ris simply r=r/hatwider=rer. (3.37) The relations between the unit vectors in the spherical coordinates and those in the Cartesian coordinates can be seen from Fig. 3.3b. In the plane AOB, we have drawn erandeθfrom the origin. It can be seen er=s i nθe/rho1+c o sθk, eθ=c o s θe/rho1−sinθk, where e/rho1is a unit vector along OB. In Fig. 3.3c, e/rho1andeθare drawn from the origin, clearly 124 3 Curved Coordinates e/rho1=c o s ϕi+si nϕj, eϕ=−sinϕi+c o s ϕj. Thus, er=s i nθcosϕi+si nθsinϕj+c o s θk, eθ=c o s θcosϕi+co sθsinϕj−sinθk, (3.38) eϕ=−sinϕi+c o s ϕj. The inverse relations can either be read from the same ﬁgures, or be solved fori,j,kfrom the above equations, i=s i nθcosϕer+c o sθcosϕeθ−sinϕeϕ, j=s i nθsinϕer+c o sθsinϕeθ+c o s ϕeϕ, (3.39) k=c o s θer−sinθeθ. It can easily be veriﬁed that ( er,eθ,eϕ) form an orthonormal basis set and satisfy the following relations er·er=eθ·eθ=eϕ·eϕ=1, er·eθ=eθ·eϕ=eϕ·er=0, (3.40) er×eθ=eϕ,eθ×eϕ=er,eϕ×er=eθ. Any vector can be expressed in terms of them A=Arer+Aθeθ+Aϕeϕ, (3.41) where Ar,Aθ,Aϕare the radial, polar, and azimuthal components of A.The derivatives of the unit vectors are easily obtained from (3.38): ∂er ∂r=∂eθ ∂r=∂eϕ ∂r=∂eϕ ∂θ=0, (3.42) ∂er ∂θ=c o s θcosϕi+co sθsinϕj−sinθk=eθ, (3.43) ∂eθ ∂θ=−sinθcosϕi−sinθsinϕj−cosθk=−er, (3.44) ∂er ∂ϕ=−sinθsinϕi+s i nθcosϕj=s i nθeϕ, (3.45) ∂eθ ∂ϕ=−cosθsinϕi+c o s θcosϕj=c o s θeϕ, (3.46) ∂eϕ ∂ϕ=−cosϕi−sinϕj=−(sinθer+c o s θeθ). (3.47) 3.2 Spherical Coordinates 125 3.2.1 Diﬀerential Operations Gradient. We are ready to express the gradient operator ∇in the spherical coordinates. Using (3 .39), we have ∇=i∂ ∂x+j∂ ∂y+k∂ ∂z=( s i n θcosϕer+c o sθcosϕeθ−sinϕeϕ)∂ ∂x +(sin θsinϕer+co sθsinϕeθ+c o s ϕeϕ)∂ ∂y+ (cos θer−sinθeθ)∂ ∂z =er/bracketleftbigg sinθcosϕ∂ ∂x+s i nθsinϕ∂ ∂y+c o s θ∂ ∂z/bracketrightbigg +eθ/bracketleftbigg cosθcosϕ∂ ∂x+c o s θsinϕ∂ ∂y−sinθ∂ ∂z/bracketrightbigg +eϕ/bracketleftbigg −sinϕ∂ ∂x+c o s ϕ∂ ∂y/bracketrightbigg . (3.48) The quantities in the brackets can be recognized if we use (3.33) and the chain rule of derivatives ∂ ∂r=∂x ∂r∂ ∂x+∂y ∂r∂ ∂y+∂z ∂r∂ ∂z =s i nθcosϕ∂ ∂x+s i nθsinϕ∂ ∂y+c o s θ∂ ∂z, (3.49) ∂ ∂θ=∂x ∂θ∂ ∂x+∂y ∂θ∂ ∂y+∂z ∂θ∂ ∂z =rcosθcosϕ∂ ∂x+rcosθsinϕ∂ ∂y−rsinθ∂ ∂z, (3.50) ∂ ∂ϕ=∂x ∂ϕ∂ ∂x+∂y ∂ϕ∂ ∂y+∂z ∂ϕ∂ ∂z =−rsinθsinϕ∂ ∂x+rsinθcosϕ∂ ∂y. (3.51) Thus (3.48) can be written as ∇=er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ. (3.52) It follows that ∇r=er,∇θ=1 reθ,∇ϕ=1 rsinθeϕ. (3.53) 126 3 Curved Coordinates Divergence. The divergence of a vector in the spherical coordinates is ∇·V=∇·(Vrer+Vθeθ+Vϕeϕ) =∇Vr·er+Vr∇·er+∇Vθ·eθ+Vθ∇·eθ +∇Vϕ·eϕ+Vϕ∇·eϕ. (3.54) Although the divergence in the spherical coordinates can be worked out just as we did for in the cylindrical coordinates, it is instructive to ﬁnd the expression by using the derivatives of (3.42)–(3.47), ∇·er=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ·er =er·∂er ∂r+1 reθ·∂er ∂θ+1 rsinθeϕ·∂er ∂ϕ =1 reθ·eθ+1 rsinθeϕ·sinθeϕ=2 r, (3.55) ∇·eθ=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ·eθ =er·∂eθ ∂r+1 reθ·∂eθ ∂θ+1 rsinθeϕ·∂eθ ∂ϕ =1 reθ·(−er)+1 rsinθeϕ·cosθeϕ=1 rcosθ sinθ, (3.56) ∇·eϕ=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ·eϕ =er·∂eϕ ∂r+1 reθ·∂eϕ ∂θ+1 rsinθeϕ·∂eϕ ∂ϕ =1 rsinθeϕ·(−sinθer+c o s θeθ)=0. (3.57) Furthermore, ∇Vr·er=/parenleftbigg er∂Vr ∂r+eθ1 r∂Vr ∂θ+eϕ1 rsinθ∂Vr ∂ϕ/parenrightbigg ·er=∂Vr ∂r. (3.58) Similarly, ∇Vθ·eθ=1 r∂Vθ ∂θ,∇Vϕ·eϕ=1 rsinθ∂Vϕ ∂ϕ. (3.59) Thus, ∇·V=∂Vr ∂r+2 rVr+1 r∂Vθ ∂θ+1 rcosθ sinθVθ+1 rsinθ∂Vϕ ∂ϕ =1 r2∂ ∂r/parenleftbig r2Vr/parenrightbig +1 rsinθ∂ ∂θ(sinθVθ)+1 rsinθ∂ ∂ϕVϕ.(3.60) 3.2 Spherical Coordinates 127 Laplacian. The Laplacian in spherical coordinates can be written as ∇2Φ=∇·∇Φ=∇·/parenleftbigg er∂Φ ∂r+eθ1 r∂Φ ∂θ+eϕ1 rsinθ∂Φ ∂ϕ/parenrightbigg . Regarding ∇Φas a vector and using the expression of divergence, we have ∇·∇Φ=1 r2∂ ∂r/parenleftbigg r2∂Φ ∂r/parenrightbigg +1 rsinθ∂ ∂θ/parenleftbigg sinθ1 r∂Φ ∂θ/parenrightbigg +1 rsinθ∂ ∂ϕ/parenleftbigg1 rsinθ∂Φ ∂ϕ/parenrightbigg . (3.61) Therefore the Laplacian operator can be written as ∇2=1 r2∂ ∂r/parenleftbigg r2∂ ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂ ∂θ/parenrightbigg +1 r2sin2θ∂2 ∂ϕ2.(3.62) Curl. The curl of a vector in the spherical coordinates can be written as ∇×V=∇×(Vrer+Vθeθ+Vϕeϕ) =∇Vr×er+Vr∇×er+∇Vθ×eθ+Vθ∇×eθ +∇Vϕ×eϕ+Vϕ∇×eϕ. (3.63) Again we will derive the expression of curl in the spherical coordinates with the derivatives of (3.42)–(3.47). ∇×er=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ×er =er×∂er ∂r+1 reθ×∂er ∂θ+1 rsinθeϕ×∂er ∂ϕ =1 reθ×eθ+1 rsinθeϕ×sinθeϕ=0, (3.64) ∇×eθ=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ×eθ =er×∂eθ ∂r+1 reθ×∂eθ ∂θ+1 rsinθeϕ×∂eθ ∂ϕ =1 reθ×(−er)+1 rsinθeϕ×cosθeϕ=1 reϕ, (3.65) ∇×eϕ=/parenleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/parenrightbigg ×eϕ =er×∂eϕ ∂r+1 reθ×∂eϕ ∂θ+1 rsinθeϕ×∂eϕ ∂ϕ =1 rsinθeϕ×(−sinθer+c o s θeθ)=−1 reθ+1 rcosθ sinθer.(3.66) 128 3 Curved Coordinates Furthermore ∇Vr×er=/parenleftbigg er∂Vr ∂r+eθ1 r∂Vr ∂θ+eϕ1 rsinθ∂Vr ∂ϕ/parenrightbigg ×er =−1 r∂Vr ∂θeϕ+1 rsinθ∂Vr ∂ϕeθ, (3.67) ∇Vθ×eθ=/parenleftbigg er∂Vθ ∂r+eθ1 r∂Vθ ∂θ+eϕ1 rsinθ∂Vθ ∂ϕ/parenrightbigg ×eθ =∂Vθ ∂reϕ−1 rsinθ∂Vθ ∂ϕer, (3.68) ∇Vϕ×eϕ=/parenleftbigg er∂Vϕ ∂r+eθ1 r∂Vϕ ∂θ+eϕ1 rsinθ∂Vϕ ∂ϕ/parenrightbigg ×eϕ =−∂Vϕ ∂reθ+1 r∂Vϕ ∂θer. (3.69) Combining these six terms, we have ∇×V=/parenleftbigg1 r∂Vϕ ∂θ−1 rsinθ∂Vθ ∂ϕ+1 rcosθ sinθVϕ/parenrightbigg er +/parenleftbigg1 rsinθ∂Vr ∂ϕ−∂Vϕ ∂r−1 rVϕ/parenrightbigg eθ+/parenleftbigg∂Vθ ∂r−1 r∂Vr ∂θ+1 rVθ/parenrightbigg eϕ.(3.70) 3.2.2 Inﬁnitesimal Elements In spherical coordinates, the inﬁnitesimal displacement vector between a point at (r, θ, ϕ) and at ( r+dr, θ+dθ,ϕ+dϕ)i s dr=erdr+eθrdθ+eϕrsinθdϕ. (3.71) Note from Fig. 3.4 that only in the erdirection, the increment d ris an element of length. Both d θand dϕare inﬁnitesimal angles. They do not even have the units of length. The element of length in the eθdirection is rdθand in the eϕ direction is rsinθdϕ.Thus, one would expect the gradient in the spherical coordinates to be ∇=er∂ ∂r+eθ1 r∂ ∂r+eϕ1 rsinθ∂ ∂ϕ, which is indeed the case as shown in (3.52). The inﬁnitesimal volume element is the product of the three perpendicular inﬁnitesimal displacements dV=( dr)(rdθ)(rsinθdϕ)=r2sinθdrdθdϕ. (3.72) 3.2 Spherical Coordinates 129 xyz dr r q dq djjr sinqdjr sinq dj r dq Fig. 3.4. Diﬀerential elements in the spherical coordinates. The diﬀerential length in the direction of increasing θisrdθ.The diﬀerential length in the direction of increasing ϕisrsinθdϕ.The diﬀerential volume element is r2sinθdrdθdϕ The possible range of ris 0 to ∞,θfrom 0 to π,andϕfrom 0 to 2 π.Note thatθgoes from 0 to only π,and not 2 π.If it goes to 2 π,then every point would be counted twice. Example 3.2.1. Use spherical coordinates to ﬁnd ∇r,∇·r,∇rn,∇·rner,∇2rn,∇×f(r)er. (We have found them in previous chapter with Cartesian coordinates. Using spherical coordinates, the results can be easily obtained, almost by ins- pection.) Solution 3.2.1. Since these functions depend only on r,we need to retain only terms involving rvariable: ∇r=er∂ ∂rr=er=/hatwider, ∇·r=1 r2∂ ∂r/parenleftbig r2r/parenrightbig =3, ∇rn=er∂ ∂rrn=ernrn−1, ∇·rner=1 r2∂ ∂r/parenleftbig r2rn/parenrightbig =(n+2 )rn−1, ∇2rn=1 r2∂ ∂r/parenleftbigg r2∂ ∂rrn/parenrightbigg =n1 r2∂ ∂rrn+1=n(n+1 )rn−2, 130 3 Curved Coordinates ∇×f(r)er=1 rsinθ∂f(r) ∂ϕeθ+1 r∂f(r) ∂θeϕ=0. Example 3.2.2. Express r×∇in spherical coordinates. (In quantum mechan- ics, the angular momentum operator Lis deﬁned as L=r×p,wherepis the linear momentum operator, given by −i/planckover2pi1∇.) Solution 3.2.2. r×∇=rer×/bracketleftbigg er∂ ∂r+eθ1 r∂ ∂θ+eϕ1 rsinθ∂ ∂ϕ/bracketrightbigg =r/bracketleftbigg eϕ1 r∂ ∂θ−eθ1 rsinθ∂ ∂ϕ/bracketrightbigg =eϕ∂ ∂θ−eθ1 sinθ∂ ∂ϕ. 3.3 General Curvilinear Coordinate System 3.3.1 Coordinate Surfaces and Coordinate Curves In this section, we will develop the general theory of a curvilinear coordinate system. Suppose there is a one to one relationship between the Cartesian coordinate system ( x,y,z) and another curvilinear system ( u1,u2,u3). This means that ( x,y,z) can be written as functions of ui, x=x(u1,u2,u3),y=y(u1,u2,u3),z=z(u1,u2,u3), (3.73) and conversely, u1=u1(x,y,z),u2=u2(x,y,z),u 3=u3(x,y,z). (3.74) The surfaces ui= constant are referred to as coordinate surfaces and the intersections of these surfaces deﬁne the coordinate curves . For example, if the curvilinear system is the cylindrical coordinate system, then u1=ρ, u2=ϕ, u3=zas shown in Fig. 3.5. Thus, u1= constant is the surface of the cylinder, u2= constant is the vertical plane, and u3=constant is the horizontal plane shown in the ﬁgure. The intersection of the vertical plane and the horizontal plane is the u1curve which is the line shown as the ρcurve. The intersection of the horizontal plane and the surface of the cylinder is the u2curve which is the circle shown as the ϕcurve. The intersection of the surface of the cylinder and the vertical plane is the u1curve which is the vertical line shown as the zcurve. 3.3 General Curvilinear Coordinate System 131 xyz rjz curve j curve r curvez = const. j = const.r = const.P ( r, j, z) rz u1 = r curveu3 = z curve j curve u3(= z) = const.u2 (= j) = const. er = e1u1(= r) = const. ej = e 2ez = e3(a) (b) P Fig. 3.5. Coordinate surfaces and coordinate curves of cylindrical coordinate sys- tem. (a) The side surface of the cylinder ( ρ= constant), the horizontal plane ( z= constant), and the plane containing zaxis (ϕ= constant) are the coordinate surfaces. The intersections of them are the coordinate curves. ( b) Locally, the three unit vectors along the coordinate curves form an orthogonal basis set Now the position vector rcan be expressed as a function of ui, r=r(u1,u2,u3), (3.75) dr=∂r ∂u1du1+∂r ∂u2du2+∂r ∂u3du3. (3.76) The partial derivative∂r ∂u1means the rate of variation of rwithu1,while u2 andu3are held ﬁxed. So, the vector∂r ∂u1lies in the u2andu3coordinate surfaces and is, therefore, along the u1coordinate curve. This enables a unit vector e1to be deﬁned in the direction of the u1curve, e1=∂r ∂u1/h1 (3.77) where h1is the magnitude of∂r ∂u1 h1=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂u1/vextendsingle/vextendsingle/vextendsingle/vextendsingle, (3.78) known as the scale factor. The unit vectors e2ande3and the corresponding scale factors h2andh3are deﬁned in a similar way. In the case of cylindrical coordinates, e1is a unit vector along the ρcurve, which is previously deﬁned as eρ,e2is a unit vector tangent to the ϕcurve, which is previously deﬁned as eϕ, ande3is a unit vector along the zcurve, which is previously deﬁned as ez=k. 132 3 Curved Coordinates e3 e1e2drh2 du 2 h1du1h3 du3 Fig. 3.6. Volume element of an orthogonal curvilinear coordinate system. A change inuileads to a change of distance hiduiin the eidirection With the unit vectors and scale factors, the displacement vector d rof (3.76) can be written as dr=e1h1du1.+e2h2du2+e3h3du3. (3.79) If the unit vectors are orthogonal, that is ei·ej=/braceleftbigg1i=j 0i/negationslash=j, (3.80) then the coordinate curves are perpendicular to each other where they in- tersect. Such coordinate systems are known as orthogonal curvilinear coor- dinates. It will also be assumed that the coordinate system is right handed, so that e1×e2=e3,e2×e3=e1,e3×e1=e2. (3.81) Thus, locally e1,e2,e3form a set of unit orthogonal basis vectors for the co- ordinate system ( u1,u2,u3),although they may change directions from point to point. In this coordinate system, a change in uiof size d uileads to a change of distance hiduiin the eidirection. Schematically this is shown in Fig. 3.6. It follows from (3.79) and Fig. 3.6 that the arc length d sof a line element along d ris given by ds=( dr·dr)1/2=[ (h1du1)2+(h2du2)2+(h3du3)2]1/2. (3.82) The directed surface element along e1generated by the displacements d u2 and d u3is e1da=e2h2du2×e3h3du, (3.83) and similarly for surface elements e2daande3da.Finally, the volume elements d Vproduced by the displacements d u1,du2,du3are given by dV=|e1h1du1.·(e2h2du2×e3h3du)|=h1h2h3du1du2du3,(3.84) sincee1·(e2×e3)=1. 3.3 General Curvilinear Coordinate System 133 3.3.2 Diﬀerential Operations in Curvilinear Coordinate Systems Gradient. The gradient ∇Φof a scalar function is a vector perpendicular to the surface Φ= constant, deﬁned by the equation dΦ=∇Φ·dr (3.85) To ﬁnd the expression of ∇Φin a curvilinear coordinate system, let us assume ∇Φ=f1e1+f2e2+f3e3. (3.86) Since dr=h1du1e1+h2du2e2+h3du3e3, it follows that ∇Φ·dr=f1h1du1+f2h2du2+f3h3du3. (3.87) On the other hand dΦ=∂Φ ∂u1du1+∂Φ ∂u2du2+∂Φ ∂u3du3. (3.88) Equating the last two equations, we have f1h1=∂Φ ∂u1,f2h2=∂Φ ∂u2,f3h3=∂Φ ∂u3. (3.89) Thus (3.86) becomes ∇Φ=e11 h1∂Φ ∂u1+e21 h2∂Φ ∂u2+e31 h3∂Φ ∂u3. (3.90) Therefore the del operator in curvilinear coordinates can be written as ∇=e11 h1∂ ∂u1+e21 h2∂ ∂u2+e31 h3∂ ∂u3. (3.91) In particular, ∇u1=e11 h1∂u1 ∂u1+e21 h2∂u1 ∂u2+e31 h3∂u1 ∂u3. Since u1,u2,u3are independent variables, ∂u1 ∂u1=1,∂u1 ∂u2=0,∂u1 ∂u3=0. Therefore ∇u1=e11 h1. (3.92) Similarly, ∇u2=e21 h2,∇u3=e31 h3. (3.93) 134 3 Curved Coordinates Divergence. The expression of the divergence of a vector ﬁeld A=A1e1+ A2e2+A3e3in a curvilinear coordinates can be found by direct calculation using the del operator. ∇·A=∇·(A1e1+A2e2+A3e3), ∇·A1e1=∇·A1(e2×e3)=∇·A1h2h3(∇u2×∇u3) =(∇A1h2h3)·(∇u2×∇u3)+A1h2h3∇·(∇u2×∇u3). The term ∇·(∇u2×∇u3)=∇×∇u2·∇u3−∇×∇u3·∇u2vanishes because ∇×∇f=0.Thus, ∇·A1e1=(∇A1h2h3)·(∇u2×∇u3) =(∇A1h2h3)·e2×e3 h2h3=(∇A1h2h3)·e1 h2h3. Using the del operator of (3.91), we have ∇(A1h2h3)=e11 h1∂(A1h2h3) ∂u1+e21 h2∂(A1h2h3) ∂u2+e31 h3∂(A1h2h3) ∂u3, (∇A1h2h3)·e1 h2h3=1 h1h2h3∂(A1h2h3) ∂u1. Therefore, ∇·A1e1=1 h1h2h3∂(A1h2h3) ∂u1. (3.94) With similar expressions ∇·A2e2=1 h1h2h3∂(A2h3h1) ∂u2, ∇·A3e3=1 h1h2h3∂(A3h1h2) ∂u3, we obtain ∇·A=1 h1h2h3/bracketleftbigg∂(A1h2h3) ∂u1+∂(A2h3h1) ∂u2+∂(A3h1h2) ∂u3/bracketrightbigg . (3.95) Laplacian. The Laplacian follows from its deﬁnition ∇2Φ=∇·∇Φ. Since the ∇Φis given by ∇Φ=e11 h1∂Φ ∂u1+e21 h2∂Φ ∂u2+e31 h3∂Φ ∂u3, 3.3 General Curvilinear Coordinate System 135 the divergence of this vector is ∇·∇Φ=1 h1h2h3/bracketleftbigg∂ ∂u1/parenleftbigg1 h1∂Φ ∂u1h2h3/parenrightbigg +∂ ∂u2/parenleftbigg1 h2∂Φ ∂u2h3h1/parenrightbigg +∂ ∂u3/parenleftbigg1 h3∂Φ ∂u3h1h2/parenrightbigg/bracketrightbigg . Hence ∇2Φ=1 h1h2h3/bracketleftbigg∂ ∂u1/parenleftbiggh2h3 h1∂Φ ∂u1/parenrightbigg +∂ ∂u2/parenleftbiggh3h1 h2∂Φ ∂u2/parenrightbigg +∂ ∂u3/parenleftbiggh1h2 h3∂Φ ∂u3/parenrightbigg/bracketrightbigg . (3.96) Curl. The curl of a vector ﬁeld in a curvilinear coordinates can also be cal- culated directly, ∇×A=∇×(A1e1+A2e2+A3e3), ∇×A1e1=∇×A1h1∇u1=∇(A1h1)×∇u1+A1h1∇×∇u1. Since ∇×∇u1=0, ∇×A1e1=∇(A1h1)×∇u1=∇(A1h1)×e1 h1 Now ∇(A1h1)=e11 h1∂(A1h1) ∂u1+e21 h2∂(A1h1) ∂u2+e31 h3∂(A1h1) ∂u3, ∇(A1h1)×e1 h1=−e31 h2h1∂(A1h1) ∂u2+e21 h3h1∂(A1h1) ∂u3. Therefore, ∇×A1e1=−e31 h2h1∂(A1h1) ∂u2+e21 h3h1∂(A1h1) ∂u3. With similar expressions for ∇×A2e2and∇×A3e3,we have ∇×A=e1/bracketleftbigg1 h2h3∂(A3h3) ∂u2−1 h2h3∂(A2h2) ∂u3/bracketrightbigg +e2/bracketleftbigg1 h1h3∂(A1h1) ∂u3−1 h1h3∂(A3h3) ∂u1/bracketrightbigg +e3/bracketleftbigg1 h2h1∂(A2h2) ∂u1−1 h2h1∂(A1h1) ∂u2/bracketrightbigg . This expression can be put in a more symmetrical form, which is easier to remember, 136 3 Curved Coordinates ∇×A=h1e1 h1h2h3/bracketleftbigg∂(A3h3) ∂u2−∂(A2h2) ∂u3/bracketrightbigg +h2e2 h1h2h3/bracketleftbigg∂(A1h1) ∂u3−∂(A3h3) ∂u1/bracketrightbigg +h3e3 h1h2h3/bracketleftbigg∂(A2h2) ∂u1−∂(A1h1) ∂u2/bracketrightbigg =1 h1h2h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleh1e1h2e2h3e3 ∂ ∂u1∂ ∂u2∂ ∂u3 A1h1A2h2A3h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (3.97) Example 3.3.1. For the cylindrical coordinates, x=ρcosϕ, y=ρsinϕ, z=z. With u1=ρ, u2=ϕ, u 3=z,(a) ﬁnd the scale factors h1,h2,andh3, (b) ﬁnd the gradient, divergence, Laplacian, and curl in the cylindrical coor- dinates from the general formulas derived in this section. Solution 3.3.1. (a) Since r=xi+yj+zkandx,y,z are functions of u1,u2,u3,so ∂r ∂ui=∂x ∂uii+∂y ∂uij+∂z ∂uik, hi=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂ui/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂ui·∂r ∂ui/vextendsingle/vextendsingle/vextendsingle/vextendsingle1/2 =/bracketleftBigg/parenleftbigg∂x ∂ui/parenrightbigg2 +/parenleftbigg∂y ∂ui/parenrightbigg2 +/parenleftbigg∂z ∂ui/parenrightbigg2/bracketrightBigg1/2 . Now ∂x ∂u1=∂x ∂ρ=c o s ϕ,∂y ∂u1=∂y ∂ρ=s i nϕ,∂z ∂u1=∂z ∂ρ=0, ∂x ∂u2=∂x ∂ϕ=−ρsinϕ,∂y ∂u2=∂y ∂ϕ=ρcosϕ,∂z ∂u2=∂z ∂ϕ=0, ∂x ∂u3=∂x ∂z=0,∂y ∂u3=∂y ∂z=0,∂z ∂u3=∂z ∂z=1. h1=/parenleftbig cos2ϕ+s i n2ϕ/parenrightbig1/2=1, h2=/parenleftbig ρ2cos2ϕ+ρ2sin2ϕ/parenrightbig1/2=ρ, h3= (1)1/2=1. (b) ∇Φ=e11 h1∂Φ ∂u1+e21 h2∂Φ ∂u2+e31 h3∂Φ ∂u3 =eρ∂Φ ∂ρ+eϕ1 ρ∂Φ ∂ϕ+ez∂Φ ∂z. 3.3 General Curvilinear Coordinate System 137 ∇·A=1 h1h2h3/bracketleftbigg∂(A1h2h3) ∂u1+∂(A2h3h1) ∂u2+∂(A3h1h2) ∂u3/bracketrightbigg =1 ρ/bracketleftbigg∂(Aρρ) ∂ρ+∂(Aϕ) ∂ϕ+∂(Azρ) ∂z/bracketrightbigg =1 ρ∂ ∂ρ(ρAρ)+1 ρ∂Aϕ ∂ϕ+∂Az ∂z. ∇2Φ=1 h1h2h3/bracketleftbigg∂ ∂u1/parenleftbiggh2h3 h1∂Φ ∂u1/parenrightbigg +∂ ∂u2/parenleftbiggh3h1 h2∂Φ ∂u2/parenrightbigg +∂ ∂u3/parenleftbiggh1h2 h3∂Φ ∂u3/parenrightbigg/bracketrightbigg =1 ρ/bracketleftbigg∂ ∂ρ/parenleftbigg ρ∂Φ ∂ρ/parenrightbigg +∂ ∂ϕ/parenleftbigg1 ρ∂Φ ∂ϕ/parenrightbigg +∂ ∂z/parenleftbigg ρ∂Φ ∂z/parenrightbigg/bracketrightbigg =∂2Φ ∂ρ2+1 ρ∂Φ ∂ρ+1 ρ2∂2Φ ∂ϕ2+∂2Φ ∂z2. ∇×A=1 h1h2h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleh1e1h2e2h3e3 ∂ ∂u1∂ ∂u2∂ ∂u3 A1h1A2h2A3h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 ρ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleeρρeϕez ∂ ∂ρ∂ ∂ϕ∂ ∂z AρρAρAz/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/parenleftbigg1 ρ∂Az ∂ϕ−1 ρ∂ ∂z(ρAϕ)/parenrightbigg eρ+/parenleftbigg∂Aρ ∂z−∂Az ∂ρ/parenrightbigg eϕ +/parenleftbigg1 ρ∂ ∂ρ(ρAϕ)−1 ρ∂Aρ ∂ϕ/parenrightbigg ez. Example 3.3.2. For the spherical coordinates, u1=r, u2=θ, u3=ϕ,and x=rsinθcosϕ, y =rsinθsinϕ, z=rcosθ.(a) Find the scale factors h1,h2,andh3,(b) ﬁnd the gradient, divergence, Laplacian, and curl in the spherical coordinates from the general formulas derived in this section. Solution 3.3.2. (a) hi=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂ui/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂ui·∂r ∂ui/vextendsingle/vextendsingle/vextendsingle/vextendsingle1/2 =/bracketleftBigg/parenleftbigg∂x ∂ui/parenrightbigg2 +/parenleftbigg∂y ∂ui/parenrightbigg2 +/parenleftbigg∂z ∂ui/parenrightbigg2/bracketrightBigg1/2 . Now ∂x ∂u1=∂x ∂r=s i nθcosϕ,∂y ∂u1=∂y ∂r=s i nθsinϕ,∂z ∂u1=∂z ∂r=c o s θ, ∂x ∂u2=∂x ∂θ=rcosθcosϕ,∂y ∂u2=∂y ∂θ=rcosθsinϕ,∂z ∂u2=∂z ∂θ=−rsinθ, ∂x ∂u3=∂x ∂ϕ=−rsinθsinϕ,∂y ∂u3=∂y ∂ϕ=rsinθcosϕ,∂z ∂u3=∂z ∂ϕ=0. 138 3 Curved Coordinates h1=( s i n2θcos2ϕ+s i n2θsin2ϕ+c o s2θ)1/2=1, h2=/parenleftbig r2cos2θcos2ϕ+r2cos2θsin2ϕ+r2sin2θ/parenrightbig1/2=r. h3=/parenleftbig r2sin2θsin2ϕ+r2sin2θcos2ϕ/parenrightbig1/2=rsinθ. (b) ∇Φ=e11 h1∂Φ ∂u1+e21 h2∂Φ ∂u2+e31 h3∂Φ ∂u3 =er∂Φ ∂r+eθ1 r∂Φ ∂θ+eϕ1 rsinθ∂Φ ∂ϕ. ∇·A=1 h1h2h3/bracketleftbigg∂(A1h2h3) ∂u1+∂(A2h3h1) ∂u2+∂(A3h1h2) ∂u3/bracketrightbigg =1 r2sinθ/bracketleftbigg∂(Arr2sinθ) ∂r+∂(Aθrsinθ) ∂θ+∂(Aϕr) ∂ϕ/bracketrightbigg =1 r2∂ ∂r/parenleftbig r2Ar/parenrightbig +1 rsinθ∂ ∂θ(sinθAθ)+1 rsinθ∂ ∂ϕ(Aϕ). ∇2Φ=1 h1h2h3/bracketleftbigg∂ ∂u1/parenleftbiggh2h3 h1∂Φ ∂u1/parenrightbigg +∂ ∂u2/parenleftbiggh3h1 h2∂Φ ∂u2/parenrightbigg +∂ ∂u3/parenleftbiggh1h2 h3∂Φ ∂u3/parenrightbigg/bracketrightbigg =1 r2sinθ/bracketleftbigg∂ ∂r/parenleftbigg r2sinθ∂Φ ∂r/parenrightbigg +∂ ∂θ/parenleftbigg sinθ∂Φ ∂θ/parenrightbigg +∂ ∂ϕ/parenleftbigg1 sinθ∂Φ ∂ϕ/parenrightbigg/bracketrightbigg =1 r2∂ ∂r/parenleftbigg r2∂ ∂r/parenrightbigg +1 r2sinθ∂ ∂θ/parenleftbigg sinθ∂ ∂θ/parenrightbigg +1 r2sin2θ∂2 ∂ϕ2. ∇×A=1 h1h2h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleh1e1h2e2h3e3 ∂ ∂u1∂ ∂u2∂ ∂u3 A1h1A2h2A3h3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 r2sinθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleerreθrsinθeϕ ∂ ∂r∂ ∂θ∂ ∂ϕ ArrAθrsinθAϕ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =1 r2sinθ/parenleftbigg∂ ∂θ(rsinθAϕ)−∂ ∂ϕ(rAθ)/parenrightbigg er +1 rsinθ/parenleftbigg∂ ∂ϕ(Ar)−∂ ∂r(rsinθAϕ)/parenrightbigg eθ+1 r/parenleftbigg∂ ∂r(rAθ)−∂ ∂θ(Ar)/parenrightbigg eϕ. 3.4 Elliptical Coordinates There are many coordinate systems. In the classical text of Morse and Feshbach, “Methods of Theoretical Physics,” no less than 13 coordinate sys- tems are discussed. Each of them is particularly convenient for certain special 3.4 Elliptical Coordinates 139 problems. However, because of the development of high-speed computers, the need for most of them has diminished. In this section, we will introduce only the elliptical coordinate system as one more example of special coordinate systems. The elliptical coordinate system is important in dealing with the two center problems in diatomic molecules. 3.4.1 Coordinate Surfaces Elliptical coordinates are families of confocal ellipses and hyperbolas in two dimensions. Rotating them around the major axis of the ellipses, surfaces of prolate spheroids and hyperboloids are generated. These surfaces together with the planes containing the major axis form a three dimensional coordinate system which is commonly known as the elliptical coordinates. The coordinate surfaces are shown in Fig. 3.7a. Let r1andr2be the distances from the two focal points which are separated by a distance 2 c on the zaxis as shown in Fig. 3.7b. A point in space is determined by r1and r2,the distances from the two focal points, and the angle ϕaround the zaxis. The coordinates of the point are λ,µandϕwith λ=r2+r1 2c, (3.98) µ=r2−r1 2c. (3.99) xF Fjrz (a) y j = const.l = const.P (λ, µ, j) j = const. 0 Brz yxA F Fcrr1 r2z(b) rz BA 0 Fig. 3.7. Elliptical Coordinate System. ( a) The coordinate surfaces generated by an ellipse, two hyperbolas and a plane containing the major axis of the ellipse. (b) The confocal ellipse given by r2+r1= constant, and the confocal hyperbolas given by r2−r1= constant 140 3 Curved Coordinates Forλ= constant, (3.98) maps out a prolate spheroid in space, on any ϕ= constant plane, it is just an ellipse as shown in Fig. 3.7b. This can be seen as follows: r2=/bracketleftBig (z+c)2+ρ2/bracketrightBig1/2 , (3.100) r1=/bracketleftBig (z−c)2+ρ2/bracketrightBig1/2 , (3.101) r2+r1=2cλ. (3.102) Square both sides of r2=2cλ−r1,and collect terms, it becomes z=cλ2−λr1. Square both sides again, we ﬁnd (λ2−1)z2+λ2ρ2=c2λ2/parenleftbig λ2−1/parenrightbig . This equation can be written in the standard form of an ellipse, z2 c2λ2+ρ2 c2/parenleftbig λ2−1/parenrightbig=1, (3.103) which cuts the zaxis at ±cλ,theρaxis at ±c/parenleftbig λ2−1/parenrightbig1/2.The range of λis clearly ∞≥λ≥1.When λ=1,the ellipse reduces to the line between the two focal points. Starting with r2−r1=2cµ and following the same procedure, we get z2 c2µ2+ρ2 c2(µ2−1)=1, which is of the same form as the ellipse. However, in this case it is clear from Fig. 3.7b that r1+2c≥r2, which simply says that the sum of two sides of a triangle must be greater than the third side. It follows that 2c≥r2−r1=2cµ. Therefore 1 ≥µ.Thus the equation is seen to be in the form of a hyperbola: z2 c2µ2−ρ2 c2(1−µ2)=1, (3.104) 3.4 Elliptical Coordinates 141 which cuts the zaxis at ±cµ.There are two sheets of hyperbola, one corres- ponds to positive value of µand the other, negative µ.Therefore the range ofµis 1≥µ≥−1.When µ=0,the hyperbola reduces to a straight line perpendicular to the zaxis through the origin. When µ=1,it reduces to a line from z=calong the zaxis to ∞.When µ=−1,it reduces to a line from z=−calong the zaxis to −∞. Surfaces of hyperboloids are generated by rotating this family of hyper- bolas around the zaxis. The range of the angle of rotation ϕis of course, 0≤ϕ≤2π. 3.4.2 Relations with Rectangular Coordinates The transformation between ( x,y,z) and ( λ,u,ϕ ) can be seen from (3.100) and (3.101): r2 2=z2+2zc+c2+ρ2, r2 1=z2−2zc+c2+ρ2. It follows that r2 2−r2 1=4zc. Since r2 2−r2 1=(r2−r1)(r2+r1)=( 2 cµ)(2cλ), therefore 4 zc=4c2µλwhich gives z=cµλ. (3.105) Putting this into (3.103), we have c2µ2λ2 c2λ2+ρ2 c2/parenleftbig λ2−1/parenrightbig=1, which gives ρ2=c2/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig . (3.106) Now, from Fig. 3.7a x=ρcosϕ, y =ρsinϕ. (3.107) Therefore x=c/bracketleftbig/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig/bracketrightbig1/2cosϕ, y=c/bracketleftbig/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig/bracketrightbig1/2sinϕ, (3.108) z=cµλ. From the position vector r=x(λ,µ,ϕ )i+y(λ,µ,ϕ )j+z(λ,µ,ϕ )k, (3.109) 142 3 Curved Coordinates we can ﬁnd the unit vectors along the λ,µ,ϕ coordinate curves. The three unit vectors are deﬁned as eλ=∂r ∂λ/hλ,eµ=∂r ∂µ/hµ,eϕ=∂r ∂ϕ/hϕ. (3.110) Since ∂r ∂λ=∂x ∂λi+∂y ∂λj+∂x ∂λk=cλ/parenleftbig 1−µ2/parenrightbig /bracketleftbig/parenleftbig λ2−1/parenrightbig (1−µ2)/bracketrightbig1/2cosϕi +cλ/parenleftbig 1−µ2/parenrightbig /bracketleftbig/parenleftbig λ2−1/parenrightbig (1−µ2)/bracketrightbig1/2sinϕj+cµk, (3.111) ∂r ∂µ=∂x ∂µi+∂y ∂µj+∂x ∂µk=−cµ/parenleftbig λ2−1/parenrightbig /bracketleftbig/parenleftbig λ2−1/parenrightbig (1−µ2)/bracketrightbig1/2cosϕi +−cµ/parenleftbig λ2−1/parenrightbig /bracketleftbig/parenleftbig λ2−1/parenrightbig (1−µ2)/bracketrightbig1/2sinϕj+cλk, (3.112) ∂r ∂ϕ=∂x ∂ϕi+∂y ∂ϕj+∂x ∂ϕk=−c/bracketleftbig/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig/bracketrightbig1/2sinϕi +c/bracketleftbig/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig/bracketrightbig1/2cosϕj, (3.113) the scale factors are seen to be hλ=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/bracketleftBigg/parenleftbigg∂x ∂λ/parenrightbigg2 +/parenleftbigg∂y ∂λ/parenrightbigg2 +/parenleftbigg∂z ∂λ/parenrightbigg2/bracketrightBigg1/2 =/bracketleftBigg c2/parenleftbig λ2−µ2/parenrightbig λ2−1/bracketrightBigg1/2 , (3.114) hµ=/bracketleftBigg c2/parenleftbig λ2−µ2/parenrightbig 1−µ2/bracketrightBigg1/2 ,h ϕ=/bracketleftbig c2/parenleftbig λ2−1/parenrightbig/parenleftbig 1−µ2/parenrightbig/bracketrightbig1/2. (3.115) It can be readily veriﬁed that eλ×eϕ=eµ,eϕ×eµ=eλ,eµ×eλ=eϕ. (3.116) Therefore eλ,eϕ,eµform an orthogonal basis set. Note that in the right-hand convention, the sequence is ( eλ,eϕ,eµ) and not ( eλ,eµ,eϕ). The volume element in this system is dV=hλhϕh µdλdϕdµ=c3/parenleftbig λ2−µ2/parenrightbig dλdϕdµ. (3.117) 3.4 Elliptical Coordinates 143 Example 3.4.1. Use the elliptical coordinates to ﬁnd the volume of the prolate spheroid generated by rotating the ellipse z2 a2+ρ2 b2=1 around its major axis z. Solution 3.4.1. In terms of elliptical coordinates, the ellipse is given by z2 c2λ2+ρ2 c2(λ2−1)=1, where 2 cis the distance between the two focal points. To ﬁnd the upper limit ofλ,we note a2=c2λ2,or λ=a/c Furthermore, b2=c2(λ2−1) =c2[(a/c)2−1] =a2−c2. The volume of the prolate spheroid is V=/integraldisplay/integraldisplay/integraldisplay dV=/integraldisplay2π 0/integraldisplay1 −1/integraldisplaya/c 1c3/parenleftbig λ2−u2/parenrightbig dλdµdϕ =2πc3/bracketleftBigg/integraldisplay1 −1dµ/integraldisplaya/c 1λ2dλ−/integraldisplaya/c 1dλ/integraldisplay1 −1µ2dµ/bracketrightBigg , /integraldisplay1 −1dµ/integraldisplaya/c 1λ2dλ=2 3/bracketleftbigg/parenleftBiga c/parenrightBig3 −1/bracketrightbigg ,/integraldisplaya/c 1dλ/integraldisplay1 −1µ2dµ=2 3/bracketleftBiga c−1/bracketrightBig . V=4π 3c3/bracketleftbigg/parenleftBiga c/parenrightBig3 −a c/bracketrightbigg =4π 3a/parenleftbig a2−c2/parenrightbig =4π 3ab2. Example 3.4.2. Evaluate the following integral over all space I=/integraldisplay/integraldisplay/integraldisplay e−r1e−r2dV, where r1andr2are distances from two ﬁxed points separated by a distance R. (This happens to be the overlap integral of the H+ 2molecular ion.) Solution 3.4.2. I=/integraldisplay/integraldisplay/integraldisplay e−r1e−r2dV=/integraldisplay/integraldisplay/integraldisplay e−(r1+r2)dV. 144 3 Curved Coordinates Using elliptical coordinates r1+r2=2cλ=Rλ, I=/parenleftbiggR 2/parenrightbigg3/integraldisplay2π 0/integraldisplay1 −1/integraldisplay∞ 1e−Rλ(λ2−µ2)dλdµdϕ =/parenleftbiggR 2/parenrightbigg3 2π/bracketleftbigg/integraldisplay1 −1dµ/integraldisplay∞ 1e−Rλλ2dλ−/integraldisplay∞ 1e−Rλdλ/integraldisplay1 −1µ2dµ/bracketrightbigg =π(1 +R+1 3R2)e−R. 3.4.3 Prolate Spheroidal Coordinates The transformation (3.108) can be expressed in a more compact form with still another change of variables. Taking advantage of the identities sin2θ=1−cos2θ,sinh2η=1+c o s h2η, we can set λ=c o s h η, µ =c o s θ. (3.118) With this set of variables, the transformation (3.108) becomes x=csinhηsinθcosϕ, y=csinhηsinθsinϕ, (3.119) z=ccoshηcosθ. The set of coordinates ( η,θ,ϕ ) is known as the prolate spheroidal coordi- nate system. The range of ηis 0≤η<∞,the range of θis 0≤θ≤π.The scale factors for this system is hη=c/parenleftbig sinh2η+s i n2θ/parenrightbig1/2, (3.120) hθ=hη,h ϕ=csinhηsinθ. (3.121) The volume element in this system is dV=c3/parenleftbig sinh3ηsinθ+s i n3θsinhη/parenrightbig dηdθdϕ. Note that hηhθhϕ/negationslash=hλhϕhµ,since d λdµis not equal to d ηdθ. 3.5 Multiple Integrals So far we have seen how to ﬁnd surface and volume elements for a multiple integral in an orthogonal coordinate system. In this section, we will show that following the same line of reasoning, this method can also be used for any change of variables in multiple integrals, regardless whether the new coordi- nates are orthogonal or not. 3.5 Multiple Integrals 145 3.5.1 Jacobian for Double Integral Consider the double integral in the Cartesian coordinates/integraltext/integraltext Sf(x,y)dawhere the area element d ais of course just d xdy.Very often the variables of integration ( x,y) are not the most convenient for evaluating the integral. It is desirable to deﬁne double integrals in terms of a general pair of curvilinear coordinates. Let the curvilinear coordinates be ( u,v),and there be a one-to-one trans- formation between ( x,y) and ( u,v): x=x(u,v),y =y(u,v). (3.122) The position vector from the origin to a point inside Sis r=x(u,v)i+y(u,v)j. (3.123) Therefore, rcan also be considered as a function of the curvilinear coordinates, that is r=r(u,v).Thus, dr=∂r ∂udu+∂r ∂vdv. (3.124) Now ( ∂r/∂u)duis an inﬁnitesimal vector along the line where vinr(u,v)i s kept constant and ( ∂r/∂v)dvis an inﬁnitesimal vector along the line where u is kept constant. While they may not be orthogonal, the area of the paralle- logram formed by these two vectors is still given by their cross product, da=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂udu×∂r ∂vdv/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂r ∂u×∂r ∂v/vextendsingle/vextendsingle/vextendsingle/vextendsingledudv. (3.125) It follows from (3.123), ∂r ∂u=∂x ∂ui+∂y ∂uj, (3.126) ∂r ∂v=∂x ∂vi+∂y ∂vj. (3.127) Thus the cross product of these two vectors is ∂r ∂u×∂r ∂v=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleij k ∂x ∂u∂y ∂u0 ∂x ∂v∂y ∂v0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=k/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂u∂y ∂u ∂x ∂v∂y ∂v/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (3.128) The last determinant is called Jacobian determinant (or simply as Jacobian) written as∂(x,y) ∂(u,v), 146 3 Curved Coordinates J=∂(x,y) ∂(u,v)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂u∂y ∂u ∂x ∂v∂y ∂v/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (3.129) It follows from (3.125) that the area element is equal to the absolute value of the Jacobian times d udv da=∂(x,y) ∂(u,v)dudv. (3.130) Therefore the double integral can be written as /integraldisplay/integraldisplay Sf(x,y)dxdy=/integraldisplay/integraldisplay Sf(x(u,v),y(u,v))∂(x,y) ∂(u,v)dudv. (3.131) The integrand on the right-hand side is a function of uandv.Now suppose we want to change it to an integral over xandy,we should have /integraldisplay/integraldisplay Sf(x(u,v),y(u,v)∂(x,y) ∂(u,v)dudv=/integraldisplay/integraldisplay Sf(x,y)∂(x,y) ∂(u,v)∂(u,v) ∂(x,y)dxdy. The right-hand side of this equation must be identical to the left-hand side of (3.131). Therefore ∂(x,y) ∂(u,v)∂(u,v) ∂(x,y)=1. (3.132) This is a useful relation. Often we need∂(x,y) ∂(u,v),but∂(u,v) ∂(x,y)is much easier to calculate. In that case, we simply set ∂(x,y) ∂(u,v)=/bracketleftbigg∂(u,v) ∂(x,y)/bracketrightbigg−1 . Now we must be careful not to assert that d xdyis equal to∂(x,y) ∂(u,v)dudv. They are equal only in the sense that under the integral sign the area element dxdycan be changed to∂(x,y) ∂(u,v)dudv,provided the area Scovered by ( x,y) is the same as covered by ( u,v).Locally they cannot be equal. From ( 3.122), we have dx=∂x ∂udu+∂x ∂vdv, (3.133) dy=∂y ∂udu+∂y ∂vdv. (3.134) If we multiply d xby dy,it is certainly not equal to∂(x,y) ∂(u,v)dudv. 3.5 Multiple Integrals 147 Incidentally, the transformation of the diﬀerentials can be written as /parenleftbiggdx dy/parenrightbigg =⎛ ⎜⎝∂x ∂u∂x ∂v ∂y ∂u∂y ∂v⎞ ⎟⎠/parenleftbiggdu dv/parenrightbigg =(J)/parenleftbiggdu dv/parenrightbigg , (3.135) where ( J) is known as Jacobian matrix . It is the matrix associated with the Jacobian determinant∂(x,y) ∂(u,v).Jacobian determinant and Jacobian matrix are named after the German mathematician Carl Jacobi (1804–1851). Both are very useful, but we must not get confused by the two. 3.5.2 Jacobians for Multiple Integrals The deﬁnition of the triple integral/integraltext/integraltext/integraltext Vf(x,y,z)dVover a given region Vis entirely analogous to the deﬁnition of a double integral. If x,y,z are rectangular coordinates, then d V=dxdydz.Just as in double integral, often the triple integral is much easier to evaluate with a set of curvilinear coordinates u1,u2,u3.Again, let x=x(u1,u2,u3),y=y(u1,u2,u3),z=z(u1,u2,u3), (3.136) and the position vector be r=x(u1,u2,u3)i+y(u1,u2,u3)j+z(u1,u2,u3)k, (3.137) then dr=∂r ∂u1du1+∂r ∂u2du2+∂r ∂u3du3. (3.138) The partial derivative ( ∂r/∂u1) is the rate of variation of rwithu2andu3held ﬁxed. Therefore ( ∂r/∂u1)du1is an inﬁnitesimal vector along the u1coordinate curve. Similarly, ( ∂r/∂u2)du2and (∂r/∂u3)du3are, respectively, inﬁnitesimal vectors along the u2andu3coordinate curves. Regardless whether they are orthogonal or not, the volume of parallelepiped formed by these three vectors is equal to the scalar triple product of them dV=∂r ∂u1du1·/parenleftbigg∂r ∂u2du2×∂r ∂u3du3/parenrightbigg . (3.139) It follows from (3.137) that ∂r ∂u1=∂x ∂u1i+∂y ∂u1j+∂z ∂u1k. (3.140) With similar expressions for ∂r/∂u2and∂r/∂u3,the scalar triple product can be written as 148 3 Curved Coordinates dV=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂u1∂y ∂u1∂z ∂u1∂x ∂u2∂y ∂u2∂z ∂u2∂x ∂u3∂y ∂u3∂z ∂u3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledu1du2du3. (3.141) Again the determinant is known as the Jacobian determinant, written as ∂(x,y,z) ∂(u1,u2,u3)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂u1∂y ∂u1∂z ∂u1∂x ∂u2∂y ∂u2∂z ∂u2∂x ∂u3∂y ∂u3∂z ∂u3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (3.142) Thus, if the region covered by x,y,z and by u1,u2,u3is the same, then /integraldisplay/integraldisplay/integraldisplay Vf(x,y,z)dxdydz=/integraldisplay/integraldisplay/integraldisplay VF(u1,u2,u3)∂(x,y,z) ∂(u1,u2,u3)du1du2du3, (3.143) where F(u1,u2,u3)=f(x(u1,u2,u3),y(u1,u2,u3),z(u1,u2,u3)). We mention in passing that it can shown by induction that a multiple integral of nvariables can be similarly transformed, that is /integraldisplay/integraldisplay ···/integraldisplay Vf(x1,x2,...,x n)dx1dx2···dxn =/integraldisplay/integraldisplay ···/integraldisplay VF(u1,u2,...,u 3)∂(x1,x2,...,x n) ∂(u1,u2,...,u n)du1du2···dun.(3.144) Example 3.5.1. Evaluate the integral I=/integraldisplay/integraldisplay x2y2dxdy over the interior of the ellipse x2 a2+y2 b2=1. Solution 3.5.1. Parametrically, the coordinates of a point on the ellipse can be written as x=acosθ, y =bsinθ, since x2 a2+y2 b2=a2cos2θ a2+b2sin2θ b2=1. 3.5 Multiple Integrals 149 ab θy xa cos θ b sin θ Fig. 3.8. Parametric form of an ellipse. Parametrically an ellipse can be written as x=acosθ, y=bsinθ This is shown in Fig. 3.8. Any point inside the ellipse can be expressed as x=γacosθ, y =γbsinθ, withγ<1.Therefore, we can take γandθas curvilinear coordinates. (Note that the ellipse of γ= constant, and the straight line of θ= constant are not orthogonal unless a=b.) Thus, the integral can be written as I=/integraldisplay/integraldisplay (γacosθ)2(γbsinθ)2∂(x,y) ∂(γ,θ)dγdθ where the Jacobian is given by ∂(x,y) ∂(γ,θ)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂γ∂y ∂γ ∂x ∂θ∂y ∂θ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingleacosθbsinθ −γasinθγ bcosθ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=γab. Therefore I=/integraldisplay1 0/integraldisplay2π 0(γacosθ)2(γbsinθ)2γabdγdθ =a3b3/integraldisplay1 0γ5dγ/integraldisplay2π 0cos2θsin2θdθ=a3b3/parenleftbigg1 6/parenrightbiggπ 4=π 24a3b3. 150 3 Curved Coordinates Example 3.5.2. Evaluate the integral I=/integraldisplay∞ 0/integraldisplay∞ 0x2+y2 1+(x2−y2)2exp(−2xy)dxdy by making a change of variable u=x2−y2,v=2xy. Solution 3.5.2. First note the range of uis from −∞to∞, I=/integraldisplay∞ 0/integraldisplay∞ −∞x2+y2 1+(x2−y2)2exp(−2xy)∂(x,y) ∂(u,v)dudv. The Jacobian∂(x,y) ∂(u,v)is not easy to calculate directly, but ∂(u,v) ∂(x,y)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂u ∂x∂u ∂y ∂v ∂x∂v ∂y/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle2x−2y 2y2x/vextendsingle/vextendsingle/vextendsingle/vextendsingle=4/parenleftbig x2+y2/parenrightbig . Therefore, ∂(x,y) ∂(u,v)=/bracketleftbigg∂(u,v) ∂(x,y)/bracketrightbigg−1 =1 4(x2+y2). Thus I=/integraldisplay∞ 0/integraldisplay∞ −∞x2+y2 1+(x2−y2)2exp(−2xy)1 4(x2+y2)dudv =1 4/integraldisplay∞ 0/integraldisplay∞ −∞1 1+(x2−y2)2exp(−2xy)dudv =1 4/integraldisplay∞ 0/integraldisplay∞ −∞1 1+u2exp(−v)dudv=1 4/integraldisplay∞ 0exp(−v)dv×2/integraldisplay∞ 01 1+u2du =1 4[−exp(−v)]∞ 02/bracketleftbig tan−1u/bracketrightbig∞ 0=π 4. Exercises 1. Express the vector v=2xi−zj+ykin cylindrical coordinates. Ans.v=/parenleftbig 2ρcos2ϕ−zsinϕ/parenrightbig eρ−(2ρcosϕsinϕ+zcosϕ)eϕ+ρsinϕez. 2. Find the curl of Awhere A=ezln (1/ρ) in cylindrical coordinates. Ans.∇×A=eϕ1 ρ.(The magnetic vector potentialof a long wire carrying 3.5 Multiple Integrals 151 a current Iin the zdirection is A=ezµI 2πln (1/ρ).The magnetic ﬁeld is given by B=∇×A=eϕµI 2πln (1/ρ).) 3. Show that ln ρsatisﬁes the Laplace’s equation/parenleftbig ∇2lnρ=0/parenrightbig , (a) use cylin- drical coordinates, (b) use spherical coordinates ( ρ=rsinθ),(c) use Cartesian coordinates/parenleftBig ρ=/parenleftbig x2+y2/parenrightbig1/2/parenrightBig . 4. Show that 1 /rsatisﬁes the Laplace’s equation/parenleftbig ∇2(1/r)=0/parenrightbig forr/negationslash=0,(a) use cylindrical coordinates/parenleftBig r=/parenleftbig ρ2+z2/parenrightbig1/2/parenrightBig , (b) use spherical coordi- nates,(c) use Cartesian coordinates/parenleftBig r=/parenleftbig x2+y2+z2/parenrightbig1/2/parenrightBig . 5. (a) Show that in cylindrical coordinates dr dt=eρdρ dt+eϕρdϕ dt+ezdz dt, ds dt=/bracketleftBigg/parenleftbiggdρ dt/parenrightbigg2 +/parenleftbigg ρdϕ dt/parenrightbigg2 +/parenleftbiggdz dt/parenrightbigg2/bracketrightBigg1/2 , where d sis the diﬀerential arc length. (b) Find the length of the spiral described parametrically by ρ=a, ϕ=t, z=btfromt=0t o t=5. Ans. 5( a2+b2)1/2. 6. With the vector ﬁeld Agiven by A=ρeρ+ezin cylindrical coordinates, (a) show that ∇×A=0.(b) Find a scalar potential Φ,such that ∇Φ=A. Ans.1 2ρ2+z 7. Use the inﬁnitesimal volume element ∆ Vof Fig. 3.2 and the deﬁnition of the divergence ∇·F=1 ∆V/integraldisplay/integraldisplay /circlecopyrt SF·nda to derive the expression of the divergence in the cylindrical coordinate system. Hint: Find the surface elements of the six sides of ∆ V, then add pairwise the surface integrals of opposite sides. For example, /integraldisplay/integraldisplay leftF·nda+/integraldisplay/integraldisplay rightF·nda=−Fρ(ρ,ϕ,z )ρdϕdz +Fρ(ρ+dρ,ϕ,z )(ρ+dρ)dϕdz=∂ ∂ρ(ρFρ)dρdϕdz. With the othertwo pairs, the result is seen identical to (3.23). 152 3 Curved Coordinates 8. A particle is moving in space. Show that the spherical coordinate compo- nents of its velocity and acceleration are given by vr=/squaresmallsolidr, v θ=r/squaresmallsolid θ, v ϕ=rsinθ/squaresmallsolidϕ, ar=..r−r. θ2 −rsin2θ.ϕ2, aθ=r.. θ+2.r. θ−rcosθsinθ.ϕ2, aϕ=rsinθ..ϕ+2.rsinθ.ϕ+2rcosθ. θ.ϕ. 9. Starting with the expression of ∇Φin spherical system, express er,eθ,eϕ in terms of i,j,k,then equate it with ∇Φin rectangular coordinates. In this way, verify that ∂Φ ∂x=s i nθcosϕ∂Φ ∂r+c o s θcosϕ1 r∂Φ ∂θ−sinϕ rsinθ∂Φ ∂ϕ, ∂Φ ∂y=s i nθsinϕ∂Φ ∂r+c o s θsinϕ1 r∂Φ ∂θ+cosϕ rsinθ∂Φ ∂ϕ, ∂Φ ∂z=c o s θ∂Φ ∂r−sinθ1 r∂Φ ∂θ. 10. Use the inﬁnitesimal volume element ∆ Vof Fig. 3.4 and the deﬁnition of the divergence ∇·F=1 ∆V/integraldisplay/integraldisplay /circlecopyrt SF·nda to derive the expression of the divergence in the spherical coordinate system. 11. Find the expression of the Laplacian ∇2in spherical coordinates by directly transforming ∇2=∂2/∂x2+∂2/∂y2+∂2/∂z2into spherical coor- dinates using the results of the last problem. 12. Show that the following three forms of ∇2Φ(r) are equivalent: (a)1 r2d dr/bracketleftbigg r2d drΦ(r)/bracketrightbigg ,(b)d2 dr2Φ(r)+2 rd drΦ(r),(c)1 rd2 dr2[rΦ(r)]. 13. (a) Show that the vector ﬁeld F=/parenleftbigg A−B r3/parenrightbigg cosθer−/parenleftbigg A+B 2r3/parenrightbigg sinθeθ is irrotational ( ∇×F=0). (b) Find a scalar potential Φsuch that ∇Φ=F. (c) Show that Φsatisﬁes the Laplace equation ∇2Φ=0. Ans. Φ=/parenleftbigg Ar+B 2r2/parenrightbigg cosθ. 3.5 Multiple Integrals 153 14. Use spherical coordinates to evaluate the following integrals over a sphere of radius Rcentered at the origin, (a)/integraldisplay/integraldisplay/integraldisplay dV,(b)/integraldisplay/integraldisplay/integraldisplay x2dV,(c)/integraldisplay/integraldisplay/integraldisplay y2dV,(d)/integraldisplay/integraldisplay/integraldisplay r2dV. Ans.4π 3R3,4π 15R3,4π 15R3,4π 5R3. 15. Let L=−i/parenleftbigg eϕ∂ ∂θ−eθ1 sinθ∂ ∂ϕ/parenrightbigg , show that (a)ez·L=−i∂ ∂ϕ, (b)L·L=−/bracketleftBig 1 sinθ∂ ∂θ/parenleftbig sinθ∂ ∂θ/parenrightbig +1 sin2θ∂2 ∂ϕ2/bracketrightBig . (These are quantum mechanical Lz,L2angular momentum operators with /planckover2pi1=1.) 16. Find the area of the Earth’s surface which lies further north than the 45◦N latitude. Assume the Earth is a sphere of radius R. Ans.πR2(2−√ 2),which is only about 15% of the total surface area of the earth (4 πR2). 17. Use spherical coordinates to verify the divergence theorem /integraldisplay/integraldisplay/integraldisplay V∇·FdV=/integraldisplay/integraldisplay /circlecopyrt SF·nda with F=r2cosθer+r2cosϕeθ−r2cosθsinϕeϕ over a sphere of radius R. Ans. Both sides equal to 0 . 18. Use elliptical coordinates to evaluate the following integral over all space I=/integraldisplay/integraldisplay/integraldisplay1 r2exp(−2r1)dV, where r1andr2are the distances from two ﬁxed points which are sepa- rated by a distance R.(This integral happens to be the so-called Coulomb integral for the H 2molecule.) Ans.π R/bracketleftbigg1 R−exp (−2R)/parenleftbigg 1+1 R/parenrightbigg/bracketrightbigg . Hint: r2=1 2R(λ+µ),r1=1 2R(λ−µ). 154 3 Curved Coordinates 19. Parabolic coordinates ( u,v,w ) are related to Cartesian coordinates ( x,y,z) by the relation x=2uv, y =u2−v2,z=w.(a) Find the scale factors hu,hv,hw.(b) Show that the ( u,v,w ) coordinate system is orthogonal. Ans. 2/parenleftbig u2+v2/parenrightbig1/2,2/parenleftbig u2+v2/parenrightbig1/2,1. 20. Show that in terms of prolate spheroidal coordinates, the Laplace equation/parenleftbig ∇2Φ=0/parenrightbig is given by 1/parenleftbig sinh2η+s i n2θ/parenrightbig/bracketleftbigg∂2 ∂η2Φ+ coth η∂ ∂ηΦ+∂2 ∂θ2Φ+c o t θ∂ ∂θΦ/bracketrightbigg +1 sinh2ηsin2θ∂2 ∂ϕ2Φ=0. 21. An orthogonal coordinate system ( u1,u2,u3) is related to Cartesian coor- dinates ( x,y,z)b y x=x(u1,u2,u3),y=y(u1,u2,u3),z=z(u1,u2,u3). Show that (a)∂r ∂u1·∂r ∂u2×∂r ∂u3=h1h2h3,(b)∇u1·∇u2×∇u3=1 h11 h21 h3, (c)∂r ∂u1·∂r ∂u2×∂r ∂u3=∂(x, y, z) ∂(u1,u2,u3),(d)∇u1·∇u2×∇u3=∂(u1,u2,u3) ∂(x, y, z). 22. Use the transformation x+y=u, x−y=vto evaluate the double integral I=/integraldisplay/integraldisplay (x2+y2)dxdy within a square whose vertices are (0 ,0),(1,1),(2,0),(1,−1). Ans. 8/3. Hint: Recall the Jacobian is the absolute value of the determinant /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂x ∂u∂y ∂u ∂x ∂v∂y ∂v/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle.Draw the square and show that the four sides of the square arev=0,u=2,v=2,u=0. 4 Vector Transformation and Cartesian Tensors The universal validity of physical laws is best expressed in terms of mathematical quantities that are independent of any reference frame. Yet physical problems governed by these laws can be solved, in most cases, only if the relevant quantities are resolved into their components in some coordinate system. For example, if we consider a block sliding on an inclined plane, the motion of the block is of course governed by Newton’s second law of dynamics F=ma,in which no coordinates appear. However, to get the actual values of velocity, acceleration, etc. of the block, we have to set a coordinate system. We will obtain the correct answer no matter how we orient the axes, although some orientations are more convenient than others. It is possible to take the x-axis horizontal or along the incline. The components of Fandain the x andydirections are of course diﬀerent in these two cases, but they will com- bine to give the same correct results. In other words, if the coordinate system is rotated, the components of a vector will of course change. But they must change in a speciﬁc way in order for the vector equation to remain valid. For this reason, the vector ﬁeld is best deﬁned in terms of the behavior of its components under axes rotation. When the coordinate system is rotated, the transformation of the compo- nents of the position vector rcan be expressed in terms of a rotation matrix. We will use this rotation matrix to deﬁne all other vectors. The properties of this rotation matrix will be used to analyze a variety of ways of combining the components of two or more vectors. This approach to vector analysis can be easily generalized to vectors of dimensions higher than three. It also naturally leads to tensor analysis. Many physical quantities are neither vectors nor scalars. For example, the electric current density Jﬂowing in a material is linearly related to the electric ﬁeld Ethat drives it. If the material is isotropic, the three components ofJandEare related by the same constant σin the Ohm’s law Ji=σEi, 156 4 Vector Transformation and Cartesian Tensors where σis known as the conductivity. However, if the material is anisotropic (nonisotropic), the direction of the current is diﬀerent from the direction of the ﬁeld. In that case, the Ohm’s law is described by Ji=3/summationdisplay j=1σijEj, where σijis the conductivity tensor. It is a tensor of rank two because it has two subscripts iandj, each of them runs from 1 to 3. All together it has nine components. The deﬁning property of a tensor is that, when the coordinate axes are rotated, its components must change according to certain transformation rules, analogous to the vector transformation. In fact, a vector, with one subscript attached to its components, is a tensor of rank one. A tensor ofnth rank has nsubscripts. In this chapter, the mathematics of tensors will be restricted to Cartesian coordinate systems, therefore the name Cartesian tensors. 4.1 Transformation Properties of Vectors 4.1.1 Transformation of Position Vector The coordinate frame we use to describe positions in space is of course entirely arbitrary, but there is a speciﬁc transformation rule for converting vector components from one frame to another. For simplicity, we will ﬁrst consider a simple case. Suppose the rectangular coordinate system is rotated counterclockwise about the z-axis through an angle θ.The point P is at the position ( x,y,z) before the rotation. After the rotation, the same posi- tion becomes ( x/prime,y/prime,z/prime),as shown in Fig. 4.1. Therefore the position vector r expressed in the original system is r=xi+yj+zk, (4.1) and expressed in the rotated system is r=x/primei/prime+y/primej/prime+z/primek/prime, (4.2) where ( i,j,k) and ( i/prime,j/prime,k/prime) are the unit vectors along the three axes of the original and the rotated coordinates, respectively. The relationship between primed and unprimed systems can be easily found, since x/prime=i/prime·r=i/prime·(xi+yj+zk)=(i/prime·i)x+(i/prime·j)y+(i/prime·k)z =xcosθ+ycos/parenleftBigπ 2−θ/parenrightBig =xcosθ+ysinθ (4.3) 4.1 Transformation Properties of Vectors 157 Y9 X9 XY rP Q OAyy9 i9 ijj9x x9 B qq q Fig. 4.1. The coordinate system is rotated around z-axis. The primed quantities are those in the rotated system and the unprimed quantities are those in the original system and y/prime=j/prime·r=j/prime·(xi+yj+zk)=(j/prime·i)x+(j/prime·j)y+(i/prime·k)z =xcos/parenleftBigπ 2+θ/parenrightBig +ycosθ=−xsinθ+ycosθ. (4.4) Sincek=k/prime, z/prime=z. Of course, these relations are also geometrical statements of the rotation. It is seen in Fig. 4.1 x/prime=O A+A B=OQ cosθ+PAsinθ=OQ cosθ+( P Q −AQ) sin θ =x cosθ+(y−xtanθ)sinθ=x/parenleftbigg1 cosθ−sin2θ cosθ/parenrightbigg +ysinθ =xcosθ+ysinθ, y/prime=P Ac o s θ=( P Q −AQ) cos θ=(y−xtanθ)cosθ =ycosθ−xsinθ, which are identical to (4 .3) and (4 .4). With matrix, these relations can be expressed as ⎛ ⎝x/prime y/prime z/prime⎞ ⎠=⎛ ⎝cosθsinθ0 −sinθcosθ0 00 1⎞ ⎠⎛ ⎝x y z⎞ ⎠. (4.5) The 3×3 matrix is known as rotation matrix. 158 4 Vector Transformation and Cartesian Tensors 4.1.2 Vector Equations Vector equations are used to express physical laws which should be indepen- dent of the reference frame. For example, Newton’s second law of dynamics F=ma, (4.6) relates the force Fon the particle of mass mand the acceleration aof the par- ticle. No coordinates appear explicitly in the equation, as it should be since the law is universal. However, often we ﬁnd it easier to set up a coordinate system and work with the individual components. In any particular coordi- nate system, each vector is represented by three components. When we change the refrence frame, these components will change. But they must change in a speciﬁc way in order for (4 .6) to remain true. The coordinates will be changed by either a translation and/or a rotation of the axes. A translation changes the origin of the coordinate system, resulting in some additive constants in the components of r.Since the derivative of a constant is zero, the translation will not aﬀect vectors Fanda.Therefore the important changes are due to the rotation of the axes. First we note that if (4 .6) holds in one coordinate system it holds in all, for the equation may be written as F−ma=0, (4.7) and under axes rotation, the zero vector will obviously remain zero in the new system. In terms of its components in Cartesian coordinate system, (4 .7) can be written as (Fx−max)i+(Fy−may)j+(Fz−maz)k=0, (4.8) which leads to Fx=max,F y=may,F z=maz. (4.9) Now if the system is rotated counterclockwise about z-axis through an angle θas indicated in Fig. 4.1, (4 .7) becomes (F/prime x/prime−ma/prime x/prime)i/prime+(F/prime y/prime−ma/prime y/prime)j/prime+(F/prime z/prime−ma/prime z/prime)k/prime=0, (4.10) where by deﬁnition a/prime x/prime=d2 dt2x/prime=d2 dt2(xcosθ+ysinθ) =axcosθ+aysinθ, (4.11) a/prime y/prime=d2 dt2y/prime=d2 dt2(−xsinθ+ycosθ) =−axsinθ+aycosθ, (4.12) 4.1 Transformation Properties of Vectors 159 a/prime z/prime=d2 dt2z/prime=d2 dt2z=az. (4.13) Each component of (4 .10) must be identically equal to zero. This gives F/prime x/prime=ma/prime x/prime=m(axcosθ+aysinθ), F/prime y/prime=ma/prime y/prime=m(−axsinθ+aycosθ), F/prime z/prime=ma/prime z/prime=maz. Using (4 .9), we have F/prime x/prime=Fxcosθ+Fysinθ, F/prime y/prime=−Fxsinθ+Fycosθ, F/prime z/prime=Fz. Written in the matrix form, these relations are expressed as ⎛ ⎝F/prime x/prime F/prime y/prime F/prime z/prime⎞ ⎠=⎛ ⎝cosθsinθ0 −sinθcosθ0 00 1⎞ ⎠⎛ ⎝Fx Fy Fz⎞ ⎠. (4.14) Comparing (4 .5) and (4 .14),we see that the rotation matrix is exactly the same. In other words, the components of the vector Ftransform in the same way as those of the position vector r. In physical applications, it means that in order for a quantity to be considered as a vector, the measured values of its components in a rotated system must be related in this way to those in the original system. The orientation between two coordinate systems is of course not limited to a single rotation around a particular axis. If we know the relative orientation of the systems, we can follow the procedure of (4 .3) to establish the relation ⎛ ⎝x/prime y/prime z/prime⎞ ⎠=⎛ ⎜⎝(i/prime·i) (j/prime·i) (k/prime·i)(i/prime·j) (j/prime·j) (k/prime·j)(i/prime·k) (j/prime·k) (k/prime·k)⎞ ⎟⎠⎛ ⎝x y z⎞ ⎠. In Sect. 4.1.3 we consider the actual rotation that will bring ( i,j,k)i n t o/parenleftbig i/prime,j/prime,k/prime/parenrightbig . 4.1.3 Euler Angles Often we need to express the transformation matrix in terms of concrete rota- tions which bring the coordinate axes into a speciﬁed orientation. In general, the rotation can be regarded as a combination of three rotations, performed successively, about three diﬀerent directions in space. The most useful descrip- tion of this kind is in terms of Euler’s angles α,β,γ, which we now deﬁne. 160 4 Vector Transformation and Cartesian Tensors Z/H11032 XZ β line of nodesαγYY/H11032 X/H11032(a) X line of nodesαY X1(b) α Y1 Z1 β line of nodesY11(c) βY1Z line of nodesγY11 X/H11032(d) Y/H11032 β X1X11X11γ γZ1Z Z/H11032 Z11Z/H11032α Fig. 4.2. Euler angles. ( a) Relative orientation of two rectangular coordinate systems XYZangX/primeY/primeZ/primew i t hac o m m o no r i g i ni ss p e c i ﬁ e db yt h r e eE u l e ra n g l e s α, β, γ. The line of nodes is the intersection of XY and X/primeY/primeplanes. The transforma- tion matrix is the product of the matrices representing the following three rotations. (b) First rotate αalong the Zaxis, bring Xaxis to coincide with the line of nodes. (c) Rotate βalong the line of nodes. ( d) Finally rotate γalong Z/primeaxis The two-coordinate systems are shown in Fig. 4.2a. Let XYZbe the axes of the initial system, X/primeY/primeZ/primebe the axes of the ﬁnal system. The intersection of XY plane and X/primeY/primeplane is known as the line of nodes. The relative orientation of the two systems is speciﬁed by the three angles α,β,γ. As shown in Fig. 4.2a, αis the angle between Xaxis and the line of nodes, βis the angle between ZandZ/primeaxes, and γis the angle between the line of nodes and X/primeaxis. The transformation matrix from XYZ to X/primeY/primeZ/primecan be obtained by writing it as the product of the separate rotations, each of which has a relative simple rotation matrix. First rotate the initial axes XYZ, by an angle αcounterclock- wise about the Zaxis, bring the Xaxis to coincide with the line of nodes. The resultant coordinate system is labeled the X1,Y1,Z1axes, as shown in Fig. 4.2b. In the second stage the intermediate axes are rotated about X1axis counterclockwise by an angle βto produce another intermediate X11,Y11,Z11 axes, as shown in Fig. 4.2c. Finally the X11,Y11,Z11axes are rotated counter- clockwise by an angle γabout the Z11axis to produce the desired X/primeY/primeZ/prime axes, as shown in Fig. 4.2d. 4.1 Transformation Properties of Vectors 161 After the ﬁrst rotation, the coordinates ( x,y,z)o frin the initial system becomes ( x1,y1,z1)i nt h e X1Y1Z1system. They are related by a rotation matrix,⎛ ⎝x1 y1 z1⎞ ⎠=⎛ ⎝cosαsinα0 −sinαcosα0 00 1⎞ ⎠⎛ ⎝x y z⎞ ⎠. (4.15) The second rotation is about the X1axis. After the rotation, ( x1,y1,z1)b e - comes ( x11,y11,z11) with the relation ⎛ ⎝x11 y11 z11⎞ ⎠=⎛ ⎝10 0 0c o s βsinβ 0−sinβcosβ⎞ ⎠⎛ ⎝x1 y1 z1⎞ ⎠. (4.16) After the ﬁnal rotation about Z11axis, the coordinates of rbecomes ( x/prime,y/prime,z/prime) which is given by ⎛ ⎝x/prime y/prime z/prime⎞ ⎠=⎛ ⎝cosγsinγ0 −sinγcosγ0 00 1⎞ ⎠⎛ ⎝x11 y11 z11⎞ ⎠. (4.17) It is clear from (4 .15) to (4 .17) that ⎛ ⎝x/prime y/prime z/prime⎞ ⎠=(A)⎛ ⎝x y z⎞ ⎠, (4.18) where (A)=⎛ ⎝cosγsinγ0 −sinγcosγ0 00 1⎞ ⎠⎛ ⎝10 0 0c o s βsinβ 0−sinβcosβ⎞ ⎠⎛ ⎝cosαsinα0 −sinαcosα0 00 1⎞ ⎠.(4.19) Hence the 3 ×3 matrix ( A) is the rotation matrix of the complete transfor- mation. Multiplying the three matrices out, one can readily ﬁnd the elements ofA, (A)=⎛ ⎜⎝cosγcosα−sinγcosβsinα −sinγcosα−cosγcosβsinα sinβsinα cosγsinα+s i nγcosβcosα −sinγsinα+c o s γcosβcosα −sinβcosαsinγsinβ cosγsinβ cosβ⎞ ⎟⎠ (4.20) 162 4 Vector Transformation and Cartesian Tensors It is not diﬃcult to verify that the product of matrix ( A) and its transpose/parenleftbig AT/parenrightbig is the identity matrix ( I),(The transpose of ( A) is the matrix ( A) with row and column interchanged.) (A)/parenleftbig AT/parenrightbig =(I). Therefore the inverse ( A)−1is given by the transpose ( A)T, ⎛ ⎝x y z⎞ ⎠=/parenleftbig AT/parenrightbig⎛ ⎝x/prime y/prime z/prime⎞ ⎠. These are general properties of a rotation matrix which we shall prove in Sect. 4.1.4 It should be noted that diﬀerent authors deﬁne Euler angles in slightly diﬀerent ways, because the sequence of rotations used to deﬁne the ﬁnal orien- tation of the coordinate systems is to some extent arbitrary. We have adopted the deﬁnition used in most textbooks on classical mechanics. 4.1.4 Properties of Rotation Matrices To study the general properties of vector space, it is convenient to use a more systematic notation. Let ( x,y,z)b e( x1,x2,x3); (i,j,k)b e(e1,e2,e3),and (Vx,Vy,Vz)b e(V1,V2,V3).The quantities in the rotated system are similarly labeled as prime quantities. One of the advantages of the new notation is that it permits us to use the summation symbol Σto write the equations in a more compact form. The orthogonality of ( i,j,k) is expressed as (ei·ej)=(e/prime i·e/prime j)=δij, where the symbol δij, known as the Kronecker delta, is deﬁned as δij=/braceleftbigg1i=j 0i/negationslash=j. In general, the same position vector r, expressed in two diﬀerent coordinate systems can be written as r=3/summationdisplay j=1x/prime je/prime j=3/summationdisplay j=1xjej. (4.21) Taking the dot product e/prime i·r,w eh a v e e/prime i·3/summationdisplay j=1x/prime je/prime j=3/summationdisplay j=1(e/prime i·e/prime j)x/prime j=3/summationdisplay j=1δijx/prime j=x/prime i. (4.22) 4.1 Transformation Properties of Vectors 163 The same dot product from (4 .21) gives e/prime i·3/summationdisplay j=1xjej=3/summationdisplay j=1(e/prime i·ej)xj.=3/summationdisplay j=1aijxj. (4.23) It follows from (4 .22) and (4 .23) that x/prime i=3/summationdisplay j=1(e/prime i·ej)xj=3/summationdisplay j=1aijxj, (4.24) where aij=(e/prime i·ej) (4.25) is the direction cosine between e/prime iandej.Note that iin (4.24) remains as a parameter which gives rise to three separate equations when it is set to 1 ,2, and 3.In matrix notation, (4 .24) is written as ⎛ ⎝x/prime 1 x/prime 2 x/prime 3⎞ ⎠=⎛ ⎝a11a12a13 a21a22a23 a31a32a33⎞ ⎠⎛ ⎝x1 x2 x3⎞ ⎠. (4.26) If, instead of e/prime i·r,we take ei·rand follow the same procedure, we will obtain xi=3/summationdisplay j=1(ei·e/prime j)x/prime j. Since ( ei·e/prime j) is the cosine of the angle between eiande/prime jwhich can be expressed just as well as ( e/prime j·ei),and by deﬁnition of (4 .25) (e/prime j·ei)=aji, therefore xi=3/summationdisplay j=1ajix/prime j, (4.27) or ⎛ ⎝x1 x2 x3⎞ ⎠=⎛ ⎝a11a21a31 a12a22a32 a13a23a33⎞ ⎠⎛ ⎝x/prime 1 x/prime 2 x/prime 3⎞ ⎠. (4.28) Comparing (4 .26) and (4 .28) we see that the inverse of the rotation matrix is equal to its transpose (aij)−1=(aji)=(aij)T. (4.29) Any transformation that satisﬁes this condition is known as an orthogonal transformation. Renaming the indices iandj,we can write (4 .27) as xj=3/summationdisplay i=1aijx/prime i. (4.30) 164 4 Vector Transformation and Cartesian Tensors It is thus clear from (4 .24) and the last equation that aij=∂x/prime i ∂xj=∂xj ∂x/prime i. (4.31) We emphasize this relation is true only in the Cartesian coordinate system. The nine elements of the rotation matrix are not independent of each other. One way to derive the relationships between them is to note that if the two coordinate systems have the same origin then the length of the position vector should be the same in both systems. This requires r·r=3/summationdisplay i=1x/prime2 i=3/summationdisplay j=1x2 j. (4.32) Using (4 .24),we have 3/summationdisplay i=1x/prime2 i=3/summationdisplay i=1x/prime ix/prime i=3/summationdisplay i=1⎛ ⎝3/summationdisplay j=1aijxj⎞ ⎠/parenleftBigg3/summationdisplay k=1aikxk/parenrightBigg =3/summationdisplay j=13/summationdisplay k=1xjxk3/summationdisplay i=1aijaik=3/summationdisplay j=1x2 j. This can be true for all points if and only if 3/summationdisplay i=1aijaik=δjk. (4.33) This relation is known as the orthogonality condition. Any matrix whose elements satisfy this condition is called an orthogonal matrix. The rotation matrix is an orthogonal matrix. With all possible values of iandj,(4.33) consists of a set of six equations. This set of equations is equivalent to 3/summationdisplay i=1ajiaki=δjk (4.34) which can be obtained in the same way from (4 .32),but starting from right to left with the transformation of (4 .27). Example 4.1.1. Show that the determinant of an orthogonal transformation is equal to either +1 or −1. Solution 4.1.1. Let the matrix of the transformation be ( A).Since ( A) (A−1)=(I),the determinant of the identity matrix is of course equal to one,/vextendsingle/vextendsingleAA−1/vextendsingle/vextendsingle=1.For an orthogonal transformation A−1=AT,so/vextendsingle/vextendsingleAAT/vextendsingle/vextendsingle=1. Since/vextendsingle/vextendsingleAAT/vextendsingle/vextendsingle=|A|/vextendsingle/vextendsingleAT/vextendsingle/vextendsingleand|A|=/vextendsingle/vextendsingleAT/vextendsingle/vextendsingle,it follows that |A|2=1.Therefore |A|=±1. 4.1 Transformation Properties of Vectors 165 Example 4.1.2. Show that the determinant of a rotation matrix is equal to +1 . Solution 4.1.2. Express e/prime iin terms of {ek}:e/prime i=/summationtext3 i=1bikek. (e/prime i·ej)=3/summationdisplay i=1bik(ek·ej)=3/summationdisplay i=1bikδkj=bij. But (e/prime i·ej)=aij,therefore bij=aij.So e/prime 1=a11e1+a12e2+a13e3, e/prime 2=a21e1+a22e2+a23e3, e/prime 3=a31e1+a32e2+a33e3. As we have shown in 1.2.7, the scalar triple product is equal to the determinant of the components e/prime 1·(e/prime 2×e/prime 3)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, which is the rotation matrix. On the other hand, e/prime 1·(e/prime 2×e/prime 3)=e/prime 1·e/prime 1=+ 1. Therefore /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=+ 1. 4.1.5 Deﬁnition of a Scalar and a Vector in Terms of Transformation Properties Now we come to the reﬁned algebraic deﬁnition of a scalar and a vector. Under a rotation of axes, the coordinates of the position vector in the original system xitransform to x/prime iin the rotated system according to x/prime i=/summationdisplay jaijxj (4.35) with /summationdisplay iaijaik=δjk. (4.36) 166 4 Vector Transformation and Cartesian Tensors If under such a transformation, a quantity ϕis unaﬀected, then ϕis called a scalar. This means if ϕis a scalar, then ϕ(x1,x2,x3)=ϕ/prime(x/prime 1,x/prime 2,x/prime 3). (4.37) Note that after the coordinates are transformed, the functional form may be changed (therefore ϕ/prime),but as long as ( x1,x2,x3) and ( x/prime 1,x/prime 2,x/prime 3) represent the same point, their value is the same. If a set of quantities ( A1,A2,A3) in the original system is transformed into (A/prime 1,A/prime 2,A/prime 3) in the rotated system according to A/prime i=/summationdisplay jaijAj, (4.38) then the quantity A=(A1,A2,A3) is called a vector. Since ( aij)−1=(aji), (4.38) is equivalent to Ai=/summationdisplay jajiA/prime j. (4.39) This deﬁnition is capable of generalization and ensures that vector equa- tions are independent of coordinate system. Example 4.1.3. Suppose AandBare vectors. Show that the dot product A·Bis a scalar. Solution 4.1.3. SinceAandBare vectors, under a rotation their compo- nents transform according to A/prime i=/summationdisplay jaijAj;B/prime i=/summationdisplay jaijBj. To show the dot product A·B=/summationdisplay iAiBi is a scalar, we must show that its value in the rotated system is the same as its value in the original system. (A·B)/prime=/summationdisplay iA/prime iB/prime i=/summationdisplay i⎛ ⎝/summationdisplay jaijAj⎞ ⎠/parenleftBigg/summationdisplay kaikAk/parenrightBigg =/summationdisplay j/summationdisplay k/parenleftBigg/summationdisplay iaijaik/parenrightBigg AjAk=/summationdisplay j/summationdisplay kδjkAjBk=/summationdisplay jAjBj=A·B. SoA·Bis a scalar. 4.1 Transformation Properties of Vectors 167 Example 4.1.4. Show that if ( A1,A2,A3) is such that/summationtext iAiBiis a scalar for every vector B, then ( A1,A2,A3) is a vector. Solution 4.1.4. Since/summationtext iAiBiis a scalar and Bis a vector, /summationdisplay iAiBi=/summationdisplay iA/prime iB/prime i=/summationdisplay iA/prime i/summationdisplay jaijBj. Now both iandjare running indices, we can rename iasj,andjasi.So /summationdisplay iAiBi=/summationdisplay jA/prime j/summationdisplay iajiBi=/summationdisplay i/summationdisplay jajiA/prime jBi. It follows /summationdisplay i⎛ ⎝Ai−/summationdisplay jajiA/prime j⎞ ⎠Bi=0. Since this identity holds for every B, we must have Ai=/summationdisplay jajiA/prime j. Therefore A1,A2,A3are components of a vector. Example 4.1.5. Show that, in Cartesian coordinates, the gradient of a scalar function ∇ϕis a vector function. Solution 4.1.5. As a scalar it must have the same value at a given point in space, independent of the orientation of the coordinate system ϕ/prime(x/prime 1,x/prime 2,x/prime 3)=ϕ(x1,x2,x3). (4.40) Diﬀerentiating with respect to x/prime iand using the chain rule, we have ∂ ∂x/prime iϕ/prime(x/prime 1,x/prime 2,x/prime 3)=∂ ∂x/prime iϕ(x1,x2,x3)=/summationdisplay j∂ϕ ∂xj∂xj ∂x/prime i. (4.41) It follows from (4 .31) that in Cartesian coordinates ∂xj ∂x/prime i=aij, therefore ∂ϕ/prime ∂x/prime i=/summationdisplay jaij∂ϕ ∂xj. (4.42) 168 4 Vector Transformation and Cartesian Tensors Now the components of ∇ϕare /parenleftbigg∂ϕ ∂x1,∂ϕ ∂x2,∂ϕ ∂x3/parenrightbigg . They transform under a rotation of coordinates in exactly the same way as the components of a vector, therefore ∇ϕis a vector function. A vector whose components are just numbers is a constant vector. All constant vectors behave like a position vector. When the axes are rotated, the components change into a new set of numbers in accordance with the transformation rule. Therefore, any set of three numbers can be considered as a constant vector. For vector ﬁelds, the components are functions of ( x1,x2,x3) themselves. Under a rotation, not only ( x1,x2,x3) will change to ( x/prime 1,x/prime 2,x/prime 3), the appea- rances of the component functions may also change. This leads to some complications. Mathematically the transformation rules place little restriction on what we can call a vector. We can make any set of three functions the components of a vector ﬁeld by simply deﬁning, in a rotated system, the corresponding functions obtained from the correct transformation rules as the components of the vector in that system. However, if we are discussing a physical entity, we are not free to deﬁne its components in various systems. They are determined by physical facts. As we stated earlier, all properly formulated physical laws must be independent of the coordinate system. In other words, the appearance of the equation describing physical laws must be the same in all coordinate systems. If vector functions maintain the same appearances in rotated systems, equations writ- ten in terms of them will be automatically invariant under rotation. Therefore we include in the deﬁnition of a vector ﬁeld, an additional condition that the transformed components must look the same as the original components. For example, many authors describe /parenleftbigg V1 V2/parenrightbigg =/parenleftbigg x2 x1/parenrightbigg (4.43) as a vector ﬁeld in a two-dimensional space (see, for example, E.M. Purcell, “Electricity and Magnetism”, McGraw-Hill Book Co. (1965), page 36; D.A. McQuarrie, “Mathematical Methods for Scientists and Engineers,” University Science Books, (2003), page 301), still many others would say that (4.43) cannot be called a vector ﬁeld (see, for example, G. Arfken, “Mathemat- ical Methods for Physicists”, Academic Press, (1968), page 8; P.C. Mathews, “Vector Calculus”, Springer, Berlin Heidelberg New York (2002), page 118). 4.2 Cartesian Tensors 169 If we consider (4 .43) as a vector, then the components of this vector in the system where the axes are rotated by an angle θ,are given by /parenleftbiggV/prime 1 V/prime 2/parenrightbigg =/parenleftbiggcosθsinθ −sinθcosθ/parenrightbigg/parenleftbiggV1 V2/parenrightbigg =/parenleftbiggcosθsinθ −sinθcosθ/parenrightbigg/parenleftbiggx2 x1/parenrightbigg . Furthermore, the coordinates have to change according to /parenleftbigg x1 x2/parenrightbigg =/parenleftbiggcosθ−sinθ sinθcosθ/parenrightbigg/parenleftbiggx/prime 1 x/prime 2/parenrightbigg . One can readily show that /parenleftbiggV/prime 1 V/prime 2/parenrightbigg =/parenleftbigg2x/prime 1sinθcosθ+x/prime 2(cos2θ−sin2θ) x/prime 1(cos2θ−sin2θ)−2x/prime 2sinθcosθ/parenrightbigg . (4.44) Mathematically one can certainly deﬁne (4 .44) as the components of the vector in the rotated system, but they do not look like (4 .43). On the other hand, consider a slightly diﬀerent expression /parenleftbiggV1 V2/parenrightbigg =/parenleftbiggx2 −x1/parenrightbigg . (4.45) With the same transformation rules, we obtain /parenleftbigg V/prime 1 V/prime 2/parenrightbigg =/parenleftbiggx/prime 2(cos2θ+s i n2θ) −x/prime 1(cos2θ+s i n2θ)/parenrightbigg =/parenleftbigg x/prime 2 −x/prime 1/parenrightbigg , (4.46) which has the same form as (4 .45).In this sense, we say that (4 .45) is invariant under rotations Under our deﬁnition, (4 .45) is a vector and (4 .43) is not. 4.2 Cartesian Tensors 4.2.1 Deﬁnition The deﬁnition of a vector can be extended to deﬁne a more general class of objects called tensors, which may have more than one subscript. If in the rectangular coordinate system of three-dimensional space, under a rotation of coordinates x/prime i=3/summationdisplay j=1aijxj, 170 4 Vector Transformation and Cartesian Tensors the 3Nquantities Ti1,i2,···,iN(where each of i1,i2,···,iNruns independently f r o m1t o3 )t r a n s f o r ma c c o r d i n gt ot h er u l e T/prime i1,i2,···,iN=3/summationdisplay j1=13/summationdisplay j2=1···3/summationdisplay jN=1ai1j1ai2j2···aiNjNTj1,j2,···,jN, (4.47) thenTi1,i2,···,iNare the components of a Nth rank Cartesian tensor. Since we are going to restrict our discussion to Cartesian tensors, unless explicitly otherwise speciﬁed, we shall drop the word Cartesian from here on. The rank of a tensor is the number of free subscripts. Tensor of zeroth rank has only one (30= 1) component. So it can be thought as a scalar. A tensor of ﬁrst rank has three components (31= 3). The rule of transformation of these components under a rotation is the same as the rule for a vector. So a vector is a tensor of rank one. The most useful other case is the tensor of second rank. It has nine com- ponents (32= 9), Tijwhich obey the transformation rule T/prime ij=3/summationdisplay l=13/summationdisplay m=1ailajmTlm. (4.48) The components of a second rank tensor may be conveniently expressed as a 3×3 matrix: Tij=⎛ ⎝T11T12T13 T21T22T23 T31T32T33⎞ ⎠. However, this does not mean that any 3 ×3 matrix forms a tensor. The essential condition is that its components satisfy the transformation rule. As a matter of terminology, a second-rank tensor in three-dimensional space is a collection of nine components Tij.However, very often Tijis referred to as “tensor” instead of “tensor components” for simplicity. In other words, Tijis used to mean the totality of the components as well as the individual component. The context will make its meaning clear. Another often used symbol for tensors is a double bar over a letter, such as T. Example 4.2.1. Show that in a two-dimensional space, the following quantity is a second-rank tensor Tij=/parenleftbigg x1x2−x2 1 x2 2−x1x2/parenrightbigg . Solution 4.2.1. In a two-dimensional space, a second rank tensor has four (22= 4) components. If it is a tensor, in a rotated system it must look like T/prime ij=/parenleftbigg x/prime 1x/prime 2−x/prime2 1 x/prime2 2−x/prime 1x/prime 2/parenrightbigg , 4.2 Cartesian Tensors 171 where/parenleftbiggx/prime 1 x/prime 2/parenrightbigg =/parenleftbigga11a12 a21a22/parenrightbigg/parenleftbiggx1 x2/parenrightbigg =/parenleftbiggcosθsinθ −sinθcosθ/parenrightbigg/parenleftbiggx1 x2/parenrightbigg . Now we must check if each component satisﬁes the transformation rule. T/prime 11=x/prime 1x/prime 2= (cos θx1+s i nθx2)(−sinθx1+c o s θx2) =−cosθsinθx2 1+c o s2θx1x2−sin2θx2x1+s i nθcosθx2 2. This is to be compared with T/prime 11=2/summationdisplay l=12/summationdisplay m=1a1la1mTlm =a11a11T11+a11a12T12+a12a11T21+a12a12T22 =c o s2θx1x2−cosθsinθx2 1+s i nθcosθx2 2−sin2θx1x2. It is seen that these two expressions are identical. The same process will show that other components will also satisfy the transformation rule. Therefore Tij is a second rank tensor in the two-dimensional space. This transformation property is not to be taken for granted. In the above example, if one algebraic sign is changed, the transformation rule will not be satisﬁed. For example if T22is changed to x1x2, Tij=/parenleftbigg x1x2−x2 1 x2 2x1x2/parenrightbigg , (4.49) then T/prime 11/negationslash=2/summationdisplay l=12/summationdisplay m=1a1la1mTlm. Therefore (4 .49) is not a tensor. 4.2.2 Kronecker and Levi-Civita Tensors Kronecker delta tensor. The Kronecker delta which we have already encoun- tered, δij=/braceleftbigg 1i=j 0i/negationslash=j, is a second rank tensor. To prove this, consider the transformation δ/prime ij=3/summationdisplay l=13/summationdisplay m=1ailajmδlm=3/summationdisplay l=1ailajl=δij. (4.50) 172 4 Vector Transformation and Cartesian Tensors Therefore δ/prime ij=/braceleftbigg 1i=j 0i/negationslash=j. Thusδijobeys the tensor transformation rule and is invariant under rotation. In addition, it has a special property. The numerical values of its components are the same in all coordinate systems. A tensor with this property is known as an isotropic tensor. Since /summationdisplay kDikδjk=Dij, the Kronecker delta tensor is also known as the substitution tensor . It is also called a unit tensor, because of its matrix representation: δij=⎛ ⎝100 010 001⎞ ⎠. Levi-Civita Tensor . The Levi-Civita symbol εijk, εijk=⎧ ⎨ ⎩1i f ( i,j,k) is an even permutation of (1 ,2,3) −1i f ( i,j,k) is an odd permutation of (1 ,2,3) 0 if any index is repeated, which we used for the deﬁnition of a third-order determinant, is a third rank isotropic tensor. This is also known as the alternating tensor .T op r o v et h i s , recall the deﬁnition of the third-order determinant 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1a1la2ma3nεlmn=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (4.51) Now if the row indices (1, 2, 3) are replaced by ( i,j,k), we have 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleai1ai2ai3 aj1aj2aj3 ak1ak2ak3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (4.52) This relation can be demonstrated by writing out the nonvanishing terms of both sides. It can also be proved by noting the following. First for i=1 ,j=2 , k=3,it reduces to (4 .51). Now consider the eﬀect of interchanging iandj. The left-hand side changes sign because 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ajlaimaknεlmn=3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ajmailaknεmln =−3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn. 4.2 Cartesian Tensors 173 The right-hand side also changes sign because it is an interchange of two rows of the determinant. If two indices of i,j,k are the same, then both sides are equal to zero. The left-hand side is zero, since the quantity is equal to its negative. The right-hand side is equal to zero since two rows of the determinant are identical. This suﬃces to prove the result since all permutations of i,j,k can be achieved by a sequence of interchanges. It follows from the properties of determinants and the deﬁnition of εijk that /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleai1ai2ai3 aj1aj2aj3 ak1ak2ak3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=εijk/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, (4.53) and (4 .52) becomes 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn=εijk/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (4.54) This relation is true for any determinant. Now if aijare elements of a rotation matrix, /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1, as shown in example 4.1.2. To decide if εijkis a tensor, we should look at its value in a rotated system. The tensor transformation rules require ε/prime ijk=3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn. But 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn=εijk/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12a13 a21a22a23 a31a32a33/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=εijk. Hence ε/prime ijk=εijk. (4.55) Therefore, εijkis indeed a third-rank isotropic tensor. Relation between δijandεijk. There is an interesting and important rela- tion between Kronecker delta and Levi-Civita tensors 3/summationdisplay i=1εijkεilm=δjlδkm−δjmδkl. (4.56) After summed over i,there are four free subscripts j,k,l,m . Therefore (4 .56) represents 81 (34= 81) equations. Yet it is not diﬃcult to prove (4 .56), when we make the following observations. 174 4 Vector Transformation and Cartesian Tensors 1. If either j=korl=m,both sides of (4 .56) are equal to zero. If j=k, the left-hand side is zero, because εikk=0.The right-hand side is also equal to zero because δklδkm−δkmδkl=0.The same result is obtained forl=m.Therefore we only need to check the cases for which j/negationslash=kand l/negationslash=m. 2. For the left-hand side not to vanish, i, j, k have to be diﬀerent. Therefore given j/negationslash=k, iis ﬁxed. Consider εilm,sinceiis ﬁxed and i, l, m have to be diﬀerent for nonvanishing εilm,therefore, l=j, m=korl=k, m=j are the only two nonvanishing options. 3. For l=jandm=k, εijk=εilm.Thus εijkandεilmmust have the same sign. (Either both equal to −1,or both equal to +1). Therefore on the left-hand side of (4 .56),εijkεilm=+ 1.On the right-hand side of (4 .56), it is also equal to +1 ,sincel/negationslash=m, δjlδkm−δjmδkl=δllδmm−δlmδml=1−0=1. 4. For l=kandm=j, εijk=εiml=−εilm.Thus εijkandεilmhave opposite sign. (one equal to −1 and the other +1 ,or vice versa). Therefore on the left-hand side of (4 .56),εijkεilm=−1.On the right-hand side of (4.56),it is also equal to −1,sincel/negationslash=m, δjlδkm−δjmδkl=δmlδlm−δmmδll=0−1=−1. This covers all 81 cases. In each case the left-hand side is equal to the right-hand side. Therefore (4 .56) is established. 4.2.3 Outer Product IfSi1i2···iNis a tensor of rank NandTj1j2···jMi sat e n s o ro f Mth rank, then Si1i2···iNTj1j2···jMis a tensor of rank (N+M). This is known as the outer product theorem. (Outer product is also known as direct product.) This theorem can be easily demonstrated. First it certainly has 3N+Mcomponents. Under a rotation S/prime i1i2···iN=/summationdisplay k1···kNai1k1···aiNkNSk1···kN, T/prime j1j2···jM=/summationdisplay l1···lNaj1l1···ajMlMTl1···lM, where we have written3/summationtext j1=13/summationtext j2=1····3/summationtext jN=1as/summationtext j1···jN, (Si1i2···iNTj1j2···jM)/prime=S/prime i1i2···iNT/prime j1j2···jM =/summationdisplay k1···kN/summationdisplay l1···lNai1k1···aiNkNaj1l1···ajMlMSk1···kNTl1···lM, (4.57) which is how a ( M+N)th rank tensor should transform. 4.2 Cartesian Tensors 175 For example, the outer product of two vectors is a second rank tensor. Let (A1,A2,A3) and ( B1,B2,B3) be vectors, so they are ﬁrst rank tensors. Their outer product AiBjis a second rank tensor. Its nine components can be displayed as a matrix AiBj=⎛ ⎜⎝A1B1A1B2A1B3 A2B1A2B2A2B3 A3B1A3B2A3B3⎞ ⎟⎠. SinceAandBare vectors, A/prime i=3/summationdisplay k=1aikAk;B/prime j=3/summationdisplay l=1ajlBl, it follows A/prime iB/prime j=3/summationdisplay k=13/summationdisplay l=1aikajlAkBl, which shows AiBjis a second rank tensor, in agreement with the outer product theorem. We mention in passing, the second rank tensor formed by two vectors A andBis sometimes denoted as AB(without anything between them). When written in this way, it is called a dyad. A linear combination of dyads is a dyadic. Since everything that can be done with vectors and dyadics can also be done by tensors and matrices, but not the other way around, we will not discuss dyadics any further. Example 4.2.2. Use the outer product theorem to show that the expression in example 4.2.1 Tij=/parenleftbigg x1x2−x2 1 x2 2−x1x2/parenrightbigg is a second rank tensor in a two-dimensional space. Solution 4.2.2. A two-dimensional position vector is given by /parenleftbiggA1 A2/parenrightbigg =/parenleftbiggx1 x2/parenrightbigg . We have also shown in (4 .46) that /parenleftbigg B1 B2/parenrightbigg =/parenleftbigg x2 −x1/parenrightbigg is a vector in the two-dimensional space. The outer product of these two vectors is a second rank tensor AiBj=/parenleftbigg x1x2−x2 1 x2 2−x1x2/parenrightbigg =Tij. 176 4 Vector Transformation and Cartesian Tensors 4.2.4 Contraction We can lower the rank of any tensor through the following theorem. IfTi1i2i3···iNis a tensor of Nthrank, then Si3···iN=/summationdisplay i1/summationdisplay i2δi1i2Ti1i2i3···iN is a tensor of rank (N−2). To prove this theorem, we ﬁrst note that Si3···iNhas 3N−2components. Next we have to show that S/prime i3···iN=/summationdisplay i1/summationdisplay i2δ/prime i1i2T/prime i1i2i3···iN(4.58) satisﬁes the tensor transformation rule. With δ/prime i1i2=δi1i2, T/prime i1,i2,···iN=/summationdisplay j1···jNai1j1ai2j2···aiNjNTj1,j2,···,jN, (4.58) becomes S/prime i3···iN=/summationdisplay i1i2δi1i2/summationdisplay j1···jNai1j1ai2j2···aiNjNTj1,j2,···,jN =/summationdisplay j1···jN/parenleftBigg/summationdisplay i1i2δi1i2ai1j1ai2j2/parenrightBigg ai3j3···aiNjNTj1,j2,···,jN. Now /summationdisplay i1i2δi1i2ai1j1ai2j2=/summationdisplay i1ai1j1ai1j2=δj1j2, so S/prime i3···iN=/summationdisplay j1···jNδj1j2ai3j3···aiNjNTj1,j2,···,jN =/summationdisplay j3···jN⎛ ⎝/summationdisplay j1j2δj1j2Tj1,j2,···,jN⎞ ⎠ai3j3···aiNjN =/summationdisplay j3···jNai3j3···aiNjNSj3···jN. (4.59) Therefore Si3···iNis a (N−2)th rank tensor. The process of multiplying by δi1i2and summing over i1andi2is called contraction . For example, we have shown that AiBj,the outer product of two 4.2 Cartesian Tensors 177 vectors AandB, is a second rank tensor. The contraction of this second-rank tensor is a zeroth rank tensor, namely a scalar. /summationdisplay ijδijAiBj=/summationdisplay iAiBi=A·B. Indeed, this zeroth rank tensor is the dot product of AandB. Simply stated, a new tensor of rank ( N−2) will be obtained if two of the indices of a Nth rank tensor are set equal to each other and summed over. (The German word for contraction is verj¨ungung, which can be translated as rejuvenation.) If the rank of the tensor is 3 or higher, we can contract over any two indices. In general we get diﬀerent ( N−2)th rank tensors if we contract over diﬀerent pairs of indices. For example, in the third rank tensor Ti1i2i3=Ai1Bi2Ci3,ifi1andi2are contracted, we obtain a vector /summationdisplay iTiii3=/summationdisplay iAiBiCi3=(A·B)Ci3, (remember Ci3could represent a particular component of C,it also could represent the totality of the components, namely the vector Citself). On the other hand, if i2andi3are contracted, we obtain another vector /summationdisplay iTi1ii=/summationdisplay iAi1BiCi=Ai1(B·C). So contracting the ﬁrst two indices we get a scalar times vector C,and con- tracting the second and third indices we get another scalar times vector A. Contraction is one of the most important operations in tensor analysis and is worth remembering. 4.2.5 Summation Convention The summation convention (invented by Einstein) gives tensor analysis much of its appeal. We note that in the deﬁnition of tensors (4 .47), all indices over which the summation is to be carried out are repeated in the expression. Moreover, the range of the index (such as from 1 to 3) is already known from the context of the discussion. Therefore, without loss of information, we may drop the summation sign with the understanding that the repeated subscripts imply that the term is to be summed over its range. The repeated subscript is referred to as a “dummy” subscript. It must appear no more than twice in a term. The choice of dummy subscript does not matter. Replace one dummy index by another is one of the most useful tricks in tensor analysis one should learn. For example, the dot product A·Bcan be equally represented either byAiBiorAkBk, since both of them mean the same thing, namely AkBk=3/summationdisplay i=1AiBi=3/summationdisplay j=1AjBj=A1B1+A2B2+A3B3=A·B. 178 4 Vector Transformation and Cartesian Tensors Example 4.2.3. Express the expression AiBjCiwith summation convention in terms of ordinary vector notation. Solution 4.2.3. AiBjCi=/parenleftBigg3/summationdisplay i=1AiCi/parenrightBigg Bj=(A·C)Bj. Note that it is the subscripts that indicate which vector is dotted with which, not the ordering of the components of the vectors. The ordering is immaterial. So ( A·C)Bj=AkCkBjis equally valid. The letter jis a free subscript, it can be replaced by any letter other than the dummy subscript. However, if the term is used in a equation, then the free subscript of every term in the equation must be represented by the same letter. From now on when we write down a quantity with Nsubscripts, if all N subscripts are diﬀerent, it will be assumed that it is a good Nth rank tensor. If any two of them are the same, it is a contracted tensor of rank N−2. Example 4.2.4. (a) What is the rank of the tensor εijkAlBm? (b) what is the rank of the tensor εijkAjBk? (c) Express εijkAjBkin terms of ordinary vector notation. Solution 4.2.4. (a) Since εijkis a third rank tensor and AlBmis a tensor of rank 2, so by the outer product theorem εijkAlBmis a tensor of rank 5 . (b)εijkAjBkis twice contracted, so it is a ﬁrst rank (5 −4 = 1) tensor. (c) εijkAjBk=/summationdisplay j/summationdisplay kεijkAjBk. Ifi=1,the only nonzero terms come from j=2or3,sinceεijkis equal to zero if any two indices are equal. If j=2,kcan only equal to 3 .Ifj=3,k has to be 2 .So ε1jkAjBk=ε123A2B3+ε132A3B2=A2B3−A3B2=(A×B)1. Similarly, ε2jkAjBk=(A×B)2,ε 3jkAjBk=(A×B)3. Therefore εijkAjBk=(A×B)i. 4.2 Cartesian Tensors 179 Example 4.2.5. Show that εijkAiBjCk=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleA1A2A3 B1B2B3 C1C2C3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Solution 4.2.5. εijkAiBjCk=(εijkAiBj)Ck=(A×B)kCk=(A×B)·C =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleC1C2C3 A1A2A3 B1B2B3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleA1A2A3 B1B2B3 C1C2C3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. With the summation convention, (4 .56) is simply εijkεilm=δjlδkm−δjmδkl. (4.60) Many vector identities can be quickly and elegantly proved with this equation. Example 4.2.6. Show that A×(B×C)=(A·C)B−C(A·B) Solution 4.2.6. (B×C)i=εijkBjCk [A×(B×C)]l=εlmnAm(B×C)n=εlmnAmεnjkBjCk=εnlmεnjkAmBjCk εnlmεnjkAmBjCk=(δljδmk−δlkδmj)AmBjCk =AkBlCk−AjBjCl=(A·C)Bl−(A·B)Cl. Since the corresponding components agree, the identity is established. Example 4.2.7. Show that (A×B)·(C×D)=(A·C)(B·D)−(A·D)(B·C). Solution 4.2.7. (A×B)·(C×D)=εkijAiBjεklmClDm =(δilδjm−δimδjl)AiBjClDm=AlBmClDm−AmBlClDm =(A·C)(B·D)−(A·D)(B·C). 4.2.6 Tensor Fields A tensor ﬁeld of Nth rank, Ti1···iN(x1,x2,x3) is the totality of 3Nfunctions which for any given point in space ( x1,x2,x3) constitute a tensor of Nth rank. 180 4 Vector Transformation and Cartesian Tensors A scalar ﬁeld is a tensor ﬁeld of rank zero. We have shown in example 4.1.5 that the gradient of a scalar ﬁeld is a vector ﬁeld. There is a corresponding theorem for tensor ﬁelds. IfTi1···iN(x1,x2,x3)is a tensor ﬁeld of rank N,then ∂ ∂xiTi1···iN(x1,x2,x3) is a tensor ﬁeld of rank N+1. The proof of this theorem goes as follows. /parenleftbigg∂ ∂xiTi1···iN(x1,x2,x3)/parenrightbigg/prime =∂ ∂x/prime iT/prime i1···iN(x/prime 1,x/prime 2,x/prime 3) =∂ ∂x/prime i/summationdisplay j1···jNai1j1···aiNjNTj1···jN(x1,x2,x3). By chain rule and (4 .31) ∂ ∂x/prime i=/summationdisplay j∂xj ∂x/prime i∂ ∂xj=/summationdisplay jaij∂ ∂xj, so /parenleftbigg∂ ∂xiTi1···iN(x1,x2,x3)/parenrightbigg/prime =/summationdisplay j/summationdisplay j1···jNaijai1j1···aiNjN∂ ∂xjTj1···jN(x1,x2,x3), (4.61) which is how a tensor of rank N+ 1 transforms. Therefore the theorem is proven. To simplify the writing, we introduce another useful notation. From now on, the diﬀerential operator ∂/∂x iis denoted by ∂i.For example, ∂ ∂xiϕ=∂iϕ, which is the ith component of ∇ϕ.Again it can also represent the totality of /parenleftbigg∂ϕ ∂x1,∂ϕ ∂x2,∂ϕ ∂x3/parenrightbigg . Thus ∂iϕis a vector. (Note it has one subscript.) Similarly, ∇×A=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglee1e2e3 ∂1∂2∂3 A1A2A3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, ∇·A=/summationdisplay i∂Ai ∂xi=/summationdisplay i∂iAi=∂iAi, 4.2 Cartesian Tensors 181 (∇×A)i=/summationdisplay j,kεijk∂ ∂xjAk=εijk∂jAk ∇2ϕ=∇·∇ϕ=∂i(∇ϕ)i=∂i∂iϕ. With these notations, vector ﬁeld identities can be easily established. Example 4.2.8. Show that ∇·(∇×A)=0. Solution 4.2.8. ∇·(∇×A)=∂i(∇×A)i=∂iεijk∂jAk=εijk∂i∂jAk =εijk∂j∂iAk (∂i∂j=∂j∂i) =−εjik∂j∂iAk =−εijk∂i∂jAk (rename iandj) =−∇·(∇×A) Thus 2 ∇·(∇×A)=0.Hence ∇·(∇×A)=0. Example 4.2.9. Show that ∇×(∇ϕ)=0. Solution 4.2.9. [∇×∇ϕ]i=εijk∂j(∇ϕ)k=εijk∂j∂kϕ=−εikj∂j∂kϕ =−εijk∂j∂kϕ (rename jandk) =−[∇×∇ϕ]i. It follows that ∇×∇ϕ=0. Example 4.2.10. Show that ∇×(∇×A)=∇(∇·A)−∇2A. Solution 4.2.10. [∇×(∇×A)]i=εijk∂j(∇×A)k=εijk∂jεklm∂lAm =εkijεklm∂j∂lAm =(δilδjm−δimδjl)∂j∂lAm =∂m∂iAm−∂l∂lAi=∂i∂mAm−∂l∂lAi =[∇(∇·A)]i−(∇2A)i. Hence ∇×(∇×A)=∇(∇·A)−∇2A,since corresponding components from both sides agree. 182 4 Vector Transformation and Cartesian Tensors Example 4.2.11. Show that ∇·(A×B)=(∇×A)·B−A·(∇×B). Solution 4.2.11. ∇·(A×B)=∂i(A×B)i=∂iεijkAjBk=εijk∂i(AjBk) =εijk(∂iAj)Bk+εijkAj(∂iBk)=(εijk∂iAj)Bk−Aj(εjik∂iBk) =(∇×A)kBk−Aj(∇×B)j=(∇×A)·B−A·(∇×B). Example 4.2.12. Show that ∇×(A×B)=(∇·B)A−(∇·A)B+(B·∇)A−(A·∇)B. Solution 4.2.12. [∇×(A×B)]i=εijk∂j(A×B)k=εijk∂jεklmAlBm =εkijεklm∂j(AlBm)=εkijεklm(Bm∂jAl+Al∂jBm) =(δilδjm−δimδjl)(Bm∂jAl+Al∂jBm) =Bm∂mAi−Bi∂lAl+Ai∂mBm−Al∂lBi =[ (B·∇)A−(∇·A)B+(∇·B)A−(A·∇)B]i. Since the corresponding components agree, the two sides of the desired equation must be equal. 4.2.7 Quotient Rule Another way to determine if a quantity with two subscripts is a second rank tensor is to use the following quotient rule. If for an arbitrary vector B, the result of summing over jof the product KijBjis another vector A Ai=KijBj, (4.62) and (4 .62) holds in all Cartesian coordinate systems, then Kijis a true second rank tensor. To prove the quotient rule, we examine the components of Ain a rotated system, A/prime i=ailAl=ailKlmBm. SinceBis a vector, Bm=ajmB/prime j. 4.2 Cartesian Tensors 183 It follows A/prime i=ailKlmBm=ailKlmajmB/prime j=ailajmKlmB/prime j. But since (4 .62) holds for all systems, A/prime i=K/prime ijB/prime j. Subtracting the last two equations, /parenleftbig K/prime ij−ailajmKlm/parenrightbig B/prime j=0. Since B/prime jis arbitrary, K/prime ij=ailajmKlm. Therefore Kijis a second rank tensor. With a similar procedure, one can show that if an Mth rank tensor is linearly related to an Nth rank tensor through a quantity TwithM+N subscripts, and the relation holds for all systems, then Tis a tensor of rank M+N. Example 4.2.13. IfTijxixjis equal to a scalar S, show that Tijis a second rank tensor. Solution 4.2.13. Since xixjis the outer product of two position vectors, so it is a second rank tensor. The scalar Sis a zeroth rank tensor, therefore by quotient rule Tijis a second rank (2 + 0 = 2) tensor. It is instructive to demonstrate this directly by looking at the components in a rotated system. S=Tijxixj=Tlmxlxm.x l=ailx/prime i;xm=ajmx/prime j. S=Tlmailx/prime iajmx/prime j=ailajmTlmx/prime lx/prime j, S/prime=T/prime ijx/prime ix/prime j,S/prime=S. Therefore (T/prime ij−ailajmTlm)x/prime ix/prime j=0 ; T/prime ij=ailajmTlm. Hence Tijis a second rank tensor. 4.2.8 Symmetry Properties of Tensors A tensor Sijk···is said to be symmetric in the indices iandjif Sijk···=Sjik···. 184 4 Vector Transformation and Cartesian Tensors A tensor Aijk···is said to be antisymmetric in the indices iandjif Aijk···=−Ajik···. For example, the outer product of rwith itself xixjis a second rank symmet- rical tensor, the Kronecker delta δijis also a second rank symmetrical tensor. On the other hand, the Levi-Civita symbol εijkis antisymmetric with respect to any two of its indices, since εijk=−εjik. Symmetry is a physical property of tensors. It is invariant to coordinate transformation. For example, if Sijis a symmetrical tensor in certain coordi- nate system, in a rotated system S/prime lm=aliamjSij=amjaliSji=S/prime ml. Therefore Sijis also a symmetric tensor in a new system. Similar results hold also for antisymmetric tensors. A symmetric second rank tensor Sijcan be written in the form Sij=⎛ ⎝S11S12S13 S12S22S23 S12S23S33⎞ ⎠, while an antisymmetric second rank tensor Aijis of the form Aij=⎛ ⎝0A12A13 −A120A23 −A13−A230⎞ ⎠. Thus a symmetric second rank tensor has six independent components, while an antisymmetric second tensor has only three independent components. Any second rank tensor Tijcan be represented as the sum of a symmetric tensor and an antisymmetric tensor. Given Tij,one can construct Sij=1 2(Tij+Tji),A ij=1 2(Tij−Tji). Clearly Sijis symmetric and Aijis antisymmetric. Furthermore Tij=Sij+Aij. Therefore any second rank tensor has a symmetric part and an antisymmetric part. As shown in the theory of matrices, the six independent elements of a symmetric matrix can be represented by a quadratic surface. In the same way, a symmetric second rank tensor can be represented uniquely by an ellipsoid Tijxixj=±1, where the sign is that of the determinant of |Tij|. The three independent components of an antisymmetric second rank tensor can also be represented geometrically by a vector. 4.2 Cartesian Tensors 185 4.2.9 Pseudotensors One of the reasons why tensors are useful is that they make it possible to formulate the laws of physics in a way that is independent of any preferred direction in space. One might also expect these laws to be independent of whether we choose a right-handed system or a left-handed system of axes. However, under a transformation from right-handed axes to left-handed axes not all tensors transform in the same way. So far our discussion has been restricted to rotations within right-handed systems. The right-handed system is deﬁned by naming the three basis vectors e1,e2,e3in such a way that the thumb of your right hand is pointing in the direction of e3if the other four ﬁngers can curl from e1toward e2 without passing through negative e2.A right-handed system can be rotated into another right-handed system. The determinant of the rotation matrix is equal to 1 as shown in example 4.1.2. Now consider the eﬀect of inversion. The three basis vectors e1,e2,e3are changed to e/prime 1,e/prime 2,e/prime 3such that e/prime 1=−e1,e/prime 2=−e2,e/prime 3=−e3. This new set of coordinate axes is a left-handed system. This is called a left- handed system because only if you use your left hand, can the thumb point in the positive e/prime 3direction while other four ﬁngers curl from e/prime 1toward e/prime 2 without passing through negative e/prime 2.Note that one cannot rotate a right- handed system into a left-handed system. If we use the same right hand rule for the deﬁnition of vector cross products in all systems, then with right-handed axes (e1×e2)·e3=1, and with left-handed axis (e/prime 1×e/prime 2)·e/prime 3=−1. The position vector r=x1e1+x2e2+x3e3 expressed in the inverted system becomes r/prime=−x1e/prime 1−x2e/prime 2−x3e/prime 3. In other words, it is the same vector r=r/prime,except in the prime system the coeﬃcients become negative since the axes are inverted. Vectors behaving this way when the coordinates are changed from a right-handed system to a left-handed system are called polar vectors . They are just regular vectors. 186 4 Vector Transformation and Cartesian Tensors A fundamental diﬀerence appears when we encounter the cross product of two polar vectors. The components of C=A×Bare given by C1=A2B3−A3B2, and so on. Now when the coordinates axes are inverted, Aigoes to −Ai,Bi changes to −BibutCigoes to + Cisince it is the product of two negative terms. It does not behave like polar vectors under inversion. To distinguish, the cross product is called a pseudo vector , also known as Axial vector . In addition to inversion, reﬂection (reversing one axis) and interchanging of two axes also transform a right-handed system into a left-handed system. The transformation matrices of these right-left operations are as follows. Inversion: ⎛ ⎝x/prime 1 x/prime 2 x/prime 3⎞ ⎠=⎛ ⎝−10 0 0−10 00 −1⎞ ⎠⎛ ⎝x1 x2 x3⎞ ⎠. Reﬂection with respect to the x2x3plane: ⎛ ⎝x/prime 1 x/prime 2 x/prime 3⎞ ⎠=⎛ ⎝−100 01 0 00 1⎞ ⎠⎛ ⎝x1 x2 x3⎞ ⎠. Interchange of x1andx2axes: ⎛ ⎝x/prime 1 x/prime 2 x/prime 3⎞ ⎠=⎛ ⎝010 100 001⎞ ⎠⎛ ⎝x1 x2 x3⎞ ⎠. These equations can be written in the form x/prime i=aijxj. It is obvious that the determinants |aij|of these transformation matrices are all equal to −1.A left-handed system can be rotated into another left-handed system with Euler angles in the same way as in the right-handed systems. Hence, if a matrix ( aij) transforms a right-handed system into a left-handed system, or vice versa, the determinant |aij|is always equal to −1.Furthermore, we can show, in the same manner as with the rotation matrix, that its elements also satisfy the orthogonality condition: aikajk=δij. Therefore it is also an orthogonal transformation. Thus the orthogonal transformations can be divided into two classes: proper transformation for which the determinant |aij|is equal to one, and improper transformation for which the determinant is equal to negative one. If the transformation is a rotation, it is a proper transformation. If the 4.2 Cartesian Tensors 187 transformation changes a right-handed system into a left-handed system, the transformation is improper. A pseudovector can now be deﬁned as satisfying the transforming rule V/prime i=|aij|aijVj. If the transformation is proper, polar vectors and pseudovectors transform in the same way. If the transformation is improper, polar vectors transform as a regular vector, but pseudovectors will ﬂip direction. Pseudotensors are deﬁned in the same way. The components of a Nth rank pseudotensor transform according to the rule: T/prime i1···iN=|aij|ai1j1···aiNjNTj1···jN, (4.63) which is exactly the same as a regular tensor, except for the determinant |aij|. It follows from this deﬁnition that: 1. The outer product of two pseudotensors of rank MandNis a regular tensor of rank M+N. 2. The outer product of a pseudotensor of rank Mand a regular tensor of rankNgives a pseudotensor of rank M+N. 3. The contraction of a pseudotensor of rank Ngives a pseudotensor of rank N−2. A zeroth rank pseudotensor is a pseudoscalar, which changes sign un- der inversion, whereas a scalar does not. An example of pseudoscalar is the scalar triple product ( A×B)·C. The cross product of A×Bis a ﬁrst rank pseudotensor, the polar vector Cis a regular ﬁrst rank tensor. Hence (A×B)i·Ci=(A×B)·Cis the contraction of a second rank pseudotensor (A×B)i·Cj,therefore a pseudoscalar. If A,B,Care the three sides of a parallelepiped, ( A×B)·Cis the volume of the parallelepiped. Deﬁned this way, volume is actually a pseudoscalar. We have shown εijkis an isotropic third rank tensor under rotations. We will now show that if εijkis to mean the same thing in both right-handed and left-handed systems, then εijkmust be regarded as third rank pseudoten- sor. This is because if we want ε/prime ijk=εijkunder both proper and improper transformations, then εijkmust be expressed as ε/prime ijk=|aij|ailajmankεlmn. (4.64) Since ailajmaknεlmn=εijk|aij|, as shown in (4 .54),it follows from (4 .64) that ε/prime ijk=|aij|εijk|aij|=|aij|2εijk=εijk. Therefore εijkis a third rank pseudotensor. 188 4 Vector Transformation and Cartesian Tensors Example 4.2.14. LetT12,T13,T23be the three independent components of an antisymmetric tensor, show that T23,−T13,T13can be regarded as the com- ponents of a pseudovector. Solution 4.2.14. Since εijkis a third rank pseudotensor, Tjkis a second rank tensor, after contracting twice, Ci=εijkTjk the result Ciis a ﬁrst rank pseudotensor, which is just a pseudovector. Since Tij=−Tji,so T21=−T12,T31=−T13,T32=−T23. Now C1=ε123T23+ε132T32=T23−T32=2T23, C2=ε213T13+ε231T31=−T13+T31=−2T13, C3=ε312T12+ε321T21=T12−T21=2T12. Since ( C1,C2,C3) is a pseudovector, ( T23,−T13,T12) is also a pseudovector. Example 4.2.15. Use the fact that εijkis a third rank pseudotensor to show thatA×Bis a pseudovector. Solution 4.2.15. LetC=A×B,so Ci=εijkAjBk. Expressed in a new coordinate system where A,B,Care, respectively, trans- formed to A/prime,B/prime,C/prime,the cross product becomes C/prime=A/prime×B/primewhich can be written in the component form C/prime l=ε/prime lmnA/prime mB/prime n. Since εijkis a pseudotensor, ε/prime lmn=|aij|aliamjankεijk, soC/prime lbecomes C/prime l=|aij|aliamjankεijkA/prime mB/prime n. But Aj=amjA/prime m,B k=ankB/prime n, thus C/prime l=|aij|aliεijkAjBk=|aij|aliCi. Therefore C=A×Bis a pseudovector. 4.3 Some Physical Examples 189 Since mathematical equations describing physical laws should be inde- pendent of coordinate systems, we cannot equate tensors of diﬀerent rank because they have diﬀerent transformation properties under rotation. Like- wise, in classical physics, we cannot equate pseudotensors to tensors because they transform diﬀerently under inversion. However, surprisingly nature under the inﬂuence of weak interaction can distinguish a left-handed system from a right-handed system. By introducing a pseudoscalar, the counterintuitive events that violate parity conservation can be described. (See, T.D. Lee in “Thirty Years Since Parity Nonconservation”, Birkh¨ auser, 1988, page 158). 4.3 Some Physical Examples 4.3.1 Moment of Inertia Tensor One of the most familiar second rank tensors in physics is the moment of inertial tensor. It relates the angular momentum Land the angular velocity ωof the rotational motion of a rigid body. The angular momentum Lof a rigid body rotating about a ﬁxed point is given by L=/integraldisplay r×vdm, where ris the position vector from the ﬁxed point to the mass element d m, andvis the velocity of d m.We have shown v=ω×r, therefore L=/integraldisplay r×(ω×r)dm =/integraldisplay [(r·r)ω−(r·ω)r]dm. Written in tensor notation with the summation convention, the ith component ofLis Li=/integraldisplay [r2ωi−xjωjxi]dm. Since r2ωi=r2ωjδij, Li=ωj/integraldisplay [r2δij−xjxi]dm=Iijωj, where Iij, known as the moment of inertia tensor, is given by Iij=/integraldisplay [xkxkδij−xjxi]dm. (4.65) 190 4 Vector Transformation and Cartesian Tensors Since δijandxixjare both second rank tensors, Iijis a symmetric second rank tensor. Explicitly the components of this tensor are Iij=⎛ ⎝/integraltext (x2 2+x2 3)dm−/integraltext x1x2dm−/integraltext x1x3dm −/integraltext x2x1dm/integraltext (x2 1+x2 3)dm−/integraltext x2x3dm −/integraltext x3x1dm−/integraltext x3x2dm/integraltext (x2 2+x2 1)dm⎞ ⎠. (4.66) 4.3.2 Stress Tensor The name tensor comes from the tensile force in elasticity theory. Inside a loaded elastic body, there are forces between neighboring parts of the material. Imagine a cut through the body, the material on the right exerts a force F on the material to the left, and the material on the left exerts an equal and opposite force −Fon the material to the right. Let us examine the force across a small area ∆ x1∆x3, shown in Fig. 4.3, in the imaginary plane perpendicular to the x2axis. If the area is small enough, we expect the force is proportional to the area. So we can deﬁne the stress P2 as the force per unit area. The subscript 2 indicates that the force is acting on a plane perpendicular to the positive x2axis. The components of P2along x1,x2,x3axes are denoted as P12,P22,P32,respectively. Next, we can look at a small area on the imaginary plane perpendicular to the x1axis, and deﬁne the stress components P11,P21,P31.Finally, imagine the cut is perpendicular to the x3axis, so the stress components P13,P23,P33can be similarly deﬁned. Ife1,e2,e3are the unit bases vectors, these relations can be expressed as Pj=Pijei. (4.67) Thus the stress has nine components Pij=⎛ ⎝P11P12P13 P21P22P23 P31P32P33⎞ ⎠. (4.68) ∆x1∆x 3 P12P22P32 x1x2x3 P2 = ∆x1 ∆x3F*2 Fig. 4.3. StressP2,deﬁned as force per unit area, on a small surface perpendicular to the x2axis. Its components along the three axes are, respectively, P12,P22,P32 4.3 Some Physical Examples 191 P12 P22P32 x 1x2x3 P12P13P23P33 P32P22 Fig. 4.4. The nine components of a stress tensor at a point can be represented as normal and tangential forces on the surfaces of an inﬁnitesimal cube around the point The ﬁrst subscript in Pijindicates the direction of the force component, the second subscript indicates the direction of the normal to the surface on which the force is acted upon. The physical meaning of Pijis as follows. Imagine an inﬁnitesimal cube around a point inside the material, Pijare the forces per unit area on the faces of this cube, as shown in Fig. 4.4. For clarity, forces are shown only on three surfaces. There are both normal forces Pii(tensions shown but they could be pressures with arrows reversed) and tangential forces Pij(i/negationslash=j, shears). Note that, in equilibrium, the forces on the opposite faces must be equal and opposite. Furthermore, (4 .67) is symmetric Pij=Pji,because of rotational equilibrium. For example, the shearing force on the top surface in thex2direction is P23∆x1∆x2.The torque around x1axis due to this force is (P23∆x1∆x2)∆x3.The opposite torque due to the shearing force on the right surface is ( P32∆x3∆x1)∆x2.Since the net torque around x1axis must be zero, therefore (P23∆x1∆x2)∆x3=(P32∆x3∆x1)∆x1, and we have P23=P32. Similar argument will show that in general Pij=Pji,therefore (4 .68) is symmetric. Now we are going to show that the nine components of (4 .68) is a tensor, known as the stress tensor. For this purpose, we construct an inﬁnitesimal tetrahedron with its edges directed along the coordinate axes as shown in Fig. 4.5. Let ∆ a1,∆a2,∆a3denote the areas of the faces perpendicular to the axes x1,x2,x3,respectively, the forces per unit area on these faces are −P1,−P2,−P3,since these faces are directed in the negative direction of the axes. Let ∆ andenote the area of the inclined face with an unit exterior normal n,andPnbe the force per unit area on this surface. The total force on these 192 4 Vector Transformation and Cartesian Tensors ∆a3∆a2∆an ∆a1n x1x2x3 Pn Fig. 4.5. Forces on the surfaces of an inﬁnitesimal tetrahedron. The equilibrium conditions show that the stress must be a second rank tensor four surfaces must be zero, even if there are body forces, such as gravity. The body forces will be proportional to the volume, whereas all the surface forces are proportional to the area. As dimensions shrink to zero, the body forces have one more inﬁnitesimal, and can always be neglected compared with the surface forces. Therefore Pn∆an−P1∆a1−P2∆a2−P3∆a3=0. (4.69) Since ∆ a1is the area of ∆ anprojected on the x2x3plane, therefore ∆a1=(n·e1)∆an. With similar expressions of ∆ a2and ∆ a3,we can write (4 .69) in the form Pn∆an=P1(n·e1)∆an+P2(n·e2)∆an+P3(n·e3)∆an, or Pn=(n·ek)Pk. (4.70) What we want to ﬁnd is the stress components in a rotated system with axes e/prime 1,e/prime 2,e/prime 3.Now without loss of generality, we can assume that the jth axis of the rotated system is directed along n,that is n=e/prime j. Therefore PnisP/prime jin the rotated system. Thus (4 .70) becomes P/prime j=/parenleftbig e/prime j·ek/parenrightbig Pk. (4.71) In terms of their components along the respective coordinate axes, as shown in (4.67), P/prime j=P/prime ije/prime i,Pk=Plkel 4.3 Some Physical Examples 193 the last equation can be written as P/prime ije/prime i=/parenleftbig e/prime j·ek/parenrightbig Plkel. Take the dot product with e/prime ion both sides, P/prime ij(e/prime i·e/prime i)=/parenleftbig e/prime j·ek/parenrightbig Plk(el·e/prime i), we have P/prime ij=(e/prime i·el)/parenleftbig e/prime j·ek/parenrightbig Plk. Since ( e/prime m·en)=amn,we see that P/prime ij=ailajkPlk. (4.72) Therefore the array of the stress components (4 .68) is indeed a tensor. 4.3.3 Strain Tensor and Hooke’s Law Under imposed forces, an elastic body will deform and exhibit strain. The deformation is characterized by the change of distances between neighboring points. Let PatrandQatr+∆rbe two nearby points, as shown in Fig. 4.6. When the body is deformed, point Pis displaced by the amount u(r)t o P/prime,andQbyu(r+∆r)t oQ/prime.If the displacements of these neighboring points are the same, that is if u(r)=u(r+∆r),the relative positions of the points are not changed. That part of the body is unstrained, since the distances PQandP/primeQ/primewill be the same. Therefore the strain is associated with the variation of the displacement vector u(r).The change of u(r)c a n be written as ∆u=u(r+∆r)−u(r). Q 0PP9Q9 ∆r r r + ∆ru(r) u(r + ∆r) r + u(r)r + ∆r + u(r + ∆r)∆r’ Fig. 4.6. T h es t r a i no fad e f o r m e de l a s t i cb o d y .T h eb o d yi ss t r a i n e di ft h er e l a t i v e distance of two nearby points is changed. The strain tensor depends on the variation of the displacement vector uwith respect to the position vector r 194 4 Vector Transformation and Cartesian Tensors Neglecting second and higher terms, the components of ∆ucan be written as ∆ui=ui(x1+∆x1,x2+∆x2,x3+∆x3)−ui(x1,x2,x3) =∂ui ∂x1∆x1+∂ui ∂x2∆x2+∂ui ∂x3∆x3=∂ui ∂xj∆xj. Since uiis a vector and ∂/∂x jis a vector operator, ∂ui/∂xjis the outer product of two ﬁrst rank tensors. Therefore ∂ui/∂xjis a second rank tensor. It can be decomposed into a symmetric part and an antisymmetric part ∂ui ∂xj=1 2/parenleftbigg∂ui ∂xj+∂uj ∂xi/parenrightbigg +1 2/parenleftbigg∂ui ∂xj−∂uj ∂xi/parenrightbigg . (4.73) We can also divide ∆uinto two parts ∆u=∆us+∆ua, where ∆us i=1 2/parenleftbigg∂ui ∂xj+∂uj ∂xi/parenrightbigg ∆xj, ∆ua i=1 2/parenleftbigg∂ui ∂xj−∂uj ∂xi/parenrightbigg ∆xj. The antisymmetric part of (4 .73) does not alter the distance between Pand Qbecause of the following. Let the distance P/primeQ/primebe∆r/prime.It is clear from Fig. 4.6 that ∆r/prime=[∆r+u(r+∆r)]−u(r). It follows that ∆r/prime−∆r=u(r+∆r)−u(r)=∆u=∆us+∆ua. Now ∆ua·∆r=∆ua i∆xi=1 2/parenleftbigg∂ui ∂xj−∂uj ∂xi/parenrightbigg ∆xj∆xi=0. This is because in this expression, both iandjare summing indices and can be interchanged. Thus ∆uais perpendicular to ∆r,and it can be considered as the inﬁnitesimal arc length of a rotation around the tail of ∆r,as shown in the following sketch. ∆r∆r′ ∆u ∆ua∆us Fig. 4.7. The change of distance between two nearby points in an elastic body. It is determined by the symmetric strain tensor 4.3 Some Physical Examples 195 Therefore ∆uadoes not change the length ∆r.The change of distance between two nearby points of an elastic body is uniquely determined by the symmetric part of (4 .73) Eij=1 2/parenleftbigg∂ui ∂xj+∂uj ∂xi/parenrightbigg . (4.74) This quantity is known as the strain tensor. The stain tensor plays an important role in the elasticity theory because it is a measure of the degree of deformation. Since in one-dimension the elastic force in a spring is given by the Hooke’s lawF=−kx,one might expect that in three-dimensional elastic media the strain is proportional to the stress. For most solid materials with a relative strain of a few percent, this is indeed the case. The linear relationship between the strain tensor and the stress tensor is given by the generalized Hooke’s law Pij=cijklEkl, (4.75) where cijklis known as the elasticity tensor. Since PijandEklare both second rank tensors, by the quotient rule, cijklmust be a fourth rank tensor. While there are 81 components of a fourth rank tensor, but various symmetry con- siderations will show that the number of independent components in a general crystalline body is only 21. If the body is isotropic, the elastic constants are further reduced to only two. While we are not going into these details which are the subjects of books on elasticity, we only want to show that concepts of tensor are crucial in describing these physical quantities. Exercises 1. Find the rotation matrix for (a) a rotation of π/2a b o u t zaxis, (b) a rotation of πabout xaxis. Ans.⎛ ⎝01 0 −100 00 1⎞ ⎠;⎛ ⎝10 0 0−10 00−1⎞ ⎠. 2. With the transformation matrix ( A) given by (4 .20),show that (A)/parenleftbig AT/parenrightbig =(I), where ( I) is the identity matrix. 3. With the transformation matrix ( A) given by (4 .20),explicitly verify that 3/summationdisplay i=1aijaik=δjk, for (a) j=1,k=1,(b)j=1,k=2,(c)j=1,k=3. 4. With the transformation matrix ( A) given by (4 .20), show explicitly that the determinant Ais equal to 1 . 196 4 Vector Transformation and Cartesian Tensors 5. Show that there is no nontrivial isotropic ﬁrst rank tensor. Hint: (1) Assume there is an isotropic ﬁrst rank tensor ( A1,A2,A3). Under a rotation, A/prime 1=A1,A/prime 2=A2,A/prime 3=A3,since it is isotropic .(2) Consider a rotation of π/2a b o u t x3axis, and show that A1=0,A2=0.(3) A rotation about x1axis will show that A3=0.(4) Therefore only the zero vector is a ﬁrst rank isotropic tensor. 6. Let Tij=⎛ ⎝102 021 123⎞ ⎠,A i=⎛ ⎝3 2 1⎞ ⎠. Find the following contractions (a)Bi=TijAj, (b)Cj=TijAi, (c)S=TijAiAj. Ans. (a) Bi=( 5,5,10),(b)Ci=( 4,6,11),(c)S=3 5. 7. Let AijandBijbe two second rank tensors, and let Cij=Aij+Bij. Show that Cijis also a second rank tensor. 8. The equation of an ellipsoid centered at the origin is of the form Aijxixj=1. Show that Aijis a second rank tensor. Hint: In a rotated system, the equation of the surface is A/prime ijx/prime ix/prime j=1. 9. Show that Ai=/parenleftbigg −x2 x1/parenrightbigg is a two-dimensional vector. 10. Show that the following 2 ×2 matrices represent second rank tensors in two-dimensional space: (a)/parenleftBigg −x1x2x2 1 −x2 2x1x2/parenrightBigg ,(b)/parenleftBigg x2 2−x1x2 −x1x2x2 1/parenrightBigg , (c)/parenleftBigg −x1x2−x2 2 x2 1x1x2/parenrightBigg ,(d)/parenleftBigg x2 1x1x2 x1x2x2 2/parenrightBigg . Hint: Show that they are various outer products of the position vector and the vector in the last question. 4.3 Some Physical Examples 197 11. Explicitly show that 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleai1ai2ai3 aj1aj2aj3 ak1ak2ak3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle by (a) writing out all nonzero terms of 3/summationdisplay l=13/summationdisplay m=13/summationdisplay n=1ailajmaknεlmn, and (b) expand the determinant /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleai1ai2ai3 aj1aj2aj3 ak1ak2ak3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle over the elements of the ﬁrst row. 12. Show that εijkcan be written as εijk=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleδi1δi2δi3 δj1δj2δj3 δk1δk2δk3/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Hint: Show that the determinant has all the properties εijkhas. 13. Show that (a)/summationtext ijεijkδij=0 ; (b)/summationtext jkεijkεljk=2δil; (c)/summationtext ijkεijkεijk=6. 14. The following equations are written with summation convention, verify them (a)δijδjkδki=3, (b)εijkεklmεmni=εjnl. Hint: (a) Recall δijis a substitution tensor, (b) Use (4 .60). 15. With the summation convention, show that (a)Aiδij=Aj (b)Bjδij=Bi (c)δ1jδj1=1 (d)δijδji=δii=3 (e)δijδjl=δil 16. With subscripts and summation convention, show that (a)∂ixj=δij (b)∂i(xjxj)1/2=1 (xjxj)1/2xi. Hint: (a) x1,x2,x3are independent variables. (b) ∂i(xjxj)=2xj∂ixj. 198 4 Vector Transformation and Cartesian Tensors 17. The following equations are written with summation convention (a)∂ixi=3,(b)∂i(xjxj)1/2=(xjxj)−1/2xi,translate them into ordi- nary vector notation. Ans. (a) ∇·r=3,(b)∇r=r/r. 18. The following expressions are written with summation convention (a)ViAjBiej;( b ) cAiBjδij; (c)AlBjεijkδliek;( d ) εijkεlmkAiBjClDm, translate them into ordinary vector notation. Ans. (a) ( V·B)A,(b)cA·B,(c)A×B,(d) (A×B)·(C×D). 19. Use the Levi-Civita tensor technique to prove the following identities: (a)A×B=−B×A, (b)A·(B×C)=(A×B)·C. 20. Use the Levi-Civita tensor technique to prove the following identity ∇×(φA)=φ(∇×A)+(∇φ)×A. 21. Let Tij=⎛ ⎝123 045 006⎞ ⎠. Find the symmetric part Sijand the antisymmetric part Aijof the tensor Tij. Ans.Sij=⎛ ⎝11 1 .5 14 2 .5 1.52.56⎞ ⎠,A ij=⎛ ⎝01 1 .5 −10 2 .5 −1.5−2.50⎞ ⎠. 22. If Sijis a symmetric tensor and Aijis an antisymmetric tensor, show that SijAij=0. 23. Let ϕbe a scalar, Vibe a pseudovector, Tijbe a second rank tensor, and let Aijk=εijkϕ, B ij=εijkVk,C i=εijkTjk. Show that Aijkis a third rank pseudotensor, Bijis a second rank tensor, andCiis a pseudovector. 24. Find the strain tensor for an isotropic elastic material when it is subjected to (a) A stretching deformation u=( 0,0,αx3); (b) A shearing deformation u=(βx3,0,0). Ans. (a) Eij=⎛ ⎝000 000 00α⎞ ⎠;( b ) Eij=⎛ ⎝00β/2 000 β/20 0⎞ ⎠. Part II Diﬀerential Equations and Laplace Transforms 5 Ordinary Diﬀerential Equations The laws of physics that govern important and signiﬁcant problems in engineering and sciences are most often expressed in the form of diﬀeren- tial equations. A diﬀerential equation is an equation involving derivatives of an unknown function that depends upon one or more independent variables. If the unknown function depends on only one independent variable, then the equation is called an ordinary diﬀerential equation. In this chapter, after a review of the standard methods for solving ﬁrst-order diﬀerential equations, we will present a comprehensive treatment of linear diﬀerential equations with constant coeﬃcients, in terms of which a great many physical problems are formulated. We will use mechanical vibrations and electrical circuits as illustrative examples. Then we will dis- cuss systems of coupled diﬀerential equations and their applications. Series solutions of diﬀerential equations will be discussed in the chapter on special functions. Another important method of solving diﬀerential equation is the Laplace transformation, which we will discuss in the next chapter. 5.1 First-Order Diﬀerential Equations To solve a diﬀerential equation is to ﬁnd a way to eliminate the derivatives in the equation so that the relation between the dependent and the independent variables can be exhibited. For a ﬁrst-order diﬀerential equation, this can be achieved by carrying out an integration. The simplest type of diﬀerential equations is dy dx=f(x), (5.1) where f(x) is a given function of x.We know from calculus that y(x)=/integraldisplayx af(x/prime)dx/prime(5.2) 202 5 Ordinary Diﬀerential Equations is a solution. Equation (5 .1) contains only the ﬁrst derivative of y,and is called a ﬁrst-order diﬀerential equation. The order of a diﬀerential equation is equal to the order of the highest derivative in the equation. The solution (5 .2) is known as a general solution, and contains an arbitrary integration constant. If the integral in (5 .2) exists, then by deﬁnition there is a function F(x),such that d dxF(x)=f(x),dF(x)=f(x)dx and y(x)=/integraldisplayx adF(x/prime)=F(x)+F(a). In this sense, we often use the notation of the indeﬁnite integral y(x)=/integraldisplay f(x)dx+C where C=F(a) is an arbitrary constant. If we know that ytakes the value y0when x=x0,then the constant is determined. This condition “ y=y0 when x=x0” is called either “initial condition” or “boundary condition.” To satisfy both the equation and the boundary condition, we can carry out the following deﬁnite integrals: /integraldisplayy y0dy/prime=/integraldisplayx x0f(x/prime)dx/prime, which can be written in the form of y(x)=/integraldisplayx x0f(x/prime)dx/prime+y0. This is known as the speciﬁc solution. (The term “particular solution” is often used, however, this may cause confusion, since “particular solution” is also used in the solution of nonhomogeneous equations, which we shall discuss a little later.) In most physical applications, it is the speciﬁc solution that is of interest. A physical problem, when formulated in the mathematical language, usually consists of a diﬀerential equation and an appropriate number of boundary and/or initial conditions. The problem is solved only after the speciﬁc solution is found. 5.1.1 Equations with Separable Variables If an equation can be written in the form f(x)dx+g(y)dy=0 5.1 First-Order Diﬀerential Equations 203 the solution can be immediately obtained in the form of /integraldisplay f(x)dx+/integraldisplay g(y)dy=C. This method is called solution by the separation of variables and is one of the most commonly used methods. For example, the diﬀerential equation dy dx=−x y can be solved by noting that the equation can be written as ydy+xdx=0. Therefore the solution is given by /integraldisplay ydy+/integraldisplay xdx=C or 1 2y2+1 2x2=C. This general solution can be written as y(x)=(C/prime−x2)1/2 or equivalently as F(x,y)=C/prime with F(x,y)=x2+y2. Clearly this general solution represents a family of circles with radius√ C/prime centered at the origin. If it is speciﬁed that x=5,y=0,is a point on the circle, then the speciﬁc solution is x2+y2=2 5. This speciﬁc solution can also be obtained from the deﬁnite integral /integraldisplayy 0y/primedy/prime+/integraldisplayx 5x/primedx/prime=0, which gives the same result by way of 1 2y2+1 2x2−1 252=0. 204 5 Ordinary Diﬀerential Equations 5.1.2 Equations Reducible to Separable Type Certain equations of the form g(x,y)dy=f(x,y)dx (5.3) that are not separable can be made separable by a change of variable. This can always be done, if the ratio of f(x,y)/g(x,y) is a function of y/x. Letu=y/xand the function of y/xbeh(u),so that dy dx=f(x,y) g(x,y)=h(u). Since yis function of x,so isu.It follows that y(x)=xu(x)a n d dy dx=u+xdu dx. Thus the diﬀerential equation can be written as u+xdu dx=h(u), or xdu dx=h(u)−u Clearly it is separable du h(u)−u=dx x. So we can solve for u(x) and the solution of the original diﬀerential equation is simply y=xu(x). For example, if dy dx=y2+xy x2=y2 x2+y x, then with u=y/x, h (u)=u2+u.Since h(u)−u=u2,so du u2=dx x and /integraldisplaydu u2=/integraldisplaydx x, which gives −1 u+C=l nx. Since u=y/x,the solution of the original diﬀerential equation is therefore given byx y+l nx=C. 5.1 First-Order Diﬀerential Equations 205 5.1.3 Exact Diﬀerential Equations Suppose we want to ﬁnd a diﬀerential equation that represents the following family of curves: F(x,y)=C. First let us look at two nearby points ( x,y) and ( x+∆x,y+∆y),both on a speciﬁc curve of this family. If the curve is characterized by C=k,then F(x,y)=k, F (x+∆x,y+∆y)=k. Clearly, the diﬀerence between the two is equal to zero ∆F=F(x+∆x,y+∆y)−F(x,y)=0. This diﬀerence can be written in the form of F(x+∆x,y+∆y)−F(x,y)=F(x+∆x,y+∆y)−F(x,y+∆y) +F(x,y+∆y)−F(x,y). (5.4) With the understanding that ∆ xand ∆ yare approaching zero as a limit, we can use the deﬁnition of partial derivative F(x+∆x,y+∆y)−F(x,y+∆y)=∂F ∂x∆x, F(x,y+∆y)−F(x,y)=∂F ∂y∆y to write (5 .4) as ∆F=∂F ∂x∆x+∂F ∂y∆y or dF=∂F ∂xdx+∂F ∂ydy. This is known as the total diﬀerential. Since ∆ F=0,so we have ∂F ∂xdx+∂F ∂ydy=0. (5.5) This is the diﬀerential equation representing the family of curves F(x,y)=C. In other words, the solution of the diﬀerential equation in the form of (5 .5) is given by F(x,y)=C. Now let ∂F ∂x=f(x,y),∂F ∂y=g(x,y) (5.6) 206 5 Ordinary Diﬀerential Equations so ∂2F ∂y∂x=∂ ∂y∂F ∂x=∂ ∂yf(x,y), ∂2F ∂x∂y=∂ ∂x∂F ∂y=∂ ∂xg(x,y). Since the order of diﬀerentiation can be interchanged as long as the function has continuous partial derivatives ∂2F ∂y∂x=∂2F ∂x∂y, one has ∂ ∂yf(x,y)=∂ ∂xg(x,y). (5.7) Any diﬀerential equation of the form f(x,y)dx+g(x,y)dy=0 that satisﬁes (5 .7) is known as an exact equation. An exact equation can be expressed as d F=0,where d Fis a total diﬀerential and F(x,y)=Cis the solution. The function F(x,y) can be obtained by integrating the two equations of (5 .6). For example, the diﬀerential equation dy dx+xy2 2+x2y=0 can be written in the form (2 +x2y)dy+xy2dx=0. Since ∂ ∂x(2 +x2y)=2xy,∂ ∂y(xy2)=2xy are equal, the diﬀerential equation is exact. Therefore, we can ﬁnd the general solution in the form of F(x,y)=C with ∂F(x,y) ∂y=2+ x2y,∂F(x,y) ∂x=xy2. 5.1 First-Order Diﬀerential Equations 207 The ﬁrst equation yields F(x,y)=2y+1 2x2y2+p(x). The second equation requires ∂F(x,y) ∂x=xy2+d dxp(x)=xy2. Therefored dxp(x)=0,p (x)=k. Thus the solution is F(x,y)=2y+1 2x2y2+k=C. Combining the two constants, we can write the solution as 2y+1 2x2y2=C/prime. 5.1.4 Integrating Factors A multiplying factor which will convert a diﬀerential equation that is not exact into an exact one is called an integrating factor. For example, the equation ydx+(x2y3+x)dy= 0 (5.8) is not exact. If, however, we multiply it by ( xy)−2,the resulting equation 1 x2ydx+(y+1 xy2)dy=0 is exact, since ∂ ∂y/parenleftbigg1 x2y/parenrightbigg =−1 x2y2, ∂ ∂x/parenleftbigg y+1 xy2/parenrightbigg =−1 x2y2. Hence by deﬁnition, ( xy)−2is an integrating factor. By the method of exact diﬀerential equation, we have ∂ ∂xF(x,y)=1 x2y, F(x,y)=−1 xy+q(y) 208 5 Ordinary Diﬀerential Equations and ∂ ∂yF(x,y)=1 xy2+d dyq(y)=y+1 xy2, d dyq(y)=y, q (y)=1 2y2. Therefore the solution of the original equation is −1 xy+1 2y2=C. It is sometimes possible to ﬁnd an integrating factor by inspection. For example, one may rearrange (5 .8) into (ydx+xdy)+x2y3dy=0 and recognize ydx+xdy=d (xy).Then it is readily seen that the equation d(xy)+x2y3dy=0 can be solved by multiplying by a factor of ( xy)−2,since it will change the equation to d(xy) (xy)2+ydy=0, which immediately gives the result of −1 xy+1 2y2=C. Theoretically an integrating factor exists for every diﬀerential equation of the form f(x,y)dx+g(x,y)dy=0.Unfortunately no general rule is known to ﬁnd it. For certain special type of diﬀerential equations, integrating factors can be found systematically. We assume that f(x,y)dx+g(x,y)dy=0 is not an exact diﬀerential equation. We wish to ﬁnd an integrating factor µ, so that µf(x,y)dx+µg(x,y)dy=0 is exact. For this equation to be exact, it must satisfy the condition ∂ ∂y(µf)=∂ ∂x(µg) which gives µ/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =∂µ ∂xg−∂µ ∂yf. (5.9) 5.1 First-Order Diﬀerential Equations 209 Now we consider the following possibilities. The integrating factor µis a function of xonly.In this case, (5 .9) becomes 1 g/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =1 µ∂µ ∂x. If the left-hand side of this equation is also a function of xonly 1 g/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =G(x), then clearly dµ µ=G(x)dx, which gives lnµ=/integraldisplay G(x)dx or µ=e/integraltext G(x)dx. For example, the diﬀerential equation (3xy+y2)dx+(x2+xy)dy=0 is not exact. Written in the form of f(x,y)dx+g(x,y)dy=0,we see that ∂f ∂y=∂ ∂y(3xy+y2)=3x+2y, ∂g ∂x=∂ ∂x(x2+xy)=2x+y are not equal. But 1 g/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =3x+2y−(2x+y) x2+xy=x+y x(x+y)=1 x is a function of xonly. Therefore the integrating factor is given by µ=e/integraltext 1 xdx=elnx=x. Multiply the original diﬀerential equation by x,it becomes (3x2y+xy2)dx+(x3+x2y)dy=0. This equation is exact, since ∂ ∂y(3x2y+xy2)=∂ ∂x(x3+x2y). 210 5 Ordinary Diﬀerential Equations Integrating these two equations ∂F ∂x=3x2y+xy2,∂F ∂y=x3+x2y. we ﬁnd F(x,y)=x3y+1 2x2y2. Therefore the solution is x3y+1 2x2y2=C. The integrating factor µis a function of yonly.In this case, (5 .9) becomes 1 f/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =−1 µ∂µ ∂y. If the left-hand side of this equation is also a function of yonly 1 f/parenleftbigg∂f ∂y−∂g ∂x/parenrightbigg =−H(y), then clearly dµ µ=H(y)dy, which gives lnµ=/integraldisplay H(y)dy or µ=e/integraltext H(y)dy. 5.2 First-Order Linear Diﬀerential Equations A special type of ﬁrst-order diﬀerential equation of some importance is of the form dy dx+p(x)y=q(x), (5.10) in which both the dependent variable yand its ﬁrst derivative y/primeare of the ﬁrst degree, and p(x)a n d q(x) are continuous functions of the independent variable x.This type of equation is called linear diﬀerential equation of the ﬁrst-order. In what follows, we will derive a general solution for this equation. First, if q(x)=0,we have dy dx=−p(x)y. 5.2 First-Order Linear Diﬀerential Equations 211 In this case y(x)=e−/integraltextxp(x)dx. For the general case, we introduce a variable coeﬃcient y(x)=f(x)e−/integraltextxp(x)dx. With this trial solution, (5 .10) becomes df(x) dxe−/integraltextxp(x)dx−p(x)f(x)e−/integraltextxp(x)dx+p(x)f(x)e−/integraltextxp(x)dx=q(x), or df(x) dxe−/integraltextxp(x)dx=q(x). Thus f(x)=/integraldisplay q(x)e/integraltextxp(x)dxdx+C. Hence the solution of the ﬁrst-order linear diﬀerential equation (5 .10) is given by y(x)=f(x)e−/integraltextxp(x)dx =e−/integraltext p(x)dx/integraldisplay e/integraltext p(x)dxq(x)dx+Ce−/integraltext p(x)dx. (5.11) To use this formula, it is important to remember to put the diﬀerential equa- tion in the form of y/prime+p(x)y=q(x).In other words, the coeﬃcient of the derivative must be one. This solution enables us to see that µ(x)=e/integraltext p(x)dx is the integrating factor of the equation. In terms of µ(x),(5.11) can be written as µ(x)y=/integraldisplay µ(x)q(x)dx+C, which is a solution of the diﬀerential equation d dx[µ(x)y]=µ(x)q(x). (5.12) Furthermore d dx[µ(x)y]=µ(x)dy dx+/bracketleftbiggd dxµ(x)/bracketrightbigg y, d dxµ(x)=d dxe/integraltext p(x)dx=e/integraltext p(x)dx[p(x)] =µ(x)p(x). 212 5 Ordinary Diﬀerential Equations Hence (5 .12) becomes µ(x)dy dx+µ(x)p(x)y=µ(x)q(x), which clearly shows that µ(x) is a integrating factor of the original equation. Thus, an easier way to make use of the complicated formula of (5 .11) is to write it in terms of the integrating factor y(x)=1 µ(x)/bracketleftbigg/integraldisplay µ(x)q(x)dx+C/bracketrightbigg with µ(x)=e/integraltextxp(x)dx. Example 5.2.1. Find the general solution of the following diﬀerential equation: xdy dx+( 1+ x)y=ex. Solution 5.2.1. This is a linear diﬀerential equation of ﬁrst-order dy dx+1+x xy=ex x. The integrating factor is given by µ(x)=e/integraltext 1+x xdx. Since /integraldisplayx1+x xdx=/integraldisplay/parenleftbigg1 x+1/parenrightbigg dx=l nx+x, µ(x)=elnx+x=xex. It follows that: y=1 xex/bracketleftbigg/integraldisplay xexex xdx+C/bracketrightbigg =1 xex/bracketleftbigg/integraldisplay e2xdx+C/bracketrightbigg =1 xex/bracketleftbigge2x 2+C/bracketrightbigg . Therefore the solution is given by y=ex 2x+Ce−x x. 5.2 First-Order Linear Diﬀerential Equations 213 5.2.1 Bernoulli Equation The type of diﬀerential equations dy dx+p(x)y=q(x)yn is known as Bernoulli equations, named after Swiss mathematician James Bernoulli (1654–1705). This is a nonlinear diﬀerential equation if n/negationslash=0or1. However, it can be transformed into a linear equation by multiplying both sides with a factor (1 −n)y−n (1−n)y−ndy dx+( 1−n)p(x)y1−n=( 1−n)q(x). Since (1−n)y−ndy dx=d dx(y1−n), the last equation can be written as d dx(y1−n)+( 1 −n)p(x)y1−n=( 1−n)q(x), which is a ﬁrst-order linear equation in terms of y1−n.This equation can be solved for y1−n,from which the solution of the original equation can be obtained. Example 5.2.2. Find the solution of dy dx+1 xy=x2y3 with the condition y(1) = 1 . Solution 5.2.2. This a Bernoulli equation of n=3.Multiplying this equation by (1−3)y−3,we have −2y−3dy dx−21 xy−2=−2x2, which can be written as d dxy−2−2 xy−2=−2x2. This equation is ﬁrst-order in y−2and can be solved by multiplying it with an integrating factor µ, µ=e/integraltext (−2 x)dx=e−2l nx=1 x2. 214 5 Ordinary Diﬀerential Equations Thus1 x2y−2=/integraldisplay1 x2(−2x2)dx+C=−2x+C. Atx=1,y=1,therefore 1=−2+C, C =3. Hence the speciﬁc solution of the original nonlinear linear diﬀerential equa- tion is y−2=−2x3+3x2 or written as y(x)=( 3 x2−2x3)−1/2. 5.3 Linear Diﬀerential Equations of Higher Order A great many physical problems can be formulated in terms of linear diﬀeren- tial equations. A second-order diﬀerential equation is called linear if it can be written d2 dx2y(x)+p(x)d dxy(x)+q(x)y(x)=h(x) (5.13) and nonlinear if it cannot be written in this form. To simplify the notation, this equation is also written as y/prime/prime+p(x)y/prime+q(x)y=h(x). The characteristic feature of this equation is that it is linear in the unknown function yand its derivatives. For example: y/prime2=xis not linear because of the term y/prime2.The equation yy/prime= 1 is also not linear because of the product yy/prime.The functions pandqare called coeﬃcients of the equation. Ifh(x) = 0 for all xconsidered, the equation becomes y/prime/prime+p(x)y/prime+q(x)y=0 and is called homogeneous. If h(x)/negationslash=0,it is called nonhomogeneous. Another convenient way of writing a diﬀerential equation is based on the so-called operator notation. The symbol of diﬀerentiationd dxis replaced by D: dy dx=Dy,d2y dx2=D2y, and so on. Therefore (5.13) can be written as D2y+p(x)Dy+q(x)y=h(x), 5.3 Linear Diﬀerential Equations of Higher Order 215 or [D2+p(x)D +q(x)]y=h(x). If we deﬁne f(D) = D2+p(x)D +q(x), then the equation is simply f(D)y=h(x). A fundamental theorem about homogeneous linear diﬀerential equation is the following. If f(D) is second-order, then there are two linearly independent solutions y1andy2.Furthermore, any linear combination of y1andy2is also a solution. This means that if f(D)y1=0,f(D)y2=0, (5.14) then with any two arbitrary constants c1andc2 f(D)(c1y1+c2y2)=0. This is very easy to show, f(D)(c1y1+c2y2)=f(D)c1y1+f(D)c2y2 =c1f(D)y1+c2f(D)y2=0. For the last step we have used (5.14). It is important to remember that this theorem does not hold for nonlinear or nonhomogeneous linear diﬀerential equations. To discuss the general solution of the nonhomogeneous diﬀerential equa- tionf(D)y=h(x),we deﬁne a complementary function ycand a particular solution yp.The complementary function is the solution of the corresponding homogeneous equation, that is f(D)yc=0. If this is an nth order equation, then ycwill contain narbitrary constants. The particular solution is a function when it is substituted into the original nonhomogeneous equation, the result is an identity f(D)yp(x)=h(x). The particular solution can be found by various methods as we shall discuss in later sections. There is no arbitrary constant in the particular solution. The most general solution of the nonhomogeneous diﬀerential equation is the sum of the complementary function and the particular solution y(x)=yc(x)+yp(x). (5.15) 216 5 Ordinary Diﬀerential Equations It is a solution since f(D)[yc(x)+yp(x)] =f(D)yc(x)+f(D)yp(x)=h(x). It is a general solution since the arbitrary constants, necessary for satisfying boundary or initial conditions, are contained in the complementary function. These general statements are also true for ﬁrst-order linear equations. For example, we have found that y=ex 2x+Ce−x x is the general solution of xdy dx+( 1+ x)y=ex. It can be readily veriﬁed that /bracketleftbigg xd dx+( 1+ x)/bracketrightbigge−x x=0, /bracketleftbigg xd dx+( 1+ x)/bracketrightbiggex 2x=ex. Therefore e−x/xis the complementary function and ex/(2x) is the particular solution. 5.4 Homogeneous Linear Diﬀerential Equations with Constant Coeﬃcients We will now focus our attention on linear homogeneous diﬀerential equations with constant coeﬃcients. In searching for the solution of a homogeneous diﬀerential equation such as y/prime/prime−5y/prime+6y=0, (5.16) it is natural to try y=emx where mis a constant, because all its derivatives have the same functional form. Substituting into (5.16) and using the fact that y/prime=memxandy/prime/prime= m2emx,we have emx(m2−5m+6 )=0 . This is the condition to be satisﬁed if y=emxis to be a solution. Since emx can never be zero, it is thus necessary that m2−5m+6=0 . 5.4 Homogeneous Linear Diﬀerential Equations with Constant Coeﬃcients 217 This purely algebraic equation is known as the characteristic or auxiliary equation of the diﬀerential equation. The roots of this equation are m=2 andm=3.Therefore y1=e x p ( 2 x)a n d y2=e x p ( 3 x) are two solutions of (5.16). The general solution is then given by a linear combination of these two functions y=c1e2x+c2e3x. (5.17) In other words, all solutions of (5.16) can be written in this form. For a second-order linear diﬀerential equation, the general solution contains two arbitrary constants c1andc2.These constants can be used to satisfy the initial conditions. For example, suppose it is given that at x=0,y=0a n d y/prime=2, then y(0) = c1+c2=0, y/prime(0) = 2 c1+3c2=2. Thus c1=−2,c2=2.So the speciﬁc solution for the diﬀerential equation together with the given initial conditions is y(x)=−2e2x+2 e3x. To facilitate further discussion, we will repeat this process in the operator notation. If we deﬁne f(D) = D2−5D + 6 , (5.18) then (5.16) can be written as f(D)y=0.Substituting y=emxinto this equation, we obtain f(m)emx=0,where f(m)=m2−5m+6. The characteristic equation f(m) = 0 has two roots; m=2,3.So the solution is given by (5 .17).Although obvious, it is useful to remember that to get the characteristic equation, we need only to change D in the operator function of (5.18) to mand set it to zero. 5.4.1 Characteristic Equation with Distinct Roots Clearly this line of reasoning can be applied to any homogeneous linear diﬀer- ential equation with constant coeﬃcients, regardless of its order. If f(D)y=0 is anth-order diﬀerential equation, then the characteristic equation f(m)=0 is anth-order algebraic equation. It has n roots, m=m1,m2,····mn.If they are all distinct (diﬀerent from each other), exactly nindependent solutions exp(m1x),exp(m2),····exp(mnx) of the diﬀerential equation are so obtained and the general solution is y=c1em1x+c2em2x+····+cnemnx. However, if one or more of the roots are repeated, less than nindependent solutions are obtained in this way. Fortunately, it is not diﬃcult to ﬁnd the missing solutions. 218 5 Ordinary Diﬀerential Equations 5.4.2 Characteristic Equation with Equal Roots To ﬁnd the solutions when two or more roots of the characteristic equation are the same, we ﬁrst consider the following identities: (D−a)xneax=Dxneax−axneax =(nxn−1eax+axneax)−axneax =nxn−1eax. If we apply (D −a) once more to both side of this equation, we have (D−a)2xneax=( D−a)nxn−1eax =n(n−1)xn−2eax. It follows that: (D−a)nxneax=n!eax. Applying (D −a) once more (D−a)n+1xneax=n!(D−a)eax=0. Clearly, if we continue to apply (D −a) to both the sides of the last equation, they will all be equal to zero. Therefore (D−a)lxneax=0 for l > n. This means that exp( ax),xexp(ax),····xn−1exp(ax) are solutions of the diﬀerential equation (D −a)ny=0.In other words, if the roots of the charac- teristic equation are repeated ntimes, and the common root is a,then the general solution of the diﬀerential equation is y=c1eax+c2xeax+····+cnxn−1eax. 5.4.3 Characteristic Equation with Complex Roots If the coeﬃcients of the diﬀerential equation are real and the roots of the characteristic equation have an imaginary part, then from the theory of alge- braic equations, we know that the roots must come in conjugate pairs such as a±ib.So the general solution corresponding to these two roots is y=c1e(a+bi)x+c2e(a−bi)x. (5.19) There are two other very useful equivalent forms of (5 .19).Since e(a±bi)x=eaxe±ibx=eax(cosbx±isinbx), 5.4 Homogeneous Linear Diﬀerential Equations with Constant Coeﬃcients 219 we can write (5.19) as y=eax[c1cosbx+ic1sinbx+c2cosbx−ic2sinbx] =eax[(c1+c2)cosbx+( ic1−ic2)sinbx]. Since c1andc2are arbitrary constants, we can replace c1+c2and ic1−ic2 by two new arbitrary constants AandB.Therefore y=eax(Acosbx+Bsinbx). (5.20) We can write (5 .20) in still another form. Recall Ccos(bx−φ)=Ccosbxcosφ+Csinbxsinφ. If we put Ccosφ=A, C sinφ=B, thenC=(A2+B2)1/2andφ= tan−1(B/A),and (5 .20) becomes y=Ceaxcos(bx−φ). (5.21) Therefore (5.19–5.21) are all equivalent. They all contain two arbitrary constants. One set of constants can be easily transformed into another set. However, there is seldom any need to do this. In solving actual problems we simply use the form that seems best for the problem at hand, and determine the arbitrary constants in that form from the given conditions. We summarize in Table 5.1 the relationships between the roots of the characteristic equation and the solution of the diﬀerential equation. Table 5.1. Relationship between the roots of the characteristic equation and the general solution of the diﬀerential equation my (x) 0 c1 0,0 c1+c2x 0,0,0 c1+c2x+c3x2 ··· ··· ac 1eax a, a c 1eax+c2xeax ··· ··· ±ibc 1cosbx+c2sinbx ±ib,±ib (c1+c2x)c osbx+(c3+c4x)si nbx ··· ··· a±ib eax(c1cosbx+c2sinbx) a±ib, a±ib eax[(c1+c2x)c osbx+(c3+c4x)si nbx] ··· ··· 220 5 Ordinary Diﬀerential Equations Example 5.4.1. Find the general solution of the diﬀerential equation y/prime/prime/prime=0. Solution 5.4.1. We can write the equation as D3y=0. The characteristic equation is m3=0. The three roots are 0 ,0,0.Therefore the general solution is given by y=c1e0x+c2xe0x+c3x2e0x =c1+c2x+c3x2. This seems to be a trivial example. Obviously the result can be obtained by inspection. Here, we have demonstrated that by using the general method, we can ﬁnd all the linear independent terms. Example 5.4.2. Find the general solution of the diﬀerential equation y/prime/prime/prime−6y/prime/prime+9y/prime=0. Solution 5.4.2. We can write the equation as (D3−6D2+9 D ) y=0. The characteristic equation is m3−6m2+9m=m(m2−6m+9 ) =m(m−3)2=0. The three roots are 0 ,3,3.Therefore the general solution is y=c1+c2e3x+c3xe3x. Again, for this third-order diﬀerential equation, the general solution has three arbitrary constants. To determine these constants, we need three conditions. Example 5.4.3. Find the solution of y/prime/prime+9y=0, satisfying the boundary conditions y(π/2) = 1 and y/prime(π/2) = 2 . 5.4 Homogeneous Linear Diﬀerential Equations with Constant Coeﬃcients 221 Solution 5.4.3. We can write the equation as (D2+9 )y=0. The characteristic equation is m2+9=0 . The two roots of this equation are m=±3i.Therefore the general solution a c c o r d i n gt o( 5 .20) is y(x)=Acos 3x+Bsin 3x. We also need y/primeto determine AandB y/prime(x)=−3Asin 3x+3Bcos 3x. The initial conditions require that y/parenleftBigπ 2/parenrightBig =Acos3π 2+Bsin3π 2 =−B=1, y/prime/parenleftBigπ 2/parenrightBig =−3Asin3π 2+3Bcos3π 2 =3A=2. Thus A=2/3,andB=−1.Therefore the solution is y=2 3cos 3x−sin 3x. We will get the same solution if we use either (5 .19) or (5 .21) instead of (5 .20). Example 5.4.4. Find the general solution of y/prime/prime+y/prime+y=0. Solution 5.4.4. The characteristic equation is m2+m+1=0 . The roots of this equation are m=1 2(−1±√ 1−4) =−1 2±i√ 3 2. Therefore the general solution is y=e−x/2/parenleftBigg Acos√ 3 2x+Bsin√ 3 2x/parenrightBigg . 222 5 Ordinary Diﬀerential Equations Example 5.4.5. Find the general solution of the diﬀerential equation (D4+8 D2+ 16)y=0. Solution 5.4.5. The characteristic equation is m4+8m2+1 6=( m2+4 )2=0. The four roots of this equation are m=±2i,±2i.The two indepen- dent solutions associated with the roots ±2i are cos 2 x,sin 2x.The other two independent solutions corresponding to the repeated roots ±2i are xcos 2x, xsin 2x.Therefore the general solution is given by y=Acos 2x+Bsin 2x+x(Ccos 2x+Dsin 2x). 5.5 Nonhomogeneous Linear Diﬀerential Equations with Constant Coeﬃcients 5.5.1 Method of Undetermined Coeﬃcients We will use an example to illustrate that the general solution of nonhomo- geneous diﬀerential equation is given by (5 .15), namely the sum of the com- plementary function and the particular solution. For an equation, such as (D2+5 D+6 ) y=e3x(5.22) it is not diﬃcult to ﬁnd the particular function yp(x).Because of e3xin the right-hand side, we try yp(x)=ce3x. (5.23) Replace ybyyp(x),(5.22) becomes (D2+5 D+6 ) ce3x=e3x. Since (D2+5 D+6 ) ce3x=( 9+5 ×3+6 ) ce3x=3 0ce3x, thusc=1/30.The function yp(x)o f( 5 .23) with c=1/30 is therefore the particular solution ypof the nonhomogeneous diﬀerential equation, that is yp=1 30e3x. Other than this particular solution, the nonhomogeneous equation has many more solutions. In fact the general solution of (5 .22) should also have two 5.5 Nonhomogeneous Linear Diﬀerential Equations 223 arbitrary constants. To ﬁnd the general solution, let us ﬁrst solve the corres- ponding homogeneous diﬀerential equation (D2+5 D+6 ) yc=0. The general solution of this homogeneous equation is known as the comple- mentary function ycof the nonhomogeneous equation. Following the rules of solving homogeneous equation, we ﬁnd yc=c1e−2x+c2e−3x. Thus y=yc+yp =c1e−2x+c2e−3x+1 30e3x(5.24) is the general solution of the nonhomogeneous equation (5 .22). First it is certainly a solution, since (D2+ 5D + 6)( yc+yp)=( D2+5 D+6 ) yc+( D2+5 D+6 ) yp =0+e3x=e3x. Furthermore, it has two arbitrary constants c1andc2.This means that every possible solution of the nonhomogeneous equation (5 .22) can be obtained by assigning suitable values to the arbitrary constants c1andc2in (5.24). Clearly this principle is not limited to this particular problem. As we have already learnt how to solve homogeneous equations, we shall now discuss some systematical ways of ﬁnding particular solutions. Let us consider another nonhomogeneous equation (D2−5D + 6) y=e3x. (5.25) Since this equation is only slightly diﬀerent from (5 .22),to ﬁnd the particular solution ypof this equation, we may again try yp=ce3x. Putting it into (5 .25),we ﬁnd (D2−5D + 6) ce3x=( 9−15 + 6) ce3x=0. Obviously no value of ccan make it equal to e3x.Clearly, we need a more general method. The idea is that if we can transform the nonhomogeneous equation into a homogeneous equation, then we know what to do. First we note that (D−3)e3x=0. 224 5 Ordinary Diﬀerential Equations Applying (D −3) to both side of (5 .25),we have (D−3)(D2−5D + 6) y=( D−3)e3x=0. (5.26) Then we note that the general solution yc+ypof the nonhomogeneous equation (5.25) must also satisfy the newly formed homogeneous equation (5 .26), since (D−3)(D2−5D + 6)( yc+yp)=( D −3)(0 + e3x)=0, (5.27) w h e r ew eh a v eu s e d (D2−5D + 6) yc= 0 (5.28) and (D2−5D + 6) yp=e3x. (5.29) This means we can obtain the general solution yc+ypof the original nonhomo- geneous equation (5 .25) by assigning certain speciﬁc values to some constants in the general solution of the newly formed homogeneous equation (5 .26).For example, since the roots of the characteristic equation ( m−3)(m2−5m+6) = 0 of the newly formed homogeneous equation are m=2,3,3, the general solution is y=c1e2x+c2e3x+c3xe3x. (5.30) But the complementary function given by (5 .28) is yc=c1e2x+c2e3x, we see that the particular solution ypcan be obtained by assigning an appro- priate value to c3,since the general solution of the original nonhomogeneous equation yc+ypis a special solution of the newly formed homogeneous equa- tion. To determine c3,we substitute c3xe3xinto (5 .29) (D2−5D + 6) c3xe3x=e3x. Since (D2−5D + 6) c3xe3x=c3[D(e3x+3xe3x)−5(e3x+3xe3x)+6xe3x] =c3e3x, clearly c3= 1. Thus the particular solution is yp=xe3x. Hence the general solution of the nonhomogeneous diﬀerential equation (5 .25) is 5.5 Nonhomogeneous Linear Diﬀerential Equations 225 y=c1e2x+c2e3x+xe3x. Now we summarize the general procedure of solving a nonhomogeneous diﬀ- erential equation f(D)y(x)=h(x). 1. The general solution is y=yc+yp, where ycis the complementary function and ypis the particular solu- tion, and f(D)yc=0,f(D)yp=h(x). 2. Let the roots of f(m)=0b e m=m1,m2...... Complementary function ycis given by the linear combination of all the linear independent functions arising from m. 3. To ﬁnd yp,we ﬁrst ﬁnd another equation such that g(D)h(x)=0. Applying g(D) to both sides of f(D)y(x)=h(x),we have g(D)f(D)y=0. 4. The general solution yc+ypof the original nonhomogeneous diﬀeren- tial equation f(D)y(x)=h(x) is a special solution of the newly formed homogeneous diﬀerential equation g(D)f(D)y=0,since g(D)f(D)(yc+yp)=g(D)[f(D)yc+f(D)yp] =g(D)[0 + h(x)] = 0 . 5. The general solution of g(D)f(D)y= 0 is associated with the roots of the characteristic equation g(m)f(m)=0. Since the roots of f(m) = 0 lead to yc,the roots of g(m)=0m u s tb e associated with yp. 6. Let the roots of g(m)=0b e m=m/prime 1,m/prime 2.... . If there is no duplication between m/prime 1,m/prime 2,···andm1,m2,···.... ., then the particular solution ypis given by the linear combination of all the linear independent functions arising from m/prime 1,m/prime 2,···. If there is duplication between m/prime 1,m/prime 2,···andm1,m2,···,then the func- tions arising from m/prime 1,m/prime 2,···must be multiplied by the lowest positive integer powers of xwhich will eliminate all such duplications. 226 5 Ordinary Diﬀerential Equations 7. Determine the arbitrary constants in the functions arising from m/prime 1,m/prime 2··· from f(D)yp=h(x). This procedure may seem to be complicated. Once understood, the imple- mentation is actually relative simple. This we illustrate with following examples. Example 5.5.1. Find the general solution of (D2+5 D+6 ) y=3 e−2x+e3x. Solution 5.5.1. The characteristic equation is f(m)=m2+5m+6=0 its roots are m=−2,−3.(m1=−2,m2=−3). Therefore yc=c1e−2x+c2e−3x. For g(D)(3e−2x+e3x)=0, the roots of g(m)=0 must be m=−2,3(m/prime 1=−2,m/prime 2=3 ). Since m/prime 1repeats m1,the term arising from m/prime 1=−2 must be multiplied byx.Therefore yp=c3xe−2x+c4e3x. Since (D2+ 5D + 6)( c3xe−2x+c4e3x)=c3e−2x+3 0c4e3x, (D2+5 D+6 ) yp=3 e−2x+e3x, we have c3=3 ; c4=1 30. Thus y=c1e−2x+c2e−3x+3xe−2x+1 30e3x. 5.5 Nonhomogeneous Linear Diﬀerential Equations 227 Example 5.5.2. Find the general solution of (D2+1 )y=x2. Solution 5.5.2. The roots of the characteristic equation f(m)=m2+1=0 are m=±i, therefore yc=c1cosx+c2sinx. Forg(D)x2=0,we can regard x2asc1+c2x+c3x2withc1=0,c2=0,c3=1. The roots of g(m)=0,as seen in Table 5.1, are m=0,0,0. Thus yp=a+bx+cx2. Substituting into the original equation (D2+ 1)(a+bx+cx2)=x2, we have 2c+a+bx+cx2=x2. Thus 2c+a=0,b=0,c=1. It follows that a=−2a n d yp=−2+x2. Finally y=c1cosx+c2sinx−2+x2. Example 5.5.3. Find the general solution of (D2+4 D+5 ) y=3 e−2x. Solution 5.5.3. The roots of f(m)=m2+4m+5=0a r e m=−2±i. Therefore yc=e−2x(c1cosx+c2sinx). Forg(D)3e−2x=0,g(D) = D + 2 .Obviously, the root of g(m)=0i s m=−2. 228 5 Ordinary Diﬀerential Equations Thus yp=Ae−2x. With (D2+4 D+5 ) Ae−2x=3 e−2x we have A=3. The general solution is therefore given by y=e−2x(c1cosx+c2sinx)+3 e−2x. Example 5.5.4. Find the general solution of (D2−2D + 1) y=xex−ex. Solution 5.5.4. The roots of m2−2m+1=0a r e m=1,1(m1=1,m2=1 ) Hence yc=c1ex+c2xex. The characteristic equation of the diﬀerential equation for which xex−exis the solution g(D)(xex−ex)=0 is of course g(m)=0.We can regard xex−exasaex+bxex(with a=−1, b= 1). We see from Table 5.1 that the roots of g(m)=0a r e m=1,1(m/prime 1=1,m/prime 2=1 ). Therefore yp=Ax2ex+Bx3ex. With (D2−2D + 1)( Ax2ex+Bx3ex)=xex−ex we ﬁnd A=−1 2,B =1 6. The general solution is therefore given by y=c1ex+c2xex−1 2x2ex+1 6x3ex. 5.5 Nonhomogeneous Linear Diﬀerential Equations 229 Example 5.5.5. Find the general solution of (D2+1 )y=s i nx. Solution 5.5.5. The roots of m2+1=0a r e m=±i. Therefore yc=c1cosx+c2sinx. To ﬁnd g(D)sin x=0, we can regard sin xasacosx+bsinxwitha=0a n d b=1.From Table 5.1, we see that the roots of g(m)=0a r e m=±i(m/prime=±i). Thus yp=Axcosx+Bxsinx. With (D2+ 1)(Axcosx+Bxsinx)=s i n x we ﬁnd A=−1 2,B =0. Therefore y=c1cosx+c2sinx−1 2xcosx. 5.5.2 Use of Complex Exponentials In applied problems, the function h(x) is very often a sine or a cosine rep- resenting alternating voltage in an electric circuit or a periodic force in a vibrating system. The particular solution ypcan be found more eﬃciently by replacing sine or cosine by the complex exponential form. In the last example, we can replace sin xby eixand solving the equation (D2+1 )Y=eix. (5.31) The solution Ywill also be complex. Y=YR+iYI.Since eix=c o s x+is i n x, the equation is equivalent to (D2+1 )YR=c o s x, (D2+1 )YI=s i nx. Since the second equation is exactly the same as the original equation, we see that to ﬁnd yp,w ec a ns o l v e( 5 .31) for Yand take its imaginary part. Following the procedure of the last section, we assume Y=Axeix, 230 5 Ordinary Diﬀerential Equations so DY=Aeix+iAxeix, D2Y=2 iAeix−Axeix, (D2+1 )Y=2 iAeix−Axeix+Axeix=2 iAeix=eix. Thus A=1 2i. Taking the imaginary part of Y=1 2ixeix=−1 2ix(cosx+is i n x), we have yp=−1 2xcosx, which is, of course, the same as obtained in the last example. 5.5.3 Euler–Cauchy Diﬀerential Equations An equation of the form anxndny dxn+an−1xn−1dn−1y dxn−1+···a1xdy dx+a0y=h(x), (5.32) where the aiare constants, is called Euler, or Cauchy, or Euler–Cauchy dif- ferential equation. By a change of variable, it can be transformed into an equation with constant coeﬃcients which can then be solved. If we set x=ez,z=l nx, thendz dx=1 x=e−z. With the notation D =d dz,we can write dy dx=dy dzdz dx=e−zDy, d2y dx2=d dxdy dx=d dx(e−zDy)=d dz(e−zDy)dz dx =(−e−zDy+e−zD2y)e−z=e−2zD(D−1)y, d3y dx3=d dz/bracketleftbig e−2zD(D−1)y/bracketrightbigdz dx =/bracketleftbig −2e−2zD(D−1)y+e−2zD2(D−1)y/bracketrightbig e−z =e−3zD(D−1)(D−2)y. 5.5 Nonhomogeneous Linear Diﬀerential Equations 231 Clearly dny dxn=e−nzD(D−1)(D−2)···(D−n+1 )y. (5.33) Substituting (5 .33) into (5 .32) and using xn=enz,we have a diﬀerential equation with constants coeﬃcients, anD(D−1)(D−2)···(D−n+1 )y+···a1Dy+a0y=h(ez). If the solution of this equation is denoted y=F(z), then the solution of the original equation is given by y=F(lnx). The following example will make this procedure clear. Example 5.5.6. Find the general solution of x2d2y dx2+xdy dx−y=xlnx. Solution 5.5.6. With x=ez,this equation becomes [D(D−1) + D −1]y=zez. The complementary function comes from m(m−1) +m−1=m2−1=0,m =1,−1, which gives yc=c1ez+c2e−z. Forg(D)zez=0,g(D) must be (D −1)2.The characteristic equation is then (m/prime−1)2=0,m/prime=1,1. Therefore the particular solution is of the form yp=c3zez+c4z2ez. Substituting it back into the diﬀerential equation [D(D−1) + D −1] (c3zez+c4z2ez)=zez, we ﬁnd c3=−1 4,c 4=1 4. 232 5 Ordinary Diﬀerential Equations Therefore y=c1ez+c2e−z−1 4zez+1 4z2ez. For the general solution of the original equation, we must change zback to x. With z=l nx y(x)=c1x+c21 x−1 4xlnx+1 4x(lnx)2, which can be readily veriﬁed that this is indeed the general solution with two arbitrary constants. For a homogeneous Euler–Cauchy equation, the following procedure is perhaps simpler. For example, to solve the equation x2d2y dx2+xdy dx−y=0, we can simply start with the solution y(x)=xm. So x2m(m−1)xm−2+xmxm−1−xm=0 or [m(m−1) +m−1]xm=0. Thus m(m−1) +m−1=m2−1=0, m=1,m =−1. It follows that: y(x)=c1x+c21 x. 5.5.4 Variation of Parameters The method of undetermined coeﬃcients is simple and has important phys- ical applications, but it applies only to constant coeﬃcient equations with special forms of the nonhomogeneous term h(x).In this section, we discuss the method of variation of parameters, which is more general. It applies to equations (D2+p(x)D +q(x))y=h(x), (5.34) where p, q,andhare continuous functions of xin some interval. Let yc=c1y1(x)+c2y2(x) be the solution of the corresponding homogeneous equation 5.5 Nonhomogeneous Linear Diﬀerential Equations 233 (D2+p(x)D +q(x))yc=0. The method of variation of parameters involves replacing the parameters c1 andc2by functions uandvto be determined so that yp=u(x)y1(x)+v(x)y2(x) is the particular solution of (5 .34).Now this expression contains two unknown functions uandv,but the requirement that ypsatisﬁes (5 .34) imposes only one condition on uandv.Therefore, we are free to impose a second arbitrary condition without loss of generality. Further calculation will show that it is convenient to require u/primey1+v/primey2=0. (5.35) Now Dyp=u/primey1+uy/prime 1+v/primey2+vy/prime 2. With the imposed condition (5 .35), we are left with y/prime p=uy/prime 1+vy/prime 2. It follows that: D2yp=u/primey/prime 1+uy/prime/prime 1+v/primey/prime 2+vy/prime/prime 2. Substituting them back into the equation (D2+p(x)D +q(x))yp=h(x) and collecting terms, we have u(y/prime/prime 1+py/prime 1+qy1)+v(y/prime/prime 2+py/prime 2+qy2)+u/primey/prime 1+v/primey/prime 2=h. Since y1andy2satisfy the homogeneous equation, the quantities in the paren- thesis are equal to zero. Thus u/primey/prime 1+v/primey/prime 2=h. This equation together with the imposed condition (5 .35) can be solved for u/primeandv/prime u/prime=−hy2 y1y/prime 2−y2y/prime 1,v/prime=hy1 y1y/prime 2−y2y/prime 1. (5.36) The quantity in the denominator, known as the Wronskian Wofy1andy2, W=/vextendsingle/vextendsingle/vextendsingle/vextendsingley1y2 y/prime 1y/prime 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle=y1y/prime 2−y2y/prime 1 234 5 Ordinary Diﬀerential Equations will not be equal to zero as long as y1andy2are linearly independent. Inte- gration of (5 .36) will enable us to determine uandv u=−/integraldisplayhy2 Wdx, v =/integraldisplayhy1 Wdx. The particular solution is then yp=−y1/integraldisplayhy2 Wdx+y2/integraldisplayhy1 Wdx. Example 5.5.7. Use the variation of parameters method to ﬁnd the general solution of (D2+4 D+4 ) y=3xe−2x Solution 5.5.7. This equation can be solved easily by the method of unde- termined coeﬃcients. But we want to show that the solution can also be found by the variation of parameters method. Since m2+4m+4=( m+2 )2=0,m =−2,−2 the two independent solutions of the homogeneous equation are y1=e−2x,y 2=xe−2x. The Wronskian of y1andy2is W=/vextendsingle/vextendsingle/vextendsingle/vextendsinglee−2x −2e−2xxe−2x e−2x−2xe−2x/vextendsingle/vextendsingle/vextendsingle/vextendsingle=e−4x(1−2x)+2xe−4x=e−4x andu/primeandv/primeare given by u/prime=−3xe−2xy2 W=−3xe−2xxe−2x e−4x=−3x2, v/prime=3xe−2xy1 W=3xe−2xe−2x e−4x=3x. It follows that: u=−/integraldisplay 3x2dx=−x3+c1,v=/integraldisplay 3xdx=3 2x2+c2. Therefore y=(−x3+c1)e−2x+/parenleftbigg3 2x2+c2/parenrightbigg xe−2x =1 2x3e−2x+c1e−2x+c2xe−2x. This is the general solution. It is seen that this solution includes the comple- mentary function yc=c1e−2x+c2xe−2x 5.6 Mechanical Vibrations 235 and the particular solution yp=1 2x3e−2x. Example 5.5.8. Find the general solution of (D2+2 D+1 ) y=h(x),h(x)=2 x2e−x. Solution 5.5.8. This equation cannot be solved by the method of undeter- mined coeﬃcients, therefore we seek the solution by variation of parameters. The complementary function is obtained from m2+2m+1=( m+1 )2=0,m =−1,−1 to be yc=c1y1+c2y2=c1e−x+c2xe−x. Let yp=uy1+vy2. From the Wronskian of y1andy2 W=/vextendsingle/vextendsingle/vextendsingle/vextendsinglee−xxe−x −e−xe−x−xe−x/vextendsingle/vextendsingle/vextendsingle/vextendsingle=e−2x(1−x)+xe−2x=e−2x, we have u=−/integraldisplayy2h W=−/integraldisplayxe−x2x−2e−x e−2xdx=−2lnx, v=/integraldisplayy1h W=/integraldisplaye−x2x−2e−x e−2xdx=−2 x. Thus yp=−2e−xlnx−2e−x. Therefore the general solution is y=c1e−x+c2xe−x−2e−xlnx. Note that we have dropped the term −2e−x,since it is absorbed in c1e−x. 5.6 Mechanical Vibrations There are countless applications of diﬀerential equations in engineering and physical sciences. As illustrative examples, we will ﬁrst discuss mechanical vibrations. Any motion that repeats itself after certain time interval is called vibration or oscillation. A simple model is the spring–mass system shown in Fig. 5.1. The block of mass mis constrained to move on a frictionless table and 236 5 Ordinary Diﬀerential Equations k m xSmooth x = 0 + xm −xm Fig. 5.1. A simple harmonic oscillator. The block moves on a frictionless table. The equilibrium position of the spring is at x=0.Att=0,the block is released from rest at x=xm is fastened to a spring with spring constant k. The block is pulled a distance xmfrom its equilibrium position at x= 0 and released from rest. We want to know the subsequent motion of the block. This system is simple enough for us to demonstrate the following steps in mathematical physics: – Formulate the physical problem in terms of mathematical language, usu- ally in the form of a diﬀerential equation. – Solve the mathematical equation. – Understand the physical meaning of the mathematical solution. 5.6.1 Free Vibration The ﬁrst step is to observe that the only horizontal force on the block is coming from the spring. According to the Hooke’s law, the force is proportional to the displacement but opposite in sign, that is F=−kx. The motion of the block is governed by the Newton’s dynamic equation F=ma. Since the acceleration is equal to the second derivative of the displacement a=d2x dt2, therefore md2x dt2=−kx. This is a second-order linear homogeneous diﬀerential equation. Since the block is released from rest at x=xm,the velocity of the block, which is the ﬁrst derivative of the displacement, is zero at t= 0. Therefore the initial conditions are x(0) = xm,v(0) =dx dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0. 5.6 Mechanical Vibrations 237 With the diﬀerential equation and the initial conditions, the mathematical problem is uniquely deﬁned. The second step is to solve this equation. Since the coeﬃcients are con- stants, the solution of the diﬀerential equation is of the exponential form, x=e x p ( αt) with αdetermined by the characteristic equation mα2=−k. Clearly the roots of this equation are α=±/radicalbigg −k m=±iω0,ω 0=/radicalbigg k m. Thus the general solution of the diﬀerential equation is given by x(t)=Aeiω0t+Be−iω0t. The initial conditions requires that x(0) = A+B=xm, dx dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=iω0A−iω0B=0. Therefore A=B=1 2xm,and x(t)=xmcosω0t. The third step is to interpret this solution. The cosine function varies between 1 and −1, it repeats itself when its argument is increased by 2 π. Therefore the block oscillates between xmand−xm.The period Tof the oscillation is deﬁned as the time required for the motion to repeat itself, this means x(t+T)=x(t). Thus cos(ω0t+ω0T)=c o s ( ω0t). Clearly ω0T=2π. Therefore the period is given by T=2π ω0=2π/radicalbiggm k. The frequency fis deﬁned as the number of oscillations in one second, that is f=1 T=ω0 2π=1 2π/radicalbigg k m. 238 5 Ordinary Diﬀerential Equations Thus the block oscillates with a frequency that is prescribed by kandm. Since ω0=2πf, ω 0is called angular frequency. Often ω0is referred simply as the natural frequency with the understanding that it is actually the angular frequency. From the solution of the diﬀerential equation, we can derive all attributes of the motion, such as the velocity of the block at any given time. This type of periodic motion is called simple harmonic motion. The mass–spring system is known as a harmonic oscillator. 5.6.2 Free Vibration with Viscous Damping In practical systems, the amplitude of the oscillation gradually decreases due to friction. This is known as damping. For example, if the system is vibrating in a ﬂuid medium, such as air, water, oil, the resisting force oﬀered by the viscosity of the ﬂuid is generally proportional to the velocity of the vibrating body. Therefore with viscous damping, there is an additional force Fv=−cdx dt where cis the coeﬃcient of viscous damping and the negative sign indicates that the damping force is opposite to the direction of velocity. Thus the equa- tion of motion of the mass–spring system becomes md2x dt2=−kx−cdx dt or md2x dt2+cdx dt+kx=0. This equation can be written in the form d2x dt2+2βdx dt+ω2 0x=0, where β=c/2m, ω2 0=k/m. With x=e x p ( αt),αmust satisfy the equation α2+2βα+ω2 0=0. The roots of this equation are α1=−β+/radicalBig β2−ω2 0,α 2=−β−/radicalBig β2−ω2 0. The solution is therefore given by x(t)=A1eα1t+A2eα2t. (5.37) Depending on the strength of damping, this solution takes the following three forms. 5.6 Mechanical Vibrations 239 Over damping. Ifβ2>ω2 0,then the values of α1andα2are both negative. Thus both terms in xexponentially go to zero as t→∞.In this case, the damping force represented by βoverpowers the restoring force represented by ω0and hence prevents oscillation. The system is called overdamped. Critical damping. In this case β2=ω2 0,the characteristic equation has a double root at α=−βtwice. Hence the solution is of the form x(t)=(A+Bt)e−βt. Since β>0,both e−βtandte−βtgo to zero as t→∞.The motion dies out with time and is not qualitatively diﬀerent from the overdamped motion. In this case the damping force is just as strong as the restoring force, therefore the system is called critically damped. Under damping. Ifβ2<ω2 0,then the roots of the characteristic equation are complex α1,α2=−β±iω, where ω=/radicalBig ω2 0−β2. Therefore the solution x=e−βt(Aeiωt+Be−iωt) can be written in the form of x(t)=Ce−βtcos(ωt+ϕ). Because of the cosine term in the solution, the motion is oscillatory. Since the maximum value of cosine is one, the displacement xmust lie between the curves x(t)=±Ce−βt.Hence it resembles a cosine curve with decreasing amplitude. In this case, the damping force represented by βis weaker than the restoring force represented by ωand thus cannot prevent oscillation. For this reason, the system is called under damped. These three cases are illustrated in the following example with speciﬁc parameters. Example 5.6.1. The displacement x(t) of a damped harmonic oscillator satis- ﬁes the equation d2x dt2+2βdx dt+ω2 0x=0. Let the initial conditions be x(0) = x0;v(0) =dx dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0. Findxas a function of time t,ifω0=4,and (a)β=5,(b)β=4,(c)β=1. Show a sketch of the solutions of these three cases. 240 5 Ordinary Diﬀerential Equations Solution 5.6.1. (a) Since β=5a n d ω0=4,the motion is overdamped. The roots of the characteristic equation are α1,α2=−5±√ 25−16 =−2,−8. Thus x(t)=A1e−2t+A2e−8t. The initial conditions require A1andA2to satisfy A1+A2=x0,−2A1−8A2=0. Therefore x(t)=x0/parenleftbigg4 3e−2t−1 3e−8t/parenrightbigg . The graph of this function is shown as the dotted line in Fig. 5.2. (b) Since β=4a n d ω0=4,the motion is critically damped. The roots of the characteristic equation are −βtwice α1,α2=−β=−4. Thus x(t)=(A+Bt)e−4t. From the initial conditions, we ﬁnd A=x0,B =4x0. Therefore x(t)=x0(1 + 4 t)e−4t. The graph of this function is shown as the line of open circles in Fig. 5.2. (c) For β=1a n d ω0=4,the motion is under damped. The solution can be written as tx0 4 3 2 101 0.5 0 −0.5 −1 Fig. 5.2. Free vibrations with viscous damping. Initially the block is at x0and released from rest. The dotted line is for the over damped motion, the line of open circles is for the critically damped motion, the solid line is for the under damped motion. The damped amplitude is shown as dashed lines 5.6 Mechanical Vibrations 241 x(t)=Ce−βtcos(ωt−φ), where ω=/radicalBig ω2 0−β2=√ 15. From the initial conditions x(0) = Ccos(−φ)=x0, dx dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=−βCcos(−φ)−ωCsin(−φ)=0, we ﬁnd tanφ=β ω=1√ 15,C =1 cosφx0. Since cos φ= (1 + tan2φ)−1/2, C=/radicalbig ω2+β2 ωx0=4√ 15x0. Therefore x(t)=4√ 15x0e−tcos/parenleftbigg√ 15t−tan−11√ 15/parenrightbigg . The graph of this function is shown as the solid line in Fig. 5.2. The damped amplitude Ce−βtis also shown as the dashed line. In the over damped and critically damped cases, if there is a large negative initial velocity, it is possible for the block to overshoot the equilibrium posi- tion. In that case, it will come to a temporarily stop in a negative xposition. After that, it will return to x= 0 in a monotonically decreasing way. There- fore even if it overshoots the equilibrium position, it can do that only once. Hence the motion is not oscillatory. The under damped motion is oscillatory although its amplitude is app- roaching zero as time goes to ∞.The damped frequency/radicalBig ω2 0−β2is always less than the natural frequency ω0. 5.6.3 Free Vibration with Coulomb Damping From the ﬁrst course of physics, we all learned that the friction force of a block sliding on a plane is proportional to the normal force acting on the plane of contact. This friction force acts in a direction opposite to the direction of velocity and is given by Fc=µN where Nis the normal force and µis the coeﬃcient of friction. When the motion is damped by this friction force, it is known as Coulomb damp- ing. Charles Augustin Coulomb (1736–1806) ﬁrst proposed this relationship, 242 5 Ordinary Diﬀerential Equations but he is much better known for his law of electrostatic force. His name is also remembered through the unit of electric charge. Coulomb damping is also known as constant damping, since the magnitude of the damping force is independent of the displacement and velocity. However, the sign of the friction force changes with the direction of the velocity, and we need to consider the motion in two directions separately. When the block moves from right to left, the friction force is pointing toward the right and has a positive sign. With this friction force, the equation of motion is given by md2x dt2=−kx+µN. This equation has a constant nonhomogeneous term. The solution is x(t)=Acosω0t+Bsinω0t+µN k, (5.38) where ω0=/radicalbig k/m, which is the same as the angular frequency of the undamped oscillator. When the block is moving from left to right, the friction force is pointing toward the left and has a negative sign, and the equation of motion becomes md2x dt2=−kx−µN. The solution of this equation is x(t)=Ccosω0t+Dsinω0t−µN k. (5.39) The constants A,B,C,D are determined by the initial conditions. For example, if the block is released from rest at a distance x0to the right of the equilibrium position, then x(0) = x0and the velocity v, which is the ﬁrst derivative of x,att= 0 is zero. In this case, the motion starts from right to left. Using (5 .38),we have v(t)=dx dt=−ωAsinω0t+ωBcosω0t. Thus the initial conditions are x(0) = A+µN k=x0,v(0) = ω0B=0. Therefore A=x0−µN/k, B =0,and x(t)=(x0−µN k)cosω0t+µN k. This equation represents a simple harmonic motion with the equilibrium position shifted from zero to µN/k. However, this equation is valid only for 5.6 Mechanical Vibrations 243 the ﬁrst half of the ﬁrst cycle. When t=π/ω0,the velocity of the block is equal to zero and the block is at its extreme left position x1which is x1=/parenleftbigg x0−µN k/parenrightbigg cosω0π ω0+µN k=−x0+2µN k. In the next half cycle, the block moves from left to right, so we have to use (5 .39).To determine CandD,we use the fact x(t=π/ω0)=x1and v(t=π/ω0)=0.With these conditions, we have x/parenleftbigg t=π ω0/parenrightbigg =Ccosω0π ω0+Dsinω0π ω0−µN k=−C−µN k=x1, v/parenleftbigg t=π ω0/parenrightbigg =−ω0Csinω0π ω0+ω0Dcosω0π ω0=−ω0D=0. Therefore C=−x1−µN k=x0−3µN k,D=0,and (5 .39) becomes x(t)=/parenleftbigg x0−3µN k/parenrightbigg cosω0t−µN k. This is also a simple harmonic motion with the equilibrium position shifted to−µN/k. This equation is valid for π/ω0≤t≤2π/ω0.At the end of this half cycle, the velocity is again equal to zero v/parenleftbigg t=2π ω0/parenrightbigg =−ω0/parenleftbigg x0−3µN k/parenrightbigg sinω02π ω0=0 and the block is at x2which is x2=/parenleftbigg x0−3µN k/parenrightbigg cosω02π ω0−µN k=x0−4µN k. These become the initial conditions for the third half cycle, and the procedure can be continued until the motion stops. The displacement xas a function of timetof this motion is shown in Fig. 5.3. It is to be noted that the frequency of a Coulomb damped vibration is the same as that of the free vibration without damping. This is to be contrasted with the viscous damping. Furthermore, if xnis a local maximum, then xn−xn−2=−4µN k. This means that in each successive cycle, the amplitude of the motion is reduced by 4µN/k in a time interval of 2 π/ω0.Therefore the maxima of the oscillation all fall on a straight line. The slope of this straight enveloping line is −4µN/k 2π/ω0=−2µNω 0 πk. 244 5 Ordinary Diﬀerential Equations x (t) t4µN kx0 − 2µN kx0 −x0 −µN kµN k 2π ω04π ω0 Fig. 5.3. The displacement xof the block as a function of time tin the mass–spring system with Coulomb damping. Initially the block is at x=x0a n di sr e l e a s e df r o m rest Similarly, all local minima must fall on a straight line with a slope of 2µNω 0/πk.These characteristics are also shown in Fig. 5.3. The motion stops when the velocity becomes zero and the restoring force kxof the spring is equal or less than the friction force µN.Thus the number of half cycles n0that elapse before the motion ceases can be found from the condition k/parenleftbigg x0−n02µN k/parenrightbigg ≤µN. Ifx0≤µN/k, the motion will not even start. For µN/k < x 0<2µN/k, the block will stop before it reaches the equilibrium position. The ﬁnal position of the block is usually diﬀerent from the equilibrium position and represents a permanent displacement. 5.6.4 Forced Vibration without Damping A dynamic system is often subjected to some type of external force. In this section, we shall consider the response of a spring-mass system under the external force of the form F0cosωt.First suppose there is no damping, then the equation of motion is given by md2x dt2+kx=F0cosωt or d2x dt2+ω2 0x=F0 mcosωt, 5.6 Mechanical Vibrations 245 where ω0=(k/m)1/2is the natural frequency of the system. The general solution is the sum of the complementary function xcand the particular solution xp.The complementary function satisﬁes the homogeneous equation d2xc dt2+ω2 0xc=0 and is given by xc(t)=c1cosω0t+c2sinω0t. Since the nonhomogeneous term has a frequency ω,the particular solution takes the form xp(t)=Acosωt+Bsinωt. Substituting it into the equation d2xp dt2+ω2 0xp=F0 mcosωt we ﬁnd B=0,andA=F0/[m(ω2 0−ω2)].Thus xp(t)=F0 m(ω2 0−ω2)cosωt and the general solution, given by xc+xp x(t)=c1cosω0t+c2sinω0t+F0 m(ω2 0−ω2)cosωt is the sum of two periodic motions of diﬀerent frequencies. Beats. Suppose the initial conditions are x(0) = 0 and v(0) = 0 ,thenc1and c2are found to be c1=−F0 m(ω2 0−ω2),c2=0. Thus the solution is given by x(t)=F0 m(ω2 0−ω2)(cosωt−cosω0t) =2F0 m(ω2 0−ω2)sinω0−ω 2tsinω0+ω 2t. Let the forcing frequency ωbe slightly less than the natural frequency ω0so thatω0−ω=2/epsilon1,where /epsilon1is a small positive quantity. Then ω0+ω≈2ωand the solution becomes x(t)≈F0 2mω/epsilon1sin/epsilon1tsinωt. Since /epsilon1is small, the function sin /epsilon1tvaries slowly. Thus the factor F0 2mω/epsilon1sin/epsilon1t 246 5 Ordinary Diﬀerential Equations 2π ω F0 2πtx(t) Fig. 5.4. Beats produced by the sum of two waves with approximately the same frequencies can be regarded as the variable amplitude of the vibration whose period is 2π/ω.The oscillation of the amplitude has a large period of 2 π//epsilon1,which is called the period of beats. This kind of motion is shown in Fig. 5.4. Resonance. In the case that the frequency of the forcing function is the same as the natural frequency of the system, that is ω=ω0,then the particular solution takes the form xp(t)=Atcosω0t+Btsinω0t. Substituting it into the original nonhomogeneous equation, we ﬁnd A= 0,B=F0/(2mω0).Thus the general solution is given by x(t)=c1cosω0t+c2sinω0t+F0 2mω0tsinω0t. Because of the presence of the term tsinω0t,the motion will become unboun- ded as t→∞.This is known as resonance. The phenomenon of resonance is characterized by xpwhich is shown in Fig. 5.5. If there is damping, the t xp(t)F0t 2mw 0 2p w 0 Fig. 5.5. Resonance without damping. When the forcing frequency coincide with the natural frequency, the motion will become unbounded 5.6 Mechanical Vibrations 247 motion will remain bounded. However, there may still be a large response if the damping is small and ωis close to ω0. 5.6.5 Forced Vibration with Viscous Damping With viscous damping, the equation of motion of the spring–mass system under a harmonic forcing function is given by md2x dt2+cdx dt+kx=F0cosωt. (5.40) The solution of this equation is again the sum of the complementary func- tion and the particular solution. The complementary function satisﬁes the homogeneous equation md2xc dt2+cdxc dt+kxc=0, which represents free vibrations with damping. As discussed earlier, the free vibration dies out with time under all possible initial conditions. This part of the solution is called transient. The rate at which the transient motion decays depends on the system parameters m,k,c. The general solution of the equation eventually reduces to the particular solution which represents the steady-state vibration. The particular solution is expected to have the same frequency as the forcing function, we can write the solution in the following form: xp(t)=Acos(ωt−φ). (5.41) Substituting it into the equation of motion, we ﬁnd A/bracketleftbig (k−mω2)cos(ωt−φ)−cωsin(ωt−φ)/bracketrightbig =F0cosωt. Using the trigonometric identities cos(ωt−φ) = cos ωtcosφ+s i nωtsinφ, sin(ωt−φ)=s i n ωtcosφ−cosωtsinφ and equating the coeﬃcients of cos ωtand sin ωton both sides of the resulting equation, we obtain A/bracketleftbig (k−mω2)cosφ+cωsinφ/bracketrightbig =F0, (5.42a) A/bracketleftbig (k−mω2)sinφ−cωcosφ/bracketrightbig =0. (5.42b) It follows from (5.42b) that: (k−mω2)sinφ=cωcosφ, 248 5 Ordinary Diﬀerential Equations but sin2φ=1−cos2φ,so (k−mω2)2(1−cos2φ)=(cωcosφ)2. Therefore cosφ=k−mω2 [(k−mω2)2+(cω)2]1/2. It follows that: sinφ=cω [(k−mω2)2+(cω)2]1/2. Substituting cos φand sin φinto (5 .42a),we ﬁnd A=F0 [(k−mω2)2+(cω)2]1/2=F0 [m2(ω2 0−ω2)2+c2ω2]1/2, (5.43) where ω2 0=k/m. The particular solution xp(t) is, therefore, given by xp(t)=F0cos(ωt−φ) /bracketleftbig m2(ω2 0−ω2)2+c2ω2/bracketrightbig1/2, (5.44) where φ= tan−1 cω m(ω2 0−ω2). Notice that m2(ω2 0−ω2)2+c2ω2is never zero, even for ω=ω0.Hence with damping, the motion is always bounded. However, if the damping is not strong enough, the amplitude can still get to be very large. To ﬁnd the maximum amplitude, we take the derivative of Awith respect toω,and set it to zero. This shows that the frequency that makesdA dω=0 must satisfy the equation 2m2(ω2 0−ω2)−c2=0. Therefore the maximum amplitude occurs at ω=/radicalbigg ω2 0−c2 2m2. (5.45) Note that for c2>2m2ω2 0,no real ωcan satisfy this equation. In that case, there will not be any maximum for ω/negationslash= 0. The amplitude is a monotonically decreasing function of the forcing frequency. However, if c2<2m2ω2 0,then there will be a maximum. Substituting (5.45) into the expression of A,we obtain the maximum amplitude Amax=2mF0 c(4m2ω2 0−c2)1/2. 5.7 Electric Circuits 249 2 1.5 1 0.5 04 3 2 1g = 0.1 g = 0.3 g = 0.5 g = 1.0 g = 2.0 g = 4.0F0 mw02 w w0 Fig. 5.6. Forced vibration with viscous damping. Amplitude of the steady-state as a function of ω/ω0,γ=(c/mω 0)2represents the strength of damping. For γ≥2, there is no maximum To see the relation between the amplitude Aand the forcing frequency ω,it is convenient to express Aof (5.43) as A=F0 mω2 0/bracketleftBig/parenleftbig 1−/parenleftbig ω/ω0/parenrightbig2/parenrightbig2+γ/parenleftbig ω/ω0/parenrightbig2/bracketrightBig1/2, where γ=c2/m2ω2 0.The graphs of Ain units of F0/mω2 0as functions of ω/ω0 are shown in Fig. 5.6 for several diﬀerent values of γ.Forγ=0,it is the forced vibration without damping and the motion is unbound at ω=ω0.For a small γ,the amplitude still has a sharp peak at a frequency slightly less than ω0. Asγgets larger, the peak becomes smaller and wider. When γ≥2,there is no longer any maximum. In designing structures, we want to include suﬃcient amount of damping to avoid resonance which can lead to disaster. On the other hand, if we design a device to detect periodic force, we would want to choose m,k,c to satisfy (5.45) so that the response of the device to such a force is maximum. 5.7 Electric Circuits As a second example of application of theory of linear second-order diﬀerential equations with constant coeﬃcients, we consider the simple electric circuit shown in Fig. 5.7. It consists of three kinds of circuit elements; a resistor with a resistance Rmeasured in ohms, an inductor with an inductance Lmeasured in henries, and a capacitor with capacitance Cmeasured in farads. They are connected in series with a source of electromotive force (emf) that supplies at time t 250 5 Ordinary Diﬀerential Equations + − εI RL CQ Q Fig. 5.7. An oscillatory electrical circuit with resistance, inductance, and capacitance a voltage V(t) measured in volts. The capacitor is a device to store electric charges Q,measured in coulombs. If the switch is closed, there will be current I(t), measured in amperes, ﬂowing in the circuit. In elementary physics, we learned that the voltage drop acrose the resistor is equal to IR,the voltage drop across the inductor is LdI dt,and the voltage acrose the capacitor is1 CQ. The sum of these is equal to the applied voltage. Therefore LdI dt+RI+1 CQ=V(t). Furthermore, the rate of increase of the charge Qon the capacitor is, by deﬁnition, equal to the current dQ dt=I. With this relation, we obtain the following second-order linear nonhomoge- neous equation for Q: Ld2Q dt2+RdQ dt+1 CQ=V(t). Suppose the circuit is driven by a generator with a pure cosine wave oscillation, V(t)=V0cosωt,then the equation becomes Ld2Q dt2+RdQ dt+1 CQ=V0cosωt. (5.46) 5.7.1 Analog Computation We see that the equation describing an LRC circuit is exactly the same as (5.40),the equation describing the forced vibration of a spring–mass system with viscous damping. The fact that the same diﬀerential equation serves to describe two entirely diﬀerent physical phenomena is a striking example of the 5.7 Electric Circuits 251 Table 5.2. The analogy between mechanical and electrical systems Mechanical Property Electrical Property md2x dt2+cdx dt+kx=F0cosωt Ld2Q dt2+RdQ dt+1 CQ=V0cosωt displacement x charge Q velocity v=dx dtcurrent I=dQ dt mass m inductance L spring constant k inverse capacitance 1 /C damping coeﬃcient c resistance R applied force F0cosωt applied voltage V0cosωt resonant frequency ω2 0=k mresonant frequency ω2 0=1 LC unifying role of mathematics in natural sciences. With appropriate substitu- tions, the solution of (5 .40) can be applied to electric circuits. The correspon- dence between the electrical and mechanical cases are shown in Table 5.2. The correspondence between mechanical and electrical properties can also be used to construct an electrical model of a given mechanical system. This is a very useful way to predict the performance of a mechanical system, since the electrical elements are inexpensive and electrical measurements are usually very accurate. The method of computing the motion of a mechanical system from an electrical circuit is known as analog computation. By directly converting xp(t)o f( 5 .44) into its electrical equivalent, the steady-state solution of (5 .46) is found to be Q(t)=V0cos(ωt−φ) /bracketleftBig/parenleftbig1 C−ω2L/parenrightbig2+/parenleftbig ωR/parenrightbig2/bracketrightBig1/2, φ= tan−1ωR 1 C−ω2L= tan−1R 1 ωC−ωL. Generally, it is the current that is of primary interests, so we diﬀerentiate Q with respect to tto get the steady-state current I(t)=dQ dt=−ωV0sin(ωt−φ) /bracketleftBig/parenleftbig1 C−ω2L/parenrightbig2+/parenleftbig ωR/parenrightbig2/bracketrightBig1/2=−V0sin(ωt−φ) /bracketleftBig/parenleftbig1 ωC−ωL/parenrightbig2+R2/bracketrightBig1/2. To see more clearly the phase relation between the current I(t) and the applied voltage V(t) = cos ωt,we would like to express the current also in terms of a cosine function. This can be done by noting that tanφ=R 1 ωC−ωL can be expressed geometrically in the following triangle. 252 5 Ordinary Diﬀerential Equations R2 +1 wC− wL2 1 wC− wLRa f)) It is clear that φ=π 2−αand tanα=1 ωC−ωL R. Since sin( ωt−φ)=s i n ( ωt−π 2+α)=−cos(ωt+α),it follows that: I(t)=V0cos(ωt+α) /bracketleftBig/parenleftbig1 ωC−ωL/parenrightbig2+R2/bracketrightBig1/2. For reasons that will soon be clear, often I(t) is written in still another form: I(t)=V0cos(ωt−β) /bracketleftBig/parenleftbig1 ωC−ωL/parenrightbig2+R2/bracketrightBig1/2, (5.47) where β=−α,and tanβ= tan( −α)=−tanα=ωL−1 ωC R. 5.7.2 Complex Solution and Impedance The particular solution of (5 .46) can be found by the complex exponential method. This method oﬀers some computational and conceptual advantages. We can replace V0cosωtbyV0eiωand solve the equation Ld2Qc dt2+RdQc dt+1 CQc=V0eiωt. (5.48) The real part of the solution Qcis the charge Q,and the real part of Ic,deﬁned asd dtQc,is the current I.The time dependence of charges and currents must also be in the form of eiωt Qc=/hatwideQeiωt,I c=/hatwideIeiωt, 5.7 Electric Circuits 253 where/hatwideQand/hatwideIare complex but independent of t.Since dQc dt=iωQc,d2Qc dt2=−ω2Qc, the diﬀerential equation (5 .48) becomes the algebraic equation (−ω2L+iωR+1 C)Qc=V0eiωt. Clearly Qc=V0eiωt −ω2L+iωR+1 C, and Ic=dQc dt=iωV0eiωt −ω2L+iωR+1 C=V0eiωt R+i (ωL−1 ωC). (5.49) Writing the denominator in the polar form R+i (ωL−1 ωC)=/bracketleftbigg R2+(ωL−1 ωC)2/bracketrightbigg1/2 eiβ, β= tan−1ωL−1 ωC R, We see that Ic=V0eiωt /bracketleftbig R2+(ωL−1 ωC)2/bracketrightbig1/2eiβ=V0ei(ωt−β) /bracketleftbig R2+(ωL−1 ωC)2/bracketrightbig1/2. The real part of Icis I=V0cos(ωt−β) /bracketleftbig R2+(ωL−1 ωC)2/bracketrightbig1/2, which is identical to (5 .47). In electrical engineering, it is customary to deﬁne V0eiωtas the complex voltage Vc,and to deﬁne Z=R+iωL+1 iωC as the complex impedance Z. With these notations, (5 .49) can be written in the form Ic=Vc Z. Note that if the circuit element had consisted of the resistance Ralone, the impedance would be equal simply to R,so this relation would resembles Ohm’s law for a direct current circuit: V=RI.Thus the role the impedance plays 254 5 Ordinary Diﬀerential Equations in an alternating circuit with a sinusoidal voltage is exactly the same as the resistor in a direct current circuit. It is a simple matter to show that if the circuit element consists only of the inductance L,the impedance is simply i ωL.Similarly, with only capacitance C,the impedance is just 1 /(iωC).Thus we see that when electrical elements are connected in series, the corresponding impedances combine just as simple resistances do. In a similar way, we can show that when electrical elements are connected in parallel, the corresponding impedances also combine just as simple resis- tances do. For example, if R,L,C are connected in parallel, the complex cur- rent can be found by dividing the complex voltage by the simple impedance Zdeﬁned by the relation. 1 Z=1 R+1 iωL+iωC. The real part of the result is the current in this AC circuit. This makes it very easy to determine the steady state behavior of an electrical system. 5.8 Systems of Simultaneous Linear Diﬀerential Equations In many applications, it is necessary to simultaneously consider several dependent variables, each depending on the same independent variable, usually timet. The mathematical model is generally a system of linear diﬀerential equations. The elementary approach of solving systems of diﬀerential equa- tions is to eliminate the dependent variables one by one through combining pairs of equations, until there is only one equation left containing one dependent variable. This equation will usually be of higher order, and can be solved by the methods we have discussed. Once this equation is solved, the other dependent variables can be found in turn. This method is similar to the solution of systems of simultaneous algebraic equations. A closely related method is to ﬁnd the eigenvalues of the matrix formed by the diﬀerential equations. This method provides a mathematical framework for the discussion of normal frequencies of the system, which are physically important. 5.8.1 The Reduction of a System to a Single Equation Let us solve the following system of equations with two dependent variables x(t)a n d y(t): dx dt=−2x+y, (5.50a) dy dt=−4x+3y+1 0c o s t (5.50b) 5.8 Systems of Simultaneous Linear Diﬀerential Equations 255 with the initial conditions x(0) = 0 ,y(0) =−1. From the ﬁrst equation ,we have y=dx dt+2x,dy dt=d2x dt2+2dx dt. Substitute them into the second equation d2x dt2+2dx dt=−4x+3 (dx dt+2x) + 10cos t or d2x dt2−dx dt−2x=1 0c o s t. (5.51) This is an ordinary second-order nonhomogeneous diﬀerential equation, the complementary function xc(t) and the particular solution xp(t) are found to be, respectively, c1e−t+c2e2tand−3cost−sint.Therefore x=xc+xp=c1e−t+c2e2t−3cost−sint. The solution for y(t) is then given by y=dx dt+2x=c1e−t+4c2e2t−7cost+s i nt. The constants c1,c2are determined by the initial conditions x(0) = c1+c2−3=0, y(0) = c1+4c2−7=−1, which gives c1=2,c2=1.Thus x(t)=2 e−t+e2t−3cost−sint, y(t)=2 e−t+4 e2t−7cost+s i nt. If the number of coupled equations is small (2 or 3), the simplest method of solving the problem is this kind of direct substitution. However, for a larger system, one may prefer the more systematic approach of Sect. 5.8.2. 5.8.2 Cramer’s Rule for Simultaneous Diﬀerential Equations We will use the same example of the last section to illustrate this method. First, use the notation D to representd dt,and write the set of equations (5 .50) as (D + 2) x−y=0, (5.52a) 4x+( D−3)y=1 0c o s t. (5.52b) 256 5 Ordinary Diﬀerential Equations Recall that for a system of algebraic equations a11x+a12y=b1 a21x+a22y=b2, the solution can be obtained by the Cramer’s rule x=/vextendsingle/vextendsingle/vextendsingle/vextendsingleb1a12 b2a22/vextendsingle/vextendsingle/vextendsingle/vextendsingle /vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12 a21a22/vextendsingle/vextendsingle/vextendsingle/vextendsingle,y =/vextendsingle/vextendsingle/vextendsingle/vextendsinglea11b1 a21b2/vextendsingle/vextendsingle/vextendsingle/vextendsingle /vextendsingle/vextendsingle/vextendsingle/vextendsinglea11a12 a21a22/vextendsingle/vextendsingle/vextendsingle/vextendsingle. We can use the same formalism to solve a system of diﬀerential equations. That is, x(t) of (5.52) can be written as x=/vextendsingle/vextendsingle/vextendsingle/vextendsingle0 −1 10cos t(D + 3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle /vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) −1 4( D −3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle, or /vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) −1 4( D −3)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=/vextendsingle/vextendsingle/vextendsingle/vextendsingle0 −1 10cos t(D−3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Expanding the determinant, we have [(D + 2)(D −3) + 4] x=1 0c o s t. This means (D2−D−2)x=1 0c o s t, which is identical to (5 .51) of the last section. Proceeding in exactly the same way as in the last section, we ﬁnd x(t)=xc+xp=c1e−t+c2e2t−3cost−sint. Substituting it into the original diﬀerential equation, y(t) is found to be y(t)=c1e−t+4c2e2t−7cost+s i nt. An alternative way of ﬁnding y(t) is to note that y=/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) 0 41 0 c o s t/vextendsingle/vextendsingle/vextendsingle/vextendsingle /vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) −1 4( D −3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle 5.8 Systems of Simultaneous Linear Diﬀerential Equations 257 or /vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) −1 4( D −3)/vextendsingle/vextendsingle/vextendsingle/vextendsingley=/vextendsingle/vextendsingle/vextendsingle/vextendsingle(D + 2) 0 41 0 c o s t/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Expanding the determinant, we have (D2−D−2)y=2 0c o s t−10sin t. The solution of this equation is y(t)=yc+yp=k1e−t+k2e2t−7cost+s i nt. Note that the complementary functions xcandycsatisfy the same homoge- neous diﬀerential equation (D2−D−2)xc=0 ( D2−D−2)yc=0. Since we have already written xc=c1e−t+c2e2t,we must avoid using c1 andc2as the constants in yc.That is, in yc=k1e−t+k2e2t,k1andk2 are not necessarily equal to c1andc2,because there is no reason that they should be equal. To ﬁnd the relationship between them, we have to substitute x(t)a n d y(t) back into one of the original diﬀerential equations. For example, substituting them back into (D + 2) x=y,we have c1e−t+4c2e2t−7cot + sin t=k1e−t+k2e2t−7cost+s i nt. Therefore k1=c1,k2=4c2. Thus we obtain the same result as before. It is seen that after the ﬁrst dependent variable x(t) is found from Cramer’s rule, it is simpler to ﬁnd the second dependent variable y(t) by direct substi- tution. If we continue to use Cramer’s rule to ﬁnd y(t),we will introduce some additional constants which must be eliminated by substituting both x(t)a n d y(t) back into the original diﬀerential equation. 5.8.3 Simultaneous Equations as an Eigenvalue Problem A system of simultaneous diﬀerential equations can be solved as an eigenvalue problem in matrix theory. We will continue to use the same example to illus- trate the procedures of this method. First write the set of equations (5 .50) in the following form: −2x+y=x/prime, −4x+3y=y/prime−10cos t. With matrix notation, they become /parenleftbigg −21 −43/parenrightbigg/parenleftbigg x y/parenrightbigg =/parenleftbigg x/prime y/prime−10cos t/parenrightbigg . 258 5 Ordinary Diﬀerential Equations Let x=xc+xp,y=yc+yp. The complementary functions xcandycsatisfy the equation /parenleftbigg −21 −43/parenrightbigg/parenleftbigg xc yc/parenrightbigg =/parenleftbigg x/prime c y/prime c/parenrightbigg , (5.53) and the particular solutions xpandypsatisfy the equation /parenleftbigg−21 −43/parenrightbigg/parenleftbiggxp yp/parenrightbigg =/parenleftbiggx/prime p y/prime p−10cos t/parenrightbigg . (5.54) Since these are linear equations with constant coeﬃcients, we assume xc=c1eλt,y c=c2eλt, so x/prime=dxc dt=λc1eλt,y/prime c=dyc dt=λc1eλt. It follows that the matrix equation for the complementary functions is given by:/parenleftbigg−21 −43/parenrightbigg/parenleftbiggc1eλt c2eλt/parenrightbigg =/parenleftbiggλc1eλt λc2eλt/parenrightbigg . This is an eigenvalue problem /parenleftbigg−21 −43/parenrightbigg/parenleftbiggc1 c2/parenrightbigg =λ/parenleftbiggc1 c2/parenrightbigg with eigenvalue λand eigenvector/parenleftbigg c1 c2/parenrightbigg .Therefore, λmust satisfy the secular equation /vextendsingle/vextendsingle/vextendsingle/vextendsingle−2−λ1 −43−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 or (−2−λ)(3−λ)+4=0 . The two roots λ1,λ2of this equation are easily found to be λ1=−1,λ 2=2. Corresponding to each λi,there is an eigenvector/parenleftbigg ci 1 ci 2/parenrightbigg .The coeﬃcients ci 1 andci 2are not independent of each other, they must satisfy the equation /parenleftbigg −21 −43/parenrightbigg/parenleftbigg ci 1 ci 2/parenrightbigg =λi/parenleftbigg ci 1 ci 2/parenrightbigg . 5.8 Systems of Simultaneous Linear Diﬀerential Equations 259 It follows from this equation that for λ1=−1,c1 2=c1 1,and for λ2=2,c2 2= 4c2 1.Therefore, other than some multiplicative constants, the eigenvector for λ=−1i s/parenleftbigg 1 1/parenrightbigg ,and for λ=2i s/parenleftbigg 1 4/parenrightbigg . The complementary functions xcandycare given by the linear combina- tions of the these two sets of solutions, /parenleftbigg xc yc/parenrightbigg =c1/parenleftbigg 1 1/parenrightbigg e−t+c2/parenleftbigg 1 4/parenrightbigg e2t. For the particular solution, because of the nonhomogeneous term 10cos t, we can assume xp=Acost+Bsintandyp=Ccost+Dsint.However, it is less cumbersome to make use of the fact that 10cos tis the real part of 10eit.We can assume xpis the real part of Aceitandypis the real part of Bceit,where AcandBcare complex numbers. With these assumptions, (5 .54) becomes /parenleftbigg−21 −43/parenrightbigg/parenleftbiggAceit Bceit/parenrightbigg =/parenleftbiggiAceit iBceit−10eit/parenrightbigg . Thus −2Ac+Bc=iAc, −4Ac+3Bc=iBc−10, which yields Ac=−3+i,B c=−7−i. Therefore xp=R e ( Aceit)=−3cost−sint, yp=R e ( Bceit)=−7cost+s i nt. Finally, we have the general solution /parenleftbiggx y/parenrightbigg =/parenleftbiggxc yc/parenrightbigg +/parenleftbiggxp yp/parenrightbigg =/parenleftbiggc1e−t+c2e2t−3cost−sint c1e−t+4c2e2t−7cost+s i nt/parenrightbigg , which is what we had before. 5.8.4 Transformation of an nth Order Equation into a System of n First-Order Equations We have seen that a system of equations can be reduced to a single equation of higher order. The reverse is also true. Any nth-order diﬀerential equation can always be transformed into a simultaneous nﬁrst-order equations. Let us 260 5 Ordinary Diﬀerential Equations use this method to solve the second-order diﬀerential equation for the damped harmonic oscillator d2x dt2+c mdx dt+k mx=0. Let x1=x, x 2=dx1 dt. It follows: dx2 dt=d2x1 dt2=−c mdx1 dt−k mx1. Thus the second-order equation can be written as a set of two ﬁrst-order equations: x2=dx1 dt, −k mx1−c mx2=dx2 dt. With matrix notation, we have /parenleftBigg0 −k m1 −c m/parenrightBigg/parenleftbigg x1 x2/parenrightbigg =/parenleftbigg x/prime 1 x/prime 2/parenrightbigg . Since the coeﬃcients are constants, we can assume /parenleftbigg x1 x2/parenrightbigg =/parenleftbigg c1 c2/parenrightbigg eλt, thus we have the eigenvalue problem /parenleftBigg0 −k m1 −c m/parenrightBigg/parenleftbigg c1 c2/parenrightbigg =λ/parenleftbigg c1 c2/parenrightbigg . The eigenvalues λcan be found from the characteristic equation /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle0−λ −k m1 −c m−λ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0. The two roots of this equation are λ1,λ2=1 2m(−c±/radicalbig c2−4km). Thus the general solution of the original problem is given by x(t)=x1=c1eλ1t+c2eλ2t. This result is identical to (5 .37). 5.8 Systems of Simultaneous Linear Diﬀerential Equations 261 The fact that a linear diﬀerential equation of nth-order can be transformed into a system of ncoupled ﬁrst-order equations is of some importance because mathematically one can show that there is an unique solution for a linear ﬁrst- order system, provided the initial conditions are speciﬁed. However, we are not too concerned with uniqueness and existence because in physical applications, the mathematical model, if formulated correctly, must have a solution. 5.8.5 Coupled Oscillators and Normal Modes The motion of a harmonic oscillator is described by a second-order diﬀerential equation. Its solution shows that the motion is characterized by a single nat- ural frequency. A real physical system usually has many diﬀerent characteristic frequencies. A vibration with any of these frequencies is called a normal mode of the system. The motion of the system is generally a linear combination of these normal modes. A simple example is the system of two coupled oscillators shown in Fig. 5.8. The system consists of two identical mass–spring oscillators of mass m and spring constant k.The two masses rest on a frictionless table and are connected by a spring with spring constant K.When the displacements xA andxBare zero, the springs are neither stretched or compressed. At any moment, the connecting spring is stretched an amount xA−xBand therefore pulls or pushes on A and B with a force whose magnitude is K(xA−xB). Thus the magnitude of the restoring force on A is −kxA−K(xA−xB). The force on B must be −kxB+K(xA−xB). Therefore the equations of motion for A and B are −kxA−K(xA−xB)=md2xA dt2, −kxB+K(xA−xB)=md2xB dt2. K k xBxABA k m m Fig. 5.8. Two coupled oscillators 262 5 Ordinary Diﬀerential Equations With matrix notation, these equations can be written as ⎛ ⎜⎝−k m−K mK mK m−k m−K m⎞ ⎟⎠/parenleftbigg xA xB/parenrightbigg =⎛ ⎜⎜⎝d2xA dt2 d2xB dt2⎞ ⎟⎟⎠. To simplify the writing, let ω2 0=k/mandω2 1=K/m. With the assumption xA=aeλtandxB=beλt,the last equation becomes /parenleftbigg −ω2 0−ω2 1ω2 1 ω2 1−ω2 0−ω2 1/parenrightbigg/parenleftbigg a b/parenrightbigg eλt=λ2/parenleftbigg a b/parenrightbigg eλt. This can be regarded as an eigenvalue problem. The secular equation /vextendsingle/vextendsingle/vextendsingle/vextendsingle−ω2 0−ω2 1−λ2ω2 1 ω2 1 −ω2 0−ω2 1−λ2/vextendsingle/vextendsingle/vextendsingle/vextendsingle=0 shows that λ2satisﬁes the equation (−ω2 0−ω2 1−λ2)2−ω4 1=0 or λ2=−ω2 0−ω2 1±ω2 1. Thus λ2=−ω2 0,λ2=−(ω2 0+2ω2 1). The four roots of λare λ1,λ2=±iω0,λ 3,λ4=±iωc, where ωc=/radicalBig ω2 0+2ω2 1. These frequencies, ω0andωcare known as the normal frequencies of the system. The amplitudes aandbare not independent of each other, since they must satisfy the equation /parenleftbigg−ω2 0−ω2 1−λ2ω2 1 ω2 1 −ω2 0−ω2 1−λ2/parenrightbigg/parenleftbigga b/parenrightbigg =0. Thus for λ=λ1,λ2=±iω0,soλ2=−ω2 0,the amplitudes aandbmust satisfy /parenleftbigg −ω2 0−ω2 1+ω2 0 ω2 1 ω2 1 −ω2 0−ω2 1+ω2 0/parenrightbigg/parenleftbigg a b/parenrightbigg =0. It follows that a=b. Similarly, for λ=λ3,λ4=±iωc,the relation between aandbis given by 5.8 Systems of Simultaneous Linear Diﬀerential Equations 263 /parenleftbigg−ω2 0−ω2 1+ω2 0+2ω2 1 ω2 1 ω2 1 −ω2 0−ω2 1+ω2 0+2ω2 1/parenrightbigg/parenleftbigga b/parenrightbigg =0, which gives b=−a. The displacements xAandxBare given by a linear combinations of these four solutions, xA=a1eλ1t+a2eλ2t+a3eλ3t+a4eλ4t, xB=a1eλ1t+a2eλ2t−a3eλ3t−a4eλ4t, where we have substituted a1,a2forb1,b2,a n d −a3,−a4forb3,b4.Since λ1=iω0,λ2=−iω0, a1eλ1t+a2eλ2t=a1eiω0t+a2e−iω0t=Ccos(ω0t+α). Similarly a3eλ3t+a4eλ4t=Dcos(ωct+β). Thus xAandxBcan be written as xA=Ccos(ω0t+α)+Dcos(ωct+β), xB=Ccos(ω0t+α)−Dcos(ωct+β). The four constants a1,a2,a3,a4(orC,α,D,β ) depend on the initial condi- tions. It is seen that both xAandxBare given by some combination of the vibrations of two normal frequencies ω0andωc. Suppose the motion is started when we pull both A and B toward the same direction by an equal amount x0and then release them from rest. The distance between A and B equals the relaxed length of the coupling spring and therefore it exerts no force on each mass. Thus A and B will oscillate in phase with the same natural frequency ω0as if they were not coupled. Mathematically, we see that is indeed the case. With the initial conditions xA(0) = xB(0) = x0anddxA dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0,dxB dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0, one can easily show that C=x0,D=0,α=β=0.Therefore xA=x0cosω0t, x B=x0cosω0t. This represents a normal mode of the coupled system. Once the system is vibrating with a normal frequency, it will continue to vibrate with that fre- quency. Suppose initially we pull A and B in opposite direction by the same amount xmand then release them. The symmetry of the arrangement tells us that A and B will be mirror images of each other. They will vibrate with certain frequency, which we might expect it to be ωc,since ωcis the only 264 5 Ordinary Diﬀerential Equations other normal frequency of the system. This is indeed the case. Since with the initial conditions xA(0) =−xB(0) = xm,anddxA dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0,dxB dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=0, one can show that C=0,D=xm,α=β=0.Therefore xA=xmcosωct, x B=−xmcosωct. They oscillate with the same frequency ωcbut they are always 180◦out of phase. This constitutes the second normal mode of the system. The general motion is a linear combination of these two modes. For a real molecule or crystal, there will be many normal modes. Each normal mode corresponds to a certain symmetry of the structure. The fact that these modes can be excited by their corresponding normal frequencies is widely used in scientiﬁc applications. 5.9 Other Methods and Resources for Diﬀerential Equations Many readers probably had previously taken a course in ordinary diﬀerential equations. Here we just give a review so that even those who did not have previous exposure can gain enough background to continue. The literature of the theory and applications of diﬀerential equations is vast. Our discussion is far from complete. Among the methods we have not yet discussed are the Laplace transform, Fourier analysis and power series solutions. The Laplace transform is especially useful in solving problems with nonhomogeneous terms of a discontinuous or impulsive nature. In Chap. 6 we will study these problems in detail. If the nonhomogeneous term is periodic but not sinusoidal, Fourier series method is particularly convenient. We will discuss this method after we study the Fourier series. In general, a diﬀerential equation with variable coeﬃcients cannot be solved by the methods of this chapter. The usual procedure for such equa- tions is to obtain solutions in the form of inﬁnite series. This is known as series method. Some most important equations in physics and engineering lead us to this type equations. The series so obtained can be taken as deﬁn- itions of new functions. Some important ones are named and tabulated. We shall study this method in the chapter on special functions. In addition, diﬀerential equations can be solved numerically. Sometimes this is the only way to solve the equation. Digital computers have made numerical solutions readily available. There are several computer programs for the integration of ordinary diﬀerential equations in “Numerical Recipes” 5.9 Other Methods and Resources for Diﬀerential Equations 265 by William H. Press, Brian P. Flannery, Saul A. Teukolsky and William T. Vetterling (Cambridge University Press, 1986). For a discussion of the numerical methods, see R.J. Rice, “Numerical Methods, Software and Analy- sis” (McGraw-Hill, New York, 1983). Finally it should be mentioned that a number of commercial computer packages are available to perform algebraic manipulations, including solv- ing diﬀerential equations. They are called computer algebraic systems, some prominent ones are Matlab, Maple, Mathematica, MathCad and MuPAD. This book is written with the software “Scientiﬁc WorkPlace”, which also provides an interface to MuPAD. (Before version 5, it also came with Maple). Instead of requiring the user to adhere to a rigid syntax, the user can use natural mathematical notations. For example, to solve the diﬀerential equation d2y dx2+dy dx=x+y all you have to do is (1) type this equation in the math-mode, and (2) click on the “Compute” button, and (3) click on the “Solve ODE” button in the pull-down menu, and (4) click on the “Exact” button in the submenu. The program will return with Exact solution is: C1ex(1 2√ 5−1 2)−x+C2ex(−1 2√ 5−1 2)−1. Unfortunately, not every problem can be solved by a computer algebraic system. Sometimes it fails to ﬁnd the solution. Even worse, for a variety of reasons, the intention of the user is sometimes misinterpreted, and the com- puter returns with an answer to a wrong problem without the user knowing it. Therefore these systems must be used with caution. Exercises 1. Find the general solutions of the following separable diﬀerential equations: (a)xy/prime+y+3=0 , (b) 2yy/prime+4x=0. Ans. (a) x(y+3 )= c,(b) 2x2+y2=c. 2. Find the speciﬁc solutions of the following separable diﬀerential equations: (a)dy dx=x(1 +y2) y(1 +x2),y(0) = 1, (b)yex+ydy=dx, y (0) = 0 . Ans. (a) 1 + y2= 2(1 + x2),(b) (1 −y)ey=e−x. 266 5 Ordinary Diﬀerential Equations 3. Change the following equations into separable diﬀerential equations and ﬁnd the general solutions: (a)xyy/prime=y2−x2, (b)dy dx=x−y x+y. Ans. (a) ln x+y2 2x2=c,(b)y2+2xy−x2=c. 4. Show that the following diﬀerential equations are exact and ﬁnd the gen- eral solutions: (a) (2 xy−cosx)dx+(x2−1)dy=0 , (b) (2 x+ey)dx+xeydy=0 . Ans. (a) x2y−sinx−y=c,(b)x2+xey=c. 5. Solve the following diﬀerential equations by ﬁrst ﬁnding an integrating factor: (a) 2( y3−2)dx+3xy2dy=0 , (b) (b) ( y+x4)dx−xdy=0 . Ans. (a) µ=x, x2y3−2x2=c,(b)µ=1/x2,x3 3−y x=c. 6. Find the general solutions of the following ﬁrst-order linear diﬀerential equations: (a)y/prime+y=x, (b)xy/prime+( 1+ x)y=e−x. Ans. (a) y=x−1+c e−x,(b)y=e−x+ce−x/x. 7. Find the speciﬁc solutions of the following ﬁrst-order linear diﬀerential equations: (a)y/prime−y=1−x, y (0) = 1. (b)y/prime+1 xy=3x2,y(1) = 5. Ans. (a) y=x+ex,(b)y=3 4x3+17 4x−1. 8. The RL circuit is described by the equation Ldi dt+Ri=Acost, i(0) = 0 where iis current. Find the current ias a function of time t. Ans.i(t)=AR R2+L2/bracketleftbig cost+L Rsint−e−Rt/L/bracketrightbig . 9. Find the general solutions of the following homogeneous second-order dif- ferential equations: (a)y/prime/prime−k2y=0 , (b)y/prime/prime−(a+b)y/prime+aby=0 , (c)y/prime/prime+2ky/prime+k2y=0 . 5.9 Other Methods and Resources for Diﬀerential Equations 267 Ans. (a) y(x)=c1ekx+c2e−kx,(b)y(x)=c1eax+c2ebx,(c)y(x)= c1e−kx+c2xe−kx. 10. Find the speciﬁc solutions of the following homogeneous second-order dif- ferential equations: (a)y/prime/prime−2ay/prime+(a2+b2)y=0,y(0) = 0 ,y/prime(0) = 1. (b)y/prime/prime+4y=s i nx, y (0) = 0 ,y/prime(0) = 0. (c)y/prime/prime/prime+y/prime=e2z,y(0) = y/prime(0) = y/prime/prime(0) = 0. Ans. (a) y(x)=1 beaxsinbx,(b)y(x)=1 3sinx−1 6sin2x, (c)y(x)=1 10e2x−1 5sinx+2 5cosx−1 2. 11. Find the general solutions of the following nonhomogeneous diﬀerential equations: (a)y/prime/prime+k2y=a, (b)y/prime/prime−4y=x, (c)y/prime/prime−2y/prime+y=3x2−12x+7 . Ans. (a) y(x)=c1sinkx+c2coskx+a/k2,(b)y(x)=c1e2x+c2e−2x−1 4x, (c)y(x)=(c1+c2x)ex+3x2+1. 12. Find the general solutions of the following nonhomogeneous diﬀerential equations: (a)y/prime/prime−3y/prime+2y=e2x, (b)y/prime/prime−6y/prime+9y=4e3x, (c)y/prime/prime+9y= cos(3 x). Ans. (a) y=c1ex+c2e2x+xe2x,(b)y=c1e3x+c2xe3x+2x2e3x, (c)y=c1cos(3x)+c2sin(3x)+1 6xsin(3x). 13. Find the speciﬁc solutions of the following nonhomogeneous diﬀerential equations: (a)y/prime/prime+y/prime=x2+2x, y (0) = 4 ,y/prime(0) =−2. (b)y/prime/prime−4y/prime+4y=6s i n x−8cosx, y(0) = 3 ,y/prime(0) = 4. (c)y/prime/prime−4y=8 e2x,y(0) = 4 ,y/prime(0) = 6. Ans. (a) y=1 3x3+2 e−x+2,(b)y=( 3−4x)e2x+2s i n x,(c)y= 3e2x+e−2x+2xe2x. 14. Solve the following diﬀerential equations with the method of variation of parameters: (a)y/prime/prime+y=s e c x, (b)y/prime/prime+4y/prime+4y=e−2x x2. Ans. (a) y=c1cosx+c2sinx+c o s xln|cosx|+xsinx, (b)y=c1e−2x+c2xe−2x−e−2xlnx. 15. Solve the following set of simultaneous linear diﬀerential equation: 268 5 Ordinary Diﬀerential Equations y/prime(x)−z/prime(x)−2y(x)+2z(x)=1−2x, y/prime/prime(x)+2z/prime(x)+y(x)=0, y(0) = z(0) = y/prime(0) = 0 . Ans.y(x)=−2e−x−2xe−x+2 ; z(x)=−2e−x−2xe−x+2−x. 16. Solve the following set of simultaneous linear diﬀerential equation: y/prime(x)+z/prime(x)+y(x)+z(x)=1, y/prime(x)−y(x)−2z(x)=0, y(0) = 1 ,z(0) = 0 . Ans.y(x)=2−e−x;z(x)=e−x−1. 17. In strength of materials, you will encounter the equation d4y dx4=−4a4y where ais a positive constant. Find the general solution of this equation. (y(x) with four constants). Ans. y(x)=eax(c1cosax+c2sinax)+e−ax(c3cosax+c4sinax) 18. Solve y/prime/prime(t)+y(t)=/braceleftbigg 1−t2 π2if 0≤t≤π, 0i f t>π ,/bracerightbigg y(0) = y/prime(0) = 0 . This may be interpreted as an undamped system on which a force acts during some interval of time, for instance, the force acting on a gun barrel when a shell is ﬁred, the barrel being braked by heavy springs (and then damped by a dashpot which we disregard for simplicity). Hint: at t=π bothyandy/primemust be continuous. Ans.y(t)=/braceleftBigg −(1 +2 π2)cost+( 1+2 π2)−1 π2t2for 0 ≤t≤π /bracketleftbig 1−2(1 +2 π2)/bracketrightbig cost+2 πsintfor t≥π./bracerightBigg 19. If Na(t),Nb(t),Nc(t) represent the number of nuclei of three radioactive substances which decay according to the scheme a→b→c with decay constants λaandλb,the substance cis considered stable. Then the functions are known to obey the system of diﬀerential equations 5.9 Other Methods and Resources for Diﬀerential Equations 269 dNa dt=−λaNa, dNb dt=−λbNb+λaNa, dNc dt=λbNb. Assuming that Na(0) = N0,andNb(0) = Nc(0) = 0 ,ﬁndNa(t),Nb(t) andNc(t) as functions of time t. Ans. Na=N0e−λat;Nb=N0/bracketleftBig λa λb−λae−λat−λa λb−λae−λbt/bracketrightBig ;Nc= N0/bracketleftBig 1−λb λb−λae−λat+λa λb−λae−λbt/bracketrightBig . 20. Show that the particular solution of md2x dt2+cdx dt+kx=F0cosωt can be written in the form of xp(t)=C1cosωt+C2sinωt, where C1=(k−mω2)F0 (k−mω2)2+(cω)2,C 2=(cω)F0 (k−mω2)2+(cω)2. 21. Show that the result of previous problem can be put in the form of xp(t)=Acos(ωt−φ) where A=F0 [(k−mω2)2+(cω)2]1/2,φ= tan−1 cω m(ω2 0−ω2). 22. For two identical undamped oscillators, A and B, each of mass m,and nat- ural frequency ω0,show that each of them is governed by the diﬀerential equation md2x dt2+mω2 0x=0. They are coupled in such a way that the coupling force exerted on A is αm(d2xB/dt2),and the coupling force on B is αm(d2xA/dt2),where αis the coupling constant with a magnitude less than one. Find the normal frequencies of the system. Ans.ω=ω0(1±α)−1/2. 6 Laplace Transforms Among the tools that are very useful in solving linear diﬀerential equations is the Laplace transform method. The idea is to use an integral to trans- form the diﬀerential equation into an algebraic equation, from the solution of this algebraic equation we get the desired function through the inverse transform. The Laplace transform is named after the eminent French mathe- matician Pierre Simon Laplace (1749–1827), who is also remembered for the Laplace equation which is one of the most important equations in mathemati- cal physics. Laplace ﬁrst studied this method in 1782. However, the power and useful- ness of this method was not recognized until 100 years later. The techniques described in this chapter are mainly due to Oliver Heaviside (1850–1925), an innovative British electrical engineer, who also made signiﬁcant contributions to electromagnetic theory. The Laplace transform is especially useful in solving problems with nonhomogeneous terms of a discontinuous or impulsive nature. Such prob- lems are common in physical sciences but are relatively awkward to handle by the methods previously discussed. In this chapter certain properties of Laplace transforms are investigated and relevant formulas are tabulated in such a way that the solution of ini- tial value problems involving linear diﬀerential equations can be conveniently obtained. 6.1 Deﬁnition and Properties of Laplace Transforms 6.1.1 Laplace Transform – A Linear Operator The Laplace transform L[f] of the function f(t) is deﬁned as L[f]=/integraldisplay∞ 0e−stf(t)dt=F(s), (6.1) 272 6 Laplace Transforms we assume that this integral exists. One of the reasons that Laplace transform is useful is that scan be chosen large enough that (6.1) converges even if f(t) does not go to zero as t→∞.Of course, there are functions that diverge faster than est.For such functions, the Laplace transform does not exit. Fortunately such functions are of little physical interests. Note that the transform is a function of s.The transforms of the functions of our concern not only exist, but also go to zero ( F(s)→0) ass→∞. It follows immediately from the deﬁnition that the Laplace transform is a linear operator, that is L[af(t)+bg(t)] =/integraldisplay∞ 0e−st[af(t)+bg(t)]dt =a/integraldisplay∞ 0e−stf(t)dt+b/integraldisplay∞ 0e−stg(t)dt =aL[f]+bL[g]. (6.2) For simple functions, the integral of the Laplace transform can be readily carried out. For example: L[1] =/integraldisplay∞ 0e−stdt=/bracketleftbigg −1 se−st/bracketrightbigg∞ 0=1 s. (6.3) It is also very easy to evaluate the transform of an exponential function L[eat]=/integraldisplay∞ 0e−steatdt=/integraldisplay∞ 0e−(s−a)tdt=/bracketleftbigg −1 s−ae−(s−a)t/bracketrightbigg∞ 0. As long as s>a , the upper limit vanishes and the lower limit gives 1 /(s−a). Thus L[eat]=1 s−a. (6.4) Similarly L[e−at]=1 s+a. (6.5) With these relations, the Laplace transforms of the following hyperbolic functions coshat=1 2(eat+e−at), sinhat=1 2(eat−e−at) are easily obtained. Since Laplace transform is linear, L[coshat]=1 2/braceleftbig L[eat]+L[e−at]/bracerightbig =1 2/parenleftbigg1 s−a+1 s+a/parenrightbigg =s s2−a2. (6.6) 6.1 Deﬁnition and Properties of Laplace Transforms 273 Similarly L[sinhat]=a s2−a2. (6.7) Now the parameter adoes not have to be restricted to real numbers. If a is purely imaginary a=iω,we will have L[eiωt]=1 s−iω. Since1 s−iω=1 s−iω×s+iω s+iω=s s2+ω2+iω s2+ω2, and L[eiωt]=L[cosωt+is i n ωt]=L[cosωt]+iL[sinωt], equating the real part to real part and imaginary part to imaginary part we have L[cosωt]=s s2+ω2, (6.8) L[sinωt]=ω s2+ω2. (6.9) The deﬁnition of L[cosωt] is, of course, still L[cosωt]=/integraldisplay∞ 0e−stcosωtdt. (6.10) With integration by parts, we can evaluate this integral directly, /integraldisplay∞ 0e−stcosωtdt=/bracketleftbigg −1 se−stcosωt/bracketrightbigg∞ 0−/integraldisplay∞ 01 se−stωsinωtdt =1 s−ω s/integraldisplay∞ 0e−stsinωtdt, /integraldisplay∞ 0e−stsinωtdt=/bracketleftbigg −1 se−stsinωt/bracketrightbigg∞ 0+/integraldisplay∞ 01 se−stωcosωtdt =ω s/integraldisplay∞ 0e−stcosωtdt. Combine these two equations, /integraldisplay∞ 0e−stcosωtdt=1 s−ω2 s2/integraldisplay∞ 0e−stcosωtdt. Move the last term to the left-hand side, /parenleftbigg 1+ω2 s2/parenrightbigg/integraldisplay∞ 0e−stcosωtdt=1 s 274 6 Laplace Transforms or /integraldisplay∞ 0e−stcosωtdt=s s2+ω2, which is exactly the same as (6 .8),as it should be. In principle, the Laplace transform can be obtained directly by carrying out the integral. However, very often it is much simpler to use the properties of the Laplace transform, rather than direct integration, to obtain the transform, as shown in the last example. The Laplace transform has many interesting properties, they are the reasons that the Laplace transform is a powerful tool of mathematical analysis. We will now discuss some of them, and use them to generate more transforms as illustrations. 6.1.2 Laplace Transforms of Derivatives The Laplace transform of a derivative is by deﬁnition L[f/prime]=/integraldisplay∞ 0e−stdf(t) dtdt=/integraldisplay∞ 0e−stdf(t). If we let u=e−stand d v=df(t),then d u=−se−stdtandv=f.With integration by parts, we have udv=d (uv)−vdu, so L[f/prime]=/integraldisplay∞ 0{d[(e−stf(t)] +f(t)se−stdt} =/bracketleftbig e−stf(t)/bracketrightbig∞ 0+s/integraldisplay∞ 0e−stf(t)dt=−f(0) + sL[f].(6.11) Clearly L[f/prime/prime]=L[(f/prime)/prime]=−f/prime(0) + sL[f/prime] =−f/prime(0) + s(−f(0) + sL[f]) =−f/prime(0)−sf(0) + s2L[f].(6.12) Naturally this result can be extended to higher derivatives L[f(n)]=−f(n−1)(0)−····− sn−1f(0) + snL[f]. (6.13) These properties are crucial in solving diﬀerential equations. Here we will use them to generate L[tn]. First let f(t)=t,then f/prime=1 a n d f(0) = 0 .By (6.11) L[1] =−0+sL[t], rearranging and using (6.3), we have L[t]=1 sL[1] =1 s2. (6.14) 6.1 Deﬁnition and Properties of Laplace Transforms 275 If we let f(t)=t2,then f/prime=2tandf(0) = 0 .Again by (6.11) L[2t]=−0+sL[t2]. Thus, with (6.14) L[t2]=1 sL[2t]=2 sL[t]=2 s3. (6.15) Clearly this process can be repeated L[tn]=n! sn+1. (6.16) 6.1.3 Substitution: s-Shifting If we know the Laplace transform F(s) of the function f(t),we can get the transform of eatf(t) by replacing swiths−ainF(s).This can be easily shown. By deﬁnition F(s)=/integraldisplay∞ 0e−stf(t)dt=L[f(t)], clearly F(s−a)=/integraldisplay∞ 0e−(s−a)tf(t)dt=/integraldisplay∞ 0e−steatf(t)dt=L/bracketleftbig eatf(t)/bracketrightbig .(6.17) This simple relation is sometimes known as s-shifting (or ﬁrst shifting) theorem. With the help of the s-shifting theorem, we can derive the transforms of many more functions without carrying out the integration. For example, it follows from (6.16) and (6 .17) that L[e−attn]=n! (s+a)n+1. (6.18) It can also be easily shown that L[e−atcosωt]=/integraldisplay∞ 0e−ste−atcosωtdt=/integraldisplay∞ 0e−(s+a)tcosωtdt.(6.19) Compare the integrals in (6.10) and (6.19), the only diﬀerence is that sis replaced by s+a.Therefore the last integral must equal to the right-hand side of (6.8) with schanged to s+a,that is L[e−atcosωt]=s+a (s+a)2+ω2. (6.20) Similarly L[e−atsinωt]=ω (s+a)2+ω2. (6.21) 276 6 Laplace Transforms 6.1.4 Derivative of a Transform If we diﬀerentiate the Laplace transform F(s) with respect to s,we get d dsF(s)=d dsL[f(t)] =d ds/integraldisplay∞ 0e−stf(t)dt =/integraldisplay∞ 0de−st dsf(t)dt=/integraldisplay∞ 0e−st(−t)f(t)dt=L[−tf(t)].(6.22) Continuing this process, we have dn dsnL[f(t)] =L[(−t)nf(t)]. (6.23) Many more formulas can be derived by taking advantage of this relation. For example, diﬀerentiating both sides of (6 .9) with respect to s,we have d dsL[sinωt]=d dsω s2+ω2. Since d dsL[sinωt]=d ds/integraldisplay∞ 0e−stsinωtdt=−/integraldisplay∞ 0te−stsinωtdt=−L[tsinωt], d dsω s2+ω2=−2sω (s2+ω2)2, therefore L[tsinωt]=2sω (s2+ω2)2. (6.24) Similarly we can show L[tcosωt]=s2−ω2 (s2+ω2)2. (6.25) 6.1.5 A Short Table of Laplace Transforms Since Laplace transform is a linear operator, two transforms can be combined to form a new one. For example: L[1−cosωt]=1 s−s s2+ω2=ω2 s(s2+ω2), (6.26) L[ωt−sinωt]=ω s2−ω s2+ω2=ω3 s2(s2+ω2), (6.27) L[sinωt−ωtcosωt]=ω s2+ω2−ω(s2−ω2) (s2+ω2)2=2ω3 (s2+ω2)2, (6.28) 6.1 Deﬁnition and Properties of Laplace Transforms 277 Table 6.1. A short table of Laplace transforms, in each case sis assumed to be suﬃciently large that the transform exists f(t) F(s)=L[f(t)] f(t) F(s)=L[f(t)] 11 sδ(t)1 t1 s2δ(t−c)e−sc tn n! sn+1δ/prime(t−c) se−sc eat 1 s−au(t−c)1 se−sc teat 1 (s−a)2(t−c)nu(t−c)n! sn+1e−sc tneat n! (s−a)n+1(t−c)nea(t−c)u(t−c)n! (s−a)n+1e−sc sinωtω s2+ω2sinω(t−c)u(t−c)ω s2+ω2e−sc cosωts s2+ω2cosha(t−c)u(t−c)s s2−a2e−sc sinhata s2−a2sinωtof periodπ ωω s2+ω2cothsπ 2ω coshats s2−a2tof period p1−(1 +ps)e−ps ps2(1−e−ps) eatsinωtω (s−a)2+ω21 t(ebt−eat)l ns−a s−b eatcosωts−a (s−a)2+ω22 t(1−coshat)l ns2−a2 s2 tsinωt2ωs (s2+ω2)22 t(1−cosωt)l ns2+ω2 s2 1−cosωtω2 s(s2+ω2)sinωt ttan−1ω s ωt−sinωtω3 s2(s2+ω2)ta(a>−1)Γ(a+1 ) sa+1 sinωt−ωtcosωt2ω3 (s2+ω2)2t−1/2/radicalBigπ s sinωt+ωtcosωt2ωs2 (s2+ω2)2t1/2 1 2√π s3/2 cosat−cosbt/parenleftbig b2−a2/parenrightbig s (s2+a2)(s2+b2)J0(at)1 (s2+a2)1/2 278 6 Laplace Transforms L[sinωt+ωtcosωt]=ω s2+ω2+ω(s2−ω2) (s2+ω2)2=2ωs2 (s2+ω2)2, (6.29) L[cosat−cosbt]=s s2+a2−s s2+b2=(b2−a2)s (s2+a2)(s2+b2). (6.30) There are extensive tables of Laplace transforms (For example, F. Oberherttinger and E. Badii, Tables of Laplace Transforms, Springer, New York, 1973). A short list of some simple Laplace transforms is given in Table 6.1. The items in the left-hand side of the table are the ones we have shown so far. Items in the right-hand side are relations we are going to derive in the following sections. 6.2 Solving Diﬀerential Equation with Laplace Transform 6.2.1 Inverse Laplace Transform In solving diﬀerential equation with Laplace transform, we encounter the inverse problem of determining the unknown function f(t) which has a given transform F(s).The notation of L−1[F(s)] is conventionally used for the inverse Laplace transform of F(s).That is, if F(s)=L[f(t)] =/integraldisplay∞ 0e−stf(t)dt, (6.31) then f(t)=L−1[F(s)]. (6.32) Since f(t)=L−1[F(s)] =L−1[L[f(t)]] =I[f(t)], it follows that L−1Lis the identity operator I. The inverse transforms are of great practical importance and there are a variety of ways to get them. In this section, we will ﬁrst study the transform in the form of a quotient of two polynomials F(s)=p(s) q(s), where p(s)a n d q(s) have real coeﬃcients and no common factors. Since lim s→∞F(s) = lim s→∞/integraldisplay∞ 0e−stf(t)dt→0, it is clear that the degree of p(s) is lower than that of q(s).There are several closely related methods to get the inverse of such a transform. For the sake of clarity, we list them separately. 6.2 Solving Diﬀerential Equation with Laplace Transform 279 By Inspection. If the expression is simple enough, one can get the inverse directly from the table. This is illustrated in the following examples. Example 6.2.1. Find (a) L−1/bracketleftbigg1 s4/bracketrightbigg ;( b )L−1/bracketleftbigg4 (s+4 )3/bracketrightbigg ;( c )L−1/bracketleftbigg1 s2+4/bracketrightbigg . Solution 6.2.1. (a) Since L[t3]=3! s4=6 s4,t3=L−1/bracketleftbigg6 s4/bracketrightbigg , we have L−1/bracketleftbigg1 s4/bracketrightbigg =1 6L−1/bracketleftbigg6 s4/bracketrightbigg =1 6t3. (b) Since L[e−4tt2]=2 (s+4 )3,e−4tt2=L−1/bracketleftbigg2 (s+4 )3/bracketrightbigg , so L−1/bracketleftbigg4 (s+4 )3/bracketrightbigg =2L−1/bracketleftbigg2 (s+4 )3/bracketrightbigg =2 e−4tt2. (c) Since L[sin 2t]=2 s2+4,sin2t=L−1/bracketleftbigg2 s2+4/bracketrightbigg , so L−1/bracketleftbigg1 s2+4/bracketrightbigg =1 2L−1/bracketleftbigg2 s2+4/bracketrightbigg =1 2sin 2t. Example 6.2.2. Find (a) L−1/bracketleftbigg1 s2+2s+5/bracketrightbigg ;( b ) L−1/bracketleftbigg2s+1 s2+2s+5/bracketrightbigg . Solution 6.2.2. (a) First we note that 1 s2+2s+5=1 (s+1 )2+4=1 2×2 (s+1 )2+4. Since L[e−tsin 2t]=2 (s+1 )2+4,e−tsin 2t=L−1/bracketleftbigg2 (s+1 )2+4/bracketrightbigg , so L−1/bracketleftbigg1 s2+2s+5/bracketrightbigg =1 2L−1/bracketleftbigg2 (s+1 )2+4/bracketrightbigg =1 2e−tsin2t. 280 6 Laplace Transforms (b) Recall L[e−tsin 2t]=2 (s+1 )2+4, L[e−tcos 2t]=s+1 (s+1 )2+4, so we write 2s+1 s2+2s+5=2(s+1 )−1 (s+1 )2+4=2(s+1 ) (s+1 )2+4−1 22 (s+1 )2+4. Thus L−1/bracketleftbigg2s+1 s2+2s+5/bracketrightbigg =2L−1/bracketleftbigg(s+1 ) (s+1 )2+4/bracketrightbigg −1 2L−1/bracketleftbigg2 (s+1 )2+4/bracketrightbigg =2L−1[L[e−tcos 2t]]−1 2L−1[L[e−tsin 2t]] =2 e−tcos 2t−1 2e−tsin 2t. Partial Fraction Decomposition. Take the partial fractions of F(s) and then take the inverse of each term. Most probably you are familiar with partial fractions. We will use the following examples for review. Example 6.2.3. FindL−1/bracketleftbiggs−1 s2−s−2/bracketrightbigg . Solution 6.2.3. First we note s−1 s2−s−2=s−1 (s−2)(s+1 ) =a (s−2)+b (s+1 )=a(s+1 )+ b(s−2) (s−2)(s+1 ). The following are three diﬀerent ways to determine aandb. •First note that s−1=a(s+1 )+ b(s−2) must hold for all s.One way is to set s=2,then it follows that a=1 3. Similarly if we set s=−1,we see immediately that b=2 3. •Another way is to collect the terms with the same powers in s,and require the coeﬃcients of the corresponding terms on both sides of the equation be equal to each other. That is, s−1=(a+b)s+(a−2b). This means a+b=1a n d a−2b=−1.Thus a=1 3andb=2 3. 6.2 Solving Diﬀerential Equation with Laplace Transform 281 •Still another way is to note that lim s→2/braceleftbigg (s−2)s−1 (s−2)(s+1 )/bracerightbigg = lim s→2/braceleftbigg (s−2)/bracketleftbigga (s−2)+b (s+1 )/bracketrightbigg/bracerightbigg , this means lim s→2/braceleftbiggs−1 (s+1 )/bracerightbigg = lim s→2/braceleftbigg a+(s−2)b (s+1 )/bracerightbigg =a. We see immediately that a=1 3.Similarly lim s→−1/braceleftbigg (s+1 )s−1 (s−2)(s+1 )/bracerightbigg = lim s→−1/braceleftbigg (s+1 )/bracketleftbigga (s−2)+b/bracketrightbigg/bracerightbigg =b givesb=2 3. In some problems, one way is much simpler than others. Anyway, in this problem L−1/bracketleftbiggs−1 s2−s−2/bracketrightbigg =L−1/bracketleftbigg1 31 (s−2)+2 31 (s+1 )/bracketrightbigg =1 3L−1[L[e2t]] +2 3L−1[L[e−t]] =1 3e2t+2 3e−t. Example 6.2.4. FindL−1/bracketleftbigg1 s(s2+4 )/bracketrightbigg . Solution 6.2.4. There are two ways for us to take partial fractions. •If we use complex roots, 1 s(s2+4 )=a s+b s−2i+c s+2 i. Multiplying by sand taking the limit with s→0,we have a= lim s→01 s2+4=1 4. Multiplying by s−2i and taking the limit with s→2i,we have b= lim s→2i1 s(s+ 2i)=−1 8. Multiplying by s+ 2i and taking the limit with s→−2i,we have c= lim s→−2i1 s(s−2i)=−1 8. 282 6 Laplace Transforms Therefore L−1/bracketleftbigg1 s(s2+4 )/bracketrightbigg =1 4L−1/bracketleftbigg1 s/bracketrightbigg −1 8L−1/bracketleftbigg1 s−2i/bracketrightbigg −1 8L−1/bracketleftbigg1 s+2 i/bracketrightbigg =1 4−1 8e2it−1 8e−2it=1 4−1 4cos 2t. •Another way to take partial fractions is to note that b s−2i+c s+2 i=b(s+ 2i) + c(s−2i) (s−2i)(s+ 2i)=(b+c)s+ 2i(b−c) s2+4. If we let b+c=b/primeand 2i( b−c)=c/prime,then 1 s(s2+4 )=a s+b/primes+c/prime s2+4. An important point we should note is that if the denominator is second order in s,the numerator must be allowed the possibility of being ﬁrst order in s.In other words, it will not be possible for us to get the correct answer if b/primeterm is missing. With this understanding, the partial fractions can be taken directly as 1 s(s2+4 )=a s+bs+c s2+4=a/parenleftbig s2+4/parenrightbig +(bs+c)s s(s2+4 ) =as2+4a+bs2+cs s(s2+4 )=(a+b)s2+cs+4a s(s2+4 ). Therefore 1=(a+b)s2+cs+4a. The coeﬃcients of smust be equal term by term. That is, a+b=0,c=0, 4a=1.This gives a=1/4,b=−1/4,c=0.Thus L−1/bracketleftbigg1 s(s2+4 )/bracketrightbigg =L−1/bracketleftbigg1 4s−1 4s s2+4/bracketrightbigg =1 4−1 4cos 2t. Example 6.2.5. FindL−1/bracketleftbigg1 s3(s−1)/bracketrightbigg . Solution 6.2.5. 1 s3(s−1)=a s+b s2+c s3+d (s−1) =as2(s−1) +bs(s−1) +c(s−1) +ds3 s3(s−1) =(a+d)s3+(b−a)s2+(c−b)s−c s3(s−1). This requires a+d=0,b−a=0,c−b=0,−c=1.Thus c=−1, b=−1,a=−1,d= 1. Hence 6.2 Solving Diﬀerential Equation with Laplace Transform 283 L−1/bracketleftbigg1 s3(s−1)/bracketrightbigg =L−1/bracketleftbigg−1 s−1 s2−1 s3+1 (s−1)/bracketrightbigg =−1−t−1 2t2+et. Example 6.2.6. FindL−1/bracketleftbigg2ω (s2+ω2)2/bracketrightbigg . Solution 6.2.6. 2ω (s2+ω2)2=2ω [(s−iω)(s+iω)]2 =a (s−iω)+b (s−iω)2+c (s+iω)+d (s+iω)2. Multiplying both sides by ( s−iω)2and take the limit as s→iω,we have lim s→iω/braceleftbigg2ω (s+iω)2/bracerightbigg = lim s→iω/braceleftbigg (s−iω)a+b+(s−iω)2c (s+iω)+(s−iω)2d (s+iω)2/bracerightbigg . Clearly b=2ω (2iω)2=−1 2ω. If after multiplying both sides by ( s−iω)2, we take the derivative ﬁrst and then go to the limit s→iω,we have lim s→iω/braceleftbiggd ds2ω (s+iω)2/bracerightbigg = lim s→iω/braceleftbigg a+d ds/bracketleftbigg(s−iω)2c (s+iω)+(s−iω)2d (s+iω)2/bracketrightbigg/bracerightbigg . This leads to a= lim s→iω/braceleftbigg−4ω (s+iω)3/bracerightbigg =1 2ω2i. Similarly, we can show d=−1 2ω,c=−1 2ω2i. Thus we have L−1/bracketleftbigg2ω (s2+ω2)2/bracketrightbigg = L−1/bracketleftbigg1 2ω2i1 (s−iω)−1 2ω1 (s−iω)2−1 2ω2i.1 (s+iω)−1 2ω1 (s+iω)2/bracketrightbigg = 1 2ω2iL−1/bracketleftbigg1 (s−iω)−1 (s+iω)/bracketrightbigg −1 2ωL−1/bracketleftbigg1 (s−iω)2+1 (s+iω)2/bracketrightbigg = 1 2ω2i/parenleftbig eiωt−e−iωt/parenrightbig −1 2ω/parenleftbig teiωt+te−iωt/parenrightbig =1 ω2sinωt−1 ωtcosωt. 284 6 Laplace Transforms The Heaviside Expansion. The Heaviside expansion is essentially a system- atic way of taking partial fractions. In the partial fraction decomposition of p(s)/q(s), an unrepeated factor ( s−a)o fq(s) gives rise to a single fraction of the form A/(s−a).Thus F(s) can be written as F(s)=p(s) q(s)=A s−a+G(s), (6.33) where G(s) is simply the rest of the expression. Multiplication by ( s−a) gives (s−a)p(s) q(s)=A+(s−a)G(s). If we let sapproach a,the second term in the right-hand side vanishes, since G(s) has no factor that could cancel ( s−a).Therefore A= lim s→a(s−a)p(s) q(s). (6.34) Since q(a)=0,because ais an unrepeated root of q(s)=0,the limit in (6 .34) is an indeterminant of the form of 0 /0.With the L’Hospital’s rule, we have A= lim s→ap(s)+(s−a)p/prime(s) q/prime(s)=p(a) q/prime(a). (6.35) Thus the constants in the partial fraction decomposition can be quickly determined. Example 6.2.7. Use the Heaviside expansion to ﬁnd L−1/bracketleftbiggs−1 s2−s−2/bracketrightbigg . Solution 6.2.7. The roots of s2−s−2=0a r e s=2a n d s=−1,and d ds(s2−s−2) = 2 s−1.Therefore s−1 s2−s−2=a (s−2)+b (s+1 ), a= lim s→2s−1 2s−1=1 3,b= lim s→−1s−1 2s−1=2 3. Thus L−1/bracketleftbiggs−1 s2−s−2/bracketrightbigg =1 3L−1/bracketleftbigg1 s−2/bracketrightbigg +2 3L−1/bracketleftbigg1 s+1/bracketrightbigg =1 3e2t+2 3e−t. Example 6.2.8. Use the Heaviside expansion to ﬁnd L−1/bracketleftbigg2s+1 s2+2s+5/bracketrightbigg . Solution 6.2.8. The roots of s2+2s+5=0 a r e s=−1±2i,andd ds(s2+ 2s+2 )=2 s+2.Thus 6.2 Solving Diﬀerential Equation with Laplace Transform 285 2s+1 s2+2s+5=a s−(−1 + 2i)+b s−(−1−2i), a= lim s→−1+2i2s+1 2s+2=1+i 4,b= lim s→−1−2i2s+1 2s+2=1−i 4. Therefore L−1/bracketleftbigg2s+1 s2+2s+5/bracketrightbigg =/parenleftbigg 1+1 4i/parenrightbigg L−1/bracketleftbigg1 s−(−1 + 2i)/bracketrightbigg +/parenleftbigg 1−1 4i/parenrightbigg L−1/bracketleftbigg1 s−(−1−2i)/bracketrightbigg . Recall L−1/bracketleftbigg1 s−c/bracketrightbigg =ect, we have L−1/bracketleftbigg1 s−(−1 + 2i)/bracketrightbigg =e(−1+2i) t=e−tei2t=e−t(cos 2t+is i n2 t), and L−1/bracketleftbigg1 s−(−1−2i)/bracketrightbigg =e−te−i2t=e−t(cos 2t−isin2t). Hence L−1/bracketleftbigg2s+1 s2+2s+5/bracketrightbigg =/parenleftbigg 1+1 4i/parenrightbigg e−t(cos 2t+is i n2 t) +/parenleftbigg 1−1 4i/parenrightbigg e−t(cos 2t−isin2t) =2 e−tcos 2t−1 2e−tsin 2t. In general, if q(s) is a polynomial with unrepeated roots, the Heaviside expansion is the most eﬃcient way in partial fraction decomposition. If q(s) is already in the form of a product of factors ( s−a1)(s−a2)···(s−an), then other methods of partial fraction may be equally or more eﬃcient. In any case, if the complex roots are used, it is useful to keep in mind that if the original function is real, the ﬁnal result must also be real. If there is an imaginary term in the ﬁnal result, then there must be a mistake somewhere. Ifq(s) has repeated roots, we can write it as p(s) q(s)=Am (s−a)m+Am−1 (s−a)m−1+···+A1 (s−a). With a similar argument, one can show that Ak=1 (m−k)!lim s→adm−k dsm−k/bracketleftbigg(s−a)mp(s) q(s)/bracketrightbigg ,k =1,...,m . (6.36) 286 6 Laplace Transforms Unfortunately, in practice this formula is not necessarily simpler than other partial fraction methods, such as the one shown in Example 6.2.6. In fact, problems of that nature are best solved by using the derivatives of a transform. Using Derivatives of the Transform. In Example 6.2.6, we used the partial fraction to ﬁnd L−1[1/(s2+a2)2].A simpler way to handle such problems is to make use of the properties of derivatives. The procedures are illustrated in the following examples. Example 6.2.9. Find (a) L−1/bracketleftbigg1 (s2+a2)2/bracketrightbigg ,(b)L−1/bracketleftbiggs (s2+a2)2/bracketrightbigg , (c)L−1/bracketleftbiggs2 (s2+a2)2/bracketrightbigg ,(d)L−1/bracketleftbiggs3 (s2+a2)2/bracketrightbigg . Solution 6.2.9. (a) Taking the derivative d daa s2+a2=1 s2+a2−2a2 (s2+a2)2, we can write 1 (s2+a2)2=1 2a2/parenleftbigg1 s2+a2−d daa s2+a2/parenrightbigg . Since L[sinat]=a s2+a2, we have 1 (s2+a2)2=1 2a2/parenleftbigg1 aL[sinat]−d daL[sinat]/parenrightbigg =1 2a3L[sinat]−1 2a2d da/integraldisplay∞ 0e−stsinatdt =1 2a3L[sinat]−1 2a2/integraldisplay∞ 0e−sttcosatdt =1 2a3L[sinat]−1 2a2L[tcosat]. Therefore L−1/bracketleftbigg1 (s2+a2)2/bracketrightbigg =L−1/bracketleftbigg1 2a3L[sinat]−1 2a2L[tcosat]/bracketrightbigg =1 2a3sinat−1 2a2tcosat. (b) Take derivative with respect to s d dsa s2+a2=−2as (s2+a2)2, 6.2 Solving Diﬀerential Equation with Laplace Transform 287 so s (s2+a2)2=−1 2ad dsa s2+a2=−1 2ad dsL[sinat] =−1 2ad ds/integraldisplay∞ 0e−stsinatdt=1 2a/integraldisplay∞ 0e−sttsinatdt =1 2aL[tsinat]. Therefore L−1/bracketleftbiggs (s2+a2)2/bracketrightbigg =L−1/bracketleftbigg1 2aL[tsinat]/bracketrightbigg =1 2atsinat. (c) It follows from the result of (b), sL/bracketleftbigg1 2atsinat/bracketrightbigg =ss (s2+a2)2=s2 (s2+a2)2. Recall L/bracketleftbiggdf dt/bracketrightbigg =sL[f]−f(0),s L[f]=L/bracketleftbiggdf dt/bracketrightbigg +f(0). Letf=1 2atsinat,sodf dt=1 2asinat+1 2tcosat;f(0) = 0 , we have sL/bracketleftbigg1 2atsinat/bracketrightbigg =L/bracketleftbigg1 2asinat+1 2tcosat/bracketrightbigg . Therefore L−1/bracketleftbiggs2 (s2+a2)2/bracketrightbigg =L−1/bracketleftbigg sL/bracketleftbigg1 2atsinat/bracketrightbigg/bracketrightbigg =1 2asinat+1 2tcosat. (d) From the result of (c) s3 (s2+a2)2=ss2 (s2+a2)2=sL/bracketleftbigg1 2asinat+1 2tcosat/bracketrightbigg . This time, let f=1 2asinat+1 2tcosat, sodf dt=c o s at−a 2tsinat;f(0) = 0 , thus sL/bracketleftbigg1 2asinat+1 2tcosat/bracketrightbigg =L/bracketleftBig cosat−a 2tsinat/bracketrightBig , L−1/bracketleftbiggs3 (s2+a2)2/bracketrightbigg =L−1/bracketleftBig L/bracketleftBig cosat−a 2tsinat/bracketrightBig/bracketrightBig =c o s at−a 2tsinat. 288 6 Laplace Transforms Relation satisfied by ƒ(t) (like a DE)Relation satisfied by F(s) (hopefully simple) ƒ(t) is foundStep 1: Transform. F(s) is found Step 3: Invert.Step 2: Solve for F(s). (Direct solution difficult) Fig. 6.1. Steps of using Laplace transform to solve diﬀerential equations 6.2.2 Solving Diﬀerential Equations The idea of using Laplace transform to solve diﬀerential equation is expressed in Fig. 6.1. Suppose we have a diﬀerential equation in which the unknown function is f(t).The ﬁrst step is to apply the Laplace transform to this dif- ferential equation. The result is a relation satisﬁed by F(s)=L[f].Generally this is an algebraic equation. The second step is to ﬁnd F(s) by solving this algebraic equation. The third and ﬁnal step is to ﬁnd the unknown function f(t) by taking the inverse of the Laplace transform F(s). A few example will make this procedure clear. Example 6.2.10. Find the solution of the diﬀerential equation y/prime/prime+y=s i n2 t, satisfying the initial conditions y(0) = 0 ,y/prime(0) = 1 . Solution 6.2.10. Applying Laplace transform to the equation , L[y/prime/prime+y]=L[sin 2t], we have s2L[y]−sy(0)−y/prime(0) + L[y]=2 s2+4. With the initial values of y(0) and y/prime(0),this equation can be written as (s2+1 )L[y]=1+2 s2+4. This algebraic equation can be easily solved to give L[y]=s2+6 (s2+1 )(s2+4 ). 6.2 Solving Diﬀerential Equation with Laplace Transform 289 Thus y(t)=L−1/bracketleftbiggs2+6 (s2+1 )(s2+4 )/bracketrightbigg . Using methods of the last section, we ﬁnd y(t)=5 3sint−1 3sin 2t. Example 6.2.11. Find the solution of the diﬀerential equation y/prime/prime+4y=s i n2 t, y (0) = 10 ,y/prime(0) = 0 . Solution 6.2.11. Applying the Laplace transform to both sides of the equation L[y/prime/prime+4y]=L[sin 2t], we have s2L[y]−sy(0)−y/prime(0) + 4 L[y]=2 s2+4. With the initial values of y(0) and y/prime(0),this equation can be written as (s2+4 )L[y]=1 0 s+2 s2+4. Therefore y=L−1/bracketleftbigg10s (s2+4 )+2 (s2+4 )2/bracketrightbigg . This leads to y=1 0c o s2 t+1 8sin2t−1 4tcos 2t. Example 6.2.12. Find the solution of the diﬀerential equation y/prime/prime+4y/prime+4y=t2e−2t,y (0) = 0 ,y/prime(0) = 0 . Solution 6.2.12. Applying the Laplace transform to the equation L[y/prime/prime+4y/prime+4y]=L[t2e−2t], With the initial values of y(0) and y/prime(0),we have s2L[y]+4sL[y]+4L[y]=2 (s+2 )3. Collecting terms (s2+4s+4 )L[y]=(s+2 )2L[y]=2 (s+2 )3, 290 6 Laplace Transforms or L[y]=2 (s+2 )5. The solution is the inverse transform y=2 4!L−1/bracketleftbigg4! (s+2 )5/bracketrightbigg =1 12t4e−2t. Example 6.2.13. Find the solution of the set of the diﬀerential equations y/prime−2y+z=0, z/prime−y−2z=0, satisfying the initial conditions y(0) = 1 ,z(0) = 0 . Solution 6.2.13. Applying the Laplace transform to each of the equations L[y/prime−2y+z]=L[0], L[z/prime−y−2z]=L[0], we obtain sL[y]−y(0)−2L[y]+L[z]=0, sL[z]−z(0)−L[y]−2L[z]=0. After substituting the initial conditions and collecting terms, we have (s−2)L[y]+L[z]=1, L[y]−(s−2)L[z]=0. This set of algebraic equations can be easily solved to give L[y]=s−2 (s−2)2+1, L[z]=1 (s−2)2+1. Thus y=L−1/bracketleftbiggs−2 (s−2)2+1/bracketrightbigg =e2tcost, z=L−1/bracketleftbigg1 (s−2)2+1/bracketrightbigg =e2tsint. 6.3 Laplace Transform of Impulse and Step Functions 291 6.3 Laplace Transform of Impulse and Step Functions Some of the most useful and interesting applications of the Laplace transform method occur in the solution of linear diﬀerential equations with discontinuous or impulsive nonhomogeneous functions. Equations of this type frequently arise in the analysis of the ﬂow of current in electric circuits or the vibrations of mechanical systems, where voltages or forces of large magnitude act over very short time intervals. To deal eﬀectively with functions having jump discontinuities, we ﬁrst introduce two functions known as delta function and step function. 6.3.1 The Dirac Delta Function The delta function, δ(t),was ﬁrst proposed in 1930 by Dirac in the develop- ment of the mathematical formalism of quantum mechanics. He required a function which is zero everywhere, except at a single point, where it is discon- tinuous and behaved like an inﬁnitely high and inﬁnitely narrow spike of unit area. Mathematicians were quick to point out that, strictly speaking, there is no function which has these properties. But Dirac supposed there was, and proceeded to use it so successfully that a new branch of mathematics was developed to justify its use. This area of mathematics is called the theory of distribution or of generalized functions. While it is nice to know that the mathematical foundation of the delta function has been established in com- plete details, for applications in physical sciences we need only to know its operational deﬁnition. Deﬁnition of δFunction. The delta function is a sharply peaked function deﬁned as δ(t−t0)=/braceleftbigg 0t/negationslash=t0 ∞t=t0,(6.37) but such that the integral of δ(t−t0) is normalized to unity: /integraldisplay+∞ −∞δ(t−t0)dt=1. (6.38) Clearly the limits −∞and∞may be replaced by t0−/epsilon1andt0+/epsilon1as long as/epsilon1>0,since δ(t−t0) is equal to zero for t/negationslash=t0.We can think of it as an inﬁnitely high and inﬁnitely narrow function shown in Fig. 6.2, where h→∞ andτ→0 in such a way that the area under it is equal to one. Mathematically, the δfunction is deﬁned by how it behaves inside an integral. In fact the ﬁrst operation where Dirac used the delta function is the integration/integraldisplay+∞ −∞f(t)δ(t−t0)dt, where f(t) is a continuous function. This integral can be evaluated by the following argument. Since δ(t−t0)i sz e r of o r t/negationslash=t0,the limit of integration 292 6 Laplace Transforms h t0tt Fig. 6.2. A sharply peaked function. If h→∞ andτ→0 in such a way that the area under it is equal to 1, then this function becomes a delta function δ(t−t0) may be changed to t0−/epsilon1andt0+/epsilon1, where /epsilon1is a small positive number. Moreover, since f(x) is continuous at t=t0,its values within the interval (t0−/epsilon1,t0+/epsilon1) will not diﬀer much from f(t0) and we can claim, approximately, that /integraldisplay+∞ −∞f(t)δ(t−t0)dt=/integraldisplayt0+/epsilon1 t0−/epsilon1f(t)δ(t−t0)dt≈f(t0)/integraldisplayt0+/epsilon1 t0−/epsilon1δ(t−t0)dt with the approximation improving as /epsilon1approaches zero. However, /integraldisplayt0+/epsilon1 t0−/epsilon1δ(t−t0)dt=1 for all values of /epsilon1.It appears then that letting /epsilon1→0,we have exactly /integraldisplay+∞ −∞f(t)δ(t−t0)dt=f(t0). (6.39) This integral is sometimes referred to as the shifting property of the delta function: δ(t−t0) acts as a sieve, selecting from all possible values of f(t) its value at the point t=t0. Delta Function with Complicated Arguments. In general the argument of the delta function can be any function of the independent variable. It turns out that such a function can always be rewritten as a sum of delta functions of simple argument. Here are some examples. •δ(−t) Lett/prime=−t,then d t=−dt/prime.We can write /integraldisplay+∞ −∞f(t)δ(−t)dt=−/integraldisplay−∞ +∞f(−t/prime)δ(t/prime)dt/prime=/integraldisplay+∞ −∞f(−t/prime)δ(t/prime)dt/prime=f(0). 6.3 Laplace Transform of Impulse and Step Functions 293 Since /integraldisplay+∞ −∞f(t)δ(t)dt=f(0), therefore δ(−t)=δ(t). (6.40) This result is almost self evident. •δ(at) Lett/prime=at,then d t=dt/prime/a.Hence, if a>0, /integraldisplay+∞ −∞f(t)δ(at)dt=/integraldisplay+∞ −∞f/parenleftbiggt/prime a/parenrightbigg δ(t/prime)1 adt/prime=1 a/integraldisplay+∞ −∞f/parenleftbiggt/prime a/parenrightbigg δ(t/prime)dt/prime =1 af/parenleftbigg0 a/parenrightbigg =1 af(0). Since /integraldisplay+∞ −∞f(t)1 aδ(t)dt=1 a/integraldisplay+∞ −∞f(t)δ(t)dt=1 af(0), therefore δ(at)=1 aδ(t). Since δ(−at)=δ(at), we can write δ(at)=1 |a|δ(t). (6.41) •δ(t2−a2) The argument of this function goes to zero when t=aandt=−a,which seems to imply two δfunctions. There can be contributions to the integral /integraldisplay+∞ −∞f(t)δ/parenleftbig t2−a2/parenrightbig dt=/integraldisplay+∞ −∞f(t)δ[(t−a)(t+a)]dt only at the zeros of the argument of the delta function. That is /integraldisplay+∞ −∞f(t)δ/parenleftbig t2−a2/parenrightbig dt=/integraldisplay−a+/epsilon1 −a−/epsilon1f(t)δ/parenleftbig t2−a2/parenrightbig dt+/integraldisplaya+/epsilon1 a−/epsilon1f(t)δ/parenleftbig t2−a2/parenrightbig dt. Near the two zeros, t2−a2can be approximated as t2−a2=(t−a)(t+a)=/braceleftbigg (−2a)(t+a)t→−a (+2a)(t−a)t→+a. 294 6 Laplace Transforms In the limit as /epsilon1→0,the integral becomes /integraldisplay+∞ −∞f(t)δ/parenleftbig t2−a2/parenrightbig dt=/integraldisplay−a+/epsilon1 −a−/epsilon1f(t)δ((−2a)(t+a)) dt +/integraldisplaya+/epsilon1 a−/epsilon1f(t)δ((2a)(t−a)) dt=1 |2a|/integraldisplay−a+/epsilon1 −a−/epsilon1f(t)δ(t+a)dt +1 |2a|/integraldisplaya+/epsilon1 a−/epsilon1f(t)δ(t−a)dt=/integraldisplay+∞ −∞f(t)1 |2a|[δ(t+a)+δ(t−a)]dt. Therefore δ/parenleftbig t2−a2/parenrightbig =1 |2a|[δ(t+a)+δ(t−a)]. (6.42) The Laplace Transform of the Delta Function and Its Derivative. It follows from the deﬁnitions of the Laplace transform and the delta function that the Laplace transform of the delta function is given by L[δ(t−a)] =/integraldisplay∞ 0e−stδ(t−a)dt=e−sa. (6.43) The Laplace transform of the derivative of a delta function can be evalu- ated using integration by parts: L[δ/prime(t−a)] =/integraldisplay∞ 0e−std dtδ(t−a)dt=/integraldisplay∞ 0e−std(δ(t−a)) =/bracketleftbig e−stδ(t−a)/bracketrightbig∞ 0−/integraldisplay∞ 0δ(t−a)d dte−stdt. Since δ(t−a) vanishes everywhere except at t=a,at both upper and lower limits the integrated part is equal to zero. Therefore L[δ/prime(t−a)] =s/integraldisplay∞ 0δ(t−a)e−stdt=se−sa. (6.44) In dealing with phenomena of an impulsive nature, this is a very useful expression. 6.3.2 The Heaviside Unit Step Function Deﬁnition of the Step Function. The Heaviside unit step function u(t−c)c a n be deﬁned from the integration of the delta function δ(t/prime−c) u(t−c)=/integraldisplayt −∞δ(t/prime−c)dt/prime. (6.45) The delta function is identically equal to zero if t/prime<c .The upper limit of the integration variable t/primeist.Iftis less than c,then all t/primewill be less than c. 6.3 Laplace Transform of Impulse and Step Functions 295 cty 1 Fig. 6.3. The Heaviside unit step function u(t−c) The integral is equal to zero. If tis greater than c,the integral is equal to one by the deﬁnition of the delta function. Thus u(t−c)=/braceleftbigg 0 t<c 1 t>c. (6.46) The step function can be deﬁned directly with (6 .46) without referring to (6.45).However, with (6 .45),it is immediately clear that d dtu(t−c)=δ(t−c). (6.47) A plot of the Heaviside unit step function y=u(t−c) is shown in Fig. 6.3. Interestingly, the function does not get its name because it is heavy on one side, but, rather from the British engineer Oliver Heaviside. Very often this function is simply called step function. Very often we have to deal with a pulse of a ﬁnite duration. These step functions are very convenient in such situation. For example, the square pulse y(t)=⎧ ⎨ ⎩00 <t<π 1 π<t< 2π 02 π<t< ∞ c a nb ee x p r e s s e da s y(t)=u(t−π)−u(t−2π). The sketch of this function is shown in Fig. 6.4. Shifting Operation. In some problems a system which becomes active at t=0, because of some initial disturbance, is subsequently acted upon by another disturbance beginning at a later time t=c.In this situation, the analytical description is greatly facilitated by the function y=f(t−c)u(t−c), which represent a shifting operation. First, f(t−c) represents a translation off(t) by a distance cin the positive t-direction. Multiplying u(t−c) has the 296 6 Laplace Transforms y 1 π 2π 3πt Fig. 6.4. The square impulse u(t−π)−u(t−2π) yy ƒ(0) ƒ(0) t tc(a) (b) Fig. 6.5. A translation of a given function. ( a)y=f(t); (b)y=f(t−c)u(t−c) eﬀect of “cutting oﬀ” or making everything vanish to the left of c.This is shown in Fig. 6.5. Laplace Transform Involving Step Function. The Laplace transform of the step function is easily determined: L[u(t−c)] =/integraldisplay∞ 0e−stu(t−c)dt=/integraldisplay∞ ce−stdt =1 se−sc. (6.48) The step function is particularly important in transform theory because of the following relationship between the transform of f(t) and that of its translation f(t−c)u(t−c). L[f(t−c)u(t−c)] =/integraldisplay∞ 0e−stf(t−c)u(t−c)dt=/integraldisplay∞ ce−stf(t−c)dt. Making a change of variable t/prime=t−c,we have /integraldisplay∞ ce−stf(t−c)dt=/integraldisplay∞ 0e−s(t/prime+c)f(t/prime)dt/prime=e−sc/integraldisplay∞ 0e−st/primef(t/prime)dt/prime =e−scL[f(t)]. Therefore L[f(t−c)u(t−c)] = e−scL[f(t)]. (6.49) 6.4 Diﬀerential Equations with Discontinuous Forcing Functions 297 Its inverse is of considerable importance. f(t−c)u(t−c)=L−1[e−scL[f(t)]]. (6.50) This relationship is sometimes referred as t-shifting (or second shifting) theorem. 6.4 Diﬀerential Equations with Discontinuous Forcing Functions In this section we turn our attention to some examples in which the nonho- mogeneous term, or forcing function, is discontinuous. We start with the simplest case. A particle of mass minitially at rest is set into motion by a sudden blow at t=t0.Assuming no friction, we wish to ﬁnd the position as a function of time. Such a common every-day occurrence is rather “awkward” for “ordinary” mathematics to deal with. However, with Laplace transform and delta function, it becomes very easy. In the Newton’s dynamic equation md2x dt2=F, (6.51) let us express the force of the sudden blow by the delta function F=Pδ(t−t0). (6.52) The initial conditions are x(0) = 0 ,x/prime(0) = 0 . (6.53) Applying the Laplace transform to the diﬀerential equation L[mx/prime/prime]=L[Pδ(t−t0)], (6.54) we obtain ms2L[x]=Pe−st0. (6.55) So x(t)=P mL−1/bracketleftbigge−st0 s2/bracketrightbigg =P mL−1[e−st0L[t]] =P m(t−t0)u(t−t0). (6.56) This means x(t)=/braceleftbigg0 t<t 0 P m(t−t0)t>t 0. (6.57) 298 6 Laplace Transforms The result says that the particle will stay put until t0,after that the distance will increase linearly with time. The velocity of the particle is given by v=dx dt=P m, (6.58) which is a constant. In fact we see that the amplitude Pof the delta function is equal to mvwhich is the momentum. This shows that what the sudden blow did is to impart a momentum Pto the particle. This momentum stays the same with the particle thereafter. Example 6.4.1. Let us consider the damped, driven, harmonic oscillator. The mass mis driven by an applied force F(t).It also experiences the spring force−kx(t) and a friction force −bx/prime(t),proportional to its velocity. The diﬀerential equation describing the motion is mx/prime/prime+bx/prime+kx=F(t). If it is at rest initially x(0) = 0 ,x/prime(0) = 0 , and the force function is an ideal impulse peaked at t0,that is F(t)=P0δ(t−t0), ﬁnd the displacement xas a function of time t. Solution 6.4.1. Applying the Laplace transform to both sides of the equation L/bracketleftbigg x/prime/prime+b mx/prime+k mx/bracketrightbigg =P0 mL[δ(t−t0)] leads to s2L[x]+b msL[x]+k mL[x]=P0 me−st0. Therefore L[x]=P0 m1 s2+b ms+k me−st0. Let us write s2+b ms+k m=s2+b ms+/parenleftbiggb 2m/parenrightbigg2 −/parenleftbiggb 2m/parenrightbigg2 +k m =/parenleftbigg s+b 2m/parenrightbigg2 +k m−/parenleftbiggb 2m/parenrightbigg2 , 6.4 Diﬀerential Equations with Discontinuous Forcing Functions 299 and simplify the notation with α=b 2m,ω2=k m−/parenleftbiggb 2m/parenrightbigg2 , so we have L[x]=P0 mωω (s+α)2+ω2e−st0. Three diﬀerent cases arise: (a) The oscillatory case, ω2>0. x(t)=L−1/bracketleftbiggP0 mωω (s+α)2+ω2e−st0/bracketrightbigg =P0 mωL−1/bracketleftbigg e−st0ω (s+α)2+ω2/bracketrightbigg =P0 mωL−1/bracketleftbig e−st0L[e−αtsinωt]/bracketrightbig =P0 mωe−α(t−t0)sinω(t−t0)u(t−t0). (b) The over-damped case ω2<0.Letβ2=−ω2, x(t)=L−1/bracketleftbiggP0 mββ (s+α)2−β2e−st0/bracketrightbigg =P0 mβL−1/bracketleftbigg e−st0β (s+α)2−β2/bracketrightbigg =P0 mβL−1[e−st0L[e−αtsinhβt]] =P0 mβe−α(t−t0)sinhβ(t−t0)u(t−t0). (c) The critically damped case ω2=0. x(t)=L−1/bracketleftbiggP0 m1 (s+α)2e−st0/bracketrightbigg =P0 mL−1/bracketleftbigg e−st01 (s+α)2/bracketrightbigg =P0 mL−1[e−st0L[e−αtt]] =P0 me−α(t−t0)(t−t0)u(t−t0). Notice in all three cases, x(t) is equal to zero before t=t0,as you would expect, because the system cannot respond until after the impulse has oc- curred. This kind of behavior is often referred to as being causal. Causality, a characteristic of solutions involving time, requires that there can be no re- sponse before the application of a drive. It is interesting to note that Newton’s equation is invariant under the transformation of t→−t.Thus clearly causality is not implied by Newton’s equation. The causality shown here is the result of the deﬁnition of the Laplace transform. The fact that this important physical requirement is built in the Laplace transformation is another reason that this method is so useful. Example 6.4.2. A mass m= 1 is attached to spring with constant k=4,and there is no friction, b=0.The mass is released from rest with x(0) = 3 .At the instant t=2πthe mass is struck with a hammer, providing an impulse P0=8.Determine the motion of the mass. 300 6 Laplace Transforms 035 4p 2p 6px t Fig. 6.6. The plot of x(t) = 3cos 2 t+4s i n2 ( t−2π)u(t−2π) Solution 6.4.2. We need to solve the initial value problem x/prime/prime+4x=8δ(t−2π);x(0) = 3 ,x/prime(0) = 0 . Apply the Laplace transform to get s2L[x]−3s+4L[x]=8 e−2πs, so (s2+4 )L[x]=3s+8 e−2πs. Therefore L[x]=3s (s2+4 )+8e−2πs (s2+4 ), hence x(t)=L−1/bracketleftbigg3s s2+4/bracketrightbigg +L−1/bracketleftbigg8e−2πs s2+4/bracketrightbigg =3L−1/bracketleftbiggs s2+4/bracketrightbigg +4L−1/bracketleftbigg2e−2πs s2+4/bracketrightbigg =3L−1[L[cos 2t]] + 4L−1[e−2πsL[sin 2t]] =3c o s2 t+4s i n2 ( t−2π)u(t−2π) or x(t)=/braceleftbigg3cos2 tt < 2π 3cos2 t+4s i n2 ( t−2π)t>2π. As 3cos 2 t+4s i n2 t= 5cos(2 t−θ)a n d θ= tan−1(4/3),we see the eﬀect of the impulse at t=2π.It instantaneously increases the amplitude of the oscillations from 3 to 5. Although the frequency is still the same, there is a discontinuity in velocity. The plot of x(t) is shown in Fig. 6.6. Example 6.4.3. Consider the RLC series circuit shown in Fig. 6.7 with R= 110 Ω ,L=1H ,C=0.001 F ,and a battery supplying an emf of 90 V . Initially there is no current in the circuit and no charge on the capacitor. At t= 0 the switch is closed and at t=T(T= 1 s)the battery is removed from the circuit in such a way that the RLC circuit is still closed but without emf. Find the current i(t) as a function of time. 6.4 Diﬀerential Equations with Discontinuous Forcing Functions 301 εoRL C t = 0 t = Ti(t) q(t) −q(t) Fig. 6.7. A RLC circuit. The open circuit without charge on the capacitor is closed att=0.Att=T,the battery is removed from the circuit in such a way that the circuit is closed but without emf Solution 6.4.3. The circuit equation is given by Li/prime+Ri+1 Cq=e(t) i=dq dt and the initial conditions are i(0) = 0 ,q(0) = 0 . In this problem e(t) = 90[ u(t)−u(t−1)]. Apply the Laplace transform to the two diﬀerential equations to get LsL[i]+RL[i]+1 CL[q]=L[e(t)] L[i]=sL[q]. Combine the two equations we obtain LsL[i]+RL[i]+1 CsL[i]=L[e(t)]. Putting in the R, L, C, ande(t) values in, we have sL[i] + 110 L[i]+1 0.001sL[i]=L[90[u(t)−u(t−1)]] =9 01−e−s s. 302 6 Laplace Transforms Therefore L[i]=9 01−e−s s2+ 110 s+ 1000. Since90 s2+ 110 s+ 1000=1 s+1 0−1 s+ 100, so we have i(t)=L−1/bracketleftbigg1 s+1 0−1 s+ 100−e−s/parenleftbigg1 s+1 0−1 s+ 100/parenrightbigg/bracketrightbigg =e−10t−e−100t−(e−10(t−1)−e−100(t−1))u(t−1). 6.5 Convolution Another important general property of the Laplace transform has to do with the products of transforms. It often happens that we are given two transforms F(s)a n d G(s) whose inverses f(t)a n d g(t) we know, and we would like to calculate the inverse of the product F(s)G(s) from those known inverses f(t) andg(t). The inverse is called the convolution of f(t)a n d g(t). In order to understand the meaning of the mathematical formulation, we will ﬁrst consider a speciﬁc example. 6.5.1 The Duhamel Integral Let us once again consider the damped driven oscillator mx/prime/prime+bx/prime+kx=f(t) (6.59) withx(0) = 0 ,x/prime(0) = 0 .Applying the Laplace transform we get L[x]=L[f(t)] m[(s+α)2+ω2], (6.60) where α=b 2m,ω2=k m−(b 2m)2.Iff(t) is a unit impulse at time τ, f(t)=δ(t−τ), (6.61) then we have L[x]=e−sτ m[(s+α)2+ω2]. (6.62) Therefore x(t)=L−1[e−sτL[1 mωe−αtsinωt]] =1 mωe−α(t−τ)sinω(t−τ)u(t−τ). (6.63) 6.5 Convolution 303 Fort>τ, x(t)=1 mωe−α(t−τ)sinω(t−τ)u(t−τ). (6.64) If we designate the solution as g(t) in the particular case where τis equal to zero, g(t)=1 mωe−αtsinωt, (6.65) so in general if τis not equal to zero, x(t)=g(t−τ). (6.66) If the force function is f(t)=Pδ(t−τ), (6.67) the solution (or the response function) is clearly x(t)=Pg(t−τ). (6.68) Now we consider the response of the system under a general external force function shown in Fig. 6.8. This force may be assumed to be made up of a series of impulses of vary- ing magnitude. As we have discussed, the impulse is actually momentum P imparted. Since the change of momentum is equal to force (∆ P/∆t=f),the impulse imparted during a short time interval is equal to force multiplied by the time duration. Assuming that at time τ,the force f(τ) acts on the system for a short period of time ∆ τ,the impulse acting at t=τis given by f(τ)∆τ.At any timet, the elapsed time since the impulse is t−τ,so the response of the system at time tdue to this impulse is ∆x(t)=f(τ)∆τg(t−τ). (6.69) ƒ(t) tO tƒ(τ) ∆τ τ + ∆τ τ Fig. 6.8. An arbitrary forcing function 304 6 Laplace Transforms The total response at time tcan be found by summing all the responses due to the elementary impulses acting all times x(t)=/summationdisplay f(τ)g(t−τ)∆τ. (6.70) Letting ∆ τ→0 and replacing the summation by the integration, we obtain x(t)=/integraldisplayt 0f(τ)g(t−τ)dτ (6.71) or x(t)=1 mω/integraldisplayt 0f(τ)e−α(t−τ)sinω(t−τ)dτ. (6.72) This result is known as the Duhamel integral. In many cases the function f(τ) has a form that permits an explicit integration. In the case such integra- tion is not possible, it can be evaluated numerically without much diﬃculty. 6.5.2 The Convolution Theorem The Duhamel integral can also be viewed in the following way. Since g(t)=1 mωe−αtsinωt=L−1/bracketleftbigg1 m[(s+α)2+ω2]/bracketrightbigg , (6.73) and by (6 .60) L[x(t)] =L[f(t)]1 m[(s+α)2+ω2] =L[f(t)]L[g(t)], (6.74) it follows that x(t)=L−1[L[f(t)]L[g(t)]]. (6.75) On the other hand x(t)=/integraldisplayt 0f(τ)g(t−τ)dτ, (6.76) therefore /integraldisplayt 0f(τ)g(t−τ)dτ=L−1[L[f(t)]L[g(t)]]. (6.77) It turns out, as long as these transforms exist, this relationship L/bracketleftbigg/integraldisplayt 0f(τ)g(t−τ)dτ/bracketrightbigg =L[f(t)]L[g(t)]] (6.78) is generally true for any arbitrary functions of fandg.It is known as the convolution theorem. If this is true, then L/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg =L[f(t)]L[g(t)]] (6.79) 6.5 Convolution 305 must also be true, since the roles played by fandgin the equation are symmetric. This can be easily demonstrated directly by a change of variable. Letλ=t−τ,then /integraldisplayt 0f(τ)g(t−τ)dτ=/integraldisplay0 tf(t−λ)g(λ)d(−λ) =/integraldisplayt 0f(t−λ)g(λ)dλ. (6.80) The proof of the convolution theorem goes as follows. By deﬁnition L/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg =/integraldisplay∞ 0e−st/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg dt. (6.81) Now with u(t−λ)=/braceleftbigg 1 λ<t 0 λ>t(6.82) and f(t−λ)g(λ)u(t−λ)=/braceleftbigg f(t−λ)g(λ)λ<t 0 λ>t.(6.83) We can write /integraldisplay∞ 0f(t−λ)g(λ)u(t−λ)dλ=/integraldisplayt 0f(t−λ)g(λ)u(t−λ)dλ +/integraldisplay∞ tf(t−λ)g(λ)u(t−λ)dλ,(6.84) the second term on the right-hand side is equal to zero because the lower limit ofλist,soλ>t . In the ﬁrst term on the right-hand side, the range of λis between 0 and t,soλ<t , using (6 .83) we have /integraldisplay∞ 0f(t−λ)g(λ)u(t−λ)dλ=/integraldisplayt 0f(t−λ)g(λ)dλ. (6.85) Putting (6 .85) into (6 .81) L/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg =/integraldisplay∞ 0e−st/bracketleftbigg/integraldisplay∞ 0f(t−λ)g(λ)u(t−λ)dλ/bracketrightbigg dt,(6.86) and changing the order of integration /integraldisplay∞ 0e−st/bracketleftbigg/integraldisplay∞ 0f(t−λ)g(λ)u(t−λ)dλ/bracketrightbigg dt =/integraldisplay∞ 0g(λ)/bracketleftbigg/integraldisplay∞ 0e−stf(t−λ)u(t−λ)dt/bracketrightbigg dλ, (6.87) 306 6 Laplace Transforms we obtain L[/integraldisplayt 0f(t−λ)g(λ)dλ]=/integraldisplay∞ 0g(λ)/bracketleftbigg/integraldisplay∞ 0e−stf(t−λ)u(t−λ)dt/bracketrightbigg dλ.(6.88) Because of the presence of u(t−λ), the integrand of the inner integral is identically zero for all t<λ . Hence the inner integration eﬀectively starts not att= 0 but at t=λ.Therefore L/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg =/integraldisplay∞ 0g(λ)/bracketleftbigg/integraldisplay∞ λe−stf(t−λ)dt/bracketrightbigg dλ. (6.89) Now in the inner integral on the right, let t−λ=τand d t=dτ.Then L/bracketleftbigg/integraldisplayt 0f(t−λ)g(λ)dλ/bracketrightbigg =/integraldisplay∞ 0g(λ)/bracketleftbigg/integraldisplay∞ 0e−s(τ+λ)f(τ)dτ/bracketrightbigg dλ =/integraldisplay∞ 0e−sλg(λ)/bracketleftbigg/integraldisplay∞ 0e−sτf(τ)dτ/bracketrightbigg dλ =/bracketleftbigg/integraldisplay∞ 0e−sτf(τ)dτ]/bracketrightbigg/bracketleftbigg/integraldisplay∞ 0e−sλg(λ)dλ/bracketrightbigg =L[f(t)]L[g(t)] (6.90) as asserted. A common notation is to designate the convolution integral as /integraldisplayt 0f(t−λ)g(λ)dλ=f(t)∗g(t). (6.91) So the convolution theorem is often written as L[f]L[g]=L[f∗g]. (6.92) Example 6.5.1. Use convolution to ﬁnd L−1/bracketleftbigg1 s2(s−a)/bracketrightbigg . Solution 6.5.1. Since L[t]=1 s2,L[eat]=1 s−a, we can write L−1/bracketleftbigg1 s2(s−a)/bracketrightbigg =L−1/bracketleftbigg1 s2·1 s−a/bracketrightbigg =L−1[L[t]L[eat]]. Therefore L−1/bracketleftbigg1 s2(s−a)/bracketrightbigg =teat=/integraldisplayt 0τea(t−τ)dτ=1 a2/parenleftbig eat−at−1/parenrightbig . 6.6 Further Properties of Laplace Transforms 307 6.6 Further Properties of Laplace Transforms 6.6.1 Transforms of Integrals From the property of the transform of a derivative, one can derive a formula for the transform of an integral. L/bracketleftbigg/integraldisplayt 0f(x)dx/bracketrightbigg =/integraldisplay∞ 0e−st/bracketleftbigg/integraldisplayt 0f(x)dx/bracketrightbigg dt. Let g(t)=/integraldisplayt 0f(x)dx, then g/prime(t)=f(t)a n d g(0) = 0 . Since L[g/prime(t)] =sL[g(t)]−g(0), we have L[f(t)] =sL/bracketleftbigg/integraldisplayt 0f(x)dx/bracketrightbigg . (6.93) Thus if F(s)=L[f(t)], L/bracketleftbigg/integraldisplayt 0f(x)dx/bracketrightbigg =L[f(t)] s=1 sF(s). (6.94) This formula is very useful in ﬁnding the inverse transform of a fraction that has the form of p(s)/[snq(s)].For example: L−1/bracketleftbigg1 s(s−a)/bracketrightbigg =L−1/bracketleftbigg1 sL/bracketleftbig eat/bracketrightbig/bracketrightbigg =L−1L/bracketleftbigg/integraldisplayt 0eaxdx/bracketrightbigg =/integraldisplayt 0eaxdx=1 a(eat−1). Similarly L−1/bracketleftbigg1 s2(s−a)/bracketrightbigg =L−1/bracketleftbigg1 sL/bracketleftbigg1 a(eat−1)/bracketrightbigg/bracketrightbigg =/integraldisplayt 01 a(eax−1)dx=1 a2(eat−at−1). This method is often more convenient than the method of partial fractions. 6.6.2 Integration of Transforms Diﬀerentiation of F(s) corresponds to multiplication of f(t)b y−t.It is natural to expect that integration of F(s) will correspond to division of f(t)b yt.This is indeed the case, provided the limits of integration are appropriately chosen. 308 6 Laplace Transforms IfF(s/prime) is the Laplace transform of f(t),then /integraldisplay∞ sF(s/prime)ds/prime=/integraldisplay∞ s/bracketleftbigg/integraldisplay∞ 0e−s/primetf(t)dt/bracketrightbigg ds/prime=/integraldisplay∞ 0/bracketleftbigg/integraldisplay∞ se−s/primetf(t)ds/prime/bracketrightbigg dt =/integraldisplay∞ 0f(t)/bracketleftbigg/integraldisplay∞ se−s/primetds/prime/bracketrightbigg dt=/integraldisplay∞ 0f(t)/bracketleftbigg −1 te−s/primet/bracketrightbigg∞ s/prime=sdt =/integraldisplay∞ 0f(t)1 te−stdt=/integraldisplay∞ 0e−st1 tf(t)dt=L/bracketleftbiggf(t) t/bracketrightbigg .(6.95) This relationship, namely L/bracketleftbiggf(t) t/bracketrightbigg =/integraldisplay∞ sL[f(t)] ds/prime is useful if L[f(t)] is known. Example 6.6.1. Find (a) L/bracketleftbigg1 t(e−at−e−bt)/bracketrightbigg ,(b)L/bracketleftbiggsint t/bracketrightbigg . Solution 6.6.1. (a) L/bracketleftbigg1 t(e−at−e−bt)/bracketrightbigg =/integraldisplay∞ sL/bracketleftbig e−at−e−bt/bracketrightbig ds/prime=/integraldisplay∞ s/parenleftbigg1 s/prime+a−1 s/prime+b/parenrightbigg ds/prime = [ln( s/prime+a)−ln(s/prime+b)]∞ s/prime=s=/bracketleftbigg lns/prime+a s/prime+b/bracketrightbigg∞ s/prime=s =l n1 −lns+a s+b=l ns+b s+a. (b) L/bracketleftbiggsint t/bracketrightbigg =/integraldisplay∞ sL[sint]ds/prime=/integraldisplay∞ s1 s/prime2+1ds/prime=/bracketleftbig tan−1s/prime/bracketrightbig∞ s/prime=s =π 2−tan−1s=c o t−1s= tan−11 s. 6.6.3 Scaling IfF(s)=L[f(t)] =/integraltext∞ 0e−stf(t)dtis already known, then L[f(at)] can be easily obtained by a change of scale. By deﬁnition L[f(at)] =/integraldisplay∞ 0e−stf(at)dt=1 a/integraldisplay∞ 0e−(s/a)atf(at)d(at). (6.96) Lett/prime=at,the integral becomes /integraldisplay∞ 0e−(s/a)atf(at)d(at)=/integraldisplay∞ 0e−(s/a)t/primef(t/prime)d(t/prime), 6.6 Further Properties of Laplace Transforms 309 which is just the Laplace transform of fwith the parameter sreplaced by s/a.Therefore L[f(at)] =1 aF/parenleftBigs a/parenrightBig . (6.97) Example 6.6.2. IfL[f(t)] is known to be1 s(1 + 2 s),FindL[f(2t)]. Solution 6.6.2. L[f(2t)] =1 21 (s/2)[1 + 2( s/2)]=1 s(1 +s). Example 6.6.3. FindL/bracketleftbiggsinωt t/bracketrightbigg . Solution 6.6.3. Since L/bracketleftbiggsint t/bracketrightbigg = tan−11 s,then L/bracketleftbiggsinωt ωt/bracketrightbigg =1 ωtan−1ω s. Therefore L/bracketleftbiggsinωt t/bracketrightbigg = tan−1ω s. 6.6.4 Laplace Transforms of Periodic Functions Very often the input functions in physical systems are periodic functions. A function is said to be periodic if there is a number psuch that f(t+p)=f(t). The least value of pis called the period of f.A periodic function is one that has the characteristic f(t)=f(t+p)=f(t+2p)=···f(t+np)···. (6.98) The Laplace transform of f(t) is a series of integrals L[f]=/integraldisplay∞ 0e−stf(t)dt =/integraldisplayp 0e−stf(t)dt+/integraldisplay2p pe−stf(t)dt···/integraldisplay(n+1)p npe−stf(t)dt···.(6.99) 310 6 Laplace Transforms It follows from a change of variable t=τ+npthat /integraldisplay(n+1)p npe−stf(t)dt=/integraldisplayp 0e−s(τ+np)f(τ+np)dτ=e−snp/integraldisplayp 0e−sτf(τ)dτ. The dummy integration variable τcan be set equal to t,thus L[f]=/integraldisplayp 0e−stf(t)dt+e−sp/integraldisplayp 0e−stf(t)dt+···+e−snp/integraldisplayp 0e−stf(t)dt+··· =( 1+e−sp+e−2sp+···+e−nsp+···)/integraldisplayp 0e−stf(t)dt. (6.100) With the series expansion, 1 /(1−x)=1+ x+x2+···,this equation becomes L[f]=1 1−e−sp/integraldisplayp 0e−stf(t)dt. (6.101) Example 6.6.4. Half-wave rectiﬁer: Find the Laplace transform of the periodic function (shown in Fig. 6.9) whose deﬁnition over one period is: f(t)=⎧ ⎪⎨ ⎪⎩sinωtif 0<t<π ω 0i fπ ω<t<2π ω. 1 p 2p 3p 4p 5πf(t) wt Fig. 6.9. Half-wave rectiﬁer. The deﬁnition over one period is f(t)=s i n ωtfor 0<t<π / ω andf(t)=0f o r π/ω < t < 2π/ω Solution 6.6.4. L[f]=1 1−e−s2π/ω/integraldisplay2π/ω 0e−stf(t)dt=1 1−e−s2π/ω/integraldisplayπ/ω 0e−stsinωtdt. The integral can be evaluated with integration by parts. However, it is easier to note that the integral is the imaginary part of /integraldisplayπ/ω 0e−steiωtdt=/bracketleftbigg1 −s+iωe−st+iωt/bracketrightbiggπ/ω 0=1 −s+iω/parenleftBig e−sπ/ω+iπ−1/parenrightBig =−s−iω s2+ω2/parenleftBig −e−sπ/ω−1/parenrightBig . 6.6 Further Properties of Laplace Transforms 311 Thus L[f]=1 1−e−s2π/ωω(1 + e−sπ/ω) s2+ω2=ω (s2+ω2)(1−e−sπ/ω). Example 6.6.5. Full-wave rectiﬁer: Find the Laplace transform of the periodic function (shown in Fig. 6.10) whose deﬁnition over one period is: f(t)=|sinωt|0<t<π ω. 1 p 2p 3p 4p 5pf(t) wt Fig. 6.10. Full-wave rectiﬁer. The deﬁnition over one period is f(t)=|sinωt|,for 0<t<π / ω Solution 6.6.5. In this case, the period is just π/ω.Therefore L[f]=1 1−e−sπ/ω/integraldisplayπ/ω 0e−stsinωtdt=ω(1 + e−sπ/ω) (s2+ω2)(1−e−sπ/ω). This is a perfectly good result. It can be simpliﬁed somewhat by multiplying both numerator and denominator exp(sπ 2ω) L[f]=ω(esπ/(2ω)+e−sπ/(2ω)) (s2+ω2)(esπ/(2ω)−e−sπ/(2ω))=ω (s2+ω2)cothsπ 2ω. 6.6.5 Inverse Laplace Transforms Involving Periodic Functions Any Laplace transform F(s) that either has a factor (1 −e−sp)−1,or can be written in a form with such a factor as in the last example, indicates that its inverse transform is a periodic function. However, its period may be a multiple ofp. This is illustrated in the following example. Example 6.6.6. Find the inverse and its period of the Laplace transform F(s)=s (s2+ 1)(1 −e−sπ). 312 6 Laplace Transforms Solution 6.6.6. f(t)=L−1/bracketleftbiggs (s2+ 1)(1 −e−sπ)/bracketrightbigg =L−1/bracketleftbiggs (s2+1 )(1 + e−sπ+e−2sπ+e−3sπ+···)/bracketrightbigg =c o s t+u(t−π)cos(t−π)+u(t−2π)cos(t−2π) +u(t−3π)cos(t−3π)+u(t−4π)cos(t−4π)··· =c o s t−u(t−π)cost+u(t−2π)cost−u(t−3π)cost+··· =[ 1−u(t−π)] cost+[u(t−2π)−u(t−3π)] cost+···. Therefore f(t) is a periodic function with period 2 π,whose deﬁnition over one period is f(t)=/braceleftbigg costif 0<t<π 0i f π<t< 2π. 6.6.6 Laplace Transforms and Gamma Functions The Laplace transform of tnis deﬁned as L[tn]=/integraldisplay∞ 0e−sttndt. (6.102) If we make a change of variable and let st=x,the integral becomes L[tn]=/integraldisplay∞ 0e−x/parenleftBigx s/parenrightBign d/parenleftBigx s/parenrightBig =1 sn+1/integraldisplay∞ 0e−xxndx. (6.103) The last integral is known as the gamma function ofn+1,written as Γ(n+1). Gamma function occurs frequently in practice. It is given by Γ(n)=/integraldisplay∞ 0e−xxn−1dx, (6.104) this is well deﬁned as long as nis not zero or a negative integer. For n=1, Γ(1) =/integraldisplay∞ 0e−xdx=/bracketleftbig e−x/bracketrightbig∞ 0=1. (6.105) With integration by parts, one can easily show Γ(n+1 )=/integraldisplay∞ 0e−xxndx=/bracketleftbig −xne−x/bracketrightbig∞ 0+n/integraldisplay∞ 0e−xxn−1dx=nΓ(n). (6.106) 6.7 Summary of Operations of Laplace Transforms 313 Thus if nis a positive integer, Γ(n+1 )= nΓ(n)=n(n−1)···1Γ(1) = n!, (6.107) and according to (6 .103) L[tn]=Γ(n+1 ) sn+1=n! sn+1(6.108) in agreement with the result we obtained before. Since Γ(n) is a tabulated function, as long as n>−1,L[tn] can still be evaluated even if nis not an integer. For example L/bracketleftbigg1√ t/bracketrightbigg =Γ(1 2) s1 2=√π s1/2, (6.109) L/bracketleftBig√ t/bracketrightBig =Γ(1 +1 2) s1+1 2=1 2Γ(1 2) s3 2=1 2√π s3/2, (6.110) where we have used the well-known result Γ(1 2)=√π. Example 6.6.7. Find the Laplace transform of eat(1 + 2 at)/√ πt. Solution 6.6.7. Uses-shifting property L/bracketleftbiggeat √ πt/bracketrightbigg =1 (s−a)1/2, L/bracketleftbiggeat2at√ πt/bracketrightbigg =2a√πL/bracketleftBig eat√ t/bracketrightBig =a (s−a)3/2. Thus L/bracketleftbiggeat(1 + 2 at)√ πt/bracketrightbigg =1 (s−a)1/2+a (s−a)3/2=s (s−a)3/2. 6.7 Summary of Operations of Laplace Transforms The properties of the Laplace transform are not diﬃcult to understand. How- ever, because there are so many of them, it is not easy to decide which one to use for a speciﬁc problem. In Table 6.2 we summarize these operations. In the last column we give a simple example and in the ﬁrst column we give a name to characterize the operation. This classiﬁcation is helpful in remembering the details of each operation. In Sect. 6.2.1, we discussed the inverse of the transform F(s)i nt h ef o r m of a quotient of two polynomials. If F(s) is not in that form, sometimes we can use the properties of the Laplace transforms to obtain the inverse. 314 6 Laplace Transforms Table 6.2. Summary of Laplace transform operations Name h(t) L[h(t)] Example: Let f(t)=t Deﬁnition f(t) F(s) L[t]=/integraltext∞ 0e−sttdt=1 s2=F(s) Multiply tt f (t) −d dsF(s) L[t·t]=−d ds1 s2=2 s3 Divide tf(t) t/integraltext∞ sF(ς)dς L/bracketleftBigt t/bracketrightBig =/integraltext∞ s1 ς2dς=1 s Derivative f/prime(t) sF(s)−f(0) L/bracketleftBigdt dt/bracketrightBig =s1 s2−0=1 s Integral/integraltextt 0f(τ)dτF(s) sL/bracketleftBig/integraltextt 0τdτ/bracketrightBig =1/s2 s=1 s3 Shifting – seatf(t) F(s−a) L/bracketleftbig eatt/bracketrightbig =1 (s−a)2 Shifting – tu(t−a)f(t−a)e−saF(s) L[u(t−a)(t−a)] = e−sa1 s2 Scaling f(at)1 aF/parenleftBigs a/parenrightBig L[at]=1 a1 (s/a)2=a1 s2 Period – p periodic f (t)/integraltextp 0e−stf(t)dt 1−e−psL[f]=1−(1 +ps)e−ps s2(1−e−ps) Convolution/integraltextt 0f(τ)g(t−τ)dτF(s)·G(s)L e t g(t)=f(t),G(s)=F(s) L/bracketleftBig/integraltextt 0τ(t−τ)dτ/bracketrightBig =1 s21 s2=1 s4 Example 6.7.1. FindL−1/bracketleftbigg lns+a s−b/bracketrightbigg . Solution 6.7.1. The transform is not in the form of a quotient of two poly- nomials, but its derivative is. Let L[f(t)] = lns+a s−b,f (t)=L−1/bracketleftbigg lns+a s−b/bracketrightbigg . Since L[tf(t)] =−d dsL[f(t)] =−d dslns+a s−b =−d dsln(s+a)+d dsln(s−b)=1 s−b−1 s+a, 6.7 Summary of Operations of Laplace Transforms 315 and tf(t)=L−1/bracketleftbigg −d dsL[f(t)]/bracketrightbigg =L−1/bracketleftbigg1 s−b−1 s+a/bracketrightbigg , therefore L−1/bracketleftbigg lns+a s−b/bracketrightbigg =f(t)=1 tL−1/bracketleftbigg1 s−b−1 s+a/bracketrightbigg =1 t/parenleftbig ebt−e−at/parenrightbig . Example 6.7.2. FindL−1/bracketleftbigg lns2−a2 s2/bracketrightbigg . Solution 6.7.2. L−1/bracketleftbigg lns2−a2 s2/bracketrightbigg =1 tL−1/bracketleftbigg −d dslns2−a2 s2/bracketrightbigg =−1 tL−1/bracketleftbigg2s s2−a2−2s s2/bracketrightbigg =2 t(1−coshat). Example 6.7.3. FindL−1/bracketleftbigg lns2+ω2 s2/bracketrightbigg . Solution 6.7.3. L−1/bracketleftbigg lns2+ω2 s2/bracketrightbigg =1 tL−1/bracketleftbigg −d dslns2+ω2 s2/bracketrightbigg =−1 tL−1/bracketleftbigg2s s2+ω2−2s s2/bracketrightbigg =2 t(1−cosωt). Example 6.7.4. FindL−1/bracketleftbigg tan−11 s/bracketrightbigg . Solution 6.7.4. L−1/bracketleftbigg tan−11 s/bracketrightbigg =1 tL−1/bracketleftbigg −d dstan−11 s/bracketrightbigg =−1 tL−1/bracketleftbigg1 (1/s)2+1d ds1 s/bracketrightbigg =1 tL−1/bracketleftbigg1 1+s2/bracketrightbigg =1 tsint. 316 6 Laplace Transforms 6.8 Additional Applications of Laplace Transforms 6.8.1 Evaluating Integrals Many integrals from 0 to∞can be evaluated by the Laplace transform method. By direct substitution. Integrals involving e−atcan be obtained from the Laplace transformation with a simple substitution. /integraldisplay∞ 0e−atf(t)dt=/braceleftbigg/integraldisplay∞ 0e−stf(t)dt/bracerightbigg s=a={L[f(t)]}s=a. (6.111) Example 6.8.1. Find/integraltext∞ 0e−3tsintdt. Solution 6.8.1. /integraldisplay∞ 0e−3tsintdt={L[sint]}s=3=/braceleftbigg1 s2+1/bracerightbigg s=3=1 10 Example 6.8.2. Find/integraldisplay∞ 0e−2ttcostdt. Solution 6.8.2. Since /integraldisplay∞ 0e−2ttcostdt={L[tcost]}s=2, {L[tcost]}=−d dsL[cost]=−d dss s2+1=s2−1 (s2+1 )2, therefore /integraldisplay∞ 0e−2ttcostdt=/braceleftbiggs2−1 (s2+1 )2/bracerightbigg s=2=3 25. Use integral of the transform. In Sect. 6.6.2 we have shown L/bracketleftbiggf(t) t/bracketrightbigg =/integraldisplay∞ sF(s/prime)ds/prime, where L/bracketleftbiggf(t) t/bracketrightbigg =/integraldisplay∞ 0e−stf(t) tdt, F (s)=L[f(t)] =/integraldisplay∞ 0e−stf(t)dt. Setting s=0,we obtain an equally important formula /integraldisplay∞ 0f(t) tdt=/integraldisplay∞ 0L[f(t)] ds. (6.112) This formula can be used if the integral on left side is diﬃcult to do directly. 6.8 Additional Applications of Laplace Transforms 317 Example 6.8.3. Find/integraldisplay∞ 0/bracketleftbigge−t−e−3t t/bracketrightbigg dt. Solution 6.8.3. /integraldisplay∞ 0/bracketleftbigge−t−e−3t t/bracketrightbigg dt=/integraldisplay∞ 0L/bracketleftbig e−t−e−3t/bracketrightbig ds=/integraldisplay∞ 0/parenleftbigg1 s+1−1 s+3/parenrightbigg ds = [ln( s+1 )−ln(s+ 3)]∞ 0=/bracketleftbigg lns+1 s+3/bracketrightbigg∞ 0 =l n1 −ln1 3=l n3 . Example 6.8.4. Find/integraldisplay∞ 0sint tdt. Solution 6.8.4. /integraldisplay∞ 0sint tdt=/integraldisplay∞ 0L[sint]ds=/integraldisplay∞ 01 s2+1ds =/bracketleftbig tan−1s/bracketrightbig∞ 0=π 2. (6.113) Use double integrals. We can solve the problem of the last example by a double integral. Starting with L[1] =/integraldisplay∞ 0e−stdt=1 s, (6.114) if we rename tasx,andsast,w eh a v e /integraldisplay∞ 0e−txdx=1 t. (6.115) So /integraldisplay∞ 0sint tdt=/integraldisplay∞ 0sint/bracketleftbigg1 t/bracketrightbigg dt=/integraldisplay∞ 0sint/bracketleftbigg/integraldisplay∞ 0e−txdx/bracketrightbigg dt. (6.116) Interchanging the order of integration, we have /integraldisplay∞ 0sint tdt=/integraldisplay∞ 0/bracketleftbigg/integraldisplay∞ 0e−txsintdt/bracketrightbigg dx. (6.117) The integral in the bracket is recognized as the Laplace transform of sin twith the parameter sreplaced by x,thus /integraldisplay∞ 0sint tdt=/integraldisplay∞ 01 1+x2dx=/bracketleftbig tan−1x/bracketrightbig∞ 0=π 2. (6.118) This method can be applied to more complicated cases. 318 6 Laplace Transforms Example 6.8.5. Find/integraldisplay∞ 0sin2t t2dt. Solution 6.8.5. First we note sin2t=1 2(1−cos 2t), then write /integraldisplay∞ 0sin2t t2dt=1 2/integraldisplay∞ 0(1−cos 2t)/bracketleftbigg1 t2/bracketrightbigg dt. With 1 t2=/integraldisplay∞ 0e−txxdx, we have /integraldisplay∞ 0sin2t t2dt=1 2/integraldisplay∞ 0(1−cos 2t)/bracketleftbigg/integraldisplay∞ 0e−txxdx/bracketrightbigg dt =1 2/integraldisplay∞ 0/bracketleftbigg/integraldisplay∞ 0e−tx(1−cos 2t)dt/bracketrightbigg xdx. Since /integraldisplay∞ 0e−tx(1−cos 2t)dt=[L(1−cos 2t)]s=x =1 x−x x2+4=4 x(x2+4 ), /integraldisplay∞ 0sin2t t2dt=1 2/integraldisplay∞ 0/bracketleftbigg4 x(x2+4 )/bracketrightbigg xdx =/integraldisplay∞ 02 (x2+4 )dx=/bracketleftBig tan−1x 2/bracketrightBig∞ 0=π 2. Use inverse Laplace transform. If the integral is diﬃcult to do, we can ﬁrst ﬁnd its Laplace transform and then take the inverse. Example 6.8.6. Find/integraldisplay∞ 0cosx x2+b2dx Solution 6.8.6. In order to use Laplace transform to evaluate this integral, we change cos xto cos tx,and at the end we set t=1.Let I(t)=/integraldisplay∞ 0costx x2+b2dx. 6.8 Additional Applications of Laplace Transforms 319 L[I(t)] =/integraldisplay∞ 01 x2+b2L[costx]dx=/integraldisplay∞ 01 x2+b2s s2+x2dx =s s2−b2/integraldisplay∞ 0/bracketleftbigg1 x2+b2−1 s2+x2/bracketrightbigg dx =s s2−b2/braceleftbigg/bracketleftbigg1 btan−1x b/bracketrightbigg∞ 0−/bracketleftbigg1 stan−1x s/bracketrightbigg∞ 0/bracerightbigg =s s2−b2/braceleftBigπ 2b−π 2s/bracerightBig =π 2b1 s+b. I(t)=L−1/bracketleftbiggπ 2b1 s+b/bracketrightbigg =π 2be−bt. Thus /integraldisplay∞ 0cosx x2+b2dx=I(1) =π 2be−b. 6.8.2 Diﬀerential Equation with Variable Coeﬃcients Iff(t) in the formula L[tf(t)] =−d dsL[f(t)] is taken to be the nth derivative of y(t),then L/bracketleftBig ty(n)(t)/bracketrightBig =−d dsL/bracketleftBig y(n)(t)/bracketrightBig =−d ds/braceleftbig snL[y(t)]−sn−1y(0)···−yn−1(0)/bracerightbig .(6.119) This equation can be used to transform a linear diﬀerential equation with variable coeﬃcients into a diﬀerential equation involving the transform. This procedure is useful if the new equation can be readily solved. Example 6.8.7. Find the solution of ty/prime/prime(t)−ty/prime(t)−y(t)=0,y(0) = 0 ,y/prime(0) = 2 . Solution 6.8.7. L[ty/prime/prime(t)] =−d ds/braceleftbig s2L[y(t)]−sy(0)−y/prime(0)/bracerightbig . LetL[y(t)] =F(s),withy(0) = 0 we have L[ty/prime/prime(t)] =−2sF(s)−s2F/prime(s), L[ty/prime(t)] =−F(s)−sF/prime(s). 320 6 Laplace Transforms Therefore L[ty/prime/prime(t)−ty/prime(t)−y(t)] =−s(s−1)F/prime(s)−2sF(s)=0. It follows dF(s) F(s)=−2ds s−1, lnF(s)=l n ( s−1)−2+l nC F(s)=C (s−1)2 y(t)=L−1[F(s)] =L−1/bracketleftbiggC (s−1)2/bracketrightbigg =Cett. Since y/prime(t)=Cett+Cet, y/prime(0) = C=2, therefore y(t)=2 ett. It can be easily veriﬁed that this is indeed the solution, since it satisﬁes both the equation and the initial conditions. Example 6.8.8. Zeroth Order Bessel Function: Find the solution of ty/prime/prime(t)+y/prime(t)+ty(t)=0,y(0) = 1 ,y/prime(0) = 0 . Solution 6.8.8. With L[y(t)] =F(s)a n d y(0) = 1 ,y/prime(0) = 0 , L[ty/prime/prime(t)+y/prime(t)+ty(t)] =−d ds/braceleftbig s2F(s)−s/bracerightbig +sF(s)−1−d dsF(s)=0. Collecting terms (s2+1 )d dsF(s)+sF(s)=0, or dF(s) F(s)=−sds s2+1=−1 2ds2 s2+1. It follows lnF(s)=−1 2ln(s2+1 )+l n C, F(s)=C (s2+1 )1/2. To ﬁnd the inverse of this Laplace transform, we expand it in a series in the cases>1, 6.8 Additional Applications of Laplace Transforms 321 F(s)=C s/bracketleftbigg 1+1 s2/bracketrightbigg−1/2 =C s/bracketleftbigg 1−1 2s2+1·3 22·2!1 s4···+(−1)n(2n)! (2nn!)21 s2n+···/bracketrightbigg . Inverting term by term, we have y(t)=L−1[F(s)] =C∞/summationdisplay n=0(−1)nt2n (2nn!)2. Since y(0) = 1 ,therefore C=1.It turned out this series with C=1i sk n o w n as the Bessel function of zeroth order J0(t),that is , J0(t)=∞/summationdisplay n=0(−1)nt2n (2nn!)2, which we will discuss in more detail in the chapter on Bessel functions. Thus the solution of the equation is y(t)=J0(t). Furthermore, this development shows that L[J0(t)] =1 (s2+1 )1/2. With the scaling property of the Laplace transform, we also have (for a>0) L[J0(at)] =1 a1 [(s/a)2+1 ]1/2=1 (s2+a2)1/2. 6.8.3 Integral and Integrodiﬀerential Equations Equations in which the unknown function appears under the integral are called integral equations. If the derivatives are also in the equation, then they are called integrodiﬀerential equations. They are often diﬃcult to solve. But if the integrals are in the form of a convolution, then Laplace transform can be used to solve them. The following example will make the procedure clear. Example 6.8.9. Solve the integral equation y(t)=t+/integraldisplayt 0y(τ)sin(t−τ)dτ. 322 6 Laplace Transforms Solution 6.8.9. L[y(t)] =L[t]+L/bracketleftbigg/integraldisplayt 0y(τ)sin(t−τ)dτ/bracketrightbigg =L[t]+L[y(t)]L[sint]=1 s2+L[y(t)]1 s2+1. Solving for L[y(t)]/parenleftbigg 1−1 s2+1/parenrightbigg L[y(t)] =1 s2. So L[y(t)] =s2+1 s4=1 s2+1 s4 and y(t)=L−1/bracketleftbigg1 s2+1 s4/bracketrightbigg =t+1 6t3. Example 6.8.10. Find the solution of y/prime(t)−3/integraldisplayt 0e−2(t−τ)y(τ)dτ=z(t),y (0) = 4 , z(t)=/braceleftbigg4e−2t1<t 00 <t< 1. Solution 6.8.10. First note L[z(t)] =L/bracketleftbig 4e−2tu(t−1)/bracketrightbig =L/bracketleftBig 4e−2e−2(t−1)u(t−1)/bracketrightBig =4 e−21 s+2e−s, L/bracketleftbigg 3/integraldisplayt 0e−2(t−τ)y(τ)dτ/bracketrightbigg =31 s+2L[y]. Applying the Laplace transform to both sides of the equation leads to sL[y]−4−31 s+2L[y]=4 e−21 s+2e−s, collecting terms s2+2s−3 s+2L[y]=4+4 e−21 s+2e−s, or L[y]=4(s+2 )+4 e−2e−s s2+2s−3 =3 s−1+1 s+3+e−2/parenleftbigg1 s−1−1 s+3/parenrightbigg e−s. Thus y(t)=3 et+e−3t+e−2/parenleftBig e(t−1)−e−3(t−1)/parenrightBig u(t−1). 6.9 Inversion by Contour Integration 323 6.9 Inversion by Contour Integration As we have seen that to be able to invert a given transform F(s)t oﬁ n d the function f(t) is the key of solving diﬀerential equations with the Laplace Transform. For those familiar with the complex contour integration, we present in this section a universal technique of ﬁnding the inverse of the Laplace Transform. The derivation is highly imaginative, but the result is elegantly simple. First let us extend the Laplace transform to the complex domain. The function F(z) is the same function as F(s)except with sreplaced by z.In the complex plane, F(z) will have some singular points. Let us choose a line x=b in the complex plane such that all singular points of F(z) are in the left-hand side of this line. Then F(z) is analytic on the line x=band in the entire half plane to the right of this line. If sis any point in this half plane, we can choose a semicircular contour C=C1+C2, as shown in Fig. 6.11 ,and apply the Cauchy’s integral formula, F(s)=1 2πi/contintegraldisplay CF(z) z−sdz =1 2πi/integraldisplayb−iR b+iRF(z) z−sdz+1 2πi/integraldisplay C1F(z) z−sdz. (6.120) Now if we let Rgo to inﬁnity, (6 .120) is still valid, but all values of zon the semicircle C1are inﬁnitely large. Since F(z)→0a sz→∞, lim R→∞/integraldisplay C1F(z) z−sdz=0. b SC1 C 2Ry x Fig. 6.11. The ﬁrst contour used to obtain the complex inversion of the Laplace transform 324 6 Laplace Transforms Therefore in this limit (6 .120) becomes F(s)=1 2πi/integraldisplayb−i∞ b+i∞F(z) z−sdz=1 2πi/integraldisplayb+i∞ b−i∞F(z) s−zdz. In the last step we have changed the sign of the integrand and interchanged the upper and lower limits of the integral. Taking the inverse Laplace transform, we have L−1[F(s)] =L−1/bracketleftBigg 1 2πi/integraldisplayb+i∞ b−i∞F(z) s−zdz/bracketrightBigg . Since the inverse Laplace operator L−1refers only to the variable s,we can write L−1[F(s)] =1 2πi/integraldisplayb+i∞ b−i∞F(z)L−1/bracketleftbigg1 s−z/bracketrightbigg dz. Since L−1/bracketleftbigg1 s−z/bracketrightbigg =ezt we have L−1[F(s)] =1 2πi/integraldisplayb+i∞ b−i∞F(z)eztdz. (6.121) This procedure is called the Mellin inversion. This integral is from b−i∞to b+i∞along C2.Usually the evaluation of this integral is accomplished by the residue theorem. To use the residue theorem, we must have a closed contour. To close the contour, we have to add a returning line integral from b+i∞to b−i∞in such a way that the value of the integral is not changed. This can be done with the semicircular contour C3in the left half-plane as shown in Fig. 6.12, since lim R→∞/integraldisplay C3F(z)eztdz= 0. (6.122) This can be understood from the fact that with a positive t,the integrand F(z)ezt=F(z)ext+iyt goes to zero as zgoes to inﬁnity. The factor eiytis oscillatory with a maximum value of 1. For xto change from bto−∞as on C3, the factor extis always less than ebt.Therefore F(z)eztwill go to zero as long as F(z) is going to zero. Note that this will not be the case in the right half plane where xwill go to positive inﬁnite and eztwill blow up. Thus with C3,we have L−1[F(s)] =1 2πi/integraldisplayb+i∞ b−i∞F(z)eztdz. =1 2πilim R→∞/bracketleftbigg/integraldisplay C2F(z)eztdz+/integraldisplay C3F(z)eztdz/bracketrightbigg =1 2πi/contintegraldisplay CF(z)eztdz, (6.123) 6.9 Inversion by Contour Integration 325 y xbC3 C2R Fig. 6.12. The contour used in evaluating the complex inversion integral where C=C2+C3as shown in Fig. 6.12 with R→∞.This contour is also called the Bromwich contour. Since bis on the right of all singular points of F(z),the contour Cencloses all singular points of eztF(z). Therefore by the residue theorem L−1[F(s)] =1 2πi/contintegraldisplay CF(z)eztdz =/summationdisplay all residues of F(z)ezt. (6.124) Example 6.9.1. Using the complex inversion integral to ﬁnd L−1/bracketleftBigg 1 (s+a)2+b2/bracketrightBigg . Solution 6.9.1. Since 1 (s+a)2+b2=1 [s−(−a+ib)][s−(−a−ib)], the residues of ezt (z+a)2+b2 at the two singular points are r1= lim z→−a+ib[z−(−a+ib)]ezt [z−(−a+ib)][z−(−a−ib)] =e(−a+ib)t 2ib 326 6 Laplace Transforms and r2= lim z→−a−ib[z−(−a−ib)]ezt [z−(−a+ib)][z−(−a−ib)] =e(−a−ib)t −2ib. Therefore L−1/bracketleftBigg 1 (s+a)2+b2/bracketrightBigg =e(−a+ib)t 2ib+e(−a−ib)t −2ib =1 be−at1 2i/parenleftbig eibt−e−ibt/parenrightbig =1 be−atsinbt. This is a familiar result. This example shows that the complex inversion integral is indeed another way of ﬁnding the inverse Laplace transform. In more diﬃcult applications, the use of the complex inversion integral and the contour integration is either the only way or the simplest way of ﬁnding the inverse Laplace transform. 6.10 Computer Algebraic Systems for Laplace Transforms Before closing this chapter, we should mention that a number of commercial computer packages are available to perform algebraic manipulations, includ- ing Laplace transforms. They are called computer algebraic systems, some prominent ones are Matlab, Maple, Mathematica, MathCad, and MuPAD. This book is written with the software “Scientiﬁc WorkPlace,” which also provides an interface to MuPAD. (Before version 5, it also came with Maple). Instead of requiring the user to adhere to a rigid syntax, the user can use natural mathematical notations. For example, to compute L(cost−2sint), all you have to do is (1) type cos t−2sintin the math-mode, (2) click on the “Compute” button, (3) click on the “Transforms” button in the pull-down menu, and (4) click on the “Laplace” button in the submenu. The program will return with cost−2sint,Laplace transform is :s s2+1−2 s2+1. The program also recognizes the Laplace transform symbol. An alternative way to do the same problem is to choose, from the “Miscellaneous Symbols” panel, the Laplace transform symbol and type 6.10 Computer Algebraic Systems for Laplace Transforms 327 L(cost−2sint) in the math-mode, and click on “Compute” and then choose “Evaluate.” The program will return with L(cost−2sint)=s s2+1−2 s2+1 Similarly, one can compute the inverse of the Laplace transform. For example, type L−1/parenleftbiggs−2 s2+1/parenrightbigg and click on “Compute” and then on “Evaluate,” the program will return with L−1/parenleftbiggs−2 s2+1/parenrightbigg =cost−2sint To compute the Laplace transform of a derivative, one has to deﬁne the function ﬁrst. For example, to ﬁnd L(y/prime/prime/prime),ﬁrst type y(t) and click on “Compute”, then on “Deﬁnition” in the menu, then on “New Deﬁnition” in the submenu. Then type L(y/prime/prime/prime) and click on “Compute” and then on “Evaluate”. The program will return with L(y/prime/prime/prime)=s3L(y)−sy/prime(0)−s2y(0)−y/prime/prime(0). The program can also use Laplace transform to solve ordinary diﬀerential equations. For example, to solve the equation y/prime+y=x+s i nx, ﬁrst type this equation in math-mode, then click on “Compute.” In the pull- down menu, click on “Solve ODE,” then click on “Laplace” in the submenu. The program returns with a question asking which is the independent variable. Type xand click on “OK.” The program will return with Laplace solution is : x−1 2cosx+1 2sinx+e−x/parenleftbigg y(0) +3 2/parenrightbigg −1. Unfortunately, not every problem can be solved by a computer algebraic system. Sometimes it fails to ﬁnd the solution. Even worse, for a variety of reasons, the intention of the user is sometimes misinterpreted, and the com- puter returns with an answer to a wrong problem without the user knowing it. One should be aware of these pitfalls. Computer algebraic systems are no substitute for the knowledge of the subject matter, but they are useful supplements. 328 6 Laplace Transforms Exercises 1. Find the Laplace transformation of each of the following functions by direct integration. (a)1 2t2,(b) e3t,(c) 3sin (3 t). Ans. (a)1 s3,(b)1 s−3,(c)9 s2+9. 2. Find the Laplace transformation of each of the following functions by using the “Multiply t” operation (see Table 6.2). (a)tet,(b)tcost,(c)t2cost. Ans. (a)1 (s−1)2,(b)s2−1 (s2+1)2,(c)2s(s2−3) (s2+1)3. 3. Find the Laplace transformation of each of the following functions by using the “Divide t” operation (see Table 6.2). (a)1 t/parenleftbig e2t−e−2t/parenrightbig ,(b)2 t(1−cos (2t)),(c)1 tsin (4t). Ans. (a) ln/parenleftBig s+2 s−2/parenrightBig ,(b) ln/parenleftBig s2+4 s2/parenrightBig ,(c)π 2−tan−1/parenleftbigs 4/parenrightbig . 4. Find the Laplace transformation of each of the following functions by using the “Shifting – s” operation (see Table 6.2). (a) eatsin 3t,(b) e−2ttsinat,(c) sinh tcost. Ans. (a)3 (s−a)2+9,(b)2a(s+2) [(s+2)2+a2]2,(c)s2−2 s4+4. 5. Use the deﬁnition of Laplace transformation to show (a)L[f/prime]=sL[f]−f(0); use (a) to show (b)L[f/prime/prime]=s2L[f]−sf(0)−f/prime(0). 6. Use the results of previous problem and the fact thatd2 dt2cosat= −a2cosatandd2 dt2sinat=−a2sinatto show (a)L[cosat]=s s2+a2,(b)L[sinat]=a s2+a2. 7. Diﬀerentiate both sides of part (b) of the previous problem with respect toa,and show that (a)L[tcosat]=1 s2+a2−2a2 (s2+a2)2, Diﬀerentiate both sides of part (a) of the previous problem with respect tos,and show that (b)L[−tcosat]=1 s2+a2−2s2 (s2+a2)2. 8. Use the results of problems 6 and 7 to show 6.10 Computer Algebraic Systems for Laplace Transforms 329 (a)L−1/bracketleftbigg1 (s2+a2)2/bracketrightbigg =1 2a3(sinat−atcosat), (b)L−1/bracketleftbiggs2 (s2+a2)2/bracketrightbigg =1 2/parenleftbig tcosat+1 asinat/parenrightbig . 9. Do problem 8 with convolution theorem. Hint: you may need the following integral /integraldisplayt 0sinaτcosaτdτ=1 4a(1−cos 2at); /integraldisplayt 0sin2aτdτ=1 4a(2at−sin 2at). 10. If f(t)=tn,g(t)=tm,n > −1,m > −1, (a) show that /integraldisplayt 0τn(t−τ)mdτ=tn+m+1/integraldisplay1 0yn(1−y)mdy. (b) By using the convolution theorem, show that /integraldisplay1 0yn(1−y)mdy=n!m! (n+m+ 1)!. Hint: (a) let τ=yt,(b) use convolution theorem to evaluate/integraltextt 0τn (t−τ)mdτ. 11. Find the Laplace transformation of each of the following functions by direct integration. (a) sin( t−a)u(t−a), (b)f(t)=/braceleftbigg cos (t−π)t>π 0 t<π, (c)f(t)=⎧ ⎨ ⎩00≤t<5 15≤t<10 01 0 ≤t. Ans. (a) e−as1 s2+1,(b) e−πss s2+1,(c)1 s/parenleftbig e−5s−e−10s/parenrightbig . 12. Do the previous problem by using the “Shifting – t” operation (see Table 6.2). 13. Use the partial fraction to ﬁnd the inverse Laplace transform of the fol- lowing expressions. (a)L−1/bracketleftBig 4 s2−4s/bracketrightBig ,(b)L−1/bracketleftBig 1 s(s2+1)/bracketrightBig ,(c)L−1/bracketleftBig 1 s2(s2+1)/bracketrightBig . Ans. (a) e4t−1,(b) 1−cost,(c)t−sint. 330 6 Laplace Transforms 14. Do the previous problem by using the formula L/bracketleftbigg/integraldisplayt 0f(τ)dτ/bracketrightbigg =1 sL[f(t)]. 15. Use the Heaviside expansion to solve the previous problem. 16. Use the Laplace transform to solve the following diﬀerential equations (a)y/prime/prime+2y/prime+y=1,y(0) = 2 ,y/prime(0) =−2, (b) (b)y/prime/prime+y=s i n( 3 t),y(0) = y/prime(0) = 0 . Ans. (a) y(t)=1+( 1 −t)e−t,(b)y(t)=3 8sint−1 8sin 3t. 17. Use the Laplace transform to solve the following set of equations dy dt=2y−3z, dz dt=−2y+z, y(0) = 8 ,z(0) = 3 . Ans.y(t)=3 e4t+5 e−t,z(t)=5 e−t−2e4t. 18. Find the solution of the integrodiﬀerential equation y/prime(t)−/integraldisplayt 0y(τ)c o s(t−τ)dτ=0,y(0) = 1 . Ans.y(t)=1+1 2t2. 19. Solve the following equations with the initial conditions at t=0,bothy and all its derivatives are equal to zero. (a)y/prime/prime+2y/prime+y=Aδ(t−t0), (b)y/prime/prime/prime/prime−y=Aδ(t−t0). Ans. (a) y(t)=A(t−t0)e−(t−t0)u(t−t0), (b)y(t)=1 2A[sinh(t−t0)−sin (t−t0)]u(t−t0). 20. Consider a resistance Rand an inductance Lconnected in series with a voltage V(t). The equation governs the current is Ldi dt+Ri=V(t). Suppose i(0) = 0 and V(t) is a voltage impulse at t=t0given by V(t)=Aδ(t−t0). 6.10 Computer Algebraic Systems for Laplace Transforms 331 Find the current by the Laplace transform method. Ans.i(t)=A Le−R(t−t0)/Lu(t−t0). 21. The damped harmonic oscillator is governed by mx/prime/prime+bx/prime+kx=f(t); with x(0) = x/prime(0) = 0 . (a) Find the solution by convolution. (Express x(t) as an integral). (b) If f(t)=Pδ(t−t0),ﬁnd the solution by evaluating the convolution integral. (c) If b=0,andf(t)=F0sinω0twhere ω0=/radicalbig k/m, solve the problem by Laplace transformation. (d) If b=0,andf(t)=F0u(t−t0) where u(t−t0) is the step function, solve the problem. Ans. (a) x(t)=1 mω/integraltextt 0f(τ)e−α(t−τ)sinω(t−τ)dτ where α=b 2m,ω2=k m−/parenleftbigb 2m/parenrightbig2, (b)x(t)=P mωe−α(t−t0)sinω(t−t0)u(t−t0), (c)x(t)=F0 2mω2 0(sinω0t−ω0tcosω0t), (d)x(t)=F0 mω2 0[1−cosω0(t−t0)]u(t−t0). 22. Using the complex inversion integral, ﬁnd the inverses of the following Laplace transforms (a)1 (s+1)(s+3),(b)1 (s+2)2,(c)1 (s2+9)(s2+4). Ans. (a)1 2(e−t−e−3t),(b)te−2t,(c)1 30(3sin 2 t−2sin3 t). References This bibliograph includes the references cited in the text and a few other books and tables that might be useful. 1. M. Abramowitz, I.A. Stegun: Handbook of Mathematical Functions (Dover, New York 1970) 2. G.B. Arfken, H.J. Weber: Mathematical Methods for Physicists, 5th edn. (Academic Press, San Diego, 2001) 3. M. L. Boas: Mathematical Methods in the Physical Sciences, 3rd edn. (Wiley, New York 2006) 4. W.E. Boyce, R.C. DiPrima: Elementary Diﬀerential Equations and Boundary Value Problems, 4th edn. (Wiley, New York 1986) 5. T.C. Bradbury: Mathematical Methods with Applications to Problems in the Physical Sciences (Wiley, New York 1984) 6. E. Butkov: Mathematical Physics (Addison-Wesley, Reading 1968) 7. F.W. Byron, Jr., R.W. Fuller: Mathematics of Classical and Quantum Physics (Dover, New York 1992) 8. T.L. Chow: Mathematical Methods for Physicists: A Concise Introduction (Cambridge University Press, Cambridge 2000) 9. R.V. Churchill: Operational Mathematics, 3rd edn. (McGraw-Hill, New York 1972) 10. H. Cohen: Mathematics for Scientists and Engineeers (Prentice-Hall, Englewood Cliﬀs 1992) 11. R.E. Collins: Mathematical Methods for Physicists and Engineers (Reinhold, New York 1968) 12. C.H. Edwards Jr., D.E. Penney: Diﬀerential Equations and Boundary Value Problems (Prentice-Hall, Englewood Cliﬀs 1996) 13. A. Erd´ elyi, W. Magnus, F. Oberhettinger, F. Tricomi: Tables of Integral Trans- forms, Vol. 1 (McGraw-Hill, New York 1954) 14. L.R. Ford: Diﬀerential Equations (McGraw-Hill, New York 1955) 15. M.D. Greenberg: Advanced Engineering Mathematics, 2nd edn. (Prentice Hall, Upper Saddle River 1998) 16. I.S. Gradshteyn, I.M. Ryzhik: Table of Integrals, Series and Products (Academic Press, Orlando 1980) 17. D.W. Hardy, C.L. Walker: Doing Mathematics with Scientiﬁc WorkPlace and Scientiﬁc Notebook, Version 5 (MacKichan, Poulsbo 2003) 334 References 18. S. Hasssani: Mathematical Methods: For Students of Physics and Related Fields (Springer, New York 2000) 19. F.B. Hilderbrand: Advanced Calculus for Applications, 2nd edn. (Prentice-Hall, Englewood Cliﬀs 1976) 20. E.L. Ince: Ordinary Diﬀerential Equations (Dover, New York 1956) 21. H. Jeﬀreys, B.S. Jeﬀreys: Mathematical Physics (Cambridge University Press, Cambridge 1962) 22. D.E. Johnson and J.R. Johnson: Mathematical Methods in Engineering Physics (Prentice-Hall, Upper Sadddle River 1982) 23. D.W. Jordan, P. Smith: Mathematical Techniques: An Introduction for the Engineering, Physical, and Mathematical Sciences, 3rd edn. (Oxford University Press, Oxford 2002) 24. E. Kreyszig: Advanced Engineering Mathematics, 8th edn. (Wiley, New York 1999) 25. B.R. Kusse, E.A. Westwig: Mathematical Physics: Applied Mathematics for Scientists and Engineers, 2nd edn. (Wiley, New York 2006) 26. S.M. Lea: Mathematics for Physicists (Brooks/Cole, Belmont 2004) 27. T.D. Lee: “ Reminiscences ”. InThirty Years Since Parity Nonconservation ed. by R. Novick (Birkh¨ auser, Boston 1988) pp. 153–165 28. H. Margenau, G.M. Murphy: Methods of Mathematical Physics (Van Nostrand, Princeton 1956) 29. J. Mathew, R.L. Walker: Mathematical Methods of Physics, 2nd edn. (Benjamin, New York 1970) 30. P.C. Matthews: Vector Calculus (Springer, London 1998) 31. D.A. McQuarrie: Mathematical Methods for Scientists and Engineers (University Science Books, Sausalito 2003) 32. P.M. Morse, H. Feshbach: Methods of Theoretical Physics (McGraw-Hill, New York 1953) 33. G.M. Murphy: Ordinary Diﬀerential Equations and Their Solutions (Van Nostrand, Princeton 1960) 34. H.E. Newell, Jr.: Vector Analysis (McGraw-Hill, New York 1955) 35. F. Oberhettinger, E. Badii: Tables of Laplace Transforms (Springer, New York 1973) 36. A.D. Polyanin, V.F. Zaitsev: Handbook of Exact Solutions for Ordinary Diﬀerential Equations (CRC Press, Boca Raton 1995) 37. M.C. Potter, J.L. Goldber, E.F. Aboufadel: Advanced Engineering Mathematics, 3rd edn. (Oxford University Press, New York 2005) 38. W.H. Press, S.A. Teukolsky, W.T. Vettering, B.P. Flannery: Numerical Recipes, 2nd edn. (Cambridge University Press, Cambridge 1992) 39. R.J. Rice: Numerical Methods, Software and Analysis (McGraw-Hill, New York 1983) 40. K.F. Riley, M.P. Hobson, S.J. Bence: Mathematical Methods for Physics and Engineering, 2nd edn. (Cambridge University Press, Cambridge 2002) 41. H.M. Schey: Div, Grad, Curl, and All That: An Informal Text on Vector Calculus, 4th edn. (Norton, New York 2004) 42. K.A. Stroud, D.J. Booth: Advanced Engineering Mathematics, 4th edn. (Indus- trial Press, New York 2003) 43. C.R. Wylie, L.C. Barrett: Advanced Engineering Mathematics, 5th edn. (McGraw-Hill, New York 1982) 44. D. Zwillinger: Handbook of Diﬀerential Equations (Academic Press, San Diego 1998) Index Acceleration vector, 36 Alternating Tensor, 172 Analog computation, 250 Angular frequency, 238 Angular velocity vector, 37 Arfken, G., 168 Axial vector, 186 Beats in forced vibration, 245 Bernoulli equation, 213 Bernoulli, James, 213 Binormal vector, 49 Bound vector, 4 Cartesian tensor, 169 Cauchy diﬀerential equation, 230 Cauchy’s integral formula, 323 Centripetal acceleration, 40, 47 Connectivity of space, 77 Conservative vector ﬁeld, 89 Continuity equation, 69 Contraction, 176 Coordinate curves, 130 Coordinate surfaces, 130 Coriolis acceleration, 47 Coulomb damping free vibration, 241 Coulomb gauge, 95 Coupled oscillation, 261 Coupled oscillators normal modes, 261 Cross product, 13 Curl curl curl identity, 82fundamental theorem of, 74 in curvilinear coordinate systems, 135 in cylindrical coordinates, 119 in spherical coordinates, 127 of the gradient of a scalar function, 81 Curl of a vector, 70 Cylindrical coordinates, 113 curl, 119 divergence, 117 gradient, 116 inﬁnitesimal elements, 120 Laplacian, 117 Damping viscous damping in free vibration, 238 Damping Coulomb damping in free vibration, 241 Delta function as a divergence, 98 deﬁnition, 291 with complicated arguments, 292 Diﬀerential equation Bernoulli equation, 213 Euler-Cauchy equation, 230 ﬁrst-order exact equation, 206 reducible to separable type, 204 separable variables, 203 336 Index Diﬀerential equation ( Continued ) force vibration without damping, 244 forced bibration with viscous damping, 247 homogeneous linear constant coeﬃcients, 216 homogeneous linear equation, 215 linear ﬁrst-order, 210 nonhomogeneous variation of parameters, 232 nonhomogeneous linear costant coeﬃcients, 222 simultaneous equations Cramer’s rule, 256 system of equations reduction to a single equation, 254 system of simultaneous linear equation, 254 with discontinuous forcing function, 297 Diﬀerential equation Laplace transform method, 278 Diﬀerentiation in noninertial reference, 42 Dirac delta function, 291 Dirac, P.A.M., 291 Direct Product, 174 Dirichlet boundary condition, 106 Distance between two skew lines, 24 Divergence fundamental theorem of, 67 in curvilinear coordinate systems, 134 in cylindrical coordinates, 117 in spherical coordinates, 126 of curl of a vector function, 82 Divergence of a vector, 61 Divergence theorem, 65 alternative forms, 86 Dot product, 6 Duhamel integral, 302 Dyad, 175 Dyadic, 175 Electrical circuit complex solution impedance, 252 LRC circuit, 249Elliptical coordinates, 138 coordinate surfaces, 139 relation with rectangular coordinates, 141 Equation with separable variables, 202 Euler angles, 159 Euler diﬀerential equation, 230 Euler-Cauchy diﬀerential equation, 230 Exact diﬀerential equation, 205 Feshbach, Herman, 138 First-order diﬀerential equation, 201 Bernoulli equation, 213 exact equation, 205 integrating factor, 207 reducible to separable type, 204 separable variables, 202 First-order linear diﬀerential equation, 210 Flannery, Brian P., 265 Flux of a vector ﬁeld, 62 Forced vibration beats, 245 resonance, 246 with viscous damping, 247 without damping, 244 Free vector, 4 Free vibration, 236 Coulomb damping, 241 viscous damping critical damping, 239 over damping, 239 under damping, 239 Frenet-Serrect formulas, 49 Frequency angular frequency, 238 Frequency of simple harmonic motion, 238 Fundamental theorem of curls, 74 Gamma function, 312 Gauge transformation, 95 Gauss’ theorem, 65 General curvilinear coordinates, 130 Gradient fundamental theorem of, 58 geometrical interpretation, 53 in curvilinear coordinate systems, 133 in cylindrical coordinates, 116 Index 337 in spherical coordinates, 125 of a scalar function, 51 Gradient operator, 51 Green’s lemma, 85 Green’s theorem, 85 symmetrical form, 85 Heaviside expansion, 284 Heaviside unit step function, 294 Heaviside, Oliver, 271 Helmholtz’s theorem, 101 Homogeneous diﬀerential equation Euler-Cauchy equation, 232 Homogeneous linear diﬀerential equation fundamental theorem, 215 with constant coeﬃcient, 216 Hooke’s Law generalized, 193 Integrating factor, 207 Inverse Laplace transform, 278 by contour integration, 323 by partial fraction, 280 of periodic function, 311 using derivative of the transform, 286 Inverse Laplace transformation Heaviside expansion, 284 Irrotational ﬁeld, 70 Irrotational vector ﬁeld, 89 Jacobi, Carl, 147 Jacobian for double integral, 145 Jacobian for multiple integral, 147 Jacobian matrix, 147 Kronecker Delta Tensor, 171 Kronecker tensor, 171 Lagrange identity, 16 Laplace transform computer algebraic systems, 326 convolution, 302 convolution theorem, 304 deﬁnition, 271 derivative, 274 derivative of a transform, 276 diﬀerential equation with variable coeﬃcients, 319evaluating integrals, 316 Gamma function, 312 integration of transforms, 307 inverse, 278 of delta function and its derivative, 294 of impulsive function, 291 of integrals, 307 s-shifting, 275 scaling, 308 solving diﬀerential equation, 288 solving integro-diﬀerential equation, 321 step function, 296 table of , 276 Laplace transform of periodic function, 309 Laplace transforms, 271 Laplace’s equation, 104 LaPlace, Pierre Simon, 271 Laplacian, 82 in curvilinear coordinate systems, 134 in cylindrical coordinates, 117 in spherical coordinates, 127 Law of cosines, 8 Law of sine, 16 Lee, T.D., 189 Levi-Civita Tensor, 172 Line integral of a gradient vector, 56 Linear diﬀerential equation ﬁrst-order, 210 Linear diﬀerential equation higher order, 214 Linear homogeneous diﬀerential equation with constant coeﬃcients characteristic equation complex conjugate roots, 218 distinct roots, 217 equal roots, 218 Maple, 265, 326 MathCad, 265, 326 Mathematica, 265 Mathews, P.C., 168 Matlab, 265, 326 McQuarrie, D.A., 168 Mechanical vibrations, 235 Mellin inversion, 324 338 Index Moebius surface, 72 Moment of Inertia Tensor, 189 Morse, Philip M., 138 Multiple integrals, 144 Multiply connected region, 77 MuPAD, 265, 326 Neumann boundary condition, 106 Nonhomogeneous diﬀerential equation variation of parameters, 232 Nonhomogeneous linear diﬀerential equation method of complex exponential, 229 with constant coeﬃcients method of undertermined coeﬃ- cients, 222 Noninertial reference system diﬀerentiation, 42 Numerical Methods, Software and Analysis, 265 Numerical Recipes, 265 Ordinary diﬀerential equations, 201 Orientable surface, 72 Osculating plane, 49 Outer Product, 174 Partial fraction decomposition, 280 Path integral, 57 Period of oscillation, 237 Planes in space, 27 Poisson’s equation, 104 Polar vector, 185 Position vector, 23 transformation of, 156 Potential vector, 92 Potential scalar, 89 Press, William H., 265 Prolate Spheroidal coordinates, 144 Pseudoscalar, 187 Pseudotensor, 185 Pseudovector, 186 Purcell, E.M., 168 Reciprocal vectors, 22 Resonance in forced vibration, 246 Rice, R.J., 265Rotation Euler angles, 159 of coordinates, 156 Rotation matrix, 157 properties of , 162 Scalar, 3 deﬁnition in terms of transformation properties, 165 Scalar potential, 89 Scalar triple product, 17 Scientiﬁc WorkPlace, 265, 326 Shifting operation, 295 Simply connected region, 77 Simultaneous diﬀerential equations as an eigenvalue problem, 257 Cramer’s rule, 255 Simultaneous linear diﬀerential equations, 254 Singularities in the ﬁeld, 69 Solenoidal ﬁeld, 92 Spherical coordinates, 122 curl, 127 divergence, 126 gradient, 125 inﬁnitesimal elements, 128 Laplacian, 127 Spherical polar coordinate system, 122 Step function deﬁnition, 294 Stokes’ theorem, 71 alternative forms, 87 Straight lines, 23 Strain Tensor, 193 Stress Tensor, 190 Substitution Tensor, 172 Summation convention, 177 Tensor cartesian, 169 contraction, 176 deﬁnition, 169 moment of inertia tensor, 189 outer product, 174 quotient rule, 182 rank , 170 Strain tensor, 193 stress tensor, 190 Index 339 summation convention, 177 symmetry properties, 183 Tensor components, 170 Tensor ﬁeld, 179 Teukolsky, Saul A., 265 Theory of space curve, 47 Theory of vector ﬁeld, 95 Torsion of the curve, 49 Transverse acceleration, 47 Uniqueness theorem, 105 Unit Tensor, 172 Unit vector, 5 Vector addition, 5 cross product, 13 curl, 70 deﬁnition in terms of transformation properties, 165 divergence, 61 dot product, 6 multiplication by a scalar, 5subtraction, 5 triple product, 17 Vector components, 10 time derivative, 36 Vector calculus, 35 Vector equation, 158 Vector equation for lines, 23 Vector equation for planes, 23 Vector ﬁeld, 35 Vector potential, 92 Vector triple product, 18 Vectors, 3 bound, 4 free, 4 product rules, 79 transformation properties, 156 Velocity vector, 36 Velocity vector ﬁeld, 41 Vetterling, William T., 265 Viscous Damping free vibration, 238 Wronskian, 233