Home / Math and Physics Files / Math / Math Book Downloads / PDE

Boundary-Value-Problems-and-Partial-Differential-Equations-5th-ed--by-David-L-Powers

PDF · 515 pages · 3.2 MB

A published undergraduate and introductory graduate textbook by David L. Powers of Clarkson University, not Phil's own work. It covers ODE review, Fourier series and integrals, the heat, wave and potential equations by separation of variables, Bessel and Legendre functions, Laplace transforms, and numerical methods, with exercises and answers to odd-numbered problems.

AI-written summary; may contain errors.

Extracted text (machine-read; may contain errors)

BOUNDARY VALUE PROBLEMS AND PARTIAL DIFFERENTIAL EQUATIONS es oF .7 Z DAVID L.POWERS BOUNDARY VALUE PROBLEMS FIFTH EDITION This page intentionally left blank BOUNDARY VALUE PROBLEMS AND PARTIAL DIFFERENTIAL EQUATIONS DAVID L. POWERS Clarkson University FIFTH EDITION Amsterdam Boston Heidelberg London New York Oxford Paris San Diego San Francisco Singapore Sydney Tokyo Acquisitions Editor Tom Singer Project Manager Jeff Freeland Marketing Manager Linda Beattie Cover Design Eric DeCicco Interior Printer The Maple Vail Book Manufacturing Group Elsevier Academic Press 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 Theobald’s Road, London WC1X 8RR, UK This book is printed on acid-free paper. /circlecopyrt∞ Copyright © 2006, Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any informationstorage and retrieval system, without permission in writing from the publisher. Permissions may be sought directly from Elsevier’s Science & T echnology Rights Department in Oxford, UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333,e-mail: [email protected]. Y ou may also complete your request on-line viathe Elsevier homepage (http://elsevier.com), by selecting “Customer Support” and then “Obtaining Permissions.” Library of Congress Cataloging-in-Publication Data Application submitted British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN 13: 978-0-12-563738-1 ISBN 10: 0-12-563738-1 For all information on all Elsevier Academic Press publications visit our Web site at www.books.elsevier.com Printed in the United States of America 0 50 60 70 80 91 0987654321 Contents Preface ix CHAPTER 0 Ordinary Differential Equations 1 0.1 Homogeneous Linear Equations 1 0.2 Nonhomogeneous Linear Equations 14 0.3 Boundary Value Problems 26 0.4 Singular Boundary Value Problems 38 0.5 Green’s Functions 43 Chapter Review 51 Miscellaneous Exercises 51 CHAPTER 1 Fourier Series and Integrals 59 1.1 Periodic Functions and Fourier Series 591.2 Arbitrary Period and Half-Range Expansions 64 1.3 Convergence of Fourier Series 731.4 Uniform Convergence 79 1.5 Operations on Fourier Series 85 1.6 Mean Error and Convergence in Mean 90 1.7 Proof of Convergence 95 1.8 Numerical Determination of Fourier Coefﬁcients 100 1.9 Fourier Integral 106 1.10 Complex Methods 1131.11 Applications of Fourier Series and Integrals 117 1.12 Comments and References 124 Chapter Review 125 Miscellaneous Exercises 125 v vi Contents CHAPTER 2 The Heat Equation 135 2.1 Derivation and Boundary Conditions 135 2.2 Steady-State Temperatures 143 2.3 Example: Fixed End Temperatures 149 2.4 Example: Insulated Bar 157 2.5 Example: Different Boundary Conditions 1632.6 Example: Convection 170 2.7 Sturm–Liouville Problems 175 2.8 Expansion in Series of Eigenfunctions 181 2.9 Generalities on the Heat Conduction Problem 184 2.10 Semi-Inﬁnite Rod 188 2.11 Inﬁnite Rod 193 2.12 The Error Function 199 2.13 Comments and References 204 Chapter Review 206Miscellaneous Exercises 206 CHAPTER 3 The Wave Equation 215 3.1 The Vibrating String 2153.2 Solution of the Vibrating String Problem 218 3.3 d’Alembert’s Solution 227 3.4 One-Dimensional Wave Equation: Generalities 233 3.5 Estimation of Eigenvalues 236 3.6 Wave Equation in Unbounded Regions 2393.7 Comments and References 246 Chapter Review 247 Miscellaneous Exercises 247 CHAPTER 4 The Potential Equation 255 4.1 Potential Equation 2554.2 Potential in a Rectangle 259 4.3 Further Examples for a Rectangle 264 4.4 Potential in Unbounded Regions 270 4.5 Potential in a Disk 275 4.6 Classiﬁcation and Limitations 2804.7 Comments and References 283 Chapter Review 285 Miscellaneous Exercises 285 CHAPTER 5 Higher Dimensions and Other Coordinates 295 5.1 Two-Dimensional Wave Equation: Derivation 295 5.2 Three-Dimensional Heat Equation 298 5.3 Two-Dimensional Heat Equation: Solution 303 Contents vii 5.4 Problems in Polar Coordinates 308 5.5 Bessel’s Equation 3115.6 Temperature in a Cylinder 316 5.7 Vibrations of a Circular Membrane 321 5.8 Some Applications of Bessel Functions 3295.9 Spherical Coordinates; Legendre Polynomials 3355.10 Some Applications of Legendre Polynomials 3455.11 Comments and References 353 Chapter Review 354 Miscellaneous Exercises 354 CHAPTER 6 Laplace Transform 363 6.1 Deﬁnition and Elementary Properties 3636.2 Partial Fractions and Convolutions 3696.3 Partial Differential Equations 3766.4 More Difﬁcult Examples 3836.5 Comments and References 389 Miscellaneous Exercises 389 CHAPTER 7 Numerical Methods 397 7.1 Boundary Value Problems 3977.2 Heat Problems 403 7.3 Wave Equation 4087.4 Potential Equation 4147.5 Two-Dimensional Problems 4207.6 Comments and References 428 Miscellaneous Exercises 428 Bibliography 433 Appendix: Mathematical References 435Answers to Odd-Numbered Exercises 441 Index 495 This page intentionally left blank Preface This text is designed for a one-semester or two-quarter course in partial dif- ferential equations given to third- and fourth-year students of engineering andscience. It can also be used as the basis for an introductory course for graduatestudents. Mathematical prerequisites have been kept to a minimum — calculus and differential equations. Vector calculus is used for only one derivation, and necessary linear algebra is limited to determinants of order two. A reader needsenough background in physics to follow the derivations of the heat and waveequations. The principal objective of the book is solving boundary value problems involving partial differential equations. Separation of variables receives thegreatest attention because it is widely used in applications and because it pro- vides a uniform method for solving important cases of the heat, wave, and potential equations. One technique is not enough, of course. D’Alembert’s so-lution of the wave equation is developed in parallel with the series solution,and the distributed-source solution is constructed for the heat equation. Inaddition, there are chapters on Laplace transform techniques and on numeri-cal methods. The second objective is to tie together the mathematics developed and the student’s physical intuition. This is accomplished by deriving the mathemati-cal model in a number of cases, by using physical reasoning in the mathemat-ical development, by interpreting mathematical results in physical terms, andby studying the heat, wave, and potential equations separately. In the service of both objectives, there are many fully worked examples and now about 900 exercises, including miscellaneous exercises at the end of each chapter. The level of difﬁculty ranges from drill and veriﬁcation of details to development of new material. Answers to odd-numbered exercises are in ix x Preface the back of the book. An Instructor’s Manual is available both online and in print (ISBN: 0-12-369435-3), with the answers to the even-numbered prob-lems. A Student Solutions Manual is available both online and in print (ISBN:0-12-088586-7), that contains detailed solutions of odd-numbered problems. There are many ways of choosing and arranging topics from the book to provide an interesting and meaningful course. The following sections formthe core, requiring at least 14 hours of lecture: Sections 1.1–1.3, 2.1–2.5, 3.1–3.3, 4.1–4.3, and 4.5. These cover the basics of Fourier series and the solutionsof heat, wave, and potential equations i n ﬁnite regions. My choice for the next most important block of material is the Fourier integral and the solution ofproblems on unbounded regions: Sections 1.9, 2.10–2.12, 3.6, and 4.4. These require at least six more lectures. The tastes of the instructor and the needs of the audience will govern the choice of further material. A rather theoretical ﬂavor results from including:Sections 1.4–1.7 on convergence of Fourier series; Sections 2.7–2.9 on Sturm–Liouville problems, and the sequel, Section 3.4; and the more difﬁcult parts ofChapter 5, Sections 5.5–5.10 on Bessel functions and Legendre polynomials.On the other hand, inclusion of numerical methods in Sections 1.8 and 3.5 and Chapter 7 gives a very applied ﬂavor. Chapter 0 reviews solution techniques and theory of ordinary differential equations and boundary value problems. Equilibrium forms of the heat andwave equations are derived also. This material belongs in an elementary differ-ential equations course and is strictly optional. However, many students haveeither forgotten it or never seen it. For this ﬁfth edition, I have revised in response to students’ changing needs and abilities. Many sections have been rewritten to improve clarity, provideextra detail, and make solution processes more explicit. In the optional Chap-ter 0, free and forced vibrations are major examples for solution of differentialequations with constant coefﬁcients. In Chapter 1, I have returned to derivingthe Fourier integral as a “limit” of Fourier series. New exercises are included for applications of Fourier series and integrals. Solving potential problems on a rectangle seems to cause more difﬁculty than expected. A new section 4.3 gives more guidance and examples as well as some information about the Poissonequation. New exercises have been added and old ones revised throughout.In particular I have included exercises based on engineering research publica-tions. These provide genuine problems with real data. A new feature of this edition is a CD with auxiliary materials: animations of convergence of Fourier series; animations of solutions of the heat and wave e q u a t i o n sa sw e l la so r d i n a r yi n i t i a lv a l u ep r o b l e m s ;c o l o rg r a p h i c so fs o l u - tions of potential problems; additional exercises in a workbook style; reviewquestions for each chapter; text materi al on using a spreadsheet for numerical methods. All ﬁles are readable with just a browser and Adobe Reader, availablewithout cost. Preface xi I wish to acknowledge the skillful work of Cindy Smith, who was the LaT eX compositor and corrected many of my mistakes, the help of Academic Presseditors and consultants, and the guidance of reviewers for this edition: Darryl Y ong, Harvey Mudd College Ken Luther, Valparaiso University Alexander Kirillov, SUNY at Stony Brook J a m e sV .H e r o d ,G e o r g i aT e c hU n i v e r s i t yHilary Davies, University of Alaska AnchorageCatherine Crawford, Elmhurst CollegeAhmed Mohammed, Ball State University I also wish to acknowledge the guidance of reviewers for the previous edi- tion: Linda Allen, T exas T ech University Ilya Bakelman, T exas A&M University Herman Gollwitzer, Drexel UniversityJames Herod, Georgia Institute of T echnologyRobert Hunt, Humboldt State UniversityMohammad Khavanin, University of North DakotaJeff Morgan, T exas A&M UniversityJim Mueller, California Polytechnic State University Ron Perline, Drexel University William Royalty, University of IdahoLawrence Schovanec, T exas T ech UniversityAl Shenk, University of California at San DiegoMichael Smiley, Iowa State UniversityMonty Strauss, T exas T ech University Kathie Yerion, Gonzaga University David L. Powers This page intentionally left blank Ordinary Differential Equations CHAPTER0 0.1 Homogeneous Linear Equations The subject of most of this book is partial differential equations: their physical meaning, problems in which they appear, and their solutions. Our principal solution technique will involve separating a partial differential equation intoordinary differential equations. Therefore, we begin by reviewing some factsabout ordinary differential equations and their solutions. We are interested mainly in linear differential equations of ﬁrst and second orders, as shown here: du dt=k(t)u+f(t), (1) d2u dt2+k(t)du dt+p(t)u=f(t). (2) In either equation, if f(t)is 0, the equation is homogeneous .( A n o t h e rt e s t :I f the constant function u(t)≡0 is a solution, the equation is homogeneous.) In the rest of this section, we review homogeneous linear equations. A. First-Order Equations The most general ﬁrst-order linear homogeneous equation has the form du dt=k(t)u. (3) 1 2 Chapter 0 Ordinary Differential Equations This equation can be solved by isolating uon one side and then integrating: 1 udu dt=k(t), ln|u|=/integraldisplay k(t)dt+C, u(t)=± eCe/integraltext k(t)dt=ce/integraltext k(t)dt. (4) It is easy to check directly that the last expression is a solution of the differential equation for any value of c.T h a ti s , cis an arbitrary constant and can be used to satisfy an initial condition if one has been speciﬁed. Example. Solve the homogeneous differential equation du dt=− tu. The procedure outlined here gives the general solution u(t)=ce−t2/2 for any c. If an initial condition such as u(0)=5i ss p e c i ﬁ e d ,t h e n cmust be chosen to satisfy it (c=5). /square The most common case of this differential equation has k(t)=kconstant. The differential equation and its general solution are du dt=ku,u(t)=cekt. (5) Ifkis negative, then u(t)approaches 0 as tincreases. If kis positive, then u(t) increases rapidly in magnitude with t. This kind of exponential growth often signals disaster in physical situations, as it cannot be sustained indeﬁnitely. B. Second-Order Equations It is not possible to give a solution method for the general second-order linear homogeneous equation, d2u dt2+k(t)du dt+p(t)u=0. (6) Nevertheless, we can solve some important cases that we detail in what follows. The most important point in the general theory is the following. Chapter 0 Ordinary Differential Equations 3 Principle of Superposition. If u 1(t)and u 2(t)are solutions of the same linear homogeneous equation (6), then so is any linear combination of them: u (t)= c1u1(t)+c2u2(t). /square This theorem, which is very easy to prove, merits the name of principle be- cause it applies, with only superﬁcial changes, to many other kinds of linear, homogeneous equations. Later, we will be using the same principle on partialdifferential equations. T o be able to satisfy an unrestricted initial condition, weneed two linearly independent solutions of a second-order equation. Two so-lutions are linearly independent on an interval if the only linear combination of them (with constant coefﬁcients) that is identically 0 is the combination with 0for its coefﬁcients. There is an alternative test: Two solutions of the same linear homogeneous equation (6) are independent on an interval if and only if their Wronskian W(u 1,u2)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleu 1(t)u2(t) u/prime 1(t)u/prime 2(t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(7) is nonzero on that interval. If we have two independent solutions u 1(t),u2(t)of a linear second-order homogeneous equation, then the linear combination u(t)=c1u1(t)+c2u2(t) is a general solution of the equation: Given any initial conditions, c1andc2can be chosen so that u(t)satisﬁes them. 1. Constant coefﬁcients The most important type of second-order linear differential equation that can be solved in closed form is the one with constant coefﬁcients, d2u dt2+kdu dt+pu=0(k,pare constants ). (8) There is always at least one solution of the form u(t)=emtfor an appropriate constant m.T oﬁ n d m, substitute the proposed solution into the differential equation, obtaining m2emt+kmemt+pemt=0, or m2+km+p=0( 9 ) (since emtis never 0). This is called the characteristic equation of the differ- ential equation (8). There are three cases for the roots of the characteristicequation (9), which determine the nature of the general solution of Eq. (8).These are summarized in Table 1. This method of assuming an exponential form for the solution works for linear homogeneous equations of any order with constant coefﬁcients. In all 4 Chapter 0 Ordinary Differential Equations Roots of Characteristic General Solution of Differential Equation Equation Real, distinct: m1/negationslash=m2 u(t)=c1em1t+c2em2t Real, double: m1=m2 u(t)=c1em1t+c2tem1t Conjugate complex: u(t)=c1eαtcos(βt)+c2eαtsin(βt) m1=α+iβ,m2=α−iβ Table 1 Solutions ofd2u dt2+kdu dt+pu=0 cases, a pair of complex conjugate roots m=α±iβleads to a pair of complex solutions eαteiβt,eαte−iβt(10) that can be traded for the pair of real solutions eαtcos(βt), eαtsin(βt). (11) We include two important examples. First, consider the differential equation d2u dt2+λ2u=0, (12) where λis constant. The characteristic equation is m2+λ2=0, with roots m=± iλ. The third case of Table 1 applies if λ/negationslash=0; the general solution of the differential equation is u(t)=c1cos(λt)+c2sin(λt). (13) Second, consider the similar differential equation d2u dt2−λ2u=0. (14) The characteristic equation now is m2−λ2=0, with roots m=±λ.I fλ/negationslash=0, the ﬁrst case of Table 1 applies, and the general solution is u(t)=c1eλt+c2e−λt. (15) It is sometimes helpful to write the solution in another form. The hyperbolic sine and cosine are deﬁned by sinh(A)=1 2/parenleftbig eA−e−A/parenrightbig ,cosh(A)=1 2/parenleftbig eA+e−A/parenrightbig . (16) Thus, sinh (λt)and cosh (λt)are linear combinations of eλtand e−λt.B yt h e Principle of Superposition, they too are solutions of Eq. (14). The Wronskian Chapter 0 Ordinary Differential Equations 5 Figure 1 Mass–spring–damper system. test shows them to be independent. Therefore, we may equally well write u(t)=c/prime 1cosh(λt)+c/prime 2sinh(λt) as the general solution of Eq. (14), where c/prime 1and c/prime 2are arbitrary constants. Example: Mass–Spring–Damper System. The displacement of a mass in a mass–spring–damper system (Fig. 1) is de- scribed by the initial value problem d2u dt2+bdu dt+ω2u=0, u(0)=u0du dt(0)=v0. The equation is derived from Newton’s second law. Coefﬁcients bandω2 are proportional to characteristic constants of the damper and the spring, re- spectively. The characteristic equa tion of the differential equation is m2+bm+ω2=0, with roots −b±√ b2−4ω2 2=−b 2±/radicalBigg/parenleftbiggb 2/parenrightbigg2 −ω2. The nature of the solution, and therefore the motion of the mass, is determined by the relation between b/2a n dω. b=0:undamped. The roots are ±iωa n dt h eg e n e r a ls o l u t i o no ft h ed i f f e r - ential equation is u(t)=c1cos(ωt)+c2sin(ωt). The mass oscillates forever. 6 Chapter 0 Ordinary Differential Equations 0<b/2<ω:underdamped. T h er o o t sa r ec o m p l e xc o n j u g a t e s α±iβwith α=− b/2,β=/radicalbig ω2−(b/2)2.T h eg e n e r a ls o l u t i o no ft h ed i f f e r e n t i a le q u a - tion is u(t)=e−bt/2/parenleftbig c1cos(βt)+c2sin(βt)/parenrightbig . The mass oscillates, but approaches equilibrium as tincreases. b/2=ω:critically damped. T h er o o t sa r eb o t he q u a lt o b/2. The general solution of the differential equation is u(t)=e−bt/2(c1+c2t). The mass approaches equilibrium as tincreases and may pass through equi- librium ( u(t)may change sign) at most once. b/2>ω :overdamped. Both roots of the characteristic equation are real, say,m1and m2.T h eg e n e r a ls o l u t i o no ft h ed i f f e r e n t i a le q u a t i o ni s u(t)=c1em1t+c2em2t. The mass approaches equilibrium as tincreases, and u(t)may change sign at most once. These cases are illustrated on the CD. /square 2. Cauchy–Euler equation One of the few equations with variable coefﬁcients that can be solved in com- plete generality is the Cauchy–Euler equation: t2d2u dt2+ktdu dt+pu=0. (17) The distinguishing feature of this equation is that the coefﬁcient of the nth derivative is the nth power of t, multiplied by a constant. The style of solution for this equation is quite similar to the preceding: Assume that a solution has the form u(t)=tm,a n dt h e nﬁ n d m. Substituting uin this form into Eq. (17) leads to t2m(m−1)tm−2+ktmtm−1+ptm=0,or m(m−1)+km+p=0(k,pare constants ). (18) This is the characteristic equation for Eq. (17), and the nature of its roots de- termines the solution, as summarized in Table 2. One important example of the Cauchy–Euler equation is t2d2u dt2+tdu dt−λ2u=0, (19) Chapter 0 Ordinary Differential Equations 7 Roots of Characteristic General Solution of Differential Equation Equation Real, distinct roots: m1/negationslash=m2 u(t)=c1tm1+c2tm2 Real, double root: m1=m2 u(t)=c1tm1+c2(lnt)tm1 Conjugate complex roots: u(t)=c1tαcos(βlnt)+c2tαsin(βlnt) m1=α+iβ,m2=α−iβ Table 2 Solutions of t2d2u dt2+ktdu dt+pu=0 where λ> 0. The characteristic equation is m(m−1)+m−λ2=m2−λ2=0. The roots are m=±λ, so the ﬁrst case of Table 2 applies, and u(t)=c1tλ+c2t−λ(20) is the general solution of Eq. (19). F o rt h eg e n e r a ll i n e a re q u a t i o n d2u dt2+k(t)du dt+p(t)u=0, any point where k(t)orp(t)fails to be continuous is a singular point of the differential equation. At such a point, solutions may break down in variousw a y s .H o w e v e r ,i f t 0is a singular point where both of the functions (t−t0)k(t)and (t−t0)2p(t) (21) have Taylor series expansions, then t0is called a regular singular point .T h e Cauchy–Euler equation is an example of an important differential equation having a regular singular point (at t0=0). The behavior of its solution near that point provides a model for more general equations. 3. Other equations Other second-order equations may be solved by power series, by change ofvariable to a kind already solved, or by sheer luck. For example, the equation t 4d2u dt2+λ2u=0, (22) which occurs in the theory of beams, can be solved by the change of variables t=1 z,u(t)=1 zv(z). 8 Chapter 0 Ordinary Differential Equations Here are the details. The second derivative of uhas to be replaced by its ex- pression in terms of v, using the chain rule. Start by ﬁnding du dt=d dz/parenleftbiggv z/parenrightbigg ·dz dt. Since t=1/z,a l s o z=1/t,a n d dz/dt=− 1/t2=− z2.T h u s du dt=− z2/parenleftbiggzv/prime−v z2/parenrightbigg =− zv/prime+v. Similarly we ﬁnd the second derivative d2u dt2=d dz/parenleftbiggdu dt/parenrightbiggdz dt=d dz(−zv/prime+v)/parenleftbig −z2/parenrightbig =− z2(−zv/prime/prime−v/prime+v/prime)=z3v/prime/prime. Finally, replace both terms of the differential equation: /parenleftbigg1 z/parenrightbigg4 z3v/prime/prime+λ2v z=0, or v/prime/prime+λ2v=0. This equation is easily solved, and the solution of the original is then found by reversing the change of variables: u(t)=t/parenleftbig c1cos(λ/t)+c2sin(λ/t)/parenrightbig . (23) C. Second Independent Solution Although it is not generally possible to solve a second-order linear homoge- neous equation with variable coefﬁcients, we can always ﬁnd a second inde- pendent solution if one solution is known. This method is called reduction of order . Suppose u1(t)is a solution of the general equation d2u dt2+k(t)du dt+p(t)u=0. (24) Assume that u2(t)=v(t)u1(t)is a solution. We wish to ﬁnd v(t)so that u2 is indeed a solution. However, v(t)must not be constant, as that would not supply an independent solution. A s traightforward substitution of u2=vu1 into the differential equation leads to v/prime/primeu1+2v/primeu/prime 1+vu/prime/prime 1+k(t)(v/primeu1+vu/prime 1)+p(t)vu1=0. Chapter 0 Ordinary Differential Equations 9 Now collect terms in the derivatives of v. The preceding equation becomes u1v/prime/prime+/parenleftbig 2u/prime 1+k(t)u1/parenrightbig v/prime+/parenleftbig u/prime/prime 1+k(t)u/prime 1+p(t)u1/parenrightbig v=0. However, u1is a solution of Eq. (24), so the coefﬁcient of vis 0. This leaves u1v/prime/prime+/parenleftbig 2u/prime 1+k(t)u1/parenrightbig v/prime=0, (25) which is a ﬁrst-order linear equation for v/prime. Thus, a nonconstant vcan be found, at least in terms of some integrals. Example. Consider the equation /parenleftbig 1−t2/parenrightbig u/prime/prime−2tu/prime+2u=0,−1<t<1, which has u1(t)=tas a solution. By assuming that u2=v·tand substituting, we obtain /parenleftbig 1−t2/parenrightbig (v/prime/primet+2v/prime)−2t(v/primet+v)+2vt=0. After collecting terms, we have /parenleftbig 1−t2/parenrightbig tv/prime/prime+(2−4t2)v/prime=0. From here, it is fairly easy to ﬁnd v/prime/prime v/prime=4t2−2 t(1−t2)=−2 t+1 1−t−1 1+t (using partial fractions), and then lnv/prime=− 2l n(t)−ln(1−t)−ln(1+t). Finally, each side is exponentiated to obtain v/prime=1 t2(1−t2)=1 t2+1/2 1−t+1/2 1+t, v=−1 t+1 2ln/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+t 1−t/vextendsingle/vextendsingle/vextendsingle/vextendsingle. /square D. Higher-Order Equations Linear homogeneous equations of order higher than 2 — especially order 4 — occur frequently in elasticity and ﬂuid mechanics. A general, nth-order homo- geneous linear equation may be written u(n)+k1(t)u(n−1)+···+ kn−1(t)u(1)+kn(t)u=0, (26) 10 Chapter 0 Ordinary Differential Equations Root Multiplicity Contribution m(real) 1 cemt m(real) k (c1+c2t+···+ cktk−1)emt m,m(complex) 1 (acos(βt)+bsin(βt))eαt m=α+iβ m,m(complex) k(a1+a2t+···+ aktk−1)cos(βt)eαt +(b1+b2t+···+ bktk−1)sin(βt)eαt Table 3 Contributions to general solution in which the coefﬁcients k1(t),k2(t), etc., are given functions of t.T h et e c h - niques of solution are analogous to those for second-order equations. In par- ticular, they depend on the Principle of Superposition, which remains validfor this equation. That principle allows us to say that the general solutionof Eq. (26) has the form of a linear combination of nindependent solutions u 1(t),u2(t),..., un(t)with arbitrary constant coefﬁcients, u(t)=c1u1(t)+c2u2(t)+···+ cnun(t). Of course, we cannot solve the general nth-order equation (26), but we can indeed solve any homogeneous linear equation with constant coefﬁcients, u(n)+k1u(n−1)+···+ kn−1u(1)+knu=0. (27) We must now ﬁnd nindependent solutions of this equation. As in the second- order case, we assume that a solution has the form u(t)=emt,a n dﬁ n dv a l u e s ofmfor which this is true. That is, we substitute emtforuin the differen- tial equation (27) and divide out the common factor of emt. The result is the polynomial equation mn+k1mn−1+···+ kn−1m+kn=0, (28) called the characteristic equation of the differential equation (27). Each distinct root of the characteristic equation contributes as many inde- pendent solutions as its multiplicity, which might be as high as n.R e c a l la l s o that the polynomial equation (28) may have complex roots, which will occurin conjugate pairs if — as we assume — the coefﬁcients k 1,k2, etc., are real. When this happens, we prefer to have real solutions, in the form of an expo- nential times sine or cosine, instead of complex exponentials. The contributionof each root or pair of conjugate roots of Eq. (28) is summarized in Table 3.Since the sum of the multiplicities of the roots of Eq. (28) is n, the sum of the contributions produces a solution with nterms, which can be shown to be the general solution. Chapter 0 Ordinary Differential Equations 11 Example. Find the general solution of this fourth-order equation u(4)+3u(2)−4u=0. The characteristic equation is m4+3m2−4=0, which is easy to solve because it is a biquadratic. We ﬁnd that m2=− 4 or 1, and thus the roots are m=± 2i, ±1, all with multiplicity 1. From Table 3 we ﬁnd that acos(2t)+bsin(2t)cor- responds to the complex conjugate pair, m=± 2i,w h i l e etand e−tcorrespond tom=1a n d m=− 1. Thus we build up the general solution, u(t)=acos(2t)+bsin(2t)+c1et+c2e−t. /square Example. Find the general solution of the fourth-order equation u(4)−2u(2)+u=0. The characteristic equation is m4−2m2+1=0, whose roots, found as in the preceding, are ±1, both with multiplicity 2. From Table 3 we ﬁnd that each of the roots contributes a ﬁrst-degree pol ynomial times an exponential. Thus, we assemble the general solution as u(t)=(c1+c2t)et+(c3+c4t)e−t. With sinh (t)=(et−e−t)/2 and cosh (t)=(et+e−t)/2, the terms of the pre- ceding combination can be rearranged to give the general solution in a differ- ent form, u(t)=(C1+C2t)cosh(t)+(C3+C4t)sinh(t). /square Some important equations and their solutions. 1.du dt=ku (kis constant), u(t)=cekt. 2.d2u dt2+λ2u=0, u(t)=acos(λt)+bsin(λt). 3.d2u dt2−λ2u=0, u(t)=acosh(λt)+bsinh(λt)or u(t)=c1eλt+c2e−λt. 4.t2u/prime/prime+tu/prime−λ2u=0, u(t)=c1tλ+c2t−λ. 12 Chapter 0 Ordinary Differential Equations EXERCISES In Exercises 1–6, ﬁnd the general solution of the differential equation. Be care- ful to identify the dependent and independent variables. 1.d2φ dx2+λ2φ=0. 2.d2φ dx2−µ2φ=0. 3.d2u dt2=0. 4.dT dt=−λ2kT. 5.1 rd dr/parenleftbigg rdw dr/parenrightbigg −λ2 r2w=0. 6.ρ2d2R dρ2+2ρdR dρ−n(n+1)R=0. In Exercises 7–11, ﬁnd the general solution. In some cases, it is helpful to carry out the indicated differentiation, in others it is not. 7.d dx/parenleftbigg (h+kx)dv dx/parenrightbigg =0( h,kare constants). 8.(exφ/prime)/prime+λ2exφ=0. 9.d dx/parenleftbigg x3du dx/parenrightbigg =0. 10.r2d2u dr2+rdu dr+λ2u=0. 11.1 rd dr/parenleftbigg rdu dr/parenrightbigg =0. 12.Compare and contrast the form of the solutions of these three differential equations and their behavior as t→∞ . a.d2u dt2+u=0; b.d2u dt2=0; c.d2u dt2−u=0. In Exercises 13–15, use the “exponential guess” method to ﬁnd the general solution of the differential equations ( λis constant). 13.d4u dx4−λ4u=0. 14.d4u dx4+λ4u=0. Chapter 0 Ordinary Differential Equations 13 15.d4u dx4+2λ2d2u dx2+λ4u=0. In Exercises 16–18, one solution of the differential equation is given. Find a second independent solution. 16.d2u dt2+2adu dt+a2u=0, u1(t)=e−at. 17.t2d2u dt2+(1−2b)tdu dt+b2u=0, u1(t)=tb. 18.d dx/parenleftbigg xdu dx/parenrightbigg +4x2−1 4xu=0, u1(x)=cos(x)√x. In Exercises 19–21, use the indicated change of variable to solve the differential equation. 19.d dρ/parenleftbigg ρ2dR dρ/parenrightbigg +λ2ρ2R=0, R(ρ)=u(ρ) ρ. 20.d dρ/parenleftbigg ρdφ dρ/parenrightbigg +4λ2ρ2−1 4ρφ=0,φ(ρ)=v(ρ)√ρ. 21.t2d2u dt2+ktdu dt+pu=0, x=lnt,u(t)=v(x). 22.Solve each initial value problem. Assuming that the solution represents the displacement of a mass in a mass–spring–damper system, as in the text, describe the motion in words. a.d2u dt2+4u=0, u(0)=1,du dt(0)=0; b.d2u dt2+2du dt+2u=0, u(0)=1,du dt(0)=1; c.d2u dt2+2du dt+u=0, u(0)=1,du dt(0)=1; d.d2u dt2+2du dt+0.75u=0, u(0)=0,du dt(0)=1. 23.Sheet metal is produced by repeatedly feeding the sheet between steel rollers to reduce the thickness. In the article “On the characteristics and mechanism of rolling instability and chatter” [Y.-J. Lin et al., J. of Manu- facturing Science and Engineering ,125(2003): 778–786], the authors ﬁnd that the distance between rollers is well approximated by h+y,w h e r e h is the nominal output thickness and yis the solution of the differential equation y/prime/prime+2αy/prime+σ2y=0. The elasticity of the sheet and the rollers provides the restoring force, and the plastic deformation of the sheet ef- fectively provides damping. 14 Chapter 0 Ordinary Differential Equations For high-speed operation, the system is underdamped. Solve the initial value problem consisting of the differential equation and the initial con-ditions y(0)=− 0.001h,y /prime(0)=0. 24.(Continuation) For an input speed of 25.4 m/s, it is observed that σ∼= 600 Hz or 1200 πradians/s and α=0.103σ. Using these values, obtain a graph of the solution of the preceding exercise, over the range 0 <t< 0.02 s. How far does the sheet move in 0.02 s? 25.(Continuation) The damping constant αreferred to in the previous ex- ercises appears to depend on v, the speed of the sheet into the rollers, according to the relation α/σ=A/v,w h e r e Ais a constant. From the in- formation given previously, the value of Ais about 2.62. Assuming this is correct, ﬁnd the speed vat which damping is critical. 0.2 Nonhomogeneous Linear Equations In this section, we will review methods for solving nonhomogeneous linear equations of ﬁrst and second orders, du dt=k(t)u+f(t), d2u dt2+k(t)du dt+p(t)u=f(t). Of course, we assume that the inhomogeneity f(t)is not identically 0. The simplest nonhomogeneous equation is du dt=f(t). (1) This can be solved in complete generality by one integration: u(t)=/integraldisplay f(t)dt+c. (2) We have used an indeﬁnite integral and have written cas a reminder that there is an arbitrary additive constant in the general solution of Eq. (1). A moreprecise way to write the solution is u(t)=/integraldisplay t t0f(z)dz+c. (3) Here we have replaced the indeﬁnite integral by a deﬁnite integral with vari- able upper limit. The lower limit of integration is usually an initial time. Note 0.2 Nonhomogeneous Linear Equations 15 that the name of the integration variable is changed from tto something else (here, z) to avoid confusing the limit with the dummy variable of integration. The simple second-order equation d2u dt2=f(t) (4) can be solved by two successive integrations. The two theorems that follow summarize some properties of linear equa- tions that are useful in constructing solutions. Theorem 1. The general solution of a nonhomogeneous linear equation has the form u (t)=up(t)+uc(t),w h e r eu p(t)is any particular solution of the nonho- mogeneous equation and u c(t)is the general solution of the corresponding homo- geneous equation. /square Theorem 2. If u p1(t)and u p2(t)are particular solutions of a differential equation with inhomogeneities f 1(t)and f 2(t),r e s p e c t i v e l y ,t h e nk 1up1(t)+k2up2is a par- ticular solution of the differential equation with inhomogeneity k 1f1(t)+k2f2(t) (k1,k2are constants). /square Example. Find the solution of the differential equation d2u dt2+u=1−e−t. The corresponding homogeneous equation is d2u dt2+u=0, with general solution uc(t)=c1cos(t)+c2sin(t)(found in Section 1). A par- ticular solution of the equation with the inhomogeneity f1(t)=1, that is, of the equation d2u dt2+u=1, isup1(t)=1. A particular solution of the equation d2u dt2+u=e−t isup2(t)=1 2e−t. (Later in this section, we will review methods for constructing these particular solutions.) Then, by Theorem 2, a particular solution of thegiven nonhomogeneous Eq. (5) is u p(t)=1−1 2e−t. Finally, by Theorem 1, the 16 Chapter 0 Ordinary Differential Equations Inhomogeneity, f(t) Form of Trial Solution, up(t) (a0tn+a1tn−1+···+ an)eαt(A0tn+A1tn−1+···+ An)eαt (a0tn+···+ an)eαtcos(βt) +(b0tn+···+ bn)eαtsin(βt)(A0tn+···+ An)eαtcos(βt) +(B0tn+···+ Bn)eαtsin(βt) Table 4 Undetermined coefﬁcients g e n e r a ls o l u t i o no ft h eg i v e ne q u a t i o ni s u(t)=1−1 2e−t+c1cos(t)+c2sin(t). If two initial conditions are given, then c1andc2are available to satisfy them. Of course, an initial condition applies to the entire solution of the given dif- ferential equation, not just to uc(t). /square Now we turn our attention to methods for ﬁnding particular solutions of nonhomogeneous linear differential equations. A. Undetermined Coefﬁcients This method involves guessing the form of a trial solution and then ﬁnding the appropriate coefﬁcients. Naturally, it is limited to the cases in which we canguess successfully: when the equation has constant coefﬁcients and the inho-mogeneity is simple in form. Table 4 offers a summary of admissible inhomo-geneities and the corresponding forms for particular solution. The parametersn,α,β and the coefﬁcients a 0,..., an,b0,..., bnare found by inspecting the given inhomogeneity. The table compresses several cases. For instance, f(t)in line 1 is a polynomial if α=0o ra ne x p o n e n t i a li f n=0a n d α/negationslash=0. In line 2, both sine and cosine must be included in the trial solution even if one is absent from f(t);b u tα=0 is allowed, and so is n=0. Example. Find a particular solution of d2u dt2+5u=te−t. W eu s el i n e1o fT a b l e4 .E v i d e n t l y , n=1a n d α=− 1. The appropriate form for the trial solution is up(t)=(A0t+A1)e−t. When we substitute this form into the differential equation, we obtain (A0t+A1−2A0)e−t+5(A0t+A1)e−t=te−t. 0.2 Nonhomogeneous Linear Equations 17 Now, equating coefﬁcients of like terms gives these two equations for the coef- ﬁcients: 6A0=1(coefﬁcient of te−t), 6A1−2A0=0(coefﬁcient of e−t). T h e s ew es o l v ee a s i l yt oﬁ n d A0=1/6,A1=1/18. Finally, a particular solution is up(t)=/parenleftbigg1 6t+1 18/parenrightbigg e−t. /square A trial solution from Table 4 will not work if it contains any term that is a so- lution of the homogeneous differential equation. In that case, the trial solutionhas to be revised by the following rule. Revision Rule. Multiply by the lowest positive integral power of t such that no term in the trial solution satisﬁes the corresponding homogeneous equation. /square Example. Table 4 suggests the trial solution up(t)=(A0t+A1)e−tfor the differential equation d2u dt2−u=te−t. However, we know that the solution of the corresponding homogeneous equa- tion, u/prime/prime−u=0, is uc(t)=c1et+c2e−t. The trial solution contains a term (A1e−t)that is a solution of the homoge- neous equation. Multiplying the trial solution by teliminates the problem. T h u s ,t h et r i a ls o l u t i o ni s up(t)=t(A0t+A1)e−t=/parenleftbig A0t2+A1t/parenrightbig e−t. Similarly, the trial solution for the differential equation d2u dt2+2du dt+u=te−t has to be revised. The solution of the corresponding homogeneous equation isuc(t)=c1e−t+c2te−t. The trial solution from the table has to be multiplied byt2to eliminate solutions of the homogeneous equation. /square Example: Forced Vibrations. The displacement u(t)of a mass in a mass–spring–damper system, starting 18 Chapter 0 Ordinary Differential Equations Figure 2 Mass–spring–damper system with an external force. from rest, with an external sinusoidal force (see Fig. 2) is described by this initial value problem: d2u dt2+bdu dt+ω2u=f0cos(µt), u(0)=0,du dt(0)=0. See the Section 1 example on the mass–spring–damper system. The coefﬁ- cient f0is proportional to the magnitude of the force. There are three impor- tant cases. b=0,µ/negationslash=ω:undamped, no resonance. The form of the trial solution is up(t)=Acos(µt)+Bsin(µt). Substitution and simple algebra lead to the particular solution u0(t)=f0 ω2−µ2cos(µt) (that is, B=0 ) .T h eg e n e r a ls o l u t i o no ft h ed i f f e r e n t i a le q u a t i o ni s u(t)=f0 ω2−µ2cos(µt)+c1cos(ωt)+c2sin(ωt). Applying the initial conditions determines c1and c2. Finally, the solution of the initial value problem is u(t)=f0 ω2−µ2/parenleftbig cos(µt)−cos(ωt)/parenrightbig . 0.2 Nonhomogeneous Linear Equations 19 b=0,µ=ω:resonance. Now, since ω=µ, the trial solution must be revised to up(t)=Atcos(µt)+Btsin(µt). Substitution into the differential equation and simple algebra give A=0,B= f0/2µ,o r up(t)=f0 2µtsin(µt). The general solution of the differential equation is u(t)=f0 2µtsin(µt)+c1cos(µt)+c2sin(µt). (Remember that b=0a n d ω=µ.) The initial conditions give c1=c2=0, so the solution of the initial value problem is u(t)=f0 2µtsin(µt). The presence of the multiplier tmeans that the amplitude of the oscillation is increasing. This is the phenomenon of resonance. b>0:damped motion. The ideas are straightforward applications of the techniques developed earlier. The trial solution is a combination of cos (µt) and sin (µt).S o m e w h a tl e s ss i m p l ea l g e b r ag i v e s up(t)=f0 /Delta1/parenleftbig (ω2−µ2)cos(µt)+µbsin(µt)/parenrightbig , where /Delta1=(ω2−µ2)2+µ2b2.T h eg e n e r a ls o l u t i o no ft h ed i f f e r e n t i a le q u a t i o n may take different forms, depending on the relation between bandω.( S e e Section 1.) Assuming the underdamped case holds, we have u(t)=f0 /Delta1/parenleftbig (ω2−µ2)cos(µt)+µbsin(µt)/parenrightbig +e−bt/2/parenleftbig c1cos(γt)+c2sin(γt)/parenrightbig for the general solution of the differential equation. Here, γ=/radicalbig ω2−(b/2)2 is real because we assumed underdamping. Applying the initial conditions gives, after some nasty algebra, c1=−f0 /Delta1/parenleftbig ω2−µ2/parenrightbig ,c2=−f0 /Delta1b γω2+µ2 2. 20 Chapter 0 Ordinary Differential Equations Notice that, as tincreases, the terms that come from the complementary solution approach 0, while the terms that come from the particular solutionpersist. These cases are illustrated with animation on the CD. /square B. Variation of Parameters Generally, if a linear homogeneous differential equation can be solved, the cor-responding nonhomogeneous equation can also be solved, at least in terms ofintegrals. 1. First-order equations Suppose that uc(t)is a solution of the homogeneous equation du dt=k(t)u. (5) Then to ﬁnd a particular solution of the nonhomogeneous equation du dt=k(t)u+f(t), (6) we assume that up(t)=v(t)uc(t). Substituting upin this form into the differ- ential equation (6) we have dv dtuc+vduc dt=k(t)vuc+f(t). (7) However, u/prime c=k(t)uc,s oo n et e r mo nt h el e f tc a n c e l sat e r mo nt h er i g h t , leaving dv dtuc=f(t), ordv dt=f(t) uc(t). (8) The latter is a nonhomogeneous equation of simplest type, which can be solved for v(t)in one integration. Example. Use this method to ﬁnd a solution of the nonhomogeneous equation du dt=5u+t. We should try the form up(t)=v(t)·e5t,b e c a u s e e5tis a solution of u/prime=5u. Substituting the preceding form for up,w eﬁ n d dv dt·e5t+v·5e5t=5ve5t+t, 0.2 Nonhomogeneous Linear Equations 21 or, after canceling 5 ve5tfrom both sides and simplifying, we ﬁnd dv dt=e−5tt. This equation is integrated once (by parts) to ﬁnd v(t)=/parenleftbigg −t 5−1 25/parenrightbigg e−5t. From here, we obtain up(t)=v(t)·e5t=−/parenleftbig1 5t+1 25/parenrightbig . /square 2. Second-order equations T o ﬁnd a particular solution of the nonhomogeneous second-order equation d2u dt2+k(t)du dt+p(t)u=f(t), (9) we need two independent solutions, u1(t)and u2(t),o ft h ec o r r e s p o n d i n gh o - mogeneous equation d2u dt2+k(t)du dt+p(t)u=0. (10) Then we assume that our particular solution has the form up(t)=v1(t)u1(t)+v2(t)u2(t), (11) where v1andv2are functions to be found. If we simply insert upin this form i n t oE q .( 9 ) ,w eo b t a i no n ec o m p l i c a t e ds e c o n d - o r d e re q u a t i o ni nt w ou n -known functions. However, if we impose the extra requirement that dv 1 dtu1+dv2 dtu2=0, (12) then we ﬁnd that u/prime p=v/prime 1u1+v/prime 2u2+v1u/prime 1+v2u/prime 2=v1u/prime 1+v2u/prime 2, (13) u/prime/prime p=v/prime 1u/prime1+v/prime 2u/prime2+v1u/prime/prime 1+v2u/prime/prime 2, (14) and the equation that results from substituting Eq. (11) into Eq. (9) becomes v/prime 1u/prime1+v/prime 2u/prime2+v1(u/prime/prime 1+k(t)u/prime 1+p(t)u1)+v2(u/prime/prime 2+k(t)u/prime 2+p(t)u2)=f(t). This simpliﬁes further: The multipliers of v1andv2are both 0, because u1and u2satisfy the homogeneous Eq. (10). 22 Chapter 0 Ordinary Differential Equations Thus, we are left with a pair of simultaneous equations, v/prime 1u1+v/prime 2u2=0, (12/prime) v/prime 1u/prime1+v/prime 2u/prime2=f(t), (15) in the unknowns v/prime 1andv/prime 2. The determinant of this system is /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleu 1u2 u/prime 1u/prime2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=W(t), (16) the Wronskian of u 1and u2. Since these were to be independent solutions of Eq. (10), their Wronskian is nonzero, and we may solve for v/prime 1(t)andv/prime 2(t)and hence for v1andv2. Example. Use variation of parameters to solve the nonhomogeneous equation d2u dt2+u=cos(ωt). Assume a solution in the form up(t)=v1cos(t)+v2sin(t), because sin (t)and cos (t)are independent solutions of the corresponding ho- mogeneous equation u/prime/prime+u=0. The assumption of Eq. (12) is v/prime 1cos(t)+v/prime 2sin(t)=0. (17) Then our equation reduces to the following, corresponding to Eq. (15): −v/prime 1sin(t)+v/prime 2cos(t)=cos(ωt). (18) Now we solve Eqs. (17) and (18) simultaneously to ﬁnd v/prime 1=− sin(t)cos(ωt), v/prime 2=cos(t)cos(ωt). (19) T h e s ee q u a t i o n sa r et ob ei n t e g r a t e dt oﬁ n d v1andv2,a n dt h e n up(t). /square Finally, we note that v1(t)andv2(t)can be found from Eqs. (12) and (15) in general: v/prime 1=−u2f W,v/prime 2=u1f W. (20) Integrating these two equations, we ﬁnd that v1(t)=−/integraldisplayu2(t)f(t) W(t)dt,v 2(t)=/integraldisplayu1(t)f(t) W(t)dt. (21) 0.2 Nonhomogeneous Linear Equations 23 Now, Eq. (11) may be used to form a particular solution of the nonhomoge- neous equation (9). We may also obtain v1andv2by using deﬁnite integrals with variable upper limit: v1(t)=−/integraldisplayt t0u2(z)f(z) W(z)dz,v 2(t)=/integraldisplayt t0u1(z)f(z) W(z)dz. (22) T h el o w e rl i m i ti su s u a l l yt h ei n i t i a lv a l u eo f t,b u tm a yb ea n yc o n v e n i e n t value. The particular solution can now be written as up(t)=− u1(t)/integraldisplayt t0u2(z)f(z) W(z)dz+u2(t)/integraldisplayt t0u1(z)f(z) W(z)dz. Furthermore, the factors u1(t)and u2(t)can be inside the integrals (which are notwith respect to t), and these can be combined to give a tidy formula, as follows. Theorem 3. Let u 1(t)and u 2(t)be independent solutions of d2u dt2+k(t)du dt+p(t)u=0( H ) with Wronskian W (t)=u1(t)u/prime 2(t)−u2(t)u/prime 1(t).T h e n up(t)=/integraldisplayt t0G(t,z)f(z)dz is a particular solution of the nonhomogeneous equation d2u dt2+k(t)du dt+p(t)u=f(t), (NH) where G is the Green’s function deﬁned by G(t,z)=u1(z)u2(t)−u2(z)u1(t) W(z). (23) /square EXERCISES In Exercises 1–10, ﬁnd the general solution of the differential equation. 24 Chapter 0 Ordinary Differential Equations 1.du dt+a(u−T)=0. 3.du dt+au=e−at. 5.d2u dt2+u=cos(t). 7.d2u dt2+3du dt+2u=cosh(t). 9.1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =− 1.2.du dt+au=eat. 4.d2u dt2+u=cos(ωt)( ω /negationslash=1). 6.d2u dx2−γ2(u−U)=0 (U,γ2are constants ). 8.1 rd dr/parenleftbigg rdu dr/parenrightbigg =− 1. 10.d2u dt2=− 1. 11.Leth(t)be the height of a parachutist above the surface of the earth. Con- sideration of forces on his body leads to the initial value problem for h: Md2h dt2+Kdh dt=− Mg, h(0)=h0,dh dt(0)=0 (M=mass, g=acceleration of gravity, K=parachute constant). Solve the problem, taking g=32 ft/s2and K/M=0.1/s. 12.Solve this initial value problem for forced vibrations, d2u dt2+ω2u=f0sin(µt), u(0)=0,du dt(0)=0, in two cases: (a) µ/negationslash=ω,( b )µ=ω. In Exercises 13–19, use variation of parameters to ﬁnd a particular solution ofthe differential equation. Be sure that th e differential equation is in the correct form. 13.du dt+au=e−at,uc(t)=e−at. 14.tdu dt=− 1, uc(t)=1. 15.d2y dx2+y=tan(x),y1(x)=cos(x),y2(x)=sin(x). 16.d2y dx2+y=sin(x),y1(x)=cos(x),y2(x)=sin(x). 0.2 Nonhomogeneous Linear Equations 25 17.d2u dt2=− 1, u1(t)=1, u2(t)=t. 18.1 rd dr/parenleftbigg rdu dr/parenrightbigg =− 1, u1(r)=1, u2(r)=ln(r). 19.t2d2u dt2+tdu dt−u=1, u1(t)=t,u2(t)=1 t. In Exercises 20–22, use Theorem 3 to develop the formula shown for a partic- ular solution of the differential equation. 20.d2u dt2+γ2u=f(t),up(t)=1 γ/integraldisplayt 0sin/parenleftbig γ(t−z)/parenrightbig f(z)dz. 21.du dt+au=f(t),up(t)=/integraldisplayt 0e−a(t−z)f(z)dz. 22.d2u dt2−γ2u=f(t),up(t)=1 γ/integraldisplayt 0sinh/parenleftbig γ(t−z)/parenrightbig f(z)dz. 23.In “Model for temperature estimation of electric couplings suffering heavy lightning currents” [A.D. Polykriti et al., IEE Proceedings — Generation, Transmission and Distribution ,151(2004): 90–94], the authors model the temperature rise above ambient in a coupling with this initial value prob- lem: ρcdT dt=i2(t)R(1+αT), T(0)=0. Parameters: ρis density, cis speciﬁc heat, i(t)is the current due to a light- ening strike, Ris the resistance of the coupling at ambient temperature, and the factor (1+αT)shows how resistance increases with temperature. Simplify the differential equ ation algebraically to get dT dt=Ki2(t)(β+T), T(0)=0, and identify βand Kin terms of the other parameters. 24.(Continuation) The authors model the lightning current with the func- tion i(t)=Imax(e−λt−e−µt)/n,w h e r e nis a factor to make Imaxthe ac- tual maximum. Obtain graphs of this function and the simpler function i(t)=Imaxe−λt, using these values: Imax=100 kA, n=0.93,λ=2.1, µ=150. The unit for time is milliseconds. Graph for tfrom 0 to 2 ms, which is the range of interest. 25.(Continuation) Solve the initial value problem using the simpler functionfor current. (Don’t forget to square.) Graph the result for tfrom 0 to 2 ms, usingβ=0.26 and K=13. 26 Chapter 0 Ordinary Differential Equations 0.3 Boundary Value Problems Aboundary value problem in one dimension is an ordinary differential equa- tion together with conditions involving values of the solution and/or its deriv-atives at two or more points. The number of conditions imposed is equal to the order of the differential equation. Usually, boundary value problems of any physical relevance have these characteristics: (1) The conditions are imposedat two different points; (2) the solution is of interest only between those twopoints; and (3) the independent variable is a space variable, which we shallrepresent as x. In addition, we are primarily concerned with cases where the differential equation is linear and of second order. However, problems in elas-ticity often involve fourth-order equations. In contrast to initial value problems, even the most innocent looking boundary value problem may have exactly one solution, no solution, or aninﬁnite number of solutions. Exercise 1 illustrates these cases. When the differential equation in a boundary value problem has a known general solution, we use the two boundary conditions to supply two equationsthat are to be satisﬁed by the two constants in the general solution. If the dif- ferential equation is linear, these are two linear equations and can be easily solved, if there is a solution. In the rest of this section we examine some physical examples that are natu- rally associated with boundary value problems. Example: Hanging Cable. First we consider the problem of ﬁnding the shape of a cable that is fastened at each end and carries a distributed load. The cables of a suspension bridge provide an important example. Let u(x)denote the position of the centerline of the cable, measured upward from the x-axis, which we assume to be hori- zontal. (See Fig. 3.) Our objective is to ﬁnd the function u(x). The shape of the cable is determined by the forces acting on it. In our analy- sis, we consider the forces that hold a small segment of the cable in place. (SeeFig. 4.) The key assumption is that the cable is perfectly ﬂexible. This means that force inside the cable is always a tension and that its direction at every point is the direction tangent to the centerline. Figure 3 The hanging cable. 0.3 Boundary Value Problems 27 We suppose that the cable is not moving. Then by Newton’s second law, the sum of the horizontal components of the forces on the segment is 0, andlikewise for the vertical components. If T(x)andT(x+/Delta1x)are the magnitudes of the tensions at the ends on the segment, we have these two equations: T(x+/Delta1x)cos/parenleftbig φ(x+/Delta1x)/parenrightbig −T(x)cos/parenleftbig φ(x)/parenrightbig =0 (Horizontal ),(1) T(x+/Delta1x)sin/parenleftbig φ(x+/Delta1x)/parenrightbig −T(x)sin/parenleftbig φ(x)/parenrightbig −f(x)/Delta1x=0(Ver t ical ).(2) In the second equation, f(x)is the intensity of the distributed load, measured in force per unit of horizontal length, so f(x)/Delta1xis the load borne by the small segment. From Eq. (1) we see that the horizontal component of the tension is the same at both ends of the segment. In fact, the horizontal component of tension hasthe same value — call it T— at every point, including the endpoints where the cable is attached to solid supports. By simple algebra we can now ﬁnd thetension in the cable at the ends of our segment, T(x+/Delta1x)=T cos/parenleftbig φ(x+/Delta1x)/parenrightbig,T(x)=T cos/parenleftbig φ(x)/parenrightbig, and substitute these into Eq. (2), which becomes T cos/parenleftbig φ(x+/Delta1x)/parenrightbigsin/parenleftbig φ(x+/Delta1x)/parenrightbig −T cos/parenleftbig φ(x)/parenrightbigsin/parenleftbig φ(x)/parenrightbig −f(x)/Delta1x=0 or T/parenleftbig tan/parenleftbig φ(x+/Delta1x)/parenrightbig −tan/parenleftbig φ(x)/parenrightbig/parenrightbig −f(x)/Delta1x=0. Before going further we should note (Fig. 4) that φ(x)measures the angle between the tangent to the centerline of the cable and the horizontal. As theposition of the centerline is given by u(x),t a n(φ(x))is just the slope of the cable at x. From elementary calculus we know tan/parenleftbig φ(x)/parenrightbig =du dx(x). Figure 4 Section of cable showing forces acting on it. The angles are α=φ(x), β=φ(x+/Delta1x). 28 Chapter 0 Ordinary Differential Equations Substituting the derivative for the slope and making some algebraic adjust- ments, we obtain T/parenleftbig u/prime(x+/Delta1x)−u/prime(x)/parenrightbig =f(x)/Delta1x. Dividing through by /Delta1xyields Tu/prime(x+/Delta1x)−u/prime(x) /Delta1x=f(x). In the limit, as /Delta1xapproaches 0, the difference quotient in the left member becomes the second derivative of u, and the result is the equation Td2u dx2=f(x), (3) which is valid for xin the range 0 <x<a, where the cable is located. In addi- tion, u(x)must satisfy the boundary conditions u(0)=h0,u(a)=h1. (4) For any particular case, we must choose an appropriate model for the load- ing, f(x). One possibility is that the cable is hanging under its own weight of wunits of weight per unit length of cable. Then in Eq. (2), we should put f(x)/Delta1x=w/Delta1s /Delta1x/Delta1x, where srepresents arc length along the cable. In the limit, as /Delta1xapproaches 0, /Delta1s//Delta1xhas the limit lim /Delta1x→0/Delta1s /Delta1x=/radicalBigg 1+/parenleftbiggdu dx/parenrightbigg2 . Therefore, with this assumption, the boundary value problem that determines the shape of the cable is d2u dx2=w T/radicalBigg 1+/parenleftbiggdu dx/parenrightbigg2 ,0<x<a, (5) u(0)=h0,u(a)=h1. (6) Notice that the differential equation is nonlinear. Nevertheless, we can ﬁnd its general solution in closed form and satisfy the boundary conditions by appro-priate choice of the arbitrary constants that appear. (See Exercises 4 and 5.) Another case arises when the cable supports a load uniformly distributed in the horizontal direction, as given by f(x)/Delta1x=w/Delta1x. 0.3 Boundary Value Problems 29 Figure 5 Cylinder of heat-conducting material. This is approximately true for a suspension bridge. The boundary value prob- lem to be solved is then d2u dx2=w T,0<x<a, u(0)=h0,u(a)=h1. (7) The general solution of the differential equation (7) can be found by the procedures of Sections 1 and 2. It is u(x)=/parenleftbiggw 2T/parenrightbigg x2+c1x+c2, where c1and c2are arbitrary. The two boundary conditions require u(0)=h0:c2=h0, u(a)=h1:/parenleftbiggw 2T/parenrightbigg a2+c1a+c2=h1. T h e s et w oa r es o l v e df o r c1and c2in terms of given parameters. The result, after some beautifying algebra, is u(x)=w 2T/parenleftbig x2−ax/parenrightbig +h1−h0 ax+h0. (8) Clearly, this function speciﬁes the cab le’s shape as part of a parabola opening upward. /square Example: Heat Conduction in a Rod. A long rod of uniform material and cross section conducts heat along its axial direction (see Fig. 5). We assume that the temperature in the rod, u(x),d o e s not change in time. A heat balance (“what goes in must come out”) applied toa slice of the rod between xand x+/Delta1x(Fig. 6) shows that the heat ﬂow rate q, measured in units of heat per unit time per unit area, obeys the equation q(x)A+g(x)A/Delta1x=q(x+/Delta1x)A, (9) 30 Chapter 0 Ordinary Differential Equations Figure 6 Section cut from heat-conducting cylinder showing heat ﬂow. in which Ais the cross-sectional area and gis the rate at which heat enters the slice by means other than conduction through the two faces. For instance, if heat is generated in the slice by an electric current I, we might have g(x)A/Delta1x=I2R/Delta1x, (10) where Ris the resistance of the rod per unit length. If heat is lost through the cylindrical surface of the rod by convection to a surrounding medium attemperature T,t h e n g(x)would be given by “Newton’s law of cooling,” g(x)A/Delta1x=− h(u(x)−T)C/Delta1x, (11) where Cis the circumference of the rod and his the heat transfer coefﬁcient. (This minus sign appears because, if u(x)>T, heat actually leaves the rod.) Equation (9) may be altered algebraically to read q(x+/Delta1x)−q(x) /Delta1x=g(x), and application of the limiting process leaves dq dx=g(x). (12) The unknown function u(x)does not appear in Eq. (12). However, a well- known experimental law (Fourier’s law) says that the heat ﬂow rate through aunit area of material is directly proportional to the temperature difference andinversely proportional to thickness. In the limit, this law takes the form q=−κdu dx. (13) The minus sign expresses the fact that heat moves from hotter toward cooler regions. Combining Eqs. (12) and (13) gives the differential equation −κd2u dx2=g(x), 0<x<a, (14) where ais the length of the rod and the conductivity κis assumed to be con- stant. 0.3 Boundary Value Problems 31 If the two ends of the rod are held at constant temperature, the boundary conditions on uwould be u(0)=T0,u(a)=T1. (15) On the other hand, if heat were supplied at x=0 (by a heating coil, for in- stance), the boundary condition there would be −κAdu dx(0)=H, (16) where His measured in units of heat per unit time. Example. Solve the problem −κd2u dx2=− hu(x)C A,0<x<a, (17) u(0)=T0,u(a)=T0. (18) (Physically, the rod is losing heat to a surrounding medium at temperature 0, w h i l eb o t he n d sa r eh e l da tt h es a m et e m p e r a t u r e T0.) If we designate µ2= hC/κA, the differential equation becomes d2u dx2−µ2u=0,0<x<a, with general solution u(x)=c1cosh(µx)+c2sinh(µx). Application of the boundary condition at x=0g i v e s c1=T0; the second boundary condition requires that u(a)=T0:T0=T0cosh(µa)+c2sinh(µa). Thus c2=T0(1−cosh(µa))/sinh(µa)and u(x)=T0/parenleftbigg cosh(µx)+1−cosh(µa) sinh(µa)sinh(µx)/parenrightbigg . /square It should be clear now that solving a boundary value problem is not sub- stantially different from solving an initial value problem. The procedure is (1)ﬁnd the general solution of the differential equation, which must contain somearbitrary constants, and (2) apply the boundary conditions to determine val-ues for the arbitrary constants. In our examples the differential equations have 32 Chapter 0 Ordinary Differential Equations Figure 7 Column carrying load P. Figure 8 Section of column showing forces and moments. been of second order, causing the appearance of two arbitrary constants, which are to be determined by the boundary conditions. The next example is somewhat different in spirit from the others. Instead of just ﬁnding the solution of a boundary value problem, we will be looking forparameter values that permit the existence of solutions of special form. Example: Buckling of a Column. A long, slender column whose bottom end is hinged carries an axial load as shown in Fig. 7. The upper end of the column can move up or down but not sideways. The displacement of the column’s centerline from a vertical referenceline is given by u(x).I ft h ec o l u m nw e r ec u ta ta n yp o i n t x, an upward force P and a clockwise moment Pu(x)w o u l dh a v et ob ea p p l i e dt ot h eu p p e rp a r tt o keep it in equilibrium (see Fig. 8). This force and moment must be suppliedby the lower part of the column. 0.3 Boundary Value Problems 33 It is known that the internal bending moment (positive when counterclock- wise) in a column is given by the product EId2u dx2, where Eis Y oung’s modulus and Iis the moment of inertia of the cross- sectional area. (The moment I=b4/12 for a column whose cross section is as q u a r eo fs i d e b.) Thus equating the external moment to the internal mo- ment gives the differential equation EId2u dx2=− Pu,0<x<a, (19) which, together with the boundary conditions u(0)=0,u(a)=0, (20) determines the function u(x). In order to study this problem more conveniently, we set P EI=λ2 so that the differential equation becomes d2u dx2+λ2u=0,0<x<a. (21) Now, the general solution of this differential equation is u(x)=c1cos(λx)+c2sin(λx). Asu(0)=0, we must choose c1=0, leaving u(x)=c2sin(λx). The second boundary condition requires that u(a)=0:c2sin(λa)=0. If sin(λa)is not 0, the only possibility is that c2=0. In this case we ﬁnd that the solution is u(x)≡0,0<x<a. Physically, this means that the column stands straight and transmits the load to its support, as it was probably intended to do. Something quite different happens if sin (λa)=0, for then any choice of c2gives a solution. The physical manifestation of this case is that the column assumes a sinusoidal shape and may then collapse, or buckle, under the axialload. Mathematically, the condition sin (λa)=0m e a n st h a t λais an integer 34 Chapter 0 Ordinary Differential Equations multiple of π,s i n c es i n (π)=0, sin(2π)=0, etc., and integer multiples of π are the only arguments for which the sine function is 0. The equation λa=π, in terms of the original parameters, is /radicalbigg P EIa=π. It is reasonable to think of E,I,a n d aas given quantities; thus it is the force P=EI/parenleftbiggπ a/parenrightbigg2 , called the critical or Euler load, that causes the buckling. The higher critical loads, corresponding to λa=2π,λa=3π, etc., are so unstable as to be of no physical interest in this problem. /square The buckling example is one instance of an eigenvalue problem. The gen- eral setting is a homogeneous different ial equation containing a parameter λ and accompanied by homogeneous boundary conditions. Because both dif- ferential equations and boundary condi tions are homogeneous, the constant function 0 is always a solution. The question to be answered is: What valuesof the parameter λallow the existence of nonzero solutions? Eigenvalue prob- lems often are employed to ﬁnd the dividing line between stable and unstablebehavior. We will see them frequently in later chapters. EXERCISES 1.Of these three boundary value problems, one has no solution, one hasexactly one solution, and one has an inﬁnite number of solutions. Whichis which? a.d2u dx2+u=0, u(0)=0, u(π)=0; b.d2u dx2+u=1, u(0)=0, u(1)=0; c.d2u dx2+u=0, u(0)=0, u(π)=1. 2.Find the Euler buckling load of a steel column with a 2 in. ×3i n .r e c t a n - gular cross section. The parameters are E=30×106lb/in.2,I=2i n .4, a=10 ft. 3.Find all values of the parameter λfor which these homogeneous boundary value problems have a solution other than u(x)≡0. a.d2u dx2+λ2u=0, u(0)=0,du dx(a)=0; 0.3 Boundary Value Problems 35 b.d2u dx2+λ2u=0,du dx(0)=0, u(a)=0; c.d2u dx2+λ2u=0,du dx(0)=0,du dx(a)=0. 4.Verify, by differentiating and substituting, that u(x)=c/prime+1 µcosh/parenleftbig µ(x+c)/parenrightbig is the general solution of the differential equation (5). (Here µ=w/T. The graph of u(x)is called a catenary .) 5.Find the values of candc/primefor which the function u(x)in Exercise 4 satisﬁes the conditions u(0)=h,u(a)=h. 6.A beam that is simply supported at its ends carries a distributed lateral load of uniform intensity w(force/length) and an axial tension load T (force). The displacement u(x)of its centerline (positive down) satisﬁes the boundary value problem here. Find u(x). d2u dx2−T EIu=−w EILx−x2 2,0<x<L, u(0)=0,u(L)=0. 7.The temperature u(x)in a cooling ﬁn satisﬁes the differential equation d2u dx2=hC κA(u−T), 0<x<a, and boundary conditions u(0)=T0,−κdu dx(a)=h/parenleftbig u(a)−T/parenrightbig . That is, the temperature at the left end is held at T0>Twhile the surface of the rod and its right end exchange heat with a surrounding medium at temperature T.F i n d u(x). 8.Calculate the limit as atends to inﬁnity of u(x), the solution of the prob- lem in Exercise 7. Is the result physically reasonable? 9.In an electrical heating element, the temperature u(x)satisﬁes the bound- ary value problem that follows. Find u(x). d2u dx2=hC κA(u−T)−I2R κA,0<x<a, u(0)=T,u(a)=T. 36 Chapter 0 Ordinary Differential Equations Figure 9 Poiseuille ﬂow. 10.Verify that the solution of the problem given in Eqs. (17) and (18) can also be written as follows, with µ2=Ch/Aκ: u(x)=T0cosh(µ(x−a/2)) cosh(µa/2). 11.(Poiseuille ﬂow) A viscous ﬂuid ﬂows steadily between two large paral- lel plates so that its velocity is parallel to the x-axis. (See Fig. 9.) The x-component of velocity of the ﬂuid at any point (x,y)is a function of yonly. It can be shown that this component u(x)satisﬁes the differential equation d2u dy2=−g µ,0<y<L, where µis the viscosity and −gis a constant, negative pressure gradi- ent. Find u(y), subject to the “no-slip” boundary conditions, u(0)=0, u(L)=0. 12.If the beam mentioned in Exercise 6 is subjected to axial compression in- stead of tension, the boundary value problem for u(x)becomes the one here. Solve for u(x). d2u dx2+P EIu=−w EILx−x2 2,0<x<L, u(0)=0,u(L)=0. 13.For what value(s) of the compressive load Pin Exercise 12 does the prob- lem have no solution or inﬁnitely many solutions? 14.The pressure p(x)in the lubricant under a plane pad bearing satisﬁes the problem d dx/parenleftbigg x3dp dx/parenrightbigg =− K,a<x<b, p(a)=0,p(b)=0. 0.3 Boundary Value Problems 37 Find p(x)in terms of a,b,a n d K(constant). Hint: The differential equa- tion can be solved by integration. 15.In a nuclear fuel rod, nuclear reaction constantly generates heat. If we treata rod as a one-dimensional object, the temperature u(x)in the rod might satisfy the boundary value problem d 2u dx2+g κ=hC κA(u−T), 0<x<a, u(0)=T,u(a)=T. Here, gis the heat generation rate or power density, and the terms on the right-hand side represent heat transfer by convection to a surrounding medium, usually pressurized water. Find u(x). 16.Sketch the solution of Exercise 15 and determine the maximum temper- ature encountered. Typical values for the parameters are g=300 W/cm3, T=325◦C,κ=0.01 cal/cm s◦C,a=2.9m , C/A=4/cm, h= 0.035 cal/cm2s◦C. It will be useful to know that 1 W =0.239 cal/s. 17.An assembly of nuclear fuel rods is housed in a pressure vessel shaped roughly like a cylinder with ﬂat or hemispherical ends. The temperature in the thick steel wall of the vessel affects its strength and thus must be studied for design and safety. Treating the vessel as a long cylinder (thatis, ignoring the effects of the ends), it is easy to derive this differentialequation in cylindrical coordinates for the temperature u(r)in the wall: 1 rd dr(rd u) dr=0,a<r<b, where aand bare the inner and outer radii, respectively. The boundary conditions both involve convection, with hot pressurized water at the in- ner radius and with air at the outer radius: −κu/prime(a)=h0/parenleftbig Tw−u(a)/parenrightbig , κu/prime(b)=h1/parenleftbig Ta−u(b)/parenrightbig . Find u(r)in terms of the parameters, carefully checking the dimensions. 18.If a beam of uniform cross section is simply supported at its ends and carries a distributed load w(x)along its length, then the displacement u(x) of its centerline satisﬁes the boundary value problem d4u dx4=w(x) EI,0<x<a, u(0)=0,u/prime/prime(0)=0,u(a)=0,u/prime/prime(a)=0. 38 Chapter 0 Ordinary Differential Equations (Here, Eis Y oung’s modulus and Iis the second moment of the cross sec- tion.) Solve this problem if w(x)=w0, constant. 19.If the beam of Exercise 18 is built into a wall at the left end and is unsup- ported at the right end, the boundary conditions become u(0)=0,u/prime(0)=0,u/prime/prime(a)=0,u/prime/prime/prime(a)=0. Solve the same differential equation subject to these conditions. 0.4 Singular Boundary Value Problems A boundary value problem can be singular in two different ways. In one case, an endpoint of the interval of interest is a singular point of the differentialequation. In the other, the interval is inﬁnitely long. Regular Singular Point Recall that a point x0is a (regular) singular point of the differential equation u/prime/prime+k(x)u/prime+p(x)u=f(x) if the products (x−x0)k(x), ( x−x0)2p(x) both have Taylor series expansions centered at x0but either k(x)orp(x)or both become inﬁnite as x→x0. For example, the point x0=1 is a regular singular point of the differential equation (1−x)u/prime/prime+u/prime+xu=0. In standard form, the equation is u/prime/prime+1 1−xu/prime+x 1−xu=0. Since both k(x)=1 1−xand p(x)=x 1−x become inﬁnite at x=1, but (x−1)k(x)and(x−1)2p(x)both have Taylor series expansions about the center x=1, the point x0=1 is a regular singular point. Another convenient example is provided by the Cauchy–Euler equationof Section 1, which has a regular singular point at the origin. This situation typically arises when a boundary point is a mathematical boundary without being a physical boundary. For instance, a circular disk of 0.4 Singular Boundary Value Problems 39 radius cmay be described in polar (r,θ)coordinates as occupying the region 0≤r≤c. The origin, at r=0, is a mathematical boundary, yet physically this point is in the interior of the disk. At a singular point, one cannot specify a value for u(x0), the solution of the differential equation, or for its deri vative. However, it is usually necessary to require that both u(x0)and u/prime(x0)be ﬁnite, or bounded. Tacitly, we always require that the solution and its derivative be ﬁnite at every point of the interval where we are solving a differential equation. But when a singular point is aboundary point of that interval, we enforce the condition explicitly. In theexample that follows we shall see how these conditions act so as to make thesolution of a boundary value problem unique. Example: Radial Heat Flow. Suppose a long cylindrical bar, surrounded by a medium at temperature T, carries an electrical current. If heat ﬂows in the radial direction much faster than in the axial direction, the temperature u(r)in the rod may be described by the problem 1 rd dr/parenleftbigg rdu dr/parenrightbigg =− H,0≤r<c, (1) u(c)=T. (2) Here, cis the radius of the rod, ris a polar coordinate, and H(constant) is proportional to the electrical power being converted into heat. In this problem, only the physical boundary condition has been noted. The mathematical boundary r=0 is a singular point, as is clear from the differen- tial equation in the form d2u dr2+1 rdu dr=− H. Thus, at this point we will require that uand du/drbe ﬁnite: u(0), u/prime(0)ﬁnite. (3) Now the differential equation (1) is easy to solve. Multiply through by rand integrate once to ﬁnd that rdu dr=− Hr2 2+c1. Divide through this equation by rand integrate once more to determine that u(r)=− Hr2 4+c1ln(r)+c2. 40 Chapter 0 Ordinary Differential Equations Application of the special condition, that u(0)and u/prime(0)be ﬁnite, immediately t e l l su st h a t c1=0; for both, ln (r)and its derivative 1 /rbecome inﬁnite as r approaches 0. The physical boundary condition, Eq. (2), says that u(c)=− Hc2 4+c2=T. Hence, c2=Hc2/4+T, and the complete solution is u(r)=H(c2−r2) 4+T. (4) /square From this example, it is clear that the “artiﬁcial” boundary condition, boundedness of u(r)at the singular point r=0, works just the way an or- dinary boundary condition works at an ordinary (not singular) point. It givesone condition to be fulﬁlled by the unknown constants c 1and c2,w h i c ha r e then completely determined by the second boundary condition. Semi-Inﬁnite and Inﬁnite Intervals Another type of singular boundary value problem is one for which the inter-val of interest is inﬁnite. (Of course, this is always a mathematical abstractionthat cannot be realized physically.) For instance, on the interval 0 <x<∞, sometimes called a semi-inﬁnite interval , as it does have one ﬁnite endpoint, a boundary condition would normally be imposed at x=0. At the other “end,” no boundary condition is imposed, because no boundary exists. However, we normally require that both u(x)and u /prime(x)remain bounded as xincreases. In precise terms, we require that there exist constants Mand M/primefor which /vextendsingle/vextendsingleu(x)/vextendsingle/vextendsingle≤M and/vextendsingle/vextendsingleu/prime(x)/vextendsingle/vextendsingle≤M/prime are both satisﬁed for all x, no matter how large. We never identify MorM/prime, and the entire condition is usually written u(x)and u/prime(x)bounded as x→∞. Example: Cooling Fin. A long cooling ﬁn has one end held at a constant temperature T0and ex- c h a n g e sh e a tw i t ham e d i u ma tt e m p e r a t u r e Tthrough convection. The tem- perature u(x)in the ﬁn satisﬁes the requirements d2u dx2=hC κA(u−T), 0<x, (5) u(0)=T0 (6) 0.4 Singular Boundary Value Problems 41 (see Section 3). As the problem has been posed for a semi-inﬁnite interval (because the ﬁn is very long and, perhaps, to mask our ignorance of what ishappening at the other physical end), we must also impose the condition u(x), u /prime(x)bounded as x→∞. (7) Now, the general solution of the differential equation (5) is u(x)=T+c1cosh(µx)+c2sinh(µx), where µ=/radicalbig hC/κA. The boundary condition at x=0r e q u i r e st h a t u(0)=T0:T+c1=T0. The boundedness condition, Eq. (7), requires that c2=− c1. The reason for this is that of all the linear combinations of cosh and sinh, the only one that is bounded as x→∞ is cosh(µx)−sinh(µx)=e−µx, and its constant multiples. The ﬁnal solution is easily found to be u(x)=T+(T0−T)/parenleftbig cosh(µx)−sinh(µx)/parenrightbig . /square Satisfying the boundedness condition in the example would have been sim- pler if we had expressed the general solution of the differential equation (5) as u(x)=T+c/prime 1eµx+c/prime 2e−µx. We would have seen immediately that choosing c/prime 1=0 is the only way to satisfy the boundedness condition. We summarize the observation as a rule of thumb : the solution of d2u dx2−µ2u=0 on an interval Iis best expressed as u(x)=/braceleftbiggc1cosh(µx)+c2sinh(µx),if Iis ﬁnite, c1eµx+c2e−µx, if Iis inﬁnite. EXERCISES 1.Put each of the following equations in the form u/prime/prime+ku/prime+pu=f 42 Chapter 0 Ordinary Differential Equations and identify the singular point(s). a.1 rd dr/parenleftbigg rdu dr/parenrightbigg =u; c.d dφ/parenleftbigg sin(φ)du dφ/parenrightbigg =sin(φ)u;b.d dx/parenleftbigg/parenleftbig 1−x2/parenrightbigdu dx/parenrightbigg =0; d.1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =−λ2u. 2.The temperature ui nal a r g eo b j e c th a v i n gah o l eo fr a d i u s cin the middle may be said to obey the equations 1 rd dr/parenleftbigg rdu dr/parenrightbigg =0,r>c, u(c)=T. Solve the problem, adding the appropriate boundedness condition. 3.Compact kryptonite produces heat at a rate of Hcal/s cm3.I fas p h e r e (radius c) of this material transfers heat by convection to a surrounding medium at temperature T,t h et e m p e r a t u r e u(ρ)in the sphere satisﬁes the boundary value problem 1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =−H κ,0<ρ< c, −κdu dρ(c)=h/parenleftbig u(c)−T/parenrightbig . Supply the proper boundedness condition and solve. What is the tempera- ture at the center of the sphere? 4.(Critical radius) The neutron ﬂux uin a sphere of uranium obeys the dif- ferential equation λ 31 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg +(k−1)Au=0 in the range 0 <ρ< a,w h e r e λis the effective distance traveled by a neu- tron between collisions, Ais called the absorption cross section, and kis the number of neutrons produced by a collision during ﬁssion. In addition, the neutron ﬂux at the boundary of the sphere is 0. Make the substitutionu=v/ρ and 3(k−1)A/λ=µ 2, and determine the differential equation satisﬁed by v(ρ) . See Section 0.1, Exercise 19. 5.Solve the equation found in Exercise 4 and then ﬁnd u(ρ) that satisﬁes the boundary value problem (with boundedness condition) stated in Ex-ercise 4. For what radius ais the solution not identically 0? 0.5 Green’s Functions 43 6.Inside a nuclear fuel rod, heat is constantly produced by nuclear reaction. At y p i c a lr o di sa b o u t3ml o n ga n da b o u t1c mi nd i a m e t e r ,s ot e m p e r a t u r ev a r i a t i o na l o n gt h el e n g t hi sm u c hl e s st h a na l o n gar a d i u s .T h u s ,w et r e a tthe temperature in such a rod as a function of the radial variable alone. Find this temperature u(r), which is the solution of the boundary value problem 1 rd dr/parenleftbigg rdu dr/parenrightbigg =−g κ,0<r<a, u(a)=T0. 7.For the problem of Exercise 6, ﬁnd the temperature at the center of the rod, u(0), using these values for the parameters: a=0.5c m ,t h ep o w e r density g=418 W/cm3=100 cal/s cm3,c o n d u c t i v i t y κ=0.01 cal/s cm◦C, and the surface temperature T0=325◦C. 8.A model for microwave heating of food uses this equation for the tempera- ture u(x)in a large solid object: d2u dx2=− Ae−x/L,0<x. Here, Ais a constant representing the strength of the radiation and prop- erties of the object, and Lis a characteristic length, known as penetration depth, that depends on frequency of the radiation and properties of the ob- ject. (Typically, Lis about 12 cm in frozen raw beef or 2 cm thawed.) Show that the boundary condition u/prime(0)=0 is incompatible with the condition that u(x)be bounded as xgoes to inﬁnity. [See C.J. Coleman, The mi- crowave heating of frozen substances, Applied Math. Modeling ,14(1990): 439–443.] 9.Solve the differential equation in Exercise 8 subject to the conditions u(0)=T0,u(x)bounded . 0.5 Green’s Functions The most important features of the solution of the boundary value problem,1 d2u dx2+k(x)du dx+p(x)u=f(x), l<x<r, (1) αu(l)−α/primeu/prime(l)=0, (2) βu(r)+β/primeu/prime(r)=0, (3) 1The primes on the constants α/prime,β/primeare not to indicate differentiation, of course, but to show that they are coefﬁcients of derivatives. 44 Chapter 0 Ordinary Differential Equations can be developed by using the variation-of-parameters solution of the differ- ential equation (1), as presented in Section 2. T o begin, we need to have twoindependent solutions of the homogeneous equation d 2u dx2+k(x)du dx+p(x)u=0,l<x<r. (4) Let us designate these two solutions as u1(x)and u2(x). It will simplify algebra later if we require that u1satisfy the boundary condition at x=land u2the condition at x=r; αu1(l)−α/primeu/prime 1(l)=0, (5) βu2(r)+β/primeu/prime 2(r)=0. (6) According to Theorem 3 of Section 2, the general solution of the differential equation (1) can be written as u(x)=c1u1(x)+c2u2(x)+/integraldisplayx l/parenleftbig u1(z)u2(x)−u2(z)u1(x)/parenrightbigf(z) W(z)dz. (7) Recall that in the denominator of the integrand, we have the Wronskian of u1 and u2, W(z)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleu 1(z)u2(z) u/prime 1(z)u/prime 2(z)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, (8) which is nonzero because u 1and u2are independent. We will need to know the following derivative of the function in Eq. (7): du dx=c1u/prime 1(x)+c2u/prime 2(x)+/integraldisplayx l/parenleftbig u1(z)u/prime 2(x)−u2(z)u/prime 1(x)/parenrightbigf(z) W(z)dz. (See Leibniz’s rule in the Appendix.) Now let us apply the boundary conditio n ,E q .( 2 ) ,t ot h eg e n e r a ls o l u t i o n u(x).F i r s t ,a t x=lwe have αu(l)−α/primeu/prime(l)=c1/parenleftbig αu1(l)−α/primeu/prime 1(l)/parenrightbig +c2/parenleftbig αu2(l)−α/primeu/prime 2(l)/parenrightbig =0. (9) Note that the integrals in uand u/primeare both 0 at x=l. Because of the boundary condition (5) imposed on u1,E q .( 9 )r e d u c e st o c2/parenleftbig αu2(l)−α/primeu/prime 2(l)/parenrightbig =0, (10) and we conclude that c2=0. 0.5 Green’s Functions 45 Second, the boundary condition at x=rbecomes βu(r)+β/primeu/prime(r)=c1/parenleftbig βu1(r)+β/primeu/prime(r)/parenrightbig +/integraldisplayr l/bracketleftbig u1(z)/parenleftbig βu2(r)+β/primeu/prime 2(r)/parenrightbig −u2(z)/parenleftbig βu1(r)+β/primeu/prime 1(r)/parenrightbig/bracketrightbigf(z) W(z)dz=0. (11) Now, the boundary condition (6) on u2atx=reliminates one term of the integrand, leaving c1/parenleftbig βu1(r)+β/primeu/prime 1(r)/parenrightbig −/integraldisplayr lu2(z)/parenleftbig βu1(r)+β/primeu/prime 1(r)/parenrightbigf(z) W(z)dz=0. (12) The common factor of βu1(r)+β/primeu/prime 1(r)can be canceled from both terms, and we then ﬁnd c1=/integraldisplayr lu2(z)f(z) W(z)dz. (13) Now we have found c1and c2so that u(x)in Eq. (7) satisﬁes both boundary conditions. If we use the values of c1and c2as found, we have u(x)=u1(x)/integraldisplayr lu2(z)f(z) W(z)dz +/integraldisplayx l/parenleftbig u1(z)u2(x)−u2(z)u1(x)/parenrightbigf(z) W(z)dz. (14) The solution becomes more compact if we break the interval of integration at xin the ﬁrst integral, making it /integraldisplayr lu2(z)f(z) W(z)dz=/integraldisplayx lu2(z)f(z) W(z)dz+/integraldisplayr xu2(z)f(z) W(z)dz. (15) When the integrals on the range ltoxare combined, there is some cancella- tion, and our solution becomes u(x)=/integraldisplayx lu1(z)u2(x)f(z) W(z)dz+/integraldisplayr xu1(x)u2(z)f(z) W(z)dz. (16) Finally, these two integrals can be combined into one. We ﬁrst deﬁne the Green’s function for the problem (1), (2), (3) as G(x,z)=  u1(z)u2(x) W(z),l<z≤x, u1(x)u2(z) W(z),x≤z<r.(17) 46 Chapter 0 Ordinary Differential Equations Then the formula given in Eq. (16) for usimpliﬁes to u(x)=/integraldisplayr lG(x,z)f(z)dz. (18) Example. Solve the problem that follows by constructing the Green’s function. d2u dx2−u=− 1,0<x<1, u(0)=0,u(1)=0. First, we must ﬁnd two independent solutions of the homogeneous differential equation u/prime/prime−u=0 that satisfy the boundary conditions as required. The g e n e r a ls o l u t i o no ft h eh o m o g e n e ous differential equation is u(x)=c1cosh(x)+c2sinh(x). Asu1(x)is required to satisfy the condition at the left, u1(0)=0, we take c1=0,c2=1 and conclude u1(x)=sinh(x). The second solution is to sat- isfyu2(1)=0. We may take u2(x)=sinh(1)cosh(x)−cosh(1)sinh(x)=sinh(1−x). The Wronskian of the two solutions is W(x)=/vextendsingle/vextendsingle/vextendsingle/vextendsinglesinh(x) sinh(1−x) cosh(x)−cosh(1−x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=− sinh(1). Now, by Eq. (17), the Green’s function for this problem is G(x,z)=  sinh(z)sinh(1−x) −sinh(1),0<z≤x, sinh(x)sinh(1−z) −sinh(1),x≤z<1. Furthermore, since f(x)=− 1, the solution, by Eq. (18), is the integral u(x)=/integraldisplay1 0−G(x,z)dz. T o actually carry out the integration, we must break the interval of integration atx,t h u sr e v e r t i n gi ne f f e c tt oE q .( 1 6 ) .T h er e s u l t : 0.5 Green’s Functions 47 u(x)=/integraldisplayx 0sinh(z)sinh(1−x) sinh(1)dz+/integraldisplay1 xsinh(x)sinh(1−z) sinh(1)dz =sinh(1−x) sinh(1)cosh(z)/vextendsingle/vextendsingle/vextendsingle/vextendsinglex 0+sinh(x) sinh(1)/parenleftbig −cosh(1−z)/parenrightbig/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 x =sinh(1−x) sinh(1)/parenleftbig cosh(x)−1/parenrightbig +sinh(x) sinh(1)/parenleftbig cosh(1−x)−1/parenrightbig =sinh(1−x)cosh(x)+sinh(x)cosh(1−x) sinh(1)−sinh(1−x)+sinh(x) sinh(1) =1−sinh(1−x)+sinh(x) sinh(1). This, ﬁnally, is easily seen to be the correct solution. In this instance, there are much quicker ways to arrive at the same result. The advantage of theGreen’s function is that it shows how the solution of the problem dependson the inhomogeneity f(x). It is an efﬁcient way to obtain the solution in some cases. /square Now let us look back over the calculations and see if there is some place they might fail. Aside from the possibility that the coefﬁcients k(x)orp(x)in the differential equation might not be continuous, it seems that division by 0 isthe only possibility of failure. Quantities canceled or divided by were W(x)=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleu 1(x)u2(x) u/prime 1(x)u/prime 2(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, αu 2(l)−α/primeu/prime 2(l), βu1(r)+β/primeu/prime 1(r) in Eqs. (7), (10), and (12), respectively. It can be shown that all three of these are 0 if any one of them is 0, and, in that case, u1(x)andu2(x)are proportional. We summarize in a theorem. Theorem. Let k(x),p(x),a n df (x)be continuous, l ≤x≤r. The boundary value problem d2u dx2+k(x)du dx+p(x)u=f(x), l<x<r, αu(l)−α/primeu/prime(l)=0, (i) βu(r)+β/primeu/prime(r)=0, (ii) 48 Chapter 0 Ordinary Differential Equations has one and only one solution, unless there is a nontrivial solution of d2u dx2+k(x)du dx+p(x)u=0,l<x<r, that satisﬁes (i)and (ii). When a unique solution exists, it is given by Eqs. (17) and (18). /square Example. The boundary value problem d2u dx2+u=− 1,0<x<π , u(0)=0,u(π)=0, does not have a unique solution, according to the theorem, because u(x)= sin(x)is a nontrivial solution of the problem d2u dx2+u=0,0<x<π , u(0)=0,u(π)=0. Indeed, if we try to follow through the construction, we ﬁnd that u1(x)= sin(x)and also u2(x)=sin(x)(or a multiple thereof), and so all three quanti- ties in Eq. (19) are 0. On the other hand, suppose we try to obtain a solution by the usual method. The general solution of the differential equation is u(x)=− 1+c1cos(x)+c2sin(x). However, application of the boundary conditions leads to the contradictory requirements −1+c1=0a n d −1−c1=0. Thus, in this case, there simply is no solution to the problem stated. /square If the differential equation (1) has a singular point at x=lorx=r(or both), a Green’s function may still be constructed. The boundary condition (2)or (3) would be replaced by a boundedness condition, which would also apply tou 1oru2as the case may be. Example. Construct Green’s function for the problem 1 xd dx/parenleftbigg xdu dx/parenrightbigg =f(x), 0<x<1, u(0)bounded ,u(1)=0. 0.5 Green’s Functions 49 The general solution of the corresponding homogeneous equation is u(x)= c1+c2ln(x).T h u s ,w ew o u l dc h o o s e u1(x)=1,u2(x)=ln(x) so that u1(x)is bounded at x=0a n d u2(x)is 0 at x=1. The Green’s function is thus G(x,z)=/braceleftbiggzln(x),0<z≤x, zln(z), x≤z<1. A similar procedure is followed if the interval l<x<ris inﬁnite in length. /square EXERCISES In Exercises 1–8, ﬁnd the Green’s function for the problem stated. 1.d2u dx2=f(x),0<x<a, u(0)=0, u(a)=0. 2.d2u dx2=f(x),0<x<a, u(0)=0,du dx(a)=0. 3.d2u dx2−γ2u=f(x),0<x<a, du dx(0)=0, u(a)=0. 4.1 rd dr/parenleftbigg rdu dr/parenrightbigg =f(r),0≤r<c, u(c)=0, u(r)bounded at r=0. 5.1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =f(ρ),0≤ρ< c, u(c)=0, u(ρ)bounded at ρ=0. 6.d2u dx2+1 xdu dx−1 4x2u=f(x),0≤x<a, u(a)=0, u(x)bounded at x=0. 7.d2u dx2−γ2u=f(x),0<x, u(0)=0, u(x)bounded as x→∞ . 50 Chapter 0 Ordinary Differential Equations 8.d2u dx2−γ2u=f(x),−∞<x<∞, u(x)bounded as x→± ∞ . 9.Use the Green’s function of Exercise 5 to solve the problem 1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =1,0≤ρ< c, u(c)=0, and compare with the solution found by integrating the equation directly. 10.Use the Green’s function of Exercise 8 to solve the problem d2u dx2−γ2u=−γ2,−∞<x<∞, u(x)bounded as x→± ∞ , and compare with the result found directly. 11.Use the Green’s function of Exercise 1 to solve the problem stated there, if f(x)=/braceleftbigg0,0<x<a/2, 1,a/2<x<a. 12.In conﬁrmation of the theorem, show that the homogeneous problem a has a nontrivial solution; problem bh a sn os o l u t i o n( e x i s t e n c ef a i l s ) ;a n d problem chas inﬁnitely many solutions (uniqueness fails). a.u/prime/prime+u=0, u(0)=0, u(π)=0, b.u/prime/prime+u=− 1, u(0)=0, u(π)=0, c.u/prime/prime+u=π−2x,u(0)=0, u(π)=0. 13.Considering zto be a parameter (l<z<r), deﬁne the function v(x)= G(x,z)with Ga si nE q .( 1 7 ) .S h o wt h a t vhas these four properties, which a r es o m e t i m e su s e dt od e ﬁ n et h eG r e e n ’ sf u n c t i o n . (i)vsatisﬁes the boundary conditions, Eqs. (2) and (3), at x=land r. (ii)vis continuous, l<x<r.( T h ep o i n t x=zneeds to be checked.) (iii)v/primeis discontinuous at x=z,a n d lim h→0+/parenleftbig v/prime(z+h)−v/prime(z−h)/parenrightbig =1. (iv)vsatisﬁes the differential equation v/prime/prime+k(x)v/prime+p(x)v=0 for l< x<zand z<x<r. Miscellaneous Exercises 51 14.Show that the boundary value problem d2u dx2+λ2u=f(x), 0<x<a, u(0)=0,u(a)=0, will have no solution or inﬁnitely many solutions if λis an eigenvalue of d2u dx2+λ2u=0, u(0)=0,u(a)=0. Chapter Review See the CD for review questions. Miscellaneous Exercises In Exercises 1–15, solve the given boundary value problem, supplying bound- edness conditions where necessary. 1.d2u dx2−γ2u=0, 0 <x<a, u(0)=T0,u(a)=T1. 2.d2u dx2−r=0, 0 <x<a(ris constant), u(0)=T0,du dx(a)=0. 3.d2u dx2=0, 0 <x<a, u(0)=T0,du dx(a)=0. 4.d2u dx2−γ2u=0, 0 <x<a, du dx(0)=0, u(a)=T1. 5.1 rd dr/parenleftbigg rdu dr/parenrightbigg =− p,0<r<a, u(a)=0. 52 Chapter 0 Ordinary Differential Equations 6.1 rd dr/parenleftbigg rdu dr/parenrightbigg =0, a<r<b, u(a)=T0,u(b)=T1. 7.1 ρ2d dρ/parenleftbigg ρ2du dρ/parenrightbigg =− H,0<ρ< a, u(a)=T0. 8.1 rd dr/parenleftbigg rdu dr/parenrightbigg =0, a<r<∞, u(a)=T. 9.d2u dx2−γ2(u−T)=0, 0 <x<a, du dx(0)=0, u(a)=T1. 10.d2u dx2−γ2u=0, 0 <x<∞, u(0)=T. 11.d2u dx2=γ2(u−T0),0<x<∞, u(0)=T. 12.d dx/parenleftbigg x3du dx/parenrightbigg =− k,a<x<b(kis constant), u(a)=0, u(b)=0( N o t e : 0 <a.) 13. In this problem, his the groundwater level between two trenches in which water is held at constant levels. Solve for h(x). Note that the equa- tion is nonlinear. d dx/parenleftbigg hdh dx/parenrightbigg +e=0,0<x<a, h(0)=h0,h(a)=h1. 14. Solve for u(x). d4u dx4=w, 0<x<a(wis constant ), u(0)=0,u(a)=0,d2u dx2(0)=0,d2u dx2(a)=0. Miscellaneous Exercises 53 15.Solve for u(x).N o t et h ei n t e r v a l . d4u dx4+k EIu=w, 0<x<∞(wis constant ), u(0)=0,d2u dx2(0)=0. 16.Show that any two of the four functions sinh (λx),s i n h(λ(a−x)), cosh(λx),c o s h(λ(a−x))are independent solutions of the differential equation φ/prime/prime−λ2φ=0. 17.In this problem, uis the temperature in a wall composed of two sub- stances. Find u(x). d2u dx2=0,0<x<αaand αa<x<a, u(0)=T0,u(a)=T1, κ1du dx(αa−)=κ2du dx(αa+), u(αa−)=u(αa+). The last two conditions say that the heat ﬂow rate and the temperature are both continuous across the interface at x=αa. 18.Find the general solution of the differential equation 1 x2d dx/parenleftbigg x2du dx/parenrightbigg +ku=0 for the cases k=λ2and k=− p2.( H i n t :L e t u(x)=v(x)/xand ﬁnd the equation that v(x)satisﬁes.) 19.Find the solution of the boundary value problem exd dx/parenleftbigg exdu dx/parenrightbigg =− 1,0<x<a, u(0)=0,u(a)=0. 20.Solve the boundary value problem 1 rd dr/parenleftbigg rdu dr/parenrightbigg =− rk,0<r<a, u(0)bounded and u(a)=0. 54 Chapter 0 Ordinary Differential Equations 21. Solve the differential equation d2u dx2=p2u,0<x<a, subject to the following sets of boundary conditions. a.u(0)=0, u(a)=1; b.u(0)=1, u(a)=0; c.u/prime(0)=0, u(a)=1; d.u(0)=1, u/prime(a)=0; e.u/prime(0)=1, u/prime(a)=0; f.u/prime(0)=0, u/prime(a)=1. 22. Solve the integro-differential boundary value problem d2u dx2=γ2/parenleftbigg u−/integraldisplay1 0u(x)dx/parenrightbigg ,0<x<1, du dx(0)=0,u(1)=T. Hint: Look for a solution in the form u(x)=Acosh(γx)+Bsinh(γx)+C. 23. Use a variation of parameters to ﬁnd a second independent solution of the following differential equation. One solution is given in parentheses. d2u dx2−2x 1−x2du dx+2 1−x2u=0(u=x). 24. By applying the method of variation of parameters, derive this formula for a particular solution of the differential equation d2u dx2−γ2u=f(x), u(x)=/integraldisplayx 0f(x/prime)sinhγ(x−x/prime) γdx/prime. 25. The absolute temperature u(x)in a cooling ﬁn that radiates heat to a medium at absolute temperature Tobeys the differential equation u/prime/prime= γ2(u4−T4). Solve the special version in the boundary value problem Miscellaneous Exercises 55 that follows, which can be done in closed form. d2u dx2=γ2u4,0<x, u(0)=U,lim x→∞u(x)=0. 26.A uniform, straight shaft exhibits violent behavior at certain frequen- cies of rotation. Let the x-axis between 0 and arepresent the undeﬂected centerline of the shaft, and let u(x)be the displacement of the actual cen- terline of the shaft measured from the x-axis. Centrifugal force provides a transverse loading on the shaft when uis not identically equal to zero. The equation for the displacement is d4u dx4−ω2w EIgu=0,0<x<a, where wis the weight per unit length of the shaft, gis the acceleration of gravity, Eis Y oung’s modulus, Iis the second moment of the cross- sectional area of the shaft, and ωis the angular velocity. If the shaft is held in narrow bearings at the ends, these can be interpreted as simplesupports, leading to boundary conditions u(0)=0,u /prime/prime(0)=0,u(a)=0,u/prime/prime(a)=0. Find a formula for those values of angular velocity (critical values or whirling speeds) that permit the existence of nonzero solutions to thisboundary value problem. 27.Find the lowest critical value for the angular velocity of a steel shaft with these speciﬁcations: diameter 1.5 in.; length 48 in.; w=0.5l b / i n . ; E= 30×10 6lb/in.2,I=0.5i n .4. 28.Sulphur dioxide (SO 2) is a common air pollutant that reacts with water to form sulphuric acid. If the water is airborne, the result is acid rain; ifthe water is in snow, the result is acid runoff when the snow melts. Use ananalysis similar to that of Section 3 to obtain a boundary value problemfor the concentration u(x)(in units of mass per unit volume) of sulphur dioxide in the air included in a layer of snow. Introduce q(x),t h eﬂ o w rate of sulphur dioxide (in units of mass per unit time per unit of cross- sectional area.) There are two important physical facts: (1) Diffusion is governed by Fick’s law (similar to Fourier’s law) q(x)=− Ddu dx, 56 Chapter 0 Ordinary Differential Equations where Dis the diffusion constant ; and (2) when the sulphur dioxide reacts with water, it “disappears” at a rate proportional to its concentration, sayku(x)(in units of mass per unit time per unit volume). 29. The sulphur dioxide concentration in the air in a deep layer of snow satisﬁes this boundary value problem in equilibrium conditions: d 2u dx2−a2u=0,0<x, u(0)=C0. Here, C0is the concentration in freely circulating air. Add an appropriate boundedness condition and solve for u(x). 30. In “Mechanical properties of thin ﬁlms from the load deﬂection of long clamped plates” [V . Ziebart et al., J. of Microelectromechanical Systems ,7 (1998): 320–327] this boundary value problem is studied: d4w dx4−γ2d2w dx2=P,−1 2<x<1 2, w/parenleftbigg ±1 2/parenrightbigg =0,dw dx/parenleftbigg ±1 2/parenrightbigg =0. The variables are wdeﬂection, xdistance measured across the short di- mension; and the parameters are Ppressure beneath the plate and γ2 effective stress, all dimensionless. Find the general solution of the differ- ential equation. 31. (Continuation) Solve the boundary value problem in Exercise 30. 32. (Continuation) The parameter γ2is related to stress, which is related to deﬂection. It must satisfy the equation γ2=S0+/integraldisplay1/2 0/parenleftbiggdw dx/parenrightbigg2 dx. Use your solution to ﬁnd a single explicit equation that γsatisﬁes. 33. (Continuation) If S0is negative in the equation of Exercise 32, γ2might be negative, say, γ2=−λ2. If this is the case, is there a value of λfor which the solution breaks down? 34. Suppose that u(t)is a function, not identically 0, for which u/prime/prime u=constant >0. Miscellaneous Exercises 57 Show that this relation is a differential equation and solve it. (Call the constant p2.) Prove that exactly one of the following three possibilities holds: (i)u(t)=0 for one value of tand u/prime(t)is never 0; (ii) u/prime(t)=0 for one value of tand u(t)is never 0; (iii) neither u(t)noru/prime(t)is ever 0. This page intentionally left blank Fourier Series and Integrals CHAPTER1 1.1 Periodic Functions and Fourier Series Af u n c t i o n fis said to be periodic with period p >0 if: (1) f(x)has been de- ﬁned for all x; and (2) f(x+p)=f(x)for all x. The familiar functions sin (x) and cos (x)are simple examples of periodic functions with period 2 π,a n dt h e functions sin (2πx/p)and cos (2πx/p)are periodic with period p. A periodic function has many periods, for if f(x)=f(x+p)then also f(x)=f(x+p)=f(x+2p)=···= f(x+np), where nis any integer. Thus sin( x)h a sp e r i o d s2 π,4π,..., 2nπ,... .T h ep e - riod of a periodic function is generally taken to be positive, but the periodicity condition holds for negative as well as positive changes in the argument. That is to say, f(x−p)=f(x)for all x,s i n c e f(x)=f(x−p+p)=f(x−p).A l s o , f(x)=f(x−p)=f(x−2p)=···= f(x−np). The deﬁnition of periodic says essentially that functional values repeat themselves. This implies that the graph of a periodic function can be drawnfor all xby making a template of the graph on any interval of length pand then copying the graph from the template up and down the x-axis (see Fig. 1). Many of the functions that occur in engineering and physics are periodic in space or time — for example, acoustic waves — and in order to understandthem better it is often desirable to represent them in terms of the very simple periodic functions 1, sin (x),c o s(x),s i n(2x),c o s(2x), and so forth. Allof these 59 60 Chapter 1 Fourier Series and Integrals Figure 1 A periodic function of period p. functions have the common period 2 π, although each has other periods as well. Iffis periodic with period 2 π, then we attempt to represent fin the form of an inﬁnite series f(x)=a0+∞/summationdisplay n=1/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig . (1) Each term of the series has period 2 π, so if the sum of the series exists, it will be a function of period 2 π. There are two questions to be answered: (a) What values must a0,an,bnhave? (b) If the appropriate values are assigned to the coefﬁcients, does the series actually represent the given function f(x)? On the face of it, the ﬁrst question is tremendously difﬁcult, for Eq. (1) rep- resents an equation in an inﬁnite number of unknowns. But a reasonable an-swer can be found easily by using the orthogonality 1relations shown in Table 1. We may summarize those relations by saying: The deﬁnite integral (over theinterval −πtoπ) of the product of any two different functions from the series in Eq. (1) is zero. The fundamental idea is that if the equality proposed in Eq. (1) is to be a real equality, then both sides must give the same result after the same operation. The orthogonality relations then suggest operations that simplify the right-hand side of Eq. (1). Namely, we multiply both sides of the proposed equationby one of the functions that appears there and integrate from −πtoπ.( W e must assume that the integration of the series can be carried out term by term.This is sometimes difﬁcult to justify, but we do it nonetheless.) Multiplying both sides of Eq. (1) by the constant 1 ( =cos(0x))a n di n t e - grating from −πtoπ,w eﬁ n d /integraldisplay π −πf(x)dx=/integraldisplayπ −πa0dx+∞/summationdisplay n=1/integraldisplayπ −π/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig dx. 1The word orthogonality should not be thought of in the geometric sense. Chapter 1 Fourier Series and Integrals 61 /integraldisplayπ −πsin(nx)dx=0 /integraldisplayπ −πcos(nx)dx=/braceleftbigg0, n/negationslash=0 2π, n=0 /integraldisplayπ −πsin(nx)cos(mx)dx=0 /integraldisplayπ −πsin(nx)sin(mx)dx=/braceleftbigg0,n/negationslash=m π, n=m /integraldisplayπ −πcos(nx)cos(mx)dx=/braceleftbigg0,n/negationslash=m π, n=m/negationslash=0. Table 1 Orthogonality relations Each of the terms in the integrated series is zero, so the right-hand side of this equation reduces to 2 π·a0,g i v i n g a0=1 2π/integraldisplayπ −πf(x)dx. (In words, a0is the mean value of f(x)over one period.) Now multiplying each side of Eq. (1) by sin (mx),w h e r e mis a ﬁxed integer, and integrating from −πtoπ,w eﬁ n d /integraldisplayπ −πf(x)sin(mx)dx=/integraldisplayπ −πa0sin(mx)dx+∞/summationdisplay n=1/integraldisplayπ −πancos(nx)sin(mx)dx +∞/summationdisplay n=1/integraldisplayπ −πbnsin(nx)sin(mx)dx. All terms containing a0orandisappear, according to the orthogonality rela- tions. Furthermore, of all those containing a bn, the only one that is not zero is the one in which n=m.( N o t i c et h a t nis a summation index and runs through all the integers 1 ,2,....W ec h o s e mto be a ﬁxed integer, so n=monce.) We now have the formula bm=1 π/integraldisplayπ −πf(x)sin(mx)dx. By multiplying both sides of Eq. (1) by cos (mx)(mis a ﬁxed integer) and integrating, we also ﬁnd am=1 π/integraldisplayπ −πf(x)cos(mx)dx. 62 Chapter 1 Fourier Series and Integrals We can now summarize our results. In order for the proposed equality f(x)=a0+∞/summationdisplay n=1/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig (2) to hold, the a’s and b’s must be chosen according to the formulas a0=1 2π/integraldisplayπ −πf(x)dx, (3) an=1 π/integraldisplayπ −πf(x)cos(nx)dx, (4) bn=1 π/integraldisplayπ −πf(x)sin(nx)dx. (5) When the coefﬁcients are chosen this way, the right-hand side of Eq. (1) is called the Fourier series off.T h e a’s and b’s are called Fourier coefﬁcients .W e have not yet answered question (b) about equality, so we write f(x)∼a0+∞/summationdisplay n=1/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig to indicate that the Fourier series corresponds tof(x). See the CD for an ani- mated example. Example. Suppose that f(x)is periodic with period 2 πand is given by the formula f(x)=xin the interval −π< x<π (see Fig. 2). According to our formulas, a0=1 2π/integraldisplayπ −πf(x)dx=1 2π/integraldisplayπ −πxd x=0, an=1 π/integraldisplayπ −πf(x)cos(nx)dx=1 π/integraldisplayπ −πxcos(nx)dx =1 π/bracketleftbiggcos(nx) n2+xsin(nx) n/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ −π=0, bn=1 π/integraldisplayπ −πf(x)sin(nx)dx=1 π/integraldisplayπ −πxsin(nx)dx =1 π/bracketleftbiggsin(nx) n2−xcos(nx) n/bracketrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingleπ −π =1 π(−2π)cosnπ n=2 n(−1)n+1. Chapter 1 Fourier Series and Integrals 63 Figure 2 f(x)=x,−π< x<π,fperiodic with period 2 π. Thus, for this function, we have f(x)∼∞/summationdisplay n=12(−1)n+1 nsin(nx) ∼2/parenleftbigg sin(x)−1 2sin(2x)+1 3sin(3x)−···/parenrightbigg . /square The Appendix contains some integration formulas that are convenient for ﬁnding Fourier coefﬁcients. It is also useful to know these special values ofsines and cosines that come up frequently in Fourier series. sin(nπ)=0,cos(nπ)=(−1) n,forn=0,±1,±2,..., sin/parenleftbigg(2n−1)π 2/parenrightbigg =(−1)n+1,cos/parenleftbigg(2n−1)π 2/parenrightbigg =0, forn=0,±1,±2,.... Note that the second line involves only odd multiples of π/2. Even multiples ofπ/2 are included in the ﬁrst line. EXERCISES 1.Find the Fourier coefﬁcients of the functions given in what follows. All are supposed to be periodic with period 2 π. Sketch the graph of the function. a.f(x)=x,−π< x<π; b.f(x)=|x|,−π< x<π; c.f(x)=/braceleftBig0,−π< x<0, 1,0<x<π; d.f(x)=|sinx|. 2.Sketch for at least two periods the graphs of the functions deﬁned by: a.f(x)=x,−1<x≤1,f(x+2)=f(x); 64 Chapter 1 Fourier Series and Integrals b.f(x)=/braceleftBig0,−1<x≤0, x,0<x<1,f(x+2)=f(x); c.f(x)=/braceleftBig0,−π< x≤0, 1,0<x≤2π,f(x+3π)=f(x); d.f(x)=/braceleftBig0, −π< x≤0, sinx,0<x≤π,f(x+2π)=f(x). 3.Show that the constant function f(x)=1 is periodic with every possible period p>0. 4.Carry out the details of deriving the equation for am. 5.Suppose f(x)has period p. Show that for any c, the following equation holds. Hint: Think of the integral as the net signed area. /integraldisplayc+p cf(x)dx=/integraldisplayp 0f(x)dx. 6.Suppose f(x),g(x)are periodic with a common period p. Show that af(x)+ bg(x)and f(x)·g(x)also are periodic with period p(a,bare constants). 7.Find the Fourier series of each of the following periodic functions. Integra- tion is not necessary: Use trigonometric identities. a.f(x)=cos2(x); b.f(x)=sin(x−π/6); c.f(x)=sin(x)cos(2x). 8.Ver ify that sin (πx/a)and cos (πx/a)are periodic with period 2 a. 1.2 Arbitrary Period and Half-Range Expansions In Section 1 we found a way to represent a periodic function of period 2πwith a Fourier series. It is not necessary to restrict ourselves to this pe- riod. In fact, we may broaden the idea of Fourier series to include func- tions of any period by a simple rescaling of the variables. Let us supposethat a function fis periodic with period 2 a.( W eu s e2 ain place of pfor later convenience.) Then we may relate ft oas e r i e so ft h ef u n c t i o n s1 , sin(πx/a),cos(πx/a),sin(2πx/a),cos(2πx/a),... ,a l lh a v i n gp e r i o d2 a,i n the form f(x)∼a 0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg . The coefﬁcients of this Fourier series may be determined either by scaling from the formulas of Section 1 or through the concept of orthogonality. In either 1.2 Arbitrary Period and Half-Range Expansions 65 case, the coefﬁcients are a0=1 2a/integraldisplaya −af(x)dx,an=1 a/integraldisplaya −af(x)cos/parenleftbiggnπx a/parenrightbigg dx, bn=1 a/integraldisplaya −af(x)sin/parenleftbiggnπx a/parenrightbigg dx.(1) Example. Find the Fourier series of f(x)=|sin(πx)|. Solution: This function is periodic with period 1, so a=p/2=1/2. T o do the integrals for the Fourier coefﬁcients, we need to get rid of the absolute value signs: f(x)=/braceleftbiggsin(πx), 0<x<1, −sin(πx),−1<x<0. Then, it is easy to calculate a0=1 1/integraldisplay1/2 −1/2/vextendsingle/vextendsinglesin(πx)/vextendsingle/vextendsingledx=/integraldisplay0 −1/2−sin(πx)dx+/integraldisplay1/2 0sin(πx)dx =cos(πx) π/vextendsingle/vextendsingle/vextendsingle/vextendsingle0 −1/2−cos(πx) π/vextendsingle/vextendsingle/vextendsingle/vextendsingle1/2 0=1 π−−1 π=2 π. (Recall that cos (±π/2)=0.) The other coefﬁcients are found similarly: an=2 1/integraldisplay1/2 −1/2/vextendsingle/vextendsinglesin(πx)/vextendsingle/vextendsinglecos(2nπx)dx =2/bracketleftbigg/integraldisplay0 −1/2−sin(πx)cos(2nπx)dx+/integraldisplay1/2 0sin(πx)cos(2nπx)dx/bracketrightbigg =−4 π·1 4n2−1. And bnis found to be 0 for all n. Consequently, the Fourier series of the function is /vextendsingle/vextendsinglesin(πx)/vextendsingle/vextendsingle∼2 π−4 π∞/summationdisplay n=11 4n2−1cos(2nπx). We will see later that |sin(πx)|is equal to its series. /square It is often necessary to use a Fourier series to represent a function that has been deﬁned only in a ﬁnite interval. We can justify such a representation bymaking the given function part of a per iodic function. If the given function fis deﬁned on the interval −a<x<a, we may construct ¯f,t h e periodic extension 66 Chapter 1 Fourier Series and Integrals of period 2 a, by using the following deﬁnitions: ¯f(x)=f(x), −a<x<a, ¯f(x)=f(x+2a),−3a<x<−a, ¯f(x)=f(x−2a),a<x<3a and so on, up and down the x-axis. Notice that the argument of fon the right- hand side always falls in the interval −a<x<a,w h e r e fwas originally given. Graphically, this kind of extension amounts to making a template of the graph offon−a<x<aand then copying from the template in abutting intervals of length 2 a. For the extended function with period 2 a, the formulas for the Fourier co- efﬁcients become a0=1 2a/integraldisplaya −a¯f(x)dx, an=1 a/integraldisplaya −a¯f(x)cos/parenleftbiggnπx a/parenrightbigg dx, bn=1 a/integraldisplaya −a¯f(x)sin/parenleftbiggnπx a/parenrightbigg dx.(2) If we are concerned with f(x)only in the interval −a<x<awhere it was originally given, the process of periodic extension is strictly formal, becausethe formulas for the coefﬁcients involve fonly on the original interval. Thus, we may write f(x)∼a 0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg ,−a<x<a. The inequality for xdraws attention to the fact that fwas deﬁned only on the interval −atoa. Example. Suppose f(x)=xin the interval −1<x<1. The graph of its periodic exten- sion (with period 2) is seen in Fig. 3, and the Fourier coefﬁcients are a0=0,an=0, bn=/integraldisplay1 −1xsin(nπx)dx=−2c o s(nπ) nπ=2 π(−1)n+1 n. /square The sine and cosine functions that appear in a Fourier series have some special symmetry properties that are useful in evaluating the coefﬁcients. Thegraph of the cosine function is symmetric about the vertical axis, and that ofthe sine is antisymmetric. We formalize these properties with a deﬁnition. 1.2 Arbitrary Period and Half-Range Expansions 67 Figure 3 f(x)=x,−1<x<1,fperiodic with period 2. Deﬁnition Af u n c t i o n g(x)iseven ifg(−x)=g(x);h(x)isodd ifh(−x)=− h(x).N o t e that a function must be deﬁned on a symmetric interval, say −c<x<c (where cmight be ∞), in order to qualify as even or odd. An even function is often said to be symmetric about the vertical axis, and an odd function is said to be symmetric in the origin. Many familiar functions areeither even or odd. For example, sin (kx),x,x 3, and any other odd power of x are all odd functions deﬁned on the interval −∞<x<∞. Similarly, cos (kx), |x|,1(=x0),x2, and any other even power of xare even functions over the same interval. Most functions are neither even nor odd, but any function that is deﬁned on a symmetric interval can be written as a sum of an even and anodd function: f(x)=1 2/parenleftbig f(x)+f(−x)/parenrightbig +1 2/parenleftbig f(x)−f(−x)/parenrightbig . It is easy to show that the ﬁrst term is an even function and the second is odd. Even and odd functions preserve their symmetries in some algebraic opera- tions, as summarized here: even+even=even,odd+odd=odd, even×even=even,odd×odd=even,odd×even=odd. We are also concerned with deﬁnite integrals of even and odd functions over symmetric intervals. The symmetry properties lead to important simpliﬁca-tions in our calculations. Theorem 1. Let g(x)be an even function deﬁned in a symmetric interval −a<x<a. Then /integraldisplaya −ag(x)dx=2/integraldisplaya 0g(x)dx. 68 Chapter 1 Fourier Series and Integrals Let h(x)be an odd function deﬁned in a symmetric interval −a<x<a. Then /integraldisplaya −ah(x)dx=0. /square Suppose now that gis an even function in the interval −a<x<a. Since the sine function is odd and the product g(x)sin(nπx/a)is odd, bn=1 a/integraldisplaya −ag(x)sin/parenleftbiggnπx a/parenrightbigg dx=0. That is, all the sine coefﬁcients are zero. Also, since the cosine is even, so is g(x)cos(nπx/a),a n dt h e n an=1 a/integraldisplaya −ag(x)cos/parenleftbiggnπx a/parenrightbigg dx=2 a/integraldisplaya 0g(x)cos/parenleftbiggnπx a/parenrightbigg dx. Thus the cosine coefﬁcients can be computed from an integral over the interval from 0 to a. Parallel results hold for odd functions: the cosine coefﬁcients are all zero and the sine coefﬁcients can be simpliﬁed. We summarize the results. Theorem 2. If g(x)is even on the interval −a<x<a(g(−x)=g(x)),t h e n g(x)∼a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg ,−a<x<a, where a0=1 a/integraldisplaya 0g(x)dx,an=2 a/integraldisplaya 0g(x)cos/parenleftbiggnπx a/parenrightbigg dx. If h(x)is odd on the interval −a<x<a(h(−x)=− h(x)),t h e n h(x)∼∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg ,−a<x<a, where bn=2 a/integraldisplaya 0h(x)sin/parenleftbiggnπx a/parenrightbigg dx. /square Very frequently, a function given in an interval 0 <x<amust be repre- sented in the form of a Fourier series. There are inﬁnitely many ways of do-ing this, but two ways are especially simple and useful: extending the givenfunction to one deﬁned on a symmetric interval −a<x<aby making the extended function either odd or even. 1.2 Arbitrary Period and Half-Range Expansions 69 (a) (b) (c) (d) Figure 4 A function is given in the interval 0 <x<a(heavy curve). The ﬁgure shows: (a) the odd extension; (b) the even extension; (c) the odd periodic exten-sion; and (d) the even periodic extension. Deﬁnition Letf(x)be given for 0 <x<a.T h e odd extension offis deﬁned by fo(x)=/braceleftbigg f(x), 0<x<a, −f(−x),−a<x<0. The even extension offis deﬁned by fe(x)=/braceleftbigg f(x), 0<x<a, f(−x),−a<x<0. Notice that if −a<x<0, then 0 <−x<a, so the functional values on the right are known from the given functions. Graphically, the even extension is made by reﬂecting the graph in the vertical axis. The odd extension is made by reﬂecting ﬁrst in the vertical axis and thenin the horizontal axis (see Fig. 4). Now the Fourier series of either extension may be calculated from the for- mulas in Theorem 2. Since f eis even and fois odd, we have fe(x)∼a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg ,−a<x<a, fo(x)∼∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg ,−a<x<a. 70 Chapter 1 Fourier Series and Integrals If the series on the right converge, they actually represent periodic functions with period 2 a. The cosine series would represent the even periodic extension off— the periodic extension of fe; and the sine series would represent the odd periodic extension of f. When the problem at hand is to represent the function f(x)in the interval 0<x<a, where it was originally given, we may use either the Fourier sine series or the cosine series because both feand focoincide with fin the interval. Thus we may summarize by saying: If f(x)is given for 0 <x<a,t h e n f(x)∼a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg ,0<x<a, a0=1 a/integraldisplaya 0f(x)dx,an=2 a/integraldisplaya 0f(x)cos/parenleftbiggnπx a/parenrightbigg dx and f(x)∼∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg ,0<x<a, bn=2 a/integraldisplaya 0f(x)sin/parenleftbiggnπx a/parenrightbigg dx. These two representations are called half-range expansions , and the series are called the Fourier cosine and Fourier sine series of f,r e s p e c t i v e l y .W es h a l l need these, more than any other kind of Fourier series, in the applications wemake later in this book. Example. Let us suppose that the function fhas the formula f(x)=x,0<x<1. Then the odd periodic extension of fis as shown in Fig. 5, and the Fourier sine coefﬁcients of fare bn=2/integraldisplay1 0xsin(nπx)dx=−2 nπcos(nπ). The even periodic extension of fis shown in Fig. 6. The Fourier cosine co- efﬁcients are a0=/integraldisplay1 0xd x=1 2, an=2/integraldisplay1 0xcos(nπx)dx=−2 n2π2/parenleftbig 1−cos(nπ)/parenrightbig . /square 1.2 Arbitrary Period and Half-Range Expansions 71 Figure 5 Odd periodic extension (period 2) of f(x)=x,0<x<1. Figure 6 Even periodic extension (period 2) of f(x)=x,0<x<1. The following six correspondences (we will later show them to be equalities) follow from the ideas of this section. Note that the inequalities showing theapplicable range of xare crucial. ∞/summationdisplay n=1−2c o s(nπ) nπsin(nπx)∼  f(x)=x,0<x<1, fo(x)=x,−1<x<1, ¯fo(x), −∞<x<∞, 1 2−∞/summationdisplay n=12(1−cos(nπ)) n2π2cos(nπx)∼  f(x)=x, 0<x<1, fe(x)=|x|,−1<x<1, ¯fe(x), −∞<x<∞. EXERCISES 1.Find the Fourier series of each of the following functions. Sketch the graph of the periodic extension of ffor at least two periods. a.f(x)=|x|,−1<x<1; b.f(x)=/braceleftBig−1,−2<x<0, 1, 0<x<2; c.f(x)=x2,−1 2<x<1 2. 72 Chapter 1 Fourier Series and Integrals 2.Show that the functions cos (nπx/a)and sin (nπx/a)satisfy orthogonality relations similar to those given in Section 1. 3.Suppose a Fourier series is needed for a function deﬁned in the interval0<x<2a. Show how to construct a periodic extension with period 2 a, and give formulas for the Fourier coefﬁcients that use only integrals from 0t o2 a. (Hint: See Exercise 5, Section 1.) 4.Show that the formula ex=cosh(x)+sinh(x) gives the decomposition of the function exinto a sum of an even and an odd function. 5.Identify each of the following as being even, odd, or neither. Sketch on a symmetric interval. a.f(x)=x; c.f(x)=|cos(x)|; e.f(x)=xcos(x);b.f(x)=|x|; d.f(x)=arc sin (x); f.f(x)=x+cos(x+1). 6.Iff(x)i sg i v e ni nt h ei n t e r v a l0 <x<a, what other ways are there to extend it to a function on −a<x<a? 7.Find the Fourier series of these functions. a.f(x)=x,−1<x<1; b.f(x)=1,−2<x<2; c.f(x)=/braceleftBigg x, −1 2<x<1 2, 1−x,1 2<x<3 2. 8.Is it true that if all the sine coefﬁcients of a function fdeﬁned on −a <x<aare zero, then fis even? 9.We know that if f(x)is odd on the interval −a<x<a, its Fourier se- ries is composed only of sines. What additional symmetry condition on f will make the sine coefﬁcients with even indices be zero? Give an exam- ple. 10.Sketch both the even and odd extensions of these functions. a.f(x)=1, 0 <x<a; c.f(x)=sin(x),0<x<1;b.f(x)=x,0 <x<a; d.f(x)=sin(x),0<x<π. 11.Find the Fourier sine series and cosine series for the functions given in Exercise 10. Sketch the even and odd periodic extensions for several peri-ods. 1.3 Convergence of Fourier Series 73 12.Prove the orthogonality relations /integraldisplaya 0sin/parenleftbiggnπx a/parenrightbigg sin/parenleftbiggmπx a/parenrightbigg dx=/braceleftbigg0, n/negationslash=m, a/2,n=m, /integraldisplaya 0cos/parenleftbiggmπx a/parenrightbigg cos/parenleftbiggnπx a/parenrightbigg dx=/braceleftBigg0, n/negationslash=m, a/2,n=m/negationslash=0, a, n=m=0. 13.Iff(x)is continuous on the interval 0 <x<a, is its even periodic ex- tension continuous? What about the odd periodic extension? Check espe-cially at x=0a n d±a. 14.Justify Theorem 1 by considering the integral as a sum of signed areas. See Fig. 4 for typical even and odd functions. 15.Justify or prove these statements. a.Ifh(x)is an odd function, then |h(x)|is an even function. b.Iff(x)is deﬁned for all positive x,t h e n f(|x|)is an even func- tion. c.Iff(x)is deﬁned for all xand g(x)is any even function, then f(g(x))is even. d.Ifh(x)is an odd function, g(x)is even, and g(x)is deﬁned for all x, then g(h(x))is an even function. 1.3 Convergence of Fourier Series Now we are ready to take up the second question of Section 1: Does the Fourier series of a function actually represent that function? The word represent has many interpretations, but for most practical purposes we really want to knowthe answer to this question: If a value of xis chosen, the numbers cos (nπx/a) and sin (nπx/a)are computed for each nand inserted into the Fourier series off, and the sum of the series is calculated, is that sum equal to the functional value f(x)? In this section we shall state, without proof, some theorems that answer the question (a proof of the convergence theorem is given in Section 7). But ﬁrstwe need a few deﬁnitions about limits and continuity. The ordinary limit lim x→x0f(x)can be rewritten as lim h→0f(x0+h).H e r e hmay approach zero in any manner. But if his required to be positive only, we g e tw h a ti sc a l l e dt h e right-hand limit offatx0,d e ﬁ n e db y f(x0+)=lim h→0+f(x0+h)=lim h→0 h>0f(x0+h). 74 Chapter 1 Fourier Series and Integrals (a) (b) (c) Figure 7 Three functions with different kinds of discontinuities at x=1. (a)f(x)=(x−x2)/(1−x)has a removable discontinuity. (b) f(x)=xfor 0<x<1a n d f(x)=x−1f o r1 <x; this function has a jump discontinuity. (c)f(x)=− ln(|1−x|)has a “bad” discontinuity. The left-hand limit is deﬁned similarly: f(x0−)=lim h→0−f(x0+h)=lim h→0 h<0f(x0+h)=lim h→0+f(x0−h). Note that f(x0+)and f(x0−)need not be values of the function f. If both left- and right-hand limits exist and are equal, the ordinary limit exists and is equal to the one-handed limits. It is quite possible that the left- and right-handed limits exist but are different. This happens, for instance, atx=0 for the function f(x)=/braceleftBig1, 0<x<π, −1,−π< x<0. In this case, the left-hand limit at x 0=0i s−1, whereas the right-hand limit is+1. A discontinuity at which the one-handed limits exist but do not agree is called a jump discontinuity . It is also possible that at some point both limits exist and agree but that the function is not deﬁned at that point or its value is not equal to the limit. Ins u c hac a s e ,af u n c t i o ni ss a i dt oh a v ea removable discontinuity .I ft h ev a l u eo f the function at the troublesome point is redeﬁned to be equal to the limit, thefunction will become continuous. For example, the function f(x)=sin(x)/x has a removable discontinuity at x=0. The discontinuity is eliminated by re- deﬁning f(x)=sin(x)/x(x/negationslash=0),f(0)=1. Removable discontinuities are so simple that we may assume they have been removed from any function under discussion. Other discontinuities are more serious. They occur if one or both of the one-handed limits fail to exist. Each of the functions sin (1/x),e 1/x,1/xhas a discontinuity at x=0t h a ti sn e i t h e rr e m o v a b l en o raj u m p( s e eF i g .7 ) .T a b l e2 summarizes continuity behavior at a point. 1.3 Convergence of Fourier Series 75 Name Criterion Continuity f(x0+)=f(x0−)=f(x0) Removable discontinuity f(x0+)=f(x0−)/negationslash=f(x0) Jump discontinuity f(x0+)/negationslash=f(x0−) “Bad” discontinuity f(x0+)orf(x0−)or both fail to exist Table 2 Types of continuity behavior at x0 Figure 8 Typical sectionally continuous function made up of four continuous “sections.” We shall say that a function is sectionally continuous (also called piecewise continuous )o na ni n t e r v a l a<x<bif it is bounded and continuous, ex- cept possibly for a ﬁnite number of jumps and removable discontinuities. (SeeFig. 8.) A function is sectionally continuous (without qualiﬁcation) if it is sec-tionally continuous on every interval of ﬁnite length. For instance, if a periodicfunction is sectionally continuous on any interval whose length is one period or more, then it is sectionally continuous. Examples. 1.The square wave ,d e ﬁ n e db y f(x)=/braceleftBig1, 0<x<a, −1,−a<x<0,f(x+2a)=f(x), is sectionally continuous. There are jump discontinuities at x=0,±a, ±2a,e t c . 2.The function f(x)=1/xcannot be sectionally continuous on any interval that contains 0 or even has 0 as an endpoint, because the function is notbounded at x=0. 3.Iff(x)=x,−1<x<1, then fis continuous on that interval. Its periodic extension (see Fig. 3) is sectionally continuous but not continuous. /square 76 Chapter 1 Fourier Series and Integrals The examples clarify a couple of facts about the meaning of sectional con- tinuity. Most important is that a sect ionally continuous function must not “blow up” at any point — even an endpoint — of an interval. Note also that afunction need not be deﬁned at every point in order to qualify as sectionally continuous. No value was given for the square-wave function at x=0,±a, but the function remains sectionally continuous, no matter what values areassigned for these points. Af u n c t i o ni s sectionally smooth (also, piecewise smooth )i na ni n t e r v a l a< x<bif:fis sectionally continuous; f /prime(x)exists, except perhaps at a ﬁnite number of points; and f/prime(x)is sectionally continuous. The graph of a section- ally smooth function then has a ﬁnite number of removable discontinuities, jumps, and corners. (The derivative will not exist at these points.) Between these points, the graph will be continuous, with a continuous derivative. Novertical tangents are allowed, for these indicate that the derivative is inﬁnite. Examples. 1.f(x)=|x|1/2is continuous but not sectionally smooth in any interval that contains 0, because |f/prime(x)|→∞ asx→0. 2.T h es q u a r ew a v ei ss e c t i o n a l l ys m o o t hb u tn o tc o n t i n u o u s . /square Most of the functions useful in math ematical modeling are sectionally smooth. Fortunately we can also give a positive statement about the Fourier series of such functions. Theorem. If f(x)is sectionally smooth and periodic with period 2a, then at each point x the Fourier series corresponding to f converges, and its sum is a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg =f(x+)+f(x−) 2. /square See an animated example on the CD. This theorem gives an answer to the question at the beginning of the section. Recall that a sectionally smooth function has only a ﬁnite number of jumpsand no bad discontinuities in every ﬁnite interval. Hence, f(x−)=f(x+)=1 2/parenleftbig f(x+)+f(x−)/parenrightbig =f(x), e x c e p tp e r h a p sa taﬁ n i t en u m b e ro fp o i n t so na n yﬁ n i t ei n t e r v a l .F o rt h i sr e a - son, if fsatisﬁes the hypotheses of the theorem, we write fe q u a l to its Fourier series, even though the equality may fail at jumps. In constructing the periodic extension of a function, we never deﬁned the values of f(x)at the endpoints. Since the Fourier coefﬁcients are given by in- tegrals, the value assigned to f(x)at one point cannot inﬂuence them; in that 1.3 Convergence of Fourier Series 77 sense, the value of fatx=+ ais unimportant. But because of the averaging features of the Fourier series, it is reasonable to deﬁne f(a)=f(−a)=1 2/parenleftbig f(a−)+f(−a+)/parenrightbig . That is, the value of fat the endpoints is the average of the one-handed limits at the endpoints, each limit taken from the interior. For instance, if f(x)= 1+x,0<x<1, and f(x)=0,−1<x<0, then f(±1)should be taken to be 1, and f(0)should be1 2. Examples. 1.The square-wave function f(x)=/braceleftBig1, 0<x<1, −1,−1<x<0 is sectionally smooth; therefore the corresponding Fourier series con- verges to /braceleftBigg1, for 0<x<1, −1,for−1<x<0, 0, forx=0,1,−1 and is periodic with period 2. 2.For the function f(x)=|x|1/2,−π< x<π ,f(x+2π)=f(x),t h ep r e - ceding theorem does not guarantee convergence of the Fourier series atany point, even though the function is continuous. Nevertheless, the se-ries does converge at any point x! This shows that the conditions in the theorem are perhaps too strong. (But they are useful.) /square EXERCISES 1.F o re a c hf u n c t i o ng i v e n ,i fi ti sn o ts e c t i o n a l l ys m o o t ho nt h ei n t e r v a l ,e x - plain why not. Sketch. a.f(x)=|x|−| 1−x|,−1<x<2; b.f(x)=√|x|,−1<x<1; c.f(x)=ln/parenleftbig 2c o s(x/2)/parenrightbig ,−π< x<π; d.f(x)=tan(x), 0<x<π/ 2; e.f(x)=tan(x), 0<x<π. 2.Check each function described in what follows to see whether it is section- ally smooth. If it is, state the value to which its Fourier series converges ateach point xin the interval and at the endpoints. Sketch. 78 Chapter 1 Fourier Series and Integrals a.f(x)=|x|+x,−1<x<1; b.f(x)=xcos(x),−π 2<x<π 2; c.f(x)=xcos(x),−1<x<1; d.f(x)=/braceleftBigg0,1<x<3, 1,−1<x<1, x,−3<x<−1. 3.T ow h a tv a l u ed o e st h eF o u r i e rs e r i e so f fconverge if fis a continuous , sectionally smooth, periodic function? Give an example. 4.State convergence theorems for the Fourier sine and cosine series that arisefrom half-range expansions. 5.A function is given on the interval 0 <x<2 by the formula f(x)=/braceleftBigx, 0<x<1, 1−x,1<x<2. a.Sketch the odd periodic extension ¯f0(x)for−4<x<4. b.Explain why ¯f0(x)is sectionally smooth. c.Determine the value that the sine series of fconverges to at these points: x=1,x=2,x=9.6,x=− 3.8. 6.For the same function given in Exercise 5, answer the same questions for ¯fe(x), the even periodic extension of fand its cosine series. 7.The series ∞/summationdisplay n=1(−1)n n2cos(nx) converges to a function f(x)whose formula on the interval −π< x<π is f(x)=A+Bx+Cx2. Determine A,B,a n d C. 8.The series ∞/summationdisplay n=11 n3sin(nx) converges to a continuous periodic function. On the interval 0 <x<2π, this function coincides with a polynomial p(x)of degree 3. Find the polyno- mial. Hint: Determine points xon the interval 0 <x<2πwhere p(x)=0. Use this information to get a form for p(x). 1.4 Uniform Convergence 79 9.The function f(x)is periodic with period 2. Its graph for −1<x<1i sa semicircle with radius 1 centered at the origin. a.Find the equation of f(x)for−1<x<1. b.Determine the value of the coefﬁcient a0in its Fourier series. (This is the only cosine coefﬁcient that can be found in closed form.) c.Isf(x)sectionally smooth? d.What does the theorem tell us about the convergence of the Fourier se- ries of f(x)? 1.4 Uniform Convergence The theorem of the preceding section treats convergence at individual points of an interval. A stronger kind of convergence is uniform convergence in an interval. Let SN(x)=a0+N/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg be the partial sum of the Fourier series of a function f.T h em a x i m u md e v i a - tion between the graphs of SN(x)and f(x)is δN=max/vextendsingle/vextendsinglef(x)−SN(x)/vextendsingle/vextendsingle,−a≤x≤a, where the maximum2is taken over all xin the interval, including the end- points. If the maximum deviation tends to zero as Nincreases, we say that the series converges uniformly in the interval −a≤x≤a. Roughly speaking, if a Fourier series converges uniformly, then the sum of a ﬁnite number Nof terms gives a good approximation — to within ±δN—o f the value of f(x)atanyand every point of the interval. Furthermore, by taking al a r g ee n o u g h N, one can make the error as small as necessary. There are two important facts about uniform convergence. If a Fourier series converges uniformly in a period interval, then (1) it must converge to a con- tinuous function, and (2) it must converge to the (continuous) function thatgenerates the series. Thus, a function that has a nonremovable discontinuitycannot have a uniformly convergent Fourier series. (And not all continuous functions have uniformly convergent Fourier series.) Figure 9 presents graphs of some partial sums of a square-wave function. It is easy to see that for every Nthere are points near x=0a n d x=±πwhere 2Iffis not continuous, the maximum must be replaced by the supremum, or least upper bound. 80 Chapter 1 Fourier Series and Integrals Figure 9 Partial sums of the square-wave function. Convergence is notuniform. 1.4 Uniform Convergence 81 Figure 10 Partial sums of a sawtooth function. Convergence is uniform. 82 Chapter 1 Fourier Series and Integrals |f(x)−SN(x)|i sn e a r l ye q u a lt o1 ,s oc o n v e r g e n c ei s notuniform. (Inciden- tally, the graphs in Fig. 9 also show the partial sums of f(x)overshooting their mark near x=0. This feature of Fourier series is called Gibbs’ phenomenon and always occurs near a jump.) On the other hand, Fig. 10 shows graphs of a “sawtooth” function and the partial sums of its Fourier series. The maximum deviation always occurs at x=0, and the convergence is uniform. One of the ways of proving uniform convergence is by examining the coef- ﬁcients. Theorem 1. If the series/summationtext∞ n=1(|an|+| bn|)converges, then the Fourier series a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg converges uniformly in the interval −a≤x≤a and, in fact, on the whole interval −∞<x<∞. /square Example. For the function f(x)=|x|,−π< x<π , the Fourier coefﬁcients are a0=π 2,an=2 πcos(nπ)−1 n2,bn=0. Since the series/summationtext∞ n=11/n2converges, the series of absolute values of the coefﬁ- cients converges, and so the Fourier series converges uniformly on the interval−π≤x≤πto|x|. The Fourier series converges uniformly to the periodic extension of f(x)on the whole real line (see Fig. 10). /square Another way of proving uniform convergence of a Fourier series is by exam- ining the function fthat generates it. Theorem 2. If f is periodic and continuous and has a sectionally continuous derivative, then the Fourier series corresponding to f converges uniformly to f (x) on the entire real axis. /square While this theorem is stated for a periodic function, it may be adapted to a function f(x)given on the interval −a<x<a.I ft h e periodic extension off satisﬁes the conditions of the theorem, then the Fourier series of fconverges uniformly on the interval −a≤x≤a. Example. Consider the function f(x)=x,−1<x<1. 1.4 Uniform Convergence 83 Although f(x)is continuous and has a continuous derivative in the interval −1<x<1, the periodic extension of fisnotcontinuous. The Fourier series cannot converge uniformly in any interval containing 1 or −1b e c a u s et h ep e - riodic extension of fhas jumps there, but uniform convergence must produce ac o n t i n u o u sf u n c t i o n . On the other hand, the function f(x)=|sin(x)|, periodic with period 2 π,i s continuous and has a sectionally continuous derivative. Therefore, its Fourierseries converges uniformly to f(x)everywhere. /square Here is a restatement of Theorem 2 for a function given on the interval −a<x<a. The condition at the endpoints replaces the condition of conti- nuity of the periodic extension of f. Theorem 3. If f(x)is given on −a<x<a, if f is continuous and bounded and has a sectionally continuous derivative, and if f (−a+)=f(a−), then the Fourier series of f converges uniformly to f on the interval −a≤x≤a.(The series con- verges to f (a−)=f(−a+)at x=± a.) /square If an odd periodic function is to be continuous, it must have value 0 at x=0 and at the endpoints of the symmetric period-interval. Thus, the odd periodicextension of a function given in 0 <x<amay have jump discontinuities even though it is continuous where originally given. The even periodic extension causes no such difﬁculty, however. Theorem 4. If f(x)is given on 0<x<a, if f is continuous and bounded and has a sectionally continuous derivative, and if f (0+)=f(a−)=0, then the Fourier sine series of f converges uniformly to f in the interval 0≤x≤a.(The series converges to 0at x=0and x=a.) /square Theorem 5. If f(x)is given on 0<x<a and if f is continuous and bounded and has a sectionally continuous derivative, then the Fourier cosine series of fc o n v e r g e su n i f o r m l yt of i nt h ei n t e r v a l 0≤x≤a.(The series converges to f (0+) at x=0and to f (a−)at x=a.) /square EXERCISES 1.Determine whether the Fourier series of the following functions converge uniformly or not. Sketch each function. a.f(x)=ex,−1<x<1; b.f(x)=sinh(x),−π< x<π; c.f(x)=sin(x),−π< x<π; 84 Chapter 1 Fourier Series and Integrals d.f(x)=sin(x)+|sin(x)|,−π< x<π; e.f(x)=x+|x|,−π< x<π; f.f(x)=x(x2−1),−1<x<1; g.f(x)=1+2x−2x3,−1<x<1. 2.The Fourier series of the function f(x)=sin(x) x,−π< x<π , converges at every point. T o what value does the series converge at x=0? atx=π? The convergence is uniform. Why? 3.Determine whether the sine and cosine series of the following functions converge uniformly. Sketch. a.f(x)=sinh(x), 0<x<π; b.f(x)=sin(x), 0<x<π; c.f(x)=sin(πx), 0<x<1 2; d.f(x)=1/(1+x), 0<x<1; e.f(x)=1/(1+x2), 0<x<2. 4.Ifanand bntend to zero as ntends to inﬁnity, show that the series a0+∞/summationdisplay n=1e−αn/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig converges uniformly (α > 0). 5.For each of the following coefﬁcients, use Theorem 1 to decide whether convergence of the associate Fourier series is uniform. a.an=sin2(nπ/2) n2π2,bn=0; b.an=0,bn=1−cos(nπ) nπ; c.a1=0,an=2(1+cos(nπ)) n2−1(n≥2), bn=0; d.an=0,bn=1 cosh(nπ/2). 1.5 Operations on Fourier Series 85 1.5 Operations on Fourier Series In the course of this book we shall have to perform certain operations on Fourier series. The purpose of this section is to ﬁnd conditions under which they are legitimate. Two things must be noted, however. First, the theoremsstated here are not the best possible: There are theorems with weaker hypothe-ses and the same conclusions. Second, in applying mathematics, we often carryout operations formally, legitimate or not. The results must then be checkedfor correctness. Throughout this section we shall state results about functions and Fourier series with period 2 π, for typographic convenience. The results remain true when the period is 2 ainstead. For functions deﬁned only on a ﬁnite interval, the periodic extension must fulﬁll the hypotheses. We shall refer to a functionf(x)with the series shown: f(x)∼a 0+∞/summationdisplay n=1ancos(nx)+bnsin(nx). (1) Theorem 1. The Fourier series of the function cf (x)has coefﬁcients ca 0,c a n,a n d cbn(ci sc o n s t a n t ). /square This theorem is a simple consequence of the fact that a constant passes through an integral. The fact that the integral of a sum is the sum of the inte-grals leads to the following. Theorem 2. The Fourier coefﬁcients of the sum f (x)+g(x)are the sums of the corresponding coefﬁcients of f (x)and g(x). /square These two theorems are so natural that the reader has probably used them already without thinking about it. The theorems that follow are much moredifﬁcult to prove, but they are extremely important. Theorem 3. If f(x)is periodic and sectionally continuous, then the Fourier series of f may be integrated term by term: /integraldisplayb af(x)dx=/integraldisplayb aa0dx+∞/summationdisplay n=1/integraldisplayb a/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig dx. (2) /square Theorem 4. If f(x)is periodic and sectionally continuous and if g (x)is sectionally continuous for a ≤x≤b, then 86 Chapter 1 Fourier Series and Integrals /integraldisplayb af(x)g(x)dx=/integraldisplayb aa0g(x)dx +∞/summationdisplay n=1/integraldisplayb a/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig g(x)dx. (3) /square In Theorems 3 and 4, the function f(x)is only required to be sectionally continuous. It is not necessary that the Fourier series of f(x)converge at all. Nevertheless, the theorems guarantee that the series on the right converges andequals the integral on the left in Eqs. (2) and (4). One important application of Theorem 4 was the derivation of the formulas for the Fourier coefﬁcients in Section 1. An application of Theorems 3 and 4 is given in what follows. Example. The periodic function g(x)whose formula in the interval 0 <x<2πis g(x)=x,0<x<2π has the Fourier series g(x)∼π−2∞/summationdisplay n=1sin(nx) n. By applying Theorems 1 and 2, we ﬁnd that the function f(x)deﬁned by f(x)= [π−g(x)]/2 has the series f(x)∼∞/summationdisplay n=1sin(nx) n. This manipulation would be simple algebra if the correspondence ∼were an equality. The function f(x)satisﬁes the hypotheses of Theorem 3. Thus we may inte- grate the preceding series from 0 to bto obtain /integraldisplayb 0f(x)dx=∞/summationdisplay n=11−cos(nb) n2. Theorem 3 guarantees that this equality holds for any b.I nt h ei n t e r v a lf r o m0 to 2πwe have the formula f(x)=(π−x)/2. Hence /integraldisplayb 0f(x)dx=πb 2−b2 4=∞/summationdisplay n=11−cos(nb) n2,0≤b≤2π. 1.5 Operations on Fourier Series 87 Now, replacing bbyx,w eh a v e x(2π−x) 4=∞/summationdisplay n=11 n2−∞/summationdisplay n=1cos(nx) n2,0≤x≤2π. (4) Outside the indicated interval, the periodic extension of the function on the left equals the series on the right. It is worthwhile to mention that the series on the right of Eq. (4) is the Fourier series of the function on the left. That is to say, 1 2π/integraldisplay2π 0x(2π−x) 4dx=∞/summationdisplay n=11 n2, (5) 1 π/integraldisplay2π 0x(2π−x) 4cos(nx)dx=−1 n2, (6) 1 π/integraldisplay2π 0x(2π−x) 4sin(nx)dx=0. (7) Equations (6) and (7) can be veriﬁed directly, of course, but Theorem 4, to- gether with the orthogonality relations of Section 1, also guarantees them. In addition, Eq. (5) gives us a way to evaluate the series on the right. /square Although the uniqueness property stated in the following theorem is so very natural that we tend to assume it is true without checking, it really is a conse-quence of Theorem 4. Theorem 5. If f(x)is periodic and sectionally continuous, its Fourier series is unique. /square That is to say, only one series can correspond to f(x).W eo f t e nm a k eu s eo f uniqueness in this way: If two Fourier series are equal (or correspond to thesame function), then the coefﬁcients of like terms must match. The last operation to be discussed is differentiation, one that plays a princi- pal role in applications. Theorem 6. If f(x)is periodic, continuous, and sectionally smooth, then the dif- ferentiated Fourier series of f (x)converges to f/prime(x)at every point x where f/prime/prime(x) exists: f/prime(x)=∞/summationdisplay n=1/parenleftbig −nansin(nx)+nbncos(nx)/parenrightbig . (8) /square The hypotheses on f(x)itself imply (see Section 4) that the Fourier series of f(x)converges uniformly. If f(x)(or its periodic extension) fails to be contin- 88 Chapter 1 Fourier Series and Integrals uous, it is certain that the differentiated series of f(x)will fail to converge, at some points at least. Example. Letfbe the function that is periodic with period 2 πand has the formula f(x)=|x|,−π< x<π . This function is indeed continuous and sectionally smooth and is equal to its Fourier series, f(x)=π 2−4 π/parenleftbigg cos(x)+cos(3x) 9+cos(5x) 25+···/parenrightbigg . According to Theorem 5, the differentiated series 4 π/parenleftbigg sin(x)+sin(3x) 3+sin(5x) 5+···/parenrightbigg converges to f/prime(x)at any point xwhere f/prime/prime(x)exists. Now, the derivative of the sawtooth function f(x)(see Fig. 10) is the square-wave function f/prime(x)=/braceleftBig1, 0<x<π, −1,−π< x<0(9) (see Fig. 9). Moreover, we know that the foregoing sine series is the Fourier s e r i e so ft h es q u a r ew a v e f/prime(x)a n dt h a ti tc o n v e r g e st ot h ev a l u e sg i v e nb y E q .( 9 ) ,e x c e p ta tt h ep o i n t sw h e r e f/prime(x)has a jump. These are precisely the points where f/prime/prime(x)does not exist. /square Later on, it will frequently happen that we know a function only through its Fourier series. Thus, it will be important to obtain properties of the function by examining its coefﬁcients, as the next theorem does. Theorem 7. If f is periodic, with Fourier coefﬁcients a n,bn,a n di ft h es e r i e s ∞/summationdisplay n=1/parenleftbig/vextendsingle/vextendsinglenkan/vextendsingle/vextendsingle+/vextendsingle/vextendsinglenkbn/vextendsingle/vextendsingle/parenrightbig converges for some integer k ≥1, then f has continuous derivatives f/prime,..., f(k) whose Fourier series are differentiated series of f . /square Example. Consider the function deﬁned by the series f(x)=∞/summationdisplay n=1e−nαcos(nx), 1.5 Operations on Fourier Series 89 in which αis a positive parameter. For this function we have a0=0,an=e−nα, bn=0. By the integral test, the series/summationtextnke−nαconverges for any k. Therefore fhas derivatives of all orders. The Fourier series of f/primeand f/prime/primeare f/prime(x)=∞/summationdisplay n=1−ne−nαsin(nx), f/prime/prime(x)=∞/summationdisplay n=1−n2e−nαcos(nx). /square EXERCISES 1.Evaluate the sum of the series/summationtext∞ n=11/n2by performing the integration indicated in Eq. (5). 2.Sketch the graphs of the periodic extension of the function f(x)=π−x 2,0<x<2π, and of its derivative f/prime(x)and of F(x)=/integraldisplayx 0f(t)dt. 3.Suppose that a function has the formula f(x)=x,0<x<π. What is its derivative? Can the Fourier sine series of fbe differentiated term by term? What about the cosine series? 4.Verify Eqs. (6) and (7) by integration. 5.Suppose that a function f(x)is continuous and sectionally smooth in the interval 0 <x<a. What additional conditions must f(x)satisfy in order to guarantee that its sine series can be differentiated term by term? thecosine series? 6.Is the derivative of a periodic function periodic? Is the integral of a peri-odic function periodic? 7.It is known that the equality ln/parenleftbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle2c o s/parenleftbiggx 2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg =∞/summationdisplay n=1(−1)n+1 ncos(nx) is valid except when xis an odd multiple of π. Can the Fourier series be differentiated term by term? 90 Chapter 1 Fourier Series and Integrals 8.Use the series that follows, together with integration or differentiation, to ﬁnd a Fourier series for the function p(x)=x(π−x),0<x<π. x=2∞/summationdisplay n=1(−1)n+1 nsin(nx), 0<x<π . 9.Letf(x)be an odd, periodic, sectionally smooth function with Fourier sine coefﬁcients b1,b2,.... Show that the function deﬁned by u(x,t)=∞/summationdisplay n=1bne−n2tsin(nx), t≥0, has the following properties: a.∂2u ∂x2=∞/summationdisplay n=1−n2bne−n2tsin(nx), t>0; b.u(0,t)=0,u(π,t)=0,t>0; c.u(x,0)=1 2/parenleftbig f(x+)+f(x−)/parenrightbig . 10.Letfbe as in Exercise 9, but deﬁne u(x,y)by u(x,y)=∞/summationdisplay n=1bne−nysin(nx), y>0. Show that u(x,y)has these properties: a.∂2u ∂x2=∞/summationdisplay n=1−n2bne−nysin(nx), y>0; b.u(0,y)=0,u(π,y)=0,y>0; c.u(x,0)=1 2/parenleftbig f(x+)+f(x−)/parenrightbig . 1.6 Mean Error and Convergence in Mean While we can study the behavior of inﬁnite series, we must almost always use ﬁnite series in practice. Fortunately, Fourier series have some properties thatmake them very useful in this setting. Before going on to these properties, we shall develop a useful formula. Suppose fis a function deﬁned in the interval −a<x<a, for which /integraldisplay a −a/parenleftbig f(x)/parenrightbig2dx 1.6 Mean Error and Convergence in Mean 91 is a ﬁnite number. Let f(x)∼a0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg and let g(x)have a ﬁnite Fourier series g(x)=A0+N/summationdisplay 1Ancos/parenleftbiggnπx a/parenrightbigg +Bnsin/parenleftbiggnπx a/parenrightbigg . Then we may perform the following operations: /integraldisplaya −af(x)g(x)dx=/integraldisplaya −af(x)/bracketleftBigg A0+N/summationdisplay 1Ancos/parenleftbiggnπx a/parenrightbigg +Bnsin/parenleftbiggnπx a/parenrightbigg/bracketrightBigg dx =A0/integraldisplaya −af(x)dx+N/summationdisplay 1An/integraldisplaya −af(x)cos/parenleftbiggnπx a/parenrightbigg dx +N/summationdisplay 1Bn/integraldisplaya −af(x)sin/parenleftbiggnπx a/parenrightbigg dx. We recognize the integrals as multiples of the Fourier coefﬁcients of fand rewrite 1 a/integraldisplaya −af(x)g(x)dx=2a0A0+N/summationdisplay 1(anAn+bnBn). (1) Now suppose we wish to approximate f(x)by a ﬁnite Fourier series. The difﬁculty here is deciding what “approximate” means. Of the many ways wecan measure approximation, the one that is easiest to use is the following: E N=/integraldisplaya −a/parenleftbig f(x)−g(x)/parenrightbig2dx. (2) (Here gis the function with a Fourier series containing terms up to and in- cluding cos (Nπx/a).) Clearly, ENcan never be negative, and if fand gare “close,” then ENwill be small. Thus our problem is to choose the coefﬁcients ofgso as to minimize EN.( W ea s s u m e Nﬁxed.) To com p u te EN, we ﬁrst expand the integrand: EN=/integraldisplaya −af2(x)dx−2/integraldisplaya −af(x)g(x)dx+/integraldisplaya −ag2(x)dx. (3) The ﬁrst integral has nothing to do with g; the other two integrals clearly de- pend on the choice of gand can be manipulated so as to minimize EN.W e 92 Chapter 1 Fourier Series and Integrals already have an expression for the middle integral. The last one can be found by replacing fwith gin Eq. (1): /integraldisplaya −ag2(x)dx=a/bracketleftBigg 2A2 0+N/summationdisplay 1A2n+B2 n/bracketrightBigg . (4) Now we have a formula for ENin terms of the variables A0,An,Bn: EN=/integraldisplaya −af2(x)dx−2a/bracketleftBigg 2A0a0+N/summationdisplay 1Anan+Bnbn/bracketrightBigg +a/bracketleftBigg 2A2 0+N/summationdisplay 1A2n+B2 n/bracketrightBigg . (5) The error ENtakes its minimum value when all of the partial derivatives with respect to the variables are zero. We must then solve the equations ∂EN ∂A0=− 4aa0+4aA0=0, ∂EN ∂An=− 2aan+2aAn=0, ∂EN ∂Bn=− 2abn+2aBn=0. These equations require that A0=a0,An=an,Bn=bn.T h u s gshould be chosen to be the truncated Fourier series of f, g(x)=a0+N/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg in order to minimize EN. Now that we know which choice of A’s and B’s minimizes EN,w ec a nc o m - pute that minimum value. After some algebra, we see that min(EN)=/integraldisplaya −af2(x)dx−a/bracketleftBigg 2a2 0+N/summationdisplay 1a2n+b2 n/bracketrightBigg . (6) E v e nt h i sm i n i m u me r r o rm u s tb eg r e a t e rt h a no re q u a lt oz e r o ,a n dt h u sw e have the Bessel inequality 1 a/integraldisplaya −af2(x)dx≥2a2 0+N/summationdisplay 1a2n+b2 n. (7) 1.6 Mean Error and Convergence in Mean 93 This inequality is valid for any Nand therefore is also valid in the limit as N tends to inﬁnity. The actual fact is that, in the limit, the inequality becomesParseval’s equality : 1 a/integraldisplaya −af2(x)dx=2a2 0+∞/summationdisplay 1a2n+b2 n. (8) Another very important consequence of Bessel’s inequality is that the two series/summationtexta2 nand/summationtextb2 nmust converge if the left-hand side of Eqs. (7) and (8) is ﬁnite. Thus, the numbers anand bnmust tend to 0 as ntends to inﬁnity. By comparing Eqs. (6) and (8), we get a different expression for the mini- mum error: min(EN)=a∞/summationdisplay N+1a2 n+b2 n. This quantity decreases steadily to zero as Nincreases. Since min (EN)is, ac- cording to Eq. (2), a mean deviation between fand the truncated Fourier se- ries of f, we often say, “The Fourier series of fconverges to fin the mean.” (Another kind of convergence!) Summary Iff(x)has been deﬁned in the interval −a<x<aand if /integraldisplaya −af2(x)dx is ﬁnite, then: 1.Among all ﬁnite series of the form g(x)=A0+N/summationdisplay 1Ancos/parenleftbiggnπx a/parenrightbigg +Bnsin/parenleftbiggnπx a/parenrightbigg the one that best approximates fin the sense of the error described by Eq. (2) is the truncated Fourier series of f: a0+N/summationdisplay 1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg . 2.1 a/integraldisplaya −af2(x)dx=2a2 0+∞/summationdisplay 1a2n+b2 n. 94 Chapter 1 Fourier Series and Integrals 3.an=1 a/integraldisplaya −af(x)cos/parenleftbiggnπx a/parenrightbigg dx→0a s n→∞ ; bn=1 a/integraldisplaya −af(x)sin/parenleftbiggnπx a/parenrightbigg dx→0a s n→∞ . 4.The Fourier series of fconverges to fin the sense of the mean. Properties 2 and 3 are very useful for checking computed values of Fourier coefﬁcients. EXERCISES 1.Use properties of Fourier series to evaluate the deﬁnite integral 1 π/integraldisplayπ −π/parenleftbigg ln/vextendsingle/vextendsingle/vextendsingle/vextendsingle2c o s/parenleftbiggx 2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg2 dx. (Hint: See Section 10, Eq. (4), and Section 5, Eq. (5).) 2.Verify Parseval’s equality for these functions: a.f(x)=x,−1<x<1; b.f(x)=sin(x),−π< x<π. 3.What can be said about the behavior of the Fourier coefﬁcients of the fol- lowing functions as n→∞ ? a.f(x)=|x|1/2,−1<x<1; b.f(x)=|x|−1/2,−1<x<1. 4.How do we know that ENhas a minimum and not a maximum? 5.If a function fdeﬁned on the interval −a<x<ahas Fourier coefﬁcients an=0,bn=1√n, what can you say about /integraldisplaya −af2(x)dx? 6.Show that, as n→∞ , the Fourier sine coefﬁcients of the function f(x)=1 x,−π< x<π , tend to a nonzero constant. (Since this is an odd function, we can take the cosine coefﬁcients to be zero, although strictly speaking they do not exist.) 1.7 Proof of Convergence 95 Use the fact that/integraldisplay∞ 0sin(t) tdt=π 2. 1.7 Proof of Convergence In this section we prove the Fourier convergence theorem stated in Section 3. Most of the proof requires nothing more than simple calculus, but there arethree technical points that we state here. Lemma 1. For all N =1,2,..., 1 π/integraldisplayπ −π/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy=1. /square Lemma 2. For all N =1,2,..., 1 2+N/summationdisplay n=1cos(ny)=sin/parenleftbig (N+1 2)y/parenrightbig 2s i n(1 2y). /square Lemma 3. Ifφ(y)is sectionally continuous, −π< y<π,t h e ni t sF o u r i e rc o e f - ﬁcients tend to 0 with n: lim n→∞1 π/integraldisplayπ −πφ(y)cos(ny)dy=0, lim n→∞1 π/integraldisplayπ −πφ(y)sin(ny)dy=0. /square In Exercises 1 and 2 of this section, you are asked to verify Lemmas 1 and 2 (also see Miscellaneous Exercise 17 at the end of this chapter). Lemma 3 wasproved in Section 6. The theorem we are going to prove is restated here for easy reference. Period 2πis used for typographic convenience; we have seen that any other period can be obtained by a simple change of variables. Theorem. If f(x)is sectionally smooth and periodic with period 2π, then the Fourier series corresponding to f converges at every x, and the sum of the series is a0+∞/summationdisplay n=1ancos(nx)+bnsin(nx)=1 2/parenleftbig f(x+)+f(x−)/parenrightbig . (1) /square 96 Chapter 1 Fourier Series and Integrals Proof: Let the point xbe chosen; it is to remain ﬁxed. T o begin with, we assume that fiscontinuous atx, so the sum of the series should be f(x). Another way to say this is that lim N→∞SN(x)−f(x)=0, where SNis the partial sum of the Fourier series of f, SN(x)=a0+N/summationdisplay n=1ancos(nx)+bnsin(nx). (2) Of course, the a’s and b’s are the Fourier coefﬁcients of f, a0=1 2π/integraldisplayπ −πf(z)dz, an=1 π/integraldisplayπ −πf(z)cos(nz)dz, bn=1 2π/integraldisplayπ −πf(z)sin(nz)dz.(3) The integrals have zas their variable of integration, but that does not affect their value. Part 1. Transformation of SN(x). In order to show a relationship between SN(x)and f, we replace the co- efﬁcients in Eq. (2) by the integrals that deﬁne them and use elementaryalgebra on the results: S N(x)=1 2π/integraldisplayπ −πf(z)dx+N/summationdisplay n=1/bracketleftbigg1 π/integraldisplayπ −πf(z)cos(nz)dzcos(nx) +1 π/integraldisplayπ −πf(z)sin(nz)dzsin(nx)/bracketrightbigg (4) =1 2π/integraldisplayπ −πf(z)dx+N/summationdisplay n=1/bracketleftbigg1 π/integraldisplayπ −πf(z)cos(nz)cos(nx)dz +1 π/integraldisplayπ −πf(z)sin(nz)sin(nx)dz/bracketrightbigg (5) =1 2π/integraldisplayπ −πf(z)dx+N/summationdisplay n=1/bracketleftbigg1 π/integraldisplayπ −πf(z)/parenleftbig cos(nz)cos(nx) +sin(nz)sin(nx)/parenrightbig dz/bracketrightbigg (6) 1.7 Proof of Convergence 97 =1 π/integraldisplayπ −πf(z)/parenleftBigg 1 2+N/summationdisplay n=1cos(nz)cos(nx)+sin(nz)sin(nx)/parenrightBigg dz (7) =1 π/integraldisplayπ −πf(z)/parenleftBigg 1 2+N/summationdisplay n=1cos/parenleftbig n(z−x)/parenrightbig/parenrightBigg dz. (8) In this very compact formula for SN(x), we now change the variable of in- tegration from ztoy=z−x: SN(x)=1 π/integraldisplayπ+x −π+xf(x+y)/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy. (9) Note that both factors in the integrand are periodic with period 2 π.T h e interval of integration can be any interval of length 2 πwith no change in the result. (See Exercise 5 of Section 1.) Therefore, SN(x)=1 π/integraldisplayπ −πf(x+y)/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy. (10) Part 2. Expression for SN(x)−f(x). Since we must show that the difference SN(x)−f(x)goes to 0, we need to have f(x)in a form compatible with that for SN(x).R e c a l lt h a t xis ﬁxed (although arbitrary), so f(x)is to be thought of as a number. Lemma 1 suggests the appropriate form, f(x)=f(x)·1 π/integraldisplayπ −π/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy =1 π/integraldisplayπ −πf(x)/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy. (11) Now, using Eq. (10) to represent SN(x),w eh a v e SN(x)−f(x)=1 π/integraldisplayπ −π/parenleftbig f(x+y)−f(x)/parenrightbig/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy. (12) Part 3. The limit. T h en e x ts t e pi st ou s eL e m m a2t or e p l a c et h es u mi nE q .( 1 2 ) .T h er e s u l tis S N(x)−f(x)=1 π/integraldisplayπ −π/parenleftbig f(x+y)−f(x)/parenrightbigsin/parenleftbig (N+1 2)y/parenrightbig 2s i n(1 2y)dy. (13) 98 Chapter 1 Fourier Series and Integrals The addition formula for sines gives the equality sin/parenleftbigg/parenleftbigg N+1 2/parenrightbigg y/parenrightbigg =cos(Ny)sin/parenleftbigg1 2y/parenrightbigg +sin(Ny)cos/parenleftbigg1 2y/parenrightbigg . Substituting it in Eq. (13) and using simple properties of integrals, we ob- tain SN(x)−f(x)=1 π/integraldisplayπ −π/parenleftbig f(x+y)−f(x)/parenrightbig1 2cos(Ny)dy +1 π/integraldisplayπ −π/parenleftbig f(x+y)−f(x)/parenrightbigcos(1 2y) 2s i n(1 2y)sin(Ny)dy.(14) T h eﬁ r s ti n t e g r a li nE q .( 1 4 )c a nb er e c o g n i z e da st h eF o u r i e rc o s i n ec o e f ﬁ - cient of the function ψ(y)=1 2/parenleftbig f(x+y)−f(x)/parenrightbig . (15) Since fis a sectionally smooth function, so is ψ, and the ﬁrst integral has limit 0 as Nincreases, by Lemma 3. The second integral in Eq. (14) can also be recognized, as the Fourier sine coefﬁcient of the function φ(y)=f(x+y)−f(x) 2s i n(1 2y)cos/parenleftbigg1 2y/parenrightbigg . (16) T o proceed as before, we must show that φ(y)is at least sectionally contin- uous,−π≤y≤π. The only difﬁculty is to show that the apparent division by 0 at y=0d o e sn o tc a u s e φ(y)to have a bad discontinuity there. First, if fis continuous and differentiable near x,t h e n f(x+y)−f(x)is continuous and differentiable near y=0. Then L’Hôpital’s rule gives lim y→0f(x+y)−f(x) 2s i n(1 2y)=lim y→0f/prime(x+y) cos(1 2y)=f/prime(x). (17) Under these conditions, the function φ(y)of Eq. (16) has a removable dis- continuity at y=0 and thus is sectionally continuous. Second, if fis continuous at xbut has a corner there, then f(x+y)−f(x) is continuous with a corner at y=0. In this case, L’Hôpital’s rule applies with the one-sided limits, which show lim y→0+f(x+y)−f(x) 2s i n(1 2y)=lim y→0+f/prime(x+y) cos(1 2y)=f/prime(x+), (18) lim y→0−f(x+y)−f(x) 2s i n(1 2y)=lim y→0−f/prime(x+y) cos(1 2y)=f/prime(x−). (19) 1.7 Proof of Convergence 99 Under these conditions, the function φ(y)of Eq. (16) has a jump disconti- nuity at y=0 and again is sectionally continuous. In either case, we see that the second integral in Eq. (14) is the Fourier sine coefﬁcient of a sectionally continuous function. By Lemma 3, then, it too has limit 0 as Nincreases, and the proof is complete for every xwhere fis continuous. Part 4. Iffis not continuous at x. Now let us suppose that fhas a jump discontinuity at x.I nt h i sc a s e ,w e must return to Part 2 and express the proposed sum of the series as 1 2/parenleftbig f(x+)+f(x−)/parenrightbig =1 π/integraldisplayπ 0f(x+)/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy +1 π/integraldisplay0 −πf(x−)/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy.(20) Here, we have used the evenness of the integrand in Lemma 1 to write 1 π/integraldisplayπ 0/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy=1 π/integraldisplay0 −π/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy=1 2.(21) Next, we have a convenient way to write the quantity to be limited: SN(x)−1 2/parenleftbig f(x+)+f(x−)/parenrightbig =1 π/integraldisplayπ 0/parenleftbig f(x+y)−f(x+)/parenrightbig/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy +1 π/integraldisplay0 −π/parenleftbig f(x+y)−f(x−)/parenrightbig/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg dy. (22) The interval of integration for SN(x)a ss h o w ni nE q .( 1 0 )h a sb e e ns p l i ti n half to conform to the integrals in Eq. (20). The last step is to show that each of the integrals in Eq. (22) approaches 0 asNincreases. Since the technique is the same as in Part 3, this is left as an exercise. Let us emphasize that the crux of the proof is to show that the function from Eq. (16), φ(y)=f(x+y)−f(x) 2s i n(1 2y)cos/parenleftbigg1 2y/parenrightbigg (23) ( o ras i m i l a rf u n c t i o nt h a ta r i s e sf r o mt h ei n t e g r a n d si nE q .( 2 2 ) ) ,d o e sn o t have a bad discontinuity at y=0. /square 100 Chapter 1 Fourier Series and Integrals EXERCISES 1.Verify Lemma 2. Multiply through by 2 sin (1 2y). Use the identity sin/parenleftbigg1 2y/parenrightbigg cos(ny)=1 2/parenleftbigg sin/parenleftbigg/parenleftbigg n+1 2/parenrightbigg y/parenrightbigg −sin/parenleftbigg/parenleftbigg n−1 2/parenrightbigg y/parenrightbigg/parenrightbigg . Note that most of the series then disappears. (T o see this, write out the result for N=3.) 2.Verify Lemma 1 by integrating the sum term by term. 3.Letf(x)=f(x+2π)and f(x)=|x|for−π< x<π .N o t et h a t fis con- tinuous and has a corner at x=0. Sketch the function φ(y)as deﬁned in Eq. (16) if x=0. Find φ(0+)andφ(0−). 4.Letfbe the odd periodic extension of the function whose formula is π−x for 0<x<π .I nt h i sc a s e , fhas a jump discontinuity at x=0. Taking x=0, sketch the functions φR(y)=f(x+y)−f(x+) 2s i n(1 2y)cos/parenleftbigg1 2y/parenrightbigg (y>0), φL(y)=f(x+y)−f(x−) 2s i n(1 2y)cos/parenleftbigg1 2y/parenrightbigg (y<0). (These functions appear if the integrands in Eq. (22) are developed as in Part 3 of the proof.) 5.Consider the function fthat is periodic with period 2 πand has the formula f(x)=|x|3/4for−π< x<π. a.Show that fis continuous at x=0 but is not sectionally smooth. b.Show that the function φ(y)(from Eq. (16), with x=0) is sectionally continuous, −π< x<π, except for a bad discontinuity at y=0. c.Show that the Fourier coefﬁcients of φ(y)tend to 0 as nincreases, de- spite the bad discontinuity. 1.8 Numerical Determination of Fourier Coefﬁcients There are many functions whose Fourier coefﬁcients cannot be determinedanalytically because the integrals involved are not known in terms of easilyevaluated functions. Also, it may happen that a function is not known explic-itly but that its value can be found at some points. In either case, if a Fourier 1.8 Numerical Determination of Fourier Coefﬁcients 101 series is to be found for the function, some numerical technique must be em- ployed to approximate the integrals that give the Fourier coefﬁcients. It turnsout that one of the crudest numerical integration techniques is the best. Any periodic, sectionally smooth function can be reduced by the procedure illustrated in Fig. 11 to the sum of some functions f 1(x)and f2(x), whose series can be found by integration, and another function that is continuous ,p e r i o d i c , and sectionally smooth. This last funct ion’s Fourier coefﬁcients will approach 0 rapidly with n. Suppose then that f(x)is continuous, sectionally smooth, and periodic with period 2 a. We wish to ﬁnd its Fourier coefﬁcients numerically. For instance, a0=1 2a/integraldisplaya −af(x)dx. The integral is approximated using the trapezoidal rule. First, cut up the inter- val−a<x<ainto requal subintervals with endpoints x0,x1,..., xrwhere xk=− a+k/Delta1x,/Delta1 x=2a r. Next, evaluate the sum a0∼=1 2a/parenleftbigg1 2f(x0)+f(x1)+···+ f(xr−1)+1 2f(xr)/parenrightbigg /Delta1x. (1) Since x0=− a,xr=a,a n d fis periodic with period 2 a,w eh a v e f(x0)=f(xr): The two terms with1 2multipliers can be combined. Thus, our approxima- tion is a0∼=1 2a/parenleftbig f(x1)+f(x2)+···+ f(xr)/parenrightbig ·2a r. The occurrences of 2 acancel, and the computed value is just the average of the functional values. We use a caret over the usual coefﬁcient name to designate approximations. Other Fourier coefﬁcients are approximated in a similar way. Summary Letf(x)be continuous, sectionally smooth and periodic with period 2 a.A p - proximate Fourier coefﬁcients of f(x)are ˆa0=1 r/parenleftbig f(x1)+···+ f(xr)/parenrightbig , (2) ˆan=2 r/parenleftbigg f(x1)cos/parenleftbiggnπx1 a/parenrightbigg +···+ f(xr)cos/parenleftbiggnπxr a/parenrightbigg/parenrightbigg , (3) ˆbn=2 r/parenleftbigg f(x1)sin/parenleftbiggnπx1 a/parenrightbigg +···+ f(xr)sin/parenleftbiggnπxr a/parenrightbigg/parenrightbigg . (4) 102 Chapter 1 Fourier Series and Integrals Figure 11 Preparation of a function for numerical integration of Fourier co- efﬁcients. (a) Graph of sectionally smooth function f(x)given on −a<x<a. (b) Graph of f1(x), which has jumps of the same magnitude and position as f(x). Coefﬁcients can be found analytically. (c) Graph of f(x)−f1(x).T h i sf u n c t i o nh a s no jumps in −a<x<a.( d )G r a p ho f f2(x). The periodic extensions of f2(x)and off(x)−f1(x)have jumps of the same magnitude at x=± a, and so forth. The co- efﬁcients of f2can be found analytically. (e) Graph of f3(x)=f(x)−f1(x)−f2(x). The Fourier series of f3(x)converges uniformly (the coefﬁcients tend to zero rapidly). 1.8 Numerical Determination of Fourier Coefﬁcients 103 Ifris odd, Eqs. (3) and (4) are valid for n=1,2,...,( r−1)/2, giving a total of rcoefﬁcients. If ris even, Eq. (4) gives ˆbr/2=0, and Eq. (3) has to be modiﬁed: ˆar/2=1 r/parenleftbigg f(x1)cos/parenleftbiggrπx1 2a/parenrightbigg +···+ f(xr)cos/parenleftbiggrπxr 2a/parenrightbigg/parenrightbigg . (3/prime) We again get rvalid coefﬁcients. /square The formulas in Eqs. (2)–(4) were derived for the case in which x0,x1,..., xr are equally spaced points in the interval −a≤x≤a. However, they remain valid for equally spaced points on the interval 0 ≤x≤2a.T h a ti s , x0=0,x1=2a r,x2=4a r, ..., xr=2a. (5) Note also that when f(x)is given in the interval 0 ≤x≤aand the sine or co- sine coefﬁcients are to be determined, the formulas may be derived from those already given here. Let the interval be divided into sequal subintervals with endpoints 0 =x0,x1,..., xs=a(in general, xi=ia/s). Then the approximate Fourier cosine coefﬁcients for for its even extension are ˆa0=1 s/parenleftbigg1 2f(x0)+f(x1)+···+ f(xs−1)+1 2f(xs)/parenrightbigg , ˆan=2 s/parenleftbigg1 2f(x0)+f(x1)cos/parenleftbiggnπx1 a/parenrightbigg +···+1 2f(xs)cos/parenleftbiggnπxs a/parenrightbigg/parenrightbigg , n=1,..., s−1, ˆas=1 s/parenleftbigg1 2f(x0)+f(x1)cos/parenleftbiggsπx1 a/parenrightbigg +···+1 2f(xs)cos/parenleftbiggsπxs a/parenrightbigg/parenrightbigg .(6) Similarly, the approximate Fourier sine coefﬁcients for for its odd extension are ˆbn=2 s/parenleftbigg f(x1)sin/parenleftbiggnπx1 a/parenrightbigg +···+ f(xs−1)sin/parenleftbiggnπxs−1 a/parenrightbigg/parenrightbigg , n=1,2,..., s. (7) An important feature of the approximate Fourier coefﬁcients is this: If F(x)=ˆa0+ˆa1cos/parenleftbiggπx a/parenrightbigg +ˆb1sin/parenleftbiggπx a/parenrightbigg +··· is a ﬁnite Fourier series using a total of rapproximate coefﬁcients calculated from Eqs. (3) and (4), then F(x)actually interpolates the function f(x)at x1,x2,..., xr.T h a ti s , F(xi)=f(xi), i=1,2,..., r. 104 Chapter 1 Fourier Series and Integrals ix i cosxi cos 2 xicos 3 xisin(xi)/xi 00 1 .01 .01 .01 .0 1π 60.86603 0 .50 0 .95493 2π 30.5 −0.5−1.00 .82699 3π 20 −1.00 0 .63662 42π 3−0.5 −0.51 .00 .41350 55π 6−0.86603 0 .50 0 .19099 6π−1.01 .0−1.00 .0 Table 3 Numerical information n ˆan an Error 00 .58717 0 .58949 0 .00232 10 .45611 0 .45141 0 .00470 2−0.06130 −0.05640 0 .00490 30 .02884 0 .02356 0 .00528 Table 4 Approximate coefﬁcients of sin(x)/x Thus the graph of F(x)cuts the graph of f(x)at the points xi,i=1,2,..., r. Example. Calculate the approximate Fourier coefﬁcients of f(x)=sin(x)/xin−π< x<π.S i n c e fis even, it will have a cosine series. We simplify computation by using the half-range formulas and making seven. We take s=6,x0=0,x1= π/6,..., x5=5π/6,x6=π. The numerical information is given in Table 3. The results of the calculation are given in Table 4. On the left are the approx- imate coefﬁcients calculated from the table. On the right are the correct values(to ﬁve decimals), obtained with the aid of a table of the sine integral (see Ex-ercise 2). Figure 12 shows the difference between f(x)andF(x)(the sum of the Fourier series using the approximate coefﬁcients through ˆa 6). /square For hand calculation, choosing sto be a multiple of 4 makes many of the cosines “easy” numbers such as 1 and 0.5. When the calculation is done by digital computer, this is not a consideration. EXERCISES 1.Since Table 3 gives sin (x)/xfor seven points, seven cosine coefﬁcients can be calculated. Find ˆa6. 1.8 Numerical Determination of Fourier Coefﬁcients 105 Figure 12 Graph of the difference between f(x)=sin(x)/xand F(x), the sum of the Fourier series using the approximate coefﬁcients ˆa0through ˆa6. 2.Express the Fourier cosine coefﬁcients of the example in terms of integrals of the form Si/parenleftbig (n+1)π/parenrightbig =/integraldisplay(n+1)π 0sin(t) tdt. This is the sine integral function and is tabulated in many books, especially Handbook of Mathematical Functions , Abramowitz and Stegun, 1972. 3.Each entry in the list that follows represents the depth of the water in Lake Ontario (minus the low-water datum of 242.8 feet) on the ﬁrst of the cor-responding month. Assuming that the water level is a periodic functionof period one year, and that the observations are taken at equal intervals, compute the Fourier coefﬁcients ˆa 0,ˆa1,ˆb1,ˆa2,ˆb2, thus identifying the mean level, and ﬂuctuations of period 12 months, 6 months, 4 months, and so forth. Take x0as January, ...,x11as December, and x12as January again. Jan. 0.75 July 2.35 Feb. 0.60 Aug. 2.15 Mar. 0.65 Sept. 1.75Apr. 1.15 Oct. 1.05May 1.80 Nov. 1.00June 2.25 Dec. 0.90 4.The numbers in the table that follows represent the monthly precipi-t a t i o n( i ni n c h e so fw a t e r )i nL a k eP l a c i d ,N Y ,a v e r a g e do v e rt h ep e -riod 1950–1959. Find the approximate Fourier coefﬁcients ˆa 0,...,ˆa6and ˆb1,...,ˆb5. 106 Chapter 1 Fourier Series and Integrals Jan. 2.751 July 3.861 Feb. 2.004 Aug. 4.088 Mar. 3.166 Sept. 4.093 Apr. 2.909 Oct. 3.434May 3.215 Nov. 2.902June 3.767 Dec. 3.011 1.9 Fourier Integral In Sections 1 and 2 of this chapter we developed the representation of a pe- riodic function in terms of sines and cosines with the same period. Then, by means of periodic extension, we obtaine d series representations for functions deﬁned only on a ﬁnite interval. Now we must deal with nonperiodic functionsdeﬁned for xbetween −∞ and∞. Can such functions also be represented in terms of sines and cosines? We make some transformations that suggest ananswer. Suppose f(x)is deﬁned for −∞<x<∞and is sectionally smooth in every ﬁnite interval. Then for any positive a,f(x)can be represented in the interval −a<x<aby its Fourier series: f(x)=a 0+∞/summationdisplay n=1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg ,−a<x<a, a0=1 2a/integraldisplaya −af(x)dx,an=1 a/integraldisplaya −af(x)cos/parenleftbiggnπx a/parenrightbigg dx, bn=1 a/integraldisplaya −af(x)sin/parenleftbiggnπx a/parenrightbigg dx. (1) Example 1. Let f(x)=/braceleftbigg e−x,0<x, 0, x<0. For any a>0, we have the Fourier series for f(x)on the interval −a<x<a: f(x)=a0+∞/summationdisplay 1ancos/parenleftbiggnπx a/parenrightbigg +bnsin/parenleftbiggnπx a/parenrightbigg ,−a<x<a, a0=1−e−a 2a,an=1−e−acos(nπ) a(1+(nπ/a)2), bn=(1−e−acos(nπ))nπ a2(1+(nπ/a)2). (2) (The series converges to 1 /2a tx=0a n dt o e−a/2a tx=a.) /square 1.9 Fourier Integral 107 Now we modify Eq. (1). Let λn=nπ/aand deﬁne two functions Aa(λ)=1 π/integraldisplaya −af(x)cos(λx)dx,Ba(λ)=1 π/integraldisplaya −af(x)sin(λx)dx.(3) Notice that an=π aAa(λn), bn=π aBa(λn). Because of this, the Fourier series Eq. (1) becomes f(x)=a0+∞/summationdisplay n=1/bracketleftbig Aa(λn)cos(λnx)+Ba(λn)sin(λnx)/bracketrightbig ·/Delta1λ, −a<x<a,(4) where /Delta1λ=π/a=λn+1−λn. The form in which Eq. (4) is written is chosen to suggest an integral with respect to λover the interval 0 <λ< ∞. We may imagine aincreasing to inﬁnity, so /Delta1λ→0a n d Aa(λ)→A(λ)=1 π/integraldisplay∞ −∞f(x)cos(λx)dx, (5) Ba(λ)→B(λ)=1 π/integraldisplay∞ −∞f(x)sin(λx)dx, (6) and a0→0. Then Eq. (4) suggests f(x)=/integraldisplay∞ 0/bracketleftbig A(λ)cos(λx)+B(λ)sin(λx)/bracketrightbig dλ,−∞<x<∞. (7) Example 1 (continued). Forf(x)as in Example 1, we ﬁnd A(λ)=1 π/integraldisplay∞ 0e−xcos(λx)dx=1 π(1+λ2), B(λ)=1 π/integraldisplay∞ 0e−xsin(λx)dx=λ π(1+λ2), and therefore we expect that /integraldisplay∞ 0/bracketleftbigg1 π(1+λ2)cos(λx)+λ π(1+λ2)sin(λx)/bracketrightbigg dx=/braceleftbigg e−x,0<x, 0, x<0. /square The foregoing derivation is not a proof, but it does suggest the following theorem. 108 Chapter 1 Fourier Series and Integrals Fourier Integral Representation Theorem. Let f(x)be sectionally smooth on every ﬁnite interval, and let/integraltext∞ −∞|f(x)|dx be ﬁnite. Then at every point x, /integraldisplay∞ 0/parenleftbig A(λ)cos(λx)+B(λ)sin(λx)/parenrightbig dλ=1 2/parenleftbig f(x+)+f(x−)/parenrightbig , −∞<x<∞, (8) where A(λ)=1 π/integraldisplay∞ −∞f(x)cos(λx)dx,B(λ)=1 π/integraldisplay∞ −∞f(x)sin(λx)dx. (9) /square Equation (8) is called the Fourier integral representation of f (x);A(λ)andB(λ) in Eq. (9) are the Fourier integral coefﬁcient functions off(x). The right-hand s i d eo fE q .( 8 )w a sa l s os e e ni nt h eF o u r i e rs e r i e sc o n v e r g e n c et h e o r e m .S i n c ef(x)is sectionally smooth, the expression in Eq. (8) is the same as f(x)almost everywhere, so we often write Eq. (7) instead of Eq. (8). Example 2. The function f(x)=/braceleftbigg1,|x|<1, 0,|x|>1 has the Fourier integral coefﬁcient functions A(λ)=1 π/integraldisplay∞ −∞f(x)cos(λx)dx=1 π/integraldisplay1 −1cos(λx)dx=2s i n(λ) πλ, B(λ)=0. Since f(x)is sectionally smooth, the Fourier integral representation is legit- imate, and we write f(x)=/integraldisplay∞ 02s i n(λ) πλcos(λx)dλ. (Actually the integral equals1 2atx=± 1, so equality is not strictly correct at these two points.) /square Example 3. Find the Fourier integral representation of f(x)=exp(−|x|). Solution: Direct integration gives A(λ)=1 π/integraldisplay∞ −∞exp/parenleftbig −|x|/parenrightbig cos(λx)dx, (10) 1.9 Fourier Integral 109 A(λ)=2 π/integraldisplay∞ 0e−xcos(λx)dx, (11) A(λ)=2 πe−x(−cos(λx)+λsin(λx)) 1+λ2/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞ 0=2 π1 1+λ2. (12) B(λ)=0, because exp (−|x|)is even. Since exp (−|x|)is continuous and sec- tionally smooth, we may write exp(−|x|)=2 π/integraldisplay∞ 0cos(λx) 1+λ2dλ,−∞<x<∞. /square These two examples illustrate the fact that, in general, one cannot evaluate the integral in the Fourier integral representation. It is the theorem stated in the preceding that allows us to write the equality between a suitable functionand its Fourier integral. Iff(x)is deﬁned only in the interval 0 <x<∞, one can construct an even or odd extension whose Fourier integral contains only cos (λx)or sin(λx). These are called the Fourier cosine and sine integral representations off,r e - spectively. Letf(x)be deﬁned and sectionally smooth for 0 <x<∞,a n dl e t/integraltext∞ 0|f(x)|dx<∞.T h e nw ew r i t e : Fourier cosine integral representation f(x)=/integraldisplay∞ 0A(λ)cos(λx)dλ, 0<x<∞ with A(λ)=2 π/integraldisplay∞ 0f(x)cos(λx)dx, Fourier sine integral representation f(x)=/integraldisplay∞ 0B(λ)sin(λx)dλ, 0<x<∞ with B(λ)=2 π/integraldisplay∞ 0f(x)sin(λx)dx. Example 4. Find the Fourier sine and cosine integral representations of f(x)given for 0 <x by f(x)=/braceleftbigg sin(x),0<x<π, 0,π < x. 110 Chapter 1 Fourier Series and Integrals Since f(x)=0 for x>π, the integral for B(λ)reduces to one over the interval 0<x<π: B(λ)=2 π/integraldisplay∞ 0f(x)sin(λx)dx=2 π/integraldisplay∞ 0sin(x)sin(λx)dx =2 π/bracketleftbiggsin((λ−1)x) 2(λ−1)−sin((λ+1)x) 2(λ+1)/bracketrightbiggπ 0 =2 π/bracketleftbiggsin((λ−1)π) 2(λ−1)−sin((λ+1)π) 2(λ+1)/bracketrightbigg . This expression can be simpliﬁed by using the fact that sin/parenleftbig (λ±1)π/parenrightbig =sin(λπ±π)=− sin(λπ). Then, creating a common denominator, we obtain B(λ)=−2s i n(λπ) π(λ2−1). Hence the Fourier sine integral representation of f(x)is f(x)=/integraldisplay∞ 0−2s i n(λπ) π(λ2−1)sin(λx)dλ, 0<x. Since f(x)is continuous for 0 <x, the equality holds at every point. Similarly, we can compute the cosine coefﬁcient function A(λ)=−2(1+cos(λπ)) π(λ2−1), and the cosine integral representation of f(x)is f(x)=/integraldisplay∞ 0−2(1+cos(λπ)) π(λ2−1)cos(λx)dλ, 0<x. Note that both A(λ)and B(λ)have removable discontinuities at λ=1. /square It seems to be a rule of thumb that if the Fourier coefﬁcient functions A(λ) and B(λ)can be found in closed form for some function f(x), then the inte- gral in the Fourier integral representation cannot be carried out by elementarymeans, and vice versa. (See Exercise 3.) Rules for operations on Fourier integrals generally follow the lines men- tioned in Section 1.5 for Fourier series. In particular: If f(x)is continuous and if both f(x)and f /prime(x)have Fourier integral representations, then f(x)=/integraldisplay∞ 0/bracketleftbig A(λ)cos(λx)+B(λ)sin(λx)/bracketrightbig dλ, 1.9 Fourier Integral 111 f/prime(x)=/integraldisplay∞ 0/bracketleftbig −λA(λ)sin(λx)+λB(λ)cos(λx)/bracketrightbig dλ. Example 5. Letf(x)=exp(−|x|)as in Example 3. Then its derivative is f/prime(x)=/braceleftbigg −e−x,0<x, ex, x<0. Clearly, f(x)is continuous, and both f(x)and f/prime(x)have Fourier integral rep- resentations. The one for f(x)is in Example 3. Thus we have f/prime(x)=/integraldisplay∞ 02 π−λ 1+λ2sin(λx)dλ. /square EXERCISES 1.Sketch the even and odd extensions of each of the following functions, and ﬁnd the Fourier cosine and sine integrals for f.E a c hf u n c t i o ni sg i v e ni n the interval 0 <x<∞. a.f(x)=e−x; b.f(x)=/braceleftBig1,0<x<1, 0,1<x; c.f(x)=/braceleftBigπ−x,0<x<π, 0,π < x. 2.Find the Fourier integral representation of the following function ft(x). This is sometimes called a “window” because it is “open” for t−h<x< t+h. ft(x)=/braceleftbigg 1,|x−t|<h, 0,|x−t|>h. 3.Find the Fourier integral representation of each of the following functions. a.f(x)=1 1+x2; b.f(x)=sin(x) x. (Hint: T o evaluate the Fourier integral coefﬁcient functions, consult the Fourier integral representations found in the examples.) 4.In Exercise 3b, the integral/integraltext∞ −∞|f(x)|dxis not ﬁnite. Nevertheless, A(λ) and B(λ)do exist (B(λ)=0). Find a rationale in the convergence theo- rem for saying that this function can be represented by its Fourier integral.(Hint: See Example 1.) 5.Find the Fourier integral representation of each of these functions: 112 Chapter 1 Fourier Series and Integrals a.f(x)=/braceleftbigg sin(x),−π< x<π, 0, |x|>π; b.f(x)=/braceleftbigg sin(x),0<x<π, 0, otherwise; c.f(x)=/braceleftbigg |sin(x)|,−π< x<π, 0, otherwise. 6.Show that if kand Kare positive, then the following are true: a./integraldisplay∞ 0e−kxsin(x)dx=1 1+k2; b./integraldisplay∞ 01−e−Kx xsin(x)dx=tan−1(K); c./integraldisplay∞ 0sin(x) xdx=π 2. (Part (a) by direct integration, (b) by integration of (a) with respect to k over the interval 0 to K,( c )b yl i m i to f( b )a s K→∞ .) 7.Starting from Exercise 6c, show that /integraldisplay∞ 0sin(λz) λdλ=/braceleftBiggπ/2, 0<z, 0, z=0, −π/2,z<0. Is this the Fourier integral of some function? 8.Change the variable of integration in the formulas for Aand B, and justify each step of the following string of equalities. (Do not worry about chang- ing order of integration.) f(x)=1 π/integraldisplay∞ 0/integraldisplay∞ −∞f(t)/parenleftbig cos(λt)cos(λx)+sin(λt)sin(λx)/parenrightbig dt dλ =1 π/integraldisplay∞ −∞f(t)/integraldisplay∞ 0cos/parenleftbig λ(t−x)/parenrightbig dλdt =1 π/integraldisplay∞ −∞f(t)/bracketleftbigg lim ω→∞sin(ω(t−x)) t−x/bracketrightbigg dt =lim ω→∞1 π/integraldisplay∞ −∞f(t)sin(ω(t−x)) t−xdt. The last integral is called Fourier’s single integral .S k e t c ht h ef u n c t i o n sin(ωv)/v as a function of vfor several values of ω. What happens near 1.10 Complex Methods 113 v=0? Sometimes notation is compressed and, instead of the last line, we write f(x)=/integraldisplay∞ −∞f(t)δ(t−x)dt. Although δis not, strictly speaking, a function, it is called Dirac’s delta func- tion. 1.10 Complex Methods Fourier series Suppose that a function f(x)equals its Fourier series f(x)=a0+∞/summationdisplay n=1ancos(nx)+bnsin(nx). (W e use period 2 πfor simplicity only.) A famous formula of Euler states that eiθ=cos(θ)+isin(θ), where i2=− 1. Some simple algebra then gives the exponential deﬁnitions of the sine and cosine: cos(θ)=1 2/parenleftbig eiθ+e−iθ/parenrightbig ,sin(θ)=1 2i/parenleftbig eiθ−e−iθ/parenrightbig . By substituting the exponential forms into the Fourier series of fwe arrive at the alternate form f(x)=a0+1 2∞/summationdisplay n=1an/parenleftbig einx+e−inx/parenrightbig −ibn/parenleftbig einx−e−inx/parenrightbig =a0+1 2∞/summationdisplay n=1(an−ibn)einx+(an+ibn)e−inx. We are now led to deﬁne complex Fourier coefﬁcients forf: c0=a0,cn=1 2(an−ibn), c−n=1 2(an+ibn), n=1,2,3,.... In terms of these two coefﬁcients, we have f(x)=c0+∞/summationdisplay n=1/parenleftbig cneinx+c−ne−inx/parenrightbig =∞/summationdisplay −∞cneinx. (1) 114 Chapter 1 Fourier Series and Integrals This is the complex form of the Fourier series for f.I ti se a s yt od e r i v et h e universal formula cn=1 2π/integraldisplayπ −πf(x)e−inxdx, (2) which is valid for all integers n, positive, negative, or zero. The complex form is used especially in physics and electric al engineering. Sometimes the function corresponding to a Fourier series can be recognized by use of the complexform. Example. The series ∞/summationdisplay n=1(−1)n+1 ncos(nx) may be considered the real part of ∞/summationdisplay n=1(−1)n+1 neinx=∞/summationdisplay n=1(−1)n+1 n/parenleftbig eix/parenrightbign(3) because the real part of eiθis cos(θ). The series on the right in Eq. (3) is recog- nized as a Taylor series, ∞/summationdisplay n=1(−1)n+1 n/parenleftbig eix/parenrightbign=ln/parenleftbig 1+eix/parenrightbig . Some manipulations yield 1+eix=eix/2/parenleftbig eix/2+e−ix/2/parenrightbig =2eix/2cos/parenleftbiggx 2/parenrightbigg , ln/parenleftbig 1+eix/parenrightbig =ix 2+ln/parenleftbigg 2c o s/parenleftbiggx 2/parenrightbigg/parenrightbigg . The real part of ln (1+eix)is ln(2c o s(x/2))when−π< x<π.T h u s ,w ed e r i v e the relation ln/parenleftbigg 2c o s/parenleftbiggx 2/parenrightbigg/parenrightbigg ∼∞/summationdisplay n=1(−1)n+1 ncos(nx),−π< x<π . (4) (The series actually converges except at x=±π,±3π,... .) /square 1.10 Complex Methods 115 Fourier integral The Fourier integral of a function f(x)deﬁned in the entire interval −∞< x<∞can also be cast in complex form: f(x)=/integraldisplay∞ −∞C(λ)eiλxdλ. (5) The complex Fourier integral coefﬁcient function is given by C(λ)=1 2π/integraldisplay∞ −∞f(x)e−iλxdx. (6) It is simple to show that C(λ)=1 2/parenleftbig A(λ)−iB(λ)/parenrightbig , (7) where Aand Bare the usual Fourier integral coefﬁcients. The complex Fourier integral coefﬁcient is often called the Fourier transform of the function f(x). Example. Find the complex Fourier integral representation of f(x)=/braceleftbigg1,−a<x<a, 0,x<|a|. The coefﬁcient function (or transform) of fis C(λ)=1 2π/integraldisplaya −ae−iλxdx=1 2πe−iλx −iλ/vextendsingle/vextendsingle/vextendsingle/vextendsinglea −a =1 2πeiλa−e−iλa iλ=sin(λa) πλ. The representation of fis f(x)=/integraldisplay∞ −∞sin(λa) πλeiλxdλ,−∞<x<∞. Of course, at x=± a,t h ei n t e g r a lc o n v e r g e st o1 /2. /square This example brings out a fact about symmetry: If f(x)is even, C(λ)is real; iff(x)is odd, C(λ)is imaginary. The Fourier integral or transform may be used to solve differential equations on the interval −∞<x<∞, in much the same way that Laplace transform is used. 116 Chapter 1 Fourier Series and Integrals EXERCISES 1.Use the complex form an−ibn=1 π/integraldisplayπ −πf(x)e−inxdx,n/negationslash=0, to ﬁnd the Fourier series of the function f(x)=eαx,−π< x<π . 2.Find the complex Fourier series for the “square wave” with period 2 π: f(x)=/braceleftBig1, 0<x<π, −1,−π< x<0. 3.Find the complex Fourier integral representation of the following func- tions: a.f(x)=/braceleftbigg e−x,x>0, 0, x<0; b.f(x)=/braceleftbigg sin(x),0<x<π, 0, elsewhere. 4.Find the complex Fourier integral for a.f(x)=/braceleftbigg xe−x,0<x, 0, x<0; b.f(x)=e−α|x|sin(x). 5.Relate the functions and series that follow by using complex form and Tay- lor series. a.1+∞/summationdisplay n=1rncos(nx)=1−rcos(x) 1−2rcos(x)+r2,0≤r<1; b.∞/summationdisplay n=1sin(nx) n!=ecos(x)sin/parenleftbig sin(x)/parenrightbig . 6.Show by integrating that /integraldisplayπ −πeinxe−imxdx=/braceleftbigg 0, n/negationslash=m, 2π, n=m, and develop the formula for the complex Fourier coefﬁcients using this idea of orthogonality. 1.11 Applications of Fourier Series and Integrals 117 7.Find the function f(x)whose complex Fourier coefﬁcient function is given. a.C(λ)=/braceleftBig1,−1<λ< 1, 0,otherwise; b.C(λ)=e−|λ|. 8.Show that the complex Fourier coefﬁcient of f(x)=e−x2is C(λ)=e−λ2/4 2√π. Use a change of variable in the exponent. Y ou need to know that /integraldisplay∞ −∞e−z2dz=√π. 1.11 Applications of Fourier Series and Integrals Fourier series and integrals are among the most basic tools of applied mathe- matics. In what follows, we give just a few applications that do not fall withinthe scope of the rest of this book. A. Nonhomogeneous Differential Equation Many mechanical and electrical systems may be described by the differentialequation ¨y+α˙y+βy=f(t). The function f(t)is called the “forcing function,” βythe “restoring term,” and α˙ythe “damping term.” It is known (see Section 0.2) that: (1) a sine or cosine inf(t)will cause functions of the same period in y(t);( 2 )i f f(t)is broken down as a sum of simpler functions, y(t)c a nb eb r o k e nd o w ni nt h es a m ew a y . Suppose that f(t)is periodic with period 2 π, and let its Fourier series be f(t)=a 0+∞/summationdisplay n=1ancos(nt)+bnsin(nt). Then a particular solution y(t)will be periodic with period 2 π; it and its deriv- atives have Fourier series y(t)=A0+∞/summationdisplay n=1Ancos(nt)+Bnsin(nt), 118 Chapter 1 Fourier Series and Integrals ˙y(t)=∞/summationdisplay n=1−nAnsin(nt)+nBncos(nt), ¨y(t)=∞/summationdisplay n=1−n2Ancos(nt)−n2Bnsin(nt). Then the differential equation can be written in the form βA0+∞/summationdisplay n=1/parenleftbig −n2An+αnBn+βAn/parenrightbig cos(nt) +∞/summationdisplay n=1/parenleftbig −n2Bn−αnAn+βBn/parenrightbig sin(nt)=a0+∞/summationdisplay n=1ancos(nt)+bnsin(nt). The A’s and B’s are now determined by matching coefﬁcients βA0=a0, /parenleftbig β−n2/parenrightbig An+αnBn=an, −αnAn+/parenleftbig β−n2/parenrightbig Bn=bn. When these equations are solved for the A’s and B’s, we ﬁnd An=(β−n2)an−αnbn /Delta1,Bn=(β−n2)bn+αnan /Delta1, where /Delta1=/parenleftbig β−n2/parenrightbig2+α2n2. Now, given the function f,t h e a’s and b’s can be determined, thus giving the A’s and B’s. The function y(t)represented by the series found is the periodic part of the response. Depending on the initial conditions, there may also be atransient response, which dies out as tincreases. Example. Consider the differential equation ¨y+0.4˙y+1.04y=r(t). Ifr(t)=sin(nt), the corresponding particular solution is y(t)=−0.4ncos(nt)+(1.04−n2)sin(nt) (1.04−n2)2+(0.4n)2. 1.11 Applications of Fourier Series and Integrals 119 Next, suppose that r(t)is a square-wave function with Fourier series r(t)=∞/summationdisplay n=12(1−cos(nπ)) nπsin(nt). The corresponding response is y(t)=∞/summationdisplay n=12(1−cos(nπ)) nπ·−0.4ncos(nt)+(1.04−n2)sin(nt) (1.04−n2)2+(0.4n)2. Note that the term for n=1 has a small denominator, causing a large response. /square B. Boundary Value Problems By way of introduction to the next chapter, we apply the idea of Fourier seriesto the solution of the boundary value problem d 2u dx2+pu=f(x), 0<x<a, u(0)=0,u(a)=0. First, we will assume that f(x)is equal to its Fourier sine series, f(x)=∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg ,0<x<a. And second, we will assume that the solution u(x), which we are seeking, equals its Fourier sine series, u(x)=∞/summationdisplay n=1Bnsin/parenleftbiggnπx a/parenrightbigg ,0<x<a, and that this series may be differentiated twice to give d2u dx2=∞/summationdisplay n=1−/parenleftbiggn2π2 a2Bn/parenrightbigg sin/parenleftbiggnπx a/parenrightbigg ,0<x<a. When we insert the series forms for u,u/prime/prime,a n d f(x)into the differential equation, we ﬁnd that ∞/summationdisplay n=1/parenleftbigg −n2π2 a2Bn+pBn/parenrightbigg sin/parenleftbiggnπx a/parenrightbigg =∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg ,0<x<a. 120 Chapter 1 Fourier Series and Integrals Since the coefﬁcients of like terms in the two series must match, we may con- clude that /parenleftbigg p−n2π2 a2/parenrightbigg Bn=bn,n=1,2,3,.... If it should happen that p=m2π2/a2for some positive integer m,t h e r ei s no value of Bmthat satisﬁes /parenleftbigg p−m2π2 a2/parenrightbigg Bm=bm unless bm=0 also, in which case any value of Bmis satisfactory. In summary, we may say that Bn=bn p−n2π2/a2 and u(x)=∞/summationdisplay n=1a2bn a2p−n2π2sin/parenleftbiggnπx a/parenrightbigg , with the agreement that a zero denominator must be handled separately. Example. Consider the boundary value problem d2u dx2−u=− x,0<x<1, u(0)=0,u(1)=0. We have found previously that −x=∞/summationdisplay n=12(−1)n πnsin(nπx), 0<x<1. Thus, by the preceding development, the solution must be u(x)=∞/summationdisplay n=12 π(−1)n+1 n(n2π2+1)sin(nπx), 0<x<1. Although this particular series belongs to a known function, one would not, in general, know any formula for the solution u(x)other than its Fourier sine series. /square 1.11 Applications of Fourier Series and Integrals 121 C. The Sampling Theorem One of the most important results of information theory is the sampling the- orem, which is based on a combination of the Fourier series and the Fourierintegral in their complex forms. What the electrical engineer calls a signal isjust a function f(t)deﬁned for all t. If the function is integrable, there is a Fourier integral representation for it: f(t)=/integraldisplay ∞ −∞C(ω)exp(iωt)dω, C(ω)=1 2π/integraldisplay∞ −∞f(t)exp(−iωt)dt. As i g n a li sc a l l e d band limited if its Fourier transform is zero except in a ﬁnite interval, that is, if C(ω)=0,for|ω|>/Omega1. Then/Omega1is called the cutoff frequency. If fis band limited, we can write it in the form f(t)=/integraldisplay/Omega1 −/Omega1C(ω)exp(iωt)dω (1) because C(ω)is zero outside the interval −/Omega1<ω</Omega1 .W ef o c u so u ra t t e n t i o n on this interval by writing C(ω)as a Fourier series: C(ω)=∞/summationdisplay −∞cnexp/parenleftbigginπω /Omega1/parenrightbigg ,−/Omega1<ω</Omega1. (2) The (complex) coefﬁcients are cn=1 2/Omega1/integraldisplay/Omega1 −/Omega1C(ω)exp/parenleftbigg−inπω /Omega1/parenrightbigg dω. The point of the sampling theorem is to observe that the integral for cnac- tually is a value of f(t)at a particular time. In fact, from the integral Eq. (1), we see that cn=1 2/Omega1f/parenleftbigg−nπ /Omega1/parenrightbigg . Thus there is an easy way of ﬁnding the Fourier transform of a band-limited function. We have C(ω)=1 2/Omega1∞/summationdisplay −∞f/parenleftbigg−nπ /Omega1/parenrightbigg exp/parenleftbigginπω /Omega1/parenrightbigg 122 Chapter 1 Fourier Series and Integrals =1 2/Omega1∞/summationdisplay −∞f/parenleftbiggnπ /Omega1/parenrightbigg exp/parenleftbigg−inπω /Omega1/parenrightbigg ,−/Omega1<ω</Omega1. By utilizing Eq. (1) again, we can reconstruct f(t): f(t)=/integraldisplay/Omega1 −/Omega1C(ω)exp(iωt)dω =1 2/Omega1∞/summationdisplay −∞f/parenleftbiggnπ /Omega1/parenrightbigg/integraldisplay/Omega1 −/Omega1exp/parenleftbigg−inπω /Omega1/parenrightbigg exp(iωt)dω. Carrying out the integration and using the identity sin(θ)=(eiθ−e−iθ) 2i, we ﬁnd f(t)=∞/summationdisplay −∞f/parenleftbiggnπ /Omega1/parenrightbiggsin(/Omega1t−nπ) /Omega1t−nπ. (3) This is the main result of the sampling theorem. It says that the band-limited function f(t)may be reconstructed from the samples of fatt=0,±π//Omega1,... . It is difﬁcult to determine what functions are actually band limited. However,the process usually works quite well. In practice, we must use a ﬁnite series to approximate the function f(t)∼=N/summationdisplay −Nf/parenleftbiggnπ /Omega1/parenrightbiggsin(/Omega1t−nπ) /Omega1t−nπ. (4) Since the sampled values all come from the interval −Nπ//Omega1 toNπ//Omega1 ,t h e series cannot attempt to approximate the function outside that interval. An animation on the CD shows the effects of choosing Nand/Omega1. Example. The function f(t)=t2+2t (1+t2)2 is not band limited but can be approximated satisfactorily from a ﬁnite por- tion of the sum as in Eq. (4). Figure 13 shows results for N=100 (that is, 201 terms) and /Omega1=4 and 10. The target function is dashed. Notice the improvement. /square 1.11 Applications of Fourier Series and Integrals 123 Figure 13 Graphs of approximation using sampling: Eq. (4) with N=100 and /Omega1=4 and 10. EXERCISES 1.Use the method of Part A to ﬁnd a particular solution of d2u dt2+0.4du dt+1.04u=r(t), where r(t)is periodic with period 4 πand r(t)=t 4π,0<t<4π. 2.In the solution of Exercise 1, calculate the magnitude of the coefﬁcients of the Fourier series of u(t)(periodic part). 3.A simply supported beam of length Lhas a point load win the middle and axial tension T. (See Exercises in Section 0.3.) Its displacement u(x)satisﬁes the boundary value problem d2u dx2−T EIu=w EIh(x), 0<x<L, u(0)=0,u(L)=0, where h(x)is the “triangle function” h(x)=/braceleftbigg2x/L, 0<x<L/2, 2(L−x)/L,L/2<x<L. Use the method of Part B to ﬁnd u(x)as a sine series. 4.The inhomogeneity in the differential equation in Exercise 3 has a dis- continuous derivative. Find another way to solve the differential equation.Hint: Both u(x)and u /prime(x)must be continuous for 0 <x<L. 124 Chapter 1 Fourier Series and Integrals 5.U s et h es o f t w a r et oa p p r o x i m a t et h ef u n c t i o n f(t)=e−t2by the Sampling Theorem. Try /Omega1=4,N=2. 6.Simplify the ﬁnal formula for sampling to f(t)=sin(/Omega1t)∞/summationdisplay −∞f/parenleftbiggnπ /Omega1/parenrightbigg(−1)n /Omega1t−nπ. 1.12 Comments and References The ﬁrst use of trigonometric series occurred in the middle of the eighteenth century. Euler seems to have originated the use of orthogonality for the de-termination of coefﬁcients. In the earl y nineteenth century Fourier made ex- tensive use of trigonometric series in studying problems of heat conduction(see Chapter 2). His claim, that an arbitrary function could be represented asa trigonometric series, led to an extensive reexamination of the foundations of calculus. Fourier seems to have been among the ﬁrst to recognize that a function might have different analytical expressions in different places. Dirichlet established sufﬁcient conditions (similar to those of our conver- gence theorem) for the convergence of Fourier series around 1830. Later, Rie-mann was led to redeﬁne the integral as part of his attempt to discover condi-tions on a function necessary and sufﬁcient for the convergence of its Fourier series. This problem has never been solved. Many other great mathematicians have founded important theories (the theory of sets, for one) in the course ofstudying Fourier series, and they continue to be a subject of active research. Anentertaining and readable account of the history and uses of Fourier series isinThe Mathematical Experience , by Davis and Hersh. (See the Bibliography.) Historical interest aside, Fourier series and integrals are extremely impor- tant in applied mathematics, physics, and engineering, and they merit further study. A superbly written and organized book is T olstov’s Fourier Series .I t s mathematical prerequisites are not too high. Fourier Series and Boundary Value Problems by Churchill and Brown is a standard text for some engineering ap- plications. About 1960 it became clear that the numerical computation of Fourier co- efﬁcients could be rearranged to achieve dramatic reductions in the amountof arithmetic required. The result, called the fast Fourier transform ,o rF F T ,h a s revolutionized the use of Fourier series in applications. See The Fast Fourier Transform by James S. Walker. The sampling theorem mentioned in the last section has become bread and butter in communications engineering. For extensive information on this aswell as the FFT, see Integral and Discrete Transforms with Applications and Error Analysis , by A.J. Jerri. Miscellaneous Exercises 125 Chapter Review See the CD for review questions. Miscellaneous Exercises 1.Find the Fourier sine series of the trapezoidal function given for 0<x<π by f(x)=/braceleftBiggx/α, 0<x<α, 1,α < x<π−α, (π−x)/α, π −α< x<π. 2.Show that the series found in Exercise 1 converges uniformly. 3.When αapproaches 0, the function of Exercise 1 approaches a square wave. Do the sine coefﬁcients found in Exercise 1 approach those of asquare wave? 4.Find the Fourier cosine series of the function F(x)=/integraldisplay x 0f(t)dt, where fdenotes the function in Exercise 1. Sketch. 5.Find the Fourier sine series of the function given in the interval 0 <x<a by the formula ( αis a parameter between 0 and 1) f(x)=  hx αa, 0<x<αa, h(a−x) (1−α)a,α a<x<a. 6.Sketch the function of Exercise 5. T o what does its Fourier sine series converge at x=0? at x=αa?a tx=a? 7.Suppose that f(x)=1, 0<x<a. Sketch and ﬁnd the Fourier series of the following extensions of f(x): a.even extension; b.odd extension; c.periodic extension (period a); d.even periodic extension; 126 Chapter 1 Fourier Series and Integrals e.odd periodic extension; f.the one corresponding to f(x)=x,−a<x<0. 8.Perform the same task as in Exercise 7, but f(x)=0, 0<x<a. 9.Find the Fourier series of the function given by f(x)=/braceleftBig0,−a<x<0, 2x,0<x<a. Sketch the graph of f(x)and its periodic extension. T o what values does the series converge at x=− a,x=− a/2,x=0,x=a,a n d x=2a? 10. Sketch the odd periodic extension and ﬁnd the Fourier sine series of the function given by f(x)=/braceleftbigg1,0<x<π 2, 1 2,π 2<x<π. T o what values does the series converge at x=0,x=π/2,x=π,x= 3π/2, and x=2π? 11. Sketch the even periodic extension of the function given in Exercise 10. Find its Fourier cosine series. T o what values does the series converge at x=0,x=π/2,x=π,x=3π/2, and x=2π? 12. Find the Fourier cosine series of the function g(x)=/braceleftBig1−x,0<x<1, 0, 1<x<2. Sketch the graph of the sum of the cosine series. 13. Find the Fourier sine series of the function deﬁned by f(x)=1−2x, 0<x<1. Sketch the graph of the odd periodic extension of f(x),a n d determine the sum of the sine series at points where the graph has ajump. 14. Following the same requirements as in Exercise 13, use the cosine series and the even periodic extension. 15. Find the Fourier series of the function given by f(x)=  0, −π< x<− π 2, sin(2x),−π 2<x<π 2, 0,π 2<x<π. Sketch the graph of the function. Miscellaneous Exercises 127 16.Show that the function given by the formula f(x)=(π−x)/2, 0<x< 2π, has the Fourier series f(x)=∞/summationdisplay 1sin(nx) n,0<x<2π. Sketch f(x)and its periodic extension. 17.Use complex methods and a ﬁnite geometric series to show that N/summationdisplay n=1cos(nx)=sin/parenleftbig (N+1 2)x/parenrightbig −sin(1 2x) 2s i n(1 2x). Then use trigonometric identities to identify N/summationdisplay n=1cos(nx)=sin(1 2Nx)cos/parenleftbig1 2(N+1)x/parenrightbig sin(1 2x). 18.Identify the partial sums of the Fourier series in Exercise 16 as SN(x)=N/summationdisplay n=1sin(nx) n. The series of Exercise 17 is S/prime N(x). Use this information to locate the max- ima and minima of SN(x)in the interval 0 ≤x≤π.F i n dt h ev a l u eo f SN(x)at the ﬁrst point in the interval 0 <x<π where S/prime N(x)=0 for N=5. Compare to (π−x)/2a tt h a tp o i n t . 19.Find the Fourier sine series of the function given by f(x)=/braceleftBigg sin/parenleftbiggπx a/parenrightbigg ,0<x<a, 0, a<x<π, assuming that 0 <a<π. 20.Find the Fourier cosine series of the function given in Exercise 19. 21.Find the Fourier integral representation of the function given by f(x)=/braceleftBig1,0<x<a, 0,x<0o r x>a. 128 Chapter 1 Fourier Series and Integrals 22. Find the Fourier sine and cosine integral representations of the function given by f(x)=/braceleftBigga−x a,0<x<a, 0, a<x. 23. Find the Fourier sine integral representation of the function f(x)=/braceleftbigg sin(x),0<x<π, 0,π < x. 24. Find the Fourier integral representation of the function f(x)=/braceleftBig1//epsilon1, α < x<α+/epsilon1, 0, elsewhere. 25. Use integration by parts to establish the equality /integraldisplay∞ 0e−λcos(λx)dλ=1 1+x2. 26. T h ee q u a t i o ni nE x e r c i s e2 5i sv a l i df o ra l l x. Explain why its validity implies that 2 π/integraldisplay∞ 0cos(λx) 1+x2dx=e−λ,λ > 0. 27. Integrate both sides of the equality in Exercise 25 from 0 to tto derive the equality /integraldisplay∞ 0e−λsin(λt) λdλ=tan−1(t). 28. Does the equality in Exercise 27 imply that 2 π/integraldisplay∞ 0tan−1(t)sin(λt)dt=e−λ λ? 29. F r o mE x e r c i s e2 7d e r i v et h ee q u a l i t y /integraldisplay∞ 01−e−λ λsin(λx)dλ=π 2−tan−1(x), x>0. 30. Without using integration, obtain the Fourier series (period 2 π)o fe a c h of the following functions: Miscellaneous Exercises 129 a.2+4s i n(50x)−12 cos(41x); c.sin(4x+2); e.cos3(x);b.sin2(5x); d.sin(3x)cos(5x); f.cos(2x+1 3π). 31.Let the function f(x)be given in the interval 0 <x<1 by the formula f(x)=1−x. Find (a) a sine series, (b) a cosine series, (c) a sine integral, and (d) a cosine integral that equals the given function for 0 <x<1. In each case, sketch the function to which the series or integral converges in the inter-val−2<x<2. 32.Verify the Fourier integral /integraldisplay ∞ 0cos(λq)exp/parenleftbig −λ2t/parenrightbig dλ=/radicalbiggπ 4texp/parenleftbigg −q2 4t/parenrightbigg ,t>0, by transforming the left-hand side according to these steps: (a) Con- vert to an integral from −∞ to∞by using the evenness of the inte- grand; (b) replace cos (λq)by exp (iλq)(justify this step); (c) complete the square in the exponent; (d) change the variable of integration; (e) usethe equality /integraldisplay ∞ −∞exp/parenleftbig −u2/parenrightbig du=√π. 33.Approximate the ﬁrst seven cosine coefﬁcients (ˆa0,ˆa1,...,ˆa6)of the function f(x)=1 1+x2,0<x<1. 34.Use Fourier sine series representations of u(x)and of the function f(x)= x,0<x<a, to solve the boundary value problem d2u dx2−γ2u=− x,0<x<a, u(0)=0,u(a)=0. 35–43. For each of these exercises, a.ﬁnd the Fourier cosine series of the function; b.determine the value to which the series converges at the given values ofx; 130 Chapter 1 Fourier Series and Integrals c.sketch the even periodic extension of the given function for at least two periods. 44–52. For each of these exercises, a.ﬁnd the Fourier sine series of the function; b.determine the value to which the series converges at the given values ofx; c.sketch the odd periodic extension of the given function for at least two periods. 35. & 44. f(x)=  0, 0<x<a 3, x−a 3,a 3<x<2a 3,x=0,a 3,a,−a 2, a 3,2a 3<x<a. 36. & 45. f(x)=  1 2,0<x<a 2,x=a 2,2a,0,−a, 1,a 2<x<a. 37. & 46. f(x)=  2x a, 0<x<a 2,x=0,a 2,a,3a 2, (3a−2x) 2a,a 2<x<a. 38. & 47. f(x)=  x,0<x<a 2,x=0,a,−a 2, a 2,a 2<x<a. 39. & 48. f(x)=(a−x) a,0<x<a,x=0,a,−a 2. 40. & 49. f(x)=  0,0<x<a 4, 1,a 4<x<3a 4,x=0,a 4,a 2,a,−3a 4, 0,3a 4<x<a. 41. & 50. f(x)=x(a−x), 0<x<a,x=0,−a,−a 2. 42. & 51. f(x)=ekx,0<x<a,x=0,a 2,a,−a. Miscellaneous Exercises 131 43. & 52. f(x)=  0,0<x<a 2,x=− a,a 2,a, 1,a 2<x<a. 53–58. For each of these exercises, a.ﬁnd the Fourier cosine integral representation of the function; b.sketch the even extension of the function. 59–64. For each of these exercises, a.ﬁnd the Fourier sine integral representation of the function; b.sketch the odd extension of the function. 53. & 59. f(x)=e−x,0<x. 54. & 60. f(x)=/braceleftbigg e−x,0<x<a, 0, a<x. 55. & 61. f(x)=/braceleftbigg 1,0<x<b, 0,b<x. 56. & 62. f(x)=/braceleftbigg cos(x),0<x<π, 0,π < x. 57. & 63. f(x)=/braceleftBig1−x,0<x<1, 0, 1<x. 58. & 64. f(x)=/braceleftBigg1, 0<x<1, 2−x,1<x<2, 0, 2<x. 65.(Cesaro summability.) Let f(x)be a periodic function with period 2 π whose Fourier coefﬁcients are a0,a1,b1,.... Then, the partial sum SN(x)=a0+N/summationdisplay n=1ancos(nx)+bnsin(nx) is an approximation to f(x)iffis sectionally smooth and Nis large enough. The average of these approximations is σN(x)=1 N/parenleftbig S1(x)+···+ SN(x)/parenrightbig . It is known that σN(x)converges uniformly to f(x)iffis continuous. Show that σN(x)=a0+N/summationdisplay n=1N+1−n N/parenleftbig ancos(nx)+bnsin(nx)/parenrightbig . 132 Chapter 1 Fourier Series and Integrals 66. In analogy to Lemma 2 of Section 7, prove that N−1/summationdisplay n=0sin/parenleftbigg/parenleftbigg n+1 2/parenrightbigg y/parenrightbigg =sin2(1 2Ny) sin(1 2y). 67. Following the lines of Section 7, show that σN(x)−f(x)=1 2Nπ/integraldisplayπ −π/bracketleftbig f(x+y)−f(x)/bracketrightbig/parenleftbiggsin(1 2Ny) sin(1 2y)/parenrightbigg2 dy. This equality is the key to the proof of uniform convergence mentioned in Exercise 65. 68. In a study of river freezing, E.P . Foltyn and H.T. Shen [St. Lawrence River freeze-up forecast, Journal of Waterway, Port, Coastal and Ocean Engi- neering ,112 (1986): 467–481] use data spanning 33 years to ﬁnd this Fourier series representation of the air temperature in Massena, NY: T(t)=a0+∞/summationdisplay n=1ancos(2nπt)+bnsin(2nπt). Here Tis temperature in◦C,tis time in years, and the origin is Oct. 1. The ﬁrst coefﬁcients were found to be a0=6.638,a1=5.870,b1=− 13.094,a2=0.166,b2=0.583, and the remaining coefﬁcients were all less than 0.3 in absolute value. The authors decided to exclude all the terms from a2and b2up, so their approximation could be written T(t)∼=a0+Asin(2πt+θ). a.Find the average temperature in Massena. b.Find A, the amplitude of the annual variation, and the phase angle θ. c.Find the approximate date when the minimum temperature occurs. d.Find the dates when the approximate temperature passes through 0. e.Discuss the effect on the answer to part dif the next two terms of the series were included. 69. In each part that follows, a function is equated to its Fourier series as justiﬁed by the Theorem of Section 3 . By evaluating both sides of the equality at an appropriate value of x, derive the second equality. Miscellaneous Exercises 133 a.|x|=1 2−4 π2∞/summationdisplay k=01 (2k+1)2cos/parenleftbig (2k+1)πx/parenrightbig ,−1<x<1, π2 8=1+1 9+1 25+··· ; b.4 π∞/summationdisplay k=01 2k+1sin/parenleftbig (2k+1)πx/parenrightbig =/braceleftBig1, 0<x<1, −1,−1<x<0, π 4=1−1 3+1 5−1 7+··· ; c.|sin(x)|=2 π−4 π∞/summationdisplay n=11 4n2−1cos(2nx), 1 2=1 3+1 15+1 35+··· . This page intentionally left blank The Heat Equation CHAPTER2 2.1 Derivation and Boundary Conditions As the ﬁrst example of the derivation of a partial differential equation, we consider the problem of describing the temperature in a rod or bar of heat-conducting material. In order to simplify the problem as much as possible, weshall assume that the rod has a uniform cross section (like an extrusion) and that the temperature does not vary from point to point on a section. Thus, if we use a coordinate system as suggested in Fig. 1, we may say that the temper-ature depends only on position xand time t. The basic idea in developing the partial differential equation is to apply the laws of physics to a small piece of the rod. Speciﬁcally, we apply the law ofconservation of energy to a slice of the rod that lies between xand x+/Delta1x (Fig. 2). The law of conservation of energy states that the amount of heat that enters a region plus what is generated inside is equal to the amount of heat that leavesplus the amount stored. The law is equally valid in terms of rates per unit timeinstead of amounts. Now let q(x,t)be the heat ﬂux at point xand time t. The dimensions of q are 1[q]=H/tL2,a n d qis taken to be positive when heat ﬂows to the right. T h er a t ea tw h i c hh e a ti se n t e r i n gt h es l i c et h r o u g ht h es u r f a c ea t xisAq(x,t), where Ais the area of a cross section. The rate at which heat is leaving the slice through the surface at x+/Delta1xisAq(x+/Delta1x,t). 1Square brackets are used to symbolize “dimension of.” H=heat energy, t=time, T= temperature, L=length, m=mass, and so forth. 135 136 Chapter 2 The Heat Equation Figure 1 R o do fh e a t - c o n d u c t i n gm a t e r i a l . Figure 2 Slice cut from rod. The rate of heat storage in the slice of material is proportional to the rate of change of temperature. Thus, if ρis the density and cis the heat capacity per unit mass ([c]=H/mT), we may approximate the rate of heat storage in the slice by ρcA/Delta1x∂u ∂t(x,t), where u(x,t)is the temperature. There are other ways in which heat may enter (or leave) the section of rod we are looking at. One possibility is that heat is transferred by radiation orconvection from (or to) a surrounding medium. Another is that heat is con-verted from another form of energy — for instance, by resistance to an elec- trical current or by chemical or nuclear reaction. All of these possibilities we lump together in a “generation rate.” If the rate of generation per unit volume isg,[g]=H/tL 3, then the rate at which heat is generated in the slice is A/Delta1xg. (Note that gmay depend on x,t,a n de v e n u.) We have now quantiﬁed the law of conservation of energy for the slice of rod in the form Aq(x,t)+A/Delta1xg=Aq(x+/Delta1x,t)+A/Delta1xρc∂u ∂t. (1) After some algebraic manipulation, we have q(x,t)−q(x+/Delta1x,t) /Delta1x+g=ρc∂u ∂t. The ratio q(x+/Delta1x,t)−q(x,t) /Delta1x Chapter 2 The Heat Equation 137 should be recognized as a difference quotient. If we allow /Delta1xto decrease, this quotient becomes, in the limit, lim /Delta1x→0q(x+/Delta1x,t)−q(x,t) /Delta1x=∂q ∂x. The limit process thus leaves the law of conservation of energy in the form −∂q ∂x+g=ρc∂u ∂t. (2) We are not ﬁnished, since there are two dependent variables, qand u,i nt h i s equation. We need another equation relating qand u.T h i sr e l a t i o ni sF o u r i e r ’ s law of heat conduction, which in one dimension may be written q=−κ∂u ∂x. In words, heat ﬂows downhill ( qis positive when ∂u/∂xis negative) at a rate proportional to the gradient of the temperature. The proportionality factor κ, called the thermal conductivity ,m a yd e p e n do n xif the rod is not uniform and also may depend on temperature. However, we will usually assume it to be aconstant. Substituting Fourier’s law in the heat balance equation yields ∂ ∂x/parenleftbigg κ∂u ∂x/parenrightbigg +g=ρc∂u ∂t. (3) Note that κ,ρ,a n d cmay all be functions. If, however, they are independent ofx,t,a n d u,w em a yw r i t e ∂2u ∂x2+g κ=ρc κ∂u ∂t. (4) The equation is applicable where the rod is located and after the experiment starts: for 0 <x<aand for t>0. The quantity κ/ρcis often written as k and is called the thermal diffusivity . Table 1 shows approximate values of these constants for several materials. For some time we will be working with the heat equation without genera- tion, ∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, (5) w h i c h ,t or e v i e w ,i ss u p p o s e dt od e s c r i b et h et e m p e r a t u r e uin a rod of length awith uniform properties and cross section, in which no heat is generated and whose cylindrical surface is insulated. Some qualitative features can be obtained from the partial differential equa- tion itself. Suppose that u(x,t)satisﬁes the heat equation, and imagine a graph 138 Chapter 2 The Heat Equation c ρκ k=κ ρc Material/parenleftbigcal g◦C/parenrightbig/parenleftbigg cm3/parenrightbig/parenleftbigcal sc m◦C/parenrightbig/parenleftbigcm2 s/parenrightbig Aluminum 0 .21 2 .70 .48 0 .83 Copper 0 .094 8 .90 .92 1 .1 Steel 0 .11 7 .80 .11 0 .13 Glass 0 .15 2 .60 .0014 0 .0036 Concrete 0 .16 2 .30 .0041 0 .011 Ice 0 .48 0 .92 0 .004 0 .009 Table 1 Typical values of constants ofu(x,t∗), with t∗a ﬁxed time. If a portion of the graph is shaped like U,Jor backwards J, the graph is concave there — that is, ∂2u/∂x2is positive. Then by the heat equation, ∂u/∂tmust be positive as well. Vice versa, when the graph is convex, ∂2u/∂x2and hence ∂u/∂tmust be negative. Thus, a solution of the heat equation tends to straighten out. This equation alone is not enough information to completely specify the temperature, however. Each of the functions u(x,t)=x2+2kt, u(x,t)=e−ktsin(x) satisﬁes the partial differential equation, and so do their sum and difference. Clearly this is not a satisfactory situation either from the mathematical or physical viewpoint; we would like the temperature to be uniquely determined. More conditions must be placed on the function u. The appropriate additional conditions are those that describe the initial temperature distribution in therod and what is happening at the ends of the rod. The initial condition is described mathematically as u(x,0)=f(x), 0<x<a, where f(x)is a given function of xalone. In this way, we specify the initial temperature at every point of the rod. The boundary conditions may take a variety of forms. First, the temperature at either end may be held constant, for instance, by exposing the end to an ice-water bath or to condensing steam. We can describe such conditions by theequations u(0,t)=T 0,u(a,t)=T1,t>0, where T0andT1may be the same or different. More generally, the temperature at the boundary may be controlled in some way, without being held constant.Ifx 0symbolizes an endpoint, the condition is u(x0,t)=α(t), (6) Chapter 2 The Heat Equation 139 where αis a function of time. Of course, the case of a constant function is included here. This type of boundary condition is called a Dirichlet condition orcondition of the ﬁrst kind . Another possibility is that the heat ﬂow rate is controlled. Since Fourier’s law associates the heat ﬂow rate and the gradient of the temperature, we can write ∂u ∂x(x0,t)=β(t), (7) where βis a function of time. This is called a Neumann condition orcondition of the second kind .W em o s tf r e q u e n t l yt a k e β(t)to be identically zero. Then the condition ∂u ∂x(x0,t)=0 corresponds to an insulated surface, for this equation says that the heat ﬂow is zero. Still another possible boundary condition is c1u(x0,t)+c2∂u ∂x(x0,t)=γ(t), (8) called third kind or a Robin condition . This kind of condition can also be real- ized physically. If the surface at x=ais exposed to air or other ﬂuid, then the heat conducted up to that surface from inside the rod is carried away by con-vection. Newton’s law of cooling says that the rate at which heat is transferredfrom the body to the ﬂuid is proportional to the difference in temperaturebetween the body and the ﬂuid. In symbols, we have q(a,t)=h/parenleftbig u(a,t)−T(t)/parenrightbig , (9) where T(t)is the air temperature. After application of Fourier’s law, this be- comes −κ∂u ∂x(a,t)=hu(a,t)−hT(t). (10) This equation can be put into the form of Eq. (8). (Note: his called the con- vection coefﬁcient or heat transfer coefﬁcient; [ h]=H/L2tT.) A l lo ft h eb o u n d a r yc o n d i t i o n sg i v e ni nE q s .( 6 ) ,( 7 ) ,a n d( 8 )i n v o l v et h e function uand/or its derivative at one point. If more than one point is in- volved, the boundary condition is called mixed . For example, if a uniform rod is bent into a ring and the ends x=0a n d x=aare joined, appropriate bound- ary conditions would be 140 Chapter 2 The Heat Equation u(0,t)=u(a,t), t>0, (11) ∂u ∂x(0,t)=∂u ∂x(a,t), t>0, (12) both of mixed type. Many other kinds of boundary conditions exist and are even realizable, but the four kinds already mentioned here are the most commonly encountered.A ni m p o r t a n tf e a t u r ec o m m o nt oa l lf o u rt y p e si st h a tt h e yi n v o l v ea linear operation on the function u. The heat equation, an initial condition, and a boundary condition for each e n df o r mw h a ti sc a l l e da n initial value–boundary value problem . For instance, o n ep o s s i b l ep r o b l e mw o u l db e ∂ 2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (13) u(0,t)=T0, 0<t, (14) −κ∂u ∂x(a,t)=h/parenleftbig u(a,t)−T1/parenrightbig ,0<t, (15) u(x,0)=f(x), 0<x<a. (16) Notice that the boundary conditions may be of different kinds at different ends. Although we shall not prove it, it is true that there is one, and only one, solution to a complete initial value–boundary value problem. We have derived the heat equation (4) as a mathematical model for the tem- perature in a “rod,” suggesting an object that is much longer than it is wide. The equation applies equally well to a “slab,” an object that is much wider thanit is thick. The important feature is that we may assume in either case that the temperature varies in only one space direction (along the length of the rodor the thickness of the slab). In Chapter 5, we derive a multidimensional heatequation. It may come as a surprise that the partial differential equations of this sec- tion have another completely different but equally important physical inter-pretation. Suppose that a static medium occupies a region of space betweenx=0a n d x=a(a slab!) and that we wish to study the concentration u,m e a - sured in units of mass per unit volume, of another substance, whose moleculesor atoms can move, or diffuse, through the medium. We assume that the con-centration is a function of xand tonly and designate q(x,t)to be the mass ﬂux([q]=m/tL 2). Then the principle of conservation of mass may be applied to a layer of the medium between xand x+/Delta1xto obtain the equation q(x,t)+/Delta1xg=q(x+/Delta1x,t)+/Delta1x∂u ∂t(x,t). (17) Chapter 2 The Heat Equation 141 When we rearrange Eq. (17) and take the limit as /Delta1xapproaches 0, it becomes −∂q ∂x+g=∂u ∂t. (18) In these equations, gis a “generation rate” ([g]=m/tL3),af u n c t i o nt h a ta c - counts for any gain or loss of the substance from the layer by means other thanmovement in the x-direction. For example, the substance may participate in a chemical reaction with the medium at a rate proportional to its concentration(a ﬁrst-order reaction) so that in this case the generation rate is g=− ku(x,t). (19) The concentration and the mass ﬂux are linked by a phenomenological re- lation called Fick’s ﬁrst law , written in one dimension as q=− D∂u ∂x. (20) In words, the diffusing substance moves toward regions of lower concentra- tion at a rate proportional to the gradient of the concentration. The coefﬁcientof proportionality D, usually constant, is called the diffusivity . By combining Fick’s law with Eq. (18) arising from the conservation of mass, we obtain thediffusion equation ∂ 2u ∂x2+g D=1 D∂u ∂t. (21) At a boundary of the medium, the concentration of the diffusing substance may be controlled, leading to a condition like Eq. (6), or the ﬂux of the sub- stance may be controlled, leading via Fick’s law to a condition like Eq. (7); an impermeable surface corresponds to zero ﬂux. If a boundary is covered with apermeable ﬁlm, then the ﬂux through the ﬁlm is usually taken to be propor-tional to the difference in concentrations on the two sides of the ﬁlm. Supposethat the surface in question is at x=a. Then these statements may be expressed symbolically as q(a,t)=h/parenleftbig u(a,t)−C(t)/parenrightbig , (22) where the proportionality constant his called the ﬁlm coefﬁcient and Cis the concentration outside the medium. Using Fick’s law here leads to the equation −D∂u ∂x(a,t)=hu(a,t)−hC(t). (23) This equation is analogous to Eq. (10) and can be put into the form of Eq. (8). 142 Chapter 2 The Heat Equation EXERCISES 1.Give a physical interpretation for the problem in Eqs. (13)–(16). 2.Verify that the following functions are solutions of the heat equation (5): u(x,t)=exp/parenleftbig −λ2kt/parenrightbig cos(λx), u(x,t)=exp/parenleftbig −λ2kt/parenrightbig sin(λx). 3.Suppose that the rod exchanges heat through the cylindrical surface by con- vection with a surrounding ﬂuid at temperature U(constant). Newton’s law of cooling says that the rate of heat transfer is proportional to exposed area and temperature difference. What is gin Eq. (1)? What form does Eq. (4) take? 4.Suppose that the end of the rod at x=0 is immersed in an insulated con- tainer of water or other ﬂuid; that the temperature of the ﬂuid is the sameas the temperature of the end of the rod; that the heat capacity of the ﬂuidisCunits of heat per degree. Show that this situation is represented math- ematically by the equation C∂u ∂t(0,t)=κA∂u ∂x(0,t), where Ais the cross-sectional area of the rod. 5.Put Eq. (10) into Eq. (8) form. Notice that the signs still indicate that heat ﬂows in the direction of lower temperature. That is, if u(a,t)>T(t),t h e n q(a,t)is positive and the gradient of uis negative. Show that, if the surface atx=0 (left end) is exposed to convection, the boundary condition would read κ∂u ∂x(0,t)=hu(0,t)−hT(t). Explain the signs. 6.Suppose the surface at x=ais exposed to radiation. The Stefan–Boltzmann law of radiation says that the rate of radiation heat transfer is proportional to the difference of the fourth powers of the absolute temperatures of the bodies: q(a,t)=σ/parenleftbig u4(a,t)−T4/parenrightbig . Use this equation and Fourier’s law to obtain a boundary condition for radiation at x=ato a body at temperature T. 7.The difference cited in Exercise 6 may be written u4−T4=(u−T)/parenleftbig u3+u2T+uT2+T3/parenrightbig . 2.2 Steady-State Temperatures 143 Under what conditions might the second factor on the right be taken ap- proximately constant? If the factor were constant, the boundary conditionwould be linear. 8.Interpret this problem in terms of diffusion. Be sure to explain how theboundary conditions could arise physically. D∂ 2u ∂x2+K=∂u ∂t, 0<x<a,0<t, u(0,t)=C,∂u ∂x(a,t)=0,0<t, u(x,0)=0, 0<x<a. 2.2 Steady-State Temperatures Before tackling a complete heat conduction problem, we shall solve a simpli- ﬁed version called the steady-state or equilibrium problem. We begin with this problem: ∂2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (1) u(0,t)=T0, 0<t, (2) u(a,t)=T1, 0<t, (3) u(x,0)=f(x), 0<x<a. (4) We may think of u(x,t)as the temperature in a cylindrical rod, with insulated lateral surface, whose ends are held at constant temperatures T0and T1. Experience indicates that after a long time under the same conditions, the variation of temperature with time dies away. In terms of the function u(x,t) that represents temperature, we thus expect that the limit of u(x,t),a sttends to inﬁnity, exists and depends only on x, lim t→∞u(x,t)=v(x), and also that lim t→∞∂u ∂t=0. The function v(x),c a l l e dt h e steady-state temperature distribution ,m u s ts t i l l satisfy the boundary conditions and the heat equation, which are valid for allt>0. 144 Chapter 2 The Heat Equation Example. For the preceding problem, v(x)should be the solution to the problem d2v dx2=0,0<x<a, (5) v(0)=T0,v ( a)=T1. (6) On integrating the differential equation twice, we ﬁnd dv dx=A,v ( x)=Ax+B. The constants Aand Ba r et ob ec h o s e ns ot h a t v(x)satisﬁes the boundary conditions: v(0)=B=T0,v ( a)=Aa+B=T1. When the two equations are solved for Aand B, the steady-state distribution becomes v(x)=T0+(T1−T0)x a. (7) Of course, Eqs. (5) and (6), which together form the steady-state problem corresponding to Eqs. (1)–(4), could have been derived from scratch, as was done in Chapter 0, Section 3. Here, however, we see it as part of a more com- prehensive problem. /square We can establish this rule for setting up the steady-state problem corre- sponding to a given heat conduction prob lem: Take limits in all equations that are valid for large t(the partial differential equation and the boundary condi- tions), replacing uand its derivatives with respect to xbyvand its derivatives, and replacing ∂u/∂tby 0. Example. Find the steady-state problem and solution of Eqs. (13)–(16) of Section 2.1, which were ∂2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (8) u(0,t)=T0, 0<t, (9) −κ∂u ∂x(a,t)=h/parenleftbig u(a,t)−T1/parenrightbig ,0<t, (10) u(x,0)=f(x), 0<x<a. (11) 2.2 Steady-State Temperatures 145 When the rule given here is applied to this problem, we are led to the following equations: d2v dx2=0,0<x<a, v(0)=T0,−κv/prime(a)=h/parenleftbig v(a)−T1/parenrightbig . The solution of the differential equation is v(x)=A+Bx. The boundary con- ditions require that Aand Bsatisfy v(0)=T0:A=T0, −κv/prime(a)=h/parenleftbig v(a)−T1/parenrightbig :−κB=h(A+Ba−T1). Solving simultaneously, we ﬁnd A=T0,B=h(T1−T0) κ+ha. Thus the steady-state solution of Eqs. (8)–(11) is v(x)=T0+xh(T1−T0) κ+ha. (12) /square In both of these examples, the steady-state temperature distribution has been uniquely determined by the differential equation and boundary condi-tions. This is usually the case, but not always. Example. For the problem ∂2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (13) ∂u ∂x(0,t)=0, 0<t, (14) ∂u ∂x(a,t)=0, 0<t, (15) u(x,0)=f(x), 0<x<a, (16) which describes the temperature in an insulated rod that also has insulated ends, the corresponding steady-state problem for v(x)=lim t→∞u(x,t)is d2v dx2=0,0<x<a, dv dx(0)=0,dv dx(a)=0. 146 Chapter 2 The Heat Equation It is easy to see that v(x)=T(any constant) is a solution to this problem. However, there is no information to tell what value Tshould take. Thus, this boundary value problem has inﬁnitely many solutions. /square It should not be supposed that every steady-state temperature distribution has a straight-line graph. This is certainly not the case in the problem of Exer-cise 1. While the steady-state solution gives us some valuable information about the solution of an initial value–boundary value problem, it also is importantas the ﬁrst step in ﬁnding the complete solution. We now isolate the “rest” ofthe unknown temperature u(x,t)by deﬁning the transient temperature distri- bution , w(x,t)=u(x,t)−v(x). The name transient is appropriate because, according to our assumptions about the behavior of ufor large values of t,w ee x p e c t w(x,t)to tend to zero asttends to inﬁnity. In general, the transient also satisﬁes an initial value–boundary value prob- lem that is similar to the original one but is distinguished by having a homo-geneous partial differential equation and boundary conditions. T o illustratethis point, we shall treat the problem stated in Eqs. (1)–(4) whose steady-statesolution is given by Eq. (7). By using the equality u(x,t)=w(x,t)+v(x)and what we know about v— that is, Eqs. (5) and (6) — we make the original problem for u(x,t)into a new problem for w(x,t),a ss h o w ni nw h a tf o l l o w s . ∂ 2u ∂x2=∂2w ∂x2+d2v dx2by rules of calculus , =∂2w ∂x2because of Eq. (5) , ∂u ∂t=∂w ∂t+dv dtby rules of calculus , =∂w ∂tv(x)does not depend on t, ∂2w ∂x2=1 k∂w ∂tby substituting into Eq. (1) , u(0,t)=w(0,t)+v(0) by deﬁnition of the transient , T0=w(0,t)+T0 from Eqs. (2) and (6) , u(a,t)=w(a,t)+v(a) by deﬁnition of the transient , T1=w(a,t)+T1 from Eqs. (3) and (6) , u(x,0)=w(x,0)+v(x) by deﬁnition of the transient , f(x)=w(x,0)+T0+(T1−T0)x/afrom Eqs. (4) and (7) . 2.2 Steady-State Temperatures 147 Now we collect and simplify these transformations of Eqs. (1)–(4) to get an initial value–boundary value problem for w: ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, (17) w(0,t)=0, 0<t, (18) w(a,t)=0, 0<t, (19) w(x,0)=f(x)−/bracketleftbigg T0+(T1−T0)x a/bracketrightbigg (20) ≡g(x), 0<x<a. (21) In the last line, we have just renamed the combination of f(x)andv(x)in Eq. (20). I nt h en e x ts e c t i o n ,w es h a l ls e eh o wt h ep r o b l e mf o rt h et r a n s i e n tt e m p e r - ature can be solved. The mathematical purpose of setting up the steady-state problem and then the transient problem is that the transient problem is ho- mogeneous .Y o uc a nt e s tt h i sb yt r y i n g w(x,t)≡0: This function satisﬁes the partial differential equation (17) and the boundary conditions (18) and (19). Itis crucially important for the method we will develop to have a homogeneouspartial differential equation and boundary conditions. EXERCISES S e ee x t r ae x e r c i s e so nt h eC D . 1.State and solve the steady-state problem corresponding to ∂2u ∂x2−γ2(u−U)=1 k∂u ∂t, 0<x<a,0<t, u(0,t)=T0,u(a,t)=T1,0<t, u(x,0)=0,0<x<a. Also ﬁnd a physical interpretation of this problem. (See Exercise 3, Sec- tion 1.) 2.State the problem satisﬁed by the transient temperature distribution corre-sponding to the problem in Exercise 1. 3.Obtain the steady-state solution of the problem ∂2u ∂x2+γ2(u−T)=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T,u(a,t)=T,0<t, u(x,0)=T1x a,0<x<a. 148 Chapter 2 The Heat Equation Can you think of a physical interpretation of this problem? Note the differ- ence between the partial differential equation in this exercise and in Exer-cise 1. What happens if γ=π/a? 4.State the initial value–boundary value problem satisﬁed by the transient temperature distribution corresponding to Eqs. (8)–(11). 5.Find the steady-state solution of the problem ∂ ∂x/parenleftbigg κ∂u ∂x/parenrightbigg =cρ∂u ∂t,0<x<a,0<t, u(0,t)=T0,u(a,t)=T1,0<t if the conductivity varies in a linear fashion with x:κ(x)=κ0+βx,w h e r e κ0andβare constants. 6.Find and sketch the steady-state solution of ∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t together with boundary conditions a.∂u ∂x(0,t)=0, u(a,t)=T0; b.u(0,t)−∂u ∂x(0,t)=T0,∂u ∂x(a,t)=0; c.u(0,t)−∂u ∂x(0,t)=T0,u(a,t)+∂u ∂x(a,t)=T1. 7.Find the steady-state solution of this problem, where ris a constant that represents heat generation. ∂2u ∂x2+r=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,∂u ∂x(a,t)=0,0<t. 8.Find the steady-state solution of ∂2u ∂x2+γ2/parenleftbig U(x)−u/parenrightbig =1 k∂u ∂t,0<x<a,0<t, u(0,t)=U0,∂u ∂x(a,t)=0,0<t, where U(x)=U0+Sx(U0,Sare constants). 2.3 Example: Fixed End Temperatures 149 9.This problem describes the diffusion of a substance in a medium that is moving with speed Sto the right. The unknown function u(x,t)is the con- centration of the diffusing substance. Write out the steady-state problemand solve it. ( D,U,a n d Sare constants.) D∂ 2u ∂x2=∂u ∂t+S∂u ∂x, 0<x<a,0<t, u(0,t)=U,u(a,t)=0,0<t, u(x,0)=0, 0<x<a. 2.3 Example: Fixed End Temperatures In Section 1 we saw that the temperature u(x,t)in a uniform rod with insu- lated material surface would be determined by the problem ∂2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (1) u(0,t)=T0, 0<t, (2) u(a,t)=T1, 0<t, (3) u(x,0)=f(x), 0<x<a (4) if the ends of the rod are held at ﬁxed temperatures and if the initial tempera- ture distribution is f(x). In Section 2 we found that the steady-state tempera- ture distribution, v(x)=lim t→∞u(x,t), satisﬁed the boundary value problem d2v dx2=0,0<x<a, (5) v(0)=T0,v ( a)=T1. (6) I nf a c t ,w ew e r ea b l et oﬁ n d v(x)explicitly: v(x)=T0+(T1−T0)x a. (7) We also deﬁned the transient temperature distribution as w(x,t)=u(x,t)−v(x) 150 Chapter 2 The Heat Equation a n dd e t e r m i n e dt h a t wsatisﬁes the boundary value–initial value problem ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, (8) w(0,t)=0, 0<t, (9) w(a,t)=0, 0<t, (10) w(x,0)=f(x)−v(x)≡g(x), 0<x<a. (11) Our objective is to determine the transient temperature distribution, w(x,t), and — since v(x)is already known — the unknown temperature will be u(x,t)=v(x)+w(x,t). (12) The problem in wcan be attacked by a method called product method, sepa- ration of variables ,o rFourier’s method . For this method to work, it is essential to have homogeneous partial differential equation and boundary conditions.Thus, the method may be applied to the transient distribution wbutnotto the original function u. Of course, because both the partial differential equa- tion and the boundary conditions satisﬁed by w(x,t)are homogeneous, the function w≡0 satisﬁes them. Because this solution itself is obvious and is of no help in satisfying the initial condition, it is called the trivial solution .W e are seeking the unobvious, nontrivial solutions, so we shall avoid the trivialsolution at every turn. The general idea of the method is to assume that the solution of the par- tial differential equation has the form of a product: w(x,t)=φ(x)T(t).W e require that neither of the factors φ(x)and T(t)be identically 0, since that would lead back to the trivial solution. Now, each of the factors depends ononly one variable, so we have ∂ 2w ∂x2=φ/prime/prime(x)T(t),∂w ∂t=φ(x)T/prime(t). The partial differential equation becomes φ/prime/prime(x)T(t)=1 kφ(x)T/prime(t), and on dividing through by φTwe ﬁnd φ/prime/prime(x) φ(x)=T/prime(t) kT(t),0<x<a,0<t. Here is the key argument : The ratio on the left contains functions of xalone and cannot vary with t. On the other hand, the ratio on the right contains functions of talone and cannot vary with x. Since this equality must hold for 2.3 Example: Fixed End Temperatures 151 allxin the interval 0 <x<aand for all t>0, the common value of the two sides must be a constant, varying neither with xnort: φ/prime/prime(x) φ(x)=p,T/prime(t) kT(t)=p. Now we have two ordinary differential equations for the two factor functions: φ/prime/prime−pφ=0,T/prime−pkT=0. (13) The two boundary conditions on wmay also be stated in the product form: w(0,t)=φ(0)T(t)=0,w ( a,t)=φ(a)T(t)=0. There are two ways these equations can be satisﬁed for all t>0. Either the function T(t)≡0 for all t, which is forbidden, or the other factors must be zero. Therefore, we have φ(0)=0,φ ( a)=0. (14) Our job now is to solve Eqs. (13) and satisfy the boundary conditions (14) while avoiding the trivial solution. Case 1 :I fp>0, the solutions of Eqs. (13) are φ(x)=c1cosh/parenleftbig√px/parenrightbig +c2sinh/parenleftbig√px/parenrightbig ,T(t)=cepkt. Now we apply the boundary conditions: φ(0)=0:c1=0, φ(a)=0:c2sinh/parenleftbig√pa/parenrightbig =0. Because the sinh function is 0 only when its argument is 0 — clearly not true of√pa—w eh a v e c1=c2=0a n dφ(x)≡0, which is not acceptable. Case 2 :I fw et a k e p=0, the solutions of the differential equations (13) are φ(x)=c1+c2x,T(t)=c. The boundary conditions require φ(0)=0:c1=0, φ(a)=0:c2a=0. Again we have φ(x)≡0. Case 3 : We now try a negative constant. Replacing pby−λ2in Eqs. (13) gives us the two equations φ/prime/prime+λ2φ=0,T/prime+λ2kT=0, whose solutions are φ(x)=c1cos(λx)+c2sin(λx), T(t)=cexp/parenleftbig −λ2kt/parenrightbig . 152 Chapter 2 The Heat Equation Ifφhas the form given in the preceding, the boundary conditions require that φ(0)=c1=0, leaving φ(x)=c2sin(λx).T h e n φ(a)=c2sin(λa)=0. We now have two choices: either c2=0, making φ(x)≡0 for all values of x, or sin(λa)=0. We reject the ﬁrst possibility, for it leads to the trivial solution w(x,t)≡0. In order for the second possibility to hold, we must have λ= nπ/a,w h e r e n=± 1,±2,±3,....T h en e g a t i v ev a l u e so f ndo not give any new functions, because sin (−θ)=− sin(θ).H e n c ew ea l l o w n=1,2,3,... only. We shall set λn=nπ/a. Incidentally, because the differential equations (13) and the boundary con- ditions (14) for φ(x)are homogeneous, any constant multiple of a solution is still a solution. We shall therefore remember this fact and drop the constant c2 inφ(x). Likewise, we delete the cinT(t). T o review our position, we have, for each n=1,2,3,...,af u n c t i o n φn(x)= sin(λnx)and an associated function Tn(t)=exp(−λ2 nkt).T h ep r o d u c t wn(x,t) =s i n(λnx)exp(−λ2 nkt)has these properties: 1.∂2wn ∂x2=−λ2 nwn;∂w n ∂t=−λ2 nkwn; and therefore wnsatisﬁes the heat equa- tion. 2.wn(0,t)=sin(0)e−λ2 nkt=0 for any nand t; and therefore wnsatisﬁes the boundary condition at x=0. 3.wn(a,t)=sin(λna)e−λ2 nkt=0 for any nand tbecause λna=nπand sin(nπ)=0. Therefore wnsatisﬁes the boundary condition at x=a. Now we call on the Principle of Superposition in order to continue. Principle of Superposition. Ifu1,u2,... are solutions of the same linear, homogeneous equations, then so is u=c1u1+c2u2+···. /square In fact, we have inﬁnitely many solutions, so we need an inﬁnite series to combine them all: w(x,t)=∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (15) Using an inﬁnite series brings up questions about convergence that we are go- ing to ignore. However, it is easy to verify that the function deﬁned by theseries does satisfy the boundary conditions: At x=0a n da t x=a,e a c ht e r m is 0, so the sum is 0 as well. T o check the partial differential equation, we haveto differentiate w(x,t)by differentiating each term of the series. This done, it is easy to see that terms match and the heat equation is satisﬁed. 2.3 Example: Fixed End Temperatures 153 Notice that the choice of the coefﬁcients bndoes not enter into the check- ing of the partial differential equation and the boundary conditions. Thus,Eq. (15) plays the role of a general solution of Eqs. (8)–(10). Of the four parts of the original problem, only the initial condition has not yet been satisﬁed. At t=0, the exponentials in Eq. (15) are all unity. Thus the initial condition takes the form w(x,0)=∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg =g(x), 0<x<a. (16) We immediately recognize a problem in Fourier series, which is solved by choosing the constants bnaccording to the formula bn=2 a/integraldisplaya 0g(x)sin/parenleftbiggnπx a/parenrightbigg dx. (17) If the function gis continuous and sectionally smooth, we know that the Fourier series actually converges to g(x)in the interval 0 <x<a,s ot h es o l u - tion that we have found for w(x,t)actually satisﬁes all requirements set on w. Even if gdoes not satisfy these conditions, it can be shown that the solution we have arrived at is the best we can do. Once the transient temperature has been determined, we ﬁnd the original unknown u(x,t)as the sum of the transient and the steady-state solutions, u(x,t)=v(x)+w(x,t). Example. Suppose the original problem to be ∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,0<t, u(a,t)=T1,0<t, u(x,0)=0, 0<x<a. The steady-state solution is v(x)=T0+(T1−T0)x a. The transient temperature, w(x,t)=u(x,t)−v(x),s a t i s ﬁ e s ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, w(0,t)=0, 0<t, 154 Chapter 2 The Heat Equation w(a,t)=0, 0<t, w(x,0)=− T0−(T1−T0)x a≡g(x), 0<x<a. According to the preceding calculations, whas the form w(x,t)=∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig (18) and the initial condition is w(x,0)=∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg =g(x), 0<x<a. The coefﬁcients bnare given by bn=2 a/integraldisplaya 0/bracketleftbigg −T0−(T1−T0)x a/bracketrightbigg sin/parenleftbiggnπx a/parenrightbigg dx =2T0 acos(nπx/a) (nπ/a)/vextendsingle/vextendsingle/vextendsingle/vextendsinglea 0 −2 a2(T1−T0)sin(nπx/a)−(nπx/a)cos(nπx/a) (nπ/a)2/vextendsingle/vextendsingle/vextendsingle/vextendsinglea 0 =−2T0 nπ/parenleftbig 1−(−1)n/parenrightbig +2(T1−T0) nπ(−1)n bn=−2 nπ/parenleftbig T0−T1(−1)n/parenrightbig . Now the complete solution (see Fig. 3) is u(x,t)=w(x,t)+T0+(T1−T0)x a, where w(x,t)=−2 π∞/summationdisplay n=1T0−T1(−1)n nsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (19) T h es o l u t i o no ft h i sp r o b l e mi ss h o w na sa na n i m a t i o no nt h eC D . /square We can discover certain features of u(x,t)by examining the solution. First, u(x,0)really is zero (0<x<a)because the Fourier series converges to −v(x) att=0. Second, when tis positive but very small, the series for w(x,t)will almost equal −T0−(T1−T0)x/a.B u ta t x=0a n d x=a, the series adds up to zero (and w(x,t)is a continuous function of x); thus u(x,t)satisﬁes the boundary conditions. Third, when tis large, exp (−λ2 1kt)is small, and the 2.3 Example: Fixed End Temperatures 155 Figure 3 The solution of the example with T1=100 and T0=20. The function u(x,t)is graphed as a function of xf o rf o u rv a l u e so f t, chosen so that the dimen- sionless time kt/a2has the values 0.001, 0.01, 0.1, and 1. For kt/a2=1, the steady state is practically achieved. See the CD. other exponentials are still smaller. Then w(x,t)may be well approximated by the ﬁrst term (or ﬁrst few terms) of the series. Finally, as t→∞ ,w(x,t) disappears completely. EXERCISES Also see Separation of Variables Step by Step on the CD. 1.Write out the ﬁrst few terms of the series for w(x,t)in Eq. (19). 2.Ifk=1c m2/s,a=1 cm, show that after t=0.5 s the other terms of the series for ware negligible compared with the ﬁrst term. Sketch u(x,t)for t=0,t=0.5,t=1.0, and t=∞ .T a k e T0=100, T1=300. 3.We can see from Eq. (19) that the dimensionless combinations x/aand kt/a2appear in the sine and exponential functions. Reformulate the partial differential equation (8) in terms of the dimensionless variables. ξ=x/a, τ=kt/a2.S e t u(x,t)=U(ξ,τ) . 4.Sketch the functions φ1,φ2,a n dφ3, and verify that they satisfy the bound- ary conditions φ(0)=0,φ(a)=0. In Exercises 5–8, solve the problem ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, w(0,t)=0,w ( a,t)=0,0<t, w(x,0)=g(x), 0<x<a for the given function g(x). 156 Chapter 2 The Heat Equation 5.g(x)=T0(constant). 6.g(x)=βx(βis constant). 7.g(x)=β(a−x)(βis constant). 8.g(x)=  2T0x a, 0<x<a 2, 2T0(a−x) a,a 2<x<a. 9.A.N. Virkar, T.B. Jackson, and R.A. Cutler [Thermodynamic and kinetic ef- fects of oxygen removal on the thermal conductivity of aluminum nitride,Journal of the American Ceramic Society ,72(1989): 2031–2042] use the fol- lowing boundary value problem to study the kinetics of oxygen removal from a grain of aluminum nitride by diffusion: ∂C ∂t=D∂2C ∂x2,0<x<a, 0<t, C(0,t)=C1,C(a,t)=C1,0<t, C(x,0)=C0,0<x<a. In these equations, Cis the oxygen concentration, Dis the diffusion con- stant, ais the thickness of a grain, C0and C1are known concentrations. a.Find the steady-state solution, v(x). b.State the problem (partial differential equation, boundary conditions and initial condition) for the transient, w(x,t)=C(x,t)−v(x). c.Solve the problem for w(x,t), and write out the complete solution C(x,t). d.The concentration in the center of the grain, C(a/2,t),v a r i e sf r o m C0 at time t=0t o w a r d C1astincreases. Suppose we want to ﬁnd out how long it takes for this concentration to complete 90% of the change it will make from C0toC1; that is, we want to solve this equation for t: C/parenleftbigga 2,t/parenrightbigg −C0=0.9(C1−C0). Show that this equation is equivalent to the equation w/parenleftbigga 2,t/parenrightbigg =− 0.1(C1−C0). Find an approximate formula for the solution by using just the ﬁrst term of the series for w(x,t). 2.4 Example: Insulated Bar 157 e.Use the formula in dto ﬁnd texplicitly for a=5×10−6m,D= 10−11cm2/s. Be careful to check dimensions. 2.4 Example: Insulated Bar We shall consider again the uniform bar that was discussed in Section 1. Let us suppose now that the ends of the bar at x=0a n d x=aare insulated in- stead of being held at constant temperatures. The boundary value–initial valueproblem that describes the temperature in this rod is: ∂ 2u ∂x2=1 k∂u ∂t,0<x<a, 0<t, (1) ∂u ∂x(0,t)=0,∂u ∂x(a,t)=0,0<t, (2) u(x,0)=f(x), 0<x<a, (3) where f(x)is supposed to be a given function. We saw in Section 2 that the solution of the steady-state problem is not unique. However, the mathematical purpose behind ﬁnding the steady-statesolution is to pave the way for a homogeneous problem (partial differentialequation and boundary conditions) for the transient. In this example the par-tial differential equation and boundary conditions are already homogeneous.Thus, we do not need the steady-state solution or the transient problem. We m a yl o o kf o r u(x,t)directly. Assume that uhas the product form u(x,t)=φ(x)T(t), with neither factor identically 0. The heat equation becomes φ /prime/prime(x)T(t)=1 kφ(x)T/prime(t), and the variables are separated by dividing through by φT,l e a v i n g φ/prime/prime(x) φ(x)=T/prime(x) kT(t),0<x<a,0<t. In order that a function of xequal a function of t,t h e i rm u t u a lv a l u em u s t be a constant. If that constant were positive, Twould be an increasing expo- nential function of time, which would be unacceptable. It is also easy to showthat if the constant were positive, φcould not satisfy the boundary conditions without being identically zero. Assuming then a negative constant, we can write φ /prime/prime(x) φ(x)=−λ2=T/prime(t) kT(t) 158 Chapter 2 The Heat Equation and separate these equalities into two ordinary differential equations linked by the common parameter λ2: φ/prime/prime+λ2φ=0, 0<x<a, (4) T/prime+λ2kT=0,0<t. (5) The boundary conditions on ucan be translated into conditions on φ,b e - cause they are homogeneous conditions. The boundary conditions in productform are ∂u ∂x(0,t)=φ/prime(0)T(t)=0,0<t, ∂u ∂x(a,t)=φ/prime(a)T(t)=0,0<t. T o satisfy these equations, we must have the function T(t)always zero (which would make u(x,t)≡0), or else φ/prime(0)=0,φ/prime(a)=0. The second alternative avoids the trivial solution. We now have a homogeneous differential equation for φtogether with ho- mogeneous boundary conditions: φ/prime/prime+λ2φ=0,0<x<a, (6) φ/prime(0)=0,φ/prime(a)=0. (7) A problem of this kind is called an eigenvalue problem . We are looking for those values of the parameter λ2for which nonzero solutions of Eqs. (6) and (7) may exist. Those values are called eigenvalues , and the corresponding solutions are called eigenfunctions . Note that the signiﬁcant parameter is λ2,n o tλ.T h e square is used only for convenience. It is worth mentioning that we already saw an eigenvalue problem in Section 3 and in the Euler buckling problem ofChapter 0. The general solution of the differential equation in Eq. (6) is φ(x)=c 1cos(λx)+c2sin(λx). Applying the boundary condition at x=0, we see that φ/prime(0)=c2λ=0, giv- ingc2=0o rλ=0. We put aside the case λ=0a n da s s u m e c2=0, so φ(x)=c1cos(λx). Then the second boundary condition requires that φ/prime(a)= −c1λsin(λa)=0. Once again, we may have c1=0o rs i n (λa)=0. But c1=0 makes φ(x)≡0. We choose therefore to make sin (λa)=0b yr e s t r i c t i n g λto 2.4 Example: Insulated Bar 159 the values π/a,2π/a,3π/a,.... We label the eigenvalues with a subscript: λ2 n=/parenleftbiggnπ a/parenrightbigg2 ,φ n(x)=cos(λnx), n=1,2,.... Notice that any constant multiple of an eig enfunction is still an eigenfunction; thus, we may take c1=1 for simplicity. Returning to the case λ=0, we see that Eqs. (6) and (7) become φ/prime/prime=0,0<x<a, φ/prime(0)=0,φ/prime(a)=0. The solution of the differential equation is φ(x)=c1+c2x. Both boundary conditions say c2=0. Therefore φ(x)is a (any) constant. Thus 0 is an eigen- value of the problem Eqs. (6) and (7), and we designate λ2 0=0,φ 0(x)=1. Let us summarize our ﬁndings by saying that the eigenvalue problem, Eqs. (6) and (7), has the solution   λ2 0=0,φ 0(x)=1, λ2 n=/parenleftbiggnπ a/parenrightbigg2 ,φ n(x)=cos(λnx),n=1,2,.... Now that the numbers λ2 nare known, we can solve Eq. (5) for T(t),ﬁ n d i n g T0(t)=1,Tn(t)=exp/parenleftbig −λ2 nkt/parenrightbig . The products φn(x)Tn(t)give solutions of the partial differential equation (1) that satisfy the boundary conditions, Eq. (2): u0(x,t)=1,un(x,t)=cos(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (8) Because the partial differential equation and the boundary conditions are all linear and homogeneous, the principle of superposition applies, and any linearcombination of solutions is also a solution. The solution u(x,t)of the whole system may therefore have the form u(x,t)=a 0+∞/summationdisplay 1ancos(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (9) There is only one condition of the original set remaining to be satisﬁed, the initial condition Eq. (3). For u(x,t)in the form of Eq. (9), the initial condi- tion is u(x,0)=a0+∞/summationdisplay 1ancos(λnx)=f(x), 0<x<a. 160 Chapter 2 The Heat Equation Figure 4 The solution of the example, u(x,t), as a function of xfor several times. The initial temperature distribution is f(x)=T0+(T1−T0)x/a. For this illustra- tion, T0=20,T1=100, and the times are chosen so that the dimensionless time kt/a2takes the values 0.001, 0.01, 0.1, and 1. The last case is indistinguishable from the steady state. See the CD also. Because λn=nπ/a, we recognize a problem in Fourier series and can imme- diately cite formulas for the coefﬁcients: a0=1 a/integraldisplaya 0f(x)dx,an=2 a/integraldisplaya 0f(x)cos/parenleftbiggnπx a/parenrightbigg dx. (10) When these coefﬁcients are computed and substituted in the formulas for u(x,t), that function becomes the solution to the initial value–boundary value problems, Eqs. (1)–(3). Notice that when t→∞ , all other terms in the sum- mation for u(x,t)disappear, leaving lim t→∞u(x,t)=a0=1 a/integraldisplaya 0f(x)dx. Example. Find the complete solution of Eqs. (1)–(3) for the initial temperature dis- tribution f(x)=T0+(T1−T0)x/a.I tr e q u i r e sn oi n t e g r a t i o nt oﬁ n dt h a t a0=(T1+T0)/2. The remaining coefﬁcients are an=2 a/integraldisplaya 0/parenleftbigg T0+(T1−T0)x a/parenrightbigg cos/parenleftbiggnπx a/parenrightbigg dx =2(T1−T0)cos(nπ)−1 n2π2. Thus the solution is given by Eq. (9) with these coefﬁcients for a0and an. Ag r a p ho f u(x,t)as a function of xis shown in Fig. 4 and as an animation on the CD. /square 2.4 Example: Insulated Bar 161 EXERCISES 1.Using the initial condition u(x,0)=T1x a,0<x<a, ﬁnd the solution u(x,t)of Eqs. (1)–(3). Sketch u(x,0),u(x,t)for some t>0 (using the ﬁrst three terms of the series), and the steady-state solu- tion. 2.Repeat Exercise 1 using the initial condition u(x,0)=T0+T1/parenleftbiggx a/parenrightbigg2 ,0<x<a. 3.Same as Exercise 1, but with initial condition u(x,0)=  2T0x a, 0<x<a 2, 2T0(a−x) a,a 2<x<a. 4.Solve Eqs. (1)–(3) using the initial condition u(x,0)=f(x),w h e r e f(x)=  T1,0<x<a 2, T2,a 2<x<a. 5.Consider this heat problem, which is related to Eqs. (1)–(3): ∂2u ∂x2=1 k∂u ∂t,0<x<a, 0<t, ∂u ∂x(0,t)=S0,∂u ∂x(a,t)=S1,0<t, u(x,0)=f(x), 0<x<a. a.Show that the steady-state problem has a solution if and only if S0=S1, and give a physical reason why this should be true. (Recall that the heatﬂux qis proportional to the derivative of uwith respect to x.) Find the steady-state solution if this condition is met. b.I ft h es t e a d y - s t a t es o l u t i o n v(x)exists, show that the “transient,” w(x,t)=u(x,t)−v(x), has the boundary conditions ∂w ∂x(0,t)=0,∂w ∂x(a,t)=0,0<t. 162 Chapter 2 The Heat Equation c.Show that the function u(x,t)=A(kt+x2/2)+Bxsatisﬁes the heat equation for arbitrary AandBand that AandBcan be chosen to satisfy the boundary conditions ∂u ∂x(0,t)=S0,∂u ∂x(a,t)=S1,0<t. What happens to u(x,t)astincreases if S0/negationslash=S1? 6.Ver ify that un(x,t)in Eq. (8) satisﬁes the partial differential equation (1) and the boundary conditions, Eq. (2). 7.State the eigenvalue problem associated with the solution of the heat prob-lem in Section 3. Also state its solution. 8.Suppose that the function φ(x)satisﬁes the relation φ/prime/prime(x) φ(x)=p2>0. Show that the boundary conditions φ/prime(0)=0,φ/prime(a)=0, then force φ(x) to be identically 0. Thus, a positive “separation constant” can only lead to the trivial solution. 9.Refer to Eqs. (9) and (10), which give the solution of the problem stated in Eqs. (1)–(3). If fis sectionally continuous, the coefﬁcients an→0a s n→∞ .F o r t=t1>0, ﬁxed, the solution is u(x,t1)=a0+∞/summationdisplay 1anexp/parenleftbig −λ2 nkt1/parenrightbig cos(λnx) and the coefﬁcients of this cosine series are An(t1)=anexp/parenleftbig −λ2 nkt1/parenrightbig . Show that An(t1)→0 so rapidly as n→∞ that the series given in the preceding converges uniformly 0 ≤x≤a. (See Chapter 1, Section 4, The- orem 1.) Show the same for the series that represents ∂2u ∂x2(x,t1). 10.Sketch the functions φ1,φ2,a n dφ3and verify graphically that they satisfy the boundary conditions of Eq. (7). 11.The boundary conditions Eq. (2) require that ∂u ∂x(0,t)=0,0<t, and similarly at x=a.D o e st h i sm e a nt h a t uis constant at x=0? 2.5 Example: Different Boundary Conditions 163 12.This table gives values of u(0,t)for the function ufound in the example and shown in Fig. 4. Make a graph of u(0,t)and describe the graph in words. kt/a2:0.001 0.003 0.01 0.03 0.1 0.3 1 u(0,t):22.9 24.929.035.647.958.360.0 13.Check that the partial differential equation and boundary conditions are satisﬁed by the series in Eq. (9). 2.5 Example: Different Boundary Conditions In many important cases, boundary conditions at the two endpoints will bedifferent kinds. In this section we shall solve the problem of ﬁnding the tem-perature in a rod having one end insulated and the other held at a constant temperature. The boundary value–initial value problem satisﬁed by the tem- perature in the rod is ∂ 2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (1) u(0,t)=T0, 0<t, (2) ∂u ∂x(a,t)=0, 0<t, (3) u(x,0)=f(x), 0<x<a. (4) It is easy to verify that the steady-state solution of this problem is v(x)=T0. Using this information, we can ﬁnd the boundary value–initial value problemsatisﬁed by the transient temperature w(x,t)=u(x,t)−T 0: ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, (5) w(0,t)=0,∂w ∂x(a,t)=0,0<t, (6) w(x,0)=f(x)−T0=g(x), 0<x<a. (7) Since this problem is homogeneous, we can attack it by the method of sep- aration of variables. The assumption that w(x,t)has the form of a product, w(x,t)=φ(x)T(t), and insertion of win that form into the partial differential equation (5) lead, as before, to φ/prime/prime(x) φ(x)=T/prime(t) kT(t)=constant . (8) 164 Chapter 2 The Heat Equation The boundary conditions take the form φ(0)T(t)=0,0<t, (9) φ/prime(a)T(t)=0,0<t. (10) As before, we conclude that φ(0)andφ/prime(a)should both be zero: φ(0)=0,φ/prime(a)=0. (11) By trial and error we ﬁnd that a positive or zero separation constant in Eq. (8) forces φ(x)≡0. Thus we take the constant to be −λ2. The separated equations are φ/prime/prime+λ2φ=0, 0<x<a, (12) T/prime+λ2kT=0,0<t. (13) Now, the general solution of the differential equation (12) is φ(x)=c1cos(λx)+c2sin(λx). The boundary condition, φ(0)=0, requires that c1=0, leaving φ(x)=c2sin(λx). The boundary condition at x=anow takes the form φ/prime(a)=c2λcos(λa)=0. The three choices are: c2=0, which gives the trivial solution; λ=0, which should be investigated separately (Exercise 2), and cos (λa)=0. The third al- ternative — the only acceptable one — requires that λabe an odd multiple of π/2, which we may express as λn=(2n−1)π 2a,n=1,2,.... (14) Thus, we have found that the eigenvalue problem consisting of Eqs. (11) and (12) has the solution λn=(2n−1)π 2a,φ n(x)=sin(λnx), n=1,2,3,.... (15) With the eigenfunctions and eigenvalues now in hand, we return to the dif- ferential equation (13), whose solution is Tn(t)=exp/parenleftbig −λ2 nkt/parenrightbig . 2.5 Example: Different Boundary Conditions 165 As in previous cases, we assemble the general solution of the homogeneous problem expressed in Eqs. (5)–(7) by forming a general linear combination ofour product solutions, w(x,t)=∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (16) T h ec h o i c eo ft h ec o e f ﬁ c i e n t s , bn,m u s tb em a d es oa st os a t i s f yt h ei n i t i a lc o n - dition, Eq. (8). Using the form of wgiven by Eq. (16), we ﬁnd that the initial condition is w(x,0)=∞/summationdisplay n=1bnsin/parenleftbigg(2n−1)πx 2a/parenrightbigg =g(x), 0<x<a. (17) A routine Fourier sine series for the interval 0 <x<awould involve the func- tions sin (nπx/a), rather than the functions we have. By one of several means (Exercises 10–12), it may be shown that the series in Eq. (17) represents thefunction g(x),p r o v i d e dt h a t gis sectionally smooth and that we choose the coefﬁcients by the formula b n=2 a/integraldisplaya 0g(x)sin/parenleftbigg(2n−1)πx 2a/parenrightbigg dx. (18) Now the original problem is completely solved. The solution is u(x,t)=T0+∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (19) It should be noted carefully that the T0term in Eq. (19) is the steady-state solution in this case; it is not part of the separation-of-variables solution. Example. Find the solution of Eqs. (1)–(4) with the initial condition u(x,0)=T1,0<x<a. Then g(x)=T1−T0,0<x<a, and the coefﬁcients as determined by Eq. (18) are bn=(T1−T0)4 π(2n−1). Therefore, the complete solution of the boundary value–initial value problem with initial condition u(x,0)=T1would be u(x,t)=T0+(T1−T0)4 π∞/summationdisplay n=11 2n−1sin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . (20) See Fig. 5 for graphs and an animation on the CD. /square 166 Chapter 2 The Heat Equation Figure 5 Solution of the example, Eq. (20): u(x,t)is shown as a function of x for various times, which are chosen so that the dimensionless time kt/a2takes the values 0.001, 0.01, 0.1, 1.0. For the illustration, T0has been chosen equal to 0 and T1=100. N o wt h a tw eh a v eb e e nt h r o u g ht h r e em a j o re x a m p l e s ,w ec a no u t l i n et h e method we have been using to solve linear boundary value–initial value prob- lems. Up to this moment we have seen only homogeneous partial differential equations, but a nonhomogeneity that is independent of tcan be treated by the same technique. Summary of Separation of Variables Prepare If the partial differential equation or a boundary condition or both are nothomogeneous, ﬁrst ﬁnd a function v(x), independent of t, that satisﬁes the partial differential equation and the boundary conditions. Since v(x)does not depend on t, the partial differential equation applied to v(x)becomes an ordi- nary differential equation. Finding v(x)is just a matter of solving a two-point boundary value problem. Determine the initial value–boundary value problem satisﬁed by the “tran- sient solution” w(x,t)=u(x,t)−v(x).T h i sm u s tb ea homogeneous problem . That is, the partial differential equation and the boundary conditions (but notusually the initial condition) are sa tisﬁed by the constant function 0. Separate Assuming that w(x,t)=φ(x)T(t), with neither factor 0, separate the partial differential equation into two ordinary differential equations, one for φ(x)and one for T(t), linked by the separation constant, −λ2. Reduce the boundary conditions to conditions on φalone. 2.5 Example: Different Boundary Conditions 167 Solve Solve the eigenvalue problem for φ. That is, ﬁnd the values of λ2for which the eigenvalue problem has nonzero solutions. Label the eigenfunctions andeigenvalues φ n(x)andλ2 n. Solve the ordinary differential equation for the time factors, Tn(t). Combine and Satisfy Remaining Condition Form the general solution of the homogeneous problem as a sum of constant multiples of the product solutions: w(x,t)=/summationdisplay cnφn(x)Tn(t). Choose the cnso that the initial condition is satisﬁed. This may or may not b ear o u t i n eF o u r i e rs e r i e sp r o b l e m .I fn o t ,a no r t h o g o n a l i t yp r i n c i p l em u s t be used to determine the coefﬁcients. (We shall see the theory in Sections 7 and 8.) Check Form the solution of the original problem u(x,t)=v(x)+w(x,t) and check that all conditions are satisﬁed. EXERCISES See Common Eigenvalue Problems on the CD. 1.Find the steady-state solution of the problem stated in Eqs. (1)–(4). 2.Determine whether 0 is an eigenvalue of the eigenvalue problem stated in Eqs. (11) and (12). That is, take λ=0 and see whether the solution is nonzero. 3.Solve the problem stated in Eqs. (1)–(4), taking f(x)=Tx/a. 4.Solve the problem stated in Eqs. (1)–(4) if f(x)=/braceleftbiggT0,0<x<a/2, T1,a/2<x<a. 5.Solve the nonhomogeneous problem ∂2u ∂x2=1 k∂u ∂t−T a2, 0<x<a,0<t, 168 Chapter 2 The Heat Equation u(0,t)=T0,∂u ∂x(a,t)=0,0<t, u(x,0)=T0, 0<x<a. 6.Solve this problem for the temperature in a rod in contact along the lateral surface with a medium at temperature 0. ∂2u ∂x2=1 k∂u ∂t+γ2u, 0<x<a,0<t, u(0,t)=0,∂u ∂x(a,t)=0,0<t, u(x,0)=T0, 0<x<a. 7.Solve the problem ∂2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, ∂u ∂x(0,t)=0,u(a,t)=T0,0<t, u(x,0)=T1, 0<x<a. 8.Compare the solution of Exercise 7 with Eq. (20). Can one be turned into the other? 9.Solve the problem in Exercise 7 taking T0=0a n d u(x,0)=T1cos/parenleftbiggπx 2a/parenrightbigg ,0<x<a. 10. a. Show that the eigenfunctions found in this section are orthogonal. That is, prove that /integraldisplaya 0sin(λnx)sin(λmx)dx=/braceleftBigg0(m/negationslash=n),a 2(m=n) whenλn=(2n−1)π 2a. b.Use the orthogonality relation in part ato justify the formula in Eq. (18). 11.T o justify the expansion of Eq. (17), for an arbitrary sectionally smoothg(x), ∞/summationdisplay n=1bnsin/parenleftbigg(2n−1)πx 2a/parenrightbigg =g(x), 0<x<a, 2.5 Example: Different Boundary Conditions 169 construct the function G(x)with these properties: G(x)=g(x), 0<x<a, G(x)=g(2a−x), a<x<2a. Show that G(x)corresponds to the series G(x)∼∞/summationdisplay N=1BNsin/parenleftbiggNπx 2a/parenrightbigg ,0<x<2a. 12.Show that the BNof the series in the preceding equation satisfy BN=0(Neven), BN=2 a/integraldisplaya 0g(x)sin/parenleftbiggNπx 2a/parenrightbigg dx(Nodd). 13. a. Solve this problem over the interval 0 <x<2a. ∂2u ∂x2=1 k∂u ∂t, 0<x<2a,0<t, u(0,t)=T0,u(2a,t)=T0,0<t, u(x,0)=g(x), 0<x<2a. Af u n c t i o n fis given over the interval 0 <x<a,a n d gis an extension offdeﬁned by g(x)=/braceleftbigg f(x), 0<x<a, f(2a−x),a<x<2a. b.Explain why the solution of the problem comprising Eqs. (1)–(4) is exactly the same as the solution of the problem in part a. 14.In the ceramics industry, the following problem has to be analyzed for parameter measurement. A cylindrical rod of uniform porous material issuspended vertically so that its lower end is immersed in water. The cylin-drical surface and the upper end are sealed — with wax, for example. Theconcentration of water in the rod (weight per unit volume) is a functionC(x,t)that satisﬁes the boundary value problem D∂ 2C ∂x2=∂C ∂t,0<x<L, 0<t, C(0,t)=C0,∂C ∂x(L,t)=0,0<t, C(x,0)=0,0<x<L. 170 Chapter 2 The Heat Equation In these equations, Dis the diffusion constant, Lis the length of the cylin- der, and C0is the saturation concentration, which depends on the porosity of the material. Find C(x,t). 15.Use the solution of Exercise 14 to ﬁnd an expression for the total weight of water absorbed by the rod, W(t)=A/integraldisplayL 0C(x,t)dx. 16.A plot of W(t)/C0as a function of s=/radicalbig Dt/L2over the range 0 to 2 resembles a slanted line segment joined by a curve to a horizontal linesegment. The slope of the slanted line segment in this graph is approx-imately 1. Experimenters plot measured values of W(t)/C 0vs√tto get a similar graph. Then they use the slope of the slanted line segment toﬁnd D.E x p l a i nh o w . 2.6 Example: Convection We have seen three examples in which boundary conditions speciﬁed either uor∂u/∂x. Now we shall study a case where a condition of the third kind is involved. The physical model is conduction of heat in a rod with insulatedlateral surface whose left end is held at constant temperature and whose rightend is exposed to convective heat transfer. The boundary value–initial valueproblem satisﬁed by the temperature in the rod is ∂ 2u ∂x2=1 k∂u ∂t, 0<x<a,0<t, (1) u(0,t)=T0, 0<t, (2) −κ∂u ∂x(a,t)=h/parenleftbig u(a,t)−T1/parenrightbig ,0<t, (3) u(x,0)=f(x), 0<x<a. (4) We found in Section 2 that the steady-state solution of this problem is v(x)=T0+xh(T1−T0) κ+ha. (5) Now, since the original boundary conditions were nonhomogeneous, we form the problem for the transient solution w(x,t)=u(x,t)−v(x).B yd i r e c t substitution it is found that ∂2w ∂x2=1 k∂w ∂t, 0<x<a,0<t, (6) 2.6 Example: Convection 171 w(0,t)=0,hw(a,t)+κ∂w ∂x(a,t)=0,0<t, (7) w(x,0)=f(x)−v(x)≡g(x), 0<x<a. (8) The solution for w(x,t)can now be found by the product method. On the assumption that whas the form of a product φ(x)T(t), the variables can be separated exactly as before, giving two ordinary differential equations linkedby a common parameter λ 2: φ/prime/prime+λ2φ=0, 0<x<a, T/prime+λ2kT=0,0<t. Also, since the boundary conditions are linear and homogeneous, they can be translated directly into conditions on φ: w(0,t)=φ(0)T(t)=0, κ∂w ∂x(a,t)+hw(a,t)=/bracketleftbig κφ/prime(a)+hφ(a)/bracketrightbig T(t)=0. Either T(t)is identically zero (which would make w(x,t)identically zero) or φ(0)=0,κ φ/prime(a)+hφ(a)=0. Combining the differential equation and boundary conditions on φ,w eg e t the eigenvalue problem φ/prime/prime+λ2φ=0,0<x<a, (9) φ(0)=0,κ φ/prime(a)+hφ(a)=0. (10) The general solution of the differential equation is φ(x)=c1cos(λx)+c2sin(λx). The boundary condition at x=0r e q u i r e st h a t φ(0)=c1=0, leaving φ(x)= c2sin(λx). Now, at the other boundary, κφ/prime(a)+hφ(a)=c2/parenleftbig κλcos(λa)+hsin(λa)/parenrightbig =0. Discarding the possibilities c2=0a n d λ=0, which both lead to the trivial solution, we are left with the equation κλcos(λa)+hsin(λa)=0,or tan (λa)=−κ hλ. (11) 172 Chapter 2 The Heat Equation A n 0.2500 0 .5000 1 .0000 2 .0000 4 .0000 12 .5704 2 .2889 2 .0288 1 .8366 1 .7155 25 .3540 5 .0870 4 .9132 4 .8158 4 .7648 38 .3029 8 .0962 7 .9787 7 .9171 7 .8857 41 1 .3348 11 .1727 11 .0855 11 .0408 11 .0183 51 4 .4080 14 .2764 14 .2074 14 .1724 14 .1548 Table 2 First ﬁve positive solutions of the equation tan (x)=− Ax Figure 6 Graphs of tan (λa)and−λκ/h. The points of intersection are solutions of tan(λa)=−λκ/h, eigenvalues of the problem Eqs. (9)–(10). The intersection atλ=0 corresponds to the trivial solution. From sketches of the graphs of tan (λa)and−κλ/h(Fig. 6), we see that there is an inﬁnite number of solutions, λ1,λ2,λ3,..., and that, for very large n,λnis given approximately by λn∼=2n−1 2π a. Table 2 shows the ﬁrst ﬁve values of the product λafor several different values of the dimensionless parameter κ/ha. (More solutions are tabulated in Hand- book of Mathematical Functions by Abramowitz and Stegun.) Thus we have for each n=1,2,... an eigenvalue λ2 nand an eigenfunction φn(x), which satisﬁes the eigenvalue problem Eqs. (9) and (10). Accompanying φn(x)is the function Tn(t)=exp/parenleftbig −λ2 nkt/parenrightbig 2.6 Example: Convection 173 that makes wn(x,t)=φn(x)Tn(t)a solution of the partial differential equa- tion (6) and the boundary conditions Eq. (7). Since Eqs. (6) and (7) are lin-ear and homogeneous, any linear combination of solutions is also a solution.Therefore, the transient solution will have the form w(x,t)= ∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig , and the remaining condition to be satisﬁed, the initial condition Eq. (8), is w(x,0)=∞/summationdisplay n=1bnsin(λnx)=g(x), 0<x<a. (12) Thus the constants bnare to be chosen so as to make the inﬁnite series equal g(x). Although Eq. (12) looks like a Fourier series problem, it is not, because λ2, λ3, and so forth are not all integer multiples of λ1.I fw ea t t e m p tt ou s et h ei d e a of orthogonality, we can still ﬁnd a way to select the bn, for it may be shown by direct computation that /integraldisplaya 0sin(λnx)sin(λmx)dx=0,ifn/negationslash=m. (13) Then if we multiply both sides of the proposed Eq. (12) by sin (λmx)(where mis ﬁxed) and integrate from 0 to a,w eh a v e /integraldisplaya 0g(x)sin(λmx)dx=∞/summationdisplay n=1bn/integraldisplaya 0sin(λnx)sin(λmx)dx, where we have integrated term by term. According to Eq. (13), all the terms of the series disappear except the one in which n=m, yielding an equation forbm: bm=/integraltexta 0g(x)sin(λmx)dx/integraltexta 0sin2(λmx)dx. (14) By this formula, the bmmay be calculated and inserted into the formula for w(x,t). Then we may put together the solution u(x,t)of the original problem Eqs. (1)–(4): u(x,t)=v(x)+w(x,t) =T0+xh(T1−T0) (κ+ha)+∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig . In Fig. 7 are graphs of u(x,t)for two different values of the parameter κ/ha; both have initial conditions u(x,0)=0 .S e ea n i m a t i o n so nt h eC D . 174 Chapter 2 The Heat Equation (a) (b) Figure 7 Solution of Eqs. (1)–(4) with T0=20,T1=100, and f(x)=0. Graphs (a) and (b) correspond to κ/ha=0.1a n d κ/ha=1.0, respectively. In each case, u(x,t)is graphed as a function of xfor times chosen so that the dimensionless time kt/a2takes on the values 0.001, 0.01, 0.1, 1. Note that both the temperature and its slope at the right end (x=a)change with time, so the boundary condition Eq. (3) is satisﬁed. See animations on the CD. EXERCISES 1.Sketch v(x)a sg i v e ni nE q .( 5 )a s s u m i n g a.T1>T0; b.T1=T0; c.T1<T0. 2.IfT1>T0,a si nF i g .7 ,w h a ti st h em a x i m u mv a l u eo ft h et e m p e r a t u r e u(x,t)on the interval 0 ≤x≤aat any ﬁxed time t?T h es o l u t i o nw i l lb ea function of T0,T1and z=κ/ha. 3.W h yh a v ew ei g n o r e dt h en e g a t i v es o l u t i o n so ft h ee q u a t i o n tan(λa)=−κλ h? 4.Derive the formula Eq. (12) for the coefﬁcients bm. 5.Sketch the ﬁrst two eigenfunctions of this example taking κ/h=0.5(λ1= 2.29/a,λ2=5.09/a). 6.Ver ify that /integraldisplaya 0sin2(λmx)dx=a 2+κ hcos2(λma) 2. 2.7 Sturm–Liouville Problems 175 7.Find the coefﬁcients bmcorresponding to g(x)=1,0<x<a. 8.Using the solution of Exercise 7, write out the ﬁrst few terms of the solu- tion of Eqs. (6)–(8), where g(x)=T,0<x<a. 9.Same as Exercise 7 for g(x)=x,0<x<a. 10.Verify the orthogonality integral by direct integration. It will be necessary to use the equation that deﬁnes the λn: κλncos(λna)+hsin(λna)=0. 2.7 Sturm–Liouville Problems At the end of the preceding section, we saw that ordinary Fourier series are not quite adequate for all the problems we can solve. We can make some gen-eralizations, however, that do cover most cases that arise from separation of variables. In simple problems, we often ﬁnd eigenvalue problems of the form φ /prime/prime+λ2φ=0,l<x<r, (1) α1φ(l)−α2φ/prime(l)=0, (2) β1φ(r)+β2φ/prime(r)=0. (3) It is not difﬁcult to determine the eigenvalues of this problem and to show the eigenfunctions orthogonal by direct calculation, but an indirect calculation isstill easier. Suppose that φ nandφmare eigenfunctions corresponding to different eigenvalues λ2 nandλ2 m.T h a ti s , φ/prime/prime n+λ2 nφn=0,φ/prime/prime m+λ2 mφm=0, and both functions satisfy the boundary conditions. Let us multiply the ﬁrst differential equation by φmand the second by φn,s u b t r a c tt h et w o ,a n dm o v e the terms containing φnφmto the other side: φ/prime/prime nφm−φ/prime/prime mφn=/parenleftbig λ2 m−λ2 n/parenrightbig φnφm. The right-hand side is a constant (nonzero) multiple of the integrand in the orthogonality relation /integraldisplayr lφn(x)φm(x)dx=0,n/negationslash=m, 176 Chapter 2 The Heat Equation which is proved true if the left-hand side is zero: /integraldisplayr l/parenleftbig φ/prime/prime nφm−φ/prime/prime mφn/parenrightbig dx=0. This integral is integrable by parts: /integraldisplayr l/parenleftbig φ/prime/prime nφm−φ/prime/prime mφn/parenrightbig dx =/bracketleftbig φ/prime n(x)φm(x)−φ/prime m(x)φn(x)/bracketrightbig/vextendsingle/vextendsingler l−/integraldisplayr l/parenleftbig φ/prime nφ/prime m−φ/prime mφ/prime n/parenrightbig dx. T h el a s ti n t e g r a li so b v i o u s l yz e r o ,s ow eh a v e /parenleftbig λ2 m−λ2 n/parenrightbig/integraldisplayr lφn(x)φm(x)dx=/bracketleftbig φ/prime n(x)φm(x)−φ/prime m(x)φn(x)/bracketrightbig/vextendsingle/vextendsingler l. Bothφnandφmsatisfy the boundary condition at x=r, β1φm(r)+β2φ/prime m(r)=0, β1φn(r)+β2φ/prime n(r)=0. These two equations may be considered simultaneous equations in β1andβ2. At least one of the numbers β1andβ2is different from zero; otherwise, there would be no boundary condition. Hence the determinant of the equationsmust be zero: φ m(r)φ/prime n(r)−φn(r)φ/prime m(r)=0. A similar result holds at x=l.T h u s /bracketleftbig φ/prime n(x)φm(x)−φ/prime m(x)φn(x)/bracketrightbig/vextendsingle/vextendsingler l=0, and, therefore, we have proved the orthogonality relation /integraldisplayr lφn(x)φm(x)dx=0,n/negationslash=m, for the eigenfunctions of Eqs. (1)–(3). We may make a much broader generalization about orthogonality of eigen- functions with very little trouble. Consider the following model eigenvalueproblem, which might arise from separation of variables in a heat conductionproblem (see Section 9): /bracketleftbig s(x)φ /prime(x)/bracketrightbig/prime−q(x)φ(x)+λ2p(x)φ(x)=0,l<x<r, α1φ(l)−α2φ/prime(l)=0, β1φ(r)+β2φ/prime(r)=0. 2.7 Sturm–Liouville Problems 177 Let us carry out the procedure used in the preceding with this problem. The eigenfunctions satisfy the differential equations /parenleftbig sφ/prime n/parenrightbig/prime−qφn+λ2 npφn=0, /parenleftbig sφ/prime m/parenrightbig/prime−qφm+λ2 mpφm=0. Multiply the ﬁrst by φmand the second by φn, subtract (the terms containing q(x)cancel), and move the term containing pφnφmto the other side: /parenleftbig sφ/prime n/parenrightbig/primeφm−/parenleftbig sφ/prime m/parenrightbig/primeφn=/parenleftbig λ2 m−λ2 n/parenrightbig pφnφm. (4) Integrate both sides from ltor, and apply integration by parts to the left-hand side: /integraldisplayr l/bracketleftbig/parenleftbig sφ/prime n/parenrightbig/primeφm−/parenleftbig sφ/prime m/parenrightbig/primeφn/bracketrightbig dx =/bracketleftbig sφ/prime nφm−sφ/prime mφn/bracketrightbig/vextendsingle/vextendsingler l−/integraldisplayr l/parenleftbig sφ/prime nφ/prime m−sφ/prime nφ/prime m/parenrightbig dx. The second integral is zero. From the boundary conditions we ﬁnd that φ/prime n(r)φm(r)−φ/prime m(r)φn(r)=0, φ/prime n(l)φm(l)−φ/prime m(l)φn(l)=0 by the same reasoning as before. Hence, we discover the orthogonality relation /integraldisplayr lp(x)φn(x)φm(x)dx=0,λ2 n/negationslash=λ2 m for the eigenfunctions of the problem stated. During these operations, we have made some tacit assumptions about in- tegrability of functions after Eq. (4). In individual cases, where the coefﬁcientfunctions s,q,a n d pand the eigenfunctions themselves are known, one can easily check the validity of the steps taken. In general, however, we would like to guarantee the existence of eigenfunctions and the legitimacy of computa-t i o n sa f t e rE q .( 4 ) .T od os o ,w en e e dt h ef o l l o w i n g . Deﬁnition The problem (sφ/prime)/prime−qφ+λ2pφ=0,l<x<r, (5) α1φ(l)−α2φ/prime(l)=0, (6) β1φ(r)+β2φ/prime(r)=0( 7 ) 178 Chapter 2 The Heat Equation is called a regular Sturm–Liouville problem if the following conditions are ful- ﬁlled: a.s(x),s/prime(x),q(x),a n d p(x)are continuous for l≤x≤r; b.s(x)>0a n d p(x)>0 for l≤x≤r; c. The α’s and β’s are nonnegative, and α2 1+α2 2>0,β2 1+β2 2>0. d. The parameter λoccurs only where shown. Condition a and the ﬁrst condition b guarantee that the differential equation has solutions with continuous ﬁrst and second derivatives. Notice that s(l)and s(r)must both be positive (not zero). Condition c just says that there are two boundary conditions: α2 1+α2 2=0o n l yi f α1=α2=0, which would be no condition. The other requirements contribute to the desired properties in waysthat are not obvious. We are now ready to state the theorems that contain necessary information about eigenfunctions. Theorem 1. The regular Sturm–Liouville problem has an inﬁnite number of eigenfunctions φ1,φ2,...,e a c hc o r r e s p o n d i n gt oad i f f e r e n te i g e n v a l u e λ2 1,λ22,.... If n/negationslash=m, the eigenfunctions φnandφmare orthogonal with weight function p (x): /integraldisplayr lφn(x)φm(x)p(x)dx=0,n/negationslash=m. /square The theorem is already proved, for the continuity of coefﬁcients and eigen- functions makes our previous calculations legitimate. It should be noted thatany constant multiple of an eigenfunction is also an eigenfunction; but asidefrom a constant multiplier, the eigenfunctions of a Sturm–Liouville problemare unique. A number of other properties of the Sturm–Liouville problem are known. We summarize a few here. Theorem 2. (a)The regular Sturm–Liouville problem has an inﬁnite number of eigenvalues, and λ2 n→∞ as n→∞ . (b)I ft h ee i g e n v a l u e sa r en u m b e r e di no r d e r , λ2 1<λ22<···, then the eigen- function corresponding to λ2 nhas exactly n −1zeros in the interval l <x<r (endpoints excluded ). (c)If q(x)≥0andα1,α2,β1,β2, are all greater than or equal to zero, then all t h ee i g e n v a l u e sa r en o n n e g a t i v e . /square Examples. 1. We note that the eigenvalue problems in Sections 3–6 of this chapter are all regular Sturm–Liouville problems, as is the problem in Eqs. (1)–(3) of 2.7 Sturm–Liouville Problems 179 this section. In particular, the problem φ/prime/prime+λ2φ=0,0<x<a, φ(0)=0,hφ(a)+κφ/prime(a)=0 is a regular Sturm–Liouville problem, in which s(x)=p(x)=1,q(x)=0,α 1=1,α 2=0, β1=h,β 2=κ. All conditions of the deﬁnition are met. 2. A less trivial example is (xφ/prime)/prime+λ2/parenleftbigg1 x/parenrightbigg φ=0,1<x<2,φ ( 1)=0,φ ( 2)=0. We identify s(x)=x,p(x)=1/x,q(x)=0. This is a regular Sturm– Liouville problem. The orthogonality relation is /integraldisplay2 1φn(x)φm(x)1 xdx=0,n/negationslash=m. The conclusions of Theorems 1 and 2 hold for both examples. /square EXERCISES 1.The general solution of the differential equation in Example 2 is φ(x)=c1cos/parenleftbig λln(x)/parenrightbig +c2sin/parenleftbig λln(x)/parenrightbig . Find the eigenvalues and eigenfunctions, and verify the orthogonality rela- tion directly by integration. 2.Check the results of Theorem 2 for the problem consisting of φ/prime/prime+λ2φ=0,0<x<a, with boundary conditions a.φ(0)=0,φ(a)=0; b.φ/prime(0)=0,φ/prime(a)=0. In case b,λ2 1=0. 3.Find the eigenvalues and eigenfunctions, and sketch the ﬁrst few eigenfunc- tions of the problem φ/prime/prime+λ2φ=0,0<x<a, 180 Chapter 2 The Heat Equation with boundary conditions a.φ(0)=0,φ/prime(a)=0, b.φ/prime(0)=0,φ(a)=0, c.φ(0)=0,φ(a)+φ/prime(a)=0, d.φ(0)−φ/prime(0)=0,φ/prime(a)=0, e.φ(0)−φ/prime(0)=0,φ(a)+φ/prime(a)=0. 4.In Eqs. (1)–(3), take l=0,r=a, and show that a.The eigenfunctions are φn(x)=α2λncos(λnx)+α1sin(λnx). b.The eigenvalues must be solutions of the equation −tan(λa)=λ(α 1β2+α2β1) α1β1−α2β2λ2. 5.Show by applying Theorem 1 that the eigenfunctions of each of the follow- ing problems are orthogonal, and state the orthogonality relation. a.φ/prime/prime+λ2(1+x)φ=0,φ(0)=0,φ/prime(a)=0; b.(exφ/prime)/prime+λ2exφ=0,φ(0)−φ/prime(0)=0,φ(a)=0; c.φ/prime/prime+/parenleftbiggλ2 x2/parenrightbigg φ=0,φ(1)=0,φ/prime(2)=0; d.φ/prime/prime−sin(x)φ+exλ2φ=0,φ/prime(0)=0,φ/prime(a)=0. 6.Consider the problem (sφ/prime)/prime−qφ+λ2pφ=0,l<x<r, φ(r)=0, in which s(l)=0,s(x)>0 for l<x≤r,b u t pandqsatisfy the conditions of a regular Sturm–Liouville problem. Require also that both φ(x)andφ/prime(x) have ﬁnite limits as x→l+. Show that the eigenfunctions (if they exist) are orthogonal. 7.The following problem is not a regular Sturm–Liouville problem. Why? Solve, and show that the eigenfunctions are notorthogonal. φ/prime/prime+λ2φ=0,0<x<a, φ(0)=0,φ/prime(a)−λ2φ(a)=0. 2.8 Expansion in Series of Eigenfunctions 181 8.S h o wt h a t0i sa ne i g e n v a l u eo ft h ep r o b l e m (sφ/prime)/prime+λ2pφ=0,l<x<r, φ/prime(l)=0,φ/prime(r)=0, where sand psatisfy the conditions of a regular Sturm–Liouville problem. 9.Find all values of the parameter µfor which there is a nonzero solution of this problem: φ/prime/prime+µφ=0, φ(0)+φ/prime(0)=0,φ ( a)+φ/prime(a)=0. One solution is negative. Does this contradict Theorem 2? 2.8 Expansion in Series of Eigenfunctions We have seen that the eigenfunctions that arise from a regular Sturm–Liouville problem (sφ/prime)/prime−qφ+λ2pφ=0,l<x<r, (1) α1φ(l)−α2φ/prime(l)=0, (2) β1φ(r)+β2φ/prime(r)=0( 3 ) are orthogonal with weight function p(x): /integraldisplayr lp(x)φn(x)φm(x)dx=0,n/negationslash=m, (4) and it should be clear, from the way in which the question of orthogonality arose, that we are interested in expressing functions in terms of eigenfunction series. Suppose that a function f(x)is given in the interval l<x<rand that we wish to express f(x)in terms of the eigenfunctions φn(x)of Eqs. (1)–(3). That is, we wish to have f(x)=∞/summationdisplay n=1cnφn(x), l<x<r. (5) The orthogonality relation Eq. (4) clearly tells us how to compute the coef- ﬁcients. Multiplying both sides of the proposed Eq. (5) by φm(x)p(x)(where 182 Chapter 2 The Heat Equation mis a ﬁxed integer) and integrating from ltoryields /integraldisplayr lf(x)φm(x)p(x)dx=∞/summationdisplay n=1cn/integraldisplayr lφn(x)φm(x)p(x)dx. The orthogonality relation says that all the terms in the series, except that one in which n=m, must disappear. Thus /integraldisplayr lf(x)φm(x)p(x)dx=cm/integraldisplayr lφ2 m(x)p(x)dx gives a formula for choosing cm. We can now cite a convergence theorem for expansion in terms of eigen- functions. Notice the similarity to the Fourier series convergence theorem. Of course, the Fourier sine or cosine series are series of eigenfunctions on a regu-lar Sturm–Liouville problem in which the weight function p(x)is 1. Theorem. Letφ1,φ2,... be eigenfunctions of a regular Sturm–Liouville problem Eqs. (1)–(3),i nw h i c ht h e α’s andβ’s are not negative. If f(x)is sectionally smooth on the interval l <x<r, then ∞/summationdisplay n=1cnφn(x)=f(x+)+f(x−) 2,l<x<r, (6) where cn=/integraltextr lf(x)φn(x)p(x)dx/integraltextr lφ2n(x)p(x)dx. Furthermore, if the series ∞/summationdisplay n=1|cn|/bracketleftbigg/integraldisplayr lφ2 n(x)p(x)dx/bracketrightbigg1/2 converges, then the series Eq. (6)converges uniformly, l ≤x≤r. /square EXERCISES 1.Ver ify that λ2 n=/parenleftbiggnπ ln(b)/parenrightbigg2 ,φ n=sin/parenleftbig λnln(x)/parenrightbig are the eigenvalues and eigenfunctions of 2.8 Expansion in Series of Eigenfunctions 183 (xφ/prime)/prime+λ2/parenleftbigg1 x/parenrightbigg φ=0,1<x<b, φ(1)=0,φ ( b)=0. Find the expansion of the function f(x)=xin terms of these eigenfunc- tions. T o what values does the series converge at x=1a n d x=b? 2.Ifφ1,φ2,... are the eigenfunctions of a regular Sturm–Liouville problem and are orthogonal with weight function p(x)onl<x<rand if f(x)is sectionally smooth, then /integraldisplayr lf2(x)p(x)dx=∞/summationdisplay n=1anc2 n, where an=/integraldisplayr lφ2 n(x)p(x)dx and cnis the coefﬁcient of fas given in the theorem. Show why this should be true, and conclude that cn√an→0a s n→∞ . 3.Verify that the eigenvalues and eigenfunctions of the problem (exφ/prime)/prime+exγ2φ=0,0<x<a, φ(0)=0,φ ( a)=0 are γ2 n=/parenleftbiggnπ a/parenrightbigg2 +1 4,φ n(x)=exp/parenleftbigg −x 2/parenrightbigg sin/parenleftbiggnπx a/parenrightbigg . Find the coefﬁcients for the expansion of the function f(x)=1, 0<x<a, in terms of the φn. 4.Ifφ1,φ2,... are eigenfunctions of a regular Sturm–Liouville problem, the numbers√anare called normalizing constants , and the functions ψn= φn/√anare called normalized eigenfunctions . Show that /integraldisplayr lψ2 n(x)p(x)dx=1,/integraldisplayr lψn(x)ψm(x)p(x)dx=0,n/negationslash=m. 5.Find the formula for the coefﬁcients of a sectionally smooth function f(x) in the series f(x)=∞/summationdisplay n=1bnψn(x), l<x<r, where the ψnare normalized eigenfunctions. 184 Chapter 2 The Heat Equation 6.Show that, for the function in Exercise 5, /integraldisplayr lf2(x)p(x)dx=∞/summationdisplay n=1b2 n. 7.What are the normalized eigenfunctions of the following problem? φ/prime/prime+λ2φ=0,0<x<1, φ/prime(0)=0,φ/prime(1)=0. 2.9 Generalities on the Heat Conduction Problem On the basis of the information we have about the Sturm–Liouville problem, we can make some observations on a fairly general heat conduction problem.We take as a physical model a rod whose lateral surface is insulated. In orderto simplify slightly, we will assume that no heat is generated inside the rod. Since material properties may vary with position, the partial differential equation that governs the temperature u(x,t)in the rod will be ∂ ∂x/parenleftbigg κ(x)∂u ∂x/parenrightbigg =ρ(x)c(x)∂u ∂t,l<x<r,0<t. (1) Any of the three types of boundary conditions may be imposed at either boundary, so we use as boundary conditions α1u(l,t)−α2∂u ∂x(l,t)=c1,t>0, (2) β1u(r,t)+β2∂u ∂x(r,t)=c2,t>0. (3) If the temperature is ﬁxed, the coefﬁcient of ∂u/∂xis zero. If the boundary is insulated, the coefﬁcient of uis zero, and the right-hand side is also zero. If there is convection at a boundary, both coefﬁcients will be positive, and thesigns will be as shown. We already know that in the case of two insulated boundaries, the steady- state solution has some peculiarities, so we set this aside as a special case. As-sume, then, that either α 1orβ1or both are positive. Finally we need an initial condition in the form u(x,0)=f(x), l<x<r. (4) Equations (1)–(4) make up an initial value–boundary value problem. 2.9 Generalities on the Heat Conduction Problem 185 Assuming that c1and c2are constants, we must ﬁrst ﬁnd the steady-state solution v(x)=lim t→∞u(x,t). The function v(x)satisﬁes the boundary value problem d dx/parenleftbigg κ(x)dv dx/parenrightbigg =0,l<x<r, (5) α1v(l)−α2v/prime(l)=c1, (6) β1v(r)+β2v/prime(r)=c2. (7) Since we have assumed that at least one of α1orβ1is positive, this problem can be solved. In fact, it is possible to give a formula for v(x)in terms of the function (see Exercise 1) /integraldisplayx ldξ κ(ξ)=I(x). (8) Before proceeding further, it is convenient to introduce some new functions. Let¯κ,¯ρ,a n d¯cindicate average values of the functions κ(x),ρ(x),a n d c(x). We shall deﬁne dimensionless functions s(x)and p(x)by κ(x)=¯κs(x), ρ( x)c(x)=¯ρ¯cp(x). Also, we deﬁne the transient temperature to be w(x,t)=u(x,t)−v(x). By direct computation, using the fact that v(x)is a solution of Eqs. (5)–(7), we can show that w(x,t)satisﬁes the initial value–boundary value problem ∂ ∂x/parenleftbigg s(x)∂w ∂x/parenrightbigg =1 kp(x)∂w ∂t, l<x<r,0<t, (9) α1w(l,t)−α2∂w ∂x(l,t)=0, 0<t, (10) β1w(r,t)+β2∂w ∂x(r,t)=0, 0<t, (11) w(x,0)=f(x)−v(x)=g(x), l<x<r, (12) which has homogeneous boundary conditions. The constant kis deﬁned to be ¯κ/¯ρ¯c. Now we use our method of separation of variables to ﬁnd w.I fwhas the formw(x,t)=φ(x)T(t), the differential equation becomes T(t)/parenleftbig s(x)φ/prime(x)/parenrightbig/prime=1 kp(x)φ(x)T/prime(t), 186 Chapter 2 The Heat Equation and, on dividing through by pφT,w eﬁ n dt h es e p a r a t e de q u a t i o n (sφ/prime)/prime pφ=T/prime kT,l<x<r,0<t. As before, the equality between a function of xand a function of tcan hold only if their common value is constant. Furthermore, we expect the constantto be negative, so we put (sφ /prime)/prime pφ=T/prime kT=−λ2 and separate two ordinary equations T/prime+λ2kT=0, 0<t, (sφ/prime)/prime+λ2pφ=0,l<x<r. The boundary conditions, being linear and homogeneous, can also be changed into conditions of φ.F o ri n s t a n c e ,E q .( 1 0 )b e c o m e s /bracketleftbig α1φ(l)−α2φ/prime(l)/bracketrightbig T(t)=0,0<t, and, because T(t)≡0m a k e s w(x,t)≡0, we take the other factor to be zero. We have, then, the eigenvalue problem (sφ/prime)/prime+λ2pφ=0,l<x<r, (13) α1φ(l)−α2φ/prime(l)=0, (14) β1φ(r)+β2φ/prime(r)=0. (15) Since sand pare related to the physical properties of the rod, they should be positive. We suppose also that s,s/prime,a n d pare continuous. Then Eqs. (13)–(15) comprise a regular Sturm–Liouville problem, and we know the following. 1.There is an inﬁnite number of eigenvalues 0<λ2 1<λ22<···. 2.T o each eigenvalue corresponds just one eigenfunction (give or take a con- stant multiplier). 3.The eigenfunctions are orthogonal with weight p(x): /integraldisplayr lφn(x)φm(x)p(x)dx=0,n/negationslash=m. The function Tn(t)that accompanies φn(x)is given by Tn(t)=exp/parenleftbig −λ2 nkt/parenrightbig . 2.9 Generalities on the Heat Conduction Problem 187 We now begin to assemble the solution. For each n=1,2,3,...,w n(x,t) =φn(x)Tn(t)satisﬁes Eqs. (9)–(11). As these are all linear homogeneous equa- tions, any linear combination of solutions is again a solution. Thus the tran-sient temperature has the form w(x,t)=∞/summationdisplay n=1anφn(x)exp/parenleftbig −λ2 nkt/parenrightbig . The initial condition Eq. (12) will be satisﬁed if we choose the anso that w(x,0)=∞/summationdisplay n=1anφn(x)=g(x), l<x<r. The convergence theorem tells us that the equality will hold, except possibly at a ﬁnite number of points, if f(x)— and therefore g(x)— is sectionally smooth. Thusw(x,t)is the solution of its problem, if we choose an=/integraltextr lg(x)φn(x)p(x)dx/integraltextr lφ2n(x)p(x)dx. Finally, we can write the complete solution of Eqs. (1)–(4) in the form u(x,t)=v(x)+∞/summationdisplay n=1anφn(x)exp/parenleftbig −λ2 nkt/parenrightbig . (16) Working from the representation Eq. (16) we can draw some conclusions about the solution of Eqs. (1)–(4). 1.Since all the λ2 nare positive, u(x,t)does tend to v(x)ast→∞ . 2.For any t1>0, the series for u(x,t1)converges uniformly in l≤x≤r because of the exponential factors; therefore u(x,t1)is a continuous func- tion of x. Any discontinuity in the initial condition is immediately elimi- nated. 3.For large enough values of t, we can approximate u(x,t)by v(x)+a1φ1(x)exp/parenleftbig −λ2 1kt/parenrightbig . (T o judge how large tmight be, we need to know something about the an and the λn.) Because φ1(x)is of one sign on the interval l<x<r(that is,φ1(x)>0o rφ1(x)<0 for all xbetween land r), the graph of our approximation will lie either above or below the graph of v(x)but will not cross it (provided that a1/negationslash=0). 188 Chapter 2 The Heat Equation EXERCISES 1.Find the explicit form for v(x)in terms of the function in Eq. (8) assuming a.α1=β1=0, c1=c2=0; b.α1>0o rβ1>0, and no coefﬁcient negative. Why are these two cases separate? 2.Justify each of the conclusions. 3.Derive the general form of u(x,t)if the boundary conditions are ∂u/∂x=0 at both ends. In this case, λ2=0i sa ne i g e n v a l u e . 2.10 Semi-Inﬁnite Rod Up to this point we have seen only problems over ﬁnite intervals. Frequently, however, it is justiﬁable and useful to assume that an object is inﬁnite in length.(Sometimes this assumption is used to disguise ignorance of a boundary con- dition or to suppress the inﬂuence of a complicated condition.) Thus, if the rod we have been studying is very long, we may treat it as semi-inﬁnite —t h a ti s ,a s extending from 0 to ∞. If properties are uniform and there is no “generation,” the partial differential equation governing the temperature u(x,t)remains ∂ 2u ∂x2=1 k∂u ∂t,0<x,0<t. Let us suppose that at x=0 the temperature is held constant, say, u(0,t)=0 in some temperature scale. In the absence of another boundary, there is no other boundary condition. However, it is desirable that u(x,t)remain ﬁnite — less than some ﬁxed bound — as x→∞ . Thus, our mathematical model is ∂2u ∂x2=1 k∂u ∂t, 0<x<∞,0<t, (1) u(0,t)=0, 0<t, (2) u(x,t)bounded as x→∞, (3) u(x,0)=f(x), 0<x. (4) The heat equation (1) and the boundary condition (2) are homogeneous. The boundedness condition (3) is also homogeneous in an important way:A (ﬁnite) sum of bounded functions is bounded. Thus, we can attack Eqs. (1)–(3) by separation of variables. Assume that u(x,t)=φ(x)T(t),s ot h ep a r t i a l 2.10 Semi-Inﬁnite Rod 189 differential equation can be separated into two ordinary equations as usual: φ/prime/prime(x) φ(x)=T/prime(t) kT(t)=const. (5) There is just one boundary condition on u, which requires that φ(0)=0. The boundedness condition also requires that φ(x)remain ﬁnite as x→∞ .I ti s easy to check (see Exercises) that a positive separation constant produces func-tionsφ(x)that cannot fulﬁll both the boundary and boundedness conditions without being identically 0. Thus, we must choose a negative separation con- stant,−λ 2. The differential equation, together with the boundary and bound- edness conditions, forms a singular eigenvalue problem (singular because of the semi-inﬁnite interval), φ/prime/prime+λ2φ=0,0<x, (6) φ(0)=0,φ ( x)bounded as x→∞. (7) The general solution of the differential equation is φ(x)=c1cos(λx)+c2sin(λx), which is bounded for any choice of the constants and for any value of λ.T h e boundedness condition told us to use a negative constant in Eq. (5) and nowcontributes nothing further. Applying the boundary condition at x=0s h o w st h a t c 1=0, leaving φ(x)= c2sin(λx). In this singular eigenvalue problem, there are no “special” values ofλ:Any value produces a nonzero solution of the differential equation that also satisﬁes the boundary and boundedness conditions. (But negative valuesofλproduce no new solutions.) Recalling that any constant multiple of a so- lution of a homogeneous problem is still a solution, we choose c 2=1a n d summarize the solution of the singular eigenvalue problem as φ(x;λ)=sin(λx), λ > 0. (8) The solution of Eq. (5) for T(t), with constant −λ2,i s T(t)=exp/parenleftbig −λ2kt/parenrightbig . For any value of λ2,t h ef u n c t i o n u(x,t;λ)=sin(λx)exp/parenleftbig −λ2kt/parenrightbig satisﬁes Eqs. (1)–(3). Equation (1) and the boundary condition Eq. (2) are ho- mogeneous, and Eq. (3) is homogeneous in effect; therefore any linear combi-nation of solutions is a solution. Since the parameter λmay take on any value, 190 Chapter 2 The Heat Equation we must use an integral — the continuous analogue of a sum or series — to include all possibilities. Thus ushould have the form u(x,t)=/integraldisplay∞ 0B(λ)sin(λx)exp/parenleftbig −λ2kt/parenrightbig dλ. (9) (We need not include negative values of λ. They give no new solutions.) The initial condition will be satisﬁed if B(λ)is chosen to make u(x,0)=/integraldisplay∞ 0B(λ)sin(λx)dλ=f(x), 0<x. We recognize this as a Fourier integral; B(λ)is to be chosen as B(λ)=2 π/integraldisplay∞ 0f(x)sin(λx)dx. (10) IfB(λ)exists, then Eq. (9) is the solution of the problem. Notice that when t>0, the exponential function makes the improper integral in Eq. (9) con- verge very rapidly. Some care must be taken in the interpretation of our solution. If the rod re- ally is ﬁnite (say, length L) the expression in Eq. (9) is, of course, meaningless forxgreater than L. The presence of a boundary condition at x=Lwould inﬂuence temperatures nearby, so Eq. (9) can be considered a valid approxi-mation only for x/lessmuchL. Example. Solve the problem in Eqs. (1)–(4) using the initial temperature distribution f(x)=/braceleftbigg T0,0<x<b, 0, b<x. This means that a section of length bat the left end of the rod starts out at temperature T0, different from the temperature of the long right end, which is at the same temperature as the left boundary. (We assume T0>0.) The solution is given by Eq. (9), with B(λ)calculated from Eq. (10): B(λ)=2 π/integraldisplay∞ 0f(x)sin(λx)dx =2 π/integraldisplayb 0T0sin(λx)dx =2T0 λπ/parenleftbig 1−cos(λb)/parenrightbig . 2.10 Semi-Inﬁnite Rod 191 Figure 8 Graphs of the solution of the example, u(x,t)as a function of xover the interval 0 <x<3b,w h e r e b=1a n d T0=100 for convenience. The times have been chosen so that the dimensionless time kt/b2takes the values 0.001, 0.01, 0.1, and 1. When kt/b2=0.01, the temperature near x=b/2 has not changed noticeably from its initial value. Therefore, the complete solution is u(x,t)=2 πT0/integraldisplay∞ 01−cos(λb) λsin(λx)exp/parenleftbig −λ2kt/parenrightbig dλ. In Fig. 8 are graphs of u(x,t)as a function of xf o rv a r i o u sv a l u e so f t;a n animation can be seen on the CD. /square EXERCISES 1.Find the solution of Eqs. (1)–(3) if the initial temperature distribution is given by f(x)=/braceleftBigg0,0<x<a, T,a<x<b, 0,b<x. 2.Ver ify that u(x,t)as given by Eq. (9) is a solution of Eqs. (1)–(3). What is t h es t e a d y - s t a t et e m p e r a t u r ed i s t r i b u t i o n ? 3.Find the solution of Eqs. (1)–(4) if f(x)=T0e−αx,x>0. 4.Find a formula for the solution of the problem 192 Chapter 2 The Heat Equation ∂2u ∂x2=1 k∂u ∂t, 0<x,0<t, ∂u ∂x(0,t)=0, 0<t, u(x,0)=f(x), 0<x. 5.Determine the solution of Exercise 4 if f(x)is the function given in Exer- cise 1. 6.Penetration of heat into the earth. Assume that the earth is ﬂat, occupying the region 0 <x(so that xmeasures distance down from the surface). At the surface, the temperature ﬂuctuates according to season, time of day, etc.We cover several cases by taking the boundary condition to be u(0,t)= sin(ωt), where the frequency ωc a nb ec h o s e na c c o r d i n gt ot h ep e r i o do f interest. a.Show that u(x,t)=e−pxsin(ωt−px)satisﬁes the boundary condition and is a solution of the heat equation if p=/radicalbig ω/2k. b.Sketch u(x,t)as a function of tforx=0,1, and 2 m, taking ω=2× 10−7rad/s (approximately one cycle per year) and k=0.5×10−6m2/s. c.Withωas in part b, ﬁnd the depth (as a function of k)a tw h i c hs e a s o n s are reversed. 7.Consider the problem ∂2u ∂x2=1 k∂u ∂t, 0<x,0<t, u(0,t)=T0, 0<t, u(x,0)=f(x), 0<x. Show that, for our method of solution to work, it is necessary to have T0= lim x→∞f(x). Find a formula for u(x,t)if this is the case. 8.If the separation constant in Eq. (5) were positive (say, p2), we would at- tempt to solve φ/prime/prime−p2φ=0 subject to the conditions, Eq. (7). Solve the differential equation, and show that any nonzero, bounded solution is not0a tx=0 and that any solution that is 0 at x=0 is not bounded. 9.R.C. Bales, M.P . Valdez and G.A. Dawson [Gaseous deposition to snow, 2: Physical-chemical model for SO 2deposition, Journal of Geophysical Re- search ,92(1987): 9789–9799] develop a mathematical model for the trans- port of SO 2gas into snow by molecular diffusion. The governing partial 2.11 Inﬁnite Rod 193 differential equation is ∂C ∂t=D/parenleftbigg∂2C ∂x2−a2C/parenrightbigg , where Cis the concentration of SO 2as a function of x(depth into the snow) and time and Dis a diffusion constant. The term containing Cappears be- cause the SO 2takes part in a chemical reaction with water in the snow, forming sulphuric acid, H 2SO 4.T h ec o e f ﬁ c i e n t a2depends on pH, temper- ature, and other circumstances; we treat it as a constant. The problem is tobe solved for a wide range of values for the parameters. If the snow is deep, the authors believe that it is reasonable to use a semi- inﬁnite interval for xand to add the condition C(x,t)→0a s x→∞ .I n addition, a natural boundary condition at the snow surface is that con- centration in the snow match that in the air: C(0,t)=C 0.F u r t h e r m o r e , if the snow is fresh, we can assume that the concentration throughout isinitially 0, C(x,0)=0,0<x. a.Find a steady-state solution v(x)that satisﬁes the partial differential equation and the boundary conditions. b.State the problem (partial differential equation, boundary condition atx=0, condition as x→∞ , and initial condition) to be satisﬁed by the transient w(x,t)=C(x,t)−v(x). c.Solve the problem for the transient. Note that the condition as x→∞ must be relaxed to: w(x,t)bounded as x→∞ .I n d i v i d u a lp r o d u c ts o - lutions do not approach 0 as xincreases. 2.11 Inﬁnite Rod If we wish to study heat conduction in the center of a very long rod, we may as- sume that it extends from −∞ to∞. Then there are no boundary conditions, and the problem to be solved is ∂2u ∂x2=1 k∂u ∂t, −∞<x<∞,0<t, (1) u(x,0)=f(x), −∞<x<∞, (2) /vextendsingle/vextendsingleu(x,t)/vextendsingle/vextendsinglebounded as x→± ∞ . (3) 194 Chapter 2 The Heat Equation Using the same techniques as before, we look for solutions in the form u(x,t)=φ(x)T(t)so that the heat equation (1) becomes φ/prime/prime(x) φ(x)=T/prime(t) T(t)=constant . As in the previous section, the constant must be nonpositive (say, −λ2)i n order for the solutions to be bounded. Thus, we have the singular eigenvalueproblem φ /prime/prime+λ2φ=0,−∞<x<∞, φ(x)bounded as x→± ∞ . It is easy to see that every solution of φ/prime/prime/φ=−λ2is bounded. Thus, our fac- torsφ(x)and T(t)are φ(x;λ)=Acos(λx)+Bsin(λx), T(t;λ)=exp/parenleftbig −λ2kt/parenrightbig . We combine the solutions φ(x)T(t)i nt h ef o r mo fa ni n t e g r a lt oo b t a i n u(x,t)=/integraldisplay∞ 0/parenleftbig A(λ)cos(λx)+B(λ)sin(λx)/parenrightbig exp/parenleftbig −λ2kt/parenrightbig dλ. (4) At time t=0, the exponential factor becomes 1, and the initial condition is /integraldisplay∞ 0/parenleftbig A(λ)cos(λx)+B(λ)sin(λx)/parenrightbig dλ=f(x),−∞<x<∞. As this is clearly a Fourier integral problem, we must choose A(λ)and B(λ)to be the Fourier integral coefﬁcient functions, A(λ)=1 π/integraldisplay∞ −∞f(x)cos(λx)dx,B(λ)=1 π/integraldisplay∞ −∞f(x)sin(λx)dx.(5) Then the function u(x,t)in Eq. (4) satisﬁes the partial differential equation (1) and the initial condition (2), provided that fis sectionally smooth and |f(x)| has a ﬁnite integral. It can be proved that the boundedness condition (3) is also satisﬁed, provided that the initial value f(x)is bounded as x→± ∞ . Example. Solve the problem posed in Eqs. (1)–(3) with f(x)=/braceleftBigg0, x<−a, T0,−a<x<a, 0, a<x. 2.11 Inﬁnite Rod 195 Figure 9 Solution of example problem. At t=0, the temperature is T0>0f o r −a<x<aand is 0 in the rest of the rod; u(x,t)is shown as a function of x on the interval −3a<x<3afor three times. The times are chosen so that the dimensionless time kt/a2takes the values 0.01, 1, and 10 (to get a clear picture of the changes in u). Note that u(x,t)is positive everywhere for any t>0. The values T0=100 and a=1 have been used for convenience. Also see the CD. In words, the rod has a center section of length 2 awhose temperature is differ- ent from that of the long sections to the left and right. We must compute thecoefﬁcient functions A(λ)and B(λ). The latter is identically 0 because f(x)is an even function; and A(λ)=1 π/integraldisplay∞ −∞f(x)cos(λx)dx =1 π/integraldisplaya −aT0cos(λx)dx =2T0 λπsin(λa). Thus, the solution of the problem is u(x,t)=2T0 π/integraldisplay∞ 0sin(λa) λcos(λx)exp/parenleftbig −λ2kt/parenrightbig dλ. (6) This function is graphed as a function of xfor several values of tin Fig. 9 and animated on the CD. The ﬁgure suggests that u(x,t)ispositive for all x when t>0. This is indeed true and illustrates an interesting property of the solutions of the heat equation: the instantaneous transmission of information.T h e“ h o t ”s e c t i o ni nt h ei n t e r v a l −a<x<ainstantly raises the temperature everywhere else from the initial value of 0 to a positive value. /square Starting from the general form of a solution in Eq. (4), we can derive some very interesting results. Change the variable of integration in Eq. (5) to x/primeand 196 Chapter 2 The Heat Equation substitute the formulas for A(λ)and B(λ)into Eq. (4): u(x,t)=1 π/integraldisplay∞ 0/bracketleftbigg/integraldisplay∞ −∞f(x/prime)cos(λx/prime)dx/primecos(λx) +/integraldisplay∞ −∞f(x/prime)sin(λx/prime)dx/primesin(λx)/bracketrightbigg exp/parenleftbig −λ2kt/parenrightbig dλ. Combining terms we ﬁnd u(x,t)=1 π/integraldisplay∞ 0/integraldisplay∞ −∞f(x/prime)/bracketleftbig cos(λx/prime)cos(λx)+sin(λx/prime)sin(λx)/bracketrightbig dx/prime ×exp/parenleftbig −λ2kt/parenrightbig dλ =1 π/integraldisplay∞ 0/integraldisplay∞ −∞f(x/prime)cos/parenleftbig λ(x/prime−x)/parenrightbig dx/primeexp/parenleftbig −λ2kt/parenrightbig dλ. If the order of integration may be reversed, we may write u(x,t)=1 π/integraldisplay∞ −∞f(x/prime)/integraldisplay∞ 0cos/parenleftbig λ(x/prime−x)/parenrightbig exp/parenleftbig −λ2kt/parenrightbig dλdx/prime. The inner integral can be computed by complex methods of integration. It is known to be (Miscellaneous Exercises 32, Chapter 1) /integraldisplay∞ 0cos/parenleftbig λ(x/prime−x)/parenrightbig exp/parenleftbig −λ2kt/parenrightbig dλ=/radicalbiggπ 4ktexp/bracketleftbigg−(x/prime−x)2 4kt/bracketrightbigg ,t>0. This gives us, ﬁnally, a new form for the temperature distribution: u(x,t)=1√ 4kπt/integraldisplay∞ −∞f(x/prime)exp/bracketleftbigg−(x/prime−x)2 4kt/bracketrightbigg dx/prime. (7) Using this form, we ﬁnd the solution of the example problem solved earlier ( s e eE q .( 6 ) )t ob e u(x,t)=T0√ 4πkt/integraldisplaya −aexp/bracketleftbigg−(x/prime−x)2 4kt/bracketrightbigg dx/prime. (8) Of the two formulas, Eqs. (4) and (7), for the solution u(x,t), each has its advantages. For simple problems we may be able to evaluate the coefﬁcientsA(λ)and B(λ)in Eq. (5). However, it is a rare case indeed when the integral in Eq. (4) can be evaluated analytically. The same is true for the integral inE q .( 7 ) .T h u s ,i ft h ev a l u eo f uat a speciﬁc xand tis needed, either integral would be calculated numerically. For large values of kt, the exponential factor 2.11 Inﬁnite Rod 197 in the integrand of Eq. (4) will be nearly zero, except for small λ.T h u s ,E q .( 4 ) is approximately u(x,t)∼=/integraldisplay/Lambda1 0/parenleftbig A(λ)cos(λx)+B(λ)sin(λx)/parenrightbig exp/parenleftbig −λ2kt/parenrightbig dλ for/Lambda1not large, and the right-hand side may be found to a high degree of accuracy with little effort. On the other hand, if ktis small, the exponential in the integrand of Eq. (7) will be nearly zero, except for x/primenear x. The approximation u(x,t)∼=1√ 4kπt/integraldisplayx+h x−hf(x/prime)exp/bracketleftbigg−(x/prime−x)2 4kt/bracketrightbigg dx/prime is satisfactory for hnot large, and again numerical techniques are easily applied to the right-hand side. The expression in Eq. (7) also has a number of other advantages. It requires no intermediate integrations (compare Eq. (5)). It shows directly the inﬂuenceof initial conditions on the solution. Moreover, the function f(x)need not satisfy the restriction /integraldisplay ∞ −∞/vextendsingle/vextendsinglef(x)/vextendsingle/vextendsingledx<∞ in order for Eq. (7) to satisfy the original problem. EXERCISES See the exercise Common Singular Eigenvalue Problems on the CD. 1.Find the solution of Eqs. (1)–(3) using the form given in Eq. (7) if the initial temperature distribution is f(x)=/braceleftbigg T0,x<0, T1,0<x. 2.Find the solution of Eqs. (1)–(3) using the form given in Eq. (4) if f(x)=/braceleftbigg T0/parenleftbig a−|x|/parenrightbig ,−a<x<a, 0, otherwise. 3.Same task as in Exercise 2, with f(x)=T0e−|x/a|for all x. 4.Show that the solution of the problem studied in Section 10, ∂2u ∂x2=1 k∂u ∂t, 0<x,0<t, 198 Chapter 2 The Heat Equation u(0,t)=0, 0<t, u(x,0)=f(x), 0<x, can be expressed as u(x,t)=1√ 4πkt/integraldisplay∞ 0f(x/prime)/bracketleftbigg exp/parenleftbigg−(x/prime−x)2 4kt/parenrightbigg −exp/parenleftbigg−(x/prime+x)2 4kt/parenrightbigg/bracketrightbigg dx/prime. Hint: Start from the problem of this section with initial condition u(x,0)=fo(x),−∞<x<∞, where fois the odd extension of f.T h e nu s eE q .( 7 ) ,a n ds p l i tt h ei n t e r v a l of integration at 0. 5.Verify by differentiating that the function u(x,t)=1√ 4kπtexp/bracketleftbigg −x2 4kt/bracketrightbigg is a solution of the heat equation ∂2u ∂x2=1 k∂u ∂t,0<t,−∞<x<∞. What can be said about uatx=0? at t=0+?W h a ti sl i m t→0+u(0,t)? Sketch u(x,t)f o rv a r i o u sﬁ x e dv a l u e so f t. 6.Suppose that f(x)is an odd periodic function with period 2 a. Show that u(x,t)deﬁned by Eq. (7) also has these properties. 7.Iff(x)=1 for all x, the solution of our heat conduction problem is u(x,t)=1. Use this fact together with Eq. (7) to show that 1=1√ 4πkt/integraldisplay∞ −∞exp/bracketleftbigg−(x/prime−x)2 4kt/bracketrightbigg dx/prime. 8.Solve the problem that follows using Eq. (7). ∂2u ∂x2=1 k∂u ∂t,−∞<x<∞,0<t, u(x,0)=/braceleftBig1, x>0, −1,x<0. 9.Can Exercise 8 be solved in the form of Eq. (4)? Note that 2 π/integraldisplay∞ 0sin(λx) λdλ=/braceleftBig1, 0<x, −1,x<0. 2.12 The Error Function 199 Figure 10 Graph of the error function erf (z)for−3<z<3. 2.12 The Error Function I nS e c t i o n1 1w em a d et r a n s f o r m a t i o n so faF o u r i e ri n t e g r a lt oo b t a i nt h es o - lution of the heat problem ∂2u ∂x2=1 k∂u ∂t,−∞<x<∞,0<t, (1) u(x,0)=f(x),−∞<x<∞, (2) in the form of a single integral, u(x,t)=1√ 4πkt/integraldisplay∞ −∞f(x/prime)e−(x−x/prime)2/4ktdx/prime. (3) Even for the simplest functions f, this integration cannot be carried out in closed form, mainly because the indeﬁnite integral/integraltext e−x2dxis not an elemen- tary function. We can improve our understanding of the solution Eq. (3) if we introduce the error function ,d e ﬁ n e da s erf(z)=2√π/integraldisplayz 0e−y2dy. (4) Ag r a p ho fe r f (z)is shown in Fig. 10. Convenient tables, together with approx- imations to the error function, will be found in Handbook of Mathematical Functions , by Abramowitz and Stegun. Several important properties of the error function follow immediately from the deﬁnition. First, it is clear that erf (0)=0, and it is easy to show that erf is an odd function (Exercise 1). Second, by the fundamental theorem of calculus,the derivative of the error function is d dzerf(z)=2√πe−z2. (5) 200 Chapter 2 The Heat Equation And ﬁnally, the error function supplies the integral /integraldisplayb ae−y2dy=√π 2/parenleftbig erf(b)−erf(a)/parenrightbig . (6) The reason for the choice of the constant in front of the integral in Eq. (4) is to make lim z→∞erf(z)=1. (7) T o see that this is true, deﬁne A=/integraldisplay∞ 0e−y2dy. We are going to show that A=√π/2. First write A2as the product of two integrals, A2=/integraldisplay∞ 0e−y2dy/integraldisplay∞ 0e−x2dx. Remember that the name of the variable of integration in a deﬁnite integral is immaterial. This expression for A2can be interpreted as an iterated double integral over the ﬁrst quadrant of the x,y-plane, equivalent to A2=/integraldisplay∞ 0/integraldisplay∞ 0e−(x2+y2)dx dy. Now change to polar coordinates. The ﬁrst quadrant is described by the in- equalities 0 <r<∞,0<θ<π / 2, and the element of area in polar coordi- nates is rd rdθ.T h u s ,w eh a v e A2=/integraldisplayπ/2 0/integraldisplay∞ 0e−r2rd rdθ. (8) This integral, which can be evaluated by elementary means (see Exercise 2), has value π/4. Hence A=√π/2 and Eq. (7) is validated. Many workers also use the complementary error function ,e r f c(z),d e ﬁ n e da s erfc(z)=2√π/integraldisplay∞ ze−y2dy. (9) By using Eq. (7) we obtain the identity erfc(z)=1−erf(z). (10) Some properties of the complementary error function are found in Exercise 3. 2.12 The Error Function 201 We are interested in the error function because of its role in solving the heat equation. First we shall show that the solution of the problem ∂2u ∂x2=1 k∂u ∂t, −∞<x<∞,0<t, (11) u(x,0)=sgn(x),−∞<x<∞ (12) is u(x,t)=erf(x/√ 4kt).( R e c a l lt h a ts g n (x)has the value −1i fxis negative or +1i fxis positive.) The easy way to prove this statement is to verify it directly. (See Exercises 4 and 5.) Here, we shall arrive at the same conclusion, startingfrom Eq. (3), u(x,t)=1 √ 4πkt/integraldisplay∞ −∞sgn(x/prime)e−(x−x/prime)2/4ktdx/prime. (13) First, change the variable of integration to y=(x/prime−x)/√ 4kt.T h e n dy= dx/prime/√ 4kt,a n d u(x,t)=1√π/integraldisplay∞ −∞sgn/parenleftbig x+y√ 4kt/parenrightbig e−y2dy. (14) Now the function e−y2is even, and the sgn function changes from −1t o+1a t y=− x/√ 4kt. Thus, the integrand of Eq. (14) is as shown in Fig. 11. The tail to the left of −x/√ 4kthas the same area as the tail to the right of x/√ 4ktbut opposite sign. These two areas cancel, leaving u(x,t)=1√π/integraldisplayx/√ 4kt −x/√ 4kte−y2dy. Finally, use the symmetry of the integrand to halve the interval of integra- tion and double the result: u(x,t)=2√π/integraldisplayx/√ 4kt 0e−y2dy=erf/parenleftbig x/√ 4kt/parenrightbig . This is the result we wanted to arrive at. Figure 12 shows graphs of u(x,t)= erf(x/√ 4kt)as a function of xfor several values of kt. Because erf (0)=0, the function u(x,t)=erf(x/√ 4kt)must also be the so- lution of the problem ∂2u ∂x2=1 k∂u ∂t,0<x,0<t, u(0,t)=0, 0<t, u(x,0)=1, 0<x. 202 Chapter 2 The Heat Equation (a) (b) Figure 11 (a) Graph of exp (−y2)and (b) graph of sgn (x+y√ 4kt)exp(−y2). The tails beyond ±x/√ 4kthave the same areas, with opposite signs. Figure 12 Graphs of the solution of the problem in Eqs. (11) and (12), u(x,t)=erf(x/√ 4kt),f o r xin the range −2t o2a n df o r kt=0.01, 0.1, 1, and 10. Asktincreases, the graph of u(x,t)collapses toward the x-axis. A simple modiﬁcation leads to the conclusion that the complementary error function, u(x,t)=erfc(x/√ 4kt)is the solution of this problem with zero ini- tial condition and constant boundary condition, ∂2u ∂x2=1 k∂u ∂t,0<x,0<t, u(0,t)=1, 0<t, u(x,0)=0, 0<x. 2.12 The Error Function 203 EXERCISES 1.Show that erf (−z)=− erf(z),t h a ti s ,t h a te r fi sa no d df u n c t i o n . 2.Carry out the integration indicated in Eq. (8). 3.Verify these properties of the complementary error function: a.d dzerfc(z)=− e−z22√π; b.erfc(0)=1; c.lim z→∞erfc(z)=0; d.lim z→−∞erfc(z)=2; e.erfc(z)is neither even nor odd. 4.Verify by differentiating that u(x,t)=erf(x/√ 4kt)satisﬁes the heat equa- tion (1). 5.Ver ify that u(x,t)=erf(x/√ 4kt)satisﬁes the initial condition u(x,t)=/braceleftBig1, 0<x, −1,x<0. 6.In probability and statistics, the normal ,o rGaussian , probability density function is deﬁned as f(z)=1√ 2πe−z2/2,−∞<z<∞, and the cumulative distribution function is /Phi1(x)=/integraldisplayx −∞f(z)dz. Show that the cumulative distributio n function and the error function are related by /Phi1(x)=[1+erf(x/√ 2)]/2. 7.Express this integral in terms of the error function: I(x)=/integraldisplaye−x √xdx. 8.Use error functions to solve the problem ∂2u ∂x2=1 k∂u ∂t,0<x,0<t, 204 Chapter 2 The Heat Equation u(0,t)=Ub,0<t, u(x,0)=Ui,0<x. (Hint: What conditions does u(x,t)−Ubsatisfy?) 9.Assuming that Ub<0a n d Ui>0, the problem in Exercise 8 might be interpreted as representing the temperature in a freezing lake. (Think of xas measuring depth from the surface.) Deﬁne x(t)as the depth of the ice–water interface; then u(x(t),t)=0. Now ﬁnd x(t)explicitly. 10.Use error functions to solve the problem ∂2u ∂x2=1 k∂u ∂t,−∞<x<∞,0<t, u(x,0)=f(x),−∞<x<∞, where f(x)=U0forx<0a n d f(x)=U1forx>0. 2.13 Comments and References In about 1810, Fourier made an intensive study of heat conduction problems, in which he used the product method of solution and developed the idea ofFourier series. Sturm and Liouville made their clear and simple generalizationof Fourier series in the 1830s. Among modern works, Conduction of Heat in Solids , by Carslaw and Jaeger, is the standard reference. The Mathematics of Diffusion ,b yC r a n k ,a n d The Heat Equation , by D.V . Widder, are also useful references. (See the Bibliography.) Although we have motivated our study in terms of heat conduction and, to a lesser extent, by diffusion, many other physical phenomena of interest inengineering are described by the heat/diffusion equation: for example, voltage and current in an inductance-free cable and vorticity transport in ﬂuid ﬂow. The heat/diffusion equation and allied equations are being employed in biol- ogy to model cell physiology, chemical reactions, nerve impulses, the spread ofpopulations, and many other phenomena. Two good references are Differen- tial Equations and Mathematical Biology , by D.S. Jones and B.D. Sleeman, and Mathematical Biology ,b yJ . D .M u r r a y . The diffusion equation also turns up in some classical problems of probabil- ity theory, especially the description of Brownian motion. Suppose a particle moves exactly one step of length /Delta1xin each time interval /Delta1t.T h es t e pm a y be either to the left or to the right, each equally likely. Let u i(m)denote the probability that, at time m/Delta1t, the particle is at point i/Delta1x(m=0,1,2,..., i=0,±1,±2,...). In order to arrive at point i/Delta1xat time (m+1)/Delta1t,t h e particle must have been at one of the adjacent points (i±1)/Delta1xat the preced- ing time m/Delta1tand must have moved toward i/Delta1x. From this, we see that the 2.13 Comments and References 205 i m−3 −2−10 12 3 00 0 0 1 0 0 10 0 0 .50 0 .50 0 20 0 .25 0 0 .50 0 .25 0 30.125 0 0 .375 0 0 .375 0 0 .125 Table 3 Random-walk probabilities probabilities are related by the equation ui(m+1)=1 2ui−1(m)+1 2ui+1(m). The u’s are completely determined once an initial probability distribution is given. For instance, if the particle is initially at point zero (u0(0)=1,ui(0)=0, fori/negationslash=0), the ui(m)are formed by successive applications of the difference equation, as shown in Table 3. The equation may be transformed into a close relative of the heat equation. First, subtract ui(m)from both sides: ui(m+1)−ui(m)=1 2/parenleftbig ui+1(m)−2ui(m)+ui−1(m)/parenrightbig . Next divide by /Delta1x2/2o nt h er i g h ta n db y /Delta1t·(/Delta1x2/2/Delta1t)on the left to obtain ui(m+1)−ui(m) /Delta1t2/Delta1t (/Delta1x)2=ui+1(m)−2ui(m)+ui−1(m) (/Delta1x)2. If both the time interval /Delta1tand the step length /Delta1xare small, we may think ofui(m)as being the value of a continuous function u(x,t)atx=i/Delta1x,t= m/Delta1t. In the limit, the difference quotient on the left approaches ∂u/∂t.T h e right-hand side, being a difference of differences, approaches ∂2u/∂x2.T h e heat equation thus results if, in the simultaneous limit as /Delta1xand/Delta1ttend to zero, the quantity 2 /Delta1t/(/Delta1x)2approaches a ﬁnite, nonzero limit. In this context, the heat equation is called the Fokker–Planck equation .M o r ed e t a i l s and references may be found in Feller, Introduction to Probability Theory and Its Applications . We have used the term linear partial differential equation several times. The most general such equation, of second order in two independent variables, canbe put in the form A∂ 2u ∂x2+B∂2u ∂x∂t+C∂2u ∂t2+D∂u ∂x+E∂u ∂t+Fu+G=0, where A,B,..., Gare known — perhaps functions of xand tbut not of uor its derivatives. If Gis identically zero, the equation is homogeneous. Of course, 206 Chapter 2 The Heat Equation the ordinary heat equation has this form if we take A=1,E=− 1/kand all other coefﬁcients equal to zero. Some astute students will have wondered why we should seek solutions in product form. The simplest answer is that in many cases it works. A more subtle rationale is that of seeking solutions that are geometrically similar func- tions of xat different times. The idea of similarity — related to dimensional analysis — has been most fruitful in the mechanics of ﬂuids. Chapter Review See the CD for Review Questions. Miscellaneous Exercises Also see Review Questions on the CD. In Exercises 1–16, ﬁnd the steady-state solution, the associated eigenvalue problem, and the complete solution for each problem. 1.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,u(a,t)=T0,0<t, u(x,0)=T1,0<x<a. 2.∂2u ∂x2−γ2u=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,u(a,t)=T0,0<t, u(x,0)=T1,0<x<a. 3.∂2u ∂x2+r=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,u(a,t)=T0,0<t, u(x,0)=T1,0<x<a(ris constant). 4.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,∂u ∂x(a,t)=0, 0 <t, u(x,0)=T1x a,0<x<a. 5.∂2u ∂x2−γ2u=1 k∂u ∂t,0<x<a,0<t, Miscellaneous Exercises 207 ∂u ∂x(0,t)=0,∂u ∂x(a,t)=0, 0 <t, u(x,0)=T1x a,0<x<a. 6.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=0, u(a,t)=T0,0<t, u(x,0)=0, 0 <x<a. 7.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,u(a,t)=T0, u(x,0)=T0,0<x<a. 8.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, ∂u ∂x(0,t)=/Delta1T a,∂u ∂x(a,t)=/Delta1T a,0<t, u(x,0)=T0,0<x<a. 9.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,∂u ∂x(a,t)=0, 0 <t, u(x,0)=T1,0<x<a. 10.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, ∂u ∂x(0,t)=0,∂u ∂x(a,t)=0, 0 <t, u(x,0)=  T0,0<x<a 2, T1,a 2<x<a. 11.∂2u ∂x2=1 k∂u ∂t,0<x<∞,0<t, u(0,t)=T0,0<t, u(x,0)=T0/parenleftbig 1−e−αx/parenrightbig ,0<x. 12.∂2u ∂x2=1 k∂u ∂t,0<x<∞,0<t, 208 Chapter 2 The Heat Equation u(0,t)=T0,0<t, u(x,0)=/braceleftBig0, 0<x<a, T0,a<x. 13.∂2u ∂x2=1 k∂u ∂t,0<x<∞,0<t, ∂u ∂x(0,t)=0, 0 <t, u(x,0)=/braceleftbigg T0,0<x<a, 0, a<x. 14.∂2u ∂x2=1 k∂u ∂t,−∞<x<∞,0<t, u(x,0)=exp/parenleftbig −α|x|/parenrightbig ,−∞<x<∞. 15.∂2u ∂x2=1 k∂u ∂t,−∞<x<∞,0<t, u(x,0)=/braceleftBigg0,−∞<x<0, T0,0<x<a, 0, a<x<∞. 16.∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, ∂u ∂x(0,t)=0, u(a,t)=T0,0<t, u(x,0)=T0+S(a−x),0<x<a. 17. Give a physical interpretation for this problem and thus explain why u(x,t)should increase steadily as tincreases. (Assume that Sis a pos- itive constant.) ∂2u ∂x2=1 k∂u ∂t,0<x<a,0<t, ∂u ∂x(0,t)=0,∂u ∂x(a,t)=S,0<t, u(x,0)=0,0<x<a. 18. Show that v(x,t)=(S/2a)(x2+2kt)satisﬁes the heat equation and the boundary conditions of the problem in Exercise 17. Also ﬁnd w(x,t), deﬁned by u(x,t)=v(x,t)+w(x,t). 19. Show that the four functions u0=1,u1=x,u2=x2+2kt,u3=x3+6kxt Miscellaneous Exercises 209 are solutions of the heat equation. (These are sometimes called heat polynomials.) Find a linear combination of them that satisﬁes theboundary conditions u(0,t)=0,u(a,t)=t. 20.Suppose that u(x,t)is a positive function that satisﬁes ∂ 2u ∂x2=∂u ∂t. Show that the function w(x,t)=−2 u∂u ∂x satisﬁes the nonlinear partial differential equation called Burgers’ equa- tion: ∂w ∂t+w∂w ∂x=∂2w ∂x2. 21.Find a solution of the Burgers’ equation that satisﬁes the conditions w(0,t)=0,w ( 1,t)=0,0<t, w(x,0)=1,0<x<1. 22.Taking the function u(x,t)given here as a solution of the heat equation (with k=1), ﬁnd a solution wof Burgers’ equation. Verify that wsatis- ﬁes Burgers’ equation. u(x,t)=1√ 4πtexp/parenleftbigg−x2 4t/parenrightbigg . 23.Consider a solid metal bar surrounded by a ﬁnite quantity of water con- ﬁned in a water jacket. If the bar and the water are at different tempera-tures, they will exchange heat. Let u 1and u2be the temperatures in the bar and in the water, respectively. Heat balances for the water and the bar give these two equations: c1du1 dt=h(u2−u1), c2du2 dt=h(u1−u2). Here, c1and c2are the heat capacities of the bar and the water, respec- tively, and his the product of the convection coefﬁcient with the area of the bar–water interface. Find temperatures u1and u2assuming initial conditions u1(0)=T0,u2(0)=0. 210 Chapter 2 The Heat Equation 24. Solve the eigenvalue problem by setting φ(ρ)=ψ(ρ)/ρ : 1 ρ2/parenleftbig ρ2φ/prime/parenrightbig/prime+λ2φ=0,0<ρ< a, φ(0)bounded ,φ ( a)=0. Is this a regular Sturm–Liouville problem? Are the eigenfunctions or- thogonal? 25. Solve this problem for heat conduction in a sphere. (Hint: Let u(ρ,t)= v(ρ, t)/ρ.) 1 ρ2∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg =1 k∂u ∂t,0<ρ< a,0<t, u(0,t)bounded ,u(a,t)=0,0<t, u(ρ,0)=T0,0<ρ< a. 26. State and solve the eigenvalue problem associated with e−x∂ ∂x/parenleftbigg ex∂u ∂x/parenrightbigg =1 k∂u ∂t,0<x<a,0<t, u(0,t)=0,∂u ∂x(a,t)=0. 27. Find the steady-state solution of the problem ∂2u ∂x2+γ2/parenleftbig T(x)−u/parenrightbig =1 k∂u ∂t,0<x<a,0<t, u(0,t)=T0,∂u ∂x(a,t)=0,0<t, where T(x)=T0+Sx. 28. Determine whether or not λ=0i sa ne i g e n v a l u eo ft h ep r o b l e m φ/prime/prime+λ2xφ=0,0<x<a, φ/prime(0)=0,φ ( a)=0. 29. Same question as Exercise 28, but with boundary conditions φ/prime(0)=0,φ/prime(a)=0. Miscellaneous Exercises 211 30.Prove the following identity: 1√ 4πkt/integraldisplaya bexp/bracketleftbigg −(ξ−x)2 4kt/bracketrightbigg dξ=1 2/bracketleftbigg erf/parenleftbiggb−x√ 4kt/parenrightbigg −erf/parenleftbigga−x√ 4kt/parenrightbigg/bracketrightbigg . 31.In Exercise 6 of Section 10, it was shown that the function w(x,t;ω)=e−pxsin(ωt−px), where p=/radicalbig ω/2k, satisﬁes the heat equation and also the boundary con- dition w(0,t;ω)=sin(ωt). Show how to choose the coefﬁcient B(ω)so that the function u(x,t)=/integraldisplay∞ 0B(ω)e−pxsin(ωt−px)dω satisﬁes the boundary condition u(0,t)=f(t), 0<t for a suitable function t. 32.Use the idea of Exercise 31 to ﬁnd a solution of ∂2u ∂x2=1 k∂u ∂t, 0<x,0<t, u(0,t)=h(t), 0<t, where h(t)=/braceleftbigg 1,0<t<T, 0,T<t. 33.S.E. Serrano and T.E. Unny develop probabilistic mathematical models for groundwater ﬂow under uncertain conditions [Predicting groundwa- t e rﬂ o wi nap h r e a t i ca q u i f e r , Journal of Hydrology ,95(1987): 241–268], and compare the results to measurements. One of the models uses thisnonlinear Boussinesq equation, S∂y ∂t−∂ ∂x/parenleftbigg Kh∂y ∂x/parenrightbigg =I+φ, 0<x<L,0<t, together with the conditions y(0,t)=y1(t), y(L,t)=y2(t), 0<t, y(x,0)=y0(x), 0<x<L. 212 Chapter 2 The Heat Equation In these equations, y(x,t)is the water table elevation above sea level, h(x,t)is water table elevation above bedrock, Kis hydraulic conductiv- ity (in meters per day, m/da), Sis the aquifer speciﬁc yield, Iis a function representing input by percolation from the aquifer, and φ(x,t)is a ran- dom function that accounts for uncertainty in input. The partial differential equation is nonlinear because hand yrepresent the same thing relative to two different references. We assume that thebedrock elevation has constant slope a,s oy=h+ax. Then the equation can be written in terms of halone as S∂h ∂t−∂ ∂x/parenleftbigg Kh∂h ∂x/parenrightbigg −a∂ ∂x(Kh)=I+φ. N e x t ,t h i se q u a t i o ni sl i n e a r i z e d .A s s u m et h a t Kis constant and that h c a nb eb r o k e nd o w na s h=¯h+h/prime,w h e r e ¯his a constant mean value of h, h/primeis a ﬂuctuation much smaller than ¯h. (In this case, ¯his about 150 m and h/primeis less than 1 m.) Then the product Khis approximately equal to K¯h=T(transmissivity) and, as a coefﬁcient in the second term, can be t r e a t e da sac o n s t a n t .T h ee q u a t i o ni sn o wl i n e a ri n h/prime(we drop the prime for convenience): S∂h ∂t−T∂2h ∂x2−aK∂h ∂x=I+φ, 0<x<L,0<t. a.Treating Ias a constant, ﬁnd a steady-state solution v(x)for the sta- tistical mean value of h, which is obtained by replacing φ(x,t)with 0. The boundary and initial conditions are h(0,t)=h1,h(L,t)=h2,0<t, h(x,0)=h0(x), 0<x<L. b.State the problem (partial differential equation, boundary conditions, and initial condition) to be satisﬁed by the mean transient, w(x,t)= h(x,t)−v(x). (Again, the statistical mean corresponds to φ≡0.) c.Solve the problem in b. d.Values for the parameters are: a=0.0292 m/m, K=17.28 m/da, T= 218.4m2/da, S=0.15,L=116.25 m. Find the eigenvalues. 34. A ﬂat enzyme electrode can be visualized by imagining it seen from the side. The electrode itself lies to the left of x=0 (its thickness is unimpor- tant); a gel-containing enzyme lies in a layer between x=0a n d x=L; and the test solution lies to the right of x=L. When the substance to be detected is introduced into the solution, it diffuses into the gel and re-acts with the enzyme, yielding a product. The electrode responds to theproduct with a measurable electric potential. Miscellaneous Exercises 213 P .W. Carr [Fourier analysis of the transient response of potentiomet- ric enzyme electrodes, Analytical Chemistry ,49(1977): 799–802] stud- ied the transient response of such an electrode via two partial differentialequations that describe the concentrations, Sand P, of the substance be- ing detected and the enzyme-reaction product as they diffuse and react in the gel: ∂S ∂t=D∂2S ∂x2−VS K+S,0<x<L,0<t, (1∗) ∂P ∂t=D∂2P ∂x2+VS K+S,0<x<L,0<t. (2∗) In these equations, Vis the speciﬁc enzyme activity (mol/ml s), Kis a constant related to reaction rate, and Dis the diffusion constant (cm2/s), assumed to be the same for both substance and product. Reasonable boundary conditions are ∂S ∂x(0,t)=0,∂P ∂x(0,t)=0,0<t, (3∗) representing no reaction or penetration at the electrode surface, and S(L,t)=S0,P(L,t)=0,0<t, (4∗) where the gel meets the test solution. Initially, we assume S(x,0)=0,P(x,0)=0,0<x<L. (5∗) Equation (1 ∗) is nonlinear because the unknown function Sappears in the denominator of the last term. However, if Sis much smaller than K, we may replace K+SbyK,a n dE q .( 1 ∗)b e c o m e s ∂S ∂t=D∂2S ∂x2−V KS,0<x<L,0<t. (6∗) a.State and solve the steady-state problem for this equation, subject to the boundary conditions on Sin Eqs. (3 ∗)a n d( 4 ∗). b.Find the transient solution and then the complete solution S(x,t). 35.Refer to Exercise 34. Equation (2 ∗), though linear, is not easy to solve. However, if Eqs. (1 ∗)a n d( 2 ∗)a r ea d d e dt o g e t h e r ,t h en o n l i n e a rt e r m s cancel, leaving this homogeneous linear equation for the sum of the con-centrations: ∂(S+P) ∂t=D∂2(S+P) ∂x2,0<x<L,0<t. 214 Chapter 2 The Heat Equation Deﬁning u=S+P, ﬁnd the boundary and initial conditions for u,a n d solve completely. Then ﬁnd P(x,t)asu(x,t)−S(x,t). 36. Refer to Exercises 34 and 35. In order to determine the response time of the enzyme electrode, one wants to know the function P(0,t).A p - proximate this, using in your solution only steady-state terms and the ﬁrst term of each inﬁnite series. Sketch. Find the “time constants,” the multipliers of tin the exponential functions. 37. Consider a steel plate that is much larger in length and width ( x-a n d z- directions) than in thickness ( y-direction), and suppose the plate is free to expand or contract under the effects of heating. Assume that the tem-perature Tin the plate is a function of yand tonly. Timoshenko and Goodier ( Theory of Elasticity , pp. 399–403) derive the following expres- sion for the stresses due to thermal effects: σ x=σz=−αTE 1−ν+1 2c(1−ν)/integraldisplay+c −cαTE dy +3y 2c3(1−ν)/integraldisplay+c −cαTEy dy . The parameters, and their values for steel are as follows: αis the co- efﬁcient of expansion, 6 .5×10−6per degree F; Eis Y oung’s modulus, 28×106lb/in.2;νis Poisson’s ratio, 0 .7; and 2 cis the thickness of the plate. Note that the origin is located so that the plate lies between y=c and y=− c. a.Show that if T(y)=T0+Sy,w h e r e T0and Sare constants, then the thermal stress is 0. (This is a typical steady-state temperature distri- bution.) b.Suppose that the plate is initially at temperature 500◦F throughout and that the temperature on the face y=cis suddenly changed to 200◦w h i l et h et e m p e r a t u r ea t y=− cremains at 500◦.F i n d T(y,t). c.Assume the initial and boundary conditions given in b. Use your un- derstanding of the function T(y,t)to explain why the thermal stress near the face y=cis large just after time 0. The Wave Equation CHAPTER3 3.1 The Vibrating String A simple and historically important example of a problem that includes the wave equation is provided by the study of the vibration of a string, like a violinor guitar string. We set up a coordinate system as shown in Fig. 1. The un- known is the transverse displacement, u(x,t),m e a s u r e du pf r o mt h e x-axis. The situation is similar to that of the hanging cable discussed in Chapter 0,but here the string is taut, and of course motion is allowed. In order to ﬁnd theequation of motion of the string, we consider a short piece whose ends are atxand x+/Delta1xand apply Newton’s second law of motion to it. First, we must analyze the nature of the forces on the string. We assume that the only external force is the attraction of gravity, acting perpendicular to the x-direction. Internal forces are exerted on the segment by the rest of the string. We will assume that the string is perfectly ﬂexible and offers no resistance to bending. Then the only force that can be transmitted by the string is a pullor tension, which acts in a direction tangential to the centerline of the string.Its magnitude we denote by T(x,t). The forces on the small segment of string are shown in Fig. 2. We shall fur- ther assume that each point on the string moves only in the vertical direction . Thus, the horizontal component of acceleration is zero. Application of New-ton’s second law for the horizontal direction to the segment leads to the equa-tion −T(x,t)cos/parenleftbig φ(x,t)/parenrightbig +T(x+/Delta1x,t)cos/parenleftbig φ(x+/Delta1x,t)/parenrightbig =0, 215 216 Chapter 3 The Wave Equation Figure 1 String ﬁxed at the ends. Figure 2 Section of string showing forces exerted on it. The angles are α=φ(x,t)andβ=φ(x+/Delta1x,t). or T(x,t)cos/parenleftbig φ(x,t)/parenrightbig =T(x+/Delta1x,t)cos/parenleftbig φ(x+/Delta1x,t)/parenrightbig . (1) This says that the horizontal component of tension in the string is the same at every point: T(x,t)cos/parenleftbig φ(x,t)/parenrightbig =T(x+/Delta1x,t)cos/parenleftbig φ(x+/Delta1x,t)/parenrightbig =T, independent of x. If the string is taut, Tcan vary only slightly with t,s ow ew i l l assume that Tis constant. In the absence of external forces other than gravity, Newton’s second law for the vertical direction yields −T(x,t)sin/parenleftbig φ(x,t)/parenrightbig +T(x+/Delta1x,t)sin/parenleftbig φ(x+/Delta1x,t)/parenrightbig −mg=m∂2u ∂t2(x,t). (2) (Because u(x,t)measures displacement in the vertical direction, its second partial derivative with respect to tis the vertical acceleration.) The mass of the short piece of string we are examining is proportional to its length, m=ρ/Delta1x, where ρis the linear density, measured in units of mass per unit length. Now we use Eq. (1) to solve for the tensions at the ends of the segment of string as T(x,t)=T cos(φ(x,t)),T(x+/Delta1x,t)=T cos(φ(x+/Delta1x,t)). Chapter 3 The Wave Equation 217 When these expressions are substituted into Eq. (2), we have −Ttan/parenleftbig φ(x,t)/parenrightbig +Ttan/parenleftbig φ(x+/Delta1x,t)/parenrightbig −ρ/Delta1xg=ρ/Delta1x∂2u ∂t2. (3) Recall from elementary calculus that tan (φ(x,t))is the slope of the string at (x,t)and hence may be expressed in terms of the (partial) derivative with respect to x: tan/parenleftbig φ(x,t)/parenrightbig =∂u ∂x(x,t), tan/parenleftbig φ(x+/Delta1x,t)/parenrightbig =∂u ∂x(x+/Delta1x,t). Substituting these into Eq. (3), we have T/parenleftbigg∂u ∂x(x+/Delta1x,t)−∂u ∂x(x,t)/parenrightbigg =ρ/Delta1x/parenleftbigg∂2u ∂t2+g/parenrightbigg . On dividing through by /Delta1x, we see a difference quotient on the left: T /Delta1x/parenleftbigg∂u ∂x(x+/Delta1x,t)−∂u ∂x(x,t)/parenrightbigg =ρ/parenleftbigg∂2u ∂t2+g/parenrightbigg . In the limit as /Delta1x→0, the difference quotient becomes a partial derivative with respect to x, leaving Newton’s second law in the form T∂2u ∂x2=ρ∂2u ∂t2+ρg, (4) or ∂2u ∂x2=1 c2∂2u ∂t2+1 c2g, (5) where c2=T/ρ.I fc2is very large (usually on the order of 105m2/s2), we neglect the last term, giving the equation of the vibrating string ∂2u ∂x2=1 c2∂2u ∂t2,0<x<a,0<t. (6) This equation is called the wave equation in one dimension. Two- and three- dimensional versions will be treated in Chapter 5. In describing the motion of an object, one must specify not only the equa- tion of motion, but also both the initial position and velocity of the object. The initial conditions for the string, then, must state the initial displacementof every particle — that is, u(x,0)— and the initial velocity of every particle, ∂u/∂t(x,0). For the vibrating string as we have described it, the boundary conditions are zero displacement at the ends, so the boundary value–initial value problem for 218 Chapter 3 The Wave Equation the string is ∂2u ∂x2=1 c2∂2u ∂t2, 0<x<a, 0<t, (7) u(0,t)=0, u(a,t)=0,0<t, (8) u(x,0)=f(x), 0<x<a, (9) ∂u ∂t(x,0)=g(x), 0<x<a, (10) under the assumptions noted plus the assumption that gravity is negligible. EXERCISES 1.Find the dimensions of each of the following quantities, using the facts that force is equivalent to mL/t2, and that the dimension of tension is F(force): u,∂2u/∂x2,∂2u/∂t2,c,g/c2. Check the dimension of each term in Eq. (5). 2.Suppose a distributed vertical force F(x,t)(positive upwards) acts on the string. Derive the equation of motion: ∂2u ∂x2=1 c2∂2u ∂t2−1 TF(x,t). The dimension of a distributed force is F/L. (If the weight of the string is considered as a distributed force and is the only one, then we would haveF(x,t)=−ρg. Check dimensions and signs.) 3.Find a solution v(x)of Eq. (5) with boundary conditions Eq. (8) that is independent of time. (This corresponds to a “steady-state solution,” butthe term steady-state is no longer appropriate. Equilibrium solution is more accurate.) 4.Suppose that the string is located in a medium that resists its movement, such as air. The resistance is expressed as a force opposite in direction andproportional in magnitude to velocity. Thus it affects only Eq. (2). Proceedto derive the equation that replaces Eq. (7) for this case. 3.2 Solution of the Vibrating String Problem The initial value–boundary value problem that describes the displacement ofthe vibrating string, 3.2 Solution of the Vibrating String Problem 219 ∂2u ∂x2=1 c2∂2u ∂t2, 0<x<a, 0<t, (1) u(0,t)=0, u(a,t)=0,0<t, (2) u(x,0)=f(x), 0<x<a, (3) ∂u ∂t(x,0)=g(x), 0<x<a, (4) contains a linear, homogeneous partial differential equation and linear, homo- geneous boundary conditions. Thus we may apply the method of separation ofvariables with hope of success. If we assume that 1u(x,t)=φ(x)T(t), Eq. (1) becomes φ/prime/prime(x)T(t)=1 c2φ(x)T/prime/prime(t). Dividing through by φT,w eo b t a i n φ/prime/prime(x) φ(x)=T/prime/prime(t) c2T(t),0<x<a,0<t. For the equality to hold, both members of this equation must be constant. We write the constant as −λ2and separate the preceding equation into two ordinary differential equations linked by the common parameter λ2: T/prime/prime+λ2c2T=0,0<t, (5) φ/prime/prime+λ2φ=0,0<x<a. (6) The boundary conditions become φ(0)T(t)=0,φ ( a)T(t)=0,0<t and, since T(t)≡0 gives a trivial solution for u(x,t),w em u s th a v e φ(0)=0,φ ( a)=0. (7) T h ee i g e n v a l u ep r o b l e mE q s .( 6 )a n d( 7 )i se x a c t l yt h es a m ea st h eo n ew e have seen and solved before. (See Chapter 2, Section 3.) We know that theeigenvalues and eigenfunctions are λ 2 n=/parenleftbiggnπ a/parenrightbigg2 ,φ n(x)=sin(λnx), n=1,2,3,.... Equation (5) is also of a familiar type, and its solution is known to be Tn(t)=ancos(λnct)+bnsin(λnct), 1Tno longer symbolizes tension. 220 Chapter 3 The Wave Equation where anandbna r ea r b i t r a r y .( I no t h e rw o r d s ,t h e r ea r et w oi n d e p e n d e n ts o l u - tions.) Note, however, that there is a substantial difference between the Tthat arises here and the Tthat we found in the heat conduction problem. The most important difference is the behavior as ttends to inﬁnity. In the heat conduc- tion problem, T(t)tends to 0, whereas here T(t)has no limit but oscillates periodically in agreement with our intuition. For each n=1,2,3,...,w en o wh a v ep r o d u c ts o l u t i o n s un(x,t)=sin(λnx)/bracketleftbig ancos(λnct)+bnsin(λnct)/bracketrightbig . (8) Such solutions are called standing waves .F o rap a r t i c u l a r anand bn,un(x,t) maintains the same shape with a variable, periodic amplitude. For any choiceofa nand bn,un(x,t)is a solution of the homogeneous partial differential equation (1) and also satisﬁes the boundary conditions Eq. (2). Some standingwaves are shown animated on the CD. By the Principle of Superposition, linear combinations of the u n(x,t)also satisfy both Eqs. (1) and (2). In making our linear combinations, we need no new constants because the anand bnare arbitrary. We have, then, u(x,t)=∞/summationdisplay n=1sin(λnx)/bracketleftbig ancos(λnct)+bnsin(λnct)/bracketrightbig . (9) The initial conditions, which remain to be satisﬁed, have the form u(x,0)=∞/summationdisplay n=1ansin/parenleftbiggnπx a/parenrightbigg =f(x), 0<x<a, ∂u ∂t(x,0)=∞/summationdisplay n=1bnnπ acsin/parenleftbiggnπx a/parenrightbigg =g(x), 0<x<a. (Here we have assumed that ∂u ∂t(x,t)=∞/summationdisplay n=1sin(λnx)/bracketleftbig −anλncsin(λnct)+bnλnccos(λnct)/bracketrightbig . In other words, we assume that the series for umay be differentiated term by term.) Both initial conditions take the form of Fourier series problems: A givenfunction is to be expanded in a series of sines. In each case, then, the constantmultiplying sin (nπx/a)must be the Fourier sine coefﬁcient for the given func- tion. Thus we determine that a n=2 a/integraldisplaya 0f(x)sin/parenleftbiggnπx a/parenrightbigg dx, (10) 3.2 Solution of the Vibrating String Problem 221 and bnnπ ac=2 a/integraldisplaya 0g(x)sin/parenleftbiggnπx a/parenrightbigg dx or bn=2 nπc/integraldisplaya 0g(x)sin/parenleftbiggnπx a/parenrightbigg dx. (11) If the functions f(x)andg(x)are sectionally smooth on the interval 0 <x< a, then we know that the initial conditions really are satisﬁed, except possibly at points of discontinuity of forg. By the nature of the problem, however, one would expect that f, at least, would be continuous and would satisfy f(0)= f(a)=0. Thus we expect the series for fto converge uniformly. Example. If the string is lifted in the middle and then released, appropriate initial condi- tions are u(x,0)=f(x)=  h·2x a, 0<x<a 2, h/parenleftbigg 2−2x a/parenrightbigg ,a 2<x<a, ∂u ∂t(x,0)=g(x)≡0,0<x<a. Then bn=0 for n=1,2,3,...,a n d an=2 a/bracketleftbigg/integraldisplaya/2 0h·2x asin/parenleftbiggnπx a/parenrightbigg dx+/integraldisplaya a/2h/parenleftbigg 2−2x a/parenrightbigg sin/parenleftbiggnπx a/parenrightbigg dx/bracketrightbigg =8h π2sin(nπ/2) n2. Therefore the complete solution is u(x,t)=8h π2∞/summationdisplay n=1sin(nπ/2) n2sin/parenleftbiggnπx a/parenrightbigg cos/parenleftbiggnπct a/parenrightbigg . (12) The CD shows an animated version of this solution. /square Although the solution in the example can be considered valid, it is difﬁcult to see, in the present form, what shape the string will take at various times.However, because of the simplicity of the sines and cosines, it is possible torewrite the solution in such a way that u(x,t)may be determined without summing a series. 222 Chapter 3 The Wave Equation By applying the trigonometric identity sin(A)cos(B)=1 2/bracketleftbig sin(A−B)+sin(A+B)/bracketrightbig we can express u(x,t)as u(x,t)=1 2/bracketleftBigg 8h π2∞/summationdisplay n=1sin(nπ/2) n2sin/parenleftbiggnπ(x−ct) a/parenrightbigg +8h π2∞/summationdisplay n=1sin(nπ/2) n2sin/parenleftbiggnπ(x+ct) a/parenrightbigg/bracketrightBigg . We know that the series 8h π2∞/summationdisplay n=1sin(nπ/2) n2sin/parenleftbiggnπx a/parenrightbigg actually converges to the odd periodic extension, with period 2 a,o ft h ef u n c - tion f(x). Let us designate this extension by ¯fo(x)a n dn o t et h a ti ti sd e ﬁ n e df o r all values of its argument. Using this observation, we can express u(x,t)more simply as u(x,t)=1 2/bracketleftbig¯fo(x−ct)+¯fo(x+ct)/bracketrightbig . (13) In this form, the solution u(x,t)can easily be sketched for various values oft.T h eg r a p ho f ¯fo(x+ct)has the same shape as that of ¯fo(x)but is shifted ct units to the left. Similarly, the graph of ¯fo(x−ct)is the graph of ¯fo(x)shifted ctunits to the right. When the graphs of ¯fo(x+ct)and¯fo(x−ct)are drawn on the same axes, they may be averaged graphically to get the graph of u(x,t). In Fig. 3 are graphs of ¯fo(x+ct),¯fo(x−ct),a n d u(x,t)=1 2/bracketleftbig¯fo(x+ct)+¯fo(x−ct)/bracketrightbig for the particular example discussed here and for various values of t.T h ed i s - placement u(x,t)is periodic in time, with period 2 a/c. During the second half-period (not shown), the string returns to its initial position through the positions shown. The horizontal portions of the string have a nonzero veloc-ity. Equation (12) can also be used to ﬁnd u(x,t)for any given xand t.F o r instance, if we take x=0.2aand t=0.9a/c,w eﬁ n dt h a t u/parenleftbigg 0.2a,0.9a c/parenrightbigg =1 2/bracketleftbig¯fo(−0.7a)+¯fo(1.1a)/bracketrightbig =1 2/bracketleftbig (−0.6h)+(−0.2h)/bracketrightbig =− 0.4h. 3.2 Solution of the Vibrating String Problem 223 Figure 3 On the left are the graphs of ¯fo(x+ct)(solid) and ¯fo(x−ct)(dashed) for the given value of ct. On the right is the graph of u(x,t)for 0<x<a,m a d e by averaging the graphs on the left. The function values can be read directly from a graph of f(x). The manipula- tions with the series solution in the example can be done for any f(x).T h e r e - fore the formula of Eq. (13) is a solution of Eqs. (1)–(4) for any f(x),p r o v i d e d that g(x)≡0. We will generalize in later sections. Frequencies of Vibration The product solutions in Eq. (8) provide important information about pos- sible frequencies of vibration. The multipliers λncin the sines and cosines of 224 Chapter 3 The Wave Equation tare frequencies, in radians per unit time; λnc/2πare frequencies in cycles per unit time (or Hertz, if the time unit is seconds). For the vibrating stringproblem, the possible frequencies of vibration are (nπ/a)c 2π=nπc 2a. The fact that these form an arithmetic sequence guarantees a common period for all the un(x,t)in Eq. (8), and thus u(x,t)in Eq. (9) is a function that is periodic in time. EXERCISES 1.Verify that the product solution un(x,t)=sin(λnx)/bracketleftbig ancos(λnct)+bnsin(λnct)/bracketrightbig satisﬁes the wave equation (1) and the boundary conditions, Eq. (2). 2.Sketch u1(x,t)andu2(x,t)as functions of xfor several values of t.A s s u m e a1and a2=1,b1and b2=0. (The solutions un(x,t)are called standing waves .) In Exercises 3–5, solve the vibrating string problem, Eqs. (1)–(4), with the ini- tial conditions given. 3.f(x)=0, g(x)=1, 0 <x<a. 4.f(x)=sin/parenleftbiggπx a/parenrightbigg ,g(x)=0, 0 <x<a. 5.f(x)=/braceleftbiggU0,0<x<a/2, 0, a/2<x<a, g(x)=0, 0 <x<a. (This initial condition is difﬁcult to justify for a vibrating string, but it may be reasonable where the unknown function is pressure in a pipe with a membrane at the midpoint. See Miscellaneous Exercise 18 of this chapter for some derivations.) 6.If ∞/summationdisplay n=1ansin/parenleftbiggnπx a/parenrightbigg =¯fo(x),∞/summationdisplay n=1bncos/parenleftbiggnπx a/parenrightbigg =¯Ge(x), show that u(x,t)a sg i v e ni nE q .( 9 )m a yb ew r i t t e n 3.2 Solution of the Vibrating String Problem 225 u(x,t)=1 2/parenleftbig¯fo(x−ct)+¯fo(x+ct)/parenrightbig +1 2/parenleftbig¯Ge(x+ct)−¯Ge(x−ct)/parenrightbig . Here,¯fo(x)and¯Ge(x)are periodic with period 2 a. 7.The pressure of the air in an organ pipe satisﬁes the equation ∂2p ∂x2=1 c2∂2p ∂t2,0<x<a,0<t, with the boundary conditions ( p0is atmospheric pressure) a.p(0,t)=p0,p(a,t)=p0if the pipe is open, or b.p(0,t)=p0,∂p ∂x(a,t)=0 if the pipe is closed at x=a. Find the eigenvalues and eigenfunctions associated with the wave equation for each of these sets of boundary conditions. 8.Find the lowest frequency of vibration of the air in the organ pipes referredto in Exercise 7aandb. 9.If a string vibrates in a medium that resists the motion, the problem for the displacement of the string is ∂2u ∂x2=1 c2∂2u ∂t2+k∂u ∂t, 0<x<a,0<t, u(0,t)=0,u(a,t)=0,0<t plus initial conditions. Find eigenfunctions, eigenvalues, and product so- lutions for this problem. (Assume that kis small and positive.) 10.For the problem in Exercise 9, ﬁnd frequencies of vibration and show that they do notform an arithmetic sequence. If we form a series solution, will it be periodic? What happens to u(x,t)ast→∞ ? 11.The displacements u(x,t)of a uniform thin beam satisfy ∂4u ∂x4=−1 c2∂2u ∂t2,0<x<a,0<t. If the beam is simply supported at the ends, the boundary conditions are u(0,t)=0,∂2u ∂x2(0,t)=0,u(a,t)=0,∂2u ∂x2(a,t)=0. Find product solutions to this problem. What are the frequencies of vibra- tion? 12.Write out formulas for the ﬁrst four frequencies of vibration for a thinbeam (Exercise 11) and for a string (text). Then ﬁnd their values, assum-ing that parameters cand ahave values that make the lowest frequency of 226 Chapter 3 The Wave Equation Figure 4 Shapes of car antenna. e a c he q u a lt o2 5 6c y c l e sp e rs e c o n d .T h ed i f f e r e n c ei nt h es e to ff r e q u e n - cies accounts for some of the difference between the sound of a stringedinstrument and that of a xylophone or glockenspiel. 13.My car’s antenna vibrates in the wind under various conditions in one of the two shapes shown in Fig. 4. If the antenna is modeled as a uniform thin beam with centerline displacement u(x,t),t h e n usatisﬁes the equation ∂4u ∂x4=−1 c2∂2u ∂t2+f(x,t), 0<x<a,0<t, where fis a “forcing function” that represents the effect of wind or other distributed forces. Because the base of the antenna is built into the car, theboundary conditions at the base are zero displacement and slope: u(0,t)=0,∂u ∂x(0,t)=0,0<t. The top of the antenna is free to move. There, the internal moment and shear are both zero, leading to the conditions ∂2u ∂x2(a,t)=0,∂3u ∂x3(a,t)=0,0<t. (These four boundary conditions are standard for a cantilevered beam.) It can be shown that the solution of the foregoing problem, together with initial conditions on uand ut, can be represented as a series of products of the form /parenleftbig ancos/parenleftbig λ2 nct/parenrightbig +bnsin/parenleftbig λ2 nct/parenrightbig +Fn(t)/parenrightbig φn(x), where Fn(t)comes from the forcing function and φn(x)andλnare related through the eigenvalue problem φ/prime/prime/prime/prime−λ4φ=0,0<x<a, φ(0)=0,φ/prime(0)=0,φ/prime/prime(a)=0,φ/prime/prime/prime(a)=0. This arises in the obvious way from the boundary conditions and the ho- mogeneous partial differential equation. 3.3 d’Alembert’s Solution 227 Solve the eigenvalue problem, sketch the ﬁrst two eigenfunctions, and compare them to the ﬁgure. In Exercises 14–16, ﬁnd a solution by separation of variables. 14.∂2u ∂x2=1 c2/parenleftbigg∂2u ∂t2+2k∂u ∂t/parenrightbigg ,0<x<a,0<t, u(0,t)=0, u(a,t)=0, 0 <t, u(x,0)=f(x),∂u ∂t(x,0)=0, 0 <x<a, where f(x)is as in Eq. (11). (Assume that kis small and positive.) 15.∂2u ∂x2=1 c2∂2u ∂t2+γ2u,0<x<a,0<t, u(0,t)=0, u(a,t)=0, 0 <t, u(x,0)=h,∂u ∂t(x,0)=0, 0 <x<a, where handγ2are constants. 16.∂2u ∂x2=1 c2∂2u ∂t2,0<x<a,0<t, u(0,t)=0,∂u ∂x(a,t)=0, 0 <t, u(x,0)=0,∂u ∂t(x,0)=1, 0 <x<a. 17.Does the series in Eq. (12) converge uniformly? 18.In the text, we assumed that the ratio φ/prime/prime/φhad to be a negative con- stant. Show that, if φ/prime/prime/φ=p2>0 (or equivalently, if φ/prime/prime−p2φ=0), then the only function that also satisﬁes the boundary conditions, Eq. (7), is φ(x)≡0. 3.3 d’Alembert’s Solution In Section 2 we saw that, in some cases, we could express the solution of the wave equation directly in terms of the initial data. From this evidence we mightsuspect that there is something special about x+ctandx−ct.T ot e s tt h i si d e a , we change variables and see what the wave equation looks like. Let w=x+ct, z=x−ct,a n d u(x,t)=v(w, z). Then a calculation using the chain rule shows that the wave equation becomes (see Exercise 11) ∂ 2v ∂z∂w=0. 228 Chapter 3 The Wave Equation It is actually possible to ﬁnd the general solution of this last equation. Put in another form it says ∂ ∂z/parenleftbigg∂v ∂w/parenrightbigg =0, which means that ∂v/∂w is independent of z,o r ∂v ∂w=θ(w). Integrating this equation, we ﬁnd that v=/integraldisplay θ(w) dw+φ(z). Here, φ(z)plays the role of an integration “constant.” Since the integral of θ(w) is also a function of w,w em a yw r i t et h e general solution of the partial differential equation foregoing as v(w, z)=ψ(w) +φ(z), where ψandφarearbitrary functions with continuous derivatives. Trans- forming back to our original variables, we obtain u(x,t)=ψ(x+ct)+φ(x−ct) (1) as a form for the general solution of the one-dimensional wave equation. This is known as d’Alembert’s solution or the traveling wave solution . It represents the solution as the superposition of two waves, one moving to the left and theother to the right, with propagation speed c. Now let us look at the vibrating string problem: ∂ 2u ∂x2=1 c2∂2u ∂t2, 0<x<a, 0<t, (2) u(0,t)=0, u(a,t)=0,0<t, (3) u(x,0)=f(x), 0<x<a, (4) ∂u ∂t(x,0)=g(x), 0<x<a. (5) We already know a form for u. The problem is to choose ψandφin such a way that the initial and boundary conditions are satisﬁed. We assume then that u(x,t)=ψ(x+ct)+φ(x−ct). The initial conditions are ψ(x)+φ(x)=f(x), 0<x<a, cψ/prime(x)−cφ/prime(x)=g(x),0<x<a.(6) 3.3 d’Alembert’s Solution 229 If we divide through the second equation by cand integrate, it becomes ψ(x)−φ(x)=G(x)+A,0<x<a, (7) where G(x)stands for G(x)=1 c/integraldisplayx 0g(y)dy (8) andAis an arbitrary constant. Equations (6) and (7) can now be solved simul- taneously to determine ψ(x)=1 2/parenleftbig f(x)+G(x)+A/parenrightbig ,0<x<a, φ(x)=1 2/parenleftbig f(x)−G(x)−A/parenrightbig ,0<x<a. These equations give ψandφonly for values of the argument between 0 and a.B u t x±ctmay take on any value whatsoever, so we must extend these functions to deﬁne them for arbitrary values of their arguments. Let us desig-nate them as ψ(x)=1 2/parenleftbig˜f(x)+˜G(x)+A/parenrightbig , φ(x)=1 2/parenleftbig˜f(x)−˜G(x)−A/parenrightbig , where ˜fand˜Gare some extensions of fandG.( T h a ti s ˜f(x)=f(x)and˜G(x)= G(x)for 0<x<a.) However we choose these extensions, the wave equation and the initial conditions are satisﬁed. Thus, they must be determined by theboundary conditions, u(0,t)=ψ(ct)+φ(−ct)=0,t>0, (9) u(a,t)=ψ(a+ct)+φ(a−ct)=0,t>0. (10) The ﬁrst of these equations says that ˜f(ct)+˜G(ct)+A+˜f(−ct)−˜G(−ct)−A=0, or ˜f(ct)+˜f(−ct)+˜G(ct)−˜G(−ct)=0. As these equations must be true for arbitrary functions fand G(because the two functions are not interdependent), we must have individually ˜f(ct)=−˜f(−ct),˜G(ct)=˜G(−ct). (11) That is, ˜fis odd and ˜Gis even. 230 Chapter 3 The Wave Equation At the second endpoint, a similar calculation shows that ˜f(a+ct)+˜f(a−ct)+˜G(a+ct)−˜G(a−ct)=0. Once again, the independence of fand Gimplies that ˜f(a+ct)=−˜f(a−ct),˜G(a+ct)=˜G(a−ct). (12) The oddness of ˜fand evenness of ˜Gcan be used to transform the right-hand sides. Then ˜f(a+ct)=˜f(−a+ct),˜G(a+ct)=˜G(−a+ct). These equations say that ˜fand˜Gare both periodic with period 2 a,b e c a u s e changing their arguments by 2 adoes not change the functional value. Thus we want ˜fto be the odd periodic extension of fand˜Gto be the even periodic extension of G. In the notation we used in Chapter 1, the explicit expressions forφandψare ψ(x+ct)=1 2/parenleftbig¯fo(x+ct)+¯Ge(x+ct)+A/parenrightbig , φ(x−ct)=1 2/parenleftbig¯fo(x−ct)−¯Ge(x−ct)−A/parenrightbig . Finally, we arrive at an expression for the solution u(x,t): u(x,t)=1 2/bracketleftbig¯fo(x+ct)+¯fo(x−ct)/bracketrightbig +1 2/bracketleftbig¯Ge(x+ct)−¯Ge(x−ct)/bracketrightbig .(13) The CD shows an animated version of Fig. 3 (Section 3.2) using this form of the solution. This form of the solution of Eqs. (2)–(5) allows us to see directly how the initial data inﬂuence the solution at later times. From a practical point of view,it also permits us to calculate u(x,t)at any xand tand even to sketch uas a function of one variable for a ﬁxed value of the other. The following procedureis helpful in sketching u(x,t)as a function of xfor a ﬁxed t=t ∗, when the initial condition (5) has g(x)≡0. It is easily adapted to other cases. 1.Sketch the odd periodic extension of f;c a l lt h i s ¯fo(x). 2.Sketch ¯fo(x+ct∗)against x;t h i si sj u s tt h eg r a p ho f ¯foshifted ct∗units to the left. 3.Sketch ¯fo(x−ct∗)against xon the same axes. This graph is the same as that of ¯fobut shifted ct∗units to the right. 4.Average graphically the graphs made in the two preceding steps. Check that the boundary conditions are satisﬁed. 3.3 d’Alembert’s Solution 231 Similarly, if f(x)≡0, sketch G(x)and its even periodic extension ¯Ge(x). Then sketch the graphs of ¯Ge(x+ct∗)(same shape as ¯Ge(x)but shifted ct∗ units to the left) and −¯Ge(x−ct∗)(graph of ¯Ge(x)shifted ct∗units to the right and reﬂected in the horizontal axis). These two are then averaged graph- ically to obtain the graph of u(x,t∗). Check that the boundary conditions are satisﬁed. EXERCISES 1.Letu(x,t)be a solution of Eqs. (2)–(5), with g(x)≡0a n d f(x)af u n c t i o n whose graph is an isosceles triangle of width aand height h.F i n d u(x,t) forx=0.25aand 0.5aand for t=0, 0.2a/c,0.4a/c,0.8a/c,1.4a/c. 2.Sketch u(x,t)of Exercise 1 as a function of xfor the times given. Compare your results with Fig. 3. 3.Letu(x,t)be a solution of Eqs. (2)–(5), with f(x)≡0a n d g(x)=αc,0< x<a.F i n d u(x,t)at:x=0,t=0.5a/c;x=0.2a,t=0.6a/c;x=0.5a, t=1.2a/c. 4.Sketch u(x,t)of Exercise 3 as a function of xfor times t=0, 0.25a/c, 0.5a/c,a/c. 5.Find the function G(x)corresponding to (see Eq. (8)) g(x)=/braceleftBigg0,0<x<0.4a, 5c,0.4a<x<0.6a, 0,0.6a<x<a. 6.Justify this alternate description of the function G(x),t h a ti ss p e c i ﬁ e di n Eq. (8): Gis the solution of the initial value problem dG dx=1 cg(x), 0<x, G(0)=0. 7.Using Eq. (8) or Exercise 6, sketch the function G(x)of Exercise 5. 8.Letu(x,t)be the solution of the vibrating string problem, Eqs. (2)–(5), with f(x)≡0a n d g(x)as in Exercise 5. Sketch u(x,t)as a function of xfor times ct=0, 0.2a,0.4a,0.5a,a,1.2a. Hint: Sketch ¯Ge(x+ct)and −¯Ge(x−ct); then average them graphically. 232 Chapter 3 The Wave Equation 9.Sketch the solution of the vibrating string problem, Eqs. (2)–(5), at times ct=0, 0.1a,0.3a,0.4a,0.5a,0.6a,i fg(x)=0a n d f(x)=  0, 0<x<0.4a, 10h(x−0.4a),0.4a<x<0.5a, 10h(0.6a−x),0.5a<x<0.6a, 0, 0.6a<x<a. 10.Ver ify direc tly that u(x,t)as given by Eq. (1) is a solution of the wave equation (2) if φandψhave at least two derivatives. 11.Use the change of variables at the beginning of this section to transform t h ew a v ee q u a t i o n .Y o un e e dt ou s et h ec h a i nr u l ee x t e n s i v e l y — f o ri n -stance, ∂u ∂t=∂v ∂w∂w ∂t+∂v ∂z∂z ∂t=∂v ∂w·c+∂v ∂z·(−c). In this way, ﬁnd expressions for the second derivatives of u, and then sub- stitute into the wave equation, ∂2u ∂x2=1 c2∂2u ∂t2. 12.The equation for the forced vibrations of a string is ∂2u ∂x2−1 c2∂2u ∂t2=−1 TF(x,t) (∗) (see Section 1, Exercise 2). Changing variables to w=x+ct,z=x−ct,u(x,t)=v(w, z), f(w,z)=F(x,t), this equation becomes ∂2v ∂w∂ z=−1 4Tf(w,z). Show that the general solution of this equation is v(w, z)=−1 4T/integraldisplay/integraldisplay f(w,z)dwdz+ψ(w) +φ(z). 13.Find the general solution of Eq. ( ∗)i nE x e r c i s e1 2i nt e r m so f xand t,i f F(x,t)=Tcos(t). 3.4 One-Dimensional Wave Equation: Generalities 233 3.4 One-Dimensional Wave Equation: Generalities As for the one-dimensional heat equation, we can make some comments for a generalized one-dimensional wave equation. For the sake of generality, weassume that some nonuniform properties are present. For the sake of simplic-ity, we assume that the equation is homogeneous and free of u. Our initial value–boundary value problem will be ∂ ∂x/parenleftbigg s(x)∂u ∂x/parenrightbigg =p(x) c2∂2u ∂t2, l<x<r,0<t, (1) α1u(l,t)−α2∂u ∂x(l,t)=c1,0<t, (2) β1u(r,t)+β2∂u ∂x(r,t)=c2,0<t, (3) u(x,0)=f(x), l<x<r, (4) ∂u ∂t(x,0)=g(x), l<x<r. (5) We assume that the functions s(x)and p(x)are positive for l≤x≤r,b e - cause they represent physical properties, that s,s/prime,a n d pare all continuous, and that sand phave no dimensions. Also, suppose that none of the coefﬁ- cients α1,α2,β1,β2are negative. T o obtain homogeneous boundary conditions we can write u(x,t)=v(x)+w(x,t), just as before. In the wave equation, however, neither of the names “steady- state solution” nor “transient solution” is appropriate; for, as we shall see, thereis no steady state, or limiting case, nor is there a part of the solution that tendsto zero as ttends to inﬁnity. Nevertheless, vrepresents an equilibrium solu- tion, and, more important, it is a useful mathematical device to consider uin the form provided in the preceding equation. The function v(x)is required to satisfy the conditions (sv /prime)/prime=0,l<x<r, α1v(l)−α2v/prime(l)=c1, β1v(r)+β2v/prime(r)=c2. Thusv(x)is exactly equivalent to the “steady-state solution” discussed for the heat equation. 234 Chapter 3 The Wave Equation The function w(x,t), being the difference between u(x,t)andv(x),s a t i s ﬁ e s the initial value–boundary value problem ∂ ∂x/parenleftbigg s(x)∂w ∂x/parenrightbigg =p(x) c2∂2w ∂t2,l<x<r,0<t, (6) α1w(l,t)−α2∂w ∂x(l,t)=0,0<t, (7) β1w(r,t)+β2∂w ∂x(r,t)=0,0<t, (8) w(x,0)=f(x)−v(x), l<x<r, (9) ∂w ∂t(x,0)=g(x), l<x<r. (10) Since the equation and the boundary conditions are homogeneous and lin- ear, we attempt a solution by separation of variables. If w(x,t)=φ(x)T(t),w e ﬁnd in the usual way that the factor functions φand Tmust satisfy T/prime/prime+c2λ2T=0,0<t, (11) /parenleftbig s(x)φ/prime/parenrightbig/prime+λ2p(x)φ=0,l<x<r, (12) α1φ(l)−α2φ/prime(l)=0, (13) β1φ(r)+β2φ/prime(r)=0. (14) The eigenvalue problem represented in the last three lines is a regular Sturm–Liouville problem, because of the assumptions we have made about s,p, and the coefﬁcients. We know that there are an inﬁnite number of non- negative eigenvalues λ2 1,λ22,... and corresponding eigenfunctions φ1,φ2,... that have the orthogonality property /integraldisplayr lφn(x)φm(x)p(x)dx=0,n/negationslash=m. The solution of the equation for Tis Tn(t)=ancos(λnct)+bnsin(λnct). From here it is clear that the frequencies of vibration that occur in the solution of Eqs. (1)–(3) are λnc/2π(cycles per time unit). Thus, it is the eigenvalues coming from Eqs. (12)–(14) that determine these frequencies. Having solved the subsidiary problems that arose after separation of vari- ables, we can begin to assemble the solution. The function wwill have the form w(x,t)=∞/summationdisplay n=1φn(x)/parenleftbig ancos(λnct)+bnsin(λnct)/parenrightbig , (15) 3.4 One-Dimensional Wave Equation: Generalities 235 and its two initial conditions, yet to be satisﬁed, are w(x,0)=∞/summationdisplay n=1anφn(x)=f(x)−v(x), l<x<r, ∂w ∂t(x,0)=∞/summationdisplay n=1bnλncφn(x)=g(x), l<x<r. By employing the orthogonality of the φn, we determine that the coefﬁcients anand bnare given by an=1 In/integraldisplayr l/bracketleftbig f(x)−v(x)/bracketrightbig φn(x)p(x)dx, (16) bn=1 Inλnc/integraldisplayr lg(x)φn(x)p(x)dx, (17) where In=/integraldisplayr lφ2 n(x)p(x)dx. (18) Finally, u(x,t)=v(x)+w(x,t)is the solution of the original problem, and each of its parts is completely speciﬁed. From the form of w(x,t),w ec a nm a k e certain observations about u. 1.u(x,t)does not have a limit as t→∞ . Each term of the series form of w is periodic in time and thus does not die away. 2. Except in very special cases, the eigenvalues λ2 nare not closely related to each other. So in general, if ucauses acoustic vibrations, the result will not be musical to the ear. (A sound would be musical if, for instance, λn=nλ1, as in the case of the uniform string.) 3. In general, u(x,t)is not even periodic in time. Although each term in the series for wis periodic, the terms do not have a common period (except in special cases), and so the sum is not periodic. EXERCISES 1.Verify the formulas for the anand bn. Under what conditions on fand g can we say that the initial conditions are satisﬁed? 2.Check the statement that v(x)is the same for the heat conduction problem and for the problem considered here. 3.Identify the period of Tn(t)and the associated frequency. 236 Chapter 3 The Wave Equation 4.Although u(x,t)has no limit as t→∞ , show that the following general- ized limit is valid: v(x)=lim T→∞1 T/integraldisplayT 0u(x,t)dt. (Hint: Do the integration and limiting term by term.) 5.Formally solve the problem ∂ ∂x/parenleftbigg s(x)∂u ∂x/parenrightbigg =p(x) c2/parenleftbigg∂2u ∂t2+γ∂u ∂t/parenrightbigg +q(x)u,l<x<r,0<t, with the boundary conditions Eqs. (2) and (3) and initial conditions Eqs. (4) and (5), taking γto be constant. 6.Ver ify that w(x,t)as given in Eq. (15) satisﬁes the differential equation and the boundary conditions Eqs. (6)–(8). 7.In reference to the observations at the end of the section, prove the follow-ing statement: The product solutions of the problem in Eqs. (6)–(8) all havea common period in time if the eigenvalues of the problem in Eqs. (12)– (14) satisfy the relation λ n=α(n+β), where βis a rational number. 8.Find a separation-of-variables solution of the problem ∂u2 ∂x2=1 c2/parenleftbigg∂2u ∂t2+γ2u/parenrightbigg , 0<x<a,0<t, u(0,t)=0,u(a,t)=0, 0<t, u(x,t)=f(x),∂u ∂t(x,0)=g(x), 0<x<a. Is this an instance of the problem in this section? Which of the observations at the end of the section are valid for the solution of this problem? 3.5 Estimation of Eigenvalues In many instances, one is interested not in the full solution to the wave equa- tion, but only in the possible frequencies of vibration that may occur. For ex-ample, it is of great importance that bridges, airplane wings, and other struc-tures not vibrate; so it is important to know the frequencies at which a struc-ture can vibrate, in order to avoid them. By inspecting the solution of the gen-eralized wave equation, which we investigate in the preceding section, we can 3.5 Estimation of Eigenvalues 237 see that the frequencies of vibration are λnc/2π,n=1,2,3,....T h u sw em u s t ﬁnd the eigenvalues λ2 nin order to identify the frequencies of vibration. Consider the following Sturm–Liouville problem: /parenleftbig s(x)φ/prime/parenrightbig/prime−q(x)φ+λ2p(x)φ=0,l<x<r, (1) φ(l)=0,φ ( r)=0, (2) where s,s/prime,q,a n d pare continuous and sand pare positive for l≤x≤r.( N o t e that we have a rather general differential equation but very special boundary conditions.) Ifφ1is the eigenfunction corresponding to the smallest eigenvalue λ2 1,t h e n φ1satisﬁes Eq. (1) for λ=λ1.A l t e r n a t i v e l y ,w ec a nw r i t e −/parenleftbig sφ/prime 1/parenrightbig/prime+qφ1=λ2 1pφ1,l<x<r. Multiplying through this equation by φ1and integrating from ltor,w eo b t a i n /integraldisplayr l−/parenleftbig sφ/prime 1/parenrightbig/primeφ1dx+/integraldisplayr lqφ2 1dx=λ2 1/integraldisplayr lpφ2 1dx. If the ﬁrst integral is integrated by parts, it becomes −sφ/prime 1φ1/vextendsingle/vextendsingler l+/integraldisplayr lsφ/prime 1φ/prime 1dx. Butφ1(l)=φ1(r)=0, so the ﬁrst term vanishes and we are left with the equal- ity /integraldisplayr ls/bracketleftbig φ/prime 1/bracketrightbig2dx+/integraldisplayr lqφ2 1dx=λ2 1/integraldisplayr lpφ2 1dx. Because p(x)is positive for l≤x≤r, the integral on the right is positive and we may deﬁne λ2 1as λ2 1=/integraltextr ls[φ/prime 1]2dx+/integraltextr lqφ2 1dx/integraltextr lpφ2 1dx=N(φ1) D(φ1). (3) It can be shown that, if y(x)is any function with two continuous derivatives (l≤x≤r)that satisﬁes y(l)=y(r)=0, then λ2 1≤N(y) D(y). (4) By choosing any convenient function ythat satisﬁes the boundary conditions, we obtain from the ratio N(y)/D(y)an upper bound on λ2 1. Usually this bound is quite a good estimate. One should keep in mind that the graph of the eigen-function φ 1(x)does not cross the x-axis between land r,s ot h eg r a p ho f y(x) should not cross the axis either. 238 Chapter 3 The Wave Equation Example. Estimate the ﬁrst eigenvalue of φ/prime/prime+λ2φ=0,0<x<1, φ(0)=φ(1)=0. Let us try y(x)=x(1−x), which satisﬁes the boundary conditions. Then y/prime(x)=1−2xand N(y)=/integraldisplay1 0/bracketleftbig y/prime(x)/bracketrightbig2dx=/integraldisplay1 0(1−2x)2dx=1 3, D(y)=/integraldisplay1 0y2(x)dx=/integraldisplay1 0x2(1−x)2dx=1 30. Therefore, N(y)/D(y)=10. We know, of course, that φ1(x)=sin(πx),a n d N(φ1)=/integraldisplay1 0π2cos2(πx)dx=π2 2,D(φ1)=/integraldisplay1 0sin2(πx)dx=1 2, soN(φ1)/D(φ1)=λ2 1=π2<10, conﬁrming Eq. (4). /square Example. Estimate the ﬁrst eigenvalue of (xφ/prime)/prime+λ21 xφ=0,1<x<2, φ(1)=φ(2)=0. The integrals to be calculated are N(y)=/integraldisplay2 1x(y/prime)2dx,D(y)=/integraldisplay2 11 xy2dx. The tabulation gives results for several trial functions. It is known that the ﬁrst eigenvalue and eigenfunction are λ2 1=/parenleftbiggπ ln 2/parenrightbigg2 ∼=20.5423, φ1(x)=sin/parenleftbiggπlnx ln 2/parenrightbigg . The error for the best of the trial functions is about 1.44% y(x)√x(2−x)(x−1) (2−x)(x−1)(2−x)(x−1) x N(y) D(y)23.7500 22.1349 20.8379 /square 3.6 Wave Equation in Unbounded Regions 239 This method of estimating the ﬁrst eigenvalue is called Rayleigh’s method , and the ratio N(y)/D(y)is called the Rayleigh quotient .I ns o m em e c h a n i c a l systems, the Rayleigh quotient may be interpreted as the ratio between poten-tial and kinetic energy. There are many other methods for estimating eigenval- ues and for systematically improving the estimates. EXERCISES 1.Using Eq. (3), show that if q≥0, then λ2 1≥0a l s o . 2.Verify the results for at least one of the trial functions used in the second example. 3.Estimate the ﬁrst eigenvalue of the problem φ/prime/prime+λ2(1+x)φ=0,0<x<1, φ(0)=φ(1)=0. 4.Verify that the general solution of the following differential equation is axcos(λ/x)+bxsin(λ/x), and then solve the eigenvalue problem. φ/prime/prime+λ2 x4φ=0,1<x<2, φ(1)=0,φ ( 2)=0. 5.Estimate the lowest eigenvalue of the problem in Exercise 4 using y= (x−1)(2−x). 6.Estimate the lowest eigenvalue of the problem φ/prime/prime+λ2xφ=0,0<x<1, φ(0)=0,φ ( 1)=0. Use the trial function xb(1−x), and minimize the Rayleigh quotient with respect to b. 3.6 Wave Equation in Unbounded Regions When the wave equation is to be solved for 0 <x<∞or for −∞< x<∞, we can proceed as we did for the solution of the heat equation in these unbounded regions. That is to say, we separate variables and use a Fourierintegral to combine the product solutions. 240 Chapter 3 The Wave Equation Consider the problem ∂2u ∂x2=1 c2∂2u ∂t2, 0<t,0<x, (1) u(x,0)=f(x), 0<x, (2) ∂u ∂t(x,0)=g(x), 0<x, (3) u(0,t)=0, 0<t. (4) We require in addition that the solution u(x,t)be bounded as x→∞ . On separating variables, we make u(x,t)=φ(x)T(t)and ﬁnd that the fac- tors satisfy T/prime/prime+λ2c2T=0,0<t, φ/prime/prime+λ2φ=0, 0<x, φ(0)=0,|φ(x)|bounded . The solutions are easily found to be φ(x;λ)=sin(λx), T(t;λ)=Acos(λct)+Bsin(λct), and we combine the products φ(x;λ)T(t;λ)in a Fourier integral u(x,t)=/integraldisplay∞ 0/parenleftbig A(λ)cos(λct)+B(λ)sin(λct)/parenrightbig sin(λx)dλ. (5) The initial conditions become u(x,0)=f(x)=/integraldisplay∞ 0A(λ)sin(λx)dλ, 0<x, ∂u ∂t(x,0)=g(x)=/integraldisplay∞ 0λcB(λ)sin(λx)dλ, 0<x. Both of these equations are Fourier integrals. Thus the coefﬁcient functions are given by A(λ)=2 π/integraldisplay∞ 0f(x)sin(λx)dx,B(λ)=2 πλc/integraldisplay∞ 0g(x)sin(λx)dx. It is sufﬁcient to demand that/integraltext∞ 0|f(x)|dxand/integraltext∞ 0|g(x)|dxboth be ﬁnite in order to guarantee the existence of Aand B. The deﬁciency of the Fourier integral form of the solution given in Eq. (5) is that the formula gives no idea of what u(x,t)looks like. The d’Alembert 3.6 Wave Equation in Unbounded Regions 241 solution of the wave equation can come to our aid again here. We know that the solution of Eq. (1) has the form u(x,t)=ψ(x+ct)+φ(x−ct). The two initial conditions boil down to ψ(x)+φ(x)=f(x), 0<x, ψ(x)−φ(x)=G(x)+A,0<x. As in the ﬁnite case, we have deﬁned G(x)=1 c/integraldisplayx 0g(y)dy and Ais any constant. From the two initial conditions we obtain ψ(x)=1 2/parenleftbig f(x)+G(x)+A/parenrightbig ,x>0, φ(x)=1 2/parenleftbig f(x)−G(x)−A/parenrightbig ,x>0. Both fand Gare known for x>0. Thus ψ(x+ct)=1 2/parenleftbig f(x+ct)+G(x+ct)+A/parenrightbig is deﬁned for all x>0a n d t≥0. But φ(x−ct)is not yet deﬁned for x−ct<0. That means that we must extend the functions fand Gin such a way as to deﬁne φfor negative arguments and also satisfy the boundary condition. The sole boundary condition is Eq. (4), which becomes u(0,t)=0=ψ(ct)+φ(−ct). In terms of ˜fand˜G, extensions of fand G,t h i si s 0=f(ct)+G(ct)+A+˜f(−ct)−˜G(−ct)−A. Since fand Gare not dependent on each other, we must have individually f(ct)+˜f(−ct)=0,G(ct)−˜G(−ct)=0. That is, ˜fisfo, the odd extension of f,a n d˜GisGe, the even extension of G. Finally, we arrive at a formula for the solution: u(x,t)=1 2/bracketleftbig fo(x+ct)+Ge(x+ct)/bracketrightbig +1 2/bracketleftbig fo(x−ct)−Ge(x−ct)/bracketrightbig . (6) 242 Chapter 3 The Wave Equation Figure 5 Solution of Eqs. (1)–(4) with g(x)≡0. On the left are graphs of fo(x+ct)(solid) and of fo(x−ct)(dashed) at the times shown. On the right are the graphs of u(x,t)for 0<x, made by averaging the graphs on the left. Now, given the functions f(x)and g(x), it is a simple matter to construct fo andGeand thus to graph u(x,t)as a function of either variable or to evaluate it for speciﬁc values of xandt. By way of illustration, Fig. 5 shows the solution of Eqs. (1)–(4) as a function of xat various times, for f(x)as shown and g(x)≡0. Another interesting problem that can be treated by the d’Alembert method is one in which the boundary condition is a function of time. For simplicity, we take zero initial conditions. Our problem becomes ∂2u ∂x2=1 c2∂2u ∂t2,0<t,0<x, (7) u(x,0)=0, 0<x, (8) ∂u ∂t(x,0)=0, 0<x, (9) 3.6 Wave Equation in Unbounded Regions 243 u(0,t)=h(t), 0<t. (10) Asuis to be a solution of the wave equation, it must have the form u(x,t)=ψ(x+ct)+φ(x−ct). (11) The two initial conditions, Eqs. (8) and (9), can be treated exactly as in the ﬁrst problem. Of course, G(x)≡0, and the constant A, being arbitrary, may be taken as 0. The conclusion is that ψ(x)=0,φ ( x)=0,0<x. Because both xand ta r ep o s i t i v ei nt h i sp r o b l e m ,w es e et h a t ψ(x+ct)=0 always, so Eq. (11) may be simpliﬁed to u(x,t)=φ(x−ct). (12) The boundary condition Eq. (10) will tell us how to evaluate φfor negative arguments. The equation is u(0,t)=φ(−ct)=h(t), 0<t. (13) We now put together what we know of the function φ: φ(q)=  0, q>0, h/parenleftbigg −q c/parenrightbigg ,q<0.(14) The argument qis a dummy, used to avoid association with either xort.E q u a - tions (12) and (14) now specify the solution u(x,t)completely. Example. Take h(t)as shown in Fig. 6. The major steps to construct the graph of φ(q). Note that the graph of φfor negative argument is that of h,r e ﬂ e c t e d .I no t h e r words, to make the graph of φ(q), start from the graph of h(t):( 1 )G r a p ht h e even extension he(t); (2) replace the right half (from 0 up) with 0; (3) adjust scales so that q=− cwhere t=− 1, etc. The graphs in Fig. 7 show u(x,t)as a function of xf o rv a r i o u sv a l u e so f t.I t is clear from both the graphs and the formula that the disturbance caused bythe variable boundary condition arrives at a ﬁxed point xat time x/c.T h u st h e disturbance travels with the velocity c, the wave speed. An example is animated on the CD. /square A wave equation accompanied by nonzero initial conditions and time- varying boundary conditions can be solved by breaking it into two prob-lems, one like Eqs. (1)–(4) with zero boundary condition, and the other likeEqs. (7)–(10) with zero initial conditions. 244 Chapter 3 The Wave Equation Figure 6 Graphs of h(t)andφ(q)for semi-inﬁnite string with time-varying boundary condition. Figure 7 Graphs of u(x,t)versus xfor the semi-inﬁnite string under the time-varying boundary condition u(0,t)=h(t),w i t h h(t)a ss h o w ni nF i g .6 .T h e shape seen in the last drawing will continue to travel to the right. 3.6 Wave Equation in Unbounded Regions 245 EXERCISES 1.Derive a formula similar to Eq. (6) for the case in which the boundary con- dition Eq. (4) is replaced by ∂u ∂x(0,t)=0,0<t. 2.Derive Eq. (6) from Eq. (5) by using trigonometric identities for the prod- uct sin (λx)·cos(λct), and so forth, and recognizing certain Fourier inte- grals. 3.Sketch the solution of Eqs. (1)–(4) as a function of xat times t=0, 1/2c, 1/c,2/c,3/c,i fg(x)=0e v e r y w h e r ea n d f(x)is the rectangular pulse f(x)=/braceleftBigg0,0<x<1, 1,1<x<2, 0,2<x. 4.Same as Exercise 3, but f(x)=0a n d g(x)is the rectangular pulse g(x)=/braceleftBigg0,0<x<1, c,1<x<2, 0,2<x. 5.Sketch the solution of Eqs. (7)–(10) at times t=0,π/2,π,3π/2, 2π,5π/2, ifh(t)=sin(t).T a k e c=1. 6.Sketch the solution of Eqs. (7)–(10) at times t=0, 1/2, 3/2, and 5 /2, if c=1a n d h(t)=/braceleftBigg0,0<t<1, 1,1<t<2, 0,2<t. 7.Use the d’Alembert solution of the wave equation to solve the problem ∂2u ∂x2=1 c2∂2u ∂t2,−∞<x<∞,0<t, u(x,0)=f(x), −∞<x<∞, ∂u ∂t(x,0)=g(x),−∞<x<∞. 8.The solution of the problem stated in Exercise 7 is sometimes written u(x,t)=1 2/parenleftbig f(x+ct)+f(x−ct)/parenrightbig +1 2c/integraldisplayx+ct x−ctg(z)dz. 246 Chapter 3 The Wave Equation Show that this is correct. Y ou will need Leibniz’s rule (see the Appendix) to differentiate the integral. 3.7 Comments and References T h ew a v ee q u a t i o ni so n eo ft h eo l d e s te quations of mathematical physics. Euler, Bernoulli, and d’Alembert all solved the problem of the vibrating stringabout 1750, using either separation of variables or what we called d’Alembert’smethod. This latter is, in fact, a very special case of the method of characteris-tics, in essence a way of identifying new independent variables having specialsigniﬁcance. Street’s Analysis and Solution of Partial Differential Equations has a chapter on characteristics, including their use in numerical solutions. Wan’s Mathematical Models and Their Analysis g i v e sa p p l i c a t i o n st ot r a f ﬁ cﬂ o wa n d also discusses other wave phenomena. (See the Bibliography.) Because many physical phenomena described by the wave equation are part of our everyday experience — the sounds of musical instruments, forinstance — they are often featured in popular expositions of mathematical physics. The book of Davis and Hersch ( The Mathematical Experience )e x - plains standing waves (product solutions) and superposition in an elemen-tary way. Of course, many other phenomena are described by the wave equa-tion. Among the most important for modern life are electrical and mag-netic waves, which are solutions of special cases of the Maxwell ﬁeld equa-tions. These and other kinds of waves (including water waves) are studied inMain’s Vibrations and Waves in Physics ; both exposition and ﬁgures are ﬁrst rate. The potential difference Vbetween the interior and exterior of a nerve axon can be modeled approximately by the Fitzhugh–Nagumo equations, ∂V ∂t=∂2V ∂x2+V−1 3V3−R, ∂R ∂t=k(V+a−bR). Here Rrepresents a restoring effect and a,b,a n d kare constants. At ﬁrst glance, one would expect Vto behave like the solution of a heat equation. But a trav- eling wave solution, V(x,t)=F(x−ct), R(x,t)=G(x−ct), of these equations can be found that shows many important features of nerve impulses. This system and many other exciting biological applications ofmathematics are reported by Murray’s excellent book, Mathematical Biology . Miscellaneous Exercises 247 More information about the Rayleigh qu otient and estimation of eigenval- ues is in Boundary and Eigenvalue Problems in Mathematical Physics , by Sagan. The classic reference for eigenvalues, and indeed for the partial differentialequations of mathematical physics in general, is the work by Courant and Hilbert, Methods of Mathematical Physics . Chapter Review See the CD for Review Questions. Miscellaneous Exercises E x e r c i s e s1 – 5r e f e rt ot h ep r o b l e m ∂2u ∂x2=1 c2∂2u ∂t2, 0<x<a,0<t, u(0,t)=0,u(a,t)=0, 0<t, u(x,0)=f(x),∂u ∂t(x,0)=g(x), 0<x<a. 1.Take f(x)=1, 0<x<a,a n d g(x)≡0 .( T h i si sr a t h e ru n r e a l i s - tic if u(x,t)is the displacement of a vibrating string.) Find a series (separation-of-variables) solution. 2.Sketch u(x,t)of Exercise 1 as a function of xat various times throughout one period. 3.The solution u(x,t)of Exercise 1 takes on only the three values 1, 0, and −1. Make a sketch of the region 0 <x<a,0<t, and locate the places where ut a k e so ne a c ho ft h ev a l u e s . 4.Take g(x)=0a n d f(x)to be this function: f(x)=  3hx 2a, 0<x<2a 3, 3h(a−x) a,2a 3<x<a. The graph of fis triangular, with peak at x=2a/3. Find a series solution foru(x,t). 5.Sketch u(x,t)of Exercise 4 as a function of xat times 0 to a/cin steps of a/6c. 248 Chapter 3 The Wave Equation 6.Find an analytic (integral) solution of this wave problem ∂2u ∂x2=1 c2∂2u ∂t2, −∞<x<∞,0<t, u(x,0)=f(x),∂u ∂t(x,0)=g(x),−∞<x<∞, with g(x)=0a n d f(x)=/braceleftbiggh,|x|</epsilon1, 0,|x|>/epsilon1. 7.Sketch the solution of the problem in Exercise 6 at times t=0,/epsilon1/c,2/epsilon1/c, 3/epsilon1/c. 8.Same as Exercise 7, but f(x)=0a n d g(x)=/braceleftbiggc,|x|</epsilon1, 0,|x|>/epsilon1. 9.Letu(x,t)be the solution of the problem ∂2u ∂x2=1 c2∂2u ∂t2, 0<x,0<t, u(0,t)=0, 0<t, u(x,0)=f(x),∂u ∂t(x,0)=g(x), 0<x. Sketch the solution u(x,t)as a function of xat times t=0,a/6c,a/2c, 5a/6c,7a/6c.U s e g(x)≡0a n d f(x)=  3hx 2a, 0<x<2a 3, 3h(a−x) a,2a 3<x<a, 0, a<x. 10. Same task as in Exercise 9 but f(x)=/braceleftbiggsin(x),0<x<π, 0,π < x and g(x)=0. Sketch at times t=0,π/4c,π/2c,3π/4c,π/c,2π/c. 11. Letu(x,t)be the solution of the wave equation on the semi-inﬁnite in- terval 0 <x<∞, with both initial conditions equal to zero but with the Miscellaneous Exercises 249 time-varying boundary condition u(0,t)=  sin/parenleftbiggct a/parenrightbigg ,0<t<πa c, 0,πa c<t. Sketch u(x,t)as a function of xat various times. 12.Same as Exercise 11, but the boundary condition is u(0,t)=h, for all t>0. 13.Same as Exercise 11, but the boundary condition is u(0,t)=  hct a, 0<t<a c, h(2a−ct) a,a c<t<2a c, 0,2a c<t. 14.Estimate the lowest eigenvalue of the problem /parenleftbig eαxφ/prime/parenrightbig/prime+λ2eαxφ=0,0<x<1, φ(0)=0,φ ( 1)=0. (This problem can be solved exactly.) 15.Estimate the lowest eigenvalue of the problem φ/prime/prime−xφ+λ2φ=0,0<x<1, φ(0)=0,φ ( 1)=0. 16.Show that the nonlinear wave equation ∂u ∂t+u∂u ∂x+∂3u ∂x3=0 (the Korteweg–deVries equation) has, as one solution, u(x,t)=12a2sech2/parenleftbig ax−4a3t/parenrightbig . A wave of this form is called a soliton or solitary wave. 17.T h es o l u t i o ni nE x e r c i s e1 6i so ft h ef o r m u(x,t)=f(x−ct). What is the function f, and what is the wave speed c? 250 Chapter 3 The Wave Equation 18. Fort<0, water ﬂows steadily through a long pipe connected at x=0 to a large reservoir and open at x=ato the air. At time t=0, a valve at x=ais suddenly closed. Reasonable expressions for the conservation of momentum and of mass are ∂u ∂t=−∂p ∂x, 0<x<a,0<t, (A) ∂p ∂t=− c2∂u ∂x,0<x<a,allt, (B) where pis gauge pressure and uis mass ﬂow rate. If the pipe is rigid, c2=K/ρ, the ratio of bulk modulus of water to its density. Show that both pand usatisfy the wave equation. The phenomenon described here is called water hammer . 19. Introduce a function vwith the deﬁnition u=∂v/∂ x,p=−∂v/∂ t. Show that (A) becomes an identity and that (B) becomes the wave equa- tion for v. 20. Reasonable boundary and initial conditions for uand pare u(x,0)=U0(constant ), 0<x<a, p(x,0)=0, 0<x<a, p(0,t)=0, allt, u(a,t)=0, t>0. Restate these as conditions on v. Show that the ﬁrst and third equations may be replaced by v(x,0)=U0x,0<x<a, v(0,t)=0, allt. 21. Solve the problem in Exercise 20; ﬁnd a series form for v(x,t). 22. In many problems involving ﬂuid ﬂow, the combination ∂u ∂t+V∂u ∂x (called the Stokes derivative )a p p e a r s .H e r e Vis the speed of the ﬂuid in thex-direction. If Vequals uor otherwise depends on u,t h i so p e r a t o ri s nonlinear and difﬁcult to work with. Let us assume that Vis a constant, so that the operator is linear, and deﬁne new variables ξ=x+Vt,τ=x−Vt,u(x,t)=v(ξ,τ). Miscellaneous Exercises 251 Show that ∂u ∂t+V∂u ∂x=2V∂v ∂ξ. 23.Assume that u(x,y,t)has the product form shown in what follows. Sep- arate the variables in the given partial differential equation. u(x,y,t)=ψ(x+Vt)φ(x−Vt)Y(y), ∂2u ∂y2=1 k/parenleftbigg∂u ∂t+V∂u ∂x/parenrightbigg . 24.A ﬂuid ﬂows between two parallel plates held at temperature 0. At the inlet, ﬂuid temperature is T0and initially the ﬂuid is at temperature T1. IfVis the speed of the ﬂuid in the x-direction, a problem describing the temperature u(x,y,t)is ∂2u ∂y2=1 k/parenleftbigg∂u ∂t+V∂u ∂x/parenrightbigg ,0<y<b, 0<x,0<t, u(x,0,t)=0, u(x,b,t)=0,0<x,0<t, u(0,y,t)=T0, 0<y<b, 0<t, u(x,y,0)=T1, 0<x, 0<y<b. Make a separation of variables as in Exercise 23. State and solve the eigen- value problem for Y. Show that un(x,y,t)=φn(x−Vt)exp/parenleftbig −λ2 nk(x+Vt)/2V/parenrightbig sin(λny) satisﬁes the partial differential equation and boundary conditions at y= 0a n d y=b, without restriction of φn(except differentiability). 25.Show how to satisfy the initial and inlet conditions in the problem of Ex- ercise 24, by forming a sum of product solutions and correctly choosing theφn. 26.Find all functions φsuch that u(x,t)=φ(x−ct)is a solution of the heat equation, ∂2u ∂x2=1 k∂u ∂t. 27.Take the constant c=(1+i)/radicalbig ωk/2 in Exercise 26 and show that the functions e−pxsin(ωt−px), e−pxcos(ωt−px) 252 Chapter 3 The Wave Equation can be obtained from φ(x−ct)andφ(x−¯ct).( H e r e p=/radicalbig ω/2kand¯cis the complex conjugate of c. Refer to Exercise 6 in Chapter 2, Section 10.) 28. Some nonlinear equations can also result in “traveling wave solutions,” u(x,t)=φ(x−ct). Show that Fisher’s equation , ∂2u ∂x2=∂u ∂t−u(1−u), has a solution of this form if φsatisﬁes the nonlinear differential equa- tion φ/prime/prime+cφ/prime+φ(1−φ)=0. 29. Show that the function uis a solution of the problem ∂2u ∂x2=1 c2∂2u ∂t2, 0<x<a,0<t, u(0,t)=0,u(a,t)=sin(ωt), 0<t, provided that the parameters are such that sin (ωa/c)/negationslash=0. u(x,t)=sin(ωx/c)sin(ωt) sin(ωa/c). 30. Ifω=πc/a, the denominator of the function in Exercise 29 is 0. Show that, for this value of ω, a function satisfying the wave equation and the given boundary condition is u(x,t)=−ct asin/parenleftbiggπx a/parenrightbigg cos/parenleftbiggπct a/parenrightbigg −x acos/parenleftbiggπx a/parenrightbigg sin/parenleftbiggπct a/parenrightbigg . 31. A string in a musical instrument is typically not as ﬂexible as assumed in Section 1. For such a string, the displacement umay satisfy the partial differential equation ∂2u ∂x2−/epsilon1∂4u ∂x4=1 c2∂2u ∂t2,0<x<a,0<t, where c2=T/ρ,a si nS e c t i o n1 ,a n d /epsilon1=EI/T, with E=Yo u n g’s m o d - ulus for the material, I=second moment of area. (See an elasticity ref- erence.) Assuming that u(x,t)=φ(x)T(t), carry out a separation of variable and ﬁnd the eigenvalue problem for φ. Take the boundary conditions to be Miscellaneous Exercises 253 u(0,t)=0,u(a,t)=0, ∂2u ∂x2(0,t)=0,∂2u ∂x2(a,t)=0,0<t. 32.Show that φ(x)=sin(µx)satisﬁes the eigenvalue problem found in Ex- ercise 31, provided that µ=nπ/aand λ2=µ2+/epsilon1µ4, where −λ2=T/prime/prime(t)/c2T(t). 33.The values of λ2are related to the frequencies of vibration of the string mentioned in Exercise 31. Show that λnapproaches nπ/afor any ﬁxed n as/epsilon1approaches 0. 34.The longitudinal vibration of a thin rod has been described by Love (see Bibliography) with the equation ∂2u ∂x2=1 c2/parenleftbigg∂2u ∂t2−/epsilon1∂4u ∂x2∂t2/parenrightbigg ,0<x<a,0<t. Here, u(x,t)is the displacement of the points that are at xwhen there is no motion; /epsilon1=νK2,w h e r e νis Poisson’s ratio and Kis the radius of gyration of a cross section of the rod. Take boundary conditions u(0,t)=0,u(a,t)=0, and separate the variables by assuming u(x,t)=φ(x)T(t).I tw i l lb eu s e - ful to name φ/prime/prime/φ=−λ2. 35.The eigenvalue problem in Exercise 34 is routine. Once that is solved, ﬁnd the differential equation for T(t), solve it, and determine the fre- quencies of vibration. This page intentionally left blank The Potential Equation CHAPTER4 4.1 Potential Equation The equation for the steady-state temperature distribution in two dimensions (see Chapter 5) is ∂2u ∂x2+∂2u ∂y2=0. The same equation describes the equilibrium (time-independent) displace- ments of a two-dimensional membrane , and so is an important common part of both the heat and wave equations in two dimensions. Many other physicalphenomena — gravitational and electrostatic potentials, certain ﬂuid ﬂows —and an important class of functions are described by this equation, thus mak- ing it one of the most important of mathematics, physics, and engineering.The analogous equation in three dimensions is ∂ 2u ∂x2+∂2u ∂y2+∂2u ∂z2=0. Either equation may be written ∇2u=0 and is commonly called the potential equation orLaplace’s equation . The solutions of the potential equation (called harmonic functions )h a v e many interesting properties. An important one, which can be understood in-tuitively, is the maximum principle: If ∇ 2u=0i nar e g i o n ,t h e n ucannot have a relative maximum or minimum inside the region unless uis constant. (Thus, if ∂u/∂xand∂u/∂ya r eb o t hz e r oa ts o m ep o i n t ,i ti sas a d d l ep o i n t . ) Ifuis thought of as the steady-state temperature distribution in a metal plate, 255 256 Chapter 4 The Potential Equation it is clear that the temperature cannot be greater at one point than at all other nearby points. For if such were the case, heat would ﬂow away from the hotpoint to cooler points nearby, thus reducing the temperature at the hot point.But then the temperature would not be unchanging with time. We return to this matter in Section 4. A complete boundary value problem consists of the potential equation in a region plus boundary conditions. These may be of any of the three types ugiven,∂u ∂ngiven,orαu+β∂u ∂ngiven along any section of the boundary. (By ∂u/∂n,w em e a nt h ed i r e c t i o n a ld e r i v - ative in the direction normal, or perpendicular, to the boundary.) When uis speciﬁed along the whole boundary, the problem is called Dirichlet’s problem ; if∂u/∂nis speciﬁed along the whole boundary, it is Neumann’s problem .T h e solutions of Neumann’s problem are not unique, for if uis a solution, so is u plus a constant. It is often useful to consider the potential equation in other coordinate sys- tems. One of the most important is the polar coordinate system, in which thevariables are r=/radicalbig x2+y2,θ=tan−1/parenleftbiggy x/parenrightbigg , x=rcos(θ), y=rsin(θ). By convention we require r≥0. We shall deﬁne u(x,y)=u/parenleftbig rcos(θ), rsin(θ)/parenrightbig =v(r,θ) and ﬁnd an expression for the Laplacian of u, ∇2u=∂2u ∂x2+∂2u ∂y2, in terms of vand its derivatives by using the chain rule. The calculations are elementary but tedious. (See Exercise 7.) The results are ∂2u ∂x2=cos2(θ)∂2v ∂r2−2s i n(θ)cos(θ) r∂2v ∂θ∂r+sin2(θ) r2∂2v ∂θ2 +sin2(θ) r∂v ∂r+2s i n(θ)sin(θ) r2∂v ∂θ, ∂2u ∂y2=sin2(θ)∂2v ∂r2+2s i n(θ)cos(θ) r∂2v ∂θ∂r+cos2(θ) r2∂2v ∂θ2 +cos2(θ) r∂v ∂r−2s i n(θ)sin(θ) r2∂v ∂θ. Chapter 4 The Potential Equation 257 From these equations we easily ﬁnd that the Laplacian in polar coordinates is ∇2v=∂2v ∂r2+1 r∂v ∂r+1 r2∂2v ∂θ2=1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg +1 r2∂2v ∂θ2. In cylindrical (r,θ,z)coordinates, the Laplacian is ∇2v=1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg +1 r2∂2v ∂θ2+∂2v ∂z2. EXERCISES 1.Find a relation among the coefﬁcients of the polynomial p(x,y)=a+bx+cy+dx2+exy+fy2 that makes it satisfy the potential equation. Choose a speciﬁc polynomial that satisﬁes the equation, and show that, if ∂p/∂xand∂p/∂yare both zero at some point, the surface there is saddle shaped. 2.Show that u(x,y)=x2−y2andu(x,y)=xyare solutions of Laplace’s equa- tion. Sketch the surfaces z=u(x,y). What boundary conditions do these f u n c t i o n sf u l ﬁ l lo nt h el i n e s x=0,x=a,y=0,y=b? 3.If a solution of the potential equation in the square 0 <x<1, 0<y<1 has the form u(x,y)=Y(y)sin(πx), of what form is the function Y?F i n da function Ythat makes u(x,y)satisfy the boundary conditions u(x,0)=0, u(x,1)=sin(πx). 4.Find a function u(x), independent of y, that satisﬁes the potential equation. 5.What functions v(r), independent of θ, satisfy the potential equation in polar coordinates? 6.Show that rnsin(nθ)and rncos(nθ)both satisfy the potential equation in polar coordinates (n=0,1,2,...) . 7.Find expressions for the partial derivatives of uwith respect to xand yin terms of derivatives of vwith respect to randθ. 8.Ifuandvare the x-a n d y-components of the velocity in a ﬂuid, it can be shown (under certain assumptions) that these functions satisfy the equa-tions ∂u ∂x+∂v ∂y=0, (A) ∂u ∂x−∂v ∂x=0. (B) 258 Chapter 4 The Potential Equation (a) (b) (c) (d) Figure 1 (a)uis displacement of a membrane; the graph of f(x)is an isosceles triangle. (b) uis the temperature on a cross section of a long bar. (c) uis voltage in a rectangular sheet of conducting material. (d) φis a velocity potential (see Exercise 8). What are x-a n d y-velocities on the boundaries? Show that the deﬁnition of a velocity potential function φby the equations u=−∂φ ∂x,v=−∂φ ∂y causes (B) to be identically satisﬁed and turns (A) into the potential equa- tion. (See Section 4.7, Comments and References, at the end of this chap-ter.) 9.For each of the diagrams in Fig. 1, (a) write out the problem in mathemat-ical form (partial differential equation and boundary conditions); (b) pro-vide an interpretation in words of the boundary conditions for the given interpretation of the unknown function. 4.2 Potential in a Rectangle 259 4.2 Potential in a Rectangle One of the simplest and most important problems in mathematical physics is Dirichlet’s problem in a rectangle. T o take an easy case, we consider a problemwith just two nonzero boundary conditions: ∂ 2u ∂x2+∂2u ∂y2=0,0<x<a,0<y<b, (1) u(x,0)=f1(x), 0<x<a, (2) u(x,b)=f2(x), 0<x<a, (3) u(0,y)=0, 0<y<b, (4) u(a,y)=0, 0<y<b. (5) It is not immediately clear that separation of variables will work. However, we have a homogeneous partial differential equation and some homogeneous boundary conditions, so we can try the method. If we assume that u(x,y)has a product form u=X(x)Y(y),t h e nE q .( 1 )b e c o m e s X/prime/prime(x)Y(y)+X(x)Y/prime/prime(y)=0. This equation can be separated by dividing through by XYto yield X/prime/prime(x) X(x)=−Y/prime/prime(y) Y(y). (6) The nonhomogeneous conditions Eqs. (2) and (3) will not, in general, become conditions on XorY, but the homogeneous conditions Eqs. (4) and (5), as usual, require that X(0)=0,X(a)=0. (7) Now, both sides of Eq. (6) must be constant, but the sign of the constant is not obvious. If we try a positive constant (say, µ2), Eq. (6) represents two ordinary equations: X/prime/prime−µ2X=0,Y/prime/prime+µ2Y=0. The solutions of these equations are X(x)=Acosh(µx)+Bsinh(µx), Y(y)=Ccos(µy)+Dsin(µy). In order to make Xsatisfy the boundary conditions Eq. (7), both AandBmust be zero, leading to a solution u(x,y)≡0. Thus we try the other possibility for sign, taking both members in Eq. (6) to equal −λ2. 260 Chapter 4 The Potential Equation Under the new assumption, Eq. (6) separates into X/prime/prime+λ2X=0,Y/prime/prime−λ2Y=0. (8) The ﬁrst of these equations, along with the boundary conditions, is recogniz- able as an eigenvalue problem, whose solutions are Xn(x)=sin(λnx), λ2 n=/parenleftbiggnπ a/parenrightbigg2 . The functions Ythat accompany the X’s are Yn(y)=ancosh(λny)+bnsinh(λny). The a’s and b’s are for the moment unknown. We see that Xn(x)Yn(y)is a solution of the (homogeneous) potential Eq. (1), which satisﬁes the homogeneous conditions Eqs. (4) and (5). A sum of thesefunctions should satisfy the same conditions and equation, so umay have the form u(x,y)=∞/summationdisplay n=1/parenleftbig ancosh(λny)+bnsinh(λny)/parenrightbig sin(λnx). (9) The nonhomogeneous boundary conditions Eqs. (2) and (3) are yet to be satisﬁed. If uis to be of the form of Eq. (9), the boundary condition Eq. (2) becomes u(x,0)=∞/summationdisplay n=1ansin/parenleftbiggnπx a/parenrightbigg =f1(x), 0<x<a. (10) We recognize a problem in Fourier series immediately. The anmust be the Fourier sine coefﬁcients of f1(x), an=2 a/integraldisplaya 0f1(x)sin/parenleftbiggnπx a/parenrightbigg dx. The second boundary condition reads u(x,b)=∞/summationdisplay n=1/parenleftbig ancosh(λnb)+bnsinh(λnb)/parenrightbig sin/parenleftbiggnπx a/parenrightbigg =f2(x), 0<x<a. This also is a problem in Fourier series, but it is not as neat. The constant ancosh(λnb)+bnsinh(λnb) 4.2 Potential in a Rectangle 261 must be the nth Fourier sine coefﬁcient of f2.S i n c e anis known, bncan be determined from the following computations: ancosh(λnb)+bnsinh(λnb)=2 a/integraldisplaya 0f2(x)sin(λnx)dx=cn, bn=cn−ancosh(λnb) sinh(λnb). If we use this last expression for bnand substitute into Eq. (9), we ﬁnd the solution u(x,y)=∞/summationdisplay n=1/braceleftbigg cnsinh(λny) sinh(λnb) +an/bracketleftbigg cosh(λny)−cosh(λnb) sinh(λnb)sinh(λny)/bracketrightbigg/bracerightbigg sin(λnx). (11) Notice that the function multiplying cnis 0 at y=0a n di s1a t y=b. Similarly, the function multiplying anis 1 at y=0a n d0a t y=b. An easier way to write this latter function is sinh(λn(b−y)) sinh(λnb), as can readily be found from hyperbolic identities. Example. Suppose f1and f2are both given by f1(x)=f2(x)=  2x a, 0<x<a 2, 2/parenleftbigga−x a/parenrightbigg ,a 2<x<a. Then cn=an=8 π2sin(nπ/2) n2. The solution of the potential equation for these boundary conditions is u(x,y)=8 π2∞/summationdisplay n=1sin/parenleftbignπ 2/parenrightbig n2sinh/parenleftbignπ ay/parenrightbig +sinh/parenleftbignπ a(b−y)/parenrightbig sinh/parenleftbignπb a/parenrightbig sin/parenleftbiggnπx a/parenrightbigg .(12) In Fig. 2 is a graph of some level curves, u(x,y)=constant, for the case where a=b, and also a view of the surface z=u(x,y). Also see color ﬁgures on the CD. /square 262 Chapter 4 The Potential Equation (a) (b) Figure 2 (a) Level curves of the solution u(x,y)of the example problem (see Eq. (12)) for the case b=a=1. Each curve is part of the locus of points that satisfy u(x,y)=constant for constants 0 to 0.9 in steps of 0.1. For some constants, the locus consists of more than one connected curve. (b) Perspective view of thesurface z=u(x,y). Now we have seen a solution of Dirichlet’s problem in a rectangle with ho- mogeneous conditions on two parallel sides. In general, of course, the bound-ary conditions will be nonhomogeneous on all four sides of the rectangle. But this more general problem can be broken down into two problems like the one we have solved. Consider the problem ∇ 2u=0, 0<x<a,0<y<b, (13) u(x,0)=f1(x), 0<x<a, (14) u(x,b)=f2(x), 0<x<a, (15) u(0,y)=g1(y), 0<y<b, (16) u(a,y)=g2(y), 0<y<b. (17) Letu(x,y)=u1(x,y)+u2(x,y). We will put conditions on u1and u2so that they can readily be found, and from them ucan be put together. The most obvious conditions are the following: ∇2u1=0, ∇2u2=0, u1(x,0)=f1(x), u2(x,0)=0, u1(x,b)=f2(x), u2(x,b)=0, u1(0,y)=0, u2(0,y)=g1(y), u1(a,y)=0, u2(a,y)=g2(y). 4.2 Potential in a Rectangle 263 It is evident that u1+u2is the solution of the original problem Eqs. (13)– (17). Also, each of the functions u1and u2has homogeneous conditions on parallel boundaries. We already have determined the form of u1. The other function would be of the form u2(x,y)=∞/summationdisplay n=1sin(µny)Ansinh(µnx)+Bnsinh(µn(a−x)) sinh(µna), (18) where µn=nπ/band An=2 b/integraldisplayb 0g2(y)sin(µny)dy, Bn=2 b/integraldisplayb 0g1(y)sin(µny)dy. In the individual problems for u1and u2, the technique of separation of variable works because the homogeneous conditions on parallel sides of therectangle can be translated into conditions on one of the factor functions. When the boundary conditions are not complicated functions, it may be possible to satisfy some of them with a polynomial function. (See Exercises 1and 2 of Section 4.1.) Then the difference between uand the polynomial is a solution of the potential equation that satisﬁes some homogeneous boundaryconditions. EXERCISES 1.Show that sinh (λy)and sinh (λ(b−y))are independent solutions of Y/prime/prime−λ2Y=0 with λ/negationslash=0. Thus a combination of these two functions may replace a combination of sinh and cosh as the general solution of this dif- ferential equation. 2.Show that the solution of the example problem may be written u(x,y)=8 π2∞/summationdisplay n=1sin/parenleftbignπ 2/parenrightbig n2cosh/parenleftbignπ a(y−1 2b)/parenrightbig cosh/parenleftbignπb 2a/parenrightbig sin/parenleftbiggnπx a/parenrightbigg . 3.Use the form in Exercise 2 to compute uin the center of the rectangle in the three cases b=a,b=2a,b=a/2. (Hint: Check the magnitude of the terms.) 4.Verify that each term of Eq. (9) satisﬁes Eqs. (1), (4), and (5). 264 Chapter 4 The Potential Equation 5.Solve the problem ∇2u=0, 0<x<a,0<y<b, u(0,y)=0,u(a,y)=0, 0<y<b, u(x,0)=0,u(x,b)=f(x),0<x<a, where fis the same as in the example. Sketch some level curves of u(x,y). 6.Solve the potential problem on the rectangle 0 <x<a,0<y<b,s u b j e c t to the boundary conditions u(a,y)=1, 0<y<b,a n d u=0o nt h er e s to f the boundary. 7.Solve the problem of the potential equation in the rectangle 0 <x<a,0< y<b, for each of the following sets of boundary conditions. Before solving, make a pictorial version of the problem as in Exercise 9 of Section 4.1. a.u(x,b)=100, 0 <x<a;u=0 on the other three sides of the rectangle. b.u(x,b)=100, 0 <x<a;u(a,y)=100, 0 <y<b;u=0 on the other two sides of the rectangle. c.u(x,b)=bx,0<x<a;u(a,y)=ay,0<y<b;u=0 on the other two sides of the rectangle. 8.Solve the problem for u2.( T h a ti s ,d e r i v eE q .( 1 8 ) . ) 4.3 Further Examples for a Rectangle In Section 4.2, we solved Dirichlet problems with separation of variables. The same method applies to problems with other types of boundary conditions, as shown in the following. Example 1. In this problem, the unknown function might be a voltage in a conductor. Theleft and right sides are electrically insulated. ∂ 2u ∂x2+∂2u ∂y2=0,0<x<a, 0<y<b, ∂u ∂x(0,y)=0,∂u ∂x(a,y)=0, 0<y<b, u(x,0)=0, u(x,b)=V0x/a,0<x<a. We have homogeneous conditions on the facing sides at x=0a n d x=a.I f we look for solutions in the product form u(x,y)=X(x)Y(y),w eﬁ n d( a se x - 4.3 Further Examples for a Rectangle 265 pected) that X/prime/prime(x) X(x)=−Y/prime/prime(y) Y(y)=constant . The conditions at x=0a n d x=abecome X/prime(0)=0,X/prime(a)=0. If we make the separation constant −λ2, we ﬁnd a familiar eigenvalue problem forXwhose solution is X0(x)=1,λ 0=0, Xn(x)=cos(λnx), λ n=nπ/a,n=1,2,.... For the factor Y(y), the differential equation is Y/prime/prime 0=0,or Y/prime/prime n−λ2 nYn=0 with solution Y0(y)=a0+b0yor Yn(y)=ancosh(λny)+bnsinh(λny). Thus, the principle of superposition leads to the series solution u(x,y)=a0+b0y+∞/summationdisplay n=1/parenleftbig ancosh(λny)+bnsinh(λny)/parenrightbig cos(λnx). The boundary condition at y=0b e c o m e s a0+∞/summationdisplay 1ancos(λnx)=0,0<x<a, from which we see that all the a’s are 0. Then at y=bwe have b0b+∞/summationdisplay n=1/parenleftbig bnsinh(λnb)/parenrightbig cos(λnx)=V0x a,0<x<a. This is a slightly disguised cosine series. The coefﬁcients are b0b=1 a/integraldisplaya 0V0/parenleftbiggx a/parenrightbigg dx, bnsinh(λnb)=2 a/integraldisplaya 0V0/parenleftbiggx a/parenrightbigg cos(λnx)dx. See a color graphic of the solution on the CD. /square 266 Chapter 4 The Potential Equation We have seen that the success of the separation of variables method depends on having homogeneous boundary conditions at the ends of one of the inter-vals involved. In Section 4.2 we mentioned splitting up a Dirichlet problem,if necessary, to achieve this. The same splitting technique applies in problems where boundary condition of other kinds are used. The principle is to zero conditions on two facing sides of the region and to copy the rest. Example 2. This problem may describe the temperature u(x,y)in a thin plate between insulating sheets. ∂2u ∂x2+∂2u ∂y2=0,0<x<a, 0<y<b, ∂u ∂x(0,y)=0, u(a,y)=Sy, 0<y<b, ∂u ∂y(x,0)=S, u(x,b)=Sbx a,0<x<a. Since we have nonhomogeneous conditions on adjacent sides, we must split the problem in order to solve by separation of variables. Here are the two prob-lems: ∂ 2u1 ∂x2+∂2u1 ∂y2=0,∂2u2 ∂x2+∂2u2 ∂y2=0, ∂u1 ∂x(0,y)=0,u1(a,y)=0,∂u2 ∂x(0,y)=0,u2(a,y)=Sy, ∂u1 ∂y(x,0)=S,u1(x,b)=Sbx a,∂u2 ∂y(x,0)=0,u2(x,b)=0. The solution of the original problem is the sum u=u1+u2.H e r ei st h e reasoning in detail. 1.The potential equation is linear and homogeneous. By the Principle of Superposition, the sum of solutions is a solution. 2.Atx=awe have u(a,y)=0+Sy,a n da t y=bwe have u(x,b)= Sbx/a+0. Both conditions are satisﬁed. 3.From elementary calculus, we know ∂u ∂x=∂u1 ∂x+∂u2 ∂x,∂u ∂y=∂u1 ∂y+∂u2 ∂y. 4.3 Further Examples for a Rectangle 267 Then at the left and bottom boundaries, we have ∂u ∂x(0,y)=0+0,∂u ∂y(x,0)=S+0. These are satisﬁed as well. Thus, it remains to solve the two problems for u1and u2. (See the Exercises.) Here are product solutions. For u1: cos(λnx)/parenleftbig ancosh(λny)+bnsinh(λny)/parenrightbig ,λ n=/parenleftbigg n−1 2/parenrightbiggπ a,n=1,2,.... Foru2: cos(µny)/parenleftbig Ancosh(µnx)+Bnsinh(µnx)/parenrightbig ,µ n=/parenleftbigg n−1 2/parenrightbiggπ b,n=1,2,.... /square The simple polynomial solutions that we found in Section 4.1, Exercise 1, can be very useful in reducing the number of series needed for a solution. Ifnonhomogeneous conditions are given on adjacent sides and these are con-stants or ﬁrst-degree polynomials in one variable, then a polynomial may beable to satisfy enough of them to simplify the work. Example 3. Refer to the problem in Example 2. The polynomial v(y)=Sysatisﬁes the potential equation and several of the boundary conditions: ∂v ∂x(0,y)=0,v ( a,y)=Sy,0<y<b, ∂v ∂y(x,0)=S,v ( x,b)=Sb,0<x<a. Thus, we may set u(x,y)=v(y)+w(x,y)and determine that wmust be the solution of this problem, similar to the problem for u2in Example 2: ∂2w ∂x2+∂2w ∂y2=0,0<x<a, 0<y<b, ∂w ∂x(0,y)=0,w ( a,y)=0, 0<y<b, ∂w ∂y(x,0)=0,w ( x,b)=Sb(x−a) a,0<x<a. T h es o l u t i o ni sl e f ta sa ne x e r c i s e . /square 268 Chapter 4 The Potential Equation Poisson Equation Many problems in engineering and physics require the solution of the Poisson equation, ∇2u=− H in a region R. Here are three examples of such problems. (1) uis the deﬂection of a membrane that is fastened at its edges, so u=0o n the boundary of R;His proportional to the pressure difference across the membrane. (See Section 5.1.) (2) uis the steady-state temperature in a cross section of a long cylindrical r o dt h a ti sc a r r y i n ga ne l e c t r i c a lc u r r e n t ; His proportional to the power in resistance heating. (See Section 5.2.) (3) uis the stress function on the cross section Rof a cylindrical bar or rod in torsion (the shear stresses are proportional to the partial derivativesofu);His proportional to the rate of twist and to the shear modulus of the material; u=0 on the boundary of R. IfHis a constant, a polynomial of the form P(x,y)=A+Bx+Cy+Dx2+Exy+Fy2 is a solution of Poisson’s equation, provided that 2(D+F)=− H. The other coefﬁcients are arbitrary and may be chosen for convenience in sat- isfying boundary conditions. Example 4. Find the deﬂection uof a membrane that is modeled by this problem. The constant is H=p/σ,w h e r e pis the pressure difference (below to above) and σis the surface tension in the membrane. A polynomial can be chosen that satisﬁes the partial differential equation and boundary conditions on facing sides. For instance, v(x)=Hx(a−x) 2 satisﬁes the Poisson equation and two boundary conditions, v(0)=0,v ( a)=0. 4.3 Further Examples for a Rectangle 269 Thus, we may set u(x,y)=v(x)+w(x,y)and determine that wis a solution of the problem ∂2w ∂x2+∂2w ∂y2=0,0<x<a, 0<y<b, w(0,y)=0,w ( a,y)=0, 0<y<b, w(x,0)=−v(x), w( x,b)=−v(x),0<x<a. The CD has color graphics of the solution. /square In general, if His a polynomial in xand y, a solution can be found in the form of a polynomial of total degree 2 higher than H.I fHis a more gen- eral function, it may be expressed as a double Fourier series (see Chapter 5),and the partial differential equation can be solved following the idea of Sec-tion 1.11B. EXERCISES 1.Solve the problem consisting of the potential equation on the rectangle 0<x<a,0<y<bwith the given boundary conditions. Two of the three are very easy if a polynomial is subtracted from u. a.∂u ∂x(0,y)=0; u=1o nt h er e m a i n d e ro ft h eb o u n d a r y . b.∂u ∂x(0,y)=0,∂u ∂x(a,y)=0; u(x,0)=0, u(x,b)=1. c.∂u ∂x(x,0)=0, u(x,b)=0; u(0,y)=1, u(a,y)=0. 2.Same task as Exercise 1. a.u(x,b)=100; the outward normal derivative is 0 on the rest of the boundary. b.u(x,b)=100, u(0,y)=0, u(a,y)=100,∂u ∂y(x,0)=0. 3.Finish the work for Example 1: Find the bn, form the series, and check that all conditions are satisﬁed. 4.In Example 2, check that the given product solution for u1(x,y)satisﬁes the conditions and determine the coefﬁcients anand bn. 5.In Example 2, check that the given product solution for u2(x,y)satisﬁes the conditions and determine the coefﬁcients Anand Bn. 270 Chapter 4 The Potential Equation 6.Explain the difference between the cosine series in Example 1 and the co- sine series for u1(x,y)in Example 2. What is the source of the difference? 7.Finish the work for Example 3. That is, ﬁnd w(x,y)as a series and check that the boundary conditions are all satisﬁed. 8.Compare the amount of work involved in solving the problem of Exam-ple 2 (including Exercises 4 and 5) with the work for Example 3 (including Exercise 7). 9.Finish the work of Example 4: Find the solution, as a series, for w(x,y). Form u(x,y)and use the ﬁrst term of the series for w(x,y)to obtain an expression for the value of u(a 2,b 2). This would be the maximum deﬂection of the membrane. 10.Find the condition on the coefﬁcients so that the following generalsecond-degree polynomial is a solution of the Poisson equation, ∇ 2p= −H,w h e r e His constant: p(x,y)=A+Bx+Cy+Dx2+Exy+Fy2. 11.Same task as Exercise 10, but H=K(x2+y2)and pis this part of the general fourth-degree polynomial p(x,y)=Ax4+Bx3y+Cx2y2+Dxy3+Ey4, ∂2u ∂x2+∂2u ∂y2=H,0<x<a, 0<y<b, u(0,y)=0, u(a,y)=0,0<x<a, u(x,0)=0, u(x,b)=0,0<y<b. 4.4 Potential in Unbounded Regions The potential equation, as well as the heat and wave equations, can be solved in unbounded regions. Consider the following problem, in which the regioninvolved is half a vertical strip, or a slot: ∂ 2u ∂x2+∂2u ∂y2=0,0<x<a,0<y, (1) u(x,0)=f(x), 0<x<a, (2) u(0,y)=g1(y), 0<y, (3) u(a,y)=g2(y), 0<y. (4) As usual, we required that u(x,y)remain bounded as y→∞ . 4.4 Potential in Unbounded Regions 271 In order to make the separation of variables work, we must break this up into two problems. Following the model of Sections 4.2 and 4.3 we setu(x,y)=u 1(x,y)+u2(x,y)and require that the parts satisfy these two solv- able problems: ∇2u1=0, ∇2u2=0, 0<x<a,0<y, u1(x,0)=f(x), u2(x,0)=0, 0<x<a, u1(0,y)=0, u2(0,y)=g1(y), 0<y, u1(a,y)=0, u2(a,y)=g2(y), 0<y. We attack the problem for u1by assuming the product form and separating variables: u1(x,y)=X(x)Y(y),X/prime/prime(x) X(x)=−Y/prime/prime(y) Y(y)=−λ2. The sign of the constant −λ2is determined by the boundary conditions at x=0a n d x=a, which become homogeneous conditions on the factor X(x): X(0)=0,X(a)=0. (5) (We also can see that the condition to be satisﬁed along y=0d e m a n d sf u n c - tions of xthat permit a representation of an arbitrary function.) The boundary conditions, Eq. (5), together with the differential equation X/prime/prime+λ2X=0( 6 ) that comes from the separation of variables, constitute a familiar eigenvalue problem, whose solution is Xn(x)=sin/parenleftbiggnπx a/parenrightbigg ,λ2 n=/parenleftbiggnπ a/parenrightbigg2 ,n=1,2,3,.... The equation for Yis Y/prime/prime−λ2Y=0,0<y. In addition to satisfying this differential equation, Ymust remain bounded as y→∞ . The solutions of the equation are eλyand e−λy. Of these, the ﬁrst is unbounded, so Yn(y)=exp(−λny). Finally, we can write the solution of the ﬁrst problem as u1(x,y)=∞/summationdisplay n=1ansin/parenleftbiggnπx a/parenrightbigg exp/parenleftbigg−nπy a/parenrightbigg . (7) The constants ana r et ob ed e t e r m i n e df r o mt h ec o n d i t i o na t y=0. 272 Chapter 4 The Potential Equation The solution of the second problem is somewhat different. Again we seek solutions in the product form u2(x,y)=X(x)Y(y). The homogeneous bound- ary condition at y=0 and the boundedness condition become conditions on Y(y): Y(0)=0,Y(y)bounded as y→∞. Then the potential equation becomes X/prime/prime(x) X(x)+Y/prime/prime(y) Y(y)=0, (8) and both ratios must be constant. If Y/prime/prime/Yis positive, the auxiliary conditions force Yto be identically 0. Thus, we take Y/prime/prime/Y=−µ2,o rY/prime/prime+µ2Y=0, and ﬁnd that the solution that satisﬁes the auxiliary conditions is Y(y)=sin(µy), for any µ> 0. Then the general solution of the equation X/prime/prime/X=µ2is X(x)=Asinh(µx) sinh(µa)+Bsinh(µ(a−x)) sinh(µa). We have chosen this special form on the basis of our experience in solving the potential equation in the rectangle. Sinceµis a continuous parameter, we combine our product solutions by means of an integral, ﬁnding u2(x,y)=/integraldisplay∞ 0/bracketleftbigg A(µ)sinh(µx) sinh(µa)+B(µ)sinh(µ(a−x)) sinh(µa)/bracketrightbigg sin(µy)dµ. (9) The nonhomogeneous boundary conditions at x=0a n d x=aare satisﬁed if u2(0,y)=/integraldisplay∞ 0B(µ)sin(µy)dµ=g1(y), 0<y, u2(a,y)=/integraldisplay∞ 0A(µ)sin(µy)dµ=g2(y), 0<y. Obviously these two equations are Fourier integral problems, so we know how to determine the coefﬁcients A(µ)and B(µ). An example of this kind of prob- lem is shown on the CD. The potential equation can also be solved in a strip (0 <x<a,−∞<y <∞), a quarter-plane (0 <x,0<y), or a half-plane (0 <x,−∞<y<∞). Along each boundary line, a boundary condition is imposed, and the solutionis required to remain bounded in remote portions of the region considered. Ingeneral, a Fourier integral is employed in the solution, because the separationconstant is a continuous parameter, as in the second problem here. 4.4 Potential in Unbounded Regions 273 EXERCISES 1.Find a formula for the constants anin Eq. (7). 2.Ver ify that u1(x,y)in the form given in Eq. (7) satisﬁes the potential equa- tion and the homogeneous boundary conditions. 3.Find formulas for A(µ)and B(µ)of Eq. (9). 4.Solve the potential equation in the slot, 0 <x<a,0<y, for each of these sets of boundary conditions. a.u(0,y)=0, u(a,y)=0, 0<y; u(x,0)=1, 0<x<a; b.u(0,y)=0, u(a,y)=e−y,0<y; u(x,0)=0, 0 <x<a; c.u(0,y)=f(y)=/braceleftbigg 1,0<y<b, 0,b<y,u(a,y)=0, 0 <y; u(x,0)=0, 0 <x<a. 5.Solve the potential equation in the slot, 0 <x<a,0<y, for each of these sets of boundary conditions. a.∂u ∂x(0,y)=0, u(a,y)=0, 0 <y; u(x,0)=1, 0 <x<a; b.∂u ∂x(0,y)=0, u(a,y)=e−y,0<y; u(x,0)=0, 0 <x<a; c.u(0,y)=0, u(a,y)=f(y)=/braceleftbigg 1,0<y<b, 0,b<y; ∂u ∂y(x,0)=0, 0 <x<a. 6.Show that if the separation constant had been chosen as −µ2instead of µ2in solving for u2(leading to Y/prime/prime−µ2Y=0), then Y(y)≡0i st h e only function that satisﬁes the differential equation, satisﬁes the conditionY(0)=0, and remains bounded as y→∞ . 7.Solve the problem of potential in a slot under the boundary conditions u(x,0)=1,u(0,y)=u(a,y)=e−y. 8.Show that the function v(x,y)given here satisﬁes the potential equation and the boundary conditions on the “long” sides in Exercise 7, provided 274 Chapter 4 The Potential Equation that cos (a/2)/negationslash=0: v(x,y)=cos/parenleftbig x−1 2a/parenrightbig cos/parenleftbig1 2a/parenrightbige−y. What partial differential equation and boundary conditions are satisﬁed byw(x,y)=u(x,y)−v(x,y)ifuis the function of Exercise 7? 9.Solve the potential equation in the slot 0 <y<b,0<xfor each of the following sets of boundary conditions: a.u(0,y)=0,u(x,0)=0,u(x,b)=f(x)=/braceleftBig1,0<x<a, 0,a<x; b.u(0,y)=0,u(x,0)=e−x,u(x,b)=0. 10.Find product solutions of this potential problem in a strip: ∂2u ∂x2+∂2u ∂y2=0,0<x<a,−∞<y<∞, subject to the boundedness condition u(x,y)bounded as y→± ∞ . 11.Solve the potential problem consisting of the equation and boundedness conditions from Exercise 10 and the boundary conditions u(0,y)=0,u(a,y)=e−|y|,−∞<y<∞. 12.Show how to solve the potential problem of Exercise 10 together with the boundary conditions u(0,y)=g1(y), u(a,y)=g2(y),−∞<y<∞, where g1and g2are suitable functions. 13.Find product solutions of this potential problem in the quarter-plane: ∂2u ∂x2+∂2u ∂y2=0,0<x,0<y, u(0,y)=0,u(x,0)=f(x). Note that u(x,y)must remain bounded as x→∞ and as y→∞ . 14.Solve the potential equation in the quarter-plane, x>0,y>0, subject to the boundary conditions u(0,y)=e−y,y>0; u(x,0)=e−x,x>0. 4.5 Potential in a Disk 275 15.Find product solutions of the potential equation in the half-plane y>0: ∂2u ∂x2+∂2u ∂y2=0,−∞<x<∞,0<y<∞, u(x,0)=f(x), −∞<x<∞. What boundedness conditions must u(x,y)satisfy? 16.Convert your solution of Exercise 15 into the following formula (see Ex- ercise 8 of Section 4.4 and Section 2.11): u(x,y)=1 π/integraldisplay∞ −∞f(x/prime)y y2+(x−x/prime)2dx/prime. 17.Use the formula in Exercise 16 to solve the potential problem in the upper half-plane, with boundary condition u(x,0)=f(x)=/braceleftBig1,0<x, 0,x<0. 18.Solve the problem stated in Exercise 15 if the boundary function is f(x)=/braceleftbigg1,|x|<a, 0,|x|>a. 19.Show that u(x,y)=xis the solution of the potential equation in a slot under the boundary conditions f(x)=x,g1(y)=0,g2(y)=a. Can this solution be found by the method of this section? 4.5 Potential in a Disk If we need to solve the potential equation in a circular disk x2+y2<c2,i ti s n a t u r a lt ou s ep o l a rc o o r d i n a t e s r,θ, in terms of which the disk is described by 0<r<c. We found in Section 4.1 that the potential equation in polar coordinates is 1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg +1 r2∂2v ∂θ2=0. There are some special features of this coordinate system. First, it is clear that some coefﬁcients of the Laplacian are negative powers of r. Thus, we must enforce a boundedness condition at r=0. Second, θandθ+2πrefer to the same angle. Therefore, we must require that the function v(r,θ)be periodic with period 2 πinθ. 276 Chapter 4 The Potential Equation The Dirichlet problem on a disk can now be stated as 1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg +1 r2∂2v ∂θ2=0,0≤r<c, (1) v(c,θ)=f(θ), (2) v(r,θ+2π)=v(r,θ), 0<r<c, (3) v(r,θ)bounded as r→0+. (4) By assuming v(r,θ)=R(r)Q(θ)we can separate variables. The potential equa- tion becomes 1 r/parenleftbig rR/prime/parenrightbig/primeQ+1 r2RQ/prime/prime=0. Separation is effected by dividing through by RQ/r2: r(rR/prime(r))/prime R(r)+Q/prime/prime(θ) Q(θ)=0. As usual, both terms must be constant. We know that if Q/prime/prime/Qis a posi- tive constant, then Qwill be exponential, not periodic. Therefore we choose Q/prime/prime/Q=−λ2and obtain this singular eigenvalue problem: Q/prime/prime+λ2Q=0, (5) Q(θ+2π)=Q(θ). (6) The accompanying equation for R(r)is r/parenleftbig rR/prime/parenrightbig/prime−λ2R=0, (7) R(r)bounded as r→0+. (8) The general solution of Eq. (5) (if λ> 0) is Q(θ)=Acos(λθ)+Bsin(λθ). This function is periodic for all λ, but the period is 2 πonly if λis an inte- ger. Thus we have λn=n,n=1,2,....I na d d i t i o n ,i f λ=0, we have a peri- odic solution that is any constant. Thus, the solution of the singular eigenvalueproblem of Eqs. (5) and (6) is λ 0=0,Q0(θ)=1, λn=n,Qn(θ)=Acos(nθ)+Bsin(nθ), n=1,2,3,.... The novelty here is that we have two eigenfunctions for each eigenvalue n= 1,2,.... 4.5 Potential in a Disk 277 Knowing that λ2 n=n2,w ec a ne a s i l yﬁ n d R(r). The equation for Rbecomes r2R/prime/prime+rR/prime−n2R=0,0<r<c, when the indicated differentiations are carried out. This is a Cauchy–Euler equation, whose solutions are known to have the form R(r)=rα,w h e r e α is constant. Substituting R=rα,R/prime=αrα−1,a n d R/prime/prime=α(α−1)rα−2into it leaves /parenleftbig α(α−1)+α−n2/parenrightbig rα=0,0<r<c. Because rαis not zero, the constant factor in parentheses must be zero — that is,α=± n. The general solution of the differential equation is any combina- tion of rnand r−n.T h el a t t e r ,h o w e v e r ,i su n b o u n d e da s rapproaches zero, so we discard that solution, retaining Rn(r)=rn.I nt h es p e c i a lc a s e n=0, the two solutions are the constant function 1 and ln (r). The logarithm is discarded because of its behavior at r=0. Now we reassemble our solution. The functions r0·1=1,rncos(nθ), rnsin(nθ) (9) are all solutions of the potential equation, so a general linear combination of these solutions will also be a solution. Thus v(r,θ)may have the form v(r,θ)=a0+∞/summationdisplay n=1anrncos(nθ)+∞/summationdisplay n=1bnrnsin(nθ). (10) At the true boundary r=c, the boundary condition reads v(c,θ)=a0+∞/summationdisplay n=1cn/parenleftbig ancos(nθ)+bnsin(nθ)/parenrightbig =f(θ), −π<θ ≤π. This is a Fourier series problem, as in Section 1.1, solved by choosing a0=1 2π/integraldisplayπ −πf(θ)dθ, an=1 πcn/integraldisplayπ −πf(θ)cos(nθ)dθ, bn=1 πcn/integraldisplayπ −πf(θ)sin(nθ)dθ. (11) Example. Consider the problem consisting of Eqs. (1)–(4) with v(c,θ)=f(θ)=/braceleftBigg0,−π<θ< −π/2, 1,−π/2<θ<π/ 2, 0,π / 2<θ<π . 278 Chapter 4 The Potential Equation T h es o l u t i o ni sg i v e nb yE q .( 1 0 ) ,p r o v i d e dt h a tt h ec o e f ﬁ c i e n t sa r ec h o s e n a c c o r d i n gt oE q .( 1 1 ) .S i n c e f(θ)is an even function, bn=0, and a0=1 π/integraldisplayπ 0f(θ)dθ=1 2, an=2 πcn/integraldisplayπ 0f(θ)cos(nθ)dθ=2s i n(nπ/2) nπcn. Therefore, the solution of the problem is v(r,θ)=1 2+∞/summationdisplay n=12s i n(nπ/2) nπrn cncos(nθ). (12) The level curves of this function are all arcs of circles that pass through the boundary points r=c,θ=±π/2, where f(θ)jumps between 0 and 1. Along thex-axis, the function has the simple closed form 1 /2+(2/π)tan−1(x/c). The CD has a color graphic of the solution. /square Properties of the Solution Now that we have the form Eq. (10) of the solution of the potential equation, we can see some important properties of the function v(r,θ).I np a r t i c u l a r ,b y setting r=0w eo b t a i n v(0,θ)=a0=1 2π/integraldisplayπ −πf(θ)dθ=1 2π/integraldisplayπ −πv(c,θ)dθ. This says that the solution of the potential equation at the center of a disk is equal to the average of its values around the edge of the disk. It is easy to show also that v(0,θ)=1 2π/integraldisplayπ −πv(r,θ)dθ (13) for any rbetween 0 and c! This characteristic of solutions of the potential equa- tion is called the mean value property . From the mean value property, it is just a step to prove the maximum principle mentioned in Section 4.1, for the meanvalue of a function lies between the minimum and the maximum and cannotequal either unless the function is constant. An important consequence of the maximum principle — and thus of the mean value property — is a proof of the uniqueness of the solution of theDirichlet problem. Suppose that uandvare two solutions of the potential equation in some region Rand that they have the same values on the bound- ary of R. Then their difference, w=u−v, is also a solution of the potential equation in Rand has value 0 all along the boundary of R.B yt h em a x i m u m 4.5 Potential in a Disk 279 principle, whas maximum and minimum values 0, and therefore wis identi- cally 0 throughout R.I no t h e rw o r d s , uandvare identical. EXERCISES 1.Solve the potential equation in the disk 0 <r<cif the boundary condi- tion is v(c,θ)=|θ|,−π<θ ≤π. 2.Same as Exercise 1 if v(c,θ)=θ,−π<θ<π . Is the boundary condition satisﬁed at θ=±π? 3.Same as Exercise 1, with boundary condition v(c,θ)=f(θ)=/braceleftbigg cos(θ), −π/2<θ<π/ 2, 0, otherwise. 4.Find the value of the solution at r=0 for the problems of Exercises 1, 2, and 3. 5.If the function f(θ)in Eq. (2) is continuous and sectionally smooth and satisﬁes f(−π+)=f(π−), what can be said about convergence of the series for v(c,θ)? 6.Show that v(r,θ)=a0+∞/summationdisplay n=1r−n/parenleftbig ancos(nθ)+bnsin(nθ)/parenrightbig is a solution of Laplace’s equation in the region r>c(exterior of a disk) and has the property that |v(r,θ)|is bounded as r→∞ . 7.If the condition v(c,θ)=f(θ)is given, what are the formulas for the a’s and b’s in Exercise 6? 8.The solution of Eqs. (1)–(4) can be written in a single formula by the following sequence of operations: a.Replace θbyφin Eq. (11) for the a’s and b’s; b.replace the a’s and b’s in Eq. (10) by the integrals in part a; c.use the trigonometric identity cos(nθ)cos(nφ)+sin(nθ)sin(nφ)=cos/parenleftbig n(θ−φ)/parenrightbig ; d.take the integral outside the series; 280 Chapter 4 The Potential Equation e.add up the series (see Section 1.10, Exercise 5a). Then v(r,θ)is given by the single integral (Poisson integral formula) v(r,θ)=1 2π/integraldisplayπ −πf(φ)c2−r2 c2+r2−2rccos(θ−φ)dφ. 9.Solve Laplace’s equation in the quarter-disk 0 <θ<π / 2, 0<r<c,s u b - ject to the boundary conditions v(r,0)=0,v(r,π/2)=0,v(c,θ)=1. 10.Generalize the results of Exercise 9 by solving this problem: 1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg +1 r2∂2v ∂θ2=0,0<θ<α π , 0<r<c, v(r,0)=0,v ( r,απ)=0,0<r<c, v(c,θ)=f(θ), 0<θ<α π . Here,αis a parameter between 0 and 2. 11.Suppose that α> 1 in Exercise 10. Show that there is a product solution with the property that∂v ∂r(r,θ)is not bounded as r→0+. 4.6 Classiﬁcation of Partial Differential Equations and Limitations of the Product Method By this time, we have seen a variety of equations and solutions. We have con- centrated on three different, homogeneous equations (heat, wave, and poten-tial) and have found the qualitative features summarized in the following table: Equation Features Heat Exponential behavior in time. Existence of a limiting (steady-state) solution. Smooth graph for t>0. Wave Oscillatory (not always periodic) behavior in time. Retention of discontinuities fort>0. Potential Smooth surface. Maximum principle. Mean value property. These three two-variable equations are the most important representatives of the three classes of second-order linear partial differential equations in twovariables. The most general equation that ﬁts this description is A∂ 2u ∂ξ2+B∂2u ∂ξ∂η+C∂2u ∂η2+D∂u ∂ξ+E∂u ∂η+Fu+G=0, 4.6 Classiﬁcation and Limitations 281 where A,B,C,and so forth are, in general, functions of ξandη.( W eu s eG r e e k letters for the independent variables to avoid implying any relations to spaceor time.) Such an equation can be classiﬁed according to the sign of B 2−4AC: B2−4AC<0: elliptic , B2−4AC=0: parabolic , B2−4AC>0: hyperbolic . Because A,B,and Care functions of ξandη(not of u), the classiﬁcation of an equation may vary from point to point. It is easy to see that the heat equationis parabolic, the wave equation is hyper bolic, and the potential equation is elliptic. The classiﬁcation of an equatio n determines important features of the solution and also dictates the method of attack when numerical techniques are used for solution. The question naturally arises whether separation of variables works on all equations. The answer is no. For instance, the equation /parenleftbig ξ+η 2/parenrightbig∂2u ∂ξ2+∂2u ∂η2=0 does not admit separation of variables. In general, it is difﬁcult to say just which equations can be solved by this method. However, it is necessary to have B≡0. The region in which the solution is to be found also limits the applicability o ft h em e t h o dw eh a v eu s e d .T h er e g i o nm u s tb ea generalized rectangle .B y this we mean a region bounded by coordinate curves of the coordinate systemof the partial differential equation. Put another way, the region is describedby inequalities on the coordinates, whose endpoints are ﬁxed quantities. For instance, we have worked in regions described by the following sets of inequal- ities: 0<x<a, 0<t, 0<x, 0<t, −∞<x<∞,0<t, 0<x<a, 0<y<b, 0<r<c, −π<θ ≤π. All of these are generalized rectangles, but only one is an ordinary rectangle. An L-shaped region is not a generalized rectangle, and our methods wouldbreak down if applied to, for instance, the potential equation there. There are, as we know, restrictions on the kinds of boundary conditions that can be handled. From the examples in this chapter it is clear that we need 282 Chapter 4 The Potential Equation homogeneous or “homogeneous-like” con d i t i o n so no p p o s i t es i d e so fag e n e r - alized rectangle. Examples of “homogeneous-like” conditions are the require-ment that a function remain bounded as some variable tends to inﬁnity, or theperiodic conditions at θ=±π(see Section 4.5). The point is that if two or more functions satisfy the conditions, so does a sum of those functions. In spite of the limitations of the method of separation of variables, it works well on many important problems in two or more variables and provides in-sight into the nature of their solutions. Moreover, it is known that in thosecases where separation of variables can be carried out, it will ﬁnd a solution ifone exists. EXERCISES 1.Classify the following equations. a.∂2u ∂x∂y=0; b.∂2u ∂x2+∂2u ∂x∂y+∂2u ∂y2=2x; c.∂2u ∂x2−∂2u ∂x∂y+∂2u ∂y2=2u; d.∂2u ∂x2−2∂2u ∂x∂y+∂2u ∂y2=∂u ∂y; e.∂2u ∂x2−∂2u ∂y2−∂u ∂y=0. 2.S h o wt h a t ,i np o l a rc o o r d i n a t e s ,a na n n u l u s ,as e c t o r ,a n das e c t o ro fa n annulus are all generalized rectangles. 3.I nw h i c ho ft h ee q u a t i o n si nE x e r c i s e1c a nt h ev a r i a b l e sb es e p a r a t e d ? 4.Sketch the regions listed in the text as generalized rectangles. 5.Solve these three problems and compare the solutions. a.∂2u ∂x2+∂2u ∂y2=0, 0 <x<1, 0 <y, u(x,0)=f(x),0<x<1, u(0,y)=0, u(1,y)=0, 0 <y; b.∂2u ∂x2=∂2u ∂y2,0<x<1, 0 <y, 4.7 Comments and References 283 u(x,0)=f(x),∂u ∂y(x,0)=0, 0 <x<1, u(0,y)=0, u(1,y)=0, 0 <y; c.∂2u ∂x2=∂u ∂y,0<x<1, 0 <y, u(x,0)=f(x),0<x<1, u(0,y)=0, u(1,y)=0, 0 <y. 6.Show that if f1,f2,... all satisfy the periodic boundary conditions f(−π)=f(π), f/prime(−π)=f/prime(π), then so does the function c1f1+c2f2+··· ,w h e r et h e c’s are constants. 7.Longitudinal waves in a slender rod may be described by this partial differ- ential equation: ∂2u ∂x2=∂2u ∂t2−/epsilon1∂4u ∂x2∂t2. Show how to separate the variables. 8.Deﬂections of a thin plate and slow ﬂow of a viscous ﬂuid may both be described by the biharmonic equation ∂4u ∂x4+2∂4u ∂x2∂y2+∂4u ∂y4=0. Assume that u(x,y)=X(x)Y(y)and show that the variables don’t separate. Show that, under the additional assumption X/prime/prime/X=−λ2, a differential equation for Yresults. 4.7 Comments and References While the potential equation describes many physical phenomena, there is one that makes the solution of the Dirichlet problem very easy to visualize. Sup-pose a piece of wire is bent into a closed curve or frame. When the frame isheld over a level surface, its projection onto the surface is a plane curve Cen- closing a region R. If one forms a soap ﬁlm on the frame, the height u(x,y)of the ﬁlm above the level surface is a function that satisﬁes the potential equa- tion approximately, if the effects of gravity are negligible (see Chapter 5). Theheight of the frame above the curve Cgives the boundary condition on u.F o r example, Fig. 2(b) shows the surface corresponding to the problem solved inS e c t i o n4 . 2 .Ag r e a td e a lo fi n f o r m a t i o na b o u ts o a pﬁ l m si si nt h eb o o k The S c i e n c eo fS o a pF i l m sa n dS o a pB u b b l e s by C. Isenberg (see the Bibliography). 284 Chapter 4 The Potential Equation Figure 3 Streamlines (solid) and equipotential curves (dashed) for ﬂow in a cor- ner. The streamlines are described by the equation 2 xy=constant, with a differ- ent constant for each one. Similarly, the equipotential curves are described by theequation x 2−y2=constant. It turns out that the potential equation (but not all elliptic equations) is best studied through the use of complex variables. A complex variable may bewritten z=x+iy,w h e r e xand yare real and i 2=− 1; similarly a function of zis denoted by f(z)=u(x,y)+iv(x,y),uandvbeing real functions of real variables. If fhas a derivative with respect to z,t h e nb o t h uandvsatisfy the potential equation. Easy examples, such as polynomials and exponentials, leadto familiar solutions: z 2=(x+iy)2=x2−y2+i2xy, ez=ex+iy=exeiy=excos(y)+iexsin(y), ln(z)=1 2ln/parenleftbig x2+y2/parenrightbig +itan−1/parenleftbiggy x/parenrightbigg . (See Section 4.1, Exercises 1, 2; Section 4.4, Exercises 13, 14; Miscellaneous Exercise 18 in this chapter.) Knowing these elementary solutions often helps in simplifying a problem. In certain idealized ﬂuid ﬂows (steady, irrotational, two-dimensional ﬂow of an inviscid, incompressible ﬂuid) the velocity vector is given by V=− gradφ, where the velocity potential φis a solution of the potential equation. The streamlines along which the ﬂuid ﬂows are level curves of a related function ψ, called the stream function , which also is a solution of the potential equation. The two functions φandψare, respectively, the real and imaginary parts of a function of the complex variable z. The level curves φ=constant and ψ=constant form two families of orthogonal curves called a ﬂow net .T h e ﬂow net in Fig. 3, for φ=x2−y2andψ=2xy(the real and imaginary parts of the function f(z)=z2), illustrates ﬂow near a corner formed by two walls. Many other ﬂow nets are shown in the book Potential Flows :Computer Graphic Solutions by R.H. Kirchhoff. Civil engineers sometimes sketch a ﬂow net by eye Miscellaneous Exercises 285 to get a rough graphical solution of the potential equation for hydrodynamics problems. Where a physical boundary is formed by an impervious wall, the velocity vector Vmust be parallel to the boundary. This fact leads to two boundary conditions. First, the wall must coincide with a streamline; thus ψ=constant along a boundary. Second, the component of Vthat is normal to the wall must be zero there, because no ﬂuid passes through it; thus the normal derivative ofφis zero, ∂φ/∂ n=0, at a boundary. See Miscellaneous Exercises 30–32. Chapter Review See the CD for Review Questions. Miscellaneous Exercises 1.Solve the potential equation in the rectangle 0 <x<a,0<y<bwith the boundary conditions u(0,y)=1,u(a,y)=0,0<y<b, u(x,0)=0,u(x,b)=0,0<x<a. 2.Ifa=bin Exercise 1, then u(a/2,a/2)=1/4. Use symmetry to explain this fact. 3.Solve the potential equation on the rectangle 0 <x<a,0<y<bwith the boundary conditions u(0,y)=1, u(a,y)=1, 0<y<b, ∂u ∂y(x,0)=0,∂u ∂y(x,b)=0,0<x<a. 4.Same as Exercise 3, but the boundary conditions are u(0,y)=1,∂u ∂x(a,y)=0,0<y<b, u(x,0)=1,∂u ∂y(x,b)=0,0<x<a. 286 Chapter 4 The Potential Equation 5.Same as Exercise 3, but the boundary conditions are u(0,y)=1, u(a,y)=1,0<y<b, ∂u ∂y(x,0)=0,u(x,b)=0,0<x<a. 6.Same as Exercise 3, but the boundary conditions are u(0,y)=1,u(a,y)=0,0<y<b, u(x,0)=1,u(x,b)=0,0<x<a. 7.Same as Exercise 3, but the region is a square (b=a)and the boundary conditions are u(0,y)=f(y), u(a,y)=0,0<y<a, u(x,0)=f(x), u(x,a)=0,0<x<a, where fis a function whose graph is an isosceles triangle of height hand width a. 8.Solve the potential equation in the region 0 <x<a,0<ywith the boundary conditions u(x,0)=1, 0<x<a, u(0,y)=0,u(a,y)=0,0<y. 9.Find the solution of the potential equation on the strip 0 <y<b, −∞<x<∞, subject to the conditions that follow. Supply bounded- ness conditions as necessary. u(x,0)=/braceleftbigg1,−a<x<a, 0,|x|>a, u(x,b)=0,−∞<x<∞. 10. Show that the function u(x,y)=tan−1(y/x)is a solution of the potential equation in the ﬁrst quadrant. What conditions does usatisfy along the positive x-a n d y-axes? 11. Solve the potential problem in the upper half-plane, ∂2u ∂x2+∂2u ∂y2=0,−∞<x<∞,0<y, u(x,0)=f(x),−∞<x<∞, taking f(x)=exp(−α|x|). Miscellaneous Exercises 287 12.Apply the following formula (see Section 4.4, Exercise 16) for the solu- tion of the potential problem in the upper half-plane if the boundarycondition is u(x,0)=f(x),w h e r e f(x)=/braceleftBig0,x<0, 1,x>0, u(x,y)=1 π/integraldisplay∞ −∞f(x/prime)y y2+(x−x/prime)2dx/prime. 13.Apply the formula in Exercise 12 to the case where f(x)=1,−∞<x <∞. The solution of the problem should be u(x,y)≡1. 14. a. Find the separation-of-variables solution of the potential problem in a disk of radius 1 if the boundary condition is u(1,θ)=f(θ),w h e r e f(θ)=/braceleftBig−π−θ,−π<θ< 0, π−θ, 0<θ<π . b.Show that the function given in polar and Cartesian coordinates by u(r,θ)=2t a n−1/parenleftbiggrsin(θ) 1−rcos(θ)/parenrightbigg =2t a n−1/parenleftbiggy 1−x/parenrightbigg satisﬁes the potential equation (use the Cartesian coordinates) and the boundary condition. The following identity is useful: sin(θ) 1−cos(θ)=tan/parenleftbiggπ−θ 2/parenrightbigg . c.Sketch some level curves of the solution inside the circle of radius 1. 15.Solve the potential equation in a disk of radius cwith boundary condi- tions u(c,θ)=/braceleftBig1,0<θ<π , 0,−π<θ< 0. 16.What is the value of uat the center of the disk in Exercise 15? 17.Same as Exercise 15, but the boundary condition is u(c,θ)=/vextendsingle/vextendsinglesin(θ)/vextendsingle/vextendsingle. 288 Chapter 4 The Potential Equation 18. For the potential problem on an annular ring 1 r∂ ∂r/parenleftbigg r∂2u ∂r/parenrightbigg +1 r2∂2u ∂θ2=0,a<r<b, show that product solutions have the form A0+B0ln(r), ( C0+D0θ)ln(r), or rn/parenleftbig Ancos(nθ)+Bnsin(nθ)/parenrightbig +r−n/parenleftbig Cncos(nθ)+Dnsin(nθ)/parenrightbig . 19. Solve the potential problem on the annular ring as stated in Exercise 18 with boundary conditions u(a,θ)=1,u(b,θ)=0. 20. Find product solutions of the potential equation on a sector of a disk with zero boundary conditions on the straight edges. ∇2u=0, 0≤r<c,0<θ<α , u(r,0)=0,u(r,α)=0. 21. Solve the potential problem in a slit disk: ∇2u=0, 0≤r<c,0<θ< 2π, u(r,0)=0, u(r,2π)=0, u(r,θ)=f(θ), 0<θ< 2π. 22. Show that the function u(x,y)=sin(πx/a)sinh(πy/a)satisﬁes the po- tential problem ∇2u=0, 0<x<a, 0<y, u(0,y)=0,u(a,y)=0,0<y, u(x,0)=0,0<x<a. This solution is eliminated if it is also required that u(x,y)be bounded asy→∞ . 23. Solve the potential equation in the rectangle 0 <x<a,0<y<b, with the boundary conditions u(0,y)=0,∂u ∂x(a,y)=0,0<y<b, u(x,0)=0,u(x,b)=x, 0<x<a. Miscellaneous Exercises 289 24.Find a polynomial of second degree in xand y, v(x,y)=A+Bx+Cy+Dx2+Exy+Fy2, that satisﬁes the potential equation and these boundary conditions: v(0,y)=0,0<y<b, v(x,0)=0,v ( x,b)=x,0<x<a. 25.Find the problem (partial differential equation and boundary condi- tions) satisﬁed by w(x,y)=v(x,y)−u(x,y),w h e r e uandvare the so- lutions of the problems in Exercises 23 and 24. Solve the problem. Is thisproblem easier to solve than the one in Exercise 23? 26.Solve the potential equation in the quarter-plane 0 <x,0<y,s u b j e c tt o the boundary conditions u(x,0)=f(x), 0<x, u(0,y)=f(y), 0<y. The function fthat appears in both boundary conditions is given by the equation f(x)=/braceleftBig1,0<x<a, 0,a<x. 27.(Flow past a plate) A ﬂuid occupies the half-plane y>0 and ﬂows past (left to right, approximately) a plate located near the x-axis. If the xandy components of velocity are U 0+u(x,y)andv(x,y),r e s p e c t i v e l y( U0= constant free-stream velocity), under certain assumptions, the equations of motion, continuity, and state can be reduced to ∂u ∂y=∂v ∂x,/parenleftbig 1−M2/parenrightbig∂u ∂x+∂v ∂y=0, valid for all xand y>0.Mis the free-stream Mach number. Deﬁne the velocity potential φby the equations u=∂φ/∂ xandv=∂φ/∂ y.S h o w that the ﬁrst equation is automatically satisﬁed and the second is a partialdifferential equation that is elliptic if M<1 or hyperbolic if M>1. 28.If the plate is wavy — say, its equation is y=/epsilon1cos(αx)— then the boundary condition, that the vector velocity be parallel to the wall, is v/parenleftbig x,/epsilon1cos(αx)/parenrightbig =−/epsilon1αsin(αx)/parenleftbig U 0+u/parenleftbig x,/epsilon1cos(αx)/parenrightbig/parenrightbig . This equation is impossible to use, so it is replaced by v(x,0)=−/epsilon1αU0sin(αx) 290 Chapter 4 The Potential Equation on the assumption that /epsilon1is small and uis much smaller than U0. Using this boundary condition and the condition that u(x,y)→0a s y→∞ , set up and solve a complete boundary value problem for φ,a s s u m i n g M<1. 29. By superposition of solutions ( αranging from 0 to ∞) ﬁnd the ﬂow past aw a l lw h o s ee q u a t i o ni s y=f(x). Hint: Use the boundary condition v(x,0)=U0f/prime(x)=/integraldisplay∞ 0/bracketleftbig A(α)cos(αx)+B(α)sin(αx)/bracketrightbig dα. 30. In hydrodynamics, the velocity vector in a ﬂuid is V=− grad(u),w h e r e uis a solution of the potential equation. The normal component of ve- locity, ∂u/∂n, is 0 at a wall. Thus the problem ∇2u=0, 0<x<1, 0<y<1, ∂u ∂x(0,y)=0,∂u ∂x(1,y)=− 1,0<y<1, ∂u ∂y(x,0)=0,∂u ∂y(x,1)=1, 0<x<1, represents a ﬂow around a corner: ﬂow inward at the top, outward at the right, with walls at left and bottom. Explain why, in a ﬂuid ﬂow problem,it must be true that/integraldisplay C∂u ∂nds=0( ∗) ifuis a solution of the potential equation in a region R,∂u/∂nis the outward normal derivative, Cis the boundary of the region, and sis arc length. 31. Under the conditions stated in Exercise 30, prove the validity of ( ∗). Hint: Use Green’s theorem. 32. The Neumann problem consists of the potential equation in a region R and conditions on ∂u/∂nalongC, the boundary of R. Show (a) that /integraldisplay C∂u ∂nds=0 is a necessary condition for a solution to exist, and (b) if uis a solution of the Neumann problem, so is u+c(cis constant). 33. Show that u(x,y)=1 2(y2−x2)is a solution of the problem in Exercise 30. 34. a. Show that the given function is a sol ution of the potential equation. Miscellaneous Exercises 291 b.Find the gradient of uand plot some vectors V=− grad(u)near the origin. u(x,y)=− tan−1/parenleftbiggy x/parenrightbigg . T h eﬂ o wﬁ e l d( s e eE x e r c i s e3 0 )g i v e nb yt h i sf u n c t i o ni sc a l l e da n irrotational vortex . 35.S a m et a s k sa si nE x e r c i s e3 4 .T h eﬂ o wﬁ e l dg i v e nb yt h i sf u n c t i o ni sc a l l e d a source at the origin. u(x,y)=− ln/parenleftbig/radicalbig x2+y2/parenrightbig . 36.Solve this potential problem in a half-annulus (sketch the region). At some point, it may be useful to make the substitution s=ln(r). 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg +1 r2∂2u ∂θ2=0,1<r<e, 0<θ<π , u(1,θ)=0, u(e,θ)=0,0<θ<π , u(r,0)=0, u(r,π)=1,1<r<e. 37.Solve the Poisson equation, ∇2u=− f, in polar coordinates by ﬁnding a function that depends only on rfor: a.f(r,θ)=1; b.f(r,θ)=1 r2. 38.In “An improved transmission line structure for contact resistivity mea- surements” [L.P . Floyd et al., Solid-State Electronics ,37(1994): 1579– 1584], a strip of conducting material is carrying a current in the direc-tion of its length. A second, long conducting strip of width Lis placed at right angles to the ﬁrst, forming a cross. A voltage is to be measured bya probe on the second strip some distance from the ﬁrst. In the secondstrip, the voltage V(x,y)satisﬁes the boundary value problem ∂ 2V ∂x2+∂2V ∂y2=0, 0<x<L,0<y, ∂V ∂x(0,y)=0,∂V ∂x(L,y)=0,0<y, V(x,0)=f(x), 0<x<L. In this problem, xis in the direction of current ﬂow in the lower strip; yis in the direction of the length of the second strip; y=0a tt h ee d g e 292 Chapter 4 The Potential Equation Figure 4 Exercise 38. of the lower strip (see Fig. 4). Of course, Vis bounded as y→∞ .S o l v e this boundary value problem for Vin terms of f(x). 39. The authors of the article cited in Exercise 38 say that any measurement ofV(x,y)made at a distance ygreater than 5 Lis independent of x.E x - plain this statement, and determine what value (in terms of f) would be measured. 40. In the article “A production-planning and design model for assessing the thermal behavior of thick steel strip during continuous heat treatment”[W.D. Morris, Journal of Process Engineering (2001): 53–63] the author models the temperature T(x,y)of a long steel strip that comes out of an oven at x=0, moving to the right, where it is exposed to coolant air. These equations enter into the modeling (see Table 1 and Fig. 5): a.Conservation of energy/steady-state heat equation, derived by con- sidering conservation of energy for a rectangle of dimensions /Delta1xby /Delta1ythat is ﬁxed in space (see Sections 5.1 and 5.2): v k∂T ∂x=∂2T ∂x2+∂2T ∂y2,0<x,−b<y<b; b.Symmetry condition: ∂T ∂y(x,0)=0,0<x; c.Cooling by convection at the surface (see Section 2.1, Eq. (10)): −κ∂T ∂y(x,b)=h/parenleftbig T(x,b)−Ta/parenrightbig ,0<x; Miscellaneous Exercises 293 h convection coefﬁcient (W/m2K) κ thermal conductivity of steel (W/mK) L length of cooling line T(x,y)temperature in the strip Ta temperature of coolant T0 temperature of the strip at entry to cooling line v strip speed (m/s) k thermal diffusivity of steel (m2/s) B Biot number (dimensionless) Table 1 Table of Notation. Figure 5 Exercise 40. d.Condition at entry to cooling line (at x=0): T(0,y)=T0,−b<y<b. The author treats the strip as inﬁnite. In fact, typical dimensions are 100 m in the x-direction and 2 cm in the y- d i r e c t i o n ,s ot h er a t i oo f xtoylengths is on the order of 104. Next, these dimensionless variables are introduced: θ=T−Ta T0−Ta,Y=y b,X=x b, and these dimensionless parameter combinations appear in the equa- tions: γ=bv k,B=hb κ. 294 Chapter 4 The Potential Equation The problem in terms of dimensionless variables and parameters is: γ∂θ ∂X=∂2θ ∂X2+∂2θ ∂Y2, 0<X,0<Y<1, ∂θ ∂Y(X,0)=0, 0<X, ∂θ ∂Y(X,1)=− Bθ(X,1),0<X, θ(0,Y)=1, 0<Y<1. S o l v et h ep r o b l e mf o rt h ec a s eo fah i g hB i o tn u m b e r , B→∞ ,w h i c h means that θ(X,1)=0, 0<X. 41. (Continuation) Solve the problem in Exercise 40 for the case of a low Biot number, B≈0, which means that∂θ ∂Y(X,1)=0, 0<X. Higher Dimensions and Other Coordinates CHAPTER5 5.1 Two-Dimensional Wave Equation: Derivation For an example of a two-dimensional wave equation, we consider a membrane that is stretched taut over a ﬂat frame in the xy-plane (Fig. 1). The displace- ment of the membrane above the point (x,y)at time tisu(x,y,t).W ea s s u m e that the surface tension σ(dimensions F/L) is constant and independent of position. We also suppose that the membrane is perfectly ﬂexible; that is, itdoes not resist bending. (A soap ﬁlm satisﬁes these assumptions quite accu-rately.) Let us imagine that a small rectangle (of dimensions /Delta1xby/Delta1yaligned with the coordinate axes) is cut out of the membrane, and then apply Newton’slaw of motion to it. On each edge of the rectangle, the rest of the membrane ex-erts a distributed force of magnitude σ(symbolized by the arrows in Fig. 2a); these distributed forces can be resolved into concentrated forces of magnitude σ/Delta1xorσ/Delta1y, according to the length of the segment involved (see Fig. 2b and Fig. 3). Figure 1 Frame in the xy-plane. 295 296 Chapter 5 Higher Dimensions and Other Coordinates (a) (b) Figure 2 (a) Distributed forces. (b) Concentrated forces. Figure 3 Forces on a piece of membrane. (a) (b) Figure 4 Forces (a) in the xu-plane; (b) in the yu-plane. Looking at projection on the xu-a n d yu-planes (Figs. 4a, 4b), we see that the sum of forces in the x-direction is σ/Delta1y(cos(β)−cos(α)) ,a n dt h es u mo f forces in the y-direction is /Delta1x(cos(δ)−cos(γ)) . It is desirable that both these sums be zero or at least negligible. Therefore we shall assume that α,β,γ,a n d δare all small angles. Because we know that tan(α)=∂u ∂x,tan(γ)=∂u ∂y and so forth, when the derivatives are evaluated at some appropriate point near (x,y), we are assuming that the slopes ∂u/∂xand∂u/∂yof the membrane are very small. Chapter 5 Higher Dimensions and Other Coordinates 297 Adding up forces in the vertical direction and equating the sum to the mass times acceleration (in the vertical direction) we obtain σ/Delta1y/parenleftbig sin(β)−sin(α)/parenrightbig +σ/Delta1x/parenleftbig sin(δ)−sin(γ)/parenrightbig =ρ/Delta1x/Delta1y∂2u ∂t2, where ρis the surface density [ m/L2]. Because the angles α,β,γ ,a n dδare small, the sine of each is approximately equal to its tangent: sin(α)∼=tan(α)=∂u ∂x(x,y,t), and so forth. With these approximations used throughout, the preceding equa- tion becomes σ/Delta1y/parenleftbigg∂y ∂x(x+/Delta1x,y,t)−∂u ∂x(x,y,t)/parenrightbigg +σ/Delta1x/parenleftbigg∂u ∂y(x,y+/Delta1y,t)−∂u ∂y(x,y,t)/parenrightbigg =ρ/Delta1x/Delta1y∂2u ∂t2. On dividing through by /Delta1x/Delta1y, we recognize two difference quotients in the left-hand member. In the limit they become partial derivatives, yieldingthe equation σ/parenleftbigg∂ 2u ∂x2+∂2u ∂y2/parenrightbigg =ρ∂2u ∂t2, or ∂2u ∂x2+∂2u ∂y2=1 c2∂2u ∂t2, ifc2=σ/ρ. This is the two-dimensional wave equation. If the membrane is ﬁxed to the ﬂat frame, the boundary condition would be u(x,y,t)=0 for (x,y)on the boundary . Naturally, it is necessary to give initial conditions describing the displacement and velocity of each point on the membrane at t=0: u(x,y,0)=f(x,y), ∂u ∂t(x,y,0)=g(x,y). EXERCISES 1.Suppose that the frame is rectangular, bounded by segments of the lines x=0,x=a,y=0,y=b. Write an initial value–boundary value problem, complete with inequalities, for a membrane stretched over this frame. 298 Chapter 5 Higher Dimensions and Other Coordinates 2.Suppose that the frame is circular and that its equation is x2+y2=a2. Write an initial value–boundary value problem for a membrane on a circu-lar frame. (Use polar coordinates.) 3.What should the three-dimensional wave equation be? 5.2 Three-Dimensional Heat Equation: Vector Derivation T o illustrate a different technique, we are going to derive the three-dimensional heat equation using vector methods. Suppose we are investigating the temper- ature in a body that occupies a region Rin space. (See Fig. 5.) Let Vbe a subregion of Rbounded by the surface S. The law of conservation of energy, applied to V,s a y s net rate of heat in +rate of generation inside =rate of accumulation. Our next job is to quantify this statement. The heat ﬂow rate at any point inside Ris a vector function, q,m e a s u r e di nJ / m2s or similar units. The rate of heat ﬂow through a small piece of the surface Swith area /Delta1Ais approximately ˆn·q/Delta1A(see Fig. 6), where ˆnis the outward unit normal. This quantity is positive for outward ﬂow, so the inﬂow is its negative. The net inﬂow over theentire surface Sis a sum of quantities like this, which becomes, in the limit as /Delta1Ashrinks, the integral /integraldisplay/integraldisplay S−q·ˆndA. The term “rate of generation inside” in the energy balance is intended to in- clude conversion of energy from other forms (chemical, electrical, nuclear) to thermal. We assume that it is speciﬁed as an intensity gmeasured in J/m3so r Figure 5 A solid body occupying a region Rin space and a subregion Vwith boundary S. 5.2 Three-Dimensional Heat Equation 299 Figure 6 The heat ﬂow rate through a small section of surface with area /Delta1Ais q·ˆn/Delta1A. similar units. Then the rate at which heat is generated in a small region of vol- ume Vcentered on point Pis approximately g(P,t)/Delta1V. These contributions are summed over the whole subregion V;a s/Delta1Vshrinks, their total becomes the integral /integraldisplay/integraldisplay/integraldisplay Vg(P,t)dV. The rate at which heat is stored in a small region of volume /Delta1Vcentered on point Pis proportional to the rate at which temperature changes there. That is, the storage rate is ρc/Delta1Vu t(P,t). The storage rate for the whole subregion Vis the sum of such contributions, which passes to the integral /integraldisplay/integraldisplay/integraldisplay Vρc∂u ∂t(P,t)dV. Now the heat balance equation in mathematical terms becomes /integraldisplay/integraldisplay S−q·ˆndA+/integraldisplay/integraldisplay/integraldisplay Vgd V=/integraldisplay/integraldisplay/integraldisplay Vρc∂u ∂tdV. (1) At this point, we call on the divergence theorem, which states that the integral over a surface Sof the outward normal component of a vector function equals the integral over the volume bounded by Sof the divergence of the function. Thus/integraldisplay/integraldisplay Sq·ˆndA=/integraldisplay/integraldisplay/integraldisplay V∇·qdV (2) and we make this replacement in Eq. (1). Next collect all terms on one side of the equation to ﬁnd /integraldisplay/integraldisplay/integraldisplay V/bracketleftbigg −∇ · q+g−ρc∂u ∂t/bracketrightbigg dV=0. (3) 300 Chapter 5 Higher Dimensions and Other Coordinates Because the subregion Vwas arbitrary, we conclude that the integrand must be 0 at every point: −∇ · q+g−ρc∂u ∂t=0i n R,0<t. (4) The argument goes this way. If the integrand were not identically 0, we could ﬁnd some subregion of Rthroughout which it is positive (or negative). The integral over that subregion then would be positive (or negative), contradict- ing Eq. (3), which holds for any subregion. The vector form of Fourier’s law of heat conduction says that the heat ﬂow rate in an isotropic solid (same properties in all directions) is negatively pro-portional to the temperature gradient, q=−κ∇u. (5) Again, the minus sign makes the heat ﬂow “downhill” — from hotter to colder regions. Assuming that the conductivity κis constant, we ﬁnd, on substituting Fourier’s law into Eq. (4), the three-dimensional heat equation, κ∇ 2u+g=ρc∂u ∂tinR,0<t. (6) Of course, we must add an initial condition of the form u(P,0)=f(P)forPinR. (7) I na d d i t i o n ,a te v e r yp o i n to ft h es u r f a c e Bbounding the region R,s o m e boundary condition must be speciﬁed. Commonly we have conditions suchas those that follow, any one of which may be given on Bor some portion of it,B/prime. (1) T emperature speciﬁed, u(P,t)=h1(P,t), for Pany point in B/prime,w h e r e h1is a given function. (2) Heat ﬂow rate speciﬁed. The outward heat ﬂow rate through a small portion of surface surrounding point PonB/primeisq(P,t)·ˆntimes the area. If this is controlled, then by Fourier’s law ∇u·ˆnis controlled. But this dot product is just the directional derivative of uin the outward normal direction at the point P. Thus, this type of boundary condition takes the form ∂u ∂n(P,t)=h2(P,t)forPonB/prime, (8) where h2is a given function. ( 3 )C o n v e c t i o n .I fap a r to ft h es u r f a c ei se x p o s e dt oaﬂ u i da tt e m p e r a t u r e T(P,t), then an accounting of energy passing through a small piece of surface centered at Pleads to the equation q(P,t)·ˆn=h/parenleftbig u(P,t)−T(P,t)/parenrightbig forPonB/prime. 5.2 Three-Dimensional Heat Equation 301 Again using Fourier’s law, we obtain the boundary condition κ∂u ∂n(P,t)+hu(P,t)=hT(P,t)forPonB/prime. (9) As an example, we set up the three-dimensional problem for a solid in the form of a rectangular parallelepiped. In this case, Cartesian coordinates areappropriate, and we may describe the region Rby the three inequalities 0 < x<a,0<y<b,0<z<c. Assuming no generation inside the object, we have the partial differential equation ∂2u ∂x2+∂2u ∂y2+∂2u ∂z2=1 k∂u ∂t,0<x<a,0<y<b,0<z<c,0<t. (10) Suppose that on the faces at x=0a n d a, the temperature is controlled, so the boundary condition there is u(0,y,z,t)=T0,u(a,y,z,t)=T1,0<y<b,0<z<c,0<t. (11) Furthermore, assume that the top and bottom surfaces are insulated. Then theboundary conditions at z=0a n d care ∂u ∂z(x,y,0,t)=0,∂u ∂z(x,y,c,t)=0,0<x<a,0<y<b,0<t. (12) (The outward normal directions on the top and bottom are the positive andnegative z-directions, respectively.) Finally, assume that the faces at y=0a n d aty=bare exposed to a ﬂuid at temperature T 2, so they transfer heat by convection there. The resulting boundary conditions are −κ∂u ∂y(x,0,z,t)+hu(x,0,z,t)=hT2, κ∂u ∂y(x,b,z,t)+hu(x,b,z,t)=hT2, 0<x<a,0<z<c,0<t. (13) Finally, we add an initial condition, u(x,y,z,0)=f(x,y,z), 0<x<a,0<y<b,0<z<c. (14) A full, three-dimensional problem is complicated to solve, so we often look for ways to reduce it to two or even one dimension. In the example problem of Eqs. (10)–(14), we might eliminate zby ﬁnding the temperature averaged over the interval 0 <z<c, v(x,y,t)=1 c/integraldisplayc 0u(x,y,z,t)dz. 302 Chapter 5 Higher Dimensions and Other Coordinates Because differentiation with respect to x,y,o r tgives the same result inside or outside the integral with respect to z, and because of the boundary condi- tion (12), we ﬁnd that 1 c/integraldisplayc 0/parenleftbigg∂2u ∂x2+∂2u ∂y2+∂2u ∂z2/parenrightbigg dz=∂2v ∂x2+∂2v ∂y2, andvsatisﬁes the two-dimensional heat equation, ∂2v ∂x2+∂2v ∂y2=1 k∂v ∂t,0<x<a,0<y<b,0<t. (See the exercises for details and for boundary and initial conditions.) Ifz-variation cannot be ignored, we could try to get rid of the y-variation by introducing an average in that direction, w(x,z,t)=1 b/integraldisplayb 0u(x,y,z,t)dy. From the boundary condition (13) we ﬁnd that /integraldisplayb 0∂2u ∂y2dy=∂u ∂y(x,b,z,t)−∂u ∂y(x,0,z,t) =/parenleftbiggh κ/parenrightbigg/bracketleftbig/parenleftbig T2−u(x,b,z,t)/parenrightbig +/parenleftbig T2−u(x,0,z,t)/parenrightbig/bracketrightbig . Ifbis small — the parallelepiped is more like a plate — we may accept the ap- proximation u(x,b,z,t)+u(x,0,z,t)≡2w(x,z,t),w h i c hw o u l dm a k e ,f r o m the preceding expression, 1 b/integraldisplayb 0∂2u ∂y2dy≡/parenleftbigg2h bκ/parenrightbigg/parenleftbig T2−w(x,z,t)/parenrightbig . After applying the averaging process to Eqs. (10), (11), (12), and (14) we ob- tain the following two-dimensional problem for w: ∂2w ∂x2+∂2w ∂z2+2h bκ(T2−w)=1 k∂w ∂t,0<x<a,0<z<c,0<t, (15) w(0,z,t)=T0,w ( a,z,t)=T1, 0<z<c,0<t,( 16) ∂w ∂z(x,0,t)=0,∂w ∂z(x,c,t)=0, 0<x<a,0<t,( 17) w(x,z,0)=1 b/integraldisplayb 0f(x,y,z)dy, 0<x<a,0<z<c.( 18) 5.3 Two-Dimensional Heat Equation: Solution 303 EXERCISES 1.For the function u(x,y,z,t)that satisﬁes Eqs. (10)–(14), show that /integraldisplayc 0∂2u ∂z2dz=0. 2.Find the initial and boundary conditions satisﬁed by the function v(x,y,t)=1 c/integraldisplayc 0u(x,y,z,t)dz, where usatisﬁes Eqs. (10)–(14). 3.In Eqs. (15)–(18), suppose that w(x,z,t)→W(x,z)ast→∞ .S t a t ea n d solve the boundary value problem for W. (This problem is much easier than it appears, because there is no variation with z.) 4.Find the dimensions of ρ,c,κ,q,a n d g, and verify that the dimensions of the right and left members of the heat equation are the same. 5.Suppose the plate lies in the rectangle 0 <x<a,0<y<b.S t a t eac o m p l e t e initial value–boundary value problem for temperature in the plate if: thereis no heat generation; the temperature is held at T 0along x=aand y=0; the edges at x=0a n d y=bare insulated. 5.3 Two-Dimensional Heat Equation: Double Series Solution In order to see the technique of solution for a two-dimensional problem, we shall consider the diffusion of heat in a rectangular plate of uniform, isotropicmaterial. The steady-state temperature distribution is a solution of the poten- tial equation (see Exercise 6). Suppose that the initial value–boundary value problem for the transient temperature u(x,y,t)is ∂ 2u ∂x2+∂2u ∂y2=1 k∂u ∂t, 0<x<a,0<y<b,0<t,( 1) u(x,0,t)=0,u(x,b,t)=0,0<x<a,0<t,( 2) u(0,y,t)=0,u(a,y,t)=0,0<y<b,0<t,( 3) u(x,y,0)=f(x,y), 0<x<a,0<y<b.( 4) This problem contains a homogeneous partial differential equation and ho- mogeneous boundary conditions. We may thus proceed with separation of 304 Chapter 5 Higher Dimensions and Other Coordinates variables by seeking solutions in the form u(x,y,t)=φ(x,y)T(t). On substituting uin product form into Eq. (1), we ﬁnd that it becomes /parenleftbigg∂2φ ∂x2+∂2φ ∂y2/parenrightbigg T=1 kφT/prime. Separation can be achieved by dividing through by φT,w h i c hl e a v e s /parenleftbigg∂2φ ∂x2+∂2φ ∂y2/parenrightbigg1 φ=T/prime kT. We may argue, as usual, that the common value of the members of this equa- tion must be a constant, which we expect to be negative (−λ2).T h ee q u a t i o n s that result are T/prime+λ2kT=0, 0<t, (5) ∂2φ ∂x2+∂2φ ∂y2=−λ2φ, 0<x<a,0<y<b. (6) In terms of the product solutions, the boundary conditions become φ(x,0)T(t)=0,φ ( x,b)T(t)=0, φ(0,y)T(t)=0,φ ( a,y)T(t)=0. In order to satisfy all four equations, either T(t)≡0 for all torφ=0o nt h e boundary. We have seen many times that the choice of T(t)≡0w i p e so u to u r solution completely. Therefore, we require that φsatisfy the conditions φ(x,0)=0,φ ( x,b)=0,0<x<a, (7) φ(0,y)=0,φ ( a,y)=0,0<y<b. (8) We are not yet out of difﬁculty, because Eqs. (6)–(8) constitute a new prob- lem, a two-dimensional eigenvalue problem. It is evident, however, that thepartial differential equation and the boundary conditions are linear and ho-mogeneous; thus separation of variables may work again. Supposing that φ has the form φ(x,y)=X(x)Y(y), we ﬁnd that the partial differential equation (6) becomes X /prime/prime(x) X(x)+Y/prime/prime(y) Y(y)=−λ2,0<x<a,0<y<b. 5.3 Two-Dimensional Heat Equation: Solution 305 The sum of a function of xand a function of ycan be constant only if those two functions are individually constant: X/prime/prime X=constant ,Y/prime/prime Y=constant . Before naming the constants, let us look at the boundary conditions on φ=XY: X(x)Y(0)=0,X(x)Y(b)=0,0<x<a, X(0)Y(y)=0,X(a)Y(y)=0,0<y<b. If either of the functions XorYis zero throughout the whole interval of its variable, the conditions are certainly satisﬁed, but φis identically zero. We therefore require each of the functions Xand Yto be zero at the endpoints of its interval: Y(0)=0,Y(b)=0, (9) X(0)=0,X(a)=0. (10) Now it is clear that each of the ratios X/prime/prime/Xand Y/prime/prime/Yshould be a negative constant, designated by −µ2and−ν2, respectively. The separate equations for Xand Yare X/prime/prime+µ2X=0,0<x<a, (11) Y/prime/prime+ν2Y=0,0<y<b. (12) Finally, the original separation constant −λ2is determined by λ2=µ2+v2. (13) Now we see two independent eigenvalue problems: Eqs. (9) and (12) form one problem and Eqs. (10) and (11) the other. Each is of a very familiar form;the solutions are X m(x)=sin/parenleftbiggmπx a/parenrightbigg ,µ2 m=/parenleftbiggmπ a/parenrightbigg2 ,m=1,2,..., Yn(y)=sin/parenleftbiggnπy b/parenrightbigg ,ν2 n=/parenleftbiggnπ b/parenrightbigg2 , n=1,2,.... Notice that the indices nand mare independent. This means that φwill have a double index. Speciﬁcally, the solutions of the two-dimensional eigenvalueproblem Eqs. (6)–(8) are φ mn(x,y)=Xm(x)Yn(y), λ2 mn=µ2 m+ν2 n, 306 Chapter 5 Higher Dimensions and Other Coordinates and the corresponding function Tis Tmn=exp/parenleftbig −λ2 mnkt/parenrightbig . We now begin to assemble the solution. For each pair of indices m,n(m= 1,2,3,..., n=1,2,3,...) there is a function umn(x,y,t)=φmn(x,y)Tmn(t) =sin/parenleftbiggmπx a/parenrightbigg sin/parenleftbiggnπy b/parenrightbigg exp/parenleftbig −λ2 mnkt/parenrightbig that satisﬁes the partial differential equation (1) and the boundary conditions Eqs. (2) and (3). We may form linear combinations of these solutions to get other solutions. The most general linear combination would be the doubleseries u(x,y,t)=∞/summationdisplay m=1∞/summationdisplay n=1amnφmn(x,y)Tmn(t), (14) and any such combination should satisfy Eqs. (1)–(3). There remains the ini- tial condition Eq. (4) to be satisﬁed. If uhas the form given in Eq. (14), then the initial condition becomes ∞/summationdisplay m=1∞/summationdisplay n=1amnφmn(x,y)=f(x,y), 0<x<a,0<y<b. (15) The idea of orthogonality is once again applicable to the problem of selecting the coefﬁcients amn. One can show by direct computation that /integraldisplayb 0/integraldisplaya 0φmn(x,y)φpq(x,y)dx dy=  ab 4ifm=pand n=q, 0,otherwise .(16) Thus, the appropriate formula for the coefﬁcients amnis amn=4 ab/integraldisplayb 0/integraldisplaya 0f(x,y)sin/parenleftbiggmπx a/parenrightbigg sin/parenleftbiggnπy b/parenrightbigg dx dy. (17) Iffis a sufﬁciently regular function, the series in Eq. (15) will converge and equal f(x,y)in the rectangular region 0 <x<a,0<y<b.W em a yt h e ns a y that the problem is solved. It is reassuring to notice that each term in the seriesof Eq. (14) contains a decaying exponential, and thus, as tincreases, u(x,y,t) tends to zero, as expected. Example. Let us take the speciﬁc initial condition f(x,y)=xy,0<x<a,0<y<b. 5.3 Two-Dimensional Heat Equation: Solution 307 T h ec o e f ﬁ c i e n t sa r ee a s i l yf o u n dt ob e amn=4ab π2cos(mπ)cos(nπ) mn=4ab π2(−1)m+n mn, so the solution to this problem is u(x,y,t)=4ab π2∞/summationdisplay m=1∞/summationdisplay n=1(−1)m+n mnsin/parenleftbiggmπx a/parenrightbigg sin/parenleftbiggnπy b/parenrightbigg exp/parenleftbig −λ2 mnkt/parenrightbig . (18) This solution is shown animated on the CD. /square T h ed o u b l es e r i e st h a ta p p e a rh e r ea r eb e s th a n d l e db yc o n v e r t i n gt h e mi n t o single series. T o do this, arrange the terms in order of increasing values of λ2 mn. Then the ﬁrst terms in the single series are the most signiﬁcant, those thatdecay least rapidly. For example, if a=2b,s ot h a t λ 2 mn=(m2+4n2)π2 a2, then the following list gives the double index (m,n)in order of increasing values of λ2 mn: (1,1),(2,1),(3,1),(1,2),(2,2),(4,1),(3,2),.... EXERCISES 1.Write out the “ﬁrst few” terms of the series of Eq. (18). By “ﬁrst few,” we mean those for which λ2 mnis smallest. (Assume a=bin determining relative magnitudes of the λ2.) 2.Provide the details of the separation of variables by which Eqs. (9)–(13) are derived. 3.Find the frequencies of vibration of a rectangular membrane. See Sec- tion 5.1, Exercise 1. 4.Ver ify that umn(x,y,t)satisﬁes Eqs. (1)–(3). 5.Show that Xm(x)=cos(mπx/a)(m=0,1,2,...) if the boundary condi- tions Eq. (3) are replaced by ∂u ∂x(0,y,t)=0,∂u ∂x(a,y,t)=0,0<y<b,0<t. What values will the λ2 mnhave, and of what form will the solution u(x,y,t) be? 308 Chapter 5 Higher Dimensions and Other Coordinates 6.S u p p o s et h a t ,i n s t e a do fb o u n d a r yc o n d i t i o n sE q s .( 2 )a n d( 3 ) ,w eh a v e u(x,0,t)=f1(x), u(x,b,t)=f2(x), 0<x<a,0<t,( 2/prime) u(0,y,t)=g1(y),u(a,y,t)=g2(y),0<y<b,0<t.( 3/prime) Show that the steady-state solution involves the potential equation, and indicate how to solve it. 7.Solve the two-dimensional heat conduction problem in a rectangle if thereis insulation on all boundaries and the initial condition is a.u(x,y,0)=1; b.u(x,y,0)=x+y; c.u(x,y,0)=xy. 8.Verify the orthogonality relation in Eq. (16) and the formula for amn. 9.Show that the separation constant −λ2must be negative by showing that −µ2and−ν2must both be negative. 10.Show that the function umn(x,y,t)=sin(µmx)sin(νny)cos(λmnct), where µm,νn,a n d λmna r ea si nt h i ss e c t i o n ,i sas o l u t i o no ft h et w o - dimensional wave equation on the rectangle 0 <x<a,0<y<b, with u=0 on the boundary. The function umay be thought of as the displace- ment of a rectangular membrane (see Section 5.1). 11.The places where umn(x,y,t)=0 for all tare called nodal lines .D e s c r i b e the nodal lines for (m,n)=(1,2),(2,3),(3,2),(3,3). 12.Determine the frequencies of vibration for the functions umnof Exer- cise 10. Are there different pairs (m,n)that have the same frequency if a=b? 5.4 Problems in Polar Coordinates We found that the one-dimensional wave and heat problems have a great deal in common. Namely, the steady-state or time-independent solutions and theeigenvalue problems that arise are identical in both cases. Also, in solvingproblems in a rectangular region, we have seen that those same features areshared by the heat and wave equations. 5.4 Problems in Polar Coordinates 309 If we consider now the vibrations of a circular membrane or heat conduc- tion in a circular plate, we shall see common features again. In what follows,these two problems are given side by side for the region 0 <r<a,0<t. Wave Heat ∇ 2v=1 c2∂2v ∂t2∇2v=1 k∂v ∂t v(a,θ,t)=f(θ) v( a,θ,t)=f(θ) v(r,θ,0)=g(r,θ) v( r,θ,0)=g(r,θ) ∂v ∂t(r,θ,0)=h(r,θ) In both problems we require that vbe periodic in θwith period 2 π: v(r,θ,t)=v(r,θ+2π,t), as in Section 4.5. Although the interpretation of the function vis different in the two cases, we see that the solution of the problem ∇2v=0,v ( a,θ)=f(θ) is the rest-state or steady-state solution for both problems, and it will be needed in both problems to make the boundary condition at r=ahomo- geneous. Let us suppose that the time-independent solution has been foundand subtracted; that is, we will replace f(θ)by zero. Then we have ∇ 2v=1 c2∂2v ∂t2∇2v=1 k∂v ∂t v(a,θ,t)=0 v(a,θ,t)=0 plus the appropriate initial conditions. If we attempt to solve by separation of variables, setting v(r,θ,t)=φ(r,θ)T(t),i nb o t hc a s e sw ew i l lﬁ n dt h a t φ(r,θ) must satisfy 1 r∂ ∂r/parenleftbigg r∂φ ∂r/parenrightbigg +1 r2∂2φ ∂θ2=−λ2φ, (1) φ(a,θ)=0, (2) φ(r,θ+2π)=φ(r,π), (3) φbounded as r→0. (4) Now we shall concentrate on the solution of this two-dimensional eigen- value problem. We can separate variables again by assuming that φ(r,θ)= 310 Chapter 5 Higher Dimensions and Other Coordinates R(r)Q(θ). After some algebra, we ﬁnd that (rR/prime)/prime rR+Q/prime/prime r2Q=−λ2, (5) R(a)=0, (6) Q(θ)=Q(θ+2π), (7) R(r)bounded as r→0. (8) The ratio Q/prime/prime/Qmust be constant; otherwise, λ2could not be constant. Choos- ingQ/prime/prime/Q=−µ2, we get a familiar, singular eigenvalue problem: Q/prime/prime+µ2Q=0, (9) Q(θ+2π)=Q(θ). (10) We found (in Chapter 4) that the solutions of this problem are µ2 0=0, Q0(θ)=1, µ2 m=m2,Qm(θ)=cos(mθ) and sin (mθ),(11) where m=1,2,3,.... There remains a problem in R: /parenleftbig rR/prime/parenrightbig/prime−µ2 rR+λ2rR=0,0<r<a, (12) R(a)=0, (13) R(r)bounded as r→0. (14) Equation (12) is called Bessel’s equation , and we shall solve it in the next sec- tion. EXERCISES 1.State the full initial value–boundary value problems that result from theproblems as originally given when the steady-state or time-independent so-lution is subtracted from v. 2.Verify the separation of variables that leads to Eqs. (1) and (2). 3.Substitution of v(r,θ,t)in the form of a product led to the problem of Eqs. (1)–(4) for the factor φ(r,θ). What differential equation is to be satis- ﬁed by the factor T(t)? 4.Solve Eqs. (9)–(11) and check the solutions given. 5.5 Bessel’s Equation 311 5.Suppose the problems originally stated were to be solved in the half-disk 0<r<a,0<θ<π , with additional conditions: v(r,0,t)=0,0<r<a,0<t, v(r,π,t)=0,0<r<a,0<t. What eigenvalue problem arises in place of Eqs. (9)–(11)? Solve it. 6.Suppose that the boundary condition ∂v ∂r(a,θ,t)=0,−π<θ ≤π, 0<t were given instead of v(a,θ,t)=f(θ). Carry out the steps involved in sepa- ration of variables. Show that the only change is in Eqs. (6) and (13), which become R/prime(a)=0. 7.One of the consequences of Green’s theorem is the integral relation /integraldisplay/integraldisplay R/parenleftbig f∇2g−g∇2f/parenrightbig dA=/integraldisplay C/parenleftbigg f∂g ∂n−g∂f ∂n/parenrightbigg ds, where Ris a region in the plane, Cis the closed curve that bounds R,a n d ∂f/∂nis the directional derivative in the direction normal to the curve C. Use this relation to show that eigenfunctions of the problem ∇2φ=−λ2φinR, φ=0o n C are orthogonal if they correspond to different eigenvalues. (Hint: Use f=φk,g=φm,m/negationslash=k.) 8.Same problem as Exercise 7, except the boundary condition is φ+λ∂φ ∂n=0o n C. 5.5 Bessel’s Equation In order to solve the Bessel equation, /parenleftbig rR/prime/parenrightbig/prime−µ2 rR+λ2rR=0, (1) we apply the method of Frobenius. Assume that R(r)has the form of a power series multiplied by an unknown power of r: R(r)=rα/parenleftbig c0+c1r+···+ ckrk+···/parenrightbig . (2) 312 Chapter 5 Higher Dimensions and Other Coordinates When the differentiations in Eq. (1) are carried out and the equation is multi- plied by r,i tb e c o m e s r2R/prime/prime=α(α−1)c0rα+(α+1)αc1rα+1+(α+2)(α+1)c2rα+2+··· +(α+k)(α+k−1)ckrα+k+··· rR/prime= αc0rα+(α+1)c1rα+1+(α+2)c2rα+2+··· +(α+k)ckrα+k+··· −µ2R=− µ2c0rα−µ2c1rα+1−µ2c2rα+2−··· −µ2ckrα+k+··· λ2r2R= λ2c0rα+2+··· +λ2ck−2rα+k+··· The expression for λ2r2Ris jogged to the right to make like powers of rline up vertically. Note that the lowest power of rpresent in λ2r2Risrα+2. Now we add the tableau vertically. The sum of the left-hand sides is, accord- ing to the differential equation, equal to zero. Therefore 0=c0/parenleftbig α2−µ2/parenrightbig rα+c1/bracketleftbig (α+1)2−µ2/bracketrightbig rα+1 +/bracketleftbig c2/parenleftbig (α+2)2−µ2/parenrightbig +λ2c0/bracketrightbig rα+2 +···+/bracketleftbig ck/parenleftbig (α+k)2−µ2/parenrightbig +λ2ck−2/bracketrightbig rα+k+···. Each term in this power series must be zero in order for the equality to hold. Therefore, the coefﬁcient of each term must be zero: c0/parenleftbig α2−µ2/parenrightbig =0, c1/parenleftbig (α+1)2−µ2/parenrightbig =0, ... ck/parenleftbig (α+k)2−µ2/parenrightbig +λ2ck−2=0,k≥2. As a bookkeeping agreement, we take c0/negationslash=0. Thus α=±µ. Let us study the caseα=µ≥0. The second equation becomes c1/parenleftbig (µ+1)2−µ2/parenrightbig =0 and this implies c1=0. Now, in general the relation ck=−λ2ck−2 (µ+k)2−µ2=−λ2ck−2 k(2µ+k),k≥2, (3) says that ckcan be found from ck−2.I np a r t i c u l a r ,w eﬁ n d c2=−λ2 2(2µ+2)c0, c4=−λ2 4(2µ+4)c2=λ4 2·4·(2µ+2)(2µ+4)c0, 5.5 Bessel’s Equation 313 and so forth. All c’s with odd index are zero, since they are all multiples of c1. The general formula for a coefﬁcient with even index k=2mis c2m=(−1)m m!(µ+1)(µ+2)···(µ+m)/parenleftbiggλ 2/parenrightbigg2m c0. (4) For integral values of µ,c0is chosen by convention to be c0=/parenleftbiggλ 2/parenrightbiggµ ·1 µ!. Then the solution of Eq. (1) that we have found is called the Bessel function of the ﬁrst kind of order µ: Jµ(λr)=/parenleftbiggλr 2/parenrightbiggµ∞/summationdisplay m=0(−1)m m!(µ+m)!/parenleftbiggλr 2/parenrightbigg2m . (5) This series serves us for evaluating the function and for obtaining its proper- ties. (See the Exercises.) However, from now on, we consider the Bessel functions of the ﬁrst kind to be as well known as sines and cosines , although less familiar. There must be a second independent solution of Bessel’s equation, which can be found by using variation of parameters. This method yields a solutionin the form J µ(λr)·/integraldisplaydr rJ2µ(λr). (6) In its standard form, the second solution of Bessel’s equation is called the Bessel function of second kind of order µand is denoted by Yµ(λr). The most important feature of the second solution is its behavior near r=0. When ris very small, we can approximate Jµ(λr)by the ﬁrst term of its series expansion: Jµ(λr)∼=/parenleftbiggλ 2/parenrightbiggµ1 µ!rµ,r/lessmuch1. The solution Eq. (6) then can be approximated by constant ×rµ/integraldisplaydr r1+2µ=constant ×/braceleftbigg ln(r),ifµ=0, r−µ, ifµ> 0. In either case, it is easy to see that /vextendsingle/vextendsingleYµ(λr)/vextendsingle/vextendsingle→∞ asr→0. Both kinds of Bessel functions have an inﬁnite number of zeros. That is, t h e r ei sa ni n ﬁ n i t en u m b e ro fv a l u e so f α(andβ) for which Jµ(α)=0,Yµ(β)=0. 314 Chapter 5 Higher Dimensions and Other Coordinates Figure 7 Graphs of Bessel functions of the ﬁrst kind. Also see the CD. (a) J0and J1,( b ) Y0and Y1. n m 12 3 4 0 2.405 5.520 8.654 11.792 1 3.832 7.016 10.173 13.324 2 5.136 8.417 11.620 14.796 3 6.380 9.761 13.015 16.223 Table 1 Zeros of Bessel functions. The values αmn satisfy the equation Jm(αmn)=0. Also, as r→∞ ,b o t h Jµ(λr)and Yµ(λr)tend to zero. Figure 7 gives graphs of several Bessel functions, and Table 1 provides values of their zeros. Further information can be found in most books of tables. The modiﬁed Bessel equation differs from the Bessel equation only in the sign of one term. It is /parenleftbig rR/prime/parenrightbig/prime−µ2 rR−λ2rR=0. (7) 5.5 Bessel’s Equation 315 Using the same method as in the preceding, an inﬁnite series can be developed for the solutions (see Exercise 8). The solution that is bounded at r=0, in standard form, is called the modiﬁed Bessel function of the ﬁrst kind of order µ, designated Iµ(λr), and its series is Iµ(λr)=/parenleftbiggλr 2/parenrightbiggµ∞/summationdisplay m=01 m!(µ+m)!/parenleftbiggλr 2/parenrightbigg2m . Summary The differential equation d dr/parenleftbigg rdR dr/parenrightbigg −µ2 rR+λ2rR=0 is called Bessel’s equation. Its general solution is R(r)=AJµ(λr)+BYµ(λr) (Aand Bare arbitrary constants). The functions Jµand Yµare called Bessel functions of order µof the ﬁrst and second kinds, respectively. The Bessel function of the second kind is unbounded at the origin. EXERCISES 1.Find the values of the parameter λfor which the following problem has a nonzero solution: 1 rd dr/parenleftbigg rdφ dr/parenrightbigg +λ2φ=0,0<r<a, φ(a)=0,φ ( 0)bounded . 2.Sketch the ﬁrst few eigenfunctions found in Exercise 1. 3.Show that d drJµ(λr)=λJ/prime µ(λr), where the prime denotes differentiation with respect to the argument. 4.Show from the series that d drJ0(λr)=−λJ1(λr). 316 Chapter 5 Higher Dimensions and Other Coordinates 5.By using Exercise 4 and Rolle’s theorem, and knowing that J0(x)=0 for an inﬁnite number of values of x, show that J1(x)=0 has an inﬁnite number of solutions. 6.Using the inﬁnite series representations for the Bessel functions, verify theformulas d dx/parenleftbig x−µJµ(x)/parenrightbig =− x−µJµ+1(x), d dx/parenleftbig xµJµ(x)/parenrightbig =xµJµ−1(x). 7.Use the second formula in Exercise 6 to derive the integral formula /integraldisplay xµJµ(x)xd x=xµ+1Jµ+1(x). 8.Use the method of Frobenius to obtain a solution of the modiﬁed Bessel equation (7). Show that the coefﬁcients of the power series are just thesame as those for the Bessel function of the ﬁrst kind, except for signs. 9.Use the modiﬁed Bessel function to solve this problem for the temperaturein a circular plate when the surface is exposed to convection: 1 rd dr/parenleftbigg rdu dr/parenrightbigg −γ2(u−T)=0,0<r<a, u(a)=T1. 10.Using the result of Exercise 4, solve the eigenvalue problem 1 rd dr/parenleftbigg rdφ dr/parenrightbigg +λ2φ=0,0<r<a, dφ dr(a)=0,φ ( 0)bounded . 5.6 Temperature in a Cylinder In Section 5.4, we observed that both the heat and wave equations have a great deal in common, especially the equilibrium solution and the eigenvalue prob- lem. T o reinforce that observation, we will solve a heat problem and a wave problem with analogous conditions so that their similarities may be seen.These examples illustrate another important point: Problems that would betwo-dimensional in one coordinate system (rectangular) may become one-dimensional in another system (polar). In order to obtain this simpliﬁcation,we will assume that the unknown function, v(r,θ,t), is actually independent 5.6 Temperature in a Cylinder 317 of the angular coordinate θ.( W ew r i t e v(r,t)then.) As a consequence of this assumption, the two-dimensional Laplacian operator becomes ∇2v=1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg . Suppose that the temperature v(r,t)in a large cylinder (radius a)s a t i s ﬁ e s the problem 1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg =1 k∂v ∂t,0<r<a,0<t, (1) v(a,t)=0, 0<t, (2) v(r,0)=f(r), 0<r<a. (3) Because the differential equation (1) and boundary condition (2) are ho- mogeneous, we may start the separation of variables by assuming v(r,t)= φ(r)T(t). Using this form for v, we ﬁnd that the partial differential equa- tion (1) becomes 1 r(rφ/prime)/primeT=1 kφT/prime. After dividing through this equation by φT, we arrive at the equality (rφ/prime(r))/prime rφ(r)=T/prime(t) kT(t). (4) The two members of this equation must both be constant; call their mutual value−λ2. Then we have two linked ordinary differential equations, T/prime+λ2kT=0,0<t, (5) (rφ/prime)/prime+λ2rφ=0,0<r<a. (6) The boundary condition, Eq. (2), becomes φ(a)T(t)=0, 0<t.I tw i l lb es a t - isﬁed by requiring that φ(a)=0. (7) We can recognize Eq. (6) as Bessel’s equation with µ=0. (See Summary, Section 5.5.) The general solution, therefore, has the form φ(r)=AJ0(λr)+BY0(λr). IfB/negationslash=0,φ(r)must become inﬁnite as rapproaches zero. The physical impli- cations of this possibility are unacceptable, so we require that B=0. In effect we have added the boundedness condition /vextendsingle/vextendsinglev(r,t)/vextendsingle/vextendsinglebounded at r=0, (8) which we shall employ frequently. 318 Chapter 5 Higher Dimensions and Other Coordinates The function φ(r)=J0(λr)is a solution of Eq. (6), and we wish to choose λ so that Eq. (7) is satisﬁed. Then we must have J0(λa)=0 or λn=αn a,n=1,2,..., where αnare the zeros of the function J0. Thus the eigenfunctions and eigen- values of Eqs. (6), (7), and (8) are φn(r)=J0(λnr), λ2 n=/parenleftbiggαn a/parenrightbigg2 . (9) T h e s ea r es h o w no nt h eC D . Returning to Eq. (5), we determine that the time factors Tnare Tn(t)=exp/parenleftbig −λ2 nkt/parenrightbig . We may now assemble the general solution of the partial differential equa- tion (1), under the boundary condition (2) and boundedness condition (8),as a general linear combination of our product solutions: v(r,t)=∞/summationdisplay n=1anJ0(λnr)exp/parenleftbig −λ2 nkt/parenrightbig . (10) It remains to determine the coefﬁcients anso as to satisfy the initial condi- tion (3), which now takes the form v(r,0)=∞/summationdisplay n=1anJ0(λnr)=f(r), 0<r<a. (11) While this problem is not a routine exercise in Fourier series or even a reg- ular Sturm–Liouville problem (see Secti on 2.7, especially Exercise 6 there), it is nevertheless true that the eigenfunctions of Eqs. (6) and (7) are orthogonal, as expressed by the relation /integraldisplaya 0φn(r)φm(r)rd r=0(n/negationslash=m) or/integraldisplaya 0J0(λnr)J0(λmr)rd r=0(n/negationslash=m). More importantly, the following theorem gives us justiﬁcation for Eq. (11). 5.6 Temperature in a Cylinder 319 Theorem. If f(r)is sectionally smooth on the interval 0<r<a, then at every point r on that interval, ∞/summationdisplay n=1anJ0(λnr)=f(r+)+f(r−) 2,0<r<a, where the λnare solutions of J 0(λa)=0and an=/integraltexta 0f(r)J0(λnr)rd r/integraltexta 0J2 0(λnr)rd r. (12) /square The CD shows an animation of a Bessel series converging. Now we may proceed with the problem at hand. If the function f(r)in the initial condition (3) is sectionally smooth, the use of Eq. (12) to chose thecoefﬁcients a nguarantees that Eq. (11) is satisﬁed (as nearly as possible), and hence the function v(r,t)=∞/summationdisplay n=1anJ0(λnr)exp/parenleftbig −λ2 nkt/parenrightbig (13) satisﬁes the problem expressed by Eqs. (1), (2), (3), and (8). By way of example, let us suppose that the function f(r)=T0,0<r<a.I t is necessary to determine the coefﬁcients anby formula (12). The numerator is the integral /integraldisplaya 0T0J0(λnr)rd r. This integral is evaluated by means of the relation (see Exercise 6 of Sec- tion 5.5) d dx/parenleftbig xJ1(x)/parenrightbig =xJ0(x). (14) Hence, we ﬁnd /integraldisplaya 0J0(λnr)rd r=1 λnrJ1(λnr)/vextendsingle/vextendsingle/vextendsingle/vextendsinglea 0 =a λnJ1(λna)=a2 αnJ1(αn). (15) The denominator of Eq. (12) is known to have the value (Exercise 5) /integraldisplaya 0J2 0(λnr)rd r=a2 2J2 1(λna) =a2 2J2 1(αn). (16) 320 Chapter 5 Higher Dimensions and Other Coordinates n αn J1(αn)2 αnJ1(αn) 1 2.405 +0.5191 +1.6020 2 5.520 −0.3403 −1.0647 3 8.654 +0.2715 +0.8512 4 11.792 −0.2325 −0.7295 Table 2 Values for Eq. (18) Figure 8 Graphs of the solution of the example problem. The function v(r,t)as g i v e ni nE q .( 1 8 )i ss h o w nv s rfor times chosen so that kt/a2takes the values 0, 0.01, 0.1, and 0 .5.T0=100 and a=1. Putting together the numerator and denominator from Eqs. (15) and (16), we ﬁnd that the coefﬁcients we need are an=2T0 αnJ1(αn). (17) Thus, the solution to the heat conduction problem is v(r,t)=T0∞/summationdisplay n=12 αnJ1(αn)J0(λnr)exp/parenleftbig −λ2 nkt/parenrightbig . (18) I nT a b l e2a r el i s t e dt h eﬁ r s tf e wv a l u e so ft h er a t i o2 /[αnJ1(αn)]. Figure 8 shows graphs of v(r,t)as a function of rfor several times. Also, see Exercise 1. An animation is shown on the CD. 5.7 Vibrations of a Circular Membrane 321 EXERCISES 1.Use Eq. (18) to ﬁnd an expression for the function v(0,t)/T0.E v a l u a t et h e function for kt a2=0.1,0.2,0.3. (The ﬁrst two terms of the series are sufﬁcient.) 2.Write out the ﬁrst three terms of the series in Eq. (18). 3.Solve the heat problem consisting of Eqs. (1)–(3) if f(r)is f(r)=  T0,0<r<a 2, 0,a 2<r<a. 4.Letφ(r)=J0(λr)so that φ(r)satisﬁes Bessel’s equation of order 0. Multiply through the differential equation by rφ/primeand conclude that d dr/bracketleftbig (rφ/prime)2/bracketrightbig +λ2r2d dr/bracketleftbig φ2/bracketrightbig =0. 5.Assuming that λis chosen so that φ(a)=0, integrate the equation in Exer- cise 4 over the interval 0 <r<ato ﬁnd /integraldisplaya 0φ2(r)rd r=1 2λ2/parenleftbig aφ/prime(a)/parenrightbig2. 6.Use Exercise 5 to validate Eq. (16). 5.7 Vibrations of a Circular Membrane We shall now attempt to solve the problem of describing the displacement of a circular membrane that is ﬁxed at its edge. Symmetric Vibrations T o begin with, we treat the simple case in which the initial conditions are in-dependent of θ. Thus the displacement v(r,t)satisﬁes the problem 1 r∂ ∂r/parenleftbigg r∂v ∂r/parenrightbigg =1 c2∂2v ∂t2,0<r<a,0<t, (1) v(a,t)=0, 0<t, (2) 322 Chapter 5 Higher Dimensions and Other Coordinates v(r,0)=f(r), 0<r<a, (3) ∂v ∂t(r,0)=g(r), 0<r<a. (4) We start immediately with separation of variables, assuming v(r,t)= φ(r)T(t). The differential equation (1) becomes 1 r(rφ/prime)/primeT=1 c2φT/prime/prime, and the variables may be separated by dividing by φT.T h e nw eﬁ n d (rφ/prime(r))/prime rφ(r)=T/prime/prime(t) c2T(t). The two sides must both be equal to a constant (say, −λ2), yielding two linked, ordinary differential equations T/prime/prime+λ2c2T=0,0<t, (5) (rφ/prime)/prime+λ2rφ=0,0<r<a. (6) The boundary condition Eq. (2) is satisﬁed if φ(a)=0. (7) Of course, because r=0 is a singular point of the differential equation (6), we add the requirement /vextendsingle/vextendsingleφ(r)/vextendsingle/vextendsinglebounded at r=0, (8) which is equivalent to requiring that |v(r,t)|be bounded at r=0. We recognize that Eq. (6) is Bessel’s equation, of which the function φ(r)= J0(λr)is the solution bounded at r=0. In order to satisfy the boundary con- d i t i o nE q .( 7 ) ,w em u s th a v e J0(λa)=0, which implies that λn=αn a,n=1,2,..., (9) where αnare the zeros of the function J0. Thus the eigenfunctions and eigen- values of Eqs. (6)–(8) are φn(r)=J0(λnr), λ2 n=/parenleftbiggαn a/parenrightbigg2 . 5.7 Vibrations of a Circular Membrane 323 The rest of our problem can now be dispatched easily. Returning to Eq. (5), we see that Tn(t)=ancos(λnct)+bnsin(λnct), and then for each n=1,2,... w eh a v eas o l u t i o no fE q s .( 1 ) ,( 2 ) ,a n d( 8 ) : vn(r,t)=φn(r)Tn(t). The most general linear combination of the vnwould be v(r,t)=∞/summationdisplay n=1J0(λnr)/bracketleftbig ancos(λnct)+bnsin(λnct)/bracketrightbig . (10) The initial conditions Eqs. (3) and (4) are satisﬁed if v(r,0)=∞/summationdisplay n=1anJ0(λnr)=f(r), 0<r<a, ∂v ∂t(r,0)=∞/summationdisplay n=1bnλncJ0(λnr)=g(r),0<r<a. As in the preceding section, the coefﬁcients of these series are to be found through the integral formulas an=1 Dn/integraldisplaya 0f(r)J0(λnr)rd r,bn=1 λncDn/integraldisplaya 0g(r)J0(λnr)rd r, Dn=/integraldisplaya 0/bracketleftbig J0(λnr)/bracketrightbig2rd r. With the coefﬁcients determined by these formulas, the function given in Eq. (10) is the solution to the vibrating membrane problem that we startedwith. General Vibrations Having seen the simplest case of the vibrations of a circular membrane, wereturn to the more general case. The full problem was 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg +1 r2∂2u ∂θ2=1 c2∂2u ∂t2,0<r<a,0<t. (11) u(a,θ,t)=0, 0<t, (12) /vextendsingle/vextendsingleu(0,θ,t)/vextendsingle/vextendsinglebounded , 0<t, (13) u(r,θ+2π,t)=u(r,θ,t), 0<r<a,0<t, (14) 324 Chapter 5 Higher Dimensions and Other Coordinates u(r,θ,0)=f(r,θ), 0<r<a, (15) ∂u ∂t(r,θ,0)=g(r,θ), 0<r<a. (16) Following the procedure suggested in Section 5.4, we assume that uhas the product form u=φ(r,θ)T(t) and we ﬁnd that Eq. (11) separates into two linked equations: T/prime/prime+λ2c2T=0, 0<t,( 17) 1 r∂ ∂r/parenleftbigg r∂φ ∂r/parenrightbigg +1 r2∂2φ ∂θ2=−λ2φ, 0<r<a.( 18) If we separate variables of the function φby assuming φ(r,θ)=R(r)Q(θ), Eq. (18) takes the form 1 r/parenleftbig rR/prime/parenrightbig/primeQ+1 r2RQ/prime/prime=−λ2RQ. The variables will separate if we multiply by r2and divide by RQ. Then the preceding equation may be put in the form r(rR/prime)/prime R+λ2r2=−Q/prime/prime Q=µ2. Finally we obtain two problems for Rand Q: Q/prime/prime+µ2Q=0,−π<θ ≤π, (19a) Q(θ+2π)=Q(θ), (19b) /parenleftbig rR/prime/parenrightbig/prime−µ2 rR+λ2rR=0,0<r<a, (20) /vextendsingle/vextendsingleR(0)/vextendsingle/vextendsinglebounded , R(a)=0. As we observed before, the problem (19) has the solutions µ2 0=0, Q0=1, µ2 m=m2,Qm=cos(mθ) and sin (mθ), m=1,2,3,.... Also, the differential equation (20) will be recognized as Bessel’s equation, the general solution of which is (using µ=m) R(r)=CJm(λr)+DY m(λr). 5.7 Vibrations of a Circular Membrane 325 In order for the boundedness condition in Eq. (20) to be fulﬁlled, Dmust be zero. Then we are left with R(r)=Jm(λr). (Because any multiple of a solution is another solution, we can drop the con- stant C.) The boundary condition of Eq. (20) becomes R(a)=Jm(λa)=0, implying that λamust be a root of the equation Jm(α)=0. (See Table 1.) For each ﬁxed integer m,αm1,αm2,αm3,... are the ﬁrst, second, third, . . . solutions of the preceding equation. The values of λfor which Jm(λr) solves the differential equation and satisﬁes the boundary condition are λmn=αmn a,m=0,1,2,..., n=1,2,3,.... Now that the functions Rand Qare determined, we can construct φ.F o r m=1,2,3,... and n=1,2,3,..., both of the functions Jm(λmnr)cos(mθ), Jm(λmnr)sin(mθ) (21) are solutions of the problem Eq. (18), both corresponding to the same eigen- valueλ2 mn.F o r m=0a n d n=1,2,3,...,w eh a v et h ef u n c t i o n s J0(λ0nr), (22) which correspond to the eigenvalues λ2 0n. (Compare with the simple case.) The function T(t)that is a solution of Eq. (17) is any combination of cos (λmnct) and sin (λmnct). Now the solutions of Eqs. (11)–(14) have any of the forms Jm(λmnr)cos(mθ)cos(λmnct), Jm(λmnr)sin(mθ)cos(λmnct),(23)Jm(λmnr)cos(mθ)sin(λmnct), Jm(λmnr)sin(mθ)sin(λmnct) form=1,2,3,... and n=1,2,3,.... In addition, there is the special case m=0, for which solutions have the form J0(λ0nr)cos(λ0nct), J0(λ0nr)sin(λ0nct). (24) The CD shows a few of these “standing waves” animated. The general solution of the problem Eqs. (11)–(14) will thus have the form of a linear combination of the solutions in Eqs. (23) and (24). We shall use 326 Chapter 5 Higher Dimensions and Other Coordinates several series to form the combination: u(r,θ,t)=/summationdisplay na0nJ0(λ0nr)cos(λ0nct) +/summationdisplay m,namnJm(λmnr)cos(mθ)cos(λmnct) +/summationdisplay m,nbmnJm(λmnr)sin(mθ)cos(λmnct) +/summationdisplay nA0nJ0(λ0nr)sin(λ0nct) +/summationdisplay m,nAmnJm(λmnr)cos(mθ)sin(λmnct) +/summationdisplay m,nBmnJm(λmnr)sin(mθ)sin(λmnct). (25) When t=0, the last three sums disappear, and the cosines of tin the ﬁrst three s u m sa r ea l le q u a lt o1 .T h u s u(r,θ,0)=/summationdisplay na0nJ0(λ0nr)+/summationdisplay m,namnJm(λmnr)cos(mθ) +/summationdisplay m,nbmnJm(λmnr)sin(mθ) =f(r,θ), 0<r<a,−π<θ ≤π. (26) We expect to fulﬁll this equality by choosing the a’s and b’s according to some orthogonality principle. Since each function present in the series is an eigenfunction of the problem ∇2φ=−λ2φ, 0<r<a, φ(a,θ)=0, φ(r,θ+2π)=φ(r,θ), 0<r<a, we expect it to be orthogonal to each of the others (see Section 5.4, Exercise 7). This is indeed true: Any function from one series is orthogonal to all of thefunctions in the other series and also to the rest of the functions in its ownseries. T o illustrate this orthogonality, we have /integraldisplay/integraldisplay RJ0(λ0nr)Jm(λmnr)cos(mθ)dA =/integraldisplaya 0J0(λ0nr)Jm(λmnr)/integraldisplayπ −πcos(mθ)dθrd r=0,m/negationslash=0. (27) 5.7 Vibrations of a Circular Membrane 327 There are two other relations like this one involving functions from two dif- ferent series. We already know that the functions within the ﬁrst series are orthogonal to each other:/integraldisplayπ −π/integraldisplaya 0J0(λ0nr)J0(λ0qr)rd rdθ=0,n/negationslash=q. Within the second series we must show that, if m/negationslash=porn/negationslash=q,t h e n 0=/integraldisplayπ −π/integraldisplaya 0Jm(λmnr)cos(mθ)Jp(λpqr)cos(pθ)rd rdθ. (28) (Recall that rd rdθ=dAin polar coordinates.) Integrating with respect to θ ﬁ r s t ,w es e et h a tt h ei n t e g r a lm u s tb ez e r oi f m/negationslash=p, by the orthogonality of cos(mθ)and cos (pθ).I fm=p, the preceding integral becomes π/integraldisplaya 0Jm(λmnr)Jm(λmqr)rd r after the integration with respect to θ.F i n a l l y ,i f n/negationslash=q,t h i si n t e g r a li sz e r o ;t h e demonstration follows the same lines as the usual Sturm–Liouville proof. (SeeSection 2.7.) Thus the functions within the second series are shown orthogonalto each other. For the functions of the last series, the proof of orthogonality issimilar. Equipped now with an orthogonality relation, we can determine formulas for the a’s and b’s. For instance, a 0n=/integraltextπ −π/integraltexta 0f(r,θ)J0(λ0nr)rd rdθ 2π/integraltexta 0J2 0(λ0nr)rd r. (29) The A’s and B’s are calculated from the second initial condition. It should now be clear that, while the computation of the solution to the original problem is possible in theory, it will be very painful in practice. Worseyet, the ﬁnal form of the solution Eq. (25) does not give a clear idea of what u looks like. All is not wasted, however. We can say, from an examination of theλ’s, that the tone produced is not musical — that is, uis not periodic in t.A l s o we can sketch some of the fundamental modes of vibration of the membranecorresponding to some low eigenvalues (Fig. 9). The curves represent points for which displacement is zero in that mode (nodal curves). EXERCISES 1.Verify that each of the functions in the series in Eq. (10) satisﬁes Eqs. (1), (2), and (8). 2.Derive the formulas for the a’s and b’s of Eq. (10). 328 Chapter 5 Higher Dimensions and Other Coordinates Figure 9 Nodal curves: The curves in these graphs represent solutions of φmn(r,θ)=0. Adjacent regions bulge up or down, according to the sign. Only those φ’s containing the factor cos (mθ)have been used. (See the cover photo- graph.) 3.List the ﬁve lowest frequencies of vibration of a circular membrane. 4.Sketch the function J0(λnr)forn=1,2,3. 5.What boundary conditions must the function φof Eq. (18) satisfy? 6.Justify the derivation of Eqs. (19) and (20) from Eqs. (12)–(14) and (18). 7.Show that /integraldisplaya 0Jm(λmnr)Jm(λmqr)rd r=0,n/negationslash=q, if Jm(λmsa)=0,s=1,2,.... 8.Sketch the nodal curves of the eigenfunctions Eq. (21) corresponding to λ31,λ32,a n dλ33. 9.In the simple case of symmetric vibrations, we found the eigenfunctions φ0n(r,θ)=J0(λ0nr),w h e r e J0(λ0na)=0 for n=1,2,3....T h en o d a l curves of φ03are concentric circles. What are their radii (as multiples 5.8 Some Applications of Bessel Functions 329 Figure 10 Exercise 10. ofa)? What are the radii of the circles that are the nodal curves of φ0n(r,θ) for general n? 10.The nodal curves of φmn(r,θ)are shown in Fig. 10. a.By examining the ﬁgure, determine what values mand nhave. b.What is the numerical value of the eigenvalue λmn(as a multiple of a) for this eigenfunction? c.What is the formula for the function φmn(r,θ)whose nodal curves are shown? d.What is the frequency of vibration for the drumhead when it is vibrat- ing in this mode? (“In this mode” means “so that the displacement u equals a product solution in which this eigenfunction is a factor.”) 5.8 Some Applications of Bessel Functions After the elementary functions, the Bessel functions are among the most useful in engineering and physics. One reason for their usefulness is they solve a fairly general differential equation. The general solution of φ/prime/prime+1−2α xφ/prime+/bracketleftbigg/parenleftbig λγxγ−1/parenrightbig2−p2γ2−α2 x2/bracketrightbigg φ=0( 1 ) is given by φ(x)=xα/bracketleftbig AJp(λxγ)+BYp(λxγ)/bracketrightbig . 330 Chapter 5 Higher Dimensions and Other Coordinates Several problems in which the Bessel functions play an important role follow. The details of separation of variables, which should now be routine, are keptto a minimum. A. Potential Equation in a Cylinder The steady-state temperature distribution in a circular cylinder with insulatedsurface is determined by the problem 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg +∂2u ∂z2=0,0<r<a,0<z<b, (2) ∂u ∂r(a,z)=0, 0<z<b, (3) u(r,0)=f(r), 0<r<a, (4) u(r,b)=g(r), 0<r<a. (5) Here we are considering the boundary conditions to be independent of θ,s ou is independent of θalso. Assuming that u=R(r)Z(z)we ﬁnd that /parenleftbig rR/prime/parenrightbig/prime+λ2rR=0,0<r<a, (6) R/prime(a)=0, (7) /vextendsingle/vextendsingleR(0)/vextendsingle/vextendsinglebounded , (8) Z/prime/prime−λ2Z=0. (9) C o n d i t i o n( 8 )h a sb e e na d d e db e c a u s e r=0 is a singular point. The solution of Eqs. (6)–(8) is Rn(r)=J0(λnr), (10) where the eigenvalues λ2 nare deﬁned by the solutions of R/prime(a)=λJ/prime 0(λa)=0. (11) Because J/prime 0=− J1,t h eλ’s are related to the zeros of J1. The ﬁrst three eigen- values are 0, (3.832/a)2,a n d(7.016/a)2.N o t et h a t R(0)=J0(0)=1. The solution of the problem Eqs. (2)–(5) may be put in the form u(r,z)=a0+b0z+∞/summationdisplay n=1J0(λnr)/bracketleftbigg ansinh(λnz) sinh(λnb)+bnsinh(λn(b−z)) sinh(λnb)/bracketrightbigg .(12) 5.8 Some Applications of Bessel Functions 331 The a’s and b’s are determined from Eqs. (4) and (5) by using the orthogonality relation/integraldisplaya 0J0(λnr)J0(λmr)rd r=0,n/negationslash=m. B. Spherical Waves In spherical (ρ,θ,φ) coordinates (see Section 5.9), the Laplacian operator ∇2 becomes ∇2u=1 ρ2∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 ρ2sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg +1 ρ2sin2(φ)∂2u ∂θ2. Consider a wave problem in a sphere when the initial conditions depend only on the radial coordinate ρ: 1 ρ2∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg =1 c2∂2u ∂t2,0<ρ< a,0<t,( 13) u(a,t)=0, 0<t,( 14) u(ρ,0)=f(ρ), 0<ρ< a,( 15) ∂u ∂t(ρ,0)=g(ρ), 0<ρ< a.( 16) Assuming u(ρ,t)=R(ρ)T(t), we separate variables and ﬁnd T/prime/prime+λ2c2T=0, (17) /parenleftbig ρ2R/prime/parenrightbig/prime+λ2ρ2R=0,0<ρ< a, (18) R(a)=0, (19) /vextendsingle/vextendsingleR(0)/vextendsingle/vextendsinglebounded . (20) Again, the condition (20) has been added because ρ=0 is a singular point. Equation (18) may be put into the form R/prime/prime+2 ρR/prime+λ2R=0, and comparison with Eq. (1) shows that α=− 1/2,γ=1, and ρ=1/2; thus the general solution of Eq. (18) is R(ρ)=ρ−1/2/bracketleftbig AJ1/2(λρ)+BY1/2(λρ)/bracketrightbig . We know that near ρ=0, J1/2(λρ)∼const×ρ1/2, Y1/2(λρ)∼const×ρ−1/2. 332 Chapter 5 Higher Dimensions and Other Coordinates Thus in order to satisfy Eq. (20), we must have B=0. It is possible to show that J1/2(λρ)=2 πsin(λρ)√λρ,Y1/2(λρ)=−2 πcos(λρ)√λρ. Our solution to Eqs. (18) and (20) is, therefore, R(ρ)=sin(λρ) ρ, (21) and Eq. (19) is satisﬁed if λ2 n=(nπ/a)2. The solution of the problem of Eqs. (13)–(16) can be written in the form u(ρ,t)=∞/summationdisplay n=1sin(λnρ) ρ/bracketleftbig ancos(λnct)+bnsin(λnct)/bracketrightbig . (22) The a’s and b’s are, as usual, chosen so that the initial conditions Eqs. (15) and (16) are satisﬁed. C. Pressure in a Bearing The pressure in the lubricant inside a plane-pad bearing satisﬁes the problem ∂ ∂x/parenleftbigg x3∂p ∂x/parenrightbigg +x3∂2p ∂y2=− 1,a<x<b,−c<y<c,( 23) p(a,y)=0, p(b,y)=0,−c<y<c,( 24) p(x,−c)=0,p(x,c)=0,a<x<b.( 25) (Here aand care positive constants and b=a+1.) Equation (23) is ellip- tic and nonhomogeneous. T o reduce this equation to a more familiar one, letp(x,y)=v(x)+u(x,y),w h e r e v(x)satisﬁes the problem /parenleftbig x 3v/prime/parenrightbig/prime=− 1,a<x<b, (26) v(a)=0,v ( b)=0. (27) Then, when vis found, umust be the solution of the problem ∂ ∂x/parenleftbigg x3∂u ∂x/parenrightbigg +x3∂2u ∂y2=0,a<x<b,−c<y<c,( 28) u(a,y)=0, u(b,y)=0,−c<y<c,( 29) u(x,±c)=−v(x), a<x<b.( 30) If we now assume that u(x,y)=X(x)Y(y), the variables can be separated: 5.8 Some Applications of Bessel Functions 333 /parenleftbig x3X/prime/parenrightbig/prime+λ2x3X=0,a<x<b, (31) X(a)=0, X(b)=0, (32) Y/prime/prime−λ2Y=0, −c<y<c. (33) Equation (31) may be put in the form X/prime/prime+3 xX/prime+λ2X=0,a<x<b. By comparing to Eq. (1) we ﬁnd that α=− 1,γ=1, and p=1 and that the g e n e r a ls o l u t i o no fE q .( 3 1 )i s X(x)=1 x/parenleftbig AJ1(λx)+BY1(λx)/parenrightbig . Because the point x=0 is not included in the interval a<x<b,t h e r ei sn o problem with boundedness. Instead we must satisfy the boundary conditionsEq. (32), which after some algebra have the form AJ 1(λa)+BY1(λa)=0, AJ1(λb)+BY1(λb)=0. Not both Aand Bmay be zero, so the determinant of these simultaneous equations must be zero: J1(λa)Y1(λb)−J1(λb)Y1(λa)=0. Some solutions of the equation are tabulated for various values of b/a.F o r instance, if b/a=2.5, the ﬁrst three eigenvalues λ2are /parenleftbigg2.156 a/parenrightbigg2 ,/parenleftbigg4.223 a/parenrightbigg2 ,/parenleftbigg6.307 a/parenrightbigg2 . We now can take Xnto be Xn(x)=1 x/parenleftbig Y1(λna)J1(λnx)−J1(λna)Y1(λnx)/parenrightbig , (34) and the solution of Eqs. (28)–(30) has the form u(x,y)=∞/summationdisplay n=1anXn(x)cosh(λny) cosh(λnc). (35) The a’s are chosen to satisfy the boundary conditions Eq. (30), using the or- thogonality principle/integraldisplayb aXn(x)Xm(x)x3dx=0,n/negationslash=m. Notice that Eqs. (31) and (32) make up a regular Sturm–Liouville problem. 334 Chapter 5 Higher Dimensions and Other Coordinates EXERCISES 1.Find the general solution of the differential equation /parenleftbig xnφ/prime/parenrightbig/prime+λ2xnφ=0, where n=0,1,2,.... 2.Find the solution of the equation in Exercise 1 that is bounded at x=0. 3.Find the solutions of Eq. (9), including the case λ2=0, and prove that Eq. (12) is a solution of Eqs. (2)–(4). 4.Show that any function of the form u(ρ,t)=1 ρ/parenleftbig φ(ρ+ct)+ψ(ρ−ct)/parenrightbig is a solution of Eq. (13) if φandψhave at least two derivatives. 5.Find functions φandψsuch that u(ρ,t)a sg i v e ni nE x e r c i s e4s a t i s ﬁ e s Eqs. (14)–(16). 6.Give the formula for the a’s and b’s in Eq. (12). 7.What is the orthogonality relation for the eigenfunctions of Eqs. (18)– (20)? Use it to ﬁnd the a’s and b’s in Eq. (22). 8.Sketch the ﬁrst few eigenfunctions of Eqs. (18)–(20). 9.Find the function v(x)that is the solution of Eqs. (26) and (27). 10.Use the technique of Example C to change the following problem into a potential problem: ∂2u ∂x2+∂2u ∂y2=− f(x), 0<x<a,0<y<b, u=0 on all boundaries. 11.In Exercise 10, will the same technique work if f(x)is replaced by f(x,y)? 12.Verify that Eqs. (31) and (32) form a regular Sturm–Liouville problem. Show the eigenfunctions’ orthogonality by using the orthogonality of theBessel functions. 13.Find a formula for the anof Eq. (35). 14.Verify that Eq. (34) is a solution of Eqs. (28)–(30). 5.9 Spherical Coordinates; Legendre Polynomials 335 5.9 Spherical Coordinates; Legendre Polynomials After the Cartesian and cylindrical coordinate systems, the one most fre- quently encountered is the spherical system (Fig. 11), in which x=ρsin(φ)cos(θ), y=ρsin(φ)sin(θ), z=ρcos(φ). The variables are restricted by 0 ≤ρ,0≤θ< 2π,0≤φ≤π.I nt h i sc o o r d i - nate system the Laplacian operator is ∇2u=1 ρ2/braceleftbigg∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg +1 sin2(φ)∂2u ∂θ2/bracerightbigg . From what we have seen in other cases, we expect solvable problems in spherical coordinates to reduce to one of the following. Problem 1. ∇2u=−λ2uinR, plus homogeneous boundary conditions. Problem 2. ∇2u=0i nR, plus homogeneous boundary conditions on facing sides (where Ris a generalized rectangle in spherical coordinates). Problem 1 would come from a heat or wave equation after separating out the time variable. Problem 2 is a part of the potential problem. The complete solution of either of these problems is very complicated, but a number of special cases are simple, important, and not uncommon. We havealready seen Problem 1 solved (Section 5.8) when uis a function of ρonly. A second important case is Problem 2, when uis independent of the variable θ. We shall state a complete boundary value problem and solve it by separation Figure 11 Spherical coordinates. 336 Chapter 5 Higher Dimensions and Other Coordinates of variables: 1 ρ2/braceleftbigg∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg/bracerightbigg =0, 0<ρ< c,0<φ<π , (1) u(c,φ)=f(φ), 0<φ<π . (2) From the assumption u(ρ,φ) =R(ρ)/Phi1(φ) , it follows that (ρ2R/prime(ρ))/prime R(ρ)+(sin(φ)/Phi1/prime(φ))/prime sin(φ)/Phi1(φ)=0. Both terms are constant, and the second is negative, −µ2, because the bound- ary condition at ρ=cwill have to be satisﬁed by a linear combination of func- tions of φ. The separated equations are /parenleftbig ρ2R/prime/parenrightbig/prime−µ2R=0, 0<ρ< c,( 3)/parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime+µ2sin(φ)/Phi1=0,0<φ<π . ( 4) Neither equation has a boundary condition. However, ρ=0 is a singular point of the ﬁrst equation, and both φ=0a n dρ=πare singular points of the sec- ond equation. (At these points, the coefﬁcient of the highest-order derivativeis zero, while some other coefﬁcient is nonzero.) At each of the singular points,we impose a boundedness condition: R(0)bounded ,/Phi1 ( 0)and /Phi1(π) bounded . Equation (4) can be simpliﬁed by the change of variables x=cos(φ), /Phi1(φ)=y(x).( O fc o u r s e , xisnotthe Cartesian coordinate.) By the chain rule, the relevant derivatives are d/Phi1 dφ=− sin(φ)dy dx, d dφ/parenleftbigg sin(φ)d/Phi1 dφ/parenrightbigg =sin3(φ)d2y dx2−2s i n(φ)cos(φ)dy dx. The differential equation becomes sin2(φ)d2y dx2−2c o s(φ)dy dx+µ2y=0, or, in terms of xalone, /parenleftbig 1−x2/parenrightbig y/prime/prime−2xy/prime+µ2y=0,−1<x<1. (5) In addition, we require that y(x)be bounded at x=± 1. 5.9 Spherical Coordinates; Legendre Polynomials 337 Solutions of the differential equation are usually found by the power series method. Assume that y(x)=a0+a1x+···+ akxk+··· .T h et e r m so ft h e differential equations are then y/prime/prime=2a2+3·2a3x+4·3a4x2+··· + (k+2)(k+1)ak+2xk+··· −x2y/prime/prime=− 2a2x2−··· − k(k−1)akxk+··· −2xy/prime=− 2a1x−2a2x2−··· − 2kakxk−··· µ2y=µ2a0+µ2a1x+µ2a2x2+··· + µ2akxk+···. When this tableau is added vertically, the left-hand side is zero, according to the differential equation. The right-hand side adds up to a power series, eachof whose coefﬁcients must be zero. We therefore obtain the following relations: 2a 2+µ2a0=0, 6a3+/parenleftbig µ2−2/parenrightbig a1=0, (k+2)(k+1)ak+2+/bracketleftbig µ2−k(k+1)/bracketrightbig ak=0. The last equation actually includes the ﬁrst two, apparently special, cases. We may write the general relation as ak+2=k(k+1)−µ2 (k+2)(k+1)ak, valid for k=0,1,2,.... Suppose for the moment that µ2is given. A short calculation gives the ﬁrst few coefﬁcients: a2=−µ2 2a0, a3=2−µ2 6a1, a4=6−µ2 12a2, a5=12−µ2 20a3, =6−µ2 12·−µ2 2a0,=12−µ2 20·2−µ2 6a1. It is clear that all the a’s with even index will be multiples of a0and those with odd index will be multiples of a1.T h u s y(x)equals a0times an even function plus a1times an odd function, with both a0and a1arbitrary. It is not difﬁcult to prove that odd and even series produced by this process diverge at both x=± 1, for general µ2.H o w e v e r ,w h e n µ2has one of the spe- cial values µ2=µ2 n=n(n+1), n=0,1,2,..., one of the two series turns out to have all zero coefﬁcients after an.F o ri n - stance, if µ2=3·4, then a5=0, and all subsequent coefﬁcients with odd 338 Chapter 5 Higher Dimensions and Other Coordinates P0(x)=1 P1(x)=x P2(x)=(3x2−1)/2 P3(x)=(5x3−3x)/2 P4(x)=(35x4−30x2+3)/8 Table 3 Legendre polynomials Figure 12 Graphs of the ﬁrst ﬁve Legendre polynomials. index are also zero. Hence, one of the solutions of /parenleftbig 1−x2/parenrightbig y/prime/prime−2xy/prime+12y=0 is the polynomial a1(x−5x3/3).T h eo t h e rs o l u t i o ni sa ne v e nf u n c t i o nu n - bounded at both x=± 1. Now we see that the boundedness conditions can be satisﬁed only if µ2is one of the numbers 0 ,2,6,..., n(n+1),... . In such a case, one solution of the differential equation is a polynomial (naturally bounded at x=± 1). When normalized by the condition y(1)=1, these are called Legendre polynomials , written Pn(x). Table 3 provides the ﬁrst ﬁve Legendre polynomials. Figure 12 shows their graphs. 5.9 Spherical Coordinates; Legendre Polynomials 339 Since the differential equation (5) is easily put into self-adjoint form, /parenleftbig (1−x2)y/prime/parenrightbig/prime+µ2y=0,−1<x<1, it is routine to show that the Legendre polynomials satisfy the orthogonality relation /integraldisplay1 −1Pn(x)Pm(x)dx=0,n/negationslash=m. By direct calculation, it can be shown that /integraldisplay1 −1P2 n(x)dx=2 2n+1. (6) A compact way of representing the Legendre polynomials is by means of Ro- drigues’ formula, Pn(x)=1 n!2ndn dxn/bracketleftbig (x2−1)n/bracketrightbig . (7) Elementary algebra and calculus show that the nth derivative of (x2−1)nis ap o l y n o m i a lo fd e g r e e n. Substituting this polynomial into the differential equation (5), with µ2=n(n+1), shows that it is a solution — bounded, of course. Therefore, it is a multiple of the Legendre polynomial Pn(x).T h r o u g h Rodrigues’ formula or otherwise, it is possible to prove the following two for-mulas, which relate three consecutive Legendre polynomials: (2n+1)P n(x)=P/prime n+1(x)−P/prime n−1(x), (8) (n+1)Pn+1(x)+nPn−1(x)=(2n+1)xPn(x). (9) In order to use Legendre polynomials in boundary value problems, we need to be able to express a given function f(x)in the form of a Legendre series, f(x)=∞/summationdisplay n=0bnPn(x),−1<x<1. From the orthogonality relation and the integral, Eq. (6), it follows that the coefﬁcient in the series must be bn=2n+1 2/integraldisplay1 −1f(x)Pn(x)dx. (10) The convergence theorem for Legendre series is analogous to the one for Fourier series in Chapter 1. 340 Chapter 5 Higher Dimensions and Other Coordinates Theorem. If f(x)is sectionally smooth on the interval −1<x<1,t h e na te v e r y point of that interval the Legendre series of f is convergent, and ∞/summationdisplay n=0bnPn(x)=f(x+)+f(x−) 2. /square From Eq. (10) for the coefﬁcient of a Legendre series and from the fact that the Legendre polynomials are odd or even, we see that an odd function will have only odd-indexed coefﬁcients that are nonzero, and an even function will have only even-indexed coefﬁcients that are nonzero. Furthermore, if a func-tion fi sg i v e no nt h ei n t e r v a l0 <x<1, then its odd and even extensions have odd and even Legendre series, and fis represented by either in that interval: f(x)=/summationdisplay nevenbnPn(x), 0<x<1, bn=(2n+1)/integraldisplay1 0f(x)Pn(x)dx(neven), (11) f(x)=/summationdisplay noddbnPn(x), 0<x<1, bn=(2n+1)/integraldisplay1 0f(x)Pn(x)dx(nodd). (12) Because the Pn(x)are polynomials, the integral equation (10) for any spe- ciﬁc coefﬁcient can be done in closed form for a variety of functions f(x). However, getting anas a function of nis not so easy. Fortunately, some ele- mentary integrals can be done using the differential equation /parenleftbig (1−x2)P/prime n/parenrightbig/prime+n(n+1)Pn=0. (1) First, separate the two terms of the differential equation and integrate: n(n+1)/integraldisplay Pn(x)dx=/integraldisplay −/parenleftbig (1−x2)P/prime n/parenrightbig/primedx =−/parenleftbig 1−x2/parenrightbig P/prime n(x). This equation may be solved for the integral if n/negationslash=0. (2) Now multiply through the differential equation by x, separate terms, and integrate: n(n+1)/integraldisplay xPn(x)dx=/integraldisplay −x/parenleftbig (1−x2)P/prime n/parenrightbig/primedx =− x/parenleftbig 1−x2/parenrightbig P/prime n+/integraldisplay/parenleftbig 1−x2/parenrightbig P/prime ndx =− x/parenleftbig 1−x2/parenrightbig P/prime n+/parenleftbig 1−x2/parenrightbig Pn−/integraldisplay (−2x)Pndx. 5.9 Spherical Coordinates; Legendre Polynomials 341 Next, move the last term to the left-hand member of the equation to ﬁnd (n+2)(n−1)/integraldisplay xPn(x)dx=/parenleftbig 1−x2/parenrightbig/parenleftbig Pn(x)−xP/prime n(x)/parenrightbig . This equation can be solved for the integral on the left, provided that n/negationslash=1. (For n=1, the integration is done directly.) Summary /integraldisplay Pn(x)dx=−(1−x2) n(n+1)P/prime n(x), (13) /integraldisplay xPn(x)dx=(1−x2) (n+2)(n−1)/parenleftbig Pn(x)−xP/prime n(x)/parenrightbig (14) These integration formulas are useful if we can evaluate Pn(x)and P/prime n(x) easily for any x. The relations in Eqs. (8) and (9) are useful for this purpose. We illustrate by ﬁnding Pn(0).F i r s t ,n o t et h a t Pn(0)=0 for odd values of n, because the Legendre polynomials with odd index are odd functions of x.F o r odd n,E q .( 9 )g i v e s (n+1)Pn+1(0)+nPn−1(0)=0, or Pn+1(0)=−n n+1Pn−1(0). Because P0(0)=1, we ﬁnd successively that P2(0)=−1 2,P4(0)=1·3 2·4,P6(0)=−1·3·5 2·4·6, or in general Pn(0)=(−1)n/21·3···(n−1) 2·4···n,n=2,4,6,... Pn(0)=0, n=1,3,5,....(15) Similarly, but not as easily, Eq. (8) can be used to ﬁnd the values of P/prime n(0).I t is simpler to use the relation P/prime n(0)=nPn−1(0), (16) which can be derived from Eqs. (8) and (9). 342 Chapter 5 Higher Dimensions and Other Coordinates Example. Let f(x)=/braceleftBig−1,−1<x<0, 1, 0<x<1. The Legendre series will contain only odd-indexed polynomials, and their co- efﬁcients are bn=(2n+1)/integraldisplay1 0Pn(x)dx(nodd) =−2n+1 n(n+1)/bracketleftbig/parenleftbig 1−x2/parenrightbig P/prime n(x)/bracketrightbig1 0 =2n+1 n(n+1)P/prime n(0)=2n+1 n+1Pn−1(0) =(−1)(n−1)/21·3·5···(n−2) 2·4·6···(n−1)·2n+1 n+1(n=3,5,7,...). Speciﬁcally we ﬁnd b1=3/2 (by a separate calculation), b3=− 7/8,b5= 11/16,....B e c a u s e f(x)is indeed sectionally smooth, f(x)=3 2P1(x)−7 8P3(x)+11 16P5(x)−···. See Fig. 13 for graphs of the partial sums of this series. /square (a) (b) Figure 13 Graphs of a function and a partial sum of its Legendre series: (a) through P9(x)for the function f(x)in the example; (b) through P6(x)for f(x)=|x|,−1<x<1. Compare with the partial sums of the Fourier series, Figs. 9 a n d1 0o fC h a p t e r1 . 5.9 Spherical Coordinates; Legendre Polynomials 343 Figure 14 The nodal curves of the zonal harmonics are the parallels (φ=constant) on a sphere, where Pn(cos(φ))=0. The nodal curves are shown in projection for n=1,2,3,4. See the CD for color versions. Summary The solution of the eigenvalue problem /parenleftbig (1−x2)y/prime/parenrightbig/prime+µ2y=0,−1<x<1, y(x)bounded at x=− 1a n da t x=1, isy(x)=Pn(x),µ2 n=n(n+1),n=0,1,2,.... The solution of the eigenvalue problem /parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime+µ2sin(φ)/Phi1=0, /Phi1(φ) bounded at φ=0a n da t φ=π, is/Phi1(φ)=Pn(cos(φ)) ,µ2 n=n(n+1),n=0,1,2,.... T h eL e g e n d r ep o l y n o m i a l s Pn(cos(φ)) are often called zonal harmonics be- cause their nodal lines (loci of solutions of Pn(cos(φ))=0) divide a sphere into zones, as shown in Fig. 14. EXERCISES 1.Equation (4) may be solved by assuming /Phi1(φ)=1 2a0+∞/summationdisplay 1akcos(kφ). Find the relations among the coefﬁcients akby computing the terms of the equation in the form of series. Use the identities sin(φ)sin(kφ)=1 2/bracketleftbig cos/parenleftbig (k−1)φ/parenrightbig −cos/parenleftbig (k+1)φ/parenrightbig/bracketrightbig , sin(φ)cos(kφ)=−1 2/bracketleftbig sin/parenleftbig (k−1)φ/parenrightbig +sin/parenleftbig (k+1)φ/parenrightbig/bracketrightbig . 344 Chapter 5 Higher Dimensions and Other Coordinates Show that the coefﬁcients are all zero after anifµ2=n(n+1). 2.Derive the formula for the coefﬁcients bn, as shown in Eq. (10). 3.Find P5(x), ﬁrst from the formulas for the a’s and second by using Eq. (9) with n=4. 4.Verify Eqs. (6) and (7) for n=0,1,2 and Eq. (9) for n=2,3. 5.One of the solutions of (1−x2)y/prime/prime−2xy/prime=0i sy(x)=1(µ2=0).F i n d another independent solution of this differential equation. 6.Show that the orthogonality relation for the eigenfunctions /Phi1n(φ)= Pn(cos(φ)) is /integraldisplayπ 0/Phi1n(φ)/Phi1 m(φ)sin(φ)dφ=0,n/negationslash=m. 7.Obtain the relation P/prime n+1(x)=(n+1)Pn(x)+xP/prime n(x) by differentiating Eq. (9) and eliminating P/prime n−1between that and Eq. (8). Note that Eq. (16) follows from this relation. 8.LetF=(x2−1)n. Show that Fsatisﬁes the differential equation /parenleftbig x2−1/parenrightbig F/prime=2nxF. 9.Differentiate both sides of the preceding equation n+1 times to show that thenth derivative of Fsatisﬁes Legendre’s equation (5). Use Leibniz’s rule for derivatives of a product. 10.Obtain Eq. (6) by these manipulations: a.Multiply through Eq. (9) by Pn+1,i n t e g r a t ef r o m −1t o1 ,a n du s et h e orthogonality of Pn+1with Pn−1. b.Replace (2n+1)Pnby means of Eq. (8). c.Pn+1is orthogonal to xP/prime n−1, which is a polynomial of degree n. d.Solve what remains for the desired integral. 11.Find the Legendre series for the function f(x)=|x|,−1<x<1. 12.Find the Legendre series for the following function. Note that f(x)−1/2 is an odd function. f(x)=/braceleftBig0,−1<x<0, 1,0<x<1. 5.10 Some Applications of Legendre Polynomials 345 5.10 Some Applications of Legendre Polynomials In this section we follow through the details involved in solving some problems in which Legendre polynomials are used. First, we complete the problem statedin the previous section. A. Potential in a Sphere We consider the axially symmetric potential equation — that is, with no vari-a t i o ni nt h el o n g i t u d i n a l -o r θ-direction. The unknown function umight rep- resent an electrostatic potential, steady-state temperature, etc. 1 ρ2/braceleftbigg∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg/bracerightbigg =0, 0<ρ< c,0<φ<π , (1) u(c,φ)=f(φ), 0<φ<π . (2) Of course, the function uis to be bounded at the singular points φ=0,φ=π, andρ=0. The assumption that uhas the product form, u(ρ,φ) =/Phi1(φ) R(ρ), allows us to transform the partial differential equation into (ρ2R/prime(r))/prime R(r)+(sin(φ)/Phi1/prime(φ))/prime sin(φ)/Phi1(φ)=0. F r o mh e r ew eo b t a i ne q u a t i o n sf o r Rand/Phi1individually, /parenleftbig ρ2R/prime/parenrightbig/prime−µ2R=0, 0<ρ< c, (3) /parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime+µ2sin(φ)/Phi1=0,0<φ<π . (4) In Section 5.9 we found the eigenfunctions of Eq. (4), subject to the bound- edness conditions at φ=0a n d π,t ob e /Phi1n(φ)=Pn(cos(φ)) , corresponding to the eigenvalues µ2 n=n(n+1). We must still solve Eq. (3) for R.A f t e rt h e differentiation has been carried out, the problem for Rbecomes ρ2R/prime/prime n+2ρR/prime n−n(n+1)Rn=0,0<ρ< c, Rnbounded at ρ=0. The equation is of the Cauchy–Euler type, solved by assuming R=ραand determining α. Two solutions, ρnandρ−(n+1), are found, of which the sec- ond is unbounded at ρ=0. Hence Rn=ρn, and our product solutions of the potential equation have the form un(ρ,φ) =Rn(ρ)/Phi1 n(φ)=ρnPn/parenleftbig cos(φ)/parenrightbig . 346 Chapter 5 Higher Dimensions and Other Coordinates The general solution of the partial differential equation that is bounded in the region 0 <ρ< c,0<φ<π is thus the linear combination u(ρ,φ) =∞/summationdisplay n=0bnρnPn/parenleftbig cos(φ)/parenrightbig . (5) Atρ=c, the boundary condition becomes u(c,φ)=∞/summationdisplay n=0bncnPn/parenleftbig cos(φ)/parenrightbig =f(φ), 0<φ<π . (6) The coefﬁcients bnare then found to be bn=2n+1 2cn/integraldisplayπ 0f(φ)Pn/parenleftbig cos(φ)/parenrightbig sin(φ)dφ. (7) B. Heat Equation on a Spherical Shell The temperature on a spherical shell satisﬁes the three-dimensional heat equa- tion. If initially there is no dependence on θ, then there will never be such de- pendence. Furthermore, if the shell is thin (thickness much less than averageradius R), we may also assume that temperature does not vary in the radial direction. The heat equation then becomes one-dimensional: 1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg =R2 k∂u ∂t,0<φ<π , 0<t, (8) u(φ,0)=f(φ), 0<φ<π . (9) Naturally, we require boundedness of uatφ=0a n dφ=π. The assumption of a product form for the solution, u(φ,t)=/Phi1(φ) T(t), leads to the conclusion that (sin(φ)/Phi1/prime(φ))/prime sin(/Phi1(φ))=R2T/prime(t) kT(t)=−µ2. Thus, we have the eigenvalue problem /parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime+µ2sin(φ)/Phi1=0,0<φ<π , /Phi1(0)and /Phi1(π) bounded . The solution of this problem was found in Section 5.9 to be µ2=n(n+1)and /Phi1n(φ)=Pn/parenleftbig cos(φ)/parenrightbig ,n=0,1,2,.... Obviously, the other factor in a product solution must be Tn(t)=exp/parenleftbig −n(n+1)kt/R2/parenrightbig . 5.10 Some Applications of Legendre Polynomials 347 Now, a series of constant multiples of product solutions is the most general solution of our problem: u(φ,t)=∞/summationdisplay n=0bnPn/parenleftbig cos(φ)/parenrightbig e−n(n+1)kt/R2. (10) The initial condition, Eq. (9), now takes the form of a Legendre series, ∞/summationdisplay n=0bnPn/parenleftbig cos(φ)/parenrightbig =f(φ), 0<φ<π . (11) From the information in Section 5.9, we know that the coefﬁcients bnmust be chosen to be bn=2n+1 2/integraldisplayπ 0Pn/parenleftbig cos(φ)/parenrightbig f(φ)sin(φ)dφ. Then if f(φ)is sectionally smooth for 0 <φ<π , the series of Eq. (11) actually equals f(φ), and thus the function u(φ,t)in Eq. (10) satisﬁes the problem originally posed. For instance, if f(φ)=T0in the northern hemisphere (0<φ<π / 2)and f(φ)=− T0in the southern (π/2<φ<π) , then the coefﬁcients are bn=2n+1 2/integraldisplayπ 0f(φ)Pn/parenleftbig cos(φ)/parenrightbig sin(φ)dφ =2n+1 2/bracketleftbigg/integraldisplay1 −1f/parenleftbig cos−1(x)/parenrightbig Pn(x)dx/bracketrightbigg =T02n+1 n+1Pn−1(0), as found in the previous section. Figure 15 shows graphs of u(φ,t)as a func- tion of φin the interval 0 <φ<π for various times. The CD shows an ani- mated version of the solution. C. Spherical Waves In Section 5.8, we solved the wave equation in spherical coordinates for thecase where the initial conditions depend only on the radial variable ρ.N o ww e consider the case where the variable φis also present. A full statement of the 348 Chapter 5 Higher Dimensions and Other Coordinates Figure 15 Graphs of the solution of the example problem, with u(φ,0)positive in the north and negative in the south. The function u(φ,t)is shown as a function ofφin the range 0 to πfor times chosen so that the dimensionless time kt/R2 takes the values 0, 0 .01, 0.1, and 1; for convenience, T0=100. problem is 1 ρ2∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 ρ2sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg =1 c2∂2u ∂t2, 0<ρ< a,0<φ<π , 0<t, u(a,φ,t)=0, 0<φ<π , 0<t, u(ρ,φ, 0)=f(ρ,φ), 0<ρ< a,0<φ<π , ∂u ∂t(ρ,φ, 0)=g(ρ,φ), 0<ρ< a,0<φ<π .(12) As usual, we require in addition that ube bounded as ρ→0a n da s φ→0 andφ→π. First, we seek solutions in the product form u(ρ,φ, t)=R(ρ)/Phi1(φ) T(t). Inserting uin this form into the partial differential equation (12) and manip- ulating, we ﬁnd 1 ρ2/parenleftbigg(ρ2R/prime)/prime R+(sin(φ)/Phi1/prime)/prime sin(φ)/Phi1/parenrightbigg =T/prime/prime c2T. (13) Both sides of this equation must have the same and constant value, say, −λ2. Thus, we must have (ρ2R/prime)/prime R+(sin(φ)/Phi1/prime)/prime sin(φ)/Phi1=−λ2ρ2. 5.10 Some Applications of Legendre Polynomials 349 Again we see that the ratio containing /Phi1must be constant, say, −µ2.H e n c e , we have two separate problems for the functions /Phi1and R: /parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime+µ2sin(φ)/Phi1=0,0<φ<π , /Phi1(φ) bounded at φ=0,π ,/parenleftbig ρ2R/prime/parenrightbig/prime−µ2R+λ2ρ2R=0, 0<ρ< a, R(a)=0, R(ρ)bounded at 0 . The ﬁrst of these problems is now quite familiar, and we know its solution to be µ2 n=n(n+1), /Phi1 n(φ)=Pn/parenleftbig cos(φ)/parenrightbig ,n=0,1,2,.... The second problem is less familiar. In standard form, the differential equa- tion is R/prime/prime+2 ρR/prime−µ2 ρ2R+λ2R=0. Comparison with the four-parameter form of Bessel’s equation (Eq. (1) of Sec- tion 5.8) shows α=− 1/2,γ=1, and p2=µ2+α2.S i n c e µ=n(n+1), p2=n2+n+1 4,a n dt h e n p=n+1 2. Thus, the general solution of the differ- ential equation is Rn(ρ)=ρ−1/2/bracketleftbig AJn+1/2(λρ)+BYn+1/2(λρ)/bracketrightbig . The fact that the Bessel functions of the second kind, Yp(λρ) ,a r eu n - bounded at ρ=0 allows us to discard them from the solution, leaving Rn(ρ)=ρ−1/2Jn+1/2(λρ) as the bounded solution. These functions occur frequently in problems in spherical coordinates. Sometimes the functions jn(z)=/radicalbiggπ 2zJn+1/2(z), called spherical Bessel functions of the ﬁrst kind of order n ,a r ei n t r o d u c e d .A s noted in Section 5.8, there is a relation to sines and cosines: j0(z)=sin(z)/z, j1(z)=/parenleftbig sin(z)−zcos(z)/parenrightbig /z2, j2(z)=/parenleftbig (3−z2)sin(z)−3zcos(z)/parenrightbig /z3. We have yet to satisfy the boundary condition Rn(a)=0. This cannot be done by formula, except for n=0. In this case, R0(a)=0c o m e sd o w nt o 350 Chapter 5 Higher Dimensions and Other Coordinates sin(λa)/λa=0, soλm=mπ/aform=1,2,....F o ro t h e r n’s, solutions of Jn+1/2(λa)=0 must be found numerically. For instance, for n=1t h ee q u a t i o n is sin(λa)−λacos(λa)=0, with solutions λa=4.493, 7.725, 10 .904,....( S e e Handbook of Mathematical Functions by Abramowitz and Stegun, listed in the Bibliography.) Finally we can put together some product solutions. Clearly the factor func- tion T(t)will be a sine or cosine of λct. Thus our product solutions have the form ρ−1/2Jn+1/2(λnmρ)Pn/parenleftbig cos(φ)/parenrightbig sin(λnmct), ρ−1/2Jn+1/2(λnmρ)Pn/parenleftbig cos(φ)/parenrightbig cos(λnmct). The solution u(ρ,φ, t)will be an inﬁnite series of constant multiples of these functions. We will not write it out. Let us summarize some of the information we have obtained. First, the fre- quencies of vibration of a sphere are λmnc(radians per unit time), where λmn is the mth positive solution of Jn+1/2(λa)=0. Second, the nodal surfaces (loci of poin ts where a product solution is 0 for all time) are the values of ρandφfor which Jn+1/2(λmnρ)Pn/parenleftbig cos(φ)/parenrightbig =0. One or the other factor must be 0, so these surfaces are either concentric spheres, ρ=const., determined by Jn+1/2(λmnρ)=0, or else cones φ=const., determined by Pn(cos(φ))=0. Finally, let us observe that, because P0(cos(φ))≡1, the product solutions with n=0 are precisely what we found as product solutions of the problem in Section 5.8, Part B. EXERCISES 1.Solve the potential equation in the sphere 0 <ρ< 1, 0<φ<π with the boundary condition u(1,φ)=/braceleftbigg1,0<φ<π/ 2, 0,π / 2<φ<π , together with appropriate boundedness conditions. 5.10 Some Applications of Legendre Polynomials 351 2.Solve the potential equation in a hemisphere, 0 <ρ< 1, 0<φ<π / 2, subject to boundedness conditions at ρ=0a n dφ=0, and the boundary conditions u(1,φ)=1, 0<φ<π/ 2, u(ρ,π/ 2)=0,0<ρ< 1. Hint: Use odd-order Legendre polynomials. 3.Solve this heat problem with convection on a spherical shell of radius R: 1 R2sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg −γ2(u−T)=1 k∂u ∂t, 0<φ<π , 0<t, u(φ,0)=0,0<φ<π . Think carefully about the physical situ ation before attempting a solution. 4.Solve this heat problem on a hemispherical shell of radius R: 1 R2sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg =1 k∂u ∂t,0<φ<π/ 2,0<t, ∂u ∂φ(π/2,t)=0, 0<t, u(φ,0)=cos(φ), 0<φ<π/ 2. 5.Solve the eigenvalue problem 1 ρ2/bracketleftbigg∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg/bracketrightbigg =−λ2u, 0<ρ< a,0<φ<π/ 2, u(a,φ)=0, 0<φ<π/ 2, u(r,π/2)=0,0<r<a, subject to boundedness conditions at ρ=0a n da t φ=0. 6.In Part C of this section we mention nodal surfaces (i.e., surfaces where the function is 0). Find the nodal surfaces of the function ρ−1/2J3/2(λρ) P1/parenleftbig cos(φ)/parenrightbig ifλis the second positive solution of J3/2(λ)=0. 7.Describe in words the nodal surfaces for ρ−1/2J5/2(λρ) P2/parenleftbig cos(φ)/parenrightbig 352 Chapter 5 Higher Dimensions and Other Coordinates ifλis the second positive solution of J5/2(λ)=0. 8.Solve the potential problem in the exterior of a sphere. 1 ρ2/bracketleftbigg∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg/bracketrightbigg =0, R<ρ, 0<φ<π , u(R,φ)=f(φ), 0<φ<π . 9.L.M. Chiappetta and D.R. Sobel [T emp erature distribution within a hemi- sphere exposed to a hot gas stream, SIAM Review ,26(1984): 575–577] analyze the steady-state temperature in the rounded tip of a combustion-gas sampling probe. The tip is approximately hemispherical in shape. Itsouter surface is exposed to hot gases at temperature T G, and its base is cooled by water at temperature TWcirculating inside the probe. If T(ρ,φ) is the temperature inside the tip, it should satisfy the condi- tions 1 ρ2/bracketleftbigg∂ ∂ρ/parenleftbigg ρ2∂T ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂T ∂φ/parenrightbigg/bracketrightbigg =0, 0<ρ< R,0<φ<π 2, T(ρ,π/ 2)=TW, 0<ρ< R, k∂T ∂ρ(R,φ)=h/bracketleftbig TG−T(R,φ)/bracketrightbig ,0<φ<π 2 together with boundedness conditions at ρ=0a n da t φ=0. The authors then change the variables to simplify the problem. Let r=ρ/R,u(r,φ)=T(ρ,φ) −TW, and show that the problem for ube- comes ∂ ∂ρ/parenleftbigg ρ2∂u ∂ρ/parenrightbigg +1 sin(φ)∂ ∂φ/parenleftbigg sin(φ)∂u ∂φ/parenrightbigg =0,0<r<1,0<φ<π/ 2, u(r,π/2)=0, 0<r<1, K∂u ∂r(1,φ)+u(1,φ)=D, 0<φ<π/ 2, where K=k/hRand D=TG−TW. 10.Solve the problem in Exercise 9. Hint: Use odd-indexed Legendre polyno- mials to satisfy the boundary condition at φ=π/2. 5.11 Comments and References 353 5.11 Comments and References We have seen just a few problems in two or three dimensions, but they are sufﬁcient to illustrate the complications that may arise. A serious drawbackto the solution by separation of variables is that double and triple series tendto converge slowly, if at all. Thus, if a numerical solution to a two- or three-dimensional problem is needed, it may be advisable to sidestep the analyticalsolution by using an approximate numerical technique from the beginning. One advantage of using special coordinate systems is that some problems that are two-dimensional in Cartesian coordinates may be one-dimensional in another system. This is the case, for instance, when distance from a point(rin polar or ρin spherical coordinates) is the only signiﬁcant space variable. Of course, nonrectangular systems may arise naturally from the geometry of aproblem. As Sections 5.3 and 5.4 point out, solving the two-dimensional heat or wave equation in a region Rof the plane depends on being able to solve the eigen- value problem ∇2φ=−λ2φinRwithφ=0 on the boundary. The solution of this problem in a region bounded by coordinate curves (that is, in a gen-eralized rectangle) is known for many coordinate systems. We have discussedthe most common cases; others can be found in Methods of Theoretical Physics by Morse and Feshbach. Information about the special functions involved isavailable from the Handbook of Mathematical Functions by Abramowitz and Stegun and also from Special Functions of Mathematics for Engineers by L.C. Andrews. Eigenfunctions and eigenvalues are known for a few regions that arenot generalized rectangles. (See Miscellaneous Exercises 20 and 21 in the textthat follows.) Eigenvalues of the Laplacian in a region can be estimated by a Rayleigh quo- tient, much as in Section 3.5. Furthermore, we have theorems of the follow- ing type. Let λ 2 1be the lowest eigenvalue of ∇φ=−λ2φinRwithφ=0 on the boundary. Let ¯λ2 1have the same meaning for another region, ¯R.I f¯R ﬁts inside R,t h e n ¯λ2 1≥λ2 1. [The smaller the region, the larger the ﬁrst eigen- value. For further information, see Methods of Mathematical Physics , Vol. 1, by Courant and Hilbert. In the famous article “Can one hear the shape of adrum?,” American Mathematical Monthly ,73(1966): 1–23], Mark Kac shows that one can ﬁnd the area, perimeter, and connectivity of a region from theeigenvalues of the Laplacian for that region. However, Kac’s title question has been answered negatively. In the Bulletin of the American Mathematical Soci- ety,27(1992): 134–138, authors C. Gordon, D.L. Webb, and S. Wolpert display two plane regions, or “drums,” of different shapes, on which the Laplacian hasexactly the same eigenvalues. The nodal curves of a membrane shown in Fig. 9 can be realized physically. Photographs of such curves, along with an explanation of the physics of the 354 Chapter 5 Higher Dimensions and Other Coordinates vibrating membrane, will be found in The Physics of Musical Instruments ,b y Fletcher and Rossing. Chapter Review See the CD for review questions and special exercises. Miscellaneous Exercises 1.Solve the heat problem ∇2u=1 k∂u ∂t, 0<x<a,0<y<b,0<t, ∂u ∂x(0,y,t)=d0,∂u ∂x(a,y,t)=0,0<y<b,0<t, u(x,0,t)=0,u(x,b,t)=0, 0<x<a,0<t, u(x,y,0)=Tx a, 0<x<a,0<y<b. 2.Same as Exercise 1, but the initial condition is u(x,y,0)=Ty b,0<x<a,0<y<b. 3.Letu(x,y,t)be the solution of the heat equation in a rectangle as stated here. Find an expression for u(a/2,b/2,t).W r i t eo u tt h eﬁ r s tt h r e e nonzero terms for the case a=b. ∇2u=1 k∂u ∂t, 0<x<a,0<y<b,0<t, u=0 on all boundaries , u(x,y,0)=T,0<x<a,0<y<b. 4.Find the nodal lines of the square membrane. These are loci of points satisfying φmn(x,y)=0, where φmnsatisﬁes ∇2φ=−λ2φin the square andφ=0 on the boundary. 5.Find the solution of the boundary value problem 1 rd dr/parenleftbigg rdu dr/parenrightbigg =− 1,0<r<a, u(0)bounded ,u(a)=0, Miscellaneous Exercises 355 both directly and by assuming that both u(r)and the constant function 1 h a v eB e s s e ls e r i e so nt h ei n t e r v a l0 <r<a: u(r)=∞/summationdisplay n=1CnJ0(λnr),0<r<a, 1=∞/summationdisplay n=1cnJ0(λnr), 0<r<a. /parenleftbigg Hint:1 rd dr/parenleftbigg rdJ0(λr) dr/parenrightbigg =−λ2J0(λr)./parenrightbigg 6.Suppose that w(x,t)andv(y,t)are solutions of the partial differential equations ∂2w ∂x2=1 k∂w ∂t,∂2v ∂y2=1 k∂v ∂t. Show that u(x,y,t)=w(x,t)v(y,t)satisﬁes the two-dimensional heat equation ∂2u ∂x2+∂2u ∂y2=1 k∂u ∂t. 7.Use the idea of Exercise 6 to solve the problem stated in Exercise 1. 8.Letw(x,y)andv(z,t)satisfy the equations ∂2w ∂x2+∂2w ∂y2=0,∂2v ∂z2=1 c2∂2v ∂t2. Show that the product u(x,y,z,t)=w(x,y)v(z,t)satisﬁes the three- dimensional wave equation ∂2u ∂x2+∂2u ∂y2+∂2u ∂z2=1 c2∂2u ∂t2. 9.Find the product solutions of the equation 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg =1 k∂u ∂t,0<r,0<t, that are bounded as r→0+and as r→∞ . 10.Show that the boundary value problem /parenleftbig (1−x2)φ/prime/parenrightbig/prime=− f(x),−1<x<1, φ(x)bounded at x=± 1, 356 Chapter 5 Higher Dimensions and Other Coordinates h a sa si t ss o l u t i o n φ(x)=/integraldisplayx 01 1−y2/integraldisplay1 yf(z)dz dy, provided that the function fsatisﬁes /integraldisplay1 −1f(z)dz=0. 11. Suppose that the functions f(x)andφ(x)in the preceding exercise have expansions in terms of Legendre polynomials f(x)=∞/summationdisplay k=0bkPk(x),−1<x<1, φ(x)=∞/summationdisplay k=0BkPk(x),−1<x<1. What is the relation between Bkand bk? 12. By applying separation of variables to the problem ∇2u=0,0<ρ< a,0≤φ<π , with ubounded at φ=0,πand uperiodic (2π)inθ,d e r i v et h ef o l l o w - ing equation for the factor function /Phi1(φ) : sin(φ)/parenleftbig sin(φ)/Phi1/prime/parenrightbig/prime−m2/Phi1+µ2sin2(φ)/Phi1=0, where m=0,1,2,... comes from the factor /Theta1(θ) . 13. Using the change of variables x=cos(φ),/Phi1(φ)=y(x)on the equation of Exercise 12, derive a differential equation for y(x). 14. Solve the heat conduction problem 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg =1 k∂u ∂t, 0<r<a,0<t, ∂u ∂r(a,t)=0, 0<t, u(r,0)=T0−/Delta1/parenleftbiggr a/parenrightbigg2 ,0<r<a. Miscellaneous Exercises 357 15.Solve the following potential problem in a cylinder: 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg +∂2u ∂z2=0, 0<r<a,0<z<b, u(a,z)=0, 0<z<b, u(r,0)=0,u(r,b)=U0,0<r<a. 16.Find the solution of the heat conduction problem 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg =1 k∂u ∂t,0<r<a,0<t, u(a,z)=T0, 0<t, u(r,0)=T1, 0<r<a. 17.Find some frequencies of vibration of a cylinder by ﬁnding product so- lutions of the problem 1 r∂ ∂r/parenleftbigg r∂u ∂r/parenrightbigg +∂2u ∂z2=1 c2∂2u ∂t2,0<r<a,0<z<b,0<t, u(r,0,t)=0,u(r,b,t)=0,0<r<a,α < t, u(a,z,t)=0, 0<z<b,0<t. 18.Derive the given formula for the solution of the following potential equa- tion in a spherical shell: ∇2u=0,a<ρ< b, 0<φ<π , u(a,φ)=f/parenleftbig cos(φ)/parenrightbig ,u(b,φ)=0,0<φ<π , u(ρ,φ) =∞/summationdisplay n=0Anb2n+1−ρ2n+1 b2n+1−a2n+1/parenleftbigga ρ/parenrightbiggn+1 Pn/parenleftbig cos(φ)/parenrightbig , An=2n+1 2/integraldisplay1 −1f(x)Pn(x)dx. 19.Show that the function φ(x,y)=sin(πx)sin(2πy)−sin(2πx)sin(πy) is an eigenfunction for the triangle Tbounded by the lines y=0,y=x, x=1. That is, ∇2φ=−λ2φinT, φ=0 on the boundary of T. What is the eigenvalue λ2associated with φ? 358 Chapter 5 Higher Dimensions and Other Coordinates 20. Observe that the function φin Exercise 19 is the difference of two differ- ent eigenfunctions of the 1 ×1 square (see Section 5.3) corresponding to the same eigenvalue. Use this idea to construct other eigenfunctions forthe triangle Tof Exercise 19. 21. LetTbe the equilateral triangle in the xy-plane whose base is the interval 0<x<1o ft h e x-axis and whose sides are segments of the lines y=√ 3x and y=√ 3(1−x). Show that for n=1,2,3,...,t h ef u n c t i o n φn(x,y)=sin/parenleftbig 4nπy/√ 3/parenrightbig +sin/parenleftbig 2nπ/parenleftbig x−y/√ 3/parenrightbig/parenrightbig −sin/parenleftbig 2nπ/parenleftbig x+y/√ 3/parenrightbig/parenrightbig is a solution of the eigenvalue problem ∇2φ=−λ2φinT,φ=0o nt h e boundary of T. What are the eigenvalues λ2 ncorresponding to the func- tionφnthat is given? [See “The eigenvalues of an equilateral triangle,” SIAM Journal of Mathematical Analysis ,11(1980): 819–827, by Mark A. Pinsky.] 22. In Comments and References, Section 5.11, a theorem is quoted that re- lates the least eigenvalue of a region to that of a smaller region. Conﬁrmthe theorem by comparing the solution of Exercise 19 with the smallesteigenvalue of one-eighth of a circular disk of radius 1: 1 r∂ ∂r/parenleftbigg r∂φ ∂r/parenrightbigg +1 r2∂2φ ∂θ2=−λ2φ, 0<θ<π 4, 0<r<1, φ(r,0)=0,φ/parenleftbigg r,π 4/parenrightbigg =0,0<r<1, φ(1,θ)=0, 0<θ<π 4. 23. Same task as Exercise 22, but use the triangle of Exercise 21 and the smallest eigenvalue of one-sixth of a circular disk of radius 1. 24. Show that u(ρ,t)=t−3/2e−ρ2/4tis a solution of the three-dimensional heat equation ∇2u=∂u ∂t, in spherical coordinates. 25. For what exponent bisu(r,t)=tbe−r2/4ta solution of the two- dimensional heat equation ∇2u=∂u ∂t? (Use polar coordinates.) 26. Suppose that an estuary extends from x=0t o x=a, where it meets the open sea. If the ﬂoor of the estuary is level but its width is proportional tox, then the water depth u(x,t)satisﬁes 1 x∂ ∂x/parenleftbigg x∂u ∂x/parenrightbigg =1 gU∂2u ∂t2,0<x<a,0<t, Miscellaneous Exercises 359 where gis the acceleration of gravity and Uis mean depth. The tidal motion of the sea is represented by the boundary condition u(a,t)=U+hcos(ωt). Find a bounded solution of the partial differential equation that satisﬁes the boundary condition by setting u(x,t)=U+y(x)cos(ωt). (See Lamb, Hydrodynamics , pp. 275–276.) 27.Is there any combination of parameters for which the solution of Exer- cise 26 does not exist in the form suggested? 28.If the estuary of Exercise 26 has uniform width but variable depth h=Ux/a, then the equation for uis ∂ ∂x/parenleftbigg x∂u ∂x/parenrightbigg =a gU∂2u ∂t2,0<x<a,0<t, subject to the same boundary condition as in Exercise 26. Find a bounded solution in the form suggested. (See Eq. (1) of Section 5.8.) 29.The equation for radially symmetric waves in n-dimensional space is 1 rn−1∂ ∂r/parenleftbigg rn−1∂u ∂r/parenrightbigg =1 c2∂2u ∂t2 where ris distance to the origin. Find product solutions of this equation that are bounded at the origin. 30.Show that the equation of Exercise 29 has solutions of the form u(r,t)=α(r)φ(r−ct) forn=1a n d n=3. [See “A simple proof that the world is three- dimensional” by T om Morley, SIAM Review ,27(1985): 69–71.] 31.A certain kind of chemical reactor contains particles of a solid catalyst and a liquid that reacts with a gas bubbled through it. M. Chidambaran [“Catalyst mixing in bubble column slurry reactors,” Canadian Journal of Chemical Engineering ,67(1989): 503–506] uses the following problem to model the catalyst concentration Cin a cylindrical reactor: Dr1 r∂ ∂r/parenleftbigg r∂C ∂r/parenrightbigg +Dz∂2C ∂z2+U∂C ∂z=0,0<r<R,0<z<L, ∂C ∂r(R,z)=0,0<z<L. 360 Chapter 5 Higher Dimensions and Other Coordinates Here, Drand Dzare the diffusion constants in the radial and axial di- rections, respectively. The term containing ∂C/∂zrepresents physical movement of particles at speed U. Show that the change of variables ρ=r/R,ζ=z/L,u(ρ,ζ) =C(r,z) leads to the equivalent equations b1 ρ∂ ∂ρ/parenleftbigg ρ∂u ∂ρ/parenrightbigg +∂2u ∂ζ2+p∂u ∂ζ=0,0<ρ< 1,0<ζ< 1, ∂u ∂ρ(1,ζ)=0,0<ζ< 1, and identify the parameters band p. 32. When a boundedness condition at ρ=0i sa d d e d ,p r o d u c ts o l u t i o n so f the foregoing equation are found to have the form u(ρ,ζ) =R(ρ)Z(ζ): R0(ρ)=1,Z0(ζ)=/braceleftbigg e−pζ 1, Rn(ρ)=J0(λnρ), Zn(ζ)=/braceleftbigg em1ζ em2ζ, where m1<0<m2are the roots of the equation m2+pm−λ2 nb=0a n d λnis chosen to satisfy J/prime 0(λn)=0. a.Check the details of the solution. b.Show that the λ’s also satisfy J1(λn)=0. 33. The solution of the problem in Exercise 31 has the form u(ρ,ζ) =a0e−pζ+b0+∞/summationdisplay n=1/parenleftbig anem1ζ+bnem2ζ/parenrightbig J0(λnρ). The coefﬁcients would normally be found by applying boundary condi- tions u(ρ,0)=f(ρ), u(ρ,1)=g(ρ), 0<ρ< 1. In this case, however, information is scarce. The author suggests discard- ing the solutions that do not approach 0 as ζ→∞ . The justiﬁcation is that g(ρ)is approximately 0. The solution then becomes u(ρ,ζ) =a0e−pζ+∞/summationdisplay n=1anem1ζJ0(λnρ), Miscellaneous Exercises 361 a n dt h ec o e f ﬁ c i e n t ss h o u l db ed e t e r m i n e db y a0=2/integraldisplay1 0f(ρ)ρ dρ, an=/integraltext1 0f(ρ)J0(λnρ)ρdρ /integraltext1 0J2 0(λnρ)ρ dρ. The function fis known only roughly through experiment. Use the numbers in the following table to ﬁnd a0and a1by the trapezoidal rule of numerical integration. ρ 0 0.1 0.2 0.3 0.4 f(ρ) 8.8 8.9 9.2 9.8 10.3 J0(λ1ρ) 10 0.964 0.858 0.696 0.493 ρ 0.5 0.6 0.7 0.8 0.9 1.0 f(ρ) 11.2 12.0 13.1 14.1 14.8 15 J0(λ1ρ) 0.273 0.056 −0.135 −0.281 −0.373 −0.403 34.In the article “Asymptotic analysis of intraparticle diffusion in GAC batch reactors” [D.A. Lyn, Journal of Environmental Engineering ,122 (1996): 1013–1022], the author analyzes chemical diffusing into a spher- ical particle, with a view to determining some parameter. The concen-tration qis modeled in dimensionless variables by 1 r2∂ ∂r/parenleftbigg r2∂q ∂r/parenrightbigg =∂q ∂t,0<r<1,0<t, q(r,t)bounded as r→0, ∂q ∂t(1,t)=− D∂q ∂r(1,t). Separate the variables and ﬁnd the eigenvalue problem, assuming that q(r,t)=R(r)T(t). 35.The eigenvalue problem that comes from Exercise 34 has a peculiar boundary condition that prevents the eigenfunctions from being orthog- onal. However, the author needs only the ﬁrst terms of a series solution. Find an equation for the eigenvalues. Conﬁrm that λ=0i sas o l u t i o n . Find the next one numerically for D=0,1,10. This page intentionally left blank Laplace Transform CHAPTER6 6.1 Deﬁnition and Elementary Properties The Laplace transform serves as a device for simplifying or mechanizing the solution of ordinary and partial differential equations. It associates a functionf(t)with a function of another variable F(s)from which the original function can be recovered. Letf(t)be sectionally continuous in every interval 0 ≤t<T.T h eL a p l a c e transform of f, written L(f)orF(s),i sd e ﬁ n e db yt h ei n t e g r a l L(f)=F(s)=/integraldisplay∞ 0e−stf(t)dt. (1) We use the convention that a function of tis represented by a lowercase letter and its transform by the corresponding capital letter. The variable smay be real or complex, but in the computation of transforms by the deﬁnition, sis usually assumed to be real. Two simple examples are L(1)=/integraldisplay∞ 0e−st·1dt=1 s, L/parenleftbig eat/parenrightbig =/integraldisplay∞ 0e−steatdt=−e−(s−a)t s−a/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞ 0=1 s−a. Not every sectionally continuous function of thas a Laplace transform, for the deﬁning integral may fail to converge. For instance, exp (t2)has no trans- form. However, there is a simple sufﬁcient condition, as expressed in the fol- lowing theorem. 363 364 Chapter 6 Laplace Transform Theorem. Let f(t)be sectionally continuous in every ﬁnite interval 0≤t<T. If, for some constant k, it is true that lim t→∞e−ktf(t)=0, then the Laplace transform of f exists for Re(s)>k. /square A function that satisﬁes the limit condition in the hypotheses of the theorem is said to be of exponential order . The Laplace transform inherits two important properties from the integral used in its deﬁnition: L/parenleftbig cf(t)/parenrightbig =cL/parenleftbig f(t)/parenrightbig ,cconstant , (2) L/parenleftbig f(t)+g(t)/parenrightbig =L/parenleftbig f(t)/parenrightbig +L/parenleftbig g(t)/parenrightbig . (3) By exploiting these properties, we easily determine that L/parenleftbig cosh(at)/parenrightbig =L/bracketleftbigg1 2/parenleftbig eat+e−at/parenrightbig/bracketrightbigg =1 2/parenleftbigg1 s−a+1 s+a/parenrightbigg =s s2−a2, L/parenleftbig sin(ωt)/parenrightbig =L/bracketleftbigg1 2i/parenleftbig eiωt−e−iωt/parenrightbig/bracketrightbigg =1 2i/parenleftbigg1 s−iω−1 s+iω/parenrightbigg =ω s2+ω2. Notice that the linearity properties work with complex constants and func- tions. Because of the factor e−stin the deﬁnition of the Laplace transform, expo- nential multipliers are easily han dled by the “shifting theorem”: L/parenleftbig ebtf(t)/parenrightbig =/integraldisplay∞ 0e−stebtf(t)dt =/integraldisplay∞ 0e−(s−b)tf(t)dt=F(s−b), where F(s)=L(f(t)). For instance, since L(sin(ωt))=ω/(s2+ω2), L/parenleftbig ebtsin(ωt)/parenrightbig =ω (s−b)2+ω2=ω s2−2sb+b2+ω2. The real virtue of the Laplace transform is seen in its effect on derivatives. Suppose f(t)is continuous and has a sectionally continuous derivative f/prime(t). Chapter 6 Laplace Transform 365 Then by deﬁnition L/parenleftbig f/prime(t)/parenrightbig =/integraldisplay∞ 0e−stf/prime(t)dt. Integrating by parts, we get L/parenleftbig f/prime(t)/parenrightbig =e−stf(t)/vextendsingle/vextendsingle∞ 0−/integraldisplay∞ 0(−s)e−stf(t)dt. Iff(t)is of exponential order, e−stf(t)must tend to 0 as ttends to inﬁnity (for large enough s), so the foregoing equation becomes L/parenleftbig f/prime(t)/parenrightbig =− f(0)+s/integraldisplay∞ 0e−stf(t)dt =− f(0)+sL/parenleftbig f(t)/parenrightbig . (Iff(t)has a jump at t=0,f(0)is to be interpreted as f(0+).) Similarly, if fand f/primeare continuous, f/prime/primeis sectionally continuous; and if all three functions are exponential order, then L/parenleftbig f/prime/prime(t)/parenrightbig =− f(0)+sL/parenleftbig f/prime(t)/parenrightbig =− f(0)−sf(0)+s2L/parenleftbig f(t)/parenrightbig . An easy generalization extends this formula to the nth derivative, L/bracketleftbig f(n)(t)/bracketrightbig =− f(n−1)(0)−sf(n−2)(0)−···− sn−1f(0)+snL/parenleftbig f(t)/parenrightbig , (4) on the assumption that fand its ﬁrst n−1 derivatives are continuous, f(n)is sectionally continuous, and all are of exponential order. We may apply Eq. (4) to the function f(t)=tk,kbeing a nonnegative inte- ger. Here we have f(0)=f/prime(0)=···= f(k−1)(0)=0,f(k)(0)=k!,f(k+1)(t)=0. Thus, Eq. (4) with n=k+1y i e l d s 0=− k!+sk+1L/parenleftbig tk/parenrightbig , or L/parenleftbig tk/parenrightbig =k! sk+1. A different application of the derivative rule is used to transform integrals. If f(t)is sectionally continuous, then/integraltextt 0f(t/prime)dt/primeis a continuous function, equal to zero at t=0, and has derivative f(t).H e n c e L/parenleftbig f(t)/parenrightbig =sL/bracketleftbigg/integraldisplayt 0f(t/prime)dt/prime/bracketrightbigg , 366 Chapter 6 Laplace Transform L(f)=F(s)=/integraldisplay∞ 0e−stf(t)dt L(cf(t))=cL(f(t)) L(f(t)+g(t))=L(f(t))+L(g(t)) L(f/prime(t))=− f(0)+sF(s) L(f/prime/prime(t))=− f/prime(0)−sf(0)+s2F(s) L(f(n)(t))=− f(n−1)(0)−sf(n−2)(0)−···− sn−1f(0)+snF(s) L(ebtf(t))=F(s−b) L/parenleftBig/integraldisplayt 0f(t/prime)dt/prime/parenrightBig =1 sF(s)L/parenleftBig1 tf(t)/parenrightBig =/integraltext∞ sF(s/prime)ds/prime L(tf(t))=−dF ds Table 1 Properties of the Laplace transform or L/bracketleftbigg/integraldisplayt 0f(t/prime)dt/prime/bracketrightbigg =1 sL/parenleftbig f(t)/parenrightbig . (5) Differentiation and integration with respect to smay produce transforma- tions of previously inaccessible functions. We need the two formulas −de−st ds=te−st,/integraldisplay∞ se−s/primetds/prime=1 te−st to derive the results L/parenleftbig tf(t)/parenrightbig =−dF(s) ds,L/parenleftbigg1 tf(t)/parenrightbigg =/integraldisplay∞ sF(s/prime)ds/prime. (6) (Note that, unless f(0)=0, the transform of f(t)/twill not exist.) Examples of the use of these formulas are L/parenleftbig tsin(ωt)/parenrightbig =−d ds/parenleftbiggω s2+ω2/parenrightbigg =2sω (s2+ω2)2, L/parenleftbiggsin(t) t/parenrightbigg =/integraldisplay∞ sds/prime s/prime2+1=π 2−tan−1(s)=tan−1/parenleftbigg1 s/parenrightbigg . Signiﬁcant properties of the Laplace transform are summarized in Table 1. When a problem is solved by use of Laplace transforms, a prime difﬁculty is computation of the corresponding function of t. Methods for computing the “inverse transform” f(t)=L−1(F(s))include integration in the complex plane, convolution, partial fractions (discussed in Section 6.2), and tables of Chapter 6 Laplace Transform 367 f(t) F(s) f(t) F(s) 00 tkk! sk+1 11 sebtcos(ωt)s−b s2−2bs+b2+ω2 eat1 s−aebtsin(ωt)ω s2−2bs+b2+ω2 cosh(at)s s2−a2ebttkk! (s−b)k+1 sinh(at)a s2−a2eat−1a s(s−a) cos(ωt)s s2+ω2tcos(ωt)s2−ω2 (s2+ω2)2 sin(ωt)ω s2+ω2tsin(ωt)2sω (s2+ω2)2 t1 s2 Table 2 Laplace transforms transforms. The last method, which involves the least work, is the most popu- lar. The transforms in Table 2 were all calculated from the deﬁnition or by useof formulas in this section. EXERCISES 1.By using linearity and the transform of eat, compute the transform of each of the following functions. a.sinh(at); d.sin(ωt−φ);b.cos(ωt); e.e2(t+1);c.cos2(ωt); f.sin2(ωt). 2.Use differentiation with respect to tto ﬁnd the transform of a.teatfromL(eat), b.sin(ωt)fromL(cos(ωt)), c.cosh(at)fromL(sinh(at)). 3.Compute the transform of each of the following directly from the deﬁni- tion. a.f(t)=/braceleftBig0,0<t<a, 1,a<t; 368 Chapter 6 Laplace Transform b.f(t)=/braceleftBigg0,0<t<a, 1,a<t<b, 0,b<t; c.f(t)=/braceleftBigt,0<t<a, a,a<t. 4.The Heaviside step function is deﬁned by the formula Ha(t)=/braceleftBig1,t>a, 0,t<a. Assuming a≥0, show that the Laplace transform of Hais L/parenleftbig Ha(t)/parenrightbig =e−as s. 5.Use completion of square and the shifting theorem to ﬁnd the inverse trans- form of a.1 s2+2s, b.s+1 s2+2s+2, c.1 s2+2as+b2,b>a. 6.Find the Laplace transform of the square-wave function f(t)=/braceleftBig1,0<x<a, 0,a<x<2a,f(x+2a)=f(x). Hint: Break up the integral as shown in the following, evaluate the integrals, and add up a geometric series: F(s)=∞/summationdisplay n=0/integraldisplay2(n+1)a 2naf(t)e−stdt. 7.Use any method to ﬁnd the inverse transform of the following. a.1 (s−a)(s−b); c.s2 (s2+ω2)2;b.s (s2−a2)2; d.1 (s−a)3; e.1−e−s s. 8.Use any theorem or formula to ﬁnd the transform of the following. a.1−cos(ωt) t; c.t2e−at;b./integraldisplayt 0sin(at/prime) t/primedt/prime; d.tcos(ωt); e.sinh(at)sin(ωt). 9.Find the inverse transform of these functions of sby any method. 6.2 Partial Fractions and Convolutions 369 a.1 (s2+ω2)2; c.s2 (s2+ω2)2;b.s (s2+ω2)2; d.s3 (s2+ω2)2. 6.2 Partial Fractions and Convolutions Because of the formula for the transform of derivatives, the Laplace transform ﬁnds important application to linear differential equations with constant co- efﬁcients, subject to initial conditions. In order to solve the simple problem u/prime+au=0,u(0)=1, we transform the entire equation, obtaining L(u/prime)+aL(u)=0 or sU−1+aU=0, where U=L(u). The derivative has been “transformed out,” and Uis deter- mined by simple algebra to be U(s)=1 s+a. By consulting Table 2 we ﬁnd that u(t)=e−at. E q u a t i o n so fh i g h e ro r d e rc a nb es o l v e di nt h es a m ew a y .W h e nt r a n s - formed, the problem u/prime/prime+ω2u=0,u(0)=1,u/prime(0)=0 becomes s2U−s·1−0+ω2U=0. Note how both initial conditions have been incorporated into this one equa- tion. Now we solve the transformed equation algebraically to ﬁnd U(s)=s s2+ω2, the transform of cos (ωt)=u(t). 370 Chapter 6 Laplace Transform In general we may outline our procedure as follows: Original problem LSolution of original problem Transformed problemSolution of transformed problemL−1 In the step marked L−1, we must compute the function of tto which the solu- tion of the transformed problem corresponds. This is the difﬁcult part of the process. The key property of the inverse transform is its linearity, as expressed by L−1/parenleftbig c1F1(s)+c2F2(s)/parenrightbig =c1L−1/parenleftbig F1(s)/parenrightbig +c2L−1/parenleftbig F2(s)/parenrightbig . This property allows us to break down a complicated transform into a sum of simple ones. A simple mass–spring–damper system leads to the initial value problem u/prime/prime+au/prime+ω2u=0,u(0)=u0,u/prime(0)=u1, whose transform is s2U−su0−u1+a(sU−u0)+ω2U=0. Determination of Ugives it as the ratio of two polynomials: U(s)=su0+(u1+au0) s2+as+ω2. Although this expression is not in Table 2, it can be worked around to a func- tion of s+a/2, whose inverse transform is available. The shift theorem then gives u(t).T h e r ei s ,h o w e v e r ,ab e t t e rw a y . The inversion of a rational function of s(that is, the ratio of two polynomi- als) can be accomplished by the technique of “partial fractions.” Suppose wew i s ht oc o m p u t et h ei n v e r s et r a n s f o r mo f U(s)=cs+d s2+as+b. The denominator has two roots, r1and r2, which we assume for the moment to be distinct. Thus s2+as+b=(s−r1)(s−r2), U(s)=cs+d (s−r1)(s−r2). 6.2 Partial Fractions and Convolutions 371 The rules of elementary algebra suggest that Ucan be written as a sum, cs+d (s−r1)(s−r2)=A1 s−r1+A2 s−r2, (1) for some choice of A1and A2. Indeed, by ﬁnding the common denominator form for the right-hand side and matching powers of sin the numerator, we obtain cs+d (s−r1)(s−r2)=A1(s−r2)+A2(s−r1) (s−r1)(s−r2), c=A1+A2,d=− A1r2−A2r1. When A1and A2are determined, the inverse transform of the right-hand side of Eq. (1) is easily found: L−1/parenleftbiggA1 s−r1+A2 s−r2/parenrightbigg =A1exp(r1t)+A2exp(r2t). For a speciﬁc example, suppose that U(s)=s+4 s2+3s+2. The roots of the denominator are r1=− 1a n d r2=− 2. Thus s+4 s2+3s+2=A1 s+1+A2 s+2=(A1+A2)s+(2A1+A2) (s+1)(s+2). We ﬁnd A1=3,A2=− 2. Hence L−1/parenleftbiggs+4 s2+3s+2/parenrightbigg =L−1/parenleftbigg3 s+1−2 s+2/parenrightbigg =3e−t−2e−2t. A little calculus takes us much further. Suppose that Uhas the form U(s)=q(s) p(s), where pand qare polynomials and the degree of qis less than the degree of p. Assume that phas distinct roots r1,..., rk: p(s)=(s−r1)(s−r2)···(s−rk). We tr y to write Uin the fraction form U(s)=A1 s−r1+A2 s−r2+···+Ak s−rk=q(s) p(s). 372 Chapter 6 Laplace Transform The algebraic determination of the A’s is very tedious, but notice that (s−r1)q(s) p(s)=A1+A2s−r1 s−r2+···+ Aks−r1 s−rk. Ifsis set equal to r1, the right-hand side is just A1. The left-hand side becomes 0/0, but L’Hôpital’s rule gives lim s→r1(s−r1)q(s) p(s)=lim s→r1(s−r1)q/prime(s)+q(s) p/prime(s)=q(r1) p/prime(r1). Therefore A1and all the other A’s are given by Ai=q(ri) p/prime(ri). Consequently, our rational function takes the form q(s) p(s)=q(r1) p/prime(r1)1 s−r1+···+q(rk) p/prime(rk)1 s−rk. From this point we can easily obtain the inverse transform, as expressed in the conclusion of the following theorem. Theorem 1. Let p and q be polynomials, q of lower degree than p, and let p have only simple roots, r 1,r2,..., rk.T h e n L−1/parenleftbiggq(s) p(s)/parenrightbigg =q(r1) p/prime(r1)exp(r1t)+···+q(rk) p/prime(rk)exp(rkt). (2) (Equation (2)is known as Heaviside’s formula. ) /square Let us apply the theorem to the example in which q(s)=s+4,p(s)=s2+ 3s+2,p/prime(s)=2s+3. Then L−1/parenleftbiggs+4 s2+3s+2/parenrightbigg =−1+4 2(−1)+3e−t+−2+4 2(−2)+3e−2t=3e−t−2e−2t. In nonhomogeneous differential equations also, the Laplace transform is a useful tool. T o solve the problem u/prime+au=f(t), u(0)=u0, we again transform the entire equation, obtaining sU−u0+aU=F(s), U(s)=u0 s+a+1 s+aF(s). 6.2 Partial Fractions and Convolutions 373 The ﬁrst term in this expression is recognized as the transform of u0e−at.I f F(s)is a rational function, partial fractions may be used to invert the second term. However, we can identify that term by solving the problem another way.For example, e at(u/prime+au)=eatf(t), /parenleftbig ueat/parenrightbig/prime=eatf(t), ueat=/integraldisplayt 0eat/primef/parenleftbig t/prime/parenrightbig dt/prime+c, u(t)=/integraldisplayt 0e−a(t−t/prime)f/parenleftbig t/prime/parenrightbig dt/prime+ce−at. The initial condition requires that c=u0. On comparing the two results, we see that L/bracketleftbigg/integraldisplayt 0e−a(t−t/prime)f/parenleftbig t/prime/parenrightbig dt/prime/bracketrightbigg =1 s+aF(s). Thus the transform of the combination of e−atand f(t)on the left is the prod- uct of the transforms of e−atand f(t). This simple result can be generalized in the following way. Theorem 2. If g(t)and f(t)have Laplace transforms G (s)and F(s),r e s p e c t i v e l y , then L/bracketleftbigg/integraldisplayt 0g/parenleftbig t−t/prime/parenrightbig f/parenleftbig t/prime/parenrightbig dt/prime/bracketrightbigg =G(s)F(s). (3) (This is known as the convolution theorem. ) /square The integral on the left is called the convolution ofgand f, written g(t)∗f(t)=/integraldisplayt 0g/parenleftbig t−t/prime/parenrightbig f/parenleftbig t/prime/parenrightbig dt/prime. It can be shown that the convolution follows these rules: g∗f=f∗g, (4a) f∗(g∗h)=(f∗g)∗h, (4b) f∗(g+h)=f∗g+f∗h. (4c) The convolution theorem provides an important device for inverting Laplace transforms, which we shall apply to ﬁnd the general solution of thenonhomogeneous problem u /prime/prime−au=f(t), u(0)=u0,u/prime(0)=u1. 374 Chapter 6 Laplace Transform The transformed equation is readily solved, yielding U(s)=su0+u1 s2−a+1 s2−aF(s). Because 1 /(s2−a)is the transform of sinh (√at)/√a, we easily determine that uis u(t)=u0cosh/parenleftbig√at/parenrightbig +u1√asinh/parenleftbig√at/parenrightbig +/integraldisplayt 0sinh(√a(t−t/prime))√af/parenleftbig t/prime/parenrightbig dt/prime.(5) A slightly different problem occurs if the mass in a spring–mass system is struck while the system is in motion. The mathematical model of the systemmight be u /prime/prime+ω2u=f(t), u(0)=u0,u/prime(0)=u1, where f(t)=F0fort0<t<t1and f(t)=0 for other values. The transform of uis U(s)=su0+u1 s2+ω2+1 s2+ω2F(s). The inverse transform of U(s)is then u(t)=u0cos(ωt)+u1sin(ωt) ω+/integraldisplayt 0sin(ω(t−t/prime)) ωf/parenleftbig t/prime/parenrightbig dt/prime. The convolution in this case is easy to calculate: /integraldisplayt 0sin(ω(t−t/prime)) ωf/parenleftbig t/prime/parenrightbig dt/prime =  0, t<t 0, F01−cos(ω(t−t0)) ω2, t0<t<t1, F0cos(ω(t−t1))−cos(ω(t−t0)) ω2,t1<t. EXERCISES 1.Solve these initial value problems. a.u/prime−2u=0, u(0)=1; b.u/prime+2u=0, u(0)=1; c.u/prime/prime+4u/prime+3u=0, u(0)=1, u/prime(0)=0; d.u/prime/prime+9u=0, u(0)=0, u/prime(0)=1. 6.2 Partial Fractions and Convolutions 375 2.Solve the initial value problem u/prime/prime+2au/prime+u=0,u(0)=u0,u/prime(0)=u1 in these three cases: 0 <a<1,a=1,a>1. 3.Solve these nonhomogeneous problems with zero initial conditions. a.u/prime+au=1; c.u/prime/prime+4u=sin(t); e.u/prime/prime+2u/prime=1−e−t;b.u/prime/prime+u=t; d.u/prime/prime+4u=sin(2t); f.u/prime/prime−u=1. 4.Complete the square in the denominator and use the shift theorem [F(s−a)=L(eatf(t))]t oi n v e r t U(s)=su0+(u1+2au0) s2+2as+ω2. There are three cases, corresponding to ω2−a2>0,=0,< 0. 5.Use partial fractions to invert the following transforms. a.1 s2−4; c.(s+3) s(s2+2);b.1 s2+4; d.4 s(s+1). 6.Prove properties (4a) and (4c) of the convolution. 7.Compute the convolution f∗gfor a.f(t)=1,g(t)=sin(t); b.f(t)=et,g(t)=cos(ωt); c.f(t)=t,g(t)=sin(t). 8.Demonstrate the following properties of convolution either directly or by using Laplace transform. a.1∗f/prime(t)=f(t)−f(0); b.(t∗f(t))/prime/prime=f(t); c.(f∗g)/prime=f/prime∗g=f∗g/prime,i ff(0)=g(0)=0. 376 Chapter 6 Laplace Transform 6.3 Partial Differential Equations In applying the Laplace transform to partial differential equations, we treat variables other than tas parameters. Thus, the transform of a function u(x,t) is deﬁned by L/parenleftbig u(x,t)/parenrightbig =/integraldisplay∞ 0e−stu(x,t)dt=U(x,s). For instance, we easily ﬁnd the transforms L/parenleftbig e−atsin(πx)/parenrightbig =1 s+asin(πx), L/parenleftbig sin(x+t)/parenrightbig =ssin(x)+cos(x) s2+1. The transform Unaturally is a function not only of sbut also of the “untrans- formed” variable x. We assume that derivatives or integrals with respect to the untransformed variable pass through the transform L/parenleftbigg∂u ∂x/parenrightbigg =/integraldisplay∞ 0∂u(x,t) ∂xe−stdt =∂ ∂x/integraldisplay∞ 0u(x,t)e−stdt=∂ ∂x/parenleftbig U(x,s)/parenrightbig . I fw ew i s ht of o c u so nt h er o l eo f xas a variable and keep sin the background as a parameter, we might use the symbol for the ordinary derivative: L/parenleftbigg∂u ∂x/parenrightbigg =dU dx. The rule for transforming a derivative with respect to tcan be found, as before, with integration by parts: L/parenleftbigg∂u ∂t/parenrightbigg =sL/parenleftbig u(x,t)/parenrightbig −u(x,0). If the Laplace transform is applied to a boundary value–initial value prob- lem in xand t, all time derivatives disappear, leaving an ordinary differential equation in x. We shall illustrate this technique with some trivial examples. I n c i d e n t a l l y ,w ea s s u m ef r o mh e r eo nt h a tp r o b l e m sh a v eb e e np r e p a r e d( f o rexample, by dimensional analysis) so as to eliminate as many parameters aspossible. 6.3 Partial Differential Equations 377 Example 1. ∂2u ∂x2=∂u ∂t, 0<x<1,0<t, u(0,t)=1,u(1,t)=1,0<t, u(x,0)=1+sin(πx), 0<x<1. The partial differential equation and the boundary conditions (that is, every- thing that is valid for t>0) are transformed, while the initial condition is incorporated by the transform d2U dx2=sU−/parenleftbig 1+sin(πx)/parenrightbig ,0<x<1, U(0,s)=1 s,U(1,s)=1 s. This boundary value problem is solved to obtain U(x,s)=1 s+sin(πx) s+π2. We direct our attention now to Uas a function of s. Because sin (πx)is a con- stant with respect to s,t a b l e sm a yb eu s e dt oﬁ n d u(x,t)=1+sin(πx)exp/parenleftbig −π2t/parenrightbig . /square Example 2. ∂2u ∂x2=∂2u ∂t2, 0<x<1,0<t, u(0,t)=0,u(1,t)=0,0<t, u(x,0)=sin(πx), 0<x<1, ∂u ∂t(x,0)=− sin(πx), 0<x<1. Under transformation the problem becomes d2U dx2=s2U−ssin(πx)+sin(πx), 0<x<1, U(0,s)=0,U(1,s)=0. The function Uis found to be U(x,s)=s−1 s2+π2sin(πx), 378 Chapter 6 Laplace Transform from which we ﬁnd the solution, u(x,t)=/parenleftbigg cos(πt)−1 πsin(πt)/parenrightbigg sin(πx). /square Example 3. Now we consider a problem that we know to have a more complicated solu- tion: ∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=1,u(1,t)=1,0<t, u(x,0)=0,0<x<1. The transformed problem is d2U dx2=sU, 0<x<1, U(0,s)=1 s,U(1,s)=1 s. The general solution of the differentia le q u a t i o ni sw e l lk n o w nt ob eac o m b i - nation of sinh (√sx)and cosh (√sx). Application of the boundary conditions yields U(x,s)=1 scosh/parenleftbig√sx/parenrightbig +(1−cosh(√s))sinh(√sx) ssinh(√s) =sinh(√sx)+sinh(√s(1−x)) ssinh(√s). This function rarely appears in a table of transforms. However, by extending the Heaviside formula, we can compute an inverse transform. When Uis the ratio of two transcendental functions (not polynomials) of s, we wish to write U(x,s)=/summationdisplay An(x)1 s−rn. In this formula, the numbers rnare values of sfor which the “denominator” of Uis zero, or, rather, for which |U(x,s)|becomes inﬁnite; the Anare functions ofxbut not s.F r o mt h i sf o r mw ee x p e c tt od e t e r m i n e u(x,t)=/summationdisplay An(x)exp(rnt). This solution should be checked for convergence. The hyperbolic sine (also the cosh, cos, sin, and exponential functions) is not inﬁnite for any ﬁnite value of its argument. Thus U(x,s)becomes inﬁnite 6.3 Partial Differential Equations 379 only where sor sinh (√s)is zero. Because sinh (√s)=0 has no real root besides zero, we seek complex roots by setting√s=ξ+iη(ξandηreal). The addition rules for hyperbolic and trigonometric functions remain valid for complex arguments. Furthermore, we know that cosh(iA)=cos(A), sinh(iA)=isin(A). By combining the addition rule and these identities we ﬁnd sinh(ξ+iη)=sinh(ξ)cos(η)+icosh(ξ)sin(η). This function is zero only if both the real and imaginary parts are zero. Thus ξ andηmust be chosen to satisfy simultaneously sinh(ξ)cos(η)=0,cosh(ξ)sin(η)=0. Of the four possible combinations, only sinh (ξ)=0 and sin (η)=0p r o d u c e solutions. Therefore ξ=0a n dη=± nπ(n=0,1,2,...); whence √s=± inπ, s=− n2π2. Recall that only the value of s,n o tt h ev a l u eo f√s, is signiﬁcant. Finally then, we have located r0=0, and rn=− n2π2(n=1,2,...). We proceed to ﬁnd the An(x)by the same method used in Section 6.2. The com- putations are done piecemeal and then the solution is assembled. Part a. (r0=0.) In order to ﬁnd A0we multiply both sides of our proposed partial fractions development U(x,s)=∞/summationdisplay n=0An(x)1 s−rn bys−r0=sand take the limit as sapproaches r0=0. The right-hand side goes to A0. On the left-hand side we have lim s→0ssinh(√sx)+sinh(√s(1−x)) ssinh(√s)=x+1−x=1=A0(x). Thus the part of u(x,t)corresponding to s=0i s1·e0t=1, which is easily recognized as the steady-state solution. Part b. (rn=− n2π2,n=1,2,....) For these cases, we ﬁnd An=q(rn) p/prime(rn), 380 Chapter 6 Laplace Transform where qandpare the obvious choices. We take√rn=+ inπin all calculations: p/prime(s)=sinh/parenleftbig√s/parenrightbig +s1 2√scosh/parenleftbig√s/parenrightbig , p/prime(rn)=1 2inπcosh(inπ)=1 2inπcos(nπ), q(rn)=sinh(inπx)+sinh/parenleftbig inπ(1−x)/parenrightbig =i/bracketleftbig sin(nπx)+sin/parenleftbig nπ(1−x)/parenrightbig/bracketrightbig . Hence the portion of u(x,t)that arises from each rnis An(x)exp(rnt)=2sin(nπx)+sin(nπ(1−x)) nπcos(nπ)exp/parenleftbig −n2π2t/parenrightbig . Part c. On assembling the various pieces of the solution, we get u(x,t)=1+2 π∞/summationdisplay 1sin(nπx)+sin(nπ(1−x)) ncos(nπ)exp/parenleftbig −n2π2t/parenrightbig . The same solution would be found by separation of variables but in a slightly different form. /square Example 4. Now consider the wave problem ∂2u ∂x2=∂2u ∂t2,0<x<1, 0<t, u(0,t)=0,∂u(1,t) ∂x=0,0<t, u(x,0)=0,∂u(x,0) ∂t=x,0<x<1. The transformed problem is d2U dx2=s2U−x,0<x<1, U(0,s)=0,U/prime(1,s)=0, and its solution (by undetermined coefﬁcients or otherwise) gives U(x,s)=sxcosh(s)−sinh(sx) s3cosh(s). The numerator of this function is never inﬁnite. The denominator is zero at s=0a n d s=± i(2n−1)π/2(n=1,2,...). We shall again use the Heaviside formula to determine the inverse transform of U. 6.3 Partial Differential Equations 381 Part a. (r0=0.) The limit as sapproaches zero of sU(x,s)may be found by L’Hôpital’s rule or by using the Taylor series for sinh and cosh. From the latter, sU(x,s)=sx/parenleftbig 1+s2 2+···/parenrightbig −/parenleftbig sx+s3x3 6+···/parenrightbig s2/parenleftbig 1+s2 2+···/parenrightbig =s3/parenleftbigx 2−x3 6−···/parenrightbig s2/parenleftbig 1+s2 2+···/parenrightbig→0. Thus, in spite of the formidable appearance of s3in the denominator, s=0i s not really a signiﬁcant value and contributes nothing to u(x,t). Part b. It is convenient to take the remaining roots in pairs. We label ±i(2n−1)π 2=± iρn. The derivative of the denominator is p/prime(s)=3s2cosh(s)+s3sinh(s), p/prime(±iρn)=± i3ρ3 nsinh(±iρn) =ρ3 nsin(ρn) since sinh (iρ)=isin(ρ)and(±i)4=1. The contribution of these two roots together may be calculated using the exponential deﬁnition of sine: q(iρn) p/prime(iρn)exp(iρnt)+q(−iρn) r/prime(−iρn)exp(−iρnt) =−sinh(iρnx)exp(iρnt)+sinh(iρnx)exp(−iρnt) ρ3nsin(ρn) =sin(ρnx) ρ3nsin(ρn)i/parenleftbig −exp(iρnt)+exp(−iρnt)/parenrightbig =2s i n(ρnx)sin(ρnt) ρ3nsin(ρn). Part c. The ﬁnal form of u(x,t), found by adding up all the contributions from Part b, is the same as would be found by separation of variables u(x,t)=2∞/summationdisplay 1sin(ρnx)sin(ρnt) ρ3nsin(ρn). /square 382 Chapter 6 Laplace Transform EXERCISES 1.Find all values of s, real and complex, for which the following functions are zero. a.cosh(√s); c.sinh(s); e.cosh(s)+ssinh(s).b.cosh(s); d.cosh(s)−ssinh(s); 2.Find the inverse transforms of the following functions in terms of an inﬁ- nite series. a.1 stanh(s); b.sinh(sx) scosh(s). 3.Find the transform U(x,s)of the solution of each of the following prob- lems. a.∂2u ∂x2=∂u ∂t,0<x<1,0<t, u(0,t)=0,u(1,t)=t,0<t, u(x,0)=0,0<x<1; b.∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=e−t,0<t, u(x,0)=1,0<x<1. 4.Solve each of the problems in Exercise 3, inverting the transform by means of the extended Heaviside formula. 5.Solve each of the following problems by Laplace transform methods. a.∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=1,0<t, u(x,0)=0,0<x<1; b.∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=0,0<t, u(x,0)=1,0<x<1. 6.4 More Difﬁcult Examples 383 6.4 More Difﬁcult Examples The technique of separation of variables, once mastered, seems more straight- forward than the Laplace transform. However, when time-dependent bound-ary conditions or inhomogeneities are present, the Laplace transform offersa distinct advantage. Following are some examples that display the power of transform methods. Example 1. A uniform insulated rod is attached at one end to an insulated container of ﬂuid. The ﬂuid is circulated so well that its temperature is uniform and equalto that at the end of the rod. The other end of the rod is maintained at a con-stant temperature. A dimensionless initial value–boundary value problem thatdescribes the temperature in the rod is ∂ 2u ∂x2=∂u ∂t, 0<x<1, 0<t, ∂u(0,t) ∂x=γ∂u(0,t) ∂t,u(1,t)=1,0<t, u(x,0)=0, 0<x<1. The transformed problem and its solution are d2U dx2=sU,0<x<1, dU dx(0,s)=sγU(0,s), U(1,s)=1 s, U(x,s)=cosh(√sx)+√sγsinh(√sx) s(cosh(√s)+√sγsinh(√s))=q(s) p(s). Aside from s=0, the denominator has no real zeros. Thus we again search for complex zeros by employing√s=ξ+iη. The real and imaginary parts of the denominator are to be computed b y using the addition formulas for cosh and sinh. The requirement that both real and imaginary parts be zero leads tothe equations /parenleftbig cosh(ξ)+ξγsinh(ξ)/parenrightbig cos(η)−ηγcosh(ξ)sin(η)=0, (1) ηγsinh(ξ)cos(η)+/parenleftbig sinh(ξ)+ξγcosh(ξ)/parenrightbig sin(η)=0. (2) We may think of these as simultaneous equations in sin (η)and cos (η).B e c a u s e sin 2(η)+cos2(η)=1, the system has a solution only when its determinant is zero. Thus after some algebra, we arrive at the condition /parenleftbig 1+ξ2γ2+η2γ2/parenrightbig sinh(ξ)cosh(ξ)+ξγ/parenleftbig sinh2(ξ)+cosh2(ξ)/parenrightbig =0. 384 Chapter 6 Laplace Transform The only solution occurs when ξ=0, for otherwise both terms of this equa- tion have the same sign. After setting ξ=0, we ﬁnd that Eq. (1) reduces to tan(η)=1 ηγ, for which there is an inﬁnite number of solutions. We shall number the posi- tive solutions η1,η2,.... Now, we have found that the roots of p(s)arer0=0 and rk=(iηk)2=−η2 k. The computation of the inverse transform follows. Part a. (r0=0.) The limit of sU(x,s)asst e n d st oz e r oi se a s i l yf o u n dt ob e 1. Thus this root contributes 1 ·e0t=1t o u(x,t). Part b. (rk=−η2 k.) First, we compute p/prime(s)=cosh/parenleftbig√s/parenrightbig +√sγsinh/parenleftbig√s/parenrightbig +1 2√s(1+γ)sinh/parenleftbig√s/parenrightbig +1 2γscosh/parenleftbig√s/parenrightbig . Using the fact that cosh (√rk)+√rkγsinh(√rk)=0, we may reduce the fore- going to p/prime(rk)=−1 2γ/parenleftbig 1+γ+η2 kγ2/parenrightbig cos(ηk). Hence the contribution to u(x,t)ofrkis q(rk) p/prime(rk)exp(rkt)=− 2γcos(ηkx)−ηkγsin(ηkx) (1+γ+η2 kγ2)cos(ηk)exp/parenleftbig −η2 kt/parenrightbig . Part c. The construction of the ﬁnal solution is left to the reader. We note that an attempt to solve this problem by separation of variables would ﬁnddifﬁculties, for the eigenfunctions are notorthogonal. /square Example 2. Sometimes one is interested not in the complete solution of a problem, but only in part of it. For example, in the problem of heat conduction in a semi- inﬁnite solid with time-varying boundary conditions, we may seek that part of the solution that persists after a long time. (This may or may not be a steady-state solution.) Any initial condition that is bounded in xgives rise only to transient temperatures; these being of no interest, we assume a zero initial con-dition. Thus, the problem to be studied is ∂ 2u ∂x2=∂u ∂t, 0<x<∞,0<t, u(0,t)=f(t), 0<t, u(x,0)=0, 0<x<∞. 6.4 More Difﬁcult Examples 385 The transformed equation and its general solution are d2U dx2=sU,0<x,U(0,s)=F(s), U(x,s)=Aexp/parenleftbig −√sx/parenrightbig +Bexp/parenleftbig√sx/parenrightbig . We make two further assumptions about the solution: ﬁrst, that u(x,t)is bounded as xtends to inﬁnity and, second, that√smeans the square root ofsthat has a nonnegative real part. Under these two assumptions, we must choose B=0a n d A=F(s),m a k i n g U(x,s)=F(s)exp/parenleftbig −√sx/parenrightbig . In order to ﬁnd the persistent part of u(x,t), we apply the Heaviside inver- sion formula to those values of shaving nonnegative real parts, because a value ofswith negative real part corresponds to a function containing a decaying exponential — a transient. T o understand this fact, consider the pair f(t)=1−e−βt+αsin(ωt), F(s)=1 s−1 s+β+αω s2+ω2. Now we return to the original problem with this choice for f(t).T h ev a l u e s ofsfor which U(x,s)=F(s)exp(−√sx)becomes inﬁnite are 0, ±iω,−β.T h e last value is discarded, because it is negative. Thus, the persistent part of thesolution is given by A 0e0t+A1eiωt+A2e−iωt, and the coefﬁcients are found from A0=lim s→0/bracketleftbig sF(s)exp/parenleftbig −√sx/parenrightbig/bracketrightbig =1, A1=lim s→iω/bracketleftbig (s−iω)F(s)exp/parenleftbig −√sx/parenrightbig/bracketrightbig =α 2iexp/parenleftbig −√ iωx/parenrightbig , A2=lim s→− iω/bracketleftbig (s+iω)F(s)exp/parenleftbig −√sx/parenrightbig/bracketrightbig =−α 2iexp/parenleftbig −√ −iωx/parenrightbig . We also need to know that the roots of ±iwith positive real part are √ i=1√ 2(1+i),√ −i=1√ 2(1−i). Thus the function we seek is 386 Chapter 6 Laplace Transform 1+1 2iexp/bracketleftbigg iωt−/radicalbiggω 2(1+i)x/bracketrightbigg −α 2iexp/bracketleftbigg −iωt−/radicalbiggω 2(1−i)x/bracketrightbigg =1+αexp/parenleftbigg −/radicalbiggω 2x/parenrightbigg sin/parenleftbigg ωt−/radicalbiggω 2x/parenrightbigg . /square Example 3. If a steel wire is exposed to a sinusoidal magnetic ﬁeld, the boundary value– initial value problem that describes its displacement is ∂2u ∂x2=∂2u ∂t2−sin(ωt), 0<x<1, 0<t, u(0,t)=0, u(1,t)=0, 0<t, u(x,0)=0,∂u ∂t(x,0)=0,0<x<1. The nonhomogeneity in the partial diffe rential equation represents the effect of the force due to the ﬁeld. The transformed equation and its solution are d2U dx2=s2U−ω s2+ω2,0<x<1, U(0,s)=0,U(1,s)=0, U(x,s)=ω s2(s2+ω2)cosh(1 2s)−cosh/parenleftbig s(1 2−x)/parenrightbig cosh(1 2s). Several methods are available for the inverse transformation of U.A no b v i - ous one would be to compute v(x,t)=L−1/parenleftbiggcosh(1 2s)−cosh/parenleftbig s(1 2−x)/parenrightbig s2cosh(1 2s)/parenrightbigg and write u(x,t)as a convolution u(x,t)=/integraldisplayt 0sin/parenleftbig ω/parenleftbig t−t/prime/parenrightbig/parenrightbig v/parenleftbig x,t/prime/parenrightbig dt/prime. The details of this development are left as an exercise. We could also use the Heaviside formula. The application is now rou- tine, except in the interesting case where cosh (iω/2)=0, that is, where ω= (2n−1)π, one of the natural frequencies of the wire. Let us suppose ω=π,s o U(x,s)=π s2(s2+π2)cosh(1 2s)−cosh/parenleftbig s(1 2−x)/parenrightbig cosh(1 2s). 6.4 More Difﬁcult Examples 387 At the points s=0,s=± iπ,s=±(2n−1)iπ,n=2,3,...,U(x,s)becomes undeﬁned. The computation of the parts of the inverse transform correspond-ing to the points other than ±iπis easily carried out. However, at these two troublesome points, our usual procedure will not work. Instead of expecting a partial-fraction decomposition containing A −1 s+iπ+A1 s−iπ and other terms of the same sort, we must seek terms like A−1(s+iπ)+B−1 (s+iπ)2+A1(s−iπ)+B1 (s−iπ)2, whose contribution to the inverse transform of Uwould be A−1e−iπt+B−1te−iπt+A1eiπt+B1teiπt. One can compute A1and B1, for example, by noting that B1=lim s→iπ/bracketleftbig (s−iπ)2U(x,s)/bracketrightbig , A1=lim s→iπ/braceleftbigg (s−iπ)/bracketleftbigg U(x,s)−B1 (s−iπ)2/bracketrightbigg/bracerightbigg and similarly for A−1andB−1. The limit for B1is not too difﬁcult. For example, B1=lim s→iπ/braceleftbiggπ s2(s+iπ)cosh(1 2s)−cosh/parenleftbig s(1 2−x)/parenrightbig /parenleftbig cosh(1 2s)/parenrightbig /(s−iπ)/bracerightbigg =π −π2(2iπ)cosh(1 2iπ)−cosh/parenleftbig iπ(1 2−x)/parenrightbig 1 2sinh(1 2iπ) =−1 π2cos/parenleftbigg π/parenleftbigg1 2−x/parenrightbigg/parenrightbigg =−1 π2sin(πx). The limit for A1is rather more complicated but may be computed by L’Hôpital’s rule. Nevertheless, because B−1=B1,w ea l r e a d ys e et h a t u(x,t) contains the term B1teiπt+B−1te−iπt=−2t π2sin(πx)cos(πt), whose amplitude increases with time. This, of course, is the expected reso- nance phenomenon. /square 388 Chapter 6 Laplace Transform EXERCISES 1.Find the persistent part of the solution of the heat problem ∂2u ∂x2=∂u ∂t, 0<x<1, 0<t, ∂u ∂x(0,t)=0,∂u ∂x(1,t)=1,0<t, u(x,0)=0, 0<x<1. 2.Verify that the persistent part of the solution to Example 2 actually satisﬁes the heat equation. What boundary condition does it satisfy? 3.Find the function v(x,t)whose transform is cosh(1 2s)−cosh/parenleftbig s(1 2−x)/parenrightbig s2cosh(1 2s). What boundary value–initial value problem does v(x,t)satisfy? 4.Solve ∂2u ∂x2=∂2u ∂t2,0<x<1, 0<t, u(0,t)=0,u(1,t)=0, 0<t, u(x,0)=0,∂u ∂t(x,0)=1,0<x<1. 5. a. Solve for ω/negationslash=π: ∂2u ∂x2=∂2u ∂t2−sin(πx)sin(ωt), 0<x<1,0<t, u(0,t)=0,u(1,t)=0, 0<t, u(x,0)=0,∂u ∂t(x,0)=0, 0<x<1. b.Examine the special case ω=π. 6.Obtain the complete solution of Example 1 and verify that it satisﬁes the boundary conditions and the heat equation. 6.5 Comments and References 389 7. a. Solve ∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=1−e−at,0<t, u(x,0)=0,0<x<1. b.Examine the special case where a=n2π2for some integer n. 6.5 Comments and References The real development of the Laplace transform began in the late nineteenth century, when engineer Oliver Heaviside invented a powerful, but unjustiﬁed,symbolic method for studying the ordinary and partial differential equations of mathematical physics. By the 1920s, Heaviside’s method had been legitima- tized and recast as the Laplace transform that we now use. Later generalizationsare Schwartz’s theory of distributions (1940s) and Mikusinski’s operationalcalculus (1950s). The former seems to be the more general. Both theories givean interpretation of F(s)=1, which is not the Laplace transform of any func- tion, in the sense we use. There are a number of other transforms, under the names of Fourier, Mellin, Hänkel, and others, similar in intent to the Laplace transform, in which some other function replaces e −stin the deﬁning integral. Operational Mathematics , by Churchill, has more information about the applications of transforms. Ex-tensive tables of transforms will be found in Tables of Integral Transforms by Erdelyi et al. (See the Bibliography.) Miscellaneous Exercises 1.Solve the heat conduction problem ∂2u ∂x2−γ2(u−T)=∂u ∂t,0<x<1, 0<t, ∂u ∂x(0,t)=0,∂u ∂x(1,t)=0,0<t, u(x,0)=T0, 0<x<1. 390 Chapter 6 Laplace Transform 2.Find the “persistent part” of the solution of ∂2u ∂x2=∂u ∂t, 0<x<1,0<t, ∂u ∂x(0,t)=0,u(1,t)=t,0<t, u(x,0)=0, 0<x<1. 3.Find the complete solution of the problem in Exercise 2. 4.A solid object and a surrounding ﬂuid exchange heat by convection. The temperatures u1and u2are governed by the following equations. Solve them by means of Laplace transforms. du1 dt=−β1(u1−u2), du2 dt=−β2(u2−u1), u1(0)=1,u2(0)=0. 5.Solve the following nonhomogeneous problem with transforms. ∂2u ∂x2=∂u ∂t−1,0<x<1, 0<t, u(0,t)=0, u(1,t)=0,0<t, u(x,0)=0, 0<x<1. 6.Find the transform of the solution of the problem ∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=1,0<t, u(x,0)=0,0<x<1. 7.Find the solution of the problem in Exercise 6 by using the extended Heaviside formula. 8.Solve the heat problem ∂2u ∂x2=∂u ∂t, 0<x,0<t, u(0,t)=0, 0<t, u(x,0)=sin(x),0<x. Miscellaneous Exercises 391 9.Find the transform of the solution of ∂2u ∂x2=∂u ∂t,0<x,0<t, u(0,t)=0,0<t, u(x,0)=1,0<x, u(x,t)bounded as x→∞. 10.At the end of Section 2.12, the problem in Exercise 9 was solved by other means. Use this fact to identify 1 s/parenleftbig 1−e−√sx/parenrightbig =L/bracketleftbigg erf/parenleftbiggx√ 4t/parenrightbigg/bracketrightbigg and 1 se−√sx=L/bracketleftbigg 1−erf/parenleftbiggx√ 4t/parenrightbigg/bracketrightbigg . (The latter function is called the complementary error function ,d e ﬁ n e d by erfc (q)≡1−erf(q).) 11.Find the function of twhose Laplace transform is F(s)=e−x√s. 12.Using the deﬁnition of sinh in terms of exponentials and a geometric series, show that sinh(√sx) sinh(√s)=∞/summationdisplay n=0/parenleftbig e−√s(2n+1−x)−e−√s(2n+1+x)/parenrightbig . 13.Use the series in Exercise 12 to ﬁnd a solution of the problem in Exer- cise 6 in terms of complementary error functions. 14.Show the following relation by using Exercise 11 and differentiating with respect to s. L/bracketleftbigg1√πtexp/parenleftbigg−k2 4t/parenrightbigg/bracketrightbigg =1√se−k√s. 15.Find the Laplace transform of the odd periodic extension of the function f(t)=π−t,0<t<π , by transforming its Fourier series term by term. 392 Chapter 6 Laplace Transform 16. Suppose that the function f(t)is periodic with period 2 a. Show that the Laplace transform of fis given by the formula F(s)=G(s) 1−e−2as, where G(s)=/integraldisplay2a 0f(t)e−stdt. (Hint: See Section 6.1, Exercise 6.) 17. Apply the extended Heaviside method to the inversion of a transform with the form F(s)=G(s) 1−e−2as, where G(s)does not become inﬁnite for any value of s. 18. Show that for a periodic function f(t)the quantities cn=1 2aG/parenleftbigginπ a/parenrightbigg (G(s)is deﬁned in Exercise 16) are the complex Fourier coefﬁcients. 19. How is it possible to determine that a Laplace transform F(s)corre- sponds to a periodic f(t)? 20. Is this function the transform of a periodic function? F(s)=1 s2+a2. 21. Use the method of Exercise 16 to ﬁnd the transform of the periodic ex- tension of f(t)=/braceleftBig1, 0<t<π, −1,π < t<2π. 22. Same as Exercise 21, but use the function of Exercise 15. 23. Use the method of Exercise 16 to ﬁnd the transform of f(t)=/vextendsingle/vextendsinglesin(t)/vextendsingle/vextendsingle. 24. Find the transform of the solution of the problem Miscellaneous Exercises 393 ∂2u ∂x2=∂2u ∂t2, 0<x,0<t, u(0,t)=h(t), 0<t, u(x,0)=0,∂u ∂t(x,0)=0,0<x, u(x,t)bounded as x→∞. Use the solution of the same problem as found in Section 3.6, to verify the rule L−1/parenleftbig e−sxH(s)/parenrightbig =/braceleftbigg h(t−x),t>x, 0, t<x. 25.Solve this wave problem with time-varying boundary condition, assum- ingω/negationslash=nπ,n=1,2,.... ∂2u ∂x2=∂2u ∂t2,0<x<1, 0<t, u(0,t)=0,u(1,t)=sin(ωt),0<t, u(x,0)=0,∂u ∂t(x,0)=0, 0<x<1. 26.S o l v et h ep r o b l e mi nE x e r c i s e2 5i nt h es p e c i a lc a s e ω=π. 27.Certain techniques for growing a crystal from a solution or a melt may cause striations — variations in the concentration of impurities. AuthorsR.T. Gray, M.F. Larrousse, and W.R. Wilcox [Diffusional decay of stria-tions, Journal of Crystal Growth ,92(1988): 530–542] use a material bal- ance on a slice of a cylindrical ingot to derive this boundary value prob-lem for the impurity concentration, C: ∂ ∂x/parenleftbigg D(x)∂C ∂x/parenrightbigg −V∂C ∂x=∂C ∂t,0<x,0<t, C(0,t)=Ca+Asin/parenleftbigg2πt tC/parenrightbigg , 0<t, C(x,0)=Ca, 0<x. Here, Vis the crystal growth rate, tCis the striation period, Cais the average concentration in the solid, and D(x)is the diffusivity of the im- purity at distance xfrom the growth face (which is located at x=0). Of course, C(x,t)is bounded as x→∞ . 394 Chapter 6 Laplace Transform Next, the equations are made dimensionless by introducing new vari- ables: ¯C=C−Ca A,¯x=Vx D(0),¯t=V2t D(0). The new problem is ∂ ∂¯x/parenleftbiggD(¯x) D(0)∂¯C ∂¯x/parenrightbigg −∂¯C ∂¯x=∂¯C ∂¯t,0<¯x,0<¯t, ¯C/parenleftbig 0,¯t/parenrightbig =sin/parenleftbig ω¯t/parenrightbig , 0<¯t, ¯C(¯x,0)=0, 0<¯x, where ω=2πD(0)/V2tC. Because D(¯x)depends in a complicated way on ¯x,an u m e r i c a ls o l u t i o n was used. T o check the numerical solution, the authors wished to ﬁnd ananalytical solution of the problem corresponding to constant diffusivity,D(¯x)=D(0).L e t ube the solution of ∂ 2u ∂¯x2=∂u ∂¯x+∂u ∂¯t,0<¯x,0<¯t, u/parenleftbig 0,¯t/parenrightbig =sin/parenleftbig ω¯t/parenrightbig ,0<¯t, u(¯x,0)=0, 0<¯x, ubounded as x→∞. Find the Laplace transform of the solution of this problem. 28. The authors of the paper mentioned in Exercise 27 were particularly in- terested in the persistent part of the solution. Use the methods of Sec-tion 6.4 to show that the persistent part of the solution is u 1=1 2i/parenleftbig f(iω)−f(−iω)/parenrightbig , where f(iω)=exp/parenleftBigg/parenleftBigg 1 2−/radicalbigg 1 4+iω/parenrightBigg ¯x+iω¯t/parenrightBigg . 29. Find the square root required in the foregoing expression by setting /radicalbigg 1 4+iω=α+iβ Miscellaneous Exercises 395 so that α2−β2+2iαβ=1 4+iω, or/braceleftBigg α2−β2=1 4, 2αβ=ω. (T o solve these equations: (i) solve the second for β; (ii) substitute the expression found into the ﬁrst; (iii) solve the resulting biquadratic for α.) 30.Noting that the persistent part is u1=Im(f(iω)) (see Exercise 28), de- termine u1/parenleftbig ¯x,¯t/parenrightbig =e(1 2−α)¯xsin/parenleftbig ω¯t−β¯x/parenrightbig , where αandβa r ea si nE x e r c i s e2 9 . This page intentionally left blank Numerical Methods CHAPTER7 7.1 Boundary Value Problems More often than not, signiﬁcant practical problems in partial — and even or- dinary — differential equations cannot be solved by analytical methods. Dif-ﬁculties may arise from variable coefﬁcients, irregular regions, unsuitableboundary conditions, interfaces, or just overwhelming detail. Now that ma-chine computation is cheap and easily accessible, numerical methods provide reliable answers to formerly difﬁcult problems. In this chapter we examine a few methods that are simple and equally adaptable to machine or manualcomputation. Implementation of some of these methods with a spreadsheetprogram is explained and carried out on the CD. If we cannot ﬁnd a simple analytic formula for the solution of a boundary value problem, we may be satisﬁed with a table of (approximate) values of thesolution. For instance, the solution of the problem d 2u dx2−12xu=− 1,0<x<1, (1) u(0)=1, u(1)=− 1, (2) may be written out in terms of Airy functions, but the values of ushown in Table 1 are more informative for most of us. One way to obtain such a tableis to replace the original analytical problem by an arithmetical problem, as described in what follows. 397 398 Chapter 7 Numerical Methods x:0 .00.20 .40 .60 .81 .0 u(x):1.00.643 0 .302−0.026−0.406−1.0 Table 1 Approximate solution of Eqs. (1) and (2) Differential equation Boundary condition u(x)→ui u(0)→u0 d2u dx2(x)→ui+1−2ui+ui−1 (/Delta1x)2du dx(0)→u1−u−1 2/Delta1x du dx(x)→ui+1−ui−1 2/Delta1xu(1)→un f(x)→f(xi)du dx(1)→un+1−un−1 2/Delta1x Table 2 Constructing replacement equations First, the values of xfor the table will be uniformly spaced across the interval 0≤x≤1, which we assume to be the interval of the boundary value problem xi=i/Delta1x,/Delta1 x=1 n. These are called meshpoints . The numbers approximating the values of uare ui∼=u(xi), i=0,1,..., n. These numbers are required to satisfy a set of equations obtained from the boundary value problem by making the replacements shown in Table 2. The entry f(x)refers to any coefﬁcient or inhomogeneity in the differential equa- tion. Example. The boundary value problem in Eqs. (1) and (2) would be replaced by the algebraic equations ui+1−2ui+ui−1 (/Delta1x)2−12xiui=− 1,i=1,2,..., n−1, (3) u0=1,un=− 1. (4) Equation (3) holds for i=1,..., n−1, so the unknowns u1,..., un−1would be determined by this set of equations. The equations become speciﬁc when wechoose n.L e tu st a k e n=5, so/Delta1x=1/5, and the four (i=1,2,3,4)versions Chapter 7 Numerical Methods 399 of Eq. (3) are 25(u2−2u1+u0)−12 5u1=− 1, 25(u3−2u2+u1)−24 5u2=− 1, 25(u4−2u3+u2)−36 5u3=− 1, 25(u5−2u4+u3)−48 5u4=− 1.(5) When we use the boundary conditions u0=1,u5=− 1( 6 ) and collect coefﬁcients, the foregoing equations become −52.4u1+ 25u2 =− 26 25u1−54.8u2+ 25u3 =− 1 25u2−57.2u3+ 25u4=− 1 25u3−59.6u4=24.(7) This system of four simultaneous equations can be solved manually by elimi- nation or by software. The result will be a set of numbers giving the approxi-mate values of uat the points x 1=0.2,..., x4=0.8. The numbers in Table 1 were obtained by a similar process, but using n=100 instead of n=5. /square Example. T o see how to handle derivative boundary conditions, we solve the problem d2u dx2−10u=f(x), 0<x<1, (8) u(0)=1,du dx(1)=− 1, (9) f(x)=  0, 0<x<1 2, −50, x=1 2, −100,1 2<x<1. The replacement equations for this problem are easily obtained by using Ta- ble 2. They are ui+1−2ui+ui−1 (/Delta1x)2−10ui=f(xi), (10) u0=1,un+1−un−1 2/Delta1x=− 1. (11) 400 Chapter 7 Numerical Methods We need to know u0,u1,..., un. The derivative boundary condition at x=1 forces us to include un+1among the unknowns, so we will need to use Eq. (10) fori=1,2,..., nin order to have enough equations to ﬁnd all the unknowns. Since we have no use for un+1, the usual practice is to solve the boundary- condition replacement for un+1, un+1=un−1−2/Delta1x, (12) and then to use this expression in the version of Eq. (10) that corresponds to i=n. Thus, the equation un+1−2un+un−1 (/Delta1x)2−10un=f(xn) is combined with Eq. (12) to get 2un−1−2/Delta1x−2un (/Delta1x)2−10un=f(xn). (13) Then Eq. (10) for i=1,..., n−1 and Eq. (13) give nequations that determine unknowns u1,u2,..., un. To b e s p e c i ﬁ c , l e t u s t a ke n=4, so/Delta1x=1/4. The three (i=1,2,3)versions of Eq. (10) are 16(u2−2u1+u0)−10u1=0 (i=1), 16(u3−2u2+u1)−10u2=− 50(i=2), 16(u4−2u3+u2)−10u3=− 100(i=3) and Eq. (13) adapted to n=4i s 16/parenleftbigg 2u3−1 2−2u4/parenrightbigg −10u4=− 100. When these equations are cleaned up and the boundary condition u0=1i s applied, the result is the following system of four equations: −42u1+16u2 =− 16 16u1−42u2+16u3 =− 50 16u2−42u3+16u4=− 100 32u3−42u4=− 92.(14) In Table 3 are shown the values of uiobtained by solving Eq. (14) and also more exact values found by using n=100. /square Elimination is not the only way to get the solution of a system like Eqs. (7) or (14). An alternative is an iterative m ethod, which generates a sequence of approximate solutions. For one such method, we solve algebraically the ith Chapter 7 Numerical Methods 401 x:0 0 .25 0 .50 .75 1 u(n=4): 1 2 .174 4 .707 7 .057 7 .567 u(n=100): 1 2 .155 4 .729 7 .125 7 .629 Table 3 Approximate solution of Eqs. (8) and (9) equation for the ith unknown. In the resulting set of equations there are “cir- cular references”: The equation for u2refers to u1and u3, while the equations for these refer to u2, etc. We may start with some guessed values for the u’s, feed them through the equations to get improved values for the u’s, and repeat the process until the values settle down. This method requires a lot of arith-metic but no strategy, while elimination is just the reverse. It may also workwith nonlinear equations, where elimination cannot. So far, we have given no justiﬁcation for the procedure of constructing re- placement equations. The explanation is not difﬁcult; it depends on the fact that certain difference quotients approximate derivatives. If u(x)is a function with several derivatives, then u(x i+1)−u(xi−1) 2/Delta1x=u/prime(xi)+(/Delta1x)2 6u(3)(¯xi), (15) u(xi+1)−2u(xi)+u(xi−1) (/Delta1x)2=u/prime/prime(xi)+(/Delta1x)2 12u(4)(¯¯xi), (16) where ¯xiand¯¯xiare points near xi. Now suppose that u(x)is the solution of the boundary value problem d2u dx2+k(x)du dx+p(x)u(x)=f(x), 0<x<1, (17) αu(0)−α/primeu/prime(0)=a,β u(1)+β/primeu/prime(1)=b. (18) Ifu(x)has enough derivatives, then at any point xi=i/Delta1xit satisﬁes the dif- ferential equation (17) and thus also satisﬁes the equation u(xi+1)−2u(xi)+u(xi−1) (/Delta1x)2+k(xi)u(xi+1)−u(xi−1) 2/Delta1x+p(xi)u(xi)=f(xi)+δi, (19) where δi=(/Delta1x)2 12u(4)(¯¯xi)+k(xi)(/Delta1x)2 6u(3)(¯xi). Because δiis proportional to (/Delta1x)2, it is very small when /Delta1xis small. The replacement equation for Eq. (17) is, according to Table 2, ui+1−2ui+ui−1 (/Delta1x)2+k(xi)ui+1−ui−1 2/Delta1x+p(xi)ui=f(xi). (20) 402 Chapter 7 Numerical Methods Thus, the values of uatx0,x1,..., xn, which satisfy Eq. (19) exactly, will nearly satisfy Eq. (20); vice versa, the numbers u0,u1,..., un, which satisfy the re- placement equations (20), nearly satisfy Eq. (19). It can be proved that thecalculated numbers u 0,u1,..., undo indeed approach the appropriate values ofu(xi)as/Delta1xapproaches 0 (under continuity and other conditions on k(x), p(x),f(x)). EXERCISES 1.Set up and solve replacement equations with n=4 for the problem d2u dx2=− 1,0<x<1, u(0)=0, u(1)=1. 2.Solve the problem of Exercise 1 analytically. On the basis of Eqs. (15) and (16), explain why the numerical solution agrees exactly with the ana- lytical solution. 3.Set up and solve replacement equations with n=4 for the problem d2u dx2−u=− 2x,0<x<1, u(0)=0, u(1)=1. 4.Solve the problem in Exercise 3 analytically, and compare the numerical results with the true solution. 5.Set up and solve replacement equations with n=4 for the problem d2u dx2=x, 0<x<1, u(0)−du dx(0)=1,u(1)=0. 6.Solve the problem in Exercise 5 analytically, and compare the numerical results with the true solution. 7.Set up and solve replacement equations for the problem d2u dx2+10u=0,0<x<1, u(0)=0, u(1)=− 1. 7.2 Heat Problems 403 Use n=3a n d n=4. Sketch the results and explain why they vary so much. In Exercises 8–11, set up and solve replacement equations for the problemstated and the given value of n. If a computer is available, also solve for ntwice as large, and compare results. 8.d2u dx2−32xu=0, 0 <x<1, u(0)=0, u(1)=1( n=4). 9.d2u dx2−25u=− 25, 0 <x<1, u(0)=2, u(1)+u/prime(1)=1( n=5). 10.d2u dx2+1 1+xdu dx=− 1, 0<x<1, u(0)=0, u(1)=0( n=3). 11.d2u dx2+du dx−u=− x, du dx(0)=0, u(1)=1( n=3). 12.Use the Taylor series expansion u(x+h)=u(x)+hu/prime(x)+h2 2u/prime/prime(x)+h3 6u(3)(x)+h4 24u(4)(x)+··· with x=xiand h=±/Delta1x(xi+/Delta1x=xi+1,xi−/Delta1x=xi−1)t oo b t a i nr e p - resentations similar to Eqs. (15) and (16). 7.2 Heat Problems In heat problems, we have two independent variables xand t,a s s u m e dt ob e in the range 0 <x<1, 0<t. A table for a function u(x,t)should give values at equally spaced points and times, xi=i/Delta1x,tm=m/Delta1t, fori=0,1,..., nand m=0,1,....H e r e , /Delta1x=1/n, as before. We will use a subscript to denote position and a number in parentheses to denote the timelevel for the approximation to the solution of a problem. That is, u i(m)∼=u(xi,tm). 404 Chapter 7 Numerical Methods The spatial derivatives in a heat problem will be replaced by difference quo- tients, as before: ∂2u ∂x2(xi,tm)→ui+1(m)−2ui(m)+ui−1(m) (/Delta1x)2, (1) ∂u ∂x(xi,tm)→ui+1(m)−ui−1(m) 2/Delta1x. (2) For the time derivative, there are several possible replacements. We limit our- selves to the forward difference ∂u ∂t(xi,tm)→ui(m+1)−ui(m) /Delta1t, (3) which will yield explicit formulas for computing. Now, to solve numerically the simple heat problem ∂2u ∂x2=∂u ∂t, 0<x<1, 0<t, (4) u(0,t)=0, u(1,t)=0,0<t, (5) u(x,0)=f(x), 0<x<1, (6) we set up replacement equations according to Eqs. (1)–(3). Those equations are ui−1(m)−2ui(m)+ui+1(m) (/Delta1x)2=ui(m+1)−ui(m) /Delta1t, (7) supposed valid for i=1,2,..., n−1a n d m=0,1,2,.... The point of using a forward difference for the time derivative is that these equations may be solved for ui(m+1): ui(m+1)=rui−1(m)+(1−2r)ui(m)+rui+1(m), (8) where r=/Delta1t/(/Delta1x)2.T h u se a c h ui(m+1)is calculated from u’s at the preced- ing time level. Because the initial condition gives each ui(0),t h ev a l u e so ft h e u’s at time level 1 can be calculated by setting m=0 in Eq. (8): ui(1)=rui−1(0)+(1−2r)ui(0)+rui+1(0). Then the values of the u’s at time level 2 can be found from these, and so on into the future. Of course, rhas to be given a numerical value ﬁrst, by choosing /Delta1xand/Delta1t. It is convenient to display the numerical values of ui(m)in a table, making columns correspond to different meshpoints x0,x1,..., xnand making rows correspond to the different time levels t0,t1,....S e eT a b l e4 . 7.2 Heat Problems 405 i m 0 123 4 0 00 .25 0 .50 .75 1 1 0 0.25 0 .50 .75 0 2 0 0.25 0 .50 .25 0 3 0 0.25 0 .25 0 .25 0 4 0 0.125 0 .25 0 .125 0 5 0 0.125 0 .125 0 .125 0 Table 4 Numerical solution of Eqs. (4)–(6) Example. Solve Eqs. (4)–(6) with /Delta1x=1/4a n d r=1/2, making /Delta1t=1/32. The equa- tions giving the u’s at time level m+1a r e u1(m+1)=1 2/parenleftbig u0(m)+u2(m)/parenrightbig , u2(m+1)=1 2/parenleftbig u1(m)+u3(m)/parenrightbig , u3(m+1)=1 2/parenleftbig u2(m)+u4(m)/parenrightbig .(9) Recall that the boundary conditions of this problem specify u0(m)=0a n d u4(m)=0 for m=1,2,3,.... Thus we ﬁll in the columns of the table that correspond to points x0and x4with 0’s (shown in italics in Table 4). Also the initial condition speciﬁes ui(0)=f(xi),s ot h et o pr o wo ft h et a b l ec a nb e ﬁlled. In this example we take f(x)=x, and the corresponding values appear in italics in the top row of Table 4. The initial condition, u(x,0)=x,0<x<1, suggests that u(1,0)should be 1, while the boundary condition suggests that it should be 0. In fact, nei- ther condition speciﬁes u(1,0), nor is there a hard and fast rule telling what to do in case of conﬂict. Fortunately, it does not matter much, either. (See Exer- cise 1.) /square Stability T h ec h o i c ew em a d eo f r=1/2 in the Example seems natural, perhaps, because it simpliﬁes the computation. It might also seem desirable to take a larger value ofr(signifying a larger time step) to get into the future more rapidly. For example, with r=1(/Delta1t=1/16)the replacement equations take the form ui(m+1)=ui−1(m)−ui(m)+ui+1(m). In Table 5 are values of ui(m)computed from this formula. No one can believe that these wildly ﬂuctuating values approximate the solution to the heat prob- 406 Chapter 7 Numerical Methods i m 01 2 34 0 00 .25 0 .50 0 .75 1 1 0 0.25 0 .50 0 .75 0 2 0 0.25 0 .50−0.25 0 3 0 0.25−0.50 0 .75 0 4 0−0.75 1 .50−1.25 0 5 0 2.25−3.50 2 .75 0 Table 5 Unstable solution lem in any sense. Indeed, they suffer from numerical instability due to using a time step too long relative to the mesh size. The analysis of instability requiresfamiliarity with matrix theory, but there are some simple rules of thumb thatguarantee stability. First, write out the equations for each u i(m+1): ui(m+1)=aiui−1(m)+biui(m)+ciui+1(m). The coefﬁcients must satisfy two conditions 1. No coefﬁcient may be negative. 2. The sum of the coefﬁcients is not greater than 1. In the example, the replacement equations were u1(m+1)=ru0(m)+(1−2r)u1(m)+ru2(m), u2(m+1)=ru1(m)+(1−2r)u2(m)+ru3(m), u3(m+1)=ru2(m)+(1−2r)u3(m)+ru4(m). The second requirement is satisﬁed automatically, because r+(1−2r)+r=1. But the ﬁrst condition is satisﬁed only for r≤1/2. Thus the ﬁrst choice of r=1/2 corresponded to the longest stable time step. Example. Different problems give different maximum values for r. For the heat conduc- tion problem ∂2u ∂x2=∂u ∂t,0<x<1, 0<t, (10) u(0,t)=1,∂u ∂x(1,t)+γu(1,t)=0,0<t, (11) u(x,0)=0,0<x<1 (12) 7.2 Heat Problems 407 the replacement equations are found to be (for n=4) u1(m+1)=ru0(m)+(1−2r)u1(m)+ru2(m), u2(m+1)=ru1(m)+(1−2r)u2(m)+ru3(m), u3(m+1)=ru2(m)+(1−2r)u3(m)+ru4(m), u4(m+1)=2ru3(m)+/parenleftbig 1−2r−1 2rγ/parenrightbig u4(m).(13) (Remember that u(1,t), corresponding to u4, is an unknown. The boundary condition has been incorporated into the equation for u4(m+1).) Again, the second stability requirement is satisﬁ ed automatically; but the ﬁrst rule re- quires that 1−2r−1 2rγ≥0o r r≤1 2+1 2γ. (14) /square EXERCISES 1.Solve Eqs. (4)–(6) numerically with f(x)=x,a si nt h et e x t (/Delta1x=1/4, r=1/2), but take u4(0)=0. Compare your results with Table 4. 2.Solve Eqs. (4)–(6) numerically with f(x)=x,/Delta1x=1/4,u4(0)=1, as in the text, but use r=1/4. Compare your results with Table 4. Be sure to compare results at corresponding times. 3.For the problem in Eqs. (10)–(12), ﬁnd the longest stable time step when γ=1, and compute the numerical solution with the corresponding value ofr. 4.Solve the problem in Eqs. (10)–(12) with /Delta1x=1/4,r=1/2a n dγ=0, for mup to 5. For each problem in the following exercises, set up the replacement equationsforn=4, compute the longest stable time step, and calculate the numerical solution for a few values of m. 5.∂2u ∂x2=∂u ∂t, u(0,t)=u(1,t)=t,u(x,0)=0. 6.∂2u ∂x2−u=∂u ∂t,u(0,t)=u(1,t)=1, u(x,0)=0. 7.∂2u ∂x2=∂u ∂t−1, u(0,t)=u(1,t)=0, u(x,0)=0. 408 Chapter 7 Numerical Methods 8.∂2u ∂x2=∂u ∂t,u(0,t)=0,∂u ∂x(1,t)+u(1,t)=1, u(x,0)=0. 9.∂2u ∂x2=∂u ∂t,∂u ∂x(0,t)=0, u(1,t)=1, u(x,0)=x. 7.3 Wave Equation The simple vibrating string problem we studied in Chapter 3, ∂2u ∂x2=∂2u ∂t2, 0<x<1, 0<t, (1) u(0,t)=0, u(1,t)=0, 0<t, (2) u(x,0)=f(x),∂u ∂t(x,0)=g(x), 0<x<1, (3) rarely needs treatment by numerical methods, because the d’Alembert solu- tion provides a simple and direct means of calculating the solution u(x,t)for arbitrary xand t. However, if the partial differential equation contains uor an inhomogeneity or if the boundary conditions are more complex, a series solu- tion or a solution of the d’Alembert type may not be practical. In many suchcases, simple numerical techniques are quite rewarding. In order to convert the wave equation (1) into a suitable difference equa- tion, we ﬁrst designate points x i=i/Delta1x(/Delta1x=1/n)and times tm=m/Delta1tfor which the approximation to uwill be found: u(xi,tm)∼=ui(m). Then the par- tial derivatives with respect to both xand tare replaced by central differences: ∂2u ∂x2→ui+1(m)−2ui(m)+ui−1(m) (/Delta1x)2, ∂2u ∂t2→ui(m+1)−2ui(m)+ui(m−1) (/Delta1t)2. The wave equation (1) becomes this partial difference equation ui+1(m)−2ui(m)+ui−1(m) (/Delta1x)2=ui(m+1)−2ui(m)+ui(m−1) (/Delta1t)2 or, with ρ=/Delta1t//Delta1x, ui(m+1)−2ui(m)+ui(m−1)=ρ2/parenleftbig ui+1(m)−2ui(m)+ui−1(m)/parenrightbig . The replacement equations may be solved for the unknowns ui(m+1), yielding the equation ui(m+1)=ρ2ui−1(m)+2(1−ρ2)ui(m)+ρ2ui+1(m)−ui(m−1), (4) 7.3 Wave Equation 409 valid for i=1,2,..., n−1. Naturally, the boundary conditions, Eq. (2), carry over as u0(m)=0,un(m)=0. It is obvious that Eq. (4) requires us to know the approximate solution at time levels mandm−1i no r d e rt oﬁ n di ta tt i m el e v e l m+1. In other words, to get ui(1)we need ui−1(0),ui(0),ui+1(0)—w h i c ha r e available from the initial condition — and also ui(−1)!O fc o u r s e ,w eh a v en o t yet applied the second initial condition, ∂u ∂t(x,0)=g(x), 0<x<1. If we replace the time derivative by a central difference approximation, this equation translates into ui(1)−ui(−1) 2/Delta1t=g(xi) (5) fori=1,2,..., n−1. Equation (5), together with a slightly modiﬁed version of Eq. (4) (with m=0a n d ui(0)=f(xi)), yields the system ui(1)+ui(−1)=ρ2f(xi−1)+2/parenleftbig 1−ρ2/parenrightbig f(xi)+ρ2f(xi+1), ui(1)−ui(−1)=2/Delta1tg(xi),(6) which we can easily solve for the u’s at the ﬁrst time level: ui(1)=1 2ρ2f(xi−1)+/parenleftbig 1−ρ2/parenrightbig f(xi)+1 2ρ2f(xi+1)+/Delta1tg(xi). (7) T h u s ,i no r d e rt os o l v et h ep r o b l e mi nE q s .( 1 ) – ( 3 )n u m e r i c a l l y ,w eu s et h e initial condition, ui(0)=f(xi), to ﬁll the ﬁrst line of our table, use the starting equation (7) to ﬁll the next line, and continue with the running equation (4) to ﬁll subsequent lines. Example. Let us now attempt to solve a simple problem. Suppose that g(x)≡0 for 0 < x<1a n dt h a t f(x)is given by f(x)=/braceleftbigg2x, 0<x<1 2, 2(1−x),1 2<x<1.(8) Also, we shall choose n=4a n d ρ=1 for convenience. (That is, /Delta1t=/Delta1x= 1/4.) Our rule for calculation, Eq. (4), is then ui(m+1)=ui−1(m)+ui+1(m)−ui(m−1). (9) In Table 6 are the calculated values of ui(m). Entries in italics are given data. It is easy to check that this numerical solution is identical with the d’Alembertsolution of this particular problem. (See Exercise 6.) However, if the initial 410 Chapter 7 Numerical Methods i m 0 123 4 0 00 .51 0 .50 1 0 0.50 .50 .5 0 2 0 000 0 3 0−0.5−0.5−0.5 0 4 0−0.5−1 −0.5 0 5 0−0.5−0.50 .5 0 6 0 000 0 Table 6 Numerical solution of Eqs. (1)–(3) velocity were not identically zero, the numerical solution would in general be only an approximation to the true solution. /square Stability In our study of the heat equation, Section 7.2, we saw that the choice of /Delta1xand /Delta1twas not free. The same is true for the wave equation. Suppose we attempt to solve the same problem as earlier, but with ρ2=(/Delta1t//Delta1x)2chosen to be 2. Then Eq. (4) becomes ui(m+1)=2/parenleftbig ui−1(m)−ui(m)+ui+1(m)/parenrightbig −ui(m−1), and the “solution” corresponding to this rule of calculation is shown in Table 7 (again, entries in italics are given data). Of course, the results bear no resem-blance to the solution of the wave equation. They suffer from the same sort of instability as that observed in Section 7.2. There is a rule of thumb, similar to t h eo n et ob ef o u n dt h e r e ,a p p l i c a b l et ot h ew a v ee q u a t i o n . First, write out the equations for each u i(m+1)in terms of the u’s at time levels mand m−1: ui(m+1)=aiui−1(m)+biui(m)+ciui+1(m)−ui(m−1). The coefﬁcients must satisfy two conditions: 1. None of the coefﬁcients ai,bi,cimay be negative. 2. The sum of the coefﬁcients is not greater than 2: ai+bi+ci≤2. Of course, ui(m−1)appears with a coefﬁcient of −1; nothing can be done about that, nor does it enter into the aforementioned rules. 7.3 Wave Equation 411 i m 0 123 4 0 00 .51 0 .50 1 0 0.50 0 .5 0 2 0−1.51 −1.5 0 3 0 4.5−84 .5 0 4 0−23.53 3 −23.5 0 Table 7 Unstable numerical solution In Eq. (4) we see that both conditions are met when ρ=/Delta1t//Delta1xis less than or equal to 1; in other words, the time step must not exceed the space step.However, using ρ 2=1 when acceptable often provides the best accuracy. We conclude with one more example, illustrating how numerical results can be obtained easily in some cases that might be puzzling analytically. Example. S u p p o s ew ea r et os o l v et h ep r o b l e m ∂2u ∂x2=∂2u ∂t2−16 cos(πt), 0<x<1, 0<t, (10) u(0,t)=0, u(1,t)=0, 0<t, (11) u(x,0)=0,∂u ∂t(x,0)=0,0<x<1. (12) We replace the partial derivatives as before, obtaining ui+1(m)−2ui(m)+ui−1(m) (/Delta1x)2 =ui(m+1)−2ui(m)+ui(m−1) (/Delta1t)2−16 cos(πtm). When this is solved for ui(m+1),w eﬁ n d ui(m+1)=(2−2ρ2)ui(m)+ρ2ui+1(m)+ρ2ui−1(m) −ui(m−1)+16(/Delta1t)2cos(πm/Delta1t). (13) Let us take /Delta1x=/Delta1t=1/4a g a i n ,s o ρ=1 and Eq. (13) simpliﬁes to ui(m+1)=ui+1(m)+ui−1(m)−ui(m−1)+cos/parenleftbiggmπ 4/parenrightbigg . (14) This is our running equation. The starting equation comes from combining Eqs. (14) for m=0, ui(1)=− ui(−1)+1 412 Chapter 7 Numerical Methods i m 0 123 4 0 0 000 0 1 0 0.50 .50 .5 0 2 0 1.21 1 .71 1 .21 0 3 0 1.21 1 .91 1 .21 0 4 0 0.00 0 .00 0 .00 0 5 0−2.21 −2.91 −2.21 0 6 0−3.62 −5.12 −3.62 0 7 0−2.91 −4.33 −2.91 0 Table 8 Numerical solution of Eqs. (10)–(12) (note ui(0)=0), with the replacement initial condition ui(1)−ui(−1) 2/Delta1t=0, or ui(1)=ui(−1)=1 2 fori=1,2,3. Now we have the top two lines of Table 8, and the rest are ﬁlled using Eqs. (14) (with cos (π/4)/similarequal0.71, and so forth). Entries in italics are given data. T h ec o m p l e t ea n a l y t i c a ls o l u t i o no ft h i sp r o b l e mi s u(x,t)=32 π2tsin(πt)sin(πx) +32 π3∞/summationdisplay n=31−cos(nπ) n(n2−1)/parenleftbig cos(πt)−cos(nπt)/parenrightbig sin(nπx). Atx=1/2, the sum of the inﬁnite series is 0, so u/parenleftbigg1 2,t/parenrightbigg =32 π2tsin(πt). Comparison of the values of this function at times tmwith the middle column of Table 8 shows the numerical solution off by a few percent. Note that the growth in u(x,t)is due to resonance in the physical system, not to numerical instability. /square EXERCISES 1.Obtain an approximate solution of Eqs. (1), (2), and (3) with f(x)≡0a n d g(x)≡1. Take /Delta1x=1/4,ρ=1. 7.3 Wave Equation 413 2.Compare the results of Exercise 1 with the d’Alembert solution. 3.Obtain an approximate solution of Eqs. (1), (2), and (3) with f(x)≡0a n d g(x)=sin(πx).T a k e /Delta1x=1/4,ρ=1. 4.Compare the results of Exercise 3 with the exact solution u(x,t)= (1/π)sin(πx)sin(πt). 5.Obtain an approximate solution of Eqs. (1), (2), and (3) with g(x)≡0a n d f(x)a si nE q .( 8 ) .U s e /Delta1x=1/4a n dρ2=1/2. 6.Compare the entries of Table 6 with the d’Alembert solution. 7.Obtain an approximate solution of this problem with a time-varying boundary condition, using /Delta1x=/Delta1t=1/4. ∂2u ∂x2=∂2u ∂t2,0<x<1, 0<t, u(0,t)=0,u(1,t)=h(t), 0<t, u(x,0)=0,∂u ∂t(x,0)=0,0<x<1, h(t)=/braceleftBig1, 0<t<1, −1,1<t<2 and h(t+2)=h(t),h(0)=h(1)=0. 8.Same task as Exercise 7 but h(t)=sin(πt). Use sin (π/4)∼=0.7 instead of√ 2/2. 9.Find starting and running equations for the following problem. Using /Delta1x=1/4, ﬁnd the longest stable time step and compute values of the approximate solution for mup to 8. ∂2u ∂x2=∂2u ∂t2+16u,0<x<1, 0<t, u(0,t)=0, u(1,t)=0, 0<t, u(x,0)=f(x),∂u ∂t(x,0)=0,0<x<1, where f(x)is given in Eq. (8). 10.Using /Delta1x=1/4a n d ρ2=1/2, compare the numerical solution of the problem in Exercise 9 with and without the 16 uterm in the partial differ- ential equation. 414 Chapter 7 Numerical Methods 7.4 Potential Equation In this section, we will be concerned with approximate solutions of the po- tential equation and related equations in a region Rof the xy-plane. For the sake of simplicity, we will limit ourselves to regions whose boundaries can bemade to coincide with the lines on a sheet of graph paper with square divi-sions. Thus, we admit such shapes as rectangles, L’s and T’s, but not circles or triangles. The graph paper provides us with a ready-made mesh of points in the region Rand on its boundary, at which we wish to know the solution of o u rp r o b l e m .T h e s ep o i n t sa r et ob en u m b e r e di ns o m ef a s h i o n—u s u a l l yl e f tto right and bottom to top. On such a mesh, the replacement for the Laplacian operator is the following: ∂ 2u ∂x2+∂2u ∂y2→uW−2ui+uE (/Delta1x)2+uN−2ui+uS (/Delta1y)2, (1) where the subscripts E,Wstand for the indices of the mesh points to the left and right of point iand the subscripts N,Sstand for those above and below (see Fig. 1). The result is sometimes called the ﬁve-point approximation to the Laplacian . Because we are assuming that /Delta1x=/Delta1y,w eo b t a i naf u r t h e rs i m - pliﬁcation in the replacement: ∂2u ∂x2+∂2u ∂y2→uN+uS+uE+uW−4ui (/Delta1x)2. (2) Example. Solve this problem numerically (see Chapter 4 for the analytical solution): ∂2u ∂x2+∂2u ∂y2=0,0<x<1, 0<y<1, (3) u(0,y)=0, u(1,y)=0, 0<y<1, (4) u(x,0)=f(x), u(x,1)=f(x), 0<x<1, (5) f(x)=/braceleftbigg2x, 0<x<1 2, 2(1−x),1 2≤x<1.(6) Let us take /Delta1x=/Delta1y=1/4 and number the mesh points inside the 1 ×1 square as shown in Fig. 2. At each of the nine mesh points, we will have the replacement equation uN+uS+uE+uW−4ui=0. (7) T ogether, these make up a system of nine equations in the nine unknowns u1,u2,..., u9. Referring to Fig. 2, where the values of uat boundary points are 7.4 Potential Equation 415 Figure 1 Point ion a square mesh and its four neighbors. Figure 2 Numbering for mesh points, and values on boundary. shown, we can write down the equations to be solved: u2+u4+1 2−4u1=0 u1+u3+u5+1−4u2=0 u2+u6+1 2−4u3=0 u1+u5+u7−4u4=0 u2+u4+u6+u8−4u5=0 u3+u5+u9−4u6=0 u4+u8+1 2−4u7=0 u5+u7+u9+1−4u8=0 u6+u8+1 2−4u9=0.(8) This is simply a system of simultaneous equations. It can be solved by elim- ination to obtain the results shown in Fig. 3. In this particular case, there arenumerous symmetries in the problem, so u 1=u3=u7=u9,u2=u8,a n d u4=u6. Thus, only u1,u2,u4,a n d u5need to be found. The system can be reduced to four equations in these four unknowns, which can even be solved manually. /square 416 Chapter 7 Numerical Methods Figure 3 Numerical solution of Eqs. (3)–(6). Example. Set up the replacement equations for the problem ∂2u ∂x2+∂2u ∂y2=16(u−1), 0<x<1, 0<y<1, (9) u(x,0)=0, u(x,1)=0,0<x<1, (10) u(0,y)=0, u(1,y)=0,0<y<1. (11) We may use the same numbering as in the ﬁrst example (Fig. 2). At each mesh point, the replacement is uN+uS+uE+uW−4ui (/Delta1x)2=16(ui−1). (12) Because /Delta1x=1/4,(1//Delta1x)2=16, and the typical replacement equation be- comes uN+uS+uE+uW−4ui=ui−1, or uN+uS+uE+uW−5ui=− 1. (13) Finally, we may write out the equations to be solved. The ﬁrst four of the nine equations, corresponding to Eq. (13) with i=1,2,3,4, are u2+u4−5u1=− 1 u1+u3+u5−5u2=− 1 u2+u6−5u3=− 1 u1+u5+u7−5u4=− 1.(14) The solution of this problem is left as an exercise. /square On more complicated regions, the replacement for the Laplacian operator h a se x a c t l yt h es a m ef o r m ,s i n c ew es t i l lu s et h e“ g r a p h - p a p e rm e s h . ”T h es y s - 7.4 Potential Equation 417 Figure 4 Mesh numbering for L-shaped region. tem of equations to be solved will be rather less regular than that for a rectan- gle. Example. Consider the problem ∂2u ∂x2+∂2u ∂y2=− 16 in R, (15) u=0 on the boundary of R, (16) where Ris an L-shaped region formed from a 1 ×1s q u a r eb yr e m o v i n ga1 /4× 1/4 square from the upper right corner. The general replacement equation is uN+uS+uE+uW−4ui=− 1. (17) With the numbering shown in Fig. 4, the eight equations to be solved are u2+u4−4u1=− 1 u1+u3+u5−4u2=− 1 u2+u6−4u3=− 1 u1+u5+u7−4u4=− 1 u2+u4+u6+u8−4u5=− 1 u3+u5−4u6=− 1 u4+u8−4u7=− 1 u5+u7−4u8=− 1.(18) The results, rounded to three digits, are shown in Eq. (19). Note the equalities, which arise from symmetries in the problem: u1=0.656 u2=u4=0.813 u3=u7=0.616 u5=0.981 u6=u8=0.649.(19) /square 418 Chapter 7 Numerical Methods Iterative Methods Systems of up to 10 equations, such as those in the foregoing examples, can readily be solved by elimination. It is easy to see, however, that we might wellwant a ﬁner mesh to get better accuracy and that a ﬁner mesh will increase thenumber of equations dramatically. For example, if we use /Delta1x=/Delta1y=1/10 in a numerical solution of Eqs. (3)–(5), the system to be solved contains 81 unknowns (or 25 if we use symmetry). Problems involving many thousands of unknowns are quite common. These large systems of simultaneous equationsare almost always solved by iterative methods , which generate a sequence of approximate solutions. Consider again the potential problem in Eqs. (3)–(6). Let us take a mesh with/Delta1x=/Delta1y=1/Nand number the points of the mesh with a double index so that u(x i,yj)∼=ui,j. (20) Then the replacement equations for the potential equation are ui+1,j−2ui,j+ui−1,j (/Delta1x)2+ui,j+1−2ui,j+ui,j−1 (/Delta1y)2=0, or, using /Delta1x=/Delta1yand some algebra, ui,j=1 4(ui+1,j+ui−1,j+ui,j+1+ui,j−1), (21) valid for iand jranging from 1 to N−1. (This is the same as Eq. (7).) The boundary conditions, Eqs. (4) and (5), determine u0,j=0, uN,j=0, j=0,..., N, (22) ui,0=f(xi), ui,N=f(xi), i=0,..., N. (23) The simplest iterative method, called the Gauss–Seidel method, works this way. We sweep through the array of u’s, replacing each ui,jby the combination ofu’s given on the right-hand side of Eq. (21). After several sweeps through the array, the numbers no longer change much. When the new and old valuesofu i,jat each point agree closely enough, we stop. The result is a set of numbers that satisfy Eq. (21) approximately. Since the exact solution of the replacement equat ions is still just an approximation to the solution of the original problem in Eqs. (3)–(6), it is not urgent to get thatexact solution of the replacement equations. An iterative method such as the Gauss–Seidel method is very easy to imple- ment on a spreadsheet without programming. (See the CD.) 7.4 Potential Equation 419 Figure 5 Regions and mesh numbering for Exercises 5–9. EXERCISES Set up and solve replacement equations for each of the following problems. Use symmetry to reduce the number of unknowns. 1.∇2u=− 1, 0<x<1, 0<y<1,u=0 on the boundary. /Delta1x=/Delta1y=1/4. 2.Same as Exercise 1 with /Delta1x=/Delta1y=1/8. Compare the solutions. 3.∇2u=0, 0<x<1, 0<y<1,u(0,y)=0,u(x,0)=0,u(1,y)=y, u(x,1)=x./Delta1x=/Delta1y=1/4. 4.Same as Exercise 3 with /Delta1x=/Delta1y=1/8. 5.The region Ris a square of side 1 from the center of which a similar square of side 1 /7h a sb e e nr e m o v e d ; ∇2u=0i nR,u=0 on the outside bound- ary, and u=1 on the inside boundary; /Delta1x=/Delta1y=1/7. See Fig. 5. 6.Same as Exercise 5, but the partial differential equation is ∇2u=− 1, and the boundary condition is u=0 on all boundaries. See Fig. 5. 7.The region Rhas the shape of a T, made by removing strips from the cor- ners of a 1 ×1 square. The partial differential equation is ∇2u=− 25 inR, andu=0 on the boundary. Take /Delta1x=/Delta1y=1/5. See Fig. 5 for numbering of mesh points. 8.The region is a rectangle, 2 units wide and 1 unit high. The potential equa-t i o nh o l d si nt h ei n t e r i o r ; u=1 on the upper half of the boundary (the top 420 Chapter 7 Numerical Methods and the upper halves of the vertical sides), and u=0 on the lower half. Take /Delta1x=/Delta1y=1/3. See Fig. 5. 9.The region, as seen in Fig. 5, is shaped like an upside-down Uand is formed by removing a small (1×2)rectangle from the bottom of a larger (5×4) one. In the interior of the region, ∇2u=0. The boundary conditions are: u=1 on the left and right sides and the top of the rectangle; u=0o nt h e bottom and on the boundary formed by the removal of the small rectangle.Use/Delta1x=/Delta1y=1. 7.5 Two-Dimensional Problems Separation of variables and other analytical methods produce satisfactory so- lutions to two-dimensional problems in only the nicest cases. However, simplenumerical methods work quite well on two-dimensional problems. In this ele-mentary exposition, we will limit ourselves to the heat and wave equations on two-dimensional regions that “ﬁt on graph paper,” as in Section 7.4. We will compute an approximation to the solution of a problem, denot- ing space position with one or two subscripts and time level with an index inparentheses. Both heat and wave problems will require the replacement of theLaplacian operator. We use the same replacement as in Section 7.4, ∂ 2u ∂x2+∂2u ∂y2→uE(m)−2ui(m)+uW(m) (/Delta1x)2+uN(m)−2ui(m)+uS(m) (/Delta1y)2. Because we are using a square mesh, with /Delta1x=/Delta1y, the replacement simpliﬁes to ∂2u ∂x2+∂2u ∂y2→uN(m)+uS(m)+uE(m)+uW(m)−4ui(m) (/Delta1x)2, (1) where N,S,E,Wstand for the indices of the four grid points adjacent to the point with the index i. Heat Problems Now let us consider this heat problem on a rectangle: ∂2u ∂x2+∂2u ∂y2=∂u ∂t,0<x<1.25, 0<y<1, 0<t, (2) u(0,y,t)=0, u(1.25,y,t)=0,0<y<1, 0<t, (3) u(x,0,t)=0, u(x,1,t)=0, 0<x<1.25,0<t, (4) u(x,y,0)=1, 0<x<1.25, 0<y<1. (5) 7.5 Two-Dimensional Problems 421 Figure 6 Mesh numbering for numerical solution of Eqs. (2)–(5). We take /Delta1x=/Delta1y=1/4 and number the interior points of the region as shown in Fig. 6. Then we will be computing the approximations u1(m)∼=u/parenleftbigg1 4,1 4,tm/parenrightbigg ,u2(m)∼=u/parenleftbigg1 2,1 4,tm/parenrightbigg ,u3(m)∼=u/parenleftbigg3 4,1 4,tm/parenrightbigg ,... (6) and so forth, for m=1,2,.... The replacement equations are obtained using Eq. (1) for the Laplacian and a forward difference to replace the time deriva-tive. The typical equation is u N(m)+uS(m)+uE(m)+uW(m)−4ui(m) (/Delta1x)2=ui(m+1)−ui(m) /Delta1t.(7) W h e nw es o l v et h i se q u a t i o nf o r ui(m+1),w eo b t a i n ui(m+1)=r[uN(m)+uS(m)+uE(m)+uW(m)]+(1−4r)ui(m), (8) in which r=/Delta1t /Delta1x2=/Delta1t /Delta1y2=16/Delta1t. The stability considerations of Section 7.2 are still important, and the rules of thumb are still valid. We must limit rby the requirement that 1 −4r≥0, or, in this case, /Delta1t≤1/64. We shall take the longest acceptable time step, /Delta1t=1/64, r=1/4, which makes the equations a little simpler. Atm=0 ,a l lt e m p e r a t u r e sa r eg i v e na s1 .F o r m≥1, all the boundary tem- peratures are zero and the ui(m)are all found to equal 1. For m=2, we calcu- late u1(2)=1 4/parenleftbig u2(1)+u5(1)+0+0/parenrightbig =1 2, u2(2)=1 4/parenleftbig u1(1)+u3(1)+u6(1)+0/parenrightbig =3 4, 422 Chapter 7 Numerical Methods i m 1256 0 1111 11 23 43 41 23 89 161 213 16 317 647 1625 6439 64 Table 9 Numerical solution of Eqs. (2)–(5) ... u5(2)=1 4/parenleftbig u1(1)+u6(1)+u9(1)+0/parenrightbig =3 4, u6(2)=1 4/parenleftbig u2(1)+u5(1)+u7(1)+u10(1)/parenrightbig =1. The 0’s in these equations stand for boundary temperatures. An alert calculator will notice that only the unknowns u1,u2,u5,u6need be calculated, since, in this example, the others will be given at each time step bysymmetry u 1(m)=u4(m)=u9(m)=u12(m), u5(m)=u8(m), u6(m)=u7(m), u2(m)=u3(m)=u10(m)=u11(m). I nT a b l e9a r ec o m p u t e dv a l u e so ft h es i g n i ﬁ c a n t u’s at a few times. Now consider this heat problem, which is not solvable by separation of vari- ables: ∂2u ∂x2+∂2u ∂y2=∂u ∂tinR, (9) u=f(t)onC, (10) u=0i n Ratt=0. (11) Here,Ris an L-shaped region and Cis its boundary. The function fwe take to be f(t)=t, but more complicated functions can be used. T o start the numerical solution, we set up a square grid, as shown in Fig. 7. The spacing is /Delta1x=/Delta1y=1/5 and the numbering of the points is shown. The typical replacement equation is just as given in Eqs. (7) and (8). We must bearin mind, however, that some points are adjacent to boundary points where thetemperature is given by f(t). 7.5 Two-Dimensional Problems 423 Figure 7 Mesh numbering for numerical solution of Eqs. (9)–(11). Because /Delta1x=/Delta1y=1/5, the parameter rin Eq. (8) is r=/Delta1t /Delta1x2=25/Delta1t. Clearly, the longest stable time step is /Delta1t=1/100, corresponding to r=1/4. Using this value of rsimpliﬁes the typical replacement equation to ui(m+1)=1 4/parenleftbig uN(m)+uS(m)+uE(m)+uW(m)/parenrightbig . (12) Speciﬁcally, we have u1(m+1)=1 4/parenleftbig u2(m)+u5(m)+2f(tm)/parenrightbig , u2(m+1)=1 4/parenleftbig u1(m)+u3(m)+u6(m)+f(tm)/parenrightbig , and so on. The f(tm)terms enter because point 1 is adjacent to two boundary points and point 2 to one boundary point. Note that symmetry about the line through points 4 and 7 makes it unnecessary to compute u8(m),..., u12(m). Table 10 contains calculated values of ufor the ﬁrst four time levels. Wave Problems In solving two-dimensional wave problems, we replace the Laplacian, as in the foregoing, and use a central difference for the time derivative, as we did inSection 7.3: ∂ 2u ∂t2→ui(m+1)−2ui(m)+ui(m−1) /Delta1t2. (13) 424 Chapter 7 Numerical Methods i m 1234567 f(tm) 0 0000000 0 1 0000000 1 .0 20 .50 .25 0 .25 0 .50 .50 .25 0 2 .0 31 .22 0 .75 0 .69 0 .75 1 .22 0 .69 0 .25 3 .0 41 .99 1 .40 1 .19 1 .84 1 .98 1 .30 0 .69 4 .0 Table 10 Numerical solution of Eqs. (9)–(11). Entries are 100 ×ui(m) As an example, let us consider the vibrations of a square membrane, as de- scribed by the problem ∂2u ∂x2+∂2u ∂y2=∂2u ∂t2, 0<x<1, 0<y<1,0<t, (14) u(x,0,t)=0, u(x,1,t)=0,0<x<1,0<t, (15) u(0,y,t)=0, u(1,y,t)=0,0<y<1,0<t, (16) u(x,y,0)=f(x,y), 0<x<1, 0<y<1, (17) ∂u ∂t(x,y,0)=g(x,y), 0<x<1, 0<y<1. (18) A typical replacement for the wave equation (14) is constructed using Eq. (1) for the Laplacian and Eq. (13) for the time derivative: ui(m+1)−2ui(m)+ui(m−1) (/Delta1t)2 =uN(m)+uS(m)+uE(m)+uW(m)−4ui(m) (/Delta1x)2. (19) As usual we solve for ui(m+1), using the abbreviation ρ=/Delta1t//Delta1x.T h er e s u l t is ui(m+1)=ρ2/bracketleftbig uE(m)+uW(m)+uN(m)+uS(m)/bracketrightbig +(2−4ρ2)ui(m)−ui(m−1). (20) The stability rules given earlier still apply. Thus we must choose ρ2≤1/2i n order to get a sensible solution. Let us now be speciﬁc. We shall take /Delta1x=/Delta1y=1/4,ρ2=1/2 (that is, /Delta1t=1/4√ 2), and suppose that the initial data from Eqs. (11) and (12) are f(x,y)=/braceleftbigg 1n e a r x=1 4,y=1 4, 0e l s e w h e r e , g(x,y)≡0. 7.5 Two-Dimensional Problems 425 The running equation is Eq. (20), which, with ρ2=1/2, simpliﬁes to ui(m+1)=1 2/bracketleftbig uE(m)+uW(m)+uN(m)+uS(m)/bracketrightbig −ui(m−1). (21) T o ﬁnd the starting equation we solve Eq. (21) with m=0 together with the re- placement equation for the initial-velocity condition, Eq. (18). The equations are ui(1)+ui(−1)=1 2/bracketleftbig uE(0)+uW(0)+uN(0)+uS(0)/bracketrightbig , (22) ui(1)−ui(−1)=2/Delta1tgi. (23) Because g(x,y)=0 in this instance, we ﬁnd ui(1)=1 4/bracketleftbig uE(0)+uW(0)+uN(0)+uS(0)/bracketrightbig as the starting equation; the right-hand side contains known values of uonly. In Fig. 8 are representations of the numerical solution at various time levels. T h es i m p l en u m e r i c a lt e c h n i q u ew eh a v ed e v e l o p e dc a nb ea d a p t e de a s i l y to treat inhomogeneities, boundary conditions involving derivatives of u,o r time-varying boundary conditions. Even nonrectangular regions can be han- dled, provided they ﬁt neatly on a rectangular grid. Several exercises illustratethese points. EXERCISES In Exercises 1–5, set up replacement equations using the given space mesh andthe numbering shown in the ﬁgure cited. Then ﬁnd the u i(m)for a few values ofmusing the largest stable value of r. Let boundary conditions override the initial condition if there is a disagreement. 1.∇2u=∂u ∂t,0 <x<1, 0 <y<0.75, 0 <t, u(0,y,t)=0, u(1,y,t)=0, 0 <y<0.75, 0 <t, u(x,0,t)=0, u(x,0.75,t)=1, 0 <x<1, 0 <t, u(x,y,0)=0, 0 <x<1, 0 <y<0.75, /Delta1x=/Delta1y=1/4. (See Fig. 9a.) 2.∇2u=∂u ∂tinR,0<t u=0o nb o u n d a r y ,0 <t u=1i nR,t=0. 426 Chapter 7 Numerical Methods Figure 8 Displacements of the square membrane. Numbers shown are ui(m)×64. The region Ris an inverted T: Starting with a rectangle of width 1 and height 3 /4, remove a 1 /4×1/4s q u a r ef r o mt h eu p p e rl e f ta n dr i g h tc o r - ners. Take /Delta1x=/Delta1y=1/4. (See Fig. 9b.) 7.5 Two-Dimensional Problems 427 Figure 9 Regions for Exercises 1–3. 3.Same as Exercise 2, except that the region is a cross. (See Fig. 9c.) 4.Same as Eqs. (9)–(11), except that the boundary condition is u=1o nt h e bottom (y=0)and u=0 elsewhere. (See Fig. 7.) 5.∇2u=∂u ∂t,0 <x<1, 0 <y<1, 0 <t, u(0,y,t)=0, u(1,y,t)=1, 0 <y<1, 0 <t, u(x,0,t)=0, u(x,1,t)=1, 0 <x<1, 0 <t, u(x,y,0)=0, 0 <x<1, 0 <y<1, /Delta1x=/Delta1y=1/4. (See Fig. 2.) 6.Find a numerical solution of the heat problem on a 1 ×1 square with /Delta1x= /Delta1y=1/4. Initially u=0 and on the outside boundary u=0. There is a tiny hole in the center of the square, so u(1/2,1/2,t)=1,t>0. (Actually, the region is a punctured square.) 7.Solve numerically Eqs. (14)–(18) with /Delta1x=/Delta1y=1/4,ρ2=1/2. Take f(x,y)≡0a n d g(x,y)=/braceleftbigg 4√ 2a t/parenleftbig1 2,1 2/parenrightbig , 0e l s e w h e r e . Physically, udescribes the vibrations of a square membrane struck in the middle. 8.Obtain an approximate solution of Eqs. (14)–(18) with f(x,y)≡0a n d g(x,y)=4√ 2. Take /Delta1x=/Delta1y=1/4a n dρ2=1/2. 9.Same as Exercise 8, but f(x,y)≡1a n d g(x,y)≡0i nt h es q u a r e . 10.Obtain an approximate numerical solution to the wave equation on an L- shaped region (a 1 ×1 square with a 1 /4×1/4s q u a r er e m o v e df r o mt h e upper right corner). Assume initial displacement =1 in the lower right corner, initial velocity equal to 0, and zero displacement on the boundary.Take/Delta1x=/Delta1y=1/4a n dρ 2=1/2. 428 Chapter 7 Numerical Methods 11.Approximate the solution of the wave equation in a semi-inﬁnite strip 3 units wide. Assume u=0 on all boundaries, zero initial velocity, and an initial value for uthat is 1 in a corner and 0 elsewhere. Take /Delta1x=/Delta1y=1 andρ2=1/2. 7.6 Comments and References Our objective in this chapter has been to survey some elementary numerical methods for problems like those we attacked analytically in earlier chapters.We have only had enough space to touch on the central topics: obtaining re- placement equations, solving linear systems of equations by direct and iterative methods, numerical stability, and order of error. The methods we have introduced are satisfactory for a ﬁrst introduction and for learning something about partial differential equations, but they are notadequate for any serious problem solving. New techniques for these problemsare superior in speed, accuracy and stability but are also more complicated. Ofthe many texts available, two excellent ones are Numerical Analysis by Burden and Faires, for general methods, and Numerical Solution of Partial Differential Equations by Smith. (See the Bibliography.) Almost all numerical methods for linear partial differential equations rely on the symbolism and theory of matrices. Two outstanding texts on matrixtheory are Applied Linear Algebra , 3rd ed., by Noble and Daniel, and Matrices by Barnett. Miscellaneous Exercises 1.Set up and solve replacement equations for this boundary value problem. Use/Delta1x=1/3. d2u dx2−√ 24xu=0,0<x<1, du dx(0)=1,u(1)=1. 2.Use the change of variables x=(r−a)/(b−a)andv(r)=u(x)to con- vert the equation 1 rd dr/parenleftbigg rdv dr/parenrightbigg −q(r)v=f(r), a<r<b, to an equation in uon the interval 0 <x<1. Miscellaneous Exercises 429 3.By means of the transformation mentioned in Exercise 2, a heat problem on an annular ring is converted to d2u dx2+1 1+xdu dx=−(1+x), 0<x<1, u(0)=1,u(1)=0. Set up and solve replacement equations for this problem using /Delta1x= 1/4. 4.The boundary value problem 1 rd dr/parenleftbigg rdv dr/parenrightbigg −γ2v=0,a<r<b, v(a)=1,v ( b)=0, can be transformed into the problem d2u dx2+1 α+xdu dx−γ2L2u=0,0<x<1, u(0)=1,u(1)=0, where L=b−aandα=a/L. Set up and solve replacement equations using/Delta1x=1/4,α=1,γL=1. 5.Set up replacement equations for the heat problem in the following, and solve for tup to 1 /4, using /Delta1x=1/4,/Delta1t=1/32. ∂2u ∂x2=∂u ∂t, 0<x<1,0<t, u(0,t)=u(1,t)=1−e−t,0<t, u(x,0)=0, 0<x<1. 6.Same as Exercise 5, but use u(0,t)=u(1,t)=1−e−32(ln 2)tso that u(0,tm)=1−(0.5)m. 7.Compare the numerical solution of the problem ∂2u ∂x2=∂u ∂t,0<x<1, 0<t, u(0,t)=0,u(1,t)=0,0<t, u(x,0)=1,0<x<1, 430 Chapter 7 Numerical Methods with the solution of the problem consisting of the equation ∂2u ∂x2−16u=∂u ∂t,0<x<1,0<t, with the same initial and boundary conditions. Use /Delta1x=1/4,/Delta1t= 1/48 in both cases. 8.In Exercise 7, what is the longest stable time step for each of the two problems? 9.Solve for several time levels using /Delta1x=1/5a n d r=1/2. What is /Delta1t? ∂2u ∂x2=∂u ∂t, 0<x<1, 0<t, u(0,t)=25t,u(1,t)=0,0<t, u(x,0)=0, 0<x<1. 10. Same as Exercise 9, except the second boundary condition is∂u ∂x(1,t)=0. 11. This problem describes the displacement of a string whose end is jerked: ∂2u ∂x2=∂2u ∂t2,0<x<1, 0<t, u(0,t)=0, u(1,t)=1, 0<t, u(x,0)=0,∂u ∂t(x,0)=0,0<x<1. Solve numerically through one period (until t=2) with /Delta1x=/Delta1t= 1/4. 12. Same problems as Exercise 11, except the right-hand boundary condi- tion is u(1,t)=h(t),0<t,w h e r e h(t)=/braceleftBig1,0<t≤1, 0,1<t≤2, and h(t+2)=h(t). Solve numerically with /Delta1x=/Delta1t=1/4 for enough values of tso that resonance becomes noticeable. 13. Using /Delta1x=/Delta1y=1/4, ﬁnd a numerical solution of this problem: ∂2u ∂x2+∂2u ∂y2=0,0<x<1, 0<y<1, u(x,0)=0, u(x,1)=2 πtan−1/parenleftbigg1 x/parenrightbigg ,0<x<1, u(0,y)=1, u(1,y)=2 πtan−1(y), 0<y<1. Miscellaneous Exercises 431 14.The analytical solution of the problem in Exercise 13 is u(x,y)= (2/π)tan−1(y/x). Compare your numerical results with the exact so- lution. 15.Using /Delta1x=/Delta1y=1/4a n d r=1/4, ﬁnd a numerical solution for this problem: ∇2u=∂u ∂t, 0<x<1,0<y<1,0<t, u(x,0,t)=u(x,1,t)=0,0<x<1,0<t, u(0,y,t)=u(1,y,t)=0,0<y<1,0<t, u(x,y,0)=1, 0<x<1,0<y<1. 16.T h ea n a l y t i c a ls o l u t i o no ft h ep r o b l e mi nE x e r c i s e1 5i s u(x,y,t)=∞/summationdisplay n=1∞/summationdisplay n=14(1−cos(nπ))( 1−cos(mπ)) π2mn ×sin(nπx)sin(nπy)e−(m2+n2)π2t. Using just the term m=n=1 of this solution, compare the ratio R=u/parenleftbig1 2,1 2,tm+1/parenrightbig u/parenleftbig1 2,1 2,tm/parenrightbig and the ratio of the corresponding u’s computed in Exercise 15. This page intentionally left blank Bibliography Abramowitz, M., and I. Stegun (eds). Handbook of Mathematical Functions , 10th ed. Washington, DC, National Bureau of Standards, 1972. (Reprinted by Dover, 1974.) Andrews, L.C. Special Functions of Mathematics for Engineers ,2 n de d .B e l l i n g - ham, WA, SPIE — The International Society for Optical Engineering, 1997. Barnett, S. Matrices: Methods and Applications . New Y ork, Oxford University Press, 1990. Burden, R.L., and J.D. Faires. Numerical Analysis , 7th ed. Belmont, CA, Brooks/Cole, 2000. Carslaw, H.S., and J.C. Jaeger. Conduction of Heat in Solids ,2 n de d .N e wY o r k , Oxford University Press, 1986. Churchill, R.V ., and J.W. Brown. Fourier Series and Boundary Value Problems , 6th ed. New Y ork, McGraw-Hill, 2000. Churchill, R.V . Operational Mathematics ,3 r de d .N e wY o r k ,M c G r a w - H i l l , 1972. Courant, R., and D. Hilbert. Methods of Mathematical Physics , Vol. 1. New Y ork, Wiley-Interscience, 1953/1989. Crank, J. The Mathematics of Diffusion ,2 n de d .N e wY o r k ,O x f o r dU n i v e r s i t y Press, 1980. Davis, P .J., and R. Hersh. The Mathematical Experience .B o s t o n ,H o u g h t o n Mifﬂin, 1999. Erdelyi, A., W. Magnus, F. Oberhettinger, and F. Tricomi. Tables of Inte g ral Transforms , Vols. 1 and 2. New Y ork, McGraw-Hill, 1954. Feller, W. Introduction to Probability Theory and Its Applications , Vol. 1, 3rd ed. New Y ork, Wiley, 1968. 433 434 Bibliography Fletcher, N.H., and T.D. Rossing. The Physics of Musical Instruments ,2 n de d . New Y ork, Springer-Verlag, 1998. Isenberg, C. The Science of Soap Films and Soap Bubbles . Avon, UK, Tieto Ltd, 1978. (Reprinted by Dover, 1992.) Jerri, A.J. Integral and Discrete Transforms with Applications and Error Analysis . New Y ork, Marcel Dekker, 1991. Jones, D.S., and B.D. Sleeman. Differential Equations and Mathematical Biol- ogy. San Francisco, Harper-Collins, 1983. Kirchhoff, R.H. Potential Flows: Computer Graphic Solutions .N e wY o r k ,M a r c e l Dekker, 1998. Lamb, H. Hydrodynamics , 6th ed. Cambridge, UK, Cambridge University Press, 1971. Love, A.E.H. A Treatise on the Mathematical Theory of Elasticity ,4 t he d .N e w Y ork, Dover, 1944. Main, I.G. Vibrations and Waves in Physics ,3 r de d .C a m b r i d g e ,U K ,C a m b r i d g e University Press, 1993. Morse, P .M., and H. Feshbach. Methods of Theoretical Physics, Part I .N e wY o r k , McGraw-Hill, 1953. Murray, J.D. Mathematical Biology . New Y ork, Springer-Verlag, 1993. Noble, B., and J.W. Daniel. Applied Linear Algebra , 3rd ed. Englewood Cliffs, NJ, Prentice-Hall, 1988. Sagan, H. Boundary and Eigenvalue Problems in Mathematical Physics .N e w Y ork, Wiley, 1966. (Reprinted by Dover, 1989.) Smith, G.D. Numerical Solution of Partial Differential Equations ,3 r de d .N e w Y ork, Oxford University Press, 1986. Street, R.L. Analysis and Solution of Partial Differential Equations .M o n t e r e y , CA, Brooks/Cole, 1973. Timoshenko, S., and J.N. Goodier. Theory of Elasticity ,2 n de d .N e wY o r k , McGraw-Hill, 1951. T olstov, G.P . Fourier Series . Englewood Cliffs, NJ, Prentice-Hall, 1962. (Reprinted by Dover, 1976.) Walker, J.S. The Fast Fourier Transform , 2nd ed. Boca Raton, FL, CRC Press, 1996. Wan, F.Y.M. Mathematical Models and Their Analysis .N e wY o r k ,H a r p e r& Row, 1989. Widder, D.V. The Heat Equation . New Y ork, Academic Press, 1975. Appendix: Mathematical References Trigonometric Functions sin(A±B)=sin(A)cos(B)±cos(A)sin(B) cos(A±B)=cos(A)cos(B)∓sin(A)sin(B) sin(A)+sin(B)=2s i n/parenleftbiggA+B 2/parenrightbigg cos/parenleftbiggA−B 2/parenrightbigg sin(A)−sin(B)=2c o s/parenleftbiggA+B 2/parenrightbigg sin/parenleftbiggA−B 2/parenrightbigg cos(A)+cos(B)=2c o s/parenleftbiggA+B 2/parenrightbigg cos/parenleftbiggA−B 2/parenrightbigg cos(A)−cos(B)=2s i n/parenleftbiggA+B 2/parenrightbigg sin/parenleftbiggA−B 2/parenrightbigg sin(A)sin(B)=1 2/parenleftbig cos(A−B)−cos(A+B)/parenrightbig sin(A)cos(B)=1 2/parenleftbig sin(A−B)+sin(A+B)/parenrightbig cos(A)cos(B)=1 2/parenleftbig cos(A−B)+cos(A+B)/parenrightbig cos(A)=1 2/parenleftbig eiA+e−iA/parenrightbig ,sin(A)=1 2i/parenleftbig eiA−e−iA/parenrightbig cos2(A)+sin2(A)=1,1+tan2(A)=sec2(A) 435 436 Appendix: Mathematical References Hyperbolic Functions cosh(A)=1 2/parenleftbig eA+e−A/parenrightbig ,sinh(A)=1 2/parenleftbig eA−e−A/parenrightbig dcosh(u)=sinh(u)du,dsinh(u)=cosh(u)du sinh(A±B)=sinh(A)cosh(B)±cosh(A)sinh(B) cosh(A±B)=cosh(A)cosh(B)±sinh(A)sinh(B) sinh(A)+sinh(B)=2s i n h/parenleftbiggA+B 2/parenrightbigg cosh/parenleftbiggA−B 2/parenrightbigg sinh(A)−sinh(B)=2c o s h/parenleftbiggA+B 2/parenrightbigg sinh/parenleftbiggA−B 2/parenrightbigg cosh(A)+cosh(B)=2c o s h/parenleftbiggA+B 2/parenrightbigg cosh/parenleftbiggA−B 2/parenrightbigg cosh(A)−cosh(B)=2s i n h/parenleftbiggA+B 2/parenrightbigg sinh/parenleftbiggA−B 2/parenrightbigg sinh(A)sinh(B)=1 2/parenleftbig cosh(A+B)−cosh(A−B)/parenrightbig sinh(A)cosh(B)=1 2/parenleftbig sinh(A+B)+sinh(A−B)/parenrightbig cosh(A)cosh(B)=1 2/parenleftbig sinh(A+B)+cosh(A−B)/parenrightbig cosh2(A)−sinh2(A)=1,1−tanh2(A)=sech2(A) Calculus 1.Derivative of a product (uv)/prime=u/primev+uv/prime (uv)/prime/prime=u/prime/primev+2u/primev/prime+uv/prime/prime (uv)(n)=u(n)v+/parenleftbiggn 1/parenrightbigg u(n−1)v/prime+···+/parenleftbiggn n−1/parenrightbigg uv(n−1)+uv(n) In this formula,/parenleftbign k/parenrightbig =n! (n−k)!k!is a binomial coefﬁcient. 2.Rules of integration a./integraldisplayb a/parenleftbig c1f1(x)+c2f2(x)/parenrightbig dx=c1/integraldisplayb af1(x)dx+c2/integraldisplayb af2(x)dx Appendix: Mathematical References 437 b./integraldisplaya af(x)dx=0 c./integraldisplayb af(x)dx=−/integraldisplaya bf(x)dx d./integraldisplayb af(x)dx=/integraldisplayc af(x)dx+/integraldisplayb cf(x)dx 3.Derivatives of integrals a.d dt/integraldisplayb af(x,t)dx=/integraldisplayb a∂f ∂t(x,t)dx b.d dt/integraldisplayt af(x)dx=f(t)(Fundamental theorem of calculus; ais constant) c.d dt/integraldisplayv(t) u(t)f(x,t)dx=f/parenleftbig v(t),t/parenrightbig v/prime(t)−f/parenleftbig u(t),t/parenrightbig u/prime(t) +/integraldisplayv(t) u(t)∂f ∂t(x,t)dx (Leibniz’s rule) 4.Integration by parts a./integraldisplay uv/primedx=uv−/integraldisplay vu/primedx b./integraldisplay uv/prime/primedx=v/primeu−vu/prime+/integraldisplay vu/prime/primedx 5.Functions deﬁned by integrals a.Natural logarithm ln(x)=/integraldisplayx 1dz z b.Sine-integral function Si(x)=/integraldisplayx 0sin(z) zdz c.Normal probability distribution function /Phi1(x)=1√ 2π/integraldisplayx −∞e−z2/2dz d.Error function erf(x)=2√π/integraldisplayx 0e−z2dz Note: erf (x)=2/Phi1/parenleftbig√ 2x/parenrightbig −1 438 Appendix: Mathematical References e.Integrated Bessel function IJ(x)=/integraldisplayx 0J0(z)dz Table of Integrals Any letter except xrepresents a constant. The integration constants have been left off. 1.Rational functions 1.1/integraldisplaydx h+kx=1 kln|h+kx| 1.2/integraldisplaydx x2+a2=1 atan−1/parenleftbiggx a/parenrightbigg 1.3/integraldisplayxd x x2+a2=1 2ln/parenleftbig x2+a2/parenrightbig 1.4/integraldisplaydx x2−a2=1 2aln/vextendsingle/vextendsingle/vextendsingle/vextendsinglex−a x+a/vextendsingle/vextendsingle/vextendsingle/vextendsingle 1.5/integraldisplayxd x x2−a2=1 2ln/vextendsingle/vextendsinglex2−a2/vextendsingle/vextendsingle 2.Radicals 2.1/integraldisplaydx√ x2+a2=ln/parenleftbig x+/radicalbig x2+a2/parenrightbig or sinh−1/parenleftbiggx a/parenrightbigg 2.2/integraldisplayxd x√ x2+a2=/radicalbig x2+a2 2.3/integraldisplaydx√ x2−a2=ln/parenleftbig x+/radicalbig x2−a2/parenrightbig (x>a) 2.4/integraldisplayxd x√ x2−a2=/radicalbig x2−a2 2.5/integraldisplaydx√ a2−x2=sin−1/parenleftbiggx a/parenrightbigg (|x|<a) 2.6/integraldisplayxd x√ a2−x2=−/radicalbig a2−x2(|x|<a) 3.Exponentials and hyperbolic functions 3.1/integraldisplay ekxdx=ekx k Appendix: Mathematical References 439 3.2/integraldisplay xekxdx=kx−1 k2ekx 3.3/integraldisplay sinh(kx)dx=cosh(kx) k 3.4/integraldisplay cosh(kx)dx=sinh(kx) k 3.5/integraldisplay xsinh(kx)dx=xcosh(kx) k−sinh(kx) k2 3.6/integraldisplay xcosh(kx)dx=xsinh(kx) k−cosh(kx) k2 4.Sines and cosines 4.1/integraldisplay sin(λx)dx=−cos(λx) λ 4.2/integraldisplay cos(λx)dx=sin(λx) λ 4.3/integraldisplay xsin(λx)dx=sin(λx) λ2−xcos(λx) λ 4.4/integraldisplay xcos(λx)dx=cos(λx) λ2+xsin(λx) λ 4.5/integraldisplay x2sin(λx)dx=2xsin(λx) λ2+(2−λ2x2)cos(λx) λ3 4.6/integraldisplay x2cos(λx)dx=2xcos(λx) λ2+(λ2x2−2)sin(λx) λ3 4.7/integraldisplay sin(λx)sin(µx)dx=sin(µ−λ)x 2(µ−λ)−sin(µ+λ)x 2(µ+λ)(λ/negationslash=µ) 4.8/integraldisplay sin(λx)cos(µx)dx=cos(µ−λ)x 2(µ−λ)−cos(µ+λ)x 2(µ+λ)(λ/negationslash=µ) 4.9/integraldisplay cos(λx)cos(µx)dx=sin(µ−λ)x 2(µ−λ)+sin(µ+λ)x 2(µ+λ)(λ/negationslash=µ) 4.10/integraldisplay sin2(λx)dx=x 2−sin(2λx) 4λ 4.11/integraldisplay sin(λx)cos(λx)dx=sin2(λx) 2λ 4.12/integraldisplay cos2(λx)dx=x 2+sin(2λx) 4λ 440 Appendix: Mathematical References 4.13/integraldisplay ekxsin(λx)dx=ekx(ksin(λx)−λcos(λx)) k2+λ2 4.14/integraldisplay ekxcos(λx)dx=ekx(kcos(λx)+λsin(λx)) k2+λ2 4.15/integraldisplay sinh(kx)sin(λx)dx=kcosh(kx)sin(λx)−λsinh(kx)cos(λx) k2+λ2 4.16/integraldisplay sinh(kx)cos(λx)dx=kcosh(kx)cos(λx)+λsinh(kx)sin(λx) k2+λ2 4.17/integraldisplay cosh(kx)sin(λx)dx=ksinh(kx)sin(λx)−λcosh(kx)cos(λx) k2+λ2 4.18/integraldisplay cosh(kx)cos(λx)dx=ksinh(kx)cos(λx)+λcosh(kx)sin(λx) k2+λ2 5.Bessel functions 5.1/integraldisplay xJ0(λx)dx=xJ1(λx) λ 5.2/integraldisplay x2J0(λx)dx=x2J1(λx) λ+xJ0(λx) λ2−1 λ3IJ(λx)1 5.3/integraldisplay J1(λx)dx=−J0(λx) λ 5.4/integraldisplay xn+1Jn(λx)dx=xn+1Jn+1(λx) λ 5.5/integraldisplay Jn(λx)dx xn−1=−Jn−1(λx) λxn−1 5.6/integraldisplay J2 0(λx)xd x=x2 2/bracketleftbig J2 0(λx)+J2 1(λx)/bracketrightbig 5.7/integraldisplay J2 n(λx)xd x=x2 2/bracketleftbig J2 n(λx)−Jn−1(λx)Jn+1(λx)/bracketrightbig =x2 2/bracketleftbig J/prime n(λx)/bracketrightbig2+/parenleftbiggx2 2−n2 2λ2/parenrightbigg/bracketleftbig Jn(λx)/bracketrightbig2 6.Legendre polynomials 6.1/integraldisplay Pn(x)dx=−−(1−x2) n(n+1)P/prime n(x) 6.2/integraldisplay xPn(x)dx=(1−x2) (n+2)(n−1)/parenleftbig Pn(x)−xP/prime n(x)/parenrightbig . 1See Calculus 5e. Answers to Odd-NumberedExercises Chapter 0 Section 0.1 1.φ(x)=c1cos(λx)+c2sin(λx). 3. The equation has constant coefﬁcients k=0,p=0;u(t)=c1+c2t. 5.w(r)=c1rλ+c2r−λ. 7. Integrate, solve for dv/dx, and integrate again: v(x)=c1+c2ln|h+kx|. 9.u(x)=c1+c2/x2. 11.u(r)=c1+c2ln(r). 13. Characteristic polynomial m4−λ4=0; roots m=±λ,±iλ. General so- lution u(x)=c1cos(λx)+c2sin(λx)+c3cosh(λx)+c4sinh(λx). 15. Characteristic polynomial (m2+λ2)2=0; roots m=± iλ(double). General solution u(x)=(c1+c2x)cos(λx)+(c3+c4x)sin(λx). 17.v(t)=ln(t)and u2(t)=tbln(t). 19.u/prime/prime+λ2u=0;R(ρ)=(acos(λρ)+bsin(λρ))/ρ . 21.t2d2u/dt2=v/prime/prime−v/prime;td u/dt=v/prime;v/prime/prime+(k−1)v/prime+pv=0 (constant coefﬁcients). 441 442 Answers to Odd-Numbered Exercises 23. Roots of characteristic equation: m=−α±iβ, β =/radicalbig σ2−α2. Solution of differential equation: y(t)=e−αt/parenleftbig c1cos(βt)+c2sin(βt)/parenrightbig . Initial conditions give: c1=− 0.001h,c2=(α/β) c1. 25.v=2.62 m/s. Section 0.2 1.u(t)=T+ce−at. 3.u(t)=te−at+ce−at. 5.u(t)=1 2tsin(t)+c1cos(t)+c2sin(t). 7.u(t)=1 12et+1 2te−t+c1e−t+c2e−2t. 9.u(ρ)=−1 6ρ2+c1 ρ+c2. 11.h(t)=− 320t+c1+c2e−0.1t,c1=h0+3200, c2=− 3200. 13.v(t)=t,up(t)=te−at. 15.v1(x)=sin(x)−ln|sec(x)+tan(x)|,v2=− cos(x); up(x)=− cos(x)ln|sec(x)+tan(x)|. 17.v1(t)=t2/2,v2(t)=− t;up(t)=− t2/2. 19.v1(t)=− 1/2t,v2(t)=− t/2,up(t)=− 1. 23.β=1/α,K=Rα/ρc. 25.T=β(exp(KI2 max(1−e−2λt)/2λ)−1). Section 0.3 1. a. u(x)=c2sin(x),c2arbitrary; b.u(x)=1−cos(x)−1−cos(1) sin(1)sin(x)(unique); c. No solution exists. 3. a. and b. λ=±(2n−1)π 2a,n=1,2,...; Answers to Odd-Numbered Exercises 443 c.λ=±nπ a,n=0,1,2,.... 5.c=− a/2,c/prime=h−1 µcosh/parenleftbiggµa 2/parenrightbigg . 7.u(x)=T+c1cosh(γx)+c2sinh(γx),w h e r e γ=/radicalbigg hC κAand c1=T0−T,c2=−κγsinh(γa)+hcosh(γa) κγcosh(γa)+hsinh(γa)c1. 9.u(x)=T+H/parenleftbigg 1−cosh(γx)−1−cosh(γa) sinh(γa)sinh(γx)/parenrightbigg , where H=I2R hCandγ=/radicalbigg hC κA. 11.u(y)=y(L−y)g/2µ. 13.P=EI(nπ/L)2,n=1,2,.... 15.u(x)=T+A/parenleftbigg 1−cosh(γx)−1−cosh(γa) sinh(γa)sinh(γx)/parenrightbigg , A=g/κγ2,a n dγ=/radicalbigg hC κA. 17.u(r)=c1ln(r/a)+c2,c1=h0h1(Ta−TW)/D,c2=[h0(κ/b+ h1ln(b/a))TW+(κ/a)h1Ta]/D,D=h1κ/a+h0κ/b+h0h1ln(b/a). 19.u(x)=w0 EI/parenleftbiggx4 24−ax3 6+a2x2 2/parenrightbigg . Section 0.4 1. a. u/prime/prime+1 ru/prime−u=0,r=0; b.u/prime/prime−2x 1−x2u/prime=0,x=± 1; c.u/prime/prime+cot(φ)u/prime−u=0,φ=0,±π,±2π,... ; d.u/prime/prime+2 ρu/prime+λ2u=0,ρ=0. 3.u(0)bounded; u(ρ)=H 6κ/parenleftbig c2−ρ2/parenrightbig +Hc 3h+T. 5.u(ρ)=1 ρ/parenleftbig c1cos(µρ)+c2sin(µρ)/parenrightbig , u(ρ)≡0u n l e s s µa=π,2π,... . The critical radius is a=π µ. 444 Answers to Odd-Numbered Exercises 7.u(r)=325+104(0.25−r2)/4;u(0)=950. 9.u(x)=T0+AL2(1−e−x/L). Section 0.5 1.G(x,z)=/braceleftbigg z(a−x)/(−a),0<z≤x, x(a−z)/(−a),x≤z<a. 3.G(x,z)=/braceleftbigg cosh(γz)sinh(γ(a−x))/(−γcosh(γa)), 0<z≤x, cosh(γx)sinh(γ(a−z))/(−γcosh(γa)), x≤z<a. 5.G(ρ,z)=  (c−ρ)/ρ −c/z2,0≤z<ρ, (c−z)/z −c/z2,ρ ≤z<c. 7.G(x,z)=  sinh(γz)e−γx −γ,0<z≤x, sinh(γx)e−γz −γ,x≤z. 9.u(ρ)=(ρ2−c2)/6. 11.u(x)=/integraldisplaya 0G(x,z)f(z)dz=/integraldisplayx 0z(a−x) −af(z)dz+/integraldisplaya xx(a−z) −af(z)dz. There are two cases: (i)x≤a/2, so u(x)=/integraldisplaya a/2x(a−z) −adz; and (ii)x>a/2, so u(x)=/integraldisplaya xx(a−z) −adz. Results: u(x)=/braceleftbigg−ax/8, 0<x<a/2, −x(a−x)2/2a,a/2<x<a. 13. (i) At the left boundary, x=l<z, so the second line of Eq. (17) holds. The boundary condition (2) is satisﬁed by vbecause it is satisﬁed by u1. At the right boundary, use the ﬁrst line of Eq. (17). (ii) At x=z,b o t hl i n e so fE q .( 1 7 )g i v et h es a m ev a l u e . (iii)v/prime(z+h)−v/prime(z−h)=u1(z)u/prime 2(z+h)−u/prime 1(z−h)u2(z) W(z). Ashapproaches 0, the numerator approaches W(z). Answers to Odd-Numbered Exercises 445 (iv) This is true because u1(x)and u2(x)are solutions of the homoge- neous equation. Chapter 0 Miscellaneous Exercises 1.u(x)=T0cosh(γx)+/parenleftbig T1−T0cosh(γa)/parenrightbigsinh(γx) sinh(γa). 3.u(x)=T0. 5.u(r)=p(a2−r2)/4. 7.u(ρ)=H(a2−ρ2)/6+T0. 9.u(x)=T+(T1−T)cosh(γx)/cosh(γa). 11.u(x)=T0+(T−T0)e−γx. 13.h(x)=/radicalbig ex(a−x)+h2 0+(h2 1−h2 0)(x/a). 15.u(x)=w(1−e−γxcos(γx))EI/k,w h e r e γ=(k/4EI)1/4. 17.u(x)=/braceleftbigg T0+Ax, 0<x<αa, T1−B(a−x), α a<x<a, A=κ2 κ1(1−α)+κ2αT1−T0 a,B=κ1 κ2A. 19.u(x)=1 2/parenleftbig 1−e−2x/parenrightbig −1 2/parenleftbig 1−e−2a/parenrightbig1−e−x 1−e−a. 21. a. u(x)=sinh(px)/sinh(pa); b.u(x)=cosh(px)−cosh(pa) sinh(pa)sinh(px)=sinh/parenleftbig p(a−x)/parenrightbig /sinh(pa); c.u(x)=cosh(px)/cosh(pa); d.u(x)=cosh(p(a−x))/cosh(pa); e.u(x)=− cosh(p(a−x))/psinh(pa); f.u(x)=cosh(px)/psinh(pa). 23.u(x)=x 2ln/vextendsingle/vextendsingle/vextendsingle/vextendsingle1+x 1−x/vextendsingle/vextendsingle/vextendsingle/vextendsingle−1. 25. Multiply by u/primeand integrate:1 2(u/prime)2=1 5γ2u5+c1.S i n c e u(x)→0 asx→∞ ,a l s o u/prime(x)→0; thus c1=0. Now u/prime=−/radicalbig 2γ2/5u5/2or u−5/2u/prime=−/radicalbig 2γ2/5( t h en e g a t i v er o o tm a k e s udecrease) can be in- tegrated to result in (−2/3)u−3/2=−/radicalbig 2γ2/5x+c2. The condition at x=0g i v e s c2=(−3/2)U−3/2. 446 Answers to Odd-Numbered Exercises Finally u(x)=(U−3/2+(3/2)/radicalbig 2γ2/5x)−2/3. 27. 459 .77 rad/s. 29.u(x)=C0e−ax. 31.w(x)=P 2γ2/bracketleftbigg1 4−x2+cosh(γx)−cosh(γ/2) γsinh(γ/2)/bracketrightbigg . 3 3 . T h es o l u t i o nb r e a k sd o w n( b u c k l i n go c c u r s )i ft a n (λ/2)=γ/2. Chapter 1 Section 1.1 1. a. 2/parenleftbigg sin(x)−1 2sin(2x)+1 3sin(3x)−···/parenrightbigg ; b.π 2−4 π/parenleftbigg cos(x)+1 9cos(3x)+1 25cos(5x)+···/parenrightbigg ; c.1 2+2 π/parenleftbigg sin(x)+1 3sin(3x)+1 5sin(5x)+···/parenrightbigg ; d.2 π−4 π/parenleftbigg1 3cos(2x)+1 15cos(4x)+1 35cos(6x)+···/parenrightbigg . 3.f(x+p)=1=f(x)for any pand all x. 5. If cis a multiple of p,t h eg r a p ho f f(x)between cand c+pis the same as that between 0 and p.O t h e r w i s e ,l e t kbe the integer such that kplies between cand c+p: /integraldisplayc+p cf(x)dx=/integraldisplaykp cf(x)dx+/integraldisplayc+p kpf(x)dx=/integraldisplayp c∗f(x)dx+/integraldisplayc∗ 0f(x)dx, where c∗=c−(k−1)p. 7. a. cos2(x)=1 2+1 2cos(2x); b. sin/parenleftbigg x−π 6/parenrightbigg =cos/parenleftbiggπ 6/parenrightbigg sin(x)−sin/parenleftbiggπ 6/parenrightbigg cos(x); c. sin(x)cos(2x)=−1 2sin(x)+1 2sin(3x). Section 1.2 1. a.1 2−4 π2/bracketleftbigg cos(πx)+1 9cos(3πx)+1 25cos(5πx)+···/bracketrightbigg ; Chapter 1 447 b.4 π/bracketleftbigg sin/parenleftbiggπx 2/parenrightbigg +1 3sin/parenleftbigg3πx 2/parenrightbigg +1 5sin/parenleftbigg5πx 2/parenrightbigg +···/bracketrightbigg ; c.1 12−1 π2/bracketleftbigg cos(2πx)−1 4cos(4πx)+1 9cos(6πx)−···/bracketrightbigg . 3.¯f(x)=f(x−2na),2na<x<2(n+1)a, f(x)∼a0+∞/summationdisplay 1ancos(nπx/a)+bnsin(nπx/a), a0=1 2a/integraldisplay2a 0f(x)dx,an=1 a/integraldisplay2a 0f(x)cos(nπx/a)dx, bn=1 a/integraldisplay2a 0f(x)sin(nπx/a)dx. 5. Odd: (a), (d), (e); even: (b), (c); neither: (f). 7. a.2 π/parenleftbigg sin(πx)−1 2sin(2πx)+···/parenrightbigg ; b. This function is its own Fourier series; c.4 π2/parenleftbigg sin(πx)−1 9sin(3πx)+1 25sin(5πx)−···/parenrightbigg . 9. If f(−x)=− f(x)and f(x)=f(a−x)for 0<x<a, sine coefﬁcients with even indices are zero. Example: square wave. 11. a. f(x)=1=2 π∞/summationdisplay 11−cos(nπ) nsin/parenleftbiggnπx a/parenrightbigg ; b.f(x)=a 2−2a π2∞/summationdisplay 11−cos(nπ) n2cos/parenleftbiggnπx a/parenrightbigg =2a π∞/summationdisplay 1−cos(nπ) nsin/parenleftbiggnπx a/parenrightbigg ; c.f(x)=∞/summationdisplay 1(−1)n+1sin(1)2nπ (nπ)2−1sin(nπx), 0<x<1 =∞/summationdisplay 1/parenleftbig (−1)ncos(1)−1/parenrightbig 2 (nπ)2−1cos(nπx), 0<x<1; d.f(x)=2 π/bracketleftBigg 1−∞/summationdisplay 11+cos(nπ) n2−1cos(nx)/bracketrightBigg =sin(x). 13. Even, yes. Odd, yes only if f(0)=f(a)=0. 448 Answers to Odd-Numbered Exercises Section 1.3 1. a. sectionally smooth; b, c, d, e are not; b: vertical tangent at 0; c: vertical asymptote at ±π/2; d, e: vertical asymp- tote at π/2. 3. T o f(x)everywhere. 5. b. Graph consists of straight-line segments. c. x=1, sum =1/2;x=2, sum=0;x=9.6, sum =− 0.6;x=− 3.8, sum =0.2. Use periodicity. 7.B=0,A=−π2/12,C=1/4. 9. a.√ 1−x2;b . a0=π/4; c. No; d. nothing. Section 1.4 1. (c), (d), (f), (g) have uniformly convergent Fourier series. 3. All of the cosine series converge uniformly. The sine series converges uni- formly only in case (b). 5. (a), (b), (d) converge uniformly; (c) does not. Section 1.5 1.∞/summationdisplay n=11 n2=π2 6. 3.f/prime(x)=1, 0<x<π. The sine series cannot be differentiated, because the odd periodic extension of fis not continuous. But the cosine series can be differentiated. 5. For the sine series: f(0+)=0a n d f(a−)=0. For the cosine series no additional condition is necessary. 7. No. The function ln |2c o s(x 2)|is not even sectionally continuous. 9. Since fis odd, periodic, and sectionally smooth, (c) follows, and also bn→0a s n→∞ .T h e n/summationtext∞ n=1|nkbne−n2t|converges for all integers k (t>0) by the comparison test and ratio test: /vextendsingle/vextendsinglenkbne−n2t/vextendsingle/vextendsingle≤Mnke−n2tfor some M and M(n+1)ke−(n+1)2t Mnke−n2t=/parenleftbiggn+1 n/parenrightbiggk e−(2n+1)t→0 asn→∞ . Then by Theorem 7, (a) is valid. Property (b) follows by direc- tion substitution. Chapter 1 449 Section 1.6 1.1 π/integraldisplayπ −π/parenleftbigg ln/vextendsingle/vextendsingle/vextendsingle/vextendsingle2c o s/parenleftbiggx 2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg 2 dx=∞/summationdisplay n=11 n2=π2 6. 3. a. Coefﬁcients tend to zero. b. Coefﬁcients tend to zero, although/integraldisplay1 −1|x|−1dxis inﬁnite. 5. The integral must be inﬁnite, because∞/summationdisplay n=1a2 n+b2 n=∞ . Section 1.7 1 . T h ee q u a l i t yt ob ep r o v e di s 2s i n/parenleftbigg1 2y/parenrightbigg/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg =sin/parenleftbigg/parenleftbigg N+1 2/parenrightbigg y/parenrightbigg . The left-hand side is transformed as follows: 2s i n/parenleftbigg1 2y/parenrightbigg/parenleftBigg 1 2+N/summationdisplay n=1cos(ny)/parenrightBigg =sin/parenleftbigg1 2y/parenrightbigg +N/summationdisplay n=12s i n/parenleftbigg1 2y/parenrightbigg cos(ny) =sin/parenleftbigg1 2y/parenrightbigg +N/summationdisplay n=1/parenleftbigg sin/parenleftbigg/parenleftbigg n+1 2/parenrightbigg y/parenrightbigg −sin/parenleftbigg/parenleftbigg n−1 2/parenrightbigg y/parenrightbigg/parenrightbigg =sin/parenleftbigg1 2y/parenrightbigg +N/summationdisplay n=1sin/parenleftbigg/parenleftbigg n+1 2/parenrightbigg y/parenrightbigg −N−1/summationdisplay n=0sin/parenleftbigg/parenleftbigg n+1 2/parenrightbigg y/parenrightbigg =sin/parenleftbigg/parenleftbigg N+1 2/parenrightbigg y/parenrightbigg because all other terms cancel. 3.φ(0+)=1,φ(0−)=− 1. See Fig. 1. 5. a. f/prime(x)=3 4x−1/4for 0<x<π (and f/primeis an odd function). Thus, fhas a vertical tangent at x=0, although it is continuous there. b.φ(y)=|y|3/4 2s i n(1 2y)cos/parenleftbigg1 2y/parenrightbigg ,−π< y<π 450 Answers to Odd-Numbered Exercises Figure 1 Graph for Exercise 3, Section 1.7. is a product of continuous functions and is therefore continuous, except perhaps where the denominator is 0. At y=0, cos(1 2y)∼=1, 2 sin (1 2y)∼=y, soφ(y)∼=|y|3/4/y=± | y|−1/4near y=0. c. Now,/integraltextπ −πφ2(y)dyis ﬁnite, so the Fourier coefﬁcients of φapproach zero. Section 1.8 1.ˆa6=− 0.00701, a6=− 0.00569. 3.ˆa0=1.367, ˆa1=− 0.844,ˆb1=− 0.043, ˆa2=0.208, ˆb2=− 0.115, ˆa3=0.050, ˆb3=− 0.050, ˆa4=0.042, ˆb4=0.00, ˆa5=− 0.0064,ˆb5=0.043, ˆa6=0.0167. Section 1.9 1. Each function has the representations (for x>0) f(x)=/integraldisplay∞ 0A(λ)cos(λx)dλ=/integraldisplay∞ 0B(λ)sin(λx)dλ. a.A(λ)=2/π(1+λ2),B(λ)=2λ/π( 1+λ2); b.A(λ)=2s i n(λ)/πλ ,B(λ)=2(1−cos(λ))/πλ ; c.A(λ)=2(1−cos(λπ))/λ2π,B(λ)=2(πλ−sin(λπ))/πλ2. 3. a.1 1+x2=/integraldisplay∞ 0e−λcos(λx)dλ; Chapter 1 451 b.sin(x) x=/integraldisplay∞ 0A(λ)cos(λx)dλ,w h e r e A(λ)=/braceleftBig1,0<x<1, 0,1<x. 5. a. A(λ)≡0,B(λ)=2s i n(λπ) π(1−λ2); b.A(λ)=1+cos(λπ) π(1−λ2),B(λ)=sin(λπ) π(1−λ2); c.A(λ)=2(1+cos(λπ)) π(1−λ2),B(λ)≡0. 7. Change variable from xtoλwith x=λz. Section 1.10 1.eαx=2sinh(απ) π/parenleftBigg 1 2α+∞/summationdisplay n=1(−1)n α2+n2/parenleftbig αcos(nx)−nsin(nx)/parenrightbig/parenrightBigg . 3.f(x)=/integraldisplay∞ −∞C(λ)eiλxdλ. a.C(λ)=1 2π(1+iλ);b . C(λ)=1+e−iλπ 2π(1−λ2). 5. a. 1 +∞/summationdisplay n=1rncos(nx)=Re∞/summationdisplay 0/parenleftbig reix/parenrightbign=Re1 1−reix; b.∞/summationdisplay n=1sin(nx) n!=Im∞/summationdisplay n=1einx n!=Im exp/parenleftbig eix/parenrightbig . 7. a. f(x)=2s i n(x) x;b . f(x)=2 1+x2. Section 1.11 1.u(t)=A0+∞/summationdisplay n=1Ancos(nt/2)+Bnsin(nt/2), A0=1 2.08,An=0.4/π (1.04−n2)2+(0.4n)2, Bn=−1 nπ1.04−n2 (1.04−n2)2+(0.4n)2. 3.u(x)=∞/summationdisplay n=1Bnsin(nπx/L),Bn=8Ksin(nπ/2) ((nπ/L)2+γ2)n2π2, K=w/EI,γ2=T/EI. 452 Answers to Odd-Numbered Exercises Chapter 1 Miscellaneous Exercises 1.f(x)=∞/summationdisplay n=1bnsin(nx), bn=/braceleftBigg0, neven, 4s i n(nα) παn2,nodd. 3. Yes. As α→0, sin(nα)/nα→1. 5.f(x)=∞/summationdisplay n=1bnsin(nπx/a), bn=2h π2sin(nπα) n2/parenleftbigg1 α+1 1−α/parenrightbigg . 7. a. bn=0,an=0,a0=1; b.∞/summationdisplay n=1bnsin(nπx/a),bn=2(1−cos(nπ)) nπ; c. and d. same as a; e. same as b; f.a0+∞/summationdisplay n=1ancos(nπx/a)+bnsin(nπx/a), a0=1 2,an=0,bn=1−cos(nπ) nπ. 9.f(x)=a0+∞/summationdisplay n=1ancos(nπx/a)+bnsin(nπx/a), a0=1 2a,an=−2a(1−cos(nπ)) n2π2,bn=−2acos(nπ) nπ, x=− a,−a/2,0,a,2a, sum=a, 0, 0,a,0. 11.f(x)=a0+∞/summationdisplay n=1ancos(nx), a0=3 4,an=sin(nπ/2) nπ, x=0,π / 2,π , 3π/2,2π, sum=1,3 4,1 2,3 4, 1. Chapter 1 453 13.f(x)=∞/summationdisplay n=1bnsin(nπx),bn=2/parenleftbig 1+cos(nπ)/parenrightbig /nπ. 15.f(x)=∞/summationdisplay n=1bnsin(nx), b2=1 2,o t h e r bn=4s i n(nπ/2) π(4−n2). 17.N/summationdisplay 1cos(nx)=ReN/summationdisplay 1einx=Reeix−eiNx 1−eix=Reeix/2−ei(2N−1)x/2 e−ix/2−eix/2. The denominator is now −2isin(x/2). 19.f(x)=∞/summationdisplay n=1bnsin(nx),bn=2asin(na+π) n2a2−π2. 21.f(x)=/integraldisplay∞ 0/parenleftbiggsin(λa) λπcos(λx)+1−cos(λa) λπsin(λx)/parenrightbigg dλ. 23.f(x)=/integraldisplay∞ 02s i n(λπ) π(1−λ2)sin(λx)dλ(x>0). 29. Use/integraldisplay∞ 0sin(λt) λdλ=π 2. 31. These answers are not unique. a.∞/summationdisplay n=1bnsin(nπx),bn=2/nπ; b.a0+∞/summationdisplay n=1ancos(nπx),a0=1 2,an=2/parenleftbig 1−cos(nπ)/parenrightbig /n2π2; c./integraldisplay∞ 0B(λ)sin(λx)dλ,B(λ)=2/parenleftbig λ−sin(λ)/parenrightbig/slashbig/parenleftbig πλ2/parenrightbig ; d./integraldisplay∞ 0A(λ)cos(λx)dλ,A(λ)=2/parenleftbig 1−cos(λ)/parenrightbig/slashbig/parenleftbig πλ2/parenrightbig . The integrals of parts c. and d. converge to 0 for x>1. 33. Use s=6i nE q .( 7 )o fS e c t i o n8 . ˆa0=0.78424,ˆa4=− 0.00924, ˆa1=0.22846,ˆa5=0.00744, ˆa2=− 0.02153,ˆa6=− 0.00347, ˆa3=0.01410. 454 Answers to Odd-Numbered Exercises 35.a0=a 6,an=2a n2π2/parenleftbigg cos/parenleftbigg2nπ 3/parenrightbigg −cos/parenleftbiggnπ 3/parenrightbigg/parenrightbigg . 37.a0=5 8,an=2 n2π2/parenleftbigg 3c o s/parenleftbiggnπ 2/parenrightbigg −2−cos(nπ)/parenrightbigg . 39.a0=1 2,an=2 n2π2/parenleftbig 1−cos(nπ)/parenrightbig . 41.a0=a2 6,an=−2a2 n2π2/parenleftbig 1+cos(nπ)/parenrightbig . 43.a0=1 2,an=−1 nπ2s i n/parenleftbiggnπ 2/parenrightbigg . 45.bn=1+cos(nπ/2)−2c o s(nπ) nπ. 47.bn=a/parenleftbigg2s i n(nπ/2) n2π2−cos(nπ) nπ/parenrightbigg . 49.bn=2 nπ/parenleftbigg cos/parenleftbiggnπ 4/parenrightbigg −cos/parenleftbigg3nπ 4/parenrightbigg/parenrightbigg . 51.bn=2nπ(1−ekacos(nπ)) (a2k2+n2π2). 53.A(λ)=2 π(1+λ2). 55.A(λ)=2s i n(λb) πλ. 57.A(λ)=2(1−cos(λ)) πλ2. 59.B(λ)=2λ π(1+λ2). 61.B(λ)=2(1−cos(λb)) λπ. 63.B(λ)=2(λ−sin(λ)) λ2π. 65. The term ancos(nx)+bnsin(nx)appears in Sn,Sn+1,..., SN,a n dt h u s N+1−ntimes in σN. 67. Use Eq. (13) of Section 7 and the identity in Exercise 66.69. a. Use x=0; b. x=1/2; c. x=0. Chapter 2 455 Chapter 2 Section 2.1 1. One possibility: u(x,t)is the temperature in a rod of length awhose lat- eral surface is insulated. The temperature at the left end is held constantatT 0. The right end is exposed to a medium at temperature T1.I n i t i a l l y the temperature is f(x). 3.A/Delta1xg=hC/Delta1x(U−u(x,t)),w h e r e his a constant of proportionality and Cis the circumference. Eq. (4) becomes ∂2u ∂x2+hC κA(U−u)=1 k∂u ∂t. 5. If∂x ∂u(0,t)is positive, then heat is ﬂowing to the left, so u(0,t)is greater than T(t). 7. The second factor is approximately constant if Tis much larger than uor ifTand uare approximately equal. Section 2.2 1.v/prime/prime−γ2(ν−U)=0, 0<x<a, v(0)=T0,v(a)=T1, v(x)=U+Acosh(γx)+Bsinh(γx), A=T0−U,B=(T1−U)−(T0−U)cosh(γa) sinh(γa). One interpretation: uis the temperature in a rod, with convective heat transfer from the cylindrical surface to a medium at temperature U. 3.v(x)=T. Heat is being generated at a rate proportional to u−T.I fγ= π/a, the steady-state problem does not have a unique solution. 5.v(x)=Aln(κ0+βx)+B,A=(T1−T0)/ln(1+aβ/κ 0), B=T0−Aln(κ0). 7.v(x)=T0+r(2a−x)x/2. 9.Du/prime/prime−Su/prime=0, 0<x<a;u(0)=U,u(a)=0, u(x)=U(eSx/D−eSa/D)/(1−eSa/D). 456 Answers to Odd-Numbered Exercises Section 2.3 1.w(x,t)=−2 π(T0+T1)sin/parenleftbiggπx a/parenrightbigg exp/parenleftbigg −π2kt a2/parenrightbigg −2 π/parenleftbiggT0−T1 2/parenrightbigg sin/parenleftbigg2πx a/parenrightbigg exp/parenleftbigg −4π2kt a2/parenrightbigg −··· 3. The partial differential equation is ∂2U ∂ξ2=∂U ∂τ,0<ξ< 1,0<τ. 5.w(x,t)=∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg exp/parenleftbig −n2π2kt/a2/parenrightbig ,bn=T02(1−cos(nπ)) πn. 7.w(x,t)as in the answer to Exercise 5, with bn=2βa π·1 n. 9. a.v(x)=C1; b.∂w ∂t=D∂2w ∂x2,0<x<a,0<t, w(0,t)=0,w(a,t)=0, 0 <t, w(x,0)=C0−C1; c.C(x,t)=C1+∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg exp/parenleftbig −n2π2kt/a2/parenrightbig , bn=(C0−C1)2(1−cos(nπ)) πn; d.t=−a2 Dπ2ln/parenleftbiggπ 40/parenrightbigg ; e.t=6444 s =107.4m i n . Section 2.4 1.a0=T1/2,an=2T1(cos(nπ)−1)/(nπ)2. 3.u(x,t)a sg i v e ni nE q .( 9 ) ,w i t h λn=nπ/a,a0=T0/2, and an= 4T0(2c o s(nπ/2)−1−cos(nπ))/ n2π2. 5. a. The general solution of the steady-state equation is v(x)=c1+c2x. The boundary conditions are c2=S0,c2=S1; thus there is a solution ifS0=S1. If heat ﬂux is different at the ends, the temperature cannot approach a steady state. If S0=S1,t h e n v(x)=c1+S0x,c1undeﬁned. Chapter 2 457 c.A=(S1−S0)/a,B=S0.I fS0/negationslash=S1,t h e n∂u ∂t=kAfor all t. 7.φ/prime/prime+λ2φ=0, 0<x<a, φ(0)=0,φ(a)=0. Solution: φn=sin(λnx),λn=nπ/a(n=1,2,...) . 9. The series∞/summationdisplay n=1|An(t1)|converges. 11. No. u(0,t)is constant if ut(0,t)=0. Section 2.5 1.v(x,t)=T0. 3.u(x,t)=T0+∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig ,λn=(2n−1)π/2a, bn=8T(−1)n+1 π2(2n−1)2−4T0 π(2n−1). 5. The steady-state solution is v(x)=T0−Tx(x−2a)/2a2. The transient satisﬁes Eqs. (5)–(8) with g(x)=T0−v(x)=Tx(x−2a) 2a2. 7.u(x,t)=T0+∞/summationdisplay n=1cncos(λnx)exp/parenleftbig −λ2 nkt/parenrightbig , λn=(2n−1)π/2a,cn=4(T1−T0)(−1)n+1 π(2n−1). 9.u(x,t)=T1cos(πx/2a)exp/parenleftbigg −/parenleftbiggπ 2a/parenrightbigg2 kt/parenrightbigg . 11. The graph of Gin the interval 0 <x<2ais made by reﬂecting the graph ofgin the line x=a( l i k ea ne v e ne x t e n s i o n ) . 13. a. u(x,t)=T0+∞/summationdisplay n=1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig ,λn=(2n−1)π/2a, bn=1 a/integraldisplay2a 0g(x)sin/parenleftbiggnπx 2a/parenrightbigg dx. 458 Answers to Odd-Numbered Exercises In the integral for bn, break the interval of integration at a; in the second integral, make the change of variable y=2a−x. The two integrals cancel ifnis even, and the coefﬁcient is the same as Eq. (18) if nis odd. b. In the solution of Eqs. (1)–(4), the eigenfunction φ(x)=sin((2n− 1)πx/2a)has the property φ(2a−x)=φ(x),s ot h es u mo ft h es e r i e s has the same property. This implies 0 derivative at x=a. 15.W(t)=C0LA/bracketleftBigg 1−∞/summationdisplay n=12e−(λ2 nDt) (n−1/2)2π2/bracketrightBigg . Section 2.6 1. The graph of v(x)is a straight line from T0atx=0t o T∗atx=a,w h e r e T∗=T0+ha k+ha(T1−T0). In all cases, T∗is between T0and T1. 3. Negative solutions provide no new eigenfunctions. 7.bm=2(1−cos(λma)) λm[a+(κ/h)cos2(λma)]. 9.bm=−2(κ+ah)cos(λma) λm(ah+κcos2(λma)). Section 2.7 1.λn=nπ/ln 2,φn=sin(λnln(x)). 3. a. sin (λnx),λn=(2n−1)π/2a; b. cos(λnx),λn=(2n−1)π/2a; c. sin(λnx),λnas o l u t i o no ft a n (λa)=−λ; d.λncos(λnx)+sin(λnx),λna solution of cot (λa)=λ; e.λncos(λnx)+sin(λnx),λnas o l u t i o no ft a n (λa)=2λ/(λ2−1). 5. The weight functions in the orthogonality relations and limits of integra- tion are: a. 1+x,0t o a;b . ex,0t o a;c .1 x2,1t o2 ; d . ex,0t o a. 7. Because λappears in a boundary condition. 9. The negative value of µdoes not contradict Theorem 2 because the coef- ﬁcient α2is not positive. Chapter 2 459 Section 2.8 1.x=∞/summationdisplay n=1cnφn,1<x<b;cn=2nπ1−bcos(nπ) n2φ2+ln2(b). 3. 1=∞/summationdisplay n=1cnφn,0<x<a;cn=2nπ1−ea/2cos(nπ) n2π2+a2/4. (Hint: Find the sine series of ex/2.) 5.bn=/integraldisplayr lf(x)ψn(x)p(x)dx. 7. 1 and√ 2c o s(nπx),n=1,2,.... Section 2.9 1. a.v(x)=constant; b. v(x)=AI(x)+B. 3. If∂u/∂x=0 at both ends, then the steady-state problem is indeterminate. But Eqs. (1)–(3) are homogeneous, so separation of variables applies di-rectly. Note that λ 0=0a n d φ0=1. The constant term in the series for u(x,t)is a0=/integraltextr lp(x)f(x)dx/integraltextr lp(x)dx. Section 2.10 1. The solution is as in Eq. (9), with B(λ)=2T(cos(λa)−cos(λb))/λπ . 3.u(x,t)is given by Eq. (6) with B(λ)=2T0λ π(α2+λ2). 5.u(x,t)=/integraldisplay∞ 0A(λ)cos(λx)exp/parenleftbig −λ2kt/parenrightbig dλ; A(λ)=2T πλ/parenleftbig sin(λb)−sin(λa)/parenrightbig . 7.u(x,t)=T0+/integraldisplay∞ 0B(λ)sin(λx)exp/parenleftbig −λ2kt/parenrightbig dλ; B(λ)=2 π/integraldisplay∞ 0/parenleftbig f(x)−T0/parenrightbig sin(λx)dx. 9. a.v(x)=C0e−ax; b.∂w ∂t=D/parenleftbigg∂2w ∂x2−a2w/parenrightbigg ,0<x,0<t, 460 Answers to Odd-Numbered Exercises w(0,t)=0, 0<t, w(x,0)=− C0e−ax,0<x; c.w(x,t)=e−a2Dt/integraldisplay∞ 0B(λ)sin(λx)e−λ2Dtdλ, B(λ)=− 2C0λ//parenleftbig π/parenleftbig λ2+a2/parenrightbig/parenrightbig . Section 2.11 1. Break the interval of integration at x/prime=0. 3.B(λ)=0,A(λ)=2T0a π(1+λ2a2). 5. The function u(x,t), as a function of x, is the famous “bell-shaped” curve. The smaller tis, the more sharply peaked the curve. 7. In Eq. (3) replace both f(x/prime)and u(x,t)by 1. 9. Using the integral given, obtain u(x,t)=2 π/integraldisplay∞ 01 λsin(λx)e−λ2ktdλ. Note, however, that B(λ)=2/λπ isnotfound using the usual formulas for Fourier coefﬁcient functions. Section 2.12 5. As t→0+,x/√ 4πkt→/braceleftbigg +∞ ifx>0, −∞ ifx<0, so erf(x/√ 4πkt)→/braceleftbigg +1i f x>0, −1i f x<0. 7. Make the substitution x=y2.T h e n I(x)=√πerf(√x)+c. 9. Let zbe deﬁned by erf (z)=− Ub/(Ui−Ub).T h e n x(t)=z√ 4kt. Chapter 2 Miscellaneous Exercises 1. SS: v(x)=T0,0<x<a. EVP:φ/prime/prime+λ2φ=0,φ(0)=0,φ(a)=0,λn=nπ/a,φn=sin(λnx), n=1,2,.... Chapter 2 461 u(x,t)=T0+∞/summationdisplay 1bnsin(λnx)e−λ2 nkt, bn=2 a/integraldisplaya 0(T1−T0)sin/parenleftbiggnπx a/parenrightbigg dx. 3. SS: v(x)=T0+r 2x(x−a),0<x<a. EVP:φ/prime/prime+λ2φ=0,φ(0)=0,φ(a)=0,λn=nπ/a,φn=sin(λnx), n=1,2,.... u(x,t)=T0−r 2x(x−a)+∞/summationdisplay 1bnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig , bn=2 a/integraldisplaya 0/bracketleftbigg T1−T0+r 2x(x−a)/bracketrightbigg sin/parenleftbiggnπx a/parenrightbigg dx. 5. SS: not needed. (Hint: Put −γ2uon the other side of the equation. Separation of vari- ables gives φ/prime/prime/φ=γ2+T/prime/kT=−λ2.) EVP:φ/prime/prime+λ2φ=0,φ/prime(0)=0,φ/prime(a)=0,λ0=0,φ0=1;λn=nπ/a, φn=cos(λnx),n=1,2,.... u(x,t)=e−γ2kt/parenleftBig a0+/summationdisplay ancos(λnx)exp/parenleftbig −λ2 nkt/parenrightbig/parenrightBig . a0=T1/2,an=− 2T1/parenleftbig 1−cos(nπ)/parenrightbig /n2π2. 7.u(x,t)=T0. 9.u(x,t)=T0+∞/summationdisplay n=1cnsin(λnx)exp/parenleftbig −λ2 nkt/parenrightbig , λn=(2n−1)π 2a,cn=(T1−T0)·4 (2n−1)π. 11.u(x,t)=T0+/integraldisplay∞ 0B(λ)sin(λx)exp/parenleftbig −λ2kt/parenrightbig dλ,B(λ)=−2λT0 π(α2+λ2). 13.u(x,t)=/integraldisplay∞ 0A(λ)cos(λx)exp/parenleftbig −λ2kt/parenrightbig dλ,A(λ)=2T0sin(λa) πλ. 15.u(x,t)=/integraldisplay∞ 0/parenleftbig A(λ)cos(λx)+B(λ)sin(λx)/parenrightbig exp/parenleftbig −λ2kt/parenrightbig dλ, A(λ)=T0sin(λa) πλ,B(λ)=T0(1−cos(λa)) πλ or 462 Answers to Odd-Numbered Exercises u(x,t)=T0√ 4πkt/integraldisplaya 0exp/parenleftbigg −(x/prime−x)2 4kt/parenrightbigg dx/prime =T0 2/bracketleftbigg erf/parenleftbigga−x√ 4kt/parenrightbigg +erf/parenleftbiggx√ 4kt/parenrightbigg/bracketrightbigg . 17. Interpretation: uis the temperature in a rod with insulation on the cylin- drical surface and on the left end. At the right end, heat is being forcedinto the rod at a constant rate (because q(a,t)=−κ ∂u ∂x(a,t)=−κS,s o heat is ﬂowing to the left, into the rod). The accumulation of heat energyaccounts for the steady increase of temperature. 19.(1/6ka)u 3−(a/6k)u1satisﬁes the boundary conditions. 21.w(x,t)=−2 u∂u ∂x,w h e r e u(x,t)=a0+/summationdisplay ancos(nπx)exp/parenleftbig −n2π2t/parenrightbig , where a0=2/parenleftbig 1−e−1/2/parenrightbig and an=1−e−1/2cos(nπ) 1 4+(nπ)2. 23.u2=T0β2V β1+β2,u1=T0/parenleftbigg 1−β1V β1+β2/parenrightbigg , where V=1−exp(−(β1+β2)t)andβi=h/ci. 25.u(ρ,t)=1 ρ∞/summationdisplay n=1bnsin(λnρ)exp/parenleftbig −λ2 nkt/parenrightbig , λn=nπ/a,bn=2 a/integraldisplaya 0ρT0sin(λnρ)dρ. 27.v(x)=T0+Sx−Ssinh(λx) γcosh(γa). 29. If λ=0, the differential equation is φ/prime/prime=0 with general solution φ(x)= c1+c2x. The boundary conditions require c2=0b u ta l l o w c1/negationslash=0. Thus, this value of λpermits the existence of a nonzero solution, and therefore λ=0i sa ne i g e n v a l u e . 31. Choose B(ω)=2 π/integraltext∞ 0f(t)sin(ωt)dt.I ffhas a Fourier integral represen- tation, then this choice of Bwill make u(0,t)=f(t),0<t. 33. a. v(x)=− Ix/aK+c1+c2(1−e−aKx/T), c1=h1,c2=(h2−h1+IL/aK)/(1−e−aKL/T). b.∂2w ∂x2+µ∂w ∂x=1 k∂w ∂t,0<x<L,0<t, w(0,t)=0,w(L,t)=0, 0<t, w(x,0)=h0(x)−v(x),0<x<L, where µ=aK/T,k=T/S. Chapter 3 463 c.w(x,t)=/summationdisplay cnφn(x)e−λ2 nkT,φn(x)=e−µx/2sin(nπx/L), λ2 n=/parenleftbiggnπ L/parenrightbigg2 +µ2 4; d.λ2 n=(7.30n2+0.0133)×10−4m−1. 35. a.∂u ∂t=D∂2u ∂x2,0<x<L,0<t, ∂u ∂x(0,t)=0, u(L,t)=S0,0<t; u(0,t)=0, 0 <x<L; b.u(x,t)=S0+∞/summationdisplay n=1cncos(λnx)exp/parenleftbig −λ2 nDt/parenrightbig , cn=4S0(−1)n/(2n−1). 37.T(y,t)=300−150y/c+/summationtextbnsinλn(y+c)exp(−λ2 nkt),λn=nπ/2c, bn=(400 cos (nπ)+1000)/nπ. c. Just before time t=0, the three terms a d dt o0 .J u s ta f t e rt i m e t=0, the integrated terms do not change sensi- bly, but in the ﬁrst term, near y=c,T(y,t)changes suddenly. Chapter 3 Section 3.1 1. [u]=L,[c]=L/t. 3.v(x)=(x2−ax)g 2c2. Section 3.2 3.u(x,t)=∞/summationdisplay n=1bnsin/parenleftbiggnπx a/parenrightbigg sin/parenleftbiggnπct a/parenrightbigg , bn=2a(1−cos(nπ)) n2π2c. 5.u(x,t)=∞/summationdisplay n=1ancos/parenleftbiggnπct a/parenrightbigg sin/parenleftbiggnπx a/parenrightbigg ,an=2U01−cos(nπ/2) nπ. 7. a. sin/parenleftbiggnπx a/parenrightbigg ;b . s i n/parenleftbigg2n−1 2πx a/parenrightbigg . 464 Answers to Odd-Numbered Exercises 9. Product solutions are φn(x)Tn(t),w h e r e φn(x)=sin(λnx), Tn(t)=exp/parenleftbig −kc2t/2/parenrightbig ×/braceleftbigg sin(µnt) cos(µnt), λn=nπ a,µ n=/radicalbigg λ2nc2−1 4k2c4. 11. Product solutions are φn(x)Tn(t),w h e r e φn(x)=sin/parenleftbiggnπx a/parenrightbigg , Tn(t)=sin or cos/parenleftbiggn2π2ct a2/parenrightbigg . Frequencies n2π2c/a2. 13. The general solution of the differential equation is φ(x)=Acos(λx)+ Bsin(λx)+Ccosh(λx)+Dsinh(λx). Boundary conditions at x=0r e - quire A=− C,B=− D;t h o s ea t x=alead to C/D=−(cosh(λa)+ cos(λa))/(sinh(λa)−sin(λa))and 1 +cos(λa)cosh(λa)=0. The ﬁrst eigenvalues are λ1=1.875/a,λ2=4.693/a, and the eigenfunctions are similar to the functions shown in the ﬁgure. 15.u(x,t)=∞/summationdisplay n=1/parenleftbig ancos(µnt)+bn(sinµnt)/parenrightbig sin(λnx):λn=nπ/a, µn=/radicalbig λ2n+γ2c,an=2h/parenleftbig 1−cos(nπ)/parenrightbig /nπ,bn=0,n=1,2,.... 17. Convergence is uniform because/summationtext|bn|converges. Section 3.3 1. Table shows u(x,t)/h. t x 00 .2a/c0.4a/c0.8a/c1.4a/c 0.25a 0.5 0.5 0.2 −0.5−0.2 0.5a 1.0 0.6 0.2 −0.6−0.2 3.u(0,0.5a/c)=0;u(0.2a,0.6a/c)=0.2αa;u(0.5a,1.2a/c)=− 0.2αa. (Hint: G(x)=αx,0<x<a.) 5.G(x)=/braceleftBigg0, 0<x<0.4a, 5(x−0.4a),0.4a<x<0.6a, a, 0.6a<x<a. Notice that Gis a continuous function whose graph is composed of line segments. Chapter 3 465 Figure 2 Solution for Exercise 7, Section 3.3. Figure 3 Solution for Exercise 9, Section 3.3. 7. See Fig. 2. 9. See Fig. 3. 11. By the chain rule we calculate ∂u ∂x=∂v ∂w∂w ∂x+∂v ∂z∂z ∂x=∂v ∂w+∂v ∂z, ∂2u ∂x2=∂ ∂w/parenleftbigg∂v ∂w+∂v ∂z/parenrightbigg∂w ∂x+∂ ∂z/parenleftbigg∂v ∂w+∂v ∂z/parenrightbigg∂z ∂x =∂2v ∂w2+2∂2v ∂w∂ z+∂2v ∂z2 466 Answers to Odd-Numbered Exercises and similarly ∂2u ∂t2=c2/parenleftbigg∂2v ∂w2−2∂2v ∂z∂w+∂2v ∂z2/parenrightbigg . (We have assumed that the two mixed partials ∂2v/∂z∂wand∂2v/∂w∂ z are equal.) If u(x,t)satisﬁes the wave equation, then ∂2u ∂x2=1 c2∂2u ∂t2. In terms of the function vand the new independent variables this equa- tion becomes ∂2v ∂w2+2∂2v ∂z∂w+∂2v ∂z2=∂2v ∂w2−2∂2v ∂z∂w+∂2v ∂z2 or, simply, ∂2v ∂z∂w=0. 13.u(x,t)=− c2cos(t)+φ(x−ct)+ψ(x+ct). Section 3.4 1. If fand gare sectionally smooth and fis continuous. 3. The frequency is cλnrads/sec, and the period is 2 π/cλnsec. 5. Separation of variables leads to the following in place of Eqs. (11) and (12): T/prime/prime+γT/prime+λ2c2T=0, (11/prime) /parenleftbig s(x)φ/prime/parenrightbig/prime−q(x)φ+λ2p(x)φ=0. (12/prime) The solutions of Eq. (11/prime) all approach 0 as t→∞ ,i fγ> 0. 7. The period of Tn(t)=ancos(λnct)+bnsin(λnct)is 2π/λ nc.A l l Tn’s have a common period pif and only if for each nthere is an integer msuch that m(2π/λ nc)=p,o r m=(pc/2π)λ nis an integer. For λnas shown andβ=q/r,w h e r e qand rare integers, this means m=/parenleftbiggpc 2π/parenrightbigg α/parenleftbigg n+q r/parenrightbigg or m=/parenleftbiggpc 2π/parenrightbiggα r(rn+q). Chapter 3 467 Figure 4 Solution for Exercise 3, Section 3.6. Given α,pcan be adjusted so that mis an integer whenever nis an integer. Section 3.5 1. If q≥0, the numerator in Eq. (3) must also be greater than or equal to 0, sinceφ1(x)cannot be identically 0. 3. 2π2/3i so n ee s t i m a t ef r o m y=sin(πx). 5./integraldisplay2 1(y/prime)2dx=1 3,/integraldisplay2 1y2 x4dx=25 6−6l n2 ; N(y)/D(y)=42.83;λ1≤6.54. Section 3.6 1.u(x,t)=1 2[fe(x+ct)+Go(x+ct)]+1 2[fe(x−ct)−Go(x−ct)], where fe is the even extension of fand Gois the odd extension of G. 3. See Fig. 4.5. See Fig. 5. 7.u(x,t)=1 2/bracketleftbig f(x+ct)+f(x−ct)/bracketrightbig +1 2c/integraldisplayx+ct x−ctg(y)dy. Chapter 3 Miscellaneous Exercises 1.u(x,t)=∞/summationdisplay 1bnsin(λnx)cos(λnct),bn=2/parenleftbig 1−cos(nπ)/parenrightbig /nπ,λn=nπ/a. 468 Answers to Odd-Numbered Exercises Figure 5 Solution for Exercise 5, Section 3.6. Figure 6 Solution of Miscellaneous Exercise 3, Chapter 3. 3. See Fig. 6. 5. See Fig. 7.7. See Fig. 8.9. See Fig. 9. 11. See Fig. 10. Chapter 3 469 Figure 7 Solution of Miscellaneous Exercise 5, Chapter 3. Figure 8 Solution of Miscellaneous Exercise 7, Chapter 3. 13. See Fig. 11. 15. Using y(x)=x(1−x),ﬁ n dλ2 1≤10.5. 17.f(q)=12a2sech2(aq),c=4a2. 470 Answers to Odd-Numbered Exercises Figure 9 Solution of Miscellaneous Exercise 7, Chapter 3. Figure 10 Solution for Miscellaneous Exercise 11, Chapter 3. Figure 11 Solution of Miscellaneous Exercise 13, Chapter 3. 21.v(x,t)=∞/summationdisplay n=1/parenleftbig ancos(λnct)+bnsin(λnct)/parenrightbig sin(λnx), λn=(2n−1)π/2a, Chapter 4 471 an=8aU0(−1)n+1 π2(2n−1)2,bn=0. 23.Y/prime/prime Y=2V kψ/prime ψ.T h ef u n c t i o n φ(x−Vt)cancels from both sides. 25.φn(−Vt)=T0exp(λ2 nkt/2)bn,t>0, φn(x)=T1exp(λ2 nkx/2V)bn,x>0, where∞/summationdisplay n=1bnsin(λny)=1, 0<y<b. 27.φ(x−ct)=e−c(x−ct)/k=e(c2t−cx)/k.T h eg i v e n csatisﬁes c2=iωk, soφ(x−ct)=eiωt−(1+i)px=e−pxei(ωt−px). Now form1 2(φ(x−ct)+ φ(x−ct))=e−pxcos(ωt−px)and so forth. 29. Differentiate and substitute. 31.φ(2)−/epsilon1φ(4)+λ2φ=0, φ(0)=0,φ(a)=0, φ/prime/prime(0)=0,φ/prime/prime(a)=0. 33.λn=nπ a/radicalBigg 1+/epsilon1/parenleftbiggnπ a/parenrightbigg2 ∼=nπ a. Chapter 4 Section 4.1 1.f+d=0. 3.Y(y)=Asinh(πy),A=1/sinh(π). 5.v(r)=aln(r)+b. 7.∂u ∂x=∂v ∂rcos(θ)−∂v ∂θsin(θ) r, ∂u ∂y=∂v ∂rsin(θ)+∂v ∂θcos(θ) r. 9. a.∂2u ∂x2+∂2u ∂y2=0, 0 <x<a,0<y<b, u(0,y)=0, u(a,y)=0, 0 <y<b, u(x,0)=f(x),u(x,b)=f(x),0<x<a. 472 Answers to Odd-Numbered Exercises Membrane is attached to a frame that is ﬂat on the left and right but has the shape of the graph of f(x)at top and bottom. b.∂2u ∂x2+∂2u ∂y2=0, 0 <x<a,0<y<b, ∂u ∂x(0,y)=0, u(a,y)=0, 0 <y<b, u(x,0)=0, u(x,b)=100, 0 <x<a. The bar is insulated on the left; the temperature is ﬁxed at 100 on the top, at 0 on the other two sides. c.∂2u ∂x2+∂2u ∂y2=0, 0 <x<a,0<y<b, u(0,y)=0, u(a,y)=100, 0 <y<b, ∂u ∂y(x,0)=0,∂u ∂y(x,b)=0, 0 <x<a. The sheet is electrically insulated at top and bottom. The voltage is ﬁxed at 0 on the left and 100 on the right. d.∂2φ ∂x2+∂2φ ∂y2=0, 0 <x<a,0<y<b, ∂φ ∂x(0,y)=0,∂φ ∂x(a,y)=− a,0<y<b, ∂φ ∂y(x,0)=0,∂φ ∂y(x,b)=b,0<x<a. The velocities, given by V=− ∇ φ,a r e Vx=a,Vy=0o nt h er i g h t , Vx=0,Vy=− bo nt h et o p ;a n dw a l l so nt h eo t h e rt w os i d e sm a k ev e - locities 0 there. Section 4.2 1. Show by differentiating and substituting that both are solutions of the differential equation. The Wronskian of the two functions is /vextendsingle/vextendsingle/vextendsingle/vextendsinglesinh(λy) sinh(λ(b−y)) λcosh(λy)−λcosh(λ(b−y))/vextendsingle/vextendsingle/vextendsingle/vextendsingle=−λsinh(λb)/negationslash=0. 3. In the case b=a, use two terms of the series: u(a/2,a/2)=0.32. 5.u(x,y)= ∞/summationdisplay 1bnsin/parenleftbiggnπx a/parenrightbiggsinh(nπy/a) sinh(nπb/a),bn=8 n2π2sin/parenleftbiggnπ 2/parenrightbigg . Chapter 4 473 7. a. See Eq. (11). an=0,cn=200(1−cos(nπ))/ nπ; b.u(x,y)=u1(x,y)+u2(x,y),u1(x,y)is the solution to Part a, u2(x,y)=∞/summationdisplay n=1cnsinh(µnx) sinh(µna)sin(µny), µn=nπ/b,cn=200(1−cos(nπ))/ nπ. c.u(x,y)=u1(x,y)+u2(x,y),w h e r e u1(x,y)=∞/summationdisplay n=1cnsinh(λny) sinh(λnb)sin(λnx), u2(x,y)=∞/summationdisplay n=1cnsinh(µnx) sinh(µna)sin(µny). In both series, cn=2ab(−1)n+1/nπ. Also note u(x,y)=xy. Section 4.3 1. a. u(x,y)=1, but the form found by applying the methods of this sec- tion is u(x,y)=∞/summationdisplay n=1ansinh(λny)+sinh(λn(b−y)) sinh(λnb)cos(λnx) +∞/summationdisplay n=1bncosh(µnx) cosh(µna)sin(µny), where λn=(2n−1)π 2a,an=4s i n/parenleftbig(2n−1)π 2/parenrightbig π(2n−1), µn=nπ b,bn=2(1−cos(nπ)) nπ. b.u(x,y)=y/b, and this is found by the methods of this section. In this case, 0 is an eigenvalue. c.4 π∞/summationdisplay 1(−1)n+1cos(λny) (2n−1)sinh(λn(a−x)) sinh(λna),λn=/parenleftbigg2n−1 2π b/parenrightbigg . 3.b0b=V0 2,bnsinh(λnb)=2V0(cos(nπ)−1) n2π2. 474 Answers to Odd-Numbered Exercises 5. Check zero boundary conditions by substituting. At x=a,ﬁ n d Ancosh(µna)=2 b/integraldisplayb 0Sycos(µny)dy. 7.w(x,y)=∞/summationdisplay n=1ancosh(λny)cos(λnx). From the condition at y=b, ancosh(λnb)=2 a/integraldisplaya 0Sb a(x−a)cos(λnx)dx. 9.w(x,y)=∞/summationdisplay n=1cnsinh(λny)+ansinh(λn(b−y)) sinh(λnb)sin(λnx), an=cn=−2 a/integraldisplaya 0Hx(a−x)sin(λnx)dx=− 2Ha21−cos(nπ) n3π3. 11. 12 A+2C=− K,1 2E+2C=− K. There are many solutions. Section 4.4 1.an=2 a/integraldisplaya 0f(x)sin/parenleftbiggnπx a/parenrightbigg dx. 3.A(µ)=2 π/integraldisplay∞ 0g2(y)sin(µy)dy. 5. a. u(x,y)=/summationdisplay cncos(λnx)exp(−λny),λn=(2n−1)π/2a, cn=4(−1)n+1/π(2n−1); b.u(x,y)=/integraldisplay∞ 0B(λ)cosh(λx)sin(λy)dλ,B(λ)=2λ π(λ2+1)cosh(λa); c.u(x,y)=/integraldisplay∞ 0A(λ)cos(λy)sinh(λx)dλ,A(λ)=2s i n(λb) πλsinh(λa). 7.u(x,y)=∞/summationdisplay 1bnsin(λnx)exp(−λny) +/integraldisplay∞ 0/parenleftbigg A(µ)sinh(µx) sinh(µa)+B(µ)sinh(µ(a−x)) sinh(µa)/parenrightbigg sin(µy)dµ, λn=nπ/a,bn=2(1−cos(nπ))/ nπ,A(µ)=B(µ)=2µ/π(µ2+1). Also see Exercise 8. 9. a. u(x,y)=2 π/integraldisplay∞ 01−cos(λa) λsin(λx)sinh(λy) sinh(λb)dλ; Chapter 4 475 b.u(x,y)=2 π/integraldisplay∞ 0λ 1+λ2sin(λx)sinh(λ(b−y)) sinh(λb)dλ. 11.u(x,y)=/integraldisplay∞ 02 π(1+λ2)sinh(λx) sinh(λa)cos(λy)dλ. 13.e−λysin(λx),λ> 0. 15.e−λysin(λx),e−λycos(λx),λ> 0. 17.u(x,y)=1 π/bracketleftbiggπ 2+tan−1(x/y)/bracketrightbigg . 19. This solution is unbounded as xtends to inﬁnity and cannot be found by the method of this section. Section 4.5 1.v(r,θ)is given by Eq. (10) with bn=0,a0=π/2, an=− 2( 1−cos(nπ))/π n2cn. 3. The solution is as in Eq. (10) with bn=0,a0=1/π,a1=1/2, and an=2s i n((n−1)π/2) π(n2−1)forn/negationslash=1. 5. Convergence is uniform in θ. 7.a0=1 2π/integraldisplayπ −πf(θ)dθ,an=cn π/integraldisplayπ −πf(θ)cos(nθ)dθ, bn=cn π/integraldisplayπ −πf(θ)sin(nθ)dθ. 9.2 π∞/summationdisplay n=11−cos(nπ) nc2nr2nsin(2nθ)=v(r,θ). 11.vn(r,θ)=rn/αsin(nθ/α) has∂v/∂ runbounded as r→0+,i fn=1. Section 4.6 1. Hyperbolic (a) and (e); elliptic (b) and (c); parabolic (d). 3. Only (e). 5. a. u(x,y)=∞/summationdisplay 1ansin(nπx)e−nπy; b.u(x,y)=∞/summationdisplay 1ansin(nπx)cos(nπy); 476 Answers to Odd-Numbered Exercises c.u(x,y)=∞/summationdisplay 1ansin(nπx)exp(−n2π2y), an=2/integraldisplay1 0f(x)sin(nπx)dx. 7.X/prime/prime/X=−λ2,T/prime/prime/T=−λ2/(1+/epsilon1λ2). Chapter 4 Miscellaneous Exercises 1.u(x,y)=∞/summationdisplay 1bnsinh(λn(a−x)) sinh(λna)sin(λny), λn=nπ/b,bn=2(1−cos(nπ))/ nπ. 3.u(x,y)=1 .N o t et h a t0i sa ne i g e n v a l u e . 5.u(x,y)=∞/summationdisplay n=1ansinh(λnx)+bnsinh(λn(a−x)) sinh(λna)cos(λny), λn=(2n−1)π/2b,an=bn=4(−1)n+1/π(2n−1). 7.u(x,y)=w(x,y)+w(y,x),w h e r e w(x,y)=∞/summationdisplay n=1bnsinh(λn(a−y)) sinh(λna)sin(λnx), λn=nπ/a,bn=8h n2π2sin/parenleftbiggnπ 2/parenrightbigg . 9.u(x,y)=/integraldisplay∞ 0A(λ)sinh(λ(b−y)) sinh(λb)cos(λx)dλ,A(λ)=2s i n(λa)/λπ . 11.u(x,y)=/integraldisplay∞ 0A(λ)cos(λx)e−λydλ,A(λ)=2α/π/parenleftbig α2+λ2/parenrightbig . 13.u(x,y)=−1 πtan−1/parenleftbiggx−x/prime y/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞ −∞=1 π/bracketleftbiggπ 2−/parenleftbigg −π 2/parenrightbigg/bracketrightbigg . 15.u(r,θ)=a0+∞/summationdisplay n=1/parenleftbiggr c/parenrightbiggn/parenleftbig ancos(nθ)+bnsin(nθ)/parenrightbig , a0=1 2,an=0,bn=1−cos(nπ) nπ. 17. Same form as Exercise 15, but a0=2/π, an=2(1+cos(nπ))/( 1−n2),bn=0( a n d a1=0). 19.u(r,θ)=(ln(r)−ln(b))/(ln(a)−ln(b)). Chapter 4 477 21.u(r,θ)=∞/summationdisplay 1bn/parenleftbiggr c/parenrightbiggn/2 sin(nθ/2),bn=1 π/integraldisplay2π 0f(θ)sin(nθ/2)dθ. 23.u(x,y)=/summationdisplay cnsinh(λny)sin(λnx),λn=(2n−1)π/2a, cn=2s i n(λna)/(aλ2 nsinh(λnb)). 25.wsatisﬁes the potential equation in the rectangle with boundary condi- tions w(0,y)=0,wx(a,y)=ay/b,0<y<b, w(x,0)=0,w(x,b)=0, 0<x<a. w(x,y)=∞/summationdisplay n=1bnsin(λny)cosh(λnx), λn=nπ/b,bn=2a(−1)n+1/n2π2cosh(λna). 27. The equations become ∂2φ ∂y∂x=∂2φ ∂x∂y,/parenleftbig 1−M2/parenrightbig∂2φ ∂x2+∂2φ ∂y2=0. 29.φ(x,y)=/integraldisplay∞ 0/parenleftbig A(α)cos(αx)+B(α)sin(αx)/parenrightbig e−βydα+c, where β=α√ 1−M2,cis an arbitrary constant, and A(α) B(α)/bracerightbigg =−U0 βπ/integraldisplay∞ −∞f/prime(x)/braceleftbiggcos(αx) sin(αx)/bracerightbigg dx. 31. If (x(s),y(s))is the parametric representation for the boundary curve C, then the vector y/primei−x/primejis normal to C,a n d /integraldisplay C∂u ∂nds=/integraldisplay C∂u ∂xdy−∂u ∂ydx. By Green’s theorem, /integraldisplay C∂u ∂xdy−∂u ∂ydx=/integraldisplay/integraldisplay R/parenleftbigg∂2u ∂x2+∂2u ∂y2/parenrightbigg dA, which is 0, since usatisﬁes the potential equation in R. 33. Substitute directly.35.−∇u=−(xi+yj)/(x 2+y2). 478 Answers to Odd-Numbered Exercises 37. a. u=−r2 4+c1ln(r)+c2; b.u=−(ln(r))2 2+c1ln(r)+c2. 39.V(x,y)∼=a0=1 a/integraldisplaya 0f(x)dxify>5L(because e−λ1.5L∼=0). 41. The solution is θ(X,Y)=1. The low Biot number, B=0, means a very large value for conductivity κ, so very little or no cooling takes place. Chapter 5 Section 5.1 1.∂2u ∂x2+∂2u ∂y2=1 c2∂2u ∂t2,0<x<a,0<y<b,0<t, u(x,0,t)=0, u(x,b,t)=0, 0 <x<a,0<t, u(0,y,t)=0, u(a,y,t)=0, 0 <y<b,0<t, u(x,y,0)=f(x,y),∂u ∂t(x,y,0)=g(x,y),0<x<a,0<y<b. 3.∂2u ∂x2+∂2u ∂y2+∂2u ∂z2=1 c2∂2u ∂t2. Section 5.2 1./integraldisplayc 0∂2u ∂z2dz=∂u ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsinglec 0=0 by Eq. (12). 3.W/prime/prime+(2h/bκ)(T2−W)=0, 0<x<a,W(0)=T0,W(a)=T1. W(x)=T2+Acosh(µx)+Bsinh(µx),w h e r e µ2=2h/bκ, A=T0−T2,B=(T1−T2−Acosh(µa))/sinh(µa). 5.∇2u=1 k∂u ∂t,0<x<a,0<y<b,0<t, ∂u ∂x(0,y,t)=0, u(a,y,t)=T0,0<y<b,0<t, u(x,0,t)=T0,∂u ∂y(x,b,t)=0, 0<x<a,0<t, u(x,y,0)=f(x,y),0<x<a,0<y<b. Chapter 5 479 Section 5.3 1. If a=b, the lowest eigenvalues are those with indices (m,n),i nt h i so r - der:(1,1);(1,2)=(2,1);(2,2);(3,1)=(1,3);(3,2)=(2,3);(1,4)= (4,1);(3,3). 3. Frequencies are λmnc/2π(Hz), where λ2 mnare the eigenvalues found in the text. 5.λ2 mn=(mπ/a)2+(nπ/b)2, for m=0,1,2,...,n=1,2,3,.... 7. a. u(x,y,t)=1. For b and c the solution has the form u(x,y,t)=/summationdisplay m,namncos/parenleftbiggmπx a/parenrightbigg cos/parenleftbiggnπy b/parenrightbigg exp/parenleftbig −λ2 mnkt/parenrightbig , where λ2 mn=(mπ/a)2+(nπ/b)2and mand nrun from 0 to ∞. b.a00=(a+b) 2,am0=−2b(1−cos(mπ)) m2π2, a0n=−2a(1−cos(nπ)) n2π2,amn=0o t h e r w i s e ; c.a00=ab 4,am0=−ab(1−cos(mπ)) m2π2,a0n=−ab(1−cos(nπ)) n2π2, amn=4ab(1−cos(nπ))( 1−cos(mπ)) m2n2π4 ifmand nare greater than zero. 9. The choice of a positive constant for either X/prime/prime/XorY/prime/prime/Y,u n d e rt h e boundary conditions in Eqs. (9) and (10), will lead to the trivial solution. 11. The nodal lines form a grid: umn(x,y,t)=0a t x=0,a/m,2a/m,..., a and at y=0,b/n,2b/n,..., b. Section 5.4 1. The partial differential equations are the same, the boundary conditions become homogeneous, and in the initial conditions g(r,θ)is replaced by g(r,θ)−v(r,θ). 3. In the heat problem, T/prime+λ2kT=0 .I nt h ew a v ep r o b l e m , T/prime/prime+λ2c2T=0. 5. The boundary conditions Eqs. (10) and (11) would be replaced by Q(0)=0,Q(π)=0. Solutions are Q(θ)=sin(nθ),n=1,2,.... 480 Answers to Odd-Numbered Exercises 7. Taking the hint and using the fact that ∇2φ=−λ2φ, the left-hand side becomes /parenleftbig λ2 k−λ2 m/parenrightbig/integraldisplay/integraldisplay Rφkφm while the right-hand side is zero, because of the boundary condition. Section 5.5 1.λn=αn/a,w h e r e αnis the nth zero of the Bessel function J0.T h es o l u t i o n s areφn(r)=J0(λnr)or any constant multiple thereof. 3. This is just the chain rule.5. Rolle’s theorem says that if a differentiable function is zero in two places, its derivative is zero somewhere between. From Exercise 4 it is clear that J 1must be zero between consecutive zeros of J0. Check Fig. 7 and Table 1. 7. Use the second formula of Exercise 6, after replacing µbyµ+1o nb o t h sides. 9.u(r)=T+(T1−T)I0(γr)/I0(γa). Section 5.6 1.v(0,t)/T0∼=1.602 exp (−5.78τ)−1.065 exp (−30.5τ),w h e r e τ=kt/a2. 3.v(r,t)=∞/summationdisplay n=1anJ0(λnr)exp(−λ2 nkt),λn=αn/a.U s eE q .( 1 3 )a n do t h e r st o ﬁnd an=T0J1(αn/2)/α nJ2 1(αn). 5. Integration leads to the equality /integraldisplaya 0/parenleftbig rφ/prime2/parenrightbig/primedr+λ2/integraldisplaya 0r2/parenleftbig φ2/parenrightbig/primedr=0. The ﬁrst integral is evaluated directly. The second must be integrated by parts. Section 5.7 1. Use 1 rd dr/parenleftbigg rd drJ0(λr)/parenrightbigg =−λ2J0(λr). 3. The frequencies of vibration are λmnc=αmnc/a.T h eﬁ v el o w e s tv a l u e so f αmn, in order, have subscripts (0,1),(1,1),(2,1),(0,2),a n d(3,1).S e e Table 1 in Section 5.6. Chapter 5 481 5.φ(a,θ)=0a n dφ(r,θ)=φ(r,θ+2π). 7. Set Jm(λmnr)=φn.T h e n (rφ/prime n)/prime=−λmnrφnand(rφ/prime q)/prime=−λmqrφqare the equations satisﬁed by the functions in the integrand. Follow the proof inSection 2.7. 9. Generally, radii for φ 0narer=λ0m/λ0nform=1,2,..., n.F o r n=2, r=2.405/5.520=0.436 and 1. Section 5.8 1.φ(x)=xα[AJp(λx)+BYp(λx)], where α=(1−n)/2,p=|α|. 3. For λ2=0,Z=A+Bz. 5.φ(ρ+ct)=¯Fo(ρ+ct)+¯Ge(ρ+ct), ψ(ρ−ct)=¯Fo(ρ−ct)−¯Ge(ρ−ct), where ¯Fo(x)is the odd periodic extension with period 2 aofxf(x)/2a n d ¯Ge(x)is the even periodic extension with period 2 aof/integraltext (x/2c)g(x)dx. 7. The weight function is ρ2and the interval is 0 to a. 9.v(x)=(b−x)(x−a)/(a+b)x2. 11. No. The idea is to ﬁnd a solution of the partial differential equation that depends on only one variable. That is impossible if fdepends on both x and y. 13.an=−/integraltextb av(x)Xn(x)x3dx /integraltextb aX2n(x)x3dx. Section 5.9 1. [k(k+1)−µ2]ak+1−[k(k−1)−µ2]ak−1=0, valid for k=1,2,.... 3.P5=1 8/parenleftbig 63x5−70x3+15x/parenrightbig . 5.y=Aln/parenleftbigg1+x 1−x/parenrightbigg . 7. Differentiate Eq. (9) and add to it ntimes Eq. (8). 9. Leibniz’s rule states that (uv)(k)=k/summationdisplay r=0/parenleftbigg k r/parenrightbigg u(k−r)v(r). 482 Answers to Odd-Numbered Exercises (A superscript kin parentheses means kth derivative.) The right-hand side looks like the binomial theorem. In the case at hand, at most twoterms are not zero. 11.b n=  0( nodd), −(−1)n/2(2n+1) (n+2)(n−1)1·3·5···(n−1) 2·4·6···n(neven). Section 5.10 1. The solution is as given in Eq. (5) with coefﬁcients as shown in Eq. (7). The integration yields (see Section 5.9) b0=1 2;bn=0 for neven; b1=3/4a n d bn=(−1)(n−1)/2 2·cn1·3·5···(n−2) 2·4·6···(n−1)·2n+1 n+1,n=3,5,7,.... 3.u(φ,t)=T−/summationtextbnPn(cos(φ)) exp(−(µ2+n(n+1))kt/R2),w h e r e µ2= γ2R2,nis odd, and bnis as at the end of Part B, with T0=T. 5. The eigenfunctions are as in Part C, except that nmust be odd in order to satisfy the boundary condition. 7. The nodal surfaces are: a sphere at ρ=0.634 and two naps of a cone given byφ=0.304πandρ=0.696π. Chapter 5 Miscellaneous Exercises 1.u(x,y,t)=∞/summationdisplay m=1amsin(µmy)exp/parenleftbig −µ2 mkt/parenrightbig +∞/summationdisplay n=1amncos(λnx)sin(µmy)exp/parenleftbig −/parenleftbig µ2 m+λ2 n/parenrightbig kt/parenrightbig , µm=mπb,λn=nπ/a, am=T1−cos(mπ) mπ,amn=4T π3(cos(nπ)−1)(1−cos(mπ)) n2m. 3.u(a/2,b/2,t)=∞/summationdisplay n=1bmnsin/parenleftbiggnπ 2/parenrightbigg sin/parenleftbiggmπ 2/parenrightbigg exp/parenleftbig −/parenleftbig λ2 n+µ2 m/parenrightbig kt/parenrightbig , where λn=nπ/a,µm=mπ/b,a n d bmn=4T π2(1−cos(mπ))( 1−cos(nπ)) mn. Chapter 5 483 The ﬁrst three nonzero terms are, for a=b, those with (m,n)= (1,1),(1,3)=(3,1),(3,3). All terms with an even index are 0. u(a/2,a/2,t)∼=16T π2/parenleftbigg e−2τ−2 3e−10τ+1 9e−18τ/parenrightbigg , where τ=ktπ2/a2. 5.u(r)=/parenleftbig a2−r2/parenrightbig /2a n d u(r)=∞/summationdisplay 1CnJ0(λnr), with Cn=2a2 α3nJ1(αn). 7.w(x,t)=a0+∞/summationdisplay n=1ancos(λnx)exp/parenleftbig −λ2 nkt/parenrightbig , v(y,t)=/summationdisplay bmsin(µmy)exp/parenleftbig −µ2 mkt/parenrightbig , where µm=mπ/b,λn=nπ/a, and initial conditions are v(y,0)=1,0<y<b;w(x,0)=Tx/a,0<x<a. 9.J0(λr)exp(−λ2kt). 11.Bk=bk/k(k+1)fork=1,2,...;b0must be 0, and B0is arbitrary. 13./parenleftbig/parenleftbig 1−x2/parenrightbig y/prime/parenrightbig/prime−m2 1−x2y+µ2y=0. 15.u(r,z)=∞/summationdisplay n=1ansinh(λnz) sinh(λnb)J0(λnr), where λn=αn aand an=2U0 αnJ1(αn). 17.u(r,z,t)=sin(µz)J0(λr)sin(νct)is a product solution if µ=mπ/b,λ= αn/a,a n dν=/radicalbig µ2+λ2. The frequencies of vibration are therefore νc or c/radicalBigg/parenleftbiggmπ b/parenrightbigg2 +/parenleftbiggαn a/parenrightbigg2 . 19. Each of the two terms satisﬁes ∇2φ=−(5π2)φ.O n y=0a n d x=1, both terms are 0; on y=xthey are obviously equal in value, opposite in sign. 21. Each term satisﬁes ∇2φ=−(16π2/3)φ. Ony=0,φ=sin(2nπx)−sin(2nπx); ony=√ 3x,φ=sin(4nπx)+0−sin(2nπ·2x); 484 Answers to Odd-Numbered Exercises ony=√ 3(1−x),φ=sin(4nπx)+sin(2nπ(1−2x))−sin 2 nπ. 23. For a sextant, φn=J3n(λr)sin(3nθ)and J3n(λ)=0. Thus λ1=6.380, which is less than/radicalbig 16π2/3=7.255. 25.b=− 1. 27. Since y(x)=hJ0(kx)/J0(ka),w h e r e k=ω/√gU, the solution cannot have this form if J0(ka)=0. 29.u(r,t)=R(t)T(t);R(r)=r−mJm(λr),w h e r e m=(n−2)/2;T(t)= acos(λct)+bsin(λct). 31.b=DrL2 DzR2,ρ=UL Dz. 33.a0=12.77,a1=− 4.88. 35. tan (λ)=Dλ/(D+λ2),λ=π(D=0), 3.173 ( D=1), 4.132 ( D=10). Chapter 6 Section 6.1 1. c.s2+2ω2 s(s2+4ω2);d .ωcos(φ)−ssin(φ) s2+ω2; e.e2 s−2;f .2ω2 s(s2+4ω2). 3. a.e−as s;b .e−as−e−bs s;c .1−e−as s2. 5. a. e−tsinh(t);b . e−tcos(t);c . e−atsin(√ b2−a2t)√ b2−a2. 7. a.eat−ebt a−b;b .t 2asinh(at);d .t2eat 2; e.f(t)=/braceleftBig1,0<t<1, 0,1<t. 9. a. [sin (ωt)−ωtcos(ωt)]/2ω2; b. See Table 2;c. See 7c;d. [cos (ωt)−ωtsin(ωt)]/2. Chapter 6 485 Section 6.2 1. a. e2t;b . e−2t;c .3e−t−e−3t 2;d .sin(3t) 3. 3. a.1−e−at a;b . t−sin(t);c .sin(t)−1 2sin(2t) 3; d.(sin(2t)−2tcos(2t))/16; e. −3 4+1 2t+e−t−1 4e−2t;f . c o s h (t)−1. 5. a.e2t−e−2t 4;b .1 2sin(2t); c.3 2+i√ 2−3 4exp/parenleftbig −i√ 2t/parenrightbig −i√ 2−3 4exp/parenleftbig i√ 2t/parenrightbig ;d . 4/parenleftbig 1−e−t/parenrightbig . 7. a. 1 −cos(t);b .et−cos(ωt)+ωsin(ωt) ω2+1;c . t−sin(t). Section 6.3 1. a. s=−/parenleftbigg2n−1 2π/parenrightbigg2 ,n=1,2,...; b.s=± i2n−1 2π,n=1,2,...; c.s=± inπ,n=0,1,2,...; d.s=iη,w h e r et a n η=−1 η; e.s=iη,w h e r et a n η=1 η. 3. a.sinh(√sx) s2sinh(√s);b .1 s−cosh/parenleftbig√s(1 2−x)/parenrightbig s(s+1)cosh(√s/2). 5. a. u(x,t)=x+∞/summationdisplay n=12s i n(nπx) nπcos(nπ)exp/parenleftbig −n2π2t/parenrightbig ; b.u(x,t)is 1 minus the solution of Example 3. Section 6.4 1.t+x2 2. 3.v(x,t)=4 π2∞/summationdisplay 1cos/parenleftbig (2n−1)(1 2−x)/parenrightbig sin((2n−1)πt) (2n−1)2sin/parenleftbig2n−1 2π/parenrightbig . 486 Answers to Odd-Numbered Exercises 5. a.ω ω2−π2/parenleftbigg1 πsin(πt)−1 ωsin(ωt)/parenrightbigg sin(πx); b.1 2π2/parenleftbig sin(πt)−πtcos(πt)/parenrightbig sin(πx). 7. a. u(x,t)=x−sin(√ax) sin(√a)e−at+2a π∞/summationdisplay 1sin(nπx)exp(−n2π2t) n(a−n2π2)cos(nπ); b. The term −xcos(nπx) cos(nπ)exp/parenleftbig −n2π2t/parenrightbig arises. Chapter 6 Miscellaneous Exercises 1.U(s)=T0 γ2+s+γ2T s(γ2+s), u(x,t)=T0exp/parenleftbig −γ2t/parenrightbig +T/parenleftbig 1−exp/parenleftbig −γ2t/parenrightbig/parenrightbig . 3.U(s)=cosh(√sx) s2cosh(√s), u(x,t)=t−1−x2 2+∞/summationdisplay n=12c o s(ρnx) ρ3nsin(ρn)exp/parenleftbig −ρ2 nt/parenrightbig , where ρn=(2n−1)π 2. 5.u(x,t)=x(1−x) 2−∞/summationdisplay n=14c o s/parenleftbig ρn(x−1 2)/parenrightbig ρnsin(ρn/2)exp/parenleftbig −ρ2 nt/parenrightbig , where ρn=(2n−1)π. 7.u(x,t)=x+∞/summationdisplay 12s i n(nπx) nπcos(nπ)exp/parenleftbig −n2π2t/parenrightbig . 9.U(x,s)=1 s/parenleftbig 1−exp/parenleftbig −√sx/parenrightbig/parenrightbig . 11.f(t)=x√ 4πt3exp/parenleftbigg−x2 4t/parenrightbigg . 13.u(x,t)=∞/summationdisplay n=0/bracketleftbigg erfc/parenleftbigg2n+1−x√ 4t/parenrightbigg −erfc/parenleftbigg2n+1+x√ 4t/parenrightbigg/bracketrightbigg . 15.F(s)=2∞/summationdisplay n=11 s2+n2. Chapter 7 487 17.f(t)=∞/summationdisplay −∞1 2aG/parenleftbigginπ a/parenrightbigg einπt/a. 19.F(s)must be of the form F(s)=G(s)/H(s),w h e r e G(s)is never inﬁnite. The solutions of H(s)=0 must form an arithmetic sequence of purely imaginary numbers, and H/prime(s)/negationslash=0i fH(s)=0. 21.F(s)=(1−e−πs)2 s(1−e−2πs)=1−e−πs s(1+e−πs). 23.F(s)=1+e−πs (s2+1)(1−e−πs). 25.u(x,t)=sin(ωx)sin(ωt) sin(ω)+∞/summationdisplay n=1(−1)n+12ω ω2−n2π2sin(nπx)sin(nπt). 27.U(x,s)=ω s2+ω2emx,m=1 2−/radicalbigg 1 4+s. 29.α2=1 2/parenleftBigg 1 4±/radicalBigg/parenleftbigg1 4/parenrightbigg2 +ω2/parenrightBigg .S i n c e αmust be real, take the +sign. Chapter 7 Section 7.1 1. 16(ui+1−2uij+ui−1)=− 1,i=1,2,3,u0=0,u4=1. Solution: u1= 11/32,u2=5/8,u3=27/32. 3. 16(ui+1−2ui+ui−1)−ui=−1 2i,i=1,2,3,u0=0,u4=1. Solution: u1=0.285, u2=0.556, u3=0.800. 5. 16(ui+1−2ui+ui−1)=1 4i,i=0,1,2,3,u0−2(u1−u−1)=1,u4=0. Solution: u0=0.422, u1=0.277, u2=0.148, u3=0.051. 7.n=3:u1=4.76,u2=4.24;n=4:u1=6.65,u2=9.14,u3=5.92. The actual solution, u(x)=−sin(√ 10x) sin(√ 10), has a maximum of about 50. The boundary value problem is nearly singular. 488 Answers to Odd-Numbered Exercises 9. 25(ui+1−2ui+ui−1)−25ui=− 25,i=1,2,3,4,5;u0=2,u5+(u6− u4)/(2/5)=1. When the equation for i=5 and the boundary condition are combined, they become 2 u4−3.4u5=− 1.4. Solution: u1=1.382, u2=1.146, u3=1.057, u4=1.023, u5=1.014. 11. 9(ui+1−2ui+ui−1)+(3/2)(ui+1−ui−1)−ui=−(1/3)i,i=0,1,2; u3=1,(u1−u−1)/(2/3)=0. When u−1is eliminated and coefﬁcients are collected, the equations to solve are −19u0+18u1=0, 71 2u0−19u1+101 2u2=−1 3, 71 2u1−19u2=− 111 6. Solution: u0=0.795,u1=0.839,u2=0.919. Section 7.2 1. Line mof the solution should be exactly the same as line m+1o fT a b l e4 . 3.r=2/5,/Delta1t=1/40. i m 01 2 3 4 0 0 0000 1 1 0. 0. 0. 0. 2 1 0.40 . 0. 0. 3 1 0.48 0 .16 0 . 0. 4 1 0.56 0 .224 0 .064 0 . 5 1 0.6016 0 .2944 0 .1024 0 .0512 5./Delta1t=1/32. Remember that u4(m)=u0(m)=m/Delta1t.A l ln u m b e r si nt h e table should be multiplied by /Delta1t. i m 01 2 3 4 0 00 0 0 0 1 1 000 1 2 2 1/20 1 /2 2 3 3 11 /21 3 4 4 7/41 7 /4 4 5 5 5/27/45/2 5 7./Delta1t=1/32. All numbers in this table should be multiplied by /Delta1t. Chapter 7 489 i m 01 2 3 4 0 00 0 0 0 1 0 111 0 2 0 3/22 3 /2 0 3 0 25 /22 0 4 0 9/43 9 /4 0 ... ∞ 0 343 0 9./Delta1t=1 32. Remember u−1=u1. All entries in this table should be multi- plied by /Delta1x=1 4. i m 012 34 0 012 34 11 1 2 3 4 21 3 /22 3 4 33 /23 /29 /43 4 43 /21 5 /89/42 5 /8 4 51 5 /81 5 /85/22 5 /8 4 Section 7.3 1. i m 0 1234 0 00 0 0 0 1 0 1/41 /41 /4 0 2 0 1/41 /21 /4 0 3 0 1/41 /41 /4 0 4 0 000 0 5 0−1/4−1/4−1/4 0 3. In this table, α=1/√ 2. i m 01 2 3 4 0 00 0 0 0 1 0α/41 /4α/40 2 0 1/4α/21 /4 0 3 0α/41 /4α/40 4 0 000 0 5 0−α/4−1/4−α/40 490 Answers to Odd-Numbered Exercises 5. i tm m 01 2 3 4 00 01 /21 1 /20 0.177 1 0 1/23 /41 /2 0 0.354 2 0 3/81 /43 /8 0 0.530 3 0 0 −1/80 0 0.707 4 0−7/16−3/8−7/16 0 0.884 5 0−5/8−11/16−5/8 0 7. i m 01 2 3 4 0 0 0000 1 0 000 1 2 0 001 1 3 0 011 1 4 0 111 0 5 0 110 −1 6 0 00 −1−1 7 0−1−2−1−1 8 0−2−2−2 0 9. Run: ui(m+1)=(2−2ρ2−16/Delta1t2)ui(m)+ρ2ui−1(m)+ρ2ui+1(m)− ui(m−1).S t a r t : ui(1)=1 2((2−2ρ2−16/Delta1t2)ui(0)+ρ2ui−1(0)+ ρ2ui+1(0)). Longest stable time step: /Delta1t=1/√ 24(ρ2=2/3). i m 01 2 34 0 00 .50 1 .00 0 .50 0 1 0 0.33 0 .33 0 .33 0 2 0−0.28−0.56−0.28 0 3 0−0.70−0.70−0.70 0 4 0−0.19−0.38−0.19 0 5 0 0.45 0 .45 0 .45 0 6 0 0.49 0 .98 0 .49 0 7 0 0.21 0 .21 0 .21 0 8 0−0.35−0.71−0.35 0 Section 7.4 1. At(1/4,1/4),1 1/256; at (1/2,1/4),1 4/256; at (1/2,1/2),1 8/256. 3. In both this exercise and Exercise 4, the exact solution is u(x,y)=xy,a n d the numerical solutions are exact. Chapter 7 491 5. Coordinates and values of the corresponding uiare:(1/7,1/7),5α; (2/7,1/7),1 0α;(3/7,1/7),1 4α;(1/7,2/7),2 1α;(2/7,2/7),3 2α.H e r e α=19/1159. 7.u1=0.670, u2=0.721, u3=0.961, u4=1.212, u5=0.954, u6=0.651. The remaining values are found by symmetry. 9.u1=0.386, u2=0.542, u3=0.784, u4=0.595. The remaining values are found by symmetry. Section 7.5 1. Use Eq. (8) with r=1/4. i m 1234 5 6 0 0000 0 0 10 0 0 1 /41 /41 /4 21 /16 1 /16 1 /16 5 /16 3 /85 /16 33 /32 1 /83 /32 23 /64 27 /64 23 /64 3. Note that u1=u2=u4=u5; replacement equations become u1(m+1)=u3(m)/4,u3(m+1)=u1(m). i m 13 0 11 11 /41 21 /41 /4 31 /16 1 /4 41 /16 1 /16 5. Use Eq. (8) with r=1/4. Note that u4=u2,u7=u3,u8=u6. i m 1235 6 9 0 0000 0 0 10 0 1 /40 1 /41 /2 20 1 /16 5 /16 1 /87/16 5 /8 31 /32 7 /64 3 /81 /41 7 /64 23 /32 7. Use the same numbering as for Exercise 5. Note that u1=u3=u7=u9 and u2=u4=u6=u8. The running equations become u1(m+1)=1 2u2(m)−u1(m−1), 492 Answers to Odd-Numbered Exercises u2(m+1)=1 2u1(m)+1 4u5(m)−u2(m−1), u5(m+1)=u2(m)−u5(m−1). m 012 3 4 5 6 7 u10 00 1 /80 −5/16 0 15 /32 u20 01/40 −3/80 5 /16 0 u50 10 −3/40 3 /80 −1/16 9. i m 125 0 111 11 /23 /41 2−1/40 1 /2 3−1/2−3/4−3/2 4−1/2−5/4−2 5−3/4−3/4−1 11. See Fig. 12 below for numbering of points. i m 123 4 5 6 0 100 0 0 0 10 1 /41 /40 0 0 2−3/40 0 1 /41/80 30 −1/2−7/16 0 0 3 /16 Figure 12 Solution of Exercise 11, Section 7.5. Chapter 7 493 Chapter 7 Miscellaneous Exercises 1.ui+1−2ui+ui−1 (/Delta1x)2−√24xiui=0,i=0,1,2, u1−u−1 2/Delta1x=1,u3=1; −18u0+18u1=6, 9u0−20.83u1+9u2=0, 9u1−22u2=− 9, u0=− 0.248, u1=0.08,u2=0.44. 3.ui+1−2ui+ui−1 (/Delta1x)2+1 1+xiui+1−ui−1 2/Delta1x=−(1+xi), i=1,2,3;u0=1,u4=0; −32u1+17.60u2=− 15.65, 14.67u1−32u2+17.33u3=− 1.5, 14.86u2−32u3=− 1.75, u1=0.822, u2=0.606, u3=0.335. 5.ui(m+1)=(ui−1(m)+ui+1(m))/2. Note that u3(m)=u1(m)and u4(m)=u0(m). i m 01 2 0 00 0 10 . 0 3 0 0 2 0.06 0.015 0 3 0.09 0.03 0.15 4 0.12 0.053 0.03 5 0.14 0.075 0.053 6 0.17 0.1 0.075 7 0.20 0.122 0.10 8 0.22 0.15 0.122 7. First problem: ui(m+1)=(ui+1(m)+ui(m)+ui−1(m))/3; second prob- lem: ui(m+1)=(ui+1(m)+ui−1(m))/3. 494 Answers to Odd-Numbered Exercises First Problem Second Problem ii m 0 1234 0 1234 0 0 1110 0 1110 1 0 2/31 2 /3 00 1/32 /31 /3 0 2 0 5/97 /95 /9 00 2/92 /92 /9 0 3 0 14/27 17 /27 14 /27 00 2/27 4 /27 2 /27 0 4 0 31/81 45 /81 31 /81 00 4/81 4 /81 4 /81 0 9.ui(m+1)=(ui+1(m)+ui−1(m))/2. i m 01 2 3 4 5 0 00 0 0 0 0 1 1/2 00 00 0 2 1 1/40 0 0 0 3 3/2 1/21 /80 0 0 4 2 13/16 1 /41 /16 0 0 5 5/2 9/87 /16 1 /81 /32 0 6 3 87/32 5 /81 5 /64 1 /16 0 11. i m 01234 0 00000 1 0 000 1 2 0 001 1 3 0 011 1 4 0 111 1 5 0 111 1 6 0 011 1 7 0 001 1 8 0 000 1 13. Let uij∼=u(xi,yj).T h e n u11=u22=u33=0.5,u12=0.698, u13=0.792, u21=0.302, u23=0.624, u21=0.209, u32=0.376. 1 5 . N u m b e ra si nC h a p t e r7 ,F i g .4 .T h e n u1=u3=u7=u9and u2=u4= u6=u8. m 01 2 3 4 5 u11 1/23/81/43/16 1 /8 u21 3/41/23/81/43 /16 u51 13 /41/23/81 /4 Index A acoustic vibrations, 235 aluminum nitride, oxygen removal from, 156 antenna vibrations, 226 approximation by Fourier series, 91–94 arbitrary periods, 64–65 B bad discontinuities, 74–75, 79 band-limited functions, 121–123 bars, insulated, 157–161. See also heat conduction problems beam vibrations, 225 Bessel functions, integrals of, 440Bessel inequality, 92–93 boundary conditions. See also initial value–boundary value problems of the ﬁrst kind. See entries at Dirichlet limitations of product method, 281–282 mixed, 139 potential equation. Seepotential equationof the second kind. See entries at Neumann of the third kind, 139 wave equation, 217–220, 229 boundary value problems, 26–34. See also initial value–boundary value problems Fourier series applications with, 119–120 Green’s functions, 23, 43–49 potential equation, 256–257 singular, 38–41 boundedness, 40–41Boussinesq equation, 211 Brownian motion, 204 buckling of a column, 32–34 Burger’s equation, 209 C cable, hanging, 26–29 calculus, 436–438 cantilevered beam, 226 car antenna vibrations, 226catenaries, 35 Cauchy–Euler equation, 6–7, 38, 277 Cesaro summability, 131 chain rule, 232 495 496 Index characteristic equation, 3, 10 characteristics, method of, 246 classiﬁcations of partial differential equations, 280–282 coefﬁcients of Fourier series. SeeFourier series column buckling, 32–34 complementary error function, 200, 202 complex coefﬁcients, potential equation analysis, 284 complex Fourier coefﬁcients, 113–115 sampling theorem, 121–124 conditions. Seeboundary conditions; initial condition conduction of heat. Seeheat conduction problems conservation of energy, law of, 135–136 continuity behavior, 73–76 convection, 170–174. See also heat conduction problems convergence expansions in series of eigenfunctions, 182 Fourier series, 73–77, 124 in the mean, 93–94 uniform, 79–83 cooling, Newton’s law of, 30, 139cooling ﬁns, 40–41 cosine function, 66–68 Fourier cosine integral representation, 109–110 hyperbolic, 4 integrals of, 439 critical radius, 42 cutoff frequency, 121 D d’Alembert’s method (traveling wave), 227–231, 252 damping term, 117 delta functions, 113diffusion equations. Seeheat conduction problems diffusion of sulphur dioxide, 55–56, 192–193 diffusivity, 141dimensions of heat ﬂux, 135Dinac’s delta function, 113 Dirichlet conditions, 139 Dirichlet’s problem, 256. See also potential equation in a disk, 275–279 in a rectangle, 259–269 soap ﬁlms, 283 discontinuity, 74 disk, potential equation in, 275–279 E eigenfunctions, 158–159 expansions in series of, 181–182 orthogonality of, 175–177 eigenvalue problems, 34, 158–159 estimating eigenvalues for wave equations, 236–239 one-dimensional wave equation, 234 singular, 189 Sturm–Liouville problems, 178–179 expansions in series of eigenfunctions, 181–182 generalizations on heat conduction problems, 184–187 one-dimensional wave equation, 234 elliptic equations, 281 endpoints of periodic extensions, 76–77 energy conservation, law of, 135–136 enzyme electrodes, 212–214 equilibrium problems. Seesteady-state problems error function, heat conduction problems, 199–202 even functions, 67 continuity behavior of, 83 extensions of functions, 69–71 exponential functions, integrals of, 438–439 exponential growth, 2 extensions of periodic functions, 65–71 endpoints of, 76–77 uniform convergence, 82–83 F fast Fourier transform (FFT), 124Fick’s law, 55, 141 Index 497 ﬁnite Fourier series, 91–94 ﬁrst-order equations homogeneous, 1–2 nonhomogeneous, 20–21 Fisher’s equation, 252Fitzhugh–Nagumo equations, 239–244ﬁxed end temperatures (heat equation), 149–155 ﬂat enzyme electrodes, 212–214ﬂow (ﬂuid), 284–285, 289ﬂuid ﬂows, 284–285, 289Fokker–Planck equation, 205 forced vibrations of strings, 232 forced vibrations system, 17–20forcing function, 117, 226Fourier integrals, 106–111, 124, 190, 194 applications of, 117–123coefﬁcient functions, 108complex coefﬁcients. Seecomplex Fourier coefﬁcients Fourier transforms, 115Fourier’s single integral, 112–113history of, 124 representational theorem, 108 wave equation in unbounded regions, 239–244 Fourier series, 62–63, 124 applications of, 117–123 arbitrary periods, 64–65complex coefﬁcients. Seecomplex Fourier coefﬁcients convergence, 73–77 proof of, 95–99uniform convergence, 79–83 cosine integral representation, 109history of, 124 means of, 90–94 numerical determination of coefﬁcients, 100–104 operations on, 85–89 periodic extensions, 65–71 endpoints of, 76–77uniform convergence, 82–83 potential in rectangle, 260–261sine integral representation, 109 Fourier transforms, 115Fourier’s law, 30 Fourier’s method (separation of variables), 150, 166–167 freezing lake, temperature of, 204 frequencies of vibration, 223–224, 234 functions. See speciﬁc function by name G Gaussian probability density function, 203 general solutions boundary value problems, 26 homogeneous differential equations, 158 nonhomogeneous linear equations, 15 one-dimensional wave equation, 228 second-order homogeneous equations, 3 second-order linear partial differential equations, 205, 280–281 generalized rectangles, 281 generation rate functions, 141 Gibbs’ phenomenon, 82 Green’s functions, 23, 43–49 groundwater ﬂow, 52, 211–212 H half-range extensions, 70–71hanging cable system, 26–29 harmonic functions, 255. See also potential equation heat conduction problems, 29–31, 135–206, 280 convection, 170–174 cooling ﬁns, 40–41 derivation of, 135–141 different end conditions (example), 157–161 error function, 199–202 ﬁxed end temperatures (example), 149–155 generalizations on, 184–187 insulated ends (example), 157–161 radial heat ﬂow, 39–40 steady-state temperatures, 143–147 498 Index higher-order equations, homogeneous, 9–11 homogeneous boundary conditions, as requirement, 282 homogeneous linear equations, 1–11 ﬁrst-order, 1–2 higher-order, 9–11second-order, 2–9 hyperbolic equations, 281 hyperbolic functions, 4, 436, 438–439 I inﬁnite intervals, 40–41inﬁnite rods, 193–197initial conditions. See also initial value–boundary value problems wave equation, 217, 220–221, 228 initial value–boundary value problems, 140 heat conduction problems, 138. See also heat conduction problems convection, 170–174different end conditions (example), 163–166 ﬁxed end temperatures (example), 149–155 generalizations on, 184–187 inﬁnite rods, 193–197insulated ends (example), 157–161 semi-inﬁnite rods, 184–187 wave equation, 215–247, 280 d’Alembert’s method, 227–231, 252 estimating eigenvalues for, 236–239frequencies of vibration, 223–224, 234 one-dimensional, in general, 233–235 in unbounded regions, 239–244 vibrating string problem, 215–224 insulated bars, 157–161. See also heat conduction problems insulated surfaces, 139integrals, table of, 438–440 integro-differential boundary value problems, 54 irrotational vortex, 291J jump discontinuities, 74 K kryptonite, 42 L Lake Ontario, 105 Lake Placid, 105 Laplace’s equation. Seepotential equation Laplacian operator, 256–257law of conservation of energy, 135–136 law of cooling (Newton), 30, 139 law of radiation (Stefan–Boltzmann), 142 left-hand limits, 73–74 Legendre polynomials, integrals of, 440 level curves, 261–262 L’Hospital’s rule, 98 linear density, 216 linear differential equations homogeneous, 1–11 ﬁrst-order, 1–2 higher-order, 9–11 second-order, 2–9 nonhomogeneous, 14–23 Fourier series applications with, 117–119 undetermined coefﬁcients, 16–20 variation of parameters, 20–23 linear operations, 140 linear partial differential equations general form, 205heat. Seeheat conduction problems potential. Seepotential equation wave. Seewave equation linearly independent solution, 3 M Massena, New Y ork, 132mass–spring–damper system, 5 forced vibrations system, 17–20 maximum principle, 255, 278mean error, 90–94 Index 499 mean value property, 278 periodic functions, 61 membrane displacement, 255. See also potential equation method of characteristics, 246 mixed boundary conditions, 139 N Neumann conditions, 139Neumann’s problem, 256, 290. See also potential equation Newton’s law of cooling, 30, 139 nonhomogeneous linear equations, 14–23 Fourier series applications with, 117–119 undetermined coefﬁcients, 16–20 variation of parameters, 20–23 nonremovable discontinuities, 74–75, 79 normal probability density function, 203 normalized eigenfunctions, 183 normalizing constants, 183 nuclear fuel rods. Seeheat conduction problems numerical determination of Fourier coefﬁcients, 100–104 O odd functions, 67 continuity behavior of, 83 extensions of functions, 69–71 ODEs (ordinary differential equations), 1–51 boundary value problems, 26–34 Green’s functions, 43–49 homogeneous, 1–11 ﬁrst-order, 1–2 higher-order, 9–11second-order, 2–9 nonhomogeneous, 14–23 Fourier series applications with, 117–119 undetermined coefﬁcients, 16–20 variation of parameters, 20–23singular boundary value problems, 38–41 one-dimensional wave equation, 233–235 ordinary differential equations, 1–51 boundary value problems, 26–34 Green’s functions, 43–49 homogeneous, 1–11 ﬁrst-order, 1–2 higher-order, 9–11 second-order, 2–9 nonhomogeneous, 14–23 Fourier series applications with, 117–119 undetermined coefﬁcients, 16–20 variation of parameters, 20–23 singular boundary value problems, 38–41 ordinary limits, 73–74 organ pipes, 225 orthogonality, 60–61, 73 of eigenfunctions, 175–177 oxygen removal from aluminum nitride, 156 P parabolic equations, 281 Parseval’s equality, 92–93 partial differential equation classiﬁcations, 280–282 particular solutions, nonhomogeneous linear equations, 15–23 penetration of heat into earth, 192 periodic functions, 59–63 arbitrary periods, 64–65 extensions of, 65–71 endpoints of, 76–77 uniform convergence, 82–83 piecewise continuous functions, 75–76 piecewise smooth functions, 76 plate, ﬂow past, 289 Poiseuille ﬂow, 36 Poisson equation, 268–269 polar coordinates, potential equation in, 256–257, 275–279 polynomial solution for potential equation, 256 500 Index potential equation, 255–285 in disk, 275–279 limitations of product method, 280–282 Poisson equation, 268–269 polynomial solution for, 256 in rectangle, 259–269 soap ﬁlms, 283 solutions to (harmonic functions), 255 in unbounded regions, 270–272 principle of superposition, 3, 10, 152 wave equation and standing waves, 220 probability density function, 203 product method (separation of variables), 150, 166–167 limitations of, 280–282 potential in rectangle, 259–261, 266 R radial heat ﬂow, 39–40 radiation, 142 radical functions, integrals of, 438 rational functions, integrals of, 438 Rayleigh method, 239 Rayleigh quotient, 239 rectangle, potential equation in, 259–269 reduction or order, 8–9 regular singular points, 7, 38–40 regular Sturm–Liouville problems, 178–179 convergence theorem, 182 one-dimensional wave equation, 234 removable discontinuities, 74, 79 restoring term, forcing functions, 117 Revision Rule, 17 right-hand limits, 73–74 Robin conditions, 139 rod vibrations, 253 rods of heat-conducting material. See heat conduction problems S sampling theorem, 121–124sawtooth function, 81–82second-order equations general form, 205 heat. Seeheat conduction problems homogeneous, 2–9nonhomogeneous, 21–23potential. Seepotential equation wave. Seewave equation sectionally continuous functions, 75–76sectionally smooth functions, 76semi-inﬁnite intervals, 40–41 semi-inﬁnite rods, 188–191 separation of variables (product method), 150, 166–167 limitations of, 280–282 potential in rectangle, 259–261, 266 sine function, 66–68 Fourier sine integral representation, 109–110 hyperbolic, 4 integrals of, 439 singular boundary value problems, 38–41 singular eigenvalue problems, 189 singular points, 7, 38–40soap ﬁlms, 283soliton (solitary) waves, 249 solutions, general boundary value problems, 26homogeneous differential equations, 158 nonhomogeneous linear equations, 15 one-dimensional wave equation, 228 second-order equations, 3, 205, 280–281 solutions, particular, 15–23 square-wave function, 75–77, 79–80standing waves, 220 steady-state problems. See also potential equation temperature (heat conduction), 143–147 convection, 170–174different end conditions (example), 157–161 ﬁxed end temperatures (example), 149–155 Index 501 generalizations on, 184–187 insulated ends (example), 157–161 semi-inﬁnite rods, 184–187, 193–197 wave equation, 218, 232 Stefan–Boltzmann law of radiation, 142 Stokes derivative, 250 stream function, 284 stresses due to thermal effects, 214 string, vibrating, 215–224. See also wave equation frequencies of vibration, 223–224, 234 one-dimensional wave equation, 233–235 Sturm–Liouville problems, 178–179 expansions in series of eigenfunctions, 181–182 generalizations on heat conduction problems, 184–187 one-dimensional wave equation, 234 sulphur dioxide, diffusion of, 55–56, 192–193 superposition, principle of, 3, 10, 152 wave equation and standing waves, 220 surfaces, insulated, 157–161. See also heat conduction problems suspension bridge (hanging cable system), 26–29 symmetry of sine and cosine functions, 66–68 T table of integrals, 438–440 Taylor series, 114 temperature (heat conduction) steady-state, 143–147 three-dimensional steady-state solution. Seepotential equation two-dimensional steady-state equation, 255 term, forcing functions, 117 thermal conductivity, 137 thermal diffusivity, 137 thermal stresses, 214 transient temperature distribution, 146ﬁxed end temperatures, 149–155 transverse displacement. Seewave equation trapezoidal function, 125traveling wave solution (d’Alembert’s method), 227–231, 252 triangle function, 123trigonometric functions, 435trigonometric series, history of, 124 trivial solutions, 150 truncated Fourier series, 92 U unbounded conditions potential equation, 270–272wave equation, 239–244 undetermined coefﬁcients, nonhomogeneous linear equations, 16–20 uniform convergence, 79–83 V variation of parameters, 20–23velocity potential function, 258, 284 vibrating string problem, 215–224. See also wave equation frequencies of vibration, 223–224, 234 one-dimensional wave equation, 233–235 W water hammers, 250 wave equation, 215–247, 280 d’Alembert’s method, 227–231, 252estimating eigenvalues for, 236–239 frequencies of vibration, 223–224, 234 one-dimensional, in general, 233–235 in unbounded regions, 239–244 vibrating string problem, 215–224, 234 whirling speeds, 55 windows, 111Wronskian, 3 This page intentionally left blank