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FOURIER SERIES AND BOUNDARY VALUE PROBLEMS Donated by Mrs. Yemuna Bappn to The Indian Institute of Astrophysics from the personal collection of Dr. M. K. V. Bappu FOURIER SERIES AND BOUNDARY VALUE PROBLEMS BY RUEL V. CHURCHILL Associate Professor of Mathematics University of Michigan First Edition Seventh Impression McGRAW-HILL BOOK COMPANY, Inc. NEW YORK AND LONDON 1941 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS Copyright, 1941, by the McGraw-Hill Book Company, Ino. PRINTED IN THE UNITED STATES OF AMERICA All rights reserved. This book , or parts thereof , may not be reproduced in any form without permission of the publishers. PREFACE This is an introductory treatment of Fourier series and their application to the solution of boundary value problems in the partial differential equations of physics and engineering. It is designed for students who have had an introductory course in ordinary differential equations and one semester of advanced calculus, or an equivalent preparation. The concepts from the field of physics which are involved here are kept on an elementary level. They are explained in the early part of the book, so that no previous preparation in this direction need be assumed. The first objective of this book is to introduce the reader to the concept of orthogonal sets of functions and to the basic ideas of the use of such functions in representing arbitrary functions. The most prominent special case, that of represent- ing an arbitrary function by its Fourier series, is given special attention. The Fourier integral representation and the repre- sentation of functions by scries of Bessel functions and Legendre polynomials are also treated individually, but somewhat less fully. The material covered is intended to prepare the reader for the usual applications arising in the physical sciences and to furnish a sound background for those who wish to pursue the subject further. The second objective is a thorough acquaintance with the classical process of solving boundary value problems in partial differential equations, with the aid of those expansions in series of orthogonal functions. The boundary value problems treated here consist of a variety of problems in heat conduction, vibra- tion, and potential. Emphasis is placed on the formal method of obtaining the solutions of such problems. But attention is also given to the matters of fully establishing the results as solutions and of investigating their uniqueness, for the process cannot be properly presented without some consideration of these matters. The book is intended to be both elementary and mathe- matically sound. It has been the author's experience that careful attention to the mathematical development, in contrast PREFACE vi to more formal procedures, contributes much to the student's interest as well as to his understanding of the subject, whether he is a student of pure or of applied mathematics. The few theorems that are stated here without proofs appear at the end of the discussion of the topics concerned, so they do not reflect upon the completeness of the earlier part of the development. Illustrative examples are given whenever new processes are involved. The problems form an essential part of such a book. A rather generous supply and wide variety will be found here. Answers are given to all but a few of the problems. The chapters on Bessel functions and Legendre polynomials (Chaps. VIII and IX) are independent of each other, so that they can be taken up in either order. The continuity of the subject matter will not be interrupted by omitting the chapter on the uniqueness of solutions of boundary value problems (Chap. VII) or by omitting certain parts of other chapters. This volume is a revision and extension of a planographed form developed by the author in a course given for many years to students of physics, engineering, and mathematics at the Uni- versity of Michigan. It is to be followed soon by a more advanced book on further methods of solving boundary value problems. The selection and presentation of the material for the present volume have been influenced by the works of a large number of authors, including Carslaw, Courant, Byerly, B6cher, Riemann and Weber, Watson, Hobson, and several others. To Dr. E. D. Rainville and Dr. R. C. F. Bartels the author wishes to express his gratitude for valuable suggestions and for their generous assistance with the reading of proof. In the preparation of the manuscript he has been faithfully assisted by his daughter, who did most of the typing, and by his wife and son. Ann Arbor, Mich., January, 1941. Ruel V. Churchill. CONTENTS Page Preface v Chapter I Section INTRODUCTION 1. The Two Related Problems 1 2 . Linear Differential Equations 2 3. Infinite Series of Solutions 5 4. Boundary Value Problems 6 Chapter II PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 5. Gravitational Potential 10 6 . Laplace’s Equation 12 7. Cylindrical and Spherical Coordinates 13 8 . The Flux of Heat 15 9. The Heat Equation 17 10 . Other Cases of the Heat Equation 19 11. The Equation of the Vibrating String 21 12 . Other Equations. Types 23 13. A Problem in Vibrations of a String 24 14. Example. The Plucked String 28 15. The Fourier Sine Series 29 16. Imaginary Exponential Functions 31 Chapter III ORTHOGONAL SETS OF FUNCTIONS 17. Inner Product of Two Vectors. Orthogonality 34 18. Orthonormal Sets of Vectors 35 19. Functions as Vectors. Orthogonality 37 20 . Generalized Fourier Scries 39 21 . Approximation in the Mean 40 22 . Closed and Complete Systems 42 23. Other Types of Orthogonality 44 24. Orthogonal Functions Generated by Differential Equations. ... 46 25. Orthogonality of the Characteristic Functions 49 Chapter IV FOURIER SERIES 26. Definition 53 27. Periodicity of the Function. ^Example ’ ’ 55 28. Fourier Sine Scries. Cosine Series 57 vii viii CONTENTS Section Page 29. Illustration 59 30. Other Forms of Fourier Series 61 31. Sectionally Continuous Functions 64 32. Preliminary Theory 67 33. A Fourier Theorem 70 34. Discussion of the Theorem 72 35. The Orthonormal Trigonometric Functions 74 Chapter V FURTHER PROPERTIES OF FOURIER SERIES; FOURIER INTEGRALS 36. Differentiation of Fourier Series 78 37. Integration of Fourier Series 80 38. Uniform Convergence 82 39. Concerning More General Conditions 85 40. The Fourier Integral 88 41. Other Forms of the Fourier Integral 91 Chapter VI SOLUTION OF BOUNDARY VALUE PROBLEMS BY THE USE OF FOURIER SERIES AND INTEGRALS 42. Formal and Rigorous Solutions 94 43. The Vibrating String 95 44. Variations of the Problem 98 45. Temperatures in a Slab with Faces at Temperature Zero 102 46. The Above Solution Established. Uniqueness 105 47. Variations of the Problem of Temperatures in a Slab 108 48. Temperatures in a Sphere 112 49. Steady Temperatures in a Rectangular Plate 114 50. Displacements in a Membrane. Fourier Series in Two Variables 116 51. Temperatures in an Infinite Bar. Application of Fourier Integrals 120 52. Temperatures in a Semi-infinite Bar 122 53. Further Applications of the Series and Integrals 123 Chapter VII UNIQUENESS OF SOLUTIONS 54. Introduction 127 55. Abel’s Test for Uniform Convergence of Series 127 56. Uniqueness Theorems for Temperature Problems 130 57. Example 133 58. Uniqueness of the Potential Function 134 59. An Application 137 Chapter VIII BESSEL FUNCTIONS AND APPLICATIONS 60. Derivation of the Functions J n (x) 143 61. The Functions of Integral Orders 145 CONTENTS ix Section Page 62. Differentiation and Recursion Formulas 148 63. Integral Forms of J n (x) 149 64. The Zeros of J n (x) 153 65. The Orthogonality of Bessel Functions 157 66. The Orthonormal Functions 161 67. Fourier-Bessel Expansions of Functions 162 68. Temperatures in an Infinite Cylinder 165 69. Radiation at the Surface of the Cylinder 168 70. The Vibration of a Circular Membrane 170 Chapter IX LEGENDRE POLYNOMIALS AND APPLICATIONS 71. Derivation of the Legendre Polynomials 175 72. Other Legendre Functions 177 73. Generating Functions for P„ (s) 179 74. The Legendre Coefficients 181 75. The Orthogonality of P n (x). Norms 183 76. The Functions P n (x ) as a Complete Orthogonal Set 185 77. The Expansion of x m 187 78. Derivatives of the Polynomials 189 79. An Expansion Theorem 191 80. The Potential about a Spherical Surface 193 81. The Gravitational Potential Due to a Circular Plate 198 Index 203 FOURIER SERIES AND BOUNDARY VALUE PROBLEMS CHAPTER I INTRODUCTION 1. The Two Related Problems. We shall be concerned here with two general types of problems: (a) the expansion of an arbitrarily given function in an infinite series whose terms are certain prescribed functions and (6) boundary value problems in the partial differential equations of physics and engineering. These two problems are so closely related that there are many advantages, especially to those interested in applied mathematics, in an introductory treatment that deals with both of them together. In fact an acquaintance with the expansion theory is neces- sary for the study of boundary value problems. The expansion problem can be treated independently. It is an interesting problem in pure mathematics, and its applications are not con- fined to boundary value problems. But it gains in unity and interest when presented as a problem arising in the solution of partial differential equations. The series in the problem type (a) is a Fourier series when its terms are certain linear combinations of sines and cosines. Fourier encountered this expansion problem, and made the first extensive treatment of it, in his development of the mathe- matical theory of the conduction of heat in solids.* Before Fourier's work, however, the investigations of others, notably D. Bernoulli and Euler, on the vibrations of strings, columns of air, elastic rods, and membranes, and of Legendre and Laplace on the theory of gravitational potential, had led to expansion * Fourier, “Thdorie analytique do la (dm, lour,” 1822. A translation of this book by Freeman appeared in 1878 under the title “The Analytical Theory of Heat.” 1 2 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 2 problems of the kind treated by Fourier as well as the related problems of expanding functions in series of Bessel functions, Legendre polynomials, and spherical harmonic functions. These physical problems which led the early investigators to the various expansions are all examples of boundary value problems in partial differential equations. Our plan of pres- entation here is in agreement with the historical development of the subject. The expansion problem as presented here will stress the development of functions in Fourier series. But we shall also consider the related generalized Fourier development of an arbi- trary function in series of orthogonal functions, including the important series of Bessel functions and Legendre polynomials. 2. Linear Differential Equations. An equation in a function of two or more variables and its partial derivatives is called a partial differential equation. The order of a partial differential equation, as in the case of an ordinary differential equation, is that of the highest ordered derivative appearing in it. Thus the equation d 2 u 0 du a) + = is one of the second order. A partial differential equation is linear if it is of the first degree in the unknown function and its derivatives. The equation ro \ d 2 u 9 du . f2) w +xy -d~y = 3x y is linear; equation (1) is nonlinear. If the equation contains only terms of the first degree in the function and its derivatives, it is called a linear homogeneous equation. Equation (2) is nonhomogeneous, but the equation d 2 u dx 2 + xy' du dy = 0 is linear and homogeneous. Thus the general linear partial differential equation of the second order, in two independent variables x and y, is Sec. 21 INTRODUCTION 3 where the letters A, B, • m m , Q. represent functions of x and y. If F is identically zero, the equation is homogeneous. The following theorem is sometimes referred to as the principle of superposition of solutions. Theorem 1. Any linear combination of two solutions of a linear homogeneous differential equation is again a solution . The proof for the ordinary equation (3) y" + Py ' + Qy = 0, where P and Q may be functions of x , will show how the proof can be written for any linear homogeneous differential equation, ordinary or partial. Let y = yi(x) and y = y%(x) be two solutions of equation (3). Then (4) y” + Py[ + Qyx = 0, (5) 2/2 + Py'i + Qyi — 0- It is to be shown that any linear combination of yi and y 2 — namely, Ayi + By 2) where A and B are arbitrary constants — is a solution of equation (3). By multiplying equations (4) by A and (5) by B and adding, the equation Ay" + By'f + P(Ay[ + By 2 ) + Q(Ayi + Byf) == 0 is obtained. This can be written d 2 d ^2 (Ay i + Byf) + P — (Ayi + Byf) + Q(Ayi + By 2 ) = 0, which is a statement that Ayi + By* is a solution of equation (3). For an ordinary differential equation of order n, a solution containing n arbitrary constants is known as the general solu- tion. But a partial differential equation of order n has in general a solution containing n arbitrary functions. These are functions of k — 1 variables, where k represents the number of independent variables in the equation. On those few occa- sions here where we consider such solutions, we shall refer to them as “general solutions” of the partial differential equations. But the collection of all possible solutions of a partial differential equation is not simple enough to be represented by just this “general solution” alone.* * See, for instance, Courant and Hilbert, “ Mcthoden der nmthermitiHohen. Physik,” Vol. 2, Chap. I; or Forsyth, “Theory of Differential Equations,” Vols. 5 and 6. 4 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 2 Consider, for example, the simple partial differential equation in the function u{x, y ) : ^ = 0 . dx According to the definition of the partial derivative, the solution IS u = f{y), where /(y) is an arbitrary function. Similarly, when the equation dx 1 is written —It—) — 0, its general solution is seen to be dx \dxj u = xf(y) + g(y), where f(y) and g(y) are arbitrary functions. PROBLEMS 1. Prove Theorem 1 for Laplace's equation d 2 ^ d 2 u dhx _ dx 2 + dy 2 + dz 2 “ 2. Prove Theorem 1 for the heat equation du _ / dhi dhi d 2 iA dt ~ \d# 2 ”* d?/ 2 dz 2 y Note that k may be a function of x } y, z , and t here. 3. Show by means of examples that the statement in Theorem 1 is not always true when the differential equation is nonhomogeneous. 4. Show that y = f(x + at) and y = g(x — at) satisfy the simple wave equation d*y 0 d 2 y dt 2 a dx 2 ’ where a is a constant and / and g are arbitrary functions, and hence that a general solution of that equation is V =f(x+ at) + g{x - at). 5. Show that e-» s ‘ sin nx is a solution of the simple heat equation du dhi Sec. 3] INTRODUCTION 5 If A i, A 2 , • * * , Ax are constants, show that the function N u=%A n e~ nH sin nx 71 — 1 is a solution having the value zero at x = 0 and x = r, for all t. 3. Infinite Series of Solutions. Let u n (n = 1, 2, 3, • • • ) be an infinite set of functions of any number of variables such that the series -f- U 2 + * * * + u n -f- ■ ■ • converges to a function u. If the series of derivatives of u ny with respect to one of the variables, converges to the same derivative of u , then the first series is said to be termwise differ- entiable with respect to that variable. Theorem 2. If each of the functions u 2j • • • , u n , • ■ • , is a solution of a linear homogeneous differential equation , the function u u n is also a solution provided this infinite series converges and is termwise differentiable as far as those derivatives which appear in the differential equation are concerned. Consider the proof for the differential equation (i) d 2 u d 2 u dx 2 ^ dx dt + qu 0 , where p and q may be functions of x and t. Let each of the functions u n (x, t) (n = 1, 2, * • • ) satisfy equation (1). The series 00 is assumed to be convergent and termwise differentiable; hence if u(x , t) represents its sum, then 00 00 00 du _ "^1 du n dhi _ d'hi 7l d 2 U __ d 2 U n dx dx ’ dx 2 dr' 1 ’ dx dt dx dt l l l Substituting these, the left-hand number of equation (1) becomes i l l ( 2 ) 6 FOURIER SERIES AND BOUNDARY PROBLEMS I* and if this quantity vanishes, the theorem is true. Now c.xpi » s sion (2) can be written 2(^ +p Ss + s “”)’ since the series obtained by adding three convejmt series term by term converges to the sum of the t ice ui , • sented by those series. Since u n is a solution o cqi dhc n dx 2 + V d 2 Un dx dt + qUn = 0 (n « 1, 2, • • * h and so expression (2) is equal to zero. Hence u(x , t) satisfies equation (1). , , This proof depends only upon the fact that the diffcien la equation is linear and homogeneous. It can clearly be app u ( to any such equation regardless of its order or number of vm in > < 4. Boundary Value Problems. In applied problems in dif- ferential equations a solution which satisfies some specified con- ditions for given values of the independent variables is usnnlh sought. These conditions are known as the boundary eondit ions. The differential equation together with these boundary con- ditions constitutes a boundary value problem. The student is familiar with such problems in ordinary differential equat ions. Consider, for example, the following problem. A body moves along the x-axis under a force of attraction toward the origin proportional to its distance from the origin. If it is initially in the position x = 0 and its position one. second later is x = 1, find its position x(t) at every instant. The displacement z(t) must satisfy the conditions ( 1 ) ( 2 ) d 2 x dt? x = 0 when t = 0 = —k 2 x, , x = 1 when t = 1, where k is a constant. The boundary value problem here con- sists of the equation (1) and the boundary conditions (2), which assign values to the function x at the extremities (or on the boundary) of the time interval from t = 0 to t = 1. The general solution of equation (1) is x = C\ cos kt + C% sin kt . Sec. 4] INTRODUCTION 7 ■ According to the conditions (2), Ci — 0 and C 2 = 1/sin so the solution of the problem is sin kt x = - — r * sin k From this the initial velocity which makes x = 1 when t = 1 can be written dx _ k dt sin k when t = 0. This condition could have been used in place of either of the conditions (2) to form another boundary value problem with the same solution. In general, the boundary conditions may contain conditions on the derivatives of the unknown function as well as on the function itself. The method corresponding to the one just used can sometimes be applied in partial differential equations. Consider, for instance, the following boundary value problem in u(x, y ) : (3) (4) w(0, y ) = y~, u( 1, y) = 1. Here the values of u are prescribed on the boundary, consisting of the lines x = 0 and x = 1, of the infinite strip in the ary-plane between those lines. The general solution of equation (3) is u(x, y ) = xf(y) + g(y), where /(y) and g(y) are arbitrary functions. Tho conditions (4) require that ( 5 ) ff(y) = y 2 , f(y) + g(y) = 1, so f(y) — 1 — y 2 , and the solution of the problem is u(x y y) = x(l - ?/) + y\ But it is only in exceptional cases that problems in partial differential equations can be solved by the above method. The general solution of the partial differential equation usually cannot be found in any practical form. But even when a gen- 8 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 4 eral solution is known, the functional equations, corresponding to equations (5), which are given by the boundary conditions arc often too difficult to solve. A more powerful method will bo developed in the following chapters — a method of combining particular solutions with the aid of Theorems 1 and 2. It is, of course, limited to problems possessing a certain linear character. The number and character of the boundary conditions which completely determine a solution of a partial differential equation depend upon the character of the equation. In the physical applications, however, the interpretation of the problem will indicate what boundary conditions are needed. If, after a solution of the problem is established, it is shown that only one solution is possible, the problem will have been shown to be com- pletely stated as well as solved. PROBLEMS 1. Solve the boundary value problem dhc dxdy = 0; u @,y)=y, 2. Solve the boundary value problem dhb tedy = 2z; 14(0, y ) = 0, u(x , 0) = sin x . Ans. u = y sin x. u(x , 0) = x 2 . o q * -rv n . Ans. u = xhy + x 2 . the condition ° * W 6n the Second bou ndary condition is replaced by du(x t 0) dx = z 2 . 4. By substituting the new independent variables U * V + ***‘ X = * + at, [x = x — at, show that the wave equation d W = «*0 V«**) becomes dt* -23L _ n d\ dn ~ °> ion of the w* problem = V(x, 0) = F(x), 0 ) _ n fit " u > 6. Solve thfLuXy S iT u r P rtbt e m WaVe eqUati ° n (Pr ° b ' 4 ’ ^ 2K Sec. 4] INTRODUCTION 9 where F(x) is a given function defined for all real x. Ans. y = i[F(x + at) +• F(x — at)]. 6. Solve Prob. 5 if the boundary conditions are replaced by y{x, 0) = 0, = G(x). = G(x). Also show that the solution under the more general conditions y(x, 0) = Fix), = G(x) is obtained by adding the solution just found to the solution of Prob. 5. Ans. V -(l /2a) f*** G(0 d!-. CHAPTER II PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 5. Gravitational Potential. According to the universal law of gravitation, the force of attraction exerted by a particle of mass m at the point (re, y, z) upon a unit mass at (X, Y , Z ) is directed along the line joining the two points, and its magnitude and sense are given by the equation where k is a positive constant and r is the distance between the two masses: r = V(X -xy + (F - y y + (Z =r sp. The positive sense is taken from the point (x, y, z), called Q, toward the point P (X, Y, Z). The gravitational potential V at any point P due to the mass m at Q is defined to be the function y __ km r So the derivative of this function is the force: dV __ km dr r 2 Let Q be fixed and consider V as a function of X, F, and Z. It will now be shown that the directional derivative of the potential in any direction gives the projection of the force F in that direction. . First let the direction be parallel to the X-axis. Then dV _ dV dr dX “ Hr dX kmX — x r 2 r = F cos a = F x , where cos a is the first direction cosine of the radius vector r, 10 Sec. 5] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS and F x is the projection of F on the X-axis. ( 1 ) Similarly, 11 Now if s is the directed distance along any line through P hav- ing the direction angles a', /3', -y' ; the directional derivative of V can be written (2) W = dV6X dVdY dV dZ ds dX ds ^ dY ds dZ ds = F x cos a' + F v cos /S' + F z cos y'. The last expression is the projection of the force in the direction of the line along which s is measured. , extension to the potential and force due to a continuous distribution of mass is quite direct. The potential function due to a mass of density S(x, y, z) distributed throughout a volume r, at a point P not occupied by mass, is defined to be (3) F(X, Y.Z) — h f f f s .( x > Ui z ) dx dy dz ’ ) J J Jr [(* - *)■ + (F -yY + (Z - 2) 2 l i ‘ This integral can be differentiated with respect to X, Y or Z inside the integral. Thus ' (4) This is the total component F x of the gravitational forces exerted by all the elements of mass in r upon a unit mass at P. Like- wise the total, components F y and F z satisfy relations (1), so that the directional derivative has the same form as in equation ( 2 ). Hence the projection, along any direction, of the force exerted by a mass distribution upon a unit mass at (X, Y, Z) is given by the directional derivative, along that direction, of the poten- tial function (3) ; that is, A force which can be derived in this manner from a potential function is known as a conservative force. 12 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 6 Let s be the arc length along any curve joining two points (Xi, Y h Z x ) and (X 2 , 7 2 , 7 2 ), at which s = si and s = s 2 , respectively. Then, according to formula (5), f>,ds = F(X 2 , F 2j Zi) - V(X h Y i, Zx). J8l That is, the difference between the values of the potential V at two points represents the work done by the gravitational force upon a unit mass which is moved from one of these points to the other. The amount of work depends upon the positions of the points, but not upon the path along which the unit mass moves. 6. Laplace’s Equation. The potential V(X, Y, Z) due to any distribution of mass will now be seen to satisfy an important- partial differential equation. Upon differentiating both members of equation (4), Sec. 5, with respect to X, we find that fx* - ~ k J/I 5 dxd V dz - Likewise '■-‘in 327 dZ' I _ 3(7 - y) 2 pS y.5 1 _ 3(z - z y : 8 dx dy dz, 8 dx dy dz . The sum of the terms inside the three brackets is zero; so &V dW d 2 V dX 2 + dY 2 + dZ~ 2 = °- This is Laplace’s equation. It is often written V 2 F = 0, T"*”- V ' squared,” V s = ~ 1 3 I d 2 3X 2 ‘ r dY 2 + dZ ~ 2 ^ -L V ZJ - We have ^ust r ^ presen *’ n severa -l other important equations. by the tlvitadon^ LaplaCe ’ S e ^ Uation is satisfied gravitational potential at points in space not occupied Sec. 7] PARTIAL DIFFERENTIAL EQ UA TIONS OF PHYSICS 13 by mass. It is satisfied as well by static electric or magnetic potential at points free from electric charges or magnetic poles, since the law of attraction or repulsion and the definition of the potential function in these cases are the same, except for constant factors, as in the case of gravitation. Other important functions in the applications satisfy Laplace's equation. One of them is the velocity potential of the irrota- tional motion of an incompressible fluid, used in hydrodynamics and aerodynamics. Another is the steady temperature at points in a homogeneous solid; this will be shown further on in this chapter. The gravitational potential at points occupied by mass of density 8 can be shown to satisfy Poisson’s equation: V 2 V = -4t rS, a nonhomogeneous equation. The equations of Laplace and Poisson, like most of the important partial differential equations of physics, are linear and of the second order. 7. Cylindrical and Spherical Coordinates. Since cylindrical and spherical surfaces occur frequently in the boundary value problems of physics, it is important to have expressions for the Laplacian operator in terms of cylindrical and spherical coordinates. The cylindrical coordinates (r, <p, z) determine a point P (Fig. 1) whose rectangular coordinates are (1) x = r cos <p, y — r sin <p, z = z. These relations can be written (2) r = \A 2 + y 2 , tp = arctan -> z = z, x 14 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 7 provided it is observed that the quadrant of the angle p is determined by the signs of x and y , not by the ratio y/x alone. Let u be a function of r , p } and z. In view of relations (2) it is also a function of x, y , z, and according to the formula for differentiating a composite function, du __ du dr du dp dx dr dx d<p dx Therefore du x dr \/x 2 + V 2 du y dip x 2 + y 2 d 2 u __ f x\ _ (v\ \ ^ y d ( eu\ dx 2 dr dx\r ) dcp dx \r 2 ) r dx \dr ) r 2 dx \d<p) The last two indicated derivatives can be written a_/3i lA _ d 2 u X d 2 u y cte \di * ) dr 2 r dr dtp r 2 ’ u\ _ d 2 U X d 2 u y \d \ pj dp dr r Vr 1 ' Substituting and simplifying, we find that d 2 u __ y 2 du , 2 xy du x 2 d 2 u 2 xy d 2 u y 2 d 2 u dx 2 r z dr r 4 dip r 2 dr 2 r z dr dip r 4 dtp 2 Similarly, it is found that d 2 u __ ^ du __ 2xy du y 2 d 2 u 2 xy d 2 u x 2 d 2 u dy 2 r 3 dr r A dp ^ r 2 dr 2 ~r*' dr dip + r 4 dp 2 ’ so that the Laplacian of u in cylindrical coordinates is (3) \7 2 w = — 4- I — -L L ^ I dHt dr 2 ^ r dr f r 2 <V dz 2 * It is simpler to transform the right-hand member of equation (3) into rectangular coordinates. This operation furnishes a verification of equation (3). The spherical coordinates (r, p, 6) of a point P (Fig. 2), also called polar coordinates, are related to the rectangular coordinates as follows: (4) x = r sin 6 cos p, y = r sin 6 sin p, z = r cos 6 . Sec. 8] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 15 The Lapladan of a function u in spherical coordinates is (5) = 4 \r f- 2 ( ru ) + ~ ± ( s i n d — ' L dr 2 ; J sm S dd \ dd J + 1 d 2 u sm 2 6 d<p 2 J The derivation or verification of this formula can be carried out in the same manner as that of the corresponding formula (3) for cylindrical coordinates. It is left as an exercise. PROBLEMS 1. Derive the expression given above for dhi/dy- in cylindrical coordinates, and thus complete the derivation of formula (3). 2. Verify formula (3) by transforming its right-hand member into rectangular coordinates. 3. Verify formula (5) by transforming its right-hand member into rectangular coordinates. 4. Write the formulas which give the spherical coordinates in tenm of x, y, z. 5 Derive formula (5) for the Laplacian in terms of spherical coordi- 8. The Flux of Heat. Consider an infinite slab of homogene- ous solid material bounded by the pianos x = 0 and x = L Let the faces x = 0 and * = L be kept at fixed uniform temperatures ui and u„ respectively. After the temperatures have become steady, the amount of heat per unit time which flows from the surface x = 0 to the surface x = L, per unit area, is JT ^2 U 1 A j—, where the constant K is known as the thermal conductivity Ihis statement is essentially a definition of the conductivity K. 16 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 8 The time rate of flow of heat per unit area through a surface is called the flux of heat. For the flux F through any isothermal surface (a surface at uniform temperature), the natural extension of the above definition is Here u is the temperature as a function of position, n o is the distance measured along a directed normal to the isotherm, and the positive sense of the flux F is that of the normal. In formula (1) the conductivity K may be variable, and the solid nonhomogeneous. Fig. 3. To indicate the extension of this formula to the flux F n normal to an arbitrary surface in a solid at a point P, let coordinate axes be chosen with origin at P so that the xy - plane is tangent to the isotherm through P (Fig. 3). Let X, y, v be the direction cosines of the normal n of the given surface. Now let the surface be displaced parallel to itself through a distance p, so that its tangent plane and the coordinate planes bound an elementary volume in the form of a tetrahedron. If A A is the area of the face QRS made by the tangent plane, vA A is the area of the face in the xy- plane. As p approaches zero, the rate of flow of heat into the element through one of these faces must approach the rate of flow out through the other: F n AA = F t vAA, where F z is the flux through the isotherm. The remaining two faces are perpendicular to the isotherm, so that the flux of heat through them is zero. Sec. 9] PARTIAL DIFFERENTIAL EQ UA TIONS OF PHYSICS 17 According to formula (1), F z = —K du/dz , so that But according to the formula for the directional derivative, du_^du du du _ dn d# ^ <9y v 62 dw d2 ? since du/dx and du/dy are both zero, owing to the fact that x and y are distances along the isothermal surface. It follows that ( 2 ) jp _ Fn ~ ~ K Tn' that is, the flux of heat through any surface in the direction of the normal to that surface is 'proportional to the rate of change of the temperature with respect to distance along that normal. In the derivation of relation (2) it was assumed that there is no source of heat in the neighborhood of the point P, and that the derivatives of the temperature function u exist. Further- more, our argument involved approximations, such as the use of tangent planes in place of surfaces, the validity of which use was not examined. We shall not attempt to make the derivation of relation (2) precise. In a rigorous development of the mathematical theory of heat conduction, relation (2) can be postulated instead of (1). The results which follow from (2) — in particular, the heat equation derived in the next section — have long been known to agree with experimental measurements. It should be observed that the temperature u serves as the potential function from which the flux of heat is obtained by finding its directional derivative. In the case of the gravitational or electrical potential, the directional derivative gives, respec- tively, the gravitational force or the flux of electricity in that direction; the flux of electricity is the current per unit area of surface normal to the direction. 9. The Heat Equation. Let u(x, y , 2, t ) represent the tem- perature at a point P (x y z) of a solid at time t, and let K be the thermal conductivity of this solid, where K may be a func- tion of x, y, z and t, or of u . Suppose that the point P is enclosed 18 FOURIER SERIES AND BOUNDARY PROBLEMS [Sac. 9 by any surface S lying entirely within the solid, and let n repre- sent the outward-drawn normal to the closed surface S. Then accor ing to the formula for the flux in Sec. 8, the time rate of flow of heat into the volume V enclosed by S, through the surface ( 1 ) K d ^dS. dn Now if 6 is the density of the solid and c its specific heat, or tile amount of heat required to raise the temperature of a unit mass of the solid 1 degree, another expression for the rate of increase of heat in the volume V is (2) If X, n, v are the direction cosines of the normal n, the integral (1) can be written /X( xk 5+**S + -*S)<» This can be transformed, according to Green’s theorem, into the volume integral J JX [1 0 1) +!(*!)+ 1 (* £)] ^ which must be equal to the integral (2), so that (3) + dz ( K ^) C3 - ) are^rmt^™ 11 ^ f ^ ^ * erms * n *be brackets in equation he int, eZT 0US ^ °f “ a nei ^borhood of P Since the integral m equation (3) vanishes for every volume V its SrTfnT* r iSi P ' For if were peak™ V tkL ZhiTT r°'‘ d o 5 ‘ 1U "' a small volume l pSv! The T* \ “ d throush0l ' t «“> integrand P toe ' The y »ould then be positive, in Sec. 10] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 19 contradiction to equation (3). Similarly if the integrand were negative. Therefore, at P, fA\ d /^chA , <9 / v du\ d ( ^ du\ (4) , *Tt-Tx\ K irx) + Ty\ K T,) + Tii\ K Tz) This is a general form of the equation of conduction of heat, or the heat equation. It should be noted that we have assumed in the derivation that no sources of heat exist in the neighborhood of the point (z, V> z). 10. Other Cases of the Heat Equation. If the conductivity K is constant, or does not depend upon the coordinates, the heat equation becomes (1) du _ j ( d 2 U d 2 U d 2 u \ Tt 3F/ where the coefficient k , called the diffusivity , is defined thus: The equation appears most frequently in the form (1), or the abbreviated form (2) g = kVht. The right-hand member can be expressed in terms of other coordinates by using the results of Sec. 7. The heat equation is also called the equation of diffusion. It is satisfied by the concentration u of any substance which penetrates a porous solid by diffusion. It was shown above that the temperature u everywhere within a solid satisfies the heat equation. To determine u as a definite function of x , y } 2 , and /, it is of course necessary to use, in addition to the heat equation, boundary conditions which describe the thermal state of the surface of the solid and the initial temperature. All these conditions make up the boundary value problem in the conduction of heat. There are several special cases and simple generalizations of the heat equation which arc important. First there are the cases in which the temperature is independent of one or more of the four independent variables, which consist of the space coordinates 20 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 10 and time t. If the temperatures are “ steady ” — that is, if u does not change with time — u satisfies Laplace's equation. This is approximately the case, for example, if the temperature distribution on the surface of a solid has been kept the same for a long period of time. If conditions are such that there can be no flow of heat in the direction of the z-axis, the heat equation for “ two-dimensional flow” applies: du _ , ( SHt d 2 lA at “ * \dx* + W Similarly for one-dimensional flow. Continuous sources of heat may exist within a solid. If at each point (x, y, z) they supply heat at the rate of F(x, y , z, t) units per unit time per unit volume, the heat equation becomes nonhomogeneous. Tor the case of one-dimensional flow, where the strength F of the source is a function of x and t , the equation becomes (3) ci t,-fX K f^)+ F ^ e >- This follows readily from the derivation in Sec. 9. Equation (3) may apply, for instance, to the temperature u in a wire which carries an electric current. PROBLEMS 1. The lateral surface of a homogeneous prism is insulated against the flow of heat. The initial temperature is zero throughout, and the end x = 0 is kept at temperature zero while the end x = L is kept at To, a constant temperature. Write the heat equation for this case. Ans. du/dt = k(d 2 u/dx 2 ). 2. Find the steady temperature in Prob. 1, after the conditions given there have been maintained for a very long time. What is the flux through one end during the steady state? Ans. u = (T 0 /L)x; flux - KT 0 /L. 3. State a physical problem whose solution is represented by the finite series in Prob. 5, Sec. 2. 4. Show that the temperature u in a uniform circular disk whose entire surface is insulated, and whose initial temperature is a function only of the distance r from the axis of the disk, satisfies the equation du z / d 2 u 1 du\ dt ~ * \dr* + r TrJ Sec. 11] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 21 5. If the initial temperature of a homogeneous sphere is a function only of the distance r from the center, and the surface is insulated, show that the temperature u of points inside satisfies the equation du _ ( dhc 2 du\ Bt “ * \dr* + r dr)' 11. The Equation of the Vibrating String. The transverse displacements of the points of a stretched string satisfy an important partial differential equation. Let the string be stretched between two fixed points on the a>axis and then given a displacement or velocity parallel to the y- axis. Its subsequent motion, with no external forces acting on it, is to be considered; this is described by finding the displacement y as a function of x and t. It will be assumed that 5, the mass per unit length, is uniform over the entire length of the string, and that the string is perfectly flexible, so that it can transmit tension but not bending or shearing forces. It will also be assumed that the displacements are small enough so that t he square of the inclination dy/dx can be neglected in comparison to 1; hence, if 8 is distance measured along the string at any instant, approximately. The length of each part of the string therefore remains essentially unaltered, and hence the tension is approxi- mately constant. Consider the vertical components of the forces exerted by the string upon any element A s of its length, lying between x and x + Ax (Fig. 4). The ^-component of the tensile force P exerted upon the element, at the end (x, y) is _ p dy = __ P d JL d Jt = ds dx ds bx 22 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 11 approximately. The corresponding force at the end whose abscissa is x + Ax is P^f + P^y Ax + R(Axy, dx dx 2 where R is the usual factor in the remainder in Taylor’s formula. Setting the sum of these forces equal to the product of the mass of the element and the acceleration in the ^-direction, we have pg Ax + R(Ax)^5Ax-^- By div iding by Ax and letting Ax approach zero, it follows that d>y . d*v (i) w = a a? where This is the equation of the vibrating string; it is also called the simple wave equation, since it is a special case of the wave equation of theoretical physics. . . If an external force parallel to the y-axis acts along the string, it is easily seen that the equation becomes ( 2 ) g-*3+n*o. where SF(x, t) is the force per unit length of string. In case the weight of the string is to be considered, for instance, the function F becomes the constant g, the acceleration of gravity. If the transverse displacements arc not confined to tlio ay-plane, two equations of type (2) are found, one involving the y, the other the z, of the points of the string, while the acceleration r is replaced by the y and z components of the external acceleration in those two equations, respectively. Equation (1) is also satisfied by the longitudinal displacements in a homogeneous elastic bar; y is then the displacement along the bar of any point from its position of equilibrium. A column of air may be substituted for the bar, and the equation becomes one of importance in the theory of sound. The equation also applies to the torsional displacements in a right circular cylinder. Sec. 12] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 23 PROBLEMS 1. Derive equation (2) above. 2. State Prob. 5, Sec. 4, as a problem of displacements in a stretched string of infinite length. Show that the motion given by the result of that problem can be described as the sum of two displacements, obtained by separating the initial displacement into two equal parts, one of which moves to the left along the string with the velocity a, and the other to the right with the same velocity. 3. If a damping force proportional to the velocity, such as air resist- ance, acts upon the string, show that the equation of motion has the form d 2 y 0 d 2 y dy dt 2 " a dx* dt’ where b is a positive constant. 12. Other Equations. Types. Some further partial differ- ential equations of importance in the applications will be described briefly at this point. For their derivation and complete descrip- tion, the reader should refer to books on the subjects involved. A natural generalization of equation (1) of the last section is the equation of the vibrating membrane: Here the position of equilibrium of the stretched membrane is the rry-plane, so that z is the transverse displacement of any point from that position. Assumptions similar to those in the case of the string are necessary. The membrane is assumed to be thin and perfectly flexible, with uniform mass 5 per unit area. The tensile stress P, or tension per unit length across any line, is assumed to be large, and the displacements small. The constant a 2 is then the ratio P/8. The telegraph equation , (2) ~ = KL g + (RK + SL) + RSv, is satisfied by either the electric potential or the current in a long slender wire with resistance P, the electrostatic capacity K , the leakage conductance S , and the self-inductance L, all per 24 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 13 unit length of wire. The simple wave equation is a special case of this. The transverse displacements y(x, t) of a uniform beam satisfy the fourth-order equation (3) d*y dt 2 + c 5 d 4 y dx 4 = o, where the constant c 2 depends upon the stiffness and mass of the beam.* Airy’s stress function tp(x } y) , used in the theory of elasticity, satisfies the fourth-order equation (4) , O dV , gv = 0 dx 4 ^ dx 2 dy 2 ^ dy 4 often written VV = 0. It serves in a sense as a potential function from which shearing and normal stresses within an elastic body can be derived. The form (4) assumes that no deformations exist in the 2 -direction. Linear partial differential equations of the second order with two independent variables x ) y are classified into three types in the theory of these equations. If the terms of second order, when collected on one side of the equation, are + B d 2 u dx dy , c d 2 U + L -r-T y 7 dy- where A, B , C are constants, the equation is of elliptic , parabolic , or hyperbolic type according as (J5 2 — 4 AC) is negative, zero, or positive. In the study of boundary value problems it will be observed that these three types require different kinds of bound- ary conditions to completely determine a solution. Note that Laplace’s equation in x and y is elliptic, while the heat equation and the simple wave equation in x and t arc parabolic and hyperbolic, respectively. The telegraph equation is also hyperbolic, if KL ^ 0. 13. A Problem in Vibrations of a String. When the differen- tial equation is linear and the boundary conditions consist of linear equations, the boundary value problem itself is called linear . *See, for instance, Timoshenko, “Vibration Problems in Engineering,” p. 221. Sec. 13] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 25 A method which can be used to solve a large class of such prob- lems will now be illustrated. It will be seen that the process leads naturally to a problem in Fourier series. A formal solution of the following problem will be given. Find the transverse displacements y(x, t) in a string of length L stretched between the points (0, 0) and (L, 0) if it is displaced initially into a position y = f(x) and released from rest at this position with no external forces acting. The required function y is the solution of the following bound- ary value problem : (1) «>0,0<*<i), (2) 2/(0, 0 = 0, y(L, <)=0 (< £ 0), (3) y(x, 0) = fix) (o ^ X g L), (4) =0 (0 g x g L). ot Our method consists of finding particular solutions of the partial differential equation (1) which satisfy the homogeneous boundary conditions (2) and (4), and then of determining a linear combination of those solutions which satisfies the non- homogeneous boundary condition (3). Particular solutions of equation (1) of the type ( 5 ) y = XT, where X is a function of x alone and T a function of t alone, can easily he found by means of ordinary differential equations. According to equation (5), dy/dx = X'T, dy/dt = XT', etc., where the prime denotes the ordinary derivative with respect to the only independent variable involved in the function. Substituting into equation (1), we find XT" = a*X"T, or, upon separating the variables by dividing by a 2 XT, X"(x) T"(t) X(xf a ir T(t ) " Since the member on the left is a function of x alone, it cannot vary with t; it is equal to a function of t alone, however, and 26 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 13 thus it cannot vary with x. Hence both members must be equal to a constant, say y, so that (6) X"(x) - yX(x) = 0, (7) T”{t) - ya?T(t) = 0. If our particular solution is to satisfy conditions (2), XT must vanish when x = 0 and when x = L, for all values of t involved. Therefore (8) X(0) = 0, X{L) = 0. Similarly, if it is to satisfy condition (4), (9) T'( 0) = 0. Equations (6) and (7) are linear homogeneous ordinary differential equations with constant coefficients. The auxiliary equation corresponding to (6), m 2 — 7 = 0, has the roots m, = ± Vy. The general solution of equation (6) is therefore X = C,e x ^ + C 2 e~* where C 1 and C 2 are arbitrary constants. But if 7 is positive, it is easily seen that there are no values of C\ and C 2 for which this function X satisfies both of conditions (8). Suppose 7 is negative, and write 7 = -£ 2 . The general solution of equation (6) can then be written X = A sin fix + B cos where A and B arc arbitrary constants. If X(0) = 0, the constant B must vanish. Then A must be different from zero, since we are not interested in the trivial solution X(x) = 0. So if X{L) = 0, we must have sin fiL — 0. Hence there is a discrete set of values of /3 } namely, (» = 1 , 2 ,--*), Sec. 13] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 27 for which the system consisting of equation (6) and conditions (8) has solutions. These solutions are X = A sin mrx Note that no new solutions are obtained when n = — 1, —2, - 3 , • • ■ . Substituting — nV 2 /L 2 for y in differential equation (7) and applying condition (9), we find that T = C cos mrat where C is an arbitrary constant. Therefore all the functions (10) A n sin mrat mrx T cos T (» = 1 , 2 , • • • ) are solutions of our partial differential equation (1) and satisfy the linear homogeneous conditions (2) and (4), when A J? A 2 , • • • are arbitrary constants. Any finite linear combination of these solutions will also satisfy the same conditions (Theorem 1, Chap. I); but when t = 0, it will reduce to a finite linear combination of the functions sin ( mrx/L ). Thus condition (3) will not be satisfied unless the given function f(x) has this particular character. Consider an infinite series of functions (10), oo /11N . . mrx mrat (11) V = An sin -j~ cos • This satisfies equation (1) provided it converges and is termwise differentiable (Theorem 2, Chap. I); it also satisfies conditions (2) and (4). It will satisfy the nonhomogeneous condition (3) provided the numbers A n can be so determined that ( 12 ) 90 f(x) = 2 An . mrx sm - T - * It will be shown in Sec. 15 that if such an expansion of f(x) is possible, the numbers A n must have the values 28 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 14 Equation (11) with coefficients (13) is formally the solution of the boundary value problem (l)-(4). The series on the right of equation (12) with the coefficients defined by (13) is called the Fourier sine series of the function f(x). In a later chapter it will be shown that this series actually converges to the function f(x) in the interval 0 x ^ L, pro- vided /(x) satisfies certain moderate conditions — conditions which are almost always satisfied by functions which arise in the applications. Other questions are left unsettled at this point in the treatment of this problem. Series (11) has not been shown to be conver- gent, or to represent a continuous function, or to be termwise differentiable twice with respect to either x or t. It has not been shown that series (11) is the only solution of the problem (l)-(4). Questions of this character are to be treated later on. 14. Example. The Plucked String. As a special case of the problem just treated, let the string be stretched between the points (0, 0) and (2, 0), and suppose its mid-point is raised to a height h above the rc-axis. The string is then released from rest in this broken-line position (Fig. 5). The function f(x) which describes the initial position can be written, in this case, f(x) = hx when 0 ^ x g 1, = — hx + 27i when 1 ^ x ^ 2. The coefficients in solution (11), Sec. 13, are, according to formula (13), Sec. 13, A n = f(x) sin dx = h x sin V ^‘ dx + h ( — x + 2) sin dx. Sec. 15] PARTIAL DIFFERENTIAL EQUA TIONS OF PHYSICS 29 After integrating and simplifying, we find that A 8h . mr An = -r-r, sin ; 7 r 2 n 2 2 7 so that the displacement y(x, t ) in this case of the plucked string is given by the formula V = 8h 1 . rnr . mrx V>2jn> sm Y- sm - T cos mrat 8 h( 7T 2 \ tx irat 1 . 2nrx sm — cos g sin — cos St at ~2~ , 1 ■ 5 tx 57 rat \ + 23 sm — COS -g- - • • . y Another form of this solution will be obtained later [formula (4), Sec. 43]. 15. The Fourier Sine Series. In the solution of the problem of Sec. 13 it was necessary to determine the coefficients A „ so that the series of sines would converge to f(x). Assuming that, an expansion of the type needed there, namely, (1) Six) = A i sin ^ + A« sin ~ x • D L , A . UirX i A n sin - T -j- Jj is possible when 0 ^ i 1 1, and that the scries can be integrated term by term after being multiplied by sin (mrx/L), it is easy to see what values the coefficients must have. It is necessary to recall that sin sin «•* . i &.-»>» _ „„ (» + oJ T1 2 m ' 7TJC 1 ( i 2ni7rx\ h,n L -2\ l -™»- L )> and hence, when m and n arc integers, ( 2 ) r' J . nnr.r . riTX , Jo K1U ' // H,n '// dx = 0 if L 2 in n, if ni = ?i. The functions sin (mrx/L) (a = 1, 2, • • • ) therefore form an orthogonal system m the interval 0 < x < /,; that is, the integral 30 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 15 over that interval of the product of any two distinct functions of the system is zero. Now let all terms in equation (1) be multiplied by sin (rnrx/L) and integrated between 0 and L. The first term on the right becomes This is zero unless n = 1, according to the orthogonality property (2) . Likewise all terms on the right except the nth one become zero; so the process gives (* ft \ • '0/KX -j A ‘ 7 A-nfi I f[x) sm -j— ax = A n I sin 2 -j- dx = according to property (2). Hence the coefficients in equation (1) must have the values (3) A n = ^ f(x) sin ^ dx. The Fourier sine series corresponding to f(x) can be written (4) where the sign ^ is used here to denote correspondence. It is to be shown later on that the series does converge to f(x) in general. PROBLEMS 1. Show that the Fourier sine series corresponding to fix) = 1 in the interval 0 < x < tt is 1 ^sin x + ^ sin Sx + i sin 5x -j- ■ • • j. 2. Show that the sine series for f(x) = x in the interval 0 < x < 1 is x sin mrx. 3. Find the solution of the problem of the string in Sec. 13 if the initial displacement is f(x) = A sin (ttx/L). Discuss the motion. Arts, y — A sin (tx/L) cos (tt at/L), Sec. 16 ] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 31 16. Imaginary Exponential Functions. According to the power series expansion of e z , e ix ( ix) n n\ t 2 2! 1 4! where i = V"— 1- So — 1 ” 1 2 ! ' 4 ! + * (* - , x* 3! + 6! (1) e ix = cos x + i sin x. This is usually taken as the definition of the exponential function with imaginary exponents. Then e -ix — cos x — { sin x f t and by first eliminating cos x and then sin x between this equa- tion and equation (1), we find that (2) i sin x = ^ = si n h (ix), nix I ix (3) cos x = ^ = cosh (ix). When the coefficients of a linear homogeneous differential equation are constant, particular solutions in the form of exponen- tial functions can be found. ^ To illustrate the use of exponential functions in partial differ- ential equations, consider again the problem in Sec. 13. The function y = fOcx+01^ where a and 13 are constants, is clearly a solution of the equation (4) provided that Hence the functions dt 2 dx 1 ’ /3 2 = a 2 « 2 . (\ ax (> i C~ ax C aat are solutions. Except for a constant factor, the difference between the two products just written is the only linear combination which 32 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 16 vanishes at x = 0 [condition (2), Sec. 13]. Thus the functions e±aoU si n h aX satisfy that condition as well as equation (4). The linear com- bination of these two functions which satisfies the condition that dy/dt = 0 when t = 0 is the sum, or (6) sinh ax cosh aat But here a must be imaginary, since our function is to vanish when x = L, because the hyperbolic sine of a real argument vanishes only when the argument is zero. According to equations (2) and (3), when a = in the function (5) can be written, except for a constant factor, as sin ixx cos j uat. This vanishes when x = L if n == mr/L. Thus we again have the particular solutions , . utx meat A n sm -j- cos -j— of equation (4), which satisfy the homogeneous conditions in the problem in Sec. 13. From this point on the procedure is the same as in that section. As another application of imaginary exponential functions, llote that N N 2(cos 0 + cos 20 + • • • + cos N9) = ^ e in9 + ^ e~ in9 . Summing the finite geometric series on the right, this becomes N 2 cos nO = l e i6 (l - e iN9 ) e~ i9 (l - e~ iN9 ) 1 — e l9 1 — e~ 19 — JV4-$) _|_ g— \iQ g—iO(N+}f) q— bid This can be written at once in the form which is known as Lagrange’s trigonometric identity . This identity will be useful in the theory of Fourier series. Sec. 16] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 33 PROBLEMS 1. Use exponential functions to determine particular solutions of the simple heat equation du d 2 w dt = dF 2 which vanish when x = 0 and x = x. (Compare Prob. 5, Sec. 2.) 2. Use exponential functions to determine particular solutions of the equation d' l u dhi _ dx* dy 2, ~ such that u = 0 when y = 0, and du/dx = 0 when x = 0 and x = 1. Ans. u — An cos mrx sinh mry ( n = 0, 1, 2, • ■ CHAPTER III ORTHOGONAL SETS OF FUNCTIONS 17. Inner Product of Two Vectors. Orthogonality. The concept of an orthogonal set of functions is a natural generaliza- tion of that of an orthogonal set of vectors, that is, a set of mutually perpendicular vectors. In fact, a function can be , considered as a generalized vector, so that the fundamental properties of the set of functions ar$ suggested by the analogous properties of the set of vectors. In the following discussion of simple vectors, the terminology and notation which apply to the generalized case will be used whenever it seems advanta- geous for the later generalizations. Let either g or g(r) denote a vector in ordinary three-dimen- sional space whose rectangular components are the three numbers #(1), <7(2), and <7(3). It is the radius vector of the point having these numbers as rectangular cartesian coordinates. The square of the length of this vector, called its norm , will be written N(g)] it is the sum of the squares of the components of g: (1) N(g) = g\ 1) + g\ 2 ) + (7*0) = g\r). r = 1 If N (g) = 1, g is a unit vector, also called a normed or a normal- ized vector. Let 6 be the angle between two vectors g x (r) and g 2 (r ). Since the components £i(l), 0i(2), g x {Z) are proportional to the direc- tion cosines of the vector g Xj and similarly for g 2 , the formula from analytic geometry for cos 6 can be written n = ffi(l)ff2(l) + gi(2)g 2 (2) + <7i(3)gr 2 (3) The numerator on the right is called the inner product (or scalar product) of the vectors g x and g 2) denoted by the symbol (g h g 2 ) ; thus 34 Sec. 18] ORTHOGONAL SETS OF FUNCTIONS 35 3 ( 2 ) (01, gt ) = 5 ) 01M02M T = 1 = VN( gi ) v%) cos 6. When JV(0 2 ) = 1, (g 1, #2) is the projection of the vector <71 in the direction of g 2 . The condition that the vectors <71 and g 2 be orthogonal, or perpendicular to each other, can be written (3) (g h 02) = 0 or, in terms of components, (4) X gi(r)gt(r) = 0. r = 1 Note also that expression (1) for the norm of g can be written N(g) = (0, 0). 18. Orthonormal Sets of Vectors. Given an orthogonal set of three vectors g n (n = 1, 2, 3), a set of unit vectors <p n having the same directions can be formed by dividing each component of g n by the length of g n . The components of <p h for instance, are <pi(r) = 0i(r)[JV(0i)]-* (r = 1, 2, 3). This set of mutually perpendicular unit vectors <p n , obtained by normalizing the mutually perpendicular vectors g n , is called an orihonormal set. Such a set can be described by means of inner products by writing (1) (<Pm, <pn) = & mn (w, U = 1, 2, 3), where 5 mn , called Kronecker’s <5, is 0 or 1 according as m and n are different or equal: &rnn — 0 if 771 7^ 72, = 1 if rn = n. The condition (1) therefore requires that each vector of the set <p h <p 2 , <ps is perpendicular to every other one in that set, and that each has unit length. The symbol {<?„} will be used to denote an orthonormal set whose vectors arc <p h and <^ 3 . The simplest example of such a set is that consisting of the unit vectors along the three coordinate axes. 36 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 18 Every vector / in the space considered can be expressed as a linear combination of the vectors (pi, (pi, and <ps. That is, three numbers Ci, c 2 , c% can be found for which (2) f(r) = Ci<pi(r) + c 2 (pi(r) + c z <pz(r) (r = 1, 2, 3), when the components /( 1), /( 2), /( 3) are given. To find the number Ci in a simple way, consider equation (2) as a vector equation and take the inner product of both its members by (pi. This gives (/, ^l) — (pi) + Ci(<pi, (pi) + Cz((pz, <pl) = Cl, since ((pi, <pi) = 1 and (<pi, (pi) = (<Pz, <pi) = 0, according to condition (1). Similarly c 2 and c z are found by taking the inner product of the members of equation (2) by (pi and (p z , respectively. The coefficients are therefore (3) c n = (/, <p n ) = X f(r)ip n (r) (n = 1, 2, 3). r — 1 The representation (2) can then be written (4) f(r) = (/, <pi)<pi(r) + (/, <Pi)<Pi{r) + (/, = X n = 1 The representation (2) or (4) may be called an “expansion” of the arbitrary vector / in a finite series of the orthonormal reference vectors <pi, (pi , and <p 3 . These orthogonal reference vectors were assumed to be normalized only as a matter of convenience, in order to obtain the simple formulas (3) for the coefficients in the expansion. The normalization is not neces- sary, of course. The definitions and results just given can be extended immedi- ately to vectors in a space of k dimensions. In this case the index r , which indicates the component, has values from 1 to k, instead of 1 to 3; similarly the indices m and n, which distinguish the different vectors of an orthonormal set, run from 1 to k. The definition of the inner product of the vectors g\ and g% in this space, for instance, becomes k (g i, gt) = X s'iMffaW. r = l ( 5 ) Sec. 19] ORTHOGONAL SETS OF FUNCTIONS 37 The formal extension to vectors in a space of a countably infinite number of dimensions (k = °o ) is also possible. In this case the numbers g(r) (r = 1, 2, • • * ) which define a vector g would be so restricted that the infinite series involved, such as the series in (5) with k infinite, would converge. The possibility of the representation corresponding to (2) would have to be examined, of course. A generalization of another sort is also possible. The units of length on the rectangular coordinate axes, with respect to which the components of vectors are measured, may vary from one axis to another. In such a case the scalar product of two vectors gi and g 2 in three-dimensional space has the form 3 (?i, 0*) = X r = 1 The “ weight numbers” p(l), p( 2), and p( 3) here depend upon the units of length used along the three axes. 19. Functions as Vectors. Orthogonality. A vector g(r) in three dimensions was described above by the numbers g( 1), g(2) y g( 3). Any function g(r) which has real values when r = 1, 2, 3 will represent a vector if it is agreed that these values are the components of the vector. This function may not be defined for any other values of r, in which case its graph would consist only of three points. The function g(r) will represent a vector in space of k dimen- sions if it has real values when r = 1, 2, * • • , /c, which are considered as the components of the vector. If g(r) is defined only at these points, it is determined by the vector; graphically it is represented by k points whose abscissas are r = 1, 2, * * • , kj and whose ordinates arc the corresponding components of the vector. Now let g(x) be a function defined for all values of x in an interval a ^ x ^ b. To consider this function as a vector, the components should consist of all the ordinates of its graph in the interval. The argument x, which has replaced r here, has as many values as there are points in the interval, so that the number of components is not only infinite but uncountable. It is therefore impossible to sum with respect to x as we do with the index r. The natural process now is to sum by integration. 38 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 19 The norm of the function or vector g(x), or the sum of the squares of its components, is therefore defined as the number (1) N(g) = f*\g(x)]*dx. The inner product of two functions g m (x) and g n (x) is defined as the number ( 2 ) (i g m , g n ) = f b g m (x)g n (x) dx , Ja in analogy to equation (5), Sec. 18. The condition that the two functions be orthogonal is written (Qm, gn) = 0 , or (3) gm(x)g n (x) dx = 0. Just as before, definition (1) can be written N{g) = ( g , g). A set (or system) of functions {<7n(z)} (n = 1, 2, • ■ ■ ) is orthogonal in the interval (a, b) if condition (3) is true when m t* n for all functions of the set. The functions of the set are normed by dividing each function g n (x) by [2V (£«)]*, thus forming a set {<p n (a;)} (n = 1, 2, ■ • • ), which is normed and orthogonal, or orthonormal . An orthonormal system in (a, b ) is then characterized as follows: (4) <Pn) = <5mn (^, XI == 1, 2, * * * ); where 5 m n is Kronecker's 5, defined in Sec. 18. Written in full, equation (4) becomes (5) r <Pm{x)ip n {x) dx = 0 if m n, = 1 if m = ft, (m, n = 1, 2, • • • ). The interval (a, b) over which the functions and their inner products are defined is called the fundamental interval. Func- tions for which the integrals representing the inner product and the norm fail to exist must, of course, be excluded. Throughout this book, only functions which are bounded and integrable in the fundamental interval , and whose norms are not zero , will be considered. The aggregate of all such functions for the given interval makes up the function space being considered, Sec. 20] ORTHOGONAL SETS OF FUNCTIONS 39 in just the same way that the three-dimensional vector space con- sists of all vectors with three components g(r) (r = 1, 2, 3). An example of an orthogonal set of functions has already been given in Sec. 15; namely, the functions sm mrx (n = 1, 2, • • • ). The fundamental interval is the interval (0, L). The norm of all these functions is the same number, L/2, so the orthonormal set consists of the functions f 2 . mrx , % Vi sin it (« = i, 2, • ■ • The set {sin (nirx/L)} is also orthogonal in the interval ( — L, L ); the normalizing factor is easily seen to be l/\/L in this case. PROBLEMS 1. Show that the set of functions j cos nx} (n = 0, 1, 2, • • ■ ) is orthogonal in the interval (0, 7r). What is the corresponding ortho- normal set? Ans. {l/V7r, V2/7T cos nx\ (n = 1, 2, • • • ). 2. Show that the set (sin x, sin 2x, sin 3$, ••■,], cos x 1 cos 2x, • - * } is orthogonal in the interval (—7 r, t). Normalize this set. 20. Generalized Fourier Series. Given a countably infinite orthonormal set of functions {<£>«,(#) 1 (n = 1, 2, • • ■ ), it may be possible to represent an arbitrary function in the fundamental interval as a linear combination of the functions <p, t (r), (1) f(x) = Ci<pi(x) + C 2 <p*(x) + • • * + Cn<Pn(x) + * * * (a < x < b). This corresponds to representation (2), Sec. IS, of any vector in terms of the vectors of an orthonormal set. If the series in equation (1) converges and if, after being mul- tiplied by <p„(:r), it can be integrated term by term over the funda- mental interval (a, b), the coefficients c n can be found in the same way as before. Writing tin 1 , inner product of both members of equation (1) by <p n (x) — that is, multiplying (1) by <p n and integrat- ing over (a, b) — we have (/, <Pn) = ipn) + C 2 (^2, <Pn) + * * ' + C n (<p n , <Pn) + * * * 40 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 21 since ( <p m , , <p n ) — 8 mn . That is, c n is the projection of the vector / on the unit vector p*. These numbers c n are called the Fourier constants of f(x) corresponding to the orthonormal system { <p n (x) } ; they can be written (2) Cn = £ f(x)<p n (x) dx (n = 1, 2, • • • ). The series in (1) with these coefficients is called the generalized Fourier series corresponding to f(x), written (3) f(x) ~ ^ C n (p n (x) = tp n {x) £ b f (£) <Pn(£) d%. The above correspondence between J{x) and its series will not always be an equality. This can be anticipated at once by considering the case of vectors in three dimensions. In that case if only two vectors <pi(r), ^ 2 (r), make up the orthonormal system, any vector not in the plane of those two could not be represented in the form Ci<pi(r) + c 2 <p 2 (r). The reference system here is not complete, in the sense that there is a vector in the three-dimensional space which is perpendicular to both of its vectors <pi and <p 2 . Likewise in formula (3), if f(x) is orthogonal to every member <p n (x) of the system, every term in the series on the right is zero, and so the series does not represent f(x). If there is no function in the space considered which is orthog- onal to every <p n (x), the system {vn(,x)\ is called complete . So the system must necessarily be complete if all functions are to be represented by their generalized Fourier series with respect to that system. PROBLEMS 1. Show that the set \\^2/L cos ( mrx/L )} {n = 1, 2, • • * ) is ortho- normal in the interval (0, L), but not complete without the addition of a function corresponding to n = 0. 2. Show that the system {sin nirx] (n = 1, 2, * • • ) is orthogonal but not complete in the interval ( — 1, 1). 21. Approximation in the Mean. Let K m (x) represent a finite linear combination of m functions of an orthonormal set { <Pn(x)\ (n = 1, 2, ■ • • ; a ^ x ^ 6); that is, (1) K m {x) = ym(x) + 72 ^ 2 ^) + * * * + 7 Sec. 21] ORTHOGONAL SETS OF FUNCTIONS 41 The values of the constants y n can easily be found for which Km(x) is the best approximation in the mean to any given function f(x ) ; this means the best approximation in the sense that the value of the integral (2) J = £ [f(x) - K n (x)Y dx is to be as small as possible; it is also the approximation in the sense of least squares. Writing c n for the Fourier constants of f(x) with respect to <Pn( X )> Cn = £ f(x)<p n (x) dx, the integral J can be written J — f a lf ( x ) ~ yi<pi(z) ~ 72^2(2) — • • • — r m<p?n(x)] 2 dx = fa + Tl + 7i + ■ ' ' + jl - 2 T ,Ci - 2 -y id — • • • - 27 m C m . Completing the squares here by adding and subtracting rf, cl, ' ' ' A, gives ( 3 ) J = f b [f(x)] 2 dx - c\ - 4 - • • • - d + ( Tl - o,)2 + (72 — c 2 ) 2 + • • • + (7,,, - c m y-. It is clear from (2) that J ^ 0, so it follows from equation (3) that J has its least value when 71 = c lt y 2 = * * • y m = c m . The result can be stated as follows: Theorem 1. The Fourier constants of a function f(x) with respect to the functions <p 1 (r), <p' 2 (x), • • • , (p m (x) of an ortho normal set are those coefficients for which a linear combination K m (x) of these functions is the best approximation in the mean to f(x), in the fundamental interval (a, b). Since / S 0, it follows from equation (3), by taking y n = c ni that W cl + 4 + ■■■ + d S £ [f(x)]* dx. This is known as Bessel’s inequality. The number on the right is independent of m\ so it follows that the scries of squares of th© 42 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 22 Fourier constants of any function c? + d+ • • • + c* + • • • =X C * 1 always converges; and its sum is not greater than the norm of /c*o, (5) = It follows that the Fourier constants of every function correspond- ing to any orthonormal system {<p n ) approach zero as n tends to infinity: (6) lim c n = 0; n— > oo because a necessary condition for the convergence of the series in (5) is that its general term c\ approaches zero as n becomes infinite. 22. Closed and Complete Systems. Let S m (x) be the sum of m terms of the generalized Fourier series corresponding to f(x), with respect to an orthonormal set of functions {tp n ) (n = 1, 2, * • ■ ) ; that is, m (1) " S m (x) = 5} C n <Pn{x). This is the sum K m (x) in the last section when y n = c n . The sum S n (x) is said to converge in the mean to the function m if (2) lim p [ f{x ) — *S m (a;)] 2 dx = 0. m — * so This is also written l.i.m. S m (x) = f(x), m—> oo where the abbreviation l.i.m. stands for limit in the mean . If the relation (2) is true for each f(x) in the function space considered, the system is said to be closed .* According to Theorem 1, then, the system is closed if every function can be * The definitions of the terms closed and complete (Sec. 20) given here are those most commonly used today. Many German writers use the term closed ( abgeschlossen ) to denote what we have called complete, and complete (vollstandig) for our concept of closed. Sec. 22] ORTHOGONAL SETS OF FUNCTIONS 43 approximated arbitrarily closely in the mean by some linear combination of the functions <p n (x ). By expanding the integrand in equation (2) and keeping the definition of c n in mind, we have i m m I fa dX ” 2 2/ C « + X C *[ == °* Hence for every closed systern it is true that ( 3 ) X c » = Jf r /(*)] 2 dx - This is known as ParsevaVs theorem. When written in the form (4) X & = N ^’ T it identifies the sum of the squares of the components of /, with respect to the reference vectors <p n , with the* norm of /. Suppose B(x) is a function which is orthogonal to every func- tion of the closed set. Substituting it for / in equation (4) gives N(8) = 0, so that 0(.r) cannot belong to the function space; and thus it is shown that the set is complete (Sec. 20). The following theorem is therefore established: Theorem 2. If the set {(Pn(z)\ i* closed , it is complete. It is an immediate consequence that if there is a function which is orthogonal to every member of the set, the set cannot be closed. This is only a bare introduction to a general theory which has been developed extensively in recent years. To carry it further (even to prove the converse of Theorem 2), a broader class of functions and the idea of the Lebesgue integral are needed. But the term “closed” was defined hero with respect to con- vergence in the mean, and this type of convergence does not guarantee ordinary convergence at any point. That is, the statement (2) is quite different from the statement of ordinary convergence : lim & m (x) = f(x) (a g x ^ b). m— > oo It is this ordinary convergence, and the concept of closed orthog- 44 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 23 onal sets with respect to it, which are usually needed in the applications. No general tests of a practical nature exist for showing that a set of functions is closed. That is another reason for deserting the general theory at this point. 23. Other Types of Orthogonality. Some of the important extensions of the concept of orthogonal sets of functions should be noted. a. A set {0n(a;)} (n = 1, 2, • • • ) is orthogonal in an interval (a, b ) with respect to a given weight function p{x) y where it is usually supposed that p{x) ^ 0 in (a, b ), if (1) p(x)g m (x)g n (x) dx = 0 whenra n (m, n = 1, 2, • - • ). The integral on the left represents the inner product (g m , g n ) with respect to the weight function, a generalization of the inner product of vectors in terms of components with respect to axes along which different units of length are used (Sec. 18). The norm of g n {x) in this case is, of course, N(g n ) = ( g n , g n ) = JT p(x)[g n (x )] 2 dx (n = 1, 2, • • • ). By multiplying each function g n of the set by the normalizing factor [N(g n )]-* } the corresponding orthonormal set is obtained. This type of orthogonality can be reduced at once to the ordi- nary type having the weight function 1. It is only necessary to use the products y/ p(x)g n (x) as the functions of the system; then equation (1) shows that the system so formed has ordinary orthogonality in the interval (a, b). An important instance of orthogonality with respect to weight functions will be seen in the study of Bessel functions later on. The Tchebichef polynomials , (2) T n (x) = 2^1 cos ( n arccos x ), T 0 (x) = 1 (n = 1, 2, ■ - ■ also form a set of this type. This set is orthogonal in the interval ( — 1, 1) with respect to the weight function p(x) = (1 — x 2 )~l. Sjuc. 23] ORTHOGONAL SETS OF FUNCTIONS 45 This is easily verified by integration; thus, J 1 dx 1 C r T m (x)T n (x) = J o cos me cos nd de = 0 if m ^ n. 6. Another extension of orthogonality applies to a system of complex functions of a real variable x (a g x g 6). A system consisting of the functions gJx), where g n (x) = w„(a;) + iv n (x), is said to be orthogonal in the Hermitian sense if (3) fa ffm(x)g n (x) dx = 0 when m 9 ^ n, where g n (x) = u n (x) — iv n (x), the conjugate of g n . The system is normed if J[ 6 9n(x)g n (x) dx = 1; that is, if £ Wl{x) + vl(x)] dx = 1 for every n. When the functions are real, v n (x) = 0, and this type reduces to the ordinary orthogonality. Imaginary exponential functions furnish the most important examples of such systems. For instance, the functions (4) e inx — cos nx + i sin nx (n -= 0, ±1, ±2, • • • ) form a system which is orthogonal on the interval (— tt, tt) in the above sense. The proof is left as a problem. c. Extensions to cases in which the fundamental interval is infinite in length arc obtained by replacing a by — 00 or & by 00 , or both. d. For systems [g n (x, y)} (n = 1, 2, • • • ) of functions of two variables, the fundamental interval is replaced by a region in the £ 2 /-plane, and the integrations are carried out over this region. Similar extensions apply when three or more variables are present. Weight functions may be introduced in such cases too, as well as in case c. 46 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 24 PROBLEMS 1. By using the binomial expansion and equating real parts in the well-known formula (cos 6 + i sin d) n = cos nd + i sin nd , obtain the identity cos nd = cos n 6 — cos n_2 6 sin 2 6 -f cos n ” 4 6 sin 4 6 — * * • + A n , where are the binomial coefficients, and A n i n sin 71 6 if n is even, A n = ni n ~ l cos 0 sin 71- * 1 6 if n is odd. Hence show that the functions T n (x) defined by equations (2) above are actually polynomials in x of degree n. 2. Prove that the system of exponential functions in equation (4) of this section is orthogonal on the interval ( — t, t) in the Hermitian sense. 3. Prove that the system ^exp / 2 mrix\ \ [b-a)j (n 0 , ± 1 , ± 2 , • • * ), where exp(w) denotes e u , is orthogonal in the Hermitian sense on the interval (a, b). 24. Orthogonal Functions Generated by Differential Equa- tions. In solving the problem of displacements in a stretched string in Sec. 13, we used particular solutions of the partial differential equation of motion which vanished when x = 0 and x — L. In order that y = X(x)T{t) be such a solution, it was found that the function X{x) must satisfy the conditions (1) X"(x) + \X(x) = 0, (2) X(0) = 0, X(L) = 0, for some constant value of X, denoted there by —7. Equations (1) and (2) form a homogeneous boundary value problem in ordinary differential equations containing \ as a parameter. Since the solution of equation (1) that vanishes when x = 0 is X = C sin \/Xx, the problem has solutions not identically zero only if X satisfies the equation (3) sin V^L = 0. Therefore X = n 2 7r 2 /L 2 (n = 1 , 2, • • • ), and the corresponding solutions of the problem (l)-(2) for these values of X are, Sec. 24] ORTHOGONAL SETS OF FUNCTIONS 47 except for a constant factor, sin ( mrx/L ). These functions were shown to form an orthogonal set on the interval (0, L). Corresponding results can be found in much more general cases. When applied to a more general partial differential equation, separation of variables will yield an equation in X(x) of the type X" +Mx)X' + [Mx) + \f*(x)]X = 0. Here/ 1 ,/ 2 , and/ 3 are known functions involved in the coefficients of the partial differential equation, and X is the constant which arises upon separation of variables. When the last equation is multiplied through by the factor r(x ), where r(x) = eJViWrf* it takes the form (4) Tx [ r{x) 2] + [«(*> + x ^ x = °> known as the Sturm-Liouville equation. The boundary conditions on X(x) may have the form (5) aiX(a) + a 2 X\a) = 0, b.Xib) + b 2 X'(b) = 0, where a Xj a 2 , b i, and b 2 arc constants. The problem composed of the differential equation (4) and the boundary conditions (5) is called a Sturm-Liouvitte problem or system , in honor of the two mathematicians who made the first extensive study of that problem. * Under rather general conditions on the functions p , q, and r, it can be shown that there is a discrete set of values Xj, X 2 , * * * of the parameter X for which the system (4)-(5) has solutions not identically zero. These numbers X n are called the character- istic numbers of the system. In the above special case — equations (1) and (2) — they are the numbers nV 2 /L 2 , the roots of the char- acteristic equation (3). The solutions X n (x) (n = 1, 2, • • • ), obtained when X = X n in equation (4), are the characteristic functions of the Sturm- Liouville problem. These are the functions sin (mrx/L) in the special case. * Papons by Liouvillo and Sturm on this problem will be found in the first three volumes of Journal de matk&maliquc. , 183G-1838. 48 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 24 It will be showx in the following section that the set of func- tions jZnOc)} (n = 1, 2, • • • ) is orthogonal on the interval (a, b) with respect to the weight function p(x). Moreover, it can be shown that any function f(x ), defined in the interval (a, b ) and satisfying certain restrictions as to its continuity and differentiability, is represented by its generalized Fourier series corresponding to that orthogonal set of functions. That is, if (p n (x ) is the function obtained by normalizing X n (x), the series 2) C n <p n (x), where P (?&)f (**0 Vn (%) dx, converges to the function f(x) in the interval (a, b). It should be noted that the normalizing factor for X n here is the number (£vxid X y. When r(b ) = r(a), the statements made above are also true when the boundary conditions (5) are replaced by the conditions (7) X(a) = X(b), X'(a) = X'(b), called the periodic boundary conditions. Conditions of this sort frequently arise when x represents a coordinate such as the angle 0 in polar coordinates, or cos 0. The proof that series (6) converges to the function f(x) is quite long and involved, as we may well expect in view of the fact that the coefficients in differential equation (4) are arbitrary functions of x. The proofs generally make use of the theory of functions of complex variables, or the comparison of the expan- sion with a Fourier series, or both. The development of a general expansion theorem, along with other interesting and useful results in the general theory of Sturm-Liouville systems, is beyond the scope of the present volume.* The expansions considered in the chapters to follow are all special cases of the general theory. But in two of the important cases, those of Bessel and Legendre functions, the equations are * -A- treatment of these topics is included in a companion volume, now being prepared by the author, on further methods of solving partial differen- tial equations. Also see Ince, “Ordinary Differential Equations,” pp. 235#., 1927; and the references listed at the end of the present chapter. Sec. 25] ORTHOGONAL SETS OF FUNCTIONS 49 singular cases of equation (4) which must be treated separately even in the general theory. Hence the plan of presentation followed here is not especially inefficient. Many of the important sets of orthogonal functions are gen- erated in the above manner, as solutions of a homogeneous differential system involving a parameter. The expansion theorem shows that these sets are closed with respect to ordinary convergence, rather than convergence in the mean, so that we have an important advantage over the general theory discussed in the preceding sections. 25. Orthogonality of the Characteristic Functions. Two theorems from the general theory of Sturm-Liouvillc systems can easily be established here. They will be useful in the following chapters. The first shows the orthogonality of the characteristic functions, and the second shows that the char- acteristic numbers arc real. The existence of such functions and numbers will be established in each case treated later on, of course, by actually finding them. Theorem 3. Let the coefficients p, q, and r in the Sturm-Liouville problem be continuous in the interval a ^ x g b, and let X w , \ n be any two distinct characteristic numbers , and X m (x), X n (x) the corresponding characteristic functions , whose derivatives X' m (x) y X' n (z ) are continuous. Then X m (x) and X n (x) are orthogonal on the interval (a, b ), with respect to the weight function p(x). Furthermore , in case r(a) = 0, the first of the conditions (5), Sec. 24, can be dropped from the problem , and if r(b) = 0 the second of those conditions can be dropped. If r(b) = r(a) f those conditions can be replaced by the periodic conditions (7), Sec. 24. Since X m and A r n arc solutions of equation (4), Sec. 24, when X = A w and X = X n , respectively, (rX'J + (q + \ m p)X m = 0, ~ (rXJ + (q + Kp)X n = 0. Multiplying the first equation by X n and the second by X m , and subtracting, gives (X m - \ n )pX m X n = X, A (rX'J - X. (rXJ = [(rXJX,„ - (rXJXJ. 50 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 25 Integrating both members over the interval (a, V), (1) (Xm - X„) £ pX m X n dx = [r(X m X' n - XnX'J ]*. In the special case when = b% = 0, the boundary conditions (5), Sec. 24, become ( 2 ) X{a) = 0, X(b) = 0. Since both X m (x) and X n (x) then satisfy these conditions, it is evident that the right-hand member of equation (1) vanishes. But X m — X n 0, so that (3) £ p(x)X m (x)X n (x) dx = 0, which is the statement of orthogonality between X m and X n . In case r(a) = 0, it is clear that (3) follows from (1) without the use of the first of the conditions (2). Similarly, if r(b) = 0, the second condition is not needed. The proof that equation (3) follows when the general boundary conditions (5), Sec. 24, or the periodic boundary conditions, are substituted for (2), will be left for the problems. Theorem 4. If in addition to the conditions stated in Theorem 3 the coefficient p(x) does not change sign in the interval ( a , b), then every characteristic number of the Sturm-Liouville problem is real . * Suppose there is a complex characteristic number X, where X = cx. + i^ Let X(x) = u(x) + iv(x) be the corresponding characteristic function. Substituting this in the Sturm-Liouville equation, we have d fa ( ™ ' + irv ') + (q + ap + i$p) (u + iv) = 0. Equating the real and imaginary parts to zero, separately, fa (™0 + (? + ap)u — /Spy = 0, d fa ( rv ') + (Q + ap)v + Ppu = 0, * The functions p , q , and r are assumed to be real. Sec. 25] ORTHOGONAL SETS OF FUNCTIONS 51 and, upon multiplying the first of these equations by v and the second by u , and subtracting, it follows that — P(u 2 + v 2 )p = u ^ ( rv' ) — v ( ru ') = ^ [(rv')u - ( ru')v ]. Consequently, an integration gives the relation (4) — (u 2 + v 2 )p dx = ( uv ' — Again let us complete the proof here when the boundary condi- tions of the Sturm-Liouville problem have the special form (2). Since our characteristic function u + iv satisfies (2), its real and imaginary parts must each vanish when x = a and x = b. The right-hand member of (4) therefore vanishes. But if the function p(x) in the integral does not change sign in the interval (a, b) 7 the integrand itself cannot change sign, and so the integral cannot vanish. It follows that ft = 0, and therefore the char- acteristic number X is real. As before, if r(a) = 0, the first of conditions (2) is not needed, and if r(b) — 0, the second can be dropped. The argument is not essentially different when the more general boundary conditions (5), Sec. 24, or t.he periodic con- ditions are used. This matter is left for the problems. PROBLEMS 1. Complete the proof of Theorem 3 when the boundary conditions are (a) the conditions (5), Sec. 24; (/>) the periodic conditions (7), Sec. 24, assuming that r(b) = r(a). 2. Complete the proof of Theorem 4 when the boundary conditions are (a) the conditions (5), Sec. 24; (/>) the periodic conditions (7), Sec. 24, assuming that r(b) = r(a). Find the characteristic functions of each of the following special cases of the Sturm-Liouville problem. Also note the interval and weight function in the orthogonality relation ensured by Theorem 3, and find the normalizing factors. 3. X" + XX = 0; X'(0) = 0, X f (L) = 0. A ns. X n = cos ( mrx/L ) (n = 0, 1, 2, • • ■ ). 4. X" + XA r = 0; X ((»_= 0, X'(L) = 0. Am. ip n = V2/Lsin [(2 n — \)ttx/( 2L)] (n = 1, 2, • * - ). 52 FOURIER SERIES AND BOUNDARY PROBLEMS (Sec. 25 5. X" + XX = 0; X(tt) - X(-tt), X'(tt) = X'(-tt). Ans. {1, cos a, cos 2a;, • • • , sin a;, sin 2a;, • * * }. 6. X" + XX = 0; X(0) = 0, X'(l) + hX(l) = 0, where h is a con- stant. Show in this case that X n = sin a n x, where a n represents the positive roots of the equation tan a = —a/h, an equation whose roots can be approximated graphically. Also show that X n is normalized by multiplying it by V2 h/(h + cos 2 a n ). 7. (d/dx)(x*X r ) + \xX = 0; X(l) = 0, X(e) = 0. Note that the equation here reduces to one of the Cauchy type after the indicated differentiation is carried out. Ans. (p n = (a/ 2 / x ) sin (mr log x) (n — 1, 2, ■ • • ). REFERENCES 1. Courant, R., and D. Hilbert: “Methoden der mathematischen Physik,” Vol. 1, 1931. 2. Mises, R. v.: “Die Differential- und Integralgleichungen der Mechanik und Physik” (Riemann- Weber), Vol. 1, 192,5. CHAPTER IV FOURIER SERIES 26. Definition. The trigonometric series (1) |a 0 + (ai cos x + b i sin x) + (a 2 cos 2x + b 2 sin 2x) + * • * + (a n cos nx + b n sin nx) + ■ • ■ is a Fourier series provided its coefficients are given by the formulas ( 2 ) (In b, = 1 T /(*) e If J — v .-if T J-7T cos nx dx f(x) sin nx dx (n = 0, 1, 2, (n = 1, 2, ), ), where f(x) is some function defined in the interval (—7 r, 7r). In particular, series (1) with the coefficients (2) is called the Fourier series corresponding to f(x) in the interval (— t, 7r), written (3) /(*) + 2^ (a n cos nx + b n sin nx) (— 7T < X < 7r). Formulas (2) for the coefficients arc special cases of those for the generalized Fourier series in the chapter preceding. The functions 1, cos x, sin x, cos 2x , sin 2x } • • * constitute an orthog- onal (but not normalized) set in (— tt, 1 r). This was noted in Prob. 5, Sec. 25; but we can easily show it here independently. For if m, n = 0, 1 , 2, • • • , then J* cos mx cos nx dx = 0, J sin mj; sin nx dx = 0, if m 5* n f and whether m and n are distinct or not, cos mx sin nx dx = 0. 53 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 26 When, m = n the first two integrals become cos 2 nxdx = r = 2 T sin 2 nx dx = t. if ft 0, if ft = 0; Considering (3) as an equality and multiplying first by and integrating therefore gives formally cos nx fl T fi x ) cos nx dx = 7m„ (n = 0, 1, 2, • • ■ ). Similarly, multiplying by sin nx and integrating gives fl T fi x ) sin nx dx = irb n (» = 1, 2, • • • )• These are formulas (2) for the Fourier coefficients. Again, the corresponding orthonormal set of functions is 1 cos x sin x cos ‘lx sin 2x V2t Vir Vir ’ Vir ’ ’ and the Fourier constants c„ of f(z) corresponding to these functions are the integrals of the products, or the inner products, ? \ . eSe ^ unc ^ ons by f(x). So the Fourier series, with respect to this set, corresponding to Six), is /(*) v&L. '\/2t f(x') dx' cos nx' , , cos nx V? vT + r kx') s ^dx' s ^ i J-ir V 7T V 7T J where x' is used for the variable of integration. This can be written (4) Six) ~ g- j_J(x') dx' + i 2 [cos nx J* Six') cos nx' dx' + sin nx L Six') sin nx' d*'l; which is the same as series (3) with the coefficients (2). Note that the constant term is the mean value of f(x) over the interval (— x, vr). 27. Periodicity of the Function. Example. Every term in the above series is periodic with the period 2i r. Consequently, if the series converges to f(x) in the interval (-t, tt), it must converge to a periodic function with period 27 r for all values of x. Thus it would represent /(x) for every finite value of x , provided Fig. (>. the definition of /(x) is extended to include all values of x by the periodicity relation f(x + 2tt) = /(*). Thus the Fourier scries may conceivably serve either of two purposes: (a) to represent, a function defined in the interval (— 7r, 7 r), for values of x in that interval, or (/;) to represent a periodic function, with period 2tt, for all values of x. It clearly cannot represent a function for all values of x if that function is not periodic. The particular interval (—t, tt) was introduced only as a matter of convenience. We shall soon see that it is easy to change to any other finite interval. It is not necessary that f(x) be described by a single analytic expression, or that it be continuous, in order to determine the coefficients in its Fourier series. Of course the mere fact that the series can be written does not ensure its convergence or, if convergent, that its sum will be /(x). Conditions for this are to be established in Sec. 33. 56 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 27 Example. Write the Fourier series corresponding to the func- tion /(s) defined in the interval — r < x < * as follows: f(x) = 0 = x when —t<x^0, when 0 g x < w. The graph of this function is indicated by the heavy lines in *lg. b. ihe Fourier coefficients are do dn 1 C T i p* ~ Z I f( x ) COS nxdx = - I r J - T * JO cos nx dx cos nx + nx sin nx cos mr n (- 1 )* . sin nx x cos nx dx r ]o = (C0S nx - 1 )’ ~ t I s ' n nx dx = - I x sin nx dx J ~ T * Jo = ^2 |^ s i n n x — nx cos nx j = The series is therefore (i) /(*)^; + >br(-i)--i w ^2 cosnx- = | + (sin x - 2 cos a;) - i s in 2x + (| sin 3a: - |- cos 3a:) - i sin 4a: + • • • . If converges to /(*) when - r < x < T , it also converges or all other values of x to the periodic function represented by the dotted lines in the figure. Note that this periodic func- sZelbvZ tmU0U \ at V ±T ’ ±3t ’ • * • ’ the vaIue ^pre- sented by the senes at such points will be found later. As an indication of the convergence of the series (1) to f(x) it is instructive to sum a few terms of the series by composition t o™ be ,0U " d - ,0r tata “ ce ' 7T e 2 = j + sin x — - cos x * r 1 • o 2 sm Sec. 28] FOURIER SERIES 57 is a wavy approximation to the curve shown in the figure. The addition of more terms from the series generally improves the approximation. PROBLEMS Write the Fourier series corresponding to each of the following func- tions defined for In a few of the problems, sum a few terms of the series graphically; also show graphically the periodic func- tion which is represented by the series provided the series converges to the given function. 1. fix) = x when —t < x <tt. (Also note the sum of the series when x = ±i r.) 2. f(x) = e x when —7 r < x < ir. 2 sinh 7 r A ns 2 n sin nx. A ns. \ + 2 r+T 2 ( c,oB nx “ n sin l 3. fix) = 1 when —tt<x<0; fix) = 2 when 0 < x < ir. a 3 , 1 X? 1 r- ( — l)"' . Am - 2 + t^J sinr 1 4. f{x) = 0 when —tt<x< 0; /(;r) = sin x when 0 < x < r. 00 1,1. 2 cos 2 nx _ + - sm * - - 2j 4^-“T 1 A ns. 28. Fourier Sine Series. Cosine Series. When /( — #) = is called an odd function; its graph is symmetric with respect to the origin, and its integral from —w to ir is zero. When f( — x) = f{x), the function is wen; its graph is symmetric to the axis of ordinates, and Jy /O) dx = 2 J['/(.r) dx. As examples, the funetions x, x : \ and r- sin lex are odd, while 1, x 2 , cos lex, and x sin lex are even. Although most funetions are neither even nor odd, every function can bo written as the sum of an even and an odd one by means of the identity f(x) = i[f(x) +/(-*)! + *[/(*) - /(-*)]• ( 1 ) 58 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 28 When f(x) is an odd function defined in ( — t, 7 r), formulas (2), Sec. 26, for its Fourier coefficients become a n = 0 (n = 0, 1, 2, • • • )» 2 f T = - I f(x) sin nx dx (n = 1, 2, • • • ). Hence its Fourier series reduces to (2) /(a;) ~ ~ sin na: J* /(a; 7 ) sin ra;' da;'. The series in (2) is known as the Fourier sine series. It can clearly be written when fix) is any function defined in the interval (0, t), provided the integrals representing its coefficients exist. Furthermore, when f(x) is defined only in the interval (0, 7r), an odd function exists in (—7 r, 7 r) which is identical with f(x) in (0, 7r) . If that odd function is represented by its Fourier series, so is fix) in (0, t). Thus the question of convergence of the sine series to fix) in (0, t) depends directly upon the conditions of convergence of the series in the last section. Similarly, when fix) is an even function defined in the interval (— 7r, 7 r), the coefficients in its Fourier series are 2 r x a n = - I fix) cos nx dx (n = 0, 1, 2 • • ■ ), t jo b, = 0 (n = 1, 2, • • • ) ; and the series becomes the Fourier cosine series (3) /(s) ~ ^ f(x ') dx' + ^ cos nx J f(x') cos nx’ dx'. When /Or) is defined only in (0, tt), this series can be written, in general, and again the conditions under which it converges to /(a;) will be known when the conditions are found for the more general series in the last section. For functions defined in the interval (0, if), then, both the sine series and the cosine series representations can be considered. As indicated earlier, these are the series corresponding to /(a-) with respect to two different sets of functions, sin nx], and {l/Vw, y/2/x cos nx) (n = 1, 2, • • • ), each of which is Sec. 29 ] FOURIER SERIES 59 orthonormal in (0, t). The series (2) and (3) can be written more easily from this viewpoint; but in the theory of these series it is important to consider them as special cases of the series in Sec. 26. Every term of the sine series is an odd function. So if the series converges when 0 < x < w, it must converge to an odd function with the period 2t for all values of x. Similarly, if the cosine series converges, it must represent an even periodic function with period 2t. 29. Illustration. Let us write («) the Fourier sine series, and ( b ) the Fourier cosine series, corresponding to the function /(x), defined in the interval 0 < x < r as follows: f(x) = x when 0 ^ x < £j 7T = 0 when - < x ^ w. Z a. The coefficients in the sine series are b n — - I /Or) sin nx <lx = - I x sin nx dx 7T jo n Jo 1 . nir ht\ = r t l 2 Sill -X UT COS 7T 1; 7 Tti" \ Z Z / so the series is /W 2 / 2 . 'Hit it rnr\ . t \n 2 KIU "2" ~ n <! ° 1 * * * S 2 ) sm UX \ ( 7 f ^ /i = - ( 2 sin a; + x sin 2a: — - sin 3a; — T sin 4a; + 7T \ Z 2 9 7T 4 > If this sine series converges to our function f(x :), it must also represent the odd periodic extension of f(x) shown in Fig. 7. 60 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 29 b. The coefficients in the cosine series are 2 CLq = — T X dx = - A y 4 = - I f(x) cos nx dx = - I x cos nx dx * JO 7T JO 1 / 0 nir .nr 0 \ = iro* \ 2 C0S T + nir sm "2 - 2 /‘ /./ N 7T 1 / 2 n 7 r. 7 r.n 7 r 2 \ /W~8 + i2i^ cos T + ; m T-^j Therefore «■ , 1 \ , 8 ir L T _ 2) cos x — cos 2x -(i+D cos nx cos Sx + Assuming its convergence to f(x) } this cosine series would converge for all x to the even periodic function shown in Fig. 8. PROBLEMS Find (a), the Fourier sine series, and ( b ) the Fourier cosine series, corresponding to each of the following functions defined in the interval (0, 7r). Assuming that each series represents its function within that interval, show what function it represents outside the interval. 1. f(x) = x when 0 < x < it. (Compare Prob. 1, Sec. 27.) 00 J0 Ans. (a) 2 "S’ f-iu+i ^ nx - fh\ Z _ i ^ cos (2n - l)x ^ } n ’ w 2 T Zx ~(2n - 1)* " 2. f(x) = sin x when 0 < x < tt. 3. f(x) = cos x when 0 Ans. (a) sin x; (b) - - - 7 r 7T 1 cos 2nx 4 ri l — 1 < X < 7T. Ans. n sin 2 nx „ , ~ 4 n 2 _ 1 ; ( 0 ) cos 3 . Sec. 30] FOURIER SERIES 4 . = 7 i- — x when 0 < x < tt. , „ „ sin nx , n ir , 4 ^ cos (2n - 1)* Am. (a) 2 ^ — fr ~, + - 1)2 • l i 5. f(x) = 1 when 0 < x < x/2, f(x) = 0 when x/2 < x < x. . / \ 2 rwr\ sin Tia; Ans. (a) - 2i y - cos t) ~r~ ; 1 6. fix) = x when 0 < x < ir/2, fix) = tt — x when tt/2 < x <tt. , \ a ^A^ ,ia 8 ^ # cos (4?^ 2)s (a) (Compare Sec. 14); (6) j ^ " 7. /(a;) = e* when 0 < x <ir. , * 2 ^ ri , nN 1 u sin na; s * ^ x ^ n 2 + 1 ; l ... e* - — 1 2 "ST* ri ... .cos nx (b) 8. Obtain series (4), Sec. 2(5, for any function in ( — 7r, tt) from series (2) and (3), Sec. 28, for odd and even functions, respectively, using identity (1), Sec. 28. 30. Other Forms of Fourier Series. The Fourier series cor- responding to any function F(z), defined in the interval 7T < Z < 7T, J* F{z') dz' + ~ J^coh nz J F{z r ) cos nz r dz r + sin nz l /(*') • sin nz f dz r . Substituting the new variable x and the new variable of inte- gration x f throughout, where Lz . Lz ' X = ; X = 7 7 T 7T and writing fix) for F(irx/L) } ihe above correspondence becomes 62 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 30 ( 1 ) /(*)■ , 1 "sh r nirz C L + r^L c “ s — J_, fix') cos dx' •L h i • n irx C L . rnrx ' , ,1 + sin —j~ J L f\ x ) sin — cfo' . The series in (1) is the Fourier series on the interval ( —L, L ) corresponding to any function f(x) defined in that interval. The same substitution changes the sine series to one cor- responding to a function /(a;) defined in the interval (0, L), or an odd function in (— L, L): ( 2 ) f(x) ~ sin J f{x r ) sin dx'. It also changes the cosine series to the form (3) f(x) ~ 1 £ f(x’) <b'+|2“ B T J[V) cos dx', corresponding to a function f(x) defined in the interval (0, L), or an even function in (-L, L). The substitution simply changes the unit of length on the 3 -axis. Of course the forms (1) to (3) can also be written by noting the orthogonality of the sine and cosine functions involved there in the interval (— L, L ) or (0, L). Let us obtain the series for any interval (a, b) in this manner. Upon integrating, it will be found that f 6 ( 2mirix\ ( 2nirix \ , J. exp [b^J ,xp * That is, the set of complex functions / (2mrix\\ (»- 0 , ± 1 , ± 2 , ■ • ■ ), = 0 if m t* —n y = b — a if m = ~n. is orthogonal on the interval (a, &), in the Hermitian sense. Sec. 30] FOURIER SERIES 63 Assuming a series representation of f(x) in terms of those functions, /(*) = 2 C - exp (K) o<*< *>. the coefficients C„ can be found formally by multiplying through by exp [—2 mirix/{b — a)] and integrating. In view of the above orthogonality property, this gives J fix) exp dx = (b — a)C m . Thus, the exponential form of the Fourier series corresponding to a function fix) defined in the interval (a, b) is Grouping the terms for which the indices n differ only in sign, (4) takes the trigonometric form, <«> 2mr(x' — x) + of the Fourier scries corresponding to jf(rr) in (a, b). This can be obtained as well from the earlier form (5), Sec. 26, for the interval (— 7 r, tt) by making a linear substitution in the variables x and x These additional forms of the series, therefore, arise from the original form for the interval (— t, 1 r) by changing the origin and the unit of length on the rr-axis. So it is only necessary to develop the theory of convergence of the series for the interval (-7T, tt); the results will then be evident for the other forms. Form (5) contains the earlier forms as special cases. The series represents a periodic function with period ( b - a), if it converges. Therefore it can be considered as a possible expansion of either a function which is periodic with period (b - a), or a function which is defined only in the interval (a, b). Both types of applications are important. In the second case, 64 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 31 however, there may be many Fourier series representations of the function; for the function can be defined at pleasure in any extension of the interval, and the new series in the extended interval may still represent its function. It would then represent the given function in (a, b ) . # PROBLEMS 1. Write formula ( 1 ) in the form corresponding to ( 3 )-( 2 ), Sec. 26; also in the form corresponding to (5), Sec. 26. 2 . Write formula (5) when a = 0 , b = 2 L, and compare it with for- mula ( 1 ). Write the Fourier series corresponding to each of the following functions. 3. fix) = — 1 when —L < x < 0, fix) — 1 when 0 < x < L. Ans oo 4 X? 1 x 2 n — 1 • (^ n iy& sin 4. f{x) = \x\ when — L < x < L; that is, fix) = —x in (— L, 0) and (2 n — 1 )xx _J (2n — l) 2 cos l f(x) = * in ( 0 , L). Ans. \ - § ^ ( 2 ^ 5 . /(as) = x 2 when — L < x < L. , L 2 , 4L 2 '^(-1)“ mrx Ans - 1 + 1 ? 2i~^ cos ~T' 1 2 1 . —5 cos nxx — - sm mrx irn 2 n 6 . fix) = x + x 2 when — 1 < x < 1. i x 7. /(x) = 0 when —2 < x < 1, /(x) = 1 when 1 < x. < 2. , 1 1 XI if. 7VTT Ans. -r > - sm it 4 x n *2 l L ■> nxx , cos — + / nx\ . nxxl l cos nx — cos ~2 1 sin * 8 . fix) = 1 when 0 < x < 1, fix) = 2 when 1 < x < 3, and fix + 3) = fix) for all x. A„. 3 x n l L 2 nx 2 nxx , ( . 2mr\ . 2 nxx sin -TjT COS — + { l-COS-g-j sui- 3 - 9. fix) = e* when 0 < x < 1 , using the exponential form of the Fourier series. 31. Sectionally Continuous Functions. At this point let us introduce some special classes of functions, the use of which will Sbc. 31] FOURIER SERIES gg keep the theory which is to follow on a fairly elementary level These classes will include most of the functions which arise in the applications; but they are rather old-fashioned classes. As we shall point out from time to time, our principal results can be obtained for a considerably broader class of functions by using somewhat more advanced methods of modern analysis. A function is sectionally continuous , or 'piecewise continuous, in a finite interval if that interval can be subdivided into a finite number of intervals in each of which the function is continuous and has finite limits as the variable approaches either end point from the interior. Any discontinuities of such a function are of the type known as ordinary points of discontinuity. Every such function is bounded and integrable over the interval, its integral being the sum of a finite number of integrals of continu- ous functions. The symbol /(a* + 0) denotes the limit of f(x) as x approaches x 0 from the right. For /(*# - 0) the approach is from the left That is, if X is positive, f(x 0 + 0) = lim f(x 0 + X), X— 0 /(.To — 0) = lim f(x„ — X). A— >0 We define the right-hand derivative, or derivative from the right, of /(a:) at xo as the following limit: lim ^ Xo Q ~ f( Xa + 0) x-o X ' where X is positive, provided of course that this limit exists. Similarly, the left-hand derivative is lim /if l.~ . 0) ~ /fa ~ X) x— ►o X ' where X is again a positive variable. , ^ f°ll° WK c,f once that if f(x) has an ordinary derivative f (x) at x 0 , then its derivatives from the right and left both exist there and have the common value f'(x„). But a function may ave one-sided derivatives without having an ordinary derivative. For example, if f(x) = x 2 when x ^ 0, — sin x when x 0, 66 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 31 then /'( 0) does not exist, but at the point x = 0 the derivatives from the right and left have the values 1 and 0, respectively. Again, for the step function J{x) — 0 when x < 0, = 1 when x > 0, /'( 0) does not exist, but its one-sided derivatives have the common value zero. All the functions described in the problems and examples in this book are sectionally continuous and have one-sided derivatives. If two functions fix) and fix) have derivatives from the right at a point x = x 0 , so does their product. For the right- hand derivative of their product is the limit, as X approaches zero through positive values, of the ratio fijxo -j- \)fjXQ + X) — fijxp + 0)fjXQ + 0) X This can be written Mx* + x) /^° + x ) . -/^° . + °) A + + 0 ) fi(xo + X) — fijxo + 0) The limit of fi(xo + X) exists, and the limits of the two frac- tions exist, since they represent the right-hand derivatives of fix) and fiix) at the point x Q . Hence the limit of the ratio representing the right-hand derivative of the product / 1/2 exists. In the same manner it can be seen that the left-hand derivative of the product exists at each point where the two factors have left-hand derivatives. One further property will be useful, in connection with our theorem on the differentiation of Fourier series (Chap. V). Let f(x) be a function which is continuous in an interval a ^ x ^ 6, and whose derivative f'{x) exists and is continuous at all interior points of that interval. Also let the limits /'(a + 0) and f r (b — 0) exist. Then the right-hand derivative of fix) exists at x = a, and the left-hand derivative exists at x = 6, and these have the values /'(a 0) and/'(b — 0), respectively. Sec. 32] FOURIER SERIES 67 Since f(x) is continuous, and differentiable when a < x < b, the law of the mean applies. So, for every X (0 < X < b — a), a number 0 (0 < 0 < 1) exists such that /( a + X) - jf(fl) X = f'(a + 0X). Since f'(a + 0) exists, the limit, as X approaches zero, of the function on the right exists and has that same value. The function on the left must have the same limit; that is, the deriva- tive from the right at x = a has the value f (a + 0). Similarly for the derivative from the left at x = b. It follows at once that if f(x) and f{x) are sectionally continu- ous , the one-sided derivatives of f(x) exist at every 'point. 32. Preliminary Theory. In order to establish conditions under which a Fourier series converges to its function, a few preliminary theorems, or lemmas, on limits of trigonometric integrals are useful. The integrals involved in these lemmas are known as Dirichlct\s integrals. The lemmas here will be so formulated that they can also be used in the theory of the Fourier integral (Chap. V). There it is essential that the parameter fc used in the first lemma be permitted to vary continuously rather than just through the positive integers. In the latter case ( [k = n) the limit in Lemma 1 would follow quite easily from equation (6), Sec. 21. Lemma 1. If F(x) is sectionally continuous in the interval a ^ x ^ bj then (1) lim f b F(x) sin kx dx = 0. Jfe-> oo Let the interval (a, b) be divided into a finite number of parts in each of which F(x) is continuous, and let ( g , h) represent any one of those parts. Then, if it is shown that (2) lim f h F(x) sin kx dx = 0, the lemma will be proved. Divide the interval (g, h) into r equal parts by the points z 0 = g, x h x 2 , • • * , x r = h. Then the integral in equation (2) can be written r ~ 1 2 J F(x) sin kx dx , 60 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 32 or r — 1 2 0 1 £‘ +l F{ ^ sin kx dx + r ^ '*) - F &)] sin kxdx\- Carrying out the first integration and using the fact that the absolute value of an integral is not greater than the integral ot the absolute value of the integrand, we find that ( 3 ) I C K r -Z\ c i | F{x) sin kx dx S I F(x x ) 0 iZ o 1 I cos kxi— cos kxi+i k + J |[^ (z) — F( Xi )] sin kx\ dx The oscillation of F(x) in the interval ( Xi , Xi¥l ) is the difference between the greatest and least values of the function in that interval. Let Vr be the greatest oscillation of F(x) in any of the r intervals (xi, x^), so that \F(x) - F( Xi )\ g Vr in each interval. Also let M be the greatest value of |F(s)| in the interval (g, h). Then according to (3), sin kx dx 2M, / £ ~r Vr[Xi + i — - 2M J + 1J r (h — g ). Now let r be selected as the largest integer which does not exceed y/k. Then 2M r z <L k ~ y/k and this approaches zero as k tends to infinity. But r tends to m mty with k, and so Vr approaches zero too, because the osci ation of a continuous function approaches zero uniformly in all intervals of length (h - g)/ r as r becomes infinite. Hence lim £ F(x) sin kx dx = 0; so relation (2) is true, and the lemma is established. Lemma 2. If F(z) is sectionally continuous in the interval u — x = b and has a right-hand derivative at x = 0, then Sec. 32] FOURIER SERF EE 69 (4) l™JV W ^*_I P(+0) . The integral in (4) can ho writ, ten as the sum («) IV°> =r* * + X" *, Consider the first of these integrals. We can write lim J fl F(+ 0) ^ tf, = /'’(+0) lun = IFC+O), A: since f>jnj* rfw * Jo M 2 The function — /'’( +0 )]/.t in the second integral in (5) is scctionally continuous in the interval (0, h) since F(x) itself is, and since ' 1 lim W r—> f-o .r exists because F(r) has a right -hand derivative at * = 0. Lemma 1 therefore applies to the second integral in (5) ; giving rtijiLT. «.+«!> „ hlt , rf , k~> °a J 0 .r The limit of expression (5) is therefore F (- f 0)r/2; hence (4) is true and the lemma is proved. Lemma 3. If F(x) is scctionally continuous in the interval (a, b) and has derivatives from the right and left at a point x = Xo where a < x 0 < b, then 0)]. <*> & J> * - l Wr. + 0) + ffe - I h(^ integral in ((}) can bo written as the sum .TV) d.r + f /-■(,) dx _ Substituting x f = .r (l — .r in tin* first of those integrals, and x = x — Xo in the s(*cond, wo can write* their sum as 70 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 33 a F(xo - x') dx' + X °F(. x" + Xo) Sin J X " dx". Lemma 2 applies to each of the integrals here, and since and lim F(x o — x') = F(xq — 0), x f — >+0 lim F(x" + a; 0 ) = F(x 0 + 0), x n — ► +0 the limit of their sum is [F{x 0 — 0) + F(x 0 + 0)]t/2 . State- ment (6) in the lemma is therefore true. 33. A Fourier Theorem. A theorem which gives conditions under which a Fourier series corresponding to a function con- verges to that function is called a Fourier theorem. One such theorem will now be established. The conditions are only sufficient for the representation; necessary and sufficient con- ditions are not known. It will be convenient to consider the function as periodic with period 2w. Theorem 1. Let fix) satisfy these conditions: ( a ) fix + 2t) =f(x) for all values of and ( b ) fix) is sectionally continuous in the interval (— 7r, tt). Then the Fourier series (1) i a ° + ( a n cos nx + b n sin nx) 7 where i r a n = - I fix) cos nx dx (n = 0, 1, 2, • ■ • ), K J-7T (2) l c* b n = - fix) sin nx dx ( n =1,2, • TT J-t • • ), converges to the value m* + o) +/(*- o)] at every point where fix) has a right- and left-hand derivative . Condition ( h ) ensures the existence of the Fourier coefficients defined by equations (2), since the products fix) cos nx and fix) sin nx are continuous by segments and therefore integrable. It was pointed out in Sec. 26 that series (1) with coefficients (2) can be put in the form Sec. 33] FOURIER SERIES 71 cos [m(x r — a;)] dx f . The sum S n {x) of the first n + 1 terms of the series can there- fore be written s„0) = i J_ /(x') || + 2 cos NC®' - ®)]| dx '- Applying Lagrange’s trigonometric identity (Sec. 16) to the sum of cosines here, we have S n (x) = i TT sin [(n + - s)] , , 2 sin [|(x' — a:)] The integrand here is a periodic function of x' with period 2t ; hence its integral over every interval of length 2 tt is the same. Let us integrate over the interval (a, a + 2i r), where the number a has been selected so that the point x is in the interior of that interval; that is, a < x < a + 2t. Introducing the factor (x f - x) in both the numerator and the denominator of the integrand, we have 1 si (3) S n (x) = - F(x') - K J a sin [( n + i)(af ~ ®)] dx> x — X where Now (4) Moreover, F{x') is written as the product of two functions each of which is sectionally continuous in every interval and has a derivative from the right and left at the point x f = x , This was assumed in the theorem for the first factor f{x') } and it is easily verified for the second. Therefore, F{x f ) is sectionally continuous, and, according to Sec. 31, its derivatives from the right and left exist at x f = x. Therefore F(x') satisfies the conditions of Lemma 3 in which xq = x and k = n + •£. Applying that lemma to the integral in equation (3), we have 72 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 34 lim S n (x) = W(x + 0) + F(x - 0 )]. 71— ► « But according to equation (4), Fix + 0 ) = f(x + 0), F(x - 0 ) = fix - 0), and therefore lim S n ix ) = i[f(x + 0 ) + fix — 0 )]. 71— > 80 This is the same as the statement in the theorem. 34. Discussion of the Theorem. At any point where the periodic function f(x) is continuous, fix + 0) = fix - 0) = fix ) ; hence at such a point the mean value of the limits of the function, from the right and left, is the value of the function. If the one-sided derivatives of f(x) exist there, the Fourier series converges to f(x). Suppose f(x) is defined only in the interval (— 7 r, 7r). Then it is the periodic extension of this function which is referred to in Theorem 1 . Consequently, if j{x) is sectionally continuous, its Fourier series converges to the value «/(* + 0 ) +/(* “ 0 )] at each interior point where both one-sided derivatives exist. But at both the end 'points x = ±7 r the series converges to the value M/fr — 0) +/(— 7r + 0)], provided f(x) has a right-hand derivative at x = —t and a left-hand derivative at x = t, because that is the mean value of the periodic function at those points. It follows that if the series is to converge to /(— t +0)when x — —x, or to /(x — 0 ) when x = x, it is necessary that the function have equal limiting values at the end points of its interval; that is, /( — x + 0) = /(x — 0). It was pointed out that the other forms of Fourier series (Sec. 30) arise from the form used in Theorem 1 by changing the unit or the origin of the variable x . The sine series and cosine series are special cases arising when f(x) is an odd or an Sec. 34] FOURIER SERIES 73 even function. Consequently the Fourier theorem applies to these series at once with the quite obvious modifications neces- sary because of the changes in the interval. For the series corresponding to the interval (~L, L), for example, the theorem becomes Corollary 1. Let f(x + 2 L) = f(x) for all x , and let f{x) be sectionally continuous in the interval (— L, L). Then at any point where f(x) has a right - and left-hand derivative , it is true that (1) |[/(* + 0 )+/(*- 0 )] where = 2 ao + ?(- cos nrx + b n sin mrx\ J f(x) cos dx J ^ f(x) sin ~~ dx (n = 0, 1, 2, • • • ), (n = 1, 2, ■ • • ). It should be observed here, as well as in Theorem 1, that the existence of the one-sided derivatives is not required at all points of the interval, but only at those points where representation (1) is used. The function \/x~ 2 in the interval (— L, L), for instance, does not have one-sided derivatives at x = 0. But, according to our expansion theorem, the Fourier series corresponding to this function must converge to \Zx^ at all points for which — L g x <0 or 0 < x S L. At x = 0 the convergence is not ensured by our theorem. Again, if f{x) is defined in the interval (0, L ) and is sectionally continuous there, its Fourier sine series 00 /o\ V , • nirx (2) >sm where b n = y f f(x) sin dx (n = 1, 2, • • • ), U Jo U converges to i-[f(x + 0) + /(: v — 0)] at each point, a: (0 < x < L) where f(x ) lias one-sided derivatives. Series (2) obviously always converges to zero when x = 0 and when x = L. 74 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 35 Under the same conditions f(x) is represented by its Fourier cosine series in the interval (0, L ) : (3) i[/(* + 0)+/(3-0)] = la 0 + 2 a " cos ^ (0 <x <L), where a n = j- f(x) cos dz (ft = 0, 1, 2, • • * ). But in view of the even periodic function represented by the cosine series, this series converges to /(+0) at the point x = 0 when the derivative from, the right exists at that point. It converges to f(L — 0) at the point x = L when fix) has a left-hand derivative at that point. Broader conditions than these, under which the Fourier series converges to its function, will be stated in the next chapter. 35. The Orthonormal Trigonometric Functions. Let us denote by 4i the aggregate or space of all functions defined in the interval (— L, L) which are sectionally continuous there and which possess right- and left-hand derivatives at all points, except the end points, of the interval. At the end points let the derivatives from the interior exist. Also let every function of the class A L be defined at each point x of discontinuity to have the value ■§[/(# + 0) +/(a; — 0)], and at the end points x = ± L to have the value %[f(L — 0) +/(—L + 0)]. Then, according to Corollary 1, for every function /(x) belong- ing to the class A h there is a series (the Fourier series) of the functions sin (nirx/L), cos ( mrx/L ) which converges in the ordi- nary sense to fix). This can be stated as follows in the termi- nology of Chap. III. Corollary 2. In the function space A L) the orthonormal set consisting of all the functions ( 1 ) _1 1 _ V2l ; Vl mrx 1 . nirx C0S ~L ! VX 8m ~L~ (n 1 , 2 , ), is closed with respect to ordinary convergence. It is also complete. The proof of completeness is left for the problems. Similar statements can be made for functions defined in the interval (0, L), with respect to either the set of sine functions or the set of cosine functions. Sec. 35] FOURIER SERIES 75 Note that the last corollary is a statement about functions whose one-sided derivatives exist at all points of the interval, a condition which is not used in Corollary 1. Let us observe finally how the conditions of our Fourier theorem apply to our examples. The function in the example treated in Sec. 27 ; namely, fix) = 0 when — r < x ^ 0, = x when 0 g x < t, is continuous in the interval (— ir, r). It has one-sided deriva- tives at all points. Series (1), Sec. 27, therefore converges to fix) at all points in the interval — r < x < r, according to Theorem 1. At the points x = ±r it converges to the value 7r/2, since /(— t + 0) =0 and /( r — 0) = r. The graph of the periodic function shown there (Fig. 6) would be a complete representation of the function represented by the series if the points (±7r, 7r / 2) , (±37t, 7t/2), • • • were inserted. In Sec. 29 the cosine and sine series were found for the function fix ) = x when 0 ^ x < = 0 when ^ < x g r. This function is sectionally continuous in the interval (0, t), and its one-sided derivatives exist there. The sine series there- fore converges to fix) when 0 g x S tt, except »at x — 7t/2, where it converges to 7t/4. At x = 0 and x = t it converges to jT ( — | — 0) and fir — 0), since these are both zero. The cosine series for this function converges in just the same manner in the interval (0, r). PROBLEMS 1. Show that each of the functions described in Probs. 1 to 4, Sec. 27, satisfies the conditions under which the series found there converges to the function, except possibly at certain points. What is the sum of the series at those points? Ans. Prob. 1: x = ±r; sum = 0; Prob. 2: x = ±r; sum = cosh7r; Prob. 3: x =0, ±w; sum — f. 2. Solve Prob. 1 above for each of the functions in Probs. 1 to 7, Sec. 29. 76 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 35 3. Solve Prob. 1 above for each of the functions in Probs. 3 to 9, Sec. 30. 4. If f(x) = 0 when -1 < x < 0, fix) = cos tx when 0 < x < 1, /(0) = §■>/(!) = ~h, and fix + 2) — fix) for ! all x, show that 1 * fix) = g cos tx + - ^ 4w2 ”_ l sin 2mrx Ti for all values of x. 6. If fix) = c/4 - x when OiiS c/2, fix) = x — 3c/4 when c/2 S i S c, show that COS (4 n — 2 )ttx for all x in the interval 0 ^ x ^ c. 6. If /( x) = x 2 when —1 < as ^ 0, f(x ) = 0 when 0 ^ a? < 1, /(!•) = I? and J{x + 2) = f(x) for all x, find its Fourier series and show that it converges to f(x) for all values of x. 7. Prove that the orthonormal set of functions in Corollary 2 is complete in the function space A Lt (Compare Sec. 22; show that any function in A L which is orthogonal to every member of the set must be identically zero.) 8. State and prove the corollary, corresponding to Corollary 2, for functions defined in the interval (0, L ), with respect to the orthonormal set of functions { V2/Z sin (rnrx/L) } . 9. Show that the series L . mrx \ f(x) sin -g- ax J of squares of the coefficients in the Fourier sine series converges when- ever f(x) is bounded and integrate on the interval (0, L), and that 2 - 1 J 0 t/wi 2 dx - [See formula (5), Sec. 21.] 10. Show that the series 1 2 a 2 0 mrx _ cos — f- dx ) involving the squares of the coefficients in the Fourier cosine series, Sec. 35] FOURIER SERIES 77 converges whenever f(x) is bounded and integrable in the interval (0, L), and that 2 a ° + - L I dx - l c/O (Compare Prob. 9.) 11. If f(x) is bounded and integrable in the interval (— L, L), show that the series ’ ’ §«0 + 2 ) al + where a n and b n are the coefficients in Fourier series (1), Sec. 34, con- verges to a sum not greater than X f [/(*)] 2 dx. (Compare Prob. 9.) 12. For every function which is bounded and integrable in the interval (-L, L), the Fourier coefficients a n and b n in series (1), Sec. 34, approach zero as n tends to infinity. Show how this follows from Prob. 11. When the function is sectionally continuous, show that the result for b n follows also from Lemma 1. 13. The coefficients (i n and b, l} in Corollary 1, are those for which the sum of any fixed finite number of terms of the series written there will be the best approximation in the mean to f(x), in the interval (-L, L). Show how this follows as a special case of Theorem 1, Chap. III. 14. Find the values of A lf A 2 , and A* such that the function y = Ai sin y + A 2 sin + A [t sin will be the best approximation in the mean to the function f(x) = 1, over the interval (0, 2) (compare Prob. 13). Also draw the graph of y [ using the coefficients found, and compare it to the graph of f(x). Ans. A , = 4/j r, A, = 0, A z = 4/(3tt). 16. Show that it follows from the expansion in Prob. 5, Sec. 30, by setting x = L, that Similarly, show that CHAPTER V FURTHER PROPERTIES OF FOURIER SERIES; FOURIER INTEGRALS 36. Differentiation of Fourier Series. We have seen that the Fourier series representation of the function f(x) = x is valid in the interval —x < x < x; thus (Prob. 1, Sec. 29) x = 2(sin x — i sin 2x + i sin 3x — * • • ) when — x < x < 7T. But the series obtained by differentiating this series term by term, namely, 2(cos x — cos 2x + cos 3x — • • • ), does not converge to the derivative of x in the interval (— x, x). The term cos nx does not approach zero as n tends to infinity; hence the series does not converge. For all values of x, the above series for the function f(x) = x represents a periodic function with discontinuities at the points x = ±x, ±3x, • • * . We shall see that the continuity of the periodic function is an important condition for the termwise differentiation of a Fourier series. A complete set of sufficient conditions can be stated as follows: Theorem 1. Let f(x) be a continuous function in the interval “X # = 7r such that /(x) = /(— x), and let its derivative f'(x ) be sectionally continuous in that interval. Then the One-sided derivatives of f(x) exist (Sec. 31), and hence f(x) is represented by its Fourier series 00 (1) f(x) — %a 0 + ^ (a n cos nx + b n sin nx) (— x ^ x ^ x), where 1 i C* (2) a» = - I f(x) cos nx dx , b n = - I f{x) sin nx dx , xJ_ T irj-f and at each point where f{x) has a derivative that series can be differentiated termwise; that is, 78 Sec. 36] FURTHER PROPERTIES OF FOURIER SERIES 79 (3) fix) = ^ n( — a n sin nx + b n cos nx) (— 7r < x < x). Since fix) satisfies the conditions of our Fourier theorem, it is represented by its Fourier series at each point where its deriva- tive/"^) exists. At such a point fix) is continuous, so that (4) f(x) = ^a' 0 + 2) i a h cos nx + b r n sin nx ), where if 7 " 1 r ,r (5) a' - - I /'(a) cos no; ire, b' n = - I /'(x) sin nx da;. T J-7T These integrals can be integrated by parts, since fix) is con- tinuous and fix) is scctionally continuous. Therefore a' — - (6) fix) cos nx j + ~ J fix) sin nx dx LfM ~ /( — tt)] + nb». 7T cos n7r This reduces to n& n because of our condition that/( 7 r) = /(— 7 r), Furthermore, aj = 0. Likewise, b^, = i £/(x) w ^ u j — - /(x) cos nx dx = —na n . Substituting these values of a' and b r n into equation (4), we have 0O fix) = ^ (nbn cos nx — na n sin nx). This is the equation (3) which was obtained by differentiating (1) term by term; hence the theorem is proved. It is important to observe that, according to equation (6), the Fourier scries for fix) docs not reduce to series (3) obtained by termwise differentiation if the function fails to satisfy the condition fW) = fi~i r). This condition ensures the continuity of the periodic extension of/(x) at the points x = ±7r, and therefore at all points, in view of the continuity of fix) in the interval (— x, 7 r). 80 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 37 At a point where f(x) has a derivative from the right and from the left, but no ordinary derivative, we can easily see from the above proof that termwise differentiation is still valid in the sense that 00 i[f(x + 0) + f'(x — 0)] = V n( — a n sin nx + b n cos nx). i Since this is true for the periodic extension of fix), the derived series converges at the points x = ±tt to the value «f(- 7T + 0) + /'(*•- 0)] if fix) has a right-hand derivative at -tt and a left-hand deriva- tive at t. We are assuming the continuity of the periodic extension of f(x) at all points, of course. Theorem 1 applies with the usual changes to the other forms of Fourier series. PROBLEMS 1. Show that the series in Prob. 4, Sec. 27, can be differentiated term by term, and state what function is represented by the derived series. (Compare Prob. 4, Sec. 35.) 2. In the problems, Sec. 29, obtain the series in Prob. 3a by differen- tiating the series in Prob. 26. Note that this is permissible according to Theorem 1; but we cannot reverse the process and*obtain the latter series by differentiating the former. 3. In Probs. 1 to 7, Sec. 29, which of the series can be differentiated termwise? _ Am. 1(6); 2(a), (6); 3(6); 4(6); 6(a), (6); 7(6). 4 . Show that in Probs. 4 and 5, Sec. 30, the series are termwise differentiable. 5. Show that the Fourier coefficients a n and b n for the function f(x), described in the first sentence of Theorem 1, satisfy the relations lim na n = 0, lim n6„ = 0. 71 — ► oo 7 i — > oo 37. Integration of Fourier Series. Termwise integration of a Fourier series is possible under much more general conditions than those for differentiation. This is to be expected, because an integration introduces a factor n in the denominator of the general term. It will be shown in the following theorem that it is not even essential that the original series converge to its function, in order that the integrated series converge to the Sec. 37] FURTHER PROPERTIES OF FOURIER SERIES 81 integral of the function. Of course, the integrated series is not a Fourier series if a 0 ^ 0, for it contains a term a Q x/ 2. Theorem 2. Let f(x) be sectionally continuous in the interval (-“ir, tt). Then whether the Fourier series corresponding to f(x), c© (1) f(%) ~ ^ (a n cos nx + b n sin nx), converges or not, the following equality is true : , ( 2 ) /(*) dx = — (x + tt) + ^ - [a w sin nx — b n (cos nx — cos nx)], 7 n when —ir^x^T. The latter series is obtained by integrating the former one term by term . Since /(x) is sectionally continuous, the function F(x), where (3) F(x) = J* r f(&) dx — ioox, is continuous; moreover F'(x) = f(x) - except at points where /(x) is discontinuous, and even there F(x) has right- and left-hand derivatives. Also, F( tt) = J^/(x) dx — = a 0 7r — -ja ( )7r = ^-aoTr, and F(— 7r) = ^a ( )7r; hence /'’(t) = According to our Fourier theorem then, for all x in the interval — 7 r 5^ x ^ 7 r, it is true that 00 F(x) = ^ (A„ cos nx + B n sin nx), where \ C T i A n = - I F(x) cos nx dx , B n = - I F(x) sin nx dx. J -TT IT J -TT Since F{x) is continuous and F'(x) is sectionally continuous, the integrals for A n and B n can be integrated by parts. Thus if n y* 0, 82 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 38 A n = -J- F(x) sin nx — ( F'(x) sin nx dx nr L J-* nr J_ T = f_ w [ f(x) ~ i \ sin dx = ~l b - Similarly, B n = a n /n; hence 1 00 1 (4) F(x) = g A 0 + - (a n sin nx — 6* cos no;). l But since F(r ) = ^oott, - a 0 7T = ^ A 0 — i b n cos n7r. 1 Substituting the value of A 0 given here in equation (4), F( x ) 1“ + - [a n sin nx — b n ( cos nx — cos nir)]. l - In view of equation (3), equation (2) follows at once. The theorem can be written for the integral from x 0 to x, when —ir ^ x Q ^ r and —it ^ x ^ tt, by noting that j£7(*) = /_V^) dx - <f*. The other forms of Fourier series can be integrated termwise under like conditions, of course. Still more general conditions under which the Fourier series can be integrated term by term will be noted in Sec. 39. PROBLEMS 1. By integrating the expansion found in Prob. 4, Sec. 27, from —tt to x , obtain the expansion N X 1 1 1 F(x) = - + 9 — 9 COS X T L A 7 r 1 sin 2 nx n4n 2 — 1 * where and F(x) = 0 when —ir^x^O, F(x) = l — C os x when 0 ^ x ^ x. 2. Integrate the series obtained in Probs. 1 and 3, Sec. 27, from 0 to x , and describe the functions represented by the new series. 38. Uniform Convergence. If A n and B n (n = 1, 2, • * • ; m) represent any real numbers, the equation Sec. 38] FURTHER PROPERTIES OF FOURIER SERIES 83 m m hi m 2, ( A » X + 5 ») 2 = X 2 X A l + 2* 2 A ' B » + 2 B * = 0 1 111 cannot have distinct real roots. In fact, if it has a real root x = x Q , then A n x 0 + B n = 0 for all n, and the ratio B n /A n must be independent of n. The discriminant of the quadratic equa- tion in x is therefore negative or zero; that is, ( m \ 2 in m X A -Bn) g X A l X Bl With the help of this relation, known as Cauchy 1 s inequality , we can readily show that the convergence of the Fourier series to the function f(x) described in Theorem 1 is absolute and uniform. Broader conditions for uniform convergence will be cited in the next section. But it should be noted that a Fourier series cannot converge uniformly in any interval containing a dis- continuity of its function, since a uniformly convergent series of continuous functions always converges to a continuous function. Theorem 3. Let f(x) be a continuous function in the interval — 7 r S x ^ t such that /(t) = /(— 7 r), and let its derivative f'(x) be sectionally continuous in that interval. Then the Fourier series for the function f{x) converges absolutely and uniformly in the interval ( — 7 r , tt). The theorem will be proved if we can show that for each positive number e an integer m 0 , independent of x , can be found such that m' ^ | a n cos nx + b n sin nx\ < e m when m > m 0 , for all m' > m. The term between the absolute value signs represents, of course, the general term in the Fourier series corresponding to f(x). Since it can be written as \fa\ + bl cos ( nx — 0) ^0 = arctan it is clear that |a n cos nx + b n sin nx | ^ + 84 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 38 so it will suffice to show that ( 2 ) m X VoJ + b\ < € ( m > mo, ml > m). In the proof of Theorem 1 we found that ( 3 ) «» = £>'„ = —na n , where a'„ and fc' are the Fourier coefficients of the function Therefore wt m' 2 = 2 Applying inequality (1) to the last sum, we have , m* m’ v * ( 4 ) 2 = { 2^2 [(<)2 + w i] \ ■ Bessel's inequality (4), Sec. 21, applies to the bounded inte- grate function f'(x) 3 with respect to the orthonormal set of functions (6> cos ^ sfa “} • • ■ >■ giving the relation 2 K <) 2 + "l 2 dx for every integer m'. Let M denote the member on the right here ; then the second sum on the right of inequality (4) does not exceed the number M. Now the series converges; so for any positive number e 2 /M an integer m 0 can be found such that m Sec. 39] FURTHER PROPERTIES OF FOURIER SERIES 85 when m > mo, for all rr\I > m; and mo is clearly independent of x. For this choice of mo, the right-hand member of inequality (4) is less than €, so that inequality (2) is established and the theorem is proved. But in view of inequality (2) we have also shown that, under the conditions in Theorem 3, the series 00 2) Val + bl always converges . Consequently each of the following series converges : 5) W, | \bn\. It is of interest to note that the Parse val relation (3), Sec. 22, applies to the class of functions described in Theorem 3 with respect to the orthonormal set of trigonometric functions (5). This follows by multiplying the Fourier series expansion of f(x) by f(x), thus leaving it still uniformly convergent, and integrat- ing, to obtain [ f ( x )] 2 dx = -g-ao dx + ^a n f* v f(x) cos nx dx f(x ) sin nx ctaj. In view of the definitions of a n and h n , this can be written (6) S-* dX= * [i a ° + X (“n + K) ] - This is the Parse val relation. PROBLEM Show that if a class of functions satisfies the Parseval relation, the orthonormal set is closed with respect to the limit in the mean (Sec. 22). Hence deduce that set (5) is closed in that sense, for the class of all functions satisfying the conditions in Theorem 3. 39. Concerning More General Conditions. The theory of Fourier series developed above will be sufficient for our purposes. Let us note at this point, however, a few of the many more 86 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 39 general results which are known. These will be stated without proof, since our purpose is only to inform the reader of the existence of such theorems. They cannot be stated in their most general form, usually, without introducing Lebesgue integrals in place of the Riemann integrals considered here. a. Fourier Theorem . Let f{x) denote here a periodic func- tion with period 2 x, and let f(x) dx exist. If the integral is improper, let it be absolutely convergent. Then the Fourier series corresponding to /(x) converges to the value M* * * § + o) +/(*-o)] at each point x which is interior to an interval in which /(x) is of bounded variation.* b. Uniform Convergence . If the periodic function /(x) described under ( a ) is continuous and of bounded variation in some interval (a, b) then its Fourier series converges to /(x) uniformly in any interval interior to (a, b). f We have noted earlier that the partial sums S n (x ) of a Fourier series cannot approach the function f(x) uniformly over any interval containing a point of discontinuity of /(x). The nature of the deviation of S n (x ) from /(x) in such an interval is known as the Gibbs phenomenon c. Integration. The Parseval relation (1) \ j [/(z)P dx = i a 2 0 + ^ ( a * + h n), is true whenever jf(x) is bounded and integrable in the interval ( — x, 7r).§ That is, the series of squares of the Fourier coefficients of /(x) on the right of equation (1) converges to the number on the left. Now let a n and ($ n be the Fourier coefficients of a function <p(x), bounded and integrable in the interval (—7 r, tt). Then (a n + <*n) and ( b n + £ n ) are the coefficients of the function U + <?)> an d according to equation (1) we have * See first the proof in Ref. 2 at the end of this chapter. t For a proof, see Ref. 2. t See Ref. 1. § See first the proof given in Ref. 2. Sue. 39] FURTHER PROPERTIES OF FOURIER SERIES 87 l f [/(*) + <?(x)] 2 dx T J ~T GO = 2 ( a ° a ° ) 2 ” t “ [C + <*») 2 +' (&» + j 8 ») 2 ]. I - Likewise <p(x)] 2 dx to = i (do — ao) 2 + 2 [(“» ~ a ") 2 + (&» ~ 0») 2 ], and by adding the last two equations we find that ( 2 ) r/_ + b n fi n ). In form (2) of the Parseval formula suppose that <p(x) = g(x) when — ir < x < t, = 0 when t < x < ir (— 7r ^ t ^ 7r), where {/(x) is bounded and integrable in the interval (— 7r, tt). Then 1 f #(x) cos nx dx, f$, ,-ir 7 T J-x g(x) sin no; dx, and form (2) becomes (3) J f(x)g(:r) dx = £a<> *7(x) dx -T J IT + ?[-f < 7 (x) cos no: do: + 7) "J- < 7 (x) sin no: dx So it follows from statement (c) that if the Fourier series cor- responding to any bounded integrable function /(x) is multiplied by any other function of the same class and then integrated term by term, the resulting series converges to the integral of the product f(x)g(x). When g(x) = 1, we have a general theorem for the termwise integration of a Fourier series. 88 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 40 PROBLEM Assuming statement (c), show that it follows that the set of func- tions (5), Sec. 38, is closed, in the sense of convergence in the mean, with respect to the class of bounded integrable functions in the interval (— 7 r, 7r). (Compare the problem at the end of Sec. 38.) 40. The Fourier Integral. The Fourier series (Sec. 30) cor- responding to fix) in the interval (— L, L) can be written It converges to \ [fix + 0) + fix — 0)] when — L < x < L, provided f(x) is sectionally continuous and has right- and left- hand derivatives in the interval (— L, L). If fix) satisfies those conditions in every finite interval, then L may be given any fixed value, arbitrarily large but finite, in order that we may obtain a representation of f(x) in a large interval. But this series representation cannot be valid outside that interval unless fix) is periodic with the period 2 L, since series (1) represents only such functions. To indicate a representation which may be valid for all real x when fix) is not periodic, it is natural to try to extend series (1) to the case L = oo . The first term would then vanish, assuming that J f ^ fix) dx converges. Putting Aa = t/L, the remaining terms can be written £ 2 j C ° S [t ^ X ' ~ *)] dX ' 60 = ^ A a j f(x') cos [nAa(V — a;)] dx'. 1 —L OO The last series has the form ^F(nAo:)Aa, where Fia) = J^ L fi x ') cos [aix' — x)] dx'; hence when Aa is small, it may be expected to approximate the integral F(a) da . (Note, however, that its limit as A a 3ec. 40] FURTHER PROPERTIES OF FOURIER SERIES 89 approaches zero is not the definition of this integral; further- more, when A a. approaches zero, L becomes infinite, so F(a) itself changes.) But if the process were sound, when L becomes infinite series (1) would become 1 f* 00 f* 00 - I da I f(x') cos [a(x r — a;)] dx r . * Jo J- « This is the Fourier integral of f{x). Its convergence to f(x) for all finite values is suggested but by no means established by the above argument. It will now be shown that this repre- sentation is valid whenf(x) satisfies the conditions in the follow- ing Fourier integral theorem: Theorem 4. Let f(x) be sectionally continuous in every finite interval (a, b), and let * \f(x)\dx converge . Then at every 'point x ( — °o < x < °o), where f{x) has a right- and left-hand deriva- tive , f(x) is represented by its Fourier integral as follows: (2) i[/(x + 0) +f(x - 0)] = - I (la I fix' ) cos [a (x' — x)] dx'. In every interval (a, b),f(x) satisfies the conditions of Lemma 3, Sec. 32, so that (3) | [f(x + 0) +f(x - 0)] = lim C f(x') a jllJ0xL - g)j dx > at, any point x (a < x < b), where f(x) has a right- and left-hand derivative. Now (4) fix') A [<*(*' - *)] dx> i: Whenever a < x, r f(x') sin _ x — X and the latter integral converges because |/(a;)| dx does. 90 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 40 Similarly for the last integral in equation (4), when b > x. Hence for any e > 0 a positive number N can be found such that ^ a < and b > N, the first and last integrals on the right of equation (4) will each be numerically less than e/3. The second integral there can be made to differ from the value + 0) + f(x — 0)] by an amount numerically less than <:/3 by taking a sufficiently large, according to equation (3). Hence the integral (4) differs numerically from the above value by an amount less than e for all a greater than some fixed number; that is, (6) M 8ln 5 - , - 1 w* + 0) +f(„ - 0)]. Writing the fraction in the integrand as an integral, and divid- ing by 7 r, this becomes 2 f f( x + 0) + f(x — 0)] = 1™ l J m fix') dx' cos [o' (a/ - *)] da' = l l d “' /_ _ fix') COS [a'(x' - x)] dx'. The inversion of order of integration in the last step is valid because the integrand does not exceed \f(x')\ in absolute value, so that the integral f_ " fix ' ) cos W(x' - x)] dx' converges uniformly for all «'.* The last equation is the same as statement (2) in the theorem. Fourier integral theorems with somewhat broader conditions on f(x) are also known. The more modem theorems take advantage of the use of Lebesgue integration. PROBLEMS 1. Verify the Fourier integral theorem directly for the function f(x) = 1 when -1 < * < 1, fix) = 0 when x < - 1 and when x > 1. ihe following integration formula, usually established in advanced calculus, will be useful: * See, for instance, p. 199 of Ref. 1. U FURTHER PROPERTIES OF FOURIER SERIES 91 f °° sinfcx dx =Z ii k > 0, Jo * i 2 = 0 if k = 0, = ~ ilk <0. how that the function f(x) = 0 when x < 0, f(x) = e~~ x when ■ /(0) = is represented by its Fourier integral; hence show that ;egral X cos ax + a sin ax 1 + a 2 da y value 0 if x < 0, tt/2 if x = 0, and we"* if a; > 0. how that the Fourier integral of the function f(z) = 1 does not ge. Other Forms of the Fourier Integral. Let f(x) be an odd :m which satisfies the conditions of Theorem 4. Then (x') cos [a(x' — x)] dx f f* OO /• 00 J ^ jf(x') cos [a (s' — *)] dx' +1 f( — y) cos [a(y + x)] dy j /(x') cos [a(x' — x)] d,x' — J /(x') cos [a(x' + x)] dx' f* oo 5 sin ax I f(x r ) sin ax' dx' . Jo the Fourier integral formula becomes [f(x + 0) + f(x — 0)] sin ax da sin ax' dx'. i is the Fourier sine integral, corresponding to the Fourier M’ies. If f(x) is defined only when x > 0, formula (1) <1 provided f(x) is piecewise continuous in each finite i,l in x ^ 0 and has a right- and left-hand derivative point x (x > 0), and provided £ |/(:r)| dx converges. Iarly if f(x ) is an even function satisfying the conditions orem 4, it is represented by its Fourier cosine integral: 92 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 41 (2) g[/(* + 0) +/(*- 0)] = T Jo C0S aX Jo ft*') C0S aX> dx '- Under the conditions just given for the sine integral, formula (2) is also valid if f(x) is defined only when x > 0. Moreover, the integral converges at x = 0 to /(+0) provided fix) has a right- hand derivative there. “ By writing cos [a(x' — x)] in terms of imaginary exponential functions, the integral formula of Theorem 4 can be reduced to (3) 2 [/(* + 0) + fix — 0)] “ ^ J m eiax da J- " e~ iax ' fix') dx'. This is the exponential form of the Fourier integral of the func- tion f{x) defined for all real values of x. If gi.a) is a known function when a > 0, note that the integral equation (4) fix') sin ax' dx' = gia), can be solved easily for the unknown function fix) (x > 0), provided that function is one of the class for which the Fourier integral formula (1) is true. For by multiplying equation (4) through by y/2 sin ax and integrating with respect to a over the interval (0, oo) W e have, in view of formula (1), (5) fix) = g(a) sin ax da ix > 0). Of course this formula would give the mean value of fix) at a point of discontinuity. K The integral in equation (4) is called the Fourier sine transform of fix) Formula (5), which gives fix) in terms of its transform Qi<x), has precisely the same form as equation (4). In view of formula (2), the sine functions can clearly be replaced by cosines in equations (4) and (5). PROBLEMS 1. Show that the formula in Theorem 4 reduces to formula (2) when fix) is an even function. Sec. 41] FURTHER PROPERTIES OF FOURIER SERIES 93 2. Transform the formula in Theorem 4 to the exponential form (3). 3. Apply formula (2) to the function /(as) = 1 when 0 g* x < 1, /(as) — 0 when x > 1, and hence show that sin a cos ax , ir , A . ^ - da = - when 0 ^ x < 1. a 2 — | when a; — 1, = 0 when a; > 1. 4. Apply formula (1) to the function /(as) = cos as, and thus show that I 15 sin ax , t _ ^ A -r— — da = - e~ x cos as if xb > 0. x 4 + 4 2 6. By applying formula (1) to the function /(as) = sin as when 0 £ x /(as) = 0 when x > t, show that X sin ax sin 7 ro : 1 — a 1 da t . = - sin x = 0 if 0 ^ as ^ 7 r, if as > 7 r. 6. Apply formula (2) to the function /(as) of Prob. 5 and obtain another integration formula. 7. Show that the solution of the integral equation sin ax dx = g(a), where g{a) = 1 when 0 < a <7 r, g (a) =0 when a > ir, is . 2 1 — cos7ras m = - 5 — 8. Show that the integral equation (x > 0). has the solution /(as) cos ax dx = e~ a ft*) = l ttv* (as > 0). REFERENCES 1. Carslaw, H. S.: ‘‘Fourier’s Series and Integrals,” 1930. 2. Whittaker, E. T., and G. N. Watson: “ Modern Analysis,” Chap. 9, 1927. 3. Titchmarsh, E. C.: “Theory of Functions,” Chap, 13, 1939. This is more advanced. CHAPTER VI SOLUTION OF BOUNDARY VALUE PROBLEMS BY THE USE OF FOURIER SERIES AND INTEGRALS 42. Formal and Rigorous Solutions. In an introductory treatment of boundary value problems in the partial differential equations of physics, it seems best to follow to some extent the plan used in introductory courses in ordinary differential equations; that is, to stress the method of obtaining a solution of the problem as stated, and give less attention to the precise statement of the problem that would ensure that the solution found is the only one possible. But it is important that the student be aware of the shortcomings of this sort of treatment; hence some discussion of the rigorous statement and solution of problems will be given. The subject of boundary value problems in partial differential equations is still under development; in particular, the uniqueness of the solutions of some of the impor- tant types of problems has not yet been satisfactorily investigated. In ordinary differential equations, the solution for all x ^ 0 of the simple boundary value problem y'{x) = 2, 2/(0) = 0, would generally be given as y = 2x, because it is understood that y(x) must be continuous. Without such an agreement, however, the function y = 2x + c when x > 0, y = 0 when x = 0, is a solution for every constant c; that is, the solution is not unique. Even when the boundary condition is written 2/(+0) = 0, the solution could be written, for instance, as y = 2x when 0^x^ a, y = 2x + c when x > a, unless y(x) is required to be continuous for all 3 ^ 0. Such tacit agreements necessary for the existence of just one solution are not nearly so evident in partial differential equations. Furthermore, if the result is found only in the form of an infinite series or integral, it is sometimes quite difficult 94 Sec. 43] SOLUTION OF BOUNDARY VALUE PROBLEMS 95 to determine the precise conditions under which that series qr integral converges and represents even one possible solution. The treatment of an applied boundary value problem is only a formal one unless it is shown (a) that the result found is actu- ally a solution of the differential equation and satisfies all the boundary conditions, and ( b ) that no other solution is possible. The physical problem will require that there should be only one solution; hence the mathematical statement of the problem is not strictly complete unless the uniqueness condition ( b ) is satisfied. 43. The Vibrating String. The formula for the displacements y(x, t) in a string stretched between the points (0, 0) and (L, 0) and given an initial displacement y = f{x) was found in Sec. 13 to be 00 .... . . mrx rnr at (1) 2 / = > | A n sin -j- cos -£-> “i where . 2 f L \ . nirx , A n = j I f(x ) sin -j-dx. The function fix) must of course be continuous in the interval 0 ^ x ^ L and vanish when x = 0 and x — L. In addition, let, fix) be required to have a right- and left-hand derivative at each point. Then the Fourier sine series obtained when t = 0 in formula (1) does converge to/(z); hence this initial condition is actually satisfied. Thus an important improvement in the formal solution is made possible by the theory of Fourier series. The nature of the problem requires the solution y(x, t) to he continuous with respect to x and t. Since y(x, t) is to satisfy the equation of motion ( 2 ) d' l y „ d' 2 y W ~ n ~ a * 5 it > 0, 0 < x < L), and all the boundary conditions 2 /( 0 , t ) = 0, yiL, 0 = 0, , o, o) - /<*>, some conditions relative to the existence of its derivatives must also be satisfied. We shall now examine the function 96 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 43 defined by formula (1) to see if it is actually a solution of our problem. The Solution Established. It is possible to sum the series in formula (1); that is, to write the result in a closed, or finite, form. This will make it much easier to examine the function y(x, t). Since 2 sin cos ^ = sin [*£ (* - at)] + sin [*£ (* + at)], equation (1) can be written (3) y = \ 2 An sin [t ~ a< ^] + j2 A ” sin [x (a: + a<) ' The two series here are those obtained by substituting (x — at) and (x + at), respectively, for the variable x in the Fourier sine series for f(x). Since the sine series represents an odd periodic function, the last equation can be written (4) y = $F(x - at) + F(x + at)], where the function F(x') is defined for all real values of x ’ as the odd periodic extension of f(x') ; that is, F(x r ) = fix') if 0 ^ a ■/ £L, F(-x') = -F{x'), and Fix' + 2 L) = Fix') for all x'. The function jix) is continuous in the interval (0, L) and vanishes at the end points; hence Fix') is continuous for all x . According to our Fourier theorem, the two sine series in equation (3) converge to the functions in equation (4) whenever fix) has one-sided derivatives. The same function yix, t) is then repre- sented by each of the three formulas (1), (3), and (4); moreover, according to (4), yix, t) is a continuous function of x and t for all values of these variables. Sec. 43] SOLUTION OF BOUNDARY VALUE PROBLEMS 97 By differentiating equation (4) we can easily see that y(x, t ) satisfies differential equation (2) whenever the derivative F"(x') exists. When it is observed that F'(x') and F"(x') are even and odd functions, respectively, it can be seen that the second derivative exists for all x' provided fix) has a second derivative whenever 0 < x < L, and provided that the one-sided derivatives of f'(x) at the end points x = 0 and x = L exist and have the value zero. Under these rather severe conditions on f(x), then, our func- tion y(x f t) satisfies the equation of motion for all x and t } and it is also evident from equation (4) that dy/dt is continuous and vanishes when t = 0. The remaining boundary conditions are clearly satisfied, in view of either equation (1) or (4); hence y(Xj t) is established as a solution. If we permit /'(rc) and f"(x) to be only sectionally continuous, or if the one-sided second derivatives of fix) do not vanish at the points x = 0 and x = L, then at each instant t there will be a finite number of points x at which the second derivatives of y(x, t) fail to exist. Except at these points, differential equation (2) will still be satisfied. In this case we have a solution of our problem in a broader sense. In cither case an examination of the uniqueness of the solution found would be necessary to make the treatment of the problem complete. An Approximate Solution. Except for the nonhomogeneous boundary condition (5) y(x, 0) = J{x) 9 our boundary value problem is satisfied by the sum of any finite number of terms of the scries in equation (1), say (6) Vn — N mrx L cos mrat ~77 where N is some integer. In place of condition (5) this function satisfies the condition N , x , . UTX (7) ysix, 0) = An sm ~7“‘ i The function y„(r, t) has continuous derivatives of all orders. 98 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 44 The sum in condition (7) is that of the first N terms of the Fourier sine series for f(x). According to Theorem 3, Chap. V, that series converges uniformly to f(x) provided f(x) is con- tinuous by segments. Hence, by taking N sufficiently large, the sum can be made to approximate f{x) arbitrarily closely for all values of x in the interval 0 ^ x ^ L. The function y N {x , t) is therefore established as a solution of the “approximating problem,” obtained by replacing condition (5) of the original boundary problem by condition (7). Similar approximations can be made to the problems to be considered later on. But the remarkable feature in the present case is that the approximating function y N (x, t) does not deviate from the actual displacement y(x, t) by more than the maximum deviation of ynix, 0) from/(x). This is true because ynix, t ) can be written Vn = 5 An sin (x - at) J + ^ A « sin [77 (* + «oJ|> and each sum here consists of the first N terms of the sine series for the odd periodic extension of /($), except for substitutions of new variables. But the greatest deviation of the first sum from F(x — at), or of the second from F(x + at), is the same as the greatest deviation of ynix, 0) from jf(x). PROBLEMS 1. Show that the motion of every point of the string in the above problem is periodic in t with the period 2 L/a. 2 . The position of the string at any time t can be found by moving the curve y = iF( x ) to the right with the velocity a and an identical curve to the left at the same rate and adding the ordinates, in the interval 0 ^ x ^ L, of the two curves so obtained at the instant t. Show how this follows from formula (4). 3. Plot a few positions of the plucked string of Sec. 14, using the method of Prob. 2 above. 44. Variations of the Problem. If each point of the string is given an initial velocity in its position of equilibrium, the bound- ary value problem in the displacement y(x, t) is the following: Sec. 44] SOLUTION OF BOUNDARY VALUE PROBLEMS 99 y(0, t) = 0, y(L, t) = 0, y(x, 0) = 0, = g(x). As before, functions of the form y = X(x)T(t) which satisfy the differential equation and all the homogeneous boundary conditions can be found. Writing the series of these particular solutions, we have 00 2 , . nirx . mrat An sin -j- sm ~j ~ * i The final condition, that dy/dt = g(x) when t = 0, shows that the numbers mraA n /L should be the Fourier sine coefficients of g{x ) ; hence the solution of the problem becomes a) sin mrx r ~L~ dx'. By the method of the last section, dy/dt can be written here in terms of the odd periodic extension G(x r ) of the function g(x f ). This leads to the closed forms ( 2 ) y = 2 J 0 “ at ') + + a ^)] dt' of solution (1). The details of these derivations are left for the problems. Superposition of Solutions. If the string is given both an initial displacement and initial velocity, the last two boundary conditions become (3) y(x, 0) = f(x), = d( x )- All the other conditions of the linear boundary problem are homogeneous. They are satisfied by the solution of the problem of the preceding section and by solution (2) above, and therefore by the sum of those two functions, namely 1 1 (4) y = 2 l F ( x ~ at ) + F ( x + ®01 + 2a J ( dx '- 100 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 44 When t = 0, the first function of the sum becomes f(x) and the second vanishes; hence the first of conditions (3) is satisfied. Likewise it is seen that the second of those conditions is satisfied, and therefore equation (4) is the required solution. In general the solution of a linear problem containing more than one nonhomogeneous boundary condition can be written as the sum of solutions of problems each of which contains only one nonhomogeneous condition. Of course we cannot always find the solutions of the simpler problems which are to be super- posed in this way. Units . It is often possible and advantageous to select units so that some of the constants in our problem become unity. For example, if we write r for the product (at), the equation of motion of the string reduces to d^y = cPy dr 2 dx 2 Such changes sometimes help to bring out reductions in compu- tation, or general properties of the solution. Since the boundary problem of the last section, for example, does not involve the number a when the problem is written in terms of r and x , its solution must be a function only of x, L, and the product (at). This conclusion is possible without our knowing the formula for the solution. But a 2 is proportional to the tension in the string; hence if y i(x, t) and y 2 (x, t) are the displacements when the tension has the values Pi and P 2 , respec- tively, then (5) yi(x, h) = y 2 (x, t 2 ) if k\/Pi = £ 2 a/P^ That is, the same set of instantaneous positions is assumed by the string whether the tension is Pi or P 2 , but the tim es U a nd t 2 required to reach any one position are in the ratio \/P 2 /Pi. Nonhomogeneous Differential Equations. The substitution of a new unknown function sometimes reduces a linear differential equation which is not homogeneous to one which is homogeneous, so that our method of solution can be employed. To illustrate this, consider the problem of displacements in a stretched string upon which an external force acts proportional to the distance from one end. If the initial displacement and velocity are zero, the units for t and x can be so selected that the Sec. 44] SOLUTION OF BOUNDARY VALUE PROBLEMS 101 problem becomes ^ = H + Az (0 < * < 1, t > 0), 2/(0, t) = 0, 2/(1, t) = 0, 2 /(*, o) = o, = 0. In terms of the new function F, where 2/(x, 0 = Y(x , 0 + ^(x), and iA(x) is to be determined later, the differential equation becomes d 2 Y d 2 Y = -^2 + ^" 0*0 + Ax (0 < x < 1 , £ > 0 ). This will be homogeneous if (6) ^"(x) = —Ax (0 < x < 1). The first pair of boundary conditions on F are m o + m = o, f(i, t) + ^d) = o; hence these arc homogeneous if (7) *(0) = 0, *(1) = 0. In view of conditions (6) and (7), (8) i (z) = 4 (* - .x 3 ) (0 < x < 1), and with this choice of ^ the problem in F becomes a special case of the problem in the preceding section; for the initial conditions arc Y{x, 0) = -i(x), = 0. The solution of our problem in forced vibrations therefore can be written , (9) y = yjy{x) - - t) + x + /)], where ^(x') is the odd periodic extension of the function ^(x') defined by equation (8) in the interval (0, 1). 102 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 45 PROBLEMS* 1. Carry out the details in the derivation of formula (1). 2. Write out the steps used in deriving formulas (2) from (1). 3. Show that relation (5) fails to hold between the displacements of a given string under different tensions if the initial velocity is the same in both cases and not zero. What change in initial velocity must accom- pany an increase in tension to cause a more rapid vibration with the same amplitude? 4. A string is stretched between the points (0, 0) and (1,0). If it is initially at rest on the z-axis, find its displacements under a constant external force proportional to sin ttx at each point. Verify your solu- tion by showing that it satisfies the equation of motion and all boundary conditions’. Ans. y = A/(j 2 a 2 ) sin7rx(l — cos 7 rat). 5. A wire stretched between two fixed points of a horizontal line is released from rest while it lies on that line, its subsequent motion being due to the force of gravity and the tension in the wire. Set up and solve the boundary value problem for its displacements. Show that its solution can be written in the form (9), if a = 1 and \p(x) — (: x 2 — Lx)g / 2 in the interval (0, L), where g is the acceleration of gravity. 45. Temperatures in a Slab with Faces at Temperature Zero. Let a slab of homogeneous material bounded by the planes x = 0 and x = ir have an initial temperature u = f(x), varying only with the distance from the faces, and let its two faces be kept at temperature zero. The formula for the temperature u at every instant and at all points of the slab is to be determined. In this problem it is clear that the temperature is a function of the variables x and t only; hence at each interior point this function u(x } t) must satisfy the heat equation for one-dimen- sional flow, /i\ du , d 2 u ^ (1) -jfi = k (0 < x < V, t > 0). In addition, it must satisfy the boundary conditions (2) t*(+0, t) = 0, w(tt - 0, 0 = 0 (t> 0), (3) u(x, +0) = f(x) (0 < x < tt). The boundary value problem (l)-(3) is also the problem of temperatures in a right prism or cylinder whose length is ir * Only formal solutions of the boundary value problems here and in the sets of problems to follow are expected, unless it is expressly stated that the solution is to be completely established. Sec. 45] SOLUTION OF BOUNDARY VALUE PROBLEMS 103 (taken so for convenience in the computation), provided its lateral surface is insulated. Its ends x = 0 and x = t are held at temperature zero and its initial temperature is f(x ). To find particular solutions of equation (1) that satisfy conditions (2), we write u = X(x)T(t). When substituted in equation (1), this gives XT' = kX"T , or X" _ V ~x~W Since the function on the left can vary only with x and the one on the right only with t, they must both equal a constant a; that is, (4) X" - aX = 0, T' - akT = 0. Moreover, if the function XT' is to satisfy conditions (2), then (5) X(0) = 0, X(r) = 0, provided X(x) is a continuous function. The solution of the first of differential equations (4) that satisfies the first of conditions (5) is X = C\ sinh X's/a, and this can satisfy the second of conditions (5) only if a = — ft 2 (ft = 1, 2, • • - ). Then X = C 2 sin nx. The solution of the second of equations (4) is, then, T — CV n2 K Hence the solutions of equations (1) and (2) of the form u — XT are (6) b n (~ n2kt sin nx (?i = 1, 2, ■ • • ), where the constants b n are arbitrary. Clearly no sum of a finite number of functions (6) can satisfy the nonhomogencous condition (3) unless f(x) happens to be a linear combination of sines of multiples of x. But the infinite series of those functions, 0Q (7) u(x, /.) = ^ b n c~ n * kt sin nx , does in general reduce to f(x) in (0, 7 r) when t = 0, provided the coefficients b n are those of the Fourier sine series for /(a:); namely, .-5 r 7T JO f{x) sin nx dx (n = 1,2, b 104 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 45 More precisely, if f(x) is sectionally continuous and has one- sided derivatives at all points in (0, r), then QO u(x, o ) = 2} bn sin nx = i[f(x + 0) + fix — 0)] (0 < x < 7 r), and this represents f(x) at each interior point where f(x) is continuous. With those mild restrictions on f(x) } then, the solution of the problem is (8) u(x } t) = ^ e~ n * kt sin nx J fix') sin nx 1 dx r , provided this series converges to a function u(x, t) such that u(x, +0) = u(x , 0) when 0 < x < tt, ^(+0, t) = u( 0 , t) and u(t — 0, t) = u(tt, t) when t > 0, and provided the series can be differentiated termwise once with respect to t and twice with respect to x when t > 0 and 0 < x < tt. It will be shown in the next section that the series does satisfy those conditions. PROBLEMS 1. Solve the above problem if the faces of the slab are the planes x — 0 and x — L. a / ,v 2 ( nVki\ . mrx C L . mrx' Ans. u(x, t) = Tj Za exP ^ jJ~) sm irj 0 ^ sin ~TT dx> ' 2. Find the formula for the temperatures in a slab of width L which is initially at the uniform temperature u 0 , if its faces are kept at tem- perature zero. Ans ‘ u(x, 0-^r2 2^=1 ex P (2 n - 1 )Vkt IJ sin (2 n — 1)7 rx L 3. The initial temperature in a bar with ends x = 0 and x = tt is u = sin x. If the lateral surface is insulated and the ends are held at zero, find the temperature u(x, t) . V erify your result completely. How does the temperature distribution vary with time? Ans. u = e~ kt sin x. 4. Write the solution of Prob. 1 if f(x) = A when 0 < x < L /2 fix) = 0 when L/2 < x < L. Ans. _ M sin 2 (mr/4) / nVkt\ . mrx tt n exp L 2 J sm ~L~ Sec. 46] SOLUTION OF BOUNDARY VALUE PROBLEMS 105 5. Two slabs of iron, each 20 cm. thick, one at temperature 100°C. and the other at temperature 0°C. throughout, are placed face to face in perfect contact, and their outer faces are kept at 0°C. (compare Prob. 4). Given that 7c = 0.15 c.g.s. (centimeter-gram-second) unit, find to the nearest degree the temperature 10 min. after contact was made, at a point on their common face and at points 10 cm. from it. Am. 37°C.;33°C.; 19°C. 6. If the slabs in Prob. 5 are made of concrete with k = 0.005 c.g.s. unit, how long after contact will it take the points to reach the same temperatures found in the iron slabs after 10 min.? Am. 5 hr. 46. The Above Solution Established. Uniqueness. It is not difficult to show that the series found in Sec. 45, namely 00 1 (1) ^ b n e~~ n ' kt sin nx , represents a function u(x, t) which satisfies all the conditions of the boundary value problem, provided the initial temperature function /(x) is sectionally continuous in the interval (0, x) and has one-sided derivatives at all interior points of that interval. For the sake of convenience, we define the value of J{x) to be %[f(x + 0) +f(x — 0)] at each point x where the function is discontinuous. Since |/(x)| is bounded, |6 n | =-\ \ f(x) sin nx dx ^ - I |/(x)| dx < M, 7T | Jo T Jo where M is a fixed number independent of n. Consequently, for each to > 0, | b n e~ n2kt sin nx\ < Me- n * kto when t ^ t Q . The series of the constant terms c- nHto converges; hence, accord- ing to the Weierstrass M-test, series (1) converges uniformly with respect to x and t when t ^ to, 0 ^ x ^ x. Also, the terms of series (1) are continuous with respect to x and t, so that the function u(x, t) represented by the series is continuous for those values of x and t ; consequently, whenever t > 0, u(+ 0, 0 = u( 0, t) = 0, u(x — 0, l) = u(ir , t) = 0. 106 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 46 The terms of the series obtained by differentiating (1) with respect to t satisfy the inequality | —kb n n 2 e~- n2kt sin nx\ < kMn 2 e~ n2kt ° when t 5; to. Since the series whose terms are n 2 e ~ nm ° also converges, accord- ing to the ratio test, that differentiated series is uniformly convergent for all t ^ to- Hence series (1) can be differentiated termwise; that is, oo = 2 ^ ( b ^~ nakt sin nx ) <t > o). "i In just the same way it follows that the series can be differenti- ated twice with respect to x whenever t > 0, and since each term of series (1) satisfies the heat equation, the function u(x, t ) must do so whenever t > 0 (Theorem 2, Chap. I). It only remains to show that u(x , t) satisfies the initial condition (2) u{x j +0) = f{x) (0 < x < 7 r). This can be shown with the aid of a test, essentially due to Abel, for the uniform convergence of a series. At this time let us show how the test applies to the present problem, and defer the general statement of the test and its proof to the following chapter (see Theorem 1, Chap. VII). oo For each fixed x (0 < x < ir), the series b n sin nx con- verges to f(x). According to Abel’s test, the new series formed by multiplying the terms of a convergent series by the cor- responding members of a bounded sequence of functions of t, such as e~ n * k \ whose functions never increase in value with n, converges uniformly with respect to t. Series (1) therefore con- verges uniformly with respect to t when 0 ^ t ^ h, 0 < x < w, for every positive t\. The terms of series (1) are continuous functions of t ; hence the function u(x, t) represented by that series is continuous with respect to t when t ^ 0 and 0 < x < r. Therefore u(x, +0) = u(x, 0), and condition (2) is satisfied because u(x, 0) = f(x) (0 < x <i r). The function u{x, t ) is now completely established as a solution of the boundary value problem (l)-(3), Sec. 45. Sec. 46] SOLUTION OF BOUNDARY VALUE PROBLEMS 107 It is necessary to add to the statement of the problem some further restrictions as to the properties of continuity of the function sought, before we can prove that we have the only solution possible. We illustrate this by stating one complete form of the problem. For the sake of simplicity, we shall impose rather severe conditions of regularity on the functions involved. A Complete Statement of the Problem . Let the function u(x, t) be required to satisfy the heat equation and boundary conditions as given by equations (1) to (3), Sec. 45, in which the function /( x) is now supposed continuous in the interval 0 ^ x g 7 r. We also assume that /( 0) = f(r) = 0, and that f'(x) is sectionally continuous in the interval (0, it) . In addition let it be required that u(x , t) be continuous with respect to the two variables x , t together when 0 ^ x ^ t, t ^ 0, and that the derivative du/dt be continuous in the same manner whenever t > 0. We can show that there is just one possible solution of this problem, and that solution is the function represented by series (1) * It was shown above that that function satisfies the heat equa- tion and boundary conditions; also, that the series for du/dt converges uniformly with respect to x and t together when 0 g x g tt, t t o (/<o > 0). Since the terms of the derived series arc continuous functions of x and t together, it follows that du/dt is continuous with respect to both variables together whenever t > 0, The continuity of the function when 0 ^ x ^ tt and t ^ 0 follows again from our form of Abel’s test. For the conditions 00 on /(;r) ensure the uniform convergence of the series ^ 6 n sin nx. In this case the introduction of the factors e~ n * kt into the terms of that series produces a series which is uniformly convergent with respect to x and t together, when 0 ^ x ^ t, 0 g t S h, for every positive h. Hence scries (1) has this uniform con- vergence, and the continuity follows as before. The function defined by series (1) therefore satisfies all the conditions of the problem. Of course, the derivative d 2 u/dx 2 is continuous in the same sense as du/dt , since these two deriva- tives differ only by the factor k. * Concerning the continuity of a series with respect to more than one variable, see the remarks preceding Theorem 1, Chap. VII. 108 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 47 It is not difficult to show that two distinct functions, satisfying all the requirements made upon u( x, t) in the above statement of the problem, cannot exist. The complete statement and proof of this uniqueness theorem will be given later (Theorem 2 Chap. VII). If we accept this statement for the present, the only possible solution of our problem has been found. 47. Variations of the Problem of Temperatures in a Slab. With only slight modifications in the method, the temperature distribution can be found for the slab of Sec. 45 when the faces are subject to certain other conditions, or when the heat equation is modified. a. One Face at Temperature A. To find the temperature u(x, t) in a slab with initial temperature f(x) when the face x = 0 is held at zero and the face x = % at constant temperature A, a simple transformation can be used to obtain the result from that of Sec. 45. Here u(x, t ) must be a solution of the boundary value problem du _ , d 2 U Ht ~ “dx* m(+0, t) = 0, u(tt — 0, t ) U(x, +0) = f(x). (0 < x < it, t > 0), = A, It follows that the function (1) v(x, t) = u(x, t ) — — x, 7T must satisfy the conditions dv _ d 2 v dt ~ & dx 2 (0 < z < r, t > 0), *K+o» t) = o, v (x — 0, t) = 0, v(x, +0) = fix) --X. 7T This is the boundary value problem of Sec. 45 with fix) replaced by fix) — Ax/t, so that its solution is vix, t) sin nx Ax r 7T sin nx r dx r . Substituting this for v(x } t) in equation (1) and carrying out part of the integration, we obtain the following solution of the Sec. 47 ] SOLUTION OF BOUNDARY VALUE PROBLEMS 109 problem: u(x, t) = — x 7 r + ^ 2 e ~ nik ‘ s * n nx j^( — 1 )’‘ ~ + J* f ( x 0 sin nx> cfo'l. 6. Insulated Faces. Find the temperature u(x, t) in a slab with initial temperature fix) if the faces x = 0 and x = ir are thermally insulated. Since the flux of heat through those faces is proportional to the values of du/dx there, the boundary value problem can be written ( 2 ) du _ , d 2 u ~dt ~ W 2 (0 < x < T, t > 0), (3) (4) drt(+0, t) _ dx 0 , du(£ — 0, f) _ 0 dx u(x, +0) = f(x) it > 0), (0 < x < it). Setting u = X{x)T(t), it is found that the functions a n e~ ntkt cos nx (n = 0, 1, 2, ) satisfy the homogeneous conditions (2) and (3). The infinite series of those functions satisfies condition (4) as well, provided the coefficients a n arc those in the Fourier cosine series correspond- ing to fix). So if fix) satisfies the conditions of our Fourier theorem, the solution of the problem is (5) uix, t) c. One Face Insulated. If the face x = 0 is held at tempera- ture zero and the face x = x is insulated, the problem can be reduced to one in which both faces are held at zero. Let the slab be extended to x = 2ir with the face x = 2ir held at temperature zero, and let the initial temperature of the new slab be symmetric with respect to the plane x = x. Then, when ir < x < 2ir, the. initial temperature is f(2ir — x), where fix) is the initial temperature of the original slab. In the 110 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 47 physical problem the symmetry indicates clearly that no heat will flow through the plane x = tt. When the solution is found, it can be verified that du/ dx = 0 when x = x. According to Prob. 1, Sec. 45, the temperature in the extended slab is ,n . nx' , , x') sm dx' . By substituting a new variable of integration in the second integral, this can be reduced to 2 C* i(x, t) = - ^ e -m\kt s i n mnX /(s') sin m n x' dx', n = 1 where m n = (2n — l)/2. When 0 ^ x ^ t, this is the solution of problem c. d. The Radiating Wire . Suppose the diameter of a wire or bar is small enough so that the variation of temperature over every cross section can be neglected. If the lateral surface is exposed to surroundings at temperature zero and loses or gains heat according to Newton’s law, the heat equation takes the form ( 6 ) du _ , d*u ~dt ” dx* hu. where x is the distance along the wire and h is a positive constant. Newton’s law of surface heat transfer is an approximate law of radiation and convection according to which the flux of heat through the surface of a solid is proportional to the differ- ence between the temperature of the surface and that of the surroundings. It is generally valid only for small temperature differences; but it has the advantage over the more exact laws of being a linear relation. That the heat equation does take the form (6) when such surface heat transfer is present can be seen from the derivation of the heat equation (Sec. 9). When the ends x = 0, x = 7r, of the radiating wire are kept at temperature zero and the initial temperature is f(x), the temperature function can be found by the method of Sec. 45. Sec. 47] SOLUTION OF BOUNDARY VALUE PROBLEMS 111 The result is (7) u(x , t) = e~ ht Ui(x, t), where Ui(x, t) is the function u in equation (S), Sec. 45. When the ends are insulated, the result is (8) u(x, t) = e~ ht u 2 (x, t), where u 2 (x, t) is the function u in equation (5) above. PROBLEMS 1. Derive the solution of the problem in Sec. 476 above when the faces are x — 0 and x = L. A ns. u = i f(x') dx' + 12' nWJct L‘i COS IT jc Kx ' ] cos ■ UTX dx', 2. Show that the result of Prob. 1 can be completely established as a solution of the boundary value problem by the method of Sec. 46. 3. Solve the problem in Sec. 47c above for a slab of width L with the face x — L insulated. It will be instructive to carry out the solution directly by obtaining particular solutions u = XT, without using the method of extension, noting the orthogonal functions generated by the differential equation in A r and its boundary conditions (compare Sec. 25). A ns. u -12' w = 1 P >l n X Jo f(x') sin m n x' dx', where m n — (n — l/2)ir/L. 4. Derive formula (7). 5. Derive formula (8). 6. Use the substitution v = uc ht to simplify equation (6) and, by writing the boundary value problem in terms of v{x , t), obtain formulas (7) and (S) from known results. 7. For a wire in which heat is being generated at a constant rate, while the lateral surface is insulated, the heat equation takes the form du _ dhi dt ~ k Ox * + B, where B is a positive constant. If the ends x = 0 and x = tt are kept at temperature zero and the initial temperature is f(x), set up the boundary value problem for u(x, t) and solve it. Note the result when 112 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 48 f(x) = Bx(i r — x)/(2 k). Suggestion : Apply the method used in Sec. 44 to reduce the nonhomogeneous differential equation to a homogeneous one. Ans . u = ^ (tt — x) + 12 e -n*kt gi n nx c* r Bx r , , Jo L» (I ■)+/(*') sin nx' dx f . is insulated, instead of being 8. Solve Prob. 7 when the end x = i r kept at temperature zero. 9. A wire radiates heat into surroundings at temperature zero. The ends x = 0 and x = t are kept at temperatures zero and A , respectively, and the initial temperature is zero. Set up and solve the boundary value problem for the temperature u(x, t). Suggestion: Substitute v — u + yp{x) : then determine ^ so that hf/" — hf/ = 0 and \[/(fl) = 0, ^(tt) = -A. Ans. u = A sinh xVh/l + 2Ak e _ ht sinh iry/h/k ir n( — l) n h + kn 2 e -n*kt s i n nx. 10. The face x = 0 of a slab is kept at temperature zero and heat is supplied or extracted at a constant rate at the face x = tt, so that du/dx = A when x - tt. If the initial temperature is zero, derive the formula u “ Ax + 2 (i -H)2 e (n i,! ‘ sin _ i)*. for the temperatures in the slab, where the unit of time has been so chosen that h = 1. 48. Temperatures in a Sphere. Let the initial temperature in a homogeneous solid sphere of radius c be a function /(r) of the distance from the center, and let the surface r = c be kept at temperature zero. The temperature is then a function u{r , t), of r and t only, and the heat equation in spherical coordinates becomes du _ h d 2 (ru) dt r dr 2 The boundary conditions are u(c — 0, t) = 0 utr, +0) = f(r) (t > 0), (0 < r < c). Sec. 48] SOLUTION OF BOUNDARY VALUE PROBLEMS 113 If we set v(rj t) = ru(r , t ), the boundary value problem here can be written dv 7 d 2 v di = k d?’ t) = 0, v(c — 0, t) — 0, v (r, +0) = rf(r), where the condition v(+0, t) = 0 is included because u(r, t ) must be bounded at r = 0. Except for the presence of r instead of x and r/(r) instead of fix), this problem is that of the tempera- tures in a slab of width c. Hence the temperature formula for the sphere can be written at once (Prob. 1, Sec. 45). It is ( 1 ) n 2 r 2 kt sin mrr J r'f(r') sin dr'. PROBLEMS 1. Find the temperatures in a sphere if the initial temperature is zero throughout and the surface r — c. is kept at constant temperature A. Ans. u(r, t) — A + 2 Ac TV (-PV n nhr 2 kt sin ■ 2. Prove that the sum of the temperature function found in Prob. 1 and the function given by formula (1) above represents the temperature in a sphere whose initial temperature is f(r) and whose surface is kept at temperature A. 3. An iron sphere with radius 20 cm., initially at the temperature 100°C. throughout, is cooled by keeping its surface at 0°C. Find to the nearest degree the temperature at its center 10 min. after the cooling begins, taking k = 0.15 e.g.s. unit. Ans. 22°C. 4. Solve Prob. 3, assuming that the sphere is made of concrete with k = 0.005 e.g.s. unit. Ans. 100°C. 5. The surfaces r = b and r — c of a solid in the form of a hollow sphere are kept at temperature zero. The initial temperature of the solid is f(r) {b < r < c). Derive the following formula for the tem- peratures u(r , t) in the solid: nVkt (c - by . mr(r — b) sin — — i c — b mr(r — b) r/(r) sin - — dr. where 114 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 49 6. Show that when the surface of a sphere is insulated, the solution of the temperature problem no longer involves the expansion of rf(r) in a Fourier series, but an expansion in a series of the functions sin a n r, where a n are the roots of the mixed equation tan ac — ac . Show why these f un ctions form an orthogonal set in the interval 0 < r < c (Sec. 25). 49. Steady Temperatures in a Rectangular Plate. Let u{x, y) be the steady temperature at points in a plate with insulated faces, the edges of the plate being the lines (or planes) x = 0, x = a, y = 0, and y — b. Let three of the edges be kept at temperature zero and the fourth at a fixed temperature distribu- tion. Then u(x , y) is the solution of the following problem: (1) S + 0 =O (0<*<a,0 <y<b), (2) w(+0, y) = 0, u(a - 0, y) = 0, (0 < y < b ), (3) u(x, b — 0) = 0, u(x, +0) = f{x), (0 < x < a). Since special case (1) of the heat equation is also a case of Laplace’s equation, the function u(x, y) is also the potential in the rectangular region when the potential on the edges is prescribed by conditions (2) and (3). The region also may be considered as an infinitely long rectangular prism, or the right section of any prism in which the potential or steady temperature depends only upon x and y. Settings = X(x)Y(y), the functions sin ^ sinh ^ (y - C) J (n = 1, 2, • • • ) are found as solutions of (1) which satisfy conditions (2), for every constant C. If C = b, they also satisfy the first of con- ditions (3), and the series u = A n sin sinh (y — b) j satisfies the nonhomogeneous condition in (3) provided fix) = — 2 A n sinh sin (0 < x < a). i According to the Fourier sine series, this is true if the coeffi- cients A n are determined so that Sec. 49] SOLUTION OF BOUNDARY VALUE PROBLEMS 115 — A n sinh mrb a . mcx Sill — a dx . So the formal solution of the problem can be written “<*■*> - ; % - 1 * ” j>> * ¥ *■ Our result can be completely established as a solution of the problem (l)-(3) by the method used in See. 46. But in this case let us defer that part of the discussion, along with a com- plete statement of the problem which ensures just one solution^ until a later time when the necessary tests have been derived (Sec. 59). PROBLEMS 1. Find the solution of the above problem if u(x, y) is zero on all edges except x = a, and u(a, y) = g(y). Am. sinh (ft7 rx/b) . mr ; y sinh (mra/b) SlU b g{y') sin 2. When the temperature distributions on all four edges are given, show how the formula for the steady temperatures in the plate can be written by combining results already found. 3. What is the steady temperature at the center of a square plate with insulated faces, (a) if three edges arc kept at ()°C. and the fourth at 100°C.; (6) if two adjacent edges arc kept at 0°C. and the others at 100°C.? Suggestion: Superpose the solutions of like problems here to obtain the obvious case in which all four edges are kept at 100°C. Am. (a) 25°C.; ( b ) 50°C. 4. A square plate has its faces and its edge y — 0 insulated. Its edges x - 0 and x — x are kept at temperature zero, and its edge y = x at temperature fix). Derive the formula for its steady temperature. CO Am. u(x, y) - ^ ■ ° S J 1 tl]f sin nx I f K x') sin ?ix' dx'. x cosh nx J () 5. Derive the formula for the electric potential V{x, y) in the space 0 ^ x ^ L, y ^ 0, if the planes x = 0 and x — L are kept at zero potential and the points of the plane y — 0 at the potential f(x), if V(x , y) is to be bounded as y becomes infinite. Ans. V(x, y) = j -r2- niry " L sin r . mrx' fix') sin dx', 116 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 50 6. Find the electric potential in Prob. 5 if the planes x = 0 and x = L, instead of being kept at potential zero, are insulated, so that the electric force normal to those planes is zero; that is, dV/dx = 0. Also state this problem as a temperature problem. Am. Vix^y) = ^ J fix') dx' 2 nTX f L ,\ nwx' , + ^ e L cos J /(* ) cos — jj- ax . 7. Solve Prob. 5 if the electric potential is zero on the plane x = 0 and the electric force normal to the plane x = L is zero. 8. Find the steady temperatures in a semi-infinite strip whose faces are insulated and whose edges x = 0 and x = ir are kept at temperature zero, if the base y = 0 is kept at temperature 1 (Prob. 5). Ans. u(x, y) = - ^e-*' sin x + | e~ Zy sin 3x + jr e~ hy sin ox + * ■ ■ 9. In the power series expansion of [log (1 + z) — log (1 — z)] (|s| < 1), set z = re l(p and equate imaginary parts to find the sum of the series S = r sin cp + \r z sin 3<p + ir 5 sin 5<p + • • • ; also note that log [p( cos d + i sin 6)] - log ( pe 1 6 ) = log p + i&, and therefore show that S = 1 , 2r sin <p 2 arctan ^ * Thus show that the answer to Prob. 8 can be written in closed form as follows: u{x i y) — ~ arctan sin x sinh y Verify the answer in this form. Also trace some of the isotherms, u(x , y) ~ a constant. 50. Displacements in a Membrane. Fourier Series in Two Variables. Let z represent the transverse displacement at each point (x ) y) at time t in a membrane stretched across a rigid rectangular frame in the zy-plane. Let the boundaries of the rectangle be the lines x = 0, x = xo, y = 0, and y = yo. If the initial displacement z is a given function f(x, y) } and the mem- brane is released from rest after that displacement is made, the Sec. 50] SOLUTION OF BOUNDARY VALUE PROBLEMS 117 boundary value problem in z(x, y, t ) is the following: / 1 \ d 2 z _ „ / d 2 z dh\ ^ dt 2 a \dx 2 + dy V’ zfro, y, t ) = 0; z{x, 2 / 0 , <) = 0; = 0, z(x, y, 0) = f(x, y). s(0, y, t ) = 0, z(x, 0, t) = 0, dzQc, y, Q) dt In order that the product z = X(x)Y{y)T(t) be a solution of equation (1), its factors must satisfy the equation TIL - TL _l XL a 2 T X + 7* All three terms in this equation must be constant, since they are functions of x ) y , and t separately. Write then T" = - a\a 2 + /3 2 )7 7 . The solutions of these three equations, for which z = XYT satis- fies all the homogeneous boundary conditions, are X = sin ax, Y = sin @y , T = cos (aV a 2 + j B 2 t), where a = mw/x o, and (3 = mr/ijo (m, n = 1, 2, • • • ). So the function ( 2 ) / . fm 2 . n 2 \ • . n7r?/ cos I 7r at . — *• q — r ) sin sin — - V \ y\) ^ yo satisfies equation (1) and all the boundary conditions, formally, provided the coefficients A mn can be determined so that z = f(x, y) when t = 0; that is, provided 22 * rn = 1 // = 1 . mirx . mry sin sin xq y o (3) f(x, y) m = 1 n = l (0 ^ x ^ Xo, 0 ^ y ^ yo). By formally grouping the terms of the series, equation (3) can be written < 4 ) rn*l 'n»l ' 118 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 50 For each fixed y between zero and y 0 this series is the Fourier sine series of the function /(x, y) of the variable x (0 g j provided the coefficients of sin ( nnrx/xo ) are those of the Fourier sine series. So equation (4) is true in general if (5) sm ft = i nr y 2/o -5 J. . m ttx , . sm dx r . x 0 Again, using the formula for the Fourier sine coefficients of the functions F m (y) f where r% pxo / Fm(v) = - J o fix', y ) sin dx’ (0 ^ y ^ y 0 ), expansion (5) is valid if A mn = - C° F m (y’) sin VElI dy’ . 2/o Jo 2/o The series in equation (3) is then a Fourier sine series in two variables for /(x, y ) provided its coefficients have the values ( 6 ) A 4 p° . r*° . . . mrx . nry , A mn = — I dy f(x,y) sm sm — ^ dx. Zo2/o Jo Jo x 0 2/0 The formal solution of the membrane problem is then given by equation (2) with the coefficients defined by equation (6). According to equation (2), the displacement z is not in general periodic in t, since the numbers [(m 2 /xg) + (n 2 /?/§)]* do not change by multiples of any fixed number as the integers m and n change. Consequently the vibrating membrane, in contrast to the vibrating string, will not generally give a musical note. It can be made to do so, however, by giving it the proper initial displacement. If, for instance, for any fixed integers M and N, z(x, y, 0) = A sin Mrx x 0 . Nry sm — 2/o then the displacement (2) is given by a single term : / \ / m ata z \ x , V, 0 = A cos ( rat H 2 * ) sin ; \ \ x o 2/o / In this case z is periodic in t with the period Mrx Xo sin Nry ~y ~ (2/a)(Myxl + Ny v i)-y Sec. 50] SOLUTION OF BOUNDARY VALUE PROBLEMS 119 PROBLEMS 1. Solve the above problem if the membrane starts from the position of equilibrium, 3 = 0, with an initial velocity at each point; that is, - "<*. »>• 2. The four edges of a plate tv units square are kept at temperature zero and the faces are insulated. If the initial temperature is/(x, y ), derive the following formula for the temperature u{x, y, t) : SO 00 u — X X A mn exp [ — A:(m 2 + n 2 )t] sin mx sin ny, m = 1 n = 1 where 4 C T f 6 7 " A mn = — , I sin ny dy I f(x, y) sin mxdx. T Jo Jo 3. When fix, y) = Ax, show that the solution of Prob. 2 is u = 1/1(2, t)u 2 (y, t), where Ml .+1 — C (-1)" n nZkt sin nXj >-n*kt s i n n y ' Show that ?/i and u 2 represent temperatures in cases of one-dimensional (low of heat with initial temperatures Ax and 1, respectively. 4. Solve Prob. 2 if, instead of being kept at temperature zero, the edges arc insulated. Note the result when f(x, y) = 1. 5. If the faces x = 0, x = tt, y = 0, y = tt, z = 0, 2 = tt of a cube are kept at temperature zero and the initial temperature is given at each point u(x, y, z, 0) = fix, y, z), show that the temperature function is u(x, //, 2, 0 = X XX Amnv Hitl mX Sln Uy Sin PZe * (w2+, * S+pl) h m “ 1 7i—l /> = 1 where I I I /(•£, y, z) sin ms sin ny sin ps ds dp dz. 7r ’ Jo Jo Jo 6. When f{x, ?/, 2) = 1, show that the solution of Prob. 5 reduces to u = u 2 (x , 0^2(3/, £)m 2 (z, 0, where the function u 2 is defined in Prob. 3. 120 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 51 51. Temperatures in an Infinite Bar. Application of Fourier Integrals. Let the length of a homogeneous cylinder or prism be so great that it can be considered as extending the entire length of the z-axis. If the lateral surface is insulated and the initial temperature is given as a function f(x) of position along the bar, the temperature u(x , t) is the solution of the following boundary value problem: (1) Tt =h M (-*<*< (2) u(x, +0) = f(x) (—00 < X < oo). Particular solutions of equation (1) which are bounded for all x and t (t ^ 0) are found by the usual method to be (3) e~ a2kt cos [a(x + C)], where a and C are arbitrary constants. Any series of these functions, formed in the usual manner by taking a as multiples of a fixed number, would clearly reduce to a 'periodic function of x when t = 0. But f(x) is not assumed periodic, and con- dition (2) is to be satisfied for all values of x; hence it is natural to try to use the Fourier integral here in place of the Fourier series. Since function (3) is a solution of equation (1), so is the function - f{x r )e~ a * kt cos [a(x r — x)] 7 7 r where the parameters x' and a are independent of x and t. The integral of this with respect to these parameters, 1 f* o° f* oo (4) u(x , t) = - I da I f(x')e- a * kt cos [a(x' — z)] dx f , * J0 J- oo is then a solution of equation (1) provided this integral can be differentiated twice with respect to x and once with respect to t inside the integral signs. When t = 0, the right-hand member of equation (4) becomes the Fourier integral corresponding to fix). Hence if f(x ) satisfies the conditions of the Fourier integral theorem, and if the function u{x , t) defined by equation (4) is such that u(x , 0) == u(x, +0), then <x, +0) =*[/(* + 0)+ /(a -0)], and this is condition (2) at each point where f(x) is continuous. Sec. 51] SOLUTION OF BOUNDARY VALUE PROBLEMS 121 The solution of the problem is therefore given, at least formally, by equation (4) . By inverting the order of integration and using the integration formula J 0 cos «(*' - x) da = exp [ - (x ~ f >* ] <t > 0), equation (4) becomes (6) <x ' 0 ” 5^7S( “ xp [“ t]* 1 This can be still further reduced by using a new variable of integration £, where x' — x * - this gives (6) u( x, t) = -L V? r When fix) is bounded for all values of x and integrable in every finite interval, it can be shown that the function defined by equation (5) satisfies equation (1) and condition (2).* Under those conditions, then, the required solution is given either by equation (5) or by equation (6) . a PROBLEMS J* fix + 2 Vkt £)<r-* s d£. 1. Derive the temperature function for the above bar if fix) is periodic with period 2tt. 1 C* Ans. u(x, t) — ^ I fix') dx' 00 +£2* 1 2. If fix) = 0 when x < 0, and fix) = 1 when x > 0, show that the temperature formula for the infinite bar becomes, for t > 0, ,% 1C fix') cos [nix' — x)] dx'. 2 y/kt 1 , _1 r a: x 1 2 3 _ 2 s/ir 2 Vkt 3(2 Vkty 5 • 2!(2 \/kty * For a proof, see p. 31 of Ref. 1 at the end of this chapter. 122 FOURIER SERIES AND BOUNDARY PROBLEMS (Sec. 52 3. One very thick layer of rock at 100°C. is placed upon another of the same material at 0°C. If k = 0.01 c.g.s. unit, find the temperatures to the nearest degree at points 60 cm. on each side of the plane of con- tact, 100 hr. after contact is made. Am . 76°C.; 24°C. 52. Temperatures in a Semi-infinite Bar. If the bar of the foregoing section extends only along the positive half of the z-axis and the end x = 0 is kept at temperature zero, the bound- ary value problem in the temperature function u(x, t ) becomes the following: (1) du , d 2 u dt~ dx 2 (x > 0 , t > 0), (2) «(+o, t) = 0 (t > 0), (3) u(x, +0) = f(x) (x > 0). The solution can be formed from the function e ~ a * kt sin ax , which satisfies conditions (1) and (2). Multiplying this by (2/7 r)f(x l ) sin ax' and integrating with respect to the parameters a and x', which are independent of x and t, the function 2 r 00 r 00 (4) u(x, t) = - I e~ aikt sin ax da I f(x') sin ax' dx' n Jo Jo is found. When t = 0, the integral on the right reduces to the Fourier sine integral of f(x), which represents f(x) when 0 < x < 00. If we write 2 sin ax sin ax' — cos [ot(x' — re)] — cos [a(x' + a?)], the integration formula used in the foregoing section can be applied to reduce formula (4) to the form <5) 4 kt jy- -U (x - A 1 kt when t > 0. This can be written r (a? + s') 2 ]) j , (6) v.ix, () = --L f e~ p f(x + 2 f) V* J -X da 2"\/kt -J e~Pf(—x + 2 \, / Ft £) da iVki Sec. 53] SOLUTION OF BOUNDARY VALUE PROBLEMS 123 These results can also be found directly from those of the last section by making f(x) there an odd function. Under the conditions stated in the preceding section, function (5) then satisfies all the conditions of the problem. PROBLEMS 1. When f(x) = 1, prove that the temperature in the semi-infinite bar, or in a semi-infinite solid x S 0, with its boundary x = 0 at zero, is X o r 2 kt u(x, t) = —pz I <?-i 2 d£ Jo 2 r x x 3 x 5 ~ Vv 1.2 Vkt ~ 3(2 Vfe) 3 + 5 - 2!(2 s/ktf ~ 2. When the end x = 0 is kept at temperature A and the initial temperature of the bar is zero, show that X 3. Show that when a semi-infinite solid initially at a uniform temper- ature throughout is cooled or heated by keeping its plane boundary at a constant temperature, the times required for any two points to reach the same temperature are proportional to the squares of their distances from the boundary plane. 4. Show that the function x ~ — U\ — ^ e 4kt satisfies all conditions of the boundary value problem consisting of equations (1) to (3) when f(x) — 0. Hence this function can be multi- plied by any constant and added to the solution obtained above, to obtain as many solutions of that problem as we please. But also show that U[ is not bourn led at x = t = 0; this can be seen by letting x vanish while x 2 = t. 63. Further Applications of the Series and Integrals. Many other boundary value problems, arising frequently as problems in engineering or geology, can be solved by the methods of this chapter. A few will be slated at this point. The derivation of the results given here can be left as problems for the student. a. Electric Potential between Parallel Planes. The plane y = 0 is kept at electric potential V = 0, and the plane y = b at the 124 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 53 potential V =f(x). Assuming that the space between those planes is free of charges, the potential V(x , y) in the space is to be determined. It can be shown that Problems of this type are idealizations of problems arising in the design of vacuum tubes. They are also problems in steady temperatures, or steady diffusion, in solids ; hence their applica- tions are quite broad. The following problem is another of the same type. b. Potential in a Quadrant. A medium free of electric charges has the planes x = 0 and y = 0 as its boundaries. If those planes are kept at electric potential 7 = 0 and 7 = f(x), respectively, and if the potential V(x, y) is bounded for all x and y (x ^ 0, y ^ 0), the formula for V(x, y) is to be found. The result can be written, when y > 0, as (2) 7 -fJM i 2 + (x' - x) 1 y 2 + O' + x) 2 dx r . When f{x) = 1, this formula becomes (3) V = — arctan — • tv y In this case the equipotential surfaces are the planes x = cy , where the constant c has the value tan (wV/2). c . Angular Displacements in a Shaft. Let 6(x, t) be the angular displacement or twist in a shaft of circular cross section with its axis along the z-axis. If the ends x = 0 and x = L of the shaft are free, the displacements 0(x, t) due to an initial twist 9 = f(x) must satisfy the boundary value problem cPd dt 2 M(0, 0 = 0 dx 0) _ n dt a 2 d 2 e dx 2 ’ dJiL, t) = Q dx 9(x, 0) = /(*), where a is a constant. Sec. 53] SOLUTION OF BOUNDARY VALUE PROBLEMS 125 The solution of this problem can be written ,.v - 1 . nirx rnr at (4) 6 = 2^0 + ^ a n cos -j- cos —j-* i where a n (n = 0, 1, 2, • • • ) are the coefficients in the Fourier cosine series for f(x) in the interval (0, L). d. The Simply Supported Beam. The differential equation for the transverse displacements y(x , t) in a homogeneous beam or bar was given in Sec. 12. At an end which is simply supported or hinged, so that both the displacement and the bending moment are zero there, it can be shown that d*y/dx 2 must vanish as well as y . The displacements are to be found in a beam of length L with both ends simply supported, when the initial displacement is y = /($), and the initial velocity is zero. The result is /kn 2 . nirx nVct C L /N . nirx r , (5) 2 / = ^ > i sm "TT cos ~U~ ^ x ' sm ~TT dx 7 where c is the constant appearing in the differential equation. PROBLEMS 1. Write the boundary value problem in Sec. 53a above, and derive solution (1). 2. Write the boundary value problem in Sec. 536, and derive solution ( 2 ). 3 . Obtain solution (3) from (2), and show that the function (3) satisfies all the conditions of the boundary value problem when /(a) = 1. 4. Derive the solution (4) of Sec. 53c. Also show how this formula can be written in closed form in terms of the even periodic extension of the function f(x). 6. Set up the boundary value problem in Sec. 53d, and derive solution (5). 6. Derive the formula for the temperatures u(x, t) in the semi- infinite solid x ^ 0, if the initial temperature is f{x) and the boundary x = 0 is kept insulated. 7 . Find the formula for the displacements y(x, t) in a string stretched between the points (0, 0) and (t, 0), if the string starts from rest in the position y = f(x) and is subject to air resistance proportional to the velocity at each point. Let the unit of time be selected so that the equation of motion becomes &V _ &V _ o h d]L dt 2 “ dx* m dt ’ 126 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 53 where h is a positive constant. A ns. y = e~ ht b n (cos Kj + sin Kj^j si sin nx , where K n = Vn 2 - h 2 , and b n are the coefficients in the Fourier sine series for f(x) in the interval (Q, t). 8 . Let V(r, p) be the electric potential in the space inside the cylin- drical surface r = 1, when the potential on this surface is a given function f(p) of (p alone. Note that V (r, p) must be periodic in (p with period 27 t; it must also be a continuous function within the cylinder, since the space is supposed free of charges. Derive the following formula for V (r, <p) : V = ia Q + ^ r n (a n cos np +■ b n sin np), where a n and b n are the Fourier coefficients of f(p) for the interval (-7T, tt). 9. In Prob. 8, suppose /(<p) = —1 when —tt < p < 0, and f(p) = 1 when 0 < <p < and show in this case that the potential formula can be written in the closed form V = 2 , 2 r sin p - arctan -= 5 - with the aid of the result found in Prob. 9, Sec. 49. 10. From the infinite solid cylinder bounded by the surface r = c a wedge is cut by the axial planes ip = 0 and p = <p 0 . Find the steady temperatures u(r, cp) in this wedge if u — 0 on the surfaces p — 0 and p = <po, and u — f(p) (0 < p < po) on the convex surface of the wedge. oo mr Ans. u — ^ b n (r/cY Q sin ( nirp/p 0 ), where b n are the coefficients in the Fourier sine series for f{p) in the interval (0, p Q ). 11. If in Prob. 10, f(p) = A, where A is a constant, show that the formula for u{r, p) can be written in closed form with the aid of the result found in Prob. 9, Sec. 49. REFERENCES 1 . Carslaw, H. S.: “ Mathematical Theory of the Conduction of Heat in Solids/' 1921. 2. Ingersoll, L. R., and O. J. Zobel: “Mathematical Theory of Heat Conduc- tion with Engineering and Geological Applications," 1913. 3. Timoshenko, S.: “Vibration Problems in Engineering," 1937. CHAPTER VII UNIQUENESS OF SOLUTIONS 54. Introduction. For the most part, our solutions of the boundary value problems in the last chapter were formal, in that we did not usually attempt to establish our result completely, or to find conditions under which the formula obtained represents the only possible solution. We shall develop a few theorems here which will furnish the reader interested in such matters with a mathematically complete treatment of many of our problems. A multiplicity of solutions may actually arise when the problem is incompletely stated. Also, it is generally not a simple matter to transcribe a physical problem completely into its mathe- matical form as a boundary value problem. Consequently, the precise treatment of such problems is of practical as well as theoretical interest. Our first theorem (Abel’s test) enables us to establish the continuity of many of our results obtained in the form of series. The continuity property is useful both in demonstrating that our result is actually a solution of the boundary value problem, and in showing that it is the only solution. The remaining theorems give conditions under which not more than one solution is possible. It will be evident that they can be applied only to specific types of problems. But no “general” uniqueness theorem exists in the theory of boundary value problems in partial differential equations, in the sense that the same theorem applies to temperature problems, potential problems, etc. The uniqueness theorems given below arc again special in that they require a high degree of regularity of the functions involved. But. they will make possible a complete treatment of many of the problems considered in this book. 55. Abel’s Test for Uniform Convergence of Series. We now establish a test for the uniform convergence of infinite series whose terms arc products of specified types of functions. Appli- cations of this test have already been made in the foregoing 127 128 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 55 chapter, to establish the continuity of the solution of a boundary value problem (Sec. 46). The function represented by a uniformly convergent series of continuous functions is continuous. This is true regardless of the number of independent variables, as will be evident upon our recalling the method of proof for a single variable.* It is to be understood that the terms of the series are continuous with respect to all the independent variables taken together, in some region. The uniform convergence of the series in this region then ensures the same type of continuity of the sum of the series. A sequence of functions T n (t) (n = 1,2, - - - ) is said to be uni- formly bounded for all values of t in an interval if a constant K, independent of n, exists for which ( 1 ) \Tn(t)\<K for every n and all values of t in the interval. The sequence is monotone with respect to n if either (2) ZWO ^ T n (t) for every t in the interval and for every n, or else ( 3 ) Tn+i(t) T n (t) for every t and n. The following somewhat generalized form of a test due to Abel shows that when the terms of a uniformly convergent series are multiplied by functions T n (t) of the type just described, the new series is uniformly convergent. Theorem 1. The series X n (x)T n (t) converges uniformly with respect to the two variables x and t together , in a closed region R of the xt-plane, provided that ( a ) the series 00 X n (x ) converges uniformly with respect to x in R, and ( b ) for v all t in R the functions T n (t ) {n = 1, 2, ■ • • ) are uniformly bounded and monotone with respect to n. Let S n denote the partial sum of our series, S n (x, t) = Xi(x)Ti (0 +X a (*)r,(0 + • • • + X n (x)T n (t). * See, for instance, Sokolnikoff, “ Advanced Calculus,” p. 256, 1939. Sec. 55] UNIQUENESS OF SOLUTIONS 129 We are to prove that, given any positive number e, an integer N independent of x and t can be found such that | S m (x, t) - SnO, t) | < e if n > N, for all integers m = n + l,n + 2, • * • , and for all t in the region R. If we write Sn(x) = Xi(x) + X 2 (^) +■'*-+- X n (x), then for every pair of integers m, n (m > n), we have (4) S m - Sn = Xn+lTn+l + Xn+^Tn+2 + * * * + X m T m — (Sn-fl — 7 i-|-i H~ (Sn-4-2 S71+1) T * ' * H” (,^m Sm— l) Tm = (S n+ 1 “ S n )(T n+ 1 — T n + 2 ) + (Sn+2 “ S n )(T n +2 — T n +z) “b * * * "b (Sw— 1 $n)(T m -l "l - Sr^'Tm* Suppose now that the functions T n arc nonincreasing, with respect to n, so that they satisfy relation (2). Also let K be an upper bound of their absolute values, so that condition (1) is true. Then the factors (7 7 „ + i — 7 1 „+2), (3P»+2 — T n+3 ), * * • , in equation (4) arc non-negative, and |7\»| < K. Since the series oo V X n (x) is uniformly convergent, an integer N can be found for x which |s n+p - s„| < when n> N, for all integers p, where e is any given positive number and N is independent of x. For this choice of N it follows from equa- tion (4) that |/S„, — Sn I < [(jf’n-l-1 — Tn+i) + (Tn +2 — T n+ . s) + • • • + \T m II = 3 ^ [Tn +I - T m + \T m \\, and therefore |jS„ — <S„| < e, when n > N (m > n). The proof of the theorem is similar when it is supposed that the functions 7’» arc; of the nondecreasing type (3), with respect to n. 130 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 56 When the variable x is kept fixed, or when the functions X n (x) are constants, the theorem shows that the series with terms X n T n is uniformly convergent with respect to L The only requirement on the series in X n in this case is that the series shall converge. Extensions of the theorem to the case in which the functions X n involve the variable t as well as x , or where X n and T n are functions of any number of variables, become evident when it is observed that our proof rests on the uniform convergence of the Xn-series and the bounded monotone character, with respect to n, of the functions T n . 56. Uniqueness Theorems for Temperature Problems. Let R denote the region interior to a solid bounded by a closed surface and let R + S denote the closed region consisting of the points within the solid and upon its surface. If u(x, y, z, t) represents the temperature at any point in the solid at time t, a rather general problem in the distribution of temperatures in an arbitrary solid is represented by the following boundary value problem; (!) + <p{x, y, z, t ) (t > 0), at all points (z, y, z) in R; ( 2 ) » = /(*, y, z) in R, when t = 0; (3) u = g(x, y, z, t) (f > 0), when ( x , y, z) is on S. This is the problem of determining the temperatures u in a solid with prescribed initial temperatures /(x, y , z) and surface temperatures g(x, y, z, t). A continuous source of heat, whose strength is proportional to <p(x } y ) z, t), may be present in the solid. Suppose there are two solutions u = ui(x, y, z, t), u = u 2 (x, y , z, t), of this problem, where both U\ and u 2 are continuous functions of Xj y } z y t } together, in the region R + S when t ^ 0, while dui/dty du 2 /dt, and all* the derivatives of u\ and u 2 once or twice UNIQUENESS OF SOLUTIONS Sec. 56] 131 with respect to either x, y , or z are continuous functions when (x, y , 2 ) is in R + S and t > 0. Since ui and u 2 satisfy each of the linear conditions (1) to (3), it follows at once that their difference w, w(x, y , 2 , t) = ui(x, y, z, t ) - u 2 (x, y , 2 , t), satisfies the following linear homogeneous problem: (4) ^ = kV*w at in R (t >0); (5) w = 0 when t = 0, in R ; (6) w = 0 on S (t > 0). Moreover, w and its derivatives appearing in equation (4) must have the continuity properties required above of u y and u 2 and their derivatives. We shall show now that w must vanish at all points of R for all t > 0, so that the two solutions Ui and u 2 are identical. It follows that not more than one solution of the problem (1)- (3) can exist if the solution is required to satisfy the continuity conditions stated above. Since the function w is continuous in R + S, the integral where dV = dx dy dz , is a continuous function of t when t ^ 0. According to condition (5), J( 0) = 0. I 11 view of the continuity of dw/dt when t > 0, we can write -*///• dw JTr w — dV R at wV 2 w dV (t > 0). Since the second derivatives of w with respect to each of the coordinates art; continuous functions in R + S when t > 0, we can use Green’s theorem to write ■in. dw w — dS dn wV 2 w dV + 132 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 56 Here n is the outward-drawn normal to the surface S. But according to condition (6), w = 0 on 8, and so -- k Si X [(£)■ + (£)' + (£)'] ^ # > »>• Since the integrand here is never negative, • J'(t) g 0 when t > 0. The mean-value theorem applies to J (t) to give J(t) ~ J(0) = U'iff) (0 < h < t), and since J (0) = 0, it follows that J(t) ^ 0 whenever t > 0. However, the definition of the integral J shows that J(t) ^ 0 (t £ 0). Therefore J(t) = 0 (t ^ 0); and so the integrand w 2 of the integral J cannot be positive in R. Consequently w(x, y, z,t) = 0 throughout R + S, when 0. This completes the proof of the following uniqueness theorem: Theorem, 2. Let u(x, y, z, t) satisfy these conditions of regu- larity: (a) it is a continuous function of x, y, z, t, taken together, when (x, y, z) is in the region R S and t S: 0; (6) those derivatives of u which are present in the heat equation (1) exist in R and are continuous in the same manner when t > 0. Then if u is a solu- tion of the boundary value problem (l)-(3), it is the only possible solution satisfying the conditions (a) and (6). Our proof required only that the integral in Green’s theorem be zero or negative. The integral vanished, since w = 0 on S because of condition (3) j but it is never positive if (3) is replaced by the condition (8) ^ + hu = g(x, y, z, t ) on S, where h is a non-negative constant or function. So our theorem can be modified as follows : Sec. 57] UNIQUENESS OF SOLUTIONS 133 Theorem 3. The statement in Theorem 2 is true if boundary condition (3) is replaced by condition (8), or if (3) is satisfied on part of the surface S, and (8) on the remainder of S. The condition that u be continuous when t = 0 makes our uniqueness test somewhat limited. This condition is clearly not satisfied, for instance, if the initial temperature function is discontinuous in R + S, where the initial temperature on S is taken as the surface temperature. If the regularity conditions (a) and ( b ) in Theorem 2 are added to the requirement that u must satisfy the heat equation and boundary conditions, our temperature problem will be completely stated provided it has a solution. For that will be the only possible solution. 57. Example. In the problem of temperatures in a slab with insulated faces and initial temperature f{x) (Sec. 475), suppose f(x) is continuous when 0 ^ x g 7r, and f'(x) is sectionally continuous in that interval. Then the Fourier cosine series for f(x) converges uniformly in the interval. Let u(x, t) denote the function defined by the series (1) a n e~ nVit cos nx, l which was obtained in Sec. 47 as the formal solution, a n being the coefficients in the Fourier cosine series for f{x). Series (1) converges uniformly with respect to x and t together when and t ^ 0, according to Theorem 1. In any interval throughout which t > 0, the series obtained by differ- entiating series (1) term by term, any number of times with respect to either variable, is uniformly convergent according to the Wcicrstrass M-tcst. It readily follows that u{x , t) not only satisfies all the conditions of the boundary value problem (compare Sec. 46), but that it is also continuous when 0 ^ x ^ tt, t ^ 0, and its derivatives du/dt , d 2 u/dx 2 are continuous when 0 g x ^ tt, t > 0. That is, u(x, t) satisfies our conditions of regularity. The temperature problem for a slab is just the same as the problem for a cylindrical bar with its lateral surface insulated ( du/dn — 0); hence the region R, can be considered here as a finite cylinder. Theorem 3 therefore applies, showing that the 134 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 58 function defined by series (1) is the only possible solution which satisfies the above regularity conditions. PROBLEMS 1. In Prob. 7, Sec. 47, let fix) be continuous, and fix) sectionally continuous, in the interval (0, t), and suppose /(0) = fiw) = 0. Show that the solution found is the only one possessing the regularity prop- erties stated above. 2. Make a complete statement of Prob. 8, Sec. 47, so that it has one and only one solution. 3. Establish the solution of Prob. 10, Sec. 47, and show that it is the only possible solution satisfying the regularity properties stated above. 58. Uniqueness of the Potential Function. A function of x, y, z is said to be harmonic in a closed region R + S, where S is a closed surface bounding a region R, if it is continuous in R + S and if its second ordered derivatives with respect to x, y , and z are continuous in R and satisfy Laplace's equation there. Let Uix , y, z) be a harmonic function whose derivatives of the first order are continuous in R + S. Then since ( 1 ) V*U = 0 throughout R, Green's formula (7), Sec. 56, can be written as follows: This formula is valid for our function [/, even though we have not required the continuity of the second ordered derivatives of U in the closed region R + S. We shall not stop here to prove that, since V 2 E/ = 0, this modification of the usual condi- tions in Green's theorem is possible.* If JJ = 0 at all points on S , the first integral in equation (2) is zero, so the second integral must vanish. But the integrand of the second integral is clearly non-negative. It is also con- tinuous in R, So it must vanish at all points of R ; that is, (3) dU dx * The proof is not difficult, of this chapter. dU = dU ' dy dz See, for instance, p. 119 of Ref. 1 at the end UNIQUENESS OF SOLUTIONS Sec. 58] 135 so that U is constant in R. But U is zero on 8 and continuous in R + S, and therefore U = 0 throughout R + S. Suppose that dU/dn, instead of U, vanishes on 8; or, to make the condition more general, suppose (4) + hU = 0 on S, where h ^ 0, and h can be either a constant or function of x , y, and z. Then on S> = —hU 2 ^ 0, cm ’ so that the integral on the left in equation (2) is not positive. But the integral on the right is not negative. Both integrals therefore vanish and again condition (3) follows, so that U is constant throughout R. Of course U may vanish over part of S and satisfy condition (4) over the rest of the surface, and our argument still shows that U is constant in R. In this case the constant must be zero. Now suppose that the function V(x, y, z), together with its derivatives of the first order, is continuous in R + S, and let its derivatives of the second order be continuous in R. Also let V(x, y, z) be required to satisfy these conditions: (5) V 2 7 = / when (x, y, z ) is in R ; ( 6 ) P^~+hV = g when (x, y, z) is on the surface S. The prescribed quantities f, p, h, and g may be functions of (x 7 y , z); but it is assumed that p ^ 0 and h ^ 0. We have made boundary condition (6) general enough to include various cases of importance. When p = 0 on S, or on part of S, the value of V is assigned there; and when h = 0, the value of dV/dn is assigned. Of course p and h must not vanish simultaneously. If V = V i and V = V<> arc two solutions of this problem, then their difference, U = V x - V 2 136 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 58 satisfies Laplace’s equation (1) in R , and the condition + hU = 0 on S. Since U is harmonic and has continuous derivatives of the first order in R + S, we have shown that U must be constant throughout R + S. Moreover, if p = 0 at any point of S , so that U vanishes there, then U = 0 throughout R •+ S. We therefore have the following uniqueness theorem for problems in potential or steady temperatures, and other prob- lems in which the differential equation is that of Laplace or Poisson. Theorem 4. Let the function V ( x , y , z ) be required to satisfy these conditions of regularity: (a) it is continuous , together with its partial derivatives of the first order , in the region R + S; and (6) its derivatives d 2 V/dx 2 , d 2 V/dy 2 , and d 2 V/dz 2 are continuous functions in R. Then if V is a solution of the. boundary value problem (5)-(6), it is the only possible solution satisfying the conditions of regularity , except for an arbitrary additive constant . If, in condition (6), p = 0 at any point of S , then the additive constant is zero and the solution is unique. It is possible to show that this theorem also applies when R is the infinite region outside the closed surface S, provided V satisfies the additional requirement that the absolute values of pV, } dV J dy } t dV ' dz shall be bounded for all p greater than some fixed number, where p is the distance from the point ( x , y, z) to any fixed point.* Since V is required to approach zero as p becomes infinite, the additive constant in this case is always zero. But note that even here S is a closed surface, so that this extension of our unique- ness theorem docs not apply, for instance, to the infinite region between two planes or the infinite region inside a cylinder. The regularity requirement (a) in Theorem 4 is quite severe. It will not be satisfied, for instance, in problems in which V is prescribed on the boundary as a discontinuous function, or as a function with a discontinuous derivative of the first order. * For the proof, see Ref. 1. Sec. 59] UNIQUENESS OF SOLUTIONS 137 For problems in which V is prescribed on the entire boundary S [that is, p = 0 in condition (6)] of the finite region R, it is possible to relax the conditions of regularity so as to require only the continuity of V itself in R + 8. The derivatives of the first and second order are only required to be continuous in R. This follows directly from a remarkable theorem in potential theory: that if a function is harmonic in R + S, and not constant, its maximum and minimum values will be assumed at points on Sj never in R.* But this uniqueness theorem is limited in its applications to boundary value problems, because it does not permit such a condition as d V/dn = 0 on any part of S, a condi- tion which is often present or implied in the problem. This will be illustrated in the example to follow. 59. An Application. To illustrate the use of the theorem in the preceding section, consider the problem, in Sec. 49, of deter- mining the steady temperature u(x, y ) in a rectangular plate with three edges kept at temperature zero and with an assigned temperature distribution on the fourth. The faces of the plate are kept insulated. For the purpose of illustration it will be sufficient to consider here only the case of the square plate with edge 7 r units long. We also observe that as long as du/dn = 0 on the faces, the thickness of the plate does not affect the problem. We may as well consider this as a problem in the potential V(x , y) in the finite region R bounded by the planes x = 0, x — 7T, y = 0, y = 7r, and any two planes z = Z\, z = z 2 . Then our boundary value problem can be written (1) aw dW dx 2 dy‘ l (2) V(0, y) = 0, V(t, y ) = 0, (3) V(x, 0) = f(x), o' 11 and of course, dV/dz = 0 on z = Z\ and z = z 2 . The given function /(x) will be required here to be continuous, together with its first derivative, in the interval (0, n r). It is also supposed that/"(x) is seetionally continuous in that interval; and finally, we require /(x) to satisfy the conditions /( 0) = /( x) = 0. * The proofs of those theorems will he found quite interesting, and not difficult to follow. See Kefs. 1 and 2. 138 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59 Then, according to our theory of Fourier series, the sine series for /(*), ao _ (4) 2 bn sin nx L" = ^ Jo ^ sin nx dx ’ converges uniformly, and so does the cosine series for f(x), oo (5) ^ nb n cos nx, obtained by differentiating the sine series termwise. In demon- strating the uniform convergence of the Fourier series in Sec. 38, however, we proved that the series of the constants + b\ converges. In the case of series (5), in which the sine coefficients are zero and the cosine coefficients are nb n , this means that the series 2) \ nb *\ is convergent. Since the absolute values of the terms of the series 00 (6) V nb n sin nx 1 are not greater than \rib n \, it follows from the Weierstrass test that the series (6) also converges uniformly. In addition to the conditions (1) to (3), let the unknown function V(x , y) be required to satisfy the regularity conditions (a) and ( b ) of Theorem 4. That is, V, dV /dx, and dV/dy must be continuous in the closed region 0 ^ x S 0 ^ y ^ tt, while d 2 V/dx 2 and d 2 V/dy 2 are required to be continuous at all interior points of the region. We shall call this a complete statement of the problem of determining the function V(x, y). For accord- ing to Theorem 4, this problem cannot have more than one solution, and we shall now prove that it does have a solution. The series derived in Sec. 49 as the formal solution of our problem can be written here as sinh n{j — y) sinh 7i7T sin nx . (7) Sec. 59 ] UNIQUENESS OF SOLUTIONS 139 Let us show that this represents a function y ) which satis- fies all the requirements made upon V in the complete statement of our problem, so that V = \p (x, y ) is the unique solution of that problem. To examine the uniform convergence of series (7), let us first show that the sequence of the functions /on sinh u(t - y) sinh mr ’ which appear as factors in the terms of that series, is mono- tone nonincreasing as n increases, for every y in the interval 0 ^ y ^ t. This is evident when y = 0 and y = t. It is true when 0 < y < 7r, provided that the function T(t) = sinh bt sinh at always decreases in value as t grows, when t > 0 and a > b > 0. Now 2T'{t) sinh 2 at = 2b sinh at cosh bt — 2 a sinh bt cosh at = —(a — b) sinh ( a + b)t + (a + b) sinh ( a — b)t [sinh ( a + b)t sinh (a — b)t~\ T^b = -(a 2 - b>) L a b = -(a 2 - b-) 2 K« + b)°- n - (a - £2n-H ( 2 »+ 1)1 The terms of this series are positive, so that T'(t) < 0, and T(t) decreases as t increases. Therefore functions (8) never increase as n grows. Likewise the functions (9) cosh n (it — y ) sinh mr (0 g y ^ tt) never increase in value when n grows; because the squares of these functions can be written as the sum sinh 2 n(ir — y ) sinh 2 mr ’ ( 10 ) sinh 2 mr 140 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59 and as n grows the first term of the sum clearly decreases, while the second was just shown to be nonincreasing. Functions (8) are clearly positive and not greater than unity for all values of y and n involved. Functions (9) are also uni- formly bounded; this is evident from the expression (10) for their squares. Therefore the sequence of functions (8), or of functions (9), can be used in our form of Abel’s test for uniform convergence. So, from the uniform convergence of the sine and cosine series (4), (5), and (6), when 0 ^ x ^ 7r, we conclude not only that our series (7) converges uniformly with respect to x, y in the region 0 ^ x ^ t, 0 ^ y S but also that this uniform con- vergence holds true for the series 2 , sinh n(r — y) nl> - smhn„ obtained by differentiating series (7) with respect to x } and for the series oo -2 Tibn cosh n(w — y) sinh mr sm nx, obtained by differentiating series (7) with respect to y. Consequently series (7) converges to a function i p(x, y) which, together with its partial derivatives of the first order, is con- tinuous in the closed region 0 ^ x ^ w, 0 y S v. The func- tion \j/ clearly satisfies boundary conditions (2) and (3). When differentiated twice with respect to either x or y, the terms of series (7) have absolute values not greater than the numbers ( 11 ) n 2 \b n \ sinh n(7r — y 0 ) sinh mr for all x and y in the region 0 ^ x ^ tt, y 0 ^ y ^ x, where y 0 is any positive number less than t. Since the series of the constants (11) converges, the series of the second derivatives of the terms of series (7) converges uniformly in the region specified. Hence series (7) can be differentiated termwise in this respect whenever 0 < y < 7r; also the derivatives d^jdx 2 , d^/dy 2 are continuous whenever Q^x^w, 0<y^T. Sec. 59] UNIQUENESS OF SOLUTIONS 141 Thus i p(x } y ) satisfies the regularity conditions. It only remains to note that it satisfies Laplace's equation (1) in R . This is true because the terms of series (7) satisfy that equation, and series (7) is termwise differentiable twice with respect to x and to y in R , so that Theorem 2, Chap. I, applies. The only solution of our completely stated problem is therefore V = 00 sinh n(ir — y) sinh mr sin nx. In particular, note that we have shown that our complete problem, which includes the condition that dV/dz = 0 on the boundaries z = Zi and z = z 2 , has no solution which varies with z. In the formal treatment of the problem given earlier, the absence of the variable z was regarded as physically evident. In the present section we have omitted the term d 2 V/dz 2 in Laplace's equation, and at other times have neglected writing the variable z, only as a matter of convenience. PROBLEMS 1. Show that the formal solution found in Sec. 49 can be completely established as one possible solution of the boundary value problem written there, provided the function /(x) is sectionally continuous in the interval (0, a) and has one-sided derivatives there, and fix) is defined to have the value [fix + 0) + fix - 0)]/2 at each point x of discontinuity (0 < # < a). 2. Make a complete statement of the boundary value problem for the steady temperatures in a square plate with insulated faces, if the edges x = 0, x = 7 r, and y = 0 arc insulated, and the edge y = t is kept at the temperature u = fix). Assume that fix) is continuous when 0 ^ x Z 7 T, and that /'(()) = /'(x) = 0. Show that your problem has the unique solution cosh ny cosh mr cos nx 2 r (In = ™ * Jo fix) cos nx dx 3. Establish the result found in Prob. 5, See. 49, as a solution (but not as the only possible one) of the boundary value problem, when the function fix) there is represented by its Fourier sine series. 4. In Prob. 8, Sec. 53, let the infinite cylinder be replaced by a finite cylinder bounded by the surfaces r — 1, z = z\, z = on the last two of which dV/dz = 0. Also let the periodic function f(<p) have a con- 142 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59 tinuous second derivative. Then show that the result found there is actually a solution, and that it is the only possible solution of the problem satisfying our conditions of regularity. REFERENCES 1. Kellogg, 0. D .: “ Foundations of Potential Theory,” 1929. 2. Courant, R., and D. Hilbert: “Methoden der Mathematischen Physik,” Vol. 2, 1937. CHAPTER VIII BESSEL FUNCTIONS AND APPLICATIONS 60. Derivation of the Functions Any solution of the differential equation (1) * 2 § + *i + (x2_n2)2/ = 0 ’ known as Bessel 1 s equation , is called a Bessel function or cylindrical function. It will be shown later on how this equation arises in the process of obtaining particular solutions of the partial differential equations of physics, written in cylindrical coordi- nates. We shall let the parameter n be any real number. A particular solution of Bessel’s equation in the form of a power series multiplied by x l \ where p is not necessarily an integer, can always be found. Let a () be the coefficient of the first non- vanishing term in such a series, so that a 0 ^ 0. Then our pro- posed solution has the form 00 00 (2) y = x* ]£ ajX* = X ap***. 3=0 3=0 If the series here can be differentiated termwise, twice, the coefficients aj can be determined so that the series is a solution of equation (1). For upon differentiating and substituting in equation (1), we obtain the equation X + j) (p +j — 1) + (p + j) + (x 2 - n.*)]ajX*+i = 0. 3=0 Dividing through by :v p and collecting the coefficients of the powers of x ) we can write the equation in the form (p 2 — n 2 )a 0 + [(p + l) 2 - n 2 ]a x x + X ! Kp + j) 2 - n 2 ]a s + a,,-2)xf = 0. J- 2 143 144 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 60 This is to be an identity in x , so that the coefficient of each power of x must vanish. The constant term vanishes only if p = ±n. The second term vanishes if ai = 0; and the coeffi- cients of the second and higher powers of x all vanish if [(p + j) 2 - + Oj-s = 0 (i = 2, 3, • • • ); that is, if (p - n +j)(p + n + f)a, = -a,-_ 2 (j = 2, 3, • • • )• This is a recursion formula for a iy giving each coefficient in terms of one appearing earlier in the series. Let us make the choice V = n, so that the recursion formula becomes (3) j(2n +j)a i = -o,-_ 2 O' = 2, 3, • • • ). Since a L = 0, it follows that a 3 = 0; hence a 5 = 0, etc.; that is, (4) a 2fc _i = 0 (Jc = 1, 2, • • - ), provided n is such that 2n + j ^ 0 in formula (3). But even if 2 n + j does vanish for some integer j, coefficients (4) still satisfy formula (3). Since this is all that is required to find a solution, we can take all the coefficients a 2k - 1 as zero regardless of the value of n. Replacing j by 2 j in formula (3), we can write a 23 ' = 2 2 j(n + J) a2j “ 2 ^ = ^ provided n is not a negative integer. Replacing j by j — 1 here, we have -1 a2l '~ 2 ~ 2*0' - 1)(» + 3 ~ 1) a2l '“ 4 ’ so that (- 1) 2 a2i 200 - l)(n+i)(»+j - I) 023 '" 4 ' Continuing in this manner, it follows that a 2 j = ( — l) fc a2 3 --2fc/[2 2fc j(j — 1) • ■ * (j — k + 1) (n + j) (n +j - 1) • • • (n +j ~ fc + 1)]; Skc. 6 . 1 ] BESSEL FUNCTIONS AND APPLICATIONS 145 so that when k = j, we have the formula for a v in terms of a Q : ( — 1 ) 3 CLq . 0/27 ~ 2 23 j\{n +j)(n +7 - 1) ' * • (n + 1) ^ = 1 » 2 > ' ' * )• The coefficient a 0 is left as an arbitrary constant. Let its value be assigned as follows : 1 a ° 2T (n + 1)' Recalling that the Gamma function has the factorial property kT(k) = r(fc + 1 ), it follows that (n + j)(n + j - 1) • • • (n + 2 )(n + l)r(n + 1) = r(n + i + 1). Our formula for a 2? - can therefore be written (5) a 2 j = + j _|_ i)2^+2/ 0’ = ‘ " ‘ )> where j! = 1 if j = 0. The function represented by series (2) with coefficients (4) and (5) is called a Bessel f unction of the first kind of order n: ( 6 ) (- 0 ' j\T(n + j + 1) _ r x 2 2 n T(n + 1) [_ 2(2 n + 2) + 2 • 4(2 n -j- 2)(2n -|- 4) The series in brackets is absolutely convergent for all values of x ) according to the ratio test. It is a power series, so that the termwise differentiation employed above is valid, and hence function (6) is a solution of Bessel's equation. Of course, when n is not a positive integer, J n (x) or its derivatives beyond a certain order will not exist at x = 0, because of the factor x n . 61. The Functions of Integral Orders. When n = 0, the important case of the function of order zero is obtained: Jo(x) = 1 x 2 x A 22 + 2 2 • 4 2 2 2 • 4 2 * 6 2 a + 146 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 61 When n is a negative integer, the choice p = -n can be made and the recursion formula (3) of the last section gives the coeffi- cients just as before. If, in this case, do 2 n r(-n + 1)’ the solution of Bessel’s equation will be found to be ( 1 ) 2 (-l) a ' _ v j!r(-n + j + 1) J-n(x). Now if we define l/r(j?) to be zero when p = 0, 1, 2 • • • formula (6) of the last section can be used to define a function J n {x) even when n is a negative integer. For if n = -m where m is a positive integer, that formula becomes J —m Summing with written (- 1 )* x \-m+2f (x) 2jfr(-m+j + l) ^2, respect to k, where k = -m + j, this can be (-1)* /x\ f^(k + m)\T(k + 1) \2/ r (k + m) !r(& + 1 ) -(-u- 2 ^ j^kWim + k + 1 ) But the last series represents J m (x); hence for functions of integral order, (2) J-m(x) = (-1 ) m J m (x) ( m = i, 2, 3, • ■ • ). According to solution (1) now, the function y = ( — iw ( x ) and hence the function y = J n (x) is a solution of Bessel’s equa- tion when .n is a negative integer; hence the junction dejined by equation (6), Sec. 60, is a solution for every real n. When n is neither a positive nor a negative integer, nor zero, it can be shown that the particular solution /_ n (z) obtained by aiang p - ms not a constant times the solution J n (x) ; hence the general solution of Bessel’s equation in this case is 1/ ~ AJ n(x) -)- BJ-. n (x), where A and B are arbitrary constants. Sec. 61] BESSEL FUNCTIONS AND APPLICATIONS 147 When n is an integer, the general solution of Bessel’s equation is V = AJ n (x) + BY n (x), where F»( x) is a Bessel function of the second kind of order n. These functions will not be used here. For their derivation and properties, as well as for a more extensive treatment- of the theory of Bessel functions of the first kind than we can give here, the reader should consult the references at the end of this chapter. There are several other ways of defining the functions J n (x). When n is zero or a positive or negative integer, the generating function exp [£x(f — I/O], is often used for this purpose. By multiplying the two series it can be seen that (3) exp | (i - D = 2 J &)t m L n — — vo = Jq(x) + Ji(x)t + Jv(x)t 2 + • • • + J-\(x)t~ l -b + * * * , for all values of x and t except t = 0. Hence J n (x) can be defined as the coefficients in this expansion. It is on the basis of this definition that the above choice of the constant a 0 was made. PROBLEMS 1. Prove tlmt J -if*) 2. Prove that J\{x) 3. Derive solution (1) when n is a negative integer. 4. Carry out the derivation of formula (8). 5. Show that, for every n, Jn(-X) = (-1 )»J n ( X ). 148 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 62 62. Differentiation and Recursion Formulas. By differenti- ating the series <» - I. W+j'+D it follows that (2) xJ' n (x) 2 (-iy(n + 2j) /x\ n+ii 00 = nJ n (x ) + X 2 Tfz: (- 1 V J-l (j - iy.T(n+j + 1) (I) n+2j—l That is, (3) TtJ nix) X A!r(n + A + 2) n+l+2fc •oj n(^0 XiJ nix') xj n+l(x). Similarly, if we write n + 2j = 2(n + j) — n in the sec- ond member of equation (2), and replace r(n + j + 1) by (n + j)T(n -+- j), we obtain the relation xJ'nix) - — nJ„( x) + X that is, (4) xJ' n {x) = —nJ n {x) + zJ n-i{x). Elimination of J n {x) between equations (3) and (4) gives the formula (®) ‘^'J'ni 3 ') = J n— \(x) / n+1 (z)j and the elimination of J' n (x ) between the same two equations gives the formula (6) ~ J n ix) = J n _,ix) + J n+ i(x). (xy~ ,_ u i !r ( n +3) W The recursion formula (6) gives the function Jn+i(x) of any order in terms of the functions J n(x) and J n -\{x) of lower orders. By multiplying equation (4) by x n ~\ and equation (3) by ar"- 1 , we can write these formulas, respectively, as d \% n Jn(%y\ = X n J n -l(x), ^[a :~ n J n (x)] = —x^Jn+iix). Sec. 63] BESSEL FUNCTIONS AND APPLICATIONS 149 The following consequences of the above formulas should be noted: A(x) = -J i(x) = (7) rJo(r) dr = xJi(x ). PROBLEMS 1. Obtain formula (7) above. 2. Prove that *V"(x) = [»(n - 1) - *V.(*) + xJ n+1 (x). 3. With the aid of Probs. 1 and 2, Sec. 61, prove that - S(^- cosx ) 4. When n is half an odd integer, show that J n (x) can always be written in closed form in terms of sin x, cos x, and powers of \/\fx. 63. Integral Forms of J n (z). Let us first recall that the Beta function is defined by the formula B(n + i, j + i) = 2 £ sin 2n 0 cos 2 * d dd (n > — j > — £). Let i be zero or a positive integer. Then B(n + j + ■*) = sin 2n 0 cos 2 * 6 d6 (n > — ■£). This function is given in terms of the Gamma function by the formula B j r (n + j)r(j + 1) r(n + j + 1) ' and as a consequence we shall be able to write the general term of our series for J n (x) in terms of trigonometric integrals. Our formula for J n (x ) can be written Jn(x) Now 1 2 2j j‘!r(n + j + 1) © n 'sh ( — 2j2*>jlT(n+j + 1)’ j = U j • ^ • I • • • 0 ~ nrCD (2j)!r(n +j + l)r(i) r(j + 1) = £(«■ + j + j) ( 2 j!)r(n +i + l)r(i) (2i)!r(i)r(n + l) 150 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 63 Therefore (1) Jn(x) = Cn 2 XU Jo sin2 " 6 C0s2) 9 de, where ( 2 ) When n ^ 0, the series (-i) _ (s/2)» ” r(*)r<» + i)‘ 3-0 m x 23 sin 271 0 cos 23 0 converges uniformly with respect to 0 in the interval (0, it); because the absolute values of its terms are not greater than the corresponding terms of the convergent series i- o and the terms here are independent of 0. The first series can therefore be integrated termwise with respect to 0 over the interval (0, 7 r). In other words, the integral sign in formula (1) can be written either before or after the summation sign. Therefore, sin 2n 0 3 =0 (-iy m (x cos 0) 23 dd. Since the series in the integrand represents cos (x cos 0), (3) J n (x) = C n sin 2n 0 cos ( x cos 0) dd, where C n is defined by formula (2). Formula (3) gives one of LommeVs integral forms of J n (x ). Although the above derivation holds only for n ^ 0, form (3) is valid when n > This can be shown by writing the first term in series (1) separately and integrating the remaining terms by parts to obtain the equation Jn(x) = Cn sin 2n 0 dd ( — l) 3 ’# 23 ' 2j - 1 (2 j ) ! 2n + 1 I si sin 27l+2 0 cos 23 ’” 2 0 dd f Sec. 63] BESSEL FUNCTIONS AND APPLICATIONS 151 if n > —Jr- Here again the second integral sign can be written before the summation sign, and the series in the integrand can be seen to represent the function sin 2n+ i 0 d / cos (x cos 0) — 1 2n + 1 dB \ cos 0 The details here are left to the reader. Integrating this by parts and adding the first integral in brackets gives formula (3) for n > — i* When n = 0, formula (3) becomes 1 C T J 0 (x) = - I cos ( x cos 0) d$. T Jo When ft = 0, 1, 2, • • ■ , the following integral form is valid: i r (4) J n (x) = - I cos {nd — x sin 0) dQ (n — 0, 1, 2, • ■ ■ ). This is known as Bessel's integral form. By writing the integrand as cos nd cos (x sin '0) + sin nd sin ( x sin 0), it can be seen that formula (4) reduces to 1 f * (5) Jn(x) = — I cos nd cos (a: sin 0) dd if n = 0, 2, 4, • • • : rr Jo (6) Jn(x) = “ nd s * n ( x 0) dd if n = 1, 3, 5, * ■ • . These forms can be obtained from formula (3), Sec. 61. By substituting t — e i0 in that formula, we find that (7) cos (x sin 0) + i sin ( x sin 0) 00 00 = J„(z) + 2 2) Jin(r) cos 2nd + 2 i V J in _^x) sin (2 n — 1)0. n = 1 n = 1 Equating real parts and imaginary parts separately here, and multiplying the resulting equations by sin nd or cos nd and integrating, using the orthogonality of these functions in the interval 0 < 0 < 7r, we get formulas (5) and (6). Formula (4) follows by the addition of the right-hand members of formulas (5) and (6). The details are left for the problems. 152 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 63 From formula (4) the important property of boundedness \Jn(x)\ £ 1 (» = 0, 1 , 2, • • • ), follows at once. It also follows from the same formula that each derivative of J n (x ) is bounded for all x: d k Jn(%) = 1 (n = 0, 1 , 2, • * • j k = 1 , 2, • • • ). According to formula (5), 2«7 2 n(^r) = a> 2 n (n — 1, 2, • • * where a n denotes the coefficients in the Fourier cosine series, with respect to 0, of the function cos (x sin 0). Similarly if b n denotes the coefficients in the Fourier sine series of the function sin ( x sin 0), formula (6) shows that 2J r 2»— i(x) = &2n— i (n = 1 , 2 , ■ • • ). Since the Fourier coefficients of every bounded integrable function tend to zero as n tends to infinity, it follows that for every x the Bessel functions of integral orders have the property lim J n (x) = 0. n — > oo As to the behavior of the functions J n (x) for large values of x, it can be shown that (8) lim J n (x) = 0 (n = 0, 1, 2, • • • ). X — > 00 The proof is left to the problems. PROBLEMS 1. Use the Lommel integral form of J n (x ) to prove that 2. Prove in different ways that Jn(-x) = ( — 1 yj n (x) (n = 0 , 1 , 2, • • ■ ), and hence that J n (x) is an even or odd function of x according as n is an even or odd integer. Also deduce that J 2n— 1(0) = 0 (n - 1, 2, ■ • • 8ec. 64] BESSEL FUNCTIONS AND APPLICATIONS 153 3. Prove in different ways that /o(0) = 1. 4. Obtain formula (7) above by the method indicated there, and follow the process outlined to derive Bessel's integral forms (5) and (6), and thence (4). 6. Deduce from formula (5) that 7T 2 7*2 Jinix) = - I cos 2nd cos ( x sin 0) d6 (n = 0, 1, 2, • ■ • ). X Jo 6. Deduce from formula (6) that J 2n-l(x) 2 7 r J~ 2 sin (2 n — 1)0 sin ( x sin 0) dQ (n 1, 2, • • • ). 7. Write the integral in Prob. 5 as the sum of the integrals over the intervals (0, tt/2 — rj) and (• tt/2 — 77, t/2), where rj > 0, and thus show that \JUx)\ s l 1 7*2 v cos 2n0 5 J 0 cos e cos (x sin 0) d(x sin 0) V- By integration by parts, show that the absolute value of the integral appearing here is not greater than a positive number M V1 independent of x. Hence, given any small positive number e, by first selecting 17 sufficiently small and then x large, show that |*/2»0c)( < € when x > x Q . This establishes formula (S) when n = 0, 2, 4, • • ■ , there. 8. Apply the procedure of Prob. 7 to the formula in Prob. 6, and thus complete the proof of formula (8). 9. Note that the functions cos ( x sin 0) and sin (x sin 0), of the vari- able 0, satisfy the conditions in our theorem in Sec. 38; also, since they are even and odd functions, respectively, the series of absolute values of their Fourier coefficients converges. Deduce that the series 00 X •'»(*) n =0 is absolutely convergent for every x. 64. The Zeros of J n (x). The following theorem gives further information of importance in the applications of Bessel functions to boundary value problems. 154 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 64 Theorem 1. For any given real n the equation J n (x ) = 0 has an infinite number of real positive roots Xi, x 2 , * • * , x m , • * • , which become infinite with m. This will first be proved when — i < n ^ The proof for every real n will then follow from Rolled theorem. For if J n (%) vanishes when x = xi and x ~ x 2) for any real n, then so do x n J n(x) and x~ n J n (x), and hence their derivatives vanish at least once between xi and x 2 . But it was shown in Sec. 62 that these derivatives are x n J n ~i(x) and —zr n J lH _ 1 (x), respectively. Therefore between two zeros of J n (x ) there is at least one zero of Jn- i(x), and one of J n+1 (x). So if there is an infinite number of zeros of J n (x) when — \ < n ^ the same is true when n is diminished or increased by unity, and repetitions of the argument show the same for any real n. For the proof when — £ < n ^ ■£, we shall use the Lommel formula derived in the last section; namely, r , \ (a/2 Y r (1) Jn(z) - r(^)r(n~+ J) J 0 sm2n 6 cos & cos °) de - Now suppose that x is confined to the alternate intervals of length 7r/2 on the positive axis; that is, x = mir + g ty where m = 0, 1, 2, * * • , and 0 S t S 1. Also let a new variable of integration X, where A * \ cos d = — X, 2x 1 be introduced into the integral in formula (1). Then the integral becomes 2x 7T r 7T COS ( 7tX/2) d\ 2 X 2 x [1 - (ir\/2xy 2 ]-"+i ; and except for a factor which is always positive, this can be written ( 2 ) X 2/n-j-i cos (ttX/ 2) dX [(2m + t) % — X 2 ]“ w+ * Sec. 64] BESSEL FUNCTIONS AND APPLICATIONS 155 The sign of J n (x) is therefore the same as the sign of integral (2) . That integral can be broken up into this sum of integrals: (3) —I\ + I 2 — ^3 + * • * + (—1 ) m I m + ( — l) m H m , where rzj cos (ttX/2) d\ (i = 1 , 2 , H. 2 [(2m + ty - X 2 ]-“+i 2m+t cos (xX/2) r/rfc„ 2m , m), = (-1)” P J 2 n [(2m + ty — X 2 ]~ n+ * Now let 7j be broken up into the sum of two integrals, one over the interval (2 j — 2, 2j — 1) and the other over the interval (2j — 1, 2 j). By substituting a new variable of integration /z into these integrals, where in the first, and A = 2j - 1 - ju A - 2j - 1 + ji in the second, it will be found that U = j Q l Fi(») sin ^dn, where FA?) = [(2m + ty - (2j - 1 + m) 2 ] 71 "* - [(2m + 0 2 - (2 j - 1 - m) 2 ] 71 "*. Since n — % = 0, the function F,(ix) is never negative. By letting j assume continuous values and differentiating F 7 (/z) with respect to j, wo find that this function always increases in value with j. The integral /,• is therefore a positive increasing function of j] that is, 0 ~ I\ = I 2 = * ‘ * = Furthermore H m is not negative; because the numerator cos (t\/2) in the integral there can be written as ( — l)"* cos (7171/2) when X = 2m + /z, and cos (7171/2) is positive. 156 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 64 Now the sum (3) can be written ( — + (Irn — I m — l) + (Im-2 “ I m -z) +***], where the final term in the brackets is (I 2 — /i) if m is even, and 1 1 if m is odd. The quantity in the brackets is therefore positive, and consequently the sign of J n (mi r + irt/2) is that of ( — l) m ; that is J n (win + g ^ > 0 if m = 0, 2, 4, • • • , <0 if m = 1 , 3, 5, • • • . Since J n (x) is a continuous function of x ) its graph therefore crosses the rr-axis between x = ir/2 and x = t, and again between x = 3?r/2 and x = 2tt, and so on, when — J- < n g That is, J nix) vanishes at an infinite number of points x = x Xj x^ • ■ • , where x m tends to infinity with m. The theorem is therefore proved. Fig. 9. It follows at once from Rolle’s theorem that the equation J'nix ) = 0 also has an infinite number of positive roots x' m (m = 1 , 2, ■ • • ), and x' m tends to infinity with m. It should be observed that whenever x m is a zero of J n (x), the number — x m is also a zero. This is true for any w, as is evident from our series for J n (x ), Sec. 60. The difference between successive roots of J n (x) = 0 can be shown to approach i r as the roots become larger. Tables of numerical values of J n (x ), and of the zeros of these functions, will be found in the references at the end of this chapter.* We list below the values, correct to four significant * See Refs. 1, 3, and 4. Sec. 65] BESSEL FUNCTIONS AND APPLICATIONS 157 figures, of the first five zeros of Jo(x), and the corresponding values of Ji(x). J b(Xvi) — 0 m 1 2 3 4 5 Xm Jl(Xm) 2.405 0.5191 5.520 - 0.3403 8.654 0.2715 11.79 - 0.2325 14.93 0.2065 The graphs of the functions J Q (x) and Ji(x) are shown in Fig. 9. PROBLEMS 1. Draw the graph of (See Prob. 2, Sec. 61.) 2. Draw the graph of (See Prob. 1, Sec. 61.) 3. Draw the graph of / 2 (^) by composition of ordinates, using recur- sion formula (6), Sec. 62, and the graphs of J Q (x) and Ji(x). 65. The Orthogonality of Bessel Functions. Since J n (r ) satis- fies BessePs equation, we can write + rJ' n (r) + (r 2 - n 2 )J n (r) = 0. Substituting the new variable x , where r = \x and X is a constant, it follows that d 2 dx 1 2 JnO^x) + x - Jr k (\x) + (\ 2 x 2 — n 2 )J n (\x ) = 0; that is, J n (\x) satisfies BessePs equation in the form ( 1 ) d_ dx X ~ J n (\x ) + (\*x - J n (\x) = 0. For each fixed n this form is a special case of the Sturm-Liou- ville equation li. [ r< ^ ] + M*) + M*)]-* = °» with the parameter written as X 2 instead of X (Sec. 24). The function r(x) = x here; hence it vanishes v,hen x = 0. It follows from Theorem 3, Sec. 25, that those solutions of equation (1) in an interval 0 < x < c, which satisfy the boundary condition Jn(Xc) = 0, 158 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 65 form an orthogonal set of functions on that interval, with respect to the weight function p(x) = x . It will be observed that in this case q(x) = —n 2 /x, so it is discontinuous when x = 0, unless n = 0; but this does not affect the proof of the theorem. Now J n (\x) is a solution of equation (1), and it was shown in the last section that J n (\c) — 0 if Xc = xj (j = 1, 2, • • • ), where Xj are the positive zeros, infinite in number, of J n (x). Let Xj denote the corresponding values of X, so that Then the functions J n (\jx) ( j = 1, 2, • • • ), are orthogonal provided their derivatives J„(\jx) are continuous. This is true except possibly at x = 0, and if n ^ 0 an inspection of the series for J n shows that xJ' n (\jx) vanishes at x = 0, which is all that is necessary in the theorem. The result can be stated as follows : Theorem 2. Let X,- (j = 1, 2, • * • ) be the positive roots of the equation (2) J n (\c) = 0 , where n is fixed and n ^ 0. Then the functions J n (X ix), J n (\ 2 x)j • • • form an orthogonal set in the interval (0, c) with respect to the weight function x; that is , (3) r xJ n (Xjx)J n (\j c x) dx = 0 if k 7* j. No new functions of the set are obtained by using the negative roots of equation (2), for an inspection of the scries for J n shows that Jn( — \jX) = ( — 1 ) n J n (\jX). It should be carefully noted that n is the same for all functions of the set; hence an infinite number of sets has been determined here , one for each n (n ^ 0). Another boundary condition at x = c which gives still other orthogonal systems can be seen by examining the proof of the above theorem. For any two distinct real values Xy and of the parameter X, the functions J n (\,x) and J n (M x ) satisfy Sec. 65 ] BESSEL FUNCTIONS AND APPLICATIONS 159 equation (1 ) ; hence d_ dx A dx % dx ^ X ^ Jn(\kX) + + (\%x - J »(X,<c) = 0, (xfc - ^ J n (\ k x ) = 0. Multiplying the first of these equations by J n (\ k x) and the second by Jnfax), then subtracting and integrating, we find that (X? - xi) f xJ J n(X/i;iC) dx = J q |/„(X,a:) J„(Xifea;)j - J n (\nx) ^ jj*^ J„(X,-a:)j| da: — ^ ~(lx 7i(Xj3/) n(X^3/) n(Xfc2/) «^"n(X j3/) dx. When n ^ 0, both terms in the brackets in the last expression vanish when x = 0; hence (4) (A? - Afc) J^ C xJ n (\jx)J n (\ k x) dx = cKkJ n(Xjc)*/ n (\&c) cXjtT n (X^c)t/" n (\jC), where /^(Xc) denotes the value of ( d/dr)J n (r ) when r = Xc. Since X? — Aj? ^ 0, the orthogonality (3) exists whenever the right-hand member of equation (4) vanishes. This will be the case when Ay and \ k are two distinct values of X which satisfy the equation (5) XcJ'(Xc) = hJ 7i (Xc) , where h is any constant, including zero. The result can be written thus: Theorem 3. For any fixed n (n ^ 0), the functions J n (\ jx) (j = 1, 2, • • * ) form an orthogonal set in the interval (0, c) with respect to the weight function x, when Ay are the non-negative roots of equation (5). Here again, for every root Xy there is a root — X 7 . This can be seen by writing equation (5) in the form (6) (n + h)J n (\c ) — Xc/n+i(Xc) = 0. Consequently the negative roots introduce no new characteristic functions. The details here can be left to the problems. If n + h ^ 0, equation (6) has no purely imaginary roots. This is easily seen by examining our series for J n (x ). From now 160 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 65 on let us assume that h ^ 0, as is usually the case in the applica- tions; then n + h S 0. Similarly, equation (2) has no purely imaginary roots. Now equation (5) can be written (7) c [J„(Xz)] + hJJ\x) = 0 when x = c (h 3: 0); hence it is a boundary condition of the type introduced earlier. It involves a linear combination of the dependent variable in Bessel’s equation and the derivative of that variable. Consider the Sturm-Liouville problem, consisting of Bessel’s equation (1) and either one of the boundary conditions (2) or (7). A boundary condition at x = 0 is not involved because the function r(x ) in the general Sturm-Liouville equation is the independent variable x in this case, and it vanishes at x = 0. The characteristic functions here, J n (\ ; x), are continuous in the interval (0, c), since ti^O. Likewise for their first ordered derivatives, except possibly at the point x = 0; but the product x(djdx) J n (\jx) is continuous and vanishes at the point x = 0, which is all that matters. Finally, note that the function p(x) is also x itself here, and therefore it does not change sign in the interval (0, c). Hence according to Theorem 4, Sec. 25, the characteristic numbers Xf are all real. According to equation (6), X = 0 is a root of equation (5) only if either /„(0) = 0 or n + h = 0. In the first case the char- acteristic function J n (Xx) vanishes, so that the root X = 0 can contribute a characteristic function only if n = h = 0. A root X = 0 of equation (2) can never contribute a characteristic function. We state our results as follows : Theorem 4. When n S: 0 and h ^ 0, equations (2) and (5) have only real roots X ; . For either equation we use only the non- negative roots, since no new characteristic Junctions correspond to the negative roots. The root X = 0 is used only in the case of equation (5) with n = h = 0. PROBLEMS 1. Derive form (6) of equation (5). 2. Prove that when X,- is any root of equation (6), -X, is also a root Sec. 66] BESSEL FUNCTIONS AND APPLICATIONS 161 66. The Orthonormal Functions. It can be shown that, when n § 0 , the function J n (x ) is, except for a constant factor, the only solution of BessePs equation that is bounded at x = 0. Hence it follows from the results of the last section that the functions J n (\x) (j = 1 , 2, • • • ) represent all the characteristic functions of the Sturm-Liouville problem involved there, on the interval (0, c). We can therefore anticipate an expansion of an arbitrary function in series of the functions of this set. It should be observed that the orthogonality here with respect to the weight function x is the same as the ordinary orthogonality of the set of functions {VxJJX-x)} ( j = 1 , 2, • • • )• Let us now find the value of the norm, Nnj = x[J n (\jX)] 2 dX , of the functions J n (\-x)] these functions can then be normalized by multiplying them by the factors (N n j)~K If we multiply the terms in BessePs equation, by the factor (2 xd/dx)J n (\z), we can write the equation as i i-'-M 1 ’ - »■ Integrating, and using integration by parts in the second term, we find that js + ( X2x ' 2 Since rJ’n(r) it follows that - n 2 )[J n (\x)} 2 C _o = nJ n (r) — rj n+ i(r ), 2X 2 jT° x[J n (^x)] 2 dx = [{nJ.(X*) - \xJ„ +[ (\x)V> + (XV - n 2 ){/„(Xx)} 2 ]° c ; 162 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 67 or, since n ^ 0, (!) J Q 4Jn(\x)]*dx = J {[/ n (Xc)] 2 + [/^(Xc)] 2 } TIC «L>(Xc)t/ n +i(Xc). Hence, when X,- represents the roots of the equation J„(Xc) = 0, (2) z[J*(X 3 -a:)] 2 da; = 1 [/„ +1 (X,c)j 2 (j = 1, 2, • • • )- FAen X 3 - represents the roots of equation (5), £ec. 65, we have seen that \jCj n+l(XyC) = (n + A)t/n(XyC), and hence formula (1) reduces to (3) £ a, - - »■ (J - 1 , 2 , ■ • • ). The normalized functions <p n ,(x) can now be written Vni(x) = 'f n(XjZ) VIKt O' = 1, 2, ), where the norms N nj are given for the two types of boundary conditions by equations (2) and (3). The set of functions Wnj{x)\ is orthonormal in the interval (0, c) with a; as a weight function; that is, for each fixed n (n ^ 0), f 0 x<p n} (x)<p nk (x) dx = 0 if j y£ k, = 1 if j = k (j, k = 1, 2, • • • ). 67. Fourier-Bessel Expansions of Functions. Let c B/ be the Fourier constants of a function /( x) with respect to the functions <f> n ,{x) of our orthonormal set, where fix) is defined in the interval (0, c). Then °ni = x<p n ;(x')f(x) dx (i = l, 2, • • • ), Sec. 67] BESSEL FUNCTIONS AND APPLICATIONS 163 and the generalized Fourier series corresponding to f(x) can be written here as <W*) = 2 £ x ' J ^')f( x ') dx', when n ^ 0, and 0 < x < c. In view of the formulas given in the last section for N n j, this series can be written (1) X (* ^ 0), i=i where the coefficients A, are defined as follows: < 2 > A < - dx ' when Xi, X 2 , • * • are the positive roots, in ascending order of magnitude, of the equation (3) J n(Xc) = 0; but 2x? f c (4) A, ~ = (Xfc 2 + ¥ - n 2 )[J„(XjC)]' 2 Jo xJ ^ X i x )f( x ) dx, when Xi, X 2 , • * * arc the positive roots of the equation (5) XcJ'(Xc) + hJ n (\c) =0 (h ^ 0, n ^ 0). However, in the special case where h = n = 0, Xi is to be taken as zero , and the first term of the scries is the constant It can be shown that, when 0 < x < c, the series here does converge to f(x) under the conditions given earlier for the repre- sentation of this function by its Fourier series. Let us state one such theorem here explicitly, and accept it without proof for the purposes of the present volume.* Theorem 5. Let f(x) he arty function defined in the interval (0, c)j such that ^/x f(x) dx is absolutely convergent. Then at each point x (0 < x < c) which is interior to an interval in which * A proof, using contour integrals in the complex plane, will be found in Ref. 1. 164 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 67 fix) is of bounded variation, series (1) converges to #/(* + 0 ) +/(* - 0 )]; that is, oo (7) i lf(. x ”1" 0) + f( x 0)] ~ AjJ n (\jx) (0 <C x <C c), i =i where the coefficients Aj are defined by equation (2) or (4), and n ^ 0, h ^ 0. The theorem holds true for the special case h = n — 0, men- tioned above, if A i is defined by formula (6). It can be shown that all conditions here on fix) are satisfied everywhere, so that formula (7) is true for every x (0 < x < c) when f{x) and its derivative f' (x) are sectionally continuous in the interval (0, c). These conditions are narrower, but perhaps more practical for us, than those stated in the theorem. Expansion (7) is usually called the Fourier-Bessel expansion; but when X ? - represents the roots of equation (5), the expansion is sometimes referred to as Dini’s. Other expansion formulas in terms of the Bessel functions J n are known. There is, for instance, an integral representation of an arbitrary function which corresponds to the Fourier integral representation. Suppose the interval /0, c) is replaced by some interval (a, b) in the Sturm-Liouville problem with Bessel's equation, where (a, b) does not contain the point x = 0. Then a boundary condition is required at each end point x = a and x = b, and the problem is no longer a singular case, but an ordinary special case, of the Sturm-Liouville problem. Hence the expansion in this case will be another one in series of Bessel functions; but here the functions of the second kind may be involved together with the functions J n . PROBLEMS 1. Expand the function fix) = 1, when 0 < x < c, in series of the functions Jq(X/:c), where X,- are the positive roots of the equation An,. <»<*<'>' 2. In the expansion of fix) = 1 (0 < x < c) in series of Jo(X/c), where X,- are the non-negative roots of «/J(Xc) = 0, show that A,- = 0 when; = 2, 3, * • • , and Ai — 1. Sec. 68] BESSEL FUNCTIONS AND APPLICATIONS 165 3. Expand the function fix) = 1 when 0 < x < 1, f(x) = 0 when 1 < x < 2, /Cl) = in series of J 0 (X,-a;) where X,- are the roots of J o(2X) = 0. Ans. ^ 2 2 Xj[Ji(2X,)] 2 (° < x < 2 )- 3 = 1 4. Expand f(x) = x (0 < x < 1) in series of J i(K,<c), where X,- are the positive roots of Ji(K) = 0. Also note the function represented by the series in the interval —1 < x ^ 0. Ans . x — 2 ^ ^i(X,a;)/[Xj/ 2 (X J -)] (“1 < a; < 1). i = i 68. Temperatures in an Infinite Cylinder. Let the convex sur- face r = c of an infinitely long solid cylinder, or a finite cylinder with insulated bases, be kept at temperature zero. If the initial temperature is a function /(r), of distance from the axis only, the temperature at any time t will be a function u(r , t ). This function is to be found. The heat equation in cylindrical coordinates, and the boundary conditions, are /i\ du ( d 2 U 1 dlA ^ (1) = (0 = r < c ’ * > °)> (2) u(c - 0, t) = 0 (< > 0), ( 3 ) + 0 ) = f(r) (0 < r < c ). It will be supposed that /(?•) and f'(r) are sectionally continuous in the interval (0, c) and, for convenience, that f{r) is defined to have the value l[f(r -f- 0) +/(r — 0)] at each point r where it is discontinuous. Particular solutions of equation (1) can be found by separation of variables. The function u = R(r)T(t) is a solution, provided that is, if Since the member on the left is a function of t alone, and that on the right is a function of r alone, they must be equal to a constant; say, —X 2 . Hence we have the equations rR" + R' + XbR = 0 , T + k\*T = 0 . 166 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 68 The equation in R here is Bessel’s equation (Sec. 65), in which n = 0. If the function RT is to satisfy the condition (2), then R(r ) must satisfy the condition S(c) = 0. According to Theorem 4, there are only real values of the param- eter X in Bessel’s equation for which a solution exists and satisfies this condition, and the positive values alone yield all possible solutions. We are supposing that the function R(r) and its derivative of the first order are continuous functions when 0 ^ r ^ c. It can be shown that the second fundamental solution, R = Fo(Xr), of Bessel’s equation, or Bessel’s function of the second kind, is infinite when r = 0. Therefore the only functions R(r) which satisfy the required conditions are Jo(X,r), where X, are the positive roots of the equation (4) Jq(\c) = 0. The only particular solutions u = RT of the heat equation (1) which satisfy the homogeneous boundary condition (2) are therefore (except for a constant factor) u = Jo(\jr)e~ kXiH , where X, are the positive roots of equation (4). A series of these solutions, (5) u(r, 0=5) AjJ o(X ? r)e~ fcx »% i= i will formally satisfy the heat equation (1) and the condition (2); it will also satisfy the initial condition (3) provided the coefficients Aj can be determined so that f(r) = j? AjJ o (\r) (0 < r < c). j = 1 This is true, according to the Fourier-Bcssel expansion, if (6) A * = cWXjC)V Jo r/(r)/o(v) dr (i = 1, 2 , • • • ). The formal solution of the boundary value problem is there- fore represented by series (5) with coefficients (6), where X 3 - are the positive roots of equation (4). That is, our solution can 167 Sec. 68] BESSEL FUNCTIONS AND APPLICATIONS be written u( r> 0 = | 2 j = l [JiO*)]' i e~ k ^ H r f f(r')Jo(\jr') dr*. This result can be fully established as a solution of the boundary value problem stated here, by following the method used in Sec. 46. For it can be shown that the numbers l/[X 3 *JJ(X 3 c)] are bounded for all the roots X 3 -.* Consequently the numbers Aj/\j are bounded for all j (j = 1, 2, • • ■ ), because f(r) and Jo(\r) are bounded. Hence for each positive number to, the absolute values of the terms of series (5) are less than the constant terms M\j exp ( — kXfto) for all r (0 ^ r ^ c) and all t (t ^ to), where M is a constant. The series of these constant terms converges, since X/4-1 — X 3 approaches t as j increases. Series (5) therefore converges uniformly when t > 0, and so the function u(r, t) represented by it is continuous with respect to r when r = c. But u(c , t ) is clearly zero; hence condition (2) is satisfied. Since the derivatives of /o(X 3 r) arc also bounded, it follows in just the same way that the differentiated series converge uniformly when t > 0, and hence that result (5) satisfies the heat equation (1). Finally, owing to the convergence of series (5) when t = 0, Abel's test applies to show that u(r, +0) = u(r, 0), when 0 < r < c; hence the condition (3) is satisfied. To determine conditions under which our solution is unique, we should need information about the uniform convergence of the Fourier-Bessel expansion. This matter is beyond the scope of our introductory treatment. PROBLEMS 1. Write the solution of the above problem when the initial tempera- ture f(r) is a constant A, and c = 1. Give the approximate numerical values of the first few coefficients in the series. Ans. u = 2vl[O.SO/ 0 (2 4r)e- 6 - % ‘ - .53 / 0 (5.5r)e- 30 ** + .43*/o(8.6r)e _7r,fci -•••]. * This can he seen, for instance, from the asymptotic formulas for \j and J\(x) developed in Ref. 1. 168 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 69 2. Over a long solid cylinder of radius 1 at temperature A throughout is tightly fitted a long hollow cylinder of the same material, with thick- ness 1 and temperature B throughout. The outer surface of the latter is then kept at temperature 1 5. • Find the temperatures in the composite cylinder of radius 2 so formed. This is a heat problem in shrunk fittings. (Note that it becomes a case of the problem in this section when B is subtracted from all temperatures.) 00 Am. u(r, t) = B + — 2 - ^ where X,, X 2 , * • • are the positive roots of J r 0 (2X) = 0. 3. Derive the formula for the potential in a cylindrical space bounded by the surfaces r = c, z = 0, and z = b, when the first two surfaces are kept at potential zero and the third at potential V = f(r). 00 Ans. V(r, z) = ^ A,J 0 (X J r)(sinh Xyz/sinh X,6), where X/ are the 3=1 positive roots of equation (4), and the coefficients A f are given by equation (6). 4. Derive the formula for the steady temperatures u(r, z) in the solid cylinder bounded by the surfaces r = 1, z = 0, and 2=1, when the first surface is kept at temperature u = 0, the last at u = 1, and the surface z = 0 is insulated. 69. Radiation at the Surface of the Cylinder. Let the surface of the infinite cylinder of the last section, instead of being kept at temperature zero, undergo heat transfer into surroundings at temperature zero, according to Newton’s law. The flux of heat through the surface r = c is then proportional to the temperature of the surface; that is, —K = Eu when r = c. Hr J where K is the conductivity of the material in the cylinder and E is the external conductivity, or emissivity. Let us write h = cE/K. The boundary value problem for the temperature u(r, t) can be written as follows: ( 1 ) ( 2 ) ( 3 ) = k(— co <■ dt \dr 2 r dr ) du(c — 0, t) , . N c — ^ ; = -hu(c - 0, i) u(r, +0) = f(r) r < c, t > 0), (t > 0 ), (0 < r < c ). 169 Sec. 69] BESSEL FUNCTIONS AND APPLICATIONS The particular solution of equation (1) found before, u = Jo(Xr)e~ AXa *, will satisfy condition (2) provided X is any root X,* of the equation d c^/°(Xr) = — hJo(\r ) when r = c , that is, of the equation (4) Xc/' 0 (Xc) = —hJ Q (\c). Hence the solution of the problem (l)-(3) can be written (5) u(r, t) = V A 3 M\r)e~W where X, are the positive roots of equation (4), and where, accord- ing to Theorem 5, ox? C c Aj ' = (X?c 2 + /i 2 )[Jo(X,-c)p Jo r/o(X,T ^ (r) dr Cj - i, 2, * * ‘ )• If h = 0, then Xi = 0, and the first term of the series in formula (5) is the constant 4i, where ^ J^ r/(r) dr. This is the case if the surface r = c is thermally insulated. PROBLEMS 1. Find the steady temperatures u(r, z) in a solid cylinder bounded by the surfaces r — 1, z — 0, and z = L if the first surface is insulated, the second kept at temperature zero, and the last at temperature u = f(r). Ans. u = ~ f r'/(r') dr' Jo >/p (X jV ) s inh (Xyg) [</ o(X,)] 2 sinh (X y L) r'/ o(X,*r')/(r') dr', where X 2 , X 3 , * * • are the positive roots of Ji(X) - 0. 2. Find the steady temperatures in a semi-infinite cylinder bounded by the surfaces r = 1 (z ^ 0) and z = 0, if there is surface transfer of heat at r = 1 into surroundings at temperature zero, and the base z = 0 is kept at temperature u — 1. 170 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70 3. Show that the answer to Prob. 1 reduces to u = Az/L when f(r) is a constant A. 70. The Vibration of a Circular Membrane. A membrane, stretched over a fixed circular frame r = c in the plane z = 0, is given an initial displacement z = f(r, p) and released from rest in that position. Its displacement z(r, p, t), where (r, p, z) are cylindrical coordinates, will be found as the continuous solution of the following boundary value problem: m * _ JdH ,1^,1^ ( ' i ' a \<9r 2 + r + r 2 d<p *)’ (2) z(c, <p, t ) = 0, (3) --- gf - - ' = 0, *(r, ft 0) = /(r, *). The function z = R(r)$(p)T(t ) satisfies equation (1) if where —X 2 is any constant, according to the usual argument. Hence T — cos (a\t) if the first of conditions (3) is to be satis- fied. Also R and <£ must satisfy the equations I (r'R" + rR ') + XV 2 = ~ = where m 2 is any constant, since the member on the left cannot vary with either p or r. Hence $ = A cos fji<p + B sin up. But z must be a periodic function of p with period 2t; hence H = n (n = 0, 1, 2, • • • ). The equation in B then becomes Bessel's equation with the parameter X, r 2 R" + rR' + (X 2 r 2 - n 2 )R = 0, and so R = J*(Xr). The solution 2 = R$T will satisfy condi- tion (2) if X is any of the roots X n; * of the equation (4) J n (\c) =0 (n = 0, 1, 2, - • * ). Therefore if A n ; and B n3 - are constants, the functions J n (^njr)(A n j cos up + B n / sin np) cos ( a\ nj t ) Sec. 70] BESSEL FUNCTIONS AND APPLICATIONS 171 are solutions of (1) which satisfy all but the last of conditions (3). The function (5) z{r, <p,t) 00 00 = X X cos rap + B n j sin rap) cos (a\ nj t) n= 0 j = 1 satisfies this last condition also, provided the coefficients are such that (6) /(r, <p) = j £ J COS ritp -f - ^ BnjJ n (\ n jV) J Sin when — 7r < <p ^ t, O^r^c. For each fixed r, the right-hand member of equation (6) is the Fourier series for /(r, <p), in the interval — tt < <p < tt, provided the coefficients of cos n<p and sin rap are the Fourier coefficients; that is, if 1 f 71 " - f( r > <p) cos rap dtp (n = 1, 2, • • * ), VJ-ir d(p (n = 0), i r ,r - I f(r, (p) sin n<p dtp n J-i r (n = 1, 2, • • • )• But the left-hand member of equation (7) is a series of Bessel functions which must represent the function of r on the right when 0 S r ^ c. It is the Fourier-Besscl expansion of that function, provided (9) A n j = — ----- *-- 7 — . 7T C“[tf «+j(X n yC ( 10 ) Aoj = S' — via I rJ n (\ nj r) dr j f(r, <p) cos rup dip AnjC)\ z Jo J-r (n = 1,2, ■■■ ), L-^ f\j,M dr <,) d r . X J 7rc 2 [./i(\(i;c)]' 2 Jo . Similarly, according to equation (8), 2 C c (11) B n j — -7— v I tJ rt(X n y/’) ch 7rc 2 [j n+1 (x n yc)j- Jo J fir, ip) si sin nip dtp. 172 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70 The displacement is therefore given by formula (5) when the coefficients have the values given by equations (9), (10), and (11), and \nj are the positive roots of equation (4), provided series (5) has the necessary properties of convergence, differenti- ability, and continuity. PROBLEMS 1. Derive the formula for the displacement of the above membrane if the initial displacement is f(r), a function of r only. Also show that when/(r) = AJ 0 (X*r), where X* is a root of J Q (\c) = 0, the displacement of the membrane is periodic in time, so that the membrane gives a musical note. Ans. z(r - t] = 1 2 «/o(X,-r) cos ( a\jt ) lUhc) ]* s: r'J 0 (\jr')f(r') dr', where X,- are the positive roots of J 0 (Xc) = 0. 2. Find the displacements in the above membrane if at t = 0 every point within the boundary of the membrane has the velocity dz/dt = 1 in the position 2 = 0. This is the case if the membrane and its frame are moving as a rigid body with unit velocity and the frame is suddenly brought to rest. Ans. z(r, t) = 00 2 sin dk,i XjJifoc) Jo (X 3 r), where X,- are the positive roots of / 0 (Xc) = 0. 3. Derive the following formula for the temperatures in a solid cylinder with insulated bases, if the initial temperature is u = f(r , <p), and the surface r = c is kept at temperature zero: u(r, <p,t) = 2 j X /»(Xnir)(A n ,* cos rup + B ni sin n<p)e- k ^ n i t , n—Q j=l where A ni and B nj have the values given by equations (9), (10), and (11), and X nj - are the positive roots of equation (4). 4. Derive the formula for the temperature u(r , 2 , t) in a solid cylinder of radius c and altitude L whose entire surface is kept at temperature zero and whose initial temperature is A, a constant. Show that the formula can be written u(r, 2 , t) = Aui{z, t)u 2 (r , t), where U\{z , t) is the temperature in a slab whose faces 2 = 0 and 2 = L are kept at temperature ui = 0, and whose initial temperature is Ui = 1; while u 2 (r, t ) is the temperature in an infinite cylinder whose surface Sue. 70] BESSEL FUNCTIONS AND APPLICATIONS 173 r = c is kept at «2 = 0, and whose initial temperature is w 2 = 1. That is, e -k\,k r /-v \ ° • 5. Derive the following formula for the temperatures in an infinitely long right-angled cylindrical wedge bounded by the surface r = c and the planes <p = 0 and <p = tt/2, when these three surfaces are kept at temperature zero and the initial temperature is u = /(r, <p) : u(r, cp, t) = XX A n jJ 2 n(X TlJ r) sin (: 2n<p)e - n = 1 j= 1 where X«, are the positive roots of J 2 n(Xc) = 0, and A n f are given by the formula irc 2 [J r 2 »+i(X n /c)] 2 A»/ = 8 jT sin 2mp dtp Jj r/ 2n (X n yr)/(r, <p) dr. 6. If the planes of the wedge in Prob. 5 are = 0 and <p — <p 0 , show that the formula for the temperature will in general involve Bessel functions of nonintegral orders. Derive the formula for u(r , <p , t) in this case. 7. Solve Prob. 5 if the planes <p = 0 and <p = 7t/2 are insulated, instead of being kept at temperature zero. When f(r, ip) — 1, show that your formula reduces to = 2 ^ /o(Xjr c X y Ji(X y « T ^ 0 » where Xy are the positive roots of ,/ 0 (Xc) = 0; thus w is independent of the angle (p . 8. Solve Prob. 5 if all three surfaces r = c, = 0, and y? = 7 t/ 2, are insulated instead of being kept at temperature zero. 9. Let w(r, 0 be the temperature in a thin circular plate whose edge, r = 1, is kept at temperature u — 0, and whose initial temperature is u = 1, when there is surface heat transfer from the circular faces to surroundings at temperature zero. The heat equation can then be written du d*u 1 du , Tt = d? + ~rd7~ hu ’ 174 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70 where h is a positive constant. Derive the following formula for u(r, t): u = ^e~ ht 00 2 ■i = 1 / o(X,t) X^xCX,) e~ x * 2< , vhere X 3 - are the positive roots of J o(\) = 0. 10 . Solve Prob. 9 if there is also surface heat transfer at the edge r = 1, instead of a fixed temperature there, so that — —hou when r = 1. 11 . Derive the following formula for the potential 7(r, z) in the semi- infinite cylindrical space, r ^ 1, 2 ^ 0, if the surface r = 1 is kept at potential V = 0, and the base 2 = 0 at 7 = 1 : V = 2 2 /0 M X,/ x(X/) j = i where \ 3 are the positive roots of / 0 (X) = 0. 12 . Let 7(r, 2 ) be the potential in the space inside the cylinder r = c, when the surface r = c is kept at the potential V — f(z), where the given function f(z) is defined for all real 2 . Derive the following formula for V(r, 2 ): r ■ * J. " ^ I". £§ m ^ *' where i = *\/— T. 13 . Let 7(r, 2 ) be the potential in the semi-infinite cylindrical space r ^ 1, 2 ^ 0, if dV/dz = 0 on the surface 2 = 0 and if, on the surface r = 1, V = 1 when 0 < 2 < 1, while 7 = 0 when z > 1. Show that V{r, z) Joiiar) . — t < cos az sin a da . aJ oyia) REFERENCES 1. Watson, G. N.: “ A Treatise on the Theory of Bessel Functions,” 1922. 2. Bowman, F.: “Introduction to Bessel Functions,” 1938. 3. Gray, A., G. B. Mathews, and T. M. MacRobert: “A Treatise on Bessel Functions and Their Applications to Physics,” 1931. 4. Jahnke, E., and F. Emde: “Tables of Functions,” 1933. CHAPTER IX LEGENDRE POLYNOMIALS AND APPLICATIONS 71. Derivation of the Legendre Polynomials. Any solution of the differential equation (1) (1 - z 2 ) -2x^ + n(n + 1 )y = 0, known as Legendre’s equation , is called a Legendre function . Later on we shall see how this equation arises in the process of obtaining particular solutions of Laplace’s equation in spherical coordinates, when x is written for cos 6. We shall consider here only the cases in which the parameter ?i is zero or a positive integer. To find a solution which can be represented by a power series, if any such exist, we substitute 00 (2) V = X into equation (1) and determine the coefficients a,-. The substi- tution gives 00 X Ll’O' — l)a,x»' -2 (l - x 2 ) — 2ja i x i + n(n + 1 )a j x i ] = 0 3 =0 or (3) X + 1) ~ j(j + 1 + j(j ~ l)fl,-x 3 '- 2 } = 0. Since this must be an identity in x if our scries is to be a solution, the coefficient of each power of x in series (3) must vanish. Setting the total coefficient of x 3 in this scries equal to zero gives (j + 2)(j + + [n(n + 1) — j(j + 1 )]dj = 0 (j = 0, 1, 2, • • * ), 175 176 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 71 which is a recursion formula giving each coefficient in terms of the second one preceding it, except for do and ai. It can be written (4) 2 — (n - j)(n + j ± 1) O' + 1 )(j + 2) a,- U = 0 , 1 , 2 , )■ The power series (2) is therefore a solution of Legendre’s equation within its interval of convergence, provided its coeffi- cients satisfy relation (4); this leaves a Q and cu as arbitrary constants. But since n is an integer, it follows from relation (4) that dn +2 = 0, and consequently dn-j-4 — &n4-6 — * — 0. Also, when a 0 = 0, then a 2 = a 4 = * * * =0; and when a 1 = 0, then a 3 = a 5 = • • • =0. Hence if n is odd and a Q is taken as zero, the series reduces to a polynomial of degree n containing only odd powers of x. If n is even and a\ is set equal to zero, the series reduces to a polynomial of degree n containing only even powers of x. So there is always a polynomial solution of equation (1), and for it no question of convergence arises. These polynomials can be written explicitly in descending powers of x whether n is even or odd. All the nonvanishing coefficients can be written in terms of a n by means of recursion formula (4); thus d>n — -2 n(n — 1) '2(2 n - 1) a n > (n — 2 )(n — 3) a "- 4 “ 4(2 n- 3) an_!! _ n(n — 1 ){n — 2 ){n — 3) 2 • 4(2n - 1)(2 n - 3) and so on. Hence the polynomial (5) y = & n X n — ^ ~ 0 a-n-2 2(2 n - 1) . n(n — l)(n — 2 )(n — 3) + 2 • 4(2« — l)(2n — 3) x n-i _ 1 is a particular solution of Legendre’s equation. Sho. 72] LEGENDRE POLYNOMIALS AND APPLICATIONS 177 Here the coefficient a n is arbitrary. It turns out to be con- venient to give it the value „ _ (2 n- 1)(2 n - 3) ~ do = 1 . 3 • 1 if n = 1, 2, With this choice of a n functions (5) are known as the LeQcndTs ; polynomials : (6) P n (x) = - 1 ) (2n - 3) • • • 1 r _ n(n - 1) _ 2 n\ L 2(2n - 1) x . ~ 1)(« ~ 2 )(n - 3) _ 4 2 • 4(2 n - 1)(2 n - 3) The function P n (x) is a polynomial in x of degree n, containing only even powers of x if n is even and only odd powers if n is odd. It is therefore an even or odd function according as n is even or odd; that is, Pn(-x) = (-1)-P n (*). The first few polynomials are as follows: Po(x) = 1, P l ( x ) = a:, P 2 (x) = - 1), p *( x ) = i(&* 9 ~ 3x), P 4 ( x ) = %(35x 4 - 30x 2 + 3), Pb(x) = 1 ( 63^ 6 - 70a; 3 + 15x). PROBLEMS 1. Show that formula ((>) for P n (x) can be written in the following compact form: (~l)*(2n ~ 2 j)l 2 n jKn - j)\(n - 2j ) ! where m = n/2 if n is even, and m = (n — 1 ) /2 if n is odd. 2. With the aid of the formula in Prob. 1, chow that A„(0) = (-l) n ^L=(— ,)n P 2 n-l(0) = 0, 3 ■ 5 • • • (2n - 1) 2-A (2n) ’ (» = 1, 2, • • • ). 72. Other Legendre Functions. When n is a positive integer or zero, wc obtained the solution y = P n (x) of Legendre’s equa- tion by setting one of the two arbitrary constants a 0 or a, in the 178 FOURIER SERIES AND BOUNDARY PROBLEMS [Sac. 72 series solution equal to zero. If these constants are left arbitrary, it is easily seen that the general solution of Legendre’s equation can be written (1) y = AP n (*) + BQ n (x), where A and B are arbitrary constants. The functions Q n (x) here, called Legendre’s functions of the second kind, are defined by the following series when \x\ <1: Q n (x) = ai x- ^ ^ - "V if n is even; and (» - 1 )(» + 2 ) 3! (re - 1)(» - 3)(» + 2)(» + 4) . i — ■* - • Q »(*) = ao [ 1 n ( n + 1) _« , n(n - 2) (to + l)(n + 3) „ 2! _l 4! x if n is odd; and where ]■ «i = (-1) ao = ( — 1) 1 2-4 • • • n 1 ■ 3 • 5 • • • (re - 1)’ 2 - 4 • • • (re - 1) 1 • 3 • 5 • • • re Of course P»(x) is a solution for all x. But when |x| > 1, the above series for Q n (x) do not converge. To obtain a second fundamental solution in that case, a series of descending powers of x is used. The following solution so obtained is taken as the definition of Q n (x ) when \x\ > 1 : Qn(x) = re! 1-3-5 (2re + 1) 2 — »— 1 + ( W + ] )^ n + 2 ) x -n-i ^ 2(2re + 3) * I (n + l)(n + 2)(re + 3)(re + 4) ' r 2 -4(2re +3)(2re + 5) + Both P n (x) and Q„(x) are special cases of the function known as the hypergeometric function. When re is not an integer, the two fundamental solutions of Legendre’s equation can be written as infinite series. These Sec. 73] LEGENDRE POLYNOMIALS AND APPLICATIONS 179 are both power series when \x\ < 1; but when |x| > 1, they are series in descending powers of x . Of these various Legendre functions the polynomials P n (x) are by far the most important. Let us now continue with the study and application of those polynomials. 73. Generating Functions for P n (x ). If — 1 g x ^ 1, the function (1 - 2xz + z : and its derivatives of all orders with respect to z exist when \z\ < 1. For these functions are infinite only when 1 - 2xz + z 2 = 0, that is, if z — x ± '\fx 1 — 1 = cos 6 + i sin 6, where we have written cos 6 for x. But this shows that |z| = 1. It is shown in the theory of functions of complex variables that such regular functions of z arc always represented by their Maclaurin series within the region of regularity (|z| < 1, in this case). It will now be shown that the coefficients of the powers of z in that series representation of the above function are the Legendre polynomials; that is, when — 1 ^ x ^ 1 and \z\ < 1, (1) (1 - 2xz + z 2 )~* = Pq(x) + Pi(x)z + Pi(x)z ' 2 + • • ■ + P n {x)z n + • • ■ . To find the coefficients, it is best to write the expansion by means of the binomial series: [1 - z(2x - «)]-* = 1 + 2x - z) + z*{2x - zY + + - (2 m - 1) 2"n\ z'*( 2x — z) n + The terms in z n come from (.ho form containing z"(2x — z) n and preceding terms, so that the total coefficient of z n is 1 -3 (2m - 1 2’*n ! + 1 - 1 • *3 -- (2.r)'‘ - (2?i - 3) (n - 1) 3 2“ -1 (n - 1)! (2m - 5) (n - 2 )(n - 3) 2’- 2 (m - 2) 2! 1 ! (2x)“- 4 - (2x) n ~ 180 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 73 This can be written 1-3 ■ ( 2 n - 1 ) n\ ’ „ _ "(» ~ 1) „«-2 * 2(2 n - 1) , n(n - l)(n - 2) (ft - 3) 4 " r 2 • 4(2» - 1)(2» - 3) But this is P n (x) ; hence relation (1) is established. Incidentally, this shows the reason for the value assigned to a n in our defi- nition of P n (x) (Sec. 71). For 2 = 1, expansion (1) becomes i>n(l)s* = (1 ~ z ] )“ 1 = or 0 Consequently Pn(l) = 1 (n = o, 1, 2, Likewise, putting 2 = 0 gives • * )• = (i+ * 2 )-* = i - 1 * 2 + H ~ + (-i) n 1-3 (2 n — 1) 2-4 and therefore (2n) + P 2 „(0) = (-1) P 2n— 1(0) ' = 0 (2 n — 1) (n - 1, 2, ■ • • By differentiating equation (1) with respect to z and multiply- ing the resulting equation by (1 — 2 xz + z 2 ), the following identity in z is found: 00 (x - z)( 1 - 2 xz + z 2 ) -5 = (x — z) Vp„(r)z» 0 00 = (1 — 222 + 2 2 ) V nP n (x)z n ~\ 0 Equating the coefficients of 2 n in the last two expressions, it follows that (n + l)P n+ i( 2 ) - (2 n + l) 2 P n ( 2 ) + nP n _ 1 ( 2 ) = 0 (n = 1, 2, • ■ • this is a recursion formula for P n ( 2 ). It is valid for all values of 2 . Sec. 74] LEGENDRE POLYNOMIALS AND APPLICATIONS 181 The result of integrating polynomial (6), Sec. 71, n times from 0 to x is (2 n - l)(2n - 3) ( 2n)\ — x 2n — nx ■2n-2 . »(» ~ 1 ) r 2n — 1 . 2! and the expression in brackets differs from ( x 1 2 — l) n only by a polynomial of degree less than n. By differentiating n times, then, it follows that Pn(x) = JLJ1 {X 2- x)» 2 n n ! dx n x J This is Rodrigues 1 formula for the Legendre polynomials. PROBLEMS 1. Show that the derivatives of Legendre polynomials have the properties PL( 0 ) = 0 ; PL+M = ( — '• ( 2 ” 2 w ) 1) - The latter can be found by differentiating equation (1) with respect to x and setting x = 0. 2. Carry out the details of the derivation of Rodrigues’ formula. 3 . Using Rodrigues’ formula, show that K + i(a) - PL iW = (2 n -1- 1 )P n (x) (n = 1, 2, • ■ • ). 4 . Using the formula in Prob. 3, obtain the integration formula ^ P n (x) dx = 2n \ ^»+iWl (ft = J j 2, * • * ). 74. The Legendre Coefficients. When —1 ^ x ^ 1, we have just shown that P n (:r) is the coefficient of z n in the expansion of the generating function (1 — 2 xz + z - 2 )“* in powers of z. When x = cos 6 = %(c i0 + c"' 16 )) this generating function can be written [1 — z(e i0 + e~ i0 ) + 2 2 ]~i = (1 — ze ie )~*( 1 — and therefore as the product of the scries 1 + i«j« + 1 2-4 2«2 id z z e 1 • 3 • • • (2n - 1) 2 - 4 • • • (2 n) + • • • 182 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 74 and the series 1-8 • ■ • ( 2 .- 1 ) ^ 2 • 4 • • ■ (2 n) 2 6 The coefficient of z n in this product is 1 • 3 ■ • • (2n - 1) 2 • 4 • - • (2 n) (e in> + e~ in6 ) + \ 2^1 (e<(B_2>9 + e ~ i{n -™) + hence this is P n (cos 0 ). Thus we have the following formula for this function: (1) P n ( cos 0) = ^!2^ L C ° S + 1 • (2ra — 1) C0S (n “ W + i-2(2n-lKL-3) C0S (n~W + ■ ■ ■ + where the final term T n is the term containing cos 0 if n is odd; but it is half the constant term indicated if n is even. These functions are called the Legendre coefficients. Tables of their numerical values will be found in some of the more extensive books of mathematical tables, or in Ref. 3 at the end of this chapter. According to formula (1), the first few functions are P 0 (cos 0) = 1, Pi (cos 0) = cos 6, P 2 ( cos 0) = i(S cos 20 + 1), P 3 (cos 6) = |(5 cos 30 + 3 cos 0), p 4 (cos 0) = -g^(35 cos 40 + 20 cos 20 + 9). The coefficients of the cosines in formula (1) are all positive. Consequently P n (cos 0) has its greatest value when 0 = 0. Since P n (l) = 1, it follows that P n ( cos 0) ^ 1. Also, each cosine is greater than or equal to -1, so that P n ( cos 0)^-1. That is, the Legendre coefficients are uniformly bounded as follows: |Pn(cos 0)| ^ 1 (n = 0, 1, 2, • • • ), for all real values of 0. Sec. 75] LEGENDRE POLYNOMIALS AND APPLICATIONS 1S3 75. The Orthogonality of P n (x). Norms. Legendre’s equa- tion can be written in the form ( 1 ) d_ dx (1 + 1 ) 2 / = 0 . It is clearly a special case of the Sturnx-Liouville equation (Sec. 24), in which the parameter X has been assigned the values (2) X = n(n + 1) (n = 0, 1, 2, • • • ). In this case r(x) = 1 — x 2 , q(x) = 0, and the weight function p(x) = 1. Since r(x) = 0 when x = ±1, no boundary conditions need accompany the differential equation to form the Sturm-Liouville problem on the interval ( — 1,1). It is only required that the characteristic functions and their first ordered derivatives be continuous when — 1 ^ x ^ 1. But the polynomials P n (x) are solutions of equation (1), and, of course, they have these required continuity properties. The Legendre polynomials P n (x) are therefore the characteristic functions of the Sturm-Liouville problem here, corresponding to the characteristic numbers (2). According to Sec. 25, then, the functions P n (x) form an orthogonal set in the interval ( — 1, 1), with respect to the weight f unction p(x) — 1 ; that is, (.3) Tj Pm(r)Pn(r) dx = 0 if m ^ n ( m , n = 0, 1, 2, • • ■ Furthermore, there can be no characteristic functions of the Sturm-Liouville problem here which correspond to complex values of the parameter X, because p(x) does not change sign. We shall soon see that the functions P n (x) and the numbers (2) are the only possible characteristic functions and numbers of the problem. To find the norm of P n (x), that is, the value of the integral in (3) when m = n, a simple method consists first of squaring both members of equation (1), Sec. 73, to obtain the formula (1 — 2 xz + s 2 )" 1 = [X • We now integrate both members here with respect to x over the interval (—1, 1) and observe that the product terms on the right 184 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 75 vanish in view of the orthogonality property (3). Thus, dx 1 — 2 xz + z 2 P [Pn{x)YdX 0 j — l ( 1*1 < !)• The integral omthe left has the value - i log (1 - 2 xz + z 2 ) -l 1 . 1+2 "I 10 *— * - 2 ( 1+ I + I + ’ " + 5^TT + " ') <W<1) - By equating the coefficients of z 2n in the last two series, we have the following formula for the norm of P n (%) : (4) JP [Pn(x)] 2 dx = (» - o, l, 2, • • • ). The orthonormal set of functions here in the interval ( — 1, 1) is therefore {<p n (x)} ) where <Pn(x) = Vn + i Pn(x) (n - 0, 1, 2, • • - ). Since [P^Oc)] 2 and the product P m (x)P n (x ), in which m and n are both even or both odd, are even functions of .r, it follows from formulas (4) and (3) that the polynomials of even degree, (5) \Z2rT+lP n (x) (ft = 0,2,4, ■ ■ • \ form an orthonormal set of functions in the interval (0, 1) ; and the same is true for the polynomials of odd degree , represented by (5) when ft = 1, 3, 5, • * * . PROBLEMS 1. Establish the orthogonality property (3) by using Rodrigues’ formula for P n (x) and successive integration by parts. 2. State why it is true that dx = 0 (ft - 1, 2, 3, ■ ■ • 3. Use the method of Prob. 1 to obtain formula (4) for the norm of F»(s), Sec. 76] LEGENDRE POLYNOMIALS AND APPLICATIONS 185 76. The Functions P n (x) as a Complete Orthogonal Set. Let. us now prove the following theorem: Theorem 1. In the interval (-1, 1) the orthogonal set of func- tions consisting of all the Legendre polynomials P n (x) (n = 0, 1, 2, • ■ • ) is complete with respect to the class of all functions which , together with their derivatives of the first order , are sectionally continuous in (-1, 1). We are to prove that if \p(x) is a function of this class which is orthogonal to each of the functions P n (x), then \p(x) = 0 except at a finite number of points in the interval. Let us suppose, then, that (1) f\Pn(x)\ls(x) dx = 0 (n = 0, 1, 2, • ■ • ). According to our recursion formula (Sec. 73), (2n + 1 )xP n (x) = (n + l)P n+ i(x) + nP n ^(x) (n = 1 , 2 , ■ • * ); and this formula can be replaced by the formula xP 0 (x) = P Y (x) when n = 0. When vve multiply its terms by and integrate from —1 to 1, the integrals in the right-hand member vanish, so that (2) P n (x)xt(x) dx = 0 (n = 0, 1, 2, • • • ). If we suppose that the orthogonality property (1) is true when ^(.r) then* is replaced by x k \p(x), the method just employed clearly shows that, property (1) is true when xf^(x) is replaced by x k+ '\f/(x). In view of equation (2), then, we conclude by induction that, for every integer j, f'j l > n(-r).rty(x) dx = 0 (n, j = 0, 1 , 2, ■ ■ ■ ). As a consequence, wo have Pntx)t(x) ( mirx )- (nnrx ) 4 2! H 4! dx = 0; because the power series in the brackets, representing cos mwx , is uniformly convergent. Moreover, the series obtained by multi- 186 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 76 plying all terms of this series by the sectionally continuous func- tion P n (x)\p(x) is also uniformly convergent, so it can be integrated termwise. Thus J l 1 P n (x)\p(x) cos mTX dx = 0 (m = 0, 1, 2, • * * ); and in just the same manner it follows that j* 1 l Pn(z)ft(z) sin mTxdx = 0 (m = 1 , 2 , • ■ * ). All the coefficients in the Fourier series corresponding to the function P n (x)\p(x ) in the interval ( — 1, 1) therefore vanish. But this function and its first derivative are sectionally con- tinuous; hence it is represented by its Fourier series except at the points of discontinuity of the function. Except possibly at a finite number of points, then, i(x) =0 (-1 S s S 1), and the theorem is proved. There is an interesting consequence of the above theorem. Suppose that for some real value of A other than n(n + 1), Legendre’s equation ( 3 ) d dx ( 1 -*’>g] + x 1 ,-0 has a solution y = y<,(x), where y' 0 ( x) is continuous in the interval - 1 fk x ^ 1 . Then, according to Sec. 25, y 0 (x) is orthogonal to all the characteristic functions P n (x) already found and cor- responding to X = n(n + 1). But this is impossible according to Theorem 1, unless j/ 0 = 0. Since we have already shown that X must bo real if equation (3) is to have such a regular solution, we have the following result: Theorem 2. The only values of X for which the Legendre equa- tion (3) can have a non-zero solution with a continuous derivative of the first order, in the interval — 1 ^ x g 1, are \ = n(n + 1) (n = 0, 1, 2, • • ■ ). It can be shown that the Legendre functions of the second kind, Qn(x), which also satisfy equation (3) when X = n(n + 1), become infinite at x = ±1. Consequently, the polynomials P n (x) Sec. 77] LEGENDRE POLYNOMIALS AND APPLICATIONS 187 are , except for constant factors, the only solutions of equation (3) which have continuous first derivatives in the interval — l^x<l. 77. The Expansion of x m . Without the use of a general expan- sion theorem, we can easily show how every integral power of x and therefore every polynomial, can be expanded in a finite series of the polynomials P n (x). It will be clear that these important expansions are valid for all x, not just for the values of x in the interval (—1, 1). According to its definition, the polynomial P m (x) has the form (1) Pm(x) = ax m + bx 7> '- 2 + cx m - 4 + * • * , where a, 6, • • • are constants depending on the integer m. Therefore X m = ]-P m (x) - - T'"~“ - -*»-•* - • • - a a a That is, every integral power x m of x can be written as a constant times P m (x ) plus a polynomial in x of degree m — 2. Applying this rule to x m ~ - in the last equation, we see that x m is a linear combination of P m (x), P n ^(x), and a polynomial of degree m - 4. Continuing in this way, and noting that only the alternate exponents m, m - 2, m - 4, • • ■ appear in the poly- nomials here, it is clear that there is a finite series for x m of the following form : (2) x m = A„P m (x) + A„ 1 _ 2 P„,_ 2 (a:) + • • • -f T, where the final term T is a constant A „ if m is oven, and T = A,P,(a 0 if m is odd. To find the value of any coefficient A wo multiply all terms of equation (2) by P,„_ 2 ,(t) and integrate over the interval ( — 1,1), In view of the orthogonality of the functions P„(x), this give's f i v( x ) dec = A m -»j [Pra-s/fr)]* dx. But the integrand on the left is an even function of x for every integer m; and the integral on the right, the norm of P m _ 2j (x), has the value 2 /(2m - 4j + I). Therefore, (3) A„,- 2 j = (2m — 4j + 1) x m P m - 2 j(x) dx. 188 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 77 We shall develop here the following integration formula , valid for every real number r greater than — 1 : (4) (x) dx = r(r — 1) (r — n + 2) (r + n + 1) (r + n — 1) • ■ • (r — n + 3) (n = 2, 3, ■ • • In view of this formula, the integral in equation (3) has the value m(m - 1) • ■ • (2j + 2) (2m — 2j + l)(2m — 2j — 1) (2j + 3) ml 1-3-5 • (2m - 2j + 1)2 ijl The values of the coefficients A m - % are therefore determined, and we can write expansion (2) for any integral power of x as follows : xm = 1 • 3 • 5 • • • (2m + 1 ) } 2m + + ( 2 m - 3 ) Pm _ 2 ( x ) + (2m - 7) gg- + ‘><f> ~ » P_, (l ) +...]. For the first few values of m, we have 1 = Po(*), X = P l(x) 7 X 2 = •§p 2 (#) + 'S'Pn(^), * 3 = | -Pz(x) + fPi(z), Z 4 = + fP 2 (z) + *Pn(s). Derivation of Formula (4). To obtain the integration formula (4), let us first observe that, in view of expression (1), (5) x'P n ( x ) dx -£ (ax r+n + bx r+n ~ 2 + + ) dx + r+w + 1 1 r n — 1 ' r + n — 3 as long as r > — 1. The last member can be written Sir) , (r + n + 1 ) (r + n — 1) (r + n — 3) ■ ■ • + ( 6 ) Sec. 78] LEGENDRE POLYNOMIALS AND APPLICATIONS 189 where fir) = a(r + n — l)(r + n — 3) • ■ ■ + b(r f n + l)(r + n — 3) •••+•■■ . We can see that/(r) is a polynomial in r of degree nf 2 or (n — 1) / 2 , according as n is even or odd. Now the product x n ~ 2j P n (x ) is an even function of x for every n, when j = 1, 2, • • ■ ; so it is evident from equation (2) that jf 1 x n-up n (x) dx = 0 (j = 1, 2, • • • ; n ^ 2j). Therefore our integral (5) vanishes when r is replaced by n — 2, n — 4, ft — 6, etc., down to zero or unity, and so does the poly- nomial f(r). Also, the coefficient of the highest power of r in fir) is a + b + c + * * * , which is P n (l) or unity. Hence, when n = 2, 3, • ■ ■ , the factors of fir) can be shown as follows: f(r) = (r — n + 2)(r — n + 4) • • • r, if n is even; and f(r) = (r - n + 2)(r - n + 4) • • • (r - 1), if n is odd. In either case the fraction (6) can be written as r(r — l)(r — 2) * • - (r — n + 2) (r + + l)(r + n — 1) • ~ (r — n + 3) (n = 2, 3, ■ • • ; r > -1). This is the value of our integral; hence formula (4) is established. 78. Derivatives of the Polynomials. The derivative P r n {x) is a polynomial of degree n — 1 containing alternate powers of x , namely, a*'* -1 , x n ~ [ \ • • • . It can therefore be written as a finite series of Legendre polynomials: = A n_./Vl(>) + An-lPn-zix) + ’ # • To find the coefficient, Aj (j — n — 1, w — 3, • • • ), we multi})ly all terms by Pj(x) and integrate; thus A i = 2j -z 1 f x FMKW dx. When integrated by parts, the integral here becomes \pMPJx) 1 - f 1 P.^P^x) dx, L J i %) l 190 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 78 and this last integral vanishes because Pj(x) is a linear combina- tion of the polynomials P,_ i(x), P/_ 3 (rr), etc., each of which is of lower degree than P n (x). Therefore A l = %±±[ i -P 3 .(-l)P n (-l)] = ^i[l - (-1)*"] = 2j + l, since j + n = 2n — 1, 2n — 3, • * * . Consequently we have the following expansion, valid for all x: (1) P' n (x) = (2n - ^Pn-tCa:) + (2ft - 5)P n _ 3 (z) + (2n — 9)P n— c(^r) + • • • , ending with 3Pi(z) if n is even, and with P 0 (x) if n is odd. . When — 1 g x ^ 1, we have seen that \P n (x)\ ^ 1; hence for these values of x it follows from expansion (1) that \Pin( x )\ = (4ft ” 1) + (4ft — 5) + * * * + 3 = n(2n + 1); and similarly, l^ 2 n+l(^)| (ft + l)(2n + 1). Therefore, \P 2n(. x ) | = (2ft) 2 , iP^n+iWI = (2ft + l) 2 ; that is, for all x in the interval — 1 g x g 1, ( 2 ) l^nWI ^ ft 2 ( n = 0, 1, 2, • • • ). Differentiating both members of expansion (1) and noting that ^ ft 2 , |P^_ 3 (3)| ^ ft 2 , etc., we see by the method used above that (3) \P"(x)\ ^ n 4 (-1 g x ^ 1, n = 0, 1, 2, • • • ). Similarly, for derivatives of higher order, | P^(x)\ ^ n~ k . Let us collect our properties on the order of magnitude of the Legendre coefficients and their derivatives as follows: Theorem 3. For all x in the interval — 1 g x g 1, and for 71 = 1, 2, 3, • ■ ■ , the values of the functions i p "Wi> ^ «(*)!, ^ \p’:(x)\, • • • can never exceed unity . Sec. 79] LEGENDRE POLYNOMIALS AND APPLICATIONS 191 79. An Expansion Theorem. The normalized Legendre poly- nomials were found in Sec. 75 to be <Pn{x) = y/n + ^Pn(x) (n = 0, 1, 2, • • * ). The Fourier constants, corresponding to the orthonormal set here, for a function f(x) defined in the interval ( — 1, 1), are Cn = dx = \/n + % ^ f(x)P n (x) dx . The generalized Fourier series corresponding to f{x) is therefore oo oo 2 C ^ X) = 2 Pn ^ dx'. This can be written W XAJ> n (x), 0 where (2) An = 1 j_J(x)P n (x) dx (n = 0, 1, 2, • • • ). The series (1) with the coefficients (2) is called Legendre's series corresponding to the function /(z). It was shown above that if f(x) is any polynomial, this series contains only a finite number of terms and represents f(x) for all values of x. It can be shown that, when -1 < x < 1, Legendre's series converges to f(x ) under any of the conditions given earlier for the representation of this function by its Fourier series. We now state explicitly a fairly general theorem on such expansions, and accept it without proof for the purposes of the present volume.* Theorem 4. Let f(x) be bounded and mtcgrable in the interval ( 1, 1). I hen at each point , x ( — 1 < x <! 1) which is interior to an interval in which f(x) is of bounded variation , the Legendre series corresponding to f(x) converges to \[f(x + 0) + f(x — 0)]; that TrSy (3) i[f(x + 0) +S(x - 0)] = V AJP n (x) ( - 1 < a < 1) 0 where the coefficients A n are given by formula (2). * The theorem stated here is a special ease of a theorem proved in Chap. VII of Kef. 1. The proof is lengthy and involves more advanced concepts than wc employ in this book. 192 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 79 In particular, if f(x) is sectionally continuous in the interval ( — 1, 1), and if its derivative f'(x) is sectionally continuous in every interval interior to ( — 1, 1), then expansion (3) is valid whenever — 1 < x < 1. For it can be shown that the condi- tions in the theorem are then satisfied at all points. If /(re) is an even function, the product f(x)P n (x) is even or odd according as n is an even or odd integer. Hence A n = 0 if n is odd, and (4) A n = (2 to + 1) f f(x)PJx) dx (to = 0, 2, 4, • • • ); so that expansion (3) becomes (5) Mrta + 0) +/(*-<>)] = ^4.o + AJPzix) + AJP±(x) + * * ■ , where the coefficients are defined by formula (4). Similarly, if f(x) is an odd function , the expansion becomes (6) Mf(a + 0) + f(x - 0)] = AiPi(x) + AzP%(x) + A’tJPzix) + * ’ * j where (7) An = (2 TO + 1) jT 1 f(x)P n (x) dx (to = 1, 3, 5, ■ • • ). In the interval (0, 1), either one of the expansions (5) or (6) can be used , provided of course that f{x) satisfies the conditions of the theorem in that interval. For if f(x) is defined only in (0, 1), it can be defined in ( — 1, 0) so as to make it either even or odd in ( — 1, 1). It was pointed out earlier (Sec. 75) that the polynomials Pz n {x), and the polynomials P 2 n-\(x),. appearing in expansions (5) and (6), respectively, form two sets of orthogonal functions on the interval (0, 1). When x = cos 6 , expansion (3) can of course be written oo F(6 ) = 2) AnPn(C0S e) (0 < 9 < r) 0 at points where F($) is continuous, where A n = — F(9)P n ( cos 9) sin 9 dB (n = 0, 1, 2, - • ■ PROBLEMS 1. If f(x) = 0 when —1 < x <0, f(x) = 1 when 0 < x <1, and ./(0) = i, obtain the following expansion for f(x) when — 1 < x < 1 : Sbc. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 193 /(*) 2 2 [Ps»(0) — P 2 » + 2 ( 0 )]P 2n+1 (a:) 0 1 3 ~2 + 4 x + ^i ( _1 )” 4ra + 3 1 • 3 • • • (2n - 1) _ in + 4 2 • 4 • • • (2 n) p ^+i(x). Suggestion: See Prob. 4, Sec. 73. 2. If /(*) = 0 when -1 < x g 0, and f(x) = x when 0 show that, when —1 < x < 1, x < 1, fix) = gi + ip,(*) + | P 2 (x) - , 13 - 4 ! „ , , + 2 7 ■ 4121 ^ W - • • • . 3. Expand the function /(a) = *, when 0 g x < 1, in series of Legendre polynomials of even order, in the interval (0, 1). 80. The Potential about a Spherical Surface. Let a spherical surface be kept at a fixed distribution of electric potential V - F(6), where r, <p, d are spherical coordinates with the origin at the center of the sphere. The potential at all points in the space, assumed to be free of charges, interior to and exterior to the surface is to be determined. It will clearly be independent of <p; hence it must satisfy the following case of Laplace’s equation in spherical coordinates: ( 1 ) (rV) + — J sm g so = o. The potential V{r , 6) will also be required to be continuous, together with its second-order derivatives, in every region not containing a point ol the surface, and to vanish at points infinitely far from the {surface. The boundary conditions are therefore ( 2 ) Iim V(r, 6) = b\6) (O<0<tt), T — >C * ' where c is the radius of the spherical surface, and (3) lini V(r, 0) — 0. r — > oo Particular solutions of equation (1) can be found by the usual method. Setting V = R(r)0(6), equation (1) becomes 194 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 80 - — (rJt\ = 1 d f de\ Rdr 2 ^ 6 sin 9 d9 \ Sln 9 dd ) Both members here must be equal to a constant, say X; hence we have the equations d 2 r~(rR)=\R, 1 d / l — cos 2 9 de \ _ sin 9dd\ sin 6 d9 ) “ °‘ The first of these is Cauchy’s linear equation, r 2 R" + 2 rR' - \R = 0 , which can be reduced to one with constant coefficients by substi- tuting r = e‘. Its general solution is R = + Br -i~VY+i' Writing — + v/X + -J- = n, so that X = n(n + 1), we have R(r) = Ar» + A, where n is any constant. Writing x for cos 0, the equation in 9 becomes, in terms of the new parameter n, !r] +n(n + !)e = 0, which is Legendre’s equation. We have seen that the solution of this equation can be continuous, together with its first ordered derivative, in the interval — 1 g x ^ 1, or 0 ^ 0 ^ tt, only if n is an integer. The solutions are then the Legendre poly- nomials, which have continuous derivatives of all orders. Hence and n = 0, 1, 2, • 9 = P n (x) = P n { cos 0). Thus two sets of particular solutions RQ of Laplace's equation (1) have been determined: (4) r n P n ( cos0); r- n “lP„(cos 0) in = 0 , 1 , 2 , ■ ■ • ). Sec. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 195 In the first set the functions and their derivatives of all orders with respect to r or 0 are continuous in every finite region; and in the second set they are continuous in every region, finite or infinite, not containing the origin. Then at points inside the sphere the function 00 6) = cos 0) (r < c) (T satisfies (1) and (2) formally, provided J3„ can be determined so that oo /(cos 0) = Yj BnC n Pn( COS 0) (0 < 0 < t), 0 where /(cos 0) = F(0). This is the expansion of the last section, provided B n c 11 are the coefficients A n given there; that is, if B n = ^ /(cos 0)P«( cos 0) sin 0 dd. Hence for points inside the sphere, the solution of the problem can be written (5) V(r, 0) = ^ V f 1 dx 0 J -1 [r g c; F(0) = /(cos 0)]. For points exterior to th(% sphere, the functions of the second set in (4) satisfy condition (3), and the solution can be written OO (6) V(r, 0) = ^ ^ / J «(cos 0) (r 1 c), 0 where (7) An = — J" ^ f(x)P n (z 0 da;, since the series in (6) then reduc.es to /(cos 0) when r = c. The Solution Established. To show that our formal solution does satisfy all the conditions of the problem, we use the same method hero as in earlier problems (for example, Sec. 46). We shall suppose that the given function F(0) and its derivative F'(0) are sectionally continuous in the interval (0, r). Then 196 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 80 f{x) is sectionally continuous in the interval (“1, 1), and so is fix), in every interval interior to ( — 1, 1). Now consider the function V(r, 0) represented by formula (5). When r = c, the series there converges to f(x) if —1 < x < 1. But the sequence of functions (r/c) n (n = 0, 1, 2, * - • ) is bounded, and monotone with respect to n; hence according to Abel's test the series is uniformly convergent with respect to r (0 ^ r ^ c) for each fixed x ( — 1 < x < 1). Therefore V(c — 0, 0) = F(c, 0), and so condition (2) is satisfied. The terms of the series in equation (5) can be written as the product of the three factors AJn ) P n ( cos 0), and n(r/c) n . Since the first two factors are bounded for all r, 0, and n (» = 1 , 2 , • • ■ ), and since the series whose general term is the last factor con- verges when r < c, the series in equation (5) is uniformly con- vergent when r < c. But the series of terms w*(r/c) n , for each fixed positive k , also converges when r < c; and since n~ 2 P^(x) and vr 4 P"(x) are uniformly bounded (Theorem 3), it follows easily that the series in equation (5) can be differentiated term- wise twice with respect to r and with respect to 0, when r < c. The individual terms of that series satisfy Laplace's equation (1); hence our function V(r, 0) satisfies that equation. Also, V(r, 0) and its derivatives are continuous when r < c. This establishes our solution when r < c. When r > c, solution (6) can be proved valid in the same manner. If, as a periodic function of the angle 0, F(9) is supposed continuous and F'(6) sectionally continuous, it is also possible to show that the above solutions are the only possible solutions satisfying certain regularity conditions, essentially that V(r, 0) be continuous at r = c (see Sec. 58). PROBLEMS 1. If the potential is a constant Vo on the spherical surface of radius c, show that V — Vo at all interior points, and V = 7oc/r at each exterior point. 2. Find the steady temperatures at points within a solid sphere of unit radius if one hemisphere of its surface is kept at temperature zero and the other at temperature unity; that is, /(cos 0) = 0 when ?r/2 < 0 < 7t, and /(cos 0) = 1 when 0 < 0 < tt/2. Ans. u(r, 0) = \ + f r cos 0 — | \r z P z (cos 0) + Hi i r 6 P 6 (cos 0) - • • • . Sec. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 197 3. Find the steady temperatures u(r, 0) in a solid sphere of unit radius if u = C cos 0 on the surface. Arts, u = Cr cos 0. 4. Find the potential V in the infinite region r > c, 0 ^ 0 ^ 7 t/ 2, if F = 0 on the plane portion of the boundary (0 — tt/2, r > c) and at r = oo, and F = /(cos 0) on the hemispherical portion of the boundary (r = c, 0 ^ 0 ^ tt/2). 00 4ns. 7(r, 0) - 2J (4n + 3) (c/r)^P 2n+l ( C0S 5) dx. 5. Find the steady temperatures u(r, 0) in a solid hemisphere of radius c whose convex surface is kept at temperature u = /(cos 0), if the base is insulated; that is, 1 du n A tt rdQ =0 when 6 = 2 Also write the result when /(cos 0) = 1. 00 Ans. u(r, 0) = X ( 4n + l)(r/c) 27 *P 2 n(cos 0) f }(x)P 2n (x) dx. o *' u 6. Find the steady temperatures in a solid hemisphere of unit radius if its convex surface is kept at temperature unity and its base at tem- perature zero. 7. Show that the steady temperature u(r, 0) in a hollow sphere with its inner surface r = a kept at temperature n = /(cos 0), and its outer surface r — b at u = 0, is where , u ~ H — |*2n+l fr2n \-l _ a 2n |-1 P»(cos 0), An 2 n + 2 f(x)P n (x) dx. 8. If u(x, l) represents the temperature in a nonhomogeneous bar witli ends at x = —1 and x = 1, in which the thermal conductivity is proportional to 1 — x\ and if the lateral surface of the bar is insulated, the heat equation has the form du c U = b Ox <'->£ where b is a constant, provided the thermal coefficient c5 is constant (Sec. 9). The ends x — ±1 arc? also insulated because the conductivity vanishes there. If u = f(x) when t = 0, derive the following formula 198 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 81 for u(x, t) : 00 r n(n+l)t I = — 9* When the initial temperature function in Prob. 8 is (a) f(x) — x 2 , (b) /(®) = £ 3 , show that the solution reduces to the following formulas, respectively: (a) u = | + K&c 2 — l)e _6b *, ( b ) u = + iPz(x)e~ 1%t . 81. The Gravitational Potential Due to a Circular Plate. Another type of application of Legendre polynomials to the solution of boundary value problems will be illustrated by the following problem: Find the gravitational potential due to a thin homogeneous circular plate, or disk, of mass 5 per unit area and radius c. Let the center of the disk be taken as the origin and the axis as the 2 -axis, 0 = 0, where r, <p } 6 are spherical coordinates. The potential is a function F(r, 6) independent of <p; hence it satisfies the following form of Laplace’s equation: (1) r ^ {rV)+ ^ele( sined I) = 0 ’ except at points in the disk. Its value at points on the positive axis 0=0 can be found from the definition of potential by a simple integration; thus C c 27 rxd . V(r, 0) = k J o ^ j= = = dx = 2 wkS(V? + c 2 - r), where k is the gravitational constant in the definition of poten- tial. Then F(r, 0) must be symmetric with respect to the origin and satisfy the following boundary condition in the space 0 ^ Q < tt/ 2, r > 0: (2) V(r, 0)=^ ( - r), where M is the mass of the disk. Two solutions of equation (1) were found in the last section, namely, Sue. 81] LEGENDRE POLYNOMIALS AND APPLICATIONS 199 ( 3 ) ( 4 ) V = V a„r n P n ( cos 6), T b n 1 P„(cos 0). The coefficients a n , 6 n are now to be determined, if possible, so that boundary condition (2) is satisfied. But when 0 = 0, Pn( cos 0) = 1 and the series in (3) and (4) become power series in r and in reciprocals of r, respectively. Now the binomial expansions Vr 2 + c 2 = c (l + Vr 2 + c' 1 = r ^1 + 1 r 4 1-3 r« 2 . 4 a 4 1 2 • 4 • 6 c 6 (0 g r < c) 1 c 4 1-3 c« 2 • 4 r 4 r 2 • 4 • 6 -) (r > c). are absolutely convergent in the indicated intervals, and con- vergent when r = c. Hence boundary condition (2) can be written ( 5 ) 7(r, 0) = 2 Mk / r lr 2 1 r 4 c \ c 2 c 2 2 • 4 c 4 1 * 3 \ + 2 " • 4 • 6 c” ~ ■ ■ ■ ) when 0 < r g c; 2 Mk flc 1 c* 1-3 c 5 \ c \2? 2-4r* + 2-4-6r» ) when r jjg c. The series in (3) will then satisfy (5) for r < c if its coefficients are identified with those of the first series in (5) ; thus a 0 = 2Mk/c f ai = —2 Mk/c 2 , etc. Similarly, for the case r > c the series in (4) can be used if its coefficients are taken as those in the second series in (5), namely, b 0 — Mk , b i = 0, etc. Hence the solution of the boundary value problem (l)-(2) can be written as follows, when 0 ^ 0 < 7 t/ 2: V(r, 0) = 2 Mk - 1 1 r - Pi ( cos 6) + \ ^ P»( cos e ) 2 • 4 c 7* P 4 ( COR (9) + 2 C 2 1-3 r« 2 • 4 • 6 c' Pe( cos 0) 200 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 81 if 0 < r < c; and V(r, 8) = 2 Mk c 1 c 2 r 2T4 ^ p 2( cos 9 ) + 1-3 c 2-4-6 r -g P 4 (cos 6) if r > c. When r c, the convergence of the series here follows from the absolute convergence of the series in (5) and the fact that fPn(eos.fl)l ^ 1. PROBLEMS 1. Derive the following formula for the gravitational potential due to a mass M distributed uniformly over the circumference of a circle of radius 1, when 0 ^ 8 S tt: V(r, 8) = hM [l — ^ r 2 P 2 (cos 8) + r 4 P 4 ( cos 8) — • • • J, if 0 ^ r < 1; and V(r, 8) = kM [i lP 2 (cos 8) 1 ■ 3P 4 (cos 8) 2 r 3 ' 2 • 4 7 ' if r > 1. 2. Find the gravitational potential, at external points, due to a solid sphere, taking the unit of mass as the mass of the sphere, and the unit of length as the radius, if the density of the sphere is numerically equal to the distance from the diametral plane 8 = tt/2. An,. + 1 p4(cos 8 ) 6-8 r b 1 • 3 P fi (cos 8) 6 - 8-10 3. Find the gravitational potential, at external points, due to a hollow sphere of mass M and radii a and b, if the density is proportional to the distance from the diametral plane 6 = t/2. 4. The points along the 2-axis, 8 = 0 or 8 = t, in an infinite solid are kept at temperature u = Ce~ r \ Find the steady temperature u(r, 8) at all points. Ans. u(r, 8) = <7^ (-l)»(r 2 »/w!)7 > 2 „(cos 8). 0 6. The surface 8 = t/3, r ^ 0, of an infinitely long solid cone is kept at temperature u = Ce~'. Find the steady temperatures u(r, 8) in the « Ans. u(r, 6) = C g (- l)»[r»P„(cos 8)]/[n\P n (i)]. 0 cone. Sec. 81] LEGENDRE POLYNOMIALS AND APPLICATIONS 201 6. Solve Prob. 5 if the surface temperature is u(r f tt/3 ) = C/r m , where m is a fixed positive integer. REFERENCES 1. Hobson, E. W.: “The Theory of Spherical and Ellipsoidal Harmonics,” 1931. 2. Whittaker, E. T., and G. N. Watson: “Modern Analysis,” 1927. 3. Byerly, W. E.: “Fourier’s Series and Spherical Harmonics,” 1893. Appendix. INDEX A Abel's test, 127 Absolute convergence, 83 Approximation in mean, 40 B Beam, displacements in, 24 simply supported, 125 Bessel functions, of first kind, 48, 143-174 applications of, 105-174 boundedness of, 152 differentiation of, 148 expansions in, 162 integral forms of, 149 of integral orders, 145 norms of, 161 orthogonality of, 157 recursion formulas for, 148 tables of, 156n. zeros of, 1 53 of second kind, 147 Bessel’s equation, 143, 157, 166, 170 Bessel's inequality, 41, 84 Bessel's integral form, 151 Beta function, 149 Boundary value problem, 6, 94-142, 165-174, 193-201 complete statement of, 107, 133, 138 linear, 24 units in, 100 (See also Solutions of boundary problems) C Characteristic functions, 47 Characteristic numbers, 47 Characteristic numbers, of Leg- endre's equation, 186 Closed systems, 42 of trigonometric functions, 74, 85, 88 Complete systems, 40, 42 of Legendre polynomials, 185 of trigonometric functions, 74 Conduction, equation of, 19 Conductivity, 15, 168 Convergence in mean, 42 Cylindrical coordinates, 13 Cylindrical functions ( see Bessel functions) D Derivative from right, 65 Differential equation, linear, 2 linear homogeneous, 2 (See also Partial differential equation) Differentiation, of Fourier series, 78 of series, 5, 106, 140, 196 Diffusion, equation of, 19 Diffusivity, 19 Dini's expansion, 164 Dirichlct’s integrals, 67 E Electrical potential, 13 (See also Potential) Even function, 57 Expansion, Dini’s, 164 Fourier-Bessel, 162 in Fourier integrals, 88-93 in Fourier series, 53-88 in Legendre's series, 187, 190, 191 in series of characteristic func- tions, 48, 49 203 204 FOURIER SERIES AND BOUNDARY PROBLEMS Exponential form, of Fourier inte- gral, 92 of Fourier series, 63 Exponential functions, 31, 45, 46 F Flux of heat, 15 Formal solutions, 94 Fourier-Bessel expansions, 162 Fourier coefficients, 53, 70 properties of, 76, 77, 80, 85, 86 Fourier constants, 40-42 Fourier cosine integral, 91 Fourier cosine series, 57, 62, 74 Fourier integral, 88 applications of, 120-123 convergence of, 89 forms of, 91 Fourier integral theorem, 89 Fourier series, 53-88 convergence of, 70, 86 absolute, 83 uniform, 82, 86 differentiation of, 78 forms of, 61 generalized, 39 integration of, 80, 86 in two variables, 116 Fourier sine integral, 91 application of, 122 Fourier sine series, 28, 29, 57, 62, 73 in two variables, 118 Fourier theorem, 70, 86 Function space, 38, 74 Fundamental interval, 38 G Gamma function, 145, 149 Generating function, for Bessel functions, 147 for Legendre polynomials, 179 Gibbs phenomenon, 86 Gravitational force, 10-12 Gravitational potential, 10 (See also Potential) Green's theorem, 18, 131, 134 H Heat equation, 17, 19 Heat transfer, surface, 110 I Infinite bar, temperatures in, 120 Infinite series of solutions, 5 Inner product, 34, 38 Integration, of Bessel functions, 149 of Fourier series, 80, 86 of Legendre polynomials, 181, 188 L Lagrange's identity, 32, 71 Laplace’s equation, 12, 20 Laplacian operator, 12, 14, 15 Least squares, 41 Lebesgue integral, 43, 86 Left-hand derivative, 65-67 Legendre coefficients, 181 boundedness of, 182 Legendre functions, 175 of second kind, 178 Legendre polynomials, 175-193 applications of, 193-201 bounds of, 190 complete set of, 185 derivation of, 175 derivatives of, 189 expansions in, 187, 190, 191 generating functions for, 179 integration of, 181, 188 norms of, 183 orthogonality of, 183 recursion formula for, 180 Legendre’s equation, 175, 183, 194 characteristic numbers of, 186 general solution of, 178 Legendre’s series, 191, 192 Limit in mean, 42 Linear differential equation, 2 Lommel’s integral form, 150 INDEX 205 M Membrane ( see Vibrating, mem- brane) Monotone sequence, 128, 139 N Newton’s law, 110, 168 Norm, 34, 38 O Odd function, 57 One-sided derivative, 65-67 Orthogonal functions, 29, 38 generated by differential equa- tions, 46 Orthogonal sets, 34-52 Orthogonality, 34, 37, 44, 49 of Bessel functions, 157 of characteristic functions, 49 of exponential functions, 45, 46, 62 Hermitian, 45 of Legendre polynomials, 183 of trigonometric functions, 29, 39, 40, 53 with weight function, 44 Orthonormal sets, 35, 38 of Bessel functions, 161 of Legendre functions, 184 of trigonometric functions, 54, 74 P Parseval relation, 85, 86 Parseval’s theorem, 43, 86 Partial differential equation, 2 for beam, 24 of conduction, 19 general solution of, 3 Laplace’s, 12 linear homogeneous, 2 for membrane, 23 nonliomogcneous, 100 for string, 21 types of, 24 Periodic boundary conditions, 48 Periodic extension, 96 Periodicity of function, 55 Piecewise continuous function, 65 Plucked string, 28, 98 Poisson’s equation, 13 Polar coordinates, 14 Potential, electric, in cylindrical region, 126, 141, 174 equation of, 13 between parallel planes, 115, 116, 123 in quadrant, 124 about spherical surface, 193 in square, 137 gravitational, definition of, 10 due to disk, 198 due to hollow sphere, 200 due to ring, 200 due to sphere, 200 (See also Temperature, steady) R Radiation, 110, 168, 173, 174 Right-hand derivative, 65-67 Rodrigues’ formula, 181, 184 S Schwarz inequality, 83 Sectionally continuous functions, 64 Semi-infinite bar, temperatures in, 122 Shaft, twist in, 124 Solutions of boundary problems, 94 approximate, 97 closed form of, 96, 116, 126 established, 96, 105, 133, 141, 167, 195 superposition of, 99 uniqueness of, 105, 127—142 Sources, heat, 20, 111 Spherical coordinates, 13 String (see Vibrating string) Sturm-Liouvillc equation, 47 Sturm-Liouvillc problem, 47-52 in Bessel’s equation, 160 in Legendre’s equation, 183 Superposition of solutions, 3, 99 206 FOURIER SERIES AND BOUNDARY PROBLEMS T Tchebichef polynomials, 44 Telegraph equation, 23 Temperature, steady, in cone, 200, 201 in cylinder, 168, 169 in cylindrical wedge, 126 in hemisphere, 197 in hollow sphere, 197 in infinite solid, 200 in sphere, 196, 197 variable, in bar, 104, 120-123, 197, 198 in circular plate, 173, 174 in cube, 119 in cylinder, 165, 168, 172 in cylindrical wedge, 173 in hollow sphere, 113 in infinite solid, 120-123, 125 in radiating wire, 110-112 in slab, 102-112 in sphere, 112 in square plate, 119 (See also Potential) Termwise differentiable series, 5 U Uniform convergence, of Fourier series, 82, 86 of series, 105, 127 Uniqueness of solutions, 127-142 for potential, 134, 137 for temperature, 105, 130 Units, selection of, 100 V Vectors, 34^-37 Vibrating membrane, 23 circular, 170, 172 rectangular, 116 Vibrating string, 21 with air resistance, 125 approximating problem of, 98 forced vibrations of, 100-102 problem of, 24, 28, 95-102 W Weierstrass test, 105, 133 Weight function, 44