Boundary value problems

BOUNDARY VALUE PROBLEMS

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Chi Y. Lo Michigan State University, USA

BO UNDARY VALUE PROBLEMS bWorld Scientific

X),

Sin9aPore •NewJerseY • London • Hong Kong

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to Vivien, Karen , and Natalie

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Preface

This book is designed for use in a one-year course in boundary value problems for both beginning graduate students in mathematics and advanced graduate students in other disciplines in engineering and the physical sciences. It deals mainly with the boundary value problems of linear partial differential euqations of second order, in particular, problems relating to the three fundamental equations of mathematical physics, namely, the wave equation, the heat equation, and Laplace's equation. There are two objectives in our approach. The first goal is to obtain a formal solution of a given boundary value problem either by the method of separation of variables for all applicable linear equations or by the method of d'Alembert's solution in the wave equation. This formal solution is, in general, represented either by a series or an integral. In order to obtain such a representation, the theory of Sturm-Liouville problems, Fourier series, and Fourier transforms have been developed. The second goal is to verify that the formal solution is actually a solution of the boundary value problem. The formal solution must satisfy the problem with the required smooth properties, and the resulting solution must be unique and depend continuously on the initial and boundary data. The verification of solutions depends on the concept of uniform convergence of the series or the improper integral. In turn, the uniqueness and stability of the solution depend on the method of the energy integral for the wave equation and the maximum principles for the heat equation and Laplace's equation. The first six chapters of this book cover the boundary value problems of linear partial differential equations in two independent variables with development of the relevant theory. In Chapter 7, we treat the boundary vii

viii

Preface

value problems in three independent variables. In some boundary value problems involving cylindrical and spherical domains, the method of separation of variables leads us to two singular differential equations, namely, Bessel's equation and Legendre's equation. Hence, some important properties of Bessel functions and Legendre polynomials are treated in that chapter. Some of the treatment in this book follows that of Cannon [2], Indritz [8], Tolstov [11], and Weinberger [13]. As a prerequisite for a course based on this book, the student should have completed a course in advanced calculus and an elementary course on ordinary differential equations. The problems at the end of each chapter are chosen to fit the level of the course and, in some cases, supplement the contents of the chapter.

Contents

Preface

vii

Chapter 1 Linear Partial Differential Equations 1 1.1 Linear problems . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Well-posed problems . . . . . . . . . . . . . . . . . . . . . . . . 9 1.4 Method of general solutions . . . . . . . . . . . . . . . . . . . . 12 1.5 Method of separation of variables . . . . . . . . . . . . . . . . . 14 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Chapter 2 The Wave Equation 21 2.1 The vibrating string . . . . . . . . . . . . . . . . . . . . . . . . 21 2.2 The initial value problem . . . . . . . . . . . . . . . . . . . . . 24 2.3 The nonhomogeneous wave equation . . . . . . . . . . . . . . . 29 2.4 Uniqueness of the initial value problem . . . . . . . . . . . . . . 31 2.5 Initial-boundary value problems . . . . . . . . . . . . . . . . . . 33 2.6 Initial-boundary value problems for semi-infinite string . . . . . 39 2.7 Problems . . . . . . . . . . .. .. .. . . . . . .. . . . . . . . 41 Chapter 3 Green ' s Function and Sturm-Liouville Problems 45 3.1 Solutions of second order linear equations . . . . . . . . . . . . 45 3.2 Boundary value problems and Green's function . . . . . . . . . 50 3.3 Sturm-Liouville problems . . . . . . . . . . . . . . . . . . . . . 58 3.4 Convergence in the mean . . . . . . . . . . . . . . . . . . . . . 62 3.5 Integral operator with continuous, symmetric kernel . . . . . . 66 ix

x

Contents

3.6 Completeness of eigenfunctions of Sturm-Liouville problems . . 75 3.7 Nonhomogeneous integral equation . . . . . . . . . . . . . . . . 78 3.8 Further properties of eigenvalues and eigenfunctions . . . . . . 81 3.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Chapter 4 Fourier Series and Fourier Transforms 95 4.1 Trigonometric Fourier series . . . . . . . . . . . . . . . . . . . . 95 4.2 Uniform convergence and completeness . . . . . . . . . . . . . . 99 4.3 Other types of Fourier series . . . . . . . . . . . . . . . . . . . 102 4.4 Application to the wave equation . . . . . . . . . . . . . . . . . 105 4.5 Fourier integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 109 4.6 Fourier transforms . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.7 Contour integration . . . . . . . . . . . . . . . . . . . . . . . . 117 4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Chapter 5 The Heat Equation 127 5.1 Derivation of the heat equation . . . . . . . . . . . . . . . . . . 127 5.2 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . 130 5.3 The initial-boundary value problem . . . . . . . . . . . . . . . 133 5.4 Nonhomongeneous problems and finite Fourier transform . . . 135 5.5 The initial value problem . . . . . . . . . . . . . . . . . . . . . 138 5.6 The initial value problem for the nonhomogeneous equation . . 142 5.7 Nonhomogeneous boundary conditions for initial-boundary value problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 5.8 Problems . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 147 Chapter 6 Laplace's Equation and Poisson 's Equation 151 6.1 Boundary value problems . . . . . . . . . . . . . . . . . . . . . 151 6.2 Green's identities and uniqueness theorems . . . . . . . . . . . 152 6.3 Maximum principle . . . . . . . . . . . . . . . . . . . . . . . . . 154 6.4 Laplace's equation in a rectangle . . . . . . . . . . . . . . . . . 155 6.5 Laplace's equation in a disc . . . . . . . . . . . . . . . . . . . . 158 6.6 Poisson's integral formula . . . . . . . . . . . . . . . . . . . . . 162 6.7 Green's function for Laplace's equation . . . . . . . . . . . . . 166 6.8 Poisson's equation in a disc . . . . . . . . . . . . . . . . . . . . 173 6.9 Finite Fourier transform for Poisson's equation . . . . . . . . . 177 6.10 Dirichlet problem in the upper-half plane . . . . . . . . . . . . 181 6.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

Xi

Contents

Chapter 7 Problems in Higher Dimensions 191 7.1 Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 7.2 Double Fourier series . . . . . . . . . . . . . . . . . . . . . . . . 194 7.3 Laplace's equation in a cube . . . . . . . . . . . . . . . . . . . . 200 7.4 The two-dimensional wave equation in a rectangular domain . 202 7.5 Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 7.6 Singular Sturm-Liouville problem for Bessel's equation . . . . . 209 7.7 The two-dimensional wave equation in a circular domain . . . . 212 7.8 Initial-boundary value problems for the heat equation . . . . . 217 7.9 Legendre's equation . . . . . . . . . . . . . . . . . . . . . . . . 221 7.10 Properties of Legendre polynomials . . . . . . . . . . . . . . . . 224 7.11 Legendre series and boundary value problems . . . . . . . . . . 229 7.12 Laplace's equation in a sphere . . . . . . . . . . . . . . . . . . . 232 7.13 Poisson's integral formula in space . . . . . . . . . . . . . . . . 235 7.14 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 Appendix A Ascoli' s Theorem 243 Appendix B Answers for Selected Problems 245 Bibliography Index

255

256

Chapter 1

Linear Partial Differential Equations

1.1 Linear problems A partial differential equation in two independent variables x and y is an equation which involves an unknown function u(x, y) and its partial derivatives. The order of the equation is the order of the highest derivative of u(x, y) in the equation. Consider a second order partial differential equation of the form L[u] = Auxx + Buxy + Cu, y+ Dux + Euy + Fu = G (1.1) where the coefficients A, B, C, D, E, F, and G are continuous functions of x and y in a given domain S2 in the xy-plane . Let C-(Q) denote the space of all functions which are continuous and possess continuous partial derivatives up to order m in Q. For any u and v in C2 ( Q) and for any scalars a and b, we have L[au + bv] = aL[u] + bL[v] for L[u] in (1.1). Thus, L is a called a linear operator which maps C2 (Q) into CO (9) = C(SZ). A (classical) solution of (1.1) is a function u = u(x, y) in C2(12) which, when substituted in (1.1), turns the equation into identity. (1.1) is homogeneous if G = 0 in S2; otherwise, it is nonhomogeneous. Examples of (1.1) are the following equations: (wave equation) Utt - c2uxx = 0, (heat equation) ut - kuxx = 0, (Laplace's equation)

uxx + uyy = 0, 1

2 Linear Partial Differential Equations

uXX + uyy = -q(x, y), (Poisson's equation) and which are very important in mathematical physics and whose boundary value problems and initial value problems will be studied extensively in the following chapters. Let {ui (x, y) }, i = 1, 2,...n be n solutions of the homegeneous equation L[u] = 0. (1.2) Because of the linearity of the operator L, any linear combination E? 1 aiui is also a solution for any scalars ai, i = 1, 2,...n. This is known as the principle of superposition. Moreover, if u is a solution of (1.2) and v is a solution of (1.1), then w = u + v is also a solution of (1.1). Let {un(x, y)} be a sequence of solutions of (1.2). Then the series cc u(x,y) = E anun (x,y)

(1.3)

n=1

is also a solution of (1.2) if termwise differentiation can be justified. Similarly, let u(x, y, A) be a solution of (1.2) for any A > 0. Then the improper integral U(x, y) = fu(x,Y,A)dA

(1.4)

is also a solution of (1.2) if differentiation under the integral can be carried. To achieve this, we need some theorems on the uniform convergence of series and improper integrals. Definition 1.1 Let {vn(x, y)} be a sequence of functions defined on a rectangle K = [a, b] x [c, d]. Then the series E°° 1 vn(x, y) converges uniformly to a function V(x,y) in K if, for e > 0, there exists a positive integer N = N(e) such that for k > N, IV(x, Y) - En =1 vn(x, y)J < e for all (x, y) in K. Definition 1.2 Let f (x, y, A) be defined for 0 < A < oo and for (x, y) on a rectangle K = [a, b] x [c, d]. Then the improper integral fo f (x, y, A)dA converges uniformly to a function F(x, y) in K if for e > 0, there exists a positive number R = R(c) such that for b > R, I F(x, y )- f b f (x, y, A)dAJ < c for all (x, y) in K. A very useful and the most simple way to determine uniform convergence of series and integrals is given by the following theorems:

Linear problems 3

(Weierstrass M-Test for series) Let {vn (x, y)} be a seTheorem 1 .1 quence of functions defined on K = [a, b] x [c, d] and let E°° 1 Mn be a convergent positive series such that Ivn (x, y)I < Mn for all (x, y) in K and all n. Then the series >°° 1 vn(x, y) converges uniformly and absolutely in

K. (Weierstrass M- Test for integrals) Let f (x, y, A) be defined Theorem 1.2 for 0 < A < oo and for (x, y) in K = [a, b] x [c, d]. If there exists a positive function g(A) in (0, co) such that If (x, y, A) I < g(A) for all (x, y) in K and if f0 g(A)dA converges, then the improper integral f °O f (x, y, A)dA converges uniformly and absolutely in K. To show continuity and differentiablity of the series and integrals, we may apply the following theorems: Theorem 1.3 Let { Vn (x, y ) } be continuous in K = [a, b] x [c, d] and let V(x, Y) _ E°_1 vn (x, y) converge uniformly in K. Then V(x, y) is continuous in K. Furthermore , if, for each n, vn(x, y ) is continuously differentiable in K and if the series En' =l avn/ 8x and E°° 1 8vn /ay converge uniformly in K, then av

00

avn

_Eax n_1 ax

aV y a a=E y n=1 00 avn

and

in K. Theorem 1.4

Let f (x, y, A) be continuous in

S2={(x,y,A) la<x 0 or < 0. a = 0 only if ^x = t;y = 0. Hence a # 0 and (1.5) is reduced to the canonical form uCC + u,,,, + ... = 0. The case that A = 0 can be treated similarly. Example 1.3 Show that the equation L[u] = uxx+yuyy = 0 is hyperbolic for y < 0 and elliptic for y > 0 and transform the equation into canonical forms in these regions. Solution: We note that A = 1, B = 0 and C = y. Thus, B2 - 4AC = -4y. Hence the equation is hyperbolic for y < 0, but elliptic for y > 0. The characteristic equations are

dy/dx = f(-y)1/2. I.y 0. analytic solutions are given by x ± 2iy1/2 = C3 where C3 is a constant. Let ^ = x, and y = 2y1/2. Then a = c = 1, b = 0 , L[^] = 0, and L[i7] = -1/77. The canonical form is ugg + u,7,1 - - u,l = 0. Example 1.4 Classify the equation 2

x2

L[u] = y2u.TX - 2xyuxy + x2uyy - x ux - y uy = 0, X>0' Y>0' and transform it into the canonical form. Solution: We note that A = y2, B = -2xy and C = x2. Then B2 - 4AC = 0. Hence the equation is parabolic. The characteristic equation is dy x dx y' and the characteristic curve is x2 + y2 = C4 where C4 is a constant. Let t ; = x and 77 = x2 + y2. Then we have b = c = 0 and a = y2. Moreover, L[l;] = -y2/x and L[7)] = 0. Thus, 2 y2 u££ - uc = 0

or ^u^E - u£ = 0.

1.3 Well-posed problems All partial differential equations have infinitely many solutions. Our goal is to single out one specific individual solution by imposing auxiliary conditions to the differential equations. These conditions are governed by the

10

Linear Partial Differential Equations

physical problems and are known as initial conditions and boundary conditions. The boundary value problems we are going to study are related to the three fundamental equations: the wave equation, the heat equation, and Laplace's equation. Some typical problems are given here. Problem I (Initial value problem for the wave equation) Determine a solution of the wave equation utt - uXX = 0,

-00<X0, (1.20)

which satisfies the initial conditions u(x, 0) = O(x), ut(x, 0) = 'fi(x),

-00 < x < oo. (1.21)

Problem II (Dirichlet problem for Laplace's equation) Determine a solution of Laplace's equation uXy + uyy = 0, x2 + y2 < 1 ,

( 1.22)

which is equal to q(x, y) on the boundary x2 + y2 = 1. Problem III (Mixed problem for the wave equation) Determine a solution of the wave equation utt - UXs = 0,

0<x0, (1.23)

which satisfies the initial conditions u(x, 0) = 0(x), ut(x, 0) _ fi(x), 0 < x < 1,

(1.24)

and the boundary conditions u(0, t) = u(1, t) = 0, t > 0.

(1.25)

Problem IV (Characteristic initial value problem for the heat equation)

Determine a solution of the heat equation ut-uSX =0,

-oo<x0,

(1.26)

which satisfies the initial condition u(x, 0) = O(x),

-oo < x < oc.

(1.27)

Well-posed problems

11

We will note in the following chapters that Problems I to IV are formulated to describe observable physical phenomena. It is essential to know whether such a problem has a solution and whether the solution is unique. Moreover, the boundary or initial data are measured approximately in a certain range, any small variation of the given data should give rise to a small change in the solution. Any problem in mathematical physics is said to be well-posed if the following requirements are satisfied: (1). Existence: There exists at least one solution. (2). Uniqueness: There is at most one solution. (3). Stability: The unique solution depends continuously on the initial and/or boundary data. We shall show that under appropriate conditions on the initial and boundary data, all the above problems are well-posed. On the other hand, any problem which does not satisfy any one of the above requirements is called an improperly posed problem. For example, consider the initial value problem uxy + u = 0,

(1.28)

u(x,0) = fi( x), uy(x,0 ) = fi(x).

(1.29)

We know y = 0 is characteristic and the differential equation gives a restriction on the initial data, i.e. 0'(x) + O(x) = 0. Hence, there is no solution for the problem (1.28)-(1.29) for arbitrary functions q(x) and b(x). Even though the elliptic equation does not have real characteristic curves, its initial value problem is not well-posed. Consider the following initial value problem for Laplace's equation: (1.30)

uxx + uyy = 0,

u( 0 ,y) = 0 , u^, ( 0 ,y) =

sin ny

(1 . 31)

n ,

where n > 0 is a constant. Its solution is given by u(x , y) =

Binh nx sin ny n2

(1.32)

As n -+ oo, the initial data (1.31) approach 0, whereas its solution (1.32) goes to infinity for x 54 0. Since u = 0 is an obvious solution of (1.30) with vanishing initial data, the stability requirement cannot be satisfied.

12

Linear Partial Differential Equations

Another example of an improperly posed problem is the Dirichlet problem for the wave equation u,y = 0,

0 < x < a, 0 < y < b,

(1.33)

u(x,b ) = g(x),

0 < x < a,

(1.34)

u(0, y) = p(y), u(a,y) = q(y), 0 < y < b.

(1.35)

u(x,0) = f(x),

The general solution for (1.33) is u(x,y) = ¢(x) +0(y). To satisfy (1.34), we have

f(x) = u(x,0) = O(x) + 0(0) and

g(x) = u(x, b) = O(x) + fi(b). This implies that f (x) -g(x) = 0(0) --+/'(b) = f (0) -g(0), a constant. Hence f and g cannot be arbitrarily described , and a similar situation exists for p(y) and q (y) from (1.35). In general, the boundary problems are associated with elliptic equations, while initial value problems and mixed problems are connected with hyperbolic and parabolic equations. In the following chapters, we will study the existence, uniqueness, and stability of appropriate problems for the wave equation, the heat equation, Laplace's equation, and Poisson's equation.

1.4 Method of general solutions The general solution for a given second order partial differential equation is a family of solutions of the equation with two arbitrary functions. A particular solution can be obtained from the general solution by imposing appropriate auxiliary conditions to the equation. We shall illustrate how certain boundary value problems can be solved by the method of general solutions. Consider the characteristic initial value problem of uyy + uy = 1,

(1.36)

Method of general solutions

u(x,0) = 0, u(0,y) = Y.

13

(1.37)

Let v = ux. Then (1.36) is changed to vy+v=1.

(1.38)

This is a first order linear equation in v(x, y) with respect to y, treating x as a parameter. Multiplying (1.38) by the integrating factor ev and integrating with respect to y, we have ux(x,y) = v(x,y) = 1 + f(x)e-y where f (x) is an arbitrary function. Thus, u(x, y) = x + p(x)e-y + q(y) which is the general solution of (1.36) with two arbitrary functions p(x) = f f (x)dx and q(y). Using (1.37), we have 0 = u(x, 0) = x + p(x) + q(0) and y = u(0, y) = p(0)e-y + q(y). This implies that p(x) = -x - q(0), q(y) = y - p(0)e-y, and p(O) = -q(0). Hence, the solution of (1.36)-(1.37) is given by u(x, y) = x - xe-y + y.

(1.39)

Another example is the initial value problem for the wave equation utt - 4uxx = 0, u(x,0) = ex,

-00 < x < oo, ut(x,0) = x2.

(1.40) (1.41)

Since the equation (1.40) is hyperbolic, its characteristic equations are dt/dx = ±1/2. Thus, x ± 2t = C, where C is a constant , are two families of characteristic curves of (1.40). We set 1;=x+2t, 77 =x-2t,

(1.42)

and u(x, t) = v(l;, 77). Then (1.40) is changed to v£,7 = 0, whose general solution is v(l;, r7) = p(^) + q(77), or u(x,t) = p(x + 2t) + q(x - 2t),

(1.43)

where p(r;) and q(i7) are arbitrary functions. Differentiating (1.43) with respect to t, we get ut(x,t) = 2p'(x + 2t) - 2q'(x - 2t),

Linear Partial Differential Equations

14

where the prime denotes differentiation with the argument of the functions. From (1.41), we have p(x) + q(x) = ex, p(x) - q(x) = 6 + K, where K is an integration constant. Thus, x3 ex K ex x3 K P(X) = 12 + 2 + 2 ' q(x) = 2 12 2 ' and the solution of (1.40)-(1.41) is given by _ (x + 2t)3 u(x,t) 12

( x - 2t)3 ex+2t ex-2t 12 + 2 + 2

The method of general solutions is very effective in solving both the initial value problem and the initial-boundary value problem for the wave equation and will be discussed fully in Chapter 2.

1.5 Method of separation of variables The method of separation of variables is used to transform certain linear homogeneous partial equations with homogeneous boundary conditions into two associated linear ordinary differential equations with corresponding boundary conditions. Using the principle of superposition, we form either a series solution or an integral solution to satisfy the nonhomogeneous boundary or initial conditions. Consider the initial-boundary value problem for the wave equation Utt - c2uxx = 0, 0 < x < L, t > 0, (1.44) u(0,t) = u(L,t) = 0, t > 0,

(1.45)

u(x, 0) = f (x), ut(x, 0) = g(x), 0 < x < L,

(1.46)

where f (x) and g(x) are given functions. We look for a particular solution of (1.44) and (1.45) of the form u(x, t) _ A(x)B(t). From (1.44), we get

-c2A"B + AB" = 0.

Method of separation of variables

15

Dividing u = AB from both sides of the above equation in the region in which u does not vanish, we have

A" B" A c2B Since the term on the left is a function of x and the term on the right is a function of t, the common value is a constant. Thus, we have

A" B" A =2B=-A, where A is a separation constant. We are then led to two ordinary differential equations

A"+AA=O, 0<x 0, 0 < 6 < 21r. Define Al/2 = pl/2eie/2 = a + io, where 3 > 0. Then I exp(iA'/2 x) I = exp(-/3x) and Iexp(-iA112x)I = exp(,3x). Both functions

Linear Partial Differential Equations

18

are unbounded in x except when 0 = 0. In this case, A = a2 and A(x) _ C1 cos ax + c2 sin ax, where a is a positive constant.

For A = a2, (1.59) has a solution given by B(t) = exp(-a2kt).

We now form an integral solution of (1.56) defined by u(x, t) =fcos(ax) [a(a) + b(a) sin(ax)] exp(-a2kt)da,

(1.60)

where a(a) and b(a) are arbitrary functions of a. To satisfy (1.57), we have [a(a) cos (ax) + b(a ) sin(ax)]da,

f (x) = u(x, 0) = j

which is called the Fourier integral of f (x) on (-oo, oo). From the theory of Fourier integral which will be studied in Chapter 4, the coefficients a(a) and b(a) are given by 00

J-00 f ( v) cos (av)dv,

a(a) = -

(1.61)

and f00 b(a) = 1

J 0, f (v) sin(av)dv.

(1.62)

(1.60)-(1.62) represent a formal solution of (1.56)-(1.57).

1.6 Problems 1. Show that 00 u(x, t) =

an exp (-kn2t) sin nx n=1

satisfies the heat equation ut - kux^, = 0 in 0 < x < ir, t > 0, if Ian < M for all n where M is a positive constant. 2. Show that u(x, y)f= f ( s)e'5 sin xs ds satisfies Laplace's equation u,1 + uy, = 0 in x > 0, y > 0 if f (x) is continuous and bounded in x > 0.

Problems

19

3. Determine the region in which the following equation is (i) hyperbolic, (ii) parabolic, (iii) elliptic: (a), (x2 - l ) u „ + 2yuxy - uyy + ux + uy - 0, (b). uxx + 2xuxy + yuyy - u — 0. In Problems 4 through 8, classify the equation and find the characteristic curve(s) of the equation through a given point. 4. yuxx + (x + y)uxy + xuyy = 0 at (1,2). 5- yuyy + (x - y)uxy - xuxx = 0 at (1,3). 6. uyy+ 2exuxy + e2xuxx = 0 at (0,1). 7. utt - tuxx = 0 at (0,1). 8. (cos2 x —sin x)utt + (2cosx)uxt + uxx + u — 0 at(7r/2,0). In Problems 9 and 10, use simple integration to obtain the general solution of the equation. 9. uxy = x/y + 2, y > 0. 10. uyy — 3y 2 sinx. In Problems 11 through 13, reduce the equation to an ordinary differential equation and obtain its general solution. 11. x2uxy + xuy = y, x > 0. 12.

13.

Uyy — XUy = X2 ,

X > 0. y

xuxy +uy = 2xe ,

x > 0.

14. Find the general solution of the equation transforming it into the canonical form.

uxx - uyy - ux + uy = 0 by

IT). Explain why the initial value problem uXT = ut, - c o < x < oo, t > 0; M(X,0) = f{x),

u ( (x.O) = g[x),

-oo < x < oo;

is improperly posed. 16. Consider the boundary value problem uu ~ 4«rx = 0, 0 < x < 7 r , 0 < t < IT/2, U = 0 on x = 0, x = IT, t = 0, and t = TT/2. Verify that u(x,t) = sin a; sin 2t is a solution of the above problem, and therefore deduce that this problem is improperly posed. 17. Let u(x,y) be a solution of uxx - uyy = 0 i n x > 0 , y > 0 and satisfy

Linear Partial Differential Equations

20

the initial conditions u(x, 0) = f ( x), uy(x, 0) = g(x), u(0 , y) = F(y), and ux(0, y ) = G(y) where f (0) = F(0), g(0) = F'(0), f'(0) = G(0). Prove that f'(x) + g(x) = F'(x) + G(x) when x > 0. 18. Determine the region in which tuxx + (x + t)uxt + xutt = 0

is hy-

perbolic and transform it to the canonical form. Determine its general solution.

In Problems 19 through 23, transform the equation into the canonical form and find its general solution. 19.

x2uxx - y2uyy - 2xux + 2yuy = 0, x > 0, y > 0.

20.

uxx - 2 sin x uxy - cost x uyy - cos x uy = 0.

21. t2utt + 2xtuxt + x2uxx = 4x2, x > 0, t > 0. 22. t2utt - x2uxx = 0, x > 0, t > 0. 23. x2uxx - y2uyy - 2yuy = 0, x>0, y > 0. In Problems 24 through 30, find a set of nontrivial bounded solutions for the problem by the method of separation of variables. 24. uxx - uyy - guy = 0, 0 < x < Tr, y > 0, ux(0, y) = u(7r, y) _ u(x,0) = 0. 25. u tt - uxx - u = 0, 0 < x < 1, t > 0; ut(x, 0) = u(0, t) = u(l, t) = 0. 26. utt + 2ut - 4uxx + u = 0, 0<x0; u(x,0) = ux(0,t) = U(1, t) = 0. u(0, t) = u(1, t) = 0. 27. Ut - t2uxx - U = 0, 0 < x < 1, t > 0; 28. utt-uxx+2ut-tux+u=0, 0<x0; ut(x,0) = u(0,t) = u(lr,t) = 0. 29. uxx + uyy = 0, -00<X O. 30. utt -uxx = 0, x>0, t>0, u(0, t) = u(x,0) = 0.

Chapter 2

The Wave Equation

2.1 The vibrating string Consider a homogeneous string of constant density p, which lies in its equilibrium position along a straight line taken as the x-axis. The string has a length L and is fastened at both ends, x=0 and x=L. Let the string be displaced from its equilibrium position, and then released. The string begins to vibrate. We assume that the vibrations are small in comparison with the length of the string, and are confined in a vertical plane.

U

x

x+Ox

x

Fig. 2.1 Small segment of vibrating string Let u(x, t) denote the displacement of the string at a given position

21

22

The Wave Equation

x and at time t. At any fixed t, the graph of the function u(x, t) in the xu-plane represents the shape of the string at that instant. Consider an element AB of the string at a fixed t which is acted upon by tension Tl and T2 along the tangent of the string at the points A and B respectively (see Fig 2.1). We assume that the tension is the same at all points of the string in the equilibrium position and does not change during the subsequent vibrations of the string. Hence, Tl and T2 have the same magnitudes as T, but have different directions. Let (x, u(x, t)) and (x + Ax, u(x + Ax, t)) be coordinates of A and B. Then the resultant force from the tension of the string acting on the element AB in the direction of the u-axis is T [sin a2 - sin al] = T [tan a2 - tan all

T [ux (x + Ax, t) - ux (x, t)]

where al and a2 are the acute angles between the direction of Tl and T2 with the horizontal axis at A and B respectively. By Newton's law of motion, we have pLx utt(x, t) = T[ux(x + Ax, t) - ux(x, t)], (2.1) where x 0, we have utt - c2uxx = 0. (2.2) Equation (2.2) is called the wave equation.

If there is an external force such as gravity acting on the string, we rewrite (2.1) as pLx utt (x, t) = T [ux (x + Ox, t) - u(x, t)] + f (x, t) Ax, x < x < x + Ox, where f (x, t) represents the amount of force per unit length of the string. As before, we divide the above equation by Ax, and let Ax -> 0. We get Utt - c2uxx = f (x, t)l p = F(x, t).

(2.3)

( 2.3) is called the nonhomogeneous wave equation. Two initial conditions are prescribed for the wave equation, i.e. u(x,0) = f(x), ut(x,0) = g(x),

(2.4)

where f (x) is the initial displacement and g(x) is the initial velocity of the string for 0 < x < L. Since the string is held fixed at both ends, we have

The vibrating string

23

the boundary conditions u(0, t) = u(L, t) = 0. (2.5) Hence, we have the following initial-boundary problem: Determine a solution of the wave equation (2.3) for 0 < x < L, t > 0 with the initial conditions (2.4) and boundary conditions (2.5). If the ends of the string are allowed to move freely with no external forces acting on them, then the boundary conditions become ux (0, t ) = ux (L, t) = 0. (2.6) If the ends are connected vertically to a spring and are under the condition of an elastic constraint, then the boundary conditions are ux (0, t) + hu(0, t) = 0, ux (L, t) + k u(L, t) = 0,

(2.7)

where h and k are constants. We can also consider the initial value problem of wave equation (2.3) for -oo < x < oo and t > 0 with initial conditions (2.4), where f and g are defined on (-oo, oo). This problem corresponds to the problem of a very long vibrating string so that its boundary conditions have negligible effect on its solution and need not be specified. The longitudinal vibration of an elastic bar of constant cross section is another application for the wave equation. Let the bar be stretched or compressed along its longitudinal axis and then released. It will start vibrating. We assume that the initial displacement and the velocity of the bar are along the bar and uniform over each cross section. Thus, all the subsequent movement of all sections of the bar is parallel to the axis of the bar which is taken as the x-axis. Let u(x, t) denote the longitudinal displacement of the section of the bar at a point x and at time t. Then u(x, t) satisfies z 7ltt - C 7/, xx = 0i

C2 c =Elp

where E is Young's modulus of elasticity of the bar and p is its density. (See Tolstov, [11, 275-2771 for its derivation.)

24

The Wave Equation

2.2 The initial value problem

Consider the initial value problem 2 Utt - C 16xx = 0,

-oo < x < oo, t > 0, (2.8)

u(x, 0) = f (x), ut(x, 0) = g(x),

-00 < x < 00. (2.9)

The equation (2.8) is hyperbolic, and its characteristic curves are x ± ct = constant. Introduce the new independent variables ^ and T as ^=x+ct, T=x - ct

(2.10)

and set u(x, t) = w(^, T). As a result, (2.8) is transformed to the canonical form w£T = 0,

(2.11)

which has the general solution w(6,7-) = p(^) + q(r) where p and q are arbitrary functions. Then the general solution of (2.8) is u(x, t) = p(x + ct) + q(x - ct).

(2.12)

Differentiating (2.12) with respect to t, we get ut(x,t) = cp'(x + ct) - cq'(x - ct) where the prime denotes differentiation with respect to the argument of the function. From (2.9), we have f(x) = p(x) + q(x)

(2.13)

and g(x) = cp'(x) - cq (x) from which we get, after integration, p(x) - q(x) = 1 f ^ g(s)ds + K 0

2.14)

where K is an integration constant. Solving p and q from (2.13) and (2.14), we have p(x)

f(x) fx K

= 2 + 2c 0 g(s)ds + 2

The initial value problem

25

and x

4(x) = f 2 - 2c f g(s)ds - K . o Then 1

x+ct

u(x, t) = 1 I f (x + ct) + f (x - C01 + 2c f -et g(s)ds. (2.15) Equation (2.15) is known as d'Alembert's formula for the solution of the initial value problem (2.8)-(2.9). We can verify by direct differentiation that if f (x) E C2(_00, oo) and g(x) E Cl (-oo, oo), then u(x, t) is twice continuously differentiable and satisfies (2.8) and (2.9). Moreover, the method of deriving (2.8) shows the uniqueness of the solution. Let (^, T) be a point in the xt-plane. From (2.15), we see that the value of u(1;, T) depends on the values of f at the two points t; ±cr and the values of g on the interval I = [l; - Cr, ^ + cr]. Any changes of f and g outside the interval I have no influence on the value of u at (t;, 7-). The interval I is called the domain of dependence of the point (^,T). The triangle D formed by the interval I and the two characteristic lines x-ct=^-cr, x+ct=e+cr

(2.16)

passing through the point (^, T) is called the characteristic triangle determined by the point (^, T) (see Fig. 2.2).

(^ -

cT, 0)

Fig. 2.2 Characteristic triangle Conversely, consider the initial data at a given point (a, 0). We draw

26

The Wave Equation

two characteristic lines

x-ct=a, x+ct=a.

(2.17)

Then, according to (2.15), the set of points S bounded by the lines (2.17) will be influenced by the values of f and g at the point (a, 0) and is called the range of influence of the point (a, 0) (see Fig. 2.3). t

Fig. 2.3 Range of influence of (a, 0) Suppose f is continuous and f and g are piecewise continuous. Then the function u(x, t) in (2.15) is still continuous and satisfies u(x, 0) = f (x) and ut(x, t) = g(x) at the points of continuity. However, a discontinuity of f or g at (a, 0) produces discontinuity of uw and ut along the characteristic lines x ± ct = a and a discontinuity of f" or g' produces discontinuity of uxx and utt along the same characteristic lines. Thus, u(x, t) may not satisfy (2.8) at the points where the derivatives of u fail to exist. Nevertheless, it may be possible to approximate both f (x) and g(x) by sequences of functions {fn(x)} and {gn(x)} such that (1) fn(x) is twice continuously differentiable and gn(x) is continuously differentiable for each n and (2) fn (x) converges uniformly to f(x) and fo gn(s)ds converges uniformly to ff g(s)ds on (-oo, oo) respectively. Then u(x, t) is the limit of a uniform convergent sequence of functions un(x, t) which satisfy (2.8) and the initial conditions

In this case, we shall call u(x, t) the generalized solution of (2.8) and (2.9).

The initial value problem, 27

The introduction of the generalized solution is important in two respects. Firstly, the conditions on the existence of the generalized solution are much weaker than those on the existence of the classical solution. Secondly, in certain physical problems, the values of the functions f (x) and g(x) are known only approximately and the corresponding solution u(x, t) given by (2.15) represents only an approximation of the exact solution of the problem. It will be shown in Section 2.4 that the solution of the initial value problem of the wave equation depends continuously on the initial data; therefore, if the functions f (x) and g(x) differ from the true initial data by a uniformly small amount, then its generalized solution also differs from the exact solution of the problem by a uniformly small amount. Example 2.1

Let u(x,t) be the generalized solution of utt - uxx = 0,

u(x, 0) =

t > 0;

j X(1-X), if 0 < x < 1; 0, otherwise; ut(x, 0) = 0.

Find u(0, 3/4), u(-1/2, 3/4) and u(1/2,1/3). Solution: u(x, t) = [f(x + t) + f(x - t)]/2. u(0, 3/4) = [f (3/4) + f (- 3/4)]/2 = [(3 /4)(1/4) + 0]/ 2 = 3/32. u(-1/2, 3/4) = [f (1/4) + f (-5/4)]/2 = [(1/4)(3/4) + 0]/2 = 3/32. u(1/2, 1/3) = [f (5/6) + f (1/6)]/2 = [( 5/6)(1/6) + ( 1/6)(5/6 )]/ 2 = 5/36. Example 2.2 Find the generalized solution u(x, t) and its partial derivative ux(x,t) in different regions in t > 0 of the initial value problem utt - 4uxx = 0,

u(x 0) = 0,

-00<X0,

1, forx 1.

28

The Wave Equation

Solution: From (2.15), we have 1 , +2t 2t g(s)ds, ux ( x, t) = 4 [g(x + 2t) - g(x - 2t)]. 4 ^x1

u(x, t)

We divide the region t > 0 into three parts by the characteristic lines x + 2t = 1 and x - 2t = 1 through the point ( 1, 0) (see Fig. 2.4). t

III x+2t=1

-2t=1

I

II

1

(1,0)

x

Fig. 2 . 4 Different regions in Example 2.2 Part I .

t>0, x+2t0,

4

[1 - 1] = 0.

x-2t>1.

1 f x+2t u(x, t) 4

Jx_2t

sds = 8 [(x + 2t)2 - (x - 2t)2] = xt;

ux(x,t) = 14 [x+2t-x+2t] =t.

The nonhomogeneous wave equation 29

Part III.

t>0, x+2t>1, x-2t 0,

(2.19)

-00 < x < 00. (2.20)

We would like to find a formula for the solution of (2.19) and (2.20) by method of Green's theorem, which states as follows: Green 's Theorem. Let C be a piecewise smooth simple closed curve in the xy-plane bounding a domain D . Let P(x, y) and Q (x, y) be continuously differentiable in D and continuous in D + C and let the integral f fD (Qx Py)dxdy be convergent. Then

C (Pdx + Qdy) = f J

J

( Q1

- PP ) dxdy

(2.21)

where the line integral along C is taken in the counterclockwise direction. Let (^, rr) be a point in t > 0 and let D be the domain bounded by the segments of the characteristic lines L1 : x+ct = ^+cr and L2 : x-ct = c-cr through the point (^, rr) and the interval I = [^ - or, ^ + cr] on the x-axis (see Fig. 2.2). We integrate (2.19) over the domain D, and apply Green's theorem to the second integral. Then we have -

IL F(x, t) dxdt = J f (c2u xx - utt)dxdt = J (utdx + c 2 uxdt) C

where C=I+L1+L2.

( 2.22)

30

The Wave Equation

On I, dt=O; on L1i dx = -cdt; and on L2, dx = cdt. Hence, we have (utdx + c2uxdt) =

_ -c

L

f1

J

/' £+CT - ut(x, 0)dx,

J

CT

(utdt+udx ) _ -c

Ll

J

du = c[u(+cr, 0)-u(, T)],

L1

'

IL

=c

J

(utdt+udx) = c

L2

2

J

du = c[u(e-cr, 0) -u(, T)].

L2

Thus, £+CT

c (utdx + c2uxdt) = -2cu(^,

J

r)

+ cf (e - cr) + cf(^ + cTr) + J g(x)dx. £-CT

(2.23) Combining ( 2.22) and ( 2.23) and interchanging the role of (x, t) and (^, T), we get 1 u(x, t) = 2 [f (x + ct) + f (x - ct

x+ct

)] + 2c f _ct g

(,)d, + 2c ff F(^, T)d^dT,

(2.24) which is a formal solution of (2.19)-(2.20). If f is twice continuously differentiable, g is continuously differentiable and F and Fx are continuous, then by direct differentiation, u(x, t) is a solution of (2.19)-(2.20). (2.24) reduces to (2.15) if F - 0. Example 2.3 Let f (x) = g(x) = 0 and F(x, t) = xt and c = 1 in (2.19)(2.20). Find u(x,t). Solution: From (2.24), we have x+t-T

u(x, t) 1 f f 6 T d^dr 2 0 -t +T

t

['Z

2J 0 [2

]x-t+T TdT = x -t^-T

3 = x

J

t(t - T)TdT = x[t22 - 3 ]p = 63.

Uniqueness of the initial value problem 31 2.4 Uniqueness of the initial value problem

Consider the initial value problem Utt - c2uyx = F(x, t),

-00 < x < co,

u(x, 0) = f (x), ut(x, 0) = g(x),

t > 0,

(2.25)

-00 < x < 00. (2.26)

Suppose u and v are two solutions of (2.25)-(2.26). Set w = u - v. Then w is a solution of Wtt - c 2 wXX = 0, w(x, 0) = wt (x, 0) = 0,

-00<x0, (2.27) -oo < x < oo. (2.28)

From Section 2.2, we know that the solution of (2.27)-(2.28) is uniquely determined by d'Alembert's formula and must be identically zero. Nevertheless, we will introduce in this section the energy method of establishing the uniqueness of the solution without assuming its existence. Let w(x, y) be a solution of (2.27)-(2.28) and let P = (^, rr) be a point in t > 0. We will show that w(t;, rr) = 0. Since (6, rr) is arbitrary, we have w(x, t) = 0, or u - v. We draw two characteristic lines L2 :x-ct=.'-cr L1 :x+ct=6+cr and through P. These two lines intersect the x-axis at the points A and B respectively. We draw another line t = t', 0 < t' < T, intersecting L2 and L1 at C and D respectively. We note that 0 = Wt(Wtt - c2wXx ) = 2 (wt + c2w^ )t - c2(wxwt) x. (2.29) Let Sl be the domain whose boundary K consists of the line segments AD, DC, CB, and BA (see Fig. 2.5). We integrate the identity (2.29) over S2 and apply Green's theorem. Then we get 0 = - f

K

2

+ c2w)dx + c2wwtdt] = -Ix.

(2.30)

32

The Wave Equation

t

P(6, 'T)

L2:x-ct=6-cr L :x+ct=^+ccr

C

D

B

A

x

Fig. 2.5 Domain of integration for uniqueness proof We now evaluate the integral on each line segment . On both BA and DC, dt = 0. Hence, we have IBA = 1 [wt + c2wx]dx 2 BA

J

and IDC = 1 f [ wc + c2w2]dx. 2 c Moreover, (wt +C 2wy ) dx - cwxwtdx] =

IAD = J

A

D2

2

[wt - cwx]2dx < 0 AD

since dx = -cdt on AD; and IC B

cut + c2w2 ) cdt + c2wxwtdt] = = I B2 ('

C

since dx = cdt on CB.

L 2

+ cw]2dt < 0 B

From ( 2.30), IBA+IAD+IDC+ICB = 0, and IBA = 0 since wt = wx = 0 on BA . Thus, 0 = IBA >- -IDC = ICD 0. This implies that ICD = 0. Since the integrand in ICD is nonnegative , wt = wx = 0 on t = t'. Hence, w (x, t') = constant for ^ - c(T - t') < x < ^ + c(T - t'), 0 < t' < T.

33

Initial-boundary value problems

Since w(x,O) = 0, it follows w(t;, T) = 0 by continuity. Hence, we have the following theorem: Theorem 2.1 tion given by

The initial value problem (2.25)-(2.26) has a unique solux+ct

1

u(x, t) = 2 [f (x + ct) + f (x - ct)] + 1 f 1 2c f _ct g(^)dl; + 2c f

F(r;, T)dl dT

(2.31) where D is the characteristic triangle determined by (x, t) if f E C2(-oe, oo), gE C1(-oo, oo), and both F and Fx are continuous in t > 0. Remark: (1). The initial value problem (2.25)-(2.26) is also stable . Let ui (x, t), i = 1, 2 be two solutions of (2.25 ) with corresponding initial data fi(x) and gj(x). Suppose max fl - f2l < e and max1gl - 921 < e for -oo < x < oo. Then for fixed T, we get from ( 2.24), maxlul - U21 :5 (1 +T)e for - oo < x < oo, 0 < t < T. (2). The energy method can be extended to show the uniqueness of the initial value problem of the equation utt-c2uxx + q(x)u = 0 where q(x) > 0. This equation describes the motion of a vibrating string under the action of a restoring force -q (x)u. The proof is similar to that in Theorem 2.1 by showing that fcD[wt + c2wX + q(x)u2]dx < fBA[wt + c2wy + q(x)u2]dx.

2.5 Initial-boundary value problems Consider the initial-boundary value problem Utt - c2

uXX = 0,

u(x,0) = f(x),

0 < x < L, t > 0, ut(x,0) = g(x), 0 < x < L,

(2.32) (2.33) (2.34)

u(0, t) = u (L, t) = 0, t > 0. As in the case of the initial value problem (2.8)-(2.9 ), we get 1

1

=+ct

u(x, t) = 2 [ f (x + ct) + f (x - ct)] +

g ( s)ds

2c-ct if 0 < x - ct < x + ct < L . Then, in the triangular region

(2.35)

34

The Wave Equation

S2 = { (x, t) I ct < x < L - ct, 0 < t < 2 }, the solution is uniquely determined by the data f and g and is independent of the boundary conditions (2.34). In order to obtain the solution for large t, we have to extend the domains of both f and g from [0, L] to (-oo, oo) so that the formula (2.35) remains valid. Let F(x) and G(x) be extensions of f (x) and g(x) on (-oo, oo) respectively, and define H(x) _ 1 fox G(s)ds which is an extension of h(x) _ 1 fox g(s)ds. Hence, u(x, t) = 2 [F(x + ct) + F(x - ct) + H(x + ct) - H(x - ct)].

(2.36)

Applying the boundary conditions (2.34) to (2.36), we get

0 = 2 [F(ct) + F(-ct) + H(ct) - H(-ct)]

(2.37)

and

0 = 2 [F(L + ct) + F(L - ct) + H(L + ct) - H(L - ct)]

(2.38)

fort>0. Since F and H are independent, we have, from (2.37), F(ct)+F(-ct) = 0 and H(ct) - H(-ct) = 0. This implies that F(-ct) = -F(ct), i.e. F is an odd extension of f and H(-ct) = H(ct), i.e. H is an even extension of h. Similarly, we get from (2.38), F(L + ct) = -F(L - ct) and H(L + ct) = H(L-ct). Since F is odd and H is even, we have F(L+ct) _ -F(L-ct) = F(ct - L), i.e. F is periodic of period 2L, and H(L + ct) = H(L - ct) = H(ct - L), i.e. H is periodic of period 2L. From G(x) = cH'(x), we see that G(x) is an odd, periodic extension of g(x) of period 2L. Hence, the initial-boundary value problem (2.32)-(2.34) has a formal solution given by +Ct 1 u(x, t) = [F(x + ct) +F(x - ct)] + 2G(s)ds

(2.39)

c Jr-ct where F(x) and G(x) are odd periodic extensions of period 2L of f (x) and g(x) respectively. We now impose the following conditions on both f and g so that u(x, t) in (2.39) is a solution of (2.32)-(2.34).

Initial-boundary value problems 35

I. II. III.

f c- C'[0, L] and gE C'[0, L1. f (0) = f (L) = g(0) = g(L) = 0. f"(0) = f"(L) = 0.

Both F(x) and G(x) are periodic extensions of f (x) and g(x) respectively. Hence, from I, F is twice continuously differentiable and G is continuously differentiable in (-oo, oo) except possibly at the points x = mL, where m is any integer. If we can show F(x), G(x), F'(x), G'(x), and F"(x) are continuous at 0 and L, then, by periodic extension, F(x) E C2(-oo, oo) and G(x) E C1(-oo, oo). We know F(-x) = -F(x)

and

F(L-x) = -F(x-L) = -F(x+L ). (2.40)

As x -* 0+, F(0-) = -F(0+) _ -f ( 0) = 0 from I and II. Similarly, as x -> 0-, F(L+) _ -F(L-) = - f (L) = 0. Hence F(x) is continuous in (-oo, oo). This is also true for G(x). Differentiating ( 2.40) yields

F'(-x) = F'(x),

and

F'(L - x) = F'(x + L).

(2.41)

As x -p 0+,F'(0-) = F'(0+) = f'(0). Thus, F'(0) = f'(0). Similarly, F'(L) = f'(L), G'(0) = g'(0), and G'(L) = g'(L). Differentiating (2.41) yields

F"(-x) = -F"(x)

and

- F"(L - x) = F"(x + L). (2.42)

As x -> 0+, F"(0-) = -F"(0+) = -f"( 0) = 0 and F"(L+) = -F"(L-) _ -f"(L) = 0. Thus, from III, F"(0) = F"(L) = 0. Since we have shown that F(x) is in C2(-oo, oo) and G(x) is in C1(-oo, oo ), u(x, t) in ( 2.39) is indeed a solution of (2.32 )-(2.34). As in the case of the initial value problem , u(x, t) is a generalized solution of (2.32)-(2.34 ) if f (x ) is continuous on [0, L] and f'(x) and g (x) are piecewise continuous on [0, L]. Example 2.4 Find the generalized solution u(x, t) of ( 2.39) with x, f(x)- ^L-x

and

for0<x 0, x > 0, x + ct < L/2;

The Wave Equation

36

-L/2 < x - ct < L/2, L/2 < x + ct < 3L/2; (B). t > 0, (C). t > 0, x - ct > L/2, x < L. (See Fig. 2.6).

Fig. 2.6 Different regions in Example 2.4 Solution: u(x, t) = [F(x + ct) + F(x - ct)]/2 where F(x) is the odd, periodic extension of period 2L of f (x). Hence, F(x) = x for -L/2 < x < L/2, and = L-x for L/2 < x < 3L/2. Region A: 0 < x + ct < L/2, -L/2 < x - ct < L/2. u(x, t) = [x - ct + x + ct]/2 = x. Region B: u(x, t) = [x - ct + L - (x + ct)]/2 = [L - 2ct]/2. Region C: L/2 < x - ct < L, L/2 < x + ct < 3L/2. u(x,t) = [L - (x - ct) + L - (x + ct)]/2 = L - x. Next, we consider the initial-boundary problem utt-c2uxx=F(x,t), 0<x0

(2.43)

with initial conditions (2.33) and boundary conditions (2.34). We can extend f (x), g(x), and F(x, t) as odd, periodic functions of x of

Initial-boundary

value problems

37

period 2L. Then 1 1 fx+ct u(x, t) = - [f(x + ct) + f(x - ct)] + — / g(s)ds

is itself an odd, periodic function in x of period 2L, where D is the characteristic triangle determined by (x, t) in t > 0, and is a formal solution of (2.43), (2.33), and (2.34). If we impose the same kind of smooth conditions on f(x) and g(x) as in the homogeneous case and assume that F(x, t) and Fx(x,t) are continuous and F(0,t) — F(L,t) = 0, then u(x,t) is also a classical solution. We can deduce the uniqueness of the problem of (2.43), (2.33) and (2.34) by the energy method. Let u and v be two solutions of the problem. Then w = u — v is a solution of Wu - c2wxx = 0, w(x,0) =wt{x,0)

0 < x < L, = 0,

w(0,t) = w(L,t) = 0,

t>0,

(2.44)

0<x0.

(2.46)

Define E(t) = lf (w?+ c2w2x)dx, t>0 (2.47) 2 Jo where w(t) is a solution of (2.44)-(2.46). (2.47) is called the energy integral of the function w(x, t) over the interval 0 < x < L. Then E'(t) = / (w t w tt + c 2 w x w xt )dx = / w t (w tt - c 2 w xx )dx + c 2 w x w t |J' Jo Jo after integrating the second term by parts. Prom (2.46), we have wt{0, t) = wt(L,t) = 0. Together with (2.44), we have E'(t) = 0. Thus, E(t) = constant for t > 0. But at t=0, wx — wt — 0; therefore, £(0) — 0 and E(t) = 0 by continuity. From (2.47), wt = wx - 0 for 0 < x < L and t > 0. Then w(x,t) = constant. But as w(0, t) = 0, we get w = 0 by continuity. Hence, u = v.

38

The Wave Equation

Example 2 .5 Find the value of the solution u(x, t) of the problem utt - uxx = x(3 - x), t>O, O<x 0 where 0, for x < 0, g(x) = x. for 0 < x < 1, 1, for x > 1. 4. Find the generalized solution of the problem

x/2 0 < x < 1/2; utt = 4uxx,

-oo < x < oo,

t > 0; u(x,0) _

1

(1 - x)/2 1/2 < x < 1; 0 elsewhere.

Evaluate u(1,1/4) and u(1/2,1/6). 5. Consider the initial value problem utt - uxx = 0, t > 0; u (x,0) = f(x), ut(x,0) = g(x) where f (x) E C2(- oo, oo ) and g (x) E C1(- oo, oo). Suppose f = g = 0 outside the interval [-A, A]. ( a). Find the largest constant T such that for t > T, u(2A, t) = 0. (b). Let a be a fixed point. Find the smallest constant T such that for t > T, u(a, t ) = U, a constant. Also, determine the value of U. 6. Let u (x, t) be a solution of utt - c2uxx = 0 and let L1 and L2 be the characteristics through a point P(a,b). Show that if ut has a discontinuity at P, then this discontinuity is propagated along L1 or L2.

42

The Wave Equation

7. Let u(x, t) be a solution of utt - uxx = 0. Show that u1- U2 +u3 - u4 = 0 where u1, u2, u3, and u4 are values of u at the four successive corners of any rectangle whose sides are characteristics. 8. Let u(x, t, r) be the solution of the problem u tt = C2Uxx,

-00 < x

Set v(x,t ) = f0

u(x, t


0;

u(x) 0,T)

= 0,

ut(x, 0,T )

=

F( x ,T).

- T,T)dT.

(a). Show that v(x, t) satisfies vtt - C2vxx = F(x, t); v(x, 0) = 0 , vt (x, 0) = 0. (b). Find u (x, t, T) by d'Alembert's formula and verify that v(x, t) obtained from (a ) is the same as v(x, t) = 2c f fD F(^, 77)d^dr1 where D is the characteristic triangle passing through the point (x, t). 9. Let u (x, t) be a solution of utt - c2uxx = 0, t > 0, which vanishes together with its derivatives as IxI -* oo for 0 < t < T, where T is a constant. Show that the energy integral E(t) = a f ^(ut + c2u2)dx is constant for 0 < t < T. 10. Find the generalized solution u(x, t) of the initial-boundary value problem utt-uxx=0, 0<x0

u(x, 0) = f(x) = x2, ut(x, 0) = 0, 0 < x < 3; u(0, t) = u(3, t) = 0, t > 0 in the different regions bounded by the characteristics x ± t =constant for 0 < x < 3 and 0 < t < 3. 11. Show that if g(x) satisfies g(-x) = -g(x) and g(2p - x) = -g(x), then f x+2p g(s)ds = 0 for any x. x 12. The following initial-boundary value problem utt-uxx=0, t > 0, 0<x 0, t > 0 for which ut = aux for x=0, t > 0 and u=f(x), ut = g(x) for t=0, x > 0 where f and g are of class C2 for x > 0, and vanish near x = 0. Show that generally no solution exists when a = -1. 23. Derive the formula for the solution of the problem utt - c2uxx = 0, x > 0, t > 0,

u(x,0) = f(x),

ut(x,0) = g(x), x > 0; ux(0,t) = 0, t > 0.

What conditions do you have to impose on f and g to insure u(x, t) is twice continuously differentiable? Justify your answer.

Chapter 3

Green's Function and Sturm-Liouville Problems

3.1

Solutions of second order linear equations

Consider the second order linear ordinary differential equation a(x)u" + b(x)u +c(x)u = F(x),

a < x < /?,

(3.1)

where the coefficients a(x), b(x), c(x), and F(x) are real and continuous on \a,0\ and a(x) ^ 0 there. (3.1) is nonhomogeneous if F(x) ^ 0; otherwise, it is homogeneous. Theorem 3.1 Let XQ € [a, (3] and let h and k be two real numbers. Then there exists a unique solution of (3.1) satisfying the initial conditions u(x0) = h,

u'(x0) = k,

(3.2)

on [a,/?]. Proof: Without loss of generality, we assume that a(x) — 1 on [a,/?]. Let y — u and z = u'. Then the initial value problem is equivalent to the system y' - z,

z = -cy -bz + F;

y(x0) = h, z(x0) = k;

(3.3)

which is in turn equivalent to y{x) = h+ T z(t)dt,

z(x) = k+ I \-c{t)y{t) - b(t)z(t) + F(t)]dt.

Jx0

Jx0

(3.4) 45

Green's Function and Sturm-Liouville

46

Problems

By Picard's method of successive approximations , we want to show that there exists a unique set of continuous solutions y(x) and z(x) for (3.4). Define y0{x)=h, yn(x) = h+f

zn^{t)dt,

z0(x) = k,

zn(x) = k+f

JXQ

(3.5)

[-c(t)y„-i(0-t(02n-i(0+^(0]*Jxo

(3.6) If we can show that the sequences {yn(x)} and {zn(x)} converge uniformly to the functions y(x) and z(x) on [a, (3} respectively, then, by letting n —> oo in (3.6), we get (3.4). Let M = \h\ + \k\ + R and N = 1 + B + C where B=max|6(x)|, C=max|c(x)|, and fl=max|F(x)| for a < x < f3. Then \yi(x) - yo(x)\ < |fc||x-x 0 | < i V M | x - x 0 | ,

(3.7)

and \zi(x) - z0{x)\ < (\h\C + \k\B + R)\x - x 0 | < MN\x - x 0 |.

(3.8)

For n > 2, \yn{x) - y„_i(x)| = | / [zn-i(t) - z n _ 2 (0]d*|, Jx0

(3.9)

and |zn(x)-zn_!(x)| = I I

{{yn-i{t)-yn^2{t))c{t)-Y{zn_l{t)-zn_2{t))b{t)\dt\.

J x0

(3.10) We will show, by induction, that |y n (x)- 2 / n _ 1 (x)| < M i V n | x - x 0 | n / n ! ,

| 2 n ( x ) - 2 n _ ! ( x ) | < M7V n |x-x 0 | n /n!. (3.11) (3.11) is true for n = 1 from (3.7) and (3.8). Assume that (3.11) is true for n =TO— 1, i.e. - ym-2(x)\

< MNm~l\x

- x0\m^/{m

- 1)!,

\zm-i{x) - zm_2(x)\

< MNm-x\x

- x0\m-l/{m

- 1)!.

\ym-i(x) and

Solutions of second order linear equations 47

Then, from (3.9 ) and (3.10 ), we have I ym(x) -

y m-1 ( x)I

MNm-1 x < (m - 1)! I f

o It - xolm -ldtl

< MNm-1 Ix - xolm < MNmIx -xoIm (m - 1)! in m!

and I zm(x)- zm

-1(x)I

< MN-- '(C (m

+

B) Ix I t-xolm-l dtl < )• -

MNmlxnl

xolm

1 Hence, ( 3.11) is true for n = m. We know that the series E°° o MNn(,13 - a)n/n ! converges by the ratio test . Let e > 0 . There exists an integer k such that for i > j > MN n(/3 - a)°/n! < c. Then both Iyi(x) - yj (x)I and k , we have +1 by c. Thus, by the Cauchy criterion, both are bounded I zi(x) - zj (x) I sequences { yn(x)} and {zn (x)} converge uniformly to functions y(x) and z(x) respectively on [a ,/3]. Since yn(x) and zn (x) are continuous on [a, /3],

En-;

y(x) and z (x) are also continuous there. It remains to prove the uniqueness of solutions of (3.4). Suppose y(x) and z(x) also satisfy (3.4). Then Iy(x)- y( x)I +Iz(x)-z(x )I

L : Izt_2t+cIYt_

t E [a, /3]. By continuity, u(t)=0. Since u'(t) exists, we have u'(t) = lim u(t) - u(ti) -0. ti-*t t-ti This implies that t is not a simple zero, a contradiction . Hence, u has only a finite number of zeros in [a,,3].

3.2 Boundary value problems and Green's function Consider a(x)u" + b(x)u' + c(x)u = F(x), a < x < /3,

(3.18)

51

Boundary value problems and Green 's function

where a ( x), b(x), c (x), and F(x) are real and continuous on [a, /3] and a(x) 0 there. Multiply ( 3.18) by [a(x)]-1 exp{ fa b(t;)[a(^ )]-1dt;} and define p(x) = exp (fa b(^)[a(^)]-ldc), q(x) = -c(x )p(x)/a(x), and f (x) = -F(x)p(x)/a(x). Then ( 3.18) becomes (3.19)

L[u]=(pu')'-qu=-f, a<x< /3,

which is in self-adjoint form where p(x), p'(x ), q(x), and f (x) are real and continuous on [a, /3] and p(x) >0 there. We now consider a boundary value problem of (3.19) with the boundary conditions: Ba(u) = µiu(a) + aiu'(a) = 0,

(3.20)

BO (u) = µ2u(/3) +

(3.21)

and

where µi, ai, i = 1, 2,

a2u (3)

= 0,

are real constants such that µ? + aS

0.

To illustrate our basic approach , we first consider the following problem u" + 4u = -f,

0 < x < 1,

u(0) = u(1 ) = 0,

(3.22) (3.23)

which is a special case of ( 3.19)-(3.21). Two linearly independent solutions of the associated homogeneous equation u" + 4u = 0 (3.24) are ul (x) = cos 2x and u2 (x) = sin 2x. Then its Wronskian W (x) = 2 and a particular solution uo(x) of (3.22) is given by (3.14)-(3.15) as uo(x) =

1

fo -2

sin2(x - f(^)d^. x

52 Green's Function and Sturm-Liouville Problems

The general solution of (3.22) is u(x) = uo (x) + cl cos 2x + c2 sin 2x. From ( 3.23), we get cl = 0 and c2 = (2 sin 2 ) -1 fo sin 2(1 - ^) f (t;) de. Then we have /^ U(X)=- 2

J0

sin 2 (x - ^) f (^) dl;

+ 21s n 2 fo

sin 2 (1 -1;)

sin 2x 1 sin 2 (1 - l;) f (1;) dl;. + 2 sin 2 f After simplification , we obtain 1

u(x) = f G (x,^)f(^)d^ where fsin 2^ sin 2(1 - x)/(2 sin 2), for l; < x; Gx ( ' sin 2x sin 2(1 - e)/(2 sin 2), for x< ^. (3.25) The function G(x, t;) will be called Green's function of the problem (3.22)(3.23). We note the following properties of G(x, ^) for a fixed (1). As a function of x, G(x, l;) satisfies (3.24) in 0 < x < ^ and in Z; < x < 1. (2). G(x, e) is continuous at x = t;. (3). dG/dxl x=£+ - dG/dxl x=g_ = -1. (4). G(0, t;) = G(1, e) = 0.

e:

(5). G(t;, x) = G(x, We note that for this particular example the solution of the problem (3.22)-(3.23) depends on the construction of Green's function. This approach can also be employed for the problem (3.19)-(3.21). Definition 3.2 Green's function G(x,l;) of (3.19)-(3.21) is a function of x, for a fixed t;, satisfying the following conditions: I. L[G]=0, x 1;

II. Bo, [G] = Bp [G] = 0 ,

(3.26)

(3.27)

Boundary value problems and Green's function

53

III. G(x, l;) is continuous at x = l;,

(3.28)

IV. dG/dxI ,, + - dG/dxlx=£_ Theorem 3.7

(3.29)

If

L[u] = 0, a < x < BQ[U] = BO [u] = 0;

(3.30)

has only the trivial solution, then the Green's function G(x,1;) exists and is unique. Proof: (I). Uniqueness Let l; be fixed and let Gl (x, l;) and G2 (x, t;) be two Green's functions of (3.26)-(3.29). Set H(x, t;) = Gl (x,) - G2 (x, ^). From (3.28), H(x, T;) is continuous in [a„3]. From (3.29),

dH/dxI x=^+ = dGl /dxI x =e+ - dG2 /dxI x =C+ = dGl/dxI x=g- -1 /p(^) - dG2/ dxI x =E- + 1 /p(E) = dH/dxI x =C-• Hence, H(x, t;) E C' (a, Q). For x # H" = -[p'H' - qH]/p. Since all the functions on the right are continuous at x = t;, H" is also continuous there. This implies that H satisfies ( 3.30). By our hypothesis, H - 0. (II). Existence Let ul (x) be a nontrivial solution of L[u] = 0 satisfying Bo [u] = 0 and let u2(x) be a nontrivial solution of L[u] = 0 satisfying B,3[u] = 0. We claim that ul(x) and u2(x) are linearly independent. Suppose there exists a nonzero constant C such that ul (x) = Cue (x). Then Ba[u1] = 0 by construction and Bp[ul] = CBp[u2] = 0. This implies that ul [x] - 0, a contradiction.

Let l; be fixed. Define G(x,) = Aul (x), Bu2(x), where A and B are constants to be determined. From (3.28) and (3.29), we have

54

Green's Function and Sturm-Liouville Problems

Bu2(6) - Au1(^) = 0

and

Bu2(6) - Aui(6) = -1/p(^).

Thus, A = -u2(6)/[p(6)W(6)] and B = -u1(6)/[p()W(6)] where W(t;) = W(ul, u2i 6) 0 since ul and u2 are linearly independent. Then

G(x -u1 (^)u2(x)/[P(^)W(^)] , x > ^-

(3.31)

We note that

(pWY = pW'+p'W = P( u1u2-

u2u1)' +p (u1u2 -u2u1 ) = n ( p u2 )'- u2(pul)'

= u1Qu2 - u2qul = 0.

Thus, P(1;)W(u1, u2; t;) is a constant . From (3.31), we note that G(x, G(t;, x). Example 3.1 Find the Green's function for u" = 0, 0 < x < 1; u(0) _ u(1) = 0 if it exists. Solution: The general solution for u" = 0 is u(x) = ax + b. Then 0 = u(0) = b; and 0 = u(1) = a. Thus, u(x) 0. The Green's function exists. Using (3.31), we get u1(x) = x,u2(x) = x - 1 and W(uiiu2;t;) = 1. Therefore, G(x,) Theorem 3.8 system

{ t; (1 - x), x>.

If the system (3.30) has only the trivial solution, then the

L [u] = -f, a < x 0. This implies that

IaIIbi 0, there is an integer N = N(€) such that for n > N, ff p(x) I Sn(x ) - f (x) I2dx < e where Sn(x) = Ek=1 akOk(x)• -

Now, let F = {On (x)} be a family of orthogonal functions on [a,0] and let f E S. Our goal is to approximate f (x) by the sum Sn(x) in the mean

64 Green's Function and Sturm-Liouville Problems

for a fixed n. We have n

n

n

En=IIf-SnII2-11f112-I: ak(cbk,f)-T«k(f,Ok)+EIakI2IIPkii2. k=1

k=1

k=1

Set Ck

= (f,Ok)/IIOkII2.

(3.59)

Then n

2

En =11f 11 +

Iak

n

- C k 12110 k 112 - E ICk12110k112.

k=1

k=1

Since En > 0, the minimum of En is achieved if we choose ak = ck. Hence n

min E. =11f 112 - E Ick12110k 112. (3.60) k=1

The coefficients { ck} are called the Fourier coefficients of f with respect to F and the series Ek 1 ckOk(x) is called the Fourier series of f with respect to F. From (3.60), we have

IIf II2 ? Ek= 1 IckI2II0kII2 for any n. Let 1n = Ek=1 1ckI2I10kII2 It is easy to see that { yn} is a monotone increasing sequence of real numbers bounded above and hence {'yn} converges and 00

111112>_IIck12110k112 (3.61) k=1

which is called Bessel's inequality. Moreover, Ick I II cbk II -4 0 as k -+ oo. Definition 3.3 A family F = {0n(x)} of orthogonal functions is complete in S if, for any f in S, 00

11f112 =1: ICk12110k112

(3.62)

k=1

holds. Remark. (3.62) is known as Parseval's equation or the completeness relation.

65

Convergence in the mean

Let { 0n(x)} be a family of orthogonal functions on [a, 0]. Then the Fourier series of f E S converges in the mean to f if and only if

Theorem 3 . 14

{ cn(x)} is complete.

Proof: We note that n

n

f IIf - E C OkIi2 = 11 112 - E ICk 12II7 k II2. k

k=1

k=1

The theorem is proved if we let n -* oo in the above identity. Remark : If {q5n(x)} is an orthonormal set, i.e. Ck = (f, Ok) and the completeness relation is

(0j, Ok) = bjk, then

00

IIf112=1: Ic k I2.

(3.63)

k=1

Theorem 3.15 Let {!p n(x)} be a complete orthonormal set on [a„Q]. Then any continuous function f in S such that (f, On) = 0 for all n must be identically zero.

Proof: From (3.63), we have 11fII2 = 0. Hence f - 0. Theorem 3 .16 Let { On(x)} be a complete orthonormal set on [a, 0] and let E°° 1 cng5 ( x) be the Fourier series off in S. Then

f (x)dx = > Cn, On(x)dx. ^a n=1 a a

Proof: I

f m a f (x)dx - E cn

J

a

n=1

Ja

p m CnOn(x)I dx On(x)dxi I f (x) - E a

m

IIf - ECnOn 1I 111111 n=1

n=1

66

Green's Function and Sturm-Liouville Problems

from Schwarz's inequality. The last term goes to 0 as m -> oo since the Fourier series of f converges to f in the mean. Let { fn (x) } be a sequence of independent functions on [a, /3].

Let Define

91(x) = fi(x). 92 (X) = f2 (x) + c21g1(x) such that ( 92, 91) = 0.

Then c21 = -(f2,g1)/ IIg1112 and 92 (x) = f2(x) - (f2,91) 91/II91112. Define

g3 (x) = f3(x)+ c3191 ( x)+C32g2 (x) such that (83,91) = (93,92) = 0•

Then C31 = -(f3,91)/ II91II2 and c32 = -(f3,92 )/1192 112; and 93 (x) = f3(x) - (f3,91 )91(x)/II91II2 - (f3,92) 92(x)/II92112.

By induction , we define gn (x) = fn (x) - E^-i (fn, 9j ) gj / I 9j 112, for n>2. Then the new system {gn (x) } is an orthogonal set. This process is called the Gram- Schmidt orthogonalization process.

3.5 Integral operator with continuous , symmetric kernel Let Ku =

J

k(x, t;)u(e)dt;

(3.64)

where k(x, ^) is a complex-valued continuous function of x and 6, such that k(x, l;) = k(6, x) but not identically zero on [a,,3] x [a,,3] and u E C[a, /3]. Definition 3.4 A family of functions F on [a, /3] is uniformly bounded if there exists a constant M such that If (x) I < M for all f E F. Definition 3.5 A family of functions F on [a, /3] is equicontinuous if for E > 0, there exists a 8 = 5(e) >0 such that I f (x) - f (y) I < C for all x, y E [a, /3] with Ix - yI < b and for all fin F. For simplicity in our presentation, we assume that p(x) = 1 in the definition of the inner product and the norm in the space S. Theorem 3 .17 The set of functions {Ku}, with IIuhI = 1, is uniformly bounded and equicontinuous in [a, /3]. Proof:

Integral operator with continuous , symmetric kernel 67

We note that k(x, t;) is continuous on a closed square. Hence, k (x, t;) is uniformly continuous there and also I k (x,t;)I < M for some constant M. By Schwarz's inequality, IKul < M fa I u(^) I dt; < M (O - a). Hence, {Ku} is uniformly bounded.

Let e > 0. Then there exists a S = S(e) > 0 such that for Ix - yI < S, I k(x, e) - k(y, l;) I < e for all x, y, and l; in [a,,3]. Let u e C[a, /3] with Hull = 1, -a). I Ku(x)-Ku(y)I = I fR [k(x, ^)-k(y, ^)]u(^)d^I E f R I u(6)I < c V(,3 Hence, {Ku} is equicontinuous. Theorem 3.18 real.

(Ku, v) = (u, Kv) for u and v E C[a„0] and (Ku, u) is

Proof:

(Ku, v) = L [L k(x, ^)u(^ )(/,X«)Xi-

(3-67)

If \\Km\\ > 0 for all m, then there exists a sequence of eigenvalues {fit} with corresponding orthonormal eigenfunctions {xi(x)} of K. From Theorem 3.21, m -► 0. Example 3.4 Find all the eigenvalues and the corresponding eigenfunc­ tions of the operator K defined by Kf=

f Jo

(l+xy)f(y)dy

for/€C[0,l]. Solution: Consider Hf = Kf = / f{y)dy + x yf(y)dy = a + bx Jo Jo where a

f(y)dv

=

and

b

Jo

= /

yf{y)dy.

Jo

(I). M = 0. a + bx = 0 for x £ [0,1]. This implies that a = b = 0. Hence H = 0 is an eigenvalue of K with a set of eigenfunctions orthogonal to 1 and x with respect to the weight function p(x) = l o n [0,1]. (II) M ¥= 0. f(x) = /x _1 (a + bx). Hence, ixa= I (a + by)dy = a 4- b/2; Jo

fib

Jo

(ay + by2)dy = a/2 + 6/3.

Integral operator with continuous, symmetric kernel 73

The above equations can be rewritten as (µ - 1)a - b/2 = 0; a/2 + (1/3 - µ)b = 0. For nontrivial solutions of a and b, we require det

C

µ 1 -1/2 _ 1/2 1/3-1z 0

i.e. µ2-4t/3+1/12=0. Thus, µ = (4 ± 13)/6. Hence, for eigenvalue µ = (4 + 13)/ 6, its eigenfunction is f (x) _ 1 + (-2 + 13) x/3; and for eigenvalue µ = (4 - 13)/ 6, its eigenfunction is f (x) = 1 - (2 + 13)x/3. 0, an # am Example 3.5 Let f (x) = 2 + EO°_1 an cos nx, an all the eigenvalues and the Find converges uniformly. for n m, which corresponding eigenfunctions of the integral equation Ku =

J

k(x, y)u (y)dy = µu(x)

where k(x, y) = f ( x - y), -ir < x, y < 7r. Solution: Let u(x) be an eigenfunction of Ku = µu corresponding to an eigenvalue µ. Then ao + E an cos n(x - y)]u(y)dy µu (x) = f f (x - y)u(y)dy = f [ 2 r n=1

7raoco

00

2 + E [7rancn cos nx + 7randn sin nx],

(3.68)

n=1

where 1 7r cn = - f u (y) cos nydy , ir 7r

1

do = - J u( y) sin nydy.

(3.69)

7r ,,

The interchange of integration and summation is valid since the series of f (x) converges uniformly to f (x).

74 Green's Function and Sturm -Liouville Problems

Case I: p=0. From (3.68), we have 00 0 = 7ra0 co + 7r [ancn cos nx + andn sin nx]. n=1

Since the set {1, cos nx, sin nx, n = 1, 2...} is an independent set, we have co = cn = do = 0 for all n. Thus u(x) - 0 and p = 0 is not an eigenvalue.

Case II: p 0. We assume that u(x) in ( 3.68) converges in the mean and substitute u(x) in (3.69). After interchanging the integration and summmation, which is permissible , and making use of the orthogonal property of the set on [-7r, 7r], we get ,uco = aoco7r,

pcn = ancn7r, udn = andn7r.

For p = a,n7r, Cm and dm are arbitrary, and cn = dn = 0 for n # m. Thus, p = p,,,. = a,,,7r, m = 0, 1, 2... are eigenvalues, and um (x) = cm cos mx + dm sin mx are eigenfunctions. Let {pi} be a sequence of eigenvalues determined by the extremal principles and {Xi(x)} be the corresponding sequence of orthonormal eigenfunctions. Theorem 3 .23 Let u E C[a, Q]. The Fourier series of Ku with respect to {Xi(x)} converges uniformly to Ku on [a, /3]. Proof: Let gm (x) = u(x)->2'° 1(u, Xi ) Xi(x) . It is easy to see that (g„t, Xi) = 0 for i = 1, ....m. From the extremal principles , IIKgm II 5 Ipm+l I Ilgm II • Since pm+l -> 0, the sequence { Kgm} converges to 0 in the mean. Thus, 00

Ku =

00

00

E(u, Xi)KXi = pi(u, Xi)Xi i=1

i=1

=

00

1:(u, K Xi)Xi

i=1

where the above convergence is in the mean. For any q > p, q

q

pi (u, Xi)Xi = K [E(u, Xi) Xi] . Y,

41 .

=

E( K u,

i=1

Xi)Xi,

Completeness of eigenfunctions of Sturm-Liouville problems 75

Since IKuI < M(,3 - a) 1/2IIuII, we have q

r

q

M(,Q-

Eµi(u,Xi)XdI i=p

a)1

/2[EI (u,Xk)I2]1 /2

i=p

which goes to zero as p, q -p oo by Bessel ' s inequality. Thus, E' 1(Ku, Xi)Xi is uniformly convergent to a continuous function on [a, 0]. Since Ku is also continuous , 0 = Ku, i.e

(3.70)

Ku = j:(Ku, Xi)Xi i=1

uniformly. We now claim that all the nonzero eigenvalues of K are obtained from the extremal principles. Suppose there is a nonzero eigenvalue µ with corresponding eigenfunction 0 distinct from all {µi}. Then Kb = po and, from (3.70), we have 00 yV1 =

Kii

00

( K O, xi) xi

= i =1

= ,L i =1

(0, xi)Xi

= 0.

This implies that V = 0, a contradiction.

3.6 Completeness of eigenfunctions of Sturm-Liouville problems Let G(x, 6) be the Green's function for the Sturm-Liouville problem L[u] = (pu')' - qu = -Au(x),

(3.71)

B. [u] = Bp [u] = 0.

(3.72)

We note that G(x, t;) is continuous and symmetric on [a, a] x [a , (3]. Define Ku =

) a G(x,e)u(e)d^.

(3.73)

From the theory of the continuous, symmetric operator in Section 3.5, there exists either a finite number of nonzero eigenvalues {µi} with corresponding normalized eigenfunctions {Xi(x)) such that Iµi+il < I µil and (Xi, Xi) = bij or a sequence of nonzero eigenvalues {µi} with corresponding eigenfunctions

76 Green's Function and Sturm- Liouville Problems

{Xi(x)} with µi -> 0. However, the first case exists only if for some m, JjKmjj = 0 where Q

Kmu = f Gm (x, ^)u(t;)dl; with Gm(x,S) =G(x,S) -EiL1, iXi(x)Xi(S)•

To rule this out, we would like to show jjKmjj # 0 for all m. By way of contradiction, suppose 11Km11 = 0 for some m. Let f E C[a,0]. Then m

0 = Kmf = Kf - > PiXi(x)(.f, Xi) i=1

and m

0 = L[O] = L [Kmf]

m

= L[Kf] - E yi(f, Xi) L [Xi] = -f +

J:(f, Xi)Xi•

Then f = Ei'' 1(f, Xj) Xi. This implies that f E C2(a„Q), a contradiction. Thus, I IKm I I > 0. Hence, if the kernel is G(x,t;), then there is a sequence of eigenfunctions {Xi(x)} with corresponding eigenvalues {µi} with µi -* 0 for the operator K. Theorem 3.24 (3.72). Then

Let f E C2 [a, 3] and satisfy the boundary conditions 00

f

=

E(f,Xi)Xi

(3.74)

i=1

where the convergence is uniform on [a, Q]. Proof: Let f E C2[a,,3] and satisfy (3.72). Then u = L[f] E C[a,,3]. From Theorem 3.8, f = -Ku. From Theorem 3.23,

^(f, Xi)Xi•

f = -Ku = - r(Ku, 00Xi)Xi 00 = i=1

i =1

It can be shown that any square integrable function can be approximated in the mean by a C2-function satisfying (3.72). (See Indritz, [8, 254-260]). Based on this property, we have the following theorem:

Completeness of eigenfunctions of Sturm,-Liouville problems 77

Theorem 3.25 {Xi (x) } is a complete orthonormal set in [a,)3]. Proof: Let f E S and let c > 0 be given. Then there exists a function g E C2 [a, ,0] such that II f - 9II < E. By the triangle inequality, we have m

m

m

IIf-T(f,Xi)XiII N. This implies that Iii - cl -1 < M for all i where M = max [21cl - 1,

I c - µi I -1, ..., I c - µNI -1].

From Schwarz's inequality, we have pibiXi(C) I Ii=m E

µi

<M I IbiIIN'iXi(x)I
(Kf, xi)xi (x) = J:(f, Kxi)xi(x) i=1 00

i=1 00

_ E µi(f, xi)xi(x) = E /tibixi(x).

i=1

i=1

Hence, Ku-cu=-K c f +K

c

00

v+

f-

v= i=1

2 +^ ai

[- C

c

-

p i a 2.] Xi (x)

+

f=f ,

i.e. u(x) is a solution of (3.80) Suppose there are two solutions u and w of (3.80). Then g = u - w is a solution of Kg = cg. But c is not an eigenvalue, so g = 0. Hence, the solution is unique. Case II: c = tj. From (3 . 82), b2 = 0, i.e.

fa

J It

f(x)Xj(x)dx = 0.

(3.86)

Further properties of eigenvalues and eigenfunctions

81

Then, using the same approach in Case I, we see 00 h(x) f(x) + l.LibiXi(x) c(µi - c) c z=1,i#i

(3.87)

is a solution of (3.80) if (3.86 ) is satisfied . If k(x ) is another solution of (3.80), then w(x) = k(x ) - h(x) satisfies Kw = µjw. This implies that w(x) = AX, (x) where A is an arbitrary constant. Then

k(x) = h(x) + AX; (x)•

(3.88)

We can now summarize the above results in the following theorem: Theorem 3 . 27 Consider the nonhomogeneous integral equation (3.80). (A). If c )4 pi for all i, then there exists one and only one solution u(x) of (3.80) given by (3.83). (B). If c = j for some j and the condition (3.86) is satisfied, then the solution of (3.80) is given by (3.88) with h (x) given by (3.87).

3.8 Further properties of eigenvalues and eigenfunctions Consider the Sturm-Liouville problem L[u] _ (pu')' - qu = -Apu, a < x
0 on [a, 0]. Problem (3.89)-(3.90) possesses all the properties of eigenvalues and eigenfunctions developed in Section 3.6. Moreover, with simpler boundary conditions and the nonnegative condition of q(x), we obtain further properties of the eigenvalues and eigenfunctions.

Theorem 3 . 28 Proof:

All the eigenvalues of (3.89)-(3.90) are positive.

Green's Function and Sturm-Liouville Problems

82

Let A be an eigenvalue and let u(x) be the corresponding eigenfunction. Multiplying (3.89) by u(x) and integrating by parts from a to )3, we have 0 = f Q u [(pu' ) ' - qu + Apu] dx = puu' 1 ^3 + j [-pu 2 - qu2 + Apu2] dx. «

«

From (3.90), we get

A = f [pu12 + qu2] dx/ f a pu2dx > 0 «

(3.91)

«

since p > 0, q > 0, and p > 0 on [a, a] . If A = 0, then, from (3.91), u is a constant. From (3.90), u - 0, a contradiction. Thus, A > 0. Since A = 0 is not an eigenvalue, the Green's function for the problem L[u] = 0; u ( a) = u(/3) = 0 exists, and the corresponding nonhomogeneous problem L[w] = -f; w( a) = w(/3) = 0

(3.92)

has a unique solution given by 0 w(x) = f G (x, l;) f (t;) dl;. «

If we set f(x) = Ap(x)u(x), then the eigenfunction u(x) of (3.89)-(3.90) with corresponding eigenvalue A satisfies

u(x) = A f a G(x, t;)p (l;)u(t;)dt;.

(3.93)

Let ,O(x) = p(x)u(x) and µ = 1/A. Then, we obtain

K

= f P vt'p

)

P(^)

d^ = z (x).

(3.94)

Hence, there exists a sequence of eigenvalues {µk} of ( 3.94) with normalized eigenfunctions {Xk(x)} such that IµkI >- I1 k +1 I and Iµkl -> 0. Since Ak = 1 /µk are eigenvalues of (3.89 )-(3.90) and Ak is positive and simple, we have 0 cnAn f P(bundx n=1

«

n=1

00

«

n=1

00

_ ^ncn > n=1

Al

= Al f > cn n=1 «

Hence, 10 (pO12 + 4o2)dx ^1 f«

f« po2dx

p02dx.

«

84

Green's Function and Sturm-Liouville Problems

Let 0 = ul. Then we have cn = 0 for n = 2, 3, ..., and A1 _ fa (pui + qui)dx f« puidx We conclude that A = min

I(0)

where

I (^) = fa (POi2 +

g02)dx

(3.95)

f, pq52dx for 0 E C2 [a, /3] with 4>(a) _ 0(0) = 0 and its minimizing function is 0=ui(x). We note that the functional I(0) is still well -defined even if ¢' is required to be piecewise continuous , and no assumptions are made on 0". Hence, it is natural to consider the class of admissible functions F of continuous, piecewise smooth functions 4>(x) such that q5(a) = 4>(/3) = 0. We have shown that among the class of C2-functions 4>(x) with 4>(a) = 0(0) = 0, Al is the minimum value for 1( 0) and ul (x) is the minimizing function . It is possible that for the new class F, we may obtain a new minimizing function which no longer satisfies L[u] = -Alpu; u(a) = u(/3) = 0.

(3.96)

However, in this particular functional, we can apply the theory of calculus of variations to show that (1). if u E C2 yields a minimum value to the functional I(0), then u is also a minimizing function in F; (2). any function u E F that gives a minimum value to I(4>) must be in C2 and satisfy (3.96). (See Courant and Hilbert [5,199-202]). Hence, in view of the above observations , we have

^1 = min 'EF

f / ( P^i2 + g4>2)dx fa a p4>2dx

with minimizing function ul (x). Next, let 0 E C2 [a,,3] with 4>(a) = ¢(0) = 0 and (0,un) = 0,

Further properties of eigenvalues and eigenfunctions

85

n = 1, ...., k. Then

fa

oo (p^bl2+qcb2)dx=

E Ancn^! Ak+1 n=k+1 n=k+1

and

fa J po2dx = «

2

00 cn. n=k+1

Hence, as before, we have fa (p012 q02)dX a k+1 =min

QEF,(O,u )=O,l,...,k

fa

p02dx

and the minimizing function is uk+1(x). As a result, we have the following theorem: The nth-eigenvalue for the Sturm-Liouville problem (3.89)Theorem 3.29 (3.90) is the minimum value of the functional f p [pui2 + qu2] dx 1(u) _

(3.97)

« fa pu2dx

for the class of continuous, piecewise smooth functions satsifying (3.90) and (u, ui) = 0, i = 1, ..., n - 1, where ui(x) are the first n-1 normalized eigenfunctions. The nth eigenfunction un(x) is the corresponding normalized minimizing function. We can determine the nth eigenvalue and the nth eigenfunction without using the preceding eigenfunctions. This method is given by the following theorem. (Courant's Theorem ) Let 01( x), ..., On_1(x ) be arbiTheorem 3.30 trary continuous functions on [ca, ,3]. Let A('1i ..., On-1) be the minimum of the functional I(u) in (3.97) where the class H of admissible functions u(x) is the space of all continuous , piecewise smooth functions u(x) in [a,'3] with u(a) = u(/3) = 0 and (u, ci) = 0 for i = 1, 2, ..., n - 1 . Then the nth eigenvalue An of (3.89)-(3.90) is A. = max A(01i ..., On-1)

86

Green's Function and Sturm-Liouville Problems

for all possible choices of the functions 01, ..., On-1 Proof: We have shown that A(ul, ...un_1 ) = An. Hence, it is sufficient to show that for all possible choices of ^1, •••^n-1, A(01, ..4n-1) < ) nConsider a given set of continuous functions 01, ...On_1 vanishing at c and 3. Our goal is to construct a nontrivial function u(x) E H in the form of n

u(x) ciui(x) i=1 where c1, ..., c,, are constants to be determined. u E H if u satisfies the following conditions: n n - 1.

0 _ (u, 4'j ) ci (ui,

(3.98)

i=1

(3.98) is a system of n-1 homogeneous linear equations in n unknowns cl, ..., cn. Such a system always has a nontrivial solution. Thus, u exists and belongs to H. We get fa

n

(3.99)

Pu2dx = ci, a

f

[pu2 + qu2]dx =

[p ccuu+ q ccjuu]dx a i=1 j=1

i=1 j=1

n

_

E ccj

J

[pu?u' + quiuj ] dx.

i=1 j=1 a

Moreover,

J a[puuj + quiuj]dx = puzuj I a - J p ujL[ui]dx a

J

a

/'

= Ai

a

pujuidx = r 0, j i,

A i,

j = i.

(3.100)

Further properties of eigenvalues and

eigenfunctions

87

Hence, I(u) = JZ"=i ^» c ?/ X]"=i c ? ^ ^n- This implies that A(0i, ..0 n -i) < A„. We are going to study the effect on the eigenvalues of the problem (3.89)-(3.90) if we change the coefficients of (3.89) or the interval [a,/?]. T h e o r e m 3.31 (Monotonicity Theorem) Let An be the nth eigenvalue of (3.89)-(3.90). Then An increases if any one of the following conditions holds: (1). either p(x) or q(x) increases. (2). p(x) decreases. (3). the interval [a,0] decreases. Proof: Let <j>\(x),....,(f)Tl_i(x) be any set of continuous functions on [a,0\. Define

fZ\pu* + qu*}dx

I(u) = —

a ' J pu2dx and let H and A(i,..., 4>n-\) D e defined as in Theorem 3.30. Then we have A(0i,...0„_i) = mmu&H I(u) and An = maxA(0i,...0 n _i). (1). Consider the problem (p'u'Y - q'u + X'pu = 0,

a<x p(x) and q*(x) > q(x). Define

/

pu2dx

Then I(u) < I*(u) where u € H. Hence, A((/>i,...,0n_i) < A*(0i,...,n_i) = min Ii(u). Consequently, An < A* = max A*(0i,...,0 n _i) for all possible choices of 0i,....$ n _i. (2).

Consider (jra')' -qu + \*p*u = 0,

a < x < /3

(3.102)

88

Green's Function and Sturm-Liouville Problems

with boundary conditions (3.90) where p*(x) < p(x). Define r ,i2 + qu2]dx IZ (u) -fa U fa p*u2dx

Then 1(u) < II(u) for u E H.

Moreover, f, p(x)q (x)u(x)dx = 0

if and only if fa p*(x)cb (x)u(x)dx = 0 for 02(x)p*(x) = p(x)c5i(x) for i = 1, ...., n- 1. This implies that A(01, ..., On-1) < A*(¢i, ..., n_1) =min 12 *(u). Since both p(x) and p*(x) are positive, the set of all continuous functions (01i ..., On-1) is the same as the set of all continuous functions ( , ••, 0*-1) Consequently, An < A*„ = max A*1, +0n-1)• (3). Consider (pu')' - qu + A * pu = 0, a* <x 0 are satisfied. Show that A > 0. 35. Find all the eigenvalues of the problem u" - au = -Au, a < x < /3; u(a) = u(/3) = 0; where a > 0 is a constant. Use this problem to illustrate the monotonicity theorem. 36. Find upper and lower bounds for the kth eigenvalue Ak of the problem ((1 + x2)u')' - xu = -A(1 + x2)u, 0 < x < 1; u(0) = u(1) = 0; by comparing this problem with two similar problems with constant coefficients. 37. Let Ak be the kth eigenvalue of u" - q(x)u + Au = 0, 0 < x < 1; u(0) = u(1) = 0 where q(x) is a nonnegative continuous function on [0,1]. Show that limk-,,,, Ak/k2 = 7r2. 38. Show that every nontrivial solution of u" + exu = 0 has infinitely many zeros on (0,oo), whereas a nontrivial solution of u" - exu = 0 has at most one zero in (0, oo). 39. Let u(x) be a nontrivial solution of u" + q(x)u = 0 where q(x) is continuous on (- 00, oo). (a). Show that u(x) has at most one zero in (0, oo) if q(x) a2 > 0 and a is a constant. (c). What can you say about the number of zeros of u(x) in (0, oo) if q(x) = 1/4x2? 40. Use Theorem 3.29 to show the first eigenfunction of (3.89)-(3.90) does not vanish in (a, /3). (Hint. Show that ju1(x)l is also an eigenfunction corresponding to A1.)

Chapter 4

Fourier Series and Fourier Transforms

4.1 Trigonometric Fourier series Consider a system of trigonometric functions S = 11, cos x, sin x, ..., cos nx, sin nx, ... } on [-1r, 7r]. These functions are orthogonal on [--7r, 7r] with respect to the weight function p(x) = 1. We note that 111112 = 21r, and II Cosnx112 = 11 sin nx 11 2 = in. Hence, we may obtain an orthonormal system F = {!pn (x) } where 1 00 (X) = (2 ) 1/2 , 02i W _ 71/2 cos jx, 02j-1(x) _ -7r

1

2 sin jx, j = 1, 2, 3...

A function f (x) is integrable on [a, b] if f (x) is bounded and Riemann integrable. For any integrable function f (x), its Fourier series with respect to F is 00

7r

1: CnOn( x), Cn = f J (x)On(x)dx.

(4.1)

n=0

If we define

1 an = -

1

J

f (x) cos nxdx,

bn = -

J

f (x) sin nxdx, (4.2)

then (4.1) can be rewritten as 00 ao + E (an cos nx + bn sin nx). (4.3) n=1

95

96

Fourier Series and Fourier Transforms

We now investigate the convergence of the Fourier series (4.3). Consider the nth partial sum of the series (4.3): N

SN(x) =

>(an

2 + n=1

cos nx + bn sin nx).

=1

Substituting (4.2) into SN(x) and rearranging, we get

SN(x) = 1

J ^[2 + cosn(x - t)] f (t)dt. (4.4) n=1

We note that N

N

[2 + E cos ny] sin (y/2) = 2{sin(y/2) + E[sin(n + 2)y - sin(n n=1

2)y]}

n=1

1 +2

= Hence,

1 N sin(N+

2 + n cos ny = n=1

2) y

2 sin (y/2) '

(4.5)

Y# 0.

(4.5) also holds for y=0 by taking the limit y -40 on both sides. Thus, SN x 1 /" 1(t) sin(N+ 2)(x-t)dt= 1 "^ f(x+T) sin(N+ 2)TT. d () 27r f sin[(x - t)/2] 27r f ,7r_x sin (y/2) The functions 1, cos nx, sin nx, and SN (x) are periodic of period 27r. Hence, it is natural to extend f (x) to be periodic of period 27r in (-oo, oo). In this way, we have SN

(x) = 1 " f(x+T) sin(N+ 1/2)Td 27r J 7 sin(y/2)

T.

(4 . 6)

Integrating (4.5) from -7r to 7r gives

sin(N + 1/2)T

2^r = ,

sin(y/2)

dT.

(4.7)

In order to establish convergence of the Fourier series, we will need the following theorem.

Trigonometric Fourier series

Theorem 4.1 [a, b]. Then

97

(Riemann-Lebesgue Theorem) Let f be integrable on

Ali ^ b sin Ax f (x)dx = 0 and a

Ali ^ 6 cos Ax f (x)dx = 0. (4.8) a

f

Proof: Let f (x) = 1. Then b cows-cosAb lim sin Ax dx = lim = 0. .X-.oo J a A-.oo

Next , let f (x) be a step function , i.e. f (X ) = O'k, xk_1 < x < xk, k = 1, 2,...n such that a = xo < xl < ... < xn = b, and Qk is a constant for k = 1, ..., n. Then n

b

sin Ax f (x) dx = > f

ok(COS Aak_ 1 - cosAak] A

k=1

which goes to zero as A - oo. Let e > 0. For any integrable function f (x), there exists a step function g(x) such that g(x) < f (x)) and

J b[f(x) - g(x)]dx < e/2. a

Then, for the step function g(x), we can choose a Ao such that for A > Ao,

fb f sin Ax g(x)dxl < e/2.

iJ Thus, for A > A0, we have

i fJ

b

b sin Ax f (x)dxl < 1b sin AxI f (x) - g(x)ldx + I

a

a

i.e. lima-,,. fa sin Axf (x)dx = 0. Similarly,

rb lim

J

xoo a

J

a

cos Ax f (x)dx = 0.

sin Ax g(x)dxi < e,

Fourier Series and Fourier

98

Transforms

Theorem 4.2 Let f(x) be periodic of period 2n and be piecewise smooth on [—7r,7rj- Then the Fourier series of / ( a ) converges to [/(a + 0) + f(x — 0)J/2 at every point x. Proof: Let N{T)

°

sin(N + 1/2)T

=

2sin(r/2) '

FVom (4.6), we have to show that

,.lim i- r/ f(xti + T)x G^(r)dT / M = ^ f{x+o)+f(x-o) i—ii i. N

It is sufficient to show that Jim - f N — o o 7T / n

f(x + r) GN(r)dr

=

/ ( X + 0)

,

(4.9)

I

and

i r° lim - /

fix - o) f(x + r) GN(r)dr = ^ — ^ .

(4.10)

Since the integrand in (4.7) is even, we have

i = i f GN(T)dr = - f GN(T)dT. 2

f JO

(4.11)

A" J-7T

Hence, 2

* Jo

Thus, to prove (4.9), we have to show that lim - / [f(x + T)-f(x

N—oo 77 JQ

+ 0)]GN(T)dT = 0.

(4.12)

Let ^

;

/ ( x + r) - / ( a + 0) 2sin(r/2)

1

/ ( g + T ) - / ( a + 0) r n T 2sin(r/2)''

FVom our hypotheses, we know / + ( a ) = \imT_o+[f(x + T) — f(x + 0)]/r exists, and lim T _o sin rjr = 1. Hence, 4>{T) remains bounded as r —» 0

Uniform

convergence

and

completeness

99

and is continuous in (0,7r] and integrable on [0,7r|. Thus, by Theorem 4.1, (4.12) holds. Similarly, we can also show that (4.10) holds. 4.2

Uniform convergence and completeness

Let /(x) be continuous on [—7r,7r] such that f(—ic) = /(TT) and let f'{x) be piecewise continuous there. Let an and bn be the Fourier coefficients of f(x) on [—7r, 7r] and let an and (3n be the Fourier coefficients of f'(x) on [-7T,7r]. Let {xo,...,Xj} be the points of jump discontinuity of f'(x) in [—7r,7r] such that -7T = X 0 < X i < ... < X , = IT.

Then TTQO = f J —it

f'{x)dx

= Y,

I "

^ _ j

f'{x)dx

JXk-\

= £ [ / ( * * ) - f{Xk-l)} = /(TT) - /("TT) = 0. fc=l

7T

/

J

/-H

f (x) cos nxdx = 2^

f'(x) cos nxdx

= y^[/(x)cosnx]J*_ i ■+ n^2

/(x) sin nxdx

= cosn7r /(TT) - cos(-n7r)/(-7r) + n /

n

/

J J

/(x)sinnxiix = n7r&n.

z-ik rxt,

/ ' ( x ) s i n n x = Y_] / •if

^ _ j

/'(x) sin nxdx JXk-i

100

Fourier Series and Fourier Transforms

k

_ If (x) sin nx

]Xk_1 - n E

k=1

k=1

J

x

f (x)

cos nxdx

Xk -1

fr = f (ir) sin n7r - f (--7r) sin(-nrr) - n

J

f (x) cos nxdx = -n7ran.

Since f'(x) is integrable on [-7r, in, we have, by Bessel's inequality, 00 (an +

1

r

)3n2)

-

J

in

n=1

r

l i2( x)dx = C2,

where C is a constant. Consider m

Sm(x) - Sn (x) =

E (ak cos kx + bk sin kx) k=n+1

where Sn(x) is the nth partial sum of the Fourier series of f. Schwarz 's inequality for real numbers, n

n

/

Using

n

I^CkdkI(-

2->

1

)"+1sinnx n

n=l

(c). Let t = 2x. Then f(x) =x = t/2= g(t), a o =

f2"

1

u

t _,

2

di=

\ f(•2* t an = ~ / - cos ntdt = Wo 2

0 < t < 2TT.

1 r 2,2*

4T^"=*■ 1 /* * / sin ntdt = 0. 2TT y 0

bn = — I -sin ntdt = — -—[tcosntln* -\ / ,0 ■n J0 2 2n7rl 2mr J0 Hence, the Fourier series of f(x) is

cos ntdt = — . n

7r ^-> sin2ni 2 ~^-j n ' n= l

4.4

Application t o the wave equation

Consider «tt-Wn=0,

t > 0, 0 < i < 7r,

u(0,f) = u(7r,*) = 0,

(4-17)

>0]

(4.18)

u(i,0) = /(x),

0<X 1/n2 converges, the series > l and converges. (B). Extend g(x) to be an odd function G(x) on [-7r, 7r] and let cn and dn be the Fourier coefficients of G(x) on [-ir, 7r]. Then Cn = 0, dn = nbn.

Applying the same approach as in (A), we show that IbnI converges. By the Weierstrass M-test, the series (4.21) converges uniformly in 0 < x < 7r and t > 0 and represents a continuous function there. Hence u(0, t) = u(7r, t) = 0 and u(x, 0) = E°° 1 an sin nx = the Fourier sine series of f (x)

107

Application to the wave equation

on [0, ir], which converges to f (x) by Theorem 4.3. Thus, u (x, t) satisfies (4.18) and (4.19). (C). To show that the series (4.21) can be differentiated term by term, we assume that f E C2 [0, ir], f (0) = f (ir) = f"(0) = f"(ir) = 0, and f (3) (x) is piecewise continuous there. As before, we extend f (x) to be an odd function F(x) on [- 7r, 7r] and let pn and qn be the Fourier coefficients of F(3) on [-7r, 7r]. Then p0=0, Yin=-n3an, qn=0.

Similarly, let g(x ) E C'[0,7r], g( 0) = g(7r ) = 0, and g"(x) be piecewise continuous there. We extend g(x) to be an odd function G(x) on [-,7r, 7r] and let un and Vn be the Fourier coefficients of G"(x) on [-7r, 7r]. Then vn = -n 3bn.

Un = 0,

By Bessel's inequality, we have 00 Ens a

2

cc

[f 7r 2f^

(3)(x)]2dx,

n=1

0

2

7r

n=1

f [9, (x)]2dx.

o

We now differentiate the series formally with respect to x and t and get 00 ux - E [an cos nt + bn sin nt] n cos nx, n=1 00

Ut - [-nan sin nt + nbn cos nt] sin nx, n=1

00 uxx

-

[an cos nt + bn sin nt]n2 (- sin nx), n=1

CO

utt - E [an cos nt + bn sin nt] (-n2) sin nx. n=1

(The symbol - means the series on the right is related to the formal differentiation of u(x, t) of the indicated variable(s) to the required order.)

108

Fourier Series and Fourier Transforms

It follows from Schwarz's inequality that all these differentiated series converge uniformly in any closed bounded subset in 0 < x < 7r, t > 0. This implies that the series u(x, t) can be differentiated term by term so that u(x, t) satisfies (4.17). We then use the uniform convergence of the sine series of g(x) on [0, 7r] to get (4.20). We would like to compare the solution u(x, t) given by (4.21) and (4.22) with the d 'Alembert solution given in Section 2.5. We rewrite u(x, t) in (4.21) as u(x, t) = v (x, t) + w(x, t) where 00 v(x, t) = an cos nt sin nx

(4.23)

n=1

and 00 w(x, t) = E bn sin nt sin nx. (4.24) n=1

Let f (x) be continuously differentiable on [0,7r ] and f (0) = f (7r) = 0. Then the Fourier sine series of f (x) converges uniformly to f (x) on [0, 7r]. Thus, 00

f (x) _ an sin nx.

(4.25)

n=1

Then ao v(x,t) =

an sinn x+t +sinn x - t x+t +f(x-t

(4.26)

n=1

Similarly, if g(x) is continuously differentiable on [0, 7r] and g(0) = g(7r)=0, then

g(x) = E 00 cn sin nx, (4.27) n=1

where cn = 2 g( x)sinnxdx. 7r 0

(4.28)

Fourier

109

integrals

Comparing (4.28) with (4.22), we have c„ = nbn. Integrating (4.27) with respect to x, we have i-X + t

/

°°

rX + t

g(s)ds = YJ Cn /

■l*-1

n=l

0°

sin nsds = — YJ 6n[cos n(x + t) — cos n(x — t)]

•'x~t

n=l

which is permissible because of the uniform convergence of the Fourier sine series. Then we have 1 °° rx+t w(x,t) = - ^ 6 „ [ c o s n ( x - t) - c o s n ( x + r.)] = 1/2 / g(s)ds. 2

n=l

(4.29)

^ - '

From (4.26) and (4.29), we see that the solution in (4.21) is the same as the d'Alembert solution. However, if we compare the conditions on f(x) and g(x), we note that we do not require both f^(x) and g"(x) to exist in the d'Alembert solution. 4.5

Fourier integrals

Let / be piecewise smooth and absolutely integrable on (—00,00). Then, on any finite interval [—L, L], the Fourier series of f(x) converges to f(x) at the point of continuity, i.e. 00

f(x) = a0/2 + y~][an cos(nirx/L) + bn sin(n7T:r/L)j

(4.30)

n=l

where 1 fL an — y f(x) cos(mrx/L) dx, Li J-L

1 fL bn = y / f(x) sin(mrx/L) dx. L J-L (4.31) At the point of discontinuity, we replace f(x) by \f(x + 0) + f(x - 0)]/2. Substituting (4.31) into (4.30) and simplifying, we get 1 f(x) = Jl]

fL

°° 1 rL f{t)dt + Y . l j /(OcosMt-iJ/LJdf.

(4.32)

Let L —> 00. The first integral goes to zero since f(x) is absolutely integrable on (—00,00). For the second integral, we set un = mr/L and Au n = u n + 1 - un = ■K/L.

Fourier Series and Fourier Transforms

110

Then we have

00

1

^{ J L f (t) cos[un (t - x)]dt } Dun. f (X) = lim L-+oo 7r

(4.33)

n=1

The sum in (4.33) looks like a Riemann sum of a function F(u, x), defined by F(u, x) =

1

f

L f (t) cos[u (t - x)]dt

(4.34)

7 L

for 0 < u < oo for a fixed x. It is natural to expect as L --> oo, (4. 33) goes to the improper double integral of the form f (x) = '

J

J

f (t) cos[u(t - x)]dtdu.

(4.35)

o .

The formula (4.35) is called the Fourier integral of f (x). It represents a function defined over (-oo, oo) in the same way as a Fourier series represents a function on a finite interval. In order to justify the validity of this representation, we need to prove the following theorem. Theorem 4.5 Let f(x) be piecewise smooth and absolutely integrable on (-oo, oo). Then f f (X + 0) 2 f (X - 0) - 1 f 00 du

J

0 f (t) cos[u(t - x)]dt.

(4.36)

00

Proof: By definition , the double integral in (4.36) can be rewritten as lim 1

/a oo du f (t) cos[u(t - x)]dt. (4.37)

J

J

A-^o0 7r 0 00

Since f (x) is absolutely integrable on (-oo, oo), the inner integral converges uniformly with respect to u on [0, A]. Hence, we can reverse the order of integration to get lim '00 7r 1 1- 00

for l;=t-x.

f (t) sin [A(t - x)] f (x + dt = lim t-x A -too 7r 00

sin(Ae) d

Fourier integrals

111

By means of the contour integration, we know

(See Churchill, Brown & Verhey, [4,193]). Thus, we have

/° ! £^

J-oo

=

fJo! i£ ^ =2 *, A>O.

In order to establish (4.36), it is sufficient to show that lim I 1° [/(' + e ) - / ( ; - 0 ) W * ) * A^oo 7T J_00

f

=

o,

(4.39)

and lim I f

[/(^0-/(^0)]sin(AQ ^

= Q

^

We split the integral (4.40) into three parts, i.e. I = Ii + I2 — h , where 7

I /• b [/(z + 0 - / ( * + 0)]si"(AQ ^ £

7T J0

1 f ~ / ( i + flsin(AQ / 2 = - / JV ' ; v " « and / ( i + 0) /°°sin(A0 7T

Jb

where 6 > 0 is arbitrary. For b sufficiently large, we have

which can be made as small as possible. Once we fix 6, then I\ —> 0 as A —> 00 by the Riemann-Lebesgue theorem. Let z = A£ in ^3. Then =

/(S+Q) r * J\b /A6

-2

112

Fourier Series and Fourier Transforms

which goes to 0 as A -4oo from (4.38). Thus, (4.40) is proved. (4.39) can be proved in a similar way. Let cos[u ( t - x)] = cos ut cos ux + sin ut sin ux. Then the Fourier integral formula ( 4.36) can be written as f (x + 0) + f ( x - 0) 2

= f "0 [A(u ) cos ux + B(u) sin ux]du o 10"0

(4.41)

where 1 °O 1 O° A(u) _ f (t) cos utdt , B(u) f (t) sin utdt. 7r _ 00 00

J

J

(4.42)

If f (x) is defined on (0, oo ), we can extend f (x) as an even (or odd) function on (-oo, oo), and obtain the cosine (or sine ) integral of f (x), i.e.

00

(A). the cosine integral of f (x) on (0, oo) : f 00 A(u) cos ux du , A(u) = -

J0

f (t) cos ut dt ,

(4.43)

(B). the sine integral of f (x) on (0, oo):

L

00 B (u) = - J f (t) sin ut dt. (4.44) 0

00

B (u) sin ux du,

Remark: The conditions for pointwise convergence of the cosine integral and the sine integral of f (x) are similar to the conditions given in Theorem 4.5. Example 4.3 Find the Fourier integral of f (x) = sin x on [0, ir]; = 0 elsewhere . Deduce that 00 cos(7ru/2) 7r

10

1-u2 du= 2

Solution: It is easy to see that f (x) is continuous , piecewise smooth, and abso-

lutely integrable on (-oo, oo ). From (4.36), we have , for -oo < x < oo, 1 °° J f (x) =

Jo

[sin(t + ut - ux ) + sin(t - ut + ux)]dtdu

Fourier transforms

113

cos(t - ut + ux) o du _ 1 00 cos (t + ut - ux ) 21r 0 H 1+u 1-u - 1 ' [cos u(7r - x) + cos ux] du 1-u2 ,

7r 0

= the Fourier integral of f (x). At x = it/2, the above integral is simplified to j oocos(lru/2) du = . 1 - u2 2

4.6 Fourier transforms The Fourier integral formula (4.35) can be written in a complex form if we + e-4u(t-x)]/2. Hence, (4.35) replace the function cos[u(t - x)] by [eu'u(t-x) becomes

f (X) =27r 1

1-00 J 00 f ( t)eiu(t-x)dtdu. (4.45)

Definition 4.1 The Fourier transform of f (t) in (-oo, oo) is defined as 0 (4.46) F[f ] = f (u) = f (t)ezutdt.

f

00

F[f] exists if f is absolutely integrable on (-oo,oo). If, in addition, f is continuous and piecewise smooth, then, by Theorem 4.5, F-1[f] = 2-

f(u)e_iuxdu = f(x), (4.47)

and F-1 [g] is called the inverse Fourier transform of g(u). Hence, Theorem 4.5 can be restated as Theorem 4.6 If f (x) is absolutely integrable, continuous, and piecewise smooth on (-oo, oc), then 00 f(u)e-'uxdu f(x) = 27r,f where f (u) is the Fourier transform of f (x).

114

Fourier Series and Fourier Transforms

Theorem 4.7 Let f (x) be absolutely integrable on (-oo, oo). Then its Fourier transform j (u) is bounded and uniformly continuous for -oo < u < 00. Proof: f (u) is bounded since l f (u) l < f. l f (t) l dt = K, a constant. Let u and v be two real numbers and let v - u = h. Then

Y (u) (u) - .f (v) I = I roo f (t)ezu,t [1- ei h t]dtl. However,

l1- ei ht l

= 1 1 - e i htIle- iht /2l =

21 sin(ht /2)1.

Hence,

l f(u) - f (v) l< 2

N

If--

+ f N + f N } l f (t) sin(ht/2) l dt. N N

.

Let e > 0. Choose a positive number N such that the first integral and the third integral are each less than e/3 since f is absolutely integrable. The second integral is less than KNlhl/2 since l sin xi < lxl. Hence, we can find a S = b(e) such that whenever lu - vl = lhl < S, the second integral is less than e/3. We have thus proven the uniform continuity of AU). Theorem 4.8 limjuj-^ j (u) = 0. Proof: 00 AU) = f

f (t) [cos ut + i sin ut] dt. oo

Consider f f (t) cos ut dt = { f -N +f N + f 00 } f (t) cos ut dt. 00

oo

N N

The first and third integrals can be made as small as we like by choosing large N because f (t) is absolutely integrable on (-oo, oo) and the second integral converges to zero as Jul -> oo because of the Riemann-Lebesgue theorem. Thus, limju^,o,, foo f (t) cos ut dt = 0. Similarly, f f (t) sin ut dt

0 as Jul -+ oo.

Fourier transforms

Corollary. integrals J_

115

Let / be absolutely integrable on (—00, 00). Then both the f(t)sinut dt and J_ f(t)cosut dt go to zero as |w| —> 00.

From Theorems 4.7 and 4.8, we know that f(u) is bounded, uniformly continuous, and limiui-xx, f(u) = 0. However, f(u) is not necessarily ab­ solutely integrable in (-00,00) as illustrated by the following example. Example 4.4

Find the Fourier transform f(u) of [0,

if x < 0.

Show that f(u) is not absolutely integrable in (—00, 00). Solution: r°°

f(u) = / J(u) ^

„(-i+tu)i

e~xeiuxdx

1 1 .„

+ = l°° = (_l+iu)lo l+u2-

Thus, / which diverges.

du

rO

OO

\f{u)\du = / J —C

■OO

Theorem 4.9 Let f(x) be continuous, absolutely integrable, and converge to zero as \x\ —► 00, and let f'(x) be absolutely integrable on f—00,00). Then F[f] = -iuF{(}.

(4.48)

Proof: For u ^ O , ;

J-00

L

t«

j

-~

;_«,

in

Hence, (4.48) holds for u / 0 . Moreover, by inspection, (4.48) also holds for u — 0. Corollary. L e t / 6 CP-1(-oo,oo),f^(x) -> 0 as \x\ — 00,fc= 0,1, ...,p1, and fW be absolutely integrable in (—00,00), j = 0,1, ...,p. Then

F[/ 2, 1 < x < 2, -2<x b > 0.

32. Let f (U) = ex J °° 2

cos ux dx.

Show that f (u) satisifies the equation y' + uy = 0 and deduce from its solution that the Fourier transform of e-r2/2 is 2,r e-"2/2

Chapter 5

The Heat Equation

5.1 Derivation of the heat equation Consider a homogeneous rod of uniform density p and constant cross section S, placed along the x- axis from x = 0 to x = L. Assume that the heat flows through the rod only in the x-direction, with the lateral sides insulated, and that the temperature of the rod is constant at any point of the cross section.

-Kau,, (x + Ax, t)

-Kvux(x, t)

x Fig. 5.1 Small section of heat-conducting rod Let u(x, t) be the temperature of the cross section of the rod at the point x and the time t and let q(x, t) be the rate of heat flow at the point x 127

128

The Heat Equation

and the time t. Consider a segment AB of the rod between the cross section at A (with abscissa x) and the cross section at B (with abscissa x + Ox) (see Fig. 5.1 ). From the theory of heat conduction , the rate of heat stored in the segment AB is proportional to the mass of the material and to the rate of change of the temperature. Thus, x +Ox

q(x,t) = J

cpvut ( s,t)ds

(5.1)

x

where c is the specific heat of the material of the rod and a is the area of the cross section . We assume that the function u(x, t) is continuously differentiable. Then x+Ox

q(x, t) = J

cpaut (s, t)ds = cpuut(^, t) Ax, x < ^ < x + Ax

(5.2)

x

where the second equality is due to the mean value theorem for integrals. On the other hand , the rate of heat flow through any cross section is proportional to the area a of the cross section and the gradient of the temperature normal to the cross section , and heat flows in the direction of decreasing temperature . Thus, the rate of heat flowing in the segment througn the cross section at x is -Kaux ( x, t), while the rate of heat flowing out of the segment through the cross section at x + Ax is -Kvux (x + Ax, t) where K is the thermal conductivity of the rod. Hence , the rate of heat stored in the segment is -Kau,, (x, t) + Kau (x + Ax, t).

(5.3)

From (5.2) and (5.3), we obtain cput(^, t) Ax = K[ux(x + Ax, t) - ux(x, t)].

(5.4)

Dividing (5.4) by cpix and letting Ox - 0, we obtain ut - kuxx = 0 (k = K/cp)

(5.5)

which is called the (one-dimensional) heat equation. Heat may be transferred by convection or radiation between the rod and the surrounding medium or from some other external source. If the rate of heat supply is f (x, t) per unit volume per unit time, then the term f x+

x

f(s,t)ds= f(,t)x,

Derivation of the heat equation

129

must be added to the right side of (5.4). As before, we get ut — kuxx — F(x, t) where F(x, t) = f(x,t)/(cp). tion.

(5.6)

(5.6) is called the nonhomogeneous heat equa-

In order to determine the temperature of the rod uniquely, we need additional information regarding the initial temperature of the rod and the boundary conditions at its two ends. Hence, we may consider the initialboundary value problem for the heat equation (5.5) or (5.6) with u(x,0) = f(x)

(5.7)

and u(0,t)=u{L,t)=0.

(5.8)

If the rod is insulated at both ends such that the heat flow is zero there, then we have ux(0,t)=ux(L,t)=0.

(5.9)

A third boundary condition at x = 0 (or at a: = L) can be given by ux(0,t) =-Xu(0,t)

or

ux(L,t)

=-nu(L,t)

(5.10)

where A and fi are constants. This condition corresponds to Newton's law of cooling at the ends of the rod with the temperature of the surrounding medium equal to zero. In studying the heat conduction in a very long rod, we may assume that —oo < x < oo, and that no boundary conditions are required. Thus, we may consider the initial value problem of (5.5) or (5.6) with initial condition u(x,0) - f(x),

- o o < x < oo.

(5.11)

Another application of the heat equation concerns the diffusion of a gas in a hollow tube. We assume that at any time t the concentration of the gas in any cross section of the tube is constant. The diffusion process can be described by a function u(x, t) which denotes the concentration of the gas in the cross section at x and at time t. Then it can be shown that u(x, t)

The Heat Equation

130

satisfies ut - kuxx = 0 where k is the coefficient of diffusion. We refer to Tychonov and Samarski [12, 156-7] for its derivation.

5.2 Maximum principle Let T be a positive constant and let 0 ={(x,t)10<x 0 be given. Consider SZ = [0, L] x [0, T]. Let SN(x,t) be the nth partial sum of the series (5.24). Let e > 0. There exists an integer NE such that for N > M > NE, we have f (x) - SN (x, 0) < e/2 or SN (x, 0) - SM (x, 0) < e. The function SN (x, t) - SM (x, t) satisfies the heat equation in (0, L) x (0, T], is continuous in S2, and vanishes at x = 0 and x = L. Hence, by Theorem 5.1, SN (x, t) - SM (x, t) I < e in Q. This means that the sequence { SN (x, t) } converges uniformly in Q. Since SN (x, t) is continuous, u (x, t) is also continuous in ci. Then u(x, 0) = f (x) and u(0, t) = u(L, t) = 0. Since T is arbitrary, u(x, t) is continuous in [0, L] x [0, oo) and (5.20) and (5.21) are satisfied. Hence, we have proved the following theorem: Theorem 5.4 Let f(x) be continuous, piecewise smooth in [0, L] and f(0)=f(L)=0. Then the initial-boundary value problem (5.19)-(5.21) has a unique solution given by (5.24 ) and (5.26).

5.4 Nonhomongeneous problems and finite Fourier transform

Consider the problem vt - v1X = F(x,t),

0 < x < L,

v(x,0)=0,

0<x0,

v(0, t) = v(L, t) = 0, t > 0.

(5.27)

(5.28)

(5.29)

Based on the homogeneous case with the same boundary conditions (5.29) in Section 5.3, we assume v(x, t) = 0" > bn(t) sin (n7rx/L) n=i

(5.30)

with L bn(t) =

L f0

v(x, t) sin (n7rx /L)dx

(5.31)

136

The Heat Equation

which is called the finite sine transform of v(x , t). We assume that F(x, t) can be represented by a Fourier sine series for 0 < x < L for a fixed t, i.e. F(x, t) = E Fn(t) sin (nirx/L)

(5.32)

n=1

where L Fn (t) = L I

F(x, t) sin (nirx/L)dx.

(5.33)

Differentiate ( 5.31) with respect to t and , by (5.27), we obtain

f b;^ (t) = L

J

0

L vt sin (n^rx/L)dx = 2k

J L vyx sin (n7rx/L)dx+Fn (t).

(5.34)

After applying integration by parts and using the boundary conditions (5.29), the first integral on the right of (5.34 ) is equal to - kAnbn (t) where An = n27r2 /L2. Together with the initial condition (5.28), we have b(, + kAnbn = Fn(t),

bn(0) = 0. (5.35)

Hence, the unique solution of (5.35) is

bn(t) =

ft

exp[-kAn (t

-T)]Fn(T)dT.

0

(5.36)

Substituting ( 5.36) into ( 5.30), we get 00 t v(x, t) = E[ exp(-kA n(t

- T ))Fn( T) dr ]

sin (n7rx/L)

(5.37)

n=1 0

which is a formal solution of (5.27)-(5.29). Verification:

(A). Let 00

vn,(t) sin (n7rx/L)

v(x, t) =

(5.38)

n=1

where

vn(t) =

ft 0

exp[-kAn (t - T)]Fn(T)dT.

(5.39)

Nonhomongeneous problems and finite Fourier transform

137

We assume that F(x,t) is continuous for 0 < x < L,t > 0. Let T be a positive constant. Denote R = [0, L] x [0,T] and let M = max/j |F(x,£)|. Then

\Fn(t)\ 0.

(5.50)

We note that (A). G is continuous at x = , and G -> 0 as IxI --> oo. (B). f 00 G(x - , t)d^ = (7r)-1/2

exp (- r72)di7 = 1

for 77 = ( - x)/(4kt)'/2. (C). G., G.., and Gt are continuous in t > 0, and G satisfies the heat equation there since X) G G x 2G G:,: = _ (^ 2kt ' Gt = kGxx 2t + ( 4kt2 In order to demonstrate that u(x, t) in ( 5.46) is a solution of (5.40)-(5.41), we assume that f (x) is continuous and bounded on (-oo, oo ). Let L, to, and T be any arbitrary positive constants and let

S= { (x, t) I -L <x < L, to < t < T }. Denote u(x,t) = F f ( ^)G(x - ^,t)d^ . 0

(5.51)

We differentiate u(x, t) formally with respect to x and t and get

I,(x,t ) = J

70

f ( ^)G. (x - ^,t)d^,

I2(x,t) = f00 f(^)Gt(x - e,t)d^ = k f- f(^)Gxx(x - ^,t)d^. 00 00 Let M = sup If (x) I, -oo < x < oo. Then, for (x, t) E S, let 77 = (^ - x)/(4kt)1/2. Using (B) and (C), we obtain Iu(x,t)I M, IIi(x,t)I <M(kirto)-1/2 f IrlIe-,2dr7, and II2(x, t) I

< M(2to )-17r-1/2 f7. e-,72di7 + Mt0

17r-1/2 f ^ ij e-"12d77.

Using Theorems 1.2 and 1.4 for the uniform convergence of improper integrals , we can show that u(x, t) is bounded in t > 0, and the integrals u (x, t), 11 (x, t), and I2(x, t) converge uniformly in S. Thus, ux(x, t) =

141

The initial value problem

Ii (x, t), ut (x, t) = kuxx (x, t) = I2 (x, t), and u(x, t) satisifies the heat equation there. Since L, to, and T are arbitrary, u(x, t) satisfies the heat equation in t > 0. It remains to show that lim(x,t )-. (x0,o) u (x, t) = f(xo ). Without loss of generality, we assume that xo = 0. Since f is continuous at x = 0, for c > 0, there exists a 5 > 0 such that for IxI < b, we have If (x)- f (0) < E. From (B) and ( 5.51), we have u(x, t) - f (O) = I3 + I4 where

I3 = [f f (0)]G(x - ^, t)dt; IeIa We note that G(x - 6, t ) > 0 for all x and 6 and t > 0. Hence, we obtain

1I31

G(x -^,t)dt;

< E

< E

J 00 G(x -^,t)dl; = E. 00

For IxI < S/2,

1141

< 2M{

8

f"OX

+

J - b -x}G(v,t)dv < 2M J

< 4M7r-1/2

G(o,t)da

lo bo/2

00

2

00 f(6k

d7j.

1/2t-1/2)/4

Since fo e-,72 drl < oo, 1I4 1 < e for t sufficiently close to 0. Hence, lim(x,t)-.(o,o) u(x, t) = f (0) and the following theorem is proved. Theorem 5.6 Let f(x) be continuous and bounded in (-oo, oo). Then the initial value problem (5.40)-(5.41) has a unique bounded solution given by (5.46).

The Heat Equation

142

5.6 The initial value problem for the nonhomogeneous equation Consider the initial value problem -oo < x < oo,

ut - kux., = h(x,t),

t > 0,

(5.52)

-oo < x < oo. (5.53)

u(x, 0) = 0,

Let 00 u(w, t= j

u(x,t)edx

and f00 h(w, t) = h(x, t)e"xdx. 00

Using the same method as in Section 5.5, we obtain at + kw2u = h,

u(w,0) = 0. (5.54)

Then the solution of (5.54) is given by (w,t) = f e_ kw2(t_T)h (

r)dT

(5.55)

Using the inverse Fourier transform for u(w, t), substituting for h(w, t), and formally interchanging the order of integration, we get

u(x, t) = ^t f' G(x - ^, t - r)h(^, T)dddr 0

(5.56)

00

where e- x2/(4kt) G(x t) = 7 (4k7rt)1/2 ( 5.56) and (5.57) give a formal solution of (5.52)-(5.53). Verification:

(5.57)

The initial value problem for the nonhomogeneous equation

143

(A). We assume that h(x, t) is continuous and bounded for all x and t > 0. Let M = sup I h(x, t) I there. Then

Iu(x, t)1 <M

ft f oo

J0 J

G(x-^,t-T)d^dr=Mt. 00

Hence, u(x, t) converges uniformly in any closed rectangle I in t > 0 and is continuous in t > 0. Furthermore, lim(x,t)-.(xo,o) u(x, t) = 0. Similarly, let o t f oo

fJ

v(x, t) =

Gx (x - ^, t - r)h(^, T)dedT,

00

which is obtained by formal differentiation of (5.56) with respect to x. Then 00

Iv(x, t)I <M

oo [2 Jf 0 f

= M

t - r)]1I-xj G (x-,t-T)ddr

t

t

J0

t (k7r) -1/2(t -T)-112dr = 2M(

1fk)1/2.

This implies that v (x, t) converges uniformly in S2 and ux(x, t ) = v(x, t) in t > 0 by Theorem 1.4. (B). To show the existence of uxx, we further assume that h(x, t) is continuously differentiable in any closed rectangle ci in t > 0. By the mean value theorem, we have

Ih(r;,T) - h(x,T)I < CIt; - xI where C = C(Sl) = maxo I hx (x, t) I. We note that f

00 - T= Gt- ,tf00 00

_,t_r

00 =

J 00 G^g(x-l;,t- r)d^=G£(x-^,t-T)I£ _00=0.

Formal differentiation of v(x, t) with respect to x yields p(x, t) =

JJO FO Gxx(x - t;, t - T)h(^,T)d1;dr 0

0

(5.58)

144

The Heat ft

=

l-OO

/ JO

Equation

Gxx(x-t,t-T)[h(t,T)-h(x,T)]d£dT.

J-oo

We also note that UXX{X

^

T)

~

2k(t-T)

+

ik2(t-T)2

■

Then, using (5.58), we get pt

\p(x,t)\0, t > 0 w(0, t) = 0, w(x, 0) = O(x) - h(0). The above problem can be formally solved by extending both -h(t) and O(x) - h(0) as odd functions in x on (-oo, oo).

5.8 Problems

1. Let v (y) = u(x, t) where y x/(2v). Find a differential equation that v(y) must satisfy if u(x, t) satisfies the heat equation ut = uxx for t > 0. 2. Let u be a solution of ut - kuxx = 0, 0 < x < L, t > 0; u(0, t) = u(L, t) = 0. (a). Show that A L L u2 (x, t) dt _< 0. (b). If u(x,0)= 0 for 0 < x < L, show that u(x,t) = 0 for 0 <x < L andt>0. 3. Show that if u(x, t) is a continuous solution of ut-kuxx = 0 for 0 < x < L and t > 0 such that ux (0, t) = 0, then the maximum of u is attained either at t = 0 or x = L. (Hint. Extend u(x, t) as an even function of x on [-L, L].) In Problems 4 through 6, find a formal solution of the initial-boundary problem. 4. ut-uxx+u=0for0<x0; u(0, t) = ux ('7r, t) = 0, u(x, 0) = x(7r - x). 5. ut-uxx+u=0, 0<x0; ux(0,t)=u(1,t)=0, t>0, u(x,0)=x2-1, 0<x 0; ux (0, t ) = ux (7r, t) = 0, u(x, 0) = f (x) where w is a constant. 8. ut - uxx=t for 0 < x < -7r and t > 0; u(x, 0) = cos 2x; ux (0, t) = ux (ir, t) = 0.

9. ut -uxx = F(x,t), 0 < x < ir, t>0; ux (0, t) = u(ir, t) = 0.

u(x,0) = f(x),

10. Find a formal solution of the initial-boundary value problem of the heat equation ut - kuxx = 0, 0 < x < oo, t>0; u(x,0) = f(x), 0 < x < 00 with the boundary condition (i) u(0, t) = 0, (ii) ux (0, t) = 0 for t > 0 respectively. 11. Find a formal solution of ut -kuxx + hu = 0, 0 < x < oo, t>0;

ux (0 , t) = 1, u(x, 0) = f (x)

where h and k are positive constants and u and ux go to zero as x -> oo by using the cosine transform w(s, t) = f °O u(x, t) cos sx dx. 12. Find a formal solution of ut - uxx + u = 0,

x > 0,

t > 0; u(x , 0) = 0,

u(0, t) = t;

where u and ux go to zero as x -> oo for t > 0 by the sine transform w(s, t) = f0O u(x, t ) sin sxdx. 13. Find a formal solution u(x, t) of ut-uxx+to=0, x>0, t>0; u(x,0)=e-x, x>0;ux(0,t)=0, t>0; such that both u and ux go to 0 for t > 0 as x -* oo by the Fourier cosine transform w(s, t) = f o' u(x, t) cos sx dx. Verify your solution. 14. (a). Find the Fourier sine transform of f (x) = xe-x, x > 0. (b). Verify that ux, t

1 -t2/2 f e -BZtsinsx JO 1 ds - 4ir a

Problems

is continuous in x > 0 , t > 0 and is a solution of ut - uxx + to = 0, x > 0, t.> 0; u(x, 0) = xe - x, x > 0, u(0, t) = 0, t > 0; where u and ux go to zero as x -> oo for t >0. 15. Let u (x, t) = O[x /(4t)1/2] where O(s) = fo e-"2dv. ( a). Show that u(x, t) satisfies the heat equation ut - uxx = 0. (b). Find lim(x,t )- (o,o) u (x, t) along the line x = t. (c). Find lim(x,t)-.(,,,O ) u(x, t) for any a 0 0. 16. Let K(x, t) = (4-7rt)- 1/2 exp (- x2/(4t)), t > 0. Show that ( a). for any a > 0, limt -. o+ K(x, t ) = 0 uniformly for all (xI a. (b). for any a > 0, limt-.o+ f xI>a K(x, t)dx = 0. ( c). lima-.o+ fo K.(x, t - ^r)dr = -1/2. ( d). lima-o_ fa Kx (x, t - T)dr = 1/2. (e). Kt - Kxx = 0, t>0. 17. Let u(x, t) = xt-3/2 exp (-x2/(4t)), t > 0. Show that ( a). u(x, t ) satisfies the heat equation ut - uxx = 0 in t > 0. ( b). limt --. o+ u(xo, t) = 0 for any fixed xo. ( c). lim(x,t )_(o,o) u(x, t) does not exist. 18. Solve ut - kuxx = e-t for t > 0 and u(x, 0) = 0.

149

Chapter 6

Laplace's Equation and Poisson's Equation

6.1 Boundary value problems In this chapter, we will study certain boundary value problems with Laplace's equation uxx + uyy = 0 (6.1) and Poisson's equation UXX + uyy = -4(x, y) (6.2) in two independent variables. These two equations also arise from physical phenomena ; in particular, they correspond to the steady state of the two-dimensional heat equations. Let D be a bounded domain in the xy-plane with continuous , piecewise smooth boundary C. Let u(x, y, t) denote the temperature distribution of a thin uniform plate in the domain D. Making the assumptions similar to those given in the conduction of heat in a homogeneous rod in Section 5.1, we can show that the temperature function u(x, y, t) satisfies the homogeneous heat equation Ut - k(u.x + uyy) = 0 (6.3) if no external heat source is present in D. However, if there exists an external heat supply of Q(x, y, t) per unit volume per unit time, then the temperature function u(x, y, t) will satisfy the nonhomogeneous equation ut - k(u^x + uyy) = Q(x,y,t)• 151

(6.4)

152

Laplace's Equation and Poisson 's Equation

If the heat flow is stationary in D, then both the source function Q and the temperature u are independent of the time t. We can write Q(x, y, t) = p(x, y) and equations (6.3) and (6.4) become (6.1) and (6.2) respectively where q(x,y) = p(x,y)/k. Another application for Poisson's equation is in the theory of an electric field. Let u(x, y) be the electrostatic potential in a region in the xy-plane. Then u satisfies Poisson's equation u.Ty + uyy = -4irp(x, y), where p(x, y) is the density function of the charge per unit area. If the region is free of electric charges, then p(x, y) = 0, and u satisfies Laplace's equation. For its derivation, please refer to Tychonov and Samarski, [12, 242-4]. There are two important boundary value problems for Laplace's equation and Poisson's equation: Dirichlet problem Determine a solution of (6.1) or (6.2) in D which is equal to a given function f (x, y) on C. Neumann problem Determine a solution of (6.1) or (6.2) in D such that its normal derivative on C is equal to a given function f (x, y).

6.2 Green ' s identities and uniqueness theorems Let D be a bounded domain with continuous , piecewise smooth boundary C. Let u and v be in C2 (D) fl Cl (D) and Av = vx,, + vyy be bounded in D. Applying Green 's theorem

f (Pdx + Qdy) = f L (Qx - Py) dxdy

(6.5)

to P = -uvy and Q = uv,., we get u(vxdy - vydx) =

fc

JL

Let C: x = x(s), y = y(s), 0 < s < L, where s is the arc length param-

Green's identities and uniqueness theorems

153

eter and L is the length of C. Then

IC u(vxdy - vydx) = fC u

8n ds (6.7)

where an is the outward normal derivative of v on C. Then (6.6) can be written as Lu 8n ds = f fD [u0v + uyv,, + uyvy]dxdy

(6.8)

which is known as Green's first identity. If we interchange u and v in (6.8) and subtract the resulting identity from (6.8), we get

f (u 8n - v ands = J f (uzv - vLu)dxdy

(6.9)

which is known as Green's second identity. There is at most one solution u E C2(D) n C'(D) of the Theorem 6.1 Dirichlet problem Au = -q(x, y) in D, u(x, y) = f (x, y) on C. Proof: Let ul and u2 be two solutions of the Dirichlet problem. Then w = ui-u2 is a solution ofLw=0inDandw=0 on C. Now letu=v=w in (6.8). We arrive at

1I

(wx + wy)dxdy = 0.

(6.10)

This implies that w1 = wy = 0 in D. Hence, w is a constant. But w = 0 on C, sou=O in D or u1= U2 in D. Theorem 6.2

Any solution u E C2(D)nC'(D) of the Neumann problem Du = -q(x, y) in D, 8u/8n = f(x, y) on C,

is unique up to an additive constant. Proof: Let ul and u2 be two solutions of the Neumann problem. Then w = U1 - u2 is a solution of Ow = 0 in D and 8w/8n = 0 on C. Let u = v = w in (6.8). We get (6.10) once more, implying w is a constant. Hence, ul and u2 differ by a constant.

154

Laplace's Equation and Poisson's Equation

6.3 Maximum principle

Theorem 6.3

(Maximum Principle) Let u be a solution of Du = -q(x, y)

(6.11)

in a bounded domain D where &,y) is continuous and nonpositive and let u be continuous in D and on its boundary C. Then mcaxu(x, y) = max u(x, y).

(6.12)

Proof: We divide the proof into two parts, A and B. (A). Let q(x, y) < 0 in D. Since u is continuous in D, u assumes its maximum there. Suppose, by contradiction, u attains its maximum at a point P in D. By elementary calculus, ux = uy = 0 and uy..,, < 0, uyy $ 0 at P. This means Au < 0 at P, a contradiction . Thus, in this case, maxc u (x, y) = maxD u(x, y). (B). We now extend the result in (A) to the case q(x, y) < 0. Let M = maxCu(x, y). Our goal is to show u < M in D. Let e > 0. Define v(x, y) = u(x, y) + e(x2 + y2).

Then Ov = -q + 4e > 0 in D. From part (A), v attains its maximum on C and v(x, y) < M + cR2 where R is the radius of the circle containing D. This implies that u < v < M + eR2. Since e > 0 is arbitrary, u < M in D. We now apply the maximum principle to establish both uniqueness and stability of the Dirichlet problem of Poisson's equation. Theorem 6.4

The Dirichlet problem

Du = -q, (x, y) E D; u = f, (x, y) E C;

(6.13)

has at most one solution u E C2(D) f1 C(D). Proof: Suppose ul and u2 are two solutions of (6.13). Then v = u1 - u2 satisfies Ov = 0 in D and v = 0 on C. By the maximum principle, v < 0

155

Laplace's equation in a rectangle

in D. Similarly, applying the maximum principle to w = u2 - U1 , we get w < 0 in D. Hence, ul = u2.

Corollary Let ul and u2 be solutions of Dui =

-q,

( x) y)

E D; ui = fi,

(x, y)

E

C

(i=1,2) respectively and let If, - f21 < e on C. Then lu1 - u2I < e in D. Proof: Apply the maximum principle to ul - u2 and u2 - ul respectively in D.

6.4 Laplace 's equation in a rectangle Consider the Dirichlet problem uXX+uYY=0,

0<x 0, there exists an integer N E such that for M > N > NE, jSM(x,b) - SN(x,b)j< e. The function SM (x, y) - SN (x, y) is equal to 0 on x = 0, x = a, and y = 0, and satisfies Laplace's equation in (0, a ) x (0, b). Then, by the maximum principle , I SM (x, y) - SN (x, y) I < e in [0 , a] x [0, b]. This means that {SN(x, y)} converges uniformly to a continuous function u(x, y) there.

158

Laplace's Equation and Poisson's Equation

Hence, u (x, y) satisfies u(x, b) = f (x), u(0, y) = u(a, y) = u (x, 0) = 0. Thus, we have the following theorem: Theorem 6.5 Let f(x) be continuous, piecewise smooth on [0, a] with f(O)=f(a)=0. Then the boundary value problem (6.14)-(6.17) has a unique solution given by (6.20) and (6.22). The more general boundary value problem 0 <x (an cos nO + bn sin n0) (6.29) n=1

which is a Fourier series of f (0) on [-7r, 7r]. Thus, an = J f (0) cos no do,

bn = J 7r f (0) sin no do. n

(6.30)

(6.28) and (6.30) give a formal solution of (6.23)-(6.24). Verification: (A). Let f (0) be integrable on [-ir, -7r] and let

= -7r f If(O)IdO. 1 Then Ianl < c and IbnI < c. Formal differentiation of (6.28) with respect to r and 0 yield 00 ur (an cos nO + bn sin nO)nrn-1, n=1 00 ue - >(-nan sin nO + nbn cos nO)rn, n=1

162

Laplace 's Equation and Poisson 's Equation

urr

(an cos n0) + bn sin nO)n(n - 1)rn -2,

00 n=2

00 u00 - >(-n2 )(anCOSn0 + bnsinnO)rn. n=1

We find that the series for u and its partial derivatives ur, uei urr, and uee are bounded by the series M > n2rn where M is a suitable constant. Thus, all the series converge uniformly in r < ro < 1, and its derivatives can be obtained from termwise differentiation of the series (6.28). It follows that u(r, 0) is in C2 for r < 1 and is a solution of (6.23). (B). Let f ( 0) be continuous and piecewise smooth in [-7r, 2r] and f (-7r) _ f (7r). Then the Fourier of f converges uniformly to f on [-7r, 7r]. Let SN(r, 0 ) be the Nth partial sum of the series ( 6.28) and let e > 0. There exists an integer NE such that I SM(1, 0 ) - SN(1, 0)I < e for M > N > NE. Since SM(r, 0 ) - SN(r, 0) satisfies Laplace's equation in r < 1 and is continuous in r < 1 , it follows by the maximum principle that I SM(r, 0) SN (r, 0) 1 < e for r < 1. Hence { SN (r, 0) } converges uniformly in r < 1 to a continuous function u(r, 0) and u(1, 0) = f (0). Hence, we have the following theorem: Theorem 6.6 Let f (0) be continuous, piecewise smooth on [-7r, ir] with f (-7r) = f (ir). Then the Dirichlet problem (6.23)-(6. 24) has a unique solution given by (6.28) and (6.30).

6.6 Poisson's integral formula In the last section, we obtained a solution u(r, 0) of the Dirichlet problem for Laplace's equation

urr,

+

1 1 Ur + -uee = 0,

r

r < 1, (6.31)

r2

u(1, 0) = f ( 0),

(6.32)

Poisson's

integral

163

formula

given by

2

n=i

where 1 /"r an = — / f() cos ncj) d,

1 f" bn = — / /() sin n$ d^.

(6.34)

We would like to put the series (6.33) in a closed form. Let S^/(r,9) be its TVth partial sum of (6.33). Substituting an and bn from (6.34) in S^(r,6), we get SN(r,6)

= - ^

/ m l + f > n c o s n ( 9 - <j>)\d<j>.

For r < 1, the series 1/2 + Y^n°=i rn c o s n ( ^ - 4>) converges uniformly in . Hence, for any point (r, 6) inside the unit circle r < 1, we have 1 /"* 1 °° u(r,6) = Jim 5 w (r,0) = - / f{4>){- + V > " c o s n ( 9 - 0. Since {uk (x, y) } converges uniformly on C, by the Cauchy Criterion, there exists an integer N = N(e) such that for k > m > N(e), luk(x, y)-um(x, y)I < e for all (x, y) on C. We know that uk(x, y)-um(x, y) is harmonic in D and continuous on D + C. Thus, for (x,y) in D + C, we have, by the maximum principle,

Iuk(x,y)- um(x,Y)I <mD Iuk(x,y)-um(x,y)I =maxluk(x,y)-um(x,y)I This implies that {uk(x, y)} converges uniformly in D to a continuous function u(x, y) in D. For any point Q=(x, y) = (r, 8) in D, we can draw a circle DR with radius R and center Q contained entirely in D. Then, by Poisson's integral formula,

uk(r, 0) =

J

uk(R, ¢)P(r, 0; R, q)dq5.

Let n -a oo. We have n u(r, 0) = u(R, O)P(r, 0; R, O)dq5. The interchange of limit and integration is permissible because {uk (R, 0) } converges uniformly to u(R, 0) for 0 in [-ir, 7r] and P(r, 0; R, 0) is continuous there. This implies that u(r, 0), which is represented by Poisson's integral formula, is harmonic in DR. Since Q is arbitrary, u is harmonic in

D.

6.7 Green 's function for Laplace's equation Let (1;, i) and (x, y) be two points in the xy-plane. We look for a solution u = u(r), where r = (x - t;)2 + (y - 77)2, of Laplace's equation Urr

+

1 1

+ - uee = 0 . r r2

-ur

(6.42)

C.

Green's function for Laplace 's equation 167

Hence, u(r) =a log r+b where a and bare constants. If we set a= -1/(27r) and b = 0, we get u(r) = 1 log(1/r) (6.43) which is called a fundamental solution of Laplace's equation. Let D be a bounded domain with continuous, piecewise smooth boundary C. Our goal is to obtain a representation formula for a function u E C2(D) in terms of its boundary values on C and a fundamental solution in (6.43). Theorem 6.10 Then

Let u(x, y) be in C2 (D) and let (t;, 77) be a point in D.

u(+ 71) f [u an - v On]ds -

fD v0u dxdy

(6.44)

where v = (21r)-1 log(1/r) with r = (x - 2;)2 + (y - 71) 2.

Proof: Let D' be the domain from D by deleting a small disc of radius e about the center ( t;, 77) with boundary Co oriented in the clockwise direction (see Fig. 6.1).

Fig. 6.1 Domain of integration for Theorem 6.10

168

Laplace 's Equation and Poisson 's Equation

Apply Green's second identity (6.9) to D' bounded by C and Co. Since Av = 0inffD', we find -

(log r) Au dxdy = 2- f[u

n (log r) - log r on ]ds. (6.45)

On C0,r = c and a^ (log r) = -1/e. Introducing the polar coordinates on CO: x = e+ecos8,y = 71+esinO, we have zir (log r)ds = lo - f l o Co an 0 JCo

u(e+ e cos 0 , 77+e sin 9)dO = -27ru(e, 77).

(6.46) Since u E C'(D), we set M = maxco jau/anj. Then z, log rdsl < M f fCo an 0

I log eIedO = 2.7rMel log el (6.47)

which goes to 0 as a -+ 0. Let e -* 0 in (6 .45). We get (6.44). Let u = 1 in (6 . 44). We have

2- f

(log r ) ds = 1.

(6.48)

an Let g = g(x, y; t;, 77 ) E Cz (D ) that depends on the pole (t;, 77) in D such that Ag = 0 in D . Apply Green 's second identity (6.9) to u and v = g . Then we get f fD g Au dxdy = 0. j(u an - g an )ds +

(6.49)

Let (1;, 77) be a fixed point in D. Then H(x, y; ^ , rl) =

2^r log(1/r ) + g(x, y; ^, 77)

(6.50)

is also a fundamental solution of Laplace 's equation with respect to D. From ( 6.44), we have - H au ds u u OH HAudxd (6.51 (,rl) f ,( an an) f fD y r ) Definition 6.2 Green's function G(x, y; l;, 77) for Laplace's equation in a given domain D is a fundamental solution of the form (6.50) with its pole (t;, 77) in D such that G = 0 on its boundary C.

Green's function for Laplace's equation

169

Let u(x, y) E C2(D) be a solution of the problem

Du = -F(x, y) in D; u = f (x, y) onC.

(6.52)

If such a solution exists, then, from (6.51), it is given by

u(x, y) L f (C 77) 8n y; x, y)ds + ffD F(^, i7)G(^, 77; x, y)dt;di (6.53) where G is a Green's function with the role of (^, 77) and (x, y) interchanged. Assuming that Green's function for Laplace's equation in D exists, we would like to show that Green's function for a bounded domain is unique. Suppose that there are two Green's functions Gl and G2 for D. The resulting difference Gl -G2 is harmonic in D , continuous in I) and vanishes on the boundary C. By the maximum principle, G1 - G2 vanshies in D. Hence, Gl = G2 in D. Next, we would like to show that the Green's function is symmetric for a bounded domain D, i.e.

G(x, y; ^, rl) = G(^, rl; x, y)•

(6.54)

Fig. 6.2 Domain of integration for symmetry of Green's function

170

Laplace 's Equation and Poisson 's Equation

Let (^, 77) and ((, T) be two distinct points in D. Consider the domain D' obtained from D by deleting two small discs of radius e about the points (^, 71) and ((, T) respectively. Let C1 and C2 be the boundary curves of these two discs taken in the clockwise direction (see Fig. 6.2). Apply Green's second identity (6.9) to the functions u(x, y) G(x, y; , 77) and v(x, y) = ( , , r ) in Y. Since Du = Av = 0 in D' and u = v = 0 on C, we have 8v eu (u -` - v -)ds = 0. n an Iif+C2 a

(6.55)

Using calculations similar to ( 6.46)-(6.47), we have (u an v an )ds = -v(e, rl)

lo l

and f

8

l o c2 (u -v On -

f

au n)ds=u(C, 7- ) .

Thus, G(^, i ; C, T) = G((, T; 6, i) and ( 6.54) is proved. We will obtain Green's functions for Laplace's equation for two special domains by the method of images. (A). Let P = (, 77) = (p, 0) denote the pole of the Green 's function G(x, y; t;, 77) in a disc D of radius R and center (0,0) and Q = (x, y) = (r, 6) is a variable point in D . Denote ro = ^PQ^ to be the distance between P and Q . For (6 , ,q) = (0, 0), let

G(x, y; 0 , 0) = 2^ log(R/r). It is easy to verify that G(x, y; 0, 0) is the Green 's function with pole (0,0). For (6, 77) # (0, 0), let P* = (R2e/ p2, R2i7 / p2) denote the inverse of the point P with respect to the circle C : r = R lying on the same half line. (i.e. ID-PI x SUP-*I = R2.) (See Fig. 6.3.) Define G(x, y; ^, rl) = where ro = IQP* 1.

27r log(Rro0

(6.56)

171

Green's function for Laplace 's equation

Fig. 6.3 Green's function in a disc It is clear that 1og(1/r0*) is harmonic inside D and when Q is on C, the triangles OPQ and OQP* are similar, and

I PQI

pJOPI I

ro o

Hence, G(x, y; 6, y) is the Green's function for the disc D. We note that ro = r2 + p2 - 2rp cos(O - 0) and ro = r2 RR4p-2 - 2rR2p-1 cos(O Thus, 1 R2 + r2p2R-2 - 2rp cos(O - 0) G(x, y; ^,'7) = G(r, 0; p, ') = 47r log ^ r2 + p2 - 2rp cos (B - 0) 1.

(6.57) We now return to (6.52 ) and (6.53 ). For F( x, y) = 0 in D, a disc of

172

Laplace's Equation and Poisson's Equation

radius R and center (0,0), we have 8G

r 2 - R2

8n I p=R _ 27rR[R2 + r2 - 2Rr cos(O - ^)] Then the solution u(r, 0) of the Dirichlet problem

Du=0, r 0. As r -+ R, u(r, 0) -* 0.

Laplace's Equation and Poisson's Equation

174

It is easy to show that G(r, 9; p, 0), as a function of r and 9, is in C2 (D) and satisfies Laplace's equation in D except at the point (r, 9) = (p, 0). In order to justify differentiation under the integral for the function u(r, 0) in (6.61), we have to show that u and its partial derivatives u,., ue, u,.,., and u99, obtained by formal differentiation from (6.61), converge uniformly in the closed set S = Jr < a < R} for any positive constant a. We note that log[R2 + r2p2R-2 - 2rp cos(9 - ¢)] behaves well for r < a whereas the function log ro where ro= r2 +p2

2rpcos(9-0)

is singular at (r, 0) = (p, 0). Hence, in order to demonstrate u(r, 0) is in C' (D), it is sufficient to show the uniform convergence of the integrals

J J (log ro)Fdv, J J (log ro)TFdv, and

J f (logro)oFdci in the region r < a where do, = pdpdO. Let DE = fro < e} C D where e is sufficiently small. We note that ro = (r-pcos(B-0))z+p2sinz(9 Thus, we have

Ipsin (9 - 0)I ro.

Ir - p cos (9 - 0)I < ro, Hence, we obtain the following estimates:

(log ro)rl=lr-pcos(9- < 1 roz ro and

(log ro)e I = I rp sin(9 - ¢) I < 0

ro

Then z IIf f (log ro)Fdal < 7rM[-ez loge + 2 DE

Poisson's equation in a disc

I

ff

175

(log ro)r Fdvl < 2Mirc,

and

I

IL (logro)eFdul < 2MRzrc, .

which all go to zero as c -40. Hence, f fD GFdo, f fD GrFdv, and f fD GBFdo, all converge uniformly in r r.

(6.87)

181

Dirichlet problem in the upper- half plane

The interchange of summation and integration can be justified if we assume that F(r, 0) is periodic of period 27r in 0 and F(r, 8) is continuously differentiable in r < R. We note that for0 0. 1r 00 y + (^ - x)

(6.94)

Verification: Let f (x) be continuous and bounded on (-oo, oo). (A). Let M = sup I f (x) I for x E (-oo, oo). We would like to show that lim(x,y )- (a,o) u (x, y) = f (a). Without loss of generality, we assume that a=0.

183

Dirichlet problem in the upper - half plane

First, let f(x)=1. Then 1 y °o f td___

°°

1 dQ = 1.

J oo (S - x ) 2 + y2 7f -00 1 + U2

Since f (x) is continuous at x = 0 , for for I xl < 6, I f (x) - f (0)I < E.

e

> 0, there exists a 6 > 0 such that

u(x,y) - f(0) = I + J,

Let where

I = y f1 f(^) - f(0) d^ it E1a (x - )2 + y2 Then

d

E

Y ,

f , ^ I<s (x -

S)2

y Ed^ +y2 C o° (x- S)2 +y2 -

E.

Now, for I xl < J /2,

1 JI

2My I F

b -y

{Ja-x + f_oo

2My 1 1 } ^2 + yz dQ < LI^5/2 2+Y2 dv

- 4M °° drj J ' z It 8/(2y) 1 + 77 Since f °°(1 +r/2)-1dr7 converges, we can choose a positive constant K such that 7rc . < 2 4M a/(2K) 1 +7

J °0 dij

Thus, for 0 < y < K,

IJI < E. Then

lu(x,y) - f(0)I

< 2E for Ixl < J/2 and 0 < y yo > 0 where yo is a constant. Thus, Fx(x, y) = P(x, y) and Fxx = Q(x, y) in y > 0. Similar estimates are obtained for F. and Fyy. This means that u(x, y) in (6.94) has continuous first and second partial derivatives in some neighborhood of (x, y) in y > 0. By direct computation, u(x, y) satisfies (6.88). To show (6.88)-(6.89) has at most one solution, we need the following theorem: Theorem 6.12 Let u(x,y) be harmonic in S2 = {y > 0} and be bounded and continuous in n. Then sup u(x, y) = sup u(x, y) fl an

where 99 = {y = 0}. Proof: Let M and m be the suprema of u(x, y) in f2 and Oci respectively. Both suprema exist because u is bounded in f2.

Dirichlet problem in the upper - half plane

185

Suppose, by contradiction, M > m. Define v(x,y) = log[x2 + (y +

1)2]1/2

which is harmonic in y > 0. Let R > 0. Consider the region I1R = {(x, y)Ix2 + (y + 1)2 < R2, y > 0}. (See Fig. 6.5).

C (0,-1) Fig. 6.5 Domain for Theorem 6.12 Let P be a point in 1 such that m < u(P) < M and let 6 = M - m > 0. Then u(P) = m + 51 where b1 < 6. Now, let e > 0 and consider the harmonic function w(x, y) = u(x, y) ev(x, y). Choose R so large that P E QR. By the maximum principle, maxw(x, y) = maxw(x, y). On C1= portion of the boundary of 8S1R in y > 0, max w < M - clog R; and on C2 = portion of the boundary of aIR in y = 0, max w < m-e log 1 = m. Hence, u (P)-ev(P) _<max[M-e log R, m] = Aoru(P) < ev(P)+A= c log a + A for some positive a.

186

Laplace's Equation and Poisson 's Equation

We would like to choose R and e such that clog a < b1 /4 and M E log R < m. This implies that e log R > M - m = S > 51 or 81 / log R < e < 81/(4 log a). Now we choose R such that log R > 8 log a. Then we get u(P) < 51/4 + m < u(P), a contradiction. Hence, M=m. Applying Theorem 6.12 to the Dirichlet problem (6.88)-(6.89), we have the uniqueness result. Thus, we have established Theorem 6 .13 Let f(x) be continuous and bounded in (-oo, oo). Then the boundary value problem (6.88)-(6.89) has a unique bounded solution given by (6.94).

6.11 Problems (In the following problems, D denotes a bounded domain in the xy-plane with continuous, piecewise smooth boundary C) 1. Show that the Dirichlet problem [(1 + x2)Uxjx + [(1 + y2)y]y + ku = q(x, y) in D; u = f (x, y) on C has at most one solution u E C2(D) n Cl(D+C) where k 0 is a constant, q E C1 (D + C), and f is continuous on C. 2. Show that if u is continuous in D + C and satisfies uxx + ex+yuyy - ( x2 + y2 + 1)u = 0 in D, u < 0 on C, then u < 0 in D. 3. Let u(x, y) be continuous in D+C and satisfy uxx+uyy-}-aux+buy+cu = 0 in D, where a, b, and c are continuous functions of x and y in D + C and c < 0 there. Show that u = 0 on C implies that u = 0 in D. Construct a nontrivial solution of uxx + uyy + 2u = 0 in D = (0, 7r) x (0, 7r) with u = 0 on C. 4. Show that the Neumann problem Au + ku = q(x, y) in D,

8u = f(x,y) on C, On

has at most one solution where q E C1(D), f E C(C), and k < 0 is a constant.

187

Problems

5. Let u(x, y) be a solution of the Dirichlet problem Du = -q(x,y)

in D, u(x,y) = f(x,y)

on C,

where f is continuous on C and q is continuous and bounded in D. Show that

Iu(x,y)I 0 is a constant. 7. Solve u.,x+uyy=-COS2x, O<x 0, y > 0; ux(0, y) = 0, y > 0, u(x,0) = f{x), x > 0 where / ( x ) is bounded and continuous on [0,oo). 18. (a). Let it be a solution of uxx + uyy = 0 in D and continuous in D and on its boundary C. Deduce from the maximum principle that max£) + c \u(x,y)\ - maxc \u(x,y)\. (b). Let G = {r = y/x2 + y 2 > 1} and let u be a harmonic function in G and continuous in G. Suppose limr_oo u(x, y) = 0. Show that the maximum of \u\ exists and is equal to its maximum on dG. 19. Find a formal bounded solution of the problem Au = 0, x > 0, 0 < y < b; u(x,0) = f(x),u(x,b) = 0 , x>0; u(0, y) = 0, 0 < y < 6; where both u and ux go to 0 as x —+ oo by the Fourier sine transform of u(x, y) in x, treating y as a parameter. 20. Find a formal solution u(x, y) of uxx +uyy = 0, 0 < x < oo, 0 < y < 7r; ux(0,y) = 0, 0 < y < 7r; u(xyn) = f(x), u(x, 0) = 0, 0 < x < oo; such that u and ux go to 0 as x —* oo by the Fourier cosine transform of u(x, y) in x, treating y as a parameter. 21. Let 4 Z-00 ssinhfs(7r — ullsinsx , v u(x,y) = ds, 0 < x < c o , 0 < y < 7 r . 2 .„ r ' 7r y0 (1 + s J rsinhs7T (a). Show that u(x,y) is continuous on [0,oo) x [0,7r] and satisfies u (0i2/) = 0> w(rc, 7r) = 0 and u(x, 0) = x e _ x . (Hint. / 0 °°xe _ I sinsxdx = 2s(l + 5 2 ) - 2 . ) (b). Show that u(x,y) satisfies uxx +uyy = 0 in (0,oo) x (0,n). (c). Show that u(x, y) and u x (x,y) go to zero a s x - ^ oo. 22. Find a formal solution u(x, y) of Uxx + Uyy = 0 , 0 < x < oo, 0 < y < 7r; u^O, y) = 0, 0 < y < 7r;

190

Laplace's Equation and Poisson's Equation

u(x,0) = exp (-x2), u(x,ir) = 0, x > 0; where both u and ux go to zero as x -> oo by the Fourier cosine transform w(s, y) = 0 °° u(x, y) cos sx dx. Verify your solution.

Chapter 7

Problems in Higher Dimensions

7.1 Classification Consider a linear second order partial differential equation

aiju.,i,;j -^ ... = 0, (7.1)

L[u] _ i,j=1

where aij are constants and the dots in (7.1) stand for terms involving u and its first partial derivatives. We assume without loss of generality that the n x n constant matrix A = [aij] is symmetric. In order to carry (7.1) into a simpler equation, we consider a linear transformation of coordinates n

yi

j = 1, 2..., n, (7.2)

= Y , cijxj, j=1

where cij are constants to be determined. Then, by partial differentiation, (7.1) becomes n L[u] = E bkluykyt + ... = 0, k,1=1

where n

bkl = T, aijCk iClj, i,j=1

191

k, l = 1, ..., n. (7.4)

Problems in Higher Dimensions

192

In order to classify (7.1) as we have done in Chapter 1, we associate (7.1) with the quadratic form n E aijSj i,j=1

j

and (7.3) with the quadratic form n bkl?7k??l i,j =1

where Si = En 1 cikrlk, i = 1, ..., n. From a theorem on the diagonalization of quadratic form in linear algebra (see Indritz, [8, 98]), the coefficients cij can be chosen so that the quadratic form (7.6) can be changed to a sum of squares, i.e. bkl = 0 for k # l and t

n t

aijSiSj =

2 bkk^k-

i,j=1 k=1

By further transformation , 'Yk = lbkk171k, we have n

m

2 2 'Yi - 'Yi ,

E aij^i^j = i,j=1

i=1 i=m+1

where m is the number of positive coefficients and p is the number of nonzero coefficients . Accordingly, (7.1) can be transformed into m

p

E uyiyi E uyiyi + ... = 0 (7.7) i=1 i=m+1

which is called the canonical form of (7.1). (7.7) is (1) of elliptic type if m=p=n; (2) of ultrahyperbolic type if p = n, m > 2, p - m > 2; (3) of hyperbolic type if m = n - 1, p = n; (4) of parabolic type if p < n. For n=3, we have the following simple examples: uXX + uyy + uz y = 0, Laplace's equation, elliptic type. uxX + uyy - Utt = 0, wave equation, hyperbolic type. uXX + uyy - Ut = 0, heat equation, parabolic type.

Classification

193

Example 7.1 Classify the equation

4u,,,,, + 7uyy, + 4uzz + 2uxy + 8uxz + 2uyz = 0.

Solution: Consider the coefficient matrix 4 1 4 A= 1 7 1 4 1 4 To obtain an appropriate linear transformation for the independent variables x, y, and z, we can determine the eigenvalues of the symmetric matrix A and their corresponding eigenvectors. Consider

det(A - AI) = 0. The characteristic equation is A3 - 15A2 + 54\ = 0. Thus, the eigenvalues are 0, 6, and 9 and the corresponding eigenvectors are 1 1 1 0 , -2 1 -1 1 1 Let e=x - z, 77 =x-2y+z, = x+y+z. Then the equation becomes 6uyy + 9uCS = 0 which is of parabolic type. In this chapter, we shall confine ourselves to the following problems: (1). the Dirichlet problem for Laplace's equation in a bounded domain SZ in space, where Sl can be a cube or a sphere. (2). the initial-boundary problem for wave equation or the heat equation in a bounded domain S in the plane, where S can be a rectangle or a disc. We will use the method of separation of variables and the theory of double Fourier series to solve the above problems.

Problems in Higher

194

7.2

Dimensions

Double Fourier series

Consider a basic trigonometric system S = {cos nx cos my, cos nx sin my, sinnx cosmy, sinnx sin my, n,m = 0,1,2,...} in two variables x and y. These functions are periodic of period 2ir in both x and y and form an orthogonal system with respect to the weight function p(x,y) = 1 on the square K — [—TT,TT] X [—7r,7r]. Let f(x,y) be integrable on K. Then f(x,y) double Fourier series with respect to S on K :

can be expanded into a

oo

2_. ^nm[o-nm cos nx cos my + bnm cos nx sin my n,m=0

+cnm sin nx cos my + dnm sin nx sin my]

(7.8)

where a „ m = —- I I f (x, y) cos nx cos my dxdy, ^ J JK bnm = —K I / f{x, y) cos nx sin my dxdy, * J JK Cnm = —

/ f(x,y) sinnx cos my dxdy,

dnm = —R \ I f{x, y) sinnx sin my dxdy, n J JK

(7.9)

and 1/4 1/2 1

for n = m = 0; f o r n / 0 , m = 0 o r n = 0 , m / 0; for n ^ 0, m ^ 0.

(7.10)

A system of orthogonal functions {nm(x,y)} is complete with respect to the weight function p(x, y) on the rectangle R = [a, b] x [c, d] if, for any

195

Double Fourier series

integrable function f(x, y) in R, / / f2(x,y)p(x1y)dxdy J JR

= Y] a2nm I I „n,m=0 t^n J JR

4>2nm(x,y)p{x,y)dxdy

where

/ IR f(x, y)Km{x, y)p(x, y)dxdy nm

/ IRlm(x,y)p{x,y)dxdy

For our system S, the above Parseval's relation is ~

f

f f2(x,y)dxdy K

J •'

= f;

Xnm(alm

+ b2nm + c2nm + d2nm).

(7.11)

n,m=0

The convergence in (7.11) is absolute and independent of the order of summation. We would like to show that the system S is complete in [—IT, TT] X [—IT, IT]. To do this, we need the following theorem: Dini Theorem Let {fn(x)} be a nondecreasing sequence of continuous functions in [a, b]. Suppose that {fn{x)} converges pointwise to a continuous function f(x) on [a,b\. Then {fn(x)} converges uniformly to f(x) in [a,b]. (Please refer to Fulks, [6, 525-526], for its proof.) Theorem 7.1 Let {(/>n(x)} be a complete orthonormal system on [a,b] ■with respect to the weight function k(x) and for each n, let {ipnm(y)} be a complete orthonormal system on [c, d] with respect to the weight function h(y). Then {unm(x,y) = n(x)ipnm(y)} is a complete orthonormal system on [a, b] x [c, d] with respect to the weight function k(x)h(y). Proof: Without loss of generality, we assume that fc(x) = h(y) = 1. It is easy to show that {unm(x, y)} is an orthonormal system on [a, b] x [c, d]. Since every integrable function g(x, y) can be approximated in the mean by a continuous function f(x,y), it is sufficient to prove the completeness relation for any

Problems in Higher Dimensions

196

continuous function f (x, y), i.e. 00

jdfb

rd f b /

f2(x,y)dxdy = anm,

anm =

"^

f (x, y )On(x) ,bnm ( y)dxdy.

J

c

n,m=O

(7.12) For a fixed y, f (x, y) is a continuous function of x on [a, b]. By the completeness property of {On (x)} on [a, b], we have f b f2d =

b

(y), b (y) =

j

f(x,

)cbn( x)dx.

(7.13)

,n=0

Then Spr(y) = En 0 bn(y) is a nondecreasing sequence of continuous function on [c, d] which converges pointwise to a continuous function F(y) _ fb a f2 (x , y) dx. Then , by the Dini theorem , {Spr(y)} converges uniformly to F(y) on [c, d]. Hence , it is permissible to integrate (7.13) term by term, and we get b

jd

00

f2 (x,y)dxdy =

fa

n=0

d

J

bn(y)dy•

c

Then we apply the completeness relation for bn(y), i.e. d

d

f bn(y) dy = c

00

m=0

a nm,

bn( y) ^' nm ( y )dy

anm =

= anm•

c

Hence, (7.12) holds. By Theorem 7.1, the system S is complete and the completeness relation (7.11) holds for any integrable function f (x, y). In other words, the Fourier series of f (x, y) converges to f (x, y) in the mean. Now we would like to investigate, under what conditions on f (x, y), does the Fourier series of f (x, y) converge uniformly to f (x, y) on K? Theorem 7.2 Let f(x,y) be continuously differentiable and be periodic of period 2ir in both x and y and let fyy be piecewise continuous in K. Then the Fourier series of f(x,y) converges uniformly to f(x,y) in K. Proof: For each fixed y, the function f (x, y) can be expanded in a uniformly convergent Fourier series with respect to the trigonometric system

197

Double Fourier series

{ 1, cos nx, sin nx n = 1, 2, ... } as 00 f (x, y) = ao(,y) + > [an (y) cos nx + bn (y) sin nx] n=1

where an(y) = 1 J f ( x, y) cos nxdx ,

bn(y) = 1 j

f (x, y) sin nxdx. IT

The coefficients an(y) and bn(y) are continuously differentiable for -7r < y < 7r and periodic of period 27r. Then they can be expanded in a uniformly convergent Fourier series with respect to the trigonometric system { 1, cos my, sin my, m = 1, 2,...} as an (y) =

ao +

00 [anm cos my + bnm sin my] M=1 00 [cnm cos my + dnm sin my]

+

bn (y) =

M=1 where anm,

bnm ,

Cnm, and dnm

are given by (7.9).

Hence, we have 00 Aom(aom cos my + bom sin my)

f (x, y) = Aooaoo + m=1 cc

+ E

Ano (ano

cos nx + cno sin nx)

n=1

0.0 + E Anm [anm cos nx cos my + bnm cos nx sin my] n,m= 1 eo

+ E

Anm [Cnm

sin nx cos my + dnm sin nx sin my] (7.14)

n,m=1

where the series converges in the given order. The series (7.14) is a Fourier series of f (x, y) with respect to the system S on K. Using integration by parts and the periodic property of f (x, y), we

198

Problems in Higher Dimensions

obtain the Fourier coefficients of ff, fy, and fey with respect to the system S as follows: Fourier coefficients of ff = { nCnm, ndnm, -nanm , - nbnm}; Fourier coefficients of fy = {mbnm, -manor , , mdnm, -mcnm}; Fourier coefficients of fey = {nmdnm, -nmcnm, -nmbnm , nmanm}.

By Parseval' s equation, we get fx2 (

T2

K

00

x, y)dxdy = I 'mmn 2 [Cnm + dnm + anm + bnm] i

(7.15)

n m =0

00

I

lK

fy (x, y) dxdy =

)^nmm2 [b nm

nm0 ,=

2m + anm + dn

+ Cnm];

(7.16)

00

I

K

.f( x,

y)dxdy =

f

)mmn2m2 [am

+ b m+Cnm +m]. (7.17)

n,m =0

Let M SNM (x, y ) =

4 + 2 ) [aom cos my + bom sin my] m=1

1 N

+ 2 > [ano cos nx + Cno sin nx] n=1 N

M

+ E [anm cos nx cos my + bnm cos nx sin my] n=1 m=1

N

M

+ E [cnm sin nx cos my + dnm sin nx sin my] n=1 m=1

be a partial sum of the Fourier series of f (x, y). Now, for N2 > N1 and M2 > M1, we apply Schwarz's inequality to SN2M2 (x, y) - SN1M1 ( x, y) and

199

Double Fourier series

get f1 M2 ISN2M2 ( x,y) - S'NiMi ( x,y)I2 C {2 E m2(aOm +bOm) m=M1+1

N2

N2

M1

+2 E n2(anO+c 0)+ >

M2

m=M1+l n=1

n=N1+l m=1

n=N1+1

M2 1 1 N2 f1 x `2 E m2 + 2 n2 + m=M1+1

N2

E n2m2hnm+ E >n2m2hnm}

N2 M1

M2 N2

n=N1+1 m=1

n=N1+1

2} Mn2-

2+ W2-M m=Mi+1 n=1

where hnm = anm + bnm + Cnm + dnm From (7.15)-(7.17), the first factor is bounded by

11r2

f 1 [fx + fy + fxy]d

x dy

K

and the second factor is the difference of two partial sums of the convergent series 1

2

00

1

1

00

1

E 2 + _E T2 m_1

00

1

E n 2 7IL 2

2 n=lnm=1

which can be made less than any e if M1, N1, M2, and N2 are sufficiently large. Thus, by the Cauchy criterion, the series (7.14) converges uniformly in K. Since the double Fourier series as an iterated series converges to f (x, y), the series also converges uniformly to f (x, y). Remark : If f (x, y) is odd in both x and y, then anm = bnm = Cnm, = 0 and the resulting series is a double sine series ; similarly, if f (x, y) is even in both x and y, then bnm, = Cnm = dnm = 0 and the resulting series is a double cosine series. Moreover, if f (x, y) is defined on (0, 7r) x (0, ir), f (x, y) can be continued as an odd (or even) function of both x and y in (-7r, 7r) x (-7r, 7r), and can be expanded as a double sine (or cosine ) series. As is true in the case of one variable, double series can also be defined on [a, b] x [c, d] for any constants a, b, c, and d.

200

Problems in Higher Dimensions

Example 7.2 Find the double series for f (x, y) = xy2 on [ -7r, 7r] x [ -7r, 7r]. Solution: We note that 7r

J

_lr

Z 3 7r u2du = 3 fudu=O,

and for n > 1, u cos nudu = 0, r

T

u2 cos nudu = I

J

27.( -

1 )n +1

u sin nudu = n

47r(-1)n n2 '

I 7r

u2 sin nudu = 0.

Thus, all the Fourier coefficients are zero except 47r2(-1)n+1 8(-1)m+n+1 Cn0 = 3n ,

Cnm =

m2n

for n > 1 and m > 1. Hence, the Fourier series of f (x, y) is 27x2 00 (-1)n+1 °O (-1)n+m+1 sin nx + 8 [\J` l sin nx cos my. n nm 2 3 n=1 n m=1

7.3 Laplace 's equation in a cube Consider the problem uyy+uyy+uZZ= 0 ,

0<x