Boundary value problems

0. Then , from (3.85), there exist an integer N such that for 1/2 < E. By the Cauchy criterion, n > m > N, we have [E i= m b?] z 0 v(x) 00 N'ibixi(x)

µi - C

i=1

converges uniformly on [ca, (0]. Since {Xi(x)} are continuous, v(x) converges to a continuous function; hence u(x) in (3.83) is continuous. We note that 00 00 00 Kv = K( iiaixi(x)) _ µiaiK xi = µiaixi(x) i=1

i=1

i=1

and 00

00

K f = >(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 <

u(a) = u(,0) = 0

(3.89)

(3.90)

with the additional condition q(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
83

Further properties of eigenvalues and eigenfunctions

and Ak -+ oo and {xk(x)} is a sequence of eigenfunctions such that

fa x^(x)dx =

J a P(x)u2(x)dx

where {uk(x)} is a sequence of normalized eigenfunctions of (3.89)-(3.90) with respect to the weight function p(x). Let O(x) E C2[a„3] with 0(a) = 0(0) = 0. Then , by Theorem 3.24, O(x) _ E°°_1 cu,-,( x) ( (uniform convergence) with

In = J^

P ( x)O(x)un( x)dx.

Now, by integration by parts, p p A f L[q ]undx = f gL[un]dx = -A n f P(x)un ( x)o(x) dx = -cn An. Since { u n (x) } is a complete orthonormal set, by Parseval's equation, 00

2 In,

f p¢2dx = «

n= 1

and 00 gL[q]dx = -f {E cnun}L[q]dx =fa (pc'2 + 402)dx R f« a n=1 00

0

00

0

00

0

In f unL[O]dx In f q L[un,]dx = > 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<0

(3.101)

with boundary conditions (3.90) where p*(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
(3.103)

with boundary conditions u(a*) = u(,3* ) = 0 where a < a * < 3* < ,Q. Define I3 (u) = f« + (1ui2 + qu2)dx f3 pudx 2

Let H* be the subspace of H such that u(x) _- 0 on [a,a*] U Then H* is also the class of admissible functions for I3 (u) and for this set, 1(u) = I3 (u). As a result, A(01i ..., On-1) < min I(u) = min I3 (u) =A*(¢i, .., On-1) since the additional condition on the class of admissible functions for the functional can increase the minimum of the functional. Thus, An < Az. Theorem 3.32 tions of

(Separation Theorem) Let u and v be nontrivial solu-

L[u] + c(x) u = 0

(3.104)

L[v] + c*(x )v = 0

(3.105)

and

respectively where p(x), p'(x), c(x), and c*( x) are continuous with c(x) > c*(x) and p ( x) > 0. Let a* and /3* be two consecutive zeros of v. Then there exists at least one zero of u in (a *,/3*).

Problems

89

Proof: Suppose, by contradiction, u(x) # 0 E (a*, /3*). Without loss of generality, we assume that u > 0 and v > 0 in (a*,/3*). Multiplying (3.104) by v and (3.105) by u and subtracting, we have vL[u] - uL[v] + (c - c*)uv = 0. Integrating by parts from a* to /3*, we arrive at [pv'u-pu'vI l0:>0. Since v(/3*) = v(a*) = 0, p(/3*)v'()3*)u(/3*) - p(a*)v'(a*)u(a*) > 0.

(3.106)

Since v(/3*) = 0 and v(/3*-) > 0, this implies that v'(/3*) < 0. Similarly, v'(a*) > 0. This contradicts (3.106). Corollary. Let A and v be two distinct eigenvalues of (3.89)-(3.90) with corresponding eigenfunctions u(x) and v(x) respectively, and let A > v. Then between two consecutive zeros of v, there exists at least one zero of U. Proof: Apply Theorem 3.32 with c(x) = Ap(x) and c*(x) = vp(x).

3.9 Problems In Problems 1 and 2, formulate an equivalent integral equation for the initial value problems and solve by Picard 's method. 1. y'=x+y ,

y(0)=0.

2. y' = x2y - x, y(0) = 03. Let u (x) and v(x) be two linearly independent solutions of u"+p(x)u'+ q(x)u = 0 where p and q are continuous on [a, b]. ( a). Show that u/v is monotone in every interval in which v:74 0. (b). Suppose , for a given point a,u (a) = v(a) > 0 and u' (a) > v'(a). 0 E (a, c]. Show that if c is the first zero of v(x) in x > a, then u(x)

90

Green's Function and Sturm -Liouville Problems

4. Let u(x) and v(x) be two solutions of u" + p(x)u' + q(x)u = 0 where p(x) and q(x) are continuous in (a, b). Suppose there exists a point c E (a, b) such that u"(c) = v"(c) = 0 and either p(x) or q(x) does not vanish at c. Show that u(x) and v(x) are linearly dependent on (a, b). 5. Solve u" - u = -f (x) with u(0) = u'(0) = 0. 6. Solve x2u"+xu'+u=logxforx>0with u(1)=1 andu'(1)=2. 7. Solve u" - u = -f (x) for 0 < x < 1 with u'(0) = u'(1) = 0. 8. Solveu" - u= 2x for 0 < x < 1 with u'(0) = u(1) = 0. 9. Consider the boundary value problem L[u] = (pu')' - qu = 0, a < x < 3; Ba [u] = a, Bp [u] = b. Let u(x) and v(x) be two linearly independent solutions of L[u] = 0. (a). Show that the boundary value problem has a unique solution if D=Ba [u] Bp [v] - B[v]B[u] is not equal to zero. (b). Suppose D=O. Find the condition(s) on a and b such that the boundary value problem has a solution. In Problems 10 through 13, find the Green's function for the system if it exists: 10.

u"+u=0,

0<x<7r;

u(0)=u'(-7r)=0.

11.

u"-u=0,

0<x<7r;

u(0)=u'(7r)=0.

12. u" = 0, 0 < x < 1; 2u(0) + u'(O) = 0, u(1) = 0. 13. u"=0, 0<x<1; u(0) -u'(0)=0, u(1)+u'(1)=0. 14. (a). Find the general solution of L[u] = x2u" + 2xu' - 2u = 0. (b). Determine whether the following problem has a nontrivial solution, and find it if it exists. (i). L[u] = 0, 1 < x < 2, u(1) = u'(1), u(2) = 0. (ii). L[u] = 0, 1 < x < 2, u(1) = u'(1), u(2) = 2u'(2). (c). Find the Green's function in (bi) and/or (bii) if it exists. 15. Formulate an equivalent integral equation for the solution of the SturmLiouville problem: (xu')' +Axu = 0, 1 < x < e; u(1) = u(e) = 0. In Problems 16 and 17, find the eigenvalues and the eigenfunctions of

Problems

91

the Sturm-Liouville problem: 16. (x3u' )' + Axu = 0; u ( 1) = u(e") = 0. 17. u"=-Au; u( 0) - u'(0) = 0, u (ir) - u'(7r) = 0. 18. Consider the problem y" _ -Ay, ay(0)+y'(0) = 0, y(1) = 0 where a is a given real constant. (a). Find all positive eigenvalues of the problem for all values of a. (b). Show that A = 0 is an eigenvalue only if a = 1. (c). If a < 1, show that all the eigenvalues are positive. (d). If a > 1, show that there is exactly one negative eigenvalue. 19. Consider the Sturm-Liouville problem L[u] = (pu')' - qu = -Au, -a < x < a; u(-a) = u(a) = 0 where p(x) >0 and p(x) and q(x) are real and continuous on [-a, a]. Moreover, both p(x) and q(x) are even on [-a, a]. Show that the eigenfunctions are either even or odd. In Problems 20 through 24, let {¢,, ,(x)}, n = 1, 2, ... be a complete orthonormal set in the space S of all real square integrable functions on [a, b] where the inner product of f and g is defined as (f, g) = fa f (x)g(x)dx. 20. Suppose there exists a continuous function f (x) in S such that (f, 01 + 0j) = 0 for j = 2, 3, ... Show that f = 0 on [a, b]. 21. Suppose there exists a continuous function f (x) in S such that (f, Oj + Oj+1) = 0 for j = 1, 2,3 ... Show that f = 0 in S. 22. Suppose the functions {0,(x)} are continuous and uniformly bounded. Can you find a nonzero continuous function f(x) such that (f, 4j + 2qj+1 ) = 0 for j = 1, 2, 3...? Justify your answer. 23. Let {ck} and {dk} be the Fourier coefficients of f and g with { 0k(x)} respectively. Show that (f, g) = Ek'= 1 ckdk. 24. Show that the system { qk(x)} is linearly independent on [a, b]. 25. Find the Fourier series of f (x) = 1 on [0, ir] in terms of the system S = {sin nx}, n = 1 , 2... Assuming S is complete on [0, ir] with respect to the weight function 1, find the value of E '1 (2n - 1)-2. In Problems 26 through 28, find the eigenvalues and the eigenfunctions of the integral operator: 26. Kf = fo k(x, y)f (y)dy, f E C[0, 1] where k (x, y) = x - y on [0,1] x [0,1].

92 Green's Function and Sturm -Liouville Problems

27. K f = f -, k(x, y) f (y)dy, f E C[-7r, ir] where 28.

k(x, y) = 1 + cos(x - y) on [-7r, ir] x [-7r, 7r].

K f= f', k(x, y) f (y)dy, f E C[-7r, 7r] where

k(x, y) = sin x cosy on [-ir, -7r] x [-7r, 7r].

29. Let K be an integral operator on C[-1, 1] defined by Ki = f 11(1 + xy)u(y)dy.

(a). Find all the nonzero eigenvalues and the corresponding eigenfunctions 'of K. (b). Show that zero is an eigenvalue of K and find two corresponding independent eigenfunctions of K which are orthogonal with respect to the weight function 1 on [-1,1]. 30. Lot Ku = f', k(x, y)u(y)dy for u E C[--7r, ir] where k(x, y) is continuoi.s on [-7r, 7r] x [-7r, 7r]. Suppose k(x, y) = k(-x, -y). Show that its eigenflinction '(x) corresponding to a nonzero simple eigenvalue of K is either' odd or even. 31. Let Ku = f ',,r k(x, y)u(y)dy for u E C[-7r, 7r] where k(x, y) is real and continuous on [-7r, 7r] x [-7r, 7r].

(a)',. Show that J1K112 < f "" f , k2(x, y)dxdy and IpI < IIKII if u is an eigenva,lue of K. (b). If k(-x, y) = k(x,y), show that the eigenfunctions of K corresponding to nonzero eigenvalues are even. (c) Let k(x, y) = x2 sin y + 1. Show that zero is an eigenvalue of K and determine the class of corresponding eigenfunctions. Give an eigenfunction which is not even.

fa

32. Let Ku = s(x, y)u(y)dy where s(x,y) _ 1 pi(x)gi(y) with pi and qi are continuous on [a, b] and the domain of K is C[a, b]. Prove that (a). I K I E 1 Ilpill Il giII.

s

(b) . JA I < I I K I I if A is an eigenvalue of K. 33. Let f (x) _ E- 1 bn sin nx, (bn # 0 and bn 54 b,,,, for n # m) which converges uniformly. Find all the eigenvalues and the corresponding eigenfunctions of the integral equation Ku = f "" k(x, y)u(y)dy = Au(x) where 'k(x, y) = f (x + y) on [-7r, ir] x [-1r, -7r].

34. Let A be an eigenvalue of

Problems

93

L[u]=(pu')'-qu=-Apu, a<x 0 on [a„ and µ1a1 < 0, µ2a2 > 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) <0. (b). Show that u(x) has infinitely many zeros 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<(YCk)1/2( dk)1/2 k=1 k=1 k=1

(4.13)

whose proof will be given at the end of this Section, we get m

I Sm (x) - Sn(x)I

[

m

)]1/2[ k-2]1/2 ( k2a k + k2bk k=n+1 k=n+1

m

(ak k

+ak)] 1 /2( J

n

00

x-2dx)1/2 < Cn-1 / 2.

(4.14)

I

From the Cauchy criterion for uniform convergence of sequence of functions, {Sn(x)} converges uniformly on [-ir, -7r ]. Since f (-7r) = f ( 7r), f (x) can be extended to (-oo, oo ) as a continuous , periodic function of period 27r, hence, by Theorem 4.2, Sn(x) converges pointwise to f (x) on [7r, 7r]. Thus, the convergence is also uniform. Let m -r oo in (4.14). Then

If (x) - Sn(x)I Cn-1/ 2. Hence, we obtain

(4.15)

101

Uniform convergence and completeness

Theorem 4.3 Let f be continuous on [-7r, 7r], f (-w) = f ( 7r), and have a piecewise continuous derivative there. Then the Fourier series of f converges uniformly to f on [-7r, 7r]. Furthermore, it satisfies the bound (4.15). Proof of Schwarz's inequality (4.13). Let A, Ck, and dk,

k = 1, 2,...n, be any real numbers. We note that

n

0

n

n

< E(ACk dk)2 = A2 ck k=1

k=1

2A

k=1

n

Ck dk +

d

k.

k=1

If Ek=1 Ck 0, we get (4.13). Otherwise , for Ek=1 l ekl2 = 0 , then ck = 0 for k = 1, ..., n, and (4.13) also holds. Remarks: (1). Any integrable function g(x) can be approximated in the mean by a piecewise smooth, continuous function on [-7r, -7r] with f (-ir) = f (7r). Hence, the system S forms a complete orthogonal set on [-7r, -7r]. (2). From Theorem 3.16, the Fourier series of any integrable function f can be integrated term by term in [a, b] where -ir < a < b < 7r, and the integrated series converges to fQ f(x)dx. Theorem 4.4 Let f(x) be continuous on [-7r, 7r], f (-ir) = f (7r), and have a piecewise continuous derivative there. Then, at each point x, where f"(x) exists, 00 f'(x) = 1: (nbn cos nx - nan sin nx) n=1

where {an} and {bn } are the Fourier coefficients of f(x). Proof: Since f'(x) satisfies the conditions of Theorem 4.2, we have 00 f'(x) = 2 + ^ (an cos nx + On sin nx) n=1

where {an} and {3n} are the Fourier coefficients of f'(x). By previous calculation, we have ao = 0, an = nbn,

Qn = -nan.

102

Fourier Series and Fourier Transforms

4.3 Other types of Fourier series

Let f (x) be integrable on [-7r, 7r]. (A). Suppose f (x) is even on [-7r, 7r]. Then It 2 an = f (x) cos nxdx, 7r 0

bn = 0

and the (Fourier) cosine series of f (x) is 00 a0 2 + E an cos nx. n=1

(B). Suppose f (x) is odd on [- 7r, 7r]. Then 2 fIt an = 0, bn = f (x) sin nxdx 7r

and the (Fourier) sine series of f (x) is 00 E bn sin nx. n=1 Now, suppose f (x) is defined on [0, 7r]. We can expand f (x) in a (Fourier) cosine series by extending f (x) to be an even function on [-7r, in. Similarly, we can expand f ( x) in a (Fourier) sine series by extending f (x) to be an odd function on [-7r, 7r].

If f (x) is integrable on [a, b], then we introduce a transformation 27rx 27ra t= -7rb-a b-a

(4.16)

which maps [a, b] onto [-7r, 7r]. Let f (x) = f (g(t)) = F(t) where g(t) is the inverse transformation of (4.16). Then we can expand F (t) in a Fourier series in [-7r, 7r] as o

00 + (an cos nt + bn sin nt n=1

-7r
Other types of Fourier series

103

where an

= 1 _7r F(t) cos ntdt, bn = IT

J

J 7r F(t) sin ntdt.

If we replace t by (4.16) in the above series, we obtain the Fourier series of f (x) in [a, b] as n( 27rbx a a ) ao 00 a a ) - 7r)], - 7r) + bn sin n( ---C2 + E [an cos

a < x < b,

n=1

where b

an = b 2 a

f (x) cos

n(27rb(x aa) - 7r) dx

a

and fb bn = b 2 a

J

f (x) sin n( 7r(6- a) - 7r)dx.

a

Remark: The conditions for pointwise convergence and uniform convergence of cosine series and sine series of f (x) on [0, 7r] and the Fourier series of f(x) on any interval [a, b] are similar to the conditions given in Theorems 4.2 and 4.3. Let f (x) = x2 on [-7r, 7r] and f (x + 27r) = f (x) for all x.

Example 4.1 Prove that

f (X) = 7r2/3 + 4 Y( -l) ' cos nx/n2 n=1

for all x and deduce that 00 E(-1)n-1/ n2 = 7r2/12 and

E 1/n2 = 7r2/6.

n=1

n=1

cc

Solution: We note that f (x) is anf even function. Thus, bn = 0, ao = 21r-1

J0 7' x2dx = 21r-1 [x3/3] o = 27r2/3,

104

Fourier Series and Fourier Transforms

and an = 27r-1 f 7r x2 cos nxdx = -4(n7r)-1 [f7r x sin nxdx = 4(-1)n/n2. f (x) satisfies the conditions in Theorem 4.2 and therefore its Fourier series converges to f (x) everywhere. Thus, 00

x2 =

W2

/3 + 4 E(-1) cos nx/n2,

- Tr < x < -7r.

n=1

At x = 7r, At x = 0,

7r2 = 7r2 /3 + 4 E°° 1 1/n2. Then E' 1 1/n2 = 7r2/6. 0 = ire/3 + 4E°O 1(- 1)n/n2 . Then E', (- l)n- '/n 2 =

7r2 /12.

Example 4.2 Let f (x) = x on [0, ir]. Find (a). its Fourier cosine series, (b). its Fourier sine series , and (c). its Fourier series on [0,,7r]. Solution: (a). bn = 0

2 0'r

fr -2

f

an x cos nxdx 7r n1r

-2

sin nxdx

cos nx]0

7r

- ( 0, if n is even; if n is odd. Hence, the Fourier cosine series of f (x) is 7r

4

°O

cos ( 2k - 1)x

2 7r ^ (2k - 1)2

(b). an = 0. r n+1 2 bn = 2 x sin nxdx = 2 [x cos nx] o + ? cos nxdx = (-1) it o n7r n7r f0 'r n

105

Application to the wave equation

Hence, the Fourier sine series of f(x) is 2y>(-

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
(4.19)

ut(x, 0) = 5(1),

0 < x < 7T.

(4.20)

(

By the method of separation of variables, we obtain a formal series solution of (4.17)-(4.20) given by 00

u(x, t) = / J ( a n cos nt + bn sin nt) sin nx

(4-21)

106

Fourier Series and Fourier Transforms

where an = .

f (x) sin nx dx, J0

bn = n g(x) sin nx dx. 0

J

(4.22)

( Compare the results in Section 1.5.) We now investigate under what conditions on f and g such that (4.21)(4.22) will yield a classical solution of (4.17)-(4.20). (A). Let f (x) be continuous on [0, 7r], f (0) = f (7r) = 0, and f'(x) be piecewise continuous there. We extend f (x) to be an odd function F(x) on [-7r, 7r]. This implies that F'(x) is even on [-7r, 7r]. Let an and Nn be the Fourier coefficients of F'(x) on [-7r, 7r]. It is easy to show that ao = 0, an = nan,

On = 0.

By Bessel's inequality, 00 ir cost nx dx < f Fi2(x)dx, an 7r n=1 r

J

or 00 n=1

21'

nta2 n < 7r

fi2(x) dx = C2,

o

where C is a constant . Then, for any positive integers M and N, N > M, we have, by Schwarz 's inequality, N

N

N

N

n )1/2( 2)1/2 < G( )1/t' 1: n2a2 lan'I < ( 2 n=M+1 n=M+1 n=M + 1 n n=M+1 n

Since > 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[/
(4.49)

116

Theorem 4.10 Then

Fourier Series and Fourier Transforms

Let f (t) and t f (t) be absolutely integrable in (-oo, oo).

du [f (u)] = F[itf(t)].

(4.50)

Proof: The integral

00

00

( eaut )

f:f(t ) a

dt

itf (t )e'ut dt = f oo

converges uniformly in -oo < u < oo since t f (t) is absolutely integrable. Thus, it is permissible to differentiate f (u) under the integral sign so that (4.50) holds. Theorem 4 . 11 Let a and b be real numbers and a # 0. (A). If g(t) = f (at - b), then

F[f (at - b)]

= jal - 1eiub/a f(u/a)

(B). If g( t) = eiat f(t), then

F [g] = j (u + a) Proof: Put g(t) into F[g] and simplify. Definition 4.2 Let f (x) be absolutely integrable in (0, oo). Then (A). the (Fourier ) cosine transform of f (x) is defined as

Fr[f]

=

J0 00 f(t)cosut dt;

(4.51)

(B). the (Fourier) sine transform of f (x) is defined as Fs [f] _ f f(t) sin ut dt.

(4.52)

If f is continuous, piecewise smooth, and absolutely integrable in (0, oo), then, by extending f (x) as either an odd or even function on (-oo, oo), we obtain 00 00 2 F3 [f]sinux du= Fc[fI cosuxdu AX) = 2 IT o

j

Contour integration 117

from Theorem 4.5.

4.7 Contour integration In this section, we would like to evaluate the Fourier transform by contour integration. We will need the following results from analytic functions of complex variables. Let f (z) be analytic in a domain except at an isolated singularity zo in D. Then the function f (z) has a Laurent series expansion 00 00 = E an(z - zo)n + bn(z - zo) (4.53) f(z) n=o

n=1

in 0 < I z - zo < r for some constant r, where

an 27ri jCC f (z)(

z - zo)-n-idz, bn =

27ri f f (z

)(z - zo)n-idz (4.54)

and C is any closed contour around zo in 0 < Iz - zol < r such that the path of integration is taken in the counterclockwise direction.

In particular,

bi 27ri

f

f (z)dz = Res[f ( z), zo]

(4.55)

c

which is called the residue of f at zo. In order to compute the Fourier transform of a certain class of absolutely integrable functions , we will need the following residue theorem: Residue Theorem . Let C be a simple closed contour within and on which f ( z) is analytic except for a finite number of isolated singular points zi, ...., zn interior to C. Then n

f (z) dz = 27ri E Res[f(z), zj]. c j=1

(See Churchill, Brown & Verhey, [4, 170-174].) Now, we have the following theorem:

Fourier Series and Fourier Transforms

118

Theorem 4.12 If f (z) is analytic for z > 0 except for a finite number of singular points z1, ..., zn with positive imaginary parts, if lim [R max If(z)I] = 0, (4.56) R-.oo lzl =R,Z'z>O if

and if f (x) is absolutely integrable in -oo < x < oo, then, for u > 0, n

F[f] = 21ri > Res[f (z)e'uz, zj]. (4.57) j=1

Proof:

Fig. 4.1 Contour integration for Theorem 4.12 Let R be a positive number so large that Izj I < R for j = 1, 2, ...n. Consider the contour CR consisting of the line segment [-R, R] on the real axis and the upper half circle I'R of radius R (see Fig. 4.1). By the residue theorem, n

f (z)e2uzdz = 27ri E Res[ f (z) e2`, zj] CR, j=1

Now, on FR, I

f f(z)eiuZdzj C r

f R If (z)IIdzI

which goes to zero as R -> oo. Hence, R

lim f

R-oo R

n

f (x)eiuxdx = 27ri E Res[f ( z)ezuz, zj]. j=1

119

Contour integration Since f (x) is absolutely integrable in (-oo, oo), (4.57) holds.

Applying the same procedure in the lower half plane, we have the following theorem: Theorem 4.13 If f (z) is analytic for tz < 0 except for a finite number of singular points z1, ..., zn with negative imaginary parts, if If (z) 1] = 0, lim [R max R--*oo IzI=R,Q^z
(4.58)

and if f (x) is absolutely integrable in -oo < x < oo, then , for u < 0, n

F[f] = - 21ri > Res[f (z)eiuz, zj]. j=1

Suppose f (z) has a pole of order m at zo, i.e.

bn(z - zo)-n. + f(z) _ > an( z - zo)n 00 m

n-o

n=1

Let O(z) = (z - zo)'m' f (z). Then, for m > 1, bl = Res[f (z), zo] =

(m-1) z ( o)

and, for m=1,

b1 = Res[f (z), zo] = lim (z - zo) f (z). z---*zo

Example 4.5

Find the Fourier transform of f (x) = (1 + x2)-1.

Solution: We note that °° 1 T_+ x2 J

_ [tan 1 x] _00 _ 7r. 00

Thus, f (x) is absolutely integrable in (-oo, oo).

Let f ( Z) = 1/(1 + z2 ). For z = Reie, R > 1, we have max If(z)I < 1 2 IzI=R R - 1

(4.59)

120

Fourier Series and Fourier Transforms

Hence, R^[R max If(z)j] =0. IzI=R

Foru>0, eiuz

AU) = 27ri Res[

eiuz

, i] = 2iri lira = Ire-u. z-.iz+i 1 +z2

For u < 0, e iuz f (u) = -2iri Res [ 1 + z2 , -i] = 7reu. Hence, Example 4.6

F[(1 + x2)-1] = Ire-Iu* Find the Fourier transform of 2 f (x) = exp(- 2 ).

Solution: Let f (z) = exp(- 2) where z = x + iy is a complex variable and let u and L be two positive constants.

z= -L+iu

z=-L

I'3

" I'1

z=L+iu

z=L

Fig. 4.2 Contour integral for Example 4.6 Let r be a closed contour in the complex plane consisting of the following paths: I71 :E3`z=0, from x=-L to x=L. r2:Rz=L , from y=0toy=u. I'3:!3,z=u, from x=L tox= -L.

121

Contour integration

r4: $2z=-L, fromy=utoy=0. (See Fig. 4.2.) We know that f (z) is an analytic function. From Cauchy's theorem, we have (4.60)

f (z) exp(iuz) dz = 0. We now evaluate (4.60) along Fi, i=1,2,3,4. fL f (z) exp(iuz) = Jr,

Jr3 f (z) exp(iuz)dz = L

J

x2

exp(iux - 2 )dx. L

exp[- ( x 2iu)2 + iu (x + iu)]dx

f

u2

L

x2

exp(-2) f exp(-2)dx. L

f f (z) exp( iuz)dz = f u exp[- (L r2 0

= i exp(-

22y)2

+ iu(L + iy)]idy

L 2 u y2 + iuL) J exp(2 - yu - iLy)dy. 2 o

Thus,

I

- L2

2

u

2

) f exp(-yu + f (z) exp( iuz)dzl < exp ( I r2 0

2 )dy 0

as L -> oo. Similarly, f(z) exp(iuz)dzl-+ 0 r4

as L ->oo. Since f (x) is absolutely integrable , we have, from ( 4.60),

= Jim i(U )

L x2

f

u2

L exp(-2 + iux)dx = exp(-2) f

x2

exp(-2)dx.

122

Fourier Series and Fourier Transforms

By elementary integration, we know

Hence, for u > 0, j (u) = 27r exp(- z ). Since the Fourier transform of an even function is even, we have z j (u) = -vf2-7r exp(- 2 )

for all u.

4.8 Problems 1. Find the Fourier series for f (x) = ex in (--7r, 7r ), and find its sum at x=7r. 2. Find the Fourier series of f (x) in (-ir, ir) where f(x) _ 0,

-7r < x < 0;

cos x, 0 < x < 7r. Discuss the pointwise convergence of the Fourier series in (-ir, 7r). 3. Find the Fourier series for f (x) = IxI on [-7r, 7r] and, from the Fourier series you obtain, find the sum of the series >°° o(2n + 1)-2. 4. Find the Fourier series of f (x) = e-' on (-7r,,7r) and evaluate E,°O 1(1 + n2)-1 from your work.

5. Find the Fourier series of f (x) = t4 on [-ir, ir] and a bound for the error made in replacing x4 by the first three terms of the series. 6. Find the sine and cosine series for the function f (x) in 0 < x < 7r where f (x) = 1 when 0 < x < 7r/2; = 0 when 7r/2 < x < 7r. 7. Expand f (x) =0 Fourier series.

for

- 1 <x < 0, = cos irx for 0 < x < 1, in a

8. Use the completeness relation to express fo f2 (x)dx in terms of its Fourier coefficients of the sine series of f (x) in (0,,7r).

Problems

123

9. Let S = {sin nx} be an orthogonal set on [0, -7r] with respect to the weight function 1. (a). Show that S is complete on [0, 7r]. (Hint. F = {1, cos nx, sin nx} is complete on [-7r, 7r] ). (b). Show that the series E°° 1(-1)nn-1/2 sinnx is not a Fourier series of an integrable function f with respect to S on [0, in. 10. Let f E C'[-7r, 7r], f (-7r) = f (7r), and f f (x)dx = 0. Show that f ",r f 2 (x)dx < f I7, f i2 ( x)dx and the equality holds if and only if f (x) _ A cos x + B sin x for any constants A and B. 11. Let f (x) be continuous on [--7r, 7r] and f'(x) be integrable there. Show that nan -; 0 and (-1)'" nbn -> [f (-7r) - f (,7r)]1-7r where an and bn are the Fourier coefficients of f (x) on [-7r, ,7r). 12. Let f (x) be continuous and periodic of period 27r, and satisfy if(x')-f(x") <Mlx'-x"Ia,0< c < 1, for any x ' and x", where M is a constant . Show that its Fourier coefficients an and bn satisfy an = O(n-01) and bn = O(n-"). 13. Find a formal solution of utt+2aut+bu-c2uxx = 0, 0 < x < in, t > 0; u(0, t) = u(7r, t) = u (x, 0) = 0, ut (x, 0) = g(x);

by separation of variables where a, b, and c are constants. 14. Find a formal solution of utt + ut - uxx = 0, 0 < x < 7r, t > 0; u(O,t) = ux(in,t) = ut(x,0) = 0, u(x,0) = f(x); by separation of variables. 15. Represent the function

f(x) _ cosx,

-7r

<x<

0, IxI >

7r;

7r;

by the Fourier integral formula and verify that for -ir < x < in, cosx = 2 it 0

J

°O

u cos ux sin u-7r 1 - u2

du.

16. Express the function f (x) = e-x cos x , x > 0 by the Fourier cosine integral formula and evaluate fo (u2 + 2) (u4 + 4)-1 cos 21rudu.

124

Fourier Series and Fourier Transforms

17. Find the Fourier integral of if x < 0 and x > 7r; f(x)0, cosx, if 0 < x < ir, and deduce from your work the value of f °° u sin(7ru)(1 - u2) -1du. 18. Find the Fourier sine integral of f(x) = 1 1, 0<x<1, 0, x> 1, and evaluate f o u-1(1 - cos u) sin xu du for all x. 19. Solve the integral equation f o00 g(u) sin ux du = e-x cos x,

x > 0.

20. Prove that

1

00

cos ux 1 + u2 du

lre-Ixl 2 , -oo < x < oo.

21. Use the method of separation of variables and the Fourier integral to find a formal solution of utt - c2uxx = 0, -oo < x < oo, t > 0, which satisfies the initial conditions u(x, 0) = f (x) and ut (x, 0) = 0. Show that the solution you obtain can be reduced to [f (x + et) + f (x - ct)] /2. 22. Show that the Fourier transform of an even function is even. 23. Find a formal solution of y" + 2y' + xy = 0 on (-oo, oo ) by the Fourier transform. 24. Find a formal solution of uxx-utt= 0,

-00<x<00, t>0;

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

-00<x<00;

by means of the Fourier transform 00 ft(w, 0 _ u(x, t) eiwx dx. -00

125

Problems

In Problems 25 and 26 , find the Fourier cosine and sine transforms of f (x) - Ix, O<x<1, 0, X>1. f (x) = xe-y, x > 0.

25.

(x) AX)

26.

27. Let

O(x) =

0, 2 - x, 2+x,

1xI > 2, 1 < x < 2, -2<x<-1,

1,

1xI < 1.

Find its Fourier transform ^(u) and show it is absolutely integrable in (-00, 00). In Problems 28 through 31, find the Fourier transform of f(x) by contour integration. f(x) 28. f(x) 29. f(x) 30. f(x) 31.

= x2/(1 + x2)2. = x2(1 + x4)-1 = ( x2 + a2 )-1(x2 + b2 )- 1, = ( x2 + 5x + 1)-1.

a> 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
52

5522 Fig. 5.2 Domain for the maximum principle Theorem 5.1

(Maximum Principle) Let u be a solution of ut - kuxx < 0

(5.12)

in SZ and be continuous in 52. Then u asssumes its maximum on 8522. Proof:

131

Maximum principle

We first prove that if a function v e C(S2)flC2(52) satisfies the inequality Vt - k vxx < 0

(5.13)

in Sl, it assumes its maximum on 8522. Suppose, by contradiction, that v assumes its maximum at P=(xo, to) in Q. Then, from elementary calculus, we have vt > 0 and vxx < 0 at P. Since k > 0, we have vt - kvxx > 0 at P, a contradiction. Since v is continuous in 1, v attains its maximum on 8522. Let M = maxac2 u(x, t) where u satisfies (5.12). Let e > 0. Define v(x, t) = u(x, t) +ex2. Then vt - kvxx = ut -kuxx - 2c = -2c < 0. Hence, v satisfies (5.13) and from the previous observation, v assumes its maximum on 8522. Thus, for any point (x, t) E 0, V = u + EX2 < M + cL2. So u < v < M + EL2. Since c > 0 is arbitrary, we have u < M. Remarks: (1). The corresponding minimum principle also holds if we apply Theorem 5.1 to -u. (2). The maximum principle also holds in any rectangular domain S={(x,t)Ia<x0,

(5.14)

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

(5.15)

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

(5.16)

We can apply the maximum principle to obtain a uniqueness theorem for the problem (5.14)-(5.16). There exists at most one solution of the problem (5.14)Theorem 5.2 (5.16) which is continuous in the region S = {(x, t) 10 < x < L, t > 0}. Proof: Let v and w be two solutions of (5.14)-(5.16) and let u=v - w. Let T be a constant. Then u is a solution of (5.14) in 0 < x < L and 0 < t < T and satisfies u(x, 0) = u(0, t) = u(L,t) = 0. By Theorem 5.1, u < 0 in

132

The Heat Equation

0<_x
ut - kuyz = F(x, t),

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

-oo < x < oo,

t > O,

(5.17)

-00 < x < 00. (5.18)

Theorem 5.3 There exists at most one solution of the problem (5.17)(5.18) which is continuous and bounded in the region R = {(x, t)^ - oo < x < oo,t > 0}.

Proof: Let v and w be two solutions of (5.17)-(5.18) such that lvJ < M and JwJ < M in R. Let u = v - w. Then u satisfies ut - kuxx = 0, u(x, 0) = 0 and Jul < 2M in R.

Let L and T be two arbitrary positive numbers. Consider the region S = {(x, t) i - L < x < L, O < t < T}. Define p(x, t) = 4M(x2/2 + kt)/L2. It is clear that p satisfies pt - kpxx = 0 in S, p(x, 0) = 2Mx2/L2 > 0 = u(x, 0), and p(±L, t) = 2M + 4Mkt/L2 > 2M > u(±L, t). Hence, by Theorem 5.1, p(x, t) > u(x, t) in S. Similarly, we obtain u(x, t) > -p(x, t). Thus, for any point (x, t) in t > 0, there exist constants L and T such that (x, t) E S and Ju(x, t) < p(x, t) = 4M(x2/2 + kt)/L2. Let L -4 oo. We have lu(x, t)l = 0. Hence, v(x,t) = w(x, t) in -oo <x < oo and 0
The initial- boundary value problem

133

5.3 The initial-boundary value problem

Consider the initial-boundary value problem

ut-kuyx=0, 0<x0,

(5.19)

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

( 5.20)

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

(5.21)

By the method of separation of variables , let u(x, t) = X (x)T (t). Hence,

X" _ T' - -A X kT ' where A is a separation constant . From ( 5.21), we obtain a Sturm-Liouville problem X"+AX=O, 0<x
(5.22)

for the function X (x). As a result, we have A = An = (n-7r/L)2 and X (x) = Xn(x) = sin(n7rx/L), n= 1,2,... For the function T (t), we have

T'+ AnkT = 0, t > 0,

(5.23)

and we get Tn(t) = exp(-kAnt). Thus, un(x, t ) = exp(-kn2ir2t / L2) sin (nirx/L), n = 1, 2... To satisfy ( 5.20), we write u(x, t)

0" bnun (x, t). = > n=1

(5.24)

For t=0, we have f (x) = u(x, 0) = > 0" bn sin(n7rx/L).

(5.25)

n=1

Hence, bn =

I 2

L

f (x) sin (n-7rx / L)dx.

(5.26)

134

The Heat Equation

(5.24) and (5.26) constitute a formal solution of (5.19)-(5.21). Verification: (A). Let f (x) be integrable on [0,L] and let

L

c= L f If(x)Idx. Thus, from (5.26), Ibnl < c. Hence, (5.24) is dominated by the series 00

c E exp (- kAnto) n=1

for 0 < x < L and t > to for some to > 0 and converges uniformly there. In order to show uxx and ut exist and are continuous in (0, L) x (0, oo), we differentiate (5.24) formally with respect to t and x respectively, and obtain 00 k Ut - E bn(- n T22 z ) exp(-knz7rzt /Lz) sin (n7rx/L), n=1

-

00

bn( T om) exp (- knz7rzt

ux

/L2) cos (n7rx/L),

n=1

00

2

2

2

uxx bn(- Lz )exp(

2 -k L2

tsin(n7rx/L).

n=1

We must show that the above series converge uniformly in [0, L] x [to, oo). However, these series are dominated by the series ao nz^z kc T Lz exp(- knz-7rzto /Lz) n=1

which converges by the ratio test. This implies that it is permissible to obtain uxx and ut by termwise differentiation of (5.24) on [0, L] x [to, oo) and verify that ut - kuxx = 0 there. Since to is arbitrary, we see that (5.19) is satisfied. (B). To show u(x, t) is continuous in [0, L] x [0, oo ), we assume that f (x) is continuous , f (0) = f (L) = 0, and f'(x) is piecewise continuous there. Then

Nonhomongeneous problems and finite Fourier transform

135

the Fourier sine series of f (x) converges uniformly to f (x) on [0, L]. Let T > 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<x
t>0,

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)\
\F(x,t)\dx<2M

and K ( 0 l < exp(-fcA {-kXnn()t) /f «exp(kXnr)\Fn(r)\dr Jo

< 2M[1

6

*P(

kXnt)]

<

f^.

Since the series £] 2M/(k\n) converges, it follows by the Weierstrass M-test that the series (5.38) converges uniformly in R. This implies that v(x, t) is continuous in R. Since T is arbitrary, v is continuous in 0 < x < L,t > 0. Hence, v(0, t) = v(L, t) = v(x, 0) = 0. (B). To show v(x,t) satisfies (5.27), we assume that F, Fx, and Fxx are continuous m 0 < x < L, £ > 0 and F(0, t) = F(L, t) = 0. By integration by parts, we have Fn{t) = -n r - 5 /

ft Jo

Fxx(x,t)sm(mrx/L)dx.

Let K = m&xR\Fxx(x,t)\. Then | F n ( 0 | < 2K/Xn. Now, differentiation of (5.39) yields v'n(t) = Fn(t) - k\nvn(t). As before, we get \vn(t)\ < 2K/(kX2n). Thus, \v'n(t)\ < AK/Xn. Now, we differentiate the series (5.38) formally with respect to t and x and obtain oo

oo

vx ~ YJ ~rvn{t)

vt ~ y~^ v'n(t) sin(mrx/L), n=l

COS(IVKX/L)

n=l 00

vxx ~

rfi-jrl

^T{—j^-)vn(t)sm(mrx/L).

n=l

Using the Weierstrass M-test, we can show all these series converge uniformly in any closed subrectangle in [0, L] x (0,oo), vxx and vt are continuous and oo

vt - kvxx = 22\Fn(t) n=l

- kXnvn(t) + kXnvn(t)}sin(nirx/L)

=

F(x,t).

138

The Heat Equation

Hence, we have proved the following theorem: Theorem 5.5 Let F(x, t), Fx(x, t) and Fxx( x, t) be continuous in [0, L] x [0, oo) and F(0, t) = F(L,t) = 0. Then the problem (5.27)-(5.29) has a unique solution given by (5.37) and (5.33). Remark : The unique solution of the initial-boundary value problem wt - kwxx = F(x, t), 0 < x < L, t>0,

w(x,0) = f(x), 0 < x < L,

w(0, t) = w(L, t) = 0, t > 0, is w(x, t) = u(x, t) + v(x, t) where u(x, t) is given by (5.24) and (5.26) and v(x, t) is given by (5.37) and (5.33) provided that f (x) and F(x, t) satisfy the conditions in Theorem 5.4 and Theorem 5.5 respectively.

5.5 The initial value problem Consider the initial value problem Ut - kuxx = 0,

- oo < x < oo,

t > 0,

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

-00 < x <

F[u] = u(w, t) =

J 00 u(x, t)e2wxdx

oo.

(5.40)

(5.41)

Let (5.42)

00

be the Fourier transform of the solution u(x, t) of (5.40) and (5.41). Apply (5.42) to (5.40) and ( 5.41). Since F[u"] = (-iw)2F[u], we have N +kw 2ii=0,

(5.43)

and ii(w, 0) = f f (x)eZ"xdx = j (w). (5.44) 00

139

The initial value problem Solving for (5.43) and (5.44), we have i(w,t) = f(

w)e-k2t

Using the inverse Fourier transform for both it(w, t) and f (w), we get ao 00

00

u(x t) = 2- it(w t)e -i"'ydw = ^ f( )[ f 0o ao 00 Using the formula

is

e-iw(x- E)- kw2t

dW. ]dt

00

e-y2/2+iuydy =

27r e-u2/2, (5.45)

from Section 4.7, the inner integral is simplified to e-(f-X)2 /(4kt)

kt Thus, 00 u(x, t) = 1 f

4k^rt

CO

f () e-(t;-=)2/(4kt )d^

(5.46)

which is a formal solution of (5.40)-(5.41). Remark: In Section 1.5, we use the method of separation of variables and the Fourier integral to obtain a formal solution of (5.40)-(5.41) given as 00 u(x, t) = f[A(a) cos(ax) + B(a ) sin(ax )] exp(-a2kt)d(5.47)

where 00

00

1 1 A(a) _ - f f (v) cos(av)dv, B(a ) = - f f (v) sin( av)dv. 7r 7r 00

(5.48) If we substitute A(a) and B(a) from (5.48) into (5.47) and formally interchange the order of the improper double integral, then we obtain 1 °° W (5.49) u(x, t) f f (v) f cos[a(v - x)] exp (- a2kt ) dadv. ^ ^ o After applying the formula (5.45) to (5.49) and simplifying, we obtain (5.46). Verification:

140

The Heat Equation

Let G(x - ^, t) = (4k7rt) -1/2 exp[-(^ - x)2/(4kt)], 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; IeI<5 and

I4 = [f (6) - f (0)]G(x - 6, t)d6. I,I>a 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)\<2C(kir)-1/2

/»00

i

/ ( i - r ) - 1 / 2 ( l + 2r?2)e-"2r/d7?(iT = 6 C ( - r ) 1 / 2 Jo Jo T^h 1 2

where rj = (£ — x)(4k(t — T))

^.

This implies that p(x, t) converges uniformly in Q, and uxx(x, t) = p(x, t) there. Since fi is arbitrary, we know that ux and be obtained directly from differentiating u(x, t) in (5.56) with respect to x in t > 0. (C). To determine ut(x, t), it is not permissible to differentiate u(x, i) with respect to t since the kernel G(x, t) —> oo as t —> 0+. Let e > 0. Consider the family of functions ft—e

ue{x,t)=

/»oo

/ Jo

G{x-Z,t-

r)h(£, T)d£dT,

J-oo

where e < t. Now, using Leibnitz's rule, we have

~dt=J_

Gfr-t'eWS't-eW

+J

J

Gt(x-t,t-T)h(t,T)dtdT.

The first integral converges to h(x,t) uniformly in fi as e -> 0 as in the proof of Theorem 5.6 and the second integral converges uniformly in £1 to kuxx by the similar arguments in (A) and (B). Thus, du lim — - = h(x, t) + c-»o at uniformly in Q,.

kd2u/dx2

Nonhomogeneous boundary conditions for initial-boundary value problems

145

For any to > 0 and t > to, we have

u€(x, t) = .t to aTE (x,T)dr + u6(

x, to).

Let e -40. Then u(x, t)

f =

J

t {h(x, T) + kuxx (x, T)]dT + u (x, to),

to

and ut (x, t) = h(x, t) + kuxx ( x, t) in 0. Since SZ is arbitrary, u(x, t) is a solution of of (5.52) in t > 0. Hence, we have the following theorem: The initial value problem (5.52)-(5.53) has a unique soTheorem 5.7 lution given by (5.56)-(5.57) if h(x, t) is continuous and bounded in t > 0 and continuously differentiable in any closed rectangle Sl in t > 0. Remark : The initial value problem ut - kuxx = h(x, t),

-oo < x < oo, t > 0; u(x, 0) = f (x), -oo < x < 00

has a unique solution u(x, t) given by

t u(x, t) = J f (S)G(x t)dl; + 00

oJ

00

G(x - t;, t - T)h(S, T)dedr

f 00

provided that f (x) and h(x, t) satisfy the conditions of Theorem 5.6 and Theorem 5.7 respectively.

5.7 Nonhomogeneous boundary conditions for initial -boundary value problems The method of separation of variables works only for homogeneous equations with homogeneous boundary conditions. In the case of nonhomogeneous boundary conditions, we have to convert the above problem into an equivalent nonhomogeneous equation with homogeneous boundary conditions. This approach will be illustrated with the following problem. Consider

ut - uxx = 0,

0 < x < L, t > 0, (5.59)

u(0, t) = h(t), u(L, t) = g(t), t > 0 ,

( 5.60)

The Heat Equation

146

u(x,0) = f(x), 0 < x < L.

(5.61)

Let v(x, t) = u(x, t) - [xg(t) + (L - x)h(t)]/L. Then v(x, t) is a solution of the following problem:

vt - vxx = -[xg'(t) + (L - x)h'(t)]/L, 0 < x < L, t > 0, v(0, t) = v(L, t) = 0, t > 0, v(x, 0) = f (x) - [xg(0) + (L - x)h(0)]/L, 0 < x < L.

(5.62) (5.63) (5.64)

Problem (5.62)-(5.64) can be solved using the remark in Section 5.4. We can also solve the problem (5.59)-(5.61) directly using the finite sine transform.

Let u(x, t) _ E 1 un(t) sin(nirx/L) where

L

un(t) = (2/L) f L u(x, t) sin(n7rx/L)dx. Then

L uxx ( x, t) sin ( nirx/L)dx = -

J0 L(n7r/L)2u(x, t ) sin(nirx/L)dx

L + [ux (x, t ) sin(n7rx /L) - n7ru(x, t) cos (n7rx /L)/L] = o _ -(nir/L) 2(L/2)un(t) - (-1)nn7rg (t)/L + n7rh(t)/L. On the other hand,

I

L

ut(x, t ) sin(nirx /L)dx = (L/2)u'n(t).

Since ut = uxx and f (x) = u(x, 0 ) = En'= l un (0) sin(nirx/L) from (5.61), we arrive at the following initial value problem u'n(t) + (nir/L)2un = -2n7rL-2(-1)ng(t) + 2nirL-2h(t), L

U"'(0)

J

=L 0

Thus, Un(t) can be determined.

f (x) sin (nzrx/L)dx.

147

Problems

Consider another problem: vt - kvxx = 0,

x > 0, t > 0 (5.65)

v(0, t) = h(t), v(x, 0) = O(x).

(5.66)

Let w(x, t) = v(x, t) - h(t). Then w(x, t) satisfies wt - kwxx = -h'(t), x>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<x<7randt>0; u(0, t) = ux ('7r, t) = 0, u(x, 0) = x(7r - x). 5. ut-uxx+u=0, 0<x<1, t>0; ux(0,t)=u(1,t)=0, t>0, u(x,0)=x2-1, 0<x<1. 6. ut-kuxx=0for 0<x0;

148

The Heat Equation

u(x,0) = f(x), u(0, t) = 0, ux(L,t) + hu(L, t) = 0 where h and k are positive constants. In Problems 7 through 9, find a formal solution of the initial -boundary value problem for the nonhomogeneous heat equation. 7. ut - kuxx = cos wt for 0 < x < 7r and t > 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
u(0, y) = u(a,y) = 0, 0 < y < b,

(6.15)

u(x, 0) = 0, 0 < x < a,

(6.16)

u(x, b) = f (x), 0 < x < a.

(6.17)

Let u(x,y) = X(x)Y(y) and substitute u in (6.14). We have X"/X = -Y"/Y = -A, a separation constant. Together with (6.15) and (6.16), we are led to two separate ordinary differential equations with appropriate boundary conditions: X" + AX = 0; X (O) = X (a ) = 0,

(6.18)

and

Y"-AY=0;

Y(0)=0.

(6.19)

(6.18) is a Sturm-Liouville problem, which has eigenvalues A = An = (nir/a)2 with corresponding eigenfunctions X (x) = Xn (x) = sin(nlrx/a), n = 1,2... Put A = (n7r/a)2 in (6.19) and get Y(y) = Yn(y) = sinh(niry/a). Hence, {un(x, y) = sin(nirx/a) sinh(n7ry/a)}

156

Laplace's Equation and Poisson's Equation

is a set of particular solutions of (6.14)-(6.16). To satisfy (6.17), we form a series solution 00 u(x, y) =

an sin (n7rx/a) sinh(niry /a).

(6.20)

n=1

By (6.17), we have 00 f (x) = u(x, b) = E an sinh(n7rb/a) sin (n7rx/a) (6.21) n=1 which is a Fourier sine series of f (x) on [0, a]. Thus, 2 an a sinh (n7rb/a)

f (x) sin(n7rx /a)dx.

(6.22)

(6.20) and (6.22) represent a formal solution of (6.14)-(6.17). Verification:

(A). Let f (x) be integrable on [0,a] and let ja c=

f (x)Idx.

We note that sinh(niry/a) - exp(niry/a) - exp(-n7ry/a) sinh(n7rb/a) exp(nirb/a) - exp(-nirb/a)

=[ 1 - exp (( 2n b/a )] [

1 -exp (- 2n^ry /a)] [1 -

Similarly, cosh (n7ry/a) < 2 exp [- rur(b - y)/a] sinh (nirb/ a) - [1 - exp(-2nirb/a)] Hence, the series ( 6.20) is dominated by the series 00 [1 - exp(- 2n7rb/a)]

xp((2n b/a)].

Laplace's equation in a rectangle

157

which converges uniformly for any y such that 0 < y < yo < b. Differentiating (6.20) formally with respect to x and y respectively, we obtain 00 ux an(al) cos(n7rx/a) sinh(n7ry/a), n=1

00

an( ^) sin(n7rx /a) cosh (n7ry/a), n=1

00

n22 an(-

uxx

n=1

a2

00

2) sin(n7r x /a) sinh ( n7ry /a),

2 )

uyy '' T an (na2

sin(n7rx / a) sinh (n7ry/a).

n=1

The above series are all dominated by a constant multiple of the series cc n2 exp[-n7r(b - y)/a] n=1

which converges uniformly for 0 < y < yo. Hence uXX and uyy can be obtained by termwise differentiation of (6.20) and ux2, + uyy = 0 for (x, y) in (0, a) x (0, b). (B). To show u(x, y) is continuous in [0, a] x [0, b], we assume that f (x) is continuous on [0, a], f (0) = f (a) = 0, and f'(x) is piecewise continuous there. Then, from the theory of uniform convergence of the Fourier series, f (X) = E 00an sinh(n7rb/ a) sin (n7rx/a) n=1

holds uniformly on [0, a]. Let SN(x, y) be the Nth partial sum of the series (6.20). Then SN(x, b) is the Nth partial sum of the series (6.21). By the Cauchy criterion for uniform convergence, for e > 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
vxx+uyy =0 ,

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

u(x,b) = f2(x),

0 < x < a,

u(O,y) = 9i ( y),

u(a,y ) = 92(y),

0 < y < b,

can be decomposed into four similar problems.

6.5 Laplace ' s equation in a disc Consider the Dirichlet problem of Laplace's equation in a unit disc: vxx + vyy = 0,

x2 + y2 < 1;

v(x,y) = 9(x,y), x2 + y2 = 1.

To determine its solution by the method of separation of variables, it is natural to introduce the polar coordinates. Let x = r cos 0 and y = r sin 0. Denote v(x, y) = v(r cos 0, r sin 0) = u(r, 0). Then Xr = cos 0,

xe = -r sin 0 ,

yr = sin 0 ,

ye = r cos 0.

By partial differentiation, we get dx = cos Odr - r sin OdO, dy = sin Odr + r cos OdO. Solving for dr and dO in terms of dx and dy, we have dr = cos 0dx + sin Ody, dO =

- sin cos O dx + 0 dy r r

Thus, sin 0 cos 0

vx = cos 0 Ur - r u0 i vy = sin O Ur + r U9,

159

Laplace's equation in a disc

7lyy = sin g 0 urr +

UrO +

r

2 sin O cos 0 r

ur8 +

r

2

2 sin 0 cos O

sine B

sine O

2 sin O cos B Uxx = COS 2 0 Urr -

U99 +

r

Ur +

cost 6 cost B ur 2 UB9 + r r

-

r

UB,

2 sin 0 cos 0 U. 2 r

Hence, Laplace's equation in polar coordinates takes the form urr + 1 -Ur + 1 uBB = 0r r2

(6.23)

We seek a solution u(r, 0) of (6.23) with r < 1 which is continuous in r < 1 and satisfies u(1, O) = f (0).

(6.24)

We require f (0) to be a periodic function of period 2ir. Hence, the solution u(r, 0) is also periodic in 6 with period 2ir. Let u(r, 0) = A(r)B(0) and substitute u in (6.23). We obtain r2A" rA' B"

A + A = -B =A,

where A is a separation constant. Since u(r, 0) is continuous in r < 1, we have r2A" + rA' - AA = O, 0 < r < 1,

(6.25)

where A(r) is bounded at r = 0, and B" + AB = 0; B(-ir) = B(7r), B'(-7r) = B'(ir).

(6.26)

If, for some value A, there exists a nontrivial solution B(0) of (6.26), then A is called an eigenvalue of the periodic boundary value problem with the corresponding eigenfunction B(0). We would like to show that all the eigenvalues are real. Let B(0) be an eigenfunction of (6.26) with eigenvalue A. Then, taking the complex-conjugates of (6.26), we get B" + aB = 0; B(-ir) = B(7r), 13'(-7r) = B'(7r).

(6.27)

Laplace 's Equation and Poisson 's Equation

160

Multiplying the equation (6.26) by B and the equation (6.27) by B, subtracting the second equation from the first one, and integrating the resulting equation on [-7r, ir], we get

(A - a) f B(0)B (0)dO = f [B"(9)B(9) - B"(O)B(0)]dO. After integrating by parts on the right side and applying the boundary conditions in (6.26 ) and (6 .27) on the integrated terms, we get

7r (A -.X) f IB(9)12d9 = 0. This implies that A = A. Hence, A is real. Now we proceed to determine all the eigenvalues of (6.26). (A). \ = 0. Then B(9) = a9 + b and B'(9) = a. From the boundary conditions, we deduce a = 0. Hence B(O) = 1 is an eigenfunction with 0 as an eigenvalue. (B). A = -µ2 (µ > 0).

Then

B(9) = a exp (pO) + b exp(-µ9)

and

B'(9) = p(aexp(µ9) - bexp(-µ9)).

Using the boundary conditions, we get a sinh (p7r) = b sinh(µir) and a sinh (p7r) = -b sinh(µnr). This implies that a = b = 0 and A < 0 is not an eigenvalue. (C)• A = p2 Then and

(µ > 0). B(O) = a cos (µ9) + b sin(pO) B'(9) = -aµ sin (µ9) + by cos(tO).

From the boundary conditions, we get a sin (p r) = b sin(ter) = 0. For nontrivial values of a and b, we set sin(y-7r) = 0. This implies that p = n, n = 1,2... Hence, A = An = n2, n = 1,2... are eigenvalues, and the corresponding eigenfunctions are Bn(0) = an cos(n9) + bn sin(nO). We note that the positive eigenvalues of the periodic boundary value problem (6.26) are not simple.

Laplace's equation in a disc

161

Set A = n2 in (6.25). We get r2A" + rA' - n 2 A = 0, 0 < r < 1. For n = 0,Ao(r) = a+ blogr; and for n > 0,An(r) = am+br -n. We require that An(r) remains bounded as r -+ 0. Hence, we set b = 0 in both cases, and An(r) = rn, n = 0, 1,2... From this, we obtain a set of particular solutions {un(r,0)} such that un (r, 0)

ao/2

n=0;

anrn cos n0 + bnr"' sin n0,

n 0.

We seek u(r, 0) in the form of 00 u(r, 0) = 2 + ^(an cos nO + bn sin nO)rn n=1

(6.28)

such that 00 f (0) = u(1, 0) = ao + > (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 - <£)]#. To evaluate the series inside the integral, we note that oo

1

n=0

Let z = r exp(z#) and separating the resulting series into real and imaginary parts, we get

E

r

„

.

COS 710 =

n=0

1 — r cos 9 ~. 1 - 2 r c o s 0 + r2

Hence, 1-r2 - + 2_^ rn cos n# = 2 ^ 2(l-2rcos<9 + r 2 ) ' Thus, the solution of (6.31)-(6.32) can be written as u(r,9)=

f f(<j>)P(r,9-l,4,)d J—n

(6.35)

164

Laplace 's Equation and Poisson 's Equation

where P(r, 0;1, 0) =

1 - r2 27r[1 + r2 - 2r cos(0 - ^))] ' (6.36)

(6.35) is called Poisson's integral formula and P(r, 0;1, 0) is called Poisson's kernel. We can verify that P(r, 0; 1, ¢), as a function of r and 0, satisfies (6.31) in r < 1. From this, we can differentiate (6.35) under the integral to show u(r, 0) is a solution of (6.31) for r < 1.

The corresponding formula for the solution of the Dirichlet problem for Laplace 's equation in a (disc of radius R is u(r, 0) = and

R2

- r2 7r .f (0) dO, r < R, 'r R2 + r2 - 2rR cos(0 - ¢)

27r

J

(6.37)

u(R, 0) = f (0).

Definition 6.1 Let D be a domain in the xy-plane. A function u E C2 (D) which satisfies Laplace's equation in D is said to be harmonic in D. Theorem 6 .7 (Mean Value Theorem) Let u be harmonic in D. Then the value of u at the center of any disc in D is equal to the average of its boundary values. Proof: Set r = 0 in (6.35) or ( 6.37) and get u(0, 0) =

21

f f r

(0) do.

We would like to use Poisson 's integral formula to remove the condition that f (0) is piecewise smooth on [-ir, ir] in Theorem 6.5 . Let f (0) = 1. By the uniqueness theorem for the Dirchlet problem, u(r, 0) = 1 . Hence, from (6.37), we have

1 = f P(r, 0; R, 0) d¢

(6.38)

where R2 - r2 ' 2ir[R2 + r2 - 2Rr cos(0 - 0)]

P(r 0• R th) =

(6.39)

Poisson's integral formula

165

For r < R, Poisson 's kernel (6.39) is always positive . Without loss of generality, we would like to show that u(r, 0) is continuous at (1,0 ). Since f is continuous at 0 = 0 , for e > 0, there is a 8 > 0 such that for 101 < 6, If (0) - f (0) 1 < E. Then, from (6.37 ) and (6 . 38), we have

u(r, 0) - f (O) = f[f() - f (0)]P(r, 0; R, )d

+ f a [f() - f (0)]P(r, 0; R, O)dO + f -b [f (0) - f (0)] P(r, 0; R, O)do. (6.40)

For

101

< 6/2 and 8 <

101

< ir,

R2 + r2 - 2Rr cos(O - 0) > R2 + r2 - 2Rr cos(8/2) > R2 + r2 cos2(8/2) - 2Rr cos(8/2) = [R - r cos(8/2)]2 > R2 [1 - cos(8/2)]2. Hence, P(r, 0; R, 0) < Let M = max I f (0)1. bounded by

R2 - r2 2irR2[(1 - cos(8/2)]2.

Then the first and third integrals of (6.40) are

4M(R2 - r2)(7r - 8) 2,7rR2[1 - cos(8/2)]2

which goes to zero as r -> R and the second integral in (6.40) is bounded by e/2. Hence, for 101 < 8/2, and r is sufficiently close to R, I u(r, 0) - f (0) 1 < E. We state our result in the following theorem: Theorem 6 . 8 Let f (0) be continuous in [-ir, ir] and f (-7r) = f (ir) . Then, the boundary value problem Du = 0,

r < R; u(R, 0 ) = f(0);

(6.41)

has a unique solution given by (6.37). Next, we will use Poisson's integral formula and the maximum principle to obtain the following result:

166

Laplace 's Equation and Poisson's Equation

Theorem 6.9 Let { uk(x, y)} be a sequence of harmonic functions in a bounded domain D and continuous in D and its boundary C. If this sequence converges uniformly on C, then the sequence also converges uniformly to a harmonic function u(x,y) in D.

Proof: Let e > 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 and let (, -r7) be its image with respect to the line y = 0 ( see Fig. 6.4).

YT,

y)

(e, 77) 0

(c, -rl )

Fig. 6.4 Green's function in a half plane

Then the Green's function is given by 1 ro G(x, y; ^, 77) = 27r log(r0 )

(6.58)

Poisson's equation in a disc

where r0 = \/(x-)2 +(y-7J)2 and ro=

173

(x-^)2+(y+i)2.

6.8 Poisson 's equation in a disc Consider the Dirichlet problem 1 1 Urr + r Ur + r2 ueB = -F(r, 0), r < R,

(6.59)

u(R, 0) = 0.

(6.60)

Using the representation formula (6.53) and the Green's function for a disc of radius R constructed in (6.57), we have a formal solution of (6.59)-(6.60) given by

u(r,0) =

J 7 J0 R G(r,0;p,q)F(p,q)p dpdO

(6.61)

where G(r, 0; p, ^) =

R2 +r2p2R -2 - 2rpcos(0 - 0) 1 7r log r2 + p2 - 2rp cos(0 - 0) 4 l

(6.62)

Theorem 6.11 Let F(r, 0) be continuously differentiable in r < R. Then u(r, 0) defined by (6.61)-(6.62) is continuous in r < R and twice continuously differentiable in r < R and is the unique solution of (6.59)-(6.60).

Proof: The uniqueness of solution is assured by the maximum principle. Let F(r, 0) = 1 in the domain D = {r < 1}. Then v(r, 0) = (R2 - r2)/4 is the unique solution of (6.59 )-( 6.60). Since v(r, 0) is in C2(D), by the representation formula (6.53), we have

R2 4 r2 =

J Jig G(r, 0; p, O)p dpdo.

(6.63)

For any other function F(x, y), let maxD JF(r, 0) 1 = M. Then, from (6.61), Ju(r, 0) 1 < M(R2 - r2)/4 since G(r, 0; p, ^) > 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
IL Gr.Fd,

(6.64)

0) =

IL GBFda.

(6.65)

ug(r,

Differentiating (6.63) with respect to r and 0 respectively, we have (6.66) and

(6.67) Multiplying ( 6.66) and (6.67) by F (r, 0) and subtracting from the corresponding equations ( 6.64) and ( 6.65) yield Ur

+ rF(2 0) =

ILD

Gr[ F (p,

^) - F(r, 9 )]dv

(6.68)

and ue =

IL G e [F(p, q ) - F(r, 0)]dc.

As before, we would like to verify that

I

D

8r {[log

ro] r[

F(p, 0) - F(r, 6)]}du

and

JD T. { [log

ro]o [F ( p, 0) - F(r, B)] }do,

(6.69)

Laplace's Equation and Poisson 's Equation

176

converge uniformly in r < a. Since F (r, 9) is in C' (D ), we have, by the mean value theorem,

I F(p, 0) - F( r, 0) I < Kro where K = maxD

IVFI;

and , as before, we obtain the following estimates I = Iro -2[r-pcos(9-0)]2 < 3

I(l ogro ) rr

I

ro

ro

and

I( l ogro)eBI

=

I

rorpcos(9 - ^) - 2r

sin2(9 - ^) I < 3R2

ro4

ro

Using the above inequalities, we get

i

^DE

8r{

(log ro ) r[ F (p,

')

- F(r, 9)]}dal

ff {(logro)j I(F(p,O)-F(r,8))I+I(logro ),I IFr(r,e)I}doj <

JJ

[2 Kro + K ] dv = 8K7re

DE ro

ro

and

I J L.

80 {( log ro )o [F(p, 0) - F

(r, 9)]}do l

IF(p,O)- F(r,0)I +I(logro)eI IFe (r,9)I}dv

f

<

f f 3R2 DE [ ro Kro + ro k] do, = 27reK ( 3R2 + R).

JJ

Thus, both integrals go to zero uniformly as c --^ 0. Therefore, we can differentiate ( 6.68) with respect to r and (6 . 69) with respect to 0 and get urr + 2 + rFT = J ) - F(r, 0)] - GTFr.}d

Finite Fourier transform, for Poisson's equation 177

and ueo = f

I

D

{Goo [F( P, 0) - F(r, 0)] - GBFo}da.

Then, from (6.66), (6.67), and the fact that G satisfies Laplace's equation, 1 1 - Ur + r2 uoo = -F(r, 0). Urr r Thus, we have completed the proof of Theorem 6.11.

6.9 Finite Fourier transform for Poisson 's equation In this section, we would like to obtain a solution u(r, 0) for the Dirichlet problem urr + -Ur 1 uoe = -F(r, 0), r < R, r r2

(6.70)

u(R, 0) = 0 ,

(6.71)

in a series of the form cos n0+ bn(r)sinnB} (6.72) u(r,0) = ao(r ) +>{an(r) 00 n=1

where an(r) = 1-f n u(r, 0) cos nOdO, 7r 7r

bn(r) = 1 f n u(r, 0) sin nOdO. Ir 7r

The set of formulas {an(r), bn(r)} is called the finite Fourier transform of u(r, 0). As in the case of Laplace's equation in a disc, we assume that F(r, 0) is periodic in 0 of period 2ir and we look for the solution u and its partial derivatives that are periodic of period 27r. Then, by integration by parts, we get an (r) = -

1

f uBo (r, 0) cos nOd0, n2,r n

7r bn(r) _ - 2 f uoo(r, 0) sin nOdO. Ir 7r

Applying the finite Fourier transform to equation (6.70), we have 1

n2

an + -an - an = -An (r), r r2

n2 ( ) 1 bn = -Bn r b bn + -;^ r r2

178 Laplace 's Equation and Poisson 's Equation

where An (r) = 1 7r

J

"

F(r, 0) cos nOdO , Bn (r) = 1 f 7r F(r, 0) sin nOdO 7 7r

which are called the finite Fourier transforms of F(r, 0). From (6.71), an(R) = bn(R) = 0 and we require both an(r) and bn(r) to remain bounded as r -* 0. To determine an (r) and bn (r), we are going to study the following singular Sturm-Liouville problem: Ln(v) = (rv')' - r v = -fn(r), 0 < r < R

(6.73)

v(r) remains bounded as r -+ 0 ,

(6.74)

v(R) = 0.

(6.75)

For a fixed p, 0 < p < R, Green's function G(r,p) of (6.73)-(6.75) is a function of r with the following properties:

(A). Ln(G) = 0, r # p (B). G(r, p) remains bounded as r -> 0 and G(R, p) = 0, (C). G(r, p) is continuous ar r = p, (D). dG/drlr=p+ - dG/drlr=P_ = -1/p. As in the case of the Sturm-Liouville problem, Green's function G(r, p) of the problem (6.73)-(6.75) exists and is unique if the homogeneous problem L(u)=0, 0 < r < R; u(R) = 0 and u remains bounded as r --* 0 (6.76) has only the trivial solution. Then the problem (6.73)-(6.75) has a unique solution given by R v(r) = f G(r, p)fn(p)dp• 0 It is easy to show the homogeneous problem (6.76) has only the trivial solution for all nonnegative integers n. Its Green's function subject to the

Finite Fourier transform for Poisson's equation

179

conditions (A)-(D) above is, for n=0, Go{r P)

=

>

\-log(r/R),

(6 ?7)

r>P.

-

For n > 1, its Green's function is given by U n\ - / -Vn)-l(r/Rr\(p/Rr l*n(r,p) - I _{2n)-i{p/m{r/RT

- (R/P)% _ {R/r)%

r

r p

V-<»)

Thus, a0(r)=

fR

Jo

G0(r,p)pA0(p)dp,

b0(r) = 0,

(6.79)

and

rR

rR

an(r) = / Gn(r,p)pAn(p)dp, bn(r) = Gn(r,p)pBn(p)dp. Jo Jo (6.80) Hence, (6.72), (6.77), (6.78), (6.79), and (6.80) give a formal solution of Hence, (6.72), (6.77), (6.78), (6.79), and (6.80) give a formal solution of (6.70)-(6.71). (6.70)-(6.71). In order to verify that u(r, 9) defined by (6.72) is a solution of (6.70)(6.71), we assume that K=

[ JO

[

F2(r, 0)rdrde < oo.

(6.81)

J-T

Since {An(r),Bn(r)} are the Fourier coefficients of F(r,8) on [—7r,7r] for a fixed r, by the completeness relation, both J0 A^rfrdr An(r)rdr and / 0 B;;(r)rdr B„(r)rdr are bounded by K/n. By direct computation, we have, rR

j

R2 - r

2

Gg(r,p)pdp=—i—+

r2 T

r

log(-),

J*G2(r,p)pdp=-j(l-^) + j\og(^),

(6.82)

(6.83)

and, for n > 2, rR

1

r2

r2n

r2n

R2

(6.84)

180

Laplace 's Equation and Poisson 's Equation

For n > 2, we apply Schwarz's inequality to both an (r) and bn(r) in (6.80), and get

Ian(r)IZ - 4n7r(n 1)1

I1)n(r)Iz - 4n7r(n 1)

Thus, the series (6.72) converges uniformly to a continuous function in r < R by the Weierstrass M-test and satisfies the boundary condition (6.71). To show that u(r, 9) satisfies (6.70) in r < R, we have to put a stronger condition on F(r, 0) to justify termwise differentiation of the series. However, it is more instructive to rearrange the formal solution given by (6.72), (6.77), (6.78), (6.79), and (6.80) to obtain the formal solution (6.61)-(6.62) given in Section 6.8. Substitute (6.77)-(6.80) into (6.72) to get u(r,0) =

2

jr Ao(p)plog( )dp+ 2

+I pr 2n[( R)n-(R) ](R )'[ r

f

RAo(P)Plog( R P )dp

A.(p) cos nO + Bn(p) sin nO)]pdp

00 +E

JTR

2n[(p)n-(R)n](R )n[An(p)cosnO +Bn(p)sinn9]pdp. (6.85)

We now substitute for An(p) and Bn(p) in (6.85) and formally interchange summation and integration to get u(r, 9) = f "

f o

R

G(r, 0; p, O)F(p, cb)pdpdO

00 1 [log( R)+E 1 [(R)n- (r)n](P )n cosn (B-0)]+ 2ir r n r R R

(6.86)

p < r,

n=1

G(r, 0; p, 0) = G(p, 0; r, 0), p > 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
n=1

z 00 zncos na

n

n-1 z t; cos na d(

= JO

n =1

Jz 1 + cos a - cos

a do

1 00

f

n - [E cos na - 1] do n =0

2 log[1 + z2 - 2z cos a]

by the following identity °O

1:

rn

cos na =

n=O

1 - rcosa 1 -2rcosa+r2

Thus, the function G(r, 0; p, 0) in (6.87) can be put into a closed form given by (6.73).

6.10 Dirichlet problem in the upper -half plane Consider the Dirichlet problem of Laplace's equation -oo < x < oo,

uxx + uyy = 0, u(x, 0) = f ( x),

(6.88)

y > 0,

-00 < x < 00. (6.89)

Let 00 u(w, y) = fux,y) ei"xdx. ao Then, formally, we have 00 ✓ x, y) eiwxdx = - f uxx( x, y)ei^ xdx uyy(w,y) = f uyy( 00 00 00

-iw ( )

00 2 f

00

iwx 2 u ( x , y) e dx=w u ( w ^

y)

6.90 ()

182

Laplace 's Equation and Poisson 's Equation

and f(x ) e"xdx = J(w).

u(w, 0) = r oo

(6.91)

Hence, the general solution of (6.90) is u(w, y) = cl (w) e"y + c2 (w) a-"y, where cl(w) and c2(w) are two arbitrary functions of w. We look for a bounded solution fc(w, y). For w > 0, we set cl (w) = 0 and for w < 0, we set c2(w) = 0. Thus, from (6.91), u(w, y) = i(Lo) e-Ply. From the inverse Fourier transform of ii(w, y), we have u(x, y) = 27r f00 u(w, y) 0o

f 00 "xdw = 2^ f (w) e-1"lye-i"xdw. (6.92) J o0

e-i

Substituting j (w) from (6.91) into (6.92) and reversing the order of the resulting double integral, we get u(x , y ) =

27r

00

00

ao

00

e-I "Iye' "( 6-x)dw] f (6)g.

(6.93)

We also know that 00

00

e-lxie2"xdx = f00 e-Ix1[coswx +isin (wx)]dx 00

=2

2

J0 00 e-x cos wxdx = 1+w2*

Let t = wy and i = ^ - x. Then the inner integral in (6.93) is reduced to 2y/(y2 +,q2 ) and the formal solution of (6.88 )-(6.89) is given by u(x, y) = y [ 00 2 f (6) 2 dl;, y > 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 E1<6 ( x - S)2 + y2 and

J=

PO - f(O) d;

y

41 >a (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
(B). Let F(x, y)

°°

f

(^)

_ F ao (x - 6)2 + y2

dt;

184

Laplace's Equation and Poisson's Equation

and differentiate F(x, y) formally under the integral with respect to x to obtain F. (x, y ) ^' -

2 (x - Of W2 2 d6 = P(x, y) J. [(x- ) +y

and ) ^ -2 00 [y2 - 3 (x ^'xx (x, y

S)22f ( ) -1. = Q (x, y) ,

^ oo

[(x - e)2 + y

and similar expressions for Fy and Fyy. Now, let s = x - t; and we have

P(x, y) 4M ^^ sds - 2M o (S2 + y2)2 - 7 and

Q(x, y) 1

6M f .

(s2s+ y2)3

+ 2My2 oo (S2 + y2)3

2M r°° (3v2 + 1) do, K1 < y3 f

oo

(

1 + o.2)3 - y3

for some constant K1. This shows that the integrals P(x,y) and Q(x,y) converge uniformly for -oo < x < oo and 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 <-maaxIf(x,y)I+ 1R2mD Iq(x,y)I where R is the radius of a circle containing D. 6. Find a formal solution of the problem Du - hu = 0 , 0 < x < 7r, O < y < 7r; u(x,0) = u (0,y) = u(ir,y ) = 0,u(x,ir ) = f(x); where h > 0 is a constant. 7. Solve u.,x+uyy=-COS2x, O<x<7r, O
r > a, u (a, 0) = f (0),

has a bounded solution given by u(r, B)

__ _ 1 7 (a2 - r2 )f (cb) d^. 27r 1, a2 + r2 - tar cos(0 - 0)

10. Consider the Neumann problem Du = 0, r < R; ur(R, 0) = f (0) where f (0) is a periodic function of period 2-7r. (a). Show that if the problem has a solution, it is necessary that f f (O)dO = 0. (b). Find, by the method of separation of variables, a formal bounded solution of the Neumann problem where f satisfies the condition in (a) and put the solution in the form of u(r, 0) = 2 - R f f (0) log[R2 + r2 - 2Rr cos(0 - O)]dO.

188

Laplace 's Equation and Poisson 's Equation

where A is an arbitrary constant. (c). If f ( 0) is continuous , verify that u(r, 0) in (b) is a solution. 11. Show that if u is harmonic in the xy-plane , then v(x, y) = u(x/r2, y/r2) is also harmonic at r # 0 , where r = x2 + y2.

12. Let u be harmonic in a domain D. Show that 1 u(x, y) _ 7ra2

s

u(e, il)d^dil

where S is a disc with center (x, y) and radius a in D. 13. (a). Let u(r, 0) be a nonnegative harmonic function for r < R and continuous in r < R. Show that, by Poisson's integral formula, u satisfies the following inequalities: u(0, 0) (R - r) u(r, 0) < u(0, 0) (R + r) R+r - - R-r for r < R. (b). Show that if u is a nonnegative harmonic function in the entire plane, then u must be a constant. 14. Let G be a bounded domain in the xy-plane. A function u(x, y), continuous in G, belongs to the class M if for every disc with center (x, y) and radius r contained in C, u(x, y) <

1

2^ 0

21r u(x + r cos 0, y + r sin O)dO.

Show that (a). If u is harmonic in G, both u and u2 belong to M. (b). If U E C2(G) and Du > 0 E G, then u belongs to M. (c). If u belongs to M and u assumes a local maximum at some point P of G, then u is constant in some neighborhood of P in G. 15. Show that the Green's function for Laplace's equation is nonnegative in a bounded domain. 16. Find, by the method of images, the Green's function for Laplace's equation (a). in the quarter plane x > 0, y > 0.

Problems

189

(b). in the upper-half disc of radius R with center at the origin and express your answer in polar coordinates. 17. Deduce from Poisson's integral formula for the half-plane the solution of the problem Au = 0, 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<7r, 0
u=0 forx=0,x=7r,y=O,y=7r andz=7r;

0
u(x,y,0)=g(x,y). (7.18) The maximum principle for Laplace's equation for any bounded domain in space also holds. It can be proved similarly as in Theorem 6.3. From the maximum principle, the problem (7.18) has at most one solution. We now proceed to find the solution of (7.18) by the method of separation of variables.

Laplace's equation in a

cube

201

Let u (x, y, z) = X(x)Y(y)Z (z). Substituting u into the equation, we get X" Y" Z" X" Y" X = cl - Y = c2, -y-+- Y = - Z = cl, where cl and c2 are separation constants.

Together with the homogeneous boundary conditions, we get

X" - c2X = 0, X(0) = X(7r) = 0;

(7.19)

Y" - (cl - c2)Y = 0, Y(0) = Y(ir) = 0;

(7.20)

Z" + c1Z = 0, Z(7r) = 0. (7.21) From our previous discussion, the eigenfunctions for (7.19) are X"(x) = sin nx with eigenvalues -c2 = An = n2, n = 1, 2... For (7.20), the eigenfunctions are Ym (y) = sin my with eigenvalues c2 - c1 = µm, = m2, m = 1,2... Set cl = -(n2+m2) in (7.21). We get Znm(z) = sinh[(n2 + m2)1/2(7r - z)]. We seek a solution of the form u(x, y, z) = anm sin nx sin my sinh[(n2 + m2)1 /2(ir - z)]. (7.22) n,m=1

Let z = 0 in (7.22). We have 00 g(x, y) = u(x, y, 0) = E anm sinh[(n2 + m2)1/ 27r] sin nx sin my n,m=1

which is a double sine series of g(x, y) on [0, -7r] x [0 , 7r]. Hence, 4 fo fo g(x, y) sinnx sin my dxdy anm ire sinh[(n2 + m2)1/27r] ( 7.22) and ( 7.23) give a formal solution of (7.18). Verification:

(A). We first assume that fo fo Ig(x, y) Idxdy = c < oo. Then 4c lanml

1r2 sinh[( n2 +

m2)1/27r].

(7.23)

202

Problems in Higher Dimensions

As a result, the series (7.22) together with all of its first and second partial derivatives converges uniformly for zo < z < 7r, 0 < x < 7r, 0 < y < 7r for any constant zo > 0. We can differentiate the series (7.22) term by term and verify that u(x, y) satisfies Laplace's equation for z > 0. Moreover, the series u(x, y) converges uniformly to a continuous function in zo < z < 7r, 0 < x < 7r, 0
(B). We further assume that g(x, y) can be extended to be an odd, periodic function of x and y of period 2ir and continuously differentiable there. Furthermore, g(x, y) is piecewise continuous there. As shown in Section 7.2, we can apply the completeness relation for the functions g., g.., and gxy to establish that the double sine series of g(x, y) converges uniformly to g(x,y) on [0, 7r] x [0, 7r]. The partial sum of the series (7.22) satisfies the Cauchy criterion for uniform convergence in [0, 7r] x [0, ir] on z = 0 and vanishes on the other sides of the cube. Then, by the maximum principle, the series (7.22) converges uniformly in the cube and is continuous on z = 0. Hence, u(x, y, 0) = g(x, y).

7.4 The two- dimensional wave equation in a rectangular domain Let D be a domain in the xy-plane bounded by a simple closed piecewise smooth curve C. Consider the problem utt - C2(uxx + uyy) = 0, (x, y) E D, t>0

u=0 or

8u -=0, (x,y)EC, t>0

u(x, y, 0) = .f (x, y), ut(x) y, 0) = 9(x, y), (x, y) E D. (7.24) The above problem corresponds to a vibration of membrane in the xyplane. A membrane is an elastic film supported by a closed plane curve C. Its equilibrium position is in the xy-plane. If the membrane is displaced from its equilibrium position, and then released, it will vibrate along the normal direction to the plane. Let u(x, y, t) denote the displacement of the

The two-dimensional wave equation in a rectangular domain

203

membrane at time t of the point (x,y) in the plane. In the case of small vibrations, we can derive the two-dimensional wave equation in the same way as we have done in the case of the vibrating string with c2 = T/p where T is the tension of the membrane and p is the surface density of the membrane. Both T and p are constant. We would like to show that there exists at most one solution of (7.24). The proof of uniqueness can be made with the energy method provided that Green's theorem is applicable. Define the energy integral as

E(t) 2

JJ

[ut + c2(ux + uy)]dxdy.

(7.25)

D

Let u(x, y, t) be twice continuously differentiable for (x, y) in D and t > 0. Then

E '(t) =

JL

tutt + c2

( uut + uut

)]dxdy.

Applying Green's first identity w an ds =

I

I

L [wOv

+ wxvx

+ wyvy]dxdy

with w = ut and v = u, we have t[u- c2udxdy + c2 E'(t) = ff u

J

ut-ds.

C

Let ul and u2 be two solutions of (7.24) and let u = ul - u2 . Since u = 0 or Ou/ On = 0 on C, E(t) = 0. This implies that E(t) = K, a constant. But at t = 0, ut = u = 0; thus , E(0) = 0. Then, by continuity, E(t) = 0 for t > 0. From this, we have ut = ux = uy = 0 in D since the integrand in E(t) is nonnegative . So u(x, y, t) is a constant. But, u (x, y, 0) = 0; then, by continuity, u(x, y, t) = 0 or u1 = u2. We proceed to solve the initial-boundary value problem for the wave equation in a rectangular domain: utt - uxx - uvy = 0, 0 < x < 7r, 0 < y < 7r, t > 0, u=0 for

x=0, x=1r, y=0, y=7r

u(x, y, 0 ) = f (x, y), ut( x, y, 0) = g (x, y), 0 < x < 7r, 0 < y < 7r. (7.26)

Problems in Higher Dimensions

204

Let u (x, y, t) = X (x)Y(y)T( t) and substitute it in the equation in (7.26). Then we have

Y/1 Y" X " T" X /F T = X + =-A; =- Y - A

X

Y

where A and a are separation constants. From the boundary conditions in (7.26), we obtain

X" + µX = 0; X (O) = X(7r) = 0. Y" + (A - y)Y = O; Y(0) = Y (7r) = 0. T"+AT=O. As before, we get µ = µn = n2, X (x) = Xn( x) = sin nx, n=1,2... ; A- µ = m2, Y(y) = Ym(y) =sinmy, m=1,2... ;

and T(t) =Tnm(t )

= awn cos gnmt+

with arbitrary constants

anm

bn msin gnmt,

gnm =

Vm2 + n2

and bn,,,,.

We seek a solution of (7.26) in the form of 00 u(x, y, t) = E (anm cos qnmt + bnm sin qnmt) sin nx sin my, (7.27) n,m=1

where the coefficients an,, and bnm are determined by the initial conditions. Then we get anm

4 " 7r / f (x, y) sin nx sinmy dxdy, = 2 o o

J

bnm = 4

J

2 gnm7r 0

I g(x, y) sin nx sin my dxdy. (7.28) o

(7.27) and (7.28) give a formal solution of (7.26). Verification: In order to obtain uniform convergence of the series (7.27) and its first and second partial derivatives, we assume that f (x, y) (continued as an odd

Bessel

functions

205

function) is in C 4 and g(x,y) (continued as an odd function) is in C 3 . The approach is similar to the proof of uniform convergence of double Fourier series of f(x, y) in Section 7.2 and the details are omitted here. 7.5

Bessel functions

Consider the problem Utt -
ut(r,9,0)

Let u(r, 6, t) = A(r)B(0)C(t). C"

A"

A'

= g(r,6),

T < 1, t > 0, u(l,B,t)=0.

(7.29)

Then, from the wave equation, we have

B"

x

r2A"

rA'

,

2

B"

where A and (i are separation constants. From the boundary condition u(l,9,t) = 0, and the continuity of u in r < 1, we arrive at the following problems: B" + nB = 0,

-n<6
B(-n)

= B(jr),

(rA1)' + (Ar - ±)A = 0, 0 < r < 1; r where yl(r) and A'(r) are bounded at r = 0.

B'(-TT)

.4(1) = 0,

C" + c2XC = 0.

=

B'(n). (7.30) (7.31)

(7.32)

We have shown that the eigenvalues of (7.30) are fi = m 2 , m = 0,1,2... and the corresponding eigenfunctions are cosm# and sinm#. Set p = m2 in (7.31). We have (rA')' + (Ar

m2 )A = 0, r

r > 0.

(7.33)

For A > 0, let t = y/Xr and A(r) = v(t). Equation (7.33) is changed to t2v" + tv' + (t2 - m2)v = 0,

t > 0.

(7.34)

Equation (7.34) is called Bessel's equation of order m and its solutions are called Bessel's functions of order m. In order to solve the boundary value

Problems in Higher Dimensions

206

problem (7.31), we would like to study the properties of the solutions of Bessel's equation. We seek a solution of (7.33), which is regular at t = 0, by the method of power series (the Frobenius method) 00 v(t) = to > antn, a0 L 0 n=0

where a and an are to be determined. We differentiate v(t) term by term to obtain v' and v" and substitute the series v, v' and v" into (7.34). Then we get 00

00

to [E an {(n + a)2 - m2 }tn + an_2tn] = 0. n=0

n=2

Hence, we have the recurrence relation for {an}: ao(a2 - m2) = 0, a1[(1 + a)2 - m2] = 0, an [(n + a)2 - m2] + an-2 = 0, n > 2. We choose a = m and ao = 2-m/m!. Then a1 = 0

and an = -an_2/[n(2m + n)]. This implies that (-1)k a2k+1 = 0, a2k =

22k+m(1n + k)!k!

k=0,1,2...

Thus, we have

()k(+k)k!

v(t)=

Jm(t)

_E(

2k (7.35)

k=0

The power series in (7.35) is absolutely convergent for all t by the ratio test. It therefore represents an analytic function and can be differentiated with respect to t any number of times. J,n (t) is called the Bessel function of the first kind of order m. We note that Jo(0) = 1 and J,n(0) = 0 for m > 1. There exists another independent solution of (7.34). This particular solution Yrn (t) is called the Bessel function of the second kind of order m and Ym(t) -> oo as t -> 0. Then the general solution of (7.34) is v(t) = c1Jm(t) + c2Ym(t)

Bessel functions

207

where c1 and c2 are arbitrary constants . (For a brief description of Y7i(t), please refer to Tolstov, [11, 203-5].) There are two recurrence formulas for Jm(t), namely, [t-m'Jm(t)] = -t-mJm+1(t),

(7.36)

and d [tmJm(t)] =

t''Jm -1(t)•

(7.37)

These formulas can be obtained by multiplying (7.35) by t-m and tm respectively and differentiating the resulting series. Let v(t) = t-1/2w( t) and substitute it in (7.34). We have t2w" + (t2 - m2 + 4 ) w = 0.

(7.38)

Hence, w(t) = t1/2Jm(t) is a solution of (7.38). In particular, w(t) = t1/2Jo( t) satisfies w"+(1+4t2)w=0.

(7.39)

We would like to show that w(t), or Jo(t), has a sequence of positive zeros, {aok}, k = 1, 2, ..., such that aok -> oo as k --4 oo. Let p(t) = sin t be a solution of p" +p = 0, (7.40) which has an infinite number of zeros at t= n7r, n = 1, 2,3... By the separation theorem (Theorem 3.32), w(t) has at least one zero in [n-7r, (n + 1) 7r], n = 1, 2... Moreover, by Theorem 3.6, w(t) (or Jo(t)) has at most a finite number of simple zeros in any finite interval and t = 0 is an isolated zero of Jo(t). Hence, the positive zeros can be arranged as an increasing sequence of numbers {aok } such that aok --> oo as k -4 oo. Theorem 7.3

For each m, m=0,1,2....., the set of positive zeros of J,n(t)

consists of a sequence of numbers {a,,,,k}, k = 1, 2..., such that a,,,,k -+ 00 ask->o0.

Proof:

Problems in Higher Dimensions

208

Suppose J,,,,(a) = 0 and Jm(b) = 0 for 0 < a < b. Then t-mJm(t) vanishes at a and b. By Rolle's theorem, its derivative vanishes at at least one value of tin (a, b). By (7.36), there exists at least one zero of Jm+1(t) between two positive zeros of Jm (t). Also Jm+1 (t) can have at most a finite number of zeros in [a, b] and t = 0 is an isolated zero. Since the positive zeros of Jo (t) constitute an unbounded sequence of numbers, it follows that the zeros of J1 (t) also form such a set. By induction, there exists a sequence of positive zeros {amk} of Jm(t) such that amk -> oo as k -> oo. Next, we would like to obtain a generating function and an integral representation for the Bessel function J,,,(t). Theorem 7.4

For real t and complex z such that 0 < IzI < 00,

00 exp[2 (z - z)] = Jn(t)zn + E(- 1)"Jn(t) z-n. (7.41) -n.=0 n=1

Proof: For I zI < oo, 00

tz

tkzk

exp[2]= Ek2k

(7.42)

k=0

while for IzI > 0, t = (-1)mtm exp[-2z]

m=0

(7.43)

m!2mzm

Both series converge absolutely in 0 < IzI < oc. Then 00 t

1

tkzk

00 (

-1)"'`t'"''

00

exp[2(z - z )] = (1: k!2k)(^ m^2mzm) = k=0 m=0 k,m=0

(-1)

mtk +mzk -m

k!m!2k+m

For the terms involving zn, n = 0, 1 , 2..., let k = m+n and the coefficient of zn is given by E,°n=o(- 1)mtn+2m/[(n+m)!m!2^`+2 m] = Jn(t). For the terms involving z-n, n = 1, 2..., let m = k + n and the coefficient of z-n is given 00 by Ek o(- 1 )n +ktn+Zk/[ k!(n+k)!2n +Zk] = (- 1)'J n(t). The rearrangement of the double sum is valid because of the absolute convergence of the series. Hence, (7.41) holds.

Singular Sturm-Liouville problem for Bessel 's equation

209

Theorem 7.5 in (t) = 1 cos(t sin 0 - nO)dO . (7.44) 7r 0"

1

Proof: For a fixed t, we treat (7.41) as a Laurent series expansion for f (t, z) = exp[t(z - 1/z)/2]. Hence, 1 f f (t, z)dz

_ A (t)

27ri C zn

1

where C : Iz I = 1. Let z = exp(iO). Then Jn(t) =

f ir exp[ t (eiO - e-'B ) - inO]dO = 2^ exp[it sin 0 - inO] dO 2,7r r n

J

[cos (tsinO - nO)+isin(tsinO-nO)]dO = If 7r cos( tsin0-nO)dO. =1 2-7r r 7r o

J

Remark : From (7.44), it is easy to obtain I d^d7k(t) (< 1 (k = 1, 2, ...).

Jn(t)I 1,

(7.45)

7.6 Singular Sturm-Liouville problem for Bessel ' s equation We now return to the problem we first considered in Section 7.5, i.e, a L[u] = (xu)' 'm - u = -Axu, 0 < x < 1 X

(7.46)

with boundary conditions u(1) = 0; u and u' are bounded at x = 0.

(7.47)

For any twice continuously differentiable functions w and v satisfying the boundary conditions (7.47), we find, by integrating by parts, 1

J

0

vL[w]dx = -

J 1 [xv'w' + m2vw]dx (7.48) 0

x

Problems in Higher Dimensions

210

and (7.49)

/ 1 vL[w]dx = fo 1 wL[v]dx.

f0

With (7.48) and (7.49), we can establish the following properties of eigenvalues and eigenfunctions of (7.46)-(7.47): (A). all eigenvalues are positive; (B). eigenfunctions corresponding to distinct eigenvalues with respect to the weight function x are orthogonal on [0,1]. The proofs of the above statements are similar to those of the regular case in Chapter 3. Let t = /x in ( 7.46). We have t2u" + tu' + (t2 - m2)u = 0, 0 < t < V/A.

(7.50)

Its general solution is given by (7.51)

u(t) = ciJm (t) + c2Ym (t),

where cl and c2 are arbitrary constants. For boundedness at t = 0, we set c2 = 0; for u(V'A-) = 0, we get = amk, where { a,,,,k} is the sequence of positive zeros of Jm(t). Hence, {Ak = amk} is a sequence of eigenvalues for (7 . 46)-(7.47) with corresponding eigenfunctions {uk(x) = Jm ( amkx)}, k = 1, 2... Accordingly, we have 1 xJm(amkx)Jm(amjx)dx = 0

for j # k.

(7.52)

10 Multiplying (7.46) by 2xu', we get [(xu )2]' + (Ax2 - m2 )[u2]' = 0.

We integrate the above equation from 0 to 1 and using integration by parts for the second term, we have f1

2A

J0

xu2dx = ^(xu')2 + (axe - m2)u2] lo.

Let u(x) = Jm(amkx) and A = amk. Since Jm(amk) = 0, J0(0) = 1, and JR,(0) = 0 for m > 1, we have

f J0

1 xJr2,,(amkx)dx = [Jm (amk)]2/2 . (7.53)

Singular Sturm-Liouville problem for Bessel's equation

211

We have shown in Section 6.9 that the Green 's function G,,, (x, t;) for the nonhomogeneous equation L[u] = -F(x), 0 < x < 1

(7.54)

with boundary conditions (7.47) is given by { Gm(x

( x_m - xm S) - 1 xm

)I (2m ),

S j x;

l ( )/( ), ^ x; for all positive integers m or is given by log(1/x), ^ < x; log(' /0, ^ > x;

Go(x,^) =

for m = 0. Then the unique solution of (7.54) and (7.47) is

u(x) = f Gm(x, y)F(y)dy• 1 0

Set F(y) = Ayu(y). We have u(x) = )

f fl

Gm(x, y)yu(y)dy.

0

(7.55)

Let O (x) = V1x_ u(x) and K (x, y) = Gm (x, y) V/y-. Then we have O(x) = A

0

K (x,y)o(y)dy•

(7.56)

From (6.82), (6.83), and ( 6.84), we have ff ff K2(x, y ) dxdy < oo. Moreover , K(x, y) is symmetric in [0, 1 ] x [0, 1]. It is shown in Stakgold [10, 218220] that all the eigenfunctions of the integral operator with symmetric, square integrable kernel are complete and orthogonal on [0,1]. This implies that the eigenfunctions of (7.46 )-( 7.47) form a complete set in [0,1] with respect to the weight function p(x) = x or that {VfXJ,,,,(a.,,,,kx)}, k = 1, 2... is complete with respect to the weight function 1 on [0,1]. For a fixed m, setting A = a,2nk, k = 1, 2,... and u(x) = Jm(amkx) in (7.55), we have 2 Jm(amkx) = am,k Gm(x,Y)Jm(amky)YdY. 0

Problems in Higher Dimensions

212

Then the Fourier series expansion for G,,, (x, y) with respect to the system {Jm(amkx)} is 00 Jm(amkx)Jm(amky) 2 I k=1 amkflmk

where 01

#mk

J

Jr2ii( amkx)xCL^x.

0

By the completeness relation, 1

00 2 Jmamkx) Gm (x,y)ydy =

0

k=1

(7.57)

amkAmk

7.7 The two- dimensional wave equation in a circular domain We now return to problem (7.29), that is, 1 1 2 Utt - C [urr + Ur + 1 U00] = 0,

r<1, t > 0

r r2

u(r, 0, 0) = f (r, 0), ut(r, 0, 0) = g(r, 0), u(1, 0, t) = 0.

(7.58)

We first assume that the initial data is independent of 0 and the solution u depends only on r and t. Then Utt - C2 (urr + 1 Ur) = 0, r < 1 r

u(r, 0) = f (r), ut(r, 0) = g(r), u ( 1, t) = 0.

(7.59)

Let u (r, t) = A(r) B(t). Then , from the wave equation in (7.59 ), we get

A" + r-1A' T" A c2T A where A is a separation constant. As a result, we get A"+1A'+AA= 0, 0
T"+c2AT= 0, t>0.

(7.61)

The two-dimensional wave equation in a circular domain

213

Equation (7.60) is the parametric form of Bessel's equation of order 0 and has the general solution given by A(r) = c1Jo (/r) + c2Yo(VfAr). For continuity at r = 0, we set c2 = 0. Also u ( 1, t) = 0 implies that 0 = A(1) = c1Jo (v/"A-). Then A = Ak = µk, k = 1, 2... where {µk} is the sequence of positive zeros of Jo(t). Set A = µ2 in ( 7.61). We have Tk(t) = ak cos cykt + bk sin cµkt.

We form a series solution of the form u(r, t) = 00 E( ak cos cµkt + bk sin cµkt) Jo(µkr ).

( 7.62)

k=1

From the initial conditions in (7.59), we have 00

f (r) = u(r, 0 ) = > akJo(/Akr) k=1

and bkcµkA(µkr)• g(r) = ut (r, 0) = 0" k=1

From the orthogonal relation for {Jo(µkr)}, we have bk - .fo rg(r)Jo(µkr)dr ak ,fo rf(r)Jo(ukr)dr 7.63 ) .fo rJo (µkr)dr cµk fo rJo (µkr)dr ( (7.62) and (7.63) give a formal solution of (7.59). We now return to the general case (7.58). For simplicity, we assume that g(r, 0) = 0. Let u(r, 0, t) = A(r)B(0)C(t). As shown in Section 7.5, we arrive at the following problems

B" + µB = 0,

(I).

-ir

< 0 < ir, B(-ir) = B(ir), B'(-7r) = B'(ir).

Then its solutions are

Bm ( 0) where

cm

= cm cos

mO

+

dm sin m 0,

and dm are constants.

m = 0, 1, 2....;

µ = m2,

214

Problems in Higher

(II). {rA'), + (Xr-m2r-1)A = 0, with both A and A' bounded at r = 0.

Dimensions

A{1) = 0,

Then, its solutions are Amk(r) = Jm(amkr),

A: = 1,2

;

where {amfc} is the sequence of positive zeros of C" + c2amkC

(III).

Then its solutions are

= 0,

\/A = amk Jm(t).

C"(0) = 0.

Tmk(t) = cos camkt.

Prom (I), (II) and (III), we form a series solution 1 °° u(r,9,t) = ^ ^ cot Jo(Qofcr) cos(caokt) k= l oo

+ ^2 (cmkcosmd + drnksinmd)Jrn(amkr)cos(carnkt).

(7.64)

m,fc=l

From the initial condition u(r, 9,0) = f{r,9), 1

oo

we have

oo

/(*", 0) = - ^2 cokJoiaokr) + ^ fc = l

{cmk cosm0 + dmk sinm0) J m (a m fcr).

m,k=l

Prom the orthogonal properties of {Jm(amkr)} [0,1] and [—7r, 7r] respectively, we get cmk = —^— /

nPmk Jo

dmk = —5— /

TTPm/c Jo

/

f{r,9) cosm.9

Jm{amkr)rdrd9,

f (r, 9) sinm9

Jm(amkr)rdrd9,

J-x

/

J--n

(7.65) and {cosm9,sinm#} on

Pmk= f Jl(amkr)rdr. Jo (7.64) and (7.66) give a formal solution of (7.58) with g(r, 9) = 0. Verification:

(7.66)

The two-dimensional wave equation in a circular domain

215

We assume that (1). f (r, 0) is twice continuously differentiable and f ( 1, 0) = 0. ( 2). f (r, 0 ) is periodic of period 27r in 0. (3). f0 f 7,, F2(r, 9) rdrd9 < oo, where F(r, 0) = frr + r-lfr + r 2f0e• Let Amk and Bmk be the Fourier coefficients of F(r, 0) with respect to the system Jm(amkr) sinm9},

H = {Jm(amkr) cosm9,

m =0,1,2.... k = 1, 2,...

which is orthogonal with respect to the weight function r on [0, 1] x [-7r, in, i.e.

f

1 //^^

Amk =

lr

J

lnNmk

f

Bmk =

l

/^1

l r

/

lnNmk

F(r, 0) cos m9 Jm(amkr) rdrd0, 7r

F (r, 0) sin m9 Jm(amkr)rdrd9. 7r

We note, by integrating by parts, that

f [fr, + rr]Jm(amkr)rdr = [frJm (amkr)r]o - pl fr[ a [Jm(amkr)r]-Jm(a mkr)]dr

ff i frr 5r [Jm( amkr )]dr Jo -[f (r,

9)r8r

[Jm (a mk r)]] Io

+ fo

f ( r, 0)

ar [rarJm(amkr )]dr

m2 = f f (r, 0) 0

fee I7r 7r

(7.67)

- r c nkr]Jm(amkr)dr.

cos mOdO = [fo cos m9]" .r + m f 7r fe sin mOdO

= MI f (r, 0) sin m9]" ,r-m2

✓ -7r

ir f (r, 0) cos mOd9.

J 7r f (r, 0) cos mOdO = -m2 J n

(7.68)

Problems in Higher Dimensions

216

Thus, from (7.67) and (7.68), we have (

2 Amk = -a mkCmk•

7.69

)

Similarly, we get Bmk = -amkdmk. (7.70) Thus, the Fourier series of F(r, 0) with respect to the system is

00

00

- - ) CokaokJo(aokr) - Y a^,,k( c,,.,,k cos mB + dmk sin mO)Jm(amkr)• k=1

m,k=1

Since the system {cos mO, sin mO, m = 0, 1, 2... } is complete with the weight function 1 on [-ir, 7r] and, for each fixed m, the system {Jm(amkr), k = 1,2 ... I is complete with respect to the weight function r on [0,1], hence, by Theorem 7.1, the system H is complete with respect to the weight function r on the unit circle. Thus, by the completeness relation, we have

00 F2( r, B)rdrd6 = 0

2

k=1

00

Cokaok/3o k + It T, (C,2,ik E m,k=1

+ d,2^,k)a4,,,,k/ mk• (7.71 )

Let SMK (u, r, 0) be the partial sum of the series (7.64): K SMK(r, 0, t) = 2 E cokJo( aokr ) cos(caokt) k=1 M

K

+ E E(Cmk cos mO + dmk sin m0)Jm(amkr) cos(camkt). m=1 k=1

For M2 > M1, K2 > K1, we apply Schwarz's inequality to the sum G(r, 0, t ) = SM2K2 (r, 0, t) - SM.K1 (r, 8, t), and find that J 1 K2 M2 K2 G2 (r, 0, t ) C 12 E

2 4 4 C0ka0 k/30k + (Cmk + dmk)amk,3mk }

k=K1+1 m=M1 + 1 k=K1+1

M x 1 K2 2 k= 1

Jo (aokr )

2 K2 Jm4(amkr) 7.72

aO kOk 7/1=M1+1 k = K1+1

amkQmk } ( )

Initial-boundary

value problems for the heat equation

217

Since the series (7.71) converges, for t > 0, there exist positive integers K and M such for M2 > Mi > M, and K2 > K\> K, the first factor on the right of (7.72) can be made less than e. We would like to show that the second factor on the right of (7.72) is bounded. From (7.57), we have

l±JJM+

£ 4 W ) * rGl{r,p)pdp+± f Gll{r,p)pdp.

(7.73) From the explicit computation of the JQ G^n(r,P)pdp in (6.82)-(6.84), we note that the series (7.73) converges uniformly and absolutely in [0,1]. Hence, the second factor of (7.72) is bounded, the series for u(r, 6, t) converges uniformly to a continuous function, and u(r, 9,0) = f(r,0). Let f{r,9) be in C4 and let / and frr + r _ 1 fr vanish at r = l . Then, using similar arguments, we can show that u is in C 2 and is a solution of the problem (7.58) with g(r, 9) = 0. 7.8

Initial-boundary value problems for the heat equation

Let D be a bounded domain in the iT/-plane with continuous, piecewise boundary curve C. Consider the initial-boundary value problem for the two-dimensional heat equation: ut - k(uxx + uyy) = 0, du u = 0 or ^ - = 0 , an u(x,y,0)

= f(x,y),

(x,y) € D, t > 0, (x,y)eC,

t>0,

(i,y) 6 D.

(7.74)

A solution of (7.74) is continuously differentiable for (x, y) in D and t > 0 and is twice continuously differentiable for (x, y) in D and t > 0. We would like to show that the above problem has at most one solution. Let u(x,y,t) and v(x,y,t) be two solutions of (7.74). Define w{x, y,t) = u(x,y,t) — v(x,y,t). Then w(x,y,t) satisfies wt - k(wxx + wyy) = 0,

(x, y) e D,

t > 0,

Problems in Higher

218

a, = 0 or - ^ = 0 , an w(x,y,0)

Dimensions

(x, y) € C, * > 0,

= 0,

(x,y)eD.

(7.75)

Define I(t) = ^ffw2(x,

y, t)dxdy.

(7.76)

Then I(t) is continuous for t > 0 and, from the initial condition in (7.75), 7(0) = 0. Now, •^'(0

=

I

wwtdxdy = k I I w(wxx +

wyy)dxdy.

Prom Green's first identity (6.8), we have / / w(wxx + wyy)dxdy

= / w—ds-

/ / {wx + w^)dxdy.

But w = 0 o r dw/dn = 0 for (x,y) on C and t > 0. Hence, I'(t) < 0. By the mean value theorem, I(t) = 7(0) + t / ' ( r ) where 0 < r < «. But 7(0) = 0 and /'(«) < 0. Thus, I(t) < 0. By (7.76), I(t) > 0. Therefore, I(t) = 0 for t > 0. FVom (7.76) again, w(x,y,t) = 0 or u(x,y,t) = v(x,y,t) for (x,y) in D and t > 0 by continuity. We now use the method of separation of variables to obtain a series solution for the above problem for both a rectangular domain and a circular domain respectively. Consider ut - k(uxx + uyy) = 0 , u=0 u{x,y,0) Let u(x,y,t) have

for = f(x,y),

= X(x)Y(y)T(t).

0 < x < a, 0 < y < b, t > 0, a? = 0, x = a, y = 0, y = b, 0 < x < a; 0
(7.77)

Then substituting u into the equation, we

Initial-boundary value problems for the heat equation

219

where A and n are separation constants. Then we obtain X" + nX = 0;

X(0) = X(a) = 0.

Y" + (A - fi)Y = 0;

Y(0) = Y(b) = 0.

T + kXT = 0. As before, we get fi — nn = n2TT2/a2, X- /i = m2n2/b2,

X(x) = Xn(x) = s'm(mrx/a), n = 1,2,...; Y(y) = Ym(y) = sin(m7n//6), m = 1,2...;

and qnm = k(n27r2/a2 + m27r2/b2).

T(t) = Tnm(t) = exp(-qnmt), We form a series solution of the form oo

u(x,y,t)

= 2_] -^nmexp(— qnmt)sinnx

sin my

(7.78)

n,m=l

where the coefficients {Anm} are determined by the initial condition. Then we get 4

Anm

f

= -r /

I

717TX

/ f{x, y) sin

77X1TU

sin —^dxdy. a

ab J0 J0

(7.79)

b

(7.78) and (7.79) give a formal solution of (7.77). Remark: The maximum principle for the heat equation in higher dimensions can be proved by using a similar approach as in Theorem 5.1. Hence, if we assume that the double sine series of f(x,y) converges uniformly to / ( x , y) in [0, a) x [0, b], then we can verify that u(x, y, t) in (7.78)(7.79) is a solution of (7.77). The approach is similar to the corresponding problem in Section 5.3, and the details for verification are omitted here. Consider ut - \urr + -ur + -zuge] = 0, r r2 u(l,6,t)

= 0,

r < 1, t > 0,

u(r,0,O) = /(r,0).

(7.80)

220 Problems in Higher Dimensions

Let u(r, 0, t) = A(r)B(O)C(t). Then, from the heat equation , we have Cl _ A" A' B" r2A" rA' B" C A +rA+r2B =-a, A+ A +Ar2=- B= a where A and µ are separation constants. From the boundary condition u(1, 0, t ) = 0 and the continuity of u in r < 1, we have the following problems: B" + µB = 0 , -ir < 0 < ir;

B(-7r) = B(ir), B'(-ir) = B'(ir).

(rA')'+(Ar- r)A=0,

0
where A(r) and A'(r) are bounded at r = 1, and C'+AC=O, t>0. As demonstrated in Section 7.7, we have B(B) = Bm(0) = cos m0 or sinmO;

µ = pm = m2, m = 0, 1, 2, ...

A(r) = Amk(r) = Jm(amkr), C(t) = Cmk( t) = exp(-cemkt ),

k = 1, 2...,

where A = Am,k = a^,,k, and {amk } is the sequence of positive zeros of Jm(t). Let 00 u(r, 0, t ) = 2 E cokJo( aokr ) exp(-ao kt) k=1 +

00 E (Cmk cos mO + dmk sin mO)Jm(amkr) exp(-a2mkt ).

( 7.81)

m,k=1

The coefficients {cmk} and {dmk } are determined from the initial condition and given by 1 1 ^r Cm,k

=

f

itI3mk 0

(r, 0) cos mOJm(amkr)rdrd0, - 7r

l 1f 7r dmk = It/3mk

J n f (r, 0) sin m0Jm (amkr ) rdrd0,

221

Legendre 's equation

Q,,,k

I

(7.82)

J,n ( a,,,kr)rdr.

(7.81) and (7.82) give a formal solution of (7.80). To verify that u(r, 0, t) is a solution of (7.80), we assume that the function f (r, 0) has the same properties as in the case of the wave equation in Section 7.7. The verification is the same as in Section 7.7 and is therefore omitted here.

7.9 Legendre 's equation Consider Laplace's equation in space: 0U = 26xx + uyy + uzz = 0.

We would like to transform Du in the spherical coordinates: x = r sin 0 cos o, y = r sin 0 sin o, z = r cos 0; Let x = p cos 0,

r > 0, 0 < 0 < 7r, 0 < 27r.

y = p sin 0, and z = z. From Section 6.5, we have Du=uPP+ lUP+ 2UOO+uzz =0.

P

Next, we set z = r cos 0,

P

p = r sin 0, and _ q. Then we have 1 1

uzz + uPP = 9drr + r Ur + 72- U00-

But Up

= urrp + Uo0p = sin 0 ur + cos r

0 ue

and pup + p2 u,q = rur + c^20ue + r2

si

2 0uOO

Then Laplace's equation in spherical coordinates is 2ur (sin 0 ue)B uOO _ urr + r + r2 sin 0 + (r sin 0)2 0. (7.83) We first assume that u is independent of 0, i.e. u = u(r, 0). Then (7.83) is reduced to 2Ur (sin 0 uo)e = 0 0. (7.84) Urr + r + r2 sin

Problems in Higher Dimensions

222

Let u(r, 0) = A(r)B(O) and subsitute u in (7.84). Then we get the following equations

r2A" + 2rA' - AA = 0, r > 0,

(7.85)

(sin 0B ')' + A sin OB = 0, 0 < 0 < ir,

(7.86)

and

where .\ is a separation constant. Let t = cos 0 in (7.86) and B(0) = P(t). Then (7.86) becomes [(1-t2)P']'+AP=0, -1
Substituting P(t) into (7.87) and collecting the terms of the same power, we get 00 E [Ck k =0

+2(k + 2)(k +

1) - ckk(k + 1) + ACk] tk = 0.

Hence, for k = 0, 1, 2...

Ck+2 = [k(k + 1) - A]ck (k + 2)(k + 1)

(7.88)

which is the recurrence relation for the coefficients {ck}. Setting cl = 0, we get P(t) _ Ek o c2kt2k and setting co = 0, we get Q(t) _ Ek o C2k+lt2k+1. We note that P(t) and Q(t) are two independent solutions of (7.87). For any real A, both series for P(t) and Q(t) converge absolutely in Itl < 1 by the ratio test. However, it can be shown that as t -+ ±1, both IP(t)I and IQ(t)l will diverge to oo. (See Sagan [9, 190-192] for the proof.) Now let A = n(n+1), where n is a nonnegative integer. Then Cn+2j = 0 for all positive integers j. Thus, when n is even, P(t) is a polynomial of

Legendre's equation

223

degree n and Q{t) is a series which diverges at t = ±1; whereas, when n is odd, Q(t) is a polynomial of degree n and P(t) is a series diverging at t = ± 1 . Hence, the only solutions of (7.87) which remain finite at t = ±1 are the polynomial solutions for A = n(n + 1) when n is a nonnegative integer. Put A = n(n + 1) in (7.88). We have {k + 2)(k + l)ck+2 (n-k)(n + k+l)'

°k

Using this formula successively, we express Cn-2j in terms of c„, i.e. (-l)>n(n-I)....(n-2j+l)c Vj\{2n - l)(2n - 3)....(2n - 2j + 1)'

Cn 2]

~

To simplify our expression, we set c„ = (2rc)!/[2"(n!)2]. Then Cn-2j

(-l)>(2n-2j)! 2"j!(n-2j)!(n-j)!

and P

"

( < )

-2^^

jKn-ajJKn-j)!

'

(7 89)

"

where m = n/2 if n is even and m = (n — l)/2 if n is odd. Since dr_2n-2j eft"

Pn{t)=

=

(2n - 2j)\ t"- 2 ^ (n - 2j)\

1 d" ™ ( - l ) J n ! t2n~2j 2 ^ ! d T ^ j!(n-j)! ^

To simplify the above expression, we note, by binomial expansion, that

where Qm(t) is a polynomial of degree less than n. Hence, the sum in Pn(t) can be extended from j = m to j = n since the nth derivative of Qm(t) is

224 Problems in Higher Dimensions

zero. It follows that n

Pn (t) = 1 d [( t2 - 1)n ], 2nn! dtn

n = 0,1,2... (7.90)

which is Rodrigues' formula for the Legendre polynomials.

7.10 Properties of Legendre polynomials Consider the singular Sturm-Liouville problem for the Legendre's equation L[u] = [(1 - t2)u']' = -Au, -1 < t < 1, (7.91) where u and u' are bounded at t = ±1. For any C2-functions v and w satisfying the boundary conditions at t = ±1, we have, by integration by parts, vL[w]dt = - 1(1 J 11 J i

- t2)v'w'dt (7.92)

and vL[w]dt = J 1 wL[v]dt. f11 i

(7.93)

As in the case of Bessel 's equation, the eigenvalues of (7.91 ) are nonnegative and the eigenfunctions corresponding to distinct eigenvalues are orthogonal on [-1,1 ] with respect to the weight function 1. We have shown in Section 7.9 that the eigenvalues are A = An = n(n + 1), n = 0, 1 , 2.... and the corresponding eigenfunctions are the Legendre polynomials n

Pn(t) - 1

2nn! dtn

[(t2

-

1)n].

(7.94)

Hence, we have

J 11 Pn(t)P,n(t)dt = 0, n m.

(7.95)

For m = n, we have

L

P,2(t)dt = i

22n(n !)2

i f[( 1 - t2)n]2dt = 222( )x)2

1 1( 1 - t2)ndt

Properties of Legendre polynomials

225

after n integrations by parts. But rl

J

(1 - t2 ) ndt =

J 1 (1 - t)n(1 + t )n'dt = (2 ^i 1 ( 1 + t)2"dt 1

1

after another n integrations by parts. Hence, after simplification, we get pl

J Theorem 7.6

P,,2(t)dt = 2

1 2n+1

(7.96)

For n > 1, Pn+1(t) - Pn-1(t) = ( 2n + 1)Pn (t). (7.97)

Proof: 1 Pn

+1(t)

=

dn+1

2n +1 (n + 1)! dtn +1

- 1)n] [2t(n + 1)(t2

n t2 - 1)n-1] = 1 d [(t2 - 1)n + 2nt2 ( 2nn! dtn

dn = Pn(t)+2n - 1(n - 1)! dtn

[( t2-1)n+(t2-1)n-1] = ( 2n+1)Pn (t)+Pn_1(t).

Theorem 7.7

Pn(1) = 1;

Pn(-1) = (-1)n.

(7.98)

Proof: dtn (t2 - 1)n = (t - 1)n n! + nit (t - 1)n dtn 1 (t + 1)n +... + n!(t + 1)n. Hence, (7.98) follows. Theorem 7.8

Pn(t) has n real simple zeros in the interval (-1,1).

Proof: For n > 1, f 11 Pn(t)dt = 0. This implies that Pn(t) changes sign at least once in (-1,1). Suppose Pn(t) has k simple zeros t1, t2i ..., tk in (-1,1).

226

Problems in Higher Dimensions

By contradiction, let k < n. Then the polynomial Q(t) = (t - t1)....(t - tk) is of degree k. We have

f1 Pn(t)o,(t)dt = 0. However, Pn(t)a(t) cannot change sign in (-1,1) since it has only zeros of even multiplicity. Thus, f 11 PP (t)o(t)dt # 0, which is a contradiction. Hence, k > n. But Pn (t) cannot have more than n zeros. Thus, k = n. Theorem 7 . 9 1,

For any t E [-1, 1], we have for any complex z with IzI < 00

1 (1 - 2tz + z2)1/2 =

Pf(t)zn.

(7.99)

n=0

Proof: Let w(t, z) = (1 - 2tz + z2)-1/2. For a fixed tin [-1,1], the roots of 1 - 2tz + z2 = 0 are z = t ± i(1 - t2)1/2. Thus, I zI = 1. Then w(t, z), as a function of z, is analytic in the disc IzI < 1. Hence, it can be expanded as a power series in IzI < 1, i.e.

w(t, z) = E Qn(t)zn n=0

where 1 - 12tz + z2 )1/2z1dz ( 7.100) Qn(t) = 211( evaluated along any closed contour C around the point z = 0 and inside the disc IzI < 1. Let 1 - uz = (1 - 2tz + z2 )1/2. Then z = 2(u - t )/( u2 - 1) and ( 7.100) is transformed into

where C' is the closed contour surrounding the point u = t. Using Cauchy's formula for derivatives of an analytic function, f (n) (t) 1 f f (u) du , n! 27ri C, (u - t)n+l '

Properties of Legendre polynomials

227

we get

Qn(t)

_ 1 dn Pn(t) (u2 - 1)nl=t = n! 2 n C

Theorem 7.10 IPn(t)I <1 for - 1 < t < 1.

(7.102)

Proof: Let t = ,cos 0 in (7.99). Then 00 (1 - 2z cos

0 + z2)-1/2

= 57, Pn(cos 0)zn. n=0

(7.103)

On the other hand, (1 - 2z cos O +

z2)-1/2 = (1 - zeze)-1/2(1 - ze-ze)-1/2 00 00

_ {Eaje'jezj}{T ake-ikezk} j=0 k=O

(7.104)

where all aj = (-1/2)(-1/2 -1)...(-1/2 - j + 1)(-1)j/j! > 0. Comparing (7.103) and (7.104), we have ajake'(j-')O .

Pn(cos 0) _ j+k=n

Thus, JPn(cos0)j <

T, ajak = Pn(1) = 1. j+k=n

Theorem 7.11 lPP(t)l < n2 (7.105) P.(t)I
(7.106)

Problems in Higher Dimensions

228

and

Pen+l(t) = (4n + 1)P2n (t) + (4n - 3 )P2n_2(t) + .... + PO(t)Thus, IP2n(t) I < (4n - 1) + (4n - 5 ) +.... + 3 = n(4n - 1 + 3)/2 < (2n)2, and (4n + 1) + (4n - 3) +.... +1 (4n + 2)(n + 1)/2

(2n

Hence, (7.105) is proved. (7.106) can be proved similarly. Before we prove the next theorem, we first need the following theorem: Let f (x) be a real continuous Weierstrass Approximation Theorem function on a finite interval [a, b]. Then , for e > 0, there exists a polynomial P(x) such that I f (x) - P(x ) I < e for all x in [a, b] uniformly. (For its proof, please refer to Courant -Hilbert, [5 , 65-68]). Theorem 7 . 12 The set of the Legendre polynomials {Pn(t)} is a complete orthogonal set with respect to the weight function 1 on [ 1, 1].

Proof: Let f be a square integrable function in [-1,1]. Let e > 0. Then there is a continuous function g in [-1,1] such that

By the Weierstrass approximation theorem, there exists a polynomial P such that [g - P]2dt < e/2. I But any polynomial can be written as a linear combination of the Legendre polynomials, i.e.

P(t) =

EakPk(t). k=0

229

Legendre series and boundary value problems

This implies that f[f(t) - > C'kPk( t)]2dt < E. 1 k=0

Let {Ak} be the Fourier coefficients of f (t) with respect to {Pk(t)} on [-1,1], Ak = 2 n+I f 11 f (t)Pk(t)dt. Then we have

that is,

[f (t) - > AkPk(t)]2dt < fi

k=0

f

1 [f(t) - E akPk(t)]2dt < E.

1

k=0

Hence, {Pn(t)} is complete on [-1,1].

7.11 Legendre series and boundary value problems For any integrable function f on [-1, 1], the Fourier series of f with respect to {PP(t)} is 00 (7.107)

f(t)Pn(t)dt.

E AnPn(t) An = 2n2

1f1

n=0

(7.107) is called the Legendre series of f . Theorem 7.13 Let f E C2[-l, 1]. verges to f uniformly on [-1,1].

Then the Legendre series of f con-

Proof: By (7.97), (7.98), and repeated integrations by parts, we have, for n > 2, 2An = (2n + 1)

f(t) Pn(t)dt = f f(t)[ Pn +1(t ) -

fl i

=

f(t)[Pn +1( t)

11

- Pn-1 (t)]I l 1

1

J

Pn- 1(t)]dt

f ( t )[ Pn+l(t) - Pn-1(t)]dt

_ - f l f,(t) [ Pn+2(t) - P,n(t) - Pn(t) - Pn-2(t) ]dt 1 2n+3 2n-1

l

f"( t)[Pn+)3 n(t) Pn(t ) 2n +

P1 2(t)]dt.

Problems in Higher Dimensions

230

By Schwarz's inequality, n 2IAI

<

Ilf"ll

[IIPn + 2II + 2 IIPnII + IIPn-2II1 (2n - 1)

where II f II2 = f11 f2(t)dt. Since IIPP II = [2 /(2j + 1)]1/2, we have, after simplification,

2IAnI < 211f /III (n - 3/2)3/2 Since IPn (t)I < 1 and >22(n -3/ 2)-3/2 converges , by the Weierstrass Mtest, the series (7.107) converges uniformly to a continuous function g(t). Since { Pn(t)} is complete on [-1,1 ], the series (7.107) also converges in the mean to the continuous function f (t) in [- 1,1]. Hence, g (t) = f (t). Theorem 7.14

The boundary value problem urr +

2ur (sin 0 ue)e _ 0 - 0, r < 1 (7.108) r +r2 sin

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

0 < 0 < 7r,

(7.109)

has a unique solution given by 00

u(r, 0) = E anrnPn ( cos 0)

( 7.110)

n=0

where

J0

2n+1 "

an = 2

f( 0)Pn(cos 0) sin 9dO ,

(7.111)

if f (0) E C2 [0, ir] • Proof: From Section 7.9, we set u(r, 0) = A(r)B(0). Then (7.108) is reduced to r 2 A" + 2rA' - AA = 0, 0
(7.112)

[(1 - t2)P']' + AP = 0, -1 < t < 1

(7.113)

and

Legendre series and boundary value problems

231

where A is a separation constant, and P(t) = P(cos&) = B(9). For P(t) and P (t) bounded at t = ± 1 , we have A = n(n + 1) and P(t) = Pn(t),n = 0,1,2... r u t A = n{n + 1) in (7.112). Ihen its bounded solutions are Ayr) = r . Hence, un(r, 9) = rnPn(cos 9). We form a series solution as given by (7.110). To satisfy (7.109), we have oo

/(0) = ] £ a n P n ( c o s 0 )

(7.114)

n=0

which is the Legendre series of f{9). Thus, an is given by (7.111), and (7.110) and (7.111) yield a formal solution of (7.108)-(7.109). Verification: (A). For r < 1, the series for ur,rrr,ug, differentiation of (7.110) are ur~yjnan7-n

1

and ugg obtained from formal

P n (cosd),

urr ~ Y j n ( n — l ) a n r n ue ~

2

Pn{cos9),

^2anrn(-sin9)Pn(cos9)

and uee ~ ^ a „ r n [ - cos6>P^(cosS) + sin2 9Pn\cos9)} respectively. We have shown that \an\ < (n - 3/2)- 3 / 2 ||/"||, |P n (t)| < 1, |-Pn(*)l — n2> ano ^ l^n'COl — n 4 - Hence, all the series for u and its partial derivatives are dominated by a convergent series of the form

Mj>5'2r£ for r < ro < 1 where M is a positive constant. Hence, it is justifiable to differentiate the series in (7.110) term by term and it is easy to verify that u(r,9) satisfies (7.108) for r < 1.

Problems in Higher Dimensions

232

(B). By Theorem 7.13, the series in (7.114) converges uniformly to f (0) on [0, ir]. Define N SN (r,

0)

= > n=0

a nrnPn (COS 0).

Let e > 0. There exists an integer N = N(e) such that for K > M > N, ISK(1, 0) - SM(1, 0) 1 < e. Since SK(r, 0) - SM (r, 0) satisfies Laplace's equation , it follows from the maximum principle that, I SK (r, 0) - SM (r, 0) l < e for r < 1. Hence, {SN (r, 0) } converges uniformly to a continuous function in r < 1 and u(1 , 0) = f (0).

7.12 Laplace 's equation in a sphere Consider the Dirichlet problem 2Ur

(sin Ouo)o uo±

Urr + + + = 0, r < 1; u(l, g, 0) = f (g, 0). r r2 sin 0 r2 sin2 0

(7.115) Let u(r, 0, q) = A(r)B(0)C(cb) and substitute u in the equation in (7.115). Then we have r2 sin2 0A" + 2r sin2 0A' + sin 0 (sin 0B')' - - C" - µ A

A

B

C

(7.116)

and r2 A" 2rA' A+

(sin 0 B')' _ (7.117) sin2 0 Sin O B

where A and µ are separation constants. For the continuity of u in r < 1, we require that u(r, 0, 0) be periodic in 0 of period 21r. Hence we have C" + µC = 0 , - ,7r < 0 < ir, C (-cb) = CM, C'(-¢) = C'M. (7.118) Thus, C(0) = cos mo or

sin mo,

M=m 2, m =0,1,2... (7.119)

233

Laplace's equation in a sphere

We also have r2A" + 2rA' - AA = 0,

0 < r < 1,

(7.120)

where A(r) and A'(r) are finite at r = 0, and z (sin 0B ')'-m B+Asin0B=0, sin 0

0<0<7r.

(7.121)

Let t = cos0 and B(0) = H(t). Then (7.121) becomes 2

[( 1 - t2)H']' -

1

+ AH = 0 ,

-1 < t < 1 .

(7 . 122)

This is a singular equation with singular points at t = ±1. We seek for values of A such that there exists a solution H(t) of (7.122) with H and H' bounded at t = ±1. We differentiate Legendre's equation [(1 - t2)P']' + AP = 0 (7.123) m times with respect to t and, after simplification, we get (1 - t2)P(m+2) - 2(m + 1)tP(-+1) + [A - m(m + 1)]P(-) = 0. Define H(t) = (1 - t2)-/2P(m)(t). The above equation becomes (7.122). Thus, H(t) is a nontrivial solution of (7.122) if P(t) is a solution of Legendre's equation unless P(t) = Pk(t) where k < m. We know that for A 54 n(n + 1), n = 0, 1, 2..., (7.123) has two linearly independent solutions P(t) and Q(t) which converge absolutely for Itl < 1, but diverge at t = ±1. Thus, their corresponding solutions H(t) = (1 - t2)m/2P(m) (t) and K(t) = (1 - t2)m/2Q(m) (t) of (7.122) also converge absolutely for Itl < 1 and diverge at t = ±1. Any values of \, other than n(n + 1), cannot be eigenvalues of (7.122). For A = n(n + 1) and n > m, their corresponding eigenfunctions are

P,n (t) 2 n! (1 - t2)m/2 dt^,+ n (t2 - 1)n, n > m (7.124) which are called the associated Legendre functions. We note, as in the case

Problems in Higher Dimensions

234

of Legendre's equation, that

J 11 P,m (t)PP (t) dt = 0 ,

n

0 k.

(7.125)

For k = n, we have, after m integrations by parts,

J

1[Pn(t)]2dt = 1[( 1-

_ (-1)m Pn (t)

t2)m p(m) ( t)]Pnm)(t)dt

dtm [(1

-

t2)m Pnm)]dt.

We note that

Gn(t)

=

m

2

m

dtm [(1 - t)

(m) Pn

(t)]

is a polynomial of degree n and since f 1l Pn(t)t''dt = 0 for k = 0, 1, 2.., n-1, it is sufficient to compute the coefficients of to in Gn(t), which is equal to (-1)mcn(n+m)!/(n-m)! where cn is the coefficient of to in the polynomial Pn(t). Hence, we get

f 1 [Pn(t)] 2dt = 1

(n + m)! J 1 Pn

(t) (cntn )dt

(n

-m)!

f 1 Pn (t) dt

_ 2(n + m)! (2n + 1)(n - m)!

(7.126)

Put A = n(n+1) in both (7.120) and (7.121). Their corresponding solutions are An (r) = rn and Bnm, (0) = Pn (cos 0) where n = 0, 1, 2, ... and m = 0, 1, 2,...n. We obtain a set of particular solutions of Laplace's equation in (7.115) given by Unm (r, 0, 0) = rnPn (cos 0) cos mo or rnP,n (cos 0) sin mq which are regular inside the sphere and are called spherical harmonics. We form a series solution of the form 00

2 anornPn(COS 0)

u(r, 0, 0) _ n =o

235

Poisson's integral formula in space

00 n + E T rn(an.. cos

MO + bnm sin mq5) PP (cos 0).

(7.127)

n=O m=1

For r=1, 00

00

n

anoPn( cos 9) +E E (anm cos mlp+ bnm sin mq)Pm (cos 0). f(0, 0) = 1) n=o n=o m=1

Hence, " +1 ) [ 2ir n o0

( 2n anm = Cnm

bnm =

J J

f (a, 0) Pn (cos a ) cos m/Q sin adad,Q

cnn' (2n + 1) IT f (a, ,6)Pn (cos a) sin m,Q sin adado 27r J IT L cnm = (n - m)!/(n + m)!.

(7.128)

(7.127) and (7.128 ) yield a formal solution of (7.115 ). Verification of the solution is carried out in the next section.

7.13 Poisson ' s integral formula in space Let () and (x, y, z) be a fixed point and a variable point in space respectively. Then a fundamental solution of u+ u+ u= 0

(7.129)

is a solution u(r) _

1 47rro '

r=

of the equation 2ur urr +

r

= 0.

of (7.129) is defined in the same way as Green's function G(x, y, z; the plane. In particular, the Green's function for a unit sphere can be given by the method of images as G(x, y, z;

*, () = 47rro 41rpro

(7.130)

236

Problems in Higher

Dimensions

where r0* = ^{x - 6 ) 2 + (y - m)2 + {z- Ci) 2 , />2 = £2 + r?2 + <2, and (£1, 771, £1) = (^p~2,rjp~2, C,p~2) is the image of (£, 77, Q with respect to the spherical surface r = 1. Let £> be a bounded domain in space with a simple, closed, and piecewise smooth boundary surface 5. For any harmonic function u(x,y,z), which is twice continuously differentiable in D+S, we have u(x,y,z) = - f J u^rjX^da

(7.131)

where da is the surface element and dG/dn is the outward normal derivative of G on S. The representation formula (7.131) can be derived in the same way as in the plane given in Sections 6.2 and 6.7. In the case of the unit sphere with (x, y, z) = (r, 6, ) and (£, 77, £) = (p,a,{3), we have (£1,771, &) = (p _ 1 ,a,/3). Then rl = r2 + p2 - 2r/?cos7,

r*02 = r2 + p~2 - 2rp~x COS7

where cos 7 = sin 9 sin a cos(<j) — /3) + cos 8 cos a.

(7.132)

From (7.130), dG, dG, a n "1/5=1 ' - 1 = -a£p-1| p" =- 1l

r2-\ 4 T T ( 1 + r 2 - 2 r cos 7) 3 /2-

Thus, from (7.131),

where d r =

smadadft.

Theorem 7.15

Let f(9,(p) be continuous on [0,7r] x [-7r,7r]. Then

<*r,0,*)-l^££

J£f>

*, r
237

Poisson's integral formula in space

is the unique solution of _ 2ur (sin 0 ug)g u'k^ 07 urr + r + r2 sin 0 + r2 sinz 0

r < 1; u(1, 0, 0 ) = f (O' 0) (7.135)

where cos y is given by (7.132).

Proof: For r < 1, we can differentiate (7.134) under the integral sign and show that u(r, 0, 0) satisifes Laplace's equation. Let ro = - 2r cos -y + r2. Set u = 1 in (7.133). We have

f

f _ z 1 r r 1 3dv, 4' s o

(7.136)

where S = {r = 1}. Let N be a fixed point on S. Then , from (7 .136), we get

r2 r f f^N

Ji

f (N) = 1 4-7r ) do. s o For any point M in r < 1, f (P) rz (N) do u ( M ) - f (N) = 147r fsS ro

3f

where P is a variable point on S. Let e > 0. Since f (P) is continuous at N, there exists a 6 > 0 such that If (P) - f (N) I < e/2 if IP- NJ < 26. Let S5 be the part of the surface on S of radius 26 with N as its center and let M be the point inside the sphere r < 1 such that IM - N < J. Then, f (P)

u(M)-f (N) = 1- rz 47r

j jSa

3f ro

(N ) dv+ l -

r2

.f (P)

3 f (N) do,

411- s-sa ro

IM - PI. The first integral is bounded by

where ro =

e 1 - z r do, < e/2; 81r ffa ro3

Problems in Higher Dimensions

238

whereas the second integral is bounded by

2K(1 3r2) 4-7rS

2K(13 r2) 3 J

do, < J Js_sa

which goes to zero as r -> 1 where K = maxs If I Hence, limM-.N u(M) = f(N). Remark: (7.134) is called Poisson 's integral formula in space. From (7. 99), we can easily derive that °O 2 E(2n + 1)rnPn(t) = 1 - r

(7.137)

n=O

We also note that IPn (cos y) I < 1. It follows that the series in (7.137) for t = cosy converges uniformly for 0 < r < a < 1. If we substitute (7.137) into (7.134) for r < a and integrate term by term, we have

u(r, 0, ^) =

4-

00 Ir f (a, 0)Pn(cos y)dv. E(2n + 1)rn n =o

J

(7.138)

Applying the additional theorem for spherical harmonics, Pn (cos y) = Pn (cos 0) Pn (cos a) n +2 (n + m^i Pn (cos B)Pn (cos a) cos m(O -'3) M=1

in (7.138), we get (7.127)-(7.128). (For the proof of the additional formula, please refer to Hobson [7, 141-143]).

7.14 Problems In Problems 1 and 2, classify the equation. 1. 13uxx + 13uyy + 10uZZ - 8u,,y + 4u2z - 4uyz = 0. 2. uyy + 4uxz = 0. In Problems 3 and 4, find the double Fourier series of f (x, y). 3. f (x, y) = x2y2 in (-7r, 7r) x (-ir, ir).

4. f (x, y) = xy in (0, 27r) x (0, 27r).

Problems

239

5. Find the double sine series for f (x, y) = (1 - y2) sin x in (0, 7r) x (0, 7r). 6. Let u(x, y, z) be a solution of Du > 0 in a bounded domain D with continuous, piecewise smooth boundary C and let u(x, y, z) be continuous in D + C. Show that the maximum of u in D + C is attained on C. 7. Find a formal solution of Du=O for

0<x<-7r,

O
u=0 for x=0, x=7r, and z=1;

uy = 0 for y = 0 and y = ir;

u (x, y, 0) = X.

8. Show that, for n > 1, (a). Je(t) = [Jn-1 - Jn+1]/2. (b). t2Jn'( t) = (n2 - n - t2 )Jn(t) + tJn+1(t). (c). d/dt{t2[JJ(t) - Jn-1(t)Jn+1(t)]} = 2tJn(t). 9. Find the expansion of 1 - x2 on (0,1) in terms of the eigenfunctions {Jo(aokx)}, k = 1, 2... of (xu')' + Axu = 0 for 0 < x < 1; u(1) = 0, and u and u' are bounded at x = 0. Express your Fourier coefficients in terms of J1(aok) and J2(0'ok)• 10. Find the Fourier series of f (x) = xm with respect to the system of functions {Jm(amkx)}, k = 1,2 ... on [0,1] and find the sum of the series E -2 k=1 amk'

11. Derive the reduction formula x snJo(s)ds = xnJ1(x) + (n - 1)xn-1Jo(x) - (n - 1)2 f sn-2Jo(s)ds 0 fo x forn>2. In Problems 12 through 15, find a formal solution of the boundary value problem. 12.

Ut

=

Urr

+ r-fur - hu,

r < 1 t > 0; u(1, t) = 0,

where h is a positive constant. 13.

urr+r-fur+Uzz=0 u(1, z) = 0, z > 0,

for 00,

u(r, 0) = f (r) and limz^^ u(r, z) = 0, 0 < r < 1. 14. urr+r-fur+uzZ=0 for r
u(r, 0) = f (r);

Problems in Higher Dimensions

240

15.

u(c,z)=0, 0 < z < b; u(r, b) = f (r) and u(r, 0) = 0, 0 < r < c. urr+r-fur+uzz-ut=0, r<1, 00; u(1, z, t) = uz(r, 0, t) = uz(r, 2, t) = 0, u(r, z, 0) = f (r, z).

16. (a). Consider the singular Sturm-Liouville problem (xu')' = -Axu, 0 < x < c, u' (c) = 0 and u and u' are bounded at x=0. Show that the eigenvalues are nonnegative and that the eigenfunctions corresponding to two distinct eigenvalues are orthogonal with respect to the weight function x on [0 , c]. (Assume that all the eigenfunctions are real.) (b). Find all the eigenfunctions { uk (x)} of (a) and evaluate ff uk(x)xdx. (c). Find a formal series solution of urr + r ur + uzz = 0, r < c, 0 < z < b;

u(r, 0) = ur (c, z) = 0, and u (r, b) = f (r). 17. Show that there is at most a polynomial Pn (t) of degree n which satisifes (i). Pn(1) = 1 and (ii ) f 11 Pn(t)tkdt = 0, fork = 0, 1, 2..., n - 1. 18. Find P3(t) by using the fact that it is of degree 3 and is orthogonal to 1, t and t2 on [-1,1] and P3(1) = 1. 19. Show that

(n + 1)Pn+1(t) + nPn_1(t) = (2n + 1)tPn(t),

n > 1.

20. Evaluate (a)• P, (1). (b)• f (c). .l 11 Pn(t)P,n(t)dt.

21. Show that the following sets of functions are orthonormal and complete with respect to the weight function 1 on (0,1) (a). { 4n + 1 P2n(x)}, n = 0,1, 2... (b). { 4n + 3 P2n+1(x)}, n = 0,1, 2... 22. Consider the equation (1 - x2)u" - 3xu' + Au = 0 in IxI < 1 where A is a real constant. Obtain two linearly independent solutions of the above equation by the power series method. For what values of A, does the above equation have a polynomial solution? 23. Let u(x) = E°° o anxn be a power series solution for the equation

Problems

241

(1 - x2)u" - xu' + Au = 0 in I x I < 1 where A is a real constant. Obtain a recurrence formula for the coefficients a, For what values of A does the above equation have a polynomial solution? Give the first four polynomial solutions for these values of A. 24. Find a formal solution of Laplace's equation in a < r < b with u(a, 0) _ F(0) and u(b, 0) = 0, 0 < 0 < ir, which is independent of the angle 0.

Appendix A

Ascoli's Theorem

Ascoli's Theorem Let {fn{x)} D e a sequence of uniformly bounded and equicontinuous functions on [a,/?]. Then it contains a subsequence {fnk(x)} which converges uniformly on [a,/3]. Proof: Since the set of rational numbers is countable in [a, /?], it may be arranged as {nt}, k = 1,2,..., in some order. The set of numbers {fn(r{)} is bounded and hence by the Bolzano-Weierstrass theroem, there exists a subsequence {/^(z)} of {fn(x)} such that {/^(ri)} is convergent. Since the set of numbers {fnfo)} is bounded, there exists a subsequence {f%(x)} of ifn(x)} s u c n t n a t {/n( r 2)} converges. Repeating this procedure, we can select a subsequence {/*(a;)} of {/„(x)} such that, for a fixed k, f£{x) coverges at r i , ...,rk. Define gk(x) = fk(x). Clearly {gk(x)} converges at every rational point in [a,/3]. We will show that the sequence {gk(x)} converges uniformly on [a, 0\. Let e > 0. For x = r», there exists an integer Ni = Ni(e,ri) such that Iflfcfa) — flm('"x)| < e for fc,m > Ni. Since {fn(x)} is equicontinuous, there exists a 5 = 6(e) > 0 such that \gk(x) — gk{y)\ < e for \x - y\ < 5 and for all gk. Divide the interval [a,/?] into a finite number of subintervals Ii,...,Ip such that the length of each Ij,j — 1, ...,p, is less than 5. Since the set of rational numbers in dense in any interval Ij, we can select a rational number fj in each Ij. Let x € [a, /?]. Then x G /j for some j and we have 243

244

Ascoli's Theorem

I9k(x)-9m (x)I C I9k(x)- 9k(f )l +I9k (fi)-9m(rj)I +I9m(rj)-9m.(x)I < 3e, provided that k, m > max [N(E, r`1), ...N( e, FP)]. This implies that (g,,(x)} converges uniformly on [a„ 3].

Appendix B

Answers for Selected Problems

Chapter 1 3. (a) x2 + y2 > 1, hyperbolic; x2 + y2 = 1, parabolic; x2 + y2 < 1, elliptic. (b) x2 > y, hyperbolic; x2 = y, parabolic; x2 < y, elliptic. 4. x y, hyperbolic ; x = y, parabolic . Characteristic curves are y = x2+3andy=x+1. 5. x + y 54 0, hyperbolic ; x + y = 0, parabolic . Characteristic curves are y=-x+4 and y=3x. 6. everywhere parabolic . Characteristic curve is y = 2 - e-x. 7. t > 0, hyperbolic ; t = 0, parabolic ; t < 0, elliptic . Characterisitic curves are 3x = ±2 ( t3/2 - 1). 8. x = n7r, n = 0, ±1, ±2,..., parabolic ; x nir, hyperbolic. Characteristic curves are t = sin x ± cos x - 1. 9. u(x, y ) = x2 log y/2 + 2xy + p(x) + q(y). 10. u(x, y ) = y4 sin x/4 + yp(x) + q(x). 11. u(x,y ) = y2 log x1 ( 2x) + p(y )/x + q(x). 12. u(x, y ) = - xy + p(x ) e-xy/x + q(x). 13. u(x, y ) = xey + p (y)/x + q(x). 14. u(x,y ) = e('+y)/2p ( x - y) + q(x + y).

245

246 Answers for Selected Problems

18. x 54 t, hyperbolic. u(x, t) = p (x2 - t2)/(x - t) + q(x - t). 19. u(x,y ) = (xy)3/2p (ylx) + q(xy). 20. u(x, y) = p(cos x + x - y ) + q(cos x - x - y). 21. u(x, t) = 2x2 + xp( x/t) + q(x/t). 22. u(x, t) = xtp(t/x) + q(xt). 23. u(x,y ) =

xlyp(xy ) + q(ylx)•

24. ul (x, y ) = e-y Binh (\y/2) cos(x/2); un(x, y ) = e-y sin ( (n - 3/2)(n + 1/2)y) cos (n - 1/2)x, n>2. 25. un (x, t) = sin ( n-7rx) cos( n2ir2 - 1 t), n = 1, 2,... 26. un (x, t) = e-t sin(2n - 1)irt cos (n - 1/2)7rx, 27. un (x, t)

= et_n2,2t3 /

n = 1, 2,...

3 sin nirx, n = 1, 2,...

28. un(x, t) = e_x-t sin nx sin

1 + n2 t, n = 1, 2, ...

29. ua (x, y) = e-ay [p(a) cos ax ) + q(a) sin ax], a > 0. 30. ua (x, t) = sin ax sin at, a > 0. Chapter 2 1 'x+' sin 7rx, 0 < x < 1; 2. u(x,t) = 2 fx_t g(s)ds where g(s) 0, otherwise. u(1/2,3/4) = 1/7r. u(5/6,1/2) = 4^ 3. u(x,t) = 0 for x + t < 0; =(x+t)2/4forx-t< 0 and0<x+t<1; =xtfor0<x-t<x+t<1; = (2x+2t-1)/4forx-t <0andx+t> 1; = -(x-t)2/4+(x+t)/2-1/4 forx+t> 1 and 0 <x-t < 1; =tforx-t>1. 4. u(x, t) = [f (x+2t)+ f (x-2t)] / 2 where f (x) = u(x, t) is given in 10 different regions: I. u=0 forx+2t<0;

Ix/2 for 0 < x < 1/2; (1 - x)/ 2 . for 1 12 <x< 1; 1 0 elsewhere.

247

II. u= (x + 4t)/4 for x - 2t < 0 and 0 < x + 2t < 1/2; III. u=x / 2for0 < x-2t<x+2t<1/2; IV. u= (1- 4t)/4 for 0 < x - 2t < 1/2 and 1/2 < x + 2t < 1; V. u= (1-x)/ 2 for 1 / 2 <x-2t <x+2t < 1;

VI. u = (1 - x + 2t)/4 for 1 /2 < x - 2t < 1 andx + 2t > 1; VII. u= Oforx-2t>1; VIII. u = (1 - x - 2t )/4 for 1 / 2 < x + 2t < 1 and x - 2t < 0; IX. u=0 for x+2t>1 and x-2t<0; X. u=(x-2t )/4forx+2t>1and0 < x-2t<1/2. u(1,1/4) = 1/8 and u(1/2,1 / 6) = 1/12.

5. (a) T=A;

(b) T = A+ Ial and U= 2 f Ag(s)ds.

-x2, -3 < x < 0; 10. u(x, t ) _ [f (x+t)+ f ( x-t)] /2 where f (x) = x2, 0 < x < 3; 1 -(x - 6)2, 3 < x < 6. [(x + t)2 - (x - t ) 2] /2, -3 < x - t < 0 < x + t < 3; ( [(x-t)2+(x+t ) 2]/2, 0 <x-t <x+t <3; u x,t) _ [-(x+t-6)2+ ( x-t)2]/2 , 0 <x-t <3 <x+t <6; [-(x + t - 6)2 - (x - t)2] /2 , - 3 < x - t < 0 , 3 < x + t < 6. 12. (b) u (1/2,2) = 3; ut ( 1/2,2) = 3/2. 13. (b) u (x, t) = 1/2 fytt g(s)ds and where -x2, -1 < x < 0; x2, 0 < x < 1; g(x) = (2 - x)2, 1 < x < 2.

ut(x, t) = [g (x + t) + g(x - t)]/2

I. u(x, t) = [(x + t)3 - (x - t)3] /6, ut(x,t) = [(x + t)2 + (x - t)2] /2 for x - t > 0, x + t < 1. II. u(x,t) = [(x + t)3 + (x - t)3]/6, ut(x,t) = [(x + t)2 - (x - t)2] /2 for x - t < 0, x + t < 1. III. u(x, t) = [2 - (x - t)3 - (2 - x - t)3]/6, ut(x,t)=[(2-x-t)2+(x-t)2]/2 for x-t>0, x+t>1. 14. (b) u(1,1/2) _ -1/4, u(3/2, 1) _ -1/4. 16. (a) u(x, t) = sin x cos t + xt + xt2 . 19. u(x, t) = (3e - 2el/2 - 1)/2.

(b) u(x, t) = x + x2t2 / 2 + t4/12.

248

Answers for Selected Problems

20. u(x, t ) = sin

7r x

( 1 - cos

7rt)/7r2.

21. (a) u ( 5,12) = -29 , (b)u(5,12) = - 30, (c)u ( 5,12) = 521/3. x+t 1/2 fx g(s)ds , if x - t > 0 ; 22. u(x , t ) = [f(x+t)+f(x-t)] 2 + _t 1 -a t-x +t ( 1-c t ) f +t g(s)ds, g( s)ds - 2 (1 +«) 2 [ 2(1+a ) ] + f0 if x-t< 0.

fo

23. u(x, t) = [F(x + ct) + F(x - ct) + 1 fy ^t G(s)ds]/2 where F(x) and G(x) are even extensions c of f (x) and g(x) on (-oo, oo) respectively.

Chapter 3 1. yn+1(x) _ Vj+2 ^; y(x) = ex - 1 - x.

2. y(x)

DD

X3-1

= - En =1 (3n-1)(3n-4)...2'

5. u(x) = -

fo

sinh(x - l;) f (6)d6.

6. u(x) = cos(log x) + sin(log x) + log x. 7. u(x) = - fox sinh(x -

+ c sh i fo cosh(1 - t;) f (1;)dt;.

2 e+1 ex - 2e a-1 8. u(x) = -2x + e +1 e +1 a-x

10. G(x, C) =

r sin x cos 6, cos x sin 6,

x < 6; x > ^.

Binh x cosh (7r - C) / cosh 7r, x <_ C; 11. G(x, ) = 1 l Binh 1; cosh ( 7r - x)/ cosh 7r, x > C.

12. G(x,

f-(1-2x)(1-C), x < C, (1-x)(1-2C), C <x.

1)(C - 2)/3, x < C, 13. G(x,^) _ (x + -(C +1)(x-2)/3, x > C. 14. (bii) u(x) = x,

8C-2)/24, (ci) G(x, l;) = -x(C -C(x - 8x-2)/24,

15. u(x) = A fi G(x, C)Cu(C)d, f log x(1 - log C), x < where G(x, 6) = log 6(1 - log x), x > C.

x < 6, x > 6.

249

16. Eigenvalues, n2 + 1; eigenfunctions, sin(n log x)/x;

n — 1,2,...

17. eigenvalues are -1 and n 2 , n = 1,2,...; and eigenfunctions are eT and ncos nx + sin nx. 18. (a), (i) Q = 0, eigenvalues are (n - l/2) 2 7r 2 ,n = 1,2,... (ii) 0 < a < 1, eigenvalue is /xjj where fi0 is the root of the equation tan /J. — n/a in (0, 7r/2). (iii)a / 0, eigenvalues are /i 2 where n„ is the root of the equation tan/i = n/a in ((n - l/2)7r,(n + l/2)7r),n = 1,2,...

22. /(x) = E~,(-irV J (x)/2^- 1 . or 25

v^oo 4sin(2m-l)i. o /o ' 2-m = l (2m-1). ■ * / 8 -

26. (i) for /i = 0, its corresponding eigenfunctions are all the functions orthogonal to 1 and x on [0,1]; (ii) for /J. = ±-TTT, their corresponding eigenfunctions are x — 1/2 ± —l~. 27. (i) for fi — 0, its corresponding functions are all the functions orthog­ onal to 1, cos x and sin x on [—7r,7r] (ii). for p. = 2n, its corresponding eigenfunction is a constant; (iii). for fi = ir, its corresponding eigenfunc­ tions are cos x and sin x. 28. for ft = 0, its corresponding eigenfunctions are all the functions orthog­ onal tO COS X On [ —7T, 7r]. 29. (a) for /x = 2, its eigenfunction is 1; for /i = 2/3, its eigenfunction is x. (b) for [i = 0, its eigenfunctions are all the functions orthogonal to 1 and x on [-1, 1]. For example, u\(x) — 3x 2 — l;ii2(x) = 5x 3 — 3x. 31. (c) the eigenfunctions are all the functions orthogonal to 1 and sin x on [—7r,7r]. u(x) = sin2x. 33. for fi = ±Trbn, their corresponding eigenfunctions are cos nx ± sin nx; for n = 0, its eigenfunction is a constant. 35. An = a + n 2 7 r 2 / ( / 3 - a ) 2 , n = l,2,... 36. n 2 ?r 2 /2 < An < 2n27r2 + 1, n = 1,2,... Chapter 4 1. sinh7r/7r + YlT=i ' X " " , "[cos nx - n sin nx]\ cosh n.

Answers for Selected Problems

250

2. 2 cos x + E00 1 ir( 4 sin 2nx. 1 4 0o cos ( 2m+1)x • 2 3. 27r - 2m=0 2m+1) 7f /8. 4.

sinh ir + coo 2(- 1n Binh 7r (cos nx + n sin nx); 7r n=1 ir(l+n )

(7r cosh 7r - 1)/2.

5. s + E°° 1(-1)n[8 - 48] cos nx; 4 6. V'°° 1 ( 1 - cos (n7r/2 )) sin nx;

1 00 2-1m 2 - rm=1 lr( 2m-1 cos (2m - 1)x.

7. 2 cos 7rx + 1 Ljn 1 4n'1 sin 2n7rx. 8>o0 2

. 2 n=1 bn

13. u(x, t) = E°1 bn sin nxBn(t) where bn = TB,'2 o fo g(x) sin nxdx with e-at sink

a2 - n2c2 - bt,

a2 - b - c2n2 > 0;

Bn(t) = to-at, a2 - b -c2n2 = 0; e-at sin b + c2n2 - a2t; a2 - b - c2n2 < 0.

14. Let pn =

n2 - n. u(x, t) = E°--1 bn sin(n - 1/2)x Bn(t) where

n=1; bn f0 f(x) sin(x/2)dx, z i and { Znµn fo f (x) sin(n - 1/2)x; n > 1; - f (2 + t)e-t/2, n=1; Bn(t) l e-t/2 (2 an cos µnt + sin Ant); n > 1. 16.

f °° 4 cos uxdu; Ire -2"/2.

^r 0 00 u[sin u(lr -x )-sin ux du; 1,7 1 f 1- u

18. 2 fo 1- us u sin uxdu; 19.

7r/2.

0 for x >1; 7r/4 for x = 1; 7r/2 for 0 < x < 1.

2u3

fo

[A(a) cos ax + B(a) sin ax] cos catda; 21. u(x, t) = B(a) = 11 f 00 f (s) sin asds. A(a) _7r f00 f (s) cos asds, 7r 23. 2^ f00 e-u2 cos(u3/3 - ux)du. 24. z^r f L[f (w) Cos wt + w-lg(w) sinwt]e-'wxdw where f (w) and g(w) are Fourier transforms of f(x) and g (x) respectively.

251 Sin u

25. FF[f]

1,

u

1_U2

26. F^[f] _

(u) 2 7. j (u)

o0;

Fs[ f]

u :A

F's[f] _

= jl

2(cos

(

=

1-cos u u

u54 0; u=0.

0,

2u 1T_u )2 T

u 0;

3, u=0.

28. 7r ( 1 - lul)e-lul/2. 29. '[cos u -sin l ],-1.1/,/-2. 30. [e-blul / b - e-alul /a].

31. 2 e-"-2lul . Chapter 5

1. v"+2yv'=0. 4. u(x,t ) = En 1[(2n 11) + n(2]e-[1

5. u(x, t )

= --1 32

1) eXp[- (2n

+( r,-1/2)2]tsin

4)2"2 -

1]t COS

( n- 2)x.

( 2n 21)7rx

6. Let un be the root of tan uL = -Ph ( 2-+1)7r for 2 2L ) ^ < N < n= 1,2... 2L,

2 u(x, t) = i°° 1 bne-kµ„ t sin µnx where bn = f L f (x) sin µnxdx/ [ 2 - sin 4µn

L]

2

7. u(x, t) = Aa + a-i sin wt + E°°1 Ane-kn t Cos nx where An = 1 f0' f (x) cos nxdx.

8. u(x, t) = 2 + e-41 Cos 2x. 9. u(x, t) = E', an exp[-(n - 1/2)2t] cos(n - 1/2)x + E 1 bn (t) cos(n - 1/2)x where an = fo f (x) cos(n - 1/2)xdx and

bn(t) _ ft j exp[-(2n-1)2(t-T)/41 cos(n- 1/2)F(C,T)d dr.

11. u(x t) fOcl hh+kay[-1 + e-(h+k32)t] cos sxds.

Answers for Selected Problems

252

12. u(x, t ) =

CO t e-('2}1)t sxds. fo [' - ( 1 + ( 2 Is sin

13. (x t) =

f oods. -t2/2 a-e2CO u e 7r 10 l+s

14. (a)l+u 1y. 15. (b) 0, (c) 4. 18. 1 - e-t. Chapter 6 6. u(x, y) _ E°° 1 bn sin nx sinh h + n2y where bn = 7r sinh 2 n2+h fo f (x) sin nxdx. 7. u(x,y )= 1-y+

cos x sinh ( 1-y) 1 cos 2x

sinh1

+

4COS2x-4sinh2[Slnh2y + sinh2 ( 1-y)].

0o 4sin ( 2k-1)xsinh ( 2k-1)(1-y) 8. u(x,y ) = 2k=1 (2k-1 )7rsinh(2k-1) En 2(-1)" [coshny - 1 + (1-cos nhninhny ] sinnx. + n=1 T,^

16. (a) 1 47rlog

[( x +l;)2 +2+(v+,l)2][(x

±

^)2+(y-n)2

0-0 L log r2+p2-2rpcos 0+0 p2r2+R4-2rpR2cos (b)) -47r +v -2rpcos ( B-0) p ++R -2rpR cos(B+O) _Z (' o0 (' 00 sinh w(b-y) sin wx sin £ x f (1;) d^dw. 19. U(x y) = 7r .10 Jo sinh bw

sinh sy cos ax oo f (s) ds where 20. u(x, y) = 7r2 f 0 sinh 7r s

j (s) is the Fourier transform of f (x). 22. u(x, y) _

1

r 00 e-'214sinh s (7r-y) $x JO sinh s7rCOS dS.

Chapter 7 1. 9(ugg + u,7,, + 2u(C) = 0, where 1; = x + 2y + 2z, S = 2x - 2y + z. Elliptic.

77 = 2x + y - 2z,

2. u£C + 2u,7,, - 2uCC = 0 , where ^ = y, y = x + z, (= x - z. osnx oo -1 cosmy 'r4 4r2 -1 - 1 "cosnx -=1, n2 + ^m=1 m ] 3. 9 + 3 [En + 16 Eon, m=1 (-1),..+ monnx.

Hyperbolic.

253 4 7r 2 - 27r oo sin nx - 2^ 00 sin m 00 sin nx sin m y En-1 n Em=1 m + 4 ^n,m=1 nm [1 + (- 1)m(7r2-1 ) - 2 [( _ 1 ) m _1 ] 5 27r 1:00 ]sinx sin my. m=1 m m

7. u(x, y, z) = 9.

2 -1) "+i

00 L

1

n Binh n sinh n(1 - z) sin nx.

okx)• I:^ 1 aok !Jl( a0k )J2 JO (a

2Jm( amkx) 1 l0. E00 k=1 amkJm+l(amk)' 4m+4

E' 1 akJo(aokr) e-(aok+h)t, where

12. u(r, t) ak = f0

f (r)Jo(ao k r)rdr/ f0 J0 (aokr)rdr.

13. u(r, z) _

Ekl ake-aokzJo(aokr),

where

ak = J1(aok fo f(r)Jo(aokr)rdr. 14. u(r, z) =

Ekl akJO( ooC

Binh(.), where

f f(r)Jo(-)rdr ak sinh (-) f 0 Jo °' )rdr.

0

15. u(r, z, t) _ En'=0 Ek=1 Anke-

(n27r2

/4+a0k )tJ0(aokr) COS(nirz/2), where

f(r, z)Jo(aokr)gn(z)rdrdz = J^ 1(0 k ) L f0 1, such that gn(z) _ n=0; 2 cos(n7rz/2), n > 1. `9nk

16. (b) uo(x) = 1, f0 uo(x)xdx = 2 ; for

k > 1, uk(x ) = Jo( ), f0 uk(x)xdx = ZJ0(alk).

(c) u(r, z ) = aoz + k l akJo(

ao

2 /c =6

f0 rf

) Binh ( r), where

2 fcrf( r)Jo(r)dr

(r)dr,

ak

= c2 sinh ( alkb/c) Jo(alk)

(b) -1 " 2n ! 20. (a) n n+l 2 ' 2 (n!) ,

(c) 1- (-1)n+- for n<m-1; Oforn>m-1. 22. u1 (x) = 1 +>k 1[(2k)(2k - 2) - A]...[4.2 - A][-A]x2k/(2k)!;

254

Answers for Selected Problems

u2(x) = x + E' 1[(2k + 1)(2k - 1) - A]...[3.1 - A]x21c+1/(2k + 1)!. A = n(n + 2), n.= 0, 1, 2,... 23. (n + 1)(n + 2)an+2 = an (n2 - A); A = n2, n = 0 , 1, 2,... To(x) = 1,T1(x) = x, T2(x) = 1 - 2x2,T3(x) = x - 4x3/3. 24. u(r, 0) _ En= 0 a,,,[! - ]Pn(COS 0) , where b- T;;Tr f' F(B)P„(cos B ) sin BdO an [I-- '-] o P, ( cos O) sin Bd9

Bibliography

1. Boyce, W. and DiPrima, R., Elementary Differential Equations and Boundary Value Problems, 5th Edition, Wiley, New York, 1992. 2. Cannon, J. R., The One-Dimensional Heat Equation, Addition-Wesley, Menlo Park, CA, 1984. 3. Churchill, R. and Brown, J., Fourier Series and Boundary Value Problems, 4th Edition, McGraw Hill, New York, 1987. 4. Churchill, R., Brown, J., and Verhey, R., Complex Variables and Applications, 3rd Edition, McGraw Hill, New York, 1974. 5. Courant, R. and Hilbert, D., Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1953.

6. Folks, W., Advanced Calculus, 3rd Edition, Wiley, New York, 1978. 7. Hobson, E.W., Theory of Spherical and Ellipsoidal Harmonics, Cambridge, London, 1931. 8. Indritz, J., Methods in Analysis, MacMillan, New York, 1963. 9. Sagan, H., Boundary and Eigenvalue Problems in Mathematical Physics, Wiley, New York, 1961. 10. Stakgold, I., Boundary Value Problems of Mathematical Physics, Vol. 1, MacMillan, New York, 1967. 11. Tolstov, G., Fourier Series, Prentice-Hall, Englewood Cliffs, NJ, 1962. 12. Tychenov A.N. and Samarski, A.A., Partial Differential Equations of Mathematical Physics, Vol. 1, Holden-Day, San Francisco, 1964 13. Weinberger, H., A First Course in Partial Differential Equations, Wiley, New York, 1965. 14. Yosida, K., Lectures on Differential and Integral Equations, Dover, New York, 1991. 15. Young, E., Partial Differential Equations, An Introduction, Allyn and Bacon, Boston, 1972.

255

Index

contour integration , 117-119 convergence in the mean, 63 cosine integral, 112 cosine series, 102 cosine transform, 116 Courant's theorem, 85-86

Ascoli's theorem, 69, 243-244 associated Legendre function, 233 Bessel function, 206-207 bounds for, 209 completeness property of, 211 derivatives of, 207 generating function for, 208 integral form of, 209 orthonormal property of, 210 zeros of, 207-208 Bessel's equation, 205 Bessel 's inequality, 64 bounded operator, 67

D'Alembert formula, 25 diffusion, 129-130 Dini theorem, 195 Dirichlet problem, 152 in a cube, 200-202 in a disc, 159-162 in a half plane, 181-184 in a rectangle, 155-158 in a sphere, 232-235 discontinuities, propagation of, 26 domain of dependence, 25 double Fourier series, 194

canonical form of second order equations, 6, 192 characteristic curve, 7 characteristic equation, 6 characteristic triangle, 25 classification of partial differential equations , 5-8, 191-192 completeness of eigenfunctions, 77 completeness of orthogonal system, 64, 195-196 continuity with respect to data for heat equation, 132 for Laplace's equation, 155 for wave equation, 33, 39

eigenfunction, 58, 61 eigenvalue, 58, 61 elliptic operator, 6 energy integral, 37, 203 extremal principles, 71-72 family of equicontinuous functions, 66 uniformly bound functions, 66 256

Index

finite Fourier transform, 177 finite sine transform, 135 Fourier coefficients, 64, 95 Fourier integral, 109-110 Fourier series, 64, 95 differentiation of, 101 mean convergence of, 101 pointwise convergence of, 98-99 uniform convergence of, 100-101 Fourier transform, 113 Frobenius method, 206 fundamental solution of Laplace's equation, 166-167 Gram-Schmidt process, 66 general solution, 12-14 generalized solution, 26-27, 35 Green's function

for a disc, 170-171 for a half plane, 172-173 for equation of Euler type, 178 for Laplace's equation, 168 for Sturm-Liouville problem, 52-53 Green's identities, 153 Green's theorem, 29 harmonic function, 164 heat equation, 127-129 hyperbolic operator, 6 improper integrals continuity of, 3 differentiation of, 3 uniform convergence of, 2 improperly posed problems, 11-12 inner product, 62 initial value problems for heat equation, 138-141 for nonhomogeneous heat equation, 142-145 for nonhomogeneous wave equation, 29-30 for wave equation, 24-25 initial-boundary value problems,

257

for heat equation, 133-135, 218-222 for nonhomogeneous heat equation, 135-137, 145-146 for nonhomogeneous wave equation, 36-37, 39-40 for wave equation, 33-35 integrable function, 95 integral equation, 61 integral operator, 61 inverse Fourier transform, 113 Laplace's equation, 151 in polar coordinates, 158-159 in spherical coordinates, 221 Laurent series, 117 Legendre polynomials, 224 bounds for, 227 completeness property of, 228-229 generating function for, 226-227 orthonormal property of, 224-225 Rodrigues' formula for, 224 zeros of, 225-226 Legendre series, 229 Legendre's equation, 222 maximum principle for heat equation, 130-131 Laplace's equation, 154 mean value theorem for harmonic function, 164 method of images, 170-173 method of successive approximation, 46-47 monotonicity theorem, 87-88 Neumann problem, 152 nonhomogeneous boundary conditions for heat equation, 145-146 for wave equation, 39 norm of function, 62 operator, 67 orthogonal function, 62

258 parabolic operator, 6 periodic b o u n d a r y value problem, 159 Poisson's equation, 151 in a disc, 173-177 Poisson's integral formula, 163-164, 238 Poisson's kernel, 164 principle of superposition, 2 range of influence, 26 residue, 117 residue theorem, 117 R i e m a n n - L e b e s g u e theorem, 97 Schwarz's inequality, 63 semi-infinite vibrating string, 39-40 separation of variables, 14-18 separation theorem, 88 series of functions, continuity of, 3 differentiation of, 3 uniform convergence of, 2 sine integral, 112 sine series, 102 sine transform, 116 singular Sturm-Liouville problems for Bessel's equation, 209 for equation of Euler type, 178 for Legendre's equation, 222 spherical harmonics, 234 additional theorem for, 238 Sturm-Liouville problem, 58 symmetric operator, 67 triangle inequality, 63 trigonometric Fourier series, 95 two dimensional heat equation in a circular domain, 219-221 in a rectangular domain, 218-219 two dimensional wave equation in a circular domain, 212-217 in a rectangular domain, 203-205 uniform convergence of

Index double Fourier series, 196-199 Fourier series, 100-101 integrals, 2 series, 2 uniqueness of solution of heat equation, 131-132, 217-218 of Laplace's equation, 153, 154-155 of ordinary differential equation, 47 of wave equation, 31-32, 37, 203 variation of p a r a m e t e r s , 49-50 vibrating m e m b r a n e , 202-203 vibrating string, 21-22 vibrations of bar, 23 wave equation, 21-22 Weierstrass approximation theorem, 228 Weierstrass M-test for integrals, 3 for series, 3 well-posed problem, 11 Wronskian, 47 zeros of Bessel function, 207-208 Legendre polynomial, 225-226 ordinary differential equation, 50