or a sum of such terms, where the allowed k™s are determined by the boundary

conditions.

Example: We have three conducting sheets, each in¬nite in the z direction.

The central one has width a, and is held at voltage V0 . The outer two extend

to in¬nity also in the y direction, and are grounded. The resulting potential

should tend to zero as |x|, |y| ’ ∞.

z

y

a

V0

O x

Conducting sheets.

The voltage in the x = 0 plane is

dk

∞

a(k)e’iky ,

•(0, y, z) = (6.121)

2π

’∞

6.4. LAPLACE™S EQUATION 171

where

2V0

a/2

eiky dy =

a(k) = V0 sin(ka/2). (6.122)

k

’a/2

Then, taking into account the boundary condition at large x, the solution to

2

• = 0 is

∞ dk

a(k)e’iky e’|k||x| .

•(x, y, z) = (6.123)

’∞ 2π

The evaluation of this integral, and ¬nding the charge distribution on the

sheets, is left as an exercise .

The Cauchy Problem is Ill-posed

Although the Laplace equation has no characteristics, the Cauchy data prob-

lem is ill-posed , meaning that the solution is not a continuous function of the

data. To see this, suppose we are given 2 • = 0 with Cauchy data on y = 0:

•(x, 0) = 0,

‚•

= sin kx. (6.124)

‚y y=0

Then

•(x, y) =

sin(kx) sinh(ky). (6.125)

k

Provided k is large enough ” even if is tiny ” the exponential growth of the

hyperbolic sine will make this arbitrarily large. Any in¬nitesimal uncertainty

in the high frequency part of the initial data will be vastly ampli¬ed, and

the solution, although formally correct, is useless in practice.

Eigenfunction Expansions

Elliptic operators are the natural analogues of the one-dimensional linear

di¬erential operators we studied in earlier chapters.

The operator L = ’ 2 is formally self-adjoint with respect to the inner

product

φ— χ dxdy.

φ, χ = (6.126)

This follows from Green™s identity

φ— (’ 2 2

φ)— χ dxdy = {φ— (’ χ) ’ (’ φ)— χ} · nds

χ) ’ (’

„¦ ‚„¦

(6.127)

172 CHAPTER 6. PARTIAL DIFFERENTIAL EQUATIONS

where ‚„¦ is the boundary of the region „¦ and n is the outward normal on

the boundary.

The method of separation of variables also allows us to solve eigenvalue

problems involving the Laplace operator. For example, the Dirichlet eigen-

value problem requires us to ¬nd the eigenfunctions and eigenvalues of the

operator

2

D(L) = {φ ∈ L2 [„¦] : φ = 0, on ‚„¦}.

L=’ , (6.128)

Suppose „¦ is the rectangle 0 ¤ x ¤ Lx , 0 ¤ y ¤ Ly . The normalized

eigenfunctions are

4 nπx mπy

φn,m (x, y) = sin sin , (6.129)

Lx Ly Lx Ly

with eigenvalues

n2 π 2 m2 π 2

»n,m = + . (6.130)

L2 L2

x y

The eigenfunctions are orthonormal,

φn,m φn ,m dxdy = δnn δmm , (6.131)

and complete. Thus, any function in L2 [„¦] can be expanded as

∞

f (x, y) = Anm φn,m (x, y), (6.132)

m,n=1

where

Anm = φn,m (x, y)f (x, y) dxdy. (6.133)

A similar formula will hold for any connected domain „¦ “ only the eigen-

functions may not be so easy to ¬nd!

Polar coordinates

We can use the separation of variables method in polar coordinates. Here,

‚ 2 χ 1 ‚χ 1 ‚2χ

2

χ= 2 + + 2 2. (6.134)

‚r r ‚r r ‚θ

6.4. LAPLACE™S EQUATION 173

Set

χ(r, θ) = R(r)˜(θ). (6.135)

2

Then χ = 0 implies

r2 ‚ 2 R 1 ‚R 1 ‚2˜

0= + +

‚r2 ˜ ‚θ2

R r ‚r

m2 m2

’

= (6.136)

Therefore,

d2 ˜

+ m2 ˜ = 0, (6.137)

2

dθ