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Solved Problems

sinh(nπy)
u(x, y) = sin(nπx)
sinh(nπ)

For = 0 the solution is u = 0. Now consider any > 0. Note that |u| ≤ for
(x, y) ∈ [0 . . . 1] × [0 . . . 1]. The solution depends continuously on the given
boundary data. This problem is well posed.

Exercise 201. Use the fundamental solutions for the Laplace equation

2
∇ G = δ(x − ξ)

in three dimensions
1
G(x|ξ) = −
4π|x − ξ|

to derive the mean value theorem for harmonic functions

Z
1
u(p) = u(ξ) dA ,
ξ
4πR 2
∂S R
that relates the value of any harmonic function u(x) at the point x = p to the
average of its value on the boundary of the sphere of radius R with center at p,
(∂S ).
R
Solution: The Green function problem for a sphere of radius R centered at the
point ξ is

∆G = δ(x − ξ), G = 0. (10.1)
|x−ξ|=R
We will solve Laplace’s equation, ∆u = 0, where the value of u is known on the
boundary of the sphere of radius R in terms of this Green function.

First we solve for u(x) in terms of the Green function.

Z Z
(u∆G − G∆u) dξ = uδ(x − ξ) dξ = u(x)
S S

∂G ∂u
Z Z
(u∆G − G∆u) dξ = u − G dA ξ
S ∂S ∂n ∂n
Z
∂G
= u dA ξ
∂S ∂n
Z
∂G
u(x) = u dA ξ
∂S ∂n
We are interested in the value of u at the center of the sphere. Let ρ = |p − ξ|

Z
∂G
u(p) = u(ξ) (p|ξ) dA ξ
∂S ∂ρ

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