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Solved Problems

Now we prove our assertion. First note that if we substitute f = v∇u in the
divergence theorem,
Z Z
∇ · f dx = f · n ds,
R ∂R
we obtain the identity,

Z Z
∂u
(v∆u + ∇v∇u) dx = v ds. (10.3)
R ∂R ∂n

Let u be a solution of Laplace’s equation subject to the Robin boundary condition
with our restrictions on a. We take v = u in Equation 10.3.

Z Z ∂u Z
2
2
(∇u) dx = u ds = − au ds
R C ∂n C
Since the first integral is non-negative and the last is non-positive, the integrals
vanish. This implies that ∇u = 0. u is a constant. In order to satisfy the boundary

condition where a is non-zero, u must be zero. Thus the unique solution in this
scenario is u = 0.

Exercise 204. Solve Laplace’s equation on the surface of a semi-infinite cylinder
of unit radius, 0 < θ < 2π, z > 0, where the solution, u(θ, z) is prescribed at
z = 0: u(θ, 0) = f(θ).
Solution: The mathematical statement of the problem is

∆u ≡ u + u zz = 0, 0 < θ < 2π, z > 0,
θθ
u(θ, 0) = f(θ).

We have the implicit boundary conditions,

u(0, z) = u(2π, z), u (0, z) = u (0, z)
θ
θ
and the boundedness condition,

u(θ, +∞) bounded.

We expand the solution in a Fourier series. (This ensures that the boundary condi-

tions at θ = 0, 2π are satisfied.)

∞
X inθ
u(θ, z) = u (z) e
n
n=−∞

We substitute the series into the partial differential equation to obtain ordinary
differential equations for the u .
n
2
00
−n u (z) + u (z) = 0
n
n
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