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The Fourier series for this function

sin 2θ sin 3θ sin 4θ
θ = 2 sin θ − + − + . . .
2 3 4
Therefore, invoking formulas (10.50)-(10.51),

2 3 4
r sin 2θ r sin 3θ r sin 4θ
u(r, θ) = 2 r sin θ − + − + . . .
2 3 4

is the desired solution.

Example 10.3 Consider a circular plate of radius c m, insulated from above and
0
0
below. The temperature on the circumference is 100 C on half the circle, and 0 C
on the other half.
The differential equation to solve is

2
2
∂ u ∂u ∂ u
ρ 2 + ρ + = 0, (10.52)
∂ρ 2 ∂ρ ∂φ2
with boundary conditions

(
100, if 0 < φ < π,
u(c, φ) = (10.53)
0, if π < φ < 2π.

There is no real boundary in the φ direction, but we introduce one, since we choose to
let φ run from 0 to 2π only. So what kind of boundary conditions do we apply? We
would like to see “seamless behaviour”, which specifies the periodicity of the solution
in φ,
∂u ∂u
u(ρ, φ + 2π) = u(ρ, φ), (ρ, φ + 2π) = (ρ, φ).
∂φ ∂φ
If we choose to put the seem at φ = −π we have the periodic boundary conditions

∂u ∂u
u(ρ, 2π) = u(ρ, 0), (ρ, 2π) = (ρ, 0).
∂φ ∂φ
We separate variables, and take, as usual

u(ρ, φ) = R(ρ)Φ(φ).

This gives the usual differential equations

00
Φ − λΦ = 0,
2
0
00
ρ R + ρR + λR = 0.
Our periodic boundary conditions gives a condition on Φ,

Φ(0) = Φ(2π), Φ(0) = Φ(2π). (10.54)

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