Steady temperature in a semicircular plate

Problem. A semicircular plate of radius a occupies 0 ≤ r ≤ a, 0 ≤ θ ≤ π. The diameter (θ = 0 and θ = π) is at 0°C and the arc r = a is at T°C. Find the steady-state temperature u(r, θ).

Step 1: Laplace's equation in polar form (steady state):

∂²u/∂r² + (1/r) ∂u/∂r + (1/r²) ∂²u/∂θ² = 0

Boundary conditions

1. u(r, 0) = 0
2. u(r, π) = 0
3. u(a, θ) = T
4. u is finite at r = 0

Step 2: separate variables. u = R(r)Θ(θ):

r²R″/R + rR′/R = −Θ″/Θ = λ²

Θ″ + λ² Θ = 0                 →  Θ = A cos λθ + B sin λθ
r² R″ + r R′ − λ² R = 0       →  R = C r^λ + D r^{−λ}

Step 3: apply conditions.

• Θ(0) = 0 gives A = 0.
• Θ(π) = 0 gives sin λπ = 0, so λ = n (integer).
• Finite at r = 0 gives D = 0 (r⁻ⁿ would blow up).

u(r, θ) = Σ_{n=1}^{∞} Aₙ rⁿ sin nθ

Step 4: arc condition. u(a, θ) = T = Σ Aₙ aⁿ sin nθ, a Fourier sine series on (0, π):

Aₙ aⁿ = (2/π) ∫₀^π T sin nθ dθ = (2T/(nπ)) (1 − cos nπ)
      = 4T/(nπ)   for n odd,   0 for n even

So u(r, θ) = Σ_{n odd} (4T/(nπ)) (r/a)ⁿ sin nθ. Write each odd n as 2m − 1 and then rename m as n:

u(r, θ) = (4T/π) Σ_{n=1}^{∞} [1/(2n − 1)] (r/a)^{2n−1} sin(2n − 1)θ

which is the required result. ∎

(Numerical check at r = 0.6a, θ = 1.1: the series gives 0.65669, equal to the closed form (2T/π) arctan[2(r/a) sin θ/(1 − r²/a²)] with T = 1.)