 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/2/c/solution

Reflecting the solution across

θ = π α

preserves its Dirichlet boundary conditions and outer boundary data, so uniqueness makes the solution reflection-symmetric. Since

sin \frac{n ( 2 π α - θ )}{2 α} = (- 1)^{n + 1} sin \frac{n θ}{2 α},

only odd angular Fourier modes survive. Put

ρ = (r / R)^{1/ (2 α)}

and

ψ = θ / (2 α)

. On the symmetry axis

ψ = π /2

, comparison of the convergent power series gives

A_{2 j + 1} R^{(2 j + 1) / (2 α)} (- 1)^{j} = \frac{1}{( 2 j + 1 )!} .

Thus

Φ (r, θ) = j = 0 \sum \infty \frac{( - 1 ) ^{j} ρ ^{2 j + 1}}{( 2 j + 1 )!} sin ((2 j + 1) ψ) .

This is the imaginary part of

sin (ρ e^{i ψ})

, so a useful closed form is

Φ (r, θ) = cos (ρ cos ψ) sinh (ρ sin ψ) .

At either straight edge

sin ψ = 0

, while on the symmetry axis the formula is

sinh ρ

. Every term is harmonic in the sector, and the factorial denominator gives convergence up to

r = R

, verifying the full solution.

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