 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/14a/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write

s = lo g r

, so the two circular boundaries are

s = 0

and

s = 2

. The zero angular mode must interpolate between

- 1

and

1

s - 1 = lo g r - 1

For the

n

th cosine mode, the radial factor has equal value

A_{n}

at both boundaries. The unique harmonic function with those data is

A_{n} \frac{c o s h ( n ( s - 1 ))}{c o s h n} cos (n θ) .

Hence the solution of the annular Dirichlet problem is

ϕ (r, θ) = lo g r - 1 + n = 1 \sum \infty A_{n} \frac{cosh ( n ( lo g r - 1 ))}{cosh n} cos (n θ) .

r = 1

and

r = e^{2}

, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

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