 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/9b/c/ii/solution

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

The radial equation is

(r^{2} ϕ^{'})^{'} = - \frac{1}{r} .

Its solutions satisfying decay at infinity have the form

ϕ (r) = \frac{l o g r + 1 + C}{r} .

The boundary value

ϕ (1) = 1

gives

C = 0

, so

ϕ (r) = \frac{1 + lo g r}{r} .

It tends to zero as

r \to \infty

and directly satisfies

\nabla^{2} ϕ = - r^{- 3}

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