 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-3/11b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Since

q = - κ \nabla T

and

\nabla \cdot q = H

\oint_{S} q \cdot d S = \int_{V} H d V .

(r^{2} κ T^{'})^{'} = - r^{2} H

. Regularity at zero and decay at infinity yield

T (r) = ⎩ ⎨ ⎧ \frac{H _{0}}{3 ( α + 1 )} + \frac{H _{0}}{3 ( 2 - α )} (1 - r^{2 - α}), \frac{H _{0}}{3 ( α + 1 )} r^{- α - 1}, r \leq 1, r > 1.

For

α \leq - 1

, the exterior solution cannot tend to zero: it is logarithmic at

- 1

and grows below it.

On the exterior domain, the energy-minimizing solution with boundary value one is

ϕ = r^{- α - 1}

. Applying the Dirichlet principle to any admissible

w

gives

\int_{1}^{\infty} r^{α + 2} (w^{'})^{2} d r \geq \int_{1}^{\infty} r^{α + 2} (ϕ^{'})^{2} d r = α + 1.

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