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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ib/paper-4/17b/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 4 / 17B / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

The insulated boundary imposes the Neumann boundary condition

T_{x_{1}} (0, x_{2}, t) = 0

. Use the method of images by reflecting the initial data evenly across

x_{1} = 0

. At the origin, the original sector and its reflection contribute equally. Using the given two-dimensional heat kernel,

T (0, 0, t) = \frac{2 T _{0}}{4 π K t} \int_{- π /4}^{π /4} \int_{1}^{2} e^{- r^{2} / (4 K t)} r d r d θ = \frac{T _{0}}{2} (e^{- 1/ (4 K t)} - e^{- 1/ (K t)}) .

Differentiation gives

\frac{d T}{d t} = \frac{T _{0}}{8 K t ^{2}} (e^{- 1/ (4 K t)} - 4 e^{- 1/ (K t)}) .

This vanishes exactly when

e^{3/ (4 K t)} = 4

. Its sign is positive before that time and negative afterwards. Moreover

T \to 0

t ↓ 0

and as

t \to \infty

. Therefore this is the unique global maximum:

t_{m a x} = \frac{3}{8 K lo g 2}, T_{m a x} = \frac{3 T _{0}}{8 4 ^{1/3}} .

The factor two from the insulated-boundary reflection is essential; applying the full-plane heat kernel only to the unreflected sector would give half the temperature.

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