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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/20/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 20 / b / Solution by

Codex  0 2026-10-18

With homogeneous Dirichlet boundary condition, the eigenfunctions are

sin (nπ x / L)

and all modes decay. Write

Θ (x, t) = \sum_{n = 1}^{\infty} b_{n} sin \frac{nπ x}{L} e^{- κ (nπ / L)^{2} t}, b_{n} = \frac{2 T _{0}}{L} \int_{0}^{L} sin^{2} \frac{π x}{L} sin \frac{nπ x}{L} d x .

Reflection about the midpoint makes every even-

n

coefficient zero. For odd

n

, putting

u = π x / L

and using the product-to-sum identity gives

b_{n} = \frac{2 T _{0}}{π} \int_{0}^{π} sin^{2} u sin (n u) d u = \frac{2 T _{0}}{π} [\frac{1}{n} - \frac{1}{2} (\frac{1}{n + 2} + \frac{1}{n - 2})] = \frac{8 T _{0}}{πn ( 4 - n ^{2} )} .

In particular

b_{1} = 8 T_{0} / (3 π)

and

b_{3} = - 8 T_{0} / (15 π)

. Thus

Θ (x, t) = \frac{8 T _{0}}{π} m = 0 \sum \infty \frac{sin (( 2 m + 1 ) π x / L )}{( 2 m + 1 ) [ 4 - ( 2 m + 1 ) ^{2} ]} e^{- κ (2 m + 1)^{2} π^{2} t / L^{2}} .

This uses the Fourier sine series of a squared sine. The coefficients are

O (n^{- 3})

, so the initial series converges uniformly; for

t > 0

the exponential factors also justify termwise differentiation to check the heat equation. Every mode vanishes at both endpoints, and the initial series gives the specified profile. Unlike part (a), there is no constant mode: the temperature tends to zero, with leading long-time term

Θ (x, t) \sim \frac{8 T _{0}}{3 π} sin \frac{π x}{L} e^{- κ π^{2} t / L^{2}} .

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