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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 14A / c / Solution

Let

g (x, t) = \frac{1}{4 π D t} exp (- \frac{x ^{2}}{4 D t}), t > 0.

Convolution with the initial Dirac delta function translates the heat kernel, while part (b) propagates the impulsive source from time one. Hence

θ_{c} (x, t) = g (x - 2 D, t) - A H (t - 1) g (x + 2 D, t - 1) .

At

(x, t) = (0, 2)

this becomes

θ_{c} (0, 2) = \frac{e ^{- 1/2}}{8 π D} - A \frac{e ^{- 1}}{4 π D} .

It vanishes exactly when

A = \frac{e ^{- 1/2}}{8 π D} \frac{4 π D}{e ^{- 1}} = \frac{e}{2} .

This is the cancellation of two heat-kernel impulses.

Solved by gpt-5.6-sol high.

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