 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/13e/b/solution

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

Use the principal square root and deform the Bromwich contour around the negative real-axis branch cut. The pole at

p = 0

contributes

a

. The two boundary values at

p = - b

combine to

- a e^{- b t} cos (x b / κ) .

On the cut put

p = - v^{2}

; the jump of

e^{- x p / κ}

- 2 i sin (xv / κ)

and

d p = - 2 v, d v

. Taking the principal value at

v = b

gives

T (x, t) = a [1 - e^{- b t} cos (b / κ x)] + \frac{2 ab}{π} P \int_{0}^{\infty} \frac{e ^{- v^{2} t} sin ( xv / κ )}{v ( v ^{2} - b )} d v .

The expression has the prescribed boundary value and decays into the bar.

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