 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2/7d/solution

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

f (p) = \int_{0}^{\infty} e^{- pt} f (t), d t

, and inversion is

f (t) = \frac{1}{2 πi} \int_{c - i \infty}^{c + i \infty} e^{pt} f (p), d p,

where the vertical Bromwich contour lies to the right of all singularities. For

t > 0

it is closed leftwards, and for

t < 0

rightwards, where the exponential decays.

Transforming the PDE and imposing boundedness as

x \to \infty

gives

U (x, p) = \frac{1}{p ( p + 1 )} + \frac{p - 1}{p ( p + 1 )} e^{- x p / D},

with the square root cut conventionally along

(- \infty, 0]

. Therefore

u (x, t) = \frac{1}{2 πi} \int_{c - i \infty}^{c + i \infty} e^{pt} U (x, p), d p .

There are poles at

p = 0, - 1

(with cancellation assessed in the complete expression) and a branch point at

0

; close left for

t > 0

and wrap the cut while including the relevant residues.

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