 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/39b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

η (k, t) = \int_{- \infty}^{\infty} η (x, t) e^{- ik x} d x, η (x, t) = \frac{1}{2 π} \int_{- \infty}^{\infty} η (k, t) e^{ik x} d k .

\partial_{t}^{2} η + S^{2} ∣ k ∣^{3} η = 0.

The transformed initial data are

η (k, 0) = 0, \partial_{t} η (k, 0) = - W \int_{- ϵ}^{ϵ} e^{- ik x} d x = - \frac{2 W s i n ( k ϵ )}{k} .

Therefore

η (k, t) = - \frac{2 W s i n ( k ϵ )}{k S ∣ k ∣ ^{3/2}} sin (S ∣ k ∣^{3/2} t) .

The integrand is even, so Fourier inversion gives the surface response to a localized impulsive velocity

η (x, t) = - \frac{2 W}{π S} \int_{0}^{\infty} \frac{sin ( k ϵ ) sin ( S k ^{3/2} t ) cos ( k x )}{k ^{5/2}} d k .

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