 Solution

ID: past-exam-of-the-waves-and-compressible-flow-course-of-the-university-of-oxford/2023/2/c/i/solution

W (k, z, t) = \int_{R} W (x, z, t) e^{- ik x} d x

, with inverse factor

1/ (2 π)

. The initial profile occupies only the vertical mode

sin (π z / h)

. Its transform is

W (k, z, 0) = πh C e^{- ϵ h ∣ k ∣} sin (π z / h), W_{t} (k, z, 0) = 0.

k

, with

ω (k) = \frac{αg k}{k ^{2} + π ^{2} / h ^{2} + α ^{2} /4} .

Therefore inversion gives

W (x, z, t) = \frac{h C}{2} sin (\frac{π z}{h}) \int_{- \infty}^{\infty} e^{- ϵ h ∣ k ∣} cos (ω (k) t) e^{ik x} d k .

Using the nonnegative even frequency instead gives the same cosine. At

t = 0

\int_{R} e^{- a ∣ k ∣} e^{ik x} d k = 2 a / (a^{2} + x^{2})

with

a = ϵ h

, recovering the specified initial profile and its zero time derivative.

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