 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-1/17b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

p_{1} = p_{2}

, at

z = ζ

\partial_{t} Φ_{i} + \frac{1}{2} ∣\nabla Φ_{i} ∣^{2} + p_{i} / ρ_{i} + g z = C_{i} (t) .

Choose the potential gauge so that

C_{i} = p_{0} / ρ_{i} + U_{i}^{2} /2

, its unperturbed value. Expanding the kinetic term as

U_{i}^{2} /2 + U_{i} \partial_{x} φ_{i}

and evaluating gravity at the displaced interface gives

p_{i} ∣_{z = ζ} = p_{0} - ρ_{i} [\partial_{t} φ_{i} + U_{i} \partial_{x} φ_{i} + g ζ]

to first order. Equality of these pressures gives

ρ_{1} [(\partial_{t} + U_{1} \partial_{x}) φ_{1} + g ζ] = ρ_{2} [(\partial_{t} + U_{2} \partial_{x}) φ_{2} + g ζ] (z = 0) .

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