 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/16d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Taking the vertical component of the curl of the momentum equation gives

ω_{t} + f \nabla \cdot u = 0.

η_{t} + h_{0} \nabla \cdot u = 0

, and hence

\frac{\partial}{\partial t} (ω - \frac{f}{h _{0}} η) = 0 .

Since initially

u = 0

and

η = η_{0}

ω = \frac{f}{h _{0}} (η - η_{0}) .

Taking the divergence of the momentum equation and using

\nabla \cdot (f \times u) = - f ω

gives

(\nabla \cdot u)_{t} - f ω = - g \nabla^{2} η .

Differentiate continuity in time and substitute the last two identities to obtain

η_{tt} - g h_{0} \nabla^{2} η + f^{2} η = f^{2} η_{0} .

L_{R} = \frac{g h _{0}}{∣ f ∣} .

The even, decaying steady solution of

- L_{R}^{2} η_{\infty}^{''} + η_{\infty} = η_{0}

with continuous value and derivative at

x = \pm a

η_{\infty} (x) = {ϵ [1 - e^{- a / L_{R}} cosh (x / L_{R})], ϵ sinh (a / L_{R}) e^{- ∣ x ∣/ L_{R}}, ∣ x ∣ < a, ∣ x ∣ > a .

u = 0, v = \frac{g}{f} \frac{d η _{\infty}}{d x} .

Thus the flow is parallel to the two edges of the raised strip, in opposite

y

-directions on the two sides; for

f > 0

, it points toward

+ y

on the left and

- y

on the right. Its magnitude is concentrated within a few deformation radii of the edges.

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