 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/38c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Write the velocity potentials in the upper, middle and lower domains as

ϕ_{+}

ϕ_{0}

and

ϕ_{-}

\nabla^{2} ϕ_{j} = 0, j \in {+, 0, -} .

\nabla ϕ_{+} ⟶ U e_{x} (y \to + \infty), \nabla ϕ_{-} ⟶ U e_{x} (y \to - \infty) .

Every shear layer is a material interface. The exact kinematic boundary condition at

y = h + η

ϕ_{+, y} = η_{t} + ϕ_{+, x} η_{x}, ϕ_{0, y} = η_{t} + ϕ_{0, x} η_{x},

and at

y = - h - η

it is

ϕ_{-, y} = - η_{t} - ϕ_{-, x} η_{x}, ϕ_{0, y} = - η_{t} - ϕ_{0, x} η_{x} .

Here each equation is evaluated on the indicated moving interface.

ϕ_{+, t} + \frac{1}{2} ∣\nabla ϕ_{+} ∣^{2} = ϕ_{0, t} + \frac{1}{2} ∣\nabla ϕ_{0} ∣^{2} (y = h + η),

ϕ_{-, t} + \frac{1}{2} ∣\nabla ϕ_{-} ∣^{2} = ϕ_{0, t} + \frac{1}{2} ∣\nabla ϕ_{0} ∣^{2} (y = - h - η),

after absorbing spatially constant functions of time into the potentials.

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