Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 3 / f

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- Solution f

Solution

Insert

p^{'} = p (z) sin (nπ y) e^{i α (x - c t)} .

The side-wall condition is automatic, and the potential-vorticity equation gives

p^{''} - B (α^{2} + n^{2} π^{2}) p = 0.

λ = \frac{1}{2} B (α^{2} + n^{2} π^{2}) .

Then

p = A cosh [2 λ (z - f r a c 12)] + C sinh [2 λ (z - f r a c 12)] .

The top and bottom conditions become

- c p^{'} (0) = p (0), (1 - c) p^{'} (1) = p (1) .

Setting the determinant of these two homogeneous equations for

A, C

to zero and simplifying gives

c = \frac{1}{2} \pm \frac{1}{2 λ} (λ coth λ - 1) (λ tanh λ - 1) .

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