Insert
$$
p'=p(z)\sin(n\pi y)e^{i\alpha(x-ct)}.
$$
The side-wall condition is automatic, and the potential-vorticity equation gives
$$
p''-B(\alpha^2+n^2\pi^2)p=0.
$$
Define
$$
\lambda=\frac12\sqrt{B(\alpha^2+n^2\pi^2)}.
$$
Then
$$
p=A\cosh[2\lambda(z-	frac12)]
+C\sinh[2\lambda(z-	frac12)].
$$
The top and bottom conditions become
$$
-cp'(0)=p(0),
\qquad
(1-c)p'(1)=p(1).
$$
Setting the determinant of these two homogeneous equations for $A,C$ to zero and simplifying gives
$$
\boxed{
c=\frac12\pm
\frac1{2\lambda}
\sqrt{(\lambda\coth\lambda-1)
(\lambda\tanh\lambda-1)}}.
$$


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-331/3/f/solution