Initially $P=-f\phi_0\operatorname{sgn}(x)/c^2$. In the final steady state,
$$
u=-\frac1f\Phi_y,
\qquad v=\frac1f\Phi_x.
$$
Equating its potential vorticity to the initial value gives
$$
\boxed{\Phi_{xx}+\Phi_{yy}-\frac{f^2}{c^2}\Phi
=-\frac{f^2}{c^2}\phi_0\operatorname{sgn}(x).}
$$
This is <geostrophic adjustment>: inertia-gravity waves remove the unbalanced part. Their group speed is at most $c$, so at any finite time only a region whose distance from the initial jump is $O(ct)$ can have reached the steady state.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 333 / 1 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-333/1/iv/solution