Expand both the forcing and response in the orthogonal vertical normal modes. Projection onto $P_n$ gives
$$
\partial_t\left(
\nabla_h^2\phi_n-\frac{f_0^2}{c_n^2}\phi_n
\right)+\beta\partial_x\phi_n
=F_n(x,y)H(t).
$$
If the horizontal forcing scale obeys $L_h\gg c_n/f_0$, the <relative vorticity> term $\nabla_h^2\phi_n$ is small compared with the stretching term. The resulting long-wave equation is
$$
\boxed{
\partial_t\phi_n-C_n\partial_x\phi_n
=-\frac{c_n^2}{f_0^2}F_n(x,y)H(t)
},
\qquad
C_n=\frac{\beta c_n^2}{f_0^2}.
$$
Thus each mode communicates the forcing westward at its long <Rossby wave> speed $C_n$, with the barotropic mode fastest because $c_0$ is largest.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 333 / 3 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-333/3/v/solution