Let
$$
c=\frac N{|m|},
\qquad
D=\frac{\omega^2}{c^2}-k^2.
$$
Solving zonal momentum and incompressibility for $\hat u$ and $\hat\phi$ in terms of $\hat v$ gives
$$
\boxed{
\hat u
=\frac{i}{D}
\left(\frac{\beta y\omega}{c^2}\hat v
-k\hat v_y\right)},
$$
$$
\boxed{
\hat\phi
=\frac{i}{D}
\left(k\beta y\hat v-\omega\hat v_y\right)}.
$$
Substitution into meridional geostrophic balance cancels the terms involving $D$ and yields
$$
\boxed{
\hat v_{yy}-\frac{\beta^2y^2m^2}{N^2}\hat v
=\frac{k\beta}{\omega}\hat v}.
$$

Set
$$
\xi=\sqrt{\frac{\beta|m|}{N}}\,y.
$$
The equation becomes
$$
\hat v_{\xi\xi}-\xi^2\hat v
=\frac{kN}{\omega|m|}\hat v.
$$
Using the <Hermite polynomial> eigenvalues gives
$$
\boxed{
\omega_n=-\frac{Nk}{(2n+1)|m|}},
\qquad n=1,2,\ldots,
$$
and
$$
\boxed{
\hat v_n(y)=V_n
H_n(\xi)e^{-\xi^2/2}}.
$$
The associated pressure amplitude is
$$
\boxed{
\hat\phi_n
=\frac{i(k\beta y\hat v_n-\omega_n\hat v_{n,y})}
{\omega_n^2m^2/N^2-k^2}}.
$$
For $n=0$, the denominator used to solve for $\hat u$ and $\hat\phi$ vanishes because $\omega_0^2m^2/N^2=k^2$. The inversion therefore assumed precisely the condition that excludes that degenerate case; it must be analyzed separately and is not a member of this <Equatorial Rossby wave> family.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-333/4/c/solution