Linearizing conservation of <three-dimensional quasi-geostrophic potential vorticity> about rest gives
$$
q'_t+\beta\psi'_x=0.
$$
For $K^2=k^2+l^2$, the plane-wave ansatz yields
$$
-\omega\left[
-K^2\hat\psi+\frac{f_0^2}{N^2}\hat\psi''
\right]+\beta k\hat\psi=0.
$$
Thus the vertical structure equation is
$$
\boxed{
\hat\psi''
-\frac{N^2}{f_0^2}
\left(K^2+\frac{\beta k}{\omega}\right)\hat\psi=0}.
$$
The rigid-boundary eigenfunctions are
$$
\hat\psi_n(z)=\cos\frac{n\pi z}{H},
\qquad n=0,1,2,\ldots.
$$
Substitution gives the <Baroclinic Rossby wave> dispersion relations
$$
\boxed{
\omega_n(k,l)=
-\frac{\beta k}
{k^2+l^2+
\dfrac{f_0^2}{N^2}\left(\dfrac{n\pi}{H}\right)^2}}.
$$
The $n=0$ member is the <Barotropic Rossby wave>; $n\geq1$ are baroclinic vertical modes.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 3 / a / ii / Solution

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A barotropic Rossby wave has no vertical shear. For horizontal wavenumbers $k,l$, its quasi-geostrophic frequency is $\omega=-\beta k/(k^2+l^2)$.


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 Incoming links: Barotropic Rossby wave