Put $v=0$ and use amplitudes proportional to $e^{i(kx+mz-\omega t)}$. Zonal momentum gives
$$
\hat u=\frac{k}{\omega}\hat\phi,
$$
while incompressibility, buoyancy evolution, and hydrostatic balance give
$$
\hat w=-\frac{k}{m}\hat u
=-\frac{\omega m}{N^2}\hat\phi.
$$
Hence
$$
\omega^2=\frac{N^2k^2}{m^2}.
$$
Meridional geostrophic balance requires
$$
\frac{d\hat\phi}{dy}
=-\frac{\beta k}{\omega}y\hat\phi,
$$
so
$$
\boxed{
\hat\phi(y)=\Phi_0
\exp\left(-\frac{\beta k}{2\omega}y^2\right)}.
$$
Decay as $|y|\to\infty$, together with $\beta>0$ and $k>0$, requires $\omega>0$. Therefore
$$
\boxed{\omega=\frac{Nk}{|m|}},
\qquad
\boxed{
\hat\phi(y)=\Phi_0
\exp\left(-\frac{\beta|m|}{2N}y^2\right)}.
$$
The negative-frequency root makes the Gaussian exponent positive and the solution diverge away from the equator. The acceptable branch is the eastward <Equatorial Kelvin wave>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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