For an incompressible perturbation, $\Delta=0$, so $\Psi=g\xi_z$. The vertical equation gives
$$
\frac{d\xi_z}{dz}=\frac{\omega^2}{g}\xi_z.
$$
Incompressibility together with the horizontal equation also gives
$$
\frac{d\xi_z}{dz}
=\frac{gk_x^2}{\omega^2-4\Omega^2}\xi_z.
$$
Equating the two expressions yields
$$
\omega^2(\omega^2-4\Omega^2)=g^2k_x^2.
$$
The positive-frequency branch is the <surface gravito-inertial wave>
$$
\boxed{\omega^2
=2\Omega^2+\sqrt{(gk_x)^2+(2\Omega^2)^2}}.
$$
Its vertical displacement decays as $e^{\omega^2z/g}$ into the atmosphere $z<0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-314/2/d/solution