Set $\delta\mathbf u=-i\omega\boldsymbol\xi$. The continuity and adiabatic equations give the Lagrangian perturbations
$$
\frac{\Delta_L\rho}{\rho}=-\Delta,
\qquad
\frac{\Delta_Lp}{p}=-\gamma\Delta,
\qquad
\Delta=ik_x\xi_x+\frac{d\xi_z}{dz}.
$$
Because the equilibrium is a <neutrally stratified polytropic atmosphere>, $p\propto\rho^\gamma$, so the displacement terms cancel and
$$
\boxed{\frac{\delta p}{p}=\gamma\frac{\delta\rho}{\rho}}.
$$
Using $dp/dz=-\rho g$ then gives
$$
\boxed{\Psi=\frac{\delta p}{\rho}
=g\xi_z-v_s^2\Delta},
\qquad
v_s^2=\frac{\gamma p}{\rho}.
$$

Eliminating $\xi_y$ from the two horizontal momentum equations and combining gravity with the vertical pressure force yields
$$
\boxed{-(\omega^2-4\Omega^2)\xi_x=-ik_x\Psi},
\qquad
\boxed{-\omega^2\xi_z=-\frac{d\Psi}{dz}}.
$$
Thus
$$
\xi_x=\frac{ik_x\Psi}{\omega^2-4\Omega^2},
$$
and the requested coupled first-order system is
$$
\boxed{
\frac{d\Psi}{dz}=\omega^2\xi_z,
\qquad
\frac{d\xi_z}{dz}
=\frac{g}{v_s^2}\xi_z
+\left(\frac{k_x^2}{\omega^2-4\Omega^2}
-\frac1{v_s^2}\right)\Psi}.
$$

Solution

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

 Solution

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