Eliminate $\xi_z$ from the coupled system to obtain
$$
v_s^2(\Psi''-k^2\Psi)-g\Psi'+\omega^2\Psi=0,
\qquad
k=\frac{k_x\omega}{\sqrt{\omega^2-4\Omega^2}}.
$$
Writing $\Psi=\psi e^{kz}$ gives
$$
\boxed{
v_s^2(\psi''+2k\psi')-g(\psi'+k\psi)+\omega^2\psi=0}.
$$
For the neutral polytrope found in part (a),
$$
v_s^2=\frac{\gamma p}{\rho}=-\frac{gz}{m}.
$$
If $\psi$ is a polynomial of degree $n$ with leading term $z^n$, the coefficient of the highest power $z^n$ in the ODE is
$$
\omega^2-gk\left(1+\frac{2n}{m}\right).
$$
A polynomial solution therefore requires
$$
\boxed{\omega^2=\left(1+\frac{2n}{m}\right)gk}.
$$
The factor $e^{kz}$ traps every mode below the free surface. The $n=0$ solution is the incompressible surface gravito-inertial or $f$ mode of part (d). The $n\geq1$ solutions are vertically structured <polytropic acoustic modes>[acoustic p modes]; neutral stratification leaves no buoyancy-driven $g$-mode family.

Solution

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

 Solution

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