Set $\Delta=0$ and take $\xi_r=Cr^{\ell-1}$. The divergence equation then fixes
$$
\xi_h=\frac{Cr^\ell}{\ell}.
$$
Hydrostatic balance gives $\delta p=-\xi_rp'=\rho g\xi_r$. The horizontal momentum equation becomes
$$
\rho\omega^2\frac{Cr^\ell}{\ell}
=\rho\Omega^2r\,Cr^{\ell-1},
$$
so
$$
\boxed{\omega_\ell^2=\ell\Omega^2}.
$$
The radial equation gives the same result because $\omega^2\xi_r=g\xi_r'+g'\xi_r$. This is an <incompressible stellar surface mode>: it changes the shape of the free surface without compressing fluid elements. For $\ell=1$ it is a rigid displacement of the star in the fixed harmonic potential.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 2 / c / Solution

An incompressible stellar surface mode has vanishing Lagrangian volume perturbation. In a harmonic fixed potential, a displacement generated by a solid spherical harmonic has a frequency fixed by its angular degree.


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