The rigid lower boundary has $w=0$. For a time-dependent normal mode this gives
$$
P'(-H)=0.
$$
The free-surface condition from part i gives
$$
P'(0)+\frac{N^2}{g}P(0)=0.
$$
Writing $q=N/c$, the lower condition selects
$$
P(z)=A\cos[q(z+H)].
$$
The upper condition therefore requires
$$
q\tan(qH)=\frac{N^2}{g}.
$$
Equivalently, with $\mu=NH/c$ and $\epsilon=N^2H/g$,
$$
\boxed{\mu\tan\mu=\epsilon}.
$$
Each positive root $\mu_n$ determines the <barotropic mode> or a <baroclinic mode> through $c_n=NH/\mu_n$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 333 / 3 / iii / Solution

A barotropic mode has horizontal velocity independent of depth. It is the zeroth <quasi-geostrophic vertical mode> and usually has the largest gravity-wave speed and <Rossby deformation radius>.

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 Incoming links: Barotropic mode