Suppose $\epsilon=N^2H/g\ll1$. The smallest root satisfies
$$
\mu_0^2\sim\epsilon,
\qquad
c_0=\frac{NH}{\mu_0}\sim\sqrt{gH},
$$
so the leading <external gravity wave> eigenfunction is depth-independent:
$$
P_0(z)\sim1.
$$
For $n\geq1$, the roots lie just above $n\pi$:
$$
\mu_n=n\pi+\frac{\epsilon}{n\pi}+O(\epsilon^2),
$$
and hence
$$
c_n=\frac{NH}{n\pi}
\left[1-\frac{\epsilon}{n^2\pi^2}+O(\epsilon^2)\right].
$$
In particular, up to an arbitrary normalization and sign,
$$
P_1(z)\sim
\cos\left[\frac{\pi(z+H)}{H}\right].
$$
The hierarchy $c_0>c_1>c_2>\cdots$ separates the fast, nearly barotropic free-surface mode from the internal baroclinic modes.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-333/3/iv/solution