 Solution

For constant $N^2$, substitution of the vertical cosine and horizontal plane wave givesThe top rigid-lid condition is satisfied by $\cos (mz)$. To select a discrete vertical mode one must also impose the corresponding no-flow, adiabatic condition at the flat bottom, ${\psi }_z^{\prime }(-H)=0$. It requires $mH=n\pi$. The mode $n=0$ is barotropic; the gravest nonzero baroclinic mode has $m=\pi /H$. HenceThe lid condition alone is insufficient: $\cos [\pi z/(2H)]$ also has zero derivative at the lid but violates the bottom condition. Constant depth motivates the additional physical boundary, which must be used to obtain the printed denominator.

Let $a_d=\pi |f_0|/(HN)=L_d^{-1}$ and set $l=0$. The zonal phase velocity and group velocity areThe phase curve is even in $k$, with limiting value $U-\beta /a_d^2$ at zero and approach to $U$ from below at large $|k|$. The group curve is also even: it starts at the same long-wave limit, equals $U$ at $|k|=a_d$, reaches $U+\beta /(8a_d^2)$ at $|k|=\sqrt{3}{a}_d$, and approaches $U$ from above at large $|k|$. The normalized curves appear in the bottom panels of the preceding figure. For $k=0$, the phase speed means the limit $\omega /k$; there is no zonal crest propagation to define directly at exactly zero wavenumber.