= Solution
For constant $N^2$, substitution of the vertical cosine and horizontal plane wave gives
$$
q'=-\left[k^2+l^2+\frac{f_0^2m^2}{N^2}\right]\psi',\qquad
\boxed{\omega=Uk-\frac{\beta k}{k^2+l^2+f_0^2m^2/N^2}.}
$$
The 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, $\psi'_z(-H)=0$. It requires $mH=n\pi$. The mode $n=0$ is barotropic; the gravest nonzero baroclinic mode has $m=\pi/H$. Hence
$$
\boxed{\omega=Uk-\frac{\beta k}{k^2+l^2+\pi^2f_0^2/(H^2N^2)}.}
$$
The 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> are
$$
\boxed{c_{px}=U-\frac{\beta}{k^2+a_d^2},\qquad
c_{gx}=\frac{\partial\omega}{\partial k}=U+\beta\frac{k^2-a_d^2}{(k^2+a_d^2)^2}.}
$$
The 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 $|k|=\sqrt3a_d$, 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.
Back to article page
