 Solution

For nonzero meridional wavenumber,At least one growing zonally varying mode can exist only ifThe upper value is a supremum approached as $k\to 0$, where the growth itself tends to zero. Thus not every permitted meridional wavenumber has the same fastest-growing zonal scale; nonzero $l$ restricts the available band of $k$ and weakens the growth factor $k/\sqrt{k^2+l^2}$ at fixed $\eta$.

Using $N=\sqrt{N^2}$, not the tabulated squared value itself, givesThe planetary radii put these into angular context: $l_{\sup }R$ is about 280 for Earth and 149 for Jupiter, corresponding to minimum meridional angular wavelengths about $1.28^{\circ }$ and $2.42^{\circ }$, respectively. They are reciprocal-length wave numbers; the dimensionless angular counts should not be substituted for $l$ in the Cartesian dispersion relation. If a channel admits a half-wave across its width, the corresponding minimum width is $\pi /l_{\sup }$ rather than a full wavelength.

The oceanic scale is mesoscale, while the much larger deformation length makes the Jovian modes thousands of kilometres across. Both are small fractions of the planetary radius, consistent with using a local patch as a first approximation. Real boundaries, planetary-vorticity gradients, varying stratification and jets can change these limits; this simple model alone does not predict the complete observed flow pattern.

Finally, the source's phrase $c^2<0$ must be interpreted in the mean-flow frame. The actual criterion is $D<0$, or $(c-\Lambda H/2{)}^2<0$. In the original frame $c$ has nonzero real part, so $c^2$ is generally complex, not an ordered negative real number. For example, at $\eta =1$ and $\Lambda H=1$, $c=0.5+0.2511i$ and $c^2\simeq 0.1870+0.2511i$. This is an unstable wave even though the literal printed inequality is inapplicable.