= Solution
For an exponentially growing <normal mode> with $k\ne0$, $\operatorname{Im}(kc)>0$. Such a complex $c$ cannot equal the real basic velocity anywhere, so the interior equation forces $\widehat q=0$ at every height. Decay then fixes the unique vertical structure $Ce^{-mz}$, and the lower-boundary condition fixes $c=-S/m$, which is real. This contradiction proves the <absence of exponential instability in the semi-infinite Eady model>:
$$
\boxed{\operatorname{Im}\omega=0\quad\text{for the regular decaying edge-wave modes}.}
$$
For $k=0$, there is no advective frequency in the linear interior equation, so an exponentially growing mode would again have $q'=0$. The wall buoyancy equation then forces $\psi'_z(0)=0$, which the nonzero decaying exponential cannot satisfy. Thus these modes do not provide a missing growing branch. This conclusion rules out exponential eigenmode growth, not every possible transient response or singular neutral <potential vorticity> disturbance.
With a second rigid boundary at $z=D$, the same uniform-shear flow is the <Eady model>. Both boundaries carry <buoyancy perturbations> and support <Boundary Rossby waves>. Their intrinsic propagation directions oppose one another relative to their respective local basic flows. At suitable wavelengths their interacting fields can phase-lock and release basic-state <potential energy>, producing <baroclinic instability>. The single-boundary system lacks that second interacting boundary wave.
For a direct comparison, write $\Lambda=\bar U_z=-S$ and $m=NK/|f_0|$. The vertical solution between the two boundaries is $A\cosh(mz)+B\sinh(mz)$, and their buoyancy conditions are
$$
(\Lambda z-c)\widehat\psi'(z)-\Lambda\widehat\psi(z)=0\qquad(z=0,D).
$$
Eliminating $A,B$ gives the <finite-depth Eady dispersion relation>
$$
\boxed{\left(c-\frac{\Lambda D}{2}\right)^2=\frac{\Lambda^2}{m^2}\left[\frac{(mD)^2}{4}-(mD)\coth(mD)+1\right].}
$$
The bracket is negative for $0<mD<2.399\ldots$, so those modes with $k\ne0$ have a growing and a decaying member. At short wavelengths the two edge waves interact weakly and the frequencies are real. In the semi-infinite limit, the lower-wave root tends to $c=\Lambda/m=-S/m$, recovering the neutral result above.
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