= Solution
The hydrostatic <buoyancy> anomaly is $b'=f_0\psi_z$, and its leading adiabatic evolution is $D_gb'/Dt+N_0^2w=0$. The small vertical <velocity> advects the background stratification; vertical advection of the perturbation <buoyancy> is higher order. Consequently
$$
\boxed{w=-\frac{D_g}{Dt}\left(\frac{f_0}{N_0^2}\psi_z\right).}
$$
The <impermeability condition> at a stationary lower boundary $z=\alpha y$ is $w=\alpha v$. Using $v_g=\psi_x$ gives the <topographic buoyancy boundary condition for quasi-geostrophic flow>
$$
\frac{D_g}{Dt}\left(\frac{f_0}{N_0^2}\psi_z\right)+\alpha\psi_x=0.
$$
For application at the reference plane, the actual small-height condition is $|\alpha|L/D\ll1$. Slopes scaled by $D/L$ are of Rossby-number order, so this topographic contribution is retained alongside the small ageostrophic vertical <velocity>. The dimensional slope should not be compared to the <Rossby number> without also specifying the aspect ratio.
The basic <quasi-geostrophic streamfunction> is $\overline\psi=-\Lambda zy$. It gives $U=\Lambda z$, basic <buoyancy> gradient $\overline b_y=-f_0\Lambda$, and spatially constant interior <quasi-geostrophic potential vorticity>. Linearizing its conservation law therefore gives
$$
\boxed{(\partial_t+\Lambda z\partial_x)q'=0,\qquad
q'=\psi'_{xx}+\psi'_{yy}+s\psi'_{zz},\quad s=f_0^2/N_0^2.}
$$
The printed equation omits the shear-advection term. This cannot be omitted for general disturbances: an interior anomaly proportional to $\cos[k(x-\Lambda zt)]$ is materially conserved but not independent of time at a fixed location. The intended zero-interior-PV <Eady model> modes do have $q'=0$ and satisfy both expressions; the ensuing mode calculation is valid in that subspace.
At either boundary, linearizing the geostrophic <material derivative> includes $v'\overline\psi_{zy}=-\Lambda\psi'_x$. Let
$$
A_0=\Lambda-\frac{N_0^2\alpha}{f_0},\qquad
A_D=\Lambda-\frac{N_0^2\gamma}{f_0},\qquad
b_0=\psi'_z(0),\quad b_D=\psi'_z(D).
$$
The two boundary equations are
$$
\boxed{\partial_tb_0=A_0\psi'_x(0),\qquad
(\partial_t+\Lambda D\partial_x)b_D=A_D\psi'_x(D).}
$$
These are exactly the respective bottom and top slope corrections to the <rigid-boundary buoyancy condition for quasi-geostrophic waves>.
Assume periodicity horizontally and first consider $q'=0$. Multiply its defining equation by $\psi'_x$ and integrate over the layer. The two horizontal Laplacian terms vanish by <integration by parts>; the remaining volume term after one vertical integration is an $x$-derivative. Therefore
$$
0=\int\psi'_xq'\,dV
=s\int dx\,dy\,[\psi'_x\psi'_z]_{0}^{D}.
$$
For nonzero $A_0,A_D$, the boundary equations give
$$
\frac{d}{dt}\int\frac{b_0^2}{2f_0A_0}\,dx\,dy=\frac1{f_0}\int b_0\psi'_x(0)\,dx\,dy,
\qquad
\frac{d}{dt}\int\frac{b_D^2}{2f_0A_D}\,dx\,dy=\frac1{f_0}\int b_D\psi'_x(D)\,dx\,dy.
$$
The top advection term integrates to zero. Subtracting proves conservation of the <boundary pseudomomentum of a sloping Eady layer>:
$$
\boxed{\frac{d}{dt}\int dx\,dy\left[\frac{\psi_z'^2(D)}{2(\Lambda f_0-N_0^2\gamma)}-\frac{\psi_z'^2(0)}{2(\Lambda f_0-N_0^2\alpha)}\right]=0.}
$$
The original PDF uses squared vertical derivatives and the upper slope $\gamma$ in the first denominator; these details are corrupted in the converted TeX. For arbitrary nonzero interior PV the same calculation instead gives $d\mathcal P/dt=(f_0s)^{-1}\int\psi'_xq'\,dV$, which need not vanish. For example, at an instant take $\psi'=\cos kx+(z/D)^2\sin kx$, independent of $y$; the zonal average of $\psi'_xq'$ is $-ks/D^2$. Thus the boundary-only conservation statement also needs the zero-interior-PV qualification.
For an exponentially growing smooth <normal mode>, this qualification is automatic: the interior equation is $ik(\Lambda z-c)\widehat q=0$, and a <phase speed> with nonzero <imaginary part> cannot equal the real basic <velocity>. Therefore $\widehat q=0$. If $A_0A_D<0$, the conserved boundary <quadratic form> is definite, so a nonzero exponentially growing boundary amplitude is impossible. Instability consequently requires
$$
\boxed{\left(1-\frac{N_0^2\gamma}{f_0\Lambda}\right)\left(1-\frac{N_0^2\alpha}{f_0\Lambda}\right)>0,}
$$
for $\Lambda\ne0$. When one $A$ vanishes, the corresponding boundary derivative is simply advected and must vanish for a growing mode. The other boundary then yields a real single-edge-wave speed; if both vanish, the zero-interior-PV Neumann problem has no nonzero growing mode. These degenerate cases are neutral and should not be handled by dividing by zero.
For equal slopes, put $\zeta=z/D$, $C=c/(\Lambda D)$, $M=N_0|k|D/|f_0|$, and $a=N_0^2\alpha/(f_0\Lambda)-1$. The normal-mode interior equation is $\Phi_{\zeta\zeta}-M^2\Phi=0$ for the zero-interior-PV modes. Thus
$$
\Phi=P\cosh(M\zeta)+Q\sinh(M\zeta),\qquad
(\zeta-C)\Phi_\zeta+a\Phi=0\quad\text{at }\zeta=0,1.
$$
The coefficient equations are
$$
aP-CMQ=0,
$$
$$
[(1-C)M\sinh M+a\cosh M]P+[(1-C)M\cosh M+a\sinh M]Q=0.
$$
Setting their determinant to zero, without dividing by $C$ or $a$, gives
$$
C(1-C)M\sinh M+a\cosh M+\frac{a^2}{M}\sinh M=0,
$$
so the <Eady model with parallel sloping boundaries> has
$$
\boxed{C=\frac12\pm\sqrt{\frac14+a\frac{\coth M}{M}+\frac{a^2}{M^2}}.}
$$
For positive $k,f_0$, $M$ is the printed $\mu$, and $a$ is the printed $\widetilde\alpha$. The expression is even in the sign of that vertical scale. For $a\ge0$ the radicand is at least $1/4$, so both speeds are real and \b[these modes have no exponential growth]. For equal slopes the earlier necessary condition is merely $a^2>0$, which is satisfied in many stable cases and is not sufficient. At $a=0$ the speeds are $0$ and $\Lambda D$; the invariant with nonzero denominators cannot be used, but the determinant still correctly gives neutral modes. No claim excluding transient amplification is needed.
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