= Solution
Let $D_g=\partial_t+\mathbf u_g\cdot\nabla_h$, $\psi=\widetilde\Phi/f_0$, and $\mathbf u_g=(-\psi_y,\psi_x,0)$. The hydrostatic and thermal equations combine to give
$$
D_g\psi_z=-\frac{N^2}{f_0}w,\qquad \boxed{N^2=\frac{RS(z)}H.}
$$
Taking the vertical <curl> of the horizontal momentum equations to leading quasi-geostrophic order gives
$$
D_g\nabla_h^2\psi+\beta\psi_x+f_0\nabla_h\cdot\mathbf u_{a,h}=0.
$$
The <log-pressure coordinate> continuity equation gives $\nabla_h\cdot\mathbf u_{a,h}=-e^{z/H}\partial_z(e^{-z/H}w)$. Substitute $w=-f_0D_g\psi_z/N^2$. In commuting the vertical derivative with $D_g$, the extra term is $\mathbf u_{g,z}\cdot\nabla_h\psi_z=(-\psi_{yz},\psi_{xz})\cdot(\psi_{xz},\psi_{yz})=0$. Therefore the <log-pressure quasi-geostrophic potential vorticity> equation is
$$
\boxed{D_g\left[\psi_{xx}+\psi_{yy}+e^{z/H}\partial_z\left(e^{-z/H}\frac{f_0^2}{N^2}\psi_z\right)\right]+\beta\psi_x=0.}
$$
The required small parameters have distinct roles. The <Rossby number> $\epsilon_R=U/(|f_0|L)$ makes inertial acceleration small relative to the leading <geostrophic balance>. The parameter $\epsilon_\beta=\beta L/|f_0|$ lets that leading balance use a single $f_0$ while retaining the beta effect in the slower vorticity evolution. Finally, $\psi\sim UL$ and the thermal equation imply $w\sim |f_0|U^2/(N^2D)$. Density-weighted continuity has vertical derivative scale $1/\delta$, where $\delta=\min(D,H)$, so $|\mathbf u_{a,h}|\sim Lw/\delta$. The <vertical-advection consistency criterion for log-pressure quasi-geostrophy> is thus
$$
\boxed{\frac{|\mathbf u_{a,h}|}{U}\sim\epsilon_a=\frac{f_0^2L^2H}{RSD\min(D,H)}\frac{U}{|f_0|L}\ll1.}
$$
It also controls the omitted vertical-advection ratio $wL/(UD)\sim\epsilon_a\delta/D$. Small Rossby number alone would not ensure small <ageostrophic flow> for every choice of stratification and vertical scales.
For constant $N$, linearization about rest gives
$$
\partial_t\left[\nabla_h^2\psi+\frac{f_0^2}{N^2}\left(\psi_{zz}-\frac1H\psi_z\right)\right]+\beta\psi_x=0.
$$
Putting $\psi=\operatorname{Re}[\widehat\psi_c e^{z/(2H)}e^{i(kx+mz-\omega t)}]$ replaces the vertical operator by $-m^2-1/(4H^2)$. Consequently the <Rossby wave in a log-pressure atmosphere> has
$$
\boxed{\omega=-\frac{\beta k}{k^2+(f_0^2/N^2)[m^2+1/(4H^2)]}.}
$$
For the forced half-space, substitution of the specified ansatz gives
$$
\widehat\chi''+m^2\widehat\chi=0,\qquad m^2=-\frac1{4H^2}-\frac{N^2}{f_0^2}\left(k^2+\frac{\beta k}{\omega}\right),\qquad \widehat\chi(0)=\Psi_b.
$$
Assume $\beta>0$ and $k>0$, and define $\omega_*=-\beta k/[k^2+f_0^2/(4N^2H^2)]<0$.
If $\omega>0$ or $\omega<\omega_*$, then $m^2<0$. Boundedness excludes the growing solution and uniquely selects $\boxed{\widehat\chi=\Psi_b e^{-\sqrt{-m^2}\,z}}$. If $\omega=\omega_*$, boundedness excludes the linear-in-$z$ solution, leaving $\widehat\chi=\Psi_b$; this is the zero-vertical-wavenumber threshold.
If $\omega_*<\omega<0$, both oscillatory solutions are bounded. \b[Boundedness alone does not give a unique propagating solution.] With forcing at the bottom and no incident wave from above, impose an upward <radiation condition>. Since
$$
\frac{\partial\omega}{\partial m}=\frac{2\beta k(f_0^2/N^2)m}{[k^2+(f_0^2/N^2)(m^2+1/(4H^2))]^2},
$$
positive $m$ has upward <group velocity>. The outgoing solution is $\boxed{\widehat\chi=\Psi_b e^{imz},\quad m=\sqrt{m^2}>0}$. The physical streamfunction includes the prescribed $e^{z/(2H)}$ density-weighting factor; it is the amplitude $\widehat\chi$ that must remain bounded.
Thus the strict range for vertically propagating waves is
$$
\boxed{-\frac{\beta k}{k^2+f_0^2/(4N^2H^2)}<\omega<0.}
$$
At $\omega=0$, the unforced steady interior equation requires $\beta\psi_x=0$, so a nonzero $k$ bottom streamfunction cannot have a time-independent solution of this ideal rest-state problem without additional dynamics.
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