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The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

The isopycnal displacement isFor the $n$th cosine mode,Its maximum magnitude is thereforeFor $H=3000\mathrm{m}$, $N=10^{-3}{\mathrm{s}}^{-1}$, and $|\eta {|}_{\max }=10^{-2}\mathrm{m}$,The first baroclinic mode therefore displaces interior density surfaces by about $10^2\mathrm{m}$ even though the surface moves only one centimetre.