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A long disturbance trapped on a depth step is sometimes called a double Kelvin wave. With $H_{+}>H_{-}$, $f>0$ and $c_{\pm }=\sqrt{g{H}_{\pm }}$, cross-step geostrophic balance and the along-step acceleration yield the transport equation shown for a background height jump ${\eta }_0\mathrm{sgn}(x)$. Its zero-initial-amplitude solution is $A={\eta }_0[\mathrm{sgn}(x)-\mathrm{sgn}(x-(c_{+}-c_{-})t)]$. The adjusted region diverts cross-step volume flux along the step; the propagating front still carries flux. This outer approximation does not resolve fast adjustment at the fronts.

Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.

Set $v=0$. For $F_{+}(x+ct)$ the equations require ${\tilde{\varphi }}_{+}=-c{\tilde{u}}_{+}$ and ${\tilde{u}}_{+}^{\prime }=(f/c){\tilde{u}}_{+}$; for $F_{-}(x-ct)$ they require ${\tilde{\varphi }}_{-}=c{\tilde{u}}_{-}$ and ${\tilde{u}}_{-}^{\prime }=-(f/c){\tilde{u}}_{-}$. Hence the channel supports the Kelvin wavesEach wave propagates along one wall and is exponentially trapped toward that wall on the deformation scale $c/|f|$.

Set $c=\sqrt{g{H}_0}$ and seekThe two remaining prognostic equations implyThe $x$ momentum equation is now geostrophic:ThereforeBoundedness for $x>0$ selects $\gamma =\mathrm{sgn}(f)c$ andBecause $F$ depends on $y+\gamma t$, the wave travels with meridional phase speed $-\gamma$. Along this western boundary it travels southward in the Northern Hemisphere and northward in the Southern Hemisphere, keeping the coast on the dynamically required side. This is a coastal Kelvin wave.

Moreover,Since $u=0$, its linear potential-vorticity disturbance is

Potential-vorticity conservation still determines the final surface throughThe unbounded problem only required decay or matching as $|y|$ and $x$ become large. The coast adds a boundary condition. In the adjusted geostrophic flow,so $\eta$ must be constant along the connected wall. Odd symmetry and the arbitrary common height datum set that constant to zero:This is a Dirichlet condition on $\eta$, even though it originated as no normal flow.

The initial disturbance does not instantaneously know this along the whole coast. A southward coastal Kelvin wave for $f>0$ carries the pressure signal and establishes the constant wall height behind its wavefront. In the final streamline sketch, the quasi-geostrophic streamfunction is proportional to ${\eta }_f$. Far from the wall the contours and current resemble the unbounded east--west front; near $x=0$ those contours bend through a right angle and run along the coast, so the incident geostrophic current turns into a southward boundary current instead of crossing the wall.