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The shallow-water approximation requires depth much less than horizontal length, a homogeneous incompressible fluid, gentle surface/bottom slopes, and frequencies slow enough that vertical acceleration is negligible compared with gravity. Continuity gives $w\sim (H/L)U$, making vertical inertial acceleration smaller than the horizontal inertial scale by the aspect ratio. The vertical momentum equation therefore reduces to $p_z=-{\rho }_{0}g$. With constant atmospheric pressure,The horizontal pressure gradient is independent of depth. Linearizing about rest and retaining the leading depth-uniform horizontal motion givesTheir horizontal curl gives ${\zeta }_t=-f(u_x+v_y)$, soHere $\zeta =v_x-u_y$ is relative vorticity; the PDF sentence identifying $\eta$ as relative vorticity is a symbol error. This is the linear shallow-water potential vorticity anomaly; advection of that perturbation is second order.

In the adjusted state take alongshore independence, decay offshore, and an impermeable coast. Steady continuity gives $u=0$, and geostrophic balance gives $v=g{\eta }_x/f$. Assume $f>0$ so the printed $a=\sqrt{gH}/f$ is a positive barotropic deformation radius. For either hemisphere the decay length is $\sqrt{gH}/|f|$, with the corresponding change in current direction. Initial rest makes the conserved anomaly $-{\eta }_{\mathrm{initial}}/H$, and thereforeThe coastal adjustment of an elevated strip uses a decaying offshore solution and a particular solution in the elevated strip are matched with continuous $\eta$ and $v$, hence continuous ${\eta }_x$, at $x=-L$. Integrating this equation over $x<0$ and conserving the initial volume per alongshore length, ${\int }_{-\infty }^0\eta dx={\eta }_{0}L$, gives $a^2{\eta }_x(0)=0$. Solving those matching conditions yieldsExpanding the hyperbolic functions gives exactly the alternative summed form in the PDF. Both matching values at $-L$ are ${\eta }_0(1-e^{-2L/a})/2$, and the wall elevation is ${\eta }_0(1-e^{-L/a})$.

The adjusted flow isThus both velocity components vanish at the coast. In this inviscid problem tangential no-slip is not generally an independently imposed wall condition; here it follows from initial rest and alongshore uniformity. Indeed $v_t+fu=0$ at the impermeable wall keeps $v(0,t)=0$. The mass constraint recovers the same adjusted-state condition.

For $L\gg a$, the surface retains nearly its initial elevation through most of the strip. The offshore edge is smoothed over width of order $a$, with half-height at the original edge, while the coast remains at nearly ${\eta }_0$. The current is concentrated near the smoothed edge and vanishes at the coast.

For $L\ll a$, the elevation spreads over a much larger offshore scale $a$, with wall height approximately ${\eta }_{0}L/a$. Away from the thin original strip, $\eta \simeq {\eta }_0(L/a)e^{x/a}$. Within it the leading elevation is nearly that reduced constant, with a small curvature required to bring $v$ to zero at the wall. Wide strips retain a broad high plateau; narrow strips spread into a low deformation-scale coastal bulge. The total anomalous volume remains ${\eta }_{0}L$ in both limits.

Use small Rossby number $U/(f_{0}l)\ll 1$, slow evolution on the advective time scale, small interface displacements relative to each layer depth, shallow hydrostatic layers, stable reduced gravity $g^{\prime }>0$, and inviscid unforced flow. On a beta plane, take $\beta l/f_0$ of the same small order as the Rossby number. The internal Burger number is retained at order unity so stratification and relative-vorticity effects can both enter the leading potential-vorticity anomaly.

The rigid-lid pressure in two-layer flow comes from neglecting the free-surface volume displacement in the rigid-lid approximation; it does not permit setting the common horizontal pressure gradient to zero. The small surface displacement multiplied by $g$ retains a finite lid-pressure multiplier. Write $h_1=H_1+\chi$, $h_2=H_2-\chi$, and let $\Pi$ denote this common pressure potential. Leading geostrophic balance giveswith ${\mathbf{u}}_i=(-{\psi }_{iy},{\psi }_{ix})$ at leading order. Literally setting $h_1+h_2$ to a spatial constant in the momentum gradients before taking the rigid-lid limit would suppress the upper-layer pressure field and fail to produce general two-layer QG dynamics.

Taking curl of each shallow-water momentum equation and using layer continuity gives material conservation of $(f+{\zeta }_i)/h_i$. Expanding it to first order and advecting the anomaly by the leading geostrophic velocity yieldsThe two-layer quasi-geostrophic potential vorticity equations areHere the common background $f_0/H_i$ has been removed and the anomaly multiplied by $H_i$. The advection term is retained at the same slow order as the time derivative even though the leading velocity/pressure balance was linear geostrophy. On an $f$ plane simply set $\beta =0$.

Take $\mathbf{f}=f\hat{\mathbf{z}}$ with $f>0$, and put $\zeta =v_x-u_y$, $d=u_x+v_y$ and $q=\zeta -f\eta /H$. Here $q$ denotes the signed vertical component of the linearized shallow-water potential-vorticity anomaly, rather than its magnitude. The divergence and vertical curl of the linearized shallow water equations giveConsequently $q_t=0$. Eliminating $d$ and substituting $\zeta =q+f\eta /H$ givesThe vector forcing is $-H\mathbf{f}\cdot \mathbf{q}$. This is potential-vorticity conservation in its linear, f-plane form: the conserved anomaly forces a stationary balanced part, while the homogeneous equation supports inertial-gravity waves.

For the initial strip, differentiating the discontinuous velocity in the sense of distributions givesThese are two oppositely signed vortex sheets. In the final geostrophic balance, $v_f=0$ and $u_f=-g{\eta }_f^{\prime }/f$. The height therefore solvesThis is the Rossby deformation radius. Decay at infinity fixes the Green function to $G(y)=-\lambda e^{-|y|/\lambda }/2$. Thus the geostrophic adjustment of a finite-width current has the particularly useful representationExpanding the exponentials inside the strip gives ${\eta }_f/H=-(U/c)e^{-a/\lambda }\sinh (y/\lambda )$; above the strip it gives $-(U/c)e^{-y/\lambda }\sinh (a/\lambda )$, and below it gives $(U/c)e^{y/\lambda }\sinh (a/\lambda )$. The height is continuous, odd and exponentially localized. Its derivative has the jumps required by the two Dirac delta functions.

Differentiating the height, rather than assuming a uniform final current, gives the complete velocity:The one-sided velocity jump is $U$ at each edge, with opposite orientations. The value exactly on an idealized vortex sheet is immaterial. Inside the strip the current remains in the original direction; outside it a return current develops.

Let $r=a/\lambda$ and $s=y/\lambda$. For $r\ll 1$, $u_f/U=1-r+O(r^2)$ over the narrow strip, while the outside return flow is approximately $-r{e}^{-|s|}$. The height varies almost linearly across the strip, $c{\eta }_f/(HU)\simeq -s$, and has extrema of magnitude $(1-e^{-2r})/2\simeq r$ at its edges. For $r\gg 1$, the central current is exponentially small: $u_f(0)/U=e^{-r}$. Each edge supports a layer of width $\lambda$, with $u_f/U\simeq \frac{1}{2}{e}^{-(a-|y|)/\lambda }$ on its inner side and $-\frac{1}{2}{e}^{-(|y|-a)/\lambda }$ outside. The corresponding height extrema approach $\pm HU/(2c)$, with almost zero height deep inside and far outside. Both requested profiles are drawn from the exact functions below; the dashed jumps represent one-sided velocity limits.

For balanced shallow-water energy as a signed potential-vorticity pairing, divide out the constant density and work per unit distance in $x$. The kinetic energy plus surface potential energy isMultiplication of the stationary height equation by ${\eta }_f$ and integration by parts, with vanishing end terms, yieldsThe same expression holds with $dA$ for a finite two-dimensional domain or a finite periodic length in $x$. The infinite strip's total energy is infinite, so both energies and their ratio are understood per unit $x$-length.

There is a sign defect in the printed energy expression: $|\mathbf{q}|$ must be replaced by the signed component $\hat{\mathbf{f}}\cdot \mathbf{q}$. Indeed, $|q_0|=|U|[\delta (y-a)+\delta (y+a)]$, so pairing it with the odd ${\eta }_f$ gives zero, although the balanced state has positive energy. The signed pairing givesSince $E_i=H{U}^{2}a$, the energy retention in finite-width geostrophic adjustment isFor $r\ll 1$ this ratio is $1-r+\frac{2}{3}{r}^2+O(r^3)$: nearly all energy remains in the narrow balanced current. For $r\gg 1$ it is approximately $1/(2r)$: only the edge regions retain balanced energy. Conservation of energy still holds for the inviscid evolution. The missing balanced energy is carried away by inertia-gravity waves; “final state” means the local balanced limit after those waves leave, rather than a loss of the total energy over the entire infinite domain.

On a sphere of radius $R$ rotating at angular speed $\Omega$, the Coriolis parameter is $f=2\Omega \sin \varphi$. Near latitude ${\varphi }_0$, write $y=R(\varphi -{\varphi }_0)$, use locally eastward and northward Cartesian coordinates, and expandThe beta plane retains the first northward variation of planetary vorticity while neglecting higher latitude dependence and metric curvature. A midlatitude local calculation assumes $L/R\ll 1$ and $|\beta |L/|f_0|\ll 1$; near the equator the distinct equatorial beta plane has $f_0=0$, so the midlatitude low-frequency reduction used below does not apply.

For a homogeneous shallow layer, the hydrostatic approximation gives $p=p_{\mathrm{atm}}+\rho g(\eta -z)$ and hence $p_x/\rho =g{\eta }_x$, $p_y/\rho =g{\eta }_y$, independently of depth. The approximation follows from the small aspect ratio $H/L$ and neglect of vertical acceleration. Differentiate the linear horizontal momentum equations in $z$. Their depth derivatives obeyStarting from rest gives $u_z=v_z=0$ initially, and the unique solution remains zero. This establishes depth independence of hydrostatic shallow-water flow; an arbitrary pre-existing shear would not be removed merely by taking the hydrostatic approximation. Integrating the continuity equation between the rigid bottom and the moving surface gives ${\eta }_t+H(u_x+v_y)=0$ at linear order. With the depth-integrated shallow-water transports $U=Hu$, $V=Hv$, one obtains

It is useful to make the coefficient approximation in the height reduction explicit. Put ${\mathcal{F}}_y={\partial }_t^2+f(y{)}^2$. Differentiating the two momentum equations in time and eliminating the other transport givesSince ${\partial }_y({\mathcal{F}}_{y}V)={\mathcal{F}}_y{V}_y+2f\beta V$, taking their horizontal divergence and using the continuity equation givesApplying ${\mathcal{F}}_y$ proves the exact variable-Coriolis shallow-water height equationAt the reference latitude, or after the usual local freezing of undifferentiated $f$ factors to $f_0$, this gives the height relation written with ${\mathcal{F}}_0={\partial }_t^2+f_0^2$. With $f=f_0+\beta y$ over a finite region, that constant-coefficient version is a local approximation, not an exact identity. Keeping ${\mathcal{F}}_y$ as above avoids silently commuting a variable Coriolis parameter through a spatial derivative.

For the slow Rossby wave branch, take $|\omega |/|f_0|\ll 1$, approximate ${\mathcal{F}}_0$ by $f_0^2$, and discard the two time-derivative terms on the right compared with $-f_0^2{\eta }_x$. This is the regular midlatitude long-time ordering, with nondegenerate zonal variation. If $a=f_0/c$, the result isThe printed low-frequency equation has the opposite right-hand sign. The minus sign follows directly from the preceding eliminated equation and is also the sign needed for the printed isofrequency-circle centre. An independent check uses geostrophic balance: $\psi =g\eta /f_0$, $(u,v)=(-{\psi }_y,{\psi }_x)$, and the shallow-water quasi-geostrophic potential vorticity is ${\nabla }_h^2\psi -a^2\psi +\beta y$. Linear potential-vorticity conservation gives $({\nabla }_h^2\psi -a^2\psi {)}_t+\beta {\psi }_x=0$.

For $\eta =A{e}^{i(kx+ly-\omega t)}$, the shallow-water Rossby-wave dispersion relation isFor fixed $l$ and $\beta >0$, it is odd in $k$, zero at $k=0$, negative for $k>0$, and tends to zero from below as $k\to +\infty$. Its minimum occurs at $k=\sqrt{l^2+a^2}$ with ${\omega }_{\min }=-\beta /[2\sqrt{l^2+a^2}]$. Thus both long and short waves have small frequency. The zonal phase velocity is westward, $\omega /k<0$, whereas the zonal group velocity isIt changes sign at the frequency minimum.

At fixed nonzero frequency, completing the square yields the Rossby-wave isofrequency circleThe radius is real only if $|\omega |\le |\beta |/(2|a|)$. For $\beta >0$, the branch with $k>0$ has $\omega <0$ and a circle centred on the positive $k$-axis. If instead the printed plus-sign wave equation is taken literally with this same Fourier convention, its dispersion is $\omega =+\beta k/(k^2+l^2+a^2)$ and its circle is centred at $(+\beta /(2\omega ),0)$. These two conventions cannot be mixed.

For reflection of a Rossby wave at a meridional wall, the stationary wall preserves frequency, and its translation invariance in $y$ preserves the tangential wavenumber. Thus ${\omega }_r={\omega }_i$, $l_r=l_i$. Both $k$ values solveTheir sum and product giveAn incident wave in $x>0$ travels toward the wall in group velocity, not necessarily in phase velocity. For $\beta >0$, $0<k_i<\sqrt{l_i^2+a^2}$ has $c_{gx,i}<0$, while its partner $k_r>\sqrt{l_i^2+a^2}$ has $c_{gx,r}>0$. The double-root case has zero normal group velocity and does not describe a wave packet incident on the wall.

At leading quasi-geostrophic approximation, the impermeability condition is $u=-g{\eta }_y/f_0=0$. For $l_i\ne 0$, the boundary condition givesThe reflected height therefore has equal amplitude and a phase change of $\pi$. Equality here concerns the height or quasi-geostrophic streamfunction amplitudes in the reduced model. At the degenerate $l_i=0$, geostrophic no-normal-flow is automatically satisfied and alone does not determine their ratio; the usual homogeneous wall-streamfunction condition supplies $A_r=-A_i$ if imposed. Retaining the small ageostrophic transport changes the boundary condition towhich tends to $-1$ in the regular low-frequency ordering but is not generally of unit modulus. In particular, $l_i=0$ gives $A_r/A_i=-k_i/k_r$ in that more complete boundary relation. Thus the printed equal-amplitude assertion requires the nondegenerate leading quasi-geostrophic approximation, or an explicit homogeneous wall condition; it is not a general exact shallow-water reflection law.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

The vertical component $q_z=\zeta -f\eta /H$ is $H$ times the first-order anomaly in shallow-water potential vorticity about rest:Thus $\mathbf{q}=q_z\hat{\mathbf{z}}$ records the combination of relative vorticity and column stretching that is conserved by the unforced linearized shallow water equations. Indeed their curl and continuity equation give $(q_z{)}_t=0$. It is the persistent forcing of the balanced component during geostrophic adjustment, while inertia-gravity waves carry away the unbalanced component.

Initially $q_z=-f{\eta }_0\mathrm{sgn}(x)/H$. In the final geostrophic flow, $u=0$, $v=g{\eta }_x/f$, and $\zeta =g{\eta }_{xx}/f$. Conservation of the linearized shallow-water potential vorticity therefore givesBoundedness on both sides and continuity of $\eta ,{\eta }_x$ at the origin determine the constants:The solution is the locally attained balanced state after outgoing inertia-gravity waves have departed; the full inviscid system need not lose its globally conserved wave energy. The signed volume flux in the $y$ direction is

For double Kelvin-wave adjustment at a depth step, put $c_{\pm }=\sqrt{g{H}^{\pm }}$ and use $R_D^{\pm }=c_{\pm }/f$ on the corresponding side of the step. The PDF's far-field wording must mean $\eta \to {\eta }_0\mathrm{sgn}(x)$, rather than ${\eta }_0$ for every $x$. In the exponentially trapped outer response, cross-step geostrophic balance gives, at $y=0$,To retain the propagation of this slowly varying trapped disturbance, the along-step acceleration in $u_t-fv=-g{\eta }_x$ must also be retained. It supplies the small cross-step ageostrophic flow. Thus the two cross-step volume fluxes areHere $d\mathrm{sgn}(x)/dx=2\delta (x)$ as a distributional derivative. Equating the fluxes and dividing by $c_{+}+c_{-}$ givesThe printed forcing is missing a factor of two. It cannot follow from the printed sign-function surface profile with the usual Dirac delta distribution normalization. Pure geostrophic balance in both horizontal directions would also omit $A_t$ and could not determine the wave evolution.

The method of characteristics with zero initial amplitude gives the corrected weak solutionBehind the eastward front, $0<x<\Delta ct$, the surface elevation at the step becomes $-{\eta }_0$, matching the upstream side. All the original cross-step transport is diverted into flow along the step once the wave has passed a fixed location: $v=0$ there, with opposing along-step velocities above and below the step. This is the complete barrier in the adjusted region, and in the long-time limit at every fixed $x$. At finite time a front still carries cross-step flux: the formulas above give $H^{\pm }{v}^{\pm }=(2{\eta }_0{c}_{+}{c}_{-}/f)\delta (x-\Delta ct)$. The discontinuous outer representation does not resolve the short-scale fast adjustment at either front.

For completeness, solving the equation literally as printed instead gives $A={\eta }_0[\Theta (x)-\Theta (x-\Delta ct)]$. Its amplitude is half that required; $\eta (x,0,t)$ retains a jump ${\eta }_0$ at the origin after passage, so that literal equation does not imply complete blocking. The corrected equation and the literal printed equation are distinguished rather than asserting an inconsistent barrier result.

Use a one-layer, hydrostatic, homogeneous fluid with reference depth $H_0$, no forcing or friction, small Rossby number $ϵ=U/(|f_0|L)\ll 1$, slow evolution, and leading geostrophic balance. For the beta plane, take $\beta L/|f_0|=O(ϵ)$; the relative free-surface displacement is also small, $\eta /H_0=O(ϵ)$, with the Rossby deformation radius retained at the chosen horizontal scale. This excludes the equatorial limit. If a bottom elevation $h_b$ is included, additionally require $h_b/H_0=O(ϵ)$ and stationary topography. Set $h_b=0$ for a flat bottom.

The quasi-geostrophic streamfunction satisfies $\psi =g\eta /f_0$, $u=-{\psi }_y$, $v={\psi }_x$, so relative vorticity is $\zeta ={\nabla }_h^2\psi$. Expanding the shallow-water potential vorticity using actual depth $H_0+\eta -h_b$ givesConsequently the depth-scaled anomaly, the shallow-water quasi-geostrophic potential vorticity, isThe leading material derivative uses the geostrophic flow. Define $J(a,b)=a_x{b}_y-a_y{b}_x$; then conservation yields $Q_t+J(\psi ,Q)=0$. The term $-\psi /R_D^2$ represents free-surface stretching, $\beta y$ the variation of planetary rotation, and $f_0{h}_b/H_0$ topographic stretching. Under a rigid lid or at scales much smaller than the Rossby deformation radius, the free-surface term is absent or negligible and the flat-bottom Barotropic Rossby wave model has $Q={\nabla }_h^2\psi +\beta y$.

Write $H_{\pm }=H_0\pm d$, with the upper sign referring to $y>0$, and use an f-plane with Coriolis parameter $f_0>0$ for the stated directions. The formulas retain the signed $f_0$; a positive Rossby deformation radius means $R_{\pm }=\sqrt{g{H}_{\pm }}/|f_0|$. Let $\rho$ be the constant fluid density.

Far from the depth step the problem is independent of $y$. The linearized shallow water equations imply conservation of $v_x-f_0\eta /H_{\pm }$. Initially this is $-f_0{\eta }_0\mathrm{sgn}(x)/H_{\pm }$. In geostrophic balance, $u_s=0$ and $v_s=g{\eta }_{s,x}/f_0$, soBoundedness and matching of height and its derivative at $x=0$ give geostrophic adjustment of a surface-height jump:The potential energy release per unit length in $y$ is the finite difference between two individually infinite energies:For comparison, the final kinetic energy is $\rho g{\eta }_0^2{R}_{\pm }/2$, so the energy radiated in inertia-gravity waves is $\rho g{\eta }_0^2{R}_{\pm }$. Integrating the depth-weighted velocity over $x$ gives the volume transportsFor $f_0,{\eta }_0>0$, $Q_{-}$ is towards the step and $Q_{+}$ is away from it. Their difference is $4g{\eta }_{0}d/f_0$; a global steady state cannot simply join the two far-field currents without additional transport along the step.

To obtain the rotating shallow-water height equation with variable depth, set $\zeta =v_x-u_y$ and $J_H={\partial }_x(Hv)-{\partial }_y(Hu)$. Direct differentiation of the original momentum and continuity equations givesConsequentlyAt the step, $H_y=2d\delta (y)$ is a Dirac delta distribution; the original matching conditions are used rather than assuming a smooth depth there. Velocity elimination raises the time order: initially ${\eta }_t=0$ and ${\eta }_{tt}=g{\nabla }_h\cdot (H{\nabla }_h\eta )$, as well as the prescribed initial height.

For a decaying harmonic disturbance, the height equation on each constant-depth side givesSolving the two momentum equations yieldsContinuity of $H\hat{v}$ at the step gives the implicit step-trapped topographic Rossby wave dispersion relationThis derivation assumes ${\omega }^2\ne f_0^2$ and positive decay rates; it describes the low-frequency trapped branch, rather than the radiating inertia-gravity waves.

In the quasi-geostrophic approximation, replace $f_0^2-{\omega }^2$ by $f_0^2$. The resulting explicit frequency, and then its leading small-$d/H_0$ form, areFor $d>0$, the phase velocity has the sign of $f_0$. Thus propagation is eastward in the Northern Hemisphere, with the shallower side on the right. Differentiating the leading dispersion relation givesIn the long-wave limit the group velocity and phase velocity coincide: