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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.

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:

To find the slow amplitude, integrate the low-frequency height equation across the step. This gives ${\partial }_t[H{\eta }_y]=f_0[H]{\eta }_x$, where brackets mean the upper-side minus lower-side value. For the given exponentially trapped outer profile, $[H{\eta }_y]=2H_{0}A/R_D$ and ${\eta }_x(x,0)=2{\eta }_0\delta (x)-A_x$. Hence long-wave transport along a depth step obeysThe method of characteristics, interpreted for the Dirac delta distribution source, givesFor $c_T>0$ this is $2{\eta }_0$ in $0<x<c_{T}t$ and zero outside: a surface depression is left in the wake of the wave. The sharp $x$-jump is a long-wave outer approximation. The fast adjustment region of width $O(R_D)$ near the origin, and the detailed wavefront, are unresolved. Keeping a smoothed background ${\eta }_s(x)$ instead would give $A_t+c_T{A}_x=c_T{\eta }_s^{\prime }(x)$ and $A={\eta }_s(x)-{\eta }_s(x-c_{T}t)$ at this order.

The along-step geostrophic flow is $u=-g{\eta }_y/f_0$. Integrating separately over the two sides givesIn the wake, $T_{-}=Q_{-}$ and $T_{+}=-Q_{+}$. The shallow-side transport is eastward and the deep-side transport westward, with magnitudes equal to the respective far-field currents. The net along-step transport is $T_{-}+T_{+}=-4g{\eta }_{0}d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.

For ${\mathbf{u}}_0=-3\Omega x{\mathbf{e}}_y/2$, advection vanishes and the Coriolis acceleration exactly cancels $3{\Omega }^{2}x{\mathbf{e}}_x$, so constant $P_0$ and ${\theta }_0=0$ complete the equilibrium.

Write perturbations $(\mathbf{v},p,\vartheta )e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ with $\mathbf{k}=(k_x,0,k_z)$. Incompressibility gives $\mathbf{k}\cdot \mathbf{v}=0$. It also makes the quadratic terms $\mathbf{v}\cdot \nabla \mathbf{v}=i(\mathbf{k}\cdot \mathbf{v})\mathbf{v}$ and $\mathbf{v}\cdot \nabla \vartheta$ vanish exactly; the background-advection terms vanish because $k_y=0$. The amplitude equations areEliminating the amplitudes yields the inertia-gravity wave dispersion relation

For $N^2=0$, $\omega =\pm \Omega k_z/k$, so the group velocity isIt is perpendicular to $\mathbf{k}$ and hence to the phase velocity. An inertial wave packet transports energy along beams lying in its phase surfaces.

If $N^2>0$ and $k_x/k_z\to 0$, incompressibility suppresses vertical motion and $\omega \to \Omega$: horizontal epicyclic motion and rotation dominate. If $k_z/k_x\to 0$, radial motion is suppressed and $\omega \to N$: vertical buoyancy oscillations dominate. Intermediate ratios give hybrid inertia-gravity waves.

If $N^2<0$, exponential growth occurs exactly whenOnly modes with sufficiently large radial wavenumber permit enough vertical displacement for unstable buoyancy to overcome rotational restoration; the other orientations remain stabilized by the Coriolis force.

Substitution of $e^{i(kx+ly-\omega t)}$ gives a homogeneous three-by-three system. Its determinant issoThe zero-frequency component is in geostrophic balance and carries all conserved $P$. The oscillatory components are inertia-gravity waves with zero $P$. For $K\ll |f|/c$, $\Omega \sim |f|$ and rotation produces an inertial oscillation; for $K\gg |f|/c$, $\Omega \sim cK$ and surface gravity dominates. The crossover length $c/|f|$ is the barotropic deformation radius.

With no $y$ dependence, the linearized shallow water equations areTheir conserved linear potential-vorticity anomaly isbecause the initial velocity jump has derivative $2v_0\delta (x)$. The final steady state is in geostrophic balance, so $u=0$ and $fv=g{\eta }_x$. HenceTaking $f>0$ for definiteness and requiring decay at infinity gives

Initially, the potential energy is zero and, per unit length in $y$,For $L\gg R_d$, the final kinetic and potential energies are equal:Thus $E_{msteady}=\rho H_0{v}_0^2{R}_d$. During geostrophic adjustment, inertia-gravity waves carry the excess energy out of $|x|<L$; energy in that finite region is therefore not conserved. Their long-wave speed is $c=\sqrt{g{H}_0}$, so the adjustment of the stated region takes

Initially $\zeta =0$, so conservation of the linear potential vorticity givesThe final flow is in geostrophic balance, henceConsequently the adjusted height solves the modified Helmholtz equationwhere $L_R$ is the Rossby deformation radius.

The initial condition is independent of $x$ and odd in $y$. The bounded solution that is continuously differentiable at $y=0$ isIt givesDuring geostrophic adjustment, the part of the initial energy incompatible with the conserved potential-vorticity distribution radiates away as inertia-gravity waves. The remaining current has width $O(L_R)$ and is geostrophically balanced.