= Solution
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 $R_\pm=\sqrt{gH_\pm}/|f_0|$. 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 $-f_0\eta_0\operatorname{sgn}(x)/H_\pm$. In <geostrophic balance>, $u_s=0$ and $v_s=g\eta_{s,x}/f_0$, so
$$
\eta_s-R_\pm^2\eta_{s,xx}=\eta_0\operatorname{sgn}(x).
$$
Boundedness and matching of height and its derivative at $x=0$ give <geostrophic adjustment of a surface-height jump>:
$$
\boxed{\eta_s=\eta_0\operatorname{sgn}(x)(1-e^{-|x|/R_\pm}),\qquad v_s=\frac{g\eta_0}{f_0R_\pm}e^{-|x|/R_\pm},\qquad u_s=0.}
$$
The <potential energy> release per unit length in $y$ is the finite difference between two individually infinite energies:
$$
\Delta P_\pm=\frac{\rho g}{2}\int_{-\infty}^{\infty}(\eta_0^2-\eta_s^2)\,dx
=\frac{\rho g\eta_0^2}{2}\int_{-\infty}^{\infty}(2e^{-|x|/R_\pm}-e^{-2|x|/R_\pm})\,dx
=\boxed{\frac32\rho g\eta_0^2R_\pm.}
$$
For comparison, the final <kinetic energy> is $\rho g\eta_0^2R_\pm/2$, so the energy radiated in <inertia-gravity waves> is $\rho g\eta_0^2R_\pm$. Integrating the depth-weighted velocity over $x$ gives the volume transports
$$
\boxed{Q_\pm=\int_{-\infty}^{\infty}H_\pm v_s\,dx=\frac{2g\eta_0H_\pm}{f_0}.}
$$
For $f_0,\eta_0>0$, $Q_-$ is towards the step and $Q_+$ is away from it. Their difference is $4g\eta_0d/f_0$; 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 gives
$$
\eta_{tt}=-f_0J_H+g\nabla_h\cdot(H\nabla_h\eta),\qquad (J_H)_t=f_0\eta_t+gH_y\eta_x.
$$
Consequently
$$
\boxed{\partial_t\left[\eta_{tt}+f_0^2\eta-g\nabla_h\cdot(H\nabla_h\eta)\right]+f_0gH_y\eta_x=0.}
$$
At 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 gives
$$
\widehat\eta(y)=\eta_c\begin{cases}e^{\kappa_-y},&y<0,\\e^{-\kappa_+y},&y>0,\end{cases}
\qquad \kappa_\pm=\left[k^2+\frac{f_0^2-\omega^2}{gH_\pm}\right]^{1/2}>0.
$$
Solving the two momentum equations yields
$$
\widehat v=\frac{ig}{f_0^2-\omega^2}(f_0k\widehat\eta+\omega\widehat\eta_y).
$$
Continuity of $H\widehat v$ at the step gives the implicit <step-trapped topographic Rossby wave> dispersion relation
$$
\boxed{\omega(H_+\kappa_++H_-\kappa_-)=2f_0dk.}
$$
This 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, are
$$
\boxed{\omega_{\rm QG}=\frac{2f_0dk}{H_+\sqrt{k^2+R_+^{-2}}+H_-\sqrt{k^2+R_-^{-2}}}
\simeq\frac{f_0d}{H_0}\frac{k}{\sqrt{k^2+R_D^{-2}}},\qquad R_D=\frac{\sqrt{gH_0}}{|f_0|}.}
$$
For $d>0$, 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> gives
$$
c_g=\frac{f_0d}{H_0}\frac{R_D^{-2}}{(k^2+R_D^{-2})^{3/2}}.
$$
In the long-wave limit the <group velocity> and <phase velocity> coincide:
$$
\boxed{c_g=c_p=c_T=\frac{f_0dR_D}{H_0}.}
$$
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, $[H\eta_y]=2H_0A/R_D$ and $\eta_x(x,0)=2\eta_0\delta(x)-A_x$. Hence <long-wave transport along a depth step> obeys
$$
\boxed{A_t+c_TA_x=2c_T\eta_0\delta(x),\qquad A(x,0)=0.}
$$
The <method of characteristics>, interpreted for the <Dirac delta distribution> source, gives
$$
\boxed{A(x,t)=\eta_0[\operatorname{sgn}(x)-\operatorname{sgn}(x-c_Tt)].}
$$
For $c_T>0$ this is $2\eta_0$ in $0<x<c_Tt$ 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 $A_t+c_TA_x=c_T\eta_s'(x)$ and $A=\eta_s(x)-\eta_s(x-c_Tt)$ at this order.
The along-step <geostrophic flow> is $u=-g\eta_y/f_0$. Integrating separately over the two sides gives
$$
\boxed{T_-\equiv\int_{-\infty}^0H_-u\,dy=\frac{gH_-A}{f_0},\qquad T_+\equiv\int_0^\infty H_+u\,dy=-\frac{gH_+A}{f_0}.}
$$
In 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 $T_-+T_+=-4g\eta_0d/f_0$, balancing their mismatch. These comparisons use the stated leading long-wave and small-depth-contrast approximations.
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