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Write $c=\sqrt{gH}$ and $R_d=c/|f|$, with $f\ne 0$. For the profiles below take $f>0$, so $\alpha =L/R_d$ agrees with the stated parameter; changing the sign of $f$ reverses the balanced meridional velocity but not the height profile. The linearized shallow water equations areTaking the horizontal curl gives $(v_x-u_y{)}_t=-f(u_x+v_y)=f{\eta }_t/H$. Hence the signed linear shallow-water potential vorticity anomaly is conserved:The initial data are independent of $y$, so the solution remains independent of $y$ and $v_x=f(\eta -{\eta }_0)/H$. Differentiating continuity in time and using the zonal momentum equation gives

For a horizontal Fourier mode with wavenumber $k$, put ${\Omega }_k^2=f^2+c^2{k}^2$. The solution splits into a balanced component and inertia-gravity waves:These waves have frequencies $|\omega |\ge |f|$, phase velocity $\omega /k$, and group velocity $c^{2}k/\omega$, whose magnitude is less than $c$. They have zero perturbation potential vorticity and carry the unbalanced part away. On the unbounded line, the localized difference between the initial and balanced profiles disperses, leaving a local steady limit. No damping is needed for this local geostrophic adjustment; in a closed periodic domain, undamped waves would persist and a literal pointwise steady limit would generally not exist.

The bounded steady height satisfies the modified Helmholtz equationOddness removes the even homogeneous solution in the central region. The exterior solution is bounded and tends to the appropriate plateau. Matching both height and its first derivative at $x=\pm L$ gives the geostrophic adjustment of a finite-width height ramp:For example, writing the interior solution as $hx/L+B\sinh (x/R_d)$, matching its slope to the decaying exterior at $L$ gives $B=-h{e}^{-\alpha }/\alpha$.

The steady meridional momentum equation gives $u_s=0$, and the zonal equation gives the geostrophic balance $f{v}_s=g{\eta }_s^{\prime }$. ThusThe height is odd and monotone, and the meridional current is even, with maximum $v_s(0)=gh(1-e^{-\alpha })/(fL)$ for $h>0$.

For $\alpha \ll 1$, the initial ramp is narrow compared with the Rossby deformation radius. To leading order,Near the origin the height rises with slope $h/R_d$, rather than $h/L$; the current has width $R_d=L/\alpha$ and peak $gh\alpha /(fL)$. For $\alpha \gg 1$, the central height remains nearly $hx/L$ and the central velocity nearly $gh/(fL)$. Around each end of the ramp there is a matching layer of width $R_d=L/\alpha$, with height correction of order $h/\alpha$; the exterior current decays over the same width. At $x=\pm L$ the current is approximately half its central value. In both limits the zonal current is zero. These are the broad rotational adjustment of a narrow disturbance and the weak edge adjustment of a broad disturbance, respectively.

Use the energy normalization specified for the height energy: the linear total energy density is $(u^2+v^2)/2+g{\eta }^2/(2H)$. The linear equations implyFor $\alpha \ll 1$, let the potential energy loss mean the positive difference between the initial and final values, per unit transverse length. Although both absolute potential energies diverge on the infinite line, their difference is finite. The leading step approximation givesThe finite ramp changes this leading expression by a relative $O(\alpha )$ correction. The initial kinetic energy is zero and the final gain isThe remainder $\Delta V-\Delta T\simeq g{h}^2{R}_d/H$ is energy carried away by the outgoing inertia-gravity waves. This explains why the ratio is less than one without invoking dissipation: conservation of energy includes the waves, not only the final balanced flow.