Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 333 / 2

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.