Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 333 / 1

The first two equations are horizontal momentum balances: local acceleration plus the Coriolis acceleration equals the pressure-gradient force. The third is linearized mass conservation: convergence raises the free surface, represented by $\varphi$, while divergence lowers it. The constant $c$ is the long-wave gravity-wave speed.

Differentiate the second momentum equation in $x$, the first in $y$, subtract, and use continuity:Hence the linear shallow-water potential vorticitysatisfies $P_t=0$.

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.

Initially $P=-f{\varphi }_0\mathrm{sgn}(x)/c^2$. In the final steady state,Equating its potential vorticity to the initial value givesThis is geostrophic adjustment: inertia-gravity waves remove the unbalanced part. Their group speed is at most $c$, so at any finite time only a region whose distance from the initial jump is $O(ct)$ can have reached the steady state.

No normal flow at $y=0,L$ requires $v=0$. Geostrophic balance therefore gives ${\Phi }_x=0$ on both walls. These conditions do not fix the $x$-independent nullspace: ifthen $\Phi +\Psi (y)$ satisfies both the adjusted equation and ${\Phi }_x=0$. Thus $\Psi =A{e}^{fy/c}+B{e}^{-fy/c}$ supplies two undetermined amplitudes.

Set $v=0$. For $F_{+}(x+ct)$ the equations require ${\tilde{\varphi }}_{+}=-c{\tilde{u}}_{+}$ and ${\tilde{u}}_{+}^{\prime }=(f/c){\tilde{u}}_{+}$; for $F_{-}(x-ct)$ they require ${\tilde{\varphi }}_{-}=c{\tilde{u}}_{-}$ and ${\tilde{u}}_{-}^{\prime }=-(f/c){\tilde{u}}_{-}$. Hence the channel supports the Kelvin wavesEach wave propagates along one wall and is exponentially trapped toward that wall on the deformation scale $c/|f|$.

Form the $x$-momentum equation plus $1/c$ times continuity, multiply by $e^{\mp fy/c}$, and integrate across the channel. Integration by parts makes the $v_y$ term cancel the Coriolis term, and $v=0$ removes the endpoint contribution. The result is the pair of Kelvin-wave channel invariants

For the initial data, the two weighted integrals equal $({\varphi }_0/c)\mathrm{sgn}(x){\int }_0^L{e}^{\mp fy/c}dy$. Following their opposite characteristics to large time at fixed $x$ gives the additional final-state conditionsThese two scalar constraints determine the two coefficients $A,B$ in the nullspace from part v and therefore make the adjusted solution unique.