Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 18C (Fluid Dynamics)

The constant $f=2\Omega \sin \phi$ is the positive Coriolis parameter at northern latitude $\phi$, $g$ is the gravitational acceleration, and $h$ is the undisturbed mean depth of the layer.

The linearized shallow water equations assume a homogeneous, incompressible, inviscid layer over a flat bed; a horizontal scale much larger than the depth, so the shallow-water approximation and hydrostatic approximation apply; depth-independent horizontal velocity; a free-surface displacement $|\eta |\ll h$ and sufficiently small velocity to neglect nonlinear products; and an f-plane on which $f$ is constant. The equations also omit viscosity, forcing, and bottom topography.

The vertical relative vorticity isDifferentiate the $v$-momentum equation with respect to $x$ and the $u$-momentum equation with respect to $y$, then subtract. The mixed pressure derivatives cancel, givingThe mass-conservation equation says ${\eta }_t=-h(u_x+v_y)$. Therefore the linearized shallow-water potential vorticitysatisfiesThus

For a steady flow independent of $y$, geostrophic balance givesHence $\zeta =v_x=(g/f){\eta }^{\prime \prime }$, and the definition of $q$ becomesEquivalently,With the Rossby deformation radius $R=\sqrt{gh}/f$, the bounded solution for the stated potential-vorticity step isContinuity of $\eta$ and $v=(g/f){\eta }^{\prime }$ at $x=0$ givesThereforeand

When $q_1>q_2$, the surface height increases smoothly from $-h{q}_1/f$ to $-h{q}_2/f$, while $v$ is a positive eastward jet with its maximum at the discontinuity and exponential decay on the Rossby-radius scale: