With no $y$ dependence, the <linearized shallow water equations> are
$$
u_t-fv=-g\eta_x,
\qquad v_t+fu=0,
\qquad \eta_t+H_0u_x=0.
$$
Their conserved linear potential-vorticity anomaly is
$$
v_x-\frac f{H_0}\eta
=2v_0\delta(x),
$$
because the initial velocity jump has derivative $2v_0\delta(x)$. The final steady state is in <geostrophic balance>, so $u=0$ and $fv=g\eta_x$. Hence
$$
\boxed{
\left(\frac{d^2}{dx^2}-\frac1{R_d^2}\right)\eta
=\frac{2fv_0}{g}\delta(x),
\qquad
R_d=\frac{\sqrt{gH_0}}{|f|}.}
$$
Taking $f>0$ for definiteness and requiring decay at infinity gives
$$
\boxed{
\eta=-v_0\sqrt{\frac{H_0}{g}}e^{-|x|/R_d},
\qquad
u=0,
\qquad
v=v_0\operatorname{sgn}(x)e^{-|x|/R_d}.}
$$

Initially, the potential energy is zero and, per unit length in $y$,
$$
E_{
m early}=\frac12\rho H_0v_0^2(2L)
=\rho H_0v_0^2L.
$$
For $L\gg R_d$, the final kinetic and potential energies are equal:
$$
E_{K,\rm steady}
=\frac12\rho H_0v_0^2R_d,
\qquad
E_{P,\rm steady}
=\frac12\rho g\int\eta^2dx
=\frac12\rho H_0v_0^2R_d.
$$
Thus $E_{
m steady}=\rho H_0v_0^2R_d$. During <geostrophic adjustment>, <inertia-gravity waves> carry the excess energy out of $|x|<L$; energy in that finite region is therefore not conserved. Their long-wave speed is $c=\sqrt{gH_0}$, so the adjustment of the stated region takes
$$
\boxed{t_{\rm adj}=O\left(\frac{L}{\sqrt{gH_0}}\right)
=O\left(\frac{L}{|f|R_d}\right).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 333 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-333/1/ii/solution