Because $v=\psi_x=(g/f_0)\eta_x$, the mean boundary-current velocity is exactly related to the height change by
$$
\boxed{
\bar v=\frac1\delta\int_0^\delta v\,dx
=\frac g{f_0\delta}\Delta\eta_{\rm WBC}}.
$$
For the e-folding convention used in part d,
$$
\bar v
=A\left[\frac L\delta(1-e^{-1})-1\right].
$$
Its robust narrow-layer scaling is
$$
\boxed{
\bar v\sim
\frac{L\,\tau'(y)}{\rho H_0^2\gamma}}.
$$
Thus weaker drag makes the current proportionally narrower and faster. Their product is independent of $\gamma$ to leading order:
$$
H_0\delta\bar v
\sim\frac{f_0L}{\beta\rho H_0}\tau'(y)
=-T_{\rm int}.
$$
<Mass conservation> requires the narrow return transport to cancel the broad <Sverdrup balance> transport. The meridional gradient of planetary <potential vorticity> makes a frictional closure possible on the western side and produces western intensification; bottom drag supplies the vorticity sink that permits fluid parcels to cross potential-vorticity contours there.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 2 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-333/2/e/solution