No normal flow makes $\psi$ constant on the connected basin boundary; set that constant to zero. Put
$$
A(y)=\frac{f_0\tau'(y)}{\beta\rho H_0^2}.
$$
The interior solution is $A(L-x)$. The decaying boundary correction that enforces $\psi(0,y)=0$ is
$$
\psi_{\rm bl}=-ALe^{-x/\delta}.
$$
Thus, to leading order in $\delta/L$,
$$
\boxed{
\psi=A\left(L-x-Le^{-x/\delta}\right)},
\qquad
\boxed{
\eta=\frac{f_0A}{g}
\left(L-x-Le^{-x/\delta}\right)}.
$$

If $\delta$ denotes the e-folding width literally, then
$$
\boxed{
\Delta\eta_{\rm WBC}
=\eta(\delta,y)-\eta(0,y)
=\frac{f_0A}{g}
\left[L(1-e^{-1})-\delta\right]}.
$$
Since a boundary-layer width is only defined up to an order-one factor, matching at any point satisfying $\delta\ll x\ll L$ gives the convention-independent leading jump
$$
\boxed{
\Delta\eta_{\rm WBC}
\sim\frac{f_0^2L}{g\beta\rho H_0^2}\tau'(y)}.
$$
It is the height difference needed to return the broad Sverdrup transport in a narrow <western boundary current>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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