Assume a steady, small-Rossby-number interior in which drag is negligible and the nonlinear advection of relative vorticity,
$$
J(\psi,\nabla^2\psi),
$$
is small compared with advection of planetary vorticity. The stretching term proportional to $\psi$ has zero Jacobian with $\psi$. On the <beta plane>,
$$
J\left(\psi,\frac f{f_0}\right)
=J\left(\psi,\frac{\beta y}{f_0}\right)
=\frac{\beta}{f_0}\psi_x.
$$
Hence the forced quasi-geostrophic equation reduces to <Sverdrup balance>
$$
\boxed{
\frac{\beta}{f_0}\psi_x
=\frac1{\rho H_0^2}
(\nabla\times\boldsymbol\tau)\mathbin{\cdot}\hat{\mathbf z}}.
$$
For the zonal <wind stress> $\boldsymbol\tau=\tau(y)\hat{\mathbf x}$,
$$
(\nabla\times\boldsymbol\tau)\mathbin{\cdot}\hat{\mathbf z}
=-\tau'(y),
$$
so
$$
\boxed{
\psi_x=-\frac{f_0}{\beta\rho H_0^2}\tau'(y)}.
$$

Solved by gpt-5.6-sol high.

Solution

Paper 333 / 2 / a / Solution

The eastern wall has no normal flow, so the <quasi-geostrophic streamfunction> is constant there. Choose $\psi(L,y)=0$. Integrating the Sverdrup relation gives
$$
\boxed{
\psi_{\rm int}(x,y)
=\frac{f_0\tau'(y)}{\beta\rho H_0^2}(L-x)}.
$$
Since <geostrophic balance> gives $\psi=g\eta/f_0$,
$$
\boxed{
\eta_{\rm int}(x,y)
=\frac{f_0^2\tau'(y)}
{g\beta\rho H_0^2}(L-x)}.
$$
If this interior solution were extended to both walls, the west-minus-east height difference would be
$$
\boxed{
\eta_{\rm int}(0,y)-\eta_{\rm int}(L,y)
=\frac{f_0^2L}{g\beta\rho H_0^2}\tau'(y)}.
$$

The northward volume transport per unit meridional distance is
$$
T=H_0\int_0^L v\,dx
=H_0\int_0^L\psi_x\,dx
=\frac{gH_0}{f_0}\,[\eta(L,y)-\eta(0,y)].
$$
Thus
$$
\boxed{
T=-\frac{f_0L}{\beta\rho H_0}\tau'(y)}.
$$
This is the basin-integrated form of <Sverdrup balance>.

Solved by gpt-5.6-sol high.

Paper 333 / 2 / b / Solution

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.

Paper 333 / 2 / e / Solution

{c}
{wiki}

Sverdrup balance equates meridional advection of planetary vorticity to wind-stress curl in the weakly frictional ocean interior.


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