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

Table of contents
- Solution b

Solution

The eastern wall has no normal flow, so the quasi-geostrophic streamfunction is constant there. Choose

ψ (L, y) = 0

. Integrating the Sverdrup relation gives

ψ_{int} (x, y) = \frac{f _{0} τ ^{'} ( y )}{βρ H _{0}^{2}} (L - x) .

Since geostrophic balance gives

ψ = g η / f_{0}

,

η_{int} (x, y) = \frac{f _{0}^{2} τ ^{'} ( y )}{g βρ H _{0}^{2}} (L - x) .

If this interior solution were extended to both walls, the west-minus-east height difference would be

η_{int} (0, y) - η_{int} (L, y) = \frac{f _{0}^{2} L}{g βρ H _{0}^{2}} τ^{'} (y) .

The northward volume transport per unit meridional distance is

T = H_{0} \int_{0}^{L} v d x = H_{0} \int_{0}^{L} ψ_{x} d x = \frac{g H _{0}}{f _{0}} [η (L, y) - η (0, y)] .

Thus

T = - \frac{f _{0} L}{βρ H _{0}} τ^{'} (y) .

This is the basin-integrated form of Sverdrup balance.

Solved by gpt-5.6-sol high.

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