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Put $D(x,y)=H(x,y)-h(x,y)$ and let $\mathbf{u}=(u,v)$ be the depth-independent interior velocity. The kinematic boundary conditions on the sloping upper and lower surfaces areIntegrating mass conservation through the layer givesThe inviscid vertical-vorticity equation isConsequently the shallow-water potential vorticityobeys the forced evolution equationPositive upward Ekman pumping removes layer thickness and raises the potential vorticity of the remaining column.

For a steady small-Rossby number flow, $q\simeq f/D$, soEquivalently,The same result follows from the integrated vortex-stretching balance

When $\nabla D=(c,0)$ with $c>0$,If the upper pumping is absent or weak and the upper surface is locally level, the impermeable-bottom condition is $w_b=-cu$. The stipulated $w_b<0$ then gives $u>0$, and in the Northern HemisphereThe steady interior flow is therefore directed northeastward, along contours of $f/D$ in the unforced limit. As a parcel moves eastward into deeper water, its vortex column stretches; a poleward displacement increases $f$ and preserves potential vorticity. Nonzero Ekman pumping drives motion across the $f/D$ contours. This is topographic potential-vorticity steering and the associated topographic Sverdrup balance.

At small Rossby number, $q=(\zeta +f_0)/H\simeq f_0/H$. For steady flow with $H=y^2$, the forced potential-vorticity equation becomessoDefine a transport streamfunction by $Hu=-{\psi }_y$ and $Hv={\psi }_x$. Choosing the eastern wall as $\psi =0$ gives the topographic Sverdrup balance solutionand thereforeIn each half-basin, the transport streamlines are the level curves $(L_X-x)y^3=\text{constant}$. They move westward and toward $y=0$. This interior solution treats the two sides of the degenerate line $H=0$ separately and requires boundary layers to enforce solid-wall conditions.