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Use $u=-{\psi }_y$, $v={\psi }_x$. Since $v=W(y)/\beta$ and the eastern boundary may be chosen as the zero streamline,The lower half-basin has southward interior flow and a northward western return current, giving a clockwise ocean gyre. The upper half has northward interior flow and a southward western return current, giving a counterclockwise gyre. The interior solution cannot satisfy no slip at every wall, so viscous layers complete the circulation.

With constant lateral viscosity $\nu$, the vorticity equation contains $\nu {\nabla }^4\psi$. At the western and eastern walls, balancing $\beta {\psi }_x$ against four $x$ derivatives gives the Munk boundary layer scaleAt the southern and northern walls, $x$ variation remains on the basin scale $L_X$ while four $y$ derivatives occur across the layer, so

With $f=f_0+\beta y$ and parabolic depth, the topographic potential-vorticity pseudovelocity is $(-\beta /H,-f{H}_x/H^2)$. For positive $f,\beta$, it points west and turns south on the western slope, north on the eastern slope. Its characteristics are the parabolas $y=Qax(L_x-x)/\beta -f_0/\beta$, where $Q=f/H$. Wind-driven ocean gyres are steered by these depth gradients; drag and boundary data determine the actual ocean transport streamfunction. The zero-depth shoreline is a singular limit that requires a separate physical regularization.