 Solution

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