= Solution
Use $u=-\psi_y$, $v=\psi_x$. Since $v=W(y)/\beta$ and the eastern boundary may be chosen as the zero streamline,
$$
\boxed{
\psi(x,y)=\frac{x-L_X}{\beta}W(y)
=-\frac{x-L_X}{\beta}\sin\frac{2\pi y}{L_Y}.}
$$
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> scale
$$
\boxed{\delta_W\sim\delta_E\sim(\nu/\beta)^{1/3}.}
$$
At the southern and northern walls, $x$ variation remains on the basin scale $L_X$ while four $y$ derivatives occur across the layer, so
$$
\boxed{\delta_S\sim\delta_N
\sim(\nu L_X/\beta)^{1/4}.}
$$
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