= Solution
{c}
Put $m=\pi/b$. With $r=0$, <Sverdrup balance> gives
$$
\boxed{v(y)=\frac{F_0}\beta\cos(my).}
$$
At an impermeable eastern wall, $u(L,y)=0$. Continuity implies
$$
u_x=-v_y=\frac{mF_0}\beta\sin(my),\qquad
\boxed{u(x,y)=\frac{mF_0}\beta(x-L)\sin(my).}
$$
Equivalently,
$$
\psi(x,y)=\frac{F_0}\beta(x-L)\cos(my)
$$
satisfies the <impermeability condition> with the boundary <streamfunction> set to zero.
At a second wall $x=0$, this solution has $u(0,y)=-mF_0L\sin(my)/\beta$, generally nonzero. A useful integral proof of the obstruction is
$$
\frac d{dy}\int_0^L v\,dx=-[u(L,y)-u(0,y)]=0
$$
for two impermeable walls, whereas the proposed interior gives $\int_0^L v\,dx=LF_0\cos(my)/\beta$, which varies with $y$. Friction or another departure from the inviscid model is needed for a return flow that closes the circulation.
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