= Solution
{c}
Seek $\psi=\Psi(x)\cos(my)$, with $m=\pi/b$. The <ordinary differential equation> is
$$
r\Psi''+\beta\Psi'-rm^2\Psi=F_0.
$$
Its constant particular solution and characteristic exponents are
$$
\boxed{A=-\frac{F_0}{rm^2}=-\frac{F_0b^2}{r\pi^2},\qquad
\lambda=\frac{\beta+\sqrt{\beta^2+4r^2m^2}}{2r},\qquad
\gamma=\frac{\sqrt{\beta^2+4r^2m^2}-\beta}{2r}.}
$$
Both $\lambda,\gamma$ are positive; the two homogeneous solutions are $e^{-\lambda x}$ and $e^{\gamma x}$. Their product satisfies $\lambda\gamma=m^2$. At small $rm/\beta$,
$$
\boxed{\lambda\simeq\frac\beta r,\qquad \gamma\simeq\frac{rm^2}\beta.}
$$
For the rectangular basin, choose zero <streamfunction> on its connected impermeable boundary. The factor $\cos(my)$ already gives zero at $y=\pm b/2$. Enforcing $\Psi(0)=\Psi(L)=0$ yields the exact result
$$
\Psi=A\left[1-C_W e^{-\lambda x}-C_E e^{\gamma(x-L)}\right],
$$
where
$$
C_W=\frac{1-e^{-\gamma L}}{1-e^{-(\lambda+\gamma)L}},\qquad
C_E=\frac{1-e^{-\lambda L}}{1-e^{-(\lambda+\gamma)L}}.
$$
The narrow <Stommel boundary layer> approximation requires
$$
\boxed{\lambda^{-1}\ll L\ll\gamma^{-1}.}
$$
Then $e^{-\lambda L}$ is negligible, $C_W\simeq\gamma L$ and $C_E\simeq1$, giving
$$
\boxed{\psi\simeq A\left[1-\gamma L e^{-\lambda x}-e^{\gamma(x-L)}\right]\cos(my).}
$$
It is an asymptotic formula: at $x=0$ the remaining mismatch is of order $A(\gamma L)^2$. Expanding the eastern exponential as well gives the familiar leading <Sverdrup balance> interior plus western correction,
$$
\psi\simeq\frac{F_0}\beta\left[(x-L)+Le^{-x/\delta_W}\right]\cos(my),\qquad \delta_W=\frac r\beta.
$$
\b[Printed inequality.] The original PDF explicitly has $\lambda^{-1}\gg L$ as well as $\gamma^{-1}\gg L$. The first sign is reversed for the requested small-friction limit. This can also be disproved directly: if $\lambda L=0.02$ and $\gamma L=0.01$, the printed approximate bracket at the western wall is $1-0.01-e^{-0.01}\simeq-4.98\times10^{-5}$, while at the eastern wall it is $-0.01e^{-0.02}\simeq-0.00980$. The latter is of the same order as the retained basin-scale signal, violating the boundary condition at leading order. The exact expression above remains available without imposing the erroneous inequalities.
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