Consider the Stommel problem forced by the linear wind profile $$ \tau^x = \frac{\tau_0}{a}\left(y - \frac{a}{2}\right) $$ This profile is special because its curl vanishes at $y = 0$ and $y = a$, so the interior solution satisfies $\psi_I = 0$ at both ends

Scaling the variables $$ \mathcal{O}\!\left(\frac{\beta \Psi / a}{\beta \Psi / a}\right) + \mathcal{O}\!\left(\frac{r \Psi / a^2}{\beta \Psi / a}\right) = \mathcal{O}\!\left(\frac{\tau_0/a}{\beta \Psi / a}\right) \Rightarrow \epsilon_S \nabla^2 \hat{\psi} + \frac{\partial \hat{\psi}}{\partial \hat{x}} = -1 $$

The interior flow obeys $$ \frac{\partial \psi_I}{\partial \hat{x}} = -1 $$ so boundary layers are needed at both the meridional and zonal boundaries. The total solution is written as the sum of five parts $$ \hat{\psi} = \psi_I + \phi_W + \phi_E + \phi_N + \phi_S $$ where each $\phi$ is a boundary layer correction

The western boundary layer correction is found as $$ \phi_W = -e^{-\hat{x}/\epsilon_S} $$

At the northern boundary ($y = 1$), introduce a stretched coordinate $ \alpha = \epsilon'^{-1} (y - 1), \quad \epsilon' \ll 1 $. Let $\phi_N = \phi_N(x,\alpha)$. Substitute $\hat{\psi} = \psi_I + \phi_W + \phi_E + \phi_N + \phi_S$ into \(\boxed{\epsilon_S \nabla^2 \hat{\psi} + \frac{\partial \hat{\psi}}{\partial \hat{x}} = -1 }\), keeping only the relevant terms $$ \epsilon_S \left( \nabla^2 \psi_I + \frac{\partial^2 \phi_N}{\partial \hat{x}^2} + \frac{1}{\epsilon'^2}\frac{\partial^2 \phi_N}{\partial \alpha^2} \right) + \frac{\partial \psi_I}{\partial \hat{x}} + \frac{\partial \phi_N}{\partial \hat{x}} = -1 $$