Substitute \(\boxed{\hat{\psi} = \underbrace{\psi_I}_{\text{Sverdrup streamfunction}} + \underbrace{\phi_W(\alpha,\hat{y})}_{\text{boundary layer correction}}}\) into \(\boxed{-\epsilon_M \nabla^4 \hat{\psi} + \frac{\partial \hat{\psi}}{\partial \hat{x}} = \text{curl}_z \hat{\tau}_T}\) and subtract the Sverdrup balance. This gives $$ -\epsilon_M\!\left(\nabla^4 \psi_I + \frac{1}{\epsilon^4}\frac{\partial^4 \phi_W}{\partial \alpha^4}\right) + \frac{1}{\epsilon} \frac{\partial \phi_W}{\partial \alpha} = 0 $$ A non-trivial balance occurs when $ \epsilon = \epsilon_M^{1/3} $ which implies a dimensional thickness $a\epsilon = (\nu/\beta)^{1/3}$

At leading order, equation \(\boxed{-\epsilon_M\!\left(\nabla^4 \psi_I + \frac{1}{\epsilon^4}\frac{\partial^4 \phi_W}{\partial \alpha^4}\right) + \frac{1}{\epsilon} \frac{\partial \phi_W}{\partial \alpha} = 0}\) reduces to $$ -\frac{\partial^4 \phi_W}{\partial \alpha^4} + \frac{\partial \phi_W}{\partial \alpha} = 0 $$