Stretch the coordinate $$ x = \epsilon \alpha \quad \text{or} \quad x - 1 = \epsilon \alpha $$ At the boundary, $\phi(x,y)$ varies rapidly with $x$

Insert $\psi = \psi_I + \phi$ into vorticity balance $$ \epsilon_S (\nabla^2 \psi_I + \nabla^2 \phi) + \frac{\partial \psi_I}{\partial x} + \frac{1}{\epsilon} \frac{\partial \phi}{\partial \alpha} = \text{curl}_z \boldsymbol{\tau}_T $$

Since $\psi_I$ already satisfies Sverdrup balance, the equation reduces to $$ \epsilon_S \left( \nabla^2 \psi_I + \frac{1}{\epsilon^2} \frac{\partial^2 \phi}{\partial \alpha^2} + \frac{\partial^2 \phi}{\partial y^2} \right) + \frac{1}{\epsilon} \frac{\partial \phi}{\partial \alpha} = 0 $$

Choose $\epsilon = \epsilon_S$ (dominant balance), this gives the leading-order equation $$ \frac{\partial^2 \phi}{\partial \alpha^2} + \frac{\partial \phi}{\partial \alpha} = 0 \Rightarrow \underbrace{\phi}_{\text{boundary layer solution}} = A(y) + B(y)e^{-\alpha} $$

Choose $C=1$ in the Sverdrup interior, and apply boundary condition at $x=0$ ($\psi = \psi_I + \phi = 0$ at $x=0$) \[ \underbrace{\psi_I}_{\substack{\text{interior} \\ \text{solution}}} = \pi (1-x)\sin \pi y \;\Rightarrow\; \underbrace{\phi}_{\substack{\text{boundary layer} \\ \text{solution}}} = -\pi \sin \pi y \, e^{-x/\epsilon_S} \;\Rightarrow\; \underbrace{\psi}_{\substack{\text{composite} \\ \text{solution}}} = (1-x-e^{-x/\epsilon_S})\pi \sin \pi y \] This is the composite (boundary layer plus interior) solution \(\boxed{(1-x-e^{-x/\epsilon_S})\pi \sin \pi y}\)

Restore dimensions $$ \psi = \frac{\tau_0 \pi}{\beta} \left(1 - \frac{x}{a} - e^{-x/(a\epsilon_S)}\right) \sin \frac{\pi y}{a} $$