= Solution
At infinity,
$$
\mathbf q=-\frac{h_0^3}{12\mu}\nabla p
=\frac{h_0^3G}{12\mu}\mathbf e_x,
$$
so
$$
\psi\sim Ar\sin\theta,\qquad
\boxed{A=\frac{h_0^3G}{12\mu}}.
$$
Continuity of $q_r=r^{-1}\psi_\theta$ gives $[\psi_\theta]^+_-=0$. Since $\nabla p=-12\mu h^{-3}\mathbf q$, continuity of tangential pressure derivative gives $[h^{-3}\psi_r]^+_-=0$.
The harmonic solutions regular at zero and with the prescribed far field are
$$
\psi_>=\left(Ar+\frac Br\right)\sin\theta,
\qquad
\psi_<=Cr\sin\theta.
$$
The interface conditions give
$$
B=Aa^2\frac{h_1^3-h_0^3}{h_0^3+h_1^3},
\qquad
C=\frac{2Ah_1^3}{h_0^3+h_1^3}.
$$
Integrating the outward flux across the appropriate semicircle gives
$$
\boxed{2aC=\frac{4Aah_1^3}{h_0^3+h_1^3}}.
$$
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