= Solution
With $B=0$, the stated <biharmonic stream function for planar Stokes flow> is
$$
\psi=A\sin^2\theta+Cr\sin\theta+\frac{D\sin^3\theta}{r}.
$$
Requiring it to vanish at $r=2a\sin\theta$ and $r=4a\sin\theta$ gives
$$
A+2aC+\frac{D}{2a}=0,
\qquad A+4aC+\frac{D}{4a}=0.
$$
Thus $A=-6aC$ and $D=8a^2C$. Writing $C=\alpha$ yields
$$
\boxed{\psi=\alpha\left(r\sin\theta-6a\sin^2\theta
+\frac{8a^2\sin^3\theta}{r}\right)
=\frac{\alpha\sin\theta}{r}
(r-2a\sin\theta)(r-4a\sin\theta).}
$$
Using $u_r=\psi_\theta/r$ and $u_\theta=-\psi_r$, differentiation shows that $u_r\cos\theta+u_\theta\sin\theta$ equals $\alpha$ on the inner boundary and $-\alpha/2$ on the outer boundary. Both wall conditions hold precisely when
$$
\boxed{\alpha=a\Omega.}
$$
This is the <Stokes flow between touching counter-rotating cylinders>.
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