= Solution
Two-dimensional <incompressibility> permits $u=\psi_y$, $v=-\psi_x$. Taking the <curl> of the Stokes momentum equation eliminates <pressure> and gives $\nabla^2(v_x-u_y)=0$. Since $v_x-u_y=-\nabla^2\psi$, the <stream function> obeys $\boxed{\nabla^4\psi=0}$.
Scale radius by $a$, <velocity> by $U$ and <stream function> by $Ua$. In <polar coordinates> $u_r=r^{-1}\psi_\theta$, $u_\theta=-\psi_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires $\psi_r=0$ at $r=1$. The uniform stream has $\psi=r\sin\theta$, specifying the far-field condition.
The boundary forcing has angular mode $\sin\theta$, so its corresponding mode must have $\psi=f(r)\sin\theta$. Define $L_1=d^2/dr^2+r^{-1}d/dr-r^{-2}$. Then $\nabla^4\psi=(L_1^2f)\sin\theta$, while $L_1r^s=(s^2-1)r^{s-2}$. Thus the <indicial equation> of $L_1^2$ is $(s^2-1)[(s-2)^2-1]=(s-3)(s-1)^2(s+1)=0$, giving
$$
\boxed{f=Ar^3+Br\log r+Cr+Dr^{-1}.}
$$
The repeated exponent one supplies the logarithmic term. The condition $f'\to1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f'(1)=C-D=2\ne0$. No additional <Fourier mode> can repair the missing <boundary condition> in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.
This is <Stokes' paradox>. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero <Reynolds number>, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.
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