= Solution
The incompressible Stokes equations are
$$
-\nabla p+\mu\nabla^2u=0,\qquad\nabla\cdot u=0.
$$
Taking the curl gives $\nabla^2\omega=0$. For a planar <stream function>, $\omega=-\nabla^2\psi$, so the <biharmonic stream function for planar Stokes flow> satisfies
$$
\boxed{\nabla^4\psi=0}.
$$
For $\psi=r^2f(\theta)$,
$$
u_r=rf',\qquad u_\theta=-2rf.
$$
The polar rate-of-strain components are
$$
e_{rr}=\partial_ru_r=f',\qquad
e_{\theta\theta}=\frac1r\partial_\theta u_\theta+\frac{u_r}{r}=-f',
$$
$$
e_{r\theta}=\frac12\left(
\frac1r\partial_\theta u_r+\partial_ru_\theta-\frac{u_\theta}{r}
\right)=\frac12f''.
$$
Hence
$$
\boxed{
e=\begin{pmatrix}f'&f''/2\\f''/2&-f'\end{pmatrix},
\qquad
\tau=2\mu e
=\mu\begin{pmatrix}2f'&f''\\f''&-2f'\end{pmatrix}.}
$$
Since
$$
\nabla^2(r^2f)=4f+f'',
$$
biharmonicity gives
$$
f^{(4)}+4f''=0,
\qquad
f=A+B\theta+C\cos2\theta+D\sin2\theta.
$$
No slip at $\theta=-\alpha$, no penetration at $\theta=0$, and the imposed tangential traction give
$$
f(-\alpha)=f'(-\alpha)=f(0)=0,\qquad
\mu f''(0)=S.
$$
Put $\Delta=\sin2\alpha-2\alpha\cos2\alpha$. Solving,
$$
f(\theta)=\frac{S}{4\mu}\left[
1-\cos2\theta+
\frac{2(1-\cos2\alpha)\theta+
(1-\cos2\alpha-2\alpha\sin2\alpha)\sin2\theta}
{\Delta}\right].
$$
The upper-surface <velocity> is radial and equals $rf'(0)$. Therefore the <similarity solution for tangentially forced Stokes wedge> gives
$$
\boxed{
U(r)=\frac{Sr}{\mu}
\frac{1-\cos2\alpha-\alpha\sin2\alpha}
{\sin2\alpha-2\alpha\cos2\alpha}}.
$$
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