= Solution
For two-dimensional <incompressible flow>, introduce the <stream function> by $u_r=r^{-1}\psi_\varphi$, $u_\varphi=-\psi_r$. The scalar <vorticity> is $\omega=-\nabla^2\psi$. Taking the curl of the <Stokes equations> $-\nabla p+\mu\nabla^2u=0$ gives $\nabla^2\omega=0$, and taking their divergence gives $\nabla^2p=0$. Thus the <biharmonic stream function for planar Stokes flow> satisfies $\nabla^4\psi=0$.
The far-field shear has stream function $\Gamma y^2/2=(\Gamma r^2/4)(1-\cos2\varphi)$. For a <cylinder in a simple shear Stokes flow>, the axisymmetric biharmonic terms $r^2,\log r,1$ and the angular terms $r^2,1,r^{-2}$ suffice to impose $\psi=\psi_r=0$ at $r=a$ without changing the leading far field. They give
$$
\boxed{\psi=\frac\Gamma4\left[r^2-2a^2\log(r/a)-a^2-\left(r^2-2a^2+\frac{a^4}{r^2}\right)\cos2\varphi\right].}
$$
Differentiation supplies the flow everywhere outside the cylinder:
$$
\boxed{\begin{aligned}u_r&=\frac\Gamma2\left(r-\frac{2a^2}r+\frac{a^4}{r^3}\right)\sin2\varphi,\\
u_\varphi&=\frac\Gamma2\left[-r+\frac{a^2}r+\left(r-\frac{a^4}{r^3}\right)\cos2\varphi\right],\\
p&=p_\infty-\frac{2\mu\Gamma a^2}{r^2}\sin2\varphi.
\end{aligned}}
$$
Both velocity components vanish at $r=a$, and their far-field values are precisely the cylindrical components of $(\Gamma y,0)$. The pressure follows by substitution into the <Stokes equations>.
The tangential surface <traction> is $2\mu e_{r\varphi}=\mu\Gamma(2\cos2\varphi-1)$. Integrating its moment about the axis gives
$$
\boxed{\mathcal T_z=a^2\int_0^{2\pi}2\mu e_{r\varphi}\,d\varphi=-2\pi\mu\Gamma a^2.}
$$
This is the fluid's torque on the stationary cylinder, clockwise for $\Gamma>0$; the external holding torque has the opposite sign.
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