Cylinder in a simple shear Stokes flow
= Cylinder in a simple shear Stokes flow
For a stationary cylinder of radius $a$ in the far-field shear $(\Gamma y,0)$, the <stream function> is $\psi=(\Gamma/4)[r^2-2a^2\log(r/a)-a^2-(r^2-2a^2+a^4/r^2)\cos2\varphi]$. It solves the <biharmonic equation> and the <no-slip boundary condition>. The fluid exerts torque $-2\pi\mu\Gamma a^2$ per unit length, because the angular part of the surface <traction> averages to zero and the mean tangential traction is $-\mu\Gamma$.