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For two-dimensional incompressible flow, introduce the stream function by $u_r=r^{-1}{\psi }_{\phi }$, $u_{\phi }=-{\psi }_r$. The scalar vorticity is $\omega =-{\nabla }^2\psi$. Taking the curl of the Stokes equations $-\nabla p+\mu {\nabla }^{2}u=0$ gives ${\nabla }^2\omega =0$, and taking their divergence gives ${\nabla }^{2}p=0$. Thus the biharmonic stream function for planar Stokes flow satisfies ${\nabla }^4\psi =0$.

The far-field shear has stream function $\Gamma y^2/2=(\Gamma r^2/4)(1-\cos 2\phi )$. 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 giveDifferentiation supplies the flow everywhere outside the cylinder: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 $2\mu e_{r\phi }=\mu \Gamma (2\cos 2\phi -1)$. Integrating its moment about the axis givesThis is the fluid's torque on the stationary cylinder, clockwise for $\Gamma >0$; the external holding torque has the opposite sign.