Biharmonic stream function for planar Stokes flow

Taking the curl of the planar Stokes equation makes vorticity harmonic. Since vorticity is minus the Laplacian of the stream function,

For a fixed wall at $\theta =0$ and a wall rotating at $\dot{\alpha }$ at $\theta =\alpha$, the quasi-steady Stokes approximation with the no-slip boundary condition givesIndeed the biharmonic equation for $r^{2}G$ is $G^{(4)}+4G^{\prime \prime }=0$; the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$. The four wall conditions are $G(0)=G^{\prime }(0)=G^{\prime }(\alpha )=0$, $G(\alpha )=1$. The outward radial volume flux per unit width is $-r^2\dot{\alpha }/2$, precisely the negative rate of increase of wedge area. The associated local pressure is $p=\mu \dot{\alpha }[2\cos \alpha /(\sin \alpha -\alpha \cos \alpha )]\log (r/r_{\ast })+C$. Its logarithmic singularity is integrable, although the ideal hinge requires a microscopic regularization.

Incompressibility and periodicity imply that the spatial mean of the normal velocity is constant with height. Zero normal flow at infinity therefore requires zero mean normal boundary velocity. In the bounded biharmonic stream function for planar Stokes flow solution, the mean streamfunction is a constant plus a term linear in height.

For $\psi =r^{2}f(\theta )$, biharmonicity gives
$f^{(4)}+4f^{\prime \prime }=0$. In the wedge $-\alpha <\theta <0$, no slip at the lower wall and no penetration plus tangential stress $S$ at the upper surface giveThe resulting surface speed is

An inner cylinder of radius $a$ centered at $(0,a)$ and an outer cylinder of radius $2a$ centered at $(0,2a)$ have polar boundaries $r=2a\sin \theta$ and $r=4a\sin \theta$. If their angular velocities are $\Omega$ and $-\Omega /4$, respectively, the no-slip Stokes flow between them has streamfunctionIts unique interior stagnation point is $(x,y)=(0,2\sqrt{2}a)$, and its nonzero streamlines are closed curves around that point.

For a stationary disk of radius $a$ in the far-field shear $u_{\infty }=\gamma y{e}_x$, using $u_r={\psi }_{\theta }/r$ and $u_{\theta }=-{\psi }_r$, the exterior solution isIt satisfies no slip at $r=a$ and approaches the imposed shear in velocity.

The surface shear stress isThe torque exerted by the fluid on the disk per unit axial length is therefore