Show that two-dimensional Stokes flow $\mathbf{u}=(u(r, \phi), v(r, \phi), 0)$ in cylindrical polar coordinates $(r, \phi, z)$ has a stream function $\psi(r, \phi)$, with $u=r^{-1} \partial \psi / \partial \phi, v=-\partial \psi / \partial r$, that satisfies the biharmonic equation

$$\nabla^{4} \psi=0$$

Give, in terms of $\psi$ and/or its derivatives, the boundary conditions satisfied by $\psi$ on an impermeable plane of constant $\phi$ which is either (a) rigid or (b) stress-free.

A rigid plane passes through the origin and lies along $\phi=-\alpha$. Fluid with viscosity $\mu$ is confined in the region $-\alpha<\phi<0$. A uniform tangential stress $S$ is applied on $\phi=0$. Show that the resulting flow may be described by a stream function $\psi$ of the form $\psi(r, \phi)=S r^{2} f(\phi)$, where $f(\phi)$ is to be found. Hence show that the radial flow $U(r)=u(r, 0)$ on $\phi=0$ is given by

$$U(r)=\frac{S r}{\mu}\left(\frac{1-\cos 2 \alpha-\alpha \sin 2 \alpha}{\sin 2 \alpha-2 \alpha \cos 2 \alpha}\right)$$

By expanding this expression for small $\alpha$ show that $U$ and $S$ have the same sign, provided that $\alpha$ is not too large. Discuss the situation when $\alpha>\alpha_{c}$, where tan $2 \alpha_{c}=2 \alpha_{c}$.

[Hint: In plane polar coordinates

$$\nabla^{2}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r} \frac{\partial}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2}}{\partial \phi^{2}}$$

and the component $\sigma_{r \phi}$ of the stress tensor takes the form

$$\sigma_{r \phi}=\mu\left(r \frac{\partial(v / r)}{\partial r}+\frac{1}{r} \frac{\partial u}{\partial \phi}\right)$$