For two Stokes flows $\mathbf{u}^{(1)}(\mathbf{x})$ and $\mathbf{u}^{(2)}(\mathbf{x})$ inside the same volume $V$ with different boundary conditions on its boundary $S$, prove the reciprocal theorem

$$\int_{S} u_{i}^{(1)} \sigma_{i j}^{(2)} n_{j} d S=\int_{S} u_{i}^{(2)} \sigma_{i j}^{(1)} n_{j} d S$$

where $\sigma^{(1)}$ and $\sigma^{(2)}$ are the stress tensors associated with the flows.

Stating clearly any properties of Stokes flow that you require, use the reciprocal theorem to prove that the drag $\mathbf{F}$ on a body translating with uniform velocity $\mathbf{U}$ is given by

$$F_{i}=A_{i j} U_{j},$$

where $\mathbf{A}$ is a symmetric second-rank tensor that depends only on the geometry of the body.

A slender rod falls slowly through very viscous fluid with its axis inclined to the vertical. Explain why the rod does not rotate, stating any properties of Stokes flow that you use.

When the axis of the rod is inclined at an angle $\theta$ to the vertical, the centre of mass of the rod travels at an angle $\phi$ to the vertical. Given that the rod falls twice as quickly when its axis is vertical as when its axis is horizontal, show that

$$\tan \phi=\frac{\sin \theta \cos \theta}{1+\cos ^{2} \theta}$$