(a) The function $u$ satisfies $\nabla^{2} u=0$ in the volume $V$ and $u=0$ on $S$, the surface bounding $V$.

Show that $u=0$ everywhere in $V$.

The function $v$ satisfies $\nabla^{2} v=0$ in $V$ and $v$ is specified on $S$. Show that for all functions $w$ such that $w=v$ on $S$

$$\int_{V} \nabla v \cdot \nabla w d V=\int_{V}|\nabla v|^{2} d V$$

Hence show that

$$\int_{V}|\boldsymbol{\nabla} w|^{2} d V=\int_{V}\left\{|\boldsymbol{\nabla} v|^{2}+|\boldsymbol{\nabla}(w-v)|^{2}\right\} d V \geqslant \int_{V}|\boldsymbol{\nabla} v|^{2} d V$$

(b) The function $\phi$ satisfies $\nabla^{2} \phi=\rho(\mathbf{x})$ in the spherical region $|\mathbf{x}|<a$, with $\phi=0$ on $|\mathbf{x}|=a$. The function $\rho(\mathbf{x})$ is spherically symmetric, i.e. $\rho(\mathbf{x})=\rho(|\mathbf{x}|)=\rho(r)$.

Suppose that the equation and boundary conditions are satisfied by a spherically symmetric function $\Phi(r)$. Show that

$$4 \pi r^{2} \Phi^{\prime}(r)=4 \pi \int_{0}^{r} s^{2} \rho(s) d s$$

Hence find the function $\Phi(r)$ when $\rho(r)$ is given by $\rho(r)=\left\{\begin{array}{ll}\rho_{0} & \text { if } 0 \leqslant r \leqslant b \\ 0 & \text { if } b<r \leqslant a\end{array}\right.$, with $\rho_{0}$ constant.

Explain how the results obtained in part (a) of the question imply that $\Phi(r)$ is the only solution of $\nabla^{2} \phi=\rho(r)$ which satisfies the specified boundary condition on $|\mathbf{x}|=a$.

Use your solution and the results obtained in part (a) of the question to show that, for any function $w$ such that $w=1$ on $r=b$ and $w=0$ on $r=a$,

$$\int_{U(b, a)}|\nabla w|^{2} d V \geqslant \frac{4 \pi a b}{a-b}$$

where $U(b, a)$ is the region $b<r<a$.