The function $\phi(x, y, z)$ satisfies $\nabla^{2} \phi=0$ in $V$ and $\phi=0$ on $S$, where $V$ is a region of $\mathbb{R}^{3}$ which is bounded by the surface $S$. Prove that $\phi=0$ everywhere in $V$.

Deduce that there is at most one function $\psi(x, y, z)$ satisfying $\nabla^{2} \psi=\rho$ in $V$ and $\psi=f$ on $S$, where $\rho(x, y, z)$ and $f(x, y, z)$ are given functions.

Given that the function $\psi=\psi(r)$ depends only on the radial coordinate $r=|\mathbf{x}|$, use Cartesian coordinates to show that

$$\nabla \psi=\frac{1}{r} \frac{d \psi}{d r} \mathbf{x}, \quad \nabla^{2} \psi=\frac{1}{r} \frac{d^{2}(r \psi)}{d r^{2}}$$

Find the general solution in this radial case for $\nabla^{2} \psi=c$ where $c$ is a constant.

Find solutions $\psi(r)$ for a solid sphere of radius $r=2$ with a central cavity of radius $r=1$ in the following three regions:

(i) $0 \leqslant r \leqslant 1$ where $\nabla^{2} \psi=0$ and $\psi(1)=1$ and $\psi$ bounded as $r \rightarrow 0$;

(ii) $1 \leqslant r \leqslant 2$ where $\nabla^{2} \psi=1$ and $\psi(1)=\psi(2)=1$;

(iii) $r \geqslant 2$ where $\nabla^{2} \psi=0$ and $\psi(2)=1$ and $\psi \rightarrow 0$ as $r \rightarrow \infty$.