In a region $R$ of $\mathbb{R}^{3}$ bounded by a closed surface $S$, suppose that $\phi_{1}$ and $\phi_{2}$ are both solutions of $\nabla^{2} \phi=0$, satisfying boundary conditions on $S$ given by $\phi=f$ on $S$, where $f$ is a given function. Prove that $\phi_{1}=\phi_{2}$.

In $\mathbb{R}^{2}$ show that

$$\phi(x, y)=\left(a_{1} \cosh \lambda x+a_{2} \sinh \lambda x\right)\left(b_{1} \cos \lambda y+b_{2} \sin \lambda y\right)$$

is a solution of $\nabla^{2} \phi=0$, for any constants $a_{1}, a_{2}, b_{1}, b_{2}$ and $\lambda$. Hence, or otherwise, find a solution $\phi(x, y)$ in the region $x \geqslant 0$ and $0 \leqslant y \leqslant a$ which satisfies:

$$\begin{aligned} &\phi(x, 0)=0, \quad \phi(x, a)=0, \quad x \geqslant 0 \\ &\phi(0, y)=\sin \frac{n \pi y}{a}, \quad \phi(x, y) \rightarrow 0 \quad \text { as } \quad x \rightarrow \infty, \quad 0 \leqslant y \leqslant a \end{aligned}$$

where $a$ is a real constant and $n$ is an integer.