Defining the function $G_{f_{3}}\left(\mathbf{r} ; \mathbf{r}_{0}\right)=-1 /\left(4 \pi\left|\mathbf{r}-\mathbf{r}_{0}\right|\right)$, prove Green's third identity for functions $u(\mathbf{r})$ satisfying Laplace's equation in a volume $V$ with surface $S$, namely

$$u\left(\mathbf{r}_{0}\right)=\int_{S}\left(u \frac{\partial G_{f_{3}}}{\partial n}-\frac{\partial u}{\partial n} G_{f_{3}}\right) d S$$

A solution is sought to the Neumann problem for $\nabla^{2} u=0$ in the half plane $z>0$ :

$$u=O\left(|\mathbf{x}|^{-a}\right), \quad \frac{\partial u}{\partial r}=O\left(|\mathbf{x}|^{-a-1}\right) \text { as }|\mathbf{x}| \rightarrow \infty, \quad \frac{\partial u}{\partial z}=p(x, y) \text { on } z=0$$

where $a>0$. It is assumed that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} p(x, y) d x d y=0$. Explain why this condition is necessary.

Construct an appropriate Green's function $G\left(\mathbf{r} ; \mathbf{r}_{0}\right)$ satisfying $\partial G / \partial z=0$ at $z=0$, using the method of images or otherwise. Hence find the solution in the form

$$u\left(x_{0}, y_{0}, z_{0}\right)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} p(x, y) f\left(x-x_{0}, y-y_{0}, z_{0}\right) d x d y$$

where $f$ is to be determined.

Now let

$$p(x, y)= \begin{cases}x & |x|,|y|<a \\ 0 & \text { otherwise }\end{cases}$$

By expanding $f$ in inverse powers of $z_{0}$, show that

$$u \rightarrow \frac{-2 a^{4} x_{0}}{3 \pi z_{0}^{3}} \quad \text { as } \quad z_{0} \rightarrow \infty .$$