Hi, I'll post the question and how far I got with it.

Using the method of images, construct the Green's function of the Neumann problem for Laplace's equation in the half-space $\displaystyle D = {(x,y,z) \in \Re^{3} : z > 0}$. Hence solve:

$\displaystyle \nabla^{2}u = 0, \ \ x\in D$

$\displaystyle \frac{\partial u}{\partial z} = 1 - x^{2} - y^{2}, \ \ x^{2} + y^{2} < 1, \ \ z = 0$

$\displaystyle \frac{\partial u}{\partial z} = 0, \ \ x^{2} + y^{2} > 1, \ \ z = 0$

and evaluate the resulting integral on the z-axis.

=========

So the first thing I did was to write down the fundamental solution to Laplace's equation in 3D, and take off the image points in the lower half-space (z<0). i.e.

$\displaystyle G(x,y,z;,x',y',z') = $ $\displaystyle -\frac{1}{4\pi} \left( \frac{1}{\sqrt{(x-x')^{2} + (y-y')^{2} + (z-z')^{2}}}\right)$ $\displaystyle -\frac{1}{4\pi} \left( \frac{1}{\sqrt{(x-x')^{2} + (y-y')^{2} + (z+z')^{2}}}\right)$

Now to solve, we use Green's identity.

Integrating $\displaystyle u\nabla^{2} G - G\nabla^{2} u \ $ over the upper half-space leaves just $\displaystyle u(x',y',z')$ on one side (since $\displaystyle \nabla^{2} u$ vanishes for $\displaystyle x\in D$).

For the other side, we can write $\displaystyle \frac {\partial}{\partial n} = -\frac {\partial}{\partial z} \ $ since the outward normal is going in the opposite direction to increasing z.

We can also say $\displaystyle \frac {\partial G}{\partial z} = 0$ at $\displaystyle z = 0 \ \ $ (as it's a Neumann problem..?)

$\displaystyle u(x',y',z') = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(x,y,0;,x',y',z')\frac{\partial u}{\partial z}\ dxdy$

We now evaluate this resulting integral on the z-axis to get:

$\displaystyle u(0,0,z') = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} G(x,y,0;,0,0,z')\frac{\partial u}{\partial z}\ dxdy$

I get stuck at this point. In the solutions, the above has been transformed into polars. So on the z-axis, G becomes (leaving the constant $\displaystyle -\frac{1}{4\pi}$ outside the integral),

$\displaystyle \left (\frac{1}{\sqrt{x^{2} + y^{2} + z'^{\ 2}}} + \frac{1}{\sqrt{x^{2} + y^{2} + z'^{\ 2}}} \right)$ = $\displaystyle \frac{2}{\sqrt{r^{2} + z'^{\ 2}}} $

And the $\displaystyle dxdy$ goes to $\displaystyle rdrd\theta$, as usual

The next line of the solution is:

$\displaystyle u(0,0,z') = -\frac{1}{4\pi}\int_{\theta = 0}^{2\pi} \int_{r = 0}^{1} \frac{2}{\sqrt{r^{2} + z'^{\ 2}}} \ rdrd\theta$

I don't understand what's happened to the $\displaystyle \frac{\partial u}{\partial z}$ - is it equal to 1? Something to do with the boundary conditions maybe?

After this step, it looks pretty straightforward, just normal integration.

Any help is much appreciated!

Thanks.