# Thread: Green's function for Neumann problem

1. ## Green's function for Neumann problem

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 $D = {(x,y,z) \in \Re^{3} : z > 0}$. Hence solve:

$\nabla^{2}u = 0, \ \ x\in D$
$\frac{\partial u}{\partial z} = 1 - x^{2} - y^{2}, \ \ x^{2} + y^{2} < 1, \ \ z = 0$
$\frac{\partial u}{\partial z} = 0, \ \ x^{2} + y^{2} > 1, \ \ z = 0$

and evaluate the resulting integral on the z-axis.

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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.

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

Now to solve, we use Green's identity.

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

For the other side, we can write $\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 $\frac {\partial G}{\partial z} = 0$ at $z = 0 \ \$ (as it's a Neumann problem..?)

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

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

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

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 $-\frac{1}{4\pi}$ outside the integral),

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

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

The next line of the solution is:

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

I don't understand what's happened to the $\frac{\partial u}{\partial z}$ - 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.

2. what's happened to the $\frac{\partial u}{\partial z}$ - is it equal to 1? Something to do with the boundary conditions maybe?
Εxactly. You have $\frac{\partial u}{\partial z}(x,y,0)=1-x^2-y^2$ for $x^2+y^2<1$, so $\frac{\partial u}{\partial z}(0,0,0)=1$.