Green's function representation of electric potential

1. Jun 25, 2012

hunt_mat

Hi,

I have the following problem, I have an electric field (which no charge) which satisfies the usual Laplace equation:
$$\frac{\partial^{2}V}{\partial x^{2}}+\frac{\partial^{2}V}{\partial y^{2}}+\frac{\partial^{2}V}{\partial z^{2}}=0$$
in the region $\mathbb{R}^{2}\times [\eta ,\infty ]$. So basically it is the upper half z-plane where the boundary is some fixed surface $\eta$, I also know that on this surface:
$$\frac{\partial V}{\partial x}=\frac{\partial\eta}{\partial x}$$

I can do this in 2D by the use of the Hilbert transform. Any suggestions?

2. Jun 25, 2012

Muphrid

Let me make sure I understand: this is a region for $z > \eta$ for some fixed $\eta$, or is $\eta$ a function?

3. Jun 25, 2012

hunt_mat

$\eta$ is a function but with further thought the region could be set to $z\geqslant 0$ and I think that this will make the problem easier.

4. Jun 25, 2012

Muphrid

And if I read what you said correctly, you only know the value of the x-component of the electric field on this surface?

5. Jun 25, 2012

hunt_mat

I think that you can also say that $V=\eta$ due to other considerations too.

6. Jun 28, 2012

hunt_mat

So I think I have solved this problem, I took $V$ and $\eta$ to be of the same size but small and reduced the complexity of the problem somewhat and the domain is now: $\mathbb{R}^{2}\times [ 0,\infty )$, using Green's second formula, I can write the solution as an integral over the boundary:
$$V(x,y,z)=\int_{\mathbb{R}^{2}}g\partial_{z}V-V\partial_{z}g\Big|_{z'=0}d\Sigma$$
Where $g$ is the Green's function for Laplaces's equation for the half space given by:
$$g(x,y,x|x',y',z')=\frac{1}{4\pi\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}}-\frac{1}{4\pi\sqrt{(x-x')^{2}+(y-y')^{2}+(z+z')^{2}}}$$
Then using the boundary condition $V=\eta$, then the solution becomes:
$$u=\int_{\mathbb{R}^{2}}u\frac{\partial g}{\partial z}d\Sigma$$

Last edited: Jun 28, 2012