# Solving Poisson's equation with the help of Greens function

1. Jan 30, 2012

### spookyfw

Hey all,
some weeks ago in a tutorial our TA solved Poissons equation with Greens functions..would be very short, but he also derived the Greens function using a Fourier transform. Two points I really don't get and he could also not explain it. Maybe you can help me? There might be even a short way..but if you could help out here and tell me what went wrong I would be really helpful :)!

Function to solve:
$\nabla^2$ $\Phi$ = -$\frac{\rho}{\epsilon}$ = f(x) = $\nabla^2$ u(x)
Having the Green's function:

$\nabla^2$ G(r,r') = $\delta$(r-r')

$\nabla^2$ G(x) = $\delta$(x)
taking the fourier transform:
∫$\nabla^2$ G(x)exp(-ikx)dV = 1

and now the big mess is starting: using Greens first identity, namely:

∫$\nabla*\nabla$G exp(-ikx)dV = ∫$\nabla$ G exp(-ikx) dS - ∫$\nabla$G $\nabla$ exp(-ikx)dV

then he states that the surface term is equal to zero. I understand that using the fourier transform we integrate from minus to plus infinity and that we have the condition that our Greens function is usually prone to go to zero at infinity, but how is he doing that step? A integration by parts?

From that he is then coming to the point that
∫G Δexp(-ikx) dV = k^2 ∫ G(r) exp(-ikx) dV
After a substitution he then concludes that G(k) = 1/k^2

To be honest...after applying Greens identity i am somewhat lost. And don't see the steps anymore ;(.Maybe I copied something wrong from the blackboard..it is just driving me nuts!

In the end when transforming back he actually integrates from zero to inf
∫sinc(u) du = pi/2

I always thought this is not possible...that the integral of sinc cannot be given analytically?

As you might have guessed..it is a big constrution site for me...would be very happy if someone could help me out!!

2. Jan 31, 2012

### Born2bwire

$$\int ( \nabla^2 G ) e^{-ikx}dV = \int (\nabla G) e^{-ikx} dS - \int \nabla G \nabla e^{-ikx}V$$
This is a simple invocation of the chain rule (which you can rearrange to make something like integration by parts) and using the divergence theorem on the first term. Then he does it again to remove the gradient on the Green's function in the remaining term on the right hand side. So from there it's a simple step to see that the k-space Green's function is something like 1/k^2. However, the following,

∫G Δexp(-ikx) dV = k^2 ∫ G(r) exp(-ikx) dV

is incorrect. The gradient of the exponential is simply -ik, you are confusing this step. Work out the integration by parts step twice to see that

$$1 = k^2 \int d\mathbf{r} G(\mathbf{r})e^{-i\mathbf{k}\cdot\mathbf{r}}$$

where we see that the integration on the right hand side is simply the Fourier transform of the spatial Green's function. Then you apply the inverse Fourier transform and convert it to spherical coordinates in the k-space (above I made the wavevector to be the 3D cartesian coordinate system). So when you do that you then arrive at the integral over the sinc function. Something like,

$$G(\mathbf{r}) = \frac{1}{2\pi^2} \int_0^\infty d k \ \rm{sinc} (k r)$$

You can then just do a change of variables to integrate over the variable k*r, which is why you pull out a factor of r in the denominator and then this integration over the sinc corrects the constants. In the end, you should get a Green's function that is something like,

$$G(\mathbf{r},\mathbf{r'}) = \frac{1}{\left| \mathbf{r}-\mathbf{r'} \right|}$$

There may be a factor of 4\pi in the denominator, looks like it will be for your system of units.

3. Feb 5, 2012

### spookyfw

Thank you very much Born2bwire! Now I also get: ∫(∇2G)e−ikxdV=∫(∇G)e−ikxdS−∫∇G∇e−ikxV.
That makes sense now, but when I apply the 'integration-by-parts' again on the last term, I have a slight problem because the divergence theorem..well is working with divergences, but here we have a grad, no? I don't see how u can simply apply it again and get grid of the grad. U would just blow up the terms and not get rid of them...
what is my mistake here?

4. Feb 5, 2012

### Born2bwire

The first is easiest by just looking at using the chain rule. You take,

$$\nabla \cdot \left[ \left(\nabla G \right) e^{-i\mathbf{k}\cdot\mathbf{r}} \right] = \left( \nabla^2 G \right) e^{-i\mathbf{k}\cdot\mathbf{r}} + \left(\nabla G \right) \nabla e^{-i\mathbf{k}\cdot\mathbf{r}} = \left( \nabla^2 G \right) e^{-i\mathbf{k}\cdot\mathbf{r}} - ik \left(\nabla G \right) e^{-i\mathbf{k}\cdot\mathbf{r}}$$

You can then rearrange it so that the Laplacian is on the LHS and the rest are on the RHS. Obviously the volume integral over the divergence because a surface integral because of the divergence theorem. Since the volume here is infinite, we know that this surface integral is zero because the gradient of the Green's function is zero at infinity (likewise the Green's function at infinity is also zero). This is deduced from physical insight of our problem where we know that we consider the potential of a point source at infinity to be zero.

Thus, we are now interested in the volume integral,

$$-ik \int d\mathbf{r} \left(\nabla G \right) e^{-i\mathbf{k}\cdot\mathbf{r}}$$

Once again we use the chain rule,

$$\nabla \left( G e^{-i\mathbf{k}\cdot\mathbf{r}} \right) = \left( \nabla G \right) e^{-i\mathbf{k}\cdot\mathbf{r}} + G \nabla e^{-i\mathbf{k}\cdot\mathbf{r}} = \left( \nabla G \right) e^{-i\mathbf{k}\cdot\mathbf{r}} - ik G e^{-i\mathbf{k}\cdot\mathbf{r}}$$

Thus, to evaluate the gradient of the LHS, we can use the fundamental rule of calculus and it just because the evaluation of the Green's function and the exponential function at the limits of integration (infinity which again evaluates to zero). Consider for example, the problem in Cartesian coordinates. In one-dimension we have,

$$\int_{-\infty}^{\infty} dx \frac{\partial}{\partial x} \left( G e^{-ik_xx} \right) = \left. G e^{-ik_xx} \right|_{-\infty}^\infty$$

A simple example. Then in two-dimensions,

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dx dy \left[ \frac{\partial}{\partial x} \left( G e^{-i(k_xx+k_yy)} \right) \hat{x} + \frac{\partial}{\partial y} \left( G e^{-i(k_xx+k_yy)} \right) \hat{y} \right] = \int_{-\infty}^{\infty} dy \left. G e^{-i(k_xx+k_yy)} \right|_{x=-\infty}^{x=\infty} \hat{x} + \int_{-\infty}^{\infty} dx \left. G e^{-i(k_xx+k_yy)} \right|_{y=-\infty}^{y=\infty} \hat{y}$$

The extension to three-dimensions is not hard to imagine. So we are just left with my second equation given in my previous post.

EDIT: Regarding my last post,

This is correct, I was misinterpreting your symbol for the Laplacian to meaning gradient.

Also, I have been a bit sloppy in the above, obviously we should be getting scalars but I seem to have accidentally omitted the dot products with the volume elements in the above when we have gradients.

Last edited: Feb 5, 2012
5. Feb 6, 2012

### spookyfw

Excellent. Now I a understand. Thank you very much and thanks for your time :)!!!