# Fourier transform solution to electrostatics Poisson equation?

1. Feb 18, 2009

### Peeter

Am just playing around, and
following examples of Fourier transform solutions of the heat equation, tried the same thing for
the electrostatics Poisson equation
$$\nabla^2 \phi &= -\rho/\epsilon_0 \\$$

With fourier transform pairs
\begin{align*} \hat{f}(\mathbf{k}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x} } d^3 x \\ {f}(\mathbf{x}) &= \frac{1}{(\sqrt{2\pi})^3} \iiint \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x} } d^3 k \\ \end{align*}

one gets

\begin{align*} \phi(\mathbf{x}) &= \frac{1}{\epsilon_0} \int \rho(\mathbf{x}') G(\mathbf{x-x'}) d^3 x' \\ G(\mathbf{x}) &= \frac{1}{(2 \pi)^3} \iiint \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k \end{align*}

Now it seems to me that this integral $G$ only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of
the
integrals
$$\int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1$$

using a an upper half plane contour the result is zero unless $k_2 = k_3 = 0$. So one is left with something loosely like

$$G(\mathbf{x}) &= \lim_{\epsilon \rightarrow 0} \frac{1}{(2 \pi)^3} \int_{k_1 = -\epsilon}^{\epsilon} dk_1 \int_{k_2 = -\epsilon}^{\epsilon} dk_2 \int_{k_3 = -\epsilon}^{\epsilon} dk_3 \frac{1}{\mathbf{k}^2} e^{ i \mathbf{k} \cdot \mathbf{x} }$$

However, from electrostatics we also know that the solution to the Poission equation means that $G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}$.
Does anybody know of a technique that would reduce the integral limit expression above for $G$ to the $1/x$ form? Am thinking something residue related, but I'm a bit rusty with my complex variables and how exactly to procede isn't obvious.

Last edited: Feb 18, 2009
2. Feb 19, 2009

### Peeter

found the answer in a book recently purchased, but not yet read (Mathematics of Classical and Quantum Physics). They cleverly introduce a pole in the upper half plane by evaluating

$$\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k$$

After a change to spherical polar coordinates, that new integral can be evaluated, and the Poisson Green's function follows by letting $a$ tend to zero.

3. Sep 13, 2010

### gn0m0n

I still don't see how to take these integrals. I've tried it a few ways.

4. Sep 13, 2010

### Peeter

Do you mean you don't see how to evaluate:

$$\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k$$

or, you don't see how to get to that point by taking the Fourier transforms?

5. Sep 13, 2010

### gn0m0n

Nevermind - I got it.

6. Sep 13, 2010

### gn0m0n

Hi, yes, it was taking that integral that I was having trouble with. Thanks.

7. Mar 1, 2011

### ajit_up

hi

I am trying to solve the electrostatic poisson's equation mentioned in the first post in 2D using Discrete Fourier Transform (I am using fftw3 library and REDFT10 / REDFT01 transforms). For my problem, I have charge densities given in a 2D plane at discrete points and I have to find out potential at those points. The range of solution, and boundary condition are :

0 < x < L1 and 0 < y < L2 and Dirichlet boundary condition.

I have taken phi to be 2D - Gaussian function (peaked at center of 2D grid) with sigma = 1.0, phi0 = 10.0, so that I am able to check the results analytically too.

I am writing below the steps:

1. I take 2D - DFT of charge densitiy
2. Solve the equation in fourier space for phi in fourier space
3. take inverse 2D - DFT of phi to get phi in x-y plane

When I compare my computational result with analytical result, they do not match.
Can anybody help me in understanding the reason and finding the right way to solve Poisson's equation using DFTs.

thanks