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Fourier transform solution to electrostatics Poisson equation?

  1. Feb 18, 2009 #1
    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
    \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 \\

    one gets

    \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

    Now it seems to me that this integral [itex]G[/itex] only has to be evaluated around a small neighbourhood of the origin. For example if one evaluates one of
    \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 [itex]k_2 = k_3 = 0[/itex]. 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 [itex]G(\mathbf{x}) = \frac{1}{4\pi\lvert{\mathbf{x}}\rvert}[/itex].
    Does anybody know of a technique that would reduce the integral limit expression above for [itex]G[/itex] to the [itex]1/x[/itex] 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. jcsd
  3. Feb 19, 2009 #2
    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 [itex]a[/itex] tend to zero.
  4. Sep 13, 2010 #3
    I still don't see how to take these integrals. I've tried it a few ways.
  5. Sep 13, 2010 #4
    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?
  6. Sep 13, 2010 #5
    Nevermind - I got it.
  7. Sep 13, 2010 #6
    Hi, yes, it was taking that integral that I was having trouble with. Thanks.
  8. Mar 1, 2011 #7

    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.

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