Fourier transform solution to electrostatics Poisson equation?

In summary, the electrostatic poisson's equation can be solved using the Discrete Fourier Transform. The charge density is taken to be a 2D - Gaussian function with a sigma of 1.0. The input to the DFT is taken to be the Fourier transform of the charge density. The output from the DFT is the Fourier transform of the phi Gaussian function. The inverse DFT is used to get the phi Gaussian function in the x-y plane. When the computational and analytical results are compared, they do not match.
  • #1
Peeter
305
3
Am just playing around, and
following examples of Fourier transform solutions of the heat equation, tried the same thing for
the electrostatics Poisson equation
[tex]
\nabla^2 \phi &= -\rho/\epsilon_0 \\
[/tex]

With Fourier transform pairs
[tex]
\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*}
[/tex]

one gets

[tex]
\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*}
[/tex]

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
the
integrals
[tex]
\int_{-\infty}^\infty \frac{1}{{k_1}^2 + {k_2}^2 + {k_3}^3 } e^{ i k_1 x_1 } dk_1
[/tex]

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

[tex]
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} }
[/tex]

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.
 
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  • #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

[tex]
\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k
[/tex]

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.
 
  • #3
I still don't see how to take these integrals. I've tried it a few ways.
 
  • #4
Do you mean you don't see how to evaluate:

[tex]
\iiint \frac{1}{\mathbf{k}^2 + a^2} e^{ i \mathbf{k} \cdot \mathbf{x} } d^3 k
[/tex]

or, you don't see how to get to that point by taking the Fourier transforms?
 
  • #5
Nevermind - I got it.
 
  • #6
Hi, yes, it was taking that integral that I was having trouble with. Thanks.
 
  • #7
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
 

What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its constituent frequencies. It is commonly used in signal processing and physics to analyze and solve differential equations.

How is the Fourier transform used in electrostatics?

The Fourier transform is used in electrostatics to solve the Poisson equation, which describes the relationship between electric charge and electric potential. By applying the Fourier transform, the Poisson equation can be transformed into an algebraic equation that is easier to solve.

What is the Poisson equation in electrostatics?

The Poisson equation is a partial differential equation that relates the electric potential to the distribution of electric charges in a given region. It is an important tool in electrostatics as it allows us to calculate the electric potential at any point in space.

What are the advantages of using the Fourier transform in solving the Poisson equation?

The Fourier transform allows for a simpler and more efficient solution to the Poisson equation compared to traditional methods. It also provides a more intuitive understanding of the relationship between charge and potential in a given system.

Are there any limitations to using the Fourier transform in electrostatics?

The Fourier transform method may not be suitable for all situations, such as when dealing with non-linear materials or boundary conditions. In these cases, other methods may be more appropriate for solving the Poisson equation in electrostatics.

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