How to Solve Poisson's Equation Using Fourier Transforms

Last Updated: June 17, 2017

Poisson’s equation is an important partial differential equation that has broad applications in physics and engineering. This article will deal with electrostatic potentials, though the techniques outlined here can be applied in general.

One way to solve this equation is to perform Fourier transforms (FT) relating the variables both in position space ${\mathbf {x}}$ and in the ${\mathbf {k}}$ space. This converts the equation into an integration problem, which is relatively easier to deal with.

Steps

Download Article

1
Begin with Poisson’s equation. Recall that the electric field ${\mathbf {E}}$ can be written in terms of a scalar potential ${\mathbf {E}}=-\nabla \phi .$ We can then use Gauss’ law to obtain Poisson’s equation as seen in electrostatics.
- $\nabla ^{{2}}\phi =-{\frac {\rho }{\epsilon _{{0}}}}$
- In this equation, it is often the case that we know the charge density $\rho ,$ called the source function, and wish to know the potential $\phi .$ Therefore, we need to find some way to invert this equation.
2
Write out the FTs and inverse FTs of the potential and charge density. Since we are dealing with three dimensions, the FTs are adjusted accordingly, with the constant factor there for normalization purposes. The bounds will differ depending on conventions on where to set the potential to 0. Although we will not explicitly write the boundaries until evaluating the integrals, we will set the potential to 0 at infinity, so that we are integrating over all space.
- ${\begin{aligned}{\tilde {\phi }}({\mathbf {k}})&={\frac {1}{(2\pi )^{{3/2}}}}\int \phi ({\mathbf {x}})e^{{-i{\mathbf {k}}\cdot {\mathbf {x}}}}{\mathrm {d}}^{{3}}{\mathbf {x}}\\\phi ({\mathbf {x}})&={\frac {1}{(2\pi )^{{3/2}}}}\int {\tilde {\phi }}({\mathbf {k}})e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}\end{aligned}}$
- ${\begin{aligned}{\tilde {\rho }}({\mathbf {k}})&={\frac {1}{(2\pi )^{{3/2}}}}\int \rho ({\mathbf {x}})e^{{-i{\mathbf {k}}\cdot {\mathbf {x}}}}{\mathrm {d}}^{{3}}{\mathbf {x}}\\\rho ({\mathbf {x}})&={\frac {1}{(2\pi )^{{3/2}}}}\int {\tilde {\rho }}({\mathbf {k}})e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}\end{aligned}}$
Advertisement
3
Relate ${\tilde {\phi }}({\mathbf {k}})$ with ${\tilde {\rho }}({\mathbf {k}})$ . The result will relate the potential and charge density in the ${\mathbf {k}}$ space, and as it will turn out, the relation is algebraic, which is considerably simpler.
- Take the Laplacian of $\phi ({\mathbf {x}}).$ We can differentiate under the integral here because the integral is being taken with respect to ${\mathbf {k}},$ and ${\mathbf {x}}$ is an independent variable.
  - $\nabla ^{{2}}\phi ({\mathbf {x}})={\frac {1}{(2\pi )^{{3/2}}}}\int -k^{{2}}e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\tilde {\phi }}({\mathbf {k}}){\mathrm {d}}^{{3}}{\mathbf {k}}={\frac {\rho ({\mathbf {x}})}{\epsilon _{{0}}}}$
- FT charge density so that it is also written in the ${\mathbf {k}}$ space.
  - ${\frac {\rho ({\mathbf {x}})}{\epsilon _{{0}}}}={\frac {1}{(2\pi )^{{3/2}}}}\int {\frac {{\tilde {\rho }}({\mathbf {k}})}{\epsilon _{{0}}}}e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}$
- By direct comparison, we see that the below relation holds.
  - $k^{2}{\tilde {\phi }}({\mathbf {k}})={\frac {{\tilde {\rho }}({\mathbf {k}})}{\epsilon _{{0}}}}$
- If we were given charge density in the ${\mathbf {k}}$ space and wanted to find potential in the same space, it would be very easy. However, we are interested in finding these quantities in the ${\mathbf {x}}$ space. Therefore, we will need to transform a second time.
4
Write $\phi ({\mathbf {x}})$ in terms of $\rho ({\mathbf {x}})$ . Inverse FT charge density and simplify the resulting expression. The prime symbols for the dummy variables in line 2 signify that we are taking a separate integral.
- ${\begin{aligned}\phi ({\mathbf {x}})&={\frac {1}{(2\pi )^{{3/2}}}}\int e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\frac {{\tilde {\rho }}({\mathbf {k}})}{k^{{2}}\epsilon _{{0}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}\\&={\frac {1}{(2\pi )^{{3/2}}}}\int e^{{i{\mathbf {k}}\cdot {\mathbf {x}}}}{\frac {{\mathrm {d}}^{{3}}{\mathbf {k}}}{k^{{2}}\epsilon _{{0}}}}\left[{\frac {1}{(2\pi )^{{3/2}}}}\int e^{{-i{\mathbf {k}}\cdot {\mathbf {x}}^{{\prime }}}}\rho ({\mathbf {x}}^{{\prime }}){\mathrm {d}}^{{3}}{\mathbf {x}}^{{\prime }}\right]\\&={\frac {1}{(2\pi )^{{3}}}}\int {\frac {e^{{i{\mathbf {k}}\cdot ({\mathbf {x}}-{\mathbf {x}}^{{\prime }})}}}{k^{{2}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}\int {\frac {\rho ({\mathbf {x}}^{{\prime }})}{\epsilon _{{0}}}}{\mathrm {d}}^{{3}}{\mathbf {x}}^{{\prime }}\end{aligned}}$
5
Evaluate the ${\mathbf {k}}$ space integral. It is easier if we change to spherical coordinates (we are using the physicist's convention). In line 5, we recognize that $\sin {a}={\frac {e^{{ia}}-e^{{-ia}}}{2i}}$ from Euler's formula, and in line 7, we recognize the integral $\int _{{0}}^{{\infty }}{\frac {\sin a}{a}}{\mathrm {d}}a={\frac {\pi }{2}}.$
- ${\begin{aligned}&{\frac {1}{(2\pi )^{{3}}}}\int _{{-\infty }}^{{\infty }}{\frac {e^{{i{\mathbf {k}}\cdot ({\mathbf {x}}-{\mathbf {x}}^{{\prime }})}}}{k^{{2}}}}{\mathrm {d}}^{{3}}{\mathbf {k}}\\=\ &{\frac {1}{(2\pi )^{{3}}}}\int _{{0}}^{{2\pi }}{\mathrm {d}}\phi \int _{{0}}^{{\infty }}k^{{2}}{\mathrm {d}}k\int _{{0}}^{{\pi }}\sin \theta {\mathrm {d}}\theta {\frac {e^{{ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|\cos \theta }}}{k^{{2}}}}\\=\ &{\frac {1}{(2\pi )^{{2}}}}\int _{{0}}^{{\infty }}{\mathrm {d}}k\int _{{0}}^{{\pi }}\sin \theta {\mathrm {d}}\theta e^{{ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|\cos \theta }},\ u=\cos \theta \\=\ &{\frac {1}{(2\pi )^{{2}}}}\int _{{0}}^{{\infty }}{\mathrm {d}}k\int _{{-1}}^{{1}}{\mathrm {d}}ue^{{ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|u}}\\=\ &{\frac {1}{(2\pi )^{{2}}}}\int _{{0}}^{{\infty }}{\mathrm {d}}k{\frac {1}{ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}(e^{{ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}-e^{{-ik|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}})\\=\ &{\frac {1}{2\pi ^{{2}}}}\int _{{0}}^{{\infty }}{\mathrm {d}}k{\frac {1}{k|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}\sin {(k|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|)},\ v=k|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|\\=\ &{\frac {1}{2\pi ^{{2}}|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}\int _{{0}}^{{\infty }}{\mathrm {d}}v{\frac {\sin v}{v}}\\=\ &{\frac {1}{4\pi |{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}\end{aligned}}$
6
Substitute into the equation of the potential $\phi ({\mathbf {x}})$ . This is the general solution to Poisson's equation up to a charge density, where $\phi (\infty )=0.$ The general solution to this equation cannot be written in closed form. Thus, we opt for the integral form, where we integrate the known charge density over all space to find the corresponding potential, though the integration for more complicated charge distributions becomes rather impractical.
- $\phi ({\mathbf {x}})={\frac {1}{4\pi \epsilon _{{0}}}}\int {\frac {\rho ({{\mathbf {x^{{\prime }}}}})}{|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}{\mathrm {d}}^{{3}}{\mathbf {x^{{\prime }}}}$

Expert Q&A

Search

Add New Question

Ask a Question

200 characters left

Include your email address to get a message when this question is answered.

Submit

Tips

For a stationary point charge $q,$ we make use of $q=\rho V$ (the integration is trivial) and recognize that $|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|$ is the distance between the source and test charge to recover the familiar equation.
- $\phi ({\mathbf {x}})={\frac {1}{4\pi \epsilon _{{0}}}}{\frac {q}{|{\mathbf {x}}-{\mathbf {x^{{\prime }}}}|}}$
Thanks
Helpful 0 Not Helpful 0
In step 3, we skip over a few additional steps into relating potential and charge density in the ${\mathbf {k}}$ space. We can set the equation to 0 by putting the integrands under the same integral, since integration is fundamentally a linear operator, and the integration is being carried out over the same object (all of space). Then we can say that the only thing whose integral is 0, is 0 itself, so the integrand can be set to 0, leading us to the derived relation.

Thanks
Helpful 0 Not Helpful 0