Laplace and Poisson Equation

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Derivation of Laplace Equation from Maxwell's Equations

Consider the first of Maxwell's Equations

\nabla\cdot\textbf{D} = \rho.

Remember that \textbf{D} = \epsilon \textbf{E}. Since, for static electric fields, we also recognise that

E = -\nabla\Phi.

Putting these two relationships together yields

\nabla\cdot\nabla\Phi = \nabla^2\Phi = -\frac{\rho}{\epsilon}

This is Poisson's Equation, which, together with suitable boundary conditions, yields the potential in terms of the distribution of charge density ρ.

In regions where the charge density is zero, Poisson's Equation becomes the Laplace Equation

\nabla^2\Phi = 0.

The Laplace Equation is very useful problems where potentials are defined on boundaries and one wishes to compute the field in a source-free region. The Poisson Equation is convenient for problems where charge distributions are used to compute fields.

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General solution of the Poisson Equation

See Dirac Delta Function.

Without sacrificing any generality, we can solve for the "impulse response" of the electrostatic system by deriving a solution for

\nabla^2\Phi_\delta = - \delta(\textbf{r}).

By integrating this equation over a spherical volume that encloses the origin and applying the Divergence Theorem, we see that

\oint_S \nabla\Phi_\delta \cdot \hat{\textbf{n}}\, dS = -1.

If the sphere is "small", the surface integral on the left-hand-side becomes

4\pi r^2 \frac{\partial \Phi_\delta}{\partial r} = -1.

Rearranging yields

\frac{\partial \Phi_\delta}{\partial r} = \frac{-1}{4\pi r^2}.

This is easily integrable in r and we see that the final result for the impulse response potential is

\Phi_\delta = \frac{1}{4\pi r} + V_c\,.

V_c\, is a constant that describes the reference potential (usually zero). Given the spherical symmetry of the delta function source, we know that this is the complete solution.

By shifting the origin to \textbf{r}^\prime\,, the impulse response can be generalised to any point in space:

\Phi_\delta = \frac{1}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert} + V_c.

It is left to the reader to show that this is the solution of

\nabla^2\Phi_\delta = -\delta(\textbf{r}-\textbf{r}^\prime).

(Hint: use the Divergence Theorem on the \nabla^2 operator when \textbf{r}=\textbf{r}^\prime.)

The next stage is to show that any distribution of charge \rho( \textbf{r}^\prime) uniquely defines, within a constant value, a potential everywhere in space.

By using Green's Theorem,

\int_V \left (\Phi_\delta \nabla^2\Phi - \Phi\nabla^2\Phi_\delta\right ) \, dV = \oint_{\partial V}\left (\Phi\nabla\Phi_\delta - \Phi_\delta\nabla\Phi\right ) \cdot\hat{\textbf{n}}\, dS,

we can illustrate what the expression for potential must look like. Assume \Phi\, is the unknown potential and \Phi_\delta\, is the "impulse response" function that we found in the first part of this chapter. All we know is that the function \Phi\, satisfies Poisson's Equation as well as any necessary boundary conditions on conductors as well as at infinity (where it must go to zero). Sources (charges) must be defined over a finite volume.

Inserting what we know and changing the integration to primed (source) coordinates yields

\int_V \left (-\frac{1}{4\pi \vert\textbf{r}- \textbf{r}^\prime\vert} \frac{\rho}{\epsilon} +\delta(\textbf{r}-\textbf{r}^\prime) \Phi\right ) \, dV^\prime = \oint_{\partial V}\left (-\Phi\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3} - \frac{1}{ 4\pi \vert\textbf{r}-\textbf{r}\vert} \nabla\Phi\right ) \cdot\hat{\textbf{n}}\, dS^\prime.

Carrying out the volume integration on the left hand side and the surface integration on the right gives

-\int_V \frac{\rho}{4\pi\epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\Phi(\textbf{r}) = -\oint_{\partial V}\Phi\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3}\cdot\hat{\textbf{n}}\, dS +\oint_{\partial V} \frac{\rho_s}{ 4\pi\epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime.

The surface charge density \rho_s\, in the final surface integral comes about if we recognise that

\textbf{D}\cdot\hat{\textbf{n}} = -\epsilon\nabla\Phi\cdot \hat{\textbf{n}} = \rho_s


The first integral on the left did not change. However, the integration of the product of the delta function and the unknown potential over the problem volume just yields back the potential at the obervation point in the second term. The surface integral is a bit trickier to understand. At infinity, contributions to the surface integral vanish. Only surfaces (like conductors, dielectrics) at finite distances in the problem domain contribute to the potential (since we assume a finite charge distribution). The first surface integral term (over surfaces where potential is fixed) actually collapses to a constant value on metallic boundaries, because the conducting boundary must be an equipotential. As a result, the surface integral

I=\oint_{\partial V}\frac{\textbf{r}-\textbf{r}^\prime}{4\pi \vert\textbf{r}-\textbf{r}^\prime\vert^3}\cdot\hat{\textbf{n}}\, dS

is -1\, if \textbf{r} (the observation point) lies on a surface source and 0 if the observation point is not on the surface.

Hence, we find that

-\int_V \frac{\rho}{4\pi\epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\Phi(\textbf{r}) = \oint_{\partial V} \frac{\rho_s}{ 4\pi\epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime,

where the first surface integral term vanishes for observation points away from surface sources.


If we rearrange, the potential anywhere in space can be written as

\Phi(\textbf{r})= \int_V \frac{\rho}{4\pi \epsilon\vert\textbf{r}- \textbf{r}^\prime\vert} \, dV^\prime +\oint_{\partial V} \frac{\rho_s}{4\pi \epsilon\vert\textbf{r}-\textbf{r}\vert} \, dS^\prime.

In source-free regions (where Laplace's equation is valid) only the surface integral term is needed. Many useful numerical methods have been developed using

\Phi = \oint_{\partial V} \frac{\rho_s}{4\pi \epsilon\vert\textbf{r}-\textbf{r}^\prime\vert} \, dS^\prime

as the basis for solving many complicated problems. Furthermore, in source-free regions, we can make the philosophically important observation that knowing the field along a surface is enough to know the field everywhere in space!

By finding the "impulse response" (known as a "Green's function") of the potential function, we can use what signal-processing people will recognise as convolution to generate solutions for general potentials based on a known distribution of charge. Special techniques can also be used to reconstruct the charge distribution from a known potential distribution. These form the basis of a wide class of computer simulation techniques known as boundary element models.

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