Poisson's Equation

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\nabla ^2 u(\mathbf{r})=\rho(\mathbf{r}) \label{pg}
 
\nabla ^2 u(\mathbf{r})=\rho(\mathbf{r}) \label{pg}
 
\end{align}
 
\end{align}
 +
Poisson's equation arises in Electrostatics by plugging in the expression $\mathbf{E}=-\nabla u(\mathbf{r})$ with the potential $u(\mathbf{r})$ into the first [[Maxwellgleichungen|Maxwell Equation]] $\nabla \mathbf{E}=\rho(\mathbf{r})$.
 
In a [[Greens Function for PDEs|related article]] we found that the solution to an inhomogenious linear PDE is
 
In a [[Greens Function for PDEs|related article]] we found that the solution to an inhomogenious linear PDE is
 
\begin{align}
 
\begin{align}
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Solving this integrals is in general difficult. Thus we employ the follwoing idea. According to Gauß's Law only charges within a volume are sources of the electric field. If we place a charge outside the volume $V$ we integrate over, it does hot add to the potential (and hence the electric field). We can add any charge $H(\mathbf{r},\mathbf{r}_0)$ outside the volume $V$ that satisfies the homogenious PDE (Laplace's equation). Green's function then takes the form
 
Solving this integrals is in general difficult. Thus we employ the follwoing idea. According to Gauß's Law only charges within a volume are sources of the electric field. If we place a charge outside the volume $V$ we integrate over, it does hot add to the potential (and hence the electric field). We can add any charge $H(\mathbf{r},\mathbf{r}_0)$ outside the volume $V$ that satisfies the homogenious PDE (Laplace's equation). Green's function then takes the form
 
\begin{align}
 
\begin{align}
G(\mathbf{r},\mathbf{r}_0)=-\frac{1}{4 \pi} \frac{1}{|\mathbf{r}-\mathbf{r}_0|}+H(\mathbf{r},\mathbf{r}_0)
+
G(\mathbf{r},\mathbf{r}_0)=F(\mathbf{r},\mathbf{r}_0)+H(\mathbf{r},\mathbf{r}_0)=-\frac{1}{4 \pi} \frac{1}{|\mathbf{r}-\mathbf{r}_0|}+H(\mathbf{r},\mathbf{r}_0)
 
\end{align}
 
\end{align}
 
The inhomogenious PDE is still satisfied since
 
The inhomogenious PDE is still satisfied since
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===Neumann Problem===
 
===Neumann Problem===
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In a Neumann Problem $\partial u (\mathbf{r}) / \partial n = f(\mathbf{r})$ is given on some surface $\mathcal{S}$. In the Dirichlet problem we choose $u(\mathbf{r})=0$ to solve \ref{gensol}, in a similar way we might choose $\partial G(\mathbf{r},\mathbf{r}_0) / \partial n=0$, but in general this is not permitted, since
 +
\begin{align}
 +
\int_\mathcal{S} \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} d\mathcal{S} = \int_\mathcal{S} \nabla G(\mathbf{r},\mathbf{r}_0) \cdot \hat{\mathbf{n}} d\mathcal{S}=\int_\mathcal{V} \nabla^2 G(\mathbf{r},\mathbf{r}_0) d\mathcal{V}=1
 +
\end{align}
 +
The easiest way to satisfy this condition is to set
 +
\begin{align}
 +
\frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n}=\frac{1}{A}
 +
\end{align}
 +
for $\mathbf{r}$ on $\mathcal{S}$ with surface area $A$. The solution \ref{gensol} now becomes
 +
\begin{align}
 +
u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV + \frac{1}{A} \oint_\mathcal{S}  u(\mathbf{r}_0) \, d\mathcal{S} - \oint_\mathcal{S} G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0)  \, d\mathcal{S}
 +
\end{align}
 +
Where the term
 +
\begin{align}
 +
\langle u(\mathbf{r}) \rangle_\mathcal{S} = \frac{1}{A} \oint_\mathcal{S}  u(\mathbf{r}_0) \, d\mathcal{S}
 +
\end{align}
 +
is nothing but the average of $u$ over the surface $d\mathcal{S}$, so the solution can be written in the form.
 +
\begin{align}
 +
u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV + \langle u(\mathbf{r}) \rangle_\mathcal{S} - \oint_\mathcal{S} G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0)  \, d\mathcal{S}
 +
\end{align}
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
----
 +
Further Reading:
 +
*Riley, Hobson, Bence - Mathamatical Methods for Physics and Engineering
 +
*John David Jackson - Klassische Elektrodynamik

Latest revision as of 09:39, 9 July 2013

In this article we discuss the solution of inhomogenious Laplace's Equation, which is called Poisson's equation. \begin{align} \nabla ^2 u(\mathbf{r})=\rho(\mathbf{r}) \tag{1} \end{align} Poisson's equation arises in Electrostatics by plugging in the expression $\mathbf{E}=-\nabla u(\mathbf{r})$ with the potential $u(\mathbf{r})$ into the first Maxwell Equation $\nabla \mathbf{E}=\rho(\mathbf{r})$. In a related article we found that the solution to an inhomogenious linear PDE is \begin{align} u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0)\rho(\mathbf{r}_0)dV(\mathbf{r}_0) \tag{2}=G(\mathbf{r},\mathbf{r}_0) * \rho(\mathbf{r}_0) \end{align}

where we integrate over some volume $V$ surrounded by the surface $S$ containing the charge distribution $\rho(\mathbf{r}_0)$. To avoid confusion I want to make clear explicitly that we need to distinguish between in/homogenious PDE's and in/homogenious boundary conditions. Inhomogenious PDE's are given by $\mathcal{L}u=\rho$, whereas homogenious PDE's have $\rho=0$. Homogenious boundary conditions mean, that there are essentially no boundary conditions, in electrostatics these are problems with e.g. charges in free space. Inhomogenious boundary conditions are those where additionally to a given charge distribution in space, the solution is constrained to have a certain potential on some region $\mathcal{S}$ surrounding the volume $\mathcal{V}$. For example a charge distribution is kept in some metal sphere, which itself is held at a certain potential. We distinguish two boundary-value problems

  • Dirichlet boundary conditions, where $u(\mathbf{r})$ is specified on some surface $\mathcal{S}$ (e.g. the metal sphere)
  • Neumann boundary conditions, where $\partial u(\mathbf{r})/ \partial n$ is given on some surface $\mathcal{S}$
Figure shows the charge distribution $\rho(\mathbf{r}_0)$ in the volume $V$ surrounded by the surface $S$, which is to distinguish from the Volume $\mathcal{V}$ and the surface $\mathcal{S}$ on which the boundary conditions are given.

Example


Find the solution $G(\mathbf{r},\mathbf{r}_0)$ for Poisson's equation (inhomogenious Laplace's equation) with homogenious boundary conditions in three dimensions that tends to zero as $|\mathbf{r}|\rightarrow \infty$. \begin{align} \nabla ^2 G(\mathbf{r},\mathbf{r}_0)=\delta(\mathbf{r}-\mathbf{r}_0) \tag{3} \end{align} In other words, find the Green's function for Poisson's Equation. First notice that the problem is spherically symmetric about $\mathbf{r}_0$. To find $G(\mathbf{r},\mathbf{r}_0)$ we integrate over the sphere with Volume $V$ and Surface $S$ centered at $\mathbf{r}_0$. \begin{align} \int_V \nabla ^2 G(\mathbf{r},\mathbf{r}_0) \, dV= \int_V \delta(\mathbf{r}-\mathbf{r}_0) \,dV =1 \tag{4} \end{align} and following Gauß's theorem \begin{align} \int_V \nabla \cdot \nabla G(\mathbf{r},\mathbf{r}_0) \, dV= \oint_S \nabla G(\mathbf{r},\mathbf{r}_0) \, d\mathbf{S} \end{align} We expect the solution to be spherically symmetric about $\mathbf{r}_0$, this means the solution is constant on spheres with a particular radius, thus $G(\mathbf{r},\mathbf{r}_0)$ must be a function of $|\mathbf{r}-\mathbf{r}_0|$ \begin{align} G(\mathbf{r},\mathbf{r}_0)=G(|\mathbf{r}-\mathbf{r}_0|)=G(r) \end{align} where $r$ is the radial distance from $\mathbf{r}_0$. Equation (4) becomes \begin{align} \oint_S \nabla_r G(r) \, d\mathbf{S}=4\pi r^2 \nabla_r G(r)=1 \end{align} Integrating we find the solution \begin{align} \int \nabla_r G(r) dr=\int \frac{1}{4\pi r^2} dr=-\frac{1}{4\pi r} \end{align} which is

Green's function for homogenious boundary conditions \begin{align} G(\mathbf{r},\mathbf{r}_0)=G(|\mathbf{r}-\mathbf{r}_0|)=-\frac{1}{4\pi}\frac{1}{|\mathbf{r}-\mathbf{r}_0|} \end{align}

The solution to a inhomogenious PDE is also called Fundamental solution.


[edit] Boundary Value Problems

To find the solution to a inhomogenious boundary value problem we use Green's second theorem \begin{align} \oint_\mathcal{S} \left[ \phi \, \frac{\partial \psi}{\partial n} -\psi \, \frac{\partial \phi}{\partial n} \right] \, d\mathcal{S}=\int_\mathcal{V} \left[ \phi \, \nabla^2 \psi-\psi \, \nabla^2 \phi \right] \,d\mathcal{V} \end{align} and choose $\phi=u(\mathbf{r})$ and $\psi=G(\mathbf{r},\mathbf{r}_0)$ \begin{align} \oint_\mathcal{S} \left[u(\mathbf{r}) \, \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} -G(\mathbf{r},\mathbf{r}_0) \, \frac{\partial u(\mathbf{r})}{\partial n}\right] \, d\mathcal{S}=\int_\mathcal{V} \left[ u(\mathbf{r}) \, \nabla^2 G(\mathbf{r},\mathbf{r}_0)-G(\mathbf{r},\mathbf{r}_0) \, \nabla^2 u(\mathbf{r})\right] \,d\mathcal{V} \end{align} with (3) and (1) the RHS becomes \begin{align} \oint_\mathcal{S} \left[ u(\mathbf{r}) \, \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} -G(\mathbf{r},\mathbf{r}_0) \, \frac{\partial u(\mathbf{r})}{\partial n} \right] \, d\mathcal{S}=\int_\mathcal{V} \left[ u(\mathbf{r}) \, \delta(\mathbf{r}-\mathbf{r}_0)-G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}) \right] \,d\mathcal{V} \end{align} With the integral $\int_V u(\mathbf{r}) \, \delta(\mathbf{r}-\mathbf{r}_0) \, dV=u(\mathbf{r}_0)$ the solution for $u(\mathbf{r}_0)$ can be written as \begin{align} u(\mathbf{r}_0)=\int_\mathcal{V} G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}) \,d\mathcal{V} +\oint_\mathcal{S} \left[ u(\mathbf{r}) \, \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} -G(\mathbf{r},\mathbf{r}_0) \, \frac{\partial u(\mathbf{r})}{\partial n}\right]\, d\mathcal{S} \end{align} Notice that following the property $G(\mathbf{r},\mathbf{r}_0)=G(\mathbf{r}_0,\mathbf{r})$, we can switch the role of $\mathbf{r}_0$ and $\mathbf{r}$ to find a solution $u(\mathbf{r})$. The vector $\mathbf{r}_0$ lies within the volume $V$, since the charge distribution outside $V$ is zero $\int_\mathcal{V} G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,d\mathcal{V}=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV$. \begin{align} u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV +\oint_\mathcal{S} \left[ u(\mathbf{r}_0) \, \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} -G(\mathbf{r},\mathbf{r}_0) \, \frac{\partial u(\mathbf{r}_0)}{\partial n}\right]\, d\mathcal{S} \tag{5} \end{align} The right most term in brackets contains $u(\mathbf{r})$ as well as $\partial u(\mathbf{r})/ \partial n$. However, a problem where both $u(\mathbf{r})$ as well as $\partial u(\mathbf{r})/ \partial n$ are specified is overdetermined and there will be no solution to the problem. Thus, for a solvable problem eather $u(\mathbf{r})$ or $\partial u(\mathbf{r})/ \partial n$ is specified on some surface $\mathcal{S}$.

[edit] Dirichlet Problem

In a Dirichlet Problem the solution of Poisson's equation $u(\mathbf{r})$ takes specific values $f(\mathbf{r})$ on the surface $\mathcal{S}$ and $G(\mathbf{r},\mathbf{r}_0)=0$ on the surface $\mathcal{S}$, so that (5) becomes \begin{align} u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV +\oint_\mathcal{S} f(\mathbf{r}) \, \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} \, d\mathcal{S} \end{align} Solving this integrals is in general difficult. Thus we employ the follwoing idea. According to Gauß's Law only charges within a volume are sources of the electric field. If we place a charge outside the volume $V$ we integrate over, it does hot add to the potential (and hence the electric field). We can add any charge $H(\mathbf{r},\mathbf{r}_0)$ outside the volume $V$ that satisfies the homogenious PDE (Laplace's equation). Green's function then takes the form \begin{align} G(\mathbf{r},\mathbf{r}_0)=F(\mathbf{r},\mathbf{r}_0)+H(\mathbf{r},\mathbf{r}_0)=-\frac{1}{4 \pi} \frac{1}{|\mathbf{r}-\mathbf{r}_0|}+H(\mathbf{r},\mathbf{r}_0) \end{align} The inhomogenious PDE is still satisfied since \begin{align} \nabla ^2 G(\mathbf{r},\mathbf{r}_0)=\nabla ^2 \left( -\frac{1}{4 \pi}\frac{1}{|\mathbf{r}-\mathbf{r}_0|} \right) +\nabla ^2 F(\mathbf{r},\mathbf{r}_0)=\delta(\mathbf{r}-\mathbf{r}_0) + 0 \end{align} and \begin{align} \nabla ^2 u(\mathbf{r})=\nabla ^2 \left( -\frac{1}{4 \pi}\frac{\rho(\mathbf{r})}{|\mathbf{r}-\mathbf{r}_0|} \right) +\nabla ^2 F(\mathbf{r},\mathbf{r}_0)=\rho(\mathbf{r}) + 0 \end{align} $H(\mathbf{r},\mathbf{r}_0)$ plays the role of distroting the field lines in order to satisfy the boundary conditions. $H(\mathbf{r},\mathbf{r}_0)$ has a physical interpretation, it is the potential arising from a so called image charge, thus this method of solving Poisson's equation with $F(\mathbf{r},\mathbf{r}_0)$ is also called the method of image charges.

[edit] Neumann Problem

In a Neumann Problem $\partial u (\mathbf{r}) / \partial n = f(\mathbf{r})$ is given on some surface $\mathcal{S}$. In the Dirichlet problem we choose $u(\mathbf{r})=0$ to solve (5), in a similar way we might choose $\partial G(\mathbf{r},\mathbf{r}_0) / \partial n=0$, but in general this is not permitted, since \begin{align} \int_\mathcal{S} \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n} d\mathcal{S} = \int_\mathcal{S} \nabla G(\mathbf{r},\mathbf{r}_0) \cdot \hat{\mathbf{n}} d\mathcal{S}=\int_\mathcal{V} \nabla^2 G(\mathbf{r},\mathbf{r}_0) d\mathcal{V}=1 \end{align} The easiest way to satisfy this condition is to set \begin{align} \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n}=\frac{1}{A} \end{align} for $\mathbf{r}$ on $\mathcal{S}$ with surface area $A$. The solution (5) now becomes \begin{align} u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV + \frac{1}{A} \oint_\mathcal{S} u(\mathbf{r}_0) \, d\mathcal{S} - \oint_\mathcal{S} G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0) \, d\mathcal{S} \end{align} Where the term \begin{align} \langle u(\mathbf{r}) \rangle_\mathcal{S} = \frac{1}{A} \oint_\mathcal{S} u(\mathbf{r}_0) \, d\mathcal{S} \end{align} is nothing but the average of $u$ over the surface $d\mathcal{S}$, so the solution can be written in the form. \begin{align} u(\mathbf{r})=\int_V G(\mathbf{r},\mathbf{r}_0) \, \rho(\mathbf{r}_0) \,dV + \langle u(\mathbf{r}) \rangle_\mathcal{S} - \oint_\mathcal{S} G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0) \, d\mathcal{S} \end{align}









Further Reading:

  • Riley, Hobson, Bence - Mathamatical Methods for Physics and Engineering
  • John David Jackson - Klassische Elektrodynamik