# Poisson's Equation

### From bio-physics-wiki

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\frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n}=\frac{1}{A} | \frac{\partial G(\mathbf{r},\mathbf{r}_0)}{\partial n}=\frac{1}{A} | ||

\end{align} | \end{align} | ||

− | for $\mathbf{r}$ on $mathcal{S}$. | + | 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 +\oint_\mathcal{S} \left[ u(\mathbf{r}_0) \frac{1}{A} - G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0)\, d\mathcal{S} | ||

+ | \end{align} |

## Revision as of 09:20, 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} 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 $\mathcal{S}$

**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**.

## 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 $\mathcal{S}$.

### 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.

### 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 +\oint_\mathcal{S} \left[ u(\mathbf{r}_0) \frac{1}{A} - G(\mathbf{r},\mathbf{r}_0) \, f(\mathbf{r}_0)\, d\mathcal{S} \end{align}