[NOTES/QM-18003] Green Function for Poisson Equation

In electromagnetic theory the electric potential satisfies the Poisson equation \begin{equation} \nabla^2\Phi=-\frac{\rho}{\varepsilon_0}, \label{E1} \end{equation} where $\rho(\vec{r})$ is the volume charge density. The Green function for the Poisson equation is defined by \begin{equation} \nabla^2G(\vec{r})= - \delta^3(\vec{r}) \label{E2} \end{equation} If $\phi_0(\vec{r})$ is a solution of the Laplace equation \begin{equation} \nabla^2 \phi_0(\vec{r}) = 0, \label{E4} \end{equation} then \begin{equation} \Phi(\vec{r})= \phi_0(\vec{r}) + \int G(\vec{r}-\vec{r}^\prime) \frac{ \rho(\vec{r}^\prime)}{\varepsilon_0} d^3r^\prime \label{E5} \end{equation} is a solution of the Poisson equation which \MkGBox{can be easily verified}{qm}{18001} by applying $\nabla^2$ on both sides of \Eqref{E5}. \begin{eqnarray} \nabla^2 \Phi(\vec{r}) &=& \nabla^2 \phi_0(\vec{r}) +\nabla^2 \int G(\vec{r}-\vec{r}^\prime) \frac{ \rho(\vec{r}^\prime)}{\varepsilon_0} d^3r^\prime\\[3mm] &=&\int \nabla^2 G(\vec{r}-\vec{r}^\prime) \frac{ \rho(\vec{r}^\prime)}{\varepsilon_0} d^3r^\prime \\[3mm] &=& -\frac{1}{\varepsilon_0}\int \delta^3(\vec{r}-\vec{r}^\prime)\rho(\vec{r}^\prime) d^3\vec{r}^\prime\\[3mm] &=& -\frac{\rho(\vec{r})}{\varepsilon_0} \end{eqnarray} It can be shown that one solution of \Eqref{E2} is \begin{equation} G(\vec{r}) = \frac{1}{4\pi r}. \label{E3} \end{equation} This Green function gives the potential due to a charge distribution subject to the condition that the potential vanishes at infinity. The exact form of the Green function and the solution $\phi_0$ of the Laplace equation is determined by the boundary conditions of the problem.