[NOTES/EM-04003] Green Function Method in Electromagnetic Theory

In electrostatics of charges in presence  of conductors, the potential obeys the  Poisson equation \begin{equation} \nabla^2 \Phi(\vec{r},t) = -\rho(\vec{r})/\epsilon_0. \end{equation} We define Green function for the Poisson equation as \begin{equation} \nabla^2 G(\vec{r}-\vec{r}^\prime)= - \delta^3(\vec{r}-\vec{r}^\prime) \end{equation} So that the solution of the Poisson equation takes the form \begin{equation} \Phi(\vec{r}) = \iiint G(\vec{r}-\vec{r}^\prime) \Phi(\vec{r}^\prime) \end{equation} We will discuss boundary conditions on potential in presence of conductors, assumed to be perfectly conducting surfaces.

1. Dirichlet and Neuman  boundary conditions

Consider the case when the Poisson equation is to be solved in a given volume \(V\) bounded by a surface \(S\). If the boundary is a conductor held at a constant potential \(\phi_0\),  the potential must reduce to constant \(\phi_0\) on the  boundary of volume \(V\).This type of boundary condition is known as Dirichlet boundary condition. Another class of problems involve  the conducting boundary carrying some charge. In this case the electric field on the boundary is related to the charge density on the surface. Specifying the electric field on the bounding surface \(S\) amounts to specifying the  value of the gradient \(\nabla \Phi\) is known of the surface \(S\). This type of boundary condition on solutions of the Poisson equation belongs to the class of boundary value problems known as Neuman boundary value problem. In addition one can have a mixed boundary condition. In this class of problems potential is specified on part of the boundary and \(\nabla \Phi\) is given on the rest of the boundary. In the following sections we shall formulate the Green function method for Dirichlet type and for Neuman type of boundary value problems.

2. Green's theorem

We first prove Green's theorem which is needed for writing the solution to the boundary value problems using appropriate Green function. We start with divergence theorem \begin{equation} \iiint_V \nabla \cdot \vec{F} d^3r=\iint_S \vec{F}\cdot \vec{n} dS. \end{equation} Here \(\hat{n}\) is the normal to surface \(S\) which encloses the volume \(V\). Taking \(V= \Phi \nabla \Psi\), we have \begin{equation} \nabla. \vec{F}= \nabla (\Phi\nabla \Psi)  = \nabla\Phi \nabla \Psi + \Phi \nabla^2 \Psi. \end{equation} Therefore we have

\begin{eqnarray}\label{EQ04}
 \iiint_V\Big( \nabla \Phi  \nabla \Psi + \Phi \nabla^2 \Psi\Big)
 &=& \iint \Phi \hat{n}\cdot\big(\nabla \Psi\big) dS.\\
&=& \iint \phi \pp[\Psi]{n} dS.
\end{eqnarray}


where the  \(\pp[\Phi]{n}\) is the normal derivative \(\hat{n}\cdot\nabla \Psi\). Interchanging \(\Phi\) and \(\Psi\) in \eqRef{EQ04} we get \begin{equation}\label{EQ06} \iiint_V\Big( \nabla \Phi \nabla \Psi +  \nabla \Psi  \nabla^2 \Phi \Big) = \iint \Psi \pp[\Phi]{n} dS. \end{equation} Subtracting \eqref{EQ04} and \eqref{EQ06}, we get \begin{equation} \iiint_V\big( \Phi\nabla^2 \Psi - \Phi\nabla^2\Psi\big)= \iint_S \Big( \Phi\pp[\Psi]{n} - \Psi\pp[\Phi]{n}\Big) dS. \end{equation}

3. Green function for Dirichlet Problems

We identify \(\Phi\) with potential and \(\Psi\) with  Green function \(G\)
and use Poisson equations
\begin{equation}
 \nabla^2 \Phi = \rho/\epsilon_0, \qquad \qquad \nabla^2 G(\vec{r},
\vec{r}\Prime= -\delta^{(3)} (\vec{r}-\vec{r}\Prime)
\end{equation}
we get

\begin{eqnarray}
 \iiint \Phi(\vec{r}\Prime)-\delta^{(3)} (\vec{r}-\vec{r}\Prime) +
  \frac{1}{\epsilon_0}\Psi(\vec{r}\Prime) \rho(\vec{r}\Prime) d^3 r\Prime
  = \iint\Big( \Phi \pp[\Psi]{n} -  \Psi \pp[\Phi]{n} \Big) dS.\\
  \Phi(\vec{r}) = \iiint{G(\vec{r}-\vec{r}\Prime)}\rho(\vec{r}\Prime)d^3
r\Prime  + \iint\Big( \Phi \pp[\Psi]{n} -  \Psi \pp[\Phi]{n} \Big) dS.
\label{EQ10}
\end{eqnarray}


Suppose now we want to solve a problem for \(\rho(\vec{r})\) in volume \(V\),
with some Dirichlet  boundary condition for the potential on \(S\), we should
choose \(Psi=0\) on \(S\). This gives
\begin{equation}
 \Phi(\vec{r}) = \frac{1}{\epsilon_0} \iiint_V\rho(\vec{r}\Prime)
G(\vec{r}-\vec{r}\Prime) d^3 r\Prime - \iint_S \Phi(\vec{r}\Prime)
\pp[G(\vec{r}-\vec{r}\Prime)]{n} d^3 r\Prime.
\end{equation}
Thus knowing the Green function on \(S\) and inside volume \(V\), the potential
can be computed.

4. Neuman boundary condition

In this case the charge density \(\rho\) is given inside volume \(V\)  and
value of \(\pp[\Phi]{n}\) is  specified  on the boundary \(S\).

In this case it is not possible to have \(\pp[\Psi]{n}=0\) on surface \(S\).
To see this we note \(\Psi\) is identified with the Green fucntion. Therefore
we would get
\begin{equation}
 \iiint_V \nabla^2\Psi d^3r\Prime = -1.
\end{equation}
However, this equation, using Gauss divergence theorem, implies
\begin{equation}
 \iint_S \nabla \Psi \cdot \hat{n} dS  = -1.
\end{equation}
which shows that the choice \(\pp[\Psi]{n}=0\) is inconsistent with the
definition of the Green function.

In this case one should solve
\begin{equation}
\vec{r}\Prime= -\delta^{(3)} (\vec{r}-\vec{r}\Prime)
\end{equation}
for  the Green function equation subject to the boundary condition condition
\begin{equation}
 \pp[\Psi]{n}\Big|_S = \frac{1}{A}
\end{equation}
where \(A\) is the area of the surface. \eqref{EQ10} then take the form

\begin{eqnarray}
 \Phi(\vec{r})
 &=&\iiint{G(\vec{r}-\vec{r}\Prime)}\rho(\vec{r}\Prime)d^3
r\Prime  + \iint\Big( \Phi \pp[\Psi]{n} -  \Psi \pp[\Phi]{n} \Big) dS\\
&=& \iiint{G(\vec{r}-\vec{r}\Prime)}\rho(\vec{r}\Prime)d^3
r\Prime  + \iint\Big( \Phi \frac{1}{A} -  \Psi \pp[\Phi]{n} \Big)
\end{eqnarray}
This gives the final form for the potential in terms of Green function
\begin{equation}
\Phi(\vec{r}) =
\frac{\langle\Phi\rangle}{A} + \iiint{G(\vec{r}-\vec{r}\Prime)}\rho(\vec{r}
\Prime)d^3
r\Prime  + \iint \Psi \pp[\Phi]{n} \Big) dS.
\end{equation}
To sum up, to obtain the correct Green's function, we need to solve the equation
\begin{equation}
 \nabla^2 G(\vec{r}-\vec{r}\Prime) = - \delta^3(\vec{r}-\vec{r}\Prime)
\end{equation}
subject to
\begin{equation}
 \pp[G]{n}\Big|_S =  -\frac{1}{A}.
\end{equation}