MM-09 :: Green Function for Poisson equation in $n>2$ dimensions

The Poisson equation in $n$ dimensions is

\[ \nabla^2\phi=\left(\dydx{^2}{x_1^2}+\cdots +\dydx{^2}{x_n^2}\right)\phi(\vv{x})=\rho(\vv{x}) \]

Its Green function $G(\vv{x})$ is defined as

\[ \left(\dydx{^2}{x_1^2}+\cdots +\dydx{^2}{x_n^2}\right)G(\vv{x})=\delta(\vv{x}).\]

If we represent both the Green function as well as the Dirac delta function as Fourier transforms, then

\begin{align}
G(\vv{x}) &= \frac{1}{(2\pi)^n}\int g(\vv{k})e^{i\vv{k}\cdot\vv{x}}\, d^n\vv{k}, \\
\delta(\vv{x})&=\frac{1}{(2\pi)^n}\int e^{i\vv{k}\cdot\vv{x}}\, d^n\vv{k}.
\end{align}

Substitution in the Poisson equation determines

\[ g(\vv{k})=-\frac{1}{k^2},\qquad k^2=|\vv{k}|^2, \]

and so, 

\[ G(\vv{x}) = -\frac{1}{(2\pi)^n}\int k^{-2}e^{i\vv{k}\cdot\vv{x}}\, d^n\vv{k}. \label{G} \]

To find $G(\vv{x})$ we calculate the slightly more general  integral, (at no extra cost)

\[ F(\vv{x})=\frac{1}{(2\pi)^n}\int k^{\lambda}e^{i\vv{k}\cdot\vv{x}}\, d^n\vv{k}, \qquad \lambda > -n.\label{integral} \]

The restriction on the real parameter $\lambda$ has been  made to avoid trouble at $k=0$. In  our case 

$\lambda=-2$ and $n>2$, therefore $\lambda=-2>-n$.  

We first note the {\em scaling property} and {\em rotational invariance} of  $F$.

If $a>0$ then replacing $\vv{x}$ by $a\vv{x}$ we see that

\[ F(a\vv{x}) = a^{-\lambda-n}F(\vv{x}), \]

because $\vv{k}\cdot a\vv{x}=a\vv{k}\cdot \vv{x}$ and we can change the variable of integration to $\vv{k}'=a\vv{k}$.

Similarly, if $R$ is a rotation in the $n$ dimensional space, the integral doesn't change because  $\vv{k}\cdot R\vv{x}=R^{-1}\vv{k}\cdot \vv{x}$ and we can change the variable of integration to $\vv{k}'=R^{-1}\vv{k}$. Thus

\[ F(R\vv{x}) = F(\vv{x}). \]

This tells us that $F$ is a function $F(r)$ of $r=\sqrt{x_1^2+\cdots+x_n^2}$ only and by using the scaling property 

$F(r)=r^{-\lambda-n}F(1)$. Thus,

\[ F(\vv{x})=C r^{-\lambda-n}, \qquad  \label{C}\]

and the remaining effort now is to calculate the constant $C$ which can only depend on $n$ and $\lambda$.

The trick to evaluate $C$ is to  multiply both sides of (\ref{integral}) by

$ \exp(-x_1^2-\cdots-x_n^2)=\exp(-r^2)$ and integrate over all $\vv{x}$ space. The left hand side is 

\begin{equation*}
\int F(\vv{x})e^{-r^2} d^n\vv{x}=C\int_0^\infty\, r^{-\lambda-n}e^{-r^2}r^{n-1}dr\, d\Omega_n 
\end{equation*}

where 

\begin{equation*}
d^n\vv{x}=r^{n-1}dr\, d\Omega_n
\end{equation*}

and $d\Omega_n$ are the $(n-1)$-fold integrations over angular variables in $n$-dimensions. The angular integrations are trivially equal to $\Omega_n$ the measure of surface of the unit sphere in $n$-dimensions.

Thus, using the definition of the Gamma function,

\[
\int F(\vv{x})\exp(-x_1^2-\cdots-x_n^2)d^n\vv{x}=\frac{1}{2}C\Omega_n
\Gamma\left(-\frac{\lambda}{2}\right).\label{lhs}
\]

On the other hand, the right hand side is equal to 

\begin{equation*}
\frac{1}{(2\pi)^n}\int k^{\lambda}e^{i\vv{k}\cdot\vv{x}-x_1^2-\cdots-x_n^2}\, d^n\vv{k}d^n\vv{x}.
\end{equation*}

Doing the $\vv{x}$ integrals, and using integrals like

\begin{equation*} \int e^{ik_1x_1-x_1^2} dx_1=\sqrt{\pi}e^{-k_1^2/4} \end{equation*}

we reduce the integration to

\[
\frac{(\pi)^{n/2}}{(2\pi)^n}\int_0^\infty k^{\lambda+n-1}e^{-k^2/4}\, dk\,d\Omega_n 
=2^{\lambda-1}(\pi)^{-n/2}\Omega_n\Gamma\left(\frac{\lambda+n}{2}\right)\label{rhs}
\]

Equating (\ref{lhs}) and (\ref{rhs}), we get the answer

\begin{equation*}
C=2^\lambda(\pi)^{-n/2}
\frac{\Gamma\left(\frac{\lambda+n}{2}\right)}{\Gamma\left(-\frac{\lambda}{2}\right)},
\end{equation*}

and so,

\[
\frac{1}{(2\pi)^n}\int k^{\lambda}e^{i\vv{k}\cdot\vv{x}}\, d^n\vv{k}=
2^\lambda(\pi)^{-n/2}
\frac{\Gamma\left(\frac{\lambda+n}{2}\right)}{\Gamma\left(-\frac{\lambda}{2}\right)}
r^{-n-\lambda}.
\]

Coming back to our original problem in (\ref{G}),  for which $\lambda=-2$, we see that 

\[ G(\vv{x})= -\frac{(\pi)^{-n/2}}{4}\Gamma\left(\frac{n}{2}-1\right)r^{-n+2} \]

is the Green's function for Poisson equation in $n$-dimensions. For three dimensions it reduces to the familiar and important result

\begin{equation*} G(\vv{x})=-\frac{1}{4\pi r}, \end{equation*}

and in four dimensions

\begin{equation*}s G(\vv{x})=-\frac{4}{\pi^2 r^2}.\end{equation*}