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# MM Solved Problem-09 Green's function for Poisson equation in $n>2$ dimensions

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Green's function for Poisson equation in $n>2$ dimensions

Author - Pankaj Sharan, Jamia Milia Islamia, New Delhi 110025

$\newcommand{\dydx}[2]{\frac{\partial{#1}}{\partial{#2}}} \newcommand{\vv}[1]{{\bf #1}} % three dim vector$

The UGC-CSIR NET examination of December 2013 contained the following question:

The expression

\begin{eqnarray*}
\left(\dydx{^2}{x_1^2}+\dydx{^2}{x_2^2}+\dydx{^2}{x_3^2}+\dydx{^2}{x_4^2}\right)
\left(\frac{1}{x_1^2+x_2^2+x_3^2+x_4^2}\right)
\end{eqnarray*}

is proportional to

$(A) \delta(x_1+ x_2 + x_3 +x_4)$
$(B) \delta(x_1)\delta(x_2)\delta(x_3)\delta(x_4)$
$(C) x_1^2+x_2^2+x_3^2+x_4^2)^{-3/2}$
$(D) (x_1^2+x_2^2+x_3^2+x_4^2)^{-2}$

The answer could be deduced from dimensional analysis alone. The given expression has dimension of $L^{-4}$ if $x_i$ have dimension $L$. Therefore only (B) or (D) could be correct because $\delta(x)$ has dimension of $1/x$. But the given expression is equal to zero for $x_1^2+x_2^2+x_3^2+x_4^2\neq 0$ as can be seen by a short calculation. Therefore (D) is ruled out, and (B) is the correct answer.

But I thought we never tell our students about the Green's function of the Poisson equation in higher dimensions, even though it is just a step from 3-dimension. Also, doing the angular integral in 3-dimensional k-space explicitly is easy, I do not know how to do in it four dimensions. So I had to go to Gelfand-Shilov to check!

The following discussion is based on \S\ 3.3, Chapter II of I. M. Gel'fand and G. E. Shilov, Generalized Functions, Volume I (Academic Press). This is the place I will turn to if I encounter any trouble with generalized functions! The student is supposed to know the definition of the Dirac delta and the Gamma functions. Just that.

### Section 1

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,

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

### Section 2

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 $\lambda$.

### Section 3

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,

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

### Section 4

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*}

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