Module 1 : A Crash Course in Vectors
Lecture 6 : Laplacian
  Example :
  Show that $\nabla^2(1/r)$ is a delta function.
  Solution :
  As $r= \sqrt{x^2+y^2+z^2}$, we have
 
\begin{displaymath}\frac{\partial r}{\partial x}= \frac{x}{r},\ \ \ \ \ \frac{\p... ...ac{y}{r},\ \ \ \ \ \ \frac{\partial r}{\partial z}= \frac{z}{r}\end{displaymath}
  using this it is easy to show that
 
\begin{displaymath}\frac{\partial^2}{\partial x^2}(\frac{1}{r}) = \frac{3x^2-r^2}{r^5}\end{displaymath}
  Thus
 
\begin{displaymath}\nabla^2\left(\frac{1}{r}\right) = \left(\frac{\partial^2}{\p... ...z^2}\right)(\frac{1}{r}) = \frac{3(x^2+y^2+z^2)-3r^2}{r^5} = 0\end{displaymath}
  However, the above is not true at the origin as $1/r$ diverges at $r=0$ and is not differentiable at that point. Interestingly, however, the integral of $\nabla^2(1/r)$ over any volume which includes the point $r=0$ is not zero. As the value of the integrand is zero everywhere excepting at the origin, the point $r=0$ has to be treated with care.
   
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