Maxwell's Eqn: Vanishing Laplacian of 1/r Explained

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SUMMARY

The Laplacian of the function 1/r, where r is the norm of a radial vector, vanishes everywhere except at r=0. This conclusion is derived using the Laplacian in spherical coordinates, specifically the formula: \nabla^2 u= \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+ \frac{1}{r^2 sin^2(\phi)}\frac{\partial^2 u}{\partial\theta^2}+ \frac{1}{r^2 sin(\phi)}\frac{\partial }{\partial\phi}\left(sin(\phi)\frac{\partial u}{\partial \phi}\right). In this case, since u=1/r, the derivatives with respect to the angles θ and φ are zero, confirming that the Laplacian indeed vanishes for all r ≠ 0.

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Master J
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In my derivation of one of Maxwell's Equations, I needed the fact that the Laplacian of 1/r vanishes everywhere except at r=0, where r is the norm of a radial vector.

I cannot see how this is? I like to be solid in the math I use for a derivation, so this would really help if someone could clear this up for me. Thanks!
 
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Putting the Laplacian in spherical coordinates, we have
[tex]\nabla^2 u= \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+ \frac{1}{r^2 sin^2(\phi)}\frac{\partial^2 u}{\partial\theta^2}+ \frac{1}{r^2 sin(\phi)}\frac{\partial }{\partial\phi}\left(sin(\phi)\frac{\partial u}{\partial \phi}\right)[/tex].
(See http://mathworld.wolfram.com/SphericalCoordinates.html)

With u= 1/r, the derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex] will be 0, of course.
 

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