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Laplacian solution(page 2) to Jackson 1.5

  1. Aug 14, 2008 #1
    Hi.

    In http://www-personal.umich.edu/~pran/jackson/P505/p1s.pdf" solution(page 2) to Jackson 1.5 it is stated that

    [tex]\nabla^2 \left(\frac{1}{r}\right)=-4\pi\delta^3(\mathbf r)[/tex].

    But why is this true?

    [tex]\nabla^2\left(\frac{1}{r}\right)=\frac{1}{r^2}\frac{d}{d r}\left(r^2\frac{d}{dr}\frac{1}{r}\right)=\frac{1}{r^2}\frac{d}{dr}(-1)[/tex]
     
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Aug 14, 2008 #2
    Re: Laplacian

    This equality is to be understood in the distributional sense. It should be read as:

    [tex] \int \frac{\Delta \phi}{|x|}\, \mathrm{d}x = -4\pi \phi (0), \qquad \forall \phi \in C^{\infty}_c (\mathbf{R}^3) [/tex]

    :)
     
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