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Proving some Dirac-Delta identity that uses Laplacian

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that ##(x^2+y^2+z^2)\nabla^2[\delta(x)\delta(y)\delta(z)]=6\delta(x)\delta(y)\delta(z)##

    2. Relevant equations
    ##\delta''(x)/2=\delta(x)/x^2##

    3. The attempt at a solution
    I have obtained this:

    ##6\delta(x)\delta(y)\delta(z) + (xy)^2/2\Bigg(\delta''(y)\delta''(z)+\delta''(x)\delta''(z)\Bigg)+(xz)^2/2\Bigg(\delta''(y)\delta''(z)+\delta''(x)\delta''(y)\Bigg)+(yz)^2/2\Bigg(\delta''(x)\delta''(z)+\delta''(x)\delta''(y)\Bigg)##

    This looks like a huge mess, hehe, sorry.

    Any ideas on how to make disappear those ugly terms and only remain with the first one?

    Thanks.
     
  2. jcsd
  3. Apr 12, 2015 #2
    Whats the identity for [itex]x\delta(x)[/itex]?
     
  4. Apr 12, 2015 #3
    ##x\delta(x)=0## Is this a hint or were you asking me this?
     
  5. Apr 12, 2015 #4
    yes, it was meant as a hint, your original expression seems a mess indeed, just do only the term [itex]\frac{\partial^2}{\partial x^2}[/itex] of the laplacian and you see easily that it contributes [itex]2\delta(x)\delta(y)\delta(z)[/itex]
     
  6. Apr 12, 2015 #5
    Oh! What a dope am I. Thanks hehe!
     
  7. Apr 12, 2015 #6
    Wait, sorry. But those terms are divided by (e.g. the x term) ##x^2## So I would get in total that ##(x^2+y^2+z^2)\nabla^2 (\delta(x)\delta(y)\delta(z))=2\delta(x)\delta(y)\delta(z)(x^-2+y^-2+z^-2)(x^2+y^2+z^2)##
     
  8. Apr 12, 2015 #7
    yes, expand it, one term of the expansion will be (for x^-2) [itex]2\delta(x)\delta(y)\delta(z)(1+\frac{y^2+z^2}{x^2})[/itex]. Now you can see that for example that [itex]\frac{y^2}{x^2}2\delta(x)\delta(y)\delta(z)[/itex] is zero .
     
  9. Apr 12, 2015 #8
    Thanks Delta^2 I think I've got it now.
     
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