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Homework Help: Laplace operator help

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Let x = (x,y,z) .
    Recall that the vector x is determined by its direction and length
    r = |x| = [itex] \sqrt{x^2 + y^2 + z^2} [/itex]

    and assume we are given a function f which depends only on the length of x

    f = f(r)

    Show that

    [tex] \Delta f = f'' + \frac{2}{r} f'[/tex]

    where [tex] f' = \frac{\partial f}{\partial r} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    [tex] u = x^2 + y^2 + z^2 [/tex]
    [tex] r = \sqrt{u} [/tex]
    [tex] \frac{\partial r}{\partial x} = \frac{\partial \sqrt{u}}{\partial u} \frac{\partial u}{\partial x} = \frac{1}{2}(\frac{1}{\sqrt{r}})(2x) = \frac{x}{\sqrt{r}} [/tex]

    [tex] \frac{\partial ^2 r}{\partial x^2} = \frac{\partial (x)}{\partial x } \frac{1}{\sqrt{u}} + x \frac{\partial \frac{1}{\sqrt{u}}}{\partial u} \frac{\partial u }{\partial x} [/tex]

    [tex] = \frac{1}{\sqrt{u}} - x^2 \frac{1}{\sqrt{u^3}} [/tex]

    Since f and u are symmetric in x,y,z

    [tex] \frac{\partial ^2 r}{\partial y^2} = \frac{1}{\sqrt{u}} - y^2 \frac{1}{\sqrt{u^3}} [/tex]

    [tex] \frac{\partial ^2 r}{\partial z^2} = \frac{1}{\sqrt{u}} - z^2 \frac{1}{\sqrt{u^3}} [/tex]

    [tex] x^2 + y^2 + z^2 = u [/tex]

    [tex] \Delta r = (\frac{1}{\sqrt{u}}) - x^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - y^2 (\frac{1}{\sqrt{u^3}}) + (\frac{1}{\sqrt{u}}) - z^2 (\frac{1}{\sqrt{u^3}}) [/tex]

    [tex] = \frac{3}{\sqrt{u}} - (x^2 + y^2 + z^2) \frac{1}{\sqrt{u^3}} [/tex]

    [tex] = \frac{2}{\sqrt{u}} [/tex]

    [tex] = \frac{2}{r} [/tex]

    I see that this is a part of the solution but I have no idea what to do to get the rest.

    Any help would be greatly appreciated because I'm lost and it has to be in tomorrow morning.
  2. jcsd
  3. Mar 1, 2009 #2

    [tex]\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right)f(r(x,y,z))[/tex]

    using the chain rule.
  4. Mar 2, 2009 #3
    I'm not really sure how to apply the chain rule in this case. Is there any general formula that I can use?
  5. Mar 2, 2009 #4
    Actually its cool, I got it.

    Thanks for the help!
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