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Laplace operator help

  • Thread starter Ed Aboud
  • Start date
197
0
1. Homework Statement

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. Homework 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.
 
316
0
Compute

[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.
 
197
0
I'm not really sure how to apply the chain rule in this case. Is there any general formula that I can use?
 
197
0
Actually its cool, I got it.

Thanks for the help!
 

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