Vector Calculus - Laplacian on Scalar Field

[Insolite]

A scalar field [itex]\psi[/itex] is dependent only on the distance [itex]r = \sqrt{x^{2} + y^{2} + z^{2}}[/itex] from the origin.

Show:

[itex] \partial_{x}^{2}\psi = \left(\frac{1}{r} - \frac{x^{2}}{r^{3}}\right)\frac{d\psi}{dr} + \frac{x^{2}}{r^{2}}\frac{d^{2}\psi}{dr^{2}}[/itex]

I've used the chain and product rules so far, but I'm unsure how to approach the problem to move onwards from this point. My working is as follows:

[itex]\frac{\partial\psi}{\partial x} = \frac{\partial r}{\partial x}\frac{d\psi}{dr}[/itex] where the identity [itex] \frac{\partial r}{\partial x} = \frac{x}{r}[/itex] is given.
[itex]\frac{\partial\psi}{\partial x} = \frac{x}{r}\frac{d\psi}{dr} (1) [/itex]
[itex]\frac{\partial^{2}\psi}{\partial^{2}x} = \frac{\partial}{\partial x}\left(\frac{x}{r}\frac{d\psi}{dr}\right)[/itex]
[itex]= \left(\frac{\partial}{\partial x}\frac{x}{r}\right)\left(\frac{d\psi}{dr}\right) + \left(\frac{\partial}{\partial x}\frac{d\psi}{dr}\right)\left(\frac{x}{r}\right)[/itex]

At this point, I can solve the left term to give me:
[itex]-\frac{x^{2}}{r^{3}}\frac{d\psi}{dr}[/itex]

But I don't know how to properly manipulate the second term. I've tried re-arranging and substituting in (1), but this didn't work out. Any hints on how to proceed would be greatly appreciated.

Thanks.

 

[LCKurtz]

A scalar field [itex]\psi[/itex] is dependent only on the distance [itex]r = \sqrt{x^{2} + y^{2} + z^{2}}[/itex] from the origin.

Show:

[itex] \partial_{x}^{2}\psi = \left(\frac{1}{r} - \frac{x^{2}}{r^{3}}\right)\frac{d\psi}{dr} + \frac{x^{2}}{r^{2}}\frac{d^{2}\psi}{dr^{2}}[/itex]

I've used the chain and product rules so far, but I'm unsure how to approach the problem to move onwards from this point. My working is as follows:

[itex]\frac{\partial\psi}{\partial x} = \frac{\partial r}{\partial x}\frac{d\psi}{dr}[/itex] where the identity [itex] \frac{\partial r}{\partial x} = \frac{x}{r}[/itex] is given.
[itex]\frac{\partial\psi}{\partial x} = \frac{x}{r}\frac{d\psi}{dr} (1) [/itex]

Good so far.

[itex]\frac{\partial^{2}\psi}{\partial^{2}x} = \frac{\partial}{\partial x}\left(\frac{x}{r}\frac{d\psi}{dr}\right)[/itex]
[itex]= \left(\frac{\partial}{\partial x}\frac{x}{r}\right)\left(\frac{d\psi}{dr}\right) + \left(\frac{\partial}{\partial x}\frac{d\psi}{dr}\right)\left(\frac{x}{r}\right)[/itex]

You might find it easier to write it as ##\frac \partial {\partial x} (x\cdot \frac{\psi'(r} r)##. Now you have a product rule and when you calculate ##\frac \partial {\partial x}## of the second term (quotient rule), just use the chain rule like you did in the beginning.

 

[Insolite]

Aah, thanks a lot for your help, I've got it now. (: