# Vector Calculus - Laplacian on Scalar Field

1. Apr 30, 2014

### Insolite

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

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

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:

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

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

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.

2. Apr 30, 2014

### LCKurtz

Good so far.

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.

3. May 1, 2014

### Insolite

Aah, thanks alot for your help, i've got it now. (: