Expansion of scalar function times laplace's equation

[vtfjg87]

Apparently,

$f \nabla^2 f = \nabla \cdot f \nabla f - \nabla f \cdot \nabla f$

where f is a scalar function.

Can someone please show me why this is step by step.
Feel free to use suffix notation.

Thanks in advance.

 

[da_nang]

Are you familiar with the product rules of the del operator?

 

[vtfjg87]

Yes i am. I think i got the answer. I believe the appropriate rule is:

$\nabla \cdot \vec{μ} f = \nabla f \cdot \vec{μ} + f \nabla \cdot \vec{μ}$

You need to start with

$<br /> \nabla \cdot f\nabla f = \nabla f \cdot \nabla f + f(\nabla \cdot \nabla f)<br />$

which becomes:
$\nabla \cdot f\nabla f - \nabla f \cdot \nabla f = f(\nabla \cdot \nabla f)$


Thank you for your help.