# Expansion of scalar function times laplace's equation

1. Aug 20, 2014

### 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.

2. Aug 20, 2014

### da_nang

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

3. Aug 20, 2014

### vtfjg87

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

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

$\nabla \cdot f\nabla f = \nabla f \cdot \nabla f + f(\nabla \cdot \nabla f)$
$\nabla \cdot f\nabla f - \nabla f \cdot \nabla f = f(\nabla \cdot \nabla f)$