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How do I show that

∇²(fg)=∇∙ (∇[fg] = f∇²g+g∇²f+2(∇f ∙∇g)

Any help would be greatly appreciated!

Regards,

Andrew

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migrated from mathoverflow.net Sep 25 '13 at 20:59

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2 Answers 2

By definition,

$\nabla^2u = \nabla \cdot \nabla u \tag{1}$

for any sufficiently differentiable function $u$; I guess taking $u \in C^2(R^n, R)$ "suffices". Then

$\nabla^2(fg) = \nabla \cdot \nabla (fg), \tag{2}$

and by the product (Leibniz) rule for derivatives, which can easily seen to apply to gradients since in fact

$\frac{\partial (fg)}{\partial s} = \frac{\partial f}{\partial s}g + f \frac{\partial g}{\partial s}, \tag{3}$

which holds for any variable $s$ upon which $f$ and $g$ may depend, we have

$\nabla(fg) = f(\nabla g) + g(\nabla f). \tag{4}$

Next, we use the formula

$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag{5}$

which holds for a differentiable function $f$ and vector field $X$. It too is a variant of the Leibniz rule, and may easily be verified by working out the relevant expressions in terms of coordinates. See also this wikipedia entry. Using (4) and (5) together yields

$\nabla \cdot \nabla (fg) = \nabla \cdot (f \nabla g + g\nabla f) = \nabla f \cdot \nabla g + f\nabla^2g + \nabla g \cdot \nabla f + g \nabla^2f, \tag{6}$

and bringing it all together yields

$\nabla^2(fg) = f \nabla^2g + g\nabla^2f + 2 \nabla f \cdot \nabla g. \tag{7}$

Hope this helps. Cheers, and

Fiat Lux!!!

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first take the gradient, $\nabla(fg)=f\nabla g+g\nabla f$, then take the divergence, $\nabla\cdot(\nabla[fg])=(\nabla f)\cdot(\nabla g)+f(\nabla\cdot\nabla g)+(\nabla g)\cdot(\nabla f)+g(\nabla\cdot\nabla f)=f\nabla^2 g+g\nabla^2 f+2(\nabla f)\cdot(\nabla g)$

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