# Divergence theorem proof

• Mar 6th 2010, 05:40 AM
chipette
Divergence theorem proof
How would I go about constructing this following proof?

Let f = f (x, y, z) be a function. Show that for any other function v = v(x, y, z)

$\displaystyle \nabla \cdot (\nabla f \times \nabla v) = 0$

I understand you have to show that it is divergence free, but would you just write each vector term in terms of a derivative? :S
• Mar 6th 2010, 07:26 AM
TheEmptySet
Quote:

Originally Posted by chipette
How would I go about constructing this following proof?

Let f = f (x, y, z) be a function. Show that for any other function v = v(x, y, z)

$\displaystyle \nabla \cdot (\nabla f \times \nabla v) = 0$

I understand you have to show that it is divergence free, but would you just write each vector term in terms of a derivative? :S

Note this is only true if the scalar function are twice continously differentiable.

Writing it out will work. Also if you know some other identities they could be helpful for example.

$\displaystyle \nabla \cdot (\vec{F} \times \vec{G})=\vec{G}\cdot( \nabla \times \vec{F})-F\cdot (\nabla \times \vec{G})$

Then for any scalar function with continous 2nd partial derviatives it is known that $\displaystyle \nabla \times (\nabla f)=0$

Thus in the above identity if $\displaystyle \vec{F}=\nabla f$ we get zero.

Or you can also use the scalar triple product

$\displaystyle \nabla \cdot (\nabla f \times \nabla g)=\begin{vmatrix} \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} & \frac{\partial g}{\partial z} \end{vmatrix}$

Just simplify from here