Proof of Identity: $\phi(P+Q)=\int_{\partial D}\phi \nabla \phi \cdot \n \ds$

[Tony11235]

Prove the identity $$\int_{\partial D}\phi \nabla \phi \cdot \n \ds = \int \int_{D} (\phi \nabla^2 \phi + \nabla \phi \cdot \nabla \phi) \dA$$
Can I just let $$\phi$$ be equal to P + Q, substitute into the left side, and try to derive the right side? This is a weird looking identity by the way.

 

[elhinnaw]

Its actually quite easy, without any vector calculus needed.
(f*gx)x= fx*gx + f*gxx

if you sum these up for y and z derivatives, you will get

Div[f*del(g)] = del(f) dot del(g) + f*del^2(g)

integrate both sides and use divergence theorem.

I apologize for having to write div for divergence, dot for dot product and del for the del operator, but i don't know how to do those fancy fonts.

 

[Tony11235]

So basically let F be equal to $$\phi \nabla \phi$$, then $$\nabla \cdot F$$ is equal to $$(\phi \nabla^2 \phi + \nabla \phi \cdot \nabla \phi)$$. What do you know it's the divergence theorem!