Green's Theorem

  • Thread starter Tony11235
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  • #1
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Prove the identity [tex]\int_{\partial D}\phi \nabla \phi \cdot \n \ds = \int \int_{D} (\phi \nabla^2 \phi + \nabla \phi \cdot \nabla \phi) \dA [/tex]
Can I just let [tex]\phi[/tex] 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.
 
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Answers and Replies

  • #2
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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 dont know how to do those fancy fonts.
 
  • #3
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So basically let F be equal to [tex] \phi \nabla \phi [/tex], then [tex] \nabla \cdot F [/tex] is equal to [tex](\phi \nabla^2 \phi + \nabla \phi \cdot \nabla \phi) [/tex]. What do you know it's the divergence theorem!
 

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