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Green's identity help.

  1. Aug 10, 2010 #1
    Harmonic function satisfies Laplace equation and have continuous 1st and 2nd partial derivatives. Laplace equation is [itex]\nabla^2 u=0[/itex].

    Using Green's 1st identity:

    [tex]\int_{\Omega} v \nabla^2 u \;+\; \nabla u \;\cdot \; \nabla v \; dx\;dy \;=\; \int_{\Gamma} v\frac{\partial u}{\partial n} \; ds [/tex]

    [tex] v=1 \;\Rightarrow\; \int_{\Omega} \nabla^2 u \; dx\;dy \;=\; \int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \;\hbox { if } \;u \;\hbox{ is a harmonic function .} [/tex]

    Why is it equal zero if u is harmonic function? Why is:

    [tex]\int_{\Omega} \nabla^2 u \; dx\;dy =0 \hbox { if } \nabla^2 u =0 [/tex]

    Or more basic question:

    What is [itex]\int_{\Gamma} 0 dxdy[/itex]? Is it not zero?
  2. jcsd
  3. Aug 10, 2010 #2
    The integral of zero is zero. I don't see a problem here.
  4. Aug 10, 2010 #3

    So you mean:

    [tex]\int_{\Omega} \nabla^2 u \; dx\;dy =0 \hbox { if } \nabla^2 u =0 [/tex]?
  5. Aug 10, 2010 #4
    Youngman, I don't want to get over my head with this or nothing, but let me try to explain since I jumped in: if the function u is harmonic, then the Laplacian is zero, not a limit which approaches zero but identically zero and surely the integral of zero is just zero.
  6. Aug 10, 2010 #5
    I was just thinking if you differentiate x twice you get zero, if integration is the reverse of differentiation, should integrating zero be something!!! I dug up books and I did not see anything about this, thats why I posted.

    Thanks for the help.
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