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Verify Harmonic function and Laplace equation.

  1. Aug 5, 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]

    [tex] \frac{\partial u}{\partial n} \;=\; \nabla u \cdot \widehat{n} \;\hbox{ where} \; \widehat{n} \;\hbox { is the outward normal of the boundary of the closed region } \Omega[/tex]

    [tex] \frac{\partial u}{\partial n} \;=\; \nabla u \cdot \widehat{n} =0 \;\Rightarrow\; \nabla u \;\hbox{ is tangent to the outward normal which means it is tengent of the boundary } \Gamma\; [/tex].

    This mean for a harmonic function [itex]u,\;\; \nabla u [/itex] on the boundary [itex] \Gamma [/itex] is tangent to the boundary. And in vector calculus term, it is total circulation.

    This mean at the points of the boundary [itex]\nabla u = \nabla \;X\; A \;\hbox { where A is some scalar function. } [/itex]

    Just want to verify with you guys.

    Thanks

    Alan
     
  2. jcsd
  3. Aug 6, 2010 #2
    Anyone please?
     
  4. Aug 7, 2010 #3
    Anyone?
     
  5. Aug 10, 2010 #4
    I think I know what I did wrong.

    [tex] \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]

    [tex] \int_{\Gamma} \frac{\partial u}{\partial n} \; ds = 0 \hbox { do not mean } \frac{\partial u}{\partial n} = 0[/tex].

    So my logic is wrong to start out.

    Thanks
     
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