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Homework Help: Prove equation in Green's function.

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Green's function [itex]G(x_0,y_0,x,y) =v(x_0,y_0,x,y) + h(x_0,y_0,x,y)[/itex] in a region [itex]\Omega \hbox { with boundary } \Gamma[/itex]. Also [itex]v(x_0,y_0,x,y) = -h(x_0,y_0,x,y)[/itex] on boundary [itex]\Gamma[/itex] and both [itex]v(x_0,y_0,x,y) \hbox { and }h(x_0,y_0,x,y)[/itex] are harmonic function in [itex]\Omega[/itex]

    [tex]v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2] [/tex]




    Let u be continuous and h is harmonic on an open disk around [itex](x_0,y_0)[/itex] in [itex]\Omega [/itex]. Show that

    [tex]_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = 0[/tex]

    Hint from the book: Both |u| and [itex] |\frac{\partial h}{\partial n}|[/itex] are bounded near [itex](x_0,y_0)[/itex], say by M. If [itex] I_r[/itex] denotes the integral in question, then [itex] |I_r| \leq 2\pi M r \rightarrow 0 \hbox { as } r\rightarrow 0[/itex]

    2. Relevant equations

    Green's 1st identity:

    [tex] \int _{\Omega} ( u\nabla^2 h + \nabla u \cdot \nabla h ) dx dy = \int_{\Gamma} u \frac{\partial h}{\partial n} ds[/tex]




    3. The attempt at a solution

    [tex] h=-v \hbox { on } \Gamma \Rightarrow\; \int _{\Omega} ( u\nabla^2 h + \nabla u \cdot \nabla h ) dx dy = -\int_{\Gamma} u \frac{\partial v}{\partial n} ds = -\int_{\Gamma} u \frac{1}{r} ds = -\int^{2\pi}_{0} u d\theta[/tex]

    h is harmonic in [itex]\Omega\;\Rightarrow \nabla^2 h = 0[/itex]

    [tex]\Rightarrow\; \int _{\Omega} \nabla u \cdot \nabla h \; dx dy = -\int^{2\pi}_{0} u d\theta[/tex]
    I really don't know how to continue, please help.
     
    Last edited: Aug 10, 2010
  2. jcsd
  3. Aug 11, 2010 #2
    h = -v on Γ. You're not integrating over Γ; you're integrating around a small circle Cr around (x0, y0). You haven't said what h is, but by the hint, presumably ∂h/∂n is bounded on some disk centered at (x0, y0). This is certainly not true if h = -v on this disk, as ∂v/∂n = 1/r is not bounded. If h has no singularity at (x0, y0), then certainly ∂h/∂n is bounded on such a disk, as h has a continuous derivative (since by assumption h is harmonic).

    Apply the hint.
     
  4. Aug 11, 2010 #3
    Thanks for you help, let me try this:

    Let u be continuous and h is harmonic on an open disk around [itex](x_0,y_0)[/itex] in [itex]\Omega [/itex]. Show that

    [tex]_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = 0[/tex]

    Hint from the book: Both |u| and [itex] |\frac{\partial h}{\partial n}|[/itex] are bounded near [itex](x_0,y_0)[/itex], say by M. If [itex] I_r[/itex] denotes the integral in question, then [itex] |I_r| \leq 2\pi M r \rightarrow 0 \hbox { as } r\rightarrow 0[/itex]

    So [itex]|u| \hbox { and } |\frac{\partial h}{\partial n}|\; \leq\; M[/itex]. As [itex] (x,y)\rightarrow\;(x_0,y_0)[/itex]

    [tex]_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = \;^+_- M^2\; \int^{2\pi}_{0} r\;d\theta \;=\; _r\stackrel{lim}{\rightarrow}_0 \; (2\pi M^2r) \;=\; 0[/tex]

    I have [itex] M^2 \hbox { because } u\frac{\partial h}{\partial n} \leq M^2[/itex]
     
  5. Aug 11, 2010 #4
    Yes. Now you have to prove that both |u| and |∂h/∂n| are bounded near that point.
     
  6. Aug 11, 2010 #5
    Thanks for your time. But I got M^2 instead of M. It does not matter because r goes to zero.
     
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