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Method of Images - Green's Function

  1. Oct 19, 2010 #1
    Use the method of images to find a Green's function for the problem in the attached image.

    Demonstrate the functions satisfies the homogenous boundary condition.
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2010 #2
    It's taken me ages, and I really am struggling to understand this, so Im not sure if this is correct but here goes....

    [tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] [tex]\frac{\partial G(x,0)}{\partial y} = 0[/tex] for [tex]y \geq 0 \ - \infty < x < \infty[/tex]

    Let [tex]r = |\underline{x} - \underline{x}_0 | = \sqrt{(x - x_0)^2 + (y - y_0)^2}[/tex]

    be the distance between [tex]\underline{x}[/tex] and [tex]\underline{x}_0[/tex]

    Then [tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] becomes

    [tex]\bigtriangledown^2 G = \frac{1}{r} \frac{\partial}{\partial r} \Big( r \frac{\partial}{\partial r} \Big) = 0[/tex]

    everywhere (although not at [tex]r = 0[/tex]) and subject to

    [tex]\iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1[/tex]

    which gives

    [tex]G(r) = A \ln r + B[/tex]

    [tex]A = \frac{1}{2 \pi}[/tex]

    Choosing [tex]B = 0[/tex]

    [tex]G = \frac{1}{2 \pi} \ln r = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|[/tex]

    I don't think this is the method of images though......(Headbang)
     
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