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Math Help - Method of Images - Green's Function

  1. #1
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    Method of Images - Green's Function

    Use the method of images to find a Green's function for the problem

    Method of Images - Green's Function-chapter-review-q17.bmp

    Demonstrate the functions satisfies the homogenous boundary condition.
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  2. #2
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    It's taken me ages, and I really am struggling to understand this, so Im not sure if this is correct but here goes....

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

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

    be the distance between \underline{x} and \underline{x}_0

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

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

    everywhere (although not at r = 0) and subject to

    \iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1

    which gives

    G(r) = A \ln r + B

    A = \frac{1}{2 \pi}

    Choosing B = 0

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

    I don't think this is the method of images though......
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  3. #3
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    <br />
\displaystyle {  G=\frac{1}{2 \pi} \; ( \; ln(\sqrt{(x-x_0)^2+(y-y_0)^2}+ln(\sqrt{(x-x_0)^2+(y+y_0)^2} \; )<br />
}<br />

    <br />
\displaystyle { \frac{ \partial G}{\partial  y}=\frac{1}{4 \pi} \; ( \; \frac{ 2(y-y_0)}{(x-x_0)^2+(y-y_0)^2}+\frac{ 2(y+y_0)}{(x-x_0)^2+(y+y_0)^2 } \; )<br />
}<br />

    <br />
\displaystyle { \frac{ \partial G}{\partial  y}|_{y=0}=0<br />
}<br />
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  4. #4
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    Is this the method of images though?
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  5. #5
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    I looked this link about method of images
    The method of images.
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