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I Kernel of Laplace equation

  1. Oct 23, 2016 #1
    Hi! Everyone. I encounter some trouble in deriving the kernel of Laplace equation with prescribed boundary conditions.

    Given the following preposition:
    $$T(x, y) = \int_{-\infty}^{\infty}dx'\frac{y/\pi}{(x-x')^{2}+y^2}F(x')......[1]$$
    satisfies the Laplace equation for ##x\in(-\infty, \infty)## and ##y\geqslant 0##, subject to the boundary conditions: ##T(\pm\infty, y)=T(x, \infty)=0## and ##T(x, 0)=F(x)##.

    For ##F(x)=\delta(x-x_{0})##, the solution ##T(x, y)## is reduced to $$T(x, y) = \frac{y/\pi}{(x-x_{0})^{2}+y^{2}} ......[2]$$, which is the kernel of the Laplace equation with the prescribed boundary conditions.


    I have to make use of the above kernel to derive the kernel of the Laplace equaiton for ##x\in(0, L)## and ##y\geqslant 0##, subject to the boundary conditions: ##T(0, y)=T(L, y)=T(x, \infty)=0## and ##T(x, 0)=\delta (x-x_{0})##.

    I try to use the method of image. It is obvious that the square in the denominator of [1] gives a second choice of ##x_{0}##, says ##x_{1}##such that ##T_{0}(0, y)=T_{1}(0, y)##. In this way, the boundary condition T(0, y) is satisfied by imposing an image ##-T_{1}(0, y)##.
    Example:
    ##-\frac{y/\pi}{(x-(-x_{0}))^{2}+y^{2}}## is imposed to satisfy the boundary condition at ##x=0##:
    $$\frac{y/\pi}{(0-x_{0})^{2}+y^{2}}-\frac{y/\pi}{(0-(-x_{0}))^{2}+y^{2}}=0$$

    ##-\frac{y/\pi}{(x-(2L-x_{0}))^{2}+y^{2}}## is imposed to satisfy the boundary condition at ##x=L##:
    $$\frac{y/\pi}{(L-x_{0})^{2}+y^{2}}-\frac{y/\pi}{(L-(2L-x_{0}))^{2}+y^{2}}=0$$

    But I have to create another image to balance the potential due to the image on the other boundary. This leads to an infinite series.

    $$T(x, y)=\frac{y}{\pi}(\frac{1}{(x-x_{0})^{2}+y^{2}}-\frac{1}{(x-(-x_{0}))^{2}+y^{2}}+\frac{1}{(x-(x_{0}-2L))^{2}+y^{2}}+\frac{1}{(x-(x_{0}+2L))^{2}+y^{2}}-\frac{1}{(x-(-x_{0}-2L))^{2}+y^{2}}-\frac{1}{(x-(-x_{0}+2L))^{2}+y^{2}}+...)$$$$=\frac{y}{\pi}\sum_{n=-\infty}^{\infty}[\frac{1}{(x-(x_{0}-2nL))^{2}+y^{2}}-\frac{1}{(x-(-x_{0}-2nL))^{2}+y^{2}}]$$

    It results in a complicated form. I have to use the new kernel and [1] to derive ##T(x, y)## for a semi-infinte strip ##(0\leqslant x \leqslant L, y\geqslant 0)## with the boundary conditions: ##T(0, y)=T(L, y)=T(x, \infty)=0## and $$T(x, 0) =
    \left\{\begin{matrix}
    x , x<L/2\\ L-x, x>L/2
    \end{matrix}\right.
    $$
    If I use the kernel above, I cannot obtain a closed form of ##T(x, y)##. I think there should be other better approach to obtain a simpler form of the kernel. What do you think?
     
  2. jcsd
  3. Oct 28, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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