I Kernel of Laplace equation

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1. Oct 23, 2016

Divergenxe

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. Oct 28, 2016