- #1
Divergenxe
- 2
- 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?
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?