How Do You Derive the Kernel of the Laplace Equation with Boundary Conditions?

[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?

 

[eljose79]

Thank you for sharing your approach to deriving the kernel of the Laplace equation with prescribed boundary conditions. I agree that using the method of images can lead to a complicated form for the kernel, especially when dealing with a semi-infinite strip.

One possible approach to simplify the kernel is to use the method of separation of variables. This involves assuming a solution of the form $T(x,y) = X(x)Y(y)$ and substituting it into the Laplace equation. This can lead to two ordinary differential equations, one for $X(x)$ and one for $Y(y)$, which can be solved separately. The solutions can then be combined to obtain the final form of the kernel.

Another approach is to use complex analysis techniques, such as the method of conformal mapping, to transform the semi-infinite strip into a simpler domain where the Laplace equation can be solved more easily. This can lead to a simpler form of the kernel.

I hope these suggestions are helpful in simplifying your derivation of the kernel. Good luck with your research!