- #1

sergiokapone

- 302

- 17

- Homework Statement
- Find 1D Green function for charged layer with distribution:

\begin{equation}

\phi=

\begin{cases}

0, x<-a \\

-\rho_0, -a<x<0 \\

\rho_0, 0<x<a \\

0, x>a

\end{cases}

\end{equation}

With boundary condition ##\phi(x \to -\infty) = 0##.

- Relevant Equations
- ##\nabla^2\phi = -4\pi\rho## and ##\nabla^2G(x,x') = -4\pi\delta(x-x')##

I came across an example of a solution to finding the potential of a charged layer using the Green function (here, pdf). The standard algorithm for finding the Green function by boundary conditions for many problems is understandable:

\begin{align*}

G_\mathrm{Left} = Ax+ B \\

G_\mathrm{Right} = Cx + D \\

G_\mathrm{Right}' - G_\mathrm{Left}' = -4\pi

\end{align*}

since the Green function is linear in 1D problems, then using boundary conditions we find constants that are functions of primed coordinates:

\begin{align*}

G_\mathrm{Left} = Ax+ B \\

G_\mathrm{Right} = Cx + D \\

G_\mathrm{Right}' - G_\mathrm{Left}' = -4\pi

\end{align*}

since the Green function is linear in 1D problems, then using boundary conditions we find constants that are functions of primed coordinates:

**But in this case I cannot understand how to find the Green function (how to determine constants) by the boundary condition ##\phi(x \to -\infty) = 0##, which, as indicated in the link, is equal ##G(x,x') = 4\pi x_{<}##. After all, the linear function does not disappear when striving to infinity. So, I need help.**