1D Green function for a charged layer

[sergiokapone]

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: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.

 

[member 553083]

A:The boundary condition $\phi \rightarrow 0$ as $x \rightarrow -\infty$ is telling you that the potential goes to zero at large negative values of $x$. This means that the Green's function must also go to zero in this limit, so you must have a solution of the form$$G(x,x') = 4\pi (x_{<} - x_0),$$where $x_0$ is some constant. Note that when $x \rightarrow -\infty$, we have $x_{<} \rightarrow x'$ and so this solution satisfies the boundary condition. The other boundary condition that you need to satisfy is continuity of the Green's function at $x = x'$. This means that you must have$$G(x',x') = 4\pi(x' - x_0) = 0.$$This implies that $x_0 = x'$, so the final solution for the Green's function is$$G(x,x') = 4\pi (x_{<} - x').$$