How to obtain Green function for the Helmholtz equation? all.
I am following Jackson's Classical Electrodynamics.
At Chapter 6.4, the book introduces how to obtain Green functions for the wave equation and the Helmholtz equation. I have a problem in fully understanding this section.
The Green function for the Helmholtz equation should satisfy
$$
(\nabla^2+k^2)G_k
=-4\pi\delta^3(\textbf{R}).\tag{6.36}
$$
Using the form of the Laplacian operator in spherical coordinates, $G_k$ satisfies
$$
\frac{1}{R}\frac{d^2}{dR^2}(RG_k)+k^2G_k
=-4\pi\delta^3(\textbf{R}).\tag{6.37}
$$
Everywhere expcept $R=0$, $RG_k$ can be given  as
$$
RG_k(R) =Ae^{ikR}+Be^{-ikR}.\tag{6.37b}
$$
Then, the book reads,

"Furthermore, the delta function has influence only at $R \rightarrow 0$. In that limit the equation reduces to the Poisson equation, since $kR<<1$."

But, I can't understand why the first Helmholtz equation with the Green function form can reduce to the Poisson equation form as $R\rightarrow0$. In other literature, condition is given as $k\rightarrow0$, instead of $R\rightarrow0$. But in this case, I can't still understand why the Delta function matters as $k\rightarrow0$.
 A: You can see from the expression
$$RG_k(R)=Ae^{ikR} + Be^{-ikR},$$
that the solution depends only on the combination $kR$. You know additionally that the Dirac delta distribution will only have an impact when $R$ goes to zero as stated. When evaluating that limit the value of $k$ becomes irrelevant (as long as it is finite) since $kR\rightarrow 0$ anyways. In particular you can make your life easier by using a specific value of $k$, namely 0, for the original differential equation. This "trick" is what leads to the Poisson equation and Jackson describes very briefly.
Observe this is simply allowing you to find the normalization by comparison with the known case from electrostatics.
A: So whenever you are dealing with delta functions, you are actually dealing with a situation where there is a discontinuity in some derivative. For example, when I write
$$\frac{\mathrm d^2y}{\mathrm dx^2} =\delta(x),$$
I am actually writing that I want to consider a situation where $y''(x)=0,x\ne 0,$ but there is a discontinuity in $y'(x)$ at $x=0$ with magnitude 1. So for example $\frac12 |x|$ is a valid solution, the slope changes from $-1/2$ to $+1/2$.
When you are dealing with Green's functions, you are trying to exploit the linearity of a differential operator, in order to write the solution as a superposition of simpler solutions. Those simpler solutions are the Green's functions themselves, and they usually correspond to a “pointlike” inhomogeneity. The reason that this term is zero everywhere except at a point, comes from this desire to make the inhomogeneity pointlike. In addition, Green's functions have a subtler requirement, which is that the usually have to have a particular boundary condition, like being zero off at infinity or being zero on some fixed surface, so that when we go to do superposition with them, we know that that point or surface stays at zero because all of our functions are zero on the boundary.
The eventual goal is, much like you can start from the scalar potential for a single point charge $-q/|\mathbf r-\mathbf r_0|$ and construct an arbitrary potential $-\int \mathrm d^3 r_0~\rho(r_0)/|\mathbf r-\mathbf r_0|$, building the continuous charge out of infinitely many point charges, so too we can do this for any other linear system with some sort of weird unknown driving force or source charge or whatever it is. That is the goal, and it helps to keep the goal in mind when working with Green's functions.
