Green's Function / Impulse Response Confusion From:
https://en.wikipedia.org/wiki/Green%27s_function
"In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential equation defined on a domain, with specified initial conditions or boundary conditions.
A Green's function, $G(x,s)$, of a linear differential operator $L = L(x)$ acting on distributions over a subset of the Euclidean space $ℝn$, at a point $s$, is any solution of
$$LG(x,s)=\delta(x-s)$$
where $δ$ is the Dirac delta function."
But it seems to me that the impulse response, $I$, would be
$$L(\delta(x-s))=I(x,s)$$
Then $I$ and $G$ are not the same.  That is, $I$ is the output from $L$ due to an impulsive input and $G$ is the input to $L$ that produces an impulsive output.
My Question: How can the impulse response be the same as the Green's function given these definitions?
 A: You are simply reading it backwards. When they say "impulse response of an inhomogeneous linear differential equation defined on a domain, with specified initial conditions or boundary conditions" they mean the function $G(x,s)$ such that $L(G(x,s)) = \delta(x-s)$. 
I know it is a little counterintuitive at first since one thinks on the differential operator $L$ as something that you feed with a function and gives back a function. But in this case the operator we are dealing with is in fact the solution operator of the differential equation. That is $S:X\rightarrow Y$, where $X$, $Y$ are function spaces such that $S(f) = u$ when 
\begin{equation}
\begin{cases}
Lu=f &\text{ in } \Omega\\
u = 0 &\text{ in } \partial\Omega 
\end{cases}
\text{ (It can also be a different boundary condition depending on the problem)}
\end{equation} 
Think of the Green functions and the $\delta$ in the following way to notice why this is useful, the $\delta$ is "kind of a base of the functions spaces" since you can "write" any function as 
\begin{gather}
f(x)"=" \sum_s f(s)\delta(x-s)\\ \text{ (It really is an integral not a sum, in fact is a convolution integral)}
\end{gather}
And, since the solution operator is linear, then to solve the problem for a general $f$ it is enough to solve it for the deltas and then you just sum using superposition. The Green functions are just the solutions of the deltas, that is
\begin{equation}
G(x,s) = S(\delta(x-s))\\
\end{equation}
so
\begin{equation}
u(x) = S(f)(x) "=" \sum_s f(s)S(\delta(x-s)) = \sum_s f(s)G(x,s)
\end{equation}
Notice that $s$ is a parameter not the variable of the delta function so $f(s)$ is a constant for the solution operator. Notice also that this is not a really formal answer, but I hope it is useful to star understanding what the Green functions are about.
