A question on Green's functions & integral operators I'm fairly new to the concept of Green's functions, but from what I understand so far, they are a powerful tool for solving PDEs with boundary conditions. 
Given a differential equation (in operator form)
$$\mathcal{L}u(x)=f(x)$$ (where $\mathcal{L}$ is a differential operator) it is possible to find a solution for $u(x)$ of the form
$$u(x)=\int G(x,\xi)f(\xi)d\xi$$
I'm struggling to understand the motivations given for the integral operator form of the solution given above. Why is it possible to represent it like this? 
I understand that the motivation partially comes from the fact that $\mathcal{L}$ will have an inverse (assuming that $\mathcal{L}$ is non-singular), and as it is a differential operator this implies that the inverse $\mathcal{L}^{-1}$ will be an integral operator. From the differential equation above we have $$u(x)=\mathcal{L}^{-1}f(x)$$ What is the motivation for introducing a so-called Green's function $G(x,\xi)$ such that the integral operator is of the form given above? Is it just that $G(x,\xi)$ is a generator of the operator? 
Also, is the reason why we end up with the defining equation for a Green's function $$\mathcal{L}G(x,\xi)=\delta (x-\xi)$$ simply because $\mathcal{L}\cdot\mathcal{L}^{-1}=\mathbf{1}$ and the fact that we can formally write (as $\mathcal{L}$ is linear) $$\mathcal{L}u(x)=\mathcal{L}\left(\int G(x,\xi)f(\xi)d\xi\right)=\int \left(\mathcal{L}G(x,\xi)\right)f(\xi)d\xi=f(x)\quad\Rightarrow\quad\mathcal{L}G(x,\xi)=\delta (x-\xi)$$
 A: The keyword here is linearity. A physicist would say that you are employing the superposition principle. You are decomposing your source term $f$ into a superposition of localized impulses: 
$$
f(x)=\int_\Omega f(y)\delta(x-y)\, dy.$$
Since the equation is linear, to solve 
$$
L_x u (x)=f(x)=\int_\Omega f(y)\delta(x-y)\, dy$$
you can solve the family of equations
$$
L_x G_y(x)=\delta(x-y), \qquad y\in \Omega$$
then multiply by $f(y)$ and superpose back:
$$\tag{1}
u(x)=\int_\Omega f(y)G_y(x)\,dy.$$
By linearity, you expect $(1)$ to be a solution to your original problem. (And of course people usually write $G(x, y)$ instead of $G_y(x)$).
To understand better, try looking in an electromagnetism book and see how it solves the equation 
$$
-\Delta \phi = \frac{\rho}{4\pi\epsilon_0},$$
where $\phi$ is the electrostatic potential and $\rho$ is the density of charge (and $\epsilon_0$ is the dielectric constant of the vacuum - I hope I have put the right constant). This is the prototypical example of Green's function.
