How to interpret the definition of a Green's function?

Question

A Green's function for a linear differential operator $L$ on a domain $\Omega \subseteq \mathbb{R}^n$ is a function $G$ on $ \overline{\Omega} \times \overline{\Omega} $ that satisfies

$$ L_xG(x, s) = \delta(x - s) $$

in the sense of distribution and

$$ G(\partial\Omega \times \Omega) = 0 $$

I'm not sure how to interpret the equation $L_xG(x, s) = \delta(x - s)$. Below I'll try my best to explain my understanding. To make things less complicated, let's ignore the boundary condition $ G(\partial\Omega \times \Omega) = 0 $.

I have no problem to interpret the RHS. Consider $\delta(x - s)$ acting on a test function $\phi(s)$ of variable $s$:

$$ \langle \delta(x - s), \phi(s) \rangle = \delta * \phi (x) = \phi(x) = \langle \delta_x, \phi \rangle $$

Thus, we can interpret the RHS as the following: for each fixed $x \in \Omega$, the RHS gives us a distribution on $\Omega$, namely, $\delta_x$.

The problem is the LHS. There are two possible interpretations.

1. For each fixed $s \in \Omega$, the function $x \mapsto G(x, s)$ has a distributional derivative $L_xG(\cdot, s)$. But then the equation $ L_xG(x, s) = \delta(x - s) $ makes no sense because in the RHS we have a distribution for each $x \in \Omega$, not for each $s \in \Omega$.
2. For each fixed $x \in \Omega$, we can consider the function $s \mapsto G(x, s)$. But then this makes even less sense because now that $x$ is fixed, we can't differentiate with respect to the variable $x$!

What is the appropriate interpretation of $L_xG(x, s) = \delta(x - s)$?

Edit:

As @Christopher A. Wong said, we can also interpret $\delta(x - s)$ as a distribution for each given $s \in \Omega$. The equation $ L_xG(x, s) = \delta(x - s) $ can be interpreted as the following: for each fixed $s \in \Omega$, the function $x \mapsto G(x, s)$ has a distributional derivative $L_xG(\cdot, s)$ equal to $\delta_s$. This works because of the symmetry $\delta(x - s) = \delta(s - x)$.

But now I have a problem to express the solution of $ Lu = f$. If we write

$$ u(x) = \int_\Omega G(x, s) f(s) ds = \langle G(x, \cdot), f \rangle $$

then we would have

$$ Lu(x) = \langle L_xG(x, \cdot), f \rangle $$

But I don't know what $L_xG(x, \cdot)$ is (we only have $L_xG(\cdot, s) = \delta_s$), so I'm unable to deduce $ Lu = f$.

Answer 1

The distribution $\delta(x - s)$ is a distribution in both $x$ and $s$. For a function $G: \Omega \times \Omega \rightarrow \mathbb{R}$ and $L = L_x$, $LG$ is a distribution on $\Omega$ for both fixed $x$ and fixed $s$. That is, both $\langle LG(x, \cdot), \phi(\cdot)\rangle$ and $\langle LG(\cdot,s), \phi(\cdot)\rangle$ are defined on test functions $\phi$.

In particular, this means that we can write $$\langle LG(x,\cdot), f(\cdot) \rangle = f(x),$$ which allows you to deduce that the Green's function can be used to construct the solution to the PDE $Lu = f$.