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

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 $$Lu = f$$.

• Interpretation 1 works. The equation does make sense, because $\delta(x - s)$ is a distribution for each fixed $s$ as well as every fixed $x$. Commented Feb 2, 2022 at 4:29
• @ChristopherA.Wong I take that you mean $L_xG(\cdot, s) = \delta_s$ for each fixed $s \in \Omega$? But then if we write $u(x) = \langle G(x, \cdot), f(\cdot) \rangle$, we would have $L_x u = \langle L_xG(x, \cdot), f(\cdot) \rangle$, which can't be reduced to $f(x)$. Commented Feb 2, 2022 at 4:39
• $L_x G(x,s) = \delta(x - s)$ precisely means $\langle L_x G(x,s), \phi(s) \rangle = \phi(x)$ for every $x$ and every test function $\phi$. Commented Feb 2, 2022 at 4:55
• OK, I see your concern. I will write out a more careful response shortly. Commented Feb 2, 2022 at 4:57

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 $$Lu = f$$.