Verifying the delta function satisfies Poisson's Equation Suppose we have the equation $\nabla^2 G= \delta(\mathbf{r-r_0})$ where $\delta$ is the Dirac Delta function in $\mathbb R^3$ and I want to verify that the solution to this equation is given by Green's Function as such (without worrying about the initial conditions):
$$G(\mathbf{r,r_0})=-\frac{1}{4\pi\mathbf{|r-r_0|}}$$
Then for $\mathbf{r \neq r_0}$ it follows $\nabla^2 G=0$ and the Dirac function is also zero by definition so this equation is satisfied as long as the denominator is not singular. However to verify the solution satisfies the equation for $\mathbf{r= r_0}$ the proofs I have seen take the form of integrating over a volume containing $\mathbf{r_0}$. Could it be explained why this method means that the Green's function here will satisfy the equation for $\mathbf{r=r_0}$ and thus all $\mathbf{r}$?
 A: The basic reason for this is that $G(\mathbf{r},\mathbf{r_0})$ is locally integrable on the whole $\mathbb{R}^n$, $n\geq 3$, i.e $G\in L^1_{loc}(\mathbb{R}^n)$. To use the terminology of V.S. Vladimirov [1, § 1.6 p. 15], $G$ is a regular generalized function so, by the linearity of and by the theorem on the passage to the limit under the sign of integral, it is represented exactly as 
$$
\langle G,\varphi \rangle= \int_{\mathbb{R}^n} G(\mathbf{r},\mathbf{r_0})\varphi(\mathbf{r}) d\mathbf{r} \quad \forall \varphi \in C^\infty_0(\mathbb{R}^n)
$$
It is therefore a somewhat natural choice to use the structure of this distribution to prove that 
$$
\langle\nabla^2 G,\varphi \rangle= \langle G,\nabla^2\!\varphi \rangle= \int_{\mathbb{R}^n} G(\mathbf{r},\mathbf{r_0}) \nabla^2\!\varphi(\mathbf{r}) d\mathbf{r}=\varphi(\mathbf{r_0}) \quad \forall \varphi \in C^\infty_0(\mathbb{R}^n)
$$
The method of proof of this equality to which user258521 alludes is simply the above formula in disguise. To see this, consider any domain $\Omega\in \mathbb{R}^n$ for which a form of Gauss's theorem holds, and a mollifier $\varphi_\varepsilon \in C^\infty_0(B_\varepsilon)$, i.e. an infinitely smooth function of compact support contained inside the ball $B_\varepsilon$ of radius $\varepsilon>0$ centered in $\mathbf{0}\in \mathbb{R}^n$ such that $\varphi_\varepsilon(\mathbf{r})\to \delta(\mathbf{r})$ in $\mathscr{D}^\prime$ for $\varepsilon\to 0$. Then 
$$
\varphi_\varepsilon\ast\chi_\Omega (\mathbf{r})= \int_{\mathbb{R}^n} \varphi_\varepsilon(\mathbf{r}-\mathbf{s})\chi_\Omega(\mathbf{s}) d\mathbf{s} \in C^\infty_0(\Omega+B_\varepsilon) \subset C^\infty_0(\mathbb{R}^n)
$$
where


*

*$\chi_\Omega:\mathbb{R}^n\to\{0,1\}$ is the indicator or characteristic function of the domain $\Omega$

*$\Omega+B_\varepsilon$ is the set formed by summing a vector of $\Omega$ and a vector of $B_\varepsilon$. 


By using the above definition for $\langle G,\varphi_\varepsilon\ast\chi_\Omega \rangle$ and letting $\varepsilon\to 0$ 
$$
\langle G,\varphi_\varepsilon\ast\chi_\Omega \rangle \to \int_\Omega G(\mathbf{r},\mathbf{r_0}) d\mathbf{r},
$$
and consequently applying Gauss's theorem:
$$
\langle \nabla^2 G,\varphi_\varepsilon\ast\chi_\Omega \rangle =
\langle G,\nabla^2\!\varphi_\varepsilon\ast\chi_\Omega \rangle\to
\begin{cases} 
1 & \mathbf{r_0}\in\Omega\\
0 & \mathbf{r_0}\notin\Omega
\end{cases}
$$


*

*Note that this kind of proof works only for domains $\Omega$ for which some kind of form of Gauss's theorem holds.

*A direct distribution theory proof, like the one offered by Vladimirov [1, §2.3.8 pp. 33-35], does not need such hypothesis.


[1] Vladimirov, V. S. (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR 2012831, Zbl 1078.46029.
