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 $\mathbf{r}$?

• My inner pedant compels me to note that $\nabla^2 G = \delta(\mathbf{r} - \mathbf{r_0})$ is not an equation that makes sense pointwise. It's an equality of distributions, so to check it one verifies that $\int G(r,r_0)\nabla^2\varphi(r)\,dr = \varphi(r_0)$ for all test functions $\varphi$. – Daniel Fischer Dec 28 '17 at 19:46
• I have been told that the Dirac function is loosely seen as a 'function' so I think you may be onto somthing here! By integrating over all test functions we deduce that the functions must be equal? The proof I have been presented with considers the integral over a volume of both sides of the 'equation' and deduces equality of this integral to show that the Green's function is valid at $\mathbf{r=r_0}$. – user258521 Dec 28 '17 at 19:58
• A distribution is a continuous linear functional on the space of test functions, so equality of distributions is by definition equality of the values on applying them to all test functions. I'm not sure how "integrating both sides over a volume" should work. Though the Dirac distribution is a quite well-behaved measure and thus "integrating it over a volume" has a fairly straightforward interpretation (as $\delta_{r_0}(V)$, being $1$ if $r_0 \in V$ and $0$ otherwise), I don't see how one should integrate $\nabla^2G$ over an open set containing $r_0$ without committing a petitio principii. – Daniel Fischer Dec 28 '17 at 20:33
• Using Gauss's Theorem we can decude the integral is $0$ if the origin is outside of the volume and $4*pi$ if the origin is within the volume is how the proof went about that. – user258521 Dec 28 '17 at 22:49
• I don't know what else you want from this question. I was going to write an answer but Daniel Fischer's has answered your question exactly. Delta 'functions' or similar things only make sense when they act on smooth test functions. – amcalde Jan 3 '18 at 10:23

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.