# Dirac's delta in 3 dimensions: proof of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$

If $T_f$ is a distribution, i.e. a linear functional, continuous according to the convergence defined here, defined on the space $K$ of the functions of class $C^\infty$ that are null outside a bounded interval (which is not the same for all functions), its derivative is defined as $$\frac{dT_f}{dx}(\varphi):=-T_f(\varphi’)$$where $\varphi’$ is the derivative of $\varphi$. The symbolic writing $T_f(\varphi)=\int_{-\infty}^\infty f(x)\varphi(x)dx$ is often used to write such a functional, since, if $g$ is (Riemann or Lebesgue) integrable on every bounded interval, then $\int_{-\infty}^\infty g(x)\varphi(x)dx$ indeed is such a continuous functional. In this context, we can symbolically define the “derivative” $f’$, for any $T_f$, even if the symbolic writing $f$ does not refer to an integrable function, according to the expression$$\int_{-\infty}^\infty f'(x)\varphi(x)dx:=-\int_{-\infty}^\infty f(x)\varphi'(x)dx=:\frac{dT_f}{dx}(\varphi).$$

Let us come to my question. While studying physics, in particular the theory of electromagnetism and the derivation of the Biot-Savart law from Ampère’s law, I always find the equality$$\nabla^2\left(\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}_0\|}\right)=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$$where $\nabla^2$ is the Laplacian$^1$. I suppose that, in the tridimensional case, with $\phi:\mathbb{R}^3\to\mathbb{R}$, $\int_{\mathbb{R}^3}\frac{\partial f(\boldsymbol{x})}{\partial x_i}\phi(\boldsymbol{x}) dx_1dx_2dx_3$ is analogously defined as $\frac{\partial T_f}{\partial x_i}(\varphi)$, which I suppose to be analogously defined, in turn, as $-\int_{\mathbb{R}^3}f(\boldsymbol{x})\frac{\partial \phi(\boldsymbol{x})}{\partial x_i} dx_1dx_2dx_3$, although I say I suppose because I have not found a rigourous definition of such derivatives on line nor in cartaceous texts; as to mathematical resources, I have studied Kolmogorov-Fomin’s Элементы теории функций и функционального анализа, which only focuses on the monodimensional $\varphi:\mathbb{R}\to\mathbb{R}$ case. Once fixed a proper definition of such derivatives, how can it be proved that $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$?

$^1$ The link contains a derivation of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ (equations (18)-(24)), but I do not understand it: I would understand it if we could apply Gauss’s divergence theorem at (20), but I know it for functions of class $C^1(\mathring{A})$, $\overline{V}\subset\mathring{A}$, only, while $\nabla\left(\frac{1}{\|\boldsymbol{x}-\boldsymbol{x}_0\|}\right)$ is not even defined for $\boldsymbol{x}=\boldsymbol{x}_0$; the other derivation of the identity $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})$ $=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ that I have found uses a “weak limit”, but it does not use the formal definiton of derivative of a distribution that I have written above. These are the two only references addressing what I am asking that I have managed to find.

#### Solutions Collecting From Web of "Dirac's delta in 3 dimensions: proof of $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$"

You are right, the partial derivatives of distributions on higher-dimensional spaces are defined as you surmised,

$$\frac{\partial T}{\partial x_i} \colon \varphi \mapsto -T\biggl[\frac{\partial \varphi}{\partial x_i}\biggr],$$

more generally

$$D^{\alpha} T \colon \varphi \mapsto (-1)^{\lvert\alpha\rvert} T[D^{\alpha}\varphi]$$

for higher derivatives.

From that you obtain $(\nabla^2 T)[\varphi] = T[\nabla^2\varphi]$, and for the locally integrable function $f(x) = \lVert x-x_0\rVert^{-1}$ we therefore have

$$(\nabla^2 T_f)[\varphi] = T[\nabla^2\varphi] = \int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3.\tag{1}$$

Now choose $R > 0$ so large that $\operatorname{supp} \varphi \subset B_R(x_0)$, and choose $0 < \varepsilon < R$. Then

$$\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = \lim_{\varepsilon \searrow 0} \int_{\varepsilon < \lVert x-x_0\rVert < R} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3$$

since $f$ is locally integrable and $\nabla^2\varphi$ is continuous.

Away from $x_0$, $f$ is smooth, and a slightly tedious computation shows that $\nabla^2 f \equiv 0$ on $\mathbb{R}^3\setminus \{x_0\}$, so

$$\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = \lim_{\varepsilon \searrow 0} \int_{\varepsilon < \lVert x-x_0\rVert < R} f(x)\cdot \nabla^2\varphi(x) – \varphi(x)\cdot \nabla^2 f(x)\,dx_1\,dx_2\,dx_3,$$

and to that integral you can apply Green’s formula to obtain

\begin{align}
\int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 &= \lim_{\varepsilon \searrow 0}\; \Biggl(\int_{\lVert x-x_0\rVert = R} f(x)\frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS\\
&\qquad\qquad + \int_{\lVert x-x_0\rVert = \varepsilon} f(x)\frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS\Biggr)\\
&= \lim_{\varepsilon \searrow 0} \int_{\lVert x-x_0\rVert = \varepsilon} f(x)\frac{\partial \varphi}{\partial \nu}(x) – \varphi(x)\frac{\partial f}{\partial\nu}(x)\,dS,
\end{align}

where $dS$ denotes the surface measure of the sphere, $\frac{\partial}{\partial\nu}$ the directional derivative in direction of the outer normal, and the first integral on the right hand side vanishes since $\varphi$ vanishes in a neighbourhood of the outer sphere. The outer normal on the sphere $\lVert x-x_0\rVert = \varepsilon$ is $-\frac{x-x_0}{\lVert x-x_0\rVert}$,

so we are left with

$$\frac{1}{\varepsilon}\int_{\lVert x-x_0\rVert = \varepsilon} f(x)\langle \nabla\varphi(x),x-x_0\rangle – \varphi(x)\langle \nabla f, x-x_0\rangle\,dS.$$

An easy estimate shows

$$\frac{1}{\varepsilon}\int_{\lVert x-x_0\rVert = \varepsilon} f(x)\langle\nabla\varphi(x),x-x_0\rangle\,dS \leqslant \frac{1}{\varepsilon^2}\int_{\lVert x-x_0\rVert = \varepsilon} \lVert\nabla\varphi\rVert\cdot\varepsilon\,dS \leqslant C\frac{4\pi\varepsilon^2}{\varepsilon} \xrightarrow{\varepsilon \searrow 0} 0.$$

For the other part of the integral, computing $\frac{\partial f}{\partial\nu}$ gives

$$\int_{\lVert x-x_0\rVert = \varepsilon} \varphi(x)\cdot \frac{1}{\varepsilon^2}\,dS,$$

and since $\varphi$ is continuous at $x_0$ we have

$$\lim_{\varepsilon\searrow 0} \int_{\lVert x-x_0\rVert = \varepsilon} \varphi(x)\cdot \frac{1}{\varepsilon^2}\,dS = 4\pi\varphi(x_0).$$

Collecting everything without mucking up the signs, we get

$$(\nabla^2 f)[\varphi] = \int_{\mathbb{R}^3} f(x)\cdot \nabla^2\varphi(x)\,dx_1\,dx_2\,dx_3 = -4\pi \varphi(x_0) = -4\pi\delta_{x_0}[\varphi].$$

That holds for all test functions $\varphi$, hence $\nabla^2 f = -4\pi\delta_{x_0}$.

For a fixed $x_0$, define $G_{x_0}=|x-x_0|^{-1}$. Then $\nabla^2 G_{x_0}=0$ for $x \ne x_0$. If $\varphi$ is a compactly supported $C^{\infty}$ function on $\mathbb{R}^3$, then
$$\nabla\cdot(G_{x_0}\nabla\varphi-\varphi \nabla G_{x_0})=G_{x_0}\nabla^2\varphi-\varphi\nabla^2G_{x_0}=G_{x_0}\nabla^2\varphi,\;\;\; x \ne x_0.$$
Integrate and apply the divergence theorem, keeping in mind that $\varphi$ vanishes outside a large sphere:
\begin{align}
\int_{\mathbb{R}^3} G_{x_0}\nabla^2\varphi dV
&=\lim_{\epsilon\downarrow 0}\int_{|x-x_0|\ge \epsilon}G_{x_0}\nabla^2\varphi dV\\
&=-\lim_{\epsilon\downarrow 0}\int_{|x-x_0| \ge \epsilon}\nabla\cdot(\varphi \nabla G_{x_0} – G_{x_0}\nabla\varphi)dV \\
& =-\lim_{\epsilon\downarrow 0}\int_{|x-x_0|=\epsilon}\left(\varphi\frac{\partial G_{x_0}}{\partial n}-G_{x_0}\frac{\partial\varphi}{\partial n}\right)dS \\
& =-\lim_{\epsilon\downarrow 0}\int_{|x-x_0|=\epsilon}\varphi\frac{\partial G_{x_0}}{\partial n}dS
\end{align}
The normal derivative is in the inward direction on $|x-x_0|=\epsilon$ because the outward normal on $\epsilon \le |x-x_0| \le R$ is where this started. $G_{x_0}(x)=1/r$ for $|x-x_0|=r$. So you end up with a normal derivative of $G_{x_0}$ equal to $1/\epsilon^2$ on $|x-x_0|=\epsilon$. Therefore, you pick up a factor equal to the area of the unit sphere:
$$\int_{\mathbb{R}^3} G_{x_0}\nabla^2\varphi dV = -\lim_{\epsilon\downarrow 0}\frac{1}{\epsilon^2}\int_{|x-x_0|=\epsilon}\varphi(x)dS = -4\pi\varphi(x_0).$$