Second derivatives of 1/r

Question

Let $r = \sqrt{x^2 + y^2 + z^2}$. From the fact that $\nabla^2 r^{-1} = -4\pi \delta^{(3)}(\vec{r})$, is it correct to say that $$ \frac{\partial^2}{\partial x^2}(r^{-1}) = \frac{3x^2 - r^2}{r^5} - \frac{4\pi}{3} \delta^{(3)}(\vec{r}) \\ \frac{\partial^2}{\partial x \partial y}(r^{-1}) = \frac{3xy}{r^5} $$

The question is, is it justified to split the Dirac delta evenly in all three directions? This seems like the most straightforward way. $$ \delta = \frac{\delta}{3} + \frac{\delta}{3} + \frac{\delta}{3} $$

Or maybe there are other ways to distribute, while still respecting symmetry, like $$ \delta = \frac{x^2 \delta}{r^2} + \frac{y^2 \delta}{r^2} + \frac{z^2 \delta}{r^2} $$

Answer 1

Let $\psi$ be the distribution $\psi(r)=\frac1r$. Then, for $\phi \in C_C^\infty$, we have

$$\begin{align} \langle \partial_{ij}\psi, \phi\rangle &=\langle \psi, \partial_{ij}\phi\rangle\\\\ &=\int_{\mathbb{R}^3}\frac1r \frac{\partial^2\phi(\vec r)}{\partial x_i\partial x_j}\,d^3\vec r\\\\ &=\underbrace{\int_{\mathbb{R}^3}\frac{\partial}{\partial x_i}\left(\frac1r \frac{\partial\phi(\vec r)}{\partial x_j}\right)\,d^3\vec r}_{=0\,\,\text{since}\,\,\phi\in C_C^\infty}-\int_{\mathbb{R}^3}\frac{\partial}{\partial x_i}\left(\frac1r \right)\left(\frac{\partial\phi(\vec r)}{\partial x_j}\right)\,d^3\vec r\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\frac{\partial}{\partial x_i}\left(\frac1r \right)\left(\frac{\partial\phi(\vec r)}{\partial x_j}\right)\,d^3 \vec r\tag1 \end{align}$$

where $B(\vec r_c,R)$ is a sphere of radius $R$ centered at $\vec r_C$. Now, integrating by parts again, we find that

$$\begin{align} \langle \partial_{ij}\psi, \phi\rangle&=-\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\frac{\partial }{\partial x_j}\left(\phi(\vec r)\frac{\partial}{\partial x_i}\left(\frac1r \right)\right)\,d^3\vec r\\\\ &+\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\phi(\vec r)\frac{\partial^2 }{\partial x_i\partial x_j}\left(\frac1r\right)\,d^3\vec r\\\\ &=\lim_{\varepsilon\to0^+}\oint_{\partial B(0,\varepsilon)}(\hat n\cdot \hat x_j)\phi(\vec r)\frac{\partial }{\partial x_i}\left(\frac1r\right)\,dS\\\\ &+\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\phi(\vec r)\frac{\partial^2 }{\partial x_i\partial x_j}\left(\frac1r\right)\,d^3\vec r\\\\ &=-\lim_{\varepsilon\to0^+}\int_0^{2\pi}\int_0^\phi (\hat r\cdot \hat x_j)\phi(\vec r)\left.\left(\frac{x_i}{r^3}\right)\right|_{r=\varepsilon}\,\varepsilon^2\,\sin(\theta)\,d\theta\,d\phi\\\\ &+\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\phi(\vec r)\frac{\partial^2 }{\partial x_i\partial x_j}\left(\frac1r\right)\,d^3\vec r\\\\ &=-\frac{4\pi}{3}\delta_{ij}\phi(0)+\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3\setminus B(0,\varepsilon)}\phi(\vec r)\frac{\partial^2 }{\partial x_i\partial x_j}\left(\frac1r\right)\,d^3\vec r\\\\ \end{align}$$

Therefore, in distribution we see that

$$\frac{\partial^2}{\partial x_i\partial x_j}\left(\frac1r\right)=-\frac{4\pi}{3}\delta_{ij}\delta(\vec r)+\text{PV}\left(\frac{3x_ix_j-\delta_{ij}r^2}{r^5}\right)$$

When $i=j$, we see that in distribution

$$\frac{\partial^2}{\partial x_i^2}\left(\frac1r\right)=-\frac{4\pi}{3}\delta(\vec r)+\text{PV}\left(\frac{3x_i^2-r^2}{r^5}\right)$$

whereupon summing over $i$ yields the familar result

$$\nabla^2 \frac1r =-4\pi \delta(\vec r)$$

Answer 2

You're completely correct that $\delta$-at-$0$ in 3D is some sort of "product" of the one-dimensional $\delta$'s in any choice of (maybe best to be orthogonal...?) coordinates.

But, although this is probably unsatisfying, the "decomposition" of it is as a tensor product $\delta_x\otimes \delta_y\otimes \delta_z$, in the (standard) coordinate variables. Thus, by a sort of product rule, applying $\Delta$, expressed as sum of second derivatives in the chosen variables, $$ \Delta(\delta) \;=\; (\partial_x^2+\partial_y^2+\partial_z^2)(\delta_x\otimes\delta_y\otimes\delta_z) $$ $$ \;=\; \delta_x''\otimes \delta_y\otimes \delta_z + \delta_x\otimes \delta_y''\otimes \delta_z + \delta_x \otimes\delta_y\otimes\delta_z'' $$ This is literally correct, but there are (perhaps unsurprisingly) some subtleties in interpreting/using it. :)