Derivatives of Norms and Absolute Values (distributions)

Question

For example we have for $x \in \mathbb{R}$, $$\frac{\partial}{\partial x}\left| x\right| = 2\Theta(x) -1 $$ and thus $$\frac{\partial^2}{\partial x^2}\left| x\right| = 2\delta(x) $$ We also have, for example for $\mathbf{r} \in \mathbb{R}^3$ $$\newcommand{\norm}[1]{\left\lVert#1\right\rVert} \boldsymbol{\nabla}^2 \frac{1}{\norm{\mathbf{r}}} = -4\pi \delta\left(\mathbf{r} \right)$$ But what about more generally? What about $\frac{\partial}{\partial x}\left| x\right|^a $ ? What about $\newcommand{\norm}[1]{\left\lVert#1\right\rVert} \boldsymbol{\nabla}^2\norm{\mathbf{r}}^a $ with $\mathbf{r} \in \mathbb{R}^N$? I'm mainly curious about $a \in \mathbb{Z}$.

Answer 1

The full answer with all details for all possible $\alpha$ is quite long, so we will make some simplifications and omit proofs.

Let $\alpha\in \Bbb Z$. The case $\alpha=0$ is trivial, we will skip it.

In $\Bbb R$ everything is quite simple. If $\alpha \ge 2 $, then $|x|^\alpha$ is a $C^2$ function, so $\partial_x |x|^\alpha$ and $\Delta |x|^\alpha$ are well-defined. If $\alpha= 1$, then you've already know the answer. However, for $\alpha\le -1$ the function $|x|^\alpha$ is not a distribution, hence we can't take derivatives. On a side note, by subtracting $\delta_0$ and its derivatives with well-chosen coefficients, we can obtain a distribution, and, therefore, have a derivative.

In $\Bbb R^2$ things are a little trickier, but still manageable. $\Delta \ln|x|=2\pi\delta_0$. for $\alpha>0$ we have $$\Delta |x|^\alpha = \alpha^2|x|^{\alpha-2},$$ and hence is well-defined in $D'$. For $\alpha = -1$ this wouldn't work, so we will have to proceed by a usual approach in the theory of distributions (see the end of this answer). For $\alpha\le -2$ the function $|x|^\alpha$ is not a distribution, we again will need some modifications.

Finally, $\Bbb R^d$. The result is a mere generalisation of the previous ones. $$\begin{cases}\text{the derivative in the classical sense},&\alpha>d-2\\ Const\times\delta_0,&\alpha=d-2\\\text{not a distribution without additional tinkering},&\alpha\le d-3.\end{cases}$$

The method of the proof for finding derivatives of functions with singularities.

Suppose that a sequence $f_n\to f$ in the sense of $D'(\Bbb R^d)$, then $\Delta f_n\to \Delta f$ in $D'(\Bbb R^d)$. Now we want to choose $f_n$ such that we know how to calculate its laplacian. In our case with the only singularity in zero we will take $$f_n=\begin{cases}r^\alpha,&r\ge \frac 1n,\\\frac{1}{n^\alpha},&r<\frac 1n.\end{cases}$$ Then we will prove the convergence of $f_n$, find its laplacian in $D'$ by integrating $f_n$ against $\Delta \phi$, $\phi$ - test function. Then study the limit $\Delta f_n$. As an exercise, one can work this approach to prove that $\Delta \ln |x|=2\pi\delta$ in $D'(\Bbb R^2)$.