Weak Laplacian of $\|x\|^\alpha$ Let $\alpha> 0$ and consider the function $\|\mathbf x\|^\alpha = (x^2 + y^2)^{\frac{\alpha}{2}}$ defined on $\mathbb R^2$. I want to compute the Laplacian $\Delta (\|\mathbf x\|^\alpha)$ in the sense of distributions.
If $\alpha \geq 2$ the function is differentiable and we get $\Delta (\|\mathbf x\|^\alpha) = \alpha^2 \|\mathbf x\|^{\alpha -2}$.
Does this formula hold in general? From the fact that $\|x\|^{\alpha}$ is smooth outside the origin the above formula should hold on $\mathbb R^2 \setminus \{0\}$, but what about the singularity at $0$?
 A: This result holds at least for $\alpha>0$. I will describe the method that allows to make an analysis even in the case of arbitrary dimension. Denote $\vec r=(x,y)$, $r=\|\vec r\|$.
First element: let the function $f_\epsilon\in L^1_{loc}(\Bbb R^2)$ be given by
$$f_\epsilon = \begin{cases}r^\alpha,&r>\epsilon,\\\epsilon^\alpha,&r\le\epsilon.
\end{cases}$$ We claim that $f_\epsilon\to r^\alpha$ in the sense of distributions. Indeed, this sequence converges in the sense of $L^1$, hence also in the sense of $D'$.  
Second element: if a sequence of distributions converge $T_n\to T$, then $\Delta T_n\to \Delta T$. This is a basic result from theory of distributions.
Third element: find $\Delta f_\epsilon$; let $\phi$ be a test function, then by definition 
$$\langle\Delta f_\epsilon,\phi\rangle=\int_{\Bbb R^2}f_\epsilon\Delta\phi =\int_{r\le \epsilon} \epsilon^\alpha\Delta\phi +\int_{r>\epsilon}r^\alpha\Delta\phi$$
First integral in the last expression rewrites as $$ \epsilon^\alpha\int_{r\le \epsilon}\Delta\phi = \epsilon^\alpha \int_{r= \epsilon}\nabla \phi\cdot \frac{\vec r}{r}.$$
Second integral in that expression rewrites $$ \int_{r>\epsilon}r^\alpha\Delta\phi =\int_{r>\epsilon}(\nabla\cdot (r^\alpha\nabla\phi)-\nabla r^\alpha\cdot\nabla\phi) $$
$$=\int_{r=\epsilon}    r^\alpha\nabla\phi\cdot\frac{-\vec r}{r}  -\int_{r>\epsilon} \nabla r^\alpha\cdot\nabla\phi = -\epsilon^\alpha \int_{r= \epsilon}\nabla \phi\cdot \frac{\vec r}{r} -\int_{r>\epsilon}( \nabla\cdot (\nabla r^\alpha\cdot \phi)-\Delta r^\alpha\phi )$$
$$=-\epsilon^\alpha \int_{r= \epsilon}\nabla \phi\cdot \frac{\vec r}{r}+\int_{r= \epsilon}\nabla r^\alpha\cdot\frac{\vec r}{r} \phi +\int_{r>\epsilon} \Delta r^\alpha\phi  $$
$$=-\epsilon^\alpha \int_{r= \epsilon}\nabla \phi\cdot \frac{\vec r}{r}+\int_{r= \epsilon}\alpha r^{\alpha-1} \phi +\int_{r>\epsilon} \alpha^2 r^{\alpha-2}\phi. $$
So, the grand total can be put as $$\langle\Delta f_\epsilon,\phi\rangle=\alpha \epsilon^{\alpha-1}\int_{r= \epsilon} \phi+\int_{r>\epsilon} \alpha^2 r^{\alpha-2}\phi.$$
Fourth point: convergence  as $\epsilon\to0$.
The first integral converges to zero. Indeed, put $\phi(\vec r) = \phi(0)+\nabla\phi(0)\cdot \vec r +\mathcal O(\epsilon^2)$, and use it in the integral to obtain
$$\alpha \epsilon^{\alpha-1}\int_{r= \epsilon} \phi(\vec r) =\alpha \epsilon^{\alpha-1}\int_{r= \epsilon} (\phi(0)+\nabla\phi(0)\cdot \vec r +\mathcal O(\epsilon^2))$$$$ = \alpha  \epsilon^{\alpha-1}(2\pi\epsilon\phi(0)+0+\mathcal O(\epsilon^3))\to 0\quad \mbox{as }\epsilon\to0. $$
edit
If $\alpha>0 $, then $ \alpha^2 r^{\alpha-2}\in L^1_{loc}(\Bbb R^2)$; indeed, if we integrate it on the compact that doesn't contain the point $(0,0)$, then it's just an integral of a continuous function over a compact. When we integrate on a compact that contains the point $(0,0)$, in order to study the existence of the integral, it's sufficient to examine the integral $$\int_{r\le 1} \alpha^2 r^{\alpha-2} dxdy = \int_{r=0}^{r=1}dr \int_0^{2\pi}d\theta \alpha^2 r^{\alpha-1} = 2\pi \alpha .  $$ Therefore, for all compact sets $K$ the integral $\int_K\alpha^2 r^{\alpha-2} $ exists, which is the definition of the space $L^1_{loc}$.
We need the convergence of the integral $\int_{r>\epsilon} \alpha^2 r^{\alpha-2}\phi$ because of the choice of workflow: we approximate $r^\alpha$ in the sense of distributions by a sequence of $f_\epsilon$, i.e. $f_\epsilon\to r^\alpha$ in the sense of $D'(\Bbb R^2)$. Then we find $\Delta f_\epsilon$ in the sense of distributions: it is equal to $$\langle\Delta f_\epsilon,\phi\rangle=\alpha \epsilon^{\alpha-1}\int_{r= \epsilon} \phi+\int_{r>\epsilon} \alpha^2 r^{\alpha-2}\phi.$$ Now when we pass to the limit in the above expression, we will obtain $\Delta r^\alpha$, because $\Delta f_\epsilon\to \Delta r^\alpha$ in the sense of $D'(\Bbb R^2)$. The first integral converges to zero, so we need to study the convergence of the second integral to something. The sufficient condition of this convergence is to demand $\alpha^2 r^{\alpha-2}\in L^1_{loc}$, i.e. to demand $\alpha>0$.
end edit
