I am interested in non-negative, "classical" solutions to the following PDE $$-\Delta u = \frac{u^p}{|x|^2}$$ in $\mathbf R^n$ with $n \geq 3$ and $p>\frac{n+6}{n-2}$. Here by a classical solution $u$ I mean $u\in C^2(\mathbf R^n \setminus \{0\}) \cap C(\mathbf R^n)$ and the PDE is satisfied pointwise in $\mathbf R^n \setminus \{0\}$.
My approach is to look for radial solutions, namely $u(x)=u(|x|)$. With this, I arrive at the following Cauchy problem $$ \left\{ \begin{aligned} -u''-\frac{n-1}ru' &= \frac{u^p}{r^2} & & \text{ for } t>0,\\ u(0)&=a > 0 \end{aligned} \right. $$ for suitable $a$.
Usually, to solve the above Cauchy problem, we look for a fixed point of $$ \mathcal F : u(t) \mapsto a - \frac 1{n-2}\int_0^t \Big[ 1 - \big( \frac s t \big)^{n-2} \Big]\frac{u^p}s ds $$ for $u$ belonging to a convex closed subset of $C[0, +\infty)$. But it seems that the coefficient $\frac 1{|x|^2}$ of the PDE is not a correct one. If we replace this by any $\frac 1{|x|^\alpha}$ with $0<\alpha<2$, then a local solution is guaranteed.
Is it possible to obtain a local existence for the case of the coefficient $\frac 1{|x|^2}$?