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Let $\Omega$ and open subset of $\mathbb{R}^n$. Let us consider the following operator: $$ \Delta_A (u)\, \, \colon= \text{div}(A \nabla u ), \qquad u \in C^{\infty}(\Omega) $$ where $A(x)$ is a matrix with smooth coefficients and uniformly elliptic on $\mathbb{R}^n$, i.e. $A$ is symmetric and there are positive constants $\lambda$ and $\Lambda$ such that for all $x \in \mathbb{R}^n$ and for all vectors $v, w \in \mathbb{R}^n$: $$ \langle v, A(x) w\rangle \le \Lambda |v||w| $$ $$ \langle v, A(x) v\rangle \ge \lambda |v|^2 $$

Consider the distance function from the origin $\rho(x) := dist(x, 0)$.

Is it possible to have an estimate of the kind:

$$ \Delta_A(\rho) \le \frac{K_1}{\rho} + K_2 \qquad \text{on } \Omega = \mathbb{R}^n \setminus\{0\} $$ where $K_1, K_2$ are constants possibly depending on $n$ and on $\lambda$ and $\Lambda$?

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  • $\begingroup$ Why the down vote? $\endgroup$
    – Onil90
    Commented May 10, 2017 at 14:17
  • $\begingroup$ Are you using $A$ as a metric? Or else why isn't $\rho(x)$ just $|x|$? $\endgroup$
    – Thompson
    Commented May 10, 2017 at 18:34
  • $\begingroup$ Yes, $\rho(x) = |x|$, probably my notation is a bit confusing.. I guess one could think $A$ as a metric, but then $\Delta_A$ does not coincide with the Laplace-Beltrami operator induced by that metric... $\endgroup$
    – Onil90
    Commented May 11, 2017 at 6:16
  • $\begingroup$ But then can't you just write out whatever $\Delta_A|x|$ is in terms of the coefficients $A_{ij}$ and their first derivatives? $\endgroup$
    – Thompson
    Commented May 11, 2017 at 13:32
  • $\begingroup$ @Thompson yeah, seems like I have somehow to assume some extra condition on the derivatives of the coefficients of the matrix.. I wanted to avoid that, but I guess there is no other way $\endgroup$
    – Onil90
    Commented May 11, 2017 at 14:35

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