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Let $A$ be a constant symmetric matrix with $\lambda < A < \Lambda$ and $0<\lambda < \Lambda$ are fixed constants. Let $u$ be a solution of $\text{div}A \nabla u = 0$. Is it true that $\Delta |A\nabla u|^2 \geq 0$? When $A$ is the identity matrix, this is indeed true, since $|\nabla u|^2$ is subharmonic when $u$ is harmonic.

More generally, I am looking to see if the following holds: $$ \Delta |A \nabla u|^2 - 2 \sum_i |A \nabla u_{x_i}|^2 \geq 0$$

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  • $\begingroup$ You do not specify that $\lambda$ is. It is positive? $\endgroup$ Commented Dec 13, 2024 at 14:57
  • $\begingroup$ Ah yes, sorry. Both $\Lambda > \lambda>0$ are two fixes constants. $\endgroup$ Commented Dec 13, 2024 at 15:14

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