Maximal principle: proof for subsolution

Question

I have recently gone through the statement that a subsolution $v$ satisfies the maximal principle: $\sup_{\Omega T} =\sup_{\partial\Omega T}$. So if $u(x,t)>0$ is a supersolution, then how can we show that $v = -\ln u$ is a subsolution?

Answer 1

$u$ is a supersolution, i.e. $\Delta u\leq0$, then $\Delta v=-\Delta\ln u=-\frac{1}{u}\Delta u+\frac{|\nabla u|^2}{u^2}\geq0$, i.e. $v$ is a subsolution.