Given the PDE, $x_t - \triangle x + c(m,t)x = 0$, on a bounded domain $\omega$ where $c(m,t)$ is a smooth function (bounded but can be negative). Show that if $x_{0}(m) = x(m,0)\geq 0$, then $x(m,t)\geq 0$ for every $x\in \omega$ and $t > 0$.
My attempt: We consider the transformation $v(m,t)=e^{\lambda\ t} x(m,t)$, so after doing some algebraic manipulation, we can rewrite the PDE:
$v_t - \triangle v + c(m,t)v = e^{\lambda t}[x_t(1-2\lambda) - \triangle x + x(-\lambda^{2} + c(m,t)] = 0$
$\Leftrightarrow x_t - \frac{\triangle x}{1-2\lambda} + x(\frac{-\lambda^{2} + \lambda + c(m,t)}{1-2\lambda}) = 0$ (assume $\lambda \neq \frac{1}{2}$)
Now, the main idea is to choose $\lambda$ such that the term $\frac{-\lambda^{2} + \lambda + c(m,t)}{1-2\lambda}\geq 0$. Since $c(m,t)$ is bounded, such $\lambda$ exists (I examined each case to find the value of $\lambda$ based on whether $\min\ (\max\ c(m,t), 1) = \max\ c(m,t)$ or $1$). Thus, by applying the weak maximum principle for the problem $x_t - \triangle x + c(m,t) x = 0$ with $c(m,t)\geq 0$, we get: $\min{x}\geq \max\ (x_{0}^{-})$ where $x_0^{-}$ denotes the negative parts of $x_0$. Since $x_0\geq 0$ by the given assumption, $\min\ (x)\geq \max\ (x_0^{-}) \geq 0$. Thus $x(m,t)\geq 0$ for every $x\in \omega$ and $t > 0$ (Q.E.D)
Concern about the solution above: In the case above, the coefficient of $\triangle x$ is not $-1$, but $\frac{-1}{1-2\lambda}$, but I think the conclusion is the same? Someone please corrects me if I'm wrong on this, as the final conclusion would be incorrect if this premise is not true.