I have this Dirichlet probem
\begin{align*} \Delta u - a^2u&=a^2\quad\text{on}\;\, \Omega \subset \mathbb{R}^n \\ u&\equiv 0\quad \, \text{ on }\partial \Omega, \end{align*} where $a^2$ is constant and $a^2>0$.
And the book says, "by the maximum principle implies $u>-1$"
I know this about the maximum principle: it says that the maximum of a function in a domain is to be found on the boundary of that domain.
But I dont understand why $u> -1$.
I try this.
$$ \dfrac{\Delta u }{a^2} - u = 1 \\ \Rightarrow u = \dfrac{\Delta u }{a^2} -1 ...$$
Following the hint by Ian, let $v=u+1$. The function $v$ satisfies the PDE $\Delta v=v$, and is positive on the boundary of $\Omega$. So, if $v$ was negative somewhere, its (negative) global minimum would be attained in the interior... but the Laplacian can't be negative at an interior minimum.