Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 13 / 4 / c / Solution

Apply the preceding result to $L=\Delta$, $u=w$ and $f=\Delta w$. Define the barrierThen $\Delta v=1$. Since $0\in \Omega$ and $d=\mathrm{diam}\Omega$, $|x|=|x-0|\le d$ on $\overline{\Omega }$, so $v\le 0$ on the boundary. All hypotheses of the barrier comparison hold, givingThe assumption that the origin belongs to the domain is what validates this particular choice of centered quadratic barrier.