Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 107 / 1 / a

The weak maximum principle for elliptic operators says that if $c\le 0$ andthenIndeed, a positive interior maximum has $Du=0$ and $\Delta u\le 0$, so the differential inequality is incompatible with a strict positive maximum. Applying this argument to a standard strictly perturbed function and then letting the perturbation tend to zero handles equality and proves the weak statement.

SetThen $v\ge A$ and, because $1\le e^{x_1+d}\le e^{2d}$ and $c\le 0$,Thus $(\Delta +c)(u-v)=f-(\Delta +c)v\ge 0$ and $u-v\le 0$ on the boundary. The weak maximum principle for elliptic operators gives $u\le v$. Applying the same argument to $-u$ gives $|u|\le v$, hence

If $u_1,u_2$ are two solutions with the same boundary values, put $w=u_1-u_2$. The mean value theorem givesTherefore $\Delta w-qw=0$ with zero boundary data. Apply the weak maximum principle for elliptic operators to $w$ and $-w$ with zeroth-order coefficient $-q\le 0$. It follows that $w=0$, proving uniqueness.