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

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