Set
$$
A=\sup_{\partial\Omega}|u|,
\qquad
B=\sup_\Omega|f|,
\qquad
v(x)=A+B(e^{2d}-e^{x_1+d}).
$$
Then $v\geq A$ and, because $1\leq e^{x_1+d}\leq e^{2d}$ and $c\leq0$,
$$
(\Delta+c)v=-Be^{x_1+d}+cv\leq-B.
$$
Thus $(\Delta+c)(u-v)=f-(\Delta+c)v\geq0$ and $u-v\leq0$ on the boundary. The <weak maximum principle for elliptic operators> gives $u\leq v$. Applying the same argument to $-u$ gives $|u|\leq v$, hence
$$
\boxed{\sup_\Omega|u|
\leq\sup_{\partial\Omega}|u|
+(e^{2d}-1)\sup_\Omega|f|.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-107/1/a/i/solution