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

With the convention $Lu\ge 0$ for a classical subsolution, the weak maximum principle for elliptic operators isHere $L$ is uniformly elliptic with bounded drift, has no zeroth-order term, and the domain is bounded. Equivalently, $Lu\le 0$ gives ${\inf }_{\Omega }u\ge {\min }_{\partial \Omega }u$.

One can see why measurable coefficients suffice for the pointwise classical statement by adding $\epsilon e^{\kappa x_1}$. Choose $\kappa$ large enough that $\lambda \kappa >\Vert b_1{\Vert }_{\infty }$; then $L(e^{\kappa x_1})=e^{\kappa x_1}({\kappa }^2{a}_{11}+\kappa b_1)>0$. The perturbed subsolution cannot attain a maximum inside: its gradient would be zero and its Hessian negative semidefinite, forcing its elliptic second-order term to be nonpositive. Its maximum is therefore on the boundary. Letting $\epsilon \downarrow 0$ proves the stated estimate. Only the symmetric part of $a$ acts on the Hessian, and ellipticity gives its positive definiteness.

Put $F={\sup }_{\Omega }{f}^{+}\ge 0$ and $m={\inf }_{\partial \Omega }u$, both finite by continuity on the compact closure. Set $z=Fv+m-u$. Constants are annihilated by $L$, soOn the boundary, $z\le m-u\le 0$, using $v\le 0$ and $F\ge 0$. The weak maximum principle for elliptic operators therefore gives $z\le 0$ inside as well. Rearrangement provesThe nonnegative multiplier is essential when multiplying the barrier inequality.

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.