Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 / 1 / ii

The weak maximum principle for elliptic operators gives the clipped boundary boundsIf $c=0$, constants solve the equation, and applying these bounds after subtracting the boundary maximum or minimum gives the sharper conclusion that both extrema occur on the boundary. For $c<0$, the clipping by zero is necessary in the general statement.

To prove the upper bound, write $B=\Vert b_1{\Vert }_{\infty }$ and $C_0=\Vert c{\Vert }_{\infty }$, and choose $\alpha >0$ large enough that $\lambda {\alpha }^2-B\alpha -C_0>0$. For $\varphi (x)=e^{\alpha x_1}$, uniform ellipticity givesPut $u_{\epsilon }=u+\epsilon \varphi$. Then $L{u}_{\epsilon }>0$. If $u_{\epsilon }$ had a positive maximum at an interior point, its gradient would vanish there and its Hessian matrix would be negative semidefinite. Only the symmetric part of $a_{ij}$ contracts with this Hessian, and its positive definiteness implies $a_{ij}{D}_{ij}{u}_{\epsilon }\le 0$. Also $c{u}_{\epsilon }\le 0$. These facts would give $L{u}_{\epsilon }\le 0$, a contradiction.

Since $\overline{\Omega }$ is compact and $u_{\epsilon }$ is continuous on it, its maximum exists. It is therefore either nonpositive or attained on the boundary, proving ${\sup }_{\Omega }{u}_{\epsilon }\le \max \{0,{\sup }_{\partial \Omega }{u}_{\epsilon }\}$. The exponential is bounded on the bounded domain; letting $\epsilon \downarrow 0$ proves the upper estimate. Apply the same argument to $-u$ for the lower estimate. This is the exponential perturbation proof of the weak maximum principle with drift and requires no continuity of the coefficient functions.