= Solution
The <weak maximum principle for elliptic operators> gives the clipped boundary bounds
$$
\boxed{\sup_\Omega u\leq\max\{0,\sup_{\partial\Omega}u\},\qquad\inf_\Omega u\geq\min\{0,\inf_{\partial\Omega}u\}.}
$$
If $c=0$, 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 $B=\|b_1\|_\infty$ and $C_0=\|c\|_\infty$, and choose $\alpha>0$ large enough that $\lambda\alpha^2-B\alpha-C_0>0$. For $\phi(x)=e^{\alpha x_1}$, <uniform ellipticity> gives
$$
L\phi=\phi(a_{11}\alpha^2+b_1\alpha+c)\geq\phi(\lambda\alpha^2-B\alpha-C_0)>0.
$$
Put $u_\varepsilon=u+\varepsilon\phi$. Then $Lu_\varepsilon>0$. If $u_\varepsilon$ 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 $a_{ij}D_{ij}u_\varepsilon\leq0$. Also $cu_\varepsilon\leq0$. These facts would give $Lu_\varepsilon\leq0$, a contradiction.
Since $\overline\Omega$ is compact and $u_\varepsilon$ is continuous on it, its maximum exists. It is therefore either nonpositive or attained on the boundary, proving $\sup_\Omega u_\varepsilon\leq\max\{0,\sup_{\partial\Omega}u_\varepsilon\}$. The exponential is bounded on the bounded domain; letting $\varepsilon\downarrow0$ 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.
Back to article page
