= Solution
Assume $c\leq0$, the coefficients are bounded, and $u$ is bounded above. Choose $\kappa$ so large that the bounded positive function $q(y)=e^{\kappa y}$ satisfies
$$
Lq=(a^{nn}\kappa^2+y\kappa+c)q\geq m>0.
$$
Also set $\psi(x)=\log(1+|x|^2)$. Boundedness of the coefficients, $x\mathbin\cdot D\psi\leq2$, and $c\leq0$ give a global upper bound $L\psi\leq C$.
For $\varepsilon>0$ and $0<\delta<\varepsilon m/C$, the function
$$
v=u+\varepsilon q-\delta\psi
$$
tends to $-\infty$ as $|x|\to\infty$ and satisfies $Lv>0$. If $v$ exceeded both zero and its values on $y=\pm1$, it would attain a positive interior maximum. At that point $Dv=0$ and $D^2v\leq0$, whence $Lv\leq cv\leq0$, a contradiction. Letting $\delta\downarrow0$ and then $\varepsilon\downarrow0$ proves
$$
\sup_\Omega u\leq\max\{0,\sup_{\partial\Omega}u\}.
$$
This is the <weak maximum principle for elliptic operators> on the slab. The boundedness or a comparable growth condition is necessary because the domain is unbounded.
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