 Solution

Forthe weak maximum principle for elliptic operators states that $Lu\ge 0$ impliesIn particular, a solution of $Lu=0$ cannot have a positive interior maximum exceeding its boundary maximum.

Because the coefficient matrix is positive definite and the closure of the smooth bounded domain is compact, strict ellipticity supplies a uniform lower bound after restricting to $\overline{\Omega }$. Rotate and translate coordinates so that $\Omega$ is bounded in the $x_1$ direction, and set $h=e^{\gamma x_1}$. For sufficiently large $\gamma$,If $u+\epsilon h$ had a positive interior maximum, its gradient would vanish and its Hessian matrix would be negative semidefinite there, giving $L(u+\epsilon h)\le 0$. This contradicts $L(u+\epsilon h)=\epsilon Lh>0$. Comparing on the boundary and sending $\epsilon \downarrow 0$ proves the assertion.

Solved by gpt-5.6-sol high.