 Solution

TakeIts principal symbol is ${\xi }^4$, so it is elliptic, and $a_4{a}_0=-1<0$. Yet $u(x)=\sin x$ satisfies $Lu=0$ on $(0,\pi )$ and vanishes at both boundary points while remaining positive inside. Thus second-order ellipticity is essential to the usual weak maximum principle for elliptic operators.