Symmetry: $a^{i j}(x)=a^{j i}(x)$ for all $i, j$ and $x \in \Omega$ (this is no serious restriction).
Ellipticity: There exists a constant $\lambda>0$ with
$$
\lambda|\xi|^{2} \leq \sum_{i, j=1}^{d} a^{i j}(x) \xi^{i} \xi^{j} \quad \text { for all } x \in \Omega, \xi \in \mathbb{R}^{d}
$$
In particular, the matrix $\left(a^{y}(x)\right)_{i, j=1}, \ldots, d$ is positive definite for all $x$, and the smallest eigenvalue is greater than or equal to $\lambda$.
Boundedness of the coefficients: There exists a constant $K$ with
$$
\left|a^{i j}(x)\right|,\left|b^{i}(x)\right|,|c(x)| \leq K \text { for all } i, j \text { and } x \in \Omega
$$
MATH0090 COURSE NOTES ：
Since at an interior maximum $x_{0}$ of $u$, we must have
$$
u_{x^{i}}\left(x_{0}\right)=0 \quad \text { for } i=1, \ldots, d
$$
and
$$
\left(u_{x^{i} x^{j}}\left(x_{0}\right)\right){i, j=1, \ldots, d} \quad \text { negative semidefinite, } $$ and thus by the ellipticity condition also $$ L u\left(x{0}\right)=\sum_{i, j=1}^{d} a^{i j}\left(x_{0}\right) u_{x^{i} x^{j}}\left(x_{0}\right) \leq 0
$$
such an interior maximum cannot occur.
Returning to the general case $L u \geq 0$, we now consider the auxiliary function
$$
v(x)=e^{\alpha x^{1}}
$$
for $\alpha>0$. Then