= Solution
The operator is <strictly elliptic operator>[strictly elliptic] when its symmetric principal matrix $A(x)=(a^{ij}(x))$ is positive definite at every point:
$$
a^{ij}(x)\xi_i\xi_j>0
\qquad(\xi\ne0).
$$
It is a <uniformly elliptic operator> when some $\lambda>0$ satisfies
$$
a^{ij}(x)\xi_i\xi_j\geq\lambda|\xi|^2
$$
for every $x$ and $\xi$. Here the least eigenvalue of the continuous matrix $A(x)$ is a positive continuous function on the compact set $\overline B_1$. It therefore has a positive minimum, which supplies $\lambda$ and proves uniform ellipticity.
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