Is there an asymmetric positive definite second-order linear differential operator? The second-order differential operator is $Lu=-\sum_{i,j=1}^n (a^{ij}(x)u_{x_i})_{x_j} +\sum_{i=1}^n b^i(x) u_{x_i} +c(x) u$. We say it's positive definite if there exsits constant $\beta>0$ such that $\langle Lu,u\rangle =\int_{\Omega} \left( \sum_{i,j=1}^n a^{ij}(x) u_{x_i}u_{x_j} +\sum_{i=1}^n b^i(x) u_{x_i} u+c(x)u^2 \right) dx\geq \beta \| u\|_{H_0^1(\Omega )}^2$ where $\Omega$ is a bounded domain of $\mathbb{R}^n$
We say it's symmetric if $\langle Lu,v\rangle =\langle Lv,u\rangle$ which is obviously equivalent to $b^i=0,\forall i$.
My question is: if some $b^i\neq 0$, is there still chance for $L$ to be positive definite?
Remark: It's quite clear if $L$ is elliptic(namely, $\exists \theta>0,\text{ s.t. }\sum_{i,j=1}^n a^{ij}(x) \xi_i \xi_j \geq \theta \| \xi\|^2$ ), $b^i=0,c\geq 0$ will lead to $L$'s positive definiteness.
Remark: It's easy to find an asymmetric positive definite matrix:$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
 A: Suppose for example that $b^i\in L^\infty(\Omega)$ and choose $\alpha$ such that $\|b^i\|_\infty\leq\alpha$ for all $i=1,...,n$.
Note that \begin{eqnarray}
 \int_\Omega\sum_{i=1}^n b^i(x)u_{x_i}u &=& -\int_\Omega \sum_{i=1}^n |b^i(x)u_{x_i}u|      \nonumber \\
   &\geq &  -\alpha\sum_{i=1}^n\|u_{x_i}\|_2\|u\|_2\nonumber \\
   &\geq & \tag{1}-\alpha n\|\nabla u\|_2\|u\|_2\end{eqnarray}
Now we use Young inequality $ab\leq\frac{a^2}{2}+\frac{b^2}{2}$, $a,b\geq 0$ to get from $(1)$ 
$$\tag{2}\int_\Omega\sum_{i=1}^n b^i(x)u_{x_i}u\geq-\alpha n\left(\frac{\|\nabla u\|_2^2}{2}+\frac{\|u\|_2^2}{2}\right)$$
Now, suppose for example that $L$ is elliptic as in your definition and assume that $c(x)\geq -\beta$ where $\beta>0$. We get from $(2)$ that 
\begin{eqnarray}
 \langle Lu,u\rangle &\geq& \theta\|\nabla u\|_2^2 -\alpha n\left(\frac{\|\nabla u\|_2^2}{2}+\frac{\|u\|_2^2}{2}\right) -\beta\|u\|_2^2   \nonumber \\
   &=& \tag{3}\left(\theta-\frac{\alpha n}{2}\right)\|\nabla u\|_2^2-\left(\frac{\alpha n}{2}+\beta\right)\|u\|_2^2  \\
   &\geq& \tag{4} \left(\theta-\frac{\alpha n}{2}\right)\|\nabla u\|_2^2-\frac{1}{\lambda_1}\left(\frac{\alpha n}{2}+\beta\right)\|\nabla u\|_2^2 \\
 &=& \tag{5}\left(\theta-\frac{\alpha n}{2}\left(1+\frac{1}{\lambda_1}\right)-\frac{\beta}{\lambda_1}\right)\|\nabla u\|_2^2
\end{eqnarray}
From $(3)$ to $(4)$ I have use Poincare's inequality and $\lambda_1$ is the first eigenvalue associated with $(-\Delta,H_0^1(\Omega)$). From $(5)$ we see that it is possible to impose some conditions on $\theta,\alpha,\beta$ in such a way that $L$ is positive definite. 
Remark: This is just a example, possibly there are more general conditions.
