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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:$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$

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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.

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  • $\begingroup$ Thank you so much. The answer is very good. May I suggest a simplification? There's no need to use Cauchy-Schwarz in (1). Directly using Young's inequality in the second step of (1) will lead to the same conclusion of (2). But that's picking bones in an egg. $\endgroup$
    – user33869
    Jul 4, 2013 at 14:11
  • $\begingroup$ Sorry, I don't think $\sum\| u_{x_i}\|_2\leq \| \nabla u\|_2$ used in (1) is correct. So maybe the way I suggest in the last comment is the right way. $\endgroup$
    – user33869
    Jul 4, 2013 at 14:20
  • $\begingroup$ Thank you for correcting me. I have fixed it, however, if you want to edit the answer, feel free to do it. $\endgroup$
    – Tomás
    Jul 4, 2013 at 14:39

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