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

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.

• 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. Jul 4, 2013 at 14:11
• 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. Jul 4, 2013 at 14:20
• Thank you for correcting me. I have fixed it, however, if you want to edit the answer, feel free to do it. Jul 4, 2013 at 14:39