It is well known that in general the Laplace operator is a dissipative operator, i.e. if you call $A$ its realization on $D(A)=H^2(\Omega) \cap H_0^1(\Omega)$ where $\Omega \subset \mathbb{R}^n$ is a bounded domain with regular boundary then we have:
$$\langle Ax,x \rangle\leq 0$$
for every $x \in D(A)$.
Now consider a general second order elliptic operator of the form
$$A=\sum_{i,j} \partial_i (a_{ij}\partial_j)+ \sum_{i} b_i \partial_i + c$$
defined always on $D(A)=H^2(\Omega) \cap H_0^1(\Omega)$.
Assume that $a_{ij},b_i,c$ are bounded functions on $\Omega$ and the operator is uniformly elliptic, i.e.
$$\sum_{ij}a_{ij}\xi_i \xi_j \geq \theta|\xi|^2$$ for every $\xi \in \Omega$.
Note that if $b_i=c=0$ than $A$ is dissipative.
My question is: when $b,c$ are present is the operator $A$ operator dissipative in general or are there some conditions under which $A$ is dissipative? can you provide references ?
Best Answer
As mentioned in MaoWao's comment, $\mathcal A$ is not always dissipative. But you can always find an explicit constant $\lambda=\lambda(\|B\|_\infty, \|c\|_\infty, \theta)$ so that $\mathcal A-\lambda$ is dissipative.
Edit : Indeed, let $\mathcal A u=\mathrm{div}(A\nabla u)+ B\cdot \nabla u+ cu$. We have by Green's formula
\begin{align*} \langle \mathcal A u, u\rangle &=-\int_\Omega A\nabla u\cdot \nabla u\,dx + \int_\Omega B\cdot\nabla u u\, dx + \int_\Omega c u^2\, dx\\ & \le - \theta \|\nabla u\|^2 + \left(\frac{\theta}{2} \|\nabla u\|^2 + \frac{\|B\|_\infty^2}{2\theta} \|u\|^2\right) + \|c\|_\infty \|u\|^2. \end{align*} So, you can take $\lambda=\frac{\|B\|_\infty^2}{2\theta}+ \|c\|_\infty$.