Let $\Omega$ be a bounded open domain and $v:\Omega\to\mathbb{R}^n$. Consider the following elliptic operator in divergence form, defined on smooth functions $u: \Omega \to \mathbb{R}$

\begin{align} L u=\operatorname{div}[\nabla u-v u] \end{align} with a reflecting boundary condition $$ \boldsymbol{n} \cdot(\nabla u-vu)=0 \text{ on } \partial \Omega. $$ What can we say about the spectrum of $L$? I have been searching for the answer for one week. Any relevant reference is greatly appreciated!

(In fact, I want to verify whether $\sigma(L) \subset \Sigma_{\omega}=\{\lambda \in \mathbb{C} ;|\arg \lambda|<\omega\},$ with some fixed angle $0<\omega<\frac{\pi}{2},$ and the resolvent satisfies the estimate $\left\|(\lambda-L)^{-1}\right\| \leq M /|\lambda|,\lambda \notin \Sigma_{\omega} $ with some constant $M \geq 1$. This is one of the assumptions for generating an evolution operator in nonautonomous parabolic equations.)

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    $\begingroup$ Do you just want the analytic semigroup property, or do you also want to guarantee that there are no eigenvalues with positive real part? In the latter case, you probably need some condition on the divergence of v. $\endgroup$ Mar 12, 2021 at 21:09
  • $\begingroup$ Dear @MichaelRenardy, thanks for your reply. I want both, but even the first case is very helpful. Do you know in what reference I can find related theorems? $\endgroup$
    – Mr_Rabbit
    Mar 22, 2021 at 13:30

1 Answer 1


To get the analytic semigroup estimate, we note first of all that the operator $M$: $Mu=Lu+(v\cdot\nabla)u$, with the same boundary condition as $L$, is self-adjoint and hence generates an analytic semigroup. We can then use results on perturbation of analytic semigroups and elliptic regularity to show that $L$ also generates an analytic semigroup. Now for the spectrum of $L$: $L$ has an obvious eigenvalue at $0$, since the range of $L$ is orthogonal to constants. Let $\exp(Lt)$ be the semigroup generated by $L$. Then $\exp(Lt)$ maps positive functions to positive functions (maximum principle), and it preserves the integral. These two facts together imply that the $L^1$ norm of $\exp(Lt)u$ is monotone decreasing. This is inconsistent with eigenvalues in the right half plane or nonzero eigenvalues on the imaginary axis.


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