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.)



Spectrum of an elliptic operator in divergence form with a reflecting boundary condition
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.


