Positivity of principal eigenvalue for $L\phi=-\triangle \phi + \nabla \cdot( u \phi )$ EDIT: This question is still unresolved as of April 18, 2014. The two answers provide useful work in the right direction, but neither resolves the question. A counterexample should have $u \in C^1(\overline{\Omega})$.
I am reading this paper and in section 2 the authors explicitly state "We do not assume in this section that the flow $u(x)$ is incompressible." They immediately proceed to consider $\mu_1[u]$ and $\eta(x)$ the principal eigenvalue and normalized positive eigenfunction of the problem
$$
-\triangle \eta - \nabla \cdot (u \eta) = \mu_1[u] \eta
$$
in $\Omega$ and $\eta=0$ on $\partial \Omega$. Here $\Omega$ is any smooth bounded domain.
The authors then make the cryptic remark: "Note that $\mu_1[u]>0$ as the operator $-\triangle + u \cdot \nabla$ has no zero order term."
I see that $-\triangle + u \cdot \nabla$ is the adjoint of the operator in question. But how exactly does this tell me that $\mu_1[u]>0$? I know that a sufficient condition is that $\nabla \cdot u \leq 0$, (e.g. here), but the authors of the paper I am reading make no mention of any sort of divergence assumption. 
 A: The following counterexample is not fully satisfying ($u$ is unbounded, and $\mu$ is not strictly negative) but perhaps serves as a start:
Take $\Omega$ to be the unit disk and set $u(r,\theta) = \left(\frac{3r^2}{1-r^3}, 0\right)$. This vector field diverges as $r\to 1$ but is differentiable on the interior of $\Omega$. But for $\eta = 1-r^3$, we have that
$$-\Delta \eta - \nabla\cdot (u\eta) = 9r - \nabla\cdot (3r^2,0) = 9r - 9r = 0,$$
so $f$ is an eigenfunction of the differential operator with nonpositive eigenvalue $\mu=0$.
A: Let us try to obtain a counterexample. We assume that $u$ is sufficiently smooth.
Let us write the variational definition of the first eigenvalue:
$$
\mu_1[u] = \inf_{\eta \in W_0^{1,2}(\Omega)}\frac{\int_\Omega |\nabla \eta|^2 \,dx - \int_\Omega \nabla \cdot (u \eta) \, \eta \,dx}{\int_\Omega |\eta|^2 \,dx}, \quad \left(  \eta \not\equiv 0 \right).
\tag 1
$$
For the second integral in the numerator we have
$$
\int_\Omega \nabla \cdot (u \eta) \, \eta \,dx = 
\int_\Omega (\nabla \eta \cdot u) \, \eta \,dx + 
\int_\Omega \eta^2 (\nabla \cdot u) \,dx.
\tag 2
$$
Auxiliary, we note that integration-by-parts formula gives us
$$
\int_\Omega \eta_{x_i} u_i \, \eta \,dx = 
-\int_\Omega \eta \, u_i \, \eta_{x_i} \,dx - \int_\Omega \eta^2 (u_i)'_{x_i} \,dx,
$$
(here we used the fact that $\eta = 0$ on $\partial \Omega$). 
Therefore,
$$
\int_\Omega \eta_{x_i} u_i \, \eta \,dx = -\frac{1}{2}\int_\Omega \eta^2 (u_i)'_{x_i} \,dx.
$$
Using this equality, for the first integral on the rhs of $(2)$ we derive
$$
\int_\Omega (\nabla \eta \cdot u) \, \eta \,dx = \sum_{i=1}^n \int_\Omega \eta_{x_i} u_{i} \, \eta \,dx = -\frac{1}{2} \int_\Omega \eta^2 \sum_{i=1}^n(u_i)'_{x_i} \,dx = 
-\frac{1}{2} \int_\Omega \eta^2 (\nabla \cdot u) \,dx.
$$
Hence, for $(2)$ we have
$$
\int_\Omega \nabla \cdot (u \eta) \, \eta \,dx = \frac{1}{2} \int_\Omega \eta^2 (\nabla \cdot u) \,dx \geq \frac{1}{2} \min_{\Omega} (\nabla \cdot u) \int_\Omega \eta^2 \,dx = C \int_\Omega \eta^2 \,dx.
\tag 3
$$
Finally, for $(1)$ we obtain
$$
\mu_1[u] \leq \inf_{\eta \in W_0^{1,2}(\Omega)} \left( \frac{\int_\Omega |\nabla \eta|^2 \,dx - C \int_\Omega \eta^2 \,dx}{\int_\Omega |\eta|^2 \,dx} \right) = 
\inf_{\eta \in W_0^{1,2}(\Omega)} \left( \frac{\int_\Omega |\nabla \eta|^2 \,dx}{\int_\Omega |\eta|^2 \,dx} - C\right) = \lambda_1 - C \leq 0
$$
for any $C \geq \lambda_1$ (it is possible, due the fact that we have no a priori assumptions on $\nabla \cdot u$). Here $\lambda_1$ is the standard first eigenvalue of the Dirichlet Laplacian in $\Omega$. 
(Note also that the first inequality is true, since it is true for any $\eta \in W_0^{1,2}(\Omega)$).
Using the inverse inequality for $(3)$ we obtain the condition for the strict positivity of $\mu_1[u]$:  $\max_{\Omega} (\nabla \cdot u) < 2 \lambda_1$. (consequently, if $\nabla \cdot u \leq 0$, as you wrote.)
P.S. Actually, I can not figure out right now is the variational characterization correct. But in any case, if we assume that the first eigenfunction is "normal", then the construction above holds (without taking $\inf$).
A: Simple Special Case: $u=\nabla\phi$
We have
$$
\nabla (e^{\pm \phi/2}\psi)=e^{\pm\phi/2}\left[\nabla\psi\pm\frac{u}{2} \psi\right].
$$
In particular, $\left[\nabla+\frac{u}{2}\right]=e^{-\phi/2}\nabla e^{\phi/2}$ is injective, and we have
$$
\mu_1e^{\phi/2}\eta=e^{\phi/2}\left[-\nabla\right]\cdot\left[\nabla+u\right]\eta=\left[-\nabla+\frac{u}{2}\right]\cdot\left[\nabla+\frac{u}{2}\right]e^{\phi/2}\eta.
$$
It follows that
$$
\mu_1\lVert e^{\phi/2}\eta\rVert^2_{L^2}=\lVert \left[\nabla+\frac{u}{2}\right]e^{\phi/2}\eta \rVert^2_{L^2}.
$$
Since the operator $\left[\nabla+\frac{u}{2}\right]$ is injective, we find
$$
\mu_1=\frac{\lVert \left[\nabla+\frac{u}{2}\right]e^{\phi/2}\eta \rVert^2_{L^2}}{\lVert e^{\phi/2}\eta\rVert^2_{L^2}}>0.
$$
General case
Let $\eta^*>0,\mu^*_1>0$ be the principal eigenfunction and principal eigenvalue of $L^*=-\Delta+u\cdot \nabla$. Since $\eta^*>0$ and $\eta>0$, we have $\langle \eta^*,\eta\rangle>0$. But then
$$
\mu_1=\frac{\langle \eta^*,L\eta\rangle}{\langle \eta^*,\eta\rangle}=\frac{\langle L^*\eta^*,\eta\rangle}{\langle \eta^*,\eta\rangle}=\mu_1^*>0.
$$
A: Let $L$ denote the operator
$$L\eta:= -\Delta \eta - {\rm div}\big( u(x) \eta \big)$$
and consider the adjoint
$$L^t\xi = -\Delta \xi + u(x) \cdot \nabla \xi.$$
Consider now the eigenvalue problems
$$
(1)_{\lambda}
\left \{
\begin{array}{rccl}
L \eta &=& \lambda \eta & \text{ in } \Omega\\
\eta &=& 0 & \text{ on } \partial\Omega
\end{array}
\right.
\quad \text{and} \quad
(2)_{\lambda}\left \{
\begin{array}{rccl}
L^{t} \xi &=& \lambda \xi & \text{ in } \Omega\\
\xi &=& 0 & \text{ on } \partial\Omega.
\end{array}
\right.
$$
By the Fredholm alternative, we have that:
(A) $(1)_{\lambda}$ has a unique solution (which is $\eta \equiv 0$) if and only if $(2)_{\lambda}$ has only the trivial solution $\xi \equiv 0$.
(B) Moreover, $(1)_{\lambda}$ admits nontrivial solutions if and only if $(2)_{\lambda}$ admits nontrivial solutions, and the dimension of the solution space of each problem coincides.
Let $\Sigma_{L^{t}} \subseteq \mathbb{C}$ denote the set of eigenvalues of $L^{t}$ (that is, the set of $\lambda$'s such that $(2)_{\lambda}$ admits nontrivial solutions).
By the above, we have that $(1)_{\lambda}$ admits nontrivial solutions if and only if $\lambda \in \Sigma_{L^t}.$
From the elliptic theory we know that $L^{t}$ has a smallest real eigenvalue $\lambda_1$ such that the solution space of $(2)_{\lambda_1}$ is one dimensional and, moreover, is generated by a positive function
(for instance, see the book of Evans:Theorem 3, section 6.5, for details).
By the maximum principle, we even know that $\lambda_1 > 0.$
Then, by the Fredholm alternative,
for
$(1)_{\lambda}$
to admit solutions, we must necessarily have ${\rm Re} \lambda \geq \lambda_1,$
and we also know that $(1)_{\lambda_1}$ has a one-dimensional space of solutions.
It remains to show that this space of solutions may be generated by a positive function.
However, I do not know how to show this at the moment (in the self-adjoint case, one uses the variational characterization of the eigenvalue, and in the non self-adjoint case the proof in Evans uses the maximum principle, which is not available for $L$)
Edit: It seems that the positivity of the principal eigenfunction already follows from the Krein-Rutman Theory. I was aware that, for $L^{t}$ with a maximum principle, the theory implied the positivity of the first eigenfunction, but Theorem 1.1. here implies that it is also the case for its adjoint $L$.
