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.

Solution 1:

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 $\mu=0$.

Solution 2:

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 $\inf$).