Strong maximum principle for a PDE with coefficient in $L^1$

Question

Let $U$ be a bounded open subset of $\mathbb{R}^n$ with smooth boundary, $n \geq 3$. Set $N = \frac{2n}{n-2}$. I am interested in the following equation: $$ -\Delta \phi + R \phi + \phi^{N-1} = 0 $$ with boundary condition $\phi \equiv 1$ on $\partial U$. I also assume that $R \in L^1(U)$, $R \geq -1$.

It turns out that we can construct $\phi \in W^{1,2}(U) \cap L^\infty(U)$ hwich is a weak solution to this equation as follows.

First a naive remark. If $R$ was regular, we see that $1$ is a supersolution while $0$ is a (sub) solution so we expect $0 \leq \phi \leq 1$.

We introduce the functional $$ I(\phi) = \int_U \left[\frac{1}{2}\left(|d\phi|^2 + R \phi^2\right) + \frac{1}{N} |\phi|^N\right] dx. $$ This functional is ill-defined over $W^{1,2}(U)$ but well-defined over the closed subset $C$ consisting of functions that are between $-2$ and $2$ ($C$ is closed for the weak topology as it can be described as the set of functions $\phi$ such that $-2 \mu(A) \leq \int_A \phi dx \leq 2 \mu(A)$ for all measurable $A \subset U$).

By classical arguments $I$ is sequentially lower semicontinuous on $C$ and, hence, admits a minimum $\phi_0$. As $I(\max\{\phi_0, 0\}) \leq I(\phi_0)$ and $I(\min\{\phi_0, 1\}) \leq I(\phi_0)$, we can assume that $0 \leq \phi_0 \leq 1$.

As we have allowed ourselves some wiggle room in the definition of $C$, we have, for each $\psi \in W^{1, 2}_0(U) \cap L^\infty(U)$, that $\phi_0 + \lambda \psi \in C$ for small enough $\lambda$ and, hence, "differentiating" the inequality $I(\phi_0 + \lambda \psi) \geq I(\phi_0)$, we get that, for all such $\psi$, $$ 0 = \int_U \left(\langle d\phi_0, d\psi\rangle + R \phi_0 \psi + \phi^{N-1} \psi\right) dx, $$ i.e. $\phi_0$ is a weak solution to our problem.

Is $\phi_0$ bounded from below by a positive constant?

With a more regular $R$, that would follow from Harnack's inequality but I could not find a proof with regularity as low as what I presented.

For the interested reader, the context in which this problem appears is the study of the mass of asymptotically hyperbolic manifolds (see e.g. https://arxiv.org/abs/math/0110035).

Answer 1

I think that the answer is no, you need more regularity to insure that $\phi_0$ is bounded from below by a positive constant (e.g. $R\in L^p$ with $p>n/2$).

We look on the radial function $\phi_0(x)=\|x||^\alpha$ on the unit Euclidean ball $U=\mathbb{B}$. Then $$-\Delta \phi_0+\phi_0^{N-1}=\left(-\frac{\alpha(\alpha+n-2)}{\| x\|^2}+\| x\|^{\frac{4\alpha}{n-2}}\right)\phi_0.$$ Letting $$R(x)=\frac{\alpha(\alpha+n-2)}{\| x\|^2}-\| x\|^{\frac{4\alpha}{n-2}},$$ we get $$-\Delta \phi_0+R\phi_0+\phi_0^{N-1}=0.$$ Clearly in dimension $n\ge 3$, $R\in L^1(\mathbb{B})$. And $V$ is bounded from below and if $\alpha(\alpha+n-2)\ge 0$ then $V(x)\ge \alpha(\alpha+n-2)-1\ge -1$ on the unit ball $\mathbb{B}$.