On stable and finite Morse index solutions of the nonlocal Hénon-Gelfand–Liouville equation

Abstract

We consider the nonlocal Hénon-Gelfand–Liouville problem

$$\begin{aligned} (-\Delta )^s u = |x|^a e^u\quad \mathrm {in}\quad \mathbb {R}^n, \end{aligned}$$

for every $s\in (0,1)$, $a>0$ and $n>2s$. We prove a monotonicity formula for solutions of the above equation using rescaling arguments. We apply this formula together with blow-down analysis arguments and technical integral estimates to establish non-existence of finite Morse index solutions when

$$\begin{aligned} \dfrac{\Gamma (\frac{n}{2})\Gamma (s)}{\Gamma (\frac{n-2s}{2})}\left( s+\frac{a}{2}\right) > \dfrac{\Gamma ^2(\frac{n+2s}{4})}{\Gamma ^2(\frac{n-2s}{4})}. \end{aligned}$$

Appendix A

The purpose of this appendix is to provide necessary details for the proof of Proposition 2.5. First, we notice that the stability condition (1.3) can be extended to $\overline{u}$. More precisely, if u is stable in $\Omega$ then

$$\begin{aligned} \int _{\mathbb {R}_+^{n+1}}t^{1-2s}|\nabla \Phi |^2dxdt\ge \kappa _s\int _{\mathbb {R}^n}|x|^a e^u\varphi ^2dx, \end{aligned}$$

(A.1)

for every $\Phi \in C_c^\infty ({\overline{\mathbb {R}_+^{n+1}}})$ satisfying $\varphi (\cdot ):=\Phi (\cdot ,0)\in C_c^\infty (\Omega )$. Indeed, if $\overline{\varphi }$ is the s-harmonic extension of $\varphi$, we have

$$\begin{aligned} \int _{\mathbb {R}_+^{n+1}}t^{1-2s}|\nabla \Phi |^2dxdt&\ge \int _{\mathbb {R}_+^{n+1}}t^{1-2s}|\nabla {\overline{\varphi }} |^2dxdt\\ {}&=\kappa _s\int _{\mathbb {R}^n}\varphi (-\Delta )^s\varphi dx\ge \kappa _s\int _{\mathbb {R}^n}|x|^a e^u\varphi ^2dx. \end{aligned}$$

Before giving the conclusion of Moser’s iteration, we present the following lemma which will be used in the later context.

Lemma A.1

Let $e^{\alpha u}\in L^1( \Omega )$ for some $\Omega \subset \mathbb {R}^n$. Then $t^{1-2s}e^{\alpha \overline{u}}\in L^1_{\mathrm {loc}}(\Omega \times [0, \infty ))$.

Proof

Let $\Omega _0\Subset \Omega$ be fixed. Since $u\in L_s(\mathbb {R}^n)$, we have for $x\in \Omega _0$ and $t\in (0,R)$

$$\begin{aligned} \overline{u}(x,t)\le C+\int _{\Omega }u(y)P(X,y)dy=C+\int _{\Omega }g(x,t)u(y)\frac{P(X,y)dy}{g(x,t)}, \end{aligned}$$

where $C \le g(x,t):=\int _{\Omega }P(X,y)dy\le 1$ for some positive constant C depending solely on R, $\Omega _0$ and $\Omega$. Therefore, by the Jensen’s inequality

$$\begin{aligned}&\int _{\Omega _0}e^{\alpha \overline{u}(x,t)}dx \le C \int _{\Omega _0}\int _{\Omega }e^{\alpha g(x,t)u(y)}P(X,y)dydx\\&\quad \le C\int _{\Omega }\max \{e^{\alpha u(y)},1\}\int _{\Omega _0}P(X,y)dxdy \le C+C\int _{\Omega }e^{\alpha u(y)}dy, \end{aligned}$$

where the constant C depends on $R,~\Omega _0$ and $\Omega$, but not on t. Hence,

$$\begin{aligned} \int _{\Omega _0\times (0,R)} t^{1-2s}e^{\alpha \overline{u}(x,t)}dxdt \le \int _0^R t^{1-2s}\int _{\Omega _0}e^{\alpha \overline{u}(x,t)}dxdt <\infty . \end{aligned}$$

This finishes the proof. $\square$

The following lemma is the main conclusion of this appendix and it is essential in the proof of Proposition 2.5.

Lemma A.2

Let $u\in L_s(\mathbb {R}^n)\cap \dot{H}^s_{\mathrm {loc}}(\mathbb {R}^n)$ be a solution to (1.1). Assume that u is stable in $\Omega \subseteq \mathbb {R}^n$. Let $\Phi \in C_c^\infty ({{\overline{\mathbb {R}_+^{n+1}}}})$ be of the form $\Phi (x,t) =\varphi (x)\eta (t)$ for some $\varphi \in C_c^\infty (\Omega )$ and $\eta \equiv 1$ on [0, 1]. Then for every $0<\alpha <2$ we have

$$\begin{aligned} \begin{aligned} (2-\alpha )\kappa _s\int _{\mathbb {R}^n}|x|^a e^{(1+2\alpha ) u}\varphi ^2dx&\le 2\int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha {\bar{u}}}|\nabla \Phi |^2dxdt\\&\quad -\frac{1}{2}\int _{\mathbb {R}_+^{n+1}}e^{2\alpha \bar{u}}\nabla \cdot [t^{1-2s}\nabla \Phi ^2]dxdt . \end{aligned} \end{aligned}$$

Proof

For $k\in \mathbb {N}$ we set $\overline{u}_k:=\min \{\overline{u},k\}$. Let $u_k$ be the restriction of $\overline{u}_k$ on $\mathbb {R}^n$. It is straightforward to see that $e^{2\alpha \overline{u}_k}\Phi ^2$ can be considered as a test function in (1.8). Therefore,

$$\begin{aligned} \begin{aligned}&\kappa _s\int _{\mathbb {R}^n}|x|^a e^u e^{2\alpha u_k}\varphi ^2dx\\&\quad =2\alpha \int _{\mathbb {R}_+^{n+1}}t^{1-2s} e^{2\alpha \overline{u}_k}\Phi ^2\nabla \overline{u}\cdot \nabla \overline{u}_kdxdt +\int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}\nabla \overline{u}\cdot \nabla \Phi ^2dxdt\\&\quad =2\alpha \int _{\mathbb {R}_+^{n+1}}t^{1-2s} e^{2\alpha \overline{u}_k}\Phi ^2\ | \nabla \overline{u}_k|^2dxdt +\int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}\nabla \overline{u}\cdot \nabla \Phi ^2dxdt. \end{aligned} \end{aligned}$$

(A.2)

Now, we assume that $t^{1-2s}e^{2(\alpha +\varepsilon )\overline{u}}\in L^1_{\mathrm {loc}}(\Omega \times [0,\infty ))$ for some $\varepsilon >0$. Then by [28, Lemma 3.5], up to a subsequence, we have

$$\begin{aligned} \kappa _s\int _{\mathbb {R}^n}|x|^a e^u e^{2\alpha u_k}\varphi ^2dx=(2\alpha +o(1))\int _{\mathbb {R}_+^{n+1}}t^{1-2s} e^{2\alpha \overline{u}_k}\Phi ^2\ | \nabla \overline{u}_k|^2dxdt+O(1). \end{aligned}$$

Taking $e^{\alpha \overline{u}_k} \Phi$ as a test function in the stability inequality (A.1) yields

$$\begin{aligned} \begin{aligned}&\kappa _s \int _{\mathbb {R}^n}|x|^a e^ue^{2\alpha u_k}\varphi ^2dx\\&\quad \le \alpha ^2 \int _{\mathbb {R}_+^{n+1}}t^{1-2s}\Phi ^2e^{2\alpha \overline{u}_k}|\nabla \overline{u}_k|^2dxdt + \int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}|\nabla \Phi |^2dxdt\\&\qquad +\frac{1}{2} \int _{\mathbb {R}_+^{n+1}}t^{1-2s} \nabla e^{2\alpha \overline{u}_k}\nabla \Phi ^2dxdt\\&\quad = \alpha ^2 \int _{\mathbb {R}_+^{n+1}}t^{1-2s}\Phi ^2e^{2\alpha \overline{u}_k}|\nabla \overline{u}_k|^2dxdt + \int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}|\nabla \Phi |^2dxdt \\&\qquad -\frac{1}{2} \int _{\mathbb {R}_+^{n+1}}e^{2\alpha \overline{u}_k} \nabla \cdot [t^{1-2s}\nabla \Phi ^2]dxdt, \end{aligned} \end{aligned}$$

(A.3)

where the last equality follows from integration by parts. Notice that the boundary term vanishes as $\eta (t)=1$ on [0, 1]. Combining the prior estimates, we obtain

$$\begin{aligned} \begin{aligned}&(2-\alpha -\varepsilon )\kappa _s\int _{\mathbb {R}^n}|x|^a e^ue^{2\alpha u_k}\varphi ^2dx \\&\quad \le (-2\alpha \varepsilon +o(1))\int _{\mathbb {R}_+^{n+1}}t^{1-2s} e^{2\alpha \overline{u}_k}\Phi ^2|\nabla \overline{u}_k|^2dxdt+O(1) \\&\quad \quad +2\int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}|\nabla \Phi |^2dx -\int _{\mathbb {R}_+^{n+1}}e^{2\alpha \overline{u}_k}\nabla \cdot [t^{1-2s}\nabla \Phi ^2]dxdt. \end{aligned} \end{aligned}$$

(A.4)

Concerning the last term in (A.4), one can notice that

$$\begin{aligned} \nabla \cdot [t^{1-2s}\nabla \Phi ^2]=t^{1-2s}\eta ^2\Delta _x\varphi ^2+\varphi ^2\partial _t(t^{1-2s}\partial _t\eta ^2). \end{aligned}$$

Again, as $\eta =1$ on [0, 1], the second term in the right-hand side of the above expression is identically zero for $0\le t\le 1$. Therefore, Lemma A.1 yields

$$\begin{aligned} \left| \int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}_k}|\nabla \Phi |^2dxdt\right| +\left| \int _{\mathbb {R}_+^{n+1}}e^{2\alpha \overline{u}_k}\nabla \cdot [t^{1-2s}\nabla \Phi ^2]dxdt\right| \le C. \end{aligned}$$

Thus,

$$\begin{aligned} (2-\alpha -\varepsilon )\int _{\mathbb {R}^n}e^ue^{2\alpha u_k}\varphi ^2dx\le C, \end{aligned}$$

provided $\int _{\Omega }e^{2(\alpha +\varepsilon )u}dx<\infty .$ Now, choosing $\alpha \in (0,\frac{1}{2})$ and $0<\varepsilon <\frac{1}{2}-\alpha$ in the above estimate and then taking $k\rightarrow \infty$ we conclude $e^{(1+2\alpha )u}\in L^1_{\mathrm {loc}}(\Omega )$. By an iteration argument we conclude that $e^{(1+2\alpha )u}\in L^1_{\mathrm {loc}}(\Omega )$ for every $\alpha \in (0,2)$.

Now, send $k\rightarrow \infty$ in (A.4) to get

$$\begin{aligned} \int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}}\Phi ^2|\nabla \overline{u}|^2dxdt<\infty . \end{aligned}$$

And, take limit in (A.2) and (A.3) as $k\rightarrow \infty$. Then, the proof of lemma follows immediately since the second term on the right-hand side of (A.2), when $k\rightarrow \infty$, can be re-written as

$$\begin{aligned} \int _{\mathbb {R}_+^{n+1}}t^{1-2s}e^{2\alpha \overline{u}}\nabla \overline{u}\cdot \nabla \Phi ^2dxdt&=\frac{1}{2\alpha } \int _{\mathbb {R}_+^{n+1}}t^{1-2s}\nabla e^{2\alpha \overline{u}}\cdot \nabla \Phi ^2dxdt\\&=-\frac{1}{2\alpha } \int _{\mathbb {R}_+^{n+1}} e^{2\alpha \overline{u}}\nabla \cdot [t^{1-2s}\nabla \Phi ^2]dxdt. \end{aligned}$$

Note that the boundary integral is zero as $\eta =1$ on [0, 1]. This completes the proof. $\square$