\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 45, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/45\hfil Asymptotic behavior of singular solutions] {Asymptotic behavior of singular solutions to semilinear fractional elliptic equations} \author[G. Lin, X. Zheng \hfil EJDE-2014/45\hfilneg] {Guowei Lin, Xiongjun Zheng} % in alphabetical order \address{Guowei Lin \newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{lgw2008@sina.cn} \address{Xiongjun Zheng \newline Department of Mathematics, Jiangxi Normal University, Nanchang, Jiangxi 330022, China} \email{xjzh1985@126.com} \thanks{Submitted September 24, 2013. Published February 10, 2014.} \subjclass[2000]{ 35R11, 35B06, 35B40} \keywords{Fractional Laplacian; radial symmetry; asymptotic behavior; \hfill\break\indent singular solution} \begin{abstract} In this article we study the asymptotic behavior of positive singular solutions to the equation $$ (-\Delta)^{\alpha} u+u^p=0\quad\text{in } \Omega\setminus\{0\}, $$ subject to the conditions $u=0$ in $\Omega^c$ and $\lim_{x\to0}u(x)=\infty$, where $p\geq1$, $\Omega$ is an open bounded regular domain in $\mathbb{R}^N$ ($N\ge2$) containing the origin, and $(-\Delta)^\alpha$ with $\alpha\in(0,1)$ denotes the fractional Laplacian. We show that the asymptotic behavior of positive singular solutions is controlled by a radially symmetric solution with $\Omega$ being a ball. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \newcommand{\norm}[1]{\|#1\|} \section{Introduction} In this article, we study the positive singular solutions to semilinear elliptic equations involving the fractional Laplacian \begin{equation}\label{e1.1} \begin{gathered} (-\Delta)^{\alpha} u+u^p=0\quad \text{in } \Omega\setminus\{0\},\\ u=0\quad \text{in } \Omega^c,\\ \lim_{x\to0}u(x)=\infty, \end{gathered} \end{equation} where $\Omega$ is an open bounded $C^2$ domain in $\mathbb{R}^N$ $(N\ge2)$ with $0\in \Omega$ and $p\geq1$. The operator $(-\Delta)^\alpha$ with $\alpha\in(0,1)$ is the fractional Laplacian defined as \begin{equation}\label{DELTA} (-\Delta)^\alpha u(x)=P.V.\int_{\mathbb{R}^N}\frac{u(x)-u(y)}{|x-y|^{N+2\alpha}}dy. \end{equation} Here $P.V$. denotes the principal value of the integral, and for notational simplicity we omit it in what follows. In recent years, nonlinear elliptic equations involving general integro-differential operators, especially, fractional Laplacian, have been studied by many authors. Various regularity issues for fractional elliptic equations have been studied, see for instance \cite{CS0,CS1,RS,S}. The existence of solutions to semilinear equations involving fractional Laplacian has been investigated by \cite{FQT,SV} and others by using variational methods. When $\alpha=1$ and $p\in(1,\frac N{N-2})$, it was made in \cite{V0} the description of the all possible singular behavior of positive solutions to equation \eqref{e1.1} with $u=0$ on $\partial\Omega$. In particular, there are only two types of singular behavior occur: \\ (i) either $u(x)\sim c_Nk|x|^{2-N}$ when $x\to 0$ and $k$ can take any positive value ($u$ is said to have a {\it weak singularity} at $0$); \\ (ii) or $u(x)\sim c_{N,p}|x|^{-\frac2{p-1}}$ when $x\to 0$ ($u$ is said to have a {\it strong singularity} at $0$). \smallskip For $\alpha\in(0,1)$, it was shown in \cite{CV2} the existence of singular solutions for \eqref{e1.1} by using Perron's method. It was studied in \cite{CV3} the existence of weak singular solutions and asymptotic behavior of solutions, and the that regularity of weak singular solutions was improved into classical solutions in \cite{CV4}. In particular, for $p\in(1+\frac{2\alpha}N,\frac{N}{N-2\alpha})$, there exist a solution $u_p$ of problem \eqref{e1.1} and some positive constant $c_1$ such that $$ \lim_{x\to0} u_p(x)|x|^{\frac{2\alpha}{p-1}}=c_1. $$ For $p\in(0,\frac{N}{N-2\alpha})$, there exists a solution $u_t$ of problem \eqref{e1.1} for each $t>0$ such that $$ \lim_{x\to0} u_t(x)|x|^{N-2\alpha}=t. $$ Moreover, if $p\in(0,1+\frac{2\alpha}N)$, the solutions $\{u_t\}$ blow up everywhere in $\Omega$ as $t\to\infty$. If $p\in(1+\frac{2\alpha}N,\frac{N}{N-2\alpha})$, the limit of $\{u_t\}$ as $t\to\infty$ is a strongly singular solution of \eqref{e1.1}, which coincides with $u_p$ for $p\in(\max\{\frac{2\alpha}{N-2\alpha},1+\frac{2\alpha}N\},\frac{N}{N-2\alpha})$. However, it does not make all the classification of the singularities of \eqref{e1.1} in \cite{CV2,CV4}. Our purpose in this paper is to describe the asymptotic behavior of positive singular solutions of \eqref{e1.1}, more precisely, any positive solution of \eqref{e1.1} with the general domain $\Omega$ is controlled by a radially symmetric solution of \eqref{e1.1} with $\Omega$ being a ball. Moreover, we show that any positive solution of \eqref{e1.1} is radially symmetric when $\Omega$ is a ball. Before stating the main theorem we make precise the notion of solution that we use in this article. We say that a continuous function $u:\mathbb{R}^N\setminus\{0\}\to\mathbb{R}$ is a classical solution of equation \eqref{e1.1} if the fractional Laplacian of $u$ is defined at any point of $\Omega\setminus\{0\}$, according to the definition given in \eqref{DELTA}, and if $u$ satisfies the equation and the external condition in a pointwise sense. In this article, we deal with the singular solution only in the classical sense, since the viscosity solution of \eqref{e1.1} (see Definition \ref{def2.1}) could be improved into classical sense by regularity results \cite[Lemma 3.1]{CV4}, \cite[Theorem 2.1]{CFQ} and even the solution in the weak sense (see \cite[Definition 1.1]{CV4}) has been improved into classical sense by \cite[Theorem 3.1]{CV4}. Now we are ready to state our main result. It will be convenient for our description to define \begin{equation}\label{r0} r_0=\max\{r>0: B_r(0)\subset\Omega\}. \end{equation} \begin{theorem}\label{thm1.1} Assume that $\alpha\in(0,1)$, $p\geq1$ and $u$ is a positive classical solution of \eqref{e1.1}. Then there exists a radially symmetric solution $u_s$ of \eqref{e1.1} with $\Omega=B_{r_0}(0)$ such that $$ u_s\le u\le u_s+c_2\quad \text{in }\quad \Omega\setminus\{0\}, $$ where $c_2=\sup_{x\in \mathbb{R}^N\setminus B_{r_0}(0)}u(x)$. \end{theorem} We remark if the domain $\Omega$ is a ball centered at$0$, then any positive solution $u$ of \eqref{e1.1} is radially symmetric. In fact, when $\Omega=B_R(0)$ with $R>0$, by the definition of $r_0$, it is obvious that $r_0=R$. Using Theorem \ref{thm1.1} and the fact of $c_2=0$, there exists a radially symmetric solution $u_s$ of \eqref{e1.1} such that $u=u_s$ in $\mathcal{B}_R:=B_R(0)\setminus\{0\}$; therefore, $u$ is radially symmetric. In precise statement, we have following corollary. \begin{corollary} \label{coro1.1} Assume that $\alpha\in(0,1)$, $p\geq1$ and $u$ is a positive classical solution of \eqref{e1.1} with $\Omega=B_{r_0}(0)$. Then $u$ is radially symmetric. \end{corollary} To prove Theorem \ref{thm1.1}, we first show that problem \eqref{e1.1} with $\Omega=B_{r_0}(0)$ admits a positive singular solution $u_s$ by Perron's method. This will be done by constructing super and sub-solutions by using the solution $u$. Next, we prove that $u_s$ is radially symmetric by the classical moving planes method which is developed in \cite{BLW,CLO2,FW,L} to obtain the symmetry results for the fractional semilinear problem. In this paper, we extend this method to obtain the symmetry property of isolated singular solutions to \begin{equation}\label{e1.00} (-\Delta)^{\alpha} u=h(u)+\eta\ \ {\rm in} \ \mathcal{B}_1,\quad u=0 \text{ in }B_1^c(0), \end{equation} where $\eta:\mathcal{B}_1\to\mathbb{R}$ is radially symmetric and decreasing function. We note that the singularity at the origin gives rise to difficulties in the procedure of moving planes. The paper is organized as follows. In section 2, we obtain the existence of solutions by Perron's Method. Section 3 is devoted to obtain symmetry property for general nonlinearity. \section{Existence of solutions} In this section, we show that problem \eqref{e1.1} with $\Omega=B_{r_0}(0)$ admits a positive singular solution $u_s$ under the hypotheses of Theorem \ref{thm1.1}. This result will be shown by the barrier method. To this purpose, we need some auxiliary lemmas. \begin{lemma}\label{thm2.1}\cite[Theorem 2.5]{CFQ} Let $p>0$ and ${\mathcal{O}}$ be an open bounded $C^2$ domain in $\mathbb{R}^N$. Suppose that $g:{\mathcal{O}^c} \to \mathbb{R}$ is in $L^1({\mathcal{O}}^c, \frac{dx}{1+|x|^{N+2\alpha}})$ and it is of class $C^2$ in $\{z\in{\mathcal{O}}^c: \operatorname{dist}(z,\partial{\mathcal{O}}) \leq \delta \}$ for some $\delta>0$ and $f:\bar{\mathcal{O}}\to \mathbb{R}$ is continuous, $f\in C^\beta_{\rm loc}({\mathcal{O}})$ with $\beta\in(0,1)$. Then there exists a classical solution $u$ of \begin{equation}\label{3.2.4} \begin{gathered} (-\Delta)^{\alpha} u(x)+|u|^{p-1}u(x)=f(x),\quad x\in{\mathcal{O}},\\ u(x)=g(x),\quad x\in{\mathcal{O}}^c, \end{gathered} \end{equation} which is continuous in $\bar{{\mathcal{O}}}$. \end{lemma} Next, we introduce the comparison principle. \begin{lemma}\label{comparison}\cite[Theorem 2.3]{CFQ} Let $u$ and $v$ be classical super-solution and sub-solution of $$ (-\Delta)^{\alpha} u+h(u)=\eta\quad \text{in } {\mathcal{O}}, $$ respectively, where ${\mathcal{O}}$ is an open, bounded domain, the function $\eta:{\mathcal{O}}\to\mathbb{R}$ is continuous and $h:\mathbb{R}\to\mathbb{R}$ is increasing. Suppose further that $u$ and $v$ are continuous in $\bar {\mathcal{O}}$ and $v(x)\le u(x)$ for all $x\in{\mathcal{O}}^c$. Then $u(x)\ge v(x)$, $x\in {\mathcal{O}}$. \end{lemma} Once we have a sub-solution and a super-solution of \eqref{e1.1}, we may find a solution of \eqref{e1.1} by the Perron's method. More precisely, we have the following result. \begin{lemma}\label{lem2.3} Let $p>0$ and $0\in \Omega$ be an open bounded $C^2$ domain. Suppose that there are super-solution $\overline U$ and sub-solution $\underline{U}$ of \eqref{e1.1} such that $\overline U$ and $\underline{U}$ are locally $C^2$ in $\Omega\setminus\{0\}$, and $$ \overline U\geq \underline{U}\text{ in } \Omega\setminus\{0\},\quad \lim_{x\to 0}\underline U(x)=+\infty,\quad \overline U\ge0\ge\underline{U}\quad \text{in }\Omega^c. $$ Then there exists at least one positive solution $u$ of \eqref{e1.1} such that $$ \underline{U}\leq u\leq \overline U\quad\text{in }\Omega\setminus\{0\}. $$ \end{lemma} Before proving the above lemma, we introduce the definition of viscosity solution to \eqref{e1.1}. \begin{definition} \label{def2.1} \rm We say that a function $u\in L^1(\mathbb{R}^N,\frac{dy}{1+|y|^{N+2\alpha}})$, continuous in $\mathbb{R}^N\setminus\{0\}$, is a viscosity super solution (sub solution) of \eqref{e1.1} if $$ u\geq 0\quad (\text{resp. } u\leq 0)\quad\text{in } \Omega^c $$ and for every point $x_0\in\Omega\setminus\{0\}$ and some neighborhood $V$ of $x_0$ with $\bar V\subset \Omega\setminus\{0\} $ and for any $\phi\in C^2(\bar V)$ such that $u(x_0)=\phi(x_0)$ and $$ u(x)\ge \phi(x)\quad (\text{resp. } u(x)\leq \phi(x))\quad\text{for all } x\in V, $$ defining \[ \tilde u=\begin{cases} \phi & \text{in } V,\\ u & \text{in }V^c, \end{cases} \] we have $$ (-\Delta)^{\alpha} \tilde u(x_0)+u^{p}(x_0)\geq 0\quad (\text{resp. } (-\Delta)^{\alpha} \tilde u(x_0)+u^{p}(x_0)\leq 0). $$ We say that $u$ is a viscosity solution of \eqref{e1.1} if it is a viscosity super solution and also a viscosity sub solution of \eqref{e1.1}. \end{definition} \begin{proof}[Proof of Lemma \ref{lem2.3}] Let $\Omega_n=\{x\in\Omega: |x|>1/n\}$. Then, $\Omega_{n}$ is of class $C^2$ for $n\ge N_0$, where $N_0$ is chosen large enough such that $$ \bar B_{1/N_0}(0)\subset \Omega\quad \text{and}\quad \underline U>0\quad\text{in } \bar B_{1/N_0}(0). $$ Since $\underline U$, $\overline U$ are locally $C^2$ in $\Omega\setminus\{0\}$, applying Lemma \ref{thm2.1} with ${\mathcal{O}} = \Omega_{n}$ and $\delta=\frac{1}{4n}$, we find a solution $u_n$ of the problem \begin{equation}\label{3.2.7} \begin{gathered} (-\Delta)^{\alpha} u(x)+|u|^{p-1}u(x)=0,\quad x\in\Omega_{n},\\ u(x)=\underline U(x),\quad x\in \bar B_{\frac1n}(0)\setminus\{0\},\\ u(x)=0, \quad x\in \Omega^c \end{gathered} \end{equation} and a solution $v_n$ of the problem \begin{equation}\label{3.2.07} \begin{gathered} (-\Delta)^{\alpha} u(x)+|u|^{p-1}u(x)=0,\quad x\in\Omega_{n},\\ u(x)=\overline U(x),\quad x\in \bar B_{\frac1n}(0)\setminus\{0\},\\ u(x)=0, \quad x\in \Omega^c. \end{gathered} \end{equation} Now we show that for any $n\ge N_0$, \begin{equation}\label{25-08-0} \underline U \le u_n\le v_n \le \overline U\quad\text{in } \mathbb{R}^N\setminus\{0\}. \end{equation} In fact, since $u_{n}$ is the solution of \eqref{3.2.7} in $\Omega_{n}$, $\underline U$ is a sub-solution of \eqref{3.2.7} in $\Omega_{n}$ and $u_n=\underline U$ in $B_{\frac1n}(0)\setminus\{0\}$, $u_n=0\ge\underline U$ in $\Omega^c$, then we apply Lemma \ref{comparison} to obtain that $\underline U \le u_n$ in $\mathbb{R}^N\setminus\{0\}$. Similarly, we have $v_{n}\le \overline U$ in $\mathbb{R}^N\setminus\{0\}$. Since $u_n$ and $v_n$ are solutions of \begin{equation}\label{14-01-14-1} (-\Delta)^{\alpha} u(x)+|u|^{p-1}u(x)=0,\quad x\in\Omega_{n} \end{equation} and $u_n=\underline U\le \overline U=v_n$ in $B_{\frac1n}$, $u_n=v_n=0$ in $\Omega^c$, by Lemma \ref{comparison}, we have $u_n\le v_n$ in $\mathbb{R}^N\setminus\{0\}$. Next, we prove that for all $n\ge N_0$, \begin{equation}\label{25-08-1} u_{n}\leq u_{n+1}\le v_{n+1}\le v_{n}\quad\text{in } \mathbb{R}^N\setminus\{0\}. \end{equation} Since $u_{n+1}\ge \underline U$ in $\mathbb{R}^N\setminus\{0\}$ and $u_{n}=\underline U$ in $\bar B_{\frac1n}(0)\setminus\{0\}$, we obtain $u_{n+1}\ge u_n$ in $\bar B_{\frac1n}(0)\setminus\{0\}$. For $u_{n+1}$ and $u_n$ being solutions of \eqref{14-01-14-1}, by Lemma \ref{comparison}, we have $u_{n}\leq u_{n+1}$ in $\mathbb{R}^N\setminus\{0\}$. Similarly, $v_{n}\geq v_{n+1}$ in $\mathbb{R}^N\setminus\{0\}$. The inequality $u_{n+1}\le v_{n+1}$ follows from \eqref{25-08-0}. Therefore, we can define a function $u$ by $$ u(x)=\lim_{n\to+\infty}u_n(x),\quad x\in \mathbb{R}^N\setminus\{0\}, $$ which satisfies \begin{equation}\label{25-08-2} \underline U(x)\le u_{N_0}\leq u(x)\leq v_{N_0}(x)\le \overline U(x),\quad x\in \mathbb{R}^N\setminus\{0\}. \end{equation} Since $\underline U$ and $\overline U$ belong to $L^1({\mathcal{O}}^c, \frac{dx}{1+|x|^{N+2\alpha}})$, then $u_n\to u$ in $L^1({\mathcal{O}}^c, \frac{dx}{1+|x|^{N+2\alpha}})$ as $n\to\infty$. By interior estimates as given in \cite[Lemma 3.1]{CV4} or \cite[Proposition 2.3]{RS}, for any compact set $K$ of $\Omega\setminus\{0\}$, there exists $N_K\ge N_0$ such that $\{u_n\}$ is uniformly bounded in $C^\theta(K)$ for $n\ge N_k$ and some $\theta\in(0,2\alpha)$. By Ascoli-Arzel\`{a} Theorem we find that $u$ is continuous in $K$ and $u_n\to u$ uniformly in $K$. Together with that $u_{N_0}=v_{N_0}=0$ in $\Omega^c$ and $u_{N_0}$, $v_{N_0}$ are continuous up to the boundary of $\Omega$, we have $u=0$ in $\Omega^c$, $u$ is continuous in $\Omega\setminus\{0\}$ and up to $\partial\Omega$. By stability theorem \cite[Theorem 2.4]{CFQ}, $u$ is a solution of \eqref{e1.1} in the viscosity sense. Applying \cite[Theorem 2.6]{CS2}, we find that $u$ is $C^\theta_{\rm loc}({\mathcal{O}})$, and using \cite[Theorem 2.1]{CFQ} we conclude that $u$ is a classical solution. \end{proof} Our main result in this section is as follows. \begin{proposition}\label{prop2.1} Let $0\in \Omega$ be an open bounded $C^2$ domain in $\mathbb{R}^N$ and $\alpha\in(0,1)$. Assume that $p>0$ and $u$ is a positive classical solution of \eqref{e1.1}. Then there exists a positive singular solution $u_s$ of \eqref{e1.1} with $\Omega=B_{r_0}(0)$ such that $$ u_s\le u\le u_s+c_2\quad \text{in } \Omega\setminus\{0\}, $$ where $c_2=\sup_{x\in \mathbb{R}^N\setminus B_{r_0}(0)}u(x)$. \end{proposition} \begin{proof} We will construct a sub-solution and a super-solution of \eqref{e1.1}. Let $u$ be a solution of \eqref{e1.1} and $ u_0(x)=u(x)-c_2$ for $x\in\mathbb{R}^N\setminus\{0\}$. We observe that $$ (-\Delta)^\alpha u_0=(-\Delta)^\alpha u\quad {\rm in } B_{r_0}(0)\setminus\{0\} $$ and $$ |u_0|^{p-1}u_0\le u^p \quad \text{in } B_{r_0}(0)\setminus\{0\}. $$ Hence, $u_0$ is a sub-solution of \eqref{e1.1} with $\Omega=B_{r_0}(0)$. Since $u$ is a super-solution of \eqref{e1.1} with $\Omega=B_{r_0}(0)$ and $u\ge u_0$ in $\mathbb{R}^N$, by Lemma \ref{lem2.3} there exists a solution $u_s$ of \eqref{e1.1} with $\Omega=B_{r_0}(0)$ such that \begin{equation}\label{14-08-1} u-c_2\le u_s\le u\quad \text{in } \Omega\setminus\{0\}. \end{equation} The proof is complete. \end{proof} \section{Symmetry of solutions} In this section, we will prove that the singular solution $u_s$ of \eqref{e1.1} with $\Omega=B_{r_0}(0)$ is radially symmetric. To this end, we investigate the radially symmetric property of positive singular solutions to more general semilinear elliptic equations \begin{equation}\label{e1} \begin{gathered} (-\Delta)^{\alpha} u=h(u)+\eta\quad \text{in } B_1(0)\setminus\{0\},\\ u=0\quad \quad \text{in } B_1^c(0),\\ \lim_{x\to0}u(x)=\infty. \end{gathered} \end{equation} We assume that $h$ and $\eta$ satisfy \begin{itemize} \item[(H0)] The function $h:[0,\infty)\to\mathbb{R}$ is Lipschitz continuous in $[0,R]$ for any $R>0$. \item[(H1)] The function $\eta: B_1(0)\setminus\{0\}\to\mathbb{R}$ is radially symmetric and decreasing in $|x|$. \end{itemize} The main results of this section reads as follows. \begin{proposition}\label{prop3.1} Suppose that {\rm (H0)} and {\rm (H1)} hold. If $u$ is a positive classical solution of \eqref{e1}, then $u$ must be radially symmetric and strictly decreasing in $r=|x|$ for $r\in(0,1)$. \end{proposition} Proposition \ref{prop3.1} will be proved by the moving plane method. However, since the solution $u$ is singular at the origin, we need the following variant of the maximum principle for small domain. \begin{lemma}\label{lem3.1} Let $\Omega$ be an open and bounded subset of $\mathbb{R}^N$. Suppose that $\varphi:\Omega\to\mathbb{R}$ is in $L^\infty(\Omega)$ and $w\in L^\infty(\mathbb{R}^N)$ is a classical solution of \begin{equation}\label{ea1} \begin{gathered} -(-\Delta)^\alpha w(x)\le \varphi(x)w(x),\quad x\in \Omega,\\ w(x)\ge 0,\quad x\in \mathbb{R}^N\setminus \Omega. \end{gathered} \end{equation} Then there is $\delta>0$ such that whenever $|\Omega^-|\le \delta$, $w$ has to be non-negative in $\Omega$, where $\Omega^-=\{x\in\Omega:w(x)<0\}$. \end{lemma} For a proof of the above lemma, see \cite[Corollary 2.1]{FW} and \cite{JW,RS0}. Now we use the moving plane method to show the radial symmetry and monotonicity of positive solutions to equation \eqref{e1}. For simplicity, we denote $\mathcal{B}_1=B_1(0)\setminus\{0\}$, \begin{gather}\label{d1} \Sigma_\lambda=\{x=(x_1,x')\in \mathcal{B}_1: x_1>\lambda\},\\ \label{d3} u_\lambda(x)=u(x_\lambda) \quad\text{and}\quad w_\lambda(x)=u_\lambda(x)-u(x), \end{gather} where $\lambda\in (0,1)$ and $x_\lambda=(2\lambda-x_1,x')$ for $x=(x_1,x')\in\mathbb{R}^N$. By the moving plane method, we will prove that $w_\lambda>0$ in $\Sigma_\lambda$ for all $\lambda\in (0,1)$. This is proved in a indirect way. Suppose, on the contrary, that $\Sigma_\lambda^-=\{x\in \Sigma_\lambda:w_\lambda(x)<0\}\not=\emptyset$ for $\lambda\in(0,1)$. Let us define \begin{gather}\label{ec} w_\lambda^+(x)=\begin{cases} w_\lambda(x),& x\in \Sigma_\lambda^-,\\ 0,& x\in \mathbb{R}^N\setminus \Sigma_\lambda^-, \end{cases}\\ \label{e4.01} w_\lambda^-(x)=\begin{cases} 0, & x\in \Sigma_\lambda^-,\\ w_\lambda(x), & x\in \mathbb{R}^N\setminus \Sigma_\lambda^-. \end{cases} \end{gather} Hence, $w_\lambda^+(x)=w_\lambda(x)-w_\lambda^-(x)$ for all $x\in\mathbb{R}^N$. It is obvious that $(2\lambda,0,\dots,0)\not\in \Sigma_\lambda^-$ for $\lambda$ small, since $\lim_{x\to0}u(x)=+\infty$. \begin{lemma}\label{lem13-08-1} For any $ \lambda\in(0,1)$ and any $x\in \Sigma_\lambda^-$, we have that \begin{equation}\label{claim1} (-\Delta)^{\alpha} w_\lambda^-(x)\le0. \end{equation} \end{lemma} \begin{proof} By direct computation, for $x\in \Sigma_\lambda^-$, we have that \begin{align*} (-\Delta)^{\alpha} w_\lambda^-(x) &= \int_{\mathbb{R}^N}\frac{w_\lambda^-(x)-w_\lambda^-(z)}{|x-z|^{N+2\alpha}}dz\\ &=-\int_{\mathbb{R}^N\setminus\Sigma_\lambda^-} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &= -\int_{(\mathcal{B}_1\setminus(\mathcal{B}_1)_\lambda) \cup ((\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1)} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &\quad -\int_{(\Sigma_\lambda\setminus\Sigma_\lambda^-) \cup (\Sigma_\lambda\setminus\Sigma_\lambda^-)_\lambda} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz -\int_{(\Sigma_\lambda^-)_\lambda}\frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &= -I_1-I_2-I_3, \end{align*} where $A_\lambda=\{x_\lambda: x\in A\}$ for any set $A$ in $\mathbb{R}^N$. We estimate these integrals separately. Since $u=0$ in $(\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1$ and $u_\lambda=0$ in $\mathcal{B}_1\setminus (\mathcal{B}_1)_\lambda$, we have \begin{align*} I_1 &= \int_{(\mathcal{B}_1\setminus(\mathcal{B}_1)_\lambda) \cup ((\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1)} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &= \int_{(\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1} \frac{u_\lambda(z)}{|x-z|^{N+2\alpha}}dz -\int_{ \mathcal{B}_1\setminus(\mathcal{B}_1)_\lambda} \frac{u(z)}{|x-z|^{N+2\alpha}}dz\\ &= \int_{(\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1}u_\lambda(z) (\frac{1}{|x-z|^{N+2\alpha}}-\frac{1}{|x-z_\lambda|^{N+2\alpha}}))dz\ge 0, \end{align*} since $u_\lambda\geq0$ and $|x-z_\lambda|>|x-z|$ for all $x\in \Sigma_\lambda^-$ and $z\in (\mathcal{B}_1)_\lambda\setminus \mathcal{B}_1$. To decide the sign of $I_2$ we observe that $w_\lambda(z_\lambda)=-w_\lambda(z)$ for any $z\in\mathbb{R}^N$. Then \begin{align*} I_2 &= \int_{(\Sigma_\lambda\setminus\Sigma_\lambda^-) \cup(\Sigma_\lambda\setminus\Sigma_\lambda^-)_\lambda} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &= \int_{\Sigma_\lambda\setminus\Sigma_\lambda^-} \frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz +\int_{\Sigma_\lambda\setminus\Sigma_\lambda^-} \frac{w_\lambda(z_\lambda)}{|x-z_\lambda|^{N+2\alpha}}dz\\ &= \int_{\Sigma_\lambda\setminus\Sigma_\lambda^-}w_\lambda(z) (\frac{1}{|x-z|^{N+2\alpha}}-\frac{1}{|x-z_\lambda|^{N+2\alpha}})dz\ge 0, \end{align*} since $w_\lambda\ge0$ in $\Sigma_\lambda\setminus\Sigma_\lambda^-$ and $|x-z_\lambda|>|x-z|$ for all $x\in \Sigma_\lambda^-$ and $z\in \Sigma_\lambda\setminus\Sigma_\lambda^-$. Finally, since $w_\lambda(z)<0$ for $z\in \Sigma_\lambda^-$, we deduce that \begin{align*} I_3 &= \int_{(\Sigma_\lambda^-)_\lambda}\frac{w_\lambda(z)}{|x-z|^{N+2\alpha}}dz\\ &=\int_{\Sigma_\lambda^-}\frac{w_\lambda(z_\lambda)}{|x-z_\lambda|^{N+2\alpha}}dz\\ &= -\int_{\Sigma_\lambda^-}\frac{w_\lambda(z)}{|x-z_\lambda|^{N+2\alpha}}dz \ge0. \end{align*} The proof is complete. \end{proof} \begin{lemma}\label{lem13-08-2} Let the function $h$ satisfy {\rm (H0)} and for $\lambda \in(0,1)$ and $x\in\Sigma_\lambda^-$, \begin{equation}\label{13-08} \varphi(x)=-\frac{h(u_\lambda(x))-h(u(x))}{u_\lambda(x)-u(x)}. \end{equation} Then there exists $C>0$ dependent of $\lambda$ such that $\norm{\varphi}_{ L^\infty(\Sigma_\lambda^-)}\le C$. \end{lemma} \begin{proof} For $x\in \Sigma_\lambda^-\subset \Sigma_\lambda\subset \mathbb{R}^N\setminus {B_\lambda(0)}$, $u_\lambda(x)0$ such that $$ \norm{u}_{L^\infty(\mathbb{R}^N\setminus {B_\lambda(0)})}\le M_\lambda. $$ Since $h$ satisfies $(H_0)$, there exists $C>0$ depending on $\lambda$ such that $$ \norm{\varphi}_{ L^\infty(\Sigma_\lambda^-)}\le C. $$ \end{proof} \begin{remark}\label{rmkxy}\rm We note that $M_\lambda\to \infty$ as $\lambda\to 0$, since $\lim_{x\to0}u(x)=\infty$. \end{remark} \begin{proof}[Proof of Proposition \ref{prop3.1}] We divide the proof into four steps. \smallskip \noindent \textbf{Step 1.} We prove that if $\lambda$ is close to $1$, then $w_\lambda>0$ in $\Sigma_\lambda$. First we show that $w_\lambda\geq0$ in $\Sigma_\lambda$, i.e. $\Sigma^-_\lambda$ is empty. By contradiction, we assume that $\Sigma^-_\lambda\not=\emptyset$. Now we apply \eqref{claim1} and linearity of the fractional Laplacian to obtain that, for $ x\in\Sigma_\lambda^-$, \begin{equation}\label{ee} (-\Delta)^{\alpha} w_\lambda^+(x)\ge (-\Delta)^{\alpha} w_\lambda(x) =(-\Delta)^{\alpha} u_\lambda(x)-(-\Delta)^{\alpha} u(x). \end{equation} Combining equation \eqref{e1} with \eqref{ee} and \eqref{ec}, for $x\in\Sigma_\lambda^-$, we have \begin{align*} (-\Delta)^{\alpha} w_\lambda^+(x) &\geq (-\Delta)^{\alpha}u_\lambda(x)-(-\Delta)^{\alpha} u(x)\\ &= h(u_\lambda(x))+\eta(x_\lambda)-h(u(x))-\eta(x)\\ &= -\varphi(x)w_\lambda^+(x)+\eta(x_\lambda)-\eta(x). \end{align*} By Lemma \ref{lem13-08-2} and assumption (H1), we have that $\eta(x_\lambda)\geq \eta(x)$ for $x\in\Sigma_\lambda^-$ and then \begin{equation}\label{ef} -(-\Delta)^{\alpha} w_\lambda^+(x)\le \varphi(x)w_\lambda^+(x),\ \quad x\in\Sigma_\lambda^-. \end{equation} Moreover, $w_\lambda^+=0$ in $(\Sigma_\lambda^-)^c$. Choosing $\lambda\in (0,1)$ close enough to $1$ we have that $|\Sigma_\lambda^-|$ is small and we apply Lemma \ref{lem3.1} to obtain that $$ w_\lambda=w_\lambda^+\geq0\quad \text{in } \Sigma_\lambda^-, $$ which is impossible. Thus, $$ w_\lambda\geq0\quad \text{in } \Sigma_\lambda. $$ To complete Step 1, we claim that for $0<\lambda<1$, if $w_\lambda\ge0$ and $w_\lambda\not\equiv0$ in $\Sigma_\lambda$, then $w_\lambda>0$ in $\Sigma_\lambda$. We assume that the claim is true for the moment. Since the function $u$ is positive in $B_1(0)$ and $u=0$ on $\partial B_1(0)$, $w_\lambda$ is positive on $\partial B_1(0)\cap \partial \Sigma_\lambda$ and then $w_\lambda\not\equiv0$ in $\Sigma_\lambda$. Now we prove the claim. Suppose on the contrary that there exists $x_0\in \Sigma_\lambda$ such that $w_\lambda(x_0)=0$, i.e., $u_\lambda(x_0)=u(x_0)$. Then \[ (-\Delta)^{\alpha} w_\lambda(x_0) = (-\Delta)^{\alpha} u_\lambda(x_0)-(-\Delta)^{\alpha} u(x_0) =\eta((x_0)_\lambda)-\eta(x_0). \] Since $x_0\in \Sigma_\lambda$, we have $|x_0|>|(x_0)_\lambda|$. By assumption (H1), we have that $\eta((x_0)_\lambda)\geq \eta(x_0)$ and then \begin{equation}\label{e7} (-\Delta)^{\alpha} w_\lambda(x_0)\geq0. \end{equation} On the other hand, let $K_\lambda=\{(x_1,x')\in\mathbb{R}^N: x_1>\lambda\}$. Noting $w_\lambda(z_\lambda)=-w_\lambda(z)$ for any $z\in\mathbb{R}^N$ and $w_\lambda(x_0)=0$, we deduce \begin{align*} (-\Delta)^{\alpha} w_\lambda(x_0) &= -\int_{K_\lambda}\frac{w_\lambda(z)}{|x_0-z|^{N+2\alpha}}dz -\int_{\mathbb{R}^N\setminus K_\lambda}\frac{w_\lambda(z)}{|x_0-z|^{N+2\alpha}}dz\\ &= -\int_{K_\lambda}\frac{w_\lambda(z)}{|x_0-z|^{N+2\alpha}}dz -\int_{K_\lambda}\frac{w_\lambda(z_\lambda)}{|x_0-z_\lambda|^{N+2\alpha}}dz\\ &= -\int_{K_\lambda}w_\lambda(z)(\frac{1}{|x_0-z|^{N+2\alpha}} -\frac{1}{|x_0-z_\lambda|^{N+2\alpha}})dz. \end{align*} The facts $|x_0-z_\lambda|>|x_0-z|$ for $z\in K_\lambda$, $w_\lambda(z)\ge0$ and $w_\lambda(z)\not\equiv0$ in $K_\lambda$ yield \begin{equation}\label{ff} (-\Delta)^{\alpha} w_\lambda(x_0)<0,\end{equation} which contradicts \eqref{e7}, completing the proof of the claim. \smallskip \noindent\textbf{Step 2.} We prove $\lambda_0:=\inf\{\lambda\in(0,1):w_\lambda>0\text{ in }\Sigma_\lambda\}=0$. Were it not true, we would have $\lambda_0>0$. Hence, $w_{\lambda_0}\ge0$ in $\Sigma_{\lambda_0}$ and $w_{\lambda_0}\not\equiv0$ in $\Sigma_{\lambda_0}$. The claim in Step 1 implies $w_{\lambda_0}>0$ in $\Sigma_{\lambda_0}$. Next we claim that if $w_\lambda>0$ in $\Sigma_\lambda$ for $\lambda\in(0,1)$, then there exists $\epsilon\in(0,\lambda/4)$ such that $w_{\lambda_\epsilon}>0$ in $\Sigma_{\lambda_\epsilon}$, where $\lambda_\epsilon=\lambda-\epsilon>3\lambda/4$. This claim directly implies that $\lambda_0=0$, which contradicts the fact $\lambda_0>0$. Now we prove the claim. Let $D_\mu=\{x\in\Sigma_\lambda:\operatorname{dist}(x,\partial\Sigma_\lambda) \ge \mu\}$ for $\mu>0$ small. Since $w_\lambda>0$ in $\Sigma_\lambda$ and $D_\mu$ is compact, there exists $\mu_0>0$ such that $w_\lambda\ge \mu_0$ in $D_\mu$. By the continuity of $w_\lambda(x)$, for $\epsilon>0$ small enough and $\lambda_\epsilon=\lambda-\epsilon$, we have that $w_{\lambda_\epsilon}(x)\ge0\text{ in } D_\mu$. Therefore, $\Sigma_{\lambda_\epsilon}^-\subset \Sigma_{\lambda_\epsilon}\setminus D_\mu$ and $|\Sigma_{\lambda_\epsilon}^-|$ is small if $\epsilon$ and $\mu$ are small. Using \eqref{claim1} and proceeding as in Step 1, we have for all $x\in \Sigma_{\lambda_\epsilon}^-$ that \begin{align*} (-\Delta)^{\alpha} w_{\lambda_\epsilon}^+(x) &= (-\Delta)^{\alpha} u_{\lambda_\epsilon}(x)-(-\Delta)^{\alpha} u(x)-(-\Delta)^{\alpha} w_{\lambda_\epsilon}^-(x)\\ &\ge (-\Delta)^{\alpha} u_{\lambda_\epsilon}(x)-(-\Delta)^{\alpha} u(x)\\ &= -\varphi(x)w_{\lambda_\epsilon}^+(x)+\eta(x_{\lambda_\epsilon})-\eta(x)\\ &\ge- \varphi(x)w_{\lambda_\epsilon}^+(x). \end{align*} By Lemma \ref{lem13-08-2}, if $\lambda_\epsilon>3\lambda/4$, $\varphi(x)$ is controlled by some constant depending on $\lambda$. Since $w_{\lambda_\epsilon}^+=0$ in $(\Sigma_{\lambda_\epsilon}^-)^c$ and $|\Sigma_{\lambda_\epsilon}^-|$ is small, for $\epsilon$ and $\mu$ small, Proposition \ref{lem3.1} implies that $w_{\lambda_\epsilon}\ge0$ in $\Sigma_{\lambda_\epsilon}$. Combining with $\lambda_\epsilon>0$ and $w_{\lambda_\epsilon}\not\equiv0$ in $\Sigma_{\lambda_\epsilon}$, we obtain $w_{\lambda_\epsilon}>0$ in $\Sigma_{\lambda_\epsilon}$. The proof of the claim complete. \smallskip \noindent\textbf{Step 3.} By Step 2, we have $\lambda_0=0$, which implies that $u(-x_1,x')\ge u(x_1,x')$ for $x_1\ge0$. Using the same argument from the other side, we conclude that $u(-x_1,x')\le u(x_1,x')$ for $ x_1\ge0$ and then $u(-x_1,x')= u(x_1,x')$ for $x_1\ge0$. Repeating this procedure in all directions we see that $u$ is radially symmetric. Finally, we prove $u(r)$ is strictly decreasing in $r\in (0,1)$. Let us consider $00\quad \text{for } x\in\Sigma_{\lambda}. $$ Then \begin{align*} 0u(\widetilde{x}_1,0,\dots,0)$. 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