\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 297, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/297\hfil Singular critical elliptic problems]
{Singular critical elliptic problems with fractional Laplacian}

\author[X. Wang, J. Yang \hfil EJDE-2015/297\hfilneg]
{Xueqiao Wang, Jianfu Yang}

\address{Xueqiao Wang \newline
Department of Mathematics,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email{wangxueqiao1989@126.com}

\address{Jianfu Yang \newline
Department of Mathematics,
Jiangxi Normal University\\
Nanchang, Jiangxi 330022, China}
\email{jfyang\_2000@yahoo.com}

\thanks{Submitted September 6, 2015. Published December 3, 2015.}
\subjclass[2010]{35J20, 35J25, 35J61}
\keywords{Fractional Laplacian; singular critical problem; 
\hfill\break\indent non-contractible domain}

\begin{abstract}
 In this article, we consider the existence of solutions of the critical
 problem with a Hardy term for fractional Laplacian
 \begin{gather*}
 (-\Delta)^s u -\mu \frac u{|x|^{2s}}=  u^{2^*_s-1} \quad \text{in }\Omega,\\
 u>0 \quad \text{in }\Omega, \\
 u=0 \quad \text{on }\partial \Omega,
 \end{gather*}
 where $\Omega\subset \mathbb{R}^N$ is a smooth bounded domain and
 $0\in\Omega$, $\mu$ is a positive parameter, $N>2s$ and $s\in(0,1)$,
 $2^*_{s} =\frac{2N}{N-2s}$ is the critical exponent.
 $(-\Delta)^s$ stands for the spectral fractional Laplacian.
 Assuming that $\Omega$ is non-contractible, we show that there exists
 $\mu_0>0$  such that $0<\mu<\mu_0$, there exists a solution.
 We also discuss a similar problem for the restricted fractional Laplacian.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

In this article, we consider the existence of solutions for the 
critical problem with a Hardy term and fractional Laplacian
\begin{equation} \label{ePsm}
\begin{gathered}
(-\Delta)^s u -\mu \frac u{|x|^{2s}} =  u^{2^*_s-1} \quad \text{in }\Omega,\\
u>0 \quad  \text{in } \Omega, \\
u=0 \quad  \text{on } \partial \Omega
\end{gathered}
\end{equation}
in a smooth bounded domain $\Omega\subset \mathbb{R}^N$ and 
$0\in\Omega$,  where $\mu$ is a positive parameter, $N>2s$ and 
$s\in(0,1)$, $2^*_{s} =\frac{2N}{N-2s}$ is the critical exponent. 
The operator $(-\Delta)^s$ is the spectral Laplacian defined in section 2.

In the case $s=1$, such a problem has been extensively studied, 
see \cite{CM, FG, GY, J, RW, T} etc. It is known that problem 
\eqref{ePsm} with $s=1$ has no nontrivial solutions if $\mu\geq 0$ and 
$\Omega$ is star shaped \cite{AA}. 
However, the situation becomes different if the domain $\Omega$ has nontrivial 
topology. 
In \cite{KW}, a nontrivial solution was found for problem
\eqref{ePsm} with $s=1$ and $\mu=0$, if $\Omega$ is an annulus.
Then it was  shown in \cite{BC} that there exists a nontrivial solution of 
\eqref{ePsm} with $s=1$ and $\mu=0$, if $\Omega$ has nontrivial topology.  
If $\mu>0$, there is a Hardy term in \eqref{ePsm} with $s=1$.
In \cite{HS}, it proved that problem \eqref{ePsm} with $s=1$
admits a solution in a non-contractible domain.
Since \eqref{ePsm} with $s=1$ is a critical problem, it involves the ground 
state solution of the problem in the whole space
\begin{equation} \label{eP1m}
\begin{gathered}
-\Delta u -\mu \frac u{|x|^{2}}=  u^{2^*-1} \quad \text{in }\mathbb{R}^N,\\
u>0 \quad  \text{in } \mathbb{R}^N.
\end{gathered}
\end{equation}
The ground state solutions of \eqref{eP1m} are  found in \cite{Ta} 
for $\mu=0$ and in  \cite{T} for $\mu\neq 0$.

Recently, Secchi et al \cite{SSS} proved that
Coron type problem admits a solution for problem \eqref{ePsm} with $\mu=0$
and the restricted fractional Laplacian; see section 2 for a definition. 
Similarly, the argument in \cite{SSS} relies on, among other things, 
the explicit form of the minimizer of the problem
\begin{equation}\label{eq:1.1}
\Lambda_s = \inf_{u\in \dot H^s(\mathbb{R}^{N}), u\not\equiv 0}
\frac{\int_{\mathbb{R}^N}|(-\Delta)^{s/2} u(x)|^2\,dx}
{\big(\int_{\mathbb{R}^N}|u(x)|^{2^*_s}\,dx\big)^{2/ 2^*_s}},
\end{equation}
where the space $\dot H^s(\mathbb{R}^{N})$ is defined as the completion 
of $C^\infty_0(\mathbb{R}^{N})$ under the norm
\begin{equation}\label{eq:1.2}
\|u\|_{\dot H^s(\mathbb{R}^{N})}^2 
= \int_{\mathbb{R}^N}|\xi|^{2s}|\hat u(\xi)|^2\,d\xi,
\end{equation}
here $\hat u$ denotes the Fourier transform of $u$. 
In $\mathbb{R}^N$, the operator $(-\Delta)^{s/2}$, $s\in \mathbb{R}$
is defined by the Fourier transform
\begin{equation}\label{eq:1.3}
(\widehat{(-\Delta)^{s/2}u})(\xi) = |\xi|^s\hat u(\xi)
 \end{equation}
for $u\in C^\infty_0(\mathbb{R}^N)$. Therefore, for $s>0$, we have
\begin{equation}\label{eq:1.4}
\|(-\Delta)^{s/2}u\|_{L^2(\mathbb{R}^N)}^2 
= \int_{\mathbb{R}^N}|\xi|^{2s}|\hat u(\xi)|^2\,d\xi.
 \end{equation}


For $N>2s$, the minimizing problem $\Lambda_s$ in \eqref{eq:1.1} 
is related to the fractional Sobolev embedding 
$\dot H^s(\mathbb{R}^{N})\hookrightarrow L^{\frac{2N}{N-2s}}(\mathbb{R}^{N})$. 
The continuity of this inclusion corresponds to the inequality
\begin{equation}\label{eq:1.5}
\|u\|^{2}_{L^{2^*_s}(\mathbb{R}^{N})}
\leq \Lambda_s^{-1}\|u\|^2_{\dot H^s(\mathbb{R}^{N})}.
 \end{equation}
The best constant $\Lambda_s$ in \eqref{eq:1.5} was computed 
in \cite{CT}. A minimizer $u$ of $\Lambda_s$ weakly solves the problem
\begin{equation}\label{eq:1.6}
(-\Delta)^s u  = |u|^{2^*_{s,\alpha}-2} u\quad\text{in } \mathbb{R}^N
 \end{equation}
up to a multiplying constant. Using the moving plane
method for integral equations, Chen et al \cite{CLO} classified the solutions of 
an integral equation, which is related to problem \eqref{eq:1.6}. 
Positive regular solutions of \eqref{eq:1.6}, and then the minimizers 
of $\Lambda_s$ are precisely given by
\begin{equation}\label{eq:1.9}
U_\varepsilon(x) = \Big(\frac {\varepsilon}{\varepsilon^2 +|x-x_0|^2}
\Big)^{\frac{N-2s}2}
\end{equation}
for $\varepsilon>0$ and $x_0\in \mathbb{R}^N$.

In this paper, we consider the existence of solutions of problem \eqref{ePsm}
 with $0<\mu<\mu_H$ and $s\in (0,1)$ in a non-contractible domain $\Omega$, 
where $\mu_H$ is the best constant in the Hardy inequality. 
Problem \eqref{ePsm} is related to the variational problem
\begin{equation}\label{eq:1.10}
\Lambda_{s,\mu} = \inf_{u\in \dot H^s(\mathbb{R}^{N}), u\not\equiv 0}
\frac{\|(-\Delta)^{s/2}u\|_{L^2(\mathbb{R}^N)}^2
- \mu \int_{\mathbb{R}^{N}}\frac{|u(x)|^2}{|x|^{2s}}\,dx}
{\big(\int_{\mathbb{R}^N}|u(x)|^{2^*_s}\,dx\big)^{\frac 2{2^*_s}}}.
\end{equation}
Although minimizers of $\Lambda_{1,\mu}$ were found explicitly in \cite{T}
for $s =1$, it is not the case for $s\in(0,1)$. So in our argument, 
we need to avoid using it. Our main result for the spectral Laplacian is as follows.

\begin{theorem}\label{thm:1.1}
Suppose $\Omega$ is not contractible. Then, there exists $0<\mu_0<\mu_H$ 
such that for each $\mu\in(0,\mu_0)$, there exists a solution of $(P_{s,\mu})$.
\end{theorem}

For the restricted fractional Laplacian $(-\Delta_{|\Omega})^s$, 
we consider the problem
\begin{equation} \label{ePsmR}
\begin{gathered}
(-\Delta_{|\Omega})^s u -\mu \frac u{|x|^{2s}}
=  u^{2^*_s-1} \quad \text{in }\Omega,\\
u>0\quad  \text{in } \Omega, \\
u=0 \quad  \text{on } \mathbb{R}^N\setminus \Omega.
\end{gathered}
\end{equation}
Similarly, we have the following result.

\begin{theorem}\label{thm:1.2}
 Suppose $\Omega$ is not contractible. Then, there exists $0<\mu_0<\mu_H$ 
such that for each $\mu\in(0,\mu_0)$, there exists a solution of \eqref{ePsmR}.
\end{theorem}

The article is organized as follows. 
After some preparations in section 2, we prove Theorems \ref{thm:1.1}
 and \ref{thm:1.2} in section 3.

\section {Sobolev-Hardy inequality}

In this section, we  develop some properties of minimizers of $\Lambda_{s,\mu}$, 
and give the definition of fractional operator $(-\Delta)^s$.
First, we define for each $s\geq 0$, the fractional Sobolev space 
\[
H^s(\mathbb{R}^N) = \{u\in L^2(\mathbb{R}^N): |\xi|^s\hat u(\xi)\in L^2(\mathbb{R}^N)\}
\]
via the Fourier transform
\[
\hat u(\xi) = \frac 1{(2\pi)^{N/2}}\int_{\mathbb{R}^N}e^{-ix\cdot\xi}u(x)\,dx .
\]
For $s\in(0,1)$, it is known from \cite{NPV} that there holds
\begin{equation}\label{eq:2.1}
\int_{\mathbb{R}^N}|\xi|^{2s}|\hat u(\xi)|^2\,d\xi 
= C_{s,N}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|u(x) 
- u(y)|^2}{|x - y|^{N+2s}}\,dx\,dy,
\end{equation}
where $C_{s,N}$ is a positive constant. This provides an alternative norm
\[
\|u\|_{\dot{H}^s(\mathbb{R}^{N})} 
= \Big(\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}} \frac{|u(x) 
- u(y)|^2}{|x-y|^{N+2s}}\,dx\,dy\Big)^{1/2}
\]
on $\dot H^s(\mathbb{R}^N)$.
If $N>2s$, the optimal constant $\mu_{s,N}$ was found in \cite{Y} 
for the fractional Hardy inequality
\begin{equation}\label{eq:2.2}
\mu_H\int_{\mathbb{R}^{N}}\frac{|u(x)|^2}{|x|^{2s}}\,dx
\leq \int_{\mathbb{R}^{N}}|\xi|^{2s}|\hat u(\xi)|^2\,d\xi,
\end{equation}
where $u\in C_0^\infty(\mathbb{R}^N)$. By a denseness argument, we have
\begin{equation}\label{eq:2.3}
\mu_H\int_{\mathbb{R}^{N}}\frac{|u(x)|^2}{|x|^{2s}}\,dx
\leq \int_{\mathbb{R}^{N}}|(-\Delta)^{s/2} u(x)|^2\,dx
\end{equation}
for $u\in \dot H^s(\mathbb{R}^N)$, where 
$\mu_H = 4^s\frac {\Gamma^2(\frac{N+2s}4)}{\Gamma^2(\frac{N-2s}4)}$. 
If $0<\mu< \mu_H$, we may verify that
\[
|u|_{\dot{H}^s(\mathbb{R}^{N})}
:= \Big(\|(-\Delta)^{s/2}u\|_{L^2(\mathbb{R}^N)}^2
 - \mu \int_{\mathbb{R}^{N}}\frac{|u(x)|^2}{|x|^{2s}}\,dx\Big)^{1/2}
\]
defines an equivalent norm on $\dot H^s(\mathbb{R}^N)$. 
Therefore, there exists $C>0$ such that for any $u\in \dot H^s(\mathbb{R}^{N})$,
\begin{equation}\label{eq:2.4}
\Big(\int_{\mathbb{R}^N}|u(x)|^{2^*_s}\,dx\Big)^{\frac 2{2^*_s}}
\leq C \Big[\int_{\mathbb{R}^N}|(-\Delta)^{s/2} u(x)|^2\,dx
 -  \mu \int_{\mathbb{R}^{N}}\frac{|u(x)|^2}{|x|^{2s}}\,dx\Big],
\end{equation}
where $2^*_s =\frac{2N}{N-2s}$. Define
\begin{equation}\label{eq:2.5}
\Lambda_{s,\mu} = \inf_{u\in \dot H^s(\mathbb{R}^{N}), u
\not\equiv 0}\frac{|u|^2_{\dot{H}^s(\mathbb{R}^{N})}}
{\big(\int_{\mathbb{R}^N}|u(x)|^{2^*_s}\,dx\big)^{\frac 2{2^*_s}}}.
\end{equation}
It is proved in \cite{GS} that $\Lambda_{s,\mu}>0$ is achieved if $\mu\geq 0$.

We remark that any minimizer of $\Lambda_{s,\mu}$ does not change sign, 
and is radially symmetric. Indeed, let $u$ be a minimizer. By formula 
(A.11) in \cite{SV},
\[
\int_{\mathbb{R}^N}|(-\Delta)^{s/2}|u||^2\,dx
\leq \int_{\mathbb{R}^N}|(-\Delta)^{s/2}u|^2\,dx.
\]
Hence, $|u|$ is also a minimizer, and we have $u>0$. Denote by $u^*$ 
the symmetric-decreasing rearrangement of $u$. By strict rearrangement 
inequalities in \cite{FS}, $u^*$ is also a minimizer of $\Lambda_{s,\mu}$. 
Therefore, by strict rearrangement inequalities again, $u(x) = u^*(x-A)$ 
for some $A\in \mathbb{R}^N$. Moreover, any minimizer $u$ of $\Lambda_{s,\mu}$ 
weakly solves \eqref{ePsm}  with $\Omega = \mathbb{R}^N$ up to multiplying a 
constant. In the case $\Omega = \mathbb{R}^N$, since both $u(x)$ and $u^*(x)$ 
solve equation \eqref{ePsm} , and \eqref{ePsm} is not translation invariant, 
we obtain that $A = 0$, that is $u = u^*$.

Next, we define fractional Laplacians in a bounded domain. 
There are two types of fractional Laplacians in bounded domainds, one is the 
spectral fractional Laplacian, another one is the restricted fractional Laplacian.

In a bounded domain $\Omega\subset \mathbb{R}^N$, we define the spectral 
fractional Laplacian  $(-\Delta)^s$ as follows.
Let $\{\lambda_k,\varphi_k\}^\infty_{k=1}$ be the eigenvalues and corresponding 
eigenfunctions of the Laplacian operator $-\Delta$ in $\Omega$ with zero 
Dirichlet boundary values on $\partial\Omega$ normalized by 
$\|\varphi_k\|_{L^2(\Omega)} = 1$, i.e.
\[
-\Delta \varphi_k = \lambda_k \varphi_k\quad\text{in } \Omega;\quad 
\varphi_k = 0\quad{\rm on}\ \partial\Omega.
\]
For any $u\in L^2(\Omega)$, we may write $$u = \sum_{k=1}^\infty u_k\varphi_k, \quad{\rm where}\quad u_k = \int_\Omega u\varphi_k\,dx.$$
 We define the space
\begin{equation}\label{eq:2.6}
H=\{u=\sum_{k=1}^\infty u_k\varphi_k\in L^2(\Omega):
\sum_{k=1}^\infty \lambda_k^{2s}u_k^2<\infty\},
\end{equation}
which is equipped with the norm
\[
\|u\|_{H} =\Big(\sum_{k=1}^\infty \lambda_k^{2s}u_k^2\Big)^{1/2}.
\]
For any $u\in H$, the spectral fractional Laplacian
$(-\Delta)^s$,  is defined by
\begin{equation}\label{eq:2.6a}
(-\Delta)^su = \sum_{k=1}^\infty \lambda_k^su_k\varphi_k.
\end{equation}

The space $H$ defined in \eqref{eq:2.6} is the interpolation
space $(H^2_0(\Omega), L^2(\Omega))_{s,2}$, see \cite{A,LM,Tar}.
It was shown in \cite{LM} that 
$(H^2_0(\Omega), L^2(\Omega))_{s,2} = H^s_0(\Omega)$ if $0<s<1$ and 
$s\not= 1/2$; while
 $(H^2_0(\Omega), L^2(\Omega))_{\frac 12 ,2} = H_{00}^{1/2}(\Omega)$, where
\[
H_{00}^{1/2}(\Omega) =\{u\in H^{1/2}(\Omega):
\int_\Omega\frac{u^2(x)}{d(x)}\,dx<\infty\},
\]
and $d(x) = \operatorname{dist}(x,\partial\Omega)$ for all $x\in\Omega$.

Now, using the idea in \cite{CS}, for any $u\in H^{s}_0(\Omega)$, 
we may define the extension $w = E_s(u)$ of  $u$ as the solution 
$w\in H^1_{0,L}(\mathcal{C}_\Omega)$ of the problem
\begin{equation}\label{eq:2.7}
\begin{gathered}
-\operatorname{div}(y^{1-2s}\nabla w) = 0, \quad
 \mathcal{C}_{\Omega}=\Omega\times(0,\infty),\\
w=0, \quad \partial_L\mathcal{C}_{\Omega}=\partial\Omega\times(0,\infty),\\
w = u , \quad \Omega\times\{0\},
 \end{gathered}
 \end{equation}
where
\[
H^1_{0,L}(\mathcal{C}_\Omega)
= \big\{w\in L^2(\mathcal{C}_\Omega): w = 0 \text{ on }
 \partial_L\mathcal{C}_\Omega,\;  
\kappa_{s}\int_{\mathcal{C}_\Omega}y^{1-2s}|\nabla w|^2\,dx\,dy<\infty\big\}
\]
is a Hilbert space with the norm
\[
\|w\|_{H^1_{0,L}(\mathcal{C}_\Omega)}^2
=\kappa_s\int_{\mathcal{C}_\Omega}y^{1-2s}|\nabla w|^2\,dx\,dy.
\]
The extension operator $E$ is an isometry between $H^{s}_0(\Omega)$
and $H^1_{0,L}(\mathcal{C}_\Omega)$. That is
\begin{equation}\label{eq:2.80}
\|E(u)\|_{H^1_{0,L}(\mathcal{C}_\Omega)} = \|u\|_{H_0^{s}(\Omega)}.
\end{equation}
It was shown in \cite{CS}, see also \cite{BCPa}, that
\begin{equation}\label{eq:2.8a}
-\kappa_s\lim_{y\to 0^+}y^{1-2s}\frac {\partial w}{\partial y} = (-\Delta)^{s}u,
\end{equation}
where $(-\Delta)^{s}$ is the spectral fractional Laplacian.
We remark that if $\Omega = \mathbb{R}^N$, for any $u\in H^s(\mathbb{R}^N)$, 
the extension $w = E(u)$ of $u$ is given by
\begin{equation}\label{eq:2.8b}
\begin{gathered}
-\operatorname{div}(y^{1-2s}\nabla w) = 0, \quad \mathbb{R}_+^{N+1},\\
w = u , \quad \mathbb{R}^{N},
 \end{gathered}
 \end{equation}
then by \cite{CS}, problem \eqref{eq:2.8b} corresponds to the fractional 
Laplacian given by the Fourier transform in \eqref{eq:1.3}, 
which also satisfies \eqref{eq:2.8a}. In this sense, the fractional
 Laplacian given in \eqref{eq:1.3} can be approached by the spectral 
fractional Laplacian through extending the domain $\Omega$ in 
\eqref{eq:2.7} to $\mathbb{R}^{N}$. In the sequel, we use the same 
notation $(-\Delta)^s$ to denote these two operators.

Using this sort of extension,  we may reformulate the nonlocal problem \eqref{ePsm}
 in a local way; that is,
\begin{equation}\label{eq:2.8}
\begin{gathered}
-\operatorname{div}(y^{1-2s}\nabla w) = 0, \quad \mathcal{C}_{\Omega},\\
w=0,\quad \partial_L\mathcal{C}_{\Omega},\\
\kappa_s y^{1-2s}\frac {\partial w}{\partial \nu} 
= \mu \frac u{|x|^{2s}} + u^{2^*_s-1}  \quad \Omega\times\{0\},
 \end{gathered}
 \end{equation}
where $\frac {\partial }{\partial \nu}$ is the outward normal derivative.

The other type of fractional Laplacian is the restricted fractional Laplacian 
defined by the following formula:
\[
(-\Delta_{|\Omega} )^su(x)= c_{ N, s  }\, 
 P.V. \int_{\mathbb R^N} \frac{ u(x)-u(y)}{|y-x|^{N+2s}}\,dy
\]
for all functions $u$ which are zero outside $\Omega$. 
Similarly, the extension problem is given as follows.
\begin{equation}
\begin{gathered}
\operatorname{div}(y^{1-2\alpha}\nabla w)=0 \quad\text{in }  \mathbb R^{N+1} , \\
 w=0  \quad\text{in } \mathbb R^N\setminus \Omega, \\
 y^{1-2\alpha}\frac{\partial w}{\partial \nu}
= \mu \frac u{|x|^{2s}} + u^{2^*_s-1} \quad\text{on }  \Omega.
 \end{gathered}\label{ePsmRb}
 \end{equation}

\section{Existence of solutions}

In this section, we first show that problem \eqref{ePsm} admits a nontrivial 
solution in a non-contractible domain.
Define the functional
\[
I_\mu(u) = \frac 12\int_\Omega\big( |(-\Delta)^{s/2}u|^2 
 -\mu\frac{u^2}{|x|^{2s}}\big)\,dx
-\frac 1{2_s^*}\int_\Omega |u|^{2_s^*}\,dx
\]
on $H^s_0(\Omega)$, where $(-\Delta)^{s}$ is the spectral factional Laplacian 
defined in \eqref{eq:2.6a}, and the functional
\[
J_\mu(u) = \frac 12\int_{\mathbb{R}^N}
\big( |(-\Delta)^{s/2}u|^2 -\mu\frac{u^2}{|x|^{2s}}\big)\,dx
-\frac 1{2_s^*}\int_{\mathbb{R}^N} |u|^{2_s^*}\,dx
\]
on $\dot H^s(\mathbb{R}^N)$, where $(-\Delta)^{s}$ is given in \eqref{eq:1.3}. 
It is discussed in section 2 the relation between the spectral 
factional Laplacian and the fractional Laplacian defined by the Fourier transform.

We consider the minimizing problem
\[
c_\mu = \inf\big\{J_\mu(u): u\in \mathcal{N}_{\mu,\mathbb{R}^N}\big\},
\]
where $\mathcal{N}_{\mu,\Omega}$ denotes the Nehari manifold
\[
\mathcal{N}_{\mu,\Omega} = \big\{u\in H^s_0(\Omega)\setminus\{0\}
: \int_\Omega\big( |(-\Delta)^{s/2}u|^2 -\mu\frac{u^2}{|x|^{2s}}\big)\,dx 
= \int_\Omega |u|^{2_s^*}\,dx\big\}.
\]
Obviously, we have
\[
c_\mu = \inf\big\{\frac sN\int_{\mathbb{R}^N}\big( |(-\Delta)^{s/2}u|^2
 -\mu\frac{u^2}{|x|^{2s}}\big)\,dx: u\in \mathcal{N}_{\mu,\mathbb{R}^N}\big\}
\]

We may verify that $c_\mu = \frac sN \Lambda_{s,\mu}$. If $u$ is a minimizer 
of $\Lambda_{s,\mu}$, then $v = \Lambda_{s,\mu}^{\frac {N-2s}{4s}} u$ 
is a minimizer of $c_\mu$, and vice verse. 
Since minimizers of $\Lambda_{s,\mu}$ are radially symmetric, 
so are minimizers of $c_\mu$.
The following result is proved in \cite{PP}.

\begin{lemma}\label{lem:3.1}
Let $0<s<\frac N2$ and $\{u_n\}\subset \dot H^s(\mathbb{R}^N)$ be a 
bounded sequence such that
\[
\inf_{n\in\mathbb{N}}\|u_n\|_{L^{2^*_s}}\geq C>0.
\]
Then up to subsequence, there exist $\{x_n\}\subset\mathbb{R}^N$,
$\lambda_n\in(0,\infty)$ such that
\[
v_n\rightharpoonup v\not\equiv 0\quad \text{in }  \dot H^s(\mathbb{R}^N),
\]
where $v_n(x): = \lambda_n^{\frac{N-2s}2}u_n(x_n+ \lambda_n x)$.
\end{lemma}

Using Lemma \ref{lem:3.1}, we have a description of $(PS)_c$ sequences. 
By a $(PS)_c$ sequences for $I_\mu$ we mean a sequence 
$\{u_n\}\subset H^s_0(\Omega)$ such that $I_\mu(u_n) \to c$ and 
$I'_\mu(u_n) \to 0$.

\begin{proposition}\label{prop:3.1}
 Let $\mu\in (0,\mu_H)$ and $\{u_n\}\subset\mathcal{N}_{\mu,\Omega} $ 
be a sequence such that for $0<c<c_0$,
\begin{equation}\label{eq:3.1}
\lim_{n\to\infty}I_\mu(u_n) = c,\quad \lim_{n\to\infty} I'_\mu(u_n) = 0.
\end{equation}
Suppose problem \eqref{ePsm} has no nontrivial solutions. 
Then, there exist $\{\lambda_n\}\subset \mathbb{R}^N_+\}$ and 
$\{x_n\}\subset\mathbb{R}^N$ such that
\[
\lim_{n\to\infty}\lambda_n = 0,\quad 
\lim_{n\to\infty}x_n = 0\quad{\rm and}\quad 
\lim_{n\to\infty}\|u_n - u^\mu_{\lambda_n,x_n}\|_{\dot H^s} = 0,
\]
where $u^\mu$ is a minimizer of $c_\mu$.
\end{proposition}

\begin{proof}
By \eqref{eq:3.1} and the Hardy inequality, we know that $\{u_n\}$ 
is bounded in $H^s_0(\Omega)$, and
\begin{equation}\label{eq:3.2}
\lim_{n\to \infty}\|u_n\|_{L^{2^*_s}(\Omega)} = \frac{cN}s>0.
\end{equation}
Therefore, we have
\[
u_n \rightharpoonup u\quad\text{in } H^s_0(\Omega), \quad 
u_n\to u \quad\text{in } L^{2^*_s-1}(\Omega), \quad
u_n\to u \quad\text{a.e. } \Omega.
\]
By the assumption that problem \eqref{ePsm} does not have nontrivial solution, 
we have $u=0$. Extend $u_n$  to be zero outside $\Omega$, then the extension 
of $u_n$ belongs to $H^s(\mathbb{R}^N)$, see 
\cite[Theorem 7.40]{A}. By \eqref{eq:3.2} and Lemma \ref{lem:3.1}, 
there exist $\{x_n\}\subset\mathbb{R}^N$, $\lambda_n\in(0,\infty)$ such that
\[
v_n\rightharpoonup v\not\equiv 0\quad \text{in }  \dot H^s(\mathbb{R}^N),
\]
where $v_n(x): = \lambda_n^{\frac{N-2s}2}u_n(x_n+ \lambda_n x)$, and 
$u_n$ has been extended to $\mathbb{R}^N$ by setting $u_n = 0$ 
outside $\Omega$. We also have $v_n\in H^s_0(\Omega_n)$, 
where $\Omega_n=\{x\in \mathbb{R}^N: x_n+\lambda_n x\in \Omega\}$. 
Moreover, $v_n$ satisfies
\begin{equation}\label{eq:3.3}
(-\Delta)^s v_n -\mu\frac {\lambda_n^{2s} v_n}{|x_n + \lambda_n x|^{2s}}
 - v_n^{2_s^*-1}\to 0
\end{equation}
and
\begin{equation}\label{eq:3.4}
\begin{split}
&\frac 12\int_{\Omega_n}\big( |(-\Delta)^{s/2}v_n|^2 -\mu\frac{\lambda_n^{2s}v_n^2}{|x_n + \lambda_n x|^{2s}}\big)\,dx
-\frac 1{2_s^*}\int_{\Omega_n} |v_n|^{2_s^*}\,dx\\
&= \frac 12\int_\Omega\big( |(-\Delta)^{s/2}u_n|^2 -\mu\frac{u_n^2}{|x|^{2s}}\big)\,dx
-\frac 1{2_s^*}\int_\Omega |u_n|^{2_s^*}\,dx\to c\\
\end{split}
\end{equation}
as $n\to\infty$.

We may assume that $\lambda_n\to\lambda_0\geq 0$. 
If $\lambda_0>0$, since $u_n\rightharpoonup 0$ in $\dot H^s(\mathbb{R}^N)$, 
we have $v_n\rightharpoonup 0$ in $\dot H^s(\mathbb{R}^N)$, which is a contradiction.

We may assume, up to a sequence, $\frac{x_n}{\lambda_n}\to x_0\in \mathbb{R}^N$ or
 $|\frac{x_n}{\lambda_n}|\to\infty$ as $n\to\infty$.
If $|\frac{x_n}{\lambda_n}|\to\infty$ as $n\to\infty$, by \eqref{eq:3.2}, 
we see that the limit function $v$ satisfies
\begin{equation}\label{eq:3.5}
(-\Delta)^s v = v^{2_s^*-1} \quad \text{in } \mathbb{R}^N.
\end{equation}
Thus, equations \eqref{eq:3.3} and \eqref{eq:3.4} imply
\[
\begin{split}
c_0 & > c = \lim_{n\to\infty}\Big\{\frac 12\int_{\Omega_n}\big( |(-\Delta)^{s/2}v_n|^2 -\mu\frac{\lambda_n^{2s}v_n^2}{|x_n + \lambda_n x|^{2s}}\big)\,dx
-\frac 1{2_s^*}\int_{\Omega_n} |v_n|^{2_s^*}\,dx\Big\}\\
&= \frac sN\lim_{n\to\infty}\int_{\Omega_n}\big( |(-\Delta)^{s/2}v_n|^2 -\mu\frac{\lambda_n^{2s}v_n^2}{|x_n + \lambda_n x|^{2s}}\big)\,dx\\
&\geq\frac sN\int_{\Omega_n} |(-\Delta)^{s/2}v|^2\,dx\geq c_0,\\
\end{split}
\]
which is impossible.

So we have $\frac{x_n}{\lambda_n}\to x_0 \in \mathbb{R}^N$, 
which yields $\lim_{n\to\infty}|x_n| = 0$. It follows that $v$ satisfies
\[
(-\Delta)^s v -\mu\frac {v}{|x_0 + x|^{2s}} = v^{2_s^*-1} \quad 
\text{in } \mathbb{R}^N.
\]
By the translation, we may assume $x_0 = 0$. Indeed, 
let $\tilde v_n(x) = v_n(x-\frac {x_n}{\lambda_n})$. 
Then $\tilde v_n\rightharpoonup \tilde v$ in $\dot H^s(\mathbb{R}^N)$. 
By \eqref{eq:3.3},  $\tilde v_n$ satisfies
\begin{equation}\label{eq:3.5a}
(-\Delta)^s\tilde v_n -\mu\frac {\tilde v_n}{|x|^{2s}} - \tilde v_n^{2_s^*-1}\to 0
\end{equation}
as $n\to\infty$ and $\tilde v$ satisfies
\[
(-\Delta)^s\tilde v -\mu\frac {\tilde v}{|x|^{2s}} 
= \tilde v^{2_s^*-1} \quad \text{in } \mathbb{R}^N.
\]
Now, we prove that
$$
\lim_{n\to 0}\int_{\mathbb{R}^N}|(-\Delta)^{s/2}(\tilde v_n - \tilde v)|^2\,dx = 0.
$$
Suppose on the contrary that
\[
\lim_{n\to \infty}\int_{\mathbb{R}^N}|(-\Delta)^{s/2}(\tilde v_n - \tilde v)|^2\,dx > 0.
\]
Noting
\begin{align*}
&\lim_{n\to \infty}\int_{\mathbb{R}^N}|(-\Delta)^{s/2}(\tilde v_n - \tilde v)|^2\,dx
 + \int_{\mathbb{R}^N}|(-\Delta)^{s/2} \tilde v)|^2\,dx\\
&= \lim_{n\to \infty}\int_{\mathbb{R}^N}|(-\Delta)^{s/2}\tilde v_n|^2\,dx,\\
\end{align*}
and by the Br\'{e}zis-Lieb lemma \cite{BL},
\begin{gather*}
\lim_{n\to\infty}\|\tilde v_n-\tilde v\|^{2^*_s} + \|\tilde v\|^{2^*_s} 
= \lim_{n\to\infty}\|\tilde v_n\|^{2^*_s}, \\
\lim_{n\to \infty}\int_{\mathbb{R}^N}\frac{|\tilde v_n - \tilde v|^2}{|x|^{2s}}\,dx 
+ \int_{\mathbb{R}^N}\frac{|\tilde v|^2}{|x|^{2s}}\,dx
= \lim_{n\to \infty}\int_{\mathbb{R}^N}\frac{|\tilde v_n|^2}{|x|^{2s}}\,dx,
\end{gather*}
we find from \eqref{eq:3.5a} that
\begin{equation}\label{eq:3.5b}
\lim_{n\to \infty}\int_{\mathbb{R}^N}\big(|(-\Delta)^{s/2}
(\tilde v_n - \tilde v)|^2 -\mu\frac{|\tilde v_n - \tilde v|^2}{|x|^{2s}} 
- |\tilde v_n-\tilde v|^{2^*_s} \big)\,dx = 0.
\end{equation}
That is, $\tilde v_n-\tilde v$ is close to the manifold 
$\mathcal{N}_{\mu,\mathbb{R}^N}$. Let $t_n$ be such that 
$t_n(\tilde v_n -\tilde v)\in \mathcal{N}_{\mu,\mathbb{R}^N}$. 
By \eqref{eq:3.5b}, we can show that $t_n\to 1$ as $n\to\infty$. 
Hence, $J_\mu(\tilde v_n-\tilde v) - J_\mu(t_n(\tilde v_n-\tilde v))\to 0$ 
as $n\to\infty$. But $J_\mu(t_n(\tilde v_n-\tilde v))\geq c_\mu$, it yields
\[
\liminf_{n\to\infty}J_\mu(\tilde v_n-\tilde v)\geq c_\mu,
\]
and then
\[
\liminf_{n\to\infty}J_\mu(\tilde v_n)
\geq \liminf_{n\to\infty}J_\mu(\tilde v_n-\tilde v) + J_\mu(\tilde v)\geq 2c_\mu.
\]
This is a contradiction. Consequently, $\tilde v_n\to \tilde v $ 
strongly in $\dot H^s(\mathbb{R}^N)$ and $J_\mu(\tilde v) = c_\mu$.
The proof is complete.
\end{proof}

In the case $\mu = 0$, we have the following result, its proof is similar 
to that of Proposition \ref{prop:3.1}.

\begin{proposition}\label{prop:3.2} 
Let$\{u_n\}\subset\mathcal{N}_{0,\Omega} $ be a sequence such that
\begin{equation}\label{eq:3.6}
\lim_{n\to\infty}I(u_n) \leq c_0,\quad \lim_{n\to\infty} I'(u_n) = 0.
\end{equation}
Then, there exist $\{\lambda_n\}\subset \mathbb{R}^N_+\}$ and 
$\{x_n\}\subset\mathbb{R}^N$ such that
\[
\lim_{n\to\infty}\lambda_n = 0,\quad \lim_{n\to\infty}x_n = 0,\quad
\lim_{n\to\infty}\|u_n - u^0_{\lambda_n,x_n}\|_{\dot H^s} = 0,
\]
where $u^0$ is a minimizer of $c_0$.
\end{proposition}

For each set $A\subset \mathbb{R}^N$ and each point $x\in \mathbb{R}^N$, 
$d(x,A)$ denotes the distance between $x$ and $A$. 
For each $d>0$, we denote $\Omega_d = \{x\in \mathbb{R}^N: d(x,\Omega)< d\}$ 
and $\Omega_d^i = \{x\in \Omega: d(x,\partial\Omega)> d\}$. 
For subsets $A, B\subset\mathbb{R}^N$, $A\cong B$ stands for that $A$ and 
$B$ are homotopy  equivalent.

Now, we choose $d>0$ so that $\Omega_d\cong \Omega$. Let
\[
\beta(u) = \frac {\int_\Omega x|(-\Delta)^{s/2} u|^2\,dx}
{\int_\Omega |(-\Delta)^{s/2} u|^2\,dx}\quad \text{for }
 u\in H^s_0(\Omega)\setminus\{0\}.
\]

\begin{lemma}\label{lem:3.2}
There exists $\mu_0\in (0, \mu_H)$ such that for each $\mu\in(0,\mu_0)$ 
and $u\in \mathcal{N}_{\mu,\Omega}$ with $I(u)< c_0$, $\beta(u)\in \Omega_d$.
\end{lemma}

\begin{proof}
Suppose on the contrary that there exist $\mu_n\in \mathbb{R}_+$ and 
$u_n\in \mathcal{N}_{\mu,\Omega}$ such that $\lim_{n\to\infty}\mu_n = 0$, 
$I_{\mu_n}(u_n)< c_0$ and $\beta(u_n)\not\in\Omega_d$ for all $n\geq 1$. 
We may assume that $\beta(u_n)\to x_0\in\mathbb{R}^N\setminus\Omega_d$. 
Since $\mu_n\to 0$, by the Hardy inequality,
\[
\mu_n\int_\Omega \frac{u_n^2}{|x|^{2s}}\,dx \to 0
\]
as $n\to \infty$. Hence,
\begin{gather*}
\lim_{n\to\infty}\int_\Omega|(-\Delta)^{s/2}u_n|^2\,dx 
= \lim_{n\to\infty}\int_\Omega|u_n|^{2^*_s}\,dx, \\
\lim_{n\to\infty} I_{\mu_n}(u_n) \leq c_0.
\end{gather*}
By Proposition \ref{prop:3.2}, there exist sequences $\{\lambda_n\}$ and 
$\{x_n\}\subset \mathbb{R}^N$ such that
\[
\lim_{n\to\infty}\lambda_n = 0\quad\text{and}\quad 
\lim_{n\to\infty}\|u_n - u^0_{\lambda_n,x_n}\|_{\dot H^s} = 0.
\]
By the assumption, we have $\lim_{n\to\infty} x_n = x_0$. 
However, $x_0\not\in\Omega_d$, we have that
$\lim_{n\to\infty}\|u_n - u^0_{\lambda_n,x_n}\|_{\dot H^s} \neq 0$. 
This is a contradiction. The assertion follows.
\end{proof}

Now, we choose $d_1>0$ such that $\Omega\cong \Omega_{d_1}^i$. Let
\[
\lambda =\inf\{\frac \mu{|x|^{2s}}: x\in \Omega_d\}.
\]
Let $\xi\in C^\infty(\mathbb{R}_+)$ be such that $\xi(t) = 1$ 
for $t\in[0,\frac {d_1}2]$ and $\xi(t) = 0$ for $t\in [d_1, \infty)$. 
For each $(\varepsilon, z)\in \mathbb{R}_+\times\mathbb{R}^N$, we define
\[
w_{\varepsilon,z}(x) = \tau_{\varepsilon,z}\xi(x-z)U_\varepsilon(x-z)
\]
for $x\in \mathbb{R}^N$, where $U_\varepsilon$ is given in \eqref{eq:1.9} 
and $\tau_{\varepsilon,z}$ is a positive constant such that $w_{\varepsilon,z}$
satisfying
\[
\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2 
-\lambda w_{\varepsilon,z}^2\big)\,dx 
= \int_\Omega |w_{\varepsilon,z}|^{2_s^*}\,dx.
\]
It is proved in \cite{BCP} that if $0\in\Omega$,
\begin{gather*}
\|\xi U_\varepsilon\|^2_{H^s_0(\Omega)} 
= \|U_\varepsilon\|^2_{H^s_0(\Omega)} + O(\varepsilon^{N-2s}); 
\\
\|\xi U_\varepsilon\|^2_{L^2(\Omega)}
=\begin{cases} 
 C\varepsilon^{2s} + O(\varepsilon^{2s}),&\text{if } N>4s, \\
 -C\varepsilon^{2s}\log\varepsilon + O(\varepsilon^{2s}),&\text{if } N=4s, 
\end{cases} 
\\
\|\xi U_\varepsilon\|^{2^*_s-1}_{L^{2^*_s-1}(\Omega)}
\geq C\varepsilon^{\frac{N-2s}2}\quad \text{if }  N>2s.
\end{gather*}
Let
\[
Q(u) = \frac 12\int_\Omega\big( |(-\Delta)^{s/2}u|^2 -\lambda u^2\big)\,dx
-\frac 1{2_s^*}\int_\Omega |u|^{2_s^*}\,dx.
\]
Then, we may verify that
\[
Q(w_{\varepsilon,z})
=\begin{cases}
c_0 - \lambda C\varepsilon^{2s} + O(\varepsilon^{2s}),&\text{if } N>4s, \\
 c_0+C\varepsilon^{2s}log\varepsilon + O(\varepsilon^{2s}),&\text{if } N=4s, 
\end{cases} 
\]
for all $z\in \Omega_{d_1}^i$, it implies that for all $z\in \Omega_{d_1}^i$,
\begin{equation}\label{eq:3.7}
Q(w_{\varepsilon,z})<c_0
\end{equation}
if $\varepsilon> 0$ small enough.


\begin{lemma}\label{lem:3.3}
Let $\mu\in(0,\mu_0)$. Then for $\varepsilon>0$ small, there holds
\[
\sup\{I(t_{w_{\varepsilon,z},\mu}w_{\varepsilon,z}):z\in\Omega_{d_1}^i\}< c_0.
\]
\end{lemma}

\begin{proof}
Let $t = t_{w_{\varepsilon,z}}$. Since $\frac\mu{|x|^{2s}}>\lambda$ 
for $x\in\Omega$,
\begin{align*}
\frac {t^2}2\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2 
 -\lambda w_{\varepsilon,z}^2\big)\,dx
&> \frac {t^2}2\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2 
 -\mu\frac{w_{\varepsilon,z}^2}{|x|^{2s}}\big)\,dx\\
&=\frac {t^{2^*_s}}{2}\int_\Omega w_{\varepsilon,z}^{2^*_s}\,dx\\
&=\frac {t^{2^*_s}}{2}\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2 
-\lambda w_{\varepsilon,z}^2\big)\,dx.
\end{align*}
Therefore, $t<1$. It results from \eqref{eq:3.7} that
\begin{align*}
I_\mu(tw_{\varepsilon,z})
& = \frac{st^2}N\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2
  -\mu\frac{w_{\varepsilon,z}^2}{|x|^{2s}}\big)\,dx\\
&\leq \frac{s}N\int_\Omega\big( |(-\Delta)^{s/2}w_{\varepsilon,z}|^2 
 -\lambda w_{\varepsilon,z}^2\big)\,dx
<c_0.
\end{align*}
The assertion follows.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm:1.1}]
  We argue by contradiction. For fixed $\mu\in(0,\mu_0)$, assume problem 
\eqref{ePsm} has no solutions. Hence, we know from \cite{St} that there 
exists a pseudo-gradient flow 
$\eta:[0,\infty)\times \mathcal{N}_{\mu,\Omega}\to \mathcal{N}_{\mu,\Omega}$ 
associated with $I_{\mu}$, such that the function $\eta$ satisfies for 
$s, t\in \mathbb{R}_+$ with $s>t$ and $u\in\mathcal{N}_{\mu,\Omega}$ that
\begin{gather*}
I_{\mu}(\eta(s,u))< I_{\mu}(\eta(t,u)), \\
\lim_{t\to\infty}I_{\mu}(\eta(t,u))>-\infty\quad 
\text{implies that}\quad
\lim_{t\to\infty}I'_{\mu}(\eta(t,u)) = 0.
\end{gather*}
For $\varepsilon$ given in Lemma \ref{lem:3.3}, we set 
$\omega = \{t_{w_{\varepsilon,z},\mu}w_{\varepsilon,z}: z\in\omega^i_{d_1}\}$. 
By the definition of $w_{\varepsilon,z}$, we have 
$w_{\varepsilon,z}\in H^s_0(\Omega)$
for each $z\in\omega$. Therefore, $\omega\subset \mathcal{N}_{\mu,\Omega}$, 
and $\sup\{I_\mu(\eta(t,u): t\geq 0)\}$ is bounded from below for each 
$u\in\omega$. By Proposition \ref{prop:3.1}, there exist 
$\{(\varepsilon_t, z_t\}\subset \mathbb{R}_+\times\Omega$ such that 
$\lim_{t\to\infty} z_t = 0$ and
\[
\lim_{t\to\infty}\|\eta(t,u) - u^\mu_{\varepsilon_t,z_t}\|_{\dot H^s} = 0.
\]
Since $u^\mu$ is radially symmetric, we have
\[
\lim_{t\to\infty}\beta(\eta(t,u)) = 0\in\Omega
\]
for all $u\in\omega$. On the other hand, by Lemma \ref{lem:3.2},
\[
\{\beta(\eta(t,u)): u\in\omega\}\subset \Omega_d.
\]
Since $\{\beta(\eta(0,u)):u\in\omega\} = \Omega^i_{d_1}$, we see that 
$\Omega^i_{d_1}$ is contractible in $\Omega_d$.
This contradicts to the assumption that $\Omega^i_{d_1}\cong\Omega\cong\Omega_d$ 
and that $\Omega$ is not contractible. This shows that problem 
\eqref{ePsm} possesses a positive solution in $\mathcal{N}_{\mu,\Omega}$. 
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:1.2}]
 We remark that Lemma \ref{lem:3.1} can be applied in this case. 
The proof of Theorem \ref{thm:1.2} is similar to that of Theorem \ref{thm:1.1}
with minor changes, we omit the details.  
\end{proof}

\subsection*{Acknowledgments} 
This work was supported by the NNSF of China: 11271170 and 11371254,
and by  the GAN PO 555 program of Jiangxi.

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\end{document}