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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 05, 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 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/05\hfil Existence of solutions]
{Existence of solutions for critical H\'enon equations in hyperbolic spaces}

\author[H. He, J. Qiu \hfil EJDE-2013/05\hfilneg]
{Haiyang He, Jing Qiu}  % in alphabetical order

\address{Haiyang He \newline
College of Mathematics and Computer Science,
Key Laboratory of High Performance Computing 
and Stochastic Information Processing (Ministry of Education of China),
Hunan Normal University, Changsha, Hunan 410081, China} 
\email{hehy917@yahoo.com.cn}

\address{Jing Qiu \newline
College of Mathematics and Computer Science,
Key Laboratory of High Performance Computing 
and Stochastic Information Processing (Ministry of Education of China),
Hunan Normal University, Changsha, Hunan 410081, China}
\email{qiujing0626@163.com}

\thanks{Submitted June 13, 2012. Published January 8, 2013.}
\subjclass[2000]{58J05, 35J60}
\keywords{ H\'enon equations; mountain pass theorem;
critical growth; \hfill\break\indent hyperbolic space}

\begin{abstract}
 In this article, we use  variational methods to prove that for a
 suitable value of $\lambda$, the problem
 \[
 -\Delta_{\mathbb{B}^N}u=(d(x))^{\alpha}|u|^{2^{*}-2}u+\lambda u,
 \quad u\geq 0,\quad  u\in H_0^1(\Omega')
 \]
 possesses at least one non-trivial  solution $u$ as  $\alpha\to 0^+$,
 where $\Omega'$ is a bounded domain in Hyperbolic space $\mathbb{B}^N$,
 $d(x)=d_{\mathbb{B}^N}(0,x)$.  $\Delta_{\mathbb{B}^N}$ denotes
 the  Laplace-Beltrami  operator  on $\mathbb{B}^N$, $N\geq 4$,
 $2^*=2N/(N-2)$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction and statement of main result}

In this article, we study the existence of non-trivial solution for
the problem
\begin{equation}\label{eq:1.1}
-\Delta_{\mathbb{B}^N}u=(d(x))^{\alpha}|u|^{2^{*}-2}u+\lambda u,
 \quad u\geq 0, u\in H_0^1(\Omega')
 \end{equation}
where $N\geq 4$,
$$
\frac{N(N-2)}{4}<\lambda<\lambda_1, \quad 2^{*}=\frac{2N}{N-2},
$$ 
$d(x)=d_{\mathbb{B}^N}(0,x)$. Here $\Delta_{\mathbb{B}^N}$ denotes
the  Laplace  Beltrami  operator  on $\mathbb{B}^N$.
 We denote by $\lambda_1$ is the first eigenvalue of the 
Laplace-Beltrami operator with Dirichlet boundary conditions. 
The domain $\Omega'$ is a bounded domain with an interior sphere condition,  
$0\in \Omega'\subset\mathbb{B}^N$, $\Omega \subset B_1(0)$ and
$\bar{\Omega} \cap \partial B_1(0)\neq 0$, where
$B_1(0)\subset \mathbb{B}^N$  is the geodesic ball  with radius $1$.

 When posed in the Euclidean space $\mathbb{R}^N$, problem \eqref{eq:1.1} 
is a generalization of the celebrated Brezis-Nirenberg problem
 \begin{equation}\label{eq:1.2}
-\Delta u=|x|^{\alpha}|u|^{2^{*}-2}u+\lambda u,  \quad 
u\geq 0, u\in H_0^1(\Omega),
 \end{equation}
see \cite{AS,BN,CFP,CW,GR} for more general and recent existence results.
 In spaces of constant curvature it has been studied by 
Bandle, Brillard and Flucher \cite{BBF}.
 The special case of $S^3$ has been treated in \cite{BB}.

 When $\alpha\neq 0$ and $\lambda=0$, problem \eqref{eq:1.2}  
is known as the H\'enon equation
 \begin{equation}\label{eq:1.3}
-\Delta u=|x|^{\alpha}|u|^{p-2}u,  \quad u\geq 0, u\in H_0^1(\Omega)
 \end{equation}
  and  the study goes  to H\'enon \cite{H}, Ni \cite{N}, Smets \cite{SSW},
 Cao-Peng \cite{CP} and others.
 Attention was focused on the existence and multiplicity of nonradial 
solutions for critical, supercritical and  slightly subcritical growth, 
symmetry properties and asymptotic behavior of ground states (for
$p\to \frac{2N}{N-2}$, or $\alpha\to \infty$).
 We refer to \cite{BS1,BW1,BW2,CPY,Hi} for more information. As far as
we know, the Brezis-Nirenberg problem for the critical H\'enon equation has been
studied only in \cite{LY}, where the authors prove that there always exists
 a solution to  \eqref{eq:1.3}, provided $\alpha$ is small enough.

In the hyperbolic space, the existence of Brezis-Nirenberg problem 
for the critical  equation
\begin{equation}\label{eq:1.4}
-\Delta_{\mathbb{B}^N}u=|u|^{2^{*}-2}u+\lambda u,  \quad 
u\geq 0, u\in H_0^1(\Omega)
 \end{equation}
has been studied in \cite{S} and  the results are very similar to the results
in the Euclidean case. However, for problem \eqref{eq:1.1}, there exists 
some difference from Euclidean space.
Firstly, the weight function $d(x)$ depends
on the Riemannian distance $r$ from a pole $o$.
Secondly, there is a lack of compactness due to the fact that the Sobolev 
imbedding $H_0^1(\Omega')\hookrightarrow L^{2^*}(\Omega')$
is noncompact, so the functional of problem \eqref{eq:1.1} cannot 
satisfy the $(PS)_{c}$ condition for all  $c >0$.
In generally,  to prove the functional of problem \eqref{eq:1.1} satisfying 
the local $(PS)_{c}$ condition,
we need to use the unique positive solution of the  problem
\begin{equation}\label{eq:1.5}
 -\Delta_{\mathbb{B}^N}u=u^{2^*-1}\quad\text{in } \mathbb{B}^N.
  \end{equation}
to control the energy of the functional. However, 
Mancini  and Sandeep \cite{BS} proved that \eqref{eq:1.5} did not have any 
positive solutions.
Thirdly, when we study the critical elliptic problem
 \begin{equation}\label{eq:1.6}
 -\Delta_{\mathbb{B}^N}u=Q(x)u^{2^*-1}+\lambda u ,\;
x\in\Omega',\quad  u=0, \; x\in\partial\Omega',
  \end{equation}
it is necessary that the function $Q(x)$ have the maximum in $\Omega'$.
But the weight function $d(x)^{\alpha}$ of problem \eqref{eq:1.1} has 
 the maximum on $\partial\Omega'$.
  So we have the difficulty to control the energy.
Our main result is as follows.

\begin{theorem}\label{tm:1.1}
There exists $ \bar{\alpha}> 0$, such that when  $0<\alpha < \bar{\alpha}$,
problem \eqref{eq:1.1}  has  at  least  one   non-trivial positive solution.
\end{theorem}

The proof of this result will be given in Section 3.
In section 2, we give  some basic facts about hyperbolic space and  
prove that the functional of problem \eqref{eq:1.1} satisfies 
the local $(PS)_{c}$ condition.

\section{Preliminaries}

A hyperbolic space, denoted by $\mathbb{H}^N$, is a complete simple 
connected Riemannian  manifold which has constant sectional curvature 
equal to $-1$.  There are several models for hyperbolic space, and we 
will use the  Poincar\'e ball model 
\[
\mathbb{B}^N=\{x=(x_1, x_2,\dots, x_n)\in \mathbb{R}^N: |x|<1\}
\]
endowed with Riemannian  metric  $g_{ij}=(p(x))^2\delta_{ij}$ where 
$p(x)=\frac{2}{1-|x|^2}$.
 We denote the hyperbolic volume by
$dV_{\mathbb{B}^N}$  and is given by $dV_{\mathbb{B}^N}=(p(x))^N \,dx$. 
The hyperbolic gradient and the Laplace Beltrami operator are:
\[
\Delta_{\mathbb{B}^N}=(p(x))^{-N}\operatorname{div}((p(x))^{N-2} \nabla u)),\quad
 \nabla_{\mathbb{B}^N} u=\frac{\nabla u}{p(x)}
\]
where $\nabla$ and div denotes the Euclidean gradient and divergence 
in $\mathbb{R}^N$, respectively.

The hyperbolic distance $d_{\mathbb{B}^N}(x,y)$ between $x, y\in \mathbb{B}^N$ 
in the Poincar\'e ball model is 
\[
d_{\mathbb{B}^N}(x,y)=\operatorname{arccosh}
(1+\frac{2|x-y|^2}{(1-|x|^2)(1-|y|^2)}).
\]
From this we immediately obtain that for $x\in \mathbb{B}^N$,
\[
d(x)=d_{\mathbb{B}^N}(0,x)=\log(\frac{1+|x|}{1-|x|}).
\]
Let us denote the energy functional corresponding to \eqref{eq:1.1} by
\begin{equation}\label{eq:2.1}
I(u)= \frac{1}{2}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}u|^2
 -\lambda u^2)dV_{\mathbb{B}^N}-
\frac{1}{2^{*}}\int_{\Omega'}|d(x)|^{\alpha}(u^+)^{2^{*}}dV_{\mathbb{B}^N}
 \end{equation}
defined on $H_0^1(\Omega')$. If $\lambda<\lambda_1$, we know that
 \[
\|u\|_\lambda:=[\int_{\Omega'}(|\nabla_{\mathbb{B}^N}u|^2
-\lambda u^2)dV_{\mathbb{B}^N}]^{1/2}
\]
 is  a norm equivalent to the  $H_0^1(\Omega')$ norm, and  it is known 
that critical points of the functional $I\in C^1(H_0^1(\Omega'), \mathbb{R})$ 
correspond to solutions of  \eqref{eq:1.1}.
 If $u$ is a nontrivial solution of \eqref{eq:1.1}, we define
\[
v(x)=\Big(\frac{2}{1-|x|^2}\Big)^{(N-2)/2} u
\] 
which is a nontrivial solution of the Euclidean equation
\begin{equation}\label{eq:1.6b}
-\Delta v+\frac{N(N-2)}{4}p^2v
=(\ln\frac{1+|x|}{1-|x|})^{\alpha}|v|^{2^{*}-2}v+\lambda p^2v , \quad
 x\in \Omega;\quad v\geq 0,\;    v\in H_0^1(\Omega),
\end{equation}
where $\Omega\subset \mathbb{R}^N$ is the stereographic projection of 
$\Omega'$  into  $\mathbb{R}^N$, and
$\bar{\Omega}\cap\partial B_{\frac{e-1}{e+1}}(0)\neq\emptyset$,  
$B_{\frac{e-1}{e+1}}(0)$ is a ball in the Euclidean space.
Let us define the energy functional corresponding to \eqref{eq:1.6b} by
\begin{equation}\label{eq:2.3}
\begin{aligned}
J(v)
& =\frac{1}{2} \int_{\Omega} |\nabla v|^2-(\lambda-\frac{N(N-2)}{4})
 (\frac{2}{1-|x|^2})^2 v^2\ dx\\
& -\frac{1}{2^*}\int_{\Omega}|\ln \frac{1+|x|}{1-|x|}|^\alpha (v^+)^{2^*}\ dx.
\end{aligned}
\end{equation}
Thus for any $u\in H^1(\Omega)$ if $\tilde{u}$ is defined as 
$\tilde{u}=(\frac{2}{1-|x|^2})^{(N-2)/2} u$, then $I(u)=J(\tilde{u})$.
Moreover $\langle I'(u), v\rangle=\langle J'(\tilde{u}), \tilde{v}\rangle$ 
where $\tilde{v}$ is defined in the same way.

Now, we want to prove that the functional $I$ satisfies the $(PS)_{c}$ condition.
It is well known that the best Sobolev constant
\[
S = \inf\big\{\int_{\mathbb{R}^N}|\nabla u|^2\,dx:
u\in\mathcal{D}^{1,2}(\mathbb{R}^N),
\int_{\mathbb{R}^N}|u|^{2^*}\,dx =1 \big\}
\]
is attained by the function
\[
U(x) = \frac{[N(N-2)]^\frac{N-2}{4}}{(1+|x|^2)^{\frac{N-2}2}},
\]
which is a solution of the problem
\begin{equation}\label{eq:4.1}
-\Delta u= |u|^{2^{*}-2} u, \quad x \in  \mathbb{R}^N
\end{equation}
with $\int_{\mathbb{R}^N}|\nabla
U|^2=\int_{\mathbb{R}^N}U^{2^*}dx=S^{N/2}$.

\begin{lemma}\label{lm:2.1}
For all $c \in(0,S^{N/2}/N)$, the function $I(u)$ satisfies
the $(PS)_{c}$ condition.
\end{lemma}

\begin{proof}
Suppose $c\in(0,S^{N/2}/N),\{u_{n}\}\subset H_0^1(\Omega')$ is the $(PS)_{c}$ 
sequence of the function $I(u)$, then $I(u_{n})\to c$ as
$n\to \infty,I'(u_{n})\to 0 $ as $n \to \infty$.
We have that
\begin{align*}
c+1+\|u_{n}\|_{\mathbb{B}^N}&\geq I(u_{n})
 -\frac{1}{2^{*}}\langle I'(u_{n}), u_{n}\rangle\\
 &=\frac{1}{2}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}u_{n}|^2
 -\lambda u_{n}^2)dV_{\mathbb{B}^N}
 -\frac{1}{2^{*}}\int_{\Omega'}d(x)^{\alpha}(u_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}\\
 &\quad -\frac{1}{2^{*}}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}u_{n}|^2
 -\lambda u_{n}^2)dV_{\mathbb{B}^N}
 +\frac{1}{2^{*}}\int_{\Omega'}d(x)^{\alpha}(u_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}\\
 &=(\frac{1}{2}-\frac{1}{2^{*}})\int_{\Omega}(|\nabla_{\mathbb{B}^N}u_{n}|^2
 -\lambda u_{n}^2)dV_{\mathbb{B}^N}\\
 &=(\frac{1}{2}-\frac{1}{2^{*}})\|u_{n}\|_{\mathbb{B}^N}^2.
 \end{align*} 
It follows that $\|u_{n}\|$ is bounded in $H_0^1(\Omega')$. It implies that
 \begin{equation}\label{2.2}
  \begin{gathered}
u_n\rightharpoonup u  \quad  \text{for } x\in\  H_0^1(\Omega'),    \\
u_n\to  u  \quad   \text{for } x\in  L^{p}(\Omega'), \, 2<p<2^*   \\
u_n\to  u   \quad\text{a.e.  on } \Omega',  
\end{gathered}
\end{equation}
From $\{u_{n}^+\}$ being bounded in ${L}^{2^{*}}(\Omega')$,
it follows that $\{(u_{n}^+)^{2^{*}-1}\}$ is bounded in 
$L^{\frac{2N}{N+2}}(\Omega')$. It follows that
$\{d(x)^{\alpha}(u_{n}^+)^{2^{*}-1}\}$ is bounded in 
${L}^{\frac{2N}{N+2}}(\Omega')$
and 
\[
d(x)^{\alpha}(u_{n}^+)^{2^{*}-1}\rightharpoonup d(x)^{\alpha}(u^+)^{2^{*}-1}, 
 \quad\text{in }  L^\frac{2N}{N+2}(\Omega').
\]
So  $u$ is the solution of problem \eqref{eq:1.1} and
\[
I(u)
=\frac{1}{2}\|u\|_{\mathbb{B}^N}^2
  -\frac{1}{2^{*}}\int_{\Omega'}d(x)^{\alpha} (u^+)^{2^{*}}dV_{\mathbb{B}^N}
=\frac{1}{N}\int_{\Omega'}d(x)^{\alpha} (u^+)^{2^{*}}dV_{\mathbb{B}^N} \geq 0.
\]
 Let us define $v_{n}=u_{n}-u $. The Br\'ezis-Lieb Lemma leads to
\[
|u_{n}|_{2^{*}}^{2^{*}}=|u_{n}-u|_{2^{*}}^{2^{*}}+|u|_{2^{*}}^{2^{*}}+o(1),
\]
and 
\[
\int_{\Omega'}d(x)^{\alpha}(u_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}
=\int_{\Omega'}d(x)^{\alpha}[(u_{n}-u)^+]^{2^{*}}dV_{\mathbb{B}^N}
+\int_{\Omega'}d(x)^{\alpha}(u^+)^{2^{*}}dV_{\mathbb{B}^N}+o(1).
\]
 So we have
\begin{align*}
I(u_{n})
&=\frac{1}{2}\|u_{n}\|_{\mathbb{B}^N}^2
 -\frac{1}{2^{*}}\int_{\Omega'}d(x)^{\alpha} (u_n^+)^{2^{*}}dV_{\mathbb{B}^N}\\
&=I(u)+\frac{1}{2}\|v_{n}\|_{\mathbb{B}^N}^2
 -\frac{1}{2^{*}}\int_{\Omega}d(x)^{\alpha} (v_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}+o(1)\to c.
\end{align*}
Since $\|u_{n}\|$ is bounded in $H_0^1(\Omega')$, and 
$\langle I'(u_{n}),u_{n}\rangle\to 0$,  
we  obtain
\[
\|v_{n}\|_{\mathbb{B}^N}^2
 -\int_{\Omega'}d(x)^{\alpha}(v_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}
 =\langle I'(u_{n}),u_{n} \rangle-\langle I'(u),u \rangle \to 
 -\langle I'(u),u \rangle =0.
 \]
It implies that
\[
\|v_{n}\|_{\mathbb{B}^N}^2
 -\int_{\Omega'}d(x)^{\alpha}(v_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}\to 0,
\]
So we can assume that
\[
\|v_{n}\|_{\mathbb{B}^N}^2\to b, \quad 
 \int_{\Omega'}d(x)^{\alpha}(v_{n}^+)^{2^{*}}dV_{\mathbb{B}^N}\to b.
\]
Since
$u_{n}\to u$,  in $L^2(\Omega')$, it follows that
$v_{n}=u_{n}-u\to 0$ in $L^2(\Omega')$,
and 
\[
\int_{\Omega'}|\nabla_{\mathbb{B}^N}v_{n}|^2\ dV_{\mathbb{B}^N} \to b.
\]

We know that when we consider the above calculation in the Hyperbolic space,
the Sobolev inequality is not satisfied;
so we should transfer it into the Euclidean space. Let us define:
\[
v_{n}(x) :=p^{-\frac{N-2}{2}}w_{n}(x),  \quad  p=\frac{2}{1-|x|^2}.
\]
Then
\begin{align*}
\int_{\Omega'}|\nabla_{\mathbb{B}^N}v_{n}|^2dV_{{\mathbb{B}^N}}
&=\int_{\Omega}|\nabla(p^{-\frac{N-2}{2}}w_{n})|^2p^{N-2}dx \\
&=\int_{\Omega}|\nabla w_{n}|^2dx+\frac{N(N-2)}{4}\int_{\Omega}p^2w_{n}^2dx
\end{align*}
and 
\[
\int_{\Omega'}d(x)^\alpha (v_{n}^+)^{2^*}dV_{{\mathbb{B}^N}}
=\int_{\Omega}|\ln \frac{1+|x|}{1-|x|}|^\alpha( w_{n}^+)^{2^*}dx\to  b\quad
\text{as }   n\to \infty.
\]
From $v_{n}\to 0$ in $L^2(\Omega')$, and 
\[
2\leq p(x)\leq \frac{(e+1)^2}{2e},
\]
we have  that
$\int_{\Omega}p^2w_{n}^2dx \to 0$ in $L^2(\Omega)$.
So
\[
\int_{\Omega}|\nabla w_{n}|^2dx\to b , \quad \text{as }  n\to \infty.
\]
By the Sobolev inequality, we have
\[
\frac{\int_{\Omega}|\nabla w_{n}|^2dx}
{(\int_{\Omega}(w_{n}^+)^{2^{*}}dx)^{2/2^*}}\geq S.
\]
Thus
\[
\int_{\Omega}|\nabla w_{n}|^2dx
\geq S\Big(\int_{\Omega}(w_{n}^+)^{2^{*}}dx\Big)^{2/2^*},
\]
which implies  $b\geq S b^{2/2^*}$. Then
either $b=0$ or $b\geq S^{N/2}$.

If $b\geq S^{N/2}$, by the above, it follows that
\[
\frac{S^{N/2}}{N}\leq (\frac{1}{2}-\frac{1}{2^{*}})b\leq c<\frac{S^{N/2}}{N},
\]
we know that is a contradiction. So $b=0$, the lemma is proved .
\end{proof}

\begin{lemma}\label{lm:2.2}
There exist $ \bar{\alpha}>0,$ and a nonnegative function 
$w\in H_0^1(\Omega)\setminus \{0\}$  such that
\[
\int_{\Omega}(|\nabla w|^2+(\frac{N(N-2)}{4}-\lambda)p^2w^2)dx\big/
\Big(\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}(w^+)^{2^{*}}dx \Big)^{2/2^*}
< S,
\]
for any $0<\alpha<\bar{\alpha}$.
\end{lemma}

\begin{proof}
When $N\geq 5$, suppose $(\frac{e-1}{e+1},0,\dots,0)\in \Omega$, 
$x_0=(\frac{e-1}{e+1}-\frac{\sqrt[4]{\varepsilon}}{2},0,\dots,0)\in R^N$. 
Fix $\varphi \in C_0^{\infty}(\Omega)$, such that
\begin{equation}\label{2.3}
\varphi(x)=  \begin{cases}
1   &\text{if } x\in B_{\frac14 \sqrt[4]{\varepsilon} }(x_0),    \\
0   &\text{if } x\in  \mathbb{R}^N \setminus 
B_{\frac14 \sqrt[4]{\varepsilon} } (x_0),
\end{cases}
\end{equation}
$0\leq \varphi(x)\leq 1$, 
$|\nabla \varphi(x)|\leq \frac{c}{\sqrt[4]{\varepsilon}}$.
Let
\[
u_{\varepsilon}(x)=\varphi(x)U_{\varepsilon}(x),\quad
\tilde{u_{\varepsilon}}(x)=p^{-\frac{N-2}{2}}u_{\varepsilon}(x),
\]
where
\[
p=\frac{2}{1-|x|^2},\quad 
 U_{\varepsilon}(x)
=\frac{[N(N-2)\varepsilon^2]^{\frac{N-2}{4}}}{[\varepsilon^2+|x-x_0|^2]
^{(N-2)/2}}.
\]
First we prove that
\begin{equation}\label{eq:2.4}
\int_{\Omega}|\nabla u_{\varepsilon}|^2dx
=\int_{\mathbb{R}^N}|\nabla U_{\varepsilon}|^2dx
+o(\varepsilon^{\frac{3N-6}{4}})=S^{N/2}+o(\varepsilon^{\frac{3N-6}{4}}).
\end{equation}
 Indeed, since
\begin{align*}
\int_{\Omega'}|\nabla u_{\varepsilon}|^2dx
  &=\int_{\Omega'}|\varphi(x)\cdot \nabla U_{\varepsilon}(x)
 +\nabla \varphi(x)\cdot U_{\varepsilon}(x)|^2dx\\
  &=\int_{B_{\frac14 \sqrt[4]{\varepsilon}}(x_0)}
 |\nabla U_{\varepsilon}(x)|^2dx\\
  &\quad +\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |\varphi(x)\cdot \nabla U_{\varepsilon}(x)+\nabla \varphi(x)
 \cdot U_{\varepsilon}(x)|^2dx,
 \end{align*}
 we have
 \begin{align*}
&\Big|\int_{\Omega}|\nabla u_{\varepsilon}|^2dx
 -\int_{\mathbb{R}^N}|\nabla U_{\varepsilon}|^2dx\Big|\\
&\leq|\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |\varphi\cdot \nabla U_{\varepsilon}+\nabla\varphi\cdot U_{\varepsilon}|^2dx
\\
&\quad +\Big|\int_{\mathbb{R}^N\setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|
 \nabla U_{\varepsilon}|^2dx\Big|
\\
&\quad +\int_{\mathbb{R}^N\setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
|\nabla U_{\varepsilon}|^2dx
\\
&\leq 2\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
\setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |\nabla U_{\varepsilon}|^2dx
 +\frac{2c}{\sqrt[4]{\varepsilon}}
 \int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|U_{\varepsilon}|^2dx
\\
&\quad +\int_{\mathbb{R}^N\setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |\nabla U_{\varepsilon}|^2dx\\
&\leq c\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 \frac{\varepsilon^{N-2}|x-x_0|^2}{[\varepsilon^2+|x-x_0|^2]^N}dx
\\
  &\quad +\frac{c}{\sqrt[4]{\varepsilon}}
 \int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}\frac{\varepsilon^{N-2}}
 {[\varepsilon^2+|x-x_0|^2]^{N-2}}dx\\ 
&\quad +c\int_{\mathbb{R}^N\setminus 
 B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 \frac{\varepsilon^{N-2}|x-x_0|^2}{[\varepsilon^2+|x-x_0|^2]^N}dx\\
&\leq c\varepsilon^{\frac{3N-6}{4}}+c\varepsilon^{\frac{3N-5}{4}}
=O(\varepsilon^{\frac{3N-6}{4}}).
\end{align*}
Therefore \eqref{eq:2.4} is proved.

Now we prove that 
\begin{equation}\label{eq:2.5}
\begin{aligned}
\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}|u_{\varepsilon}|^{2^{*}}dx
&\geq\Big(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}
 {1-(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}\Big)^{\alpha}
 \int_{\Omega}|u_{\varepsilon}|^{2^{*}}dx\\
&=\Big(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}
 {1-(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}\Big)^{\alpha}
\Big[\int_{\mathbb{R}^N}|U_{\varepsilon}|^{2^{*}}dx+O(\varepsilon^{3N/4})
 \Big]\\
&=\Big(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}{1-(\frac{e-1}{e+1}
 -\sqrt[4]{\varepsilon})}\Big)^{\alpha}\Big[S^{N/2}+O(\varepsilon^{3N/4})
\Big].
\end{aligned}
 \end{equation}
Indeed, 
\begin{align*}
&\Big|\int_{\Omega}|u_{\varepsilon}|^{2^{*}}dx
 -\int_{\mathbb{R}^N}|U_{\varepsilon}|^{2^{*}}dx\Big|\\
&=\Big|\int_{B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |u_{\varepsilon}|^{2^{*}}dx
 +\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}\setminus 
 B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|u_{\varepsilon}|^{2^{*}}dx
 -\int_{\mathbb{R}^N}|U_{\varepsilon}|^{2^{*}}dx\Big|
\\
&=\Big|\int_{B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|\varphi
 \cdot U_{\varepsilon}|^{2^{*}}dx
 +\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|\varphi
 \cdot U_{\varepsilon}|^{2^{*}}dx
 -\int_{\mathbb{R}^N}|U_{\varepsilon}|^{2^{*}}dx\Big|
\\
&\leq \Big|\int_{B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|U_{\varepsilon}|^{2^{*}}dx
 +\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}\setminus 
 B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|U_{\varepsilon}|^{2^{*}}dx
 -\int_{\mathbb{R}^N}|U_{\varepsilon}|^{2^{*}}dx\Big|
\\
 &\leq \int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}|U_{\varepsilon}|^{2^{*}}dx
 +\int_{\mathbb{R}^N\setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 |U_{\varepsilon}|^{2^{*}}dx
\\
&=c\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}\setminus 
 B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}\frac{\varepsilon^N}{[\varepsilon^2
 +|x-x_0|^2]^N}dx+c\int_{\mathbb{R}^N\setminus 
B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}\frac{\varepsilon^N}{[\varepsilon^2
+|x-x_0|^2]^N}dx
\\
&=c\varepsilon^N\int_{\frac{\sqrt[4]{\varepsilon}}{4}}
^{\frac12 \sqrt[4]{\varepsilon}}\frac{r^{N-1}}{[\varepsilon^2+r^2]^N}dr
+c\varepsilon^N\int_{\frac{\sqrt[4]{\varepsilon}}{4}}^{+\infty}
\frac{r^{N-1}}{[\varepsilon^2+r^2]^N}dr
\\
&\leq c\varepsilon^N\int_{\frac{\sqrt[4]{\varepsilon}}{4}}
 ^{\frac12 \sqrt[4]{\varepsilon}}r^{-N-1}dr
 +c\varepsilon^N\int_{\frac{\sqrt[4]{\varepsilon}}{4}}^{+\infty}r^{-N-1}dr
\\
&\leq c\varepsilon^{3N/4}+c\varepsilon^{3N/4}
=O(\varepsilon^{3N/4}).
 \end{align*}


Now we estimate $\int_{\Omega}(\frac{N(N-2)}{4}-\lambda)p^2u_{\varepsilon}^2dx$. 
We claim that
\begin{equation}\label{eq:2.6}
\int_{\Omega}p^2u_{\varepsilon}^{2^{*}}dx
\geq c\varepsilon^2+O(\varepsilon^{\frac{3N-4}{4}}).
 \end{equation}
Indeed, since $p(x)=\frac{2}{1-|x|^2}\geq p(0)=2$, 
\begin{align*}
\int_{\Omega}p^2u_{\varepsilon}^2dx
&\geq 4\int_{\Omega}u_{\varepsilon}^2dx\\
&=4\int_{\Omega}\frac{\varphi^2[N(N-2)\varepsilon^2]^{(N-2)/2}}
 {[\varepsilon^2+|x-x_0|^2]^{N-2}}dx\\
&=4\int_{B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 \frac{[N(N-2)\varepsilon^2]^{(N-2)/2}}{[\varepsilon^2+|x-x_0|^2]^{N-2}}dx\\
&\quad   +4\int_{B_{\frac12 \sqrt[4]{\varepsilon}(x_0)}
 \setminus B_{\frac14 \sqrt[4]{\varepsilon}(x_0)}}
 \frac{\varphi^2[N(N-2)\varepsilon^2]^{(N-2)/2}}
 {[\varepsilon^2+|x-x_0|^2]^{N-2}}dx\\
& \geq c\varepsilon^2+O(\varepsilon^{\frac{3N-4}{4}}).
\end{align*}
By  \eqref{eq:2.4}, \eqref{eq:2.5}, \eqref{eq:2.6}, we know that
\begin{align*}
&\frac{\int_{\Omega}|\nabla u_{\varepsilon}|^2
 +(\frac{N(N-2)}{4}-\lambda)p^2u_{\varepsilon}^2dx}
 {[\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}
 |u_{\varepsilon}|^{2^{*}}dx]^{2/2^*}}\\
&\leq \frac{1}{(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}
 {1-(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})})^{2\alpha/2^*}}
  \frac{S^{N/2}
  +O(\varepsilon^{\frac{3N-6}{4}})+(\frac{N(N-2)}{4}
 -\lambda)(c\varepsilon^2+O(\varepsilon^{\frac{3N-4}{4}}))}{[S^{N/2}
  +O(\varepsilon^{3N/4})]^{2/2^*}} 
\\ 
&=\frac{1}{(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}
 {1-(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})})^{2\alpha/2^*}}S(\varepsilon).
\end{align*}
Since 
\[
\Big(\ln\frac{1+(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}
{1-(\frac{e-1}{e+1}-\sqrt[4]{\varepsilon})}\Big)\to 1\quad\text{as }
\alpha\to  0^+,
\]
 there exists $ \varepsilon_0>0$(small enough), such that
 $S(\varepsilon_0)<S$. The case $N\geq 5$ is proved.
\smallskip

When $N=4$, let 
$x_0=(\frac{e-1}{e+1}-2\beta,0,\dots,0)\in 
\mathbb{R}^N,\varphi(x)\in C_0^{\infty}(\Omega)$,
\begin{equation}\label{2.7}
\varphi(x)=  \begin{cases}
1      &\text{if } x\in\  B_{\beta}(x_0),    \\
0      &\text{if } x\in\  \mathbb{R}^N\setminus B_{2\beta}(x_0),
\end{cases}
\end{equation}
for all $x\in \mathbb{R}^N$, $\beta>0$, 
$0\leq\varphi(x)\leq 1,|\nabla\varphi(x)|\leq c$,
and let
\begin{align*}
\tilde{u}_{\varepsilon}(x)
&=p^{-\frac{N-2}{2}} \varphi(x)
\frac{[N(N-2)\varepsilon^2]^{\frac{N-2}{4}}}{[\varepsilon^2
 +|x-x_0|^2]^{(N-2)/2}}\\
&=p^{-\frac{N-2}{2}} u_{\varepsilon}(x)\\
&=\frac{1}{p} u_{\varepsilon}(x)=\frac{1-|x|^2}{2}u_{\varepsilon}(x),
\end{align*}
that is to prove that
\[
\frac{\int_{\Omega}(|\nabla u_{\varepsilon}|^2
+(2-\lambda)p^2u_{\varepsilon}^2)dx}
{[\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}|u_{\varepsilon}|^{2^{*}}dx
]^{2/2^*}}<S.
\]
Similarly, we can prove that
\begin{itemize}
\item[(i)] 
\[
\int_{\Omega}|\nabla u_{\varepsilon}|^2dx
=\int_{\mathbb{R}^N}|\nabla U_{\varepsilon}|^2dx+O(\varepsilon^2)
=S^2+O(\varepsilon^2),
\]

\item[(ii)] 
\begin{align*}
&\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}|u_{\varepsilon}|^{2^{*}}dx\\
&\geq
\Big(\ln\frac{1+(\frac{e-1}{e+1}-4\beta)}{1-(\frac{e-1}{e+1}-4\beta)}
\Big)^{\alpha}\int_{\Omega}|u_{\varepsilon}|^{2^{*}}dx\\
&=\Big(\ln\frac{1+(\frac{e-1}{e+1}-4\beta)}{1-(\frac{e-1}{e+1}-4\beta)}
 \Big)^{\alpha}\int_{\mathbb{R}^N}| U_{\varepsilon}|^{2^{*}}dx
 +O(\varepsilon^{4})=S^2+O(\varepsilon^{4}),
\end{align*}

\item[(iii)] 
\[
\int_{\Omega}p^2u_{\varepsilon}^2dx\geq c\varepsilon^2|
\ln\varepsilon|+O(\varepsilon^2).
\]
Thus,
\begin{align*}
&\frac{\int_{\Omega}(|\nabla u_{\varepsilon}|^2
 +(2-\lambda)p^2u_{\varepsilon}^2)dx}
{[\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}
 |u_{\varepsilon}|^{2^{*}}dx]^{2/2^*}}\\
&<\frac{1}{\Big(\ln\frac{1+(\frac{e-1}{e+1}-4\beta)}{1-(\frac{e-1}{e+1}
-4\beta)}\Big)^{2\alpha/2^*}}\frac{S^2+O(\varepsilon^2)
+(2-\lambda)(c\varepsilon^2|\ln\varepsilon|+O(\varepsilon^2))}
{(S^2+O(\varepsilon^{4}))^{2/2^*}}\\
&=\frac{1}{\Big(\ln\frac{1+(\frac{e-1}{e+1}-4\beta)}
{1-(\frac{e-1}{e+1}-4\beta)}\Big)^{2\alpha/2^*}}S(\varepsilon).
\end{align*}
Since $(\ln\frac{1+(\frac{e-1}{e+1}-4\beta)}
{1-(\frac{e-1}{e+1}-4\beta)})^{\alpha}\to 1$ as
$\alpha\to 0^+$, then there exists $ \varepsilon_0>0$, such that 
$S(\varepsilon_0)<S$.
\end{itemize}
Hence,  there exists $ \bar{\alpha}>0$, such that when $0<\alpha<\bar{\alpha}$,
\[
\frac{\int_{\Omega}(|\nabla u_{\varepsilon}|^2
 +(\frac{N(N-2)}{4}-\lambda)p^2u_{\varepsilon}^2)dx}
{[\int_{\Omega}(\ln\frac{1+|x|}{1-|x|})^{\alpha}
|u_{\varepsilon}|^{2^{*}}dx]^{2/2^*}}<S.
\]
It implies that  in the hyperbolic space, we have
\[
\frac{\int_{\Omega'}(|\nabla_{\mathbb{B}^N}
\tilde{u}_{\varepsilon}|^2-\lambda\tilde{u}_{\varepsilon}^2)dV_{\mathbb{B}^N}}
{[\int_{\Omega'}d(x)^{\alpha}|\tilde{u}_{\varepsilon}|^{2^{*}}
dV_{\mathbb{B}^N}]^{2/2^*}}<S. \qedhere
\]
\end{proof}

\section{Proof of main results}

\begin{proof}[Proof of Theorem \ref{tm:1.1}]
Since the solution of \eqref{eq:1.1} is the critical point of the function $I$,
it suffices to apply the Mountain Pass theorem with a value 
$c<\frac{1}{N}S^{N/2}$.

By the Lemma  \ref{lm:2.1},  we know if  
$0<\alpha<\bar{\alpha},\exists\tilde{u}_{\varepsilon}(x)\in H_0^1(\Omega')
\setminus \{0\}$ such that
\[
\frac{\|\tilde{u}_{\varepsilon}\|_{\mathbb{B}^N}^2}{[\int_{\Omega'}
|x|^{\alpha}|\tilde{u}_{\varepsilon}|^{2^{*}}dV_{\mathbb{B}^N}]^{2/2^*}}<S.
\]
So let $v(x)=\tilde{u}_{\varepsilon}(x)$,  then
\begin{align*}
0<\max_{t\geq0}I(tv)
&= \max_{t\geq 0}[\frac{1}{2}\int_{\Omega}
 [|\nabla_{\mathbb{B}^N}(tv)|^2-\lambda(tv)^2]dV_{\mathbb{B}^N}
 -\frac{1}{2^{*}}\int_{\Omega}d(x)^{\alpha}|tv|^{2^{*}}dV_{\mathbb{B}^N}]\\
&=(\frac{1}{2}-\frac{1}{2^{*}})[\int_{\Omega}(|\nabla_{\mathbb{B}^N}v|^2
 -\lambda v^2)dV_{\mathbb{B}^N}/(\int_{\Omega}d(x)^{\alpha}|v|^{2^{*}}
 dV_{\mathbb{B}^N})]^{N/2}\\
&<\frac{1}{N}S^{N/2}.
\end{align*}
Since
\begin{align*}
I(u)
&\geq \frac{1}{2}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}u|^2
 -\lambda u^2)\ dV_{\mathbb{B}^N}-
\frac {1}{2^{*}}\int_{\Omega'}|u|^{2^{*}}\  dV_{\mathbb{B}^N} \\
&\geq \frac{1}{2}\|u\|_{\mathbb{B}^N}^2
-\frac {1}{2^{*}S(\mathbb{B}^N)^\frac{2^*}{2}} 
 \int_{\Omega'}|\nabla_{\mathbb{B}^N}u|^2\ dV_{\mathbb{B}^N},
\end{align*}
then there exists $r>0$ such that
\[
b=\inf_{\|u\|_{\mathbb{B}^N}=r} I(u)>0=I(u).
\]
For
\begin{align*}
I(t_0v)
&=\frac{1}{2}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}(t_0v)|^2
 -\lambda (t_0v)^2)dV_{\mathbb{B}^N}-\frac{1}{2^{*}}
 \int_{\Omega'}d(x)^{\alpha}|t_0v|^{2^{*}}dV_{\mathbb{B}^N}
\\
 &=\frac{t_0^2}{2}\int_{\Omega'}(|\nabla_{\mathbb{B}^N}v|^2-\lambda v^2)
 dV_{\mathbb{B}^N}-\frac{t_0^{2^{*}}}{2^{*}}\int_{\Omega'}
 d(x)^{\alpha}|v|^{2^{*}}dV_{\mathbb{B}^N}
\\
 &= \frac{t_0^2}{2}\|v\|_{\mathbb{B}^N}^2
 -\frac{t_0^{2^{*}}}{2^{*}}\int_{\Omega'}d(x)^{\alpha}|v|^{2^{*}}
dV_{\mathbb{B}^N}
 \to  -\infty\quad \text{as }t_0\to  +\infty,
\end{align*}
then there exists $t_0>0$, such that  when 
$\|t_0v\|_{\mathbb{B}^N}>r$, we have  $I(t_0v)<0$.
 Thus  
\[
\max_{t\in [0,1]} I(t\cdot(t_0v))<\frac{S^{N/2}}{N}.
\]
From Lemma \ref{lm:2.1}, Lemma \ref{lm:2.2} and the Mountain Pass theorem, 
we know that $I$ has a critical value and problem \eqref{eq:1.1}
has a  nontrivial solution $u$. Multiplying the equation by $u^{-}$ 
and integrating, we find $u^{-}=0$, and $u$ is a solution of  \eqref{eq:1.1}.
\end{proof}

\subsection*{Acknowledgments}
This work was supported  by grants 11201140 from the
National Natural Sciences Foundations of China,
and 212120 from the Key Project of Chinese Ministry of Education,
China. 

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\end{document}