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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 12, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/12\hfil Semilinear elliptic problems]
{Semilinear elliptic problems involving
Hardy-Sobolev-Maz'ya potential and Hardy-Sobolev critical exponents}

\author[R.-T. Jiang, C.-L. Tang \hfil EJDE-2016/12\hfilneg]
{Rui-Ting Jiang, Chun-Lei Tang}

\address{Rui-Ting Jiang \newline
School  of  Mathematics  and  Statistics, Southwest University,
 Chongqing 400715, China}
\email{1152373321@qq.com}

\address{Chun-Lei Tang (corresponding author)\newline
School  of  Mathematics  and  Statistics, Southwest University,
Chongqing 400715, China}
\email{tangcl@swu.edu.cn, phone +86 23 68253135, fax +86 23 68253135}

\thanks{Submitted July 14, 2015. Published January 7, 2016.}
\subjclass[2010]{35B09, 35B33, 35J75}
\keywords{Hardy-Sobolev-Maz'ya potential;  Hardy-Sobolev critical exponents;
\hfill\break\indent  positive solution; Mountain Pass Lemma;
 local Palais-Smale condition}

\begin{abstract}
 In this article, we study a class of  semilinear elliptic equations
 involving  Hardy-Sobolev critical exponents and Hardy-Sobolev-Maz'ya
 potential in a bounded domain. We obtain the existence of positive
 solutions using the  Mountain Pass Lemma.
\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 results}

  The main purpose of this article is to investigate the existence of 
positive solution to the following semilinear elliptic problem with 
Dirichlet boundary value conditions
\begin{equation}\label{1.1}
\begin{gathered}
-\Delta u -\mu \frac{u}{|y|^{2}}
 =\frac{|u|^{2^{*}(s)-2}u}{|y|^{s}}+\lambda f(x,u), \quad \text{in }   \Omega, \\
u>0, \quad \text{in }    \Omega , \\
u=0, \quad  \text{on }    \partial\Omega,
\end{gathered}
\end{equation}
where $f\in C(\overline{\Omega}\times\mathbb{R}, \mathbb{R})$, $\Omega$ is a 
smooth bounded domain in $\mathbb{R}^{N}=\mathbb{R}^{k}\times\mathbb{R}^{N-k}$ 
with $N\geq3$ and $2\leq k<N$, a point $x\in\mathbb{R}^{N}$ is 
denoted as $x=(y,z)\in\mathbb{R}^{k}\times\mathbb{R}^{N-k}$ and 
$(0,z^{0})\in\Omega$, $0\leq\mu<\bar{\mu}=\frac{(k-2)^{2}}{4}$ for $k>2$, 
$\mu =0$ for $k=2$. The so-called Hardy-Sobolev critical exponents are 
denoted by $2^{*}(s)=\frac{2(N-s)}{N-2}$ where $0\leq s<2$, 
$2^{*}=2^{*}(0)=\frac{2N}{N-2}$ are the Sobolev critical exponents. 
$F(x,t)$ is a primitive function of $f(x,t)$ defined by 
$ F(x,t)=\int_0^{t}f(x,s)ds$. $H^1_0(\Omega)$ is the Sobolev
space with norm 
$$
\|u\|=\Big(\int_{\Omega}(|\nabla u|^{2}-\mu\frac{u^{2}}{|y|^{2}})dx
\Big)^{1/2},
$$ 
which is equivalent to its general norm due to the Hardy inequality
$$ 
C_{k}\int_{\mathbb{R}^{N}}\frac{u^{2}}{|y|^{2}}dx
\leq\int_{\mathbb{R}^{N}}|\nabla u|^{2}dx,
\quad \forall u\in D^{1,2}(\mathbb{R}^{N}),
$$
where $ C_{k}=\big(\frac{k-2}{2}\big)^{2}$ is the best constant and is
not attained. Let $S_{\mu}$ be the best Sobolev constant, namely
\begin{equation}\label{1.2}
S_{\mu}=\inf_{u\in D^{1,2}(\mathbb{R}^{N}\backslash(0,z^{0})), 
u\neq0} \frac{\int_{\mathbb{R}^{N}}\big(|\nabla u|^{2}
-\mu\frac{u^{2}}{|y|^{2}}\big)dx}{\big(\int_{\mathbb{R}^{N}}
\frac{|u|^{2^{*}(s)}}{|y|^{s}}dx\big)^{\frac{2}{2^{*}(s)}}}.
\end{equation}
When $k=N$, problem \eqref{1.1} becomes
\begin{equation}\label{1.4}
\begin{gathered}
-\Delta u -\mu \frac{u}{|x|^{2}}=\frac{|u|^{2^{*}(s)-2}u}{|x|^{s}}+\lambda f(x,u), 
\quad \text{in }  \Omega, \\
u>0, \quad\text{in }   \Omega , \\
u=0, \quad  \text{on }   \partial\Omega,
\end{gathered}
\end{equation}
where $0\leq\mu<\bar{\mu}=\frac{(N-2)^{2}}{4}$. There are many papers 
concerning the Dirichlet problem with critical exponents 
(see \cite{CF,GA,RW,CW,MZ}) after the work of Breizis and Nirenberg \cite{TR}.
 When $\mu=0$ and $s=0$, problem \eqref{1.4} becomes the well-known 
Brezis-Nirenberg problem, and is studied extensively in \cite{MZ}. 
When $\mu\neq0$, the problem has its singularity at 0 and attracts 
much attention. Ding and Tang in \cite{XW} studied the existence of 
positive solutions with $N\geq3$, $0\leq s<2$ and $f(x,t)$ satisfying the 
(AR) condition in the case $\lambda=1$. Kang in \cite{CKW} showed 
the existence of positive solutions replacing $f(x,u)$ by $|u|^{q-2}u$ 
with $q>2$ for $0\leq s<2$.

 When $2\leq k<N$, the  problem has stronger singularity. 
Bhakta and Sandeep  \cite{IE} studied the regularity, P.S. characterization 
and existence of solutions with $\lambda=0$. Wang and Wang in \cite{LS} 
showed that the existence of infinitely many solutions replacing $f(x,u)$ 
by $u$ for $N>6+s$. In \cite{JX}, Yang studied \eqref{1.1} with 
Neumann boundary condition and obtained the existence of positive solution 
by the Mountain Pass Lemma. In order to estimate the mountain pass energy, 
the author use the extremal function of $S_{\mu}$ (\cite{IE}), 
which is achieved when 
\[
s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu}}. 
\]
In this article, we estimate the mountain pass energy for $0\leq s<2$ 
through $\lambda$ large enough instead of the extremal function, 
which makes the results more extensive and interested. Here is our main result.

\begin{theorem} \label{thm1} 
Suppose $N\geq3$, $2\leq k<N$ and $0\leq\mu<\bar{\mu}$. 
$f\in C(\overline{\Omega}\times \mathbb{R}^{+}, \mathbb{R})$ satisfies

\begin{itemize}
\item[(A1)] $f(x,t)=0$ for $t\leq0$ uniformly for $x\in\overline{\Omega}$. 
There exists a nonempty open  subset $\Omega_0\subset\Omega$ with
$(0,z^{0})\in\Omega_0$, such that $f(x,t)\geq0$ for almost everywhere
$x\in\Omega$ and all $t>0$, and $f(x,t)>0$ for almost $x\in\Omega_0$
and all $t>0$,

\item[(A2)]  $\lim_{t\to 0^{+}}\frac{f(x,t)}{t}=0$ and 
$\lim_{t\to +\infty}\frac{f(x,t)}{t^{2^{*}(s)-1}}=0$  uniformly for
 $x\in\overline{\Omega}$.
\end{itemize}
Then there exists $\Lambda_{*}>0$, such that problem \eqref{1.1}
admits at least one positive solution for all $\lambda\geq\Lambda_{*}$.
\end{theorem}


\begin{remark} \label{rmk1} \rm 
First, there are many functions satisfying our assumptions of Theorem \ref{thm1}. 
For instance, $f(x,t)=t^{q}$ with $1<q<2^{*}(s)-1$. 
Second, it is worth noting that, when $\mu=s=0$, problem \eqref{1.1} 
reduces to the classical semilinear elliptic problem with critical exponents, 
\cite{TR} proved the existence of positive solution for $\lambda>0$ large enough. 
In this paper, we obtain the similiar result as in \cite{TR} when $2\leq k<N$ and 
$0\leq s<2$. Our results complete the existence of positive solutions for 
elliptic problem with Hardy-Sobolev critical exponents.
\end{remark}

\section{Proof of Theorem \ref{thm1}}

In this article, we  use the following notation:
\\
$\bullet$ The dual space of a Banach space $E$ will be denoted by $E'$.
\\
$\bullet$ $L^{p}(\Omega, |y|^{-s}dx)$ denotes the weighted Sobolev space.
\\
$\bullet$ $\to $ denotes the strong convergence, while
$\rightharpoonup$ denotes the weak convergence.
\\
$\bullet$ $C$, $C_{i}$ ($i =0, 1, 2 \dots$) will denote various positive constants 
and their values can vary from line to line.

To study the positive solutions of problem \eqref{1.1}, we first consider 
the existence of nontrivial solutions to the problem
\begin{equation}\label{2.1}
\begin{gathered}
-\Delta u -\mu \frac{u}{|y|^{2}}
=\frac{(u^{+})^{2^{*}(s)-1}}{|y|^{s}}+\lambda f(x,u^{+}), \quad \text{in }
   \Omega, \\
u>0, \quad  \text{in }    \Omega , \\
u=0, \quad  \text{on }   \partial\Omega,
\end{gathered}
\end{equation}
where $u^{+}=\max\{u,0\}$. The energy functional corresponding to  \eqref{2.1}
 is 
\begin{equation}\label{2.2}
I(u)=\frac{1}{2}\int_{\Omega}\Big(|\nabla u|^{2}-\mu\frac{u^{2}}{|y|^{2}}\Big)dx
-\frac{1}{2^{*}(s)}\int_{\Omega}\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}dx\\
-\lambda\int_{\Omega}F(x,u^{+})dx,
\end{equation}
for $u\in H_0^1(\Omega)$. Clearly, $I$ is well defined and is $C^1$
smooth thanks to the Hardy-Sobolev-Maz'ya inequality \cite{YJ}
\begin{equation}\label{2.3}
\Big(\int_{\mathbb{R}^{N}}\frac{|u|^{2^{*}(s)}}{|y|^{2}}dx\Big)^{\frac{2}{2^{*}(s)}}
\leq S_{\mu}^{-1}\int_{\mathbb{R}^{N}}
\Big(|\nabla u|^{2}-\mu\frac{u^{2}}{|y|^{2}}\Big)dx,
\end{equation}
where $S_{\mu}=S(\mu,N,k,s)$ is the best constant defined in \eqref{1.2}. 
By the existence of the one to one correspondence between the critical points 
of $I$ and the weak solutions of problem \eqref{2.1}, we know that if $u$ 
is a weak solution of problem \eqref{2.1}, there holds
$$
\langle I'(u),v\rangle=\int_{\Omega}
\Big((\nabla u,\nabla v)-\mu\frac{uv}{|y|^{2}}\Big)dx
-\int_{\Omega}\frac{(u^{+})^{2^{*}(s)-1}v}{|y|^{s}}dx
-\lambda\int_{\Omega}f(x,u^{+})vdx=0,
$$
for any $v\in H^1_0(\Omega)$.

Before proving Theorem \ref{thm1}, it is necessary to prove the following lemmas.

\begin{lemma} \label{lem2.1}
 Assume that conditions {\rm (A1), (A2)} hold. 
Then for $0\leq\mu<\overline{\mu}$ and $\lambda>0$, we can deduce that
\begin{itemize}
\item[(1)] there exist $r$, $\alpha>0$ such that $I(u)\geq\alpha$ when $\|u\|=r$,

\item[(2)] there exists $u_1\in H^1_0(\Omega)$ such that $\|u_1\|>r$ and
$I(u_1)<0$.
\end{itemize}
\end{lemma}

\begin{proof}
 (1) From the continuity of embeddings
 $$ 
H_0^1(\Omega)\hookrightarrow L^{q}(\Omega)(1\leq q\leq2^{*}),\quad
H_0^1(\Omega)\hookrightarrow L^{2^{*}(s)}(\Omega,|y|^{-s}dx),
$$ 
there exist $C_1>0$ and $C_2>0$ such that
 \begin{equation}\label{2.3b}
 \int_{\Omega}|u|^{q}dx\leq C_1\|u\|^{q},\quad
\int_{\Omega}\frac{|u|^{2^{*}(s)}}{|y|^{s}}dx\leq C_2\|u\|^{2^{*}(s)}.
 \end{equation}
It follows from (A2) that for $\varepsilon>0$, there exists $C_3>0$, such that
 \begin{equation}\label{2.4}
 |F(x,t)|\leq\varepsilon|t|^{2}+C_3|t|^{2^{*}(s)},
 \end{equation}
 for all $t\in \mathbb{R}^{+}$ and $x\in\overline{\Omega}$. 
Combining \eqref{2.3b} and \eqref{2.4}, one has
\begin{align*}
 I(u)&\geq \frac{1}{2}\|u\|^{2}-\frac{1}{2^{*}(s)}
 \int_{\Omega}\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}dx
 -\lambda\varepsilon\int_{\Omega}|u|^{2}dx
 -\lambda C_3\int_{\Omega}|u|^{2^{*}(s)}dx\\
&\geq \frac{1}{2}\|u\|^{2}-\frac{C_{4}}{2^{*}(s)}\|u\|^{2^{*}(s)}
 -\lambda\varepsilon C_{5}\|u\|^{2}-\lambda C_6\|u\|^{2^{*}(s)}.
\end{align*}
Therefore for a fixed $\lambda>0$, there exists $\alpha>0$ such that 
$I(u)\geq\alpha>0$ for all $\|u\|=r$, where $r>0$ small enough.

(2) Fix $v_0\in C^{\infty}_0(\Omega)\setminus\{0\}$ with $v_0\geq0$ 
in $\Omega$ and $\|v_0\|=1$. From $(A1)$, one has
\begin{align*}
 I(tv_0)
&=\frac{1}{2}t^{2}\|v_0\|^{2}-\frac{1}{2^{*}(s)}t^{2^{*}(s)}
 \int_{\Omega}\frac{(v_0^{+})^{2^{*}(s)}}{|y|^{s}}dx
 -\int_{\Omega}F(x,tv_0)dx\\
&\leq \frac{1}{2}t^{2}\|v_0\|^{2}-\frac{1}{2^{*}(s)}t^{2^{*}(s)}
 \int_{\Omega}\frac{(v_0^{+})^{2^{*}(s)}}{|y|^{s}}dx,
\end{align*}
then $\lim_{t\to +\infty}I(tv_0)\to -\infty$. Thus we can find $t'>0$ such 
that $I(t'v_0)<0$ when $\|t'v_0\|>r$. Set $u_1=t'v_0$, the proof is complete.
\end{proof}

According to the Mountain Pass Lemma  without (PS) condition (see \cite{AL}),
 there exists a sequence $\{u_n\}\subset H^1_0(\Omega)$, such that
\[
I(u_n)\to  c_{\lambda}>\alpha>0, \quad
I'(u_n)\to 0~\text{in }~\big(H^1_0(\Omega)\big)',
\]
as $n\to \infty$, where 
\begin{gather*}
c_{\lambda}=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}I(\gamma(t)),\\
\Gamma=\{\gamma\in C([0,1], H^1_0(\Omega))| \gamma(0)=0, \gamma(1)=u_1\}.
\end{gather*}

\begin{lemma} \label{lem2.2} 
Suppose that {\rm (A1)} holds, then $\lim_{\lambda\to +\infty}c_{\lambda}=0$.
\end{lemma}

\begin{proof} 
If $v_0$ is the function given in the proof of Lemma \ref{lem2.1}, then 
one deduces that $I(tv_0)>0$ for $t>0$ small enough and 
$I(tv_0)\to -\infty$ as $t\to +\infty$. Thus there exists $t_{\lambda}>0$ 
such that $I(t_{\lambda}v_0)=\max_{t\geq0}I(tv_0)$. Hence 
$$
t_{\lambda}^{2}\|v_0\|^{2}=t_{\lambda}^{2^{*}(s)}
\int_{\Omega}\frac{(v_0^{+})^{2^{*}(s)}}{|y|^{s}}dx
+\lambda\int_{\Omega}f(x,t_{\lambda}v_0^{+})t_{\lambda}v_0^{+}dx.
$$
By (A1), one has 
\[
 t_{\lambda}^{2}\|v_0\|^{2}\geq t_{\lambda}^{2^{*}(s)}
\int_{\Omega}\frac{(v_0^{+})^{2^{*}(s)}}{|y|^{s}}dx,
\]
 which implies that $\{t_{\lambda}\}$ is bounded. 
Therefore there exist a sequence $\{\lambda_n\}$ and $t_0\geq0$,
such that $\lambda_n\to +\infty$ and $t_{\lambda_n}\to  t_0$ as $n\to \infty$.
Consequently, there is $C_6>0$ such that $t_{\lambda_n}^{2}\|v_0\|^{2}\leq C_6$
for all $n\in\mathbb{N}$, and thus
\begin{equation}\label{2.5}
 \lambda_n\int_{\Omega}f(x,t_{\lambda_n}v_0^{+})t_{\lambda_n}v_0^{+}dx
+t_{\lambda_n}^{2^{*}(s)}\int_{\Omega}\frac{(v_0^{+})
^{2^{*}(s)}}{|y|^{s}}dx\leq C_6,
\end{equation}
for all $n\in \mathbb{N}$. If $t_0>0$, by $(A1)$, one obtains
$$
\lim_{n\to \infty}\lambda_n\int_{\Omega_0}f(x,t_{\lambda_n}v_0^{+})
t_{\lambda_n}v_0^{+}dx+t_{\lambda_n}^{2^{*}(s)}\int_{\Omega_0}\frac{(v_0^{+})
^{2^{*}(s)}}{|y|^{s}}dx=+\infty,$$
then
 $$
\lim_{n\to \infty}\lambda_n\int_{\Omega}f(x,t_{\lambda_n}v_0^{+})
t_{\lambda_n}v_0^{+}dx+t_{\lambda_n}^{2^{*}(s)}\int_{\Omega}\frac{(v_0^{+})
^{2^{*}(s)}}{|y|^{s}}dx=+\infty,
$$ 
which contradicts \eqref{2.5}. Thus we conclude that $t_0=0$. 
Now, let us consider the path $\gamma_{*}(t)=tu_1$ for $t\in[0,1]$, 
which belongs to $\Gamma$, then we get the following estimate
$$
0<c_{\lambda}\leq\max_{t\in[0,1]}I(\gamma_{*}(t))
\leq I(t_{\lambda}v_0)\leq\frac{1}{2}t_{\lambda}^{2}\|v_0\|^{2}.
$$
As $\lambda\to +\infty$, from the above inequality, we get $c_{\lambda}\to 0$.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1}]
 By Lemma \ref{lem2.1}, there exists a sequence $\{u_n\}\subset H^1_0(\Omega)$
 such that
$I(u_n)\to  c_{\lambda}$ and $I'(u_n)\to 0$ as $n\to \infty$.
According to Lemma \ref{lem2.2}, one deduces that there exists 
$\Lambda_{*}>0$ such that 
\[
 0<c_{\lambda}<\frac{2-s}{2(N-s)}S_{\mu}^{\frac{N-s}{2-s}}
\]
 as $\lambda\geq\Lambda_{*}$. First, we prove that  $\{u_n\}$ is bounded.
Indeed, by $(A2)$ and the boundedness of $\Omega$, for any $\varepsilon>0$, 
there exists $M>0$, such that
\begin{gather*}
|F(x,t)|\leq\varepsilon|t|^{2^{*}(s)}, \quad x\in\Omega,\; t\geq M;\quad
|F(x,t)|\leq C_1(\varepsilon),\quad t\in(0,M];\\
|f(x,t)t|\leq\varepsilon|t|^{2^{*}(s)},\quad x\in\Omega,\; t\geq M;\quad
|f(x,t)t|\leq C_2(\varepsilon),\quad t\in(0,M].
\end{gather*}
Thus, we have
\begin{equation}\label{2.16}
|F(x,t)|\leq C_1(\varepsilon)+\varepsilon|t|^{2^{*}(s)},\quad
|f(x,t)t|\leq C_2(\varepsilon)+\varepsilon|t|^{2^{*}(s)},
\end{equation}
for any $(x,t)\in\overline{\Omega}\times \mathbb{R}^{+}$. 
Then for $\xi\in(2,2^{*}(s))$, one has
\begin{equation}\label{2.17}
F(x,t)-\frac{1}{2}f(x,t)t\leq F(x,t)-\frac{1}{\xi}f(x,t)t
\leq C_3(\varepsilon)+\varepsilon|t|^{2^{*}(s)},
\end{equation}
for any $(x,t)\in\overline{\Omega}\times \mathbb{R}^{+}$. 
Set $l(x,t):=|y|^{-s}|t|^{2^{*}(s)-1}+\lambda f(x,t)$,
 we claim that $l(x,t)$ satisfies the (AR) condition. By \eqref{2.17}, 
one easily gets
\begin{align*}
\xi L(x,t)-l(x,t)t
&= \Big(\frac{\xi}{2^{*}(s)}-1\Big)|y|^{-s}|t|^{2^{*}(s)}
 +\lambda\Big(\xi F(x,t)-f(x,t)t\Big)\\
&\leq \Big(\frac{\xi}{2^{*}(s)}-1\Big)|y|^{-s}|t|^{2^{*}(s)}
 +\xi\lambda C_{4}(\varepsilon)+\xi\lambda\varepsilon|t|^{2^{*}(s)}\\
&= \Big((\frac{\xi}{2^{*}(s)}-1)|y|^{-s}+\lambda\xi\varepsilon\Big)|t|^{2^{*}(s)}
 +\xi\lambda C_{4}(\varepsilon).
\end{align*}
Thus for a fixed $\lambda>0$ and $\varepsilon>0$ sufficiently small, 
there exists $M_{\lambda}'>0$, such that
$$
0\leq\xi L(x,t)\leq l(x,t)t,\quad t\geq M'_{\lambda},
$$
where $ L(x,t)=\int^{t}_0l(x,s)ds$. Moreover, by $(A2)$, we obtain
$$
L(x,t)-\frac{1}{\xi}l(x,t)t\leq\max_{x\in\overline{\Omega},
0\leq t\leq M_{\lambda}'}\Big(F(x,t)-
\frac{1}{\xi}f(x,t)t\Big):= M_{\lambda}.
$$
It follows from the inequalities above that
\begin{equation}\label{2.18}
L(x,t)-\frac{1}{\xi}l(x,t)t\leq M_{\lambda},\quad\text{for all }
x\in\overline{\Omega}\setminus\{(0,z^{0})\},\; t\geq0.
\end{equation}
Then, one has
\begin{align*}
c+1+o(1)\|u_n\|
&\geq I(u_n)-\frac{1}{\xi}\langle I'(u_n),u_n \rangle \\
&\geq \big(\frac{1}{2}-\frac{1}{\xi}\big) \|u_n\|^{2}
+\Big(\frac{1}{\xi}-\frac{1}{2^{*}(s)}\Big)
 \int_{\Omega}\frac{(u_n^{+})^{2^{*}(s)}}{|y|^{s}}dx\\
&\quad -\lambda\int_{\Omega}
\Big(F(x,u_n^{+})-\frac{1}{\xi}f(x,u_n^{+})u_n^{+}\Big)dx\\
&\geq \big(\frac{1}{2}-\frac{1}{\xi}\big) \|u_n\|^{2}
 -\int_{\Omega}\Big(L(x,u_n^{+})-\frac{1}{\xi}l(x,u_n^{+})u_n^{+}\Big)dx\\
&\geq \big(\frac{1}{2}-\frac{1}{\xi}\big)\|u_n\|^{2}-M_{\lambda}|\Omega|.
\end{align*}
Thus, $\{u_n\}$ is bounded. Due to the continuity of embedding
$ H_0^1(\Omega)\hookrightarrow L^{2^{*}(s)}(\Omega)$, we have
 $\int_{\Omega}|u_n|^{2^{*}(s)}dx\leq C_7<\infty$. Up to a subsequence,
 still denoted by $\{u_n\}$, there exists $u_0\in H^1_0(\Omega)$ satisfying
\begin{equation}\label{2.37}
\begin{gathered}
u_n\rightharpoonup u_0,\quad \text{weakly in }H_0^1(\Omega),\\
u_n\to  u_0,\quad \text{strongly in }L^{p}(\Omega),\; 1\leq p<2^{*},\\
u_n(x)\to  u_0(x),\quad \text{a.e. in }\Omega,\\
u_n^{2^{*}(s)-1}\rightharpoonup u_0^{2^{*}(s)-1} ,\quad
\text{weakly in }\big(L^{2^{*}(s)}(\Omega,|y|^{-s}dx)\big)',
\end{gathered}
\end{equation}
as $n\to \infty$. By (A2), for any $\varepsilon>0$ there exists 
$a(\varepsilon)>0$ such that
$$
|F(x,t)|\leq\frac{1}{2C_7}\varepsilon|t|^{2^{*}(s)}+a(\varepsilon)\quad
\text{for }(x,t)\in\Omega\times \mathbb{R}^{+}.
$$
Set $\delta=\frac{\varepsilon}{2a(\varepsilon)}>0$. When $E\subset\Omega$,
$\operatorname{meas}E<\delta$, one gets
\begin{align*}
 \big|\int_{E}F(x,u_n^{+})dx\big|
&\leq \int_{E}|F(x,u_n^{+})|dx\\
&\leq \int_{E}a(\varepsilon)dx+\frac{1}{2C_7}\varepsilon\int_{E}|u_n|^{2^{*}(s)}dx\\
&\leq a(\varepsilon)\operatorname{meas}E+\frac{1}{2C_7}\varepsilon C_7
\leq \varepsilon.
\end{align*}
Hence $\big\{\int_{\Omega}F(x,u_n^{+})dx, n\in N\big\}$ is equi-absolutely-continuous.
 It follows from  Vitali's Convergence Theorem that
\begin{equation}\label{2.38}
\int_{\Omega}F(x,u_n^{+})dx\to \int_{\Omega}F(x,u_0^{+})dx,
\end{equation}
as $n\to \infty$. Applying the same method, one has
\begin{equation}\label{2.39}
\int_{\Omega}f(x,u_n^{+})u_ndx\to \int_{\Omega}f(x,u_0^{+})u_0dx,
\end{equation}
as $n\to \infty$.
By \eqref{2.37} and \eqref{2.39} we have
\begin{equation}\label{2.40}
\begin{aligned}
 \lim_{n\to  +\infty}\langle I'(u_n),v\rangle
&=\int_{\Omega}\Big((\nabla u_0,\nabla v)-\mu\frac{u_0v}{|y|^{2}}\Big)dx
-\int_{\Omega}\frac{(u_0^{+})^{2^{*}(s)-1}v}{|y|^{s}}dx\\
&\quad -\int_{\Omega}f(x,u_0^{+})vdx=0,
\end{aligned}
\end{equation}
for all $v\in H^1_0(\Omega)$. That is, $\langle I'(u_0),v\rangle=0$ for any 
$v\in H^1_0(\Omega)$. Then $u_0$ is a critical point of $I$, thus $u_0$ is 
a solution of problem \eqref{2.1}. Now we verify that $u_0\not\equiv0$. 
Let $v=u_0$ in \eqref{2.40}, we get
\begin{equation}\label{2.41}
\|u_0\|^{2}-\int_{\Omega}\frac{(u_0^{+})^{2^{*}(s)}}{|y|^{s}}dx
-\int_{\Omega}f(x,u_0^{+})u_0dx=0.
\end{equation}
Set $w_n=u_n-u_0$, then we have
\begin{equation}\label{2.42}
\int_{\Omega}|\nabla u_n|^{2}dx
=\int_{\Omega}|\nabla u_0|^{2}dx+\int_{\Omega}|\nabla w_n|^{2}dx+o(1).
\end{equation}
And from Br\'ezis-Lieb's lemma  \cite{TR}, it follows that
\begin{gather}\label{2.43}
\int_{\Omega}\frac{u_n^{2}}{|y|^{2}}dx
=\int_{\Omega}\frac{u_0^{2}}{|y|^{2}}dx
+\int_{\Omega}\frac{w_n^{2}}{|y|^{2}}dx+o(1), \\
\label{2.44}
\int_{\Omega}\frac{(u_n^{+})^{2^{*}(s)}}{|y|^{s}}dx
=\int_{\Omega}\frac{(u_0^{+})^{2^{*}(s)}}{|y|^{s}}dx
+\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx+o(1).
\end{gather}
By \eqref{2.38} and \eqref{2.42}-\eqref{2.44}, one has
\begin{equation}\label{2.45}
I(u_n)=I(u_0)+\frac{1}{2}\|w_n\|^{2}
-\frac{1}{2^{*}(s)}\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx
=c_{\lambda}+o(1).
\end{equation}
Since $\langle I'(u_n),u_n\rangle=o(1)$, combining with \eqref{2.39} 
and \eqref{2.41}, one has
\[
\|w_n\|^{2}-\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx\\=o(1). 
\]
We may assume that $\|w_n\|^{2}\to  b$ and 
\[
\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx\to  b
\]
  as $n\to \infty$. Clearly, $b\geq0$. We now suppose that $u_0\equiv0$.
 On the one hand, if $b=0$, together with \eqref{2.45}, we get 
$c_{\lambda}=I(0)=0$, which contradicts with $c_{\lambda}>0$.
 On the other hand, if $b\neq0$, we have from the definition of $S_{\mu}$ that
\[
\|w_n\|^{2}=\int_{\Omega}\Big(|\nabla w_n|^{2}-\mu\frac{w_n^{2}}{|y|^{2}}\Big)dx
\geq S_{\mu}\Big(\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx\Big)
^{\frac{2}{2^{*}(s)}},
\]
then $b\geq S_{\mu}^{\frac{N-s}{2-s}}$, together with \eqref{2.45} we deduce
\begin{align*}
c_{\lambda}+o(1)
&= I(u_0)+\frac{1}{2}\|w_n\|^{2}
 -\frac{1}{2^{*}(s)}\int_{\Omega}\frac{(w_n^{+})^{2^{*}(s)}}{|y|^{s}}dx+o(1)\\
&\geq \frac{2-s}{2(N-s)}S_{\mu}^{\frac{N-s}{2-s}}+o(1),
\end{align*}
 which contradicts $ c_{\lambda}<\frac{2-s}{2(N-s)}S_{\mu}^{\frac{N-s}{2-s}}$. 
Therefore $u_0\not\equiv0$ and $u_0$ is a nontrivial solution of problem \eqref{2.1}.
 Then by $\langle I'(u_0),u_0^{-}\rangle$=0 where $ u_0^{-}=\min\{u_0,0\}$, 
one has $\|u_0^{-}\|=0$, which implies that $u_0\geq0$. From \eqref{2.40}, 
we get $\int_{\Omega}(\nabla u_0,\nabla v)dx\geq0$ for any $v\in H^1_0(\Omega)$,
 which means $-\triangle u_0\geq0$ in $\Omega$. By the strong maximum principle, 
we know $u_0$ is a positive solution of problem \eqref{1.1}. 
Therefore Theorem \ref{thm1} holds.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the
National Natural Science Foundation of China (No. 11471267).


\begin{thebibliography}{99}

\bibitem{CF} A. Bahri, J. M. Coron;
\emph{On a nonlinear elliptic equation involving the critical Sobolev exponents: 
the effect of the topology of the domain}, Comm. Pure Appl. Math. 41 (1988), 253--294.

\bibitem{IE} M. Bhakta, K. Sandeep;
\emph{Hardy-Sobolev-Maz'ya type equation in bounded domain}, 
J. Differential Equations. 247 (2009), 119--139.


\bibitem{TR} H. Br\'ezis, L. Nirenberg;
\emph{Positive solutions of nonlinear elliptic equations involving critical 
Sobolev exponents}, Comm. Pure. Appl. Math. 36 (1983), 437--477.

\bibitem{GA} J. Q. Chen;
\emph{Multiple positive solutions of a class of nonlinear elliptic equations}, 
J. Math. Anal. Appl. 295 (2004), 341--354.

\bibitem{XW} L. Ding, C. L. Tang;
\emph{Existence and multiplicity of solutions for semilinear elliptic equations 
with Hardy terms and Hardy-Sobolev critical exponents}, 
Appl. Math. Lett. 20 (2007), 1175--1183.

\bibitem{RW} D. S. Kang;
\emph{On the quasilinear elliptic equations with critical Sobolev-Hardy exponents
 and Hardy-terms}, Nonlinear Anal. 68 (2008), 1973--1985.

\bibitem{CKW} D. S. Kang, S. J. Peng;
\emph{Positive solutions for singular critical elliptic problems}, 
Appl. Math. Lett. 17 (2004), 411--416.

\bibitem{CW} D. S. Kang, S. J. Peng;
\emph{Solutions for semilinear elliptic problems with critical 
Sobolev-Hardy exponents and Hardy potential}, Appl. Math. Lett. 18 (2005), 1094--1100.

\bibitem{YJ} V.G. Maz'ya, Sobolev Space, Springer, Berlin, 1985.

\bibitem{AL} P. H. Rabinowitz;
\emph{Minimax Methods in Critical Point Theory with Applications 
to Differential Equations}, CBMS. 65, American Mathematical Society, 
Providence, RI, 1985.

\bibitem{MZ} M. Schechter, W. M. Zou;
\emph{On the Brezis-Nirenberg problem}, 
Arch. Ration. Mech. Anal. 197 (2010), 337--356.

\bibitem{LS} C. H. Wang, J. T. Wang;
\emph{Infinitely many solutions for Hardy-Sobolev-Maz'ya equation 
involving critical growth}, Commun. Contemp. Math. 14 (6) (2012),
 1250044, 38 pp.

\bibitem{JX} J. Yang;
\emph{Positive solutions for the Hardy-Sobolev-Maz'ya equation with 
Neumann boundary condition}, J. Math. Anal. Appl. 421 (2015), 1889--1916.

\end{thebibliography}

\end{document}