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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 172, pp. 1--27.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/172\hfil Nonhomogeneous Choquard equations]
{Multiple solutions for nonhomogeneous Choquard equations}

\author[L. Wang \hfil EJDE-2018/172\hfilneg]
{Lixia Wang}

\address{Lixia Wang \newline
School of Sciences, Tianjin Chengjian University,
Tianjin 300384, China. \newline
Center for Applied Mathematics,
Tianjin University, Tianjin 300072, China}
\email{wanglixia0311@126.com}

\dedicatory{Communicated by Marco Squassina}

\thanks{Submitted August 2, 2017. Published October 17, 2018.}
\subjclass[2010]{35J20, 35J60}
\keywords{Choquard equation; nonhomogeneous; critical exponent}

\begin{abstract}
 In this article, we consider the multiple solutions for the
 nonhomogeneous Choquard equations
 $$
 - \Delta u +u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u+h(x),
 \quad x\in \mathbb{R}^N,
 $$
 and
 $$
 - \Delta u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{2^{\ast}_{\alpha}}
 \Big)|u|^{2^{\ast}_{\alpha}-2}u+h(x), \quad x\in \mathbb{R}^N,
 $$
 where $N\geq 3$, $0<\alpha<N$,
 $2-\frac{\alpha}{N}<p<2^{\ast}_{\alpha}=\frac{2N-\alpha}{N-2}$.
 Under suitable assumptions on $h$, we obtain at least two solutions
 on the subcritical case
 $2-\frac{\alpha}{N}<p<2^{\ast}_{\alpha}$
 and on the critical case $p=2^{\ast}_{\alpha}$.
\end{abstract}

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


\section{Introduction and main results}

 In this article, we consider the nonhomogeneous Choquard equation
 \begin{equation} \label{eq1}
- \Delta u +u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u+h(x), \quad
 x\in \mathbb{R}^N,
 \end{equation}
 where $N\geq 3$, $0<\alpha<N$ and
 $2-\frac{\alpha}{N}<p\leq2^{\ast}_{\alpha}=\frac{2N-\alpha}{N-2}$.

 A special case of \eqref{eq1} is the Choquard equation
\[
- \Delta u +u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{2}\Big)u, \quad
x\in \mathbb{R}^N,
\]
which was proposed by Choquard in 1976, can be described as an approximation
to Hartree-Fock theory of a one-component plasma \cite{Lieb1,Lieb2}.
It was also proposed by Moroz, Penrose and Tod \cite{Moroz4} as a model for
the self-gravitational collapse of a quantum mechanical wave function.
In this context, Choquard equation is usually called the nonlinear
Schr\"odinger-Newton equation.
For more details on the physical aspects of the problem we refer the
readers to \cite{Choquard,Cingolani2,Cingolani3,Cingolani1,Lions,Moroz4,Pekar}
and the references therein.


Recently, the nonlinear Choquard equations has been widely studied.
 When $h\equiv 0$, the existence and multiplicity
results of system \eqref{eq1} have been discussed in many papers.
Take for instance, Lieb \cite{Lieb1} proved the existence and uniqueness of the
 ground state to \eqref{eq1} by using symmetric decreasing rearrangement
inequalities. Later, Lions \cite{Lions} showed the existence of infinitely
many radially symmetric solutions to \eqref{eq1}.
Gao and Yang \cite{GaoYang} established some existence results for
the Brezis-Nirenberg type problem for the nonlinear Choquard equation with
critical exponent. Further results for related problems may be
found in \cite{Alves3,Alves4,Alves1, Alves2,Clapp,GaoYang2,Ma,Moroz1,Moroz3}
and the references therein.

Next, we consider the nonhomogeneous case, that is $h\not\equiv 0$.
In \cite{Xie}, Xie, Xiao and Wang proved the following Choquard equation
\[
- \Delta u +V(x)u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u+h(x), \quad
x\in \mathbb{R}^N,
\]
 had two nontrivial solutions if $2-\frac{\alpha}{N}<p<\frac{2N-\alpha}{N-2}$
satisfies the compactness condition:
\begin{itemize}
\item[(A1)] $V\in C(\mathbb{R}^N,\mathbb{R^+})$ is coercive, that is
$\lim_{|x|\to +\infty}V(x)=+\infty$.
\end{itemize}
Zhang, Xu and Zhang \cite{Zhang} also considered the bound and ground states
for nonhomogeneous Choquard equations under the condition
\begin{itemize}
\item[(A2)] $\inf_{\mathbb{R}^N} V>0$, and there exists a constant $r>0$
such that, for any $M>0$,
$\operatorname{meas}\{x\in \mathbb{R}^N,|x-y|\leq r, V(x)\leq M\}\to 0$
 as $|y|\to \infty$, where meas stands for the Lebesgue measure.
\end{itemize}
Under condition(A1) or (A2), they define a new Hilbert space
\[
E:=\big\{u\in H^1(\mathbb{R}^N): \int_{\mathbb{R}^N}|\nabla u|^2+V(x)u^2dx
<\infty\big\}
\]
with the inner product
\[
\langle u,v\rangle_E
= \int_{\mathbb{R}^N}(\nabla u\cdot \nabla v+V(x)uv)dx
\]
and the norm $\|u\|_E=\langle u,u\rangle^{1/2}$. Obviously, the embedding
$E\hookrightarrow L^s(\mathbb{R}^N)$ is continuous, for any $s\in [2,2^\ast]$.
Consequently, for each $s\in [2,2^\ast]$, there exists a constant $d_s>0$ such that
\begin{align}
\label{eq2}
|u|_s\leq d_s \|u\|_E, \quad \forall u\in E.
\end{align}
Furthermore, it follows from condition(A1) (or (A2)) that the embedding
$E \hookrightarrow L^s(\mathbb{R}^N)$ are compact for any
$s\in [2,2^{\ast})$(See \cite{Bartsch}).
Other related results about nonhomogeneous equations can be found
in \cite{Cerami,Chen,Ding,Du,Jiang,Qi,Salvatore,Shen2,Tarantello,Zhangxu}
and the references therein.

Motivated by the works above, in this paper we study the existence of
multiple solutions to the nonhomogenous Choquard equation with the critical exponent
\begin{equation} \label{eq3}
- \Delta u =\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{2^{\ast}_{\alpha}}\Big)
|u|^{2^{\ast}_{\alpha}-2}u+h(x), \quad x\in \mathbb{R}^N
\end{equation}
 and the subcritical exponent
\begin{equation} \label{eq4}
- \Delta u +u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u+h(x), \quad
x\in \mathbb{R}^N,
\end{equation}
where $N\geq 3$, $0<\alpha<N$ and $2-\frac{\alpha}{N}<p<2^{\ast}_{\alpha}$.

Before giving our main results, we give some notation.
Let $H^1(\mathbb{R}^N)$ be the usual Sobolev space endowed with the standard
scalar and norm
$$
(u,v)= \int_{\mathbb{R}^N} (\nabla u \nabla v+uv)\,dx, \quad
 \|u\|^2=\int_{\mathbb{R}^N} (|\nabla u|^2+ |u|^2)dx.
$$
$D^{1,2}(\mathbb{R}^N)$ is the completion of $C_0^\infty (\mathbb{R}^N)$
with respect to the norm
\[
\|u\|_D^2:= \|u\|_{D^{1,2}(\mathbb{R}^N)}^2=\int_{\mathbb{R}^N} |\nabla u|^2\,dx.
\]
The norm on $L^s=L^s(\mathbb{R}^N)$ with $1<s<\infty $ is given by $|u|^s_s=\int_{\mathbb{R}^N}|u|^sdx$.

We use the following assumptions:
\begin{itemize}
\item[(A3)] $\|h\|_{H^{-1}}< C_{2^{\ast}_{\alpha}}
S^{2^{\ast}_{\alpha}/(2(2^{\ast}_{\alpha}-1)})_{H,L}$,
where $H^{-1}$ is the dual space of $D^{1,2}(\mathbb{R}^N)$, $S_{H,L}$
is the best constant defined by
 $$
 S_{H,L}=\inf_{u\in D^{1,2}(\mathbb{R}^N)\setminus\{0\}}\frac{\int_{\mathbb{R}^N}|\nabla u|^2dx}{(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy)^{\frac{N-2}{2N-\alpha}}}
 $$
 and
 $$
 C_{2^{\ast}_{\alpha}}=\frac{2(2^{\ast}_{\alpha}-1)}
{(2\cdot 2^{\ast}_{\alpha}-1)^{\frac{2\cdot 2^{\ast}_{\alpha}-1}
{2\cdot 2^{\ast}_{\alpha}-2}}};
 $$
\\

\item[(A4)] $h\in L^{\frac{2Np}{2N(p-1)+\alpha}}(\mathbb{R}^N)$, $h(x)\geq 0$
and $h\not\equiv 0$;

\item[(A5)] $|h|_{\frac{2Np}{2N(p-1)+\alpha}}<\varepsilon
=\varepsilon(N,p,\alpha,d_{\frac{2Np}{2N-\alpha}})$,
 where $\varepsilon $ is a positive constant,
$d_{\frac{2Np}{2N-\alpha}}$ is defined in Lemma \ref{lem4.1} below.
\end{itemize}

To our best knowledge, in the nonhomogeneous case, this is the first result
involving critical exponent, so that we think this type of problem is
worth to consider. We mentioned here that the basic idea of this paper
follows from that of \cite{Tarantello}. Our main results read as follows:

\begin{theorem} \label{thm1.1}
Assume $h\not\equiv 0$ and {\rm (A3)} hold. Then \eqref{eq3}
 has at least two solutions. One of which is a local minimum solution with
the ground state energy, and the other one has the energy which is
strictly bigger than the least energy. If additionally, we assume $h>0$ holds,
 then the two solutions are positive.
\end{theorem}

\begin{theorem} \label{thm1.2}
Assume $h\not\equiv 0$, {\rm (A4)} and {\rm (A5)} hold.
Then \eqref{eq4} has a local minimum solution with the ground state energy.
If additionally, if $h>0$, then \eqref{eq4} has at least two positive solutions.
\end{theorem}


\begin{remark} \label{rmk1} \rm
 There is a standard method for obtaining two solutions of a nonhomogeneous system.
Usually it is not difficult to obtain a negative local minimum and a positive
 mountain-pass value of the energy functional. But because of the lack of the
compactness of the embedding
$H^1(\mathbb{R}^N)\hookrightarrow L^p(\mathbb{R}^N), p\in(2,2^{\ast})$,
the Palais-Smale condition no longer holds. Especially, many authors avoid
the lack of compactness by some coercive assumptions on the potential or
by restricting the problem to the radially symmetric subspace of
$H^1(\mathbb{R}^N)$. But in this paper, these methods are not adopted.
To overcome this difficulty, we use the Brezis-Nirenberg method
\cite{Brezis1,Brezis2}, which preserve the compactness except some fixed
bad energy level. And then by estimating the asymptotic behavior of the
local minimum solution, we obtain the second solution.
\end{remark}

Throughout this paper, the letters $C_0, d, c_i$, $i=1, 2, 3\ldots$ will be used
to denote various positive constants which may vary from line to line and are
not essential to the problem.
We denote by $\rightharpoonup$ weak convergence and by $\to$ strong convergence.
Also if we take a subsequence of a sequence $\{u_n\}$, we shall denote
it again $\{u_n\}$.

This article is organized as follows. In Section 2, we introduce the
variational setting for the problem and give some related preliminaries.
In Section 3, we manage to give the existence of the solutions for the
critical case. In Section 4, we give the proof of Theorem \ref{thm1.2}.

\section{Variational setting and compactness condition}

First we give the well-known Hardy-Littlewood-Sobolev inequality.


\begin{lemma}[Hardy-Littlewood-Sobolev inequality \cite{Lieb2}] \label{lem2.1}
 Assume $f\in L^p(\mathbb{R}^N)$ and $g\in L^q(\mathbb{R}^N)$. Then
 \begin{align}
 \label{eq5}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|f(x)||g(y)|}{|x-y|^{\alpha}}\,dx\,dy
\leq C(p,q,\alpha)|f|_p|g|_q,
\end{align}
where $1<p,q<\infty$, $0<\alpha<N$, $\frac{1}{p}+\frac{1}{q}+\frac{\alpha}{N}=2$.

If $p=q=2^{\ast}_{\alpha}$, then
$$
C(p,q,\alpha)=C(N,\alpha)=\pi^{\alpha/2}
\frac{\Gamma(\frac{N-\alpha}{2})}{\Gamma(N-\frac{\alpha}{2})}
\Big\{\frac{\Gamma(\frac{N}{2})}{\Gamma(N)}\Big\}^{\frac{\alpha}{N}-1}.
$$
The equality in \eqref{eq5} holds if and only if $f$ is a constant times
$g$ and
$$
g(x)=\frac{A_1}{(C+|x-a|^2)^{(2N-\alpha)/2}}
$$
for some $A_1\in \mathbb{C}$, $0\neq C\in \mathbb{R}$ and $a\in \mathbb{R}^N$.
\end{lemma}

By the Hardy-Littlewood-Sobolev inequality, the integral
\[
B(u)=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{|u(x)|^{p}|u(y)|^{q}}{|x-y|^{\alpha}}\,dx\,dy
\]
is well defined if $|u|^p\in L^s(\mathbb{R}^N)$ for some $s>1$ satisfying
\[
\frac{2}{s}+\frac{\alpha}{N}=2.
\]
Therefore, for $u\in H^1(\mathbb{R}^N)$, by Sobolev embedding Theorem, we know that
\[
2\leq sp\leq \frac{2N}{N-2};
\]
that is
\[
\frac{2N-\alpha}{N}\leq p\leq \frac{2N-\alpha}{N-2}.
\]
Thus, $\frac{2N-\alpha}{N}$ is called the lower critical exponent and
$2^{\ast}_{\alpha}=\frac{2N-\alpha}{N-2}$ is the upper critical exponent
in the sense of the Hardy-Littlewood-Sobolev inequality.

 For all $u\in D^{1,2}(\mathbb{R^N})$, by the Hardy-Littlewood-Sobolev inequality,
we have
\[
\Big(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}
\,dx\,dy\Big)^{\frac{N-2}{2N-\alpha}}
\leq C(N,\alpha)^{\frac{N-2}{2N-\alpha}}|u|^2_{2^{\ast}}\,,
\]
$C(N,\alpha)$ is defined in Lemma \ref{lem2.1}.
 We use $S_{H,L}$ to denote best constant defined in (A3).


\begin{lemma}[\cite{GaoYang}] \label{lem2.2}
 The constant $S_{H,L}$ defined in {\rm (A3)} is achieved if and only if
\[
U(x)=C\Big(\frac{b}{b^2+|x-a|^2}\Big)^{N-2/2},
\]
where $C>0$ is a fixed constant, $a\in \mathbb{R}^N$ and $b\in (0 ,+\infty)$
are parameters. We can also obtain
\[
S_{H,L}=\frac{S}{C(N,\alpha)^{N-2/2N-\alpha}},
\]
where $S$ is the best Sobolev constant.
In particular, let
\[
U(x)=\frac{[N(N-2)]^{\frac{N-2}{4}}}{(1+|x|^2)^{\frac{N-2}{2}}}
\]
be a minimizer for $S$, $S$ is the best Sobolev constant, then
 \[
W(x)=S^{\frac{(N-\alpha)(2-N)}{4(N-\alpha+2)}}
C(N,\alpha)^{\frac{2-N}{2(N-\alpha+2)}}U(x)
\]
is the unique minimizer for $S_{H,L}$ and satisfies
\[
-\Delta u=\int_{\mathbb{R}^N}\frac{|u(y)|^{2^{\ast}_{\alpha}}}
{|x-y|^{\alpha}}dy|u|^{2^{\ast}_{\alpha}-2}u \quad in \quad \mathbb{R}^N.
\]
Moreover,
\[
\|W\|_D=\int_{\mathbb{R}^N}\frac{|W(x)|^{2^{\ast}_{\alpha}}
|W(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy
=S^{\frac{2N-\alpha}{N-\alpha+2}}_{H,L}.
\]
\end{lemma}

To prove the problem by variational methods, we define the energy functional
associated with \eqref{eq3} by
\begin{equation} \label{eq6}
I(u)=\frac{1}{2}\int_{\mathbb{R}^N} |\nabla u|^2\,dx-
\frac{1}{2\cdot 2^{\ast}_{\alpha}}\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N} \frac{|u(x)|^{2^{\ast}_{\alpha}}
|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy
-\int_{\mathbb{R}^N}h(x)u\,dx,
\end{equation}
for $u\in D^{1,2}(\mathbb{R}^N)$.

By the Hardy-Littlewood-Sobolev inequality, we know that
$I\in C^1(D^{1,2}(\mathbb{R}^N),\mathbb{R})$ and
\begin{equation} \label{eq7}
\begin{aligned}
\langle I'(u),v\rangle
&= \int_{\mathbb{R}^N} |\nabla u||\nabla v|\,dx
 -\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}-2}u(y)v(y)}
 {|x-y|^{\alpha}}\,dx\,dy \\
&\quad -\int_{\mathbb{R}^N}h(x)v\,dx
\end{aligned}
\end{equation}
for all $v\in C^{\infty}_0(\mathbb{R}^N)$. And so $u$ is a weak solution
of \eqref{eq3} if and only if $u$ is a critical point of function $I$.
We will constrain the functional $I$ on the Nehari manifold
\[
\Lambda=\{u\in D^{1,2}(\mathbb{R}^N), \langle I'(u),u\rangle=0\}.
\]
Denote $\Phi(u)=\langle I'(u),u\rangle$, so we know that
\[
\langle I'(u),u\rangle =\|u\|^2_D-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy
-\int_{\mathbb{R}^N}h(x)u\,dx,
\]
and
\[
\langle \Phi'(u),u\rangle=2\|u\|^2_D-2\cdot2^{\ast}_{\alpha}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy
-\int_{\mathbb{R}^N}h(x)u\,dx.
\]
Notice that, when $u_0$ is a local minimum solution of $I$, it holds
\[
\langle I'(u_0),u_0\rangle=0, \quad
\langle \Phi'(u_0),u_0\rangle\geq 0,
\]
which leads us to consider the following manifolds:
\begin{gather*}
\Lambda=\{u\in D^{1,2}(\mathbb{R}^N): \langle I'(u),u\rangle=0\},\\
\Lambda^+=\{u\in \Lambda: \langle \Phi'(u),u\rangle > 0\},\\
\Lambda^-=\{u\in \Lambda: \langle \Phi'(u),u\rangle< 0\},\\
\Lambda^0=\{u\in \Lambda: \langle \Phi'(u),u\rangle = 0\}.
\end{gather*}

Obviously, only $\Lambda^0$ contains the element $0$. Furthermore,
it is easy to see that $\Lambda^0\cup \Lambda^+$ and $\Lambda^0\cup \Lambda^-$
are both closed subsets of $D^{1,2}(\mathbb{R}^N)$.

To simplify calculations, for $u\in D^{1,2}(\mathbb{R}^N)$, we denote
\begin{gather*}
A=A(u)=\|u\|^2_D,\\
B=B(u)=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
 \frac{|u(x)|^{2^{\ast}_{\alpha}}|u(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy,\\
C=C(u)=\int_{\mathbb{R}^N}h(x)u\,dx.
\end{gather*}
Define the fibering map
\begin{equation} \label{eq8}
\varphi_u(t)=I(tu)=\frac{A}{2}t^2-\frac{B}{2\cdot 2^{\ast}_{\alpha}}
t^{2\cdot 2^{\ast}_{\alpha}}-Ct, \quad t>0.
\end{equation}
Therefore,
\begin{equation} \label{eq9}
 \begin{gathered}
\varphi'_u(t)=At-Bt^{2\cdot 2^{\ast}_{\alpha}-1}-C,\\
 \varphi''_u(t)=A-(2\cdot 2^{\ast}_{\alpha}-1)Bt^{2\cdot 2^{\ast}_{\alpha}-2}.
\end{gathered}
\end{equation}
Obviously, $tu\in \Lambda$ with $t>0$ if and only if $\varphi'_u(t)=0$.
 By the sign of $\varphi''_u(t)$, the stationary points of $\varphi_u(t)$
can be classified into three types, namely local minimum, local maximum and
turning point. Moreover, the set $\Lambda$ is a natural constraint for the
functional $I$. This is means that if
 the infimum is attained by $u\in \Lambda$, then $u$ is a solution
of \eqref{eq3}. However, in our case, the global maximum point of
$\varphi_u(t)$ is not unique. This leads us to partition
the set $\Lambda$ according to the critical points of $\varphi_u(t)$.
 This kind of idea was first introduced by Tarantello in \cite{Tarantello}.
Later, many mathematicians apply this idea to study other problems;
for instance, see \cite{Benmansour,Qi,Zhang,Zhangxu} and the references therein.
Now we give some properties of $\Lambda^{\pm}$ and $\Lambda^0$.


\begin{lemma} \label{lem2.3}
 (i) Assume that $h\not\equiv0 $ for $u\in D^{1,2}(\mathbb{R}^N)\backslash \{0\}$,
there is a unique $t^-=t^-(u)>0$ such that $t^-u\in \Lambda^-$.
If additionally we assume $\int_{\mathbb{R}^N}hu\,dx>0$, then there exists a
 unique $0<t^+=t^+(u)<t^-$ satisfying $t^+u\in \Lambda^+$. Moreover,
\begin{gather*}
I(t^-u)=\max_{t\geq 0}I(tu)\quad\text{for } \int_{\mathbb{R}^N}hu\,dx\leq 0;\\
I(t^-u)=\max_{t\geq t^+}I(tu),\quad I(t^+u)=\min_{0\leq t\leq t^-}I(tu)\quad
\text{for } \int_{\mathbb{R}^N}hu\,dx>0.
\end{gather*}
\end{lemma}

\begin{proof}
 Define $\varphi_u(t)=\frac{A}{2}t^2-\frac{B}{2\cdot
2^{\ast}_{\alpha}}t^{2\cdot2^{\ast}_{\alpha}}-Ct$ for all $t>0$.
In the case $\int_{\mathbb{R}^N}hu\,dx\leq 0$, there is a unique $t^->0$
such that $\varphi'_u(t^-)=0$ and $\varphi_u''(t^-)<0$.
So that
\begin{gather*}
\langle I'(t^-u),t^-u\rangle=0, \\
\|t^-u\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u)(t^-)^{2^{\ast}_{\alpha}-2}< 0.
\end{gather*}
Thus, $t^-u\in \Lambda^-$ and $I(t^-u)=\max_{t\geq 0}I(tu)$.

In the case $\int_{\mathbb{R}^N}hu\,dx>0$,
for $t_0=t_0(u)=\big[\frac{A}{(2\cdot 2^{\ast}_{\alpha}-1)B}
\big]^{\frac{1}{2\cdot 2^{\ast}_{\alpha}-2}}>0$, we have
\begin{align*}
\max_{t\geq 0}\varphi'_u(t)
&\geq At_0-Bt_0^{2\cdot 2^{\ast}_{\alpha}-1}-C\\
&=\Big[\frac{\|u\|^2_D}{(2\cdot 2^{\ast}_{\alpha}-1)B}
 \Big]^{\frac{1}{2\cdot 2^{\ast}_{\alpha}-2}}\cdot 
 \frac{2\cdot 2^{\ast}_{\alpha}-2}{2\cdot 2^{\ast}_{\alpha}-1}\|u\|^2_D
 -\int_{\mathbb{R}^N}hu\,dx\\
&\geq S^{2^{\ast}_{\alpha}/2(2^{\ast}_{\alpha}-1)}_{H,L}
 \frac{2(2^{\ast}_{\alpha}-1)}{(2\cdot 2^{\ast}_{\alpha}-1)^{2\cdot 
 2^{\ast}_{\alpha}-1/2(2^{\ast}_{\alpha}-1)}}\|u\|_D-\|h\|_{H^{-1}}\|u\|_D\\
&=S^{2^{\ast}_{\alpha}/2(2^{\ast}_{\alpha}-1)}_{H,L}C^{\ast}_{2_\alpha}
 \|u\|_D-\|h\|_{H^{-1}}\|u\|_D
>0.
\end{align*}
From $\varphi'_u(0)=-C<0$ and $\varphi'_u(t)\to -\infty$ as $t\to +\infty$,
 we know that there exist unique $ 0<t^+<t_0<t^-$ such that
$ \varphi'_u(t^-)=\varphi'_u(t^+)=0, \varphi''_u(t^-)<0<\varphi''_u(t^+) $.
Equivalently, $t^+u\in \Lambda^+$ and $t^-u\in \Lambda^-$.
Moreover, since $\frac{d}{dt}I(tu)=\varphi'_u(t)$, we can easily see that
 $I(t^-u)=\max_{t\geq t^+}I(tu)$ and
$I(t^+u)=\min_{0\leq t\leq t^-}I(tu)$. The proof is complete. 
\end{proof}

\begin{lemma} \label{lem2.4} 
 Assume  $h\not\equiv0 $ and {\rm (A3)} hold. Then 
\begin{itemize}
\item[(i)] $\Lambda^0= \{0\}$.
\item[(ii)] $\Lambda^{\pm}\neq \emptyset$ and $ \Lambda^-$ is closed.
\end{itemize}
\end{lemma}

\begin{proof} 
(i) To prove $\Lambda^0=\{0\}$, we need to prove that, for 
$u\in D^{1,2}(\mathbb{R}^N)\setminus \{0\}$, $\varphi_u(t)$ 
has no critical point that is a turning point. Set $\|u\|_D=1$, define
\begin{equation} \label{eq10}
\kappa(t)=At-Bt^{2\cdot 2^{\ast}_{\alpha}-1}.
\end{equation}
Then $\varphi'_u(t)=\kappa(t)-C, \kappa''(t)
=-B(2\cdot 2^{\ast}_{\alpha}-1)(2\cdot 2^{\ast}_{\alpha}-2)
t^{2\cdot 2^{\ast}_{\alpha}-3}<0$ for $t>0$.
So $\kappa(t)$ is strictly concave.
If $\kappa'(t_0)=0$, $t_0=(\frac{1}{(2\cdot 2^{\ast}_{\alpha}-1)B}
)^{1/(2\cdot 2^{\ast}_{\alpha}-2)}>0$, for $2^{\ast}_{\alpha}>2-\frac{\alpha}{N}>1$.
Moreover, $\lim_{t\to 0^+}\kappa(t)=0, \lim_{t\to +\infty}\kappa(t)=-\infty$ and
$\kappa(t)>0$ for $t>0$ small. Therefore, we have that $\kappa(t)$ has a unique
global maximum point $t_0$ and
\[
\kappa(t_0)=\frac{2(2^{\ast}_{\alpha}-1)}{2\cdot2^{\ast}_{\alpha}-1}
\Big(\frac{1}{(2\cdot 2^{\ast}_{\alpha}-1)B}\Big)^{1/(2\cdot 2^{\ast}_{\alpha}-2)}
:=\kappa_0.
\]
By \eqref{eq8} and \eqref{eq9}, we infer that if $0<C<\kappa_0$,
the equation $\varphi'_u(t)=0$ has exactly two points $t_1, t_2$ satisfying
$t_1<t_0<t_2$. If $C\leq 0$, the equation $\varphi'_u(t)=0$ has one roots
$t_3>t_0$. Since $\varphi''_u(t)=A-(2\cdot 2^{\ast}_{\alpha}-1)B
t^{2\cdot 2^{\ast}_{\alpha}-2}$, it follows that
$\varphi''_u(t_1)>0, \varphi''_u(t_2)<0$ and $\varphi''_u(t_3)<0$.
It follows that $t_1u\in \Lambda^+, t_2u\in \Lambda^-$
if $0<C<\kappa_0$ and $t_3u\in \Lambda^-$ if $C\leq 0$.
Since $\{u\in D^{1,2}(\mathbb{R}^N): \|u\|_D=1, 0<C<\kappa_0\}$ and
$\{u\in D^{1,2}(\mathbb{R}^N): \|u\|_D=1, C\leq 0\}$ are nonempty, we can
infer that $\Lambda^{\pm}$ are nonempty. This implies $\Lambda^0=\{0\}$.

It is suffices to prove $\kappa_0>C$.
By (A3), Lemma \ref{lem2.3} and the definition of $S_{H,L}$ we have
\begin{align*}
\kappa_0-C
&=k(t_0)-C=At_0-Bt^{2\cdot 2^{\ast}_{\alpha}-1}_0-C\\
 &=t_0[1-t^{2\cdot 2^{\ast}_{\alpha}-2}_0B]-\int_{\mathbb{R}^N}hu\,dx\\
 &\geq S^{2^{\ast}_{\alpha}/2(2^{\ast}_{\alpha}-1)}_{H,L}
\frac{2(2^{\ast}_{\alpha}-1)}{(2\cdot 2^{\ast}_{\alpha}-1)^{2\cdot 
2^{\ast}_{\alpha}-1/2(2^{\ast}_{\alpha}-1)}}-\|h\|_{H^{-1}}
 >0.
\end{align*}

(ii) Let $u\in \Lambda^-$, denote $\tilde{u}=\frac{u}{\|u\|_D}$, then 
$\|\tilde{u}\|_D=1$. By (i), we know that 
$C(\tilde{u})<\kappa_0=\frac{2(2^{\ast}_{\alpha}-1)}{2\cdot2^{\ast}_{\alpha}-1}
\left(\frac{1}{(2\cdot 2^{\ast}_{\alpha}-1)B}\right)^{1/2( 2^{\ast}_{\alpha}-1)}$ 
with $B:=B(\tilde{u})$. Furthermore, if $0<C(\tilde{u})<\kappa_0$, the equation 
$\varphi_{\tilde{u}}'(t)=0$ has exactly two roots $\tilde{t}_1, \tilde{t}_2$ 
satisfying $0<\tilde{t}_1<t_0<\tilde{t}_2$ such that 
$\tilde{t}_1\tilde{u}\in \Lambda^+, \tilde{t}_2\tilde{u}\in \Lambda^-$. 
Then $\tilde{t}_2\tilde{u}=u$ and so $\|u\|_D=\tilde{t}_2>t_0$.
 If $C\leq 0$, the equation $\varphi_{\tilde{u}}'(t)=0$ has exactly one roots 
$\tilde{t}_3>t_0$. Then $\tilde{t}_3\tilde{u}=u\in \Lambda^-$ and so 
$\|u\|_D=\tilde{t}_3>t_0$. In other words,
\[
\|u\|_D>t_0>0, \quad u\in \Lambda^-.
\]
So there exists $\tau>0$ such that
\begin{align} \label{eq11}
\|u\|_D>\tau>0, \quad \forall u\in \Lambda^-.
\end{align}
Therefore, $0\notin cl(\Lambda^-)$, where $cl(\Lambda^-)$ is the closure of 
$\Lambda^-$. On the other hand, by (i),
\[
cl(\Lambda^-)\subset \Lambda^-\cup \Lambda^0=\Lambda^-\cup \{0\}.
\]
Hence, $cl(\Lambda^-)=\Lambda^-$ and $\Lambda^-$ is closed. 
The proof is complete. 
\end{proof}

\begin{lemma} \label{lem2.5} 
 Under assumption {\rm (A3)}, for $u\in \Lambda\setminus \{0\}$, there exists
 $\epsilon>0$ and a differential function 
$t=t(w)>0, w\in D^{1,2}(\mathbb{R}^N), \|w\|<\epsilon$ such that
\begin{itemize}
\item[(1)] $t(0)=1$;
 
\item[(2)] $t(w)(u-w)\in \Lambda$, for all $w\in B_{\epsilon}(0)$;

\item[(3)] $\langle t'(0),w\rangle=\frac{2\int_{\mathbb{R}^N}
\nabla u\nabla w\,dx-2\cdot 2^{\ast}_{\alpha}
\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|u(y)|^{2^{\ast}_{\alpha}}|u(x)|^{2^{\ast}_{\alpha}-2}u(x)w(x)}
{|x-y|^{\alpha}}\,dx\,dy-\int_{\mathbb{R}^N}hw\,dx }
{\|u\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u)}$.
\end{itemize}
\end{lemma}
 
\begin{proof} 
We define $F: \mathbb{R}\times D^{1,2}\to \mathbb{R}$ by
\begin{align*}
F(t, w)&=t\|u-w\|^2_D-t^{2\cdot 2^{\ast}_{\alpha}-1}
 \int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|(u-w)(x)|^{2^{\ast}_{\alpha}}
 |(u-w)(y)|^{2^{\ast}_{\alpha}}}{|x-y|^\alpha}\,dx\,dy \\
&\quad -\int_{\mathbb{R}^N}h(u-w).
\end{align*}
Obviously, $F(1,0)=0$, $F'_t(1,0)=\varphi''_u(1)\not=0$. According to the 
implicit function theorem at point $(1,0)$, we get that there exist 
$\epsilon=\epsilon(u)>0$ and differentiable function 
$t: B_{\epsilon}(0)\to \mathbb{R}^+$ such that:
(1) $t(0)=1$;
(2) $t(w)(u-w)\in \Lambda$, for all $w\in B_{\epsilon}(0)$; and
(3) $\langle t'(0),w\rangle=-\frac{\langle\frac{\partial F}{\partial w}|_{(1,0)},
 w\rangle}{\frac{\partial F}{\partial t}|_{(1,0)}} $.
The proof is complete.
\end{proof}

\section{Local minimum solution}

Now we can  define 
\begin{equation*}
 c_0=\inf_{u\in \Lambda}I(u),\quad c_1=\inf_{u\in \Lambda^-}I(u),\quad 
c^+=\inf_{u\in \Lambda^+}I(u).
\end{equation*}

\begin{proposition} \label{prop3.1} 
Under assumption {\rm (A3)}, equation \eqref{eq3}
 has a local minimum solution with the least energy $c_0=\inf_{\Lambda}I(u)$.
\end{proposition}

\begin{proof}
 Firstly, we will show that $\|u\|_D$ is bounded from both above and below.
For any $u\in \Lambda$,
\begin{equation} \label{eq12}
\begin{split}
I(u)
&=\frac{1}{2}\|u\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u)-C(u)\\
&=\Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u\|^2_D
 -\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\int_{\mathbb{R}^N}hu\,dx\\
&\geq \frac{2^{\ast}_{\alpha}-1}{2\cdot 2^{\ast}_{\alpha}}\|u\|^2_D
 -\frac{2\cdot 2^{\ast}_{\alpha}-1}{2\cdot 2^{\ast}_{\alpha}}\|h\|_{H^{-1}}\|u\|_D\\
&\geq -\frac{(2^{\ast}_{\alpha}-1)^2}{8\cdot 2^{\ast}_{\alpha}(2^{\ast}_{\alpha}-1)}
 \|h\|^2_{H^{-1}}.
\end{split}
\end{equation}
Thus,
\begin{equation*}
c_0\geq -\frac{(2\cdot 2^{\ast}_{\alpha}-1)^2}{8\cdot 2^{\ast}_{\alpha}
(2^{\ast}_{\alpha}-1)}\|h\|^2_{H^{-1}}.
\end{equation*}
By the proof of Lemma \ref{lem2.4}, we have that if $u\in D^{1,2}(\mathbb{R}^N)$
and $\|u\|_D=1$, then
$$
0<C<\kappa_0=\frac{2(2^{\ast}_{\alpha}-1)}{2\cdot2^{\ast}_{\alpha}-1}
\Big(\frac{1}{(2\cdot 2^{\ast}_{\alpha}-1)B}\Big)^{1/2(2^{\ast}_{\alpha}-1)}.
$$
So that equation $\varphi'_u(t)=0$ has a positive solution $t_1$ satisfying
$0<t_1<t_0$ and $t_1u\in \Lambda^+$. Since
$\varphi'_u(t)=t-Bt^{2^{\ast}_{\alpha}-1}-C$, we know that
$\lim_{t\to 0}\varphi'_u(t)=-C<0$, $\varphi''_u(t)>0$ for all $t\in (0, t_0)$.
From $\varphi'_u(t_1)=0$ we have that $\varphi_u(t_1)<\lim_{t\to 0^+}\varphi_u(t)=0$
and $\varphi_u(t_1)=I(t_1u)\geq c^+$.
 Thus $c^+<0$ and $c_0=\inf_{\Lambda}I(u)\leq \inf_{\Lambda^+}I(u)=c^+<0$.


 By using the Ekeland's variational principle on $\Lambda$, we get a minimizing 
sequence $\{u_n\}\subset \Lambda$ which satisfies
\begin{equation} \label{eq13}
\begin{gathered}
I(u_n)<c_0+\frac{1}{n},\\
I(w)\geq I(u_n)-\frac{1}{n}\|u-w\|_D,\quad w\in \Lambda.
\end{gathered}
\end{equation}
Since $\{u_n\}\subset \Lambda$, it follows that $\|u_n\|^2_D=B(u_n)+C(u_n)$. 
Furthermore,  we infer from \eqref{eq13} that
\begin{equation}\label{eq14}
\begin{split}
c_0+\frac{1}{n}\geq I(u_n)
&=\Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u_n\|^2_D
 -\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&\geq \Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u_n\|^2_D
 -\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|h\|_{H^{-1}}\|u_n\|_D.
\end{split}
\end{equation}
We know that $\{u_n\}$ is bounded. We claim that $\inf_n\|u_n\|_D\geq \sigma>0$, 
which $\sigma$ is a positive constant.
Indeed, if not, by \eqref{eq14}, $I(u_n)$ would converge to zero.
 We can infer that $c_0\geq 0$ which is contradict with $c_0<0$. So we have
\begin{equation} \label{eq15}
\sigma \leq \|u_n\|_D\leq \delta.
\end{equation}

Secondly, we claim that, for a subsequence of $\{u_n\}$ (still denoted by $\{u_n\}$),
 $\|\nabla I(u_n)\|_D\to 0$ as $n\to \infty$.
In fact, if the claim were false, we could assume
 $\|\nabla I(u_n)\|_D\geq c>0$ for $n$ large enough. 
Consequently, according to Lemma \ref{lem2.5}, for $u_n$ there exist $\epsilon_n$
 and differentiable $t_n$ satisfying
\begin{equation*}
t_n(0)=1,\quad t_n(w)(u_n-w)\in \Lambda,\quad \|w\|_D<\epsilon_n
\end{equation*}
and
\begin{align*}
&\langle t_n'(0),w\rangle\\
&=\frac{2\int_{\mathbb{R}^N}\nabla u_n\nabla w\,dx
-2\cdot 2^{\ast}_{\alpha}\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N} \frac{|u_n(y)|^{2^{\ast}_{\alpha}}
 |u_n(x)|^{2^{\ast}_{\alpha}-2}u_n(x)w(x)}{|x-y|^{\alpha}}\,dx\,dy
 -\int_{\mathbb{R}^N}hw\,dx }
 {\|u_n\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u_n)}.
\end{align*}
We choose $w_n=\delta_n \nabla I(u_n)/\|\nabla I(u_n)\|_D$, 
$v_n=t_n(w_n)(u_n-w_n)$, where $0<\delta_n<\epsilon_n$ is sufficiently small 
satisfying $\delta_n\to 0$, $t_n(w_n)\to 1$ as $n\to \infty$ and
\begin{gather*}
\frac{\big|I(v_n)-I(u_n)-\langle I'(u_n), v_n-u_n \rangle \big|}
 {\|u_n-v_n\|_D}<\frac{1}{n},\\
\frac{\big|t_n(w_n)-1-\langle t_n'(0), w_n \rangle \big|}{\|w_n\|_D}<1.
\end{gather*}
From \eqref{eq13},  that $v_n\in \Lambda$ and the above,
 we deduce
\[
\frac{1}{n}\|v_n-u_n\|_D \geq I(u_n)-I(v_n)
\geq\langle I'(u_n), u_n-v_n \rangle -\frac{1}{n}\|u_n-v_n\|_D.
\]
Thus, we have
\begin{align*}
&\frac{2}{n}\|t_n(w_n)(u_n-w_n)-u_n\|_D\\
&\geq (1-t_n(w_n))\langle I'(u_n), u_n \rangle
 +t_n(w_n)\delta_n\langle I'(u_n), \frac{\nabla I(u_n)}{\|\nabla I(u_n)\|_D} \rangle,
\end{align*}
and
\[
\frac{2}{n}\Big[(|\langle t'_n(0), w_n \rangle|+\|w_n\|_D) \|u_n\|_D
+ t_n(w_n)\|w_n\|_D\Big]
\geq t_n(w_n)\delta_n\|\nabla I(u_n)\|_D.
\]
Dividing by $\delta_n>0$ on both left and right hand of the above inequality, 
we obtain
\begin{equation}\label{eq 3.5}
\frac{2}{n}\Big[(|\langle t'_n(0), \frac{\nabla I(u_n)}
{\|\nabla I(u_n)\|_D} \rangle|+1) \|u_n\|_D+ t_n(w_n)\Big]
\geq t_n(w_n)\|\nabla I(u_n)\|_D.
\end{equation}
Now, if there exists $\lambda>0$ such that
\begin{equation*}
\big|\|u_n\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u_n) \big|\geq \lambda\,,
\end{equation*}
we can get the claim. In fact,
\begin{align*}
&\Big|\langle t_n'(0),h_n\rangle\Big|\\
&=\Big|\frac{\int_{\mathbb{R}^N}(2\nabla u_n\nabla h_n-h_nu_n)dx
 -2\cdot 2^{\ast}_{\alpha}\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N}\frac{|u_n(y)|^{2^{\ast}_{\alpha}}|u_n(x)|^{2^{\ast}_{\alpha}-2}
 u_n(x)h_n(x)}{|x-y|^{\alpha}}\,dx\,dy}
{\|u_n\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u_n)}\Big|\\
&\leq \frac{C}{\lambda}.
\end{align*}
Here, $h_n=\frac{\nabla I(u_n)}{\|\nabla I(u_n)\|_D}$ and we have used the 
uniformly boundedness of $\|u_n\|_D$.
Consequently, as $n\to\infty$,
\begin{equation*}
\frac{2}{n}\Big[(|\langle t'_n(0), h_n \rangle|+1) \|u_n\|_D+ t_n(w_n)\Big]\to 0.
\end{equation*}
So that, by passing to the limit as $n\to\infty$ in \eqref{eq 3.5},
we get a contradiction which implies the claim is true.

To show the existence of positive lower bound of 
$\Big|\|u_n\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u_n) \Big|$, 
we argue indirectly  and assume
\begin{equation*}
\|u_n\|^2_D-(2\cdot 2^{\ast}_{\alpha}-1)B(u_n) =o(1), \quad n\to \infty.
\end{equation*}
Here, $\{u_n\}$ is a subsequence still denoted by the original symbol.
Combining this and \eqref{eq14}, similarly to the proof of Lemma \ref{lem2.4}(i),
we can easily get a contradiction.

So that we conclude that, for a subsequence which we still denote by $\{u_n\}$,
\[
I(u_n)\to c_0, \quad 
\|\nabla I(u_n)\|_D\to 0,
\]
as $n\to \infty$. By \eqref{eq14} we know that $\{u_n\}$ is bounded in
$D^{1,2}(\mathbb{R}^N)$, and the weak limit of $\{u_n\}$ which
we denote by $u_0$ is a weak solution of system \eqref{eq3}.
 Obviously, $u_0\in\Lambda$ and
\begin{align*}
c_0&\leq I(u_0)
=\frac{1}{2}\|u_0\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_0)-C(u_0)\\
&=\Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u_0\|^2_D
 -\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)C(u_0)\\
&\leq \liminf I(u_n)=c_0.
\end{align*}
Therefore, $u_0$ is a least energy solution.

Now, we only need to show that $u_0$ is a local minimum solution. 
We apply Lemma \ref{lem2.3} to $u_0$ and $|u_0|$.
Since
\[
\frac{d}{dt}I(tu_0)=\varphi'_{u_0}(t)>0,\quad t\in (t^+(u_0), t^-(u_0)),
\]
we know that $u_0\in\Lambda^+$. Otherwise, $u_0\in \Lambda^-$ and 
$c_0\leq I(t^+(u_0)u_0)<I(u_0)=c_0$ which is a contradiction. 
By Lemma \ref{lem2.3} and $u_0\in\Lambda^+$ we know that
\[
1=t^+(u_0)< t_0(u_0)
=\Big[\frac{\|u_0\|^2_D}{(2\cdot 2^{\ast}_{\alpha}-1)B(u_0)}
 \Big]^{1/(2\cdot 2^{\ast}_{\alpha}-2)}.
\]
Therefore
\begin{equation*}
1<\Big[\frac{\|u_0-w\|^2_D}{(2\cdot 2^{\ast}_{\alpha}-1)B(u_0-w)}
\Big]^{1/(2\cdot 2^{\ast}_{\alpha}-2)}, \quad \|w\|_D<\varepsilon,
\end{equation*}
for $\varepsilon$ small enough. Applying Lemma \ref{lem2.5}, 
we get a $t(w)>0$ such that $t(w)(u_0-w)\in\Lambda$ for $\|w\|<\varepsilon$ small.
Moreover, it holds $t(w)\to 1$ as $w\to 0$. Thus we can assume that, for $\|w\|<\varepsilon$ sufficiently small,
\[
t(w)<\Big[\frac{\|u_0-w\|^2_D}{(2\cdot 2^{\ast}_{\alpha}-1)B(u_0-w)}
\Big]^{1/(2\cdot 2^{\ast}_{\alpha}-2)}, \quad
t(w)(u_0-w)\in \Lambda^+.
\]
Then by Lemma \ref{lem2.3}, we  conclude that
\begin{equation}
\label{eq16}
I(u_0)\leq I(t(w)(u_0-w))\leq I(t(u_0-w))
\end{equation}
for $0<t<[\frac{\|u_0-w\|^2_D}{(2\cdot 2^{\ast}_{\alpha}-1)B(u_0-w)}
]^{1/(2\cdot 2^{\ast}_{\alpha}-2)}$.
Taking $t=1$ in \eqref{eq16} we have
\[
I(u_0)\leq I(u_0-w),\quad \|w\|_D<\varepsilon,
\]
which means $u_0$ is a local minimum solution.

Additionally, if we assume that $h>0$,
\begin{equation*}
\varphi'_{|u_0|}(t)<\varphi'_{u_0}(t)<0, \quad t\in [0,1).
\end{equation*}
Hence, $t^+(|u_0|)\geq 1$ and consequently,
\begin{equation*}
c_0\leq I(t^+(|u_0|)|u_0|)\leq I(|u_0|)\leq I(u_0)=c_0.
\end{equation*}
Therefore, $t^+(|u_0|)=1$ and 
$\int_{\mathbb{R}^N}h(x)|u_0|\,dx=\int_{\mathbb{R}^N}h(x)u_0\,dx$, which yields 
$u_0\geq 0$. Then by the maximum principle, we know $u_0>0$.
The proof is complete.
\end{proof}

We remark that since $c_0<0$, any solution $u_0$ of
system \eqref{eq3} with the least energy $c_0$ satisfies
\begin{align*}
c_0
&=I(u_0)=\frac{1}{2}\|u_0\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_0)-\int_{\mathbb{R}^N}h(x)u_0dx\\
&=\Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u_0\|^2_D
 -\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\int_{\mathbb{R}^N}h(x)u_0\,dx
<0.
\end{align*}
Thus, $\int_{\mathbb{R}^N}h(x)u_0dx> 0$ and consequently, $u_0\in\Lambda^+$.

\begin{proposition} \label{prop3.2} 
Assume $N\geq 3$, $0<\alpha<N$ and {\rm (A3)}. 
If $\{u_n\}$ is a $(PS)_c$ sequence of $I$ with
\begin{align}
\label{eq17}
 c<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L},
\end{align}
then $\{u_n\}$ has a convergent subsequence.
\end{proposition}

\begin{proof} 
Obviously, $\|u_n\|_D$ is bounded. In fact
\begin{align*}
c+\|u_n\|_D
&\geq I(u_n)-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\langle I'(u_n), u_n\rangle \\
&=\frac{1}{2}\|u_n\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_n)-C(u_n)\\
&\quad -\frac{1}{2\cdot 2^{\ast}_{\alpha}}\|u_n\|^2_D
 +\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_n)
 +\frac{1}{2\cdot 2^{\ast}_{\alpha}}C(u_n)\\
&\geq \frac{1}{2}(1-\frac{1}{2^{\ast}_{\alpha}})\|u_n\|^2_D
 -(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}})\|h\|_{H^{-1}}\|u_n\|_D.
\end{align*}
Thus, there exists $w\in D^{1,2}(\mathbb{R}^N)$ which satisfies 
$u_n\rightharpoonup w$ weakly in $ D^{1,2}(\mathbb{R}^N)$
and solves \eqref{eq3}.
Therefore, $w\not=0$ and $I(w)\geq c_0$. Let $u_n-w=v_n$, by Brezis-Lieb 
lemma \cite{Brezis1} and \cite[Lemmas 2.1 and 2.2]{GaoYang}, we deduce that
\begin{equation*}
\|u_n\|^2_D=\|v_n\|^2_D+\|w\|^2_D+o(1),\quad n\to \infty,
\end{equation*}
and
\begin{align*}
&\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|u_n(x)|^{2^{\ast}_{\alpha}}|u_n(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}
 \,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|v_n(x)|^{2^{\ast}_{\alpha}}|v_n(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}
 \,dx\,dy 
 +\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|w(x)|^{2^{\ast}_{\alpha}}
 |w(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy+o_n(1)
\end{align*}
as $ n\to \infty $.
So we obtain
\begin{align*}
&c\leftarrow I(u_n) \\
&=\frac{1}{2}\|u_n\|^2_D-
\frac{1}{2\cdot 2^{\ast}_{\alpha}}\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N} \frac{|u_n(x)|^{2^{\ast}_{\alpha}}|u_n(y)|^{2^{\ast}_{\alpha}}}
 {|x-y|^{\alpha}}\,dx\,dy -\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&=\frac{1}{2}\|v_n\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}
 \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|v_n(x)|^{2^{\ast}_{\alpha}}
 |v_n(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy
 -\int_{\mathbb{R}^N}h(x)v_n\,dx\\
&\quad +\frac{1}{2}\|w\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}
 \int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|w(x)|^{2^{\ast}_{\alpha}}|w(y)|^{2^{\ast}_{\alpha}}}{|x-y|^{\alpha}}\,dx\,dy 
 -\int_{\mathbb{R}^N}h(x)w\,dx+o_n(1)\\
&=I(w)+\frac{1}{2}\|v_n\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(v_n)+o_n(1).
\end{align*}
As a result, for $n$ large we have
\begin{equation}
\label{eq18}
\frac{1}{2}\|v_n\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(v_n)+o_n(1)
<\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}.
\end{equation}
On the other hand,
\[
o(1)=\langle I'(u_n),u_n\rangle=\langle I'(w),w\rangle+\|v_n\|^2_D-B(v_n)+o(1),
\]
which implies
\begin{equation} \label{eq19}
\|v_n\|^2_D-B(v_n)=o(1).
\end{equation}
If we can show that $\{v_n\}$ has subsequence converging strong to $0$, 
we have the result. Therefore, arguing indirectly, we assume
 $\|v_n\|_D\geq c>0$ for $n$ large.
According to \eqref{eq19}, the definition of $S_{H,L}$ and
$\frac{2^{\ast}_{\alpha}}{2^{\ast}_{\alpha}-1}=\frac{2N-\alpha}{N+2-\alpha}$, 
we have
\[
\|v_n\|^2_D=B(v_n)+o(1)\leq \frac{\|v_n\|^{2\cdot 2^{\ast}_{\alpha}}_D}
{S^{2^{\ast}_{\alpha}}_{H,L}}
\]
and
\begin{align*}
\frac{1}{2}\frac{N+2-\alpha}{2N-\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}
&=\frac{1}{2}\frac{2^{\ast}_{\alpha}-1}{2^{\ast}_{\alpha}}
 S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}\\
&\leq \frac{1}{2}\frac{2^{\ast}_{\alpha}-1}{2^{\ast}_{\alpha}}\|v_n\|^2_D\\
&=\frac{1}{2}\|v_n\|^2_D-\frac{1}{2}\frac{1}{2^{\ast}_{\alpha}}B(v_n)+o_n(1)\\
&<\frac{N+2-\alpha}{2(2N-\alpha)}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L},
\end{align*}
which is a contradiction.
The proof is complete.
\end{proof}

\begin{remark} \label{rmk3.3}
To find the second solution of \eqref{eq3},
the only we need to show is that
\begin{equation*}
c_0<c_1=\inf_{\Lambda^-}I(u)
<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}.
\end{equation*}
\end{remark}

By the continuity of $I$ and Lemma \ref{lem2.2}, we know that there exists 
$\delta>0$ such that
\begin{equation*}
I(u_0+tW)<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}, 
\quad 0\leq t<\delta,
\end{equation*}
where $u_0$ is the positive local minimum solution we get in Proposition \ref{prop3.1}.
For $t\geq \delta $, a directly computation shows us
\begin{align*}
I(u_0+tW)
&=\frac{1}{2}\|u_0+tW\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_0+tW)
 -\int_{\mathbb{R}^N}h(u_0+tW)dx\\
&=\frac{1}{2}\|u_0\|^2_D+t\int_{\mathbb{R}^N}\nabla u_0\nabla W\,dx
 +\frac{t^2}{2}\|W\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_0)\\
&\quad +\frac{1}{2\cdot 2^{\ast}_{\alpha}}[B(u_0)
 +B(tW)-B(u_0+tW)]-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(tW)\\
&\quad -\int_{\mathbb{R}^N}hu_0dx-\int_{\mathbb{R}^N}htW\,dx\\
&=I(u_0)+\frac{t^2}{2}[\|W\|^2_D-\frac{t^{2(2^{\ast}_{\alpha}-1)}}
{2\cdot 2^{\ast}_{\alpha}}B(W)]
+\frac{1}{2\cdot 2^{\ast}_{\alpha}}[B(u_0)+B(tW)\\
&\quad -B(u_0+tW)+2\cdot 2^{\ast}_{\alpha}\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N}\frac{|u_0(x)|^{2^{\ast}_{\alpha}}|u_0(y)
 |^{2^{\ast}_{\alpha}-2}u_0(y)}{|x-y|^{\alpha}}\,dx\,dy]\\
&<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}.
\end{align*}
Here, we use  that $\langle I'(u_0), tW\rangle =0$ and $W$ is a 
minimizer of $S_{H,L}$.


\begin{proof}[Proof of Theorem \ref{thm1.1}]
Firstly, we show that
\begin{equation*}
c_0<c_1=\inf_{\Lambda^-}I(u)<c_0+\frac{N+2-\alpha}{4N-2\alpha}
S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}.
\end{equation*}
We observe that for every $u\in D^{1,2}({\mathbb{R}^N})$ with $\|u\|_D=1$, 
there exists a unique $t^-(u)>0$ such that (see Lemma \ref{lem2.3})
\begin{equation*}
t^-(u)u\in \Lambda^-.
\end{equation*}
Similar to Lemma \ref{lem2.5}, we know that $t^-(u)$ is a continuous function of $u$. 
And consequently the manifold $\Lambda^-$ disconnects $D^{1,2}({\mathbb{R}^N})$ 
in exactly two connected components $U_1$ and $U_2$, where
\begin{gather*}
U_1=\Big\{u\in D^{1,2}({\mathbb{R}^N}): u=0 \text{ or }
 \|u\|_D<t^-\big(\frac{u}{\|u\|_D}\big)\Big\}, \\
U_2=\Big\{u\in D^{1,2}({\mathbb{R}^N}): \|u\|_D>t^-\big(\frac{u}{\|u\|_D}\big)\Big\}.
\end{gather*}
Obviously, $D^{1,2}({\mathbb{R}^N})=\Lambda^-\cup U_1 \cup U_2$. 
In particular, $u_0\in \Lambda^+ \subset U_1$.
Since
\begin{equation*}
t^-\Big(\frac{u_0+tW}{\|u_0+tW\|_D}\Big)\frac{u_0+tW}{\|u_0+tW\|_D}\in \Lambda,
\end{equation*}
we have
\begin{equation*}
0<t^-\Big(\frac{u_0+tW}{\|u_0+tW\|_D}\Big)<C_0
\end{equation*}
uniformly for $t\in \mathbb{R}$.

On the other hand, there exists $\tilde{t}>0$ such that
\begin{equation*}
\|u_0+tW\|_D\geq t\|W\|_D-\|u_0\|_D\geq C_0,\quad t\geq \tilde{t}.
\end{equation*}
So that we can fix a $t_0>0$ such that
$\|u_0+t_0W\|_D> t^-(\frac{u_0+t_0W}{\|u_0+t_0W\|_D})$. Thus,
$u_0+t_0W\in U_2$. Combining this and the fact $u_0\in U_1$, we know that
\begin{equation*}
u_0+t_1W\in \Lambda^-,
\end{equation*}
for some $0<t_1<t_0$. Consequently, by  Remark \ref{rmk3.3}, we have
\begin{equation*}
c_1=\inf_{\Lambda^-}I(u)\leq \max_{0\leq t\leq t_0}I(u_0+tW)
<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}.
\end{equation*}

Next, we show that $c_1$ is a critical value of $I$ and satisfies $c_1>c_0$.
Similarly to the proof of Proposition \ref{prop3.1}, we apply Ekeland's variational 
principle and get a minimizing sequence $\{u_n\}\subset \Lambda^-$ such that
\begin{gather*}
I(u_n)<c_1+\frac{1}{n};\\
I(w)\geq I(u_n)-\frac{1}{n}\|u-w\|_D,\quad w\in \Lambda^-.
\end{gather*}
So that 
\begin{align*}
c_1+1
&>I(u_n)=\frac{1}{2}\|u_n\|^2_D-\frac{1}{2\cdot 2^{\ast}_{\alpha}}B(u_n)-\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&\geq \Big(\frac{1}{2}-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|u_n\|^2_D
-\Big(1-\frac{1}{2\cdot 2^{\ast}_{\alpha}}\Big)\|h\|_{H^{-1}}\|u_n\|_D,
\end{align*}
which implies $\|u_n\|$ has a upper bound.
Moreover, from $\{u_n\}\subset \Lambda^-$, we know that
\[
\|u_n\|^2_D\leq (2\cdot 2^{\ast}_{\alpha}-1)
\frac{\|u_n\|^{2^{\ast}_{\alpha}}_D}{S^{2^{\ast}_{\alpha}}_{H,L}}.
\]
Thus, $\|u_n\|_D$ has a uniform positive lower bound.
 Then, analogously to the proof of Proposition \ref{prop3.1}, we know that
\[
I(u_n)\to c_1, \quad 
I'(u_n)\to 0\quad in\quad H^{-1}.
\]
By Proposition \ref{prop3.1} and 
$c_1<c_0+\frac{N+2-\alpha}{4N-2\alpha}S^{\frac{2N-\alpha}{N+2-\alpha}}_{H,L}$, 
we can conclude that there exists a subsequence of $\{u_n\}$ such that 
$u_n\to u_1$ strongly in $D^{1,2}(\mathbb{R}^N)$. Therefore $u_1$ is a critical 
point of $I$ and $I(u_1)=c_1$. Noting that $\Lambda^-$ is closed, we have 
$u_1\in\Lambda^-$.
To show $c_1>c_0$, arguing indirectly, we assume $c_1=c_0$. 
Thus by Remark \ref{rmk3.3} we have
\begin{equation*}
\int_{\mathbb{R}^N}h(x)u_1dx>0 \quad and \quad u_1\in\Lambda^+,
\end{equation*}
which leads to a contradiction.

Finally, we consider the case $h>0$. Applying Lemma \ref{lem2.3} to $u_1$ and $|u_1|$, 
we know that there exist a $t^-(|u_1|)$ such that
\begin{equation*}
t^-(|u_1|)|u_1|\in\Lambda^-.
\end{equation*}
Moreover,
\begin{equation*}
t^-(|u_1|)\geq t_0(|u_1|)=t_0(u_1)
=\Big[\frac{A(u_1)}{(2^{\ast}_{\alpha}-1)B(u_1)}\Big]^{\frac{1}{2^{\ast}_{\alpha}-2}}
\end{equation*}
Thus in both  cases $\int_{\mathbb{R}^N}h(x)u\,dx>0$ and 
$\int_{\mathbb{R}^N}h(x)u\,dx\leq 0$, we can deduce that
\begin{equation*}
c_1=I(u_1)\geq I(t^-(u_1)u_1)\geq I(t^-(|u_1|)|u_1|)=t_0(u_1)\geq c_1.
\end{equation*}
Therefore, $\int_{\mathbb{R}^N}h(x)u_1dx=\int_{\mathbb{R}^N}h(x)|u_1|dx$, 
which implies $u_1\geq 0$. According to the maximum principle we get $u_1>0$.
The proof is complete.
\end{proof}

\section{Proof of Theorem \ref{thm1.2}}

We define the energy functional associated with
\begin{align} \label{eq20}
- \Delta u +u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u+h(x), \quad 
x\in \mathbb{R}^N,
\end{align}
where $N\geq 3$, $0<\alpha<N$ and $2-\frac{\alpha}{N}<p<2^{\ast}_{\alpha}$,
 by
\begin{equation}
 \label{eq21}
J(u)=\frac{1}{2}\int_{\mathbb{R}^N} |\nabla u|^2 + u^2dx-
\frac{1}{2p}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy -\int_{\mathbb{R}^N}h(x)u\,dx,
\end{equation}
for $u\in H^1(\mathbb{R}^N)$.
By the Hardy-Littlewood-Sobolev inequality of Lemma \ref{lem2.1}, we know 
that $J\in C^1(H^1(\mathbb{R}^N),\mathbb{R})$ and
\begin{align*}
\langle J'(u),v\rangle 
&= \int_{\mathbb{R}^N} (|\nabla u||\nabla v| + uv)dx
 -\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
  \frac{|u(x)|^{p}|u(y)|^{p-2}u(y)v(y)}{|x-y|^{\alpha}}\,dx\,dy\\
&\quad -\int_{\mathbb{R}^N}h(x)v\,dx.
\end{align*}
We will constrain the functional $J$ on the Nehari manifold
\begin{equation}  \label{eq22}
\mathcal{N}=\{u\in H^1(\mathbb{R}^N), \langle J'(u),u\rangle=0\}.
\end{equation}
Denote $\Psi(u)=\langle J'(u),u\rangle$, so we know that
\begin{gather*}
\langle J'(u),u\rangle
=\|u\|^2-\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}
\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy-\int_{\mathbb{R}^N}h(x)u\,dx, \\
\langle \Psi'(u),u\rangle=2\|u\|^2-2p\int_{\mathbb{R}^N}
\int_{\mathbb{R}^N} \frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy
-\int_{\mathbb{R}^N}h(x)u\,dx.
\end{gather*}
Note that, when $u_0$ is a local minimum solution of $J$, it holds
\[
\langle J'(u),u\rangle=0,\quad
\langle \Psi'(u),u\rangle\geq 0,
\]
which leads us to consider the following manifolds:
\begin{gather*}
\mathcal{N}=\{u\in H^1(\mathbb{R}^N): \langle J'(u),u\rangle=0\},\\
\mathcal{N}^+=\{u\in\mathcal{N}: \langle \Psi'(u),u\rangle > 0\},\\
\mathcal{N}^-=\{u\in\mathcal{N}: \langle \Psi'(u),u\rangle< 0\},\\
\mathcal{N}^0=\{u\in \mathcal{N}: \langle \Psi'(u),u\rangle = 0\}.
\end{gather*}
Moreover, we define
\begin{equation*}
j_0=\inf_{\mathcal{N}}J(u);\quad j_1=\inf_{\mathcal{N}^-}J(u); \quad
j^+=\inf_{\mathcal{N}^+}J(u).
\end{equation*}
Obviously, only $\mathcal{N}^0$ contains the element $0$.
It is easy to see that $\mathcal{N}^0\cup \mathcal{N}^+$ and
$\mathcal{N}^0\cup \mathcal{N}^-$ are both closed subsets of $H^1(\mathbb{R}^N)$.

To simplify the calculation, for $u\in H^1(\mathbb{R}^N)$, we denote
 \begin{gather*}
\tilde{A}=\tilde{A}(u)=\|u\|^2,\\
\tilde{B}=\tilde{B}(u)
=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|u(x)|^{p}|u(y)|^{p}}
{|x-y|^{\alpha}}\,dx\,dy, \\
\tilde{C}=\tilde{C}(u)=\int_{\mathbb{R}^N}h(x)u\,dx.
\end{gather*}
Define the fibering map
\begin{equation}
 \label{eq23}
\psi(t)=\psi_u(t):=J(tu)=\frac{\tilde{A}}{2}t^2-\frac{\tilde{B}}{2p}t^{2p}
-\tilde{C}t,\quad t>0.
\end{equation}
Therefore,
\begin{equation} \label{eq24}
 \begin{gathered}
\psi'(t)=\tilde{A}t-\tilde{B}t^{2p-1}-\tilde{C},\\
\psi''(t)=\tilde{A}-(2p-1)\tilde{B}t^{2p-2}.
\end{gathered}
\end{equation}
Obviously, $tu\in \mathcal{N}$ with $t>0$ if and only if $\psi'(t)=0$.
By the sign of $\psi''(t)$, the stationary points of $\psi(t)$ can be
classified into three types, namely local minimum, local maximum and
turning point. Moreover, the set $\mathcal{N}$ is a natural constraint
for the functional $J$. This is means that if the infimum is attained
by $u\in \mathcal{N}$, then $u$ is a solution of \eqref{eq4}.
 However, in our case, the global maximum point of $\psi(t)$ is not unique.
This leads us to partition
the set $\mathcal{N}$ according to the critical points of $\psi(t)$.
Now we give some properties of $\mathcal{N}^{\pm}$ and $\mathcal{N}^0$.


\begin{lemma} \label{lem4.1} 
(i) Assume that $h\not\equiv 0 $ for $u\in H^1(\mathbb{R}^N)\backslash \{0\}$, 
there is a unique $\tilde{t}^-=\tilde{t}^-(u)>0$ such that 
$\tilde{t}^-u\in \mathcal{N}^-$. If additionally we assume 
$\int_{\mathbb{R}^N}hu\,dx>0$ and (A5), then there exists 
$\varepsilon=\varepsilon(N,p,\alpha,d_{\frac{2Np}{2N-\alpha}})$ and a 
unique $0<\tilde{t}^+=\tilde{t}^+(u)<\tilde{t}^-$ satisfying 
$\tilde{t}^+u\in \mathcal{N}^+$. Moreover,
\begin{gather*}
J(\tilde{t}^-u)=\max_{t\geq 0}J(tu)\quad\text{for } \int_{\mathbb{R}^N}hu\,dx\leq 0;\\
J(\tilde{t}^-u)=\max_{t\geq \tilde{t}^+}J(tu),\quad 
J(\tilde{t}^+u)=\min_{0\leq t\leq \tilde{t}^-}J(tu)\quad \text{for }
 \int_{\mathbb{R}^N}hu\,dx>0.
\end{gather*}
\end{lemma}

\begin{proof} 
Define $\psi(t)=\frac{\tilde{A}}{2}t^2-\frac{\tilde{B}}{2\cdot p}t^{2p}-\tilde{C}t$, 
for all $t>0$.
In the case $\int_{\mathbb{R}^N}hu\,dx\leq 0$, there is a unique $\tilde{t}^->0$ 
such that $\psi'(\tilde{t}^-)=0$ and $\psi''(\tilde{t}^-)<0$.
So that
\begin{gather*}
\langle J'(\tilde{t}^-u),\tilde{t}^-u\rangle=0;\\
\|\tilde{t}^-u\|^2-(2p-1)\tilde{B}(u)(\tilde{t}^-)^{2p-2}<0.
\end{gather*}
Thus, $\tilde{t}^-u\in \mathcal{N}^-$ and $J(\tilde{t}^-u)=\max_{t\geq 0}J(tu)$.

In the case $\int_{\mathbb{R}^N}hu\,dx>0$, for 
$\|u\|=1, \tilde{t}_0=\tilde{t}_0(u)
=\big[\frac{1}{(2p-1)\tilde{B}}\big]^{\frac{1}{2p-2}}>0$
and (A5), we have
\begin{align*}
\max_{t\geq 0}\psi'(t)
&\geq t_0-\tilde{B}t_0^{2p-1}-\tilde{C}\\
&=\Big[\frac{1}{(2p-1)\tilde{B}}\Big]^{\frac{1}{2p-2}}\cdot 
\frac{2p-2}{2p-1}-\int_{\mathbb{R}^N}hu\,dx\\
&\geq \Big[\frac{2p-2}{(2p-1)^{2p-1/(2p-2)}B_0^{1/2p-2}}
-|h|_{\frac{2Np}{2N(p-1)+\alpha}}d_{\frac{2Np}{2N-\alpha}}\Big]
>0.
\end{align*}
Here
\begin{gather*}
\varepsilon(N,p,\alpha,d_{\frac{2Np}{2N-\alpha}})
=\frac{2p-2}{(2p-1)^{2p-1/(2p-2)}B_0^{1/2p-2}d_{\frac{2Np}{2N-\alpha}}}, \\
B_0=\sup_{\|u\|=1}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy.
\end{gather*}
From $\psi'(0)=-\tilde{C}<0$ and $\psi'(t)\to -\infty$ as $t\to +\infty$, 
we know that there exist unique $ 0<\tilde{t}^+<\tilde{t}_0<\tilde{t}^-$ such that
$ \psi'(\tilde{t}^-)=\psi'(\tilde{t}^+)=0, \psi''(\tilde{t}^-)<0<\psi''(\tilde{t}^+) $.
Equivalently, $\tilde{t}^+u\in \mathcal{N}^+$ and $\tilde{t}^-u\in \mathcal{N}^-$.

Moreover, since $\frac{d}{dt}J(tu)=\psi'(t)$, we can easily see that
 $J(\tilde{t}^-u)=\max_{t\geq \tilde{t}^+}J(tu)$ and
$J(\tilde{t}^+u)=\min_{0\leq t\leq \tilde{t}^-}J(tu)$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem4.2} 
 Assume t $h\not\equiv0 $, {\rm (A4)} and (A5) {\rm hold}. 
Then 
\begin{itemize}
\item[(i)] $\mathcal{N}^0= \{0\}$;
\item[(ii)] $\mathcal{N}^{\pm}\neq \emptyset, \mathcal{N}^-$ is closed.
\end{itemize}
\end{lemma}

\begin{proof} (i) To prove $\mathcal{N}^0=\{0\}$, we need to prove that, 
for $u\in H^1(\mathbb{R}^N)\setminus \{0\}$, $\tilde{\varphi}(t)$ 
has no critical point that is a turning point. Set $\|u\|=1$, define
\begin{equation} \label{eq25}
k(t)=\tilde{A}t-\tilde{B}t^{2p-1}.
\end{equation}
Then $\psi'(t)=k(t)-\tilde{C}, k''(t)=-\tilde{B}(2p-1)(2p-2)t^{2p-3}<0$ for $t>0$.
So $k(t)$ is strictly concave.
If $k''(\tilde{t}_0)=0$,
$\tilde{t}_0=(\frac{\tilde{A}}{(2p-1)\tilde{B}})^{1/(2p-2)}>0$,
for $p>2-\frac{\alpha}{N}>1$. Moreover, $\lim_{t\to 0^+}k(t)=0$,
$\lim_{t\to +\infty}k(t)=-\infty$ and $k(t)>0$ for $t>0$ small.
Therefore, $k(t)$ has a unique global maximum points $t_0$ and
\[
k(t_0)=\frac{2\tilde{A}(2p-1)}{2p-1}
\Big(\frac{\tilde{A}}{(2p-1)\tilde{B}}\Big)^{1/(2p-2)}:=k_0.
\]
By \eqref{eq23} and \eqref{eq24}, we infer that if $0<\tilde{C}<k_0$,
the equation $\psi'(t)=0$ has exactly two points $\tilde{t}_1, \tilde{t}_2$
satisfying $\tilde{t}_1<\tilde{t}_0<\tilde{t}_2$.
If $\tilde{C}\leq 0$, the equation $\psi'(t)=0$ has one root
 $\tilde{t}_3>\tilde{t}_0$. Since $\psi''(t)=\tilde{A}-(2p-1)\tilde{B}t^{2p-2}$,
it follows that $\psi''(\tilde{t}_1)>0, \psi''(\tilde{t}_2)<0$ and
$\psi''(\tilde{t}_3)<0$. It follows that
$\tilde{t}_1u\in \mathcal{N}^+, \tilde{t}_2u\in \mathcal{N}^-$ if
 $0<\tilde{C}<k_0$ and $\tilde{t}_3u\in \mathcal{N}^-$ if $\tilde{C}\leq 0$.
Since $\{u\in H^1(\mathbb{R}^N): \|u\|=1, 0<\tilde{C}<k_0\}$ and
$\{u\in H^1(\mathbb{R}^N): \|u\|=1, \tilde{C}\leq 0\}$ are nonempty,
 we can infer that $\mathcal{N}^{\pm}$ are nonempty.
This implies $\mathcal{N}^0=\{0\}$.

It is suffices to prove that $k_0>\tilde{C}$. 
By (A4), (A5) and Lemma \ref{lem4.1} we have
\[
k_0-\tilde{C}=k(t_0)-\tilde{C}=\tilde{A}t_0-\tilde{B}t^{2p-1}_0-\tilde{C}>0.
\]

(ii) Let $u\in \mathcal{N}^-$ and  denote $\tilde{u}=u/\|u\|$.
Then $\|\tilde{u}\|=1$. By (i), we know that
\[
\tilde{C}(\tilde{u})<k_0=\frac{2(p-1)}{2p-1}
\Big(\frac{1}{(2p-1)\tilde{B}}\Big)^{1/(2p-1)}
\]
 with $\tilde{B}:=B(\tilde{u})$. Furthermore, if $0<\tilde{C}(\tilde{u})<\kappa_0$,
 the equation $\psi'(t)=0$ has exactly two roots $\tilde{t}_1, \tilde{t}_2$ 
satisfying $0<\tilde{t}_1<t_0<\tilde{t}_2$ such that 
$\tilde{t}_1u\in \mathcal{N}^+, \tilde{t}_2u\in \mathcal{N}^-$. 
Then $\tilde{t}_2\tilde{u}=u$ and so $\|u\|=\tilde{t}_2>\tilde{t}_0$.
 If $\tilde{C}\leq 0$, the equation $\psi'(t)=0$ has exactly one root
 $\tilde{t}_3>\tilde{t}_0$. Then $\tilde{t}_3\tilde{u}=u\in \mathcal{N}^-$ 
and so $\|u\|=\tilde{t}_3>\tilde{t}_0$. In other words, 
\[
\|u\|>\tilde{t}_0>0, \quad u\in \mathcal{N}^-.
\]
So there exists $\tau>0$ such that
\begin{align} \label{eq26}
\|u\|>\tau>0, \quad \forall u\in\mathcal{N}^-.
\end{align}
Therefore, $0\notin cl(\mathcal{N}^-)$, where $cl(\mathcal{N}^-)$ 
is the closure of $\mathcal{N}^-$. On the other hand, by (i),
\[
cl(\mathcal{N}^-)\subset \mathcal{N}^-\cup \mathcal{N}^0=\mathcal{N}^-\cup \{0\}.
\]
Hence, $cl(\mathcal{N}^-)=\mathcal{N}^-$ and $\mathcal{N}^-$ is closed. 
The proof is complete. 
\end{proof}

\begin{lemma} \label{lem4.3}
 Under assumption {\rm (A4)} and {\rm (A5)}, for $u\in \mathcal{N}\setminus \{0\}$, 
there exists $\epsilon>0$ and a differential function 
$\eta=\eta(w)>0, w\in H^1(\mathbb{R}^N), \|w\|<\epsilon$ such that
\begin{itemize}
\item[(1)] $\eta(0)=1$;
 
\item[(2)] $\eta(w)(u-w)\in \mathcal{N}$, for all $w\in B_{\epsilon}(0)$;

\item[(3)] 
\begin{align*}
\langle \eta'(0),w\rangle
&=\Big(2\int_{\mathbb{R}^N} (\nabla u\nabla w+uw)dx \\
&\quad -2p\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} \frac{|u(y)|^{2p}|u(x)|^{p-2}
u(x)w(x)}{|x-y|^{\alpha}}\,dx\,dy
-\int_{\mathbb{R}^N}hw\,dx \Big) \\
&\quad\div \big(\|u\|^2-(2p-1)\tilde{B}(u)\big).
\end{align*}
 \end{itemize}
 \end{lemma}

\begin{proposition} \label{prop4.1} 
 Assume {\rm (A4)} and {\rm (A5)} hold.
Then \eqref{eq4} has a local minimum solution with the least energy
$j_0=\inf_{\mathcal{N}}J(u)$.
\end{proposition}

\begin{proof} Firstly, we show that $\|u\|$ is bounded from both above and below:
For any $u\in \mathcal{N}$,
\begin{equation} \label{eq25b}
\begin{split}
J(u)
&=\frac{1}{2}\|u\|^2-\frac{1}{2p}\tilde{B}(u)-\tilde{C}(u)\\
&=\Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u\|^2
 -\Big(1-\frac{1}{2p}\Big)\int_{\mathbb{R}^N}h(x)u\,dx\\
&\geq \frac{p-1}{2p}\|u\|^2-\frac{2p-1}{2p}|h|_\frac{2Np}{2N(p-1)
 +\alpha}d_{\frac{2Np}{2N-\alpha}}\|u\|\\
&\geq -\frac{(2p-1)^2}{8p(p-1)}\|h\|_{\frac{2Np}{2N-\alpha}}^2.
\end{split}
\end{equation}
Thus,
\[
j_0\geq -\frac{(2p-1)^2}{8p(p-1)}\|h\|_{\frac{2Np}{2N-\alpha}}^2.
\]
Similar to the proof of Proposition \ref{prop3.1}, we can prove $j_0<0$.
By using the Ekeland's variational principle on $\mathcal{N}$,
we get a minimizing sequence $\{u_n\}\subset\mathcal{N}$ which satisfies
\begin{equation}\label{eq26b}
\begin{gathered}
J(u_n)<j_0+\frac{1}{n},\\
J(w)\geq J(u_n)-\frac{1}{n}\|u-w\|,\quad w\in \mathcal{N}.
\end{gathered}
\end{equation}
Since $\{u_n\}\subset \mathcal{N}$, it follows that
$\|u_n\|^2=\tilde{B}(u_n)+\tilde{C}(u_n)$.
Furthermore, we infer from \eqref{eq26} that
\begin{equation} \label{eq27}
\begin{split}
j_0+\frac{1}{n}
&\geq J(u_n)=\Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u_n\|^2
 -\Big(1-\frac{1}{2p}\Big)\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&\geq \Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u_n\|^2
 -\Big(1-\frac{1}{2p}\Big)|h|_\frac{2Np}
{2N(p-1)+\alpha}d_{\frac{2Np}{2N-\alpha}}\|u_n\|.
\end{split}
\end{equation}
We know that $\{u_n\}$ is bounded. We claim that $\inf_n\|u_n\|\geq \sigma_1>0$,
 which $\sigma_1$ is a positive constant.
Indeed, if not, by \eqref{eq27}, $J(u_n)$ would converge to zero.
 We can infer that $j_0\geq 0$ which is contradict with $j_0<0$. So we have
\begin{equation}\label{eq28}
\sigma_1 \leq \|u_n\|\leq \delta_1.
\end{equation}

Secondly, we claim that, for a subsequence of $\{u_n\}$ (still denoted 
by $\{u_n\}$), $\|\nabla J(u_n)\|\to 0$ as $n\to \infty$.

In fact, if the claim were false, we could assume $\|\nabla J(u_n)\|\geq d>0$ for
 $n$ large enough. Consequently, according to Lemma \ref{lem4.3}, for $u_n$ there 
exist $\epsilon_n$ and differentiable $\eta_n$ satisfying
\begin{equation*}
\eta_n(0)=1,\quad \eta_n(w)(u_n-w)\in \mathcal{N},\quad \|w\|<\epsilon_n
\end{equation*}
and
\begin{align*}
\langle \eta_n'(0),w\rangle
&=\Big(2\int_{\mathbb{R}^N}(\nabla u_n\nabla w+u_nw)dx\\
&\quad -2p\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|u_n(y)|^{p}|u_n(x)|^{p-2}u_n(x)w(x)}{|x-y|^{\alpha}}\,dx\,dy
 -\int_{\mathbb{R}^N}hw\,dx \Big)\\
&\quad\div \big(\|u_n\|^2-(2p-1)\tilde{B}(u_n)\big).
\end{align*}
We choose $w_n=\delta_n\frac{\nabla J(u_n)}{\|\nabla J(u_n)\|}$, 
$v_n=\eta_n(w_n)(u_n-w_n)$, where $0<\delta_n<\epsilon_n$ is sufficiently 
small satisfying
$\delta_n\to 0$, $\eta_n(w_n)\to 1$ as $n\to \infty$ and
\begin{gather*}
\frac{\big|J(v_n)-J(u_n)-\langle J'(u_n), v_n-u_n) \rangle \big|}
{\|u_n-v_n\|}<\frac{1}{n}, \\
\frac{\big|\eta(w_n)-1-\langle \eta_n'(0), w_n \rangle \big|}{\|w_n\|}<1.
\end{gather*}
From \eqref{eq25}, that $v_n\in \mathcal{N}$
and the above, we deduce that
\[
\frac{1}{n}\|v_n-u_n\| \geq J(u_n)-J(v_n)
\geq\langle J'(u_n), u_n-v_n \rangle -\frac{1}{n}\|u_n-v_n\|
\]
Thus, we have
\begin{align*}
&\frac{2}{n}\|\eta_n(w_n)(u_n-w_n)-u_n\|\\
&\geq (1-\eta_n(w_n))\langle J'(u_n), u_n \rangle
 +\eta_n(w_n)\delta_n\langle J'(u_n), 
\frac{\nabla J(u_n)}{\|\nabla J(u_n)\|} \rangle,
\end{align*}
and
\[
\frac{2}{n}\Big[(|\langle \eta'_n(0), w_n \rangle|+\|w_n\|) \|u_n\|
+ \eta_n(w_n)\|w_n\|\Big]\geq
\eta_n(w_n)\delta_n\|\nabla J(u_n)\|.
\]
Dividing by $\delta_n>0$ on both left and right hand of the above inequality, we get
\begin{equation}\label{eq29}
\frac{2}{n}\Big[(|\langle \eta'_n(0), 
\frac{\nabla J(u_n)}{\|\nabla J(u_n)\|} \rangle|+1) \|u_n\|+ \eta_n(w_n)\Big]
\geq \eta_n(w_n)\|\nabla J(u_n)\|.
\end{equation}
Now, if there exists $\lambda>0$ such that
\begin{equation*}
\big|\|u_n\|^2-(2p-1)\tilde{B}(u_n) \big|\geq \lambda,
\end{equation*}
we can get the claim. In fact,
\begin{align*}
&\Big|\langle \eta_n'(0),h_n\rangle\Big|\\
&=\Big|\frac{2(u_n,h_n)-\int_{\mathbb{R}^N}h_nu_n-2p\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|u_n(y)|^{p}|u_n(x)|^{p-2}u_n(x)h_n(x)}{|x-y|^{\alpha}}\,dx\,dy}
{\|u_n\|^2-(2p-1)\tilde{B}(u_n)}\Big|\\
&\leq \frac{\tilde{C}}{\lambda}.
\end{align*}
Here, $h_n=\nabla J(u_n)/\|\nabla J(u_n)\|$ and we have used the 
uniformly boundedness of $\|u_n\|$. 
Consequently, as $n\to\infty$,
\begin{equation*}
\frac{2}{n}\Big[(|\langle \eta'_n(0), h_n \rangle|+1) \|u_n\|+ \eta_n(w_n)\Big]\to 0.
\end{equation*}
So that, by passing to the limit as $n\to\infty$ in \eqref{eq29},
we get a contradiction which implies the claim is true.

To show the existence of positive lower bound of
 $\big|\|u_n\|^2-(2p-1)\tilde{B}(u_n) \big|$, we argue indirectly again and assume
\begin{equation*}
\|u_n\|^2-(2p-1)\tilde{B}(u_n) =o(1)\quad n\to \infty.
\end{equation*}
Here, $\{u_n\}$ is a subsequence still denoted by itself.
Combining this and \eqref{eq27}, similarly to the proof of Lemma \ref{lem4.2}(ii),
 we can easily get a contradiction.

So that we conclude that, for a subsequence  still denoted by $\{u_n\}$,
\[
I(u_n)\to j_0,\quad \|\nabla J(u_n)\|\to 0,
\]
as $n\to \infty$. By \eqref{eq27} we know that $\{u_n\}$
is bounded in $H^1(\mathbb{R}^N)$, and the weak limit of $\{u_n\}$ which
we denote by $u_0$ is a weak solution of system \eqref{eq4}.
Obviously, $u_0\in\mathcal{N}$ and
\begin{align*}
j_0\leq J(u_0)
&=\frac{1}{2}\|u_0\|^2-\frac{1}{2p}\tilde{B}(u_0)-\tilde{C}(u_0)\\
 &=\Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u_0\|^2
-\Big(1-\frac{1}{2p}\Big)\tilde{C}(u_0)\\
 &\leq \liminf J(u_n)=j_0.
\end{align*}
Therefore, $u_0$ is a least energy solution.

Now, we only need to show that $u_0$ is a local minimum solution. 
We apply Lemma \ref{lem4.1} to $u_0$ and $|u_0|$.
Since
\begin{equation*}
\frac{d}{dt}J(tu_0)=\psi'(t)>0,\quad t\in (\tilde{t}^+(u_0), \tilde{t}^-(u_0)),
\end{equation*}
we know that $u_0\in\mathcal{N}^+$. Otherwise, $u_0\in \mathcal{N}^-$
 and $j_0\leq J(\tilde{t}^+(u_0)u_0)<J(u_0)=j_0$ which is a contradiction. 
By Lemma \ref{lem4.1} and $u_0\in\mathcal{N}^+$ we know that
\begin{equation*}
1=\tilde{t}^+(u_0)< \tilde{t}_0(u_0)
=\Big[\frac{\|u_0\|^2}{(2p-1)\tilde{B}(u_0)}\Big]^{1/(2p-2)}.
\end{equation*}
Therefore
\begin{equation*}
1<\Big[\frac{\|u_0-w\|^2}{(2p-1)\tilde{B}(u_0-w)}\Big]^{1/(2p-2)}, \quad 
\|w\|<\varepsilon,
\end{equation*}
for $\varepsilon$ small enough. Applying Lemma \ref{lem4.3}, we get a $\eta(w)>0$ 
such that $\eta(w)(u_0-w)\in\mathcal{N}$ for $\|w\|<\varepsilon$ small.
Moreover, it holds $\eta(w)\to 1$ as $w\to 0$. Thus we can assume that, 
for $\|w\|<\varepsilon$ sufficiently small,
\begin{equation*}
\eta(w)<\Big[\frac{\|u_0-w\|^2}{(2p-1)\tilde{B}(u_0-w)}\Big]^{1/(2p-2)}, \quad
\eta(w)(u_0-w)\in \mathcal{N}^+.
\end{equation*}
Then by Lemma \ref{lem4.1}, we  conclude that
\begin{equation}
\label{eq30}
J(u_0)\leq J(\eta(w)(u_0-w))\leq J(t(u_0-w))
\end{equation}
for $0<\eta<\big[\frac{\|u_0-w\|^2}{(2p-1)\tilde{B}(u_0-w)}\big]^{1/(2p-2)}$.
Taking $\eta=1$ in \eqref{eq30} we have
\begin{equation*}
J(u_0)\leq J(u_0-w),\quad \|w\|<\varepsilon,
\end{equation*}
which means $u_0$ is a local minimum solution.

Additionally, if we assume that $h>0$, then
\begin{equation*}
\psi'_{|u_0|}(t)<\psi'_{u_0}(t)<0, \quad t\in [0,1).
\end{equation*}
Hence, $\tilde{t}^+(|u_0|)\geq 1$ and consequently,
\begin{equation*}
j_0\leq J(\tilde{t}^+(|u_0|)|u_0|)\leq J(|u_0|)\leq J(u_0)=j_0.
\end{equation*}
Therefore, $\tilde{t}^+(|u_0|)=1$ and 
$\int_{\mathbb{R}^N}h(x)|u_0|\,dx=\int_{\mathbb{R}^N}h(x)u_0\,dx$, 
which yield $u_0\geq 0$. Then by the maximum principle, we know $u_0>0$.
The proof is complete.
\end{proof}

Consider the nonlinear Schr\"odinger equation
\begin{equation}
\label{eq31}
-\Delta u+u=\Big(\frac{1}{|x|^{\alpha}}\ast |u|^{p}\Big)|u|^{p-2}u \quad
\text{in } \mathbb{R}^N.
\end{equation}
By \cite[Proposition 2.]{Moroz2}, we know that \eqref{eq31} has positive
smooth solution $V(x)$, which is also a minimizer of
\begin{equation*}
 S_{\alpha,p}=\inf_{u\in H^{1}(\mathbb{R}^N)\setminus\{0\}}
\frac{\int_{\mathbb{R}^N}|\nabla u|^2+u^2dx}
{(\int_{\mathbb{R}^N}\int_{\mathbb{R}^N}\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}
\,dx\,dy)^{1/p}}.
\end{equation*}
If $V$ is a positive solution of \eqref{eq31} if and only if $V$ is a
critical point of the energy functional
\begin{equation*}
\mathcal{J}(u)=\frac{1}{2}\int_{\mathbb{R}^N} (|\nabla u|^2 + u^2)dx
-\frac{1}{2p}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy.
\end{equation*}
We know that
\begin{equation*}
\|V\|=\tilde{B}(V)=S_{\alpha, p}^{\frac{p}{p-1}}.
\end{equation*}
From the fact that the Sobolev embedding
\begin{equation*}
H^1(\mathbb{R}^N)\hookrightarrow L^q(\mathbb{R}^N)\quad (2\leq q\leq 2^{\ast})
\end{equation*}
is not compact, the variational functional $J(u)$ fails to satisfy the (PS) 
condition. Such a failure brings us some difficulties in
applying the variational approach. In order to overcome the lack of 
compactness, we introduce the following proposition which plays a key role
in our argument.
We remark here that since $j_0<0$, any solution $u_0$ of system \eqref{eq4}
 with the least energy $j_0$ satisfies
\begin{align*}
j_0&=J(u_0)
=\frac{1}{2}\|u_0\|^2-\frac{1}{2p}\tilde{B}(u_0)-\int_{\mathbb{R}^N}h(x)u_0dx\\
&=\Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u_0\|^2
-\Big(1-\frac{1}{2p}\Big)\int_{\mathbb{R}^N}h(x)u_0\,dx
<0.
\end{align*}
Thus, $\int_{\mathbb{R}^N}h(x)u_0dx> 0$ and consequently, $u_0\in\mathcal{N}^+$.


\begin{proposition} \label{prop4.2} 
 Let $N\geq 3$, $0<\alpha<N$, {\rm (A4)} and {\rm (A5)} hold. 
If $\{u_n\}$ be a $(PS)_c$ sequence of $J$ with
\begin{equation} \label{eq32}
 c<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1},
\end{equation}
then $\{u_n\}$ has a convergent subsequence.
\end{proposition}

\begin{proof} 
Obviously, $\|u_n\|$ is bounded. Thus, there exists $w\in H^1(\mathbb{R}^N)$
 which satisfies $u_n\rightharpoonup v$ weakly
in $ H^1(\mathbb{R}^N)$ and solves \eqref{eq4}.
Therefore, $v\not=0$ and $J(v)\geq j_0$. Let $u_n-v=w_n$, by Brezis-Lieb 
lemma and \cite[Lemmas 2.1 and 2.2]{GaoYang}, we deduce
\begin{equation*}
\|u_n\|^2=\|w_n\|^2+\|v\|^2+o(1),\quad n\to \infty,
\end{equation*}
and
\begin{align*}
&\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|u_n(x)|^{p}|u_n(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy\\
&=\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|w_n(x)|^{p}|w_n(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy 
 +\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|w(x)|^{p}|w(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy+o_n(1)
\end{align*}
as $ n\to \infty $.
So we obtain
\begin{align*}
c\leftarrow J(u_n)
&=\frac{1}{2}\|u_n\|^2-
\frac{1}{2p}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
\frac{|u_n(x)|^{p}|u_n(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy 
-\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&=\frac{1}{2}\|w_n\|^2-\frac{1}{2p}\int_{\mathbb{R}^N}\int_{\mathbb{R}^N} 
 \frac{|w_n(x)|^{p}|w_n(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy 
 -\int_{\mathbb{R}^N}h(x)w_n\,dx\\
&\quad +\frac{1}{2}\|v\|^2-\frac{1}{2p}\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N} \frac{|v(x)|^{p}|v(y)|^{p}}{|x-y|^{\alpha}}\,dx\,dy
 -\int_{\mathbb{R}^N}h(x)v\,dx+o_n(1)\\
&=J(v)+\frac{1}{2}\|w_n\|^2-\frac{1}{2p}\tilde{B}(w_n)+o_n(1).
\end{align*}
As a result, for $n$ large, we have
\begin{equation} \label{eq33}
\frac{1}{2}\|w_n\|^2-\frac{1}{2p}\tilde{B}(w_n)+o_n(1)
<\frac{p-1}{2p}S_{\alpha, p}^{\frac{p}{p-1}}.
\end{equation}
On the other hand,
\[
o(1)=\langle J'(u_n),u_n\rangle
=\langle J'(v),v\rangle+\|w_n\|^2-\tilde{B}(w_n)+o(1),
\]
which implies
\begin{equation}\label{eq34}
\|w_n\|^2-\tilde{B}(w_n)=o(1).
\end{equation}
If we can show that $\{w_n\}$ has subsequence converging strong to $0$, 
we have the result. Therefore, arguing indirectly, we assume
 $\|w_n\|\geq C>0$ for $n$ large.
By \eqref{eq34}, we have
\begin{equation*}
\|w_n\|^2=\tilde{B}(w_n)\leq \frac{\|w_n\|^{2p}}{S_{\alpha, p}^{p}}
\end{equation*}
and
\begin{align*}
\frac{1}{2}\frac{p-1}{p}S_{\alpha, p}^{\frac{p}{p-1}}
&=\frac{1}{2}(1-\frac{1}{p})S^{\frac{p}{p-1}}\\
&\leq \frac{1}{2}(1-\frac{1}{p})\|w_n\|^{2}\\
&=\frac{1}{2}\|w_n\|^2-\frac{1}{2p}\tilde{B}(w_n)+o_n(1)\\
&<\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}
\end{align*}
which is a contradiction.
The proof is complete.
\end{proof}

To prove Theorem \ref{thm1.2}, the only we need to show  that
\[
j_0<j_1=\inf_{\mathcal{N}^-}J(u)<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}.
\]
Consider $V(x)$ is a minimizer for both $S_{\alpha, p}$.
 Let $u_0$ be the positive local minimum solution we get above.
By the continuity of $J$, we know that there exists $\gamma>0$ such that
\[
J(u_0+tV)<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}, \quad 0\leq t<\gamma.
\]
\begin{align*}
J(u_0+tV)
&=\frac{1}{2}\|u_0+tV\|^2-\frac{1}{2p}\tilde{B}(u_0+tV)
 -\int_{\mathbb{R}^N}h(u_0+tV)dx\\
&=J(u_0)+\frac{t^2}{2}[\|V\|^2-\frac{t^{p-2}}{p}\tilde{B}(V)]
 +\tilde{B}(u_0)+\tilde{B}(tV)-\tilde{B}(u_0+tV)\\
&<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}.
\end{align*}
For $t\geq \gamma $, a directly computation shows that
\begin{align*}
J(u_0+tV)
&=\frac{1}{2}\|u_0+tV\|^2-\frac{1}{2p}\tilde{B}(u_0+tV)
 -\int_{\mathbb{R}^N}h(u_0+tV)dx\\
&=\frac{1}{2}\|u_0\|^2+t\int_{\mathbb{R}^N}\nabla u_0\nabla V+u_0Vdx
 +\frac{t^2}{2}\|V\|^2-\frac{1}{2p}\tilde{B}(u_0)\\
&\quad +\frac{1}{2p}[\tilde{B}(u_0)+\tilde{B}(tV)-\tilde{B}(u_0+tV)]
 -\frac{1}{2p}\tilde{B}(tV)\\
&\quad -\int_{\mathbb{R}^N}hu_0dx-\int_{\mathbb{R}^N}htVdx\\
&=J(u_0)+\frac{t^2}{2}[\|V\|^2-\frac{t^{2(p-1)}}{2p}\tilde{B}(V)]
+\frac{1}{2p}[\tilde{B}(u_0)+\tilde{B}(tV)\\
&\quad -\tilde{B}(u_0+tV)+2p\int_{\mathbb{R}^N}
 \int_{\mathbb{R}^N}\frac{|u_0(x)|^{p}|u_0(y)|^{p-2}u_0(y)}{|x-y|^{\alpha}}\,dx\,dy]\\
&<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}.
\end{align*}
Here, we use  that $\langle J'(u_0), tV\rangle =0$ and $V(x)$
 is a solution of \eqref{eq31}.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
To show that
\begin{equation*}
j_0<j_1=\inf_{\mathcal{N}^-}J(u)<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}.
\end{equation*}
Firstly, we observe that for every $u\in H^1({\mathbb{R}^N})$ with $\|u\|=1$, 
there exists a unique $\eta^-(u)>0$ such that (see Lemma \ref{lem4.3})
\begin{equation*}
t^-(u)\in \mathcal{N}^-.
\end{equation*}
By Lemma \ref{lem4.3}, we know that $\eta^-(u)$ is a continuous function of $u$. 
Consequently the manifold $\mathcal{N}^-$ disconnects $H^1({\mathbb{R}^N})$ 
in exactly two connected components $U_1$ and $U_2$, where
\begin{gather*}
U_1=\Big\{u\in H^{1}({\mathbb{R}^N}): u=0 \text{ or }
 \|u\|<t^-\big(\frac{u}{\|u\|}\big)\Big\}, \\
U_2=\Big\{u\in H^{1}({\mathbb{R}^N}); \|u\|>t^-\big(\frac{u}{\|u\|}\big)\Big\}.
\end{gather*}
Obviously, $H^{1}({\mathbb{R}^N})=\mathcal{N}^-\cup U_1 \cup U_2$. 
In particular, $u_0\in \mathcal{N}^+ \subset U_2$.
Since
\begin{equation*}
t^-\Big(\frac{u_0+tV}{\|u_0+tV\|}\Big)\frac{u_0+tV}{\|u_0+tV\|}\in \mathcal{N},
\end{equation*}
we have
\begin{equation*}
0<t^-\Big(\frac{u_0+tV}{\|u_0+tV\|}\Big)<C_0
\end{equation*}
uniformly for $t\in \mathbb{R}$.

On the other hand, there exists $\tilde{t}>0$ such that
\[
\|u_0+tU\|\geq t\|V\|-\|u_0\|\geq C_0,\quad t\geq \tilde{t}.
\]
So that we can fix a $t_0>0$ such that
$\|u_0+t_0V\|> t^-(\frac{u_0+t_0V}{\|u_0+t_0V\|})$. Thus,
$u_0+t_0V\in U_2$. Combining this and the fact $u_0\in U_1$, we know that
\[
u_0+t_1V\in \Lambda^-,
\]
for some $0<t_1<t_0$. Consequently, by Remark \ref{rmk3.3},
we have
\[
j_1=\inf_{\Lambda^-}J(u)\leq \max_{0\leq t\leq t_0}J(u_0+tV)
<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}.
\]

Next, we show that $j_1$ is a critical value of $J$ and satisfies $j_1>j_0$.
Similarly to the proof of Proposition \ref{prop4.1}, we apply Ekeland's variational 
principle and get a minimizing sequence $\{u_n\}\subset \mathcal{N}^-$ such that
\begin{gather*}
J(u_n)<j_1+\frac{1}{n},\\
J(w)\geq J(u_n)-\frac{1}{n}\|u-w\|,\quad w\in \mathcal{N}^-.
\end{gather*}
So that we have
\begin{align*}
j_1+1
&>J(u_n)
 =\frac{1}{2}\|u_n\|^2-\frac{1}{2p}\tilde{B}(u_n)-\int_{\mathbb{R}^N}h(x)u_n\,dx\\
&\geq \Big(\frac{1}{2}-\frac{1}{2p}\Big)\|u_n\|^2
 -\Big(1-\frac{1}{2p}\Big)|h|_{\frac{2Np}{2N(p-1)+\alpha}}
d_{\frac{2Np}{2N-\alpha}}\|u_n\|
\end{align*}
which applies $\|u_n\|$ has a upper bound.
Moreover, from $\{u_n\}\subset \mathcal{N}^-$, we know that
\begin{equation*}
\|u_n\|^2\leq (2p-1)\frac{\|u_n\|^{p}}{S_{\alpha, p}^p}.
\end{equation*}
Thus, $\|u_n\|$ has a uniform positive lower bound. 
Then, analogously to the proof of Proposition \ref{prop4.1}, we know that
\[
J(u_n)\to j_1,\quad J'(u_n)\to 0\quad\text{in } H^{-1}.
\]
From Proposition \ref{prop4.2} and $j_1<j_0+\frac{p-1}{2p}S_{\alpha, p}^\frac{p}{p-1}$,
 we can conclude that there exists a subsequence of $\{u_n\}$ such that
 $u_n\to u_1$ strongly in $H^1(\mathbb{R}^N)$. 
Therefore $u_1$ is a critical point of $J$ and $J(u_1)=j_1$. 
Noting that $\mathcal{N}^-$ is closed, we have $u_1\in\mathcal{N}^-$.
To show $j_1>j_0$, arguing indirectly, we assume $j_1=j_0$. 
Thus by Remark \ref{rmk3.3} we have
\begin{equation*}
\int_{\mathbb{R}^N}h(x)u_1dx>0 \quad\text{and} \quad u_1\in\mathcal{N}^+,
\end{equation*}
which leads to a contradiction.

Finally, we consider the case $h>0$. Applying Lemma \ref{lem4.3} to $u_1$ and $|u_1|$, 
we know that there exist a $\eta^-(|u_1|)$ such that
$\eta^-(|u_1|)|u_1|\in\mathcal{N}^-$.
Moreover,
\[
\eta^-(|u_1|)\geq \eta_0(|u_1|)=\eta_0(u_1)
=\Big[\frac{\tilde{A}(u_1)}{(p-1)\tilde{B}(u_1)}\Big]^{\frac{1}{p-2}}
\]
Thus both in the case $\int_{\mathbb{R}^N}h(x)u\,dx>0$ and 
$\int_{\mathbb{R}^N}h(x)u\,dx\leq 0$, we can deduce that
\[
j_1=J(u_1)\geq J(\eta^-(u_1)u_1)\geq J(\eta^-(|u_1|)|u_1|)=\eta_0(u_1)\geq j_1.
\]
Therefore, $\int_{\mathbb{R}^N}h(x)u_1dx=\int_{\mathbb{R}^N}h(x)|u_1|dx$, 
which implies $u_1\geq 0$. By the maximum principle we get $u_1>0$.
The proof is complete.
\end{proof}

\subsection*{Acknowledgments}
L. Wang is partially supported by Postdoctoral Science
Foundation of China (2017M611159) and Research Fund of National Natural
Science Foundation of China (11801400,11571187).

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