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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 63, pp. 1--21.\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-2018/63\hfil Nonlinear Choquard equation]
{Existence of positive solutions to the nonlinear Choquard equation
with competing potentials}

\author[J. Wang, M. Qu, L. Xiao \hfil EJDE-2018/63\hfilneg]
{Jun Wang, Mengmeng Qu, Lu Xiao}

\address{Jun Wang (crresponding author) \newline
Faculty of Science,
Jiangsu University,
Zhenjiang, Jiangsu 212013, China}
\email{wangmath2011@126.com}

\address{Mengmeng Qu  \newline
Faculty of Science,
Jiangsu University,
Zhenjiang, Jiangsu 212013, China}
\email{qumengmengab@126.com}

\address{Lu Xiao \newline
School of management,
Jiangsu University,
Zhenjiang, Jiangsu 212013, China}
\email{hnlulu@126.com}

\dedicatory{Communicated by Paul H. Rabinawitz}

\thanks{Submitted October 11, 2017. Published March 7, 2018.}
\subjclass[2010]{35J61, 35J20, 35Q55, 49J40}
\keywords{Positive solutions; Choquard equation; competing coefficients;
\hfill\break\indent variational methods}

\begin{abstract}
 This article concerns the existence of positive solutions of the
 nonlinear Choquard equation
 \begin{equation*}
 -\Delta u+a(x)u=b(x)\Big(\frac{1}{|x|}*|u|^2\Big)u,\quad
  u\in H^{1}({\mathbb R}^3),
 \end{equation*}
 where the coefficients $a$ and $b$ are positive functions such that
 $a(x)\to\kappa_\infty$ and $b(x)\to \mu_\infty$ as
 $|x|\to\infty$. By comparing the decay rate of the
 coefficients $a$ and $b$, we prove the existence of positive ground
 and bound stat solutions of Choquard equation.
\end{abstract}

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

\section{Introduction}

In this article studies the existence of positive solution of the
nonlinear Choquard equation
\begin{equation}\label{1-1}
-\Delta u+a(x)u=b(x)\Big(\frac{1}{|x|}*|u|^2\Big)u,\quad
  u\in H^{1}({\mathbb R}^3),
\end{equation}
where the coefficients $a(x)$ and $b(x)$ are positive functions such
that $\lim_{|x|\to\infty}a(x) =\kappa_{\infty}>0$ and
$\lim_{|x|\to\infty}b(x)=\mu_{\infty}>0$.


Equation \eqref{1-1} is  called the nonlinear Choquard or
Choquard-Pekar equation. It has several physical origins.
Equation \eqref{1-1} first appeared  as early as in 1954, in
a work by  Pekar describing the quantum mechanics of a polaron
at rest \cite{Pekar-1954}. In 1976,  Choquard used \eqref{1-1} to
describe an electron trapped in its own hole, in a certain
approximation to Hartree-Fock theory of one component plasma
\cite{Lieb-1976-Stud-math}. In 1996, Penrose proposed \eqref{1-1} as
a model of self-gravitating matter, in a programme in which quantum
state reduction is understood as a gravitational phenomenon
\cite{Moroz-Penrose-Tod-1998}. In this context equation of type
\eqref{1-1} is usually called the nonlinear Schrodinger-Newton
equation. In general, many mathematicians study the existence of the
solitary solutions of the nonlinear generalized Choquard equation
\begin{equation}\label{1-2}
i\psi_t-\Delta\psi+K(x)\psi-b(x)\Big(\frac{1}{|x|^{\alpha}}*|\psi|^p\Big)
|\psi|^{p-2}\psi=0,\
(x,t)\in\mathbb{R}^{N}\times\mathbb{R}_+.
\end{equation}
where $N\geq 1$, $\alpha \in (0,N)$,
$\frac{N-2}{N+\alpha}<\frac{1}{p}<\frac{N}{N+\alpha}$. To
obtain the solitary solutions of \eqref{1-2}, we set
$\psi(t,x)=u(x)e^{i\omega t}$ ($\omega$ is a constant) in \eqref{1-2}
and get the stationary equation of the form
\begin{equation}\label{1-3}
-\Delta
u+a(x)u=b(x)\Big(\frac{1}{|x|^{\alpha}}*|u|^p\Big)|u|^{p-2}u,\quad
u\in H^{1}(\mathbb R^N),
\end{equation}
where $a=K(x)-\omega$. Obviously, if $N=3, \alpha=2$ and $p=2$ the
equation \eqref{1-3} reduces to \eqref{1-1}. In recent years, many
papers are concerned with the existence of solutions of \eqref{1-3}.
 Lieb \cite{Lieb-1976-Stud-math} proved the existence and
uniqueness of the ground state to \eqref{1-2}.  Lions
\cite{Lions-1980-NA} obtained the existence of a sequence of
radially symmetric solutions for \eqref{1-3} by using variational
methods. Papers \cite{Ackermann-2004-MZ,Wei-Winter-2009-JMP}
showed the existence of multi-bump solutions of \eqref{1-3}.
Recently,  papers
\cite{Lenzmann-2009-Ann, Ma-Zhao-2010,Wang-Yi-2017-AA} showed some
partial uniqueness of the positive solutions of \eqref{1-3}. 
Papers \cite{Ghimenti-Schaftingen-2016-JFA, Moro-Schaftingen-2013}
showed the existence of positive and nodal solution of \eqref{1-3}.
For more results on this direction one can refer to
\cite{Bonheure-Cingolani-2017-JFA, Cingolani-Secchi-2015-Proc-A,
Cingolani-Clapp-2013-DCDS, Cingolani-Clapp-2012-ZAMP,
Cingolani-Secchi-2010-ProcA,
Cingolani-secchi-2015-JDE,
Clapp-Salazar-2013-JMAA,
Moroz-Schaftingen-2015-TAMS,
Moroz-Schaftingen-2015-CCM,
Moroz-Schaftingen-2017-JFPTA,
Schaftingen-2013-JDE,Moroz-Schaftingen-2015-CVPDE} and the references therein.

It is worth to point out that in most of the papers mentioned above, the
search for the positive ground state solutions to \eqref{1-3}. In
the present paper we consider a nonautonomous situation that has to
be studied in a different way. We will find the positive solution
which different from positive ground state solution. Here a solution
$u$ of \eqref{1-1} is nontrivial if $u\neq0$. A solution of
\eqref{1-1} is a \emph{nontrivial bound state solution} if $u$ is a
nontrivial solution. A solution $u$ with $u>0$ is called a
positive solution. A solution is called a \emph{nontrivial ground
state solution} (or \emph{positive ground state solution}) if its
energy is minimal among all the nontrivial solutions (or all the
positive solutions) of \eqref{1-1}. Here the energy functional
corresponding to \eqref{1-1} is defined by
\begin{equation}\label{1-4}
I_{\lambda}(u)=\frac{1}{2}\int_{\mathbb R^3}\left(|\nabla
u|^2+a(x)u^2\right)-\frac{1}{4}\int _{\mathbb
R^3}b(x)\phi_uu^{2 },\ \ u\in H^{1}(\mathbb{R}^3).
\end{equation}
We set
\begin{equation}\label{1-5}
 a(x)=\kappa_{\infty}+\lambda\kappa(x),\quad
 b(x)=\mu_{\infty}+\mu(x),
\end{equation}
where $\lambda\in\mathbb R^{+}$ and we assume
\begin{itemize}
 \item[(A1)] $\kappa\in L^{3/2}( \mathbb R^3)$, $\kappa\geq0$,
 $\kappa\neq0$,  $\lim_{|x|\to\infty}\kappa(x)=0$;
 \item[(A2)] $\mu\in L^2(\mathbb R^3)$,  $\mu\geq0$, $\mu\neq0$,
$\lim_{|x|\to\infty}\mu(x)=0$.
\end{itemize}
Hence,  equation \eqref{1-1} can be rewritten as
\begin{equation} \label{auto-2}
 -\Delta u+(\kappa_{\infty}+\lambda\kappa(x))u
=(\mu_{\infty}+\mu(x))\phi_u(x)u,\ u\in H^{1}({\mathbb R}^3).
\end{equation}
The purpose of this paper is to describe some phenomena that can
occur when the coefficients are competing. For each
$\lambda\in[0,\infty)$, we prove the existence of positive ground
state solution of \eqref{auto-2} if $\kappa(x)$ decays faster than
$\mu(x)$. Conversely, if $\mu(x)$ decays faster than $\kappa(x)$, we
find the threshold value $\lambda^*>0$ such that \eqref{auto-2} has
a ground state solution if $\lambda\in[0,\lambda^*)$, and no ground
state solution for $\lambda\in[\lambda^*,\infty)$. Furthermore, we
find the positive bound state solution of \eqref{auto-2} if
$\lambda\in[\lambda^*,\infty)$. Our study mainly motivated by the
recent works \cite{Gioanna-Alessio-2017,Giovanna-Riccardo-2016},
while the authors study the existence of positive solutions of
Schr\"odinger equation and Schr\"odinger-Poisson system with
competing coefficients. Comparing to the previous works
\cite{Gioanna-Alessio-2017,Giovanna-Riccardo-2016}, we encounter new
difficulty in finding the positive solutions of \eqref{auto-2}.
Precisely, we let $u_0$ denote the sign-changing solution of the
Sch\"odinger equation
\begin{equation}\label{1-7}
 -\Delta u+\kappa_{\infty}u=\mu_{\infty}|u|^{p-2}u,\quad u\in H^{1}({\mathbb R}^N).
\end{equation}
It is easy to check that $J_{\infty}(u_0)\geq2k_\infty$
(see \cite{Gioanna-Alessio-2017}), where
\begin{equation}\label{1-8}
 k_\infty=\inf J_{\infty}\quad\text{and}\quad
 J_{\infty}(u)=\frac{1}{2}\int_{\mathbb{R}^N}(|\nabla u|^2+\kappa_{\infty}u^2)
 -\frac{\mu_{\infty}}{p}\int_{\mathbb{R}^N}|u|^p.
\end{equation}
In \cite{Gioanna-Alessio-2017,Giovanna-Riccardo-2016}, this fact was
play an important role in recovering the compactness and finding the
bound state solution. However, the situation is totally different in
our case. In fact, consider the Choquard equation
\begin{equation}\label{1-9}
 -\Delta u+\kappa_{\infty}u=\mu_{\infty}\phi_u(x)u,\quad
 u\in H^{1}({\mathbb R}^3).
\end{equation}
According to \cite{Ghimenti-Schaftingen-2016-JFA}, we know that the
energy of sign-changing solution of \eqref{1-9} is strictly less
than two times the least energy level of \eqref{1-9}. This brings
the difficulty in recovering the compactness. We shall use the idea
of \cite{Ma-li-1997,J.Wang-2017} and consider our problem in convex
set $H_+^1(\mathbb{R}^3)$ to overcome this difficult, where
$H_+^1(\mathbb{R}^3):=\{u\in H^1(\mathbb{R}^3): u\geq0\}$.

Now we are ready to give the main results of the paper. We first
state the results when $\kappa(x)$ decays faster than $\mu(x)$.

\begin{theorem}\label{theorem1-1}
For $\tau\in(0,1)$, we assume that {\rm (A1), (A2)} hold and
\begin{itemize}
\item [(A3)] $\lim_{|x|\to\infty}\kappa(x)|x|
 e^{\frac{2\tau}{1-\tau}\sqrt{\kappa_\infty}|x|}=0$,\quad
$\lim_{|x|\to\infty}\mu(x)e^{\frac{2\tau}{1+\tau}\sqrt{\kappa_\infty}|x|}=+\infty$.
\end{itemize}
 Then for all $\lambda\in \mathbb R^{+}$, Equation \eqref{auto-2} always has 
a positive ground  state  solution.
\end{theorem}

Next we study the case when $\mu(x)$ decays faster than $\kappa(x)$.

\begin{theorem}\label{theorem1-2}
Assume that {\rm (A1), (A2)} hold, and for some $\tau\in(0,1)$,
$\sigma\in(0,\kappa_\infty)$, and $c_1, c_2>0$, we have
\begin{itemize}
\item [(A4)] $\lim\inf_{|x|\to\infty}\kappa(x)e^{\frac{4\tau}
{1+\tau}\sqrt{\sigma}|x|}\geq c_1$ and
$\lim\sup_{|x|\to\infty}\mu(x)e^{\frac{4\tau}{1-\tau}\sqrt{\sigma}|x|}\leq c_2$.
\end{itemize}
 Then there exist a number $\lambda^{*}>0$, such that for all
$\lambda\in[0,\lambda^{*})$, Equation \eqref{auto-2}  has a positive
ground state  solution, while if
$\lambda\in[\lambda^{*},+\infty)$, Equation \eqref{auto-2} has
no positive ground state  solution.
Additionally, if we assume that
\begin{itemize}
\item [(A5)] $\lim\sup_{|x|\to\infty}\kappa(x)|x|^2e^{2\sqrt{\kappa_\infty}|x|}
\leq c_{3}$ for some $c_3>0$,
\end{itemize}
then for $\lambda\in[\lambda^{*},+\infty)$,
Equation \eqref{auto-2} has a positive solution.
\end{theorem}

\begin{remark} \rm
To the best our knowledge, this is the first results on the
existence of positive solution of Choquard equation \eqref{auto-2}
with competing coefficients. We believe our arguments can also work
on the generalized Choquard equation \eqref{1-3} and other nonlocal
problems. This is an interesting issue that can be pursued in the
future.
\end{remark}

\section{Preliminary results}

Throughout this article we shall use the following notation.
\begin{itemize}
 \item The scalar product in $H^1(\mathbb{R}^3)$ is
defined by
\[
(u,v)= \int_{\mathbb{R}^3}[\nabla u \nabla v+\kappa_{\infty}u v]
\]
 and the norm is defined by
$\|u\|=\sqrt{(u,u)}$, where $\kappa_{\infty}>0$ is given in
\eqref{auto-2};

\item the norm of $D^{1,2}(\mathbb{R}^3)$ defined by
$\|u\|_{D^{1,2}}^2=\int_{\mathbb R^3}|\nabla u|^2$;

\item $c^{*}$ or $c,c_i$  denote different positive constants;

\item  the norm in $L^{p}(\mathbb{R}^3)$ defined by
$|u|_{p}^{p}=\int_{\mathbb R^3}|u|^{p}$.
\end{itemize}

In this part we  given some basic knowledge which will be used
in the later. Considering for all $u\in H^{1}(\mathbb{R}^3)$, the
linear functional ${J}_u$ defined in $D^{1,2}(\mathbb{R}^3)$ by
\begin{equation*}
J_u(v)=\int_{\mathbb{R}^3}u^2v.
\end{equation*}
We infer from the H\"older inequality that
\begin{equation}\label{b-2-1}
|J_u(v)|\leq C|u|_{12/5}^2\|v\|_{D^{1,2}}.
\end{equation}
By the Lax-Milgram theorem, we know that there exists unique
$\phi_u\in D^{1,2}(\mathbb{R}^3)$ such that
\begin{equation}\label{b-2-2}
\int_{\mathbb{R}^3}\nabla\phi_u\nabla v
=\int_{\mathbb{R}^3}u^2v \quad \forall v\in D^{1,2}(\mathbb{R}^3).
\end{equation}
So, $\phi_u$ is a weak solution of $-\Delta\phi=u^2$ and the
following formula holds
\begin{equation}\label{b-2-3}
\phi_u(x)=\int_{\mathbb{R}^3}\frac{u^2(y)}{|x-y|}dy=\frac{1}{|x|}\ast u^2.
\end{equation}
Moreover, $\phi_u>0$ when $u\neq0$.

We recall the following classical Hardy-Littlewood-Sobolev
inequality (see \cite[Theorem 4.3]{Lieb-Loss-2001}). Assume that
$f\in L^p(\mathbb{R}^3)$ and $g\in L^q(\mathbb{R}^3)$. Then one has
\begin{equation}\label{2-4}
\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\frac{f(x)g(y)}{|x-y|^{t}}dx
dy \leq c(p,q,t)|f|_{p}|g|_{q},
\end{equation}
where $1<p,q<\infty$, $0<t<N$ and
$\frac{1}{p}+\frac{1}{q}+\frac{t}{3}=2$. By \eqref{2-4} we know
that
\begin{equation}\label{2-5}
\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\frac{u^2(x)u^2(y)}{|x-y|}dxdy
\leq c|u|_{12/5}^{4}\leq c\|u\|^{4}.
\end{equation}
It is well-known that solutions of \eqref{auto-2} correspond to
critical points of the energy functional
\begin{equation}\label{2-6}
I_{\lambda}(u)=\frac{1}{2}\int_{\mathbb R^3}\left(|\nabla
u|^2+(\kappa_{\infty}+ \lambda\kappa(x))u^2\right)
-\frac{1}{4}\int _{\mathbb R^3}(\mu_{\infty}+\mu(x))\phi_uu^{2 },
\end{equation}
for $u\in H^{1}(\mathbb{R}^3)$.
From \eqref{2-5}, we know that $I_{\lambda}$ is well defined, and
that
\begin{equation}\label{2-7}
I'_{\lambda}(u)[v]=\int_{\mathbb R^3}[\nabla u\nabla v+
(\kappa_{\infty}+ \lambda\kappa(x))uv]-\int _{\mathbb
R^3}(\mu_{\infty}+\mu(x))\phi_uuv,
\end{equation}
for all $v\in H^{1}(\mathbb{R}^3)$.
We define the operator $\Phi:H^{1}(\mathbb{R}^3)\to
D^{1,2}(\mathbb{R}^3)$ as
\begin{equation*}%\label{a-1}
\Phi[u]=\phi_u.
\end{equation*}
From \cite[Proposition 2.2-2.3]{Giovanna-Riccardo-2016} we know that
$\Phi$ has the following properties.
\begin{lemma}\label{Lem-2-1}
\begin{enumerate}
  \item [(1)] $\Phi$ is continuous;
  \item [(2)] $\Phi$ maps bounded sets into bounded sets;
  \item [(3)] $\Phi[tu]=t^2\Phi[u]$  for all $t\in\mathbb{R}$;
  \item [(4)] If $u_n\rightharpoonup u \in H^{1}(\mathbb{R}^3)$ then
 $\Phi[u_n]\to\Phi[u]\ in\ D^{1,2}(\mathbb{R}^3)$. 
Moreover, 
\begin{gather*}
\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}(x)u_n^2
 \to\int_{\mathbb{R}^3}\mu(x)\phi_u(x)u^2, \\
\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}(x)u_n\phi\to\int_{\mathbb{R}^3}\mu(x)
 \phi_u(x)u\phi, 
\end{gather*}
for all $\phi\in H^{1}(\mathbb{R}^3)$.
\end{enumerate}
\end{lemma}

It is  easy to verify that, whatever $ \lambda\in\mathbb R $
 is, the function $I_{\lambda}$ is bounded neither from above nor
 from below. Hence, it is convenient to consider $I_{\lambda}$
restricted to a natural constraint, the Nehari  manifold. We set
\begin{equation}\label{2-8}
\mathscr{N}_{\lambda}:=\left\{u\in
H^{1}(\mathbb{R}^3)\backslash\{0\}:I'_{\lambda}(u)u=0\right\}.
\end{equation}
The next lemma contains the statement of the main properties of
$\mathscr{N}_{\lambda}$.

\begin{lemma}\label{lemma2-2}
Assume that {\rm (A1), (A2)} hold. Then for all
$\lambda\in\mathbb{R}^+$, we have
\begin{itemize}
  \item [($i$)] $\mathscr{N}_{\lambda}$ is a $C^{1}$ regular manifold
 diffeomorphic to the sphere of  $ H^{1}(\mathbb{R}^{N})$;
  \item [($ii$)] $I_{\lambda}$ is bounded from below on $\mathscr{N}_{\lambda}$
by a positive constant;
  \item [($iii$)]u is a free critical point of $I_{\lambda}$ if and only if u
  is a critical point of $I_{\lambda}$ constrained on $\mathscr{N}_{\lambda}$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) Let $u\in H^{1}(\mathbb{R}^3)\backslash \{0\}$ be such that
$\|u\|=1$. We claim that there exists a unique
 $t\in(0,+\infty)$ for which $tu\in\mathscr{N}_{\lambda}$.
In fact, considering the equation
\begin{equation}\label{2-9}
I'_{\lambda}(tu)[tu]=t^2[\int_{\mathbb{R}^3}|\nabla u|^2+
(k_{\infty}+\lambda k(x))u^2
-t^2\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))\phi_u u^2]=0.
\end{equation}
It is clear that it admits a unique positive solution $t_{\lambda}(u)>0$
and that corresponding point
 $t_{\lambda}(u)u\in\mathscr{N}_{\lambda}$, the projection of $u$ on
$\mathscr{N}_{\lambda}$, is such that
\begin{equation*} %\label{a-9}
I_{\lambda}(t_{\lambda}(u)u=\max_{t\geq0}I_{\lambda}(tu).
\end{equation*}
Similar to the proof of \eqref{2-5}, we infer from
$u\in\mathscr{N}_{\lambda}$ and $(A_1)$-$(A_2)$ that
\begin{equation}\label{2-10}
\|u\|^2\leq\int_{\mathbb{R}^3}|\nabla u|^2+(k_{\infty}
+\lambda k(x))u^2=\int_{\mathbb{R}^3}(\mu_{\infty}
+\mu(x))\phi_uu^2\leq c\|u\|^{4}.
\end{equation}
This implies that
\begin{equation}\label{2-11}
\|u\|\geq c>0.
\end{equation}
Set $G_{\lambda}(u):=I'_{\lambda}(u)[u]$.
 By the regularity of $I_{\lambda}$ we know that
$G_{\lambda}\in C^{1}(H^{1}(\mathbb{R}^3),\mathbb{R})$. Moreover, by using
\eqref{2-11}, we obtain that
\begin{equation}\label{2-12}
G'_{\lambda}(u)[u]=-2\int_{\mathbb{R}^3}|\nabla
u|^2+(k_{\infty}+\lambda k(x))u^2\leq-2 c<0.
\end{equation}

(ii) For all $u\in\mathscr{N}_{\lambda}$, one sees that
\begin{equation}\label{2-13}
I_{\lambda}(u)=\frac{1}{4}\int_{\mathbb{R}^3}|\nabla u|^2
 +(k_{\infty}+\lambda k(x))u^2\geq\frac{1}{4}\|u\|^2\geq C> 0.
\end{equation}

(iii) If $u\neq0$ is a critical point of $I_{\lambda}$, then
$I'_{\lambda}(u)=0$ and then
 $u\in\mathscr{N}_{\lambda}$. On the other hand, if $u$ is a critical
point of $I'_{\lambda}$ constrained on $\mathscr{N}_{\lambda}$,
then there exist $\ell\in\mathbb{R}$ such that
\begin{equation*} %\label{a-9}
0=I'_{\lambda}(u)[u]=G_{\lambda}(u)=\ell G'_{\lambda}(u)[u].
\end{equation*}
We infer from \eqref{2-12} that $\ell=0$.
\end{proof}

Next we consider the limit functional $I_{\infty}:
H^{1}(\mathbb{R}^3)\to\mathbb{R}$, defined as
\begin{equation*}
I_{\infty}(u)=\frac{1}{2}\int_{\mathbb{R}^3}\left(|\nabla u|^2+
 k_{\infty}u^2\right)-\frac{1}{4}\int_{\mathbb{R}^3}\mu_{\infty}\phi_u(x)u^2,\quad
 u\in H^{1}(\mathbb{R}^3).
\end{equation*}
and the related natural constraint
\begin{equation*}
\mathscr{N}_{\infty}:=\left\{u\in H^{1}(\mathbb{R}^3)\backslash\{0\}:
I'_{\infty}(u)u=0\right\}.
\end{equation*}
Obviously, critical points of $I_{\infty}$ are solutions of the
limit problem at infinity
\begin{equation}\label{b-2-14}
\begin{gathered}
 -\Delta u+\kappa_{\infty}u=\mu_{\infty}\phi_u(x)u,\quad
  \text{in } {\mathbb R}^3,\\
-\Delta \phi=u^2,\quad  u\in H^{1}({\mathbb R}^3) .
\end{gathered}
\end{equation}
Clearly, the conclusions of  Lemma \ref{lemma2-2} hold  for
$I'_{\infty}$ and $\mathscr{N}_{\infty}$. Furthermore,
for any $u\in H^{1}(\mathbb{R}^3)\backslash\{0\}$,
it is easy to see that there
exists unique $t(u)>0$ such that $t(u)u\in\mathscr{N}_{\infty}$. Set
\begin{equation}\label{b-2-15}
m_{\infty}:=\inf\{I_{\infty}(u),u\in\mathscr{N}_{\infty}\}.
\end{equation}
From \cite{Moro-Schaftingen-2013,Ma-Zhao-2010}, we know that
$m_{\infty}$ is achieved by a radially symmetric function $w$,
unique up to translations, and decreasing when the radial coordinate
increases. Precisely, there exists a constant $c^{*}>0$ such that
\begin{equation}\label{b-2-16}
\lim_{|x|\to+\infty}|w(x)\|x|^{1-c^{*}}e^{\sqrt{\kappa_\infty}|x|}
=\text{constant}.
\end{equation}
 In what follows, for any $y\in\mathbb{R}^3,$ we use the  translation symbol
\begin{equation}\label{b-2-17}
w_{y}:=w(\cdot-y).
\end{equation}
Set
\begin{equation}\label{b-2-18}
m_{\lambda}:=\inf \{I_{\lambda}(u),u\in\mathscr{N_{\lambda}}\}.
\end{equation}
Then the following properties of $m_{\lambda}$ and $m_{\infty}$ hold.

\begin{lemma}\label{Proposition 2-3}
 Suppose that {\rm (A1), (A2)}  hold. Then for $\lambda\geq0$ we have
\begin{equation}\label{b-2-19}
0<m_{\lambda}\leq m_{\infty}.
\end{equation}
\end{lemma}

\begin{proof}
Let $\lambda\geq0$ be fixed. The first inequality of \eqref{b-2-19}
is a straight consequence of \eqref{2-13}. In order to show the
second inequality we should construct a sequence
$\{u_n\}\subset\mathscr{N_{\lambda}}$ and
$\lim_nI_{\lambda}(u_n)=m_{\infty}$. To this aim, let us
 consider $(y_n)_n$, with $y_n\in\mathbb{R}^3,
|y_n|\to+\infty,$ as $n\to+\infty$ and we set $u_n=t_nw_{y_n}$,
where $w_{y_n}$  is defined in \eqref{b-2-17} and
$t_n=t_{\lambda}(w_{y_n})$ such that
 $u_n=t_nw_{y_n}\in \mathscr N_{\lambda}$. We observe that
\begin{equation}\label{2-19}
\begin{aligned}
&I_{\lambda}(u_n) \\
&= \frac{t_n^2}{2} \int_{\mathbb{R}^3}|\nabla w_{y_n}|^2
 +(\kappa_\infty+\lambda\kappa (x))w_{y_n}^2
 -\frac{t_n^{4}}{4}\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))
 \phi_{w_{y_n}}(x)w_{y_n}^2\\
&=\frac{t_n^2}{2}\left[\|w\|^2
 +\lambda\int_{\mathbb{R}^3}\kappa(x+y_n))w^2\right]-
 \frac{t_n^{4}}{4}\int_{\mathbb{R}^3}(\mu_{\infty}
 +\mu(x+y_n))\phi_{w}(x+y_n)w^2.
\end{aligned}
\end{equation}
Moreover, from $t_nw_{y_n}\in \mathscr N_{\lambda}$ it follows
that
\begin{equation*} %\label{a-9}
t^2_n=\frac{\|w\|^2+\lambda\int_{\mathbb{R}^3}\kappa(x+y_n)w^2}
{\int_{\mathbb{R}^3}\mu_{\infty}\phi_{w}w^2
+\int_{\mathbb{R}^3}\mu(x+y_n)\phi_{w}(x+y_n)w^2}.
\end{equation*}
It is clear that
\begin{gather*}
\lim_{n\to\infty}\int_{\mathbb{R}^3}\kappa(x+y_n)w^2=0, \\
 \lim_{n\to\infty}\int_{\mathbb{R}^3}\mu(x+y_n)\phi_{w}(x+y_n)w^2=0.
\end{gather*}
 Thus, we infer that
\begin{equation}\label{b-2-21}
t_n\to 1,\quad I_{\lambda}(u_n)\to m_{\infty},\quad \text{as } n\to +\infty.
\end{equation}
\end{proof}


By apply the well-known  concentration-compactness
principle\cite{P.L.Lions-1984} and maximum
principle\cite{Willem-1996}, we have the following results for
$m_\lambda$.

\begin{lemma}\label{Lemma-2-4}
If the strictly inequality
\begin{equation}\label{b-2-21b}
m_{\lambda}< m_{\infty}
\end{equation}
holds, then $m_{\lambda}$ is achieved by a positive function.
Moreover, all the minimizing sequences are relatively compact.
\end{lemma}

\begin{lemma}\label{Proposition 2-5}
Assume that $\lambda=0$, {\rm (A1), (A2)} hold. Then
\eqref{auto-2} has a positive ground state solution.
\end{lemma}
\begin{proof}
Note that
\begin{gather*}
I_0(u)=\frac{1}{2}\|u\|^2-\frac{1}{4}\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))
\phi_u(x)u^2, \\
I_{\infty}(u)=\frac{1}{2}\|u\|^2-\frac{1}{4}\int_{\mathbb{R}^3}
\mu_{\infty}\phi_u(x)u^2.
\end{gather*}
Thus, we infer that
$m_0\leq m_{\infty}$.
To complete the proof we only need to show that
$ m_0< m_{\infty}$.
 Assume, by contradiction, that $w\in\mathscr{N}_\infty$ and
$I_{\infty}(w)= m_{\infty}=m_0$. Then there exists $t_n>0$ such that
$t_nw_{y_n}\in\mathscr{N}_0$, and $t_n\to1$, as
$n\to\infty$. This implies that
\begin{equation*}%\label{a-9}
m_{\infty}=m_0\leq I_0(t_nw_{y_n})<I_{\infty}(w)=m_{\infty}.
\end{equation*}
This is impossible.
\end{proof}

The next lemma analyzes the behavior of some sequences of
$\{u_n\}\subset\mathscr{N}_{\lambda_n}$.

\begin{lemma}\label{lemma2-6}
Suppose that {\rm (A1), (A2)} hold.
Let $(\lambda_n)_n$ be a sequence of positive numbers,
for all $n\in\mathbb{N}$, and $u_n\in\mathscr{N}_{\lambda_n}$ be such that
$I_{\lambda_n}(u_n)\leq C$. Then $\{u_n\}_n$ is bounded in
$H^{1}(\mathbb{R}^3)$.
\end{lemma}

\begin{proof}
We infer from $\{u_n\}\subset\mathscr{N}_{\lambda_n}$ that
\begin{equation}\label{b-2-22}
I_{\lambda_n}(u_n)=\frac{1}{4}\Big(\|u_n\|^2
+\lambda_n\int_{\mathbb{R}^3}\kappa(x)u_n^2\Big)
\leq C.
\end{equation}
Thus the conclusion holds.
\end{proof}

\begin{lemma}\label{lemma2.7}
Assume that {\rm (A1), (A2)} hold. Let
$u_n\in\mathscr{N}_{\lambda_n}$ be such that
$I_{\lambda_n}(u_n)\leq C$, and
$\lambda_n\to\infty$, as $n\to\infty$. Then, for all $R>0$,
\begin{equation} \label{b-2-23}
\begin{gathered}
u_n|_{B_{R}}\to 0, \text{ in } L^2(B_{R}),\quad
\lambda_n\int_{\mathbb{R}^3}\kappa(x)u_n^2\leq C,\\
\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}u_n^2\to~0, \quad\text{as }
n\to +\infty.
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
Since $\{u_n\}$ satisfies the inequality \eqref{b-2-22}, one can
check that the first two conclusions of \eqref{b-2-23} are true.
Next we prove the third one. By $(A_2)$, we know that for any
$\varepsilon>0$, there exists $R>0$ such that for all
$x\in\mathbb{R}^3\setminus B_{R}$, $\mu(x)<\varepsilon$. Thus we
infer from Lemma \ref{lemma2-6} that
\begin{equation}\label{b-2-24}
\int_{\mathbb{R}^3\setminus
B_{R}}\mu(x)\phi_{u_n}u_n^2\leq\varepsilon\int_{\mathbb{R}^3\setminus
B_{R}}\phi_{u_n}(x)u_n^2\leq \varepsilon C.
\end{equation}
On the other hand, by Hardy-Littlewood-Sobolev inequality, one sees
that
 \begin{equation}\label{b-2-25}
\int_{B_{R}}\mu(x)\phi_{u_n}u_n^2\leq
C\int_{B_{R}}\phi_{u_n}u_n^2\leq
C\left(\int_{B_{R}}u_n^{\frac{12}{5}}\right)^{\frac{5}{6}}\to0,~~~as~~
n\to\infty.
\end{equation}
From \eqref{b-2-24}-\eqref{b-2-25}, we know that the conclusions
hold.
\end{proof}


\section{Properties of the map $\lambda\mapsto m_{\lambda}$}

In this section we shall show that the monotonicity property of the
map $\lambda\mapsto m_{\lambda}$.

\begin{lemma}\label{lemma-3-1}
Assume that {\rm (A1), (A2)} hold. Then  the map
 $\lambda\mapsto m_{\lambda}$ is nondecreasing.
\end{lemma}

\begin{proof}
For $u\in H^{1}(R^3)\backslash\{0\}$, $\lambda\in\mathbb{R}^{+}$
and $t_{\lambda}(u)u\in\mathscr{N}_{\lambda}$, we have
\begin{equation*}%\label{a-9}
[t_{\lambda}(u)]^2=\frac{\|u\|^2+\lambda\int_{\mathbb{R}^3}
\kappa(x)u^2}{\int_{\mathbb{R}^3}[\mu_{\infty}+\mu(x)]\phi_uu^2}.
\end{equation*}
If $\lambda_1, \lambda_2\in\mathbb{R}^{+}$ such that
$\lambda_1<\lambda_2$, then
 $t_{\lambda_1}(u)\leq t_{\lambda_2}(u)$. Moreover,
$t_{\lambda_1}(u)=t_{\lambda_2}(u)$ if and only if
$\int_{\mathbb{R}^3}\kappa(x)u^2=0$. So, we obtain
\begin{equation}\label{3-1}
\begin{aligned}
I_{\lambda_1}(t_{\lambda_1}u)
&=\frac{t_{\lambda_1}^2}{4}\Big(\|u\|^2
 +\int_{\mathbb{R}^3}\lambda_1\kappa(x)u^2\Big)\\
&\leq\frac{t_{\lambda_2}^2}{4}
 \Big(\|u\|^2+\int_{\mathbb{R}^3}\lambda_2\kappa(x)u^2\Big)\\
&=I_{\lambda_2}(t_{\lambda_1}u).
\end{aligned}
\end{equation}
Therefore, by the arbitrariness of $u$, we obtain that
$m_{\lambda_1}\leq m_{\lambda_2}$.
\end{proof}

\begin{remark}\label{rem-3-2}\rm
We observe that if $u\in H^{1}(R^3)\backslash\{0\}$, and
$\lambda_1<\lambda_2$, $\lambda_1$, $\lambda_2\in
\mathbb{R}^{+}$, then
\begin{equation*} %\label{a-9}
t_{\lambda_1}(u)=t_{\lambda_2}(u)~\Longleftrightarrow~
\int_{\mathbb{R}^3}\kappa(x)u^2=0.
\end{equation*}
\end{remark}

Next we prove some properties for $m_{\lambda}$ and $m_\infty$.

\begin{lemma}\label{lemma-3-3}
Assume that {\rm (A1), (A2)} hold. If there exists
$\nu\in\mathbb{R}^{+}$ such that $m_{\nu}=m_{\infty}$, then we have
$m_{\lambda}=m_{\infty}$ for all $\lambda>\nu$. Moreover,
$m_{\lambda}$ is not achieved.
\end{lemma}

\begin{proof}
By Lemmas \ref{Lemma-2-4} and \ref{lemma-3-1}, we deduce that
\begin{equation*}%\label{a-9}
m_{\infty}=m_{\nu}\leq m_{\lambda}\leq m_{\infty}.
\end{equation*}
Thus, we obtain $m_{\lambda}=m_{\infty}$.

Next we shall prove that $m_{\lambda}$ is not achieved. Arguing by
contradiction, we assume that there exists
$u_{\lambda}\in\mathscr{N}_{\lambda}$ such that
$I_{\lambda}(u_{\lambda})=m_{\lambda}=m_{\infty}$. Let
$t_{\nu}=t_{\nu}(u_{\lambda})>0$ be such that
$t_{\nu}u_{\lambda}\in\mathscr{N}_{\nu}.$ By using the same
arguments as in Lemma \ref{lemma-3-1} and Remark \ref{rem-3-2}, we
can get $t_{\nu}<1$, and thus
\begin{equation*} %\label{a-9}
m_{\infty}=m_{\nu}\leq
I_{\nu}(t_{\nu}u_{\lambda})<I_{\lambda}(u_{\lambda})=m_{\infty}.
\end{equation*}
This is a contradiction.
\end{proof}

As a consequence of Lemma \ref{lemma-3-3}, we have the following
results.

\begin{corollary}\label{corollary-3-5}
Assume that {\rm (A1), (A2)} hold. Then there exists at most
one number $\nu\in\mathbb{R}^{+}$ such that $m_{\nu}=m_{\infty}$ and
it is achieved.
\end{corollary}

Let us define
\begin{equation}\label{3-2}
\lambda^{*}:=\sup\{\lambda\in\mathbb{R}^{+}:m_{\lambda}<m_{\infty}\}.
\end{equation}
Then the following lemma states the role of $\lambda^{*}$.

\begin{proposition}\label{Proposition 3-3}
Suppose that $\lambda^{*}<+\infty$. Then
\begin{equation}\label{3-3}
m_{\lambda^{*}} = m_{\infty}.
\end{equation}
 and
 \begin{equation}\label{3-4}
\sup\{\lambda\in\mathbb{R}^{+}:m_{\lambda}<m_{\infty}\}=\min
\{\lambda\in\mathbb{R}^{+}:m_{\lambda}=m_{\infty}\}.
\end{equation}
\end{proposition}

\begin{proof}
We use the contradiction method.
If $m_{\lambda^{*}}<m_{\infty}$,
then there exists $u^{*}\in\mathscr{N}_{\lambda^{*}}$ such that
$I_{\lambda^{*}}(u_{\lambda^{*}})=m_{\lambda^{*}}$. Let
$(\lambda_n)_n$ be a sequence of number such that
$\lambda_n\searrow\lambda^{*}$. By Lemma \ref{lemma-3-3}, we know
that $m_{\lambda_n}=m_{\infty}$. Moreover, there exist
$t_n:=t_{\lambda_n}(u_{\lambda^{*}})$ such that
$t_nu_{\lambda^{*}}\in\mathscr{N}_{\lambda_n}$. By the
definition of $t_n$, we obtain $t_n\to1$, as
$n\to\infty$. So, we obtain
\begin{gather*} %\label{a-9}
m_{\infty}=m_{\lambda_n}\leq I_{\lambda_n}(t_nu_{\lambda^{*}})=
\frac{1}{4}t_n^2[\|u_{\lambda^{*}}\|^2+\int_{\mathbb{R}^3}
\lambda_n\kappa(x)u_{\lambda^{*}}^2], \\
%\label{a-9}
\overrightarrow{n\to+\infty}\frac{1}{4}
[\|u_{\lambda^{*}}\|^2
+\int_{\mathbb{R}^3}\lambda^{*}\kappa(x)u_{\lambda^{*}}^2]
=I_{\lambda^{*}}(u_{\lambda^{*}})=m^{*}<m_{\infty}.
\end{gather*}
This is a contradiction. Finally, \eqref{3-4} is easily obtain from
\eqref{3-3}.
\end{proof}

Next we show the continuity of the map $\lambda\mapsto m_{\lambda}$.

\begin{lemma}\label{lemma-3-6}
Let {\rm (A1), (A2)} be satisfied. Then the map $\lambda\mapsto m_{\lambda}$
is continuous for $\lambda\in\mathbb{R}^+$.
\end{lemma}

\begin{proof}
We divide into the following two cases to prove the results.
\smallskip

\noindent\textbf{Case 1.}
 $\lambda^*=\infty$. By the definition of $\lambda^*$,
we infer that for each $\lambda\in\mathbb{R}^{+}$,
$m_{\lambda}<m_{\infty}$. By Lemma \ref{Lemma-2-4}, there exists
$u_{\lambda}\in\mathscr{N}_{\lambda}$ such that
$I_{\lambda}(u_{\lambda})=m_{\lambda}.$
 Let $\{\lambda_n\}$ be such that $\lambda_n\to\lambda$, and
$t_n:=t_{\lambda_n}(u_{\lambda})$ be such that
 $t_nu_{\lambda}\in\mathscr{N}_{\lambda_n}$.
Then, by using the definition of $t_n$,
 we obtain that $t_n\to1$ as $n\to\infty$. Hence, one
 sees that
 \begin{equation}\label{3-5}
\begin{aligned}
 m_{\lambda_n}\leq I_{\lambda_n}(t_nu_{\lambda_n})
&= \frac{1}{4} t_n^2[\|u_{\lambda}\|^2+\lambda_n\int_{\mathbb{R}^3}
 \kappa(x)u_{\lambda_n}^2]\\
&\overrightarrow{n\to+\infty}~\frac{1}{4}~[\|u_{\lambda}\|^2
 +\lambda\int_{\mathbb{R}^3}\kappa(x)u_{\lambda}^2] \\
&= I_{\lambda}(u_{\lambda})=m_{\lambda}.
\end{aligned}
\end{equation}
This implies
\begin{equation}\label{3-6}
\limsup_n m_{\lambda_n}\leq m_{\lambda}.
\end{equation}
Since $m_{\lambda_n}< m_{\infty}$ for all $n\in\mathbb{N}$, we
deduce from Lemma \ref{Lemma-2-4} that there exist
$u_n\in\mathscr{N}_{\lambda_n}$ such that
$I_{\lambda_n}(u_n)=m_{\lambda_n}$. Moreover, one infers from
Lemma \ref{lemma2.7} that the sequence $\{u_n\}$ is bounded in
$H^{1}({\mathbb{R}^3})$.

 Let $\tilde{t}_n:=t_{\lambda}(u_n)$ be such that
$\tilde{t}_nu_n\in \mathscr{N}_{\lambda}$. Since
\begin{equation*} %\label{a-9}
1=\frac{\|u_n\|^2+\lambda_n\int_{\mathbb{R}^3}
\kappa(x)u_n^2}{\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))\phi_{u_n}u_n^2},\quad
(\tilde{t}_n)^2=\frac{\|u_n\|^2+\lambda\int_{\mathbb{R}^3}
\kappa(x)u_n^2}{\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))\phi_{u_n}u_n^2}.
\end{equation*}
we deduce that $\tilde{t}_n\to1 $ and
$|I_{\lambda} (\tilde{t}_nu_n)-m_{\lambda_n}|\to0$, as
$n\to+\infty$. Thus, we obtain
\begin{equation}\label{3-7}
 m_{\lambda}\leq \limsup_n~m_{\lambda_n}.
\end{equation}
Combining \eqref{3-6} and \eqref{3-7}, we obtain the conclusion as
required.
\smallskip

\noindent\textbf{Case 2.}
$\lambda^*\in\mathbb{R}^{+}$. For any
$\lambda\in(0,\lambda^{*})$, we can use the same arguments as Case 1
to obtain the conclusion. On the other hand, for any
$\lambda\in(\lambda^{*},+\infty)$, $\lambda\mapsto m_{\lambda}$ is a
constant map. Therefor, we just need to prove the continuity when
$\lambda=\lambda^{*}$.

 Let $\{\lambda_n\}$ be a sequence of number and  $\lambda_n\to\lambda^{*}$
By \eqref{3-3}, if $\lambda_n\searrow\lambda^{*}$, the result is
trivial. So, in the following we study the case
$\lambda_n\nearrow\lambda^{*}$. By the definition of $\mathcal
{N}_{\lambda^{*}}$, for fixed $\varepsilon>0$, there exists
  $u_{\varepsilon}\in\mathcal
{N}_{\lambda^{*}}$ such that
 $I_{\lambda^{*}}(u_{\varepsilon})<m_{\lambda^{*}}+\varepsilon$.
Let $t_{n,\varepsilon}: =t_{\lambda_n}(u_{\varepsilon})$ be such that
 $t_{n,\varepsilon}u_{\varepsilon}\in\mathscr{N}_{\lambda_n}$.
On can easy to deduce that   $t_{n,\varepsilon}\to1$, as $n\to+\infty$.
Moreover, we have
\begin{align*} %\label{a-9}
m_{\lambda_n}
&\leq I_{\lambda_n}(t_{n,\varepsilon}u_{\varepsilon})
 =  \frac{1}{4}t_{n,\varepsilon}^2[\|u_{\varepsilon}\|^2+
 \lambda_n\int_{\mathbb{R}^3}\kappa(x)u_{\varepsilon}^2]\\
&\overrightarrow{n\to\infty}\;\frac{1}{4}[\|u_{\varepsilon}\|^2+
\lambda^{*}\int_{\mathbb{R}^3} \kappa(x)u_{\varepsilon}^2]
=I_{\lambda^{*}}(u_{\varepsilon})<m_{\lambda^{*}}+\varepsilon.
\end{align*}
Thus, we obtain
\begin{equation*}
\limsup_n m_{\lambda_n}\leq m_{\lambda^{*}}+\varepsilon.
\end{equation*}
By the arbitrariness of $\varepsilon$, we can obtain
\begin{equation*}
\limsup_n~m_{\lambda_n}\leq m_{\lambda^{*}}.
\end{equation*}
On the other hand, for all $n$, $m_{\lambda_n}<m_{\infty}$, we can
use the same arguments as Case 1 to showing that
\begin{equation*}
\liminf_n~m_{\lambda_n}\leq m_{\lambda^{*}}.
\end{equation*}
This completes the proof.
\end{proof}

\begin{remark}\label{rem-3-7} \rm
By the continuity of the map $\lambda\mapsto m_{\lambda}$ and the
fact that $m_0<m_{\infty}$, we infer that $\lambda^{*}>0$.
\end{remark}

\section{Two kinds of possible situations for $\lambda^*$}

In this section we study the properties of $\lambda^*$ according to
the decay of the functions $\kappa(x)$ and $\mu(x)$. Let us first
consider the case when $\kappa(x)$ decays faster than $\mu(x)$.

\begin{lemma}\label{Proposition 4-1}
Assume that {\rm (A1)--(A3)} hold. Then we have that
$\lambda^{*}=+\infty$, where $\lambda^{*}$ is defined in
\eqref{3-2}.
\end{lemma}

\begin{proof}
First, we infer from Lemma \ref{Proposition 2-5} that
$m_0<m_{\infty}$. So, in the following we only need to consider
the case $\lambda>0$. For fixed $\lambda>0$, we choose $t_n$ such
that $u_n=t_nw_{y_n}\in\mathscr{N}_{\lambda}$, where $y_n$
and $t_n$ are chose as in the proof of Lemma \ref{Proposition 2-3}.
Moreover, as in \eqref{b-2-21}, we infer that $t_n\geq c>0$.
Thus, we obtain that
\begin{equation}\label{4-1}
\begin{aligned}
m_{\lambda}&\leq I_{\lambda}(u_n)
=I_{\infty}(t_nw)+\frac{t_n^2}{2}\Big[\lambda\int_{\mathbb{R}^3}
 \kappa(x+y_n))w^2-\frac{t_n^2}{2}\int_{\mathbb{R}^3}\mu(x+y_n))
 \phi_{w}w^2\Big]\\
 &\leq I_{\infty}(w)+\frac{t_n^2}{2}\Big[\lambda\int_{\mathbb{R}^3}
 \kappa(x+y_n))w^2-c\int_{\mathbb{R}^3}\mu(x+y_n))\phi_{w}w^2\Big]\\
 &=m_{\infty}+\frac{t_n^2}{2}\Big[\lambda\int_{\mathbb{R}^3}
 \kappa(x+y_n))w^2-c\int_{\mathbb{R}^3}\mu(x+y_n))\phi_{w}w^2\Big].
\end{aligned}
\end{equation}
Hence, we obtain the conclusion if we show that, for large $n$,
\begin{equation}\label{4-2}
\int_{\mathbb{R}^3}[\lambda\kappa(x+y_n)w^2-C\mu(x+y_n)\phi_{w}w^2]<0.
\end{equation}
This is equivalent to prove that, for large $n$,
\begin{align*} %\label{a-9}
I_1&:=\int_{\mathbb{R}^3 \backslash
B_{\tau|y_n|}}\Big[\frac{\lambda}{2}\kappa(x+y_n)w^2-C
\mu(x+y_n)\phi_{w}w^2\Big] \\
&< I_2:= \int_{B_{\tau|y_n|}}
\Big[C\mu(x+y_n)\phi_{w}w^2-\frac{\lambda}{2}\kappa(x+y_n)w^2\Big]
\end{align*}
To estimate $I_1$, from \eqref{b-2-16} we have
\begin{equation}\label{d-4-3}
\begin{aligned}
I_1&<\int_{\mathbb{R}^3\backslash
B_{\tau|y_n|}}\lambda\kappa(x+y_n)w^2 \\
&<\lambda\Big[\int_{\mathbb{R}^3 \backslash
B_{\tau|y_n|}}|\kappa(x+y_n)|^{\frac{3}{2}}\Big]^{2/3}
\Big[\int_{\mathbb{R}^3\backslash B_{\tau|y_n|}}|w|^6\Big]^{1/3} \\
&<ce^{-2\tau\sqrt{\kappa_\infty}|y_n|}.
\end{aligned}
\end{equation}
Now we estimate the $I_2$ term. By (A3), for all
$\varepsilon>0$ and $M>0$, there exists $n_0\geq1$ such that, for
all $n\geq n_0$ and for all $x\in B_{\tau|y_n|}$,
\begin{equation}\label{d-4-3b}
\kappa(x+y_n)\leq \varepsilon(1-\tau)^{-1}|y_n|^{-1}
 e^{-2\tau\sqrt{\kappa_\infty}|y_n|}, \quad
 \mu(x+y_n)\geq Me^{-2\tau\sqrt{\kappa_\infty}|y_n|}.
\end{equation}
By \cite[Lemmas 2.3 and 2.6]{Li-Ni-1988-Duke}, we know that
\begin{equation}\label{d-4-4}
\phi_{w}(x)=\int_{\mathbb{R}^3}\frac{w^2(y)}{|x-y|}dy\sim
\frac{1}{|x|},\quad \text{as}\ |x|\to\infty.
\end{equation}
Thus, we infer that, for $n$ sufficiently large and for all
$x\in B_{\tau|y_n|}$,
\begin{equation*}
C\phi_{w}(x)-\frac{\lambda\kappa(x+y_n)} {\mu(x+y_n)}> \frac{C}{2}\phi_{w}(x).
\end{equation*}
Hence, one sees that
\begin{equation}\label{d-4-6}
\begin{aligned}
I_2
&=\int_{B_{\tau|y_n|}}\mu(x+y_n)w^2[C\phi_{w}
 (x)-\frac{\lambda\kappa(x+y_n)}{\mu(x+y_n)}]\\
&>\frac{C}{2}\int_{B_{\tau|y_n|}}\mu(x+y_n)\phi_{w}w^2
 >C Me^{-2\tau\sqrt{\kappa_\infty}|y_n|}\int_{B_1}\phi_{w}w^2\\
 &>CMe^{-2\tau\sqrt{\kappa_\infty}|y_n|}.
\end{aligned}
\end{equation}
Combining \eqref{d-4-3}-\eqref{d-4-6}, together with the
arbitrariness of $M$, we can conclude that $I_1<I_2$. This
completes the proof.
\end{proof}

Next we consider the case when $\kappa(x)$ decays slower than
$\mu(x)$.

\begin{lemma}\label{Proposition 4-2}
Suppose that {\rm (A1), (A2), (A4)} hold.
Then $\lambda^{*}\in \mathbb{R}^{+}$, where $\lambda^{*}$ is defined
in \eqref{3-2}.
\end{lemma}

\begin{proof}
We use the contradiction method. Assume that for all
$\lambda\in\mathbb{R}^{+}$, $m_{\lambda}<m_{\infty}$. By proposition
\ref{Proposition 3-3}. Let $\{\lambda_n\}$ be a diverging
sequence. From Lemma \ref{Lemma-2-4}, there exist $\{u_n\}$ such
that for all $n\in N$,
\begin{equation*}
u_n>0,\quad u_n\in\mathscr{N}_{\lambda_n},\quad
 I_{\lambda_n}(u_n)=m_{\lambda_n}<m_{\infty},\quad
I'_{\lambda_n}(u_n)=0.
\end{equation*}
We infer from Lemma \ref{lemma2-6} that $\{u_n\}$ is bounded in
$H^{1}(\mathbb{R}^3)$. Let $\theta_n=\theta_{u_n}$ be such
that $\theta_nu_n\in\mathscr{N}_{\infty}$. A direct computation
show that
\begin{equation}\label{d-4-7}
\theta_n^2=\frac{\|u_n\|
^2}{\mu_{\infty}\int_{\mathbb{R}^3}\phi_u(x)u^2}.
\end{equation}
We claim that
\begin{equation}\label{d-4-8}
I_{\lambda_n}(\theta_nu_n)<I_{\infty}(\theta_nu_n).
\end{equation}
Otherwise, one sees that
\begin{equation}\label{d-4-9}
m_{\infty}\leq I_{\infty}(\theta_nu_n)\leq I_{\lambda_n}
(\theta_nu_n)\leq
I_{\lambda_n}(u_n)=m_{\lambda_n}<m_{\infty}.
\end{equation}
This is impossible. So, the claim \eqref{d-4-8} holds. Moreover, we
deduce from the boundedness of $\|u_n\|$ that there exist two
numbers $c,C>0$ such that
\begin{equation}\label{d-4-10}
c\leq\theta_n\leq C.
\end{equation}
From \eqref{d-4-8} we infer that
\begin{equation}\label{d-4-11}
\frac{\lambda_n}{2}\int_{\mathbb{R}^3}\kappa(x)u^2_n- \frac
{\theta_n^2}{4}\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}(x)u^2_n<0.
\end{equation}
We deduce from \eqref{b-2-23} and \eqref{d-4-10} that
\begin{equation}\label{d-4-12}
\begin{aligned}
\frac{\lambda_n}{2}\int_{\mathbb{R}^3}\kappa(x)u^2_n\to 0, \quad
\text{as } n\to\infty.
\end{aligned}
\end{equation}
Since $u_n\in \mathscr{N}_{\lambda_n}$ and
$\theta_nu_n\in\mathscr{N}_{\infty}$, we deduce from \eqref{2-5}
that
\begin{equation}\label{d-4-13}
\begin{gathered}
c\Big(\int_{\mathbb{R}^3}\phi_{u_n}(x)u^2_n\Big)^{1/2}
\leq\|u_n\|^2=\mu_\infty\int_{\mathbb{R}^3}\phi_{u_n}(x)u^2_n, \\
\mu_{\infty}(\theta_n^2-1)\int_{\mathbb{R}^3}\phi_{u_n}(x)u^2_n=
\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}(x)u^2_n-\lambda_n
\int_{\mathbb{R}^3}\kappa(x)u^2_n=o(1).
\end{gathered}
\end{equation}
These together with \eqref{b-2-23} imply that
\begin{equation}\label{d-4-14}
 \lim_n\theta_n=1.
\end{equation}
Hence, one infers from \eqref{b-2-23} and \eqref{d-4-12} that
\begin{equation}\label{d-4-15}
\begin{aligned}
m_{\infty}
&>I_{\lambda}(u_n)=I_{\lambda_n}(\theta_nu_n)+o(1)\\
&=I_{\infty}(\theta_nu_n)+\frac{\theta_n\lambda_n}{2}
\int_{\mathbb{R}^3}\kappa(x)u^2_n-\frac{\theta_n^2}{4}
\int_{\mathbb{R}^3}\mu(x)\phi_{u_n}(x)u^2_n\\
&= I_{\infty}(\theta_nu_n)+o(1)\\
&\geq m_{\infty}+o(1).
\end{aligned}
\end{equation}
This implies
\begin{equation}\label{d-4-16}
I_{\infty}(\theta_nu_n)\to m_{\infty},\quad \text{as } n\to+\infty.
\end{equation}
By the uniqueness of the family of minimizers of $I_{\infty}$ on
$\mathscr{N}_{\infty}$,
 there exists sequence $\{y_n\}$ such that $y_n\in\mathbb{R}^3$ and
\begin{equation*} %\label{a-9}
\theta_nu_n-w_{y_n}\to0\quad \text{in}\
H^{1}(\mathbb{R}^3)\quad \text{as}\ n\to+\infty,
\end{equation*}
where $w$ is given by \eqref{b-2-16}.
 Set $v_n(x)=u_n(x+y_n)$.
We infer from \eqref{d-4-14} that
\begin{equation*}
v_n\to w\quad \text{in } H^{1}(\mathbb{R}^3)\quad
\text{as } n\to+\infty.
\end{equation*}
Since, for all $n$, $v_n$ is a solution of
\begin{equation}\label{d-4-17}
 -\Delta u+(\kappa_{\infty}+\lambda \kappa(x+y_n)u
=(\mu_{\infty}+\mu(x+y_n)\phi_u(x)u.
\end{equation}
By  the Schauder interior(see \cite{J.Serrin-1964}), we
know that $v_n\to w$ in $C_{\rm loc}^2(\mathbb{R}^3)$.
Moreover, from the decay estimates (see
\cite{Moro-Schaftingen-2013}), one deduces that for some
$\sigma\in(0,\sqrt{\kappa_\infty})$
\begin{equation}\label{d-4-18}
|v_n(x)|\leq ce^{-\sqrt{\sigma}|x|}.
\end{equation}

By \eqref{d-4-11}, it suffices to show that for  $n$ large enough,
\begin{equation}\label{d-4-19}
\begin{aligned}
\frac{\lambda_n}{2}\int_{\mathbb{R}^3}\kappa(x+y_n)v_n^2-
\frac{3}{8}\int_{\mathbb{R}^3}\mu(x+y_n)\phi_{v_n}v_n^2>0.
\end{aligned}
\end{equation}
That is, we need to prove that for $\tau\in(0,1)$,
\begin{equation}\label{d-4-20}
\begin{aligned}
I_1&:= \int_{B_{\tau|y_n|}}[\frac{\lambda_n}{2}
\kappa(x+y_n)v_n^2-\frac{3}{8}
\mu(x+y_n)\phi_{v_n}v_n^2] \\
&< I_2:= \int_{\mathbb{R}^3 \backslash
B_{\tau|y_n|}}[\frac{3}{8}\mu(x+y_n)\phi_{v_n}v_n^2
-\frac{\lambda_n}{2}\kappa(x+y_n)v_n^2]
\end{aligned}
\end{equation}
To  estimate $I_1$, from (A4) we have that
for all $x\in B_{\tau|y_n|}$,
\begin{equation*}
\kappa(x+y_n)\geq c_1e^{{-4\tau}\sqrt{\sigma}|y_n|}\quad
\text{and}\quad \mu(x+y_n)\leq
c_2e^{{-4\tau}\sqrt{\sigma}|y_n|}.
\end{equation*}
Thus, for any $C>0$ and $x\in B_{\tau|y_n|}$, we infer that if $n$
is  large enough,
\begin{equation*} %\label{a-9}
\frac{\lambda_n}{2}-\frac{c\kappa(x+y_n)}{\mu(x+y_n)}>\frac{\lambda_n}{4}.
\end{equation*}
Since $v_n\to w$ in $C_{\rm loc}^2(\mathbb{R}^3)$, it follows that
\begin{equation}\label{d-4-21}
\begin{aligned}
I_1&\geq\int_{B_{\tau|y_n|}}\kappa(x+y_n)v_n^2
\Big(\frac{\lambda_n}{2}-\frac{c\mu(x+y_n)}{\kappa(x+y_n)}\Big)\\
&>c\lambda_n\int_{B_{\tau|y_n|}}\kappa(x+y_n)v_n^2\\
&>c\lambda_ne^{{-4\tau}\sqrt{\sigma}|y_n|}\int_{B_1}v_n^2
 >c\lambda_ne^{{-4\tau}\sqrt{\sigma}|y_n|}.
\end{aligned}
\end{equation}
On the other hand, from \eqref{d-4-18} one infers that
\begin{equation}\label{d-4-22}
\begin{aligned}
I_2<\int_{\mathbb{R}^3 \backslash B_{\tau|y_n|}}
\frac{3}{8}\mu(x+y_n)\phi_{v_n}v_n^2\leq
c\Big(\int_{\mathbb{R}^3\backslash
B_{\tau|y_n|}}v_n^{\frac{12}{5}}\Big)^{5/3}
<ce^{{-4\tau}\sqrt{\sigma}|y_n|}.
\end{aligned}
\end{equation}
Hence, by the divergence of $\lambda_n$, for large $n$, we can
conclude that $I_2<I_1$. This completes the proof.
\end{proof}


\begin{proof}[Proof of Theorem\ref{theorem1-1}]
By Lemmas \ref{Lemma-2-4} and \ref{Proposition 4-1}, we know that
the conclusions of Theorem \ref{theorem1-1} hold.
\end{proof}

\section{Proof of Theorem \ref{theorem1-2}}

As we already pointed out in the introduction, new difficulty arises
here. That is, according to \cite{Ghimenti-Schaftingen-2016-JFA}, we
know that any sign-changing solution $u$ of \eqref{b-2-14} such that
$I_{\infty}(u)<2m_\infty$. From this we can not prove that
$I_\lambda$ satisfies the $(PS)_c$-condition for $c\in(m_\infty,
2m_\infty)$. Motivated by \cite{Ma-li-1997,J.Wang-2017}, we shall
consider our problem in convex set $H_+^1(\mathbb{R}^3)$ to
overcome the difficult, where $H_+^1(\mathbb{R}^3):=\{u\in
H^1(\mathbb{R}^3): u\geq0\}$.

For any point $u\in H_+^1(\mathbb{R}^3)$, we define
\begin{equation}\label{e-5-1}
J(u)=\sup_{u_1\in H_+^1(\mathbb{R}^3),~\|u-u_1\|<1}\langle\
I'(u),u-u_1\rangle.
\end{equation}
It is easy to check that $J$ is continuous on $H_+^1(\mathbb{R}^3)$.
We define
\begin{gather}\label{e-5-2}
\mathscr{N}_{\lambda}^{+}=\mathscr{N_{\lambda}}\cap H_+^1(\mathbb{R}^3), \\
\label{e-5-3}
d=\inf_{u\in\mathscr{N_{\lambda}}}I(u),\quad
 d^+=\inf_{u\in\mathscr{N_{\lambda}^{+}}}I(u).
\end{gather}
Next we study the properties of the Palais-Smale sequence
 of \eqref{auto-2} on $H_+^1(\mathbb{R}^3)$ at level $c$,
for $c\in(m_{\infty},2m_{\infty})$.

\begin{lemma}\label{lem--5.1}
Suppose that {\rm (A1), (A2)} hold. Let
$\{u_n\}\subseteq\mathscr{N_{\lambda}^{+}}$ be a sequence such
that
$I_{\lambda}(u_n)$  is bounded, and
$J(u_n)\to 0$ strongly in  $H _+^1(\mathbb{R}^3)$.
Then, up to a subsequence, there exist a solution $\bar{u}$ of
$(I_{\lambda})$, a number $k\in\mathbb{N}\cup\{0\}$,
 $k$ functions  $u^{1},\dots ,u^{k}$ of $H_+^1(\mathbb{R}^3)$ and
$k$ sequence of points $(y_n^j),y_n^j\in\mathbb{R}^3$,
$0\leq j\leq k$ such that, as $n\to+\infty$,
\begin{gather*}
u_n-\sum^{k}_{j=1}u^j(\cdot-y_n^j) \to\bar{u} \text{ in }
H_+^1(\mathbb{R}^3),\quad
I_{\lambda}(u_n)\to\sum^{k}_{j=1}
 I_{\infty}(u^j)+I_{\lambda}(\bar{u}),\\
|y_n^j|\to+\infty,\quad
|y_n^j-y_n^j|\to+\infty\; (\text{if }i\neq j),\\
\text{and $u^j$ are weak solutions of \eqref{b-2-14}}.
\end{gather*}
Moreover, we notice that in the case $k=0$,
 the above holds without $u^j$.
\end{lemma}

\begin{proof}
We claim that $I'(u_n)\to0$ as $n\to\infty$. To
prove this we first prove that $\{u_n\}$ is bounded in
$H^1(\mathbb{R}^3)$. We assume that $\|u_n\|\to\infty$ as
$n\to\infty$.  It is very easy to see that
\begin{equation}\label{e-4}
y_n=u_n\pm\frac{u_n}{1+\|u_n\|}\in H_+^1(\mathbb{R}^3).
\end{equation}
So, we infer from $J(u_n)\to 0$ as $n\to\infty$, that
\begin{equation}\label{e-5}
 \langle I'(u_n),\frac{u_n}{1+\|u_n\|}
 \rangle\to0\quad \text{as } n\to\infty .
\end{equation}
Thus, we obtain
\begin{equation}\label{e-6}
\begin{aligned}
0\leftarrow\frac{I(u_n)-\frac{1}{4}\left\langle
I'(u_n),u_n\right\rangle}{1+\|u_n\|}
&=\frac{\frac{1}{4}\|u_n\|^2+\frac{1}{4}\int_{\mathbb{R}^3}
\lambda\kappa(x)u_n^2}{1+\|u_n\|}\\
 &\geq\frac{\frac{1}{4}\|u_n\|^2}{1+\|u_n\|}\to\infty.
\end{aligned}
\end{equation}
as $n\to\infty$. This is contradiction. So, $\|u_n\|$ is
bounded. Moreover, as in \eqref{e-4}-\eqref{e-6} we obtain that
$I'_{\lambda}(u_n)u_n\to 0$ as $n\to\infty$.

We now use an idea from \cite[Theorem 7]{Ma-li-1997} to claim that
$I'(u_n)\to0$ as $n\to\infty$. Since $\|u_n\|$ is
bounded, without loss of generality we assume that
$u_n\rightharpoonup u_0$ in $H^{1}(\mathbb{R}^3)$,
$u_n\to u_0$ in $L_{\rm loc}^{p}(\mathbb{R}^{N})(\forall
p\in(2,2^{*}))$, and $u_n(x)\to u_0(x)$ a.e., in
$\mathbb{R}^3$, where $u_0\geq0$. From the assumption we can
  infer that $J(u_n)=o_n(1)$, where $o_n(1)\to0$ as $n\to\infty$. Let $\varepsilon_n>0$ be such that
 $\lim_{n\to\infty}\varepsilon_n=0$ and
  $\lim_{n\to\infty}o_n(1)\varepsilon_n^{-1}=0$.
For any $g_1\in C_0^\infty(\mathbb{R}^3)$, we set
\begin{equation}\label{e-7}
u_{1,n}=u_n+\varepsilon_n g_1+g_{1,\varepsilon_n}\in H_+^1(\mathbb{R}^3),
\end{equation}
where $g_{1,\varepsilon_n}=-\min\{0,u_n+\varepsilon_n g_1\}\geq0$.
 By the definition of $J$ we know that
\begin{equation}\label{e-8}
\langle I'(u_n),u_n-u_{1,n}\rangle\leq J(u_n)=o_n(1).
\end{equation}
So,
\begin{equation}\label{e-9}
\langle I'(u_n),g_1\rangle\geq-\varepsilon_n^{-1}\langle I'(u_n),
g_{1,\varepsilon_n}\rangle+\varepsilon_n^{-1}o_n(1).
\end{equation}
By a direct computations we can show that
\begin{equation}\label{b-23}
\begin{split}
&-\langle I'(u_n),g_{1,\varepsilon_n}\rangle \\
&=-\int_{\mathbb{R}^3}\left(\nabla u_n\nabla
g_{1,\varepsilon_n}+(\kappa_{\infty}+\lambda\kappa(x))u_n
g_{1,\varepsilon_n}+
(\mu_{\infty}+\mu(x))\phi_{u_n}u_ng_{1,\varepsilon_n}\right)\\
&=\int_{\Omega_n}\left(\nabla u_n\nabla (u_n+\varepsilon_n
g_1)+(\kappa_{\infty}+\lambda\kappa(x))u_n (u_n+\varepsilon_n
g_1)\right)\\
&\quad-\int_{\Omega_n} \left(\mu_{\infty}+\mu(x))
 \phi_{u_n}u_n(u_n+\varepsilon_n g_1\right)\\
&\geq\varepsilon_n\int_{\Omega_n}\left(\nabla u_n\nabla g_1
 + (\kappa_{\infty}+\lambda\kappa(x))u_n g_1)
 -(\mu_{\infty}+\mu(x))\phi_{u_n}u_ng_1\right)\\
&\quad-\int_{\Omega_n}(\mu_{\infty}+\mu(x))\phi_{u_n}u_n^2\\
&\geq\varepsilon_n\int_{\Omega_n}\left(\nabla u_n\nabla g_1
 + (\kappa_{\infty}+\lambda\kappa(x))u_n g_1)
 - (\mu_{\infty}+\mu(x))\phi_{u_n}u_ng_1\right)\\
&\quad-\varepsilon_n^2\int_{\Omega_n}(\mu_{\infty}
 +\mu(x))\phi_{u_n}g_1^2.
\end{split}
\end{equation}
where $\Omega_n:=\{x\in\mathbb{R}^3: u_n(x)+\varepsilon_n
g_1<0\}$. Form $\|u_n\|$ is bounded, we infer that
$|\int_{\Omega_n}(\mu_{\infty}+\mu(x))$ $\phi_{u_n}g_1^2|$
and
\[
|\int_{\Omega_n}\left(\nabla u_n\nabla g_1+
(\kappa_{\infty}+\lambda\kappa(x))u_n
g_1)-(\mu_{\infty}+\mu(x))\phi_{u_n}u_ng_1\right)|
\]
are bounded. Moreover, since $|\Omega_n|\to0$ as
$n\to\infty$, we can obtain that:
\begin{equation}\label{e-11}
-\langle I'(u_n),g_{1,\varepsilon_n}\rangle\geq o(\varepsilon_n).
\end{equation}
By letting $n\to\infty$, we infer from \eqref{e-9} and
\eqref{e-11} that
\begin{equation}\label{e-12}
\lim_{n\to\infty}\langle I'(u_n),g_1\rangle\geq0,
\quad \forall g_1\in C_0^\infty(\mathbb{R}^3).
\end{equation}
Reversing the sign of $g_1$ and since $C_0^\infty(\mathbb{R}^3)$
is dense in  $H^1(\mathbb{R}^3)$, We infer that
  $\lim_{n\to\infty}\langle I'(u_n),g_1\rangle=0$,
for all $g_1\in H^1(\mathbb{R}^3)$. So, $I'(u_n)\to0$ as
$n\to\infty$ and the claim holds.

The rest of proof is similar to
\cite[Theorem 4.1]{A.Ambrosetti-G.Cerami-D.Ruiz-2008}, and we omit the details
here.
\end{proof}

Now we are ready to prove the compactness condition for the
functional $I_{\lambda}$.

\begin{lemma}\label{lem--5.2}
Assume that {\rm (A1), (A2)} hold on
 $H_+^1(\mathbb{R}^3)$. If $m_{\lambda}=m_{\infty}$,
 then the functional $I_{\lambda}$
 satisfies the (PS) condition at level $c$, for
$c\in(m_{\infty},2m_{\infty})$.
\end{lemma}

\begin{proof}
Let $\{u_n\}$ be a Palais-Smale sequence of $I_{\lambda}$
constrained on  $\mathscr{N_{\lambda}^{+}}$
  at level $c$, for $c\in(m_{\infty},2m_{\infty})$.
Applying Lemma \ref{lem--5.1} we can get that for any solution of
\eqref{b-2-14} satisfies $u\geq0$ and $I_{\infty}\geq m_{\infty}$.
Moreover, any critical point $\bar{u}$ of $(I_{\lambda})$ is such
that $I_{\lambda}(\bar{u})\geq m_{\lambda}=m_{\infty}$. Thus, we
know that $k$ must be zero, and the conclusion of this lemma holds.
\end{proof}

Let us now recall the barycenter definition of a function $u\in
H_+^1(\mathbb{R}^3)\setminus\{0\}$,
 which has introduced in \cite{G.Cerami-D.Passase-2003}. Set
\[
\widehat{\mu}(u)(x)=\frac{1}{|B_1(0)|}\int_{|B_1(x)|}|u(y)|,
\]
which belongs to $L^{\infty}(\mathbb{R}^3)$ and is
continuous; and set
\[
\widehat{u}(x)=\Big[\widehat{\mu}(u)(x)-\frac{1}{2}\max\widehat{\mu}(x)\Big]^{+},\quad
\widehat{u}\in C_0(\mathbb{R}^3).
\]
We define that $\beta:H_+^1(\mathbb{R}^3)
\setminus\{0\}\to\mathbb{R}^3$ as
\begin{equation*}
\beta(u)=\frac{1}{|\widehat{u}|_1}
\int_{\mathbb{R}^3} x\widehat{u}(x)\in\mathbb{R}^3.
\end{equation*}
Since $\widehat{u}$ has compact support, $\beta$ is well defined.
Moreover, the following properties hold
\begin{itemize}
  \item[(a)] $\beta$ is continuous in $H_+^1(\mathbb{R}^3)$;
  \item[(b)] if u is a radial function, $\beta(u)=0 $;
  \item[(c)] for all $t\neq 0$ and $u\in H_+^1(\mathbb{R}^3)\setminus\{0\}$,
$\beta(tu)=\beta(u)$;
  \item[(d)] given $z\in\mathbb{R}^3$ and setting $u_{z}(x)=u(x-z)$,
$\beta(u_{z})=\beta(u)+z$.
\end{itemize}
Let
\begin{equation*}
\mathscr{B}_0^{\lambda}:= \inf\{I_{\lambda}(u):u\in\mathscr{N_{\lambda}^{+}},
\beta(u)=0\}.
\end{equation*}

\begin{lemma}\label{lem--5.3}
Assume that {\rm (A1), (A2)} hold. If $\lambda\geq0$ be fixed,
and let $m_{\lambda}=m_{\infty}$ be not achieved. then
\begin{equation*}
m_{\lambda}=m_{\infty}<\mathscr{B}_0^{\lambda}.
\end{equation*}
\end{lemma}

\begin{proof}
We use the contradiction method. Let
$\{u_n\}\subseteq\mathscr{N_{\lambda}^{+}}$
 be such that $\beta(u_n)=0$ and $I_{\lambda}(u_n)=m_{\infty}+o_n(1)$.
From the Ekeland  variational principle(see \cite{Willem-1996}
or \cite{Ma-li-1997}), we can obtain there exist
  a sequence of functions $\{v_n\}$ such that
\begin{equation}\label{e-13}
\begin{gathered}
v_n\in\mathscr{N_{+}},\quad  I_{\lambda}(v_n)=m_{\infty}+o_n(1),\quad
J(v_n)\to0,\\
 |\beta(v_n)-\beta(u_n)|=o_n(1).
\end{gathered}
\end{equation}
Since $m_{\lambda}$ is not achieved, $(v_n)_n$ can not be
relatively compact, by Lemma \ref{lem--5.1}, the equality
\begin{equation*}
v_n=w_{y_n}+o(1).
\end{equation*}
must be true with $|y_n|\to+\infty$, which contradicts
\eqref{e-13}.
\end{proof}

Let $\xi\in\mathbb{R}^3$ with $|\xi|=1$ and
$\Sigma=\partial B_2(\xi)$. We define
\begin{equation}\label{e-14}
\mathbf{w}=\frac{w}{\big(\int_{\mathbb{R}^3}\phi_{w}w^2\big)^{1/4}}
\end{equation}
and for any $y\in\mathbb{R}^3$, $\mathbf{w}_y=\mathbf{w}(\cdot-y)$.
 Observing that $\mathbf{w}$ satisfies
\begin{equation}\label{e-15}
-\Delta\mathbf{w}+\kappa_{\infty}\mathbf{w}=\mathbf{M}\phi_{\mathbf{w}}\mathbf{w},
\end{equation}
and by a direct computation we obtain that
\begin{equation}\label{e-16}
\mathbf{M}=2m_{\infty}^{1/2}\mu_{\infty}^{1/2}.
\end{equation}
For any $\rho>0$ and $(z,s)\in\Sigma\times[0,1]$, we define
\begin{equation*}
\psi_{\rho}(z,s)=(1-s)\mathbf{w}_{\rho z}+s \mathbf{w}_{\rho\xi}.
\end{equation*}
Let $\Psi_{\rho}:\Sigma\times[0,1]\to
\mathscr{N_{\lambda}^{+}}$ be defined by
\begin{equation*}
\Psi_{\rho}(z,s)=t_{z,s}^{\lambda}\psi_{\rho}(z,s),
\end{equation*}
where $t_{z,s}^{\lambda}>0$ be such that
$t_{z,s}^ {\lambda}\psi_{\rho}(z,s)\in\mathscr{N_{\lambda}^{+}}$. Then we have
the following results to describe the property of
$\mathscr{B}_0^{\lambda}$.

\begin{lemma}\label{lem--5.4}
Assume that {\rm (A1), (A2)} hold and let $\lambda>0$ be
fixed. Then for all $\rho>0$ we have
\begin{equation*}
\mathscr{B}_0^{\lambda}\leq\mathscr{T}_{\rho}^{\lambda}
:=\max_{\Sigma\times[0,1]}I_{\lambda}(\Psi_{\rho}(z,s)).
\end{equation*}
\end{lemma}

\begin{proof}
Since $\beta(\Psi_{\rho}(z,0))=\rho z$, we assert that
 $\beta\circ\Psi_{\rho}(\Sigma\times\{0\})$ is homotopically
 equivalent in $\mathbb{R}^3\backslash\{0\}$ to $\rho\Sigma$, then,
 we can find $(\overline{z},\overline{s})\in\Sigma\times[0,1]$
 and satisfied  $\beta(\Psi_{\rho}(\overline{z},\overline{s}))=0$,
and, naturally,
\begin{equation*}
\mathscr{B}_0^{\lambda}\leq I_{\lambda}(\Psi_{\rho}
(\overline{z},\overline{s}))\leq\mathscr{T}_{\rho}^{\lambda}.
\end{equation*}
This completes the proof.
\end{proof}

\begin{lemma}\label{lem--5.5}
Let assumptions {\rm (A1), (A2), (A4)} hold. Then
there exist $\rho_0>0$ such that for $\rho>\rho_0$,
\begin{equation*}
\mathscr{T}_{\rho}^{\lambda}<2m_{\infty}.
\end{equation*}
\end{lemma}

\begin{proof}
The idea of the proof is similar to that used in
\cite{Giovanna-Riccardo-2003,Giovanna-Riccardo-2010}, and we just
sketch it here for reader's convenience. Observing that
\begin{equation*}
I_{\lambda}(\Psi_{\rho}(z,s))=\frac{1}{4}
\Big\{\frac{\|\psi_{\rho}(z,s)\|^2+
 \lambda\int_{\mathbb{R}^3}\kappa(x)\psi^2_{\rho}(z,s)}
{[\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))
\phi_{\psi_{\rho}(z,s)}(x)
\psi^2_{\rho}(z,s)]^{1/2}}\Big\}^2
\end{equation*}
 Let us first evaluate
\begin{align*}
&\mathscr{N}^{\lambda}_{\rho}(z,s)\\
&:=\|\psi_{\rho}(z,s)\|^2+
\lambda\int_{\mathbb{R}^3}\kappa(x)\psi^2_{\rho}(z,s)\\
&=(1-s)^2\|\mathbf{w}_{\rho z}\|^2+2s(1-s)(\mathbf{w}_{\rho z},
 \mathbf{w}_{\rho \xi})_{H_{+}^{1}}+
 s^2\|\mathbf{w}_{\rho \xi}\|^2\\
&\quad +\lambda\Big[(1-s)^2\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}^2_{\rho z}+2s(1-s)
 \int_{\mathbb{R}^3}\kappa(x)\mathbf{w}_{\rho z}\mathbf{w}_{\rho \xi}+s^2
 \int_{\mathbb{R}^3}\kappa(x)\mathbf{w}_{\rho \xi}^2\Big].
\end{align*}
Since $\mathbf{w}$ satisfies \eqref{e-15}, it follows that
$\|\mathbf{w}_{\rho z}\|^2 =
\|\mathbf{w}_{\rho \xi}\|^2=\mathbf{M}$, and
\begin{equation*}
(\mathbf{w}_{\rho z}, \mathbf{w}_{\rho \xi})_{H^{1}}=
\mathbf{M}\int_{\mathbb{R}^3}\phi_{\mathbf{w}_{\rho
z}}\mathbf{w}_{\rho z}\mathbf{w}_{\rho
\xi}=\mathbf{M}\int_{\mathbb{R}^3}\phi_{\mathbf{w}_{\rho \xi}}
\mathbf{w}_{\rho \xi}\mathbf{w}_{\rho z}.
\end{equation*}
Then, from \cite[Proposition 1.2]{A.Bahri-Y.-Y.Li-1990} or
\cite[Lemma 3.7]{ Ambrosetti-Colorado-Ruiz-2007-CVPDE}, and
\eqref{d-4-4} and $(A_{5})$ and the facts $|z|\geq1$ and $c^*>0$, we
infer that
\begin{gather*}
\varepsilon_{\rho}
=\int_{\mathbb{R}^3}
 \phi_{\mathbf{w}_{\rho z}}\mathbf{w}_{\rho z}\mathbf{w}_{\rho \xi}
=\int_{\mathbb{R}^3}\phi_{\mathbf{w}_{\rho \xi}} \mathbf{w}_{\rho
\xi}\mathbf{w}_{\rho
z}\sim|2\rho|^{2c^*-1}e^{-2\rho\sqrt{\kappa_{\infty}}},
\\
\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}^2_{\rho z}\leq c|\rho
z|^{2c^*-2}\log|\rho z| e^{-2|\rho z|\sqrt{\kappa_{\infty}}}
 <c\log(3\rho)\rho^{2c^*-2}e^{-2\rho\sqrt{\kappa_{\infty}}}=o(\varepsilon_{\rho}),
\\
\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}^2_{\rho \xi} \leq c|\rho
\xi|^{2c^*-2}\log|\rho\xi|e^{-2|\rho \xi|\sqrt{\kappa_{\infty}}}
 <c\rho^{2c^*-2} e^{-2\rho\sqrt{\kappa_{\infty}}}\log\rho=o(\varepsilon_{\rho}),
\\
\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}_{\rho z}\mathbf{w}_{\rho \xi}
\leq c\left(\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}^2_{\rho z}
+ \int_{\mathbb{R}^3}\kappa(x)\mathbf{w}^2_{\rho\xi}\right)
 =o(\varepsilon_{\rho}).
\end{gather*}
So
\begin{equation*}
\mathscr{N}^{\lambda}_{\rho}(z,s)=[(1-s)^2+s^2]
\mathbf{M}+2s(1-s)\mathbf{M}\varepsilon_{\rho}+o(\varepsilon_{\rho}).
\end{equation*}
Moreover, by \cite[lemma 2.7]{G.Cerami-D.Passaseo-1995}, we obtain
\begin{align*}
\mathscr{D}_{\rho}^{\lambda}(z,s)
&:= \int_{\mathbb{R}^3} (\mu_{\infty}+
\mu(x))\phi_{\psi_{\rho}(z,s)}(x)\psi^2_{\rho}(z,s)\\
&\geq[(1-s)^{4}+s^{4}]\mu_{\infty}+ 3[(1-s)^3s+(1-s)s^3]
\mu_{\infty}\varepsilon_{\rho}.
\end{align*}
Hence
\begin{equation*}
\frac{\mathscr{N}^{\lambda}_{\rho}(z,s)}
{(\mathscr{D}_{\rho}^{\lambda}(z,s))^{1/2}} \leq\frac{1}{\mu_{\infty}^{1/2}}
\Big\{\frac{[(1-s)^2+s^2]\mathbf{M}}{[(1-s)^{4}+s^{4}] ^{1/2}}+
2\gamma(s)\mathbf{M}\varepsilon_{\rho} +o(\varepsilon_{\rho})\Big\}
\end{equation*}
where
\begin{equation*}
\gamma(s)=\frac{(1-s)s}{[(1-s)^{4}+s^{4}] \frac{1}{2}}
\Big(\frac{1}{4}-\frac{3s^2(1-s)^2}{2(1-s)^{4}+2s^{4}}\Big).
\end{equation*}
By a direct computation we obtain that $\gamma(1/2)<0$,
 hence, there exists $\mathscr{I}_{\frac{1}{2}}$, neighborhood of
  $1/2$, satisfied $\gamma(1/2)<c<0$ for all
$t\in\mathscr{I}_{\frac{1}{2}}$. Hence, for $\rho$ enough large,
\begin{equation*}
\max\Big\{\frac{\mathscr{N}^{\lambda}_{\rho}(z,s)}
{\mathscr{D}_{\rho}^{\lambda}(z,s)^{1/2}}|z\in\Sigma,s\in\mathscr{I}_{\frac{1}{2}}
\Big\}
\leq\frac{\frac{1}{4}\mathbf{M}+2c\mathbf{M}\varepsilon_{\rho}+
 o(\varepsilon_{\rho})} {\mu_{\infty}^{1/2}}
<\frac{1}{4}\mu_{\infty}^{-1/2}\mathbf{M};
\end{equation*}
On the another hand, we have
\begin{align*}
&\lim_{\rho\to+\infty}\max\Big\{\frac{\mathscr{N}^{\lambda}_{\rho}(z,s)}
{\mathscr{D}_{\rho}^{\lambda}(z,s)^{1/2}}|z\in\Sigma,
s\in[0,1]\backslash\mathscr{I}_{\frac{1}{2}} \Big\}\\
&\leq \mu_{\infty}^{-1/2}\mathbf{M}\max\Big\{\frac{[(1-s)^2+s^2]}{[(1-s)^{4}+s^{4}]
 ^{1/2}}|s\in[0,1]\backslash\mathscr{I}_{\frac{1}{2}} \Big\}\\
&<\frac{1}{4}\mu_{\infty}^{-1/2}\mathbf{M}.
\end{align*}
 When $\rho$ is large enough,
\begin{equation*}
\max_{\Sigma\times[0,1]}\frac{\mathscr{N}^{\lambda}_{\rho}(z,s)}
{\mathscr{D}_{\rho}^{\lambda}(z,s)^{1/2}}<\frac{1}{4}\mu_{\infty}^{-1/2}\mathbf{M}.
\end{equation*}
By \eqref{e-16}, we have
\begin{equation*}
\mathscr{T}_{\rho}^{\lambda}<\frac{1}{4}(2^{1/2}
\mu_{\infty}^{-1/2}\mathbf{M})^2=2m_{\infty}.
\end{equation*}
This completes the proof.
\end{proof}

\begin{lemma}\label{lem--5.6}
Let the assumptions of lemma \ref{lem--5.3} hold. Then for $\rho>0$
sufficiently large,
\begin{equation*}
\mathscr{A}_{\rho}^{\lambda}:=\max_{\Sigma} I_{\lambda}(\Psi_{\rho}(z,0))
<\mathscr{B}_0^{\lambda}.
\end{equation*}
\end{lemma}

\begin{proof}
From \eqref{e-14}, \eqref{e-15} and \eqref{e-16}, we have that for
sufficiently large $\rho$,
\begin{align*}
I_{\lambda}(\Psi_{\rho}(z,0))
&=\frac{1}{4}\Big\{\frac{\|\mathbf{w}_{\rho z}\|^2
 + \lambda\int_{\mathbb{R}^3}\kappa(x)\mathbf{w}_{\rho z}^2}
{[\int_{\mathbb{R}^3}(\mu_{\infty}+\mu(x))
\phi_{\mathbf{w}_{\rho z}}\mathbf{w}_{\rho z}^2]^{1/2}}\Big\}^2\\
&=\frac{1}{4}\left[\mu_{\infty}^{-1/2}\mathbf{M}+o_{\rho}(1)\right]^2\\
&=m_{\infty}+o_{\rho}(1).
\end{align*}
From lemma \ref{lem--5.3} the conclusion follows.
\end{proof}

\begin{proof}[Proof of Theorem\ref{theorem1-2}]
Let $\lambda^{*}$ be the number which has defined in \eqref{3-2}. We
infer from Proposition \ref{Proposition 4-2} that $\lambda^{*}\in
\mathbb{R}^{+}$. Then, we deduce from Proposition \ref{Proposition
2-5} that if $\lambda<\lambda^{*}$, then $m_{\lambda}<m_{\infty}$.
Furthermore, $m_{\lambda}$ is achieved.

Next we consider the case $\lambda>\lambda^{*}$. From Lemma
\ref{lemma-3-3} and Proposition \ref{Proposition 3-3}, one deduces
that $m_{\lambda}=m_{\infty}$, and $m_{\lambda}$ is not achieved.
Thus, we can not use minimization to solve \eqref{auto-2}. However,
we can prove that \eqref{auto-2} has a higher energy than
$m_{\infty}$ exists. For any $c\in\mathbb{R}$, we let
$I_{\lambda}^{c}:=\{u\in\mathscr{N_{\lambda}^{+}}:I_{\lambda}(u)\leq
c\}$. By Lemmas \ref{lem--5.4}-\ref{lem--5.6}, we have the following
inequalities
\begin{equation*}
m_{\infty}\leq\mathscr{A}_{\rho}^{\lambda}<
\mathscr{B}_0^{\lambda}\leq\mathscr{T}_{\rho}^{\lambda}<2m_{\infty}.
\end{equation*}
We end the proof by showing that there exists a number
$c^{*}\in[\mathscr{B}_0^{\lambda},\mathscr{T}_{\rho}^{\lambda}]$
which is a critical level of
$I_{\lambda}|_{\mathscr{N_{\lambda}^{+}}}$. We use the contradiction
arguments. Assume that this is not the case. Then the Palais-Smale
condition holds in $(m_\infty,2m_\infty)$ by Lemma \ref{lem--5.2}.
We can apply usual deformation arguments(see \cite{Willem-1996}) and
assert the existence of a number $\delta>0$ and a continuous
function $\eta:I_{\lambda}^{\mathscr{T}_{\rho}^{\lambda}}\to
I_{\lambda}^{\mathscr{B}_0^{\lambda}-\delta}$ such that
$\mathscr{B}_0^{\lambda}-\delta>\mathscr{A}_{\rho}^{\lambda}$ and
$\eta(u)=u$ for all $u\in
I_{\lambda}^{\mathscr{B}_0^{\lambda}-\delta}$. Thus, we see that
\begin{equation}\label{e-17}
0\notin\beta\circ\eta\circ\Psi_{\rho}(\Sigma,[0,1]).
\end{equation}
On the other hand, since $\Psi_{\rho}(\Sigma,[0,1])\subset
I_{\lambda}^{\mathscr{A}_{\rho}^{\lambda}}$,
$\beta\circ\eta\circ\Psi_{\rho}(\Sigma,[0,1])$ is homeomorphic to
$\rho\Sigma$ in $\mathbb{R}^3\backslash\{0\}$. So, one has
\begin{equation*}
0\in\beta\circ\eta\circ\Psi_{\rho}(\Sigma,[0,1]),
\end{equation*}
which contradicts \eqref{e-17}.

Finally, because for any $\lambda\in\mathbb{R}^{+}$,
 we can find a solution $u_{\lambda}$ of
\eqref{auto-2} with $I_{\lambda}(u_{\lambda})<2m_{\infty}$.
Moreover, since we find the second solution in
$H_+^1(\mathbb{R}^3)$, we conclude that it is positive.
\end{proof}



\subsection*{Acknowledgements} This work was supported by NNSFC
(Grants 11571140, 11671077), by the Fellowship of Outstanding Young Scholars of
Jiangsu Province (BK20160063), by the Six big talent peaks project in
Jiangsu Province (XYDXX-015), and by the NSF of Jiangsu Province
(BK20150478).

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\end{document}