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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 35, pp. 1--22.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/35\hfil Existence of two positive solutions]
{Existence of two positive solutions for indefinite Kirchhoff equations 
in $\mathbb{R}^3$}

\author[L. Ding, Y.-J. Meng, S.-W. Xiao, J.-L. Zhang \hfil EJDE-2016/35\hfilneg]
{Ling Ding, Yi-Jie Meng, Shi-Wu Xiao, Jin-Ling Zhang}

\address{Ling Ding \newline
School of Mathematics and Computer Science,
 Hubei University of Arts and Science, Hubei 441053, China}
\email{dingling1975@qq.com, dingling19750118@163.com}

\address{Yi-Jie Meng \newline
School of Mathematics and Computer Science,
Hubei University of Arts and Science, Hubei 441053, China}
\email{245330581@qq.com}

\address{Shi-Wu Xiao \newline
School of Mathematics and Computer Science,
Hubei University of Arts and Science, Hubei 441053, China}
\email{xshiwu@sina.com}

\address{Jin-Ling Zhang \newline
School of Mathematics and Computer Science,
Hubei University of Arts and Science, Hubei 441053, China}
\email{jinling48@163.com}

\thanks{Submitted August 20, 2015. Published January 25, 2016.}
\subjclass[2010]{35J60, 35B38, 35A15}
\keywords{Indefinite Kirchhoff equation; concentration compactness lemma;
\hfill\break\indent  (PS) condition;  Ekeland's variational principle}

\begin{abstract}
 In this article we study the Kirchhoff type equation
 \begin{gather*}
 -\Big(1+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+u
 =k(x)f(u)+\lambda h(x)u,\quad x\in \mathbb{R}^3, \\
 u\in H^{1}(\mathbb{R}^3),
 \end{gather*}
 involving a linear part $-\Delta u+u-\lambda h(x)u$ which is coercive
 if $0<\lambda<\lambda_1(h)$ and is noncoercive if $\lambda>\lambda_1(h)$,
 a nonlocal nonlinear term $-b\int_{\mathbb{R}^3}|\nabla u|^2dx\Delta u$
 and a sign-changing nonlinearity of the form $k(x)f(s)$, where $ b>0$,
 $\lambda>0$ is a real parameter and $\lambda_1(h)$ is the first eigenvalue
 of $-\Delta u+u=\lambda h(x)u$. Under  suitable assumptions on
 $f$ and $h$, we obtain positives solution for $\lambda\in(0,\lambda_1(h))$ 
 and two positive solutions with a condition on $k$.
\end{abstract}

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


\section{Introduction and statement of main results}

In this paper, we consider the Kirchhoff equation
\begin{equation} \label{1}
\begin{gathered}
-\Big(1+b\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)\Delta u+u
 =k(x)f(u)+\lambda h(x)u,\quad x\in \mathbb{R}^3, \\
u\in H^{1}(\mathbb{R}^3),
\end{gathered}
\end{equation}
where $b$ is a positive constant, $\lambda>0$ is a real parameter, $k(x)$ 
is sign changing in $\mathbb{R}^3$ which is why
we call problem \eqref{1} indefinite Kirchhoff equation, 
$f: \mathbb{R}^+\to \mathbb{R}$ is a continuous function
and $h(x)$ is a positive function.

When $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$, the problem
\begin{equation} \label{2}
\begin{gathered}
-\Big(a+b\int_{\Omega}|\nabla u|^2dx\Big)\Delta u=g(x,u),\quad  x\in \Omega, \\
 u=0,\quad x\in \partial\Omega,
\end{gathered}
\end{equation}
is related to the stationary analogue of the Kirchhoff equation
which was proposed by Kirchhoff in 1883 (see \cite{1}) as
an generalization of the well-known d'Alembert's wave equation
$$
 \rho \frac{\partial ^2u}{\partial t^2}
-\Big(\frac{P_0}{h}+ \frac{E}{2L}\int_0^L|\frac{\partial u}{\partial
x}|^2dx\Big)\frac{\partial ^2u}{\partial x^2}=g(x,u)
$$
for free vibrations of elastic strings. Kirchhoff's model takes into 
account the changes in length of the string produced by transverse vibrations. 
Here, $L$ is the length of the string, $h$ is the area of the cross section, 
$E$ is the Young modulus of the material, $\rho$ is the mass density 
and $P_0$ is the initial tension.

In \cite{2}, it was pointed out that the problem \eqref{2} models several 
physical systems, where $u$ describes a process which depends on the 
average of itself. Nonlocal effect also finds its applications in biological 
systems. A parabolic version of equation \eqref{1} can, in theory, 
be used to describe the growth and movement of a particular species. 
The movement, modeled by the integral term, is assumed to be dependent 
on the energy of the entire system with $u$ being its population density.
 Alternatively, the movement of a particular species may be subject to 
the total population density within the domain (for instance, the spreading 
of bacteria) which gives rise to equations of the type 
$u_t-a(\int_\Omega udx)\Delta u =h$. Some early classical studies of 
Kirchhoff's equation were those of Bernstein \cite{3} and Poho\v{z}aev \cite{4}. 
However, equation \eqref{2} received great attention only after that
Lions \cite{5} proposed an abstract framework for the problem. 
Some interesting results  for problem \eqref{2} can be found in \cite{6,7,8}
and the references therein.

Some interesting studies by variational methods can be found 
in \cite{2} and \cite{10}-\cite{22} 
for Kirchhoff-type problem \eqref{2} in a bounded domain $\Omega$
 of $\mathbb{R}^N$. Very recently, some authors had studied the multiplicity 
of solutions for the Kirchhoff equation on the whole space $\mathbb{R}^N$.
 Jin and Wu \cite{23} obtained the existence of infinitely many radial 
solutions for problem (1) in $\mathbb{R}^N$ using the Fountain Theorem. 
Wu \cite{24} obtained four new existence results for nontrivial solutions 
and a sequence of high energy  solutions for  \eqref{1} in $\mathbb{R}^N$ 
which was obtained by
using the Symmetric Mountain Pass Theorem.  Azzollini, d'Avenia 
and Pomponio \cite{25} obtained a multiplicity result concerning the critical 
points of  a class of functionals involving local and nonlocal nonlinearities, 
then they apply their result to the nonlinear elliptic
 Kirchhoff equation \eqref{1} in $\mathbb{R}^N$ assuming that the local nonlinearity 
satisfies the general hypotheses introduced by Berestycki and Lions \cite{26}. 
He and Zou \cite{27} study the existence, multiplicity and concentration
 behavior of positive solutions for the nonlinear Kirchhoff type problem. 
They relate the number of solutions with the topology of the set.
 Alves and Figueiredo in \cite{29} study a periodic Kirchhoff equation 
in $\mathbb{R}^N$, they get the nontrivial solution when the nonlinearity 
is in subcritical case and critical case. Liu and He \cite{30} get multiplicity 
of high energy solutions for superlinear
Kirchhoff equations in $\mathbb{R}^3$. Recently, Chen in \cite{56} obtained 
the existence result of a positive solution for any $\lambda\in(0,\lambda_1(h))$ 
and the multiplicity result of two positive solutions for any 
$\lambda\in(\lambda_1(h),\lambda_1(h)+\widetilde{\delta})$ for problem \eqref{1}
 with the indefinite nonlinearity $k(x)f(s)=k(x)|s|^{p-2}s$ ($4<p<6$).
 Especially, inspired by \cite{56,55}, we shall consider the general $f(s)$
 instead of power nonlinearity $f(s)=|s|^{p-2}s$ in the indefinite 
nonlinearity $k(x)f(s)$ like as problem \eqref{1}.

In this article, we shall prove that \eqref{1} has multiple positive solutions 
for suitable range $\lambda$. We assume that functions $k, f, h$ satisfy 
the following hypotheses:
\begin{itemize}
\item[(H1)]  $k\in C(\mathbb{R}^3)$ and $k(x)$ changes sign, i.e.,
$\Omega^+\not=\emptyset,\ \Omega^-\not=\emptyset$, where
 $\Omega^+=\{x\in \mathbb{R}^3|k(x)>0\}$, 
$\Omega^-=\{x\in \mathbb{R}^3|k(x)<0\}$. Moreover,
let $\Omega^0=\{x\in \mathbb{R}^3|k(x)=0\}$.

\item[(H2)] There exist positive constants $R_0$, $K_0$ and $M$
such that $k(x)<-K_0$ and $|k(x)|\leq M$ if $|x|>R_0$.

\item[(H3)]  $f\in C(\mathbb{R}^+,\mathbb{R})$, $f(s)>0$ for any $s>0$.

\item[(H4)]  $\lim_{s\to0}\frac{f(s)}{s^{p-1}}=1$, $4<p<6$

\item[(H5)]  $|f(s)|=|s|^{q-1}+O(|s|^\alpha)$ as $|s|\to\infty$ for any
$4<q<6$ and some $\alpha\in[0,1)$.

\item[(H6)] $h\in L^{3/2}(\mathbb{R}^3)$, $h(x)\geq0$ for any
$x\in \mathbb{R}^3$ and $h\not\equiv0$.
\end{itemize}
Furthermore, without loss of generality, we will assume that 
$f(s)$ is defined for all $s\in \mathbb{R}$ as an odd function.

 Before stating our main results, we give some notations and remarks. 
For any $1\leq t\leq+\infty$, we denote by $\|\cdot\|_t$ the usual norm of
the Lebesgue space $L^t(\mathbb{R}^3)$.
Define the function space
$$
H^1(\mathbb{R}^3):=\{u\in L^2(\mathbb{R}^3): \nabla u\in L^2(\mathbb{R}^3)\}
$$
 with the standard product and norm
$$
(u,v)=\int_{\mathbb{R}^3}(\nabla u\cdot\nabla v+uv)dx,\quad
\|u\|:=\Big(\int_{\mathbb{R}^3}(|\nabla u|^2+|u|^2)dx\Big)^{1/2}.
$$
The functional corresponding to problem \eqref{1} is defined in 
$H^{1}(\mathbb{R}^3)$ by
\begin{align*}
I_\lambda(u)
&=\frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u|^2+| u|^2)dx
+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2 \\
&\quad -\int_{\mathbb{R}^3}k(x) F(u)dx
 -\frac{\lambda}{2}\int_{\mathbb{R}^3}h(x)u^2dx,
\end{align*}
where $F(s)=\int_0^sf(\sigma)d\sigma$.
Clearly, $I_\lambda$ is well defined and is of class $C^1$ 
in $H^1(\mathbb{R}^3)$ by  (H1)--(H3) and (H5)--(H6). Moreover,
\begin{align*}
\langle I_\lambda'(u),\varphi\rangle
&=\int_{\mathbb{R}^3}(\nabla u\cdot\nabla\varphi+u\varphi) dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla u\cdot\nabla\varphi dx\\
&\quad -\int_{\mathbb{R}^3}k(x)f(u)\varphi dx
 -\lambda\int_{\mathbb{R}^3}h(x)u\varphi dx.
\end{align*}
There is one to one correspondence between the solution of problem \eqref{1} 
and the critical point of the functional $I_\lambda$. 
So in order to seek for the positive solution of \eqref{1}, we only 
need to study the existence of the positive critical point of 
$I_\lambda$. Furthermore, if (H6) holds, then for every $u\in H^1(\mathbb{R}^3)$,
there exists a unique $w\in H^1(\mathbb{R}^3)$ such that
$$
-\Delta w+w=h(x)u.
$$
Moreover, the operator $\mathcal{F}_h: H^1(\mathbb{R}^3)\to H^1(\mathbb{R}^3)$ 
defined by $\mathcal{F}_h(u)=w$ is positive and compact. Using this and the 
spectral theory of compact symmetric operator on Hilbert space, 
there exists a sequence of eigenvalues $\lambda_n(h)$ going to infinity for problem
$$
-\Delta u+u=\lambda h(x)u,\ u\in H^{1}(\mathbb{R}^3)
$$
with $0<\lambda_1(h)<\lambda_2(h)\leq\dots \leq\lambda_n(h)\leq\dots $
and each eigenvalue being of finite multiplicity. The first eigenvalue 
$\lambda_1(h)$ is simple, has a positive eigenfunction. 
The associated normalized eigenfunctions to sequence of eigenvalues are 
denoted by $e_1, e_2,\dots$ with $\|e_i\|=1$. In addition, we have the 
following variational characterization of $\lambda_n(h)$:
\begin{equation}
\lambda_1(h)=\inf_{u\in H^{1}(\mathbb{R}^3) \setminus\{0\} }
\frac{\|u\|^2}{\int_{\mathbb{R}^3}h(x)u^2dx},\quad
\lambda_n(h)=\inf_{u\in S_{n-1}^{\perp}\setminus\{0\}) }
\frac{\|u\|^2}{\int_{\mathbb{R}^3}h(x)u^2dx},
\label{63}
\end{equation}
where $S_{n-1}^{\perp}=\{\operatorname{span}\{e_1, e_2,\dots ,e_{n-1}\}\}^{\perp}$.


Let us consider the closed subspace of $H^{1}(\mathbb{R}^3)$ defined by
$$
H^{1}\Big(\frac{1}{\sqrt{1+a}}\Omega^0\Big)
=\big\{u\in H^{1}(\mathbb{R}^3)|u(x)=0\text{ a. e. in }
 \mathbb{R}^3\setminus \frac{1}{\sqrt{1+a}}\Omega^0\big\}
$$ 
for some $a>0$.
Let $u(x)=w(x/\sqrt{1+a})$, then the  equation
$$
-(1+a)\Delta u+u=\lambda h(x)u,\ x\in \Omega^0
$$
becomes
$$
-\Delta w+w=\lambda h(\sqrt{1+a}x)w,\ x\in \frac{1}{\sqrt{1+a}}\Omega^0,
$$
which has a sequence of eigenvalues $\lambda_n(a,h)$ with 
$0<\lambda_1(a,h)<\lambda_2(a,h)\leq\dots \leq\lambda_n(a,h)\leq\dots $, 
each eigenvalue being of finite multiplicity and 
$$
\lambda_1(a, h)=\inf_{u\in H^{1}(\mathbb{R}^3)\setminus\{0\} }
\frac{\int_{\mathbb{R}^3}[(1+a)|\nabla u|^2+|u|^2]dx}
{\int_{\mathbb{R}^3}h(x)u^2dx}.
$$ 
Clearly, we have
$$
\lambda_1(a, h)>\lambda_1( h).
$$
Especially, $\lambda_1(a, h)=\lambda_1( h)$ if $a=0$.
 Let $\delta_*=\lambda_1(a, h)-\lambda_1( h)$. 
If $\overline{\delta}\in[0,\delta_*)$, then 
$$
\lambda\not\in\sigma(-\Delta,\frac{1}{\sqrt{1+a}}\Omega^0,h(\sqrt{1+a}x) )\quad
\text{if } \lambda\in(0,\lambda_1(h)+\overline{\delta}),
$$
where $\sigma(-\Delta,\frac{1}{\sqrt{1+a}}\Omega^0,h(\sqrt{1+a}x) )$ denotes
 by the collection of eigenvalues of $-(1+a)\Delta+Id$ in 
$H_0^1(\frac{1}{\sqrt{1+a}}\Omega^0)$.
 If the Lebesgue measure of $\Omega^0$ is zero, i.e., $|\Omega^0|=0$, and 
$a=0$, then $\sigma(-\Delta,\Omega^0,h(x) )=\emptyset$. 
If $|\Omega^0|\neq0$ and $H^{1}(\frac{1}{\sqrt{1+a}}\Omega^0)\neq\{0\}$, 
it follows that $\sigma(-\Delta,\frac{1}{\sqrt{1+a}}\Omega^0,h(\sqrt{1+a}x) )$ 
is discrete set and the equation
 $$
\int_{\Omega^0}[(1+a)\nabla u\cdot\nabla\varphi+u\varphi] dx
=\lambda\int_{\Omega^0}h(x)u\varphi dx,\quad \forall\varphi\in H^{1}(\Omega^0) 
$$
has a nontrivial solution $u\in H_0^{1}(\Omega^0)$ if and only if 
$\lambda\in \sigma(-\Delta,\frac{1}{\sqrt{1+a}}\Omega^0,h(\sqrt{1+a}x) )$.

Our main result is as follows.

\begin{theorem} \label{thm1.1}
Suppose that {\rm (H1)--(H6)} hold. Then
\begin{itemize}
\item[(1)] for $0<\lambda\leq\lambda_1(h)$, problem \eqref{1} has at least one 
positive solution in $H^{1}(\mathbb{R}^3)$;

\item[(2)] there exists $\widetilde{\delta}>0$, for 
$\lambda_1(h)<\lambda<\lambda_1(h)+\widetilde{\delta}$, 
problem \eqref{1} has at least two positive solutions in $H^{1}(\mathbb{R}^3)$.
\end{itemize}
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
Theorem \ref{thm1.1} generalizes \cite[Theorem 1.1]{56} with 
$f(s)=|s|^{p-2}s$ to general form $f(s)$ satisfying (H3)-(H5).
Furthermore, if $|\Omega^0|\neq0$, Theorem \ref{thm1.1} still holds in this paper,
 which is not considered in \cite{56}, and conditions of (H1) and (H2) are weaker
than the corresponding ones in \cite{56} because the existence of limit 
of $k(x)$ at $|x|\to\infty$ is not necessary. Moreover, 
it is not difficult to find some functions $f$ satisfying (H3)-(H5).
The typical example is that $f(s)=|s|^{p-2}s$. More generally,
taking $\psi\in C_0^\infty(\mathbb{R}^3,[0,1])$ such that $\psi(x)=1$ 
if $|x|<1$ and $\psi(x)=0$ if $|x|>2$. Let 
$$
f(s)=\psi(x)s^{p-1}+(1-\psi(x))\big[s^{q-1}+P(x)/\widetilde{P}(x)s^{\alpha}\big]
\quad \text{for } s>0,
$$ 
where $P(x)$ and $\widetilde{P}(x)$ are two polynomials with the same degree.
Clearly, $f$ satisfies (H3)-(H5).
\end{remark}

\begin{remark} \label{rmk1.2}\rm
 For elliptic equations with indefinite nonlinearity, Alama and Tarantello
 \cite{57} also have studied the existence of multiple positive solutions of
$$
-\Delta u-\widetilde{\lambda} u=W(x)f(u),\quad u\in H_0^1(\Omega)
$$
under the suitable assumptions $f$ behaving like $|t|^{p-1}t$($p\in(2,2N/(N-2))$) 
near zero with $\int_\Omega W(x)\widetilde{e}_1^pdx<0$,
 where $\Omega$ is a smooth bounded domain of $\mathbb{R}^N(N\geq3)$,
 $W$ is sign changing on $\Omega$ and $\widetilde{e}_1$ is the positive 
eigenfunction corresponding to the first eigenvalue $\widetilde{\lambda}_1$ 
of the problem $-\Delta u=\widetilde{\lambda} u,\ u\in H_0^1(\Omega)$.
 Costa and Tehrani \cite{55} obtained existence results of positive solutions 
for 
$$
-\Delta u-\widehat{\lambda}h(x) u=A(x)g(u),\quad u\in D^{1,2}(\mathbb{R}^N)
$$ 
for suitable $h$, sign changing $A$ and $g$ with 
$\int_{\mathbb{R}^N} A(x)\overline{e}_1^pdx<0$, where $\overline{e}_1$ 
is the positive eigenfunction corresponding to the first eigenvalue 
$\overline{\lambda}_1$ of problem 
$-\Delta u=\overline{\lambda} h(x)u,\ u\in D^{1,2}(\mathbb{R}^N)$ and 
$A$ has \lq\lq thick" zero set. But for Kirchhoff equations with indefinite 
nonlinerity like as \eqref{1}, this kind of condition such as
 $\int_{\mathbb{R}^N} a(x)\overline{e}_1^pdx<0$ and 
$\int_\Omega W(x)\widetilde{e}_1^pdx<0$ and so on is not necessary,
because the nonlocal nonlinear term 
$-b\int_{\mathbb{R}^3}|\nabla u|^2dx\Delta u$ dominated over indefinite 
nonlinear term $k(x)f(s)$(see \cite{56}). Furthermore, this yields 
that the condition: $A$ has ``thick'' zero set in \cite{55} or 
$|\Omega^0|=0$ like as in \cite{56} is also not necessary.
\end{remark}

To obtain our result, we have to overcome various difficulties. 
First of all, since the equation is considered in the whole space 
$\mathbb{R}^3$ and the Sobolev embedding 
$H^1(\mathbb{R}^3)\hookrightarrow L^s(\mathbb{R}^3)$($2\leq s<6$) 
is no longer compact, the concentration-compactness lemma 
in \cite{52} is applied to restore compactness properties to prove 
that $I_\lambda$ satisfies $(PS)$ condition by constructing sequences 
of ``almost critical points" at those energy levels where compactness 
is available. On the other hand, because of the general term $f$ 
in indefinite nonlinearity $k(x)f(s)$, we use the concentration-compactness 
lemma in \cite{52} and not use Brezis-Lieb Lemma to prove $(PS)$ 
condition like as in \cite{56}.
 When $\lambda\in(0,\lambda_1(h))$, the linear part 
$-\Delta u+u-\lambda h(x)u$ of problem \eqref{1} is coercive, 
we can use standard variational techniques to find that zero is a local 
minimizer of the corresponding functional $I_\lambda$.
But when $\lambda\in(\lambda_1(h),\lambda_1(h)+\widetilde{\delta})$,
the linear part $-\Delta u+u-\lambda h(x)u$ of problem \eqref{1}
 is not coercive, this case with indefinite nonlinearity $k(x)f(s)$ 
involving general $f$ and the nonlocal nonlinear term 
$-b\int_{\mathbb{R}^3}|\nabla u|^2dx\Delta u$ makes us to face more 
difficult than the case of $\lambda\in(0,\lambda_1(h))$ such as the 
proofs of the boundedness of (PS) sequence and the mountain pass geometry 
of $I_\lambda$. To overcome these difficulties, we need more analysis 
technical to delicately analyze the behavior of the nonlocal nonlinear 
term $-b\int_{\mathbb{R}^3}|\nabla u|^2dx\Delta u$ and the indefinite 
nonlinear term $k(x)f(s)$.

This paper is organized as follows. 
In section 2, we  prove a (PS) condition. 
In section 3, we obtain the proof of our main result.
In the following discussion, we denote various positive
 constants by $C$ or $C_i(i=1, 2, 3,\dots )$ for convenience.


\section{Palais-Smale condition}

In this section, we shall prove that the functional $I_\lambda$ satisfies 
the (PS) condition, that is, any $(PS)_c$ sequence has a convergent subsequence 
in $H^{1}(\mathbb{R}^3)$, where $(PS)_c$ sequence for the functional $I_\lambda$ 
is referred to a sequence $\{u_n\}\subset H^{1}(\mathbb{R}^3)$ such that 
$I_\lambda(u_n)\to c$
and $I'_\lambda(u_n)\to 0$ in $H^{-1}(\mathbb{R}^3)$ for $c\in \mathbb{R}$.
We need the following Lemmas.

\begin{lemma}[\cite{51}]  \label{lem2.1}
 Suppose that {\rm (H6)} holds. Then the functional defined by
$u\in H^{1}(\mathbb{R}^3)\mapsto \int_{\mathbb{R}^3}h(x)u^2dx$ 
is weakly continuous.
\end{lemma}

\begin{lemma} \label{lem2.2}
 Suppose that {\rm (H1)--(H6)} hold.
Then for every $c\in \mathbb{R}$, the $(PS)_c$ sequence is bounded 
in $H^{1}(\mathbb{R}^3)$ if $\lambda\in(0,\lambda_1(h)+\overline{\delta})$.
\end{lemma}

\begin{proof} 
 Let $\{u_n\}\subset H^{1}(\mathbb{R}^3)$ be a $(PS)_c$ sequence for 
$I_\lambda$ at the level $c$, i. e.,
\begin{equation}
\begin{aligned}
I_\lambda(u_n)&=\frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx
 +\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2 \\
&\quad -\int_{\mathbb{R}^3}k(x)F( u_n)dx
 -\frac{\lambda}{2}\int_{\mathbb{R}^3}h(x)u_n^2dx\to c
\end{aligned}\label{80}
\end{equation}
and
\begin{equation}
\begin{aligned}
\langle I_\lambda'(u_n),\varphi\rangle
&=\int_{\mathbb{R}^3}(\nabla u_n\cdot\nabla\varphi+u_n\varphi) dx
+ b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla u_n\cdot\nabla\varphi dx \\
&\quad -\int_{\mathbb{R}^3}k(x)f( u_n)\varphi dx
 -\lambda\int_{\mathbb{R}^3}h(x)u_n\varphi dx=o(1)\|\varphi\|
\end{aligned} \label{3}
\end{equation}
for any $\varphi\in H^{1}(\mathbb{R}^3) $ as $n\to\infty$.
Arguing by contradiction, we assume that $t_n=\|u_n\|$ and 
$t_n\to\infty$ as $n\to\infty$.
Denote $v_n:=u_n/t_n$. Then, we have that $\|v_n\|=1$ for each $n$. 
Going to a subsequence, if necessary, we may assume that there is 
$v\in H^{1}(\mathbb{R}^3)$ such that for each bounded domain 
$\Omega\subset \mathbb{R}^3$,
\begin{equation}
\begin{gathered}
 v_n\rightharpoonup v \quad \text{in } H^{1}(\mathbb{R}^3), \\
 v_n(x)\to v(x) \quad \text{a. e. in } \mathbb{R}^3,\\
 v_n\to v \quad  \text{in } L^{t}(\Omega) \text{ for } 2\leq t<6,\\
 |v_n(x)|\leq w(x) \quad  \text{for some } w\in L^{t}(\Omega).
\end{gathered}\label{4}
\end{equation}
Hence, for any $\varphi\in H^{1}(\mathbb{R}^3)$, we have that
\begin{equation}
\int_{\mathbb{R}^3}\nabla v_n\cdot\nabla\varphi dx\to
\int_{\mathbb{R}^3}\nabla v\cdot\nabla\varphi dx,\quad
\int_{\mathbb{R}^3} v_n\varphi dx\to
 \int_{\mathbb{R}^3}v\varphi dx.\label{5}
\end{equation}
\smallskip

\noindent\textbf{Step I:}
 We claim $v(x)=0$ a. e. in $\mathbb{R}^3$.
In fact, since $u_n=t_nv_n$, \eqref{3} becomes
\begin{equation}
\begin{aligned}
&\int_{\mathbb{R}^3}(\nabla v_n\cdot\nabla\varphi+v_n\varphi) dx
+ bt_n^2\int_{\mathbb{R}^3}|\nabla v_n|^2dx\int_{\mathbb{R}^3}
 \nabla v_n\cdot\nabla\varphi dx \\
&-|t_n|^{q-2}A_n(\varphi)-\lambda\int_{\mathbb{R}^3}h(x)v_n\varphi dx\\
&=\frac{o(1)\|\varphi\|}{t_n}=o(1),
\end{aligned}\label{6}
\end{equation}
where $A_n(\varphi)=\int_{\mathbb{R}^3}h_n(x)dx$ and
 $h_n(x):=k(x)|v_n|^{q-2}v_n\frac{f(t_n v_n)}{|t_n v_n|^{q-2}t_n v_n}\varphi$.
Note that
$$
|t_n|^{q-2}A_n(\varphi)=\int_{\mathbb{R}^3}k(x)
\frac{f(t_n v_n)}{t_n v_n}v_n\varphi dx.
$$
On the set $\{x|v(x)\not=0\}$, we have $|t_n v_n|\to+\infty$, and then, 
\eqref{4} and (H5) imply
\begin{align*}
h_n(x)
&= k(x)|v_n|^{q-2}v_n\frac{f(t_n v_n)}{|t_n v_n|^{q-2}t_n v_n}\varphi\\
&\to k(x)|v|^{q-2}v\varphi;
\end{align*}
On the set $\{x:v(x)=0\}$, we have $v_n(x)\to0$, so, by (H1)--(H2),
$\varphi\in H^1(\mathbb{R}^3)$, (H3) and (H5), we obtain
\[
|h_n(x)|=\big|k(x)\frac{f(t_n v_n)}{t_n ^{q-1}}\varphi \big| 
\leq\frac{C_1(1+|t_n|^{q-1}|v_n(x)|^{q-1})}{|t_n| ^{q-1}}\to 0.
\]
This with \eqref{4} involving $t=q-1$ yield that
$$
|h_n(x)|\leq C_1(1+|w(x)|^{q-1})\in L^1(\Omega)
$$
where $\Omega=\operatorname{supp}(\varphi)$.
From the discussion above, by the Lebesgue dominated convergence theorem, 
we conclude
\begin{equation}
\begin{aligned}
A_n(\varphi)
&:=\int_{\mathbb{R}^3}h_n(x) dx=\int_{\{x|v(x)\neq0\}}h_n(x) dx
 +\int_{\{x|v(x)=0\}}h_n(x) dx \\
&\to \int_{\mathbb{R}^3}k(x)|v|^{q-2}v\varphi dx
\end{aligned}\label{9}
\end{equation}
as $n\to\infty$.
 Divided \eqref{6} by $t_n^{q-2}$ and passing to limit, together with 
$q\in(4,6)$, $t_n\to\infty$ and \eqref{9}, we obtain
\begin{equation}
A_n(\varphi)\to\int_{\mathbb{R}^3}k(x)|v|^{q-2}v\varphi dx=0.\label{10}
\end{equation}
Since $\varphi\in H^1(\mathbb{R}^3)$ is arbitrary, \eqref{10} implies 
\begin{equation}
v(x)=0 \text{ if }  x\in \Omega^+\cup\Omega^-.\label{78}
\end{equation}

 Taking any $\varphi\in C_0^\infty(\mathbb{R}^3)$, passing to limit in
 \eqref{6}, by \eqref{78}, \eqref{10} and the definition of $\Omega^0$ in (H1),
 we obtain
\begin{equation}
(1+b\|\nabla u_n\|_2^2)\int_{\mathbb{R}^3}\nabla v_n\cdot\nabla\varphi dx
+\int_{\mathbb{R}^3}v_n\varphi dx
-\lambda\int_{\mathbb{R}^3}h(x)v_n\varphi dx\to0\label{86}
\end{equation}
as $n\to\infty$.
If $\{\|\nabla u_n\|_2^2\}$ is bounded, then there exist a convergent 
subsequence (still denoted by $\|\nabla u_n\|_2^2$)
and some $a>0$ such that $\|\nabla u_n\|_2^2\to a/b$ as $n\to\infty$. 
Together with \eqref{5}, Lemma \ref{lem2.1} and \eqref{78}, \eqref{86} can become to
\begin{align*}
\int_{\Omega_0}(1+a)\nabla v\cdot\nabla\varphi dx
+\int_{\Omega_0}v\varphi dx-\lambda\int_{\Omega_0}h(x)v\varphi dx=0
\end{align*}
as $n\to\infty$. Since $\lambda\in(0,\lambda_1(h)+\overline{\delta})$, 
it follows that $v=0$ a.e. on $\Omega_0$.
If $\|\nabla u_n\|_2^2\to\infty$, divided \eqref{86} by $1+b\|\nabla u_n\|_2^2$,
passing to limit, we obtain
$$
\int_{\Omega_0}\nabla v\cdot\nabla\varphi dx=0.
$$
Together with $v\in H^1(\mathbb{R}^3)$ and any 
$\varphi\in C_0^\infty(\mathbb{R}^3)$, we have $v=0$ a.e. on $\Omega_0$.
Thus, $v=0$ a.e. in $\mathbb{R}^3$.
The proof of the claim is completed.
\smallskip

\noindent\textbf{Step II:}
 We shall prove that $u_n$ is bounded in $H^1(\mathbb{R}^3)$. Indeed,
Suppose that $\int_{\mathbb{R}^3}|\nabla v_n|^2dx\to \beta\geq0$ as 
$n\to\infty$.
If $\beta=0$, since $\|v_n\|=1$, then 
$\int_{\mathbb{R}^3}| v_n|^2dx\to1$ as $n\to\infty$.
This contradicts to $v_n\to 0$ a. e. in $\mathbb{R}^3$ which follows 
from Step I. Therefore, we conclude that $\|u_n\|$ is bounded.
If $\beta>0$, we need the following arguments.

Divided \eqref{80} by $t_n^2=\|u_n\|^2$ and \eqref{3} by $t_n=\|u_n\|$, we obtain
\begin{equation}
\frac{1}{2}+\frac{b}{4}\int_{\mathbb{R}^3}|\nabla u_n|^2dx
\int_{\mathbb{R}^3}|\nabla v_n|^2dx
-\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}dx
-\frac{\lambda}{2}\int_{\mathbb{R}^3}h(x)v_n^2dx\to 0 \label{15}
\end{equation}
and
\begin{equation}
\begin{aligned}
\frac{1}{t_n}\langle I_\lambda'(u_n),\varphi\rangle
&=\int_{\mathbb{R}^3}(\nabla v_n\nabla\varphi+v_n\varphi) dx
+ b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla v_n\nabla\varphi dx \\
&\quad -\int_{\mathbb{R}^3}k(x)\frac{f( u_n)}{t_n}\varphi dx
 -\lambda\int_{\mathbb{R}^3}h(x)v_n\varphi dx\to 0
\end{aligned} \label{14}
\end{equation}
as $n\to\infty$.
Moreover, if we localize and take $\varphi=v\xi$ in \eqref{14} with 
$\xi\in C_0^\infty(\mathbb{R}^3)$, since 
$\langle I_\lambda'(u_n),v\xi\rangle=
\langle I_\lambda'(u_n),v_n\xi\rangle-\langle I_\lambda'(u_n),(v_n-v)\xi\rangle$ 
and \eqref{4}, passing to limit, we obtain
\begin{equation}
\begin{aligned}
&\int_{\mathbb{R}^3}(|\nabla v_n|^2\xi+v_n^2\xi+v_n\nabla v_n\nabla\xi)dx+b
\int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla v_n|^2\xi dx
 \\
&+b \int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx
-\int_{\mathbb{R}^3}k(x)\frac{f( u_n)u_n}{t_n^2}\xi dx \\
&-\lambda\int_{\mathbb{R}^3}h(x)v_n^2\xi dx\to0.
\end{aligned} \label{20}
\end{equation}
Since $v_n\rightharpoonup v$ in $H^1(\mathbb{R}^3)$, $v=0$ a.e. in 
$\mathbb{R}^3$, \eqref{4} and Lemma \ref{lem2.1}, we obtain
\begin{gather}
\int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx\to0,\quad
 \int_{\mathbb{R}^3}| v_n|^2\xi dx\to0,  \label{21} \\
\int_{\mathbb{R}^3}h(x)v_n^2\xi dx\to0,\quad
 \lim_{n\to\infty}\int_{\mathbb{R}^3}h(x)v_n^2dx=0. \label{22}
\end{gather}
as $n\to\infty$.
Inserting \eqref{22} into \eqref{15}, we have
\begin{equation}
\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}dx
=\frac{1}{2}+\frac{b}{4}\int_{\mathbb{R}^3}|\nabla u_n|^2dx
\int_{\mathbb{R}^3}|\nabla v_n|^2dx+o(1). \label{81}
\end{equation}
Inserting \eqref{21} and \eqref{22} into \eqref{20}, we obtain
\begin{equation}
\begin{aligned}
\int_{\mathbb{R}^3}k(x)\frac{f( u_n)u_n}{t_n^2}\xi dx
&=\Big(1+b \int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}|\nabla v_n|^2\xi dx \\
&\quad +b \int_{\mathbb{R}^3}|\nabla u_n|^2dx
 \int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx+o(1).
\end{aligned} \label{88}
\end{equation}

Now, we claim that
\begin{equation}
\int_{\mathbb{R}^3}k(x)\frac{qF( u_n)}{t_n^2}\xi dx
=\int_{\mathbb{R}^3}k(x)\frac{f( u_n)u_n}{t_n^2}\xi dx+o(1). \label{17}
\end{equation}
Indeed, by (H3), we know that $|qF(s)-f(s)s|\leq C_2$ for $|s|\leq M$.
 From $t_n\to\infty$ and (H1)--(H2), we clearly have
$$
\int_{[|u_n|\leq M]}|k(x)|\big|\frac{qF( u_n)-f( u_n)u_n}{t_n^2}\big||\xi|dx=o(1).
$$
Also, (H5) implies that $|qF(s)-f(s)s|=O(|s|^{\alpha+1})$ as $|s|\to\infty$,
where $1\leq\alpha+1<2$, together with (H1) and (H2), we have
$$
\int_{[|u_n|\geq M]}|k(x)|\big|\frac{qF( u_n)-f( u_n)u_n}{t_n^2}\big||\xi|dx
\leq C_3\int_{[supp\xi]\cap[|u_n|\geq M]}\frac{|u_n|^{\alpha+1}}{t_n^2}dx\to0
$$
as $n\to\infty$. Therefore, claim \eqref{17} is proved.

Choosing $\xi\in C_0^\infty(\mathbb{R}^3)$ such that 
$\xi\in[0,1]$, $\xi(x)=1$ if $|x|<R_0$ and $\xi(x)=0$ if $|x|>2R_0$, 
from \eqref{17}$, \eqref{81}$, \eqref{88} and \eqref{21}, we obtain
\begin{equation}
\begin{aligned}
&\liminf_{n\to\infty}\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}(1-\xi )dx
 \\
&= \liminf_{n\to\infty}\Big[\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}dx
-\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}\xi dx\Big]
 \\
&=\liminf_{n\to\infty}\Big[\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}dx
 -\frac{1}{q}\int_{\mathbb{R}^3}k(x)\frac{f( u_n)u_n}{t_n^2}\xi dx\Big] 
\\
&=\liminf_{n\to\infty}\Big[\frac{1}{2}+\frac{b}{4}\int_{\mathbb{R}^3}
 |\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla v_n|^2dx\\
&\quad -\frac{1}{q}\Big(1+b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}|\nabla v_n|^2\xi dx  
 -\frac{b}{q}\int_{\mathbb{R}^3}|\nabla u_n|^2dx
\int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx\Big] 
\\
&\geq \liminf_{n\to\infty}\Big[\frac{1}{2}-\frac{1}{q}+
b\big(\frac{1}{4}-\frac{1}{q}\big)
 \int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla v_n|^2dx\\
&\quad -\frac{b}{q}\int_{\mathbb{R}^3}|\nabla u_n|^2dx
\int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx\Big] 
\\
&\geq \frac{1}{2}-\frac{1}{q}+b\liminf_{n\to\infty}
 \int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big[\big(\frac{1}{4}-\frac{1}{q}\big)
 \int_{\mathbb{R}^3}|\nabla v_n|^2dx-\frac{1}{q}
 \int_{\mathbb{R}^3}v_n\nabla v_n\nabla\xi dx\Big] \\
&\geq \frac{1}{2}-\frac{1}{q}>0
\end{aligned} \label{25}
\end{equation}
because $\int_{\mathbb{R}^3}|\nabla v_n|^2dx\to \beta>0$ as $n\to\infty$ and $q>4$. 
Moreover, since $f$ is an odd function, by the definition $\xi$ and (H2),
we have
 $$
\liminf_{n\to\infty}\int_{\mathbb{R}^3}k(x)\frac{F( u_n)}{t_n^2}(1-\xi )dx
=\liminf_{n\to\infty}\int_{|x|\geq R_0}k(x)\frac{F( u_n)}{t_n^2}(1-\xi )dx\leq0,
$$
which contradicts to \eqref{25}. Thus, we conclude that $t_n=\|u_n\|$ is bounded. 
\end{proof}


\begin{lemma}[\cite{52}] \label{lem2.3}
 Let $\{\rho_n\}$ be a sequence in $L^1(\mathbb{R}^N)$ satisfying $\rho_n\geq0$ 
and $\int_{\mathbb{R}^N}\rho_ndx=\overline{\lambda}>0$. Then, there exists 
a subsequence $\{\rho_{n_k}\}$ for which one of the three possibilities holds:

Vanishing: $\lim_{k\to\infty}\sup_{y\in \mathbb{R}^N}\int_{y+B_R}\rho_{n_k}dx=0$ 
for all $R>0$;

Dichotomy: There exists $0<\alpha<\overline{\lambda}$ such that, for any given 
$\varepsilon>0$, there is a sequence $\{y_n\}\subset \mathbb{R}^N$, 
a number $R>0$ and a sequence $\{R_n\}\subset\mathbb{R}_+$, with 
$R<R_1$, $R_n<R_{n+1}\to+\infty$, such that, if we set 
$\rho_n^1=\rho_n\chi_{[|x-y_n|\leq R]}$ and
$\rho_n^2=\rho_n\chi_{[|x-y_n|\geq R_n]}$, then we have
$$
\|\rho_n-\rho_n^1-\rho_n^2\|_1\leq\varepsilon,\quad
 \big|\int_{\mathbb{R}^N}\rho_n^1dx-\alpha\big|\leq\varepsilon,\quad
\big|\int_{\mathbb{R}^N}\rho_n^2dx-(\overline{\lambda}-\alpha)\big|
 \leq\varepsilon;
$$

Compactness: There exists $y_k\in\mathbb{R}^N$ such that $\rho_{n_k}(\cdot+y_k)$ 
is tight, i.e.
for each $varepsilon>0$ there exists $R>0$ such that
\[
\int_{y_k+B_R}\rho_{n_k}dx\geq\overline{\lambda}-\varepsilon.
\]
\end{lemma}

\begin{lemma} \label{lem2.4}
 Suppose that {\rm (H1)--(H6)} hold, then
$I_\lambda$ satisfies the (PS) condition if
 $\lambda\in(0,\lambda_1(h)+\overline{\delta})$.
\end{lemma}

\begin{proof}
 By Lemma \ref{lem2.2}, we know that a $(PS)_c$ sequence $\{u_n\}$ is bounded in
 $H^1(\mathbb{R}^3)$. Without loss of generality, we may assume that
 \begin{equation}
C_0\geq\|u_n\|^2\geq C_4>0.\label{50}
\end{equation}
Therefore, by considering the sequence of $L^1({\mathbb{R}^3})$ functions
$$
\rho_n=|\nabla u_n|^2+|u_n|^2,
$$
we have(by passing to a subsequence, if necessary) that
$$
\int_{\mathbb{R}^3}\rho_ndx\to\overline{\lambda}>0.
$$
We shall use the concentration-compactness (Lemma \ref{lem2.3}) 
to show that $\{u_n\}$ has a convergent subsequence in $H^1(\mathbb{R}^3)$. 
In fact, we will rule out vanishing and dichotomy for the $L^1$ sequence 
of $\{\rho_n\}$.


Vanishing: In our situation vanishing can not occur. 
Indeed, if vanishing happens, i.e.,
$\lim_{n\to\infty}\sup_{y\in \mathbb{R}^3}\int_{y+B_t}\rho_{n}dx=0$ 
for all $t>0$, then we have $u_n\rightharpoonup0$ weakly in $H^1(\mathbb{R}^3)$, 
therefore, $u_n\to0$ in $L_{\rm loc}^s(\mathbb{R}^3)$ for any $2\leq s\leq 6$.
Together with \eqref{80} and \eqref{3}, we have
\begin{align*}
\frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx
+\frac{b}{4}\left(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\right)^2
-\int_{\mathbb{R}^3}k(x)F( u_n)dx\to c\label{26}
\end{align*}
and
\begin{equation}
\begin{aligned}
&\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx
+ b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2 \\
&-\int_{|x|\leq R_0}k(x)f( u_n)u_n dx
 -\int_{|x|> R_0}k(x)f( u_n)u_n dx\to0
\end{aligned} \label{27}
\end{equation}
as $n\to\infty$.
From (H3)-(H5), there exist $C_i(i=5,\dots ,12)$ and $\delta_0>0$ such that
\begin{gather}
|f(s)|\leq C_5|s|^{p-1}+C_6|s|^{q-1},\quad
|F(s)|\leq C_7|s|^{p}+C_8|s|^{q},\label{28}
\\
C_9|s|^{p}\leq |F(s)|,\quad C_{10}|s|^{p}\leq f(s)s\quad 
\text{if } |s|\leq\delta_0, \nonumber %\label{29}
\\
C_{11}|s|^{q}\leq |F(s)|,\quad C_{12}|s|^{q}\leq f(s)s\quad \text{if }
 |s|\geq\delta_0. \nonumber %\label{30}
\end{gather}
Since $u_n\to0$ in $L_{\rm loc}^s(\mathbb{R}^3)$ for any $2\leq s\leq 6$, 
by (H1)--(H2) and \eqref{28}, we obtain
$$
\big|\int_{[|x|\leq R_0]}k(x)f( u_n)u_n dx\big|
+\big|\int_{[|x|\leq R_0]}k(x)F( u_n) dx\big|=o(1),
$$
where $R_0$ is given in (H2). Then \eqref{27} yields
\begin{align*}
\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2
-\int_{|x|> R_0}k(x)f( u_n)u_n dx\to0
\end{align*}
as $n\to\infty$. By (H2), we deduce that
$$
\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx=o(1),\quad
\int_{\mathbb{R}^3}k(x)f( u_n)u_n dx=o(1).
$$
Hence, $\|u_n\|\to 0$ as $n\to\infty$, which is a contradiction.
\smallskip

Dichotomy: If dichotomy occurs, then there exists 
$0<\alpha<\overline{\lambda}$ such that, for any given $\varepsilon>0$ and 
$\overline{R}\geq1$, there are sequences $\{y_n\}\subset \mathbb{R}^3$, 
$\{R_n\}\subset\mathbb{R}_+$ and $\widehat{R}>\overline{R}$ satisfying
 $R_0<\widehat{R}<\frac{1}{2}R_1$, $R_n<R_{n+1}\to+\infty$ and
 \begin{gather}
\alpha-\varepsilon<\int_{|x-y_n|\leq \frac{1}{2}\widehat{R}}
(|\nabla u_n|^2+|u_n|^2)dx<\alpha+\varepsilon, \nonumber %\label{40}
\\
\overline{\lambda}-\alpha-\varepsilon<\int_{|x-y_n|\geq 3R_n}
(|\nabla u_n|^2+|u_n|^2)dx<\overline{\lambda}-\alpha+\varepsilon.
 \label{42}
\end{gather}
In particular, we have
\begin{equation}
\int_{\frac{1}{2}\widehat{R}<|x-y_n|\leq3 R_n}
(|\nabla u_n|^2+|u_n|^2)dx<2\varepsilon.
 \label{43}\end{equation}
Note that we also have
 \begin{equation}
\int_{\widehat{R}<|x-y_n|\leq 2 R_n}|u_n|^6dx<C_{13}\varepsilon^{3}.
 \label{44}\end{equation}
In fact, if $\xi_n\in C_0^\infty(\mathbb{R}^3)$ is such that 
$\xi_n(x)=0$ if $|x|\leq\frac{1}{2}\widehat{R}$ and $|x|\geq 3R_n$,
 $\xi_n=1$ if $\widehat{R}\leq |x|\leq 2R_n$ and 
$|\nabla\xi_n|\leq\frac{1}{R_n-\widehat{R}}$, then 
$\xi_nu_n\in H^1(\mathbb{R}^3)$, by the Sobolev's inequality, 
\eqref{50} and \eqref{43}, we have
\begin{align*}
&\int_{\widehat{R}<|x-y_n|\leq2 R_n}|u_n|^6dx\\
&\leq \int_{\frac{1}{2}\widehat{R}<|x-y_n|\leq3 R_n}|\xi_nu_n|^6dx\\
&\leq \int_{\mathbb{R}^3}|\xi_nu_n|^6dx\\
&\leq  C_{14}\Big(\int_{\mathbb{R}^3}(|\nabla (\xi_nu_n)|^2+|\xi_nu_n|^2)dx\Big)^{3} \\
&\leq C_{15}\Big(\int_{\mathbb{R}^3}|\nabla \xi_n|^2u_n^2dx
 +\int_{\frac{1}{2}\widehat{R}<|x-y_n|\leq3 R_n}(|\nabla u_n|^2+|u_n|^2)dx\Big)^{3}\\
&\leq  C_{13}\varepsilon^3
 \end{align*}
as $n\to\infty$.

 Next, let us take $\zeta\in C_0^\infty(\mathbb{R}^3)$ such that 
$\zeta(x)=1$ for $|x|\leq1$,
$\zeta(x)=0$ for $|x|\geq2$, as well as $\eta(x)=1-\zeta(x)$, and set
$$
u_n^1=\zeta\Big(\frac{\cdot-y_n}{\widehat{R}}\Big)u_n:=\zeta_nu_n,\quad
u_n^2=\eta\Big(\frac{\cdot-y_n}{R_n}\Big)u_n:=\eta_nu_n.
$$ 
Clearly, $u_n^1u_n^2=0$.
\smallskip

\noindent\textbf{Case 1:}
 If $\{y_n\}$ is bounded. Then, the support of the sequence $\{u_n^2\}$ 
approaches infinity, let $R_n\to\infty$,
then $u_n^2\rightharpoonup0$ in $H^1(\mathbb{R}^3)$.
 By \eqref{44}, then we have
\begin{align*}
\big|\int_{\mathbb{R}^3}h(x)u_n^2(u_n^2-u_n)dx\big|
&\leq \int_{R_n\leq|x-y_n|\leq2R_n}|h(x)||u_n^2||u_n^2-u_n|dx\\
&\leq \int_{R_n\leq|x-y_n|\leq2R_n}|h(x)||\zeta_n\eta_n||u_n|^2dx\\
&\leq C_{16}\|h\|_{3/2}\varepsilon:= O(\varepsilon)
 \end{align*}
In particular, letting $\mu(\varepsilon)$ denote a function which goes 
to zero as $\varepsilon\to0$, we have
 \begin{equation}
\int_{\mathbb{R}^3}h(x)u_n^2u_ndx
=\int_{\mathbb{R}^3}h(x)|u_n^2|^2dx+\mu(\varepsilon).
 \label{45}\end{equation}
Similarly, using \eqref{43} and argue as above, it is easy to see that
 \begin{equation}
\int_{\mathbb{R}^3}\nabla u_n\cdot\nabla u_n^2dx
=\int_{\mathbb{R}^3}|\nabla u_n^2|^2dx+\mu(\varepsilon),\quad
\int_{\mathbb{R}^3}u_nu_n^2dx= \int_{\mathbb{R}^3}|u_n^2|^2dx+\mu(\varepsilon).
 \label{46}
\end{equation}
Furthermore, combining (H1)--(H2), \eqref{43}, \eqref{44}, \eqref{28}
 and Sobolev's inequality, we obtain
\begin{align*}
&\big|\int_{\mathbb{R}^3}k(x)(f(u_n)-f(u_n^2))u_n^2dx\big|\\
&\leq C_{17}\int_{R_n\leq|x-y_n|\leq2R_n}|u_n|^pdx
 +C_{18}\int_{R_n\leq|x-y_n|\leq2R_n}|u_n|^qdx\\
&\leq C_{19}\Big(\int_{R_n\leq|x-y_n|\leq2R_n}|u_n|^6dx\Big)^{p/6}
 + C_{20}\Big(\int_{R_n\leq|x-y_n|\leq2R_n}|u_n|^6dx\Big)^{q/6}\\
&\leq C_{21}\varepsilon^{p/2}+C_{22}\varepsilon^{q/2}=\mu(\varepsilon).
\end{align*}
This yields
\begin{equation}
\int_{\mathbb{R}^3}k(x)f(u_n)u_n^2dx
=\int_{\mathbb{R}^3}k(x)f(u_n^2)u_n^2dx+\mu(\varepsilon).\label{48}
\end{equation}
By \eqref{45}-\eqref{48}, we obtain
\begin{align*}
&o(1)\\
&=\langle I'(u_n), u_n^2\rangle\\
&=\int_{\mathbb{R}^3}(\nabla u_n\nabla u_n^2+u_n u_n^2) dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla( u_n-u_n^2+u_n^2)\nabla u_n^2dx \\
&\quad -\int_{\mathbb{R}^3}k(x)f( u_n)u_n^2 dx
 -\lambda\int_{\mathbb{R}^3}h(x)u_n u_n^2 dx\\
&=\int_{\mathbb{R}^3}(|\nabla u_n^2|^2+| u_n^2|^2) dx
+b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla u_n^2|^2dx \\
&\quad -\int_{\mathbb{R}^3}k(x)f( u_n^2)u_n^2 dx
 -\lambda\int_{\mathbb{R}^3}h(x)|u_n^2|^2 dx+\mu(\varepsilon).
\end{align*}
Since $R_n\to\infty$, $\{y_n\}$ is bounded, $u_n^2\rightharpoonup0$ weakly in 
$H^1(\mathbb{R}^3)$ and $\int_{\mathbb{R}^3}h(x)|u_n^2|^2 dx=o(1)$ as
 $n\to\infty$, we deduce
\begin{align*}
&\int_{\mathbb{R}^3}(|\nabla u_n^2|^2+| u_n^2|^2) dx
+ b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla u_n^2|^2dx
-\int_{\mathbb{R}^3}k(x)f( u_n^2)u_n^2 dx\\
&=o(1)+\mu(\varepsilon),
\end{align*}
which implies that
\begin{gather*}
\int_{\mathbb{R}^3}(|\nabla u_n^2|^2+| u_n^2|^2) dx=o(1)+\mu(\varepsilon),\\
b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\int_{\mathbb{R}^3}|\nabla u_n^2|^2dx
=o(1)+\mu(\varepsilon),\\
-\int_{\mathbb{R}^3}k(x)f( u_n^2)u_n^2 dx=o(1)+\mu(\varepsilon)
\end{gather*}
 because (H2) as $n\to\infty$.
Then we have a contradiction, indeed, by \eqref{42}, we have
$$
\int_{\mathbb{R}^3}(|\nabla u_n^2|^2+| u_n^2|^2) dx
\geq\int_{|x-y_n|\geq3R_n}(|\nabla u_n^2|^2+| u_n^2|^2) dx
>\overline{\lambda}-\alpha-\varepsilon.
$$
\smallskip

\noindent\textbf{Case 2:}
 If $\{y_n\}$ is not bounded. Then, passing to a subsequence if necessary, 
we can assume that $|y_n|\to\infty$. In this case, the support of the 
sequence $\{u_n^1\}$ approcahes infinity, we can apply the same arguments 
above to $u_n^1$ to get a contradiction.

Compactness: Since we have ruled out vanishing and dichotomy, 
it follows that compactness necessarily take place, i.e. there exists 
$y_n\in\mathbb{R}^3$ such that
for each $\varepsilon>0$ there exists $R>0$ such that
\[
\int_{y_n+B_R}(|\nabla u_n|^2+|u_n|^2)dx\geq\overline{\lambda}-\varepsilon.
\]
In particular, we have
 \begin{equation} 
\int_{|x-y_n|\geq R}(|\nabla u_n|^2+|u_n|^2)dx<\varepsilon.\label{60}
\end{equation}
By \eqref{60} and a similar method as above, we claim that $\{y_n\}$
 must remain bounded. In fact, if not, then
\eqref{60} implies that $u_n\rightharpoonup 0$ weakly in 
$H^1(\mathbb{R}^3)$, together with (H1)--(H6), we have
\begin{align*}
o(1)&=\langle I_\lambda'(u_n),u_n\rangle\\
&=\int_{\mathbb{R}^3}(|\nabla u_n|^2+|u_n|^2) dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2
 -\int_{\mathbb{R}^3}k(x)f(u_n)u_ndx+o(1)\\
&= \int_{\mathbb{R}^3}(|\nabla u_n|^2+|u_n|^2) dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2
 -\int_{|x|\leq R_0}k(x)f(u_n)u_ndx\\
&\quad -\int_{|x|> R_0}k(x)f(u_n)u_ndx+o(1)\\
&=\int_{\mathbb{R}^3}(|\nabla u_n|^2+|u_n|^2) dx
+b\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2\\
&\quad -\int_{|x|> R_0}k(x)f(u_n)u_ndx+o(1)+\mu(\varepsilon).
\end{align*}
This yields
\begin{gather*}
\int_{\mathbb{R}^3}(|\nabla u_n|^2+|u_n|^2) dx=o(1)+\mu(\varepsilon), \quad
\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2=o(1)+\mu(\varepsilon), \\
\int_{\mathbb{R}^3}k(x)f(u_n)u_ndx=o(1)+\mu(\varepsilon).
\end{gather*}
Therefore, 
\begin{align*}c+o(1)
&=I_\lambda(u_n)\\
&=\frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u_n|^2+| u_n|^2)dx
 +\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)^2 \\
&\quad -\int_{\mathbb{R}^3}k(x)F( u_n)dx
 -\frac{\lambda}{2}\int_{\mathbb{R}^3}h(x)u_n^2dx\\
&= o(1)+\mu(\varepsilon).
\end{align*}
This is a contradiction. Thus, $\{y_n\}$ is bounded in $\mathbb{R}^3$.
The boundedness of $\{y_n\}$ and \eqref{60} imply
\begin{equation} 
\int_{|x|\geq R}(|\nabla u_n|^2+|u_n|^2)dx<\varepsilon.\label{61}
\end{equation}
Since $\{u_n\}$ is bounded, we have $u_n\rightharpoonup u$ weakly 
in $H^1(\mathbb{R}^3)$ and
 $u_n\to u$ strongly in $L^t(\Omega)$ for any $2\leq t<6$, where $\Omega$ 
is bounded. From  \eqref{61}, we obtain 
\begin{equation}
u_n\to u\quad \text{strongly in $L^t(\mathbb{R}^3)$ for } 2\leq t<6.\label{62}
\end{equation}
Equations \eqref{50}, \eqref{62}, (H1)--(H6) imply
\begin{gather*}
\int_{\mathbb{R}^3}k(x)(f(u_n)-f(u))(u_n-u)dx\to0,\quad
\int_{\mathbb{R}^3}h(x)(u_n-u)^2dx\to0, \\
\int_{\mathbb{R}^3}k(x)f(u_n)(u_n-u)dx\to0,\quad
\int_{\mathbb{R}^3}h(x)u_n(u_n-u)dx\to0
\end{gather*}
as $n\to\infty$. This yields 
\begin{align*}
o(1)&= \langle I_\lambda'(u_n),u_n-u\rangle\\
&=\Big(1+b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla u_n\nabla( u_n-u) dx
  +\int_{\mathbb{R}^3}u_n( u_n-u) dx\\
&\quad -\int_{\mathbb{R}^3}k(x)f(u_n)(u_n-u)dx
 -\lambda\int_{\mathbb{R}^3}h(x)u_n(u_n-u)dx\\
&=\Big(1+b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
\int_{\mathbb{R}^3}\nabla u_n\nabla( u_n-u) dx+o(1).
\end{align*}
From the boundedness of $\{u_n\}$ in $H^1(\mathbb{R}^3)$, we obtain
$$
\int_{\mathbb{R}^3}\nabla u\nabla(u_n-u)dx\to0
$$
as $n\to\infty$. Moreover, we have
\begin{align*}
&\langle I_\lambda'(u_n)-I_\lambda'(u),u_n-u\rangle\\
&= \Big(1+b\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}|\nabla( u_n-u)|^2 dx+\int_{\mathbb{R}^3}| u_n-u|^2 dx\\
&\quad +b\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx
 -\int_{\mathbb{R}^3}|\nabla u_n|^2dx\Big)
 \int_{\mathbb{R}^3}\nabla u\nabla(u_n-u))dx\\
&\quad -\int_{\mathbb{R}^3}k(x)(f(u_n)-f(u))(u_n-u)dx
 -\lambda\int_{\mathbb{R}^3}h(x)(u_n-u)^2dx\\
&\geq \int_{\mathbb{R}^3}(|\nabla( u_n-u)|^2 +| u_n-u|^2) dx+o(1).
\end{align*}
This yields 
$$
\|u_n-u\|^2\leq \langle I_\lambda'(u_n)-I_\lambda'(u),u_n-u\rangle+o(1)\to0
$$
as $n\to\infty$. Thus, $u_n\to u$ strongly in $H^1(\mathbb{R}^3)$.
\end{proof}

\section{Existence of positive solutions}

In this section, we shall prove our main result. Firstly, we obtain 
the local minimum of $I_\lambda$ for $\lambda\in(0,\lambda_1(h))$ and 
prove $I_\lambda$ has the mountain pass structure. Then we prove the 
existence and multiplicity of positive solution for \eqref{1} by the mountain
 pass theorem and Ekeland's variational principle, respectively.

\begin{lemma} \label{lem3.1}
Suppose that {\rm (H1)--(H6)} hold.
\begin{itemize}
\item[(a)]  If $\lambda\in(0,\lambda_1(h))$, then $u=0$ is a local minimum of 
$I_\lambda$;

\item[(b)] There exist $\widetilde{\delta}$, $\rho$ and $\alpha$ such that, 
for any $\lambda\in[\lambda_1(h),\lambda_1(h)+\widetilde{\delta})$, 
$I_\lambda(u)\geq\alpha>0$ if $\|u\|=\rho$;

\item[(c)]  There exists $w\in H^1(\mathbb{R}^3)$ with $\|w\|>\rho$ such that 
$I_\lambda(w)<0$ for any $\lambda>0$.
\end{itemize}
\end{lemma}

\begin{proof} (a) By \eqref{63}, (H1)--(H2), \eqref{28} and the Sobolev 
inequality, we have
\begin{align*}
I_\lambda(u)
&= \frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u|^2+| u|^2)dx
 +\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2
 -\int_{\mathbb{R}^3}k(x) F(u)dx\\
&\quad -\frac{\lambda}{2}\int_{\mathbb{R}^3}h(x)u^2dx\\
&\geq \frac{1}{2}\Big(1-\frac{\lambda}{\lambda_1(h)}\Big)\|u\|^2
 -\int_{|x|>R_0}k(x) F(u)dx-\int_{|x|\leq R_0}k(x) F(u)dx\\
&\geq \frac{1}{2}\Big(1-\frac{\lambda}{\lambda_1(h)}\Big)\|u\|^2
 -\int_{|x|\leq R_0}k(x) F(u)dx\\
&\geq \frac{1}{2}\Big(1-\frac{\lambda}{\lambda_1(h)}\Big)\|u\|^2
 -C_{23}\int_{\mathbb{R}^3}|F(u)|dx\\
&\geq \frac{1}{2}\Big(1-\frac{\lambda}{\lambda_1(h)}\Big)\|u\|^2
 -C_{24}\|u\|_p^p-C_{24}\|u\|_q^q\\
&\geq \frac{1}{2}\Big(1-\frac{\lambda}{\lambda_1(h)}\Big)\|u\|^2
 -C_{25}\|u\|^p-C_{26}\|u\|^q\\
&\geq C_{27}\|u\|^2
\end{align*}
for $\|u\|$ suitable small. Hence $u=0$ is a local minimizer of $I_\lambda$. 
Thus, (a) holds.

(b) For any $u\in H^1(\mathbb{R}^3)$, we decompose $u$ as $u=te_1+v$, 
where $t\in \mathbb{R}$ and $v\in\{span\{e_1\}\}^\perp$. Clearly, we have
\begin{gather*}
\|u\|^2=t^2+\|v\|^2,\ \lambda_1(h)\int_{\mathbb{R}^3}h(x)e_1^2dx=\|e_1\|^2=1,\\
\lambda_2(h)\int_{\mathbb{R}^3}h(x)|v|^2dx
 \leq\|v\|^2,\quad
\lambda_1(h)\int_{\mathbb{R}^3}h(x)e_1vdx
 =\int_{\mathbb{R}^3}(\nabla e_1\nabla v+e_1v)=0.
\end{gather*}
Using this decomposition, we also know that
$$
\int_{\mathbb{R}^3}h(x)u^2dx=t^2\int_{\mathbb{R}^3}h(x)e_1^2dx
+\int_{\mathbb{R}^3}h(x)v^2dx,
$$
\begin{align*}
& \Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2\\
&=\Big(t^2\int_{\mathbb{R}^3}|\nabla e_1|^2dx
 +2t\int_{\mathbb{R}^3}\nabla e_1\nabla vdx
 +\int_{\mathbb{R}^3}|\nabla v|^2dx\Big)^2\\
&= t^4\Big(\int_{\mathbb{R}^3}|\nabla e_1|^2dx\Big)^2
 +4t^3\int_{\mathbb{R}^3}|\nabla e_1|^2dx\int_{\mathbb{R}^3}\nabla e_1\nabla vdx\\
&\quad +2t^2\int_{\mathbb{R}^3}|\nabla e_1|^2dx\int_{\mathbb{R}^3}|\nabla v|^2dx
 +4t^2 \Big(\int_{\mathbb{R}^3}\nabla e_1\nabla vdx\Big)^2\\
&\quad +4t\int_{\mathbb{R}^3}\nabla e_1\nabla vdx\int_{\mathbb{R}^3}|\nabla v|^2dx
 +\Big(\int_{\mathbb{R}^3}|\nabla v|^2dx\Big)^2\\
&\leq t^4\Big(\int_{\mathbb{R}^3}|\nabla e_1|^2dx\Big)^2
 +\Big(\int_{\mathbb{R}^3}|\nabla v|^2dx\Big)^2
 +C_{28}|t|^3\|v\|+C_{29}|t|^2\|v\|^2+C_{30}|t|\|v\|^3,
\end{align*}
and
\begin{align*}
& -\int_{\mathbb{R}^3}k(x) F(u)dx\\
&= -\int_{|x|\leq R_0}k(x) F(u)dx-\int_{|x|>R_0}k(x) F(u)dx\\
&= -\frac{1}{p}\int_{|x|\leq R_0}k(x) |te_1|^pdx
 -\int_{|x|\leq R_0}k(x) \big[F(te_1)-\frac{1}{p} |te_1|^p\big]dx\\
&\quad -\int_{|x|\leq R_0}k(x) [F(te_1+v)-F(te_1)]dx
 -\int_{|x|>R_0}k(x) F(te_1+v)dx
\\
&\geq -\frac{1}{p}\int_{|x|\leq R_0}k(x) |te_1|^pdx
 -\int_{|x|\leq R_0}k(x) \big[F(te_1)-\frac{1}{p} |te_1|^p\big]dx\\
& -C_{31}\int_{|x|\leq R_0}k(x) f(te_1+\theta v)vdx-\int_{|x|>R_0}k(x) F(te_1+v)dx\\
&\geq -C_{32}|t|^p-\int_{|x|\leq R_0}k(x) 
 \big[F(te_1)-\frac{1}{p} |te_1|^p\big]dx+K_0\int_{|x|>R_0}F(te_1+v)dx\\
& -C_{33}\Big[\int_{|x|\leq R_0}k(x) |te_1+\theta v|^{p-1}vdx
 +\int_{|x|\leq R_0}k(x) |te_1+\theta v|^{q-1}vdx\Big]\\
&\geq -C_{32}|t|^p-\int_{|x|\leq R_0}k(x) 
 \big[F(te_1)-\frac{1}{p} |te_1|^p\big]dx+K_0\int_{|x|>R_0}F(te_1+v)dx\\
& -C_{34}\Big[|t|^{p-1}\Big(\int_{|x|\leq R_0}|v|^pdx\Big)^{1/p}
 +|t|^{q-1}\Big(\int_{|x|\leq R_0}|v|^qdx\Big)^{1/q} \\
& \quad +\int_{|x|\leq R_0} |v|^{p}dx+\int_{|x|\leq R_0} |v|^{q}dx\Big]
\\
&\geq -C_{32}|t|^p-\int_{|x|\leq R_0}k(x)\big[F(te_1)
 -\frac{1}{p} |te_1|^p\big]dx+K_0\int_{|x|>R_0}F(te_1+v)dx\\
& -C_{35} [|t|^{p-1}\|v\|+|t|^{q-1}\|v\|+\|v\|^p+\|v\|^q]\\
&\geq -C_{32}|t|^p+o(1)|t|^{p}-C_{36}\left[|t|^{p-1}\|v\|+\|v\|^p\right],
\end{align*}
because (H1)--(H3), \eqref{28}, (H5) and the odd nature of $f$ as $|t|$ 
and $\|u\|$ small enough, where $\theta\in[0,1]$. Then
\begin{equation}
\begin{aligned}
I_{\lambda_1(h)}(u)
&= \frac{1}{2}\int_{\mathbb{R}^3}(|\nabla u|^2+| u|^2)dx
+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla u|^2dx\Big)^2
 -\int_{\mathbb{R}^3}k(x) F(u)dx \\
& -\frac{\lambda_1(h)}{2}\int_{\mathbb{R}^3}h(x)u^2dx \\
&\geq \frac{1}{2}\Big(1-\frac{\lambda_1(h)}{\lambda_2(h)}\Big)\|v\|^2
 +\frac{b}{4}\theta_0 t^4+\frac{b}{4}\Big(\int_{\mathbb{R}^3}|\nabla v|^2dx\Big)^2
 -C_{28}|t|^3\|v\| \\
& -C_{29}|t|^2\|v\|^2-C_{30}|t|\|v\|^3-C_{32}|t|^p+o(1)|t|^{p} \\
& -C_{36}[|t|^{p-1}\|v\|+\|v\|^p],
\end{aligned} \label{55}
\end{equation} as $|t|$ and $\|u\|$ small enough.
Furthermore, by the Young inequality, 
\begin{gather}
|t|^2\|v\|^2\leq\frac{2}{p}|t|^p+\frac{p-2}{p}\|v\|^{\frac{2p}{p-2}},
\label{56}\\
|t|\|v\|^3\leq\frac{1}{p}|t|^p+\frac{p-1}{p}\|v\|^{\frac{3p}{p-1}},
\label{57}\\
|t|^3\|v\|\leq\frac{q_0-1}{q_0}|t|^{\frac{3q_0}{q_0-1}}+\frac{1}{q_0}\|v\|^{q_0}
\label{58}
\end{gather} 
for some $q_0$ with $q_0\in(2,4)$.
Inserting \eqref{56}-\eqref{58} to \eqref{55}, letting $\|v\|$ and $|t|$ 
small enough, we obtain
$$
I_{\lambda_1(h)}(u)\geq C_{37}\|v\|^2+C_{38}|t|^4
$$
as $|t|$ and $\|u\|$ small enough, because $p>4$, $4>q_0>2$, 
$\frac{3q_0}{q_0-1}>4$, $\frac{2p}{p-2}>2$ and $\frac{3p}{p-1}>2$. 
Therefore, there exists $ \widetilde{\rho}>0$ and $\widetilde{\alpha}>0$ 
such that
\begin{equation}
I_{\lambda_1(h)}(u)\geq \widetilde{\alpha}\|u\|^4\quad
 \text{for } \|u\|\leq\widetilde{\rho}.\label{59}
\end{equation}
Taking 
\[
\widetilde{\delta}
=\min\Big\{\frac{\lambda_1(h)}{2}\widetilde{\alpha}\widetilde{\rho},
\overline{\delta},\lambda_2(h)-\lambda_1(h)\Big\}.
\]
 Note that, for any $\lambda\in[\lambda_1(h),\lambda_1(h)+\widetilde{\delta})$, 
we obtain
\begin{align*}
I_{\lambda}(u)
&= I_{\lambda_1(h)}(u)+\frac{1}{2}(\lambda_1(h)-\lambda)
 \int_{\mathbb{R}^3}h(x)u^2dx\\
&\geq \widetilde{\alpha}\|u\|^4-\frac{\lambda-\lambda_1(h)}{2\lambda_1(h)}\|u\|^2\\
&\geq \|u\|^2\Big(\widetilde{\alpha}\|u\|^2
 -\frac{\lambda-\lambda_1(h)}{2\lambda_1(h)}\Big)\\
&\geq \|u\|^2\Big(\frac{\widetilde{\alpha}\widetilde{\rho}}{2}
  -\frac{\lambda-\lambda_1(h)}{2\lambda_1(h)}\Big)\\
&\geq \|u\|^2\Big(\frac{\widetilde{\alpha}\widetilde{\rho}}{2}
  -\frac{1}{4}\widetilde{\alpha}\widetilde{\rho}\Big)\\
&= \frac{1}{4}\widetilde{\alpha}\widetilde{\rho}\|u\|^2
\end{align*}
for $\frac{\widetilde{\rho}}{2}\leq\|u\|\leq\widetilde{\rho}$. 
Choosing $\rho\in[\frac{\widetilde{\rho}}{2},\widetilde{\rho}]$ and 
$\alpha=\frac{1}{8}\widetilde{\alpha}\widetilde{\rho}^2$, we obtain (b).


(c) Choose $\psi\in H^1(\mathbb{R}^3)$ with $\operatorname{supp}\psi\subset\Omega^+$
 such that $\psi(x)\geq0$ for all $x\in\Omega^+$ and $\psi=t_0e_1+v$ 
with $t_0\neq0$. Then for any $s>0$ large such 
$\|s\psi\|\geq \max\{\delta_0, \rho\}$, by \eqref{30}, we have
\begin{align*}
&I_\lambda(s\psi)\\
&= \frac{s^2}{2}\|\psi\|^2+\frac{bs^4}{4}
 \Big(\int_{\mathbb{R}^3}|\nabla \psi|^2dx\Big)^2
 -\int_{\Omega^+}k(x) F(s\psi)dx
 -\frac{\lambda s^2}{2}\int_{\mathbb{R}^3}h(x)\psi^2dx\\
&\leq \frac{s^2}{2}\|\psi\|^2+\frac{bs^4}{4}
 \Big(\int_{\mathbb{R}^3}|\nabla \psi|^2dx\Big)^2
 -Cs^q\int_{\Omega^+}k(x) |\psi|^qdx
 -\frac{\lambda s^2}{2}\int_{\mathbb{R}^3}h(x)\psi^2dx\\
&\to -\infty
\end{align*}
as $s\to+\infty$, because $q>4$. From the choice of $\psi$, 
take $w=s\psi$ with $s$ large enough, then $I_\lambda(w)<0$, (c) is proved.
\end{proof}

\begin{lemma} \label{lem3.2}
 Suppose that {\rm (H1)--(H6)} hold. 
Then problem \eqref{1} has at least one positive solution 
$u_\lambda$ with $I_\lambda(u_\lambda)>0$ for
 $0<\lambda<\lambda_1(h)+\widetilde{\delta}$.
\end{lemma}

\begin{proof}
 From Lemma \ref{lem3.1} and the Mountain Pass Theorem, then there exists  a
$(PS)_{c_\lambda}$ sequence $\{u_n\}$ such that 
$I_\lambda(u_n)\to c_\lambda>0$ and $I'_\lambda(u_n)\to 0$ in 
$H^{-1}(\mathbb{R}^3)$,
where
$$
c_\lambda=\inf_{g\in\Gamma}\max_{u\in g[0,1]}I_\lambda(u)\quad \text{with }
\Gamma=\{g\in C([0,1],H^{1}(\mathbb{R}^3)):g(0)=0, g(1)=w\}.
$$
Then by Lemma \ref{lem2.4}, we know that $I_\lambda$ satisfies $(PS)$ condition. 
Thus, the mountain pass theorem implies
$c_\lambda$ is a critical value of $I_\lambda$, $c_\lambda>0$ and 
$u_\lambda$ is a critical point of $I_\lambda$.
Since $I_\lambda(u)=I_\lambda(|u|)$ for any $u\in H^{1}(\mathbb{R}^3)$, 
by using an idea from \cite{60,57}, for every $n\in \mathbb{N}$, 
there exists $g_n(t)\in\Gamma$ with
$g_n(s)\geq0$ for all $s\in[0,1]$ such that
\[ 
c_\lambda\leq \max_{s\in[0,1]} I_\lambda(g_n(s)) 
<c_\lambda+\frac{1}{n}. % \label{83}
\]
By using Ekeland's variational principle in \cite{51}, there exists 
$g_n^*\in\Gamma$ satisfying the following properties:
\begin{equation}
\begin{gathered}
c_\lambda\leq \max_{s\in[0,1]} I_\lambda(g_n^*(s))
 \leq \max_{s\in[0,1]} I_\lambda(g_n(s))< c_\lambda+\frac{1}{n},\\
\max_{s\in[0,1]} \|g_n-g_n^*\|\leq\frac{1}{\sqrt{n}},\quad 
\|I'_\lambda(w_n)\|\leq \frac{1}{n},
\end{gathered}\label{84}
\end{equation}
and there exists $s_n\in[0,1]$ such that $ z_n=g_n^*(s_n)$ satisfying 
\begin{equation}
I_\lambda(z_n)=\max_{s\in[0,1]}I_\lambda(g_n^*(s)),\quad
\|I'_\lambda(z_n)\|\leq\frac{1}{\sqrt{n}}.\label{85}
\end{equation}
This inequality implies that $\{z_n\}\subset H^1(\mathbb{R}^3)$ is a 
$(PS)_{c_\lambda}$ sequence, by Lemma \ref{lem2.4}, there exists a convergent 
subsequence (still denoted by $\{z_n\}$ and $z\in H^1(\mathbb{R}^3)$ 
satisfying $z_n\to z$ as $n\to\infty$. Thus $g_n(s_n)\to z$ in 
$H^1(\mathbb{R}^3)$ as $n\to\infty$ by \eqref{84} and \eqref{85}. 
It follows that $z\geq0$ from $g_n(s)\geq0$ a. e. in $\mathbb{R}^3$ with 
$I_\lambda(z)>0$. Let $u_\lambda=z\geq0$. By \eqref{50}, we know that
 $0<C\leq\|u_\lambda\|\leq C_0$, which implies that there exists $M_0>0$ 
such that $u_\lambda\leq M_0$ 
a. e. in $\mathbb{R}^3$. Moreover, by (H1)--(H5), there exists a constant 
$C(M_0)>0$ depending $M_0$ such that
 $$
|k(x)f(u_\lambda)|\leq C(M_0)u_\lambda.
$$
 Together with (H6) and $\lambda>0$, we have
 \begin{align*}
 -\Big(1+b\int_{\mathbb{R}^3}|\nabla u_\lambda|^2dx\Big)\Delta u_\lambda+u_\lambda 
&=k(x)f(u_\lambda)+\lambda h(x)u_\lambda\\
&\geq k(x)f(u_\lambda)\geq -C(M_0)u_\lambda.
 \end{align*}
This yields
 $$ 
-\Delta u_\lambda+L u_\lambda\geq0, 
$$
 where $L=(1+C(M_0))/(1+b\int_{\mathbb{R}^3}|\nabla u_\lambda|^2dx)$.
 Then by the maximum principle that $ u_\lambda>0$ in $\mathbb{R}^3$,
 then it is a positive solution of problem \eqref{1}.
\end{proof}

\begin{lemma} \label{lem3.3} 
Suppose that {\rm (H1)--(H6)} hold. 
Then  \eqref{1} has at least one positive solution $\omega_\lambda$
 with $I_\lambda(\omega_\lambda)<0$ for 
$\lambda_1(h)<\lambda<\lambda_1(h)+\widetilde{\delta}$.
\end{lemma}


\begin{proof}
 By (b) of Lemma \ref{lem3.1} and its proof, there exist $\widetilde{\delta}$, 
$\rho>0$ with $\rho\to0$, $\alpha>0$ if 
$\lambda\in(\lambda_1(h),\lambda_1(h)+\widetilde{\delta})$ and
\[
I_\lambda(u)\geq \alpha>0\ \text{if}\ \|u\|=\rho.\label{82}
\]
Let 
$m_\lambda:=\inf_{B_{\rho}}I_\lambda(u)$,
where $B_{\rho}:=\{u\in H^1{\mathbb{R}^3}:\ \|u\|\leq\rho\}$ with $\rho$ 
as in Lemma \ref{lem3.1}.
It is clear $m_\lambda>-\infty$. Next, we prove that $m_\lambda<0$.
In fact, given $R>0$, define $\kappa_R\in C_0^\infty(\mathbb{R}^3)$ 
with $0\leq \kappa_R\leq1$ and $|\nabla\kappa_R |\leq\frac{2}{R}$ 
for any $x\in\mathbb{R}^3$ and
$$
\kappa_R(x)= \begin{cases}
1, &  |x|\leq R, \\
0, &  |x|\geq2 R.
\end{cases}
$$
Then $\kappa_R e_1\in H^1({\mathbb{R}^3})$ and we have
\begin{equation}
\begin{aligned}
& I_\lambda(t\kappa_Re_1) \\
&= \frac{t^2}{2}\int_{\mathbb{R}^3}|\nabla (\kappa_Re_1)|^2dx
 +\frac{t^2}{2}\int_{\mathbb{R}^3}|\kappa_Re_1|^2dx
 +\frac{bt^4}{4} \Big(\int_{\mathbb{R}^3}|\nabla (\kappa_Re_1)|^2dx\Big)^2 \\
&\quad -\int_{\mathbb{R}^3}k(x) F(t\kappa_Re_1)dx
 -\frac{\lambda t^2}{2}\int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx \\
&= \frac{t^2}{2}\int_{\mathbb{R}^3}(\kappa_\mathbb{R}^2|\nabla e_1|^2
 +|\nabla\kappa_R|^2 e_1^2+2\kappa_Re_1\nabla e_1\nabla\kappa_R)dx
 +\frac{t^2}{2}\int_{\mathbb{R}^3}|\kappa_Re_1|^2dx \\
& +\frac{bt^4}{4}
\Big(\int_{\mathbb{R}^3}|\nabla (\kappa_Re_1)|^2dx\Big)^2
 -\int_{\mathbb{R}^3}k(x) F(t\kappa_Re_1)dx
 -\frac{\lambda t^2}{2}\int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx.
\end{aligned} \label{70}
\end{equation}
Multiplying both sides of the equation $-\Delta e_1+e_1=\lambda_1(h)h(x)e_1$
 by $\kappa_\mathbb{R}^2e_1$ and integrate by parts, we obtain 
\begin{equation}
\begin{aligned}
&2\int_{\mathbb{R}^3}\kappa_Re_1\nabla e_1\nabla\kappa_Rdx
 +\int_{\mathbb{R}^3}\kappa_\mathbb{R}^2e_1^2dx
 +\int_{\mathbb{R}^3}\kappa_\mathbb{R}^2|\nabla e_1|^2dx\\
&=\lambda_1(h)\int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx.
\end{aligned} \label{71}
\end{equation}
Inserting \eqref{71} to \eqref{70}, we obtain
\begin{equation}
\begin{aligned}
I_\lambda(t\kappa_Re_1)
&= \frac{t^2}{2}\int_{\mathbb{R}^3} |\nabla\kappa_R|^2 e_1^2dx
+\frac{ t^2(\lambda_1(h)-\lambda)}{2}
 \int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx \\
&\quad +\frac{bt^4}{4}
\Big(\int_{\mathbb{R}^3}|\nabla (\kappa_Re_1)|^2dx\Big)^2
 -\int_{\mathbb{R}^3}k(x) F(t\kappa_Re_1)dx.
\end{aligned} \label{72}
\end{equation}
Moreover, by the definition of $\kappa_R$, the H\"{o}lder and Sobolev 
inequalities, we deduce
\begin{equation}
\begin{aligned}
\int_{\mathbb{R}^3} |\nabla\kappa_R|^2 e_1^2dx
&= \int_{R\leq|x|\leq2R} |\nabla\kappa_R|^2 e_1^2dx \\
&\leq \Big(\int_{R\leq|x|\leq2R}e_1^6dx\Big)^{1/3}
 \Big(\int_{R\leq|x|\leq2R} |\nabla\kappa_R|^3dx\Big)^{2/3}
 \\
&\leq \Big(\int_{R\leq|x|\leq2R}e_1^6dx\Big)^{1/3}
 \Big(\big(\frac{2}{R}\big)^3\int_{R\leq|x|\leq2R} dx\Big)^{2/3}
 \\
&\leq  C_{39}\Big(\int_{R\leq|x|\leq2R}e_1^6dx\Big)^{1/3}
\to 0
\end{aligned} \label{73}
\end{equation}
as $R\to\infty$ because $\|e_1\|=1$. Then, multiplying both sides of 
the equation $-\Delta e_1+e_1=\lambda_1(h)h(x)e_1$ by $e_1$ and 
integrate by parts, we obtain the identity
\begin{equation}
\int_{\mathbb{R}^3}h(x)e_1^2dx
=\frac{1}{\lambda_1(h)}\|e_1\|^2
=\frac{1}{\lambda_1(h)}. \label{74}
\end{equation}
Furthermore, by choosing $R$ sufficiently large, the definition of 
$\kappa_R$ and \eqref{74}, we obtain that
\begin{equation}
\int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx
 \geq\int_{|x|\leq R}h(x)\kappa_\mathbb{R}^2e_1^2dx
= \int_{|x|\leq R}h(x)e_1^2dx\geq\frac{1}{2\lambda_1(h)}.\label{75}
\end{equation}
Therefore, by choosing $R_1\geq1$ sufficiently large, by \eqref{73} and \eqref{75},
 we obtain
\begin{equation}
\int_{\mathbb{R}^3} |\nabla\kappa_R|^2 e_1^2dx
\leq\frac{\lambda-\lambda_1(h)}{2}\int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx
\label{76}
\end{equation}
for all $R\geq R_1$ and $\lambda>\lambda_1(h)$. Inserting \eqref{76} to \eqref{72}, 
by \eqref{28} and (H2), we have
\begin{align*}
I_\lambda(t\kappa_Re_1)
&\leq \frac{ t^2(\lambda_1(h)-\lambda)}{4}
 \int_{\mathbb{R}^3}h(x)\kappa_\mathbb{R}^2e_1^2dx
+\frac{bt^4}{4} \Big(\int_{\mathbb{R}^3}|\nabla (\kappa_Re_1)|^2dx\Big)^2\\
&\quad -\int_{\mathbb{R}^3}k(x) F(t\kappa_Re_1)dx\\
&\leq -C_{40}t^2+C_{41}t^4+C_{42}t^p+C_{43}t^q,
\end{align*}
for all $R\geq R_1$. This yields that $I_\lambda(t\kappa_Re_1)<0$ 
for all $t>0$ small enough because that $p,\ q>4$. 
Then there exists $t_0$ such that $\|t_0\kappa_Re_1\|\leq\rho\to0$ 
such that $I_\lambda(t_0\kappa_Re_1)<0$.
Thus, $m_\lambda=\inf_{B_{\rho}}I_\lambda(u)<0$.

Since $m_\lambda=\inf_{B_{\rho}}I_\lambda(u)<0$ and 
$I_\lambda(u)=I_\lambda(|u|)$, there exists a minimizing sequence 
$\{w_n^*\}$ for $m_\lambda$ with $w_n^*\geq0$ and $\|w_n^*\|\leq \rho$ such that
$$
m_\lambda\leq I_\lambda(w_n*)<m_\lambda+\frac{1}{n}.
$$
Then Ekeland's variational principle in \cite{51} yields another 
sequence $\{w_n\}$ with $\|w_n\|\leq \rho$ such that
\begin{equation}
\begin{gathered}
m_\lambda\leq I_\lambda(w_n)\leq I_\lambda(w_n^*)<m_\lambda+\frac{1}{n}, \\
\|w_n-w_n^*\|\leq\frac{1}{\sqrt{n}}, quad 
\|I'_\lambda(w_n)\|\leq \frac{1}{n}.
\end{gathered} \label{77}
\end{equation}
Thus, the sequence $\{w_n\}$ satisfies
$$
I_\lambda(w_n)\to m_\lambda\ \text{and}\ I'_\lambda(w_n)\to0
$$
as $n\to\infty$. By lemma \ref{lem2.4}, there exists a minimizer 
$w_\lambda\in B_{\rho}$ such that $w_n\to w_\lambda$ in 
$H^1(\mathbb{R}^3)$ and $I_\lambda(w_\lambda)=m_\lambda$. 
By \eqref{77}, $w_n^*\to w_\lambda$. Together with $w_n^*\geq0$, 
we obtain that $w_\lambda\geq0$ a.e. in $\mathbb{R}^3$ with 
$I_\lambda(w_\lambda)<0$ and $w_\lambda$ is a solution of problem \eqref{1}. 
Since $w_\lambda\in B_{\rho}$ and $\rho\to0^+$, by (H4), we have
\begin{align*}
-(1+b\rho)(\Delta w_\lambda+w_\lambda)
&\geq-\Big(1+b\int_{\mathbb{R}^3}|\nabla w_\lambda|^2dx\Big)
 \Delta w_\lambda+w_\lambda\\
&=k(x)f(w_\lambda)+\lambda h(x)w_\lambda\geq0.
\end{align*}
 Then we deduce that $w_\lambda>0$ in $\mathbb{R}^3$ by strong maximum principle. 
The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Clearly, the first conclusion of Theorem \ref{thm1.1} is deduced directly from 
Lemma \ref{lem3.2}. The second conclusion  is obtained from
 Lemmas \ref{lem3.2} and \ref{lem3.3}.
In fact, from Lemmas \ref{lem3.2} and \ref{lem3.3}, there is a positive solution 
$u_\lambda$ of \eqref{1} with $ I_\lambda(u_\lambda)>0$ 
and a positive solution $w_\lambda$ of \eqref{1} with 
$ I_\lambda(w_\lambda)<0$. Clearly, $u_\lambda\neq w_\lambda$. 
Hence, the second statement of Theorem \ref{thm1.1} is proved.
\end{proof}

\subsection*{Acknowledgment} 
The authors are very grateful to the anonymous referees for the useful 
comments and remarks.
The first author expresses their thanks to Professor S. M. Sun 
for his valuable help and suggestion during her visit to Virginia Tech. 
This research is supported by the Key Project in Science and Technology 
Research Plan of the Education Department of Hubei Province (No. D20142602).
This article is dedicated to Professor Bolin Guo on his 80-th birthday.

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\end{document}