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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 275, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/275\hfil Multiple positive solutions]
{Multiple positive solutions for a nonlocal problem involving critical exponent}

\author[Y. Wang, H.-M. Suo, C.-Y. Lei \hfil EJDE-2017/275\hfilneg]
{Yue Wang, Hong-Min Suo, Chun-Yu Lei}

\address{Yue Wang \newline
School of Data Science and Information Engineering,
Guizhou Minzu University,
Guiyang 550025, China}
\email{wyeztf@gmail.com}

\address{Hong-Min Suo (corresponding author)\newline
School of Data Science and Information Engineering,
Guizhou Minzu University,
Guiyang 550025, China}
\email{11394861@qq.com}

\address{Chun-Yu Lei \newline
School of Data Science and Information Engineering,
Guizhou Minzu University,
Guiyang 550025, China}
\email{leichygzu@sina.cn}

\dedicatory{Communicated by Paul H. Rabinowitz}

\thanks{Submitted August 15, 2017. Published November 5, 2017.}
\subjclass[2010]{35A15, 35B09, 35B33}
\keywords{Multiple positive solutions; nonlocal problem; critical exponent}

\begin{abstract}
 This article concerns the nonlocal problem
 \begin{gather*}
 -\Big(a-b\int_{\mathbb{R}^4}|\nabla u|^2\,dx\Big)\Delta u=|u|^2u+\mu f(x),
 \quad\text{in }\mathbb{R}^4,\\
 u\in \mathcal{D}^{1,2}(\mathbb{R}^4),
 \end{gather*}
 where $a, b$ are positive constants, $\mu$ is a non-negative parameter,
 $f(x)\in L^{4/3}(\mathbb{R}^4)$ is a non-negative function.
 By using the variational method, the existence of multiple positive solutions
 are obtained.
\end{abstract}

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

\section{Introduction and main results}

In this article, we focus on multiple positive solutions to the nonlocal problem
\begin{equation}\label{wt1}
\begin{gathered}
-\Big(a-b\int_{\mathbb{R}^4}|\nabla u|^2\,dx\Big)\Delta u=|u|^2u+\mu f(x),
\quad\text{in }\mathbb{R}^4,\\
u\in \mathcal{D}^{1,2}(\mathbb{R}^4),
\end{gathered}
\end{equation}
here $a, b$ are positive constants, $\mu$ is a parameter, 
$f(x)\in L^{4/3}(\mathbb{R}^4)$ is a non-negative function. 
The problem \eqref{wt1} is related to the stationary problem
\begin{align}\label{wt0}
\varrho h \frac{\partial^2u}{\partial t^2} +\delta \frac{\partial u}{\partial t}
+f_1\big(\frac{\partial u}{\partial t}\big)
=\Big( p_0+\frac{Eh}{2L}\int_0^L\Big|\frac{\partial u}{\partial x}\Big|^2\,dx \Big)
\frac{\partial^2 u}{\partial x^2}+ f_2(x,u)
\end{align}
with $0<x<L$ and $t\geq0$. Where $u=u(x,t)$ is the lateral displacement,
$\varrho$ the mass density, $E$ the Young modulus,
$h$ the cross-section area, $L$ the length, $\delta$ the resistance modulus,
$p_0$ the initial axial tension, $f_1$ and $f_2$ the external forces.
More precisely, this problem as an extension of the classical d'Alembert's 
wave equation for free vibrations of elastic strings and first proposed by 
Kirchhoff \cite{k1} when $f_1=f_2=0$. The equation \eqref{wt0} with external
forces is considered for analyzing phenomena in real world and it is studied
by many researchers (see for instance \cite{o1,t2} 
and the references therein).

The distinguishing feature of \eqref{wt0} is that
the equation contains a nonlocal coefficient
$\big( p_0+\frac{Eh}{2L}\int_0^L\big|\frac{\partial u}{\partial x}\big|^2\,dx \big)$
which depends on the average
$\frac{1}{2L}\int_0^L\big|\frac{\partial u}{\partial x}\big|^2\,dx$
of the Kinetic energy
$\frac{1}{2L}\big|\frac{\partial u}{\partial x}\big|^2$
on $[0,L]$, \eqref{wt0} is no longer a pointwise identity and therefore 
it is often called nonlocal problem.
Restating that \eqref{wt0} received much attention after the abstract 
functional analysis framework was proposed by Lions \cite{l7}.

It is worth paying more concerns for Young's modulus, which is also known 
as the elastic modulus, is a measure of the
sensitivity of the variable to the independent variable.
It allows the elastic modulus to be sign-changing in others fields,
because of elasticities may be change sign
(e.g. the price elasticities of demand \cite{f1}).
Young's modulus can also be used in computing tension, where the atoms 
are pulled apart instead of squeezed together. In those cases, the strain 
is negative because the atoms are stretched instead of compressed, this 
leads to minus Young's modulus.
Indeed, for example, an elastic meta-material which exhibits simultaneously 
negative effective mass density and bulk modulus with a single unit 
structure made of solid materials was presented in \cite{l10}, 
authors of \cite{w1,w2} got the Young's modulus of the 
nanoplate exhibits a negative temperature coefficient, the meta-material 
model that possess simultaneously negative effective mass density and 
negative effective Young's modulus were proposed in \cite{h1,s1}.
Therefore, problem \eqref{wt0} with $E<0$ is still an interesting model.

Recently, the  Kirchhoff type problem
\begin{equation*}
\begin{array}{ll}
- \Big(a+b \int_{\Omega} |\nabla u|^2 \,dx \Big)\Delta u=f(x,u) ,&{\rm in~} \Omega
\end{array}
\end{equation*}
with $a,b\geq0,a+b>0$, $\Omega=\mathbb{R}^N$ or
$\Omega$ is a smooth bounded domain in $\mathbb{R}^N$ has been studied by 
many researchers; we refer the reader to  \cite{b2,d1,d2,l4,m1,t1,z5}
with sub-critical growth, and 
\cite{h2,l1,l5,l6,l8,l9,p1,x1,y1,y2,z1,z2,z3,z4,z6} with critical cases.
Particularly, \cite{h2,l5,l6,z3}
$N=4$ and some show interesting results.
Only a few authors mentioned problem of the form
\begin{equation}\label{wt2}
- \Big(a-b \int_{\Omega}|\nabla u|^2\,dx \Big)\Delta u = f(x,u).
\end{equation}
Yin and Liu \cite{y3} researched  problem \eqref{wt2}
when $f(x,u)=|u|^{p-2}u$ (where $2<p<2^*=\frac{2N}{N-2}$ as 
$N\geq3$ and $2^*=+\infty$ as $N=1,2$) and they got \eqref{wt2}
 has at least a nontrivial non-negative solution and a nontrivial non-positive
 solution with Dirichlet's boundary condition. Lei et al \cite{l2} studied 
\eqref{wt2}  assuming $f(x,u)=f_{\lambda}(x)|u|^{q-2}u$ $(1<q<2)$ with $N\geq3$,
 with the assumption $f_{\lambda}(x)\in L^{\infty}(\Omega)$, they concluded 
that \eqref{wt2} has at least two positive solutions.
Also Lei et al \cite{l3} obtained many solutions for $f(x,u)=u^{-\gamma}$ with 
$1<q < 2$ and $0<\gamma<1$.

To the best of our knowledge, there is no result for equation \eqref{wt1}.
From \cite[pp.7]{b1},
$\mathcal{D}^{1,2}(\mathbb{R}^4)\hookrightarrow L^{4}(\mathbb{R}^4)$ 
continuously but this embedding is never compact. 
Motivated by \cite{l2,l3,y3}, since the typical difficulty 
is the lack of compactness of the embedding 
$D^{1,2}(\mathbb{R}^4)\hookrightarrow L^4(\mathbb{R}^4)$, we overcome the 
difficulty by using the methods from \cite{h2,l5,l6,z3}.
Our main results can be stated as follows:

\begin{theorem} \label{Cor.thm}
Problem \eqref{wt1} has infinitely many positive solutions when $\mu=0$.
\end{theorem}

\begin{theorem} \label{thm1.1}
Assume that $f(x)\in L^{4/3}(\mathbb{R}^4)$ is a positive function, 
then there exists $\mu_*>0$ such that problem \eqref{wt1} has at least two 
positive solutions when $\mu\in(0,\mu_{*}]$.
\end{theorem}

\section{Preliminaries}

In this section, we give some notation and definitions.
All results are based on
$\mathcal{D}^{1,2}(\mathbb{R}^4)=\big\{u\in L^4(\mathbb{R}^4)\big|
\frac{\partial u}{\partial x_i}\in L^2(\mathbb{R}^4),i=1,\dots,4\big\}$.
 For $u, v\in \mathcal{D}^{1,2}(\mathbb{R}^4)$, the inner product is
$\langle u,v\rangle= \int_{\mathbb{R}^4} \nabla u \nabla v \,dx$
and the norm is
\[
\|u\|=\langle u,u\rangle^{1/2}
= \Big(\int_{\mathbb{R}^4}|\nabla u|^2\,dx \Big)^{1/2}.
\]
We recall that a function $u\in \mathcal{D}^{1,2}(\mathbb{R}^4)$ is called 
a solution of problem \eqref{wt1} if
$$
 \big(a-b\|u\|^2\big)\int_{\mathbb{R}^4}\nabla u \nabla v \,dx
=\int_{\mathbb{R}^4}|u|^2uv\,dx +\mu\int_{\mathbb{R}^4}fv\,dx
$$
hold for all $v\in \mathcal{D}^{1,2}(\mathbb{R}^4)$.
Throughout this paper, we denote by $\|\cdot\|_s$ the usual $L^s$-norm 
and $\to$ (resp. $\rightharpoonup$) the strong (resp. weak) convergence. Set
\begin{align}\label{eb-S}
S= \inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^4)\backslash\{0\}}
\frac{\|u\|^2}{ \left(\int_{\mathbb{R}^4}u^4\,dx\right)^{1/2}}.
\end{align}
It is well known, for any $\varepsilon >0$ and $y\in\mathbb{R}^4$, 
all positive solutions for the problem
\begin{gather*}
- \Delta u = u^3,\quad x\in \mathbb{R}^4, \\
u \in \mathcal{D}^{1,2}(\mathbb{R}^4)
\end{gather*}
can be expressed as
\begin{equation}\label{first}
u_{\varepsilon,y}: =\frac{2\sqrt{2}\varepsilon}{\varepsilon^2+|x-y|^2},
\end{equation}
as a consequence, $S$ can be archive by \eqref{first} and
$\|u_{\varepsilon,y}\|^2=\|u_{\varepsilon,y}\|_4^4=S^2$.

Because of that we are looking for positive solution, for equation 
\eqref{wt1}, set the energy 
$I: \mathcal{D}^{1,2}(\mathbb{R}^4)\mapsto \mathbb{R}$ be the functional 
defined by
\begin{equation}\label{e-fun}
I(u)=\frac{a}{2}\|u\|^2-\frac{b}{4} \|u\|^4
 - \frac{1}{4}\int_{\mathbb{R}^4}(u^+)^4\,dx-\mu\int_{\mathbb{R}^4}fu\,dx,
\end{equation}
here $u^+=\max \{0, u\}$. It is able to verify 
$I(u)\in C^1(\mathcal{D}^{1,2}(\mathbb{R}^4),\mathbb{R})$, and for all 
$v\in \mathcal{D}^{1,2}(\mathbb{R}^4)$, $I$ has the G\^{a}teaux derivative given by
\begin{equation}\label{d-e-fun}
\langle I'(u),v \rangle
=  (a-b\|u\|^2)\int_{\mathbb{R}^4}\nabla u \nabla v\,dx  
-\int_{\mathbb{R}^4}(u^+)^3v\,dx-\mu\int_{\mathbb{R}^4}fv\,dx.
\end{equation}

\section{Main Lemmas}

\begin{lemma} \label{lem r rho mu*}
Assume that $f(x)\in L^{4/3}(\mathbb{R}^4)$ is a positive function, then, 
there exist $r, \rho, \mu_1>0$ such that, for any $\mu\in(0,\mu_1]$, one has
\begin{itemize}
\item[(i)] $I(u)\geq\rho$ with $\|u\|=r$;
\item[(ii)] $\inf I(u)<0$ with $\|u\|<r$;
\item[(iii)] There exists $e\in \mathcal{D}^{1,2}(\mathbb{R}^4)$ 
which satisfies $I(e)< 0$ with $\|e\|>r$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) From \eqref{e-fun} and \eqref{eb-S}, we obtain
\begin{align*} %\label{eqq.1}
I(u)&= \frac{a}{2}\|u\|^2 - \frac{b}{4}\|u\|^4 - \frac{1}{4}\int_{\mathbb{R}^4}(u^+)^4\,dx-\mu\int_{\mathbb{R}^4}fu\,dx \\
&\geq \frac{a}{2}\|u\|^2 - \frac{b}{4}\|u\|^4 -
\frac{1}{4S^2}\|u\|^4 -\frac{\mu}{\sqrt{S}} \|f\|_{4/3}\|u\|\\
&= \|u\| \Big(\frac{a}{2}\|u\| - \frac{bS^2+1}{4S^2}\|u\|^3 
- \frac{\mu}{\sqrt{S}} \|f\|_{4/3} \Big).
\end{align*}
Set $g_1(t,\mu):=\frac{a}{2}t - \frac{bS^2+1}{4S^2}t^3
- \frac{\mu}{\sqrt{S}} \|f\|_{4/3} $ for all
$t\geq0$,
we can see that there exist constants $r=\sqrt{\frac{2aS^2}{3(bS^2+1)}}>0$ and
$\mu_1=\frac{aS}{4\|f\|_{4/3}}\sqrt{\frac{aS}{bS^2+1}}>0$
such that
\begin{align*}
\max_{t>0} g_1(t,\mu)
& =g_1(r,\mu)
= \frac{aS}{3}\sqrt{\frac{2a}{3(bS^2+1)}}
 -\frac{\mu}{\sqrt{S}}\|f\|_{4/3} \\
&\geq \frac{aS}{3}\sqrt{\frac{2a}{3(bS^2+1)}}
 -\frac{\mu_1}{\sqrt{S}}\|f\|_{4/3}
\end{align*}
for any $\mu\in(0,\mu_1]$. Particularly, we have $I(u)\geq rg(r,\mu_1)$
when $\|u\|=r$. Thus
\[ % \label{eqq.3}
I(u) \geq  \sqrt{\frac{2aS^2}{3(bS^2+1)}} 
\Big(\frac{aS}{3}\sqrt{\frac{2a}{3(bS^2+1)}} 
- \frac{\mu_1}{\sqrt{S}}\|f\|_{4/3} \Big)
> \frac{a^2S^2}{72(bS^2+1)} :=\rho.
\]
Therefore, there exist $r, \rho, \mu_1>0$ such that $I(u)\geq\rho>0$.

(ii) For $u_0\in\mathcal{D}^{1,2}(\mathbb{R}^4)$ with $\|u_0\|=r$ such that 
$\int_{\mathbb{R}^4} f u_0 \,dx>0$, then
\begin{align*}
\lim_{t\to 0^+} \frac{I(tu_0)}{t} = - \mu\int_{\mathbb{R}^4} f u_0 \,dx<0.
\end{align*}
Hence, there exists some $u\in\mathcal{D}^{1,2}(\mathbb{R}^4)$ such that 
$I(u)<0$ when $\|u\|$ enough small. Therefore,
$ c_1:=\inf_{\|u\|<r}I(u)<0$ is well defined.

(iii) For any $t\in\mathbb{R}$ and a $u_0\in \mathcal{D}^{1,2}(\mathbb{R}^4)$ 
is fixed with $\|u_0\|=r$, we have
\[
\lim_{t\to\infty}\frac{I(tu_0)}{t^4}
=-\frac{b}{4}\|u_0\|^4 -\frac{1}{4}\int_{\mathbb{R}^4}(u_0^+)^4\,dx
\leq -\frac{br^4}{4} < 0,
\]
so, there is $t_e>1$ satisfies $I(t_eu_0)<0$.
Let $e:=t_eu_0\in \mathcal{D}^{1,2}(\mathbb{R}^4)$, then
$ I(e) < 0$ and $\|e\|=t_er>r$.
For example, take $e\in \mathcal{D}^{1,2}(\mathbb{R}^4)$ with
$\|e\|^2=\frac{4a}{b}+4(\frac{\mu}{b\sqrt{S}} \|f\|_{4/3})^{2/3}$,
we can verify $\|e\|>r$ and $ I(e) < 0$.
The proof is complete.
\end{proof}


\begin{lemma} \label{lem.PSc}
Assume that $\mu>0$ and $f(x)\in L^{4/3}(\mathbb{R}^4)$ is a positive function, 
then $I$ satisfies the $(PS)_c$ condition with
\[
c<\frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3},\quad 
\Lambda=\Big( \frac{4bS^2}{bS^2+1}\Big)^{-1/3}  \|f\|_{4/3}^{4/3}.
\]
\end{lemma}

\begin{proof}
Let $\{u_n\}\subset \mathcal{D}^{1,2}(\mathbb{R}^4)$ is a $(PS)_c$ 
sequence such that $I(u_n) \to c$, $I'(u_n) \to 0$ as $n\to\infty$. 
So by the H\"older and Sobolev inequalities, for $n$ large enough, one has
\[
c+o(\|u_n\|)\geq I(u_n)-\frac{1}{4}\langle I'(u_n), u_n \rangle
\geq \frac{a}{4}\|u_n\|^2 - \frac{3\mu}{4\sqrt{S}} \|f\|_{4/3}\|u_n\|.
\]
This means that $\{u_n\}$ is bounded in $\mathcal{D}^{1,2}(\mathbb{R}^4)$.
That is, there exist a subsequence (still denoted by $\{u_n\}$) and 
$u_0$ in $\mathcal{D}^{1,2}(\mathbb{R}^4)$ such that
$u_n\rightharpoonup u_0$ in $\mathcal{D}^{1,2}(\mathbb{R}^4)$,
$u_n\to u_0$ in $L_{\rm loc}^p$ $(1\leq p<4)$,
$u_n(x)\to u_0(x)$ in $\mathbb{R}^4$ as $n\to\infty$.

Set $\omega_n:=u_n-u_0$, then $\|\omega_n\|\to0$ as $n\to\infty$. 
Otherwise $\|\omega_n\|\not\to0$. Through of contradiction, we can assume 
there is a subsequence (still denoted by $\{\omega_n\}$) such that
$\lim_{n\to\infty}\|\omega_n\|=l>0$. For $v\in \mathcal{D}^{1,2}(\mathbb{R}^4)$,
it holds that
\begin{equation*}
\big(a-b\|u_n\|^2\big)\int_{\mathbb{R}^4} \nabla u_n \nabla v \,dx
-\int_{\mathbb{R}^4}(u_n^+)^3v\,dx
- \mu\int_{\mathbb{R}^4}fv\,dx=o(1)
\end{equation*}
as $n\to\infty$.
Lebesgue's dominated convergence theorem (see \cite[pp.27]{r1}) leads to
\begin{align}\label{dominated convergence}
\int_{\mathbb{R}^4} f u_n\,dx 
= \int_{\mathbb{R}^4} f u_0\,dx + o(1).
\end{align}
Using the Br\'ezis-Lieb lemma (see \cite[Theorem 1]{b3}) and
\eqref{dominated convergence}, it satisfies
\begin{equation}\label{PS-eq.2}
\Big( a-(bl^2 + b\|u_0\|^2)\Big)\int_{\mathbb{R}^4}\nabla u_0 \nabla v\,dx
- \int_{\mathbb{R}^4}(u_0^+)^3v\,dx
- \mu\int_{\mathbb{R}^4}fv\,dx=0.
\end{equation}
Particularly, take $v=u_0$ in \eqref{PS-eq.2}, there is
\begin{equation}\label{PS-eq.3}
\big(a-bl^2-b\|u_0\|^2\big)\|u_0\|^2
- \int_{\mathbb{R}^4}(u_0^+)^4\,dx
- \mu\int_{\mathbb{R}^4}fu_0\,dx=0.
\end{equation}
Furthermore, as $n\to\infty$, it holds
\begin{equation*}
\langle I'(u_n),u_n\rangle=a\|u_n\|^2-b\|u_n\|^4
- \int_{\mathbb{R}^4}(u_n^+)^4\,dx - \mu\int_{\mathbb{R}^4}fu_n\,dx=o(1).
\end{equation*}
Using the Br\'ezis-Lieb lemma again, we get
\begin{equation} \label{PS-eq.4}
\begin{aligned}
o(1)
&=a\|\omega_n\|^2+a\|u_0\|^2 -2b\|\omega_n\|^2\|u_0\|^2-b\|u_0\|^4
 -b\|\omega_n\|^4  \\
&\quad -\int_{\mathbb{R}^4}(u_0^+)^4\,dx
 - \int_{\mathbb{R}^4}(\omega_n^+)^4\,dx
 - \mu\int_{\mathbb{R}^4}fu_0\,dx
\end{aligned}
\end{equation}
Cutting \eqref{PS-eq.3} out of \eqref{PS-eq.4}, we have
\begin{equation}\label{PS-eq.5}
a\|\omega_n\|^2-b\|\omega_n\|^4-b\|\omega_n\|^2\|u_0\|^2
=\int_{\mathbb{R}^4}(\omega_n^+)^4\,dx + o(1).
\end{equation}
Noting that $\int_{\mathbb{R}^4}(\omega_n^+)^4\,dx
\leq\int_{\mathbb{R}^4}\omega_n^4\,dx$, we obtain
$0\leq l^2(a-b\|u_0\|^2-bl^2)\leq \frac{l^4}{S^2}$, $l>0$. So that
\begin{equation}\label{l^2}
l^2\geq\frac{S^2(a-b\|u_0\|^2)}{bS^2+1}>0.
\end{equation}
On the one hand, applying \eqref{PS-eq.5}--\eqref{l^2},
it holds that
\begin{equation} \label{PS-eq.6}
\begin{aligned}
I(u_0)
&= \frac{a}{2}\|u_0\|^2-\frac{b}{4}\|u_0\|^4
 - \frac{1}{4}\int_{\mathbb{R}^4}(u_0^+)^4\,dx
 - \mu\int_{\mathbb{R}^4}fu_0\,dx   \\
&= c- \frac{a}{2}\|\omega_n\|^2
+ \frac{b}{4}\|\omega_n\|^4
+ \frac{b}{2}\|\omega_n\|^2\|u_0\|^2
+ \frac{1}{4}\int_{\mathbb{R}^4}(\omega_n^+)^4\,dx+ o(1)  \\
&= c- \frac{a}{2}l^2 + \frac{b}{4}l^4
+ \frac{b}{2}l^2\|u_0\|^2 + \frac{1}{4}\left(al^2-bl^4
- bl^2\|u_0\|^2 \right)  \\
&= c- \frac{a-b\|u_0\|^2}{4}l^2  \\
&\leq c- \frac{a^2S^2}{4(bS^2+1)}
+\frac{abS^2}{2(bS^2+1)}\|u_0\|^2
-\frac{b^2S^2}{4(bS^2+1)}\|u_0\|^4  \\
&< -\Lambda\mu^{4/3}
+\frac{abS^2}{2(bS^2+1)}\|u_0\|^2
-\frac{b^2S^2}{4(bS^2+1)}\|u_0\|^4 .
\end{aligned}
\end{equation}
On the other hand, from \eqref{PS-eq.3} it follows that
\begin{equation}\label{PS-eq.7}
 a\|u_0\|^2
= b\|u_0\|^4 + bl^2\|u_0\|^2
+\int_{\mathbb{R}^4}(u_0^+)^4\,dx
+\mu\int_{\mathbb{R}^4}fu_0\,dx.
\end{equation}
Moreover, H\"older and Yang's inequalities lead to
$\frac{\mu}{2}\int_{\mathbb{R}^4}fu_0\,dx
\leq \frac{\mu}{2\sqrt{S}}\|f\|_{4/3}\|u_0\|$ and
\begin{align}\label{PS-eq.8}
 \|f\|_{4/3}\|u_0\|
&\leq  \frac{\sqrt{S}b}{2\mu(bS^2+1)}\|u_0\|^4
+ \Big(\frac{\sqrt{S}b}{2\mu(bS^2+1)}\Big)^{-1/3} \|f\|_{4/3}^{4/3}.
\end{align}
Therefore, from \eqref{e-fun}, \eqref{PS-eq.7} and \eqref{PS-eq.8}, we get
\begin{equation} \label{PS-eq.9}
\begin{aligned}
I(u_0)
&=  \frac{a}{2}\|u_0\|^2-\frac{b}{4}\|u_0\|^4
 - \frac{1}{4}\int_{\mathbb{R}^4}(u_0^+)^4\,dx
 - \mu\int_{\mathbb{R}^4}fu_0\,dx    \\
&=  \frac{bl^2}{2}\|u_0\|^2
+ \frac{b}{4}\|u_0\|^4
+ \frac{1}{4}\int_{\mathbb{R}^4}(u_0^+)^4\,dx
- \frac{\mu}{2}\int_{\mathbb{R}^4}fu_0\,dx  \\
&\geq  \frac{b}{2}\|u_0\|^2\cdot \frac{S^2(a-b\|u_0\|^2)}{bS^2+1}
+ \frac{b}{4}\|u_0\|^4
- \frac{\mu}{2}\int_{\mathbb{R}^4}fu_0\,dx  \\
&\geq  \frac{abS^2}{2(bS^2+1)}\|u_0\|^2
-\frac{b^2S^2}{4(bS^2+1)}\|u_0\|^4
- \Big( \frac{4bS^2}{bS^2+1}\Big)^{-1/3}
  \|f\|_{4/3}^{4/3}\mu^{4/3}.   \\
&=  \frac{abS^2}{2(bS^2+1)}\|u_0\|^2
-\frac{b^2S^2}{4(bS^2+1)}\|u_0\|^4
- \Lambda\mu^{4/3}.
\end{aligned}
\end{equation}
Which is a contradiction by comparing the calculations from \eqref{PS-eq.6}
 with \eqref{PS-eq.9}. Hence $l=0$. As a consequence, we get $u_n\to u_0$
in $\mathcal{D}^{1,2}(\mathbb{R}^4)$. This proof is complete.
\end{proof}

By \eqref{first}, we can obtain the following estimate for the mountain 
pass level.

\begin{lemma} \label{sup I>=c2}
There exists $\mu_{*}\in(0,\mu_1]$ such that 
$\sup_{t\geq 0} I(tu_{\varepsilon,y})<\frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}$
with $\mu\in(0,\mu_{*}]$
($\mu_1$ is defined in the Lemma \ref{lem r rho mu*}).
\end{lemma}

\begin{proof}
Set $g(t)=I(tu_{\varepsilon,y})$ and
$h(t)=I(tu_{\varepsilon,y})
+\mu t\int_{\mathbb{R}^4}fu_{\varepsilon,y}\,dx$
with $t\geq 0$, then
\begin{align*}
g(t)&= \frac{a}{2}\|tu_{\varepsilon,y}\|^2
- \frac{b}{4}\|tu_{\varepsilon,y}\|^4
- \frac{1}{4}\int_{\mathbb{R}^4}(tu_{\varepsilon,y})^4\,dx
-\mu\int_{\mathbb{R}^4}f\cdot tu_{\varepsilon,y}\,dx \\
&= \frac{aS^2}{2}t^2 - \frac{bS^4}{4}t^4
-\frac{S^2}{4}t^4 -\mu t\int_{\mathbb{R}^4}f u_{\varepsilon,y}\,dx
\end{align*}
and
\[
h(t)= \frac{aS^2}{2}t^2 - \frac{bS^4}{4}t^4 - \frac{S^2}{4}t^4.
\]
So, there exists $t_1=\sqrt{\frac{a}{bS^2+1}}$ such that
$\max_{t>0} h(t)=h(t_1)=\frac{a^2S^2}{4(bS^2+1)}$. For any
$\mu\in(0,\mu_1)$ and $t\in(0,t_1)$, noticing
$\mu_1=\frac{aS}{4\|f\|_{4/3}}
\sqrt{\frac{aS}{bS^2+1}}$, we can see that
\[
 \mu t\int_{\mathbb{R}^4}fu_{\varepsilon,y}\,dx
< \mu_1 t_1\int_{\mathbb{R}^4} fu_{\varepsilon,y}\,dx
\leq \frac{\mu_1 t_1}{\sqrt{S}}\|f\|_{4/3}\|u_{\varepsilon,y}\|
=\frac{a^2S^2}{4(bS^2+1)}
=\max_{t>0} h(t).
\]
Therefore, $\max_{t>0} g(t)>0$. Take $\mu_2\in(0,\mu_1]
\cap \Big(0, \frac{aS^2(a^2b)^{1/4}}{2(bS^2+1)\|f\|_{4/3}}\Big)$,
then
\begin{equation}\label{PSc>0}
\frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}
> \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu_2^{4/3}>0
\end{equation}
for all $\mu\in(0,\mu_2)$. Thus there exists $t_2\in(0, t_1)$
such that
\begin{align*}
\max_{0\leq t\leq t_2}g(t)
&\leq \max_{0\leq t\leq t_2}
\big\{\frac{aS^2}{2}t^2-\frac{bS^4}{4}t^4 \big\} \\
&\leq \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu_2^{4/3}
\leq \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}.
\end{align*}
for all $\mu\in(0,\mu_2)$. Choose $\mu_{*}\in(0,\mu_2)$ such that, 
for any $\mu\in(0,\mu_{*}]$, it holds
\[
\mu t_2\int_{\mathbb{R}^4}fu_{1,0}\,dx > \Lambda\mu^{4/3}.
\]
Hence for all $\mu\in(0,\mu_{*}]$, one has
\begin{align*}
\sup_{t\geq t_2}g(t)
\leq \sup_{t\geq t_2}h(t)- \mu t_2\int_{\mathbb{R}^4}fu_{1,0}\,dx
< h(t_1)- \Lambda\mu^{4/3}
= \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}.
\end{align*}
Consequently, 
\begin{align*}
c^+:= \sup_{t\geq 0}I(tu_{1,0})
= \sup_{t\geq 0}g(t)
< \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}.
\end{align*}
Thus the proof is complete.
\end{proof}

\section{Proof Theorem \ref{Cor.thm}}

\begin{proof}
For any $\lambda>0$, let $v_{\varepsilon,y}=\lambda^{1/2} u_{\varepsilon,y}$,
 then
$u_{\varepsilon,y}=\lambda^{-1/2}v_{\varepsilon,y}$ by \eqref{first}. So
\begin{equation} \label{character1}
-\lambda\Delta v_{\varepsilon,y}
= -\lambda\Delta \lambda^{1/2}u_{\varepsilon,y}
= -\lambda^{\frac{3}{2}}\Delta u_{\varepsilon,y}
=\lambda^{\frac{3}{2}}(\lambda^{-1/2}v_{\varepsilon,y})^3
= v_{\varepsilon,y}^3.
\end{equation}
Noting that there are infinitely many $u_{\varepsilon,y}$,
we can verify all $v_{\varepsilon,y}$ are infinitely many and its are
positive solutions of \eqref{character1} for any $\lambda>0$.
Considering the equation
\begin{align}\label{character2}
\lambda=a-b\|v_\lambda\|^2=a-b\lambda S^2.
\end{align}
Obviously, the solution of equation \eqref{character2} is
$\lambda_0=\frac{a}{bS^2+1}$. As a consequence, we have
\begin{equation} \label{character3}
v_{\lambda_0}
= \lambda_0^{1/2}u_{\varepsilon,y}
= \big(\frac{a}{bS^2+1}\big)^{1/2} u_{\varepsilon,y}.
\end{equation}
Therefore, for equation
\begin{equation}\label{mu=0}
-\Big(a-b\int_{\mathbb{R}^4}|\nabla u|^2\,dx\Big)\Delta u=|u|^2u,
\quad  \text{in }\mathbb{R}^4,
\end{equation}
we can verify that all $v_{\lambda_0} $ are positive solutions of \eqref{mu=0}
 easily by \eqref{character1}--\eqref{character3}.
Thus equation \eqref{mu=0} has infinitely positive solutions
$\big(\frac{a}{bS^2+1}\big)^{1/2} u_{\varepsilon,y}$ when $\mu=0$.
\end{proof}


\section{Proof of Theorem \ref{thm1.1}}

\begin{proof}[Existence of the first positive solution]
Taking ${B}_{r}:=\{u\in \mathcal{D}^{1,2}(\mathbb{R}^4):  \|u\| \leq r\}$
and $\mu_*$ from the Lemma \ref{sup I>=c2}, where 
$r=\sqrt{\frac{2aS^2}{3(bS^2+1)}}$.
Reason by the Lemma \ref{lem r rho mu*}, there exists $\mu_1>0$ such that
$\inf I({B}_{r})<0$ for any $\mu\in(0,\mu_*]\subset(0,\mu_1)$. 
By the Ekeland variational principle (see\cite[Lemma 1.1]{e1}),
there exists a sequence $\{u_n\}\subset{B}_{r}$ such that
\begin{equation} \label{ek1}
I(u_n)\leq \inf I({B}_{r})+\frac{1}{n}   \quad\text{and}\quad
I(u)\geq I(u_n)-\frac{1}{n}\|u-u_n\|
\end{equation}
for all $n\in \mathbb{N}$ and for any $u\in \overline{B_r}$.
Therefore, we get $I(u_n) \to c_1$ and $I'(u_n)\to 0$ in dual space
of $\mathcal{D}^{1,2}(\mathbb{R}^4)$ as $n\to \infty$.
Noting that $c_1<0<\frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}$
(see Lemma \ref{lem r rho mu*} and inequality \eqref{PSc>0}),
by Lemma \ref{lem.PSc}, there exist a subsequence (still denoted by $ \{u_n\}$)
and $u_*\in B_r$ such that  $u_n \to u_*$ as $n\to \infty$.
Then, $I(u_*) = c_1<0$ and $I'(u_*)=0$.
Which implies that $u_*$ is a local minimizer for $c_1$.
Consequently, $u_*$ is a solution of problem \eqref{wt1}.
Define $u_*^- = \max \{0,-u_*\}$, then $u_*^- = 0$ by both
$\|u_*\|<\sqrt{\frac{2aS^2}{3(bS^2+1)}}$ and
$\langle I'(u_*), u_*^- \rangle =0$, which deduces $u_* \geq 0$.
By the strong maximum principle, we obtain $u_* > 0$. The proof is complete.
\end{proof}

\begin{proof}[Existence of the second positive solution]
We shall divide it into three steps.
\smallskip

\noindent\textbf{Step 1.}
There exists a critical point $u_{**}$ with $I(u_{**})>0$.
By Lemma \ref{lem r rho mu*}, the functional $I$ has mountain pass 
geometry. Set
$$
\Gamma^+=\Big\{\tau(t)\in C^1\big([0, 1],
\mathcal{D}^{1,2}(\mathbb{R}^4)\big); \tau(0)=0, \tau(1)=e\Big\}.
$$
By \eqref{e-fun}--\eqref{d-e-fun}, $I\big(\tau(t)\big)$ has continuity. 
Besides, $I\big(\tau(0)\big)=0$, $I\big(\tau(1)\big)\leq 0$ and 
$I\big(\tau(t)\big)>\rho$ with some $t\in (0, 1)$.
Moreover, by Lemma \ref{sup I>=c2}, there is a $\mu_{*}\in(0,\mu_1)$, such that
\[
 0<\rho \leq c_2:=\inf_{\tau\in \Gamma^+} \sup_{t\in[0,1]} I\big(\tau(1)\big)
\leq c^+< \frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}.
\]
hold for all $\mu\in(0,\mu_{*}]$. Via Lemma \ref{lem.PSc} and the mountain 
pass theorem (see \cite[Theorem 2.1--2.4]{a1})
implies that for $I$, there exist $u_{**}$ and a sequence $\{u_k\}$
in $\mathcal{D}^{1,2}(\mathbb{R}^4)$ such that $u_k\to u_{**}$ in
$\mathcal{D}^{1,2}(\mathbb{R}^4)$, $I(u_k) \to c_2=I(u_{**})$ and
$I'(u_k) \to 0=I'(u_{**})$ in dual space of 
$\mathcal{D}^{1,2}(\mathbb{R}^4)$.
Hence $u_{**}$ is a solution of problem \eqref{wt1} with
$\|u_{**}\| \geq \sqrt{\frac{2aS^2}{3(bS^2+1)}}$. Because of
$I(u_{*})<0<I(u_{**})$, we get $u_{**}\neq u_{*}$.
\smallskip

\noindent\textbf{Step 2.}
The critical point $u_{**}$ satisfies $\|u_{**}\|^2<a/b$.
Note that $u_{**}$ is a critical point of $I$. Relying on
$\langle I'(u_{**}),u_{**} \rangle =0$, one has
\begin{align}\label{solution}
(a-b\|u_{**}\|^2)\|u_{**}\|^2
=\int_{\mathbb{R}^4}(u_{**}^+)^4\,dx +\mu\int_{\mathbb{R}^4} f u_{**}\,dx.
\end{align}
Obviously $\|u_{**}\|^2<\frac{a}{b}$ if $u_{**}$ is trivial one.
Without loss of generality, we can suppose that there satisfies 
$\|u_{**}\|^2 \geq\frac{a}{b}$,
then $(a-b\|u_{**}\|^2)\|u_{**}\|^2 \leq 0$,
which implies $\int_{\mathbb{R}^4} f u_{**}\,dx\leq 0$ from \eqref{solution}.
By $I(u_{**})=c$ and $I'(u_{**})=0$, there is
\begin{align}\label{c value}
\frac{a^2S^2}{4(bS^2+1)} - \Lambda\mu^{4/3}
> I(u_{**})-\frac{1}{4}I'(u_{**})
= \frac{a}{4}\|u_{**}\|^2 - \frac{3\mu}{4}\int_{\mathbb{R}^4}fu_{**}\,dx
\geq \frac{a^2}{4b}.
\end{align}
This is a contradiction. So $\|u_{**}\|^2<\frac{a}{b}$.
\smallskip

\noindent\textbf{Step 3.}
$u_{**}$ is a positive critical point of $I$.
Define $u^-=\max \{0,-u\}$, then $u=u^+-u^-$.
By $I'(u_{**}) =0$, we have
\begin{align*}
0 &= \langle I'(u_{**}), u_{**}^- \rangle \\
&= (a-b\|u_{**}\|^2)\int_{\mathbb{R}^4} \nabla u_{**}  \nabla u_{**}^- \,dx
- \int_{\mathbb{R}^4}(u_{**}^+)^3 u_{**}^-\,dx
- \mu\int_{\mathbb{R}^4} f u_{**}^-\,dx\\
&= (a-b\|u_{**}\|^2)\|u_{**}^-\|^2 - \mu\int_{\mathbb{R}^4} f u_{**}^-\,dx \\
&\geq (a-b\|u_{**}\|^2)\|u_{**}^-\|^2,
\end{align*}
which implies $\|u_{**}^-\|=0$. Hence $u_{**}$ is non-negative. 
According to the
Lemma \ref{sup I>=c2} that $\|u_{**}\|\geq\sqrt{\frac{2aS^2}{3(bS^2+1)}}$,
we have $u_{**}\not\equiv 0$. By the strong maximum principle, we obtain $u_{**}>0$.
Hence the problem \eqref{wt1} has a positive solution $u_{**}$ which different 
with $u_{*}$. Therefore, the problem \eqref{wt1} has at least two positive 
solutions. The proof is complete.
\end{proof}


\subsection*{Acknowledgments}
The authors would like to thank the anonymous referees and the editors 
for their helpful comments and suggestions, which led to an improvement 
of the original manuscript. This work was supported by the National
 Natural Science Foundation of China (No. 11661021), 
by the Innovation Group Major Program of Guizhou Province (No. KY[2016]029) 
and by the Natural Science Foundation of Education of Guizhou Province 
(No. KY[2016]163, No. KY[2013]405).

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\end{document}