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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 282, pp. 1--13.\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-2017/282\hfil Least energy sign-changing solutions]
{Least energy sign-changing solutions for the nonlinear
Schr\"odinger-Poisson system}

\author[C. Ji, F. Fang, B. Zhang \hfil EJDE-2017/282\hfilneg]
{Chao Ji, Fei Fang, Binlin Zhang}

\address{Chao Ji \newline
Department of Mathematics,
East China University of Science and Technology,
Shanghai 200237, China}
\email{jichao@ecust.edu.cn}

\address{Fei Fang \newline
Department of Mathematics,
Beijing Technology and Business University,
 Beijing 100048, China}
\email{fangfei68@163.com}

\address{Binlin Zhang (corresponding author) \newline
Department of Mathematics,
Heilongjiang Institute of Technology,
Harbin 150050, China}
\email{zhangbinlin2012@163.com}

\dedicatory{Communicated by Vicentiu D. Radulescu}

\thanks{Submitted September 9, 2017. Published November 13, 2017.}
\subjclass[2010]{35J20, 35J60}
\keywords{Schr\"odinger-Poisson system; sign-changing solutions;
\hfill\break\indent constraint variational method; quantitative  deformation lemma}

\begin{abstract}
 This article  concerns the existence of the least energy sign-changing
 solutions for the Schr\"odinger-Poisson system
 \begin{gather*}
 -\Delta u+V(x)u+\lambda\phi(x)u=f(u),\quad \text{in } \mathbb{R}^3,\\
 -\Delta\phi=u^2,\quad \text{in } \mathbb{R}^3.
 \end{gather*}
 Because the so-called nonlocal term $\lambda\phi(x)u$ is involved in the
 system, the variational functional of the above system has totally different
 properties from the case of $\lambda=0$. By constraint variational method
 and quantitative  deformation lemma, we prove that the above problem has one
 least energy sign-changing solution. Moreover, for any $\lambda>0$,
 we show that the energy of a sign-changing solution is strictly larger than
 twice of the ground state energy.  Finally, we consider $\lambda$ as a
 parameter and study the convergence property of  the least energy sign-changing
 solutions as $\lambda\searrow 0$.
\end{abstract}

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

\section{Introduction}\label{intro}

In this article, we are interested in the existence, energy property of
sign-changing solution $u_{\lambda}$ and a convergence property of
 $u_{\lambda}$ as $\lambda\searrow 0$ for the nonlinear
Schr\"odinger-Poisson system
\begin{equation} \label{e1.1}
\begin{gathered}
-\Delta u+V(x)u+\lambda\phi(x)u=f(u),\quad \text{in } \mathbb{R}^3,\\
-\Delta\phi=u^2,\quad \text{in } \mathbb{R}^3,
\end{gathered}
\end{equation}
where $\lambda>0$ is a parameter.  We assume that
$f\in C^{1}(\mathbb{R}, \mathbb{R})$ and satisfies the following hypotheses:
\begin{itemize}
\item[(H1)] $f(t)=o(| t|)$ as $t\to  0$.

\item[(H2)] $| f(t)|\leq C(1+| t|^{p})$
 for all $ t\in \mathbb{R}$ and $3<p<5$.

\item[(H3)] $\lim_{t\to  \infty} F(t)/t^{4}=+\infty$,
 where $ F(t)=\int_0^{t}f(s)ds$.

\item[(H4)] $f(t)/| t|^3$ is an increasing function of $t$ on
$\mathbb{R}\setminus\{0\}$.
\end{itemize}
We assume the potential $V(x)$, satisfies
\begin{itemize}
\item[(H5)] $V(x)\in C(\mathbb{R}^3, \mathbb{R})$,
$\inf_{x\in \mathbb{R}^3} V(x)>0$ and $\lim_{|x|\to \infty} V(x)=+\infty$.
\end{itemize}
We define the Sobolev space
\begin{equation*}
H=\big\{u\in H^{1}(\mathbb{R}^3): \int_{\mathbb{R}^3}V(x)u^2dx< \infty\big\}
\end{equation*}
with the norm
\begin{equation*}
\| u\|=\Big(\int_{\mathbb{R}^3}(| \nabla u|^2+V(x)u^2)dx\Big)^{1/2}, \quad
 \forall u\in H.
\end{equation*}
By (H5), it follows that for $2\leq q<6$, the embedding
 $H\hookrightarrow L^{q}(\mathbb{R}^3)$ is compact, see \cite{rBW1}.
In fact, the condition (H5)
may be weaken, for example, we refer to \cite{rbpw, rBW1} for more details.

In recent years, there has been a great deal work dealing with problem \eqref{e1.1},
specially on the existence of positive solutions, ground states and
semiclassical states, see for examples,
\cite{rA, rAD, rAP, rI, rJ, rL, rR, RR, rR1, rR2, rWZ1, rZZ},
etc. To the best of our knowledge, there are very few results
about the existence of sign-changing solutions for problem \eqref{e1.1}.
Recently, in \cite{rLWZ}, the authors study the infinitely many sign-changing
solutions for the nonlinear Schr\"odinger--Poisson
system. And in \cite{rWZ2}, the authors studied the existence of sign-changing
solutions for a Schr\"odinger--Poisson system with pure power nonlinearity
$| u|^{p-1}u$, moreover, only when $\lambda>0$ is small enough, the
authors showed that the energy of any sign-changing solution is strictly
larger than the least energy. However, their method strongly depends on the
fact that the nonlinearity is homogeneous, so it is difficult to apply their
method to our problem.

For $u\in H$. Let $\phi_{u}$ be unique solution of $-\Delta \phi=u^2$
in $D^{1, 2}(\mathbb{R}^3)$, then
\begin{equation*}
\phi_{u}(x)=\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{u^2(y)}{| x-y|}dy.
\end{equation*}
The weak solutions to problem \eqref{e1.1} are the critical points of 
the functional defined by
\begin{equation*}
I_{\lambda}(u)=\frac{1}{2}\int_{\mathbb{R}^3}(| \nabla u|^2+V(x)|  u|^2)dx
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{u}u^2dx-\int_{\mathbb{R}^3}F(u)dx.
\end{equation*}
Then  $I_{\lambda}\in C^{1}(H, \mathbb{R})$  and for any $\psi\in H$,
\begin{equation*}
\langle I_{\lambda}'(u), \psi\rangle
=\int_{\mathbb{R}^3}(\nabla u\nabla\psi+V(x)u\psi)dx
+\lambda\int_{\mathbb{R}^3}\phi_{u}u\psi dx-\int_{\mathbb{R}^3}f(u)\psi dx.
\end{equation*}
We call $u$ a least energy sign-changing solution to problem \eqref{e1.1} if $u$ 
is a solution of problem \eqref{e1.1} with $u^{\pm}\neq 0$ and
\begin{equation*}
I_{\lambda}(u)=\inf \{I_{\lambda}(v): v^{\pm}\neq 0, I'_{\lambda}(v)=0\},
\end{equation*}
where $u^{+}=\max\{u(x), 0\}$ and $u^{-}=\min\{u(x), 0\}$.

When $\lambda=0$, problem \eqref{e1.1} does not depend on the nonlocal
term $\phi_{u}(x)$ any more, that is, it becomes to the following semilinear 
local equation
\begin{equation} \label{e1.2}
-\Delta u+V(x)u=f(u),\quad \text{in } \mathbb{R}^3.
\end{equation}
There are several ways in the literature to obtain sign-changing solutions 
for equation \eqref{e1.2}, see for instance 
\cite{rBLW, rBW2, rCCN, rMR, rNW,  Re, rZ, ZZX, ZZR}. 
However, all these methods heavily relay on the following decompositions:
\begin{gather} \label{e1.3}
 I_0(u)=  I_0(u^{+})+ I_0(u^{-}), \\
\label{e1.4}
\langle I_0'(u), u^{+}\rangle=\langle I_0'(u^{+}), u^{+}\rangle
\quad \text{and}\quad
\langle I_0'(u), u^{-}\rangle=\langle I_0'(u^{-}), u^{-}\rangle,
\end{gather}
where
\begin{equation*}
I_0(u)=\frac{1}{2}\int_{\mathbb{R}^3}(| \nabla u|^2+V(x)|  u|^2)dx
-\int_{\mathbb{R}^3}F(u)dx.
\end{equation*}
Furthermore, \eqref{e1.3} and \eqref{e1.4} imply that the energy of any 
sign-changing solution to \eqref{e1.2} is larger than two times the 
least energy in $H$. However, for the case $\lambda>0$, 
due to the effect of the nonlocal term, the functional  $I_{\lambda}$ 
no longer possesses the same decompositions as \eqref{e1.3}, \eqref{e1.4}. 
Indeed, we have
\begin{gather} \label{e1.5}
I_{\lambda}(u)=I_{\lambda}(u^{+})+ I_{\lambda}(u^{-})
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx, \\
 \label{e1.6}
\langle I_{\lambda}'(u), u^{+}\rangle=\langle I_{\lambda}'(u^{+}), u^{+}\rangle
+\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx, \\
 \label{e1.7}
\langle I_{\lambda}'(u), u^{-}\rangle=\langle I_{\lambda}'(u^{-}), u^{-}\rangle
+\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx.
\end{gather}
So the methods to obtain sign-changing solutions of the local problem \eqref{e1.2} 
and to estimate the energy of the sign-changing solutions seem not suitable 
for our nonlocal one \eqref{e1.1}.

To obtain a sign-changing solution for problem \eqref{e1.1},
 borrowing the idea in \cite{rS}, we first try to seek a minimizer of the 
energy functional $I_{\lambda}$ over the following constraint:
\begin{equation*}
\mathcal{M}_{\lambda}=\{u\in H: u^{\pm}\neq 0, \langle I_{\lambda}'(u), 
u^{+}\rangle=\langle I_{\lambda}'(u), u^{-}\rangle=0\}
\end{equation*}
and then we show that the minimizer is a sign-changing solution of \eqref{e1.1}. 
Note that the paper \cite{rBW2} is concerned with equation \eqref{e1.2}, 
but in our problem \eqref{e1.1} the nonlocal term is involved such that the 
properties \eqref{e1.3}, \eqref{e1.4} fail, and it is rather difficult 
to show that $\mathcal{M}_{\lambda}\neq \emptyset$. 
To prove it, in \cite{rSW}, the authors used the parametric method and 
implicit function theorem, this makes the problem very complicated, 
here we use Miranda's Theorem in \cite{rM}, which was first used 
in \cite{rAS} for the least energy sign-changing solution to Schr\"odinger-Poisson
system on bounded domain and can greatly simplify the proof in \cite{rSW}.
 To show that the minimizer of the constrained problem is a sign-changing 
solution, we will use the quantitative deformation lemma and degree theory.

The following are the main results of this article.

\begin{theorem}\label{thm1.1} 
Let {\rm (H1)--(H5)} hold. Then for any $ \lambda>0$, problem \eqref{e1.1}
 has a least energy sign-changing solution $u_{\lambda}$, which has precisely two
nodal domains.
\end{theorem}

In \cite{rWZ2} the authors proved that the energy of any sign-changing solution 
is strictly larger than the least energy only when $\lambda>0$ is small enough, 
here we improve it to the case for any $\lambda>0$. In order to describe 
our result, some notations are needed.  Let
\begin{gather} \label{e1.8}
\mathcal{N}_{\lambda}:=\{u\in H\setminus \{0\}: 
 \langle I_{\lambda}'(u), u\rangle=0\}, \\
\label{e1.9}
c_{\lambda}:=\inf_{u\in \mathcal{N}_{\lambda}} I_{\lambda}(u)
\end{gather}
Let $u_{\lambda}\in H$ be a sign-changing solution of problem \eqref{e1.1}, 
it is clear from \eqref{e1.6} and \eqref{e1.7} that 
$u_{\lambda}^{\pm}\not\in \mathcal{N}_{\lambda}$.

\begin{theorem}\label{thm1.2}
Under the assumptions of Theorem \ref{thm1.1}, $c_{\lambda}>0$ is achieved and
 $I_{\lambda}(u_{\lambda})>2c_{\lambda}$, where
$u_{\lambda}$ is the least energy sign-changing solution obtained in 
Theorem \ref{thm1.1}. In particular, $c_{\lambda}>0$ is achieved
 either by a positive or a negative function.
\end{theorem}

It is evident that the energy of the sign-changing solution $u_{\lambda}$ 
obtained in Theorem \ref{thm1.1} depends on $\lambda$. As a by-product of this paper, 
we give a convergence property of $u_{\lambda}$  as $\lambda\searrow 0$, 
which reflects some relationship between $\lambda>0$ and $\lambda=0$ 
in problem \eqref{e1.1}.

\begin{theorem}\label{thm1.3} 
If the assumptions of Theorem \ref{thm1.1} hold, then for any sequence $\lambda_{n}$ 
with $\lambda_{n}\searrow 0$ as $n\to  \infty$, there exists a subsequence, 
still denoted by $\lambda_{n}$, such that $u_{\lambda_{n}}\to  u_0$
strongly in $H$ as $n\to  \infty$, where $u_0$ is a least energy
sign-changing solution of the problem
\begin{equation} \label{e1.10}
\begin{gathered}
-\Delta u+V(x)u=f(u),\quad \text{in } \mathbb{R}^3,\\
u\in H,
\end{gathered}
\end{equation}
which has precisely two nodal domains.
\end{theorem}

This paper is organized as follows. In Section2, we
present some preliminary lemmas which are essential for this paper. 
In Section 3, we give the proofs of Theorems \ref{thm1.1}--\ref{thm1.3}
 respectively.

\section{Some technical lemmas}\label{prel}

In the sequel, we will use constraint minimization on 
$\mathcal{M}_{\lambda}$ to look for a critical point of $I_{\lambda}$. 
For this, we start with this section by claiming that the set 
$\mathcal{M}_{\lambda}$ is nonempty in $H$.

\begin{lemma} \label{lem2.1}
Assume that {\rm (H1)--(H5)} hold, if $u\in H$ with $u^{\pm}\neq 0$, 
then there exists a unique pair 
$(s_{u}, t_{u})\in \mathbb{R}_{+}\times \mathbb{R}_{+}$ such that 
$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}_{\lambda}$.
\end{lemma}

\begin{proof}
Fixed $u\in H$ with $u^{\pm}\neq 0$. We first establish the existence of 
$s_{u}$, $t_{u}$. Let
\begin{equation} \label{e2.1}
\begin{aligned}
g(s, t)&=\langle I'_{\lambda}(su^{+}+tu^{-}), su^{+}\rangle \\
& =s^2\| u^{+}\|^2+s^{4}\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx
 +s^2t^2\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx\\
&\quad -\int_{\mathbb{R}^3}f(su^{+})su^{+} dx,
\end{aligned}
\end{equation}
\begin{equation} \label{e2.2}
\begin{aligned}
h(s, t)&=\langle I'_{\lambda}(su^{+}+tu^{-}), tu^{-}\rangle \\
&=t^2\| u^{-}\|^2+t^{4}\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{-})^2dx
 +s^2t^2\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx \\
&\quad -\int_{\mathbb{R}^3}f(tu^{-})tu^{-} dx.
\end{aligned}
\end{equation}
 From $(f_{1})$ and (H3), it is easy to obtain that $g(s, s)=0$, $h(s, s)>0$
for $s>0$ small and $g(t, t)<0$, $h(t, t)>0$ for $t>0$ large.
 Hence there exist $0<r<R$ such that
 \begin{equation} \label{e2.3}
 g(r, r)>0,\quad\ h(r, r)>0,\quad g(R, R)<0,\quad h(R, R)<0.
\end{equation}
 From \eqref{e2.1}, \eqref{e2.2} and \eqref{e2.3}, we have
\begin{gather*}
 g(r, \beta)>0, \quad g(\beta, R)<0,\quad \forall  \beta\in[r, R], \\
 h(\alpha, r)>0, \quad h(R, \alpha)<0,\quad \forall  \alpha\in[r, R].
\end{gather*}
By Miranda's Theorem  \cite{rM}, there exists some point $(s_{u}, t_{u})$ 
with $\alpha<s_{u}, t_{u}<\beta$ such that
 $g(s_{u}, t_{u})=h(s_{u}, t_{u})=0$. So 
$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}_{\lambda}$.

Now we show that the pair $(s_{u}, t_{u})$ is unique and consider it in two cases.
 If $u\in \mathcal{M}_{\lambda}$, then $u^{+}+u^{-}=u\in \mathcal{M}_{\lambda}$. 
It means that
\begin{equation*}
 \langle I_{\lambda}'(u), u^{+}\rangle=\langle I_{\lambda}'(u), u^{-}\rangle=0;
\end{equation*}
that is,
\begin{equation} \label{e2.4}
\| u^{+}\|^2+\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx
=\int_{\mathbb{R}^3}f(u^{+})u^{+} dx,
\end{equation}
and
\begin{equation} \label{e2.5}
\| u^{-}\|^2+\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{-})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx
=\int_{\mathbb{R}^3}f(u^{-})u^{-} dx.
\end{equation}
We show that $(s_{u}, t_{u})=(1, 1)$ is the unique pair of numbers such 
that $s_{u}u^{+}+t_{u}u^{-}\in\mathcal{M}_{\lambda}$.

Assume that $(\tilde{s}_{u}, \tilde{t}_{u})$ is another pair of numbers such 
that $\tilde{s}_{u}u^{+}+\tilde{t}_{u}u^{-}\in\mathcal{M}_{\lambda}$.
 By the definition of $\mathcal{M}_{\lambda}$, we have
\begin{gather} \label{e2.6}
\begin{aligned}
&\tilde{s}_{u}^2\| u^{+}\|^2+\tilde{s}_{u}^{4}
 \lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx
 +\tilde{s}_{u}^2\tilde{t}_{u}^2\lambda\int_{\mathbb{R}^3}
 \phi_{u^{-}}(u^{+})^2dx \\
&=\int_{\mathbb{R}^3}f(\tilde{s}_{u}u^{+})\tilde{s}_{u}u^{+} dx,
\end{aligned} \\
 \label{e2.7}
\begin{aligned}
&\tilde{t}_{u}^2\| u^{-}\|^2+\tilde{t}_{u}^{4}
 \lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{-})^2dx
 +\tilde{s}_{u}^2\tilde{t}_{u}^2\lambda\int_{\mathbb{R}^3}
 \phi_{u^{+}}(u^{-})^2dx \\
&=\int_{\mathbb{R}^3}f(\tilde{t}_{u}u^{-})\tilde{t}_{u}u^{-} dx.
\end{aligned}
\end{gather}
Without loss of generality, we may assume that $0<\tilde{s}_{u}\leq \tilde{t}_{u}$. 
Then, from \eqref{e2.6}, we have
\begin{align*}
\tilde{s}_{u}^2\| u^{+}\|^2+\tilde{s}_{u}^{4}\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx+\tilde{s}_{u}^{4}\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx
\leq\int_{\mathbb{R}^3}f(\tilde{s}_{u}u^{+})\tilde{s}_{u}u^{+} dx,
\end{align*}
Moreover, we have
\begin{equation} \label{e2.8}
\tilde{s}_{u}^{-2}\| u^{+}\|^2+\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx
\leq\int_{\mathbb{R}^3}\frac{f(\tilde{s}_{u}u^{+})
\tilde{s}_{u}}{ \tilde{s}_{u}^3}u^{+} dx,
\end{equation}
By \eqref{e2.8} and \eqref{e2.4}, one has
\begin{equation} \label{e2.9}
(\tilde{s}_{u}^{-2}-1)\| u^{+}\|^2\leq\int_{\mathbb{R}^3}
\Big(\frac{f(x,\tilde{s}_{u} u^{+})}{(\tilde{s}_{u}u^{+})^3}
- \frac{f(x, u^{+})}{(u^{+})^3}\Big)(u^{+})^{4}dx.
\end{equation}
It follows from (H4) and \eqref{e2.9} that 
$1\leq \tilde{\alpha}_{u}\leq \tilde{\beta}_{u}$. 
By the same method,  we may get $\tilde{\beta}_{u}\leq 1$ by (H4), \eqref{e2.5} 
and \eqref{e2.7}, which shows
that $\tilde{\alpha}_{u}=\tilde{\beta}_{u}=1$.

If $u\not\in \mathcal{M}_{\lambda}$, then there exists a pair of positive numbers 
$(\alpha_{u}, \beta_{u})$ such that
$\alpha_{u}u^{+}+\beta_{u}u^{-}\in\mathcal{M}_{\lambda}$. 
Suppose that there exists another pair of positive numbers 
$(\alpha'_{u}, \beta'_{u})$
such that $\alpha'_{u}u^{+}+\beta'_{u}u^{-}\in\mathcal{M}_{\lambda}$. 
Set $v:=\alpha_{u}u^{+}+\beta_{u}u^{-}$ and 
$v':=\alpha'_{u}u^{+}+\beta'_{u}u^{-}$, we have
\begin{equation*}
\frac{\alpha'_{u}}{\alpha_{u}}v^{+}+\frac{\beta'_{u}}{\beta_{u}}v^{-}
=\alpha'_{u}u^{+}+\beta'_{u}u^{-}=v'\in \mathcal{M}_{\lambda}.
\end{equation*}
Since $v\in \mathcal{M}_{\lambda}$, we obtain that $\alpha_{u}=\alpha'_{u}$ 
and $\beta_{u}=\beta'_{u}$, which implies that $(\alpha_{u}, \beta_{u})$
is the unique pair of numbers such that 
$\alpha_{u}u^{+}+\beta_{u}u^{-}\in\mathcal{M}_{\lambda}$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.2}
Assume that {\rm (H1)--(H5)} hold. For a fixed $u\in H$ with 
$u^{\pm}\neq 0$. If $g_{1}(1, 1)\leq 0$ and $h_{1}(1, 1)\leq 0$, 
then there exists a unique pair $(s_{u}, t_{u})\in (0, 1]\times (0, 1]$ 
such that $g_{1}(s_{u}, t_{u})=h_{1}(s_{u}, t_{u})=0$.
\end{lemma}

\begin{proof}
Suppose that $s_{u}\geq t_{u}>0$. By Lemma \ref{lem2.1}, we know that 
$s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}_{\lambda}$, then
\begin{equation} \label{e2.10}
\begin{aligned} 
&s_{u}^2\| u^{+}\|^2+s_{u}^{4}\lambda\int_{\mathbb{R}^3}
 \phi_{u^{+}}(u^{+})^2dx+s_{u}^{4}\lambda\int_{\mathbb{R}^3}
 \phi_{u^{-}}(u^{+})^2dx \\
&\geq s_{u}^2\| u^{+}\|^2+s_{u}^{4}\lambda\int_{\mathbb{R}^3}
 \phi_{u^{+}}(u^{+})^2dx+s_{u}^2t_{u}^2
 \lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx \\
&= \int_{\mathbb{R}^3}f(s_{u}u^{+})s_{u}u^{+} dx.
\end{aligned}
\end{equation}
Moreover, $g_{1}(1, 1)\leq 0$ implies that
\begin{equation} \label{e2.11}
\| u^{+}\|^2+\lambda\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{+})^2dx
\leq \int_{\mathbb{R}^3}f(u^{+})u^{+} dx.
\end{equation}
Combining \eqref{e2.4} with \eqref{e2.5}, we have
\begin{equation*}
\Big(\frac{1}{s_{u}^2}-1\Big)\| u^{+}\|^2
\geq \int_{\mathbb{R}^3}\Big(\frac{f(s_{u}u^{+})}{(s_{u}u^{+})^3}
-\frac{f(u^{+})}{(u^{+})^3}\Big)| u^{+}|^{4}dx.
\end{equation*}
If $s_{u}>1$, the left-hand side of this inequality  is negative. 
But from (H4), the right-hand side of this inequality is positive, 
so we have $s_{u}\leq 1$.
The proof is thus complete.
\end{proof}

\begin{lemma} \label{lem2.3}
For a fixed $u\in H$ with $u^{\pm}\neq 0$, then $(s_{u}, t_{u})$ 
obtained in Lemma \ref{lem2.1} is the unique maximum point of the
function $\phi:\mathbb{R}_{+}\times \mathbb{R}_{+}\to  \mathbb{R}$ 
defined as $\phi(s, t)=I_{\lambda}(su^{+}+tu^{-})$.
\end{lemma}

\begin{proof}
From the proof of Lemma \ref{lem2.1}, we know that $(s_{u}, t_{u})$ is the unique critical 
point of $\phi$ in $\mathbb{R}_{+} \times \mathbb{R}_{+}$.
By (H3), we conclude that $\phi(s, t)\to  -\infty$ uniformly as
$| (s, t)|\to \infty$, so it is
sufficient to show that a maximum point cannot be achieved on the boundary of 
$(\mathbb{R}_{+}, \mathbb{R}_{+})$. If we assume that
$(0, \bar{t})$ is a maximum point of $\phi$. Then since
\begin{align*}
\phi(s, \bar{t})&=I_{\lambda}(su^{+}+\bar{t}u^{-})\\
&=\frac{s^2}{2}\| u^{+}\|^2+\frac{\lambda}{4}s^{4}
 \int_{\mathbb{R}^3}\phi_{u^{+}}(u^{+})^2dx-\int_{\mathbb{R}^3}F(su^{+})dx\\
&\quad +\frac{\lambda}{4}\Big(s^2\bar{t}^2\int_{\mathbb{R}^3}
 \phi_{u^{-}}(u^{+})^2dx+
s^2\bar{t}^2\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx\Big)\\
&\quad +\frac{\bar{t}^2}{2}\| u^{-}\|^2
 +\frac{\lambda}{4}\bar{t}^{4}\int_{\mathbb{R}^3}\phi_{u^{-}}(u^{-})^2dx
 -\int_{\mathbb{R}^3}F(\bar{t}u^{-})dx
\end{align*}
is an increasing function with respect to $s$ if $s$ is small enough, the pair 
$(0, \bar{t})$ is not a maximum point of $\phi$ in
$\mathbb{R}_{+}\times \mathbb{R}_{+}$. The proof is now finished.
\end{proof}

By Lemma \ref{lem2.1}, we define the minimization problem
\begin{equation*}
m_{\lambda}:=\inf\Big\{I_{\lambda}(u): u\in \mathcal{M}_{\lambda}\Big\}.
\end{equation*}

\begin{lemma} \label{lem2.4}
Assume that {\rm (H1)--(H5)} hold, then $m_{\lambda}>0$ can be achieved for any 
$\lambda>0$.
\end{lemma}

\begin{proof}
For every $u\in \mathcal{M}_{\lambda}$, we have $\langle I'_{\lambda}(u), u\rangle=0$.
 From $(f_{1})$, $(f_2)$, for any
$\epsilon>0$, there exists $C_{\epsilon}>0$ such that
\begin{equation} \label{e2.12}
f(s)s\leq \epsilon s^2+C_{\epsilon}| s|^{p+1} \quad \text {for all }
 s\in \mathbb{R}.
\end{equation}
By Sobolev embedding theorem, we obtain
\begin{equation} \label{e2.13}
\begin{aligned}
\| u\|^2
&\leq \int_{\mathbb{R}^3}(| \nabla u|^2+V(x)| u|^2)dx
 +\lambda\int_{\mathbb{R}^3}\phi_{u}u^2dx=\int_{\mathbb{R}^3}f(u)udx \\
&\leq \epsilon \int_{\mathbb{R}^3}| u|^2dx
 +C_{\epsilon}\int_{\mathbb{R}^3}| u|^{p+1}dx \\
&\leq C_2\epsilon \| u\|^2+C_{\epsilon}'\| u\|^{p+1}.
\end{aligned}
\end{equation}
Pick $\epsilon=1/(2C_2)$. So there exists a constant $\alpha>0$ 
such that $\| u\|^2>\alpha$.
By \eqref{e2.3}, we have
\begin{equation*}
f(s)s-4F(s)\geq 0.
\end{equation*}
Then
\begin{equation*}
 I_{\lambda}(u)= I_{\lambda}(u)-\frac{1}{4}\langle I'_{\lambda}(u), 
u\rangle\geq \frac{\| u\|^2}{4}\geq \frac{\alpha}{4}.
\end{equation*}
This implies that $I_{\lambda}(u)$ is coercive in $\mathcal{M}_{\lambda}$ and 
$m_{\lambda}\geq \frac{\alpha}{4}>0$.

Let $\{u_{n}\}_n\subset \mathcal{M}_{\lambda}$ be such that 
$ I_{\lambda}(u_{n})\to  m_{\lambda}$. Then $\{u_{n}\}_n$ is bounded in $H$ 
and there exists
$u_{\lambda}\in H$ such that $u_{n}^{\pm}\rightharpoonup u_{\lambda}^{\pm}$ 
weakly in $H$. Since $u_{n}\in \mathcal{M}_{\lambda}$, we have
$\langle I'_{\lambda}(u_{n}), u_{n}^{\pm}\rangle=0$, that is
\begin{equation*}
\| u_{n}^{\pm}\|^2+\lambda\int_{\mathbb{R}^3}
\phi_{u_{n}^{\pm}}(u_{n}^{\pm})^2dx+\lambda\int_{\mathbb{R}^3}
\phi_{u_{n}^{\mp}}(u_{n}^{\pm})^2dx-
\int_{\mathbb{R}^3}f(u_{n}^{\pm})u_{n}^{\pm} dx=0.
\end{equation*}
Similar as \eqref{e2.7} we also have $\| u_{n}^{\pm}\|^2\geq \beta$
for all $n\in N$, where $\beta$ is a constant.

Since $u_{n}\in \mathcal{M}_{\lambda}$, by \eqref{e2.6} again, we have
\begin{equation*}
\beta\leq\| u_{n}^{\pm}\|^2<
\int_{\mathbb{R}^3}f(u_{n}^{\pm})u_{n}^{\pm} dx
\leq \epsilon\int_{\mathbb{R}^3}| u_{n}^{\pm}|^2dx+
C_{\epsilon}\int_{\mathbb{R}^3}| u_{n}^{\pm}|^{p+1}dx.
\end{equation*}
In view of the boundedness of $\{u_{n}\}_n$, there is $C_2>0$ such that
\begin{equation*}
\beta\leq\epsilon C_2+C_{\epsilon}\int_{\mathbb{R}^3}| u_{n}^{\pm}|^{p+1}dx.
\end{equation*}
Choosing $\epsilon=\beta/(2C_2)$, we obtain
\begin{equation} \label{e2.14}
\int_{\mathbb{R}^3}| u_{n}^{\pm}|^{p+1}dx\geq \frac{\beta}{2\bar{C}}.
\end{equation}
where $\bar{C}$ is a positive constant, in fact, 
$\bar{C}=C_{\frac{\beta}{2C_2}}$.

By \eqref{e2.8} and the compact embedding $H\hookrightarrow L^{q}(\mathbb{R}^3)$ 
for $2\leq q< 6$,  we obtain
\begin{equation*}
\int_{\mathbb{R}^3}| u_{\lambda}^{\pm}|^{p+1}dx\geq \frac{\beta}{2\bar{C}}.
\end{equation*}
Thus, $u_{\lambda}^{\pm}\neq 0$. By $(f_{1})$, $(f_2)$, the compact embedding 
and \cite[Theorem A.4]{rW},
\begin{equation} \label{e2.15}
\begin{gathered}
\lim_{n\to \infty} \int_{\mathbb{R}^3}f(u_{n}^{\pm})u_{n}^{\pm} dx
=\int_{\mathbb{R}^3}f(u_{\lambda}^{\pm})u_{\lambda}^{\pm} dx,\\
\lim_{n\to \infty} \int_{\mathbb{R}^3}F(u_{n}^{\pm}) dx
=\int_{\mathbb{R}^3}F(u_{\lambda}^{\pm}) dx.
\end{gathered}
\end{equation}
By the weak semicontinuity of norm and Fatou's Lemma, we have
\begin{align*}
&\| u_{\lambda}^{\pm}\|^2+\lambda\int_{\mathbb{R}^3}
 \phi_{u_{\lambda}^{\pm}}(u_{\lambda}^{\pm})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u_{\lambda}^{\mp}}(u_{\lambda}^{\pm})^2dx \\
&\leq \underset{n\to  \infty}{\lim\inf}\Big\{\| u_{n}^{\pm}\|^2
 +\lambda\int_{\mathbb{R}^3}\phi_{u_{n}^{\pm}}(u_{n}^{\pm})^2dx+\lambda\int_{\mathbb{R}^3}\phi_{u_{n}^{\mp}}(u_{n}^{\pm})^2dx\Big\}.
\end{align*}
From \eqref{e2.9}, we have
\begin{equation} \label{2.16e}
\| u_{\lambda}^{\pm}\|^2+\lambda\int_{\mathbb{R}^3}\phi_{u_{\lambda}^{\pm}}
(u_{\lambda}^{\pm})^2dx
+\lambda\int_{\mathbb{R}^3}\phi_{u_{\lambda}^{\mp}}(u_{\lambda}^{\pm})^2dx
\leq \int_{\mathbb{R}^3}f(u_{\lambda}^{\pm})u_{\lambda}^{\pm} dx
\end{equation}
From \eqref{e2.10} and Lemma \ref{lem2.2}, there exists 
$(s_{u_{\lambda}}, t_{u_{\lambda}})\in (0, 1]\times (0, 1]$ such that
\begin{equation*}
\bar{u}_{\lambda}:=s_{u_{\lambda}}u_{\lambda}^{+}
+t_{u_{\lambda}}u_{\lambda}^{-}\in \mathcal{M}_{\lambda}.
\end{equation*}
Condition (H4) implies that $H(s):=sf(s)-4F(s)$ is a non-negative function, 
increasing in $| s|$, so we have
\begin{align*}
m_{\lambda}
&\leq I_{\lambda}(\bar{u}_{\lambda})
 =I_{\lambda}(\bar{u}_{\lambda})-\frac{1}{4}\langle  I'_{\lambda}(\bar{u}_{\lambda}),
  \bar{u}_{\lambda}\rangle\\
&=\frac{1}{4}\| \bar{u}_{\lambda}\|^2+\frac{1}{4}\int_{\mathbb{R}^3}
\Big(f(\bar{u}_{\lambda})\bar{u}_{\lambda}-4F(\bar{u}_{\lambda})\Big)dx\\
&= \frac{1}{4}\| s_{u_{\lambda}}u_{\lambda}^{+}\|^2+\frac{1}{4}
\int_{\mathbb{R}^3}\Big(f(s_{u_{\lambda}}u_{\lambda}^{+})
s_{u_{\lambda}}u_{\lambda}^{+}
-4F(s_{u_{\lambda}}u_{\lambda}^{+})\Big)dx\\
&\quad+ \frac{1}{4}\| t_{u_{\lambda}}u_{\lambda}^{-}\|^2
 +\frac{1}{4}\int_{\mathbb{R}^3}\Big(f(t_{u_{\lambda}}
u_{\lambda}^{-})t_{u_{\lambda}}u_{\lambda}^{-}
-4F(t_{u_{\lambda}}u_{\lambda}^{-})\Big)dx\\
&\leq\frac{1}{4}\| u_{\lambda}\|^2+\frac{1}{4}
 \int_{\mathbb{R}^3}\Big(f(u_{\lambda})u_{\lambda}-4F(u_{\lambda})\Big)dx\\
&\leq \underset{n\to  \infty}{\lim\inf}\Big[I_{\lambda}(u_{n})
 -\frac{1}{4}\langle  I'_{\lambda}(u_{n}), u_{n}\rangle\Big]=m_{\lambda}.
\end{align*}
Then we conclude that $s_{u_{\lambda}}=t_{u_{\lambda}}=1$. 
Thus, $\bar{u}_{\lambda}=u_{\lambda}$ and $I_{\lambda}(u_{\lambda})=m_{\lambda}$.
\end{proof}

\section{Proof of main results} \label{pf1}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
We first prove that the minimizer $u_{\lambda}$ for the minimization problem 
is indeed a sign-changing solution of problem \eqref{e1.1};
 that is, $I'_{\lambda}(u_{\lambda})=0$. For it, we will use the 
quantitative deformation lemma.

It is obvious that $I'_{\lambda}(u_{\lambda})u_{\lambda}^{+}=
0=I'_{\lambda}(u_{\lambda})u_{\lambda}^{-}$. From Lemma \ref{lem2.3}, 
for any $(s, t)\in \mathbb{R}_{+}\times \mathbb{R}_{+}$ and
$(s, t)\neq (1, 1)$,
\begin{equation*}
I_{\lambda}(s u_{\lambda}^{+}+tu_{\lambda}^{-})
<I_{\lambda}( u_{\lambda}^{+}+u_{\lambda}^{-})=m_{\lambda}.
\end{equation*}
If $I'_{\lambda}(u_{\lambda})\neq 0$, then there exist $\delta>0$ and $\kappa>0$ 
such that
\begin{equation*}
\| I'_{\lambda}(v)\|\geq \kappa \quad \text{for all } \| v-u_{\lambda}\|\leq 3\delta.
\end{equation*}
Let $D:=(\frac{1}{2}, \frac{3}{2})\times (\frac{1}{2}, \frac{3}{2})$ and 
$g(s, t):=su_{\lambda}^{+}+tu_{\lambda}^{-}$. From Lemma \ref{lem2.3}, we also have
\begin{equation*}
\bar{m}_{\lambda}:=\max_{\partial D} I_{\lambda}\circ g<m_{\lambda}.
\end{equation*}
For $\epsilon:=\min\{(m_{\lambda}-\bar{m}_{\lambda})/2, \kappa\delta/8\}$ and
 $S:=B(u_{\lambda}, \delta)$, there is a deformation $\eta$ such that
\begin{itemize}
\item[(a)] $\eta(1, u)=u$ if $u\not\in I_{\lambda}^{-1}([m_{\lambda}-2\epsilon,
 m_{\lambda}+2\epsilon])\cap S_{2\delta}$;
\item[(b)] $\eta(1, I_{\lambda}^{m_{\lambda}+\epsilon}\cap S)
 \subset I_{\lambda}^{m_{\lambda}-\epsilon}$;
\item[(c)] $I_{\lambda}(\eta(1, u)))\leq I_{\lambda}(u)$ for all 
$u\in H$.
\end{itemize}
See \cite{rW} for more details. It is clear that
\begin{equation*}
\max_{(s, t)\in \bar{D}} I_{\lambda}(\eta(1, g(s, t))))<m_{\lambda}.
\end{equation*}
 Now we prove that $\eta(1, g(D))\cap \mathcal{M}_{\lambda}\neq \emptyset$
 which contradicts to the definition of $m_{\lambda}$. Let us define
 $h(s, t)=\eta(1, g(s, t)))$ and
\begin{equation*}
\Psi_0(s, t):=\Big(I'_{\lambda}(g(s, t))u_{\lambda}^{+}, I'_{\lambda}
(g(s, t))u_{\lambda}^{-}\Big)=\Big(I'_{\lambda}(su_{\lambda}^{+}
+tu_{\lambda}^{-})u_{\lambda}^{+}, I'_{\lambda}(su_{\lambda}^{+}
+tu_{\lambda}^{-})u_{\lambda}^{-}\Big),
\end{equation*}
\begin{equation*}
\Psi_{1}(s, t):=\Big(\frac{1}{s}I'_{\lambda}(h(s, t))h^{+}(s, t), 
\frac{1}{t}I'_{\lambda}(h(s, t))h^{-}(s, t)\Big).
\end{equation*}
Lemma \ref{lem2.1} and the degree theory imply that $\deg (\Psi_0, D, 0)=1$.
It follows from that $g=h$ on $\partial D$. Consequently,
we obtain
\begin{equation*}
\deg (\Psi_{1}, D, 0)=\deg (\Psi_0, D, 0)=1.
\end{equation*}
 Thus, $\Psi_{1}(s_0, t_0)=0$ for some $(s_0, t_0)\in D$, so that
 \begin{equation*}
\eta(1, g(s_0, t_0)))=h(s_0, t_0)\in \mathcal{M}_{\lambda},
\end{equation*}
which is a contradiction. From this, $u_{\lambda}$ is a critical point of
$I_{\lambda}$, moreover, it is a sign-changing solution for problem \eqref{e1.1}.

Now we prove that $u_{\lambda}$ has exactly two nodal domains. 
By contradiction, we assume that $u_{\lambda}$ has at least three nodal domains
$\Omega_{1}$, $\Omega_2$, $\Omega_{3}$. Without loss generality, 
we may assume that $u_{\lambda}>0$ a.e. in $\Omega_{1}$ and $u_{\lambda}<0$ 
a.e. in $\Omega_2$.
Set
\begin{equation*}
{u_{\lambda_{i}}}:=\chi_{\Omega_{i}}u_{\lambda}, \quad\quad i=1, 2, 3,
\end{equation*}
where
\begin{equation*}
\chi_{\Omega_{i}}=\begin{cases}
1& x\in \Omega_{i},\\
0& x\in \mathbb{R}^{N}\setminus \Omega_{i}.
\end{cases}
\end{equation*}
Then ${u_{\lambda_{i}}}\neq 0$ and 
$\langle I'(u_{\lambda}), {u_{\lambda_{i}}}\rangle=0$ for $i=1, 2, 3$, so we have
 \begin{equation*}
\langle I'({u_{\lambda_{1}}}+{u_{\lambda_2}}), ({u_{\lambda_{1}}}
+{u_{\lambda_2}})^{\pm}\rangle<0.
\end{equation*}
By Lemma \ref{lem2.2}, there exists $(s_{v}, t_{v})\in (0, 1]\times (0, 1]$ such that 
$s_{v}{u_{\lambda_{1}}}+t_{v}{u_{\lambda_2}}\in \mathcal{M}_{\lambda}$.
Since
\begin{align*}
0&=\frac{1}{4}\langle  I'_{\lambda}(u_{\lambda}), {u_{\lambda_{3}}}\rangle\\
&=\frac{1}{4}\| {u_{\lambda_{3}}}\|^2+\frac{\lambda}{4}\int_{\mathbb{R}^3}
 \phi_{u_{\lambda}}{u_{\lambda_{3}}}^2dx
-\frac{1}{4}\int_{\mathbb{R}^3}f({u_{\lambda_{3}}}){u_{\lambda_{3}}}dx \\
&\leq \frac{1}{4}\| {u_{\lambda_{3}}}\|^2+\frac{\lambda}{4}
 \int_{\mathbb{R}^3}\phi_{u_{\lambda}}{u_{\lambda_{3}}}^2dx
-\frac{1}{4}\int_{\mathbb{R}^3}F({u_{\lambda_{3}}})dx \\
&< I_{\lambda}({u_{\lambda_{3}}})+\frac{\lambda}{4}
 \int_{\mathbb{R}^3}\phi_{{u_{\lambda}}_{1}}{u_{\lambda_{3}}}^2dx+
\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_2}}}{u_{\lambda_{3}}}^2dx.
\end{align*}
From (H4), we have
\begin{align*}
m_{\lambda}
&\leq I_{\lambda}(s_{v}{u_{\lambda_{1}}}+t_{v}{u_{\lambda_2}})\\
&= I_{\lambda}(s_{v}{u_{\lambda_{1}}}+t_{v}{u_{\lambda_2}})
 -\frac{1}{4}\langle I'_{\lambda}(s_{v}{u_{\lambda_{1}}}+t_{v}{u_{\lambda_2}}),
 s_{v}{u_{\lambda_{1}}}+t_{v}{u_{\lambda_2}}\rangle\\
&=\frac{s_{v}^2\| {u_{\lambda_{1}}}\|^2+t_{v}^2\| {u_{\lambda_2}}\|^2}{4}
 +\int_{\mathbb{R}^3}\Big(\frac{1}{4}f(s_{v}{u_{\lambda_{1}}})s_{v}{u_{\lambda_{1}}}
 -F(s_{v}{u_{\lambda_{1}}})\Big)dx\\
&\quad +\int_{\mathbb{R}^3}\Big(\frac{1}{4}f(t_{v}{u_{\lambda_2}})t_{v}{u_{\lambda_2}}
 -F(t_{v}{u_{\lambda_2}})\Big)dx\\
&\leq \frac{\| {u_{\lambda_{1}}}\|^2+\| {u_{\lambda_2}}\|^2}{4}+
\int_{\mathbb{R}^3}\Big(\frac{1}{4}f({u_{\lambda_{1}}}){u_{\lambda_{1}}}
 -F({u_{\lambda_{1}}})\Big)dx\\
&\quad +\int_{\mathbb{R}^3}\Big(\frac{1}{4}f({u_{\lambda_2}}){u_{\lambda_2}}
 -F({u_{\lambda_2}})\Big)dx\\
&=I_{\lambda}({u_{\lambda_{1}}})+I_{\lambda}({u_{\lambda_2}})
 +\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_2}}}{u_{\lambda_{1}}}^2dx
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_{3}}}}{u_{\lambda_{1}}}^2dx\\
&\quad+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_{1}}}}{u_{\lambda_2}}^2dx
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_{3}}}}{u_{\lambda_2}}^2dx\\
&<I_{\lambda}({u_{\lambda_{1}}})+I_{\lambda}({u_{\lambda_2}})
 +I_{\lambda}({u_{\lambda_{3}}})+
\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_2}}}{u_{\lambda_{1}}}^2dx\\
&\quad+\frac{\lambda}{4}\int_{\mathbb{R}^3}
 \phi_{{u_{\lambda_{3}}}}{u_{\lambda_{1}}}^2dx
\quad +\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_{1}}}}{u_{\lambda_2}}^2dx
+\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_{3}}}}{u_{\lambda_2}}^2dx\\
&\quad+\frac{\lambda}{4}\int_{\mathbb{R}^3}
 \phi_{{u_{\lambda_{1}}}}{u_{\lambda_{3}}}^2dx+
\frac{\lambda}{4}\int_{\mathbb{R}^3}\phi_{{u_{\lambda_2}}}{u_{\lambda_{3}}}^2dx\\
&=I_{\lambda}(u_{\lambda})=m_{\lambda},
\end{align*}
which is impossible, so $u_{\lambda}$ has exactly two nodal domains.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
As in the proof of Lemma \ref{lem2.4}, for each $\lambda>0$, we can get a 
$v_{\lambda}\in \mathcal{N}_{\lambda}$ such that
$I_{\lambda}(v_{\lambda})=c_{\lambda}>0$, where 
$\mathcal{N}_{\lambda}$ and $c_{\lambda}$ are defined by \eqref{e1.8} 
and \eqref{e1.9}, respectively.
Moreover, the critical points of $I_{\lambda}$ on $\mathcal{N}_{\lambda}$ 
are the critical points of $I_{\lambda}$ in $H$. Thus, $v_{\lambda}$
is a ground state solution of problem \eqref{e1.1}.

From Theorem \ref{thm1.1}, we know that problem \eqref{e1.1} has a least energy 
sign-changing solution $u_{\lambda}$ which changes sign only once. Suppose that
$u_{\lambda}=u_{\lambda}^{+}+u_{\lambda}^{-}$.  
As the proof of Step 1 in Lemma \ref{lem2.1}, there exist unique $s_{u_{\lambda}^{+}}>0$ 
and $t_{u_{\lambda}^{-}}>0$ such that
\begin{equation*}
s_{u_{\lambda}^{+}}u_{\lambda}^{+}\in \mathcal{N}_{\lambda},
\quad 
t_{u_{\lambda}^{-}}u_{\lambda}^{-}\in \mathcal{N}_{\lambda}.
\end{equation*}
From \eqref{e1.6} and \eqref{e1.7}, we have
\begin{equation*}
\langle I_{\lambda}'(u_{\lambda}^{+}), u_{\lambda}^{+}\rangle<0,\quad
 \langle I_{\lambda}'(u_{\lambda}^{-}), u_{\lambda}^{-}\rangle<0.
\end{equation*}
So, by {\rm (H1)--(H4)}, one has $s_{u_{\lambda}^{+}}\in (0, 1)$ and 
$t_{u_{\lambda}^{-}}\in (0, 1)$.
Then, by Lemma \ref{lem2.3}, we obtain
\begin{equation*}
2c_{\lambda}\leq I_{\lambda}(s_{u_{\lambda}^{+}}u_{\lambda}^{+})
+I_{\lambda}(t_{u_{\lambda}^{-}}u_{\lambda}^{-})
\leq I_{\lambda}(s_{u_{\lambda}^{+}}u_{\lambda}^{+}
+t_{u_{\lambda}^{-}}u_{\lambda}^{-})<I_{\lambda}(u_{\lambda}^{+}+u_{\lambda}^{-})
=m_{\lambda},
\end{equation*}
that is $I_{\lambda}(u_{\lambda}) >2c_{\lambda}$, which implies that 
$c_{\lambda}>0$ can not be achieved by a sign-changing function. 
This completes the proof.
\end{proof}

Now we prove Theorem \ref{thm1.3}.  In the following, we regard $\lambda>0$ 
as a parameter in problem \eqref{e1.1}. We shall study the convergence 
property of $u_{\lambda}$ as $\lambda\searrow 0$.

\begin{proof}[Proof of Theorem \ref{thm1.3}]
For any $\lambda>0$, let $u_{\lambda}\in H$ be the least energy 
sign-changing solution of problem \eqref{e1.1} obtained in Theorem \ref{thm1.1}, 
which has exactly two nodal domains.
\smallskip

\noindent\textbf{Step 1.} We show that, for any sequence $\{\lambda_{n}\}_n$ 
with $\lambda_{n} \searrow 0$ as $n\to  \infty$, $\{{u_{\lambda_{n}}}\}_n$
is bounded in $H$. Choose a nonzero function 
$\varphi \in C_0^{\infty}(\mathbb{R}^3)$ with $\varphi^{\pm}\neq 0$.
From $f(s)s-4F(s)\geq 0$,
for $s\neq 0$, we have
$f(s)s>4F(s)$. Then, (H3) implies that, for any $\lambda\in[0, 1]$, 
there exists a pair $(\theta_{1}, \theta_2)\in(\mathbb{R}_{+}\times \mathbb{R}_{+})$,
 which does not depend on $\lambda$, such that
\begin{gather*}
\begin{aligned}
&\theta_{1}^2\| \varphi^{+}\|^2+\theta_{1}^{4}\lambda\int_{\mathbb{R}^3}
 \phi_{\varphi^{+}}(\varphi^{+})^2dx+\theta_{1}^2\theta_2^2
\lambda\int_{\mathbb{R}^3}\phi_{\varphi^{-}}(\varphi^{+})^2dx \\
&-\int_{\mathbb{R}^3}f(\theta_{1}\varphi^{+})\theta_{1}\varphi^{+} dx<0,
\end{aligned}\\
\begin{aligned}
&\theta_2^2\| \varphi^{-}\|^2+\theta_2^{4}\lambda
 \int_{\mathbb{R}^3}\phi_{\varphi^{-}}(\varphi^{-})^2dx+\theta_2^2\theta_{1}^2
\lambda\int_{\mathbb{R}^3}\phi_{\varphi^{+}}(\varphi^{-})^2dx \\
&-\int_{\mathbb{R}^3}f(\theta_2\varphi^{-})\theta_2\varphi^{-} dx<0.
\end{aligned}
\end{gather*}
In view of Lemmas \ref{lem2.1} and \ref{lem2.2}, for any $\lambda\in[0, 1]$, 
there is a unique pair $(s_{\varphi}(\lambda), t_{\varphi}(\lambda) )
\in(0, 1]\times(0, 1]$ such that 
$\bar{\varphi}:=s_{\varphi}(\lambda)\theta_{1}\varphi^{+}
+t_{\varphi}(\lambda)\theta_2\varphi^{-}\in \mathcal{M}_{\lambda}$. 
Thus, for all $\lambda\in[0, 1]$, we have
\begin{align*}
 I_{\lambda}(u_{\lambda})
&\leq  I_{\lambda}(\bar{\varphi})
= I_{\lambda}(\bar{\varphi})-\frac{1}{4}\langle I'_{\lambda}(\bar{\varphi}), 
 \bar{\varphi}\rangle\\
&=\frac{\| \bar{\varphi}\|^2}{4}+
\int_{\mathbb{R}^3}\Big(\frac{1}{4}f(\bar{\varphi})\bar{\varphi}
 - F(\bar{\varphi}) \Big)dx\\
&\leq \frac{\| \bar{\varphi}\|^2}{4}+
\int_{\mathbb{R}^3}\Big(C_{3}\bar{\varphi}^2+ C_4| \bar{\varphi}|^{p+1}\Big)dx\\
&\leq \frac{\| \theta_{1}\varphi^{+}\|^2}{4}+\frac{\| \theta_2\varphi^{-}\|^2}{4}
 +\int_{\mathbb{R}^3}\Big(C_{3}(\theta_{1}\varphi^{+})^2
 + C_4| \theta_{1}\varphi^{+}|^{p+1} \\
&\quad +C_{3}(\theta_2\varphi^{-})^2
 + C_4| \theta_2\varphi^{-}|^{p+1}\Big)dx\\
=C_0.
\end{align*}
Moreover, for $n$ large enough, we obtain
\begin{equation*}
C_0+1\geq I_{\lambda_{n}}(u_{\lambda_{n}})
=I_{\lambda_{n}}(u_{\lambda_{n}})
-\frac{1}{4}\langle I'_{\lambda_{n}}(u_{\lambda_{n}}),
 u_{\lambda_{n}}\rangle\geq \frac{1}{4}\| u_{\lambda_{n}}\|^2.
\end{equation*}
So $\{{u_{\lambda}}_{n}\}_n$ is bounded in $H$.
\smallskip

\noindent\textbf{Step 2.} There exists a subsequence of $\{\lambda_{n}\}_n$,
 still denoted by $\{\lambda_{n}\}_n$, such that $u_{\lambda_{n}}\rightharpoonup u_0$
 weakly in $H$. Then, $u_0$ is a weak solution of \eqref{e1.10}. 
Since $u_{\lambda_{n}}$ is the least energy sign-changing solution of \eqref{e1.1} 
with $\lambda=\lambda_{n}$, then by the compactness of the embedding 
$H\hookrightarrow L^{q}(\mathbb{R}^3)$ for $2\leq q< 2^{*}$, we obtain 
that $u_{\lambda_{n}}\to   u_0$ strongly in $H$ as $n\to \infty$. In fact,
\begin{align*}
\| u_{\lambda_{n}}-u_0\|^2
&= \langle  I'_{\lambda_{n}}(u_{\lambda_{n}})-I'_0(u_0), u_{\lambda_{n}}
 -u_0\rangle-\lambda_{n}\int_{\mathbb{R}^3}
 \phi_{u_{\lambda_{n}}}{u_{\lambda_{n}}}(u_{\lambda_{n}}-u_0)dx\\
&\quad +\int_{\mathbb{R}^3}f(u_{\lambda_{n}})(u_{\lambda_{n}}-u_0)dx
 -\int_{\mathbb{R}^3}f(u_0)(u_{\lambda_{n}}-u_0)dx.
\end{align*}
Then $u_0\neq 0$ and $u_0$  has exactly two nodal domains.
\smallskip

\noindent\textbf{Step 3.} Suppose that $v_0$ is a least energy 
sign-changing solution of \eqref{e1.10}, we may refer to \cite{rBWW} 
for the existence of $v_0$.
 By Lemma \ref{lem2.1}, for each $\lambda_{n}>0$, there is a unique pair 
$(s_{\lambda_{n}}, t_{\lambda_{n}})\in \mathbb{R}_{+}\times \mathbb{R}_{+}$ such that
$s_{\lambda_{n}}v_0^{+}+t_{\lambda_{n}}v_0^{-}\in \mathcal{M}_{\lambda_{n}}$. 
So we have
\begin{gather*}
 \begin{aligned}
&s_{\lambda_{n}}^2\| v_0^{+}\|^2+s_{\lambda_{n}}^{4}\lambda_{n}\int_{\mathbb{R}^3}
\phi_{v_0^{+}}(v_0^{+})^2dx+ s_{\lambda_{n}}^2t_{\lambda_{n}}^2\lambda_{n}
\int_{\mathbb{R}^3}\phi_{v_0^{-}}(v_0^{+})^2dx \\
&=\int_{\mathbb{R}^3}f(s_{\lambda_{n}}v_0^{+})s_{\lambda_{n}}v_0^{+} dx,
\end{aligned}\\
\begin{aligned}
&t_{\lambda_{n}}^2\| v_0^{-}\|^2+t_{\lambda_{n}}^{4}\lambda_{n}\int_{\mathbb{R}^3}
\phi_{v_0^{-}}(v_0^{-})^2dx+ s_{\lambda_{n}}^2t_{\lambda_{n}}^2\lambda_{n}
\int_{\mathbb{R}^3}\phi_{u^{+}}(u^{-})^2dx \\
&=\int_{\mathbb{R}^3}f(t_{\lambda_{n}}v_0^{-})t_{\lambda_{n}}v_0^{-} dx.
\end{aligned}
\end{gather*}
We know that $v_0^{\pm}$ satisfies 
 $\| v_0^{\pm}\|^2=\int_{\mathbb{R}^3}f(v_0^{\pm})v_0^{\pm}dx$. It is easy to check
that
\begin{equation} \label{e2.17}
(s_{\lambda_{n}}, t_{\lambda_{n}})\to  (1, 1), \quad \text{as } n\to \infty.
\end{equation}
From this limit and Lemma \ref{lem2.3}, we have
\begin{align*}
I_0(v_0)&\leq I_0(u_0)
=\lim_{n\to \infty} I_{\lambda_{n}}(u_{\lambda_{n}})
=\lim_{n\to \infty} I_{\lambda_{n}}(u_{\lambda_{n}}^{+}+u_{\lambda_{n}}^{-})\\
&\leq \lim_{n\to \infty} I_{\lambda_{n}}(s_{\lambda_{n}}u_{\lambda_{n}}^{+}
+t_{\lambda_{n}}u_{\lambda_{n}}^{-})\\
&=I_0(v_0).
\end{align*}
This means that $u_0$ is a least energy sign-changing solution of 
 \eqref{e1.10} which has precisely two
nodal domains. The proof is complete.
\end{proof}

\subsection*{Acknowledgements}
C. Ji is supported by NSFC (grant No. 11301181) and China Postdoctoral 
Science Foundation funded project. F. Fang
is supported by NSFC (grant No. 11626038) and Young Teachers Foundation 
 of BTBU (No. QNJJ2016-15). B. Zhang is supported by Research 
Foundation of Heilongjiang Educational Committee (No. 12541667).


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\end{document}