\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 230, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/230\hfil Existence of infinitely many radial solutions]
{Existence of infinitely many radial solutions for quasilinear Schr\"odinger equations}

\author[G. Bao, Z.-Q. Han \hfil EJDE-2014/230\hfilneg]
{Gui Bao, Zhi-Qing Han}  % in alphabetical order

\address{Gui Bao \newline
 School of Mathematics and Statistics Science,
 Ludong University, Yantai, Shandong 264025,  China}
\email{baoguigui@163.com}

\address{Zhiqing Han \newline
 School of Mathematical Sciences, Dalian University of Technology,
Dalian 116024, China}
\email{hanzhiq@dlut.edu.cn}

\thanks{Submitted September 1, 2014. Published October 27, 2014.}
\subjclass[2000]{37J45, 58E05, 34C37,70H05}
\keywords{Quasilinear elliptic equations; variational methods; radial solutions}

\begin{abstract}
 In this article we prove the existence of radial solutions with arbitrarily
 many sign changes for quasilinear Schr\"odinger equation
 $$
 -\sum_{i,j=1}^{N}\partial_j(a_{ij}(u)\partial_iu)
 +\frac{1}{2}\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju+V(x)u
 =|u|^{p-1}u,~x\in\mathbb{R}^N,
 $$
 where $N\geq3$, $p\in(1,\frac{3N+2}{N-2})$. The proof is accomplished
 by using minimization under a constraint.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

We consider the  quasilinear elliptic problem
\begin{equation}\label{1.1}
-\sum_{i,j=1}^{N}\partial_j(a_{ij}(u)\partial_iu)
+\frac{1}{2}\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju+V(x)u
=|u|^{p-1}u, \quad x\in\mathbb{R}^N,
\end{equation}
where $N\geq3$, $1<p<2(2^*)-1=\frac{3N+2}{N-2}$, $2^*=\frac{2N}{N-2}$
is the critical Sobolev constant, $a_{ij}\in C^{1,\alpha}(\mathbb{R})$
is a symmetric matrix function, $\alpha\in(0,1)$ and
$a'_{ij}(u)=\frac{d}{du}a_{ij}(u)$.

For $a_{ij}(u)=(1+u^2)\delta_{ij}$, Equation \eqref{1.1} is reduced to the
well known  Modified Nonlinear Schr\"odinger Equation
\begin{equation}\label{1.2}
-\Delta{u}+V(x)u-\frac{1}{2}u\Delta(u^2)=|u|^{p-1}u,\quad x\in\mathbb{R}^N.
\end{equation}
This type of equations arise from the study of steady states and standing
wave solutions of time-dependent nonlinear Schr\"odinger equations,
and are derived as models in various branches of
mathematical physics; see \cite{BG,BE1,BE2,BL,HW,KI,KS,MF,RB}.

In the literature several papers have considered problem \eqref{1.2}.
For example, the existence of positive ground state solution of \eqref{1.2}
was proved by Poppenberg, Schmitt and Wang \cite{PS} by using a
constrained minimization argument. Liu et al \cite{LIU1}, by a change of variables,
transformed the quasilinear problem into a semilinear one, and used
an Orlicz space was the working space. The authors proved the existence of
soliton solutions of \eqref{1.2} for a Lagrange multiplier $\lambda>0$.
Colin and Jeanjean \cite{ML} also  used the change variables but work in
the Sobolev space $H^1(\mathbb{R}^N)$, they proved the existence of positive
solution for \eqref{1.2} with a Lagrange multiplier appears in the equation.
The same method of changing variables was also used recently to obtain
the existence of infinitely many solutions of problem \eqref{1.2} in \cite{FS}.
See also \cite{JM2} for the existence of positive solutions of problem \eqref{1.2}
for the case of critical growth.

The main mathematical difficulties with problem \eqref{1.2} are caused by
the term $\int_{\mathbb{R}^N}u^2|\nabla u|^2\,{\rm d}x$ which is not convex.
A further problem is caused by usual lack of compactness since these
problems are dealt with in the whole $\mathbb{R}^N$.

In this article, we consider a  general problem \eqref{1.1}. Under a certain
constraint, we prove that \eqref{1.1} possess infinitely many sign-changing
solutions for $p\in(1,\frac{3N+2}{N-2})$. As far as we know, besides \cite{LWW},
there are very few results for the existence of sign-changing solutions for
\eqref{1.1}.
However, we point out that in \cite{LWW}, solutions are founded in the case
$p\geq3$.


Throughout this article, we denote the positive constants (possibly different)
by $C, C_1, C_2, \dots$.
First we state the following assumptions.
\begin{itemize}
\item[(V1)] $V(x)\in C^{\alpha}(\mathbb{R^N})$ is a radially symmetric function
and satisfies
$$
0<V_0\leq V(x)\leq\lim_{|x|\to+\infty}V(x)=V_\infty<+\infty,\quad
  \forall x\in\mathbb{R}^N.
$$

\item[(V2)] The function $x\longmapsto x\cdot\nabla V(x)$ belongs to
 $L^{\infty}(\mathbb{R}^N)$ and $\|x\cdot\nabla V(x)\|_{\infty}\leq C_0<(p-1)V_0$.

\item[(V3)] The map $s\longmapsto s^{N+2}V(sx)$ is concave for any
$x\in\mathbb{R}^N$, $s\in\mathbb{R}$.

\item[(A1)] There exist constants $C_1>0,C_2>0$, such that for all
$\xi\in\mathbb{R}^N$ and $ s\in\mathbb{R}$,
 $$
C_1(1+s^2)|\xi|^2\leq\sum_{i,j=1}^N a_{ij}(s)\xi_i\xi_j\leq C_2(1+s^2)|\xi|^2.
$$

\item[(A2)] There exists constant $b>0$ such that for all $\xi\in\mathbb{R}^N$
and $s\in\mathbb{R}$ such that
$$
(b-2)\sum_{i,j=1}^Na_{ij}(s)\xi_i\xi_j\leq s\sum_{i,j=1}^Na'_{ij}(s)\xi_i\xi_j
\leq(p-1)\sum_{i,j=1}^Na_{ij}(s)\xi_i\xi_j-b|\xi|^2.
$$

\item[(A3)]  $|s|^{N-1}\sum_{i,j=1}^N(a_{ij}(s)
+\frac{1}{N}sa'_{ij}(s))\xi_i\xi_j$ is decreasing in $s\in(0,+\infty)$
and increasing in $s\in(-\infty,0)$.

\end{itemize}
Here is our main result.

\begin{theorem}\label{Th}
Assume {\rm (V1)--(V3), (A1)--(A3)}.
Then for any $k\in\{0,1,2,\dots\}$, there exists a pair of radial solutions
$u_k^{\pm}$ of \eqref{1.1} with the following properties:
\begin{itemize}
\item[(i)] $u_k^-(0)<0<u_k^+(0)$;
\item[(ii)] $u_k^{\pm}$ possess exactly $k$ nodes $r_l$ with
$0<r_1<r_2<\dots<r_k<+\infty$, and $u_k^{\pm}(x)|_{|x|=r_l}=0$, $l=1,2,\dots,k$.
\end{itemize}
\end{theorem}


We shall prove Theorem 1.1 under a convenient constraint, which is not
of Nehari-type; instead, we use a Pohozaev identity. This kind of argument
can be found in \cite{RD1}, see also \cite{AP,RD2,SJ} for different applications.
 Moreover, the main idea to prove Theorem 1.1 can be found in \cite{DENG},
see also \cite{BW,CZ}. However, since we deal with a more general case and
$p\in(1,\frac{3N+2}{N-2})$, there are more difficulties.

This article is organized as follows:
Section 2 is devoted to establish some preliminary results and
useful lemmas.
Theorem 1.1 will be proved in Section 3.

\section{Preliminary lemmas}

Set $H^1_r(\mathbb{R}^N)=\{u\in H^1(\mathbb{R}^N):u(x)=u(|x|)\}$,
and $X=\{u\in H^1_r(\mathbb{R}^N):
\int_{\mathbb{R}^N}|\nabla u|^2|u|^2\,{\rm d}x<+\infty\}$,
 where $H^1(\mathbb{R}^N)$ is the usual Sobolev space and
$\|u\|^2_{H^1}=\int_{\mathbb{R}^N}(|\nabla u|^2+V|u|^2)\,{\rm d}x$.
 $X$ is a complete metric space with distance:
$$
d_X(u,v)=\|u-v\|_{H^1}+\|\nabla u^2-\nabla v^2\|_{L^2}.
$$
Then, $u\in X$ is a weak solution of \eqref{1.1} if for all
$\phi\in C_0^{\infty}(\mathbb{R}^N)$,
\begin{equation}\label{e20}
\int_{\mathbb{R}^N}\Big(\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_j\phi
+\frac{1}{2}\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju\phi
+V(x)u\phi-|u|^{p-1}u\phi\Big)\,{\rm d}x=0.
\end{equation}
The corresponding functional is
\[
 I(u)=\frac{1}{2}\int_{\mathbb{R}^N}
\Big(\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_ju+V(x)u^2\Big)
\,{\rm d}x-\frac{1}{p+1}\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x.
\]
Given $u\in X$ and $\phi\in C_0^{\infty}(\mathbb{R}^N)$, the
 G\^ateaux derivative of $I$ in the direction $\phi$ at $u$,
denoted by $\langle I'(u),\phi\rangle$
is defined as $\lim_{t\to0^+}\frac{I(u+t\phi)-I(u)}{t}$.
It is easy to check that
\begin{align*}
&\langle I'(u),\phi\rangle\\
&=\int_{\mathbb{R}^N}
\Big(\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_j\phi
+\frac{1}{2}\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju\phi
+V(x)u\phi-|u|^{p-1}u\phi\Big)\,{\rm d}x.
\end{align*}
Hence, $u$ is a weak solution of problem \eqref{1.1} if this derivative
is zero in every direction $\phi\in C_0^{\infty}(\mathbb{R}^N)$.

 From \cite{PS}, we have the following two lemmas.

\begin{lemma}\label{l21}
For $N\geq2$, there is a constant $C=C(N)>0$ such that
\[
|u(x)|\leq C|x|^{\frac{1-N}{2}}\|u\|_{H^1},
\]
for any $|x|\geq1$ and $u\in H_r^1(\mathbb{R}^N)$.
\end{lemma}

\begin{lemma}
Let $\{u_n\}\subset H_r^1(\mathbb{R}^N)$ satisfy $u_n\rightharpoonup u$
in $H^1(\mathbb{R}^N)$. Then
\[
   \liminf_n\int_{\mathbb{R}^N}|\nabla u_n|^2|u_n|^2\,{\rm d}x
\geq\int_{\mathbb{R}^N}|\nabla u|^2|u|^2\,{\rm d}x.
\]
\end{lemma}

\begin{lemma}[\cite{St}] \label{imb}
Let $N\geq2$ and $2<q<2^*$. Then the imbedding
\[
    H_r^1(\mathbb{R}^N)\hookrightarrow L^q(\mathbb{R}^N)
\]
is compact.
\end{lemma}


\begin{lemma}(Br$\acute{e}$zis-Lieb lemma \cite{Bre})\label{BL} Let $\{u_n\}\subset L^q(\mathbb{R}^N)$ be a bounded sequence, where $1\leq q<+\infty$, such that $u_n\to u$ almost everywhere in $\mathbb{R}^N$. Then
\[
   \lim_{n\to+\infty}(|u_n|_q^q-|u_n-u|_q^q)=|u|_q^q.
\]
\end{lemma}

\begin{lemma}[\cite{LWW}]
 Let $u$ be a weak solution of \eqref{1.1}. Then $u$ and $\nabla u$ are bounded.
Moreover, $u$ satisfies the following exponential decay at infinity
\[
  |u(x)|\leq Ce^{-\delta R},\quad
|x|=R,\quad
\int_{\mathbb{R}^N\setminus B_R}(|\nabla u|^2+|u|^2)\,{\rm d}x\leq Ce^{-\delta R},
\]
for some positive constants $C,\delta$.
\end{lemma}

Let $\Omega$ be one of the following three types of domains:
\begin{equation}\label{e21}
\begin{gathered}
\{x\in\mathbb{R}^N| |x|<R_1\}, \\
\{x\in\mathbb{R}^N| 0<R_2\leq|x|<R_3<+\infty\},\\
\{ x\in\mathbb{R}^N| |x|\geq R_4>0\} .
\end{gathered}
\end{equation}
Set
\begin{gather*}
H_{0,r}^1(\Omega)=\{u\in H_0^1(\Omega)|u(x)=u(|x|)\},\\
X(\Omega)=\{u\in H^1_{0,r}(\Omega)|\int_{\Omega}
|\nabla u|^2u^2\,{\rm d}x<+\infty\}.
\end{gather*}
Now we consider the following equation on $\Omega$:
\begin{equation}\label{e22}
\begin{gathered}
   -\sum_{i,j=1}^{N}\partial_j(a_{ij}(u)\partial_iu)
+\frac{1}{2}\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju+V(x)u=|u|^{p-1}u,
\quad x\in\Omega,\\
   u|_{\partial\Omega}=0.
   \end{gathered}
\end{equation}
The corresponding functional is
\[
 I_{\Omega}(u)=\frac{1}{2}\int_{\Omega}
\Big(\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_ju+V(x)u^2\Big)\,{\rm d}x
-\frac{1}{p+1}\int_{\Omega}|u|^{p+1}\,{\rm d}x.
\]
Similarly we can define the G\^ateaux derivative of $I_{\Omega}$ at
 $u\in X(\Omega)$ and weak solution of problem \eqref{e22}.

We extend any $u\in X(\Omega)$ to $X$ by setting $u\equiv0$ on
$x\in\mathbb{R}^N\backslash{\Omega}$. Hereafter denote by $u_t$ the map:
\[
   \mathbb{R}^+\ni t\mapsto u_t\in X,~u_t(x)=tu(t^{-1}x),
\]
and consider
\begin{align*}
 f_u(t):=I(u_t)
    &=\frac{t^N}{2}\int_{\mathbb{R}^N}\sum_{i,j=1}^{N}a_{ij}(tu)
\partial_iu\partial_ju\,{\rm d}x\\
&\quad +\frac{t^{N+2}}{2}\int_{\mathbb{R}^N}V(tx)u^2\,{\rm d}x
-\frac{t^{N+p+1}}{p+1}\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x.
    \end{align*}
By  conditions (V1) and (A1), and the fact that $p+1>2$,
it is easy to see that $f_u(t)$ is positive for small $t$ and tends to
 $-\infty$ if $t\to+\infty$.
This implies that $f_u(t)$ attains its maximum. Moreover, thanks to (V2),
$f_u:\mathbb{R}^+\to\mathbb{R}$ is $C^1$, and
\begin{align*}
f'_u(t)&=\frac{N}{2}t^{N-1}\int_{\mathbb{R}^N}\sum_{i,j=1}^{N}a_{ij}(tu)
\partial_iu\partial_ju\,{\rm d}x
+\frac{t^N}{2}\int_{\mathbb{R}^N}u\sum_{i,j=1}^{N}a'_{ij}(tu)\partial_iu\partial_ju\,{\rm d}x\\
&\quad +\frac{{N+2}}{2}t^{N+1}\int_{\mathbb{R}^N}V(tx)u^2\,{\rm d}x
 +\frac{t^{N+2}}{2}\int_{\mathbb{R}^N}\nabla V(tx)\cdot xu^2\,{\rm d}x\\
&\quad -\frac{N+p+1}{p+1}t^{N+p}\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x.
\end{align*}
Let
$$
M(\Omega)=\{u\in X(\Omega)\setminus\{0\}:J_{\Omega}(u)=0\},
$$
where $J_{\Omega}:X(\Omega)\to\mathbb{R}$ is defined as
\begin{align*}
J_{\Omega}(u)
&= \frac{N}{2}\int_{\Omega}\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_ju\,{\rm d}x
 +\frac{1}{2}\int_{\Omega}u\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju\,{\rm d}x\\
&\quad +\frac{{N+2}}{2}\int_{\Omega}V(x)u^2\,{\rm d}x+\frac{1}{2}\int_{\Omega}
\nabla V(x)\cdot xu^2\,{\rm d}x-\frac{N+p+1}{p+1}\int_{\Omega}|u|^{p+1}\,{\rm d}x.
\end{align*}
In other words, $M(\Omega)$ is the set of functions $u\in X(\Omega)$
such that $f'_u(1)=0$. Moreover, $M(\Omega)\neq\emptyset$
(actually, given any $u\neq0$, there exists $t>0$ such that $u_t\in M(\Omega)$
(cf. \cite{RD1})).

In the appendix of \cite{LWW}, by using Moser and De Giorgi iterations,
the authors proved that weak solutions of \eqref{1.1} are bounded
in $L^{\infty}(\mathbb{R}^N)$. Their arguments work also for $p\in(1,3)$.
A density argument show that weak formulation \eqref{e20} holds also for
test functions in $H^1(\mathbb{R}^N)\cap L^{\infty}(\mathbb{R}^N)$.
By \cite[theorems 5.2 and 6.2 in chapter 4]{LU} it follows that
$u\in C^{1,\alpha}$. From Schauder theory we conclude that $u\in C^{2,\alpha}$
 is a classical solution of \eqref{1.1}. Moreover, if $u\in X$ is a solution,
$u,Du,D^2u$ have an exponential decay as $|x|\to+\infty$
(see \cite{LWW}). By \cite{PPS}, assume that $u\in X$ is a $C^2$ solution
of \eqref{1.1}.
Then, for all $a\in\mathbb{R}$, we have the  identity
\begin{equation} \label{po}
\begin{aligned}
&(\frac{N-2}{2}-a)\int_{\mathbb{R}^N}\sum_{i,j=1}^Na_{ij}(u)\partial_iu\partial_ju\,{\rm d}x-\frac{a}{2}\int_{\mathbb{R}^N}u\sum_{i,j=1}^Na'_{ij}(u)\partial_iu\partial_ju\,{\rm d}x \\
&+(\frac{N}{2}-a)\int_{\mathbb{R}^N}V(x)u^2\,{\rm d}x+\frac{1}{2}\int_{\mathbb{R}^N}\nabla V(x)\cdot xu^2\,{\rm d}x \\
&+(a-\frac{N}{p+1})\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x=0.
\end{aligned}
\end{equation}
Observe also that $M(\Omega)$ is nothing but the set of functions
 $u\in X(\Omega)$ such
that the identity \eqref{po} holds for $a =-1$. Then, all solutions belong
to $M(\Omega)$.


\begin{lemma}\label{inf}
For any $u\in X(\Omega)$, the map $f_u$ attains its maximum at exactly one
point $t^u$. Moreover, $f_u$ is positive and increasing for $t\in[0,t^u]$
and decreasing for $t>t^u$. Also,
\[
   c:=\inf_{M(\Omega)}I_\Omega=\inf_{u\in X(\Omega),u\neq0}\max_{t>0}I(u_t).
\]
\end{lemma}

\begin{proof}
We employ a similar argument as in \cite[Lemma 3.1]{RD1}. Set
\[
    g(t)=\frac{t^N}{2}\int_{\mathbb{R}^N}\sum_{i,j=1}^{N}a_{ij}(tu)
\partial_iu\partial_ju\,{\rm d}x-\frac{t^{N+p+1}}{p+1}\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x.
\]
Let $t_1\in\mathbb{R}^+$, $t_2\in\mathbb{R}^+$, $t_1\neq t_2$, then we have
\begin{align*}
g'(t_1)-g'(t_2)
&=\frac{N}{2}\int_{\mathbb{R}^N}t^{N-1}_1
 \Big(\sum_{i,j=1}^{N}a_{ij}(t_1u)+\frac{1}{N}t_1u\sum_{i,j=1}^{N}a'_{ij}(t_1u)\Big)
 \partial_iu\partial_ju \,{\rm d}x \\
&\quad -\frac{N}{2}\int_{\mathbb{R}^N}t_2^{N-1}\Big(\sum_{i,j=1}^{N}a_{ij}(t_2u)+\frac{1}{N}t_2u\sum_{i,j=1}^{N}a'_{ij}(t_2u)\Big)\partial_iu\partial_ju\,{\rm d}x \\
&\quad -\frac{N+p+1}{p+1}(t_1^{N+p}-t_2^{N+p})\int_{\mathbb{R}^N}|u|^{p+1}\,{\rm d}x.
\end{align*}
By using (A3) we obtain
\[
(g'(t_1)-g'(t_2))(t_1-t_2)\leq0.
\]
This implies that $g(t)$ is a concave function. Then by assumption (V3),
 \begin{align*}
    f_u(t)=g(t)+\frac{t^{N+2}}{2}\int_{\mathbb{R}^N}V(tx)u^2\,{\rm d}x
\end{align*}
is a concave function. We already know that it attains its maximum. 
Let $t^u$ be the unique point at which this maximum is achieved. 
Then $t^u$ is the unique critical point of $f_u$ and $f_u$ is positive and 
increasing for $0<t<t^u$ and decreasing for $t>t^u$.

In particular, for any $u\in X(\Omega)\setminus\{0\}$, $t^u\in\mathbb{R}$ 
is the unique value such that $u_{t^u}$ belongs to $M(\Omega)$, 
and $I(u_t)$ reaches a global maximum for $t=t^u$. 
\end{proof}

Similar to \cite[Proposition 3.3]{RD1}, we can prove the coercivity 
of $I_{\Omega}\mid_{M(\Omega)}$.

\begin{proposition}\label{pro1}
There exists $C>0$ such that for any $u\in M(\Omega)$,
\[
I_{\Omega}(u)\geq C\int_{\Omega}(u^2+|\nabla u|^2+u^2|\nabla u|^2)\,{\rm d}x.
\]
\end{proposition}

\begin{proof}
 Take $u\in M(\Omega)$ and extend $u$ to $X$ by setting $u\equiv0$ on 
$\mathbb{R}^N\setminus\Omega$. Choose $t\in(0,1)$, then
\begin{align*}
I(u_t)-t^{N+p+1}I(u)
&=\int_{\mathbb{R}^N}\sum_{i,j=1}^{N}\Big(\frac{t^N}{2}a_{ij}(tu)
 \partial_iu\partial_ju-\frac{t^{N+p+1}}{2}a_{ij}(u)\partial_iu\partial_ju\Big)\,{\rm d}x\\
&\quad +\int_{\mathbb{R}^N}\Big(\frac{t^{N+2}}{2}V(tx)-\frac{t^{N+p+1}}{2}
V(x)\Big)u^2\,{\rm d}x.
\end{align*}
Observe that $V(tx)\geq V_0\geq\delta V_{\infty}\geq\delta V(x)$, 
for some positive $\delta\in(0,1)$ depending only on $V_0$ and $V_{\infty}$. 
By choosing a smaller $t$, if necessary, we obtain
\[
    \frac{t^{N+2}}{2}V(tx)-\frac{t^{N+p+1}}{2}V(x)
\geq\Big(\delta\frac{t^{N+2}}{2}-\frac{t^{N+p+1}}{2}\Big)V(x)\geq\gamma_0,
\]
for a fixed constant $\gamma_0>0$. Since $u\in M(\Omega)$, from Lemma \ref{inf} 
we obtain that $I(u_t)\leq I(u)$. By choosing 
$t\in\left(0,(\frac{C_1}{C_2})^{\frac{1}{p-1}}\right)$ small enough, 
from (A1) we have
\begin{align*}
&(1-t^{N+p+1})I(u)\\
&\geq I(u_t)-t^{N+p+1}I(u)\\
&\geq \int_{\mathbb{R}^N}\Big(\frac{t^N}{2}C_1(1+(tu)^2)|\nabla u|^2
 -\frac{t^{N+p+1}}{2}C_2(1+u^2)|\nabla u|^2\Big)\,{\rm d}x
 +\gamma_0\int_{\mathbb{R}^N}u^2\,{\rm d}x\\
&= \frac{t^N}{2}\int_{\mathbb{R}^N}\Big((C_1-t^{p+1}C_2)+\big(C_1-C_2t^{p-1}\big)
 (tu)^2\Big)|\nabla u|^2\,{\rm d}x+\gamma_0\int_{\mathbb{R}^N}u^2\,{\rm d}x\\
&\geq \frac{t^N}{2}(C_1-C_2t^{p-1})\int_{\mathbb{R}^N}(1+t^2u^2)|\nabla u|^2
 \,{\rm d}x+\gamma_0\int_{\mathbb{R}^N}u^2\,{\rm d}x\\
&\geq \frac{t^{N+2}}{2}(C_1-C_2t^{p-1})\int_{\mathbb{R}^N}(1+u^2)|\nabla u|^2
 \,{\rm d}x+\gamma_0\int_{\mathbb{R}^N}u^2\,{\rm d}x.
\end{align*}
Note that $u\equiv0$ on $\mathbb{R}^N\setminus\Omega$, we conclude by defining 
\[
C=\min\big\{\frac{C_1 t^{N+2}-C_2t^{N+p+1}}{2 (1-t^{N+p+1})}
 \frac{\gamma_0}{1-t^{N+p+1}}\big\}.
\]
\end{proof}

\begin{lemma}\label{le27}
Suppose that the domain $\Omega$ is one of the forms of \eqref{e21}. 
Then $c=\inf_{M(\Omega)}I_{\Omega}(u)$ can be achieved by some positive 
function $u$ which is a solution of problem \eqref{e22}.
 Moreover, $\int_{\Omega}u^2|\nabla\phi|^2\,{\rm d}x<+\infty$, 
$\int_{\Omega}\phi^2|\nabla u|^2\,{\rm d}x<+\infty$.
\end{lemma}

\begin{proof}
 We divide the proof into three steps.

\noindent\textbf{Step 1.} $c$ is attained.
By the definition of $c$, there exists a sequence $\{u_n\}\subset M(\Omega)$ 
such that
\[
    I_{\Omega}(u_n)=c+o(1), \quad J_{\Omega}(u_n)=0.
\]
By Proposition \ref{pro1}, $\{u_n\}$ is bounded in $X(\Omega)$.
 Hence, by Lemma \ref{imb}, we can extract a subsequence of $\{u_n\}$ 
(still denoted by $\{u_n\}$), such that
\begin{gather*}
  u_n\rightharpoonup u  \quad\text{in } X(\Omega),\\
  u_n\to u  \quad \text{in } L^q(\Omega), \; 2<q<2^*.
\end{gather*}
Since $\nabla(u_n)^2$ is uniformly bounded in $L^2(\Omega)$, 
by Sobolev's inequality we have $|u_n^2|_{2^*}\leq C$,
 which gives $|u_n|_{22^*}\leq C$. By H\"older's inequality we have
\[
    u_n\to u \quad \text{in  }L^q(\Omega),\; 2<q<22^*.
\]
Taking the limit in $n$, it follows from $J_{\Omega}(u_n)=0$ that
\begin{align*}
J_{\Omega}(u)
&=\frac{N}{2}\int_{\Omega}\sum_{i,j=1}^{N}a_{ij}(u)\partial_iu\partial_ju\,{\rm d}x
 +\frac{1}{2}\int_{\Omega}u\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju\,{\rm d}x\\
&\quad +\frac{{N+2}}{2}\int_{\Omega}V(x)u^2\,{\rm d}x
 +\frac{1}{2}\int_{\Omega}\nabla V(x)\cdot xu^2\,{\rm d}x
 -\frac{N+p+1}{p+1}\int_{\Omega}|u|^{p+1}\,{\rm d}x\\
&\leq 0.
\end{align*}
By Lemma \ref{inf}, there exists $t>0$ such that $J_{\Omega}(u_t)=0$. 
Extend $u_n$ and $u$ to $X$ by setting $u_n\equiv0$ and $u\equiv0$ on 
$\mathbb{R}^N\setminus\Omega$.
In the following we just need to recall the expression of $I((u_n)_t)$,
\[
  c=\lim_{n\to+\infty}I_{\Omega}(u_n)
=\lim_{n\to+\infty}I(u_n)\geq\lim_{n\to+\infty}\inf I((u_n)_t)
\geq I(u_t),\quad \forall t>0.
\]
So $\max_t I(u_t)=c$. Then, by Lemma \ref{inf}, there exists $t_0>0$ 
such that $u_{t_0}\in M(\Omega)$, which implies that $c$ is attained.
\smallskip

\noindent\textbf{Step 2.} $u$ is a radial solution of \eqref{e22}.
We use an indirect argument which is based on a general idea used 
in \cite{LWW}. Suppose that $u\in M(\Omega)$, $I_{\Omega}(u)=c$ but 
$I'_{\Omega}(u)\neq0$. In such a case, we can find a function 
$\phi\in X(\Omega)$ with the property that 
$\int_{\Omega}u^2|\nabla\phi|^2\,{\rm d}x<+\infty$, 
$\int_{\Omega}\phi^2|\nabla u|^2\,{\rm d}x<+\infty$ but
\[
   \langle I'_{\Omega}(u),\phi\rangle\leq-1.
\]
Extend $u\in X(\Omega)$ to $X$ as above and choose $\varepsilon>0$ small 
enough such that
\[
   \langle I'(u_t+\sigma\phi),\phi\rangle\leq-\frac{1}{2},\quad 
\forall |t-1|+|\sigma|\leq\varepsilon.
\]
Let $\eta$ be a cut-off function, 
\[
    \eta(t)=\begin{cases}
    1, &|t-1|\leq\frac{1}{2}\varepsilon,\\
    0, &|t-1|\geq\varepsilon.
    \end{cases}
\]
Define
\[
    \gamma(t)=\begin{cases}
    u_t, &|t-1|\geq\varepsilon,\\
    u_t+\varepsilon\eta(t)\phi, &|t-1|<\varepsilon.
    \end{cases}
\]
Next we estimate $\sup_tI(\gamma(t))$. If $|t-1|\leq\varepsilon$, then
\begin{equation}\label{e23}
\begin{split}
  I(\gamma(t))
&=I(u_t+\varepsilon\eta(t)\phi)\\
&=I(u_t)+\int_0^1\langle I'(u_t+\sigma\varepsilon\eta(t)\phi),
\varepsilon\eta(t)\phi\rangle\,{\rm d}\sigma\\
& \leq I(u_t)-\frac{1}{2}\varepsilon\eta(t).
\end{split}
\end{equation}
If $|t-1|\geq\varepsilon$, then $\eta(t)=0$, and the above estimate is trivial. 
Now since $u\in M(\Omega)$, for $t\neq1$ we get $I(u_t)<I(u)$. 
Hence it follows from \eqref{e23} that
\begin{equation}\label{e24}
    I(u_t+\varepsilon\eta(t)\phi)\leq\begin{cases}
    I(u_t)<I(u), &t\neq1,\\
    I(u)-\frac{1}{2}\varepsilon\eta(1)=I(u)-\frac{1}{2}\varepsilon, &t=1.
    \end{cases}
\end{equation}
In any case we have $I(\gamma(t))<I(u)=c$.

To conclude observe that $J(\gamma(1-\varepsilon))>0$ and $J(\gamma(1+\varepsilon))<0$. 
As a result, we can find $t_0\in(1-\varepsilon,1+\varepsilon)$ such that $J(\gamma(t_0))=0$, 
which implies that $\gamma(t_0)=u_{t_0}+\varepsilon\eta(t_0)\phi\in M(\Omega)$. 
However, it follows from \eqref{e24} that $I_{\Omega}(\gamma(t_0))<c$. 
This is a contradiction.
\smallskip

\noindent\textbf{Step 3.} $u>0$.
Consider $u\in M(\Omega)$ a minimizer of $I_{\Omega}|_{M(\Omega)}$. 
Then the absolute value $|u|\in M(\Omega)$ is also a minimizer. 
By the classical maximum principle and the fact that solutions are $C^2$, $|u|>0$.
\end{proof}

\section{Proof of Theorem \ref{Th}}
For given $k+2$ numbers $r_l$ $(l=0,1,\dots,k+1)$ such that 
$0=r_0<r_1<\dots<r_k<r_{k+1}=+\infty$, denote
\[
   \Omega^1=\{x\in\mathbb{R}^N:|x|<r_1\},\quad
   \Omega^l=\{x\in\mathbb{R}^N:r_{l-1}<|x|<r_{l}\}.
\]
We will always extend $u_l\in X(\Omega^l)$ to $X$ by setting $u\equiv0$ 
on $x\in\mathbb{R}^N\backslash{\Omega^l}$ for every 
$u_l\in X(\Omega^l),l=1,2,\dots,k+1$. In this sense, we use $I(u_l)$ 
to replace $I_{\Omega^l}(u_l)$ and $J(u_l)$ to replace $J_{\Omega^l}(u_l)$.
Define
\begin{align*}
Y_k^\pm(r_1,r_2,\dots,r_{k+1})
=\Big\{&u\in X:u=\pm\sum_{l=1}^{k+1}(-1)^{l-1}u_l,\;
   u_l\geq0,\\
&u_l\not\equiv0,u_l\in X(\Omega^l),\; l=1,2,\dots,k+1\Big\},
 \end{align*}
\begin{align*}
    M_k^{\pm}=\Big\{&u\in X:\exists0<r_1<r_2<\dots<r_k<r_{k+1}=+\infty,
\text{\ such  that }\\
    & u\in Y_k^\pm(r_1,r_2,\dots,r_k,r_{k+1})\text{ and }
u_l\in M(\Omega^l),l=1,2,\dots,k+1\Big\}.
\end{align*}
Note that $M_k^\pm\neq\emptyset$, $k=1,2,\dots$. In the following we  
always refer to $M_k$ and we  drop the $"+"$. For $M_k^-$, everything 
could be done exactly in the same way. 
By Lemma \ref{inf}, it is easy to verify that for all $u$,
\begin{equation}\label{e31}
u=\sum_{l=1}^{k+1}(-1)^{l-1}u_l\in M_k\Leftrightarrow I(u)
=\max_{\substack{\alpha_l>0\\ 1\leq l\leq k+1}}
I\Big(\sum_{l=1}^{k+1}(-1)^{l-1}(u_l)_{\alpha_l}\Big).
\end{equation}
Set
\[
c_k=\inf_{M_k}I(u),\quad ~k=1,2,\dots.
\]

\begin{lemma}\label{le31}
$c_k$ is attained, $k=0,1,2,\dots$.
\end{lemma}

\begin{proof} 
By induction we prove that for each $k$ there exists $u_k\in M_k$ such that
\[
 I(u_k)=c_k.
\]
The case that $k=0$ can be deduced by setting $\Omega=\mathbb{R}^N$ 
in Lemma \ref{le27}. We suppose the claim is true for $k-1$ and discuss 
the case $k\geq1$ in the following. For convenience, we divide the proof 
of the rest proof into four steps.
\smallskip

\noindent\textbf{Step 1.}
 $I$ is bounded from below on $M_k$ by a positive constant.
Since
\[
I(u)=I\Big(\sum_{l=1}^{k+1}(-1)^{l-1}u_l\Big)
=\sum_{l=1}^{k+1}I_{\Omega^l}(u_l), \quad \forall u\in M_k.
\]
We just need to prove that, for $l=1,2,\dots,k+1$, $I_{\Omega^l}$ 
is bounded from below on $M(\Omega^l)$ by a positive constant.

For any $u_l\in M(\Omega^l)$, we extend it to $X$ by setting $u_l\equiv0$ 
on $\mathbb{R}^N\setminus\Omega^l$. By (V1) and (A1) we have
\[
I(u_l)\geq\frac{1}{2}\int_{\mathbb{R}^N}(C_1(1+u_l^2)|\nabla u_l|^2
+V_0u_l^2)\,{\rm d}x-\frac{1}{p+1}\int_{\mathbb{R}^N}|u_l|^{p+1}\,{\rm d}x.
\]
Let
\[
   \bar{I}(u_l) =\frac{1}{2}\int_{\mathbb{R}^N}(C_1(1+u_l^2)|\nabla u_l|^2
+V_0u_l^2)\,{\rm d}x-\frac{1}{p+1}\int_{\mathbb{R}^N}|u_l|^{p+1}\,{\rm d}x.
\]
Obviously,
\[
    \bar{c}:=\inf_{u_l\in X(\Omega^l),u_l\neq0}\max_{t>0}\bar{I}((u_l)_t)
\leq\inf_{u_l\in X(\Omega^l),u_l\neq0}\max_{t>0}I((u_l)_t)=c.
\]
Let us define
\[
    \bar{M}(\Omega^l)=\{u_l\in X(\Omega^l)\setminus\{0\}:g'_{u_l}(1)=0\} \quad 
\text{ where }g_{u_l}(t)=\bar{I}((u_l)_t).
\]
Similar to Lemma \ref{inf}, we know that
\[
   \bar{c}=\inf_{u_l\in\bar{M}(\Omega^l)}\bar{I}_{\Omega^l}(u_l).
\]
For any $u_l\in\bar{M}(\Omega^l)$,
\begin{align*}
 &\frac{N+2}{2}V_0\int_{\Omega^l}u_l^2\,{\rm d}x 
 + \frac{C_1(N+2)}{2}\int_{\Omega^l}|\nabla u_l|^2u_l^2\,{\rm d}x\\
&\leq \frac{N+p+1}{p+1}\int_{\Omega^l}|u_l|^{p+1}\,{\rm d}x\\
&\leq \frac{N+2}{2}V_0\int_{\Omega^l}u_l^2\,{\rm d}x
 +C\int_{\Omega^l}|u_l|^{\frac{4N}{N+2}}\,{\rm d}x,
\end{align*}
for a suitable constant $C>0$. So, by using the Sobolev's inequality,
\[
  \frac{C_1(N+2)}{2}\int_{\Omega^l}|\nabla u_l|^2u_l^2\,{\rm d}x
\leq C\int_{\Omega^l}|u_l|^{\frac{4N}{N+2}}\,{\rm d}x
\leq C'\Big(\int_{\Omega^l}|\nabla u_l|^2u_l^2\,{\rm d}x\Big)^{\frac{N}{N-2}},
\]
this shows that $\int_{\Omega^l}|\nabla u_l|^2u_l^2\,{\rm d}x$ is bounded away 
from zero on $\bar{M}(\Omega^l)$. Since the functional $\bar{I}_{\Omega^l}$ 
restricted to $\bar{M}(\Omega^l)$ has the expression
\begin{align*}
\bar{I}_{\Omega^l}(u_l)
&=\frac{C_1}{2}\frac{p+1}{N+p+1}\int_{\Omega^l}|\nabla u_l|^2\,{\rm d}x
   +\frac{V_0}{2}\frac{p-1}{N+p+1}\int_{\Omega^l}u_l^2\,{\rm d}x\\
&\quad +\frac{C_1(p-1)}{N+p+1}\int_{\Omega^l}|\nabla u_l|^2|u_l|^{2}\,{\rm d}x.
\end{align*}
We obtain that $\bar{c}>0$, and hence $c>0$. This implies that 
$d_X(M(\Omega^l),0)>0$. Then by Proposition \ref{pro1}, we get that 
$I_{\Omega^l}$ is bounded from below on $M(\Omega^l)$ by a positive constant.
\smallskip

\noindent\textbf{Step 2.} 
We suppose $\{u_m\}_{m\geq1}$ be a minimizing sequence of $c_k$ in $M_k$; that is
\[
  \lim_{m\to+\infty}I(u_m)=c_k,\quad u_m\in M_k,\quad  m=1,2,\dots.
\]
${u_m}$ corresponds to $k$ nodes, $r_m^1,r_m^2,\dots,r_m^k$ with 
$0<r_m^1<r_m^2<\dots<r_m^k<+\infty$. By Proposition \ref{pro1},
 we know that $\{u_m\}$ is bounded in $X$. Set
\begin{gather*}
   \Omega_m^l=\{x\in\mathbb{R}^N : r_m^{l-1}<|x|<r_m^l\},\\
 u_m^l=\begin{cases}
   u_m, & x\in \Omega_m^l,\\
   0,   & x\not\in \Omega_m^l.
   \end{cases}
\end{gather*}
By selecting a subsequence, we may assume that $\lim_{m\to+\infty}r_m^l=r^l$, 
and clearly $0\leq r^1\leq r^2\leq\dots\leq r^k\leq+\infty$.

Next we  prove that $r^l\neq r^{l-1}, l=1,2,\dots,k$. Here we denote $r^0=0$. 
If there exists some $l\in\{1,2,\dots,k\}$ such that $r^l=r^{l-1}$, then 
$\lim_{m\to+\infty}r_m^l=\lim_{m\to+\infty}r_m^{l-1}$. 
We denote the measure of $\Omega_m^l$ by $|\Omega_m^l|$, so that 
$|\Omega_m^l|\to0$ as $m\to+\infty$. From (A1) and the fact that $\{u_m\}$ 
is bounded in $X$, we have
\begin{equation} \label{31'}
\begin{aligned}
I(u_m^l)
&= \frac{1}{2}\int_{\Omega_m^l}\Big(\sum_{i,j=1}^{N}a_{ij}(u_m^l)
 \partial_iu_m^l\partial_ju_m^l+V(u_m^l)^2\Big)\,{\rm d}x
 -\frac{1}{p+1}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x \\
&\leq \frac{1}{2}\int_{\Omega_m^l}\Big(C_2(1+(u_m^l)^2)|\nabla u_m^l|^2
 +V_{\infty}(u_m^l)^2\Big)\,{\rm d}x
 -\frac{1}{p+1}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x \\
&\leq C-\frac{1}{p+1}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x.
\end{aligned}
\end{equation}
By using H\"older's inequality,
\[
\int_{\Omega_m^l}|u_m^l|^2\,{\rm d}x
\leq\Big(\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x\Big)^{\frac{2}{p+1}}
|\Omega_m^l|^{1-\frac{2}{p+1}},
\]
i.e.,
\begin{equation}\label{32}
\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x
\geq\Big(\int_{\Omega_m^l}|u_m^l|^2\,{\rm d}x\Big)^{\frac{p+1}{2}}
|\Omega_m^l|^{\frac{1-p}{2}}.
\end{equation}
Since $u_m^l\in M_k$,
\begin{equation} \label{3.4}
\begin{aligned}
&\frac{N+p+1}{p+1}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x \\
&=\frac{N}{2}\int_{\Omega_m^l}\sum_{i,j=1}^Na_{ij}(u_m^l)
 \partial_iu_m^l\partial_ju_m^l\,{\rm d}x
 +\frac{1}{2}\int_{\Omega_m^l}u_m^l\sum_{i,j=1}^Na'_{ij}(u_m^l)
 \partial_iu_m^l\partial_ju_m^l\,{\rm d}x \\
&\quad +\frac{N+2}{2}\int_{\Omega_m^l}V(x)(u_m^l)^2\,{\rm d}x
+\frac{1}{2}\int_{\Omega_m^l}\nabla V(x)\cdot x(u_m^l)^2\,{\rm d}x \\
&\geq \frac{C_1(N+b-2)}{2}\int_{\Omega_m^l}(1+(u_m^l)^2)|\nabla u_m^l|^2\,{\rm d}x
 +\frac{N+2}{2}V_0\int_{\Omega_m^l}(u_m^l)^2\,{\rm d}x \\
&\quad -\frac{1}{2}C_0\int_{\Omega_m^l}(u_m^l)^2\,{\rm d}x \\
&\geq \frac{C_1(N+b-2)}{2}\int_{\Omega_m^l}(u_m^l)^2|\nabla u_m^l|^2\,{\rm d}x
 +\frac{(N+3-p)V_0}{2}\int_{\Omega_m^l}(u_m^l)^2\,{\rm d}x.
\end{aligned}
\end{equation}
On the other hand,
\[
\frac{N+p+1}{p+1}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x
\leq \frac{(N+3-p)V_0}{2}\int_{\Omega_m^l}|u_m^l|^{2}\,{\rm d}x
 +C\int_{\Omega_m^l}|u_m^l|^{\frac{4N}{N+2}}\,{\rm d}x,
\]
for a suitable $C>0$. So by using Sobolev's inequality
\begin{align*}
\frac{C_1(N+b-2)}{2}\int_{\Omega_m^l}|u_m^l|^{2}|\nabla u_m^l|^2\,{\rm d}x
&\leq C\int_{\Omega_m^l}|u_m^l|^{\frac{4N}{N+2}}\,{\rm d}x\\
&\leq C'\int_{\Omega_m^l}|u_m^l|^{2}|\nabla u_m^l|^2\,{\rm d}x.
\end{align*}
This shows that $\int_{\Omega_m^l}|u_m^l|^{2}|\nabla u_m^l|^2\,{\rm d}x$
is bounded away from zero on $M_k$.  This implies that
\[
 \int_{\Omega_m^l}|u_m^l|^2\,{\rm d}x\geq\delta>0.
\]
Then from \eqref{32} we obtain
\[
\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x
\geq\Big(\int_{\Omega_m^l}|u_m^l|^2\,{\rm d}x\Big)^{\frac{p+1}{2}}|\Omega_m^l
 |^{\frac{1-p}{2}}
\geq \delta^{\frac{p+1}{2}}|\Omega_m^l|^{\frac{1-p}{2}} .
\]
Note that $|\Omega_m^l|\to0$ as $m\to+\infty$ and $p>1$, we have
$$
\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x\to+\infty, \quad {\rm as}~m\to+\infty.
$$
This and \eqref{31'} implies that $I(u_m^l)\to-\infty$ as $m\to+\infty$,
which contradicts Step 1. Thus $r^l\neq r^{l-1}, l=1,2,\dots,k$.
\smallskip


\noindent\textbf{Step 3.} 
$r^k<+\infty$.
If $r^k=+\infty$, then $\lim_{m\to+\infty}r_m^k=+\infty$. 
Since $u_m^k\in M(\Omega_m^k)$, from (V1), (V2), (A1) and (A2) we have
\begin{equation} \label{e32}
\begin{aligned}
&I(u_m^k) \\
&= \frac{1}{2}\int_{\Omega_m^k}\Big(\sum_{i,j=1}^{N}a_{ij}(u_m^k)
 \partial_iu_m^k\partial_ju_m^k+V(x)(u_m^k)^2\Big)\,{\rm d}x
 -\frac{1}{p+1}\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x \\
&= \frac{1}{2}\int_{\Omega_m^k}\Big(\sum_{i,j=1}^{N}a_{ij}(u_m^k)
 \partial_i u_m^k\partial_j u_m^k+V(x)(u_m^k)^2\Big)\,{\rm d}x \\
&\quad -\frac{1}{N+p+1}\Big(\frac{N}{2}\int_{\Omega_m^k}
 \sum_{i,j=1}^{N}a_{ij}(u_m^k)\partial_i u_m^k\partial_j u_m^k\,{\rm d}x \\
&\quad +\frac{1}{2}\int_{\Omega_m^k}u_m^k\sum_{i,j=1}^{N}a'_{ij}
 (u_m^k)\partial_iu_m^k\partial_ju_m^k\,{\rm d}x+\frac{N+2}{2}
 \int_{\Omega_m^k}V(x)(u_m^k)^2\,{\rm d}x \\
&\quad +\frac{1}{2}\int_{\Omega_m^k}\nabla V(x)\cdot x(u_m^k)^2\,{\rm d}x\Big) \\
&\geq \Big(\frac{1}{2}-\frac{N}{2(N+p+1)}-\frac{p-1}{2(N+p+1)}\Big)
 \int_{\Omega_m^k}\sum_{i,j=1}^{N}a_{ij}(u_m^k)\partial_iu_m^k\partial_ju_m^k\,{\rm d}x \\
&\quad +\Big(\Big(\frac{1}{2}-\frac{N+2}{2(N+p+1)}\Big)V_0-\frac{C_0}{2(N+p+1)}\Big)
 \int_{\Omega_m^k}(u_m^k)^2\,{\rm d}x \\
&\quad +\frac{b}{2(N+p+1)}\int_{\Omega_m^k}|\nabla u_m^k|^2\,{\rm d}x \\
&\geq \frac{1}{N+p+1}\int_{\Omega_m^k}C_1\big(1+(u_m^k)^2\big)
 |\nabla u_m^k|^2\,{\rm d}x+\frac{(p-1)V_0-C_0}{2(N+p+1)}
 \int_{\Omega_m^k}(u_m^k)^2\,{\rm d}x \\
&\quad +\frac{b}{2(N+p+1)}\int_{\Omega_m^k}|\nabla u_m^k|^2\,{\rm d}x \\
&\geq  C\eta^2(u_m^k),
\end{aligned}
\end{equation}
where
\[
\eta^2(u_m^k)=\int_{\Omega_m^k}\big(1+(u_m^k)^2\big)|\nabla u_m^k|^2\,{\rm d}x
+\int_{\Omega_m^k}(u_m^k)^2\,{\rm d}x.
\]
From Step 1 we know that
$\int_{\Omega_m^k}|u_m^k|^{2}|\nabla u_m^k|^2\,{\rm d}x$ is bounded away from
zero on $M(\Omega_m^k)$.
Then there exists some $\delta_0>0$ such that
\[
\int_{\Omega_m^k}|u_m^k|^2\,{\rm d}x\geq\delta_0>0.
\]
This and \eqref{3.4} imply that there exists some $\delta_1>0$ such that
\[
\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x\geq\delta_1>0.
\]
Then from \eqref{31'}, we have
\begin{equation}\label{e35'}
\begin{aligned}
I(u_m^k)
&\leq  C-\frac{1}{p+1}\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x \\
&\leq  C+C\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x \\
&\leq C\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x
\Big(\Big(\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x\Big)^{-1}+1\Big) \\
&\leq C\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x(\delta_1^{-1}+1) \\
&\leq C\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x,
\end{aligned}
\end{equation}
for some suitable $C>0$.
It follows from \eqref{e32},\eqref{e35'} and Lemma \ref{l21} that
 \begin{align*}
\eta^2(u_m^k)&\leq I(u_m^k)\\
&\leq C\int_{\Omega_m^k}|u_m^k|^{p+1}\,{\rm d}x\\
&\leq C\int_{\Omega_m^k}|u_m^k|^2|u_m^k|^{p-1}\,{\rm d}x\\
&\leq C \|u_m^k\|^{p-1}\int_{\Omega_m^k}|u_m^k|^2|x|^{\frac{(1-N)(p-1)}{2}}\,{\rm d}x\\
&\leq C \Big(\eta^2(u_m^k)\Big)^{\frac{p+1}{2}}|r_m^k|^{\frac{(1-N)(p-1)}{2}}.
\end{align*}
Thus
\begin{equation}\label{e36}
  \eta^2(u_m^k)\geq C|r_m^k|^{N-1}.
\end{equation}
From \eqref{e36} we have
\[
 \eta^2(u_m^k)\to+\infty\quad \text{as }m\to+\infty.
\]
So \eqref{e32} implies
\begin{equation}\label{e35}
    I(u_m^k)\to+\infty \quad \text{as }m\to+\infty.
\end{equation}
By the inductive assumption and \eqref{e35}, for $\varepsilon>0$ fixed we
choose $M>0$ such that
\[
  I(u_m^k)>c_k-c_{k-1}+\varepsilon,\quad |I(u_m)-c_k|<\varepsilon,\quad \text{as }m\geq M.
\]
Then we may define $\hat{u}(x)\in M_{k-1}$ by
\[
\hat{u}(x)=\begin{cases}
u_m^s(x), &x\in\Omega_m^s\text{ as } s<k,\\
0, &x\in\Omega_m^k.
\end{cases}
\]
Hence $I(\hat{u})=I(u_m)-I(u_m^k)<c_k+\varepsilon-(c_k-c_{k-1}+\varepsilon)=c_{k-1}$ as
$m\geq M$, which contradicts the fact that $c_{k-1}=\inf_{M_{k-1}}I(u)$.
Then, we  obtain $r^k<+\infty$.
\smallskip

\noindent\textbf{Step 4.} $c_k$ is attained.
By Proposition \ref{pro1} we can find a subsequence (still denoted by $\{u_m\}$) 
such that
\begin{gather*}
    u_m\rightharpoonup u \quad \text{in }X,\\
u_m\to u \quad \text{in }L^{p+1}(\mathbb{R}^N).
\end{gather*}
Set $\Omega^l=\{x\in\mathbb{R}^N|r^{l-1}<|x|<r^l\}$, for all 
$l=1,2,\dots,k+1,r^0=0$ and $r^{k+1}=+\infty$. Lemma \ref{le27} implies that
 $c=\inf_{u\in M(\Omega^l)}I(u)$ is attained by some positive function 
$\hat{u}^l$ which satisfies the  boundary-value problem
 \begin{gather*}
   -\sum_{i,j=1}^{N}\partial_j(a_{ij}(u)\partial_iu)+\frac{1}{2}
\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju+V(x)u=|u|^{p-1}u, \quad x\in\Omega^l,\\
   u|_{\partial\Omega^l}=0.
\end{gather*}
Define $u_k=\sum_{l=1}^{k+1}(-1)^{l-1}\hat{u}^l(x)$, 
$(\hat{u}^l(x)=0, x\not\in\Omega^l)$. Then, clearly, $u_k\in M_k$. 
Consider the coordinate transformations
$ \Phi_m:\mathbb{R}^N\to\mathbb{R}^N$, $m=1,2,\dots$, defined
by
\[
\Phi_m(x)=\varphi_m(|x|)\frac{x}{|x|},\quad x\in\mathbb{R}^N,
\]
where
\[
  \varphi_m(r)=\frac{(r^l-r^{l-1})(r-r_m^{l-1})}{r_m^l-r_m^{l-1}}+r^{l-1}.
\]
For any $r\in\mathbb{R}$, clearly $\Phi_m(\Omega_m^l)=\Omega^l$.
Let $y=\Phi_m(x)\in\Omega^l$, if $x\in\Omega_m^l$. 
It is easy to show that
\begin{gather}\label{e37}
    |\nabla u(y)|=(R^l_m)^{-1}|\nabla u(x)|, \\
\label{e38}
{\rm d}y=|J_m^l|\,{\rm d}x, \\
\label{e39}
a_m^l\leq\Big(\frac{\Phi_m(r)}{r}\Big)^{N-1} \leq A_m^l,
\end{gather}
where
\begin{gather*}
   R_m^l=\frac{r^l-r^{l-1}}{r_m^l-r_m^{l-1}},\quad
J_m^l=\big(\varphi_m(|x|)\big)^{N-1}\big(\varphi_m(|x|)\big)'|x|^{1-N},\\
a_m^l=\Big(\min\big\{\frac{r^l}{r^l_m},\frac{r^{l-1}}{r^{l-1}_m}\big\}\Big)^{N-1},
\quad A_m^l=\Big(\max\big\{\frac{r^l}{r^l_m},\frac{r^{l-1}}{r^{l-1}_m}\big\}
\Big)^{N-1}.
\end{gather*}
Clearly,
\begin{equation}\label{e310}
a_m^lR_m^l\leq|J_m^l|\leq A_m^lR_m^l,
\end{equation}
and
\begin{equation}\label{e311}
R_m^l\to1,\quad a_m^l\to1,\quad A_m^l\to1,\quad J_m^l\to1, \quad 
\text{as }m\to+\infty.
\end{equation}
Let
\begin{align*}
   \gamma(t)&=\frac{N}{2}t^{N-1}\int_{\Omega^l}\sum_{i,j=1}^{N}a_{ij}(tu_m^l)
\partial_iu_m^l\partial_ju_m^l\,{\rm d}y\\
&\quad +\frac{t^N}{2}\int_{\Omega^l}u_m^l
\sum_{i,j=1}^{N}a'_{ij}(tu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}y
 +\frac{{N+2}}{2}t^{N+1}\int_{\Omega^l}V(ty)(u_m^l)^2\,{\rm d}y\\
&\quad +\frac{t^{N+2}}{2}\int_{\Omega^l}\nabla V(ty)\cdot y(u_m^l)^2\,{\rm d}y
 -\frac{N+p+1}{p+1}t^{N+p}\int_{\Omega^l}|u_m^l|^{p+1}\,{\rm d}y.
\end{align*}
From Lemma \ref{inf}, there exists some $t_m^l>0$, such that $\gamma(t_m^l)=0$, 
thus $(u_m^l)_{t_m^l}\in M(\Omega^l)$. Now we claim that
\begin{equation}\label{e312}
  t_m^l\to1 \quad \text{as }m\to+\infty,\; l=1,2,\dots,k.
\end{equation}
Indeed, since $\gamma(t_m^l)=0$, we have
\begin{equation}\label{e313}
\begin{aligned}
&\frac{N}{2}(t_m^l)^{N-1}\int_{\Omega^l}
\sum_{i,j=1}^{N}a_{ij}(t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}y \\
&+\frac{(t_m^l)^N}{2}\int_{\Omega^l}u_m^l\sum_{i,j=1}^{N}a'_{ij}
(t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}y \\
&+\frac{{N+2}}{2}(t_m^l)^{N+1}\int_{\Omega^l}V(t_m^ly)(u_m^l)^2\,{\rm d}y
 +\frac{(t_m^l)^{N+2}}{2}\int_{\Omega^l}\nabla V(t_m^ly)\cdot y(u_m^l)^2\,{\rm d}y \\
&-\frac{N+p+1}{p+1}(t_m^l)^{N+p}\int_{\Omega^l}|u_m^l|^{p+1}\,{\rm d}y=0.
\end{aligned}
\end{equation}
We can prove that there exists a constant $\tilde{t}>0$ such that
\[
   0<t_m^l\leq\tilde{t}<+\infty.
\]
By selecting a subsequence, we may assume that $\lim_{m\to+\infty}t_m^l=t_*^l$.
Using \eqref{e37}-\eqref{e311}, we have
\begin{gather}\label{e314}
\begin{aligned}
&\lim_{m\to+\infty}\int_{\Omega^l}\sum_{i,j=1}^{N}a_{ij}(t_m^lu_m^l)
 \partial_iu_m^l\partial_ju_m^l\,{\rm d}y \\
&=\lim_{m\to+\infty}\int_{\Omega_m^l}\sum_{i,j=1}^{N}a_{ij}(t_m^lu_m^l)
 \partial_iu_m^l\partial_ju_m^l\,{\rm d}x,
\end{aligned}\\
\label{e315}
\begin{aligned}
&\lim_{m\to+\infty}\int_{\Omega^l}u_m^l\sum_{i,j=1}^{N}a'_{ij}
 (t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}y \\
&=\lim_{m\to+\infty}\int_{\Omega_m^l}u_m^l\sum_{i,j=1}^{N}a'_{ij}
 (t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x,
\end{aligned}\\
\label{e316}
\lim_{m\to+\infty}\int_{\Omega^l}V(t_m^ly)(u_m^l)^2\,{\rm d}y
=\lim_{m\to+\infty}\int_{\Omega_m^l}V(t_m^lx)(u_m^l)^2\,{\rm d}x,\\
\label{e317}
\lim_{m\to+\infty}\int_{\Omega^l}\nabla V(t_m^ly)\cdot y(u_m^l)^2\,{\rm d}y
=\lim_{m\to+\infty}\int_{\Omega_m^l}\nabla V(t_m^lx)\cdot x(u_m^l)^2\,{\rm d}x,\\
\label{e318}
\lim_{m\to+\infty}\int_{\Omega^l}|u_m^l|^{p+1}\,{\rm d}y
=\lim_{m\to+\infty}\int_{\Omega_m^l}|u_m^l|^{p+1}\,{\rm d}x.
\end{gather}
Substituting \eqref{e314}-\eqref{e318} in \eqref{e313} we find that
\begin{align}
&\lim_{m\to+\infty}\Big(\frac{N}{2}(t_m^l)^{N-1}
\int_{\Omega^l_m}\sum_{i,j=1}^{N}a_{ij}(t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x 
 \nonumber \\
&+\frac{(t_m^l)^N}{2}\int_{\Omega^l_m}u_m^l
 \sum_{i,j=1}^{N}a'_{ij}(t_m^lu_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x \nonumber \\
&+\frac{{N+2}}{2}(t_m^l)^{N+1}\int_{\Omega^l_m}V(t_m^lx)(u_m^l)^2\,{\rm d}x
 +\frac{(t_m^l)^{N+2}}{2}\int_{\Omega^l_m}\nabla V(t_m^lx)\cdot x(u_m^l)^2\,{\rm d}x 
\nonumber \\
&-\frac{N+p+1}{p+1}(t_m^l)^{N+p}\int_{\Omega^l_m}|u_m^l|^{p+1}\,{\rm d}x\Big)=0.
 \label{e319}
\end{align}
But for $u_m^l(x)\in M(\Omega_m^l)$, we know that
\begin{equation} \label{e320}
\begin{aligned}
&\frac{N}{2}\int_{\Omega^l_m}\sum_{i,j=1}^{N}a_{ij}(u_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x+\frac{1}{2}\int_{\Omega^l_m}u_m^l\sum_{i,j=1}^{N}a'_{ij}(u_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x \\
&+\frac{{N+2}}{2}\int_{\Omega^l_m}V(t_m^lx)(u_m^l)^2\,{\rm d}x+\frac{1}{2}\int_{\Omega^l_m}\nabla V(x)\cdot x(u_m^l)^2\,{\rm d}x \\
&-\frac{N+p+1}{p+1}\int_{\Omega^l_m}|u_m^l|^{p+1}\,{\rm d}x=0.
\end{aligned}
\end{equation}
Set
\begin{equation}\label{e321}
\begin{aligned}
h(s)
&= \frac{N}{2}s^{N-1}\int_{\Omega^l_m}\sum_{i,j=1}^{N}a_{ij}(s u_m^l)\partial_iu_m^l\partial_ju_m^l\,{\rm d}x \\
&\quad +\frac{s^N}{2}\int_{\Omega^l_m}u_m^l\sum_{i,j=1}^{N}a'_{ij}(s u_m^l)
 \partial_iu_m^l\partial_ju_m^l\,{\rm d}x \\
&\quad +\frac{{N+2}}{2}s^{N+1}\int_{\Omega^l_m}V(s x)(u_m^l)^2\,{\rm d}x
  +\frac{s^{N+2}}{2}\int_{\Omega^l_m}\nabla V(s x)\cdot x(u_m^l)^2\,{\rm d}x\\
&\quad -\frac{N+p+1}{p+1}s^{N+p}\int_{\Omega^l_m}|u_m^l|^{p+1}\,{\rm d}x.
\end{aligned}
\end{equation}
From the proof of Lemma \ref{inf}, we know that $h(s)$ has only one zero on
$(0,+\infty)$.
So, from \eqref{e319}-\eqref{e321} we get that $t_*^l=1$. Moreover,
\[
  \lim_{m\to+\infty}I((u_m^l)_{t_m^l})=\lim_{m\to+\infty}I(u_m^l).
\]
On the other hand, since $I(\hat{u}^l)=\inf_{M(\Omega^l_m)}I(u)$ and
$(u_m^l)_{t_m^l}\in M(\Omega_m^l)$, we obtain
\[
    I(\hat{u}^l)\leq I((u_m^l)_{t_m^l}).
\]
Hence $\lim_{m\to+\infty}I((u_m^l)_{t_m^l})\geq I(\hat{u}^l)$, $l=1,2,\dots, k+1$.
Thus
\[
  c_k=\lim_{m\to+\infty}I(u_m)=\lim_{m\to+\infty}\sum _{l=1}^{k+1}I(u^l_m)
\geq \sum _{l=1}^{k+1}I(\hat{u}^l)=I(u_k).
\]
Since $u_k\in M_k$, we have that $c_k=I(u_k)$, which means that $c_k$ is attained.
\end{proof}

\begin{proof}[Proof of Theorem \ref{1.1}]
By Lemma \ref{le31}, there exists $u_k\in M_k$ which attains $c_k$. 
We will prove that $u_k$ is indeed a solution to problem \eqref{1.1}. 
For convenience, we denote $u:=u_k$. Thus we get $k$ nodes: 
$r_1,r_2,\dots,r_k$, $0<r_1<r_2<\dots<r_k<+\infty$. Clearly, $u$ satisfies 
\eqref{1.1} in $E=\{x\in\mathbb{R}^N: |x|\neq r_l,l=1,2,\dots,k+1\}$. 
We know already that $u$ is of class $C^2$ on $E$ and satisfies, for $x\in E$
\begin{equation}\label{e322}
    -\sum_{i,j=1}^{N}\partial_j(a_{ij}(u)\partial_iu)+\frac{1}{2}
\sum_{i,j=1}^{N}a'_{ij}(u)\partial_iu\partial_ju+V(x)u=|u|^{p-1}u.
\end{equation}
We will prove that $u$ satisfies \eqref{e322} for all $x\in\mathbb{R}^N$.

We use an indirect argument. Assume that for some $l=1,2,\dots,k$, 
there exists $x_0\in\mathbb{R}^N, |x_0|=r_l$ such that \eqref{e322} does not hold. 
To complete the proof, it suffices to show that for $a_{ij}(u)=(1+u^2)\delta_{ij}$, 
there exists a contradiction.

The existence of the contradiction can be proved similar to that as in \cite{DENG}, 
by a slight modification, their arguments worked also for $p\in(1,3]$. 
We just sketch the proof. We set $r:=|x|$ and treat the special case 
$a_{ij}(u)=(1+u^2)\delta_{ij}$ as an ordinary differential equation: 
$$
-(1+u^2)(r^{N-1}u')'=r^{N-1}(|u|^{p-1}-V+|u'|^2)u,
$$
where $'$ denotes $\frac{d}{dr}$. Then our assumption becomes to 
$$
u'_+=\lim_{r\to r^+_l}u'(r)\neq\lim_{r\to r^-_l}u'(r)=u'_-.
$$
Firstly, we construct some $w$ such that $w\in M_k$. Let 
$$
\psi(h)=\int_{r_{l-1}}^{r_{l+1}}
\Big(\frac{1}{2}(h'^2+Vh^2+h^2h'^2)-\frac{1}{p+1}|h|^{p+1}\Big)r^{N-1}\,{\rm d}r.
$$
Then, according to the definition of $u$, there holds 
$$
\psi(u)\leq\psi(w).
$$
However, under the assumption $u'_+\neq u'_-$, we can prove that 
$\psi(w)<\psi(u)$ (cf. \cite{DENG}). 
This is a contradiction.
As a result, we complete the proof.
\end{proof}

\subsection*{Acknowledgments}
 Z.-Q. Han was supported by NSFC 11171047.

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\end{document}
