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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 152, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/152\hfil Multiple positive solutions]
{Multiple positive solutions for second-order three-point
boundary-value problems with sign changing nonlinearities}

\author[J. Liu, Z. Zhao \hfil EJDE-2012/152\hfilneg]
{Jian  Liu, Zengqin Zhao}  % in alphabetical order

\address{Jian Liu \newline
School of Mathematics and Quantitative Economics,
Shandong University of Finance and Economics, 
Jinan, Shandong, 250014, China}
\email{liujianmath@163.com}

\address{Zengqin Zhao \newline
School of Mathematical Sciences, Qufu Normal
University, Qufu, Shandong, 273165, China}
\email{zqzhao@mail.qfnu.edu.cn}

\thanks{Submitted March 14, 2012. Published September 7, 2012.}
\subjclass[2000]{34B15, 34B25}
\keywords{Multiple positive solutions; sign changing; fixed-point theorem}

\begin{abstract}
 In this article, we study the second-order three-point boundary-value
 problem
 \begin{gather*}
 u''(t)+a(t)u'(t)+f(t,u)=0,\quad  0 \leq t \leq 1,   \\
 u'(0)=0,\quad u(1)=\alpha u(\eta),
 \end{gather*}
 where $0<\alpha$, $\eta<1$, $a\in C([0,1],(-\infty, 0))$ and $f$ is
 allowed to change sign. We show that there exist two positive
 solutions by using Leggett-Williams fixed-point theorem.
\end{abstract}

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

\section{Introduction}

The study of multi-point boundary-value problems for
linear second-order ordinary differential equations was initiated
by Kiguradze and Lomtatidze \cite{Kiguradze}, Lomtatidze \cite{Lomtatidze},
 Il'in and Moviseev \cite{Il'in1,Il'in2}, Agarwal and Kiguradze \cite{AK}, 
Lomtatidze and  Malaguti \cite{LM}. Motivated by the study of \cite{Il'in1,Il'in2},
 Gupta \cite{Gupta1} studied certain three-point boundary-value problems for
nonlinear ordinary differential equations. Since then,  more
general nonlinear multi-point boundary-value problems have been
studied by several authors. We refer the reader to  Gupta \cite{Gupta2}, 
Li, Liu and Jia \cite{LiLiuJia}, Liu \cite{Liu}, Ma\cite{Ma1,Ma2} for some
references along this line. Using results from fixed point theory,
such as the fixed-point theorems by Banach, Krasnosel'skii,
Leggett-Williams etc., in studying second-order dynamic systems is a standard
and useful tool (see, e.g. \cite{Anderson2,Anderson1,BS}).

Recently, Ma \cite{Ma1} studied  the   three-point boundary-value problem
(BVP)
\begin{gather*}
u''(t)+a(t)f(u)=0,\quad  0 \leq t \leq 1, \\
 u(0)=0,\quad u(1)=\alpha u(\eta),
\end{gather*}
where $0<\eta<1$, $\alpha$ is a positive constant, $a\in C[0,1]$,
$f\in C([0,+\infty),[0,+\infty))$ and there exists $x_0\in(0,1)$
such that $a(x_0)>0$. Author got the existence  and multiplicity
of positive solutions theorems under the condition that $f$ is
either superlinear or sublinear by using Krasnoselskii's fixed
point theorem.

In 2001, Ma\cite{Ma2} considered $m$-point boundary-value problem 
\begin{gather*}
u''(t)+h(t)f(u)=0,\quad 0 \leq t \leq 1,   \\
 u(0)=0,\quad u(1)=\sum _{i=1}^{m-2}\beta_{i} u(\xi_{i}),
\end{gather*}
where
$\beta_{i}>0$ ($i=1,2,\dots,m-2$), $0<\xi_{1}<\xi_{2}<\dots<\xi_{m-2}<1$,
 $h\in C([0,1],[0,+\infty))$ and $f\in  C([0,+\infty),[0,+\infty))$.
 Author established the existence of positive solutions 
 under the condition that $f$ is either superlinear or sublinear.

In \cite{LiLiuJia}, the authors studied  the  three-point boundary-value problem
\begin{gather*}
u''(t)+a(t)u'(t) +\lambda f(t, u)=0,\quad  0 \leq t \leq 1,   \\
 u'(0)=0,\quad u(1)=\alpha u(\eta),
\end{gather*}
where $0<\eta<1$, $\alpha$ is a positive constant, 
$a\in C([0,1],(-\infty, 0))$, $ f\in C([0,1]\times \mathbb{R^+},\mathbb{R})$ 
and there exists $M>0$ such that $f(t,u)\geq -M$ for
 $(t,u)\in [0,1]\times \mathbb{R^+} $.
They obtained the existence of one positive solution by using
Krasnoselskii's fixed point theorem.

Motivated by the results mentioned above, in this paper, we study
the  existence of positive solutions of three-point boundary-value problem
\begin{equation}
\begin{gathered}
u''(t)+a(t)u'(t)+f(t,u)=0,\quad 0 \leq t \leq 1,   \\
u'(0)=0,\quad u(1)=\alpha u(\eta),
\end{gathered} \label{e1.1}
\end{equation}
where $0<\alpha$, $\eta<1$, $a\in C([0,1],(-\infty, 0))$ and $f$ is
allowed to change sign. We show that there exist  two positive
solutions by using Leggett-Williams fixed-point theorem. Our ideas
are similar those used in  \cite{LiLiuJia}, but a little different.
By applying Leggett-Williams fixed-point theorem, we get the new
results, which are different from the previous results and the
conditions are easy to be checked. In particular, we do not need
that $f$ be either superlinear or sublinear which was required in
\cite{Gupta2,LiLiuJia,Ma1,Ma2}.

In the rest of the paper, we make the following assumptions
\begin{itemize}
\item[(H1)] $0<\alpha$, $\eta<1$;

\item[(H2)] $a\in C([0,1],(-\infty, 0))$;

\item[(H3)] $f: [0,1]\times \mathbb{R^+}\to \mathbb{R}$  is continuous  and
there exists $M>0$ such that $f(t,u)\geq -M$ for 
$(t,u)\in [0,1]\times \mathbb{R^+}$.

\end{itemize}
By a positive solution of  \eqref{e1.1}, we understand a function $u$
which is positive on $(0,1)$ and satisfies the differential
equations as well as the boundary  conditions in \eqref{e1.1}.

\section{Preliminaries}

In this section, we give some definitions  and  lemmas.

\begin{definition} \label{def2.1} \rm
 Let $E$ be a real Banach space. A nonempty closed set $P\subset E$ 
is said to be a cone provided that
\begin{itemize}
\item[(i)]  $u\in P$ and $a\geq0$  imply $a u\in P$;

\item[(ii)]  $u,\ -u \in P$ implies $u=0$;

\item[(ii)] $u,v\in P$ implies $u+v\in P$.
\end{itemize}
\end{definition}

\begin{definition} \label{def2.2}\rm
Given a cone $P$ in a real Banach space $E$, an operator $\psi: P\to P$ is said to be
increasing on $P$, provided $\psi(x)\leq\psi(y)$, for all 
$x,y\in P$ with $x\leq y$.

A functional $\alpha: P\to [0,\infty)$ is said
to be nonnegative continuous concave on $P$, provided
$\alpha(tx+(1-t)y)\geq t\alpha(x)+(1-t)\alpha(y)$, for all 
$x,y\in P$ with $t\in[0,1]$.
\end{definition}

Let $a,  b,  r>0$ be constants with $P$ and $\alpha$ as defined
above, we note 
$$
P_r=\{y\in P:  \|y\|<r\},\quad   
P\{\alpha, a, b\}=\{y\in P:  \alpha(y)\geq a,\,  \|y\|\leq b\}.
$$
The main tool of this paper is the following well known
Leggett-Williams fixed-point theorem.

\begin{theorem}[\cite{Henderson,Leggett}] \label{thm2.1}
Assume $E$ be a real Banach space, $P\subset E$ be a cone. 
Let $A:\overline{P}_c\to\overline{P}_c$ be completely continuous and
$\alpha$ be a nonnegative continuous concave functional on $P$
such that $\alpha(y)\leq \|y\|$,  for $y\in \overline{P}_c$.
Suppose that there exist $0<a<b<d\leq c$ such that

\noindent(i)\ \ $\{y\in P(\alpha, \ b,\ d)|\ \
\alpha(y)>b\}\neq\emptyset$ and $\alpha(Ay)>b$, \ for all $y\in
P(\alpha, \ b,\ d) $;
\begin{itemize}
\item[(ii)] $\|Ay\|<a$,   for all $\|y\|\leq a$;

\item[(iii)] $\alpha(Ay)>b$   for all $y\in P(\alpha, b,c)$ with $\|Ay\|>d$.
\end{itemize}
 Then  $A$ has at least three fixed points $y_1, y_2,y_3$ satisfying
$$ 
\|y_1\|<a,\quad  b<\alpha(y_2),\quad 
\|y_3\|>a,\quad \alpha(y_3)<b.
$$
\end{theorem}

\begin{lemma}[\cite{LiLiuJia}] \label{lem2.1}
 Assume that {\rm (H1), (H2)} hold. 
Then for any  $y\in C[0,1]$ the BVP
\begin{equation}
\begin{gathered}
u''(t)+a(t)u'(t)+y(t)=0,\quad  0 \leq t \leq 1,   \\
u'(0)=0,\quad u(1)=\alpha u(\eta),
\end{gathered}\label{e2.1}
\end{equation}
has unique solution
\begin{equation}
\begin{aligned}
u(t)
&=-\int_0^{t}\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)y(\tau)d\tau\Big)ds
 +\frac{1}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)y(\tau)d\tau\Big)ds\\
&\quad -\frac{\alpha}{1-\alpha}\int_0^{\eta}
\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)y(\tau)d\tau\Big)ds,
\end{aligned}\label{e2.2}
\end{equation}
where
$p(t)=\exp\big(\int_0^{t}a(\tau)d\tau\big)$.
\end{lemma}

\begin{lemma}[\cite{LiLiuJia}] \label{lem2.2} 
 Assume that {\rm (H1), (H2)} hold. Let $y\in C[0,1]$ and $y(t)\geq 0$ 
for all $t\in[0,1]$, then the unique solution of  \eqref{e2.1} satisfies 
$u(t)\geq0$.
\end{lemma}


\begin{lemma}[\cite{LiLiuJia}] \label{lem2.3}
 Let {\rm (H1),  (H2)} hold.
If $y\in C[0,1]$ and $y(t)\geq 0$ for all  $t\in [0,1]$, the unique solution 
of  \eqref{e2.1} satisfies
 $$
\min_{t\in[0,1]}u(t)\geq\gamma\|u\|,
$$
 where
 $\gamma=\frac{\alpha(1-\eta)} {1-\alpha\eta}$.
\end{lemma}

\begin{lemma}[\cite{LiLiuJia}] \label{lem2.4}
Let $\omega$ be the unique solution of the initial-value problem
\begin{equation} \begin{gathered}
u''(t)+a(t)u'(t)+1=0,\quad  0\leq t \leq 1,   \\
 u'(0)=0,\quad u(1)=\alpha u(\eta).
\end{gathered}\label{e2.3}
\end{equation}
Then $\omega(t)\leq\Gamma\gamma$ for $t\in[0,1]$,
where
$\gamma=(\alpha(1-\eta))/(1-\alpha\eta)$ and
$$
\Gamma=\Big(\int_0^{\eta}(\frac{1}{p(s)}
\int_0^{s}p(\tau)d\tau)ds+\frac{1}{1-\alpha}
\int_{\eta}^1(\frac{1}{p(s)}\int_0^{s}p(\tau)d\tau)ds\Big)\frac{1}{\gamma}.
$$
\end{lemma}

 \section{Main Results}

For convenience, we let 
$\gamma=\frac{\alpha(1-\eta)} {1-\alpha\eta}$, 
\[
l=\int_0^1\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)d\tau\Big)ds, \quad
h=\frac{1}{1-\alpha}\int_0^1\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)d\tau\Big)ds.
\]
Let $E=C[0,1]$, then $E$ is Banach space, with  the
norm $\|u\|=\sup_{t\in[0,1]}|u(t)|$. We define a cone in $E$ by
$$
P=\{u\in E:u\geq0,\min_{t\in[0,1]}u(t)\geq\gamma\|u\|\}.
$$
Our main results are the following theorems.

\begin{theorem} \label{thm3.1}
Suppose conditions {\rm (H1)--(H3)} hold and there exist positive constants
$e, b, c, N $ with $M\Gamma<e<e+M\Gamma\gamma<b<\gamma^2c$,
$\frac{1}{\gamma}<N<\frac{cl}{bh}$ such that
\begin{itemize}
\item[(A1)]  $f(t,u)<\frac{e}{h}-M$ for $t\in [0,1]$, $0\leq u\leq e$;

\item[(A2)] $f(t,u)\geq\frac{b}{l}N-M$ for $t\in [0,1]$,
$b-M\Gamma\gamma\leq u\leq \frac{b}{\gamma^2}$;

\item[(A3)] $f(t,u)\leq\frac{c}{h}-M$ for $t\in [0,1]$,
$0\leq u\leq c$,
\end{itemize}
where the number $\Gamma$ is defined
in Lemma \ref{lem2.4}. Then  \eqref{e1.1} has at least two positive solutions.
\end{theorem}

\begin{proof}
 Let $\omega$ be a solution of \eqref{e2.3} and $z=M\omega$. 
By Lemma \ref{lem2.4} we have $z(t)=M\omega(t)\leq M\Gamma\gamma<e\gamma$. 
It is easy to see that  \eqref{e1.1} has a positive solution $u$ if and only if
$u+z=\overline{u}$ is a solution of the boundary-value problem
\begin{equation}
\begin{gathered}
u''(t)+a(t)u'(t)=-g(t,u-z),\quad  0\leq t \leq 1,   \\
u'(0)=0,\quad u(1)=\alpha u(\eta),
\end{gathered}\label{e3.1}
\end{equation}
and $\overline{u}>z$ for $t\in(0,1)$, where
 $g:[0,1]\times R \to [0, +\infty)$ is defined by
$$
g(t,y)=\begin{cases}
f(t,y)+M, & (t,y)\in[0,1]\times[0,+\infty),   \\
f(t,0)+M, & (t,y)\in[0,1]\times(-\infty,0).
\end{cases}
$$
For $v\in P$, define the operator
\begin{align*}
Tv(t)&=-\int_0^{t}\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)
 -z(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{1-\alpha}\int_0^1
\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\quad -\frac{\alpha}{1-\alpha}\int_0^{\eta}
\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds,
\end{align*}
where $p$ is defined in Lemma \ref{lem2.1}.
By Lemmas \ref{lem2.1},  \ref{lem2.2} and  \ref{lem2.3}, we can  check
$T(P)\subseteq P$. It is easy to check $T$ is completely continuous
by Arzela-Ascoli theorem.

In the following, we show that all the conditions of Theorem \ref{thm2.1}
are satisfied. Firstly, we define the nonnegative, continuous
concave functional $\alpha: P\to[0,\infty) $ by
$$
\alpha(v)=\min_{t\in [0,1]}v(t),
$$
Obviously, for every $v\in P$,
$\alpha(v)\leq \|v\|$.

We first show
that $T(\overline{P}_c)\subseteq\overline{P}_c$. Let
 $v\in \overline{P}_c$ and $t\in[0,1]$ be arbitrary. 
When $v(t)\geq z(t)$, we have $0\leq v(t)-z(t)\leq v(t)\leq c$ and thus
 $g(t,v(t)-z(t))=f(t,v(t)-z(t))+M\geq0$. By (A3) we have
$$
g(t,v(t)-z(t))\leq\frac{c}{h}.
$$
 When $v(t)<z(t)$, we have $v(t)-z(t)<0$ and then
$g(t,v(t)-z(t))=f(t, 0)+M\geq0$. Again by (A3) we have
$$
g(t,v(t)-z(t))\leq\frac{c}{h}.
$$ 
Therefore, we have proved that,  if $v\in \overline{P}_c$, then
$g(t,v(t)-z(t))\leq\frac{c}{h} $  for $ t\in [0,1]$. 
 Then,
\begin{align*} 
\|Tv\|&=Tv(0)=\frac{1}{1-\alpha}
 \int_0^1\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\quad -\frac{\alpha}{1-\alpha}\int_0^{\eta}
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\leq\frac{1}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\leq\frac{c}{h}\frac{1}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)d\tau\Big)ds
 =c.
\end{align*}
  Thus $Tv\in\overline{P}_c$.
Therefore,  we have $T(\overline{P}_c)\subseteq\overline{P}_c$.
Especially, if $v\in\overline{P}_e$, then assumption (A1)
yields  $ g(t,v(t)-z(t))\leq\frac{e}{h} $  for $ t\in [0,1]$.  So,
we have $T:\overline{P}_e\to P_e$, i.e., the assumption (ii) of
Theorem \ref{thm2.1} holds.

To verify condition (i) of Theorem \ref{thm2.1}, let
$v(t)=\frac{b}{\gamma^2}$, then $v\in P$,
$\alpha(v)=b/\gamma^2>b$. That is 
$\{v\in P(\alpha, b, \frac{b}{\gamma^2}): \alpha(v)>b\}\neq\emptyset$.
Moreover, if $v\in P(\alpha,  b, \frac{b}{\gamma^2})$, then
$\alpha(v)\geq b$, so $b\leq \|v\|\leq \frac{b}{\gamma^2}$.
Thus, $0<b-M\Gamma\gamma\leq v(t)-z(t)\leq v(t)\leq\frac{b}{\gamma^2},
t\in [0,1]$. From assumption (A2) we obtain
$g(t,v(t)-z(t))\geq\frac{b}{l}N$ for $t\in [0,1]$. 
By the definition of $\alpha$ and above-proved inclusion $T(P)\subseteq P$, 
we have
\begin{align*} 
\alpha(Tv)&=\min_{t\in[0,1]}Tv(t)
 \geq \gamma\|Tv\|=\gamma Tv(0)\\
&=\gamma\Big(\frac{1}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\quad -\frac{\alpha}{1-\alpha}\int_0^{\eta}
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\Big)\\
&\geq\gamma\Big(\frac{1}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\\
&\quad -\frac{\alpha}{1-\alpha}\int_0^1
 \Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))d\tau\Big)ds\Big)\\
&=\gamma\int_0^1\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)g(\tau,v(\tau)-z(\tau))
 d\tau\Big)ds\\
&\geq\gamma\frac{b}{l} N\int_0^1\Big(\frac{1}{p(s)}\int_0^{s}p(\tau)d\tau\Big)ds\\
&=\gamma Nb >b.
\end{align*}
Therefore, condition (i) of Theorem  \ref{thm2.1} is satisfied with $d=b/\gamma^2$.

Finally, we address condition (iii) of Theorem  \ref{thm2.1}. For this we
choose $v\in P(\alpha, b,  c)$ with
$\|Tv\|>b/\gamma^2$. Then from above-proved inclusion $T(P)\subseteq P$, we have
$$
\alpha(Tv)=\min_{t\in[0,1]}Tv(t)\geq\gamma\|Tv\|\geq\frac{b}{\gamma}>b.
$$
Hence, condition (iii) of Theorem  \ref{thm2.1} holds with $\|Tv\|>b/\gamma^2$.

To sum up, all the hypotheses of Theorem \ref{thm2.1}  are satisfied. Hence
$T$ has at least three positive fixed points $v_1$, $v_2$ and
$v_3$ such that
$$ 
\|v_1\|<e,\quad b<\alpha(v_2),\quad
\|v_3\|>e,\quad \alpha(v_3)<b.
$$ 
Further, $u_i=v_i-z$ ($i=1,2,3$) are solutions of  \eqref{e3.1}. Moreover,
\begin{gather*}
v_2(t)\geq \gamma \|v_2\|\geq \gamma \alpha(v_2)>\gamma b>\gamma M\Gamma\geq z(t), 
\quad  t\in [0,1],\\
v_3(t)\geq \gamma \|v_3\|>\gamma e>\gamma M\Gamma\geq z(t), \quad t\in [0,1].
\end{gather*}
So $u_2= v_2-z$, $u_3=v_3-z$ are two positive solutions of 
\eqref{e1.1}. This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.2}
  Suppose {\rm  (H1)--(H3)} hold, and there exist positive constants
$a_i, b_i, N $ with
$M\Gamma<a_i<a_i+M\Gamma\gamma<b_i<\gamma^2a_{i+1}$,
$\frac{1}{\gamma}<N<\frac{a_{i+1}l}{b_ih}$, $(i=1,2,\dots, n-1)$
such that
\begin{itemize}
\item[(A4)]  $f(t,u)<\frac{a_i}{h}-M$ for $t\in [0,1]$,
$0\leq u\leq a_i$ $(i=1,2,\dots,n)$;

\item[(A5)] $f(t,u)\geq\frac{b_i}{l}N-M$ for $t\in [0,1]$,
$b_i-M\Gamma\gamma\leq u\leq \frac{b_i}{\gamma^2}$ $(i=1,2,\dots,n-1)$.
\end{itemize}
 Then,  \eqref{e1.1} has at least $2(n-1)$ positive
solutions.
\end{theorem}

\begin{proof}
When $n=2$, the assumptions of Theorem \ref{thm3.1} hold (with $c=a_2$), 
so we can get at least two positive solutions $u_2$ and $u_3$ such 
that $a_{1 }< u_{2}\leq a_{2}$ and $u_{3} \leq a_{2}$. 
Following the identical fashion, by the induction
method we immediately complete the proof.
\end{proof}

Our results are different from those in \cite{LiLiuJia}, 
in particular, the following condition that was used in \cite{LiLiuJia},
is not needed in this article
$$
\lim_{u\to\infty}\frac{f(t,u)}{u}=+\infty\quad \text{uniformly on } [0,1].
$$

\subsection*{Acknowledgments}
The first author is supported by grants ZR2012AQ024 from the Natural Science
Foundation of Shandong  Province of China,
and J10LA62 from the University Science and Technology  Foundation of
Shandong Provincial Education Department.
The second author is supported by grant 10871116 from the National
Natural Science Foundation of China,
ZR2010AM005 from the Natural Science Foundation of Shandong Province of
China,  and 200804460001 from the Doctoral Program Foundation of Education
Ministry of China.


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\end{thebibliography}

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