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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 95, pp. 1--11.\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-2012/95\hfil Multiple symmetric positive solutions]
{Multiple symmetric positive solutions to four-point boundary-value problems
of differential systems with p-Laplacian}

\author[H. Feng, D. Bai, M. Feng \hfil EJDE-2012/95\hfilneg]
{Hanying Feng, Donglong Bai, Meiqiang Feng}  

\address{Hanying Feng \newline
 Department of Mathematics,
Shijiazhuang  Mechanical Engineering College\\
Shijiazhuang 050003, China}
\email{fhanying@yahoo.com.cn}

\address{Donglong Bai  \newline
 Department of Mathematics,
Shijiazhuang  Mechanical Engineering College\\
Shijiazhuang 050003, China}
\email{baidonglong@yeal.net}

\address{Meiqiang Feng \newline
 School of Applied Science,
 Beijing Information Science and Technology University\\
 Beijing 100092,  China}
\email{meiqiangfeng@sina.com}


\thanks{Submitted March 9, 2012. Published June 10, 2012.}
\thanks{Supported by grants 10971045 from the NNSF, and 
A2012506010 from  HEBNSF of China}
\subjclass[2000]{34B10, 34B15, 34B18}
\keywords{Four-point boundary-value problem; differential system;
\hfill\break\indent  fixed point theorem; symmetric positive solution;
 one-dimensional $p$-Laplacian}

\begin{abstract}
 In this article,  we study the four-point boundary-value problem
 with the one-dimensional $p$-Laplacian
 \begin{gather*}
 (\phi_{p_i}(u_i'))'+q_i(t)f_i(t,u_1,u_2)=0,\quad
 t\in(0,1),\quad i=1,2;\\
 u_i(0)-g_i(u_i'(\xi))=0,\quad
 u_i(1)+g_i(u_i'(\eta))=0, \quad i=1,2.
 \end{gather*}
 We obtain sufficient conditions such that by means of a fixed point
 theorem on a cone,  there exist  multiple  symmetric positive solutions
 to the above boundary-value problem. As an application, we give an example
 that we illustrates  our results.
\end{abstract}

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

\section{Introduction}

In this article, we discuss  the  existence of multiple symmetric
positive solutions to the four-point boundary-value problem
 (BVP) for a differential system with the one-dimensional $p$-Laplacian,
 \begin{gather}\label{e1.1}
(\phi_{p_i}(u_i'))'+q_i(t)f_i(t,u_1,u_2)=0,\quad
  t\in(0,1),\; i=1,2; \\
\label{e1.2}
 u_i(0)-g_i(u_i'(\xi))=0,\quad
u_i(1)+g_i(u_i'(\eta))=0, \quad i=1,2,
\end{gather}
where $\phi_{p_i}(s)=|s|^{p_i-2}s$, $p_i>1$, $0<\xi<1/2$,
$\xi+\eta=1$, and the functions $q_i$, $f_i$, $g_i$, $i=1,2$ satisfy
the following conditions:
\begin{itemize}
\item[(H1)]  $q_i\in L^1[0,1]$ is nonnegative symmetric on $[0,1]$
(i.e., $q_i(t)=q_i(1-t)$,\ $t\in[0,1]$)
and $q_i(t)\not\equiv0$ on any subinterval of $[0,1]$;

\item[(H2)] $ f_i\in C ([0,1]\times [0,+\infty)\times
[0,+\infty),(0,+\infty))$
is symmetric on $[0,1]$ (i.e., $f_i(t,u_1,u_2)=f_i(1-t,u_1,u_2)$,
for $t\in[0,1]$);

\item[(H3)] $ g_i\in C ((-\infty,+\infty),
(-\infty,+\infty))$ is strictly increasing, odd
 and satisfies the condition that there exists $m_i>0$ such that $0\leq
g_i(s)\leq m_is$ for all $s\geq0$.
\end{itemize}


Multipoint boundary-value problems for ordinary differential equations and systems
arise in a variety of  areas of applied mathematics and
physics. The study of multipoint BVPs for linear second-order
ordinary differential equations was initiated by Il'in and Moiseev
\cite{il}. Since then, many authors studied more general nonlinear
multi-point boundary-value problems by Leray-Schauder continuation
theorem, coincidence degree theory, the method of lower and upper
solutions, monotone iterative technique, fixed point theorem in
cones and so on. We refer readers to \cite{ba,fw1,gu,gu1,mr,wp,wf}
and the references cited therein. On the other hand, the existence
of symmetric positive solutions of second or higher order boundary-value
problems have received more and more attention in the recent
literature. The results of existence of symmetric positive solutions
were obtained by some authors, see
\cite{av1,fh,fh1,gr,gr1,ko,ko1,sy,sy1,yq}.

In recent years, there were many works done for a variety of
nonlinear second order ordinary differential systems with different
boundary conditions, see \cite{ag,fi,ll,mr1,wz,ya}. However, to the
best of our knowledge, there were only a few results on the
existence of multiple positive solutions to boundary-value problems
for differential systems with the one-dimensional $p$-Laplacian.
Especially, there were few papers on the existence of symmetric
positive solutions.

Recently, Liu \cite{lb} studied the existence of positive solutions
of singular boundary value systems with $p$-Laplacian,
\begin{gather*}
(\phi_{p}(x'))'+a_1(t)f(t,x(t),y(t))=0,\quad t\in(0,1),\\
 (\phi_{p}(y'))'+a_2(t)g(t,x(t),y(t))=0,\quad t\in(0,1),\\
x(0)-\beta_1x'(0)=0, \quad x(1)+\delta_1x'(1)=0,\\
 y(0)-\beta_2y'(0)=0, \quad y(1)+\delta_2y'(1)=0.
\end{gather*}
By using  fixed-point index theory, the existence of one and
multiple positive solutions for the boundary value systems under
some conditions were established.

 Liu and zhang \cite{lz} considered the existence of positive solutions
for the nonlinear system
\begin{gather*}
(\varphi_1(x'))'+a(t)f(x,y)=0,\ (\varphi_2(y'))'+b(t)g(x,y)=0,\quad
  t\in(0,1),\\
 \alpha\varphi_1(x(0))-\beta\varphi_1(x'(0))=0,\quad
 \alpha\varphi_2(y(0))-\beta\varphi_2(y'(0))=0,\\
 \alpha\varphi_1(x(1))-\beta\varphi_1(x'(1))=0,\quad
  \alpha\varphi_2(y(1))-\beta\varphi_2(y'(1))=0,
\end{gather*}
where $\varphi_1, \varphi_2$ are the increasing homeomorphism and positive
homomorphism and  $\varphi_1(0)=0, \varphi_2(0)=0$.
They showed the sufficient conditions for the existence of positive solutions
by means of the norm type cone expansion-expression fixed point theorem.


Recently, Ji, Feng and Ge \cite{ji} discussed the existence of
symmetric positive solutions for the  boundary-value system
with $p$-Laplacian,
\begin{gather*}
(\phi_{p_1}(u'))'+a_1(t)f(u,v)=0,\quad t\in(0,1),\\
 (\phi_{p_2}(v'))'+a_2(t)g(t,u,v)=0, \quad t\in(0,1),\\
u(0)-\alpha u'(\xi)=0, \quad u(1)+\alpha u'(\eta)=0,\\
 v(0)-\alpha u'(\xi)=0,\quad v(1)+\alpha v'(\eta)=0,
\end{gather*}

Motivated by the above works, our purpose in this paper is to give
some conditions that guarantee the  existence of multiple symmetric
positive solutions for boundary value systems \eqref{e1.1},
\eqref{e1.2}.

The main tool of this article is the fixed point index 
theorem in cones.


\begin{lemma}[\cite{gd,la}] \label{lem1.1}
 Let $K$ be a cone in a Banach space $X$. Let $D$ be
an open bounded subset of $X$ with $D_k = D \cap K \neq \phi$
and $\overline{D}_k\neq K$. Assume that $A:
\overline{D}_k\to K$ is a compact map such that
$x \neq Ax$ for $x \in \partial{D}_k$. Then the following results hold:
\begin{itemize}
\item[(1)] If $\|Ax\|\leq \|x\|$, $x \in\partial D_k$, then
$i_k(A,D_k) = 1$;

\item[(2)] If there exists $e\in K\backslash \{0\}$ such that $x \neq Ax
+ \lambda e$ for all $x \in\partial{D}_k$ and all $\lambda > 0$,
then $i_k(A,{D}_k) = 0$;

\item[(3)] Let $U$ be open in $X$ such that $\overline{U} \subset
{D}_k$. If $i_k(A,{D}_k) = 1$ and $i_k(A,U_k) = 0$, then
$A$ has a fixed point in ${D}_k\backslash \overline{U}_k$. The
same results holds if $i_k(A,{D}_k) = 0$ and $i_k(A,U_k) =1$.
\end{itemize}
\end{lemma}

\section{Preliminaries}

Let $E=C[0,1]\times C[0,1]$, then $E$ is a Banach space with the norm
$\|(u,v)\|=\|u\|+\|v\|$,
 where $\|u\|=\max_{t\in [0,1]}\mid u(t)\mid$, $\|v\|=\max_{t\in [0,1]}\mid
 v(t)\mid$.

\begin{definition}\label{def3.1} \rm
We define a partial ordering in $E$. 
For $(u_1, u_2), (v_1,v_2)\in E$:
 $(u_1,u_2)\leq (v_1,v_2)$
 if and only if
$u_i(t)\leq v_i(t),\ t\in[0,1],\ i=1,2$.
\end{definition}

\begin{definition}\label{def3.2} \rm
$(u_1,u_2)\in E$ is concave and symmetric on
$[0,1]$ if and only if $u_i(t),\ i=1,2$, is concave and symmetric
on $[0,1]$.
\end{definition}

 So, define a cone $K\subset E\times E$ by
$$
K=\{(u_1,u_2)\in E\times E : (u_1,u_2)
\text{ is nonnegative, concave and symmetric on } [0,1]\}.
$$
For  $h_i\in L^1[0,1]$, let $(u_1,u_2)$ be a solution
of the BVP
\begin{gather}\label{e3.1}
(\phi_{p_i}(u_i'))'+h_i(t)=0, \quad t\in(0,1),\;i=1,2,\\
\label{e3.2}
 u_i(0)-g_i(u_i'(\xi))=0,\quad u_i(1)+g_i(u_i'(\eta))=0, \quad i=1,2.
\end{gather}
By integrating  \eqref{e3.1}, it follows that
\begin{gather*}
u_i'(t)=\phi_{p_i}^{-1}\Big(A_{h_i}-\int_0^{t}h_i(\tau)d\tau\Big),\\
u_i(t)=u_i(0)+\int_0^{t}\phi_{p_i}^{-1}\Big(A_{h_i}-\int_0^sh_i(\tau)
d\tau\Big)ds,\\
u_i(t)=u_i(1)-\int_{t}^1\phi_{p_i}^{-1}\Big(A_{h_i}-\int_0^sh_i(\tau)
d\tau\Big)ds.
\end{gather*}

Using the boundary condition \eqref{e3.2}, we can easily obtain
\[
u_i(t)=g_i\circ\phi_{p_i}^{-1}
\Big(A_{h_i}-\int_0^{\xi}h_i(\tau)d\tau\Big)+
\int_0^{t}\phi_{p_i}^{-1}\Big(A_{h_i}-\int_0^sh_i(\tau)
d\tau\Big)ds
\]
or
\[
u_i(t)=-g_i\circ\phi_{p_i}^{-1}
\Big(A_{h_i}-\int_0^{\eta}h_i(\tau)d\tau\Big)
-\int_{t}^1\phi_{p_i}^{-1}\Big(A_{h_i}-\int_0^sh_i(\tau)d\tau\Big)ds,
\]
where $A_{h_i}$ satisfies \eqref{e3.3}.
\begin{equation}\label{e3.3}
\begin{split}
&g_i\circ\phi_{p_i}^{-1}(A_{h_i}-\int_0^{\xi}h_i(\tau)
d\tau)+g_i\circ\phi_{p_i}^{-1}(A_{h_i}-\int_0^{\eta}h_i(\tau)
d\tau )\\
&+\int_0^1 \phi_{p_i}^{-1}(A_{h_i}-\int_0^s h_i(\tau)d\tau)ds=0,
 \quad i=1,2.
\end{split}
\end{equation}


\begin{lemma} \label{lem2.1}
If  $h_i\in L^1[0,1]$ is nonnegative on $[0,1]$ and
$h_i(t)\not\equiv0$ on any subinterval of $[0,1]$, then there
exists a unique $A_{h_i} \in (-\infty, +\infty)$ satisfying
\eqref{e3.3}. Moreover, there is a unique $\sigma_{h_i} \in (0,1)$
such that $A_{h_i}=\int_0^{\sigma_{h_i}}h_i(\tau)d\tau$
 for $i=1,2$.
\end{lemma}

\begin{proof}
 For any  $h_i(t)$ in Lemma \ref{lem2.1}, define
\begin{align*}
H_{h_i}(c)
&= g_i\circ\phi_{p_i}^{-1}\Big(c-\int_0^{\xi}h_i(\tau)
d\tau\Big)+g_i\circ\phi_{p_i}^{-1}\Big(c-\int_0^{\eta}h_i(\tau)
d\tau \Big) \\
&\quad +\int_0^1 \phi_{p_i}^{-1}(c-\int_0^s
h_i(\tau)d\tau)ds, \quad i=1,2.
\end{align*}
So, $H_{h_i}: R\to R$ is continuous and strictly
increasing since $g_i$ is strictly increasing. It is easy to see
that $H_{h_i}(0)<0$,
$H_{h_i}(\int_0^1 h_i(\tau)d\tau)>0$. Therefore, there exists a unique
$A_{h_i}\in (0, \int_0^1h_i(\tau)d\tau)\subset(-\infty, +\infty)$ satisfying
\eqref{e3.3}.
Furthermore, if  $F_i(t)=\int_0^{t}h_i(\tau)d\tau$, then $F_i(t)$ is
continuous and strictly increasing on $[0,1]$,  $F_i(0)=0$, and
$F_i(1)= \int_0^1h_i(\tau)d\tau$. Thus,
$$
0=F_i(0)<A_{h_i}< F_i(1)=\int_0^1h_i(\tau)d\tau.
$$
Therefore, the intermediate value theorem guarantees that there exists
a unique $\sigma_{h_i} \in(0,1)$ such that
$A_{h_i}=\int_0^{\sigma_{h_i}}h_i(\tau)d\tau$,
$i=1,2$.
\end{proof}

\begin{remark} \label{rmk2.1} \rm
By Lemma \ref{lem2.1}, if $(u_1, u_2)$ is the unique solution
 of \eqref{e3.1}-\eqref{e3.2}, then $u_i$ $(i=1,2)$ can be rewritten
as
\begin{equation}\label{e3.6}
 u_i(t)= \begin{cases}
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{\sigma_{h_i}}h_i(\tau)d\tau\Big)+
\int_0^{t}\phi_{p_i}^{-1}\Big(\int_{s}^{\sigma_{h_i}}
h_i(\tau) d\tau\Big)ds,
 & 0\leq t\leq \sigma_{h_i},
\\
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\sigma_{h_i}}^{\eta}h_i(\tau)d\tau\Big)
+\int_{t}^1\phi_{p_i}^{-1}\Big(\int_{\sigma_{h_i}}^s
h_i(\tau)d\tau\Big)ds,
 & \sigma_{h_i}\leq t\leq 1,
\end{cases}
\end{equation}
for $i=1,2$.
\end{remark}

\begin{lemma} \label{lem3.2}
If  $h_i\in L^1[0,1]$, $i=1,2$, is nonnegative symmetric
on $[0,1]$ and $h_i(t)\not\equiv0$ on any subinterval of $[0,1]$,
then the unique solution $(u_1, u_2)$ of  \eqref{e3.1}-\eqref{e3.2}
is nonnegative, concave and symmetric.
\end{lemma}

\begin{proof}
Suppose that $(u_1, u_2)$ is the solution of \eqref{e3.1}-\eqref{e3.2}.
From the fact that
$$
(\phi_{p_i}(u_i'))'(t) =-h_i(t)\leq 0,\quad i=1,2,
$$
we know that $\phi_{p_i}(u_i'(t))$ is non-increasing. It follows
that $u_i'(t)$ is also non-increasing. Thus, we know that the graph
of $u_i$ is concave down on $[0,1]$.

It is easy to know that
$H_{h_i}(\int_0^{\frac{1}{2}}h_i(\tau)d\tau)=0$;
i.e., $ \sigma_{h_i}=1/2$ from the symmetry of $h_i(t)$. So,
from \eqref{e3.6} and for $t\in[0,1/2]$, by the transformation
$\tau=1-\widehat{\tau}$, we have
\begin{align*}
u_i(t)&=g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}h_i(\tau)d\tau\Big)+
\int_0^{t}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}h_i(\tau) d\tau\Big)ds\\
&=-g_i\circ\phi_{p_i}^{-1}
\Big(\int_{1-\xi}^{1/2}h_i(\widehat{\tau})d\widehat{\tau}\Big)
-\int_0^{t}\phi_{p_i}^{-1}\Big(\int_{1-s}^{1/2}
h_i(\widehat{\tau})d\widehat{\tau}\Big)ds.
\end{align*}
Again, let $s=1-\widehat{s}$; then
\begin{align*}
u_i(t)&=-g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\eta}^{1/2}h_i(\widehat{\tau})d\widehat{\tau}\Big)
+\int_1^{1-t}\phi_{p_i}^{-1}\Big(\int_{\widehat{s}}^{1/2}
h_i(\widehat{\tau})d\widehat{\tau}\Big)d\widehat{s}\\
&=g_i\circ\phi_{p_i}^{-1}
\Big(\int_{1/2}^{\eta}h_i(\tau)d\tau\Big)
+\int_{1-t}^1\phi_{p_i}^{-1}\Big(\int_{1/2}^s
h_i(\tau)d\tau\Big)ds\\
&=u_i(1-t),\quad i=1,2.
\end{align*}
So, $u_i$ is symmetric on $[0,1]$.

Combining the concavity and symmetry of $u_i$, we have
$u_i(1/2)=\max_{0\leq t\leq1}u_i$ and $u_i'(1/2)=0$.
So, $u_i'(t)\geq0$ for $t\in(0,1/2)$ and  $u_i'(t)\leq0$ for
$t\in(1/2,1)$. It follows that
 $u_i(0)=g(u_i'(\xi))=-g(u_i'(\eta))=u_i(1)\geq0$. Therefore,
 $u_i\geq0$ on $[0,1]$, $i=1,2$.
\end{proof}

\begin{lemma} \label{lem3.3}
Let $(u_1,u_2)\in K$, and $\delta\in(0,\frac{1}{2})$, then
$\min_{\delta\leq t\leq1-\delta}(u_1(t)+u_2(t))\geq
\delta(\|u_1\|+\|u_2\|)$, $t \in [\delta, 1-\delta]$.
\end{lemma}

 The proof of the above lemma  uses standard arguments only;  we omit
it here.

\begin{lemma} \label{lem3.4}
Assume that {\rm (H1)--(H3)} hold. For any $(u_1,u_2)\in K$,
define an operator
$$
T(u_1,u_2)(t)=(T_1(u_1,u_2), T_2(u_1,u_2))(t),
$$
where
\[
T_i(u_1,u_2)(t)= \begin{cases}
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))d\tau\Big)\\
+\int_0^{t}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))
d\tau\Big)ds,
& 0\leq t\leq 1/2, \\
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{1/2}^{\eta}q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))d\tau\Big)\\
+\int_{t}^1\phi_{p_i}^{-1}\Big(\int_{1/2}^s
q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))d\tau\Big)ds,
 & 1/2\leq t\leq 1,
\end{cases}
\]
for $i=1,2$.
Then $T: K\to K$ is completely continuous.
\end{lemma}

\begin{proof}
We first verify that $T: K\to K$. To do
so, let $(u_1,u_2)\in K$. According to the definition of $T$ and
Lemma \ref{lem3.2}, it follows
 $(\phi_{p_i} (T_i(u_1,u_2))')'(t)=-q_i(t)f_i(t,u_1(t),u_2(t))\leq0$
this implies $T_i(u_1,u_2)$ is concave on $[0,1]$. Again, from
the definition of symmetry of $f_i$ and $q_i$, it is easy to
know that $T_i(u_1,u_2)(t)=T_i(u_1,u_2)(1-t)$ for
$t\in[0,1/2]$, that is $T_i(u_1,u_2)$ is symmetric on $[0,1]$.
So, indeed $TK\subset K$ from definition \ref{def3.2}.

Next we show that  $T: K\to K$ is completely continuous.

(1) We prove $T$ is compact.
 Let $U\subset K$ is a bounded subset,
then there exists a constant $D>0$, such that $\|(u_1,u_2)\|\leq
D$ for any $(u_1,u_2)\in U$. By the discussion about $T_i\
(i=1,2)$ and condition (H3), for any $(u_1,u_2)\in U$ and
$0\leq t\leq1/2$ (the case $1/2\leq t\leq1$ can be proved
similarly), we have
\begin{displaymath}
\begin{aligned}
\|T_i(u_1,u_2)\|
&= T_i(u_1,u_2)(\frac{1}{2})\\
&= g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))d\tau\Big)\\
&\quad +\int_0^{\frac{1}{2}}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau)) d\tau\Big)ds,\\
&\leq \Big[m_i\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)d\tau\Big)+
\int_0^{\frac{1}{2}}\phi_{p_i}^{-1}(\int_{s}^{1/2}
q_i(\tau) d\tau)ds\Big]\\
&\quad \times\phi_{p_i}^{-1}\Big(\sup\{f_i(t,u_1,u_2):t\in[0,1],\
(u_1,u_2)\in U\}\Big).
\end{aligned}\quad\quad\ \
\end{displaymath}
Thus $T_i(U)$ $(i=1,2)$ is bounded, this implies $T(U)$ is bounded.

Next, for any $(u_1,u_2)\in U$ and $0\leq t\leq1/2$, we have
\begin{align*}
\|T_i'(u_1,u_2)\|
&\leq \phi_{p_i}^{-1}(\int_0^{1/2} q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau)) d\tau)\\
&\leq \phi_{p_i}^{-1}(\int_0^{1/2} q_i(\tau) d\tau)
\phi_{p_i}^{-1}(\sup\{f_i(t,u_1,u_2):t\in[0,1],\ (u_1,u_2)\in U\}).
\end{align*}
Then $T(U)$ is equicontinuous; that is, $T(U)$ is a relatively
compact set according to the Ascoli-Arzela theorem.

(2) We show that $T$ is continuous.
Let $(u_1^{(n)},u_2^{(n)})\in U$ and converge uniformly to
$(u_1^{(0)},u_2^{(0)})$, then
\begin{align*}
&T_i(u_1^{(n)},u_2^{(n)})(t)\\
&\leq T_i(u_1^{(n)},u_2^{(n)})(\frac{1}{2})\\
&= \begin{cases}
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1^{(n)}(\tau),u_2^{(n)}(\tau))d\tau\Big)\\
+\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1^{(n)}(\tau),u_2^{(n)}(\tau)) d\tau\Big)ds,
& 0\leq t\leq 1/2,\\
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{1/2}^{\eta}q_i(\tau)f_i(\tau,u_1^{(n)}(\tau),u_2^{(n)}(\tau))d\tau\Big)\\
+\int_{1/2}^1\phi_{p_i}^{-1}\Big(\int_{1/2}^s
q_i(\tau)f_i(\tau,u_1^{(n)}(\tau),u_2^{(n)}(\tau))d\tau\Big)ds,
& 1/2\leq t\leq1,
\end{cases} 
\\
&\leq  \begin{cases}
\Big[m_i\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)d\tau\Big)+
\int_0^{\frac{1}{2}}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau) d\tau\Big)ds\Big]\\
\times\phi_{p_i}^{-1}(\sup\{f_i(t,u_1,u_2):t\in[0,1],\;
(u_1,u_2)\in U\}),
 & 0\leq t\leq 1/2,\\
\Big[m_i\phi_{p_i}^{-1}
\Big(\int_{1/2}^{\eta}q_i(\tau)d\tau\Big)+
\int_{\frac{1}{2}}^1\phi_{p_i}^{-1}\Big(\int_{1/2}^s
q_i(\tau) d\tau\Big)ds\Big]\\
\times\phi_{p_i}^{-1}(\sup\{f_i(t,u_1,u_2):t\in[0,1],\
(u_1,u_2)\in U\})
 & 1/2\leq t\leq1.
\end{cases} 
\end{align*}
Thus, by the dominated convergence theorem, we can get the limit
\begin{align*}
&\lim_{n\to\infty}T_i(u_1^{(n)},u_2^{(n)})(t)\\
&= \begin{cases} 
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1^{(0)}(\tau),u_2^{(0)}(\tau))d\tau\Big)\\
+\int_0^{t}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1^{(0)}(\tau),u_2^{(0)}(\tau))
d\tau\Big)ds,
& 0\leq t\leq 1/2,\\
g_i\circ\phi_{p_i}^{-1}
\Big(\int_{1/2}^{\eta}q_i(\tau)f_i(\tau,u_1^{(0)}(\tau),u_2^{(0)}(\tau))d\tau\Big)\\
+\int_{t}^1\phi_{p_i}^{-1}\Big(\int_{1/2}^s
q_i(\tau)f_i(\tau,u_1^{(0)}(\tau),u_2^{(0)}(\tau))d\tau\Big)ds,
& 1/2\leq t\leq1;
\end{cases}
\end{align*}
i.e.,
$\lim_{n\to\infty}T_i(u_{n},u_{n})(t)=T_i(u_0,u_0)(t)$.
So $T_i\ (i=1,2)$ is continuous on $U$. It follows that $T(U)$ is
continuous on $U$. Hence we complete the proof of Lemma \ref{lem3.4}.
\end{proof}

\section{Existence of multiple symmetric positive solutions 
to \eqref{e1.1}-\eqref{e1.2}}

Now for convenience we use the following notation. Let
\[
\overline{\gamma}_i
=\frac{\delta \int_{\delta}^{1/2}\phi_{p_i}^{-1}
\big(\int_{s}^{1/2} q_i(\tau)d\tau\big)ds}
{m_i\phi_{p_i}^{-1}\big(\int_{\xi}^{1/2}q_i(\tau)d\tau\big)
+\int_0^{1/2}\phi_{p_i}^{-1}\big(\int_{s}^{1/2}
q_i(\tau)d\tau\big)ds},\quad i=1,2,
\]
$\gamma_i=\delta\overline{\gamma}_i$,
$\gamma=\min\{\gamma_1,\gamma_2\}$,
$K_{\rho}=\{(u_1,u_2)\in K: \|(u_1,u_2)\|<\rho\}$,
\begin{align*}
\Omega_{\rho}&=\{(u_1,u_2)\in K: \min_{\delta\leq
t\leq1-\delta}(u_1(t)+u_2(t))<\gamma\rho\},\\
&=\{(u_1,u_2)\in K,\ \gamma\|(u_1,u_2)\|\leq
\min_{\delta\leq t\leq1-\delta}(u_1(t)+u_2(t))<\gamma\rho\},
\end{align*}
\begin{gather*}
 f_{i [\gamma \rho,\rho]}=\min\{\min_{t\in
[\delta,1-\delta]}\frac{f_i(t,u_1,u_2)}{\phi_{p_i}(\rho)}:u_1+u_2\in[\gamma
\rho,\rho]\},
\\
 f_i^{[0,\rho]}=\max\{\max_{t\in
[0,1]}\frac{f_i(t,u_1,u_2)}{\phi_{p_i}(\rho)}: u_1+u_2\in[0,\rho]\},
\\
 f_{i\alpha}=\liminf_{(u_1, u_2)\to\alpha} \min_{t\in
[\delta,1-\delta]}\frac{f_i(t,u_1,u_2)}{\phi_{p_i}(u_1+u_2)},
\\
 f_i^{\alpha}=\limsup_{(u_1, u_2)\to\alpha}
\max_{t\in [0,1]}\frac{f_i(t,u_1,u_2)}{\phi_{p_i}(u_1+u_2)},\quad
 (\alpha:=\infty,\text{ or }0),
\\
\frac{1}{N_i}=2\Big[m_i\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)d\tau\Big)
+\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)d\tau\Big)ds\Big],
\\
\frac{1}{M_i}=2\delta
\int_{\delta}^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2} q_i(\tau)d\tau\Big)ds,
\end{gather*}
where $i=1,2$ and $(u_1, u_2)\to\alpha$ if and only if 
$\|u_1\|+\|u_2\|\to\alpha$

\begin{remark} \label{rmk3.1} \rm
By (H1) it is to show that $0<N_i,M_i<\infty$ and
$M_i\gamma\leq M_i\gamma_i=M_i\delta\overline{\gamma}_i=\delta
N_i<N_i$, $i=1,2$.
\end{remark}

\begin{lemma}[\cite{la}] \label{lem3.1}
The set $\Omega_{\rho}$ defined above has the following properties:
\begin{itemize}
\item[(a)] $\Omega_{\rho}$ is open relative to $K$;
\item[(b)] $K_{\gamma\rho}\subset\Omega_{\rho}\subset K_{\rho}$;
\item[(c)] $(u_1,u_2)\in\partial\Omega_{\rho}$ if and only if
$\min_{\delta\leq t\leq1-\delta}(u_1(t)+u_2(t))= \gamma\rho$;
\item[(d)] If $(u_1,u_2)\in\partial\Omega_{\rho}$, then
$\gamma\rho\leq u_1+u_2\leq\rho$ for $t\in[\delta,1-\delta]$.
\end{itemize}
\end{lemma}

We are now ready to apply Lemma \ref{lem1.1} to the operator $T$ to
give sufficient conditions for the existence of multiple symmetric
positive solutions to \eqref{e1.1}-\eqref{e1.2}.


\begin{theorem} \label{thm3.1}
 Assume that {\rm (H1)--(H3)} hold,
and suppose that $f_i$ satisfies the following conditions:
\begin{itemize}
\item[(H4)] There exist $\rho_1, \rho_2, \rho_3\in (0, \infty)$,
with  $\rho_1<\gamma\rho_2<\rho_2<\rho_3$ such that
$$
f_i^{[0,\rho_1]}<\phi_{p_i}(N_i),\quad
f_{i [\gamma\rho_2,\rho_2]}>\phi_{p_i}(\gamma M_i),\quad
f_i^{[0,\rho_3]}\leq\phi_{p_i}(N_i),\quad i=1,2.
$$ 
\end{itemize}
Then \eqref{e1.1}-\eqref{e1.2} has three symmetric positive solutions
in $K$.
\end{theorem}

\begin{proof}
 Recall that  \eqref{e1.1}-\eqref{e1.2} has a
solution $(u_1,u_2)$ if and only if the operator $T$ has a fixed
point. Thus we set out to verify that the operator $T$ satisfies
Lemma \ref{lem1.1} which will prove the existence of three fixed
points of $T$ which satisfies the conclusion of the theorem.

Firstly, we show that $i_k(T, K_{\rho_1})=1$.
In fact, by the definition of $T$ and
$f_i^{[0,\rho_1]}<\phi_{p_i}(N_i)$, for 
$(u_1,u_2)\in \partial K_{\rho_1}$, we have
\begin{align*}
\|T_i(u_1,u_2)\|
&= \max _{0\leq t\leq 1}|T_i(u_1,u_2)(t)|
=T_i(u_1,u_2)(1/2)\\
&= g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau))d\tau\Big)\\
&+\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1(\tau),u_2(\tau)) d\tau\Big)ds\\
&< m_i\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)d\tau\phi_{p_i}(\rho_1N_i)\Big)\\
&\quad +\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)d\tau\phi_{p_i}(\rho_1N_i)\Big)ds\\
&\leq \rho_1N_i\Big[m_i\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)d\tau\Big)
+\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)d\tau\Big)ds\Big]\\
&=\frac{\rho_1}{2}
=\frac{\|(u_1,u_2)\|}{2},\quad i=1,2.
\end{align*}
Thus, 
$$
\|T(u_1u_2)\|=\|(T_1(u_1u_2),T_2(u_1u_2))\|
=\|(T_1(u_1u_2)\|+\|T_2(u_1u_2))\|< \|(u_1,u_2)\|,
$$
by Lemma\ref{lem1.1} (1), one has $i_k(T, K_{\rho_1})=1$.

Secondly, we show that $i_k(T, \Omega_{\rho_2})=0$.
Let
$(e_1(t),e_2(t))\equiv(\frac{1}{2},\frac{1}{2})$
for $t\in[0,1]$, then $(e_1(t),e_2(t))\in \partial K_1$. We
claim that
\begin{align*}
(u_1,u_2) \neq T(u_1,u_2) + \lambda (e_1,e_2) 
&= (T_1(u_1,u_2),T_2 (u_1,u_2)) + \lambda (e_1,e_2)\\
&= (T_1(u_1, u_2)+ \lambda e_1, T_2 (u_1,u_2) +
\lambda e_2);
\end{align*}
that is,  $u_i\neq T_i(u_1, u_2)+ \lambda e_i$, for
$(u_1,u_2)\in \partial\Omega_{\rho_2}$, $\lambda>0$, $i=1,2$.

In fact, if not, there exist 
$(u_1^{0},u_2^{0})\in \partial\Omega_{\rho_2}$, $\lambda_0>0$ such that
$(u_1^{0},u_2^{0}) = T (u_1^{0},u_2^{0})+ \lambda_0(e_1,e_2)$. 
From Lemma \ref{lem3.3} and 
$f_{i [\gamma \rho_2,\rho_2]}>\phi_{p_i}(\gamma M_i)$, we have
\begin{align*}
u_i^{0}(t)&= T_i(u_1^{0},u_2^{0})(t)+\lambda_0e_i(t)
\\
&\geq  \delta\|T_i(u_1^{0},u_2^{0})\|+\frac{\lambda_0}{2}
=\delta T_i(u_1^{0},u_2^{0})(\frac{1}{2})+\frac{\lambda_0}{2}\\
&= \delta\Big[g_i\circ\phi_{p_i}^{-1}
\Big(\int_{\xi}^{1/2}q_i(\tau)f_i(\tau,u_1^{0}(\tau),u_2^{0}(\tau))d\tau\Big)\\
&\quad +\int_0^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1^{0}(\tau),u_2^{0}(\tau)) d\tau\Big)ds\Big]
+\frac{\lambda_0}{2}\\
&\geq \delta\int_{\delta}^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)f_i(\tau,u_1^{0}(\tau),u_2^{0}(\tau))
d\tau\Big)ds+\frac{\lambda_0}{2}\\
&>\delta\int_{\delta}^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)d\tau\phi_{p_i}(\gamma\rho_2 M_i)\Big)ds+\frac{\lambda_0}{2}\\
&=  \delta\gamma\rho_2
M_i\int_{\delta}^{1/2}\phi_{p_i}^{-1}\Big(\int_{s}^{1/2}
q_i(\tau)d\tau\Big)ds+\frac{\lambda_0}{2}\\
&= \frac{\gamma \rho_2+\lambda_0}{2},\ i=1,2.
\end{align*}
Thus  $u_1^{0}(t)+u_2^{0}(t)>\gamma \rho_2+\lambda_0$. This
implies that $\gamma \rho_2>\gamma \rho_2+\lambda_0$, which is
a contradiction. Hence by Lemma \ref{lem1.1} (2), it follows that
$i_k(T, \Omega_{\rho_2})=0$.

Finally, similar to the proof of $i_k(T, K_{\rho_1})=1$, we can
obtain that $i_k(T, K_{\rho_3})=1$.
Therefore, it follows from Lemma \ref{lem1.1} that $T$ has three
fixed points $u_1\in K_{\rho_1}$,
$u_2\in \Omega_{\rho_2}\backslash\overline{K}_{\rho_1}$ and 
$u_3\in K_{\rho_3}\backslash\overline{\Omega}_{\rho_2}$.
\end{proof}

\begin{theorem} \label{thm3.2} 
Assume that {\rm (H1)--(H3)} hold,
and suppose that $f_i$ satisfies the  conditions:
\begin{itemize}
\item[(H5)] There exist $\rho_1, \rho_2, \rho_3\in (0, \infty)$, with
$\rho_1<\rho_2<\gamma\rho_3$ such that
$$
f_{i[\gamma\rho_1,\rho_1]}>\phi_{p_i}(\gamma M_i),\quad 
f_i^{[0,\rho_2]}<\phi_{p_i}(N_i),\quad 
f_{i[\gamma\rho_3,\rho_3]}\geq\phi_{p_i}(\gamma M_i), \quad
i=1,2.
$$ 
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has  two symmetric
positive solutions in $K$.
\end{theorem}

 The proof of the above theorem is similar to that of Theorem
\ref{thm3.1}; we omit it here.
 As a special case of Theorem
\ref{thm3.1}, we obtain the following result.

\begin{corollary} \label{cor3.3} 
Assume that {\rm (H1)--(H3)} hold.
In addition, if there exists $\rho\in(0,\infty)$ such that 
\begin{itemize} 
\item[(H6)] $ 0\leq f_i^{0}<\phi_{p_i}(N_i)$,
$ f_{i [\gamma\rho,\rho]}>\phi_{p_i}(\gamma M_i)$,
$0\leq f_i^{\infty}<\phi_{p_i}(N_i)$, $i=1,2$.
\end{itemize}
 Then \eqref{e1.1}-\eqref{e1.2} has  three symmetric positive solutions
in $K$.
\end{corollary}

\begin{proof} 
We show that (H6) implies (H4). It is easy
to verify that $0\leq f_i^{0}<\phi_{p_i}(N_i)$ implies that
there exists $\rho_1\in(0,\gamma \rho)$ such that
$f_i^{[0,\rho_1]}<\phi_{p_i}(N_i)$. Let $k_i\in
(f_i^{\infty},\phi_{p_i}(N_i))$, then there exists $r>\rho$
such that $\max_{0\leq t\leq1}f_i(t,u_1,u_2)\leq
k_{_i}\phi_{p_i}(u_1+u_2)$ for $u_1+u_2\in [r,\infty)$
since $0\leq f_i^{\infty}<\phi_{p_i}(N_i)$.
Let
\begin{gather*}
\beta_i=\max\big\{\max_{0\leq t\leq1}f_i(t,u_1,u_2): 0\leq u_1+u_2\leq r\big\},\\
\rho_3>\max\big\{\phi_{p_1}^{-1}(\frac{\beta_1}{\phi_{p_1}(N_1)-k_1}),
\phi_{p_2}^{-1}(\frac{\beta_2}{\phi_{p_2}(N_2)-k_2}),
\rho\big\}.
\end{gather*}
Then 
$$
\max_{0\leq t\leq1}f_i(t,u_1,u_2)\leq
k_i\phi_{p_i}(u_1+u_2) +\beta_i\leq
k_i\phi_{p_i}(\rho_3)
+\beta_i<\phi_{p_i}(N_i)\phi_{p_i}(\rho_3)
$$
for $u_1+u_2\in [0, \rho_3]$.  This implies
$f_i^{[0,\rho_3]}\leq\phi_{p_i}(N_i)$ and (H4) holds.
\end{proof}

Similarly, as a special case of Theorem \ref{thm3.2}, we obtain the
following result.

\begin{corollary} \label{cor3.4} 
Assume that {\rm (H1)--(H3)} hold.
In addition, if there exists $\rho\in(0,\infty)$
such that the following conditions hold
\begin{itemize}
\item[(H7)] $\phi_{p_i}(M_i)<f_{i 0}\leq \infty$, 
$f_{i}^{[0,\rho]}<\phi_{p_i}(N_i)$,
 $\phi_{p_i}(M_i)<f_{i\infty}\leq \infty$, $i=1,2$.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has  two symmetric positive
solutions in $K$.
\end{corollary}

By an argument similar to that of Theorem \ref{thm3.1}, we can
obtain the following results.

\begin{theorem} \label{thm3.5}
 Assume that {\rm (H1)--(H3)} hold.
In addition, one of  the following two condition holds
\begin{itemize}
\item[(H8)] There exist $\rho_1, \rho_2\in (0,\infty)$ with
$\rho_1<\gamma\rho_2$ such that
$$ 
f_{i}^{[0 ,\rho_1]}\leq\phi_{p_i}(N_i),\quad 
f_{i[\gamma \rho_2,\rho_2]}\geq\phi_{p_i}(\gamma M_i)\quad
i=1,2;
$$

\item[(H9)] There exist $\rho_1, \rho_2\in (0,\infty)$
with $\rho_1<\rho_2$ such that
$$
f_{i [\gamma \rho_1,\rho_1]}\geq\phi_{p_i}(\gamma M_i), \quad 
f_{i}^{[0,\rho_2]}\leq\phi_{p_i}(N_i) \quad i=1,2.
$$ 
\end{itemize}
Then \eqref{e1.1}-\eqref{e1.2} has  one symmetric positive solution in
$K$.
\end{theorem}

\begin{corollary} \label{cor3.6} 
Assume that {\rm (H1)--(H3)} hold.
In addition, one of  the following two conditions holds
\begin{itemize}
\item[(H10)] $0\leq f_i^{0}<\phi_{p_i}(N_i)$,
$\phi_{p_i}(M_i)<f_{i \infty}\leq \infty$, $i=1,2$.

\item[(H11)] $0\leq f_i^{\infty}<\phi_{p_i}(N_i)$,
$\phi_{p_i}(M_i)<f_{i 0}\leq \infty$, $i=1,2$.
\end{itemize}
Then \eqref{e1.1}-\eqref{e1.2} has one symmetric positive solution
in $K$.
\end{corollary}

\section{Example}

Let $p_i=3$, $q_i(t)=2$, $i=1,2$, in \eqref{e1.1} and
 $\xi=1/3$, $\eta=2/3$, and $g_i$ satisfies (H3) with $m_i=1$, $i=1,2$, in
\eqref{e1.2}. 
We consider the boundary-value problem
\begin{gather}\label{e5.1}
(|u_i'|u_i')'(t)+q_i(t)f_i(t,u_1,u_2)=0,\quad t\in(0,1),\quad i=1,2;\\
\label{e5.2}
 u_i(0)-g_i(u_i'(\frac{1}{3}))=0,\quad
u_i(1)+g_i(u_i'(\frac{2}{3}))=0, \quad  i=1,2,
\end{gather}
where
\begin{gather*}
f_1(t,u_1,u_2)=\begin{cases}
 t(1-t)(u_1+u_2)^{14}+\frac{1}{1000}, & 0\leq u+v\leq3, \\
 t(1-t)\cdot3^{14}+\frac{1}{1000}, & u+v>3,
\end{cases} \\
f_2(t,u_1,u_2)=\begin{cases}
 \sqrt{t(1-t)}(u_1+u_2)^{13}+\frac{1}{1000}, & 0\leq u+v\leq3, \\
 \sqrt{t(1-t)}\cdot3^{13}+\frac{1}{1000}, & u+v>3,
\end{cases} 
\end{gather*}
Choose $\rho_1=1$, $\rho_2=64(\sqrt{6}+\sqrt{2})$,
$\rho_3=1500$, $\delta=\frac{1}{4}$. we note that
$$
M_i=24,\quad N_i=\frac{3(\sqrt{6}-\sqrt{2})}{4},\quad
 \gamma=\frac{\sqrt{6}-\sqrt{2}}{128}.
$$
Consequently, $f_i(t,u_1,u_2)$, $i=1,2$, satisfies
\begin{gather*}
f_1^{[0,\rho_1]}=0.25<\phi_{p_1}(N_1)= 0.60, \quad
f_2^{[0,\rho_1]}=0.5<\phi_{p_1}(N_1)= 0.60,\\
f_{1 [\gamma \rho_2,\rho_2]} = 0.05>\phi_{p_1}(\gamma M_1)=0.04,\quad 
f_{2 [\gamma \rho_2,\rho_2]}= 0.06>\phi_{p_1}(\gamma M_1)= 0.04,\\
f_1^{[0,\rho_3]}= 0.53<\phi_{p_1}(N_1)= 0.60,\quad 
f_2^{[0,\rho_3]}= 0.35<\phi_{p_1}(N_1)= 0.60.
\end{gather*}
Then all the conditions for Theorem \ref{thm3.1} hold. Thus,
 \eqref{e1.1}-\eqref{e1.2} has  three symmetric
positive solutions in $K$.


\subsection*{Acknowledgements}
The authors would like to thank the anonymous referee for his or her 
helpful suggestions.

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