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

\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2007(2007), No. 23, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/23\hfil Multiple positive solutions]
{Multiple positive solutions for fourth-order three-point
$p$-Laplacian boundary-value problems}

\author[H. Feng, M. Feng, M. Jiang, W. Ge\hfil EJDE-2007/23\hfilneg]
{Hanying Feng, Meiqiang Feng, Ming Jiang, Weigao Ge}  % in alphabetical order

\address{Hanying Feng \newline
 Department of Mathematics, Beijing Institute of Technology \\
 Beijing 100081, China.\newline
 Department of Mathematics, Shijiazhuang,
 Mechanical Engineering College,
 Shijiazhuang 050003, China}
\email{fhanying@yahoo.com.cn}

\address{Meiqiang Feng \newline
 Department of Mathematics\\
 Beijing Institute of Technology\\
 Beijing 100081, China\newline
 Department of Fundamental Sciences,
 Beijing Information Technology Institute,
 Beijing, 100101, China}
\email{meiqiangfeng@sina.com}

\address{Ming Jiang \newline
Department of Mathematics,
Shijiazhuang Mechanical Engineering College\\
Shijiazhuang 050003, China} 
\email{jiangming27@163.com}

\address{Weigao Ge \newline
 Department of Mathematics\\
 Beijing Institute of Technology\\
 Beijing 100081, China}
\email{gew@bit.edu.cn}



\thanks{Submitted November 30, 2006. Published February 4, 2007.}
\thanks{Supported by grants 10671012 from  NNSF, and 20050007011
 from  SRFDP of China}
\subjclass[2000]{34B10, 34B15, 34B18} 
\keywords{Fourth-order boundary-value problem; one-dimensional $p$-Laplacian;
\hfill\break\indent
 five functional fixed point theorem; positive solution}

\begin{abstract}
 In this paper, we study the three-point  boundary-value problem
 for a fourth-order one-dimensional $p$-Laplacian differential
 equation
 $$
 \big(\phi_p(u''(t))\big)''+ a(t)f\big(u(t)\big)=0,
 \quad t\in (0,1),
 $$
 subject to the nonlinear boundary conditions:
\begin{gather*}
 u(0)=\xi u(1),\quad u'(1)=\eta u'(0),\\
 (\phi _{p}(u''(0))' =\alpha _{1}(\phi _{p}(u''(\delta))',
 \quad u''(1)=\sqrt[p-1]{\beta  _{1}}u''(\delta),
\end{gather*}
 where $\phi_{p}(s)=|s|^{p-2}s$, $p>1$. Using the five functional
 fixed point theorem due to Avery, we obtain sufficient conditions
 for the existence of at least three positive  solutions.
\end{abstract}

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

\section{Introduction}

This paper concerns the existence of three positive solutions for
the fourth-order three-point boundary-value problem (BVP for short)
consisting of the $p$-Laplacian differential equation
\begin{equation}\label{e1.1}
\big(\phi_p(u''(t))\big)''+a(t)f(u(t))=0,\quad t\in (0,1),
\end{equation}
with  the nonlinear boundary conditions
\begin{equation}\label{e1.2}
\begin{gathered}
u(0)=\xi u(1),\quad u'(1)=\eta u'(0),\\
(\phi _{p}(u''(0))' =\alpha_{1}(\phi _{p}(u''(\delta))',\quad
 u''(1)=\sqrt[p-1]{\beta_{1}}u''(\delta),
\end{gathered}
\end{equation}
where $f:R\to[0,+\infty)$ and $a:(0,1)\to[0,+\infty)$ are continuous
functions, $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $\alpha_{1}$,
$\beta_{1}\geq0$, $\xi\neq1$, $\eta\neq1$ and $0<\delta<1$.

Two-point boundary-problems for differential equation are used to
describe a number of physical, biological and chemical phenomena.
For additional background and results, we refer the reader to the
monograph by Agawarl, O'Regan and Wong \cite{ag} as well as to the
recent contributions by \cite{av1,el,gr,gr1,mr}.

Boundary-value problems for $n$-th order differential equation
\cite{he,lb,pa} and even-order can arise, especially for
fourth-order equations, in applications, see
\cite{bo,bo1,bo2,bo4,ch} and references therein.

Recently, three-point boundary-value problems of the differential
equations were presented and studied by many authors, see \cite{
fw1,gu,gu1,mr1} and the references cite there. However, three-point
BVP \eqref{e1.1}, \eqref{e1.2} have not received as much attention
in the literature as Lidstone condition BVP
\begin{equation}\label{e1.3}
  \begin{gathered}
  u''''(t)=a(t)f(u(t)),\quad t\in (0,1),\\
  u(0)=u(1)=u''(0)=u''(1)=0,
\end{gathered}
  \end{equation}
and the three-point BVP for the second-order differential equation
\begin{equation}\label{e1.4}
  \begin{gathered}
  u''(t)+a(t)f(u(t))=0,\quad t\in (0,1),\\
  u(0)=0,\quad u(1)=\alpha u(\eta ),
  \end{gathered}
  \end{equation}
that were extensively considered,  in \cite{gr,gr1,mr} and
\cite{mr1}, respectively. The results of existence of positive
solutions of BVP \eqref{e1.1}, \eqref{e1.2} are relatively scarce.

Most recently, Liu and Ge studied two class of  four-order
four-point BVPs successively in \cite{ly, ly1}. They proved that
existence of at least two or three positive solutions. To the best
of our knowledge, existence results of multiple positive solutions
for fourth-order three-point BVP \eqref{e1.1}, \eqref{e1.2} have not
been found in literature. Motivated by the works in \cite{ly, ly1},
the purpose of this paper is to establish the existence of at least
three positive solutions of \eqref{e1.1}, \eqref{e1.2}.

For the remainder of the paper, we assume that:
\begin{itemize}
\item[(i)] $0 < \int_{0}^{1}a(s)ds < \infty$;

\item[(ii)] $q$ satisfies $\frac{1 }{p} + \frac{1}{q}=1$ and
  $\phi_q(z) = |z|^{q-2}z$.
\end{itemize}


\section{Background and definitions}

For the convenience of the reader, we provide some background
material from the theory of cones in Banach spaces. We also state in
this section a fixed point theorem by Avery.


\begin{definition} \label{de2.1} \rm
Let $X$ be a real Banach space. A nonempty closed set $P \subset X$
is said to be a cone provided that
\begin{itemize}
 \item[(i)] $x\in P$ and $\lambda\geq0$ implies $\lambda x\in X$,
 and

\item[(ii)] $x\in P$ and $-x \in P$ implies $x=0$.
\end{itemize}
\end{definition}

 Every cone $P\subset X$ induces an ordering in $X$ given by
 $x\leq y$  if and only if  $y-x\in P$.

\begin{definition} \label{2.2} \rm
The map $\psi$ is said to be a nonnegative continuous concave
functional on a cone $P$ of a real Banach space $E$ provided that
$\psi: P\to [0, \infty)$ is continuous and
$$
\psi(tx+(1-t)y)\geq t\psi(x)+(1-t)\psi(y)
$$
for all $x, y\in P$ and  $0 \leq t \leq 1$. Similarly, we say the
map $\beta$ is a nonnegative continuous convex functional on a cone
$P$ of a real Banach space $E$ provided that $\beta: P\to [0,
\infty)$ is continuous and
$$
\beta(tx+(1-t)y)\leq t\beta(x)+(1-t)\beta(y)
$$
for all $x, y\in P$  and $0 \leq t\leq1$.
\end{definition}

Let $\gamma, \beta, \theta$ be nonnegative, continuous, convex
functionals on $P$ and $\alpha, \psi$ be nonnegative, continuous,
concave functionals on $P$. Then for nonnegative numbers $h, a, b,
d$ and $c$ we define the following sets:
\begin{gather*}
P(\gamma,c)=\{x\in P:\gamma(x)<c\},\\
P(\gamma,\alpha,a,c)=\{x\in P:a\le \alpha (x),\; \gamma(x)\le
c\},\\
 Q(\gamma,\beta,d,c)=\{x\in P: \beta (x)\le d,\; \gamma (x)\le
c\},\\
 P(\gamma ,\theta ,\alpha,a,b,c)=\{x\in P:\ a\le \alpha (x),\;
\theta (x)\le b,\; \gamma (x)\le c\}, \\
Q(\gamma,\beta,\psi,h,d,c)=\{x\in P: h\le \psi(x),\; \beta (x)\le
d,\; \gamma(x)\le c\} .
\end{gather*}

To prove our results, we need the following Five Functionals Fixed
Point Theorem due to Avery \cite{av} which is a generalization of
the Leggett-Williams fixed point theorem.

\begin{theorem} \label{thm2.1}
Suppose $X$ is a real Banach space and $P$ is a cone of $X$,
$\gamma, \beta, \theta$ are three nonnegative, continuous, convex
functionals and $\alpha, \psi$ are nonnegative, continuous, concave
functionals such that
$$
\alpha (x)\le \beta(x),\quad ||x||\le M\gamma (x)
$$
for $x\in \overline{P(\gamma,c)}$ and some positive numbers $c, M$.
Again, assume that
$$
T: \overline{P(\gamma,c)}\to\overline{P(\gamma,c)}
$$
 be a completely continuous operator
 and  there are positive numbers $h, d, a, b$ with $0<d<a$ such that
\begin{itemize}
\item[(i)] $\{x\in P(\gamma,\theta,\alpha ,a,b,c):\alpha (x)>a\}\neq
\emptyset$ and  $x\in P(\gamma,\theta ,\alpha,a,b,c)$ implies
$\alpha (Tx)>a$.

\item[(ii)] $\{x\in Q(\gamma,\beta,\psi ,h,d,c): \beta (x)<d\}\neq
\emptyset$
  and $x\in Q(\gamma,\beta ,\psi,h,d,c)$ implies $\beta(Tx)<d$.

\item[(iii)] $x\in P(\gamma,\alpha ,a,c)$ with $\theta (Tx)>b$
implies $\alpha (Tx)>a$.

\item[(iv)] $x\in Q(\gamma,\beta ,d,c)$ with $\psi(Tx)<h$  implies
$\beta (Tx)<d$.
\end{itemize}
Then $T$ has at least three fixed points $x_1,x_2,x_3\in
\overline{P(\gamma,c)}$ such thah
$$
\beta(x_1)<d,\quad a<\alpha(x_2),\quad d<\beta (x_3),\quad
\mbox{with} \quad \alpha (x_3)<a.
$$
\end{theorem}


\section{Related lemmas}

\begin{lemma}[\cite{ly}] \label{lm3.1}
Suppose $f\in C(R,R)$, then the  three-point BVP
\begin{equation}\label{e3.1}
\begin{gathered}
-y''=f(t),\quad t\in (0,1)\\
y'(0)=\alpha_1y'(\delta),\quad y(1)=\beta_1y(\delta)
\end{gathered}
\end{equation}
has a unique solution
$$
y(t)=\int_0^1 g(t,s)f(s)ds,\quad t\in (0,1),
$$
where $M=(1-\alpha_1)(1-\beta_1)\neq0$ and
$$
g(t,s)=\frac{1}{M}
\begin{cases}
1-\beta_1\delta-t+\beta_1t,
&\text{if }\ 0\le s\le t<\delta<1\\
&\text{or }\ 0\le s\le \delta\le t\le 1,\\[3pt]
1-\beta_1\delta+(1-\beta_1)(\alpha_1s-s-\alpha_1t),
&\text{if }\ 0\le t\le s\le \delta<1,\\[3pt]
1-\alpha_1-\beta_1s+\alpha_1\beta_1s-t\\
+\alpha_1t+\beta_1t+\alpha_1\beta_1t,
&\text{if }\ 0\le \delta\le s\le t\le 1,\\[3pt]
(1-s)(t-\alpha_{1}),
&\text{if }\ 0<\delta\le t\le s\le 1\\
&\textrm{or }\ 0\le t<\delta\le s\le 1.
\end{cases}
$$
\end{lemma}

\begin{lemma}[\cite{md}] \label{lm3.2}
Suppose $f\in C(R,R)$, then the  two-point BVP
\begin{equation}\label{e3.2}
\begin{gathered}
-y''=f(t),\quad \quad t\in (0,1)\\
y(0)=\xi y(1),\quad y'(1)=\eta y'(0)
\end{gathered}
\end{equation}
has a unique solution
$$
y(t)=\int_0^1 h(t,s)f(s)ds,\quad t\in[0,1],
 $$
where $M_1=(1-\xi)(1-\eta)\neq 0$ and
$$
h(t,s)=\frac{1}{M_1} \begin{cases}
s+\eta(t-s)+\xi\eta(1-t), &0\leq s\leq t\leq1, \\
t+\xi(s-t)+\xi\eta(1-s), &0\leq t\leq s\leq  1.
\end{cases}
$$
\end{lemma}

\begin{remark} \label{rmk3.1} \rm
 It is easy to check that if $\alpha_{1}<1$, and
$0\leq\beta_{1}<1$, then $g(t,s)\geq0$ for
$(t,s)\in[0,1]\times[0,1]$. If $\xi,\eta\geq0$ and
$M_1=(1-\xi)(1-\eta)\geq 0$, then $h(t,s)\geq0$ for
$(t,s)\in[0,1]\times[0,1]$

If $u(t)$ is a solution of BVP \eqref{e1.1} and \eqref{e1.2}. By
Lemma \ref{lm3.1} and \eqref{e3.1}, one has
\begin{equation}\label{e3.3}
\phi_p(u''(t))=-\int_0^1g(t,s)a(s)f(u(s))ds.
\end{equation}
By Lemma \ref{lm3.2} and \eqref{e3.2}, one obtains
\begin{equation}\label{e3.4}
u(t)=\int_0^1h(t,s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big) ds.
\end{equation}
\end{remark}

\begin{lemma}[\cite{md}] \label{lm3.3}
 Suppose $0\leq\xi,\eta<1$, $0<t_1<t_2<1$
and $\delta\in(0,1)$.  Then, for $s\in[0,1]$,
\begin{gather}\label{e3.5}
\frac{h(t_1,s)}{h(t_2,s)}\geq \frac{t_1}{t_2},\\
\label{e3.6} \frac{h(1,s)}{h(\delta,s)}\leq\frac{1}{\delta}.
\end{gather}
\end{lemma}

\begin{lemma}[\cite{md}] \label{lm3.4}
Suppose $\xi,\eta>1$, $0<t_1<t_2<1$ and $\delta\in(0,1)$. Then, for
$ s\in[0,1]$,
\begin{gather}\label{e3.7}
\frac{h(t_2,s)}{h(t_1,s)}\geq \frac{1-t_2}{1-t_1}, \\
\label{e3.8} \frac{h(0,s)}{h(\delta,s)}\leq\frac{1}{1-\delta}.
\end{gather}
\end{lemma}


\section{Triple positive solutions to \eqref{e1.1}, \eqref{e1.2}}

Now let the Banach space $E=C[0,1]$ be endowed with the maximum norm
$\|u\| =\max_{ t\in [0,1]}|u(t)|$. Let the two cones
$P_{1},P_{2}\subset X$ defined by
\begin{gather*}
P_1=\left\{u\in X: u(t)\textrm{ is nonnegative, concave,
nondecreasing
on } (0,1)\right\},\\
P_2=\left\{u\in X: u(t) \textrm{ is nonnegative, concave,
nonincreasing on } (0,1)\right\}.
\end{gather*}
Next, choose $t_1 ,t_2,t_3\in (0,1)$ and $t_1<t_2$. Define
nonnegative, continuous, concave functions $\alpha, \psi$ and
nonnegative, continuous, convex functions $\beta, \theta, \gamma$ on
$P_{1}$ by
\begin{gather*}
\gamma (u)=\max_{t\in [0,t_3]}u(t)=u(t_3),\quad u\in P_1,\\
 \psi (u)=\min_{t\in[\delta,1]}u(t)=u(\delta),\quad u\in P_1,\\
 \beta(u)=\max_{t\in[\delta,1]}u(t)=u(1),\quad u\in P_1,\\
 \alpha (u)=\min_{t\in [t_1,t_2]}u(t)=u(t_1),\quad u\in P_1,\\
 \theta  (u)=\max_{t\in [t_1,t_2]}u(t)=u(t_2),\quad u\in P_1.
\end{gather*}
It is easy to verify that $ \alpha (u)=u(t_1)\le u(1)=\beta (u) $
 and $\|u\|=u(1)\le \frac{1}{t_3}u(t_3)=\frac{1}{t_3}\gamma (u) $
 for $u\in P_1$.

\begin{theorem} \label{thm4.1}
Suppose $0\leq\xi$, $\eta<1$, $\alpha_{1}<1$, $0\leq\beta_{1}<1$ and
there exist numbers $0<a<b<c$ such that $ 0<a<b<
\frac{t_{2}}{t_{1}}b\le c $ and $f(w)$ satisfies the following
conditions:
\begin{gather}\label{e4.1}
 f(w)< \phi_p(\frac{a}{C }),\quad 0\le w\le  a,\\
\label{e4.2}
f(w)> \phi_p(\frac{b}{B}),\quad b\le w\le \frac{t_2}{t_1}b,\\
\label{e4.3}
 f(w)\leq \phi_p(\frac{c}{A }),\quad 0\le w\le\frac{1}{t_{3}}c ,
\end{gather}
where
\begin{gather*}
A=\int_0^1h(t_{3},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds,\\
B=\int_0^1h(t_{1},s)
\phi_q\Big(\int_{t_{1}}^{t_{2}}g(s,\tau)a(\tau)d\tau\Big)ds,\\
C=\int_0^1h(1,s) \phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds.
\end{gather*}
Then  \eqref{e1.1}, \eqref{e1.2} has at least three positive
solutions $u_{1}, u_{2}, u_{3}\in\overline{P_{1}(\gamma,c)}$  such
that
\begin{equation} \label{e4.4}
 u_{1}(t_{1})>b,\quad u_{2}(1)<a,\quad u_{3}(t_{1})<b
\end{equation}
with $u_{3}(1)>a$, $u_{i}(\delta)\leq c$ for $i=1,2,3$.
\end{theorem}

\begin{proof} We begin by defining the completely continuous
operator $T: P_{1}\to X $ by  \eqref{e3.4} as
\begin{equation*}(Tu)(t)=\int_0^1h(t,s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds
\end{equation*}
for $u\in P_{1}$. It is easy to prove that \eqref{e1.1},
\eqref{e1.2} has a positive solution $u=u(t)$ if and only if the
operator $T$ has a fixed point on $P_{1}$.

Firstly, we prove $T: \overline{P_{1}(\gamma,c)}\subset
\overline{P_{1}(\gamma,c)}$. For $u\in P_{1}$, by Remark 3.1, it is
easy to check that $Tu\geq0$. Moreover,
\begin{align*}
(Tu)'(t)&=(1-\xi)\Big(\eta\int_0^t
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&\quad +\int_t^1 \phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau
\Big)ds\Big)\geq 0
\end{align*}
and
\begin{equation*}
(Tu)''(t)=-\phi_q\Big(\int_0^1g(t,s)a(s)f(u(s))ds\Big)\leq0.
\end{equation*}
So, we have $TP_{1}\subset P_{1}$.

For $u\in \overline{P_{1}(\gamma,c)}$, $0\leq u(t)\leq\|u\|\leq
\frac{1}{t_3}\gamma (u)\leq\frac{1}{t_3}c$. By \eqref{e4.3}, it
follows that
\begin{align*}
\gamma(Tu)&=\max _{0\leq t\leq t_{3}}(Tu)(t)=(Tu)( t_{3})\\
&=\int_0^1h(t_{3},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&\leq \int_0^1h(t_{3},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\ \phi_{p}(c/A)\Big)ds\\
&=\frac{c}{A}\int_0^1h(t_{3},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds\\
&=\frac{c}{A}A=c.
\end{align*}
Thus, $T: \overline{P_{1}(\gamma,c)}\subset
\overline{P_{1}(\gamma,c)}$.

Secondly, by taking
\begin{gather*}
u_{1}(t)=b+\varepsilon_{1}\quad \mbox{for } 0<\varepsilon_{1}<
\frac{t_{2}}{t_{1}}b-b,\\
u_{2}(t)=a-\varepsilon_{2}\ \mbox{for}\ 0<\varepsilon_{2}< a-\delta
a,
\end{gather*}
It is immediate that
\begin{gather*}
u_{1}(t)\in\{ P(\gamma,\theta,\alpha ,b,\frac{t_{2}}{t_{1}}b,c):\
\alpha (u)>b\}\neq \emptyset,\\
u_{2}(t)\in\{Q(\gamma,\beta,\psi ,\delta a,a,c):\ \beta (u)<a\}\neq
\emptyset.
\end{gather*}
In the following steps, we verify the remaining conditions of
Theorem \ref{thm2.1}. Now the proof is divided into four steps.

Step 1:  We prove that
\begin{equation}\label{e4.5}
u\in P(\gamma,\theta,\alpha ,b,\frac{t_{2}}{t_{1}}b,c)\ \quad
\mbox{implies}\quad  \alpha(Tu)>b.
\end{equation}
In fact, $u(t)\geq u(t_{1})=\alpha(u)\geq b$ for $t_{1}\leq t\leq
t_{2}$, and $u(t)\leq u(t_{2})=\theta(u)\leq\frac{t_{2}}{t_{1}}b$
for $t_{1}\leq t\leq t_{2}$. Thus using \eqref{e4.2}, one gets
\begin{align*}
 \alpha(Tu)
&=\min _{t_{1}\leq t \leq t_{2}}(Tu)(t)=(Tu)(t_{1})\\
&=\int_0^1h(t_{1},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&\geq\int_0^1h(t_{1},s)
\phi_q\Big(\int_{t_{1}}^{t_{2}}g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&>\int_0^1h(t_{1},s)
\phi_q\Big(\int_{t_{1}}^{t_{2}}g(s,\tau)a(\tau)d\tau\
\phi_{p}(b/B)\Big)ds\\
&=\frac{b}{B}\int_0^1h(t_{1},s)
\phi_q\Big(\int_{t_{1}}^{t_{2}}g(s,\tau)a(\tau)d\tau\Big)ds\\
&=\frac{b}{B}B=b.
\end{align*}
Step 2: We show that
\begin{equation}\label{e4.6}
u\in Q(\gamma,\beta,\psi ,\delta a,a,c)\quad \mbox{implies}\quad
 \beta (Tu)<a.
\end{equation}
In fact, $0\leq u(t)\leq u(1)=\beta(u)\leq a$ for $0\leq t\leq 1$,
Thus using \eqref{e4.1}, one arrives at
\begin{align*}
\beta(Tu)
&=\max_{\delta\leq t \leq1}(Tu)(t)=(Tu)(1)\\
&=\int_0^1h(1,s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&<\int_0^1h(1,s)
\phi_q\Big(\int_{0}^{1}g(s,\tau)a(\tau)d\tau\ \phi_{p}(a/C)\Big)ds\\
&=\frac{a}{C}\int_0^1h(1,s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds\\
&=\frac{a}{C}C=a.
\end{align*}
Step 3: We verify that
\begin{equation}\label{e4.7}
u\in Q(\gamma,\beta,a,c)\quad\mbox{with}\quad\psi(Tu)< \delta a\quad
\mbox{implies}\quad \beta(Tu)<a.
\end{equation}
By Lemma \ref{lm3.3},
\begin{align*}
\beta(Tu)&=\max_{\delta\leq t \leq1}(Tu)(t)=(Tu)(1)\\
&=\int_0^1h(1,s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&=\int_0^1\frac{h(1,s)}{h(\delta,s)}h(\delta,s)
\phi_q\Big(\int_{0}^{1}g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&\leq\frac{1}{\delta}(Tu)(\delta)=\frac{1}{\delta}\psi(Tu)<a.
\end{align*}
Step 4: We prove that
\begin{equation}\label{e4.8}
u\in P(\gamma,\alpha,b,c)\quad \mbox{with}\quad
\theta(Tu)>\frac{t_{2}}{t_{1}}b\quad  \mbox{implies}\quad
\alpha(Tu)>b.
\end{equation}
By Lemma \ref{lm3.3},
\begin{align*}
\alpha(Tu)
&=\min_{t_{1}\leq t \leq t_{2}}(Tu)(t)=(Tu)(t_{1})\\
&=\int_0^1h(t_{1},s)
\phi_q\Big(\int_0^1g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&=\int_0^1\frac{h(t_{1},s)}{h(t_{2},s)}h(t_{2},s)
\phi_q\Big(\int_{0}^{1}g(s,\tau)a(\tau)f(u(\tau))d\tau\Big)ds\\
&\geq\frac{t_{1}}{t_{2}}(Tu)(t_{2})=\frac{t_{1}}{t_{2}}\theta(Tu)>b.
\end{align*}
Therefore, the hypotheses of Theorem \ref{thm2.1} are satisfied and
there exist three positive solutions $x_{1}, x_{2}, x_{3}$ for BVP
 \eqref{e1.1},  \eqref{e1.2} satisfying  \eqref{e4.4}.
\end{proof}

Similarly, choose $t_1 ,t_2,t_3\in (0,1)$ and $t_1<t_2$. Define
nonnegative, continuous, concave functions $\alpha, \psi$ and
nonnegative, continuous, convex functions $\beta, \theta, \gamma$ on
$P_{2}$ by
\begin{gather*}
\gamma (u)=\max_{t\in [t_3,1]}u(t)=u(t_3),\quad x\in P_2,\\
 \psi (x)=\min_{t\in[0,\delta]}u(t)=u(\delta),\quad u\in P_2,\\
 \beta(u)=\max_{t\in[0,\delta]}u(t)=u(0),\quad u\in P_2,\\
 \alpha(u)=\min_{t\in [t_1,t_2]}u(t)=u(t_2),\quad u\in P_2,\\
 \theta(u)=\max_{t\in [t_1,t_2]}u(t)=u(t_1),\quad u\in P_2.
\end{gather*}
It is easy to verify that $ \alpha (u)=u(t_2)\le u(0)=\beta (u) $
 and $\|u\|=u(0)\le \frac{1}{t_3}u(t_3)=\frac{1}{t_3}\gamma (u) $
 for $u\in P_2$. So, we obtain the following result.


\begin{theorem} \label{thm4.2} Suppose $\xi,\;\eta>1$,
$\alpha_{1}<1$, $0\leq\beta_{1}<1$ and there exist numbers $0<a<b<c$
such that $ 0<a<b<\frac{1-t_1}{1-t_2}b\le c $ and $f(w)$ satisfy the
following conditions:
\begin{gather}\label{e4.9}
 f(w)< \phi_p(\frac{a}{C }),\quad 0\le w\le  a,\\
\label{e4.10}
f(w)> \phi_p(\frac{b}{B}),\quad b\le w\le \frac{1-t_1}{1-t_2}b,\\
\label{e4.12}
 f(w)\leq \phi_p(\frac{c}{A }),\quad 0\le w\le\frac{1}{t_{3}}c,
\end{gather}
where
\begin{gather*}
A=\int_0^1h(t_{3},s) \phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds,\\
B=\int_0^1h(t_{2},s)
\phi_q\Big(\int_{t_{1}}^{t_{2}}g(s,\tau)a(\tau)d\tau\Big)ds,\\
C=\int_0^1h(0,s)\phi_q\Big(\int_0^1g(s,\tau)a(\tau)d\tau\Big)ds.
\end{gather*}
Then  \eqref{e1.1}, \eqref{e1.2} has at least three positive
solutions $u_{1}, u_{2}, u_{3}\in\overline{P_{1}(\gamma,c)}$  such
that
\[ % %\label{e4.4}
 u_{1}(t_{2})>b,\quad u_{2}(0)<a,\quad u_{3}(t_{2})<b,
\]
with $u_{3}(0)>a$ and $u_{i}(\delta)\leq c$ for $i=1,2,3$.
\end{theorem}

Since the proof of the above theorem is similar to that of Lemma
\ref{lm3.4}, we omit it.

\subsection*{Acknowledgments}
The authors wish to thank the referee for his (or her) valuable
corrections to the original manuscript.

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\end{document}