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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 22, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/22\hfil Monotone positive solutions]
{Monotone positive solutions for $p$-Laplacian equations with
sign changing coefficients and multi-point boundary conditions}

\author[J. Xia, Y. Liu \hfil EJDE-2010/22\hfilneg]
{Jianye Xia, Yuji Liu}

\address{Jianye Xia \newline
Department of Mathematics, Guangdong University of Finance \\
Guangzhou 510320, China}
\email{JianyeXia@sohu.com}

\address{Yuji Liu \newline
Department of Mathematics,
Hunan Institute of Science and Technology\\
 Yueyang 414000, China}
\email{liuyuji888@sohu.com}


\thanks{Submitted October 26, 2009. Published February 4, 2010.}
\thanks{Supported by grants 06JJ5008 and 7004569 from the Natural Science
Foundation of \hfill\break\indent Hunan and Guangdong provinces, China}
\subjclass[2000]{34B10, 34B15, 35B10}
\keywords{Second order differential equation; positive solution
\hfill\break\indent multi-point boundary value problem}

\begin{abstract}
 We prove the existence of three monotone positive solutions
 for the second-order multi-point boundary value problem,
 with sign changing coefficients,
 \begin{gather*}
 [p(t)\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
  x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\
 x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^mb_ix(\xi_i)
 -\sum_{i=1}^mc_ix'(\xi_i).
 \end{gather*}
 To obtain these results, we use a fixed point theorem  for
 cones in Banach spaces. Also we present an example that illustrates
 the main results.
\end{abstract}

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

\section{Introduction}

As is well known, a differential equation defined on the interval
$a\leq t\leq b$ having the form
\begin{gather*}
[p(t)x'(t)]'+(q(t)+\lambda r(t))x(t)=0,\\
a_1x(a)+a_2x'(a)=a_3x(b)+a_4x'(b)=0,
\end{gather*}
is called a Sturm-Liouville boundary-value problem or
Sturm-Liouville system.
Here $p(t)>0$, $q(t)$, the weighting
function $r(t)$, the constants $a_1, a_2, a_3, a_4 $ are given,
and the eigenvalue $\lambda $ is an unspecified parameter.

 Sturm-Liouville boundary-value problems for nonlinear
second-order $p$-Laplacian differential equations have been studied
extensively; see for example \cite{a1,a2,b1,d1,g1,g2,h1,l1,l2,l3,l4,m1,m2}.
The study of existence of positive
solutions for such  problems is complicated since there is
no Green's function for the p-Laplacian $(p\neq  2)$.

Some authors have extend Sturm-Liouville boundary conditions to
nonlinear cases. For example, He, Ge and Peng \cite{h1}, by means of
the Leggett-Williams fixed-point theorem, established criteria for
the existence of at least three positive solutions to the
one-dimensional p-Laplacian boundary-value problem
\begin{equation}  \label{e1}
\begin{gathered}
(\phi(y'))' + g(t)f(t,y) = 0,\quad t\in (0,1),\\
 y(0) - B_0(y'(0)) = 0,\\
y(1) + B_1(y'(1)) = 0,
\end{gathered} \end{equation}
 where $\phi(v)=|v|^{p-2}v$ with $v>1$, under the assumptions
$xB_0(x)\geq 0$, $xB_1(x)\geq 0$ and that there exist constants $M_i>0$ such
that
\begin{equation}  \label{e2}
|B_i(x)|\leq M_i|x|,\quad x\in \mathbb{R}.
\end{equation}
In \cite{l2,l3,l4,h1}, the authors extended \eqref{e1} to a more general
case. They established some existence results of at least
one positive solution of Sturm-Liouville boundary value problems of
higher order differential equations.  Liu \cite{l5}
studied the boundary-value problem
\begin{equation}  \label{e3}
\begin{gathered}{}
[\phi(x^{(n-1)}(t))]'=f(t,x(t),x'(t),\dots,x^{(n-2)}(t)),\quad 0<t<1,\\
x^{(i)}(0)=0\quad\text{for }i=0,1,\dots,n-3,\\
 x^{(n-2)}(0)-B_0(x^{(n-1)}(0))=0,\\
B_1(x^{(n-2)}(1))+x^{(n-1)}(1)=0.
\end{gathered}
\end{equation}
which is a general case of \eqref{e1}.
Liu \cite{l5} established existence  of at least one positive
solution of \eqref{e3} without assumption \eqref{e2}.

The Sturm-Liouville boundary conditions have been extended to
multi-point cases. For example, Ma \cite{m1,m2} studied the problem
\begin{equation} \label{e4}
\begin{gathered}{}
[p(t)x'(t)]'-q(t)x(t)+f(t,x(t))=0,\quad t\in (0,1),\\
\alpha x(0)-\beta p(0)x'(0)=\sum_{i=1}^ma _ix(\xi_i),\\
 \gamma x(1)+\delta p(1)x'(1)=\sum_{i=1}^mb_ix(\xi_i),
\end{gathered}
\end{equation}
where $0<\xi_1<\dots<\xi_m<1$, $\alpha,\beta,\gamma,\delta\geq 0$,
$a_i,b_i\geq 0$ with $\rho=\gamma\beta+\alpha\gamma+\alpha\delta>0$.
By using Green's functions (which is complicate for studying \eqref{e1})
and Guo-Krasnoselskii fixed point theorem \cite{d1,g2}, the
existence and multiplicity of positive solutions for \eqref{e4} were
given. There is no paper discussing  the existence of multiple
positive solutions of \eqref{e4} by using Leggett-Williams fixed point
theorem. Liu in \cite{l3,l4} also studied some  Sturm-Liouville
type multi-point boundary value problems.

In recent papers \cite{l1,p1}, the authors studied the
four-point boundary-value problem
\begin{equation} \label{e5}
\begin{gathered}
(\phi(x'))' + f(t,x(t),x'(t)) = 0,\quad t\in (0,1),\\
 \alpha x(0) -\beta x'(\xi) = 0,\\
\gamma x(1) + \delta x'(\eta) = 0,
\end{gathered}
\end{equation}
where $\phi(x)=|x|^{p-2}x$, $p > 1$, $\phi^{-1}(x)=|x|^{q-2}x$ with
$1/p + 1/q = 1$, $\alpha>0$, $\beta\geq 0$, $\gamma>0$, $\delta\geq 0$,
$0<\xi<\eta<1$, $f$ is continuous and nonnegative. When
$\xi\to 0$ and $\eta\to 1$, \eqref{e5} converges to
a Sturm-Liouville boundary-value problem. So \eqref{e5} can be
seen as a generalized Sturm-Liouville boundary-value problem.

Xu \cite{x1} proved the existence of at least one or two positive
solutions of the differential equation
\begin{equation} \label{e6}
\begin{gathered}{}
[\phi(x'(t))]'+a(t)f(x(t))=0,\quad t\in (0,1),\\
 x'(0)=\sum_{i=1}^ma _ix'(\xi_i),\\
x(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i)
-\sum_{i=s+1}^mc_ix'(\xi_i),
\end{gathered}
\end{equation}
 where $0<\xi_1<\dots<\xi_m<1$, $a_i,b_i,c_i\geq 0$, $a$
and $f$ are continuous functions, $\phi(x)=|x|^{p-2}x$ with $p>1$.

Motivated by above mentioned papers, we investigate the
 boundary-value problem
\begin{equation} \label{e7}
\begin{gathered}{}
[p(t)\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
 x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\
x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^mb_ix(\xi_i)
-\sum_{i=1}^mc_ix'(\xi_i),
\end{gathered}
\end{equation}
 where
\begin{itemize}
\item $0<\xi_1<\dots<\xi_m<1$,
 $\beta\geq 0$, $1\leq k,l\leq m$ and $
 a_i\geq 0,b_i\ge0,c_i\geq 0$ for all $i=1,\dots,m$;


\item $f$ is defined on $[0,1]\times \mathbb{R}\times \mathbb{R}$, continuous,
 nonnegative with $f(t,0,0)\not\equiv 0$ on each subinterval of $[0,1]$;


\item $p$ is defined on $[0,1]$, continuous and positive;

\item $\phi$ is called $p$-Laplacian, $\phi(x)=|x|^{p-2}x$ with $p>1$,
  its inverse function is denoted by $\phi^{-1}(x)$ with
$\phi^{-1}(x)=|x|^{q-2}x$ with $1/p+1/q=1$.
\end{itemize}

Sufficient conditions for the existence of at least three monotone
positive solutions of \eqref{e7} are established by using a fixed point
theorem for cones in Banach spaces.

We remark that the fixed point theorem used here
is different form the one in \cite{x1}.
Our results improve the the results in \cite{x1} since:
Three positive solutions are obtained, while
one or two positive solutions were obtained in \cite{x1};
the nonlinearity in \eqref{e7} depends on $t,x, x'$, while
the nonlinearity in \cite{x1} depends only on $t,x$.
The main result for this article is presented in Section 2,
and an example is given in Section 3.

\section{Main Results}

In this section, we first present some background definitions
and state an important fixed point theorem. Then
the main results are given and proved.

\begin{definition} \label{def2.1} \rm
Let $X$ be a semi-ordered real Banach
space. The nonempty convex closed subset $P$ of $X$ is called a cone
in $X$ if $x+y\in P$ and $ax\in P$ for all $x,y\in P$ and $a\geq 0$,
and $x\in X$ and $-x\in X$ imply $x=0$.
\end{definition}


\begin{definition} \label{def2.2} \rm
 Let $X$ be a semi-ordered real Banach space and $P$ a cone
in $X$. A map $ \psi :P\to [0,+\infty)$ is a nonnegative
continuous concave (or convex) functional map provided $\psi $ is
nonnegative, continuous and satisfies
$$
 \psi (tx+(1-t)y)\geq \text{ (or $\leq$) }
t\psi (x)+(1-t)\psi(y) \quad\text{for all }x,y\in P,t\in [0,1].
$$
\end{definition}


\begin{definition} \label{def2.3} \rm
 Let $X$ be a semi-ordered real Banach
space. An operator $T;X\to X$ is completely continuous if it
is continuous and maps bounded sets into pre-compact sets.
\end{definition}

Let $a_1,a_2,a_3,a_4>0$ be positive constants, $\psi$ be a
nonnegative continuous functional on the cone $P$. Define the sets
as follows:
\begin{gather*}
P(\beta_1;a_4)=\{x\in P:\beta_1(x)<a_4\},\\
P(\beta_1,\alpha_1;a_2,a_4)
 =\{x\in P:\alpha_1(x)\geq a_2,\;\beta_1(x)\leq a_4\},\\
P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)
 =\{x\in P:\alpha_1(x)\geq a_2,\;\beta_2(x)\leq a_3,\;\beta_1(x)\leq a_4\}\
\end{gather*}
and a closed set
$$
R(\beta_1,\psi;a_1,a_4)=\{x\in P:\psi(x)\geq a_1,\beta_1(x)\leq a_4\}.
$$

\begin{theorem}[\cite{b1}] \label{thm2.1}
Let $P$ be a cone in a real Banach space $X$
with the norm $\|\cdot\|$. Suppose that
\begin{itemize}
\item[(1)] $T:P\to P$ is completely continuous;

\item[(2)] $\beta_1$ and $\beta_2$ be nonnegative continuous
convex functionals on $P$, $\alpha_1$ be a nonnegative continuous
concave functional on $P$, and $\psi$  be a nonnegative continuous
functional on $P$ satisfying $\psi(\lambda x)\leq \lambda\psi(x)$ for
all $x\in P$ and $\lambda\in [0,1]$, and $\alpha_1(x)\leq \psi(x)$
and there exists a constant $M>0$ such that $\|x\|\leq M\beta_1(x)$
for all $x\in P$;

\item[(3)] there exist positive numbers $a_1<a_2,a_3$ and $a_4$
such that
\begin{itemize}
\item[(E1)]
 $T(\overline{P(\beta_1;a_4)})\subseteq\overline{P(\beta_1;a_4)}$;

\item[(E2)] $\alpha_1(Tx)>a_2$ for all $x\in
P(\beta_1,\alpha_1;a_2,a_4)$ with $\beta_2(Tx)>a_3$;
\item[(E3)] $\{x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4):
 \alpha_1(x)>a_2\}\neq \emptyset$
and $\alpha_1(Tx)>b$ for all
$x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$;


\item[(E4)] $0\not\in R(\beta_1,\psi;a_1,a_4)$ and
$\psi(Tx)<a_1$ for all $x\in R(\beta_1,\psi;a_1,a_4)$ with
$\psi(x)=a_1$;

\end{itemize}
\end{itemize}
 Then $T$ has at least three fixed points $x_1,
x_2, x_3\in \overline{P(\beta_1;a_4)}$ such that
\begin{gather*}
 \beta_1(x_i )\leq a_4,\quad i = 1, 2, 3;\\
\alpha_1(x_1)>a_2,\quad \psi(x_2)>a_1,\quad \alpha_1(x_2)<a_2,\quad
\psi(x_3)<a_1.
\end{gather*}
\end{theorem}

Choose $X=C^1[0,1]$.  We call $x\leq y$ for $x,y\in X$ if
$x(t)\leq y(t)$ for all $t\in [0,1]$, for $x\in X$, define its norm by
$$
\|x\|=\max\big\{\max_{t\in [0,1]}|x(t)|,\;\max_{t\in
[0,1]}|x'(t)|\big\}.
$$
It is easy to see that $X$ is a
semi-ordered real Banach space.
Let
\begin{align*}
P=\Big\{& y\in X: y(t)\geq 0 \text{ for all }t\in [0,1],\;
 y'(t)\leq 0 \text{ is decreasing on }[0,1],\\
 &y(t)\geq (1-t)y(0)\text{ for all }t\in [0,1],  \\
&x(1)+\beta p(1)x'(1)=\sum_{i=1}^kb_ix(\xi_i)
 -\sum_{i=k+1}^m b_ix(\xi_i)-\sum_{i=1}^mb_ix'(\xi_i),\\
&x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i)
\Big\}.
\end{align*}
Then $P$ is a nonempty cone in $X$. For $\sigma \in (0,1/2)$. Define
functionals from $P$ to $\mathbb{R}$ by
\begin{gather*}
\beta_1(y)=\max_{t\in [0,1]}|y'(t)|,\quad
\psi(x)=\max_{t\in [0,1]}|y(t)|,\\
\beta_2(y)=\max_{t\in [\sigma,1-\sigma]}|y(t)|,\quad
 \alpha_1(y)=\min_{t\in [\sigma,1-\sigma]}|y(t)|,\quad y\in P.
\end{gather*}

Let us list some conditions to be used in this article.
\begin{itemize}
\item[(A1)] $p:[0,1]\to (0,+\infty)$ is continuous;

\item[(A2)] $a_i\geq 0$, $b_i\geq 0,c_i\geq 0$ for all $i=1,\dots,m$
satisfying
\begin{gather*}
 \sum_{i=l+1}^ma_i\Big(\frac{p(0)}{p(\xi_i)}\Big)
 -\sum_{i=1}^la_i\frac{p(0)}{p(\xi_i)}<1,\quad
 \sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i<1, \\
 \sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}\Big)
 \geq \sum_{i=1}^l  a_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}\Big),\quad
 \sum_{i=1}^kb_i\geq \sum_{i=k+1}^mb_i;
\end{gather*}


\item[(A2')]  $a_i\geq 0$, $b_i\geq 0$ for all $i=1,\dots,m$
satisfying
\begin{gather*}
\sum_{i=l+1}^ma_i\Big(\frac{p(0)}{p(\xi_i)}\Big)<1,\quad
\sum_{i=1}^kb_i<1, \\
\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}\Big)
 \geq \sum_{i=1}^l  a_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}\Big),\quad
\sum_{i=1}^kb_i\geq \sum_{i=k+1}^mb_i;
\end{gather*}

\item[(A3)] $\beta \geq 0$;

\item[(A4)] $f:[0,1]\times [0,+\infty)\times \mathbb{R}\to
[0,+\infty)$ is continuous with $f(t,0,0)\not\equiv 0$ on
each sub-interval of $[0,1]$;

\item[(A5)] $\theta:[0,1]\to [0,+\infty)$ is a continuous
function and $\theta(t)\not\equiv 0$ on each subinterval of $[0,1]$.

\end{itemize}
It is easy to see that (A2) holds if (A2') holds.

\begin{lemma} \label{lem2.1}
 Suppose that $p:[0,1]\to (0,+\infty)$
with $p\in C^1[0,1]$, $x\in X$, $x(t)\geq 0$ for all $t\in [0,1]$ and
$[p(t)\phi(x'(t))]'\leq 0$ on $[0,1]$. Then $x$ is concave and
\begin{equation} \label{e8}
x(t)\geq \min\{t,1-t\}\max_{t\in [0,1]}x(t),\quad t\in [0,1].
\end{equation}
\end{lemma}

\begin{proof}
 Suppose $x(t_0)=\max_{t\in [0,1]}x(t)$. If $t_0<1$,
for $t\in (t_0,1)$, since $x'(t_0)\leq 0$, we have
$p(t_0)\phi(x'(t_0))\leq 0$. Then $p(t)\phi(x'(t))\leq 0$ for all
$t\in (t_0,1]$. It follows that $x'(t)\leq 0$ for all $t\in [0,1]$.
Thus $x(t_0)\geq x(t)\geq x(1)\geq 0$. Let
$$
\tau(t)=\frac{\int_{t_0}^t\phi^{-1}\big(\frac{1}{p(s)}\big)ds}
{\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds}.
$$
Then $\tau\in C([t_0,1],[0,1])$ and is increasing on $[t_0,1]$ since
$$
\frac{d\tau}{dt}=\frac{\phi^{-1}\big(\frac{1}{p(t)}\big)}
{\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds}>0,
$$
$\tau(t_0)=0$ and $\tau(1)=1$. Thus
$$
\frac{dx}{dt}=\frac{dx}{d\tau}\frac{d\tau}{dt}
=\frac{dx}{d\tau}\frac{\phi^{-1}\big(\frac{1}{p(t)}\big)}
{\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds}
$$
which implies
$$
p(t)\phi(x'(t))=\phi\Big(\frac{dx}{d\tau}\Big)
\frac{1}{\phi\Big(\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds\Big)}.
$$
Hence
$$
\phi'\Big(\frac{dx}{d\tau}\Big)\frac{\phi^{-1}\big(\frac{1}{p(t)}\big)}
{\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds}\frac{d^2x}{d\tau^2}
=\phi\Big(\int_{t_0}^1\phi^{-1}\big(\frac{1}{p(s)}\big)ds\Big)
\big[\frac{}{}p(t)\phi(x'(t))\big]'\leq 0
$$
for all $t\in [t_0,1]$.
 Note that $\phi'(x)\geq 0$. It follows that
$\frac{d^2x}{d\tau^2}\leq 0$. Together with $x''(\tau)\leq 0(\tau\in
[0,1])$. Then $x'$ is decreasing on $[0,1]$. We get that there exist
$t_0\leq \eta\leq t\leq \xi\leq 1$ such that
\begin{align*}
\frac{x(t_0)-x(1)}{t_0-1}-\frac{x(t)-x(1)}{t-1}
&= -\frac{(t-1)[x(t_0)-x(t)]+(t_0-t)[x(1)-x(t)]}
{(t-1)(t_0-1)}\\
&= -\frac{(t-1)(t_0-t)x'(\eta)+(t_0-t)(1-t)tx'(\xi)}{(t-1)(t_0-1)}\\
&\leq -\frac{(t-1)(t_0-t)x'(\xi)+(t_0-t)(1-t)tx'(\xi)}{(t-1)(t_0-1)}=0.
\end{align*}
It follows that for $t\in (t_0,1)$,
$$
x(t)\ge
x(1)+(t-1)\frac{x(t_0)-x(1)}{t_0-1}
=x(1)\Big(1-\frac{1-t}{1-t_0}\Big)+\frac{1-t}{1-t_0}x(t_0)\ge
(1-t)x(t_0).
$$
If $t_0>0$, for $t\in (0,t_0)$, similarly to above discussion, we
have
$$
x(t)\geq tx(t_0),\;t\in (0,t_0).
$$
Then one gets that $x(t)\geq \min\{t,1-t\}\max_{t\in [0,1]}x(t)$ for
all
$t\in [0,1]$. The proof is complete.
\end{proof}

Consider the boundary-value problem
\begin{equation} \label{e9}
\begin{gathered}{}
 [p(t)\phi(x'(t))]'+\theta(t)=0,\quad t\in (0,1),\\
 x'(0)=-\sum_{i=1}^la _ix'(\xi_i)+\sum_{i=l+1}^ma_ix'(\xi_i),\\
x(1)+\beta x'(1)=\sum_{i=1}^kb_ix(\xi_i)
 -\sum_{i=k+1}^sb_ix(\xi_i)-\sum_{i=s+1}^mc_ix'(\xi_i).
 \end{gathered}
\end{equation}


\begin{lemma} \label{lem2.2}
 Suppose that {\rm (A1)--(A5)} hold. If $x$ is a
solution of \eqref{e9}, then $x$ is concave, decreasing and positive on
$(0,1)$.
\end{lemma}

\begin{proof}
 Suppose $x$ satisfies \eqref{e9}. It follows from the
assumptions that $px'$ is decreasing on $[0,1]$.
Lemma \ref{lem2.1} implies that $x'$ is decreasing on $[0,1]$.
Then $x$ is concave on $[0,1]$.

First, we prove that $x'(0)\leq 0$. One sees from (A2) and the
deceasing property of $p(t)\phi(x'(t))$ that
\begin{align*}
x'(1)&= -\sum_{i=1}^la
_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)\phi^{-1}(p(\xi_i)\phi(x'(\xi_i)))
+\sum_{i=l+1}^ma_i\big(\frac{1}{p(\xi_i)}\big)
 \phi^{-1}(p(\xi_i)\phi(x'(\xi_i)))\\
&\leq -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 \phi^{-1}(p(\xi_l)\phi(x'(\xi_l)))\\
&\quad +\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 \phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))\\
&= \Big(\sum_{i=l+1}^ma_i\phi^{-1}
\big(\frac{1}{p(\xi_i)}\big)-\sum_{i=1}^la_i\phi^{-1}
\big(\frac{1}{p(\xi_i)}\big) \Big)
\phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))\\
&\quad +\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
[\phi^{-1}(p(\xi_{l+1})\phi(x'(\xi_{l+1})))-\phi^{-1}(p(\xi_l)
 \phi(x'(\xi_l)))]\\
&\leq \Big(\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 \Big)\phi^{-1}(p(\xi_{l+1}\phi(x'(\xi_{l+1})))\\
&\leq \Big(\sum_{i=l+1}^ma_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 -\sum_{i=1}^la_i\phi^{-1}\big(\frac{1}{p(\xi_i)}\big)
 \Big)\phi^{-1}(p(0)\phi(x'(0))).
\end{align*}
It follows that
$$
\Big(1-\sum_{i=l+1}^ma_i\Big(\frac{p(0)}{p(\xi_i)}\Big)
+\sum_{i=1}^la_i\frac{p(0)}{p(\xi_i)}\Big)x'(0)\leq 0.
$$
It follows that $x'(0)\leq 0$. Then $x'(t)\leq 0$ for all
$t\in [0,1]$.

Second, we prove that $x(1)\geq 0$. It follows from the
boundary conditions in \eqref{e9} that
\begin{align*}
x(1)+\beta x'(1)
&= \sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i)
 -\sum_{i=s+1}^mc_ix'(\xi_i)\\
&\geq \sum_{i=1}^kb_ix(\xi_i)-\sum_{i=k+1}^sb_ix(\xi_i)\\
&\geq \sum_{i=1}^kb_ix(\xi_k)-\sum_{i=k+1}^sb_ix(\xi_{k+1})\\
&= \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(\xi_k)
  +\sum_{i=k+1}^sb_i[x(\xi_k)-x(\xi_{k+1})]\\
&\geq \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(\xi_k)\\
&\geq \Big(\sum_{i=1}^kb_i-\sum_{i=k+1}^sb_i\Big)x(1).
\end{align*}
Then
$$
\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^sb_i\Big)x(1)+\beta x'(1)
\geq 0.
$$
We get $x(1)\geq 0$ since $x'(1)\leq 0$ and $\beta\geq 0$.
Then $x(t)> x(1)\geq 0$ for all $t\in [0,1)$. The
proof is complete.
\end{proof}

\begin{lemma} \label{lem2.3}
 Suppose that {\rm (A1),(A2'), (A3)--(A5)} hold. Let
\[
\mu =\phi\Big(\frac{1}{\sum_{i=l+1}^ma_i\phi^{-1}
\big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1.
\]
 If $y$ is a solution of \eqref{e9}, then
$$
y(t)=B_\theta-\int_t^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)}
\int_0^s\theta(u)du\Big)ds,\quad t\in [0,1],
$$
where
$$
A_\theta\in \Big[-\phi^{-1}
\Big(\frac{\int_0^1\theta(u)du}{\mu p(0)}\Big),0\Big],
$$
\begin{align*}
A_\theta
&= -\sum_{i=1}^la_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}\theta(u)du\Big)\\
&\quad +\sum_{i=l+1}^ma_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}\theta(u)du\Big)
\end{align*}
 and
\begin{align*}
B_\theta
&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}
\Big[-\beta\phi^{-1}
\Big(\frac{1}{p(1)}p(0)\phi(A_\theta)-\frac{1}{p(1)}
\int_0^1\theta(u)du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)}
\int_0^s\theta(u)du\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_\theta)-\frac{1}{p(s)}
\int_0^s\theta(u)du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_\theta)-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}\theta(u)du\Big)ds\Big].
\end{align*}
\end{lemma}

\begin{proof}
 Since $y$ is solution of \eqref{e9},
\begin{gather*}
y'(t)=\phi^{-1}\Big(\frac{1}{p(t)}p(0)\phi(y'(0))
-\frac{1}{p(t)}\int_0^t\theta(u)du\Big),\\
y(t)=y(1)-\int_t^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(y'(0))
-\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds.
\end{gather*}
 The boundary conditions in \eqref{e9} imply
\begin{align*}
y'(0)
&= -\sum_{i=1}^la_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\
&\quad +\sum_{i=l+1}^ma_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)
\end{align*}
and
\begin{align*}
&y(1)+\beta\phi^{-1}
\Big(\frac{1}{p(1)}p(0)\phi(y'(0))
-\frac{1}{p(1)}\int_0^1\theta(u)du\Big)\\
&= \sum_{i=1}^kb_i\Big(y(1)-\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(y'(0))-\frac{1}{p(s)}\int_0^s\theta(u)du
 \Big)ds\Big)\\
&\quad - \sum_{i=k+1}^mb_i
\Big(y(1)-\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(y'(0))-\frac{1}{p(s)}\int_0^s\theta(u)du
 \Big)ds\Big)\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
-\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big).
\end{align*}
It follows that
\begin{align*}
y(1)&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}
\Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(y'(0))
 -\frac{1}{p(1)}\int_0^1\theta(u)du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(y'(0))
 -\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds\\
&\quad+\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(y'(0))
 -\frac{1}{p(s)}\int_0^s\theta(u)du\Big)ds \\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)ds\Big].
\end{align*}
Lemma \ref{lem2.2} implies that $y'(0)\leq 0$. One finds that
\begin{align*}
y'(0)&= -\sum_{i=1}^la_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\
&\quad +\sum_{i=l+1}^ma_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\
&\geq \sum_{i=l+1}^ma_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(y'(0))
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big).
\end{align*}
If $y'(0)\neq 0$, one gets
 \begin{equation} \label{e10}
1-\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)
-\frac{1}{\phi(y'(0))}\frac{1}{p(\xi_i)}
\int_0^{\xi_i}\theta(u)du\Big)\leq 0.
\end{equation}
 Let
$$
G(c)=c-\sum_{i=l+1}^ma_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)\phi(c)
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big).
$$
Then
$$
\frac{G(c)}{c}=1-\sum_{i=l+1}^ma_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)-\frac{1}{\phi(c)}\frac{1}{p(\xi_i)}
\int_0^{\xi_i}\theta(u)du\Big).
$$
Note that
$$
\mu=\phi\Big(\frac{1}{\sum_{i=l+1}^ma_i\phi^{-1}
\big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1.
$$
It is easy to see that $\frac{G(c)}{c}$ is decreasing on
$(-\infty,0)$ and on $(0,+\infty)$.
First, since
$\lim_{t\to 0^+} G(c)/c =+\infty$ and
$$
\lim_{c\to +\infty}\frac{G(c)}{c}
=1-\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\Big)>0,
$$
we get that $\frac{G(c)}{c}>0$ for all $c>0$.
Second, we have
$\lim_{c\to 0^-} G(c)/c=-\infty$
and
\begin{align*}
\frac{G\Big(-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu
p(0)}\Big) \Big)}{-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu
p(0)}\Big)}
&= 1-\!\sum_{i=l+1}^ma_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)+\frac{p(0)\mu}
 {\int_0^1\theta(u)du}\frac{1}{p(\xi_i)}\int_0^{\xi_i}\theta(u)du\Big)\\
&= 1-\sum_{i=t+1}^ma_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)+\mu\frac{\int_0^{\xi_i}\theta(u)du}
 {\int_0^1\theta(u)du}\frac{1}{p(\xi_i)}p(0)\Big)\\
&\geq 1-\phi^{-1}(1+\mu)\sum_{i=l+1}^ma_i\phi^{-1}
 \Big(\frac{p(0)}{p(\xi_i)}\Big)
= 0.
\end{align*}
It follows from \eqref{e10} that $\frac{G(y'(0))}{y'(0)}\leq 0$.
We get
$$
0\geq y'(0)\ge-\phi^{-1}\Big(\frac{\int_0^1\theta(u)du}{\mu
p(0)}\Big).
$$
The proof is complete.
\end{proof}

Define the nonlinear operator $T:X\to X$ by
\[
(Tx)(t)= B_x-\int_t^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)
 -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds,
\]
for $t\in [0,1]$, where
$$
A_\sigma\in
\Big[-\phi^{-1}\Big(\frac{\int_0^1f(u,x(u),x'(u))du}
{\Big(\phi\Big(\frac{1}{\sum_{i=t+1}^ma_i\phi^{-1}
 \big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1
\Big)p(0)}\Big),0\Big],
$$
\begin{align*}
A_x&= -\sum_{i=1}^la_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)}
 \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)\\
&\quad +\sum_{i=l+1}^ma_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}f(u,x(u),x'(u))du\Big),
\end{align*}
\begin{align*}
B_x
&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
 -\frac{1}{p(\xi_i)}
\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big].
\end{align*}

\begin{lemma} \label{lem2.4}
 Suppose that {\rm (A1),(A2'), (A3), (A4)} hold. Then
\begin{itemize}
\item[(i)] the following equalities hold:
\begin{gather*}
 [p(t)\phi((Ty)'(t))]'+f(t,y(t),y'(t))=0,\quad t\in (0,1),\\
(Ty)'(0)=-\sum_{i=1}^la_i(Ty)'(\xi_i)+\sum_{i=l+1}^ma _i(Ty)'(\xi_i),\\
(Ty)(1)+\beta
(Ty)'(1)=\sum_{i=1}^kb_i(Ty)(\xi_i)-\sum_{i=k+1}^mb_i(Ty)(\xi_i)
-\sum_{i=1}^mc_i(Ty)'(\xi_i);
 \end{gather*}

\item[(ii)] $Ty\in P$ for each $y\in P$;

\item[(iii)] $x$ is a positive solution of \eqref{e7} if and only if
$x$ is a solution of the operator equation $y=Ty$ in $P$;

\item[(iv)] $T:P\to P$ is completely continuous.
\end{itemize}
\end{lemma}

\begin{proof}  The proofs of (i), (ii) and (iii) are simple. To
prove (iv), it suffices to prove that $T$ is continuous on $P$ and
$T$ is relative compact. We divide the proof into two steps:

\textbf{Step 1.} For each bounded subset $D\subset P$, and each
$x_0\in D$, since $f(t,u,v)$ is continuous, we can prove that $T$ is
continuous at $y(t)$.

\textbf{Step 2.} For each bounded subset $D\subset P$, prove that
$T$ is relative compact on $D$.
It is similar to that of the proof of Lemmas in \cite{l6} and are
omitted.
\end{proof}

\begin{lemma} \label{lem2.5}
 Suppose that {\rm (A1), (A2'), (A3), (A4)} hold. Then
there exists a constant $M>0$ such that
$$
\max_{t\in [0,1]}(Tx)(t)\leq M\max_{t\in [0,1]}|(Tx)'(t)|\quad
\text{for each }x\in P.
$$
\end{lemma}

\begin{proof}
For $x\in P$, Lemma \ref{lem2.2} implies that $(Tx)(t)\geq 0$
and $(Tx)'(t)\leq 0$ for all $t\in [0,1]$. Lemma \ref{lem2.4} implies
$$
(Tx)(1)+\beta
(Tx)'(1)=\sum_{i=1}^kb_i(Tx)(\xi_i)-\sum_{i=k+1}^mb_i(Tx)(\xi_i)
-\sum_{i=1}^mc_i(Tx)'(\xi_i).
$$
Then there exist numbers $\eta_i\in [\xi_i,1]$ such that
\begin{align*}
(Tx)(1)
&= \frac{(Tx)(1)-\sum_{i=1}^kb_i(Tx)(1)
 +\sum_{i=k+1}^mb_i(Tx)(1)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&= \frac{-\beta
(Tx)'(1)-\sum_{i=1}^mc_i(Tx)'(\xi_i)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad +\frac{\sum_{i=1}^kb_i[(Tx)(\xi_i)-(Tx)(1)]
 -\sum_{i=k+1}^mb_i[(Tx)(\xi_i)-(Tx)(1)]}
 {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&= \frac{-\beta (Tx)'(1)-\sum_{i=1}^mc_i(Tx)'(\xi_i)}
 {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad +\frac{\sum_{i=1}^kb_i(\xi_i-1)(Tx)'(\eta_i)
 -\sum_{i=k+1}^mb_i(\xi_i-1)(Tx)'(\eta_i)}
 {1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}.
\end{align*}
It follows that
\begin{align*}
&|(Tx)(t)|\\
&= |(Tx)(t)-(Tx)(1)+(Tx)(1)|\\
&\leq |(Tx)(1)|+(1-t)|(Tx)'(\xi)|\quad\text{where }\xi\in [t,1]\\
&\leq \Big(1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i)
+\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\Big)
\max_{t\in [0,1]}|(Tx)'(t)|.
\end{align*}
Then
\begin{align*}
&\max_{t\in [0,1]}|(Tx)(t)|\\
&\le\Big(1+\frac{\beta
p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i)
+\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\Big)\max_{t\in
[0,1]}|(Tx)'(t)|.
\end{align*}
It follows that there exists a constant $M>0$ such that
for all $x\in P$,
\[
\max_{t\in [0,1]}(Tx)(t)\leq M\beta_1((Tx))\,.
\]
\end{proof}

Denote
\begin{gather*}
\mu = \phi\Big(\frac{1}{\sum_{i=t+1}^ma_i\phi^{-1}
\big(\frac{p(0)}{p(\xi_i)}\big)}\Big)-1,\\
M = 1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i)
+\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i},
\\
\begin{aligned}
L_1 &= \beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\Big(\frac{1+\mu s}{\mu
p(s)}\Big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\Big(\frac{1+\mu
s}{\mu p(s)}\Big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\left(\frac{1+\mu \xi_i}{\mu
p(\xi_i)}\right)ds+\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds,
\end{aligned}\\
\begin{aligned}
L_2&= \beta\phi^{-1}\Big(\frac{1}{ p(1)}\Big)
+\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\big(\frac{s}{ p(s)}\big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{\xi_i}{p(\xi_i)}\Big)ds
+\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds.
\end{aligned}
\end{gather*}

\begin{theorem} \label{thm2.6}
 Choose $k\in (0,1)$. Suppose that
{\rm (A1), (A2'), (A3), (A4)} hold. Let $e_1,e_2,c$ be positive
numbers and
\begin{gather*}
 Q= \min\Big\{\phi\Big(\frac
{c\big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\big)}{L_1}\Big),\quad
\frac{\mu p(1)\phi(c)}{1+\mu}
\Big\};\\
W= \phi\Big(\frac
{e_2\big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\big)}{L_2}\Big);\\
 E= \phi\Big(\frac{e_1\big(1-\sum_{i=1}^kb_i
 +\sum_{i=k+1}^mb_i\big)}{L_1}\Big).
\end{gather*}
If
$Mc>e_2>\frac{e_1}{k}>e_1>0$ and
\begin{itemize}
\item[(A6)] $f(t,u,v)\leq Q$ for all $t\in [0,1],u\in [0,Mc],v\in
[-c,c]$;

\item[(A7)] $f(t,u,v)\geq W$ for all $t\in [0,k],u\in
[e_2,e_2/k],v\in [-c,c]$;

\item[(A8)] $f(t,u,v)\leq E$ for all $t\in [0,1],u\in [0,e_1],v\in
[-c,c]$;
\end{itemize}
then \eqref{e7} has at least three solutions $x_1,x_2,x_3$ such that
$$
x_1(0)<e_1,\quad x_2(k)>e_2,\quad x_3(0)>e_1,\quad x_3(k)<e_2.
$$
\end{theorem}

\begin{proof}
To apply Theorem \ref{thm2.1}, we check that all its conditions
are satisfied.
By the definitions, it is easy to see that $\alpha_1$ is a
nonnegative continuous concave functional on the cone $P$,
$\beta_1,\beta_2$ are three nonnegative continuous convex
functionals on the cone $P$, $\psi$ a nonnegative continuous
functional on the cone $P$. Lemma \ref{lem2.4} implies that $x=x(t)$ is a
positive solution of \eqref{e7} if and only if $x$ is a solution of the
operator equation $y=Ty$ and $T$ is completely continuous.

For $x\in P$, one sees that $\psi(\lambda x)\leq \lambda\psi(x)$ for
all $x\in P$ and $\lambda\in [0,1]$, and $\alpha_1(x)\leq \psi(x)$
for all $x\in P$. There exists a constant $M>0$  such that $\|x\|\le
M\beta_1(x)$ for all $x\in P$. Then (1) and (2) of
Theorem \ref{thm2.1} hold.

Corresponding to Theorem \ref{thm2.1}, we have
$a_4=c$, $a_3=\frac{e_2}{k}$, $a_2=e_2$, $a_1=e_1$.
 Now, we check that \eqref{e3} of Theorem \ref{thm2.1} holds. One sees that
$0<a_1<a_2$, $a_3>0$, $a_4>0$. The rest is divided into four steps.

\textbf{Step 1.} Prove that
$T(\overline{P_1(\beta_1;a_4)})\subseteq\overline{P_1(\beta_1;a_4)}$;
For $x\in \overline{P_1(\beta_1;a_4)}$, we have $\beta_1(x)\le
a_4=c$. Then
\begin{gather*}
0\leq x(t)\leq \max_{t\in [0,1]}x(t)\leq Mc\quad\text{for } t\in [0,1],\\
-c\leq x'(t)\leq c\quad\text{for all }t\in [0,1].
\end{gather*}
So (A6) implies that
$f(t,x(t),x'(t))< Q$ for $t\in [0,1]$.
Then Lemma \ref{lem2.3}  implies
$$
0\geq A_x\ge-\phi^{-1}\Big(\frac{\int_0^1f(u,x(u),x'(u))du}{\mu
p(0)}\Big)\geq -\phi^{-1}\Big(\frac{Q}{\mu p(0)}\Big).
$$
Since
$$
c\ge\max\big\{\frac{e_2}{\sigma_0},\;La,\;\phi^{-1}
\big(\frac{2}{p(0)}\big)a,\;\phi^{-1}\big(\frac{1}{p(1)}\big)a\big\},
$$
we obtain
$$
\phi(a)\leq \min\big\{\phi\big(\frac{c}{L}\big),\;
\frac{\phi(c)p(0)}{2},\;\phi(c)p(1)\big\}=Q.
$$
Then $\max_{t\in [0,1]}(Ty)(t)=(Ty)(0)$ implies
\begin{align*}
&(Tx)(0)\\
&= B_x-\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x)
-\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\\
&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
 -\frac{1}{p(\xi_i)}
 \int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big]\\
&\quad -\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x)
 -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\\
&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad -\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\Big(\frac{1}{p(s)}p(0)
 \phi(A_x)-\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
 -\frac{1}{p(\xi_i)}\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\
&\quad -\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\
&\leq \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}
\Big[\beta\phi^{-1}\Big(\frac{Q}{\mu p(1)}+\frac{Q}{p(1)}\Big) \\
&\quad +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\big(\frac{Q}{\mu p(s)}+\frac{Qs}{p(s)}\big)ds
 +\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\big(\frac{Q}{\mu
p(s)}+\frac{Qs}{p(s)}\big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{Q}{\mu
p(\xi_i)}+\frac{Q\xi_i}{p(\xi_i)}\big)ds
+\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}
\big(\frac{Q}{\mu p(s)}+\frac{Qs}{p(s)}\big)ds\Big]\\
&= \frac{\phi^{-1}(Q)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[\beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}
\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds
 +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1+\mu \xi_i}{\mu
p(\xi_i)}\Big)ds \\
&\quad +\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\Big]
\leq c.
\end{align*}
On the other hand, similarly to above discussion, since $(Ty)'(t)$
is decreasing and $(Ty)'(0)\leq 0$,  from Lemma \ref{lem2.3} we have
\[
(Tx)'(1)= \phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big),
\]
\begin{align*}
\max_{t\in [0,1]}|(Tx)'(t)|&= -\phi^{-1}
\Big(\frac{1}{p(1)}p(0)\phi(A_x)
-\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big)\\
&\leq \phi^{-1}\Big(\frac{Q}{\mu p(1)}+\frac{Q}{p(1)}\Big)
\leq  c.
\end{align*}
It follows that $\|Tx\|=\max\big\{\max_{t\in
[0,1]}|(Tx)(t)|,\max_{t\in [0,1]}|(Tx)'(t)|\big\}\leq c$. Then
$T(\overline{P(\beta_1;a_4)})\subseteq\overline{P(\beta_1;a_4)}$.
This completes the proof of (E1) in Theorem \ref{thm2.1}.

\textbf{Step 2.} Prove that $\alpha_1(Tx)>a_2$ for all $x\in
P(\beta_1,\alpha_1;a_2,a_4)$ with $\beta_2(Tx)>a_3$;
For $y\in P(\beta_1,\alpha_1;a_2,a_4)=P(\beta_1,\alpha_1;e_2,c)$
with $\beta_2(Ty)>a_3=\frac{e_2}{k}$, we have
\[
\alpha_1(y)=\min_{t\in [0,k]}y(t)\geq e_2, \quad
\beta_1 (y)=\max_{t\in [0,1]}|y'(t)|\leq c, \quad
\max_{t\in [0,k]}(Ty)(t)>\frac{e_2}{k}.
\]
Then
$$
\alpha_1(Ty)=\min_{t\in [0,k]}(Ty)(t)\ge
k\beta_2(Ty)>k\frac{e_2}{k}=e_2=a_2.
$$
This completes the proof of (E2) in Theorem \ref{thm2.1}.

\textbf{Step 3.}  Prove that $\{x\in
P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4):\alpha_1(x)>a_2\}\neq \emptyset$
and $\alpha_1(Tx)>b$ for all
$x\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$;
It is easy to check that $\{x\in
P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4):\alpha_1(x)>a_2\}\neq \emptyset$.
For $y\in P(\beta_1,\beta_2,\alpha_1;a_2,a_3,a_4)$, one has that
$$
\alpha_1(y)=\min_{t\in [0,k]}y(t)\geq e_2,\quad
\beta_2(y)=\max_{t\in [0,k]}y(t)\leq \frac{e_2}{k},\quad
\beta_1(y)=\max_{t\in [0,1]}| y'(t)|\leq c.
$$
Then
$$
e_2\leq y(t)\leq \frac{e_2}{k},\quad t\in [0,k],\;| y'(t)|\leq c.
$$
Thus (A7) implies
$$
f(t,y(t), y'(t))\geq W,\quad t\in [0,k].
$$
Since
$$
\alpha_1(Ty)=\min_{t\in [0,k]}(Ty)(t)\geq k\max_{t\in [0,1]}(Ty)(t),
$$
we obtain $\alpha_1(Ty)\geq k\max_{t\in [0,1]}(Ty)(t)$. Then
$$
\alpha_1(Ty)\geq k\max_{t\in [0,1]}(Ty)(t)\geq k(Ty)(0).
$$
 From $A_x\leq 0$, we obtain
\begin{align*}
\alpha_1(T_1y)
&\geq k\Big[\frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big(-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)-\frac{1}{p(1)}
 \int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du
\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
\int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}
\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\Big)\\
&\quad -\int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x)
 -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\
&= \frac{k}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\
&\quad -\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\Big)
 \int_0^1\phi^{-1}\Big(\frac{1}{p(s)}p(0)\phi(A_x)\\
&\quad -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\
&= \frac{k}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad -\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}
 \Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)-\frac{1}{p(\xi_i)}
\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\
&\quad-\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)\\
 &\quad -\frac{1}{p(s)}\int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\
&\geq \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[\beta\phi^{-1}\big(\frac{W}{p(1)}\big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\big(
\frac{Ws}{p(s)}\big)ds+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}
\big(\frac{Ws}{p(s)}\big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{W\xi_i}{p(\xi_i)}\big)ds
+\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}
\big(\frac{Ws}{p(s)}\big)ds\Big]\\
&= \frac{\phi^{-1}(W)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[\beta\phi^{-1}\big(\frac{1}{ p(1)}\big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \big(\frac{s}{ p(s)}\big)ds+\sum_{i=k+1}^mb_i
 \int_0^{\xi_i}\phi^{-1}\big(\frac{s}{ p(s)}\big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\big(\frac{\xi_i}{
p(\xi_i)}\big)ds +\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds\Big]
= e_2.
\end{align*}
 This completes the proof of (E3) in Theorem \ref{thm2.1}.

\textbf{Step 4.} Prove that $0\not\in R(\beta_1,\psi;a_1,a_4)$ and
that $\psi(Tx)<a_1$ for all $x$ in $R(\beta_1,\psi;a_1,a_4)$ with
$\psi(x)=a_1$;
It is easy to see that $0\not\in R(\beta_1,\psi;a_1,a_4)$. For $y\in
R(\beta_1,\psi;a_1,a_4)$ with $\psi(x)=a_1$, one has that
$$
\psi(y)=\max_{t\in [0,1]}y(t)= a_1,\quad
\beta_1(y)=\max_{t\in [0,1]}|y'(t)|\leq a_4.
$$
 Hence
$$
0\leq y(t)\leq a_1,\quad t\in [0,1];\quad
-c\leq y'(t)\leq c,\quad t\in [0,1].
$$
Then (A8) implies
$f(t,y(t),y'(t))\leq E$, $t\in [0,1]$.
So
\begin{align*}
\psi(Ty)&= \max_{t\in [0,1]}(Ty)(t)=(Ty)(0)\\
&= \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}\\
&\quad\times \Big[-\beta\phi^{-1}\Big(\frac{1}{p(1)}p(0)\phi(A_x)
 -\frac{1}{p(1)}\int_0^1f(u,x(u),x'(u))du\Big) \\
&\quad -\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\\
&\quad -\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}
 \Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds
\\
&\quad -\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1}{p(\xi_i)}p(0)\phi(A_x)
 -\frac{1}{p(\xi_i)}
\int_0^{\xi_i}f(u,x(u),x'(u))du\Big)ds\\
&\quad -\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}
\Big(\frac{1}{p(s)}p(0)\phi(A_x)-\frac{1}{p(s)}
 \int_0^sf(u,x(u),x'(u))du\Big)ds\Big]\\
&\leq \frac{1}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}
\Big[\beta\phi^{-1}\Big(\frac{E}{\mu p(1)}+\frac{E}{p(1)}\Big) \\
&\quad +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{E}{\mu p(s)}+\frac{Es}{p(s)}\Big)ds
+\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}\Big(\frac{E}{\mu
p(s)}+\frac{Es}{p(s)}\Big)ds
\\
&\quad +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{E}{\mu
p(\xi_i)}+\frac{E\xi_i}{p(\xi_i)}\Big)ds
+\Big(1-\sum_{i=1}^kb_i\Big)\int_0^1\phi^{-1}\Big(\frac{E}
{\mu p(s)}+\frac{Es}{p(s)}\Big)ds\Big]\\
&= \frac{\phi^{-1}(E)}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}
 \Big[\beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}
\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\\
&\quad +\sum_{i=k+1}^mb_i
 \int_0^{\xi_i}\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds
 +\sum_{i=1}^mc_i\phi^{-1}\Big(\frac{1+\mu \xi_i}{\mu
p(\xi_i)}\Big)ds \\
&\quad +\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds\Big]
\leq a_1.
\end{align*}
This completes the proof of (E4) in
Theorem \ref{thm2.1}.

Then Theorem \ref{thm2.1} implies that $T$ has at least three fixed points
$x_1$, $x_2$ and $x_3$ such that
$$
\beta(x_1)<e_1,\quad \alpha(x_2)>e_2,\quad \beta(x_3)>e_1,\quad
\alpha (x_3)<e_2.
$$
Hence \eqref{e7} has three decreasing positive solutions $x_1,x_2$ and
$x_3$ as needed in the statement of the
theorem; therefore, the proof is complete.
\end{proof}

\section{Examples}
Now, we present an example that illustrates our main results, and
that can not be solved by results in \cite{b1,l2,l3,m1,m2}.

\begin{example} \label{exa3.1} \rm
Consider the boundary-value problem
\begin{equation} \label{e11}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=-\frac{1}{4}x'(1/4)+\frac{1}{2}x'(1/2),\\
x(1)+2x'(1)=\frac{1}{4}x(1/4)-\frac{1}{4}x(1/2)
 -\frac{1}{2}x'(1/4)-\frac{1}{4}x'(1/2),
\end{gathered}
\end{equation}
where $f(t,u,v)=f_0(u)+\frac{t|v|}{25500000}$ and
\[
f_0(u)= \begin{cases}
\frac{2}{51}u & u\in [0,4],\\
\frac{8}{51} &u\in [4,12],\\
564000+\frac{564000-\frac{8}{51}}{1004-12}(u-1004) &u\in [12,1004],\\
564000 &u\in [1004,4004],\\
564000 &u\in [4004,2000004],\\
564000e^{u-2000004} &u\geq 2000004.
\end{cases}
\]
\end{example}
Corresponding to \eqref{e7}, one sees that $\phi^{-1}(x)=x$,
$\xi_1=1/4,\xi_2=1/2$, $a_1=1/4,a_2=1/2$, $b_1=1/4,b_2=1/4$,
$c_1=1/2,c_2=1/4$.
It is easy to see that (A1)--(A4) hold. Choose constants
$k=\frac{1}{2}$, $e_1=2,e_2=100$ and $c=20000$. One obtains
\begin{gather*}
M= 1+\frac{\beta p(1)+\sum_{i=1}^kb_i(1-\xi_i)+\sum_{i=k+1}^mb_i(1-\xi_i)
+\sum_{i=1}^mc_i}{1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i}=\frac{51}{4},
\\
\mu = \phi\Big(\frac{1}{\sum_{i=t+1}^ma_i\phi^{-1}
\Big(\frac{p(0)}{p(\xi_i)}\Big)}\Big)-1=1,
\\
\begin{aligned}
L_1 &= \beta\phi^{-1}\Big(\frac{1+\mu}{\mu p(1)}\Big)
+\sum_{i=1}^k b_i\int_{\xi_i}^1 \phi^{-1}
\Big(\frac{1+\mu s}{\mu p(s)}\Big) ds \\
&\quad +\sum_{i=k+1}^m b_i\int_0^{\xi_i}\phi^{-1}
\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds +\sum_{i=1}^mc_i\phi^{-1}
\Big(\frac{1+\mu \xi_i}{\mu p(\xi_i)}\Big)ds \\
&\quad +\Big(1-\sum_{i=1}^kb_i\Big)
\int_0^1\phi^{-1}\Big(\frac{1+\mu s}{\mu p(s)}\Big)ds,
\end{aligned}\\
\begin{aligned}
L_2 &= \beta\phi^{-1}\big(\frac{1}{p(1)}\big)
 +\sum_{i=1}^kb_i\int_{\xi_i}^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds\\
&\quad +\sum_{i=k+1}^mb_i\int_0^{\xi_i}\phi^{-1}
 \big(\frac{s}{ p(s)}\big)ds +\sum_{i=1}^mc_i\phi^{-1}
 \big(\frac{\xi_i}{p(\xi_i)}\big)ds \\
&\quad +\Big(1-\sum_{i=1}^kb_i\Big)
 \int_0^1\phi^{-1}\big(\frac{s}{ p(s)}\big)ds,
\end{aligned} \\
Q = \min\big\{\phi\Big(\frac
{c\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\Big)}{L_1}\Big),\;
\frac{\mu p(1)\phi(c)}{1+\mu} \big\};
\\
W= \phi\Big(\frac{e_2\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\Big)}{L_2}
\Big);\\
E = \phi\Big(\frac{e_1\Big(1-\sum_{i=1}^kb_i+\sum_{i=k+1}^mb_i\Big)}{L_1}
\Big).
\end{gather*}
It is easy to see that If
$Mc>e_2>\frac{e_1}{k}>e_1>0$ and
\begin{gather*}
f(t,u,v)\leq Q\quad\text{for all }t\in [0,1],u\in [0,Mc],v\in
[-20000,20000];\\
f(t,u,v)\geq W\quad\text{for all }t\in [0,1/2],u\in [100,200],v\in
[-20000,20000];\\
f(t,u,v)\leq E\quad\text{for all }t\in [0,1],u\in [0,2],
v\in [-20000,20000];
\end{gather*}
Then (A6), (A7) and (A8) hold. Theorem \ref{thm2.6} implies that \eqref{e11}
has at least three positive solutions such
that
$$
x_1(0)<2,\quad x_2(k)>100,\quad x_3(0)>2,\quad x_3(k)<100.
$$

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