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

\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations}, 
Vol. 2008(2008), No. 96, pp. 1--52.\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 2008 Texas State University - San Marcos.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2008/96\hfil Solutions to second order BVPs]
{Solutions to second order non-homogeneous multi-point BVPs
using a fixed-point theorem}

\author[Y. Liu\hfil EJDE-2008/96\hfilneg]
{Yuji Liu}

\address{Yuji Liu \newline
Department of Mathematics, Guangdong University of Business
Studies, Guangzhou 510320, China}
\email{liuyuji888@sohu.com}

\thanks{Submitted March 3, 2008. Published July 25, 2008.}
\thanks{Supported by grant 06JJ50008 from the Natural Science 
Foundation of  Hunan \hfill\break\indent
province and grant 7004569 from the Natural Science
 Foundation of Guangdong \hfill\break\indent
  province , China} 
\subjclass[2000]{34B10, 34B15, 35B10}
\keywords{Second order; $p$-Laplacian; positive solution;\hfill\break\indent
  mixed type multi-point boundary-value problem}

\begin{abstract}
  In this article, we study five non-homogeneous multi-point
  bound\-ary-value problems (BVPs) of second order differential
  equations with the one-dimensional $p$-Laplacian.
  These problems have a common equation (in different function domains)
  and different boundary conditions. We find conditions that
  guarantee the existence of at least three positive solutions.
  The results obtained  generalize several known ones and are illustrated
  by examples.
  It is also shown that the approach for getting three positive solutions
  by using multi-fixed-point theorems can be extended to
  nonhomogeneous BVPs.   The emphasis is on the nonhomogeneous
  boundary conditions and the nonlinear term involving first order
  derivative of the unknown. Some open problems are also proposed.
\end{abstract}

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

\section{Introduction}

Multi-point boundary-value problems (BVPs) for second
order differential equations without $p$-Laplacian have received a
wide attention because of their potential applications and BVPs are
fascinating and challenging fields of study, one may see the
textbook by Ge \cite{g}. There are four classes of such BVPs
(including their special cases) studied in known papers:
\begin{equation} \label{e1}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i),
\end{gathered}
\end{equation}
\begin{equation} \label{e2}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i),
\end{gathered}
\end{equation}
\begin{equation}\label{e3}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\;\;t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad
 x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i),
\end{gathered}
\end{equation}
and
\begin{equation}\label{e4}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad
 x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i),
\end{gathered}
\end{equation}
where $0<\xi_1<\dots<\xi_m<1$, $\alpha_i,\beta_i\in \mathbb{R}$ and $f$ is
a continuous function. The  main methods to get solutions or multiple
positive solutions of these BVPs are as follows:

(i) fixed point index theory \cite{iw1,iw2,zf,zs}, or fixed point
theorems in cones in Banach spaces \cite{d,fg2,gl,l3,m1,m2,mc},
 such as the Guo-Krasnoselskii's fixed-point theorem
\cite{chc,m4,mt,s,sczw,sz}; Leggett-Williams theorems
\cite{l1,l4,oz}, the five-functional fixed
point theorem  \cite{aah}; the fixed point theorem of Avery and
Peterson \cite{ko1}, etc.

(ii) Mawhin's continuation theorem \cite{flg,iz,ko2,ko3,l1,l2,ly1,ly2,lz,m2};

(iii) the shooting methods \cite{kw,p};


(iv) upper and lower solution methods and monotone iterative
techniques \cite{aop,cr,zm}; upper and lower solution methods and
Leray-Schauder degree theory \cite{kk1,kk2,kk3,mo}, or the approach of
a combination of nonlinear alternative of Leray-Schauder with the
properties of the associated vector field at the $(x,x')$ plane
\cite{gp};

(v) the critical point theory and variational methods \cite{la};

(vi) topological degree theory \cite{gmyz,g}; the Schauder's fixed
point theorem in suitable Banach space \cite{l4,m1,m2,m3,nm}.


In all the above papers, the boundary conditions (BCs)
are homogeneous. However, in many applications, BVPs
consist of differential equations coupled with nonhomogeneous BCs,
for example
\begin{gather*}
y''=\frac{1}{\lambda}(1+y^2)^{\frac{1}{2}},\quad t\in (a,b),\\
y(a)=a\alpha,\quad
y(b)=\beta
\end{gather*}
and
\begin{gather*}
 y''=-\frac{(1+y'(t))^2}{2(y(t)-\alpha)},\quad t\in (a,b),\\
y(a)=\alpha,\quad
y(b)=\beta
\end{gather*}
which are very well known and were  proposed in 1690 and 1696, respectively.
In 1964, the BVPs studied by Zhidkov and Shirikov \cite{zs2}
and by Lee \cite{lee} are also nonhomogeneous.

There are also several papers concerned with the existence of
positive solutions of BVPs for differential equations with
non-homogeneous BCs. Ma Ruyun \cite{m5}
studied existence of positive solutions of the following BVP
consisting of second order differential equations and three-point BC
\begin{equation} \label{e5}
\begin{gathered}
 x''(t))+a(t)f(x(t))=0,\quad t\in (0,1),\\
 x(0)=0,\quad  x(1)-\alpha x(\eta)=b,
 \end{gathered}
\end{equation}
Kwong and Kong \cite{kw} studied the BVP
\begin{equation} \label{e6}
\begin{gathered}
y''(t)=-f(t,y(t)),\quad 0<t<1, \\
\sin\theta y(0)-\cos\theta y'(0)=0,\\
y(1)-\sum_{i=1}^{m-2}\alpha_iy(\xi_i)=b\ge0,
\end{gathered}
\end{equation}
where $\xi_i\in (0,1),\alpha_i\ge 0, \theta\in [0,3\pi/4]$, $f$ is a
nonnegative and continuous function. Under some assumptions, it was
proved that there exists a constant $b^*>0$ such that:
\eqref{e6} has at least two positive solutions if $b\in (0,b^*)$;
\eqref{e6} has at least one solution if $b=0$ or $b=b^*$;
 \eqref{e6} has no positive solution if $b>b^*$.

Palamides [\cite{p}], under superlinear and/or sublinear growth rate in
$f$, proved the existence of positive solutions (and monotone in
some cases) of the  boundary-value problem
\begin{equation} \label{e7}
\begin{gathered}
y''(t)=-f(t,y(t),y'(t)),\quad 0\le t\leq1, \\
\alpha y(0)-\beta y'(0)=0,\quad
 y(1)-\sum_{i=1}^{m-2}\alpha_iy(\xi_i)=b\ge0,
\end{gathered}
\end{equation}
where $\alpha>0$, $\beta>0$, the function $f$ is continuous, and
$f(t,y,y')\ge 0$, for all $t\in[0,1]$, $y\ge 0$, $y'\in \mathbb{R}$.
The approach is based on an analysis of the corresponding
 vector field on the  face-plane and on Kneser's property for the solution's
 funnel.

 Sun, Chen, Zhang and Wang \cite{sczw} studied the existence of positive solutions
for the three-point boundary-value problem
\begin{equation} \label{e8}
\begin{gathered}
u''(t)+a(t)f(u(t))=0,\quad 0\le t\le 1,\\
u'(0)=0,\quad
u(1)-\sum_{i=1}^{m-2}\alpha_iu(\xi_i)=b\ge0,
\end{gathered}
\end{equation}
where $\xi_i\in(0,1)$, $\alpha_i\ge 0$ are given. It was proved that
there exists $b^*>0$ such that \eqref{e8} has at least one positive
solution if $b\in (0,b^*)$ and no positive solution if $b>b^*$. To
study the existence of positive solutions of BVPs \eqref{e5},
\eqref{e6}, \eqref{e7}, \eqref{e8}, the Green's functions of the
corresponding problems are established and play an important role
in the proofs of the main results.

In recent papers, using lower and upper solutions
methods, Kong and Kong \cite{kk1,kk2,kk3}  established results for solutions
and positive solutions of the following two problems
\begin{equation} \label{e9}
\begin{gathered}
x''(t))+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)-\sum_{i=1}^m\alpha_i x'(\xi_i)=\lambda_1,\quad
x(1)-\sum_{i=1}^m\beta_i x(\xi_i)=\lambda_2,
 \end{gathered}
\end{equation}
and
\begin{equation} \label{e10}
\begin{gathered}
x''(t))+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)-\sum_{i=1}^m\alpha_i x(\xi_i)=\lambda_1,\quad
x(1)-\sum_{i=1}^m\beta_i x(\xi_i)=\lambda_2,
\end{gathered}
\end{equation}
respectively. We note that the boundary conditions in \cite{kk1,kk2} are
two-parameter non-homogeneous BCs. There, the existence of
lower and upper solutions with certain relations are assumed.


In recent years, there have been many exciting results concerning the
existence of positive solutions of BVPs of second order differential
equations with $p$-Laplacian subjected to different multi-point
boundary conditions:
\begin{equation} \label{e11}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i),
\end{gathered}
\end{equation}
\begin{equation} \label{e12}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i),\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i),
\end{gathered}
\end{equation}
\begin{equation} \label{e13}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i),\quad
 x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i).
\end{gathered}
\end{equation}

These results, can be found in
\cite{fg1,fg3,fgj,flg,fpg,gmyz,jfg,jg,jgz,l5,sg,sgz,sqg,wg1,wg2,wg3,wg4,x,zwg}.
In above mentioned papers, to obtain
 positive solutions, two kinds of assumptions are
supposed. The first one is imposed on $\alpha_i,\beta_i$, the
other one called growth conditions is  imposed on the
nonlinearity $f$. To define a cone $P$ in Banach spaces and to
define the nonlinear operator $T:P\to P$ are important steps
in the the proofs of the results.

It is easy to see that
\begin{gather*}
x''(t)=-2,\quad t\in (0,1),\\
x(0)=x(1)=0
\end{gather*}
has unique positive solution $x(t)=-t^2+t$, but the BVP
\begin{gather*}
x''(t)=-2,\quad t\in (0,1),\\
x(0)=A, \quad x(1)=B
\end{gather*}
has positive solution $x(t)=-t^2+(B-A+1)t+A$ if and only if $A\ge 0$
and $B\ge 0$. It shows us that the presence of nonhomogeneous BCs
can induce nonexistence of positive solutions of a BVP.

 Motivated by the facts mentioned above, this
paper is concerned with the more generalized BVPs for second order
differential equation with $p$-Laplacian coupled with nonhomogeneous
multi-point BCs; i.e.,
\begin{equation} \label{e14}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=\sum_{i=1}^m\alpha _ix'(\xi_i)+A,\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B,
\end{gathered}
\end{equation}


\begin{equation} \label{e15}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i)+A,\quad
 x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i)+B,
\end{gathered}
\end{equation}


\begin{equation}\label{e16}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\alpha x'(0)+A,\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B,
\end{gathered}
\end{equation}

\begin{equation}\label{e17}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix'(\xi_i)+A,\quad
 x'(1)=\sum_{i=1}^m\beta_ix'(\xi_i)+B,
\end{gathered}
\end{equation}
and
\begin{equation}\label{e18}
\begin{gathered}{}
[\phi(x'(t))]'+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\sum_{i=1}^m\alpha _ix(\xi_i)+A,\quad
 x(1)=\sum_{i=1}^m\beta_ix(\xi_i)+B,
\end{gathered}
\end{equation}
 where $0<\xi_1<\dots<\xi_m<1$, $A,B\in \mathbb{R}$,
 $\alpha_i\ge0,\alpha\ge 0$,
$\beta_i\ge0$ for all $i=1,\dots,m$, $f$ is continuous and
nonnegative, $\phi$ is called $p$-Laplacian, $\phi(x)=|x|^{p-2}x$
for $x\neq 0$ and $\phi(0)=0$ with $p>1$, its inverse function is
denoted by $\phi^{-1}(x)$ with $\phi^{-1}(x)=|x|^{q-2}x$ for
$x\neq 0$ and $\phi^{-1}(0)=0 $ with $1/p+1/q=1$.


The purpose is to establish sufficient conditions for the existence
of at least three positive solutions for  \eqref{e14}--\eqref{e18}.
 The results in this paper are new
since there exists no paper concerned with the existence of at least
three positive solutions of these nonhomogeneous multi-point BVPs
even when $\phi(x)=x$. Maybe it is the first time to use the
multi-fixed-point theorem to solve these kinds of BVPs.

  The remainder of this paper is organized as follows: The main
results are presented in Section 2
(Theorems \ref{thm210}, \ref{thm214}, \ref{thm218}, \ref{thm222},
\ref{thm226}).
Some examples to show the main results are given in Section 3.


\section{Main Results}

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

\begin{definition} \label{def21}\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 $ax\in P$ for all $x\in P$ and $a\ge 0$ and $x\in X$ and
$-x\in X$ imply $x=0$.
\end{definition}

\begin{definition} \label{def22}\rm
A map $ \psi :P\to [0,+\infty)$ is a nonnegative
continuous concave or convex functional map provided $\psi $ is
nonnegative, continuous and satisfies
$$
 \psi (tx+(1-t)y)\ge t\psi (x)+(1-t)\psi(y),
$$
or
$$
 \psi (tx+(1-t)y)\le t\psi (x)+(1-t)\psi(y),
$$
for all $x,y\in P$ and $t\in [0,1]$.
\end{definition}

\begin{definition} \label{def23} \rm
 An operator $T;X\to X$ is completely continuous if it is continuous and maps bounded sets into
relative compact sets.
\end{definition}


\begin{definition} \label{def24} \rm
Let $a,b,c,d,h>0$ be positive constants,
$\alpha,\psi $ be two nonnegative continuous concave functionals on
the cone $P$, $\gamma ,\beta,\theta$ be three nonnegative continuous
convex functionals on the cone $P$. Define the convex sets:
\begin{gather*}
P_c=\{x\in P:\|x\|<c\},\\
P(\gamma,\alpha;a,c)=\{x\in P:\alpha(x)\ge a,\;\gamma(x)\le c\},\\
P(\gamma,\theta,\alpha;a,b,c)=\{x\in P:\alpha(x)\ge a,\;\theta(x)\le b,\;\gamma(x)\le c\},\\
Q(\gamma ,\beta;,d,c)=\{x\in P:\beta (x)\le d,\;\gamma(x)\le c\},\\
Q(\gamma,\beta,\psi;h,d,c)=\{x\in P:\psi(x)\ge h,\;\beta(x)\le
d,\;\gamma(x)\le c\}.
\end{gather*}
\end{definition}


\begin{lemma}[\cite{aah}] \label{lem25}
 Let $X$ be a real Banach space, $P$ be a cone
in $X$, $\alpha,\psi $ be two nonnegative continuous concave
functionals on the cone $P$, $\gamma ,\beta,\theta$ be three
nonnegative continuous convex functionals on the cone $P$. There
exist constant $M>0$ such that
$$
\alpha(x)\le \beta(x),\quad
\|x\|\le M\gamma (x)\quad\text{for all }x\in P.
$$
Furthermore, Suppose that $h,d,a,b,c>0$ are constants with $d<a$.
Let $T:\overline{P_c}\to \overline{P_c}$ be a completely
continuous operator. If
\begin{itemize}
\item[(C1)] $\{y\in
P(\gamma,\theta,\alpha;a,b,c)|\alpha(x)>a\}\neq \emptyset$ and
$\alpha(Tx)>a$ for every $x$ in $P(\gamma,\theta,\alpha;a,b,c)$;

\item[(C2)] $\{y\in Q(\gamma,\theta,\psi;h,d,c)|\beta(x)<d\}\neq \emptyset$
and $\beta(Tx)<d$ for every $x$ in $Q(\gamma,\theta,\psi;h,d,c)$;

\item[(C3)] $\alpha(T y)>a$ for $y\in P(\gamma,\alpha;a,c)$
with $\theta(Ty)>b$;

\item[(C4)]  $\beta(Tx)<d$ for each $x\in Q(\gamma ,\beta;,d,c)$
with $\psi(Tx)<h$,

\end{itemize}
then $T$ has at least three  fixed points $y_1$, $y_2$ and $y_3$
such that
$$
\beta(y_1)<d,\quad \alpha(y_2)>a,\quad \beta(y_3)>d,\quad
\alpha (y_3)<a.
$$
\end{lemma}

Choose $X=C^1[0,1]$.  We call $x\le y$ for $x,y\in X$ if
$x(t)\le y(t)$ for all $t\in [0,1]$, define the norm
$\|x\|=\max\{\max_{t\in
[0,1]}|x(t)|,\;\max_{t\in [0,1]}|x'(t)|\}$. It is easy to see that
$X$ is a semi-ordered real Banach space.

 Choose $k\in (0,1/2)$, let $\sigma_0=\min\{k,1-k\}=k$. For a cone
$P\subseteq X$ of the Banach space $X=C^1[0,1]$, define the
functionals on $P:P\to \mathbb{R}$ by
\begin{gather*}
\gamma(y)=\max_{t\in [0,1]}|y'(t)|,\;y\in P,\quad
\beta(y)=\max_{t\in [0,1]}|y(t)|,\;y\in P,\\
\theta(y)=\max_{t\in [k,1-k]}|y(t)|,\;y\in P,\quad
 \alpha(y)=\min_{t\in [k,1-k]}|y(t)|,\;y\in P,\\
 \psi(y)=\min_{t\in [k,1-k]}|y(t)|,\;y\in P.
 \end{gather*}
It is easy to see that $\alpha,\psi $ are two nonnegative continuous
concave functionals on the cone $P$, $\gamma ,\beta,\theta$ are
three nonnegative continuous convex functionals on the cone $P$ and
$\alpha(y)\le \beta(y)$ for all $y\in P$.


\begin{lemma} \label{lem26}
 Suppose that $x\in X$, $x(t)\ge 0$ for all $t\in
[0,1]$ and $x'(t)$ is decreasing on $[0,1]$. Then
\begin{equation} \label{e19}
x(t)\ge \min\{t,1-t\}\max_{t\in [0,1]}x(t),\;\;t\in [0,1].
\end{equation}
\end{lemma}

\begin{proof}
 Suppose $x(t_0)=\max_{t\in [0,1]}x(t)$.
If $t\in (0,t_0)$, we get that there exist $0\le \eta\le t\le \xi\le t_0$
such that
\begin{align*}
\frac{x(t)-x(0)}{t-0}-\frac{x(t_0)-x(0)}{t_0-0}&= -\frac{t[x(t_0)-x(t)]-(t_0-t)[x(t)-x(0)]}
{tt_0}\\
&= -\frac{t(t_0-t)x'(\xi)-(t_0-t)tx'(\eta)}{tt_0}\\
&\geq -\frac{t(t_0-t)x'(\eta)-(t_0-t)tx'(\eta)}{tt_0}=0.
\end{align*}
Then
$$
x(t)\ge \frac{t}{t_0}x(t_0)+(1-\frac{t}{t_0}x(0)\ge
\frac{t}{t_0}x(t_0)\ge tx(t_0),\quad t\in (0,t_0).
$$
Similarly
$x(t)\ge (1-t)x(t_0)$, for $t\in (t_0,1)$.
It follows that $x(t)\ge \min\{t,1-t\}\max_{t\in [0,1]}x(t)$ for all
$t\in [0,1]$. The proof is complete.
\end{proof}

\subsection{Positive Solutions of \eqref{e14}}

First, we establish an existence result for three decreasing
positive solutions of \eqref{e14}. We use the following assumptions:
\begin{itemize}

\item[(H1)] $f:[0,1]\times [h,+\infty)\times
(-\infty,0]\to [0,+\infty)$ is continuous with
$f(t,c+h,0)\not\equiv 0$ on each sub-interval of [0,1], where
$h=\frac{B}{1-\sum_{i=1}^m\beta_i}$;

\item[(H2)]  $A\le 0,B\ge 0$;


\item[(H3)] $\alpha_i\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\alpha_i<1,\;\;\sum_{i=1}^m\beta_i<1$;


\item[(H4)] $h:[0,1]\to [0,+\infty)$ is a continuous
function and $h(t)\not\equiv 0$ on each subinterval of [0,1].

\end{itemize}

Consider the boundary-value problem
\begin{equation} \label{e20}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\;\;t\in (0,1),\\
y'(0)-\sum_{i=1}^m\alpha_i
y'(\xi_i)=A,\quad
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)=0,
 \end{gathered}
\end{equation}


\begin{lemma} \label{lem27}
Suppose that {\rm (H2)--(H4)} hold. If
$y$ is a solution of \eqref{e20}, then $y$ is decreasing and positive on
$(0,1)$.
\end{lemma}

\begin{proof}
 Suppose $y$ satisfies \eqref{e20}. It follows from the
assumptions that $y'$ is decreasing on $[0,1]$. Then the BCs in \eqref{e20}
and (H3) imply
$$
y'(0)=\sum_{i=1}^m\alpha_i y'(\xi_i)+A\le \sum_{i=1}^m\alpha_i
y'(0)+A.
$$
It follows that $y'(0)\le A\big(1-\sum_{i=1}^m\alpha_i\big)^{-1}\le 0$.
We get  $y'(t)\le 0$ for $t\in [0,1]$. Then
\begin{align*}
y(1)&= \sum_{i=1}^m\beta_iy(\xi_i)\ge\sum_{i=1}^m\beta_iy(1).
\end{align*}
So $y(1)\ge 0$. Then $y(t)>y(1)\ge 0$ for $t\in (0,1)$ since
$y'(t)\le0$ on $[0,1]$. The
 proof is complete.
\end{proof}


\begin{lemma} \label{lem28}
Suppose that {\rm (H2)--(H4)} hold. If
$y$ is a solution of \eqref{e20}, then
$$
y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds,
$$
with
\begin{equation} \label{e21}
\phi^{-1}(A_h)=\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(A_h-\int_0^{\xi_i}h(s)ds\Big)+A,
\end{equation}
 and
\begin{align*}
B_h&=\frac{1}{1-\sum_{i=1}^m\beta_i}\Big[-\int_0^1\phi^{-1}
\Big(A_h-\int_0^sh(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i}
\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\Big],
\end{align*}
where
$a= \phi\left(\frac{A}{1-\sum_{i=1}^m\alpha_i}\right)$ and
$$
b= \phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
-\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds.
$$
\end{lemma}


\begin{proof}  It follows from \eqref{e20} that
$$
y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds,
$$
and BCs in \eqref{e20} imply that
\begin{equation} \label{e22}
y'(0)=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(\phi(y'(0))
-\int_0^{\xi_i}h(s)ds\Big)+A,
\end{equation}
 and
\begin{align*}
y(0)&= \frac{1}{1-\sum_{i=1}^m\beta_i}\Big[-\int_0^1\phi^{-1}
\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds \\
&\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i}
\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds\Big].
\end{align*}
Let
$$
G(c)=\phi^{-1}(c)-\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(c-\int_0^{\xi_i}h(s)ds\Big)-A.
$$
It is easy to see that
\begin{align*}
G(a)&= G\Big(\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big)\\
&\geq  \frac{A}{1-\sum_{i=1}^m\alpha_i}
-\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big) -A
= 0.
\end{align*}
On the other hand, one sees that
\begin{align*}
\frac{G(b)}{\phi^{-1}(b)}
&=1-\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(1-\frac{\int_0^{\xi_i}h(s)ds}{b}\Big)-\frac{A}{\phi^{-1}(b)}
\\
&= 1-\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(1-\frac{\int_0^{\xi_i}h(s)ds}{
\phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)-\frac{\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds}\Big)
\\
&\quad-\frac{A}{\phi^{-1}
\Big(\phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)-\frac{\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\int_0^1h(s)ds\Big)}
\\
&\ge 1-\sum_{i=1}^m\alpha_i\phi^{-1}\Big( 1+\frac{1-\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\Big)-\frac{1-\sum_{i=1}^m\alpha_i}{1+\sum_{i=1}^m\alpha_i}\\
&= 1-\sum_{i=1}^m\alpha_i\frac{2}{1+\sum_{i=1}^m\alpha_i}
-\frac{1-\sum_{i=1}^m\alpha_i}{1+\sum_{i=1}^m\alpha_i}
= 0.
\end{align*}
Hence $G(b)\le 0$. It is easy to know that
$\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on
$(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence
$G(a)\ge 0$ and $G(b)\le 0$ and
\[
\lim_{c\to 0^+}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad
\lim_{c\to +\infty}\frac{G(c)}{\phi^{-1}(c)}
= 1-\sum_{i=1}^m\alpha_i>0,
\]
we get that there exists unique constant $A_h=\phi(y'(0))\in [b,a]$
such that \eqref{e21} holds. The proof is completed.
\end{proof}

Note $h=B/(1-\sum_{i=1}^m\beta_i)$, and let $x(t)-h=y(t)$.
Then \eqref{e14} is transformed into the boundary-value problem
\begin{equation} \label{e23}
\begin{gathered}{}
[\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
y'(0)-\sum_{i=1}^m\alpha_i y'(\xi_i)=A,\quad
y(1)-\sum_{i=1}^m\beta_i y(\xi_i)=0,
 \end{gathered}
\end{equation}
Let
\begin{align*}
P_1=\Big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],\;
y'(t)\le 0 \text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t)
\text{ for all }t\in [0,1] \Big\}.
\end{align*}
Then $P_1$ is a cone in $X$.
Since
\begin{align*}
|y(t)|&=\big|\frac{\sum_{i=1}^m\beta_i y(\xi_i)-\sum_{i=1}^m\beta_i
y(1)}{1-\sum_{i=1}^m\beta_i }\big|\\
&\leq \frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\max_{t\in [0,1]}|y'(t)|\\
&=\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\gamma(y),
\end{align*}
we obtain
 $$
\max_{t\in [0,1]}|y(t)|\leq
\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\gamma(y).
$$
It is easy to see that there exists a constant $M>0$
such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$.

Define the nonlinear operator $T_1:P_1\to X$ by
\begin{align*}
(T_1y)(t)= B_y+\int_0^t\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad
y\in P_1,
\end{align*}
where $A_y$ satisfies
\begin{equation} \label{e24}
\phi^{-1}(A_y)=\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(A_y-\int_0^{\xi_i}f(s,y(s)+h,y'(s))ds\Big)+A,
\end{equation}
and $B_y$ satisfies
\begin{align*}
B_y&= \frac{1}{1-\sum_{i=1}^m\beta_i}
\Big(-\int_0^1\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&+\sum_{i=1}^m\beta_i\int_0^{\xi_i} \phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big).
\end{align*}
Then for $y\in P_1$,
\begin{align*}
(T_1y)(t)
&= -\int_t^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds.
\end{align*}

 \begin{lemma} \label{lem29}
Suppose that {\rm (H1)--(H3)} hold. It is easy to show that
\begin{itemize}
\item[(i)] the following equalities hold:
\begin{gather*}
[\phi((T_1y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_1y)'(0)-\sum_{i=1}^m\alpha_i (T_1y)'(\xi_i)=A,\quad
(T_1y)(1)-\sum_{i=1}^m\beta_i (T_1y)(\xi_i)=0;
\end{gather*}

\item[(ii)] $T_1y\in P_1$ for each $y\in P_1$;

\item[(iii)] $x$ is a solution of \eqref{e14} if and only if $x=y+h$
and $y$ is a solution of the operator equation $y=T_1y$ in $P_1$;

\item[(iv)] $T_1:P_1\to P_1$ 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 $T_1$ is continuous and $T_1$
is compact. We divide the proof into two steps:

\noindent\textbf{Step 1.} Prove that $T_1$ is continuous about $y$.
Suppose $y_n\in X$ and $y_n\to y_0\in X$. Let $A_{y_n}$ be
decided by \eqref{e21} corresponding to $y_n$ for $n=0,1,2,\dots$. We will
prove that $A_{y_n}\to A_{y_0}$ as $n$ tends to infinity.

Since $y_n\to y_0$ uniformly on [0,1] and $f$ is continuous,
we have that for $\epsilon =1$, there exists $N$, when $n>N,$ for
each $t\in [0,1]$, such that
$$
0\le f(t,y_n(t)+h,y_n'(t))\le 1+f(t,y_0(t)+h,y_0'(t))\le
1+\max_{t\in [0,1]}f(t,y_0(t)+h,y_0'(t)).
$$
Hence Lemma \ref{lem28} implies that
$A_{y_n}$ is an element in the interval
\begin{align*}
&\Big[\phi\Big(\frac{A(1+\sum_{i=1}^m\alpha_i)}
{1-\sum_{i=1}^m\alpha_i}\Big)
-\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(s,y_n(s)+h,y_n'(s))ds,\\
&\phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big]\\
&\subseteq \Big[\phi\Big(\frac{A\big(1+\sum_{i=1}^m\alpha_i\big)}
{1-\sum_{i=1}^m\alpha_i}\Big)
-\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\Big(1+\max_{t\in [0,1]}f(t,y_0(t)+h,y_0'(t))\Big),\\
&\quad \phi\Big(\frac{A}{1-\sum_{i=1}^m\alpha_i}\Big)\Big].
\end{align*}
It follows that $\{A_{y_n}\}$ is bounded. If $\{A_{y_n}\}$ does not
converge to $A_{y_0}$, then there exist two subsequences
$\{A_{y_{n_k}}^1\}$ and $\{A_{y_{n_k}}^2\}$ of $\{A_{y_n}\}$ with
$$
A_{y_{n_k}}^1\to C_1,\;\;A_{y_{n_k}}^2\to C_2,\quad
k\to +\infty,\quad C_1\neq C_2.
$$
By the construction of $A_{y_n}$,
$$
\phi^{-1}(A_{y_{n_k}}^1)=\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(A_{y_{n_k}}^1-\int_0^{\xi_i}f(s,y_{n_k}(s)+h,y_{n_k}'(s))ds\Big)+A.
$$
Since $f(t,y_n(t)+h,y_n'(t))$ is uniformly bounded, by Lebesgue's
dominated convergence theorem, letting $k\to +\infty$, we get
$$
\phi^{-1}(C_1)=\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(C_1-\int_0^{\xi_i}f(s,y_0(s)+h,y_0'(s))ds\Big)+A.
$$
Since $A_{y_0}$ satisfies
$$
\phi^{-1}(A_{y_0})=\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(A_{y_0}-\int_0^{\xi_i}f(s,y_0(s)+h,y_0'(s))ds\Big)+A,
$$
Lemma \ref{lem28} implies that $A_{y_0}=C_1$. Similarly, we can prove that
$A_{y_0}=C_2$. This contradicts to $C_1\neq C_2$. Therefore for each
$y_n\to y_0$, we have $A_{y_n}\to A_{y_0}$. It
follows that $A_y$ is continuous about $y$. So the continuity of
$T_1$ is obvious.


\noindent\textbf{Step 2.} Prove that $T_1$ is compact.
Let $\Omega\subseteq P_1$ bet a bounded set. Suppose that
$\Omega\subseteq \{y\in P_1:\|y\|\le M\}$. For $y\in \Omega$, we
have
$$
0\le \int_0^1f(s,y(s)+h,y'(s))ds\le \max_{t\in [0,1],u\in
[h,M+h],v\in [-M,M]}f(t,u,v)=:D.
$$
It follows from the definition of $T_1$ and Lemma \ref{lem28} that
\begin{align*}
&|(T_1y)(t)|\\
&= \Big|-\int_t^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big|\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du \\
&\quad +\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)\\
&\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du+\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)+\frac{D}{1-\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big)
\\
&\quad +\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i}
\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big)\\
&= \phi^{-1}
\left(E\right)+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i}
\phi^{-1}
\left(E\right),
\end{align*}
\begin{align*}
& |(T_1y)'(t)|\\
&= \big|\phi^{-1}
\left(\phi(A_y)-\int_0^tf(u,y(u)+h,y'(u))du\right)ds\big|\\
&\le \phi^{-1}
\Big(\phi\left(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\right)
\\
 &\quad+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du +\int_0^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\le \phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\Big)\\
&= \phi^{-1}(E),
\end{align*}
where
$$
E=\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{D}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}.
$$
For the uniform continuity of $\phi$ on the interval $[-E,E]$,
for each
$\epsilon>0$, there exists a $\rho>0$ such that
$$
|\phi^{-1}(Y_1)-\phi^{-1}(Y_2)|<\epsilon,\quad
Y_1,Y_2\in [-E,E],\quad |Y_1-Y_2|<\rho.
$$
Put
$$
Y_1=\phi(A_y)-\int_0^{t_1}f(u,y(u)+h,y'(u))du,\quad
Y_2=\phi(A_y)-\int_0^{t_2}f(u,y(u)+h,y'(u))du.
$$
Since $|Y_1-Y_2|=|\int_{t_1}^{t_2}f(u,y(u)+h,y'(u))du|\le
D|t_1-t_2|$, it is easy to see that there exists $\delta>0$
(independent of $\epsilon$ ) such that $|Y_1-Y_2|<\rho$ for all
$t_1,t_2$ with $|t_1-t_2|<\delta$. Hence there is $\delta>0$
(independent of $\epsilon$) such that
$$
|(Ty)'(t_1)-(Ty)'(t_2)|=\phi^{-1}(Y_1)-\phi^{-1}(Y_2)|<\epsilon,
$$
whenever $t_1,t_2\in [0,1]$ and $|t_1-t_2|<\delta$.
This shows that $(Ty)(t)$ is equi-continuous on [0,1]. The
Arzela-Askoli theorem guarantees that $T(\Omega)$ is relative
compact, which means that $T$ is compact. Hence the continuity and
the compactness of $T$ imply that $T$ is completely continuous.
\end{proof}


\begin{theorem} \label{thm210}
Suppose that {\rm (H1)--(H3)} hold
and there exist positive constants $e_1,e_2,c$,
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\left(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\right)ds\\
&\quad+\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\left(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\right)ds,
\end{aligned}\\
 Q= \min\big\{\phi\big(\frac{c}{L}\big),\;\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}}
\big\};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(s-k\right)ds}\Big);\quad
 E= \phi\big(\frac{e_1}{L}\big).
\end{gather*}
such that
$$
c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0,\quad
Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W.
$$
If
\begin{itemize}
\item[(A1)]
$f(t,u,v)<Q$ for all $t\in [0,1],u\in [h,c+h],v\in [-c,c]$;

\item[(A2)] $f(t,u,v)>W$ for all $t\in [k,1-k],u\in
[e_2+h,e_2/\sigma_0^2+h],v\in [-c,c]$;

\item[(A3)] $f(t,u,v)< E$ for all $t\in [0,1],u\in
[h,e_1/\sigma_0+h],v\in [-c,c]$;
\end{itemize}
then \eqref{e14} has at least three decreasing positive solutions
$x_1,x_2,x_3$ such that
$$
x_1(0)<e_1+h,\quad x_2(1-k)>e_2+h,\quad
x_3(0)>e_1+h,\quad x_3(1-k)<e_2+h.
$$
\end{theorem}

\begin{remark} \label{rmk1}\rm
 In paper \cite{kk1}, sufficient conditions are found
for the existence of solutions of \eqref{e9} based on the existence of
lower and upper solutions with certain relations. Using the obtained
results, under some other assumptions, the explicit ranges of values
of $\lambda_1$ and $\lambda_2$ are presented with which \eqref{e9} has a
solution, has a positive solution, and has no solution,
respectively. Furthermore, it is proved that the whole plane for
$\lambda_1$ and $\lambda_2$ can be divided into two disjoint
connected regions $\wedge E$ and $\wedge N$ such that \eqref{e9} has a
solution for $(\lambda_1,\lambda_2)\in\wedge E$ and has no solution
for $(\lambda_1,\lambda_2)\in\wedge N$. When applying Theorem \ref{thm210}
to \eqref{e9}, it shows us that \eqref{e9} has at least three decreasing
positive solutions under the assumptions $\lambda_1\le
0,\lambda_2\ge 0$ and some other assumptions.
\end{remark}

\begin{remark} \label{rmk2}\rm
 Consider the case $A\le 0$ and $B<0$, when
(H3) and (H4) hold, we can prove similarly that Lemma \ref{lem27} and
Lemma \ref{lem28} are valid. Define the same operator $T_1$ on the cone
$P_1$. Theorem \ref{thm210} shows that \eqref{e23} has at least three decreasing
and positive solutions $y_1,y_2,y_3$. Hence \eqref{e14} has at least
three decreasing solutions $x_1=y_1+h,x_2=y_2+h$ and $x_3=y_3+h$,
which need not be positive since
$h=\frac{B}{1-\sum_{i=1}^m\beta_i}<0$. In cases $A>0,B\le 0$ and
$A>0,B<0$, the author could not get the sufficient conditions
guaranteeing the existence of multiple positive solutions of
\eqref{e14}.
\end{remark}


\begin{proof}[Proof of Theorem \ref{thm210}]
To apply Lemma \ref{lem25}, we prove that its hypotheses
are satisfied. By the definitions, it is
easy to see that $\alpha,\psi $ are two nonnegative continuous
concave functionals on the cone $P_1$, $\gamma ,\beta,\theta$ are
three nonnegative continuous convex functionals on the cone $P_1$
and $\alpha(y)\le \beta(y)$ for all $y\in P_1$, there exist
constants $M>0$ such that $\|y\|\le M\gamma (y)\quad\text{for all }y\in
P_1$. Lemma \ref{lem28} implies that $x=x(t)$ is a positive solution of
\eqref{e14} if and only if $x(t)=y(t)+h$ and $y(t)$ is a solution of the
operator equation $y=T_1y$ and $T_1:P_1\to P_1$ is
completely continuous.

Corresponding to Lemma \ref{lem25},
$$
c=c,\quad h=\sigma_0e_1,\quad d=e_1,\quad a=e_2,\quad
b=\frac{e_2}{\sigma_0}.
$$
Now, we prove that all conditions of Lemma \ref{lem25} hold. One sees that $0<d<a$.
The remainder is divided into four steps.

\noindent\textbf{Step 1.} Prove that $T_1:\overline{{P_1}_c}\to
\overline{{P_1}_c}$;
For $y\in \overline{{P_1}_c}$, we have $\|y\|\le c$. Then $0\le
y(t)\le c$ for $t\in [0,1]$ and $-c\le y'(t)\le c$ for all $t\in
[0,1]$. So $(A_1)$ implies that
$$
f(t,y(t)+h,y'(t))\le Q,\quad t\in [0,1].
$$
 It follows from Lemma \ref{lem29} that $T_1y\in P_1$. Then Lemma \ref{lem28}
 implies
\begin{align*}
0&\le(T_1y)(t)\\
&= -\int_t^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad-\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\Big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}\\
&\quad\times \int_0^1f(u,y(u)+h,y'(u))du
 +\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(\phi\Big(\frac{-A(1+\sum_{i=1}^m\alpha_i)}
 {1-\sum_{i=1}^m\alpha_i}\Big)\\
 &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du+\int_0^s\! f(u,y(u)+h,y'(u))du\Big)ds\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\Big(
\frac{1+\sum_{i=1}^m\alpha_i}{2}\Big)}Q
+Qs\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)\\
 &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}Q+Qs\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(Q+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}Q+Qs\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(Q+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}Q+Qs\Big)ds
\\
&= \phi^{-1}(Q)\Big[\int_0^1\phi^{-1}
\Big(1+\frac{\phi\Big(1+\sum_{i=1}^m\alpha_i\Big)}
{\phi(2)-\phi\Big( 1+\sum_{i=1}^m\alpha_i\Big)}+s\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)
-\phi\left( 1+\sum_{i=1}^m\alpha_i\right)}+s\Big)ds\Big]\\
 &\le c.
\end{align*}
Similarly to the above discussion, we have from
Lemma \ref{lem27} that
$$
|A_y|\le \Big|\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
-\frac{1}{1-\phi\big( \frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du\Big|.
$$
Then
\begin{align*}
&|(T_1y)'(t)|\\
&\le |(T_1 y)'(0)|=|\phi^{-1} (A_y)\|\\
&\le \phi^{-1}\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
+\frac{1}{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du\Big)\\
&\le \phi^{-1}\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)+\frac{1}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}Q\Big)\\
&\le \phi^{-1}\Big(Q+\frac{\phi(2)}{\phi(2)
-\phi\big(1+\sum_{i=1}^m\alpha_i\big)}Q\Big)
\le  c.
\end{align*}
It follows that $\|T_1y\|=\max\{\max_{t\in
[0,1]}|(T_1y)(t)|,\max_{t\in [0,1]}|(T_1y)'(t)|\}\le c$. Then
$T_1:\overline{{P_1}_c}\to \overline{{P_1}_c}$.


\noindent\textbf{Step 2.} Prove that
$$\{y\in P_1(\gamma,\theta,\alpha;a,b,c)|\alpha(y)>a\}=\{y\in
P_1\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big)
|\alpha(y)>e_2\}\neq \emptyset
$$
and $\alpha(T_1y)>e_2\quad\text{for every }y\in
P_1\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big)$.

Choose
$y(t)=\frac{e_2}{2\sigma_0}$ for all $t\in [0,1]$. Then $y\in P_1$
and
$$
\alpha(y)=\frac{e_2}{2\sigma_0}>e_2,\;\theta(y)=\frac{e_2}{2\sigma_0}\le
\frac{e_2}{\sigma_0},\;\gamma(y)=0< c.
$$
It follows that $\{y\in
P_1(\gamma,\theta,\alpha;a,b,c):\alpha(y)>a\}\neq \emptyset$.
For $y\in P_1(\gamma,\theta,\alpha;a,b,c)$, one has
$$
\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2,\quad
\theta(y)=\max_{t\in [k,1-k]}y(t)\le \frac{e_2}{\sigma_0},\quad
\gamma(y)=\max_{t\in [0,1]}| y'(t)|\le c.
$$
Then
$$
e_2\le y(t)\le \frac{e_2}{\sigma_0^2},\quad t\in [k,1-k],\; |y'(t)|\le c.
$$
Thus (A2) implies
$$
f(t,y(t)+h,| y'(t)|)\ge W,\quad n\in [k,1-k].
$$
Since
$$
\alpha(T_1y)=\min_{t\in [k,1-k]}(T_1y)(t)\ge \sigma_0\max_{t\in
[0,1]}(T_1y)(t),
$$
we get
\begin{align*}
\alpha(T_1y)
&\geq \sigma_0\Big[-\int_0^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big]\\
&\ge \sigma_0\Big[-\int_0^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\Big]\\
&\ge \sigma_0\Big[\int_0^1\phi^{-1}
\left(\phi\left(\frac{-A}{1-\sum_{i=1}^m\alpha_i}\right)
+\int_0^sf(u,y(u)+h,y'(u))du\right)ds\Big]\\
&\ge \sigma_0\Big[\int_k^{1-k}\phi^{-1}
\left(\phi\left(\frac{-A}{1-\sum_{i=1}^m\alpha_i}\right)
+\int_k^sf(u,y(u)+h,y'(u))du\right)ds\Big]\\
&\ge \sigma_0\int_k^{1-k}\phi^{-1}
\big(W(s-k)\big)ds\\
 &= e_2.
\end{align*}
 This completes Step 2.

\noindent\textbf{Step 3.} Prove that
$\{y\in Q(\gamma,\theta,\psi;h,d,c):\beta(y)<d\}$ which is equal to
$\{y\in Q(\gamma,\theta,\psi;\sigma_0e_1,e_1,c):\beta(y)<e_1\}$ is not empty,
and
$$
\beta(T_1y)<e_1\quad\text{for every }
y\in Q(\gamma,\theta,\psi;h,d,c)
=Q\left(\gamma,\theta,\psi;\sigma_0e_1,e_1,c\right);
$$
Choose $y(t)=\sigma_0e_1$. Then $y\in P_1$, and
$$
\psi(y)= \sigma_0e_1\ge h,\quad
\beta(y)=\theta(y)=\sigma_0e_1<e_1=d,\quad
\gamma(y)=0\le c.
$$
It follows that $\{y\in
Q(\gamma,\theta,\psi;h,d,c)|\beta(y)<d\}\neq \emptyset$.

For $y\in Q(\gamma,\theta,\psi;h,d,c)$, one has
\begin{gather*}
\psi(y)=\min_{t\in [k,1-k]}y(t)\ge h=e_1\sigma_0,\quad
\theta(y)=\max_{t\in [k,1-k]}y(t)\le d=e_1,\\
\gamma (y)=\max_{t\in [0,1]}|y'(t)|\le c.
\end{gather*}
Hence
$0\le y(t)\le \frac{e_1}{\sigma_0}$ and $-c\le y'(t)\le c$,
for $t\in [0,1]$.
Then (A3) implies
$$
f(t,y(t)+h,|y'(t)|)\le E,\quad t\in [0,1].
$$
So
\begin{align*}
&\beta(T_1y)\\
&= \max_{t\in [0,1]}(T_1y)(t)\\
&= -\int_0^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
\\
&\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1} \Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
\\
 &\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)
 +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
 {1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
 \int_0^1f(u,y(u)+h,y'(u))du
\\
&\quad+\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)\\
 &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du
+\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)+\frac{\phi
\left(\frac{1+\sum_{i=1}^m\alpha_i}{2}\right)}{1-\phi\left(
\frac{1+\sum_{i=1}^m\alpha_i}{2}\right)}E+Es\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(\phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big)\\
 &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}
{1-\phi\big( \frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}E+Es\Big)ds\\
&\le \int_0^1\phi^{-1}
\Big(E+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}E+Es\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(E+\frac{\phi\big(\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}{1-\phi\big(
\frac{1+\sum_{i=1}^m\alpha_i}{2}\big)}E+Es\Big)ds
\\
&= \phi^{-1}(E)\Big[\int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\Big)ds \\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\Big)ds\Big]\\
  &= e_1=d.
\end{align*}
This completes Step 3.


\noindent\textbf{Step 4.} Prove that $\alpha(T_1 y)>a$ for $y\in
P_1(\gamma,\alpha;a,c)$ with $\theta(T_1y)>b$;
For $y\in P_1(\gamma,\alpha;a,c)=P_1(\gamma,\alpha;e_2,c)$ with
$\theta(T_1y)>b=\frac{e_2}{\sigma_0}$, we have that
$\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2$ and $\gamma
(y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\max_{t\in
[k,1-k]}(T_1y)(t)>\frac{e_2}{\sigma_0}$. Then
$$
\alpha(T_1y)=\min_{t\in [k,1-k]}(T_1y)(t)\ge
\sigma_0\beta(T_1y)>\sigma_0\frac{e_2}{\sigma_0}=e_2=a.
$$
This completes Step 4.

\noindent\textbf{Step 5.}
 Prove that $\beta(T_1y)<d$ for each $y\in Q(\gamma,\beta;d,c)$
with $\psi(T_1y)<h$.
For $y\in Q(\gamma ,\beta;d,c)$ with $\psi(T_1y)<d$, we have $\gamma
(y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\beta(y)=\max_{t\in
[0,1]}y(t)\le d=e_1$ and $\psi(T_1y)=\min_{t\in [k,1-k]}(T_1y)(t)<
h=e_1\sigma_0$. Then
$$
 \beta(T_1y)=\max_{t\in [0,1]}(T_1y)(t)\le
\frac{1}{\sigma_0}\min_{t\in
[k,1-k]}(T_1y)(t)<\frac{1}{\sigma_0}e_1\sigma_0=e_1=d.
$$
This completes the Step 5.


Then Lemma \ref{lem25} implies that $T_1$ has at least three  fixed points
$y_1$, $y_2$ and $y_3$ such that
$$
\beta(y_1)<e_1,\quad \alpha(y_2)>e_2,\quad
\beta(y_3)>e_1,\quad \alpha (y_3)<e_2.
$$
Hence \eqref{e14} has three decreasing positive solutions $x_1,x_2$ and
$x_3$ such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\max_{t\in [0,1]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
Hence
$x_1(0)<e_1+h$, $x_2(1-k)>e_2+h$,
$x_3(0)>e_1+h$, $x_3(1-k)<e_2+h$.
\end{proof}

\subsection{Positive Solutions of \eqref{e15}}
Now we prove the  existence of three positive increasing
solutions of \eqref{e15}. We use the assumptions:
\begin{itemize}
\item[(H5)] $f:[0,1]\times [0,+\infty)\times
[0,+\infty)\to [0,+\infty)$ is continuous with
$f(t,c+h,0)\not\equiv 0$ on each sub-interval of [0,1] for all $c\ge
0$, where $h=\frac{A}{1-\sum_{i=1}^m\alpha_i}$;

\item[(H6)] $A\ge 0,B\ge0$;


\item[(H7)] $\alpha_i\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\alpha_i\le1$, $\sum_{i=1}^m\beta_i<1$;

\end{itemize}

Consider the boundary-value problem
\begin{equation} \label{e25}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=0,\quad
y'(1)- \sum_{i=1}^m\beta_iy'(\xi_i)=B.
\end{gathered}
\end{equation}


\begin{lemma} \label{lem211}
 Suppose that {\rm (H4), (H6), (H7)} hold. If
$y$ is a solution of \eqref{e25}, then $y$ is increasing and positive on
$(0,1)$.
\end{lemma}

\begin{proof}
Suppose $y$ satisfies \eqref{e25}. It follows from the
assumptions that $y'$ is decreasing on $[0,1]$. Then the BCs in \eqref{e25}
and (H4) imply
$$
y'(1)=\sum_{i=1}^m\beta_i y'(\xi_i)+B\ge \sum_{i=1}^m\beta_i y'(1).
$$
It follows that $y'(1)\ge 0$. We get that $y'(t)\ge 0$ for $t\in
[0,1]$. Then
$$
y(0)= \sum_{i=1}^m\alpha_iy(\xi_i)\ge \sum_{i=1}^m\alpha_i y(0).
$$
It follows that $y(0)\ge 0$. Then $y(t)>y(0)\ge 0$ for $t\in (0,1)$
since $y'(t)\ge 0$ for all $t\in  [0,1]$. The
 proof is complete.
\end{proof}


\begin{lemma} \label{lem212}
Suppose that {\rm (H4), (H6), (H7)} hold. If
$y$ is a solution of \eqref{e25}, then
$$
y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds,
$$
with
\begin{equation} \label{e26}
\phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1}
\Big(A_h+\int_{\xi_i}^1h(s)ds\Big)+B,
\end{equation}
 and
$$
B_h=\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(A_h+\int_s^1h(u)du\Big)ds,$$
 where
\begin{gather*}
 a= \phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big),\\
b=\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds.
\end{gather*}
\end{lemma}

\begin{proof}  It follows from \eqref{e25} that
$$
y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds,
$$
and the BCs in \eqref{e25} imply
\begin{gather*}
y'(1)=\sum_{i=1}^m\beta_i\phi^{-1}\left(\phi(y'(1))
+\int_{\xi_i}^1h(s)ds\right)+B,
\\
y(0)= \frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.
\end{gather*}
Let
$$
G(c)=\phi^{-1}(c)-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(c+\int_{\xi_i}^1h(s)ds\Big)-B.
$$
It is easy to see that
\begin{align*}
G(a)
&= G\Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\Big)\\
&\le \frac{B}{1-\sum_{i=1}^m\beta_i}
-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\Big)-B\\
&= 0.
\end{align*}
On the other hand, one sees that
\begin{align*}
\frac{G(b)}{\phi^{-1}(b)}
&= 1-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(1+\frac{\int_{\xi_i}^1h(s)ds}{b}\Big)
-\frac{B}{\phi^{-1}(b)}
\\
&= 1-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(1+\frac{\int_{\xi_i}^1h(s)ds}{
\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1h(s)ds}\Big)
\\
&\quad -\frac{B}{\phi^{-1}
\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1h(s)ds\Big)}
\\
&\ge 1-\sum_{i=1}^m\beta_i\phi^{-1}
\Big( 1+\frac{1-\phi
\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}
{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\Big)-\frac{1-\sum_{i=1}^m\beta_i}{1+\sum_{i=1}^m\beta_i}
= 0.
\end{align*}
Hence $G(b)\ge 0$. It is easy to know that
$\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on
$(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence
$G(a)\le 0$ and $G(b)\ge 0$ and
\[
\lim_{c\to 0^-}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad \lim_{c\to
-\infty}\frac{G(c)}{\phi^{-1}(c)}= 1-\sum_{i=1}^m\beta_i>0.
\]
Then there exists unique constant $A_h=\phi(y'(1))\in [a,b]$
such that \eqref{e26} holds. The proof is complete.
\end{proof}


Note $h=\frac{A}{1-\sum_{i=1}^m\alpha_i}$, and $x(t)-h=y(t)$. Then
\eqref{e15} is transformed into the boundary-value problem
\begin{equation} \label{e27}
\begin{gathered}{}
[\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=0,\quad
y'(1)-\sum_{i=1}^m\beta_i y'(\xi_i)=B,
 \end{gathered}
\end{equation}
Let
\begin{align*}
P_2=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in [0,1]}y(t)\text{
for all }t\in [0,1]\big\}.
\end{align*}
Then $P_2$ is a cone in $X$.
Since
\begin{align*}
|y(t)|&=\big|\frac{\sum_{i=1}^m\alpha_i
y(\xi_i)-\sum_{i=1}^m\alpha_i y(0)}{1-\sum_{i=1}^m\alpha_i}\big|\\
&\le\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i
}\max_{t\in [0,1]}|y'(t)|
=\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i}\gamma(y),
\end{align*}
we have
 $$
\max_{t\in [0,1]}|y(t)|\le\frac{\sum_{i=1}^m
\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i }\gamma(y).
$$
It is easy to see that there exists a constant $M>0$
such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$.


Define the nonlinear operator $T_2:P_2\to X$ by
\[
(T_2y)(t)= B_y+\int_0^t\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_2,
\]
 where $A_y$ satisfies
\begin{equation} \label{e28}
\phi^{-1}(A_y)=\sum_{i=1}^m\beta_i\phi^{-1}\left(A_y
+\int_{\xi_i}^1f(s,y(s)+h,y'(s))ds\right)+B,
\end{equation}
and $B_y$ satisfies
\begin{align*}
B_y&= \frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \left(A_y+\int_s^1f(u,y(u)+h,y'(u))ds\right)ds.
\end{align*}
Then for $y\in P_2$,
\begin{align*}
(T_2y)(t)&= \int_0^t\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \left(A_y+\int_s^1f(u,y(u)+h,y'(u))ds\right)ds\,.
\end{align*}

\begin{lemma} \label{lem213}
 Suppose that {\rm (H5), (H6), (H7)} hold. Then
\begin{itemize}
\item[(i)]  the following equalities hold:
\begin{gather*}{}
[\phi((T_2y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_2y)(0)-\sum_{i=1}^m\alpha_i (T_2y)(\xi_i)=0,\quad
(T_2y)'(1)-\sum_{i=1}^m\beta_i (T_2y)'(\xi_i)=B;
 \end{gather*}

\item[(ii)] $T_2y\in P_2$ for each $y\in P_2$;

\item[(iii)] $x$ is a solution of \eqref{e15} if and only if $x=y+h$
and $y$ is a solution of the operator equation $y=T_2y$ in cone
$P_2$;

\item[(iv)] $T_2:P_2\to P_2$ is completely continuous.
\end{itemize}
\end{lemma}

The proofs of the above lemma is similar to that of  Lemma \ref{lem29}
and it is omitted.

\begin{theorem} \label{thm214}
Suppose that {\rm (H5)--(H7)} hold
and there exist positive constants $e_1,e_2,c$,
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds,
\end{aligned}\\
 Q= \min\big\{\phi\left(\frac
{c}{L}\right),\;\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\beta_i\right)}}
\big\};\quad
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
(1-k-s)ds}\Big);\\
 E= \phi(\frac{e_1}{L}).
\end{gather*}
such that
$$
c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0,\quad
Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad
Q>W.
$$
If (A1), (A2) and (A3) from in Theorem \ref{thm210} hold, then
\eqref{e15} has at least three increasing positive solutions
$x_1,x_2,x_3$ such that
$$
x_1(1)<e_1+h,\quad x_2(k)>e_2+h,\quad x_3(1)>e_1+h,\quad x_3(k)<e_2+h.
$$
\end{theorem}


\begin{proof}
 To apply Lemma \ref{lem25}, we prove that all its conditions
are satisfied. By the definitions, it is easy to see that
$\alpha,\psi $ are two nonnegative continuous concave functionals on
the cone $P_2$, $\gamma ,\beta,\theta$ are three nonnegative
continuous convex functionals on the cone $P_2$ and $\alpha(y)\le
\beta(y)$ for all $y\in P_2$, there exist constants $M>0$ such that
$\|y\|\le M\gamma (y)\quad\text{for all }y\in P_2$. Lemma \ref{lem213} implies
that $x=x(t)$ is a positive solution of \eqref{e15} if and only if
$x(t)=y(t)+h$ and $y(t)$ is a solution of the operator equation
$y=T_2y$ and $T_2:P_2\to P_2$ is completely continuous.

Corresponding to Lemma \ref{lem25},
$$
c=c,\quad h=\sigma_0e_1,\quad
d=e_1,\quad a=e_2,\quad b=\frac{e_2}{\sigma_0}.
$$ Now, we
prove that all conditions of Lemma \ref{lem25} hold. One sees that $0<d<a$.
The remainder is divided in four steps.

\noindent\textbf{Step 1.} Prove that $T_2:\overline{{P_2}_c}\to
\overline{{P_2}_c}$;

For $y\in \overline{{P_2}_c}$, we have $\|y\|\le c$. Then
$0\le y(t)\le c$ for $t\in [0,1]$ and $-c\le y'(t)\le c$ for all
$t\in [0,1]$. So $(A_1)$ implies that
$$
f(t,y(t)+h,y'(t))\le Q,\quad  t\in [0,1].
$$
 It follows from Lemma \ref{lem213} that $T_2y\in P_2$.
Then Lemma \ref{lem212}
 implies
\begin{align*}
0&\le(T_2y)(t)\\
&\le \int_0^1\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}
\int_0^1f(u,y(u)+h,y'(u))du \\
&\quad +\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
\\
 &\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(\phi\left(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\right)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}Q+Q(1-s)\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
\\
&\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}Q+Q(1-s)\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(Q+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}Q+Q(1-s)\Big)ds
\\
&\quad+\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}
\Big(Q+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}Q+Q(1-s)\Big)ds
\\
&= \phi^{-1}(Q)\Big[\int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds\Big]\\
 &\le c.
\end{align*}
Similarly to above discussion,  from Lemma \ref{lem212},
 we have
\begin{align*}
&|(T_2y)'(t)|\\
&\le (T_2 y)'(0)=\phi^{-1} \Big(A_y+\int_0^1f(u,y(u)+h,y'(u))du\Big)
\\
&\le \phi^{-1}\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du \\
&\quad +\int_0^1f(u,y(u)+h,y'(u))du\Big)\\
&\le \phi^{-1}
\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)+\frac{1}{1-\phi
\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}Q\Big)
\\
&\le \phi^{-1}\Big(Q+\frac{\phi(2)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\beta_i\right)}Q\Big)
\le  c.
\end{align*}
It follows that $\|T_2y\|=\max\{\max_{t\in
[0,1]}|(T_2y)(t)|,\max_{t\in [0,1]}|(T_2y)'(t)|\}\le c$. Then
$T_2:\overline{{P_2}_c}\to \overline{{P_2}_c}$.


\noindent\textbf{Step 2.} Prove that
 $$
\{y\in P_2(\gamma,\theta,\alpha;a,b,c)|\alpha(y)>a\}
=\{y\in P_2\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big):
\alpha(y)>e_2\}\neq \emptyset
$$
and $\alpha(T_2y)>e_2\quad\text{for every }y\in
P_2\big(\gamma,\theta,\alpha;e_2,\frac{e_2}{\sigma_0},c\big)$;
Choose
$y(t)=\frac{e_2}{2\sigma_0}$ for all $t\in [0,1]$. Then $y\in P_2$
and
$$
\alpha(y)=\frac{e_2}{2\sigma_0}>e_2,\quad
\theta(y)=\frac{e_2}{2\sigma_0}\le \frac{e_2}{\sigma_0},\quad
\gamma(y)=0< c.
$$
It follows that $\{y\in
P_2(\gamma,\theta,\alpha;a,b,c)|\alpha(y)>a\}\neq \emptyset$.

For $y\in P_2(\gamma,\theta,\alpha;a,b,c)$, one has
$$
\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2,\quad
\theta(y)=\max_{t\in [k,1-k]}y(t)\le \frac{e_2}{\sigma_0},\quad
\gamma(y)=\max_{t\in [0,1]}|y'(t)|\le c.
$$
Then
$$
e_2\le y(t)\le \frac{e_2}{\sigma_0^2},\quad t\in [k,1-k],\quad
|y'(t)|\le c.
$$
Thus (A2) implies
$f(t,y(t)+h,| y'(t)|)\ge W,\;\;t\in [k,1-k]$.
Since
$$
\alpha(T_2y)=\min_{t\in [k,1-k]}(T_2y)(t)\ge \sigma_0\max_{t\in
[0,1]}(T_2y)(t),
$$
from Lemma \ref{lem212}, we have
\begin{align*}
\alpha(T_2y)
&\geq \sigma_0\max_{t\in [0,1]}(T_2y)(t)\\
&= \sigma_0\Big[\int_0^1\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\quad+\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(A_h+\int_s^1f(u,y(u)+h,y'(u))ds\Big)ds\Big]\\
&\ge \sigma_0\Big[\int_0^1\phi^{-1}
\Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)
 +\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)\\
&\quad +\int_s^1f(u,y(u)+h,y'(u))ds\Big)ds\Big]
\\
&\ge \sigma_0\int_0^1\phi^{-1}
\left(\phi\Big(\frac{B}{1-\sum_{i=1}^m\beta_i}\Big)+\int_s^1f(u,y(u)+h,y'(u))du\right)ds\\
&\ge \sigma_0\int_0^1\phi^{-1}
\Big(\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\ge \sigma_0\int_k^{1-k}\phi^{-1}
\Big(\int_s^{1-k}f(u,y(u)+h,y'(u))du\Big)ds\\
&\ge \sigma_0\int_k^{1-k}\phi^{-1}
\left(W(1-k-s)\right)ds
= e_2.
\end{align*}
 This completes Step 2.

\noindent\textbf{Step 3.} Prove that the set
$\{y\in Q(\gamma,\theta,\psi;h,d,c)|\beta(y)<d\}$ which is equal to
$\{y\in Q\big(\gamma,\theta,\psi;\sigma_0e_1,e_1,c\big):\beta(y)<e_1\}$
is not empty, and
$\beta(T_2y)<e_1\quad\text{for every }y\in
Q(\gamma,\theta,\psi;h,d,c)=Q\big(\gamma,\theta,\psi;\sigma_0e_1,e_1,c\big)$.
Choose $y(t)=\sigma_0e_1$. Then $y\in P_2$, and
$$
\psi(y)= \sigma_0e_1\ge h,\quad \beta(y)=\theta(y)=\sigma_0e_1<
e_1=d,\quad \gamma(y)=0\le c.
$$
It follows that $\{y\in Q(\gamma,\theta,\psi;h,d,c):\beta(y)<d\}\neq \emptyset$.

For $y\in Q(\gamma,\theta,\psi;h,d,c)$, one has
\begin{gather*}
\psi(y)=\min_{t\in [k,1-k]}y(t)\ge h=e_1\sigma_0,\quad
\theta(y)=\max_{t\in [k,1-k]}y(t)\le d=e_1,\\\
\gamma(y)=\max_{t\in [0,1]}|y'(t)|\le c.
\end{gather*}
Hence we have
$0\le y(t)\le \frac{e_1}{\sigma_0}$ and $-c\le y'(t)\le c$ for $t\in [0,1]$.
Then (A3) implies
$f(t,y(t)+h,|y'(t)|)\le E,\quad t\in [0,1]$.
So that
\begin{align*}
\beta(T_2y)
&= \max_{t\in [0,1]}(T_2y)(t)\\
&= \int_0^1\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)\\
&\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\Big(\frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}
\int_0^1f(u,y(u)+h,y'(u))du \\
&\quad +\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)\\
&\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}
\int_0^1f(u,y(u)+h,y'(u))du\\
&\quad +\int_s^1f(u,y(u)+h,y'(u))du\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}E+E(1-s)\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1} \Big(\phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)\\
&\quad +\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big(
\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}E+E(1-s)\Big)ds
\\
&\le \int_0^1\phi^{-1}
\Big(E+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}E+E(1-s)\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}
\Big(E+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big(
\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}E+E(1-s)\Big)ds
\\
&= \phi^{-1}(E)\Big[\int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big(
\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds
\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+(1-s)\Big)ds\Big]\\
&= e_1=d.
\end{align*}
This completes Step 3.


\noindent\textbf{Step 4.} Prove that $\alpha(T_2 y)>a$ for
$y\in P_2(\gamma,\alpha;a,c)$ with $\theta(T_2y)>b$;
For $y\in P_2(\gamma,\alpha;a,c)=P_2(\gamma,\alpha;e_2,c)$ with
$\theta(T_2y)>b=\frac{e_2}{\sigma_0}$, we have that
$\alpha(y)=\min_{t\in [k,1-k]}y(t)\ge e_2$ and $\gamma
(y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\max_{t\in
[k,1-k]}(T_2y)(t)>\frac{e_2}{\sigma_0}$. Then
$$
\alpha(T_2y)=\min_{t\in [k,1-k]}(T_2y)(t)\ge
\sigma_0\beta(T_2y)>\sigma_0\frac{e_2}{\sigma_0}=e_2=a.
$$
This completes Step 4.

\noindent\textbf{Step 5.}
 Prove that $\beta(T_2y)<d$ for each $y\in Q(\gamma,\beta;d,c)$
with $\psi(T_2y)<h$.
For $y\in Q(\gamma ,\beta;d,c)$ with $\psi(T_2y)<d$, we have $\gamma
(y)=\max_{t\in [0,1]}|y(t)|\le c$ and $\beta(y)=\max_{t\in
[0,1]}y(t)\le d=e_1$ and $\psi(T_2y)=\min_{t\in [k,1-k]}(T_1y)(t)<
h=e_1\sigma_0$. Then
$$
 \beta(T_2y)=\max_{t\in [0,1]}(T_2y)(t)\le
\frac{1}{\sigma_0}\min_{t\in
[k,1-k]}(T_2y)(t)<\frac{1}{\sigma_0}e_1\sigma_0=e_1=d.
$$
This completes the Step 5.

Then Lemma \ref{lem25} implies that $T_2$ has at least three fixed points
$y_1$, $y_2$ and $y_3$ in $P_2$ such that
$$
\beta(y_1)<e_1,\quad \alpha(y_2)>e_2,\quad
\beta(y_3)>e_1,\quad \alpha(y_3)<e_2.
$$
Hence \eqref{e15} has three increasing positive solutions $x_1,x_2$ and
$x_3$ such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad \min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\max_{t\in [0,1]}x_3(t)>e_1+h,\quad \min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
Hence
$$x_1(1)<e_1+h,\quad x_2(k)>e_2+h,\quad
x_3(1)>e_1+h,\quad x_3(k)<e_2+h.
$$
The proof is complete.
\end{proof}

\begin{remark} \label{rmk3}\rm
For \eqref{e15}, when $A<0,B\ge 0$, we can also get
the existence results for three increasing solutions of \eqref{e15}
similarly, but the solutions need not be positive. By the way, it is
interesting to establish sufficient conditions guarantee the
existence of positive solutions of \eqref{e15} when $B<0$.
\end{remark}

\subsection{Positive Solutions of \eqref{e16}}

Now we prove an existence result for three positive solutions of
\eqref{e16}. We use the following there conditions:
\begin{itemize}

\item[(H8)] $f:[0,1]\times [h,+\infty)\times \mathbb{R}\to
[0,+\infty)$ is continuous with $f(t,c+h,0)\not\equiv 0$ on each
sub-interval of [0,1] for all $c\ge 0$, where
$h=\frac{B}{1-\sum_{i=1}^m\beta_i}$;

\item[(H9)] $A\ge 0,B\ge0$;

\item[(H10)] $\alpha\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\beta_i<1$ and $A\ge \frac{B}{1-\sum_{i=1}^m\beta_i}$;

\end{itemize}

Consider the following boundary-value problem
\begin{equation} \label{e29}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\quad t\in (0,1),\\
y(0)-\alpha y'(0)=D,\quad
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)=0,
 \end{gathered}
\end{equation}


\begin{lemma} \label{lem215}
 Suppose that {\rm (H4), (H10)} hold and  $D\ge 0$.
If $y$ is a solution of \eqref{e29}, then $y$ is positive on
$(0,1)$.
\end{lemma}

\begin{proof}
Suppose $y$ satisfies \eqref{e29}. It follows from
assumption (H4) that $y'$ is decreasing on $[0,1]$.

If $y'(1)>0$, the BCs in \eqref{e29} and (H4) imply  $y'(t)>0$ for
all $t\in [0,1]$. Then
$$
y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\le  \sum_{i=1}^m\beta_iy(1).
$$
Then $y(1)\le 0$. On the other hand,
$y(0)=\alpha y'(0)+D\ge 0$.
 This is a contradiction since $y'(t)>0$.

If $y'(1)\le 0$ and $y'(0)\le 0$, then we get that $y'(t)\le 0$ for
all $t\in [0,1]$. The BCs in \eqref{e29} and (H4) imply
$$
y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\ge  \sum_{i=1}^m\beta_iy(1).
$$
It follows that $y(1)\ge 0$. Then $y(t)\ge y(1)\ge 0$ for all
$t\in [0,1]$. (H4) implies  $y(t)>0$ for all $t\in (0,1)$.

If $y'(1)\le 0$ and $y'(0)>0$, then $y(0)=\alpha y'(0)+D\ge 0$. It
follows from \eqref{e29} and (H4) that
$$
y(1)=\sum_{i=1}^m\beta_iy(\xi_i)\ge
\sum_{i=1}^m\beta_i\min\{y(0),y(1)\}.
$$
If $y(1)\le y(0)$, then $y(1)\ge \sum_{i=1}^m\beta_i y(1)$ implies
that $y(1)\ge 0$. If $y(1)>y(0)$, then $y(1)>0$ since $y(0)\ge 0$.
It follows from (H4) that $y(t)\ge \min\{y(0),y(1)\}\ge 0$. Then
$y$ is positive on $(0,1)$. The proof is complete.
\end{proof}


\begin{lemma} \label{lem216}
Assume {\rm (H4), (H10)} hold and  $D\ge 0$.
If $y$ is a solution of \eqref{e29}, then
$$
y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big),\quad
t\in [0,1],
$$
where $A_h\in [b,0]$,
\begin{align*}
&\Big[\alpha\phi^{-1}\Big(A_h+\int_{\xi}^1h(u)du\Big)+D\Big]
\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&= -\Big(1-\sum_{i=1}^m\beta_i\Big)
  \int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds,
\end{align*}
\begin{gather*}
B_h=\alpha\phi^{-1}\Big(A_h+\int_{\xi}^1h(u)du\Big)+D,\\
b=-\int_0^1h(u)du-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big),\\
a= 1+\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)-\sum_{i=1}^m\beta_i\xi_i.
\end{gather*}
\end{lemma}


\begin{proof} From \eqref{e29} it follows that
\begin{gather*}
y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1(u)du\Big)ds,\\
\begin{aligned}
y(0)\Big(1-\sum_{i=1}^m\beta_i\Big)
&= -\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i}
\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds,
\end{aligned}\\
y(0)=\alpha\phi^{-1}\Big(\phi(y'(1))+\int_{0}^1h(u)du\Big)+D.
\end{gather*}
Thus
\begin{align*}
&\Big[\alpha\phi^{-1}\Big(\phi(y'(1))+\int_{0}^1h(u)du\Big)+D\Big]
\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&= -\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_0^{\xi_i}
\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.
\end{align*}
Let
\begin{align*}
G(c)
&= \Big[\alpha\phi^{-1}\Big(c+\int_{0}^1h(u)du\Big)+D\Big]
\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&\quad +\int_0^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\
&\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}
\Big(c+\int_s^1h(u)du\Big)ds\\
&= \Big[\alpha\phi^{-1}\Big(c+\int_{0}^1h(u)du\Big)+D\Big]
\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)
\int_0^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1 \phi^{-1}
\Big(c+\int_s^1h(u)du\Big)ds\\.
\end{align*}
It is easy to see that $G(c)$ is increasing on $(-\infty,+\infty)$,
$G(0)\ge 0$. Since
$$
a=1+\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)-\sum_{i=1}^m\beta_i\xi_i,
$$
we get
\begin{align*}
G(b)&=G\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)
 -\int_0^1h(u)du\Big)
\\
&= \Big[\alpha\phi^{-1}
\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)+D\Big]
\Big(1-\sum_{i=1}^m\beta_i\Big)
\\
&\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)
-\int_0^sh(u)du\Big)ds
\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)
-\int_0^sh(u)du\Big)ds
\\
&\le \Big[\alpha\phi^{-1}
\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)
+D\Big] \Big(1-\sum_{i=1}^m\beta_i\Big)\\
&\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)\Big)ds
= 0,
\end{align*}
we get  $\phi(y'(1))\ge
-\int_0^1h(u)du-\phi\Big(\frac{D\Big(1-\sum_{i=1}^m\beta_i\Big)}{a}\Big)=b$.
The proof is complete.
\end{proof}

Let
\begin{align*}
P_3=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in
[0,1]}y(t)\text{ for all }t\in [0,1] \big\}.
\end{align*}
 Then $P_3$ is a cone in $X$.
Since, for $y\in P_3$, we have
\begin{align*}
|y(t)|
&= |y(t)-y(1)+y(1)|\\
&\le |y'(\theta)|(1-t)+|y(1)|\\
&\le \max_{t\in [0,1]}|y'(t)|+\Big|\frac{\sum_{i=1}^m\beta_i
y(\xi_i)-\sum_{i=1}^m\beta_i y(1)}{1-\sum_{i=1}^m\beta_i
}\Big|\\
&\le \Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\Big)\max_{t\in [0,1]}|y'(t)|\\
&= \Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\Big)\gamma(y),
\end{align*}
we get
$$
\max_{t\in [0,1]}|y(t)|
\le\Big(1+\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\Big)\gamma(y).
$$
It is easy to see that there exists a constant
$M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_3$.

Suppose that (H10) holds. Let $x(t)-h=y(t)$. Then \eqref{e16} is
transformed into
\begin{gather*}
[\phi(y'(t))]'+f(t,y(t)+h,y'(t))=0,\quad t\in (0,1),\\
y(0)-\alpha y'(0)=A-\frac{B}{1-\sum_{i=1}^m\beta_i}\ge 0,\\
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)= 0,
\end{gather*}
Define the nonlinear operator $T_3:P_3\to X$ by
\begin{align*}
(T_3y)(t)&= B_y+\int_0^t\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_3,
\end{align*}
 where
\begin{align*}
&\Big[\alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))du\Big)
+A-\frac{B}{1-\sum_{i=1}^m\beta_i}\Big]\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&= -\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1
\phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad -\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,
\end{align*}
and
$$
B_h=\alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))duu\Big)
+A-\frac{B}{1-\sum_{i=1}^m\beta_i}.
$$
For $y\in P_3$, the definition of $A_y$ implies
\begin{align*}
(T_3y)(t)
&= \alpha\phi^{-1}\Big(A_y+\int_{0}^1f(u,y(u)+h,y'(u))du\Big)
+A-\frac{B}{1-\sum_{i=1}^m\beta_i}\\
&\quad +\int_0^t\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&= -\int_t^1\phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds\\
&\quad -\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds.
\end{align*}

\begin{lemma} \label{lem217}
 Suppose that {\rm (H8)--(H10)} hold.
Then
\begin{itemize}
\item[(i)] $T_3y\in P_3$ for each $y\in P_3$;

\item[(ii)]  $x$ is a solution of \eqref{e17} if and only if $x=y+h$
 and $y$ is a solution of the operator equation $T_3y=y$ in $P_3$;

\item[(iii)]  $T_3:P_3\to P_3$ is completely continuous;

\item[(iv)] the following equalities hold:
\begin{gather*}{}
[\phi((T_3y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_3y)(0)-\alpha (T_3y)'(\xi)=A-\frac{B}{1-\sum_{i=1}^m\beta_i},\\
(T_3y)(1)-\sum_{i=1}^m\beta_i (T_3y)(\xi_i)=0;
 \end{gather*}
\end{itemize}
\end{lemma}

 The proof of the above lemma is  similar to that of
 Lemma \ref{lem29}, so it is omitted.


\begin{theorem} \label{thm218}
 Suppose that {\rm (H8)--(H10)} hold and that there exist positive
constants $e_1,e_2,c$,
\begin{gather*}
L= \phi^{-1}(2)+\frac{\phi^{-1}(2)}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i(1-\xi_i),\\
Q= \min\big\{\phi\big(\frac {c}{L}\big), \frac{\phi(c)}{2} \big\};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
(s-k)ds}\Big);\quad
 E= \phi\big(\frac{e_1}{L}\big).
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{\big(A-\frac{B}{1-\sum_{i=1}^m\beta_i}\big)
\left(1-\sum_{i=1}^m\beta_i \right)}
 {1+\alpha-\sum_{i=1}^m\beta_i\xi_i}\Big),\quad
Q>W.
$$
If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold, then
\eqref{e16} has at least three positive solutions $x_1,x_2,x_3$ such
that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\\
\min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
\end{theorem}

The proof of the above theorem is similar to that of
 Theorem \ref{thm210}, so it is omitted.



 Assume the following three conditions:
\begin{itemize}
\item[(H11)] $f:[0,1]\times [h,+\infty)\times \mathbb{R}\to
[0,+\infty)$ is continuous with $f(t,c+h,0)\not\equiv 0$ on each
sub-interval of [0,1] for all $c\ge 0$, where $h=A$;

\item[H12)] $A\ge 0,B\ge0$;

\item[(H13)] $\alpha\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\beta_i<1$ and $B\ge
A\big(1-\sum_{i=1}^m\beta_i\big)$;
\end{itemize}

Consider the boundary-value problem
\begin{equation} \label{e30}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\quad t\in (0,1),\\
y(0)-\alpha y'(0)=0,\quad
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)=D,
\end{gathered}
\end{equation}


\begin{lemma} \label{lem219}
 Assume {\rm (H4), (H13)} and $D\ge 0$.
If $y$ is a solution of \eqref{e30}, then $y$ is positive on
$(0,1)$.
\end{lemma}

\begin{proof}  Suppose $y$ satisfies \eqref{e30}. It follows from the
assumptions that $y'$ is decreasing on $[0,1]$.

If $y'(0)<0$, the BCs in \eqref{e30} and (H4) imply that $y'(t)<0$ for
all $t\in [0,1]$ and $y(0)\le 0$. Then
$$
y(1)=\sum_{i=1}^m\beta_iy(\xi_i)+D\ge  \sum_{i=1}^m\beta_iy(1)+D.
$$
Then $y(1)\ge \frac{D}{1-\sum_{i=1}^m\beta_i}$. It follows from
$D\ge 0$ that $y(1)\ge y(0)$, a contradiction to $y'(t)<0$ for all
$t\in [0,1]$.

If $y'(0)\ge 0$ and $y'(1)>0$, we get that $y'(t)>0$ for all $t\in
[0,1]$. The BCs in \eqref{e30} and (H4) imply that $ y(0)=\alpha
y'(0)\ge 0$. Then $y(t)>y(0)\ge 0$ for all $t\in [0,1]$.

If $y'(0)\ge 0$ and $y'(1)\le0$, then $y(0)=\alpha y'(0)\ge 0$. It
follows from \eqref{e30} that
$$
y(1)=\sum_{i=1}^m\beta_iy(\xi_i)+D\ge \sum_{i=1}^m\beta_i\min\{y(0),y(1)\}.
$$
If $y(1)\le y(0)$, then $y(1)\ge \sum_{i=1}^m\beta_i y(1)$ implies
that $y(1)\ge 0$. If $y(1)>y(0)$, then $y(1)>0$ since $y(0)\ge 0$.
It follows that $y(t)\ge \min\{y(0),y(1)\}\ge 0$. Then $y$ is
positive on $(0,1)$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem220}
 Assume {\rm (H4), (H13)}, $D\ge 0$.
If $y$ is a solution of \eqref{e30}, then
$$
y(t)=B_h-\int_t^1\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big),\quad
t\in [0,1],
$$
where $A_h\in [a,b]$,
\begin{align*}
&\alpha\phi^{-1}\Big(A_h-\int_0^{\xi}h(u)du\Big)
\Big(1-\sum_{i=1}^m\beta_i\Big)\\
&\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\Big(A_h-\int_0^sh(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds-D=0,
\end{align*}
and
\begin{gather*}
B_h=\alpha\phi^{-1}\Big(\phi(y'(0))-\int_0^{\xi}h(u)du\Big),\\
a=\phi\big(\frac{D}{c}\big),\quad
b=\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du,\quad
c=\alpha\big(1-\sum_{i=1}^m\beta_i\big)
+1-\sum_{i=1}^m\beta_i\xi_i.
\end{gather*}
\end{lemma}

\begin{proof}
It follows from \eqref{e30} that
\begin{gather*}
y(t)=y(1)-\int_t^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds,\\
y(1)\Big(1-\sum_{i=1}^m\beta_i\Big)
= -\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+D,\\
y(1)-\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds=\alpha y'(0).
\end{gather*}
Thus
\begin{align*}
&\Big(1-\sum_{i=1}^m\beta_i\Big)
\Big(\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+\alpha
y'(0)\Big)\\
&= -\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+D.
\end{align*}
It follows from the proof of Lemma \ref{lem219} that $y'(0)\ge 0$. Let
\begin{align*}
G(c)&= \Big(1-\sum_{i=1}^m\beta_i\Big)
\int_0^1\phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds
+\alpha\Big(1-\sum_{i=1}^m\beta_i\Big) \phi^{-1}(c)\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds-D.
\end{align*}
It is easy to see that $G(c)$ is increasing on $(-\infty,+\infty)$
and $G(a)=G\left(\phi(\frac{D}{a})\right)\le0$. Since
\begin{align*}
G(b)&=G\Big(\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du\Big)\\
&\geq \Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\Big(\phi\big(\frac{D}{c}\big)+\int_s^1h(u)du\Big)ds\\
&\quad +\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)
\phi^{-1}\Big(\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du\Big)\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(\phi\big(\frac{D}{c}\big)+\int_s^1h(u)du\Big)ds-D\\
&\ge \Big(1-\sum_{i=1}^m\beta_i\Big)\frac{D}{c}
 +\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)
\frac{D}{c}+\sum_{i=1}^m\beta_i(1-\xi_i)\frac{D}{c}-D\\
&=  0,
\end{align*}
we get  $a\le \phi(y'(0))\le
\phi\big(\frac{D}{c}\big)+\int_0^1h(u)du$. The proof is complete.
\end{proof}

Suppose that (H13) holds. Let $x(t)-h=y(t)$. Then \eqref{e16} is
transformed into
\begin{gather*}{}
[\phi(y'(t))]'+f(t,y(t)+h,y'(t))=0,\quad t\in (0,1),\\
y(0)-\alpha y'(0)=0,\\
y(1)-\sum_{i=1}^m\beta_iy(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i}\ge
0.\end{gather*}
Let
\begin{align*}
P_3'=\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,(1-t)\}\max_{t\in
[0,1]}y(t)\text{ for all }t\in [0,1] \}.
\end{align*}
Then $P_3'$ is a cone in $X$.
Since, for $y\in P_3'$, we have
$$
|y(t)|\le|y(t)-y(0)|+|y(0)|\le \max_{t\in
[0,1]}|y'(t)|+\alpha\max_{t\in [0,1]}|y'(t)| =(1+\alpha)\gamma(y),
$$
we get $\max_{t\in [0,1]}|y(t)|\le(1+\alpha)\gamma(y)$. It is
easy to see that there exists a constant $M>0$ such that $\|y\|\le
M\gamma (y)$ for all $y\in P_3'$.

Define the nonlinear operator $T_3':P_3'\to X$ by
\[
(T_3'y)(t)= B_y-\int_t^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_3',
\]
where
\begin{align*}
&\Big(1-\sum_{i=1}^m\beta_i\Big)\Big(\int_0^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds+\alpha
\phi^{-1}(A_y)\Big)\\
&= -\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
+B-\frac{A}{1-\sum_{i=1}^m\beta_i}.
\end{align*}
 and
\[
B_y= \int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
+\alpha\phi^{-1}(A_y).
\]
Then
\begin{align*}
(T_3'y)(t)
&= \int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
+\alpha\phi^{-1}(A_y)\\
&\quad -\int_t^1\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&= \int_0^t\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds
+\alpha\phi^{-1}(A_y),\quad y\in P_3'.
\end{align*}

\begin{lemma} \label{lem221}
Assume {\rm (H11)--(H13)}.
Then the following holds:
\begin{itemize}
\item[(i)] $T_3'y\in P_3'$ for each $y\in P_3'$;

\item[(ii)] $x$ is a solution of \eqref{e16}' if and only if
$x=y+h$
 and $y$ is a solution of the operator equation $T_3'y=y$ in $P_3'$;

\item[(iii)] $T_3':P_3'\to P_3'$ is completely
continuous;

\item[(iv)] the following equalities hold:
\begin{gather*}{}
[\phi((T_3'y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_3'y)(0)-\alpha (T_3'y)(\xi)=0,\\
(T_3'y)(1)-\sum_{i=1}^m\beta_i
(T_3'y)(\xi_i)=B-\Big(1-\sum_{i=1}^m\beta_i\Big)A;
 \end{gather*}
\end{itemize}
\end{lemma}

The proof of the above lemma is similar to that  of Lemma \ref{lem29},
so it is omitted.

\begin{theorem} \label{thm222}
 Suppose that {\rm H11)--(H13)} hold and there exist positive constants
$e_1,e_2,c$,
\begin{gather*}
 Q= \min\big\{\phi\Big(\frac{c}{\int_0^1\phi^{-1}(2-s)ds+\alpha
\phi^{-1}(2)}\Big),\;\frac{\phi(c)}{2}
\big\};\\
W= \phi\Big(\frac{e_2}{\sigma_0\min\big\{\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds,\;\int_k^{1-k}\phi^{-1}(s-k)ds\big\}}\Big);\\
 E= \phi\Big(\frac{e_1}{\int_0^1\phi^{-1}(2-s)ds+\alpha\phi^{-1}(2)}\Big).
\end{gather*}
such that
$$
c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>\frac{\alpha}{
\alpha\Big(1-\sum_{i=1}^m\beta_i\Big)+1-\sum_{i=1}^m\beta_i\xi_i}
\Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big),
$$
and
$$
Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i \big)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.
$$
If {\rm (A1)--(A3)} defined in Theorem \ref{thm210} hold,
then \eqref{e16} has at least three positive solutions $x_1,x_2,x_3$
such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
\end{theorem}

The proof of the above theroem is similar to that  Theorem \ref{thm214} and
it is omitted.

\begin{remark} \label{rmk4}\rm
 Kwong and Wong \cite{kw}, Palamides \cite{p}  studied
the existence of positive solutions of \eqref{e6} and \eqref{e7}
 (the main results may be seen in Section 1). When applying
Theorem \ref{thm218} to
\eqref{e6}, we get three positive solutions  if $\theta\in
(0,\pi/2]$ and the other assumptions in Theorem \ref{thm218} hold.
\end{remark}

\subsection{Positive Solutions of \eqref{e17}}

We prove existence results for three positive solutions of
\eqref{e17}. The following assumptions are used in this sub-section.
\begin{itemize}
\item[(H14)] $f:[0,1]\times [0,+\infty)\times
[0,+\infty)\to [0,+\infty)$ is continuous with
$f(t,c+h,0)\not\equiv 0$ on each sub-interval of [0,1] for all $c\ge
0$, where $h=A$;

\item[(H15)] $\alpha_i\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\alpha_i<1$, $\sum_{i=1}^m\beta_i<1$;

\item[(H16)] $A\ge 0,B\ge 0$ with $B\ge \frac{A}{1-\sum_{i=1}^m\beta_i}$.
\end{itemize}

Suppose that {\rm (H4)--(H16)} hold and consider the
boundary-value problem
\begin{equation} \label{e31}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y'(\xi_i)=0,\quad
y'(1)- \sum_{i=1}^m\beta_iy'(\xi_i)=D,
\end{gathered}
\end{equation}

\begin{lemma} \label{lem223}
 Assume {\rm (H4), (H15), (H16)} and  $D\ge 0$. If $y$ is a
solution of \eqref{e31}, then $y$ is
positive on $(0,1)$.
\end{lemma}

\begin{proof}  Suppose $y$ satisfies \eqref{e31}. It follows from the
assumptions that $y'$ is decreasing on $[0,1]$. Then the BCs in \eqref{e31}
and (H4) imply
$$
y'(1)=\sum_{i=1}^m\beta_i y'(\xi_i)+D\ge \sum_{i=1}^m\beta_i y'(1).
$$
It follows that $y'(1)\ge 0$. We get  $y'(t)\ge 0$ for $t\in
[0,1]$. Then
$$
y(0)= \sum_{i=1}^m\alpha_iy(\xi_i)\ge \sum_{i=1}^m\alpha_i y(0).
$$
It follows that $y(0)\ge 0$. Then $y(t)>y(0)\ge 0$ for $t\in (0,1)$
since $y'(t)\ge 0$ for all $t\in  [0,1]$. The
 proof is complete.
\end{proof}

\begin{lemma} \label{lem224}
Assume {\rm (H4)--(H16)}.
 If $y$ is a solution of \eqref{e31}, then
$$
y(t)=B_h+\int_0^t\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds
$$
and  there exists unique $A_h\in [a,b]$ such that
\begin{equation} \label{e32}
\phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1}\left(A_h+\int_{\xi_i}^1h(s)ds\right)+D,
\end{equation}
and
\begin{gather*}
B_h=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(A_h+\int_{\xi_i}^1h(u)du\Big),\\
a=\phi\Big(\frac{D}{1-\sum_{i=1}^m\beta_i}\Big), \\
b=\phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)}{1-\sum_{i=1}^m\beta_i}\Big)+
  \frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi\big(
\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds.
\end{gather*}
\end{lemma}

\begin{proof}  It follows from \eqref{e31} that
$$
y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds
$$
The BCs in \eqref{e31} imply
\begin{gather*}
y'(1)=\sum_{i=1}^m\beta_i\phi^{-1}\Big(\phi(y'(1))+\int_{\xi_i}^1h(s)ds\Big)
+D,\\
y(0)=\sum_{i=1}^m\alpha_i\phi^{-1}\Big(\phi(y'(1))+\int_{\xi_i}^1h(u)du\Big).
\end{gather*}
Let
\[
G(c)= \phi^{-1}(c)-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(c+\int_{\xi_i}^1h(s)ds\Big)-D.
\]
It is easy to see that
\begin{align*}
G(a)&= \frac{D}{1-\sum_{i=1}^m\beta_i}-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(\phi\Big(\frac{D}{1-\sum_{i=1}^m\beta_i}\Big)
+\int_{\xi_i}^1h(s)ds\Big)-D\\
&\le \frac{D}{1-\sum_{i=1}^m\beta_i}-\sum_{i=1}^m\beta_i
\frac{D}{1-\sum_{i=1}^m\beta_i}-D=0.
\end{align*}
On the other hand,
\begin{align*}
\frac{G(b)}{\phi^{-1}(b)}
&=1-\sum_{i=1}^m\beta_i\phi^{-1}
\Big(1+\frac{\int_{\xi_i}^1h(s)ds}{b}\Big)-\frac{D}{\phi^{-1}(b)}\\
&= 1-\sum_{i=1}^m\beta_i\phi^{-1}
\bigg(1+\frac{\int_{\xi_i}^1h(s)ds}{
\phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1h(s)ds}\bigg)
\\
&\quad -\frac{D}{\phi^{-1}
\Big(\phi\Big(\frac{D\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big)
+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}{1-\phi
\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}\int_0^1h(s)ds\Big)}\\
&\le 1-\sum_{i=1}^m\beta_i\phi^{-1}
\Big( 1+\frac{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\Big)-\frac{1-\sum_{i=1}^m\beta_i}{1+\sum_{i=1}^m\beta_i}
= 0.
\end{align*}
Hence $G(b)\ge 0$. It is easy to know that
$\frac{G(c)}{\phi^{-1}(c)}$ is continuous and decreasing on
$(-\infty,0)$ and continuous and increasing on $(0,+\infty)$, Hence
$G(a)\le 0$ and $G(b)\ge 0$ and
\[
\lim_{c\to 0^-}\frac{G(c)}{\phi^{-1}(c)}= +\infty,\quad
\lim_{c\to -\infty}\frac{G(c)}{\phi^{-1}(c)}
= 1-\sum_{i=1}^m\beta_i>0.
\]
Then there exists unique constant $A_h=\phi(y'(1))\in [a,b]$
such that \eqref{e32} holds. The proof is complete.
\end{proof}

Suppose (H16) holds. Let $x(t)-A=y(t)$. Then \eqref{e17} is
transformed into
\begin{equation} \label{e33}
\begin{gathered}{}
[\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y'(\xi_i)=0,\\
y'(1)-\sum_{i=1}^m\beta_i
y'(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i}\ge 0,
 \end{gathered}
\end{equation}
Let
\begin{align*}
P_4=\{&y\in X:y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ y(t)\ge ty(1)\text{ for all }t\in [0,1]\}.
\end{align*}
Then $P_4$ is a cone in $X$.
For $y\in P_4$, since
$$
|y(t)|=|y(t)-y(0)|+|y(0)|\le \Big(1+\sum_{i=1}^m\alpha_i
\Big)\max_{t\in [0,1]}|y'(t)|=\Big(1+\sum_{i=1}^m\alpha_i
\Big)\gamma(y),
$$
we get
 $$
\max_{t\in [0,1]}|y(t)|\le\Big(1+\sum_{i=1}^m\alpha_i \Big)\gamma(y).
$$
It is easy to see that there exists a constant
$M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_1$.

 Define the operator $T_4:P_4\to X$ by
\[
(T_4y)(t)= B_y+\int_0^t\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_4,
\]
where
\begin{equation} \label{e34}
\phi^{-1}(A_h)=\sum_{i=1}^m\beta_i\phi^{-1}
\Big(A_h+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\Big)
+B-\frac{A}{1-\sum_{i=1}^m\beta_i},
\end{equation}
and
\[
B_y= \sum_{i=1}^m\alpha_i\phi^{-1}\left(A_y+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\right).
\]
Then
\begin{align*}
(T_4y)(t)&= \sum_{i=1}^m\alpha_i\phi^{-1}
\Big(A_y+\int_{\xi_i}^1f(u,y(u)+h,y'(u))du\Big)\\
&\quad +\int_0^t\phi^{-1} \Big(A_y+\int_s^1f(u,y(u)+h,y'(u))du\Big)ds.
\end{align*}
It follows from Lemma \ref{lem224} that
\begin{align*}
\phi\Big(\frac{\Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big)}
{1-\sum_{i=1}^m\beta_i}\Big)
& \le A_y\\
&\le \phi\Big(\frac{\Big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\Big)
\Big(1+\sum_{i=1}^m\beta_i\Big)}
{1-\sum_{i=1}^m\beta_i}\Big)\\
&\quad +  \frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
\int_0^1f(u,y(u)+h,y'(u))du.
\end{align*}


\begin{lemma} \label{lem225}
Suppose that {\rm (H14)--(H16)}. Then
\begin{itemize}
\item[(i)] the following equalities hold:
\begin{gather*}{}
[\phi((T_4y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_4y)(0)-\sum_{i=1}^m\alpha_i (T_4y)'(\xi_i)=0,\\
(T_4y)'(1)-\sum_{i=1}^m\beta_i
(T_4y)'(\xi_i)=B-\frac{A}{1-\sum_{i=1}^m\beta_i};
 \end{gather*}

\item[(ii)] $T_4y\in P_4$ for each $y\in P_4$;

\item[(iii)] $x$ is a solution of \eqref{e17} if and only if $x=y+h$
and $y$ is a solution of the operator equation $y=T_4y$ in $P_4$;

\item[(iv)] $T_4:P_4\to P_4$ is completely continuous.
\end{itemize}
\end{lemma}

The proof of the above lemma is similar to that of  Lemma \ref{lem29};
so we omit it.

\begin{theorem} \label{thm226}
Suppose that {\rm (H14)--(H16)} hold,
and there exist positive constants $e_1,e_2,c$, and
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+1-s\Big)ds\\
&\quad +\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}+1-\xi_i\Big),
\end{aligned}\\
Q= \min\big\{\phi(\frac{c}{L}),\;
\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)
-\phi\left(1+\sum_{i=1}^m\beta_i\right)}}
\big\};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds}\Big);\quad
 E= \phi\big(\frac{e_1}{L}\big).
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{\big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\big)
\left(1+\sum_{i=1}^m\beta_i \right)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.
$$
If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold,
then \eqref{e17} has at least three increasing positive solutions
$x_1,x_2,x_3$ such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
\end{theorem}

 The proof of the above theorem is similar to that of Theorem \ref{thm214};
Therefore, it is omitted.

\subsection{Positive Solutions of \eqref{e18}}

Finally, we prove an existence result for three positive solutions of
\eqref{e18}. The following assumptions will be used in the proofs
of all results in this subsection.
\begin{itemize}

\item[(H17)] $f:[0,1]\times [h,+\infty)\times \mathbb{R}\to
[0,+\infty)$ is continuous with $f(t,c+h,0)\not\equiv 0$ on each
sub-interval of [0,1] for all $c\ge0$, where
$h=\frac{A}{1-\sum_{i=1}^m\alpha_i}$;

\item[(H18)] $\alpha_i\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\alpha_i<1,\;\;\sum_{i=1}^m\beta_i<1$;

\item[(H19)] $A\ge 0,B\ge 0$ with $B\ge
\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}A$.

\end{itemize}
Consider the boundary-value problem
\begin{equation} \label{e35}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=0,\quad
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)=B,
\end{gathered}
\end{equation}

\begin{lemma} \label{lem227}
Assume {\rm (H4), (H18), (H19)}.
If $y$ is a solution of \eqref{e35}, then $y$ is positive on
$(0,1)$.
\end{lemma}

\begin{proof}
Suppose $y$ satisfies \eqref{e35}. It follows from the
assumptions that $y'$ is decreasing on $[0,1]$.
Then the BCs in \eqref{e35}
and (H4) imply that $y(t)\ge \min\{y(0),y(1)\}$ for all $t\in
[0,1]$. Then
\begin{equation} \label{e36}
y(0)\ge\sum_{i=1}^m\alpha_i \min\{y(0),y(1)\}.
\end{equation}
Similarly, we get
\begin{equation} \label{e37}
y(1)\ge\sum_{i=1}^m\beta_i \min\{y(0),y(1)\}.
\end{equation}
It follows from \eqref{e36} and \eqref{e37} that
$$
\min\{y(0),y(1)\}\ge
\min\big\{\sum_{i=1}^m\alpha_i ,\;\sum_{i=1}^m\beta_i
\big\}\min\{y(0),y(1)\}.
$$
Then (H18) implies that $\min\{y(0),y(1)\}\ge 0$. So $y(t)\ge
\min\{y(0),y(1)\}\ge 0$ for all $t\in [0,1]$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem228}
Assume {\rm (H4), (H18), (H19)}.
If $y$ is a solution of \eqref{e35}, then there exists unique
$A_h\in [0,b]$ such that
\begin{align*}
&\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\\
&\quad+\int_0^1\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\\
&\quad-\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1}
\Big(A_h-\int_0^sh(u)du\Big)ds-B=0.
\end{align*}
where
$$
b=\int_0^1h(u)du+\phi\big(\frac{B}{a}\big),\quad
a=\sum_{i=1}^m\alpha_i\xi_i +\Big(1-\sum_{i=1}^m\beta_i\Big)
+\sum_{i=1}^m\beta_i(1-\xi_i).
$$
\end{lemma}

\begin{proof}
 From \eqref{e35} it follows that
$$
y(t)=y(0)+\int_0^t\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds.
$$
The BCs in \eqref{e35} implies
$$
y(0)=y(0)\sum_{i=1}^m\alpha_i+\sum_{i=1}^m\alpha_i
\int_0^{\xi_i}\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds
$$
and
\begin{align*}
&y(0)+\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds\\
&= y(0)\sum_{i=1}^m\beta_i+\sum_{i=1}^m\beta_i\int_0^{\xi_i}
\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds+B.
\end{align*}
Then
\begin{align*}
&\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
 \sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}\Big(\phi(y'(0))
 -\int_0^sh(u)du\Big)ds\\
&\quad+\int_0^1\phi^{-1}\Big(\phi(y'(0))-\int_0^sh(u)du\Big)ds\\
&\quad -\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1}\Big(\phi(y'(0))
 -\int_0^sh(u)du\Big)ds-B=0.
\end{align*}
Let
\begin{align*}
G(c)&= \frac{1-\sum_{i=1}^m\beta_i}
{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
\Big(c-\int_0^sh(u)du\Big)ds\\
&\quad +\int_0^1\phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds\\
&\quad-\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1}
\Big(c-\int_0^sh(u)du\Big)ds-B\\
&= \frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}\Big(c-\int_0^sh(u)du\Big)ds\\
&\quad+\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
 \Big(c-\int_0^sh(u)du\Big)ds\\
&\quad+\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(c-\int_0^sh(u)du
\Big)ds-B.
\end{align*}
It is easy to see that $G(c)$ is increasing on $(-\infty,+\infty)$
and
\begin{align*}
G(0)&= -\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
 \sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}\Big(\int_0^sh(u)du\Big)ds\\
&\quad-\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
 \Big(\int_0^sh(u)du\Big)ds\\
&\quad-\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}
 \Big(\int_0^sh(u)du\Big)ds-B<0,
\end{align*}
and
\begin{align*}
G(b)&= G\Big(\int_0^1h(u)du+\phi\big(\frac{B}{a}\big)\Big)\\
 &= \frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
  \sum_{i=1}^m\alpha_i   \int_0^{\xi_i}\phi^{-1}
\Big(\phi\big(\frac{B}{a}\big)+\int_s^1h(u)du\Big)ds\\
&\quad +\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\Big(\phi\big(\frac{B}{a}\big)+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}\Big(\phi\big(\frac{B}{a}\big)+\int_s^1h(u)du\Big)ds-B\\
&\ge \frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i
  \int_0^{\xi_i}\phi^{-1}\left(\phi\big(\frac{B}{a}\big)\right)ds\\
&\quad+\Big(1-\sum_{i=1}^m\beta_i\Big)\int_0^1\phi^{-1}
\left(\phi\big(\frac{B}{a}\big)\right)ds\\
&\quad +\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}
\left(\phi\big(\frac{B}{a}\big)\right)ds-B
= 0.
\end{align*}
Then $G(0)<0$, $G(b)\ge 0$ and that $G(c)$ is increasing on
$(-\infty,+\infty)$ imply that $A_h=\phi(y'(0))\in [0,b]$ and $A_h$
satisfies
\begin{align*}
&\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\\
&+\int_0^1\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds\\
&-\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1}\Big(A_h-\int_0^sh(u)du\Big)ds-B=0.
\end{align*}
\end{proof}

Let $x(t)-h=y(t)$. Then \eqref{e18} is transformed into
\begin{equation} \label{e38}
\begin{gathered}{}
[\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\;\;t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=0,\\
y(1)-\sum_{i=1}^m\beta_i
y(\xi_i)=B-\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}A,
 \end{gathered}
\end{equation}
Let
\begin{align*}
P_5=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ y(t)\ge \min\{t,1-t\}\max_{t\in [0,1]}y(t)\text{
for all }t\in [0,1]\big\}
\end{align*}
Then $P_5$ is a cone in $X$.
For $y\in P_5$, since
\begin{align*}
|y(t)|&= |y(t)-y(0)+y(0)|\\
&\le |y'(\theta)|t+|y(0)|\\
&\le \max_{t\in [0,1]}|y'(t)|+\big|\frac{\sum_{i=1}^m\alpha_i
y(\xi_i)-\sum_{i=1}^m\alpha_i y(0)}{1-\sum_{i=1}^m\alpha_i
}\big|\\
&\le \Big(1+\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i
}\Big)\max_{t\in [0,1]}|y'(t)|\\
&= \Big(1+\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i
}\Big)\gamma(y),
\end{align*}
we get
 $$
\max_{t\in [0,1]}|y(t)|\le
\Big(1+\frac{\sum_{i=1}^m\alpha_i\xi_i}{1-\sum_{i=1}^m\alpha_i
}\Big)\gamma(y).
$$
It is easy to see that there exists a constant
$M>0$ such that $\|y\|\le M\gamma (y)$ for all $y\in P_5$.

Define the operator $T_5:P_5\to X$ by
\[
(T_5y)(t)= B_y+\int_0^t\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds,\quad y\in P_5,
\]
where
\begin{align*}
&\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}
\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&+\int_0^1\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds\\
&-\sum_{i=1}^m\beta_i\int_0^{\xi_i}\phi^{-1}
\left(A_y-\int_0^sf(u,y(u)+h,y'(u))du \right)ds-B=0.
\end{align*}
and
\[
B_y= \frac{1}
{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_0^{\xi_i}
\phi^{-1}\Big(A_y-\int_0^sf(u,y(u)+h,y'(u))du\Big)ds.
\]

\begin{lemma} \label{lem229}
Assume {\rm (H17), (H18), (H19)}. Then
\begin{itemize}
\item[(i)] the following equalities hold:
\begin{gather*}{}
[\phi((T_5y)'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
(T_5y)(0)-\sum_{i=1}^m\alpha_i (T_5y)(\xi_i)=0,\\
(T_5y)(1)-\sum_{i=1}^m\beta_i
(T_5y)(\xi_i)=B-\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}A\ge
0;
 \end{gather*}

\item[(ii)] $T_5y\in P_5$ for each $y\in P_5$;

\item[(iii)] $x$ is a solution of \eqref{e18} if and only if $x=y+h$
and $y$ is a solution of the operator equation $y=T_5y$;

\item[(iv)] $T_5:P_5\to P_5$ is completely continuous.
\end{itemize}
\end{lemma}

The proof is similar to that  of Lemma \ref{lem29}; therefore, it is  omitted.


\begin{theorem} \label{thm230}
 Suppose that {\rm (H17), (H18), (H19)} hold,
and that there exist positive constants $e_1,e_2,c$,
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big(
1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big(
1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds,
\end{aligned}
\\
 Q= \min\big\{\phi\big(\frac{c}{L}\big),\;
\frac{\phi(c)}{1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)
-\phi\big(1+\sum_{i=1}^m\beta_i\big)}}\big\};
\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds}\Big);\quad
 E= \phi\big(\frac{e_1}{L}\big).
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i\big)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad
Q>W.
$$
If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold,
then \eqref{e18} has at least three positive solutions $x_1,x_2,x_3$
such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]}x_3(t)<e_2+h.
\end{gather*}
\end{theorem}

The proof of the above theorem is similar to that of Theorem \ref{thm210};
it is omitted.


For \eqref{e18}, we have the following assumptions:
\begin{itemize}

\item[(H19)]  $f:[0,1]\times [h,+\infty)\times \mathbb{R}\to [0,+\infty)$
is continuous with $f(t,c+h,0)\not\equiv 0$ on each sub-interval of [0,1]
for all $c\ge0$, where $h=\frac{B}{1-\sum_{i=1}^m\beta_i}$;

\item[(H20)] $\alpha_i\ge 0$, $\beta_i\ge 0$ satisfy
$\sum_{i=1}^m\alpha_i<1$, $\sum_{i=1}^m\beta_i<1$;

\item[(H21)] $A\ge 0,B\ge 0$ with
$B\le \frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}A$.

\end{itemize}
Consider the boundary-value problem
\begin{equation} \label{e39}
\begin{gathered}{}
[\phi(y'(t))]'+h(t)=0,\;\;t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i y(\xi_i)=A,\quad
y(1)- \sum_{i=1}^m\beta_iy(\xi_i)=0,
\end{gathered}
\end{equation}

\begin{lemma} \label{lem231}
Assume {\rm (H4), (H20), (H21)}.
If $y$ is a solution of \eqref{e39}, then $y$ is positive on
$(0,1)$.
\end{lemma}

The proof of the above lemma is similar to that of Lemma \ref{lem221};
it is omitted.


\begin{lemma} \label{lem232}
 Assume {\rm (H4), (H20), (H21)}.
If $y$ is a solution of \eqref{e39}, then
$$
y(t)=B_h-\int_t^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds.
$$
where
\begin{gather*}
\begin{aligned}
&-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&-\int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&+\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1}
\Big(A_h+\int_s^1h(u)du\Big)ds-A=0,
\end{aligned}\\
b=-\int_0^1h(u)du-\phi\big(\frac{A}{a}\big),\\\
a=\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i(1-\xi_i)+\Big(1-\sum_{i=1}^m\alpha_i\Big)
+\sum_{i=1}^m\alpha_i\xi_i,
\\
B_h=-\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i
\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.
\end{gather*}
\end{lemma}


\begin{proof}  It follows from \eqref{e39} that
$$
y(t)=y(1)-\int_t^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.
$$
The BCs in \eqref{e39} implies
$$
y(1)=y(1)\sum_{i=1}^m\beta_i-\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds
$$
and
\begin{align*}
&y(1)-\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\
&= y(1)\sum_{i=1}^m\alpha_i-\sum_{i=1}^m\alpha_i
\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds+A.
\end{align*}
Then
\begin{align*}
&-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))
 +\int_s^1h(u)du\Big)ds\\
&-\int_0^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds\\
&+\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))
 +\int_s^1h(u)du\Big)ds-A=0.
\end{align*}
Let
\begin{align*}
G(c)&= -\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\
&\quad -\int_0^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1}
 \Big(c+\int_s^1h(u)du\Big)ds-A\\
&= -\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(c+\int_s^1h(u)du\Big)ds\\
&\quad-\Big(1-\sum_{i=1}^m\alpha_i\Big)\int_0^1\phi^{-1}
 \Big(c+\int_s^1h(u)du\Big)ds\\
&\quad-\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
 \Big(c+\int_s^1h(u)du\Big)ds-A.
\end{align*}
It is easy to see that $G(c)$ is decreasing on $(-\infty,+\infty)$
and
\begin{align*}
G(0)&= -\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(\int_s^1h(u)du\Big)ds\\
&\quad -\Big(1-\sum_{i=1}^m\alpha_i\Big)\int_0^1\phi^{-1}
 \Big(\int_s^1h(u)du\Big)ds\\
&\quad-\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
 \Big(\int_s^1h(u)du\Big)ds-A<0,
\end{align*}
and
\begin{align*}
G(b)&= G\Big(-\int_0^1h(u)du-\phi\big(\frac{A}{a}\big)\Big)\\
  &= -\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}
 \Big(-\int_0^sh(u)du-\phi\big(\frac{A}{a}\big)\Big)ds\\
&\quad-\Big(1-\sum_{i=1}^m\alpha_i\Big)\int_0^1\phi^{-1}
 \Big(-\int_0^sh(u)du-\phi\big(\frac{A}{a}\big)\Big)ds\\
&\quad -\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
 \Big(-\int_0^sh(u)du-\phi\big(\frac{A}{a}\big)\Big)ds-A\\
&\ge \frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}
 \Big(\phi\big(\frac{A}{a}\big)\Big)ds\\
&\quad+\Big(1-\sum_{i=1}^m\alpha_i\Big)\int_0^1\phi^{-1}
 \Big(\phi\big(\frac{A}{a}\big)\Big)ds\\
&\quad+\sum_{i=1}^m\alpha_i\int_0^{\xi_i}\phi^{-1}
 \Big(\phi\big(\frac{A}{a}\big)\Big)ds-A
= 0.
\end{align*}
Then $G(0)<0$, $G(b)\ge 0$ and that $G(c)$ is decreasing on
$(-\infty,+\infty)$ imply that $A_h=\phi(y'(1))\in [b,0]$ and $A_h$
satisfies
\begin{align*}
&-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}
 \Big(A_h+\int_s^1h(u)du\Big)ds\\
&-\int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&+\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1}
 \Big(A_h+\int_s^1h(u)du\Big)ds-A=0.
\end{align*}

Let $x(t)-h=y(t)$. Then \eqref{e18} is transformed into
\begin{equation} \label{e40}
\begin{gathered}{}
[\phi(y'(t))]'+f\big(t,y(t)+h,y'(t)\big)=0,\quad t\in (0,1),\\
y(0)-\sum_{i=1}^m\alpha_i
y(\xi_i)=A-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}B,\\
y(1)-\sum_{i=1}^m\beta_i y(\xi_i)=0,
\end{gathered}
\end{equation}
Let
\begin{align*}
P_6=\big\{&y\in X: y(t)\ge 0\text{ for all }t\in [0,1],
y'(t)\text{ is decreasing on }[0,1],\\
&\ \ \ \ \ \ \ \ \ \ y(t)\ge \min\{t,1-t\}\max_{t\in
[0,1]}y(t)\text{for all }t\in [0,1]\big\}
\end{align*}
Then $P_6$ is a cone in $X$.
For $y\in P_6$, since
\begin{align*}
|y(t)|&=\big|\frac{\sum_{i=1}^m\beta_i y(\xi_i)-\sum_{i=1}^m\beta_i
y(1)}{1-\sum_{i=1}^m\beta_i }\big|\\
&\le \frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i
}\max_{t\in [0,1]}|y'(t)|\\
&=\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i}\beta(y),
\end{align*}
we get
$$
\max_{t\in [0,1]}|y(t)|\le
\frac{\sum_{i=1}^m\beta_i(1-\xi_i)}{1-\sum_{i=1}^m\beta_i}\gamma(y).
$$
It is easy to see that there exists a constant $M>0$
such that $\|y\|\le M\gamma (y)$ for all $y\in P_6$.

Define the operator $T_6:P_6\to X$ by
\begin{align*}
(T_6y)(t)&= B_y-\int_t^1\phi^{-1}
\Big(A_y+\int_s^1f(u,y(u)+h',y'(u))du\Big)ds,\quad y\in P_6,
\end{align*}
where $A_h\in [b,0]$ satisfies
\begin{align*}
&-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}
\sum_{i=1}^m\beta_i\int_{\xi_i}^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&\quad -\int_0^1\phi^{-1}\Big(A_h+\int_s^1h(u)du\Big)ds\\
&\quad +\sum_{i=1}^m\alpha_i\int_{\xi_i}^1\phi^{-1}
 \Big(A_h+\int_s^1h(u)du\Big)ds-A=0,
\end{align*}
 and $B_h$ satisfies
$$
B_h=-\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i
\int_{\xi_i}^1\phi^{-1}\Big(\phi(y'(1))+\int_s^1h(u)du\Big)ds.
$$
\end{proof}

\begin{lemma} \label{lem233}
 Suppose that {\rm (H19)--(H21)} hold. Then
\begin{itemize}
\item[(i)]  the following equalities hold:
\begin{gather*}{}
[\phi((T_6y)'(t))]'+f\left(t,y(t)+h',y'(t)\right)=0,\quad t\in (0,1),\\
(T_6y)(0)-\sum_{i=1}^m\alpha_i
(T_6y)(\xi_i)=A-\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}B\ge 0,\\
(T_6y)(1)-\sum_{i=1}^m\beta_i (T_6y)(\xi_i)=0;
 \end{gather*}

\item[(ii)] $T_6y\in P_6$ for each $y\in P_6$;

\item[(iii)] $x$ is a solution of \eqref{e18} if and only if $x=y+h$
and $y$ is a solution of the operator equation $y=T_6y$;

\item[(iv)] $T_6:P_6\to P_6$ is completely continuous.
\end{itemize}
\end{lemma}

The proof of the above lemma is similar to that  of Lemma
\ref{lem29}; we omit it.


\begin{theorem} \label{thm234}
 Suppose that {\rm (H19)--(H21)} hold  and that there exist positive
constants $e_1,e_2,c$,
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)
-\phi\big(1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\
&\quad+\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\big(
1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds,
\end{aligned}\\
 Q= \min\big\{\phi\big(\frac{c}{L}\big),\;
\frac{\phi(c)}{1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\beta_i\right)}}\big\};
\\
W = \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds}\Big);\quad
 E = \phi\big(\frac{e_1}{L}\big).
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{B\big(1+\sum_{i=1}^m\beta_i
\big)} {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.
$$
If {\rm (A1)--(A3)} in Theorem \ref{thm210} hold,
then \eqref{e18} has at least three positive solutions $x_1,x_2,x_3$
such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<e_1+h,\quad
\min_{t\in [k,1-k]}x_2(t)>e_2+h,\\
\min_{t\in [k,1-k]}x_3(t)>e_1+h,\quad
\min_{t\in [k,1-k]} x_3(t)<e_2+h.
\end{gather*}
\end{theorem}

The proof of the above theorem is similar to that of Theorem \ref{thm210};
it is omitted.


\begin{remark} \label{rmk5}\rm
 In papers \cite{kk1,kk3}, sufficient conditions are
found for the existence of solutions of \eqref{e10}. It was proved that
the whole plane is divided by a ``continuous decreasing curve''
$\Gamma$ into two disjoint connected regions ĻĢ$\wedge E$ and
$\wedge N$ such that \eqref{e10} has at least one solution for
$(\lambda_1,\lambda_2)\in\Gamma$, has at least two solutions for
$(\lambda_1,\lambda_2)\in\wedge E\setminus \Gamma$, and has no
solution for $(\lambda_1,\lambda_2)\in\wedge N$. The explicit
subregions of $\wedge E$ where \eqref{e10} has at least two solutions
and two positive solutions, respectively. When applying Theorem \ref{thm226}
to \eqref{e10}, it shows us that \eqref{e10} has at least three positive
solutions under the assumptions $\lambda_2\ge
\frac{1-\sum_{i=1}^m\beta_i}{1-\sum_{i=1}^m\alpha_i}\lambda_1$ and
some other assumptions. When applying Theorem \ref{thm230} to \eqref{e10}, it
shows us that \eqref{e10} has at least three positive solutions under
the assumptions $\lambda_1\ge
\frac{1-\sum_{i=1}^m\alpha_i}{1-\sum_{i=1}^m\beta_i}\lambda_2$ and
some other assumptions.
\end{remark}

\begin{remark} \label{rmk6}\rm
 In paper \cite{wg1}, the authors studied the existence
of multiple positive solutions of \eqref{e11} under the assumption
$\alpha_i\le \beta_i$ for all $i=1,\dots,m$ and other assumptions,
when we apply Theorem \ref{thm230} and Theorem \ref{thm234}
to \eqref{e11}, the
assumptions $\alpha_i\le \beta_i$ for all $i=1,\dots,m$ are
deleted. So Theorem \ref{thm230} and Theorem \ref{thm234} generalize and improve the
theorems in \cite{wg1}.
\end{remark}

\begin{remark} \label{rmk7}\rm
 The existence problem on multiple positive
solutions of \eqref{e18} is solved in the case $A\ge 0,B\ge 0$, but such
problems remains unsolved in the cases $A\ge 0,B<0$, $A<0,B\ge 0$
and $A<0,B<0$.
\end{remark}

\section{Examples}

Now, we present three examples, whose three positive solutions can
not be obtained by theorems in known papers, to illustrate the main
results.

\subsection*{Example 3.1} Consider the boundary-value problem
\begin{equation} \label{e41}
\begin{gathered}{}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x'(0)=\frac{1}{4}x'(1/4)-2,\quad
x(1)=\frac{1}{4}x(1/2)+1.
\end{gathered}
\end{equation}
Corresponding to \eqref{e14}, one sees that $\phi(x)=x=\phi^{-1}(x)$,
$\alpha=2$, $\xi_1=1/4,\xi_2=1/2$,
$\alpha_1=1/4,\alpha_0=0,\beta_1=0,\beta_2=1/4$, $A=-2$, $B=1$.

Choose $k=1/4$, then $\sigma_0=4$, choose $e_1=10$, $e_2=50$,
$c=20000$. Then
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\beta_i}\sum_{i=1}^m\beta_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\alpha_i\big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\alpha_i\right)}+s\Big)ds=\frac{299}{120},
\end{aligned}\\
 Q= \min\big\{\phi\big(\frac{c}{L}\big),\;
\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)
-\phi\big(1+\sum_{i=1}^m\alpha_i\big)}}
\big\}=\frac{120\times 20000}{299};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(s-k\right)ds}\Big)=1600;\quad
 E= \phi\big(\frac{e_1}{L}\big)=\frac{1200}{299}.
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W.
$$
If
$$
f_0(u)=\begin{cases}
 \frac{150}{299}x,& x\in [0,4],\\
\frac{600}{299}, & x\in [4,44],\\
(x-44)\frac{4000-\frac{600}{299}}{54-44}+\frac{600}{299},&
x\in [44,54],\\
4000, & x\in [54,20004],\\
x-16004, &x\ge 20004,
\end{cases}
$$
and
$$
f(t,u,v)=f_0(u)+\frac{1+\sin t}{10000}+\frac{u^2+v^2}{2\times
10^12},
$$
it is easy to see that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{-A\big(1+\sum_{i=1}^m\alpha_i\big)}
 {1-\sum_{i=1}^m\alpha_i}\Big),\quad Q>W
$$
 and
\begin{itemize}
\item[(A1)] $f(t,u,v)<\frac{120\times 20000}{299}$ for all
$t\in [0,1],u\in [4,20004],v\in [-20000,20000]$;

\item[(A2)] $f(t,u,v)>1600$ for all $t\in [1/4,3/4],u\in
[54,804],v\in [-20000,20000]$;

\item[(A3)] $f(t,u,v)\le \frac{1200}{299}$ for all $t\in
[0,1],u\in [4,44],v\in [-20000,20000]$;
\end{itemize}
then Theorem \ref{thm210} implies that \eqref{e41} has at least three decreasing
and positive solutions $x_1,x_2,x_3$ such that
$x_1(0)<14$, $x_2(3/4)>54$, $x_3(0)>14$, $x_3(3/4)<54$.

\subsection*{Example 3.2} Consider the boundary-value problem
\begin{equation} \label{e42}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\frac{1}{2}x'(1/4)+2,\quad
x'(1)=\frac{1}{4}x'(1/4)+\frac{1}{4}x'(1/2)+5.
\end{gathered}
\end{equation}
Corresponding to \eqref{e17}, one sees that $\phi(x)=x=\phi^{-1}(x)$,
$\alpha=2$, $\xi_1=1/4$, $\xi_2=1/2$,
 $\alpha_1=1/2$, $\alpha_2=0$, $\beta_1=1/4=\beta_2$, $A=2$, $B=5$, $h=2$.

Choose $k=1/3$, then $\sigma_0=1/3$, $e_1=20$, $e_2=80,c=30000$ and
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}+1-s\Big)ds\\
&\quad +\sum_{i=1}^m\alpha_i\phi^{-1}
\Big(1+\frac{\phi\big(\frac{1+\sum_{i=1}^m\beta_i}{2}\big)}
{1-\phi\Big( \frac{1+\sum_{i=1}^m\beta_i}{2}\Big)}+1-\xi_i\Big)
 =\frac{55}{8},
\end{aligned}\\
 Q= \min\big\{\phi\left(\frac {c}{L}\right),\;
\frac{\phi(c)}{1+\frac{\phi(2)}{\phi(2)
-\phi\big( 1+\sum_{i=1}^m\beta_i\big)}}
\big\}=\frac{48000}{11};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds}\Big)=4320;\quad
 E= \phi\big(\frac{e_1}{L}\big)=\frac{32}{11}.
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{\big(B-\frac{A}{1-\sum_{i=1}^m\beta_i}\big)
\left(1+\sum_{i=1}^m\beta_i\right)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.
$$
If
\begin{itemize}
\item[(A1)] $f(t,u,v)<\frac{48000}{11}$ for all $t\in
[0,1],u\in [2,30002],v\in [-30000,30000]$;

\item[(A2)] $f(t,u,v)>4320$ for all $t\in [1/3,2/3],u\in
[82,722],v\in [-30000,30000]$;

\item[(A3)] $f(t,u,v)\le \frac{32}{11}$ for all $t\in
[0,1],u\in [2,62],v\in [-30000,30000]$;

\end{itemize}
then Theorem \ref{thm226} implies that \eqref{e42} has at least three positive
solutions $x_1,x_2,x_3$ such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<22,\quad
\min_{t\in [1/3,2/3]}x_2(t)>82,\\
\max_{t\in [0,1]}x_3(t)>22,\quad
\min_{t\in [k,1-k]}x_3(t)<82.
\end{gather*}


\subsection*{Example 3.3}
Consider the boundary-value problem
\begin{equation} \label{e43}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in (0,1),\\
x(0)=\frac{1}{2}x(1/4)+\frac{1}{3}x(1/2)+2,\\
x(1)=\frac{1}{4}x(1/4)+\frac{1}{4}x(1/2)+8.
\end{gathered}
\end{equation}
Corresponding to \eqref{e18}, one sees that $\phi(x)=x=\phi^{-1}(x)$,
$\xi_1=1/4$, $\xi_2=1/2$,
$\alpha_1=1/2$, $\alpha_2=1/3$, $\beta_1=1/4=\beta_2$, $A=2,B=8$,
$h\frac{2}{1-\sum_{I=1}^m\beta_i}=16$.

Choose $k=1/4$, then $\sigma_0=1/4$. Choose
$e_1=50$, $e_2=250$, $c=400000$ and
\begin{gather*}
\begin{aligned}
L&= \int_0^1\phi^{-1}
\Big(1+\frac{\phi\Big(1+\sum_{i=1}^m\beta_i\Big)}{\phi(2)-\phi\big(
1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^m\alpha_i}\sum_{i=1}^m\alpha_i\int_{\xi_i}^1
\phi^{-1}
\Big(1+\frac{\phi\big(1+\sum_{i=1}^m\beta_i\big)}{\phi(2)
-\phi\big(1+\sum_{i=1}^m\beta_i\big)}+1-s\Big)ds=\frac{481}{64},
\end{aligned}\\
Q= \min\big\{\phi\left(\frac
{c}{L}\right),\;\;\;\frac{\phi(c)}{1+\frac{\phi\Big(1+\sum_{i=1}^m\beta_i\Big)}{\phi(2)-\phi\left(
1+\sum_{i=1}^m\beta_i\right)}}
\big\}=\frac{25600000}{481};\\
W= \phi\Big(\frac{e_2}{\sigma_0\int_k^{1-k}\phi^{-1}
\left(1-k-s\right)ds}\Big)=8000;\quad
 E= \phi\big(\frac{e_1}{L}\big)=\frac{3200}{481}.
\end{gather*}
such that
$c\ge \frac{e_2}{\sigma_0^2}>e_2>\frac{e_1}{\sigma_0}>e_1>0$,
$$
Q\ge \phi\Big(\frac{B\left(1+\sum_{i=1}^m\beta_i\right)}
 {1-\sum_{i=1}^m\beta_i}\Big),\quad Q>W.
$$
If
\begin{itemize}
\item[(A1)] $f(t,u,v)<\frac{25600000}{481}$ for all $t\in
[0,1],u\in [4,400004],v\in [-400000,400000]$;

\item[(A2)] $f(t,u,v)>8000$ for all $t\in [1/4,3/4],u\in
[254,4004],v\in [-400000,400000]$;

\item[(A3)] $f(t,u,v)\le \frac{3200}{481}$ for all $t\in
[0,1],u\in [4,204],v\in [-400000,400000]$;

\end{itemize}
then Theorem \ref{thm230} implies that \eqref{e43} has at least three positive
solutions $x_1,x_2,x_3$ such that
\begin{gather*}
\max_{t\in [0,1]}x_1(t)<54,\quad
\min_{t\in [k,1-k]}x_2(t)>254, \\
\max_{t\in [0,1]}x_3(t)>54,\quad
\min_{t\in [k,1-k]}x_3(t)<254.
\end{gather*}


\subsection*{Acknowledgements}
The author is grateful to the anonymous referee for his/her detailed
reading and constructive comments which improve the presentation of
this article. The author also wants to thank the managing editor of
this journal for his suggestions.

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\end{document}