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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 147, pp. 1--29.\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{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/147\hfil Multiple positive solutions]
{Multiple positive solutions for singular $m$-point
boundary-value problems with nonlinearities depending on the
derivative}

\author[Y. Ma, B. Yan\hfil EJDE-2008/147\hfilneg]
{Ya Ma, Baoqiang Yan}  % in alphabetical order

\address{Department of Mathematics, 
Shandong Normal University, Jinan, 250014,  China} 
\email[Ya Ma]{maya-0907@163.com}
\email[Baoqiang Yan]{yanbqcn@yahoo.com.cn}

\thanks{Submitted August 13, 2008. Published October 24, 2008.}
\thanks{Supported by grants 10871120 from the National Natural
 Science, and Y2005A07 \hfill\break\indent
from the Natural Science of Shandong Province, China}
\subjclass[2000]{34B10, 34B15} 
\keywords{$m$-point boundary-value problem; singularity;
 positive solutions; \hfill\break\indent fixed point theorem}

\begin{abstract}
 Using the fixed point theorem in cones, this paper
 shows the existence of multiple positive solutions for the
 singular $m$-point boundary-value problem
 \begin{gather*}
 x''(t)+a(t)f(t,x(t),x'(t))=0,\quad 0<t<1,\\
 x'(0)=0,\quad x(1)= \sum_{i=1}^{m-2}a_ix(\xi_i),
 \end{gather*}
 where $0<\xi_{1}<\xi_{2}<\dots<\xi_{m-2}<1$, $a_i\in [0,1)$,
 $i = 1, 2,\dots, m-2$, with $0< \sum_{i=1}^{m-2}a_i <1 $ and
 $f$  maybe singular at $x=0$ and $x'=0$.
\end{abstract}

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

\section{Introduction}


The study of multi-point boundary-value problems (BVP) for linear
second-order ordinary differential equations was initiated by
Il'in and Moiseev \cite{h1,i1}. Since then, many authors have studied
general nonlinear multi-point BVP; see for examples \cite{g1,m3}, and
references therein.  Gupta, Ntouyas and Tsamatos \cite{g1} considered
the existence of a solution in $C^{1}[0, 1]$ for the $m$-point
boundary-value problem
\begin{equation} \label{e1.1}
\begin{gathered}
x''(t) = f(t,x(t),x'(t))+ e(t),\quad  0 < t <1, \\
 x'(0) = 0, \quad x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i),
\end{gathered}
\end{equation}
where $\xi_i\in (0, 1)$, $i = 1, 2, \dots, m-2$,
$0<\xi_{1} < \xi_{2} < \dots< \xi_{m-2} < 1$, $a_i\in \mathbb{R}$,
$i = 1, 2, \dots, m-2$, have the same sign,
$\sum_{i=1}^{m-2}a_i\neq 1$,
$e\in L^{1}[0, 1]$, $f: [0, 1] \times R^{2}\to \mathbb{R}$ is a
function satisfying Carath\`eodory conditions and a
growth condition of the form $|f(t,u, v)|\leq p_{1}(t)|u|
+ q_{1}(t)|v| + r_{1}(t)$, where $p_{1}$, $q_{1}$,
$r_{1}\in L^{1}[0, 1]$. Recently, using Leray-Schauder continuation theorem,
Ma and  O'Regan proved the existence of positive solutions of
$C^{1}[0, 1)$ solutions for the above BVP, where
$f : [0, 1] \times R^{2} \to \mathbb{R}$ satisfies the Carath\`eodory
conditions (see \cite{m3}). Khan and Webb \cite{k1}
obtained a interesting result which presents the multiplicity of
existence of at least three solutions of a second-order
three-point boundary-value problem. However, up to now, there are
a fewer results on the existence of multiple solutions to \eqref{e1.1}. In view of the importance of the research on the
multiplicity of positive solutions for differential equations
\cite{a1,a2,h1,j2,k1,l1,l2,l4,m1}, the goal of this paper is to
fill this gap in the literature.

There are main four sections in this paper.
In section 2, we give a special cone and its properties.
In section 3, using the theory of fixed point
index on a cone, we present the existence of  multiple positive
solutions to  \eqref{e1.1} with $f$ may be singular at $x'=0$ but
not at $x=0$.
In section 4, under the condition $f$ is singular at $x'=0$ and $x=0$,
we present the existence of  multiple positive solutions to \eqref{e1.1}.
In section 5, under the condition $f$ is singular at $x=0$ but not
at $x'=0$, we present the existence of  multiple positive solutions
to \eqref{e1.1}.

\section{Preliminaries}

Let $C^1[0,1]=\{x:[0,1]\to \mathbb{R}$ such that $x(t)$ be continuous on
$[0,1]$ and  $x'(t)$  continuous on  $[0,1]$ \} with norm
$\|x\| =\max\{\gamma\|x\|_1,\gamma\delta\|x\|_2 \}$,
where
\begin{gather*}
\|x\|_1=\max_{t\in[0,1]}|x(t)|,\quad \|x\|_2=\max_{t\in[0,1]}|x'(t)|, \\
\gamma=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i\xi_i},
\quad \delta=\sum_{i=1}^{m-2}a_i(1-\xi_i).
\end{gather*}
Let
$$
P=\{x\in C^1[0,1]:x(t)\geq \gamma\|x\|_1,\forall t\in[0,1],x(0)\geq
\delta\|x\|_2\}.
$$
Obviously,  $C^1[0,1]$ is a Banach space and  $P$ is a cone in  $C^1[0,1]$.

\begin{lemma} \label{lem2.1}
Let $\Omega$ be a bounded open set in real Banach space $E$,
$\theta\in\Omega $,  $P$ be a cone in  $E$  and
$A:\bar{\Omega}\cap P\to P $ be continuous and completely
continuous. Suppose
\begin{equation}
\lambda Ax\neq  x,\quad \forall  x\in \partial  \Omega \cap P,
\lambda \in(0,1] .\label{e2.1}
\end{equation}
Then $i(A,\Omega\cap P,P)=1$.
\end{lemma}

\begin{lemma} \label{lem2.2}
Let $\Omega$ be a bounded open set in real Banach space $E$,
$\theta\in\Omega $,  $P$ be a cone in  $E$  and
$A:\bar{\Omega}\cap P\to P $ be continuous and completely continuous. Suppose
\begin{equation}
 Ax \not\leq  x,\quad \forall  x\in \partial  \Omega \cap P.
 \label{e2.2}
\end{equation}
Then  $i(A,\Omega\cap P,P)=0$.
\end{lemma}

Let
$\mathbb{R}_+=(0,+\infty)$, $\mathbb{R}_-=(-\infty,0)$,
$\mathbb{R}=(-\infty,+\infty)$. The following conditions will
be used in this article.
\begin{gather}
a(t)\in C(0,1)\cap L^1[0,1],
\quad a(t)>0,\; t\in (0,1);\label{e2.3}\\
f\in C([0,1]\times  \mathbb{R}_+ \times \mathbb{R}_-,[0,+\infty));
\label{e2.4}
\end{gather}
 There exists $g\in C([0,+\infty)\times (-\infty,0],[0,+\infty))$ such that
\begin{equation}
f(t,x,y)\leq g(x,y),\forall (t,x,y)\in [0,1]\times
 \mathbb{R}_+ \times \mathbb{R}_-. \label{e2.5}
\end{equation}
 For  $x\in P$ and $t\in [0,1]$, define operator
\begin{equation}
 \begin{aligned}
(Ax)(t)&=-\int_{0}^{t}(t-s)a(s)f(s,x(s)x'(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)
  a(s)f(s,x(s),x'(s))\Big)ds\,.
\end{aligned} \label{e2.6}
\end{equation}

\begin{lemma}[\cite{m3}] \label{lem2.3}
Assume \eqref{e2.1}. Then for $y\in C[0,1]$  the problem
\begin{equation}
\begin{gathered}
 x''+y(t)=0,t\in(0,1)  \\
x'(0)=\sum_{i=1}^{m-2}b_ix'(\xi_i),\quad
x(1)=\sum_{i=1}^{m-2}a_ix(\xi_i)
\end{gathered}\label{e2.7}
\end{equation}
has a unique solution
\begin{equation}
x(t)=-\int_{0}^{t}(t-s)y(s)ds+Mt+N,\label{e2.8}
\end{equation}
where,
\begin{gather*}
M=\frac{\sum_{i=1}^{m-2}b_i\int_{0}^{\xi_i}a(s)y(s)ds}
 {\sum_{i=1}^{m-2}b_i-1},\\
\begin{aligned}
N&=\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)y(s)ds-\sum_{i=1}^{m-2}a_i
  \int_{0}^{\xi_i}(\xi_i-s)a(s)y(s)ds\\
&\quad -\frac{\sum_{i=1}^{m-2}b_i\int_{0}^{\xi_i}a(s)y(s)ds}{\sum_{i=1}^{m-2}b_i-1}
(1-\sum_{i=1}^{m-2}a_i\xi_i)\Big).
\end{aligned}
\end{gather*}
 Further, if $y\geq 0$, for all $t\in [0,1]$, $x$ satisfies
\begin{equation}
\inf_{t\in[0,1]}x(t)\geq \gamma\|x\|_1,\label{e2.9}
\end{equation}
where
$\gamma=\big(\sum_{i=1}^{m-2}a_i(1-\xi_i)\big)/
\big(1-\sum_{i=1}^{m-2}a_i\xi_i\big)$.
\end{lemma}

\begin{lemma} \label{lem2.4}
Suppose \eqref{e2.3}--\eqref{e2.5}hold. Then $A:P\to P $ is
a completely continuous operator.
\end{lemma}

\begin{proof}
For $x\in P$, from \eqref{e2.6}, one has
\begin{equation}\begin{aligned}
(Ax)(t)&\geq -\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s),x'(s))ds
\Big)\\
&\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i
 \int_{0}^{1}(\xi_i-s)a(s)f(s,x(s),x'(s))ds\\
&\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)f(s,x(s),x'(s))ds
\geq 0, \quad t\in[0,1],
\end{aligned}  \label{e2.10}
\end{equation}
\begin{equation}
\begin{aligned}
|(Ax)(t)|
&=-\int_{0}^{1}(t-s)a(s)f(s,x(s)x'(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)
 a(s)f(s,x(s),x'(s))ds \Big)\\
&\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds
\\
&\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x(s),x'(s))ds\\
&\leq \frac{1}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)ds
\max_{0\leq c\leq \|x\|,-\|x\|\leq c'\leq 0}g(c,c')
<+\infty, \quad t\in[0,1],
\end{aligned}\label{e2.11}
\end{equation}
\begin{equation}
\begin{aligned}
|(Ax)'(t)|
&=\big|-\int_{0}^{t}a(s)f(s,x(s)x'(s))ds\big|\\
&=\int_{0}^{t}a(s)f(s,x(s)x'(s))ds\\
&\leq \int_{0}^{1}a(s)f(s,x(s)x'(s))ds\\
&\leq \int_{0}^{1}a(s)ds\max_{0\leq c\leq \|x\|,
-\|x\|\leq c'\leq 0}g(c,c')
<+\infty,
\end{aligned}\label{e2.12}
\end{equation}
which implies that $A$ is well defined.
\begin{align*}
(Ax)(0)
&=\frac{1}{1-\sum_{i=1}^{m-2}a_i}
(\int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}
 (\xi_i-s)a(s)f(s,x(s),x'(s))ds)\\
&\geq \frac{1}{1-\sum_{i=1}^{m-2}a_i}(\sum_{i=1}^{m-2}a_i
  \int_{0}^{1}(1-s)a(s)f(s,x(s),x'(s))ds\\
&\quad - \sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x(s),x'(s))ds)\\
&= \sum_{i=1}^{m-2}a_i(1-\xi_i)\int_{0}^{1}a(s)f(s,x(s),x'(s))ds.
\end{align*}
On the other hand,
\begin{align*}
\|Ax\|_{2}
&=\max_{t\in[0,1]}|(Ax)'(t)|\\
&=\max_{t\in[0,1]}|-\int_{0}^{t}a(s)f(s,x(s)x'(s))ds|\\
&=\int_{0}^{1}a(s)f(s,x(s)x'(s))ds .
\end{align*}
Then
\begin{equation}
(Ax)(0)\geq \delta\|Ax\|_{2}.\label{e2.13}
\end{equation}
By Lemma \ref{lem2.3}, we have $(Ax)(t)\geq \gamma \|Ax\|_{1}.$
As a result $Ax\in P$, which implies $AP\subseteq P$.
By a standard argument, we know that $A:P\to P$ is continuous
and completely continuous.
\end{proof}

\section{Singularities at $x'=0$ but not at $x=0$}

In this section the nonlinearity $f$ may be singular at $x'=0$ but
not at $x=0$. We will assume that the following conditions hold.
\begin{itemize}
\item[(H1)] $a(t)\in C(0,1)\cap L^1[0,1]$,   $a(t)>0$, $t\in(0,1)$

\item[(H2)] $f(t,u,z)\leq h(u)[g(z)+r(z)]$, where
$f\in C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+)$,
$g(z)>0$ continuous and nondecreasing on $\mathbb{R}_-$,
$h(u)\geq 0$ continuous and nondecreasing on $\mathbb{R}_+$, $r(z)>0$
continuous and non-increasing on $(-\infty,0]$;

\item[(H3)]
$$
\sup_{c\in \mathbb{R}_+}{\frac{c
}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}I^{-1}(h(c)\int_0^1
a(s)ds)}}>1,
$$
where
$I(z)={\int_z^0{\frac{du}{g(u)+r(u)}}}$, $z\in \mathbb{R}_-$;

\item[(H4)] There exists a function
$g_1\in C([0,+\infty)\times(-\infty,0],[0,+\infty))$, such that
 $f(t,u,z)\geq g_1(u,z),\forall (t,u,z)\in [0,1]\times \mathbb{R}_+\times
\mathbb{R}_-$, and
  $\lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty$,
uniformly for $z\in \mathbb{R}_-$.

\item[(H5)] There exists a function $\Psi_H\in C([0,1],\mathbb{R}_+)$
and a constant $0\leq \delta<1$ such that
$f(t,u,z)\geq \Psi_H(t)u^\delta$,
 for all $(t,u,z)\in [0,1]\times[0,H]\times \mathbb{R}_-$.

\end{itemize}
For $n\in\{1,2,\dots\}$ and $x\in P$, define operator
\begin{equation}
\begin{aligned}
 (A_{n}x)(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s),-|x'(s)|-\frac{1}{n})ds
\Big),\quad t\in [0,1].
\end{aligned} \label{e3.1}
\end{equation}


\begin{theorem} \label{thm3.1}
Suppose {\rm (H1)--(H5)} hold. Then
\eqref{e1.1} has at least two positive solutions
 $ x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$ with
$x_{0,1}(t),x_{0,2}(t)>0$, $t \in [0,1]$.
\end{theorem}

\begin{proof}
 Choose $R_1>0$ such that
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}
I^{-1}(h(R_1)\int_0^1 a(s)ds)}}>1.\label{e3.2}
\end{equation}
 From the continuity of $I^{-1}$ and  $h$, we can choose
$\varepsilon>0$ and  $\varepsilon<R_1$ with
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}I^{-1}(h(R_1)\int_0^1
a(s)ds+I(-\varepsilon))}}>1, \label{e3.3}
\end{equation}
$n_0\in\{1,2,\dots\}$ with
${\frac{1}{n_0}} < \min\{\varepsilon,{\delta/2} \}$ and let
  $N_0 =\{n_0,n_0+1, \dots\}$.

Lemma \ref{lem2.4} guarantees that for $n\in N_0$,
$A_n:P\to P$ is a completely continuous operator.
Let
$$
\Omega_1=\{x\in C^1[0,1]:\| x \| <R_1\}.
$$
We show that
\begin{equation}
x\neq \mu A_n x,\quad \forall x\in P\cap \partial\Omega_1,\;
\mu\in(0,1],\;  n\in N_0.\label{e3.4}
\end{equation}
In fact, if there exists an $ x_0\in P\cap\partial \Omega_1$ and
$\mu_0\in(0,1]$ such that $x_0=\mu_0 A_n x_0$,
\begin{align*}
x_0(t)
&=-\mu_0\int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad +\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds
\Big),\quad t\in [0,1].
\end{align*}
Then
\begin{equation}
x'_0(t)=-\mu_0\int_0^t a(s)f(s,x_0(s),
-|x_0'(s)|-{\frac{1}{n}})ds,\quad \forall t\in[0,1].\label{e3.5}
\end{equation}
Obviously, $x'_0(t)\leq 0$,  $t\in(0,1)$, and
since $x_0(1)> 0$,  $x_0(t)>0,t\in[0,1]$.
Differentiating \eqref{e3.5},
we have
\begin{equation}
\begin{gathered}
x''_0(t)+\mu_0 a(t)f(t,x_0(t),x_0'(t)-{\frac{1}{n}})=0,\quad 0<t<1,\\
x_0'(0)=0,\quad x_0(1)=\sum_{i=1}^{m-2}a_i x_0(\xi_i).
\end{gathered}\label{e3.6}
\end{equation}
Then
\begin{gather*}
\begin{aligned}
- x''_0(t)&=\mu_0 a(t)f(t,x_0(t),x_0'(t)-{\frac{1}{n}})\\
&\leq a(t)h(x_0(t))[g(x_0'(t)-{\frac{1}{n}})+r(x_0'(t)-{\frac{1}{n}})],\quad
\forall t\in(0,1),
\end{aligned}\\
{\frac{ -x''_0(t)}{g(x_0'(t)-{\frac{1}{n}})+r(x_0'(t)-{\frac{1}{n}})}}
\leq a(t)h(x_0(t)),\quad \forall t\in(0,1).
\end{gather*}
Integrating from $0$ to $t$, we have
$$
I( x'_0(t)-{\frac{1}{n}})-I(-{\frac{1}{n}})
\leq\int_0^t a(s)h(x_0(s))ds \leq h(R_1)\int_0^t a(s)ds,
$$
and
$$
I( x'_0(t)-{\frac{1}{n}})\leq h(R_1)\int_0^t a(s)ds+I(-\varepsilon ).
$$
Then
$$
x'_0(t)\geq I^{-1}(h(R_1)\int_0^t a(s)ds+I(-\varepsilon ));
$$
that is,
\begin{equation}
- x'_0(t)\leq -I^{-1}( h(R_1)\int_0^t
a(s)ds+I(-\varepsilon )),\quad t\in(0,1).\label{e3.7}
\end{equation}
Then
\begin{align*}
x_0(0)
&={\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0-x'_0(s)ds
 -{\frac{\mu_0\sum_{i=1}^{m-2}a_i\int^{\xi_i}_{0}
 -x'_0(s)ds}{1-\sum_{i=1}^{m-2}a_i}}\\
&\leq {\frac{1}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0 -I^{-1}
\Big( h(R_1)\int_0^s a(\tau)d\tau+I(-\varepsilon )\Big)ds \\
&\quad +{\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}}
\int^{\xi_i}_{0}-I^{-1}( h(R_1)
\int_0^s a(\tau)d\tau+I(-\varepsilon ))ds\\
&\leq {\frac{1}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0 -I^{-1}
\Big( h(R_1)\int_0^1 a(\tau)d\tau+I(-\varepsilon )\Big)ds \\
&\quad +{\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}}
\int^{\xi_i}_{0} -I^{-1}( h(R_1)
\int_0^1 a(\tau)d\tau+I(-\varepsilon ))ds\\
&={\frac{1+\sum_{i=1}^{m-2}a_i\xi_i}{1-\sum_{i=1}^{m-2}a_i}}\Big(-I^{-1}
\Big( h(R_1)\int_0^1 a(s)ds+I(-\varepsilon )\Big)\Big).
\end{align*}
Since $x_0(0)\geq  x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x_0(0)
\geq \gamma\delta\|x_0\|_2,x_0(0)\geq \|x_0\|=R_1$,
\begin{equation}
{\frac{R_1}{-{\frac{1+\sum_{i=1}^{m-2}a_i\xi_i}{1-\sum_{i=1}^{m-2}a_i
} }I^{-1}(h(R_1)\int_0^1 a(s)ds+I(-\varepsilon))}}\leq
1,\label{e3.8}
\end{equation}
 which is a contradiction to \eqref{e3.3}. Then \eqref{e3.4} holds.

 From Lemma \ref{lem2.1}, for  $n\in N_0$,
\begin{equation}
i(A_n,\Omega_1\cap P,P)=1. \label{e3.9}
\end{equation}
Now we show that there exists a set $\Omega_2$ such that
\begin{equation}
A_{n}x \not\leq x,\quad \forall  x\in \partial \Omega_2 \cap P.
 \label{e3.10}
\end{equation}
Choose  $a^*$ with $0<a^*<1$. Let
$$
N^*=\Big(\frac{1}{\gamma a^{*}
\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
\int_0^1 a(s)ds}\Big)^{-1}+1 .
$$
From (H4), there exists $R_2>R_1$ such that
\begin{equation}
g_1(x,y)\geq N^{*}x ,\quad \forall x\geq R_2,\; y\in \mathbb{R}_-.
\label{e3.11}
\end{equation}
Let $\Omega_2=\{x\in C^1[0,1]:\|x\|<\frac{R_2}{a^*}\}$. Then
$$
Ax \not\leq  x,\quad \forall  x\in \partial  \Omega_2 \cap P.
$$
In fact, if  there exists $x_0\in  \partial  \Omega_2 \cap P$ with
$x_0\geq A_{n}x_0$. By the definition of the cone and Lemma  \ref{lem2.3},
one has
$$
x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x(0)\geq \gamma\delta\|x_0\|_2,\quad
x_0(t)\geq \frac{R_2}{a^*}>R_2,\quad \forall t\in[0,1],
$$
from \eqref{e3.11},
\begin{align*}
\gamma x_0(t)&\geq \gamma A_{n}x_0(t)\\
&=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
 &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds
\Big)\Big)\\
&\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds
\Big)\Big)\\
&\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{t}(t-s)a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\frac{\sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x_0(s),
  -|x_0'(s)|-\frac{1}{n})ds}{1-\sum_{i=1}^{m-2}a_i}\Big)\\
 &=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
 &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)g_1(x_0(s),-|x_0'(s)|-\frac{1}{n})ds\\
 &\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)dsN^{*} x_0(s)\\
 &\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*}
 >\frac{R_2}{a^*}.
\end{align*}
Then $\|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}$,  which is a
contradiction to $x_0\in \partial  \Omega_2 \cap P$. Then \eqref{e3.10}
holds. From Lemma \ref{lem2.2},
\begin{equation}
\begin{aligned}
i(A_n,\Omega_2\cap P,P)=0. \end{aligned} \label{e3.12}
\end{equation}

which with \eqref{e3.9} guarantee that
\begin{equation}
\begin{aligned}
i(A_n,(\Omega_2-\bar{\Omega}_1)\cap P,P)=-1. \end{aligned}
\label{e3.13}
\end{equation}
 From this equality and \eqref{e3.9}, $A_n$ has two fixed points with
$x_{n,1}\in \Omega_1\cap P,x_{n,2}\in (\Omega_2-\bar{\Omega}_1)\cap P$.

For each $n\in N_0$, there exists $x_{n,1}\in\Omega_1\cap P$ such
that $x_{n,1}=A_nx_{n,1}$; that is,
\begin{equation}
\begin{aligned}
 x_{n,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)
 f(s,x_{n,1}(s),-|x'_{n,1}(s)|-\frac{1}{n})ds
\Big).
\end{aligned} \label{e3.14}
\end{equation}
As in the proof of \eqref{e3.5}, we have
 $ x'_{n,1}(t)\leq 0$, $t \in (0,1)$ and
$$
x'_{n,1}(t)=-\int_0^t a(s)f(s,x_{n,1}(s),x_{n,1}'(s)
-{\frac{1}{n}})ds,\quad  n\in N_0,\; t\in(0,1).
$$
Now we consider $\{x_{n,1}(t)\}_{ n\in N_0}$ and
$\{x' _{n,1}(t)\}_{n\in N_0}$. Since
$\| x_{n,1}\|\leq R_1$, it follows that
\begin{gather}
\text{$\{x _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$,} \label{e3.15}\\
\text{$\{x' _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$.} \label{e3.16}
\end{gather}
Then
\begin{equation}
\text{$\{x _{n,1}(t)\}$ is equicontinuous on  $[0,1]$.} \label{e3.17}
\end{equation}
As in the  proof as \eqref{e3.6},
\begin{equation}
\begin{gathered}
x''_{n,1}(t)+a(t)f(t,x_{n,1}(t),x_{n,1}'(t)-{\frac{1}{n}})=0,\quad 0<t<1,\\
x_{n,1}'(0)=0,x_{n,1}(1)=\sum_{i=1}^{m-2} a_ix_{n,1}(\xi_i).
\end{gathered} \label{e3.18}
\end{equation}
Now we show that for all  $t_1,t_2\in[0,1]$,
\begin{equation}
|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I( x'_{n,1}(t_1)-{\frac{1}{n}})|
 \leq h(R_1)|\int _{t_1}^{t_2}a(t)dt|.\label{e3.19}
\end{equation}
 From \eqref{e3.18},
 \begin{align*}
-x''_{n,1}(t)
&=a(t)f(t,x_{n,1}(t),x'_{n,1}(t)-{\frac{1}{n}})\\
&\leq a(t)h(x_{n,1}(t))\big[g\big(x_{n,1}'(t)-{\frac{1}{n}}\big)
 +r(x_{n,1}'(t)-{\frac{1}{n}})\big],
\quad \forall t\in(0,1),
\end{align*}

 and
\begin{align*}
x''_{n,1}(t)
&=-a(t)f\big(t,x_{n,1}(t),x'_{n,1}(t)-{\frac{1}{n}}\big)\\
&\geq -a(t)h(x_{n,1}(t))
\big[g\big(x_{n,1}'(t)-{\frac{1}{n}}\big)+r(x_{n,1}'(t)-{\frac{1}{n}})\big],
\quad \forall t\in(0,1),
\end{align*}
so
\begin{equation}
 \frac{- x''_{n,1}(t)}{g(x_{n,1}'(t)-{\frac{1}{n}})+r(x_{n,1}'(t)
-{\frac{1}{n}})} \leq
a(t)h(x_{n,1}(t)),\ \forall t\in(0,1),\label{e3.20}
\end{equation}
and
\begin{equation}
 {\frac{
x''_{n,1}(t)}{g(x_{n,1}'(t)-{\frac{1}{n}})+r(x_{n,1}'(t)-{\frac{1}{n}})}}
\geq -a(t)h(x_{n,1}(t)),\quad
 \forall t\in(0,1).\label{e3.21}
\end{equation}
Then, for all  $t_1$, $t_2$ in $[0,1]$ and $t_1<t_2$,
\begin{align*}
\big|-\int_{t_1}^{t_2} \frac{1}{g(x_{n,1}'(s)-{\frac{1}{n}})
+r(x_{n,1}'(s)-{\frac{1}{n}})}d(x_{n,1}'(s)- {\frac{1}{n}})\big|
&\leq h(R_1)\int_{t_1}^{t_2}a(t)dt\\
&=h(R_1)|\int_{t_1}^{t_2}a(t)dt|,
\end{align*}
Inequality \eqref{e3.19} holds.

Since  $I^{-1}$ is uniformly continuous on $[0,I(-R_1-\epsilon)]$,
 for all  $\bar\epsilon>0$, there exists $\epsilon'>0$ such that
\begin{equation}
|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon,\quad
 \forall |s_1-s_2|<\epsilon',\;
  s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e3.22}
\end{equation}
And  \eqref{e3.19}  guarantees that for  $\epsilon'>0$, there exists
$\delta'>0$ such that
\begin{equation}
|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I(
x'_{n,1}(t_1)-{\frac{1}{n}})|<\epsilon',
 \quad \forall\ |t_1-t_2|<\delta',\; t_1,t_2\in[0,1].\label{e3.23}
\end{equation}
From this inequality and \eqref{e3.22},
\begin{equation}
\begin{aligned}
  | x'_{n,1}(t_2)- x'_{n,1}(t_1)|
&=| x'_{n,1}(t_2)-{\frac{1}{n}}-(  x'_{n,1}(t_1)-{\frac{1}{n}})|\\
&=|I^{-1}(I( x'_{n,1}(t_2)-{\frac{1}{n}}))-I^{-1}(I(  x'_{n,1}(t_1)
  -{\frac{1}{n}}))|\\
&< \bar\epsilon, \quad \forall |t_1-t_2|<\delta',\; t_1,t_2\in[0,1];
\end{aligned}  \label{e3.24}
\end{equation}
that is,
\begin{equation}
\text{$\{x' _{n,1}(t)\}$ is equicontinuous on  $[0,1]$.} \label{e3.25}
\end{equation}
 From \eqref{e3.15}--\eqref{e3.17}, \eqref{e3.25} and  the Arzela-Ascoli
Theorem, $\{x _{n,1}(t)\}$ and  $\{x'_{n,1}(t)\}$ are relatively
compact  on $C[0,1]$,   which implies there exists a subsequence
$\{x_{n_j,1}\}$ of $\{x_{n,1}\}$ and function $x_{0,1}(t)\in C[0,1]$
such that
 $$
\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad
\lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0.
$$
Since $x'_{n_j,1}(0)=0$, $x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i)$,
   $x'_{n_j,1}(t)<0$, $x_{n_j,1}(t)>0$, $t\in(0,1)$, $j\in\{1,2,\dots\}$,
\begin{equation}
x'_{0,1}(0)=0, \quad
x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i), \quad
x'_{0,1}(t)\leq 0,\quad  x_{0,1}(t)\geq 0, \quad t\in(0,1).\label{e3.26}
\end{equation}
 For $(t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n}})\in
[0,1]\times [0,R_1+\epsilon]\times (-\infty,0)$, from
(H5) there exists a function $\Psi_{R_1}\in C([0,1],\mathbb{R}_+)$
such that
$$
f(t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n_j}})ds
\geq \Psi_{R_1}(t)(x_{n_j,1}(t))^\delta,\quad  0\leq \delta<1.
$$
Then, for $n\in N_0$,
\begin{align*}
 x_{n_j,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
 &\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s),
 x'_{n_j,1}(s)-\frac{1}{n_j})ds
\Big)\\
&\geq  -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
 \Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)
 f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds \Big)\\
&\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}(\xi_i-s)a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)\Psi_{R_1}(s)(x_{n_j,1}(s))^\delta ds\\
&\geq
\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)\Psi_{R_1}(s)\gamma^{\delta}ds\|x_{n_j,1}\|_1^\delta,
\end{align*}
which implies
$$
\|x_{n_j,1}\|_1\geq  \Big(\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}
 {1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)\Psi_{R_1}(s)
 \gamma^{\delta}ds\Big)^{\frac{1}{1-\delta}},
$$
and
 $$
x_{n_j,1}(t)\geq \Big(\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}
{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)\Psi_{R_1}(s)\gamma^{\delta}ds\Big)
^{\frac{1}{1-\delta}} =a_0>0 .
$$
Thus
\begin{align*}
x_{n_j,1}'(t)
&=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s),x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\leq -\int_{0}^{t}a(s)\Psi_{R_1}(s)(x_{n_j,1}(s))^\delta ds\\
&\leq -\int_{0}^{t}a(s)\Psi_{R_1}(s) ds a_0^\delta ,\quad  t\in[0,1],
n\in N_0.
\end{align*}
Consequently,
\begin{gather*}
\inf_{j\geq 1}\min_{s\in[\frac{1}{2},t]}|x'_{n_j,1}(s)| >0,\quad
t\in[\frac{1}{2},1),\\
\inf_{j\geq 1}\min_{s\in[t,\frac{1}{2}]}|x'_{n_j,1}(s)|
 >0, \quad t\in(0,\frac{1}{2}].
\end{gather*}
 Since $$x_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}})
=- \int_{1/2}^{t} a(s)f(s,x_{n_j,1}(s),x_{n_j,1}'(s)-{\frac{1}{n_j}})ds,\quad
t\in(0,1),
$$
and
\begin{align*}
f(t,x_{n_j,1}(t),x_{n_j,1}'(t)-{\frac{1}{n_j}})
&\leq h(x_{n_j,1}(t))[g(x_{n_j,1}'(t)-{\frac{1}{n_j}})+r(x_{n_j,1}'(t)-{\frac{1}{n_j}})]\\
&\leq h(\frac{R_1}{\gamma})[g(-\int_{0}^{t}a(s)\Psi_{R_1}(s) ds
a_0^\delta)+r(-\frac{R_1}{\gamma\delta}-\epsilon)],
\end{align*}
letting $j\to +\infty$, the  Lebesgue Dominated Convergence
Theorem guarantees that
\begin{equation}
x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})= - \int_{1/2}^{t}
a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,\quad t\in(0,1). \label{e3.27}
\end{equation}
Differentiating, we have
$$
x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 0<t<1,
$$
and from \eqref{e3.26}
$x_{0,1}(t)$ is a positive solution of
  \eqref{e1.1} with $x_{0,1}\in C^1[0,1]\cap C^2(0,1)$.

For the set
$\{x_{n,2}\}_{n\in N_0}\subseteq (\Omega_2-\overline{\Omega}_1)\cap P$,
the proof is as that for the set $\{x_{n,1}\}_{n\in N_0}$.
We can obtain a convergent subsequence
$\{x_{n_i,2}\}_{n\in N_0} $ of $\{x_{n,2}\}_{n\in N_0}$ with
$\lim_{i\to +\infty}x_{n_i,2}=x_{0,2}  \in C^1[0,1]\cap C^2(0,1)$.
Moreover, $x_{0,2}$ is a positive solution to \eqref{e1.1}.
\end{proof}
\textbf{Example 3.1} \ In \eqref{e1.1}, let
$f(t,u,z)=\mu[1+(-z)^{-a}][1+u^b+u^d]$ and $a(t)\equiv 1$ with
$0<a<1,b>1,0<d<1$ and $\mu>0$. If
\begin{equation}
 \mu<\sup_{c\in \mathbb{R}_+}\frac{I(-\frac{c(1-\sum_{i=1}^{m-2}a_i)}
{1+\sum_{i=1}^{m-2}a_i\xi_i})}{1+c^b+c^d}. \label{e3.28}
\end{equation}
Then  \eqref{e1.1} has at least two positive solutions
$x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$.

We apply Theorem \ref{thm3.1} with
$g(z)=(-z)^{-a},r(z)=1,h(u)=\mu(1+u^b+u^d),\Psi(t)=\mu,g_1(u,z)=\mu u^b$.
(H1), (H2), (H4), (H5) hold. Also
\begin{align*}
&\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2}
a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1}(h(c)\int_0^1
a(s)ds)}}\\
&=\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}I^{-1}(\mu(1+c^b+c^d))}},
\end{align*}
and \eqref{e3.28} guarantees that (H3) holds.


\section{Singularities at $x'=0$ and  $x=0$}

In this section the nonlinearity $f$ may be singular at $x'=0$ and $x=0$.
We assume that the following conditions hold.
\begin{itemize}
\item[(P1)] $a(t)\in C[0,1]$,  $a(t)>0$, $t\in(0,1)$;

\item[(P2)] $f(t,u,z)\leq [h(u)+\omega(u)][g(z)+r(z)]$, where  $f\in
C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+)$, $g(z)>0$
continuous and non-increasing on $(-\infty,0]$, $\omega(u)>0$ continuous and
non-increasing on $[0,+\infty)$, $h(u)\geq 0$  continuous and
nondecreasing on $\mathbb{R}_+$, $r(z)>0$ continuous and nondecreasing on
$\mathbb{R}_-$;

\item[(P3)]
$$
\sup_{c\in \mathbb{R}_+}{\frac{c}{-{\frac{\sum_{i=1}^{m-2}
 a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in
[0,1]}a(t)(ch(c)+\int_0^c \omega(s)ds)])}}>1,
$$
where
$I(z)={\int_z^0{\frac{udu}{g(u)+r(u)}}}$, $z\in \mathbb{R}_-$,
$\int_{0}^{a}\omega(s)ds < +\infty$;

\item[(P4)] There exists a function
 $g_1\in C( [0,+\infty)\times (-\infty,0],[0,+\infty))$, such that
$f(t,u,z)\geq g_1(u,z), \forall (t,u,z)\in [0,1]\times \mathbb{R}_+\times
\mathbb{R}_-$, and
$\lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty$, uniformly for
$z\in \mathbb{R}_-$.

\item[(P5)] There exists a function $\Psi_H\in C([0,1],\mathbb{R}_+)$
with  $f(t,u,z)\geq \Psi_H(t)$, for all
$(t,u,z)\in [0,1]\times[0,H]\times[-H,0)$.

\end{itemize}
For $n\in\{1,2,\dots\}$, $x\in P$, $t\in [0,1]$, define operator
\begin{equation}
\begin{aligned}
 (A_{n}x)(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s)
+\frac{1}{n},-|x'(s)|-\frac{1}{n})ds\Big).
\end{aligned} \label{e4.1}
\end{equation}

\begin{theorem} \label{thm4.1}
Suppose {\rm (P1)--(P5)} hold. Then
\eqref{e1.1} has at least two positive solutions
 $ x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$ and
$x_{0,1}(t),x_{0,2}(t)>0$, $t \in [0,1]$.
\end{theorem}

\begin{proof}
Choose $R_1>0$ such that
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in[0,1]}a(t)(R_1
h(R_1)+\int_0^{R_1} \omega(s)ds)])}}>1.\label{e4.2}
\end{equation}
 From the continuity of $ I^{-1}$ and  $h$, we can choose
$ \epsilon>0$ and $\epsilon<R_1$ such that
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[I(-\epsilon)
-\max_{t\in[0,1]}a(t)((R_1+\epsilon)
h(R_1+\epsilon)+\int_0^{R_1+\epsilon} \omega(s)ds)])}}
\label{e4.3}
\end{equation}
is greater than $1$, $n_0\in\{1,2,\dots\}$  with ${\frac{1}{n_0}} < \min\{\epsilon,{\delta/2} \}$
and let   $N_0 =\{n_0,n_0+1, \dots\}$.
Then Lemma \ref{lem2.4} guarantees that for $n\in N_0$,
$A_n:P\to P$ is a completely continuous operator.
Let
$$
\Omega_1=\{x\in C^1[0,1]:\| x \| <R_1\}.
$$
We show that
\begin{equation}
x\neq \mu A_n x,\quad \forall x\in P\cap \partial\Omega_1,\;
\mu\in(0,1], \; n\in N_0.\label{e4.4}
\end{equation}
In fact, if there exists an $ x_0\in P\cap\partial \Omega_1$ and
$\mu_0\in(0,1]$  with $x_0=\mu_0 A_n x_0$,
\begin{align*}
x_0(t)
&=-\mu_0\int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\quad +\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}
(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)
+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds),\quad t\in [0,1].
\end{align*}
Then
\begin{equation}
x'_0(t)=-\mu_0\int_0^t a(s)f(s,x_0(s)+\frac{1}{n},
-|x_0'(s)|-{\frac{1}{n}})ds,\quad \forall t\in[0,1].\label{e4.5}
\end{equation}
Obviously, $x'_0(t)\leq 0$, $t\in(0,1)$, and
since $x_0(1)> 0$, $x_0(t)>0,t\in[0,1]$.
Differentiating \eqref{e4.5}, we have
\begin{equation}
\begin{gathered}
x''_0(t)+\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},x_0'(t)
-{\frac{1}{n}})=0,\quad  0<t<1,\\
x_0'(0)=0,\quad x_0(1)=\sum_{i=1}^{m-2}a_i x_0(\xi_i).
\end{gathered} \label{e4.6}
\end{equation}
Then, for $t\in(0,1)$,
\begin{align*}
- x''_0(t)
&=\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},x_0'(t)-{\frac{1}{n}})\\
&\leq a(t)\big[h\big(x_0(t)+\frac{1}{n}\big)+\omega(x_0(t)+\frac{1}{n})\big]
\big[g\big(x_0'(t)-{\frac{1}{n}}\big)+r(x_0'(t)-{\frac{1}{n}})\big],
\end{align*}
and
$$
{\frac{ -x''_0(t)}{g(x_0'(t)-{\frac{1}{n}})+r(x_0'(t)-{\frac{1}{n}})}}
\leq a(t)[h(x_0(t)+{\frac{1}{n}})+\omega(x_0(t)+{\frac{1}{n}})],\quad
 \forall t\in(0,1),
$$
and
\begin{equation}
\begin{aligned}
&{\frac{ -x''_0(t)(x_0'(t)-{\frac{1}{n}})}{g(x_0'(t)-{\frac{1}{n}})
 +r(x_0'(t)-{\frac{1}{n}})}}\\
&\geq a(t)[h(x_0(t)+{\frac{1}{n}})+\omega(x_0(t)+
 {\frac{1}{n}})](x_0'(t)-{\frac{1}{n}})\\
&\geq a(t)[h(R_1+\epsilon)+\omega(x_0(t)
 +\frac{1}{n}(1-t))](x_0'(t)-{\frac{1}{n}})\\
&=a(t)[h(R_1+\epsilon)(x_0'(t)-{\frac{1}{n}})+\omega(x_0(t)
 +\frac{1}{n}(1-t))(x_0'(t)-{\frac{1}{n}})]
\end{aligned} \label{e4.7}
\end{equation}
Integrating  from $0$ to $t$, we have
\begin{align*}
&I( x'_0(t)-{\frac{1}{n}})-I(-{\frac{1}{n}})\\
&\geq\int_0^t a(s)\big[h(R_1+\epsilon)(x_0'(s)-{\frac{1}{n}})
+\omega(x_0(s)+\frac{1}{n}(1-s))(x_0'(s)-{\frac{1}{n}})\big]ds\\
&\geq\max_{t\in [0,1]}a(t) h(R_1+\epsilon)
\Big(\int_0^t x_0'(s) ds -\int_0^t {\frac{1}{n}}ds\Big)
+ \int_0^t \omega(x_0(s)+\frac{1}{n}(1-s))d(x_0(s)\\
&\quad +\frac{1}{n}(1-s))\\
&\geq -\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)(R_1+\epsilon)
+\int_0^{R_1+\epsilon}\omega(s)ds),
\end{align*}
$$
I( x'_0(t)-{\frac{1}{n}})\geq
I(-\epsilon)-\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)(R_1+\epsilon)
+\int_0^{R_1+\epsilon}\omega(s)ds).
$$
Then
$$
x'_0(t)\geq I^{-1}\Big( I(-\epsilon)-\max_{t\in [0,1]}\{a(t)\}
\big(h(R_1+\epsilon)(R_1+\epsilon)+\int_0^{R_1+\epsilon}\omega(s)ds\big)\Big);
$$
that is,
\begin{equation}
- x'_0(t)\leq -I^{-1}\Big( I(-\epsilon)-\max_{t\in [0,1]}
a(t)(h(R_1+\epsilon)(R_1+\epsilon)+\int_0^{R_1+\epsilon}\omega(s)ds)\Big),
\quad t\in(0,1).\label{e4.8}
\end{equation}
Since
\begin{align*}
x_0(0)
&={\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0-x'_0(s)ds-{\frac{\mu_0\sum_{i=1}^{m-2}a_i\int^{\xi_i}_{0}
-x'_0(s)ds}{1-\sum_{i=1}^{m-2}a_i}}
\\
&\leq {\frac{1}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0 -I^{-1}
\Big( I(-\epsilon)-\max_{t\in [0,1]}
\{a(t)\}(h(R_1+\epsilon)(R_1+\epsilon)\\
&\quad +\int_0^{R_1+\epsilon}\omega(s)ds)\Big)ds
 +{\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}}
\int^{\xi_i}_{0}
-I^{-1}\Big( I(-\epsilon)\\
&\quad -\max_{t\in [0,1]}a(t)
(h(R_1+\epsilon)(R_1+\epsilon)+\int_0^{R_1+\epsilon}\omega(s)ds)\Big)ds\\
&={\frac{1+\sum_{i=1}^{m-2}a_i\xi_i}{1-\sum_{i=1}^{m-2}a_i}}
(-I^{-1}\Big( I(-\epsilon)-\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)
(R_1+\epsilon)\\
&\quad +\int_0^{R_1+\epsilon}\omega(s)ds)\Big).
\end{align*}
Since $x_0(0)\geq  x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x_0(0)\geq
\gamma\delta\|x_0\|_2$, $x_0(0)\geq \|x_0\|=R_1$.
 So
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[I(-\epsilon)-\max\{a(t)\}((R_1+\epsilon)
h(R_1+\epsilon)+\int_0^{R_1+\epsilon} \omega(s)ds)])}}\leq
1,\label{e4.9}
\end{equation}
which is a contradiction to \eqref{e4.3}. Then \eqref{e4.4} holds.

From Lemma \ref{lem2.1}, for  $n\in N_0$,
\begin{equation}
 i(A_n,\Omega_1\cap P,P)=1. \label{e4.10}
\end{equation}
Now we show that there exists a set $\Omega_2$ such that
\begin{equation}
A_{n}x \not\leq x,\quad \forall x\in \partial \Omega_2 \cap P.
 \label{e4.11}
\end{equation}
Choose  $a^*, N^{*}$ as in section 3.
Let
$$
\Omega_2=\{x\in C^1[0,1]:\|x\|<\frac{R_2}{a^*}\}.
$$
Then
$$
A_{n}x \not\leq  x,\quad \forall x\in \partial  \Omega_2 \cap P.
$$
In fact, if  there exists $x_0\in  \partial  \Omega_2 \cap P$ with
$x_0\geq A_{n}x_0 .$ By the definition of the cone and Lemma  \ref{lem2.3},
one has
$$
x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x(0)\geq \gamma\delta\|x_0\|_2,
$$
and so  $x_0(t)\geq \frac{R_2}{a^*}>R_2$, for all
$t\in [0,1]$, $x_0(t)+\frac{1}{n}>R_2$, from \eqref{e3.11},
\begin{align*}
\gamma x_0(t)
&\geq \gamma A_{n}x_0(t)\\
&=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|
  -\frac{1}{n})ds\\
&\quad+\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)
 +\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds \Big)\Big)\\
&\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|
 -\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)
 +\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds \Big)\Big)
\\
&\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{t}(t-s) a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\quad -\frac{\sum_{i=1}^{m-2} a_i\int_{0}^{1}
(\xi_i-s)a(s)f\big(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n}\big)ds}
{1-\sum_{i=1}^{m-2}a_i}\Big)\\
&=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)g_1(x_0(s)+\frac{1}{n},-|x_0'(s)|-\frac{1}{n})ds\\
&\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)dsN^{*} x_0(s)\\
&\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
  \int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*}
>\frac{R_2}{a^*}.
\end{align*}
Then $\|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}$,  which is a
contradiction to $x_0\in \partial  \Omega_2 \cap P$. Then  \eqref{e4.11}
holds.

From Lemma \ref{lem2.2},
\begin{equation}
 i(A_n,\Omega_2\cap P,P)=0. \label{e4.12}
\end{equation}
This equality and \eqref{e4.10} guarantee,
\begin{equation}
i(A_n,(\Omega_2-\bar{\Omega}_1)\cap P,P)=-1. \label{e4.13}
\end{equation}
From this equality and \eqref{e4.10}, $A_n$ has two fixed points
with $x_{n,1}\in \Omega_1\cap P,x_{n,2}\in
(\Omega_2-\overline{\Omega}_1)\cap P$. For each $n\in N_0$, there
exists $x_{n,1}\in\Omega_1\cap P$ with $x_{n,1}=A_nx_{n,1}$; that
is,
\begin{equation}
\begin{aligned}
 x_{n,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|-\frac{1}{n})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|
 -\frac{1}{n})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n,1}(s)
 +\frac{1}{n},-|x'_{n,1}(s)|-\frac{1}{n})ds
\Big).
\end{aligned}\label{e4.14}
\end{equation}
As in the same proof of \eqref{e3.5},
 $ x'_n(t)\leq 0$, $t \in (0,1)$ and
$$
x'_{n,1}(t)=-\int_0^t a(s)f(s,x_{n,1}(s)+\frac{1}{n},x_{n,1}'(s)
-{\frac{1}{n}})ds,\quad n\in N_0,\; t\in(0,1).
$$
Now we consider  $\{x_{n,1}(t)\}_{ n\in N_0}$ and
$\{x' _{n,1}(t)\}_{n\in N_0}$, since $\| x_{n,1}\|\leq R_1$, it follows that
\begin{gather}
\text{$\{x _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$,} \label{e4.15}\\
\text{$\{x' _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$.} \label{e4.16}
\end{gather}
Then
\begin{equation}
\text{$\{x _{n,1}(t)\}$ is equicontinuous on  $[0,1]$.} \label{e4.17}
\end{equation}
As in the same proof of \eqref{e3.6},
\begin{equation}
\begin{gathered}
x''_{n,1}(t)+a(t)f(t,x_{n,1}(t)+\frac{1}{n},x_{n,1}'(t)-{\frac{1}{n}})=0,
\quad 0<t<1,\\
x_{n,1}'(0)=0,\quad x_{n,1}(1)=\sum_{i=1}^{m-2}a_i x_{n,1}(\xi_i).
\end{gathered}\label{e4.18}
\end{equation}
Now we show for all  $t_1$, $t_2$ in $[0,1]$,
\begin{equation}
\begin{aligned}
&|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I( x'_{n,1}(t_1)-{\frac{1}{n}})|\\
&\leq \max_{t\in [0,1]}a(t)\big[h(R_1+\epsilon)(|x_{n,1}(t_2)-x_{n,1}(t_1)|
 +|t_2-t_1|)\\
&\quad +|\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt|\big].
\end{aligned}  \label{e4.19}
\end{equation}
 From \eqref{e4.18}, it follows that for $t\in(0,1)$,
\begin{align*}
-x''_{n,1}(t)
&=a(t)f(t,x_{n,1}(t)+\frac{1}{n},x'_{n,1}(t)-{\frac{1}{n}})\\
&\leq a(t)[h(x_{n,1}(t)+\frac{1}{n})+\omega(x_{n,1}(t)
 +\frac{1}{n})][g(x_{n,1}'(t)-{\frac{1}{n}})+r(x_{n,1}'(t)-{\frac{1}{n}})],
\end{align*}
\begin{align*}
x''_{n,1}(t)
&=-a(t)f(t,x_{n,1}(t)+\frac{1}{n},x'_{n,1}(t)-{\frac{1}{n}})\\
&\geq -a(t)[h(x_{n,1}(t)+\frac{1}{n})+\omega(x_{n,1}(t)
  +\frac{1}{n})][g(x_{n,1}'(t)-{\frac{1}{n}})+r(x_{n,1}'(t)-{\frac{1}{n}})],
\end{align*}
and so for $t\in(0,1)$,
\begin{equation}
\begin{aligned}
&\frac{-x''_{n,1}(t)(x_{n,1}'(t)-{\frac{1}{n}})}{g(x_{n,1}'(t)
-{\frac{1}{n}})+r(x_{n,1}'(t)-{\frac{1}{n}})}\\
&\geq a(t)[h(x_{n,1}(t)+{\frac{1}{n}})+\omega(x_{n,1}(t)
 +{\frac{1}{n}})](x_{n,1}'(t)-{\frac{1}{n}})\\
&\geq a(t)[h(R_1+\epsilon)(x_{n,1}'(t)-{\frac{1}{n}})
 +\omega(x_{n,1}(t)+{\frac{1}{n}}(1-t))(x_{n,1}(t)+{\frac{1}{n}}(1-t))'],
\end{aligned}\label{e4.20}
\end{equation}
\begin{equation}
\begin{aligned}
&{\frac{x''_{n,1}(t)(x_{n,1}'(t)-{\frac{1}{n}})}{g(x_{n,1}'(t)
 -{\frac{1}{n}})+r(x_{n,1}'(t)-{\frac{1}{n}})}}\\
&\leq -a(t)[h(x_{n,1}(t)+{\frac{1}{n}})+\omega(x_{n,1}(t)
 +{\frac{1}{n}})](x_{n,1}'(t)-{\frac{1}{n}})\\
&\leq -a(t)[h(R_1+\epsilon)(x_{n,1}'(t)-{\frac{1}{n}})+\omega(x_{n,1}(t)
+{\frac{1}{n}}(1-t))(x_{n,1}(t)+{\frac{1}{n}}(1-t))'].
\end{aligned}\label{e4.21}
\end{equation}
Then, for all  $t_1$, $t_2\in[0,1]$ and $t_1<t_2$,
\begin{align*}
&I( x'_{n,1}(t_2)-{\frac{1}{n}})-I( x'_{n,1}(t_1)-{\frac{1}{n}})\\
&\geq \max_{t\in
[0,1]}a(t)[h(R_1+\epsilon)(\int_{t_1}^{t_2}x'_{n,1}(s)dt
 -{\frac{1}{n}}(t_2-t_1)
+\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt]\\
&\geq -\max_{t\in [0,1]}a(t)\big[h(R_1+\epsilon)(|x_{n,1}(t_2)
 -x_{n,1}(t_1)|+|t_2-t_1|) \\
&\quad +|\int _{x_{n,1}(t_1)
 +\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)+\frac{1}{n}(1-t_2)}\omega(s)dt|\big],
\end{align*}
\begin{align*}
&I( x'_{n,1}(t_1)-{\frac{1}{n}})-I( x'_{n,1}(t_2)-{\frac{1}{n}})\\
&\leq \max_{t\in [0,1]}a(t)\big[h(R_1+\epsilon)(|x_{n,1}(t_2)-x_{n,1}(t_1)|
+|t_2-t_1|)\\
&\quad +|\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt|\big].
\end{align*}
Therefore, \eqref{e4.19} holds.
Since  $I^{-1}$ is uniformly continuous on $[0,I(-R_1-\epsilon)]$,
 for all  $\bar\epsilon>0$, there exists  $\epsilon'>0$ such that
\begin{equation}
|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon, \forall\ |s_1-s_2|<\epsilon',
  s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e4.22}
\end{equation}
Then  \eqref{e4.19} guarantees that for $\epsilon'>0$, there exists
$\delta'>0$ such that
\begin{equation}
|I( x'_{n,1}(t_2)-{\frac{1}{n}})-I(
x'_{n,1}(t_1)-{\frac{1}{n}})|<\epsilon', \quad  \forall
|t_1-t_2|<\delta',t_1,t_2\in[0,1]. \label{e4.23}
\end{equation}
 From this inequality and \eqref{e4.22},
\begin{equation}
\begin{aligned}
  | x'_{n,1}(t_2)- x'_{n,1}(t_1)|
&=| x'_{n,1}(t_2)-{\frac{1}{n}}-(
  x'_{n,1}(t_1)-{\frac{1}{n}})|\\
&=|I^{-1}(I( x'_{n,1}(t_2)-{\frac{1}{n}}))-I^{-1}(I(
  x'_{n,1}(t_1)-{\frac{1}{n}}))|\\
&< \bar\epsilon, \quad  \forall |t_1-t_2|<\delta',t_1,t_2\in[0,1];
\end{aligned}\label{e4.24}
\end{equation}
that is,
\begin{equation}
\text{$\{x' _{n,1}(t)\}$  is equicontinuous on  $[0,1]$.}\label{e4.25}
\end{equation}
 From \eqref{e4.15}--\eqref{e4.17}, \eqref{e4.25} and the
 Arzela-Ascoli Theorem, $\{x_{n,1}(t)\}$ and  $\{x'_{n,1}(t)\}$
are relatively compact on $C[0,1]$,
which implies, there exists a subsequence
$\{x_{n_j,1}\}$ of $\{x_{n,1}\}$ and
 function $x_{0,1}(t)\in C[0,1]$ such that
 $$
\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad
\lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0.
$$
Since $x'_{n_j,1}(0)=0$, $x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i)$,
$x'_{n_j,1}(t)<0$, $x_{n_j,1}(t)>0$, $t\in(0,1)$, $j\in\{1,2,\dots\}$,
\begin{equation}
x'_{0,1}(0)=0, x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i),
   x'_{0,1}(t)\leq 0, x_{0,1}(t)\geq 0,\quad
 t\in(0,1).\label{e4.26}
\end{equation}
For $(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}})\in
[0,1]\times [0,R_1+\epsilon]\times (-\infty,0)$, from
(P5) there exists a function $\Psi_{R_1}\in C([0,1],\mathbb{R}_+)$ such that
$$
f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}})ds
\geq \Psi_{R_1}(t).
$$
Then, for $n\in N_0$,
\begin{align*}
 x_{n_j,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)
 -\frac{1}{n_j})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds
\Big)\\
&\geq  -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)
 -\frac{1}{n_j})ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds
)\\
&\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)-\frac{1}{n_j})ds\\
&\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)\Psi_{R_1}(s) ds
=a_0,
\end{align*}
and
\begin{align*}
x_{n_j,1}'(t)
&=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s)
 -\frac{1}{n_j})ds\\
&\leq -\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s) ds,\quad t\in[0,1],\;n\in N_0.
\end{align*}
Thus,
\begin{gather*}
\inf_{j\geq 1}\min\{\min_{s\in[\frac{1}{2},t]}x_{n_j,1}(s),
\min_{s\in[\frac{1}{2},t]}|x'_{n_j,1}(s)|\}>0,\quad t\in[\frac{1}{2},1),\\
\inf_{j\geq 1}\min\{\min_{s\in[t,\frac{1}{2}]}x_{n_j,1}(s),
 \min_{s\in[t,\frac{1}{2}]}|x'_{n_j,1}(s)|\} >0, \quad t\in(0,\frac{1}{2}].
\end{gather*}
 From (P2), we have
\begin{align*}
&f\big(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t)-{\frac{1}{n_j}}\big)\\
&\leq [h(x_{n_j,1}(t)+{\frac{1}{n_j}})+\omega(x_{n_j,1}(t)
  +{\frac{1}{n_j}})][g(x_{n_j,1}'(t)-{\frac{1}{n_j}})+r(x_{n_j,1}'(t)
  -{\frac{1}{n_j}})]\\
&\leq [h(\frac{R_1}{\gamma}+\epsilon)+\omega(a_0)]
  [r(-\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)ds )
  +g(-\frac{R_1}{\gamma\delta}-\epsilon)].
\end{align*}
and since
$$
x_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}})
=- \int_{1/2}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x_{n_j,1}'(s)
 -{\frac{1}{n_j}})ds,\quad t\in(0,1),
$$
letting $j\to +\infty$, the  Lebesgue Dominated Convergence Theorem
guarantees that
\begin{equation}
 x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})=
- \int_{1/2}^{t}
a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,t\in(0,1).\label{e4.27}
\end{equation}
Differentiating, we have
$$
x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 0<t<1,
$$
and from \eqref{e4.26},
  $x_{0,1}(t)$ is a positive solution of
  \eqref{e1.1} with $x_{0,1}\in C^1[0,1]\cap C^2(0,1)$.

For the set
$\{x_{n,2}\}_{n\in N_0}\subseteq (\Omega_2-\overline{\Omega}_1)\cap P$,
as in proof for the set $\{x_{n,1}\}_{n\in N_0}$,
we  obtain a convergent subsequence $\{x_{n_i,2}\}_{n\in N_0} $ of
$\{x_{n,2}\}_{n\in N_0}$ with
$\lim_{i\to +\infty}x_{n_i,2}=x_{0,2} \in C^1[0,1]\cap C^2(0,1)$.
Moreover, $x_{0,2}$ is a positive solution to \eqref{e1.1}.
\end{proof}
\textbf{Example 4.1}
 \ In \eqref{e1.1}, let $f(t,u,z)=\mu[1+(-z)^{-a}][1+u^b+u^{-d}]$ and
$a(t)\equiv 1$ with $0<a<1$, $b>1$, $0<d<1$ and $\mu>0$. If
\begin{equation}
\mu<\sup_{c\in \mathbb{R}_+}\frac{I(-\frac{c(1-\sum_{i=1}^{m-2}a_i)}
{\sum_{i=1}^{m-2}a_i\xi_i+1})}
{\max_{t\in [0,1]}a(t)(c+c^{1-d}+  \frac{c^{1+b}}{1+b})}. \label{e4.28}
\end{equation}
Then equation \eqref{e1.1} has at least two positive solutions
$x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$.
We apply Theorem \ref{thm4.1} with $g(z)=1$, $r(z)=(-z)^{-a}$,
$h(u)=\mu u^{-d}$, $\omega(u)=\mu(1+u^b)$, $\Psi(t)=\mu$,
$g_1(u,z)=\mu u^b$.  Note that (P1), (P2), (P4) and
 (P5) hold, and that
\begin{align*}
&\sup_{c\in \mathbb{R}_+}{\frac{c
}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1}[\max_{t\in
[0,1]}a(t)(ch(c)+\int_0^c
\omega(s)ds)]}}\\
&=\sup_{c\in \mathbb{R}_+}{\frac{c
}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}I^{-1}
[\mu\max_{t\in [0,1]}a(t)(c+c^{1-d}+  \frac{c^{1+b}}{1+b})]}}
\end{align*}
Then \eqref{e4.28} guarantees that (P3) holds.


\section{Singularities at  $x=0$ but not  $x'=0$}

In this section the nonlinearity $f$ may be singular at $x=0$, but
 not at $x'=0$. We  assume that the following conditions hold.
\begin{itemize}
\item[(S1)] $a(t)\in C[0,1]$,  $a(t)>0$,   $t\in(0,1)$;

\item[(S2)] $f(t,u,z)\leq [h(u)+\omega(u)]r(z)$, where  $f\in
C([0,1]\times \mathbb{R}_+\times \mathbb{R}_-,\mathbb{R}_+)$,
$\omega(u)>0$ continuous and
non-increasing on $[0,+\infty)$, $h(u)\geq 0$ continuous and
nondecreasing on $\mathbb{R}_+$, $r(z)>0$ continuous and nondecreasing on
$\mathbb{R}_-$;

\item[(S3)]
$$
\sup_{c\in \mathbb{R}_+}{\frac{c
}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in
[0,1]}a(t)(ch(c)+\int_0^c \omega(s)ds)])}}>1,
$$
where
$I(z)={\int_z^0{\frac{udu}{r(u)}}}$, $z\in \mathbb{R}_-$,
$\int_{0}^{a}\omega(s)ds<+\infty$, $a\in \mathbb{R}_+$;

\item[(S4)] There exists a function
$g_1\in C([0,+\infty)\times (-\infty,0],[0,+\infty))$, such that
$f(t,u,z)\geq g_1(u,z), \forall (t,u,z)\in [0,1]
 \times \mathbb{R}_+\times \mathbb{R}_-$, and
$\lim_{u\to +\infty}\frac{g_1(u,z)}{u}=+\infty$,
uniformly for $z\in \mathbb{R}_-$.

\item[(S5)]  There exists a function $\Psi_H\in C([0,1],\mathbb{R}_+)$
and a constant $0\leq \delta$ such that
$f(t,u,z)\geq \Psi_H(t)(-z)^\delta$,  for all
$(t,u,z)\in [0,1]\times[0,H]\times[-H,0)$.

\end{itemize}
For $n\in\{1,2,\dots\}$, $x\in P$, define operator
\begin{equation}
\begin{aligned}
 (A_{n}x)(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x(s)+\frac{1}{n},-|x'(s)|)ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x(s)
 +\frac{1}{n},-|x'(s)|)ds\Big),\quad t\in [0,1].
\end{aligned} \label{e5.1}
\end{equation}

\begin{theorem} \label{thm5.1}
Suppose {\rm (S1)--(S5)}  hold. Then
\eqref{e1.1} has at least two positive solutions
 $ x_{0,1}$, $x_{0,2}$ in $C^1[0,1]\cap C^2(0,1)$ with
$x_{0,1}(t),x_{0,2}(t)>0$, $t \in [0,1]$.
\end{theorem}

\begin{proof}
Choose $R_1>0$ such that
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
 {1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in[0,1]}a(t)(R_1 h(R_1)
+\int_0^{R_1} \omega(s)ds)])}}>1.\label{e5.2}
\end{equation}
 From the continuity of $ I^{-1}$ and  $h$, we can choose
$\epsilon>0$ and $\epsilon<R_1$ such that
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
 {1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in[0,1]}a(t)(R_1 h(R_1
+\epsilon)+\int_0^{R_1+\epsilon}
\omega(s)ds)])}}>1. \label{e5.3}
\end{equation}
Let $n_0\in\{1,2,\dots\}$ with
${\frac{1}{n_0}} < \min\{\epsilon,{\delta/2} \}$
and let   $N_0 =\{n_0,n_0+1, \dots\}$.
Then Lemma \ref{lem2.4} guarantees that for $n\in N_0$,
$A_n:P\to P$ is a completely continuous operator.
Let
$$
\Omega_1=\{x\in C^1[0,1]:\| x \| <R_1\}.
$$
We show that
\begin{equation}
x\neq \mu A_n x,\quad \forall x\in P\cap \partial\Omega_1,\;
\mu\in(0,1],\; n\in N_0.\label{e5.4}
\end{equation}
In fact, if there exists an $ x_0\in P\cap\partial \Omega_1$ and
$\mu_0\in(0,1]$ with $x_0=\mu_0 A_n x_0$,
\begin{align*}
x_0(t)&=-\mu_0\int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|)ds\\
&\quad +\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},-|x_0'(s)|)ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)
+\frac{1}{n},-|x_0'(s)|)ds\Big).
\end{align*}
Then
\begin{equation}
x'_0(t)=-\mu_0\int_0^t a(s)f(s,x_0(s)+\frac{1}{n},
-|x_0'(s)|)ds,\quad \forall t\in[0,1].\label{e5.5}
\end{equation}
Obviously, $x'_0(t)\leq 0$, $t\in(0,1)$, and
since  $x_0(1)> 0$, $x_0(t)>0,t\in[0,1]$.
Differentiating \eqref{e5.5}, we have
\begin{equation}
\begin{gathered}
x''_0(t)+\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},\quad x_0'(t))=0,\;0<t<1,\\
x_0'(0)=0,\quad x_0(1)=\sum_{i=1}^{m-2}a_i x_0(\xi_i).
\end{gathered}\label{e5.6}
\end{equation}
and
\begin{align*}
- x''_0(t)
&=\mu_0 a(t)f(t,x_0(t)+\frac{1}{n},x_0'(t))\\
&\leq a(t)[h(x_0(t)+\frac{1}{n})+\omega(x_0(t)+\frac{1}{n})]r(x_0'(t)),
\quad \forall t\in(0,1).
\end{align*}
Then
$$
{\frac{ -x''_0(t)}{r(x_0'(t))}}
\leq a(t)[h(x_0(t)+\frac{1}{n})+\omega(x_0(t)+\frac{1}{n})],\quad
 \forall t\in(0,1),
$$
and

\begin{align*}
{\frac{ -x''_0(t)x_0'(t)}{r(x_0'(t))}}
&\geq a(t)[h(x_0(t)+{\frac{1}{n}})+\omega(x_0(t)+{\frac{1}{n}})]x_0'(t)\\
&\geq a(t)[h(R_1+\epsilon)+\omega(x_0(t)+\frac{1}{n})]x_0'(t).
\end{align*} %\label{e5.7}
Integrating from $0$ to $t$, we have
\begin{align*}
I( x'_0(t))
&\geq\int_0^t a(s)[h(R_1+\epsilon)x_0'(s))
 +\omega(x_0(s)+\frac{1}{n})x_0'(s)]ds\\
&\geq  \max_{t\in [0,1]}a(t) \int_0^t[h(R_1+\epsilon) x_0'(s)
 \omega(x_0(s)+\frac{1}{n}) x_0'(s)]ds\\
&\geq -\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)R_1
 +\int_0^t \omega(x_0(s)+\frac{1}{n})d(x_0(s)+\frac{1}{n}))\\
&=-\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)R_1+\int_{x_0(0)
 +\frac{1}{n}}^{x_0(t)+\frac{1}{n}}\omega(s)ds)\\
&\geq -\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)R_1+\int_{0}^{R_1
 +\epsilon}\omega(s)ds)
\end{align*}
and
 $$
I( x'_0(t))\geq-\max_{t\in
[0,1]}a(t)(h(R_1+\epsilon)R_1+\int_{0}^{R_1+\epsilon}\omega(s)ds).
$$
Then
$$
x'_0(t)\geq I^{-1}( -\max_{t\in
[0,1]}a(t)(h(R_1+\epsilon)R_1+\int_{0}^{R_1+\epsilon}\omega(s)ds));
$$
that is,
\begin{equation}
- x'_0(t)\leq -I^{-1}( -\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)R_1
+\int_{0}^{R_1+\epsilon}\omega(s)ds)),\ t\in(0,1).\label{e5.7}
\end{equation}
Since
\begin{equation}
\begin{aligned}
&x_0(0)\\
&={\frac{\mu_0}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0-x'_0(s)ds
-{\frac{\mu_0\sum_{i=1}^{m-2}a_i\int^{\xi_i}_{0}
-x'_0(s)ds}{1-\sum_{i=1}^{m-2}a_i}} \\
&\leq {\frac{1}{1-\sum_{i=1}^{m-2}a_i}}\int^1_0 -I^{-1}\Big( -\max_{t\in [0,1]}
a(t)(h(R_1+\epsilon)R_1+\int_0^{R_1+\epsilon}\omega(s)ds)\Big)ds\\
&\quad +{\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}}\int^{\xi_i}_{0}
-I^{-1}\Big( I(-\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)(R_1)
+\int_0^{R_1+\epsilon}\omega(s)ds))\Big)ds\\
&={\frac{1+\sum_{i=1}^{m-2}a_i\xi_i}{1-\sum_{i=1}^{m-2}a_i}}
(-I^{-1}\Big(-\max_{t\in [0,1]}a(t)(h(R_1+\epsilon)R_1+\int_0^{R_1
+\epsilon}\omega(s)ds)\Big),
\end{aligned}\label{e5.8}
\end{equation}
$x_0(0)\geq  x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x_0(0)\geq
\gamma\delta\|x_0\|_2,x_0(0)\geq \|x_0\|=R_1$. So
\begin{equation}
{\frac{R_1}{-{\frac{\sum_{i=1}^{m-2}a_i\xi_i+1}
{1-\sum_{i=1}^{m-2}a_i}}(I^{-1}[-\max_{t\in
[0,1]}a(t)(R_1 h(R_1+\epsilon)+\int_0^{R_1+\epsilon}
\omega(s)ds)])}}\leq 1,\label{e5.9}
\end{equation}
which is a contradiction to
\eqref{e5.3}. Then \eqref{e5.4} holds.
From Lemma \ref{lem2.1}, for  $n\in N_0$,
\begin{equation}
i(A_n,\Omega_1\cap P,P)=1. \label{e5.10}
\end{equation}
Now we show that there exists a set $\Omega_2$ such that
\begin{equation}
A_{n}x \not\leq x,\quad \forall\ x\in \partial \Omega_2 \cap P. \label{e5.11}
\end{equation}
Choose $a^*, N^{*}$ as in section 3.
Let
$$
\Omega_2=\{x\in C^1[0,1]:\|x\|<\frac{R_2}{a^*}\}.
$$
Then
$$
A_{n}x \not\leq  x,\quad \forall  x\in \partial  \Omega_2 \cap P.
$$
In fact, if  there exists $x_0\in  \partial  \Omega_2 \cap P$
with $x_0\geq A_{n}x_0 $,  by the definition of the cone and Lemma
\ref{lem2.3},
one has
$$
x_0(t)\geq \gamma\|x_0\|_1\geq \gamma x(0)\geq \gamma\delta\|x_0\|_2,
$$
$x_0(t)\geq \frac{R_2}{a^*}>R_2$ for all $t\in[0,1]$.
 Then $x_0(t)+\frac{1}{n}>R_2$.
 From \eqref{e3.11},
\begin{align*}
\gamma x_0(t)
&\geq \gamma A_{n}x_0(t)\\
&=\gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds
\Big)\Big)\\
&\geq \gamma\Big(- \int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&-\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds
\Big)\Big)\\
&\geq \gamma\Big(\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{t}(t-s)a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&\quad -\frac{\sum_{i=1}^{m-2}a_i\int_{0}^{1}(\xi_i-s)a(s)f(s,x_0(s)
+\frac{1}{n},x_0'(s)}{1-\sum_{i=1}^{m-2}a_i})\Big)\\
&=\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)f(s,x_0(s)+\frac{1}{n},x_0'(s))ds\\
&\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)g_1(x_0(s)+\frac{1}{n},x_0'(s))ds\\
&\geq \frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}a(s)dsN^{*} x_0(s)\\
&\geq a^*\frac{\gamma\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}
a_i}\int_{0}^{1}a(s)dsN^{*}\frac{R_2}{a^*}
>\frac{R_2}{a^*};
\end{align*}
that is, $\|x_0\|\geq \gamma\|x_0\|_1>\frac{R_2}{a^*}$, which is
a contradiction to $x_0\in \partial  \Omega_2 \cap P$. Then \eqref{e5.11}
holds.
From Lemma \ref{lem2.2},
\begin{equation}
i(A_n,\Omega_2\cap P,P)=0. \label{e5.12}
\end{equation}
This equality and (5.10) guarantee
\begin{equation}
i(A_n,(\Omega_2-\bar{\Omega}_1\cap P,P)=-1. \label{e5.13}
\end{equation}
From this equality and (5.10), $A_n$ has two fixed points with
$x_{n,1}\in \Omega_1\cap P,x_{n,2}\in (\Omega_2-\bar{\Omega}_1)\cap
P$.

For each $n\in N_0$, there exists $x_{n,1}\in\Omega_1\cap P$ such that
$x_{n,1}=A_nx_{n,1}$; that is, for $t\in [0,1]$,
\begin{equation} \begin{aligned}
 x_{n,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\Big(\int_{0}^{1}(1-s)a(s)f(s,x_{n,1}(s)+\frac{1}{n},-|x'_{n,1}(s)|)ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n,1}(s)
+\frac{1}{n},-|x'_{n,1}(s)|)ds\Big).
\end{aligned}\label{e5.14}
\end{equation}
As in the proof of \eqref{e3.5},
 $ x'_n(t)\leq 0$, $t \in (0,1)$ and
$$
x'_{n,1}(t)=-\int_0^t
a(s)f(s,x_{n,1}(s)+\frac{1}{n},x_{n,1}'(s))ds,\quad  n\in N_0,\;
t\in(0,1).
$$
Now we consider $\{x_{n,1}(t)\}_{ n\in N_0}$ and
$\{x' _{n,1}(t)\}_{n\in N_0}$, since
$\| x_{n,1}\|\leq R_1$,
\begin{gather}
\text{$\{x _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$,} \label{e5.15}\\
\text{$\{x' _{n,1}(t)\}$ is  uniformly bounded on $[0,1]$.} \label{e5.16}
\end{gather}
Then
\begin{equation}
\text{$\{x _{n,1}(t)\}$ is equicontinuous on  $[0,1]$.}\label{e5.17}
\end{equation}
As in the  proof of \eqref{e3.6},
\begin{equation}
\begin{gathered}
x''_{n,1}(t)+a(t)f(t,x_{n,1}(t)+\frac{1}{n},x_{n,1}'(t))=0,0<t<1,\\
x_{n,1}'(0)=0,x_{n,1}(1)=\sum_{i=1}^{m-2} a_ix_{n,1}(\xi_i).
\end{gathered} \label{e5.18}
\end{equation}
 Now we show that for all  $t_1,t_2\in[0,1]$,
\begin{equation}
\begin{aligned}
&|I( x'_{n,1}(t_2))-I( x'_{n,1}(t_1))|\\
&\leq \max_{t\in [0,1]}a(t)\big[h(R_1+\epsilon)(|x_{n,1}(t_2)-x_{n,1}(t_1)|
 +|t_2-t_1|)\\
&\quad +|\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt|\big].
\end{aligned}\label{e5.19}
\end{equation}
 From \eqref{e5.18},
\begin{align*}
-x''_{n,1}(t)
&=a(t)f(t,x_{n,1}(t)+\frac{1}{n},x'_{n,1}(t))\\
&\leq a(t)[h(x_{n,1}(t)+\frac{1}{n})+\omega(x_{n,1}(t)
  +\frac{1}{n})]r(x_{n,1}'(t)), \quad \forall t\in(0,1),
\end{align*}
\begin{align*}
x''_{n,1}(t)
&=-a(t)f(t,x_{n,1}(t)+\frac{1}{n},x'_{n,1}(t))\\
&\geq -a(t)[h(x_{n,1}(t)+\frac{1}{n})+\omega(x_{n,1}(t)
  +\frac{1}{n})]r(x_{n,1}'(t)), \quad \forall t\in(0,1),
\end{align*}
\begin{equation}
\begin{aligned}
&\frac{-x''_{n,1}(t)x_{n,1}'(t)}{r(x_{n,1}'(t))}\\
&\geq a(t)[h(x_{n,1}(t)+{\frac{1}{n}})+\omega(x_{n,1}(t)
  +{\frac{1}{n}})]x_{n,1}'(t)\\
&\geq a(t)[h(R_1+\epsilon)x_{n,1}'(t)+\omega(x_{n,1}(t)
  +{\frac{1}{n}}(1-t))x'_{n,1}(t)],\quad \forall t\in(0,1),
\end{aligned}\label{e5.20}
\end{equation}
\begin{equation}
\begin{aligned}
&\frac{ x''_{n,1}(t)x_{n,1}'(t)}{r(x_{n,1}'(t))} \\
&\leq -a(t)[h(x_{n,1}(t)+{\frac{1}{n}})+\omega(x_{n,1}(t)
  +{\frac{1}{n}})]x_{n,1}'(t)\\
&\leq -a(t)[h(R_1+\epsilon)x_{n,1}'(t)+\omega(x_{n,1}(t)
 +{\frac{1}{n}}(1-t))x'_{n,1}(t)],\quad
 \forall t\in(0,1).
\end{aligned}\label{e5.21}
\end{equation}
Since the right-hand sides of \eqref{e5.20} and
\eqref{e5.21} are positive,
for all $t_1$, $t_2\in[0,1]$ and $t_1<t_2$,
\begin{align*}
&I( x'_{n,2}(t_2))-I( x'_{n,1}(t_1))\\
&\geq \max_{t\in [0,1]}a(t)[h(R_1+\epsilon)
\Big(\int_{t_1}^{t_2}x'_{n,1}(s)dt
+\int _{t_1}^{t_2}\omega(x_{n,1}(s)+\frac{1}{n})x'_{n,1}(s)ds\Big)]\\
&\geq -\max_{t\in [0,1]}a(t)[h(R_1+\epsilon)(|x_{n,1}(t_2)-x_{n,1}(t_1)|)
 +|\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt|],
 \end{align*}
\begin{align*}
&I( x'_{n,1}(t_1))-I( x'_{n,1}(t_2))\\
&\leq \max_{t\in [0,1]}a(t)[h(R_1+\epsilon)(|x_{n,1}(t_2)-x_{n,1}(t_1)|)
 +|\int _{x_{n,1}(t_1)+\frac{1}{n}(1-t_1)}^{x_{n,1}(t_2)
 +\frac{1}{n}(1-t_2)}\omega(s)dt|]\,,
\end{align*}
 \eqref{e5.19} holds.

Since  $I^{-1}$ is uniformly continuous on  $[0,I(-R_1-\epsilon)]$,
 for all $\bar\epsilon>0$, there exists $\epsilon'>0$ such that
\begin{equation}
|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\epsilon, \quad \forall |s_1-s_2|<\epsilon',\;
  s_1, s_2\in[0,I(-R_1-\epsilon)].\label{e5.22}
\end{equation}
Also  \eqref{e5.19}  guarantees that for $\epsilon'>0$, there exists
$\delta'>0$ such that
\begin{equation}
|I( x'_{n,1}(t_2))-I( x'_{n,1}(t_1))|<\epsilon',
 \quad  \forall\ |t_1-t_2|<\delta',\; t_1,t_2\in[0,1].\label{e5.23}
\end{equation}
 From this inequality and \eqref{e5.22},
\begin{equation} \begin{gathered}
| x'_{n,1}(t_2)- x'_{n,1}(t_1)|
=|I^{-1}(I( x'_{n,1}(t_2)))-I^{-1}(I(  x'_{n,1}(t_1)))|< \bar\epsilon, \\
\forall |t_1-t_2|<\delta',\;t_1,t_2\in[0,1];
\end{gathered}  \label{e5.24}
\end{equation}
that is,
\begin{equation}
\text{$\{x' _{n,1}(t)\}$ is equi-continuous on  $[0,1]$.}\label{e5.25}
\end{equation}
From  \eqref{e5.15}--\eqref{e5.17}, \eqref{e5.25} and  the Arzela-Ascoli
Theorem, $\{x _{n,1}(t)\}$ and  $\{x'_{n,1}(t)\}$ are relatively compact on
$C^1[0,1]$. This implies, there exists a subsequence
$\{x_{n_j,1}\}$  of $\{x_{n,1}\}$ and  function
$x_{0,1}(t)\in C^1[0,1]$ such that
 $$
\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{n_j,1}(t)-x_{0,1}(t)|=0, \quad
\lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{n_j,1}(t)-x'_{0,1}(t)|=0.
$$
Since $x'_{n_j,1}(0)=0$, $x_{n_j,1}(1)=\sum_{i=1}^{m-2}a_i x_{n_j,1}(\xi_i)$,
   $x'_{n_j,1}(t)<0$, $x_{n_j,1}(t)>0$, $t\in(0,1)$,
$j\in\{1,2,\dots\}$,
\begin{equation}
x'_{0,1}(0)=0, x_{0,1}(1)=\sum_{i=1}^{m-2}a_i x_{0,1}(\xi_i),
x'_{0,1}(t)\leq 0, x_{0,1}(t)\geq 0, \;t\in(0,1).\label{e5.26}
\end{equation}
For $(t,x_{n_j,1}(t)+{\frac{1}{n_j}}$,
$x_{n_j,1}'(t))\in [0,1]\times [0,R_1+\epsilon]\times  (-\infty,0)$,
from (S5) there exists a function $\Psi_{R_1}\in C([0,1],\mathbb{R}_+)$
such that
$$
f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t))ds\geq
\Psi_{R_1}(t)(-x_{n_j,1}'(t))^{\delta},\quad 0\leq \delta<1.
$$
Then, for $n\in N_0$,
\begin{align*}
x_{n_j,1}'(t)
&=-\int_{0}^{t}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\leq -\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)(-x'_{n_j,1}(s))^\delta ds,\quad
  t\in[0,1],\; n\in N_0,
\end{align*}
which implies
$$
x_{n_j,1}'(t)\leq -(1-\delta)(\int_{0}^{t}a(s)
\Psi_{R_1+\epsilon}(s)ds)^{\frac{1}{1-\delta}} ,
$$
and
\begin{align*}
 x_{n_j,1}(t)
&=-\int_{0}^{t}(t-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds
)\\
&\geq  -\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
(\int_{0}^{1}(1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\quad -\sum_{i=1}^{m-2}a_i\int_{0}^{\xi_i}(\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds
)\\
&\geq \frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}
 (1-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\quad -\frac{\sum_{i=1}^{m-2}a_i}{1-\sum_{i=1}^{m-2}a_i}\int_{0}^{1}
 (\xi_i-s)a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&=\frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x'_{n_j,1}(s))ds\\
&\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)\Psi_{R_1+\epsilon}(s)(-x'_{n_j,1}(s))^\delta ds\\
&\geq \frac{\sum_{i=1}^{m-2}a_i(1-\xi_i)}{1-\sum_{i=1}^{m-2}a_i}
 \int_{0}^{1}a(s)\Psi_{R_1+\epsilon}(s)
((1-\delta)(\int_{0}^{s}a(\tau)\Psi_{R_1+\epsilon}(\tau)d\tau)
 ^{\frac{1}{1-\delta}})^\delta ds\\
&=:F,\quad t\in [0,1].
\end{align*}
Since
\begin{align*}
&f(t,x_{n_j,1}(t)+{\frac{1}{n_j}},x_{n_j,1}'(t))\\
&\leq [h(x_{n_j,1}(t)+{\frac{1}{n_j}})
 +\omega(x_{n_j,1}(t)+{\frac{1}{n_j}})]r(x_{n_j,1}'(t))\\
&\leq [h(\frac{R_1}{\gamma}+\epsilon)+\omega(F)]r((\delta-1)
\Big(\int_{0}^{t}a(s)\Psi_{R_1+\epsilon}(s)ds\Big)^{\frac{1}{1-\delta}})),
\end{align*}
and
$$
x_{n_j,1}'(t)-x_{n_j,1}'({\frac{1}{2}})
=-\int_{1/2}^{t} a(s)f(s,x_{n_j,1}(s)+{\frac{1}{n_j}},x_{n_j,1}'(s))ds,\
\quad t\in(0,1),
$$
letting $j\to +\infty$, the  Lebesgue Dominated
Convergence Theorem guarantees that
\begin{equation}
x_{0,1}'(t)-x_{0,1}'({\frac{1}{2}})= - \int_{1/2}^{t}
a(s)f(s,x_{0,1}(s),x_{0,1}'(s))ds,\ t\in(0,1).\label{e5.27}
\end{equation}
Differentiating, we have
$$
x''_{0,1}(t)+a(t)f(t,x_{0,1}(t),x'_{0,1}(t))=0,\quad 0<t<1\,.$$
From this equality  and from \eqref{e5.26},
$x_{0,1}(t)$ is a positive solution of
\eqref{e1.1} with $x_{0,1}\in C^1[0,1]\cap C^2(0,1)$.

For the set
$\{x_{n,2}\}_{n\in N_0}\subseteq (\Omega_2-\overline{\Omega}_1)\cap P$,
as in the proof for the set $\{x_{n,1}\}_{n\in N_0}$,
we obtain a convergent subsequence
$\{x_{n_i,2}\}_{n\in N_0} $ of $\{x_{n,2}\}_{n\in N_0}$
with $\lim_{i\to +\infty}x_{n_i,2}=x_{0,2}  \in C^1[0,1]\cap C^2(0,1)$.
Moreover, $x_{0,2}$ is a positive solution to \eqref{e1.1}.
\end{proof}

\textbf{Example 5.1}
 In \eqref{e1.1}, let $f(t,u,z)=\mu[1+(-z)^{a}][1+u^b+u^{-d}]$ and
$a(t)\equiv 1$ with $0\leq a<1,b>1,0<d<1$ and $\mu>0$. If
\begin{equation}
\mu<\sup_{c\in \mathbb{R}_+}\frac{I(\frac{c(1-\sum_{i=1}^{m-2}a_i)}
{\sum_{i=1}^{m-2}a_i\xi_i+1})}{\max_{t\in [0,1]}a(t)(c+c^{1-d}+
 \frac{c^{1+b}}{1+b})},
\label{e5.28}
\end{equation}
Then  \eqref{e1.1} has at least two positive solutions
$x_{0,1},x_{0,2}\in C^1[0,1]\cap C^2(0,1)$.


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\end{document}