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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 116, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/116\hfil Positive solutions]
{Positive solutions for singular three-point boundary-value
problems}

\author[R. P. Agarwal, D. O'Regan, B. Yan \hfil EJDE-2008/116\hfilneg]
{Ravi P. Agarwal, Donal O'Regan, Baoqiang Yan}  % in alphabetical order

\address{Ravi P. Agarwal \newline
Department of Mathematical Science, Florida Institute
of Technology, Melbourne, Florida 32901, USA}
\email{agarwal@fit.edu}

\address{Donal O'Regan \newline
Department of Mathematics, National University of Ireland,
Galway, Ireland}
\email{donal.oregan@nuigalway.ie}

\address{Baoqiang Yan \newline
 Department of Mathematics, Shandong Normal University, Ji-nan, 250014, China}
\email{yanbqcn@yahoo.com.cn}

\thanks{Submitted January 17, 2008. Published August 25, 2008}
\thanks{Supported by grants 10571111 from National Natural Science,
 and J07WH08 from \hfill\break\indent the Shandong Education Committee}
\subjclass[2000]{34B15}
\keywords{Three-point boundary value problems; singularity;\hfill\break\indent
  positive solutions; fixed point index}

\begin{abstract}
 Using the theory of fixed point index, this paper discusses the
 existence of at least one positive solution and the existence
 of multiple positive solutions for the singular three-point
 boundary value problem:
 \begin{gather*}
 y''(t)+a(t)f(t,y(t),y'(t))=0,\quad 0<t<1,\\
 y'(0)=0,\quad y(1)=\alpha y(\eta),
 \end{gather*}
 where $0<\alpha<1$, $0<\eta<1$, and $f$ may be singular at $y=0$ and
 $y'=0$.
\end{abstract}

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

\section{Introduction}

In this paper, we consider the singular three-point
boundary-value problem (BVP):
\begin{gather}
 y''(t)+a(t)f(t,y(t),y'(t))=0,\quad 0<t<1,\label{e1.1}\\
 y'(0)=0,\quad y(1)=\alpha y(\eta),\label{e1.2}
\end{gather}
where $0<\alpha <1$, $0<\eta<1$, $f$ may be singular at $y=0$ and
$y'=0$, and $a \in C((0,1), (0,\infty))$.

When $f(t,x,z)$ has no singularity at $x=0$ and $z=0$, there are
many results on the existence of solutions to \eqref{e1.1}-\eqref{e1.2} with
different boundary conditions such as $x(0)=0$, $x(1)=\delta
x(\eta)$, or $x(0)=x_0$, $x(\eta)-x(1)=x_1$
(see \cite{g1,g2,g4,h1,k1}). Also
when $f(t,x,z)=f(t,x)$ has no singularity at $x=0$, using
Krasnoselkii's fixed point theorem,  Liu cite{l1} discussed the
existence of positive solutions to \eqref{e1.1}-\eqref{e1.2}.
 In \cite{c1,y1}, the authors obtained the existence of at least one
positive solutions to \eqref{e1.1}-\eqref{e1.2} when $f(t,x,z)$ is
singular at $x=0$ and $z=0$.

The features in this article, that differ from those in \cite{c1,y1}, are
as follows. Firstly, the nonlinearity $f(t,x,z)$ may be sublinear in
$x$ at $x=+\infty$ and the degree of singularity in $x$ and $z$ may
be arbitrary; i.e., $f(t,x,z)$ contains $\frac{1}{x^{\alpha}}$,
$x^{\beta}$ and $\frac{1}{(-z)^{-\gamma}}$ for any $\alpha>0$,
$\beta>0$ and $\gamma>0$. Secondly, \eqref{e1.1}-\eqref{e1.2} may have
at least two positive solutions. Thirdly, \eqref{e1.1}-\eqref{e1.2}
may have no positive solutions.

There are main five sections in our paper. In sections 2, we discuss
a special Banach space and define a new cone in this space, and some
lemmas are proved for convenience. In section 3, we discuss the
nonexistence of positive solutions to \eqref{e1.1}-\eqref{e1.2}.
In section 4, the existence of at least one positive solution
to \eqref{e1.1}-\eqref{e1.2} is presented when $f(t,x,z)$ is singular
at $x=0$ and $z=0$. In section 5, we consider the existence of at
least two positive solutions to \eqref{e1.1}-\eqref{e1.2} when
$f(t,x,z)$ is singular at $x=0$ and $z=0$ and $f$
is suplinear at $x=+\infty$. Some of the ideas in this paper were
motivated from \cite{a1,a2,o1,o2}.


\section{Preliminaries}

Let
$$
C^1[0,1]=\{y:[0,1]\to R : \text{ $y(t)$ and $y'(t)$are continuous on
$[0,1]$}\}
$$
 with  norm $\| y\| =\max\{\max_{t\in[0,1]}|y(t)|, \max_{t\in[0,1]}|y'(t)|\}$
and
$$
P=\{y\in C^1[0,1]:y(t)\geq 0,\forall t\in[0,1]\}.
$$
Obviously $C^1[0,1]$ is a Banach space and $P$ is a cone
in $C^1[0,1]$. The
following lemmas are needed later.


\begin{lemma}[cite{g3}] \label{lem2.1}
 Let $\Omega$ be a bounded  open set
in real Banach space $E$, $P$ be a cone of $E$, $\theta\in\Omega $
and $A:\bar{\Omega}\cap P\to P $ be continuous and
compact. 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}[\cite{g3}] \label{lem2.2}
 Let $\Omega$ be a bounded  open set
in real Banach space $E$, $P$ be a cone of $E$, $\theta\in\Omega $
and $A:\bar{\Omega}\cap P\to P $ be continuous and
compact. Suppose
\begin{equation}
Ax\neq  x, \quad \forall\  x\in \partial  \Omega \cap P,\label{e2.2}
\end{equation}
then  $i(A,\Omega\cap P,P)=0$.
\end{lemma}

\begin{lemma}[\cite{l1}] \label{lem2.3}
 Let $0<\alpha<1$, $a$, $h\in C((0,1), (0,\infty))$,
 $ a, h \in L^1[0,1]$ and
\begin{align*}
 y(t)&=\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)h(\tau) \,d\tau\,ds
 -\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)h(\tau)\,d\tau\,ds\\
 &\quad -\int^t_0\int^s_0a(\tau)h(\tau)\,d\tau\,ds.
 \end{align*}
  Then
\begin{equation}
 \min_{t\in[0,1]}y(t)\geq
 \frac{\alpha(1-\eta)}{1-\alpha\eta}\max_{t\in[0,1]}|y(t)|.\label{e2.3}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.4}
 Assume that
$ f\in C((0,1)\times(0,+\infty)\times(-\infty,0), (0,+\infty))$, that
$a\in C((0,1),(0,+\infty))$,
and that for any constant $H>0$ there exists a function $\Psi_H(t)$
continuous on $(0,1)$, and positive on $(0,1)$ and a constant
$0\leq \gamma<1$ such that
\begin{equation}
f(t,x,z)\geq \Psi_H(t)(-z)^{\gamma},\quad
 \forall t \geq 0, \; 0<x \leq H,  \; z<0,\label{e2.4}
\end{equation}
where $\int_0^1a(s)\Psi_H(s)ds<+\infty$. Then there is a $c_0>0$ such that
for any positive solution $x\in C[0,1]$ with $x'(t)<0$ for all
$t\in(0,1)$ to \eqref{e1.1}-\eqref{e1.2} we have
\begin{equation}
x(t)\geq c_0, \quad  t\in[0,1].\label{e2.5}
\end{equation}
Moreover, if $x_0\in C[0,1]$ is a positive solution to
\begin{gather*}
y''(t)+a(t)f(t,\max\{c_0,y(t)\},-|y'(t)|-\frac{1}{n})=0,\quad 0<t<1,\\
y'(0)=0,\quad y(1)=\alpha y(\eta),
\end{gather*}
where $\frac{\alpha}{1-\alpha}(1-\eta)\frac{1}{n}<c_0$, $x_0$ is a
positive solution to
\begin{gather*}
y''(t)+a(t)f(t,y(t),-|y'(t)|-\frac{1}{n})=0,\quad 0<t<1,\\
y'(0)=0,\quad y(1)=\alpha y(\eta).
\end{gather*}
\end{lemma}

\begin{proof}
 Assume that $x$ is a positive solution to \eqref{e1.1}-\eqref{e1.2}
with $x'(t)<0$ for $t\in(0,1)$. Then Lemma \ref{lem2.3} implies
$\min_{t\in[0,1]}x(t)\geq
\frac{\alpha(1-\eta)}{1-\alpha\eta}\max_{t\in[0,1]}|x(t)|>0$.

Let $H=1$. Then there exists a function $\Psi_1(t)$ continuous on
$[0,1]$, and positive on $(0,1)$ a constant $0\leq\gamma<1$ such
that
$$
f(t,x,z)\geq \Psi_1(t)(-z)^{\gamma},\quad \forall t
\geq 0, \; 0<x \leq 1,  \; z<0.
$$
There are two cases to be considered:
(1) $x(t)\geq 1$ for all $t\in[0,1]$.
(2) $x(1)<1$.
Let $t_*=\inf\{t|x(t)<1$ for all $s\in[t,1]\}$. If
$t_*>0$, we have $x(t_*)=1$ and $x(0)\geq 1$. Then, Lemma \ref{lem2.3} yields
\begin{equation}
\min_{t\in[0,1]}|x(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
\max_{t\in[0,1]}|x(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}.\label{e2.6}
\end{equation}
If $t_*=0$ and $x(t_*)=1$, Lemma \ref{lem2.3} implies
\begin{equation}
\min_{t\in[0,1]}|x(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
\max_{t\in[0,1]}|x(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}\label{e2.7}
\end{equation}
also. If $t_*=0$ and $x(t_*)<1$, from \eqref{e2.4}, we have
$$
-x''(t)=a(t)f(t,x(t),x'(t))\geq
a(t)\Psi_1(t)(-x'(t))^{\gamma}, \quad t\in(0,1).
$$
Also note
$$
-\frac{x''(t)}{(-x'(t))^{\gamma}}\geq a(t)\Psi_1(t),\quad
  t\in(0,1).
$$
Integrating from $0$ to $t$, we have
$$
\frac{1}{1-\gamma}(-x'(t))^{1-\gamma}\geq\int_0^ta(s)\Psi_1(s)ds,
t\in(0,1),
$$
which implies
$$
-x'(t)\geq[(1-\gamma)\int_0^ta(s)\Psi_1(s)ds]^{\frac{1}{1-\gamma}},\quad
 t\in(0,1).
$$
Integration from $\eta$ to $1$ yields
$$
x(\eta)-x(1)\geq\int_{\eta}^1[(1-\gamma)\int_{\eta}^{1}\int_0^ta(s)
\Psi_1(s)ds]^{\frac{1}{1-\gamma}}dt.
$$
Since $x(1)=\alpha x(\eta)$, we have
\begin{equation}
x(1)\geq\frac{\alpha}{1-\alpha}\int_{\eta}^1[(1-\gamma)
\int_0^ta(s)\Psi_1(s)ds]^{\frac{1}{1-\gamma}}dt.\label{e2.8}
\end{equation}
Let
$c_0=\frac{1}{2}\min\{1,\frac{\alpha(1-\eta)}{1-\alpha\eta},
\frac{\alpha}{1-\alpha}\int_{\eta}^1[(1-\gamma)\int_0^ta(s)
\Psi_1(s)ds]^{\frac{1}{1-\gamma}}dt\}$.
Combining \eqref{e2.6}, \eqref{e2.7} and \eqref{e2.8}, we have
$$
\min_{t\in[0,1]}x(t)\geq c_0.
$$

Suppose that $x_0$ satisfies
\begin{gather*}
x''_0(t)+a(t)f(t,\max\{c_0,x_0(t)\},-|x'_0(t)|-\frac{1}{n})=0, \quad
t\in(0,1),\\
x_0'(0)=0,\quad x_0(\eta)=\alpha x_0(1).
\end{gather*}
Then $x_0''(t)<0$ and so $x_0¡¯(t)<0$  for $t\in (0,1)$.
 Then $x_0$ satisfies
\begin{gather*}
x''_0(t)+a(t)f(t,\max\{c_0,x_0(t)\},x'_0(t)-\frac{1}{n})=0, \quad
t\in(0,1),\\
x_0'(0)=0,\quad x_0(\eta)=\alpha x_0(1).
\end{gather*}
There are are two cases to be considered:

\noindent (1) $x_0(t)\geq 1$ for all $t\in[0,1]$.
In this case, since $c_0\leq 1$, we have
$$
x''_0(t)+a(t)f(t,\max\{c_0,x_0(t)\},x'_0(t)-\frac{1}{n})
=x''_0(t)+a(t)f(t,x_0(t),x'_0(t)-\frac{1}{n})=0,
$$
for $0<t<1$.

\noindent (2) $x_0(1)<1$. Let $t_*=\inf\{t|x_0(t)<1$ for all $s\in[t,1]\}$.
If $t_*>0$, we have $x_0(t_*)=1$ and $x_0(0)\geq 1$. Then
$$
\min_{t\in[0,1]}x_0(t)\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
\max_{t\in[0,1]}|x_0(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}.
$$
If $t_*=0$ and $x_0(t_*)=1$, we have
$$
\min_{t\in[0,1]}x_0(t)\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
\max_{t\in[0,1]}|x_0(t)|\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
$$
also. If $t_*=0$ and $x_0(t_*)<1$, from \eqref{e2.4}, we have
$$
-x_0''(t)=a(t)f(t,\max\{c_0,x_0(t)\},x_0'(t)-\frac{1}{n})\geq
a(t)\Psi_1(t)(-x_0'(t)+\frac{1}{n})^{\gamma}, \quad
t\in(0,1).
$$
Also note
$$
-\frac{x_0''(t)}{(-x_0'(t)+\frac{1}{n})^{\gamma}}\geq a(t)\Psi_1(t),\quad
  t\in(0,1).
$$
Integrating from $0$ to $t$, we have
$$
\frac{1}{1-\gamma}[(-x_0'(t)+\frac{1}{n})^{1-\gamma}
-({\frac{1}{n}})^{1-\gamma}]\geq\int_0^ta(s)\Psi_1(s)ds, t\in(0,1),
$$
which implies
$$
-x_0'(t)+\frac{1}{n}\geq[(1-\gamma)\int_0^ta(s)\Psi_1(s)ds
]^{\frac{1}{1-\gamma}}, t\in(0,1).
$$
Integration from $\eta$ to $1$ yields
$$
x_0(\eta)-x_0(1)\geq\int_{\eta}^1[(1-\gamma)
\int_{\eta}^{1}\int_0^ta(s)\Psi_1(s)ds]^{\frac{1}{1-\gamma}}]dt-(1-\eta)
\frac{1}{n}.
$$
Since $x_0(1)=\alpha x_0(\eta)$, we have
$$
x_0(1)\geq\frac{\alpha}{1-\alpha}\int_{\eta}^1[(1-\gamma)
\int_{\eta}^{1}\int_0^ta(s)\Psi_1(s)ds]^{\frac{1}{1-\gamma}}]dt
-\frac{\alpha}{1-\alpha}(1-\eta)\frac{1}{n}\geq c_0.
$$
Consequently, the definition of $c_0$ implies that
$x_0(t)\geq c_0$ for all $t\in[0,1]$.
Therefore,
$$
x''_0(t)+a(t)f(t,\max\{c_0,x_0(t)\},x'_0(t)-\frac{1}{n})=
x''_0(t)+a(t)f(t,x_0(t),x'_0(t)-\frac{1}{n})=0,
$$
for $0<t<1$. The proof is complete.
\end{proof}

To discuss the existence of multiple positive solutions, we
construct a new space.
Let $q(t)=1-t$, $t\in[0,1]$ and
$$
C^1_q[0,1]=\{y:[0,1]\to R: \text{$y(t)$ and $q(t)y'(t)$ are
 continuous on $[0,1]$}\}
$$
with  norm $\|y\|_q=\max\{\max_{t\in[0,1]}|y(t)|,
 \max_{t\in[0,1]}q(t)|y'(t)|\}$
and
$$
P_q=\{y\in C^1_q[0,1]:\min_{t\in[0,1]}y(t)\geq
\frac{\alpha(1-\eta)}{1-\alpha\eta}\max_{t\in[0,1]}|y(t)|\text{
and } y(0)\geq\max_{t\in[0,1]}q(t)|y'(t)|\}.
$$

\begin{lemma} \label{lem2.5}
The set $C^1_q[0,1]$ is a Banach space and $P_q$ is
cone in $C^1_q[0,1]$
\end{lemma}

\begin{proof} It is easy to see that $\|\cdot\|_q$ is a norm of the
space $C^1_q$. Now we show that $C^1_q$ is a Banach space.
Assume that $\{x_n\}_{n=1}^{\infty}\subseteq C^1_q$ is a Cauchy
sequence; i.e., for each $\varepsilon>0$, there is a $N>0$ such
that
\begin{equation}
\|x_n-x_m\|_q<\varepsilon,\quad \forall n>N,\; m>N\,.\label{e2.9}
\end{equation}
Then
$$
\max_{t\in[0,1]}|x_n(t)-x_m(t)|<\varepsilon,\quad \forall n>N,\; m>N.
$$
Thus, there is a $x_0\in C[0,1]$ such that
\begin{equation}
\lim_{n\to+\infty}\max_{t\in[0,1]}|x_n(t)-x_0(t)|=0.\label{e2.10}
\end{equation}
For  $1>\delta>0$, since
$(1-\delta)\max_{t\in[0,1-\delta]}|x'_n(t)-x'_m(t)|
\leq\max_{t\in[0,1-\delta]}q(t)|x'_n(t)-x'_m(t)|$,
we have
$$
\max_{t\in[0,1-\delta]}|x'_n(t)-x'_m(t)|\leq
\frac{1}{1-\delta}\max_{t\in[0,1-\delta]}q(t)|x'_n(t)-x'_m(t)|
<\frac{1}{1-\delta}\varepsilon,
$$
which implies that for any $\delta>0$, $x'_n(t)$ is uniformly
convergent on $[0,1-\delta]$. Hence, $x_0(t)$ is continuously
differentiable on $[0,1)$. And since $q(t)x'_{N+1}(t)$ is uniformly
continuous on $[0,1]$, there exists a $\delta'>0$ such that
$$
|q(t_1)x'_{N+1}(t_1)-q(t_2)x'_{N+1}(t_2)|<\varepsilon \quad
\text{for } |t_1-t_2|<\delta¡¯, t_1, t_2 \in [0,1).
$$
Then
\begin{align*}
&|q(t_1)x'_0(t_1)-q(t_2)x'_{0}(t_2)|\\
&=|q(t_1)x'_{0}(t_1)-q(t_1)x'_{N+1}(t_1)\\
&\quad +q(t_1)x'_{N+1}(t_1)-q(t_2)x'_{N+1}(t_2)+q(t_2)x'_{N+1}(t_2)-q(t_2)x'_{0}(t_1)|\\
&\leq|q(t_1)x'_{0}(t_1)-q(t_1)x'_{N+1}(t_1)|\\
&\quad +|q(t_1)x'_{N+1}(t_1)-q(t_2)x'_{N+1}(t_2)|+|q(t_2)x'_{N+1}(t_2)-q(t_2)x'_{0}(t_1)|\\
&<3\varepsilon,\quad \text{ for  } |t_1-t_2|<\delta',\;
 t_1, t_2\in [0,1),
\end{align*}
which implies that $\lim_{t\to 1^-}q(t)x_0'(t)$ exists. Let
$q(1)x_0(1)=\lim_{t\to 1^{-}}q(t)x'_0(t)$. Now from \eqref{e2.9}, we have
for any $t\in[0,1]$,
$$
q(t)|x_n'(t)-x_m'(t)|<\varepsilon,\quad \forall n>N, m>N.
$$
Letting $m\to+\infty$,  for all $t\in[0,1]$, we have
\begin{equation}
q(t)|x_n'(t)-x_0'(t)|\leq\varepsilon,\quad
\forall n>N .\label{e2.11}
\end{equation}
Combining \eqref{e2.10} and \eqref{e2.11} shows
$C^1_q[0,1]$ is a Banach space.

Clearly $P_q$ is a cone of $C^1_q[0,1]$. The proof is complete.
\end{proof}


\begin{lemma} \label{lem2.6}
For each $y\in P_q$, $\|y\|_q=\max_{t\in[0,1]}|y(t)|$.
\end{lemma}

\begin{proof} For $y\in P$, obviously
$\|y\|_q\geq\max_{t\in[0,1]}|y(t)|$. On the other hand, since
$y\in P_q$,
$$
\max_{t\in[0,1]}|y(t)|\geq
y(0)\geq\max_{t\in[0,1]}q(t)|y'(t)|.
$$
Then
\begin{align*}
\|y\|_q
&=\max\{\max_{t\in[0,1]}|y(t)|,\max_{t\in[0,1]}q(t)|y'(t)|\}\\
&\leq \max\{\max_{t\in[0,1]}|y(t)|,y(0)\}=\max_{t\in[0,1]}|y(t)|.
\end{align*}
Consequently, $\|y\|_q=\max_{t\in[0,1]}|y(t)|$. The proof is
complete.
\end{proof}

Now we list the following conditions to be used in this article.
\begin{itemize}
\item[(H)]
$f \in C( (0,1) \times (0,\infty) \times (-\infty,0), (0,\infty))$
and there are three functions
$g,h\in C((0,+\infty),(0,+\infty))$, $\Phi\in C((0,1),[0,+\infty))$,
with $\Phi(t)>0$ for all $t\in(0,1)$, and
\begin{equation}  \label{e2.12}
f(t,x,z)\leq\Phi(t)h(x)g(|z|)\quad \forall\,
(t,x,z)\in(0,1)\times(0,+\infty)\times (-\infty,0).
\end{equation}

\item[(H')]  For any constant $H>0$ there exists a function
$\Psi_H(t)$ continuous on $(0,1)$  and positive on
$(0,1)$,  and a constant $0\leq \gamma<1$  such that
\begin{equation} \label{e2.13}
f(t,x,z)\geq \Psi_H(t)(-z)^{\gamma},\quad  \forall t \in(0,1),
 \; 0<x \leq H,  \; z<0,
\end{equation}
 where $\int_0^1a(s)\Psi_H(s)ds<+\infty$.

\end{itemize}
 For each $n\in N=\{1,2,\dots\}$, for $y\in P$ (or $y\in P_q$),
define operators
\begin{equation}
\begin{aligned}
(A_ny)(t)
&=\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)f(\tau,\max\{c_0,y(\tau)\},
 -|y'(\tau)|-\frac{1}{n})\,d\tau\,ds\\
&\quad -\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)
  f(\tau,\max\{c_0,y(\tau)\},-|y'(\tau)|-\frac{1}{n})\,d\tau\,ds\\
&\quad -\int^t_0\int^s_0a(\tau)f(\tau,\max\{c_0,y(\tau)\},
 -|y'(\tau)|-\frac{1}{n})\,d\tau\,ds,
 \end{aligned}\label{e2.14}
\end{equation}
for $t\in[0,1]$ and $c_0>0$.

Suppose that (H) and (H') hold. A standard argument (see \cite{g3,g4})
applied to \eqref{e2.14} yields that $A_n:P\to P$ is continuous and
completely continuous for each $n\in N$.

\begin{lemma} \label{lem2.7}
Suppose {\rm (H)} and {\rm (H')} holds and
$\int_0^1a(t)\Phi(t)\sup_{\frac{1}{c}\leq u\leq
\frac{1}{c}+\frac{1}{1-t}c}g(u)dt<+\infty$ for all $c>1$. Then
$A_n:P_q\to P_q$ is a continuous and completely continuous for each
$n\in N$.
\end{lemma}

\begin{proof}
For $y\in P_q$, it is easy to see that
$|y'(t)|\leq\frac{1}{1-t}\|y\|_q$ for all $t\in[0,1)$.
Also (H) and Lemma \ref{lem2.3} yield
$$
\frac{\alpha(1-\eta)}{1-\alpha\eta}\max_{t\in[0,1]}|(A_ny)(t)|
\leq (A_ny)(t)<+\infty,\quad \forall t\in[0,1]
$$
and $(A_ny)'(t)>-\infty$ for all $t\in[0,1)$.
Moreover,  since
\begin{align*}
(A_ny)(0)
&=\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)f(\tau,\max\{c_0,y(\tau)\},
  -|y'(\tau)|-\frac{1}{n})\,d\tau\,ds\\
&\quad -\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)
  f(\tau,\max\{c_0,y(\tau)\},-|y'(\tau)|-\frac{1}{n})\,d\tau\,ds\\
&\geq \int^1_0\int^s_0a(\tau)f(\tau,\max\{c_0,y(\tau)\},
  -|y'(\tau)|)\,d\tau\,ds\\
&=\int^1_0(1-s)a(s)f(s,\max\{c_0,y(s)\},-|y'(s)|-\frac{1}{n})ds
\end{align*}
and
\begin{align*}
q(t)|(A_nx)'(t)|
&=(1-t)\int_0^ta(s)f(s,\max\{c_0,y(s)\},-|y'(s)|-\frac{1}{n})ds\\
&\leq\int^1_0(1-s)a(s)f(s,\max\{c_0,y(s)\},-|y'(s)|-\frac{1}{n})ds,
\end{align*}
we have
$$
(A_nx)(0)\geq \max_{t\in[0,1]}q(t)|(A_nx)'(t)|.
$$
Consequently, $A_nP_q\subseteq P_q$ for each $n\in N=\{1,2,\dots\}$.
Moreover, since
$$
\lim_{t\to 1^-}|(A_ny)'(t)|=\int_0^1a(s)f(s,\max\{c_0,y(s)\},|y'(s)|
-\frac{1}{n})ds,
$$
we can assume that $A_ny\in C^1[0,1]$.

Next we show that $A_n:P_q\to P_q$ is continuous and completely
continuous.
Suppose that $\{y_m\}\subseteq P_q$, $y_0\in P_q$ with
$\lim_{m\to+\infty}\|y_m-y_0\|_q=0$. Then, there is an $M>c_0$ such
that
$$
\|y_m\|_q\leq M, \|y_0\|_q\leq M, \quad m\in N.
$$
Then
$|y_m'(t)|\leq M/(1-t)$ for $m\in\{1,2,\dots\}$ and so
$$
f(t,\max\{c_0,y_m(t)\},-|y_m'(t)|-\frac{1}{n})
\leq \Phi(t)\max_{c_0\leq u\leq M}h(u)\sup_{\frac{1}{n}\leq u\leq
\frac{1}{n}+\frac{1}{1-t}M}g(u),
$$
for $t\in(0,1)$. Moreover, since
$$
\lim_{m\to+\infty}f(t,\max\{c_0,y_m(t)\},-|y_m'(t)|-\frac{1}{n})
=f(t,\max\{c_0,y_0(t)\},-|y_0'(t)|-\frac{1}{n}),
$$
for $t\in(0,1)$, the Lebesgue Dominated Convergence Theorem guarantees that
\begin{align*}
&\max_{t\in[0,1]}|(A_ny_m)(t)-(A_ny_0)(t)|\\
&\leq\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)
 \Big|f(\tau,\max\{c_0,y_m(\tau)\},-|y'_m(\tau)|-\frac{1}{n})\\
&\quad -f(\tau,\max\{c_0,y_0(\tau)\},-|y'_0(\tau)|-\frac{1}{n})\Big|\,d\tau\,ds
\\
&\quad +\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)
 \Big||f(\tau,\max\{c_0,y_m(\tau)\},-|y'_m(\tau)|-\frac{1}{n})\\
&\quad -f(\tau,\max\{c_0,y_0(\tau)\},-|y'_0(\tau)|-\frac{1}{n})Big|\,d\tau\,ds
\\
&\quad +\int\int_0^1\int_0^sa(\tau)
 \Big|f(\tau,\max\{c_0,y_m(\tau)\},-|y'_m(\tau)| \\
&\quad -\frac{1}{n})-f(\tau,\max\{c_0,y_0(\tau)\},
 -|y'_0(\tau)|-\frac{1}{n})\Big|\,d\tau\,ds
\to 0, \quad\text{as } m\to+\infty.
\end{align*}
Since $A_ny_m$, $A_ny_0\in P_q$, Lemma \ref{lem2.6} yields
$$
\lim_{m\to+\infty}\|A_ny_m-A_ny_0\|_q
 =\max_{t\in[0,1]}|(A_ny_m)(t)-(A_ny_0)(t)|=0,
$$
which implies that $A_n:P_q\to P_q$ is continuous.

Suppose $D\subseteq P_q$ is bounded. Then, there is an $M>c_0$ such
that $\|y\|_q\leq M$ for all $y\in D$.
Then
$|y'(t)|\leq M/(1-t)$ for all $y\in D$,
and so
$$
f(t,\max\{c_0,y(t)\},-|y'(t)|-\frac{1}{n})
\leq \Phi(t)\max_{c_0\leq u\leq M}h(u)\sup_{\frac{1}{n}
\leq u\leq\frac{1}{n}+\frac{1}{1-t}M}g(u), \quad
t\in(0,1).
$$
Thus
\begin{align*}
&\max_{t\in[0,1]}|(A_ny)(t)|\\
&\leq\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)|f(\tau,\max\{c_0,y(\tau)\},
  -|y'(\tau)|-\frac{1}{n})|\,d\tau\,ds\\
&\quad +\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)
  |f(\tau,\max\{c_0,y(\tau)\},-|y'(\tau)|-\frac{1}{n})|\,d\tau\,ds\\
&\quad +\int_0^1\int_0^sa(\tau)|f(\tau,\max\{c_0,y(\tau)\},
 -|y'(\tau)|-\frac{1}{n})|\,d\tau\,ds\\
&\leq\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)\Phi(s)
  \max_{c_0\leq u\leq M}h(u)\sup_{\frac{1}{n}\leq u\leq\frac{1}{n}
 +\frac{1}{1-\tau}M}g(u)\,d\tau\,ds\\
&\quad +\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0\Phi(s)a(\tau)
 \max_{c_0\leq u\leq M}h(u)\sup_{\frac{1}{n}\leq u\leq\frac{1}{n}
 +\frac{1}{1-\tau}M}g(u)\,d\tau\,ds\\
&\quad +\int_0^1\int_0^sa(\tau)\Phi(\tau)\max_{c_0\leq u\leq M}h(u)
 \sup_{\frac{1}{n}\leq u\leq\frac{1}{n}+\frac{1}{1-\tau}M}g(u)\,d\tau\,ds
\end{align*}
and
\begin{align*}
&\max_{t\in[0,1]}|(A_ny)'(t)|\\
&\leq \max_{t\in[0,1]}\int_0^ta(\tau)|f(\tau,\max\{c_0,y(\tau)\},
 -|y'(\tau)|-\frac{1}{n})|d\tau\\
&\leq\max_{t\in[0,1]}\int_0^ta(\tau)\Phi(\tau)
 \max_{c_0\leq u\leq M}h(u)\sup_{\frac{1}{n}\leq u\leq\frac{1}{n}
 +\frac{1}{1-\tau}M}g(u)d\tau.
\end{align*}
Also $A_n D$ is bounded in the norm
$\|x\|_0=\max\{\max_{t\in[0,1]}|x(t)|,\max_{t\in[0,1]}|x'(t)|\}$.
For $t_1$, $t_2\in[0,1]$, $y\in D$, we have
\begin{align*}
&|(A_ny)(t_1)-(A_ny)(t_2)|\\
&=|\int_{t_1}^{t_2}\int_0^sa(\tau)\Phi(\tau)|f(\tau,\max\{c_0,y(\tau)\},
 -|y'(\tau)|-\frac{1}{n})|\,d\tau\,ds|\\
&\leq|\int_{t_1}^{t_2}\int_0^sa(\tau)\Phi(\tau)\max_{c_0\leq u\leq M}h(u)
 \sup_{\frac{1}{n}\leq
u\leq\frac{1}{n}+\frac{1}{1-\tau}M}g(u)\,d\tau\,ds|
\end{align*}
and
\begin{align*}
|(A_ny)'(t_1)-(A_ny)'(t_2)|
&=|\int_{t_1}^{t_2}a(\tau)|f(\tau,\max\{c_0,y(\tau)\},-|y'(\tau)|
-\frac{1}{n})|d\tau |\\
&\leq|\int_{t_1}^{t_2}a(\tau)\Phi(\tau)\max_{c_0\leq u\leq
M}h(u)\sup_{\frac{1}{n}\leq
u\leq\frac{1}{n}+\frac{1}{1-\tau}M}g(u)d\tau|,
\end{align*}
which implies that $\{(A_ny)(t)|y\in D\}$ and $\{(A_ny)'(t)|y\in
D\}$ are equicontinuous on $[0,1]$.

The Arzela-Ascoli Theorem guarantees that $A_nD$ and $(A_nD)'$ are
relatively compact in $C[0,1]$.
Since
\begin{align*}
\|A_ny\|_q
&=\max\{\max_{t\in[0,1]}|(A_ny)(t)|,\max_{t\in[0,1]}(1-t)|(A_ny)'(t)|\}\\
&\leq\max\{\max_{t\in[0,1]}|(A_ny)(t)|,\max_{t\in[0,1]}|(A_ny)'(t)|\},
\end{align*}
the set $A_nD$ is relatively compact in $C^1_q[0,1]$.
Consequently, $A_n:P_q\to P_q$ is continuous and completely
continuous for each $n\in\{1,2,\dots\}$. The proof is complete.
\end{proof}


\section{Nonexistence of  positive solutions to \eqref{e1.1}-\eqref{e1.2}}

In this section, we notice that the presence of $z$ in $f(t,x,z)$
can lead to the nonexistence of positive solutions to
\eqref{e1.1}-\eqref{e1.2}.


\begin{theorem} \label{thm3.1}
 Suppose  {\rm (H)} holds and
$\int_0^z\frac{1}{g(r)}dr=+\infty$ for all $z\in(0,+\infty)$ and
$\int_0^1a(s)\Phi(s)ds<+\infty$. Then \eqref{e1.1}-\eqref{e1.2}
has no positive solution.
\end{theorem}

\begin{proof}
Suppose $x_0(t)$ is a positive solution to \eqref{e1.1}-\eqref{e1.2}.
Then
\begin{gather*}
x''_0(t)+a(t) f(t,x_0(t),x'_0(t))=0, \quad t\in(0,1)\\
x_0'(0)=0,\quad  x_0(1)=\alpha x_0(\eta),
\end{gather*}
which means that there is a $t_0\in(0,1)$ with $x_0'(t_0)<0$,
$x_0(t_0)>0$ (otherwise $x'(t) \geq 0$ for all $t\in (0,1)$
which would contradict $x(1)=\alpha x(\eta)<x(\eta)$).
Let $t_*=\inf\{t<t_0|x_0'(s)<0$ for all $s\in[t,t_0]\}$.
Obviously, $t_*\geq 0$ and $x_0'(t_*)=0$ and
$x_0'(t)<0$ for all $t\in(t_*,t_0]$. Condition (H) implies
$$
-x''_0(t)\leq a(t)f(t,x_0(t),x'_0(t))\leq a(t)\Phi(t)h(x_0(t))g(|x'_0(t)|)
,\quad \forall t\in(t_*,t_0),
$$
and so
$$
\frac{-x''_0(t)}{g(-x'_0(t))}\leq
a(t)\Phi(t)h(x_0(t))\leq a(t)\Phi(t)h(x_0(t)), \quad \forall
t\in(t_*,t_0).
$$
 Integration from $t$ to $t_0$ yields
$$
\int_{-x'_0(t)}^{-x'_0(t_0)}\frac{1}{g(r)}dr
=\int_{t}^{t_0}\frac{1}{g(-x'_0(s))}d(-x'_0(s))
\leq \max_{u\in[x_0(t_0),x_0(t)]}h(u)\int_0^1a(s)\Phi(s)ds.
$$
Letting $t\to t_*$, we have
$$
+\infty=\int_{0}^{-x'_0(t_0)}\frac{1}{g(r)}dr
\leq \max_{u\in[x_0(t_0),x_0(t_*)]}h(u)\int_0^1a(s)\Phi(s)ds<+\infty,
$$
a contradiction.
Consequently, \eqref{e1.1}-\eqref{e1.2} has no positive solution.
\end{proof}

\begin{example} \label{exa3.1} \rm
 Consider the boundary-value problem
\begin{gather*}
x''+(1-t)^a(|x'|)^{a}[x^{b}+(x+1)^{-d}+1]=0, \quad t\in(0,1),\label{e3.1}\\
x(0)=0, \quad x(1)=\frac{1}{2}x(\frac{1}{2}),\label{e3.2}
\end{gather*}
where $a\geq1$, $b>1$, $d>0$.
This problem has no positive solution.
It is easy to see that
$f(t,x,z)=(1-t)^a(|z|)^{a}[x^{b}+(x+1)^{-d}+1]$ for all
$(t,x,z)\in[0,1]\times[0,+\infty)\times(-\infty,+\infty)$.
Obviously, $g(r)=r^a$ and $\int_0^z\frac{1}{g(r)}dr=+\infty$ for all
$z\in(0,+\infty)$. Then Theorem \ref{thm3.1} guarantees that
\eqref{e3.1}-\eqref{e3.2} has no positive solution.
\end{example}

\section{Existence of at least one positive solution to
 \eqref{e1.1}-\eqref{e1.2}}

In this section our nonlinearity $f$ may be singular at $y'=0$ and
$y=0$ and $\Phi$ . Throughout this section we will assume that the
following conditions hold:
\begin{itemize}
\item[(H1)]  $a(t)\in C(0,1)$, $a(t)>0$ for all $t\in(0,1)$;

\item[(H2)] Conditions (H) and (H') hold and
$I(z)=\int^z_0\frac{1}{g(r)}dr<+\infty$ for all $z\in[0,+\infty)$
with
$$
\sup_{c_0 \leq r\leq c} h(r) \int^1_0 a(s)\Phi(s) ds
<\int^{\infty}_0 \frac{dr}{g(r)}
$$
 for all $c\in[c_0,+\infty)$ and suppose
$$
\sup_{c_0\leq c<+\infty}\frac{c}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}
(\sup_{c_0\leq r\leq c}h(r)\int_0^1a(s)\Phi(s)ds)}>1,
$$
where $c_0$ is defined in Lemma \ref{lem2.4}.
\end{itemize}

\begin{theorem} \label{thm4.1}
 Suppose that {\rm (H1)--(H2)} hold. Then
\eqref{e1.1}-\eqref{e1.2} has at least
 one positive  solution $ y_0\in C[0,1]\cap C^2(0,1)$ with $y_0(t)>0$  on
 $[0,1]$ and $y'_0(t)<0$ on
 $(0,1)$.
\end{theorem}

\begin{proof}
 Choose $R_1>0$ with
\begin{equation}
\frac{R_1 }{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r
\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds)}>1.\label{e4.1}
\end{equation}
 From the continuity of $ I^{-1}$ and $I$, we can choose
$\varepsilon>0$ and $\varepsilon<R_1$ with
\begin{equation}
\frac{R_1 }{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r\leq R_1}h(r)
\int_0^1a(s))\Phi(s)ds)+I(\varepsilon))}>1. \label{e4.2}
\end{equation}

Let $n_0\in\{1,2,\dots\}$ with $\frac{1}{n_0} <
\min\{\varepsilon,\frac{1}{2}\frac{1-\alpha}{\alpha(1-\eta)}c_0\}
\}$ and let $N_0 =\{n_0,n_0+1, \dots\}$.
Now (H1)--(H2) guarantee that for each $n\in  N_0$,
$A_n:P\to P$ is a continuous and completely continuous
operator.

Let $\Omega_1=\{y\in C^1[0,1]:\| y\| <R_1\}$.
We now show that
\begin{equation}
y\neq \mu A_n y,\quad \forall y\in P\cap \partial\Omega_1,\mu\in(0,1],
n\in N_0.\label{e4.3}
\end{equation}
 Suppose there exists a $ y_0\in P\cap\partial \Omega_1$ and
a $\mu_0\in(0,1]$ such that $y_0=\mu_0 A_n y_0$. It is easy to see
that $y_0'(t)\leq0$ and
\begin{equation}
y'_0(t)=-\mu_0\int_0^t a(s)f(s,\max\{c_0,y_0(s)\}, y_0'(s)
-\frac{1}{n})ds,t\in(0,1).  \label{e4.4}
\end{equation}
Also
\begin{gather}
y''_0(t)+\mu_0 a(t)f(t,\max\{c_0,y_0(t)\},y_0'(t)-\frac{1}{n})=0,\quad
0<t<1,\label{e4.5} \\
y_0'(0)=0,y_0(1)=\alpha y_0(\eta). \label{e4.6}
\end{gather}
Therefore,
\begin{align*}
- y''_0(t)&=\mu_0 a(t)f(t,\max\{c_0,y_0(t)\},y_0'(t)-\frac{1}{n})\\
&\leq a(t)\Phi(t)h(\max\{c_0,y_0(t)\})g(-y_0'(t)+\frac{1}{n}),
\forall t\in(0,1),
\end{align*}
which yields
$$
\frac{ -y''_0(t)}{g(-y_0'(t)+\frac{1}{n})}
\leq a(t)\Phi(t)h(\max\{c_0,y_0(t)\}),\quad \forall t\in(0,1).
$$
Integration from $0$ to $t$ yields
\begin{align*}
I(-y'_0(t)+\frac{1}{n})-I(\frac{1}{n})
&\leq\int_0^t a(s)\Phi(s)h(\max\{c_0,y_0(s)\})ds\\
&\leq \sup_{c_0\leq r\leq R_1}h(r)\int_0^ta(s)\Phi(s)ds,
\end{align*}
and so
$$
I(-y'_0(t)+\frac{1}{n})\leq \sup_{c_0\leq r\leq R_1}h(r)\int_0^t a(s)
\Phi(s)ds+I(\varepsilon).
$$
Thus
\begin{equation}
- y'_0(t)\leq I^{-1}\Big( \sup_{c_0\leq r\leq
R_1}h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon )\Big),\quad
t\in(0,1).\label{e4.7}
\end{equation}
Therefore,
\begin{equation}
y_0(t)-y_0(1)\leq (1-t)I^{-1}\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big).\label{e4.8}
\end{equation}
Let $t=\eta$ in \eqref{e4.8}.
Then
$$
y_0(\eta)-y_0(1)\leq (1-\eta)I^{-1}\Big( \sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big).
$$
Since $y_0(1)=\alpha y_0(\eta)$,
one has
$$
(\frac{1}{\alpha}-1)y_0(1)\leq (1-\eta)I^{-1}
\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big),
$$
which yields
$$
y_0(1)\leq \frac{\alpha}{1-\alpha}(1-\eta)I^{-1}
\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big).
$$
Then \eqref{e4.8} implies
\begin{equation}
\begin{aligned}
y_0(0)&\leq y_0(1)+I^{-1}\Big( \sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon)\Big)\\
&=\frac{1-\alpha\eta}{1-\alpha}I^{-1}\Big( \sup_{c_0\leq r\leq
R_1}h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon )\Big).
\end{aligned} \label{e4.9}
\end{equation}
Now \eqref{e4.7} and \eqref{e4.9} guarantee that
\begin{align*}
R_1&=\max\{\max_{t\in[0,1]}|y_0(t)|,\max_{t\in[0,1]}|y'_0(t)|\}\\
 &\leq\frac{1-\alpha\eta}{1-\alpha}I^{-1}\Big( \sup_{c_0\leq r\leq
R_1}h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon )\Big)
\end{align*}
which implies
$$
\frac{R_1}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r\leq R_1}
h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon))}\leq 1.
$$
This contradicts \eqref{e4.2}. Thus \eqref{e4.3} is true.

 From Lemma \ref{lem2.1}, for each $n\in N_0$, we have
$i(A_n,\Omega_1\cap P,P)=1$.
As a result, for each $n\in N_0$, there exists a $y_n\in\Omega_1\cap
P$, such that $y_n=A_ny_n$; i.e.,
\begin{align*}
 y_n(t)
&=\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)f(\tau,\max\{c_0,y_n(\tau)\},
 -|y'_n(\tau)|- \frac{1}{n})\,d\tau\,ds\\
 &\quad -\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)
 f(\tau,\max\{c_0,y_n(\tau)\},-|y'_n(\tau)|- \frac{1}{n})\,d\tau\,ds\\
 &\quad -\int^t_0\int^s_0a(\tau)f(\tau,\max\{c_0,y_n(\tau)\},-|y'_n(\tau)|-
 \frac{1}{n})\,d\tau\,ds.
 \end{align*} %\label{e4.9b}
It is easy to see that $ y'_n(t)<0$, and
$$
y'_n(t)=-\int_0^t a(s)f(s,\max\{c_0,y_n(s)\},y_n'(s)-\frac{1}{n})ds,\quad
n\in N_0,\; t\in(0,1).
$$
Now we consider $\{y_n(t)\}_{ n\in N_0}$ and $\{y' _n(t)\}_{n\in
N_0}$. Since $\| y_n \|\leq R_1$, one has
\begin{gather}
\text{the functions belonging to $\{y _n\}$  are uniformly bounded on
$[0,1]$,}  \label{e4.10}\\
 \text{the functions belonging to $\{y' _n\}$ are
uniformly bounded on $[0,1]$.} \label{e4.11}
\end{gather}
Thus
\begin{equation}
\text{the functions belonging to $\{y _n\}$  are
equicontinuous on $[0,1]$.}  \label{e4.12}
\end{equation}
 A similar argument to that used to show \eqref{e4.5} yields
\begin{equation}
\begin{gathered}
y''_n(t)+a(t)f(t,\max\{c_0,y_n(t)\},y_n'(t)-\frac{1}{n})=0,\quad 0<t<1,\\
y_n'(0)=0,\quad y_n(1)=\alpha y_n(\eta).
\end{gathered} \label{e4.13}
\end{equation}
By Lemma \ref{lem2.4}, we have
\begin{equation}
y_n(t)\geq c_0,\quad \forall n\in N_0. \label{e4.14}
\end{equation}
Now we claim that for any $t_1,t_2\in[0,1]$,
\begin{equation}
|I( y'_n(t_2)-\frac{1}{n})-I( y'_n(t_1)-\frac{1}{n})|
 \leq \sup_{c_0\leq r\leq  R_1}h(r))|\int _{t_1}^{t_2}a(t)\Phi(t))dt|.
\label{e4.15}
\end{equation}
Notice that
\begin{align*}
-y''_n(t)
&=a(t)f(t,\max\{c_0,y_n(t)\},y'_n(t)-\frac{1}{n})\\
&\leq a(t)|f(t,\max\{c_0,y_n(t)\},y'_n(t)-\frac{1}{n})|\\
&\leq a(t)\Phi(t)h(\max\{c_0,y_n(t)\})g(y_n'(t)-\frac{1}{n}),
\quad \forall t\in(0,1),\end{align*}
and
\begin{align*}
y''_n(t)
&=-a(t)f(t,\max\{c_0,y_n(t)\},y'_n(t)-\frac{1}{n})\\
&\leq a(t)|f(t,\max\{c_0,y_n(t)\},y'_n(t)-\frac{1}{n})|\\
&\leq a(t)\Phi(t)h(\max\{c_0,y_n(t)\})g(y_n'(t)-\frac{1}{n}),
\forall t\in(0,1),
\end{align*}
which yields
\begin{gather}
 \frac{-y''_n(t)}{g(y_n'(t)-\frac{1}{n})} \leq
a(t)\Phi(t)h(\max\{c_0,y_n(t)\}),\forall t\in(0,1),\label{e4.16}\\
\frac{y''_n(t)}{g(y_n'(t)-\frac{1}{n})} \leq
a(t)\Phi(t)h(\max\{c_0,y_n(t)\}),\forall t\in(0,1).\label{e4.17}
\end{gather}
 Note that the right hand sides are always positive in \eqref{e4.16}
and \eqref{e4.17}.
For any $t_1$, $t_2\in[0,1]$ with $t_1<t_2$, we have
$$
|\int_{t_1}^{t_2}\frac{1}{g(-y_n'(s)+\frac{1}{n})}d(-y_n'(s)+
\frac{1}{n})|\leq \sup_{c_0\leq r\leq
R_1}h(r)|\int_{t_1}^{t_2}a(t)\Phi(t)dt|;
$$
i.e., \eqref{e4.15} is true.

 Since $I^{-1}$ is uniformly continuous on $[0,I(R_1)]$,
for any $\bar\varepsilon>0$, there is a $\varepsilon'>0$ such that
\begin{equation}
|I^{-1}(s_1)-I^{-1}(s_2)|<\bar\varepsilon, \forall\ |s_1-s_2|<\varepsilon',
 \quad s_1, s_2\in[0,I(R_1)].\label{e4.18}
\end{equation}
Also \eqref{e4.15} guarantees that, for $\varepsilon'>0$, there is a
$\delta'>0$ such that
\begin{equation}
|I( y'_n(t_2)-\frac{1}{n})-I( y'_n(t_1)-\frac{1}{n})|<\varepsilon',
  \forall\ |t_1-t_2|<\delta',\quad t_1,t_2\in[0,1].\label{e4.19}
\end{equation}
Now \eqref{e4.18} and  \eqref{e4.19} yield
\begin{align*}
  | y'_n(t_2)- y'_n(t_1)|
&=|-y'_n(t_2)+\frac{1}{n}+  y'_n(t_1)-\frac{1}{n}|\\
&=|I^{-1}(I(-y'_n(t_2)+\frac{1}{n}))-I^{-1}(I(  -y'_n(t_1)+\frac{1}{n}))|\\
&< \bar\varepsilon, \quad \forall\ |t_1-t_2|<\delta',\; t_1,t_2\in[0,1],
  \end{align*}
which implies
\begin{equation}
\text{the functions belonging to $\{y' _n\}$ are equicontinuous on $[0,1]$.}
 \label{e4.20}
\end{equation}
  Consequently \eqref{e4.10}, \eqref{e4.11}, \eqref{e4.12} and \eqref{e4.20},
 the Arzela-Ascoli Theorem  guarantees
 that  $\{y _n\}$ and $\{y' _n\}$ are relatively compact in $C[0,1]$;
 i.e., there is a function $y_0 \in C^1[0,1]$, and a subsequence
$\{y_{n_j}\}$ of $\{y_n\}$ such that
 $$
\lim_{j\to +\infty}\max_{t\in[0,1]}|y_{n_j}(t)-y_0(t)|=0,\quad
\lim_{j\to +\infty}\max_{t\in[0,1]}|y'_{n_j}(t)-y'_0(t)|=0.
$$
Since $y'_{n_j}(0)=0$, $y_{n_j}(1)=\alpha y_{n_j}(\eta)$,
   $y'_{n_j}(t)<0$, $y_{n_j}(t)>0$, $t\in(0,1)$, $j\in\{1,2,\dots\}$,
   then one has
\begin{equation}
y'_0(0)=0, y_0(1)=\alpha y_0(\eta),
y'_0(t)\leq 0, y_0(t)\geq 0, t\in(0,1).\label{e4.21}
\end{equation}
Now since $\sup_{n\geq 1}\|y_n\|\leq R_1$, (H') guarantees that
there exists a $\Psi_{R_1}(t)$ continuous and $\Psi_{R_1}(t)>0$ on
$(0,1)$ such that
$$
f(t,x,z)\geq \Psi_{R_1}(t)(-z)^{\gamma},\quad t\in(0,1),\; x\in(0,R_1],\; z<0.
$$
Then
$$
-y_{n_j}''(t)=a(t)f(t,\max\{c_0,y_{n_j}(t)\},y_{n_j}'(t)-\frac{1}{n_j}))\geq
a(t)\Psi_{R_1}(t)(-y_{n_j}'(t)+\frac{1}{n_j})^{\gamma},
$$
for $t\in(0,1)$. Also note that
$$
-\frac{y''(t)}{(-y_{n_j}'(t)+\frac{1}{n_j})^{\gamma}}\geq
a(t)\Psi_{R_1}(t),\quad  t\in(0,1).
$$
Integrating from $0$ to $t$, we have
$$
\frac{1}{1-\gamma}(-y_{n_j}'(t)+\frac{1}{n_j})^{1-\gamma}
-\frac{1}{1-\gamma}(\frac{1}{n_j})^{1-\gamma}\geq\int_0^ta(s)\Psi_1(s)ds,
\quad t\in(0,1),
$$
which implies
$$
-y_{n_j}'(t)+\frac{1}{n_j}\geq[(1-\gamma)(\int_0^ta(s)\Psi_1(s)ds
+\frac{1}{1-\gamma}(\frac{1}{n_j})^{1-\gamma})]^{\frac{1}{1-\gamma}}, \quad
t\in(0,1).
$$
Letting $j\to+\infty$, we have
$$
-y_0'(t)\geq[(1-\gamma)(\int_0^ta(s)\Psi_1(s)ds)]^{\frac{1}{1-\gamma}}, \quad
t\in(0,1).
$$
Consequently,  $y'_0(t)<0$ for all $t\in(0,1)$, which
together with $y_0(1)>0$  guarantees that $y_0(t)>0$ for all $t\in[0,1]$.
Therefore,
\begin{gather*}
\min\{\min_{s\in[\frac{1}{2},t]}y_0(s),\min_{s\in[\frac{1}{2},t]}|y'_0(s)|\}
 >0,\quad \text{for all } t\in[\frac{1}{2},1),\\
 \min\{\min_{s\in[t,\frac{1}{2}]}y_0(s),\min_{s\in[t,\frac{1}{2}]}|y'_0(s)|\}
 >0,\quad \text{for all } t\in(0,\frac{1}{2}].
\end{gather*}
 Since
$$
y_{n_j}'(t)-y_{n_j}'(\frac{1}{2})
=- \int_{\frac{1}{2}}^{t} a(s)f(s,\max\{c_0,y_{n_j}(s)\},y_{n_j}'(s)
 -\frac{1}{n_j})ds,\quad t\in(0,1),
$$
letting $j\to +\infty$, one has
$$y_0'(t)-y_0'(\frac{1}{2})=
- \int_{\frac{1}{2}}^{t}
a(s)f(s,\{c_0,y_0(s)\},y_0'(s))ds,t\in(0,1).$$ Now by direct
differentiation, we have
$$
y''_0(t)+a(t)f(t,\{c_0,y_0(t)\},y'_0(t))=0,0<t<1.
$$
Now \eqref{e4.14} guarantees that $y_0(t)\geq c_0$ for all $t\in[0,1]$ and
so
$$
y''_0(t)+a(t)f(t,y_0(t),y'_0(t))=0,\quad 0<t<1.
$$
 From \eqref{e4.21}, we have $y_0\in C[0,1]\cap C^2(0,1)$ and
$y_0$ is a positive solution to   \eqref{e1.1}-\eqref{e1.2}.
\end{proof}

\begin{example} \label{exa4.1}\rm
 Consider the three-point boundary value problems
\begin{gather*}
y''+\alpha[(-y')^{\frac{1}{2}}+(-y')^{-a}][y^b
 +(\frac{1}{\alpha})^{\frac{1}{2}d}y^{-d}]=0,t\in(0,1),\label{e4.22}\\
y'(0)=0,y(1)=\frac{1}{2}y(\frac{1}{2}),    \label{e4.23}
\end{gather*}
where $\alpha>0$, $a>0$, $1>\gamma\geq 0$, $b\geq 0$ and $d>0$.
Then, there is a $\alpha_0>0$ such that \eqref{e4.22}-\eqref{e4.23} has one
positive solution $y_0\in C[0,1]\cap C^2(0,1)$ with $y_0(t)>0$  on
 $[0,1]$ and
$y_0'(t)<0$ on $(0,1)$ for all $0<\alpha<\alpha_0$.

Let $a(t)\equiv \mu$, $\Phi(t)\equiv1$ for all $t\in[0,1]$,
$h(x)=x^b+(\frac{1}{\alpha})^{\frac{1}{2}d}x^{-d}$ for
$x\in(0,+\infty)$ and $g(z)=z^{\frac{1}{2}}+z^{-a}$ for
$z\in(0,+\infty)$. From the proof of Lemma \ref{lem2.4}, we have
$c_0=\frac{7}{192}\alpha^{\frac{1}{2}}$ with $\alpha\leq1$, and then
$\alpha[y^b+(\frac{1}{\alpha})^{\frac{1}{2}d}y^{-d}]\leq\alpha
y^{b}+(\frac{192}{7})^{d}$ for all $y\in[c_0,+\infty)$. Let
$I(z)=\int_0^z\frac{1}{r^{\frac{1}{2}}+r^{-a}}dr$. Thus there exists
an $\alpha_0$ such that
$$
\frac{I(1/3)}{\alpha\sup_{c_0\leq r\leq1}h(r)}>1, \quad
 \forall \alpha\in(0,\alpha_0]
$$
and then
$$
\sup_{c_0\leq c<+\infty}\frac{c}{3I^{-1}(\sup_{c_0\leq r\leq
c}h(r)\alpha)}>1.
$$
Hence, the conditions (H1) and (H2) hold.
Thus Theorem \ref{thm4.1} guarantees that \eqref{e4.22} and \eqref{e4.23}
has at least one positive solution.
\end{example}


\section{Multiple positive solutions to \eqref{e1.1}-\eqref{e1.2}}

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

\item[(P2)] Conditions (H) and (H') hold and
$I(z)=\int^z_0\frac{1}{g(r)}dr<+\infty$ for all $z\in[0,+\infty)$
with $\sup_{c_0 \leq r\leq c} h(r) \int^1_0 a(s)\Phi(s) ds
<\int^{\infty}_0 \frac{dr}{g(r)}$ for all $c\in[c_0,+\infty)$ and
suppose
$$\sup_{c_0\leq c<+\infty}\frac{c}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r\leq
c}h(r)\int_0^1a(s)\Phi(s)ds)}>1,$$ where $c_0$ is defined by
Lemma \ref{lem2.4};

\item[(P3)]
$\lim_{u\to+\infty} f(t,u,z)/u =+\infty$  uniformly for
$(t,z)\in[\frac{1}{4},\frac{3}{4}]\times(0,+\infty)$.

\end{itemize}

\begin{theorem} \label{thm5.1}
Suppose that {\rm (P1)--(P3)} hold. Then
\eqref{e1.1}-\eqref{e1.2} has at least two positive solutions $y_{1,0}$,
$y_{1,0}\in C[0,1]\cap C^2(0,1)$ with $y_{1,0}(t)>0$, $y_{2,0}(t)>0$
on  $[0,1]$
and $y_{1,0}'(t)<0$, $y_{2,0}'(t)<0$ on $(0,1)$.
\end{theorem}

\begin{proof}
Choose $R_1>0$ with
\begin{equation}
\frac{R_1}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}
\big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds\big)}>1.\label{e5.1}
\end{equation}
 From the continuity of $ I^{-1}$ and $I$, we can choose
$\varepsilon>0$ and $\varepsilon<R_1$ with
\begin{equation}
\frac{R_1}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r\leq R_1}h(r)
\int_0^1a(s)\Phi(s)ds+I(\varepsilon))}>1.\label{e5.2}
\end{equation}
Let $n_0\in\{1,2,\dots\}$ so that $\frac{1}{n_0} <
\min\{\varepsilon,\frac{1}{2}\frac{1-\alpha}{\alpha(1-\eta)}c_0\}$
and let $N_0 =\{n_0,n_0+1, \dots\}$.
Lemma \ref{lem2.7} guarantees that for each $n\in    N_0$,
$A_n:P_q\to P_q$ is a continuous and completely
continuous operator.
 From $(P_3)$, there is a $R'>R_1$ such that
$$
f(t,x,y)\geq N^* x, \quad \forall x\geq R',
$$
where
$N^*>(\int_{1/4}^{3/4}(1-s)a(s)ds
\frac{\alpha(1-\eta)}{1-\alpha\eta})^{-1}$.
Let
$$
R_2>\max\{R',\frac{1-\alpha\eta}{\alpha(1-\eta)}R'\}.
$$
Now let
$$
\Omega_1=\{y\in C^1_q[0,1]:\| y\|_q<R_1\},\quad
\Omega_2=\{y\in C^1_q[0,1]:\| y\|_q<R_2\}.
$$
We now show that
\begin{equation}
 y\neq \mu A_n y,\quad \forall y\in P\cap
\partial\Omega_1,\;\mu\in(0,1], \;n\in N_0,\label{e5.3}
\end{equation}
and
\begin{equation}
A_nx\not \leq  x,\quad \forall\  x\in \partial  \Omega_2
\cap P,n\in N_0.\label{e5.4}
\end{equation}
  Suppose there exists a $ y_0\in P\cap\partial \Omega_1$ and
a $\mu_0\in(0,1]$ such that $y_0=\mu_0 A_n y_0$. It is easy to see
that $y_0'(t)\leq0$ and
\begin{equation}
y'_0(t)=-\mu_0\int_0^t a(s)f(s,\max\{c_0,y_0(s)\}, y_0'(s)
-\frac{1}{n})ds,t\in(0,1).  \label{e5.5}
\end{equation}
Also
\begin{gather*}
y''_0(t)+\mu_0 a(t)f(t,\max\{c_0,y_0(t)\},y_0'(t)-\frac{1}{n})=0,\quad
0<t<1,\\
y_0'(0)=0,y_0(1)=\alpha y_0(\eta).
\end{gather*}
Therefore,
\begin{align*}
- y''_0(t)&=\mu_0 a(t)f(t,\max\{c_0,y_0(t)\},y_0'(t)-\frac{1}{n})\\
&\leq a(t)\Phi(t)h(\max\{c_0,y_0(t)\})g(-y_0'(t)+\frac{1}{n}), \quad
\forall t\in(0,1).
\end{align*}
which yields
$$
\frac{-y''_0(t)}{g(-y_0'(t)+\frac{1}{n})}
\leq a(t)\Phi(t)h(\max\{c_0,y_0(t)\}),\quad \forall t\in(0,1).
$$
Integration from $0$ to $t$ yields
\begin{align*}
I(-y'_0(t)+\frac{1}{n})-I(\frac{1}{n})
&\leq\int_0^t a(s)\Phi(s)h(\max\{c_0,y_0(s)\})ds\\
&\leq \sup_{c_0\leq r\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds,
\end{align*}
 and so
$$
I(-y'_0(t)+\frac{1}{n})\leq \sup_{c_0\leq r\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds+I(\varepsilon
).
$$
Thus
\begin{equation}
-y'_0(t)\leq I^{-1}\Big( \sup_{c_0\leq r\leq R_1}
h(r)\int_0^1a(s)\Phi(s)ds+I(\varepsilon)\Big), \quad t\in(0,1).\label{e5.6}
\end{equation}
Integration from $t$ to $1$ yields
\begin{equation}
y_0(t)-y_0(1)\leq (1-t)I^{-1}
\Big( \sup_{c_0\leq r\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds+I(\varepsilon )\Big),
\quad t\in(0,1).\label{e5.7}
\end{equation}
Let $t=\eta$ in \eqref{e5.7}. Then
$$
y_0(\eta)-y_0(1)\leq (1-\eta)I^{-1}\Big( \sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big).
$$
Since $y_0(1)=\alpha y_0(\eta)$, one has
$$
(\frac{1}{\alpha}-1)y_0(1)\leq (1-\eta)
I^{-1}\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big),
$$
which yields
$$
y_0(1)\leq \frac{\alpha}{1-\alpha}(1-\eta)I^{-1}
\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon )\Big).
$$
Then \eqref{e5.7} implies
\begin{equation}
\begin{aligned}
y_0(0)
&\leq y_0(1)+I^{-1}( \sup_{c_0\leq r\leq R_1}h(r)\int_0^1
a(s)\Phi(s)ds+I(\varepsilon ))\\
&=\frac{1-\alpha\eta}{1-\alpha}I^{-1}( \sup_{c_0\leq r\leq
R_1}h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon )).
\end{aligned} \label{e5.8}
\end{equation}
Now \eqref{e5.6} and \eqref{e5.8} guarantees
\begin{align*}
R_1&=\max\{\max_{t\in[0,1]}|y_0(t)|,\max_{t\in[0,1]}(1-t)|y'_0(t)|\}\\
&\leq\frac{1-\alpha\eta}{1-\alpha}I^{-1}\Big( \sup_{c_0\leq r
\leq R_1}h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon )\Big)
\end{align*}
which implies
$$
\frac{R_1}{\frac{1-\alpha\eta}{1-\alpha}I^{-1}(\sup_{c_0\leq r\leq R_1}
h(r)\int_0^1 a(s)\Phi(s)ds+I(\varepsilon))}\leq 1.
$$
This contradicts \eqref{e5.2}. Thus \eqref{e5.3} is true.

 From Lemma \ref{lem2.1}, for each $n\in N_0$, we have
\begin{equation}
i(A_n,\Omega_1\cap P,P)=1, \quad n\in N_0.\label{e5.9}
\end{equation}
Suppose there is a $x_0\in \partial \Omega_2 \cap P$
such that $A_nx_0\leq  x_0$. Then
$\|x_0\|_q=R_2$. Also Lemma \ref{lem2.3} implies
$$
\min_{t\in[0,1]}x_0(t)\geq\frac{\alpha(1-\eta)}{1-\alpha\eta}
\max_{t\in[0,1]}|x_0(t)|=\frac{\alpha(1-\eta)}{1-\alpha\eta}\|x_0\|_q
=\frac{\alpha(1-\eta)}{1-\alpha\eta}R_2>R'.
$$
Then, we have
\begin{align*}
x_0(0)&\geq (Ax_0)(0)\\
&=\frac{1}{1-\alpha}\int^1_0\int^s_0a(\tau)f(\tau,\max\{c_0,x_0(\tau)\},-|x_0'(\tau)|-
 \frac{1}{n})\,d\tau\,ds\\
&\quad -\frac{\alpha}{1-\alpha}\int^\eta_0\int^s_0a(\tau)f(\tau,\max\{c_0,x_0(\tau)\},-|x_0'(\tau)|-
 \frac{1}{n})\,d\tau\,ds\\
&\geq\int_{1/4}^{3/4}(1-s)a(s)f(\tau,\max\{c_0,x_0(s)\},-|x_0'(s)|-
 \frac{1}{n}) ds\\
&\geq\int_{1/4}^{3/4} (1-s)a(s)N*\max\{c_0,x_0(s)\} ds\\
&\geq\int_{1/4}^{3/4} (1-s)a(s)ds N*\frac{\alpha(1-\eta)}{1-\alpha\eta}R_2\\
&>\|x_0\|_q,
\end{align*}
which is a contradiction.
Thus, \eqref{e5.4} is true. Then Lemma \ref{lem2.2} implies
\begin{equation}
i(A_n,\Omega_2\cap P,P)=0, \quad n\in N_0.\label{e5.10}
\end{equation}
 From \eqref{e5.9} and \eqref{e5.10}, we have
\begin{equation}
i(A_n,(\Omega_2-\overline{\Omega}_1)\cap P,P)=-1, \quad
 n\in N_0.\label{e5.11}
\end{equation}
By \eqref{e5.9}, \eqref{e5.11}, there is a $x_{1,n}\in \Omega_1\cap P$
and another $x_{2,n}\in \Omega_2\cap P$ such that
$$
A_nx_{1,n}=x_{1,n}, \quad A_n x_{2,n}=x_{2,n},\quad n\in N_0.
$$
Now we consider $\{x_{1,n}\}_{n\in N_0}$ and
$\{x_{2,n}\}_{n\in N_0}$. By Lemma \ref{lem2.4}, we have $x_{1,n}(t)\geq c_0$
and $x_{2,n}\geq c_0$.

We consider $\{x_{1,n}\}_{n\in N_0}$. Obviously
$\max_{t\in[0,1]}|x_{1,n}(t)|\leq R_1$ for all $n\in N_0$ and
$\max_{t\in[0,1]}(1-t)|x'_{1,n}(t)|\leq R_1$ for all $n\in N_0$.
Also  $|x'_{1,n}(t)|\leq\frac{1}{1-t}R_1$ for all $t\in[0,1)$ and
$n\in N_0$. Hence,
the functions belonging to $\{x_{1,n}\}$ are uniformly bounded
on $[0,1]$.

Since $x_{1,n}(t)$ satisfies
\begin{align*}
x''_{1,n}(t)+a(t)f(t,\max\{c_0,x_{1,n}(t)\},x_{1,n}'(t)-\frac{1}{n})=0,
\quad 0<t<1,\\
x_{1,n}'(0)=0,x_{1,n}(1)=\alpha x_{1,n}(\eta).
\end{align*}
A similar argument to that used to show \eqref{e5.6} yields that
$$
-x'_{1,n}(t)\leq I^{-1}
\Big(\sup_{c_0\leq r\leq R_1}h(r)\int_0^1a(s)\Phi(s)ds+I(\varepsilon )\Big),
\quad t\in(0,1),
$$
which implies that the functions belonging to $\{x'_{1,n}\}$
 are uniformly bounded on $[0,1]$ and so
the functions belonging to $\{x_{1,n}\}$
 are equicontinuous on $[0,1]$.

A similar argument to that to used to show \eqref{e4.15} yields
the functions belonging to $\{x'_{1,n}\}$
are equicontinuous on $[0,1]$.

Consequently, the Arzela-Ascoli Theorem  guarantees
 that  $\{x_{1,n}(t)\}$ and $\{x'_{1,n}(t)\}$ are relatively compact
in $C[0,1]$;  i.e., there is a
function $x_{1,0}\in C^1[0,1]$, and a subsequence $\{x_{1,n_j}\}$ of
$\{x_{1,n}\}$ such that
 $$
\lim_{j\to +\infty}\max_{t\in[0,1]}|x_{1,n_j}(t)-x_{1,0}(t)|=0, \quad
\lim_{j\to +\infty}\max_{t\in[0,1]}|x'_{1,n_j}(t)-x'_{1,0}(t)|=0.
$$
Similar reasoning as in the proof of Theorem \ref{thm4.1} establishes that
$x_{1,0}$ is a positive solution to \eqref{e1.1} and \eqref{e1.2}.

Similarly, there is a convergent subsequence $\{x_{2,n_k}\}$ of
$\{x_{2,n}\}$ such that
$$
\lim_{k\to+\infty}|x_{2,n_k}(t)-x_{2,0}(t)|=0,\quad
\lim_{k\to+\infty}|x_{2,n_j}'(t)-x_{2,0}'(t)|=0
$$
and $x_{2,0}$ satisfies \eqref{e1.1}-\eqref{e1.2}.

Since $\|x_{1,0}\|_q=\max_{t\in[0,1]}|x_{1,0}(t)|\leq R_1$ and
$\|x_{2,0}\|_q=\max_{t\in[0,1]}|x_{2,0}(t)|\geq R_1$, a similar
argument to that used to show \eqref{e5.3} yields that $x_{1,0}$,
$x_{2,0}\not\in P\cap\partial\Omega_1$; i.e.,
$$
\|x_{1,0}\|_q=\max_{t\in[0,1]}|x_{1,0}(t)|<R_1,\quad
\|x_{2,0}\|_q=\max_{t\in[0,1]}|x_{2,0}(t)|>R_1.
$$
Consequently, $x_{1,0}$ and $x_{2,0}$ are different positive
solutions to \eqref{e1.1}-\eqref{e1.2}.
\end{proof}

\begin{example} \label{exa5.1} \rm
 Consider the three-point boundary value problems
\begin{gather*}
y''+\alpha(1-t)^a[1+(-y')^e+(-y')^{-a}][1+y^b+y^{-d}]=0,\quad
t\in(0,1),\label{e5.12}\\
y'(0)=0,y(1)=\frac{1}{2}y(\frac{1}{2}) \label{e5.13}
\end{gather*}
where $1\geq e\geq 0$, $a>0$, $b\geq 0$, $d>0$ and $\alpha>0$. Then
there is a $\alpha_0>0$ such that \eqref{e5.11}-\eqref{e5.12} has at
least two positive solutions
 $y_{1,0}$, $y_{2,0}\in C[0,1]\cap C^2(0,1)$ with
$y_{1,0}(t)>0$, $y_{2,0}(t)>0$  on
 $[0,1]$ and  $y_{1,0}'(t)<0$,
 $y_{2,0}'(t)<0$ on (0,1) for all $0<\alpha\leq\alpha_0$.

Let $a(t)\equiv \mu$, $\Phi(t)=(1-t)^a$ for all $t\in[0,1]$,
$h(x)=1+x^b+x^{-d}$ for $x\in(0,+\infty)$ and $g(z)=1+z^{e}+z^{-a}$
for $z\in(0,+\infty)$. From the proof of Lemma \ref{lem2.4}, we have
$c_0=\frac{1}{2}\min\{\frac{1}{3},\frac{1}{a+1}(\frac{1}{2}
-\frac{1}{a+2}(\frac{1}{2})^{a+2})\}$, and then
$\alpha(1-t)^a[1+y^b+y^{-d}]\leq\alpha (1-t)^a[1+y^{b}+{c_0}^{-d}]$
for all $y\in[c_0,+\infty)$. Let
$I(z)=\int_0^z\frac{1}{1+r^{e}+r^{-a}}dr$. Thus there exists an
$\alpha_0$ such that
$$
\frac{I(\frac{1}{3})}{\alpha\frac{1}{a+1}\sup_{c_0\leq r\leq1}h(r)}>1, \quad
\forall \alpha\in(0,\alpha_0]
$$
and then
$$
\sup_{c_0\leq c<+\infty}\frac{c}{3I^{-1}(\sup_{c_0\leq r\leq
c}h(r)\int_0^1a(s)\Phi(s)ds)}>1.
$$
Hence, the conditions (P1),
(P2) and (P3) hold. Thus Theorem \ref{thm5.1} guarantees that
\eqref{e5.12}-\eqref{e5.13} has at least two positive solutions.
\end{example}


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 \end{document}