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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 63, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/63\hfil Positive solutions]
{Positive solutions for singular three-point boundary-value
problems with sign changing nonlinearities depending on $x'$}

\author[Y. Chen, B. Yan, L. Zhang\hfil EJDE-2007/63\hfilneg]
{Yun Chen, Baoqiang Yan, Lili Zhang}

\address{Yun Chen \newline
Department of Mathematics, Shandong Normal
University, Jinan, 250014,  China}
\email{chenyun001982@126.com}

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

\address{Lili Zhang \newline
Department of Mathematics, Shandong Normal
University, Jinan, 250014,  China}
\email{kuaile100@163.com}

\thanks{Submitted September 15, 2006. Published April 25, 2007.}
\thanks{Supported by grants 10571111 from the National Natural Science,
 and Y2005A07 from the \hfill\break\indent Natural Science of Shandong Province}
\subjclass[2000]{34B15, 34B10}
\keywords{Three-point boundary value problem; singularity;
        positive solutions; \hfill\break\indent fixed point theorem}

\begin{abstract}
 Using a fixed point theorem in cones, this paper shows 
 the existence of positive solutions for the singular
 three-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)=\alpha x(\eta),
 \end{gather*}
 where $0<\alpha <1$, $0<\eta<1 $, and $f$ may change sign and may
 be singular at $x=0$ and $x'=0$.
\end{abstract}

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

\section{Introduction}

The study of multi-point boundary value problem (BVP) for linear
second-order ordinary differential equations was initiated by Il'in
and Moiseev \cite{i1,i2}. Since then, many authors studied more
general nonlinear multi-point BVPs, for example \cite{g1,l1,m1}, and
references therein. Recently, Liu \cite{l1} proved the existence of
positive solutions for the three-point BVP
\begin{gather*}
y''(t)+a(t)f(y(t))=0,\quad 0<t<1,\\
y'(0)=0,\quad y(1)=\beta y(\eta),
\end{gather*} 
where $0< \beta <1$, $0<\eta<1$ and $f:[0,+\infty)\to [0,+\infty)$
has no singularity at $y=0$.
Guo and Ge \cite{g1} presented the existence of positive solutions
for the three-point BVP
\begin{gather*}
x''(t)+f(t,x,x')=0,\quad 0<t<1,\\
x(0)=0,\quad x(1)=\beta x(\eta),
\end{gather*}  
where $\beta\eta\in(0,1)$, $0<\eta<1$ and
$f\in C([0,1]\times[0,+\infty)\times R, [0,+\infty))$
 has no singularity at $t=0$, $x=0$ and $x'=0$.

Motivated by the works of \cite{i2,l1}, in this paper, we discuss the
equation
\begin{equation} \label{e1.1}
\begin{gathered}
x''(t)+a(t)f(t,x(t),x'(t))=0, \quad   0<t<1,\\
x'(0)=0,\quad x(1)=\alpha x(\eta),
\end{gathered}
\end{equation}
where $0< \alpha <1$, $0<\eta<1$, $f$ may change sign and may be
singular at $x=0$ and $x'=0$.

The  features in this article, that different from those in \cite{g1,l1},
are as follows: First, the nonlinearity  $a(t)f(t,x,x')$ may be singular
at $t=0$, $t=1$, $x=0$ and $x'=0$; also the degree of singularity in
$x$ and $x'$ may be arbitrary; i. e., if $f$ contains
$\frac{1}{x^{\alpha}}$ and $\frac{1}{(-x')^{\gamma}}$, $\alpha$ and
$\gamma$ may be big enough). Second, $f$ is allowed to change sign.

 The paper is organized as follows. In the next section, we present
some preliminaries. Section  3 is devoted to our main result,
Theorem \ref{thm3.1}. An example is also given to illustrate the main result.
Some of the idea used here come from \cite{m1,y1}.

\section{Preliminaries}

 In this paper, we assume the following conditions
\begin{itemize}
\item[(P1)] $f(t,x,y) \in C((0,1) \times (0,+\infty) \times (-
\infty,0),(-\infty,+\infty))$;

\item[(P2)] $\beta(t),a(t),k(t) \in C((0,1),(0,+\infty))$,
$F(x)\in C((0,+ \infty),(0,+\infty))$,
$G(y)\in C((-\infty,0),(0, +\infty)),a(t)k(t)\in L[0,1]$;

\item[(P3)] $0< \alpha <1$, $0<\eta<1$ and   $| f(t,x,y)| \leq k(t)F(x)G(y)$;

\item[(H1)] There  exists  $\delta>0 $ such that $ f(t,x,y)\geq
 \beta(t),y\in(-\delta,0)$;

\item[(H2)] $\sup F[z,+\infty)=\sup\{F(x),z\leq x < +\infty\}< + \infty $
for all fixed $z\in(0,+\infty)$;

\item[(H3)] $ \frac{1}{ G(y)}\not\in  L (-\infty,-1]$;
\end{itemize}

\begin{lemma}[\cite{d1}] \label{lem2.1}
 Let $E$ be a Banach space, $K $ a cone of $E$, and
$B_R=\{x\in E: \| x\| < R\}$, where $0<r<R$. Suppose that
$ F: K \cap \overline{ B_R \backslash B_r}= K_{R,r}\to K$ is a
completely continuous operator and the following two conditions are
satisfied
\begin{enumerate}
\item $\| F(x)\| \geq \| x\| $ for any $ x\in K$ with $\| x\| =r$.

\item If $x \not = \lambda F(x)$ for any $x\in K$ with $\| x\| = R$
and $0<\lambda<1$.
\end{enumerate}
 Then $F$ has a fixed point in $K_{R,r}$.
\end{lemma}

\begin{lemma} \label{lem2.2}
 For each  natural number $n>0$, there exists
$y_n(t)\in C[0,1]$ with $y_n(t) \leq  -\frac{1}{n}$ such that
\begin{equation}
y_n(t)=-\frac{1}{n}+\min\{0,-\int^t_0 a(s)f(s,Ay_n(s)
+\frac{1}{n},y_n(s))ds\},\quad  t\in[0,1]. \label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}
 For $y(t)\in P =\{y(t):y(t)\leq 0,y(t)\in C[0,1]\}$, define the operator
\begin{gather*}
Ty(t)=-\frac{1}{n}+\min\{0,-\int^t_0 a(s)f(s,Ay(s)+\frac{1}{n},
\min\{y(s),-\frac{1}{n}\})ds\},\\
A y(s)=\frac{1}{1- \alpha }\int^1_0 -y(\tau)d\tau
-\frac{\alpha}{1-\alpha} \int ^\eta_0 -y(\tau)d\tau
-\int^s_0 -y(\tau)d\tau,
\end{gather*}
where $n>0$ is a natural number. Using the equality
$\min\{c,0\}=\frac{c-|c|}{2}$ and
$$
c(y(t))=-\int^t_0 a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}\})ds,
$$
it is easy to know that
$$
Ty(t)=-\frac{1}{n}+\frac{c(y(t))-|c(y(t)|}{2}.
$$
Let $y_k(t),y(t)\in P,\| y_k-y\|  \to 0$, then there exists a
constant $h>0$, such that $\| y_k\| \leq h$ and $\| y\| \leq h$, and let
$$
c(y_k(t))=-\int^t_0 a(s)f(s,Ay_k(s)+\frac{1}{n},\min\{y_k(s),-\frac{1}{n}\})ds,
$$
which yields
\begin{align*}
|Ty_k(t)-Ty(t)|
&=\frac12 \big|c(y_k(t))-c(y(t))-|c(y_k(t))|+|c(y(t))|\big|\\
&\leq \frac 12\big|c(y_k(t))-c(y(t))+|c(y_k(t))-c(y(t))|\big|.
\end{align*}
Assumption (P1) implies that
$\{a(s)f(s,Ay_k(s)+\frac{1}{n},\min\{y_k(s),-\frac{1}{n}\}$
converges to $\{a(s)f(s,Ay(s)+\frac{1}{n},
\min\{y(s),-\frac{1}{n}\}$,
 for $s\in(0,1)$. By the Lebesgue dominated convergence theorem
(the dominated function
$a(s)k(s)F[\frac{1}{n},+\infty)G[-h-\frac{1}{n},-\frac{1}{n}]$),
$|Ty_k(t)-Ty(t)|\to 0$,
$T$ is a  continuous operator in $P$.

Let $C$ be a bounded set in $P$, i.e., there exists $h_1>0$ such
that $\| y\| \leq h_1$, for any
$y(t)\in C,t_1,t_2\in[0,1]$, $t_1<t_2$, $y(t)\in P$,
\begin{align*}
|Ty(t_2)-Ty(t_1)|
&=\frac12 \big|-\int^{t_2}_{t_1}
a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}\}\\
&\quad +\int^{t_2}_{t_1}
a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}|)ds \big|\\
&\leq \big|-\int^{t_2}_{t_1} a(s)k(s)ds
F[\frac{1}{n}+\infty)G[-h_1-\frac{1}{n},
-\frac{1}{n}]\\
&\quad +\int^{t_2}_{t_1}a(s)k(s)ds F[\frac{1}{n},
+\infty)G[-h_1-\frac{1}{n},-\frac{1}{n}]|\big| .
\end{align*}
According to the absolute continuity of the Lebesgue integral, for
any $\epsilon>0$, there exists $\delta>0$ such that, when
$|t_2-t_1|<\delta,|\int^{t_2}_{t_1}a(s)k(s)ds<\epsilon$ holds.
Therefore, $\{Ty(t),y(t)\in P\}$ is equicontinuous. Hence $T$ is a
completely continuous operator in $P$.

 By (H3), we may choose a sufficiently large $R_n>1$ to fit
$$
\int^{-1}_{-R_n}\frac{dy}{G(y)}\geq\int_0^t a(s)k(s)ds \sup F
[\frac{1}{n},+\infty).
$$
For any fixed $n$, we prove that
\begin{equation}
y(t)\neq \lambda Ty(t)=\frac{-\lambda}
{n}+\lambda\min\{0,-\int^t_0
a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}\})ds\}\label{e*}
\end{equation}
for any $y(t)\in P$ with $\| y\| = R_n$ and $0<\lambda<1$.

In fact, if there exist $y(t)\in P$ with $\| y\| = R_n$ and
$0<\lambda<1$, such that
\begin{equation}
y(t)=\frac{-\lambda}{n} +\lambda\min\{0,-\int^t_0
a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}\})ds\}. \label{e2.2}
\end{equation}
First, we prove an important fact: for $t,z\in[0,1]$,
$t>z$, $y(t)<y(z)\leq - \frac{1}{n}$,
 \begin{equation}
\int_{y(t)}^{y(z)}\frac{dy}{G(y)}\leq \int_z^t a(s)k(s)ds
\sup F[Ay(t)+\frac{1}{n},+\infty). \label{e2.3}
\end{equation}
Let $t'\in(0,t]$ such that $ y(t')=y(t),y(s) \geq y(t'),s\in(0,t']$.
We may choose $\{t_i\}(i=1,2,\dots,2m)$ to fit

$1)$ $t' = t_1 > t_2 \geq t_3> t_4 \geq t_5> \dots \geq
t_{2m-1}>t_{2m} = z\geq 0$;
\begin{enumerate}
\item  $y( t_1)=y(t')$, $y ( t_{2i} ) = y( t_{2i+1})$,
$i=1,2,\dots m-1$, $y(t_{2m})=y(z)$;

\item $y(t)$ is decreasing in $[t_{2i},t_{2i-1}]$, $i=1,2,\dots m$.
(if $y(t)$ is  decreasing in $ [0,t']$. Let $m=1$, i.e.
$[t_2,t_1]=[0,t']$.)
\end{enumerate}
Note that $y(t)<-\frac{1}{n}$, $t\in(t_{2i},t_{2i-1}]$, which implies
$$
-\int^t_0 a(s)f(s,Ay(s)+\frac{1}{n},\min\{y(s),-\frac{1}{n}\})ds<0,
\quad t\in(t_{2i},t_{2i-1}].
$$
 Differentiating \eqref{e2.2} and using (H2), we obtain
\begin{gather*}
-y'(t)= \lambda a(t)f(t,Ay(t)+\frac{1}{n},y(t))\\
\frac{ -y'(t)}{ G (y(t))} \leq a(t)k(t) \sup
F[Ay(t)+\frac{1}{n},+\infty)
\leq a(t)k(t) \sup F[\frac{1}{n},+\infty),
\end{gather*} %\label{e2.4}
for $t\in(t_{2i},t_{2i-1}]$, $i=1,2,\dots m$.
Integrating from $t_{2i}$ to $t_{2i-1}$, we have
$$
\int^{y(t_{2i})}_{y(t_{2i-1})}\frac{dy}{G(y)}\leq
\int^{t_{2i-1}}_{t_{2i}} a(s)k(s)ds\sup F[\frac{1}{n},+\infty),
     \quad  i=1,2,\dots m.
$$      % \label{e2.5}
Summing  from $m$ to $1$, we have
$$
\int_{y(t)}^{y(z)}\frac{dy}{G(y)}\leq \int_z^t a(s)k(s)ds\sup
F[\frac{1}{n},+\infty).
$$
Set $y(z)=-\frac{1}{n}$, $y(t)=-R_n$ in
\eqref{e2.3}, we have
$$
\int_{-R_n}^{-1}\frac{dy}{G(y)}\leq \int_{-R_n}^{-\frac{1}{n}}\frac{dy}{G(y)}
\leq \int^t_0 a(s)k(s)ds\sup F[\frac{1}{n},+\infty),
$$
which contradicts
$$
\int_{-R_n}^{-1}\frac{dy}{G(y)}\geq \int^t_0
a(s)k(s)ds\sup F[\frac{1}{n},+\infty).
$$
Hence \eqref{e*} holds. Put $r=\frac{1}{n}$, Lemma \ref{lem2.1} leads to the
desired result.
\end{proof}

\section{Main results}

Main result in this paper is as follows.

\begin{theorem} \label{thm3.1}
 Let {\rm (H1)--(H3)} hold. Then the three-point boundary-value
 problem \eqref{e1.1} has at least one positive solution.
\end{theorem}

\begin{proof} Put $M_n=\min\{y_n(t):t\in[0,\eta]\}$. (H1) implies
$\gamma =\sup\{M_n\}<0$.
Set $\tau=\max\{\gamma,-\delta\},n >-\frac{1}{\tau}$.

(1) First, we prove that
\begin{equation}
y_n(t)=-\frac{1}{n}-\int^t_0 a(s)f(s,Ay_n(s)+\frac{1}{n},y_n(s))ds, \quad
 t\in[0,1].  \label{e3.1}
\end{equation}
Set $y_n(t_n)=\tau,t_n\in(0,\eta] ,y_n(t)\geq\tau,t\in[0,t_n]$. We
easily check that $y_n(t)$ is decreasing in $(0,t_n]$. We only need
to prove that
\begin{equation}
y_n(t)\leq \tau,\quad t\in[t_n,1]. \label{e3.2}
\end{equation}
 If there exist $t\in(t_n,1]$ such that $y_n(t)> \tau$, then we may choose
$t',t''\in[t_n,1], t'<t''$ to fit
$y_n(t')=\tau,\tau< y_n(t)< -\frac{1}{n}, t\in(t',t'']$,
we have from \eqref{e2.1}
$$
0<\int^{t''}_{t'}a(s)f(s,Ay_n(s)+\frac{1}{n},y_n(s))ds=y_n(t')-y_n(t'')<0.
$$
This contradiction implies \eqref{e3.2}.

Using $y_n(t),1$ and $0$ in place of $y(t), \lambda$ and $z$ in
\eqref{e2.2} in Lemma  \ref{lem2.2}, we notice that
\begin{align*}
A y_n(t)+ \frac{1}{n}
&=\frac{1}{1- \alpha }\int^1_0 -y_n(\tau)d\tau
  -\frac{\alpha}{1-\alpha} \int ^\eta_0 -y_n(\tau)d\tau
  -\int^t_0-y_n(\tau)d\tau + \frac{1}{n}\\
&>\frac{\alpha}{1-\alpha}\int^1_\eta-y_n(\tau)d\tau \\
&\geq \frac{\alpha}{1-\alpha}(-\tau)(1-\eta),\quad t\in[0,1].
\end{align*}
 From \eqref{e2.3}, putting $t=t_n$, we know that
\begin{equation}
\int_{y_n(t_n)}^{-\frac{1}{n}}\frac{d y_n}{G(y_n)}
 \leq \int^{t_n}_0a(s)k(s)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty). \label{e3.3}
\end{equation}
Equation \eqref{e3.3} shows $t_0=\inf\{t_n\}>0$. Also,
 $y_n(t)$ is decreasing for $t\in(0,t_0]$ and (H1) imply that
$W(t)=\sup\{y_n(t)\}<0$, $t\in(0,t_0]$.

(2) We show that $\{y_n(t)\}$ is equicontinuous on
$[\frac{1}{3k},1-\frac{1}{3k}]$, for a natural number $k\geq 1$,
 and uniformly bounded on $[0,1]$.

 Using $y_n(t)$, $1$ and $0$ instead of
$y_(t)$, $\lambda$ and $z$ in \eqref{e2.2} in Lemma \ref{lem2.2}, we notice that
$$
A y_n(t)+ \frac{1}{n}\geq
\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),\quad t\in [0,1].
$$
We know from \eqref{e2.3},
\begin{equation}
\int_{y_n(t)}^{-\frac{1}{n}}\frac{d y_n}{G(y_n)}\leq \int^t_0a(s)k(s)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) , \quad
t\in [0,1] . \label{e3.4}
\end{equation}
Now we use (H3) and \eqref{e3.4} show that $\omega(t)=\inf \{y_n(t)\}>-\infty$
 is bounded on $[0,1]$. On the other hand, it follows from \eqref{e3.1} and
\eqref{e3.2} that
\begin{equation}
|y'_n(t)|\leq k(t)a(t)\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) \sup G
[\omega_k,\max\{\tau,W(\frac{1}{k})\}],\   (n \geq k).  \label{e3.5}
\end{equation}
Where $\omega_k =\inf\{\omega(t)$,
$t\in [\frac{1}{3k},1-\frac{1}{3k}] \}$. Thus \eqref{e3.5} and the absolute
 continuity of Lebesgue integral show that $ \{y_n(t)\}$ is
 equicontinuous on $[\frac{1}{3k},1-\frac{1}{3k}]$. Now the
 Arzela-Ascoli theorem guarantees that there exists a subsequence of
 $ \{y_n(t)\}$, which converges uniformly on $[\frac{1}{3k},1-\frac{1}{3k}]$.
When $k=1$, there exists a subsequence $\{{y_n^{(1)}}(t)\}$ of
$\{y_n(t)\}$, which converges uniformly on
$[\frac{1}{3},\frac{2}{3}]$. When $k=2$, there exists a subsequence
$\{{y_n^{(2)}}(t)\}$ of $\{{y_n^{(1)}}(t)\}$, which converges
uniformly on $[\frac{1}{6},\frac{5}{6}]$. In general, there exists a
subsequence $\{{y_n^{(k+1)}}(t)\}$ of $\{{y_n^{(k)}}(t)\}$, which
converges uniformly on $[\frac{1}{3(k+1)},1-\frac{1}{3(k+1)}]$. Then
the diagonal sequence $ \{y_k^{(k)}(t)\}$ converges everywhere in
$(0,1)$ and it is easy to verify that $ \{y_k^{(k)}(t)\}$ converges
uniformly on any interval $[c,d]\subseteq (0,1)$. Without loss of
generality, let  $ \{y_k^{(k)}(t)\}$  be itself of  $ \{y_n(t)\}$ in
the rest. Put $y(t)=\lim_{n\to\infty}y_n(t),t\in(0,1)$. Then
$y(t)$ is continuous in $(0,1)$ and $y(t)<0,t\in(0,1)$.

(3) Now \eqref{e3.4} shows that
$$
\sup\{\max\{-y_n(t),t\in[0,1]\}\}<+\infty.
$$
We have
\begin{equation}
\lim_{t\to 0+}{\sup\{\int^t_0-{{y_n}(s)}ds\}}=0, \quad
\lim_{t\to 1-}{\sup\{\int_t^1-{{y_n}(s)}ds\}}=0,  \label{e3.6}
\end{equation}
and we obtain
\begin{equation}
\begin{aligned}
A y_n(t)
&=\frac{1}{1- \alpha }\int^1_0-y_n(\tau)d\tau
-\frac{\alpha}{1-\alpha} \int ^\eta_0 -y_n(\tau)d\tau
-\int^t_0 -y_n(\tau)d\tau \\
&<\frac{1}{1- \alpha }\int^1_0 -y_n(\tau)d\tau
 < +\infty, \quad t\in[0,1].
\end{aligned}  \label{e3.7}
\end{equation}
Since \eqref{e3.6} and \eqref{e3.7} hold,  Fatou's theorem of the Lebesgue
integral implies $Ay(t)<+\infty$, for any fixed $t\in(0,1)$.

(4) $y(t)$ satisfies
$$
y(t)=-\int^t_0 a(s)f(s,Ay(s),y(s))ds, \quad    t\in(0,1). % \label{e3.8}
$$
Since $y_n(t)$ converges uniformly on $[a,b]\subset(0,1)$, \eqref{e3.6}
 leads that ${A y_n(s)}$ converges to $Ay(s)$ for any
$s\in(0,1)$. For each fixed $t\in(0,1)$, thee exists $d>0$ such
that $0<d<t$, then
$$
y_n(t)-y_n(d)=-\int^t_d a(s)f(s,Ay_n(s)+\frac{1}{n},y_n(s))ds.% \label{e3.9}
$$
for all $n>k$. Since $y_n(s)\leq \max \{\tau,W(d)\}$, $A y_n(s)+
\frac{1}{n}\geq \frac{\alpha}{1-\alpha}(-\tau)(1-\eta)$, $s\in
[d,t]$, the set $\{A y_n(s)\}$ or $\{y_n(s)\}$ is bounded and
equicontinuous on $[d,t]$. Let $n\to\infty$
\begin{equation}
y(t)-y(d)=-\int^t_d a(s)f(s,Ay(s),y(s))ds.  \label{e3.10}
\end{equation}
 Putting  $t=d$ in \eqref{e3.4}, we have
$$
\int_{y_n(d)}^{-\frac{1}{n}}\frac{d y_n}{G(y_n)}\leq \int^d_0a(s)k(s)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) . % \label{e3.11}
$$
Let $n\to\infty$ and  $d\to 0+$, we obtain
$$ y(0+)=\lim_{d\to 0+}y(d)=0.
$$
Letting $d\to0+$ in \eqref{e3.10}, we have
\begin{equation}
y(t)=-\int^t_0 a(s)f(s,Ay(s),y(s))ds ,\quad t\in(0,1), \label{e3.12}
\end{equation}
and $Ay(1)=\alpha Ay(\eta)$.
Hence $x(t)=Ay(t)$ is a positive solution of \eqref{e1.1}.
\end{proof}

\begin{corollary} \label{coro3.1}
 Suppose that {\rm (H1)-(H3)} hold , then the
set of  positive solutions of \eqref{e1.1} is  compact.
\end{corollary}

\begin{proof} Let $M=\{y\in C[0,1]:Ay(t)
\text{is a positive solution of \eqref{e1.1}} \}$. First we show that
 $M$ is compact. Note that
(1) $M$ is not empty;
(2) $M$ is relatively compact(bounded, equicontinuous).
(3) $M$ is closed.

Obviously Theorem \ref{thm3.1} implies $M$ is not empty.

First we show that $M\in C[0,1]$ is relatively compact.  For any
$y(t)\in M$, differentiating \eqref{e3.12} and using (H2), we obtain
\begin{gather*}
-y'(t)= \lambda a(t)f(t,Ay(t),y(t)) \\
\begin{aligned}\frac{ -y'(t)}{ G (y(t))}
&\leq a(t)k(t) \sup F[Ay(t),+\infty)\\
&\leq a(t)k(t) \sup F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty),
\quad t\in [0,1].
\end{aligned}
\end{gather*} %\label{e3.13}
Integrating  from $0$ to $t$, we have
\begin{equation}
\int_{y(t)}^{0}\frac{d y}{G(y)}\leq \int^1_0a(s)k(s)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) , \quad
t\in [0,1] . \label{e3.14}
\end{equation}
 Now (H3) and  \eqref{e3.14} show that for any $y(t)\in M$, there exists
 $K>0$ such that $|y(t)|<K$, for all $t\in[0,1]$. Then $M$ is bounded.

 For each $y(t)\in M$, we obtain from \eqref{e3.12},
\begin{align*}
 -y'(t)&=a(t)f(t,Ay(t),y(t))\\
&\leq a(t)|f(t,Ay(t),y(t))|\\
&\leq a(t)k(t)F [\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) G(y(t)),
\quad t\in(0,1) ,
\end{align*}
and
\begin{align*}
 y'(t)&=-a(t)f(t,Ay(t),y(t))\\
&\leq a(t)|f(t,Ay(t),y(t))|\\
&\leq a(t)k(t)F [\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) G(y(t)),
\quad t\in(0,1) ,
\end{align*}
which yields
\begin{gather}
\frac{ -y'(t)}{G(y(t))+1}\leq a(t)k(t)\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty),\quad t\in(0,1), \label{e3.15}
\\
\frac{ y'(t)}{G(y(t))+1}\leq a(t)k(t)\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty),\quad t\in(0,1).
\label{e3.16}
\end{gather}
Note that the right-hand sides of the above inequalities are always positive.
 Let $I(y(t))=\int^{y(t)}_0\frac{dy}{G(y)+1}$,
for any $t_1,t_2\in[0,1]$. Integration from $t_1$ to $t_2$ in
\eqref{e3.15} and \eqref{e3.16} yields
\begin{equation}
|I(y(t_1))-I(y(t_2))|\leq\int^{t_2}_{t_1}a(t)k(t)
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty)dt. \label{e3.17}
\end{equation}
Since $I^{-1}$ is uniformly continuous on $[I(-K),0]$, for any
$\overline{\epsilon}>0$, there is a $\epsilon'>0$ such that
\begin{equation}
|I^{-1}(s_1)-I^{-1}(s_2)|<\overline{\epsilon},\forall|s_1-s_2|
<\epsilon',s_1,s_2\in[I(-K),0]. \label{e3.18}
\end{equation}
Inequality \eqref{e3.17} guarantees that for $\epsilon'>0$, there is
 a $\delta'>0$ such that
$$
|I(y(t_1))-I(y(t_2))|<\epsilon',\forall|t_1-t_2|<\delta',\quad
t_1,t_2\in[0,1]. %\label{e3.19}
$$
This inequality and \eqref{e3.18}imply
$$
|y(t_1)-y(t_2)|=|I^{-1}(I(y(t_1))-I^{-1}(I(y(t_2))|<\overline{\epsilon},\quad
 t_1,t_2\in[0,1],
$$ %\label{e3.20}
which means that $M$ is equicontinuous. So $M$ is relatively
compact.

Second, we show that $M$ is closed. Suppose that
$\{y_n\}\subseteq M$ and
$$
\lim_{n\to+\infty}\max_{t\in[0,1]}|y_n(t)-y_0(t)|=0.
$$
Obviously $y_0\in C[0,1]$ and $\lim_{n\to+\infty}Ay_n(t)=Ay_0(t)$,
$t\in[0,1]$. Moreover,
\begin{align*}
Ay_n(t)&=\frac{1}{1- \alpha }\int^1_0-y_n(\tau)d\tau
-\frac{\alpha}{1-\alpha} \int ^\eta_0 -y_n(\tau)d\tau
-\int^t_0 -y_n(\tau)d\tau \\
&<\frac{1}{1- \alpha }\int^1_0 -y_n(\tau)d\tau \\
&< \frac{K}{1- \alpha }, \quad t\in[0,1].
\end{align*} % \label{e3.21}
For $y_n(t)\in M$, from \eqref{e3.12} we obtain
$$
y_n(t)=-\int^t_0 a(s)f(s,Ay_n(s),y_n(s))ds ,\quad
 t\in(0,1).
$$% \label{e3.22}
For fixed $t\in(0,1)$, there exists $d>0$ such that $0<d<t$ , then
$$
y_n(t)-y_n(d)=-\int^t_d a(s)f(s,Ay_n(s),y_n(s))ds.
$$% \label{e3.23}
Since $y_n(s)\leq \max \{\tau,W(d)\}$,
$A y_n(s)\geq \frac{\alpha}{1-\alpha}(-\tau)(1-\eta)$, $s\in[d,t]$,
the Lebesgue dominated convergence theorem yields
\begin{equation}
y_0(t)-y_0(d)=-\int^t_d a(s)f(s,Ay_0(s),y_0(s))ds, \quad t\in(0,1).
 \label{e3.24}
\end{equation}
 From \eqref{e3.12}, we have
$$
-y'_n(t)=a(t)f(t,Ay_n(s),y_n(s))
\leq a(t)k(t)F [\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty)
G(y_n(t)),
$$
which yields
$$
\frac{ -y'_n(t)}{G(y_n(t))}\leq a(t)k(t)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty),\quad t\in(0,1),
$$
integrating from 0 to $d$,
$$
\int_{y_n(d)}^0\frac{d y_n}{G(y_n)}\leq \int^d_0a(s)k(s)ds\sup
F[\frac{\alpha}{1-\alpha}(-\tau)(1-\eta),+\infty) .
$$ 
 Let $n\to\infty$ and  $d\to0+$, we
obtain
$y_0(0+)=\lim_{d\to 0+}y_0(d)=0$.
Letting $d\to0+$ in \eqref{e3.24}, we have
\begin{gather*}
y_0(t)=-\int^t_0 a(s)f(s,Ay_0(s),y_0(s))ds ,t\in(0,1),
Ay_0(1)=\alpha Ay_0(\eta),
\end{gather*}
 then $x_0(t)=Ay_0(t)$ is a positive solution of \eqref{e1.1}.
So $y_0(t)\in M$  and $M$ is a closed set.
Hence $\{Ay(t),y(t)\in M\}\in C^1[0,1]$ is compact.
\end{proof}

\begin{example} \label{exa3.1} \rm
 In \eqref{e1.1}, let
$$
f(t,x,y)=k(t)  [1+ x^{-\gamma} +
(-y)^{-\sigma}- (-y) \ln (- y )],a(t) =t^{-\frac{1}{3}} ,
$$
 and
$k(t) =t^{-\frac{1}{2}}$, $0<t < 1$,
where $ \gamma>0$, $\sigma \geq 0$, and let
$F(x)=1+x^{-\gamma},G(y)=1+ (-y)^{-\sigma}+(-y)\ln(-y)$. Then
\begin{gather*}
f(t,x,y)\leq k(t)F(x)  G(y), \quad \delta = -1,\quad \beta (t)= k(t),
\\
\int^{-1}_{-\infty}\frac{dy}{G(y)}=+\infty.
\end{gather*}
By Theorem \ref{thm3.1}, equation \eqref{e1.1} has at least a positive
solution and Corollary \ref{coro3.1} implies the set of solutions is compact.
\end{example}

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\end{document}