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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 38, 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}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/38\hfil Positive solutions]
{Positive solutions for second-order singular
three-point boundary-value problems with sign-changing
nonlinearities}

\author[C. Ji, B. Yan\hfil EJDE-2010/38\hfilneg]
{Caisheng Ji, Baoqiang Yan}  % in alphabetical order

\address{Caisheng Ji \newline
Department of Mathematics, Shandong Normal University,
Jinan 250014,  China}
\email{jicaisheng@163.com}

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

\thanks{Submitted July 30, 2009. Published March 14, 2010.}
\thanks{Supported by grants 10871120 from the fund of National
Natural Science, J07WH08 from  \hfill\break\indent the
 Shandong Education Committee, and Y2008A06 from the  Shandong
Natural Science} 
\subjclass[2000]{34B10, 34B15} 
\keywords{Singular three-point boundary-value problem; sign-changing;
\hfill\break\indent  nonlinearity; positive solution; fixed point}

\begin{abstract}
 In this article, we study the existence and  uniqueness
 of the positive solution for a second-order singular
 three-point boundary-value problem with sign-changing
 nonlinearities. Our main tool is a fixed-point theorem.
\end{abstract}

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

\section{Introduction}

 In this article, we consider the second-order boundary-value problem
\begin{gather}
x''(t)+f(t,x(t))=0,\quad 0<t<1,\label{e1.1}\\
x(0)=0,\quad x(1)= \alpha x(\eta), \quad 0<\eta<1,\quad
  0<\alpha<1. \label{e1.2}
\end{gather}
The singularity may appear at $t=0$, $x=0$ and the function $f$
may be superlinear at $x=\infty$ and change sign.

 Webb \cite{w1} employed the fixed-point index for compact maps to
investigate the existence of at least one positive solution
for the  second-order boundary-value problem
\begin{equation}
\begin{gathered}
x''(t)+g(t)f(x(t))=0,\quad 0<t<1,\\
x(0)=0,\quad x(1)=\alpha x(\eta),
\end{gathered}\label{e1.3}
\end{equation}
 where $0<\eta<1$, $0<\alpha \eta <1$, and
$f_0=\limsup_{ x\to 0} \frac{f(x)}{x}$,
$f_\infty =\liminf_{ x\to \infty} \frac{f(x)}{x}$ exist and $g(t)>0$.
Moreover, when $g(t)$ is a sign-changing function in $[0,1]$
and $f$ is nondecreasing and without any singular points,
using the fixed point theorem of strict-set-contractions,
Bing Liu \cite{l1} established the existence of at least two positive
solutions for  \eqref{e1.3}. When $g(t)>0$ and $f$ is a
given sign-changing function without any singular points and
any monotonicity, using the increasing operator theory and
approximation process, Xian Xu \cite{x1} showed at least three solutions
for the three-point boundary-value problem \eqref{e1.3}.

 In addition, the existence of solutions of nonlinear multi-point
boundary-value problems have been studied by many other authors;
 the readers are referred to \cite{l1,m1,x2,x3}
 and the references therein.

Motivated by \cite{h1,y1}, the purpose of this article is to examine
the existence and the uniqueness of the positive solution of
 \eqref{e1.1}-\eqref{e1.2} under the assumption that $f$ may be
singular at $t=0$, $x=0$ and be superlinear at $x=\infty$
and change sign.
There are only a few papers considering
\eqref{e1.1}-\eqref{e1.2} under this assumptions.
We try to fill this gap in the literature with this paper.

 In this article, we use the following assumptions:
\begin{itemize}
\item[(H1)] $f(t,x)\in C((0,1]\times(0,+\infty),(-\infty,+\infty))$,

\item[(H2)] $k(t), a(t), b(t)\in C((0,1],(0,+\infty))$,
 $tk(t)\in L(0,1]$,

\item[(H3)] there exist $F(x)\in C((0,+\infty), (0,+\infty))$,
 $G(x)\in C([0,+\infty),[0,+\infty))$ such that
 $f(t,x)\leq k(t)(F(x)+G(x))$.

\item[(S1)] $f(t,x)\geq a(t)$ hold for $0<x<b(t)$, $x\in C[0,1]$,

\item[(S2)] $F(x)$ is decreasing in $(0,+\infty)$,

\item[(S3)] there exist $R>1$, such that
 $\int_1^R\frac{dy}{F(y)}\cdot (1+\frac{\bar{G}(R)}{F(R)})^{-1}
 >\int_0^1 sk(s)ds$, where
 $\bar{G}(R)=\max_{s\in [0,R]} G(s)$.

\end{itemize}
This paper is organized as follows. In Section 2, we give some
preliminaries. In Section 3, we obtain the existence of at least
one positive solution for \eqref{e1.1}-\eqref{e1.2}, and show an
application of our results.

\section{Preliminaries}

\begin{lemma}[\cite{d1}] \label{lem2.1}
Let $E$ be a Banach space, $R>0$, $B_R=\{x\in E:\|x\|\leq R\}$,
$F:B_R\to E$ be a completely continuous operator.
 If $x\neq \lambda F(x)$ for any $x\in E$ with $\|x\|=R$ and
$0<\lambda<1$, then $F$ has a fixed point in $B_R$.
\end{lemma}

  Let $n>[\frac{1}{\eta}+1]$ be a natural number,
$d_n=\min\{b(t):t\in[\frac{1}{n},1]\}$,
$b_n=\min\{d_n,\frac{1}{n}\}$, $C_n=\{x: x\in C[\frac{1}{n},1]\}$
with norm $ \|x\|=\max\{|x(t)|, \frac{1}{n}\leq t\leq 1\}$. It is
easy to see that ($C_n, \| \cdot \|$) is a Banach space.

  Inspired by \cite{y1}, we define $T_n$ as
$$
(T_nx)(t)=b_n+\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s)
f(s,\max\{b_n,x(s)\})ds,\quad x\in C_n,\; t\in[\frac{1}{n},1],
$$
where
\[
G_{\frac{1}{n},1}(t,s)=\begin{cases}
G_1(t,s), &\frac{1}{n}<\eta \leq s,\\
G_2(t,s), &\frac{1}{n}\leq s\leq \eta,
\end{cases}
\]

\begin{gather*}
G_1(t,s)=\begin{cases}
\frac{1}{1- \alpha \eta-(1- \alpha) \frac{1}{n}}(1-s)
(t- \frac{1}{n}),& \frac{1}{n}\leq t\leq s\leq 1,
\\
 \frac{1}{1- \alpha \eta-(1- \alpha)
\frac{1}{n}}[ \alpha (t-s)(\eta - \frac{1}{n})-(t-1)(s- \frac{1}{n})],
 &\eta \leq s\leq t \leq 1,
\end{cases}
\\
G_2(t,s)=\begin{cases}
\frac{(1-\alpha \eta)(t-\frac{1}{n})-s(1-\alpha)
(t-\frac{1}{n})}{1- \alpha \eta-(1- \alpha) \frac{1}{n}},
&\frac{1}{n}\leq t\leq s\leq 1,\\
\frac{(1-\alpha \eta)(s-\frac{1}{n})-t(1-\alpha)
(s-\frac{1}{n})}{1- \alpha \eta-(1- \alpha) \frac{1}{n}},
&\frac{1}{n}\leq s\leq t\leq 1,
\end{cases}
\end{gather*}
and $G_{\frac{1}{n},1}(t,s)$ is Green's function to the
 boundary-value problem
\begin{gather*}
x''(t)=0,\quad \frac{1}{n}<t<1,\\
x(\frac{1}{n})=0,\quad x(1)=\alpha x(\eta),\quad
 0<\alpha<1, \quad 0<\eta<1.
\end{gather*}
By a standard argument we have the following result;
see for example \cite{w2}.

\begin{lemma} \label{lem2.2}
The operator $T_n$ is  completely continuous  from $C_n$ to $C_n$.
\end{lemma}

\begin{lemma} \label{lem2.3}
There exist $x_n\in C_n$, $b_n\leq x_n(t)\leq R$
for $t\in[\frac{1}{n},1]$ such that
\begin{equation}
x_n(t)=b_n+\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s)f(s,x_n(s))ds,\quad
 t\in[\frac{1}{n},1]. \label{e2.1}
\end{equation}
\end{lemma}

\begin{proof}
We prove that
\begin{equation}
x(t)\neq \lambda (T_nx)(t)=\lambda b_n+\lambda
\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s)
f(s,\max\{b_n,x(s)\})ds, \quad t\in[\frac{1}{n},1], \label{e2.2}
\end{equation}
for any $\|x\|=R$ and $\lambda\in (0,1)$.
In fact, if \eqref{e2.2} is not true, there exist $x\in C_n$ with
$\|x\|=R$ and $0< \lambda <1$ such that
\begin{equation}
x(t)=\lambda (T_nx)(t)=\lambda b_n+\lambda
\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s)f(s,\max\{b_n,x(s)\})ds,\quad
  t\in[\frac{1}{n},1]. \label{e2.3}
\end{equation}
It is easy to see that $x(\frac{1}{n})=\lambda b_n$,
$x(1)-\alpha x(\eta)=(1-\alpha)\lambda b_n$.

 We  first claim that $x(t)\geq \lambda b_n$ for any
$t\in[\frac{1}{n},1] $. In fact if $x(\eta)<\lambda b_n$,
we have $x(1)=\lambda b_n+\alpha x(\eta)-\alpha \lambda b_n<\lambda b_n $
and $x(\eta)<x(1)$. Since $x(\frac{1}{n})=\lambda b_n>x(1)$, we can get
a point $t_1\in (\frac{1}{n},\eta)$ such that $x(t_1)=x(1)$.
Let $\gamma=\sup \{t_1:\ t_1\in (\frac{1}{n},\eta), x(t_1)=x(1)\}$.
It follows that $x(\gamma)=x(1)$ and $x(t)<x(\gamma)=x(1)$,
$t\in(\gamma,\eta)$.
 Since $x(\eta)<x(1)<\lambda b_n$, we have two cases:\\
Case (1). There exist $t_1'\in (\eta,1)$ such that $x(1)\leq x(t_1')$.
and\\
Case (2). $x(t)<x(1)$ for all $t\in (\eta,1)$.

 In case (1), we may get a point $t_2\in (\eta,t_1')$ such
that $x(t_2)=x(1)$. Setting
$\beta=\inf \{t_2: t_2\in (\eta,1), x(t_2)=x(1)\}$, we get
$x(\beta)=x(1)$ and $x(t)<x(\beta)=x(1), t\in(\eta,\beta)$.
In case (2), setting $\beta=1$, we also get $x(\beta)=x(1)$ and
$x(t)<x(\beta)=x(1), t\in(\eta,\beta)$.
 Hence, there exist an interval
$[\gamma,\beta]\subseteq (\frac{1}{n},1](\gamma<\beta)$ such that
\begin{equation}
x(\gamma)=x(\beta)<\lambda b_n, x(t)<x(\gamma), x(t)<x(\beta), \quad
 t\in (\gamma,\beta). \label{e2.4}
\end{equation}
By \eqref{e2.3} and (S1), we have
$x''(t)=-\lambda f(t,b_n)<0$, $t\in [\gamma,\beta]$ and
$x(t)$ is concave down on $[\gamma,\beta]$, which
contradicts \eqref{e2.4}.
Hence $x(\eta)\geq \lambda b_n$, and then
$x(1)\geq \lambda b_n, x(1)\leq x(\eta)$.
  If there exist $t^{'}_2\in (\frac{1}{n},\eta)$ such that
$x(t^{'}_2)<\lambda b_n$, a similar argument as before yields an
interval $[\gamma',\beta']\subseteq [\frac{1}{n},\eta](\gamma'<\beta')$,
such that
\begin{equation}
x(t)<x(\gamma'),\quad
x(t)<x(\beta'), \quad t\in (\gamma',\beta'), \quad
x(\gamma')\leq \lambda b_n, \quad
x(\beta')\leq \lambda b_n. \label{e2.5}
\end{equation}
It follows from \eqref{e2.3} and (S1) that
$x''(t)=-\lambda f(t,b_n)<0, t\in [\gamma',\beta'] $ and
$x(t)$ is concave down on $[\gamma',\beta']$, which
contradicts \eqref{e2.5}.
So we have $x(t)\geq \lambda b_n$, $t\in [\frac{1}{n},\eta]$.
By the same argument used for $t\in [\frac{1}{n},\eta]$, we
can easily show that $x(t)\geq \lambda b_n, t\in [\eta,1]$.

 Next we  claim that: for any $z\in (\frac{1}{n},1)$, if
$b_n<x(z)<R$, we have
\begin{equation}
\int_{b_n}^{x(z)}\frac{dx}{F(x)}\leq
\big(1+\frac{\bar{G}(R)}{F(R)}\big) \int_0^z\int_t^1k(s)\,ds\,dt.
\label{e2.6}
\end{equation}
Since $x(\frac{1}{n})=\lambda b_n<R$,
$x(1)\leq x(\eta)$, there exist $t^*\in(\frac{1}{n},1)$ such
that $x(t^*)=R$,
$x'(t^*)=0$. Setting $t'=\inf\{t^*:t^*\in (\frac{1}{n},1),
x'(t^*)=0, x(t^*)=\|x\|=R\}$, we obtain
 $t'\in (\frac{1}{n},1), x'(t')=0, x(t')=\|x\|=R$.
Obviously there exist
$t''\in(\frac{1}{n},t')$ such
that $x(t'')=b_n$. Furthermore we get a countable set
$\{t_i\}$ of $(\frac{1}{n},1)$ such that
\begin{enumerate}
\item $t''=t_1<t_2\leq t_3<t_4\leq t_5<\ldots \leq t_{2m-1}<t_{2m}\leq
\ldots <1$, $t_{2m}\to t'$,
\item $x(t_1)=b_n$, $x(t_{2i})=x(t_{2i+1})$, $x'(t_{2i})=0$,
$i=1,2,3\ldots$,
\item  $x(t)$ is strictly increasing in
$[t_{2i-1},t_{2i}]$, $i=1,2,3\ldots$ (if $x(t)$ is strictly
increasing in $[t'',t']$, put $m=1$; i.e, $[t_1,t_2]=[t'',t']$).
\end{enumerate}
  Differentiating \eqref{e2.3} and using the assumptions,
we obtain easily
\begin{equation}
\begin{aligned}
-x''(t)&=\lambda f(t,x(t))
 \leq \lambda k(t)(F(x(t))+G(x(t)))\\
&=\lambda k(t)F(x(t))(1+\frac{G(x(t))}{F(x(t))})\\
&<k(t)F(x(t))(1+\frac{\bar{G}(R)}{F(x(t))})\\
&\leq k(t)F(x(t))(1+\frac{\bar{G}(R)}{F(R)}), \quad
 t\in[t_{2i-1},t_{2i}), \; i=1,2,3\ldots.
\end{aligned} \label{e2.7}
\end{equation}
Integrating  \eqref{e2.7} from $ t$ to $t_{2i}$, we have by the
decreasing property of $F(x)$,
\begin{equation}
\begin{aligned}
-\int_t^{t_{2i}}x''(s)ds
&\leq (1+\frac{\bar{G}(R)}{F(R)}) \int_t^{t_{2i}}k(s)F(x(s))ds\\
&\leq F(x(t))(1+\frac{\bar{G}(R)}{F(R)})\int_t^{t_{2i}}k(s)ds,
\end{aligned}\label{e2.8}
\end{equation}
for $t\in[t_{2i-1},t_{2i})$, $i=1,2,3\ldots$;
that is to say
\begin{equation}
x'(t)\leq F(x(t))(1+\frac{\bar{G}(R)}{F(R)})\int_t^{t_{2i}}k(s)ds , \quad
 t\in[t_{2i-1},t_{2i}), \; i=1,2,3\ldots. \label{e2.9}
\end{equation}
It follows from \eqref{e2.9} that
\begin{equation}
\frac{x'(t)}{F(x(t))}\leq (1+\frac{\bar{G}(R)}{F(R)})
\int_t^{t_{2i}}k(s)ds\leq (1+\frac{\bar{G}(R)}{F(R)})\int_t^1k(s)ds,
\label{e2.10}
\end{equation}
for $t\in[t_{2i-1},t_{2i})$, $i=1,2,3\ldots$.

On the other hand, we can choose $i_0$ and
$z'\in (\frac{1}{n},1), z'\leq z$ such that
$z'\in[t_{2i_0-1},t_{2i_0})$ and $x(z')=x(z)$.
Integrating  \eqref{e2.10} from $t_{2i-1}$ to $t_{2i},\ i=1,2,3...i_0-1$ and from
$t_{2i_0-1}$ to $z'$, we have
\begin{equation}
\int_{x(t_{2i-1})}^{x(t_{2i})}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_{t_{2i-1}}^{t_{2i}}
\int_t^1k(s)\,ds\,dt, \quad i=1,2,3\ldots i_0-1,  \label{e2.11}
\end{equation}
and
\begin{equation}
\int_{x(t_{2i_0-1})}^{x(z')}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_{t_{2i_0-1}}^{z'}
\int_t^1k(s)\,ds\,dt.  \label{e2.12}
\end{equation}
Summing  \eqref{e2.11} from 1 to $i_0-1$, we have by \eqref{e2.12} and
$x(t_{2i})=x(t_{2i+1})$, that
$$
\int_{b_n}^{x(z')}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{z'}\int_t^1k(s)\,ds\,dt
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{z}\int_t^1k(s)\,ds\,dt.
$$
Since $x(z)=x(z')$,
$$
\int_{b_n}^{x(z)}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{z}\int_t^1k(s)\,ds\,dt;
$$
i.e, \eqref{e2.6} holds.
 Letting $z\to t'$ in \eqref{e2.6}, we have
\begin{equation}
\begin{aligned}
\int_{b_n}^R\frac{dx}{F(x)}
&\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{t'}\int_t^1k(s)\,ds\,dt\\
&\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^1\int_t^1k(s)\,ds\,dt\\
&=(1+\frac{\bar{G}(R)}{F(R)})\int_0^1sk(s)ds.
\end{aligned} \label{e2.13}
\end{equation}
The inequality above contradicts
$\int_1^R\frac{dx}{F(x)}>(1+\frac{\bar{G}(R)}{F(R)})\int_0^1sk(s)ds$.
Hence \eqref{e2.2} holds.

It follows from Lemma \ref{lem2.1} and \eqref{e2.2} that $T_n$ has
a fixed point $x_n$ in $C_n$. Using $x_n$ and 1 in the place
of $x$ and $\lambda$ in \eqref{e2.2}, we obtain easily
$b_n\leq x_n(t)\leq R, t\in [\frac{1}{n},1]$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.4}
For a fixed $h\in (0,\min\{\frac{1}{2},\eta\})$, suppose
$m_{n,h}=\min\{x_n(t), t\in[h,1]\}$.
Then $m_h=\inf\{m_{n,h}\}>0$.
\end{lemma}

\begin{proof}
Since $x_n(t)\geq b_n>0$, we get $m_h\geq 0$. For any fixed natural
numbers $n$ ($n>[\frac{1}{\eta}]+1$),
let $t_n\in[h,1]$ such that
$ x_n(t_n)=\min\{x_n(t),t\in [h,1]\}$. If $m_h=0$, there exist a
countable set $\{n_i\}$ such that
\begin{equation}
\lim_{n_i\to+\infty}x_{n_i}(t_{n_i})=0 .\label{e2.14}
\end{equation}
So there exist $N$ such that
$x_{n_i}(t_{n_i})\leq \min\{b(t), t\in [\frac{h}{2},1]\}$, $n_i>N$.
Then we have two cases.

  Case 1. There exist $n_k\in \{n_i\}, n_k>N$ and
$t^*_{n_k}\in [\frac{h}{2},h]$ such
that $x_{n_k}(t^*_{n_k})\geq x_{n_k}(t_{n_k})$. By the same
argument in Lemma \ref{lem2.3}, we can get
$t'_{n_k}, t''_{n_k}\in [\frac{h}{2},1], t'_{n_k}<t''_{n_k}$ such that

$$x_{n_k}(t)\leq \min\{b(t), t\in [\frac{h}{2},1] \}, \quad
t\in [t'_{n_k},t''_{n_k}],$$
\begin{gather}
x_{n_k}(t)\leq x_{n_k}(t'_{n_k}), x_{n_k}(t)\leq x_{n_k}(t''_{n_k}),
\quad t\in (t'_{n_k},t''_{n_k}), \label{e2.15}\\
x''_{n_k}(t)=-f(t,x_{n_k}(t))<0, \quad t\in (t'_{n_k},t''_{n_k}).
\label{e2.16}
\end{gather}
Inequality \eqref{e2.15} shows that $x_{n_k}(t)$ is concave
down in $ [t'_{n_k},t''_{n_k}]$,
which contradicts \eqref{e2.16}.

 Case 2. $x_{n_i}(t)<x_{n_i}(t_{n_i})$, $t\in[\frac{h}{2},h]$ for any
$n_i\in \{n_i\}$, $n_i>N$. And so we have
\begin{equation}
\lim_{n_i\to +\infty }x_{n_i}(t)=0, \quad t\in[\frac{h}{2},h]. \label{e2.17}
\end{equation}
On the other hand for any
$t\in[\frac{h}{2},h]$,
\begin{equation}
\begin{aligned}
x_{n_i}(t)&=\frac{2}{h}\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s)
 f(s,x_{n_i}(s))ds\\
&\quad +\frac{2}{h}\int_t^h(s-\frac{h}{2})(h-t)
 f(s,x_{n_i}(s))ds+x_{n_i}(\frac{h}{2})+x_{n_i}(h)\\
&\geq  \frac{2}{h}[\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s)a(s)ds
+\int_t^h(s-\frac{h}{2})(h-t)a(s)ds]>0,
\end{aligned} \label{e2.18}
\end{equation}
which contradicts \eqref{e2.17}. The proof is complete.
\end{proof}

\section{Main Result}

\begin{theorem} \label{thm3.1}
If {\rm (S1)--(S3)} hold, the three-point boundary-value problem
\eqref{e1.1}-\eqref{e1.2} has at least one positive solution.
\end{theorem}

\begin{proof}
 For any natural numbers $n\geq [\frac{1}{\eta}+1]$, it follows
from Lemma \ref{lem2.3} that there exist $x_n\in C_n, b_n\leq x_n\leq R$
satisfying \eqref{e2.1}. Now we divide the proof into three steps.

  Step 1. There exist a convergent subsequence of $\{x_n\}$ in (0,1].
For a natural number $k\geq \max\{3,[\frac{1}{\eta}]+1\}$,
it follows from Lemma \ref{lem2.4} that
$0<m_{\frac{1}{k}}\leq x_n(t)\leq R$, $t\in [\frac{1}{k},1]$
for any natural numbers $n\geq [\frac{1}{\eta}+1]$;
i.e., $\{x_n\}$ is uniformly bounded in $[\frac{1}{k}, 1]$.  Since
$x_n$ also satisfies
\begin{align*}
x_n(t)&=-\int_{\frac{1}{k}}^t(t-s)f(s,x_n(s))ds\\
&\quad+\frac{t-\frac{1}{k}}{1-\alpha \eta-\frac{1}{k}(1-\alpha)}
[\int_{\frac{1}{k}}^1(1-s)f(s,x_n(s))ds
 -\alpha \int_{\frac{1}{k}}^{\eta}(\eta -s)f(s,x_n(s))ds]\\
&\quad +x_n(\frac{1}{k})+\frac{(t-\frac{1}{k})
 (1-\alpha)}{1-\alpha \eta-\frac{1}{k}(1-\alpha)}(b_n-x_n(\frac{1}{k})),
 \quad t\in [\frac{1}{k},1],
\end{align*}
we have
\begin{align*}
x'_n(t)&=-\int_{\frac{1}{k}}^tf(s,x_n(s))ds
 +\frac{\int_{\frac{1}{k}}^1(1-s)f(s,x_n(s))ds
 -\alpha \int_{\frac{1}{k}}^\eta
 (\eta-s)f(s,x_n(s))ds}{1-\alpha \eta-\frac{1}{k}(1-\alpha)}\\
&\quad +\frac{(1-\alpha)(b_n-x_n(t))}{1-\alpha \eta
 -\frac{1}{k}(1-\alpha)}, \quad
 t\in[\frac{1}{k},1].
\end{align*}
Obviously
\begin{equation}
|x'_n(t)|\leq \frac{3-\eta}{1-\eta}\max\{|f(t,x(t))|:
(t,x)\in [\frac{1}{k},1]\times [m_{\frac{1}{k}},R]\}
+\frac{2R}{1-\eta}, \label{e3.1}
\end{equation}
for $t\in[\frac{1}{k},1]$.
It follows from inequality \eqref{e3.1} that $\{x_n\}$ is
equicontinuous in $[\frac{1}{k},1]$. The Ascoli-Arzela theorem
guarantees that there exists a subsequence of $\{x_n(t)\} $
which converges uniformly on $[\frac{1}{k},1]$.
We may choose the diagonal sequence $\{x_k^{(k)}(t)\} $
(see more details in \cite{y2}) which converges everywhere in
$ (0,1]$ and it is easy to verify that $\{x_k^{(k)}(t)\} $
converges uniformly on any interval $[c,d]\subseteq (0,1]$.
Without loss of generality, let
$\{x_k^{(k)}(t)\}$ be  $\{x_n(t)\}$ in the rest.
Putting $x(t)=\lim_{n\to+\infty}x_n(t), t\in(0,1]$,
we have $x(t)$ is continuous in $(0,1]$ and
$x(t)\geq m_h>0, t\in(0,1]$ by Lemma \ref{lem2.4}.

 Step 2. $x(t)$ satisfies \eqref{e1.1}.
 Fixed $t\in (0,1]$, we may choose $h\in (0,\min\{\frac{1}{2},\eta\})$
such that $t\in (h,1]$ and
\begin{equation}
\begin{aligned}
x_n(t)&=-\int_h^t(t-s)f(s,x_n(s))ds\\
&\quad +\frac{t-h}{1-\alpha \eta-h(1-\alpha)}
 [\int_h^1(1-s)f(s,x_n(s))ds-\alpha \int_h^{\eta}(\eta -s)
 f(s,x_n(s))ds]\\
&\quad +x_n(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)}
 (b_n-x_n(h)), \quad t\in (h,1].\label{e3.2}
\end{aligned}
\end{equation}
Letting $n\to +\infty$ in \eqref{e3.2}, we have
\begin{equation}
\begin{aligned}
x(t)&=-\int_h^t(t-s)f(s,x(s))ds
+\frac{t-h}{1-\alpha \eta-h(1-\alpha)}\\
&\quad\times \big[\int_h^1(1-s)f(s,x(s))ds
 -\alpha \int_h^{\eta}(\eta -s)f(s,x(s))ds\big]\\
&\quad +x(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)}(-x(h)),
\quad  t\in (h,1].\label{e3.3}
\end{aligned}
\end{equation}
Differentiating \eqref{e3.3}, we get the desired result.

 Step 3. $x(t)$ satisfies \eqref{e1.2}. Let
$$
t_n=\inf \{t:x_n(t)=\|x_n\|, x'_n(t)=0, t\in[\frac{1}{n},1]\},
$$
where $\|x_n\|=\max _{\frac{1}{n}\leq t\leq 1}x_n(t)\leq R$.
Then
$$
t_n\in[\frac{1}{n},1], \quad
x_n(t_n)=\|x_n\|, \quad
x'_n(t_n)=0.
$$
Using $x_n(t), 1$ and $t_n$ in place of $x(t), \lambda$ and
$t'$ in Lemma \ref{lem2.3}, we obtain easily by \eqref{e2.13}
\begin{equation}
\int_{b_n}^{\|x_n\|}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{t_n}\int_t^1k(s)\,ds\,dt.
\label{e3.4}
\end{equation}
It follows from  \eqref{e3.4} and Lemma \ref{lem2.4} that
$0<a=\inf\{t_n\}\leq 1$.  Fixed $z\in(0,a)$, then
$b_n<x_n(z)<\|x_n\|\leq R$. By Lemma \ref{lem2.3} we easily get
\begin{equation}
\int_{b_n}^{x_n(z)}\frac{dx}{F(x)}\leq
(1+\frac{\bar{G}(R)}{F(R)})\int_0^z\int_t^1k(s)\,ds\,dt, \quad z\in
(0,a).\label{e3.5}
\end{equation}
Letting $n\to +\infty$ in \eqref{e3.5} and
noticing $b_n\to 0$, we have
\begin{equation}
\int_0^{x(z)}\frac{dx}{F(x)}\leq
(1+\frac{\bar{G}(R)}{F(R)})\int_0^z\int_t^1k(s)\,ds\,dt, \quad
 z\in (0,a).\label{e3.6}
\end{equation}
It follows from \eqref{e3.6} that $x(0)=\lim_{z\to 0^+}x(z)=0$.
 Using $1$ in place of $\lambda$ in \eqref{e2.3}, we obtain
easily
\begin{equation}
x_n(1)=\alpha x_n(\eta)+(1-\alpha)b_n.\label{e3.7}
\end{equation}
Letting $n\to +\infty$, we have $x(1)=\alpha x(\eta)$.
This complete the proof.
\end{proof}

 When $G(x)\equiv 0$ in (H3), it is easy to see that the assumption
(S3) is satisfied by the decreasing property of $F(x)$.
Then under the assumption $G(x)\equiv 0$ we get the following
corollaries to Theorem \ref{thm3.1}.

\begin{corollary} \label{coro3.1}
 Suppose {\rm (S1), (S2)} hold. Then \eqref{e1.1}-\eqref{e1.2}
has at least one positive solution
\end{corollary}

 \begin{corollary} \label{coro3.2}
Suppose the assumptions of Corollary \ref{coro3.1} hold. If further
$f(t,\cdot)$ is non-increasing in $(0,+\infty)$ for each $t\in(0,1)$,
the solution of \eqref{e1.1}-\eqref{e1.2} is unique.
\end{corollary}

\begin{proof} Suppose $x_1(t)$ and $x_2(t)$ are two solutions of
\eqref{e1.1}-\eqref{e1.2}. We need to prove that
$x_1(t)\equiv x_2(t), t\in[0,1]$. Let $z(t)=x_1(t)-x_2(t), t\in[0,1]$.
It follows that $z(0)=0, z(1)=\alpha z(\eta)$.
We first show  that $x_1(\eta)=x_2(\eta)$, which implies that
$x_1(1)=x_2(1)$. In fact, if it is not true, without loss of generality,
 we can suppose $x_1(\eta)>x_2(\eta)$. That is to say
$z(\eta)>0, 0<z(1)=\alpha z(\eta)<z(\eta)$.
 Setting $t_1=\max\{t\in(0,\eta), z(t)=z(1)\}$ and
$t_2=\min\{t\in(\eta,1), z(t)=z(1)\}$, we get
\[
z(t_1)=z(t_2)=z(1), \quad
z(t)=x_1(t)-x_2(t)>z(1)>0, \quad t\in(t_1,t_2).
\]
 Letting $s(t)=z(t)-z(1)$, we have that $s(t_1)=s(t_2)=0$ and
$s(t)>0, t\in(t_1,t_2)$.
 It follows from \eqref{e1.1} and the monotonicity of $f(t,\cdot)$
that $s''(t)=z''(t)\geq 0, t\in (t_1,t_2)$.
 An elementary form of the maximum principle implies
 $s(t)\leq 0$ for all $t\in (t_1,t_2)$ and hence a contradiction.
Then, $x_1(\eta)=x_2(\eta)$, which also yields that
 $x_1(1)=x_2(1)$. That is to say $z(0)=z(\eta)=z(1)=0$.

We next claim that $x_1(t)=x_2(t), t\in(0,\eta)$. In fact,
if it is not true, without loss of generality,
we can get $x_1(t_0)>x_2(t_0)$ for some $t_0\in(0,\eta)$.
Let $t_3=\max\{t\in(0,t_0), z(t)=0\},
t_4=\min\{t\in(t_0,\eta), z(t)=0\}$(note $z(\eta)=0)$).
Then $z(t_3)=z(t_4)=0$ and $z(t)>0, t\in(t_3,t_4)$.
 Let $s_1(t)=z_1(t)-z_2(t)$, $t\in[t_3,t_4]$. Then
$s_1(t)>0$ for all $t\in[t_3,t_4]$. On the other hand,
the monotonicity of $f(t,\cdot)$ implies that
  $s''_1(t)\geq 0, t\in (t_3,t_4)$.
An elementary form of the maximum principle implies
$s_1(t)\leq 0$ for all $t\in (t_3,t_4)$ and hence a contradiction.

  The same argument yields that $x_1(t)=x_2(t), t\in(\eta,1)$.
Hence we get $x_1(t)=x_2(t), t\in[0,1]$. Thus the result is proved.
\end{proof}

\subsection*{Example}
 Consider the second order singular three-point boundary-value problem
\begin{gather}
x''(t)+\frac{1}{4}(x^2(t)+\frac{1}{x^2(t)}-\frac{x^3(t)}{t^5}
-\frac{1}{t^2})=0,\quad 0<t<1, \label{e3.8} \\
x(0)=0,\quad x(1)=\frac{1}{3} x(\frac{1}{4}).  \label{e3.9}
\end{gather}
Set $\alpha=\frac{1}{3}$, $\eta=\frac{1}{4}$,
\begin{gather*}
f(t,x)=\frac{1}{4}(x^2+\frac{1}{x^2}-\frac{x^3}{t^5}-\frac{1}{t^2}),\quad
k(t)=\frac{1}{4},\quad
F(x)=\frac{1}{x^2}, \\
G(x)=x^2, \quad a(t)=\frac{1}{4t^2}, \quad b(t)=\frac{t}{2}.
\end{gather*}
It is easy to prove that $f(t,x)\leq k(t)(F(x)+G(x))$ and
(S1)--(S3) hold. By Theorem \ref{thm3.1}, the three-point
boundary-value problem
\eqref{e3.8}-\eqref{e3.9} has at least one positive
solution. Moreover, if
$f(t,x)=\frac{1}{x^2(t)}-\frac{x^3(t)}{t^5}$ in \eqref{e3.8}, the
three-point boundary-value problem \eqref{e3.8}-\eqref{e3.9} has only one positive
solution by Corollaries \ref{coro3.1} and  \ref{coro3.2}.

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\end{document}