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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 110, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/110\hfil Positive solutions]
{Positive solutions for nonlinear second-order $m$-point 
 boundary-value problems}

\author[J. Jiang, L. Liu \hfil EJDE-2009/110\hfilneg]
{Jiqiang Jiang, Lishan Liu}  % in alphabetical order

\address{Jiqiang Jiang \newline
School of Mathematical Sciences,
Qufu Normal University \\ 
Qufu 273165, Shandong, China}
\email{qfjjq@mail.qfnu.edu.cn, qfjjq@163.com}

\address{Lishan Liu \newline
School of Mathematical Sciences,
Qufu Normal University \\ 
Qufu 273165, Shandong, China}
\email{lls@mail.qfnu.edu.cn, Tel. 86-537-4456234, Fax 86-537-4455076}

\thanks{Submitted December 26, 2008. Published September 10, 2009.}
\thanks{Supported  by  grants:
10771117 from the National Natural Science
Foundation of China, \hfill\break\indent
20060446001 from the State Ministry of
Education Doctoral Foundation of China,
and  \hfill\break\indent
Y2007A23 from the Natural Science Foundation of
Shandong Province of China.}
\subjclass[2000]{34B15, 34B25}
\keywords{Positive solution; singular; $m$-point boundary-value
problem; cones}

\begin{abstract}
 By constructing a special cone and applying the fixed
 index theory in the cone, we prove the existence of positive solutions
 for a class of singular $m$-point boundary-value problems.
\end{abstract}

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

\section{Introduction}

This paper considers the existence of positive solutions for the
 second-order $m$-point boundary-value problem
\begin{gather} \label{e1.1}
(p(t)x'(t))'-q(t)x(t)+f(t,x(t))=0 ,\quad  t\in(0,1) ,\\
 \label{e1.2}
ax(0)-bp(0)x'(0)=\sum_{i=1}^{m-2}\alpha_i x(\xi_i),\quad
cx(1)+dp(1)x'(1)=\sum_{i=1}^{m-2}\beta_i x(\xi_i),
\end{gather}
where $a,c\in[0,+\infty)$, $b,d\in(0,+\infty)$ with $ac+ad+bc>0$,
$\xi_i\in(0,1)$, $\alpha_i,\beta_i\in[0,+\infty)$ for
$i\in\{1,2,\dots,m-2\}$ are given constants,
$p\in C^1([0,1],(0,+\infty))$, $q\in C([0,1],(0,+\infty))$ and
$f\in C((0,1)\times(0,+\infty),[0,+\infty))$, $f(t,x)$ is allowed to be
singular at $t=0$, $t=1$ and $x=0$.

If $p\equiv1$, $q\equiv0$, $\alpha_i,\beta_i=0$, (for
$i=1,2,\dots,m-2)$, then \eqref{e1.1}-\eqref{e1.2} reduces to the
two-point boundary-value problem
\begin{gather} \label{e1.3}
x''(t)+f(t,x(t))=0 ,\quad  t\in(0,1) ,\\
 \label{e1.4}
ax(0)-bx'(0)=0,\quad
cx(1)+dx'(1)=0,
\end{gather}
which has been intensively studied; see \cite{h2,l1}.

In \cite{l2}, by using the fixed index theory in a cone, positive solutions
were obtained for differential systems
\begin{gather*}
-x''(t)=f(t,y) ,\quad  t\in(0,1) ,\\
-y''(t)=g(t,x) ,\quad  t\in(0,1) ,\\
\alpha_1x(0)-\beta_1x'(0)=\gamma_1x(1)+\delta_1x'(1)=0,\\
\alpha_2y(0)-\beta_2y'(0)=\gamma_2y(1)+\delta_2y'(1)=0,
\end{gather*}
where $\alpha_i,\beta_i,\gamma_i,\delta_i\geq0$ and
$\rho_i=\alpha_i\gamma_i+\alpha_i\gamma_i+\gamma_i\beta_i>0$
($i=1,2$), $f(t,y)$ and $g(t,x)$ may be singular at $t=0$, $t=1$ and
$x=0$, $y=0$, respectively.

In recent years, singular multi-point boundary-value problems have
been extensively studied and many optimal results have been
obtained, see \cite{l1,w1,x1,z1,z2}
and references therein. In addition, many
papers investigated the existence of solutions for the nonsingular
multi-point boundary-value problems, for example,
\cite{f1,h1,h2,s1}.

Recently, Ma \cite{m1},  Ma and Thompson \cite{m2} obtained  excellent
results about the existence of positive solutions for the more
general $m$-point boundary-value problem \eqref{e1.1}-\eqref{e1.2},
but in the above papers
there are no studies  for singularity of the nonlinearity
$f(t,x)$ at the point $x = 0$. Recently, by using Nonlinear
Alternative of Leray-Schauder with the properties of the associated
vector field at the $(u,u')$ plane, Galanis and Palamides \cite{g1}
studied the problem
$$
-[\phi_p(u')]'=q(t)f(t,u(t)),\quad 0<t<1
$$
subject to
$$
u(0)-g(u'(0))=0,\quad u(1)-\beta u(\eta)=0,
$$
or to
$$
u(0)-\alpha u'(\eta)=0,\quad u(1)+g(u'(1))=0,
$$
where $f(t,u)$ is allowed to have singularity at $u = 0$, the
obtained solutions remains away from the origin and avoid the
singularity of the nonlinear term at $u = 0$.

Motivated by the above mentioned papers, we consider the existence
of positive solutions for \eqref{e1.1}-\eqref{e1.2}.
Here we allow $f(t,x)$ to have a singularity
at $t=0,1$, and  at $x=0$.
As far as we know, there were only a few works  when $f$
has singularities at $t=0,1$ and
$x=0$. This paper attempts to fill part of this gap in the
literature.

This work is organized as follows. In section 2, we present some
lemmas that are used to prove our main results. Then in section 3,
the existence of positive solution for  \eqref{e1.1}-\eqref{e1.2} will be established
by using the fixed
 point theory in the cone, which we state here for the convenience
of the reader.


 \begin{lemma}[\cite{g2}] \label{lem1.1}
Let  $P$  be  a  cone of the real Banach
space $E$, $\Omega$ be a bounded open subset of $E$  with
$\theta\in \Omega$ and $T: \overline{\Omega}\cap P\to P$ is
a completely continuous. Suppose that
 $ Tu\neq\lambda u$, for all $u\in\partial\Omega \cap P$,
$\lambda\geq 1$, then $i(T,\Omega\cap P,P)=1$.
\end{lemma}

\begin{lemma}[\cite{g2}] \label{lem1.2}
 Let  $P$  be  a  cone of the real Banach
space $E$, $\Omega$ be a bounded open subset of $E$  with
$\theta\in \Omega$ and $T: \overline{\Omega}\cap P\to P$ is a completely
continuous. Suppose that
\begin{itemize}
\item[(i)] $\inf_{u\in P\cap\partial\Omega}\|Tu\|>0$
\item[(ii)] $Tu\neq\lambda u$, for all $u\in\partial\Omega \cap P$,
$\lambda\in(0,1]$,
\end{itemize}
 then $i(T,\Omega\cap P,P)=0$.
\end{lemma}



\section{Preliminaries}

Let $E=C[0,1]$ be a real Banach space, with the norm
$\|x\|=\max_{t\in[0,1]}|x(t)|$  for
$x\in C[0,1]$. Let $P=\{x\in E: x(t)\geq  0,t\in[0,1]\}$.
Clearly $P$ is a  cone in $E$.

 The function $x$ is said to be a positive solution of \eqref{e1.1}-\eqref{e1.2}
if $x(t)$ is positive solution on $(0,1)$
and satisfies the differential equation \eqref{e1.1} and the boundary conditions
 \eqref{e1.2}.

The following lemmas play an important role when proving
 our main results.

\begin{lemma}[\cite{m1,m2}] \label{lem2.1}
 Assume
\begin{itemize}
\item[(H1)]   $p\in C^1([0,1], (0, +\infty))$,
$q\in C([0,1], (0, +\infty))$.
\end{itemize}
Let $\psi$ and $\phi$ be  the solutions of the linear problems
\begin{gather} \label{e2.1}
(p(t)\psi'(t))'(t)-q(t)\psi(t)=0, \quad t\in (0,1), \\
 \label{e2.2}
 \psi(0)=b,\quad  p(0)\psi'(0)= a,
\end{gather}
and
\begin{gather} \label{e2.3}
(p(t)\phi'(t))'(t)-q(t)\phi(t)=0, \quad  t\in (0,1), \\
 \label{e2.4}
\phi(1)=d,\quad  p(1)\phi'(1)= -c,
\end{gather}
respectively. Then
\begin{itemize}
\item[(i)]  $\psi$ is strictly increasing on $[0,1]$, and
$\psi(t)>0$ on $[0,1]$;

\item[(ii)] $\phi$ is strictly decreasing on $[0,1]$,  and
$\phi(t)>0$ on $[0,1]$.
\end{itemize}
\end{lemma}

As in \cite{m2}, set
$$
\Delta= \det \begin{pmatrix}
-\sum_{i=1}^{m-2} \alpha_i\psi(\xi_i)
  & \rho-\sum_{i=1}^{m-2} \alpha_i\phi(\xi_i) \\
\rho-\sum_{i=1}^{m-2} \beta_i\psi(\xi_i)
  & -\sum_{i=1}^{m-2} \beta_i\phi(\xi_i)
\end{pmatrix},
\quad \rho= p(t)\det\begin{pmatrix}
  \phi(t)  & \psi(t) \\
 \phi'(t)  & \psi'(t)
\end{pmatrix}.
$$
Then, by Liouville's formula, we have
$$
\rho= p(0)\det\begin{pmatrix}
  \phi(0)  & \psi(0) \\
 \phi'(0)  & \psi'(0)
\end{pmatrix}=\text{constant}.
$$
Define
\begin{equation} \label{e2.5}
G(t,s)= \frac{1}{\rho}
\begin{cases}
\phi(t)\psi(s), & 0\leq s\leq t\leq 1,\\
\phi(s)\psi(t), & 0\leq t\leq s\leq 1.
\end{cases}
\end{equation}
It is easy to see that
\begin{equation} \label{e2.6}
0\leq G(t,s)\leq G(s,s),\quad 0\leq s,\; t\leq 1.
\end{equation}

\begin{remark} \label{rmk2.1}\rm
 By \eqref{e2.5} and Lemma \ref{lem2.1}, for any
$t\in [0,1]$, we have
$$
\frac{G(t,s)}{G(s,s)}
=\begin{cases}\frac{\phi(t)}{\phi(s)}, & 0\leq s\leq t\leq 1,\\[3pt]
 \frac{\psi(t)}{\psi(s)}, & 0\leq t\leq s\leq 1,
\end{cases}
\geq \begin{cases}
\frac{d}{\phi(0)}, & 0\leq s\leq t\leq 1,\\[3pt]
 \frac{b}{\psi(1)},& 0\leq t\leq s\leq 1.
\end{cases}
$$
\end{remark}

Let $\gamma= \min\{\frac{d}{\phi(0)},\
\frac{b}{\psi(1)}\}$, then
 $ G(t,s)\geq\gamma G(s,s)$, for $t,s\in [0,1]$.

\begin{remark} \label{rmk2.2}\rm
Since $\gamma= \min\{\frac{d}{\phi(0)}, \frac{b}{\psi(1)}\}$,
according to the monotonicity of $\psi(t)$, we have
$\gamma\leq\frac{b}{\psi(1)}
=\frac{\psi(0)}{\psi(1)}\leq\frac{\psi(t)}{\psi(1)}$,
so $\psi(t)\geq\gamma\psi(1)$,  for $t\in [0,1]$. Similarly, by the
monotonicity of $\phi(t)$, we have
$\gamma\leq\frac{d}{\phi(0)}
=\frac{\phi(1)}{\phi(0)}\leq\frac{\phi(t)}{\phi(0)}$,
so $\phi(t)\geq\gamma\phi(0)$, for $t\in [0,1]$.
\end{remark}

\begin{lemma}[\cite{m1,m2}] \label{lem2.2}
Assume  {\rm (H1)} and that  $\Delta\neq 0$.
 Then for any $y\in L[0,1]$, the problem
\begin{gather} \label{e2.7}
(p(t)x'(t))'(t)-q(t)x(t)+y(t)=0, \quad  t\in (0,1), \\
 \label{e2.8}
ax(0)-bp(0)x'(0)= \sum_{i=1}^{m-2}\alpha_i x(\xi_i),\quad
cx(1)+dp(1)x'(1)= \sum_{i=1}^{m-2}\beta_i x(\xi_i),
\end{gather}
has a unique solution
\begin{equation} \label{e2.9}
x(t)=\int_0^1G(t,s)y(s)ds +A(y)\psi(t)+B(y) \phi(t),
\end{equation}
where
 \begin{gather} \label{e2.10}
 A(y)=\frac{1}{\Delta} \det\begin{pmatrix}
 \sum_{i=1}^{m-2} \alpha_i\int_0^1G(\xi_i,s)y(s)ds
  & \rho-\sum_{i=1}^{m-2} \alpha_i\phi(\xi_i) \\
\sum_{i=1}^{m-2} \beta_i\int_0^1G(\xi_i,s)y(s)ds
  & -\sum_{i=1}^{m-2} \beta_i\phi(\xi_i)
\end{pmatrix}
\\ \label{e2.11}
B(y)=\frac{1}{\Delta} \det\begin{pmatrix}
-\sum_{i=1}^{m-2} \alpha_i\psi(\xi_i)
  & \sum_{i=1}^{m-2} \alpha_i\int_0^1G(\xi_i,s)y(s)ds  \\
\rho-\sum_{i=1}^{m-2} \beta_i\psi(\xi_i)
  & \sum_{i=1}^{m-2} \beta_i\int_0^1G(\xi_i,s)y(s)ds
\end{pmatrix}.
\end{gather}
\end{lemma}

\begin{lemma}[\cite{m1,m2}] \label{lem2.3}
Assume {\rm (H1)} and
\begin{itemize}
\item[(H2)]  $\Delta<0$,
$\rho-\sum_{i=1}^{m-2}\alpha_i\phi(\xi_i)>0$,
$\rho-\sum_{i=1}^{m-2}\beta_i\psi(\xi_i)>0$.
\end{itemize}
Then for $y\in L[0,1]$ with $y\geq0$, the unique solution $x$ of
 \eqref{e2.7}-\eqref{e2.8} satisfies
$x(t)\geq0$, for $t\in[0,1]$.
\end{lemma}

Let $Q=\{x\in P :x(t)\geq\gamma\|x\|\}$. It is obvious that $Q$ is a
subcone of $P$. With Lemma \ref{lem2.2},  Problem \eqref{e1.1}-\eqref{e1.2} has a positive
solution $x=x(t)$ if and only if $x\in Q\backslash\{\theta\}$ is
a solution  of the  nonlinear integral equation
\begin{equation} \label{e2.12}
x(t)=\int_0^1G(t,s)f(s,x(s))ds +A(f(s,x(s)))\psi(t)+B(f(s,x(s)))
\phi(t),
\end{equation}
where $f$ satisfies the condition
\begin{itemize}
\item[(H3)] $f\in C((0,1)\times(0,+\infty),[0,+\infty))$ and there
exist $h\in C((0,1),[0,+\infty))$, $g\in C((0,+\infty),[0,+\infty))$
satisfying that for any $t\in (0,1)$, $u\in (0,+\infty)$ implies
\begin{gather*}
f(t,u)\leq h(t)g(u),\quad t\in(0,1),\; u\in(0,+\infty),\\
0<\int_0^1G(s,s)h(s)ds<+\infty.
\end{gather*}
\end{itemize}


Define an operator $T:Q\backslash \{\theta\}\to P$ by
\begin{equation} \label{e2.13}
(Tx)(t)=\int_0^1G(t,s)f(s,x(s))ds
+A(f(s,x(s)))\psi(t)+B(f(s,x(s)))\phi(t).
\end{equation}
It is easy to prove that the existence of solutions to  \eqref{e1.1}-\eqref{e1.2} is
equivalent to the existence of solutions to \eqref{e2.12}. That is, the
existence of a fixed point of operator $T$.

To overcome the singularity, we consider the following
approximating equation of \eqref{e2.13} with the boundary conditions
\eqref{e1.2}.
\begin{equation} \label{e2.14}
(T_nx)(t)=\int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t),
\end{equation}
where $n$ is a positive integer and
\begin{equation} \label{e2.15}
f_n(t,x)=f(t,\max\{\frac{1}{n},x\}).
\end{equation}

\begin{remark} \label{rmk2.3}\rm
By (H3), there exists
$\tau\in(0,\frac{1}{2})$ such that
$$
0<\int_{\tau}^{1-\tau}G(s,s)h(s)ds<+\infty.
$$
\end{remark}

\begin{lemma} \label{lem2.4}
Assume {\rm (H1)-(H3)}. Then
 $T_n: P\to P$ is completely continuous for any fixed natural number $n$.
\end{lemma}


\begin{proof}
 First it is easy to see that $T_n$ maps
$P$ into $P$. Then we prove that $T_n$ maps bounded sets into
bounded sets.

Suppose $D\subset P$ is an arbitrary bounded set. Then there exists
a constant $M_1>0$
 such that $\|  x\|\leq M_1$  for any $x\in D$.
 By (H1), for any $x \in D$ and $s \in [0,1]$, we have
$$\begin{aligned}
|(T_nx)(t)|
&=\int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t)\\
&\leq \int_0^1G(s,s)h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)ds\\
&\quad +A\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)\psi(1)\\
&\quad +B\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)\phi(0)\\
&\leq M_2\Big[\int_0^1G(s,s)h(s)ds+A\left(h(s)\right)\psi(1)
+B\left(h(s)\right)\phi(0)\Big]\\
&\leq M_2(1+A\psi(1)
+B\phi(0))\int_0^1G(s,s)h(s)ds,
\end{aligned}$$
where
$M_2=\sup_{x\in[\frac{1}{n},\frac{1}{n}+M_1]}g(x)$,
\begin{gather} \label{e2.16}
A=\frac{1}{\Delta} \det\begin{pmatrix}
  \sum_{i=1}^{m-2} \alpha_i & \rho-\sum_{i=1}^{m-2} \alpha_i\phi(\xi_i) \\
 \sum_{i=1}^{m-2} \beta_i & -\sum_{i=1}^{m-2} \beta_i\phi(\xi_i) \\
\end{pmatrix}, \\
 \label{e2.17}
B=\frac{1}{\Delta}  \det \begin{pmatrix}
-\sum_{i=1}^{m-2} \alpha_i\psi(\xi_i)
& \sum_{i=1}^{m-2} \alpha_i  \\
 \rho-\sum_{i=1}^{m-2} \beta_i\psi(\xi_i)
 &
 \sum_{i=1}^{m-2} \beta_i\\
\end{pmatrix}.
\end{gather}
Therefore, $T_n(D)$ is uniformly bounded.

Now we show that $T_n(D)$ is equicontinuous on $[0,1]$. For any
$\varepsilon>0$, since $G(t,s),\psi(t)$ and $\phi(t)$ are uniformly
continuous on $[0,1]\times[0,1]$ and $[0,1]$, respectively.  There
exists $\delta>0$ such that for any
$t_1,t_2\in[0,1],|t_1-t_2|<\delta$ implies that
\begin{gather*}
|G(t_1,s)-G(t_2,s)|<\frac{\varepsilon\min_{0\leq
s\leq1}G(s,s)}{3M_2\int_0^1G(s,s)h(s)ds},
\\
|\psi(t_1)-\psi(t_2)|<\frac{\varepsilon}{3M_2A\int_0^1G(s,s)h(s)ds},
\\
|\phi(t_1)-\phi(t_2)|<\frac{\varepsilon}{3M_2B\int_0^1G(s,s)h(s)ds}.
\end{gather*}
Consequently, for any $x\in D$, $t_1,t_2\in[0,1]$,
$|t_1-t_2|<\delta$, we
have
\begin{align*}
&|T_nx(t_1)-T_nx(t_2)|\\
&\leq\int_0^1|G(t_1,s)-G(t_2,s)|f_n(s,x(s))ds\\
&\quad +A(f_n(s,x(s)))|\psi(t_1)-\psi(t_2)|
  +B(f_n(s,x(s)))|\phi(t_1)-\phi(t_2)|
\\
&\leq \int_0^1|G(t_1,s)-G(t_2,s)|h(s)
 g\big(\max\{\frac{1}{n},x(s)\}\big)ds\\
&\quad +A\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)
  |\psi(t_1)-\psi(t_2)|\\
&\quad +B\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)
 |\phi(t_1)-\phi(t_2)|\\
&\leq M_2\int_0^1|G(t_1,s)-G(t_2,s)|h(s)ds\\
&\quad +A(h(s)M_2)|\psi(t_1)-\psi(t_2)|
+B(h(s)M_2)|\phi(t_1)-\phi(t_2)|\\
&\leq M_2\int_0^1|G(t_1,s)-G(t_2,s)|h(s)ds\\
&\quad +M_2A|\psi(t_1)-\psi(t_2)|\int_0^1G(s,s)h(s)ds\\
&\quad +M_2B|\phi(t_1)-\phi(t_2)|\int_0^1G(s,s)h(s)ds\\
&<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}
=\varepsilon.
\end{align*}
Thus, $T_n(D)$ is equicontinuous on $[0,1]$. According to
Ascoli-Arzela Theorem, $T_n(D)$ is a relatively compact set.

 In the end, we show $T_n$  is continuous. Suppose $x_m,x\in D,x_m\to x \
 (m\to +\infty)$. Then  there exists a constant $M_3>0$
 such that $\|x\|\leq M_3$,
$\|x_m\|\leq M_3$ ($m=1,2,\dots$).
 Since $f_n(t,x)$ is uniformly continuous on $[0,1]\times D$ for any
 fixed natural number $n$, hence,
$$
\lim_{m\to+\infty}f_n(t,x_m(t))=f_n(t,x(t)),\quad
\text{uniformly on } t\in[0,1].
$$
According to the Lebesgue
dominated convergence theorem,
$$
\lim_{m\to+\infty}\int_0^1G(s,s)|f_n(s,x_m(s))-f_n(s,x(s))|ds=0.
$$
Thus for the above $\varepsilon>0$, there exists a natural number
$M$, such that $m>M$ implies that
\begin{equation} \label{e2.18}
\int_0^1G(s,s)|f_n(s,x_m(s))-f_n(s,x(s))|ds
 <\frac{\varepsilon}{1+A\psi(1)+B\phi(0)}.
\end{equation}
 From \eqref{e2.18}, we obtain that for $m>M$,
\begin{align*}
 &\|T_nu_m-T_nu\|\\
&= \max_{0\leq t\leq1}\Big[\int_0^1G(t,s)f_n(s,x_m(s))ds
+A(f_n(s,x_m(s)))\psi(t)+B(f_n(s,x_m(s)))\phi(t) \\
&\quad -\int_0^1G(t,s)f_n(s,x(s))ds
-A(f_n(s,x(s)))\psi(t)-B(f_n(s,x(s)))\phi(t)\Big]\\
&\leq \int_0^1G(s,s)|f_n(s,x_m(s))-f_n(s,x(s))|ds\\
&\quad +A\big(|f_n(s,x_m(s))-f_n(s,x(s))|\big)\psi(1)\\
&\quad +B(|f_n(s,x_m(s))-f_n(s,x(s))|)\phi(0)\\
&\leq \big(1+A\psi(1)+B\phi(0)\big)\int_0^1G(s,s)|f_n(s,x_m(s))
 -f_n(s,x(s))|ds<\varepsilon.
\end{align*}
Therefore, $T_n:P\to P$ is  continuous. Thus $T_n:P \to P$ is a
completely  continuous operator.
\end{proof}

 \begin{lemma} \label{lem2.6}
 $T_n(Q)\subset Q$.
\end{lemma}

\begin{proof}
For any $x\in Q$, (H2) and (H3)  imply
$(T_nx)(t)\geq0$. From \eqref{e2.6}, \eqref{e2.14} and
the monotonicity of $\psi(t)$ and $\phi(t)$, we have
$$
(T_nx)(t)\leq\int_0^1G(s,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(1)+B(f_n(s,x(s)))\phi(0),
$$
which implies
\begin{equation} \label{e2.19}
 \|T_nx\|\leq\int_0^1G(s,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(1)+B(f_n(s,x(s)))\phi(0).
\end{equation}
By Remarks \ref{rmk2.1} and \ref{rmk2.2}, we have
\begin{equation} \label{e2.20}
 \begin{aligned}
(T_nx)(t)
&=\int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t)\\
&\geq \gamma\int_0^1G(s,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\gamma\psi(1)+B(f_n(s,x(s)))\gamma\phi(0)\\
&\geq\gamma\Big[\int_0^1G(s,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(1)+B(f_n(s,x(s)))\phi(0)\Big].
\end{aligned}
\end{equation}
Then, \eqref{e2.19} and \eqref{e2.20} yield
$$
(T_nx)(t)\geq\gamma\|T_nx\|.
$$
Hence $T_nx\in Q$.
\end{proof}

\section{Main results}

In this section, we present our main results as
follows.

\begin{theorem} \label{thm3.1}
 Suppose that {\rm (H1)--(H3)} hold
and there exist numbers $R>0$ and $L>0$ such that
\begin{gather} \label{e3.1}
\int_0^1G(s,s)h(s)ds<\frac{R}{\widetilde{M}(1+A\psi(1)+B\phi(0))},
\\ \label{e3.2}
L\gamma^2\int_\tau^{1-\tau}G(s,s)ds>1,\quad
\liminf_{x\to+\infty}\min_{\tau\leq
t\leq1-\tau}\frac{f(t,x)}{x}>L.
\end{gather}
Then  \eqref{e1.1}-\eqref{e1.2} has at least one positive solution, where
$\widetilde{M}=\max_{u\in[\gamma R,1+R]}g(u)$, $\gamma$ is defined
in Remark  \ref{rmk2.1} and $A,B$ are defined by \eqref{e2.16} and \eqref{e2.17},
respectively.
\end{theorem}


\begin{proof}
 Firstly, we shall prove that when $n$ is
sufficiently large, we have
\begin{equation} \label{e3.3}
T_nx\neq\lambda x,\quad x\in\partial Q_R,\; \lambda\geq1,
\end{equation}
where $Q_R=\{x\in Q:\|x\|<R\}$ for $R>0$. In fact, if there exists
$x_0\in \partial Q_R$,  and $\lambda_0\geq 1$  such that
$\lambda_0x_0=T_nx_0$, then
$x_0(t)\leq T_nx_0(t)$ for $t\in[0,1]$ and any $n$.
Choose a sufficiently large $n$ satisfying
$n>\frac{1}{\gamma R}$. Then we have
\begin{equation} \label{e3.4}
\begin{aligned}
x_0(t)&\leq(T_nx_0)(t)\\
&=\int_0^1G(t,s)f_n(s,x_0(s))ds
+A(f_n(s,x_0(s)))\psi(t)+B(f_n(s,x_0(s)))\phi(t)\\
&\leq\int_0^1G(s,s)f_n(s,x_0(s))ds
+A(f_n(s,x_0(s)))\psi(1)+B(f_n(s,x_0(s)))\phi(0)\\
&\leq(1+A\psi(1)+B\phi(0))\int_0^1G(s,s)f_n(s,x_0(s))ds\\
&\leq(1+A\psi(1)+B\phi(0))\int_0^1G(s,s)h(s)
 g\big(\max\{\frac{1}{n},x_0(s)\}\big)ds\\
&\leq(1+A\psi(1)+B\phi(0))\widetilde{M}\int_0^1G(s,s)h(s)ds<R.
\end{aligned}
\end{equation}
Therefore, by \eqref{e3.4} we have $\|x_0\|<R$, which is a contradiction
to $x_0\in\partial Q_R$. So applying Lemma \ref{lem1.1},
$i(T_n,Q_R,Q)=1$.

Next, according to \eqref{e3.2}, there exists $R_1$ such that $x>R_1$
implies
\begin{equation} \label{e3.5}
f(t,x)>Lx,\quad t\in[\tau,1-\tau].
\end{equation}
Choose $R'>\{R,\gamma^{-1}R_1\}$. When $n$ being sufficiently
large we can claim that
\begin{equation} \label{e3.6}
T_nx\neq\lambda x,\quad \forall x\in\partial Q_{R'},\; \lambda\in(0,1],
\end{equation}
where $Q_R'=\{x\in Q:\|x\|<R'\}$. Suppose \eqref{e3.6} is not true, then
there exist $x_1\in\partial Q_{R'}$ and $\lambda'\in(0,1]$ such that
$\lambda' x_1=T_nx_1$. Similarly, we choose sufficiently large $n$
satisfying that $n>\frac{1}{\gamma R'}$. Therefore, by \eqref{e3.5}
we have
\begin{align*}
x_1(t)&\geq (T_nx_1)(t)\\
&=\int_0^1G(t,s)f_n(s,x_1(s))ds
+A(f_n(s,x_1(s)))\psi(t)+B(f_n(s,x_1(s)))\phi(t)\\
&\geq\int_0^1G(t,s)f_n(s,x_1(s))ds\\
&\geq\gamma\int_\tau^{1-\tau}G(s,s)f_n(s,x_1(s))ds\\
&\geq L\gamma\int_\tau^{1-\tau}G(s,s)x_1(s)ds \\
&\geq LR'\gamma^2\int_\tau^{1-\tau}G(s,s)ds.
\end{align*}
This is a contradiction to $x_1\in\partial Q_{R'}$. Consequently,
\eqref{e3.6} holds. Furthermore, for each $x\in\partial Q_{R}$,
\begin{align*}
\|T_nx\|&\geq \int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t)\\
&\geq\int_0^1G(t,s)f_n(s,x(s))ds\\
&\geq\gamma\int_\tau^{1-\tau}G(s,s)f_n(s,x(s))ds\\
&\geq L\gamma\int_\tau^{1-\tau}G(s,s)x(s)ds \\
&\geq LR'\gamma^2\int_\tau^{1-\tau}G(s,s)ds.
\end{align*}
So $\inf_{x\in\partial Q_{R'}}\|T_nx\|>0$. Thus from Lemma \ref{lem1.2},
$i(T_n,Q_{R'},Q)=0$.

By the additivity of fixed point index, we know that
$$
i(T_n,Q_{R'}\setminus\overline{Q}_R,Q)=i(T_n,Q_{R'},Q)-i(T_n,Q_R,Q)=-1.
$$
 As a result, there exist
$x_n\in Q_{R'}\setminus\overline{Q}_R$ satisfying $T_nx_n=x_n$
provided that $n$ is sufficiently large.

Without loss of generality,
suppose $T_nx_n=x_n$, $n\geq n_0$.
Let $D=\{x_n\}_{n\geq n_0}$ be the
sequence of solutions to \eqref{e2.14}. It is not difficult to
prove that $D$ is uniformly bounded.
Next we show $\{x_n\}_{n\geq n_0}$ is
equicontinuous on $[0,1]$. It is obvious that we only need to prove
$\lim_{t\to0+}(x_n(t)-x_n(0))=0$,
$\lim_{t\to1-}(x_n(t)-x_n(1))=0$ uniformly with respect to
$n\geq n_0$ and $D$ is equicontinuous on
$[\sigma,1-\sigma]\subset(0,1)$ for $\sigma\in(0,1/2)$.


Now we prove that
\begin{equation} \label{e3.7}
\lim_{t\to0+}(x_n(t)-x_n(0))=0, \quad
\text{uniformly with respect to } n\geq n_0.
\end{equation}
According to \eqref{e2.12},
\begin{align}
 &\big|x_n(t)-x_n(0)\big| \notag \\
&=\Big|\int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t) \notag \\
&\quad -\int_0^1G(0,s)f_n(s,x(s))ds
-A(f_n(s,x(s))\psi(0)-B(f_n(s,x(s))\phi(0)\Big| \notag \\
&=\frac{1}{\rho}\phi(t)\int_0^t\psi(s)f_n(s,x(s))ds
 +\frac{1}{\rho}(\psi(t)-\psi(0))\int_t^1\phi(s)f_n(s,x(s))ds \label{e3.8} \\
&\quad -\frac{1}{\rho}\phi(0)\int_0^t\psi(s)f_n(s,x(s))ds
 +A(f_n(s,x(s)))(\psi(t)-\psi(0)) \notag \\
&\quad +B(f_n(s,x(s)))(\phi(t)-\phi(0)) \notag \\
&\leq \frac{1}{\rho}\phi(t)\int_0^t\psi(s)h(s)g
 \big(\max\{\frac{1}{n},x(s)\}\big)ds \notag \\
&\quad +\frac{1}{\rho}(\psi(t)-\psi(0))
 \int_t^1\phi(s)h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)ds \notag\\
&\quad-\frac{1}{\rho}\phi(0)\int_0^t\psi(s)h(s)
g\big(\max\{\frac{1}{n},x(s)\}\big)ds  \notag \\
&\quad +A\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)
(\psi(t)-\psi(0)) \notag\\
&+B\Big(h(s)g\big(\max\{\frac{1}{n},x(s)\}\big)\Big)(\phi(0)-\phi(t))
 \notag \\
&\leq \frac{1}{\rho}M_4\phi(t)\int_0^t\psi(s)h(s)ds+
\frac{1}{\rho}M_4(\psi(t)-\psi(0))\int_t^1\phi(s)h(s)ds \notag\\
&\quad +\frac{1}{\rho}\phi(0)M_4\int_0^t\psi(s)h(s)ds
+AM_4(\psi(t)-\psi(0))\int_0^1G(s,s)h(s)ds \notag\\
&\quad +BM_4(\phi(0)-\phi(t))\int_0^1G(s,s)h(s)ds. \notag
\end{align}
Since $\{x_n(t)\}_{n\geq n_0}$ is uniformly bounded, it follows that
$\{g(x_n)\}_{n\geq n_0}$ is bounded. Therefore, there exists a
constant $M_4$ such that $\|g(x_n)\|<M_4$ for $n\geq n_0$. This
together with (H3) and \eqref{e3.8} show that we need to prove only
that
\begin{gather} \label{e3.9}
\lim_{t\to0+}\frac{1}{\rho}\phi(t)\int_0^t\psi(s)h(s)ds=0,\\
\label{e3.10}
\lim_{t\to0+}\frac{1}{\rho}\phi(0)\int_0^t\psi(s)h(s)ds=0,\\
\label{e3.11}
\lim_{t\to0+}\frac{1}{\rho}(\psi(t)-\psi(0))\int_t^1\phi(s)h(s)ds=0,\\
\label{e3.12}
\lim_{t\to0+}(\psi(t)-\psi(0))=0, \quad
\lim_{t\to0+}(\phi(0)-\phi(t))=0.
\end{gather}
Since $\psi(t)$ and $\phi(t)$ are continuous on $[0,1]$,
\eqref{e3.12} holds. For all $\varepsilon>0$, by the absolutely
continuity of integral function and (H3), there exists
$\delta_1\in(0,\frac{1}{2})$ such that
$t_1,t_2\in[0,1],|t_1-t_2|<\delta_1$ implies
\begin{equation} \label{e3.13}
\big|\int_{t_1}^{t_2}G(s,s)h(s)ds\big|<\varepsilon.
\end{equation}
Therefore, from \eqref{e2.5} and \eqref{e3.13}, we have
\begin{gather*}
\frac{1}{\rho}\phi(t)\int_0^t\psi(s)h(s)ds\leq\int_0^tG(s,s)h(s)ds
<\varepsilon,\quad t\in(0,\delta_1], \\
\frac{1}{\rho}\phi(0)\int_0^t\psi(s)h(s)ds
 \leq\frac{\phi(0)}{\phi(\delta_1)}\int_0^tG(s,s)h(s)ds<\varepsilon,
\quad  t\in(0,\delta_1];
\end{gather*}
i.e., \eqref{e3.9} and \eqref{e3.10} hold.
\begin{align*}
&\frac{1}{\rho}(\psi(t)-\psi(0))\int_t^1\phi(s)h(s)ds\\
&\leq \frac{1}{\rho}\psi(t)\int_t^{\delta_1}\phi(s)h(s)ds+\frac{1}{\rho}(\psi(t)-\psi(0))
\int^1_{\delta_1}\phi(s)h(s)ds\\
&\leq \int_t^{\delta_1}
G(s,s)h(s)ds+\frac{\psi(t)-\psi(0)}{\psi(\delta_1)}
\int_{\delta_1}^1G(s,s)h(s)ds\leq2\varepsilon.
\end{align*}
That is, \eqref{e3.11} holds. By \eqref{e3.9}-\eqref{e3.12},
 \eqref{e3.7} holds.
Since
\begin{align*}
&|x_n(t)-x_n(1)| \\
&= \Big|\int_0^1G(t,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t)+B(f_n(s,x(s)))\phi(t)\\
&\quad -\int_0^1G(1,s)f_n(s,x(s))ds
-A(f_n(s,x(s))\psi(1)-B(f_n(s,x(s))\phi(1)\Big|\\
&\leq \frac{1}{\rho}(\phi(t)-\phi(1))\int_0^t\psi(s)f_n(s,x(s))ds+\frac{1}{\rho}\psi(t)\int_t^1\phi(s)f_n(s,x(s))ds\\
&\quad -\frac{1}{\rho}\phi(1)\int_t^1\psi(s)f_n(s,x(s))ds+A(f_n(s,x(s)))(\psi(t)-\psi(1))\\
&\quad +B(f_n(s,x(s)))(\phi(t)-\phi(1))\\
&\leq \frac{1}{\rho}M_4(\phi(t)-\phi(1))\int_0^t\psi(s)h(s)ds+
\frac{1}{\rho}M_4\psi(t)\int_t^1\phi(s)h(s)ds\\
&\quad +\frac{1}{\rho}\phi(1)M_4\int_0^t\psi(s)h(s)ds
+AM_4(\psi(1)-\psi(t))\int_0^1G(s,s)h(s)ds\\
&\quad +BM_4(\phi(t)-\phi(1))\int_0^1G(s,s)h(s)ds,
\end{align*}
similar to the above, we can easily prove that
\begin{equation} \label{e3.14}
\lim_{t\to1-}(x_n(t)-x_n(1))=0, \quad \text{uniformly with respect
to } n\geq n_0.
\end{equation}

Next we prove that $D$ is equicontinuous on $[\sigma,1-\sigma]$ for
any $\sigma\in(0,1/2)$. In fact, for $n\geq n_0$,
$t_1,t_2\in[\sigma,1-\sigma]$ with $t_2>t_1$, we have
\begin{equation} \label{e3.15}
\begin{aligned}
&|x_n(t_2)-x_n(t_1)|\\
&= \Big|\int_0^1G(t_2,s)f_n(s,x(s))ds
+A(f_n(s,x(s)))\psi(t_2)+B(f_n(s,x(s)))\phi(t_2) \\
&\quad -\int_0^1G(t_1,s)f_n(s,x(s))ds
-A(f_n(s,x(s)))\psi(t_1)-B(f_n(s,x(s)))\phi(t_1)\Big|\\
&\leq \frac{1}{\rho}(\phi(t_2)-\phi(t_1))\int_0^{t_1}\psi(s)f_n(s,x(s))ds
+\frac{1}{\rho}\phi(t_2)\int_{t_1}^{t_2}\psi(s)f_n(s,x(s))ds\\
&\quad +\frac{1}{\rho}(\psi(t_2)-\psi(t_1))\int_{t_2}^1\phi(s)f_n(s,x(s))ds-
\frac{1}{\rho}\psi(t_1)\int_{t_1}^{t_2}\phi(s)f_n(s,x(s))ds\\
&\quad +A(f_n(s,x(s)))(\psi(t_2)-\psi(t_1))+B(f_n(s,x(s)))
 (\phi(t_2)-\phi(t_1)).
\end{aligned}
\end{equation}
By \eqref{e2.5} and the monotonicity of $\psi(t)$ and $\phi(t)$, we have
\begin{gather} \label{e3.16}
\frac{1}{\rho}\int_0^{t_1}\psi(s)h(s)ds\leq\frac{1}{d}
\int_0^1G(s,s)h(s)ds, \\
 \label{e3.17}
\frac{1}{\rho}\int_{t_2}^1\phi(s)h(s)ds\leq\frac{1}{b}
 \int_0^1G(s,s)h(s)ds, \\
 \label{e3.18}
\frac{1}{\rho}\phi(t_2)\int_{t_1}^{t_2}\psi(s)h(s)ds
\leq\int_{t_1}^{t_2}G(s,s)h(s)ds, \\
 \label{e3.19}
\frac{1}{\rho}\psi(t_1)\int_{t_1}^{t_2}\phi(s)h(s)ds
\leq\int_{t_1}^{t_2}G(s,s)h(s)ds.
\end{gather}
By \eqref{e3.15}-\eqref{e3.19}, we have
\begin{align*} %\label{e3.20}
&|x_n(t_2)-x_n(t_1)|\\
&\leq M_4(\frac{1}{b}+A)(\psi(t_2)-\psi(t_1))\int_0^1G(s,s)h(s)ds \\
&\quad +M_4(\frac{1}{d}+B)(\phi(t_1)-\phi(t_2))\int_0^1G(s,s)h(s)ds
 +2M_4\int_{t_1}^{t_2}G(s,s)h(s)ds.
\end{align*}
By the above inequality, (H3), \eqref{e3.9}-\eqref{e3.12}, and
continuity of $\psi(t)$, $\phi(t)$, $D$ is equicontinuous on
$[\sigma,1-\sigma]$.

 From the above proof, we can know $D$ is equicontinuous on $[0,1]$.
It follows from Ascoli-Arzela's theorem that the sequence
$\{x_n\}_{n\geq n_0}$ has a subsequence which uniformly converges on
$[0,1]$. Without loss of generality, we assume that $\{x_n\}$ itself
uniformly converges to $x$ on $[0,1]$. According to the Lebesgue's
dominated theorem, we know that $x$ is the positive solution of
\eqref{e1.1}-\eqref{e1.2}.
\end{proof}

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\end{document}
