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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 63, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/63\hfil Existence of positive solutions]
{Existence of positive solutions for singular fractional
differential equations with integral boundary conditions}

\author[J. Jin,  X. Liu, M. Jia \hfil EJDE-2012/63\hfilneg]
{Jingfu Jin,  Xiping Liu, Mei Jia} 


\address{Jingfu Jin \newline
College of Science, University of Shanghai for Science and Technology,
Shanghai 200093, China}
\email{jinjingfu2005@126.com}

\address{Xiping Liu \newline
College of Science, University of Shanghai for Science and Technology,
 Shanghai 200093,  China}
\email{xipingliu@163.com}

\address{Mei Jia \newline
College of Science, University of Shanghai for Science and Technology,
 Shanghai 200093,  China}
\email{jiamei-usst@163.com}


\thanks{Submitted January 17, 2012. Published April 19, 2012.}
\thanks{Supported by grants 10ZZ93 from Innovation Program of Shanghai Municipal
Education \hfill\break\indent  Commission, and 11171220  from the National Natural Science
Foundation of China.}
\subjclass[2000]{34B16, 34B18, 26A33}
\keywords{Caputo derivative; fractional differential equations;
 positive solutions; \hfill\break\indent  integral boundary conditions;  
  singular differential equation}

\begin{abstract}
 This article shows the  existence of a positive solution
 for the singular fractional differential equation with integral
 boundary condition
 \begin{gather*}
 {}^C\!D^p u(t)=\lambda h(t)f(t, u(t)), \quad t\in(0, 1), \\
 u(0)-au(1)=\int^1_0g_0(s)u(s)\,\mathrm{d}s, \\
 u'(0)-b\,{}^C\!D^qu(1)=\int^1_0g_1(s)u(s)\,\mathrm{d}s, \\
 u''(0)=u'''(0)=\dots =u^{(n-1)}(0)=0,
 \end{gather*}
 where $\lambda $ is a parameter and the nonlinear term is allowed to be singular
 at $t=0, 1$ and $u=0$.
 We obtain an explicit interval for $\lambda$ such that for any $\lambda$ in
 this interval, existence of at least one positive solution is guaranteed.
 Our approach is by a fixed point theory  in cones combined with linear operator
  theory.
\end{abstract}

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

\section{Introduction}

We consider the singular integral boundary-value problem
involving Caputo fractional derivative:
\begin{equation}\label{eq1.5}
\begin{gathered}
{}^C\!D^p u(t)=\lambda h(t)f(t, u(t)), \quad t\in(0, 1), \\
u(0)-au(1)=\int^1_0g_0(s)u(s)\,\mathrm{d}s, \\
 u'(0)-b\,{}^C\!D^q u(1)=\int^1_0g_1(s)u(s)\,\mathrm{d}s, \\
u''(0)=u'''(0)=\dots =u^{(n-1)}(0)=0,
\end{gathered}
\end{equation}
where $ {}^C\!D$ is the standard Caputo derivative, $n\geq3$ is an integer,
 $p \in (n-1, n)$, $0<q<1$, $0<a<1, \ 0< b<\Gamma(2-q)$ are real numbers.
 $f\in C([0, 1]\times (0, +\infty), [0, +\infty))$, and $f(t,u)$ may be
singular at $u=0$. $g_0$, $g_1\in C[0,1]$ are given
 functions. $h\in C((0,1),[0,+\infty))$, $h(t)$ is allowed to be singular at $t=0,1$.

There exist a great number of important applications using
fractional differential equations in many areas, such as physics,
mechanics, chemistry, engineering, etc. Due to this, the study of
related problems has attracted much attention of the researchers,
especially most recently
\cite{Bai,Caballero,Zhang,Agarwal,Stanek,Feng,Jia,LXP,LXP2,LXP3,Podlubny}.
Also, as another important factor,
singularity is sometimes inevitable in the mathematical models of
modern science and technology areas.
 Among the studies of the existence of positive solutions for singular boundary-value problems,
 extensive work has been done for the singular integer order differential equations with integral
  boundary conditions; see \cite{Liu1,Liu2,Liu3,Li,Liu4,Liu5,Liu6,Jiang}, and the references
   therein.
On the other hand, for fractional differential equations, however,
most results on singular boundary-value problems are only restricted
to the two-point boundary conditions
\cite{Bai,Caballero,Zhang,Agarwal,Stanek,  Feng}.
 For example,  with the assumptions of $1<\alpha < 2$ and $f(t, x, y) $ is singular at
 $x=0$, \cite{Agarwal} discussed existence and multiplicity of positive solutions for the
 two-point boundary-value problem \eqref{E1}:
\begin{equation}\label{E1}
\begin{gathered}
D^\alpha_{0+} u(t)+f(t, u(t),D^\mu u(t))=0, \quad t\in(0, 1), \\
u(0)=u(1)=0 ,
\end{gathered}
\end{equation}
where $\mu>0$ and $\alpha-\mu\geq1$, $D^\alpha_{0+}$ is the
standard Riemann-Liouville derivative.

When $2<\alpha <3$, the two-point boundary-value problems \eqref{E2}
and \eqref{E3} are studied in \cite{Bai} and \cite{Stanek} respectively:
\begin{equation}\label{E2}
\begin{gathered}
D^\alpha_{0+} u(t)+f(t, u(t))=0, \quad t\in(0, 1), \\
u(0)=u'(1)=u''(0)=0,
\end{gathered}
\end{equation}
\begin{equation}\label{E3}
and
\begin{gathered}
D^\alpha u(t)+f(t, u(t),u'(t),D^\mu u(t))=0,  \\
u(0)=0, \ u'(0)=u'(1)=0.
\end{gathered}
\end{equation}
In \eqref{E2}, $f$ is assumed to be singular at $t=0$, and $D^\alpha_{0+}$
is the standard Caputo derivative. In \eqref{E3}, $f(t,x,y,z)$ may be singular
at the value $0$ of all variables $x, y, z$ and $D^\alpha u(t)$ is the
standard Riemann-Liouville fractional derivative.

In the literature, results on singular integral boundary-value problems of the
fractional differential equations are relatively rare. In this paper,
we first give the Green function of  boundary-value problem (BVP) \eqref{eq1.5}
and prove some of its properties. Then, applying a  fixed-point theorem with
linear operator theory analysis, we obtain some sufficient conditions
on the existence of positive solutions of \eqref{eq1.5}.  An explicit interval
for $\lambda$ is derived such that for any $\lambda$ in this interval,
the existence of at least one positive solution is guaranteed.

\section{Preliminaries}

  In this section, we introduce definitions and preliminary facts which
are used throughout this   paper.

\begin{definition}[\cite{Podlubny}] \label{def2.1} \rm
The fractional integral of order $\alpha>0$ of a function
 $y:(0, +\infty)\to \mathbb{R}$ is given by
$$
I^\alpha y(t)=\frac{1}{\Gamma(\alpha)}\int^t_0(t-s)^{\alpha-1}y(s)\,\mathrm{d}s.
$$
provided that the right side is point wise defined on $(0,
+\infty)$,  and $\Gamma$ denotes the
  Gamma function.
\end{definition}

\begin{definition}[\cite{Podlubny}] \label{def2.2} \rm
The fractional Caputo  derivative of order
  $\alpha>0$  for a function $x:(0, +\infty)\to \mathbb{R}$ is given by
$$
{}^C\!D^\alpha
x(t)=\frac{1}{\Gamma(n-\alpha)}\int^t_0\frac{x^{(n)}(s)}{(t-s)^{\alpha+1-n}}
\,\mathrm{d}s,
$$
where $n=[\alpha] + 1$, provided the right integral converges.
\end{definition}

\begin{lemma}\label{lemma1}
 Suppose that $y\in C[0, 1]$ and  $n\geq 3$ is an integer, $p\in(n-1, n)$,
$0<q<1$, $0<a<1$,
 $0<b<\Gamma(2-q)$.
 Then the integral boundary-value problem
\begin{equation}\label{eq2.1}
\begin{gathered}
{}^C\!D^p u(t)=y(t), \quad t\in(0, 1), \\
u(0)-au(1)=\int^1_0g_0(s)u(s)\,\mathrm{d}s, \\
 u'(0)-b{}^C\!D^qu(1)=\int^1_0g_1(s)u(s)\,\mathrm{d}s, \\
u''(0)=u'''(0)=\dots =u^{(n-1)}(0)=0
\end{gathered}
\end{equation}
is equivalent to the fractional integral equation
\begin{align}  \label{eq2.2}
    u(t)=
  \int^1_0G(t, s)y(s)\,\mathrm{d}s+\int^1_0\Phi(t, s)u(s)\,\mathrm{d}s,
\end{align}
where
\begin{equation}\label{eq2.3}
  G(t, s)=
\begin{cases}
     \frac{(t-s)^{p-1}}{\Gamma(p)}+\frac{a\Gamma(p-q)(\Gamma(2-q)-b)
      (1-s)^{p-1}+b\Gamma(2-q)\Gamma(p)(a+t-at)(1-s)^{p-q-1}}{(1-a)
      (\Gamma(2-q)-b)\Gamma(p-q)\Gamma(p)}, \\
\quad \text{if } 0\leq s\leq t\leq 1, \\[4pt]
   \frac{a\Gamma(p-q)(\Gamma(2-q)-b)(1-s)^{p-1}+b\Gamma(2-q)
      \Gamma(p)(a+t-at)(1-s)^{p-q-1}}{(1-a)(\Gamma(2-q)-b)\Gamma(p-q)
      \Gamma(p)}, \\
\quad \text{if } 0\leq t\leq s\leq 1.
   \end{cases}
\end{equation}
and
\begin{equation}\label{eq2.4}
  \Phi(t, s)=\frac{(a+t-at)\Gamma(2-q)g_1(s)}{(1-a)(\Gamma(2-q)-b)}
        +\frac{g_0(s)}{1-a}.
\end{equation}
\end{lemma}

\begin{proof}
 From ${}^C\!D^p  u(t)=y(t)$, $t\in(0, 1)$
and the boundary conditions $u''(0)=u'''(0)=\dots =u^{(n-1)}(0)=0$,
we have
\begin{align*}
   u(t)&= I_{t}^p y(t)+u(0)+u'(0)t+\frac{u''(0)}{2!}t^2+\dots
   +\frac{u^{(n-1)}(0)}{(n-1)!}t^{n-1} \\
       &=\frac{1}{\Gamma(p)}\int^t_0(t-s)^{p-1}y(s)\,\mathrm{d}s+u(0)+u'(0)t.
\end{align*}
By  properties of the  Caputo  derivative, we get
\begin{align*}
   {}^C\!D^q u(t)&= I_{t}^{p-q}y(t)+^C\hspace{-0.4em}D^q(u(0)+u'(0)t) \\
       &=\frac{\int^t_0(t-s)^{p-q-1}y(s)\,\mathrm{d}s}{\Gamma(p-q)}
       +\frac{u'(0)t^{1-q}}{\Gamma(2-q)}.
\end{align*}
Then
$$
  u(1)=\frac{1}{\Gamma(p)}\int^1_0(1-s)^{p-1}y(s)\,\mathrm{d}s+u(0)+u'(0),
$$
and
$$
    {}^C\!D^q u(1)=\frac{\int^1_0(1-s)^{p-q-1}y(s)\,\mathrm{d}s}{\Gamma(p-q)}
    +\frac{u'(0)}{\Gamma(2-q)}.
$$
By the boundary conditions
$u(0)-au(1)=\int^1_0g_0(s)u(s)\,\mathrm{d}s$ and
 $ u'(0)-b{}^C\!D^qu(1)=\int^1_0g_1(s)u(s)\,\mathrm{d}s$, we
have
$$
  u(0)-\frac{a}{\Gamma(p)}\int^1_0(1-s)^{p-1}y(s)\,\mathrm{d}s-au(0)-au'(0)
  =\int^1_0g_0(s)u(s)\,\mathrm{d}s,
$$
and
$$
  u'(0)-\frac{b\int^1_0(1-s)^{p-q-1}y(s)\,\mathrm{d}s}{\Gamma(p-q)}-
  \frac{bu'(0)}{\Gamma(2-q)}=\int^1_0g_1(s)u(s)\,\mathrm{d}s.
$$
Hence,
\begin{align*}
u'(0)
&=\frac{b\Gamma(2-q)}{(\Gamma(2-q)-b)\Gamma(p-q)}\int^1_0(1-s)^{p-q-1}y(s)
 \,\mathrm{d}s\\
&\quad +\frac{\Gamma(2-q)}{\Gamma(2-q)-b}\int^1_0g_{1}(s)u(s)\,\mathrm{d}s,
\end{align*}
and
\begin{align*}
  u(0)&=\frac{1}{1-a}\int^1_0g_0(s)u(s)\,\mathrm{d}s
       +\frac{a\int^1_0(1-s)^{p-1}y(s)\,\mathrm{d}s}{(1-a)\Gamma(p)}\\
&\quad  +\frac{ab\Gamma(2-q)\int^1_0(1-s)^{p-q-1}y(s)\,\mathrm{d}s}
      {(1-a)(\Gamma(2-q)-b)\Gamma(p-q)}
  +\frac{a\Gamma(2-q)\int^1_0g_{1}(s)u(s)\,\mathrm{d}s}
      {(1-a)(\Gamma(2-q)-b)}.
\end{align*}
We  can  easily obtain
\begin{align*}
  u(t)&= \frac{1}{\Gamma(p)}\int^t_0(t-s)^{p-1}y(s)\,\mathrm{d}s
      +\frac{bt\Gamma(2-q)\int^1_0(1-s)^{p-q-1}y(s)\,\mathrm{d}s}
      {(\Gamma(2-q)-b)\Gamma(p-q)}\\
&\quad  +\frac{a\int^1_0(1-s)^{p-1}y(s)\,\mathrm{d}s}{(1-a)\Gamma(p)}
  +\frac{ab\Gamma(2-q)\int^1_0(1-s)^{p-q-1}y(s)\,\mathrm{d}s}
      {(1-a)(\Gamma(2-q)-b)\Gamma(p-q)}\\
&\quad +\frac{1}{1-a}\int^1_0g_0(s)u(s)
      \,\mathrm{d}s+\frac{t\Gamma(2-q)\int^1_0g_{1}(s)u(s)\,\mathrm{d}s}
      {\Gamma(2-q)-b}\\
&\quad+\frac{a\Gamma(2-q)\int^1_0g_{1}(s)u(s)\,\mathrm{d}s}
      {(1-a)(\Gamma(2-q)-b)}\\
    &=
     \int^1_0G(t, s)y(s)\,\mathrm{d}s+\int^1_0\Phi(t, s)u(s)\,\mathrm{d}s.
\end{align*}
The proof is complete.
\end{proof}

Denote
\begin{gather*}
  k_1 =\frac{(\Gamma(2-q)-b)\Gamma(p-q)+b\Gamma(2-q)\Gamma(p)}
  {(1-a)(\Gamma(2-q)-b)\Gamma(p-q)\Gamma(p)},\\
  k_2=\frac{ab\Gamma(2-q)}{(1-a)(\Gamma(2-q)-b)\Gamma(p-q)}.
\end{gather*}

\begin{lemma}\label{lem2}
  The function $G(t, s)$ in Lemma \ref{lemma1} satisfies the following
  conditions:
\begin{itemize}
\item[(i)] $G(t, s)$ is continuous on $[0, 1]\times[0, 1]$;
\item[(ii)] $G(t, s)\leq
     k_1 (1-s)^{p-q-1}$, for any $(t, s)\in[0, 1]\times[0, 1]$;
\item[(iii)] $G(t, s)\geq
    k_2(1-s)^{p-q-1}$, for any $(t, s)\in[0, 1]\times[0, 1]$.
\end{itemize}
\end{lemma}

\begin{proof}   It is easy to check that (i) holds and
$G(t, s)\geq 0$ on  $[0, 1]\times[0, 1]$.

 (ii) For $0\leq s\leq t\leq 1$, denote
\begin{align*}
& G_1(t, s)\\
&=\frac{a\Gamma(p-q)(\Gamma(2-q)-b)
  (1-s)^{p-1}
  +b\Gamma(2-q)\Gamma(p)(a+t-at)(1-s)^{p-q-1}}{(1-a)(\Gamma(2-q)-b)
  \Gamma(p-q)\Gamma(p)}\\
&\quad +\frac{(t-s)^{p-1}}{\Gamma(p)},
\end{align*}
 and for $0\leq t\leq s \leq1$, denote
\begin{align*}
&G_2(t,s)\\
&=\frac{a\Gamma(p-q)(\Gamma(2-q)-b)(1-s)^{p-1}
  +b\Gamma(2-q)\Gamma(p)
  (a+t-at)(1-s)^{p-q-1}}{(1-a)(\Gamma(2-q)-b)\Gamma(p-q)\Gamma(p)}.
\end{align*}
For $0\leq s\leq t\leq 1$, we have
\begin{align*}
    &(1-a)(\Gamma(2-q)-b)\Gamma(p-q)\Gamma(p)G_1(t, s)\\
&\leq
    (1-a)(\Gamma(2-q)-b)\Gamma(p-q)(1-s)^{p-1}+a\Gamma(p-q)(\Gamma(2-q)-b)
    (1-s)^{p-1}\\
&\quad
    +b\Gamma(2-q)\Gamma(p)(a+t-at)(1-s)^{p-q-1}\\
&\leq
   (1-s)^{p-q-1}[(\Gamma(2-q)-b)\Gamma(p-q)(1-s)^{q}
    +b\Gamma(2-q)\Gamma(p)(a+t-at)]\\
&\leq (1-s)^{p-q-1}[(\Gamma(2-q)-b)\Gamma(p-q)+b\Gamma(2-q)\Gamma(p)].
\end{align*}
Hence, $G_1(t, s)\leq k_1 (1-s)^{p-q-1}$, for any $0\leq s\leq t\leq 1$.

For $0\leq t\leq s\leq 1$, we have
\begin{align*}
&(1-a)(\Gamma(2-q)-b)\Gamma(p-q) \Gamma(p)G_2(t, s)\\
&=   (1-s)^{p-q-1}[a\Gamma(p-q)(\Gamma(2-q)-b)(1-s)^{q}+
   b\Gamma(2-q)\Gamma(p)(a+t-at)]\\
&\leq  (1-s)^{p-q-1}[(\Gamma(2-q)-b)\Gamma(p-q)+b\Gamma(2-q)\Gamma(p)].
\end{align*}
Hence, $G_{2}(t, s)\leq k_1 (1-s)^{p-q-1}$.

Therefore, $G(t, s)\leq k_1 (1-s)^{p-q-1}$, for any $(t, s)\in[0,1]\times[0, 1]$.

(iii) It is easy to see for $(t, s)\in[0, 1]\times[0, 1]$,
\begin{align*}
  (1-a)(\Gamma(2-q)-b)\Gamma(p-q)\Gamma(p)G(t, s)
&\geq    b\Gamma(2-q)\Gamma(p)(a+t-at)(1-s)^{p-q-1}\\
&\geq   ab\Gamma(2-q)\Gamma(p)(1-s)^{p-q-1}.
\end{align*}
Therefore, $G(t, s)\geq k_2 (1-s)^{p-q-1}$, for any $(t, s)\in[0,1]\times[0, 1]$.
 \end{proof}

Denote
$$
  m_0=\min_{t,s\in[0,1]}\Phi(t,s),\quad
  M_0=\max_{t,s\in[0,1]}\Phi(t,s).
$$
Let $E=C[0,1]$ be the Banach space with the norm
$\| u\|=\max_{0\leq t\leq1}| u(t)|$,
$P=\{u\in E:u(t)\geq 0\}$ and
$K=\{u\in P: u(t)\geq \frac{k_2(1-M_0)}{k_1 }\| u \|\}$
be cones of $E$.

Denote $ K_{r}=\{u\in K: \| u\|<r\}$,
$\partial K_{r}=\{u\in K: \| u \|=r\}$, and
$\overline{K}_{r,R}=\{u\in K: r\leq\| u \|\leq R\}$,
where $0<r<R<+\infty$.

\begin{lemma}[\cite{Guo,Deimling}] \label{lem2.3}
Let $K$ be a positive cone in real Banach space $E$, $0<r<R<+\infty$,
and let $S:\overline{K}_{r,R}\to K$
be a completely continuous operator and such that
\begin{itemize}
\item[(i)] $\|Su\|\leq\|u\|$ for $u\in\partial K_{R}$;
\item[(ii)] There exists $e\in \partial K_{r}$ such that $u\neq Su+me$
for any $u\in \partial K_{r}$, and $m>0$.
\end{itemize}
Then $S$ has a fixed point in $\overline{K}_{r,R}$.
\end{lemma}

\begin{remark}\label{remark} \rm
If (i) and (ii) are satisfied for $u\in \partial K_{r}$ and $e\in
\partial K_{R}$, respectively. Then Lemma \ref{lem2.3} is still true.
\end{remark}

Define a linear operator  $A:E \to E$, by
\begin{equation}\label{eq3.1}
 Au(t)=\int^1_0\Phi(t,s)u(s)\,\mathrm{d}s.
\end{equation}

\begin{lemma}\label{lemma3.1}
Suppose $0\leq m_0\leq M_0<1$ holds. Then
\begin{itemize}
\item[(i)] $A$ is a bounded linear operator;
\item[(ii)] $A(P)\subset P$;
\item[(iii)] $(I-A)$ is invertible and
$\|(I-A)^{-1}\|\leq\frac{1}{1-M_0}$.
\end{itemize}
\end{lemma}

\begin{proof}
 (i) It is easy to see that $A$ is a linear
operator with
$$
  |Au(t)|=\big|\int^1_0\Phi(t, s)u(s)\,\mathrm{d}s\big|\leq
  M_0\|u\|.
$$
Therefore, $\|A\|\leq M_0<1$. It follows that $A$ is a bounded
linear operator.

(ii)  For each $u\in P$, we have $u\in C([0,1])$, $u(t)\geq 0$.
Since $\Phi(t, s)$ is continuous and nonnegative, it is easy to
check that $Au\in C([0,1])$, $Au(t)\geq 0$. This implies that
$A(P)\subset P$.


(iii) We have proved in (i) that $\|A\|\leq M_0<1$, which implies
that $(I-A)^{-1}$ is invertible.

To find the expression for $(I-A)^{-1}$, we use the theory of
Fredholm integral equations. We have $u(t)=(I-A)^{-1}v(t)$ if and
only if $u(t)=v(t)+Au(t)$ for each $t \in [0,1]$. The definition of
the operator $A$ implies that
\begin{equation}\label{eq3.2}
u(t)=v(t)+\int^1_0\Phi(t, s)u(s)\,\mathrm{d}s.
\end{equation}
The condition $\|A\|\leq M_0<1$ implies that 1 is not an
eigenvalue of the kernel $\Phi(t, s)$.

Hence, \eqref{eq3.2} has a unique solution $u\in E$, for each $v\in
E$. By successive substitutions in \eqref{eq3.2}, we obtain
\begin{equation}\label{eq3.3}
   u(t)=v(t)+\int^1_0 \rho(t, s)v(s)\,\mathrm{d}s,
\end{equation}
where the resolvent kernel $\rho(t, s)$ is given by
$$
  \rho(t,s)=\sum_{j=1}^{\infty} \Phi_{j}(t, s),
$$
where $\Phi_{1}(t, s)=\Phi(t, s)$,
 $\Phi_{j}(t, s)=\int^1_0\Phi(t,\tau)\Phi_{j-1}(\tau, s)\,\mathrm{d}\tau$,
$(j=2, 3, \dots)$. Since
$0\leq m_0\leq \Phi(t, s)\leq M_0<1$, we have
$m_0^{j}\leq\Phi_{j}(t, s)\leq M_0^{j}$, $(j=1, 2, 3, \dots)$.
Hence, we have
\begin{equation}\label{eq3.4}
  \frac{m_0}{1-m_0}\leq \rho(t, s)\leq \frac{M_0}{1-M_0},
\end{equation}
and $\rho(t, s)$ is continuous on $[0,1]\times[0,1]$. In view of
\eqref{eq3.3} and \eqref{eq3.4}, we obtain
\begin{align*}
|(I-A)^{-1}v(t)|&\leq |v(t)|+\int^1_0 |\rho(t,
s)v(s)|\,\mathrm{d}s\leq(1+\frac{M_0}{1-M_0})\|v\|=\frac{1}{1-M_0}\|v\|.
\end{align*}
That is, $\|(I-A)^{-1}\|\leq 1/(1-M_0)$.
 \end{proof}

Define a nonlinear operator $T: E\to E $,  by
\begin{equation}\label{eq3.5b}
  Tu(t)=\lambda\int^1_0G(t, s)h(s)f(s, u(s))\,\mathrm{d}s.
\end{equation}

 In view of \eqref{eq3.1}, \eqref{eq3.5b}, and Lemma \ref{lemma1},
we can easily prove that the  existence of solutions to  \eqref{eq1.5}
is equivalent to the existence of solutions to the  equation
\begin{equation}\label{eq3.6}
  u(t)=Tu(t)+Au(t),\quad t\in [0,1].
\end{equation}
 It follows from Lemma \ref{lemma3.1}  that $u$ is a
 solution of \eqref{eq3.6} if and only if $u$ is a solution of
 $u(t)=(I-A)^{-1}Tu(t)$. That is, $u$ is a fixed point of the operator
$S:=(I-A)^{-1}T$. By \eqref{eq3.3} and \eqref{eq3.5b}, we have
\begin{equation}\label{eq3.7}
\begin{split}
  (Su)(t)&=\lambda\int^1_0G(t, s)h(s)f(s, u(s))\,\mathrm{d}s\\
 &\quad +\lambda\int^1_0\rho(t, s)\int^1_0
  G(s, \tau)h(\tau)f(\tau, u(\tau))\,\mathrm{d}\tau\,\mathrm{d}s.
\end{split}
\end{equation}

We can prove the following lemma.

\begin{lemma}\label{lemma}
A function  $u$ is a solution of {\rm\eqref{eq1.5}} if and only if
$u$ is a fixed point of  the operator $S$.
\end{lemma}

We denote
$$
L=\int^1_0 (1-s)^{p-q-1}h(s)\,\mathrm{d}s.
$$
and assume the following conditions hold
\begin{itemize}
\item[(H)] $h\in C((0,1),[0,+\infty))$,
$\int^1_0 h(s)\,\mathrm{d}s<+\infty$ and $0<L<+\infty $.
\end{itemize}
To overcome the singularity, we consider the following approximating
equation of \eqref{eq3.7} with boundary condition of \eqref{eq1.5},
\begin{equation}\label{eq3.8}
\begin{split}
  (S_{n}u)(t)
&=\lambda\int^1_0G(t, s)h(s)f_{n}(s, u(s))\,\mathrm{d}s\\
&\quad  +\lambda\int^1_0\rho(t, s)\int^1_0G(s, \tau)h(\tau)f_{n}(\tau, u(\tau))
  \,\mathrm{d}\tau\,\mathrm{d}s.
\end{split}
\end{equation}
where $n$ is a positive integer and
$f_{n}(t,u)=f(t,\max\{1/n,u\})$.


\begin{lemma}\label{lemma3.2}
Suppose $0\leq m_0\leq M_0<1$ and {\rm (H)} holds. Then for each positive
integer $n$, we have
\begin{itemize}
\item[(i)] For any $0<r\leq R<+\infty$, the operator
$S_{n}:\overline{K}_{r,R}\to P$ is completely continuous;

\item[(ii)] $S_{n}(\overline{K}_{r,R})\subset K$.
\end{itemize}
\end{lemma}

\begin{proof}  (i) Suppose $D\subset \overline{K}_{r,R}$ is a
 bounded set. Then there exists $r_1  >0$ such that $\|u\|\leq r_1  $
for any $u\in D$. Denote
$$
M_1 =\max\{f(t,\max\{1/n,u\}):
(t,u)\in[0,1]\times[\frac{1}{n},\frac{1}{n}+r_1  ]\}.
$$
By  \eqref{eq3.4} and
Lemma \ref{lem2}, for any $u\in D$ and $t\in[0,1]$, we have
\begin{align*}
&|S_{n}u(t)|\\
&= |\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u(s))\,\mathrm{d}s
       +\lambda \int^1_0 \rho(t,s)\int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u(\tau))
       \,\mathrm{d}\tau\,\mathrm{d}s|\\
&\leq \lambda k_1 \int^1_0 (1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
&\quad+\frac{M_0\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)
       f_{n}(s,u(s))\,\mathrm{d}s \\
&=\frac{\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)
       f(s,\max\{1/n,u(s)\})\,\mathrm{d}s \\
&\leq \frac{\lambda k_1 M_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}
       h(s)\,\mathrm{d}s \\
&= \frac{\lambda k_1 M_1 L}{1-M_0}.
\end{align*}
Therefore, $S_{n}(D)$ is uniformly bounded.

We can also prove that $S_{n}(D)$ is equicontinuous.
For $t_1,t_2 \in[0,1]$ and $u\in D$, we have
\begin{align*}
&|(S_nu)(t_1)-(S_nu)(t_2)|\\
&= |\lambda \int^1_0 (G(t_1,s)-G(t_2,s))h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
&\quad +\lambda \int^1_0 (\rho(t_1,s)-\rho(t_2,s))\int^1_0 G(s,\tau)h(\tau)
      f_{n}(\tau,u(\tau))\,\mathrm{d}\tau\,\mathrm{d}s| \\
&\leq \lambda M_1 \Big(\int^1_0 |G(t_1,s)-G(t_2,s)|h(s)\,\mathrm{d}s\\
&\quad +k_1 \int^1_0 |\rho(t_1,s)-\rho(t_2,s)|\int^1_0 (1-s)^{p-q-1}h(\tau)
       \,\mathrm{d}\tau\,\mathrm{d}s\Big) \\
&\leq \lambda M_1 \Big(\int^1_0 |G(t_1,s)-G(t_2,s)|h(s)\,\mathrm{d}s
       +k_1 L\int^1_0 |\rho(t_1,s)-\rho(t_2,s)|\,\mathrm{d}s\Big).
\end{align*}
Since $G(t,s)$ and $\rho(t,s)$ are continuous on $[0,1]\times[0,1]$,
we can get $G(t,s)$ and $\rho(t,s)$ are uniformly continuous on
$[0,1]\times[0,1]$, it follows that
$|(S_{n}u)(t_{2})-(S_{n}u)(t_{1})|\to 0$ as
$|t_{2}-t_{1}|\to 0 $.
Hence, $S_{n}(D)$ is equicontinuous.
Using the Ascoli-Arzela's theorem, $S_{n}(D)$ is
relatively compact. Therefore, $S_{n}:\overline{K}_{r,R}\to
P$ is compact.

Now we show that $S_{n}$ is continuous. Suppose
$u, u_{m}\in D,\;(m=1,2,3,\dots)$ with $\|u_{m}-u\|\to 0$
as $ m\to\infty $. Then there exists $r_2  >0$ such that
$\|u_{m}\|<r_2  $ and $\|u\|<r_2  $.

For $t\in[0,1]$ ,
\begin{align*}
&|(S_n u_m)(t)-(S_nu)(t)|\\
&= |\lambda \int^1_0
    G(t,s)h(s)\big(f_{n}(s,u_m(s))-f_{n}(s,u(s))\big)\,\mathrm{d}s\\
 &\quad +\lambda \int^1_0 \rho(t,s)\int^1_0 G(s,\tau)h(\tau)
      \big(f_{n}(\tau,u_m(\tau))-f_{n}(\tau,u(\tau))\big)\,\mathrm{d}\tau\,\mathrm{d}s| \\
 &\leq     \lambda k_1 \int^1_0
     (1-s)^{p-q-1}h(s)\big|f_{n}(s,u_m(s))-f_{n}(s,u(s))\big|\,\mathrm{d}s\\
 &\quad +\frac{\lambda k_1M_0}{1-M_0} \int^1_0 (1-s)^{p-q-1}h(s)\big|f_{n}(s,u_m(s))-
     f_{n}(s,u(s))\big|\,\mathrm{d}s\\
 &=  \frac{\lambda k_1}{1-M_0} \int^1_0 (1-s)^{p-q-1}h(s)\big|f_{n}(s,u_m(s))-
     f_{n}(s,u(s))\big|\,\mathrm{d}s.
\end{align*}
Since $f_{n}(s,u)$ is continuous on
$[0,1]\times[\frac{1}{n},\frac{1}{n}+r_2  ]$, we can get
$f_{n}(s,u)$ is uniformly continuous on
$[0,1]\times[\frac{1}{n},\frac{1}{n}+r_2  ]$. Hence, we have
$$
  \lim_{m\to \infty}\|f_{n}(s,u_m(s))-f_{n}(s,u(s))\|=0.
$$
 It is easy to see
$$
  \lim_{m\to \infty}\|S_nu_m-S_nu\|=0.
$$
Therefore, $S_{n}$ is continuous on $P$.

(ii) For any $u\in \overline{K}_{r,R}$, $t\in[0,1]$, we have
\begin{align*}
&S_{n}u(t) \\
&=\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u(s))\,\mathrm{d}s+\lambda \int^1_0 \rho(t,s)
    \int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u(\tau))\,\mathrm{d}\tau\,\mathrm{d}s \\
&\leq \lambda k_1 \int^1_0 (1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
&\quad +\frac{M_0\lambda  k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s  \\
&=\frac{\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s,
         u(s))\,\mathrm{d}s.
\end{align*}
This implies that $\|S_{n}u\|\leq\frac{\lambda k_1
}{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s,
    u(s))\,\mathrm{d}s$.

On the other hand, for $t\in[0,1]$,
\begin{align*}
    S_{n}u(t) &=\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
 &\quad   +\lambda \int^1_0 \rho(t,s)\int^1_0G(s,\tau)
 h(\tau)f_{n}(\tau,u(\tau))\,\mathrm{d}
    \tau\,\mathrm{d}s \\
  &\geq     \lambda k_2\int^1_0
         (1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s \\
  &\geq    \lambda k_2\frac{1-M_0}{\lambda k_1 }\|S_{n}u\| \\
  &=    \frac{k_2}{k_1 }(1-M_0)\|S_{n}u\|.
\end{align*}
Therefore $S_{n}(\overline{K}_{r,R})\subset K$.
 \end{proof}


\section{Main results and proof}

Denote
$$
  f^{0}=\limsup_{u\to 0^+}\max_{t\in[0,1]}\frac{f(t,u)}{u},\quad
  f_{\infty}=\liminf_{u\to+\infty}\min_{t\in[0,1]}\frac{f(t,u)}{u},
$$
and
$$
  f^{\infty}=\limsup_{u\to+\infty}\max_{t\in[0,1]}\frac{f(t,u)}{u},\quad
  f_0=\liminf_{u\to0^+}\min_{t\in[0,1]}\frac{f(t,u)}{u}.
$$

\begin{theorem}\label{thm1}
Suppose $0\leq m_0\leq M_0<1$ and {\rm (H)} holds. If
\begin{equation}\label{eq3.1b}
  0<f^{0}<\frac{1-M_0}{k_1 L}\quad \text{ and }\quad
0<\frac{1}{k_2L}<f_{\infty}<+\infty,
\end{equation}
then \eqref{eq1.5} has at least one positive solution for
$ \lambda\in(\frac{1}{k_2Lf_{\infty}},\frac{1-M_0}{k_1 Lf^{0}})$.
\end{theorem}

\begin{proof}  For
$ \lambda\in(\frac{1}{k_2Lf_{\infty}},\frac{1-M_0}{k_1 Lf^{0}})$,
 there exists $\varepsilon >0$ such that
\[
 f_{\infty}-\varepsilon>0, \quad
 \frac{1}{(f_{\infty}-\varepsilon)k_2L}\leq\lambda\leq\frac{1-M_0}
 {k_1 L(f^{0}+\varepsilon)}.
\]
By \eqref{eq3.1b}, there exist $r>0$ and $R_0>0$, such that
\begin{gather}\label{3.1}
 f(t,u)\leq(f^{0}+\varepsilon)u,\quad \text{for } t\in[0,1], \;
 0<u\leq r.\\
\label{3.2}
 f(t,u)>(f_{\infty}-\varepsilon)u,\quad\text{for }t\in[0,1], \; u\geq R_0.
\end{gather}
For any $u\in\partial K_{r}$ and
$n>[\frac{k_1}{rk_2(1-M_0)}]+1=: n_0$, we have
$$
  r=\|u\|\geq
  u(t)\geq\frac{k_2(1-M_0)}{k_1}\|u\|=\frac{rk_2(1-M_0)}{k_1}>\frac{1}{n}.
$$
It follows that
\begin{equation}
 f_{n}(t,u(t))=f(t,\max\{1/n,u(t)\})=f(t,u(t))\leq(f^{0}+\varepsilon)u
\end{equation}
from \eqref{3.1}.
Hence,
\begin{align*}
    \|S_{n}u\|
&=\max_{t\in[0,1]}|\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
&\quad +\lambda   \int^1_0 \rho(t,s)\int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u(\tau))
 \,\mathrm{d}\tau\,\mathrm{d}s| \\
&\leq \lambda k_1 \int^1_0 (1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
&\quad +\frac{M_0\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s  \\
&=\frac{\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s, u(s))\,\mathrm{d}s\\
&\leq\frac{\lambda  k_1 (f^{0}+\varepsilon)}{1-M_0}\int^1_0(1-s)^{p-q-1}
h(s)u(s)\,\mathrm{d}s \\
&\leq\frac{\lambda  k_1 L(f^{0}+\varepsilon)}{1-M_0}\|u\|
\leq  \|u\|.
\end{align*}
We can get $\|S_{n}u\|\leq\|u\|$, for each $u\in\partial K_{r}$.

Let $R=\max\{2r,\frac{k_1 R_0}{k_2(1-M_0)}\}$ and $e(t)\equiv1$
for $t\in[0,1]$. Then $R>r$ and $e(t)\in K_{1}=\{u\in K: \|u\|<1\}$.
Subsequently, we can show $u\neq S_{n}u+me$, for any $m>0$ and
$u\in\partial K_{R}$.

Otherwise, there exists $u_0\in \partial K_{R}$ and $m_{1}>0$ such
 that $u_0=S_{n}u_0+m_1e$.
We notice that for any $s\in[0,1]$,
\[
  u_0(s)\geq\min_{s\in[0,1]}u_0(s)\geq\frac{k_2}{k_1 }(1-M_0)R\geq
  R_0.
\]
From \eqref{3.2}, it follows that
\[
  f_{n}(t,u_0(t))=f(t,\max\{1/n,u_0(t)\})=f(t,u_0(t))
  >(f_{\infty}-\varepsilon)u_0(t)\,.
\]
Let $\xi=\min_{t\in[0,1]}u_0(t)$.  Consequently, for any
$t\in[0,1]$, we have
\begin{align*}
u_0(t)
&=\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u_0(s))\,\mathrm{d}s\\
&\quad +\lambda \int^1_0 \rho(t,s)
    \int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u_0(\tau))\,\mathrm{d}\tau\,\mathrm{d}s
 +m_{1}e(t) \\
&\geq\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u_0(s))\,\mathrm{d}s+m_{1}e(t)\\
&\geq \lambda k_2(f_{\infty}-\varepsilon)\int^1_0
         (1-s)^{p-q-1}h(s)u_0(s)\,\mathrm{d}s+m_{1} \\
&\geq  \frac{\xi}{L}\int^1_0 (1-s)^{p-q-1}h(s)\,\mathrm{d}s+m_{1} \\
&\geq \xi+m_{1}>\xi.
\end{align*}
This implies that $\xi>\xi$, which is a contradiction.

It follows that for $n\geq n_0=[\frac{k_1}{rk_2(1-M_0)}]+1$, the operator $S_{n}$
 has a fixed point $u_{n}$ in
$K$ with $r<\|u_{n}\|<R$, from Lemma \ref{lem2.3}.
Hence,
\begin{align*}
&u_n(t)\\
&=\lambda\int^1_0G(t, s)h(s)f_{n}(s, u_n(s))\,\mathrm{d}s
  +\lambda\int^1_0\rho(t, s)\int^1_0G(s, \tau)h(\tau)f_{n}(\tau, u_n(\tau))
  \,\mathrm{d}\tau\,\mathrm{d}s,
\end{align*}
for $t\in[0,1]$. Since $u_{n}\in K$, we have
$$
  u_{n}(t)\geq\frac{k_2(1-M_0)}{k_1 }\|u_{n}\|
  =\frac{rk_2(1-M_0)}{k_1 }>\frac{1}{n}>0,
  \quad t\in[0,1],
$$
and
$$
  f_{n}(t,u_n(t))=f(t,\max\{1/n,u_n(t)\})=f(t,u_n(t)),\quad
  t\in[0,1].
$$
It is easy to see that
\begin{align*}
&u_n(t)\\
&=\lambda\int^1_0G(t, s)h(s)f(s, u_n(s))\,\mathrm{d}s
  +\lambda\int^1_0\rho(t, s)\int^1_0G(s, \tau)h(\tau)f(\tau, u_n(\tau))
  \,\mathrm{d}\tau\,\mathrm{d}s,
\end{align*}
for $t\in[0,1]$.
By Lemma \ref{lemma}, we obtain that $u_{n}$ is a positive solution of
\eqref{eq1.5}.
\end{proof}

By proof similar to the one for Theorem \ref{thm1}, we can show the
following theorem.

\begin{theorem}\label{thm2}
Suppose $0\leq m_0\leq M_0<1$ and {\rm(H)} holds. If
$$
  f^{0}=0\quad\text{and}\quad f_{\infty}=+\infty,
$$
then  \eqref{eq1.5} has at least one positive solution for
$ \lambda\in(0,+\infty)$.
\end{theorem}

\begin{remark}\label{remark2} \rm
  In Theorem \ref{thm1}, if $f^{0}=0$ or $f_{\infty}=+\infty$,
we can obtain  conclusions similar to Theorems \ref{thm1} and \ref{thm2}.
\end{remark}

\begin{theorem}\label{thm3}
Suppose $0\leq m_0\leq M_0<1$ and {\rm (H)} holds. If
\begin{equation}\label{eq3.5}
  0<f^{\infty}<\frac{1-M_0}{k_1 L}\quad\text{and}\quad
0<\frac{1}{k_2L}<f_0<+\infty,
\end{equation}
then \eqref{eq1.5} has at least one positive solution for
 $ \lambda\in\big(\frac{1}{k_2Lf_0},\frac{1-M_0} {k_1 Lf^{\infty}}\big)$.
\end{theorem}

\begin{proof}
 For $ \lambda\in\big(\frac{1}{k_2Lf_0},\frac{1-M_0} {k_1 Lf^{\infty}}\big)$,
 there exists $\varepsilon >0$ such that
\[
 f_0-\varepsilon>0, \quad
 \frac{1}{(f_0-\varepsilon)k_2L}\leq\lambda\leq\frac{1-M_0}
 {k_1 L(f^{\infty}+\varepsilon)}.
\]
By  \eqref{eq3.5}, there exist $r>0$ and $R_0>1$, such that
\begin{gather}\label{3.4}
  f(t,u)\geq(f_0-\varepsilon)u,\quad\text{for }  t\in[0,1],  \; 0<u\leq r.
\\ \label{3.5}
  f(t,u)\leq(f^{\infty}+\varepsilon)u,\quad \text{for }  t\in[0,1],\;u\geq R_0.
\end{gather}
Take $R\geq\max\{r,R_0,\frac{k_1R_0}{k_2(1-M_0)}\}$
For $u\in\partial K_{R}$ and
$n>[\frac{k_1}{Rk_2(1-M_0)}]+1=: n_0$, we have
$$
  u(t)\geq\frac{k_2(1-M_0)}{k_1}\|u\|=\frac{Rk_2(1-M_0)}{k_1}\geq R_0.
$$
From \eqref{3.5}, we have
\[
 f_{n}(t,u(t))=f(t,\max\{1/n,u(t)\})=f(t,u(t))\leq(f^{\infty}+\varepsilon)u.
\]
Hence,
\begin{align*}
    \|S_{n}u\|
&=\max_{t\in[0,1]}|\lambda \int^1_0 G(t,s)h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
 &\quad   +\lambda \int^1_0 \rho(t,s)\int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u(\tau))\,\mathrm{d}
    \tau\,\mathrm{d}s| \\
&\leq \lambda k_1 \int^1_0 (1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s\\
 &\quad   +\frac{M_0\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s,u(s))\,\mathrm{d}s  \\
&=\frac{\lambda k_1 }{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)f_{n}(s, u(s))\,\mathrm{d}s\\
&\leq\frac{\lambda
         k_1 (f^{\infty}+\varepsilon)}{1-M_0}\int^1_0(1-s)^{p-q-1}h(s)u(s)\,\mathrm{d}s \\
&\leq\frac{\lambda
         k_1 L(f^{\infty}+\varepsilon)}{1-M_0}\|u\|
\leq  \|u\|.
\end{align*}
We can get $\|S_{n}u\|\leq\|u\|$, for each $u\in\partial K_{R}$.


Let $e(t)\equiv 1$, $t\in[0,1]$. Then $e(t)\in\partial K_{1}$, and
we can prove $u\neq S_{n}u+me$, for any $m>0, \ and \ u\in K_{r}$.
Otherwise there exists $u_0\in K_{r}$ and $m_{1}>0$ such that
$u_0=S_{n}u_0+m_{1}e$. Let $\eta=\min\{u_0(t):t\in[0,1]\}$,
for $t\in[0,1]$, by \eqref{3.4}, we have
\begin{align*}
    u_0(t)
  &=
    \lambda \int^1_0 G(t,s)h(s)f_{n}(s,u_0(s))\,\mathrm{d}s\\
&\quad  +\lambda \int^1_0 \rho(t,s)\int^1_0G(s,\tau)h(\tau)f_{n}(\tau,u_0(\tau))
    \,\mathrm{d}\tau\,\mathrm{d}s+m_{1}e(t) \\
&\geq \lambda \int^1_0 G(t,s)h(s)f_{n}(s,u_0(s))\,\mathrm{d}s+m_{1} \\
&\geq  \lambda \int^1_0 k_2(1-s)^{p-q-1}h(s)(f_0-\varepsilon)u_0(s)\,\mathrm{d}s+m_{1} \\
&\geq  \frac{\eta}{L}\int^1_0(1-s)^{p-q-1}h(s)\,\mathrm{d}s+m_{1}
= \eta+m_{1}.
\end{align*}
This is a contradiction. It follows from Lemma \ref{lem2.3} that $S_{n}$ has
a fixed point $u_{n}$ in $K$ with $r<\|u_{n}\|< R$.
Hence,
\begin{align*}
u_n(t)&=\lambda\int^1_0G(t, s)h(s)f_{n}(s, u_n(s))\,\mathrm{d}s\\
 &\quad +\lambda\int^1_0\rho(t, s)\int^1_0G(s, \tau)h(\tau)f_{n}(\tau, u_n(\tau))
  \,\mathrm{d}\tau\,\mathrm{d}s,
\end{align*}
for $t\in[0,1]$.
 Since $u_{n}\in K$, for $n>\frac{k_1 }{rk_2(1-M_0)}$, we have
$$
  u_{n}(t)\geq\frac{k_2(1-M_0)}{k_1 }\|u_{n}\|
  =\frac{rk_2(1-M_0)}{k_1 }>\frac{1}{n}>0,
  \quad t\in[0,1],
$$
and
$$
  f_{n}(t,u_n(t))=f(t,\max\{1/n,u_n(t)\})=f(t,u_n(t)),\quad
  t\in[0,1].
$$
It is easy to see that
\begin{align*}
  u_n(t)&=\lambda\int^1_0G(t, s)h(s)f(s, u_n(s))\,\mathrm{d}s\\
 &\quad +\lambda\int^1_0\rho(t, s)\int^1_0G(s, \tau)h(\tau)f(\tau, u_n(\tau))
  \,\mathrm{d}\tau\,\mathrm{d}s,
\end{align*}
for $t\in[0,1]$.
By Lemma \ref{lemma}, we can get $u_{n}$ is a positive solution of
 \eqref{eq1.5}.
\end{proof}

Similarly to the proof of Theorem \ref{thm1}, we can obtain the
following theorem.

\begin{theorem}\label{thm4}
Suppose $0\leq m_0\leq M_0<1$ and {\rm (H)} holds. If
$$
  f^{\infty}=0\quad\text{and}\quad f_0=+\infty,
$$
then \eqref{eq1.5} has at least one positive solution for
 $\lambda\in(0,+\infty)$.
\end{theorem}

\begin{remark}\label{remark3}\rm
  In Theorem \ref{thm3}, if $f^{\infty}=0$ or $f_0=+\infty$, we can 
obtain similar  conclusions as those in Theorems \ref{thm3} and  \ref{thm4}.
\end{remark}

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