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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 238, pp. 1--15.\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-2012/238\hfil Existence and multiplicity of solutions]
{Existence and multiplicity of positive solutions for
 m-point nonlinear fractional differential equations on the half line}

\author[K. Ghanbari, Y. Gholami \hfil EJDE-2012/238\hfilneg]
{Kazem Ghanbari, Yousef Gholami}  % in alphabetical order

\address{Kazem Ghanbari \newline
Department of Mathematics, 
Sahand University of Technology, 
Tabriz, Iran}
\email{kghanbari@sut.ac.ir}

\address{Yousef Gholami \newline
Department of Mathematics, 
Sahand University of Technology, 
Tabriz, Iran}
\email{y\_gholami@sut.ac.ir}

\thanks{Submitted August 3, 2012. Published December 28, 2012.}
\subjclass[2000]{34A08, 34B10, 34B15, 34B18}
\keywords{Fractional derivative; fixed point theorem; positive solution}

\begin{abstract}
 In this article we find sufficient conditions for existence
 and multiplicity of positive solutions for an $m$-point nonlinear
 fractional boundary-value problem on an infinite interval.
 Moreover,  we prove that the set of positive solutions is compact.
 Nonexistence results for the boundary-value problem also are obtained.
\end{abstract}

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

\section{Introduction}

Fractional calculus has played a significant role in engineering,
science, economy, and other fields. The monographs
\cite{Kilbas,Poudlobny,Oldham,Miller} are
commonly cited for the theory of fractional derivatives and
integrals and applications to differential equations of fractional
order. Recently,  there have been some papers dealing with the
existence and multiplicity of positive solutions of nonlinear
boundary value problems of fractional order using the
techniques of nonlinear analysis (fixed point theorem,
Leray-Schauder theory, etc).
See \cite{Liang,Agrawal,Zhang,Troung,Zhao} for more details.

In this article we investigate existence and nonexistence results
for a boundary-value problem of nonlinear fractional differential
equation with $m$-point boundary conditions on an infinite interval
of the  form
\begin{gather} 
D_{{0}^{+}}^{\alpha}u(t)+\lambda
a(t)f(t,u(t))=0,\quad t\in(0,\infty),\; \alpha\in(2,3),
\label{e1.1} \\
u(0)+u'(0)=0,\quad \lim_{t\to +\infty}D_{{0}^{+}}^{\alpha-1}u(t)
=\sum_{i=1}^{m-2}\beta_iu'(\xi_i),
 \label{e1.2}\\
0<\xi_1<\xi_2<\dots <\xi_{m-2}<\infty,\quad
\beta_i\in \mathbb{R}^{+}\cup \{0\},\quad i=1,2,\dots ,m-2 \label{e1.3}
\end{gather}
where $D_{{0}^{+}}^{\alpha}$ is the  fractional
Riemann-Liouville derivative of order $\alpha>0$ and $\lambda$ is
a positive parameter. We assume the following conditions:
 \begin{itemize}
\item[(H1)]  $f\in C((0,\infty)\times[0,\infty),[0,\infty))$,
$f(t,0)\neq 0$ on any subinterval of $(0,+\infty)$, also when $u$ is bounded
$f(t,(1+t^{\alpha-1})u)$ is bounded on $[0,+\infty)$.

\item[(H2)] $a\in C((0,\infty),[0,\infty))$ and $a(t)$ is not identically zero
on any interval of the form $(t_0,\infty)$. Also assume that
$$
0<\int_0^{\infty}a(s)ds<\infty.
$$

\item[(H3)] $0<{\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}
<\Gamma(\alpha)$.
\end{itemize}

\section{Preliminaries}

In this section we introduce some fundamental tools of
fractional calculus. We also remind the well
known  fixed point theorem due to Krasnosel'skii for operators
acting on cones in Banach spaces.

\begin{definition} \label{def2.1} \rm
The  Riemann-Liouville fractional integral of order
$\alpha>0$  of a function $u:(0,\infty)\to \mathbb{R}$ is given by
$$
I_{{0}^{+}}^{\alpha}u(t)=\frac{1}{\Gamma(\alpha)}
\int_0^{t}(t-s)^{\alpha-1} u(s)ds
$$
\end{definition}

\begin{definition} \label{def2.2} \rm
The Riemann-Liouville fractional derivative of order $\alpha>0$
for a function $u:(0,\infty)\to \mathbb{R}$ is defined
by
$$
D_{{0}^{+}}^{\alpha}u(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}}
\int_0^{t}(t-s)^{n-\alpha-1}u(s)ds
$$
where $n=[\alpha]+1$.
\end{definition}

\begin{lemma}[\cite{Kilbas}] \label{lem2.3}
 Let $u\in C(0,\infty)\cap L^{1}(0,\infty)$, $\beta\geq
\alpha \geq 0$, then
$$
D_{{0}^{+}}^{\alpha}I_{{0}^{+}}^{\beta}u(t)=I_{{0}^{+}}^{\beta-\alpha}u(t)
$$
\end{lemma}

\begin{lemma}[\cite{Kilbas}] \label{lem2.4}
Let $\alpha>0$ then
\begin{itemize}
\item[(i)]
If $\mu>-1$, $\mu\neq\alpha-i$ with
$i=1,2,\dots ,[\alpha]+1$, $t>0$ then
$$
D_{{0}^{+}}^{\alpha}t^{\mu}
=\frac{\Gamma(\mu+1)}{\Gamma(\mu-\alpha+1)}t^{\mu-\alpha}.
$$

\item[(ii)]
For $i=1,2,\dots, [\alpha]+1$, we have $D_{{0}^{+}}^{\alpha}t^{\alpha-i}=0$.

\item[(iii)] For every $t\in (0,\infty)$, $u\in L^{1}(0,\infty)$
$$
D_{{0}^{+}}^{\alpha}I_{{0}^{+}}^{\alpha}u(t)=u(t),\quad
I_{{0}^{+}}^{\alpha}D_{{0}^{+}}^{\alpha}u(t)=u(t)
+\sum_{i=1}^{n}c_it^{\alpha-i},\quad c_i\in \mathbb{R},\;n=[\alpha]+1.
$$

\item[(iv)]
$D_{{0}^{+}}^{\alpha}u(t)=0$ if and only if
$u(t)={\sum_{i=1}^{n}c_it^{\alpha-i}}$, $c_i\in \mathbb{R}$, $n=[\alpha]+1$.
\end{itemize}
\end{lemma}

\begin{lemma} \label{lem2.5}
Let $h\in C[0,\infty)$ such that
$0<{\int_0^{+\infty}h(s)ds}<+\infty$, then the
fractional boundary-value problem
\begin{gather}
D_{{0}^{+}}^{\alpha}u(t)+h(t) =
0,\quad t\in(0,\infty),\; \alpha\in(2,3), \label{e2.1} \\
u(0)+u'(0)=0,\quad \lim_{t\to +\infty}D_{{0}^{+}}^{\alpha-1}u(t)
=\sum_{i=1}^{m-2}\beta_iu'(\xi_i) \label{e2.2}
\end{gather}
has a unique solution
\begin{equation} \label{e2.3}
u(t)=\int_0^{+\infty}G(t,s)h(s)ds,
\end{equation}
where
\begin{equation} \label{e2.4}
G(t,s)=H_1(t,s)+H_2(t,s)
\end{equation}
with
\begin{gather}
H_1(t,s)=\frac{1}{\Gamma(\alpha)}
\begin{cases}
t^{\alpha-1}-(t-s)^{\alpha-1},  & 0\leq s \leq t<+\infty\\
t^{\alpha-1}, & 0\leq t \leq s<+\infty
\end{cases},  \label{e2.5}
\\
H_2(t,s)=\frac{{\sum_{i=1}^{m-2}\beta_it^{\alpha-1}}}
{{\Gamma(\alpha)-\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
\frac{\partial H_1(t,s)}{\partial t}\big|_{t=\xi_i}. \label{e2.6}
\end{gather}
The function $G(t,s)$  is called Green's
function of boundary-value problem \eqref{e2.1}-\eqref{e2.2}.
\end{lemma}

\begin{proof}
 By Lemmas \ref{lem2.3} and \ref{lem2.4} and considering \eqref{e2.1}, we have
$$
u(t)=-c_1t^{\alpha-1}-c_2t^{\alpha-2}-c_3t^{\alpha-3}
-\int_0^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds.
$$
Then
$$
u'(t)=-(\alpha-1)c_1t^{\alpha-2}-(\alpha-2)c_2t^{\alpha-3}
-(\alpha-3)c_3t^{\alpha-4}-\int_0^{t}
\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds.
$$
Now by imposing the boundary condition
$u(0)+u'(0)=0$ we conclude that $c_2=c_3=0$, also using
boundary condition
 $$
\lim_{t\to +\infty}D_{{0}^{+}}^{\alpha-1}u(t)
=\sum_{i=1}^{m-2}\beta_iu'(\xi_i)
$$
we have
$$
c_1=\frac{1}{{\Gamma(\alpha)-\sum_{i=1}^{m-2}(\alpha-1)
\beta_i\xi_i^{\alpha-2}}}\Big[\sum_{i=1}^{m-2}\beta_i
\int_0^{\xi_i}\frac{(\xi_i-s)^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds
-\int_0^{+\infty}h(s)ds\Big].
$$
Thus
\begin{align*}
u(t)\
&=\frac{t^{\alpha-1}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\int_0^{+\infty}h(s)ds
\\
&\quad-\frac{t^{\alpha-1}}{{\Gamma(\alpha)-\sum_{i=1}^{m-2}(\alpha-1)
 \beta_i\xi_i^{\alpha-2}}}\Big[\sum_{i=1}^{m-2}\beta_i\int_0^{\xi_i}
 \frac{(\xi_i-s)^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds\Big]\\
&\quad-\int_0^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds
\\
&=\int_0^{+\infty}H_1(t,s)h(s)ds
 +\frac{{\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 {{\Gamma(\alpha)-\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 \int_0^{+\infty}\frac{t^{\alpha-1}}{\Gamma(\alpha)}h(s)ds
\\
&\quad-\frac{{\sum_{i=1}^{m-2}\beta_it^{\alpha-1}}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 \int_0^{\xi_i}\frac{(\xi_i-s)^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds
\\
&=\int_0^{+\infty}H_1(t,s)h(s)ds\\
&\quad +\frac{{\sum_{i=1}^{m-2}\beta_it^{\alpha-1}}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 \int_0^{\xi_i}\frac{\xi_i^{\alpha-2}
 -(\xi_i-s)^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds
\\
&\quad+\frac{{\sum_{i=1}^{m-2}\beta_it^{\alpha-1}}}
 {{\Gamma(\alpha)-\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 \int_{\xi_i}^{+\infty}\frac{\xi_i^{\alpha-2}}{\Gamma(\alpha-1)}h(s)ds
\\
&=\int_0^{+\infty}H_1(t,s)h(s)ds\\
&\quad +\frac{{\sum_{i=1}^{m-2}\beta_it^{\alpha-1}}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
 \int_0^{+\infty}\frac{\partial
H_1(t,s)}{\partial t}\big|_{t=\xi_i}h(s)ds
\\
&=\int_0^{+\infty}H_1(t,s)h(s)ds+\int_0^{+\infty}H_2(t,s)h(s)ds
\\
&=\int_0^{+\infty}G(t,s)h(s)ds
\end{align*}
where $G(t,s)$ is Green's function defined by \eqref{e2.4}.
Now by uniqueness of constants $c_1,c_2,c_3$ we conclude that
\eqref{e2.3} is the unique solution of boundary value problem
\eqref{e2.1}-\eqref{e2.2}. This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.6} 
The function $H_1(t,s)$ defined by \eqref{e2.5} has
the following properties:
\begin{itemize}
\item[(i)] $H_1(t,s)$  is a nonnegative continuous function
for $t,s \in [0,+\infty)$;

\item[(ii)] $H_1(t,s)$  is increasing
function with respect to the first variable;

\item[(iii)] $H_1(t,s)$  is a concave function with respect to the first
variable, for every $0<s<t<+\infty$.
\end{itemize}
\end{lemma}

\begin{proof}
Using \eqref{e2.5} it is easy to see that property (i)
obviously holds. Now we show that (ii) holds. Considering
(i) we know that $H_1(t,s)\in
C([0,\infty)\times[0,\infty),[0,\infty))$, hence
$$
\frac{\partial H_1(t,s)}{\partial t}
=\frac{(\alpha-1)}{\Gamma(\alpha)}\begin{cases}
t^{\alpha-2}-(t-s)^{\alpha-2}, & 0\leq s \leq t <+\infty\\
t^{\alpha-2}, & 0\leq t \leq s<+\infty;
\end{cases}
$$
thus $H_1(t,s)$ is an increasing function with respect to first
variable.

To prove (iii) we note that
$$
\frac{\partial^{2}H_1(t,s)}{\partial
t^{2}}=\frac{1}{\Gamma(\alpha-2)}\begin{cases}
t^{\alpha-3}-(t-s)^{\alpha-3}, & 0\leq s \leq t <+\infty\\
t^{\alpha-3}, &0\leq t \leq s<+\infty.
\end{cases}
$$
On the other hand $\alpha\in(2,3)$, thus
for $ 0<s<t<+\infty$,
$$
\frac{\partial^{2}H_1(t,s)}{\partial t^{2}}<0
$$
So $H_1(t,s)$ is a concave function with respect to first variable,
for $0<s<t<+\infty$. This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.7} \rm
According to definition of $H_1(t,s)$ in
\eqref{e2.5} we have for $t,s\in[0,+\infty)$,
\begin{gather*}
\frac{H_1(t,s)}{1+t^{\alpha-1}}\leq \frac{1}{\Gamma(\alpha)},\quad
\frac{G(t,s)}{1+t^{\alpha-1}}\leq L,\\
L=\frac{1}{\Gamma(\alpha)}
\Big(1+\frac{{\sum_{i=1}^{m-2}\beta_i\xi_{m-2}^{\alpha-1}}}{{\Gamma(\alpha)
-\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\Big).
\end{gather*}
\end{remark}

\begin{lemma} \label{lem2.8}
There exist positive constant $\gamma_1$ such that for every
$k>1$,
$$
\min_{1/k \leq t \leq k}\frac{H_1(t,s)}{1+t^{\alpha-1}}
\geq \gamma_1\sup_{0\leq t <+\infty}\frac{H_1(t,s)}{1+t^{\alpha-1}},
$$
where $H_1(t,s)$  is defined by \eqref{e2.5}.
\end{lemma}

\begin{proof}
Using \eqref{e2.5}, we have
$$
\frac{H_1(t,s)}{1+t^{\alpha-1}}=\frac{1}{\Gamma(\alpha)}
\begin{cases}
\frac{t^{\alpha-1}-(t-s)^{\alpha-1}}{1+t^{\alpha-1}}, & 0\leq
s \leq t<+\infty\\
\frac{t^{\alpha-1}}{1+t^{\alpha-1}}, & 0\leq t\leq s <+\infty\,.
\end{cases}
$$
 Now let
\begin{gather*}
h_1(t,s)=\frac{1}{\Gamma(\alpha)}
\frac{t^{\alpha-1}-(t-s)^{\alpha-1}}{1+t^{\alpha-1}},\quad s\leq t\\
h_2(t,s)=\frac{1}{\Gamma(\alpha)}\frac{t^{\alpha-1}}{1+t^{\alpha-1}},
\quad t\leq  s.
 \end{gather*}
First of all we must note that, $h_1$ is decreasing and
$h_2$ is increasing with respect to $t$, respectively, also
$h_1$ is increasing with respect to $s$. So by a direct
computation, we conclude that
\begin{gather*}
\min_{1/k \leq t \leq k}h_1(t,s)
\geq \frac{(k^{\alpha-1}-(k-s)^{\alpha-1})}{\Gamma(\alpha)(1+k^{\alpha-1})}
\geq h_1(k)
=\frac{k^{2(\alpha-1)}-(k^{2}-1)^{\alpha-1}}
{\Gamma(\alpha)k^{\alpha-1}(1+k^{\alpha-1})},
\\
\sup_{0\leq t<+\infty} h_1(t,s) \leq \frac{1}{\Gamma(\alpha)}
\\
\min_{1/k \leq t \leq k}h_2(t,s)\geq
h_2(1/k)=\frac{1}{\Gamma(\alpha)(1+k^{\alpha-1})},
\\
\sup_{0\leq t <+\infty}h_2(t,s)=\frac{1}{\Gamma(\alpha)}.
\end{gather*}
Now defining
$$
m_1=\min\Big\{\frac{k^{2(\alpha-1)}-(k^{2}-1)^{\alpha-1}}{\Gamma(\alpha)
k^{\alpha-1}(1+k^{\alpha-1})},\frac{1}{\Gamma(\alpha)(1+k^{\alpha-1})}
\Big\}\,, \quad
 M_1=\frac{1}{\Gamma(\alpha)},
$$
 and setting
\begin{equation} %\label{e}
\gamma_1=\frac{m_1}{M_1}=\min\Big\{\frac{k^{2(\alpha-1)}
-(k^{2}-1)^{\alpha-1}}{k^{\alpha-1}(1+k^{\alpha-1})},
\frac{1}{(1+k^{\alpha-1})}\Big\}
\end{equation}
we conclude that
$$
\min_{1/k \leq t \leq k}\frac{H_1(t,s)}{1+t^{\alpha-1}}
\geq \gamma_1\sup_{0\leq t <+\infty}\frac{H_1(t,s)}{1+t^{\alpha-1}}
$$
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.9}
For $H_2(t,s)$, defined by  \eqref{e2.6}  there exist positive
constant $\gamma_2$  such that 
$$
\min_{1/k \leq t \leq k}\frac{H_2(t,s)}{1+t^{\alpha-1}}
\geq \gamma_2\sup_{0\leq t<+\infty}\frac{H_2(t,s)}{1+t^{\alpha-1}},
k>1\,.
$$
\end{lemma}

\begin{proof}
 Considering $H_2(t,s)$ in \eqref{e2.6} we have
\begin{gather*}
\min_{1/k\leq t\leq k}\frac{H_2(t,s)}{1+t^{\alpha-1}}
=\frac{1}{1+k^{\alpha-1}}
\frac{{\sum_{i=1}^{m-2}\beta_i}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\frac{\partial
H_1(t,s)}{\partial t}\big|_{t=\xi_i}=m_2,
\\
\sup_{0\leq t<+\infty}
 \frac{H_2(t,s)}{1+t^{\alpha-1}}
=\frac{{\sum_{i=1}^{m-2}\beta_i}}{{\Gamma(\alpha)-\sum_{i=1}^{m-2}
(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\frac{\partial H_1(t,s)}{\partial t}
\big|_{t=\xi_i}=M_2.
\end{gather*}
Now setting
$$
\gamma_2=\frac{m_2}{M_2}=\frac{1}{1+k^{\alpha-1}},
$$
we conclude that
$$
\min_{1/k \leq t \leq k}\frac{H_2(t,s)}{1+t^{\alpha-1}}
\geq \gamma_2\sup_{0\leq t<+\infty}\frac{H_2(t,s)}{1+t^{\alpha-1}}.
$$
 The proof is complete.
 \end{proof}

 \begin{lemma} \label{lem2.10}
Let $k>1$ be fixed and $G(t,s)$ be defined by
 \eqref{e2.4}-\eqref{e2.6}. Then
\begin{gather*}
\min_{1/k \leq t\leq k}\frac{G(t,s)}{1+t^{\alpha-1}}
\geq \lambda(k)\sup_{0\leq t<+\infty}\frac{G(t,s)}{1+t^{\alpha-1}},
\\
\lambda(k)=\min\{\gamma_1,\gamma_2\}=\gamma_1.
\end{gather*}
 \end{lemma}

\begin{definition} \label{def2.11} \rm
 We introduce the  Banach space
$$
B=\{u\in C[0,+\infty):\| u\|<+\infty\}
$$
 which is equipped with the norm
 $$
\| u\|=\sup_{t\in[0,+\infty)}\frac{\mid u(t)\mid}{1+t^{\alpha-1}}.
$$
Also we define the cone $P\subset B$ as follows
 $$
P=\{u\in B: u(t)\geq 0,\min_{t\in[\frac{1}{k},k]}\frac{u(t)}{1+t^{\alpha-1}}
\geq \lambda(k)\| u\|\}.
$$
\end{definition}

 \begin{lemma} \label{lem2.12}
 Let conditions {\rm (H1)--(H3)} be  satisfied and define the Hammerstein
integral operator $T:P\to  B$ by
 \begin{equation} \label{e2.8}
Tu(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds.
 \end{equation}
Then $TP\subset P$.
\end{lemma}

\begin{proof}
 Let $u\in P$. Considering conditions
(H1), (H2) and Lemma \ref{lem2.6} it is clear that
$$
Tu(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds\geq 0.
$$
Also we have
\begin{align*}
\min_{1/k\leq t\leq k}\frac{Tu(t)}{1+t^{\alpha-1}}
&=\min_{1/k \leq t\leq k}
 \frac{{\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds}}{1+t^{\alpha-1}}
\\
&\geq \lambda\int_0^{+\infty}\min_{1/k \leq t\leq
k}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)f(s,u(s))ds\\
&\geq \lambda\int_0^{+\infty}\lambda(k)\sup_{0\leq
t<+\infty}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)f(s,u(s))ds
\\
&\geq \lambda \lambda(k)\sup_{0\leq t<+\infty}
 \frac{{\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds}}{1+t^{\alpha-1}}\\
&= \lambda(k)\| Tu\|.
\end{align*}
This shows that $TP\subset P$.
\end{proof}

\begin{definition}[\cite{Liang}] \label{def2.13} \rm
Let
$$
V=\{u\in B: \| u\|<l,\,l>0\}, \quad
W=\{\frac{u(t)}{1+t^{\alpha-1}}:u\in V\}.
$$
The set $W$ is called equiconvergent at infinity if
for each $\epsilon>0$ there exists $\mu(\epsilon)>0$, such that
for all $u\in W$ and all $t_1,t_2\geq \mu$, we have
$$
|\frac{u(t_1)}{1+t_1^{\alpha-1}}-\frac{u(t_2)}{1+t_2^{\alpha-1}}|<\epsilon.
$$
\end{definition}

\begin{lemma}[\cite{Liang}] \label{lem2.14}
Assume
$$
V=\{u\in B:\| u\|<l,~l>0\}, W=\{\frac{u(t)}{1+t^{\alpha-1}}\big|u\in V\}.
$$
If $V$ is equicontinuous on any compact interval of
$[0,+\infty)$ and equiconvergent at infinity, then $V$ is
relatively compact on $B$.
\end{lemma}

\begin{lemma} \label{lem2.15}
If conditions {\rm (H1)--(H3)} hold, then integral
operator $T:P\to P$ is completely continuous.
\end{lemma}

\begin{proof}
First we prove that the operator $T$ is uniformly bounded on
$P$. Considering real Banach space $B$, we choose a positive
constant $r_0$ such that for every $u\in P$,
$\| u \| <r_0$. Let
$$
B_{r_0}=\sup\{f(t,(1+t^{\alpha-1 })u):(t,u)\in[0,+\infty)\times[0,r_0]\}
$$
and $\Omega$ be a bounded subset of $P$. Thus there exist a
positive constant $r$ such that
$$
\| u\|\leq r.
$$
Using Definition \ref{def2.11}, we have
$$
\| Tu\|=\lambda\sup_{t\in[0,+\infty)}
\frac{{\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds}}{1+t^{\alpha-1}}
\leq \lambda LB_{r}\int_0^{+\infty}a(s)ds<+\infty.
$$
Thus $T\Omega$ is bounded.
Now we show that operator $T$ is continuous. We consider
$\{u_{n}\}_{n=1}^{\infty}\subset P$, such that
$u_{n}\to u$ as $n\to \infty$, so by the
Lebesgue dominated convergence theorem we find that
$$
\int_0^{+\infty}a(s)f(s,u_{n}(s))ds\to \int_0^{+\infty}a(s)f(s,u(s))ds
$$
as $n\to \infty$. Hence by \eqref{e2.8} we have
$$
\| Tu_{n}-Tu\|\leq L\lambda
\big| \int_0^{+\infty}a(s)f(s,u_{n}(s))ds
 -\int_0^{+\infty}a(s)f(s,u(s))ds\big|\to 0,
$$
as $n\to \infty$. Hence $T$ is a continuous operator.
Now we show that operator $T:P\to P$ is an
equiconvergent operator at infinity. For each $u\in \Omega$, we
have
$$
\int_0^{+\infty}a(s)f(s,u(s))ds\leq B_{r}\int_0^{+\infty}a(s)ds<+\infty.
$$
Since
$$
\lim_{t\to+\infty}\frac{1}{1+t^{\alpha-1}}
\int_0^{+\infty} H_1(t,s)a(s)f(s,u(s))ds=0
$$
and for $i=1,2,\dots ,m-2$, $\xi_i<\infty$, also by condition
(H2), we conclude that
\begin{align*}
&\lim_{t\to+\infty}|\frac{Tu(t)}{1+t^{\alpha-1}}|
\\
&=\lim_{t\to+\infty}\frac{1}{1+t^{\alpha-1}}
\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds
\\
&=\lim_{t\to+\infty}\frac{1}{1+t^{\alpha-1}}
\lambda\int_0^{+\infty}H_1(t,s)a(s)f(s,u(s))ds
\\
&\quad+\lim_{t\to+\infty}\frac{\lambda
t^{\alpha-1}}{1+t^{\alpha-1}}
 \frac{{\sum_{i=1}^{m-2}\beta_i}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\\
&\quad\times \int_0^{+\infty}\frac{\partial H_1(t,s)}{\partial
t}\big|_{t=\xi_i}a(s)f(s,u(s)ds)
\\
&=\frac{{\lambda\sum_{i=1}^{m-2}\beta_i}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}
\int_0^{+\infty}\frac{\partial H_1(t,s)}{\partial t}
 \big|_{t=\xi_i}a(s)f(s,u(s))ds.
\end{align*}
Then
\begin{equation*}
\lim_{t\to+\infty}\big|\frac{Tu(t)}{1+t^{\alpha-1}}\big|<+\infty.
\end{equation*}
Thus $T\Omega$ is equiconvergent at infinity.

Finally we prove that $T$ is an equicontinuous operator. For
every $s\in(0,+\infty)$, let $t_1,t_2\in[0,s]$, with $t_1<t_2$. Then we have
\begin{align*}
\big|\frac{Tu(t_2)}{1+t_2^{\alpha-1}}-\frac{Tu(t_1)}{1+t_1^{\alpha-1}}\big|
&\leq \lambda B_{r}\int_0^{+\infty}\big|
\frac{G(t_2,s)}{1+t_2^{\alpha-1}}-\frac{G(t_1,s)}{1+t_1^{\alpha-1}}\big|a(s)ds
\\
&\leq \lambda B_{r}\int_0^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_2^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\ 
&\quad+\lambda B_{r}\int_0^{+\infty}\big|
\frac{H_2(t_2,s)}{1+t_2^{\alpha-1}}-\frac{H_2(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&\leq \lambda B_{r}\int_0^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&\quad+\lambda B_{r}\int_0^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_2^{\alpha-1}}-\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&\quad+\lambda B_{r}\frac{{\sum_{i=1}^{m-2}\beta_i}}{{\Gamma(\alpha)
 -\sum_{i=1}^{m-2}(\alpha-1)\beta_i\xi_i^{\alpha-2}}}\\
&\quad\times \big|\frac{t_2^{\alpha-1}}{1+t_2^{\alpha-1}}
-\frac{t_1^{\alpha-1}}{1+t_1^{\alpha-1}}\big|\int_0^{+\infty}\frac{\partial
H_1(t,s)}{\partial t}\big|_{t=\xi_i}a(s)ds.
\end{align*}
On the other hand
\begin{align*}
&\int_0^{+\infty}\big| \frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}
-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|a(s)ds\\
&\leq \int_0^{t_1}\big|
\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&\quad+\int_{t_1}^{t_2}\big|
\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&\quad+\int_{t_2}^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
 a(s)ds\\
&=\int_0^{t_1}\frac{|(t_2^{\alpha-1}-t_1^{\alpha-1})+(t_1-s)^{\alpha-1}
 -(t_2-s)^{\alpha-1}|}{1+t_1^{\alpha-1}}a(s)ds
\\
&\quad+\int_{t_1}^{t_2}\frac{|(t_2^{\alpha-1}-t_1^{\alpha-1})
 -(t_2-s)^{\alpha-1}|}{1+t_1^{\alpha-1}}a(s)ds
 +\int_{t_2}^{+\infty}\frac{|(t_2^{\alpha-1}-t_1^{\alpha-1})|}
 {1+t_1^{\alpha-1}}a(s)ds.
\end{align*}
Thus when $t_1\to t_2$, we conclude that
\begin{equation} \label{e2.9}
\int_0^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}-\frac{H_1(t_1,s)}{1+t_1^{\alpha-1}}\big|
a(s)ds\to 0
\end{equation}
Similar to \eqref{e2.9}, when $t_1\to t_2$, we have
\begin{equation} \label{e2.10}
\int_0^{+\infty}\big|
\frac{H_1(t_2,s)}{1+t_2^{\alpha-1}}-\frac{H_1(t_2,s)}{1+t_1^{\alpha-1}}\big|
a(s)ds\to 0
\end{equation}
From \eqref{e2.9} and\eqref{e2.10} when $t_1\to t_2$, we obtain that
$$
\big| \frac{Tu(t_2)}{1+t_2^{\alpha-1}}-\frac{Tu(t_1)}{1+t_1^{\alpha-1}}\big|\to 0.
$$
Thus $T\Omega$ is equicontinuous on $(0,+\infty)$.
Using Lemma \ref{lem2.14} we attain that operator $T:P\to P$ is
completely continuous. This complete the proof.
\end{proof}

\begin{theorem}[\cite{Troung}] \label{thm2.16}
Let $X$  be a real Banach space and $P\subset X$ be a
cone in $X$ . Assume $\Omega_1,\Omega_2$ are two open bounded
subsets of $X$ with $0\in \Omega_1$,
$\overline \Omega_1\subset \Omega_2$ and
$T:P\cap(\overline \Omega_2\backslash \Omega_1 )\to P$ be a
completely continuous operator such that
\begin{itemize}
\item[(i)]
$\| Tu \|\leq \| u \|$, $u\in P\cap\partial\Omega_1$ and
$\| Tu \|\geq \| u \|$, $u\in P\cap\partial\Omega_2$ , or

\item[(ii)]
$\| Tu \|\leq \| u \|$, $u\in P\cap\partial\Omega_2$  and
$\| Tu \|\geq \| u \|$, $u\in P\cap\partial\Omega_1$.
\end{itemize}
Then $T$ has a fixed point in $P\cap(\overline \Omega_2\backslash \Omega_1 )$.
\end{theorem}


\section{Main Results}

We introduce the following notation:
\begin{gather*}
f_0=\lim\min_{u\to 0^{+}}\frac{(1+t^{\alpha-1})f(t,u)}{u},\quad
f_{\infty}=\lim\min_{u\to +\infty}\frac{(1+t^{\alpha-1})f(t,u)}{u},\quad
 t\in[\frac{1}{k},k]
\\
f^{0}=\limsup_{u\to 0^{+}}\frac{(1+t^{\alpha-1})f(t,u)}{u},\quad
f^{\infty}=\limsup_{u\to +\infty}\frac{(1+t^{\alpha-1})f(t,u)}{u},\quad
 t\in(0,\infty)
\\
A=\Big( L \int_0^{+\infty}a(s)ds\Big)^{-1}, \quad
B=\Big(\frac{\lambda^{2}(k)}{k^{\alpha-1}}\int_{1/k}^{k}a(s)ds\Big)^{-1}.
\end{gather*}
The following theorem rely on Theorem \ref{thm2.16} which has two
possibilities that may occur.

\begin{theorem} \label{thm3.2}
Let conditions {\rm (H1)--(H3)} hold. Then 
\eqref{e1.1}-\eqref{e1.2} has at least one
positive solution on $P$ in each one of the two cases:
\begin{itemize}
\item[(C1)] For every $\lambda\in(\frac{B}{f_0},\frac{A}{f^{\infty}})$ such
that $ f_0,f^{\infty}\in(0,\infty)$ with
$\lambda(k)f_0>f^{\infty}$, or

\item[(C2)] For every $\lambda\in(\frac{B}{f_{\infty}},\frac{A}{f^{0}})$ such
that $f_{\infty},f^{0}\in(0,\infty)$ with
$\lambda(k)f_{\infty}>f^{0}$.
\end{itemize}
\end{theorem}

\begin{proof}
Let
$$
\Omega_i=\{u\in B :\| u\|<R_i \}, \; i=1,2,\; R_1<R_2.
$$
Then $\Omega_1,\Omega_2$ are two open bounded subset of
$B$ such that $0\in \Omega_1,\overline \Omega_1\subset \Omega_2$.

\textbf{Case1:} Let $ f_0,f^{\infty}\in(0,\infty)$ and
$\lambda(k)f_0>f^{\infty}$, also
$\lambda\in(\frac{B}{f_0},\frac{A}{f^{\infty}})$. Let
$\epsilon>0$ be chosen such that
\begin{equation} \label{e3.1}
\frac{B}{f_0-\epsilon}<\lambda<\frac{A}{f^{\infty}+\epsilon}
\end{equation}
Since $ f_0\in(0, \infty)$, thus there exist a positive constant
$R_1$ such that for every $t\in[1/k,k]$ and $u\in[0,R_1]$,
\begin{equation*}
 f(t,u)=f(t,\frac{(1+t^{\alpha-1})u}{1+t^{\alpha-1}})
 \geq(f_0-\epsilon)\frac{u}{1+t^{\alpha-1}}.
 \end{equation*}
So if $u\in P$ with $\| u\|=R_1$, then
$$
f(t,u)\geq(f_0-\epsilon)\frac{u}{1+t^{\alpha-1}}
\geq \lambda(k)(f_0-\epsilon)\| u\|, \quad t\in[1/k,k]
$$
hence from \eqref{e3.1} we have
\begin{align*}
Tu(t)&=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds\\
&\geq \lambda\lambda(k)(f_0-\epsilon)\|
u\|\int_0^{+\infty}G(t,s)a(s)ds.
\end{align*}
Thus
\begin{align*}
\| Tu\|
&\geq \lambda\lambda(k)(f_0-\epsilon)\|
u\|\int_0^{+\infty}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)ds\\
&\geq \lambda\lambda(k)(f_0-\epsilon)\|
u\|\int_{1/k}^{k}\frac{H_1(t,s)}{1+t^{\alpha-1}}a(s)ds\\
&\geq \lambda(f_0-\epsilon)\|
u\|\frac{\lambda^{2}(k)}{k^{\alpha-1}}\int_{1/k}^{k}a(s)ds\\
&=\lambda(f_0-\epsilon)B^{-1}\| u\| >\| u\|.
\end{align*}
Therefore,
\begin{equation} \label{e3.2}
\| Tu\| \geq \| u\| \quad \forall u\in P\cap \partial\Omega_1.
\end{equation}
On the other hand, since $f^{\infty}\in(0,\infty)$, there exist a
positive constant $R$ such that for all $u\geq R$, we have
\[
f(t,u)=f(t,\frac{(1+t^{\alpha-1})u}{1+t^{\alpha-1}})
\leq(f^{\infty}+\epsilon)\frac{u}{1+t^{\alpha-1}}
\leq (f^{\infty}+\epsilon)\| u\|.
\]
Let $R_2=\max\{1+R_1,R\lambda^{-1}(k)\}$ and
$u\in P\cap \partial\Omega_2$. Using \eqref{e3.1} we have
\begin{align*}
Tu(t)&=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds\\
&\leq \lambda(f^{\infty}+\epsilon)\|
u\|\int_0^{+\infty}G(t,s)a(s)ds.
\end{align*}
So
\begin{align*}
\| Tu\|
&\leq \lambda(f^{\infty}+\epsilon)\|
u\|\int_0^{+\infty}\sup_{t\in[0,+\infty)}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)ds\\
&\leq \lambda(f^{\infty}+\epsilon)\| u\|
L\int_0^{+\infty}a(s)ds\\
&\leq \lambda(f^{\infty}+\epsilon)A^{-1}\| u\|
\leq \| u\|.
\end{align*}
Thus we find that
\begin{equation} \label{e3.3}
\| Tu\|\leq \| u\| \quad \forall u\in P\cap \partial\Omega_2.
\end{equation}
Hence, using the Theorem \ref{thm2.16} and \eqref{e3.2}, \eqref{e3.3} we conclude that the
boundary value problem \eqref{e1.1}-\eqref{e1.2} has at least one positive
solution in $P \cap (\overline\Omega_2\backslash \Omega_1)$.

\textbf{Case 2:} Let
$f_{\infty},f^{0}\in(0,\infty)$, $\lambda(k)f_{\infty}>f^{0}$ and
$\lambda\in(\frac{B}{f_{\infty}},\frac{A}{f^{0}})$.
Similar to the case1, let $\epsilon>0$ be chosen such that
\begin{equation} \label{e3.4}
\frac{B}{f_{\infty}-\epsilon}<\lambda<\frac{A}{f^{0}+\epsilon}.
\end{equation}
We can choose positive constants $R_2>R_1$ such that
\begin{gather} \label{e3.5}
\| Tu\|\geq \| u\| \quad \forall u\in P\cap \partial\Omega_1,\\
\| Tu\|\leq \| u\| \quad \forall u\in P\cap \partial\Omega_2. \label{e3.6}
\end{gather}
Considering Theorem \ref{thm2.16} and \eqref{e3.5},\eqref{e3.6} we conclude
that the boundary value problem \eqref{e1.1}-\eqref{e1.2} has at least one
positive solution in $P \cap (\overline\Omega_2\backslash\Omega_1)$.
\end{proof}

To prove multiplicity of positive solutions
for \eqref{e1.1}-\eqref{e1.2}, we need  following
 condition.
\begin{itemize}
\item[(H4)]
Assume that function $f(t,u)$ is nondecreasing with respect to the second
variable; i.e., for all $u_1,u_2\in B$,
if $u_1\leq u_2$ then $f(t,u_1)\leq f(t,u_2)$.
\end{itemize}

\begin{theorem} \label{thm3.4}
Let conditions {\rm (H1)--(H4)} hold. Assume
that there exist positive constants $R_2>R_1$, such that
\begin{equation} \label{e3.7}
\frac{BR_1}{{\min_{t\in[1/k,k]}f(t,\lambda(k)R_1)}}\leq
\lambda\leq
\frac{AR_2}{{\sup_{t\in[0,+\infty)}f(t,R_2)}}.
\end{equation}
Then  \eqref{e1.1}-\eqref{e1.2} has at least two
positive solutions $v_1,v_2$ such that
\begin{gather*}
R_1\leq \| v_1\| \leq R_2,\quad \lim_{n\to \infty}T^{n}u_0=v_1,\quad
u_0=R_2,\quad t\in[0,+\infty),\\
R_1\leq \| v_2\| \leq R_2,\quad
\lim_{n\to \infty}T^{n}w_0 =v_2,\quad
w_0=R_1,\quad t\in[0,+\infty).
\end{gather*}
\end{theorem}

\begin{proof}
We define
$$
P_{[R_1,R_2]}=\{u\in P: R_1\leq \| u\|\leq R_2\}.
$$
First we prove that $TP_{[R_1,R_2]}\subset
P_{[R_1,R_2]}$. Let $u\in P_{[R_1,R_2]}$, thus obviously
we have
\begin{equation} \label{e3.8}
\lambda(k)R_1\leq \lambda(k)\| u\|
\leq\frac{u(t)}{1+t^{\alpha-1}}\leq u(t)\leq
\| u\| \leq R_2,\quad t\in[1/k,k].
\end{equation}
Using (H4), \eqref{e3.7} and \eqref{e3.8}, we have
\[
Tu(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds
\leq \lambda\int_0^{+\infty}G(t,s)a(s)f(s,R_2)ds.
\]
Hence
\begin{align*}
\| Tu\|
&\leq \lambda\int_0^{+\infty}\sup_{t\in[0,+\infty)}
 \frac{G(t,s)}{1+t^{\alpha-1}}a(s)f(s,R_2)ds\\
&\leq \lambda \sup_{t\in[0,+\infty)}f(t,R_2)A^{-1}
\leq R_2.
\end{align*}
Also considering \eqref{e3.7} and \eqref{e3.8},  we have
\begin{equation*}
Tu(t)\geq \lambda\int_0^{+\infty}G(t,s)a(s)f(s,\lambda(k)R_1)ds.
\end{equation*}
Thus
\begin{align*}
\| Tu\|
 &\geq \lambda\frac{\lambda^{2}(k)}{k^{\alpha-1}}
 \int_{1/k}^{k}a(s)ds\min_{t\in[1/k,k]}f(t,\lambda(k)R_1)\\
&=\lambda \min_{t\in[1/k,k]}f(t,\lambda(k)R_1)B^{-1}
\geq R_1.
\end{align*}
This implies
$TP_{[R_1,R_2]}\subset P_{[R_1,R_2]}$.
For every $t\in(0,+\infty)$ and $u_0=R_2$, clearly
$u_0\in P_{[R_1,R_2]}$. Now we consider the sequence
$\{u_{n}\}_{n\in \mathbb{N}}$ in $P_{[R_1,R_2]}$
and define 
\begin{equation} \label{e3.9}
u_{n}=Tu_{n-1}=T^{n}u_0,\quad i=1,2,3,\dots.
\end{equation}
Since,  $T$ is completely continuous,  there exist a
subsequence $\{u_{nk}\}$ of the sequence
$\{u_{n}\}_{n\in \mathbb{N}}$ such that it converges uniformly
to $v_1\in B$. On the other hand considering
the condition (H4), we can see that the operator
$T:P_{[R_1,R_2]}\to P_{[R_1,R_2]}$, is
nondecreasing. Since for every $t\in(0,+\infty)$
$$
0\leq u_1(t)\leq \| u_1\| \leq R_2=u_0(t).
$$
Thus $Tu_1\leq Tu_0$. Considering \eqref{e3.9} we conclude that
$u_2\leq u_1$. Similarly by induction we deduce that
$u_{n+1}\leq u_{n}$. Hence $\{u_{n}\}_{n\in \mathbb{N}}$
is a decreasing sequence, such that has a subsequence
$\{u_{nk}\}$ converges to $v_1$. Thus
$\{u_{n}\}_{n\in \mathbb{N}}$  converges uniformly to
$v_1$. Letting $n\to +\infty$ in \eqref{e3.9}
yields
\begin{equation} \label{e3.10}
Tv_1=v_1.
\end{equation}
Let $w_0=R_1$ for every $t\in(0,+\infty)$.
So $w_0\in P_{[R_1,R_2]}$. Now consider the sequence
$\{w_{n}\}_{n\in \mathbb{N}}$ given by
\begin{equation} \label{e3.11}
w_{n}=Tw_{n-1}, \quad n=1,2,3,\dots
\end{equation}
From \eqref{e3.11} we have $\{w_{n}\}_{n\in
\mathbb{N}}\subset P_{[R_1,R_2]}$. Moreover, using definition
\eqref{e2.8}, we conclude that
\begin{align*}
w_1(t)=Tw_0(t)
&=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,w_0(s))ds\\
&\geq \lambda\int_0^{+\infty}G(t,s)a(s)f(s,\lambda(k)R_1)ds\\
&\geq R_1=w_0(t)\quad t\in(0,+\infty).
\end{align*}
Thus using the same argument as above, we deduce that
$\{w_{n}\}_{n\in \mathbb{N}}$ is a increasing sequence
with subsequence $\{w_{nk}\}$ such that
$\{w_{nk}\}$ converges uniformly to $v_2\in
P_{[R_1,R_2]}$. Thus $\{w_{n}\}_{n\in \mathbb{N}}$
converges uniformly to $v_2\in P_{[R_1,R_2]}$. Letting
$n\to +\infty$, from \eqref{e3.11} we
find that
\begin{equation} \label{e3.12}
Tv_2=v_2.
\end{equation}
Finally from \eqref{e3.10} and \eqref{e3.12} we conclude that the
boundary-value problem \eqref{e1.1}-\eqref{e1.2} has at least two
positive solutions $v_1,v_2$ in $P$ which completes the proof.
\end{proof}

We conclude this article with two nonexistence results stated
in the following theorems.
Moreover, we show the compactness of the solutions set.

\begin{theorem} \label{thm3.5}
Let conditions {\rm (H1)--(H3)} hold. If
$f^{0},f^{\infty}<\infty$, then there exist a positive constant
$\lambda_0$, such that for every $0<\lambda<\lambda_0$, the
boundary value problem \eqref{e1.1}-\eqref{e1.2} has no positive solution.
\end{theorem}

\begin{proof}
Since $f^{0},f^{\infty}<\infty$,  for every
$t\in(0,+\infty)$, there exist positive constants
$c_1,c_2,r_1,r_2$ with $r_1<r_2$ such that
\begin{gather*}
f(t,u)\leq c_1\frac{u}{1+t^{\alpha-1}},\quad u\in[0,r_1]\\
f(t,u)\leq c_2\frac{u}{1+t^{\alpha-1}},\quad u\in[r_2,+\infty).
\end{gather*}
Let
$$
C=\max\big\{c_1,c_2,\sup_{r_1\leq u\leq r_2}\frac{(1+t^{\alpha-1})f(t,u)}{u}
\big\}.
$$
Thus we have
$$
f(t,u)\leq C\frac{u}{1+t^{\alpha-1}},\quad u\in[0,+\infty),\;t\in(0,+\infty).
$$
Assume $w(t)$ is a positive solution of the boundary value problem
\eqref{e1.1}-\eqref{e1.2}. We will show that this leads to a contradiction
for every $0<\lambda<\lambda_0$ with
$\lambda_0=\frac{A}{C}$.
\[
w(t)=Tw(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,w(s))ds\\
\leq \lambda C\| w\|\int_0^{+\infty}G(t,s)a(s)ds.
\]
Hence
\[
\| w\|\leq \lambda C\|
w\|\int_0^{+\infty}\sup_{t\in[0,+\infty)}\frac{G(t,s)}{1+t^{\alpha-1}}a(s)ds
=\frac{\lambda C}{A}\| w\|
<\| w\|,
\]
which is a contradiction. Therefore, \eqref{e1.1}-\eqref{e1.2}
 has no positive solution.
\end{proof}

\begin{theorem} \label{thm3.6}
Assume that conditions{\rm  (H1)--(H3)} hold. If
$f_0,f_{\infty}>0$, then there exist a positive constant
$\lambda_0$, such that for every $\lambda>\lambda_0$, the
boundary value problem \eqref{e1.1}-\eqref{e1.2} has no positive solution.
\end{theorem}

\begin{proof}
Since $f_0,f_{\infty}>0$,  we conclude that for all
$t\in[1/k,k]$, there exist positive constants
$m_1,m_2,r_1,r_2$ with $r_1<r_2$ such that
\begin{gather*}
f(t,u)\geq m_1\frac{u}{1+t^{\alpha-1}},\quad u\in[0,r_1]\\
f(t,u)\geq m_2\frac{u}{1+t^{\alpha-1}},\quad u\in[r_2,+\infty).
\end{gather*}
Assume that
$$
m=\min\big\{m_1,m_2,\min_{r_1\leq u\leq r_2}\frac{(1+t^{\alpha-1})f(t,u)}{u}
\big\}.
$$
Hence we have
$$
f(t,u)\geq m\frac{u}{1+t^{\alpha-1}}\geq m\lambda(k)\| u\|,\quad
u\in[0,+\infty),\; t\in[1/k,k].
$$
Let $w(t)$ be a positive solution of \eqref{e1.1}-\eqref{e1.2}.
 We will show that this leads to a contradiction for
every $\lambda>\lambda_0$, with $\lambda_0=B/m$.
\[
w(t)=Tw(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,w(s))ds
\geq m\lambda\lambda(k)\|w\|\int_0^{+\infty}G(t,s)a(s)ds.
\]
So
\[
\| w\| \geq m\lambda\frac{\lambda^{2}(k)}{k^{\alpha-1}}\|
w\|\int_{1/k}^{k}a(s)ds
 = \frac{\lambda m}{B}\| w\|
>\| w\|,
\]
which is a contradiction. Therefore \eqref{e1.1}-\eqref{e1.2}
 has no positive solution. This completes the proof.
\end{proof}

\begin{theorem} \label{thm3.7}
Assume conditions {\rm (H1)--(H3)} hold and that
\begin{equation} \label{e3.13}
f_0,f^{\infty}\in(0,+\infty),\quad
f_0\lambda(k)>f^{\infty},\quad
\lambda\in(\frac{B}{f_0},\frac{A}{f^{\infty}}).
\end{equation}
Then the set of positive solutions of \eqref{e1.1}-\eqref{e1.2}
is nonempty and compact.
\end{theorem}

\begin{proof}
Let $S=\{u\in P : u=Tu\}$.
Theorem \ref{thm3.2} implies that $S$ is nonempty. It is sufficient to
show that $S$ is compact in $B$.
First of all we claim that $S$ is closed in $B$. Let
$\{u_{n}\}_{n\in\mathbb{N}}$ be sequence in $S$, such
that $\lim_{n\to \infty}\|u_{n}-u\|=0$. Thus for every $t\in (0,+\infty)$,
 we have
\begin{align*}
&\big| u(t)-\lambda \int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds\big|\\
&\leq |u_{n}-u|+\big|  u_{n}(t)-\lambda
\int_0^{+\infty}G(t,s)a(s)f(s,u_{n}(s))ds\big|\\
&\quad +\lambda\int_0^{+\infty}G(t,s)a(s)|f(s,u(s))-f(s,u_{n}(s))|ds.
\end{align*}
Let $n\to \infty$, using the continuity of $f$ and by dominated
convergence theorem, we deduce that for all $t\in(0,+\infty)$
$$
 u(t)=\lambda \int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds.
$$
Thus $u\in S$ and $S$ is closed in $B$.

It remains to check that $S$ is relatively compact in $B$. Let
\eqref{e3.13} hold. Choosing $\epsilon>0$ such that
$$
\lambda\in(\frac{B}{f_0-\epsilon},\frac{A}{f^{\infty}+\epsilon}),
$$
we find that there exists a positive constant $R$ such that for
every $u\in[R,+\infty)$,
$$
f(t,u)\leq(f^{\infty}+\epsilon)\frac{u}{1+t^{\alpha-1}}\leq
(f^{\infty}+\epsilon)\| u\|.
$$
Hence for $t\in(0,+\infty)$, we have
\begin{gather*}
f(t,u)\leq (f^{\infty}+\epsilon)\| u\| +\gamma ,\\
\gamma=\max\{f(t,u):t\in[1/k,k],\,u\in[0,R]\}.
\end{gather*}
Thus for every $u\in S$ and $t\in(0,+\infty)$, we have
 \begin{align*}
u(t)&=\lambda \int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds\\
&\leq\lambda\left[(f^{\infty}+\epsilon)\| u\|
+\gamma \right]\int_0^{+\infty}G(t,s)a(s)ds.
\end{align*}
Then
\[
\| u\| \leq\lambda(\frac{(f^{\infty}+\epsilon)\|u\|+\gamma}{A}).
\]
Therefore,  $S$ is bounded in $B$. Now by
 compactness of the operator $T:P\to P$ we deduce that
 $S=TS$ is relatively compact, which completes the proof.
 \end{proof}

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\end{document}