\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 $$\label{e2.3} u(t)=\int_0^{+\infty}G(t,s)h(s)ds,$$ where $$\label{e2.4} G(t,s)=H_1(t,s)+H_2(t,s)$$ 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 $01$, $$\min_{1/k \leq t \leq k}\frac{H_1(t,s)}{1+t^{\alpha-1}} \geq \gamma_1\sup_{0\leq t <+\infty}\frac{H_1(t,s)}{1+t^{\alpha-1}},$$ 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 $$%\label{e} \gamma_1=\frac{m_1}{M_1}=\min\Big\{\frac{k^{2(\alpha-1)} -(k^{2}-1)^{\alpha-1}}{k^{\alpha-1}(1+k^{\alpha-1})}, \frac{1}{(1+k^{\alpha-1})}\Big\}$$ 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 $$\label{e2.8} Tu(t)=\lambda\int_0^{+\infty}G(t,s)a(s)f(s,u(s))ds.$$ 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 $$V=\{u\in B: \| u\|0\}, \quad W=\{\frac{u(t)}{1+t^{\alpha-1}}:u\in V\}.$$ 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 $$V=\{u\in B:\| u\|0\}, W=\{\frac{u(t)}{1+t^{\alpha-1}}\big|u\in V\}.$$ 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 $u\in P$, $\| u \| f^{\infty}$, 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 $$\Omega_i=\{u\in B :\| u\|f^{\infty}, also \lambda\in(\frac{B}{f_0},\frac{A}{f^{\infty}}). Let \epsilon>0 be chosen such that $$\label{e3.1} \frac{B}{f_0-\epsilon}<\lambda<\frac{A}{f^{\infty}+\epsilon}$$ Since  f_0\in(0, \infty), thus there exist a positive constant R_1 such that for every t\in[1/k,k] and u\in[0,R_1], \begin{equation*} f(t,u)=f(t,\frac{(1+t^{\alpha-1})u}{1+t^{\alpha-1}}) \geq(f_0-\epsilon)\frac{u}{1+t^{\alpha-1}}. \end{equation*} So if u\in P with \| u\|=R_1, then$$ 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, $$\label{e3.2} \| Tu\| \geq \| u\| \quad \forall u\in P\cap \partial\Omega_1.$$ 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 $$\label{e3.3} \| Tu\|\leq \| u\| \quad \forall u\in P\cap \partial\Omega_2.$$ 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 $$\label{e3.4} \frac{B}{f_{\infty}-\epsilon}<\lambda<\frac{A}{f^{0}+\epsilon}.$$ 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 $$\label{e3.7} \frac{BR_1}{{\min_{t\in[1/k,k]}f(t,\lambda(k)R_1)}}\leq \lambda\leq \frac{AR_2}{{\sup_{t\in[0,+\infty)}f(t,R_2)}}.$$ 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 $$\label{e3.8} \lambda(k)R_1\leq \lambda(k)\| u\| \leq\frac{u(t)}{1+t^{\alpha-1}}\leq u(t)\leq \| u\| \leq R_2,\quad t\in[1/k,k].$$ 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 $$\label{e3.9} u_{n}=Tu_{n-1}=T^{n}u_0,\quad i=1,2,3,\dots.$$ 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 $$\label{e3.10} Tv_1=v_1.$$ 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 $$\label{e3.11} w_{n}=Tw_{n-1}, \quad n=1,2,3,\dots$$ 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 $$\label{e3.12} Tv_2=v_2.$$ 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_10, 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\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 $$\label{e3.13} f_0,f^{\infty}\in(0,+\infty),\quad f_0\lambda(k)>f^{\infty},\quad \lambda\in(\frac{B}{f_0},\frac{A}{f^{\infty}}).$$ 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} \begin{thebibliography}{00} \bibitem{Avery} R. I. Avery, A. C. Peterson; Three positive fixed points of nonlinear operators on ordered banach spaces, \emph{Comput. math. Appl.} \textbf{42} (2001), 313-322. \bibitem{Agrawal} R. P. Agrawal, D. O'Regon; Positive solutions for Dirichlet problem of singular nonlinear fractional differential equations, \emph{J. 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