\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 06, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/06\hfil short-title??] {Monotone iterative method for fractional differential equations} \author[Z. Bai, S. Zhang, S. Sun, C. Yin\hfil EJDE-2016/06\hfilneg] {Zhanbing Bai, Shuo Zhang, Sujing Sun, Chun Yin} \address{Zhanbing Bai (corresponding author) \newline College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao 266590, China} \email{zhanbingbai@163.com} \address{Shuo Zhang \newline College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao 266590, China} \email{18353250431@163.com} \address{Sujing Sun \newline College of Information Science and Engineering, Shandong University of Science and Technology, Qingdao 266590, China} \email{kdssj@163.com} \address{Chun Yin \newline School of Automation Engineering, University of Electronic Science and Technology of China, Chengdu 611731, China} \email{yinchun.86416@163.com} \thanks{Submitted October 14, 2015. Published January 6, 2016.} \subjclass[2010]{34B15, 34A08} \keywords{Fractional initial value problem; lower and upper solution method; \hfill\break\indent existence of solutions} \begin{abstract} In this article, by using the lower and upper solution method, we prove the existence of iterative solutions for a class of fractional initial value problem with non-monotone term \begin{gather*} D_{0+}^\alpha u(t)=f(t, u(t)), \quad t \in (0, h), \\ t^{1-\alpha}u(t)\big|_{t=0} = u_0 \neq 0, \end{gather*} where $00), \\ D_{0+}^{\alpha-1}u (0+) = u_0. \label{eq12} \end{gather} was obtained under the assumption that $f: [0, 1] \times \mathbb{R} \to \mathbb{R}$ is Lipchitz continuous, by using the Banach concentration mapping principle. In \cite{Zha}, the existence and uniqueness of solution of the initial value problem \begin{gather*} D_{0+}^\alpha u(t) = f(t, u(t)), \quad t \in (0, T], \\ t^{1-\alpha} u(t)\big|_{t=0} =u_0 \end{gather*} was discussed by using the method of lower and upper solutions and its associated monotone iterative method. In \cite{Nieto}, a new proof of the maximum principle was given by using the completely monotonicity of the Mittag-Leffler type function. We refer the readers to \cite{Dee} for other applications of monotone method to various fractional differential equations. In the previous works, the nonlinear term has to satisfy the monotone or other control conditions. In fact, the nonlinear fractional differential equation with non-monotone term can respond better to impersonal law, so it is very important to weaken control conditions of the nonlinear term. Motivated by the above references, we focus our attention on the problem \begin{gather} \label{eq17} D_{0+}^\alpha u(t)=f(t, u(t)), \quad t \in (0, h), \\ t^{1-\alpha}u(t)\big|_{t=0}=u_0, \label{eq18} \end{gather} where $f\in C([0, h]\times \mathbb{R}, \mathbb{R})$, $D_{0+}^\alpha u (t) $ is the standard Riemann-Liouville fractional derivative, $0<\alpha< 1$. The existence of the blow-up solution, that is to say $u \in C(0, h]$ and $\lim_{t \to 0+}u(t) =\infty$, is obtained by the use of the lower and upper solution method. This paper is organized as follows. In section 2, we recall briefly some notion of fractional calculus and theory of the operators for integration and differentiation of fractional order. Section 3 is devoted to the study of the existence of solution for utilizing the method of upper and lower solutions. The existence of maximal and minimal solutions is also given. \section{Preliminaries} Given $0 \le a < b <+\infty$ and $r>0$, define a set $$ C_r[a, b]=\{u : u \in C(a, b], (t-a)^r u(t) \in C[a, b]\}. $$ Clearly, $C_r[a, b]$ is a linear space with the normal multiplication and addition. Given $u \in C_r[a, b]$, define $$ \|u\| = \max_{t \in [a, b]}(t-a)^r|u(t)|, $$ then $(C_r[a, b], \|\cdot\|)$ is a normed space. Moreover, if $\{u_n\} \subset C_r[a, b]$ and $\|u_n-u\| \to 0$, then one has $u \in C_r[a, b]$. In fact, setting $v_n(t) =(t-a)^ru_n(t), ~v(t)=(t-a)^ru(t)$, then $v_n \in C[a, b]$ and $$ \|u_n - u\| \to 0 \Leftrightarrow \|v_n - v\|_\infty \to 0. $$ By the completeness of the space $C[a, b]$, one has $v \in C[a, b]$, so $u(t) = (t-a)^{-r} v(t) \in C_r[a, b]$. Thus, $(C_r[a, b], \|\cdot\|)$ is a Banach space. \begin{lemma}[\cite{KilST}] \label{lem2.1} The linear initial value problem \begin{gather*} D_{0+}^\alpha u(t) + \lambda u(t) = q(t), \\ t^{1-\alpha}u(t)\big|_{t=0}=u_0, \end{gather*} where $\lambda \ge 0$ is a constant and $q \in L(0, h)$, has the following integral representation for a solution $$ u(t) = \Gamma(\alpha) u_0t^{\alpha-1}E_{\alpha, \alpha}(-\lambda t^\alpha) + \int_0^t (t-s)^{\alpha-1}E_{\alpha, \alpha}(-\lambda (t-s)^{\alpha})q(s)ds. $$ Here, $E_{\alpha, \alpha}(t)$ is a Mittag-Leffler function. \end{lemma} \begin{lemma}\label{lem2.2} For $0 < \alpha \le 1$, the Mittag-Leffler type function $E_{\alpha, \alpha}(-\lambda t^\alpha)$ satisfies $$ 0 \le E_{\alpha, \alpha}(-\lambda t^\alpha) \le \frac{1}{\Gamma(\alpha)},\quad t \in[0, \infty), \; \lambda \ge 0. $$ \end{lemma} \begin{proof} According to \cite{Nieto, Sch}, the function $g(t):=E_{\alpha, \alpha}(-\lambda t^\alpha)$, $t \in (0, +\infty)$ is completely monotonic, that is to say that $g(t)$ possesses of derivatives $g^{(n)}(t)$ for all $n=0, 1, 2, \dots$, and $(-1)^nf^{(n)}(t) \ge 0$ for all $t \in (0, \infty)$. This combined with the fact that $E_{\alpha, \alpha}(-\lambda t^\alpha)$ is continuous on $\mathbb{R}$ and $E_{\alpha, \alpha}(0)=1/\Gamma(\alpha)$ yields the conclusion. \end{proof} \begin{lemma}[\cite{GSL}] \label{lem2.3} Suppose that $E$ is an ordered Banach space, $x_0, y_0 \in E$, $x_0 \le y_0$, $D=[x_0, y_0]$, $T: D \to E$ is an increasing completely continuous operator and $x_0 \le Tx_0, ~y_0 \ge Ty_0$. Then the operator $T$ has a minimal fixed point $x^*$ and a maximal fixed point $y^*$. If we let $$ x_n = Tx_{n-1}, \quad y_n = Ty_{n-1}, \quad n=1, 2, 3, \dots, $$ then \begin{gather*} x_0 \le x_1 \le x_2 \le \dots \le x_n \le \dots \le y_n \le \dots \le y_2 \le y_1 \le y_0,\\ x_n \to x^*, \quad y_n \to y^*. \end{gather*} \end{lemma} \begin{definition} \label{def2.2} \rm A function $v(t) \in C_{1-\alpha}[0, h]$ is called as a lower solution of \eqref{eq17}, \eqref{eq18}, if it satisfies \begin{gather} \label{l01} D_{0+}^\alpha v(t) \le f(t, v(t)), \quad t \in (0, h), \\ t^{1-\alpha}v(t)\big|_{t=0} \le u_0. \label{l02} \end{gather} \end{definition} \begin{definition} \label{def2.3} \rm A function $w(t) \in C_{1-\alpha}[0, h]$ is called as an upper solution of \eqref{eq17}, \eqref{eq18}, if it satisfies \begin{gather} \label{U01} D_{0+}^\alpha w(t) \ge f(t, w(t)), \quad t \in (0, h), \\ t^{1-\alpha} w(t)\big|_{t=0} \ge u_0. \label{U02} \end{gather} \end{definition} \section{Existence of solutions} The following assumptions will be used in our main results: \begin{itemize} \item[(A1)] $f: [0, h] \times \mathbb{R} \to \mathbb{R}$ and there exist constants $A, B \ge 0$ and $0 < r_1 \le 1 < r_2<1/(1-\alpha)$ such that for $t \in [0, h]$ \begin{equation}\label{eq333} |f(t, u) - f(t, v)| \le A|u-v|^{r_1} + B |u-v|^{r_2}, \quad u, v \in \mathbb{R}. \end{equation} \item[(A2)] Assume that $ f: [0, h] \times \mathbb{R} \to \mathbb{R}$ satisfies \begin{equation}\label{eq334} f(t, u) - f(t, v) + \lambda (u-v) \ge 0, \quad \text{for } \hat{u} \le v \le u \le \tilde{u}, \end{equation} where $\lambda \ge 0$ is a constant and $\hat{u}, \tilde{u}$ are lower and upper solutions of Problem \eqref{eq17}, \eqref{eq18} respectively. \end{itemize} \begin{remark} \label{rmk3.1} \rm Assume that $f(t, u) =a(t) g(u)$ and $g$ is a H\"{o}lder continuous function, $a(t)$ is bounded, then \eqref{eq333} holds. \end{remark} \begin{theorem} \label{thm3.1} Suppose {\rm (A1)} holds. The function $u$ solves problem \eqref{eq17}, \eqref{eq18} if and only if it is a fixed-point of the operator $T: C_{1-\alpha} [0, h] \to C_{1-\alpha} [0, h]$ defined by \begin{align*} (Tu)(t) &= \Gamma(\alpha)u_0t^{\alpha-1} E_{\alpha, \alpha}(-\lambda t^{\alpha}) \\ &\quad + \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha} (-\lambda( t-s)^{\alpha})[ f(s, u(s))+\lambda u(s)]ds. \end{align*} \end{theorem} \begin{proof} Firstly, we need to show that the operator $T$ is well defined, i.e., for every $u \in C_{1-\alpha}[0, h] $ and $t>0$, the integral $$ \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha}) [ f(s, u(s))+\lambda u(s)]ds $$ belongs to $C_{1-\alpha}[0, h]$. Under condition \eqref{eq333}, $$ |f(t, u)| \le A|u|^{r_1} + B|u|^{r_2} + C, $$ where $C= \max_{t \in [0, h]}f(t, 0)$. By Lemma \ref{lem2.2}, for $u(t) \in C_{1-\alpha}[0, h]$, we have \begin{align*} & \Big| t^{1-\alpha}\int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha})[ f(s, u(s))+\lambda u(s)]ds\Big| \\ & \le t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha})|f(s, u(s))+\lambda u(s)|ds \\ & \le t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha})\big(A |u|^{r_1} + \lambda |u| + B |u|^{r_2} + C\big)ds \\ & \le t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha}) \Big\{A s^{(\alpha-1)r_1}[s^{1-\alpha}|u(s)|]^{r_1} \\ &\quad + \lambda s^{\alpha-1} s^{1-\alpha}|u(s)| + B s^{(\alpha-1)r_2} [s^{1-\alpha}|u(s)|]^{r_2} +C \} ds \\ & \le \frac{A \|u\|^{r_1} t^{1-\alpha}}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1} s^{(\alpha-1)r_1}ds + \frac{\lambda \|u\|t^{1-\alpha}}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1} s^{\alpha-1}ds\\ &\quad + \frac{B \|u\|^{r_2} t^{1-\alpha}}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1} s^{(\alpha-1)r_2}ds + \frac{Ct}{\Gamma(\alpha+1)} \\ & \le A \|u\|^{r_1} \frac{\Gamma((\alpha-1)r_1+1)}{\Gamma((\alpha-1)r_1+\alpha+1)} t^{(\alpha-1)r_1 +\alpha +1-\alpha} + \lambda \|u\| \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} t^{\alpha}\\ &\quad +B \|u\|^{r_2} \frac{\Gamma((\alpha-1)r_2+1)}{\Gamma((\alpha-1)r_2 +\alpha+1)} t^{(\alpha-1)r_2 +\alpha +1-\alpha} + \frac{Ct}{\Gamma(\alpha+1)} \\ & \le \frac{\Gamma[(\alpha-1)r_1+1] A h^{(\alpha-1)r_1+1}} {\Gamma[(\alpha-1)r_1+\alpha+1]}\|u\|^{r_1} + \lambda \|u\| \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} h^{\alpha} \\ &\quad +\frac{\Gamma[(\alpha-1)r_2+1] B h^{(\alpha-1)r_2+1}} {\Gamma[(\alpha-1)r_2+\alpha+1]}\|u\|^{r_2} + \frac{Ch}{\Gamma(\alpha+1)}. \end{align*} That is to say that the integral exists and belongs to $C_{1-\alpha}[0, h]$. The the above inequality and the assumption $0< r_1 \le 1 < r_2 < 1/(1-\alpha)$ imply that $$ \lim_{ t\to 0+} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha})[ f(s, u(s))+\lambda u(s)]ds=0. $$ Combining with the fact that $\lim_{t \to 0+}E_{\alpha, \alpha}(-\lambda t^\alpha) = E_{\alpha, \alpha}(0)=1/\Gamma(\alpha)$ yields that $\lim_{t \to 0+} t^{1-\alpha}(Tu)(t) = u_0$. The above arguments combined with Lemma \ref{lem2.1} implies that the fixed-point of the operator $T$ solves \eqref{eq17}, \eqref{eq18}. And the vice versa. The proof is complete. \end{proof} In the following, we consider the compactness of a set of the space $C_r[0, h]$. Let $F \subset C_r[0, h]$ and $E= \{g(t)= t^r h(t) \mid h(t)\in F\}$, then $E \subset C[0, h]$. It is clear that $F$ is a bounded set of $C_r[0, h]$ if and only if $E$ is a bounded set of $C[0, h]$. Therefore, to proof that $F \subset C_r[0, h]$ is a compact set, it is sufficient to prove that $E \subset C[0, h]$ is a bounded and equicontinuous set. \begin{theorem} \label{thm3.2} Suppose {\rm (A1)} holds. Then $T$ is a completely continuous operator. \end{theorem} \begin{proof} Given $u_n \to u \in C_{1-\alpha}[0, h]$, with the definition of $T$ and condition (A1), one has \begin{align*} &\|Tu_n -Tu\| \\ &= \|t^{1-\alpha}(Tu_n - Tu)\|_\infty \\ &= \max_{0 \le t \le h}\Big|t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha}(-\lambda( t-s)^{\alpha}) [f(s, u_n)-f(s, u)+ \lambda(u_n-u)]ds\Big| \\ &\le \frac{1}{\Gamma(\alpha)} \max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1}[A|u_n-u|^{r_1} + B|u_n-u|^{r_2} + \lambda|u_n-u|]ds \\ &\le \frac{1}{\Gamma(\alpha)} \Big[A \max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-r_1(1-\alpha)} s^{r_1(1-\alpha)} |u_n-u|^{r_1}ds \\ &\quad + \lambda \max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-(1-\alpha)} s^{(1-\alpha)} |u_n-u|ds\\ &\quad + B \max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-r_2(1-\alpha)} s^{r_2(1-\alpha)} |u_n-u|^{r_2}ds \Big] \\ &\le \frac{1}{\Gamma(\alpha)} \Big[A \|u_n-u\|^{r_1}\max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-r_1(1-\alpha)}ds \\ &\quad + \lambda \|u_n-u\|\max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-(1-\alpha)} ds\\ &\quad + B \|u_n-u\|^{r_2}\max_{0 \le t \le h} t^{1-\alpha} \int_0^t (t-s)^{\alpha-1} s^{-r_2(1-\alpha)} ds \Big] \\ &\le \frac{A\|u_n-u\|^{r_1} \Gamma[1-r_1(1-\alpha)]}{\Gamma[1-r_1(1-\alpha) +\alpha]} h^{1-r_1(1-\alpha)} + \frac{\lambda\|u_n-u\| \Gamma[\alpha]}{\Gamma[2\alpha]} h^{\alpha} \\ &\quad +\frac{B\|u_n-u\|^{r_2} \Gamma[1-r_2(1-\alpha)]}{\Gamma[1-r_2(1-\alpha) +\alpha]} h^{1-r_2(1-\alpha)} \\ & \to 0, \quad (n \to \infty). \end{align*} That is to say that $T$ is continuous. Suppose that $F \subset C_{1-\alpha}[0, h]$ is a bounded set. The argument as in the proof of Theorem \ref{thm3.1} shows that $T(F) \subset C_{1-\alpha}[0, h] $ is bounded. At last, we prove the equicontinuity of $T(F)$. Let $f_1(t, u)= f(t, u) + \lambda u$. Given $\epsilon>0$, for every $u \in F $ and $t_1, t_2\in [0, h], t_1 \le t_2$, \begin{align*} & \big|[t^{1-\alpha} (Tu)(t)]_{t =t_2} - [t^{1-\alpha} (Tu)(t)]_{t =t_1}\big| \\ & \le \big[\Gamma(\alpha)u_0E_{\alpha, \alpha}(-\lambda t^{\alpha})\big]_{t_1}^{t_2} + \Big[t^{1-\alpha}\int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha} (-\lambda( t-s)^{\alpha}) f_1(s, u(s))ds\Big]_{t_1}^{t_2}\\ & \le \big[\Gamma(\alpha)u_0E_{\alpha, \alpha}(-\lambda t^{\alpha})\big]_{t_1}^{t_2} + \frac{1}{\Gamma(\alpha)}\int_{t_1}^{t_2} t_2^{1-\alpha} (t_2-s)^{\alpha-1} |f_1(s, u(s))|ds \\ &\quad + \frac{1}{\Gamma(\alpha)}\int_0^{t_1} \left[t_2^{1-\alpha}(t_2-s)^{\alpha-1} - t_1^{1-\alpha}(t_1-s)^{\alpha-1}\right] s^{\alpha-1} \left |s^{1-\alpha}f_1(s, u(s))\right|ds. \end{align*} For the first term of the above formula, by the function $E_{\alpha, \alpha}(-\lambda t^{\alpha})$ is continuous and therefore uniformly continuous on $[0, h]$, there exists $\delta_1>0$ such that when $|t_2 -t_1| <\delta_1$, there is $$ \big[\Gamma(\alpha)u_0E_{\alpha, \alpha}(-\lambda t^{\alpha})\big]_{t_1}^{t_2} < \frac{\epsilon}{3}; $$ For the second term, by the continuity of $ t_2^{1-\alpha} (t_2-s)^{\alpha-1}|f_1(s, u(s))|$, there is a positive $M_1$ such that $$ \frac{1}{\Gamma(\alpha)} | t_2^{1-\alpha} (t_2-s)^{\alpha-1}f_1(s, u(s))| < M_1, $$ Thus, letting $\delta_2=\epsilon/ (3 M_1)$, when $|t_2 -t_1| <\delta_2$, we have $$ \frac{1}{\Gamma(\alpha)} \int_{t_1}^{t_2} t_2^{1-\alpha} (t_2-s)^{\alpha-1}|f_1(s, u(s))|ds < \frac{\epsilon}{3}; $$ For the third term, $\int_0^{t_1} s^{\alpha-1}ds= (t_1)^{\alpha-1}/(\alpha-1)$. By the continuity of the function $ s^{1-\alpha}f_1(s, u(s))$, there is a positive constant $M_2$ such that $$ |s^{1-\alpha}f_1(s, u(s))| \le M_2; $$ The function $t_2^{1-\alpha}(t_2-s)^{\alpha-1} - t_1^{1-\alpha}(t_1-s)^{\alpha-1}$ is continuity and therefore uniform continuity on $[0, h]^3$, so there exists $\delta_3>0$ such that when $|t_2 -t_1| <\delta_3$, there is $$ \frac{1}{\Gamma(\alpha)} |t_2^{1-\alpha}(t_2-s)^{\alpha-1} - t_1^{1-\alpha}(t_1-s)^{\alpha-1}| < \frac{\epsilon(\alpha-1)}{3M_2h^{\alpha-1}}, $$ thus $$ \frac{1}{\Gamma(\alpha)} \int_0^{t_1} \left[t_2^{1-\alpha}(t_2-s)^{\alpha-1} - t_1^{1-\alpha}(t_1-s)^{\alpha-1}\right] s^{\alpha-1} \left |s^{1-\alpha}f(s, u(s))\right|ds< \frac{\epsilon}{3}. $$ To sum up, Given $\epsilon>0$, for every $u \in F $ and $t_1, t_2\in [0, h]$, let $\delta = \min\{\delta_1, \delta_2, \delta_3 \}$, when $|t_2 -t_1| <\delta$, there holds $$ \left|[t^{1-\alpha} (Tu)(t)]_{t =t_2} - [t^{1-\alpha} (Tu)(t)]_{t =t_1}\right| < \epsilon. $$ That is to say that $T(F)$ is equicontinuous. The proof is complete. \end{proof} \begin{theorem}\label{thm3.3} Assume {\rm (A1), (A2)} hold, and $v, w \in C_{1-\alpha}[0, h]$ are lower and upper solutions of \eqref{eq17} and \eqref{eq18} respectively, such that \begin{equation}\label{eq213} v(t) \le w(t), \quad 0\le t\leq h. \end{equation} Then, the fractional IVP \eqref{eq17}, \eqref{eq18} has a minimal solution $x^*$ and a maximal solution $y^*$ such that $$ x^* = \lim_{n \to \infty} T^nv, \quad y^* = \lim_{n \to \infty} T^nw. $$ \end{theorem} \begin{proof} Clearly, if functions $v, w$ are lower and upper solutions of IVP \eqref{eq17}, \eqref{eq18}, then there are $v \le Tv, w \ge Tw$. In fact, by the definition of the lower solution, there exist $q(t) \ge 0$ and $\epsilon \ge 0$ such that \begin{align*} D_{0+}^\alpha v(t) = f(t, v(t)) -q(t), \quad t \in (0, h), \\ t^{1-\alpha} v(t) =u_0 -\epsilon. \end{align*} Using Theorem \ref{thm3.1} and Lemma \ref{lem2.2}, one has \begin{align*} v(t) &= \Gamma(\alpha)(u_0-\epsilon)t^{\alpha-1} E_{\alpha, \alpha} (-\lambda t^{\alpha}) \\ & \quad +\int_0^t (t-s)^{\alpha-1} E_{\alpha, \alpha} (-\lambda( t-s)^{\alpha})[ f(s, v(s))+\lambda v(s)-q(s)]ds \\ & \le (Tv)(t). \end{align*} Similarly, there is $w \ge Tw$. By Theorem \ref{thm3.2} the operator $T: C_{1-\alpha}[0, h]\to C_{1-\alpha}[0, h]$ is increasing and completely continuous. Setting $D:= [v, w]$, by the use of Lemma \ref{lem2.3}, the existence of $x^*, y^*$ is obtained. 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