\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 166, 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 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/166\hfil Positive nondecreasing solutions] {Positive nondecreasing solutions for a multi-term fractional-order functional differential equation with integral conditions} \author[A. M. A. El-Sayed, E. O. Bin-Taher \hfil EJDE-2011/166\hfilneg] {Ahmed M. A. El-Sayed, Ebtisam O. Bin-Taher} \address{Ahmed M. A El-Sayed \newline Faculty of Science, Alexandria University, Alexandria, Egypt} \email{amasayed5@yahoo.com, amasayed@hotmail.com} \address{Ebtisam O. Bin-Taher \newline Faculty of Science, Hadhramout University of Sci. and Tech., Hadhramout, Yemen} \email{ebtsamsam@yahoo.com} \thanks{Submitted June 14, 2011. Published December 14, 2011.} \subjclass[2000]{34B10, 26A33} \keywords{Fractional calculus; Cauchy problem; nonlocal condition; \hfill\break\indent integral condition; functional differential equation; integral equation; deviated argument} \begin{abstract} In this article, we prove the existence of positive nondecreasing solutions for a multi-term fractional-order functional differential equations. We consider Cauchy boundary problems with: nonlocal conditions, two-point boundary conditions, integral conditions, and deviated arguments. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \allowdisplaybreaks \section{Introduction} Problems with non-local conditions have been extensively studied by several authors in the previous two decades; see for example \cite{B1}-\cite{b1}, \cite{e1}-\cite{N} and references therein. In this work we study the existence of nondecreasing solutions for the fractional differential equation \begin{equation}\label{eq1.1} x'(t)=f(t, D^{{\alpha}_1}x(m_1(t)), D^{{\alpha}_2}x(m_2t)), \dots, D^{{\alpha}_n}x(m_n(t))),\quad {\alpha}_{i}\in(0, 1), \end{equation} a.e. $t\in(0, 1)$, with the nonlocal condition \begin{equation}\label{eq1.2} \sum_{k=1}^{m} a_k x(\tau_k) =\beta \sum_{j=1}^{p} b_j x(\eta_j), \end{equation} where $ a_k, b_j>0$, $\tau_k\in(a,c)$, $\eta_j\in(d,b)$, $0 0$, is defined by (see \cite{P2}) \[ I_a^\beta f(t)=\int_a^t \frac{(t - s)^{\beta - 1}}{\Gamma(\beta)} f(s) \,ds. \] The Caputo fractional-order derivative of $f(t)$ of order $\alpha \in (0,1]$ is defined as (see \cite{P1,P2}) \[ D_a^\alpha f(t)=I_a^{1-\alpha} \frac{d}{dt} f(t) =\int_a^t \frac{ (t - s)^{- \alpha}}{\Gamma(1 - \alpha)} \frac{d}{ds}f(s)\,ds. \] \end{definition} \begin{theorem}[Schauder fixed point theorem \cite{KD}] \label{thm2.1} Let $E$ be a Banach space and $Q$ be a convex subset of $E$, and $T:Q\to Q$ is compact, continuous map, Then $T$ has at least one fixed point in $Q$. \end{theorem} \begin{theorem}[Kolmogorov compactness criterion \cite{JD}] \label{thm2.2} Let $\Omega \subseteq L^p [0,1]$, $1 \leq p < \infty$. If \begin{itemize} \item[(i)] $\Omega$ is bounded in $L^p [0,1]$, and \item[(ii)] $u_h \to u$ as $h \to 0$ uniformly with respect to $u \in \Omega$, \end{itemize} then $\Omega$ is relatively compact in $L^p [0,1]$, where \[ u_h(t)=\frac{1}{h} \int_t^{t+h} u(s)\,ds. \] \end{theorem} \section{Main results} We consider firstly the fractional-order functional integral equation \begin{equation}\label{eq3.1} y(t)=f(t, I^{1-{\alpha}_1}m_1'(t)y(m_1(t)),\dots, I^{1-{\alpha}_n}m_n'(t)y(m_n(t))). \end{equation} A function $ y $ is called a solution of the fractional-order functional integral equation \eqref{eq3.1} if $y\in L^{1}[0,1]$ and satisfies \eqref{eq3.1}. In this article, we use the following assumption: \begin{itemize} \item[(i)] $f:[0,1]\times R_{+}^{n}\to R_+ $ is a function with the following properties: \begin{itemize} \item[(a)] for each $t \in [0,1], f(t,.)$ is continuous, \item[(b)] for each $x \in R_{+}^{n}, f(.,x)$ is measurable; \end{itemize} \item[(ii)] there exists an integral function $ a\in L^1[0,1]$ and constants $q_{i}>0$, $i=1, 2$, such that \[ |f(t,x)| \leq a(t) + \sum_{i=1}^{n}q_{i} |x_{i}|, \text{for each } t\in[0,1], x\in R_n; \] \item[(iii)] $ m_i: [0,1] \to [0,1] $ are absolutely continuous functions; \item[(iv)] \[ \sum_{i=1}^{n}\frac{q_{i}}{\Gamma(2-\alpha_{i})} < 1. \] \end{itemize} \begin{theorem} \label{thm3.1} Assume {\rm (i)-(iv)}. Then \eqref{eq3.1} has at least one positive solution $ y\in L^{1}[0,1]$. \end{theorem} \begin{proof} Define the operator $ T $ associated with \eqref{eq3.1} by \[ Ty(t)=f(t, I^{1-{\alpha}_1}m_1'(t)y(m_1(t)),\dots, I^{1-{\alpha}_n}m_n'(t)y(m_n(t))). \] Let $B^+_{r}=\{ y\in R^+: \|y\|_{L^1} \leq r \}\subset L^1$, \[ r=\frac{\|a\|}{1 - \sum_{i=1}^{n}\frac{q_{i}}{\Gamma(2-\alpha_{i})}}. \] Let $ y $ be an arbitrary element in $B^+_{r}$. Then from the assumptions (i) and (ii), we obtain \begin{align*} \|Ty\|_{L_1} &= \int_0^1 |Ty(t)| dt\\ &= \int_0^1 |f(t, I^{1-{\alpha}_1}m_1'(t)y(m_1(t)),\dots, I^{1-{\alpha}_n}m_n'(t)y(m_n(t)))|dt\\ &\leq \int_0^1|a(t)|dt + \sum_{i=1}^{n}q_{i} \int_0^1\int_0^t\frac{(t-s)^{-\alpha_{i}}}{\Gamma(1-\alpha_{i})} |y(m_i(s))| dm_i(s) dt\\ &\leq \|a\|_{L^1} + \sum_{i=1}^{n}q_{i}\int_0^1 \Big( \int_s^1\frac{(t-s)^{-\alpha_{i}}}{\Gamma(1-\alpha_{i})}dt\Big) |y(m_i(s))| dm_i(s)\\ &\leq \|a\|_{L^1} + \sum_{i=1}^{n}q_{i}\int_{m_i(0)}^{m_i(1)} \frac{1}{\Gamma(2-\alpha_{i})}|y(m_i(s))| dm_i(s)\\ &\leq \|a\|_{L^1} + \sum_{i=1}^{n}q_{i}\int_0^1 \frac{1}{\Gamma(2-\alpha_{i})}|y(u)| du\\ &\leq \|a\|_{L^1}+ \sum_{i=1}^{n} \frac{q_{i}}{\Gamma(2-\alpha_{i})}\|y\|_{L_1}\leq r, \end{align*} which implies that the operator $T$ maps $B^+_{r}$ into itself. Assumption (i) implies that $T$ is continuous. Now, we will show that $T$ is compact. Let $\Omega$ be a bounded subset of $ B^+_r$. Then $T(\Omega)$ is bounded in $L^1[0,1]$; i.e., condition (i) of Theorem \ref{thm2.2} is satisfied. It remains to show that $(Ty)_h \to Ty$ in $L^1[0,1]$ as $h \to 0$, uniformly with respect to $Ty \in T \Omega$. Now \begin{align*} &\|(Ty)_h-Ty\|_{L^1}\\ &=\int_0^1|(Ty)_h(t)-(Ty)(t)| dt\\ &= \int_0^1 \big|\frac{1}{h} \int_t^{t+h} (Ty)(s)\,ds - (Ty)(t)\big| dt\\ &= \int_0^1 \Big(\frac{1}{h} \int_t^{t+h} |(Ty)(s) -(Ty)(t)|\,ds\Big) dt\\ &\leq \int_0^1\frac{1}{h}\int_t^{t+h} \Big|f(s, I^{1-{\alpha}_1}m_1'(s)y(m_1(s)),\dots, I^{1-{\alpha}_n}m_n'(s)y(m_n(s)))\\ &\quad - f(t, I^{1-{\alpha}_1}m_1'(t)y(m_1(t)),\dots, I^{1-{\alpha}_n}m_n'(t)y(m_n(t)))\Big| \,ds\, dt. \end{align*} Now, by assumption (ii), $ y\in \Omega $ implies $ f\in L^1[0,1]$; then \[ \frac{1}{h}\int_t^{t+h} \big|f(s, I^{1-{\alpha}_1}m_1'(s)y(m_1(s)),\dots) - f(t, I^{1-{\alpha}_1}m_1'(t)y(m_1(t)),\dots)\big| \,ds\,dt \to 0. \] Therefore, by Theorem \ref{thm2.2}, $T(\Omega)$ is relatively compact; that is, $T$ is compact, then the operator $T$ has a fixed point in $ B^+_{r}$, which proves the existence of positive solution $y\in B^+_r \subset L^1[0,1]$ of equation \eqref{eq3.1}. \end{proof} Let $ AC[0,1] $ be the class of absolutely continuous functions defined on $ [0,1]$. For the existence of solution for the nonlocal problem \eqref{eq1.1}-\eqref{eq1.2}, we have the following result. \begin{theorem} \label{thm3.2} Under the assumptions of Theorem \ref{thm3.1}, problem \eqref{eq1.1}-\eqref{eq1.2} has at least one solution $x \in AC[0,1]$. \end{theorem} \begin{proof} %Consider the differential equation \eqref{eq1.1}. Let $ y(t)=x'(t)$, then \begin{gather}\label{eq5} x(t)=x(0) + Iy(t), \\ x'(m_i(t))=m_i'(t) y(m_i(t)), \end{gather} and $ y $ is the solution of the fractional-order functional integral equation \eqref{eq3.1}. Let $t=\tau_k$ in equation \eqref{eq5}. We obtain \begin{gather*} x(\tau_k)=\int_0^{\tau_k} y(s)\,ds+x(0), \\ \sum_{k=1}^{m} a_kx(\tau_k)=\sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds +x(0)\sum_{k=1}^{m} a_k. \end{gather*} Let $t=\eta_j$ in equation \eqref{eq5}. We obtain \begin{gather*} x(\eta_j)=\int_0^{\eta_j} y(s)\,ds+x(0),\\ \sum_{j=1}^{p} b_jx(\eta_j) =\sum_{j=1}^{p} b{j}\int_0^{\eta_j} y(s)\,ds +x(0)\sum_{j=1}^{p} b_j. \end{gather*} From \eqref{eq1.2}, we obtain \[ \sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds+x(0)\sum_{k=1}^{m} a_k=\beta \sum_{j=1}^{p} b_j \int_0^{\eta_j} y(s)\,ds+x(0)\beta\sum_{j=1}^{p} b_j. \] Then \begin{gather*} x(0)=A \Big(\sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds -\beta \sum_{j=1}^{p} b_j\int_0^{\eta_j} y(s)\,ds\Big),\\ A=(\beta \sum_{j=1}^{p}b_j-\sum_{k=1}^{m} a_k)^{-1}, \\ \end{gather*} and \begin{equation}\label{eq6} x(t) =A \Big(\sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds -\beta \sum_{j=1}^{p} b_j \int_0^{\eta_j} y(s)\,ds\Big)+ \int_0^{t}y(s)\,ds, \end{equation} which, by Theorem \ref{thm3.1}, has at least one solution $ x \in AC(0,1)$. Now, from equation \eqref{eq6}, we have \[ x(0)=\lim_{t\to 0^{+}}x(t) =A \sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds-A\beta \sum_{j=1}^{p} b_j\int_0^{\eta_j} y(s)\,ds \] and \[ x(1)=\lim_{t\to 1^{-}}x(t)=A \sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds - A\beta \sum_{j=1}^{p} b_j\int_0^{\eta_j} y(s)\,ds + \int_0^{1}y(s)\,ds, \] from which we deduce that \eqref{eq6} has at least one solution $ x \in AC[0,1]$. Next we differentiate \eqref{eq6}, to obtain \begin{gather*} \frac{dx}{dt}=y(t),\\ D^{{\alpha}_{i}}x(m_i(t))=I^{1-{\alpha}_{i}} \frac{d}{dt}x(m_i(t))=I^{1-{\alpha}_{i}}m_i'(t)y(m_i(t)),\\ x'(t)=f(t, D^{{\alpha}_1}x(t), D^{{\alpha}_2}x(t), \dots, D^{{\alpha}_n}x(t)). \end{gather*} By direct calculation, we can prove that \eqref{eq6} satisfies the nonlocal condition \eqref{eq1.2}. This completes the proof. \end{proof} From the above theorem we have the following corollaries. \begin{corollary} \label{coro3.1} Under the assumptions of Theorem \ref{thm3.1}, the solution of \eqref{eq1.1}-\eqref{eq1.2} is nondecreasing. \end{corollary} \begin{proof} Let $ t_1, t_2 \in (0,1) $ and $ t_1 < t_2, $ then from \eqref{eq6} we have \begin{align*} x(t_1) &=A \Big(\sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s) \,ds -\beta \sum_{j=1}^{p} b_j \int_0^{\eta_j} y(s) \,ds\Big) + \int_0^{t_1}y(s) \,ds\\ & \leq A \Big(\sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds -\beta \sum_{j=1}^{p} b_j \int_0^{\eta_j} y(s)\,ds\Big) + \int_0^{t_2}y(s)\,ds\\ &=x(t_2) \end{align*} and the solution of the nonlocal problem \eqref{eq1.1}-\eqref{eq1.2} is nondecreasing. \end{proof} \begin{corollary} \label{coro3.2 } Under the assumptions of Theorem \ref{thm3.1}, problem \eqref{eq1.1} with the nonlocal condition \begin{equation}\label{n1} \sum_{k=1}^{m} a_kx(\tau_k)=0, \quad \tau_k\in (a,c)\subset (0,1). \end{equation} has at least one nondecreasing solution $ x \in AC[0,1] $, represented by \begin{equation}\label{beta=0} x(t)= \int_0^{t}y(s)\,ds - A^* \sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds, \quad A^* =(\sum_{k=1}^{m} a_k)^{-1}. \end{equation} This solution is positive in the interval $[c,1]$. \end{corollary} \begin{proof} Letting $ \beta =0$ in \eqref{eq1.2} and \eqref{eq6}, then from Theorem \ref{thm3.2} we deduce that the nonlocal problem \eqref{eq1.1} and \eqref{n1} has at least one nondecreasing solution given by \eqref{beta=0}. Let $ t\in [c,1]$, then $\tau_k < t $ and \[ A^* \sum_{k=1}^{m} a_k\int_0^{\tau_k} y(s)\,ds \leq A^* \sum_{k=1}^{m} a_k\int_0^{t} y(s)\,ds =\int_0^{t} y(s)\,ds, \] which proves that the solution \eqref{beta=0} is positive in $[c,1]$. \end{proof} \begin{corollary} \label{coro3.3} Under the assumptions of Theorem \ref{thm3.1}, the two point problem \begin{gather*} x'(t)=f(t, D^{{\alpha}_1}x(m_1(t)), D^{{\alpha}_2}x(m_2(t)), \dots, D^{{\alpha}_n}x(m_n(t))), {\alpha}_{i}\in(0, 1),\\ \text{a.e.} t\in(0, 1),\\ x(\tau)=\beta x(\eta), \quad \tau, \eta \in (a,c)\subset (0,1). \end{gather*} has at least one nondecreasing solution $ x \in AC[0,1] $ represented by \begin{equation}\label{eq} x(t) =A (\int_0^{\tau} y(s)\,ds-\beta \int_0^{\eta} y(s)\,ds)+ \int_0^{t}y(s)\,ds, A=(\beta -1)^{-1}. \end{equation} This solution is positive in the interval $ [c,1]$. \end{corollary} \section{Integral condition} Let$ x \in AC[0,1] $ be the solution of the nonlocal problem \eqref{eq1.1}-\eqref{eq1.2}. Let$ a_k=\phi(\tau_k)- \phi(\tau_{k-1})$, $t_k\in (\tau_{k-1}, \tau_k)$, $a=\tau_0 < \tau_1 < \tau_2,\dots < \tau_m=c $ and $ b_j=\psi(\eta_j)-\psi(\eta_{j-1})$, $t_j\in (\eta_{j-1}, \eta_j)$, $d=\eta_0 < \eta_1 < \eta_2,\dots < \eta_{p}=b $ then the nonlocal condition \eqref{eq1.2} will be \[ \sum_{k=1}^{m} (\phi(\tau_k) - \phi(\tau_{k-1})) x(t_k) =\beta \sum_{j=1}^{p} (\psi(\eta_j) - \psi(\eta_{j-1})) x(t_j). \] From the continuity of the solution $ x $ of \eqref{eq1.1}-\eqref{eq1.2} we can obtain \[ \lim_{m\to\infty}\sum_{k=1}^{m} (\phi(\tau_k) - \phi(\tau_{k-1})) x(t_k) = \beta\lim_{p\to\infty}\sum_{j=1}^{p} (\psi(\eta_j) - \psi(\eta_{j-1})) x(t_j). \] and the nonlocal condition \eqref{eq1.2} transformed to the integral condition \begin{equation}\label{eqm} \int_{a}^{c} x(s) d\phi(s)=\beta \int_{d}^{b} x(s) d\psi(s). \end{equation} Also from the continuity of the function $ Iy(t)$, where $ y$ is the solution of \eqref{eq3.1}, we deduce that the solution \eqref{eq6} will be \begin{align*} x(t)&=(\beta (b-d)-(c-a))^{-1} \Big(\int_{a}^{c}\int_0^s y(\theta) \,d\phi(\theta)\,ds -\beta\int_{d}^{b}\int_0^s y(\theta) d\psi(\theta)\,ds \Big)\\ &\quad + \int_0^{t}y(s)\,ds. \end{align*} \begin{theorem} \label{thm4.1} Under the assumptions of Theorem \ref{thm3.2}, there exists at least one nondecreasing solution $ x\in AC[0,1] $ of the nonlocal problem with integral condition, \begin{gather*} x'(t)=f(t, D^{{\alpha}_1}x(m_1(t)), D^{{\alpha}_2}x(m_2(t)), \dots, D^{{\alpha}_n}x(m_n(t))), {\alpha}_{i}\in(0, 1), \\ \text{a.e.} t\in(0, 1),\\ \int_{a}^{c} x(s)\,ds=\beta\int_{d}^{b} y(s)\,ds , \quad \beta (b-d)\neq(c-a). \end{gather*} \end{theorem} Letting $ \beta=0 $ in \eqref{eqm}, the we can easily prove the following corollary. \begin{corollary} \label{coro4.1} Under the assumptions of Theorem \ref{thm3.2}, the nonlocal problem \begin{gather*} x'(t)=f(t, D^{{\alpha}_1}x(m_1(t)), D^{{\alpha}_2}x(m_2(t)), \dots, D^{{\alpha}_n}x(m_n(t))),\\ {\alpha}_{i}\in(0, 1),\; \text{a.e.} t\in(0, 1),\\ \int_a^c x(s)\,ds=0, \quad (a,c) \subset (0,1) \end{gather*} has at least one nondecreasing solution $ x\in AC[0,1] $ represented by \[ x(t)=\int_0^{t}y(s)\,ds-(c-a)^{-1} \int_a^c\int_0^s y(\theta) d\theta\,ds. \] This solution is positive in the interval $ [c,1]$. \end{corollary} \section{Equations with deviated arguments} As a first example, let $ m_i(t)=\beta_i t$, $\beta_i \in (0,1)$, then $ m_i: [0,1]\to [0,1] $ is absolutely continuous deviated functions and all our results here can be applied to the multi-term fractional-order functional differential equation with deviated arguments \begin{equation} x'(t)=f(t, D^{{\alpha}_1}x(\beta_1 t), D^{{\alpha}_2}x(\beta_2 t), \dots, D^{{\alpha}_n}x(\beta_n t)), {\alpha}_{i}\in(0, 1), \quad \text{a.e. } t\in(0, 1). \end{equation} As a second example, let $ m_i(t)= t^{\gamma_i}$, $ \gamma_i \geq 1$, then $ m_i: [0,1]\to [0,1] $ is absolutely continuous deviated functions and all our results here can be applied to the multi-term fractional-order functional differential equation with deviated arguments \begin{equation} x'(t)=f(t, D^{{\alpha}_1}x(t^{\gamma_1}), D^{{\alpha}_2} x(t^{\gamma_2}), \dots, D^{{\alpha}_n}x(t^{\gamma_n})), {\alpha}_{i}\in(0, 1), \quad \text{a.e. } t\in(0, 1). \end{equation} \begin{thebibliography}{00} \bibitem{B1} Boucherif, A.; \emph{First-order differential inclusions with nonlocal initial conditions}, Applied Mathematics Letters 15 (2002), 409-414. \bibitem{B2} Boucherif, A.; \emph{Nonlocal Cauchy problems for first-order multivalued differential equations}, Electronic Journal of Differential Equations, Vol. 2002 (2002), No. 47, pp. 1-9. \bibitem{B3} Boucherif, A; Precup, R.; \emph{On The nonlocal initial value problem for first order differential equations}, Fixed Point Theory Vol. 4, No 2, (2003) 205-212. \bibitem{B4} Boucherif, A.; \emph{Semilinear evolution inclusions with nonlocal conditions}, Applied Mathematics Letters 22 (2009), 1145-1149. \bibitem{b} Benchohra, M.; Gatsori, E. 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