\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 64, 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 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/64\hfil Positive solutions] {Positive solutions for a nonlocal multi-point boundary-value problem of fractional and second order} \author[A. M. A. El-Sayed, E. O. Bin-Taher \hfil EJDE-2013/64\hfilneg] {Ahmed M. A. El-Sayed, Ebtisam O. Bin-Taher} % in alphabetical order \address{Ahmed M. A. El-Sayed \newline Faculty of Science, Alexandria University, Alexandria, Egypt} \email{amasayed@hotmail.com} \address{Ebtisam O. Bin-Taher \newline Faculty of Science, Hadhramout Univeristy of Science and Technology, Hadhramout, Yemen} \email{ebtsamsam@yahoo.com} \thanks{Submitted March 19, 2012. Published March 5, 2013.} \subjclass[2000]{34B10, 26A33} \keywords{Fractional calculus; boundary value problem; nonlocal condition; \hfill\break\indent integral condition; positive solution} \begin{abstract} In this article we study the existence of positive solutions for the nonlocal multi-point boundary-value problem \begin{gather*} u''(t)+f(t, ^{c}D^{\alpha}u(t))=0, \quad \alpha \in(0, 1), \text{ a.e. } t\in(0, 1), \\ u(0)=0, \quad u(1)=\sum_{k=1}^m a_k u(\tau_k), \quad \tau_k\in(a, b)\subset (0, 1). \end{gather*} We also consider the corresponding integral condition, and the two special cases $\alpha = 0$ and $\alpha = 1$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Problems with non-local conditions have been extensively studied by several authors in the previous two decades; see for example \cite{B0}-\cite{B2}, \cite{ee}-\cite{sk}, and the references therein. In this work we show the existence of at least one solution for the nonlocal multi-point boundary-value problem consisting of second and fractional-orders differential equation $$\label{e1.1} u''(t)+f(t, ^{c}D^{\alpha}u(t))=0, \quad \alpha \in(0, 1), \text{ a.e. } t\in(0, 1)$$ with the nonlocal conditions $$\label{e1.2} u(0)=0, \quad u(1)=\sum_{k=1}^m a_k u(\tau_k), \quad \tau_k\in(a, b)\subset (0, 1).$$ Also we deduce the same results for the two differential equations $$\label{*} u''(t)+f(t, u(t))=0, \quad \text{a.e. } t\in(0, 1)$$ and $$\label{**}% u''(t)+f(t, u'(t))=0, \quad \text{a.e. } t\in(0, 1).$$ with the nonlocal conditions \eqref{e1.2}. Also we study problems \eqref{e1.1}, \eqref{*} and \eqref{**} with an integral condition. \section{Preliminaries} Let $L^1(I)$ denote the class of Lebesgue integrable functions on the interval $I = [a,b]$,where $0\leq a 0$ is defined by $I_a^\beta f(t) = \int_a^t \frac{(t - s)^{\beta - 1}}{\Gamma(\beta)} f(s) ds,$ \end{definition} \begin{definition}[\cite{P1,P2}]\label{def1.1.2} \rm The Caputo fractional-order derivative of order $\alpha \in (0,1]$ of the absolutely continuous function $f(t)$ is defined by $D_a^\alpha f(t)=I_a^{1 - \alpha} \frac{d}{dt} f(t) =\int_a^t \frac{(t - s)^{-\alpha}}{\Gamma(1 - \alpha)} \frac{d}{ds} f(s) ds.$ \end{definition} \begin{theorem}[Schauder fixed point theorem \cite{KD}]\label{thm13} Let $E$ be a Banach space and $Q$ be a convex subset of $E$, and $T:Q\to Q$ a compact, continuous map. Then $T$ has at least one fixed point in $Q$. \end{theorem} \begin{theorem}[Kolmogorov compactness criterion \cite{JD}]\label{thm17} 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 $u \in \Omega$, 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{itemize} \end{theorem} \section{Main results} Consider the fractional-order functional integral equation \label{equa3.1} \begin{aligned} y(t)&=f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds-A\sum_{k=1}^ma_k \int_0^{\tau_k}(\tau_k-s)y(s) ds\big\}\\ &\quad -\int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}y(s) ds\Big). \end{aligned} The function $y$ is called a solution of the fractional-order functional integral equation \eqref{equa3.1}, if $y$ belongs to $L^{1}[0,1]$ and satisfies \eqref{equa3.1}. We consider the following assumptions: \begin{itemize} \item[(H1)] $f:[0,1]\times R\to R^{+}$ be a function with the following properties: \begin{itemize} \item[(a)] $u\to f(t,u)$ is continuous for almost all $t \in [0,1]$, \item[(b)] $t\to f(t,u)$ is measurable for all $u\in R$, \end{itemize} \item[(H2)]there exists an integrable function $a\in L^1[0,1]$ and constant $b,$ such that $|f(t,u)| \leq a(t)+b |u|, \textrm{a.e} t\in[0,1],$ \end{itemize} \begin{theorem} \label{thm3.1} % Theorem 3.1 Let the {\rm (H1), (H2)} be satisfied. If $$B=\frac{b}{\Gamma(3-\alpha)} < 1,$$ then \eqref{equa3.1} has at least one positive solution $y\in L^1[0,1]$. \end{theorem} \begin{proof} Define the operator $T$ associated with \eqref{equa3.1} by \begin{align*} Ty(t)&=f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\} \\ &\quad - \int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}y(s) ds\Big) \end{align*} Let $Q^{+}_{r} = \{y\in R^{+}:\|y\|0\}$, $r=\frac{\|a\|}{1 - B(1+A+A\sum_{k=1}^m a_k)}.$ Let $y$ be an arbitrary element in $Q^{+}_{r}$, then from assumptions (H1) and (H2), we obtain \begin{align*} \|Ty\|_{L^1} &=\int_0^1 |Ty(t)| dt\\ &=\int_0^1 \Big|f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\} \\ &\quad - I^{2-{\alpha}}y(t)\Big)\Big|dt \\ &\leq \int_0^1|a(t)|dt+b \int_0^{1} \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}\,dt \Big\{A\int_0^{1}(1-s) |y(s)| ds\\ &\quad +A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) |y(s)| ds\Big\} +b\int_0^1\int_0^t\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}|y(s)| \,ds\,dt \\ &=\|a\|_{L^1}+b \int_0^{1} \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}\,dt \Big\{A\int_0^{1} |y(s)| ds+A\sum_{k=1}^m a_k\int_0^{1} |y(s)| ds\Big\} \\ &\quad+ b\int_0^1\int_s^1\frac{(t-s)^{1-\alpha}}{\Gamma(2-\alpha)}dt|y(s)| ds \\ &\leq \|a\|_{L^1}+b (A+A\sum_{k=1}^m a_k)\|y\|_{L^1} \int_0^{1} \frac{t^{1-\alpha}}{\Gamma(2-\alpha)}dt\\ &\quad +b\int_0^1\frac{(t-s)^{2-\alpha}}{(2-\alpha) \Gamma(2-\alpha)}|_{s}^{1}|y(s)| \,ds \\ &\leq \|a\|_{L^1}+\frac{b (A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)} \|y\|_{L^1}+b\int_0^1\frac{1}{\Gamma(3-\alpha)}|y(s)| \,ds\\ &\leq \|a\|_{L^1}+\frac{b (A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)} \|y\|_{L^1}+\frac{b}{\Gamma(3-\alpha)}\|y\|_{L^1}\\ &\leq \|a\|_{L^1}+\frac{b (1+A+A\sum_{k=1}^m a_k)}{\Gamma(3-\alpha)} \|y\|_{L^1}= r\,, \end{align*} which implies that the operator $T$ maps $Q^{+}_{r}$ into it self. Assumption (H1) implies that $T$ is continuous. Now let $\Omega$ be a bounded subset of $Q^{+}_{r}$, then $T(\Omega)$ is bounded in $L^1[0,1]$; i.e., condition (i) of Theorem \ref{thm17} is satisfied. Let $y\in \Omega$. Then \begin{align*} \|(Ty)_h-Ty\| &=\int_0^1|(Ty)_h(t)-(Ty)(t)| dt\\\\ &= \int_0^1 |\frac{1}{h} \int_t^{t+h} (Ty)(s) ds - (Ty)(t)| dt\\\\ &\leq \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\Big(s,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s)y(s) ds \\ &\quad -A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s)y(s) ds\big\} -I^{2-{\alpha}}y(t)\Big)\\ &\quad - f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds\\ &\quad -A\sum_{k=1}^m a_k \int_0^{\tau_k}(\tau_k-s) y(s) ds\Big\} -I^{2-{\alpha}}y(t)\Big)\Big| \,ds\,dt. \end{align*} then assumption (H2) implies that $f\in L^1(0,1)$ and \begin{align*} &\frac{1}{h}\int_t^{t+h} \Big|f\Big(s,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\}\\ &- I^{2-{\alpha}}y(t)\Big) - f\Big(t,\frac{t^{1-\alpha}}{\Gamma(2-\alpha)} \big\{A\int_0^{1}(1-s) y(s) ds - A\sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\big\}\\ &- I^{2-{\alpha}}y(t)\Big)\Big| \,ds \to 0 \end{align*} Therefore, by Theorem \ref{thm17}, we have that $T(\Omega)$ is relatively compact; that is, $T$ is a compact operator. Then the operator $T$ has a fixed point $Q^{+}_{r}$, which proves the existence of positive solution $y\in L^{1}[0,1]$ for \eqref{equa3.1}. \end{proof} For the existence of solutions to the nonlocal problem \eqref{e1.1}--\eqref{e1.2}, we have the following theorem. \begin{theorem} \label{thm3.2} Under the assumptions of Theorem \ref{thm3.1}, if $0<\sum_{k=1}^m a_k \tau_k<1$, then nonlocal problem \eqref{e1.1}--\eqref{e1.2} has at least one positive solution $u\in C[0,1]$, with $u' \in AC[0,1]$. \end{theorem} \begin{proof} For the problem \eqref{e1.1}-\eqref{e1.2}, let $- y(t) = u''(t)$. Then $$\label{equa5} u(t) = tu'(0) - I^{2}y(t),$$ where $y$ is the solution of the fractional-order functional integral equation \eqref{equa3.1}. Letting $t=\tau_k$ in \eqref{equa5}, we obtain $u(\tau_k)=- \int_0^{\tau_k}(\tau_k-s) y(s) ds+\tau_k u'(0)$ and $$\label{e*} \sum_{k=1}^m a_ku(\tau_k) =- \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds + u'(0)\sum_{k=1}^m a_k \tau_k$$ From equation \eqref{e1.2} and \eqref{e*}, we obtain $- \int_0^{1}(1-s) y(s) ds+u'(0) =- \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds +u'(0)\sum_{k=1}^m a_k \tau_k\,.$ Then \begin{gather*} u'(0) =A \Big(\int_0^{1}(1-s) y(s) ds - \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds\Big), \\ A=(1 - \sum_{k=1}^m a_k \tau_k)^{-1}. \end{gather*} Then \label{e6} \begin{aligned} u(t) &=A t\int_0^{1}(1-s) y(s) ds - A t \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds - \int_0^{t}(t-s)y(s) ds \\ &=t\Big\{\int_0^{1}(1-s) y(s) ds+A \sum_{k=1}^m a_k \tau_k\int_0^{1}(1-s)y(s) ds\\ &\quad - A\sum_{k=1}^m a_k \int_0^{\tau_k}(\tau_k-s)y(s) ds\Big\} - \int_0^{t}(t-s)y(s) ds. \end{aligned} where $y$ is the solution of the fractional-order functional integral equation \eqref{equa3.1}. Hence, by Theorem \ref{thm3.1}, Equation \eqref{e6} has at least one solution $u \in C(0,1)$. Now, from equation \eqref{e6}, we have \begin{gather*} u(0)=\lim_{t\to 0^{+}}u(t)=0,\\ \begin{aligned} u(1)&=\lim_{t\to 1^{-}}u(t)\\ &=A \sum_{k=1}^m a_k\int_0^{\tau_k}(\tau_k-s) y(s) ds - A\int_0^{1}(1-s) y(s) ds- \int_0^{1}(1-s)y(s) ds \end{aligned} \end{gather*} from which we deduce that \eqref{e6} has at least one positive solution $u \in C[0,1]$. Now, \begin{align*} \sum_{k=1}^m a_k \int_0^{\tau_k}(\tau_k-s)y(s) ds &<\sum_{k=1}^m a_k\tau_k \int_0^{1}(1-\frac{s}{\tau_k})y(s) ds\\ &<\sum_{k=1}^m a_k\tau_k \int_0^{1}(1-s)y(s) ds \end{align*} and \[ \int_0^{t}(t-s)y(s) ds The authors were informed, but they have not sent any new corrections. \medskip End of addendum. \end{document}