\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 70, 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 2012 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2012/70\hfil Uniqueness of positive solutions] {Uniqueness of positive solutions for a fractional differential equation via a fixed point theorem of a sum operator} \author[C. Yang, C. Zhai\hfil EJDE-2012/70\hfilneg] {Chen Yang, Chengbo Zhai} % in alphabetical order \address{Chen Yang \newline Department of Mathematics, Business College of Shanxi University, Taiyuan 030031 Shanxi, China} \email{yangchen0809@126.com} \address{Chengbo Zhai \newline School of Mathematical Sciences, Shanxi University, Taiyuan 030006 Shanxi, China} \email{cbzhai@sxu.edu.cn, cbzhai215@sohu.com} \thanks{Submitted January 27, 2012. Published May 7, 2012.} \thanks{Supported by grant 2010021002-1 from the Youth Science Foundation of Shanxi Province} \subjclass[2000]{34B18} \keywords{Riemann-Liouville fractional derivative; positive solution; \hfill\break\indent fractional differential equation; existence and uniqueness; fixed point theorem} \begin{abstract} In this work, we study the existence and uniqueness of positive solutions for nonlinear fractional differential equation boundary-value problems. Our analysis relies on a fixed point theorem of a sum operator. Our results guarantee the existence of a unique positive solution, and can be applied for constructing an iterative scheme for obtaining the solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, engineering and biological sciences,etc; see \cite{d1,g1,g2,k2,m1,m2,m3,o1,p1,r1} for example. In the recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives, see the monographs of Miller and Ross \cite{m3}, Podlubny \cite{p1}, Kilbas et al \cite{k2}, and the articles \cite{b1,b2,j1,k1,k3,l1,l2,x1,y1,w1,z2,z3,z4,z5} and the references therein. In these papers, many authors have investigated the existence of positive solutions for nonlinear fractional differential equation boundary value problems. On the other hand, the uniqueness of positive solutions for nonlinear fractional differential equation boundary value problems has been studied by some authors, see \cite{y1,z3,z4,z5} for example. By means of a fixed point theorem for mixed monotone operators, Xu, Jiang and Yuan \cite{x1} considered the existence and the uniqueness of positive solutions for the following \begin{gathered} D^\alpha_{0+} u(t)=f(t,u(t)),\quad 00 $$is called the Riemann-Liouville fractional integral of order \alpha, where \alpha>0 and \Gamma(\alpha) denotes the gamma function. \end{definition} \begin{definition}[{\cite[page 36-37]{s1}}] \label{def2.2} \rm For a function f(x) given in the interval [0,\infty), the expression$$ D^\alpha_{0+}f(x)=\frac 1{\Gamma(n-\alpha)} \big(\frac d {dx}\big)^n \int^x_0 \frac {f(t)} {(x-t)^{\alpha-n+1}} dt, $$where n=[\alpha]+1,[\alpha] denotes the integer part of number \alpha, is called the Riemann-Liouville fractional derivative of order \alpha. \end{definition} \begin{lemma}[\cite{b1}] \label{lem2.3} Given y \in C[0,1] and 1<\alpha\leq 2, the boundary-value problem \begin{gathered} D^\alpha_{0+} u(t)+y(t)=0,\quad 0x. By \theta we denote the zero element of E. Recall that a non-empty closed convex set P\subset E is a cone if it satisfies (i) x\in P,\lambda\geq 0 implies \lambda x\in P; (ii) x\in P and -x\in P imply x=\theta. Putting \mathring{P}=\{x\in P: x\text{ is an interior point of }P\}, a cone P is said to be solid if \mathring{P} is non-empty. Moreover, P is called normal if there exists a constant N>0 such that, for all x,y\in E, \theta\leq x\leq y implies \|x\|\leq N\|y\|; in this case N is called the normality constant of P. We say that an operator A:E\to E is increasing if x\leq y implies Ax\leq Ay. For all x,y\in E, the notation x\sim y means that there exist \lambda >0  and \mu>0 such that \lambda x\leq y\leq \mu x. Clearly, \sim is an equivalence relation. Given h>\theta (i.e., h\geq \theta and h\neq \theta), we denote by P_h the set P_h=\{x\in E|\ x\sim h\}. It is easy to see that P_h\subset P. \begin{definition} \label{def2.5} \rm Let D=P or D=\mathring{P} and \beta be a real number with 0\leq \beta<1. An operator A:P\to P is said to be \beta-concave if it satisfies$$ A(tx)\geq t^\beta Ax, \forall\ t\in (0,1),\ x\in D. $$Notice that the definition of a \beta-concave operator mentioned above is different from that in \cite{g3}, because we need not require the cone to be solid in general. \end{definition} \begin{definition} \label{def2.6}\rm An operator A:E\to E is said to be homogeneous if it satisfies$$ A(\lambda x)=\lambda Ax, \quad \forall \lambda>0,\; x\in E. $$An operator A:P\to P is said to be sub-homogeneous if it satisfies$$ A(tx)\geq t Ax, \forall\ t\in (0,1),\ x\in P. $$\end{definition} In a recent paper Zhai and Anderson \cite{z1} considered the sum operator equation$$ Ax+Bx+Cx=x, $$where A is an increasing \beta-concave operator, B is an increasing sub-homogeneous operator and C is a homogeneous operator. They established the existence and uniqueness of positive solutions for the above equation, and when C is a null operator, they present the following interesting result. \begin{lemma} \label{lem2.7} Let P be a normal cone in a real Banach space E, A:P\to P be an increasing \beta-concave operator and B:P\to P be an increasing sub-homogeneous operator. Assume that \begin{itemize} \item[(i)] there is h>\theta such that Ah\in P_h and Bh\in P_h; \item[(ii)] there exists a constant \delta_0>0 such that Ax\geq \delta_0Bx, for all x\in P. \end{itemize} Then the operator equation Ax+Bx=x has a unique solution x^* in P_h. Moreover, constructing successively the sequence y_n=Ay_{n-1}+By_{n-1}, n=1,2,\dots for any initial value y_0\in P_h, we have y_n\to x^* as n\to \infty. \end{lemma} \section{Existence and uniqueness of positive solutions for \eqref{e1.4}} In this section, we apply Lemma \ref{lem2.7} to study problem \eqref{e1.4} and we obtain a new result on the existence and uniqueness of positive solutions. The method used here is new to the literature and so is the existence and uniqueness result to the fractional differential equations. In our considerations we will work in the Banach space C[0,1]=\{x: [0,1]\to \mathbf{R} \text{ is continuous}\} with the standard norm \|x\| = \sup\{|x(t)|:t\in [0,1]\}. Note that this space can be equipped with a partial order given by$$ x,y \in C[0,1], x \leq y \Leftrightarrow x(t)\leq y(t) \text{ for } t\in [0,1]. $$Set  P=\{x\in C[0,1]| x(t)\geq 0,\; t\in [0,1]\}, the standard cone. It is clear that P is a normal cone in C[0,1] and the normality constant is 1. Our main result is summarized in the following theorem using the following assumptions: \begin{itemize} \item[(H1)] f,g:[0,1]\times [0,+\infty)\to [0,+\infty) are continuous and increasing respect to the second argument, g(t,0)\not\equiv 0; \item[(H2)] g(t,\lambda x)\geq \lambda g(t,x) for \lambda\in (0,1), t\in [0,1], x\in [0,+\infty), and there exists a constant \beta\in (0,1) such that f(t,\lambda x)\geq \lambda^\beta f(t,x), for all t\in [0,1], \lambda\in (0,1), x\in [0,+\infty); \item[(H3)] there exists a constant \delta_0>0 such that f(t,x)\geq \delta_0g(t,x), t\in [0,1], x\geq 0. \end{itemize} \begin{theorem} \label{thm3.1} Under assumptions {\rm (H1)--(H3)}, problem \eqref{e1.4} has a unique positive solution u^* in P_h, where h(t)=t^{\alpha-1}(1-t), t\in [0,1]. Moreover, for any initial value u_0\in P_h, the sequence$$ u_{n+1}(t)=\int^1_0 G(t,s)f(s,u_n(s))ds+\int^1_0 G(t,s)g(s,u_n(s))ds,\quad n=0,1,2,\dots $$satisfies u_n(t)\to u^*(t) as n\to \infty. \end{theorem} \begin{proof} From \cite{b1}, problem \eqref{e1.4} has the integral formulation$$ u(t)=\int_0^1G(t,s)[f(s,u(s))+g(s,u(s))]ds, $$where G(t,s) is given as in Lemma \ref{lem2.3}. Define two operators A:P\to E and  B:P\to E by$$ Au(t)=\int_0^1G(t,s)f(s,u(s)))ds,\quad Bu(t)=\int_0^1G(t,s)g(s,u(s))ds. $$It is easy to prove that u is the solution of \eqref{e1.4} if and only if u=Au+Bu. From (H1) and Lemma \ref{lem2.4}, we know that A:P\to P and B:P\to P. In the sequel we check that A,B satisfy all assumptions of Lemma \ref{lem2.7}. Firstly, we prove that A,B are two increasing operators. In fact, by (H1) and Lemma \ref{lem2.4}, for u,v\in P with u\geq v, we know that u(t)\geq v(t), t\in [0,1] and obtain$$ Au(t)=\int_0^1G(t,s)f(s,u(s))ds\geq \int_0^1G(t,s)f(s,v(s))ds=Av(t). $$That is, Au\geq Av. Similarly, Bu\geq Bv. Next we show that A is a \beta-concave operator and B is a sub-homogeneous operator. In fact, for any \lambda \in (0,1) and u\in P, by (H_2) we obtain $A(\lambda u)(t)=\int^1_0G(t,s)f(s,\lambda u(s)) ds \geq \lambda^\beta\int^1_0G(t,s)f(s,u(s)) ds=\lambda^\beta Au(t).$ That is, A(\lambda u)\geq \lambda^\beta Au  for  \lambda \in (0,1), u\in P. So the operator A is a \beta-concave operator. Also, for any \lambda \in (0,1) and u\in P, from (H2) we know that$$ B(\lambda u)(t)=\int_0^1G(t,s)g(s,\lambda u(s))ds\geq \lambda\int_0^1G(t,s)g(s,u(s))ds=\lambda Bu(t); $$that is, B(\lambda u)\geq \lambda Bu for  \lambda \in (0,1),\ u\in P. So the operator B is sub-homogeneous. Now we show that Ah\in P_h and Bh\in P_h. Let h_{\rm max}=\max\{h(t)=t^{\alpha-1}(1-t):t\in [0,1]\}. Then h_{\rm max}>0. From (H1) and Lemma \ref{lem2.4}, \begin{gather*} Ah(t)=\int_0^1G(t,s)f(s,h(s))ds\leq\frac 1{\Gamma(\alpha)}h(t)\int_0^1(1-s)^{\alpha-2}f(s,h_{\rm max})ds , \\ Ah(t)=\int_0^1G(t,s)f(s,h(s))ds\geq \frac {\alpha-1}{\Gamma(\alpha)}h(t)\int_0^1s(1-s)^{\alpha-1}f(s,0)ds. \end{gather*} From (H1) and (H3), we have$$ f(s,h_{\rm max})\geq f(s,0)\geq \delta_0g(s,0)\geq 0. $$Since g(t,0)\not\equiv 0, we can obtain$$ \int_0^1f(s,h_{\rm max})ds\geq \int^1_0f(s,0)ds\geq \delta_0\int^1_0g(s,0)ds> 0, $$and in consequence, \begin{gather*} l_1:=\frac {\alpha-1}{\Gamma(\alpha)}\int_0^1s(1-s)^{\alpha-1}f(s,0)ds>0,\\ l_2:=\frac 1{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-2}f(s,h_{\rm max})ds>0. \end{gather*} So l_1h(t)\leq Ah(t)\leq l_2h(t), t\in [0,1]; and hence we have Ah\in P_h. Similarly,$$ \frac{\alpha-1}{\Gamma(\alpha)}h(t)\int_0^1s(1-s)^{\alpha-1}g(s,0)ds\leq Bh(t)\leq\frac 1{\Gamma(\alpha)}h(t)\int_0^1(1-s)^{\alpha-2}g(s,h_{\rm max})ds, $$from g(t,0)\not\equiv 0, we easily prove Bh\in P_h. Hence the condition (i) of Lemma \ref{lem2.7} is satisfied. In the following we show the condition (ii) of Lemma \ref{lem2.7} is satisfied. For u\in P, from (H3),$$ Au(t)=\int_0^1G(t,s)f(s,u(s))ds\geq \delta_0\int^1_0G(t,s)g(s,u(s))ds =\delta_0Bu(t). $$Then we get Au\geq \delta_0Bu, u\in P. Finally, an application of Lemma \ref{lem2.7} implies: the operator equation Ax+Bx=x has a unique solution u^* in P_h. Moreover, constructing successively the sequence y_n=Ay_{n-1}+By_{n-1}, n=1,2,\dots for any initial value y_0\in P_h, we have y_n\to u^* as n\to \infty. That is, problem \eqref{e1.4} has a unique positive solution u^* in P_h. Moreover, for any initial value u_0\in P_h, constructing successively the sequence$$ u_{n+1}(t)=\int^1_0 G(t,s)f(s,u_n(s))ds +\int^1_0 G(t,s)g(s,u_n(s))ds,\quad n=0,1,2,\dots,  we have $u_n(t)\to u^*(t)$ as $n\to \infty$. \end{proof} \begin{remark} \label{rmk3.1} \rm A simple example that illustrates Theorem \ref{thm3.1} is as follows: let $f(t,x)\equiv 2\delta$, $g(t,x)\equiv 1, \delta>0$. Then the conditions (H1)--(H3) are satisfied and \eqref{e1.4} has a unique solution $u(t)=(2\delta+1)\int^1_0G(t,s)ds$, $t\in [0,1]$. Evidently, \begin{gather*} u(t)\geq \frac{(2\delta+1)(\alpha-1)}{\Gamma(\alpha)} \int^1_0s(1-s)^{\alpha-1}ds\cdot h(t) =\frac{(2\delta+1)(\alpha-1)}{\alpha(\alpha+1)\Gamma(\alpha)}\cdot h(t), \\ u(t)\leq \frac {2\delta+1}{\Gamma(\alpha)} \int^1_0(1-s)^{\alpha-2}ds\cdot h(t)=\frac {2\delta+1}{(\alpha-1)\Gamma(\alpha)}\cdot h(t), \quad t\in [0,1]. \end{gather*} So the unique solution $u$ is a positive solution and satisfies $u\in P_h=P_{t^{\alpha-1}(1-t)}$. \end{remark} \begin{example} \label{examp3.1} \rm \begin{gathered} D^{\frac 32}_{0+}u(t)+ u^{1/2}(t)+\frac {u(t)}{1+u(t)}q(t)+t^2+a=0,\quad 00$is a constant,$q:[0,1]\to [0,+\infty)$is continuous with$q\not\equiv 0$. In this example, we have$\alpha=3/2$. Take$00$;$ f,g:[0,1]\times [0,+\infty)\to [0,+\infty)$are continuous and increasing respect to the second argument,$g(t,0)=a-b> 0$. Besides, for$\lambda\in (0,1)$,$t\in [0,1]$,$x\in [0,+\infty)$, we have \begin{gather*} g(t,\lambda x)=\frac {\lambda x}{1+\lambda x}q(t)+a-b\geq \frac {\lambda x}{1+ x}q(t)+\lambda(a-b)=\lambda g(t,x),\\ f(t,\lambda x)=\lambda^{1/2}x^{1/2}+t^2+b\geq \lambda^{1/2}(x^{1/2}+t^2+b)= \lambda^\beta f(t,x). \end{gather*} Moreover, if we take$\delta_0\in (0,\frac b{q_{\rm max}+a-b}], then we obtain \begin{align*} f(t,x)&=x^{1/2}+t^2+b\geq b=\frac b{q_{\rm max}+a-b} (q_{\rm max}+a-b)\\ &\geq \delta_0 [\frac x{1+x}q(t)+a-b] =\delta_0 g(t,x). \end{align*} Hence all the conditions of Theorem \ref{thm3.1} are satisfied. This implies that \eqref{e3.1} has a unique positive solution inP_h$, where$h(t)=t^{\alpha-1}(1-t)$,$t\in [0,1]\$. \end{example} \begin{thebibliography}{00} \bibitem{b1} Z.B. Bai, H. S. Li; Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311 (2005) 495-505. \bibitem{b2} C. Z. 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