\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 25, pp. 1--13.\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-2013/25\hfil Fractional differential equations] {Fractional differential equations with fractional non-separated boundary conditions} \author[X. Liu, Y. Liu \hfil EJDE-2013/25\hfilneg] {Xiaoyou Liu, Yiliang Liu} \address{Xiaoyou Liu \newline College of Sciences, Guangxi University for Nationalities, Nanning 530006, Guangxi Province, China. \newline School of Mathematical Science and Computing Technology, Central South University, Changsha 410075, Hunan Province, China} \email{liuxiaoyou2002@hotmail.com} \address{Yiliang Liu \newline College of Sciences, Guangxi University for Nationalities, Nanning 530006, Guangxi Province, China} \email{yiliangliu100@126.com} \thanks{Submitted September 28, 2012. Published January 27, 2013.} \subjclass[2000]{34A08, 34B15, 34B10} \keywords{Fractional differential equation; non-separated boundary condition; \hfill\break\indent integral boundary condition, existence of solutions} \begin{abstract} We study boundary-value problems of nonlinear fractional differential equations with fractional non-separated (integral) boundary conditions. Existence and uniqueness results are obtained by using fixed point theorems and examples are given to illustrate the results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} The study of fractional differential equations ranges from the theoretical aspects of existence and uniqueness of solutions to the analytic and numerical methods for finding solutions. A strong motivation for studying fractional differential equations comes from the fact that they have been proved to be valuable tools in the modeling of many phenomena in engineering and sciences such as physics, mechanics, chemistry, economics and biology, etc. \cite{BaleanuMachado1,LaksLeela1,SabatierAgrawal1}. For some recent developments on the existence results of fractional differential equations, we can refer to, for instance, \cite{agarwal1,agarwal2,Ahmad2,bai-zhanbing,baleanu1, Cernea1,Cernea2,changyongkui1,ChenAnping1,ChenTang1,Laks1,liuzhenhai1, liuzhenhai2,CFLi1,lvlinli1,Nieto1,guotaowang1, wangjinrong1,wangjinrong2,zhouyong} and the references therein. Ahmad and Nieto \cite{AhmadNieto1} investigated the existence and uniqueness of solutions for an anti-periodic fractional boundary-value problem \label{limitedcase} \begin{gathered} ^cD^qx(t)=f(t,x(t)),\quad t\in[0,T],\; T>0,\; 10,\; 10,\\ a_1x(0)+b_1x(T)=c_1,a_2(^cD^{\gamma}x(0))+b_2(^cD^{\gamma}x(T))=c_2,0<\gamma<1, \end{gathered} where $^cD^q$ denotes the Caputo fractional derivative of order $q$, $f$ is a continuous function on $[0,T]\times\mathbb{R}$ and $a_i,b_i,c_i$, $i=1,2$ are real constants such that $a_1+b_1\neq0$ and $b_2\neq0$. Then the results obtained for problem \eqref{mainequation} in this paper are extended to fractional differential equations with fractional non-separated integral boundary conditions of the form $$\label{mainequintegralform} \begin{gathered} ^cD^{\alpha}x(t)=f(t,x(t)),\quad t\in[0,T],\; T>0,\; 1<\alpha\leq2,\\ a_1x(0)+b_1x(T)=c_1\int_0^Tg(s,x(s))ds,\\ a_2(^cD^{\gamma}x(0))+b_2(^cD^{\gamma}x(T))=c_2\int_0^Th(s,x(s))ds,\quad 0<\gamma<1, \end{gathered}$$ where $g,h:[0,T]\times\mathbb{R}\to\mathbb{R}$ are given continuous functions. We remark that when $a_1=1$, $b_1=1$, $c_1=0$, $a_2=1$, $b_2=1$ and $c_2=0$, the problem \eqref{mainequation} reduces to an anti-periodic fractional boundary value problem \eqref{limitedcase} (cf.\cite{liuzhenhai3}). The paper is organized as follows: in Section \ref{preliminaries} we recall some preliminary facts that we need in the sequel, Sections \ref{existresultsfordiyiwenti}, \ref{existresuforintegralbound} are dedicated to the existence results of the problem \eqref{mainequation}, respectively, the problem \eqref{mainequintegralform}, in the final Section \ref{examples}, two examples are given to illustrate the results. \section{Preliminaries} \label{preliminaries} \begin{definition}[\cite{kilbasSriTru}] The Riemann-Liouville fractional integral of order $q$ for a function $f$ is defined as \begin{equation*} I^qf(t)=\frac{1}{\Gamma(q)}\int_0^t\frac{f(s)}{(t-s)^{1-q}}ds,\quad q>0, \end{equation*} provided the integral exists. \end{definition} \begin{definition}[\cite{kilbasSriTru}] For a continuous function $f$, the Caputo derivative of order $q$ is defined as \begin{equation*} ^cD^qf(t)=\frac{1}{\Gamma(n-q)}\int_0^t(t-s)^{n-q-1}f^{(n)}(s)ds,\quad n-10$, then the differential equation $$^cD^{\alpha}h(t)=0$$ has solutions$h(t)=c_0+c_1t+c_2t^2+\cdots+c_{n-1}t^{n-1}$and $$I^{\alpha}\,^cD^{\alpha}h(t)=h(t)+c_0+c_1t+c_2t^2+\cdots+c_{n-1}t^{n-1},$$ here$c_i\in\mathbb{R}$,$i=0,1,2,\cdots,n-1$,$n=[\alpha]+1$. \end{lemma} \begin{lemma}\label{linearproblem} For any$y\in C([0,T],\mathbb{R}), the unique solution of the fractional non-separated boundary-value problem $$\label{zhangjianfanchen} \begin{gathered} ^cD^{\alpha}x(t)=y(t),\quad t\in[0,T],\; 1<\alpha\leq2,\\ a_1x(0)+b_1x(T)=c_1,a_2(^cD^{\gamma}x(0))+b_2(^cD^{\gamma}x(T))=c_2,\quad 0<\gamma<1, \end{gathered}$$ is given by \label{integralformofsolution} \begin{aligned} x(t)&=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds -\frac{t\Gamma(2-\gamma)}{T^{1-\gamma}} \int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds \\ &\quad +\frac{t\Gamma(2-\gamma)c_2}{T^{1-\gamma}b_2}-\frac{b_1}{a_1+b_1} \Big\{\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds \\ &\quad -T^{\gamma}\Gamma(2-\gamma)\int_0^T \frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds \Big\} \\ &\quad -\frac{1}{a_1+b_1} \Big(\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{b_2}-c_1\Big). \end{aligned} \end{lemma} \begin{proof} For1<\alpha\leq2$, by Lemma \ref{lemforlemexistence}, we know that the general solution of the equation$^cD^{\alpha}x(t)=y(t)$can be written as $$\label{lemoflinear1} x(t)=I^{\alpha}y(t)-k_1-k_2t=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds-k_1-k_2t,$$ where$k_1,k_2\in\mathbb{R}$are arbitrary constants. Since$^cD^{\gamma}k=0$(k is a constant),$^cD^{\gamma}t=\frac{t^{1-\gamma}}{\Gamma(2-\gamma)}$,$^cD^{\gamma}I^{\alpha}y(t)=I^{\alpha-\gamma}y(t)(see \cite{kilbasSriTru}), from \eqref{lemoflinear1} we have \begin{equation*} ^cD^{\gamma}x(t)=I^{\alpha-\gamma}y(t)-\frac{k_2t^{1-\gamma}}{\Gamma(2-\gamma)} =\int_0^t\frac{(t-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds -\frac{k_2t^{1-\gamma}}{\Gamma(2-\gamma)}. \end{equation*} Using the boundary conditions, we obtain \begin{gather*} a_1(-k_1)+b_1\Big(\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds-k_1-k_2T\Big)=c_1, \\ a_2\times0+b_2\Big(\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds -\frac{k_2T^{1-\gamma}}{\Gamma(2-\gamma)}\Big)=c_2. \end{gather*} Therefore, we have \begin{gather*} \begin{aligned} k_1&=\frac{1}{a_1+b_1}\Big(\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{b_2}-c_1\Big)\\ &\quad +\frac{b_1}{a_1+b_1}\Big(\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} y(s)ds-T^{\gamma}\Gamma(2-\gamma) \int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds\Big), \end{aligned}\\ k_2=\frac{\Gamma(2-\gamma)}{T^{1-\gamma}}\Big(\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}y(s)ds -\frac{c_2}{b_2}\Big). \end{gather*} Substituting the values ofk_1,k_2$in \eqref{lemoflinear1}, we obtain \eqref{integralformofsolution}. This completes the proof. \end{proof} From the proof of the above Lemma, we notice that the solution \eqref{integralformofsolution} of problem \eqref{zhangjianfanchen} does not depend on the parameter$a_2$, that is to say, the parameter$a_2$is of arbitrary nature for this problem. \begin{theorem}[Schauder fixed point theorem]\label{Schauderfixedpoint} Let$U$be a closed, convex and non\-empty subset of a Banach space$X$, let$P:U\to U$be a continuous mapping such that$P(U)$is a relatively compact subset of$X$. Then$P$has at least one fixed point in$U$. \end{theorem} \begin{theorem}[Nonlinear alternative of Leray-Schauder type]\label{nonlinearalternative} Let$X$be a Banach space,$C$a closed, convex subset of$X$,$U$an open subset of$C$and$0\in U$. Suppose that$P:\overline{U}\to C$is a continuous and compact map. Then either (a)$P$has a fixed point in$\overline{U}$, or (b) there exist a$x\in\partial U$(the boundary of$U$) and$\lambda\in(0,1)$with$x=\lambda P(x)$. \end{theorem} \section{Existence results for problem \eqref{mainequation}} \label{existresultsfordiyiwenti} Let$\mathcal{C}=C([0,T],\mathbb{R})$denote the Banach space of all continuous functions from$[0,T]$into$\mathbb{R}$equipped with the norm$\|x\|=\sup_{t\in[0,T]}|x(t)|$. In view of Lemma \ref{linearproblem}, we define an operator$\mathcal{F}:\mathcal{C}\to\mathcal{C}as \label{definioperaofF} \begin{aligned} &(\mathcal{F}x)(t) \\ &=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x(s))ds-\frac{t\Gamma(2-\gamma)}{T^{1-\gamma}} \int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}f(s,x(s))ds \\ &\quad +\frac{t\Gamma(2-\gamma)c_2}{T^{1-\gamma}b_2} -\frac{b_1}{a_1+b_1}\Big\{\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x(s))ds \\ &\quad -T^{\gamma}\Gamma(2-\gamma) \int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}f(s,x(s))ds \Big\} \\ &\quad -\frac{1}{a_1+b_1}\Big(\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{b_2} -c_1\Big). \end{aligned} Observe that problem \eqref{mainequation} has solutions if and only if the operator equation\mathcal{F}x$has fixed points. We put$\mathcal{F}x=\mathcal{F}_1x+\mathcal{F}_2x$where \begin{equation*} (\mathcal{F}_1x)(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x(s))ds,\quad (\mathcal{F}_2x)(t)=-k^x_2t-k^x_1. \end{equation*} Here$k^x_1$and$k^x_2are constants given by \begin{gather*} \begin{aligned} k^x_1&=\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2}-\frac{c_1}{a_1+b_1} +\frac{b_1}{a_1+b_1}\Big\{\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x(s))ds\\ &\quad-T^{\gamma}\Gamma(2-\gamma)\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}f(s,x(s))ds \Big\}, \end{aligned} \\ k^x_2=\frac{\Gamma(2-\gamma)}{T^{1-\gamma}} \Big(\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}f(s,x(s))ds -\frac{c_2}{b_2}\Big). \end{gather*} Now we are in a position to present our main results. The methods used to prove the existence results are standard, however, their exposition in the framework of problem \eqref{mainequation} is new. \begin{theorem}\label{thm3.1} Suppose thatf:[0,T]\times\mathbb{R}\to\mathbb{R}$is continuous and satisfies \begin{equation*} |f(t,x)-f(t,y)|\leq m(t)|x-y| \end{equation*} for$t\in[0,T]$,$x,y\in\mathbb{R}$with$m\in L^{\infty}([0,T],\mathbb{R}^+)$. If $$\label{contractionconstant} \|m\|_{L^{\infty}}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big) \Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)<1,$$ then problem \eqref{mainequation} has a unique solution. \end{theorem} \begin{proof} For$x,y\in\mathcal{C}$and for each$t\in[0,T], we have \begin{align*} |(\mathcal{F}_1x)(t)-(\mathcal{F}_1y)(t)| &\leq \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \big|f(s,x(s))-f(s,y(s))\big|ds\\ &\leq \|m\|_{L^{\infty}}\|x-y\|\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \end{align*} \begin{equation*} |(\mathcal{F}_2x)(t)-(\mathcal{F}_2y)(t)|\leq T|k^x_2-k^y_2|+|k^x_1-k^y_1|, \end{equation*} \begin{align*} T|k^x_2-k^y_2| &\leq T^{\gamma}\Gamma(2-\gamma)\Big|\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)} \big(f(s,x(s))-f(s,y(s))\big)ds\Big|\\ &\leq \|m\|_{L^{\infty}}\|x-y\|\frac{\Gamma(2-\gamma)T^{\alpha}}{\Gamma(\alpha-\gamma+1)}, \end{align*} \begin{align*} |k^x_1-k^y_1|&\leq \frac{|b_1|}{|a_1+b_1|} \Big|\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} \big(f(s,x(s))-f(s,y(s))\big)ds\Big|\\ &\quad +\frac{|b_1|T^{\gamma}\Gamma(2-\gamma)}{|a_1+b_1|} \Big|\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)} \big(f(s,x(s))-f(s,y(s))\big)ds\Big|\\ &\leq \|m\|_{L^{\infty}}\|x-y\|\frac{|b_1|}{|a_1+b_1|} \Big(\frac{T^{\alpha}}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)T^{\alpha}}{\Gamma(\alpha-\gamma+1)}\Big). \end{align*} Therefore, we have \begin{equation*} \|\mathcal{F}x-\mathcal{F}y\| \leq\|m\|_{L^{\infty}}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big) \Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)\|x-y\|. \end{equation*} This together with \eqref{contractionconstant} implies that the map\mathcal{F}$is a contraction mapping. Hence the contraction mapping principle yields that$\mathcal{F}$has a unique fixed point which is the unique solution of problem \eqref{mainequation}. The proof is complete. \end{proof} \begin{corollary}\label{uniquesoluforexample1} Suppose that$f:[0,T]\times\mathbb{R}\to\mathbb{R}$is continuous and satisfies \begin{equation*} |f(t,x)-f(t,y)|\leq L|x-y|, \end{equation*} for$t\in[0,T]$,$x,y\in\mathbb{R}$and$L>0$. Then problem \eqref{mainequation} has a unique solution provided \begin{equation*} LT^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)<1. \end{equation*} \end{corollary} \begin{theorem}\label{atleastoneschauder} Let$f:[0,T]\times\mathbb{R}\to\mathbb{R}$be a continuous function. Assume that \begin{equation*} |f(t,x)|\leq m(t)+d|x|^{\rho} \end{equation*} for$t\in[0,T]$,$x\in\mathbb{R}$with$m\in L^{\infty}([0,T],\mathbb{R}^+)$,$d\geq0$and$0\leq\rho<1$. Then problem \eqref{mainequation} has at least one solution on$[0,T]$. \end{theorem} \begin{proof} Define$B_r=\{x:x\in\mathcal{C}\textrm{ and }\|x\|\leq r\}, where \begin{gather*} r\geq\max\{2A,(2Bd)^{\frac{1}{1-\rho}}\}, \\ \begin{aligned} A&=\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}+ \Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2} -\frac{c_1}{a_1+b_1}\Big| \\ &\quad +\|m\|_{L^{\infty}}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big) \Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big), \end{aligned} \end{gather*} $$\label{constantfor2bian} B=T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big).$$ It is obvious thatB_r$is a closed, bounded and convex subset of the Banach space$\mathcal{C}$. Firstly, we prove that$\mathcal{F}:B_r\to B_r$. For any$x\in B_r, we have \begin{gather*} |(\mathcal{F}_1x)(t)| \leq \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \big(m(s)+d|x(s)|^{\rho}\big)ds \leq \big(\|m\|_{L^{\infty}}+dr^{\rho}\big)\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \\ |(\mathcal{F}_2x)(t)|\leq T|k^x_2|+|k^x_1|, \\ \begin{aligned} T|k^x_2|&\leq T^{\gamma}\Gamma(2-\gamma)\Big|\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}f(s,x(s))ds -\frac{c_2}{b_2}\Big|\\ &\leq T^{\gamma}\Gamma(2-\gamma)\int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}|f(s,x(s))|ds +\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}\\ &\leq \big(\|m\|_{L^{\infty}}+dr^{\rho}\big)\frac{\Gamma(2-\gamma)T^{\alpha}}{\Gamma(\alpha-\gamma+1)}+\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}, \end{aligned} \\ \begin{aligned} |k^x_1|&\leq \Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2}-\frac{c_1}{a_1+b_1}\Big| +\frac{|b_1|}{|a_1+b_1|}\int_0^T\frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} |f(s,x(s))|ds\\ &\quad +\frac{|b_1|T^{\gamma}\Gamma(2-\gamma)}{|a_1+b_1|} \int_0^T\frac{(T-s)^{\alpha-\gamma-1}}{\Gamma(\alpha-\gamma)}|f(s,x(s))|ds\\ &\leq \big(\|m\|_{L^{\infty}}+dr^{\rho}\big)\frac{|b_1|}{|a_1+b_1|} \Big(\frac{T^{\alpha}}{\Gamma(\alpha+1)}+\frac{\Gamma(2-\gamma) T^{\alpha}}{\Gamma(\alpha-\gamma+1)}\Big)\\ &\quad +\Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2} -\frac{c_1}{a_1+b_1}\Big|. \end{aligned} \end{gather*} Hence we obtain \begin{align*} \|\mathcal{F}x\| &\leq \frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}+ \Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2} -\frac{c_1}{a_1+b_1}\Big|\\ &\quad +\|m\|_{L^{\infty}}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big) \Big(\frac{1}{\Gamma(\alpha+1)}+\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)} \Big)\\ &\quad +dr^{\rho}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big) \Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)\\ &\leq A+dr^{\rho}B\leq\frac{r}{2}+\frac{r}{2}=r. \end{align*} This implies that\mathcal{F}:B_r\to B_r$. Secondly, we show that$\mathcal{F}$maps bounded sets into equicontinuous sets. Let$\overline{B}$be any bounded subset of$\mathcal{C}$. Since$f$is continuous, we can assume without any loss of generality that there is positive constant$N$such that \begin{equation*} |f(t,x(t))|\leq N \end{equation*} for any$t\in[0,T]$and$x\in\overline{B}$. Now let$0\leq t_10$such that $$\label{forexistenceconst} \frac{M}{O+\varphi(M)Q}>1,$$ where \begin{gather*} O=\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}+ \Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2} -\frac{c_1}{a_1+b_1}\Big|, \\ Q=\|m\|_{L^{\infty}}T^{\alpha} \Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big). \end{gather*} Then problem \eqref{mainequation} has at least one solution. \end{theorem} \begin{proof} We, firstly, prove that$\mathcal{F}$maps bounded sets into bounded sets in$\mathcal{C}$. Let$\overline{B}$be a bounded subset of$\mathcal{C}$and assume that$\|x\|\leq r$for any$x\in\overline{B}. As in the proof of the above theorems, we have the following estimates \begin{gather*} |(\mathcal{F}_1x)(t)|\leq\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} |f(s,x(s))|ds\leq\varphi(r)\|m\|_{L^{\infty}}\frac{T^{\alpha}}{\Gamma(\alpha+1)}, \\ |(\mathcal{F}_2x)(t)|\leq T|k^x_2|+|k^x_1|, \\ T|k^x_2|\leq\varphi(r)\|m\|_{L^{\infty}}\frac{\Gamma(2-\gamma)T^{\alpha}}{\Gamma(\alpha-\gamma+1)} +\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}, \\ \begin{aligned} |k^x_1|&\leq \frac{|b_1|}{|a_1+b_1|}\Big(\varphi(r)\|m\|_{L^{\infty}}\frac{T^{\alpha}}{\Gamma(\alpha+1)} +\varphi(r)\|m\|_{L^{\infty}}\frac{\Gamma(2-\gamma)T^{\alpha}}{\Gamma(\alpha-\gamma+1)}\Big)\\ &\quad +\Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2}-\frac{c_1}{a_1+b_1}\Big|. \end{aligned} \end{gather*} Hence we have \begin{align*} \|\mathcal{F}x\| &\leq \frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}+ \Big|\frac{b_1c_2T^{\gamma}\Gamma(2-\gamma)}{(a_1+b_1)b_2} -\frac{c_1}{a_1+b_1}\Big|\\ &\quad +\varphi(r)\|m\|_{L^{\infty}}T^{\alpha} \Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)\\ &\leq O+\varphi(r)Q. \end{align*} This implies that\mathcal{F}(\overline{B})$is bounded in$\mathcal{C}$. Secondly, we claim that$\mathcal{F}$is equicontinuous on bounded subsets of$\mathcal{C}$. The proof of this claim is the same as the corresponding part in the proof of Theorem \ref{atleastoneschauder}. Finally, let$x=\lambda\mathcal{F}x$for some$\lambda\in(0,1)$. Then for each$t\in[0,T]$, we have \begin{equation*} |x(t)|=|\lambda(\mathcal{F}x)(t)|\leq O+\varphi(\|x\|)Q. \end{equation*} That is to say, we have \begin{equation*} \frac{\|x\|}{O+\varphi(\|x\|)Q}\leq1. \end{equation*} Due to \eqref{forexistenceconst}, we know that there exists$M$such that$\|x\|\neq M$. Let \begin{equation*} U=\{y\in\mathcal{C}:\|y\|0$ such that \begin{equation*} \frac{M}{\varphi_1(M)Q+ \varphi_3(M)\|m_3\|_{L^1}O+\frac{|c_1|}{|a_1+b_1|}\varphi_2(M)\|m_2\|_{L^1}}>1, \end{equation*} where \begin{gather*} O=\frac{T^{\gamma}\Gamma(2-\gamma)|c_2|}{|b_2|}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big), \\ Q=\|m_1\|_{L^{\infty}}T^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big). \end{gather*} Then problem \eqref{mainequintegralform} has at least one solution. \end{theorem} \section{Examples} \label{examples} In this section, we give two simple examples to illustrate the main results. subsection*{Example 1} Consider the fractional boundary-value problem $$\label{example1} \begin{gathered} ^cD^{\frac{3}{2}}x(t)=\frac{1}{(t+4)^2}(\sin x(t)+\frac{|x(t)|}{1+|x(t)|}),\quad t\in[0,1],\\ 3x(0)+\frac{1}{2}x(1)=2.5,\\ 2(^cD^{1/2}x(0))+\frac{1}{3}(^cD^{1/2}x(1))=-\frac{1}{3}, \end{gathered}$$ Here $\alpha=\frac{3}{2}$, $\gamma=\frac{1}{2}$, $a_1=3$, $b_1=\frac{1}{2}$, $c_1=2.5$, $a_2=2$, $b_2=\frac{1}{3}$, $c_2=-\frac{1}{3}$, $T=1$ and $f(t,x)=\frac{1}{(t+4)^2}(\sin x+\frac{|x|}{1+|x|})$. Since \begin{gather*} |f(t,x)-f(t,y)|\leq\frac{1}{(t+4)^2}|x-y|\leq\frac{1}{16}|x-y|,\\ \begin{aligned} <^{\alpha}\Big(1+\frac{|b_1|}{|a_1+b_1|}\Big)\Big(\frac{1}{\Gamma(\alpha+1)} +\frac{\Gamma(2-\gamma)}{\Gamma(\alpha-\gamma+1)}\Big)\\ &\approx \frac{1}{16}\times\frac{8}{7}\times(0.7523+0.8862)=0.1170<1. \end{aligned} \end{gather*} Thus, by Corollary \ref{uniquesoluforexample1}, the boundary value problem \eqref{example1} has a unique solution on $[0,1]$. \subsection*{Example 2} Let $\alpha=\frac{5}{4}$, $\gamma=\frac{1}{3}$ and $T=\pi$. Consider the fractional integral boundary-value problem $$\label{example2} \begin{gathered} ^cD^{\frac{5}{4}}x(t)=2t^3-3\ln(3+t)+(3t+1)^2\frac{|x(t)|^{1/2}}{2+\cos^2x(t)}, \quad t\in[0,\pi],\\ \frac{1}{2}x(0)+x(\pi)=\int_0^{\pi}\frac{x^{1/3}(t)}{7(1+|x(t)|)}ds,\\ 2(^cD^{1/3}x(0))+3(^cD^{1/3}x(\pi))=\int_0^{\pi}(3t^3-5+e^{-t}|x(t)|^{2/5})ds, \end{gathered}$$ Since $f(t,x)=2t^3-3\ln(3+t)+(3t+1)^2\frac{|x|^{1/2}}{2+\cos^2x}$, $g(t,x)=\frac{x^{1/3}}{7(1+|x|)}$, $h(t,x)=(3t^3-5+e^{-t}|x|^{2/5})$, $a_1=\frac{1}{2}$, $b_1=c_1=c_2=1$, $a_2=2$ and $b_2=3$, we have \begin{gather*} |f(t,x)|\leq|2t^3-3\ln(3+t)|+(3\pi+1)^2|x|^{1/2},\\ |g(t,x)|\leq\frac{1}{7}|x|^{1/3},\quad |h(t,x)|\leq|3t^3-5|+|x|^{2/5}. \end{gather*} Now it is easy to verify that all conditions of Theorem \ref{integralboundatleastNA} are satisfied. 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