\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 61, pp. 1--16.\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-2016/61\hfil Existence of solutions] {Existence of solutions to fractional differential equations with multi-point boundary conditions at resonance in Hilbert spaces} \author[H.-C. Zhou, F.-D. Ge, C.-H. Kou \hfil EJDE-2016/61\hfilneg] {Hua-Cheng Zhou, Fu-Dong Ge, Chun-Hai Kou} \address{Hua-Cheng Zhou (corresponding author)\newline Academy of Mathematics and Systems Science, Academia Sinica, Beijing 100190, China} \email{hczhou@amss.ac.cn} \address{Fu-Dong Ge \newline College of Information Science and Technology, Donghua University, Shanghai 201620, China} \email{gefd2011@gmail.com} \address{Chun-Hai Kou \newline Department of Applied Mathematics, Donghua University, Shanghai 201620, China} \email{kouchunhai@dhu.edu.cn} \thanks{Submitted August 22, 2015. Published February 29, 2016.} \subjclass[2010]{34A08, 34B10, 34B40} \keywords{Fractional differential equations; resonance; coincidence degree} \begin{abstract} This article is devoted to investigating the existence of solutions to fractional multi-point boundary-value problems at resonance in a Hilbert space. More precisely, the dimension of the kernel of the fractional differential operator with the boundary conditions be any positive integer. We point out that the problem is new even when the system under consideration is reduced to a second-order ordinary differential system with resonant boundary conditions. We show that the considered system admits at least a solution by applying coincidence degree theory first introduced by Mawhin. An example is presented to illustrate our results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In this article, we are concerned with the existence of solutions to the following fractional multi-point boundary value problems(BVPs) at resonance \begin{equation}\label{model-equ} \begin{gathered} D_{0^+}^{\alpha}x(t)=f(t,x(t),D_{0^+}^{\alpha-1}x(t)), \quad 1<\alpha\leq 2,\; t\in (0,1), \\ I_{0+}^{2-\alpha}x(t)|_{t=0}=\theta, \quad x(1)=Ax(\xi), \end{gathered} \end{equation} where $D_{0^+}^{\alpha}$ and $I_{0^+}^{\alpha}$ are the Riemann-Liouville differentiation and integration, respectively; $\theta$ is the zero vector in $l^2:=\{x=(x_1,x_2,\dots,.):\sum_{i=1}^{\infty}|x_i|^2<\infty\}$; $A:l^2\to l^2$ is a bounded linear operator satisfying $1\leq\operatorname{dim}\ker (I-A\xi^{\alpha-1})<\infty$; $\xi\in(0,1)$ is a fixed constant; $f:[0,1]\times l^2\times l^2\to l^2$ is a Carath\'eodory function; that is, \begin{itemize} \item[(i)] for each $(u,v)\in l^2\times l^2$, $t\mapsto f(t,u,v)$ is measurable on $[0,1]$; \item[(ii)] for a.e. $t\in[0,1],\;(u,v)\mapsto f(t,u,v)$ is continuous on $l^2\times l^2$; \item[(iii)] for every bounded set $\Omega\subseteq l^2\times l^2$, the set $\{f(t,u,v):(u,v)\in \Omega\}$ is a relatively compact set in $l^2$. Moreover, the function \[ \varphi_\Omega(t)=\sup\{\|f(t,u,v)\|_{l^2}:(u,v)\in \Omega\} \in L^1[0,1], \] where $\|x\|_{l^2}=\sqrt{\sum_{i=1}^{\infty}|x_i|^2}$ is the norm of $x=(x_1,x_2,\dots,\cdot)^\top$ in $l^2$. \end{itemize} System \eqref{model-equ} is said to be at resonance in $l^2$ if $\operatorname{dim}\ker (I-A\xi^{\alpha-1}) \geq1$, otherwise, it is said to be non-resonant. In the past three decades, the existence of solutions for the fractional differential equations with the boundary value conditions have been given considerable attention by many mathematical researchers. The attempts on $\operatorname{dim}\ker (I-A\xi^{\alpha-1})= 0$, non-resonance case, for fractional differential equations are available in \cite{AhmadEloe,BaiZ,JiaLiu,JiaoZhou,RehamEloe,ZhaiXu,ZhangWangS,ZhuZhangZhao}, and the attempts on $1\leq\operatorname{dim}\ker (I-A\xi^{\alpha-1})\leq2$, resonance case, can be found in \cite{Ba3,YCHen,JWH1,JWH2,Kosmatov1,Kosmatov2,Ba1,zhou1}. However, to the best of our knowledge, all results derived in these papers are for one equation with $\operatorname{dim}\ker L =0$ or $1$ and for two equations with $\operatorname{dim}\ker L =2$. Recently, the authors in \cite{PhungTruong} investigated the following second differential system \begin{equation} \begin{gathered} u''(t)=f(t,u(t),u'(t)), \quad 00$ of a function $x:(0,\infty)\to \mathbb{R}$ is given by \[ I^{\alpha}_{0^+}x(t)=\frac{1}{\Gamma(\alpha)} \int^{t}_{0}{(t-s)^{\alpha-1}x(s)ds}, \] provided that the right-hand side is pointwise defined on $(0,\infty)$. \end{definition} \begin{remark} \label{rmk2.1} \rm The notation $I^{\alpha}_{0^+}x(t)|_{t=0}$ means that the limit is taken at almost all points of the right-sided neighborhood $(0, \varepsilon) (\varepsilon > 0)$ of $0$ as follows: \[ I^{\alpha}_{0^+}x(t)|_{t=0}=\lim_{t\to 0^+}I^{\alpha}_{0^+}x(t). \] Generally, $I^{\alpha}_{0^+}x(t)|_{t=0}$ is not necessarily to be zero. For instance, let $\alpha\in(0,1)$, $x(t)=t^{-\alpha}$. Then \[ I^{\alpha}_{0^+}t^{-\alpha}|_{t=0}=\lim_{t\to0^+} \frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}s^{-\alpha}ds =\lim_{t\to0^+}\Gamma(1-\alpha)=\Gamma(1-\alpha). \] \end{remark} \begin{definition}[\cite{Kilbas}]\label{def2.2} \rm The fractional derivative of order $\alpha>0$ of a function $x:(0,\infty)\to \mathbb{R}$ is given by \[ D^{\alpha}_{0^+}x(t)=\frac{1}{\Gamma(n-\alpha)} \Big(\frac{d}{dt}\Big)^{n}\int^{t}_{0}{\frac{x(s)}{(t-s)^{\alpha-n+1}}ds}, \] where $n=[\alpha]+1$, provided that the right side is pointwise defined on $(0,\infty)$. \end{definition} \begin{lemma}[\cite{Kilbas}] \label{Lem2.1} Assume that $x\in C(0,+\infty)\cap L_{\rm loc}(0,+\infty)$ with a fractional derivative of order $\alpha>0$ belonging to $C(0,+\infty)\cap L_{\rm loc}(0,+\infty)$. Then \[ I_{0^+}^{\alpha}D_{0^+}^{\alpha}x(t)=x(t)+c_1t^{\alpha-1} +c_2t^{\alpha-2}+\dots+c_{n}t^{\alpha-n}, \] for some $c_i\in\mathbb{R},i=1,\dots,n$, where $n=[\alpha]+1$. \end{lemma} For any $x(t)=(x_1(t),x_2(t),\dots)^\top \in l^2$, the fractional derivative of order $\alpha>0$ of $x$ is defined by \[ D^{\alpha}_{0^+}x(t)=(D^{\alpha}_{0^+}x_1(t),D^{\alpha}_{0^+}x_2(t),\dots)^\top \in l^2. \] The following definitions and the coincidence degree theory are fundamental in the proof of our main result. We refer the reader to \cite{Mawhin1,Mawhin}. \begin{definition}\label{Fredhop} \rm Let $X$ and $Y$ be normed spaces. A linear operator $L:\operatorname{dom}(L)\subset X\to Y$ is said to be a Fredholm operator of index zero provided that \begin{itemize} \item[(i)] $\operatorname{im}L$ is a closed subset of $Y$, and \item[(ii)] $\operatorname{dim}\ker L=\operatorname{codim}\operatorname{im}L<+\infty$. \end{itemize} \end{definition} It follows from definition \ref{Fredhop} that there exist continuous projectors $P:X\to X$ and $Q:Y\to Y$ such that \[ \operatorname{im}P=\ker L,\quad \ker Q= \operatorname{im}L,\quad X=\ker L\oplus\ker P,\quad Y=\operatorname{im}L\oplus\operatorname{im}Q \] and the mapping $L|_{\operatorname{dom}L\cap\ker P}:\operatorname{dom} L\cap\ker P\to\operatorname{im}L$ is invertible. We denote the inverse of $L|_{\operatorname{dom}L\cap\ker P}$ by $K_P:\operatorname{im}L\to\operatorname{dom}L\cap\ker P$. The generalized inverse of $L$ denoted by $K_{P,Q}:Y\to\operatorname{dom}L\cap\ker P$ is defined by $K_{P,Q}=K_P(I-Q)$. Furthermore, for every isomorphism $J:\operatorname{im}Q\to \ker L$, we can obtain that the mapping $K_{P,Q}+JQ: Y\to \operatorname{dom}L$ is also an isomorphism and for all $x\in \operatorname{dom}L$, we know that \begin{equation}\label{isomorphism} (K_{P,Q}+JQ)^{-1}x= (L+J^{-1}P)x. \end{equation} \begin{definition}\label{lcompact} \rm Let $L$ be a Fredholm operator of index zero, let $\Omega \subseteq X$ be a bounded subset and $\operatorname{dom}L \cap \Omega \neq \emptyset$. Then the operator $N: \overline{\Omega}\to Y$ is called to be $L$-compact in $\overline{\Omega}$ if \begin{itemize} \item[(i)] the mapping $QN:\overline{\Omega} \to Y$ is continuous and $QN(\overline{\Omega}) \subseteq Y $ is bounded, and \item[(ii)] the mapping $K_{P,Q}N:\overline{\Omega} \to X$ is completely continuous. \end{itemize} \end{definition} Assume that $L$ is defined by Definition \ref{lcompact} and $N:\overline{\Omega}\to Y$ is $L$-compact. For any $x\in \overline{\Omega}$, by \eqref{isomorphism}, we shall see that \begin{align*} Lx&=(K_{P,Q}+JQ)^{-1}x-J^{-1}Px\\ &=(K_{P,Q}+JQ)^{-1}\left[Ix-K_{P,Q}J^{-1}Px-JQJ^{-1}Px\right] \\ &=(K_{P,Q}+JQ)^{-1}(Ix-Px). \end{align*} Then we can equivalently transform the existence problem of the equation $Lx=Nx,x\in \overline{\Omega}$ into a fixed point problem of the operator $P+(K_{P,Q}+JQ)N$ in $\overline{\Omega}$. This can be guaranteed by the following lemma, which is also the main tool in this paper. \begin{lemma}[\cite{Mawhin}] \label{lem2.2} Let $\Omega\subset X$ be bounded, $L$ be a Fredholm mapping of index zero and $N$ be $L$-compact on $\overline{\Omega}$. Suppose that the following conditions are satisfied: \begin{itemize} \item[(i)] $Lx\neq\lambda Nx$ for every $(x,\lambda)\in((\operatorname{dom}L\backslash\ker L)\cap\partial\Omega)\times(0,1)$; \item[(ii)] $Nx\not\in \operatorname{im}L$ for every $x\in\ker L\cap\partial\Omega$; \item[(iii)] $\deg(JQN|_{\ker L\cap\partial\Omega},\Omega\cap\ker L,0)\neq0$, with $Q:Y\to Y$ a continuous projector such that $\ker Q=\operatorname{im}L$ and $J:\operatorname{im}Q\to\ker L$ is an isomorphism. \end{itemize} Then the equation $Lx=Nx$ has at least one solution in $\operatorname{dom}L\cap\overline{\Omega}$. \end{lemma} In this paper, we use spaces $\mathbb{X}$, $\mathbb{Y}$ introduced as \[ \mathbb{X}=\big\{x(t)\in l^2:x(t)=I_{0+}^{\alpha-1}u(t), u\in C([0,1];l^2),t\in [0,1] \big\} \] with the norm $\|x\|_\mathbb{X}=\max \{\|x\|_{C([0,1];l^2)},\|D_{0+}^{\alpha-1}x\|_{C([0,1];l^2)}\}$ and $\mathbb{Y}=L^1([0,1];l^2)$ with the norm $\|y\|_{L^1([0,1];l^2)}=\int_0^1\|y(s)\|_{l^2}ds$, respectively, where $\|x\|_{C([0,1];l^2)}=\sup_{t\in[0,1]}\|x(t)\|_{l^2}$. We have the following compactness criterion on subset $F$ of $\mathbb{X}$ which is a slight modification of \cite[Lemma 2.2]{Ba4} (see also the Ascoli-Arzela theorem \cite[Theorem 1.2.5, p. 15]{GuoDJ}). \begin{lemma}\label{Lem2.3} $F\subset \mathbb{X}$ is a sequentially compact set if and only if $F(t)$ is a relatively compact set and equicontinuous which are understood in the following sense: \begin{itemize} \item[(1)] for any $t\in[0,1]$, $F(t):=\{x(t)|x\in F\}$ is a relatively compact set in $l^2$; \item[(2)] for any given $\varepsilon>0$, there exists a $\delta>0$ such that \[ \|x(t_1)-x(t_2)\|_{l^2}<\varepsilon,\;\|D_{0^+}^{\alpha-1}x(t_1) -D_{0^+}^{\alpha-1}x(t_2)\|_{l^2}<\varepsilon, \] for $t_1,t_2\in[0,1]$, $|t_1-t_2|<\delta$, for all $x\in F$. \end{itemize} \end{lemma} Now we define the linear operator $L:\operatorname{dom}L \subseteq\mathbb{X}\to\mathbb{Y}$ by \begin{equation}\label{Ldef} Lx:=D_{0^+ }^\alpha x, \end{equation} where $\operatorname{dom}L=\{x\in X: D_{0^+ }^\alpha x\in Y, x(0)=\theta, x(1)=Ax(\xi)\}$. Define $N:X\to Y$ by \begin{equation}\label{N} Nx(t):=f(t,x(t),D_{0^+}^{\alpha-1}x(t)), \quad t\in [0,1]. \end{equation} Then the problem can be equivalently rewritten as $Lx=Nx$. The next lemma plays a vital role in estimating the boundedness of some sets. \begin{lemma}\label{boundinequ} Let $z_1,z_2\geq 0$, $\gamma_1,\gamma_2\in[0,1)$ and $\lambda_i,\mu_i\geq0,i=1,2,3$, and the following two inequalities hold, \begin{equation}\label{baseineq} \begin{gathered} z_1\leq\lambda_1z_1^{\gamma_1}+\lambda_2z_2+\lambda_3,\\ z_2\leq\mu_1z_1+\mu_2z_2^{\gamma_2}+\mu_3 \end{gathered} \end{equation} Then $z_1,z_2$ is bounded if $\lambda_2\mu_1<1$. \end{lemma} \begin{proof} From \eqref{baseineq}, we have \begin{equation}\label{twoineq} \begin{gathered} z_1\leq\frac{\lambda_1z_1^{\gamma_1}+\lambda_2\mu_2z_2^{\gamma_2} +\lambda_2\mu_3+\lambda_3}{1-\lambda_2\mu_1},\\ z_2\leq\frac{\lambda_1\mu_1z_1^{\gamma_1}+\mu_2z_2^{\gamma_2} +\lambda_3\mu_1+\mu_3}{1-\lambda_2\mu_1}. \end{gathered} \end{equation} Let $z=\max\{z_1,z_2\}$, $\kappa_1=\max\{\lambda_1,\lambda_1\mu_1\}$ and $\kappa_2=\max\{\lambda_2\mu_2,\mu_2\}$. It follows from \eqref{twoineq} that \[ z\leq\frac{\kappa_1z^{\gamma_1}+\kappa_2z^{\gamma_2}+\lambda_2\mu_3 +\lambda_3\mu_1+\lambda_3+\mu_3}{1-\lambda_2\mu_1}. \] This, together with $\gamma_1,\gamma_2\in[0,1)$, yields that $z$ is bounded. \end{proof} \begin{lemma}\label{lemL} The operator $L$, defined by \eqref{Ldef}, is a Fredholm operator of index zero. \end{lemma} \begin{proof} For any $x\in \operatorname{dom}L$, by Lemma \ref{Lem2.1} and $x(0)=\theta$, we obtain \begin{equation}\label{integralequation} x(t)=I_{0+}^\alpha Lx(t)+ct^{\alpha-1}, \quad c\in l^2, \; t\in [0,1], \end{equation} which, together with $x(1)=Ax(\xi)$, yields \begin{equation}\label{KerLeq} \begin{aligned} \ker L&=\{x\in \mathbb{X}:x(t)=ct^{\alpha-1}, t\in [0,1], c\in \ker (I-A\xi^{\alpha-1})\}\\ &\backsimeq \ker (I-A\xi^{\alpha-1})t^{\alpha-1}. \end{aligned} \end{equation} Now we claim that \begin{equation}\label{ImL} \operatorname{im}L=\{y\in Y: h(y) \in \operatorname{im}(I-A\xi^{\alpha-1})\}, \end{equation} where $h:\mathbb{Y}\to l^2$ is a continuous linear operator defined by \begin{equation}\label{hdef} h(y):= \frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}y(s)ds -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}y(s)ds. \end{equation} Actually, for any $y\in \operatorname{im}L$, there exists a function $x\in \operatorname{dom}L$ such that $y=Lx$. It follows from \eqref{integralequation} that $x(t)=I_{0^+}^{\alpha}y(t)+ct^{\alpha-1}$. From this equality and $x(1)=Ax(\xi)$, we obtain \[ \frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}y(s)ds -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}y(s)ds =(I-A\xi^{\alpha-1})c,\quad c\in l^2, \] which means that $h(y)\in \operatorname{im}(I-A\xi^{\alpha-1})$. On the other hand, for any $y\in\mathbb{Y}$ satisfying $h(y)\in \operatorname{im}(I-A\xi^{\alpha-1})$, there exists a constant $c^*$ such that $h(y)=(I-A\xi^{\alpha-1})c^*$. Let $x^*(t)=I_{0^+ }^\alpha y(t)+c^*t^{\alpha-1}$. A straightforward computation shows that $x^*(0)=\theta$ and $x^*(1)=Ax^*(\xi)$. Hence, $x^*\in \operatorname{dom}L$ and $y(t)=D_{0^+ }^\alpha x^*(t)$, which implies that $y\in \operatorname{im}L$. Next, put $\rho_A=\kappa(I-A\xi^{\alpha-1})$, where \begin{equation}\label{k} \kappa=\begin{cases} 1,& \text{if (A1) holds, i.e., $A^2\xi^{2\alpha-2}=A\xi^{\alpha-1}$};\\ \frac{1}{2}, & \text{if (A2) holds, i.e., $A^2\xi^{2\alpha-2}=I$}. \end{cases} \end{equation} For $A^2\xi^{2\alpha-2}=A\xi^{\alpha-1}$, we have \begin{equation}\label{xiA1} \begin{gathered} \rho_A^2=(I-A\xi^{\alpha-1})^2=I-2A\xi^{\alpha-1}+A^2\xi^{2\alpha-2} =I-A\xi^{\alpha-1}=\rho_A, \\ \begin{aligned} (I-\rho_A)(\xi^{2\alpha-1}A-I) &=A\xi^{\alpha-1}(\xi^{2\alpha-1}A-I)=\xi^{3\alpha-2}A^2-A\xi^{\alpha-1}\\ &=(\xi^\alpha -1)A\xi^{\alpha-1}=(\xi^\alpha-1)(I-\rho_A). \end{aligned} \end{gathered} \end{equation} For $A^2\xi^{2\alpha-2}=I$, we have \begin{equation}\label{xiA2} \begin{gathered} \rho_A^2=\frac{1}{4}(I-A\xi^{\alpha-1})^2 =\frac{1}{4}(I-2A\xi^{\alpha-1}+A^2\xi^{2\alpha-2}) =\frac{1}{2}(I-A\xi^{\alpha-1})=\rho_A,\\ \begin{aligned} &(I-\rho_A)(\xi^{2\alpha-1}A-I)\\ &=\frac{1}{2}(I+A\xi^{\alpha-1})(\xi^{2\alpha-1}A-I)\\ & =\frac{1}{2}[\xi^{2\alpha-1}A-I+\xi^{3\alpha-2} A^2-A\xi^{\alpha-1}] =\frac{1}{2}(\xi^\alpha-1)(I+A\xi^{\alpha-1})\\ &=(\xi^\alpha-1)(I-\rho_A). \end{aligned} \end{gathered} \end{equation} It follows from \eqref{xiA1} and \eqref{xiA2} that $\rho_A$ satisfies the following properties \begin{equation}\label{rhorho} \rho_A^2=\rho_A,\quad (I-\rho_A)(\xi^{2\alpha-1}A-I) =(\xi^\alpha -1)(I-\rho_A). \end{equation} Furthermore, we note that if $y=ct^{\alpha-1}$, $c\in l^2$, then \begin{equation}\label{gker} \begin{aligned} h(y) &=\frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}y(s)ds -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}y(s)ds\\ &=\frac{(\xi^{2\alpha-1}A-I)c}{\Gamma(\alpha)\Gamma(2\alpha)}. \end{aligned} \end{equation} Define the continuous linear mapping $Q:\mathbb{Y}\to\mathbb{Y}$ by \begin{equation}\label{Q} Qy(t):=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(y)t^{\alpha-1}, \quad t\in [0,1],\; y\in \mathbb{Y}. \end{equation} By the first identity in \eqref{rhorho}, we obtain $(I-\rho_A)^2=(I-\rho_A)$, which together with \eqref{rhorho} implies \begin{align*} Q^2y(t) &=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(Qy(t))t^{\alpha-1}\\ &=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A) \frac{(\xi^{2\alpha-1}A-I)}{\Gamma(\alpha)\Gamma(2\alpha)} \frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(y)t^{\alpha-1}\\ &=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)^2h(y)t^{\alpha-1} =Qy(t); \end{align*} that is, $Q$ is a projection operator. The equality $\ker Q =\operatorname{im}L$ follows from the trivial fact that \begin{gather*} y\in \ker Q\Leftrightarrow h(y)\in\ker (I-\rho_A) \Leftrightarrow h(y)\in \operatorname{im} \rho_A \\ \Leftrightarrow h(y)\in \operatorname{im} (I-A\xi^{\alpha-1}) \Leftrightarrow y\in \operatorname{im}L. \end{gather*} Therefore, we get $\mathbb{Y}=\ker Q \oplus \operatorname{im} Q=\operatorname{im}L \oplus \operatorname{im}Q $. Finally, we shall prove that $\operatorname{im} Q=\ker L$. Indeed, for any $z\in \operatorname{im} Q$, let $z=Qy$, $y\in \mathbb{Y}$. By \eqref{rhorho}, we have \[ k(I-A\xi^{\alpha-1})z(t)=\rho_Az(t)=\rho_AQy(t) =\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}\rho_A(I-\rho_A)g(y)t^{\alpha-1} =\theta, \] which implies $z\in \ker L$. Conversely, for each $z\in \ker L$, there exists a constant $c^*\in \ker (I-A\xi^{\alpha-1})$ such that $z=c^*t^{\alpha-1}$ for $t\in [0,1]$. By \eqref{rhorho} and \eqref{gker}, we derive \[ Qz(t)=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1} (I-\rho_A)h(c^*t^{\alpha-1})t^{\alpha-1}=c^*t^{\alpha-1}=z(t),\quad t\in [0,1], \] which implies that $z\in \operatorname{im}Q$. Hence we know that $\operatorname{im}Q=\ker L$, i.e., the operator $L$ is a Fredholm operator of index zero. The proof is complete. \end{proof} Define the operator $P:\mathbb{X}\to\mathbb{X}$ as follows \begin{equation}\label{Popdef} Px(t)=\frac{1}{\Gamma(\alpha)}(I-\rho_A)D_{0+}^{\alpha-1}x(0)t^{\alpha-1}. \end{equation} \begin{lemma}\label{lem-Pdef} The mapping $P:\mathbb{X}\to\mathbb{X}$, defined by \eqref{Popdef}, is a continuous projector such that \[ \operatorname{im}P=\ker L, \quad \mathbb{X}=\ker L\oplus\ker P \] and the linear operator $K_P: \operatorname{im}L\to \operatorname{dom}L \cap\ker P$ can be written as \[ K_Py(t)=I_{0+}^\alpha y(t), \] also \[ K_P=(L|_{\operatorname{dom}L \cap\ker P})^{-1}, \quad \|K_Py\|_{\mathbb{X}}\leq 1/\Gamma(\alpha)\|y\|_{L^1([0,1];l^2)}. \] \end{lemma} \begin{proof} By \eqref{Popdef}, one can see that $P$ is a continuous operator. From the first identity of \eqref{rhorho}, we have $(I-\rho_A)^2=(I-\rho_A)$, which implies that the mapping $P$ is a projector. Moreover, if $v\in\operatorname{im}P$, there exists a $x\in \mathbb{X}$ such that $v=Px$. By the first identity of \eqref{rhorho} again, we see that \[ \frac{1}{\Gamma(\alpha)}(I-A\xi^{\alpha-1})(I-\rho_A)D_{0+}^{\alpha-1}x(0) =\frac{1}{k\Gamma(\alpha)}\rho_A(I-\rho_A)D_{0+}^{\alpha-1}x(0)=0, \] which gives us $v\in \ker L$. Conversely, if $v\in \ker L$, then $v(t)=c_*t^{\alpha-1}$ for some $c_*\in\ker (I-A\xi^{\alpha-1})$, and we deduce that \[ Pv(t)=\frac{1}{\Gamma(\alpha)}(I-\rho_A)D_{0+}^{\alpha-1}v(0)t^{\alpha-1} =(I-\rho_A)c_*t^{\alpha-1}=c_*t^{\alpha-1}=v(t), \quad t\in [0,1], \] which gives us $v\in\operatorname{im}P$. Thus, we get that $\ker L=\operatorname{im}P$ and consequently $\mathbb{X}={\rm ker }L\oplus \ker P$. Moreover, let $y\in \operatorname{im}L$. There exists $x\in \operatorname{dom}L$ such that $y=Lx$, and we obtain \[ K_Py(t)=x(t)+ct^{\alpha-1} \] where $c\in l^2$ satisfies $c=\xi^{\alpha-1}Ac$. It is easy to see that $K_Py\in \operatorname{dom}L$ and $K_Py\in\ker P$. Therefore, $K_P$ is well defined. Further, for $y\in \operatorname{im}L$, we have \[ L(K_Py(t))=D_{0+}^\alpha(K_Py(t))=y(t) \] and for $x\in \operatorname{dom}L\cap\ker P$, we obtain that \[ K_P(Lx(t))=x(t)+c_1t^{\alpha-1}+c_2t^{\alpha-1}, \] for some $c_1,c_2\in l^2$. In view of $x\in \operatorname{dom}L\cap \ker P$, we know that $c_1=c_2=\theta$. Therefore, $(K_PL)x(t)=x(t)$. This shows that $K_P=(L|_{\operatorname{dom}L \cap \ker P})^{-1}$. Finally, by the definition of $K_P$, we derive \begin{equation}\label{2.9} \|D_{0+}^{\alpha-1}K_Py\|_{C([0,1];l^2)} =\big\|\int_0^\cdot{y(s)}ds\big\|_{C([0,1];l^2)} \leq \|y\|_{L^1([0,1];l^2)} \end{equation} and \begin{equation}\label{2.10} \|K_Py\|_{C([0,1];l^2)} =\big\|\frac{1}{\Gamma(\alpha)}\int_0^\cdot{(\cdot-s)^{\alpha-1}y(s)}ds \big\|_{C([0,1];l^2)} \leq \frac{1}{\Gamma(\alpha)}\|y\|_{L^1([0,1];l^2)}. \end{equation} It follows from $(\ref{2.9})$ and $(\ref{2.10})$ that \begin{equation} \begin{aligned} \|K_Py\|_{\mathbb{X}} &=\max\{\|D_{0+}^{\alpha-1}K_Py\|_{C([0,1];l^2)},\|K_Py\|_{C([0,1];l^2)}\}\\ &\leq\max\big\{\|y\|_{L^1([0,1];l^2)},\frac{1}{\Gamma(\alpha)} \|y\|_{L^1([0,1];l^2)}\big\} \\ &=\frac{1}{\Gamma(\alpha)}\|y\|_{L^1([0,1];l^2)}. \end{aligned} \end{equation} This completes of the proof. \end{proof} \begin{lemma} \label{lem-N-compact} Let $f$ be a Carath\'eodory function. Then $N$, defined by \eqref{N}), is L-compact. \end{lemma} \begin{proof} Let $\Omega$ be a bounded subset in $\mathbb{X}$. By hypothesis (iii) on the function $f$, there exists a function $\varphi_\Omega(t)\in L^1[0,1]$ such that for all $x\in \Omega$, \begin{equation}\label{fphi} \|f(t,x(t),D_{0^+}^{\alpha-1}x(t))\|_{l^2}\leq \varphi_\Omega(t), \quad \text{a.e. }t\in [0,1], \end{equation} which, along with \eqref{hdef} implies \begin{equation}\label{gbound} \begin{aligned} \|h(Nx(t))\|_{l^2} &=\big\|\frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1} {f(s,x(s),D_{0^+}^{\alpha-1}x(s))}ds\\ &\quad -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1} {f(s,x(s),D_{0^+}^{\alpha-1}x(s))}ds\big\|_{l^2}\\ &\leq \frac{\|A\|+1}{\Gamma(\alpha)}\|\varphi_\Omega\|_{L^1[0,1]}. \end{aligned} \end{equation} Thus, from \eqref{Q} and \eqref{gbound} it follows that \begin{equation} \begin{aligned} \|QNx\|_{L^1([0,1];l^2)} &=\big\|\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(Nx) \big\|_{l^2}\int_0^1s^{\alpha-1}ds\\ &\leq \frac{\Gamma(2\alpha)(\|A\|+1)\|I-\rho_A\|}{|1-\xi^\alpha|} \|\varphi_\Omega\|_{L^1[0,1]}<\infty. \end{aligned} \end{equation} This shows that $QN(\overline{\Omega}) \subseteq\mathbb{Y}$ is bounded. The continuity of $QN$ follows from the hypothesis on $f$ and the Lebesgue dominated convergence theorem. Next, we shall show that $K_{P,Q}N$ is completely continuous. First, for any $x\in \Omega$, we have \begin{equation}\label{KPQdef} \begin{aligned} K_{P,Q}Nx(t) &=K_P(I-Q)Nx(t)=K_PNx(t)-K_PQNx(t)\\ &=I_{0+}^\alpha Nx(t)- \frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(Nx(t))I_{0+}^\alpha t^{\alpha-1}. \end{aligned} \end{equation} and \begin{equation}\label{KPQdefD} D_{0^+}^{\alpha-1}K_{P,Q}Nx(t)=I_{0+}^1Nx(t)- \frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1}(I-\rho_A)h(Nx(t))I_{0+}^1 t^{\alpha-1}. \end{equation} By the hypothesis on $f$ and the Lebesgue dominated convergence theorem, it is easy to see that $K_{P,Q}N$ is continuous. Since $f$ is a Carath\'eodory function, for every bounded set $\Omega_0\subseteq l^2\times l^2$, the set $\{f(t,u,v):(u,v)\in \Omega_0\}$ is relatively compact set in $l^2$. Therefore, for almost all $t\in[0,1]$, $\{K_{P,Q}Nx(t):x\in\Omega\}$ and $\{D_{0^+}^{\alpha-1}K_{P,Q}Nx(t):x\in\Omega\}$ are relatively compact in $l^2$. From \eqref{gbound}, \eqref{KPQdef} and \eqref{KPQdefD}, we derive that \begin{align*} &\|K_{P,Q}Nx\|_{C([0,1];l^2)}\\ &=\big\|I_{0+}^\alpha Nx(t)-\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1} (I-\rho_A)h(Nx(t))I_{0+}^\alpha t^{\alpha-1}\big\|_{C([0,1];l^2)}\\ &\leq \frac{1}{\Gamma(\alpha)}\|\varphi_\Omega\|_{L^1(0,1)} +\frac{\Gamma(2\alpha)\|I-\rho_A\|}{|\xi^\alpha-1|}\|h(Nx(t))\|_{l^2}\\ &\leq \frac{1}{\Gamma(\alpha)}\|\varphi_\Omega\|_{L^1(0,1)} +\frac{\Gamma(2\alpha)\|I-\rho_A\|(\|A\|+1)}{\Gamma(\alpha)|\xi^\alpha-1|} \|\varphi_\Omega\|_{L^1(0,1)}<\infty, \end{align*} and \begin{align*} & \|D_{0^+}^{\alpha-1}K_{P,Q}Nx\|_{C([0,1];l^2)}\\ &=\big\|I_{0+}^1 Nx(t)-\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1} (I-\rho_A)h(Nx(t))I_{0+}^1 t^{\alpha-1}\big\|_{C([0,1];l^2)}\\ &\leq \|\varphi_\Omega\|_{L^1(0,1)}+\frac{\Gamma(2\alpha) \|I-\rho_A\|}{|\xi^\alpha-1|}\|h(Nx(t))\|_{l^2}\\ &\leq \|\varphi_\Omega\|_{L^1(0,1)} +\frac{\Gamma(2\alpha)\|I-\rho_A\|(\|A\|+1)}{\Gamma(\alpha)|\xi^\alpha-1|} \|\varphi_\Omega\|_{L^1(0,1)}<\infty, \end{align*} which shows that $K_{P,Q}N\overline{\Omega}$ is uniformly bounded in $\mathbb{X}$. Noting that \begin{equation}\label{inequality} b^p-a^p\leq (b-a)^p \quad \text{for any } b\geq a>0, 0< p\leq1. \end{equation} for any $t_1,t_2\in [0,1]$ with $t_10$ such that for $x\in\operatorname{dom}L$, if $\|D_{0^+}^{\alpha-1}x(t)\|_{l^2}>A_1$ for all $t\in [0,1]$, then \begin{align*} & \frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}{f(s,x(s), D_{0^+}^{\alpha-1}x(s))}ds\\ &-\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}{f(s,x(s), D_{0^+}^{\alpha-1}x(s))}ds\notin\operatorname{im} (I-A\xi^{\alpha-1}). \end{align*} \item[(H3)] There exists a constant $A_2>0$ such that for any $e=\{(e_i)\}\in l^2$ satisfying $e=\xi^{\alpha-1}Ae$ and $\|e\|_{l^2}>A_2$, either \[ \langle e,QNe\rangle_{l^2}\leq 0 \quad \text{or} \quad \langle e,QNe\rangle_{l^2}\geq 0, \] where $\langle \cdot,\cdot\rangle_{l^2}$ is the inner product in $l^2$. \end{itemize} Then \eqref{model-equ} has at least one solution in space $X$ provided that \begin{equation}\label{Thcondition} \begin{gathered} \Gamma(\alpha)>\max\big\{(\|I-\rho_A\|+1)\|a_1\|_{L^1(0,1)}, (\|I-\rho_A\|+1)\|b_1\|_{L^1(0,1)}\big\},\\ \frac{(\|I-\rho_A\|+1)^2\|a_1\|_{L^1(0,1)}\|b_1\|_{L^1(0,1)}} {(\Gamma(\alpha)-(\|I-\rho_A\|+1)\|a_1\|_{L^1(0,1)}) (\Gamma(\alpha)-(\|I-\rho_A\|+1)\|b_1\|_{L^1(0,1)})}<1. \end{gathered} \end{equation} \end{theorem} \begin{proof} We shall construct an open bounded subset $\Omega$ in $X$ satisfying all assumption of Lemma \ref{lem2.2}. Let \begin{equation} \Omega_1=\big\{x\in \operatorname{dom}L\backslash\ker L:Lx=\lambda Nx \text{ for some } \lambda \in [0,1]\}. \end{equation} For any $x\in \Omega_1$, $x \notin\ker L$, we have $\lambda\neq 0$. Since $Nx\in \operatorname{im}L=\ker Q$, by \eqref{ImL}, we have $h(Nx)\in \operatorname{im} (I-A\xi^{\alpha-1})$, where \begin{equation}\label{hDef} \begin{aligned} h(Nx)&=\frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}{f(s,x(s), D_{0^+}^{\alpha-1}x(s))}ds\\ &\quad -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}{f(s,x(s), D_{0^+}^{\alpha-1}x(s))}ds. \end{aligned} \end{equation} From (H2) there exists $t_0\in [0,1]$ such that $|D_{0^+}^{\alpha-1}x(t_0)|_{l^2}\leq A_1$. Then from the equality $D_{0+}^{\alpha-1}x(0)=D_{0+}^{\alpha-1}x(t_0)-\int_0^t{D_{0+}^\alpha x(s)}ds$, we deduce that \[ \|D_{0+}^{\alpha-1}x(0)\|_{l^2} \leq A_1+\|D_{0+}^\alpha x\|_{L^1(0,1;l^2)} = A_1+\|Lx\|_1\leq A_1+\|Nx\|_{L^1(0,1;l^2)}, \] which implies \begin{equation}\label{P} \|Px\|_{\mathbb{X}} =\|\frac{1}{\Gamma(\alpha)}(I-\rho_A)D_{0+}^{\alpha-1}x(0)t^{\alpha-1} \|_{\mathbb{X}} \leq \frac{\|I-\rho_A\|}{\Gamma(\alpha)}( A_1+\|Nx\|_{L^1(0,1;l^2)}). \end{equation} Further, for $x\in \Omega_1$, since $\operatorname{im}P=\ker L, X=\ker L\oplus\ker P $, we have $(I-P)x \in \operatorname{dom}L \cap \ker P$ and $LPx=\theta$. Then \begin{equation}\label{I-P} \begin{aligned} \|(I-P)x\|_{\mathbb{X}} &=\|K_PL(I-P)x\|_{\mathbb{X}}\leq \|K_PLx\|_{\mathbb{X}}\\ &\leq \frac{1}{\Gamma(\alpha)}\|Lx\|_{L^1(0,1;l^2)} \leq \frac{1}{\Gamma(\alpha)}\|Nx\|_{L^1(0,1;l^2)}. \end{aligned} \end{equation} From \eqref{P} and \eqref{I-P}, we conclude that \begin{equation} \begin{aligned}\label{|x|} \|x\|_{\mathbb{X}} &=\|Px+(I-P)x\|_{\mathbb{X}} \leq \|Px\|_{\mathbb{X}}+\|(I-P)x\|_{\mathbb{X}}\\ &\leq\frac{\|I-\rho_A\|}{\Gamma(\alpha)}A_1+\frac{\|I-\rho_A\|+1}{\Gamma(\alpha)} \|Nx\|_{L^1(0,1;l^2)}. \end{aligned} \end{equation} Moreover, by the definition of $N$ and (H1), one has \begin{equation}\label{NX} \begin{aligned} &\|Nx\|_{L^1(0,1;l^2)}\\ &= \int_0^1{\|f(s,x(s),D_{0^+}^{\alpha-1}x(s))\|_{l^2}}dt\\ &\leq\|a_1\|_{L^1(0,1)}\|x\|_{C([0,1];l^2)} +\|b_1\|_{L^1(0,1)}\|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}\\ &\quad +\|a_2\|_{L^1(0,1)}\|x\|_{C([0,1];l^2)}^{\gamma_1} +\|b_2\|_{L^1(0,1)}\|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}^{\gamma_2} +\|c\|_{L^1(0,1)}. \end{aligned} \end{equation} Thus, \begin{equation}\label{x} \begin{aligned} \|x\|_{\mathbb{X}} &\leq \frac{\|I-\rho_A\|}{\Gamma(\alpha)}A_1+\frac{\|I-\rho_A\|+1}{\Gamma(\alpha)} \Big(\|a_1\|_{L^1(0,1)}\|x\|_{C([0,1];l^2)}\\ &\quad +\|b_1\|_{L^1(0,1)}\|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}\Big) +\frac{\|I-\rho_A\|+1}{\Gamma(\alpha)} \\ &\quad\times \Big(\|a_2\|_{L^1(0,1)}\|x\|_{C([0,1];l^2)}^{\gamma_1} +\|b_2\|_{L^1(0,1)}\|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}^{\gamma_2}\\ &\quad +\|c\|_{L^1(0,1)}\Big). \end{aligned} \end{equation} It follows from \eqref{Thcondition}, \eqref{x}, $\|x\|_{C([0,1];l^2)}\leq \|x\|_{\mathbb{X}}$, $\|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}\leq \|x\|_{\mathbb{X}}$ and Lemma \ref{boundinequ} that there exists $M>0$ such that \[ \max\{\|x\|_{C([0,1];l^2)}, \|D_{0^+}^{\alpha-1}x\|_{C([0,1];l^2)}\}\leq M, \] that is to say $\Omega_1$ is bounded. Let \begin{equation} \Omega_2=\{x \in\ker L:N x\in \operatorname{im}L\}. \end{equation} For any $x\in \Omega_2$, it follows from $x \in\ker L $ that $x=et^{\alpha-1}$ for some $e\in \ker (I-A\xi^{\alpha-1})\subset l^2$, and it follows from $N x\in \operatorname{im}L$ that $h(N x) \in \operatorname{im}(I-A\xi^{\alpha-1})$, where $h(N x)$ is defined by \eqref{hDef}. By hypothesis $(H_2)$, we arrive at $\|D_{0^+}^{\alpha-1}x(t_0)\|_{l^\infty}=\|e\|_{l^2}\Gamma(\alpha)\leq A_1$. Thus we obtain \[ \|x\|\leq \|e\|_{l^\infty}\Gamma(\alpha)\leq A_1. \] That is, $\Omega_2$ is bounded in $X$. If the first part of $(H_3)$ holds, denote \[ \Omega_3=\{x\in\ker L:-\lambda x+(1-\lambda)QNx=\theta, \; \lambda \in [0,1]\}, \] then for any $x\in \Omega_3$, we know that \[ x=et^{\alpha-1}\quad \text{with } e\in \ker (I-A\xi^{\alpha-1})\text{ and } \lambda x=(1-\lambda)QNx. \] If $\lambda=0$, we have $Nx\in\ker Q= \operatorname{im}L$, then $x\in \Omega_2$, by the argument above, we get that $\|x\| \leq A_1$. Moreover, if $\lambda\in (0,1]$ and if $\|e\|_{l^2}>A_2$, by (H3), we deduce that \[ 0<\lambda \|e\|^2_{l^2}=\lambda\langle e,e\rangle_{l^2} =(1-\lambda)\langle e,QNe\rangle_{l^2}\leq 0, \] which is a contradiction. Then $\|x\|_{\mathbb{X}}=\|et^{\alpha-1}\|_{\mathbb{X}}\leq \max\{\|e\|_{l^2}, \Gamma(\alpha)\|e\|_{l^2}\}$. That is to say, $\Omega_3$ is bounded. If the other part of (H3) holds, we take \[ \Omega_3=\{x\in \ker L:\lambda x+(1-\lambda)QNx=\theta, \; \lambda \in [0,1]\}. \] By using the same arguments as above, we can conclude that $\Omega_3$ is also bounded. Next, we show that all conditions of Lemma \ref{lem2.2} are satisfied. Assume that $\Omega$ is a bounded open subset of $\mathbb{X}$ such that $ \cup_{i = 1}^3\overline{\Omega}_i \subseteq \Omega$. It follows from Lemmas \ref{lemL} and \ref{lem-N-compact} that $L$ is a Fredholm operator of index zero and $N$ is $L$-compact on $\overline{\Omega}$. By the definition of $\Omega$ and the argument above, to complete the theorem, we only need to prove that condition (iii) of Lemma \ref{lem2.2} is satisfied. For this purpose, let \begin{equation} H(x,\lambda)=\pm \lambda x+(1-\lambda)QNx, \end{equation} where we let the isomorphism the $J:\operatorname{im}Q\to \ker L$ be the identical operator. Since $\Omega_3\subseteq \Omega$, $H(x,\lambda)\neq 0$ for $(x,\lambda)\in\ker L \cap \partial \Omega \times [0,1]$, then by homotopy property of degree, we obtain \begin{align*} \deg \left( {JQN{|_{\ker L\cap \partial \Omega }},\Omega\cap\ker L,0} \right) &= \deg \left( {H\left( \cdot,0\right),\Omega\cap\ker L,0} \right)\\ &=\deg \left( {H\left( \cdot,1\right),\Omega\cap\ker L,0} \right)\\ &=\deg \left( {\pm Id,\Omega\cap\ker L,0} \right)=\pm 1\neq 0. \end{align*} Thus (H3) of Lemma \ref{lem2.2} is fulfilled and Theorem \ref{mainresults} is proved. \end{proof} \section{Example} In this section, we shall present an example to illustrate our main result in $l^2$ with $\operatorname{dim}\ker L=2k$, which surely generalize the previous results \cite{Ba3,YCHen,JWH1,JWH2,Kosmatov1,Kosmatov2,Ba1,zhou1}, where the dimension of $\operatorname{dim}\ker L$ is only $1$ or $2$. Consider the following system with $\operatorname{dim}\ker L=2k$, $k=1,2,3,\dots$ in $l^2$. \begin{equation}\label{exampleker2} \begin{gathered} \begin{aligned} &D_{0^+}^{3/2} \begin{pmatrix} x_1(t)\\ x_2(t)\\ x_3(t)\\ x_4(t)\\ \vdots \end{pmatrix} \\ &=\frac{1}{10} \begin{pmatrix} \begin{cases} 1, &\text{if } \|D_{0^+}^{1/2}x(t)\|_{l^2}< 1\\ D_{0^+}^{1/2}x_1(t)+[D_{0^+}^{1/2}x_1(t)]^{-1}-1, &\text{if } \|D_{0^+}^{1/2}x(t)\|_{l^2}\geq 1 \end{cases} \\ \big(x_2(t)+D_{0^+}^{1/2}x_3(t)\big)/2\\ \big(x_3(t)+D_{0^+}^{1/2}x_3(t)\big)/2^2\\ \big(x_4(t)+D_{0^+}^{1/2}x_4(t)\big)/2^3\\ \vdots \end{pmatrix} \end{aligned}\\ x_i(0)=0, \quad i=1,2,\dots\\ x(1)=Ax(1/9). \end{gathered} \end{equation} Let $\alpha=3/2$, $\xi=1/9$. For all $t\in[0, 1]$, let $u=(x_1,x_2,x_3,\dots)$, $v=(y_1,y_2,y_3,\dots)\in l^2$ and $f=(f_1,f_2,\dots)^T$ with \[ f_1(t,u,v)=\begin{cases} 1/10, &\text{if } \|v\|_{l^2}< 1,\\ (y_1+y_1^{-1}-1)/10, &\text{if } \|v\|_{l^2}\geq 1, \end{cases} \] $f_2(t,u,v)=(x_2+y_3)/20$ and $f_i(t,u,v)=\frac{1}{5}\frac{x_i+y_i}{2^{i}}$, $i=3,4,\dots$. Moreover, \begin{equation}\label{Adef2x2x} A=\begin{bmatrix} {B_1}&0&{0}&{0}&{0}&{0}&\dots\\ {0}&{B_2}&0&{0}&{0}&{0}&\dots\\ \vdots& &\ddots& & & &\vdots\\ {0}&0&{0}&{B_k}&{0}&{0}&{\dots}\\ {0}&0&{0}&{0}&{0}&{0}&{\dots}\\{0}&0&{0}&{0}&{0}&{0}&{\dots} \\ \vdots& & & & & &\ddots \end{bmatrix} \quad\text{with}\quad B_i=\begin{bmatrix} {3}&{0}&{0}\\ 0&{-3}&{6}\\ {0}&{0}&3\\ \end{bmatrix}, \end{equation} $i=1,2,\dots,k$, $k\in \mathbf{N}$. Obviously, we see that $B_i^2=9I_3$ and $\operatorname{dim}\ker (I_3-\xi^{\alpha-1}B_i)=\operatorname{dim}\ker (I_3-B_i/3)=2$, $i=1,2,\dots$, where $I_3$ is the $3\times 3$ identity matrix. Then $A^2\xi^{2\alpha-2}=I$, $\operatorname{dim}\ker (I-A\xi^{\alpha-1})=2k$, $k\in \mathbf{N}$ and the problem \eqref{exampleker2}, with $A$ and $f$ defined above, has one solution if and only if problem \eqref{model-equ} admits one solution. Checking (H1) of Theorem \ref{mainresults}: For some $r\in \mathbb{R}$, $\Omega=\{(u,v)\in l^2\times l^2: \|u\|_{l^2}\leq r, \|v\|_{l^2}\leq r\}$, let $\varphi_\Omega(t)=\frac{1}{10}[(r+1/r+1)^2+\frac{4r^2}{3}]^{1/2}\in L^1[0,1]$. Since $\|A\|_{l^2}\leq 9\sqrt{k}$, letting \begin{equation}\label{abex2e} a_1(t)=b_1(t)=\frac{1}{5\sqrt{3}},\quad a_2(t)=b_2(t)=0 , \quad c(t)=\frac{r+1/r+1}{10}. \end{equation} condition (H1) is satisfied. Checking (H2) of Theorem \ref{mainresults}: From the definition of $f$ it follows that $f_1>1/10>0$ when $ \|D_{0^+}^{1/2}x(t)\|_{l^2}> 1$. This, \[ (B_1\xi^\alpha-I)\begin{pmatrix} {f_1}\\ {f_2}\\ {f_3} \end{pmatrix} =\begin{bmatrix} {-8/9}&{0}&{0}\\ 0&{-10/9}&{2/9}\\ {0}&{0}&{-8/9} \end{bmatrix} \begin{pmatrix} {f_1}\\ {f_2}\\ {f_3} \end{pmatrix}=\begin{pmatrix} {\frac{-8f_1}{9}}\\ {*}\\ {*} \end{pmatrix} \] and $\operatorname{im} (I-A\xi^{\alpha-1}) =\{(0,0,\tau_3,0,0,\tau_6,\dots,0,0,\tau_{3i},\dots):\tau_{3i}\in \mathbb{R}, i=1,2,\dots\}$ implies that condition (H2) is satisfied. Checking (H3) of Theorem \ref{mainresults}: Since $\operatorname{dim}\ker (I-A\xi^{\alpha-1}) =\operatorname{dim}\ker (I-A/3)=2k$, $k\in \mathbf{N}$, for any $e\in l^2$ satisfying $e=Ae$, $e$ can be expressed as $e=e_1+e_2+\dots+e_k$, with \[ e_i=\sigma_{i1}\varepsilon_{3i-2}+\sigma_{i2}(\varepsilon_{3i-1} +\varepsilon_{3i}), \quad \sigma_{ij} \in \mathbb{R},\quad i=1,2,\dots,k, j=1,2, \] where $\varepsilon_j=\big( {0,0, \dots 0,\mathop 1_{{\rm{j}} - th}, 0, 0,\dots } \big)\in l^2$ is a vector with all elements equaling to 0 except the $j$-th equaling to 1, $j=1,2,\dots$. In addition, for any $y\in \mathbb{Y}$, by \eqref{Q} and $\rho_A=\frac{1}{2}(I-A/3)$, we have \begin{equation}\label{Qexample11} Qy(t)=\frac{\Gamma(\alpha)\Gamma(2\alpha)}{\xi^\alpha-1} (I-\rho_A)h(y)t^{\alpha-1}=\frac{-27\sqrt{\pi}}{52}(I+A/3)h(y)t^{\alpha-1}, \end{equation} where \begin{equation} h(y)= \frac{A}{\Gamma(\alpha)}\int_0^\xi(\xi-s)^{\alpha-1}y(s)ds -\frac{I}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-1}y(s)ds. \end{equation} By \eqref{N}, let $d=t^{1/2}+\frac{\sqrt{\pi}}{2}$, we have \begin{equation}\label{Nct1ex11} N(et^{1/2})\\ =\frac{1 }{10} \begin{cases} \Big(1,\frac{d\sigma_{12}}{2}, \frac{d\sigma_{12}}{2^2}, \frac{ d\sigma_{21}}{2^3},\frac{d\sigma_{22}}{2^{4}}, \frac{d\sigma_{22}}{2^{5}},\dots,\\ \frac{d\sigma_{i1}}{2^{3i-3}},\frac{d\sigma_{i2}}{2^{3i-2}}, \frac{d\sigma_{i2}}{2^{3i-1}},\dots\Big)^\top_,\\ \quad\text{if } |\sigma_{11}|< 1,\; 2\leq i\leq k;\\[4pt] \Big(\sigma_{11}+\frac{1}{\sigma_{11}}-1, \frac{d\sigma_{12}}{2}, \frac{d\sigma_{12}}{2^2}, \frac{ d\sigma_{21}}{2^3}, \frac{d\sigma_{22}}{2^{4}}, \frac{d\sigma_{22}}{2^{5}},\dots,\\ \frac{d\sigma_{i1}}{2^{3i-3}},\frac{d\sigma_{i2}}{2^{3i-2}}, \frac{d\sigma_{i2}}{2^{3i-1}},\dots\Big)^\top_,\\ \quad\text{if }|\sigma_{11}|\geq 1,\; 2\leq i\leq k. \end{cases} \end{equation} Suppose that $|\sigma_{11}|>1$, $\sigma_{12}\neq 0$, and let $A_2=1$, $\widetilde{d}=\frac{-27\pi-208\sqrt{\pi}}{648}<0$. From \eqref{Qexample11} and \eqref{Nct1ex11} it follows that \begin{align*} Q(Net^{1/2}) &=\frac{-27\sqrt{\pi}}{520}(I+A/3)h(Net^{1/2})t^{1/2}\\ &= \frac{-27\sqrt{\pi} t^{1/2}}{520}\Big(\frac{-64}{27\sqrt{\pi}} (\sigma_{11}+\frac{1}{\sigma_{11}}-1), \frac{\widetilde{d}\sigma_{12}}{2^2}, \frac{\widetilde{d}\sigma_{12}}{2^2}, \frac{ \widetilde{d}\sigma_{21}}{2^3}, \frac{\widetilde{d}\sigma_{22}}{2^{5}}, \frac{\widetilde{d}\sigma_{22}}{2^{5}},\\ &\quad \dots, \frac{\widetilde{d}\sigma_{i1}}{2^{3i-3}}, \frac{\widetilde{d}\sigma_{i2}}{2^{3i-1}}, \frac{\widetilde{d}\sigma_{i2}}{2^{3i-1}},\dots\Big)^T \end{align*} and \begin{align*} \langle e,QNet^{1/2}\rangle &=\frac{-27\sqrt{\pi} t^{1/2}}{520} \Big[\frac{-64}{27\sqrt{\pi}}((\sigma_{11}-1/2)^2+3/4)\\ &\quad +\widetilde{d}\Big(\frac{2\sigma_{12}^2}{2^2} +\frac{\sigma_{21}^2}{2^3}+\frac{2\sigma_{22}^2}{2^5}+\dots +\frac{\sigma_{i1}^2}{2^{3i-3}}+\frac{2\sigma_{i2}^2}{2^{3i-1}} +\dots \Big)\Big]>0. \end{align*} Therefore, \eqref{exampleker2} admits at least one solution. \subsection*{Acknowledgements} This work was supported by Chinese Universities Scientific Fund No.CUSF-DH-D-2014061, the Natural Science Foundation of Shanghai (No.15ZR1400800) and by the National Natural Science Foundation of China (No. 11526164). \begin{thebibliography}{00} \bibitem{AhmadEloe} B. 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