\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 229, 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 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/229\hfil Impulsive fractional differential equations] {Integral boundary-value problem for impulsive fractional functional differential equations with infinite delay} \author[A. Chauhan, J. Dabas, M. Kumar \hfil EJDE-2012/229\hfilneg] {Archana Chauhan, Jaydev Dabas, Mukesh Kumar} % in alphabetical order \address{Archana Chauhan \newline Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad - 211 004, India} \email{archanasingh.chauhan@gmail.com} \address{Jaydev Dabas \newline Department of Applied Science and Engineering, IIT Roorkee, Saharanpur Campus, Saharanpur-247001, India} \email{jay.dabas@gmail.com} \address{Mukesh Kumar \newline Department of Mathematics, Motilal Nehru National Institute of technology, Allahabad - 211 004, India} \email{mukesh@mnnit.ac.in} \thanks{Submitted May 29, 2012. Published December 18, 2012.} \subjclass[2000]{26A33, 34K05, 34A12, 34A37} \keywords{Fractional differential equation; integral boundary condition; \hfill\break\indent impulsive conditions; infinite delay} \begin{abstract} In this article, we establish a general framework for finding solutions for impulsive fractional integral boundary-value problems. Then, we prove the existence and uniqueness of solutions by applying well known fixed point theorems. The obtained results are illustrated with an example for their feasibility. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction}\label{intro} The purpose of this article is to establish the existence and uniqueness of solution to an integral boundary-value problem for impulsive fractional functional integro-differential equation with infinite delay of the form: \begin{equation} \label{ME} \begin{gathered} ^cD_t^{\alpha}x(t)=f(t,x_t,Bx(t)),\quad t\in J=[0,T],\; t\neq t_k,\\ \Delta x(t_k)=Q_k(x(t_k^{-})),\quad k=1,2,\dots,m,\\ \Delta x'(t_k)=I_k(x(t_k^{-})),\quad k=1,2,\dots,m,\\ x(t)=\phi(t),\quad t\in(-\infty,0], \\ ax'(0)+bx'(T)=\int_0^Tq(x(s))ds, \end{gathered} \end{equation} where $T>0$, $\alpha\in(1,2)$, $a,b\in \mathbb{R}$ such that $a+b\neq 0$. $^cD_t^\alpha$ is the Caputo fractional derivative. The functions $f:J\times \mathfrak{B}_h\times X\to X$ and $q:X\to X$ are given functions that satisfy certain assumptions, where $\mathfrak{B}_h$ is a phase space defined in details in Section $2$. Here $0= t_00$, we define $\mathfrak{B}=\{\psi:[-a,0]\to X$ such that $\psi(t)$ is bounded and measurable$\}$ and equip the space $\mathfrak{B}$ with the norm $$ \|\psi\|_{[-a,0]}=\sup_{s\in[-a,0]}\|\psi(s)\|_X,\quad \forall\; \psi \in \mathfrak{B}. $$ Let us define $\mathfrak{B}_h=\{\psi:(-\infty,0]\to X$, such that for any $c>0$, $\psi |_{[-c,0]}\in \mathfrak{B}$ and $\int_{-\infty}^0 h(s)\|\psi\|_{[s,0]}ds<\infty\}$. If $\mathfrak{B}_h$ is endowed with the norm $$ \|\psi\|_{\mathfrak{B}_h}=\int_{-\infty}^0 h(s)\|\psi\|_{[s,0]}ds,\quad \forall\;\psi \in \mathfrak{B}_h, $$ then it is clear that $(\mathfrak{B}_h,\|\cdot\|_{\mathfrak{B}_h})$ is a Banach space. Now we consider the space \begin{align*} \mathfrak{B}_h' &= \{x:(-\infty,T]\to X\text{ such that } x|_{J_k}\in C(J_k,X) \text{ and there exist}\\ &x(t_k^{+})\ \mbox{ and } x(t_k^{-})\text{ with } x(t_k)=x(t_k^{-}),\; x_0=\phi \in \mathfrak{B}_h,k=1,\dots,m\}, \end{align*} where $x|_{J_k}$ is the restriction of $x$ to $J_k=(t_k,t_{k+1}],k=0,1,2,\dots,m$. Set $\|\cdot\|_{B_h'}$ to be a seminorm in $\mathfrak{B}_h'$ defined by $$ \|x\|_{\mathfrak{B}_h'}=\sup\{\|x(s)\|_X:s \in[0,T]\} +\|\phi\|_{\mathfrak{B}_h}, x\in \mathfrak{B}_h'. $$ Assume that $x\in \mathfrak{B}_h'$, then for $t\in J,\;x_t\in \mathfrak{B}_h$. Moreover, $$ l\|x(t)\|_X\leq\|x_t\|_{\mathfrak{B}_h}\leq l\sup_{0 0$, of a function $f:\mathbb{R}_+\to \mathbb{R}$ and $f\in L^1(\mathbb{R}_+,X)$ is defined by \begin{equation} \label{e2.1} J_t^0f(t)=f(t),\;J_t^{\alpha}f(t)={1\over \Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s)ds,\quad \alpha>0,\;t>0, \end{equation} where $\Gamma(\cdot)$ is the Euler gamma function. \end{definition} \begin{definition} \label{def2.2} \rm Caputo's derivative of order $\alpha$ for a function $f: [0,\infty)\to \mathbb{R}$ is defined as \begin{equation} D^\alpha_tf(t)={1\over \Gamma(n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}f^{(n)}(s)ds =J^{n-\alpha}f^{(n)}(t), \end{equation} for $n-1\leq\alpha0$, the general solution of fractional differential equation ${}^cD_t^\alpha x(t)=0$ is given by $x(t)=c_0+c_1t+c_2t^2+\dots+c_{n-1}t^{n-1}$, where $c_i\in \mathbb{R}$, $ i=0,1,2,\dots,n-1$ $(n=[\alpha]+1)$ and $[\alpha]$ denotes the integer part of the real number $\alpha$. \end{lemma} Note that \begin{equation} x(t)=x_0-ct-\int_0^\omega\frac{(\omega-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds \end{equation} is the solution of the Cauchy problem \begin{equation} \begin{gathered} {}^cD_t^\alpha x(t)= h(t),\quad t\in J,\; \alpha\in(1,2),\\ x(0)= x_0-\int_0^\omega\frac{(\omega-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds. \end{gathered} \end{equation} Now, we can obtain the following result. \begin{lemma}[{\cite[lemma 2.6]{MYJ}}] \label{lem2.5} Let $\alpha\in(1,2)$, $c\in \mathbb{R}$ and $h:J\to \mathbb{R}$ be continuous function. A function $x\in C(J,\mathbb{R})$ is a solution of the fractional integral equation \eqref{IE} \begin{equation}\label{IE} x(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds -\int_0^\omega\frac{(\omega-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds +x_0-c(t-\omega) \end{equation} if and only if $x$ is a solution of the fractional Cauchy problem \begin{equation} \begin{gathered} {}^cD_t^\alpha x(t)= h(t),\quad t\in J,\\ x(\omega)= x_0,\quad \omega\geq 0. \end{gathered} \end{equation} \end{lemma} \begin{lemma} \label{lem2.6} Let $\alpha\in(1,2)$ and $f:J\times \mathfrak{B}_h\times X\to \mathbb{R}$ be continuously differentiable function. A piecewise continuously differentiable function $x\in \mathfrak{B}_h'$ is a solution of system \eqref{ME} if and only if $x\in \mathfrak{B}_h'$ is a solution of the fractional integral equation \begin{equation}\label{int-sol} x(t)\\ =\begin{cases} \phi(t), \quad\text{if } t\in(-\infty,0], \\[3pt] \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0) -\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-}))\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_s,Bx(s))ds +\frac{t}{a+b}\int_0^Tq(x(s))ds, \\ \quad \text{if }t\in[0,t_1],\\ \dots, \\ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0) +\sum_{i=1}^k(t-t_i)I_i(x(t_i^{-}))\\ +\sum_{i=1}^kQ_i(x(t_i^{-}))-\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-}))\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_s,Bx(s))ds +\frac{t}{a+b}\int_0^Tq(x(s))ds, \\ \quad \text{if }t\in (t_k,t_{k+1}], \end{cases} \end{equation} where $k=1,\dots,m$. \end{lemma} \begin{proof} Assume $x$ satisfies \eqref{ME}. If $t\in[0,t_1]$, then \begin{equation}\label{L1} {}^cD_t^\alpha x(t)=f(t,x_t,Bx(t)),\;t\in(0,t_1],\; x(0)=\phi(0). \end{equation} By using Lemma \ref{lem2.5}, we can write the solution of \eqref{L1} as \begin{equation} x(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0)-ct. \end{equation} If $t\in(t_1,t_2]$, then \begin{equation} \begin{gathered} {}^cD_t^\alpha x(t)=f(t,x_t,Bx(t)),\;t\in(t_1,t_2],\\ x(t_1^+)=x(t_1^-)+Q_1(x(t_1^-)),\;x'(t_1^+)=x'(t_1^-)+I_1(x(t_1^-)). \end{gathered} \end{equation} Again by lemma \ref{lem2.5}, we have the following form of the solution \begin{align*} x(t) &= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds -\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds\\ &\quad +x(t_1^-)+Q_1(x(t_1^-))-d(t-t_1)\\ &= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds +\phi(0)-ct_1+Q_1(x(t_1^-))-d(t-t_1). \end{align*} Since $x'(t_1^+)=x'(t_1^-)+I_1(x(t_1^-))$, we obtain $d=c-I_1(x(t_1^-))$. Thus \begin{align*} x(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds +\phi(0)+(t-t_1)I_1(x(t_1^-))+Q_1(x(t_1^-))-ct. \end{align*} If $t\in(t_2,t_3]$, then by similar way using the lemma \ref{lem2.5}, we have \begin{align*} x(t) &= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds-\int_0^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds\\ &\quad +x(t_2^-)+Q_2(x(t_2^-))-e(t-t_2).\\ &= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0)-ct_2+(t_2-t_1)I_1(x(t_1^-))\\ &\quad +Q_1(x(t_1^-))+Q_2(x(t_2^-))-e(t-t_2). \end{align*} Since $x'(t_2^+)=x'(t_2^-)+I_2(x(t_2^-))$, we obtain $e=c-I_1(x(t_1^-))-I_2(x(t_2^-))$. Thus \begin{align*} x(t)&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds +\phi(0)+(t-t_1)I_1(x(t_1^-))\\ &\quad +(t-t_2)I_2(x(t_2^-))+Q_1(x(t_1^-))+Q_2(x(t_2^-))-ct. \end{align*} Similarly, if $t\in(t_k,t_{k+1}]$, then again from lemma \ref{lem2.5}, we have \begin{align*} x(t)&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0)+\sum_{i=1}^k(t-t_i)I_i(x(t_i^-))\\ &\quad+\sum_{i=1}^kQ_i(x(t_i^-))-ct. \end{align*} By using the integral boundary condition $ax'(0)+bx'(T)=\int_0^Tq(x(s))ds$, we obtain \begin{align*} c&= {b\over a+b}\int_0^T{(T-s)^{\alpha-1}\over\Gamma(\alpha-1)}f(s,x_s,Bx(s))ds\\ &\quad +{b\over a+b}\sum_{i=1}^mI_i(x(t_i^-))-{1\over a+b}\int_0^Tq(x(s))ds. \end{align*} Thus for $t\in[0,t_1]$, \begin{align*} x(t)&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds +\phi(0)-\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-}))\\ &\quad-\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,x_s,Bx(s))ds+\frac{t}{a+b}\int_0^Tq(x(s))ds, \end{align*} and for $t\in(t_k,t_{k+1}]$, $k=1,2,\dots,m$, we have \begin{align*} x(t) &=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0) +\sum_{i=1}^k(t-t_i)I_i(x(t_i^{-}))\\ &\quad+\sum_{i=1}^kQ_i(x(t_i^{-}))-\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-}))\\ &\quad -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,x_s,Bx(s))ds +\frac{t}{a+b}\int_0^Tq(x(s))ds. \end{align*} Conversely, assume that $x$ satisfies \eqref{int-sol}. By a direct computation, it follows that the solution given in \eqref{int-sol} satisfies the system \eqref{ME}. This completes the proof of the lemma. \end{proof} Further we introduce the following assumptions to establish our results. \begin{itemize} \item[(H1)] There exists constants $\mu_1,\mu_2>0$, such that \[ \|f(t,\varphi,x)-f(t,\psi,y)\|_X \leq\mu_1\|\varphi-\psi\|_{\mathfrak{B}_h} +\mu_2\|x-y\|_X, \] $t\in J$, $\varphi,\psi\in \mathfrak{B}_h$, $x,y\in X$. \item[(H2)] The function $q:X\to X$ is continuous and there exists constant $L_q>0$, such that \[ \|q(x(s))-q(y(s))\|_X\leq L_q\|x-y\|_X. \] \item [(H3)] For each $k=1,\dots,m$, there exists $\overline{L},L>0$, such that \begin{gather*} \|Q_k(x)-Q_k(y)\|_X\leq\overline{L}\|x-y\|_X,\quad \forall x,y\in X.\\ \|I_k(x)-I_k(y)\|_X\leq L\|x-y\|_X,\quad \forall x,y\in X. \end{gather*} \item[(H4)] The function $f:J\times \mathfrak{B}_h\times X\to X$ is continuous and there exist two continuous functions $\mu_1,\mu_2:J\to(0,\infty)$ such that $\|f(t,\psi,x)\|_X\leq\mu_1(t)\|\psi\|_{\mathfrak{B}_h}+\mu_2(t)\|x\|_X$ and $\mu_1^*=\sup_{t\in[0,T]}\mu_1(t),\mu_2^*=\sup_{t\in[0,T]}\mu_2(t)$. \item[(H5)] The functions $q:X\to X$, $I_k:X\to X$ and $Q_k:X\to X,\;k=1,\dots,m$ are continuous and there exist constants $C,\rho,\Omega$ such that $\|q(x)\|_X\leq C,\;x\in X$, $\rho=\max_{1\leq k\leq m,x\in B_r}\{\|I_k(x)\|_X\}$ and $\Omega=\max_{1\leq k\leq m,x\in B_r}\{\|Q_k(x)\|_X\}$. \end{itemize} \section{Existence and uniqueness results}\label{sec:3} \begin{theorem}\label{thm3.1} Suppose that the assumptions {\rm (H1)--(H3)} hold and \[ \Lambda=\Big[\frac{(a+(1+\alpha)b)(\mu_1l+\mu_2B^*)T^\alpha}{(a+b) \Gamma(\alpha+1)}+\frac{(a+2b)LTm+L_qT^2}{a+b} +\overline{L}m\Big]<1. \] Then \eqref{ME} has an unique solution. \end{theorem} \begin{proof} Consider the operator $N: B_h'\to B_h'$ defined by \[ Nx(t) =\begin{cases} \phi(t), \quad \text{if } t\in(-\infty,0], \\ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0) -\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-})) \\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_s,Bx(s))ds +\frac{t}{a+b}\int_0^Tq(x(s))ds,\\ \quad \text{if } t\in[0,t_1],\\ \dots, \\ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_s,Bx(s))ds+\phi(0) +\sum_{i=1}^k(t-t_i)I_i(x(t_i^{-}))\\ +\sum_{i=1}^kQ_i(x(t_i^{-}))-\frac{bt}{a+b}\sum_{i=1}^mI_i(x(t_i^{-})) \\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_s,Bx(s))ds +\frac{t}{a+b}\int_0^Tq(x(s))ds,\\ \quad \text{if } t\in (t_k,t_{k+1}], \end{cases} \] where $k=1,2,\dots,m$. Let $y(.):(-\infty,T]\to X$ be the function defined by \[ y(t)=\begin{cases}\phi(t),& t\in (-\infty,0];\\ 0,& t\in J,\end{cases} \] then $y_0=\phi$. For each $z\in C([0,T],\mathbb{R})$ with $z(0)=0$, we denote \[ \overline{z}(t)=\begin{cases} 0,& t\in (-\infty,0];\\ z(t),& t\in J. \end{cases} \] If $x(.)$ satisfies \eqref{int-sol} then we can decompose $x(\cdot)$ as $x(t)=y(t)+\overline{z}(t)$, which implies $x_t=y_t+\overline{z}_t$ for $t\in J$ and the function $z(\cdot)$ satisfies \[ z(t)=\begin{cases} \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,y_s+\overline{z}_s, B(y(s)+\overline{z}(s)))ds+\phi(0)\\ -\frac{bt}{a+b}\sum_{i=1}^mI_i(z(t_i^{-}))+\frac{t}{a+b}\int_0^T q(y(s)+\overline{z}(s))ds\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds,\; & t\in[0,t_1], \\ \dots, \\ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,y_s+\overline{z}_s, B(y(s)+\overline{z}(s)))ds+\phi(0)\\ +\sum_{i=1}^k(t-t_i)I_i(z(t_i^{-}))+\sum_{i=1}^kQ_i(z(t_i^{-}))\\ -\frac{bt}{a+b}\sum_{i=1}^mI_i(z(t_i^{-})) +\frac{t}{a+b}\int_0^Tq(y(s)+\overline{z}(s))ds\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds,\; & t\in (t_k,t_{k+1}], \end{cases} \] where $k=1,2,\dots,m$. Set $\mathfrak{B}_h''=\{z\in \mathfrak{B}_h'$ such that $z_0=0\}$ and let $\|\cdot\|_{\mathfrak{B}_h''}$ be the seminorm in $\mathfrak{B}_h''$ defined by \[ \|z\|_{\mathfrak{B}_h''}=\sup_{t\in J}\|z(t)\|_X+\|z_0\|_{\mathfrak{B}_h}= \sup_{t\in J}\|z(t)\|_X, \; z\in \mathfrak{B}_h''. \] Thus $({\mathfrak{B}_h''},\|\cdot\|_{{\mathfrak{B}_h''}})$ is a Banach space. We define the operator $P:{\mathfrak{B}_h''}\to {\mathfrak{B}_h''}$ by \[ Pz(t)=\begin{cases} \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds+\phi(0)\\ -\frac{bt}{a+b}\sum_{i=1}^mI_i(z(t_i^{-}))+\frac{t}{a+b} \int_0^Tq(y(s)+\overline{z}(s))ds\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds,\; & t\in[0,t_1], \\ \dots, \\ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds+\phi(0)\\ +\sum_{i=1}^k(t-t_i)I_i(z(t_i^{-}))+\sum_{i=1}^kQ_i(z(t_i^{-}))\\ -\frac{bt}{a+b}\sum_{i=1}^mI_i(z(t_i^{-}))+\frac{t}{a+b} \int_0^Tq(y(s)+\overline{z}(s))ds\\ -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds,\; & t\in (t_k,t_{k+1}], \end{cases} \] where $k=1,2,\dots,m$. It is clear that the operator $N$ has a unique fixed point if and only if $P$ has a unique fixed point. So let us prove that $P$ has a unique fixed point. Let $z,z^{*}\in \mathfrak{B}_h''$ and $t\in [0,t_1]$ we have \begin{align*} &\|(Pz)(t)-(Pz^{*})(t)\|_X\\ &\leq\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \|f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s))) -f(s,y_s+\overline{z}_s^*,B(y(s)\\ &\quad +\overline{z}^*(s)))\|_X\,ds +\frac{bt}{a+b}\sum_{i=1}^m\|I_i(z(t_i^{-}))-I_i(z^*(t_i^{-}))\|_X\\ &\quad +\frac{t}{a+b}\int_0^T\|q(y(s)+\overline{z}(s))-q(y(s)+\overline{z}^*(s))\|_X\,ds\\ &\quad +\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s))) -f(s,y_s+\overline{z}_s^*,B(y(s)\\ &\quad +\overline{z}^*(s))\|_X\,ds\\ &\leq\Big[\frac{(\mu_1l+\mu_B^*)(a+(1+\alpha)b)T^\alpha}{(a+b) \Gamma(\alpha+1)}+\frac{T(bLm+L_qT)}{a+b}\Big]\|z-z^*\|_{\mathfrak{B}_h''}. \end{align*} If $t\in(t_k,t_{k+1}]$, $k=1,2,\dots,m$, then \begin{align*} &\|(Pz)(t)-(Pz^*)(t)\|_X\\ &\leq \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \|f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s))) -f(s,y_s+\overline{z}_s^*,B(y(s)\\ &\quad +\overline{z}^*(s)))\|_X\,ds +\sum_{i=1}^k(t-t_i)\|I_i(z(t_i^{-}))-I_i(z^*(t_i^{-}))\|_X \\ &\quad +\sum_{i=1}^k\|Q_i(z(t_i^{-}))-Q_i(z^*(t_i^{-}))\|_X\\ &\quad +\frac{bt}{a+b}\sum_{i=1}^m\|I_i(z(t_i^{-}))-I_i(z^*(t_i^{-}))\|_X \\ &\quad +\frac{t}{a+b}\int_0^T\|q(y(s)+\overline{z}(s))-q(y(s) +\overline{z}^*(s))\|_X\,ds\\ &\quad +\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s))) -f(s,y_s+\overline{z}_s^*,B(y(s)\\ &\quad +\overline{z}^*(s)))\|_X\,ds\\ &\leq\Big[\frac{(a+(1+\alpha)b)(\mu_1l+\mu_2B^*)T^\alpha}{(a+b) \Gamma(\alpha+1)}+\frac{(a+2b)LTm+L_qT^2}{a+b} +\overline{L}m\Big]\|z-z^*\|_{\mathfrak{B}_h''}. \end{align*} Thus for all $t\in[0,T]$, we have the estimate \begin{align*} &\|P(z)-P(z^*)\|_{B_h''}\\ &\leq \Big[\frac{(a+(1+\alpha)b)(\mu_1l+\mu_2B^*)T^\alpha}{(a+b) \Gamma(\alpha+1)}+\frac{(a+2b)LTm+L_qT^2}{a+b} +\overline{L}m\Big]\|z-z^*\|_{B_h''}\\ &\leq\Lambda\|z-z^*\|_{\mathfrak{B}_h''}. \end{align*} Since $\Lambda<1$, the map $P$ is a contraction map and has a unique fixed point $z\in \mathfrak{B}_h''$, which is obviously a solution of the system \eqref{ME} on $(-\infty,T]$. This completes the proof of the theorem. \end{proof} Our second existence result is based on the following Krasnoselkii's fixed point theorem. \begin{theorem} \label{kraski} Let $B$ be a closed convex and nonempty subset of a Banach space $X$. Let $P$ and $Q$ be two operator such that (i) $Px+Qy\in B$, whenever $x,y\in B$. (ii) $P$ is compact and continuous. (iii) $Q$ is a contraction mapping. Then there exists $z\in B$ such that $z=Pz+Qz$. \end{theorem} \begin{theorem} Suppose that assumptions {\rm (H1), (H4), (H5)} are satisfied with $$ \Delta=\frac{(\mu_1l+\mu_2B^*)(a+(1+\alpha)b)T^\alpha}{(a+b) \Gamma(\alpha+1)}<1. $$ Then \eqref{ME} has at least one solution on $(-\infty,T]$. \end{theorem} \begin{proof} Choose \begin{align*} r&\geq \Big[\|\phi(0)\|+(\rho T+\Omega)m+\frac{(b\rho m+TC)T}{a+b}\\ &\quad +\frac{(\mu_1^*(\|\phi\|+lr)+\mu_2^*B^*r)(a+(1+\alpha)b)T^\alpha}{(a+b) \Gamma(\alpha+1)}\Big]. \end{align*} Define $B_r=\{z\in \mathfrak{B}_h'':\|z\|_{\mathfrak{B}_h''}\leq r\}$, then $B_r$ is a bounded, closed convex subset in $\mathfrak{B}_h''$. Let $P_1:B_r\to B_r$ and $P_2:B_r\to B_r$ be defined as \begin{align*} (P_1z)(t) &= \phi(0)+\frac{t}{a+b}\int_0^Tq(y(s)+\overline{z}(s))ds +\sum_{i=1}^kQ_i(z(t_i^-))\\ &\quad -\frac{bt}{a+b}\sum_{i=1}^mI_i(z(t_i^-)) +\sum_{i=1}^k(t-t_i)I_i(z(t_i^-)),\quad t\in J_k.\\ (P_2z)(t) &= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,y_s+\overline{z}_s,B(y(s) +\overline{z}(s)))ds\\ &\quad -\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s)))ds,\quad t\in J_k, \end{align*} where $J_0=[0,t_1]$ and $J_k=(t_k,t_{k+1}]$, $k=1,\dots,m$. Now, we proceed the proof in following steps: \textbf{Step 1.} Let $z,z^*\in B_r$, then we show that $P_1z+P_2z^*\in B_r$ for $t\in J_k,k=0,1,\dots,m$. We have \begin{equation}\label{L2l} \begin{aligned} &\|(P_1z)(t)+(P_2z^*)(t)\|_X\\ &\leq \|\phi(0)\|_X+\frac{t}{a+b}\int_0^T\|q(y(s)+\overline{z}(s))\|_X\,ds\\ &\quad +\sum_{i=1}^k\|Q_i(z(t_i^-))\|_X +\frac{bt}{a+b}\sum_{i=1}^m\|I_i(z(t_i^-))\|_X+\sum_{i=1}^k(t-t_i)\|I_i(z(t_i^-)) \|_X\\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \|f(s,y_s+\overline{z}^*_s,B(y(s)+\overline{z}^*(s)))\|_X\,ds\\ &\quad+\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|f(s,y_s+\overline{z}^*_s,B(y(s)+\overline{z}^*(s)))\|_X\,ds, \end{aligned} \end{equation} we estimate the inequality \eqref{L2l}, by using (H4) and (H5), as \begin{align*} \|(P_1z)(t)+(P_2z^*)(t)\|_X &\leq \Big[\|\phi(0)\|+(\rho T+\Omega)m+\frac{(b\rho m+TC)T}{a+b}\\ &\quad +\frac{(\mu_1^*(\|\phi\|+lr)+\mu_2^*B^*r)(a+(1+\alpha)b)T^\alpha} {(a+b)\Gamma(\alpha+1)}\Big], \end{align*} which implies that $\|P_1z+P_2z^*\|_{B_h''}\leq r$. \textbf{Step 2.} Now, we shall show that the mapping $(P_1z)(t)$ is continuous on $B_r$. For this purpose, let $\{z^n\}_{n=1}^\infty$ be a sequence in $B_r$ with $\lim z^n\to z$ in $B_r$. Then for $t\in J_k,k=0,1,\dots,m$, we have \begin{align*} &\|(P_1z^n)(t)-(P_1z)(t)\|_X\\ &\leq\frac{t}{a+b}\int_0^T\|q(y(s)+\overline{z}^n(s))-q(y(s) +\overline{z}(s))\|_X\,ds\\ &+\sum_{i=1}^k\|Q_i(z^n(t_i^-))-Q_i(z(t_i^-))\|_X +\frac{bt}{a+b}\sum_{i=1}^m\|I_i(z^n(t_i^-))-I_i(z(t_i^-))\|_X\\ &+\sum_{i=1}^k(t-t_i)\|I_i(z^n(t_i^-))-I_i(z(t_i^-))\|_X. \end{align*} Since the functions $q,Q_k, I_k$, $k=0,1,\dots,m$, are continuous, hence $\lim_{n\to\infty}P_1z^n=P_1z$ in $B_r$. Which implies that the mapping $P_1$ is continuous on $B_r$. \textbf{Step 3} $(P_1z)(t)$ is uniformly bounded follows by the following inequality. For $t\in J_k,k=0,1,\dots,m$, we have \begin{align*} \|(P_1z)(t)\|_X &\leq \|\phi(0)\|_X+\frac{t}{a+b}\int_0^T\|q(y(s)+\overline{z}(s))\|_X\,ds +\sum_{i=1}^k\|Q_i(z(t_i^-))\|_X\\ &\quad +\frac{bt}{a+b}\sum_{i=1}^m\|I_i(z(t_i^-))\|_X +\sum_{i=1}^k(t-t_i)\|I_i(z(t_i^-))\|_X\\ &\leq \|\phi(0)\|_X+\frac{T(TC+b\rho m)}{a+b}+\Omega m+\rho mT. \end{align*} \textbf{Step 4} To show that $P_1(B_r)$ is equicontinuous. Let $\tau_1,\tau_2\in J_k$, $t_k\leq \tau_1<\tau_2\leq t_{k+1}$, $k=0,1,\dots,m$, $z\in B_r$, we have \begin{align*} \|(P_1z)(\tau_2)-(P_2z)(\tau_1)\|_X &\leq (\tau_2-\tau_1)[\frac{1}{a+b}\int_0^T\|q(y(s)+\overline{z}(s))\|_X\,ds\\ &\quad +\sum_{i=1}^k\|I_i(z(t_i^-))\|_X +\frac{b}{a+b}\sum_{i=1}^m\|I_i(z(t_i^-))\|_X], \end{align*} which implies that $P_1(B_r)$ is equicontinuous. Finally, combing Step 2 to Step 4 together with the Ascol's theorem, we conclude that the operator $P_1$ is a compact. \textbf{Step 5.} Now, we show that $P_2$ is a contraction mapping. Let $z,z^*\in B_r$ and $t\in J_k,\;k=0,1,\dots,m$, we have \begin{align*} &\|(P_2z)(t)-(P_2z^*)(t)\|_X\\ &\leq\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \|f(s,y_s+\overline{z}_s,B(y(s)+\overline{z}(s))) -f(s,y_s+\overline{z}^*_s,B(y(s)\\ &\quad +\overline{z}^*(s)))\|_X\,ds +\frac{bt}{a+b}\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|f(s,y_s+\overline{z}_s,B(y(s)\\ &\quad +\overline{z}(s))) -f(s,y_s+\overline{z}^*_s,B(y(s)+\overline{z}^*(s)))\|_X\,ds\\ &\leq \frac{(\mu_1l+\mu_2B^*)(a+(1+\alpha)b)T^\alpha}{(a+b)\Gamma(\alpha+1)} \|z-z^*\|_{B_h''}\\ &\leq\Delta\|z-z^*\|_{\mathfrak{B}_h''}, \end{align*} where $$ \Delta=\frac{(\mu_1l+\mu_2B^*)(a+(1+\alpha)b)T^\alpha}{(a+b)\Gamma(\alpha+1)}. $$ As $\Delta<1$, then $P_2$ is a contraction map. Thus all the assumptions of the Theorem \ref{kraski} are satisfied and the conclusion of the Theorem \ref{kraski} implies that the system \eqref{ME} has at least one solution on $(-\infty,0]$. This completes the proof of the theorem. \end{proof} \section{Application} \label{sec:4} We consider the model \begin{equation}\label{ex} \begin{gathered} \begin{aligned} {}^CD_t^{3/2}u(t) &=\frac{1}{(t+9)^2}\int_{-\infty}^0e^{2\theta}\sin(\|u(t+\theta)\|_X)d\theta\\ &\quad +\frac{1}{(t+7)^2}\sin\big(\|\int_0^t(t-s)u(s)ds\|_X\big), \quad t\in[0,1],\; t\neq t_i, i=1,2,3, \end{aligned} \\ u(t)=\phi(t),\quad t\in(-\infty,0], \\ \Delta u(t_i)=\int_{-\infty}^0e^{2\theta}\frac{\|u(t_i+\theta)\|_X} {25+\|u(t_i+\theta)\|_X}d\theta, \\ \Delta u'(t_i)=\int_{-\infty}^0e^{2\theta}\frac{\|u(t_i+\theta)\|_X} {27+\|u(t_i+\theta)\|_X}d\theta,\\ u'(0)+u'(1)=\int_0^1\sin({1\over2}\|u(s)\|_X)ds, \end{gathered} \end{equation} where $X$ is a real Banach space, $0