\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 273, 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-2013/273\hfil Fractional integro-differential equations] {Impulsive neutral fractional integro-differential equations with state dependent delays and integral condition} \author[J. Dabas, G. R. Gautam \hfil EJDE-2013/273\hfilneg] {Jaydev Dabas, Ganga Ram Gautam} % in alphabetical order \address{Jaydev Dabas \newline Department of Applied Science and Engineering, IIT Roorkee, Saharanpur Campus \\ Saharanpur-247001, India} \email{jay.dabas@gmail.com} \address{Ganga Ram Gautam \newline Department of Applied Science and Engineering, IIT Roorkee, Saharanpur Campus \\ Saharanpur-247001, India} \email{gangaiitr11@gmail.com} \thanks{Submitted May 3, 2013. Published December 17, 2013.} \subjclass[2000]{26A33, 34K05, 34A12, 34A37, 26A33} \keywords{Fractional order differential equation; nonlocal condition; \hfill\break\indent contraction; impulsive condition} \begin{abstract} In this article, we establish the existence of a solution for an impulsive neutral fractional integro-differential state dependent delay equation subject to an integral boundary condition. The existence results are proved by applying the classical fixed point theorems. An example is presented to demonstrate the application of the results established. \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:1} Let $X$ be a Banach space and $PC_t:=PC([-d,t];X), d>0,0\le t\le T<\infty$, be a Banach space of all such functions $\phi:[-d,t]\to X$, which are continuous everywhere except for a finite number of points $t_i$, $i=1,2,\dots,m$, at which $\phi(t_i^+)$ and $\phi(t_i^-)$ exists and $\phi(t_i)=\phi(t_i^-)$, endowed with the norm $$ \|\phi\|_t=\sup_{-d\leq s\leq t}\|\phi(s)\|_X,\;\phi\in PC_t, $$ where $\|\cdot \|_X$ is the norm in $X$. In this article we study an impulsive neutral fractional integro-differential equation of the form \begin{gather} \label{1.1} \begin{aligned} &D^{\alpha}_t \Big[x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\Big]\\ &=f(t,x_{\rho(t,x_t)}, B(x)(t)), \quad t\in J=[0,T],\;T<\infty,\; t\neq t_k, \end{aligned} \\ \label{1.2} \Delta x(t_k)=I_k(x(t_k^-)),\quad \Delta x'(t_k)=Q_k(x(t_k^-)),\quad k=1,2,\dots,m,\\ \label{1.4} x(t)=\phi (t),\quad t\in[-d,0],\\ \label{1.5} ax'(0)+bx'(T)=\int_0^Tq(x(s))ds,\quad a+b\neq 0,\; b\neq0, \end{gather} where $x'$ denotes the derivative of $x$ with respect to $t$ and $D^{\alpha}_t$, $\alpha \in (1,2)$ is Caputo's derivative. The functions $f:J\times PC_0 \times X\to X$, $g:J\times PC_0\to X$, and $q:X\to X$ are given continuous functions where $PC_0=PC([-d,0],X)$ and for any $x\in PC_T=PC([-d,T],X)$, $t\in J$, we denote by $x_t$ the element of $PC_0$ defined by $x_t(\theta)=x(t+\theta),\;\theta\in[-d,0]$. In the impulsive conditions for $0= t_0 0$, of a function $f:\mathbb{R}^+\to \mathbb{R}$ and $f\in L^1(\mathbb{R}^+,X)$ is defined by \begin{equation} \label{def2.1e} 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{lemma}[ \cite{29}]\label{fr} For $\alpha >0$, the general solution of fractional differential equations $D^{\alpha}_tx(t)=0$ is given by $x(t)=c_0+c_1t+c_2t^2+c_3t^3+\dots+c_{n-1}t^{n-1}$ where $c_i\in \mathbb{R}$, $i=0,1,\dots,n-1$, $n=[\alpha]+1$ and $[\alpha]$ represent the integral part of the real number $\alpha$. \end{lemma} \begin{lemma}[{\cite[Lemma 2.6]{20}}] \label{fr1} 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 following fractional integral equation \begin{equation} x(t)=\int_0^t{(t-s)^{\alpha-1}\over \Gamma(\alpha)}h(s)ds-\int_0^w{(w-s)^{\alpha-1}\over \Gamma(\alpha)}h(s)ds+x_0-c(t-w), \end{equation} if and only if $x$ is a solution of the following fractional Cauchy problem \begin{equation} D^{\alpha}_tx(t)=h(t),\quad t\in J,\; x(w)=x_0,\;w\ge0. \end{equation} \end{lemma} As a consequence of Lemma \ref{fr} and Lemma \ref{fr1} we have the following result. \begin{lemma} Let $\alpha\in (1,2)$ and $f:J\times PC_0\times X\to X$ be continuously differentiable function. A piecewise continuous differential function $x(t):(-d,T]\to X$ is a solution of system \eqref{1.1}--\eqref{1.5} if and only if $x$ satisfied the integral equation \begin{equation} \label{sol} x(t)=\begin{cases} \phi(t), \quad t\in[-d,0],\\ \phi(0)-\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds\\ -\sum_{i=1}^kQ_i(x(t_i^-)) +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds \\ -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\} \\ +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \quad t\in[0,t_1],\\ \dots \\ \phi(0)+\sum_{i=1}^kI_i(x(t_i^-))+\sum_{i=1}^k(t-t_i)Q_i(x(t_i^-))\\ -\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds\\ -\sum_{i=1}^kQ_i(x(t_i^-)) +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds \\ -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\} \\ +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \quad t\in(t_k, t_{k+1}]. \end{cases} \end{equation} \end{lemma} \begin{proof} If $t\in [0,t_1]$, then \begin{equation} \label{t1} \begin{gathered} D^{\alpha}_t [x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds]= f(t,x_{\rho(t,x_t)}, B(x)(t)),\\ x(t)=\phi (t),\;t\in[-d,0]. \end{gathered} \end{equation} Taking the Riemann-Liouville fractional integral of \eqref{t1} and using the Lemma \ref{fr1}, we have \begin{equation} \label{1.7} \begin{gathered} x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ =a_0+b_0t+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \end{gathered} \end{equation} using the initial condition, we get $a_0=\phi(0)$, then \eqref{1.7} becomes \begin{equation} \label{g1} \begin{aligned} &x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &=\phi(0)+b_0t+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{aligned} \end{equation} Similarly, if $t\in (t_1, t_2]$, then \begin{gather} \label{1.9} D^{\alpha}_t [x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds]= f(t,x_{\rho(t,x_t)}, B(x)(t))\\ \label{g8} x(t_1^+)=x(t_1^-)+I_1(x(t_1^-)),\\ \label{g9} x'(t_1^+)=x'(t_1^-)+Q_1(x(t_1^-)). \end{gather} Again apply the Riemann-Liouville fractional integral operator on \eqref{1.9} and using the lemma \ref{fr1}, we obtain \begin{equation} \label{g10} \begin{aligned} & x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ & = a_1+b_1t+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \end{aligned} \end{equation} rewrite \eqref{g10} as \begin{equation} \label{g11} %\label{1.10a} \begin{aligned} &x(t_1^+)+\int_0^{t_1} \frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &= a_1+b_1t_1+\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x_{\rho(s,x_s)},B(x)(s))ds, \end{aligned} \end{equation} due to impulsive condition \eqref{g8} and the fact that $x(t_1)=x(t_1^-)$, we may write \eqref{g11} as \begin{equation} \label{g12} %\label{1.10a2} \begin{aligned} & x(t_1)+I_1(x(t_1^-))+\int_0^{t_1} \frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &= a_1+b_1t_1+\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,x_{\rho(s,x_s)},B(x)(s))ds. \end{aligned} \end{equation} Now from \eqref{g1}, we have \begin{equation} \label{g13} \begin{aligned} &x(t_1)+\int_0^{t_1} \frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)} g(s,x_{\rho(s,x_s)})ds\\& =\phi(0)+b_0t_1+\int_0^{t_1} \frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)},B(x)(s))ds. \end{aligned} \end{equation} From \eqref{g12} and \eqref{g13}, we get $a_1=\phi(0)+b_0t_1-b_1t_1+I_1(x(t^-_1))$, hence \eqref{g11} can be written as \begin{equation} \label{g14} \begin{aligned} & x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &=\phi(0)+b_0t_1+b_1(t-t_1)+I_1(x(t^-_1)) +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)},B(x)(s))ds. \end{aligned} \end{equation} On differentiating \eqref{g10} with respect to $t$ at $t=t_1$, and incorporate second impulsive condition \eqref{g9}, we obtain \begin{equation} \label{g15} \begin{aligned} &x'(t_1^-)+Q_1(x(t_1^-))+\int_0^{t_1} \frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds\\ &=b_1 +\int_0^{t_1}\frac{(t_1-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,x_{\rho(s,x_s)},B(x)(s))ds, \end{aligned} \end{equation} Now differentiating \eqref{g1}, with respect to $t$ at $t=t_1$, we get \begin{equation} \label{g16} \begin{aligned} & x'(t_1)+\int_0^{t_1} \frac{(t_1-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds\\ &=b_0+\int_0^{t_1}\frac{(t_1-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{aligned} \end{equation} From \eqref{g15} and \eqref{g16}, we obtain $b_1=b_0+Q_1(x(t_1^-))$. Thus, \eqref{g14} become \begin{equation} \label{1.8} \begin{aligned} & x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &=\phi(0)+b_0t+I_1(x(t_1^-)) +(t-t_1)Q_1(x(t_1^-))\\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{aligned} \end{equation} Similarly, for $t\in (t_2,t_3]$, we can write the solution of the problem as \begin{align*} %\label{1.8} & x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &=\phi(0)+b_0t+I_1(x(t_1^-))+I_2(x(t_2^-)) +(t-t_1)Q_1(x(t_1^-))+(t-t_2)Q_2(x(t_2^-))\\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{align*} In general, if $t\in (t_k, t_{k+1}]$, then we have the result \begin{equation} \label{1.8b} \begin{aligned} &x(t)+\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds\\ &=\phi(0)+b_0t+\sum_{i=1}^kI_i(x(t_i^-)) +\sum_{i=1}^k(t-t_i)Q_i(x(t_i^-)) \\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{aligned} \end{equation} Finally, we use the integral boundary condition $ax'(0)+bx'(T)=\int_0^Tq(x(s))ds$, where $x'(0)$ calculated from \eqref{g1} and $x'(T)$ from \eqref{1.8}. On simplifying, we get the following value of the constant $b_0$, \begin{align*} b_0&=\frac{b}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds -\sum_{i=1}^mQ_i(x(t_i^-)) +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds \\ &\quad -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\}. \end{align*} On summarizing, we obtain the desired integral equation \eqref{sol}. Conversely, assuming that $x$ satisfies \eqref{sol}, by a direct computation, it follows that the solution given in \eqref{sol} satisfies system \eqref{1.1}--\eqref{1.5}. This completes the proof of the lemma. \end{proof} \section{Existence result}\label{exir:3} The function $ \rho :J\times PC_0 \to [-d ,T]$ is continuous and $\phi (0)\in PC_0$. Let the function t$\to {\varphi}_t$ be well defined and continuous from the set $\Re(\rho^-)=\{\rho(s,\psi ):(s,\psi)\in [0,T]\times PC_0\}$ into $PC_0$. Further, we introduce the following assumptions to establish our results. \begin{itemize} \item[(H1)] There exist positive constants $L_{f1}, L_{f2}, L_q$ and $L_g$, such that \begin{gather*} \|f(t,\psi ,x)-f(t,\chi, y)\|_X \leq L_{f1}\|\psi -\chi\|_{PC_0}+L_{f2}\|x-y\|_X, \\ \|g(t,\psi)-g(t,\chi)\|_X \leq L_g\|\psi-\chi\|_{PC_0}, t \in J, \forall\; \psi,\chi \in PC_0,\;\forall\;x, y \in X,\\ \|q(x)-q(y)\|_X \leq L_q\|x-y\|_X,\;\forall x,y\in X. \end{gather*} \item[(H2)] There exist positive constants $L_Q, L_I, L_q$, such that \[ \|Q_k(x)-Q_k(y)\|_X\leq L_Q\|x-y\|_X,\quad \|I_k(x)-I_k(y)\|_X\leq L_I\|x-y\|_X. \] \item[(H3)] The functions $Q_k, I_k, q$ are bounded continuous and there exist positive constants $C_1, C_2, C_3$, such that \[ \|Q_k(x)\|_X\leq C_1, \quad \|I_k(x)\|_X\leq C_2,\quad \|q(x)\|_X\leq C_3,\quad \forall x\in X. \] \end{itemize} Our first result is based on the Banach contraction theorem. \begin{theorem}\label{thm} Let the assumptions {\rm (H1)--(H2)} are satisfied with \begin{align*} \triangle&= \Big\{m(L_I+TL_Q)+\frac{T^{\alpha}L_g}{\Gamma (\alpha+1)} +\frac{bT}{a+b}\Big(\frac{TL_q}{b}+mL_Q\\ &\quad +\frac{T^{\alpha-1}L_g}{\Gamma(\alpha)} +\frac{T^{\alpha-1}(L_{f1}+L_{f2}B^*)}{\Gamma (\alpha)}\Big)+\frac{T^{\alpha}(L_{f1}+L_{f2}B^*)}{\Gamma (\alpha+1)}\Big\}<1. \end{align*} Then \eqref{1.1}--\eqref{1.5} has a unique solution. \end{theorem} \begin{proof} We transform problem \eqref{1.1}--\eqref{1.5} into a fixed point problem. Consider the operator $P:PC_T\to PC_T$ defined by \[ Px(t)=\begin{cases} \phi(t), \quad t\in [-d,0],\\ \phi(0)-\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds \\ -\sum_{i=1}^mQ_i(x(t_i^-)) +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds\\ -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\}\\ +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \quad t\in[0,t_1]\\ \dots \\ \phi(0)+\sum_{i=1}^kI_i(x(t_i^-))+\sum_{i=1}^k(t-t_i)Q_i(x(t_i^-))\\ -\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds-\sum_{i=1}^mQ_i(x(t_i^-))\\ +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\}\\ +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds, \quad t\in(t_k, t_{k+1}]. \end{cases} \] Let $x, x^* \in PC_T$ and $t\in [0,t_1]$. Then \begin{align*} & \|P(x)-P(x^*)\|_X\\ &\leq \int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|g(s,x_{\rho(s,x_s)}) -g(s,x^*_{\rho(s,x^*_s)})\|_Xds\\ &\quad +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^T\|q(x(s))-q(x^*(s))\|_Xds +\sum_{i=1}^m\|Q_i(x(t_i^-))-Q_i(x^*(t_i^-))\|_X\\ &\quad +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}\|g(s,x_{\rho(s,x_s)}) -g(s,x^*_{\rho(s,x^*_s)})\|_Xds\\ &\quad +\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}\|f(s,x_{\rho(s,x_s)}, B(x)(s))-f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds\Big\} \\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|f(s,x_{\rho(s,x_s)}, B(x)(s))-f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds\\ &\leq \Big\{\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_g+\frac{bT}{a+b}\Big(\frac{T}{b}L_q+mL_Q+\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_g \\ & \quad +\frac{T^{\alpha-1}}{\Gamma (\alpha)}(L_{f1}+L_{f2}B^*)\Big)+\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}+L_{f2}B^*)\Big\}\|x-x^*\|_{PC_T}. \end{align*} In a similar way for $t\in (t_k,t_{k+1}]$, we have \begin{align*} & \|P(x)-P(x^*)\|_X\\ & \leq \sum_{i=1}^k\|I_i(x(t_i^-))-I_i(x^*(t_i^-))\|_X +\sum_{i=1}^k(t-t_i)\|Q_i(x(t_i^-))-Q_i(x^*(t_i^-))\|_X \\ &\quad\times \int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|g(s,x_{\rho(s,x_s)}) -g(s,x^*_{\rho(s,x^*_s)})\|_Xds\\ & +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^T\|q(x(s))-q(x^*(s))\|_Xds +\sum_{i=1}^m\|Q_i(x(t_i^-))-Q_i(x^*(t_i^-))\|_X \\ &\quad +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}\|g(s,x_{\rho(s,x_s)}) -g(s,x^*_{\rho(s,x^*_s)})\|_Xds\\ & \quad +\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}\|f(s,x_{\rho(s,x_s)}, B(x)(s))-f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds\Big\} \\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|f(s,x_{\rho(s,x_s)}, B(x)(s))-f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds\\ &\leq \Big\{mL_I+mTL_Q+\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_g+\frac{bT}{a+b} \Big(\frac{T}{b}L_q+mL_Q+\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_g \\ & \quad +\frac{T^{\alpha-1}}{\Gamma (\alpha)}(L_{f1}+L_{f2}B^*)\Big)+\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}+L_{f2}B^*)\Big\}\|x-x^*\|_{PC_T}\\ &\leq \Delta \|x-x^*\|_{PC_T}. \end{align*} Since $\Delta<1$, implies that the map $P$ is a contraction map and therefore has a unique fixed point $x\in PC_T$, hence system \eqref{1.1}--\eqref{1.5} has a unique solution on the interval $[-d,T]$. This completes the proof of the theorem. \end{proof} Our second result is based on 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 operators such that \begin{itemize} \item[(i)] $Px+Qy\in B$, whenever $x,y\in B$. (ii) $P$ is compact and continuous. \item[(iii)] $Q$ is a contraction mapping. Then there exists $z\in B$ such that $z=Pz+Qz$. \end{itemize} \end{theorem} \begin{theorem}\label{k2} Let the function $f, g$ be continuous for every $t \in [0, T]$, and satisfy the assumptions {(H1)--(H3)} with \begin{align*} \Delta &= \Big\{\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_g+\frac{bT}{a+b}\Big(\frac{T}{b}L_q+\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_g+\frac{T^{\alpha-1}}{\Gamma (\alpha)}(L_{f1}+L_{f2}B^*)\Big)\\ &\quad +\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}+L_{f2}B^*)\Big\}<1. \end{align*} Then system \eqref{1.1}--\eqref{1.5} has at least one solution on $[-d,T]$. \end{theorem} \begin{proof} Choose \begin{align*} r&\geq \Big[\|\phi(0)\|+mL_Ir+mTL_Qr+\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_gr+\frac{bT}{a+b}(\frac{T}{b}L_qr+mL_Qr\\ &\quad +\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_gr +\frac{T^{\alpha-1}}{\Gamma(\alpha)} (L_{f1}r+L_{f2}B^*r))+\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}r+L_{f2}B^*r)\Big]. \end{align*} Define $PC_T^r=\{x\in PC_T:\|x\|_{PC_T}\leq r\}$, then $PC_T^r$ is a bounded, closed convex subset in $PC_T$. Consider the operators $N:PC_T^r\to PC_T^r$ and $P:PC_T^r\to PC_T^r$ for $t\in J_k=(t_k, t_{k+1}]$, defined by \begin{gather} N(x)= \phi(0)+\sum_{i=1}^kI_i(x(t_i^-))+\sum_{i=1}^k(t-t_i)Q_i(x(t_i^-)) -\frac{bt}{a+b}\sum_{i=1}^mQ_i(x(t_i^-)) \\ \begin{aligned} P(x) &= \frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^Tq(x(s))ds+\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}g(s,x_{\rho(s,x_s)})ds \\ & \quad -\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds\Big\}\\ &\quad -\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s,x_{\rho(s,x_s)})ds +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x_{\rho(s,x_s)}, B(x)(s))ds. \end{aligned} \end{gather} We complete the proof in the following steps: \noindent\textbf{Step 1.} Let $x, x^* \in PC_T^r$ then, \begin{align*} \|N(x)+P(x^*)\|_X & \leq \|\phi(0)\|_X+\sum_{i=1}^k\|I_i(x(t_i^-))\|_X +\sum_{i=1}^k(t-t_i)\|Q_i(x(t_i^-))\|_X\\ &\quad +\frac{bt}{a+b}\sum_{i=1}^m\|Q_i(x(t_i^-))\|_X +\frac{bt}{a+b}\Big\{\frac{1}{b}\int_0^T\|q(x^*(s))\|_Xds \\ &\quad +\int_0^T \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|g(s,x^*_{\rho(s,x^*_s)})\|_Xds \\ &\quad +\int_0^T\frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} \|f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds\Big\} \\ &\quad +\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|g(s,x^*_{\rho(s,x^*_s)})\|_Xds \\ &\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\|f(s,x^*_{\rho(s,x^*_s)}, B(x^*)(s))\|_Xds \\ & \leq\Big[\|\phi(0)\|+mC_2+mTC_1+\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_gr+\frac{bT}{a+b}(\frac{T}{b}C_3\\ &\quad +mC_1+\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_gr +\frac{T^{\alpha-1}}{\Gamma (\alpha)}(L_{f1}r+L_{f2}B^*r))\\ &\quad +\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}r+L_{f2}B^*r)\Big]\leq r. \end{align*} Which shows that $PC_T^r$ is closed with respect to both the maps. \noindent\textbf{Step 2.} $N$ is continuous. Let $x_n \to x$ be sequence in $PC_T^r$, then for each $t\in J_k$ \begin{align*} & \|N(x_n)-N(x)\|_X\\ & \leq\sum_{i=1}^k\|I_i(x_n(t_i^-))-I_i(x(t_i^-))\|_X +\sum_{i=1}^k(t-t_i)\|Q_i(x_n(t_i^-))-Q_i(x(t_i^-))\|_X\\ &\quad +\frac{bt}{a+b}\sum_{i=1}^m\|Q_i(x_n(t_i^-))-Q_i(x(t_i^-))\|_X. \end{align*} Since the functions $Q_k$ and $I_k$, $k=1,\dots,m$, are continuous, hence $\|N(x_n)-N(x)\|\to 0$, as $n\to\infty$. Which implies that the mapping $N$ is continuous on $PC_T^r$. \noindent\textbf{Step 3.} The fact that the mapping $N$ is uniformly bounded is a consequence of the following inequality. For each $t\in J_k$, $k=0,1,\dots,m$ and for each $x\in PC_T^r$, we have \begin{align*} \|N(x)\|_X &\leq \|\phi(0)\|_X+\sum_{i=1}^k\|I_i(x(t_i^-))\|_X +\sum_{i=1}^k(t-t_i)\|Q_i(x(t_i^-))\|_X \\ &\quad +\frac{bt}{a+b}\sum_{i=1}^m\|Q_i(x(t_i^-))\|_X\\ &\leq \|\phi(0)\|+mC_2+mTC_1+\frac{bT}{a+b}mC_1. \end{align*} \noindent\textbf{Step 4.} Now, to show that $N$ is equi-continuous, let $l_1,l_2\in J_k$, $t_k\leq l_10$, and choose $PC^{\gamma}$ as $$ PC^{\gamma}=\{\phi \in PC((0,\infty], X): \lim_{t \to-d}e^{\gamma t }\phi (t)\,\text{exist}\} $$ with the norm $\|\phi\|_{\gamma}=\sup_{t \in (0, \infty]}e^{\gamma t }|\phi (t)|, \, \phi \in PC^{\gamma}$. We set \begin{gather*} \rho(t,\varphi )= t-\sigma (\varphi (0)),\quad (t,\varphi )\in J\times PC^{\gamma},\\ f(t,\varphi)= {{e^t(\varphi)}\over{25+(\varphi)^2}}, \quad (t,\varphi )\in J\times PC^{\gamma},\\ g(t,\varphi)= {\varphi\over 47}ds,\quad \varphi\in PC^{\gamma}, \\ B(x)(t)= \int _0^t\cos(t-s){x e^s\over(4+x )}ds,\quad (t,x )\in I\times PC^{\gamma}, \\ Q_k(x(t_k))= \int_{-d}^{t_i}\frac{\gamma _{i}(t_i-s)x(s)}{25}ds,\\ I_k(x(t_k))= \int_{-d}^{t_i}\frac{\gamma _{i}(t_i-s)x(s)}{9}ds. \end{gather*} We can see that all the assumptions of Theorem \ref{thm} are satisfied with \begin{gather*} |f(t,\varphi)-f(t,\chi)| \leq e^t{\|\varphi -\chi \|\over25} \quad \forall t\in J,\varphi,\chi \in PC^{\gamma}, \\ |B(x)-B(y)| \leq e^t {\|x -y \|\over4} \quad \forall t\in J,\; x,y \in PC^{\gamma}, \\ |g(t,\varphi)-g(t,\chi)| \leq \frac{1}{47}\|\varphi-\chi\|,\quad \forall t\in J,\; \varphi,\chi \in PC^{\gamma},\\ |Q_k(x(t_k))-Q_k(y(t_k))| \leq {\gamma }^*\frac{1}{25}\|x-y\|,\quad x,y\in X,\\ |I_k(x(t_k))-I_k(y(t_k))| \leq {\gamma }^*\frac{1}{9}\|x-y\|,\quad x,y\in X,\\ |q(x)-q(y)| \leq \frac{1}{4}\|x-y\|,\quad x,y\in X. \end{gather*} Further, we observe that \begin{align*} &\Big\{mL_I+mTL_Q+\frac{T^{\alpha}}{\Gamma (\alpha+1)}L_g+\frac{bT}{a+b}\Big(\frac{T}{b}L_q+mL_Q+\frac{T^{\alpha-1}}{\Gamma (\alpha)}L_g \\ &\quad +\frac{T^{\alpha-1}}{\Gamma (\alpha)}(L_{f1}+L_{f2}B^*)\Big)+\frac{T^{\alpha}}{\Gamma (\alpha+1)}(L_{f1}+L_{f2}B^*)\Big\}\\ &\approx 0.513 \gamma^*+0.534<1. \end{align*} We fix $\gamma^*=\int_{-d}^t\gamma_i(t_i-s)ds<0$, $0