\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 201, pp. 1--26.\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-2012/201\hfil Existence of solutions] {Existence of solutions to singular fractional differential systems with impulses} \author[X. Liu, Y. Liu \hfil EJDE-2012/201\hfilneg] {Xingyuan Liu, Yuji Liu} % in alphabetical order \address{Xingyuan Liu \newline Department of Mathematics, Shaoyang University \\ Hunan Shaoyang, 422000, China} \email{liuxingyuan888@sohu.com} \address{Yuji Liu \newline Department of Mathematics, Guangdong University of Business Studies\\ Guangzhou 510000, China} \email{liuyuji888@sohu.com} \thanks{Submitted March 14, 2012. Published November 15, 2012.} \subjclass[2000]{92D25, 34A37, 34K15} \keywords{Solution; singular fractional differential system; \hfill\break\indent impulsive boundary value problems; Leray-Schauder nonlinear alternative} \begin{abstract} By constructing a weighted Banach space and a completely continuous operator, we establish the existence of solutions for singular fractional differential systems with impulses. Our results are proved using the Leray-Schauder nonlinear alternative, and are illustrated with examples. \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} Fractional differential equation is a generalization of ordinary differential equation to arbitrary non integer orders. The origin of fractional calculus goes back to Newton and Leibniz in the seventieth century. Recent investigations have shown that many physical systems can be represented more accurately through fractional derivative formulation \cite{1}. Fractional differential equations, therefore find numerous applications in the field of visco-elasticity, feed back amplifiers, electrical circuits, electro analytical chemistry, fractional multipoles, neuron modelling encompassing different branches of physics, chemistry and biological sciences \cite{2}. There have been many excellent books and monographs available on this field \cite{3}, \cite{4} and \cite{5}, the authors gave the most recent and up-to-date developments on fractional differential and fractional integro-differential equations with applications involving many different potentially useful operators of fractional calculus. The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Processes with such a character arise naturally and often, for example, phenomena studied in physics, chemical technology, population dynamics, biotechnology and economics. For an introduction of the basic theory of impulsive differential equation, we refer the reader to \cite{6}. Recently, the authors in papers \cite{8,9,10,13,7,11,12} and the survey paper \cite{14} studied the existence of solutions of the different initial value problems for the impulsive fractional differential equations. In \cite{13}, the author studied the existence of solutions of the following impulsive anti-periodic boundary value problem \begin{equation} \label{e1} \begin{gathered} ^cD_{0^+}^qx(t)=f(t,x(t)),\quad 10$ of a function $g:(0,\infty)\to \mathbb{R}$ is given by $$ I_{0+}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}g(s)ds, $$ provided that the right-hand side exists. \end{definition} \begin{definition}[\cite{4}] \label{def2.2}\rm The Riemann-Liouville fractional derivative of order $\alpha>0$ of a continuous function $g:(0,\infty)\to \mathbb{R}$ is given by $$ D_{0^+}^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}} \int_{0}^{t}\frac{g(s)}{(t-s)^{\alpha-n+1}}ds, $$ where $n-1\le \alpha< n$, provided that the right-hand side is point-wise defined on $(0,\infty)$. \end{definition} \begin{definition} \label{def2.3}\rm Let $X$ and $Y$ be Banach spaces. $L:D(L)\subset X\to Y$ is called a Fredholm operator of index zero if $\operatorname{Im} L$ is closed in $X$ and $\operatorname{dim}\ker L=\operatorname{codim}\operatorname{Im} L<+\infty$. \end{definition} It is easy to see that if $L$ is a Fredholm operator of index zero, then there exist the projectors $P:X\to X$, and $Q:Y\to Y$ such that $$ \operatorname{Im} P=\ker L, \quad \ker Q=\operatorname{Im} L,\quad X=\ker L\oplus \ker P,\quad Y=\operatorname{Im} L \oplus \operatorname{Im} Q. $$ If $L:D(L)\subset X\to Y$ is called a Fredholm operator of index zero, the inverse of $$ L|_{D(L)\cap \ker P}: D(L)\cap \ker P\to \operatorname{Im} L $$ is denoted by $K_p$. \begin{definition} \label{def2.4} Suppose that $L:D(L)\subset X\to Y$ is called a Fredholm operator of index zero. The continuous map $N:X\to Y$ is called $L$-compact if both $QN(\overline{\Omega})$ and $K_p(I-Q)N:\overline{\Omega }\to X$ are compact for each nonempty open subset $\Omega$ of $X$ satisfying $D(L)\cap \overline{\Omega }\not=\emptyset$. \end{definition} To obtain the main results, we need the following abstract existence theorem, the Leray-Schauder Nonlinear Alternative. \begin{lemma}[\cite{15}] \label{lem2.1} Let $X,Y$ be Banach spaces and $L:D(L)\cap X\to Y$ a Fredholm operator of index zero with $\ker L=\{0\in X\}$, $N:X\to Y$ $L$-compact. Suppose $\Omega$ is a nonempty open subset of $X$ satisfying $D(L)\cap \overline{\Omega }\not=\emptyset$. Then either there exists $x\in \partial\Omega$ and $\theta\in (0,1)$ such that $Lx=\theta Nx$ or there exists $x\in \overline{\Omega}$ such that $Lx=Nx$. \end{lemma} \begin{definition}[\cite{16}] \label{def2.5} An odd homeomorphism $\Phi$ of the real line $\mathbb{R}$ onto itself is called a sup-multiplicative-like function if there exists a homeomorphism $\omega$ of $[0,+\infty)$ onto itself which {\it supports} $\Phi$ in the sense that for all $v_1, v_2\geq 0$ it holds \begin{equation} \label{e3} \Phi(v_1v_2)\geq\omega(v_1)\Phi(v_2). \end{equation} The function $\omega$ is called the supporting function of $\Phi$. \end{definition} \begin{remark} \label{rmk2.1}\rm Note that any sup-multiplicative function is sup-multiplicative-like function. Also any function of the form $$ \Phi(u):=\sum_{j=0}^kc_j|u|^ju, \quad u\in \mathbb{R} $$ is sup-multiplicative-like, provided that $c_j\geq 0$. Here a supporting function is defined by $\omega(u):=\min\{u^{k+1},u\}$, $u\geq 0$. \end{remark} \begin{remark} \label{rmk2.2}\rm It is clear that a sup-multiplicative-like function $\Phi$ and any corresponding supporting function $\omega$ are increasing functions vanishing at zero and moreover their inverses $\Phi^{-1}$ and $\nu$ respectively are increasing and such that \begin{equation} \label{e4} \Phi^{-1}(w_1w_2)\leq\nu(w_1)\Phi^{-1}(w_2), \end{equation} for all $w_1, w_2\geq 0$ and $\nu$ is called the supporting function of $\Phi^{-1}$. \end{remark} In this article we assume that $\Phi$ is a sup-multiplicative-like function with its supporting function $\omega$, the inverse function $\Phi^{-1}$ has its supporting function $\nu$. \begin{definition} \label{def2.6}\rm We call $K:(0,1]\times R^2\to \mathbb{R}$ an \emph{impulsive Caratheodory function} if it satisfies the following: \begin{itemize} \item[(i)] $t\to K\left(t, (t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y\right)$ is continuous on $(t_{k},t_{k+1}]$ for $k=0,1,\dots,p$, and there exist the following limits: $$ \lim_{t\to t_k^+}K\left(t, (t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y\right) \quad (k=0,1,\dots,p)\text{ for any }(x,y)\in \mathbb{R}^2, $$ \item[(ii)] $(x,y)\to K\left(t, (t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y\right)$ is continuous on $R^2$ for all $t\in (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$. \end{itemize} \end{definition} We use the Banach spaces \begin{align*} X=\big\{&x:(0,1]\to \mathbb{R}: x|_{(t_k,t_{k+1}]}\in C^0(t_k,t_{k+1}],\; k=0,1,\dots,p,\\ &\text{there exist the limits } \lim_{t\to t_k^+}(t-t_k)^{1-\alpha}x(t),k=0,1,\dots,p\big\} \end{align*} with the norm $$ \|x\|=\|x\|_\infty=\max\big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|x(t)|,\; k=0,1,\dots,p\big\}. $$ \begin{align*} Y=\big\{&y:(0,1]\to \mathbb{R}:y|_{(t_k,t_{k+1}]}\in C^0(t_k,t_{k+1}],k=0,1,\dots,p,\\ &\text{there exist the limits } \lim_{t\to t_k^+}(t-t_k)^{1-\beta}y(t),k=0,1,\dots,p \big\} \end{align*} with the norm $$ \|y\|=\|y\|_\infty=\max\big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|y(t)|, k=0,1,\dots,p\big\}. $$ $L^1[0,1]$ with the norm $$ \|u\|_1=\int_0^1|u(s)|ds. $$ Choose $E=X\times Y$ with the norm $\|(x,y)\|=\max\{\|x\|_\infty,\|y\|_\infty\}$, and choose $Z=L^1(0,1)\times L^1(0,1)\times R^{2p+2}$ with the norm \begin{align*} &\Big\|\begin{pmatrix}u\\ v\\ a\\ b\\ c_k\;(k=1,2,\dots,p)\\ d_k\;(k=1,2,\dots,p)\end{pmatrix}^T\Big\| =\|(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)\|\\ &=\max\{\|u\|_1,\|v\|_1,|a|,|b|,|c_1|,\dots,|d_p| ,|d_1|,\dots,|c_p|\}. \end{align*} Define $L$ to be the linear operator from $D(L)\cap E$ to $Z$ with $$ D(L)=\{(x,y)\in E:D_{t_k^+}^\alpha x,D_{t_k^+}^\beta y\in L^1(0,1)\} $$ and $$ L(x,y)(t)=\begin{pmatrix} D_{t_k^+}^\alpha x(t)\\ D_{t_k^+}^\beta y(t)\\ \lim_{t\to 1}t^{1-\alpha}x(t)+\lim_{t\to 0}t^{1-\alpha}x(t)\\ \lim_{t\to 1}t^{1-\beta}y(t)+\lim_{t\to 0}t^{1-\beta}y(t)\\ \lim_{t\to t_k^+}(t-t_k)^{1-\alpha}x(t)- x(t_k),k=1,\dots,p\\ \lim_{t\to t_k^+}(t-t_k)^{1-\beta}y(t)- y(t_k),k=1,\dots,p\end{pmatrix}^T $$ for $(x,y)\in E\cap D(L)$. Define $N:E\to Z$ by \[ N(x,y)(t)= \begin{pmatrix} m(t)f(t,x(t),y(t))\\ n(t)g(t,x(t),y(t))\\ \int_0^1\phi(t)F\left(t,x(t),y(t)\right)dt\\ \;\int_0^1\psi(t)G\left(t,x(t),y(t)\right)dt\\ I_k\left(t_k,x(t_k),y(t_k)\right),k=1,\dots,p\\ J_k\left(t_k,x(t_k),y(t_k)\right),k=1,\dots,p\end{pmatrix}^T \] for $(x,y)\in E$. Then \eqref{e2} can be written as $$ L(x,y)=N(x,y),\quad (x,y)\in E. $$ \begin{lemma} \label{lem2.2} Suppose that $f,g,F,G,I_k,J_k$ $(k=1,2,\dots,p)$ are impulsive Cara\-theodory functions. Then $L$ is a Fredholm operator of index zero and $N:X\to Y$ is $L$-compact. \end{lemma} \begin{proof} To prove that $L$ is a Fredholm operator of index zero, we should do the following six steps. \textbf{Step (i)} Prove that $\ker L=\{(0,0)\in E\}$. We know that $(x,y)\in \ker L$ if and only if \begin{gather*} D_{t_k^+}^\alpha x(t)=0,\quad D_{t_k^+}^\beta y(t)=0,\\ \lim_{t\to 1}t^{1-\alpha}x(t)+\lim_{t\to 0}t^{1-\alpha}x(t)=0,\\ \lim_{t\to 1}t^{1-\beta}y(t)+\lim_{t\to 0}t^{1-\beta}y(t)=0,\\ \lim_{t\to t_k^+}(t-t_k)^{1-\alpha}x(t)- x(t_k)=0,k=1,\dots,p,\\ \lim_{t\to t_k^+}(t-t_k)^{1-\beta}y(t)- y(t_k))=0,k=1,\dots,p. \end{gather*} Hence $(x,y)\in \ker L$ if and only if $x(t)=0$ and $y(t)=0$. Thus $\ker L=\{(0,0)\in E\}$. \textbf{Step (ii)} Prove that $\operatorname{Im} L=Z$. First, we have $\operatorname{Im} L\subseteq Z$. Second, we know that $(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)\in \operatorname{Im} L$ if and only if there exist $(x,y)\in D(L)\cap E$ such that \begin{equation} \label{e5} \begin{gathered} D_{t_k^+}^\alpha x(t)=u(t),\quad D_{t_k^+}^\beta y(t)=v(t),\\ \lim_{t\to 1}t^{1-\alpha}x(t)+\lim_{t\to 0}t^{1-\alpha}x(t)=a,\\ \lim_{t\to 1}t^{1-\beta}y(t)+\lim_{t\to 0}t^{1-\beta}y(t)=b,\\ \lim_{t\to t_k^+}(t-t_k)^{1-\alpha}x(t)- x(t_k)=c_k,k=1,\dots,p,\\ \lim_{t\to t_k^+}(t-t_k)^{1-\beta}y(t)- y(t_k))=d_k,k=1,\dots,p. \end{gathered} \end{equation} If $(x,y)$ satisfies \eqref{e5}, then there exist two numbers $\overline{M}_k$ $(k=0,1,\dots,p)$ such that \begin{equation} \label{e6} x(t)=\frac{1}{\Gamma(\alpha)}\int_{t_k}^t(t-s)^{\alpha-1}u(s)ds +\overline{M}_k(t-t_k)^{\alpha-1}, \end{equation} for $t\in (t_k,t_{k+1}]$, $k=0,1,\dots,p$. By the boundary condition $\lim_{t\to 1}t^{1-\alpha}x(t)+\lim_{t\to 0}t^{1-\alpha}x(t)=a$, we obtain \begin{equation} \label{e7} \int_{t_p}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds +\overline{M}_p(1-t_p)^{\alpha-1}+\overline{M}_0=a. \end{equation} By the impulse conditions $\lim_{t\to t_k^+}(t-t_k)^{1-\alpha}x(t)- x(t_k)=c_k$, we obtain \begin{equation} \label{e8} \overline{M}_k-\Big(\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds +\overline{M}_{k-1}(t_k-t_{k-1})^{\alpha-1}\Big)=c_k, \end{equation} for $ k=1,\dots,p$. It follows from \eqref{e8} that $$ \overline{M}_p-\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \overline{M}_0=\sum_{k=1}^p \Big(c_k+\int_{t_{k-1}}^{t_k}\frac{(t_k-s)^{\alpha-1}}{\gamma(\alpha)}u(s)ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}. $$ By this equality and \eqref{e7}, we obtain \begin{align*} &\overline{M}_0\\ &=\frac{a-\int_{t_p}^1\frac{(1-s)^{\alpha-1}} {\Gamma(\alpha)}u(s)ds}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\alpha-1}} \\ & +\frac{\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^p \Big(c_k+\int_{t_{k-1}}^{t_k}\frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\alpha-1}}, \end{align*} \begin{equation} \label{e9} \begin{split} \overline{M}_p &=\frac{\sum_{k=1}^p\Big(c_k+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\alpha-1}} \\ &\quad +\frac{\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \Big(a-\int_{t_p}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds\Big) }{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\alpha-1}}. \end{split} \end{equation} Then \eqref{e8} implies that \begin{equation} \label{e10} \overline{M}_k=c_k+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds +\overline{M}_{k-1}(t_k-t_{k-1})^{\alpha-1},\quad k=1,\dots,p-1. \end{equation} Hence \eqref{e6} is proved and $\overline{M}_k$ $(k=0,1,2,\dots,p)$ are given by \eqref{e9} and \eqref{e10}. Similarly we obtain \begin{equation} \label{e11} y(t)=\frac{1}{\Gamma(\beta)}\int_{t_k}^t(t-s)^{\beta-1}v(s)ds +\overline{N}_k(t-t_k)^{\beta-1}, \end{equation} for $t\in (t_k,t_{k+1}]$, $k=0,1,\dots,p$, where $\overline{N}_k$ $(k=0,1,\dots,p)$ are given by \begin{align*} &\overline{N}_0\\ &=\frac{b-\int_{t_p}^1\frac{(1-s)^{\beta-1}} {\Gamma(\beta)}v(s)ds}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\beta-1}} \\ & +\frac{\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\sum_{k=1}^p \Big(d_k+\int_{t_{k-1}}^{t_k}\frac{(t_k-s)^{\beta-1}}{\Gamma(\beta)}v(s)ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\beta-1}}, \end{align*} \begin{equation} \label{e12} \begin{split} \overline{N}_p &=\frac{\sum_{k=1}^p\Big(d_k+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\beta-1}}{\Gamma(\beta)}v(s)ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}}{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\beta-1}} \\ &\quad +\frac{\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \Big(b-\int_{t_p}^1\frac{(1-s)^{\beta-1}}{\Gamma(\beta)}v(s)ds\Big) }{1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\beta-1}}, \end{split} \end{equation} and \begin{equation} \label{e13} \overline{N}_k=d_k+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\beta-1}}{\Gamma(\beta)}v(s)ds +\overline{N}_{k-1}(t_k-t_{k-1})^{\beta-1},\quad k=1,\dots,p-1. \end{equation} It is easy to show that $(x,y)\in D(L)\cap E$. Hence $(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)\in \operatorname{Im} L$. Then $\operatorname{Im} L=Z$. On the other hand, we can prove that $(x,y)$ is a solution of \eqref{e5} if $x\in E$ satisfies \eqref{e6} and $y\in Y$ satisfies \eqref{e11}. \textbf{Step (iii)} Prove that $\operatorname{Im} L$ is closed in $X$ and $\operatorname{dim}\ker L=\operatorname{codim}\operatorname{Im} L<+\infty$. From Step (ii) $\operatorname{Im} L=Z$ is closed in $Z$. It follows from $\ker L=\{(0,0)\in E\}$ that $\operatorname{dim}\ker L=0$. Define the projector $P:E\to E$ by \begin{equation} \label{e14} P(x,y)(t)=(0,0)\quad \text{for }(x,y)\in E. \end{equation} It is easy to prove that \begin{equation} \label{e15} \operatorname{Im} P=\ker L,\quad X=\ker L\oplus \ker P. \end{equation} Define the projector $Q:Z\to Z$ by \begin{equation} \label{e16} Q(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)(t) =(0,0,0,0,0,\dots,0,0,\dots,0) \end{equation} for $(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)\in Z$. It is easy to show that \begin{equation} \label{e17} \operatorname{Im} L= \ker Q,\quad Y=\operatorname{Im} Q\oplus\operatorname{Im} L. \end{equation} From above discussion, we see that $\operatorname{dim}\ker L=\operatorname{codim}\operatorname{Im} L=0<+\infty$. So $L$ is a Fredholm operator of index zero. Now, we prove that $N$ is $L$-compact. This is divided into three steps (Steps (iv)-(vi)). \textbf{Step (iv)} We prove that $N$ is continuous. Let $(x_n,y_n)\in E$ with $(x_n,y_n)\to (x_0,y_0)$ as $n\to \infty$. We will show that $N(x_n,y_n)\to N(x_0,y_0)$ as $n\to \infty$. In fact, we have \begin{align*} &\|(x_n,y_n)\|\\ &=\sup_{n=0,1,2,\dots} \Big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|x_n(t)|, \\ &\quad \sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\beta}|y_n(t)|:k=0,1,\dots,p\Big\} =r<+\infty \end{align*} and \begin{equation} \label{e18} \begin{gathered} \max\big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|x_n(t)-x_0(t)|,\; k=0,1,\dots,p\big\}\to 0,\quad n\to \infty, \\ \max\big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\beta}|y_n(t)-y_0(t)|,\; k=0,1,\dots,p\big\}\to 0,\quad n\to \infty. \end{gathered} \end{equation} By \[ N(x_n,y_n)(t)=\begin{pmatrix} m(t)f(t,x_n(t),y_n(t))\\ n(t)g(t,x_n(t),y_n(t))\\ \int_0^1\phi(t)F\left(t,x_n(t),y_n(t)\right)dt\\ \int_0^1\psi(t)G\left(t,x_n(t),y_n(t)\right)dt\\ I_k\left(t_k,x_n(t_k),y_n(t_k)\right)\; (k=1,2,\dots,p)\\ J_k\left(t_k,x_n(t_k),y_n(t_k)\right)\; (k=1,2,\dots,p) \end{pmatrix}^T \] for $(x,y)\in E$, for any $\epsilon>0$, since $f,F,I_k$ $(k=1,\dots,p)$ are impulsive Caratheodory functions, we know that $f\left(t,(t-t_k)^{\alpha-1}u,(t-t_k)^{\beta-1}v\right)$ is continuous on $[t_{k},t_{k+1}]\times [-r,r]^2$ $(k=0,1\dots,p)$ respectively, so $f\left(t,(t-t_k)^{\alpha-1}u,(t-t_k)^{\beta-1}v\right)$ is uniformly continuous on $[t_{k},t_{k+1}]\times [-r,r]^2$ respectively. Similarly, $F,I_k$ $(k=1,\dots,p)$ are uniformly continuous on $[t_{k},t_{k+1}]\times [-r,r]^2$ respectively. Then there exists $\delta>0$ such that \begin{gather*} \left|f\left(t,(t-t_k)^{\alpha-1}u_1,(t-t_k)^{\beta-1}v_1\right) -f\left(t,(t-t_k)^{\alpha-1}u_2,(t-t_k)^{\beta-1}v_2\right)\right|<\epsilon,\\ t\in (t_k,t_{k+1}], \\ \left|F\left(t,(t-t_k)^{\alpha-1}u_1,(t-t_k)^{\beta-1}v_1\right) -F\left(t,(t-t_k)^{\alpha-1}u_2,(t-t_k)^{\beta-1}v_2\right)\right|<\epsilon,\\ t\in (t_k,t_{k+1}], \\ \begin{aligned} &\big|I_k\left(t_k,(t_k-t_{k-1})^{\alpha-1}u_1,(t_k-t_{k-1})^{\beta-1}v_1\right)\\ &-I_k\left(t_k,(t_k-t_{k-1})^{\alpha-1}u_2,(t_k-t_{k-1}) ^{\beta-1}v_2\right)\big|<\epsilon \end{aligned} \end{gather*} for all $k=0,1,\dots,p$, $|u_1-u_2|<\delta$ and $|v_1-v_2|<\delta$ with $u_1,u_2,v_1,v_2\in [-,r,r]$. From \eqref{e18}, there exists $N$ such that \begin{equation} \label{e19} \begin{gathered} (t-t_k)^{1-\alpha}|x_n(t)-x_0(t)|<\delta ,\quad t\in (t_k,t_{k+1}],\;k=0,1,\dots,p,\;n>N, \\ (t-t_k)^{1-\beta}|y_n(t)-y_0(t)|<\delta ,\quad t\in (t_k,t_{k+1}],\,k=0,1,\dots,p,\;n>N. \end{gathered} \end{equation} Hence using \eqref{e19}, we obtain \begin{align*} &\int_0^1\left|m(t)f\left(t,x_n(t),y_n(t)\right)-m(t)f\left(t,x_0(t),y_0(t)\right) \right|dt\\ &= \sum_{k=0}^p\int_{t_k}^{t_{k+1}}|m(t)f\left(t,(t-t_k)^{\alpha-1} (t-t_k)^{1-\alpha}x_n(t),(t-t_k)^{\beta-1}(t-t_k)^{1-\beta}y_n(t)\right) \\ &-m(t)f\left(t,(t-t_k)^{\alpha-1}(t-t_k)^{1-\alpha}x_0(t), (t-t_k)^{\beta-1}(t-t_k)^{1-\beta}y_0(t)\right)\big|dt\\ &<\sum_{k=0}^p\int_{t_k}^{t_{k+1}}\epsilon m(t)dt=\epsilon\int_0^1m(t)dt,n>N. \end{align*} It follows that \begin{equation} \label{e20} \left|\int_0^1m(t)f\left(t,x_n(t),y_n(t)\right)dt-\int_0^1f\left(t,x_0(t),y_0(t) \right)dt\right|<\epsilon\int_0^1m(t)dt, \end{equation} for $n>N$. Similarly, \begin{equation} \label{e21} \left|\int_0^1\phi(t)F\left(t,x_n(t),y_n(t)\right)dt -\int_0^1F\left(t,x_0(t),y_0(t)\right)dt\right|<\epsilon\int_0^1\phi(t)dt, \end{equation} for $n>N$, and \begin{equation} \label{e22} \left|I_k\left(t_k,x_n(t_k),y_n(t_k)\right) -I_k\left(t_k,x_0(t_k),y_0(t_k)\right)\right|<\epsilon,\quad n>N,\; k=1,\dots,p \end{equation} We can also show that \begin{equation} \label{e23} \big|\int_0^1n(t)g\left(t,x_n(t),y_n(t)\right)dt-\int_0^1n(t) g\left(t,x_0(t),y_0(t)\right)dt\big|<\epsilon\int_0^1n(t)dt, \end{equation} for $n>N$. Similarly, \begin{equation} \label{e24} \left|\int_0^1\psi(t)G\left(t,x_n(t),y_n(t)\right)dt -\int_0^1G\left(t,x_0(t),y_0(t)\right)dt\right|<\epsilon\int_0^1\psi(t)dt, \end{equation} for $n>N$, and \begin{equation} \label{e25} \left|J_k\left(t_k,x_n(t_k),y_n(t_k)\right) -J_k\left(t_k,x_0(t_k),y_0(t_k)\right)\right|<\epsilon,\quad n>N,k=1,\dots,p \end{equation} Then \eqref{e20}--\eqref{e25} imply that $$ \|N(x_n,y_n)-N(x_0,y_0)\|\to 0,\quad n\to\infty. $$ It follows that $N$ is continuous. Let $P:X\to X$ and $Q:Y\to Y$ be defined by \eqref{e14} and \eqref{e16}. For $(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)$ $\in \operatorname{Im} L=Z$, let \begin{equation} \label{e26} K_P(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)(t)=\left(x_1(t),y_1(t)\right), \end{equation} where \begin{gather*} x_1(t)=\int_{t_k}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}u(s)ds +\overline{M}_kt^{\alpha-1},t\in (t_k,t_{k+1}],\quad k=0,1,\dots,p;\\ y_1(t)=\int_{t_k}^t\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}v(s)ds +\overline{N}_kt^{\alpha-1},t\in (t_k,t_{k+1}],\quad k=0,1,\dots,p. \end{gather*} Here $\overline{M}_k,\overline{N}_k$ $(k=0,1,\dots,p)$ are given by \eqref{e9}, \eqref{e10}, \eqref{e11} and \eqref{e13}. One sees that $K_P(u,v,a,b,c_1,\dots,c_p,d_1,\dots,d_p)\in D(L)\cap E$ and $K_P:\operatorname{Im }L\to D(L)\cap \ker P$ is the inverse of $L:D(L)\cap\ker P\to \operatorname{Im} L$. The isomorphism $\wedge: \ker L\to Y/\operatorname{Im} L $ is given by $$ \wedge (0,0)=(0,0,0,0,0,\dots,0,0\dots,0). $$ Furthermore, one has \begin{equation} \label{e27} QN(x,y)(t)=(0,0,0,0,0,\dots,0,0\dots,0), \end{equation} and $$ K_p(I-Q)N(x,y)(t)=K_pN(x,y)(t)=\left(x_2(t),y_2(t)\right), $$ where \begin{equation} \label{e28} x_2(t)=\int_{t_k}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds +M_kt^{\alpha-1},t\in (t_k,t_{k+1}], \end{equation} for $k=0,1,\dots,p$, and \begin{equation} \label{e29} y_2(t)=\int_{t_k}^t\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}n(s)g(s,x(s),y(s))ds +N_kt^{\alpha-1},t\in (t_k,t_{k+1}], \end{equation} for $k=0,1,\dots,p$. Here $M_k,N_k$ $(k=0,1,\dots,p)$ are given by \begin{gather*} \begin{aligned} M_{0}&=\frac{1}{\lambda}\Big(\int_0^1\phi(s)F(s,x(s),y(s))ds -\int_{t_p}^{t_{p+1}}\frac{(t_{p+1}-s)^{\alpha-1}} {\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^p\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad\times \big(I_k(t_k,x(t_k),y(t_k))+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds\big)\Big), \end{aligned} \\ \begin{aligned} M_{1}&=I_1(t_1,x(t_1),y(t_1))\\ &\quad +\int_{t_0}^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}m(s) f(s,x(s),y(s))ds +(t_1-t_0)^{\alpha-1}M_{0}, \end{aligned}\\ \dots \\ \begin{aligned} M_{p-1}&=I_{p-1}(t_{p-1},x(t_{p-1}),y(t_{p-1})) +\int_{t_{p-2}}^{t_{p-1}}\frac{(t_{p-1}-s)^{\alpha-1}}{\Gamma(\alpha)}m(s) f(s,x(s),y(s))ds\\ &\quad +(t_{p-1}-t_{p-2})^{\alpha-1}M_{p-2}, \end{aligned} \\ \begin{aligned} M_{p} &=\frac{1}{\lambda }\Big[\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \Big(\int_0^1\phi(s)F(s,x(s),y(s))ds \\ &\quad -\int_{t_p}^{t_{p+1}} \frac{(t_{p+1}-s)^{\alpha-1}}{\Gamma(\alpha)} m(s)f(s,x(s),y(s))ds\Big)+\sum_{k=1}^p\Big(I_k(t_k,x(t_k),y(t_k)) \\ &\quad +\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\Big], \end{aligned} \end{gather*} and \begin{gather*} \begin{aligned} N_{0}&=\frac{1}{\lambda}\Big(\int_0^1\psi(s)G(s,x(s),y(s))ds -\int_{t_p}^{t_{p+1}}\frac{(t_{p+1}-s)^{\beta-1}} {\Gamma(\beta)}n(s)g(s,x(s),y(s))ds \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \sum_{k=1}^p\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}\times \\ &\quad \Big(J_k(t_k,x(t_k),y(t_k))+\int_{t_{k-1}}^{t_k}\frac{(t_k-s)^{\beta -1}}{\Gamma(\beta)}n(s)g(s,x(s),y(s))ds\Big) \Big), \end{aligned}\\ N_{1}=J_1(t_1,x(t_1),y(t_1))+\int_{t_0}^{t_1} \frac{(t_1-s)^{\beta-1}}{\Gamma(\beta)}n(s)g(s,x(s),y(s))ds +(t_1-t_0)^{\alpha-1}N_{0}, \\ \dots \\ \begin{aligned} N_{p-1}&=J_{p-1}(t_{p-1},x(t_{p-1}),y(t_{p-1}))+\int_{t_{p-2}}^{t_{p-1}} \frac{(t_{p-1}-s)^{\beta-1}}{\Gamma(\beta}n(s)g(s,x(s),y(s))ds \\ &\quad +(t_{p-1}-t_{p-2})^{\beta-1}N_{p-2}, \end{aligned}\\ \begin{aligned} N_{p}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \Big(\int_0^1\psi(s)G(s,x(s),y(s))ds \\ &\quad -\int_{t_p}^{t_{p+1}} \frac{(t_{p+1}-s)^{\beta-1}}{\Gamma(\beta)}n(s)g(s,x(s),y(s))ds\Big) +\sum_{k=1}^p(J_k(t_k,x(t_k),y(t_k)) \\ &\quad +\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\beta-1}}{\Gamma(\beta)}n(s)g(s,x(s),y(s))ds\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}\Big). \end{aligned} \end{gather*} Let $\Omega$ be a bounded open subset of $E$ satisfying $D(L)\cap \Omega\neq 30\emptyset$. We have \begin{equation} \label{e30} \begin{split} \|(x,y)\|&=\max\Big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|x(t)|, \\ &\quad \sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\beta}|y(t)|:k=0,1,\dots,p\Big\}\\ &=r<+\infty,\quad (x,y)\in \Omega. \end{split} \end{equation} Since $f,g,F,G,I_k,J_k$ are impulsive Caratheodory functions, together with \eqref{e30}, there exists $M>0$ such that \begin{align*} \left|f\left(t,x(t),y(t)\right)\right| &=\left|f\left(t,(t-t_k)^{\alpha-1}(t-t_k)^{1-\alpha}x(t), (t-t_k)^{\beta-1}(t-t_k)^{1-\beta}y(t)\right)\right|\\ &\le M \end{align*} holds for $t\in (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$. Hence $$ \left|f\left(t,x(t),y(t)\right)\right|\le M,\quad t\in (0,1]. $$ Similarly, \begin{gather*} \left|g\left(t,x(t),y(t)\right)\right|\le M,\\ \left|F\left(t,x(t),y(t)\right)\right|\le M\quad \text{for all }t\in (0,1],\\ \left|G\left(t,x(t),y(t)\right)\right|\le M\quad \text{for all }t\in (0,1],\\ \left|I_k\left(t_k,x(t_k),y(t_k)\right)\right|\le M,\quad k=1,2,\dots,p\\ \left|J_k\left(t_k,x(t_k),y(t_k)\right)\right|\le M,\quad k=1,2,\dots,p. \end{gather*} \textbf{Step (v)} Prove that $QN(\overline{\Omega})$ is bounded. It is easy to see from \eqref{e27} that $QN(\overline{\Omega})$ is bounded. \textbf{Step (vi)} Prove that $K_P(I-Q)N:\overline{\Omega}\to E$ is compact; i.e., prove that $K_P(I-Q)N(\overline{\Omega})$ is relatively compact. We must prove that $K_P(I-Q)N(\overline{\Omega})$ is uniformly bounded and equi-continuous on each subinterval $[e,f]\subseteq (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$, respectively and equi-convergent at $t=t_k$ $(k=0,1,\dots,p)$, respectively. \textbf{Substep (vi1)} Prove that $K_P(I-Q)N(\overline{\Omega})$ is uniformly bounded. We have \begin{equation} \label{e31} (t-t_k)^{1-\alpha}x_2(t)=(t-t_k)^{1-\alpha}\int_{t_k}^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds +M_k, \end{equation} for $t\in (t_k,t_{k+1}]$. By the definition of $M_k$, we have \begin{align*} &|M_0|\\ &\le \frac{1}{\lambda} \Big(\int_0^1|\phi(s)F(s,x(s),y(s))|ds+\int_{t_p}^{t_{p+1}} \frac{(t_{p+1}-s)^{\alpha-1}}{\Gamma(\alpha)}|m(s)f(s,x(s),y(s))|ds \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^p\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad\times \Big(|I_k(t_k,x(t_k),y(t_k))|+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}|m(s)f(s,x(s),y(s))|ds\Big)\Big) \\ &\le \frac{M}{\lambda}\Big(\|\phi\|_1+l_1t_{p+1}^{\alpha+k_1} \int_{\frac{t_p}{t_{p+1}}}^{1}\frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw \\ &\quad +\prod _{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^p \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad\times \Big(1+l_1t_k^{\alpha+k_1}\int_{\frac{t_{k-1}}{t_k}}^{1} \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw\Big) \Big)<+\infty. \end{align*} Similarly, \begin{gather*} |M_1|\le M+Ml_1t_1^{\alpha+k_1}\int_{\frac{t_0}{t_1}}^{1} \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw +(t_1-t_0)^{\alpha-1}|M_{0}|<+\infty, \\ \dots\\ \begin{aligned} |M_{p-1}| &\le M+Ml_1t_{p-1}^{\alpha+k_1} \int_{\frac{t_{p-2}}{t_{p-1}}}^{1} \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw +(t_{p-1}-t_{p-2})^{\alpha-1}|M_{p-2}|\\ &<+\infty, \end{aligned}\\ \begin{aligned} |M_p|&\le \frac{M}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \Big(\|\phi\|_1+l_1t_{p+1}^{\alpha+k_1}\int_{\frac{t_p}{t_{p+1}}}^{1} \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw\Big) \\ &\quad +\sum_{k=1}^p\Big(1+l_1t_k^{\alpha+k_1}\int_{\frac{t_{k-1}}{t_k}}^{1} \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw\Big) \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\Big)\\ &\quad <+\infty. \end{aligned} \end{gather*} First, use \eqref{e31}, for $t\in (t_0,t_{1}]$ we have \begin{align*} (t-t_{0})^{1-\alpha}|x_2(t)| &\le(t-t_0)^{1-\alpha}\int_{t_0}^t\frac{(t-s)^{\alpha-1}} {\Gamma(\alpha)}|m(s)f(s,x(s),y(s))|ds+|M_0| \\ &\le Ml_1(t_1-t_0)^{1-\alpha}t_1^{\alpha+k_1}\int_{\frac{t_0}{t}}^1 \frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw+|M_0|<+\infty. \end{align*} Second, for $t\in (t_k,t_{k+1}]$ $(k=1,\dots,p-1)$, we have \begin{align*} &(t-t_{0})^{1-\alpha}|x_2(t)|\\ &\le(t-t_k)^{1-\alpha}\int_{t_k}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} |m(s)f(s,x(s),y(s))|ds+|M_k| \\ &\le Ml_1(t_{k+1}-t_k)^{1-\alpha}t_k^{\alpha+k_1} \int_{\frac{t_{k-1}}{t_k}}^{1}\frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)} w^{k_1}dw+|M_k|<+\infty. \end{align*} Finally, for $t\in (t_p,t_{p+1}]$, we have \begin{align*} &(t-t_{p})^{1-\alpha}|x_2(t)|\\ &\le(t-t_p)^{1-\alpha}\int_{t_p}^t\frac{(t-s)^{\alpha-1}} {\Gamma(\alpha)}|m(s)f(s,x(s),y(s))|ds+|M_p| \\ &\le Ml_1(t_{p+1}-t_p)^{1-\alpha}t_{p+1}^{\alpha+k_1} \int_{\frac{t_{p}}{t_{p+1}}}^1\frac{(1-w)^{\alpha-1}}{\Gamma(\alpha)}w^{k_1}dw +|M_p|<+\infty. \end{align*} From above discussion, there exists $M_1>0$ such that $$ \|x_2\|_\infty=\max \Big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|x_2(t)|:k=0,1,\dots,p\Big\} \le M_1<+\infty. $$ Similarly, we can show that there exist $M_2>0$ such that $$ \|y_2\|_\infty=\max\Big\{\sup_{t\in (t_k,t_{k+1}]}(t-t_k)^{1-\alpha}|y_2(t)| :k=0,1,\dots,p\Big\}\le M_2<+\infty. $$ Hence $K_P(I-Q)N(\overline{\Omega})$ is uniformly bounded. \textbf{Substep (vi2)} Prove that $K_P(I-Q)N(\overline{\Omega})$ is equi-continuous on each subinterval $[e,f]\subseteq (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$, respectively. For each $[e,f]\subseteq (t_k,t_{k+1}]$, and $s_1,s_2\in [e,f]$ with $s_2\ge s_1$, use \eqref{e28}, we have \begin{align*} &|(s_1-t_k)^{1-\alpha}x_2(s_1)-(s_2-t_k)^{1-\alpha}x_2(s_2)| \\ &\le \frac{l_1M}{\Gamma(\alpha)} \left|(s_1-t_k)^{1-\alpha}-(s_2-t_k)^{1-\alpha}\right| s_1^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) \\ &\quad +\frac{l_1M}{\Gamma(\alpha)}(t_{k+1}-t_k)^{1-\alpha}s_2^{\alpha+k_1} \int_{s_1/s_2}^{1} (1-w)^{\alpha-1}w^{k_1}dw \\ &\quad +\frac{l_1M}{\Gamma(\alpha)}(t_{k+1}-t_k)^{1-\alpha} \Big(s_2^{\alpha+k_1}\int_0^1(1-w)^{\alpha-1}w^{k_1}dw\\ &\quad -s_1^{\alpha+k_1} \int_{0}^{s_2/s_1 }|(1-w)^{\alpha-1}w^{k_1}dw\Big)\to 0 \end{align*} uniformly as $ s_1\to s_2$. It follows that \begin{equation} \label{e32} |(s_1-t_k)^{1-\alpha}x_2(s_1)-(s_2-t_k)^{1-\alpha}x_2(s_2)|\to 0 \end{equation} uniformly as $ s_1\to s_2,\;s_1,s_2\in [e,f]\subseteq (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$. Similarly, we can prove that \begin{equation} \label{e33} |(s_1-t_k)^{1-\beta}y_2(s_1)-(s_2-t_k)^{1-\beta}y_2(s_2)|\to 0 \end{equation} uniformly as $ s_1\to s_2,\;s_1,s_2\in [e,f]\subseteq (t_k,t_{k+1}]$ $(k=0,1,\dots,p)$. \textbf{Substep (vi3)} Prove that $K_P(I-Q)N(\overline{\Omega})$ is equi-convergent at $t=t_k$ $(k=0,1,\dots,p)$, respectively. Since \begin{align*} &\left|(t-t_k)^{1-\alpha}x_2(t)-M_k\right| \\ &\le l_1M(t_{k+1}-t_k)^{1-\alpha}t_{k+1}^{\alpha+k_1} \int_{\frac{t_k}{t}}^1(1-w)^{\alpha-1}w^{k_1}dw\to 0 \end{align*} uniformly as $t\to t_k$. Similarly we can show that \begin{equation} \label{e34} \left|(t-t_{k})^{1-\beta}y_2(t)-N_k\right|\to 0\quad \text{uniformly as }t\to t_k\; (k=0,1,\dots,p. \end{equation} From \eqref{e33}--\eqref{e34}, we see that $K_P(I-Q)N(\overline{\Omega})$ is relatively compact. Then $N$ is $L$-compact. The proof is complete. \end{proof} \section{Main Result} Now, we prove the main theorem in this article, using the following assumptions: \begin{itemize} \item[(A)] $\Phi$ is a sup-multiplicative-like function with its supporting function $w$, the inverse function of $\Phi$ is $\Phi^{-1}$ with supporting function $\nu$. \item[(B)] $f,g,F,G,I_k,J_k$ $(k=1,2,\dots,p)$ are impulsive Caratheodory functions and satisfy that there exist nonnegative constants $c_i,b_i,a_i$ $(i=1,2)$, $C_i,B_i,A_i$ and $C_{i,k},B_{i,k},A_{i,k}$ $(i=1,2,k=1,2,\dots,p)$ such that \begin{gather*} |f(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)| \le c_1+b_1|x|+a_1\Phi^{-1}(|y|),\\ t\in (t_k,t_{k+1}],k=0,1,\dots,p, \\ |g(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le c_2+b_2\Phi(|x|)+a_2|y|,\\ t\in (t_k,t_{k+1}],k=0,1,\dots,p, \\ |F(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le C_1+B_1|x|+A_1\Phi^{-1}(|y|),\\ t\in (t_k,t_{k+1}],k=0,1,\dots,p, \\ |G(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le C_2+B_2\Phi(|x|)+A_2|y|,\\ t\in (t_k,t_{k+1}],k=0,1,\dots,p, \\ |I_k(t,(t_{k+1}-t_k)^{\alpha-1}x,(t_{k+1}-t_k)^{\beta-1}y)| \le C_{1,k}+B_{1,k}|x|+A_{1,k}\Phi^{-1}(|y|),\\ k=1,2,\dots,p, \\ |J_k(t,(t_{k+1}-t_k)^{\alpha-1}x,(t_{k+1}-t_k)^{\beta-1}y)| \le C_{2,k}+B_{2,k}\Phi(|x|)+A_{2,k}|y|,\\ k=1,2,\dots,p. \end{gather*} \end{itemize} Also we introduce the following notation. \begin{gather*} \lambda =1+\prod_{k=1}^{p+1}(t_k-t_{k-1})^{\alpha-1},\\ \begin{aligned} M_{0,1}&=\frac{1}{\lambda}\Big[C_1\|\phi\|_1+l_1c_1\mathbf{B} (\alpha,k_1+1)t_{p+1}^{\alpha+k_1} \\ &\quad +l_1c_1\mathbf{B}(\alpha,k_1+1)\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^pt_{k}^{\alpha+k_1}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pC_{1,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} +l_1c_1\mathbf{B}(\alpha,k_1+1)\Big], \end{aligned}\\ \begin{aligned} M_{0,2}&=\frac{1}{\lambda}\Big[B_1\|\phi\|_1+l_1b_1\mathbf{B} (\alpha,k_1+1)t_{p+1}^{\alpha+k_1}\\ &\quad +l_1b_1\mathbf{B}(\alpha,k_1+1)\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^pt_{k}^{\alpha+k_1}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pB_{1,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}+l_1b_1\mathbf{B}(\alpha,k_1+1)\Big], \end{aligned}\\ \begin{aligned} M_{0,3}&=\frac{1}{\lambda}[A_1\|\phi\|_1+l_1a_1\mathbf{B} (\alpha,k_1+1)t_{p+1}^{\alpha+k_1} \\ &\quad +l_1a_1\mathbf{B}(\alpha,k_1+1) \prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pt_{k}^{\alpha+k_1} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pA_{1,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}+l_1a_1\mathbf{B}(\alpha,k_1+1)\Big], \end{aligned} \\ M_{1,1}=C_{1,1}+l_1c_1t_1^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) +(t_1-t_0)^{\alpha-1}M_{0,1}, \\ M_{1,2}=B_{1,1}+l_1b_1t_1^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) +(t_1-t_0)^{\alpha-1}M_{0,2}, \\ M_{1,3}=A_{1,1}+l_1a_1t_1^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) +(t_1-t_0)^{\alpha-1}M_{0,3}, \\ \dots\\ M_{p-1,1}=C_{1,p-1}+l_1c_1t_{p-1}^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)+(t_{p-1}-t_{p-2})^{\alpha-1}M_{p-2,1}, \\ M_{p-1,2}=B_{1,p-1}+l_1b_1t_{p-1}^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) +(t_{p-1}-t_{p-2})^{\alpha-1}M_{p-2,2}, \\ M_{p-1,3}=A_{1,p-1}+l_1a_1t_{p-1}^{\alpha+k_1}\mathbf{B}(\alpha,k_1+1) +(t_{p-1}-t_{p-2})^{\alpha-1}M_{p-2,3}, \\ \begin{aligned} M_{p,1}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\|\phi\|_1 +l_1c_1\mathbf{B}(\alpha,k_1+1)t_{p+1}^{\alpha+k_1} \prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\\ &\quad +\sum_{k=1}^pC_{1,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\\ &\quad +l_1c_1\mathbf{B}(\alpha,k_1+1)\sum_{k=1}^pt_k^{\alpha+k_1} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\Big), \end{aligned}\\ \begin{aligned} M_{p,2}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\|\phi\|_1 +l_1b_1\mathbf{B}(\alpha,k_1+1)t_{p+1}^{\alpha+k_1} \prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\\ &\quad +\sum_{k=1}^pB_{1,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad +l_1b_1\mathbf{B}(\alpha,k_1+1)\sum_{k=1}^pt_k^{\alpha+k_1} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\Big), \end{aligned}\\ \begin{aligned} M_{p,3}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \|\phi\|_1+l_1a_1\mathbf{B}(\alpha,k_1+1)t_{p+1}^{\alpha+k_1} \prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \\ &\quad +\sum_{k=1}^pA_{1,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad +l_1a_1\mathbf{B}(\alpha,k_1+1)\sum_{k=1}^pt_k^{\alpha+k_1} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}\Big), \end{aligned} \end{gather*} and \begin{gather*} \sigma_{k,1}=l_1c_1(t_{k+1}-t_k)^{1-\alpha}t_{k+1}^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)+M_{k,1},\quad k=0,1,\dots,p, \\ \sigma_{k,2}=l_1b_1(t_{k+1}-t_k)^{1-\alpha}t_{k+1}^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)+M_{k,2},\quad k=0,1,\dots,p, \\ \sigma_{k,3}=l_1a_1(t_{k+1}-t_k)^{1-\alpha}t_{k+1}^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)+M_{k,3},\quad k=0,1,\dots,p, \\ \sigma_1=\max\left\{\sigma_{k,1}:k=0,1,\dots,p\right\}, \\ \sigma_2=\max\left\{\sigma_{k,2}:k=0,1,\dots,p\right\}, \\ \sigma_3=\max\left\{\sigma_{k,3}:k=0,1,\dots,p\right\}. \end{gather*} Denote \begin{gather*} \begin{aligned} N_{0,1}&=\frac{1}{\lambda}\Big[C_2\|\psi\|_1+l_2c_2 \mathbf{B}(\beta,k_2+1)t_{p+1}^{\beta+k_2} \\ &\quad +l_2c_2\mathbf{B}(\beta,k_2+1) \prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \sum_{k=1}^pt_{k}^{\beta+k_2}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\sum_{k=1}^pC_{2,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}+l_2c_2\mathbf{B}(\beta,k_2+1)\Big], \end{aligned}\\ \begin{aligned} N_{0,2}&=\frac{1}{\lambda}\Big[B_2\|\psi\|_1+l_2b_2\mathbf{B}(\beta,k_2+1) t_{p+1}^{\beta+k_2}\\ &\quad +l_2b_2\mathbf{B}(\beta,k_2+1)\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \sum_{k=1}^pt_{k}^{\beta+k_2}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\sum_{k=1}^pB_{2,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}+l_2b_2\mathbf{B}(\beta,k_2 +1)\Big], \end{aligned}\\ \begin{aligned} N_{0,3}&=\frac{1}{\lambda}\Big[A_2\|\psi\|_1+l_2a_2 \mathbf{B}(\beta,k_2+1)t_{p+1}^{\beta+k_2}\\ &\quad +l_2a_2\mathbf{B}(\beta,k_2+1)\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \sum_{k=1}^pt_{k}^{\beta+k_2}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\sum_{k=1}^pA_{2,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}+l_2a_2\mathbf{B}(\beta,k_2+1)\Big], \end{aligned}\\ N_{1,1}=C_{2,1}+l_2c_2t_1^{\beta+k_2}\mathbf{B}(\beta,k_2+1) +(t_1-t_0)^{\beta-1}N_{0,1}, \\ N_{1,2}=B_{2,1}+l_2b_2t_1^{\beta+k_2}\mathbf{B}(\beta,k_2+1) +(t_1-t_0)^{\beta-1}N_{0,2}, \\ N_{1,3}=A_{2,1}+l_2a_2t_1^{\beta+k_2}\mathbf{B}(\beta,k_2+1) +(t_1-t_0)^{\beta-1}N_{0,3}, \\ \dots\\ N_{p-1,1}=C_{2,p-1}+l_2c_2t_{p-1}^{\beta+k_2}\mathbf{B} (\beta,k_2+1)+(t_{p-1}-t_{p-2})^{\beta-1}N_{p-2,1}, \\ N_{p-1,2}=B_{2,p-1}+l_2b_2t_{p-1}^{\beta+k_2}\mathbf{B}(\beta,k_2+1) +(t_{p-1}-t_{p-2})^{\beta-1}N_{p-2,2}, \\ N_{p-1,3}=A_{2,p-1}+l_2a_2t_{p-1}^{\beta+k_2}\mathbf{B}(\beta,k_2+1) +(t_{p-1}-t_{p-2})^{\beta-1}N_{p-2,3}, \\ \begin{aligned} N_{p,1}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \|\psi\|_1+l_2c_2\mathbf{B}(\beta,k_2+1)t_{p+1}^{\beta+k_2} \prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\\ &\quad +\sum_{k=1}^pC_{2,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +l_2c_2\mathbf{B}(\beta,k_2+1)\sum_{k=1}^pt_k^{\beta+k_2} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}\Big), \end{aligned}\\ \begin{aligned} N_{p,2}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \|\psi\|_1+l_2b_2\mathbf{B}(\beta,k_2+1)t_{p+1}^{\beta+k_2} \prod_{k=1}^p(t_k-t_{k-1})^{\beta-1} \\ &\quad +\sum_{k=1}^pB_{2,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +l_2b_2\mathbf{B}(\beta,k_2+1)\sum_{k=1}^pt_k^{\beta+k_2} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}\Big), \end{aligned}\\ \begin{aligned} N_{p,3}&=\frac{1}{\lambda }\Big(\prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\|\psi\|_1 +l_2a_2\mathbf{B}(\beta,k_2+1)t_{p+1}^{\beta+k_2} \prod_{k=1}^p(t_k-t_{k-1})^{\beta-1}\\ &\quad +\sum_{k=1}^pA_{2,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1} \\ &\quad +l_2a_2\mathbf{B}(\beta,k_2+1)\sum_{k=1}^pt_k^{\beta+k_2} \prod_{s=k+1}^p(t_s-t_{s-1})^{\beta-1}\Big), \end{aligned} \end{gather*} and \begin{gather*} \mu_{k,1}=l_2c_2(t_{k+1}-t_{k})^{1-\beta}t_{k+1}^{\alpha+k_1} \mathbf{B}(\beta,k_2+1)+N_{k,1},\quad k=0,1,\dots,p, \\ \mu_{k,2}=l_2b_2(t_{k+1}-t_{k})^{1-\beta}t_{k+1}^{\alpha+k_1} \mathbf{B}(\beta,k_2+1)+N_{k,2},\quad k=0,1,\dots,p, \\ \mu_{k,3}=l_2a_2(t_{k+1}-t_{k})^{1-\beta}t_{k+1}^{\alpha+k_1} \mathbf{B}(\beta,k_2+1)+N_{k,3},\quad k=0,1,\dots,p, \\ \mu_1=\max\left\{\mu_{k,1}:k=0,1,\dots,p\right\}, \\ \mu_2=\max\left\{\mu_{k,2}:k=0,1,\dots,p\right\}, \\ \mu_3=\max\left\{\mu_{k,3}:k=0,1,\dots,p\right\}. \end{gather*} \begin{theorem} \label{thm3.1} Suppose that both {\rm (A)} and {\rm (B)} hold. Let $\mu_2,\mu_3$ and $\sigma_2,\sigma_3$ be defined above. Then \eqref{e2} has at least one solution if \begin{equation}\label{e35} \sigma_2<1,\quad \mu_2\frac{1}{w((1-\sigma_2)/(2\sigma_3))}+\mu_3<1. \end{equation} \end{theorem} \begin{proof} To apply Lemma 2.1, we should define an open bounded subset $\Omega$ of $E$ centered at zero such that all assumptions in Lemma 2.1 hold. To obtain $\Omega$. Let $ \Omega_1=\{(x,y)\in E\cap D(L)\setminus \ker L,\;L(x,y)=\theta N(x,y)\text{ for some }\theta \in (0,1)\}$. We will prove that $\Omega_1$ is bounded. For $(x,y)\in \Omega_1$, we obtain $L(x,y)=\theta N(x,y)$ and $N(x,y)\in \operatorname{Im} L$. Then \begin{equation} \label{e36} \begin{gathered} D_{t_k^+}^\alpha x(t)=\theta m(t)f(t,x(t),y(t)), \\ D_{t_k^+}^\beta y(t)=\theta n(t)g(t,x(t),y(t)), \\ \lim_{t\to 1}t^{1-\alpha}x(t)+\lim_{t\to 0}t^{1-\alpha}x(t) =\theta\int_0^1\phi(t)F\left(t,x(t),y(t)\right)dt, \\ \lim_{t\to 1}t^{1-\beta}y(t)+\lim_{t\to 0}t^{1-\beta}y(t) =\theta\int_0^1\psi(t)G\left(t,x(t),y(t)\right)dt, \\ \lim_{t\to t_k^+}(t-t_k)^{1-\alpha}u(t)- u(t_k) =\theta I_k(t_k,u(t_k),v(t_k)),k=1,2,\dots,p, \\ \lim_{t\to t_k^+}(t-t_k)^{1-\beta}v(t)-v(t_k) =\theta J_k(t_k,u(t_k),v(t_k)),k=1,2,\dots,p. \end{gathered} \end{equation} So \begin{gather} \label{e37} x(t)=\theta\int_{t_k}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} m(s)f(s,x(s),y(s))ds+\theta (t-t_k)^{\alpha-1}M_k,t\in (t_k,t_{k+1}],\\ \label{e38} y(t)=\theta\int_{t_k}^t\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}n(s) g(s,x(s),y(s))ds+\theta (t-t_k)^{\alpha-1}N_k,t\in (t_k,t_{k+1}], \end{gather} for $k=0,1,\dots,p$. Here $M_k,N_k$ $(k=0,1,\dots,p)$ are given in Step (iv) in the proof of Lemma 2.2. By the definition of $M_k$, we have \begin{align*} &|M_0|\\ &\le \frac{1}{\lambda}\Big(\int_0^1|\phi(s)F(s,x(s),y(s))|ds +\int_{t_p}^{t_{p+1}}\frac{(t_{p+1}-s)^{\alpha-1}}{\Gamma(\alpha)}| m(s)f(s,x(s),y(s))|ds \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^p\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} \\ &\quad\times \Big(|I_k(t_k,x(t_k),y(t_k))|+\int_{t_{k-1}}^{t_k} \frac{(t_k-s)^{\alpha-1}}{\Gamma(\alpha)}|m(s)f(s,x(s),y(s))|ds\Big) \\ &\le \frac{1}{\lambda}\Big[C_1\|\phi\|_1+l_1c_1t_{p+1}^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)\\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pC_{1,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}+l_1c_1t_k^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)\Big] \\ &\quad +\frac{1}{\lambda}\Big[B_1\|\phi\|_1+l_1b_1\mathbf{B}(\alpha,k_1+1)\\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1} \sum_{k=1}^pB_{1,k}\prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1} +l_1b_1\mathbf{B}(\alpha,k_1+1)\Big]\|x\|\\ &\quad +\frac{1}{\lambda}\Big[B_1\|\phi\|_1+l_1b_1\mathbf{B}(\alpha,k_1+1) \\ &\quad +\prod_{k=1}^p(t_k-t_{k-1})^{\alpha-1}\sum_{k=1}^pB_{1,k} \prod_{s=k+1}^p(t_s-t_{s-1})^{\alpha-1}+l_1b_1\mathbf{B}(\alpha,k_1+1) \Big]\Phi^{-1}(\|y\|) \\ &=M_{0,1}+M_{0,2}\|x\|+M_{0,3}\Phi^{-1}(\|y\|). \end{align*} Similarly, \begin{gather*} |M_{1}|\le M_{1,1}+M_{1,2}\|x\|+M_{1,3}\Phi^{-1}(\|y\|), \\ \dots\\ |M_{p-1}|\le M_{p-1,1}+M_{p-1,2}\|x\|+M_{p-1,3}\Phi^{-1}(\|y\|), \\ |M_{p}|\le M_{p,1}+M_{p,2}\|x\|+M_{p,3}\Phi^{-1}(\|y\|). \end{gather*} Similarly, we can prove that \begin{gather*} |N_0|\le N_{0,1}+N_{0,2}\Phi(\|x\|)+N_{0,3}\|y\|, \\ |N_{1}|\le N_{1,1}+N_{1,2}\Phi(\|x\|)+N_{1,3}\|y\|,\\ \dots\\ |N_{p-1}|\le N_{p-1,1}+N_{p-1,2}\Phi(\|x\|)+N_{p-1,3}\|y\|,\\ |N_{p}|\le N_{p,1}+N_{p,2}\Phi(\|x\|)+N_{p,3}\|y\|. \end{gather*} First, using \eqref{e37} for $t\in (t_0,t_{1}]$, we have \begin{align*} (t-t_0)^{1-\alpha}|x(t)| &\le\big|(t-t_{0 })^{1-\alpha}\int_{t_0}^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds+M_0\big| \\ &\le l_1(t_1-t_{0})^{1-\alpha}t_1^{\alpha+k_1} \mathbf{B}(\alpha,k_1+1)(c_1+b_1\|x\|+a_1\Phi^{-1}(\|y\|)), \end{align*} \[ M_{0,1}+M_{0,2}\|x\|+M_{0,3}\Phi^{-1}(\|y\|) \le \sigma_{0,1}+\sigma_{0,2}\|x\|+\sigma_{0,3}\Phi^{-1}(\|y\|) . \] For $k=1,2,\dots,p-1$, we have \begin{align*} &(t-t_k)^{1-\alpha}|x(t)|\\ &\le\big|(t-t_{k})^{1-\alpha}\int_{t_k}^t\frac{(t-s)^{\alpha-1}} {\Gamma(\alpha)}m(s)f(s,x(s),y(s))ds+M_{k}\big| \\ &\le l_1(t-t_{k})^{1-\alpha}\int_{t_k}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} s^{k_1}ds(c_1+b_1\|x\|+a_1\Phi^{-1}(\|y\|))+|M_k| \\ &\le \sigma_{k,1}+\sigma_{k,2}\|x\|+\sigma_{k,3}\Phi^{-1}(\|y\|). \end{align*} For $t\in (t_p,t_{p+1}]$, we have \begin{align*} &(t-t_p)^{1-\alpha}|x(t)|\\ &\le\big|(t-t_{p })^{1-\alpha}\int_{t_p}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} m(s)f(s,x(s),y(s))ds+M_p\big|\\ &\le l_1(t-t_{p})^{1-\alpha}\int_{t_p}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} s^{k_1}ds(c_1+b_1\|x\|+a_1\Phi^{-1}(\|y\|))+|M_p|\\ &\le \sigma_{p,1}+\sigma_{p,2}\|x\|+\sigma_{p,3}\Phi^{-1}(\|y\|). \end{align*} It follows that \begin{equation} \label{e39} \|x\|\le \sigma_1+\sigma_2\|x\|+\sigma_3\Phi^{-1}(\|y\|). \end{equation} Similarly, we can show that \begin{equation} \label{e40} \|y\|\le \mu_1+\mu_2\Phi(\|x\|)+\mu_3\|y\|. \end{equation} From \eqref{e39} and \eqref{e40}, we obtain $$ \|y\|\le \mu_1+\mu_2\Phi\Big(\frac{\sigma_1}{1-\sigma_2} +\frac{\sigma_3\Phi^{-1}(\|y\|)}{1-\sigma_2}\Big)+\mu_3\|y\|. $$ Without loss of generality, assume that $\|y\|>\Phi(\frac{\sigma_1}{\sigma_3})$. Then \eqref{e3} implies that \begin{align*} \|y\|&\le \mu_1+\mu_2\Phi\Big(\frac{2\sigma_3\Phi^{-1}(\|y\|)}{1-\sigma_2}\Big) +\mu_3\|y\| \\ &\le \mu_1+\mu_2\frac{\Phi\big(\Phi^{-1}(\|y\|)\big)}{w((1-\sigma_2)/(2\sigma_3))} +\mu_3\|y\|\\ &=\mu_1+\Big(\mu_2\frac{1}{w((1-\sigma_2)/(2\sigma_3))}+\mu_3\Big)\|y\|. \end{align*} It follows that $$ \|y\|\le \frac{\mu_1}{1-\big(\mu_2\frac{1}{w((1-\sigma_2)/(2\sigma_3))}+\mu_3\big)}. $$ Then $$ \|x\|\le \sigma_1+\sigma_2\|x\|+\sigma_3\Phi^{-1} \Big(\frac{\mu_1}{1-\big(\mu_2\frac{1}{w((1-\sigma_2)/(2\sigma_3))} +\mu_3\big)}\Big). $$ It follows that $\Omega _1$ is bounded. Now we show that all assumptions of Lemma 2.1 are satisfied. Set $\Omega$ be a open bounded subset of $X$ centered at zero such that $\Omega \supset \overline{\Omega_1}$. By Lemma 2.2, $L$ is a Fredholm operator of index zero, $\ker L=\{0\in E\}$ and $N$ is $L$-compact on $\overline{\Omega}$. By the definition of $\Omega$, we have $Lx\not=\theta Nx$ for $x\in (D(L)\cap \partial \Omega $ and $\theta \in (0,1)$. Thus by Lemma 2.1, $L(x,y)=N(x,y)$ has at least one solution in $D(L)\cap \overline{\Omega}$. Then $x$ is a solution of \eqref{e2}. The proof is complete. \end{proof} As an application of Theorem 3.1, we give the following theorem, under the assumption \begin{itemize} \item[(B')] $f,g,F,G,I_k,J_k$ $(k=1,2,\dots,p)$ are impulsive Caratheodory functions and satisfy that there exist nonnegative constants $c_i,b_i,a_i(i=1,2)$, $C_i,B_i,A_i$ $(i=1,2)$ and $C_{i,k},B_{i,k},A_{i,k}$ $(i=1,2,k=1,2,\dots,p)$ such that \begin{gather*} |f(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le c_1+b_1|x|+a_1|y|,\\ t\in (t_k,t_{k+1}],\; k=0,1,\dots,p, \\ |g(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le c_2+b_2|x|+a_2|y|,\\ t\in (t_k,t_{k+1}],\; k=0,1,\dots,p, \\ |F(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le C_1+B_1|x|+A_1|y|,\\ t\in (t_k,t_{k+1}],\; k=0,1,\dots,p, \\ |G(t,(t-t_k)^{\alpha-1}x,(t-t_k)^{\beta-1}y)|\le C_2+B_2|x|+A_2|y|,\\ t\in (t_k,t_{k+1}],\; k=0,1,\dots,p, \\ |I_k(t,(t_{k+1}-t_k)^{\alpha-1}x,(t_{k+1}-t_k)^{\beta-1}y)| \le C_{1,k}+B_{1,k}|x|+A_{1,k}|y|, \\ k=1,2,\dots,p, \\ |J_k(t,(t_{k+1}-t_k)^{\alpha-1}x,(t_{k+1}-t_k)^{\beta-1}y)| \le C_{2,k}+B_{2,k}|x|+A_{2,k}|y|,\\ k=1,2,\dots,p. \end{gather*} \end{itemize} \begin{theorem} \label{thm3.2} Assume that {\rm (B')} holds. Let $\mu_2,\mu_3$ and $\sigma_2,\sigma_3$ be defined at the beginning of this section. Then \eqref{e2} has at least one solution if $$ \sigma_2<1,\quad \mu_2\frac{2\sigma_3}{1-\sigma_2}+\mu_3<1. $$ \end{theorem} For the proof of the above theorem, choose $\Phi(x)=x$ and then we obtain $\Phi^{-1}(x)=x$. The proof follows from Theorem 3.1 and is omitted. \section{An example} Now, we present an example that illustrates Theorem 3.1, and can not be covered by known results. Consider the boundary-value problem for the impulsive fractional differential equation \begin{equation} \label{e41} \begin{gathered} D_{t_k^+}^{\frac{2}{3}}u(t)=t^{-1/4}f(t,u(t),v(t)),\quad t\in (t_k,t_{k+1}],k=0,1, \\ D_{t_k^+}^{1/2} v(t)=t^{-1/4}g(t,u(t),v(t)),\quad t\in (t_k,t_{k+1}],k=0,1, \\ \lim_{t\to 1}u(t)+\lim_{t\to 0}t^{1/3}u(t)=0,\\ \lim_{t\to 1}v(t)+ \lim_{t\to 0}t^{1/2}v(t)=0,\\ \lim_{t\to \frac{1}{2}^+}(t-\frac{1}{2})^{1/3}u(t)-u(1/2)=0 ,\\ \lim_{t\to \frac{1}{2}^+}(t-\frac{1}{2})^{1/2}v(t)-v(1/2)=0, \end{gathered} \end{equation} where \begin{gather*} f(t,x,y)=\begin{cases} c_1+b_1t^{-\frac{1}{3}}x+a_1t^{-3/2}y^3,& t\in (0,1/2],\\ c_1+b_1(t-1/2)^{-\frac{1}{3}}x+a_1(t-1/2)^{-3/2}y^3, &t\in (1/2,1], \end{cases} \\ g(t,x,y)=\begin{cases} c_2+b_2t^{-\frac{1}{3}}x^{1/3}+a_2t^{-3/2}y, &t\in (0,1/2],\\ c_2+b_2(t-1/2)^{-\frac{1}{9}}x^{1/3}+a_2(t-1/2)^{-1/2}y, &t\in (1/2,1] \end{cases} \end{gather*} with $c_i,b_i,a_i\ge 0(i=1,2)$ and $0=t_0