\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 91, pp. 1--17.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/91\hfil Fractional-order impulsive PDEs] {Impulsive discontinuous hyperbolic partial differential equations of fractional order on Banach algebras} \author[S. Abbas, R. P. Agarwal, M. Benchohra\hfil EJDE-2010/91\hfilneg] {Sa\"{\i}d Abbas, Ravi P. Agarwal, Mouffak Benchohra} % in alphabetical order \address{Sa\"{\i}d Abbas \newline Laboratoire de Math\'{e}matiques, Universit\'{e} de Sa\"{\i}da, B.P. 138, 20000, Sa\"{\i}da, Alg\'{e}rie} \email{abbas\_said\_dz@yahoo.fr} \address{Ravi P. Agarwal \newline Department of Mathematical Sciences, Florida Institute of Technology\\ Melboune, Florida, 32901-6975, USA\newline KFUPM Chair Professor, Mathematics and Statistics Department\\ King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia} \email{agarwal@fit.edu} \address{Mouffak Benchohra \newline Laboratoire de Math\'ematiques, Universit\'e de Sidi Bel-Abb\`es \\ B.P. 89, 22000, Sidi Bel-Abb\`es, Alg\'erie} \email{benchohra@univ-sba.dz} \thanks{Submitted May 6, 2010. Published July 7, 2010.} \subjclass[2000]{26A33} \keywords{Impulsive differential equations; fractional order; upper solution; \hfill\break\indent lower solution; extremal solutions; left-sided mixed Riemann-Liouville integral; \hfill\break\indent Caputo fractional-order derivative; fixed point; Banach algebra} \begin{abstract} This article studies the existence of solutions and extremal solutions to partial hyperbolic differential equations of fractional order with impulses in Banach algebras under Lipschitz and Carath\'eodory conditions and certain monotonicity conditions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} This article studies the existence of solutions to fractional order initial-value problems ($IVP$ for short), for the system \begin{gather}\label{e1} ^{c}D_{0}^{r}\Big(\frac{u(x,y)}{f(x,y,u(x,y))}\Big) =g(x,y,u(x,y)),\quad (x,y)\in J,\; x\neq x_k, \; k=1,\ldots,m, \\ \label{e2} u(x_k^+,y)=u(x_k^-,y)+I_{k}(u(x_k^-,y)), \quad y\in [0,b]; \; k=1,\dots,m, \\ \label{e3} u(x,0)=\varphi (x), \quad u(0,y)=\psi (y), \quad x\in [0,a], \; y\in [0,b], \end{gather} where $J=[0,a]\times [0,b]$, $a,b>0$, $ ^{c}D_{0}^{r}$ is the Caputo's fractional derivative of order $r=(r_1,r_2)\in (0,1]\times (0,1]$, $0=x_{0}0: \|u(x,y)\|\leq c, \text{ a.e. } (x,y)\in J\}, $$ where $\|\cdot\|$ denotes a suitable complete norm on $\mathbb{R}^{n}$. By $L^{1}(J,\mathbb{R}^{n})$ we denote the space of Lebesgue-integrable functions $u:J\to \mathbb{R}^{n}$ with the norm $$ \|u\|_{1}=\int_{0}^{a}\int_{0}^{b}\|u(x,y)\|\,dy\,dx. $$ $AC(J, \mathbb{R}^{n})$ is the space of absolutely continuous valued functions on $J$. Denote by $D ^{2}_{xy}:=\frac{\partial ^{2}}{\partial x\partial y}$ the mixed second order partial derivative. In all what follows set $$ J_{k}:=(x_{k},x_{k+1}]\times[0,b], \quad k=0,1,\ldots,m. $$ Consider the space \begin{align*} PC(J,\mathbb{R}^{n})= &\big\{u: J\to\mathbb{R}^{n}: u \in C(J_{k}, \mathbb{R}^{n}) ; k=1, \ldots,m, \text{ and there exist}\\ & u(x_{k}^-,y), u(x_{k}^+,y) ; k=1,\ldots,m, \text{ with } u(x_{k}^-,y)=u(x_{k},y) \big\}. \end{align*} This set is a Banach space with the norm $$ \|u\|_{PC}=\sup_{(x,y)\in J}\|u(x,y)\|. $$ Define a multiplication $``\,\cdot\,"$ by $$ (u\cdot v)(x,y)=u(x,y)v(x,y) \quad \text{for } (x,y)\in J. $$ Then $PC(J,\mathbb{R}^{n})$ is a Banach algebra with the above norm and multiplication. Let $a_1\in[0,a]$, $z^+=(a_1^+,0)\in J$, $J_z=[a_1,a]\times[0,b]$, $r_{1}, r_{2}>0 $ and $r=(r_{1},r_{2})$. For $u\in L^{1}(J_z,\mathbb{R}^{n})$, the expression $$ (I_{z^+}^{r}u)(x,y)=\frac{1}{\Gamma (r_{1})\Gamma (r_{2})} \int_{a_1^+}^{x} \int_{0}^{y}(x-s)^{r_{1}-1}(y-t)^{r_{2}-1}u(s,t)\,dt\,ds, $$ where $\Gamma (.)$ is the Euler gamma function, is called the left-sided mixed Riemann-Liouville integral of order $r$. \begin{definition}[\cite{ViGo}] \label{def2.1} \rm For $u\in L^{1}(J_z,\mathbb{R}^{n})$, the Caputo fractional-order derivative of order $r$ is defined by the expression $(^{c}D_{z^+}^{r}u)(x,y)=( I_{z^+}^{1-r}D ^{2}_{xy}u) (x,y)$. \end{definition} Let $X$ be a Banach algebra with norm $\|\cdot \|$. An operator $ T:X \to X$ is called \emph{compact} if $\overline {T(S)}$ is a compact subset of $ X$ for any $S\subset X$. Similarly $T: X \to X $ is called \emph{totally bounded} if $T$ maps a bounded subset of $X$ into the relatively compact subset of $X$. Finally $T: X \to X$ is called \emph{completely continuous} operator if it is continuous and totally bounded operator on $X$. It is clear that every compact operator is totally bounded, but the converse may not be true. \begin{definition}\label{Car} \rm A function $\gamma :J \times \mathbb{R}^{n} \to \mathbb{R}^{n}$ is called \emph{Carath\'eodory}'s if \begin{enumerate} \item [(i)] the function $(x,y)\to \gamma(x,y,u)$ is measurable for each $u\in \mathbb{R}^{n}$, \item [(ii)] the function $u\to \gamma(x,y,u)$ is continuous for almost each $(x,y) \in J$. \end{enumerate} \end{definition} A non-empty closed set $K$ in a Banach algebra $X$ is called a \emph{cone} if \begin{itemize} \item [(i)] $K+K\subseteq K$, \ \item [(ii)] $\lambda K\subseteq K$ for $\lambda\in \mathbb{R}, \ \lambda\ge 0$ and \item [(iii)] $\{-K\}\cap K=0$, where $0$ is the zero element of $X$. \end{itemize} The cone $K$ is called to be \emph{positive} if \begin{itemize} \item [(iv)] $K\circ K\subseteq K$, where ``$\circ$'' is a multiplication composition in $X$. \end{itemize} We introduce an order relation, $\le$, in $X$ as follows. Let $u,v\in X$. Then $u\le v$ if and only if $v-u\in K$. A cone $K$ is called to be \emph{normal} if the norm $\|\cdot\|$ is monotone increasing on $K$. It is known that if the cone $K$ is normal in $X$, then every order-bounded set in $X$ is norm-bounded. \begin{lemma}[\cite{D2}] Let $K$ be a positive cone in a real Banach algebra $X$ and let $u_{1}, u_{2}, v_{1}, v_{2}\in K$ be such that $u_{1}\le v_{1}$ and $u_{2}\le v_{2}$. Then $u_{1}u_{2}\le v_{1}v_{2}$. \end{lemma} For any $v,w\in X, v\le w$, the order interval $[v,w]$ is a set in $X$ given by $$ [v,w]=\{u\in X: v\le u\le w\}. $$ The nonlinear alternative of Schaefer type proved by Dhage \cite{D1} is embodied in the following theorem. \begin{theorem} \label{td} Let $X$ be a Banach algebra and let $A,B : X \to X$ be two operators satisfying \begin{itemize} \item [(a)] $A$ is Lipschitz with a Lipschitz constant $\alpha$, \item [(b)] $B$ is compact and continuous, and \item [(c)] $\alpha M< 1$, where $M= \|B(X)\|:=\sup\{\|Bu\| : u\in X\}$. \end{itemize} Then either \begin{itemize} \item[(i)] the equation $\lambda [Au \,Bu]=u $ has a solution for $0<\lambda <1$, or \item[(ii)] the set ${\mathcal E }=\{u \in X: \lambda [Au\,Bu] =u,\, 0<\lambda <1 \}$ is unbounded. \end{itemize} \end{theorem} We use the following fixed point theorems by Dhage \cite{D2} for proving the existence of extremal solutions for our problem under certain monotonicity conditions. \begin{theorem}\label{t21} Let $K$ be a cone in a Banach algebra $X$ and let $v,w\in X$. Suppose that $A,B: [v,w]\to K$ are two operators such that \begin{itemize} \item[(a)] $A$ is completely continuous, \item[(b)] $B$ is totally bounded, \item[(c)] $Au_1Bu_2\in [v,w]$ for all $u_1,u_2\in [v,w]$, and \item[(d)] $A$ and $B$ are nondecreasing. \end{itemize} Further if the cone $K$ is positive and normal, then the operator equation $Au\,Bu=u$ has a least and a greatest positive solution in $[v,w]$. \end{theorem} \begin{theorem} \label{t22} Let $K$ be a cone in a Banach algebra $X$ and let $v,w\in X$. Suppose that $A,B: [v,w]\to K$ are two operators such that \begin{itemize} \item[(a)] $A$ is Lipschitz with a Lipschitz constant $\alpha$, \item[(b)] $B$ is totally bounded, \item[(c)] $Au_1\,Bu_2\in [v,w]$ for all $u_1,u_2\in [v,w]$, and \item[(d)] $A$ and $B$ are nondecreasing. \end{itemize} Further if the cone $K$ is positive and normal, then the operator equation $Au\,Bu=u$ has least and a greatest positive solution in $[v,w]$, whenever $\alpha M<1$, where $M=\|B([v,w])\|:=\sup\{\|Bu\|: u\in [v,w]\}$. \end{theorem} \begin{remark} \label{rmk2.7} \rm Note that hypothesis (c) of Theorems \ref{t21} and \ref{t22} holds if the operators $A$ and $B$ are positive monotone increasing and there exist elements $v$ and $w$ in $X$ such that $v\le Av\,Bv$ and $Aw\,Bw\le w$. \end{remark} \section{Auxiliary Results} Let us start by defining what we mean by a solution of problem \eqref{e1}-\eqref{e3}. Set $J':=J\backslash\{(x_{1},y),\dots,(x_{m},y), \; y\in [0,b]\}$. \begin{definition}\label{d1} \rm A function $u\in PC(J,\mathbb{R}^n)$ whose $r$-derivative exists on $J'$ is said to be a solution of \eqref{e1}-\eqref{e3} if \begin{itemize} \item[(i)] the function $(x,y)\mapsto \frac{u(x,y)}{f(x,y,u(x,y))}$ is absolutely continuous, and \item[(ii)] $u$ satisfies $^{c}D_{0}^{r}\big(\frac{u(x,y)}{f(x,y,u(x,y))}\big)= g(x,y,u(x,y))$ on $ J'$ and conditions \eqref{e2}, \eqref{e3} are satisfied. \end{itemize} \end{definition} Let $f\in C([x_k,x_{k+1}]\times[0,b],\mathbb{R}^{n}\backslash \{0_{\mathbb{R}^{n}}\})$, $g\in L^{1}([x_k,x_{k+1}]\times[0,b],\mathbb{R}^{n}) $, $z_k=(x_k,0)$, and $$ \mu_{0,k}(x,y)=\frac{u(x,0)}{f(x,0)}+\frac{u(x_k^+,y)}{f(x_k^+,y)} -\frac{u(x_k^+,0)}{f(x_k^+,0)}, \quad k=0,\dots,m. $$ For the existence of solutions for the problem \eqref{e1}-\eqref{e3}, we need the following lemma. \begin{lemma}\label{L1} A function $u\in AC([x_k,x_{k+1}]\times[0,b],\mathbb{R}^{n})$, $k=0,\dots,m $ is a solution of the differential equation \begin{equation}\label{e1'} ^{c}D_{z_k}^{r}\Big(\frac{u}{f}\Big)(x,y)=g(x,y), \quad (x,y)\in [x_k,x_{k+1}]\times[0,b], \end{equation} if and only if $u(x,y)$ satisfies \begin{equation}\label{e2'} u(x,y)=f(x,y)\Big(\mu_{0,k}(x,y)+(I_{z_k}^{r}g)(x,y)\Big),\quad (x,y)\in [x_k,x_{k+1}]\times[0,b]. \end{equation} \end{lemma} \begin{proof} Let $u(x,y)$ be a solution of \eqref{e1'}. Then, taking into account the definition of the derivative $^{c}D_{z_k}^{r}$, we have $$ I_{z_k^+}^{1-r}( D_{xy}^{2}\frac{u}{f})(x,y)=g(x,y). $$ Hence, we obtain $$ I_{z_k^+}^{r}(I_{z_k}^{1-r}D_{xy}^{2}\frac{u}{f}) (x,y) =(I_{z_k^+}^{r}g)(x,y), $$ then $$ I_{z_k^+}^{1}(D_{xy}^{2}\frac{u}{f})( x,y) =( I_{z_k^+}^{r}g)(x,y). $$ Since $$ I_{z_k^+}^{1}( D_{xy}^{2}\frac{u}{f})(x,y) =\frac{u(x,y)}{f(x,y)} -\frac{u(x,0)}{f(x,0)} -\frac{u(x_k^+,y)}{f(x_k^+,y)} +\frac{u(x_k^+,0)}{f(x_k^+,0)} , $$ we have $$ u(x,y) =f(x,y)\Big(\mu_{0,k}(x,y)+(I_{z_k}^{r}g)(x,y)\Big). $$ Now let $u(x,y)$ satisfies \eqref{e2'}. It is clear that $u(x,y)$ satisfies \eqref{e1'}. \end{proof} \begin{corollary} \label{coro3.3} The function $u\in AC([x_k,x_{k+1}]\times[0,b],\mathbb{R}^{n})$, $k=0,\dots,m $ is a solution of the differential equation \eqref{e1} if and only if $u$ satisfies the equation \begin{equation} \label{e32} \begin{aligned} u(x,y)&=\big[f(x,y,u(x,y))\big]\Big(\mu_k(x,y)\\ &\quad +\frac{1}{\Gamma (r_{1})\Gamma (r_{2})}\int_{0}^{x} \int_{0}^{y}(x-s)^{r_{1}-1}(y-t)^{r_{2}-1}g(s,t,u(s,t))\,dt\,ds\Big), \end{aligned} \end{equation} for $(x,y) \in [x_k,x_{k+1}]\times[0,b]$, where $$ \mu_k(x,y)=\frac{u(x,0)}{f(x,0,u(x,0))} +\frac{u(x_k^+,y)}{f(x_k^+,y,u(x_k^+,y))} -\frac{u(x_k^+,0)}{f(x_k^+,0,u(x_k^+,0))}, \quad k=0,\dots,m. $$ \end{corollary} Let $\mu':=\mu_{0,0}$. \begin{lemma}\label{L2} Let $0< r_1,r_2\leq 1$ and let $ f: J \to\mathbb{R}^{n}\backslash \{0_{\mathbb{R}^{n}}\}$, $ g: J \to\mathbb{R}^{n}$ be continuous. A function $u$ is a solution of the fractional integral equation \begin{equation}\label{e4} u(x,y)=\begin{cases} f(x,y)\Big[\mu'(x,y)+\frac{1}{\Gamma(r_1)\Gamma(r_2)}\int_{0}^{x} \int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big],\\ \quad \text{if } (x,y)\in [0,x_{1}]\times[0,b], \\[3pt] f(x,y)\Big[\mu'(x,y)+\sum_{i=1}^{k} \Big(\frac{I_{i}(u(x_{i}^{-},y))}{f(x_{i}^{+},y)} -\frac{I_{i}(u(x_{i}^{-},0))}{f(x_{i}^{+},0)}\Big) \\ +\frac{1}{\Gamma(r_1)\Gamma(r_2)}\sum_{i=1}^{k} \int_{x_{i-1}}^{x_{i}}\int_{0}^{y} (x_{i}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds \\ + \frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{k}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds \Big],\\ \quad \text{if } (x,y)\in (x_{k},x_{k+1}]\times[0,b],\; k=1,\dots,m, \end{cases} \end{equation} if and only if $u$ is a solution of the fractional initial-value problem \begin{gather}\label{e5} ^{c}D^{r}(\frac{u}{f})(x,y)= g(x,y), \quad (x,y)\in J', \\ \label{e6} u(x_{k}^{+},y)= u(x_{k}^{-},y)+I_{k}(u(x_{k}^{-},y)), \ y\in [0,b], \ k=1,\dots,m. \end{gather} \end{lemma} \begin{proof} Assume $ u $ satisfies \eqref{e5}-\eqref{e6}. If $(x,y)\in [0,x_{1}]\times[0,b]$, then $$ ^{c}D^{r}(\frac{u}{f})(x,y)= g(x,y). $$ Lemma \ref{L1} implies $$ u(x,y)=f(x,y)\Big(\mu'(x,y)+\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{0}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}h(s,t)\,dt\,ds\Big). $$ If $(x,y)\in (x_{1},x_{2}]\times[0,b]$, then Lemma \ref{L1} implies \begin{align*} &u(x,y)\\ &=f(x,y)\Big(\mu_{0,1}(x,y)+\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)}+\frac{u(x_{1}^{+},y)}{f(x_{1}^{+},y)} -\frac{u(x_{1}^{+},0)}{f(x_{1}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)}+\frac{u(x_{1}^{-},y)}{f(x_{1}^{+},y)} -\frac{u(x_{1}^{-},0)}{f(x_{1}^{+},0)} +\frac{I_{1}(u(x_{1}^{-},y))}{f(x_{1}^{+},y)}-\frac{I_{1}(u(x_{1}^{-},0))}{f(x_{1}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)}+\frac{u(x_{1},y)}{f(x_{1}^{+},y)} -\frac{u(x_{1},0)}{f(x_{1}^{+},0)} +\frac{I_{1}(u(x_{1}^{-},y))}{f(x_{1}^{+},y)}-\frac{I_{1}(u(x_{1}^{-},0))}{f(x_{1}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)}+\frac{\psi(y)}{f(0,y)} -\frac{u(0,0)}{f(0,0)} +\frac{I_{1}(u(x_{1}^{-},y))}{f(x_{1}^{+},y)}-\frac{I_{1}(u(x_{1}^{-},0))}{f(x_{1}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{0}^{x_{1}}\int_{0}^{y}(x_{1}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\mu'(x,y)+\frac{I_{1}(u(x_{1}^{-},y))}{f(x_{1}^{+},y)}-\frac{I_{1}(u(x_{1}^{-},0))}{f(x_{1}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{0}^{x_{1}}\int_{0}^{y}(x_{1}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big). \end{align*} If $(x,y)\in (x_{2},x_{3}]\times[0,b]$, then from Lemma \ref{L1} we obtain \begin{align*} &u(x,y)\\ &=f(x,y)\Big(\mu_{0,2}(x,y)+\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{2}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)}+\frac{u(x_{2}^{+},y)}{f(x_{2}^{+},y)}-\frac{u(x_{2}^{+},0)}{f(x_{2}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{2}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)} +\frac{u(x_{2}^{-},y)}{f(x_{2}^{+},y)} -\frac{u(x_{2}^{-},0)}{f(x_{2}^{+},0)} +\frac{I_{2}(u(x_{2}^{-},y))}{f(x_{2}^{+},y)}-\frac{I_{2}(u(x_{2}^{-},0))}{f(x_{2}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{2}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\frac{\varphi(x)}{f(x,0)} +\frac{u(x_{2},y)}{f(x_{2}^{+},y)}-\frac{u(x_{2},0)}{f(x_{2}^{+},0)} +\frac{I_{2}(u(x_{2}^{-},y))}{f(x_{2}^{+},y)}-\frac{I_{2}(u(x_{2}^{-},0))}{f(x_{2}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{2}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big)\\ &=f(x,y)\Big(\mu'(x,y)+\frac{I_{1}(u(x_{1}^{-},y))}{f(x_{1}^{+},y)} -\frac{I_{1}(u(x_{1}^{-},0))}{f(x_{1}^{+},0)} +\frac{I_{2}(u(x_{2}^{-},y))}{f(x_{2}^{+},y)}-\frac{I_{2}(u(x_{2}^{-},0))}{f(x_{2}^{+},0)}\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{0}^{x_{1}}\int_{0}^{y}(x_{1}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\\ &\quad+\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{1}}^{x_{2}}\int_{0}^{y}(x_{2}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\\ &\quad+\frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{2}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t)\,dt\,ds\Big). \end{align*} If $(x,y)\in (x_{k},x_{k+1}]\times[0,b]$ then again from Lemma \ref{L1} we get \eqref{e4}. Conversely, assume that $u$ satisfies the impulsive fractional integral equation \eqref{e4}. If $(x,y)\in [0,x_{1}]\times[0,b]$ and using the fact that $^{c}D^{r}$ is the left inverse of $I^{r}$ we get $$ ^{c}D^{r}(\frac{u}{f})(x,y)= g(x,y), \quad \text{for each } (x,y)\in [0,x_{1}]\times[0,b]. $$ If $(x,y)\in [x_{k},x_{k+1})\times[0,b]$, $k=1,\dots,m$ and using the fact that $^{c}D^{r}C=0$, where $C$ is a constant, we get $$ ^{c}D^{r}(\frac{u}{f})(x,y)=g(x,y), \text{for each } (x,y)\in [x_{k},x_{k+1})\times[0,b]. $$ Also, we can easily show that $$ u(x_{k}^{+},y)=u(x_{k}^{-},y)+ I_{k}(u(x_{k}^{-},y)), \quad y\in [0,b], k=1,\dots,m. $$ \end{proof} Let $\mu:=\mu_{0}$. \begin{corollary}\label{c2} Let $0< r_1,r_2\leq 1$ and let $ f: J\times\mathbb{R}^{n} \to\mathbb{R}^{n}\backslash \{0_{\mathbb{R}^{n}}\}$, $ g: J\times\mathbb{R}^{n} \to\mathbb{R}^{n}$ be continuous. A function $u$ is a solution of the fractional integral equation \begin{equation}\label{e7} u(x,y)=\begin{cases} f(x,y,u(x,y))\Big[\mu(x,y)\\ +\frac{1}{\Gamma(r_1)\Gamma(r_2)}\int_{0}^{x}\int_{0}^{y} (x-s)^{r_1-1}(y-t)^{r_2-1}g(s,t,u(s,t))\,dt\,ds\Big],\\ \quad \text{if } (x,y)\in [0,x_{1}]\times[0,b], \\[4pt] f(x,y,u(x,y))\Big[\mu(x,y)+\sum_{i=1}^{k} \Big(\frac{I_{i}(u(x_{i}^{-},y))}{f(x_{i}^{+},y,u(x_{i}^{+},y))} -\frac{I_{i}(u(x_{i}^{-},0))}{f(x_{i}^{+},0,u(x_{i}^{+},0))}\Big)\\ +\frac{1}{\Gamma(r_1)\Gamma(r_2)}\sum_{i=1}^{k} \int_{x_{i-1}}^{x_{i}}\int_{0}^{y} (x_{i}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t,u(s,t))\,dt\,ds\\ + \frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{k}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1} g(s,t,u(s,t))\,dt\,ds\Big],\\ \quad \text{if } (x,y)\in (x_{k},x_{k+1}]\times[0,b],\ k=1,\dots,m, \end{cases} \end{equation} if and only if $u$ is a solution of the fractional initial-value problem \begin{gather}\label{e8} ^{c}D^{r}\Big(\frac{u(x,y)}{f(x,y,u(x,y))}\Big) = g(x,y,u(x,y)), \quad (x,y)\in J', \\ \label{e9} u(x_{k}^{+},y)= u(x_{k}^{-},y)+I_{k}(u(x_{k}^{-},y)), \ y\in [0,b], \ k=1,\dots,m. \end{gather} \end{corollary} \section{Existence of Solutions} In this section, we are concerned with the existence of solutions for the problem \eqref{e1}-\eqref{e3}. The following hypotheses will be used in the sequel. \begin{itemize} \item[(A1)] The function $f$ is continuous on $J\times \mathbb{R}^{n}$. \item[(A2)] There exists a function $\alpha\in C(J,\mathbb{R}_+)$ such that $$ \|f(x,y,u)-f(x,y,\overline{u})\|\le \alpha(x,y)\|u-\overline{u}\| ; \quad\text{for all} \ (x,y)\in J\ \text{and} \ u, \overline{u} \in \mathbb{R}^{n}.$$ \item[(A3)] The function $g$ is Carath\'eodory, and there exists $h\in L^{\infty}(J,\mathbb{R}_+)$ such that $$ \|g(x,y,u)\|\le h(x,y); \quad\text{a.e. } (x,y)\in J,\ \text{for all} \ u\in \mathbb{R}^{n}. $$ \item[(A4)] There exists a function $\beta\in C(J,\mathbb{R}_+)$ such that $$ \Big\|\frac{I_k(u)}{f(x,y,u)}\Big\|\le \beta(x,y); \quad\text{for all} \ (x,y)\in J\ \text{and} \ u\in \mathbb{R}^{n}.$$ \end{itemize} \begin{theorem}\label{tc} Assume that hypotheses {\rm (A1)--(A4)} hold. If \begin{equation}\label{e3'} \|\alpha\|_{\infty}\Big[\|\mu\|_{\infty}+ 2m\|\beta\|_{\infty}+ \frac{2a^{r_{1}}b^{r_{2}}\|h\|_{L^{\infty}}}{\Gamma(r_{1}+1) \Gamma(r_{2}+1)}\Big]<1, \end{equation} Then the initial-value problem \eqref{e1}-\eqref{e3} has at least one solution on $J$. \end{theorem} \begin{proof} Let $X:=PC(J,\mathbb{R}^{n})$. Define two operators $A$ and $B$ on $X$ by \begin{equation}\label{e33} Au(x,y)=f(x,y,u(x,y));\quad (x,y)\in J, \end{equation} and \begin{equation} \label{e34} \begin{aligned} Bu(x,y)&=\mu(x,y)+\sum_{i=1}^{m} \Big(\frac{I_{i}(u(x_{i}^{-},y))}{f(x_{i}^{+},y,u(x_{i}^{+},y))} -\frac{I_{i}(u(x_{i}^{-},0))}{f(x_{i}^{+},0,u(x_{i}^{+},0))}\Big)\\ &\quad +\frac{1}{\Gamma(r_1)\Gamma(r_2)} \sum_{i=1}^{m}\int_{x_{i-1}}^{x_{i}}\int_{0}^{y} (x_{i}-s)^{r_1-1}(y-t)^{r_2-1}g(s,t,u(s,t))\,dt\,ds \\ &\quad + \frac{1}{\Gamma(r_1)\Gamma(r_2)} \int_{x_{m}}^{x}\int_{0}^{y}(x-s)^{r_1-1}(y-t)^{r_2-1} g(s,t,u(s,t))\,dt\,ds; \end{aligned} \end{equation} with $(x,y)\in J$. Solving \eqref{e1}-\eqref{e3} is equivalent to solving \eqref{e32}, which is further equivalent to solving the operator equation \begin{equation}\label{e35} Au(x,y)\,Bu(x,y)=u(x,y), \quad (x,y)\in J. \end{equation} We show that operators $A$ and $B$ satisfy all the assumptions of Theorem \ref{td}. First we shall show that $A$ is a Lipschitz. Let $u_1, u_2\in X$. Then by (A2), \begin{align*} \|Au_1(x,y)-Au_2(x,y)\| &=\|f(x,y,u_1(x,y))-f(x,y,u_2(x,y))\|\\ &\le \alpha(x,y)\|u_1(x,y)-u_2(x,y)\|\\ &\le \|\alpha\|_{\infty}\|u_1-u_2\|_{PC}. \end{align*} Taking the maximum over $(x,y)$, in the above inequality yields $$ \|Au_1-Au_2\|_{PC}\le \|\alpha\|_{\infty}\|u_1-u_2\|_{PC}, $$ and so $A$ is a Lipschitz with a Lipschitz constant $\|\alpha\|_{\infty}$. Next, we show that $B$ is compact operator on $X$. Let $\{u_n\}$ be a sequence in $X$. From (A3) and (A4) it follows that $$ \|Bu_n\|_{PC} \le \|\mu\|_{\infty}+ 2m\|\beta\|_{\infty}+ \frac{2a^{r_{1}}b^{r_{2}}\|h\|_{L^{\infty}}}{\Gamma(r_{1}+1) \Gamma(r_{2}+1)}. $$ As a result $\{Bu_n: n\in {\mathbb N}\}$ is a uniformly bounded set in $X$. Let $(\tau_{1},y_{1}), (\tau_{2},y_{2})\in J, \,\,\, \tau_{1}<\tau_{2}$ and $y_{1} 0$ such that $\|u\|\le r$ for all $u\in [{\underline u},\overline{u}]$. As $f$ is continuous on compact set $J\times[-r,r]$, it attains its maximum, say $M$. Therefore, for any subset $S$ of $[{\underline u},\overline{u}]$ we have \begin{align*} \|A(S)\|&= \sup\{ \|Au\| : u\in S\}\\ &=\sup\Big\{ \sup_{(x,y)\in J} \|f(x,y, u(x,y))\|: u\in S\Big\}\\ & \le \sup\Big\{\sup_{(x,y)\in J}\|f(x,y, u)\| :u\in [-r,r]\Big\} \le M. \end{align*} This shows that $A(S)$ is a uniformly bounded subset of $X$. We note that the function $f(x,y,u)$ is uniformly continuous on $J\times[-r,r]$. Therefore, for any $(\tau_1,y_1),(\tau_2,y_2)\in J$ we have $$ \|f(\tau_1,y_1, u)-f(\tau_2,y_2, u)\|\to 0\quad \mbox{as } (\tau_1,y_1)\to (\tau_2,y_2), $$ for all $u\in [-r,r]$. Similarly for any $u_1,u_2\in [-r,r]$ $$ \|f(x,y,u_1)-f(x,y,u_2)\|\to 0\quad \mbox{as } u_1\to u_2, $$ for all $(x,y)\in J$. Hence for any $(\tau_1,y_1),(\tau_2,y_2)\in J$ and for any $u\in S$ one has \begin{align*} \|Au(\tau_1,y_1)-Au(\tau_2,y_2)\|&=\|f(\tau_1,y_1, u(\tau_1,y_1))-f(\tau_2,y_2, u(\tau_2,y_2))\|\\ &\le \|f(\tau_1,y_1, u(\tau_1,y_1))-f(\tau_2,y_2, u(\tau_1,y_1)\|\\ &\quad +\|f(\tau_2,y_2, u(\tau_1,y_1))-f(\tau_2,y_2, u(\tau_2,y_2))\|\\ &\to 0\quad \mbox{as } (\tau_1,y_1)\to (\tau_2,y_2). \end{align*} This shows that $A(S)$ is an equicontinuous set in $K$. Now an application of Arzel\`a-Ascoli theorem yields that $A$ is a completely continuous operator on $[{\underline u},\overline{u}]$. \end{proof} Next it can be shown as in the proof of Theorem \ref{t31} that $B$ is a compact operator on $[{\underline u},\overline{u}]$. Now an application of Theorem \ref{t21} yields that the problem \eqref{e1}-\eqref{e3} has a minimal and maximal positive solution on $J$. \section{An Example} As an application of our results we consider the following partial hyperbolic functional differential equations of the form \begin{gather}\label{ex1} ^{c}D_{0}^{r}\Big(\frac{u(x,y)}{f(x,y,u(x,y))}\Big) =g(x,y,u(x,y)),\quad (x,y)\in [0,1]\times [0,1], \\ \label{ex2} u\Big(\frac{1}{2}^+,y\Big)=u\Big(\frac{1}{2}^-,y\Big) +I_{1}\Big(u\Big(\frac{1}{2}^-,y\Big)\Big), \quad y\in [0,1], \\ \label{ex3} u(x,0)=\varphi (x), \quad x\in [0,1], \quad u(0,y)=\psi (y), \quad y\in [0,1], \end{gather} where $f,g : [0,1]\times [0,1]\times \mathbb{R}\to \mathbb{R}$ and $I_{1}:\mathbb{R}\to\mathbb{R}$ are defined by \begin{gather*} f(x,y,u)= \frac{1}{e^{x+y+10}(1+|u|)},\\ g(x,y,u)=\frac{1}{e^{x+y+8}(1+u^{2})},\\ I_{1}(u)=\frac{(8+e^{-10})^{2}}{512e^{10}(1+|u|)^{2}}. \end{gather*} The functions $\varphi, \psi:[0,1]\to \mathbb{R}$ are defined by $$ \varphi(x)=\begin{cases} \frac{x^2}{2}e^{-10}; & \text{if } x\in [0,\frac{1}{2}],\\ x^2e^{-10}; & \text{if } x\in (\frac{1}{2},1], \end{cases} $$ and $$ \psi(y)=ye^{-10}, \quad \text{for all } y\in [0,1]. $$ We show that the functions $\varphi, \psi, f, g$ and $I_1$ satisfy all the hypotheses of Theorem \ref{tc}. Clearly, the function $f$ satisfies (A1) and (A2) with $\alpha(x,y)=\frac{1}{e^{x+y+10}}$ and $$ \|\alpha\|_{\infty}=1/e^{10}. $$ Also, the function $g$ satisfies (A3) with $h(x,y)=\frac{1}{e^{x+y+8}}$ and $$ \|h\|_{L^{\infty}}=1/e^{8}. $$ Finally, condition (A4) holds with $\beta(x,y)=\frac{81e^{x+y}}{512}$ and $\|\beta\|_{\infty}=\frac{81e^{2}}{512}$. A simple computation gives $\mu(x,y)<4e$. Condition \eqref{e3'} holds. Indeed \begin{align*} &\|\alpha\|_{\infty}\Big[\|\mu\|_{\infty}+2m\|\beta\|_{\infty}+ \frac{2a^{r_{1}}b^{r_{2}}\|h\|_{L^{\infty}}}{\Gamma(r_{1}+1) \Gamma(r_{2}+1)}\Big]\\ &<\frac{1}{e^{10}} \Big[4e+\frac{81e^{2}}{256}+\frac{2}{e^{8}\Gamma(r_{1}+1)\Gamma(r_{2}+1)} \Big]<1, \end{align*} for each $(r_1,r_2)\in (0,1]\times (0,1]$. Hence by Theorem \ref{tc}, problem \eqref{ex1}-\eqref{ex3} has a solution defined on $[0,1]\times [0,1]$. \begin{thebibliography}{00} \bibitem{AbBe1} S. Abbas and M. 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