\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 241, pp. 1--15.\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-2013/241\hfil piecewise continuous mild solutions] {Existence of piecewise continuous mild solutions for impulsive functional differential equations with iterated deviating arguments} \author[P. Kumar, D. N. Pandey, D. Bahuguna \hfil EJDE-2013/241\hfilneg] {Pradeep Kumar, Dwijendra N. Pandey, Dhirendra Bahuguna} \address{Pradeep Kumar \newline Department of Mathematics and Statistics, Indian Institute of Technology Kanpur, Pin 208016, India} \email{prdipk@gmail.com} \address{Dwijendra N. Pandey \newline Department of Mathematics, Indian Institute of Technology Roorkee, Pin 247667, India} \email{dwij.iitk@gmail.com} \address{Dhirendra Bahuguna \newline Department of Mathematics and Statistics, Indian Institute of Technology Kanpur, Pin 208016, India} \email{dhiren@iitk.ac.in, Tel. +91-512-2597053, Fax +91-512-2597500} \thanks{Submitted May 20, 2013. Published October 31, 2013.} \subjclass[2000]{34K45, 34A60, 35R12, 45J05} \keywords{Impulsive functional differential equation; deviating argument; \hfill\break\indent analytic semigroup; fixed point theorem} \begin{abstract} The objective of this article is to prove the existence of piecewise continuous mild solutions to impulsive functional differential equation with iterated deviating arguments in a Banach space. The results are obtained by using the theory of analytic semigroups and fixed point theorems. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In the theory of differential equations with deviating arguments, we study differential equations involving variables (arguments) as well as unknown functions and its derivative; generally speaking, under different values of the variables (arguments). It is very important and significant branch of nonlinear analysis with numerous applications to physics, mechanics, control theory, biology, ecology, economics, theory of nuclear reactors, engineering, natural sciences, and many other areas of science and technology. A comprehensive coverage can be found in the book by El'sgol'ts and Norkin \cite{norkin}. For more details on recent works in this direction, we refer to \cite{gal,rhd,haloi12,haloi12d,haloi12j} and the references cited therein. Impulsive effects are common phenomena due to short-term perturbations whose duration is negligible in comparison with the total duration of the original process. The governing equations of such phenomena may be modelled as impulsive differential equations. In recent years, there has been a growing interest in the study of impulsive differential equations as these equations approach the simulation processes in the control theory, physics, chemistry, population dynamics, biotechnology, economics and so on. The investigation of existence and uniqueness of mild solutions for impulsive differential equations have been discussed by many authors (see \cite{cynj,dwwy,rmn,cya,bmh,mf,mgm,lzlj,wwc,ygs,jt,jt2,wgp,bdd,bdd2} and references cited therein). However, due to theoretical and practical difficulties, the study of impulsive differential equations with deviating arguments has been developed rather slowly. Recently, the study of impulsive differential equations with deviating arguments has studied by some authors. Guobing et al. \cite{ygs} established the existence solution of periodic boundary value problems for a class of impulsive neutral differential equations with multi-deviation arguments. Jankowski \cite{jt2} discussed the existence of solutions for second-order impulsive differential equations with deviating arguments (see also \cite{lzlj, wwc, jt, wgp, bdd, bdd2} and the references cited therein). Liu \cite{ljh} studied the existence of mild solutions of the following impulsive evolution equation by using semigroup theory \begin{equation} \label{e1.1} \begin{gathered} u'(t) = Au(t) + f(t, u(t)), \quad 0 < t < T_0, \; t\neq t_i, \\ \Delta u(t_i) = I_i(u(t_i))=u(t_i^+)-u(t_i^-),\quad i=1,2,\dots ,\; 0 < t_1 < t_2< \dots < T_0, \\ u(0)=u_0, \end{gathered} \end{equation} where $A$ is the infinitesimal generator of a $C_0$-semigroup, $u(t_i^+),u(t_i^-)$ represent the right and left hand limits respectively at $t=t_i$, $I_i$'s are some operators and $f$ is a suitable function. Chang et al. \cite{{cynj}} investigated the existence of mild solutions for the following impulsive problem \begin{gather} \label{e1.2} \begin{aligned} \frac{d}{dt}(u(t)-F(t,u(h_1(t)))) &\in A\Big[u(t)+\int_0^t f(t-s)u(s)ds\Big] \\ &\quad +G(t,u(h_2(t)))),\quad t\in[0,a],\; t\neq t_k, \end{aligned}\\ \Delta u|_{t=t_k}= I_k(u(t_k^-)),\quad k=1,\dots ,m, \label{e1.3}\\ u(0)+g(u)= u_0, \label{e1.4} \end{gather} where $A$ is the infinitesimal generator of a compact analytic resolvent operator $R(t)$, $t>0$ in a Banach space $H$, $u_0\in H$, $0=t_00$, we define $H_{-\alpha}=(H_\alpha)^{*}$, the dual space of $H_\alpha$, is a $\|x\|_{-\alpha}=\|A^{-\alpha}x\|$. For more details, we refer to the book by Pazy \cite{pazy}. \begin{lemma}[{\cite[pp. 72,74,195-196]{pazy}}] \label{bslem1} Suppose that $-A$ is the infinitesimal generator of an analytic semigroup $S(t)$, $t \ge 0$ with $\|S(t)\| \le M$ for $t \ge 0$ and $0 \in \rho(-A)$. Then we have the following: \begin{itemize} \item[(i)] $H_{\alpha}$ is a Banach space for $0 \le \alpha \le 1$; \item[(ii)] For any $0<\delta\le\alpha$ implies $D(A^\alpha)\subset D(A^\delta)$, the embedding $H_{\alpha} \hookrightarrow H_{\delta}$ is continuous; \item[(iii)] The operator $A^\alpha S(t)$ is bounded for every $t>0$ and $$ \|A^{\alpha}S(t)\|\leq C_{\alpha}t^{-\alpha}. $$ \item[(iv)] For $\alpha\le 0$, $A^\alpha$ is bounded. \end{itemize} \end{lemma} We define the space \begin{align*} X=\mathcal{PC}(H_\alpha) =\Big\{&u:[0,T_0]\to H_\alpha:u\in C((t_k,t_{k+1}],H_\alpha),k=0,1,\dots ,m, \\ & u(t^-_k),u(t^+_k) \text{ exist and } u(t^-_k)=u(t_k)\Big\}, \end{align*} is a Banach space endowed with the supremum norm $$ \|u\|_{\mathcal{PC}} :=\sup_{t\in I}\|u(t)\|_\alpha. $$ We shall use the following conditions on $f$ and $h_i$ in its arguments: \begin{itemize} \item [(H1)] Let $W\subset \operatorname{Dom}(f)$ be an open subset of $\mathbb{R_{+}}\times H_\alpha \times H_{\alpha-1}$, where $0\le \alpha<1$. For each $(t,u,v)\in W$, there is a neighborhood $V_1\subset W$ of $(t,u,v)$, such that the nonlinear map $f:\mathbb{R_{+}}\times H_\alpha \times H_{\alpha-1} \to H$ satisfies the condition \[ \|f(t,u,v)-f(s,u_1,v_1)\| \le L_f \{|t-s|^{\theta_1}+\|u-u_1\|_{\alpha}+ \|v-v_1\|_{\alpha-1} \} \] for all $(t,u,v),(s,u_1,v_1)\in V_1$, where $L_f=L_{f}(t,u,v,V_{1})> 0$ and $0<\theta_1\le 1$ are constants. \item[(H2)] Let $U_{h_i}\subset \operatorname{Dom}(h_i)$ be open subsets of $\mathbb{R}_{+} \times H_{\alpha -1} $, where $0\le \alpha<1$. For each $(t,u)\in U_{h_i}$, there is a neighborhood $V_{h_i}\subset U_{h_i}$ of $(t,u)$, such that $h_i(0,\cdot )=0$ for each $i=1,2,\dots ,m$, $h_i: \mathbb{R_{+}} \times H_{\alpha-1} \to \mathbb{R_{+}}$ satisfies the condition \begin{equation} \label{e2.1} |h_i(t,u)-h_i(s,v)| \le L_{h_i}\{\|u-v\|_{\alpha-1}+|t-s|^{\theta_2}\} \end{equation} for all $(t,u), (s,v)\in V_{h_i}$, $L_{h_i}=L_{h_i}(t,u,V_{h_i})> 0$ and $0<\theta_2\le 1$ are constants. \item[(H3)] The functions $I_k:H_\alpha\to H_\alpha$ are continuous and there exists $\Upsilon_k$ such that $\|I_k(u)\|_\alpha\le\Upsilon_k$, $k=1,2,\dots ,m$. \item[(H4)] There exist continuous nondecreasing $d_k:\mathbb{R}_+\to\mathbb{R}_+$ such that $\|I_k(u)-I_k(v)\|_\alpha\le d_k\|u-v\|_\alpha,k=1,2,\dots ,m$. \end{itemize} \subsection*{New concept of solutions} Here, we prove a new concept of solutions \cite{mf,ljh} for the problem \begin{equation} \begin{gathered} u'(t)+Au(t)=r(t),\quad t\in[0,T_0], \; t\neq t_k,\\ u(0)=u_0\\ u(t_k)=I_k(u(t_k^-)),\quad k=1,2\dots ,m, \end{gathered}\label{mild0} \end{equation} where $r\in H$. Let \begin{equation} \begin{gathered} v'(t)+A v(t)= r(t),\quad t\in[0,T_0],\\ v(0)= v_0, \end{gathered}\label{mild} \end{equation} and \begin{equation} \begin{gathered} w'(t)+A w(t)=0,\quad t\in[0,T_0],\; t\neq t_k,\\ w(0)=0,\\ w(t_k)=I_k(u(t_k^-)),\quad k=1,2,\dots ,m, \end{gathered}\label{mild2} \end{equation} be the decomposition of $u(.)=v(.)+w(.)$, where $v$ is the continuous mild solution of \eqref{mild} and $w$ is the piecewise continuous mild solution of \eqref{mild2}. By a mild solution for \eqref{mild}, we mean a continuous function $v:[0,T_0]\to H$ satisfying the integral equation \begin{equation} \label{e2.5} v(t)=S(t)v_0+\int_0^t S(t-s)r(s)ds,\quad t\in [0,T_0]. \end{equation} and by a piecewise continuous mild solution for \eqref{mild2}, we mean a function $w\in \mathcal{PC}([0,T_0],D(A))$ satisfying the integral equation equivalent to the system \eqref{mild2} (\cite[see Eq. (3.4)]{mf}) \begin{equation} w(t)=\begin{cases} -\int_0^t Aw(s)ds, & t\in[0,t_1],\\ I_1(u(t^-_1))-\int_0^t Aw(s)ds, & t\in(t_1,t_2],\\ \sum_{i=1}^k I_i(u(t^-_i))-\int_0^t Aw(s)ds, &t\in(t_k,t_{k+1}],\\ k=1,2,\dots ,m. \end{cases}\label{mild3} \end{equation} The above equation can be expressed as \begin{equation} w(t)=\sum_{i=1}^k\chi_i(t)I_i(w(t_i^-))-\int_0^t Aw(s)ds,\label{mild4} \end{equation} for $t\in[0,T_0]$, where \begin{equation} \label{e2.8} \chi_i(t)=\begin{cases} 0,&\text{for } t\in[0,t_1],\\ 1,&\text{for } t\in[t_k,t_{k+1}]. \end{cases} \end{equation} Taking Laplace transform of \eqref{mild4}, we obtain \[ w(p)=\sum_{i=1}^k \frac{e^{-t_i p}}{p}I_i-\frac{Aw(p)}{p}, \] this gives \begin{equation} \label{e2.9} w(p)=\sum_{i=1}^k {e^{-t_i p}}(pI+A)^{-1}I_i. \end{equation} We also note that $(pI+A)^{-1}=\int_0^\infty e^{-pt}S(t)dt$. Thus we can obtain the mild solution for \eqref{mild2}, \begin{equation} \label{e2.10} w(t)=\sum_{i=1}^k\chi_i(t)S(t-t_i)I_i(w(t_i^-)). \end{equation} Hence, the mild solution for the problem \eqref{mild0} is \begin{equation} u(t)=S(t)u_0+\sum_{i=1}^k \chi_i(t)S(t-t_i)I_i(u(t_i^-))+\int_0^t S(t-s)r(s)ds.\label{p1} \end{equation} We can rewrite equation \eqref{p1} as \begin{equation} \label{e2.12} u(t)=\begin{cases} S(t)u_0+\int_0^t S(t-s)r(s)ds,& t\in[0,t_1],\\ S(t)u_0+S(t-t_1)I_1(u(t^-_1))+\int_0^t S(t-s)r(s)ds, & t\in(t_1,t_2],\\ S(t)u_0+\sum_{i=1}^k S(t-t_i)I_i(u(t^-_i))+\int_0^t S(t-s)r(s)ds, &t\in(t_k,t_{k+1}],\\ k=1,2,\dots ,m. \end{cases} %\label{2} \end{equation} \section{Local existence of mild solutions} In this section, we prove the existence and uniqueness results concerning piecewise continuous-mild solutions for system \eqref{mainp}. For $0\le \alpha<1$, we define \[ X_1=\{u\in X: \|u(t)-u(s)\|_{\alpha-1}\le L|t-s|,\forall t,s\in (t_k,t_{k+1}],k=0,1,\dots ,m\}, \] which is needful for proving contraction principle (See \cite{gal,rhd}). Where $L$ is a suitable positive constant to be specified later. \begin{definition} \label{a}\rm A function $u:[0,T_0]\to H$ solution of the problem \eqref{mainp}, \begin{equation} \label{e3.1} u(t)=\begin{cases} S(t)u_0+\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds,& t\in[0,t_1],\\[4pt] S(t)u_0+S(t-t_1)I_1(u(t^-_1))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, &t\in(t_1,t_2],\\ \dots\\ S(t)u_0+\sum_{i=1}^k S(t-t_i)I_i(u(t^-_i))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, & t\in(t_k,t_{k+1}],\\ k=1,2,\dots ,m. \end{cases} %\label{2} \end{equation} is said to be a piecewise continuous-mild solution. \end{definition} For a fixed $R>0$, we define $$ \mathcal{W}=\{u\in X\cap X_1:u(0)=u_0, \|u-u_0\|_{\mathcal{PC}} \le R\}. $$ Clearly, $ \mathcal{W}$ is a closed and bounded subset of $X_1$ and is a Banach space. We define a map $\mathcal{G}:\mathcal{W}\to\mathcal{W}$ by \begin{equation} \label{e3.2} (\mathcal{G}u)(t)= \begin{cases} S(t)u_0+\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds,& t\in[0,t_1],\\[4pt] S(t)u_0+S(t-t_1)I_1(u(t^-_1))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, & t\in(t_1,t_2],\\ \dots\\ S(t)u_0+\sum_{i=1}^k S(t-t_i)I_i(u(t^-_i))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, &t\in(t_k,t_{k+1}],\\ k=1,2,\dots ,m. \end{cases} %\label{3} \end{equation} \begin{theorem} \label{b} Let $0\le\alpha<1$, $u_0\in H_\alpha$ and the assumptions {(H1)--(H4)} hold. Then problem \eqref{mainp} has a mild solution provided that \begin{gather} C_{\alpha}N{T_0^{1-\alpha}\over 1-\alpha}+M~\sum_{i=1}^k \Upsilon_i\le \frac{R}{2},\label{assum2} \\ \label{e3.4} C_{\alpha }L_f(2+L~L_h ){T_0^{1-\alpha}\over 1-\alpha}+M\sum_{i=1}^m d_i< 1, \end{gather} where $N$ and $L_h$ are positive constants to be specified later. \end{theorem} \begin{proof} Clearly, $\mathcal{G}:X\to X$. We begin with showing that $\mathcal{G}u\in X_1$ for each $u\in X_1$. Let $u\in X_1$, then for each $\tau_1,\tau_2\in[0,t_1]$, $\tau_1<\tau_2$ and $0\le \alpha<1$, we have \begin{equation} \begin{aligned} & \|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \\ &\leq \|[S(\tau_2)-S(\tau_1)]u_0\|_{\alpha-1} \\ &\quad +\int_0^{\tau_1}\|A^{\alpha-1}[S(\tau_2-s)-S(\tau_1-s)]\|\, \|f(s,u(s),u(w_1(s,u(s))))\|ds\\ &\quad+\int_{\tau_1}^{\tau_2}\|A^{\alpha-1}S(\tau_2-s)\|\, \|f(s,u(s),u(w_1(s,u(s))))\|ds. \end{aligned}\label{c} \end{equation} Since $f(t,u(t),u(w_1(u(t),t)))$ is continuous, by assumptions (H1), (H2), there exists a constant $N$, such that \begin{equation} \|f(t,u(t),u(w_1(t,u(t))))\|\le N,\quad u\in X,\; t\in[0,T_0].\label{b1} \end{equation} For the first term on the right-hand side of \eqref{c}, we have \begin{equation} \begin{aligned} \|A^{\alpha-1}[S(\tau_2)-S(\tau_1)]u_0\| &\le \int_{\tau_1}^{\tau_2}\|A^{\alpha-1}S'(s)u_0\|ds \\ &= \int_{\tau_1}^{\tau_2}\|A^\alpha S(s)u_0\|ds \\ &\le M\|u_0\|_\alpha(\tau_2-\tau_1). \end{aligned}\label{1} \end{equation} For the second term on the right-hand side of \eqref{c}, we have \begin{equation} \begin{aligned} \|(S(\tau_2-s)-S(\tau_1-s))\|_{\alpha-1} &\le \int_0^{\tau_2-\tau_1}\|A^{\alpha-1}S^{\prime}(l)S(\tau_1-s)\|dl \\ &= \int_0^{\tau_2-\tau_1}\|S(l)A^{\alpha}S(\tau_1-s)\|dl \\ & \le MC_{\alpha}(\tau_2-\tau_1)(\tau_1-s)^{-\alpha}. \end{aligned}\label{new2} \end{equation} Then using \eqref{new2}, we obtain the following bound for the second term on the right-hand side of \eqref{c}, \begin{equation} \begin{aligned} &\int_0^{\tau_1}\|(S(\tau_2-s)-S(\tau_1-s))A^{\alpha-1}\|\, \|f(s,u(s),u(w_1(s,u(s))))\|ds \\ &\le N M C_{\alpha}{T_0^{1-\alpha}\over {1-\alpha}}(\tau_2-\tau_1). \end{aligned}\label{new3} \end{equation} The third term on the right-hand side of \eqref{c} is estimated as \begin{equation} \int_{\tau_1}^{\tau_2}\|S(\tau_2-s)A^{\alpha-1}\| \|f(s,u(s),u(w_1(s,u(s))))\|ds \leq \|A^{\alpha-1}\|M N (\tau_2-\tau_1).\label{new4} \end{equation} Thus from inequalities \eqref{1}, \eqref{new3} and \eqref{new4}, we see that \begin{equation} \|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \le M\{\|u_0\|_\alpha +N~C_\alpha\frac{T_0^{1-\alpha}}{1-\alpha} +N \|A^{\alpha-1}\|\}(\tau_2-\tau_1). \label{5} \end{equation} For $\tau_1,\tau_2\in(t_1,t_2]$, $\tau_1<\tau_2$ and $0\le \alpha<1$, we have \begin{equation} \begin{aligned} &\|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \\ &\leq \|[S(\tau_2-S(\tau_1)]u_0\|_{\alpha-1}+\|A^{\alpha-1} [S(\tau_2-t_1)-S(\tau_1-t_1)]I_1(u(t_1^-))\| \\ &\quad+\int_0^{\tau_1}\|A^{\alpha-1}[S(\tau_2-s)-S(\tau_1-s)]\|\, \|f(s,u(s),u(w_1(s,u(s))))\|ds \\ &\quad+\int_{\tau_1}^{\tau_2}\|A^{\alpha-1}S(\tau_2-s)\|\, \|f(s,u(s),u(w_1(s,u(s))))\|ds. \end{aligned}\label{d} \end{equation} The second term on the right-hand side of \eqref{d} is estimated as \begin{equation} \begin{aligned} &\|A^{\alpha-1}[S(\tau_2-t_1)-S(\tau_1-t_1)]I_1(u(t_1^-))\|\\ &\le \int_{\tau_1}^{\tau_2} \|A^{\alpha-1}S'(l-t_1)I_1(u(t_1^-))\|dl \\ &= \int_{\tau_1}^{\tau_2}\|A^{\alpha}S(l-t_1)I_1(u(t_1^-))\|dl \\ &\le M \|I_1(u(t_1^-))\|_\alpha(\tau_2-\tau_1). \end{aligned}\label{4} \end{equation} Thus, from inequalities \eqref{1}, \eqref{new3}, \eqref{new4} and \eqref{4}, we obtain \begin{equation} \begin{aligned} & \|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \\ & \le M\Big\{\|u_0\|_\alpha+\Upsilon_1 +N C_\alpha\frac{T_0^{1-\alpha}}{1-\alpha} +N \|A^{\alpha-1}\|\Big\}(\tau_2-\tau_1). \end{aligned}\label{5b} \end{equation} Similarly, for $\tau_1,\tau_2\in(t_k,t_{k+1}]$, $\tau_1<\tau_2$, $k=1,2,\dots ,m$ and $0\le \alpha<1$, we have \begin{equation} \begin{aligned} & \|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \\ &\le M\Big\{\|u_0\|_\alpha+\sum_{i=1}^k \Upsilon_i +NC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha}+N\|A^{\alpha-1}\|\Big\} (\tau_2-\tau_1). \end{aligned}\label{5c} \end{equation} Thus, for each $\tau_1,\tau_2\in[0,T_0]$, $\tau_1<\tau_2$ and $0\le\alpha<1$, we have \begin{equation} \label{e3.16} \|(\mathcal{G}u)(\tau_2)-(\mathcal{G}u)(\tau_1)\|_{\alpha-1} \le L(\tau_2-\tau_1), \end{equation} where \[ L=\max\{M\|u_0\|_\alpha,M~\sum_{i=1}^m\Upsilon_i ,NMC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha},NM\|A^{1-\alpha}\|\}. \] Therefore, $\mathcal{G}$ is piecewise Lipschitz continuous on $[0,T_0]$ and so $\mathcal{G}:X_1\to X_1$. Next we will show that $\mathcal{G}:\mathcal{W}\to \mathcal{W}$. Let $u\in X\cap X_1$, then for each $t\in[0,t_1]$, \begin{align*} \|(\mathcal{G}u)(t)-u_0\|_{\alpha} &\le \|(S(t)-I)A^{\alpha}u_0\| \\ &\quad + \int_0^t\|S(t-s)A^{\alpha}\|\|f(s,u(s),u(w_1(s,u(s))))\|ds \\ &\le \|(S(t)-I)A^{\alpha}u_0\|+C_{\alpha}N{T_0^{1-\alpha}\over 1-\alpha}. \end{align*} Similarly, let $u\in X\cap X_1$, then for each $t\in (t_k,t_{k+1}],k=1,\dots ,m$, \begin{align*} &\|(\mathcal{G}u)(t)-u_0\|_{\alpha}\\ &\leq \|(S(t)-I)A^{\alpha}u_0\| + \int_0^t\|S(t-s)A^{\alpha}\|\|f(s,u(s),u(w_1(s,u(s))))\|ds \\ &\quad +\sum_{i=1}^{k}\|A^\alpha S(t-t_i)I_i(u(t^-_i))\| \\ &\leq \|(S(t)-I)A^{\alpha}u_0\|+C_{\alpha}N{T_0^{1-\alpha}\over 1-\alpha}+M \sum_{i=1}^k \Upsilon_i. \end{align*} Part (iii) of Lemma 2.1, implies that \begin{equation} \|(S(t)-I)A^{\alpha}(u_0)\|\leq {R\over 2}\label{assum1}. \end{equation} Thus, from \eqref{assum2} and \eqref{assum1}, it is clear that $$ \|\mathcal{G}u- u_0\|_{\mathcal{PC}}\leq R. $$ Therefore, $\mathcal{G}:\mathcal{W}\to \mathcal{W}$. Finally, we will claim that $\mathcal{G}$ is a contraction map. If $t\in [0,t_1]$ and $u, v\in \mathcal{W}$, then \begin{equation} \begin{aligned} &\|(\mathcal{G}u)(t)-(\mathcal{G}v)(t)\|_{\alpha} \\ &\le \int_0^t\|S(t-s)A^{\alpha}\|~\|f(s,u(s),u(w_1(s,u(s)))) -f(s,v(s),v(w(s,v(s))))\|ds. \end{aligned} \label{bs11+} \end{equation} Also, we note that \begin{equation} \label{e3.19} \begin{aligned} &\|f(t,u(t),u(w_1(t,u(t))))-f(t,v(t),v(w_1(t,v(t))))\| \\ & \leq L_f\{\|u(t)-v(t)\|_\alpha+\|u(w_1(t,u(t)))-v(w_1(t,v(t)))\|_{\alpha-1}\} \\ &\leq L_f\Big[\|u(t)-v(t)\|_\alpha+\|A^{-1}\|\, \|u(w_1(t,v(t)))-v(w_1(t,v(t)))\|_{\alpha} \\ &\quad +\|u(w_1(t,u(t)))-u(w_1(t,v(t)))\|_{\alpha-1}\Big] \\ &\leq L_f\big\{2\|u-v\|_{\mathcal{PC}}+ L|w_1(t,u(t))-w_1(t,v(t))|\big\}. \end{aligned} \end{equation} Now, let $$ w_i(t,u(t))=h_i(t,u(h_{i+1}(t,\dots u(t,h_m(t,u(t)))\dots ))),~~i=1,\dots ,m, $$ with $w_{m+1}(t,u(t))=t$. For more details, we refer to \cite[p. 2183]{ss}. Hence, we have \begin{align*} |w_1(t,u(t))-w_1(t,v(t))| &= |h_1(t,u(w_2(t,u(t))))-h_1(t,v(w_2(t,v(t)))) \\ &\le L_{h_1}\|u(w_2(t,u(t)))-u(w_2(t,u(t)))\|_{\alpha-1} \\ &\le L_{h_1}\Big\{\|u(w_2(t,u(t)))-u(w_2(t,v(t)))\|_{\alpha-1}\\ &\quad +\|u(w_2(t,v(t)))-v(w_2(t,v(t)))\|_{\alpha-1}\Big\} \\ &\le L_{h_1}\Big\{L|h_2(t,u(w_3(t,u(t)))-h_2(t,v(w_3(t,v(t)))|\\ &\quad +\|A\|^{-1}\|u-v\|_{\mathcal{PC}}\Big\} \\ &\dots \\ &\le ( L^{m-1}L_{h_1}\dots L_{h_m}+L^{m-2}L_{h_1}\dots L_{h_{m-1}}+ \dots\\ &\quad +LL_{h_1}L_{h_2}+ L_{h_1})\|A\|^{-1}\|u-v\|_{\mathcal {PC}}. \end{align*} Thus, we have \begin{equation} \begin{aligned} &\|f(t,u(t),u(w_1(t,u(t))))-f(t,v(t),v(w_1(t,v(t))))\| \\ &\leq L_f\Big(2+LL_h\|A^{-1}\|\Big)\|u-v\|_{\mathcal{PC}} \\ &\leq L_f(2+LL_h)\|u-v\|_{\mathcal{PC}}. \end{aligned}\label{f} \end{equation} where $L_h=( L^{m-1}L_{h_1}\dots L_{h_m}+L^{m-1}L_{h_1}\dots L_{h_{m-1}}+\dots + L_{h_1})>0$. We use \eqref{f} in \eqref{bs11+}, we obtain \begin{equation} \|(\mathcal{G}u)(t)-(\mathcal{G}v)(t)\|_{\alpha} \le C_{\alpha }L_f(2+LL_h){T_0^{1-\alpha}\over 1-\alpha}\|u-v\|_{\mathcal{PC}}.\label{bs145} \end{equation} For $t\in(t_1,t_2]$, we have \begin{equation} \|(\mathcal{G}u)(t)-(\mathcal{G}v)(t)\|_{\alpha} \le \Big[ C_{\alpha }L_f(2+LL_h){T_0^{1-\alpha}\over 1-\alpha}+Md_1\Big]\|u-v\|_{\mathcal{PC}}. \label{bs146} \end{equation} For $t\in(t_k,t_{k+1}]$, $k=1,2,\dots ,m$, we have \begin{equation} \|(\mathcal{G}u)(t)-(\mathcal{G}v)(t)\|_{\alpha} \le \Big[ C_{\alpha }L_f(2+LL_h){T_0^{1-\alpha}\over 1-\alpha}+M\sum_{i=1}^k d_i\Big]\|u-v\|_{\mathcal{PC}}. \label{bs146b} \end{equation} Thus, for each $t\in[0,T_0]$, we have \begin{equation*} \|(\mathcal{G}u)(t)-(\mathcal{G}v)(t)\|_{\alpha} \le \Big[ C_{\alpha }L_f(2+LL_h){T_0^{1-\alpha}\over 1-\alpha}+M\sum_{i=1}^m d_i \Big]\|u-v\|_{\mathcal{PC}}. \end{equation*} Therefore, the map $\mathcal{G}$ is a contraction map, hence $\mathcal{G}$ has a unique fixed point $u\in \mathcal{W}$. That is, problem \eqref{mainp} has a unique mild solution. \end{proof} \section{Further existence results} Theorem \ref{b} can be proved if we omit the hypothesis (H1). In that case the proof is based on the idea of Wang et al. \cite{mf}. \begin{theorem} \label{a1} Assume the conditions (H3)-(H4) hold, the semigroup $\{S(t),t\ge 0\}$ is compact, and $f:I\times H\times H\to H$ is continuous. For $u_0\in H_\alpha$ there exists a constant $r>0$ such that \begin{equation} M\{\|u_0\|_\alpha+\sum_{i=1}^m\Upsilon_i\}+C_\alpha N_f \frac{T_0^{1-\alpha}}{1-\alpha} \le r,\label{e1} \end{equation} where \[ N_f={\sup_{s\in[0,T_0],u\in\Omega}}\|f(s,u(s),u(w_1(s,u(s))))\|, \] and $u\in\Omega=\{v\in \mathcal{PC}(H_\alpha):\|v\|_{\mathcal{PC}}\le r\}$. Then there exists a mild solution $u\in\mathcal{PC}(H_\alpha)$ of problem \eqref{mainp}. \end{theorem} \begin{proof} Let us define a map $ \Upsilon:\mathcal{PC}(H_\alpha)\to \mathcal{PC}(H_\alpha)$, by \[ (\Upsilon u)(t)=\begin{cases} S(t)u_0+\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, & t\in[0,t_1],\\[4pt] S(t)u_0+S(t-t_1)I_1(u(t^-_1))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, & t\in(t_1,t_2],\\ \dots\\ S(t)u_0+\sum_{i=1}^k S(t-t_i)I_i(u(t^-_i))\\ +\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds, &t\in(t_k,t_{k+1}],\\ k=1,2,\dots ,m. \end{cases} \] \noindent\textbf{Step 1.} First we show that ${\Upsilon}$ is continuous. It follows from the continuity of $f$ that $$ \|f(s,u_n(s),u_n(w_1(s,u_n(s))))-f(s,u(s),u(w_1(s,u(s))))\| \le \epsilon,\quad\text{as } n\to \infty, $$ for $s\in[0,t]$, $t\in[0,T_0]$. Now, for each $t\in[0,t_1]$, we have \begin{equation} \label{e4.2} \|(\Upsilon u_n)(t)-({\Upsilon}u)(t)\|_\alpha\le C_\alpha \frac{T_0^{1-\alpha}}{1-\alpha}\epsilon\to 0, \quad\text{as } n\to\infty. \end{equation} For, $t\in(t_1,t_2]$, we have \begin{equation} \label{e4.3} \begin{aligned} &\|({\Upsilon}u_n)(t)-({\Upsilon}u)(t)\|_\alpha \\ &\le M\|I_1(u_n(t^-_1))-I_1(u(t^-_1))\|_\alpha+C_\alpha \frac{T_0^{1-\alpha}}{1-\alpha}\epsilon\to 0, \quad\text{as }n\to\infty. \end{aligned} \end{equation} Similarly, for each $t\in(t_k,t_{k+1}]$, $k=1,2,\dots ,m$, \begin{equation} \label{e4.4} \begin{aligned} &\|({\Upsilon}u_n)(t)-({\Upsilon}u)(t)\|_\alpha \\ &\le M\sum_{i=1}^k\|I_i(u_n(t^-_i))-I_i(u(t^-_i))\|_\alpha+C_\alpha \frac{T_0^{1-\alpha}}{1-\alpha}\epsilon\to 0, \quad \text{as }n\to\infty. \end{aligned} \end{equation} We have that ${\Upsilon}$ is continuous. \noindent\textbf{Step 2.} Next we show that ${\Upsilon}$ maps bounded sets into bounded sets in $\mathcal{PC}(H_\alpha)$. Let $u\in \Omega$, then for $t\in[0,t_1]$, we have \begin{equation} \label{e4.5} \|({\Upsilon}u)(t)\|_\alpha \le M\|u_0\|_\alpha+N_fC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha}, \end{equation} For, $t\in(t_1,t_2]$, we have \begin{equation} \label{e4.6} \|({\Upsilon}u)(t)\|_\alpha\le M\big\{\|u_0\|_\alpha+\Upsilon_1\big\}+N_fC_\alpha\frac{T_0^{1-\alpha}} {1-\alpha}. \end{equation} Similarly, for each $t\in(t_k,t_{k+1}]$, $k=1,2,\dots ,m$, we have \begin{equation} \label{e4.7} \|({\Upsilon}u)(t)\|_\alpha\le M\Big\{\|u_0\|_\alpha+\sum_{i=1}^k \Upsilon_i\Big\}+N_fC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha} . \end{equation} Thus, from inequality \eqref{e1}, we see that ${\Upsilon}:\Omega\to \Omega$. \noindent\textbf{Step 3.} In this step, we show that ${\Upsilon}$ maps bounded sets into equicontinuous sets in $\mathcal{PC}(H_\alpha)$. Let $\tau_1,\tau_2\in[0,t_1],\tau_1<\tau_2$, we have \begin{equation} \label{e4.8} \begin{aligned} &\|({\Upsilon}u)(\tau_2)-({\Upsilon}u)(\tau_1)\|_\alpha \\ &\le M\Big\{\|u_0\|_\alpha+N_fC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha}+ \|A^{\alpha-1}\|N_f\Big\}(\tau_2-\tau_1). \end{aligned} \end{equation} Similarly, for each $\tau_1,\tau_2\in(t_k,t_{k+1}]$, $\tau_1<\tau_2$, $k=1,2,\dots ,m$, we have \begin{equation} \label{e4.9} \begin{aligned} &\|({\Upsilon}u)(\tau_2)-({\Upsilon}u)(\tau_1)\|_\alpha \\ &\le M\Big\{\|u_0\|_\alpha+\sum_{i=1}^k\Upsilon_i +N_fC_\alpha\frac{T_0^{1-\alpha}}{1-\alpha}+ \|A^{\alpha-1}\|N_f\Big\}(\tau_2-\tau_1). \end{aligned} \end{equation} As $\tau_2\to\tau_1$ the right-hand side of the above inequality tends to zero. So, ${\Upsilon}(\Omega)$ is equicontinuous. \noindent\textbf{Step 4.} ${\Upsilon}$ maps $\Omega$ into a compact set in $H_\alpha$. For this purpose, we decompose ${\Upsilon}$ as ${\Upsilon}={\Upsilon_1}+{\Upsilon_2}$, where \[ ({\Upsilon_1}u)(t)= S(t)u_0+(\mathcal{T}_1u)(t),\quad t\in I \backslash \{t_1,\dots ,t_m\}, \] where $(\mathcal{T}_1u)(t)=\int_0^t S(t-s)f(s,u(s),u(w_1(s,u(s))))ds$. and \[ ({\Upsilon_2}u)(t)= \begin{cases} 0,& t\in[0,t_1],\\ \sum_{i=1}^k S(t-t_i)I_i(u(t^-_i)),& t\in(t_k,t_{k+1}],\; k=1,2,\dots ,m. \end{cases} \] ${\Upsilon_2}$ is a constant map and hence compact. Finally, we need to prove that for $0\le t\le T_0$, $({\Upsilon_1}u)(t)$ is relatively compact in $\Omega$. For each $t\in [0,T_0]$, the set $\{S(t)u_0\}$ is precompact in $H_\alpha$ since $\{S(t),t\ge 0\}$ is compact. For $t\in(0,T_0]$, and $\epsilon>0$ sufficiently small, we define \[ ({\mathcal T_1}^\epsilon u)(t) =S(\epsilon)\int_0^{t-\epsilon}S(t-\epsilon-s)f(s,u(s),u(w_1(s,u(s)))) ds,\quad u\in\Omega. \] The set $\{({\mathcal{T}_1}^\epsilon u)(t):u\in\Omega\}$ is precompact in $H_\alpha$ since $S(\epsilon)$ is compact. Moreover, for any $u\in\Omega$, we have \begin{align*} \|(\mathcal{T}_1 u)(t)-({\mathcal{T}_1^\epsilon}u)(t)\|_\alpha &\le \int_{t-\epsilon}^t \|A^\alpha S(t-s)\|\|f(s,u(s),u(w_1(s,u(s))))\| ds \\ &\le (C_\alpha N_f)\epsilon^{1-\alpha}. \end{align*} Therefore, $\{({\mathcal{T}_1^\epsilon}u)(t):u\in\Omega\}$ is arbitrarily close to the set $\{(\mathcal{T}_1 u)(t):u\in\Omega\},t>0$. Hence the set $\{({\Upsilon_1}u)(t):u\in\Omega\}$ is precompact in $H_\alpha$. Thus, ${\Upsilon_1}$ is a compact operator by Arzela-Ascoli theorem, and hence ${\Upsilon}$ is a compact operator. Then Schauder fixed point theorem ensures that ${\Upsilon}$ has a fixed point, which gives rise to a piecewise continuous-mild solution. \end{proof} \section{Applications} We consider the following semi-linear heat equation with a deviating argument (See also \cite{gal}). \begin{equation} \begin{gathered} \frac{\partial u}{\partial t}= \frac{\partial^2 u} {\partial x^2}+\tilde{H}(x,u(x,t))+G(t,x,u(x,t)),\\ x\in (0,1),\quad t\in(0,\frac{1}{2})\cup (\frac{1}{2},1),\\ \Delta u|_{t=\frac{1}{2}} = \frac{2u({\frac{1}{2}})^-}{2+u({\frac{1}{2}})^-} \\ u(0,t)=u(1,t)=0, \\ u(x,0)=u_0(x), \quad x\in (0,1), \end{gathered} \label{exappgal} \end{equation} where \begin{gather*} \tilde{H}(x,u(x,t))=\int _0^{x}K(x,y)u(y,P(t)) dy,\\ P(t)=g_1(t)\Big|u\Big(x,g_2(t)|u(x,\dots g_m(t)|u(x,t)|)|\Big)\Big|, \end{gather*} and the function $G: \mathbb{R_{+}} \times [0,1]\times \mathbb{R}\to \mathbb{R}$ is measurable in $x$, locally H\"{o}lder continuous in $t$, locally Lipschitz continuous in $u$, uniformly in $x$. Assume that $g_i:\mathbb{R_{+}}\to \mathbb{R_{+}}$ are locally H\"{o}lder continuous in $t$ with $g_i(0)=0$ for each $i=1,2,\dots ,m$, and $K\in C^1([0,1]\times [0,1];\mathbb{R})$. Let \begin{gather*} X=L^2((0,1);\mathbb{R}), \quad Au=\frac{d^2u}{dx^2}, \quad D(A)=H^2(0,1)\cap H^1_0(0,1),\\ X_{1/2}=D((A)^{1/2})=H^1_0(0,1), \quad X_{-1/2}=(H^1_0(0,1))^*=H^{-1}(0,1)\equiv H^1(0,1). \end{gather*} For $x\in (0,1)$, we define the function $f:\mathbb{R_{+}}\times X_{1/2}\times X_{-1/2}\to X$ by \begin{equation} f(t,u,\psi)(x)= \tilde{H}(x,\psi)+G(t,x,u),\label{gs55} \end{equation} where $\tilde{H}:[0,1]\times X_{-1/2}\to X$ is given by \begin{equation} \tilde{H}(x,\psi)= \int_0^x K(x,y)\psi(y)dy,\label{gs55b} \end{equation} and $G:\mathbb{R_+}\times [0,1]\times X_{1/2}\to X$ satisfies \begin{equation} \|G(t,x,u)\|\le Q(x,t)(1+\|u\|_{1/2}),\label{gs55c} \end{equation} with $Q(.,t)\in X$ and $Q$ continuous in its second arguments. For $u\in D(A) $ and $\lambda \in \mathbb{R}$ with $-Au=\lambda u$, we have \[ \langle -Au,u\rangle =\langle \lambda u,u\rangle,\quad \| u'\| _{L^2}=\lambda \| u\| _{L^2}; \] so we have $\lambda >0$. The solution $u$ of $-Au=\lambda u$ is \[ u( x) =D_1\cos ( \sqrt{\lambda }x) +D_2\sin ( \sqrt{\lambda }x). \] Using the boundary condition, we get $D_1=0$ and $\lambda= \lambda_n=n^2\pi^2$ for $n\in \mathbb{N}$. Thus, for $n\in \mathbb{N}$, we have $$ u_n(x)=D_2 \sin(\sqrt{\lambda_n}x). $$ Also $\langle u_n,u_{m} \rangle =0$, $m\neq n$ and $ \langle u_n,u_n \rangle =1$. So, for $u\in D(A)$ there exists a sequence $\alpha_n$ of real numbers such that $u(x)= \sum _{n\in \mathbb{N}} \alpha_nu_n(x)$ with $\sum _{n\in \mathbb{N}}(\alpha_n)^2<\infty$ and $\sum _{n\in \mathbb{N}}(\alpha_n)^2(\lambda _n)^2<\infty$. The semigroup is given by $$ S(t)u= \sum _{n\in \mathbb{N}} \exp(-n^2t)\langle u,u_{m} \rangle u_{m}. $$ Let $T>0$ be fixed. Now we will show that assumptions (H1) and (H2) are satisfied. For (H1), we will show that $\tilde{H}: [0,1]\times X_{-1/2}\to X$, where $$ \tilde{H}(x,\psi(x,t))=\int _0^{x}K(x,y)\psi(y,t)dy, $$ with $\psi(x,t)= u(x,w_1(t,u(x,t)))$. For $x\in [0,1]$, we have \begin{align*} \|\tilde{H}(x,\psi_{1}(x,\cdot))-\tilde{H}(x,\psi_2(x,\cdot))\| & \leq \int_0^{x}|K(x,y)|\|(\psi_{1}-\psi_2)(y,\cdot)\|dy \\ & \leq \| K \| _{\infty} \int_0^{x}|(\psi_{1}-\psi_2)(y,\cdot)|dy. \end{align*} Thus for $\psi_{1},\psi_2\in H^1(0,1)$ and by applying the Minkowski's integral inequality we obtain \begin{align*} \|\tilde{H}(x,\psi_{1}(x,\cdot))-\tilde{H}(x,\psi_2(x,\cdot))\|^2 & \leq \| K \| _{\infty}^2\int_0 ^1 \int_0^{y}|(\psi_{1}-\psi_2)(y,\cdot)|^2dxdy \\ & \leq \| K \| _{\infty} ^2\int_0^1 y |(\psi_{1}-\psi_2)(y,\cdot)|^2dy\\ & \leq \| K \| _{\infty} ^2\| \psi_{1}-\psi_2 \| ^2_{{-1/2}}. \end{align*} Since, \[ \frac{\partial}{\partial x}\tilde{H}(x,\psi(x,\cdot)) =K(x,x)\psi (x,\cdot)+\int_0^{x} \frac{\partial K}{\partial x}(x,y)\psi(y,\cdot), \] we obtain, in similar way, that $$ \| \frac{\partial}{\partial x}H(x,\psi_{1}(x,\cdot))-\frac{\partial} {\partial x}H(x,\psi_2(x,\cdot))\|\leq ( \| K \| _{\infty} +| |\frac{\partial K}{\partial x}\|_{\infty})\| \psi_{1}-\psi_2 \| _{{-1/2}}. $$ Thus \begin{align*} \|\tilde{H}(x,\psi_{1}(x,\cdot))-\tilde{H}(x,\psi_2(x,\cdot))\| & \leq (2\|K\|_{\infty}+\|\frac{\partial K}{\partial x}\|_{\infty}) \| \psi_{1}-\psi_2 \| _{{-1/2}} \\ & \leq (2\|K\|_{\infty}+\|\frac{\partial K}{\partial x}\|_{\infty}) \| \psi_{1}-\psi_2 \| _{{-1/2}}\\ &= C_{1} \| \psi_{1}-\psi_2 \| _{{-1/2}}, \end{align*} where $C_{1}= (2\|K\|_{\infty} +\| \frac{\partial K}{\partial x}\|_{\infty})$. Again the assumption on $G$ implies that there exist constants $C_2>0$ and $0< \gamma \leq 1 $ such that $$ \|G(t,x,u)-G(s,x,v)\|_{H^1_0(0,1)}\leq C_2 (|t-s|^{\gamma}+\|u-v\|_{{1/2}}), $$ for $t,s \in [0,T]$, $x\in (0,1)$ and $u,v \in H^1_0(0,1)$. Thus $f: [0,T] \times H^1_0(0,1)\times H^1(0,1)\to L^2(0,1)$ defined by $f=\tilde{H}+G$ satisfies the assumption (H1). Our next aim is to prove that the functions $h_i:[0,T]\times X_{-1/2}\to [0,T]$ defined by $h_i(t,\phi)=g_i(t)|\phi (x,\cdot)|$ for each $i=1,2,\dots ,m$, satisfies (H2). For $t\in [0,T]$, we obtain \[ |h_i(t,\phi)| =|g_i(t)|\,|\phi (x,\cdot)| \leq \|g_i\|_{\infty} \|\phi\|_{L^\infty(0,1)} \leq C \|\phi\|_{{-1/2}}, \] where we used the embedding $H^1_0(0,1)\subset C[0,1]$ in the last inequity, and $C$ is a constant depending on the bounds of $g_i$'s. Since $g_i$'s are locally H\"{o}lder continuous, for some constant $L_{g_i}>0$ and $0<\theta \leq 1$ we have $|g_i(t)-g_i(s)|\leq L_{g_i}|t-s|^{\theta}$ for $t,s\in [0,T]$. Moreover $t,s\in [0,T]$ and $\phi_{1},\phi_2\in X_{-1/2} $, we have \begin{align*} |h_i(t,\phi_{1})-h_i(t,\phi_2)| &=|g_i(t)(|\phi_{1} (x,\cdot)|-|\phi_2 (x,\cdot)|) +(g_i(t)-g_i(s))|\phi_2(x,\cdot)|\\ & \leq \|g\|_{\infty} \|\phi_{1}-\phi_2\|_{L^\infty(0,1)} +L_{g_i}|t-s|^{\theta}\|\phi_2\|_{L^\infty(0,1)}\\ & \leq C \|g\|_{\infty}\|\phi_{1}-\phi_2\|_{{-1/2}} +L_{g_i}|t-s|^{\theta}\|\phi_2\|_{{-1/2}}\\ & \leq \max \{ C \|g\|_{\infty},L_{g_i}\|\phi_2\|_{\infty} \}(\|\phi_{1}-\phi_2\|_{{-1/2}}+|t-s|^{\theta}). \end{align*} If $u, v\in D((-A)^{1/2})$, then $$ \|I_k(u)- I_k(v)\|_{1/ 2}\le {2\|u- v\|_{1/ 2}\over\|(2+ u)(2 + v)\|_{1/ 2}} \le{ 1\over 2} \|u-v\|_{1/ 2}. $$ Next for Theorem \eqref{a1} we define \[ f(t,u(t),\psi(t)))(x)={e^{-t}(\cos(u(x,t))+\sin(\psi(x,t))) \over(2+t^2)(e^t+e^{-t})}+e^{-t}, \] for $t\in[0,1/2)\cup(1/2,1]$, $u\in X$, and $x\in(0,1)$. Clearly, \[ \|f(t,u,\psi)\|\le{e^{-t}\over(2+t^2)(e^t+e^{-t})}+e^{-t}=m(t), \] with $m(t)\in L^\infty([0,1];\mathbb{R}_+)$. 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