\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 21, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/21\hfil Mild solutions] {Mild solutions for semilinear fractional differential equations} \author[G. M. Mophou, G. M. N'Gu\'er\'ekata\hfil EJDE-2009/21\hfilneg] {Gis\`ele M. Mophou, Gaston M. N'Gu\'er\'ekata} % in alphabetical order \address{Gis\`ele M. Mophou \newline Universit\'e des Antilles et de la Guadeloupe, D\'epartement de Math\'ematiques et Informatique, Universit\'e des Antilles et de La Guyane, Campus Fouillole 97159 Pointe-\`a-Pitre Guadeloupe (FWI)} \email{gmophou@univ-ag.fr} \address{Gaston M. N'Gu\'er\'ekata \newline Department of Mathematics, Morgan State University, 1700 E. Cold Spring Lane, Baltimore, MD 21251, USA} \email{Gaston.N'Guerekata@morgan.edu, nguerekata@aol.com} \thanks{Submitted October 28, 2008. Published January 23, 2009.} \subjclass[2000]{34K05, 34A12, 34A40} \keywords{Fractional differential equation} \begin{abstract} This paper concerns the existence of mild solutions for fractional semilinear differential equation with non local conditions in the $\alpha$-norm. We prove existence and uniqueness, assuming that the linear part generates an analytic compact bounded semigroup, and the nonlinear part is a Lipschitz continuous function with respect to the fractional power norm of the linear part. \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} During the past decades, fractional differential equations have attracted many authors (see for instance \cite{lak,lak1,lak2,gis1,gaston,gaston1,pod,wei,zhang} and references therein). This, mostly because it efficiently describes many phenomena arising in Engineering, Physics, Economy, and Science. Our aim in this paper is to discuss the existence and the uniqueness of the mild solution for fractional semilinear differential equation with nonlocal conditions : \begin{equation}\label{e1.1} \begin{gathered} D^q x(t)=-Ax(t)+f(t,x(t),Bx(t)),\quad t\in [0,T],\\ x(0)+g(x)=x_0 \end{gathered} \end{equation} where $T>0$, $01$ such that \begin{equation}\label{0.1} \|S(t)\|\leq M \end{equation} We assume without loss of generality that $0\in \rho(A)$. This allows us to define the fractional power $A^\alpha$ for $0<\alpha<1$, as a closed linear operator on its domain $D(A^\alpha)$ with inverse $A^{-\alpha}$(see [8]). We have the following basic properties $A^\alpha$. \begin{theorem}[{\cite[pp. 69-75]{pazy}}]\label{theo1} \begin{enumerate} \item $ \mathbb{X}_\alpha=D (A^\alpha)$ is a Banach space with the norm $\|x\|_\alpha := \|A^\alpha x\|$ for $x\in D(A^\alpha)$. \item $ S(t) : \mathbb{X} \to \mathbb{X}_\alpha$ for each $t > 0$. \item $ A^\alpha S(t)x = S(t)A^\alpha x $ for each $x \in D(A^\alpha)$ and $t \geq 0$. \item For every $t > 0$, $A^\alpha S(t)$ is bounded on $\mathbb{X}$ and there exist $ M_\alpha > 0$ and $\delta > 0 $ such that \begin{equation}\label{0.2} \|A^\alpha S(t)\|\leq {\frac{M_\alpha }{t^\alpha}}e^{-\delta t} \end{equation} \item $A^{-\alpha}$ is a bounded linear operator in $\mathbb{X}$ with $D(A^\alpha) = \mathop{\rm Im}(A^{-\alpha})$. \item If $0 < \alpha \leq \beta$, then $ D(A^\beta)\hookrightarrow D(A^\alpha)$. \end{enumerate} \end{theorem} \begin{remark} \label{rmk2.2} \rm Observe as in \cite{hsiang} that by Theorem \ref{theo1} (ii) and (iii), the restriction $S_\alpha(t)$ of $S(t)$ to $\mathbb{X}_\alpha$ is exactly the part of $S(t)$ in $\mathbb{X}_\alpha$. Let $x \in \mathbb{X}_\alpha$. Since $$ \|S(t)x\|_\alpha = \|A^\alpha S(t)x\| = \|S(t)A^\alpha x\| \leq \|S(t)\|\|A^\alpha x\| = \|S(t)\|\|x\|_\alpha, $$ and as $t$ decreases to $0$ $$ \|S(t)x - x\|_\alpha=\|A^\alpha S(t)x - A^\alpha x\| = \|S(t)A^\alpha x - A^\alpha x\|\to 0, $$ for all $x\in \mathbb{X}_\alpha$, it follows that $(S(t))_{t\geq 0}$ is a family of strongly continuous semigroup on $\mathbb{X}_\alpha$ and $\|S_\alpha (t)\|\leq \|S(t)\|$ for all $t\geq 0$. \end{remark} We have the the following result from \cite{hsiang}. \begin{lemma}\label{compact} $ (S_\alpha(t))_{t\geq 0}$ is an immediately compact semigroup in $\mathbb{X}_\alpha$, and hence it is immediately norm-continuous. \end{lemma} \begin{definition}[\cite{jara}] \rm A continuous function $x : I\to \mathbb{X}$ satisfying the equation \begin{equation}\label{defmild} x(t)=S(t)(x_0-g(x))+ {\frac{1}{\Gamma(q)}}\int_{0}^{t}(t-s)^{q-1} S(t-s)(f(s,x(s),Bx(s))\,ds \end{equation} for $t\in [0,T]$ is called a mild solution of the equation \eqref{e1.1} \end{definition} In the sequel, we will also use $\|f\|_p$ to denote the $L^p$ norm of $f$ whenever $f\in L^p(0, T)$ for some $p$ with $1\leq p<\infty$. We will set $\alpha \in (0,1)$ and we will denote by $\mathcal{C}_\alpha$, the Banach space $C([0, T], \mathbb{X}_\alpha)$ endowed with the supnorm given by $$ \|x\|_\infty := \sup_{t\in I}\|x\|_\alpha ,\quad \text{for } x\in \mathcal{C}. $$ \section{Main Results\label{main}} We assume the following conditions: \begin{itemize} \item[(H1)] The function $f : I \times \mathbb{X}_\alpha\times \mathbb{X}_\alpha \to \mathbb{X}$ is continuous, and there exists a positive function $\mu\in L_{\rm loc}^1(I, \mathbb{R}^+)$ such that \begin{equation}\label{0.3} \|f(t, x,y)\|\leq \mu(t), \end{equation} \item[(H2)] $g \in C(\mathcal{C}_\alpha, \mathbb{X}_\alpha)$ is completely continuous and there exist $\lambda, \gamma > 0$ such that $$ \|g(x)\|_\alpha \leq \lambda \|x\|_\infty + \gamma. $$ \end{itemize} \begin{theorem}\label{theo2} Suppose that assumptions {\rm (H1), (H2)} hold. If $x_0\in \mathbb{X}_\alpha$ and \begin{equation}\label{0.4} M\lambda<{\frac{1}{2}} \end{equation} then \eqref{e1.1} has a mild solution on $[0, T]$. \end{theorem} \begin{proof} We define the function $F:\mathcal{C}_\alpha\to \mathcal{C}_\alpha$ by $$ (Fx)(t)=S(t)(x_0-g(x)) +{\frac{1}{\Gamma(q)}}\int_0^t(t-s)^{q-1} S(t-s)f(s,x(s),Bx(s))\,ds, $$ and we choose $r$ such that $$ r\geq 2\Big( {\frac{M_\alpha T^{q-\alpha}}{(1-\alpha)\Gamma(q)}} \|\mu\|_{L^1_{\rm loc}(I,\mathbb{R}_+)} +M(\|x_0\|_\alpha+\gamma)\Big). $$ Let $B_r=\{x\in \mathcal{C}_\alpha:\|x\|_\infty\leq r\}$. Then we proceed in three steps. \noindent\textbf{Step 1.} We show that $FB_r\subset B_r$. Let $x\in B_r$. Then for $t\in I$, we have \begin{align*} &\|(Fx)(t)\|_\alpha\\ &\leq \|S(t)(x_0-g(x))\|_\alpha +{\frac{1}{\Gamma(q)}}\|\int_0^t(t-s)^{q-1} S(t-s)f(s,x(s),Bx(s)) \,ds\|_\alpha\\ &\leq \|S(t)\|\|x_0-g(x)\|_\alpha + {\frac{1}{\Gamma(q)}}\int_0^t\|(t-s)^{q-1}A^\alpha S(t-s)f(s,x(s), Bx(s))\|ds\\ &\leq \|S(t)\|(\|x_0\|_\alpha+ \lambda \|x\|_\infty+\gamma) + {\frac{T^{q-1}}{\Gamma(q)}}\int_0^t \|A^\alpha S(t-s)\| \| f(s,x(s),Bx(s)) \|\,ds , \end{align*} which according to (\ref{0.1}), (\ref{0.2}), (\ref{0.3}) and (\ref{0.4}) gives \begin{align*} &\|(Fx)(t)\|_\alpha\\ &\leq \|S(t)\|\left(\|x_0\|_\alpha+ \lambda \|x\|_\infty+\gamma\right) + {\frac{T^{q-1}}{\Gamma(q)}}\int_0^t M_\alpha (t-s)^{-\alpha} e^{-\delta(t-s)}\mu(s)\,ds\\ &\leq \|S(t)\|\left(\|x_0\|_\alpha+ \lambda \|x\|_\infty+\gamma\right) + {\frac{M_\alpha T^{q-1}}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha} \mu(s)\,ds\\ &\leq M\left(\|x_0\|_\alpha+ \lambda \|x\|_\infty+\gamma\right) + {\frac{M_\alpha T^{q-\alpha}}{(1-\alpha)\Gamma(q)}} \|\mu\|_{L^1_{\rm loc}(I,\mathbb{R}_+)} \leq r \end{align*} for $t\in I$. Hence, we deduce $ \|Fx\|_{\infty}\leq r$. \noindent\textbf{Step 2.} We prove that $F$ is continuous. Let $({x_n})$ be a sequence of $B_r$ such that $x_n \to x$ in $B_r$. Then $$ f(s, x_n(s),Bx_n(s)) \to f(s, x(s),Bx(s)), \quad n\to \infty $$ because the function $f$ is continuous on $I\times \mathbb{X}_\alpha\times \mathbb{X}_\alpha$. Now, for $t\in I$, we have \begin{align*} &\|Fx_n-Fx\|_\alpha\\ &\leq \|S(t)(g(x_n) - g(x))\|_\alpha\\ &\quad + \big\| {\frac{1}{\Gamma(q)}}\int_0^t (t-s)^{q-1} S(t - s)\left(f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\right)\,ds\big\|_\alpha, \end{align*} which in view of (\ref{0.1}) and (\ref{0.2}) gives \begin{align*} &\|Fx_n-Fx\|_\alpha\\ &\leq \|S(t)\|\|g(x_n) - g(x)\|_\alpha\\ &\quad + {\frac{T^{q-1}}{\Gamma(q)}}\int_0^t \|A^\alpha S(t - s)\|\|f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\|\,ds\\ &\leq M\|g(x_n) - g(x)\|_\alpha\\ &\quad + {\frac{M_\alpha T^{q-1}}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha} \|f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\|\,ds\\ &\leq M\|g(x_n) - g(x)\|_\alpha\\ &\quad+ {\frac{M_\alpha T^{q-1}}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha} \|f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\|\,ds \end{align*} for $t\in I$. Therefore, using on the one hand the fact that $$ \|f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\|\leq 2\mu(s) \quad \text{for } s \in I, $$ and for each $t\in I$ since $f$ satisfies (H1) and on the other hand the fact that the function $s\mapsto 2\mu(s) (t - s)^{-\alpha}$ is integrable on $I$, by means of the Lebesgue Dominated Convergence Theorem one proves that $$ \int_0^t (t-s)^{-\alpha} \|f(s, x_n(s),Bx_n(s)) - f(s, x(s),Bx(s))\|\,ds \to 0. $$ Hence, since $g( x_n) \to g(x)$ as $ n\to \infty$ because $g$ is completely continuous on $\mathcal{C}_\alpha$, it can easily been shown that $$ \lim_{n\to \infty}\|Fx_n-Fx\|_\infty = 0, \quad\text{as }n\to \infty. $$ In other words $F$ is continuous. \noindent \textbf{Step 3.} We show that $F$ is compact. To this end, we use the Ascoli-Arzela's theorem. We first prove that $\{(Fx)(t):x\in B_r\}$ is relatively compact in $\mathbb{X}_\alpha$, for all $t\in I$. Obviously, $\{(Fx)(0):x\in B_r\}$ is compact. Let $t\in (0,T]$. For each $h\in (0,t)$ and $x\in B_r$, we define the operator $F_h$ by \begin{align*} (F_{h}x)(t)&= S(t)(x_0-g(x))+ {\frac{1}{\Gamma(q)}}\int_{0}^{t-h}(t-s)^{q-1} S(t-s)f(s,x(s),Bx(s))\,ds\\ &= S(t)(x_0-g(x))+ {\frac{S(h)}{\Gamma(q)}}\int_{0}^{t-h}(t-s)^{q-1} S(t-h-s)f(s,x(s),Bx(s))\,ds. \end{align*} Then the sets $\{(F_hx)(t):x\in B_r\}$ are relatively compact in $\mathbb{X}_\alpha$ since by Lemma \ref{compact}, the operators $S_\alpha(t)$, $t\geq 0$ are compact on $X_\alpha$. Moreover, using (H1) and (\ref{0.2}), we have \begin{align*} \|(Fx)(t)-(F_{h}x)(t)\|_\alpha &\leq {\frac{1}{\Gamma(q)}}\int_{t-h}^{t}(t-s)^{q-1}\| S(t-s)f(s,x(s),Bx(s))\|_\alpha\,ds\\ &\leq {\frac{T^{q-1}}{\Gamma(q)}}\int_{t-h}^{t}\|A^\alpha S(t-s)\| \|f(s,x(s),Bx(s))\|\,ds\\ &\leq {\frac{T^{q-1} M_\alpha}{\Gamma(q)}}\|\mu\|_{L^1_{\rm loc} (I,\mathbb{R}_+)} \int_{t-h}^{t}(t-s)^{-\alpha}\,ds\\ &\leq {\frac{T^{q-1} M_\alpha \|\mu\|_{L^1_{\rm loc} (I,\mathbb{R}_+)}}{(1-\alpha)\Gamma(q)}}h^{1-\alpha} \end{align*} Therefore, we deduce that $\{(Fx)(t):x\in B_r\}$ is relatively compact in $\mathbb{X}_\alpha$ for all $t\in (0,T]$ and since it is compact at $t=0$ we have the relatively compactness in $\mathbb{X}_\alpha$ for all $t\in I$. Now, let us prove that $F(B_r)$ is equicontinuous. By the compactness of the set $g(B_r)$, we can prove that the functions $Fx, x\in B_r$ are equicontinuous a $t=0$. For $00$, we deduce that $ {\lim_{t_2\to t_1}} I_4=0$. In summary, we have proven that $F(B_r)$ is relatively compact, for $t\in I$, $ \{Fx:x\in B_r\}$ is a family of equicontinuous functions. Hence by the Arzela-Ascoli Theorem, $F$ is compact. By Schauder fixed point theorem $F$ has a fixed point $x\in B_r$. Consequently, \eqref{e1.1} has a mild solution. \end{proof} Now we make the following assumptions. \begin{itemize} \item[(H1')] $f: I\times \mathbb{X}_\alpha\times \mathbb{X}_\alpha\to \mathbb{X}$ is continuous and there exist functions $\mu_1,\mu_2\in L^1_{\rm loc}(I,\mathbb{R}^+)$ such that $$ \|f(t,x,u)-f(t,y,v)\|\leq \mu_1(t)\|x-y\|_\alpha+\mu_2(t)\|u-v\|_\alpha, $$ for all $t\in I$, $x,y,u,v\in\mathbb{X}_\alpha$. \item[(H2')] $g: \mathcal{C}_\alpha \to \mathbb{X}_\alpha$ is continuous and there exists a constant $b$ such that $$ \|g(x)-g(y)\|_\alpha\leq b\|x-y\|_{\infty}, \quad \text{for all }x, y\in \mathcal{C}_\alpha. $$ \item[(H3)] The function $\Omega_{\alpha,q}:I\to \mathbb{R}_+$, $0<\alpha,q<1$ defined by $$ \Omega_{\alpha,q}= M b + {\frac{T^{q-1} M_\alpha t^{1-\alpha}}{(1-\alpha)\Gamma(q)}} \left(\|\mu_1\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}+ B^*\|\mu_2\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}\right) $$ satisfies $0<\Omega_{\alpha,q}\leq\tau<1$, for all $t\in I$. \end{itemize} \begin{theorem}\label{theo1.1} Assume that {\rm (H1'), (H2'), (H3)} hold. If $x_0\in \mathbb{X}_\alpha$ then \eqref{e1.1} has a unique mild solution $x\in \mathcal{C}_\alpha$. \end{theorem} \begin{proof} Define the function $F:\mathcal{C}_\alpha\to \mathcal{C}_\alpha$ by $$ (Fx)(t)= S(t)(x_0-g(x)) + {\frac{1}{\Gamma(q)}}\int_0^t(t-s)^{q-1} S(t-s)f(s,x(s),Bx(s))\,ds. $$ Note that $F$ is well defined on $\mathcal{C}_\alpha$. Now take $t\in I$ and $x,y \in \mathcal{C}_\alpha$. We have \begin{align*} &\|(Fx)(t)-F(y)(t)\|_\alpha\\ &\leq \|S(t)\left(g(x)-g(y)\right)\|_\alpha\\ &\quad + {\frac{1}{\Gamma(q)}}\int_0^t(t-s)^{q-1} \|S(t-s) \left(f(s,x(s),Bx(s))-f(s,y(s),By(s))\right)\|_\alpha\,ds\\ &\leq \|S(t)\|\|g(x)-g(y)\|_\alpha\\ &\quad + {\frac{T^{q-1}}{\Gamma(q)}}\int_0^t \| A^\alpha S(t-s)\| \|f(s,x(s),Bx(s))-f(s,y(s),By(s))\| ds \end{align*} which according to (\ref{0.1}), (\ref{0.2}), (H1'), (H2') and (\ref{defb*}) gives \begin{align*} &\|(Fx)(t)-F(y)(t)\|_\alpha\\ &\leq M b\|x-y\|_\infty+ {\frac{T^{q-1} M_\alpha}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha}\mu_1(s) \|x(s)-y(s)\|_\alpha ds\\ &\quad + {\frac{T^{q-1} M_\alpha}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha}\mu_2(s) \|Bx(s)-By(s))\|_\alpha ds\\ &\leq M b\|x-y\|_\infty\\ &\quad + {\frac{T^{q-1} M_\alpha}{\Gamma(q)}}\|\mu_1 \|_{L^1_{\rm loc}(I,\mathbb{R}_+)} \Big(\int_0^t (t-s)^{-\alpha}\,ds\Big)\|x(s)-y(s)\|_\infty\\ &\quad + {\frac{T^{q-1} M_\alpha}{\Gamma(q)}}\int_0^t (t-s)^{-\alpha}\mu_2(s)\Big[ \int_{0}^sK(s,\sigma) \|A ^\alpha(x(\sigma)-y(\sigma))\| d \sigma \Big] ds\\ &\leq \Big( M b + {\frac{T^{q-1} M_\alpha t^{1-\alpha}}{(1-\alpha)\Gamma(q)}} \|\mu_1\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}\Big)\|x-y\|_\infty\\ &\quad + {\frac{T^{q-1} M_\alpha B^* t^{1-\alpha}}{(1-\alpha)\Gamma(q)}} \|\mu_2\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}\|x-y\|_\infty\\ &\leq \Big[ M b + {\frac{T^{q-1} M_\alpha t^{1-\alpha}}{(1-\alpha)\Gamma(q)}} \left(\|\mu_1\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}+ B^* \|\mu_2\|_{L^1_{\rm loc}(I,\mathbb{R}_+)}\right)\Big]\|x-y\|_\infty\\ &\leq\Omega_{\alpha,q}(t)\|x-y\|_\infty. \end{align*} So we get $$ \|(Fx)(t)-F(y)(t)\|_\infty \leq \Omega_{\alpha,q}(t) \|x-y\|_\infty. $$ Therefore, assumption (H3) allows us to conclude in view of the contraction mapping principe that, $F$ has a unique fixed point in $\mathcal{C}_\alpha$, and $$ x(t)=S(t)(x_0-g(x))+ {\frac{1}{\Gamma(q)}}\int_0^t(t-s)^{q-1} S(t-s)f(s,x(s),Bx(s))\,ds $$ which is the mild solution of \eqref{e1.1}. \end{proof} \begin{thebibliography}{00} \bibitem{aiz} S. 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