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\markboth{\hfil Existence results \hfil EJDE--2001/24}
{EJDE--2001/24\hfil Eduardo Hern\'{a}ndez M. \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations}, Vol. {\bf
2001}(2001), No. 24, pp. 1--14. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
 %
   Existence results for a class of semi-linear evolution equations
 %
\thanks{ {\em Mathematics Subject Classifications:} 35A05, 34G20, 34A09.
\hfil\break\indent {\em Key words:} Banach spaces, semigroup of
linear operators, abstract differential  equations,
\hfil\break\indent
 fractional powers of closed operators, regular solutions.
\hfil\break\indent \copyright 2001 Southwest Texas State
University. \hfil\break\indent 
Submitted December 14, 2000. Published April 10, 2001. \hfil\break\indent
Partially supported by grant 13394-5 from Fapesp Brazil } }
\date{}
%
\author{ Eduardo Hern\'{a}ndez M. }
\maketitle

\begin{abstract}
 We prove the existence of regular solutions for the
 quasi-linear evolution
 $$ \frac{d}{dt}(x(t)+g(t,x(t))=Ax(t)+f(t,x(t)),
 $$
 where $A$ is the infinitesimal generator of an analytic semigroup
 of bounded linear operators defined on a Banach space  and the functions
 $f, g$ are continuous.
 \end{abstract}

\newtheorem{Lemma}{Lemma}
\newtheorem{Theorem}{Theorem}
\newtheorem{Corollary}{Corollary}
\newtheorem{Definition}{Definition}

\renewcommand{\theequation}{\thesection.\arabic{equation}}
\catcode`@=11 \@addtoreset{equation}{section} \catcode`@=12

\section{Introduction}\label{sec1}

 The class of equations considered in this paper have the form
\begin{equation}\label{ne}
\begin{array}{c}
\frac{d}{dt} (x(t)+g(t,x(t))=Ax(t)+f(t,x(t)), \quad t>0, \\
 x(0)=x_0\,.
\end{array}
\end{equation}
We consider this system as a Cauchy problem on a Banach
space $X$, where $A$ is the infinitesimal generator of an
analytic semigroup of bounded linear operators $(T(t))_{t\geq
0}$;  $f,g:[0,T]\times \Omega \to X$ are appropriate continuous
functions and $\Omega$ is  an open subset of $X$.
 The case  $g\equiv 0$ has an extensive literature. The books of
 Pazy \cite{PA}, Krein \cite{KE}, Goldstein \cite{GO} and the references
contained therein, give a good account of important results.

Throughout this paper $X$ will be a Banach space equipped with
the norm $\|\cdot\|$ and the operator $A:D(A)\subset X\to X$ will
be the infinitesimal generator of an analytic semigroup of
bounded  linear operators $(T(t))_{t\geq 0}$ on $X$. For the
theory of strongly continuous semigroups,  refer to \cite{PA} and
\cite{GO}. We mention here  only  some notation and properties
essential to our purpose. In particular, it is well known that
there exist $\tilde{M}\geq 1$ and a real number $w$ such that
$$
\| T(t) \|\leq \tilde{M}e^{wt}, \quad t\geq 0\,.
$$
In what follows we assume that $\| T(t)\|$ is uniformly bounded by
$\tilde{M}$ and that $0 \in \rho(A)$. In this  case it is
possible to define the fractional power $(-A)^\alpha$, for
$0<\alpha<1$, as a closed linear operator with domain
$D((-A)^\alpha)$. Furthermore, the subspace $D((-A)^\alpha)$ is
dense in $X$ and the expression
$$
\| x \|_\alpha = \| (-A)^\alpha x \|
$$
defines a norm on $D((-A)^\alpha)$. Hereafter we represent by
$X_\alpha$ the space $D((-A)^\alpha)$ endowed with the norm $\|
\cdot \|_\alpha $. The following properties are well known (see
\cite{PA}).

\begin{Lemma}\label{an}
Under the above  conditions we have
\begin{enumerate}
\item If $0<\alpha\leq 1$, then $X_\alpha $ is a Banach space.
\item If $0< \beta \leq \alpha$, then $X_\alpha  \to X_{\beta}$ is
continuous and compact when the resolvent operator of $A$ is
compact.
\item For every constant $a>0$, there exists $C_{a}>0$ such that
$$
\| (-A)^\alpha T(t)\| \leq \frac{C_{a}}{t^\alpha}, \quad 0<t\leq
a.
$$
\item For every $a>0$ there exists a positive constant $C'_{a}$ such that
$$
\| (T(t)-I)(-A)^{-\alpha} \| \leq C'_{a}t^\alpha, \quad 0<t\leq a.
$$
\end{enumerate}
\end{Lemma}

 By analogy with the abstract Cauchy problem
\begin{eqnarray}\label{cle}
\dot{u}(t)=Au(t)+h(t)
\end{eqnarray}
we adopt the following definitions.

\begin{Definition} \rm
A function $x\in C([0,r):X)$ is a mild solution of the abstract
Cauchy problem (\ref{ne}) if the following holds: $x(0)=x_0$; for
each $0\leq t <r$ and $s\in [0,t)$,
 the function $AT(t-s)g(s,x(s))$ is integrable and
\begin{eqnarray}\label{smild}
x(t)&=&T(t)(x_0+g(0,x_0))-g(t,x(t))-\int_0^{t}AT(t-s)g(s,x(s))ds
\nonumber \\
&&+ \int_0^{t}T(t-s)f(s,x(s))ds\,. \nonumber
\end{eqnarray}
\end{Definition}

\begin{Definition} \rm
A  function $x\in C([0,r):X)$ is a classical solution of the
abstract Cauchy problem (\ref{ne}) if $x(0)=x_0$,  $x(t)\in
D(A)$  for all $t\in (0,r)$,  $\dot{x}$ is continuous on $(0,r)$,
and $x(\cdot)$ satisfies (\ref{ne}) on  $(0,r)$.
\end{Definition}

\begin{Definition} \rm
A function $x\in C([0,r):X)$ is an S-classical (Semi-classical)
solution of the abstract Cauchy problem (\ref{ne}) if $x(0)=x_0$,
$\frac{d}{dt}(x(t)+g(t,x(t)))$ is continuous on  $(0,r)$,
$x(t)\in D(A)$  for all $t\in (0,r)$,
 and $x(\cdot)$ satisfies (\ref{ne}) on  $(0,r)$.
\end{Definition}

This paper is organized as follows. In section
\ref{regularsolution} we discuss  the existence of S-classical
and classical solutions to the initial value problem  (\ref{ne}).
Our results are based on the properties  of analytic semigroups
and the  ideas used in \cite[chapter 5]{PA}. In  section
\ref{examples}, some applications are considered.

Throughout this paper  we assume that $X$ is an abstract Banach
space. The terminology and notations are those generally used in
operator theory. In particular, if $X,\,\, Y$ are Banach spaces,
we indicate by $\mathcal{L}$ $(X:Y)$ the Banach space of the
bounded linear
  operator of $X$ into $Y$ and we abbreviate to $\mathcal{L}$ $(X)$ whenever $X=Y$.
  In addition $B_{r}(x:X)$ will denote the closed ball in  space $X$ with center at $x$
  and radius $r$.

For a bounded function $\xi :[0,a]\to X$ and $0<t<a$ we will
employ the notation
$$
\| \xi(\cdot)\|_{t} =\sup\{\| \xi(s)\|: s \in [0,t]\}.
$$
Finally  for $x_0\in X$, we will use the notation $x(\cdot,x_0)$
for the mild solution of  (\ref{ne}).

\section{Regular Solutions}\label{regularsolution}

 The existence of mild solutions for the abstract Cauchy problem (\ref{ne})
 follows  from  \cite[Theorems 2.1, 2.2]{HH2}; for this reason we
 omit the proofs of the next two results.

\begin{Theorem}\label{mild}
Let $x_0\in X$ and assume that the following conditions hold
\begin{itemize}
\item [a)] There exist $\beta\in (0,1)$ and $L\geq 0$ such that the function g
 is $X_{\beta}$-valued and  satisfies the Lipschitz condition
$$
\|(-A)^{\beta}g(t,x)-(-A)^{\beta}g(s,y)\|\leq L (| t-s| +\| x-y\|)
$$
for every $0\leq s,t\leq T$ and $x,y\in \Omega$, and $\|
(-A)^{-\beta}\| L <1$.
\item [b)] The function $f$ is continuous and takes bounded sets into  bounded sets.
\item [c)] The semigroup $(T(t))_{t\geq 0}$ is compact.
\end{itemize}
Then there exists a mild solution $x(\cdot,x_0)$ of the abstract
Cauchy problem (\ref{ne}) defined on  $[0,r]$ for some $0<r<T$.
\end{Theorem}

\begin{Theorem}\label{lip}
Let $x_0\in X$ and assume that the following conditions hold:
\begin{itemize}
\item [a)] There exist $\beta\in (0,1)$ and $L\geq 0$ such that the function g is
 $X_{\beta}$-valued and satisfies the Lipschitz condition
$$
\| (-A)^{\beta}g(t,x)-(-A)^{\beta}g(s,y)\|\leq L ( | t-s|+\| x-y\| )
$$
for every $0\leq s,t\leq T$ and $x,y\in \Omega$ and $\|
(-A)^{-\beta}\| L <1$.
\item [b)]  The function $f$ is continuous and there exists $N>0$ such that
$$
\| f(t,x)-f(s,y)\|\leq N ( | t-s| +\| x-y\| )
$$
for every $0\leq s,t\leq T$ and $x,y\in \Omega$.
\end{itemize}

Then there exists a unique mild solution $x(\cdot,x_0)$ of the
abstract
 Cauchy problem (\ref{ne}) defined on  $[0,r]$ for some $0<r\leq T$.
\end{Theorem}

The existence of S-classical and classical solutions, requires
some
 additional assumptions  on the functions $g,f$. In  particular,
in the next result we assume that the following  assumption hold.

\paragraph{Assumptions on $f$ and $g$:}  There exist $0<\alpha<\beta<1$ and an  open set
$\Omega_\alpha \subset X_\alpha $ such  that the functions $f$
and $(-A)^{\beta}g$ are continuous on $[0,T]\times \Omega_\alpha$,
and there exist $L>0$ and $0<\gamma_1,\gamma_2<1$ such that for
every $(t,x_1),(s,x_2)\in [0,T]\times\Omega_\alpha $ we have
$$\displaylines{
\| (-A)^{\beta}g(t,x_1)-(-A)^{\beta}g(s,x_2)\| \leq L\{| t-s
|^{\gamma_1}
 + \| x_1-x_2\|_\alpha \}, \cr
 \| f(t,x_1)-f(s,x_2)\| \leq N\{| t-s |^{\gamma_2} +
\| x_1-x_2\|_\alpha \}, \cr L\| (-A)^{\alpha-\beta}\| < 1\,. }$$


\begin{Theorem}\label{nclass}
Let $x_0\in \Omega_\alpha $ and assume  that $f$ and $g$ satisfy
the above assumptions, that $g$ is $D(A)$-valued continuous and
that $1-\beta<\min\{\beta-\alpha,\gamma_1, \gamma_2\}$. Then
there exists a unique S-classical solution $x(\cdot,x_0)\in
C([0,r]:X)$  for some $0<r<T$.
\end{Theorem}

\paragraph{Proof.} Let  $0<r_1<T$ and $\delta>0$ such that
$$
V=\{(t,x)\in [0,r_1]\times X_\alpha :\,\, \| (-A)^\alpha
x-(-A)^\alpha x_0\| <\delta \}\subset [0,T)\times\Omega_\alpha .
$$
Assuming that the functions $f$ and $(-A)^{\beta}g$ are bounded
on $V$ by $C_1>0$, we choose  $0<r<r_1$ such that
$$ \displaylines{
\| (T(\cdot)-I)(-A)^\alpha x_0\|_{r} \leq
\frac{(1-\mu)\delta}{6},\cr \|(T(\cdot)-I)(-A)^\alpha
g(0,x_0)\|_{r} \leq \frac{(1-\mu)\delta}{6}, \cr \|
(-A)^{\alpha-\beta}\| Lr^{\gamma_1}+C_{1-\beta+\alpha}C_1
\frac{r^{\beta-\alpha}}{\beta-\alpha}+ C_\alpha
C_1\frac{r^{1-\alpha}}{1-\alpha}<\frac{(1-\mu)\delta}{6},\cr
LC_{1-\beta+\alpha}\frac{r^{\beta-\alpha}}{\beta-\alpha}+
NC_\alpha \frac{r^{1-\alpha}}{1-\alpha}<1-\mu, }$$ 
where $\mu=\|
(-A)^{\alpha-\beta}\| L$ and  $C_\alpha $, $C_{1-\beta+\alpha}$
are the constants in Lemma \ref{an}.

On the set
$$ S=\{y\in C([0,r]:X):y(0)=(-A)^\alpha x_0,\,\,\|
y(t)-(-A)^\alpha x_0\| \leq \delta ,\,\,t\in [0,r]\} $$
 we define the operator
\begin{eqnarray}
\Psi(y)(t) &=& T(t)(-A)^\alpha(x_0+g(0,x_0))-(-A)^\alpha
g(t,(-A)^{-\alpha}y(t)) \nonumber\\
&&+\int_0^{t}(-A)^{1-\beta+\alpha}T(t-s)(-A)^{\beta}g(s,
(-A)^{-\alpha}y(s))ds  \label{ie}\\
&&+\int_0^{t}(-A)^\alpha T(t-s)f(s,(-A)^{-\alpha}y(s))\,ds.
\nonumber
\end{eqnarray}
For the mapping  $\Psi$ we consider the decomposition
$\Psi=\Psi_1+\Psi_2$, where
\begin{eqnarray*}
\Psi_1(y)(t)&=&T(t)(-A)^\alpha(x_0+g(0,x_0))-(-A)^\alpha
g(t,(-A)^{-\alpha}y(t))  \\
&&+\int_0^{t}(-A)^{1-\beta+\alpha}T(t-s)(-A)^{\beta}g(s,
(-A)^{-\alpha}y(s))ds,
\\
\Psi_2(y)(t)&=&\int_0^{t}(-A)^\alpha
T(t-s)f(s,(-A)^{-\alpha}y(s))ds.
\end{eqnarray*}
Next we prove that $\Psi_1$ and $\Psi_2$ are well defined, that
$\Psi$ satisfies  a Lipschitz condition and that the ranges of
$\Psi$ is contained in $S$.

Since $T(\cdot)$ is   analytic, the function $s \to AT(t-s)$ is
continuous in the uniform operator topology on $[0,t)$,
consequently the function $AT(t-s)g(s,(-A)^{-\alpha}y(s))$ is
continuous on $[0,t)$. Moreover from  lemma \ref{an} we have
$$
\| (-A)^{1-\beta
+\alpha}T(t-s)(-A)^{\beta}g(s,(-A)^{-\alpha}y(s))\| \leq
\frac{C_{1-\beta +\alpha}C_1}{(t-s)^{1-\beta +\alpha}},
$$
 $s\in [0,t)$, which implies that $\| (-A)^{1-\beta+\alpha}T(t-s)g(s,(-A)^{-\alpha}y(s))\|$
 is integrable on $[0,t)$. We thus conclude
that $\Psi_2$ is well defined and with values in $C([0,r]:X)$.
It's  clear from the previous remark that $\Psi_1$ is also well
defined and with values in $C([0,r]:X)$.

 It remain to show  that the operator $\Psi$ is a contraction on   $S$. Firstly we prove
 that the range of $\Psi$ is contained in  $S$. Let $y $ be a function in $S$. Then for
  $t \in [0,r]$ we get
\begin{eqnarray*}
\lefteqn{\| \Psi(y)(t)-(-A)^\alpha x_0\|}\\
&\leq& \| (T(t)-I)(-A)^\alpha( x_0+g(0,x_0))\| \\
&&+\|(-A)^\alpha g(0,x_0)-(-A)^\alpha g(t,(-A)^{-\alpha}y(t))\|  \\
&&+\int_0^{t}\frac{C_{1-\beta+\alpha}}{(t-s)^{1-\beta+\alpha}}\|
(-A)^{\beta}g(s,(-A)^{-\alpha}y(s))\| ds  \\
&& + \int_0^{t}\frac{C_\alpha }{(t-s)^\alpha}\| f(s,(-A)^{-\alpha}y(s))\| ds \\
 &\leq& \frac{2(1-\mu)\delta}{6}\,+\|
(-A)^{\alpha-\beta}\| L \{r^{\gamma_1} +\| (-A)^\alpha x_0-y(t)\|\} \\
&&+\int_0^{t}\left(\frac{C_{1-\beta+\alpha}C_1}{(t-s)^{1-\beta+\alpha}}+
\frac{C_\alpha C_1}{(t-s)^\alpha}\right) ds \\
 &\leq& \frac{2(1-\mu)\delta}{6}+ \| (-A)^{\alpha-\beta}\| L \{ r^{\gamma_1}
 +\delta\} +C_{1-\beta+\alpha}C_1\frac{r^{\beta-\alpha}}{\beta-\alpha}+
 C_\alpha C_1\frac{r^{1-\alpha}}{1-\alpha}.
\end{eqnarray*}
>From  the choice of $r$ we  conclude  that
 $$\|\Psi(y)-(-A)^\alpha x_0\|_{r} \leq \delta $$
 so that  $\Psi(y) \in S$.


On the other hand for $x(\cdot),\, y(\cdot) \in S$ and $t \in
[0,r]$,
\begin{eqnarray*}
\lefteqn{\| \Psi(y)(t)- \Psi(x)(t)\|}\\
&\leq& \| (-A)^\alpha g(t,(-A)^{-\alpha}y(t))-(-A)^\alpha
g(t,(-A)^{-\alpha}x(t))\|\\
&+&  \int_0^{t}\frac{C_{1-\beta+\alpha}}{(t-s)^{1-\beta+\alpha}}\|
(-A)^{\beta}g(s,(-A)^{-\alpha}y(s))-(-A)^{\beta}g(s,(-A)^{-\alpha}x(s))\|
 ds \\
&+&  \int_0^{t}\frac{C_\alpha }{(t-s)^\alpha}\|
f(s,(-A)^{-\alpha}y(s))-f(s,(-A)^{-\alpha}x(s))\| ds  \\
&\leq& \| (-A)^{\alpha-\beta}\| L \| y(t)- x(t) \| +
\int_0^{t}\{\frac{LC_{1-\beta+\alpha}}{(t-s)^{1-\beta+\alpha}}+
\frac{NC_\alpha }{(t-s)^\alpha}\}\| y-x\|_{r}ds,
\end{eqnarray*}
thus $$\|  \Psi(y)-\Psi(x) \|_{r}\leq   \left( L\|
(-A)^{\alpha-\beta}\|
+LC_{1-\beta+\alpha}\frac{r^{\beta-\alpha}}{\beta-\alpha}+
NC_\alpha \frac{r^{1-\alpha}}{1-\alpha}\right) \| y-x\|_{r}. $$
 The last estimate and the choice of $r$ imply that  $\Psi$ is a contraction mapping on  $S$.
 Let $y(\cdot)$ be  the unique  fixed  point  of the operator $\Psi$ in $S$.
We affirm that $y(\cdot)$ is locally H\"{o}lder continuous. In fact,
let $\vartheta$ be a real number with $0<\vartheta  <
\min\{1-\alpha,\beta-\alpha\}$ and $\vartheta+\beta>1$, and let
 $\tilde{C}>0$ be the constant guaranteed in Lemma \ref{an}, such that for all $0\leq
s\leq t \leq T$ and $0<h<1$
   $$ \| (T(h)-I)(-A)^\alpha T(t-s)\| \leq
\frac{\tilde{C}h^{\vartheta}}{(t-s)^{\vartheta+\alpha}}, \quad 0
\leq s < t
$$
and $$\| (T(h)-I)(-A)^{1-\beta+\alpha}T(t-s)\| \leq
\frac{\tilde{C}h^{\vartheta}}{(t-s)^{1-\beta+\alpha+\vartheta}} ,
 \quad 0\leq s<t\,.
$$
For $t\in [0,r)$ and  $h>0$ sufficiently small,
\begin{eqnarray*}
\lefteqn{\| y(t+h)-y(t)\|}\\
&\leq& \| (T(h)-I)(-A)^\alpha T(t)(x_0-g(0,x_0))\| \\
&&+\| (-A)^{\alpha-\beta}\| L \{ h^{\gamma_1}+\|y(t+h)-y(t)\| \}\\
&& + \int_0^{t}\|(T(h)-I)(-A)^{1-\beta+\alpha}T(t-s)(-A)^{\beta}
g(s,(-A)^{-\alpha}y(s))\| ds\\
&& + \int_{t}^{t+h}\| (-A)^{1-\beta+\alpha}T(t+h-s)(-A)^{\beta}g(s,(-A)^{-\alpha}y(s))\| ds\\
&& + \int_0^{t}\|
(T(h)-I)(-A)^\alpha T(t-s)f(s,(-A)^{-\alpha}y(s))\| ds \\
&& + \int_{t}^{t+h}\| (-A)^\alpha
T(t+h-s)f(s,(-A)^{-\alpha}y(s))\| ds \nonumber
\\ &\leq& \frac{\tilde{C}}{t^{(\alpha +\vartheta)}}\| x_0-g(0,x_0)\|
h^{\vartheta} + L\| (-A)^{\alpha-\beta}\| \{ h^{\gamma_1}+\|
y(t+h)-y(t)\| \} \\ && +
\int_0^{t}\frac{\tilde{C}h^{\vartheta}C_1}{(t-s)^{1-\beta+\alpha+\vartheta}}ds
+
\int_{t}^{t+h}\frac{C_{1-\beta+\alpha}C_1}{(t+h-s)^{1-\beta+\alpha}}ds \nonumber\\
&&
+\int_0^{t}\frac{\tilde{C}h^{\vartheta}C_1}{(t-s)^{\alpha+\vartheta}}ds
+ \int_{t}^{t+h}\frac{C_\alpha C_1}{(t+h-s)^\alpha}ds
\\ &\leq& \frac{C(x_0)h^{\vartheta}}{t^{\vartheta+\alpha}}+ C_2h^{\gamma_1}+L\|
(-A)^{\alpha-\beta}\| \|
y(t+h)-y(t)\|+C_{3}h^{\vartheta}\\
&&+C_{4}h^{\beta-\alpha}+C_{5}h^{1-\alpha}
\end{eqnarray*}
where  the constants $C_{i}$ are independent of $t$. If
$\bar{\rho}=\min\{ \vartheta,\gamma_1\}$, the last inequality can
be rewritten  in the form
$$
\| y(t+h)-y(t)\| \leq
\frac{C(\alpha,\beta,\vartheta,t,x_0)}{1-\mu}h^{\bar{\rho}}
$$
since $\mu=L\| (-A)^{\alpha-\beta}\|<1$. Therefore the function
$y(\cdot)$ is locally $\bar{\rho}$-H\"{o}lder continuous on  $(0,r)$,
moreover, we can to assume that $\bar{\rho}+\beta>1$. Now it is
easy to show  that $s\to (-A)^{\beta}g(s,(-A)^{-\alpha}y(s))$ and
 $s\to f(s,(-A)^{-\alpha}y(s))$ are $\rho$-H\"{o}lder continuous on  $(0,r)$, where
 $\rho=\min\{\bar{\rho},\gamma_2\}$ and $\rho+\beta>1$. From this remark,
in \cite[Theorem 2.4.1]{GO} and Lemma \ref{regc} below, we infer
that the function
\begin{eqnarray}
x(t)&=& T(t)(x_0+g(0,x_0)) - g(t,(-A)^{-\alpha}y(t))\nonumber \\
 && +\int_0^{t}(-A)^{1-\beta}T(t-s)(-A)^{\beta}g(s,(-A)^{-\alpha}y(s))ds
\label{reggol} \\
&& + \int_0^{t}T(t-s)f(s,(-A)^{-\alpha}y(s))ds  \nonumber
 \end{eqnarray}
 is  $X_\alpha $-valued, that the integral terms  in  (\ref{reggol}) are functions in
$C^{1}([0,r]:X)$ and that $x(t)\in D(A)$ for all $t\in (0,r)$.
Operating on $x(\cdot)$ with  $(-A)^\alpha$, we conclude that
$(-A)^{-\alpha}y=x$ and hence  that $x(t)+g(t,x(t))$ is a $C^{1}$
function on $(0,b)$. The  proof is
completed.\hfill$\diamondsuit$\medskip

The proof of the next Lemma is analogous to the proof in
\cite[Theorem 2.4.1]{GO}. However there are some differences that
require special attention and we include the principal ideas of
this proof for completeness.

\begin{Lemma}\label{regc}
Let $0<\beta<1$  and $g\in C([0,T]:X_{1-\beta})$. Assume that
$g:[0,T]\to X$ is $\theta$-H\"older continuous in $(0,T)$ with
$\beta+\theta>1$. If $y:[0,T]\to X$ is defined by $$
y(t)=\int_0^{t}(-A)^{1-\beta}T(t-s)g(s)ds,
$$ then $y(t)\in D(A)$ for every $t\in [0,T)$ and $\dot{y} \in C([0,T):X)$.
\end{Lemma}

 \paragraph{Proof.} For $t\in [0,T)$ we rewrite $y(t)$ in the form
\begin{equation}\label{leii}
\int_0^{t}(-A)^{1-\beta}T(t-s)(g(s)-g(t))ds +
\int_0^{t}(-A)^{1-\beta}T(t-s)g(t)ds=v(t)+w(t).
\end{equation}
 Clearly, $Aw(t)=T(t)(-A)^{1-\beta}g(t)-(-A)^{1-\beta}g(t) \in C([0,T]:X)$. For $\epsilon >0$,
sufficiently small we define the function
$$
v_{\epsilon}(t):= \left\{
\begin{array}{ll}
\int_0^{t-\epsilon}(-A)^{1-\beta}T(t-s)(g(s)-g(t))ds\,, &
\hbox{for }
 t\in[\epsilon,T),\\[3pt]
0 & \hbox{for } t \in [0,\epsilon).\end{array} \right.
$$
It is clear that $v_{\epsilon}(t)\in D(A)$. Moreover for
$0<\delta_1<\delta_2$
\begin{eqnarray*}
\| Av_{\delta_2}(t)-Av_{\delta_1}(t)\| &\leq&
\int_{t-\delta_2}^{t-\delta_1}
\| (-A)^{2-\beta}T(t-s)(g(s)-g(t))\| ds \\
&\leq& C_{2-\beta}(\delta_2^{\beta+\theta
-1}-\delta_1^{\beta+\theta-1}).
\end{eqnarray*}
The last inequality proves that $Av_{\delta}$ is convergent,
$\beta+\theta>1$, and therefore
\begin{eqnarray}
A(v(t))=\int_0^{t}A^{2-\beta}T(t-s)( g(s)-g(t) ) ds
\end{eqnarray}
since $A$ is a closed operator. From the previous remark it
follows that $y(t)\in D(A)$ for $t\in [0,T]$. The continuity of
$\partial_{t}y$ follows as in \cite[Theorem 2.4.1]{GO}.
\hfill$\diamondsuit$\medskip

In the rest of this paper  for a function  $j:[0,b]\times X\to$ $
X$
 and $h\in {{\bf I\kern -1.6pt{\bf R}}}$ we denote by $\partial_{h}j$ to the function
 $$\partial_{h}j(t)=\frac{j(t+h)-j(t)}{h}.$$
Moreover, if $j$ is differentiable we  will employ
 the  decomposition:
\begin{equation}\label{descomi}
  j(t+s,y)-j(t,y)=D_1j(t,y)s+W_1(j,t,t+s,y)
\end{equation}
and
\begin{equation}\label{descomii}
j(t,y+y_1)-j(t,y)=D_2j(t,y)\cdot y_1+W_2(j,t,y,y+y_1)
\end{equation}
where
\begin{eqnarray*}
\frac{W_1(j,t,t+s,y)}{| s |}\to  0&& \mbox{as}\quad s\to 0\\
 \frac{W_2(j,t,y,y+y_1)}{\|  y_1\|}\to  0 &&\mbox{as}\quad y_1\to 0\,.
\end{eqnarray*}

To prove the next theorem, we need a preliminary result which is
interesting in its own right.

\begin{Lemma}\label{lipsolutions}
Under the assumptions  in  Theorem \ref{lip}, if $x_0\in D(A)$ and
$g(0,x_0)\in D(A)$, then $x(\cdot)=x(\cdot,x_0)$ is Lipschitz on
closed intervals.
\end{Lemma}

\paragraph{Proof.} Initially we prove  that $x(\cdot)$ is
$\beta$-H\"{o}lder continuous on a closed interval $[0,b]$. Using the
continuity of $(-A)^{\beta}g$ and $f$ we can to assert that
$(-A)^{\beta}g(s,x(s))$ and $f(s,x(s))$ are bounded by $C_1>0$ on
$[0,b]$. Employing that  $x_0\in D(A)$ and that $g(0,x_0)\in
D(A)$; for $t\in [0,b)$ and $h>0$ we have
\begin{eqnarray*}
\lefteqn{\|x(t+h)-x(t)\|}\\
 &\leq& C_2h + \| g(t+h,x(t+h))-g(t,x(t))\| \\
&&+\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{{1-\beta}}}\|
(-A)^{\beta}g(s+h,x(s+h))-(-A)^{\beta}g(s,x(s))\| ds \\
&&+\int_0^{h}\| (-A)^{1-\beta}T(t+h-s)(-A)^{\beta}g(s,x(s))\| ds
\\ && +\tilde{M}\int_0^{t}\| f(s+h,x(s+h))-f(s,x(s))\| ds
+\tilde{M}\int_0^{h}\| f(s,x(s))\| ds
\end{eqnarray*}
thus
\begin{eqnarray*}
 \| x(t+h)-x(t) \| &\leq&  C_{3}h^{\beta} +\| (-A)^{-\beta}\| L\| x(t+h)-x(t)\|  \nonumber \\ &&
+\int_0^{t}\{\frac{C_{1-\beta}L}{(t-s)^{1-\beta}}+N\tilde{M}\}\|
x(s+h)-x(s)\| ds. \nonumber
\end{eqnarray*}
 Since $\| (-A)^{-\beta}\| L<1$, the Gronwall-Bellman inequality
\cite[Lemma 5.6.7]{PA}  implies that $x(\cdot)$ is $\beta$-H\"{o}lder
continuous.  Reiterating the previous estimates and using that
$x(\cdot)$ is  $\beta$-H\"{o}lder; if  $t\in [0,T)$ and $h>0$  we get
\begin{eqnarray*}
\lefteqn{\|x(t+h)-x(t) \|}\\
&& \leq C_{4}h +\| (-A)^{-\beta}\| L\| x(t+h)-x(t)\| \\
&&+\int_0^{t}\{\frac{C_{1-\beta}L}{(t-s)^{1-\beta}}+N\tilde{M}\}\|
x(s+h)-x(s)\| ds\\
&&+\int_0^{h}\frac{C_{1-\beta}L}{(t+h-s)^{1-\beta}}\|
(-A)^{\beta}g(s,x(s))-(-A)^{\beta}g(0,x_0)\| ds  \\
&&+\int_0^{h}\| T(t+h-s)(-A)g(0,x_0)\| ds
\end{eqnarray*}
then
\begin{eqnarray*}
\| x(t+h)-x(t) \|  &\leq& C_{5} h^{2\beta} +\| (-A)^{-\beta}\| L\|
x(t+h)-x(t)\|  \nonumber \\ &&
+\int_0^{t}\{\frac{C_{1-\beta}L}{(t-s)^{1-\beta}}+N\tilde{M}\}\|
x(s+h)-x(s)\| ds.\nonumber
\end{eqnarray*}
The assumption $\| (-A)^{-\beta}\| L<1$ and  Gronwall Bellman
inequality, implies that $x(\cdot)$ is $2\beta$-H\"{o}lder
continuous. Clearly the previous routine permit  to infer  that
$x(\cdot)$ is Lipschitz continuous, this completes the proof.
\hfill$\diamondsuit$\medskip

In the next theorem we  establish   the existence  of classical
solutions using some usual   regularity assumptions on the
functions   $f$ and $(-A)^{\beta}g$.

\begin{Theorem}\label{class}
Assume that  $(-A)^{1-\beta}g(\cdot)$ and $f(\cdot)$ are
continuously differentiable functions on  $[0,T]\times \Omega$.\,
If $x_0,\, g(0,x_0) \in D(A)$ and $\| D_2 g(0,x_0)
\|_{{\mathcal{L}}(X)}<1$ then  $\dot{x}(\cdot, x_0) \in
C([0,b]:X)$ for some  $0< b< T$.
\end{Theorem}

\paragraph{Proof:} Let $x(\cdot)=x(\cdot,x_0)$ and $z(\cdot)$ be a
solution of the integral equation
\begin{eqnarray}
z(t)&=&T(t)(Ax_0+ A g(0,x_0) + f(0,x_0))+h(t) - D_2g(t,x(t))(z(t))
\nonumber\\
 &&+\int_0^{t}(-A)^{1-\beta}T(t-s)D_2(-A)^{\beta}g(s,x(s))( z(s))ds
\label{reg2}\\
 &&+\int_0^{t}T(t-s)D_2f(s,x(s))(z(s))ds \nonumber
\end{eqnarray}
 where
\begin{eqnarray}\label{inicond}
z(0)=Ax_0+A g(0,x_0)+f(0,x_0)
-D_1g(0,x_0)-D_2g(0,x_0)(z(0))\nonumber
\end{eqnarray}
and
\begin{eqnarray*}
h(t)&=& - D_1g(t,x(t))+\int_0^{t}(-A)^{1-\beta}T(t-s)D_1(-A)
^{\beta}g(s,x(s))ds \\
 &&+\int_0^{t}T(t-s)D_1f(s,x(s))ds\,.
\end{eqnarray*}
 The existence and uniqueness  of  local solution for  (\ref{reg2}), is  consequence  of the contraction
 mapping principle and the condition
 $\| D_2g(0,x_0)\|_{{\mathcal{L}}(X)}<1$;
  we omit details. In what follows we assume that $x(\cdot)$ and
   $z(\cdot)$  are defined on  $[0,2b]$   where  $0<2b<T$ and
     $\| D_2 g(\theta,x_{\theta})
   \|_{2b}<\eta<1$. Using the notations introduced
in   (\ref{descomi})-(\ref{descomii}),  for $t$ in $[0,b]$ and
$h>0$ sufficiently small, we have
\begin{eqnarray*}
\lefteqn{ \|  \xi(t,h) \| }\\
&=&\|\frac{x(t+h)-x(t)}{h}-z(t)\| \\
 &\leq& \| T(t)(\frac{T(h)-I}{h}x_0-A(x_0))\| \nonumber\\
&&+ \| \frac{1}{h}\int_0^{h}T(t+h-s)f(s,x(s))ds-T(t)f(0,x_0) \|
\nonumber\\ &&+\|
T(t)(\frac{T(h)-I}{h})g(0,x_0)+\frac{1}{h}\int_0^{h}
(-A)T(t+h-s)g(s,x(s))ds \|\nonumber \\ && + \|
D_1g(t,x(t+h))-D_1g(t,x(t))\| + \| D_2g(t,x(t))(\xi(t,h))\| \nonumber\\
 && +\| \frac{W_1(g,t,t+h,x(t+h))}{h}\| + \|\frac{W_2(g,t,x(t),x(t+h))}{h}\| \nonumber\\ &&+
\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
D_1(-A)^{\beta}g(s,x(s+h))- D_1(-A)^{\beta}g(s,x(s))\| ds\nonumber
\\ && +\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
D_2(-A)^{\beta}g(s,x(s))(\xi(s,h))\| ds \nonumber \\ &&
+\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
\frac{W_1((-A)^{\beta}g,s,s+h,x(s+h))}{h}\| ds \nonumber\\ && +
\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
\frac{W_2((-A)^{\beta}g,s,x(s),x(s+h))}{h}\| ds \nonumber \\ &&+
\int_0^{t}\tilde{M}\| D_1f(s,x(s+h))-D_1f(s,x(s))\| ds \nonumber
\\ &&+\int_0^{t}\tilde{M}\| D_2f(s,x(s))\|\|\xi(s,h)\|ds
+\int_0^{t}\tilde{M}\|\frac{W_1(f,s,s+h,x(s+h))}{h}\|ds \nonumber
\\ && +
\tilde{M}\int_0^{t}\| \frac{W_2(f,s,x(s),x(s+h))}{h}\|
ds.\nonumber
\end{eqnarray*}
On the other hand, from lemma \ref{lip} we know that $x(\cdot)$
is Lipschitz continuous; therefore,
 $$ \frac{W_2((-A)^{\beta}g,s,x(s),x(s+h))}{\|
x(s+h)-x(s)\|}\cdot \frac{\| x(s+h)-x(s)\|}{h}\to 0
\quad\mbox{as}\quad h \to 0 $$ and $$
\frac{W_2(f,s,x(s),x(s+h))}{\| x(s+h)-x(s)\|}\cdot \frac{\|
x(s+h)-x(s)\|}{h}\to 0 \quad\mbox{as} \quad h \to 0
$$
 uniformly for $s\in [0,b]$. This enables us to  rewrite the last
inequality in the form
\begin{eqnarray*}
\lefteqn{ \| \xi(t,h) \| }\\
&=&\| \frac{x(t+h)-x(t)}{h}-z(t)\| \\
&\leq& \rho (t,h)+\frac{1}{h}\int_0^{h}
\frac{C_{1-\beta}}{(t+h-s)^{1-\beta}}\|(-A)^{\beta}g(0,x_0)-(-A)^{\beta}g(s,x(s))\|ds \\
\, &&+\| D_2 g(t,x(t))( \xi(t,h))\|
+\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
D_2(-A)^{\beta}g(s,x(s))\|
\|\xi(s,h)\|ds \\
&&+\tilde{M}\int_0^{t}\| D_2f(s,x(s)\| \|\xi(s,h)\|ds \nonumber
\end{eqnarray*}
where $ \rho(t,h)\to 0$  as $h \to 0$, uniformly for $t\in
[0,b].$ Since $x(\cdot )$ is  Lipschitz and  $\|
D_2g(\cdot,x(\cdot )) \|_{b} <\eta$, follow that
\begin {eqnarray}
\| \xi(t,h) \| & \leq & \frac{1}{1-\eta}\rho (t,h)
+\frac{C_{1-\beta}LCh^{\beta}}{\beta}
\nonumber\\
 && +\frac{1}{1-\eta}\int_0^{t}\frac{C_{1-\beta}}{(t-s)^{1-\beta}}\|
 D_2(-A)^{\beta}g(s,x(s))\| \|\xi(s,h)\| ds \nonumber \\ &&+
 \frac{1}{1-\eta}\tilde{M}\int_0^{t}\| D_2f(s,x(s)\|
 \| \xi(s,h)\| ds. \nonumber
 \end{eqnarray}
Finally,   the  Gronwall's  inequality \cite[Lemma 5.6.7]{PA}
shows  that $\xi(t,h)\to 0$ as $h\to 0$. Therefore,
 $\dot{x}(\cdot ,x_0)=z(\cdot)$. This completes the
proof. \hfill$\diamondsuit$

\begin{Corollary}\label{classical}
If $g$ is a $D(A)$-valued continuous function then there exits a
unique classical solutions of (\ref{ne}) defined on $[0,b]$ for
some $0< b< T$.
\end{Corollary}

\paragraph{Proof:} From Theorem \ref{class} we know that
$x(\cdot )=x(\cdot ,x_0)\in C^{1}([0,b]:X)$ for some $0<b<T$.
Since $x(\cdot,x_0)$ is Lipschtiz continuous in $[0,b]$, from
\cite[Theorem 2.4.1]{GO} and Lemma \ref{regc} we infer that
$x(t)+g(t,x(t))\in D(A)$ for $t\in [0,b]$ and therefore that
$x(t)\in D(A)$ for $t\in [0,b]$. The proof is
complete.\hfill$\diamondsuit$

\section{Examples}\label{examples}
In this section we sketch briefly some applications.

\subsection*{ Functional Differential Equations with Unbounded Delay}
The regularity results obtained in this work  are used to prove
the existence of regular solutions, ``Classical" and
``N-Classical" solutions, for a class of quasi-linear neutral
functional differential equations with unbounded delay that can
be  modeled in the form
\begin{eqnarray}\label{example1}
\frac{d}{dt}(x(t)+F(t,x_{t})) &=& Ax(t)+G(t,x_{t}), \quad t\geq
\sigma,
\\ \label{example1'} x_{\sigma} &=& \varphi \in \mathcal{B},
\end{eqnarray}
where A is the infinitesimal generator of an analytic semigroup
of bounded linear operators $(T(t))_{t\geq 0}$ on a Banach space
$X$;  the history $x_{t}:(-\infty,0]\to X$,
$x_{t}(\theta)=x(t+\theta)$,  belongs to  some abstract phase
space $\mathcal{B}$ defined axiomatically, as in Hale and Kato
\cite{HK}, and where the axioms are establish employing  the
terminology and notations used in Hino-Murakami-Naito \cite{HMN}.
A complete reference including results of existence of mild,
strong and periodical solutions for
(\ref{example1})-(\ref{example1'}) are the papers \cite{HH1},
\cite{HH2}. The existence of "N-Classical " and " Classical"
solutions is studied in \cite{HH3}, actually in preparation.

\subsection*{Partial  Differential Equations of Sobolev Type}
There is a extensive literature on semi-linear Sobolev evolution
equations modeled in the form
\begin{eqnarray}\label{Sobolevequa}
&\frac{d}{dt}(Bu(t))=Au(t)+f(t,u(t)) \quad t>0,&\\
 \label{Sobolevequa'}
&u(0)=u_0, \quad u_0\in D(B)\,,&
\end{eqnarray}
where $A, B$  are closed linear operators on a Banach space $X$.
The literature includes different and complete results concerning
to existence, uniqueness and qualitative properties of mild,
strong and classical solutions for
(\ref{Sobolevequa})-(\ref{Sobolevequa'}) (see  \cite{Brill, ligh,
Showalter2, Showalter1}). Some usual assumptions on the operators
$A,B$ (see for example \cite{Brill, ligh})  are
\begin{itemize}
  \item $A,\,B$ are closed linear operators.
   \item $D(B)\subset D(A)$ and $B$ has a continuous inverse.
\end{itemize}
>From these assumptions and the Closed Graph  Theorem it follows that
$AB^{-1}$ is a bounded linear operator on $X$. In this case the
approach is to consider the related integral equation
\begin{equation}\label{absobolev0}
x(t)=T(t)Bx_0+ \int_0^{t}T(t-s)f(s,B^{-1}x(s))ds,
\end{equation}
where $T(t)$ with $t\geq 0$ is the semigroup generated by
$AB^{-1}$.

We shall consider the abstract  Cauchy problem
\begin{eqnarray}\label{absobolev21}
&\frac{d}{dt}(u(t)+Bu(t))=Au(t)+f(t,u(t)), \quad t>0,&\\
\label{absobolev22} & u(0)= u_0, \quad u_0\in D(B),&
\end{eqnarray}
 where  $A,B$ are closed linear operators on a Banach space $X$ and
\begin{itemize}
  \item $D(A)\subset D(B)$ and $B$ has a continuous inverse
  \item $AB^{-1}$ is the infinitesimal generator of an analytic
semigroup of bounded linear  operators on $X$.
  \end{itemize}
Under these conditions, we consider the associated system
\begin{eqnarray}\label{Sobolevequa2}
&\frac{d}{dt}(u+ B^{-1}u(t))=AB^{-1}u(t)+f(t,B^{-1}u(t)), \quad t>0,&\\
\label{Sobolevequa2'} &u(0)=B^{-1}u_0, \quad u_0\in D(B).&
\end{eqnarray}
If in addition  $B^{-1}$ is  $D(AB^{-1})$-valued continuous  and
$f$ is continuously differentiable,  the existence of classical
solutions for (\ref{Sobolevequa2})-(\ref{Sobolevequa2'}) and
consequently for (\ref{absobolev21})-(\ref{absobolev22}), follows
from Corollary \ref{classical}.


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}\end{thebibliography}

\noindent{\sc Eduardo Hern\'{a}ndez M.} \\
Departamento de Matem\'atica \\
Instituto de Ci\^encias Matem\'aticas de S\~ao Carlos \\
Universidade de S\~ao Paulo \\
Caixa Postal 668 \\
13560-970 S\~ao Carlos, SP. Brazil \\
e-mail: lalohm@icmc.sc.usp.br

\end{document}