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\markboth{\hfil On the ellipticity and solvability \hfil EJDE--2001/57}
{EJDE--2001/57\hfil Abdelillah El Haial \& Rabah Labbas \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 57, pp. 1--18. \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} \\
 %
  On the ellipticity and solvability of an abstract second-order
  differential equation 
 %
\thanks{ {\em Mathematics Subject Classifications:} 34G10, 35J25, 47D03.
\hfil\break\indent
{\em Key words:} Abstract elliptic equations.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted November 15, 1998. Published August 27, 2001.} }
\date{}
%
\author{ Abdelillah El Haial \& Rabah Labbas }
\maketitle

\begin{abstract} 
 In this work we give some new results on a complete abstract 
 second-order differential equation of elliptic type in the
 non-homogeneous case.  Existence, uniqueness, and maximal regularity 
 of the strict solution are proved under some natural assumptions 
 which imply the ellipticity of the differential equation.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

 In this paper, we study the second-order abstract
differential equation 
\begin{equation}
u''(t)+2Bu'(t)+Au(t)=f(t),\quad t\in (0,1)
\end{equation}
under the non-homogeneous boundary conditions 
\begin{equation}
u(0)=\varphi , \quad u(1)=\psi ,
\end{equation}
where $\varphi ,\psi $ and $f(t)$ belong to a complex Banach space $E$, and 
$A$, $B$ are two closed linear operators with domains $D(A)$, $D(B)$. We are
interested in the existence, uniqueness, and maximal regularity of the strict
solution $u$ when $f$ is regular (for instance $f$ \ is H\"{o}lder
continuous function). We recall that $u\in C([0,1] ;E)$ is a
strict solution of (1)-(2) if and only if 
\[
u\in C^{2}([0,1] ;E)\cap C^1([0,1] ;D(B))\cap C([0,1] ;D(A)) 
\]
and $u$ satisfies (1) and (2).

Throughout this paper we assume the following hypotheses: 
\begin{enumerate}
\item[H1)] Operator $B$  generates a strongly continuous group on $E$

\item[H2)] There exists $K>0$ such that for all $\lambda \geqslant 0$,
$\| (A-B^{2}-\lambda I)^{-1}\|_{L(E)}\leqslant K/(1+\lambda)$ 

\item[H3)] For all $\mu \in \mathbb{R}$, and all $\lambda \geqslant 0$,
$(A-B^{2}-\lambda I)^{-1}(B-\mu I)^{-1}-(B-\mu I)^{-1}(A-B^{2}
-\lambda I)^{-1}=0$ 

\item[H4)] There exists $K>0$ such that for all $\lambda \geqslant 0$,\\ 
i)\quad $\| B(A-B^{2}-\lambda I)^{-1}\|_{L(E)}\leqslant 
K/(1+\sqrt{\lambda })$, \\ 
ii)\quad $\| B^{2}(A-B^{2}-\lambda I)^{-1}\|_{L(E)}\leqslant K$,\\
iii)\quad $\| A(A-B^{2}-\lambda I)^{-1}\|_{L(E)}\leqslant K$.
\end{enumerate}

Note that H1 implies $\overline{D(B)}=E$, but the domain $D(A-B^{2})$
may be not dense. On the other hand, it is not difficult to see that
assumptions H1, H2, H3 permit us to apply the Da Prato
and Grisvard \cite{d1} sum theory and deduce that, necessarily, 
\[
(A-\lambda I) ^{-1}\in L(E) \quad\mbox{and}\quad \|(A-\lambda
I)^{-1}\|_{L(E)}\leqslant K/(1+\lambda ),\,\mbox{for all }\lambda
\geqslant 0. 
\]
(See the proof of Theorem \ref{thm1}, the end of statement i)).

Several authors have studied equation (1) when it is regarded as an
abstract Cauchy problem, that is under the following initial data 
$u(0)=\varphi $, $u'(0)=\psi $. See, for instance, Favini \cite{f1},
Neubrander \cite{n1}, Liang and Xiao \cite{l3}. The techniques used in these papers
are based on the parabolicity of (1), that is the parabolicity of the
operator pencil defined by
\[
P(\lambda ):D(A)\cap D(B)\to E;\quad x\mapsto P(\lambda
)x=(\lambda ^{2}I+2\lambda B+A)x. 
\]

In this paper we try to provide an unified treatment of some class of second
order partial differential equations when they are regarded as equations of
elliptic type. The principal part of these equations can be written in the
form 
\[
\mathcal{P}u=\partial_{t}^{2}u+2b(.)\partial_{xt}^{2}u+a(.)\partial
_{x}^{2}u, 
\]
or more generally
\[
\mathcal{P}u=\partial_{t}^{2}u+\mathcal{B}(x,\partial_{x})\partial_{t}u+
\mathcal{A}(x,\partial_{x})\partial_{x}^{2}u, 
\]
where $\mathcal{B}(x,\partial_{x})$ and $\mathcal{A}(x,\partial_{x})$ are
linear differential operators in $\mathbb{R}^{n}$ with some
natural assumptions on their coefficients of order $m$ and $l$.

So, our objective is to express the ellipticity of these concrete operators
in general abstract form, find a representation of the solution of the
problem (1)-(2) and study its maximal regularity.

Some particular equations of type (1) can be used to describe the
singularities of the solutions in elliptic problems on singular domains. Let
us give the following example of Laplacian operator 
\begin{gather*}
-\Delta u=g,\quad g\in L^{p}(Q) \\
u=0\quad \mbox{on } \partial Q,
\end{gather*}
in a conical domain $Q=\left\{ \rho \sigma ~:~\rho >0,\sigma \in
G\right\} \subset \mathbb{R}^{n}$, where $G$ is an open subset
of the sphere $S^{n-1}$. In the three dimensional domain, the boundary 
$\partial Q$ has a vertex at $O$ and also edges depending on whether the
boundary $\partial G$ has corners or not. Let us assume that the variational
solution $u$ of this Dirichlet problem exists. So, we would have 
$u=u_{r}+u_{s}$, with a regular part $u_{r}$ $\in W^{2,p}(Q) $
and a singular part $u_{s}$ in $\exp $licit form. Then, one performs the
change of variable $\rho =e^{t}$ which maps $Q$ onto the infinite cylinder 
$\mathbb{R}\times G$. The Laplacian equation in polar
coordinates is 
\[
D_{\rho }^{2}u+\frac{n-1}{\rho }D_{\rho }u+\frac{1}{\rho
^{2}}\Delta 'u=g 
\]
where $\Delta '$ denotes the Laplace-Beltrami operator on $S^{n-1}$.
 The above mentioned change of variable leads us to 
\[
D_{t}^{2}u+D_{t}u+\Delta 'u=e^{2t}g=g_{1}, 
\]
then by puttting $v(t)(\sigma )=e^{\varpi t}u(e^{t}\sigma )$ and 
$g_{2}(t)(\sigma )=e^{\varpi t}g_{1}$ \ we have 
\begin{gather*}
D_{t}^{2}v+(n-2-2\varpi ) D_{t}v+\varpi (\varpi
-n+2) v+\Delta 'v=g_{2} \\ 
v=0\quad \mbox{on }\partial (\mathbb{R}\times G)
\end{gather*}
where $\varpi =-2+n/p$, precisely the opposite of the Sobolev exponent of 
$W^{2,p}$ . This final equation can be regarded as a particular form of our
abstract general problem (1). See, for details, 
Labbas-Moussaoui-Najmi  \cite{l2}.

Equations (1)-(2) can be illustrated, for instance, by the following
simple model differential problem 
\begin{equation}
\begin{gathered}
\frac{\partial ^{2}u}{\partial t^{2}}(t,x)+2b\frac{\partial ^{2}u}{
\partial x\partial t}(t,x)+a\frac{\partial ^{2}u}{\partial x^{2}}
(t,x)=f(t,x),\quad (t,x)\in \Sigma , \\ 
u(0,x)=\varphi (x),\quad \,u(1,x)=\psi (x),\quad x\in \mathbb{R},
\end{gathered}
\end{equation}
where $\Sigma =\left] 0,1\right[ \times \mathbb{R}$ and 
\begin{equation}
a-b^{2}>0.
\end{equation}
Then we can choose in $E=L^{p}(\mathbb{R})$, $1<p<\infty $,
the operators $A$, $B$ defined by 
\begin{gather*}
D(B)=W^{1,p}(\mathbb{R}),\quad (Bu) (x)=bu'(x),~\forall u\in D(B) \\ 
D(A)=W^{2,p}(\mathbb{R}),\quad (Au) (x)=au''(x),~\forall u\in D(A).
\end{gather*}

In our study, hypotheses H2 and H4 express the ellipticity of 
(1) and generalize the one used in Krein  \cite{k1} in the case $B=0$. 
In example (3) these two assumptions are equivalent to (4). So we cannot reduce
equations (1)-(2) to some first order system. When $B=0$, Labbas  \cite{l1} has
studied problem (1)-(2) and gives necessary and sufficient conditions on 
$\varphi$, $\psi$, $f(0)$ and $f(1)$ for having a unique strict solution
when $f$ is H\"{o}lder continuous function. In this work we generalize these
results since all the hypotheses considered here, coincide with those used
in  \cite{l1} in the case $B=0$. Note that assumptions H1, H2, H3
allow us to apply the sum theory of linear operators as in Da
Prato-Grisvard \cite{d1} and give $D_{A}(\theta ;+\infty )=D_{L}(\theta ;+\infty
)\cap D_{B^{2}}(\theta ;+\infty )$ where $L=A-B^{2}$ and, for $\theta \in
]0,1[$ 
\[
D_{L}(\theta ;+\infty )=\left\{ \varphi \in E:\sup_{r>0}
r^{\theta }\|L(L-rI)^{-1}\varphi \|_{E}<\infty \right\} , 
\]
(see Grisvard  \cite{g2}).

In section 2, we build the natural representation of the solution of
(1)-(2) by using the operational calculus and the Dunford integral.

In section 3, we prove an essential lemma, which allow us to justify and
to study the optimal smoothness of the previous representation; we then give
necessary and sufficient conditions on $\varphi ,\psi $, $f(0)$ and $f(1)$
for having a strict solution when $f$ is H\"{o}lder continuous function.

Finally, in section 4, we give an example, to which our abstract results
can be applied.

\section{Construction of the solution}

 If $A$ and $B$ are two scalars such that $B^{2}-A=-\lambda $ is
strictly positive, then the solution of (1) is given by  
\[
u(t)=\frac{\sinh \sqrt{-\lambda }(1-t)}{\sinh \sqrt{-\lambda }}
e^{-tB}\varphi +\frac{\sinh \sqrt{-\lambda }t}{\sinh \sqrt{-\lambda }}
e^{(1-t)B}\psi \medskip -\int_0^1G_{\sqrt{-\lambda }}(t,s)f(s)ds, 
\]
where 
\[
G_{\sqrt{-\lambda }}(t,s)=\left\{ 
\begin{array}{c}
G_{\sqrt{-\lambda }}^{+}(t,s)=\frac{\sinh \sqrt{-\lambda }(1-t)\sinh \sqrt{
\lambda }s}{\sqrt{-\lambda }\sinh \sqrt{-\lambda }},~~0\leqslant
s\leqslant t\medskip \\ 
G_{\sqrt{-\lambda }}^{-}(t,s)=\frac{\sinh \sqrt{-\lambda }(1-s)\sinh \sqrt{
\lambda }t}{\sqrt{-\lambda }\sinh \sqrt{-\lambda }},\,~~t\leqslant
s\leqslant 1.
\end{array}
\right. 
\]
Now, it is well known that H2 implies the existence of $\delta
_0\in ]0,\pi /2[$ and $\varepsilon_0>0$ such that the resolvent set of 
$A-B^{2}$ contains the following sector of the complex plane 
\[
S(\delta_0,\varepsilon_0)=\left\{ z\in \mathbb{C}
:\left| \arg (z)\right| \leqslant \delta_0\right\} \cup B(
0,\varepsilon_0) , 
\]
where $B(0,\varepsilon_0) $ is the open ball of radius 
$\varepsilon_0$. If $\gamma $ denotes the sectorial boundary curve of 
$S(\delta_0,\varepsilon_0)$ oriented positively, then the natural
representation of the solution of (1)-(2), in the abstract case, is
given by the Dunford integral 
\begin{eqnarray}
u(t) &=&\frac{e^{-tB}}{2\pi i}\int\nolimits_{\gamma }g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}\varphi d\lambda \nonumber \\
&&+\frac{e^{(1-t)B}}{2\pi i}\int\nolimits_{\gamma }g_{\sqrt{-\lambda }
}(1-t)(L-\lambda I)^{-1}\psi d\lambda  \\
&&-\frac{1}{2\pi i}\int\nolimits_{\gamma }\int_0^1G_{\sqrt{-\lambda }
}(t,s)e^{-(t-s)B}(L-\lambda I)^{-1}f(s)ds\,d\lambda ,  \nonumber
\end{eqnarray}
where $L=A-B^{2}$, $D(L)=D(A)\cap D(B^{2})$,
\[
g_{\sqrt{-\lambda }}(t)=\frac{\sinh \sqrt{-\lambda }(1-t)}{\sinh \sqrt{
\lambda }}\quad t\in (0,1),\lambda \in \gamma .
\]
Here $\sqrt{-\lambda }$ is the analytic determination defined by 
$Re\sqrt{-\lambda }>0$. From writing
\[
\left| g_{\sqrt{-\lambda }}(t)\right| =
\big| e^{-\sqrt{-\lambda }t}\Big(
\frac{1-e^{-2\sqrt{-\lambda }(1-t)}}{1-e^{-2\sqrt{-\lambda }}}\Big)
\big| , 
\]
we deduce that there exists two constants $c_0$ and $K_0$ such that 
\begin{equation}
\begin{gathered} 
\left| g_{\sqrt{-\lambda }}(t)\right| \leqslant K_0e^{-c_0\left| \lambda
\right| ^{1/2}t}\quad \forall \lambda \in \gamma ,\; \forall t\in
]0,1], \\ 
\left| g_{\sqrt{-\lambda }}(1-t)\right| \leqslant K_0e^{-c_0\left|
\lambda \right| ^{1/2}(1-t)}\quad \forall \lambda \in \gamma ,\; \forall
t\in [0,1[,
\end{gathered}
\end{equation}
where $c_0=\cos (\pi /2-\delta_0/2)$ and 
$K_0=2/\big(1 -\exp(-2\mathbf{c}_0\sqrt{\varepsilon_0})\big)$.
On the other hand, for any $f\in C([0,1] ;E) $, we
can see that 
\begin{equation}
\|\int_0^1G_{\sqrt{-\lambda }}(t,s)f(s)ds\|_{E}\leqslant 
\frac{1}{\left| \lambda \right| \cos (\delta_0/2) }
\|f\|_{C(E)},\quad \forall \lambda \in \gamma .
\end{equation}
According to H1, H2 and estimates (6)-(7), all the
integrals in (5) converge absolutely for every $t\in ]0,1[$. 

\section{Smoothness of the solution}

Let us set, for $\varphi $ and $\psi $ in $E$, 
\begin{gather*}
U(t,-B,L)\varphi =\frac{e^{-tB}}{2\pi i}\int_\gamma g_{
\sqrt{-\lambda }}(t)(L-\lambda I)^{-1}\varphi d\lambda \quad t\in ]0,1], \\ 
U(1-t,B,L)\psi =\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{
\sqrt{-\lambda }}(1-t)(L-\lambda I)^{-1}\psi d\lambda \quad t\in [0,1[,
\end{gather*}
then we have the essential lemma

\begin{lemma} \label{lm1}
Under assumptions H1 and H2
\begin{enumerate}
\item[i)] there exists a positive constant $K=K(\varepsilon_0,\delta_0)$
such that 
\begin{gather*}
\|U(t,-B,L)\varphi \|_{E}\leqslant K\|\varphi \|
_{E},\quad \forall \varphi \in E, \forall t\in ]0,1], \\ 
\|U(1-t,B,L)\psi \|_{E}\leqslant K\|\psi \|
_{E},\quad \forall \psi \in E, \forall t\in \lbrack 0,1[,
\end{gather*}


\item[ii)]  $U(.,-B,L)\varphi $ $\in C([0,1] ;E)$ if and only  if  
$\varphi \in \overline{D(L)}$,

\item[iii)] $U(1-.,B,L)\psi \in C([0,1] ;E)$ if and only if 
$\psi \in \overline{D(L)}$.
\end{enumerate} \end{lemma}

\paragraph{Proof.} i) For $t>0$ we can write 
\begin{eqnarray*}
\lefteqn{U(t,-B,L)\varphi }\\
&=&\frac{e^{-tB}}{2\pi i}\int_{\gamma_{+}^{t}}
g_{\sqrt{-\lambda }}(t)(L-\lambda I)^{-1}\varphi d\lambda 
+\frac{e^{-tB}}{2\pi i}\int_{\gamma_{-}^{t}} g_{\sqrt{
-\lambda }}(t)(L-\lambda I)^{-1}\varphi d\lambda \\
&=&I_{+}+I_{-},
\end{eqnarray*}
where
\begin{equation}
\gamma_{+}^{t}=\left\{ z\in \gamma ;\,\left| z\right| \geqslant
1/t^{2}\right\} , \quad \gamma_{-}^{t}=\left\{ z\in \gamma
;\,\left| z\right| \leqslant 1/t^{2}\right\} .
\end{equation}
Then we have 
$$
\|I_{+}\|_{E} \leqslant K_0(\int_{1/t^{2}}^{+\infty }
\frac{e^{-c_0\left| \lambda \right| ^{1/2}t}}{\left| \lambda \right| }
d\left| \lambda \right| ) \|\varphi \|_{E} 
  \leqslant 2K_0\int_{1}^{+\infty }\frac{e^{-c_0\sigma }}{\sigma }
d\sigma \|\varphi \|_{E}\leqslant K\|\varphi \|
_{E},
$$
and
\begin{eqnarray*}
I_{-} &=&\frac{e^{-tB}}{2\pi i}\int_{\gamma_{-}^{t}} (g_{
\sqrt{-\lambda }}(t)-g_{\sqrt{-\lambda }}(0)) (L-\lambda
I)^{-1}\varphi d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_{\gamma_{-}^{t}} (L-\lambda
I)^{-1}\varphi d\lambda \\
&=&-\frac{e^{-tB}}{2\pi i}\int_{\gamma_{-}^{t}} (
\int_0^{t}\frac{\sqrt{-\lambda }\cosh \sqrt{-\lambda }(1-s)}{\sinh \sqrt{
\lambda }}ds) (L-\lambda I)^{-1}\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_{C_{1/t^2}} (L-\lambda I)^{-1}\varphi d\lambda \\
&=&I_{-}'+I_{-}'',
\end{eqnarray*}
where 
\[
\,C_{1/t^{2}}=\left\{ z : \left| \arg z\right| \leqslant \delta_0\mbox{
and }\left| z\right| =1/t^{2}\right\} . 
\]
Then 
\begin{gather*}
\|I_{-}'\|_{E}\leqslant \frac{K_0}{2\pi }
\int_{\varepsilon_0}^{1/t^{2}}\frac{\left| \lambda \right| ^{1/2}t}{
\left| \lambda \right| }d\left| \lambda \right| .\|\varphi \|
_{E}\leqslant K\|\varphi \|_{E}, \\ 
\|I_{-}''\|_{E}\leqslant \frac{1}{2\pi }
\int_{-\delta_0}^{+\delta_0}\|(L-\frac{e^{i\theta }}{
t^{2}}I) ^{-1}\varphi \|_{E}\frac{d\theta }{t^{2}}\leqslant
K\|\varphi \|_{E}.
\end{gather*}
Similarly we obtain 
\[
\|U(1-t,B,L)\psi \|_{E}\leqslant K\|\psi \|
_{E},\quad \forall \psi \in E\mbox{ , }\forall t\in [0,1[. 
\]
ii) Fix $\varepsilon >0$ and let $\varphi \in \overline{D(L)}$, then there
exists $y\in D(L)$ such that 
\begin{equation}
\|\varphi -y\| \leqslant \varepsilon .
\end{equation}
Using the identity 
\[
(L-\lambda I)^{-1}y=\frac{(L-\lambda I)^{-1}Ly}{\lambda }-\frac{y}{\lambda }, 
\]
we have 
\[
U(t,-B,L)y=\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}(t)\frac{(L-\lambda I)^{-1}}{\lambda }Lyd\lambda , 
\]
which gives 
\begin{equation}
\|U(t,-B,L)y\|_{E}\leqslant \frac{1}{2\pi }\int_{\varepsilon
_0}^{\infty }\frac{K}{\left| \lambda \right| ^{2}}d\left| \lambda \right|
\|Ly\|_{E}\leqslant K\|Ly\|_{E}~\forall 
t\in [0,1].
\end{equation}
Now, from the equality 
\[
U(t,-B,L)\varphi -\varphi =U(t,-B,L)\varphi
-U(t,-B,L)y+U(t,-B,L)y-y+y-\varphi , 
\]
and the estimates (9), (10) we deduce that 
\[
U(t,-B,L)\varphi -\varphi \to 0 \quad\mbox{as}\quad t\to 0^+. 
\]
The continuity in $t>0$ is easily verified.
Conversely, assume  $U(.,-B,L)\varphi\in C([0,1] ;E)$,
then $\lim_{t\to 0^+} U(t,-B,L)\varphi $ exists and is
necessarily equal to $\varphi$; however 
\[
g_{\sqrt{-\lambda }}(t)e^{-tB}(L-\lambda I)^{-1}\varphi \in D(L), 
\]
which implies that $U(t,-B,L)\varphi \in \overline{D(L)}$.

For statement iii) it is enough to substitute $1-t$ for $t$.
\hfill$\Box$\medskip

Let us consider, for $\theta \in \left] 0,1\right[ $, the well known real
interpolation space between $D(L)$ and $E$ characterized by 
\[
D_{L}(\theta ;+\infty )=\big\{ \varphi \in E : \sup_{r>0}
r^{\theta }\|L(L-rI)^{-1}\varphi \|_{E}<\infty \big\} . 
\]
(See Grisvard$ \cite{g1})$.

\begin{lemma} \label{lm2}
Under assumptions H1, H2, H3 and for $\theta \in
]0,1/2[ $ we have 
\[
U(.,-B,L)\varphi \in C^{2\theta }([0,1] ;E)
\quad\mbox{if and only if}\quad \varphi \in D_{L}(\theta ;+\infty ). 
\]
\end{lemma}

\paragraph{Proof.} 
Let $\varphi \in D_{L}(\theta ;+\infty )$ and $\tau ,t\in ]0,1[$ such that 
$\tau <t$, then 
\begin{eqnarray*}
\lefteqn{U(t,-B,L)\varphi -U(\tau ,-B,L)\varphi }\\
&=&\frac{e^{-tB}}{2\pi i}\int_\gamma (g_{\sqrt{-\lambda }}(t)
-g_{\sqrt{-\lambda }}(\tau)) \frac{L(L-\lambda I)^{-1}}{\lambda }
\varphi d\lambda \\
&&+\frac{(e^{-tB}-e^{-\tau B}) }{2\pi i}
\int_\gamma g_{\sqrt{-\lambda }}(\tau )\frac{L(L-\lambda I)^{-1}}{\lambda }\varphi
d\lambda \\
~ &=&I_{1}+I_{2}.
\end{eqnarray*}
$I_{1}$ may be written as 
\begin{eqnarray*}
I_{1}&=&\frac{e^{-tB}}{2\pi i}\int_{\gamma_{+}^{t-\tau }} 
(g_{\sqrt{-\lambda }}(t)-g_{\sqrt{-\lambda }}(\tau )) \frac{
L(L-\lambda I)^{-1}}{\lambda }\varphi d\lambda \\ 
&& +\frac{e^{-tB}}{2\pi i}\lim_{\gamma_{-}^{t-\tau }} 
(g_{\sqrt{-\lambda }}(t)-g_{\sqrt{-\lambda }}(\tau )) \frac{
L(L-\lambda I)^{-1}}{\lambda }\varphi d\lambda \\ 
&=&I_{1}'+I_{1}'',
\end{eqnarray*}
(where $\gamma_{+}^{t-\tau }$ and $\gamma_{-}^{t-\tau }$ are defined in 
(8)), and 
\begin{eqnarray*}
&&\|I_{1}'\|_{E}\leqslant 2K_0\int_{\gamma_{+}^{t-\tau }} 
\frac{d\left| \lambda \right| }{\left| \lambda \right|
^{\theta +1}}\|\varphi \|_{D_{L}(\theta ;+\infty )}, \\
&&\|I_{1}''\|_{E}\leqslant K_0
\lim_{\gamma_{-}^{t-\tau }} \frac{\left| t-\tau \right| }{\left| \lambda
\right| ^{\theta +1/2}}d\left| \lambda \right| \|\varphi \|
_{D_{L}(\theta ;+\infty )},
\end{eqnarray*}
from which it follows 
\[
\|I_{1}\|_{E}\leqslant K(t-\tau )^{2\theta }\|\varphi
\|_{D_{L}(\theta ;+\infty )}. 
\]
We write $I_{2}$ as 
\[
I_{2}=(t-\tau ) ^{2\theta }e^{-\tau B}(\frac{e^{-(
t-\tau ) B}\Phi_{1}-\Phi_{1}}{(t-\tau ) ^{2\theta }}
) , 
\]
with 
\[
\Phi_{1}=\frac{1}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(\tau )(L-\lambda I)^{-1}\varphi d\lambda , 
\]
then, for $\forall r>0$, the classical operational calculus gives 
\[
L(L-rI)^{-1}\Phi_{1}=\frac{1}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}(\tau )\frac{L(L-\lambda I)^{-1}\varphi }{r-\lambda }d\lambda
,\quad 
\]
which implies that
\[
\|L(L-rI)^{-1}\Phi_{1}\|_{E}\leqslant \frac{K}{r^{\theta }}
\|\varphi \|_{D_{L}(\theta ;+\infty )}; 
\]
therefore,
\[
\Phi_{1}\in D_{L}(\theta ;+\infty )\subset D_{B^{2}}(\theta ;+\infty
)=D_{B}(2\theta ;+\infty )\,.
\]
The last equality holds by using the well known reiteration theorem in
interpolation theory (see, Lions-Peetre  \cite{l4}). So, we obtain 
\[
\|I_{2}\|_{E}\leqslant K(t-\tau ) ^{2\theta
}\|\Phi_{1}\|_{D_{B}(2\theta ;+\infty )}. 
\]
The main result in this work is the following

\begin{theorem} \label{thm1}
 Assume H1,H2,H3, and H4. Let 
$\varphi \in D(L),\psi \in D(L)$ and $f\in C^{2\theta }([0,1] ;E)$
with $\theta \in ]0,1/2[\,$. Then $u$ given in (5) satisfies 
\begin{enumerate}
\item[i)] $u(t)\in D(A)$ for all $t\in [0,1]$

\item[ii)] $Au(.)$ belongs to $C([0,1] ;E)$ if and only if 
$A\varphi,A\psi ,f(0)$ and $f(1)$ belong to $\overline{D(L)}$

\item[iii)]  $Au(.)$ belongs to $C^{2\theta }([0,1] ;E)$ if and only
if $A\varphi$ ,$A\psi$, $f(0)$ and $f(1)$ belong to $D_{L}(\theta ;+\infty )$.
\end{enumerate}
Furthermore, if 
\begin{equation}
B\varphi \in D_{L}(\theta +1/2;+\infty )\mbox{ and }B\psi \in D_{L}(\theta
+1/2;+\infty ),
\end{equation}
then
\begin{enumerate}
\item[iv)] $u'(t)\in D(B)$ for all $t\in [0,1]$

\item[v)] $Bu'(.)\in C([0,1];E)$  if and only if $B^{2}\varphi $ and 
$B^{2}\psi $ belong to $\overline{D(L)}$

\item[vi)] $Bu'(.)\in C^{2\theta }([0,1];E)$  if and only if  
$B^{2}\varphi$, $B^{2}\psi$ belong to $D_{L}(\theta ;+\infty )$.
\end{enumerate}
\end{theorem}

\paragraph{Proof.}
Statement i). The study of the first and second integral
in representation (5) is identical since $t$ and $(1-t)$ have the same
role. So we can assume that $\psi =0$. Moreover in the third integral in 
(5) we write $f(s)=(f(s)-f(t))+f(t)$, then, after two integration by
parts, we get 
\begin{eqnarray*}
u(t) &=&\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(1-t)(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{1}{2\pi i}\int_\gamma \int_0^1G_{\sqrt{-\lambda }
}(t,s)e^{-(t-s)B}(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&+\frac{1}{2\pi i}\int_\gamma (B^{2}+\lambda
I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&=&\sum_{i=1}^{5}a_{i}.
\end{eqnarray*}
All these integrals converge absolutely. From Lemma \ref{lm1} and H4, we
deduce that the first integral $a_{1}$ belongs to $D(A)$. Writing 
$A=L+B^{2}=(L-\lambda I) +(B^{2}+\lambda I) $ and
using H4 we obtain 
\[
A(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)=(L-\lambda
I)^{-1}f(t)+(B^{2}+\lambda I)^{-1}f(t), 
\]
which implies, as for the first integral, that $a_{2}$ and $a_{3}$ belong to 
$D(A)$. For $a_{4}$, we use iii) of H5 which gives 
\begin{eqnarray*}
\lefteqn{\|\frac{1}{2\pi i}\int_\gamma \int_0^1G_{\sqrt{
\lambda }}(t,s)e^{-(t-s)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \|_{E} }\\
&\leqslant &K\int_{\varepsilon_0}^{\infty }(
\sup{0\leqslant t\leqslant 1} \int_0^1\left| G_{\sqrt{-\lambda }}(t,s)\right|
\left| t-s\right| ^{2\theta }ds) d\left| \lambda \right| \|
f\|_{C^{2\theta }(E)} \\
&\leqslant &K(\int_{\varepsilon_0}^{\infty }\frac{1}{\left| \lambda
\right| ^{1+\theta }}d\left| \lambda \right| ) \|f\|
_{C^{2\theta }(E)},
\end{eqnarray*}
the last estimate follows in virtue of H\"{o}lder's inequality. For $a_{5}$
we apply the Da Prato-Grisvard's \cite{d1} sums theory to $B^{2}$ and $L$. In
fact, from H1 we deduce that $B^{2}$ generates a bounded holomorphic
semigroup in $E$ (see Stone \cite{s1}); then $(B^{2}+L)$ is closable and 
\[
(\overline{B^{2}+L}) ^{-1}=S=\frac{1}{2\pi i}
\int_\gamma (B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}d\lambda , 
\]
which implies that
$Sx=A^{-1}x$ for all $x\in E$; therefore,  $a_{5}=A^{-1}f(t)$. 


Statements ii) and iii). We have 
\begin{eqnarray*}
Au(t) &=&\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}A\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(1-t)(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(1-t)(L-\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{1}{2\pi i}\int_\gamma \int_0^1G_{\sqrt{-\lambda }
}(t,s)e^{-(t-s)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&+f(t) \\
&=&\sum_{i=1}^{6}v_{i}(t)+f(t).
\end{eqnarray*}
As in lemmas \ref{lm1} and \ref{lm2}, $v_{1}(.)\in C([0,1] ;E)$ if and
only if $A\varphi \in \overline{D(L)}$ and $v_{1}(.)\in C^{2\theta }([0,1] ;E)$
 if and only if $A\varphi \in D_{L}(\theta ;+\infty )$. 
 Let us write for $0\leqslant \tau <t\leqslant 1$ 
\begin{eqnarray*}
v_{2}(t)-v_{2}(\tau ) &=&-\frac{e^{-tB}}{2\pi i}\int_\gamma 
g_{\sqrt{-\lambda }}(t)(B^{2}+\lambda I)^{-1}(f(t)-f(\tau ))d\lambda \\
&&-\frac{e^{-\tau B}}{2\pi i}\int_\gamma (g_{\sqrt{-\lambda }
}(t)-g_{\sqrt{-\lambda }}(\tau ))(B^{2}+\lambda I)^{-1}(f(\tau
)-f(0))d\lambda \\
&&-\frac{(e^{-tB}-e^{-\tau B})}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}(t)(B^{2}+\lambda I)^{-1}(f(\tau )-f(0))d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(B^{2}+\lambda I)^{-1}f(0)d\lambda \\
&&-\frac{e^{-\tau B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(\tau )(B^{2}+\lambda I)^{-1}f(0)d\lambda \\
&=&\sum_{i=1}^{5}b_{i},
\end{eqnarray*}
then, as in the proof of Lemma \ref{lm1}, we have 
\[
\|b_{1}\| \leqslant K(t-\tau )^{2\theta }\|f\|
_{C^{2\theta }(E)}, 
\]
and writing $b_{2}$ as 
\[
b_{2}=-\frac{e^{-\tau B}}{2\pi i}\int_{\tau }^{t}\int_\gamma 
\frac{\sqrt{-\lambda }\cosh \sqrt{-\lambda }(1-s)}{\sinh \sqrt{-\lambda }}
(B^{2}+\lambda I)^{-1}(f(\tau )-f(0))d\lambda ds, 
\]
we get 
\begin{eqnarray*}
\|b_{2}\|_{E} &\leqslant &K\int_{\tau }^{t}\int_{\varepsilon
_0}^{+\infty }\frac{e^{-\left| \lambda \right| ^{1/2}c_0s}\tau ^{2\theta
}}{\left| \lambda \right| ^{1/2}}d\left| \lambda \right| ds\|f\|
_{C^{2\theta }(E)} \\
&\leqslant &K\int_{\tau }^{t}\int_{\sqrt{\varepsilon_0}s}^{+\infty }\frac{
e^{-\sigma c_0}\tau ^{2\theta }}{s}d\sigma ds\|f\|
_{C^{2\theta }(E)} \\
&\leqslant &K\int_{\tau }^{t}(\int_0^{+\infty }e^{-\sigma
c_0}d\sigma ) s^{2\theta -1}ds\|f\|_{C^{2\theta }(E)} \\
&\leqslant &K(t^{2\theta }-\tau ^{2\theta })\|f\|_{C^{2\theta
}(E)} \\
&\leqslant &K(t-\tau )^{2\theta }\|f\|_{C^{2\theta }(E)};
\end{eqnarray*}
for $b_{3}$, we have 
\[
b_{3}=\frac{1}{2\pi i}\int_{\tau }^{t}\int_\gamma g_{\sqrt{
\lambda }}(t)e^{-sB}B(B^{2}+\lambda I)^{-1}(f(\tau )-f(0))d\lambda ds, 
\]
and 
\begin{eqnarray*}
\|b_{3}\|_{E} &\leqslant &K\int_{\tau }^{t}\int_{\varepsilon
_0}^{+\infty }\frac{e^{-\left| \lambda \right| ^{1/2}c_0t}\tau ^{2\theta
}}{\left| \lambda \right| ^{1/2}}d\left| \lambda \right| ds\|f\|
_{C^{2\theta }(E)} \\
&\leqslant &K\int_{\tau }^{t}\int_{\varepsilon_0}^{+\infty }\frac{
e^{-\left| \lambda \right| ^{1/2}c_0s}\tau ^{2\theta }}{\left| \lambda
\right| ^{1/2}}d\left| \lambda \right| ds\|f\|_{C^{2\theta }(E)}
\\
&\leqslant &K(t-\tau )^{2\theta }\|f\|_{C^{2\theta }(E)}.
\end{eqnarray*}
The sum $b_{4}+b_{5}$ can be estimated by the same method used in lemma $2$;
it follows that
\[
\|b_{4}+b_{5}\|_{E}\leqslant K(t-\tau )^{2\theta }\|
f(0)\|_{D_{B^{2}}(\theta ;+\infty )}, 
\]
if and only if $f(0)\in D_{B^{2}}(\theta ;+\infty )$. Similarly 
$v_{3}(.)$ belongs to $C^{2\theta }([0,1] ;E)$ if and only if 
$f(0)$ belongs to $D_{L}(\theta ;+\infty )$ and $v_{4}(.)+v_{5}(.)$ belongs
to $C^{2\theta }([0,1] ;E)$ if and only if $f(1)$ belongs to 
$D_{L}(\theta ;+\infty )$.

Finally,
\begin{align*}
&v_{6}(t)-v_{6}(\tau )\\
&= 
-\frac{1}{2\pi i}\int_\gamma \int_{\tau }^{t}G_{\sqrt{
\lambda }}^{+}(t,s)e^{-(t-s)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
-\frac{1}{2\pi i}\int_\gamma \int_0^{\tau }(G_{
\sqrt{-\lambda }}^{+}(t,s)-G_{\sqrt{-\lambda }}^{+}(\tau ,s))
e^{-(t-s)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
-\frac{1}{2\pi i}\int_\gamma \int_0^{\tau }G_{\sqrt{
\lambda }}^{+}(t,s)(e^{-(t-s)B}-e^{-(\tau -s)B})A(L-\lambda
I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
-\frac{1}{2\pi i}\int_\gamma \int_{\tau }^1G_{\sqrt{
\lambda }}^{-}(t,s)(e^{(s-t)B}-e^{(s-\tau )B})A(L-\lambda
I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
+\frac{1}{2\pi i}\int_\gamma \int_{\tau }^{t}G_{\sqrt{
\lambda }}^{-}(t,s)e^{(s-t)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
-\frac{1}{2\pi i}\int_\gamma \int_{\tau }^1(G_{
\sqrt{-\lambda }}^{-}(t,s)-G_{\sqrt{-\lambda }}^{-}(\tau ,s))
e^{(s-t)B}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&
+\frac{e^{-\tau B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(\tau )A(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}(f(t)-f(\tau ))d\lambda \\
&
+\frac{e^{(1-\tau )B}}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}(1-\tau )A(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}(f(t)-f(\tau
))d\lambda \\
&
-(f(t)-f(\tau ))\\
&=\sum_{i=1}^8 c_{i}-(f(t)-f(\tau )).
\end{align*} 
Using the fact that in the first integral  
(since $0\leqslant \tau<s<t\leqslant 1)
$\[
\left| G_{\sqrt{-\lambda }}^{+}(t,s)\right| \leqslant K\frac{e^{-\left|
\lambda \right| ^{1/2}c_0(t-s)}}{\left| \lambda \right| ^{1/2}}~, 
\]
we have 
\begin{eqnarray*}
\|c_{1}\|_{E} &\leqslant &K\int_{\tau }^{t}\int_{\varepsilon
_0}^{+\infty }\frac{e^{-\left| \lambda \right| ^{1/2}c_0(t-s)}}{\left|
\lambda \right| ^{1/2}}(t-s)^{2\theta }d\left| \lambda \right| ds\|
f\|_{C^{2\theta }(E)} \\
&\leqslant &K\int_{\tau }^{t}(t-s)^{2\theta -1}(\int_{\sqrt{
\varepsilon_0}(t-s)}^{+\infty }e^{-\sigma c_0}d\sigma ) ds\|
f\|_{C^{2\theta }(E)}\quad \\
&\leqslant &K(t-\tau )^{2\theta }\|f\|_{C^{2\theta }(E)}.
\end{eqnarray*}
Writing $c_{2}$ as 
\begin{eqnarray*}
c_{2}&=&-\frac{1}{2\pi i}\int_0^{\tau }\int\nolimits_{\gamma }\int_{\tau
}^{t}\frac{\cosh \sqrt{-\lambda }(1-\xi )\sinh \sqrt{-\lambda }s}{\sinh 
\sqrt{-\lambda }} \\
&&\times e^{-(\tau -s)B}A(L-\lambda I)^{-1}(f(s)-f(t))d\xi d\lambda
ds, 
\end{eqnarray*}
we obtain 
\begin{eqnarray*}
\|c_{2}\|_{E} &\leqslant &K\int_0^{\tau }\int_{\varepsilon
_0}^{+\infty }\int_{\tau }^{t}e^{-\left| \lambda \right| ^{1/2}c_0(\xi
-s)}(t-s) ^{2\theta }d\xi d\left| \lambda \right| ds\|
f\|_{C^{2\theta }(E)} \\
&\leqslant &K\int_0^{\tau }\int_{\tau -s}^{t-s}\int_{\varepsilon
_0}^{+\infty }e^{-\left| \lambda \right| ^{1/2}c_0\eta }(
t-s) ^{2\theta }d\left| \lambda \right| d\eta ds\|f\|
_{C^{2\theta }(E)}\quad \\
&\leqslant &K\int_0^{\tau }\int_{\tau -s}^{t-s}\frac{(t-s)
^{2\theta }}{\eta ^{2}}(\int_{\sqrt{\varepsilon_0}\eta }^{+\infty
}\sigma e^{-\sigma c_0}d\sigma ) d\eta ds\|f\|
_{C^{2\theta }(E)}\quad \\
&\leqslant &K\int_0^{\tau }(t-s) ^{2\theta }(\int_{\tau
-s}^{t-s}\frac{d\eta }{\eta ^{2}}) ds\|f\|_{C^{2\theta
}(E)} \\
&\leqslant &K\int_0^{\tau }(t-s) ^{2\theta }(\frac{1}{\tau -s}-
\frac{1}{t-s})ds\|f\|_{C^{2\theta }(E)} \\
&\leqslant &K(t-\tau )\int_0^{\tau }\frac{\left| t-s\right| ^{2\theta -1}}{
\tau -s}ds\|f\|_{C^{2\theta }(E)} \\
&\leqslant &K(t-\tau )^{2\theta }(\int_0^{+\infty }\frac{z^{2\theta
-1}}{1+z}dz) \|f\|_{C^{2\theta }(E)} \\
&\leqslant &K(t-\tau )^{2\theta }\|f\|_{C^{2\theta }(E)}.
\end{eqnarray*}
For $c_{3}+c_{4}$, we write 
\[
c_{3}+c_{4}=e^{-\tau B}(e^{-(t-\tau )B}\Phi_{2}-\Phi_{2}) 
\]
where 
\begin{eqnarray*}
\Phi_{2} &=&\frac{-1}{2\pi i}\int_\gamma \int_0^{\tau }G_{
\sqrt{-\lambda }}^{+}(t,s)e^{sB}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&-\frac{1}{2\pi i}\int_\gamma \int_{\tau }^1G_{\sqrt{
\lambda }}^{-}(t,s)e^{sB}A(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \,;
\end{eqnarray*}
using the fact that for $r>0$ and $\lambda \in \gamma 
$\[
L(L-rI)^{-1}A(L-\lambda I)^{-1}=\frac{r}{r-\lambda }A(L-rI)^{-1}-\frac{
\lambda }{r-\lambda }A(L-\lambda I)^{-1}\quad 
\]
we get 
\begin{eqnarray*}
\lefteqn{L(L-rI)^{-1}\Phi_{2}}\\
&=&\frac{1}{2\pi i}\int_\gamma \int_0^{\tau }G_{\sqrt{
\lambda }}^{+}(t,s)e^{sB}\frac{\lambda }{r-\lambda }A(L-\lambda
I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&+\frac{1}{2\pi i}\int_\gamma \int_{\tau }^1G_{\sqrt{
\lambda }}^{-}(t,s)e^{sB}\frac{\lambda }{r-\lambda }A(L-\lambda
I)^{-1}(f(s)-f(t))ds\,d\lambda ,
\end{eqnarray*}
and 
\begin{eqnarray*}
\lefteqn{\|L(L-rI)^{-1}\Phi_{2}\|_{E}} \\
&\leqslant &K\int_{\varepsilon_0}^{+\infty }(
\sup_{0\leqslant t\leqslant 1} 
\int_0^1\left| G_{\sqrt{-\lambda }}(t,s)\right|
\left| s-t\right| ^{2\theta }ds) \frac{\left| \lambda \right| }{\left|
r-\lambda \right| }d\left| \lambda \right| \|f\|_{C^{2\theta
}(E)} \\
&\leqslant &K(\int_{\varepsilon_0}^{+\infty }\frac{d\left| \lambda
\right| }{\left| \lambda \right| ^{\theta }\left| r-\lambda \right| })
\|f\|_{C^{2\theta }(E)} \\
&\leqslant &\frac{K}{r^{\theta }}\|f\|_{C^{2\theta }(E)};
\end{eqnarray*}
therefore, $\Phi_{2}\in D_{L}(\theta ;+\infty )\subset D_{B^{2}}(\theta
;+\infty )=D_{B}(2\theta ;+\infty )$. Thus we have 
\begin{eqnarray*}
\|c_{3}+c_{4}\|_{E} &\leqslant &K(t-\tau )^{2\theta }
\sup_{t-\tau >0} \|\frac{e^{-(t-\tau )B}\Phi_{2}-\Phi_{2}
}{(t-\tau )^{2\theta }}\|_{E} \\
&\leqslant &K(t-\tau )^{2\theta }\|\Phi_{2}\|_{D_{B}(2\theta
;+\infty )}.
\end{eqnarray*}
By reasoning in the same manner for the last integral we obtain 
\[
\|v_{6}(t)-v_{6}(\tau )\|_{E}=O((t-\tau )^{2\theta
}) . 
\]

Statement iv). We have 
\begin{eqnarray*}
u'(t) &=&\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}'(t)(L-\lambda I)^{-1}\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}B\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(t)(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)B(B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(1-t)(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(1-t)B(B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{1}{2\pi i}\int_\gamma \int_0^1G_{\sqrt{-\lambda }
}(t,s)e^{-(t-s)B}B(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&-\frac{1}{2\pi i}\int_\gamma \int_0^1\partial_{t}G_{
\sqrt{-\lambda }}(t,s)e^{-(t-s)B}(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&=&\sum_{i=1}^8 w_{i}(t).
\end{eqnarray*}
The derivative of the last integral in (5) is obtained by differentiating
the kernel and then writing $f(s)=(f(s)-f(t))+f(t)$ and carrying out two
integrations by parts. We verify that all these integrals converge
absolutely. The first integral $w_{1}(t)\in D(B)$. In fact, we write 
\[
\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(t)(L-\lambda I)^{-1}B\varphi d\lambda =\frac{e^{-tB}}{2\pi i}
\int_\gamma g_{\sqrt{-\lambda }}'(t)\frac{
L(L-\lambda I)^{-1}B\varphi }{\lambda }d\lambda , 
\]
then we have 
\begin{eqnarray*}
\lefteqn{\|\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda 
}}'(t)\frac{L(L-\lambda I)^{-1}B\varphi }{\lambda }d\lambda
\|_{E} }\\
&\leqslant &K(\int_{\varepsilon_0}^{+\infty }\frac{e^{-\left|
\lambda \right| ^{1/2}c_0t}}{\left| \lambda \right| ^{1+\theta }}d\left|
\lambda \right| ) \|B\varphi \|_{D_{L}(\theta
+1/2;+\infty )} \\
&\leqslant &K\|B\varphi \|_{D_{L}(\theta +1/2;+\infty )}.
\end{eqnarray*}
Due to Lemma \ref{lm1} and H4, we have $w_{2}(t)\in D(B)$. Moreover,
we can write 
\begin{eqnarray*}
\lefteqn{w_{3}(t)+w_{4}(t) }\\
&=&-\frac{e^{-tB}}{2\pi i}\int_\gamma (\sqrt{-\lambda }
g_{\sqrt{-\lambda }}(t)-g_{\sqrt{-\lambda }}'(t)) (L-\lambda
I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma \sqrt{-\lambda }g_{\sqrt{
\lambda }}(t)(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)B(B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&=&\frac{e^{-tB}}{\pi i}\int_\gamma \frac{\sqrt{-\lambda }e^{-
\sqrt{-\lambda }(2-t)}}{1-e^{-2\sqrt{-\lambda }}}(L-\lambda
I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(B-\sqrt{-\lambda }I) ^{-1}(L-\lambda I)^{-1}f(t)d\lambda ,
\end{eqnarray*}
which implies that $w_{3}(t)+w_{4}(t)\in D(B)$. Similarly, we show
that the sum $w_{5}(t)+w_{6}(t)\in D(B)$. As in the proof of statement 
i), we also obtain that $w_{7}(t)$ and $w_{8}(t)$ belong to $D(B)$.

Statements v) and vi). We have 
\begin{eqnarray*}
Bu'(t) &=&\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{
\lambda }}'(t)(L-\lambda I)^{-1}B\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)(L-\lambda I)^{-1}B^{2}\varphi d\lambda \\
&&-\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(t)B(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{-tB}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(t)B^{2}(B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&&-\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(1-t)B(L-\lambda I)^{-1}(B^{2}+\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{e^{(1-t)B}}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}(1-t)B^{2}(B^{2}+\lambda I)^{-1}(L-\lambda I)^{-1}f(t)d\lambda \\
&&+\frac{1}{2\pi i}\int_\gamma \int_0^1G_{\sqrt{-\lambda }
}(t,s)e^{-(t-s)B}B^{2}(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda \\
&&-\frac{1}{2\pi i}\int_\gamma \int_0^1\partial_{t}G_{
\sqrt{-\lambda }}(t,s)e^{-(t-s)B}B^{2}(L-\lambda I)^{-1}(f(s)-f(t))ds\,d\lambda
\\
&=&\sum_{i=1}^8 x_{i}(t),
\end{eqnarray*}
and for $0\leqslant \tau <t\leqslant 1$ , we get 
\begin{eqnarray*}
x_{1}(t)-x_{1}(\tau ) &=&-\frac{e^{-tB}}{2\pi i}\int_{\tau }^{t}
\int_\gamma g_{\sqrt{-\lambda }}(s)L(L-\lambda I)^{-1}B\varphi d\lambda ds
\\
&&+\frac{(e^{-tB}-e^{-\tau B}) }{2\pi i}
\int_\gamma g_{\sqrt{-\lambda }}'(\tau )\frac{L(L-\lambda I)^{-1}B\varphi 
}{\lambda }d\lambda \\
&=&I_{1}+I_{2},
\end{eqnarray*}
then 
\begin{eqnarray*}
\|I_{1}\|_{E} &\leqslant &K_0\int_{\tau
}^{t}\int_{\varepsilon_0}^{+\infty }\frac{e^{-\left| \lambda \right|
^{1/2}c_0s}}{\left| \lambda \right| ^{\theta +1/2}}d\left| \lambda \right|
ds\|B\varphi \|_{D_{L}(\theta +1/2;+\infty )} \\
&\leqslant &K_0\int_{\tau }^{t}\int_{\sqrt{\varepsilon_0}s}^{+\infty }
\frac{2\sigma e^{-\sigma c_0}}{(\frac{\sigma }{s}) ^{2\theta
+1}s^{2}}d\sigma ds\|B\varphi \|_{D_{L}(\theta +1/2;+\infty )}
\\
&\leqslant &K\int_{\tau }^{t}s^{2\theta -1}(\int_0^{+\infty }\frac{
e^{-\sigma c_0}}{\sigma ^{2\theta }}d\sigma ) ds\|B\varphi
\|_{D_{L}(\theta +1/2;+\infty )} \\
&\leqslant &K(t-\tau )^{2\theta }\|B\varphi \|_{D_{L}(\theta
+1/2;+\infty )}.
\end{eqnarray*}
Writing $I_{2}$ in the form 
\[
I_{2}=e^{-\tau B}(e^{-(t-\tau ) B}\Phi_{3}-\Phi
_{3}) , 
\]
with 
\[
\Phi_{3}=\frac{1}{2\pi i}\int_\gamma g_{\sqrt{-\lambda }
}'(\tau )\frac{L(L-\lambda I)^{-1}B\varphi }{\lambda }d\lambda , 
\]
we obtain 
\[
\|L(L-rI)^{-1}\Phi_{3}\|_{E}\leqslant \frac{K}{r^{\theta }}
\|B\varphi \|_{D_{L}(\theta +1/2;+\infty )}\quad \forall r>0. 
\]
Therefore, $\Phi_{3}\in D_{L}(\theta ;+\infty )\subset D_{B}(2\theta
;+\infty )$. Thus we have 
\begin{eqnarray*}
\|I_{2}\|_{E} &\leqslant &K(t-\tau )^{2\theta }
\sup_{t-\tau >0} \|\frac{e^{-(t-\tau )B}\Phi_{3}-\Phi_{3}}{(t-\tau
)^{2\theta }}\|_{E} \\
&\leqslant &K(t-\tau )^{2\theta }\|\Phi_{3}\|_{D_{B}(2\theta
;+\infty )}.
\end{eqnarray*}
The techniques for the rest of the proof is quite similar to those used in
the precedent statements. \hfill$\Box$

\paragraph{Remark}
In the same manner we have a similar result to Theorem \ref{thm1} when the second
member $f$ has a spatial smoothness, that is 
for all $t\in [0,1]$, $f(t)\in D_{L}(\sigma ,\infty )$ 
and $\sup_{t\in (0,1)} \|f(t)\|_{D_{L}(\sigma ,\infty)}<\infty$,
with $\sigma \in ]0,1[$. We omit to prove this assertion in this work.


\paragraph{Remark}
We can consider other hypotheses instead of (11) involving a comparison of 
$D(B)$ and $D((-L)^{\alpha }) $ for some $\alpha \in ]0,1[$.


\subsection*{Example}
Here we exhibit an example which can give an idea of other possible
applications. Consider $E=L^{p}(\mathbb{R})$ with $1<p<\infty $
and
\begin{gather*}
D(A)=W^{2m,p}(\mathbb{R}), \quad
A\psi =\mathcal{A}(x,\partial_{x})\psi =\sum_{0\leqslant k\leqslant 2m}
 a_{k}\psi ^{(k)}, 
 \\
D(B)=W^{m,p}(\mathbb{R}) , \quad
B\psi =\mathcal{B}(x,\partial_{x})\psi =\sum_{0\leqslant k\leqslant m}
b_{k}\psi ^{(k)},
\end{gather*}
where $(a_{2m}-b_{m}^{2})>0$. Then we know that 
\begin{eqnarray*}
D_{L}(\theta ;+\infty ) &=&(D(L);E)_{1-\theta ,\infty } \\
&=&(W^{2m,p}(\mathbb{R});L^{p}(\mathbb{R}
))_{1-\theta ,\infty } \\
&=&\mathcal{B}_{p,\infty }^{2m\theta }(\mathbb{R}).
\end{eqnarray*}
Where this last space is a Besov space defined, for instance, in Grisvard
\cite{g2}. It is not difficult to see that H1, H2, H3, and H4
are satisfied. Applying Theorem \ref{thm1} we have the following statement.

\begin{theorem} \label{thm2}
 Let $f\in C^{2\theta }([0,1];L^{p}(\mathbb{R}))$ such that for $j=0,1$ 
the mappings $x\mapsto f(j,x)$ belong to 
$\mathcal{B}_{p,\infty }^{2m\theta }(\mathbb{R})$ and assume that 
$\varphi $, $\psi $ belong to $W^{2m,p}(\mathbb{R})$ and 
$\varphi ^{(2m)}$, $\psi ^{(2m)}$ belong to 
$\mathcal{B}_{p,\infty }^{2m\theta }(\mathbb{R})$. Then the problem 
\begin{gather*}
\partial_{t}^{2}u(t,x)+\mathcal{B}(x,\partial_{x})\partial_{t}u(t,x)+
\mathcal{A}(x,\partial_{x})\partial_{x}^{2}u(t,x)=f(t,x) \\ 
u(0,x)=\varphi (x), \quad u(1,x)=\psi (x),
\end{gather*}
has a unique solution $u$ satisfying \begin{enumerate}

\item[i)] $u\in C^{2}([0,1];L^{p}(\mathbb{R}))\cap
C^1([0,1],W^{m,p}(\mathbb{R}))\cap C([0,1],W^{2m,p}(
\mathbb{R}))$,

\item[ii)] $\partial_{x}^{2m}u$, $\partial_{x}^{m}\partial_{t}u$ and $\partial
_{t}^{2}u$ belong to $C^{2\theta }([0,1];L^{p}(\mathbb{R}))$.
\end{enumerate}
\end{theorem}

\paragraph{Acknowledgments.} The authors would like to express our thanks to 
Professor Jerry L. Bona for his valuable remarks and suggestions concerning 
this work.

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\end{thebibliography}

\noindent\textsc{Abdelillah El Haial} \\
Universit\'{e} du Havre, I.U.T, \\
B.P 4006, 76610 Le Havre, France 
\smallskip

\noindent\textsc{Rabah Labbas} \\
Laboratoire de Math\'{e}matiques, Facult\'{e} des Sciences
et Techniques, \\
 Universit\'{e} du Havre, 
B.P 540, 76058 Le Havre Cedex, France\\
e-mail: Labbas@univ-lehavre.fr

\end{document}