\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
2005-Oujda International Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 191--205.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{191}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil Cauchy problems]
{Non-autonomous inhomogeneous boundary Cauchy problems}

\author[M. Filali, B. Karim \hfil EJDE/Conf/14 \hfilneg]
{Mohammed Filali,  Belhadj Karim}  % in alphabetical order

\address{Mohammed Filali \newline
D\'epartement de Math\'ematiques et Informatique \\
Facult\'e des Sciences \\
Universit\'e Mohammed 1er, Oujda, Maroc}
\email{filali@sciences.univ-oujda.ac.ma}

\address{Belhadj Karim \newline
D\'epartement de Math\'ematiques et Informatique \\
Facult\'e des Sciences \\
Universit\'e Mohammed 1er, Oujda, Maroc}
\email[B. Karim]{karim@sciences.univ-oujda.ac.ma}

\date{}
\thanks{Published ??, 2006.}
\subjclass[2000]{34G10, 47D06}
\keywords{Boundary Cauchy problem; evolution families; solution;
\hfill\break\indent well posedness; variation of constants formula}

\begin{abstract}
 In this paper we prove  existence and uniqueness of
 classical solutions for the non-autonomous inhomogeneous
 Cauchy problem
 \begin{gather*}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t), \quad 0 \leq s\leq t\leq T,   \\
    L(t)u(t)=\Phi(t)u(t)+g(t) , \quad  0\leq s\leq t\leq T,  \\
    u(s)=x.
 \end{gather*}
 The solution to this problem is obtained by a variation of
 constants formula.
\end{abstract}
\maketitle

\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

Consider the  boundary Cauchy problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=A(t)u(t) , \quad 0\leq s\leq t\leq T,\\
    L(t)u(t)=\Phi(t)u(t), \quad 0\leq s\leq t\leq T, \\
    u(s)=x.
\end{gathered} \label{e1.1}
\end{equation}
In the autonomous case ($A(t)=A$, $L(t)=L$), the
Cauchy problem \eqref{e1.1} was studied by Greiner \cite{g1}. The author
used the perturbation of domains of infinitesimal generators
to study the homogeneous boundary Cauchy problem. He
has also showed the existence of classical solution of \eqref{e1.1} via
a variation of constants formula. In the non-autonomous case,
Kellerman \cite{k2} and Lan \cite{l1}  showed the existence of an
evolution family $(U(t,s))_{0\leq s\leq t\leq T } $ which provides
classical solutions of homogeneous boundary Cauchy problems.
Filali and Moussi \cite{f1}  showed the existence and
uniqueness of classical solutions to the problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=A(t)u(t),  \quad 0\leq s \leq t \leq T,\\
    L(t)u(t)=\Phi(t)u(t)+g(t),  \quad 0\leq s \leq t \leq T,\\
    u(s)=x.
\end{gathered}\label{e1.2}
\end{equation}
In this paper, we prove existence and uniqueness of
classical solutions to the problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t),  \quad 0\leq s \leq t \leq T,\\
    L(t)u(t)=\Phi(t)u(t)+g(t),  \quad 0\leq s \leq t \leq T,\\
    u(s)=x.
\end{gathered}\label{e1.3}
\end{equation}
 Our technique consists on transforming \eqref{e1.3} into an ordinary
Cauchy problem and giving an equivalence between the two problems.
The solution is explicitly given by a variation of constants
formula.

\section{Evolution Family}

\begin{definition} \label{def2.1} \rm
A family of bounded linear operators $(U(t,s))_{0\leq s\leq t\leq
T}$ on $X$ is an evolution family if \\
(a) $U(t,r)U(r,s)=U(t,s)$ and $U(t,t)=Id $ for all
$0\leq s\leq r\leq t\leq T $; and \\
(b) the mapping $(t,s)\to U(t,s)x $ is continuous on
            $\triangle$, for all $x\in X $ with
$$
\triangle = \{(t,s)\in \mathbb{R}^{2}_{+} :0\leq s\leq t\leq T\}.
$$
\end{definition}

\begin{definition} \label{def2.2} \rm
A family of linear (unbounded) operators $ (A(t))_{0\leq t\leq T} $ on a Banach
space $ X $ is a stable family if there are  constants
$ M \geq 1$, $\omega\in \mathbb{R}$ such that
$]\omega,+\infty[\subset\rho(A(t))$ for all $0\leq t\leq T$
and
$$
\|\prod^{m}_{i=1}R(\lambda,A(t_{i}))\|\leq
M\frac{1}{(\lambda-\omega)^{m}} %\label{e2.1}
$$
 for  $\lambda >\omega $ and any finite sequence
 $0\leq t_{1} \leq t_{2}\leq\dots \leq t_{m}\leq T$.
 \end{definition}

 Let $D,X $ and $Y$ be Banach spaces, $D$ densely and continuously
embedded in $X$. Consider  families of operators $A(t)\in L(D,X)$,
 $L(t)\in L(D,Y)$,  $\Phi(t)\in L(X,Y)$  for  $0\leq t\leq T$.
In this section,
we use the operator matrices method to prove the existence of
classical solutions for the non-autonomous inhomogeneous boundary
Cauchy problem \eqref{e1.3}. We use the following theorem due to
Tanaka \cite{t1}.

\begin{theorem} \label{thm2.3}
Let $(A(t))_{0\leq t\leq T}$ be a stable family of linear operators on
a Banach space $X$ such that
\begin{itemize}
\item[(a)] the domain $D=(D(A(t),\|.\|_{D})$ is a Banach space independent
of $t$,
\item[(b)] the mapping $t\to A(t)x $ is continuously
differentiable in $X$ for every  $ x\in D$.
\end{itemize}
Then there is an evolution family $(U(t,s))_{0\leq s\leq t\leq T}$
  on  $\overline{D}$. Moreover, we have the following properties:
(1) $U(t,s)D(s) \subset D(t)$  for all  $0\leq s\leq t\leq T$, where
$$
D(r)=\{x\in D:A(r)x\in\overline{D}\}, 0\leq r\leq T;
$$
(2) the mapping $ t\to U(t,s)x $ is continuously differentiable in
$X$ on $[s,T]$ and
$$
 \frac{d}{dt}U(t,s)x=A(t)U(t,s)x
 $$
for all $x\in D(s)$ and  $t\in [0,T]$.
\end{theorem}

We will assume that the following hypotheses:
\begin{itemize}
\item[(H1)] The mapping  $t\to A(t)x$ is continuously differentiable
 for all $ x\in D$.

\item[(H2)] The family $(A_{0}(t))_{0\leq t\leq T},A_{0}(t)=A(t)/\ker L(t)$
the restriction  of $A(t)$
 to $\ker L(t)$, is stable, with  $M_{0}$  and $\omega_{0}$ constants
of stability.

\item[(H3)] The operator $L(t)$ is surjective for every $t\in [0,T]$
 and the mapping $ t\to L(t)x $ is continuously
 differentiable for all  $x\in D$.

\item[(H4)] The mapping  $ t \to \Phi(t)x $ is continuously differentiable
for all $x\in X $.

\item[(H5)] There exist constants $\gamma >0$  and $\omega_{1} \in \mathbb{R}$
such that
\begin{equation}
\|L(t)x\|_{Y}\geq \frac{\lambda-\omega_{1}}{\gamma}\|x\|_{X} \label{e2.2}
\end{equation}
 for $x\in \ker (\lambda I-A(t))$, $\omega_{1} < \lambda$ and $ t\in [0,T]$.

\end{itemize}

Note that under the above hypotheses, Lan \cite{l1}  has showed that
$A_{0}(t)$ generates an evolution family $(U(t,s))_{0\leq s\leq t\leq T } $
such that:
\begin{itemize}
\item[(a)] $U(t,r)U(r,s)=U(t,s)$ and  $U(t,t)=Id_{X}$ for  all
$0\leq s\leq t\leq T$;
\item[(b)] $(t,s)\to U(t,s)x $ is continuously differentiable on
$\Delta$  for  all  $x\in  X$  with
$\Delta=\{(t,s)\in \mathbb{R_{+}}^{2}:0\leq s\leq t\leq T\}$;
\item[(c)] there exists constants $M_{0} \geq 1$ and
$\omega_{0}\in \mathbb{R}$
such that
 $\|U(t,s)\|\leq M_{0}e^{\omega_{0}(t-s)}$.
\end{itemize}
The following results with will be used in this article.

 \begin{lemma}[\cite{g1}] \label{lem2.4}
 For $t\in [0,T]$ and $\lambda \in \rho (A_{0}(t))$,
 following properties are satisfied:
\begin{enumerate}
\item $D=D(A_{0}(t))\oplus \ker (\lambda I-A(t))$
\item $L(t)/\ker (\lambda I-A(t))$ is an isomorphism from $
 \ker (\lambda I-A(t))$ onto $ Y$
\item $ t\mapsto L_{\lambda,t}:=(L(t)/\ker (\lambda I-A(t)))^{-1}$
is strongly continuously differentiable.
\end{enumerate}
 \end{lemma}

 As a consequence of this lemma, we have
 $L(t)L_{\lambda,t}=Id_{Y}$, $L_{\lambda,t}L(t)$  and
 $(I-L_{\lambda,t}L(t))$ are the projections from $D$ onto
 $\ker (\lambda I-A(t))$ and $D(A_{0}(t))$ .

  \section{The Homogeneous Problem}

 In this section, we consider the Cauchy problem \eqref{e1.1}.
A function $ u:[s,T]\to X $ is called  classical solution
 if it is continuously differentiable,
$u(t)\in D$ for all  $0 \leq s \leq t \leq T$  and  $u $ satisfies
 \eqref{e1.1}.

We now introduce the Banach spaces $ Z=X\times Y$,
 $Z_{0}=X\times\{0\}\subset Z $
and  we consider  the  projection of $Z $ onto $X$:
$p_1(x,y)= x$.
Let $M(t)$ be the matrix-valued operator defined on $Z$ by
$$
M(t)=\begin{pmatrix}
  A(t) & 0 \\
  -L(t)+\Phi(t)& 0  \end{pmatrix}
=l(t)+\phi(t),
$$
where
$$
l(t)=\begin{pmatrix}
  A(t) & 0 \\
  -L(t)& 0 \end{pmatrix}, \quad
\phi(t) = \begin{pmatrix}
  0 & 0 \\
  \Phi(t) & 0 \end{pmatrix},
$$
and $D(M(t))=D\times \{0\}$.

Now, we consider the Cauchy  problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=M(t)u(t)  , \quad 0\leq s\leq t\leq T, \\
    u(s)=(x,0).
\end{gathered}\label{e3.1}
\end{equation}
 We start by proving the following lemma.

\begin{lemma} \label{lem3.2}
Assume that  hypothesis (H1)--(H5) hold.
Then, the family of operators $(M(t))_{ 0\leq t\leq T }$ is stable.
\end{lemma}

\begin{remark} \label{rmk3.3} \rm
 Since $ L_{\lambda,t}L(t)$ is the projection from $D$ onto
$\ker (\lambda  I-A(t))$ and $ x-L_{\lambda,t}L(t)x\in D(A_{0}(t))$,
we have
\begin{align*}
&R(\lambda,A_{0}(t))((\lambda I-A(t))x)+L_{\lambda,t}L(t)x \\
&= R(\lambda,A_{0}(t))((\lambda I-A(t))(x-L_{\lambda,t}L(t)x)+
L_{\lambda,t}L(t)x
\end{align*}
and
\begin{equation}
R(\lambda,A_{0}(t))((\lambda I-A(t))x)+L_{\lambda,t}L(t)x=x. \label{e3.2}
\end{equation}
\end{remark}

\begin{proof}[Proof of Lemma \ref{lem3.2}]
 Since $M(t)$ is a perturbation of $l(t)$ by a
linear bounded operator on $E$, hence, in view of the perturbation
result \cite[Theorem 5.2.3]{p1}, it is sufficient to show the stability
of $l(t)$.
For $\lambda >\omega_{0}$ and $\lambda \neq 0$, let
$$
R(\lambda)=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} \\
  0 & 0 \end{pmatrix}\,.
$$
We have $D(l(t))=D\times \{0\}$ and
$$
(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \end{pmatrix}
=\begin{pmatrix}
  (\lambda I-A(t))x \\
  L(t)x \end{pmatrix}
$$
for  $(x \\   0) \in D\times \{0\}$.
By Remark \ref{rmk3.3}, we obtain
 $$ R(\lambda)(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \end{pmatrix}
=\begin{pmatrix}
  R(\lambda,A_{0}(t))((\lambda I-A(t))x)+L_{\lambda,t}L(t)x \\
  0 \end{pmatrix}.
$$
So  that
\begin{equation}
R(\lambda)(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \end{pmatrix}
=\begin{pmatrix}
  x \\
  0 \end{pmatrix}.\label{e3.3}
\end{equation}
On the other hand, for $( x , y) \in X\times Y$,
 we have
\begin{equation}
\begin{aligned}
(\lambda I-l(t)) R(\lambda)\begin{pmatrix}
  x \\
  y\end{pmatrix}
=\begin{pmatrix}
  \lambda I-A(t) & 0 \\
  L(t) & \lambda \end{pmatrix}
\begin{pmatrix}
  R(\lambda,A_{0}(t))x+L_{\lambda,t}y \\
  0 \end{pmatrix}
=\begin{pmatrix}
  x \\
  y \end{pmatrix}
\end{aligned} \label{e3.4}
\end{equation}
from \eqref{e3.3} and \eqref{e3.4}, we obtain that the resolvent
 of $l(t)$ is given by
\begin{equation}
R(\lambda,l(t))=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} \\
  0 & 0 \end{pmatrix}.
\label{e3.5}
\end{equation}
By a direct computation, we  obtain
$$
\prod_{i=1}^m R(\lambda,l(t_{i}))=\begin{pmatrix}
  \prod_{i=1}^m R(\lambda,A_{0}(t_{i}))
& \prod_{i=1}^{m-1} R(\lambda,A_{0}(t_{i})L_{\lambda,t_{m}} \\
  0 & 0 \end{pmatrix}
$$
for a finite sequence  $0\leq t_{1} \leq t_{2}\leq\dots \leq t_{m}\leq T$
and we have
 $$
\prod_{i=1}^m R(\lambda,l(t_{i}))\begin{pmatrix}
  x \\
  y \end{pmatrix}
=\begin{pmatrix}
  \prod_{i=1}^m R(\lambda,A_{0}(t))x+\prod_{i=1}^{m-1}
R(\lambda,A_{0}(t))L_{\lambda,t_{m}}y \\
  0 \end{pmatrix}.
$$
 From  hypothesis (H5), we conclude that
$\|L_{\lambda,t}\|\leq \frac{\gamma}{(\lambda-\omega)}$ for all
$t\in [0,T]$ and $\lambda>\omega$ and by using (H2), we obtain
\begin{equation}
\begin{aligned}
\|\prod_{i=1}^m R(\lambda,l(t_{i}))\begin{pmatrix}
  x \\
  y \end{pmatrix}\|
& \leq \|\prod_{i=1}^m R(\lambda,A_{0}(t))x\|
+\|\prod_{i=1}^{m-1} R(\lambda,A_{0}(t))L_{\lambda,t_{m}}y\| \\
& \leq \frac{M}{(\lambda-\omega_{0})^{m}}\|x\|
  +\frac{\gamma M}{(\lambda-\omega_{0})^{m-1}}\frac{1}
          {\lambda-\omega_{1}}\|y\|.
\end{aligned} \label{e3.6}
\end{equation}
 For $ \omega_{2}=\max(\omega_{0},\omega_{1})$, we have
$$
\|\prod_{i=1}^m R(\lambda,l(t_{i})\begin{pmatrix}
  x \\
  y \end{pmatrix}\|
\leq \frac{M'}{(\lambda-\omega_{2})^{m}}(\|x\|+\|y\|),
$$
where $M'=\max(M,M \gamma)$.  On $ E = X\times Y $ equipped with the
norm $\|(x,y)\|_{1}=\|x\|+\|y\|$, we have:
$$
\|\prod_{i=1}^m R(\lambda,l(t_{i})\begin{pmatrix}
  x \\
  y\end{pmatrix}\|
\leq \frac{M'}{(\lambda-\omega_{2})^{m}}(\|(x,y)\|_{1}).
$$
\end{proof}

 In the following proposition we give the equivalence between the
boundary problem \eqref{e1.1} and the Cauchy problem \eqref{e3.1}.

\begin{proposition} \label{prop3.4}
Let $( x , 0 )\in D\times \{0\}$.
\begin{enumerate}
\item If the function $ t\to U(t)=( u_{1}(t),  0 )$ is a classical
 solution of \eqref{e3.1} with an
initial value $(x,0)$ then $ t\to u_{1}(t)$ is
a classical solution of \eqref{e1.1} with the initial value $x$.

\item Let $ u $ be a classical solution of  \eqref{e1.1} with the
initial value $x$. Then the function
$ t \to U(t)=(u(t),  0 )$ is a classical solution of \eqref{e3.1} with the
initial value $(x,  0)$.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) Since $ U(t)=(  u_{1}(t),  0 )$ is a classical solution of \eqref{e3.1},
 $ u_{1} $ is continuously  differentiable on $ [s,T] $ and $u_{1}(t)\in D $.
Moreover,
\begin{equation}
\frac{d}{dt}U(t)=\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \end{pmatrix}
=M(t)U(t) \quad\text{and}\quad
 U(s) =\begin{pmatrix}
  x \\
  0 \end{pmatrix}. \label{e3.7}
\end{equation}
Therefore,
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u_{1}(t)=A(t)u_{1}(t), \quad 0\leq s\leq t\leq T,\\
    L(t)u_{1}(t)=\Phi(t)u_{1}(t),\quad 0\leq s\leq t\leq T, \\
    u_{1}(s)=x.
\end{gathered} \label{e3.8}
\end{equation}
This implies that  $ u_{1} $ is a classical solution of \eqref{e1.1}.

\noindent(2) Let $ u $ is a classical solution of \eqref{e1.1}, then $ u $
is continuously differentiable, $ u(t)\in D $ for $ t\geq s $ and
\begin{gather*}
    \frac{d}{dt}u(t)=A(t)u(t),\quad 0\leq s\leq t\leq T,\\
    L(t)u(t)= \Phi(t)u(t),\quad 0\leq s\leq t\leq T,\\
    u(s)=x.
\end{gather*}
Hence
$$
\begin{pmatrix}
  \frac{d}{dt}u(t) \\
  0 \end{pmatrix}
=\begin{pmatrix}
  A(t) & 0\\
  -L(t)+\Phi(t) & 0
\end{pmatrix}
\begin{pmatrix}
  u(t) \\
  0 \end{pmatrix},
$$
with  $( u(s),   0)=(  x,   0 )$.
This implies that $ U(t)=(  u(t),  0 )$ is a classical solution
of \eqref{e3.2} with the initial
value $( x,  0)$.
\end{proof}

The above proposition allows us to get the aim of this section by
showing the well-posedness of the Cauchy problem \eqref{e1.1}.

\begin{theorem} \label{thm3.5}
Assume that the hypotheses (H1)--(H5) hold. Then for
every $ x\in D $, such that $ -L(s)x+\Phi(s)x=0 $, the problem
\eqref{e1.1} has a unique classical solution. Moreover, $u$ is
given by $ t\to p_{1}(U(t,s)\begin{pmatrix}
  x \\
  0\end{pmatrix}$,
where $ U(t,s)$ is the evolution family generated by
$(M(t)_{ 0\leq t\leq T})$.
\end{theorem}

\begin{proof} For the Cauchy problem \eqref{e3.1}, we have the following:
\begin{enumerate}
\item $D(M(t))=D\times \{0\} $ is independent of $t$.
\item $t \to M(t)\begin{pmatrix}
  x\\
  0 \end{pmatrix}$ is  continuously differentiable for
$( x,  0 ) \in D\times \{0\}$.

\item The family $ (M(t))_{0\leq t\leq T } $ is stable.
\end{enumerate}
Then the family $M(t)$ satisfies all conditions of Theorem \ref{thm2.3}.
Thus, there exist an evolution family $(U(t,s))_{0\leq s\leq t}$
 generated by the family $(M(t))_{0\leq t\leq T }$ such that
\begin{itemize}
\item[(a)] $U(t,t)=Id_{X\times \{0\}}$,
\item[(b)] $U(t,r)U(r,s)=U(t,s)$,    $0\leq s\leq r\leq t\leq T$,
\item[(c)] $(t,s)\to U(t,s)$ is strongly continuous,
\item[(d)] the function $t\to U(t,s)\begin{pmatrix}
  x \\
  0 \end{pmatrix}$ is continuously differentiable in $ X\times \{0\}$
on $[s,T]$, and satisfies
$$
\frac{d}{dt}U(t,s)\begin{pmatrix}
  x \\
  0 \end{pmatrix}
=M(t)U(t,s) \begin{pmatrix}
  x \\
  0 \end{pmatrix}
\quad\text{for}\quad
\begin{pmatrix}
  x \\
  0 \end{pmatrix}  \in D(s),
$$
and
\begin{equation}
 U(t,s)D(s)\subset D(t), \quad\text{for  all }
0\leq s\leq t\leq T ,\label{e3.9}
\end{equation}
where
\begin{equation}
\begin{aligned}
D(s)&=\{\begin{pmatrix}
  x \\
  0 \end{pmatrix} \in D\times\{0\}: M(s)\begin{pmatrix}
  x \\
  0 \end{pmatrix} \in X\times \{0\}\} \\
&=\ker \big(L(s)-\Phi(s)\big)\times \{0\}. \label{e3.10}
\end{aligned}
\end{equation}
\end{itemize}
Let $U(t,s)(  x ,  0 )
=(  u_{1}(t),  0 )$. We have
$$
\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \\
\end{pmatrix}
= M(t) \begin{pmatrix}
  u_{1}(t) \\
  0 \\
\end{pmatrix},
$$
and for $u(t)=(  u_{1}(t),  0 )$, we have $\frac{d}{dt}u(t)=M(t)u(t)$, with
$u(s)=(  x,  0 )$, thus $ u(t)=(  u_{1}(t),  0 )$ is a classical solution
of \eqref{e3.1} and
from Proposition  \ref{prop3.4}, we have $ u_{1}$ is a classical
solution of \eqref{e1.1} and
\begin{equation}
u_{1}(t)=  p_{1}\big(U(t,s)\begin{pmatrix}
  x \\
  0\end{pmatrix} \big).\label{e3.11}
\end{equation}
\end{proof}

\section{First Inhomogeneous Problem}

In this section, we consider the inhomogeneous Cauchy problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t) , \quad 0\leq s\leq t\leq T,\\
    L(t)u(t)=\Phi(t)u(t)           , \quad 0\leq s\leq t\leq T,\\
    u(s)=x.
\end{gathered}\label{e4.1}
\end{equation}
A function $u:[s,T]\to X$ is called classical solution
if it is continuously differentiable,
$u(t)\in D$, $t\geq s$ and $u$ satisfies \eqref{e4.1}.


Consider the Banach space $E=X \times Y \times C^{1}([0,T],X),
T>0$, where $C^{1}([0,T],X)$ is the space of continuously
differentiable functions from $[0,T]$ into $ X $ equipped with the norm
$\|f\|=\|f\|_{\infty}+\|f'\|_{\infty} $, for
$ f\in C^{1}([0,T],X)$.
Let $B(t)$ be the operator matrices defined on $E$ by
 \begin{equation}
B(t)=\begin{pmatrix}
  A(t) & 0 & \delta_{t} \\
  -L(t)+\Phi(t) & 0 & 0 \\
  0 & 0 & 0 \end{pmatrix}\label{e4.2}
\end{equation}
 with
 $D(B(t))=D\times \{0\}\times C^{1}([0,T],X)$.
Where $\delta_{t}:C^{1}([0,T],X)\to X$
is the Dirac function concentrated at the point $t$ with
$\delta_{t}(f)=f(t)$.
 To the family $B(t)$ we associate the
homogeneous Cauchy problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=B(t)u(t) ,  \quad 0\leq s\leq t\leq T , \\
    u(s)=(  x,  0,  f ).
\end{gathered} \label{e4.3}
\end{equation}
with $(  x,   0,  f) \in D\times \{0\}\times C^{1}([0,T]$.

\begin{lemma} \label{lem4.2}
Assume that  hypothesis (H1)--(H5)  hold. Then the
family  operators $(B(t))_{0 \leq t \leq T}$ is stable.
\end{lemma}

\begin{proof}
For $t\in[0,T]$, we write the operator $B(t)$ as
$B(t)=l(t)+\phi(t)$, with
$$
l(t)=\begin{pmatrix}
  A(t) & 0 & 0 \\
  -L(t) & 0 & 0 \\
  0 & 0 & 0 \end{pmatrix}
\quad\text{and}\quad  \phi(t)=\begin{pmatrix}
  0 & 0 & \delta_{t} \\
  \Phi(t) & 0 & 0 \\
  0 & 0 & 0 \end{pmatrix}.
$$
 We must show that $l(t)$ is stable
and that
\begin{equation}
R(\lambda,l(t))=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} & 0 \\
  0 & 0& 0 \\
  0 & 0 & 1/\lambda
\end{pmatrix}.\label{e4.4}
\end{equation}
For $ \lambda >\omega_{0}$, $\lambda\neq 0$,  and
  $t\in[0,T]$, let
$$
R(\lambda)=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} & 0 \\
  0 & 0& 0 \\
  0 & 0 & 1/\lambda
\end{pmatrix}.
$$
For $(  x,  y,  f ) \in X\times Y\times C^{1}([0,T],X)$, we have
$$
\begin{pmatrix}
  R(\lambda,A_{0}(t) & L_{\lambda,t} & 0 \\
  0 & 0 & 0 \\
  0 & 0 & 1/\lambda \end{pmatrix}
\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}
=\begin{pmatrix}
  R(\lambda,A_{0}(t))x+L_{\lambda,t}y \\
  0 \\
  \frac{f}{\lambda} \\
\end{pmatrix},
$$
by the Remark \ref{rmk3.3}, we obtain
\begin{equation}
(\lambda I-l(t))R(\lambda)
\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}
=\begin{pmatrix}
  (\lambda I-A(t))[R(\lambda,A_{0}(t))x+L_{\lambda,t}y] \\
  L(t)[R(\lambda,A_{0}(t))x+L_{\lambda,t}y] \\
  f \end{pmatrix}
=\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}.\label{e4.5}
\end{equation}
On the other hand, for  $(  x,  0,  f )\in D\times \{0\}\times C^{1}([0,T],X)$,
 we have
$$
(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}
=\begin{pmatrix}
  (\lambda I-A(t))x \\
  L(t)x \\
  \lambda f
\end{pmatrix},
$$
and
$$
R(\lambda)(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}
=\begin{pmatrix}
  R(\lambda,A_{0}(t))((\lambda I-A(t))x)+L_{\lambda,t}L(t)x \\
  0 \\
  f \end{pmatrix}.
$$
 From  Remark \ref{rmk3.3}, we have
 \begin{equation}
R(\lambda)(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}
=\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}.\label{e4.6}
\end{equation}
 From \eqref{e4.5} and \eqref{e4.6}, we obtain that
the resolvent of $l(t)$ is given by
  $$
R(\lambda,l(t))=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} & 0 \\
  0 & 0 & 0 \\
  0 & 0 & 1/\lambda
\end{pmatrix}.
$$
By recurrence we can obtain
$$
\prod_{i=1}^m R(\lambda,l(t_{i}))=
\begin{pmatrix}
  \prod_{i=1}^m R(\lambda,A_{0}(t_{i})) & \prod_{i=1}^{m-1} R(\lambda,A_{0}(t_{i}))L_{\lambda,t_{m}} & 0 \\
  0 & 0 & 0 \\
  0 & 0 & 1/\lambda^m
\end{pmatrix}.
$$
 For a finite sequence $0\leq t_{1}\leq t_{2}\dots \leq t_{m}\leq T$ and
for $(  x , y , f ) \in E$, we have
$$
\prod_{i=1}^m R(\lambda,l(t_{i}))
\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}
=\begin{pmatrix}
  \prod_{i=1}^m R(\lambda,A_{0}(t_{i}))x+\prod_{i=1}^{m-1}
R(\lambda,A_{0}(t_{i}))L_{\lambda,t_{m}}y \\
  0 \\
  f/\lambda^m \end{pmatrix}.
$$
Using (H5), we obtain
\begin{align*}
\|\prod_{i=1}^m R(\lambda,l(t_{i}))
\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}\|
&\leq \|\prod_{i=1}^m R(\lambda,A_{0}(t_{i}))x+\prod_{i=1}^{m-1}
R(\lambda,A_{0}(t_{i})) L_{\lambda,t_{m}}y\|
+\frac{\|f\|}{\lambda^{m}}\\
&\leq \frac{M}{(\lambda-\omega_{0})^{m}}\|x\|
+\frac{M}{(\lambda-\omega_{0})^{m-1}}\frac{\gamma
}{\lambda-\omega_{1}}\|y\|+\frac{\|f\|}{\lambda^{m}}.
\end{align*}
Define
$\omega_{2}=\max(0,\omega_{0},\omega_{1})$. Then
$$
\|\prod_{i=1}^m R(\lambda,l(t_{i}))
\begin{pmatrix}
  x \\
  y \\
  f \end{pmatrix}\|
\leq \frac{M'}{(\lambda-\omega_{2})^{m}}(\|x\|+\|y\|+\|f\|),
$$
where
 $M'=\max(M,M \gamma)$ and
\begin{equation}
\|\prod_{i=1}^m R(\lambda,l(t_{i}))\|\leq
 \frac{M'}{(\lambda-\omega_{2})^{m}}.\label{e4.7}
\end{equation}
This inequality shows that the family $l(t)$ is stable and by using
\cite[Theorem 5.2.3]{p1},  the family $B(t)$ is stable.
\end{proof}

\begin{proposition} \label{prop4.3}
Let $(  x,  0,  f) \in D\times \{0\}\times C^{1}([0,T],X)$.

\noindent(1) If the function $ t\to u(t)=(  u_{1}(t),  0,
  u_{2}(t))$ is a classical solution of \eqref{e4.3} with an initial
value $(  x,  0,  f )$ then $t\to u_{1}(t) $ is a classical solution of
\eqref{e4.1} with the initial value $x$.

\noindent(2) Let $u$ is a classical solution of  \eqref{e4.1} with the
initial value $x$ .Then, the  function
$t\to U(t)=(  u(t),  0,  f)$
is a classical solution of \eqref{e4.3} with the initial
value $(  x,  0,  f )$.
\end{proposition}

\begin{proof} (1) If $u(t)=(  u_{1}(t),  0,  u_{2}(t))$ is a classical
solution of  \eqref{e4.3}, then $u_{1}$ is
continuously differentiable  on $[s,T]$, $u_{1} \in D$ and we have
$$
\frac{d}{dt} u(t)=\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \\
  \frac{d}{dt}u_{2}(t)
\end{pmatrix}
=B(t)u(t),
$$
which implies
$$
\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \\
  \frac{d}{dt}u_{2}(t)
\end{pmatrix}
= \begin{pmatrix}
  A(t) & 0 & \delta_{t} \\
  -L(t)+\Phi(t) & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  u_{1}(t) \\
  0 \\
  u_{2}(t)
\end{pmatrix},
$$
and
$$
\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \\
  \frac{d}{dt}u_{2}(t)
\end{pmatrix}
=\begin{pmatrix}
  A(t)u_{1}(t)+\delta_{t}u_{2}(t) \\
  -L(t)u_{1}(t)+\Phi(t)u_{1}(t) \\
  0 \end{pmatrix},
$$
with
$$
u(s)=\begin{pmatrix}
  u_{1}(s) \\
  0 \\
  u_{2}(s) \end{pmatrix}
=\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}.
$$
One has $\frac{d}{dt}u_{2}(t)=0 $. This implies
$u_{2}(t)=u_{2}(s)=f$; therefore,
$\delta_{t}u_{2}(t)=\delta_{t}f=f(t) $ and we have
\begin{gather*}
    \frac{d}{dt}u_{1}(t)=A(t)u_{1}(t)+f(t),  \quad 0\leq s\leq t\leq
T , \\
    L(t)u_{1}(t)=\Phi(t)u_{1}(t), \quad 0\leq s\leq t\leq
T , \\
    u_{1}(s)=x.
\end{gather*}
Therefore, $u_{1} $ is a classical solution of
\eqref{e4.1} with the initial value $ x $.

\noindent(2) If $u$ is a classical solution of  \eqref{e4.1}, then $ u $ is
continuously differentiable, $ u(t)\in D $
and
\begin{gather*}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t),  \quad 0\leq s\leq t\leq T , \\
    L(t)u(t)=\Phi(t)u(t), \quad 0\leq s\leq t\leq T , \\
    u(s)=x.
\end{gather*}
Moreover,
$$
\begin{pmatrix}
  \frac{d}{dt}u(t) \\
  0 \\
  0 \end{pmatrix}
=\begin{pmatrix}
  A(t) & 0 & \delta_{t} \\
  -L(t)+\Phi(t) & 0 & 0 \\
  0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  u(t) \\
  0 \\
  f \end{pmatrix}.
$$
With $u(s)=x$,
 $ U(t)=(  u(t),  0,  f)$ is continuously differentiable,
$U(t) \in D(B(t))=D\times \{0\}\times C^{1}([0,T],X) $ then it
is a classical solution of
\eqref{e4.3} with the initial value
$( x,  0,  f )$.
\end{proof}

\begin{theorem} \label{thm4.4}
Let $ f\in C^{1}([0,T],X)$. Assume that the hypothesis (H1)--(H5) hold.
Then for all $x\in D$, such that
$-L(s)x+\Phi(s)x=0 $,  problem \eqref{e4.1} has a unique classical
solution solution $u$. Moreover, $u$ is given by
\begin{equation}
u(t)=U_{\Phi}(t,s)x+\int_{s}^{t}U_{\Phi}(t,s)f(r)dr,\label{e4.8}
\end{equation}
where
$U_{\Phi}(t,s)$ is an evolution family solution of the problem
\eqref{e3.1}
\end{theorem}

\begin{proof}  Consider the problem
\begin{gather*}
    \frac{d}{dt}u(t)=B(t)u(t), \quad 0\leq s\leq t\leq T , \\
    u(s)=(  x, 0, f ).
\end{gather*}
We have showed that  $(B(t))_{0\leq t\leq T} $ is a
stable family and the function $ t\to B(t)y $ is
continuously differentiable, for all
$ y\in D(B(t))=D\times \{0\}\times C^{1}([0,T],X)$ and that $D(B(t))$
is independent of $t$. Then there exist an evolution system $U(t,s)$
on $ X\times \{0\}\times C^{1}([0,T],X) $ such that
$$
U(t,s)\begin{pmatrix}
  x \\
  0 \\
  f \end{pmatrix}
=\begin{pmatrix}
  u_{1}(t) \\
  0 \\
  u_{2}(t)
\end{pmatrix} = u(t)
$$
is a classical solution of  \eqref{e4.3} and from the
Proposition \ref{prop4.3}, $ u_{1}$ is a classical solution of
\eqref{e4.1}, for
 $(  x,  0,  f) \in \ker (L(s)-\Phi(s))\times\{0\}\times C^{1}([0,T],X)$.
Let $v(r)= U_{\Phi}(t,r)u_{1}(r)$. Then $v$ is differentiable and
$$
\frac{d}{dr}v(r)= -U_{\Phi}(t,r)A_{\Phi}(r)u_{1}(r)+U_{\Phi}(t,r)
[A_{\Phi}(r)u_{1}(r)+f(r)],
$$
where $A_{\Phi}(t)=A(t)/\ker (L(t)-\Phi(t))$; therefore,
\begin{equation}
\frac{d}{dr}v(r)=U_{\Phi}(t,r)f(r). \label{e4.9}
\end{equation}
Integrating \eqref{e4.9} from $s$ to $ t$,  we obtain
$$
u_{1}(t)=U_{\Phi}(t,s)x+\int_{s}^{t}U_{\Phi}(t,r)f(r)dr,
$$
which completes the proof.
\end{proof}

\section{Second Inhomogeneous Problem}

In this section, we consider the Inhomogeneous Cauchy problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t) , \quad 0\leq s\leq t\leq T, \\
    L(t)u(t)=\Phi(t)u(t)+g(t)      , \quad 0\leq s\leq t\leq T,  \\
    u(s)=x.
\end{gathered} \label{e5.1}
\end{equation}
A function $u:[s,T]\to X $ is a classical solution
if it is continuously differentiable, $ u(t)\in D,
$ for all $ t\geq s $ and $u$ satisfies \eqref{e5.1}.

Consider the Banach space
$ E=X\times Y\times C^{1}([0,T],X)\times C^{1}([0,T],Y) $,
where $ C^{1}([0,T],X)$ and $ C^{1}([0,T],Y)$ are equipped with
the norm
$\|f\|=\|f\|_{\infty}+\|f'\|_{\infty}$
for $f$ in $C^{1}([0,T],X)$  or in $C^{1}([0,T],Y)$.
Consider the operator matrices
\begin{equation}
B(t)=\begin{pmatrix}
  A(t) & 0 & \delta_{t} & 0 \\
  -L(t)+\Phi(t) & 0 & 0 & \overline{\delta_{t}} \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 \end{pmatrix},\label{e5.2}
\end{equation}
 with
$$
D(B(t))=D\times\{0\}\times C^{1}([0,T],X)\times
C^{1}([0,T],Y)
$$
where $\delta_{t}: C^{1}([0,T],X)\to X $
such that $\delta_{t}(f)=f(t)$  and
$\overline{\delta_{t}}:C^{1}([0,T],Y)\to Y $ such that
$\overline{\delta_{t}}(g)=g(t)$.
To the family $B(t)$ ,we associate the homogeneous Cauchy problem
\begin{equation}
\begin{gathered}
    \frac{d}{dt}u(t)=B(t)u(t), \quad 0\leq s\leq t\leq T, \\
    u(s)=(  x,  0, f,  g)
\end{gathered} \label{e5.3}
\end{equation}
for $ (  x , 0 , f, g) \in D\times \{0\}\times C^{1}([0,T],X)\times
C^{1}([0,T],Y)=D_{1}$.

\begin{lemma} \label{lem5.2}
Assume that the hypothesis (H1)--(H5) hold. Then the
family operators $B(t)$ is stable.
\end{lemma}

 \begin{proof} For $t \in [0,T]$, we write the $B(t)$ defined in \eqref{e5.2}
as $B(t)=l(t)+\phi(t)$, where
$$
l(t)= \begin{pmatrix}
  A(t) & 0 & 0 & 0 \\
  -L(t) & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
\end{pmatrix} \quad \text{and}\quad
\phi(t)=\begin{pmatrix}
  0 & 0 & \delta_{t} & 0 \\
  \Phi(t) & 0 & 0 & \overline{\delta_{t}} \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
\end{pmatrix},
$$
we must show that the family $l(t)$ is stable.
Let
 $$
R(\lambda)=\begin{pmatrix}
  R(\lambda,A_{0}(t)) & L_{\lambda,t} & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 1/\lambda & 0 \\
  0 & 0 & 0 & 1/\lambda \end{pmatrix}.
$$
For $\lambda >\omega_{0}$, $\lambda \neq 0$ and $ t\in[0,T] $ we show
that $R(\lambda,l(t))=R(\lambda)$.
For $(  x,  y, f, g )\in X\times Y\times C^{1}([0,T],X)\times C^{1}([0,T],Y)$,
we have
\begin{equation}
R(\lambda)\begin{pmatrix}
  x \\
  y \\
  f \\
  g \end{pmatrix}
=\begin{pmatrix}
  R(\lambda,A_{0}(t))x+L_{\lambda,t}y \\
  0 \\
  f/\lambda \\
  g/\lambda \\
\end{pmatrix},\label{e5.4}
\end{equation}
 by the Remark \ref{rmk3.3} and with the same proof
as Lemma \ref{lem4.2} we obtain
\begin{equation}
(\lambda I-l(t))R(\lambda)
\begin{pmatrix}
  x \\
  y \\
  f \\
  g \end{pmatrix}
=\begin{pmatrix}
  x \\
  y \\
  f \\
  g \end{pmatrix}.\label{e5.5}
\end{equation}
On the other hand, for
 $(  x,  0,  f, g) \in D\times \{0\}\times C^{1}([0,T],X)\times
C^{1}([0,T],Y)$, we have
$$
(\lambda I-l(t))\begin{pmatrix}
  x \\
  0 \\
  f \\
  g \end{pmatrix}
=\begin{pmatrix}
  (\lambda I-A(t))x \\
  L(t)x \\
  \lambda f \\
  \lambda g \end{pmatrix},
$$
and
\begin{equation}
R(\lambda)(\lambda I-l(t))
\begin{pmatrix}
  x \\
  0 \\
  f \\
  g \end{pmatrix}
=\begin{pmatrix}
  R(\lambda,A_{0}(t))((\lambda I-A(t))x)+L_{\lambda,t}L(t)x \\
  0 \\
  f \\
  g
\end{pmatrix}
=\begin{pmatrix}
  x \\
  0 \\
  f \\
  g \end{pmatrix},\label{e5.6}
\end{equation}
then from \eqref{e5.5}, \eqref{e5.6} and  Remark \ref{rmk3.3}, we have
$R(\lambda)=R(\lambda I,l(t))$.
By recurrence we obtain
$$
\prod_{i=1}^m R(\lambda,l(t_{i}))=
\begin{pmatrix}
  \prod_{i=1}^m R(\lambda,A_{0}(t_{i})) & \prod_{i=1}^{m-1}
R(\lambda,A_{0}(t_{i}))L_{\lambda,t_{m}} & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 1/\lambda^m  & 0 \\
  0 & 0 & 0 & 1/\lambda^m
\end{pmatrix},
$$
for a finite sequence  $0\leq t_{1} \leq t_{2} \leq \dots \leq t_{m}\leq T$.

Now on the space $X\times Y\times C^{1}([0,T],X)\times
C^{1}([0,T],Y)$,we consider the norm
\begin{equation}
\|( x,  y,  f,  g )\|=(\|x\|+\|y\|+\|f\|+\|g\|). \label{e5.7}
\end{equation}
For $( x, y,  f, g )\in X\times Y\times C^{1}([0,T],X)\times
     C^{1}([0,T],Y)$,
we have
\begin{align*}
\|\prod_{i=1}^m R(\lambda,l(t_{i}))
\begin{pmatrix}
  x \\
  y \\
  f \\
  g
\end{pmatrix}\|
&\leq \frac{M}{(\lambda-\omega_{0})^{m}}\|x\|
+\frac{M\gamma}{(\lambda-\omega_{0})^{m-1}}\frac{1}{\lambda-\omega_{1}}\|y\|
+\frac{\|f\|}{\lambda^{m}}+\frac{\|g\|}{\lambda^{m}}\\
&\leq \frac{M'}{(\lambda-\omega_{2})^{m}}(\|x\|+\|y\|+\|f\|+\|g\|),
\end{align*}
where $\omega_{2}=\max(0,\omega_{0},\omega_{1})$ and
$M'=\max(M,M\gamma )$.
Since $B(t)$ is a perturbation of $l(t)$, by a linear operator
$\phi(t)$ on $E$; hence, in view of perturbation result
\cite[Theorem 5.2.3]{p1}, $B(t)$ is stable.
\end{proof}

\begin {proposition} \label{prop5.3}
Let $(x,  0,  f,  g) \in D\times \{0\}\times C^{1}([0,T],X)\times
C^{1}([0,T],Y)$

\noindent(1) If the function  $t\to u(t)=\big(
  u_{1}(t),  0,  u_{2}(t),  u_{3}(t) \big)$ is a classical solution
of \eqref{e5.3} with an initial
value  $(  x,  0,  f,  g )$ then $ t\to u_{1}(t)$ is a classical solution of
\eqref{e5.1} with the initial value $x$.

\noindent(2) Let $u$ is a classical solution of \eqref{e5.1} with the
initial value $x$.
Then, the function  $t\to U(t)= \big(  u(t),  0,  f,  g\big)$ is a
classical solution of \eqref{e5.3} with the
initial value $( x,  0,  f,  g )$.
\end {proposition}

\begin{proof} (1) If $u(t)=\big(  u_{1}(t),  0, u_{2}(t),  u_{3}(t)\big)$
 is a classical solution of \eqref{e5.3}, then $u_{1}$ is
continuously differentiable on  $[s,T]$ and we have
$$
\begin{pmatrix}
  \frac{d}{dt}u_{1}(t) \\
  0 \\
  \frac{d}{dt}u_{2}(t) \\
  \frac{d}{dt}u_{3}(t) \\
\end{pmatrix}
=\begin{pmatrix}
  A(t) & 0 & \delta_{t} & 0 \\
  -L(t)+\Phi(t) & 0 & 0 & \overline{\delta_{t}} \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  u_{1}(t) \\
  0 \\
  u_{2}(t) \\
  u_{3}(t) \end{pmatrix}.
$$
This implies
\begin{gather*}
    \frac{d}{dt}u_{1}(t)=A(t)u_{1}(t)+\delta_{t}u_{2}(t), \quad
0\leq s\leq t\leq T, \\
    L(t)u_{1}(t)=\Phi(t)u_{1}(t)+\overline{\delta_{t}}u_{3}(t),\quad
 0\leq s\leq t\leq T, \\
    \frac{d}{dt}u_{2}(t)=0, \\
    \frac{d}{dt}u_{3}(t)=0.
\end{gather*}
 One has  $\frac{d}{dt}u_{3}(t)=0$ which  implies
$u_{3}(t)=u_{3}(s)=g$  and $L(t)u_{1}(t)=\Phi(t)u_{1}(t)+g(t)$.
Also
$\frac{d}{dt}u_{2}(t)=0$ implies  $u_{2}(t)=u_{2}(s)=f$ and
$\frac{d}{dt}u_{1}(t)=A(t)u_{1}(t)+f(t)$.
Then
\begin{gather*}
    \frac{d}{dt}u_{1}(t)=A(t)u_{1}(t)+f(t),\quad 0\leq s\leq t\leq T, \\
    L(t)u_{1}(t)=\Phi(t)u_{1}(t)+g(t),\quad 0\leq s\leq t\leq T, \\
    u_{1}(s)=x.
\end{gather*}
Thus $u_{1}$ is a classical solution of \eqref{e5.1} with the initial
value $x$.

\noindent(2) Let $u$ is a classical solution of \eqref{e5.1}.
This implies that $u $ is continuously differentiable and
$u(t)\in D\times \{0\}\times C^{1}([0,T],X)\times C^{1}([0,T],Y) $. Moreover,
\begin{gather*}
    \frac{d}{dt}u(t)=A(t)u(t)+f(t),\quad 0\leq s\leq t\leq T \\
    L(t)u(t)=\Phi(t)u(t)+g(t),\quad 0\leq s\leq t\leq T \\
    u(s)=x.
\end{gather*}
 This implies
$$
\begin{pmatrix}
  \frac{d}{dt}u(t) \\
  0 \\
  0 \\
  0 \end{pmatrix}
=\begin{pmatrix}
  A(t) & 0 & \delta_{t} & 0 \\
  -L(t)+\Phi(t) & 0 & 0 & \overline{\delta_{t}} \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
  u(t) \\
  0 \\
  f \\
  g
\end{pmatrix},
$$
with $u(s)=x$.
Then $U(t)=(u(t),  0, f, g)$  is continuously differentiable,
$U(t)\in D\times
\{0\}\times C^{1}([0,T],X)\times C^{1}([0,T],Y) $, for all
$t \in [s,T]$ and $U(t)$ is  a classical solution of \eqref{e5.1}
with the initial value $(  x,  0, f, g )$.
\end{proof}

\begin{theorem}
Let $ f\in C^{1}([0,T],X)$ and $g\in C^{1}([0,T],Y)$. Assume that
the hypothesis (H1)--(H5) hold. Then for every
$x\in D$ such that $-L(s)x+\Phi(s)x+g(s)=0$,  problem \eqref{e5.1} has a
unique classical solution.
\end{theorem}

\begin{proof} Consider the homogenous Cauchy problem
\begin{gather*}
    \frac{d}{dt}u(t)=B(t)u(t), \quad 0\leq s\leq t\leq T, \\
    u(s)=( x ,  0 ,  f , g ).
\end{gather*}
 By Lemma \ref{lem5.2}, $ B(t)$ is a stable family and the
function $t\to B(t)y $ is continuously differentiable  for
all $y\in D_{1}=D(B(t))$  independent of $t$. Then there exist
an evolution family $U(t,s)$ on $X\times\{0\}\times
C^{1}([0,T],X)\times C^{1}([0,T],Y)$
such that
$$
U(t,s) \begin{pmatrix}
  x \\
  0 \\
  f \\
  g \end{pmatrix}
=\begin{pmatrix}
  u_{1}(t) \\
  0 \\
  u_{2}(t) \\
  u_{3}(t) \end{pmatrix}= u(t)
$$
is a classical solution of \eqref{e5.3} and from the Proposition \ref{prop5.3},
 $u_{1}$  is a classical solution of \eqref{e5.1}.
The uniqueness of $u_{1}$ comes from the uniqueness of the
solution of \eqref{e5.3} and Proposition \ref{prop5.3}.
\end{proof}

\begin{theorem} \label{thm5.5}
Let $ f\in C^{1}([0,T],X)$ and $ g\in C^{1}([0,T],Y)$. If $ u $ is
a classical solution of \eqref{e5.1} then  $u$ is given by the
variation of constants formula
\begin{equation}
 u(t)=U(t,s)(I-L_{\lambda,s}L(s))x+g(t,u(t))+\int^{t}_{s}U(t,r)[\lambda
g(r,u(r))-g(r,u(r))'+f(r)]dr, \label{e5.8}
\end{equation}
where $ U(t,s)$ is the evolution family generated by $A_{0}(t)$ and
$$
g(t,u(t))=L_{\lambda,t}(\Phi (t)u(t)+g(t)).
$$
\end{theorem}

\begin{proof}
Let now $u$ be a classical solution of \eqref{e5.1}.
Take
$$
u_{2}(t)=L_{\lambda,t}L(t)u(t)\quad\text{and}\quad
u_{1}(t)=(I-L_{\lambda,t}L(t))u(t).
$$
Then the functions
$$
u_{2}(t)=g(t,u(t))=L_{\lambda,t}(\Phi(t)u(t)+g(t)) \quad\text{and}\quad
u_{1}(t)
$$
are differentiable.
Since $u_{2}(t)\in \ker (\lambda I-A(t))$, we have
$A(t)u_{2}(t)=\lambda u_{2}(t) $ and
\begin{align*}
\frac{d}{dt}u_{1}(t)&=\frac{d}{dt}u(t)-\frac{d}{dt}u_{2}(t)\\
&=A(t)u(t)-(g(t,u(t)))'+f(t)\\
&=A(t)(u_{1}(t)+u_{2}(t))+f(t)-(g(t,u(t)))' \\
&=A(t)u_{1}(t)+\lambda (g(t,u(t))+f(t)-(g(t,u(t)))'.
\end{align*}
When we define $h(t):=\lambda g(t,u(t))+f(t)-(g(t,u(t)))'$, we get
\begin{equation}
 u_{1}(t)=U(t,s)u_{1}(s)+\int_{s}^{t}U(t,r)h(r)dr.\label{e5.9}
\end{equation}
By replacing $u_{1}(s)$ by $(I-L_{\lambda,s}L(s))x$, we obtain
\begin{equation}
u_{1}(t)=U(t,s)(I-L_{\lambda,s}L(s))x+\int_{s}^{t}U(t,r)h(r)dr,\label{e5.10}
\end{equation}
it follows that
\begin{align*}
u(t)&=u(t,s)(I-L_{\lambda,s}L(s))x+g(t,u(t))\\
 &\quad +\int^{t}_{s}u(t,r)[\lambda g(r,u(r))-(g(r,u(r)))'+f(r)]dr,
\end{align*}
which completes the proof.
\end{proof}

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\end{document}