\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 130, pp. 1--20.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/130\hfil Variation of constants formula]
{Variation of constants formula for functional  parabolic
partial differential equations}

\author[A. Carrasco, H. Leiva\hfil EJDE-2007/130\hfilneg]
{Alexander Carrasco, Hugo Leiva}  % in alphabetical order

\address{Alexander Carrasco \newline
 Universidad Centroccidental Lisandro Alvarado \\
 Decanato de Ciencias, Departamento de Matem\'atica\\
 Barquisimeto 3001, Venezuela}
\email{acarrasco@ucla.edu.ve}

\address{Hugo Leiva \newline
Universidad de Los Andes
Facultad de Ciencias, Departamento de Matem\'atica \\
M\'erida 5101, Venezuela}
\email{hleiva@ula.ve}

\thanks{Submitted July 2, 2007. Published October 5, 2007.}
\subjclass[2000]{34G10, 35B40}
\keywords{Functional partial parabolic equations;
 variation of constants formula; \hfill\break\indent
strongly continuous semigroups}

\begin{abstract}
 This paper presents a variation of constants formula for
 the  system of functional parabolic partial differential equations
 \begin{gather*}
 \frac{\partial u(t,x)}{\partial t}
 = D\Delta u+Lu_t+f(t,x), \quad t>0,\; u\in \mathbb{R}^n  \\
 \frac{\partial u(t,x)}{\partial \eta} = 0, \quad t>0, \; x\in \partial\Omega \\
 u(0,x) =  \phi(x)  \\
 u(s,x) = \phi(s,x), \quad s\in[-\tau,0),\; x\in\Omega\,.
 \end{gather*}
 Here $\Omega$ is a bounded domain in  $\mathbb{R}^n$, the
 $n\times n$ matrix $D$ is block diagonal with  semi-simple eigenvalues
 having non negative real part, the operator $L$ is bounded and linear,
 the delay in time is bounded, and the standard
 notation $u_{t}(x)(s) = u(t+s,x)$ is used.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

In this paper we find a variation of constants formula for the
 system of functional  parabolic partial differential equations
\begin{equation}\label{non-homogeneous}
 \begin{gathered}
\frac{\partial u(t,x)}{\partial t}
 = D\Delta u+Lu_t+f(t,x), \quad t>0,\; u\in  \mathbb{R}^n  \\
\frac{\partial u(t,x)}{\partial \eta} = 0, \quad t>0, \; x\in \partial\Omega \\
 u(0,x)  =  \phi(x) \\
 u(s,x)  = \phi(s,x), \quad s\in [-\tau,0),\; x\in\Omega
\end{gathered}
\end{equation}
where $\Omega$ is a bounded domain in $\mathbb{R}^N$,
the $n\times n$ matrix $D$ is non diagonal with  semi-simple
eigenvalues having non negative real part,
 and $f:\mathbb{R} \times\Omega\to \mathbb{R}^n $ is an
smooth function. The standard notation $u_{t}(x)$ defines a
function from $[-\tau,0]$ to $\mathbb{R}^n$  by
$u_{t}(x)(s) = u(t+s,x)$, $-\tau \leq  s \leq 0 $ (with $x$ fixed). Here
$\tau\geq 0$ is the maximum delay, which is suppose to be finite. We assume
the operator $L : L^{2}([-\tau,0];Z) \to Z$ is linear
and bounded with $Z = L^{2}(\Omega)$ and
$\phi_{0}\in Z$, $\phi\in L^{2}([-\tau,0];Z)$.

 The variational constant formula plays an important role
in the study of the stability, existence of bounded solutions and
the asymptotic behavior of non linear ordinary differential
equations. The variation of constants formula is well known for
the finite dimensional semi-linear ordinary differential equation
\begin{equation}\label{non-homo-ode}
\begin{gathered}
 x'(t)  =  A(t) + f(t,x), \quad x \in \mathbb{R}^n \\
 x(0) =  x_{0},
\end{gathered}
\end{equation}
and it gives the solution
$$
x(t) = \Phi(t)x_{0} + \int^{t}_{0}\Phi(t)\Phi^{-1}(s)f(s,x(s))ds
$$
where $\Phi(\cdot)$ is the fundamental matrix of the system
\begin{equation}\label{homogeneous-ode}
 x'(t) = A(t)x.
\end{equation}
Due to the importance of this formula, for semi linear ordinary
differential equations, in 1961 the Russian mathematician
Alekseev \cite{Alek} found a formula for the nonlinear ordinary
differential equation
\begin{equation}\label{sistema-semilineal}
 y'(t) = f(t,y) + g(t,y), \quad y(t_{0}) = y_{0},
\end{equation}
which is given by
$$
 y(t,t_{0},y_{0}) = x(t,t_{0},y_{0}) +
 \int^{t}_{t_{0}}\Phi(t,s,y(s))g(s,y(s))ds,
$$
where $x(t,t_{0},y_{0})$ is the solution of the initial value
problem
\begin{equation}\label{sistema-no-autonomo}
 x'(t) = f(t,x), \quad x(t_{0}) = y_{0},
\end{equation}
and
$$
\Phi(t,s,\xi) = \frac{\partial x(t,t_{0},y_{0})}{\partial y_{0}}.
$$
This formula is used to compare the solutions of
(\ref{sistema-semilineal}) with the solutions of (\ref{sistema-no-autonomo}).
In fact, it was used in \cite{LG}.

In infinite dimensional Banach spaces $Z$, we have the following
general situation. If $A$ is the infinitesimal generator of
strongly continuous semigroup $\{T(t) \}_{t \geq 0}$ in $Z$ and
$f: [0, \beta] \to Z$ is a suitable function, then the
solution of the initial value problem
\begin{equation}\label{sistemainfinito}
 \begin{gathered}
  z'(t) =   Az(t) + f(t), \quad t> 0, \; z \in Z \\
  z(0)  =  z_{0},
 \end{gathered}
\end{equation}
is given by the variation constant formula
\begin{equation}\label{milds}
z(t) = T(t)z_{0} + \int^{t}_{0}T(t - s)f(s)ds , \quad t\in[0,\infty ).
\end{equation}
Therefore, any solution of the problem (\ref{sistemainfinito}) is also
solution of the integral equation (\ref{milds}). However, the converse
may not be true, since a solution of (\ref{milds}) is not necessarily
differentiable. We shall refer to a continuous solution of
(\ref{milds})  as a mild solution of problem
(\ref{sistemainfinito}); a mild solution is thus a kind of
generalized solution. However, if $\{T(t) \}_{t \geq 0}$ is an
analytic semigroup and the function $f$ satisfies the following
H\"{o}lder condition
$$
\|f(s) -f(t) \| \leq L|s-t|^\theta, \quad s,t \in [0, \beta],
$$
with $L>0$,  $\theta \geq 1$, then the mild solution (\ref{milds})
is also solution of the initial value problem
(\ref{sistemainfinito}).

Our work and many others are motivated by the legendary paper by
Borisovic  and Turbabin \cite{BT}; there they
found a variational constants formula for the  system of
nonhomogeneous differential equation with delay
\begin{equation}\label{sistemadelay}
\begin{gathered}
 z'(t) =   Lz_t + f(t), \quad t> 0, \; z \in \mathbb{R}^n \\
 z(0)  =  z_{0},\\
 z(s)  =  \phi(s), \quad s \in [-\tau, 0),
\end{gathered}
\end{equation}
where  $f:\mathbb{R}^+ \to \mathbb{R}^n $ is a suitable function. The
standard notation $z_{t}$ defines a function from $[-\tau,0]$ to
$\mathbb{R}^n $ by $z_{t}(s) = z(t+s), -\tau \leq  s \leq 0 $. Here
$\tau\geq 0$ is the maximum delay, which is suppose to be finite.
We assume that the operator $L : L^{p}([-\tau,0];\mathbb{R}^n)
\to \mathbb{R}^n$ is linear and bounded, and $z_{0}\in
\mathbb{R}^n,\;\phi\in L^{p}([-\tau,0];\mathbb{R}^n)$.
Under some conditions they
prove the existence and the uniqueness of solutions for this
system and associate to it a strongly continuous semigroup $\{T(t)
\}_{t \geq 0}$ in the Banach space $\mathbb{M}_{p}([-\tau,0];\mathbb{R}^n)
= \mathbb{R}^n \oplus L_{p}([-\tau,0];\mathbb{R}^n)$.

Therefore, system (\ref{sistemadelay}) is
equivalent to the following system of ordinary differential
equations, in $\mathbb{M}_{p}$,
\begin{equation}\label{ecuacion3 ordinaria no homogenea}
\begin{gathered}
 \frac{dW(t)}{dt} = \Lambda W(t) + \Phi(t),\quad t>0, \\
 W(0)   = W_{0} = (z_{0},\phi(\cdot))
\end{gathered}
\end{equation}
where $\Lambda$  is the infinitesimal generator of the semigroup
$\{T(t)\}_{t\geq0}$ and $\Phi(t) = (f(t),0)$.

Hence, the solution of system (\ref{sistemadelay}) is given by the
variational constant formula or mild solution
\begin{equation}\label{variacion3 de parametro}
W(t) = T(t)W_{0} + \int_{0}^{t}T(t-s)\Phi(s)ds.
\end{equation}

Finally, the formula we found here is valid for those system of
PDEs that can be rewritten in the form
${\partial \over \partial t}u =D\Delta u$, like damped nonlinear
vibration of a string or a beam,
thermoplastic plate equation, etc. For more information about this, see
the paper by Oliveira \cite{LO2}.


 To the best of our knowledge, there are
variational constant formulas for reaction diffusion equations,
functional equations and neutral equations \cite{IV}, but for
functional partial parabolic equations we are not aware of results
similar to the one presented here. At the same time, if we change
the Neumann boundary condition by Dirichlet boundary condition,
the result follows trivially.

\section{Abstract Formulation of the Problem}

In this section we choose a Hilbert Space where system
\eqref{non-homogeneous} can be written as an abstract
functional differential equation. To this end, we consider the
following hypothesis.
\begin{itemize}
\item[(H1)]
 The matrix $D$ is semi simple (block diagonal) and
 the eigenvalues $d_{i} \in \mathbb{C}$  of $D$
satisfy  $\mathop{\rm Re}(d_{i}) \geq 0 $. Consequently, if
 $ 0 = \lambda_{1}< \lambda_{2}<\dots < \lambda_{n} \to \infty$
are the eigenvalues of $-\Delta$ with homogeneous Neumann boundary
conditions, then there exists a constant $M \geq 1$ such that : \\
 $\|e^{-\lambda_{n}Dt}\|\leq M$, \;\;$t\geq 0, \;\;\;n=1,2,3,\dots $\\
H2). For all $I>0$ and $z \in L^{2}_{\rm loc}([-\tau, 0); Z)$ we have
the following inequality
$$
\int^{t}_{0}| Lz_{s}| ds \leq M_{0}(t)| z
|_{L^{2}([-\tau,t),Z)},\quad \forall t\in[0,I],
$$
where $M_{0}(\cdot)$ is a positive continuous function on $[0, \infty)$.

\end{itemize}
 Consider $H = L^{2}(\Omega,\mathbb{R})$ and $0 =
\lambda_{1}<\lambda_{2}<\dots <\lambda_{n}\to \infty $
the  eigenvalues of $-\Delta$, each one with finite
multiplicity $\gamma_{n}$ equal to the dimension of the
corresponding eigenspace. Then
\begin{itemize}
\item[(i)] There exists a complete orthonormal set
$\{\phi_{n,k} \}$ of eigenvectors of $-\Delta$.

\item[(ii)] For all $\xi \in D(-\Delta)$ we have
\begin{equation} \label{p3}
-\Delta \xi = \sum_{n = 1}^{\infty} \lambda_{n} \sum_{k =
1}^{\gamma_n} \langle \xi, \phi_{n,k}\rangle  \phi_{n,k} =\sum_{n = 1}^{\infty}
\lambda_{n}E_{n}\xi,
\end{equation}
where $\langle\cdot, \cdot\rangle $ is the inner product in $H$ and
\begin{equation} \label{p4}
E_{n}x = \sum_{k = 1}^{\gamma_n} \langle \xi, \phi_{n,k}\rangle \phi_{n,k}.
\end{equation}
So, $\{ E_n \}$ is a family of complete orthogonal projections in
$H$ and
 $\xi = \sum_{n = 1}^{\infty}E_{n} \xi$,  $\xi \in H$.

\item[(iii)] $\Delta$ generates an analytic semigroup $\{
T_{\Delta}(t) \}$ given by
\begin{equation} \label{p5}
 T_{\Delta}(t)\xi =  \sum_{n = 1}^{\infty} e^{-\lambda_{n}t}E_{n}\xi.
\end{equation}

\end{itemize}
Now, we denote by $Z$ the Hilbert space $L^{2}(\Omega,\mathbb{R}^n )$ and
define the following operator
$$
A : D(A)\subset Z\to Z , \quad A\psi = -D\Delta\psi
$$
with $D(A)=H^{2}(\Omega,\mathbb{R}^n )\cap H^{1}_{0}(\Omega,\mathbb{R}^n )$.\\
Therefore, for all $z\in D(A)$ we obtain
\begin{gather*}
Az = \sum^{\infty}_{n = 1}\lambda_{n}DP_{n}z, \quad
z = \sum^{\infty}_{n = 1}P_{n}z , \quad
\| z \|^{2} = \sum^{\infty}_{n = 1}\| P_{n}z \|^{2} , \quad z\in Z
\end{gather*}
where
$P_{n} = \mathop{\rm diag}(E_{n},E_{n},\dots ,E_{n})$
is a family of complete orthogonal proyections in $Z$.
Consequently, system \eqref{non-homogeneous} can be written
as an abstract functional differential equation in $Z$:
\begin{equation}\label{ecuacion diferencial abstracta no homogenea}
\begin{gathered}
 \frac{dz(t)}{dt} = -Az(t) + Lz_{t} + f^{e}(t),\quad t>0 \\
  z(0) = \phi_{0} \\
 z(s) = \phi(s),\quad s\in[-\tau,0)
\end{gathered}
\end{equation}
Here $f^{e}: (0,\infty)\to Z$ is a function defined as
follows:
$$
f^{e}(t)(x) = f(t,x),\quad t>0, \; x\in\Omega.
$$

\section{Preliminaries Results}

For the rest of this article, we will use the following generalization
of lemma 2.1 from \cite{HLe}.

\begin{lemma} \label{LG}
Let $Z$ be a separable Hilbert space, $\{S_{n}(t)\}_{n\geq 1}$ a
family of strongly continuous semigroups and
$\{P_{n}\}_{n\geq 1}$ a family  of complete orthogonal projection
in $Z$ such that
$$
\Lambda_{n}P_{n} = P_{n}\Lambda_{n}, \quad n\geq 1,2,\dots
$$
where $\Lambda_{n}$ is the infinitesimal generator of  $S_{n}$.
 Define the  family of linear operators
$$
S(t)z = \sum_{n = 1}^{\infty}S_{n}(t)P_{n}z,\quad t\geq0.
$$
Then:
\begin{itemize}
\item[(a)] $S(t)$ is a linear and bounded operator if
$\|S_{n}(t) \| \leq g(t)$, $n = 1,2,\dots $, with $g(t)\geq 0$,
continuous for $t\geq0$.
\item[(b)] $\{S(t)\}_{t \geq 0}$ is an strongly continuous
semigroup in the Hilbert space $Z$ whose infinitesimal generator
$\Lambda$ is given by
$$
\Lambda z = \sum_{n = 1}^{\infty}\Lambda_{n}P_{n}z, \quad z\in
D(\Lambda)
$$
with
$$
D(\Lambda) = \big\{z\in Z \: / \: \sum_{n = 1}^{\infty}\|
\Lambda_{n}P_{n}z \|^{2} < \infty\big\}
$$
\item[(c)] the spectrum $\sigma(\Lambda)$ of $\Lambda$ is given
by
\begin{equation}\label{L6}
\sigma(\Lambda) = \overline{\cup_{n=1}^{\infty}
\sigma(\bar{\Lambda}_n)},
\end{equation}
where $\bar{\Lambda}_{n} = \Lambda_{n}P_n :\mathcal{R}(P_n)
\to \mathcal{R}(P_n)$.
\end{itemize}
\end{lemma}

\begin{proof}
 First, from Hille-Yosida Theorem,
$S_{n}(t)P_{n}=P_{n}S_{n}(t)$ since $\Lambda_{n}P_{n}= P_{n}\Lambda_{n}$.
So that $\{S_{n}(t)P_{n}z\}_{n\geq1}$ is a family of orthogonal
vectors in $Z$. Then
\begin{align*}
 \| S(t)z \|^{2}
& =  \langle S(t)z,S(t)z \rangle  \\
& =  \Big\langle \sum_{n = 1}^{\infty}S_{n}(t)P_{n}
z,\sum_{m = 1}^{\infty}S_{m}(t)P_{m}z \Big\rangle  \\
& =  \sum_{n = 1}^{\infty}\| S_{n}(t)P_{n}z \|^{2} \\
& \leq  (g(t))^{2}\sum_{n = 1}^{\infty}\| P_ {n}z \|^{2} \\
& =  (g(t)\| z \|)^{2}
\end{align*}
 Therefore, $S(t)$ is a bounded linear operator.

 Second, we have the following relations:
(i)
\begin{align*}
   S(t)S(s)z & =  \sum_{n = 1}^{\infty}S_{n}
(t)P_{n}S(s)z \\
 & =  \sum_{n = 1}^{\infty}S_{n}(t)P_{n}
\Big(\sum_{m = 1}^{\infty}S_{m}(s)P_{m}z\Big) \\
 & =  \sum_{n = 1}^{\infty}S_{n}(t + s)P_
{n}z \\
 & =  S(t + s)z
 \end{align*}
(ii)
$$
S(0)z = \sum_{n = 1}^{\infty}S_{n}(0)P_{n}z = \sum_{n =
1}^{\infty}P_{n}z = z
$$
(iii)
\begin{align*}
   \| S(t)z - z \|^{2}
& = \| \sum_{n = 1}^{\infty}S_{n}(t)P_{n}z -
\sum_{n = 1}^{\infty}P_{n}z \|^{2}   \\
& =  \sum_{n = 1}^{\infty}\|(S_{n}(t) - I)P_{n}z\|^{2}  \\
& = \sum_{n = 1}^{N}\| (S_{n} (t) - I)P_{n}z)\|^{2}
   + \sum_{n = N + 1}^{\infty}\| (S_ {n}(t) -I)P_{n}z \|^{2}\\
& \leq   \sup_{1\leq n \leq N} \| (S_{n}(t) - I)P_{n}z \|^{2}\sum_{n = 1}^{N}
 + K\sum_{n = N + 1}^{\infty}\| P_{n}z \|^{2},
\end{align*}
 where $K = \sup_{0\leq t \leq 1 ;\;n \geq 1}\| (S_{n}(t) - I)\|^{2}\leq (g(t)
+ 1)^{2}$.
Since $\{S_{n}(t)\}_ {t\geq 0}$ $(n =1,2,\dots )$ is an strongly
continuous semigroup and $\{P_n \}_{n \geq 1}$ is a complete
orthogonal projections, given an arbitrary $\epsilon >0$ we have,
for some natural number $N$ and $0<t <1$, the following estimates:
\begin{gather*}
\sum_{n = N + 1}^{\infty}\| P_{n}z\|^{2} <\frac {\epsilon}{2K}, \quad
\sup_{1 \leq n \leq N }\| (S_{n}(t) - I)P_{n}z\|^{2} \leq \frac{\epsilon}{2N},
\\
 \| S(t)z - z \|^{2} < \frac{\epsilon}{2N}\sum_{n = 1}^{N}
 +  K\frac{\epsilon}{2K}<\epsilon
\end{gather*}
Hence, $S(t)$ is an strongly continuous semigroup.

Let $\Lambda$ be the infinitesimal generator of this semigroup. By
definition,  for all  $z\in D(\Lambda)$, we have
$$
\Lambda z = \lim_{t\to 0^{+}}\frac{S(t)z
- z}{t} = \lim_{t\to 0^{+}}\sum_{n =
1}^{\infty}\frac{(S_{n}(t) - I)}{t}P_{n}z.
$$
Next,
$$
P_{m}\Lambda z = P_{m}\Big(\lim_{t\to 0^{+}}\sum_{n =
1}^{\infty}\frac{(S_{n}(t) - I)}{t}P_{n}z\Big) =
\lim_{t\to 0^{+}}\frac{S_{m}(t) - I}{t}P_{m}z =
\Lambda_{m}P_{m}z
$$
So,
$$
\Lambda z  = \sum_{n = 1}^{\infty}P_{n}\Lambda z
 =  \sum_{n = 1}^{\infty}\Lambda_{n}P_{n}z
$$
and
$$
D(\Lambda)\subset\big\{z\in Z / \sum_{n =1}^{\infty}\|
\Lambda_{n}P_{n}z\|^{2}<\infty\big\}
$$
On the other hand, if we assume that
$z\in\big\{z\in Z/ \sum_{n = 1}^{\infty}\|
\Lambda_{n}P_{n}z\|^{2}<\infty\big\}$, then
$$
\sum_{n = 1}^{\infty}\Lambda_{n}P_{n}z = y\in Z
$$
Next, making $z_{n} = \sum_{k = 1}^{n}P_{k}z$, we
obtain
$$
\lim_{t\to 0^{+}}\frac{S(t)z_{n} - z_{n}}
{t} = \sum_{k = 1}^{n}P_{k}\Lambda_{k}z<\infty.
$$
Therefore, $z_{n}\in D(\Lambda)$  and
$\Lambda z_{n} = \sum_{k= 1}^{n}P_{k}\Lambda_{k}z$.
Finally,  if $z_{n}\to z$ when $n\to \infty$  and
$\lim_{t\to 0^{+}} \Lambda z_{n} = y$,  then, since $\Lambda$  is closed, we
obtain that $z\in D(\Lambda)$  and $\Lambda z = y$.

To complete the proof of the lemma, we shall prove part (c). It is
equivalent to prove that
$$
\cup_{n=1}^{\infty} \sigma(\bar{\Lambda}_{n}) \subset
\sigma(\Lambda) \quad \mbox{and} \quad
 \sigma(\Lambda) \subset \overline{\cup_{n=1}^{\infty}
 \sigma(\bar{\Lambda}_{n})}.
$$
To prove the first part, We shall show that
$ \rho(\Lambda) \subset \bigcap_{n=1}^{\infty}\rho(\bar{\Lambda}_{n}) $.
In fact, let $\lambda$ be in $\rho(\Lambda)$. Then
$(\lambda - \Lambda)^{-1}: Z \to D(\Lambda)$ is a bounded linear
operator. We need to prove that
$$
(\lambda - \bar{\Lambda}_{m})^{-1}:\mathcal{R}(P_m) \to \mathcal{R}(P_m)
$$
exists and is bounded for $m \geq 1$. Suppose that $(\lambda -
\bar{\Lambda}_{m})^{-1}P_{m}z=0$. Then
\[
(\lambda -\Lambda)P_{m}z
 =  \sum_{n=1}^{\infty} (\lambda -\Lambda_n)P_{n}P_{m}z\\
 =  (\lambda -\Lambda_m)P_{m}z
 =(\lambda -\bar{\Lambda}_m)P_{m}z=0.
\]
Which implies that, $P_{m}z = 0$. So, $(\lambda -\bar{\Lambda}_{m})$
is one to one.

Now, given $y$ in $\mathcal{R}(P_m)$ we want to solve the equation
$(\lambda -\bar{\Lambda}_m)w=y$. In fact, since
$\lambda \in \rho(\Lambda)$ there exists $z \in Z$ such that
$$
(\lambda -\Lambda)z =  \sum_{n=1}^{\infty} (\lambda -\Lambda_n)P_{n}z=y.
$$
Then, applying $P_m$ to the both side of this equation we obtain
$$
P_{m}(\lambda -\Lambda)z = (\lambda -\Lambda_m)P_{m}z=(\lambda
-\bar{\Lambda}_m)P_{m}z= P_{m}y=y.
$$
Therefore, $(\lambda -\bar{\Lambda}_{m}):\mathcal{R}(P_m) \to
\mathcal{R}(P_m)$ is a bijection. Since $\bar{\Lambda}_{m}$ is close,
then, by the closed-graph theorem, we get
$$
\lambda \in \rho(\bar{\Lambda}_{m})= \{ \lambda \in \mathbb{C} :
(\bar{\Lambda}_{m}-\lambda I) \mbox{ is bijective }\}
 = \{ \lambda \in \mathbb{C} : (\bar{\Lambda}_{m}-\lambda I)^{-1}
\mbox{is bounded } \}
$$
for all $m \geq 1$. We have proved that
$$
\rho(\Lambda) \subset \bigcap_{n=1}^{\infty}
\rho(\bar{\Lambda}_{n}) \iff \bigcup_{n=1}^{\infty}
\sigma(\bar{\Lambda}_n) \subset \sigma(\Lambda).
$$
Now, we shall prove the other part of (c), that is to say:
$$
\sigma(\Lambda)\subset\overline{\cup_{n=1}^{\infty}\sigma
(\overline{\Lambda_{n}})}.
$$
In fact, if $ \lambda \in \sigma(\Lambda)$, then
\begin{itemize}
\item[(1)] $\lambda \in \sigma_{p}(\Lambda)= \{ \lambda \in \mathbb{C} :
(\Lambda-\lambda I) \mbox{ is not injective } \}$
\item[(2)] $\lambda \in \sigma_{r}(V)= \{ \lambda \in \mathbb{C} :
 (\Lambda-\lambda I) \mbox{ is injective , but }
  \overline{R(\Lambda-\lambda I)}\neq Z\}$
\item[(3)] $\lambda \in \sigma_{c}(\Lambda)= \{ \lambda \in \mathbb{C} :
 (\Lambda-\lambda I)  \mbox{ is injective, }
 \overline{R(\Lambda-\lambda I)} = Z,  \mbox{ but }
 R(\Lambda-\lambda I) \neq Z \}$.

\end{itemize}
(1) If $(A\Lambda-\lambda I)$ is not injective, then
there exists $z\in Z $ non zero such that: $(\Lambda-\lambda I)z=0$.
 This implies that for some $n_{0}$ we have
$$
(\overline{\Lambda_{n_{0}}}-\lambda I)P_{n_{0}}z= 0,  \quad
P_{n_{0}}z \not= 0 .
$$
>From here we obtain that
$\lambda\in \sigma(\overline{\Lambda_{n_{0}}})$, and therefore $\lambda\in
\overline{\cup_{n=1}^{\infty}\sigma(\overline{\Lambda_{n}})}$.

\noindent(2) If  $\overline{R(\Lambda-\lambda I)}\neq Z$, then
there exists $z_{0}\in Z$ non zero such that
$$
\langle z_{0},(\Lambda-\lambda I)z\rangle = 0, \ \ \forall z\in
D(A).
$$
But, $z= \sum_{n=1}^{\infty}P_{n}z$, so
$$
\langle z_{0},\sum_{n=1}^{\infty}(\overline{\Lambda_{n}}-\lambda
I)P_{n}z\rangle = 0.
$$
Now, if $z_{0}\neq 0$, then there is $n_{0}\in\mathbf{N}$ such
that
 $P_{n_{0}}z_{0} \neq0$. Hence,
\[
  0  =   \langle
z_{0},\sum_{n=1}^{\infty}(\overline{\Lambda_{n}}-\lambda
I)P_{n}z\rangle
    =   \langle z_{0},(\overline{\Lambda_{n_{0}}}-\lambda
I)P_{n_{0}}z\rangle
   =   \langle P_{n_{0}}z_{0},(\overline{\Lambda_{n_{0}}}-\lambda
I)P_{n_{0}}z\rangle
\]
So, $R(\overline{\Lambda}_{n_{0}}-\lambda I)\neq P_{n_{0}}Z$.
Therefore, $\lambda\in \sigma(\overline{\Lambda}_{n_{0}})
\subset\overline{\cup_{n=1}^{\infty}\sigma(\overline{\Lambda_{n}})}$.

\noindent (3) Assume that $(\Lambda-\lambda I)$ is injective,
$\overline{R(\Lambda-\lambda I)}= Z$ and
$R(\Lambda-\lambda I)\subseteq Z$.
For the purpose of getting a contradiction, we suppose that
$\lambda\in \left(\overline{\cup_{n=1}^{\infty}\sigma(\overline{\Lambda_
{n}})}\right)^{C}$.\\
However,
$$
\Big(\overline{\cup_{n=1}^{\infty}\sigma(\overline{\Lambda_{n}})}\Big)^{C}
 \subset \Big(\bigcup_{n=1}^{\infty}\sigma
(\overline{\Lambda_{n}})\Big)^{C} \\
  =  \bigcap_{n\geq1}\big(\sigma(\overline{\Lambda_{n}})\big)^{C} \\
  =  \bigcap_{n\geq1}\rho(\overline{\Lambda_{n}}),
$$
which implies that, $\lambda\in \rho(\overline{\Lambda_{n}})$, for
all $n\geq1$. Then we get that
$$
(\overline{\Lambda_{n}}-\lambda I):R(P_{n})\to
R(P_{n})
$$
is invertible, with $(\overline{\Lambda_{n}}-\lambda I)^{-1}$ bounded.
Hence, for all $z \in D(\Lambda)$ we obtain
$$
P_{j}(\Lambda-\lambda I)z =(\overline{\Lambda_{j}}-\lambda
I)P_{j}z, \quad j=1,2, \dots;
$$
i.e.,
$$
(\overline{\Lambda_{j}}-\lambda I)^{-1}P_{j}(\Lambda-\lambda I)z
=P_{j}z, \quad j=1,2, \dots
$$
Now, since $D(A)$ is dense in $Z$, we may extend the operator
$(\overline{\Lambda_{j}}-\lambda I)^{-1}P_{j}(\Lambda-\lambda I)$
to a bounded operator $T_j$ defined on $Z$. Therefore, it follows
that
$$
T_{j}z = P_{j}z, \quad \forall z \in Z, \; j=1,2, \dots,
$$
and
$$
\|T_{j}\| = \|P_{j}\| \leq 1, \quad j=1,2, \dots.
$$
Since $\overline{R(\Lambda-\lambda I)}= Z$, we get
\begin{equation}\label{milagro}
\|(\overline{\Lambda_{j}}-\lambda I)^{-1}\| \leq 1, \quad j=1,2, \dots.
\end{equation}
Now we shall see that ${R(\Lambda-\lambda I)}= Z$. In fact, given
$z \in Z$ we define $y$ as
$$
y= \sum_{j=1}^{\infty}(\overline{\Lambda_{j}}-\lambda
I)^{-1}P_{j}z.
$$
>From (\ref{milagro})  we get that $y$ is well defined. We shall
see now that $y \in D(\Lambda)$ and  $(\Lambda-\lambda I)y = z$.
In fact, we know that
$$
y \in D(\Lambda) \iff \sum_{j=1}^{\infty} \|\Lambda_{j}P_{j}y\|^2
< \infty.
$$
On the other hand, we have
$$
\sum_{j=1}^{\infty} \|\overline{\Lambda}_{j}P_{j}y\|^2 =
\sum_{j=1}^{\infty} \|\Lambda_{j}(\overline{\Lambda_{j}}-\lambda
I)^{-1}P_{j}z\|^2 =\sum_{j=1}^{\infty} \|\{I +
\lambda(\overline{\Lambda_{j}}-\lambda I)^{-1}\} P_{j}z\|^2.
$$
So,
$$
\sum_{j=1}^{\infty} \|\Lambda_{j}P_{j}y\|^2 \leq
\sum_{j=1}^{\infty} \|(1 + |\lambda|)^2 \|P_{j}z\|^2 = (1 +
|\lambda|)^2 \|z\|^2 < \infty.
$$
Then, $y \in D(\Lambda)$ and  $(\Lambda-\lambda I)= z$.
Therefore $R(\Lambda-\lambda I)= Z$, which is a contradiction that
came from the assumption: $\lambda\in
\big(\overline{\cup_{n=1}^{\infty}\sigma(\overline{\Lambda_{n}})}\big)^{C}$.
\end{proof}

\begin{lemma}\label{LG1}
Let $Z$ be a separable Hilbert space, $\{S_{n}(t)\}_{t\geq0}$ a
family of strongly continuous semigroups with generators
$\Lambda_{n}$ and $\{P_{n}\}_{n\geq1}$ a family of complete
orthogonal projections such that
\begin{equation}\label{conmutatividad}
\Lambda_{n}P_{m}=P_{m}\Lambda_{n},\quad n,m=1,2,\dots
\end{equation}
If the operator
$$
\Lambda z = \sum_{n=1}^{\infty}\Lambda_{n}P_{n}z,\quad z\in D(\Lambda)
$$
with
$$
D(\Lambda) = \{z\in Z: \sum_{n=1}^{\infty}\| \Lambda_{n}P_{n}z\|^{2}<\infty\}
$$
generates a strongly continuous semigroup  $\{S(t)\}_{t\geq0}$,
then
$$
S(t)z = \sum_{n=1}^{\infty}S_{n}(t)P_{n}z,\quad z\in Z.
$$
\end{lemma}

\begin{proof}
If $z_{0}\in Z$, then $P_{n}z_{0}\in D(\Lambda)$ and the mild
solution of the problem
\begin{equation}\label{conver1}
\begin{gathered}
z'(t)  =  \Lambda z(t) \\
z(0)  =  P_{n}z_{0}
\end{gathered}
\end{equation}
is given by $z_{n}(t) = S(t)P_{n}z_{0}$ and it is a classic solution.
Using (\ref{conmutatividad})  and the Hille-Yosida Theorem, we get
  $P_{n}S(t) = S(t)P_{n}$, which implies
\begin{equation}\label{conver2}
S(t)z_{0} = \sum_{n=1}^{\infty}P_{n}S(t)z_{0}=
\sum_{n=1}^{\infty}S(t)P_{n}z_{0}.
\end{equation}
On the other hand, since $z_{n}(t)$ is a classic solution of
(\ref{conver1}), we obtain
\begin{align*}
  z'_{n}(t) & =  \Lambda z_{n}(t)\\
   & =  \Lambda S(t)P_{n}z_{0}\\
   & =  \sum_{m=1}^{\infty}\Lambda_{m}P_{m}S(t)P_{n}z_{0}\\
   & =  \Lambda_{n}P_{n}S(t)P_{n}z_{0} \\
   & =  \Lambda_{n}S(t)P_{n}z_{0} = \Lambda_{n}z_{n}(t)
\end{align*}
So that, $z_{n}(t) = S_{n}(t)P_{n}z_{0} = S(t)P_{n}z_{0}$ and from
(\ref{conver2}) we get
$$
S_{n}(t)z_{0} =
\sum_{n=1}^{\infty}S_{n}(t)P_{n}z_{0}.
$$
\end{proof}

Now, applying Lemma \ref{LG} we can prove the following result.

\begin{theorem}\label{T1}
The operator $-A$ is the infinitesimal generator of a strongly
continuous semigroup $\{ T_{A}(t)\}_{t \geq 0}$ in the
space $Z$, given by
\begin{equation}\label{damp}
T_{A}(t)z =\sum_{n=1}^{\infty}e^{-\lambda_{n}D t}P_nz%
, \quad z\in Z ,  \; t \geq 0.
\end{equation}
\end{theorem}

\subsection{Existence and Uniqueness of Solutions}

In this part we study the existence and the uniqueness of the
solutions for system
(\ref{ecuacion diferencial abstracta no homogenea}) in case
$f^{e} \equiv 0$. That is, we analyze the
 homogeneous system
\begin{equation}\label{ecuacion diferencial abstracta  homogenea}
\begin{gathered}
 \frac{dz(t)}{dt} = -Az(t) + Lz_{t} ,\quad t>0  \\
 z(0) = \phi_{0}=z_0  \\
 z(s) = \phi(s),
 \quad s\in[-\tau,0)
 \end{gathered}.
\end{equation}

\begin{definition}\label{mildsoluction} \rm
A function $z(\cdot)$ define on $[-\tau, \alpha )$ is called a
Mild Solution of (\ref{ecuacion diferencial abstracta  homogenea})
if
$$
z(t) = \begin{cases}
 \phi(t) &-\tau\leq t <0, \\
 T_{A}(t)z_{0} + \int^{t}_{0}T_{A}(t - s)Lz_{s}ds ,
& t\in[0,\alpha )
 \end{cases}
$$
\end{definition}

\begin{theorem}\label{existencia}
Problem (\ref{ecuacion diferencial abstracta  homogenea})
admits only one mild solution defined on $[-\tau, \infty)$.
\end{theorem}

\begin{proof}
Consider the  initial function
\[
\varphi(s) = \begin{cases}
 \phi(s), & -\tau\leq s <0 \\
 T_{A}(s)z_{0} & s\geq 0
 \end{cases}
\]
 which belongs to $L^{2}_{\rm loc}([-\tau,\infty),Z)$.
For a moment we shall set the
problem on $[-\tau, I]$, $I>0$ and denote by $G$ the set
$$
G = \{\psi : \psi\in L^{2}[[-\tau,\alpha],Z] \quad\text{and}
\quad | \psi - \varphi |_{L^{2}}\leq \rho, \quad \rho>0\},
$$
where $\alpha  >0$ is a number to be determine. It is clear that
$G$ endowed with the norm of $L^{2}([-\tau,\alpha];Z)$ is a
complete metric space.

 Now, we consider the application $S:G \to Z$, for $z\in G$,
given by
\[
(Sz)(t) = Sz(t) = \begin{cases}
 \phi(t), & -\tau\leq t <0 \\
 T_{A}(t)z_{0} + \int^{t}_{0}T_{A}(t - s)Lz_{s}ds,
 & t\in[0,\alpha]
 \end{cases}
\]

\noindent {\bf Claim 1.} There exists $\alpha> 0$ such that
\begin{itemize}
\item[(i)] $Sz\in G$, for all $z\in G$.
\item[(ii)] $S$ is a contraction mapping.
\end{itemize}
In fact, we prove (i) as follows:
\[
  | Sz(t) -\varphi(t)|   \leq  \int^{t}_{0}| T_{A}(t - s)Lz_{s}| ds
 \leq  \int^{\alpha}_{0}M| Lz_{s}| ds
 \leq  M M_{0}(\alpha)| z |_{L^{2}([-\tau,\alpha),Z)}.
\]
 Integrating, we have
$$
| Sz - \varphi |_{L^{2}} \leq  K \alpha^{1/2}|z |_{L^{2}}
$$
where $K = max\{M M_{0}(\alpha)/ \alpha\in[0,I]\}$.
>From here we get
$$
| Sz - \varphi |_{L^{2}} \leq K\alpha^{1/2} (|\varphi |_{L^{2}} + \rho) ,
\quad z \in G.
$$
Taking
 $$
 \alpha < \Big(\frac{\rho}{K(|\varphi |_{L^{2}} + \rho)}\Big)^2
$$
we obtain that $Sz \in G$, for all $z \in G$.

To prove (ii), we use the linearity of $L$ to
obtain:
$$
| Sz - Sw|_{L^{2}} \leq  K\alpha^{1/2}| z -
w|_{L^{2}}, \quad \forall z,w \in G.
$$
Next, to prove that $S$ it is a contraction and
$S(G)\subset G$ it is sufficient to choose $\alpha$ so that
$$
\alpha <
\min\big\{\Big(\frac{1}{K}\Big)^{2},\Big(\frac{\rho}{K(|
\varphi |_{L^{2}} + \rho)}\Big)^{2}\big\}
$$
Therefore, $S$ is a contraction mapping. So, if we apply the
contraction mapping Theorem, there exists a unique point $z \in G$
such that $Sz = z$. i.e.,
\[
z(t) = Sz(t) = \begin{cases}
 \phi(t) , & -\tau\leq t <0 \\
 T_{A}(t)z_{0} + \int^{t}_{0}T_{A}(t - s)Lz_{s}ds ,
& t\in[0,\alpha],
 \end{cases}
\]
which proves the existence and the uniqueness of the mild solution
of the initial value problem
(\ref{ecuacion diferencial abstracta homogenea}) on $[-\tau,\alpha]$.

\noindent {\bf Claim 2.} $\alpha$ could be equal to $\infty$. In fact, let
$z$ be the unique mild solution define in a maximal interval
$[-\tau,\delta)(\delta\geq\alpha)$.

\noindent By contradiction, let us suppose that $\delta < \infty$.
Since $z$ is a mild solution of
(\ref{ecuacion diferencial abstracta homogenea}), we have that
$$
 z(t) =T_{A}(t)z_{0} + \int_{0}^{t}T_{A}(t - s)Lz_{s}ds,\quad t \in
[0,\delta).
$$
Consider the sequence $\{t_{n}\}$ such that $t_{n}\to \delta^{-}$.
Let us prove that  $\{z(t_{n})\}$ is a Cauchy sequence.
In fact,
\begin{align*}
&| z(t_{n}) - z(t_{m}) |\\
& =  | T_{A}(t_{n})z_{0} - T_{A}(t_
{m})z_{0}+\int_{0}^{t_{n}}T_{A}(t_{n} - s)Lz_{s}ds -
\int_{0}^{t_{m}}T_{A}(t_{m} - s)Lz_{s}ds | \\
& \leq  |(T_{A}(t_{n}) - T_{A}(t_{m}))z_{0}|+|
\int_{0}^ {t_{n}}T_{A}(t_{n}-s)Lz_{s}ds -
\int_{0}^{t_{m}}T_{A}(t_{m}-s)Lz_{s}ds |
\end{align*}
But,
\begin{align*}
&| \int_{0}^{t_{n}}T_{A}(t_{n}-s)Lz_{s}ds
- \int_{0}^{t_{m}}T_{A}(t_{m}-s)Lz_{s}ds | \\
& \leq  |\int_{0}^{t_{m}}(T_{A}(t_{n}-s)-T_{A}(t_{m}-s))Lz_{s}ds |
 + | \int_{t_{n}}^{t_{m}}T_{A}(t_{n}-s)Lz_{s}ds |
\end{align*}
Now, for $z\in L^{2}([-\tau,\delta])$ we obtain
$$
\int_{0}^{t_{m}}| (T_{A}(t_{n}-s)-T_{A}(t_{m}-s)) Lz_{s}|
ds\leq \int_{0}^{\delta}| (T_{A}(t_{n}-s)-T_{A}(t_{m}-s))
Lz_{s}| ds
$$
We know that
\begin{gather*}
\lim_{n,m \to \infty}| (T_{A}(t_{n}-s)-T_{A}(t_{m}-s))
Lz_{s}|=0, \\
|(T_{A}(t_{n}-s)-T_{A}(t_{m}-s)) Lz_{s}|\leq 2M| Lz_{s}|
\end{gather*}
But, from the hypothesis (H1), we obtain
$$
\int_{0}^{\delta}2M| Lz_{s}| ds\leq 2MM_{0}(\delta)|
z|_{L^{2}([-\tau,\delta);Z)}
$$
Therefore, applying the Lebesgue Dominated Convergence Theorem, we
obtain
$$
\lim_{n,m \to \infty}\int_{0}^{\delta}|
(T_{A}(t_{n}-s)-T_{A}(t_{m}-s)) Lz_{s}| ds=0
$$
Then, since the family $\{T_{A}(t)\}_{t\geq 0}$ is strongly
continuous and $t_{n},t_{m}\to \delta^{-}$ when
$n,m\to\infty$, the sequence $\{z(t_{n})\}$ is a
Cauchy sequence and therefore there exists $B\in Z$ such that
$$
\lim_{n\to \infty}z(t_{n}) = B.
$$
Now, for $t\in[0,\delta)$ we obtain that
\begin{align*}
 | z(t) - B | & \leq  | z(t) - z(t_{n}) |  + | z(t_{n}) - B |\\
& \leq  | (T_{A}(t) - T_{A}(t_{n}))z_{0} | \;+\; | z(t_{n}) - B | \\
& \quad + | \int_{0}^{t_{n}}T_{A}(t_{n}-s)Lz_{s}ds - \int_{0}^{t}T_{A}(t-s)
Lz_{s}ds |
\end{align*}
However,
\begin{align*}
&\big| \int_{0}^{t_{n}}T_{A}(t_{n}-s)Lz_{s}ds
- \int_{0}^{t}T_{A}(t-s)Lz_{s}ds \big| \\
& \leq  \int_{0}^{t_{n}}|(T_{A}(t-s)- T_{A}(t_{n}-s))Lz_{s}|ds
 + \int_{t}^{t_{n}}| T_{A}(t-s)Lz_{s}| ds.
\end{align*}
On the other hand, for $z\in L^{2}([-\tau,\delta])$ we get the
 estimate
$$
\int_{0}^{t_{n}}| (T_{A}(t-s)- T_{A}(t_{n}-s))Lz_{s}|
ds\leq \int_{0}^{\delta}| (T_{A}(t-s)- T_{A}(t_{n}-s))Lz_{s}|
ds
$$
Therefore, applying the Lebesgue Dominated Convergence Theorem, we
obtain
$$
\lim_{n \to \infty}\int_{0}^{\delta}| (T_{A}(t-s)-T_{A}(t_{n}-s))
Lz_{s}|=0
$$
Then, since the family $\{T_{A}(t)\}_{t\geq 0}$ is strongly
continuous and $t_{n}\to \delta^{-}$ when
$n\to\infty$, it follows that $z(t)\to B$ as
$t\to \delta^{-}$.
The function
\[
\varphi(s) = \begin{cases}
 z(s) , & \delta-\tau\leq s < \delta \\
 T_{A}(s)B , & s\geq\delta
 \end{cases}
\]
belongs to $ L^{2}_{\rm loc}([\delta-\tau,\infty),Z)$. So, if we apply
again the contraction mapping Theorem to the Cauchy problem
\begin{equation}\label{ecuacion diferencial abstracta  homogenea2}
\begin{gathered}
 \frac{dy(t)}{dt} = -Ay(t) + Ly_{t}, \quad t>\delta \\
 y(\delta) = B \\
 y(s) = z(s),\quad  s\in[\delta-\tau,\delta)
 \end{gathered}
\end{equation}
where $z(\cdot)$ is the unique solution of the system
(\ref{ecuacion diferencial abstracta homogenea}), then we get that
(\ref{ecuacion diferencial abstracta homogenea2}) admits only one
solution $y(\cdot)$ on the interval $[\delta-\tau,\delta +
\epsilon]$ with $\epsilon >0$. Therefore, the function
\[
\widetilde{z}(s) = \begin{cases}
 z(s) & -\tau \leq s < \delta \\
 y(s), & \delta \leq s < \delta + \epsilon
 \end{cases}
\]
is also a mild solution of
(\ref{ecuacion diferencial abstracta homogenea}) which is a contradiction.
So, $\delta = \infty$.
 \end{proof}

\section{The Variation Of Constants Formula}

 Now we are ready to find the formula announced in the title of
this paper for the system
(\ref{ecuacion diferencial abstracta no homogenea}), but first we
need to write this system as an abstract
ordinary differential equation in an appropriate Hilbert space. In
  fact, we consider the Hilbert space $\mathbb{M}_{2}([-\tau,0];Z) = Z
\oplus L_{2}([-\tau,0];Z)$
  with the usual innerproduct given by
$$
\Big\langle
\begin{pmatrix}
   \phi_{01} \\
 \phi_{1} \end{pmatrix},
\begin{pmatrix}
   \phi_{02} \\
 \phi_{2} \end{pmatrix}
 \Big\rangle = \langle\phi_{01},\phi_{02}\rangle_{Z} +
\langle\phi_{1},\phi_{2}\rangle_{L_{2}}.
$$
Define the  operators $T(t)$ in the space $\mathbb{M}_{2}$ for
$t\geq 0$ by
\begin{equation}\label{Operador Semigrupo}
T(t)\begin{pmatrix}
   \phi_{0} \\
 \phi(.) \end{pmatrix}
= \begin{pmatrix}
   z(t) \\
 z_t \end{pmatrix}
\end{equation}
where  $z(\cdot)$ is the only mild solution of the system
(\ref{ecuacion diferencial abstracta homogenea}).

\begin{theorem}\label{semigroup}
The family of operators $\{T(t)\}_{t\geq
0}$ defined by (\ref {Operador Semigrupo}) is an strongly
continuous semigroup on $\mathbb{M}_{2}$ such that
 \begin{equation}\label{semigrupo}
 T(t)W = \sum_{n = 1}^{\infty}T_{n}(t)Q_{n}W,\quad W\in \mathbb{M}_{2}
,\;t\geq 0,
\end{equation}
where
$$
Q_{n}  =  \begin{pmatrix}
 P_{n} & 0 \\
0 & \widetilde{P}_{n} \end{pmatrix},
$$
with $(\widetilde{P}_{n}\phi)(s)=P_{n}\phi(s)$,
$\phi\in L^{2}([-\tau,0];Z)$, $s\in[-\tau,0]$,
and $\{\{T_{n}(t)\}_{t\geq0},\ n = 1,2.3,\dots \}$
is a family of strongly
continuous semigroups on  $\mathbb{M}_{2}^{n}=Q_{n}\mathbb{M}_{2}$
given in the same way as in \cite[Theorem 2.4.4]{CP3} and
defined by
$$
T_{n}(t) \begin{pmatrix}
  w^{0}_{n} \\
  w_{n}\end{pmatrix}   =  \begin{pmatrix}
  W^{n}(t)\\
  W^{n}(t + \cdot) \end{pmatrix}, \quad
\begin{pmatrix}
  w^{0}_{n} \\
  w_{n}\end{pmatrix} \in \mathbb{M}_{2}^{n},
$$
where $W^{n}(\cdot)$ is the unique solution of the initial value
problem
\begin{equation}\label{ec. con retardo}
\begin{gathered}
 \frac{dw(t)}{dt} = -\lambda_{n}Dw(t) + L_{n}w_{t} ,\quad t>0 \\
 w(0) = w_{n}^{0} \\
 w(s) = w_{n}(s),\quad s\in[-\tau,0)
\end{gathered}
\end{equation}
and  $L_{n}=L\widetilde{P}_{n}= P_{n}L$, as it is in most the case
practical problems.
\end{theorem}

\begin{proof}[Proof of Theorem \ref{semigroup}]
 First, we shall prove that
$$
T(t)W = \sum_{n = 1}^{\infty}T_{n}(t)Q_{n}W,\quad W\in \mathbb{M}_
{2},\; t\geq0.
$$
In fact, let $W= \begin{pmatrix}
  w_{1} \\
  w_{2} \end{pmatrix}
\in \mathbb{M}_{2}$.
\begin{align*}
 & \sum_{n = 1}^{\infty}T_{n}(t)Q_{n}W\\
& =  \sum_{n = 1}^{\infty}T_{n}(t) \begin{pmatrix}
  P_{n} & 0 \\
  0 & \widetilde{P}_{n}
   \end{pmatrix}
 \begin{pmatrix}
   w_{1} \\
   w_{2} \\
 \end{pmatrix}\\
& =  \sum_{n = 1}^{\infty}T_{n}(t) \begin{pmatrix}
   P_{n}w_{1} \\
   \widetilde{P}_{n}w_{2}  \end{pmatrix}\\
& =  \sum_{n = 1}^{\infty}  \begin{pmatrix}
 z^{n}(t) \\
 z^{n}(t + \cdot)   \end{pmatrix}
  \quad z^{n}(\cdot)
\mbox{ is the only mild solution of (\ref{ec. con retardo})}\\
& =  \sum_{n = 1}^{\infty}  \begin{pmatrix}
 e^{\mathcal{A}_{n}t}P_{n}w_{1} + \int_{0}^{t}e^{\mathcal{A}_{n}(t-s)}L_{n}
(\widetilde{P}_{n}z^{n}(s + \cdot))ds \\
 (\widetilde{P}_{n}z(t + \cdot))   \end{pmatrix} \\
& =   \begin{pmatrix}
 \sum_{n = 1}^{\infty}e^{\mathcal{A}_{n}t}P_{n}w_{1} +
 \int_{0}^{t}\sum_{n = 1}^{\infty}e^{\mathcal{A}_{n}(t-s)}P_{n}
 \Big(L\sum_{m = 1}^{\infty}(\widetilde{P}_{m}z(s + \cdot)) \Big)ds \\
  \sum_{n = 1}^{\infty}(\widetilde{P}_{n}z(t + \cdot)) \end{pmatrix} \\
& =  \begin{pmatrix}
 T_{\mathcal{A}}(t)w_{1} + \int_{0}^{t}T_{\mathcal{A}}(t-s)L
z(s + \cdot)ds \\
 z(t + \cdot)  \end{pmatrix} \\
& =  \begin{pmatrix}
 z(t) \\
 z_{t}(\cdot) \end{pmatrix}, \quad
 z(\cdot)\mbox{ is the only mild solution of
(\ref{ecuacion diferencial abstracta  homogenea})} \\
 & =  T(t)W.
\end{align*}
In the same way as in \cite[Theorem 2.4.4]{CP3} we can prove
that the infinitesimal generator of $\{T_{n}(t)\}_{t\geq 0}$ is
given by
$$
\Lambda_{n}\begin{pmatrix}
   w_{n}^{0} \\
 w_{n}(\cdot) \end{pmatrix}
=\begin{pmatrix}
   -\Lambda_{n}Dw_{n}^{0} + L_{n}w_{n}(\cdot) \\
 \frac{\partial w_{n}(\cdot)}{\partial s} \end{pmatrix}
$$
with
$$
D(\Lambda_{n}) =\{
\begin{pmatrix}
   w_{n}^{0} \\
 w_{n}(\cdot)\end{pmatrix}
\in \mathbb{M}_{2}^{n}: w_{n} \text{ is  a.c., }
\frac {\partial w_{n}(\cdot)}{\partial s}\in L_{2}([-\tau,0];Q_{n}Z),\;
w_{n}(0)=w_{n}^{0}\}.
$$
Furthermore, the spectrum of $\Lambda_n$ is discrete and given by
\begin{equation}\label{spectrum}
\sigma(\Lambda_n) = \sigma_{p}(\Lambda_n)
= \{ \lambda \in \mathbb{C} : \det (A_{n}(\lambda)) = 0 \},
\end{equation}
where $A_{n}(\lambda)$ is given by
$$
\Lambda_{n}(\lambda)z =  \lambda z + \lambda_n Dz -
L_{n}e^{\lambda(\cdot)}z, \ \ z \in Z_n = P_n Z,
$$
which can be considered  a matrix since $\mbox{dim}(Z_n) <\infty$.

 On the other hand, $\{Q_n \}_{n \geq 1}$ is a family of
complete orthogonal projection on $\mathbb{M}_{2}$ and
$$
\Lambda_{n}Q_n = Q_n \Lambda_{n}, \quad n=1,2,3, \dots
$$
In fact,
\begin{align*}
\Lambda_{n}Q_n \begin{pmatrix}
   w_{n}^{0} \\
 w_{n}(\cdot)\end{pmatrix}
& = \Lambda_{n}\begin{pmatrix}
   P{n}w_{n}^{0} \\
 \widetilde{P_{n}}w_{n}(\cdot)
\end{pmatrix}
 =
\begin{pmatrix}
   -\Lambda_{n}DP_{n}w_{n}^{0} + L_{n}\widetilde{P_{n}}w_{n}(\cdot) \\
 \frac{\partial \widetilde{P_{n}}w_{n}(\cdot)}{\partial s} \end{pmatrix}
\\
& = \begin{pmatrix}
   -\Lambda_{n}DP_{n}w_{n}^{0} + L\widetilde{P_{n}}\widetilde{P_{n}}w_
{n}(\cdot) \\
 \widetilde{P_{n}}\frac{\partial w_{n}(\cdot)}{\partial s}
\end{pmatrix} \\
&= \begin{pmatrix}
   -\Lambda_{n}DP_{n}w_{n}^{0} + P_{n}L_{n}w_{n}(\cdot) \\
 \widetilde{P_{n}}\frac{\partial w_{n}(\cdot)}{\partial s}
 \end{pmatrix}\\
& =\begin{pmatrix} P_{n} & 0 \\
0 & \widetilde{P}_{n} \end{pmatrix}
\begin{pmatrix}
   -\Lambda_{n}Dw_{n}^{0} + L_{n}w_{n}(\cdot) \\
 \frac{\partial w_{n}(\cdot)}{\partial s} \end{pmatrix} \\
 &=Q_{n}\Lambda_{n}\begin{pmatrix}
   w_{n}^{0} \\
 w_{n}(\cdot)\end{pmatrix}
\end{align*}
Now, we shall check condition (a) of Lemma \ref{LG}. To this end
we need to prove the following claim.

\noindent {\bf Claim.} If $W^{n}(t)$ is the solution of
(\ref{ec. con retardo}), then the following inequalities hold
\begin{gather}\label{ineq1}
\| W^{n}(t) \|_{Z} \leq  c_ {2}e^{c_{1}t} \|w_{n}^{0} \|, \quad t \geq 0, \\
\int_{0}^{t}\| W^{n}(u) \|_{Z}du\leq
ke^{c_{2}t}\|w_{n}^{0} \|, \quad t \geq 0.
\end{gather}
In fact, if we put $M_{1}=\max\{M,\| L\|\}$, then we
get
$$
  \| W^{n}(t + \theta)\|_{Z} \; \leq \;M_{1}\|w_{n}^{0}
\|+M_{1}^{2} \int_{0}^{t}\|
  W_{s}^{n}\|_{L^{2}}ds;\quad \theta\in [-\tau,0],
$$
this implies
$$
\| W^{n}(t + \theta)\|_{Z}^{2}\leq \Big(
M_{1}\|w_{n}^{0} \| + M_{1}^{2}\int_{0}^{t}\|
W_{s}^{n}\|_{L^{2}}ds\Big)^{2}.
$$
Next,
\begin{align*}
  \int_{-\tau}^{0}\| W^{n}(t + \theta)\|_{Z}
^{2}d\theta
& \leq   \int_{-\tau}^{0}\Big(
 M_{1}\|w_{n}^{0} \|+M_{1}^{2}\int_{0}^{t}\|
 W_{s}^{n}\|_{L^{2}}ds\Big)^{2}d\theta \\
& \leq   \int_{-\tau}^{0}2^{2}\Big(
 M_{1}^{2}\|w_{n}^{0} \|^2 +M_{1}^{4}\Big(
 \int_{0}^{t}\|W_{s}^{n}\|_{L^{2}}ds\Big)^{2}\Big)d\theta \\
& =   2^{2}\tau M_{1}^{2}\|w_{n}^{0} \|^2+M_{1}^{4}
 \Big(\int_{0}^{t}\| W_{s}^{n}\|_{L^{2}}ds\Big)^{2}
 \int_{-\tau}^{0}d\theta  \\
& =   c_{2}^2 \|w_{n}^{0} \|^2 +c_{1}^2 \Big(
 \int_{0}^{t}\|W_{s}^{n}\|_{L^{2}}ds\Big)^{2} \\
& \leq   \Big( c_{2}\|w_{n}^{0} \|+c_{1}\Big(
\int_{0}^{t}\|W_{s}^{n}\|_{L^{2}}ds\Big)\Big)^{2}
\end{align*}
So that
$$
\| W_{t}^{n}\|_{L^{2}}\leq  c_{2}\|w_{n}^
{0} \|+c_{1}
\Big(\int_{0}^{t}\| W_{s}^{n}\|_{L^{2}}ds\Big)
$$
Therefore, applying  Gronwall's lemma we obtain
$$
\|
W_{t}^{n}\|_{L^{2}}\leq c_{2}e^{c_{1}t}\|w_{n}^{0} \|,\quad t\geq0.
$$
On the other hand, we obtain the estimate
\begin{align*}
  \| W^{n}(t)\|_{Z}
& \leq  \| T_{A_{n}}(t)w_{n}^{0}\|
 + \|\int_{0}^{t}T_{A_{n}}(t-s) L_{n}W^{n}(s+\cdot)ds\| \\
& \leq   M_{1}\|w_{n}^{0} \|+ M_{1}^{2}\int_{0}^{t}\| W^{n}
  (s+\cdot)ds\|\\
& \leq   M_{1}\|w_{n}^{0} \|+ M_{1}^{2}\int_{0}^{t}c_{1}e^{c_{2}t}
  \|w_{n}^{0} \|ds\\
& =  \big( M_{1} + \frac{M_{1}^{2}c_{1}}{c_{2}}e^{c_{2}t}\big)
  \|w_{n}^{0}\| \\
& \leq   ce^{c_{2} t}\|w_{n}^{0} \|,
\end{align*}
where $c= M_{1} + \frac{M_{1}^{2}c_{1}}{c_ {2}}$, $t\geq0$.
Finally, we get
$$
\int_{0}^{t}\| W^{n}(u)\|_{Z}du\leq ke^{c_{2}
t}\|w_{n}^{0} \|,\quad k=\frac{c}{c_{2}}, \;t\geq0.
$$
This completes the proof of the claim. \smallskip

 Now, we use the above inequalities:
\begin{align*}
\big\|  T_{n}(t) \begin{pmatrix}
  w^{0}_{n} \\
  w_{n}\end{pmatrix} \big\|^{2}
& =  \| W^{n}(t)\|_{Z}^{2}
  +\int_{-\tau}^{0}\| W^{n}(t + \tau)\|^{2}_{Z}d\tau \\
& =  \| W^{n}(t)\|_{Z}^{2}+ \int_{t-\tau}^{t}\| W^{n}(u)\|^{2}_{Z}du \\
& \leq  \| W^{n}(t)\|_{Z}^{2}+ \int_{0}^ {t}\| W^{n}(u)\|^{2}_{Z}du
   +\|w_{n}\|^{2}_{L^{2}}\\
& \leq  \big(c_{2}^2e^{2c_{2}t} + k^2e^{2c_{2}t} \big)\|w_{n}^{0} \|^2
  +  \| w_{n}\|^{2}_{L^{2}}\\
& \leq  g(t)2\big(\|w_{n}^{0}\|^2 + \|   w_{n}\|^{2}_{L^{2}}  \big),
\quad n\geq    1,2,\dots .
\end{align*}
Hence,
$$
\|  T_{n}(t) \| \leq g(t), \quad n\geq 1,2,\dots .
$$
Therefore, applying Lemma \ref{LG}, we obtain that $T(t)$ is
bounded and $\{T(t)\}_{t\geq0}$ is a strongly continuous semigroup
on the Hilbert space  $\mathbb{M}_{2}$, whose generator $\Lambda$
is given by
$$
\Lambda W=\sum_{n=1}^{\infty}\Lambda_{n}Q_{n}W,\quad W\in D(\Lambda),
$$
with
$$
D(\Lambda)=\big\{
W\in\mathbb{M}_{2}/\sum_{n=1}^{\infty}\|\Lambda_{n}Q_{n}
W\|^{2}<\infty \big\}
$$
and the spectrum $\sigma(\Lambda)$ of $\Lambda$ is given by
\begin{equation}\label{L66}
\sigma(\Lambda) = \overline{\cup_{n=1}^{\infty}
\sigma(\bar{\Lambda}_n)},
\end{equation}
where $\bar{\Lambda}_{n} = \Lambda_{n}Q_{n} :\mathcal{R}(Q_{n})
\to \mathcal{R}(Q_{n})$.
 \end{proof}

\begin{lemma} \label{lem4.2}
Let $\Lambda$ be the infinitesimal generator of the semi-group
$\{T(t)\}_{t\geq0}$. Then
$$
\Lambda\tilde{\varphi}(s) =
\begin{pmatrix}
   -A\varphi(0) + L\phi(s) \\
\frac{\partial\phi(s)}{\partial s}
\end{pmatrix}, \quad
-\tau\leq s \leq 0,
$$
\begin{align*}
D(\Lambda) = \big\{&\begin{pmatrix}
  \phi_{0} \\
  \phi(\cdot) \end{pmatrix}
\in \mathbb{M}_{2} : \phi_{0} \in D(A), \phi
\mbox{ is  a.c., }\frac{\partial\phi(s)}{\partial s} \in
L^{2}([-\tau,0];Z) \\
&\mbox{and }  \phi(0)= \phi_{0} \big\},
\end{align*}
and
$$
\sigma(\Lambda) = \overline{\cup_{n=1}^{\infty}\{ \lambda \in
\mathbb{C} : \mbox{det}(\Lambda_{n}(\lambda)) = 0 \}  }
$$
\end{lemma}

\begin{proof} Consider $\begin{pmatrix}
  \phi_{0} \\
  \phi(\cdot) \end{pmatrix}$ in $\mathbb{M}_{2}$. Then
\begin{align*}
  \Lambda W&=\Lambda \begin{pmatrix}
  \phi_{0} \\
  \phi(\cdot)\end{pmatrix}
  =  \sum_{n=1}^{\infty} \Lambda_{n}Q_{n}W  \\
& =  \sum_{n=1}^{\infty}\Lambda_{n}\begin{pmatrix}
  P_{n} & 0 \\
  0 & \widetilde{P}_{n}  \end{pmatrix}
 \begin{pmatrix}
  \phi_{0} \\
  \phi(\cdot) \end{pmatrix}
  = \sum_{n=1}^{\infty}\Lambda_{n}\begin{pmatrix}
  P_{n}\phi_{0} \\
  \widetilde{P_{n}}\phi(\cdot)
   \end{pmatrix} \\
& =  \sum_{n=1}^{\infty}\begin{pmatrix}
   -\Lambda_{n}D\widetilde{P_{n}}\phi(0) + L_{n}\widetilde{P_{n}}\phi\\
  \frac{\partial \widetilde{P}_{n}\phi(\cdot)}{\partial (s)}
  \end{pmatrix}\\
& =  \begin{pmatrix}
   -\sum_{n=1}^{\infty}\Lambda_{n}DP_{n}\phi(0) + L\sum_
{n=1}^{\infty}\widetilde{P}_{n}\phi\\
\frac{\partial}{\partial s}\Big(\sum_{n=1}^{\infty}
\widetilde{P}_{n}\phi(\cdot)\Big)
\end{pmatrix} \\
& = \begin{pmatrix}
  -A\phi(0) + L\phi(\cdot) \\
  \frac{\partial \phi(\cdot)}{\partial s}
\end{pmatrix}.
\end{align*}
The other part of the lemma follows from (\ref{L66})
\end{proof}

 Therefore, the systems (\ref{ecuacion diferencial abstracta homogenea})
and (\ref{ecuacion diferencial abstracta no homogenea}) are equivalent
 to the following two systems of ordinary di-fferential equations
in $\mathbb{M}_{2}$ respectively:
\begin{equation}\label{ecuacion ordinaria homogenea}
\begin{gathered}
 \frac{dW(t)}{dt} = \Lambda W(t) ,\quad t>0 \\
 W(0)  = W_{0} = (\phi_{0},\phi(\cdot))
 \end{gathered}
\end{equation}
and
\begin{equation}\label{ecuacion ordinaria no homogenea}
\begin{gathered}
 \frac{dW(t)}{dt} = \Lambda W(t) + \Phi(t),\quad t>0 \\
 W(0)   = W_{0} = (\phi_{0},\phi(\cdot)),
 \end{gathered}
\end{equation}
where $\Lambda$  is the infinitesimal generator of the semigroup
$\{T(t)\}_{t\geq0}$ and $\Phi(t) = (f^{e}(t),0)$.

 The steps we have taken to arrive here allow us to conclude
the proof of the main result of this work: The Variation of
Constants Formula for Functional Partial Parabolic Equations. This
result is presented as the final Theorem of the this work.

\begin{theorem} \label{thm4.3}
The abstract Cauchy problem in the Hilbert space
$\mathbb{M}_{2}$,
\begin{gather*}
 \frac{dW(t)}{dt} = \Lambda W(t) + \Phi(t),\quad t>0 \\
 W(0)   =  W_{0}
\end{gather*}
where $\Lambda$ is the infinitesimal generator of the semigroup
$\{T(t)\}_{t\geq0}$ and $\Phi(t) = (f^{e}(t),0)$ is a function
taking values in  $\mathbb{M}_{2}$, admits one and only one mild
solution given by
\begin{equation}\label{variacion de parametro}
W(t) = T(t)W_{0} + \int_{0}^{t}T(t-s)\Phi(s)ds
\end{equation}
\end{theorem}

\begin{corollary} \label{coro4.4}
If $z(t)$ is a solution of
\eqref{ecuacion diferencial abstracta no homogenea}, then the function
$W(t) := (z(t),z_{t}) $ is solution
of the equation \eqref{ecuacion ordinaria no homogenea}
\end{corollary}

\section{Conclusion}

As one can see, this work can be generalized to a broad class of
functional reaction diffusion equation in a Hilbert space $Z$ of
the form
\begin{equation}\label{ecuacion abstracta}
\begin{gathered}
 \frac{dz(t)}{dt} = {\mathcal{A}}z(t) + Lz_{t} + F (t),\quad t>0\\
 z(0) = \phi_{0} \\
 z(s) = \phi(s),\quad  s\in[-\tau,0),
 \end{gathered}
\end{equation}
where
\begin{equation}\label{L4}
{\mathcal{A}}z = \sum_{n=1}^{\infty} A_n P_n z, \quad
 z \in D({\mathcal{A}}),
\end{equation}
where $L : L^{2}([-\tau,0];Z) \to Z$ is linear and
bounded $F :[-\tau, \infty) \to Z$ is a suitable
function. Some examples of this class are the following well known
systems of partial differential equations with delay:


 The equation modelling a damped flexible beam:
\begin{equation}
\begin{gathered}
 {\partial^2 z \over \partial t^2}
 =-{\partial3 z \over \partial3 x} +2\alpha {\partial3 z \over
\partial t \partial2 x} + z(t-\tau,x) + f(t,x)  \quad t \geq 0, \;
0 \leq x \leq 1 \\
z(t,1)  =  z(t,0) = {\partial2 z \over
\partial2 x}(0,t)= {\partial2 z \over \partial2 x}(1,t)=0,\\
z(0,x)  =  \phi_{0}(x), \quad   {\partial z \over
\partial t}(0,x) = \psi_{0}(x), \quad 0 \leq x \leq 1 \\
z(s,x)  =  \phi(s,x), \quad {\partial z \over \partial t}(s,x) =
\psi(s,x), \quad s \in [-\tau, 0), \; 0 \leq x \leq 1 \\
\end{gathered}
\end{equation}
where $\alpha >0$, $f:\mathbb{R} \times [0,1] \to \mathbb{R}$ is a smooth
function, $\phi_{0}, \psi_{0} \in L^{2}[0,1]$ and
$\phi, \psi \in L^{2}([-\tau, 0]; L^{2}[0,1])$.


The strongly damped wave equation with Dirichlet boundary
conditions
\begin{equation}\label{W1}
\begin{gathered}
{\partial^2 w \over \partial t^2}+
\eta(-\Delta)^{1/2}{\partial w \over \partial t} + \gamma(-
\Delta) w  =  Lw_t + f(t,x),  \quad  t \geq 0, \; x \in \Omega, \\
w(t,x)  =  0, \quad t \geq 0, \; x \in  \partial \Omega.  \\
w(0,x) =\phi_{0}(x), \quad {\partial z \over \partial
t}(0,x) =\psi_{0}(x), \quad x \in \Omega, \\
w(s,x) =  \phi(s,x), \quad {\partial z \over \partial
t}(s,x) = \psi(s,x), \quad s \in [-\tau, 0), \; x \in \Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a sufficiently smooth bounded domain in $\mathbb{R}^N$,
$f:\mathbb{R} \times \Omega \to \mathbb{R}$ is a smooth function,
$\phi_{0}, \psi_{0} \in L^{2}(\Omega)$ and $\phi, \psi \in
L^{2}([-\tau, 0]; L^{2}(\Omega))$ and $\tau\geq 0$ is the maximum
delay, which is supposed to be finite. We assume that the
operators $L : L^{2}([-\tau,0];Z) \to Z$ is linear and
bounded and $Z = L^{2}(\Omega)$.


The thermoelastic plate equation with Dirichlet boundary
conditions
\begin{equation}\label{Th1}
\begin{gathered}
{\partial2 w \over \partial2 t}+ \Delta^{2}w + \alpha
\Delta \theta  = L_{1}w_t + f_{1}(t,x) \quad  t \geq 0, \; x \in \Omega, \\
{\partial \theta \over \partial t}- \beta \Delta \theta
- \alpha \Delta {\partial w \over \partial t}  =  L_{2}\theta_t + f_{2}(t,x)
\quad t \geq 0, \; x\in \Omega, \\
\theta = w  = \Delta w= 0, \quad t \geq 0, \; x \in \partial \Omega, \\
w(0,x) =\phi_{0}(x), \quad {\partial w \over \partial
t}(0,x) = \psi_{0}(x), \quad \theta(0,x) =\xi_{0}(x) \quad x \in \Omega, \\
w(s,x) =  \phi(s,x),  \quad  {\partial w \over
\partial t}(s,x) = \psi(s,x), \quad \theta(0,x) =\xi(s,x), \quad
 s \in [- \tau, 0), \; x \in \Omega,
\end{gathered}
\end{equation}
where $\Omega$ is a sufficiently smooth bounded domain in $\mathbb{R}^N$,
$f_1, f_2 :\mathbb{R} \times \Omega \to \mathbb{R}$ are smooth functions,
$\phi_{0}, \psi_{0}, \xi_{0} \in L^{2}(\Omega)$ and $\phi, \psi,
\xi \in L^{2}([-\tau, 0]; L^{2}(\Omega))$ and $\tau\geq 0$ is the
maximum delay, which is supposed to be finite. We assume that the
operators $L_1, L_2 : L^{2}([-\tau,0];Z) \to Z$ are
linear and bounded and $Z = L^{2}(\Omega)$.

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\end{document}
