\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 104, pp. 1--23.\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 2003 Texas State University-San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE--2003/104\hfil Linear FDE's with distributions]
{A linear functional differential equation
with distributions in the input}
\author[Vadim Z. Tsalyuk\hfil EJDE--2003/104\hfilneg]
{Vadim Z. Tsalyuk}
\address{Vadim Z. Tsalyuk\hfill\break
Mathematics Department, Kuban State University,
Stavropol'skaya 149, Krasnodar 350040, Russia}
\email{vts@math.kubsu.ru}
% http://public.kubsu.ru/vts (in Russian)
\date{}
\thanks{Submitted June 14, 2003. Published October 13, 2003.}
\thanks{Partialy supported by grant 03-01-00255 from the Russian
Foundation for Basic Research.}
\subjclass[2000]{26A42, 28B20, 34A60}
\keywords{Stieltjes integral, function of bounded variation,
multivalued integral, \hfill\break\indent
linear functional differential equation}
\begin{abstract}
This paper studies the functional differential equation
$$
\dot x(t) = \int_a^t {d_s R(t,s)\, x(s)} + F'(t), \quad t \in [a,b],
$$
where $F'$ is a generalized derivative, and
$R(t,\cdot)$ and $F$ are functions of bounded variation.
A solution is defined by the difference $x - F$
being absolutely continuous and satisfying the inclusion
$$
\frac{d}{dt} (x(t) - F(t)) \in \int_a^t {d_s R(t,s)\,x(s)}.
$$
Here, the integral in the right is the multivalued Stieltjes integral
presented in \cite{VTs1} (in this article we review and extend the results
in \cite{VTs1}).
We show that the solution set for the initial-value problem is
nonempty, compact, and convex.
A solution $x$ is said to have memory if there exists the
function $\bar x$ such that $\bar x(a) = x(a)$, $\bar x(b) = x(b)$,
$\bar x(t) \in [x(t-0),x(t+0)]$ for $t \in (a,b)$, and
$\frac{d}{dt} (x(t) - F(t)) = \int_a^t {d_s R(t,s)\, \bar{x}(s)}$,
where Lebesgue-Stieltjes integral is used.
We show that such solutions form a nonempty, compact, and convex
set. It is shown that solutions with memory obey the Cauchy-type
formula
$$
x(t) \in C(t,a)x(a) + \int_a^t C(t,s)\, dF(s).
$$
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem {theorem} {Theorem}[section]
\newtheorem {corollary}[theorem]{Corollary}
\newtheorem {prop}[theorem]{Proposition}
\newtheorem {lemma}[theorem]{Lemma}
\newtheorem {definition}[theorem]{Definition}
\newtheorem {remark}[theorem]{Remark}
\newtheorem {example}[theorem]{Example}
\section{Introduction: Model example}
Our purpose is to generalize the linear functional differential
equation
\begin{equation}\label{OrdFDE}
\dot x(t) = \int_a^t d_sR(t,s)\,x(s) + f(t), \quad t \in [a,b],
\end{equation}
introducing a distribution of order zero, otherwise a measure,
as $f$.
Such a kind of inputs for systems described by ordinary
differential equations were studied in many works; see for
instance \cite{Hal}.
The functions $R(t,\cdot)$ in the equation (\ref{OrdFDE}) are usually discontinuous
and have bounded variation. If $f$ is a measure then it is natural to suppose that
a solution $x$ belongs to a class of primitives for measures, i.e. it is also
discontinuous and has bounded variation on $[a,b]$.
To illustrate the question, we consider the model problem
\begin{equation}\label{model}
\begin{array}{l}
\dot x(t) = h(t-1) \cdot x(1) + \delta_1, \quad t \in [0, 2],\\
x(0) = 0,
\end{array}
\end{equation}
where $h$ is the Heaviside function ($h(t) = 0$ for $t < 0$ and $h(t) = 1$ for
$t > 0$), $\delta_1 = h'(t-1)$ is the unit impulse (so-called delta-function)
concentrated at the point 1.
The behaviour of the system for $t > 1$ essentially depends on the value $x(1)$.
Naturally, we assume the solution $x(t)$ of (\ref{model}) being equal to $0$
for $t < 1$ and having a unit jump at $t = 1$. Then what is $x(1)$?
Is it $x(1-0)=0$, or $x(1+0)=1$, or some intermediate value? The system do
not know how to behave further. Such a phenomenon frequently occurs in the
theory of functional differential equations and is known as ``hovering" \cite{Azb}.
To avoid hovering, Anokhin \cite{Ano,Azb}
assumed the solution of (\ref{model}) to be piecewise absolutely continuous,
having an unique jump at the point 1, and continuous from the right.
Then $x(1)$ in (\ref{model}) is well defined and, clearly, the initial
value problem has the unique solution
$$
x_+(t) = \begin{cases}
0, & t < 1,\\
t, & t \geq 1. \end{cases}
$$
If we want the solution to be continuous from the left, as in \cite{AnB},
then we get
$$
x_-(t) = \begin{cases}
0, & t \leq 1,\\
1, & t > 1. \end{cases}
$$
Let us introduce a perturbation of the right-hand side of (\ref{model}).
Exactly, we replace the equation by
\begin{equation}\label{model c}
\dot x(t) = h(t-c) \cdot x(c) + \delta_1, \quad c \neq 1.
\end{equation}
From the point of view of certain applications, such a perturbation looks
small as $c \to 1$ .
In the case $c > 1$, the solution of (\ref{model c}) has the form
$$
x_c(t) = \begin{cases}
0, & t < 1,\\
1, & 1 < t \leq c,\\
t+1-c, & t > c, \end{cases}
$$
and for $c < 1$ we have
$$
x_c(t) = \begin{cases}
0, & t < 1,\\
1, & t > 1. \end{cases}
$$
Now let $c$ tend to $1$. Then $x_c \to x_+$ as $c \downarrow 1$ and
$x_c \to x_-$ as $c \uparrow 1$.
So we have detected the discontinuity. In the theory of discontinuos systems,
such a phenomenon is usually treated as follows. First, the so-called
"nonideality" (the regularization) is introduced. The regularized equation
ought to be of well-known class of systems and uniquely solvable. Next,
the size of nonideality is assumed to tend to $0$. Any limit of the solutions is
said to be a solution of the initial "ideal" system. Thus, the uniqueness may
be lost in these situations. We shall follow this way.
Let us replace the unit impulse $\delta_1$ by the ``delta-like sequence"
(${\varepsilon} > 0$) \linebreak
$f_{\varepsilon}(t) \ge 0$, $f_{\varepsilon}(t) = 0$ for
$t \in (-\infty,1 - {\varepsilon}) \cup (1 + {\varepsilon} ,\infty)$,
$\int_{1 - {\varepsilon}}^{1 + {\varepsilon}} {f_{\varepsilon}(t)\,dt = 1}$,
and put $\alpha _{\varepsilon} = \int_{1 - {\varepsilon} }^1 {f_{\varepsilon}(t)\,dt}$. The solutions
of the problems
\begin{gather*}
\dot x(t) = h(t - 1)x(1) + f_{\varepsilon} (t), \\
x(0) = 0
\end{gather*}
are as follows:
$$
x_{\varepsilon} (t) = \begin{cases}
0, & 0 \le t \le 1 - \varepsilon , \\
\int_{1 - {\varepsilon} }^t {f_{\varepsilon} (s)\,ds},
& {1 - {\varepsilon} < t \le 1,} \\
{\int_{1 - {\varepsilon} }^t {f_{\varepsilon} (s)\,ds} +
\alpha _{\varepsilon} \cdot (t - 1),} & {1 < t \le 1 + {\varepsilon} ,} \\
{1 + \alpha _{\varepsilon} \cdot (t - 1),} & {t > 1 + {\varepsilon}.}
\end{cases}
$$
Let ${\varepsilon}$ tend to $0$. We see that, in order to the convergence of solutions,
it is necessary that $\alpha_{\varepsilon} \to \alpha \in [0,1]$.
On the other hand, each value in $[0,1]$ may be the limit of $\alpha_{\varepsilon}$.
So, if we want the limit of solutions to be a solution, the problem (\ref{model c})
gets a family of solutions
$$
x(t) = \begin{cases}
{0,} & {0 \le t < 1,} \\
{\alpha ,} & {t = 1,} \\
{1 + \alpha \cdot (t - 1),} & {t > 1.}
\end{cases}
$$
with arbitrary $\alpha \in [0,1]$.
Another kind of the ``nonideality" represents a model of inaccuracy of the
measuring device for $x(1)$ -- or our inperfect knowledge of this device.
Replace the operator $h(t - 1)x(1)$ in (\ref{model}) with one sufficiently close
to. Let $\alpha,\beta \ge 0$, $\alpha + \beta = 1$, and set
$$
({\bf K}_{\varepsilon} x)(t) = \begin{cases}
{0,} & {0 \le t \le 1 - \beta {\varepsilon} ^2 ,} \\
{{\varepsilon} ^{ - 2} \int_{1 - \beta {\varepsilon} ^2 }^t {x(s)\,ds} ,} &
{1 - \beta {\varepsilon} ^2 < t \le 1 + \alpha {\varepsilon} ^2 ,} \\
{{\varepsilon} ^{ - 2} \int_{1 - \beta {\varepsilon} ^2}^{1 + \alpha {\varepsilon} ^2 } {x(s)\,ds} ,} &
{t > 1 + \alpha {\varepsilon} ^2.}
\end{cases}
$$
The solution of the problem
\begin{gather*}
\dot x(t) = ({\bf K}_{\varepsilon} x)(t) + \delta_1 , \\
x(0) = 0
\end{gather*}
tends to
$$
x(t) = \begin{cases}
{0,} & {0 \le t < 1,} \\
{1 + \alpha \cdot (t - 1),} & {t > 1,}
\end{cases}
$$
as ${\varepsilon} \downarrow 0$.
Naturally we consider these functions with arbitrary $\alpha \in [0,1]$ as the
solutions of problem (\ref{model}).
Assuming that the value $\alpha \in [0,1]$ is chosen for each $t > 1$
independently, we get the solution family
$$
x(t) = \begin{cases}
{0,} & {0 \le t < 1,} \\
{1 + \int_1^t {\alpha (s)\,ds} ,} & {t > 1,}
\end{cases}
$$
with summable functions $\alpha(s) \in [0,1]$.
Eventually, in equation (\ref{model}) we have to mean the whole segment
$[0,1] ={}$ \linebreak
${[x(1 - 0),x(1 + 0)]}$ instead of $x(1)$. And solutions of (\ref{model})
ought to satisfy somewhat like inclusion.
\section{Preliminaries}
\subsection{Some notions and notation}
{\bf 1.} Functions $x$ are assumed further to have values in the
finite-dimensional space $\mathbb{R}^{N}$. The space for values of $R$ is the
space of matrices
$\mathbb{R}^{N \times N}$. Consequently products $R(t,s)x(s)$
and values of integrals will lie in the space $\mathbb{R}^{N}$.
All the norms in these spaces are denoted by $| \cdot |$ and we assume that the
norm of matrix $|g| = \sup_{|x| = 1} |gx|$.
The product sets $[a,b] \times \mathbb{R}$, $[a,b] \times \mathbb{R}^{N}$, etc, are
equipped with the metric
$$\rho \left( (t_1, x_1),\: (t_2, x_2) \right) =
\max \left( |t_1 - t_2|, |x_1 - x_2| \right).$$
{\bf 2.} The \textit{\textbf{distance}} from a point $x$ to a set $A$ is
$\rho(x,A) {\stackrel{\scriptstyle {\rm df}}{=}} \sup_{a \in A} | x-a |$.
We denote $\beta(A,B) {\stackrel{\scriptstyle {\rm df}}{=}} \sup_{a \in A} \rho(a,B)$.
The symmetric function
$$
\alpha(A,B) = \max(\beta(A,B),\beta(B,A))
$$
on the space ${\mathop{\rm Comp}(\mathbb{R}^N)}$ of nonempty compact subsets of
finite-dimensional space is the metric named Hausdorff distance. This metric
space is complete. We use also the \textit{\textbf{triangle inequality}} of
the form
$$
\beta(A,C) \leq \beta(A,B) + \beta(B,C)
$$
that holds for compact sets $A$, $B$ and $C$.
The sequence of sets $A_n$ is said to converge to a set $A$
if $\alpha(A_n, A) \to 0$ as $n\to\infty$.
For compact sets $A$, $A_n$, it is easy to see that the convergence
$$
\beta\left( A_n, A\right) \to 0
$$
is equivalent to what follows:
for every sequence $a_n \in A_n$
\begin{itemize}
\item[i)] this sequence is compact;
\item[ii)] if $\lim_{i \to \infty} a_{n_i} = a$ for subsequence $a_{n_i}$ then $a \in A$.
\end{itemize}
We denote
${\|A\| {\stackrel{\scriptstyle {\rm df}}{=}} \sup_{v \in A}|v|=\alpha(A,0)}$.
{\bf 3.} For $u$ and $v$ being points of a linear space, by $[u,v]$ we denote
the convex hull of these two points.
If $x$ is a function defined on
$[a,b] \subset \mathbb{R}$ and $t \in (a,b)$ then
$$\Delta x(t) {\stackrel{\scriptstyle {\rm df}}{=}} x(t+0)-x(t-0)$$
is a \textit{\textbf{jump}} of $x$ at the point $t$.
We say that the function $g$ of bounded variation is \textit{\textbf{proper}} if
$$g(t) \in [g(t-0),g(t+0)]$$
for all $t \in (a,b)$. In this case, in the definition
${\mathop{\rm Var}\nolimits}_a^b g = \sup\sum |g(t_i)-g(t_{i-1})|$
of the variation of $g$ we may assume that each point $t_i$ not coinciding with
the end of the segment $[a,b]$ is a point of continuity of $g$.
By $\bar{\Gamma}(g)$ denote a \textit{\textbf{completed graph}} of function $g$, i.e.
the set in $[a,b] \times \mathbb{G}$ that consists of the segments
$\{ a \} \times {[g(a),g(a+0)]}$, $\{ b \} \times {[g(b-0),g(b)]}$,
and $ \{ t \} \times {[g(t-0),g(t+0)]}$ for $t \in {(a,b)}$. It's obvious that
for function $g$ of bounded variation $\bar{\Gamma}(g)$ is closed set.
If $\bar{\Gamma}(g_1) = \bar{\Gamma}(g_2)$ then we say that the functions $g_1$
and $g_2$ of bouned variation are \textit{\textbf{equivalent}}. In this case for $x \in {{\bf C}[a,b]}$
(${{\bf C}[a,b]}$ is the space of continuous functions) we have
$\int_a^b dg_1(t) x(t) = \int_a^b dg_2(t) x(t)$. In the case of $N=1$
the norm of this functional is equal to ${\mathop{\rm Var}\nolimits}_a^b g$ with proper $g$
taken from the equivalence class of $g_1$ and $g_2$.
We denote by ${{\bf BV}[a,b]}$ the space of equivalence classes of the functions of bounded
variation. We define ${\mathop{\rm Var}\nolimits}_a^b g$ for equivalence
class using a proper element of the class. We say that $t \in (a,b)$ is a
\textit{\textbf{\mbox{(dis-)}continuity point}} of $g \in {{\bf BV}[a,b]}$ if the proper element of the
equivalence class is \mbox{(dis-)}continuous at $t$. In the points of continuity
the \textit{\textbf{value}} $g(t)$ is defined as the value of the proper element. We say
that the sequence of $g_n \in {{\bf BV}[a,b]}$ \textit{\textbf{uniformly converges}} to $g$ if the
proper representatives of $g_n$ converge to proper element of $g$ uniformly
for all $t \in [a,b]$ except for discontinuity points of $g$ in $(a,b)$.
{\bf 5.} Anywhere below the integration variable may be omitted if it does
not cause uncertainty.
{\bf 6.} Let $T$ be a measurable subset of the finite-dimensional space. The multivalued
function $X:T \to {\mathop{\rm Comp}(\mathbb{R}^N)}$ is \textit{\textbf{measurable}} if for each $A \in {\mathop{\rm Comp}(\mathbb{R}^N)}$
the set $\{ t \in T : X(t) \cap A \ne \emptyset \}$ is measurable.
The detailed review \cite{Cas} contains some other definitions of the
measurability of multifunctions, conditions for the equivalence of the
definitions, and various properties of measurable multifunctions.
See also \cite[pp. 148--150]{Bor} and \cite[pp. 205--206]{BlF}.
Particularly, $X$ is measurable if and only if there exists a countable set
of measurable singlevalued functions $x_i\,:\,T \to {\mathbb{R}}^N$ such that
$$
X(t) = \overline {\{ x_i(t)\,:\,i = 1,2, \ldots\} } \quad\mbox{a.e.}
$$
Using this, one can easily check that if
$x_0,x_1 :T \to {\mathbb{R}}^N$ are measurable singlevalued functions then
the multifunction $t \mapsto [x_0(t),x_1(t)]$ is measurable. If the series
$ X(t) = \sum_{i = 1}^\infty {X_i (t)} $ has a convergent in ${{\bf L}_1(a,b)}$ majorant
then the multifunction $X$ is measurable \cite[Theorem 21]{BlF}.
{\bf 7.} The \textit{\textbf{Aumann integral}} of the multifunction $X:[a,b] \to {\mathop{\rm Comp}(\mathbb{R}^N)}$
is, by definition, the set
$$
\int_a^b {X(t)\,dt} = \Big\{ \int_a^b {x(t)\,dt} :
x \in {{\bf L}_1(a,b)},\quad x(t) \in X(t) \mbox{ for a.e. }t \in [a,b]\Big\}.
$$
{\bf 8.} We say that multifunction $X$ is \textit{\textbf{boundedly summable}}
if it is measurable and
$\| {X(t)}\| \le \chi (t) $ a.e. on $[a,b]$ for some summable
(singlevalued) function $\chi$. If $X$ is such a multifunction then according to the
theorem of A.\,A. Lyapunov ( \cite[Theorem 1.7.10]{Bor}, \cite{BlF}) the set
$\int_a^b X(t)\,dt$ is nonepty, convex, and compact;
besides,
$$
\int_a^b X(t)\,dt = \int_a^b \overline {{\mathop{\rm co}} X(t)}\,dt
$$
(${\mathop{\rm co}} X$ is a convex hull of $X$).
{\bf 9.} The \textit{\textbf{support function}} of the set $A \in {\mathop{\rm Comp}(\mathbb{R}^N)}$ is
$$
{\mathop{\rm supp}}(\psi |\, A ) := \sup_{a \in A} \psi \cdot a,
$$
where $\psi \cdot a$ denotes the usual scalar product in ${\mathbb{R}}^N$.
Sufficiently complete review of the properties of the support functions
may be found in \cite[pp. 197--200]{BlF}.
The convex hull ${\mathop{\rm co}} A$ of $A \in {\mathop{\rm Comp}(\mathbb{R}^N)}$ is uniquely determined
by the values of the support functions ${\mathop{\rm supp}}(\psi |\, A )$
with $\left| \psi \right| = 1$:
$$
{\mathop{\rm co}} A = \bigcap_\psi {\{ x:\: \psi \cdot x \le
{\mathop{\rm supp}}(\psi |\, A )\}}.
$$
Note that it is sufficient here that $\psi$ ranges over the countable
dence subset of the unit sphere.
The multifunction $X: T \to {\mathop{\rm Comp}(\mathbb{R}^N)}$
is measurable if and only if
the mappings $t \mapsto {\mathop{\rm supp}}(\psi |\, X(t))$ are measurable
\cite[Theorem 22]{BlF}. For $X$ being boundedly summable, according
to \cite[Theorem 42]{BlF} we have
\begin{equation}\label{(8)}
{\mathop{\rm supp}}\Big( {\psi \Big| {\int_a^b {X(t)\,dt} } }
\Big) = \int_a^b {{\mathop{\rm supp}}(\psi \big| {X(t)} )\,dt}.
\end{equation}
\subsection{Auxiliary lemmas}
The next two lemmas may be known, but the author did not find a reference
to them.
\begin{lemma}[Lebesgue-type theorem on convergence of integrals]\label{L3}
Suppose $X$ and $X_n$, $n = 1,2, \ldots$ are bounded summable
multivalued functions on $[a,b]$ with values in
${\mathop{\rm Comp}(\mathbb{R}^N)}$;
$\| {X_n (t)}\| \le \chi (t)$ a.e. on $[a,b]$, $\chi \in {{\bf L}_1(a,b)}$.
If $\beta (X_n (t),X(t)) \to 0$ for a.e. $t \in [a,b]$ as $n \to \infty$
then
\begin{equation}\label{(9)}
\beta \Big( \int_a^b {X_n (t)\,dt}, \, \int_a^b
{X(t)\,dt} \Big) \to 0.
\end{equation}
\end{lemma}
\begin{proof}
Let $ | \psi| = 1 $. Obviously,
$$
{\mathop{\rm supp}}(\psi \left|\, {X_n (t)} \right.) \le {\mathop{\rm supp}}(\psi \left|\, {X(t)}
\right.) + \beta (X_n (t),X(t)).
$$
Integrating we get
$$
\int_a^b {\mathop{\rm supp}}(\psi \left|\, X_n (t) \right.)\,dt \le
\int_a^b {\mathop{\rm supp}}(\psi \left|\, {X(t)} \right.)\,dt +
\int_a^b \beta (X_n (t),X(t))\,dt
$$
(since $\beta (X_n (t),X(t)) \le \left\| {X_n (t)} \right\| +
\| X(t)\|$, the latter integral exists).
According to (\ref{(8)}) and the conventional Lebesgue theorem,
\begin{align*}
\limsup_{n \to \infty} \mathop{\rm supp}\Big(
\psi \Big|\, \int_a^b & {X_n(t)\, dt} \Big) \\
%& = \limsup_{n \to \infty }
%\int_a^b {{\mathop{\rm supp}}(\psi \left|\, {X_n (t)} \right.)\,dt} \\
&\le \limsup_{n \to \infty } \Big(
{\int_a^b {\mathop{\rm supp}}(\psi \big|\, {X(t)})\,dt +
\int_a^b {\beta (X_n (t),X(t))\,dt} } \Big)\\
& \le \int_a^b {{\mathop{\rm supp}}(\psi \left|\, {X(t)} \right.)\,dt} \\
& \le {\mathop{\rm supp}}\Big( {\psi \Big|\, {\int_a^b {X(t)\,dt} }} \Big).
\end{align*}
Consider an arbitrary sequence $v_n \in \int_a^b {X_n (t)\,dt}$.
Since
$\| {\int_a^b {X_n (t)\,dt} } \| \le \int_a^b {\chi (t)\,dt}$,
this sequence is compact. Furthermore, if
$\lim_{i \to \infty } v_{n_i} = v$ then
$$
\psi \cdot v \le \lim_{i \to \infty } \psi
\cdot v_{n_i} \le \limsup_{i \to \infty }
{\mathop{\rm supp}}\Big( {\psi \Big|\, {\int_a^b {X_{n_i} (t)\,dt} }}
\Big) \le {\mathop{\rm supp}}\Big( {\psi \Big|\, {\int_a^b {X(t)\,dt} }
} \Big)
$$
for every $\psi$ such that $|\psi| = 1$. Consequently,
$v \in \int_a^b X(t)\,dt$. These two facts combined are equivalent
to (\ref{(9)}).
\end{proof}
Note that the analoguous assertion with $\beta (A,B)$ replaced by
$\alpha (A,B)$ is cited in \cite{BlF} (Theorem 45 of the review).
\begin{lemma}[Differentiability of multivalued integrals]\label{L4}
If $X:[a,b] \to {\mathop{\rm Comp}(\mathbb{R}^N)}$ is boundedly summable then
for a.e. $t \in [a,b)$
\begin{equation}\label{(10)}
\beta \Big( {\frac{1}{{\Delta t}}\int_t^{t + \Delta t}{X(s)\,ds},
{\mathop{\rm co}} X(t)} \Big) \to 0 \quad\mbox{as } \Delta t \downarrow 0.
\end{equation}
\end{lemma}
\begin{proof}
Let $\Psi$ be a countable dense subset of the unit sphere. A convex set
$A \in {\mathop{\rm Comp}(\mathbb{R}^N)}$ is uniquely determined by
the values
${\mathop{\rm supp}}(\psi \big| A )$, $ \psi \in \Psi$:
$$
A = \bigcap_{\psi \in \Psi } {\{ x:\psi \cdot x \le
{\mathop{\rm supp}}(\psi \big| A )\} }.
$$
Since ${\mathop{\rm supp}}(\psi \big| X(t) ) \le \| X(t)\|$,
the functions $t \mapsto {\mathop{\rm supp}}(\psi \left| X(t) \right|)$ are
summ\-able. Due to (\ref{(8)}), for $t \in E$, $E \subset [a,b)$,
${\mathop{\rm mes}\nolimits}([a,b]\backslash E) = 0$, and $\psi \in \Psi$ we have
$${\mathop{\rm supp}}\Big( {\psi \Big|
\frac1{\Delta t} \int_t^{t + \Delta t} X(s)\,ds }
\Big) =
\frac1{\Delta t} \int_t^{t + \Delta t} {\mathop{\rm supp}}(\psi \big| X(s) )\,ds \to
{\mathop{\rm supp}}(\psi \big| X(t) )
$$
and
\begin{equation}\label{(11)}
\frac1{\Delta t} \int_t^{t + \Delta t} {\left\|
{X(s)} \right\| \,ds} \to \| X(t)\| \quad \mbox{as } \Delta t \downarrow 0.
\end{equation}
We assume in the sequel that $t \in E$.
Let $\Delta t_n \downarrow 0$ and
$v_n \in {\frac1{\Delta t_n}} \int_t^{t + \Delta t_n } X(s)\,ds$.
Due to (\ref{(11)})
\begin{itemize}
\item[i)] the sequence $\{ v_n \}$ is compact.
\end{itemize}
Besides,
\begin{itemize}
\item[ii)] if $v_{n_i} \to v$ as $i \to \infty$ then $v \in {\mathop{\rm co}} X(t)$.
\end{itemize}
Indeed, for $\psi \in \Psi$ we have
$$
\psi \cdot v =
\lim_{i \to \infty } \psi \cdot v_{n_i} \le
\lim_{i \to \infty } {\mathop{\rm supp}}
\Big({\psi \Big|\,{\frac{1}{{\Delta t_{n_i} }}\int_t^{t + \Delta t_{n_i}}
{X(s)\,ds} } } \Big) =
{\mathop{\rm supp}}\left( {\psi \big| X(t) } \right), $$
and therefore
$v \in \overline{{\mathop{\rm co}} X(t)} = {\mathop{\rm co}} X(t)$
(we recall that the set $X(t)$ is compact).
The assertions a) and b) above, being combined, give (\ref{(10)}).
\end{proof}
\section{The multivalued Stieltjes integral}
This section containes a review of the results of the paper \cite{VTs1}.
We need the concept of the multivalued integral
\begin{equation}\label{int1}
Ax = \int_a^b dg(s)\,x(s)
\end{equation}
for $g$ and $x$ being both discontinuous functions of bounded variation.
Note that the Riemann--Stieltjes integral
$$
Ax = \int_a^b dg(t)\,x(t) = g(b)x(b) - g(a)x(a) - \int_a^b g(t)\,dx(t)
$$
exists for {\bf continuous} function $g$. We define the integral (\ref{int1}) by
passing to the limit $g_i \to g$ where $g_i$ are continuous.
From the point of view of applications, it is natural to interpret a small error on
the moment of jump of $g$ as a small deviation of the function.
Therefore we construct the integral (\ref{int1}) using a convergence
$$
A_ix=\int_a^b dg_i(t)\,x(t) \to Ax
$$
almost similar to the pointwise convergence of functionals ${A_i \to A}$ in
the space ${{\bf C}[a,b]}$.
Focusing on the model equation (\ref{model}), we are about to costruct the
integral $Ax = \int_0^2 {dg(t)\,x(t)}$ for functions
$$
g(t) = \begin{cases}
0,& t \in [0,1), \\
1,& t \in (1,2],\end{cases}
\quad\mbox{and}\quad
x(t) = \begin{cases}
x_0, & t \in [0,1), \\
x_1, & t \in (1,2].\end{cases}
$$
For an arbitrary $a \in \mathbb{R}$ and each $i = 1,2,3, \ldots$, define
the continuous piecewise linear functions ${g_i(t)}$ through the vertices
$(0,0)$,
${(1 - \frac{1}{i},0)}$, ${(1,a)}$, ${(1 + \frac{1}{i},1)}$ and ${(2,1)}$.
Then, by the first Helly Theorem \cite{Kol}, the functionals $A_i$ pointwise
converge to $Ax = x(1)$ in ${\bf C}[0,2]$.
The function $x$ under the consideration, however, has only one-sided limits
at the point 1, and it is easy to check that $A_i x = x_1 - a(x_1 - x_0)$.
So $\lim_{i \to \infty } A_i x$ depends on the way
of construction of continuous approximations $g_i$. The common decision in such
a situation is to declare the functional $A$ to be {\bf multivalued} and the set
$Ax$ containing {\bf all} such limits. It is important for the theory
of equations that $Ax$ is a {\bf compact} set. But here, according to the
arbitrariness of $a$, we get $Ax = (-\infty, \infty)$ in the case
$x_1 \ne x_0$. Of course it's inadmissible.
Note that $\| A_i\| = {\mathop{\rm Var}\nolimits}_0^2 g_i =
|a| + {|a-1|}$ converge to ${\left\| A \right\|} = 1$ if and only if
$a \in [0,1]$. Under this restriction, we get
${\lim_{i \to \infty } A_i x} = ax_0 + {(1 - a)x_1} \in {[x_0 ,x_1]}$.
The set ${[x_0 ,x_1 ]} = {[x(1 - 0),x(1 + 0)]}$ of all such limits should be
considered as the value of $Ax = \int_0^2 {dg(t)\,x(t)}$. (It should be noted
that the functions $x$ and $g$ are not assumed to have certain values at their
discontinuity points.) This is why the convergence we use implies also the
convergence of the norms of functionals. It is natural to name it
\textit{\textbf{normal convergence}}.
On the other hand, the graphs of the functions $g_i$ converge to the graph of
$g$ completed with the ``vertical" segments at the points of discontinuity.
Such a type of convergence may also be used to define the integral (\ref{int1}).
For the completeness of the exposition, we give here the summary of
the paper \cite{VTs1} devoted to the introduction and the properties of such
a kind of integral.
\subsection{On the convergence}
\begin{definition} \rm \label{conv}
The sequence of functions $g_n \in {{{\bf BV}[a,b]}}$
$\beta$-\textit{\textbf{converges}}
on the segment $[a,b]$ to the function $g \in {{{\bf BV}[a,b]}}$ as $n \to \infty$ if
the following conditions hold:
\begin{itemize}
\item[1)] $g_n(a) \to g(a)$, $g_n(b) \to g(b)$;
\item[2)] $\beta \left(\bar{\Gamma}(g_n), \bar{\Gamma}(g) \right) \to 0$.
\end{itemize}
Such a convergence is denoted as $g_n {\stackrel {\beta} {\to} } g$.
If, additionally,
\begin{itemize}
\item[3)]
\end{itemize}
\begin{equation}\label{V to V}
{\mathop{\rm Var}\nolimits}_a^b g_n \to {\mathop{\rm Var}\nolimits}_a^b g,
\end{equation}
then we say that $g_n$ \textit{\textbf{normally converge}} to $g$ and denote
this by $g_n {\stackrel{N}{\to}} g$.
\end{definition}
\begin{lemma}\label{lemma2}
Let functions $g_n \in {{\bf BV}[a,b]}$ normally converge to a function
$g \in {{\bf BV}[a,b]}$.
\begin{itemize}
\item[a)]
For $\tau \in (a,b)$, the distance from the point $g_n(\tau)$ to the segment\\
${[g(\tau-0),g(\tau+0)]}$ tends to $0$ as $n \to \infty$.
\item[b)]
If $g$ is continuous at the point $a$ then $\lim_{n \to \infty} g_n(a+0) = g(a)$;
the analoguous statement takes place for the point $b$.
\item[c)]
If $g$ is continuous on $[a,b]$ then $g_n(t) \to g(t)$ uniformly for
$t \in [a,b]$.
\end{itemize}
\end{lemma}
\begin{lemma}\label{lemma3}
Let $g_n$ normally converge on $[a,b]$ to $g$.
If $c \in (a,b)$ is a continuity point of $g$ then
$g_n \to g$ normally on both intervals $[a,c]$ and $[c,b]$.
\end{lemma}
\begin{lemma}\label{lemma4}
Let $\left\{ f_n \right\}$ and $\left\{ g_n \right\}$ be the sequences of functions
normally converging on $[a,b]$ to $g$. Then for arbitrary $\lambda \in (0,1)$ the
functions $\lambda f_n + ({1 - \lambda })g_n$ normally converge to the function $g$.
\end{lemma}
\begin{prop}\label{quasi exist}
Any proper function of bounded variation on $[a,b]$ is a normal limit of
some sequence of continuous functions.
\end{prop}
\begin{proof}
For arbitrary $\delta > 0$ take a finite partition $a = t_0 < t_1 < \dots < t_n = b$
such that
\begin{itemize}
\item[1)]
$\sum |g(t_i)-g(t_{i-1})| > {\mathop{\rm Var}\nolimits}_a^b g - \delta$,
\item[2)]
$\max_{1 \leq i \leq n} (t_i - t_{i-1}) < \delta$,
\item[3)]
$\max_{1 \leq i \leq n} {\mathop{\rm Var}\nolimits}_{t_{i-1}+0}^{t_i-0} g < \delta$,
\item[4)]
at least one in the pair of the neighbour points $t_{i-1}$
and $t_i$ is the continuity point of $g$.
\end{itemize}
To satisfy each of the conditions 1)--4), we may add to the partition a
finite set of points so that the previous conditions stay fulfilled.
Let $l_{\delta}(t)$ be a piecewise linear function built through
the points $\left(t_i,g(t_i)\right)$. Hence
${\mathop{\rm Var}\nolimits}_a^b l_{\delta} \to {\mathop{\rm Var}\nolimits}_a^b g $ as $\delta \to 0$.
The inequality
$\beta \left( \bar{\Gamma}(l_{\delta}),\: \bar{\Gamma}(g) \right) \leq \delta$
is proved in \cite{VTs1}.
\end{proof}
\begin{remark} \rm
The approximations $l_{\delta}(t)$ constructed here are absolutely continuous.
\end{remark}
\smallskip
\subsection{Definition and properties of the multivalued Stieltjes integral}
Let $x,g \in {{\bf BV}[a,b]}$. The set $\int_a^b dg(t)\,x(t)$ is defined as
follows.
\begin{definition}\label{df3} \rm
The point $v$ belongs to the set $\int_a^b dg\,x$ if and
only if there exists a sequence of continuous functions
$g_i \in {{\bf BV}[a,b]}$ such that
\begin{itemize}
\item[1)]
the sequence normally converges to $g$, and
\item[2)]
$\lim_{i \to \infty} \int_a^b dg_i\,x = v$.
\end{itemize}
\end{definition}
\begin{theorem} \label{pr1}
The set $\int_a^b dg\,x$ is convex and compact; besides,
$$
\big\| \int_a^b dg\,x \big\| \leq
{\mathop{\rm Var}\nolimits}_a^b g \cdot \sup_{t \in [a,b]} |x(t)| \leq
{\mathop{\rm Var}\nolimits}_a^b g \cdot
\big(|x(a)| + {\mathop{\rm Var}\nolimits}_a^b x \big).
$$
\end{theorem}
\begin{example} \label{ex0} \rm
Consider the functions
$$
g(t) = \begin{cases}
g_0,& t < c,\\
g_1,& t > c, \end {cases}
\quad \mbox{and}\quad
x(t) = \begin{cases}
x_0,& t < d,\\
x_1,& t > d, \end{cases}
$$
where $c,d \in [a,b]$. In the case $c = a$, the value $g(a)$
ought to be chosen in such a way that the jump at the point $a$
belongs to $[a,b]$. Analogous assumptions are made for $c = b$
and for the function $x$.
\end{example}
First we suppose that $c = d \in (a,b)$. Let $g_i$ be continuous
functions normally converging to $g$ on ${[a,b]}$ and
$v = \lim_{i \to \infty} {\int_a^b dg_i\, x}$.
Therefore,
$$
v = {\lim_{i \to \infty} \Big(
g_i(t) x(t) \Big|_a^b - \int_a^b g_i\,dx \Big)} =
{g(b) x_1 - g(a) x_0} -
{\lim_{i \to \infty} \left(g_i(c) \cdot (x_1 - x_0)\right)},
$$
and the latter limit exists. Due to compactness we may assume that
$\lim_{i \to \infty} g_i(c)$ exists and by Lemma \ref{lemma2}
it belongs to the segment ${[g(c-0),g(c+0)]}$. On the other hand,
using piecewise linear approximations we can present each point of this
segment as a limit of values of the sequence of continuous functions $g_i \in {{\bf BV}[a,b]}$ (Proposition \ref{quasi exist}).
Hence,
\begin{align*}
{\int_a^b dg\,x} & =
{{g(b) x_1 - g(a) x_0}} - {[g(c-0),g(c+0)] \cdot (x_1 - x_0)} \\
& = {(g_1 - g_0) \cdot [x_0,x_1]}.
\end{align*}
Analogous reasonings lead to
$$
\int_a^b dg\,x = \begin{cases}
(g_1 - g_0) \cdot x_0 & \mbox{if } a \leq c < d \leq b \mbox{ or } c = d = b, \\
(g_1 - g_0) \cdot [x_0,x_1] & \mbox{if } a < c = d < b, \\
(g_1 - g_0) \cdot x_1 & \mbox{if } a \leq d < c \leq b \mbox{ or } a = c = d.
\end{cases}
$$
Denote
\begin{align*}
g \big|_a^b x
&= g(b)x(b-0) - g(b-0)x(b-0) + g(b-0)x(b)\\
&\quad- g(a)x(a+0)+ g(a+0)x(a+0) - g(a+0)x(a).
\end{align*}
Such symbol has the following properties:
\begin{itemize}
\item[a)] Symmetry: if one exchanges $g$ and $x$ and restores the right
order of factors
then the same expression appears.
\item[b)] Provided that at each end of the segment $[a,b]$ at least one of the functions $x$,
$g$ is continuous, the value of the symbol is equal to the ordinary
\textit{\textbf{double substitution}}
$g(t)x(t) \big|_a^b = g(b)x(b) - g(a)x(a)$.
\item[c)] if $x_n(t) \to x(t)$ and $g_n(t) \to g(t)$ as $n \to \infty$ uniformly
for $t \in [a,b]$ then $\lim_{n \to \infty}{g_n \big|_a^b x_n} =
{g \big|_a^b x}$. \qed
\end{itemize}
\begin{theorem}\label{th1}
The formula of integration by parts holds:
$$
\int_a^b dg\,x = g \Big|_a^b x - \int_a^b g\,dx\,.
$$
\end{theorem}
Let $g^c$ be the continuous component of the function
$g \in {{\bf BV}[a,b]}$, $g^c(a) = g(a)$
and $J(g)$ be the set of discontinuity points of $g$ in the
open interval $(a,b)$.
\begin{theorem}\label{th2}
The following two formulas are valid:
\begin{equation}\label{nineth}
\begin{aligned}
\int_a^b dg\,x &= \int_a^b dg^c\,x + (g(a+0)-g(a))\,x(a+0) \\
&\quad + (g(b)-g(b-0))\,x(b-0)
+ \sum_{\tau \in J(g)} \Delta g(\tau) \cdot [x(\tau-0),x(\tau+0)]
\end{aligned}
\end{equation}
and
\begin{equation}\label{nineth prim}
\begin{aligned}
\int_a^b dg\,x &= \int_a^b dg\,x^c + (g(b)-g(a)) \cdot (x(a+0)-x(a))\\
&\quad + g(b) \cdot \sum_{\tau \in J(x)} \Delta x(\tau)
- \sum_{\tau \in J(x)} [g(\tau-0),g(\tau+0)] \cdot \Delta x(\tau).
\end{aligned}
\end{equation}
\end{theorem}
\begin{corollary}
The integral $\int_a^b dg\,x$ is singlevalued if $x$ and $g$ have no
common discontinuity points.
\end{corollary}
\begin{remark} \label{note4} \rm
By the formula of integration by parts we conclude that the properties of the
multivalued integral as a function of $x$ and $g$ are almost symmetric
with respect to the transposition of $x$ and $g$. All the distinction is that
the set $\int_a^b dg\,x$ doesn't depend on the value $x(a)$ even for $x$ and $g$
discontunuous at $a$, but it depends on $g(a)$ in any case.
\end{remark}
Using (\ref{nineth}), by direct calculation we get the following statement.
\begin{theorem}\label{th4}
If $c \in (a,b)$ then
\begin{align*}
{\int_a^b dg\,x}\, +&\, {(g(c)-g(c-0)) x(c-0) + (g(c+0)-g(c)) x(c+0)}\\
&= \int_a^c dg\,x + \int_c^b dg\,x +
(g(c+0)-g(c-0)) \cdot [x(c-0),x(c+0)],
\end{align*}
or, in other words,
\begin{equation}\label{additive}
\int_a^b dg\,x = \int_a^c dg\,x + \int_c^b dg\,x
+ [(g(c)-g(c+0)), (g(c)-g(c-0))] \cdot \Delta x(c).
\end{equation}
\end{theorem}
It is obvious that multiplying one of the functions $x$ or $g$ by a scalar
constant we multiply the integral $\int_a^b dg\,x$ by the same constant.
Using the equality (\ref{nineth}), we get formulas for addition of functions.
\begin{theorem}\label{th3}
For $x,y,g,f \in {{\bf BV}[a,b]}$ the following inclusions hold:
\begin{equation}\label{tenth}
\int_a^b dg(t)\,(x(t)+y(t)) \subset \int_a^b dg(t)\,x(t) +
\int_a^b dg(t)\,y(t)
\end{equation}
and
\begin{equation}\label{tenthagain}
\int_a^b d(g(t) + f(t))\,x(t) \subset \int_a^b dg(t)\,x(t) +
\int_a^b df(t)\,x(t).
\end{equation}
If $J(x) \cap J(y) \cap J(g) = \emptyset$ or $J(x) \cap J(g) \cap J(f) = \emptyset$
then (\ref{tenth}) or, correspondingly, (\ref{tenthagain}) turns into the
equality.
\end{theorem}
\begin{remark} \label{note6} \rm
We can easily obtain the counterexample for the condition of no common discontinuity
points on the following way. Suppose all three functions in (\ref{tenth})
have unique common discontinuity in some point in $(a,b)$; then the right
of (\ref{tenth}) is in general multivalued.
But the jumps of addends $x$ and $y$ may eliminate one another;
thus the left of (\ref{tenth}) becomes singlevalued.
\end{remark}
As an addendum to \cite{VTs1}, we give here another condition for equality.
\begin{theorem}\label{eqjmps}
Suppose that $x(t) - \alpha y(t)$ is continuous in $(a,b)$ for some $\alpha > 0$.
Then for positive $\lambda$ and $\mu$
\begin{equation}\label{lambda mu}
\int_a^b dg(t)\,(\lambda x(t) + \mu y(t)) = \lambda \int_a^b dg(t)\,x(t) +
\mu \int_a^b dg(t)\,y(t).
\end{equation}
\end{theorem}
\begin{proof}
Rewrite (\ref{nineth prim}) in the form
$$
\int_a^b dg\,x = \int_a^b dg\,x^c + (g(b)-g(a)) \cdot (x(a+0)-x(a)) + M,
$$
where
$$
M = g(b) \cdot \sum_{\tau \in J(x)} \Delta x(\tau)
- \sum_{\tau \in J(x)} [g(\tau-0),g(\tau+0)] \cdot \Delta x(\tau).
$$
Since $\alpha y$ has the same jumps that $x$ has,
$$
\int_a^b dg\,y = \int_a^b dg\,y^c + (g(b)-g(a)) \cdot (y(a+0)-y(a))
+ \frac{1}{\alpha}M
$$
with the same set $M$.
The central point of the proof is that, due to convexity of $M$, $\lambda M + \frac{\mu}{\alpha} M =
(\lambda + \frac{\mu}{\alpha})M$. Thus
\begin{align*}
\lambda \int_a^b dg\,x + \mu \int_a^b dg\,y
&= \int_a^b dg\,(\lambda x + \mu y)^c
+ (g(b)-g(a)) \cdot (\lambda x(a+0)+\mu y(a+0)) \\
&\quad - (\lambda x(a)+\mu y(a))+(\lambda + \frac{\mu}{\alpha})M.
\end{align*}
To complete the proof, it remains to note that
$(\lambda + \frac{\mu}{\alpha})\Delta x(\tau)
= \Delta(\lambda x + \mu y)(\tau)$. \end{proof}
\begin{theorem}\label{alpha}
Suppose $x_n(t) \to x(t)$ and $g_n(t) \to g(t)$ uniformly on $[a,b]$,
at least one of the sequences
$\big\{ {\mathop{\rm Var}\nolimits}_a^b g_n \big\}$ and
$\big\{ {\mathop{\rm Var}\nolimits}_a^b x_n \big\}$ is bounded; then
$$
\alpha\Big(\int_a^b dg_n\,x_n, \int_a^b dg\,x \Big) \to 0.
$$
\end{theorem}
\begin{theorem}\label{beta}
a) Suppose that $g_n {\stackrel {\beta} {\to} } g$ and $x_n(t) \to x(t)$
uniformly on $[a,b]$. Then
\begin{equation}\label{beta-convergence}
\lim_{n \to \infty} \beta
\Big(\int_a^b dg_n\,x_n,\int_a^b dg\,x\Big)=0.
\end{equation}
b) Let $x_n {\stackrel {\beta} {\to} } x$, $g_n(t) \to g(t)$ uniformly
on $[a,b]$.
If at each end of $[a,b]$ at least one of the functions
$x$ and $g$ is continuous then (\ref{beta-convergence}) holds.
\end{theorem}
\begin{remark} \label{note2} \rm
The deviation $\beta\big( \int_a^b dg\,x\,, \int_a^b dg_n\,x\big)$
may not tend to 0 (see Example \ref{ex0}).
In other words, this means that dependence of the integral on the function $g$
is upper semicontinuous but not continuous.
\end{remark}
Next, we study the properties of the multivalued function
\begin{equation}\label{X}
X(t) = \begin{cases}
\displaystyle
\int_a^t dg\,x,& t \in (a,b], \\
\{0\}, & t=a.
\end{cases}
\end{equation}
Of course, the properties of the multivalued integral as a function of its
lower limit of integration are analogous.
\begin{definition} \label{defn3}\rm
We say that $X$ is a function of bounded \textit{\textbf{variation}} on $[a,b]$
if
$${\mathop{\rm Var}\nolimits}_a^b X {\stackrel{\scriptstyle {\rm df}}{=}}
\sup\sum_{i=1}^n \alpha(X(t_i),X(t_{i-1})) < \infty,
$$
where supremum is taken on the range of all finite partitions
$a=t_0 < t_1 < \dots < t_n=b$.
\end{definition}
\begin{theorem}\label{boundVar}
The multivalued function $X$ defined by (\ref{X}) has a bounded variation
on $[a,b]$.
\end{theorem}
\begin{theorem}\label{leftright}
The function $X$ has left-handed limit at every point in $(a,b]$ and
right-handed limits at all points in $[a,b)$.
\end{theorem}
\begin{remark}\rm
To prove the theorem, we use in \cite{VTs1} only that $X$ is a function of bounded
variation with values in a complete metric space.
\end{remark}
\section{The functional differential equation with measure input.
General assumptions and accessory results}
The main subject of this paper is linear functional differential equation
\begin{equation}\label{MesFDE}
\dot x(t) = \int_a^t d_sR(t,s)\,x(s) + F'(t), \quad t \in [a,b],
\end{equation}
where the matrix function $R:\Delta \to {\mathbb{R}}^{N \times N}$ has a bounded variation as a function of $s$, $\Delta = \{ (t,s):t \in [a,b], s \in [a,t]\}$,
$F'$ is a derivative of discontinuous function $F \in {{\bf BV}[a,b]}$ of bounded variation,
desired solution $x \in {{\bf BV}[a,b]}$ has values in ${\mathbb{R}}^N$.
Everywhere in what follows we assume that the function $R$ satisfies the following
conditions:
\begin{itemize}
\item[1)] $R$ is measurable on $\Delta$;
\item[2)] for a.e. $t \in [a,b]$, the variation ${\mathop{\rm Var}\nolimits}_a^t R(t, \cdot) \leq v(t)$,
where $v$ is summable function on $[a,b]$; the functions $R(t, \cdot)$ are proper;
\item[3)] the functions $R(\cdot,s)$ are summable on $[s,b]$ for every $s \in [a,b]$;
the mapping $t \mapsto R(t,t)$ is summable on $[a,b]$.
\end{itemize}
\begin{remark} \label{rmk6} \rm
Under these conditions, the initial value problem for equation (\ref{OrdFDE}) with
summable $f$ has a unique solution (see, for instance, \cite{Azb}).
The conditions ``functions $R(t, \cdot)$ are proper" and ``functions
$R(\cdot,s)$ are summable for {\bf every} $s$" are not required for the
solvability of (\ref{OrdFDE}). But by necessity we fulfil them easily by little
corrections of the function $R$ such that the integrals $\int_a^t d_sR(t,s)\,x(s)$
do not change for continuous functions $x$.
\end{remark}
\subsection{Functional differential inequality}
Denote $r(t,s) = {\mathop{\rm Var}\nolimits}_a^s R(t, \cdot)$. Since the functions $R(t,\cdot)$
are proper and $R(\cdot,s)$ summable, $r$ is measurable
with respect to $(t,s) \in \Delta$ and the mapping $t \mapsto r(t,t)$ is
summable. Besides, $r(t,a) = 0$ and ${\mathop{\rm Var}\nolimits}_a^t r(t, \cdot) \leq v(t)$.
Therefore, the problem
\begin{equation}\label{ProblemZ}
\dot z(t) = \int_a^t d_s r(t,s)\, z(s) + {\varphi} (t), \quad z(a) = z_0,
\end{equation}
has a unique solution (see, for instance, \cite{Azb}).
The following lemma on inequality is proved in the usual way but may
be unknown.
\begin{lemma}\label{L1}
Suppose $x$ is a solution of (\ref{OrdFDE}) with summable input $f$,
$z$ is the solution of the problem (\ref{ProblemZ})
with $z_0 > \left| {x(a)} \right|$ and summable
${\varphi}(t) \ge \left| {f(t)} \right|$. Then
$$ \left| {x(t)} \right| \le z(t), \quad t \in [a,b].$$
\end{lemma}
\begin{proof}
Assume the converse and let
$\tau = \inf\{t \in [a,b]\,:\, |x(t)| > z(t) \}$.
Then $\tau \in (a,b]$ and $|x(\tau)| = z(\tau)$. For $s \in [a,\tau]$ we
have $|x(s)| \leq z(s)$. Hence for $t \in [a,\tau]$,
$$
\frac{d}{dt}|x(t)| \leq \Big|\int_a^t d_s R(t,s)\, x(s)\Big|
+ |f(t)| \leq \int_a^t d_s r(t,s)\, z(s) + {\varphi} (t) = \dot z(t).
$$
Therefore,
$$
|x(\tau)| = |{x(a)}| + \int_a^\tau \frac{d}{dt}|x(t)|\, dt <
z(a) + \int_a^\tau \dot z(t)\, dt = z(t).
$$
This contradiction proves the lemma. \end{proof}
\subsection{Multivalued integral is summable}
\begin{lemma}\label{L2}
Let
\begin{itemize}
\item[1)] the function $R(t,s)$ be measurable in $[a,b] \times [c,d]$;
\item[2)] the variation $v(t) = {\mathop{\rm Var}\nolimits}_c^d R(t, \cdot )$ finite for a.e.
$t \in [a,b]$;
\item[3)] the function $v$, mappings $t \mapsto R(t,c)$ and
$t \mapsto R(t,d)$ summable on $[a,b]$.
\end{itemize}
Then, if $x \in {{\bf BV}[c,d]}$, the function $X(t) = \int_c^d {d_s R(t,s)\,x(s)}$
is boundedly summable on $[a,b]$.
\end{lemma}
First let $x$ be continuous.
\begin{prop}\label{Pr1}
If the conditions of the Lemma are fulfilled and $x$ is continuous on $[c,d]$
then $X \in {{\bf L}_1(a,b)}$.
\end{prop}
\begin{proof}
Since $R$ is summable on $[a,b] \times [c,d]$, the mapping
$t \mapsto R(t,s)$ is measurable for almost every $s \in [c,d]$
(this follows from the Fubini theorem).
Therefore we may assume that the integral sums are measurable and their
limit $\int_c^d {d_s R(t,s)\,x(s)}$ is measurable too.
The estimate $\left\| X(t) \right\| \le
v(t) \cdot \max_{s \in [c,d]} \left| {x(s)} \right|$
completes the proof of the Proposition. \end{proof}
\begin{proof}[Proof of Lemma \ref{L2}]
Substitute $R(t, \cdot)$ for $g$ in (\ref{nineth prim}).
Due to Proposition \ref{Pr1} the integral $\int_c^d {d_s R(t,s)\,x^c(s)}$
is measurable with respect to $t$; the sum of the series
$$
\sum_{\tau \in J(x)} [R(t,\tau-0),R(t,\tau+0)] \cdot \Delta x(\tau)
$$
is measurable according to the properties of measurable multivalued
functions mentioned above. Thus $X$ is measurable. The proof is completed by the estimate
$\| X(t)\| \le {\mathop{\rm Var}\nolimits}_c^d R(t, \cdot) \cdot
\max_{s \in [c,d]} \left| {x(s)} \right|$.
\end{proof}
\section{Solutions of general type}
Now we start the study of the functional differential equation with a
measure input
$$
\dot x(t) = \int_a^t d_sR(t,s)\,x(s) + F'(t), \quad t \in [a,b],
\eqno(\ref{MesFDE})
$$
where $F \in {{\bf BV}[a,b]}$.
\begin{definition} \label{sol def} \rm
The function $x \in {{\bf BV}[a,b]}$ is a \textit{\textbf{solution}} of the equation (\ref{MesFDE})
on $[a,b]$ if the function $y(t) := x(t) - F(t)$
is absolutely continuous and
$$
\dot y(t) \in \int_a^t {d_s R(t,s)\,x(s)}
$$
a.e. on $[a,b]$.
\end{definition}
If $F$ is absolutely continuous then the latter inclusion turns into the equation
equivalent to the ordinary functional differential equation with
aftereffect (\ref{OrdFDE}) with $f = F'$. The unique solvability of the initial value
problem for (\ref{OrdFDE}) is well known; see, for instance, \cite{Azb}.
\begin{remark} \label{rmk7} \rm
For every solutuion $\tilde x$ of (\ref{MesFDE}) there is a corresponding solution $x$
of the problem
$$
\dot x(t) = \int_a^t {d_s R(t,s)\, x(s)} + \Phi'(t), \quad
x(a) = \tilde x(a) + (F(a+0) - F(a))
$$
with the function $\Phi$ defined by the formulas $\Phi(a) = F(a+0)$, $\Phi(t) = F(t)$ for
$t > a$, and continuous at the point $a$. The solution $x$ differs from
$\tilde x$ only at the point $a$.
That is why in the sequel we can consider by necessity $F$ being continuous from
the right at $a$. \end{remark}
Besides, the function $F$ and, correspondingly, the solutions $x$ may be considered
to be proper individual functions in stead of equivalence classes.
\begin{example} \label{ex1} \rm
Consider the equation (\ref{MesFDE}) on $[0,3]$ with the data:
$$
R(t,s) = \begin{cases}
0, & s < 1, \\
1, & s > 1,\; 1 < t < 2, \\
-1, & s > 1,\; t > 2,
\end{cases}
\quad
F(t) = \begin{cases}
0, & t < 1, \\
1, & t > 1.
\end{cases}
$$
For this equation, the solutions $x$ that satisfy the initial condition
$x(0) = 0$ are given by the formula
$$
x(t) = \begin{cases}
0, & t < 1, \\
1 + \int_1^t \alpha(s)\, ds, & 1 < t < 2, \\
1 + \int_1^2 \alpha(s)\, ds - \int_2^t \alpha(s)\, ds, & t > 2,
\end{cases}
$$
with arbitrary summable $\alpha$, $\alpha(t) \in [0,1]$.
\end{example}
\begin{remark} \label{ex1 note} \rm
The \textit{\textbf{section of the integral funnel}} of the initial-value
problem
$H(t)= \{ x(t)\,:\, x \mbox{ is a solution} \}$
for this example is represented as follows:
$$
H(t) = \begin{cases}
0, & t < 1,\\
{[1, t]}, & 1 < t \leq 2,\\
{[3 - t, 2]}, & t > 2.
\end{cases}
$$
\end{remark}
\begin{theorem}\label{OtherWords}
The function $x \in {{\bf BV}[a,b]}$ is a solution of the equation
(\ref{MesFDE}) if and only if there exists ${\varphi} \in {{\bf L}_1(a,b)}$
such that for a.e. $t \in [a,b]$
$$
{\varphi}(t) \in \int_a^t {d_s R(t,s)\,F(s)},
$$
and $y(t) = x(t) - F(t)$ is a solution of the equation
$$
\dot y(t) = \int_a^t {d_s R(t,s)\,y(s)} + {\varphi}(t).
$$
\end{theorem}
\begin{proof}
Let $x$ be a solution of (\ref{MesFDE}). Due to Proposition \ref{Pr1} the
function ${\varphi}(t) = \dot y(t) - \int_a^t {d_s R(t,s)\,y(s)}$
belongs to ${{\bf L}_1(a,b)}$.
According to Theorem \ref{th3} for a.e. $t \in [a,b]$ we have
$$
{\varphi}(t) \in \int_a^t {d_s R(t,s)\,x(s)}
- \int_a^t {d_s R(t,s)\,y(s)} = \int_a^t {d_s R(t,s)\,F(s)}.
$$
Now let ${\varphi}$ satisfy stated conditions. Then the function $y$ is absolutely
continuous and for a.e. $t \in [a,b]$ the equality
$$
\dot y(t) \in \int_a^t {d_s R(t,s)\,y(s)} + \int_a^t {d_s R(t,s)\,F(s)} =
\int_a^t {d_s R(t,s)\,x(s)}
$$
holds (the latter equlity is due to Theorem \ref{th3} again). Thus $x$ is
a solution of (\ref{MesFDE}). \end{proof}
\begin{remark} \label{rmk9} \rm
In other words, the solutions of the certain initial value problem for
(\ref{MesFDE}) are in one-to-one correspondence with summable selectors
of the multivalued function $\int_a^t {d_s R(t,s)\,F(s)}$.
\end{remark}
If the latter integral is a.e. singlevalued then the selector ${\varphi}$
is defined uniquely.
\begin{corollary} \label{coro2}
Consider the set of the points $t$ such that the functions $F$ and $R(t,\cdot)$
have common points of discontinuity in $(a,t)$. If its measure is equal to 0
then the initial value problem for the equation (\ref{MesFDE}) has a unique solution.
\qed
\end{corollary}
Consider the sequence of the equations
\begin{equation}\label{xn}
\dot x(t) = \int_a^t {d_s R(t,s)\, x(s)} + F_n'(t).
\end{equation}
\begin{lemma}\label{Limit}
Let $F_n$ $\beta$-converge to $F$, $x_n$ be solutions of (\ref{xn}), and
the sequence $x_n(a)$ be bounded. Then
\begin{itemize}
\item[a)] the sequence of the functions $y_n(t) := x_n(t) - F_n(t)$ has a subsequence
uniformly convergent on $[a,b]$;
\item[b)] if $y_n(t) \to y(t)$ uniformly on $[a,b]$ then $x(t) := y(t) + F(t)$
is a solution of the equation (\ref{MesFDE}).
\end{itemize}
\end{lemma}
\begin{proof}
According to Theorem \ref{OtherWords}, absolutely continuous
functions $y_n$ satisfy the equations
$$
\dot y_n(t) = \int_a^t {d_s R(t,s)\,y_n (s)} + {\varphi}_n(t),
$$
where ${\varphi}_n(t) \in \int_a^t {d_s R(t,s)\,F_n(s)}.$
By Lemma \ref{L1} we have
$$
\left| {y_n (t)} \right| < z(t), \quad t \in [a,b].
$$
where $z$ is a unique solution of the problem (\ref{ProblemZ}) with
${\varphi}(t) = \sup_{s \in [a,b], n} |{F_n(s)}| \cdot v(t)$ and
and $z_0 > \sup_n |{y_n(a)}|$.
So the sequence $\{ y_n \}$ is uniformly bounded. Due the estimate
\begin{equation}\label{dot y}
\left| {\dot y_n (t)} \right| \leq \sup_{s \in [a,b]} z(s) \cdot v(t) + {\varphi}(t),
\end{equation}
the sequence is equicontinuous. This establishes the first assertion
of the theorem.
To prove the second assertion, note that since the absolute continuity of the
sequence $\{ y_n \}$ is equipower, $y$ is absolutely continuous. Suppose, without
loss of generality, that $F$ is continuous at the point $a$, then the solution $x$
is too. The functions $x_n$ $\beta$-converge to $x$ on the almost every
segment $[a,t]$ (Lemma \ref{lemma3}). By Theorem \ref{beta},
$$
\beta \Big(\int_a^t {d_s R(t,s)\,x_n(s)}, \:
\int_a^t {d_s R(t,s)\,x(s)} \Big) \to 0.
$$
According to Lemma \ref{L2} (on summability) and Lebesgue-type theorem on convergence
of integrals for multifunctions (Lemma \ref{L3}),
for $t_0 ,t_1 \in [a,b]$, $t_0 < t_1$, we have
$$
\beta\Big( \int_{t_0 }^{t_1}
\!\!\int_a^t {d_s R(t,s)\,x_n (s)} \,dt,\ %
\int_{t_0 }^{t_1 }
\!\!\int_a^t {d_s R(t,s)\,x(s)} \,dt
\Big) \to 0.
$$
Hence
\begin{align*}
y(t_1) - y(t_0) & = \lim_{n \to \infty} (y_n(t_1 ) - y_n(t_0 )) \\
& = \lim_{n \to \infty } \int_{t_0 }^{t_1 } {\int_a^t {d_s R(t,s)\,x_n (s)} \,dt} \\
& \in \int_{t_0}^{t_1} \!\!\int_a^t {d_s R(t,s)\,x(s)} \,dt.
\end{align*}
Therefore, due to Lemma \ref{L4}, for a.e. $t$ we have
$$ \dot y(t) \in \int_a^t {d_s R(t,s)\,x(s)}.
$$
This implies that $x$ is a solution of (\ref{MesFDE}).
\end{proof}
As a consequence we obtain the following theorem.
\begin{theorem}\label{Compactness}
Let $K$ be compact subset of $\mathbb{R}^N$. Then the set of the solutions
$x$ of (\ref{MesFDE}) with $x(a) \in K$ is compact in the sense of the uniform
convergence on $[a,b]$. \qed
\end{theorem}
Since the values of the multifunction $\int_a^t {d_s R(t,s)\,F(s)}$ are
convex sets (Proposiion \ref{th1}), Theorem \ref{OtherWords} easily implies
the following statement.
\begin{theorem}\label{Convexness}
Let $K$ be convex subset of $\mathbb{R}^N$. Then the set of solutions $x$
of the equation (\ref{MesFDE}) with $x(a) \in K$ is convex. \qed
\end{theorem}
Another way to proof this statement is to refer directly to Definition
\ref{sol def} and Theorem \ref{eqjmps}.
\section{Approximative solutions and solutions with memory}
Now we consider narrow notions of solution for equation (\ref{MesFDE}).
\subsection{Approximative solutions}
\begin{definition} \label{appro def} \rm
The function $x$ is \textit{\textbf{approximative solution}} of (\ref{MesFDE})
if there exist a sequences of absolutely continuous functions $F_n$ and
solutions $x_n$ of equations (\ref{xn}) such that
\begin{itemize}
\item[1)] $F_n {\stackrel {\beta} {\to} } F$,
\item[2)] $x_n(t) - F_n(t) \to x(t) - F(t)$ uniformly on $[a,b]$.
\end{itemize}
\end{definition}
The following example demonstrates that the solution of (\ref{MesFDE})
may not be approximative.
\begin{example} \label{ex2} \rm
Consider the following solution from Example \ref{ex1}:
$$
x(t) = \begin{cases}
0, & t < 1, \\
1, & 1 < t < 2, \\
3 - t, & t > 2, \end{cases}
$$
that is obtained by virtue of
$$
\alpha(t) = \begin{cases}
0, & t < 2, \\
1, & t > 2.\end{cases}
$$
This solution is not approximative.
Indeed, if it is approximative and $F_n$,
$F_n(0) = 0$, are absolutely continuous functions stated in
Definition \ref{appro def}
then the corresponding solutions are
\begin{equation}\label{example xn}
x_n(t) = \begin{cases}
F_n(t), & t \leq 1, \\
F_n(t) + F_n(1) (t - 1), & 1 < t \leq 2, \\
F_n(t) + F_n(1) (3 - t), & t > 2. \end{cases}
\end{equation}
In particular, $x_n(3) = F_n(3)$. Thus
$0 = \lim_{n \to \infty}(x_n(3) - F_n(3)) =
x(3) - F(3) = -1$. We have got a contradiction. \qed
\end{example}
\begin{remark} \label{ex2 note} \rm
This example demonstrates the existence of the points of the integral funnel
not attainable for approximative solutions. From (\ref{example xn})
we have that the section of the integral funnel of approximative solutions
of the initial value problem
$H_a(t) = \{x(t)\,:\, x \mbox{ is approximative solution}\}$
is as follows:
$$
H_a(t) = \begin{cases}
0, & t < 1,\\
{[1, t]}, & 1 < t \leq 2,\\
{[1, 4 - t]}, & 2 < t \leq 3,\\
{[4 - t, 1]}, & t > 3.\end{cases}
$$
\end{remark}
Proposition \ref{quasi exist} and Lemma \ref{Limit} imply the following
theorem.
\begin{theorem}\label{appro exists solut}
\begin{itemize}
\item[a)] The initial value problem for (\ref{MesFDE}) has an approximative solution.
\item[b)] The approximative solution is a solution (in general sense).
\end{itemize}
\end{theorem}
The first assertion of the next theorem follows directly from the
first assertion of the Lemma \ref{Limit}. The easy proof of the second one
is well known.
\begin{theorem}\label{appro lim}
Let $F_n \in {{\bf BV}[a,b]}$ $\beta$-converge to $F\in {{\bf BV}[a,b]}$, $x_n$ be approximative
solutions of (\ref{xn}) and the sequence $\left\{ x_n(a) \right\}$ is
bounded.
Then
\begin{itemize}
\item[a)] the sequence of the functions $y_n(t) := x_n(t) - F_n(t)$ is compact
in the sense of the uniform convergence on $[a,b]$;
\item[b)] if $y_n(t) \to y(t)$ uniformly on $[a,b]$ then $x(t) := y(t) + F(t)$
is an approximative solution of (\ref{MesFDE}). \qed
\end{itemize}
\end{theorem}
As a particular case, we have the following statement.
\begin{theorem}\label{Appro Compactness}
If $K$ is a compact subset of $\mathbb{R}^N$ then the set of approximative
solutions $x$ of the equation (\ref{MesFDE}) with $x(a) \in K$ is compact
in the sense of the uniform convergence on $[a,b]$. \qed
\end{theorem}
The convexness of the set of approximative solutions takes place too.
\begin{theorem}\label{Appro Convexness}
If $K$ is a convex subset of $\mathbb{R}^N$, then the set of approximative
solutions $x$ of the equation (\ref{MesFDE}) with $x(a) \in K$ is convex.
\end{theorem}
\begin{proof}
Let $x^0$ and $x^1$ be approximative solutions of (\ref{MesFDE}), $x_i^0$ and $x_i^1$
the solutions of the equations with inputs $F_i^0$ and, respectively, $F_i^1$
such as it is described in the Definition \ref{appro def}.
Since $F_i^0 {\stackrel {\beta} {\to} } F$ and $F_i^1 {\stackrel {\beta} {\to} } F$, due to Lemma \ref{lemma4}, the functions
$\lambda F_i^1 + (1 - \lambda)F_i^0$ ($\lambda \in (0,1)$) $\beta$-converge to
$F$ as $i \to \infty$. For continuous functions $x_i^k$, $k = 0,1$, the integral
has a linearity property. Thus, for the functions
$
y_i = \lambda x_i^1 + (1 - \lambda)x_i^0 - F =
\lambda (x_i^1 - F) + (1 - \lambda)(x_i^0 - F)
$ we get a.e.
$$
\dot y_i(t) = \int_a^t {d_s R(t,s)\,(x_i^1 + (1 - \lambda)x_i^0)(s)}
$$
and
\begin{align*}
\lim_{i \to \infty}y_i(t)
& = \lambda \lim_{i \to \infty}(x_i^1(t) - F(t)) +
(1 - \lambda)\lim_{i \to \infty}(x_i^0(t) - F(t)) \\
& = \lambda x^1(t) + (1 - \lambda)x^0(t) - F(t)
\end{align*}
uniformly on $[a,b]$.
\end{proof}
\begin{remark} \label{rmk11} \rm
If we replace $\beta$-convergence $F_n$ to $F$ by the normal convergence in
Definition \ref{appro def} then we get another definition of solution that is,
a priori, stronger. We shal see soon that as a matter of fact
this new notion coincides with the old one.
\end{remark}
\subsection{Solutions with memory}
Consider now the following property of the solution that may be important for
applications. Suppose the Stieltjes measure $d_s R(t,s)$ has an atom at the point
$s = \tau$. So the system modelled by equation (\ref{MesFDE}) has
onboard device for measuring the value $x(\tau)$ to determine the behaviour of
the system. The property considered is that at any moment of time in
the future, whenever the value $x(\tau)$ is needed, just the same obtained
value is used. The solution from Example \ref{ex2} do not has this property:
the value $x(1)$ is measured by two independent devices, the value obtained
by first one is valid during $t \in (1,2)$ and the value of the second is for
$t \in (2,3)$.
In this subsection we assume that the function $F$ is
continuous from the right at the point $a$.
\begin{definition} \rm
$x \in {{\bf BV}[a,b]}$ is \textit{\textbf{solution with memory}} of (\ref{MesFDE})
if it has a proper representative ${\bar x}$ such that the function
$y(t) := x(t) - F(t)$ satisfy a.e. the equality
\begin{equation}\label{memory}
\dot y(t) = {\rm (S)}\!\!\int_a^t {d_s R(t,s)\, \bar x(s)},
\end{equation}
where ${\scriptstyle\rm (S)}\!\int$ denotes a Lebesgue-Stieltjes integral.
\end{definition}
\begin{theorem}\label{memory appro}
For $x \in {{\bf BV}[a,b]}$, the following conditions are equivalent:
\begin{itemize}
\item[a)] $x$ is approximative solution of equation (\ref{MesFDE});
\item[b)] there exist sequences of absolutely continuous functions
$F_n$ and solutions $x_n$ of equations (\ref{xn}) such that
$F_n {\stackrel{N}{\to}} F$ and
$x_n(t) - F_n(t) \to x(t) - F(t)$ uniformly on $[a,b]$;
\item[c)] $x$ is a solutions with memory of (\ref{MesFDE}).
\end{itemize}
\end{theorem}
\begin{proof}
\textbf{c) $\Rightarrow$ b).} Let $x$ be a solution with memory. Define
$y(t) = x(t) - F(t)$ and ${\bar F}(t) = {\bar x}(t) - y(t)$.
The function $\bar F$ has just the same behaviour at the points of
discontinuity as $\bar x$ has.
Define the approximations $l_\delta$ of the function $\bar F$ as it was made
in the proof of the Proposition \ref{quasi exist}. Then we have
$l_\delta {\stackrel{N}{\to}} F$ as $\delta \to 0$. Let $x_\delta$ be solutions
of the equations
\begin{equation}\label{Eq delta}
\dot x(t) = \int_a^t {d_s R(t,s)\, x(s)} + l'_\delta(t)
\end{equation}
with the same initial value that $x$ has, and
$y_\delta = x_\delta - l_\delta$.
Now let $\delta$ tend to $0$ ranging over countable set of values.
For every point $\tau \in J(\bar F)$ we have $l_\delta(\tau) = \bar F(\tau)$
as $\delta$ is sufficiently small. Thus $x_\delta(s) \to \bar x(s)$ for
all $s \in [a,b]$. The proof of Lemma \ref{Limit} shows that
the functions $x_\delta$ are uniformly bounded. Hence
$$
{\dot y}_\delta(t) = {\rm (S)}\!\!\int_a^t {d_s R(t,s)\, x_\delta(s)} \to
{\rm (S)}\!\!\int_a^t {d_s R(t,s)\, \bar x(s)} = {\dot y}(t)
$$
for a.e. $t \in [a,b]$. The derivatives ${\dot y}_\delta(t)$ have common
summable majorant (see the proof of Lemma \ref{Limit}); therefore according
to the Lebesgue theorem $y_\delta(t) \to y(t)$ for all $t \in [a,b]$.
Due to Lemma \ref{Limit}a) we
may find such a subsequence $\delta \to 0$ that the convergence $y_\delta(t)
\to y(t)$ along this subsequence is uniform on $[a,b]$.
\noindent\textbf{b) $\Rightarrow$ a).} It is obvious by definition.
\noindent\textbf{a) $\Rightarrow$ c).} Let $x$ be an approximative solution and
$F_n$ the functions described in Definition \ref{appro def}. Then the equality
$\lim_{n \to \infty} F_n(t) = F(t)$ holds for all points of continuity
of $F$. Take $\tau \in J(F)$. According to Lemma \ref{lemma2}a) there exists a
subsequence such that being renumerated it has the limit
\begin{equation}\label{in segment}
\lim_{n \to \infty} F_n(\tau) \in [F(\tau-0), F(\tau+0)].
\end{equation}
Using diagonal process we get (\ref{in segment}) for all $\tau \in J(F)$.
Denote
\begin{equation}\label{FBar}
\bar F(t) = \lim_{n \to \infty} F_n(t), \quad
y(t) = x(t) - F(t), \quad
\bar x(t) = y(t) + \bar F(t),
\end{equation}
then $\bar x(t) = x(t)$ at the points of continuity and $\bar x(\tau) \in
[x(\tau-0), x(\tau+0)]$ for $\tau \in J(F) = J(x)$.
Let $x_n$ be the solutions of equations (\ref{xn}) such as described in
Definition \ref{appro def}. Due to (\ref{FBar}), $\bar x(s) = \lim_{n \to \infty} x_n(s)$
for all $s \in [a,b]$. Hence
$$
\int_a^t {d_s R(t,s)\, x_n(s)} \to
{\rm (S)}\!\!\int_a^t {d_s R(t,s)\, \bar x(s)}
$$
and the latter integral is summable function of $t$. Denote
$y_n(t) = x_n(t) - F_n(t)$; then $y_n(t) \to y(t)$ for $t \notin J(F)$.
If $t_2 > t_1$ then
\begin{align*}
y(t_2) - y(t_1) &
= \lim_{n \to \infty}(y_n(t_2) - y_n(t_1)) \\
&= \lim_{n \to \infty}\int_{t_1}^{t_2} \int_a^t {d_s R(t,s)\, x_n(s)} \, dt \\
&= \int_{t_1}^{t_2} {\rm (S)}\!\!\int_a^t {d_s R(t,s)\, \bar x(s)} \,dt,
\end{align*}
which implies the equality (\ref{memory}) for a.e. $t$.
Thus $x$ is a solution with memory.
\end{proof}
\section{The Cauchy formula}
\subsection{The Cauchy function}
It is well known that the solutions of the equation (\ref{OrdFDE}) with
summable $f$ obey the Cauchy formula
\begin{equation}\label{Cauchy}
x(t) = C(t,a) x(a) + \int_a^t C(t,s) f(s)\,ds,
\end{equation}
where $C(t,s)$ is the Cauchy function.
We describe the construction of the Cauchy
function following \cite{Azb, Sim}. Integration by parts in (\ref{OrdFDE}) gives
$$
\dot x(t) = \int_a^t K(t,s) \dot x(s)\, ds + K(t,a) x(a) + f(t),
$$
where $K(t,s) = R(t,t) - R(t,s)$.
Note that $K(t,t) = 0$.
The latter equation is uniquely solvable as the Volterra equation of second
kind in unknown $\dot x$. Integrating, we get the solution. In such a way the
formula (\ref{Cauchy}) is obtained, where
\begin{equation}\label{Cts}
C(t,s) = E + \int_s^t P(\tau,s)\, d\tau,
\end{equation}
$E$ is the unit matrix, $P(t,s)$ the kernel of the Volterra integral
operator ${\bf P}$,
$$
{I + \bf P} := (I - {\bf K})^{-1} = I + {\bf K} + {\bf K}^2 + \dots + {\bf K}^n
+ \dots,
\ %
({\bf K}z)(t) = \int_a^t {K(t,s)z(s)\,ds}.
$$
The kernel $P$ may be found as a sum of Neumann's series
\begin{equation}\label{H}
P(t,s) = \sum_{n=1}^\infty K_n(t,s),
\end{equation}
where iterated kernels $K_n$ are defined by equalities
\begin{gather*}
K_1(t,s) = K(t,s),\\
K_{n+1}(t,s) = \int_s^t K_n(t,\tau) K(\tau,s)\, d\tau, \quad n \geq 1.
\end{gather*}
To argue that all these calculations are well made, we bring out the
following proposition.
Denote by $\mathfrak{A}$ the subset of $[a,b]$ of comlete measure containing
all points $t$ such that the variation
${\mathop{\rm Var}\nolimits}_a^t R(t, \cdot) \leq v(t) < \infty$.
\begin{prop}\label{iter kern}
The following properties of the iterated kernels take place:
\begin{itemize}
\item[a)] the functions $K_n$ are summable on $\Delta$; the iterative formula
may be rewritten in the form
$$
K_{n+1}(t,s) = \int_s^t K(t,\tau) K_n(\tau,s)\, d\tau;
$$
\item[b)] the functions $K_n(\cdot,s)$ are summable on $[s,b]$ for
every $s \in [a,b]$; in particular, the functions $K_n(\cdot,a)$ are
summable;
\item[c)] $K_n(t,t) = 0$;
\item[d)] for $t \in \mathfrak{A}$ and $s \in [a,b]$ we have
$$
|K_n(t,s)| \leq {\frac{v(t)}{(n-1)!}} \Big(\int_s^t v(\tau)\, d\tau\Big)^{n-1};
$$
\item[e)] for $t \in \mathfrak{A}$ the variation
$$
{\mathop{\rm Var}\nolimits}_a^t K_n(t, \cdot ) \leq
{\frac{v(t)}{(n-1)!}} \Big(\int_a^t v(\tau)\, d\tau \Big)^{n-1}.
$$
\end{itemize}
\end{prop}
All the assertions in this proposition are easily verified by mathematical
induction with use of the Fubini theorem on multiple integral. \qed
Hence for a.e. $t$ the Neumann series absolutely converges uniformly with
respect to $s$. Integrating its sum as consistent with (\ref{Cts}) we obtain
the following theorem.
\begin{theorem}[see also \cite{Azb, Sim}] \label{C properties}
The following properties of the Cauchy function $C(t,s)$ take place:
\begin{itemize}
\item[ a)] $C$ is summable on $\Delta$;
\item[ b)] the function $C(\cdot,s)$ is absolutely continuous on $[s,b]$
for every $s \in [a,b]$;
\item[ c)] $C(t,t) = E$;
\item[ d)] for all $(t,s) \in \Delta$,
$$
|C(t,s)| \leq \exp\Big(\int_s^t v(\tau)\, d\tau\Big);
$$
\item[ e)] for every $t \in [a,b]$, the function $C(t,\cdot)$ has finite
variation on $[a,t]$, which is uniformly bounded respective to $t$:
$$
{\mathop{\rm Var}\nolimits}_a^t C(t, \cdot) \leq
\exp\Big(\int_a^t v(\tau)\, d\tau\Big).
$$ \qed
\end{itemize}
\end{theorem}
\subsection{The Cauchy formula for equations with measure input}
We suppose here that the function $F$ is continuous at the point $a$.
Likewise to (\ref{Cauchy}), we can write the formula
\begin{equation}\label{X(t)}
X(t) = C(t,a)\,x(a) + \int_a^t C(t,s)\, dF(s).
\end{equation}
In virtue of Theorem \ref{C properties} the integral here is well defined
multivalued Stieltjes integral. Hence the function $X$ is in general
multivalued and its values are convex compact sets.
Integrating by parts due to Theorem \ref{th1} we get
$$
X(t) = C(t,a)\,(x(a) - F(a)) + F(t) - \int_a^t d_sC(t,s)\, F(s)
$$
for a.e. $t \in [a,b]$. Thus, according to Lemma \ref{L2},
the multifunction $X$ is boundedly summable.
Let $x$ be approximative solution, $F_n$ and $x_n$ corresponding absolutely
continuous functions and solutions such as stated in Definition \ref{appro def}.
We have
\begin{equation}\label{Cauchy n}
x_n(t) = C(t,a)\,x_n(a) + \int_a^t C(t,s)\, dF_n(s).
\end{equation}
For $t \notin J(F)$), since $F_n {\stackrel {\beta} {\to} } F$ on $[a,t]$, according to
Theorem \ref{beta}, we have
$$
\rho\Big(\int_a^t C(t,s)\, dF_n(s),\: \int_a^t C(t,s)\, dF(s) \Big)
\to 0 \quad\mbox{as } n \to \infty.
$$
Therefore,
$\rho\left(x_n(t) , X(t)\right) \to 0$.
The set $X(t)$ is closed, hence
$$
x(t) = \lim_{n \to \infty} x_n(t) \in X(t)
$$
for all $t \notin J(x)$. So we get the following statement.
\begin{theorem}\label{appro Cauchy}
Let $F$ be continuous from the right at the point $a$. Then the approximative
solution $x$ of equation (\ref{MesFDE}) satisfies the inclusion
\begin{equation}\label{in Cauchy}
x(t) \in C(t,a)\,x(a) + \int_a^t C(t,s)\, dF(s)
\end{equation}
for all points $t$ of continuity of $F$. \qed
\end{theorem}
\begin{remark} \label{rmk12} \rm
The Cauchy function of Examples \ref{ex1} and \ref{ex2} is as follows:
$$
C(t,s) = \begin{cases}
1, & s < 1,\; t < 1,\\
t, & s < 1,\; 1 < t < 2,\\
4 - t,& s < 1,\; t > 2,\\
1, & s > 1. \end{cases}
$$
\end{remark}
Using this equality, we obtain from (\ref{X(t)}) that $X(t) = H_a(t)$.
The expression for $H_a(t)$ is given in Remark \ref{ex2 note}.
It was found in Remark \ref{ex1 note} that the set $H(t)$ for $t > 2$
contains points not belonging to $H_a(t)$.
Thus the inclusion (\ref{in Cauchy}) may be violated for the solutions
of general type.
The arbitrary selector $x$ of the multifunction $X$ (that is $x(t) \in X(t)$
$\forall t \in [a,b]$) such that the difference $x - F$ is absolutely
continuous is not necessarily a solution of (\ref{MesFDE}). Indeed,
the solution $x$ in Example \ref{ex1} is nondecreasing on $[1,2]$
that is not required for selectors of multifunction $H_a$.
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\end{document}