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\AtBeginDocument{{\noindent\small
2003 Colloquium on Differential Equations and Applications, Maracaibo, Venezuela.\newline
{\em Electronic Journal of Differential Equations},
Conference 13, 2005, pp. 57--63.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{57}

\begin{document}

\title[\hfilneg EJDE/Conf/13 \hfil On the solution of differential equations]
{On the solution of differential equations with delayed
and advanced arguments}

\author[V. Iakovleva,  C. J. Vanegas \hfil EJDE/Conf/12 \hfilneg]
{Valentina Iakovleva,  Carmen Judith Vanegas}  % in alphabetical order

\address{Valentina Iakovleva \hfill\break
Department of Mathematics \\
Universidad Sim\'{o}n Bol\'{\i}var \\
Sartenejas-Edo. Miranda \\
P. O. Box 89000, Venezuela}
\email{0080947@usb.ve}

\address{Carmen Judith Vanegas \hfill\break
Department of Mathematics \\
Universidad Sim\'{o}n Bol\'{\i}var \\
Sartenejas-Edo. Miranda \\
P.O. Box 89000, Venezuela}
\email{cvanegas@usb.ve}

\date{}
\thanks{Published May 30, 2005.}
\subjclass[2000]{34K15, 34K18}
\keywords{Differential difference equations; functional differential equations}

\begin{abstract}
 In this work, we construct solutions to differential
 difference equations with delayed and advanced arguments.
 We use a  step derivative so that a special condition on
 the initial function assures the existence and uniqueness
 of the solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{corollary}{Corollary}[section]
\allowdisplaybreaks

\section{Introduction}

The differential difference delayed equations and the differential
difference advanced equations have been  studied widely; see for
example \cite{elg,dr,ha,aw}. Applications of this equations can
be found in physics, biology, economy, and so on, \cite{gop,ha,hlun,gl}.
However as far as we have researched, there are only a few studies
on the differential equations with delayed and advanced arguments
\cite{l,kw}.

 From a strictly mathematical point of view, we are interested on
the study of the system of equations
\begin{equation}\label{0}
   x'(t)= A x(t-\omega) + B x(t+\omega) + C x(t),
\end{equation}
where $x(t)$ is a vector-value function in $\mathbb{R}^n$, $A, B$ and $C$
are arbitrary $n\times n$ matrices, and $\omega$ an real number.
However, in this article, we analyze the  simpler scalar
equation
\begin{equation}\label{1}
   x'(t)=x(t-1)+x(t+1).
\end{equation}
and leave the study of \eqref{0} for a future research.

To obtain a solution of \eqref{1}, we define the function $x(t)$
initially on some interval of $\mathbb{R}$. Then we construct the
solution using the step derivative method, which is an analog to the
step integration method \cite{bc,dr}. Then  we prove the existence,
uniqueness, and smoothness of the solution.

\section{Construction of the solution}

In this section we  construct the solution of the differential
difference equation \eqref{1}, using the step  derivatives method,
that provides an iterative formula.
Consider the differential difference equation \eqref{1},
with $t\geq 0$, $x:[-1,+\infty) \to \mathbb{R} $, and
$x(t)$ differentiable in $[0,+\infty)$.  Equation \eqref{1} is
rewritten as
$$
 x(t+1)=x'(t)-x(t-1)
$$
or equivalently as
\begin{equation}\label{1.2}
x(t)=x'(t-1)-x(t-2).
\end{equation}
 From this equation, it follows that in order to find the
solution  $x(t)$ on the interval $[m,m+1]$ it is necessary to know
its value on the interval $[m-2,m]$, with $m$ a positive integer.
In particular, to determine the solution on the interval $[1,2]$,
it is necessary to know it on the interval $[-1,1]$.

Accordingly  we define  $x(t)$ for $t\in [-1,1]$ as
\begin{equation}\label{1.3}
x(t)=\varphi(t)=
\begin{cases}
\varphi_1(t),  &t\in[-1,0]\\
\varphi_2(t),  &t\in[0,1],\end{cases}
\end{equation}
where the function $ \varphi$ is taken initially in the space
$C_{[-1,1]}$ (because $x(t)$ is differentiable
and therefore continuous in $[0,+\infty)$).

After a formal procedure,
for $t\in(1,2)$,
\begin{equation}\label{1.4}
x(t)=\varphi'_2(t-1)-\varphi_1(t-2)=\varphi'(t-1)-\varphi(t-2)\,.
\end{equation}
For $t\in(2,3)$,
\begin{equation}\label{1.41}
\begin{aligned}
x(t)&=x'(t-1)-x(t-2)\\
&=\frac{d}{dt}(\varphi'_2(t-2)-\varphi_1(t-3))-\varphi(t-2)\\
&=\varphi''_2(t-2)-\varphi'_1(t-3)-\varphi_2(t-2)\\
&=\varphi''(t-2)-\varphi'(t-3)-\varphi(t-2)\,.\end{aligned}
\end{equation}
For $t\in(3,4)$,
\begin{align*}
x(t)&=x'(t-1)-x(t-2)\\
&=\frac{d}{dt}(\varphi''_2(t-3)-\varphi'_1(t-4)-
 \varphi_2(t-3))-\varphi'_2(t-3)+\varphi_1(t-4)\\
&= \varphi'''_2(t-3)-\varphi''_1(t-4)-2\varphi'_2(t-3)+
\varphi_1(t-4)\\
&=\varphi'''(t-3)-\varphi''(t-4)-2\varphi'(t-3)+ \varphi(t-4),\
\end{align*}
and so forth. Due to the fact, that in each interval the solution
$x(t)$ is expressed by means of increasing order derivatives of
the function $\varphi$, it is necessary  to take  $\varphi$ in
$C^{\infty}_{[-1,1]}$.  From the above expressions for the
solution $x(t)$ it follows that this solution can be written via the
following iterative formulas, where $l$ is a natural number:
On the interval $(2l-1,2l)$,
\begin{equation}\label{1.5}
x(t)=\sum_{k=0}^{l-1}(c_{2k}\varphi^{(2k)}(t-2l)+c_{2k+1}
\varphi^{(2k+1)}(t-(2l-1))),
\end{equation}
and on the interval $(2l,2l+1)$,
\begin{equation}\label{1.6}
x(t)=\sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-2l)
+\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+1)}(t-(2l+1))\,,
\end{equation}
where $c_{2k}$, $c_{2k+1}$, $k=0,1,2,\dots l$ are constants.

The proofs of \eqref{1.5} and \eqref{1.6} are done by
induction (on l):
The case   $l=1$, i.e.,  $t\in(1,2)$ and $t\in(2,3)$ are
the already proven:  \eqref{1.4} and \eqref{1.41}.

Now we assume formulas \eqref{1.5} y \eqref{1.6} hold for $t\in(2l-1,2l)$
and  $t\in(2l,2l+1)$ respectively.

\noindent First we deal with formula \eqref{1.5} in the interval
$(2l+1,2l+2)$:
\begin{equation}\label{1.51}
x(t)=\sum_{k=0}^{l}(c_{2k}\varphi^{(2k)}(t-(2l+2))+c_{2k+1}
\varphi^{(2k+1)}(t-(2l+1)))\,.
\end{equation}
Using  \eqref{1.2}, \eqref{1.5}  and \eqref{1.6} it follows that
\begin{align*}
x(t)&=x'(t-1)-x(t-2)\\
&=\sum_{k=0}^{l}c_{2k}\varphi^{(2k+1)}(t-1-2l)+
\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+2)}(t-1-(2l+1))\\
&+\sum_{k=0}^{l-1}(c_{2k}\varphi^{(2k)}(t-2-2l)+c_{2k+1}
\varphi^{(2k+1)}(t-2-(2l-1)))\\
&=\sum_{k=0}^{l}(c_{2k}+c_{2k+1})\varphi^{(2k+1)}(t-(2l+1))
+\sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-(2l+2))\\
&\quad +\sum_{r=1}^{l}c_{2r-1}\varphi^{(2r)}(t-(2l+2))\\
&= \sum_{k=0}^{l}((c_{2k}+c_{2k-1})\varphi^{(2k)}(t-(2l+2))+
(c_{2k}+c_{2k+1})\varphi^{(2k+1)}(t-(2l+1)))
\end{align*}
where  $k+1=r$, $c_{2l+1}=c_{2l}=c_{-1}=0$. This shows  formula
\eqref{1.51}.

Now we prove formula  \eqref{1.6} in the interval
$(2l+2,2l+3)$:
\begin{equation}\label{1.61}
x(t)=\sum_{k=0}^{l+1}c_{2k}\varphi^{(2k)}(t-(2l+2))
+\sum_{k=0}^{l}c_{2k+1}\varphi^{(2k+1)}(t-(2l+3))\,.
\end{equation}
Using \eqref{1.2}, \eqref{1.51}  and \eqref{1.6} it follows that
\begin{align*}
x(t)&=x'(t-1)-x(t-2)\\
&=\sum_{k=0}^{l}c_{2k}\varphi^{(2k+1)}(t-1-(2l+2))+
\sum_{k=0}^{l}c_{2k+1}\varphi^{(2k+2)}(t-1-(2l+1))\\
&\quad+ \sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-2-2l)+
\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+1)}(t-2-(2l+1))\\
&=\sum_{k=0}^{l}c_{2k+1}\varphi^{(2k+2)}(t-(2l+2))
+\sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-(2l+2))\\
&\quad + \sum_{k=0}^{l}c_{2k}\varphi^{(2k+1)}(t-(2l+3))
+\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+1)}(t-(2l+3))\\
&=\sum_{r=1}^{l+1}c_{2r-1}\varphi^{(2r)}(t-(2l+2))
+\sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-(2l+2))\\
&\quad + \sum_{k=0}^{l}c_{2k}\varphi^{(2k+1)}(t-(2l+3))+
\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+1)}(t-(2l+3))\\
&=\sum_{k=0}^{l+1}(c_{2k-1}+c_{2k})\varphi^{(2k)}(t-(2l+2))+
\sum_{k=0}^{l}(c_{2k}+c_{2k+1})\varphi^{(2k+1)}(t-(2l+3))
\end{align*}
where $k+1=r$, $c_{2l+2}=c_{2l+1}=c_{-1}=0$. This shows formula
\eqref{1.61}.

That formulas \eqref{1.5} and  \eqref{1.6} hold and coincide  in
the boundary points of each interval $(m,m+1)$ will be proven
in the next section.

\begin{remark} \rm
The solution $x(t)$ may be extended to the left by rewriting
\eqref{1} as
$$
x(t-1)=x'(t)-x(t+1)\,.
$$
In this case we obtain expressions for $x(t)$ analogous to
\eqref{1.5} and  \eqref{1.6}.
\end{remark}

\section{Existence and uniqueness of the solution}

In this section we give necessary and sufficient conditions to
assure the existence and uniqueness of the solution to problem
\eqref{1}-\eqref{1.3}.

\begin{theorem} \label{thm3.1}
The  solution $x(t)$ of \eqref{1} satisfying the initial condition
\eqref{1.3} with  $\varphi$ in $C^{\infty}_{[-1,1]}$,
exists and is differentiable, if and only if
$$
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),
$$
for $n=0,1,2,\dots$.
\end{theorem}

\begin{proof} Since $\varphi$ belongs to
$C^{\infty}_{[-1,1]}$,  for each interval $(m,m+1)$ the
function $x(t)$ exists and is infinitely many times
differentiable. In order to prove the continuity of $x(t)$  and
the existence of its derivative in the end points $m$ and $m+1$ (and
therefore the existence of $x(t)$ in such points) it is necessary
to prove the equalities
\begin{equation}\label{1.8}
x^{(i)}(m^+)=x^{(i)}(m^-),\quad i=0,1;\; m=1,2,\dots ,
\end{equation}
where
\begin{equation}
\begin{gathered}
x^{(i)}(m^+):= \lim_{\varepsilon\to 0,\; \varepsilon>0}
 x^{(i)}(m+\varepsilon) \,,\\
x^{(i)}(m^-):= \lim_{\varepsilon\to 0,\;
\varepsilon>0} x^{(i)}(m-\varepsilon).
\end{gathered}
\end{equation}
By induction on the natural number $k$, we will prove the claim:
\begin{gather}\label{1.9}
x(m^+)=x(m^-),\, \, \,m=1,2,\dots k\,,\\
\label{1.10} x'(m^+)=x'(m^-),\, \, \,m=1,2,\dots k-1\,,
\end{gather}
if and only if
\begin{equation}\label{1.11}
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),\quad
n=1,2,\dots k-1 \,.
\end{equation}
But if \eqref{1.9} and \eqref{1.10} hold, then it follows the
equivalence
\begin{equation}\label{1.12}
x'(k^+)=x'(k^-)
\end{equation}
if and only if
\begin{equation}\label{1.13}
\varphi^{(k+1)}(0)=\varphi^{(k)}(-1)+\varphi^{(k)}(1).
\end{equation}
 For the case $k=1$, by formula \eqref{1.4} at point $m=1$ it is
valid
$$
x(1^+)=\varphi'(0^+)-\varphi(-1^+)\,.
$$
 From $x(1^-)=\varphi(1^-)=\varphi(1)$ it follows that
 $x(1^+)=x(1^-)$ is equivalent to
$$
\varphi'(0^+)=\varphi(-1^+)+\varphi(1^-).
$$
 From the derivatives of \eqref{1.4} it  follows
$$
x'(1^+)=\varphi''(0^+)-\varphi'(-1^+)\,,
$$
and from \eqref{1.3} follows $ x'(1^-)=\varphi'(1^-)$.
Therefore $x'(1^+)=x'(1^-) $ if and only if
$$
\varphi''(0^+)=\varphi'(-1^+)+\varphi'(1^-).
$$
Assume now the claim holds up to $ k=s$.  According to the
inductive assumption, formula \eqref{1.11} is satisfied for $
k=s$. Formula \eqref{1.10} for $ k=s+1$ implies $x'(s^+)=x'(s^-)$,
which is \eqref{1.12} for $ k=s$, but according to the inductive
assumption this is equivalent to \eqref{1.13} for $ k=s$, which
means $\varphi^{(s+1)}(0)=\varphi^{(s)}(-1)+\varphi^{(s)}(1) $. It
follows, that formulas \eqref{1.9} and \eqref{1.10} for $ k=s+1$
imply
$$
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),\quad
n=1,2,\dots s .
$$
Next will be verified, that \eqref{1.12} and \eqref{1.13} are
equivalent for $k=s+1$, provided that \eqref{1.9} and \eqref{1.10}
hold for $ k=s+1$. By computing the derivatives of formula
\eqref{1.2} one reaches $x'(t)=x''(t-1)-x'(t-2)$. Therefore,
$$
x'((s+1)^+)=x''(s^+)-x'((s-1)^+), \quad
x'((s+1)^-)=x''(s^-)-x'((s-1)^-).
$$
Formula \eqref{1.10} for $ m=s-1$ implies
$x'((s-1)^+)=x'((s-1)^-)$ and the late implies
\begin{equation}\label{1.14}
x'((s+1)^+)=x'((s+1)^-)\Longleftrightarrow x''(s^+)=x''(s^-)
\end{equation}
Then after the inductive assumption,
$x'(s^+)=x'(s^-)\Longleftrightarrow\varphi^{(s+1)}(0)
=\varphi^{(s)}(-1)+\varphi^{(s)}(1)$.

Formulas for $x'(t)$ and $x''(t)$ are obtained for each interval
$(m,m+1)$ by computing the simple and double derivatives of
\eqref{1.5} or \eqref{1.6} respectively; therefore the equality
$x''(s^+)=x''(s^-)$ in terms of $\varphi$ has the same shape than
$x'(s^+)=x'(s^-)$, with the only light difference, that the
derivatives of $\varphi$ appear increased in one degree. Since
$x'(s^+)=x'(s^-)$ if and only if
$\varphi^{(s+1)}(0)=\varphi^{(s)}(-1)+\varphi^{(s)}(1)$, it
follows $x''(s^+)=x''(s^-)$ if and only if
$\varphi^{(s+2)}(0)=\varphi^{(s+1)}(-1)+\varphi^{(s+1)}(1)$. Claim
\eqref{1.14} implies $x'((s+1)^+)=x'((s+1)^-)$ if and only if
$\varphi^{(s+2)}(0)=\varphi^{(s+1)}(-1)+\varphi^{(s+1)}(1)$, with
which the equivalence has been proven.

On the other hand, suppose \eqref{1.11} is true for $ k=s+1$ which
means
$$
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),\quad
n=1,2,\dots s .
$$
The formulas \eqref{1.13} and \eqref{1.11} hold for $ k=s$ ,
therefore after the inductive assumption formulas \eqref{1.9} and
\eqref{1.10} are true for $ k=s$ , hence formulas \eqref{1.12} and
\eqref{1.13} are equivalent for $ k=s$.

 From formula \eqref{1.2} we have
$$
x((s+1)^+)=x'(s^+)-x((s-1)^+), \quad
x((s+1)^-)=x'(s^-)-x((s-1)^-).
$$
Since  $x'(s^+)=x'(s^-)$ and $x((s-1)^+)=x((s-1)^-)$, then
$x((s+1)^+)=x((s+1)^-)$. Equalities $x'(s^+)=x'(s^-)$ and
$x((s+1)^+)=x((s+1)^-)$ together with \eqref{1.9} and \eqref{1.10}
for $ k=s$ imply, that \eqref{1.9} and \eqref{1.10} are both true
for $ k=s+1$.
\end{proof}

\begin{remark} \rm
In \cite{iv}, we have obtained non-trivial functions in the set
$$
\{\varphi\in C^{\infty}_{[-1,1]}:
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1), \,
n=0,1,2,\dots\}.
$$
\end{remark}

\begin{remark} \rm
 If no condition such as those in Theorem \ref{thm3.1} are imposed
 on the function  $\varphi$,  then the function $x(t)$, defined
by  \eqref{1.5} and  \eqref{1.6}, may be discontinuous at the
integer points.
\end{remark}

\begin{corollary} \label{cor3.1}
If for an initial function $ \varphi\in C^{\infty}_{[-1,1]}$
there exists a differentiable solution $x(t)$ for $t\geq 0$, of
\eqref{1} with the initial condition \eqref{1.3}, then this
solution belongs to the space $ C^{\infty}_{[-1,+\infty)}$.
\end{corollary}

\begin{proof} Since there exists a differentiable
solution to problem \eqref{1} and \eqref{1.3}, we have
$$
\varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),\quad
n=0,1,\dots .
$$
Then formula \eqref{1.11}  holds, as well as its equivalent
formulas \eqref{1.9} and \eqref{1.10}, therefore the equivalence
between \eqref{1.12} and  \eqref{1.13} follows. Since in each
interval $(m,m+1)$ the formulas for $x'(t)$ and $x^{(i)}(t)$ are
reached by differentiating formula \eqref{1.5} or \eqref{1.6}
according to the case, one or i-times respectively, then the
equality $x^{(i)}(k^+)=x^{(i)}(k^-)$ in terms of $\varphi$ has the
same shape than the equality $x'(k^+)=x'(k^-)$ in terms of
$\varphi$, but now with the derivatives of $\varphi$ increased in
an $i$-order degree. Hence
$$
\varphi^{(k+i)}(0)=
\varphi^{(k+i-1)}(-1)+\varphi^{(k+i-1)}(1)\,\,\mbox{for any natural number} \,(k+i)\,,
$$
due to the equivalence between \eqref{1.12} and \eqref{1.13}. But
this is equivalent to $x^{(i)}(k^+)=x^{(i)}(k^-)$, for all
$i=0,1,2\dots$ and each natural number $k$, which means that
$x(t)\in C^{\infty}_{[-1,+\infty)}$. \end{proof}

\begin{remark} \rm
For certain initial functions, we can define the semigroup
associated to the solutions $x(t)$ of  \eqref{1}. This is
another way to build solutions on the whole real line, which
has been developed in \cite{iv}.
\end{remark}

\begin{theorem} \label{thm3.2}
Let $ \varphi\in C^{\infty}_{[-1,1]}$. If a solution  $x(t)$ of
equation \eqref{1}-\eqref{1.3} exists and is differentiable,
then the solution is unique.
\end{theorem}

\begin{proof}
 On the open intervals the solution coincides with  \eqref{1.5} or
\eqref{1.6}, while in the integer points the solution is obtained
uniquely due to the continuity of $x(t)$.
\end{proof}

\begin{thebibliography}{00}

\bibitem{aw}
Aftabizedeh A. R., Wiener J.; {\it Oscillatory and periodic
solutions of advanced differential equations with piecewise
constant argument}, In Nonlinear Analysis and Applications. V.
Lakshmikanthan, ed. Marcel Dekker Inc., N. Y. and Basel, 31-38,
(1987).


\bibitem{bc}
Bellman R., Cooke K.; {\it Differential-difference equations},
Editorial Mir, Moscu, (1967).

\bibitem{dr} Driver R. D.; {\it Ordinary and Delay
Differential Equations}, Applied Mathematical
Sciences 20, Springer Verlag, Berlin (1977).

\bibitem{elg} Elgoltz L. E., Norkin C. B.;
{\it Introduction to the Theory of Differential
Equations with Deviating Arguments}, Nauka, Moscow (in
Russian) (1971).

\bibitem{gop} Gopalsamy  K.; {\it Stability and
Oscillations in Delay Differential Equations of
Populations Dynamics}, Kluwer, Dordrecht (1992).

\bibitem{gl}
Gyory I., Ladas G.; {\it Oscillation theory of delay differential
equations with applications}, Oxford University Press, Oxford,
(1991).

\bibitem{ha}
Hale J. K.; {\it Theory of  functional differential equations},
Springer-Verlag, New York (1977).

\bibitem{hlun} Hale J. K.,Verduyn Lunel S. M.;
{\it Introduction to Functional Differential Equations},
Springer-Verlag, New York (1993).

\bibitem{iv}
Iakovleva V., Vanegas C.; {\it Spectral analysis of strongly
continuous semigroups associated to solutions of differential
delay-advance equations}, preprint (2004).

\bibitem{kw}
Kenneth L., Wiener J.; {\it A Survey of Differential Equations
with Piecewise Continuous Arguments}. Delay Differential Equations
and Dynamical Systems, eds. S. Busenberg and M. Mantelli. Lectures
Notes in Math. 1475. Springer Verlag, Berlin, 1-15, (1991).

\bibitem{l}
Ladas G.; {\it Oscillations of equations with piecewise constant
mixed arguments}. In Differential Equations and Applications. II
Aftabizedeh, ed. Ohio University Press, Athens, 64-69, (1989).

\end{thebibliography}



\end{document}