\documentclass[reqno]{amsart} \usepackage{hyperref} \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 $$\label{0} x'(t)= A x(t-\omega) + B x(t+\omega) + C x(t),$$ 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 $$\label{1} x'(t)=x(t-1)+x(t+1).$$ 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 $$\label{1.2} x(t)=x'(t-1)-x(t-2).$$ 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 $$\label{1.3} x(t)=\varphi(t)= \begin{cases} \varphi_1(t), &t\in[-1,0]\\ \varphi_2(t), &t\in[0,1],\end{cases}$$ 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 $t\in(1,2)$, $$\label{1.4} x(t)=\varphi'_2(t-1)-\varphi_1(t-2)=\varphi'(t-1)-\varphi(t-2)\,.$$ For $t\in(2,3)$, \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} 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 $(2l-1,2l)$, $$\label{1.5} x(t)=\sum_{k=0}^{l-1}(c_{2k}\varphi^{(2k)}(t-2l)+c_{2k+1} \varphi^{(2k+1)}(t-(2l-1))),$$ and on the interval $(2l,2l+1)$, $$\label{1.6} x(t)=\sum_{k=0}^{l}c_{2k}\varphi^{(2k)}(t-2l) +\sum_{k=0}^{l-1}c_{2k+1}\varphi^{(2k+1)}(t-(2l+1))\,,$$ 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 $(2l+1,2l+2)$: $$\label{1.51} x(t)=\sum_{k=0}^{l}(c_{2k}\varphi^{(2k)}(t-(2l+2))+c_{2k+1} \varphi^{(2k+1)}(t-(2l+1)))\,.$$ 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 $(2l+2,2l+3)$: $$\label{1.61} x(t)=\sum_{k=0}^{l+1}c_{2k}\varphi^{(2k)}(t-(2l+2)) +\sum_{k=0}^{l}c_{2k+1}\varphi^{(2k+1)}(t-(2l+3))\,.$$ 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 $$\label{1.8} x^{(i)}(m^+)=x^{(i)}(m^-),\quad i=0,1;\; m=1,2,\dots ,$$ where $$\begin{gathered} x^{(i)}(m^+):= \lim_{\varepsilon\to 0,\; \varepsilon>0} x^{(i)}(m+\varepsilon) \,,\\ x^{(i)}(m^-):= \lim_{\varepsilon\to 0,\; \varepsilon>0} x^{(i)}(m-\varepsilon). \end{gathered}$$ 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 $$\label{1.11} \varphi^{(n+1)}(0)=\varphi^{(n)}(-1)+\varphi^{(n)}(1),\quad n=1,2,\dots k-1 \,.$$ But if \eqref{1.9} and \eqref{1.10} hold, then it follows the equivalence $$\label{1.12} x'(k^+)=x'(k^-)$$ if and only if $$\label{1.13} \varphi^{(k+1)}(0)=\varphi^{(k)}(-1)+\varphi^{(k)}(1).$$ 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 $$\label{1.14} x'((s+1)^+)=x'((s+1)^-)\Longleftrightarrow x''(s^+)=x''(s^-)$$ 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. 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