\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 103, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/130\hfil Asymptotic behavior of solutions] {Asymptotic behavior of solutions to a first-order non-homogeneous delay differential equation} \author[Q. Zhou \hfil EJDE-2011/103\hfilneg] {Qiyuan Zhou} \address{Qiyuan Zhou \newline College of Mathematics and Computer Science, Hunan University of Arts and Science, Changde, Hunan 415000, China} \email{zhouqiyuan65@yahoo.com.cn} \thanks{Submitted May 20, 2011. Published August 10, 2011.} \subjclass[2000]{34C12, 39A11} \keywords{Asymptotic behavior; delay differential equation} \begin{abstract} In this article, we study the asymptotic behavior of solutions to the delay differential equation $$x'(t)=f(t,x(t),x(t-r(t)))\,.$$ It is shown that every solution tends to either $\infty$ or a constant as $t\to \infty$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Consider the delay differential equation $$x'(t)=f(t,x(t),x(t-r(t))), \label{e1.1}$$ where $f\in C(\mathbb{R} \times \mathbb{R} \times \mathbb{R} )$ and $r\in C(\mathbb{R})$. In this article, we assume the following: $f(t,u,v)$ is non-increasing in $u$; the delay may be unbounded from above, but it is bounded from below by a positive constant, $0<\tau \leq r(t)$; the function $\lambda(t):=t-r(t)$ is non-decreasing, and $\lim_{t\to \infty}\lambda(t)=\infty$. Let $$\label{e1.1a} \alpha(s)=\sup\{t:\lambda(t)=s\}.$$ Then $\alpha(\lambda(t))\geq t$ and $\alpha(s)>s$, and when $\lambda(t)$ is strictly increasing, $\alpha$ is the inverse function of $\lambda$; i.e., $\alpha(\lambda(t))=t$. The initial condition for \eqref{e1.1} is a continuous function $\phi$ such that \[ x(t)=\phi(t) \quad \text{for } t\in E_{t_0}:=\{t_0\}\cup\{t-r(t)0$for all$u_2>u_1$;$q_i\in L^1[0,\infty)(i=1, 2, 3)$,$p, q_1, q_2$; and $$\int_t ^{\alpha(t)}p(s)ds\leq M,\quad t\in\mathbb{R}\,. \label{e1.5}$$ Then every solution of \eqref{e1.1} is bounded above, and tends to either a constant or to$-\infty$, as$t\to \infty$. \end{theorem} Recently, Yi \cite{y1} pointed out a mistake in \cite[Proposition 4]{c2}. Unfortunately, the similar mistake appears in \cite[Lemma 1]{c1} and \cite[Lemma 2]{c2}. Moreover, we found that \cite[condition (1.5)]{c2} should be strengthened to be $$p(t)> 0, \quad \int_t ^{\alpha(t)}p(s)ds\leq M,\quad \forall t\in\mathbb{R}\,. \label{e1.6}$$ The main purpose of this paper is to show the convergence of the solutions of \eqref{e1.1}. Our approach is quite different from the one in \cite{d1,y1}, and our conditions are weaker than those in \cite{y1}. Ofcourse, our proofs ovoid the mistakes in \cite{c1,c2}. \section{Preliminary results} We start with a well known result in differential equations. \begin{lemma}[See \cite{t1}] \label{lem2.1} Let$x_0\in\mathbb{R}$,$\beta>0$,$h\in C([x_0, x_0+\beta]\times \mathbb{R}, \mathbb{R})$, and$h$be a non-increasing in the second variable. Then the initial value problem $$\label{e2.1} \begin{gathered} \frac{dy}{dx}=h(x,y)\\ y(x_0)=y_0 \end{gathered}$$ has a unique solution on the interval$[x_0, x_0+\beta]$. \end{lemma} \begin{lemma} \label{lem2.2} Assume$\phi:E_{t_0}\to \mathbb{R}$is a continuous function. Then the initial-value problem $$\begin{gathered} x'(t)=f(t,x(t),x(t-r(t))),\quad t\geq t_0 \\ x(t)=\phi(t),\quad t\in E_{t_0} \end{gathered} \label{e2.4}$$ has a unique solution on$[t_0,\infty)$. \end{lemma} \begin{proof} We find the solution on intervals of length$\tau$, where$\tau$is the lower bound for the delay$r(t)$. For$t\in [t_0,t_0+\tau]$, let $$h(t,x(t)))=f(t,x(t),x(t-r(t)))=f(t,x(t),\phi(t-r(t))).$$ Then by Lemma \ref{lem2.1}, there exists a unique solution$x(t)$on$[t_0, t_0+\tau]$. Recursively, we can build a unique solution of \eqref{e2.4} for any interval$[t_0,T]$. The proof is complete. \end{proof} Using the same argument as in \cite[Proposition 3]{d1}, we can prove the following result. \begin{lemma} \label{lem2.3} Suppose that$G(u,u)\equiv 0$for all$ u\in\mathbb{R}$,$ G(u,v)$is non-increasing in$u$, and non-decreasing in$v$. Then the initial-value problem \begin{gather} \frac{du }{dt}=G(u,k), \label{e2.6} \\ u(t_0)=u_0, \label{e2.7} \end{gather} where$k$is constant in$\mathbb{R}$, has a unique solution$u=u(t,k)$on$[t_0,+\infty)$, and the function$\phi(k)=u(t,k)$is continuous in$k$. \end{lemma} \begin{lemma} \label{lem2.4} Suppose that$G(u,u)\equiv 0$for all$ u\in\mathbb{R}$,$ G(u,v)$is non-increasing in$u$, and non-decreasing in$v$. Consider the initial-value problem \begin{gather} \frac{du }{dt}=p(t)G(u,c+\varepsilon), \label{e2.8}\\ u(t_0)=u_0,\quad u_00, \quad \int_t ^{\alpha(t)}p(s)ds\leq M \quad \text{for all } t\in\mathbb{R}\,, \label{e2.10} where$\alpha(t)$is defined by \eqref{e1.1a}. Then there exists a positive constant$\mu$, independent of$t_0$and$\varepsilon$, such that $$u(t,t_0,\varepsilon)\leq c+\varepsilon-\mu, \quad\text{for } t_0\leq t\leq\alpha(t_0)\,.$$ \end{lemma} \begin{proof} The change of variables $$s=\int_{t_0}^{t}p(\xi)d\xi, \quad v(s)=u(t) \label{e2.11}$$ transform \eqref{e2.8}--\eqref{e2.9} into \begin{gather} \frac{d v}{ds}=G(v(s),c+\varepsilon),\quad s\geq 0, \label{e2.8p}\\ v(0)=u_0, \quad u_00$. Let $$\mu=\min_{0\leq\varepsilon\leq|c|/2} \psi(\varepsilon)>0.\label{e2.13}$$ It follows from \eqref{e2.12}, \eqref{e2.8p}, $G(u,u)=0$, and $G(u,v)$ being non-increasing in $u$, that $\frac{\partial v}{\partial s}v(s,\varepsilon) \geq 0$, and $$v(s,\varepsilon)\leq v(M,\varepsilon)\quad \text{for all }s\in[0,M],$$ which implies $$v(s,\varepsilon)\leq c+\varepsilon-\psi(\varepsilon) \leq c+\varepsilon-\mu\quad \text{for all }s\in[0,M].\label{e2.14}$$ By the relationship between the initial value problems \eqref{e2.8}-\eqref{e2.9} and \eqref{e2.8p}--\eqref{e2.9p}, it follows from \eqref{e2.10} that $$u(t,t_0,\varepsilon)\equiv v(s(t),\varepsilon) \quad \text{for } t_0\leq t\leq\alpha(t_0).\label{e2.15}$$ Again from \eqref{e2.10} and \eqref{e1.1a}, we obtain $$s(\alpha(t_0))\leq M.\label{e2.16}$$ By \eqref{e2.14}, \eqref{e2.15} and \eqref{e2.16}, we have $$u(t,t_0,\varepsilon)\leq c+\varepsilon-\mu\quad \text{for } t_0\leq t\leq\alpha(t_0).$$ Since $M$ is independent of $t_0$, and $\mu$ is independent of $t_0$ and $\varepsilon$, the proof is complete. \end{proof} By a similar argument, we can prove the following result. \begin{lemma} \label{lem2.5} Suppose that $G(u,u)\equiv 0$ for all $u\in\mathbb{R}$, $G(u,v)$ is non-increasing in $u$, and non-decreasing in $v$. Consider the initial-value problem \begin{gather} \frac{du }{dt}=p(t)G(u,c-\varepsilon), \label{e2.17}\\ u(t_0)=u_0,\quad u_0>c, \label{e2.18} \end{gather} where $c$ is a nonzero constant and $\varepsilon$ is a parameter such that $0\leq\varepsilon\leq |c|/2$. Denote by $u=u(t,t_0,\varepsilon)$ be the solution of the initial value problem. If (A1) and \eqref{e2.10} hold, then there exists a positive constant $\nu$ independent of $t_0$ and $\varepsilon$ such that $$u(t,t_0,\varepsilon)\geq (c-\varepsilon)+\nu\quad \text{for } t_0\leq t\leq\alpha(t_0).$$ \end{lemma} \begin{remark} \label{rmk2.2} \rm If (A1) holds for all $\eta\in\mathbb{R}$ and $\varepsilon\in [0, 1]$, using the method in the proof of Lemma \ref{lem2.4}, we can show that the conclusions in Lemmas \ref{lem2.4} and \ref{lem2.5} hold for any $c\in\mathbb{R}$. \end{remark} \section{Main results} \begin{theorem} \label{thm3.1} Assume that $f(t,u,v)$ non-increasing in $u$, and $$\label{e31i} f(t,u,v)\leq p(t)G(u,v)+q_1(t)u+q_2(t)v+q_{3}(t).$$ where $G(u,v)\in C(\mathbb{R}\times\mathbb{R})$ and $p,q_i\in C(\mathbb{R})$ satisfying the following conditions: $G(u,v)$ is non-increasing in $u$, and non-decreasing in $v$; $G(u,u)\equiv 0$ for all $u\in\mathbb{R}$; $q_i\in L^1[0,\infty)$ $(i=1, 2, 3)$; and $p, q_1, q_2$ are non-negative; and {\rm (A1)} and \eqref{e2.10} hold. Then every solution of \eqref{e1.1} is bounded above. Furthermore, if $\limsup_{t\to \infty} x(t)\neq 0$, then $x(t)$ tends to either a constant or to $-\infty$ as $t\to \infty$. \end{theorem} \begin{proof} We first prove that every solution of \eqref{e1.1} is bounded above. Let $$y_1(t)=\max\{\max_{t_0-r(t_0)\leq s\leq t}x(s),\,1\}, \quad S_1=\{t\geq t_0: y_1(t)=x(t)\}.$$ Let $D^{+}$ denote the upper right derivative. Then $D^{+}y_1(t)=0$ for $t\in[t_0,+\infty)\backslash S_1$, and $D^{+}y_1(t)\leq\max\{x'(t),0\}$ a.e. on $S_1$. From \eqref{e1.1} and \eqref{e31i}, $$\label{e3.1} \begin{split} x'(t) &\leq p(t)G(x(t),x(t-r(t)))+q_1(t)x(t)+q_2(t)x(t-r(t))+q_{3}(t)\\ &\leq p(t)G(x(t),y_1(t))+q_1(t)y_1(t)+q_2(t)y_1(t)+q_{3}(t)\\ &\leq p(t)G(x(t),y_1(t))+q_1(t)y_1(t)+q_2(t)y_1(t)+|q_{3}(t)| \quad \forall t\geq t_0. \end{split}$$ Since $G(u,u)\equiv 0$ for all $u\in\mathbb{R}$, and $D^{+}y_1(t)\leq\max\{x'(t),0\}$ a.e. on $[t_0,+\infty)$, we obtain $$D^{+}y_1(t)\leq q_1(t)y_1(t)+q_2(t)y_1(t)+|q_{3}(t)|\quad \text{a.e. on } [t_0,+\infty).$$ From $y_1(t)\geq1$, we have $$\frac{D^{+}y_1(t)}{y_1(t)}\leq q_1(t)+q_2(t)+|q_{3}(t)|\quad \text{a.e. on }[t_0,+\infty).$$ Again from the monotonicity of $y_1(t)$, we obtain that $y_1(t)$ is differentiable almost everywhere on $[t_0,\infty)$. Thus $$\ln\big(\frac{y_1(t)}{y_1(t_0)}\big) \leq \int_{t_0}^{+\infty}q_1(t)dt +\int_{t_0}^{+\infty}q_2(t)dt+\int_{t_0}^{+\infty}|q_{3}(t)|dt <+\infty\quad \forall t\geq t_0,$$ which implies $y_1(t)$ is bounded above; thus $x(t)$ is also bounded above. Set $A=\limsup_{t\to \infty}x(t) <\infty$. If $\limsup_{t\to \infty}x(t)=-\infty$, then $\lim_{t\to \infty}x(t)=-\infty$, which implies that Theorem \ref{thm3.1} holds. Next we assume that $A$ is a nonzero real number and show that $\lim_{t\to \infty}x(t)=A$. By contradiction, assume that $\lim_{t\to \infty}x(t)$ does not exist. For each $\mu_1\in [0, |A|/2]$, there exists let $t^*>t_0$ large enough such that \begin{gather} x(t)\leq A+\mu_1, \quad x(t-r(t))\leq A+\mu_1\quad \forall t\geq t^*, \label{e3.2}\\ \int_{t^*}^{+\infty} [(q_1(t)+q_2(t))A+\mu_1+|q_{3}(t)|]dt\leq\mu_1. \label{e3.3} \end{gather} For $t\geq t^*$, let $$n_1(t)=x(t)-\int_{t^*}^{t}[(q_1(s)+q_2(s))A+\mu_1+|q_{3}(s)|]ds\,. \label{e3.4}$$ Obviously, $n_1(t)$ is bounded above, and $\lim_{t\to \infty}n_1(t)$ does not exist. Let $B=\limsup_{t\to \infty}n_1(t)$ and $b=\limsup_{t\to \infty}n_1(t)$; Thus $bt^*$ and $t_m\to \infty$ as $m\to \infty$. It follows from \eqref{e31i} and \eqref{e3.2} that $$x'(t)\leq p(t)G(x(t),A+\mu_1)+(q_1(t)+q_2(t))A +\mu_1+|q_{3}(t)|\quad \text{for all } t\geq t^*.$$ From \eqref{e3.4}, we obtain $x(t)\geq n_1(t)$ and $$n_1'(t)\leq p(t)G(n_1(t),A+\mu_1)\quad \text{for all } t\geq t^*.\label{e3.5}$$ For each $m$, we consider the initial-value problem \begin{gathered} u'(t)=p(t)G(u(t),A+\mu_1) \\ u(t_m)=H, \quad H0$independent of$t_m$and of$\mu_1$, such that $$u(t)\leq A+\mu_1-\mu \quad \text{for } t_m\leq t\leq\alpha(t_m).$$ Then, by the comparison theorem and \eqref{e3.5}, we obtain $$n_1(t)\leq u(t)\leq A+\mu_1-\mu\quad \text{for } t_m\leq t\leq\alpha(t_m),$$ thus$ x(t)\leq A+2\mu_1-\mu$for$t_m\leq t\leq\alpha(t_m)$. Choosing$\mu_1\in (0,\mu/4]$, we have $$x(t)\leq A-\frac{\mu}{2}\quad \text{for } t_m\leq t\leq\alpha(t_m),\; m=1,2,\dots. \label{e3.7}$$ On the other hand, define $$y_2(t)=\max_{\lambda(t)\leq s\leq t}x(s),\quad S_2=\{t:t\in[t^*,\infty),y_2(t)=x(t)\}.$$ Then$D^{+}y_2(t)\leq0$for all$t\in[t^*,\infty)\backslash S_2$, and$D^{+}y_2(t)\leq\max\{x'(t),0\}$for all$t\in S_2$. Hence $$D^{+}y_2(t)\leq (q_1(t)+q_2(t))(A+\mu_1)+|q_{3}(t)|\quad \forall t\geq t^*. \label{e3.8}$$ For$t\geq t^*$, denote $$n_2(t)=y_2(t)-\int_{t^*}^{t}[(q_1(s)+q_2(s))(A+\mu_1)+|q_{3}(s)|]ds.$$ From \eqref{e3.8}, we obtain$D^{+}n_2(t)\leq0$for all$t\geq t^*$; therefore,$n_2(t)$is non-increasing. Since$\limsup_{t\to \infty}x(t)>-\infty$, $$\lim_{t\to \infty}y_2(t)=\lim_{t\to \infty}n_2(t) + \lim_{t\to \infty}\int_{t^*}^{t} [(q_1(s)+q_2(s))(A+\mu_1)+|q_{3}(s)|]ds$$ exists as real number. From the definition of$y_2$and the the fact that$\{s:\lambda(t)\leq s\leq t, t\geq t_0\}\supset [t_0,\infty)$, it follows that $$\lim_{t\to \infty}y_2(t) =\lim_{t\to \infty}\max_{\lambda(t)\leq s\leq t}x(s) =\limsup_{t\to \infty} x(t)=A. \label{e3.9}$$ Since$\lambda(t)\to+\infty$as$t\to+\infty$, for each$t_m$there exists$t_m'$such that$t_m=\lambda(t_m')$. Then$\alpha(t_m)=\alpha(\lambda(t_m'))\geq t_m'$and$t_m'\geq t_m\geq t^*$. By \eqref{e3.7}, $$y_2(t_m')\leq A-\frac{\mu}{2}\quad \text{for } m=1,2,\dots.\label{e3.10}$$ However, \eqref{e3.9} implies$\lim_{m\to+\infty}y_2(t_m')=A$which contradicts \eqref{e3.10}. Hence$\lim_{t\to \infty}x(t)$exists, and$\lim_{t\to \infty}x(t)=A$. This completes the proof. \end{proof} In a similar fashion, by using Lemma \ref{lem2.5}, we can show the following result. \begin{theorem} \label{thm3.2} Assume$f(t,u,v)$non-increasing in$u$, and $$\label{e32i} f(t,u,v)\geq p(t)G(u,v)+q_1(t)u+q_2(t)v+q_{3}(t),$$ where$G(u,v)\in C(\mathbb{R}\times\mathbb{R})$and$p,q_i\in C(\mathbb{R})$satisfying the following conditions:$G(u,v)$non-increasing in$u$, and non-decreasing in$v$;$G(u,u)\equiv 0$for all$u\in\mathbb{R}$;$q_i\in L^1[0,\infty)(i=1, 2, 3)$,$p, q_1, q_2$are nonnegative; and {\rm (A1)} and \eqref{e2.10} hold. Then every solution of \eqref{e1.1} is bounded below. Furthermore, if$\limsup_{t\to \infty} x(t)\neq 0$, then$x(t)$tends to either a constant or to$\infty$as$t\to \infty$. \end{theorem} \begin{theorem} \label{thm3.3} Consider the differential equation $$x'(t)=p(t)G(x(t),x(t-r(t)))+q_1(t)x(t )+q_2(t)x(t-r(t))+q_{3}(t), \label{e3.12}$$ where$G(u,v)\in C(\mathbb{R}\times\mathbb{R})$and$p,q_i\in C(\mathbb{R})$satisfying the following conditions:$G(u,v)$is non-increasing in$u$, and non-decreasing in$v$;$G(u,u)\equiv 0$for all$u\in\mathbb{R}$;$q_i\in L^1[0,\infty)(i=1, 2, 3)$,$p, q_1, q_2$are nonnegative; and {\rm (A1)} and \eqref{e2.10} hold. Then every solution of \eqref{e3.12} tends to a constant as$t\to\infty$. \end{theorem} The proof of the above theorem follows immediately from Theorems \ref{thm3.1} and \ref{thm3.2}. \begin{remark} \label{rmk3.1} \rm Let$G(u,v)=-u^{\theta}+v^{\theta}$, where$\theta$is the ratio of two odd positive integers. Then$G(u,v)$is strictly decreasing in$u$, and is strictly increasing in$v$. Moreover,$G(u, \eta)$is continuously differentiable when$u \neq 0$. Applying Cauchy's uniqueness and existence theorem, we conclude that assumption (A1) holds. Therefore, Theorem \ref{thm3.3} confirms the Bernfeld-Haddock conjecture. \end{remark} From Remark \ref{rmk2.2}, and using a similar argument as in the proof of Theorem \ref{thm3.1}, we can also show the following result, under the assumption \begin{itemize} \item[(A1')] For each$\eta$and$t_0$in$\mathbb{R}$, the initial-value problem$\frac{du }{dt}= G(u,\eta)$,$u(t_0)=\eta$has a unique left-hand solution. \end{itemize} \begin{theorem} \label{thm3.4} Assume {\rm (A1')}. Under the hypotheses of Theorem \ref{thm3.1}, every solution of \eqref{e1.1} is bounded above, and tends to either a constant or to$-\infty$, as$t\to \infty$. \end{theorem} \begin{theorem} \label{thm3.5} Assume {\rm (A1')}. Under the hypotheses of Theorem \ref{thm3.2}, every solution of \eqref{e1.1} is bounded below, and tends to either a constant or to$+\infty$, as$t\to \infty$. \end{theorem} The proofs of the two theorems above are similar to the proof of Theorem \ref{thm3.1}: Replace$\mu_1\in [0,|A|/2]$with$\mu_1\in [0,1]$, and then use Remark \ref{rmk2.2}. \begin{remark} \label{rmk3.3} \rm Note that the results in \cite{c1,c2} can be obtained only by assuming condition (A1'), and the strengthened condition \eqref{e2.10}. Since the function$G(u,v)$in this article satisfies weaker conditions than those in \cite{c1,c2}, their results there are special cases in this article. When$r(t)$is constant and$p(t)\$ is a bounded and positive function, \eqref{e2.10} holds naturally. Hence, our results include those in \cite{d1,y1}, and naturally extend the Bernfeld-Haddock conjecture. \end{remark} \subsection*{Acknowledgments} The author would like to thank the anonymous referees for their helpful comments and suggestions. This work was supported by the Scientific Research Fund of Hunan Provincial Natural Science Foundation of PR China (Grants No. 11JJ6006, No. 10JJ6011), and the Natural Scientific Research Fund of Hunan Provincial Education Department of PR China (Grants No. 10C1009, No. 09B072). \begin{thebibliography}{0} \bibitem{b1} S. R. Bernfeld and J. R. Haddock; \emph{A variation of Razumikhims method for retarded functional equstions}, Nonlinear Systems and Applications, An International Conference, Acadimic Press New York, 1977, 561-566. \bibitem{c1} B. S. Chen; \emph{Asymptotic behavior of a class of nonautonomous retarded differential equations} (in Chinese), Chinese Science Bulletin, 1988, 6: 413-415. \bibitem{c2} B. S. 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