\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 302, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/302\hfil Asymptotic stability] {Asymptotic stability of non-autonomous functional differential equations with distributed delays} \author[L. Hatvani \hfil EJDE-2016/302\hfilneg] {L\'aszl\'o Hatvani} \address{L\'aszl\'o Hatvani \newline University of Szeged, Bolyai Institute\\ Aradi v\'ertan\'uk tere 1, H-6720 Szeged, Hungary} \email{hatvani@math.u-szeged.hu} \thanks{Submitted October 5, 2016. Published November 25, 2016.} \subjclass[2010]{34K20, 34K27, 34D20} \keywords{Annulus argument; uniform asymptotic stability} \begin{abstract} We consider the integro differential equation \[ x'(t)=-a(t)x(t)+b(t)\int^t_{t-h} \lambda(s)x(s)\,{\rm d}s,\quad o\leq a(t),\; 0\le t<\infty, \] where $a,b:\mathbb{R}_+\to\mathbb{R}$, $\lambda:[-h,\infty)\to \mathbb{R}$ are piecewise continuous functions and $h$ is a positive constant. We establish sufficient conditions guaranteeing either asymptotic stability or uniform asymptotic stability for the zero solution. These conditions state that the instantaneous stabilizing term on the right-hand side dominates in some sense the perturbation term with delays. Our conditions not require $a$ being bounded from above. The results are based on the method of Lyapunov functionals and Razumikhin functions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Consider the scalar functional differential equation (FDE) \begin{equation} \label{I1} x'(t)=-a(t)x(t)+b(t)\int^t_{t-h} \lambda(s)x(s)\,{\rm d}s,\quad 0\leq a(t)\; t\in \mathbb{R}_+:=[0,\infty), \end{equation} where $a,b:\mathbb{R}_+\to\mathbb{R}$, $\lambda:[-h,\infty)\to \mathbb{R}$ are piecewise continuous and everywhere continuous from the right, $00$ in $C$ around $\varphi=0$. As is usual, if $x: [-h, \beta)\to \mathbb{R}$ $(\beta>0)$, then $x_t(\theta):= x(t+\theta)$ for $-h\le \theta\le 0$, $0\le t<\beta$. Let $x(\cdot; t_0, \varphi): [t_0 -h, t_0+\alpha)\to\mathbb{R}$ denote a solution of \eqref{I1} satisfying the initial condition $x_{t_0}(\cdot; t_0, \varphi) = \varphi$. It is known \cite{HJ} that for each $t_0 \in \mathbb{R}_+$ and $\varphi\in C$ there is exactly one solution $x(\cdot; t_0, \varphi): [t_0-h, \infty) \to \mathbb{R}$. Equation \eqref{I1} is the model of a system in which there act an instantaneous negative feedback stabilizing the equilibrium $x=0$ and a perturbation with distributed delays on the interval $[t-h, t]$. We look for sufficient conditions guaranteeing asymptotic stability for the zero solution of \eqref{I1}. To this end we have to suppose that the stabilizing term on the right-hand side dominates in some sense the perturbation term. In fact, if $a(t)\equiv b(t)\int^t_{t-h}\lambda$, then every constant is a solution of the equation, so the zero solution is not asymptotically stable. The parameters of the system are varying in time, so \eqref{I1} is non-autonomous. In the stability theory of the non-autonomous FDE's \begin{equation}\label{I2} \mathbf{x}'(t)=f(t,x_t) \end{equation} it is typically supposed that $f:[-h,\infty)\times C\to \mathbb{R}$ maps sets $[-h,\infty)\times C_H$ into bounded sets of $C$ (see, e.g., \cite[Theorem 5.2.1]{HJ}). If we apply theorems of this type to \eqref{I1}, then we have to require boundedness on $\mathbb{R}_+$ also of function $a$ (see a similar situation in \cite[equation (5.2.25)]{HJ}). However, in the case of \eqref{I1} this is not a natural condition because, obviously, the larger $a(t)$ is the better from the point of view of the asymptotic stability. So we refine the techniques and allow only conditions not requiring any boundedness type conditions from above of function $a$. Equation \eqref{I1} is not only a model, it is also an important ``test equation" in the stability theory: authors often choose it as an example to illustrate their general theorems on \eqref{I2} \cite{BT, BH, HL2, TW1, TW2}. In these applications the conditions on $a$, $b$, $\lambda$ are accorded with the requirements of the general theorems. In this article, equation \eqref{I1} is in the focus of the investigation, and we look for optimal conditions of asymptotic stability finding the most adequate methods and techniques. Equation \eqref{I1} is linear, but we will not use the consequences of this fact deeply; the results can be easily transformed for the more general equation \[ x'(t)=-a(t)f(x)+b(t)\int^t_{t-h} \lambda(s)g(x(s))\,{\rm d}s,\quad 0\leq a(t),\; 0\ (t\in \mathbb{R}_+) \] treated in \cite[Section 5]{BH}. The reason of the choice $f(x)\equiv g(x)\equiv x$ is that the main ideas can be demonstrated well by this special case so that the formulation of the results is essentially simpler. In Section 2 we give theorems based upon the combination of the method of Lyapunov functionals \cite{HJ, BT, BH} and the annulus argument \cite{HL1}; in Section 3 we use the Lyapunov-Razumikhin method \cite{HJ, KJ, PM, TT, WZZL, ZL}. \section{A Lyapunov functional} The following stability concepts are standard \cite{BT, HJ}. \begin{definition} \label{def2.1} \rm The zero solution of \eqref{I1} is: \begin{itemize} \item[(a)] \emph{stable} if for every $\varepsilon > 0$ and $t_0 \ge 0$ there is a $\delta(\varepsilon, t_0)>0$ such that $\|\varphi\|<\delta$, $t\ge t_0$ imply that $|x(t; t_0, \varphi)|<\varepsilon$. \item[(b)] \emph{uniformly stable} if for every $\varepsilon > 0$ there is a $\delta(\varepsilon) > 0$ such that $\|\varphi\|<\delta, t_0 \ge 0, t\ge t_0$ imply that $|x(t; t_0, \varphi)|< \varepsilon$. \item[(c)] \emph{asymptotically stable} if it is stable and for every $t_0 \ge 0$ there is a $\sigma(t_0)>0$ such that $\|\varphi\|<\sigma$ implies $\lim_{t\to \infty} x(t; t_0, \varphi) = 0$. \item[(d)] \emph{uniformly asymptotically stable} (UAS) if it is uniformly stable and there is a $D>0$ and for each $\mu>0$ there is a $T(\mu)$ such that $t_0 \in \mathbb{R}_+$, $\|\varphi\| < D$, $t\ge t_0 + T$ imply that $|x(t; t_0, \varphi)|<\mu$. \end{itemize} \end{definition} It can be seen that the zero solution is stable if and only if all solutions are bounded on $\mathbb{R}_+$, and it is asymptotically stable if and only if all solutions tend to zero as $t$ tends to infinity. A continuous functional $V:\mathbb{R}_+\times C\to \mathbb{R}_+$ which is locally Lipschitz in $\varphi$ is called a \emph {Lyapunov functional} if its right derivative with respect to system \eqref{I1} is non-positive: \[ V'_{\eqref{I1}} (t, \varphi) = V'(t,\varphi) :=\limsup_{\delta\to 0+0}\Big( \frac{1}{\delta} \big(V(t+\delta, x_{t+\delta} (\cdot, t, \varphi)) -V(t,\varphi)\big)\Big) \le 0. \] We consider the functional \begin{equation}\label{L1} \begin{split} V_1(t,\varphi):&= |\varphi(0)| + \int^0_{-h} \int^0_\theta |b(t+\vartheta-\theta)|\, |\lambda(t+\vartheta)||\varphi(\vartheta)|\,{\rm{d}}\vartheta \,{\rm{d}}\theta\\ &=V_2(t,\varphi)+V_3(t,\varphi), \quad V_2(t,\varphi):=|\varphi(0)|. \end{split} \end{equation} The following lemma says that $V_1$ is a Lyapunov functional under appropriate conditions. \begin{lemma}\label{absvalder} The derivative of the functional in \eqref{L1} satisfies the inequality \begin{equation}\label{L2} V_1'(t,\varphi)\le -\Big(a(t)-|\lambda(t)|\int_t^{t+h}|b|\Big)|\varphi(0)| \quad (t\in\mathbb{R}_+). \end{equation} \end{lemma} \begin{proof} Let us consider the solution $t\mapsto x(t)=x(t;t_0,\varphi)$. If $\varphi(0)\not = 0$, then \begin{equation}\label{L3} \begin{split} V'_2(t_0,\varphi) &=\operatorname{sign}(\varphi(0))x'(t_0)\\ &=-a(t_0)|\varphi(0)| +\operatorname{sign}(\varphi(0))b(t_0)\int _{t_0-h}^{t_0}\lambda(s)\varphi(s-t_0)\,{\rm d}s\\ &\le -a(t_0)|\varphi(0)|+|b(t_0)|\int _{t_0-h}^{t_0}|\lambda(s)||\varphi(s-t_0)|\,{\rm d}s. \end{split} \end{equation} If $\varphi(0)= 0$, then \begin{equation}\label{L4} \begin{split} V'_2(t_0,\varphi) &=\limsup_{\delta\to 0+0} \Big(\frac{1}{\delta}\left(|x(t_0+\delta)|-|x(t_0)|\right)\Big) \\ &\le \limsup_{\delta\to 0+0} \big|\frac{x(t_0+\delta)-x(t_0)}{\delta}\big|\\ &=|x'(t_0)| \le -a(t_0)|\varphi(0)|+ |b(t_0)|\int_{t_0-h}^{t_0}|\lambda(s)||\varphi(s-t_0)|\,{\rm d}s \end{split} \end{equation} because $\varphi(0)= 0$. On the other hand, \[ V_3(t,x_t)=\int^0_{-h} \int^t_{t+\theta} |b(s-\theta)|\, |\lambda(s)||x(s)| \,{\rm{d}}s\,{\rm{d}}\theta, \] therefore, \begin{align*} &V'_3(t_0,\varphi) \\ &=\limsup_{\delta\to 0+0}\Big( \frac{1}{\delta}\Big( \int^0_{-h}\Big( \int^{t_0+\delta}_{t_0} |b(s-\theta)||\lambda(s)||x(s)|\,{\rm{d}}s\hfill\\ &\quad -\int^{t_0+\theta+\delta}_{t_0+\theta} |b(s-\theta)||\lambda(s)||x(s)|\,{\rm{d}}s\Big) \,{\rm{d}}\theta\Big) \Big)\\ &=|\lambda(t_0)|\Big(\int_{-h}^0 |b(t_0-\theta)|\,{\rm d}\theta\Big)|\varphi(0)| -|b(t_0)|\int_{t_0-h}^{t_0}|\lambda(s)||\varphi(s-t_0)|\,{\rm d}s. \end{align*} From $V_1'(t_0,\varphi)\le V_2'(t_0,\varphi)+V_3'(t_0,\varphi)$ we obtain \eqref{L2}. \end{proof} Lemma \ref{absvalder} suggests the first condition on the dominance of the negative feedback: \begin{itemize} \item[(A1)] $a(t)-|\lambda(t)|\int_t^{t+h}|b|\ge 0\quad (t\in\mathbb{R}_+$). \end{itemize} \begin{lemma}\label{stab} If condition {\rm (A1)} is satisfied, then the zero solution of \eqref{I1} is stable. If, in addition, \begin{equation}\label{L5} \Big(\int^{t+h}_t|b|\Big)\Big(\int^t_{t-h} |\lambda|^2\Big)^{1/2} \le K_1\quad (t\in R_+) \end{equation} with some constant $K_1$, then the zero solution is uniformly stable. \end{lemma} \begin{proof} By Lemma \ref{absvalder} and condition (A1), for any solution $x(\cdot;t_0,\varphi)$ the function $t\mapsto V_1(t,x_t)$ is non-increasing. Since $x(t)\le V_1(t,x_t)$, every solution is bounded, which proves stability. To prove uniform stability, let us estimate functional in \eqref{L1}: \begin{equation*} \begin{split} V_1(t,\varphi) &=|\varphi(0)| + \int^0_{-h}|\lambda(t+\vartheta)| |\varphi(\vartheta)| \Big(\int^\vartheta_{-h} |b(t+\vartheta-\theta)|\,{\rm{d}}\theta\Big)\,{\rm{d}}\vartheta\\ &\le |\varphi(0)| + \Big(\int^{t+h}_t|b|\Big) \int^0_{-h} |\lambda(t+\vartheta)||\varphi(\vartheta)| \,{\rm{d}}\vartheta\\ &\le |\varphi(0)| + \Big(\int^{t+h}_t|b|\Big) \Big(\int^t_{t-h} |\lambda|^2\Big)^{1/2} \Big(\int^0_{-h} |\varphi|^2\Big)^{1/2}. \end{split} \end{equation*} For arbitrarily fixed $\varepsilon>0$ and $t_0\in \mathbb{R}_+$, let us define $\delta(\varepsilon):=\varepsilon/(1+K_1)$. If $\|x_{t_0}\|<\delta(\varepsilon)$, then \[ |x_t|\le V_1(t,x_t)\le V_1(t_0,x_{t_0})<\varepsilon \quad(t\ge t_0)), \] which proves uniformity. \end{proof} The following concept is widely used in stability theory of non-autonomous differential equations \cite{ BH, KAJ1, KAJ2, KJ, MAT, SU}. \begin{definition} \label{def2.4} \rm A locally integrable function $\eta: \mathbb{R}_+ \to \mathbb{R}_+$ is called \emph{integrally positive} if for every $\delta>0$ the inequality \begin{equation}\label{L51} \liminf_{t\to \infty} \int^t_{t-\delta} \eta> 0 \end{equation} holds. \end{definition} Now we need this concept in a more general form. \begin{definition} \label{def2.5}\rm Let $\eta,M: \mathbb{R}_+ \to \mathbb{R}_+$ be locally integrable, and for any $\nu>0$ define \begin{equation} \label{L52} \Delta_M(t,\nu):=\inf\{\tau>0: \int_{t-\tau}^{t}M=\nu\}. \end{equation} Function $\eta$ is called \emph{integrally positive with respect to $M$} if for every $\nu>0$ \begin{equation} \label{L53} \liminf_{t\to\infty}\int^t_{t-\Delta_M(t,\nu)}\eta>0, \end{equation} i.e., for every $\nu>0$ there exists $W(\nu)>0$ and $t_*(\nu)$ such that if $t>t_*(\nu)$, then $\int^t_{t-\Delta_M(t,\nu)}\eta>W(\nu)$. \end{definition} \begin{remark}\label{posdef} \rm If $M(t)\equiv c=\textrm{const.}$, then $\Delta_M(t,\nu)=\nu/c$, and $\eta$ is integrally positive with respect to $M$ if and only if it is integrally positive. Furthermore, if $t\mapsto\int_0^t M$ is uniformly continuous, then for every $\varepsilon>0$ there exists a $\delta(\varepsilon)>0$ such that $|t'-t''|<\delta(\varepsilon)$ implies $\int_{t'}^{t''} M<\varepsilon$, therefore $\Delta_M(t,\nu)\ge \delta(\nu)$. Consequently, integral positivity of $\eta$ implies integral positivity of $\eta$ with respect to $M$. It may be also interesting that every function is integrally positive with respect to itself. More generally, if $\eta\ge M$, then $\Delta_\eta(t,\nu)\le \Delta_M(t,\nu)$ for all $t$, from which it follows that $\eta$ is integrally positive with respect to $M$. Of course, the converse assertion is not true. \end{remark} The following lemma is based on the method of annulus argument \cite{HL1}. \begin{lemma}\label{annulus} Suppose that condition {\rm (A1)} is satisfied. If, in addition, either \break $M(t):=|b(t)|\int_{t-h}^t|\lambda|$ is integrable on $\mathbb{R}_+$, or function $\eta_1(t):=a(t)-|\lambda(t)|\int_t^{t+h}|b|$ is integrally positive with respect to $M$, then every solution of \eqref{I1} has a finite limit as $t$ goes to infinity. \end{lemma} \begin{proof} Suppose that the statement is not true, i.e., there exists a solution $x$ having no limit at infinity. We can suppose that $|x(t)|\le 1$ ($t\ge t_0$), so there exist $0<\varepsilon_1<\varepsilon_2$ and a sequence $\{r_i,s_i\}_{i=1}^\infty$ such that \begin{gather*} r_1\ge h\quad r_it_*(\varepsilon_2-\varepsilon_1)/2)$. At least one of the last two inequalities is satisfied for infinitely many $i$'s, which means that $v_1(t)\to-\infty$ as $t\to\infty$, but this is a contradiction. \end{proof} \begin{theorem}\label{light} Suppose that condition {\rm (A1)} is satisfied. If \begin{equation}\label{int} \int^\infty \Big(a(t)-|\lambda(t)|\int_t^{t+h}|b|\Big)\,{\rm d}t=\infty, \end{equation} and either $M(t):=|b(t)|\int_{t-h}^t|\lambda|$ is integrable on $\mathbb{R}_+$, or function $\eta_1(t):=a(t)-|\lambda(t)|\int_t^{t+h}|b|$ is integrally positive with respect to $M$, then the zero solution of \eqref{I1} is asymptotically stable. If, in addition, conditions \eqref{L5} and \begin{equation}\label{intunif} \lim_{S\to\infty}\int_t^{t+S} \Big(a(t)-|\lambda(t)|\int_t^{t+h}|b|\Big)\,{\rm d}t=\infty \quad \text{uniformly w.r.t. } t\in\mathbb{R}_+ \end{equation} are assumed, then the zero solution of \eqref{I1} is uniformly asymptotically stable. \end{theorem} \begin{proof} Lemma \ref{annulus} guaranties the existence of a finite limit for every solution of \eqref{I1}. If this limit is not equal to zero for a solution $x$, then \eqref{L2} and \eqref{int} imply $\lim_{t\to\infty}V_1(t,x_t)=-\infty$, which is a contradiction. To proof UAS we start off uniform stability; remember that $\delta(\varepsilon)$ corresponds to $\varepsilon$ in this definition. Setting $D=\delta(1)$ and fixing $\mu>0$ we are looking for $T(\mu)$ introduced in the definition of UAS. Let $t_0\ge 0, \varphi$ ($\|\varphi\|<\delta(1)$) be arbitrary. According to the definition of uniform stability it is enough to proof that there exists a $T(\mu)$ such that $\|x_{t_0+R}(\cdot;t_0,\varphi)\|<\delta(\mu)$ for some $R\in[t_0,t_0+T(\mu)]$. If this is not true, then there is a $\overline\mu>0$ such that for every $T$ there are $t_0,\varphi$ such that $\|x_{t_0+R}(\cdot;t_0,\varphi)\|\ge \delta(\overline\mu)$ for all $R\in[0,T]$. Let us choose a $T$ arbitrarily and fix such $t_0,\varphi$ to it. Then $|x(t)|$=$|x(t;t_0,\varphi)|$ takes a value not less than $\delta(\overline\mu)$ in every subinterval of length $h$ of the interval $[t_0, t_0+T]$. On the other hand, we can prove that $|x|$ often takes values less than $\delta(\overline\mu)/2$. In fact, by \eqref{intunif} there exists $T_1(\overline \mu)$ such that \[ \int_t^{t+T_1(\overline \mu)}\eta_1 >\frac{2(1+K_1)\delta(1)}{\delta(\overline\mu)}\quad (t\in \mathbb{R}_+). \] If $| x(t)|\ge \delta(\overline \mu)/2$ on an interval $[t',t'']\subset [t_0, t_0+T]$, then \begin{align*} -(1+K_1)\delta(1) &\le-v_1(t_0)\le v_1(t'')- v_1(t') \\ &\le -\int _{t'}^{t''}\eta_1(t)|x(t)|\,{\rm d}t \le -\frac{\delta(\overline \mu)}{2}\int_{t'}^{t''}\eta_1, \end{align*} hence $\int_{t'}^{t''}\eta_1\le 2(1+K_1)\delta(1)/\delta(\overline\mu)$. Therefore $|t''-t'|\le T_1(\overline \mu)$, i.e., $|x|$ takes values less than $\delta(\overline \mu)/2$ in every subinterval of length $T_1(\overline\mu)$ in the interval $[t_0, t_0+T]$. According to the latter two paragraphs there is a sequence $\{r_i,s_i\}_{i=1}^N$ such that \begin{equation} \label{L8} \begin{gathered} r_1\ge t_0,\quad r_i0$ and for every sequence $\{s_i\}$ ($s_i\in [t_i-h,t_i]$) there holds \begin{equation}\label{L82} \sum_{i=1}^\infty \int_{\max\{t_{i-1};s_i-\Delta(s_i,\varepsilon)\}}\eta_1=\infty. \end{equation} \end{itemize} Then the zero solution of \eqref{I1} is asymptotically stable. Suppose that, instead of \eqref{L81} estimate \eqref{L5} holds, and instead of \eqref{L82}, \begin{itemize} \item[(ii')] there are a sequence $\{t_i\}_{i=1}^\infty$ and a constant $K_2$ ($h\le t_{i+1}-t_i\le K_2$) such that for every $\varepsilon>0$ there is a $\kappa_1(\varepsilon)>0$ with \begin{equation}\label{L83} \int_{t-\Delta_M(t,\varepsilon)\}}^t\eta_1\ge \kappa_1(\varepsilon) \quad (t\in[t_i-h,t_i]); \end{equation} \end{itemize} moreover there is a constant $\kappa_2>0$ with \begin{itemize} \item[(iii')] $\int _{t_{i-1}}^{t_i-h}\eta\ge \kappa_2 \quad(i\in\mathbb{N})$. \end{itemize} Then the zero solution of \eqref{I1} is uniformly asymptotically stable. \end{theorem} \begin{proof} First we show that under conditions (i)-(ii) every solution $x$ tends to zero as $t\to\infty$. Since $V_1(t_i,x_{t_i})\le (1+K_1) \|x_{t_i}\|$, it is sufficient to prove $\liminf_{i\to\infty}\|x_{t_i}\|=0$. If this is not true, then we may suppose without any loss of the generality that $\|x_{t_i}\|\ge 3\varepsilon>0$ for all $i\in\mathbb{N}$ with some $\varepsilon>0$. This means that there exists a sequence $\{s_i\in [t_i-h,t_i]\}$ having the properties $v_2(s_i):=|x(s_i)|\ge 3\varepsilon$. Then for any $i$, either \begin{itemize} \item[(a)] $v_2(t)\ge\varepsilon$ in the interval $[t_{i-1}, s_i]$, or \item[(b)] there exists $r_i\in[t_{i-1}, s_i]$ such that $v_2(r_i)=\varepsilon$. \end{itemize} In case (a) we integrate \eqref{L2} and obtain \begin{equation} \label{L84} v_1(t_i)-v_1(t_{i-1})\le v_1(s_i)-v_1(t_{i-1}) \le -\varepsilon\int_{t_{i-1}}^{s_i}\eta_1 \le-\varepsilon\int_{t_{i-1}}^{t_i-h}\eta_1. \end{equation} In case (b) either we have \begin{itemize} \item[(b/1)] $\Delta_M(s_i,\varepsilon)\ge s_i-r_i$ and \begin{equation}\label{L85} v_1(t_i)-v_1(t_{i-1})\le v_1(s_i)-v_1(r_i)\le -2\varepsilon +\varepsilon=-\varepsilon<0, \end{equation} \end{itemize} or \begin{itemize} \item[(b/2)] $\Delta_M(s_i,\varepsilon)< s_i-r_i$, when \begin{equation}\label{L86} v_1(t_i)-v_1(t_{i-1})\le -\varepsilon\int_{s_i-\Delta_M(s_i,\varepsilon)}^{s_i}\eta_1. \end{equation} \end{itemize} If case (b/1) occurs infinitely many times, then $\lim_{t\to\infty}v_1(t)=-\infty$, which is a contradiction. Otherwise, there is a natural number $I$ such that for any $i>I$ either (a) or (b/2) is satisfied. Then from condition \eqref{L82} we obtain $\sum_{i=I}^\infty (v_1(t_i)-v_1(t_{i-1}))=-\infty$, which results in a contradiction again. This concludes the proof of asymptotic stability. Now turn to the proof of UAS. Conditions (A1) and \eqref{L5} imply uniform stability; let $\delta(\varepsilon)$ correspond to $\varepsilon$ in the sense of the definition of this property. For $t_0\in\mathbb{R}_+$, $\varphi$ ($\|\varphi\|<1/(1+K_1)$) we have $v_2(t)\le v_1(t)\le 1$ for all $t\ge t_0$. Let $i_0$ denote the natural number of the property $t_{i_0-1}0$ we will show the existence of an $I(\mu)\in\mathbb{N}$ such that $\max\{x(t_{i_0 +I(\mu)}+\theta):-h\le\theta\le 0\}<\delta(\mu)$. Suppose that for a fixed $\mu>0$ such an index does not exists amongst $i_0, i_0+1,\ldots,i_*$. Then one of the possibilities (a), (b/1), (b/2) occurs for every $i_0\le i\le i_* $ with $\varepsilon:=\delta(\mu)/3$. By the estimates \eqref{L84}-\eqref{L86}, now these cases have the consequences \begin{itemize} \item[(a)] $v_1(t_i)-v_1(t_{i-1})\le -\varepsilon\kappa_2(\varepsilon)$; \item[(b/1)] $v_1(t_i)-v_1(t_{i-1})\le -\varepsilon$; \item[(b/2)] $v_1(t_i)-v_1(t_{i-1})\le -\varepsilon\kappa_1(\varepsilon)$. \end{itemize} Consequently, there is a $\kappa(\varepsilon)>0$ such that \[ -1\le v_1(t_{i_*})-v_1(t_0)\le v_1(t_{i_*})-v_1(t_{i_0}) \le -\kappa(\varepsilon)(i_*-i_0). \] In other words, $i_*$ must not be arbitrarily large; namely, $i_*-i_0\le 1/\kappa(\varepsilon)$. This means that the choice $I(\mu):= [1/\kappa(\delta(\mu)/3)]$ is appropriate, where $[\alpha]$ denotes the fractional part of $\alpha\in\mathbb{R}$. \end{proof} Now we consider a modification of functional \eqref{L1}: \begin{equation}\label{L100} \begin{split} V_\beta(t,\varphi) :&= |\varphi(0)| + \beta\int^0_{-h} \int^0_\theta |b(t+\vartheta-\theta)| |\lambda(t+\vartheta)||\varphi(\vartheta)|\,{\rm{d}}\vartheta \,{\rm{d}}\theta\\ &=V_2(t,\varphi)+\beta V_3(t,\varphi), \quad V_2(t,\varphi):=|\varphi(0)|\quad (\beta>1). \end{split} \end{equation} It is a consequence of Lemma \ref{absvalder} that \begin{equation}\label{L101} \begin{split} V_\beta'(t,\varphi) &\le -\Big(a(t)-\beta|\lambda(t)|\int_t^{t+h}|b|\Big)|\varphi(0)|\\ &\quad-(\beta-1)|b(t)|\int_{-h}^0|\lambda(t+\theta)||\varphi(\theta)|\,{\rm d}\theta \quad (t\in\mathbb{R}_+). \end{split} \end{equation} To make $V_\beta$ non-increasing along the solutions of \eqref{I1}, we assume a stronger dominance condition \begin{itemize} \item[(A2)] $a(t)-\beta|\lambda(t)|\int_t^{t+h}|b|\ge 0\quad \text{with some } \beta>1\; (t\in\mathbb{R}_+$). \end{itemize} This condition alone guaranties the existence of limits of solutions. \begin{lemma}\label{annulus1} Condition {\rm (A2)} implies that every solution of \eqref{I1} has a finite limit as $t$ goes to infinity. \end{lemma} \begin{proof} Suppose that the statement is not true, and consider a solution $x$ and the sequence $\{r_i,s_i\}_{i=1}^\infty$ with properties \eqref{L53} from the proof of the analogous Lemma \ref{annulus}. By using \eqref{L3}, \eqref{L4}, and \eqref{L101}, for the function $v_\beta(t):=V_\beta(t,x_t)$ we obtain the inequality \[ v'_\beta\le-(\beta-1)v'_2(t). \] Integrating this inequality we obtain \[ v_\beta(s_i)-v_\beta(r_i)\le -(\beta-1)(\varepsilon_2-\varepsilon_1)<0 \quad (i\in\mathbb{N}), \] which means that $v_\beta(t)\to-\infty$ as $t\to\infty$, but this is a contradiction. \end{proof} \begin{theorem}\label{strong} Suppose that {\rm (A2)} is satisfied. If condition \eqref{int} also holds, then the zero solution of \eqref{I1} is asymptotically stable. If, in addition, we assume \eqref{L5} and \eqref{intunif}, then the zero solution is uniformly asymptotically stable. \end{theorem} \begin{proof} We have to repeat the proof of Theorem \ref{light} with the only modification that, instead of \eqref{L7}, now we have $v'_\beta(t)\le -v'_2(t)$. From this estimate it follows that \[ v_\beta(s_i)- v_\beta(r_i)\le -\frac{\delta(\overline \mu)}{2} \quad (i=1,2,\ldots,N). \] Now this also implies $\lim _{t\to \infty } v_\beta(t)=-\infty$, a contradiction. \end{proof} \subsection{The case of $\lambda(t)\equiv 1$} Consider the equation \begin{equation}\label{L10} x'(t)=-a(t)x+b(t)\int^t_{t-h} x(s)\,{\rm d}s,\quad a(t)\ge 0\ (t\in \mathbb{R}_+). \end{equation} This equation was investigated by Ting Xiu Wang in \cite[pp. 849--853, Theorems 3.1 and 3.2]{TW2}, where he applied his abstract theorems from \cite{TW1} and proved very useful sufficient conditions for the asymptotic stability and uniform asymptotic stability of the zero solution of \eqref{L10}. Our Theorem \ref{light} becomes more comparable to these results if we deduce a more explicit form of integral positivity for this special case. \begin{theorem}\label{light1} Suppose that condition {\rm (A1)} with $\lambda(t)\equiv 1$ is satisfied. If \begin{equation}\label{int1} \int^\infty \Big(a(t)-\int_t^{t+h}|b|\Big)\,{\rm d}t=\infty, \end{equation} and either $|b|$ is integrable on $\mathbb{R}_+$, or $t\mapsto \int_0^t|b|$ is uniformly continuous on $\mathbb{R}_+$ and for every $\kappa >0$ there exist $\gamma(\kappa)>0$ and $t_{**}(\kappa)$ such that \begin{equation}\label{L11} \frac{1}{\kappa}\int_t^{t+\kappa}a -\int_t^{t+\kappa+h}|b|\ge\gamma(\kappa)\quad (t\ge t_{**}(\kappa)), \end{equation} then the zero solution of \eqref{L10} is asymptotically stable. If, in addition, \begin{equation}\label{intunif1} \lim_{S\to\infty}\int_t^{t+S}\Big(a(t)-\int_t^{t+h}|b|\Big)\,{\rm d}t =\infty\quad \text{uniformly w.r.t. } t\in\mathbb{R}_+, \end{equation} then the zero solution of \eqref{L10} is uniformly asymptotically stable. \end{theorem} \begin{proof} The assertions follow from Theorem \ref{light} if we prove that \eqref{L11} is sufficient for the integral positivity of $t\to a(t)-\int_t^{t+h}|b|$ with respect to $|b|$. Let $\nu>0$ be arbitrarily fixed. Since $t\mapsto \int_0^t|b|$ is uniformly continuous, there exists a $\delta(\nu)>0$ such that $\Delta=\Delta_{|b|}(t,\nu)>\delta(\nu)$. Therefore, changing the order of the successive integration and using \eqref{L11} we obtain \begin{align*} &\int_t^{t+\Delta}a-\int_t^{t+\Delta}\int_s^{s+h}|b(u)|\,{\rm d}u \,{\rm d}s \\ &\ge\int_t^{t+\Delta}a-\int_t^{t+\Delta}\int_{t}^{t+\Delta+h}|b(u)|\,{\rm d}u \,{\rm d}s\\ &=\int_t^{t+\Delta}a-\Delta\int_t^{t+\Delta+h}|b| \\ &=\Delta\Big(\frac{1}{\Delta}\int_t^{t+\Delta}a-\int_t^{t+\Delta+h}|b|\Big)\\ &\ge \delta(\nu)\gamma(\Delta_{|b|}(t,\nu))=:W(\nu)>0\quad \text{for } t\ge t_{**}(\Delta_{|b|}(t,\nu))=:t_*(\nu), \end{align*} which means the desired integral positivity. \end{proof} Parts of Wang's results (Theorem 3.1 (a), Theorem 3.2 (a)) are consequences of Theorems \ref{light} and \ref{strong}, another part are independent of our theorems. Theorem \ref{light1} does not require some key conditions of Wang's theorem (for example, function $t\to a(t)-\int_t^{t+h}|b|$ is non-decreasing). \section{Lyapunov-Razumikhin method} This method uses \emph{functions} $V:\mathbb{R}_+\times\mathbb{R}\to \mathbb{R}_+$ instead of \emph{functionals} $V:\mathbb{R}_+\times C\to \mathbb{R}_+$. For example, if we define the function $V(\varphi(0)):=|\varphi(0)|$ ($\varphi \in C$) to equation \eqref{I1}, then, by \eqref{L3} and \eqref{L4}, the derivative of $V$ with respect to \eqref{I1} allows the estimate \begin{equation}\label{R0} V'(t,\varphi)\le -a(t)|\varphi(0)|+ |b(t)| \int_{t-h}^{t}|\lambda(s)||\varphi(s-t)|\,{\rm d}s. \end{equation} In the Lyapunov-Razumikhin method function $V$ is called a \emph{Lyapunov function} if \begin{equation}\label{R1} V'(t,\varphi)\le 0,\text{ provided that } V(t+\theta, \varphi(\theta))\le V(t,\varphi(0))\; (-h\le\theta\le 0). \end{equation} So $V(\varphi(0)):=|\varphi(0)|$ will be a Lyapunov function to \eqref{I1} if we require \begin{itemize} \item[(A3)] $a(t)-|b(t)|\int_{t-h}^{t}|\lambda|\ge 0$ ($t\in \mathbb{R}_+$), \end{itemize} which is a new dominance condition to equation \eqref{I1}. The following lemma can be proved easily (see, e.g., \cite[Theorem 5.4.1]{HJ}). \begin{lemma} If {\rm (A3)} is satisfied, and $x$ is a solution of \eqref{I1}, then the functional $\overline V(t,x_t):=\sup\{V(x(t+\theta)):-h\le\theta\le 0\}$ is non-increasing in $\mathbb{R}_+$, and the zero solution of \eqref{I1} is uniformly stable. \end{lemma} This lemma shows that the Lyapunov-Razumikhin method can be very effective: the dominance condition (A3) alone guaranties uniform stability without any boundedness condition of type \eqref{L5}. \begin{lemma}\label{lim} If \begin{itemize} \item[(A4)] $a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|\ge 0$ ($t\in \mathbb{R}_+$) with some constant $\beta>1$, \end{itemize} then every solution has a finite limit as $t$ appraoches infinity. \end{lemma} \begin{proof} Expression \eqref{R0} and condition (A4) imply that \begin{equation}\label{R3} V'(t,\varphi)\le -\Big(a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|\Big)|\varphi(0)| \le 0, \end{equation} provided that $V(\varphi(\theta))\le \beta V(\varphi(0))$ $(-h\le\theta\le 0)$. For an arbitrary solution $x$ introduce the notation \begin{equation}\label{R2} \begin{gathered} v(t):=V(x(t))=|x(t)|,\quad \overline v(t):=\sup\{v(t+\theta):-h\le \theta\le 0\},\\ \overline v_0:=\lim_{t\to \infty}\overline v(t). \end{gathered} \end{equation} For any $\varepsilon>0$ there is $t_*(\varepsilon)$ such that $t>t_*(\varepsilon)$ implies $\overline v_0\le \overline v(t)<\overline v_0+\varepsilon$. If $x$ has no limit, then $x(t)\not\to \overline v_0$, and there are $\varepsilon$ ($0<\varepsilon<\overline v_0(\beta-1)/(\beta+1)$) and $r$, $s$ with the properties \[ t_*(\varepsilon)< r\beta(\overline v_0-\varepsilon)>\overline v_0+\varepsilon>v(t+\theta)\quad (-h\le \theta \le 0), \end{equation} hence $v'(t)\le 0$, which is a contradiction. \end{proof} \begin{theorem}\label{asstab} If {\rm (A4)} is satisfied, and \begin{equation}\label{R5} \int^\infty\Big(a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|\Big)\,{\rm d}t=\infty, \end{equation} then the zero solution of \eqref{I1} is asymptotically stable. \end{theorem} \begin{proof} If $x$ is an arbitrary solution, then, by Lemma \ref{lim}, $\lim_{t\to\infty}|x(t)|=\overline v_0$. We have to show that $\overline v_0=0$. Suppose the contrary, i.e., $\overline v_0>0$. Then for every $\varepsilon$ ($0<\varepsilon<\overline v_0(\beta-1)/(\beta+1)$) there is a $t_*(\varepsilon)$ such that \eqref{R4} holds for all $t>t_*(\varepsilon)$. Consequently, \eqref{R3} yields \begin{gather*} v'(t)\le -\eta_4(t)(\overline v_0-\varepsilon)\quad \eta_4(t):=a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|,\\ v(t)-v(t_*)\le -(\overline v_0-\varepsilon)\int_{t_*}^t\eta_4\to-\infty \quad(t\to\infty), \end{gather*} a contradiction. \end{proof} \begin{theorem} If {\rm (A4)} is satisfied and \begin{equation}\label{intunifraz} \lim_{S\to\infty}\int_t^{t+S} \Big(a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|\Big)\,{\rm d}t =\infty\quad \text{uniformly w.r.t. } t\in\mathbb{R}_+, \end{equation} then the zero solution is uniformly asymptotically stable. \end{theorem} \begin{proof} Let $\delta(\varepsilon)$ belong to $\varepsilon$ in the sense of uniform stability and let $t_0\ge 0, \varphi$ ($\|\varphi\|<\delta(1)$) be arbitrary. It is sufficient to prove that for every $\mu>0$ there exists a $T(\mu)$ such that $\|x_{t_0+T(\mu)}(\cdot;t_0,\varphi)\|<\delta(\mu)$. Given $\mu>0$ introduce some notation. Let $N=N(\mu)$ denote the smallest natural number for which $1/\beta^N<\delta(\mu)$ holds. Condition \eqref{intunifraz} guaranties the existence of an $S_*(\mu)$ such that \begin{equation}\label{R6} \int_t^{t+S_*(\mu)}\eta_4>(\beta-1)\beta^{N-1}, \quad \eta_4(t):= a(t)-\beta|b(t)|\int_{t-h}^{t}|\lambda|. \end{equation} Finally, set $t_i:=t_0+i(S_*(\mu)+h)$ ($i=1,2,\ldots,N$). We state that \begin{equation}\label{R7} \text{if } t>t_i-h, \text{ then } v(t):=|x(t)| \le \frac{1}{\beta^{i}} \text{ for } i=0,1,\ldots,N. \end{equation} In fact, the assertion is true for $i=0$. Using the method of the mathematical induction, we assume that the assertion is true for some $i$ ($0\le it_i$ and $v(s)\le 1/\beta^{i+1}$, then $v(s+\tau)\le 1/\beta^{i+1}$ for all $\tau\ge 0$. Therefore, if $t\ge t_i$, $v(t)>1/\beta^{i+1}$, then $v(s)>1/\beta^{i+1}$ for $s\in[t_i,t]$ and \[ v'(s)\le -\eta_4(s)v(s)\le -\frac{1}{\beta^{i+1}}\eta_4(s) \le-\frac{1}{\beta^N}\eta_4(s). \] Consequently, if $t\ge t_i+S_*(\mu)$ and $v(t)>1/\beta^{i+1}$, then \[ v(t)-v(t_i)\le -\frac{1}{\beta^N}\int_{t_i}^{t_i+S_*(\mu)}\eta_4 <-\frac{1}{\beta^N}(\beta-1)\beta^{N-1} =-(1-\frac{1}{\beta}). \] On the other hand, \[ v(t)-v(t_i)\ge \frac{1}{\beta^{i+1}}-\frac{1}{\beta^{i}} =-\frac{1}{\beta^{i}}(1-\frac{1}{\beta}) \ge -(1-\frac{1}{\beta}), \] which is a contradiction. Consequently, if we define $t_{i+1}:=t_i+S_*(\mu)+h$, and if $t>t_{i+1}-h$, then $v(t)\le 1/\beta^{i+1}$, i.e., assertion \eqref{R7} is true for $i+1$. By the rule of the mathematical induction, assertion \eqref{R7} is true for $i=N$. In other words, if $t>t_N$, then $v(t)\le 1/\beta^N<\delta(\mu)$. So the definition $T(\mu):=N(\mu)(S_*(\mu)+h)$ completes the proof. \end{proof} In some applications it may happen that dominance condition (A4) is not satisfied with a uniform (independent of $t$) coefficient $\beta>1$, but it is satisfied with $\beta(t)\ge 1$. The remaining part of the section is devoted to this situation. Suppose that \begin{itemize} \item[(A5)] $a(t)-\beta(t)|b(t)|\int_{t-h}^{t}|\lambda|\ge 0$ ($t\in \mathbb{R}_+$) with some piecewise continuous function $\beta:\mathbb{R}_+\to[1,\infty)$. \end{itemize} This condition and \eqref{R0} imply \begin{equation}\label{R8} V'(t,\varphi)\le -\Big(a(t)-\beta(t)|b(t)|\int_{t-h}^{t}|\lambda|\Big)|\varphi(0)| \le 0, \end{equation} provided that $V(\varphi(\theta))\le \beta(t)V(\varphi(0))$ ($-h\le\theta\le 0$) instead of \eqref{R3}. Now we import two lemmas from Kato \cite{KJ}, in which he transformed estimate \eqref{R8} into a comparison statement. Let $U:\mathbb{R}_+\times \mathbb{R}\to\mathbb{R}$ be measurable with respect to the first variable and locally Lipschitz with respect to the second one, and denote by $u(\cdot;t_*,\alpha):[t_*-h,\infty)\to \mathbb{R}$ the solution of the initial value problem \begin{equation}\label{R9} u'=U(t,u), \quad u(t_*)=\alpha. \end{equation} If $v:[t_0-h,\infty)\to \mathbb{R}$ ($t_0\ge 0$) is a continuous function, then define \[ v'(t):=\limsup_{\delta\to 0+0}\Big(\frac{1}{\delta}(v(t+\delta)-v(t))\Big). \] \begin{lemma}[{\cite[Lemma 1]{KJ}}] \label{kato1} Suppose that \begin{equation}\label{R10} \begin{gathered} v'(t)\le U(t,v(t)) \quad \text{if }\ t\ge t_* \text{ and}\\ v(t+\theta)\le u(t+\theta; t,v(t))\quad (-h\le \theta\le 0). \end{gathered} \end{equation} Then \[ v(t)\le u(t;t_*,\alpha)\quad (t\ge t_*), \] provided that $v(t_*+\theta)\le u(t_*+\theta; t_*,\alpha)$ ($-h\le \theta\le0$). \end{lemma} As a consequence of \eqref{R8}, from \cite[Lemma 2]{KJ} we obtain the following lemma. \begin{lemma} If {\rm (A5)} is satisfied, then there exists a function $U:\mathbb{R}_+\times\mathbb{R}_+\to\mathbb{R}_-$ such that for every solution $x$ of \eqref{I1} the function $v(t):=|x(t)|$ satisfies the differential inequality \eqref{R10} with $t_*=t_0$. Namely, the choice \begin{equation} \label{R11} \begin{gathered} U(t,u):=-\min\Big\{\frac{1}{3h};\frac{2}{3h}\min\{\beta(t+\tau)-1:0\le\tau\le h \}; \eta_5(t) \Big\}u\\ (u>0) \quad \Big(\eta_5(t):=a(t)-\beta(t)|b(t)|\int_{t-h}^{t}|\lambda|\ge 0\Big) \end{gathered} \end{equation} is appropriate. \end{lemma} \begin{theorem}\label{kato2} Suppose that {\rm (A5)} is satisfied. If \begin{gather} \int^\infty \gamma=\infty\label{R12}\\ \Big(\gamma(t):=\min\big\{1;\min\{ \beta(t+\tau):0\le\tau\le h\}-1; a(t)-\beta(t)|b(t)|\int_{t-h}^{t}|\lambda|\big\}\Big), \nonumber \end{gather} then the zero solution of \eqref{I1} is asymptotically stable. If \begin{equation}\label{R13} \lim_{S\to\infty}\int_t^{t+S}\gamma =\infty \quad \text{uniformly w.r.t. } t\in\mathbb{R}_+, \end{equation} then the zero solution of \eqref{I1} is uniformly asymptotically stable. \end{theorem} \begin{proof} If $U(t,u)=-\Gamma(t)u$ ($\Gamma(t)\ge 0$) in \eqref{R11}, then the solution of the initial value problem \eqref{R9} is \[ u(t;t_*,\alpha)=\alpha\exp\big[-\int_{t_*}^t\Gamma\big]. \] It is easy to see that for arbitrary measurable functions $A,B:\mathbb{R}_+\to\mathbb{R}_+$ the divergence \[ \int^\infty\min\{A(t);B(t)\}\,{\rm d}t=\infty \] is true if and only if \[ \int^\infty\min\{A(t);qB(t)\}\,{\rm d}t=\infty \] holds for all $q>0$. Therefore \eqref{R12} implies $\lim_{t\to\infty}u(t;t_*,\alpha)=0$ for all $t_*, \alpha$. For arbitrary $t_0,\varphi$, applying Lemma \ref{kato1} to $v(t)=|x(t;t_0,\varphi)|$ with $t_*=t_0$, $\alpha=\|\varphi\|$ we obtain $|x(t;t_0,\varphi)|\le u(t;t_0,\|\varphi\|)\to0$ if $t\to\infty$, which proves asymptotic stability. For UAS, take $\delta(1)>0$ from the definition of the uniform stability. To prove UAS we have to show that for every $\mu>0$ there is a $T(\mu)$ such that $[t_0\in \mathbb{R}_+$, $\|\varphi\|<\delta(1)$, $t\ge t_0+T(\mu)]$ imply $|x(t;t_0,\varphi)|<\mu$. We show that the choice of $T(\mu)$ with the properties \[ \int_{t_0}^{t_0+T(\mu)}\gamma>\ln\frac{\delta(1)}{\mu} \] is appropriate. In fact, by Lemma \ref{kato1} we have \begin{equation} \begin{split} |x(t;t_0,\varphi)| &\le u(t;t_0,\|\varphi\|)\le u(t_0+T(\mu);t_0,\delta(1))\\ &\le\delta(1)\exp\Big[-\int_{t_0}^{t_0+T(\mu)}\gamma\Big] < \delta(1)\exp\big[-\ln\frac{\delta(1)}{\mu}\big]=\mu. \end{split} \end{equation} \end{proof} \section{Examples} \begin{example} \label{examp4.1} \rm Let a sequence $\{\overline t_i\}_{i=1}^\infty$ and a number $\delta$ ($0<\delta1$ let us define the functions \begin{equation}\label{E4} a(t):=\alpha (t+\frac{h}{2});\quad b(t):=\frac{t}{h};\quad \lambda(t):=1\quad (t\in\mathbb{R}_+). \end{equation} The zero solution of \eqref{I1} with coefficients \eqref{E4} is asymptotically sable. We have $\eta_1(t):=a(t)-\int_t^{t+h}|b|=(\alpha-1)(t+h/2)$ and $a(t)-\alpha\int_t^{t+h}|b|\equiv 0$, so Theorem \ref{strong} can be applied. Let us modify the definition of $\lambda$ so that \[\lambda(t):=\begin{cases} \alpha & \text{if } i\le t\le i+\frac{1}{i^2},\\ 1 & \text{otherwise}, \end{cases} \quad (i\in\mathbb{N}); \quad M(t):=|b(t)|\int _{t-h}^t|\lambda|\le \alpha t. \] Then we have \[ \eta_1(t):=a(t)-|\lambda(t)|\int_t^{t+h}|b| =\begin{cases} 0 & \text{if } i\le t\le i+\frac{1}{i^2},\\ (\alpha-1)(t+\frac{h}{2}) & \text{otherwise}, \end{cases} \] and $\Delta_M(t,\nu)\ge\nu/(2\alpha t)$. Applying Theorem \ref{light} we obtain that the assertion remains true. Obviously Theorem \ref{strong} cannot be applied. \end{example} \subsection*{Acknowledgments} Supported by the Hungarian National Foundation for Scientific Research (OTKA K109782) and Analysis and Stochastics Research Group of the Hungarian Academy of Sciences. \begin{thebibliography}{10} \bibitem{BT} T. A. Burton; \emph{Stability and Periodic Solutions of Ordinary and Functional Differential Equations}, Mathematics in Science and Engineering, 178. 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