\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 126, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/126\hfil Oscillation criterion] {Oscillation criterion for first-order delay differential equations with sign-changing coefficients} \author[H. Wu, C. Chen, R. Zhuang \hfil EJDE-2017/126\hfilneg] {Hongwu Wu, Churong Chen, Rongkun Zhuang} \address{Hongwu Wu \newline School of Mathematics, South China University of Technology, Guangzhou 510640, China} \email{hwwu@scut.edu.cn} \address{Churong Chen \newline School of Mathematics, South China University of Technology, Guangzhou 510640, China} \email{churchen@foxmail.com} \address{Rongkun Zhuang \newline Department of Mathematics, Huizhou University, Huizhou 516007, China} \email{zhuangrk@hzu.edu.cn} \thanks{Submitted March 8, 2017. Published May 10, 2017.} \subjclass[2010]{34K11, 34C10} \keywords{Oscillatory solution; delay differential equation; variable delay} \begin{abstract} We establish conditions so that all solutions to a first-order linear delay differential equation become oscillatory. The coefficients of this equation are allowed to have negative and positive values, which is an expansion of the results in the references. To illustrate our results we present two examples. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article, we study the oscillation of solutions to the first-order linear delay differential equation \begin{equation}\label{e1} x'(t)+p(t)x(\tau(t))=0,\quad t\geq{t_0}, \end{equation} where $p(t)$ and $\tau(t)$ are continuous on $[t_0,\infty)$, the delay argument $\tau(t)$ is strictly increasing, $\tau(t)\frac{1}{e}\quad\text{or}\quad \beta:=\limsup_{t \to \infty}\int^t_{\tau(t)}p(s)ds>1, \end{equation} then \eqref{e1} is oscillatory \cite{Kwong}. On the other hand, when $\beta\leq 1/e$, equation \eqref{e1} has a non-oscillatory solution \cite{Koplatadze}. Therefore, the interesting case happens when $\alpha\leq 1/e<\beta\leq 1$. In 1988, Erbe and Zhang \cite{Erbe3} used an upper bound of the ratio $x(\tau(t))/x(t)$ to establish sufficient conditions for the oscillation of all solutions to \eqref{e1}: \begin{equation*}\label{Stav1} p(t)\geq0,\quad 0<\alpha \leq\frac{1}{e}\,,\quad \beta >1-\frac{\alpha^2}{4}. \end{equation*} Since then several authors have improved these results by using upper bounds for $x(\tau(t))/x(t)$ when $\alpha \leq1/e$; see for example \cite{Myshkis,Tang,Wu,BG}. Let $\lambda_*\leq \lambda^*$ be the two roots of the equation $\lambda=e^{\alpha \lambda}$. The following results have been obtained for $p(t)\geq0$ and $\alpha\leq 1/e$: \begin{align*} &\text{in \cite{Chao}} \quad \beta> 1-\frac{\alpha^2}{2(1-\alpha)} \geq 0.892, \\ &\text{in \cite{Yu3}}\quad \beta >1-\frac{1-\alpha-\sqrt{1-2\alpha-\alpha^2}}{2} \geq 0.863 , \\ &\text{in \cite{Elbert}}\quad \beta >1-\Big(1-\frac{1}{\sqrt{{\lambda}_*}}\Big)^2\quad \geq0.845, \\ &\text{in \cite{Kwong}} \quad \beta >\frac{1+\ln {\lambda}_*}{{\lambda}_*}\geq 0.735, \\ &\text{in \cite{Philos}}\quad \beta >1-\frac{\alpha^2}{2(1-\alpha)}-\frac{\alpha^2}{2}{\lambda}_*\geq 0.709,\\ &\text{in \cite{Jaros}}\quad \beta >\frac{1+\ln {\lambda}_*}{{\lambda}_*} -\frac{1-\alpha-\sqrt{1-2\alpha-\alpha^2}}{2}\geq 0.599, \\ &\text{in \cite{Kon}}\quad \beta >2\alpha+\frac{2}{{\lambda}_*}-1\geq 0.471,\quad \\ &\text{in \cite{Stav1}}\quad \beta >\frac{\ln {\lambda}_*-1+\sqrt{5-2{\lambda}_* +2\alpha{\lambda}_*}}{{\lambda}_*}\geq 0.459\,. \end{align*} Note that when $\alpha=1/e$, the lowest bound for $\beta$ is $0.459$, so far. In 2004, Zhao et al \cite{Zhao} established the oscillation criterion \[ \limsup_{t \to \infty}\Big\{\min_{\tau(t)\leq s\leq t}\int^{s}_{\tau(s)}p(\xi)d\xi \Big\}>\frac{1+\ln\lambda_*}{\lambda_*} -\frac{1}{\lambda^*}. \] When $\alpha=1/e$, we have $\lambda_*=\lambda^*=e$, and condition above reduces to \[ \limsup_{t \to \infty}\Big\{\min_{\tau(t)\leq s\leq t}\int^{s}_{\tau(s)}p(\xi)d\xi \Big\}>\frac{1}{e}. \] Note that as $\alpha\to 0$, almost all conditions reduce to the condition $\beta>1$. However a condition in \cite{Stav1} leads to $\beta >\sqrt 3-1\approx 0.732$, which is an essential improvement. A natural question arises: can we get a lower bound for $\beta$ when $\alpha\leq 1/e$? Motivated by \cite{Stav1,Tang,Wu,Zhao} and other publications, our goal is to study \eqref{e1} when $p(t)$ takes negative values on infinitely many intervals. \section{Main results} Note that if $x(t)$ is a solution of \eqref{e1}, then $-x(t)$ is also a solution. Therefore, when $x(t)$ does not have zeros, we assume that $x(t)$ is positive. To use iterates of the delay, we define $\tau^0(t)=t$, $\tau^1(t)=\tau(t)$, $\tau^2(t)=\tau(\tau(t))$, etc. Also to use iterates of $\tau$ inverse, we define $\tau^{-2}(t)=\tau^{-1}(\tau^{-1}(t))$, $\tau^{-3}(t)=\tau^{-2}(\tau^{-1}(t))$, etc. Following an idea in \cite{Wu}, we define the functions \begin{equation} \label{e2} \rho_i(t)=\int ^t_{\tau(t)}p(s_1)\int ^{\tau(t)}_{\tau (s_1)}p(s_2)\cdots\int ^{\tau^{i-1}(t)}_{\tau(s_{i-1})}p(s_{i})ds_{i}\cdots ds_1\,. \end{equation} However, the functions $\rho_i(t)$ in \cite{Wu} are bounded below by non-negative constants; we do not use such assumption here. Note that computing $\rho_i(t)$ requires the evaluation of $p(s)$ for $s\in[\tau^i(t),t]$. Since $p(s)$ is defined only for $s\geq t_0$, the number of $\rho_i$'s depend on the condition $\tau^i(t)\geq t_0$. For example when $\tau(t)=t-{\rm const.}$ there are only finitely many $\rho_i$'s, and when $\tau(t)=\sqrt{t}$ with $0\leq t_0\leq 1$ there are infinitely many $\rho_i$'s. If $p(s)\geq 0$ on $[\tau^{-i}(t),t]$, then $\rho_i(t)\geq 0$. If $p(s)= 0$ on some interval $[\tau^n(t),\tau^{n-1}(t)]$, then $\rho_i(t)= 0$ for $i\geq n$. When $p(s)=p$ and $\tau(t)=t-\sigma$ with $p$ and $\sigma$ constants, \[ \rho_1(t)=\frac{p\sigma}{1!}, \quad \rho_2(t)=\frac{(p\sigma)^2}{2!}, \quad \cdots\,. \] We shall use the convention that for $j>n$, $\sum_{i=j}^np = 0$ and $\prod_{i=j}^n p =1$. \begin{lemma}\label{lem1} Let $x(t)$ be a solution of \eqref{e1}. Assume that there exit $n\geq 1$ and $t_1^*\geq t_1>t_0$ such that: $\tau^{n+2}(t_1)\geq t_0$, $p(s)\geq 0$ for $s\in [\tau^{n+1}(t_1),t_1^*]$, and $x(t)> 0$ for $t\in [\tau^{n+2}(t_1),t_1^*]$. Let $\rho_i(t)$ be defined by \eqref{e2}, and $f_1(t)=1/(1-\rho_1(t))$. Then \begin{equation} \label{e6} \frac{x(\tau(t))}{x(t)} \geq \frac{1}{1-\sum_{i=1}^n \rho_i(t)\prod_{j=1}^{i-1} f_{n-j}(\tau^j(t))} =:f_n(t)>0\quad \forall t\in[t_1,t_1^*]\,, \end{equation} where $f_2,f_3,\dots$ are defined recursively. \end{lemma} \begin{proof} Integrating \eqref{e1}, we have \begin{gather*} x(\tau(t))=x(t)+\int_{\tau(t)}^t p(s_1)x(\tau(s_1))\,ds_1,\\ x(\tau(s_1))=x(\tau(t))+\int_{\tau(s_1)}^{\tau(t)} p(s_2)x(\tau(s_2))\,ds_2\,,\\ x(\tau(s_2))=x(\tau^2(t))+\int_{\tau(s_2)}^{\tau^2(t)} p(s_3)x(\tau(s_3))\,ds_3\,,\\ \cdots\,. \end{gather*} Using repeated substitutions we have \begin{align*} x(\tau(t)) &=x(t)+\int_{\tau(t)}^t p(s_1)x(\tau(t))\,ds_1 +\int_{\tau(t)}^t p(s_1)\int_{\tau(s_1)}^{\tau (t)} p(s_2)x(\tau^2(t))\,ds_2\,ds_1 \\ &\quad +\int_{\tau(t)}^t p(s_1)\int_{\tau(s_1)}^{\tau (t)} p(s_2) \int_{\tau(s_2)}^{\tau^2 (t)} p(s_3) x(\tau(s_3))\,ds_3\,ds_2\,ds_1 +\dots \,. \end{align*} Since $p(\cdot)$ and $x(\tau(\cdot))$ are non-negative on the interval $[\tau^{n+1}(t_1),t_1^*]$, from \eqref{e1}, we have that $x'(t)\leq 0$ and hence $x(t)$ is non-increasing on the same interval. Using this property and \eqref{e2} we obtain \begin{equation} \label{e8} x(\tau(t)) \geq x(t)+\rho_1(t)x(\tau(t))+\rho_2(t)x(\tau^2(t))+\rho_3(t)x(\tau^3(t))+\dots \,. \end{equation} Considering the first two terms in the right-hand side, we have \[ (1-\rho_1(t))x(\tau(t))\geq x(t)>0\,; \] thus \[ 1-\rho_1(t)>0\quad\text{and}\quad \frac{x(\tau(t))}{x(t)} \geq \frac{1}{1-\rho_1(t)}=:f_1(t)>0\quad \text{for } t\in [\tau^{n-1}(t_1),t_1^*]\,. \] With $\tau(t)$ instead of $t$ in the above expression, we have \begin{equation} \label{e9} \frac{x(\tau^2(t))}{x(\tau(t))} \geq f_1(\tau(t))\quad \text{for } t\in [\tau^{n-2}(t_1),t_1^*]\,. \end{equation} Considering the first three terms in the right-hand side in \eqref{e8}, and using \eqref{e9}, we have \[ \Big(1-\rho_1(t)-\rho_2(t)f_1(\tau(t))\Big) x(\tau(t)) \geq x(t) >0\quad \text{for } t\in [\tau^{n-2}(t_1),t_1^*]\,. \] Thus $1-\rho_1(t)-\rho_2(t)f_1(\tau(t))>0$ and \[ \frac{x(\tau(t))}{x(t)}\geq \frac{1}{1-\rho_1(t)-\rho_2(t)f_1(\tau(t))} =:f_2(t)>0\quad \text{for } t\in [\tau^{n-2}(t_1),t_1^*]\,. \] With $\tau(t)$ instead of $t$ in the above expression, we have \begin{equation} \label{e9b} \frac{x(\tau^2(t))}{x(\tau(t))} \geq f_2(\tau(t))\quad \text{for } t\in [\tau^{n-3}(t_1),t_1^*]\,. \end{equation} With $\tau(t)$ instead of $t$ in \eqref{e9}, we have \begin{equation} \label{e10} \frac{x(\tau^3(t))}{x(\tau^2(t))} \geq f_1(\tau^2(t)) \quad \text{for } t\in [\tau^{n-3}(t_1),t_1^*]\,. \end{equation} Considering the first four terms in the right-hand side in \eqref{e8}, and using \eqref{e9b} and \eqref{e10}, we obtain \[ \Big(1-\rho_1(t)-\rho_2(t)f_2(\tau(t)) -\rho_3(t)f_2(\tau(t))f_1(\tau^2(t))\Big) x(\tau(t)) \geq x(t)>0 \] for $t\in [\tau^{n-3}(t_1),t_1^*]$. Thus $1-\rho_1(t)-\rho_2(t)f_2(\tau(t)) -\rho_3(t)f_2(\tau(t))f_1(\tau^2(t))>0$, and \[ \frac{x(\tau(t))}{x(t)} \geq \frac{1} {1-\rho_1(t)-\rho_2(t)f_2(\tau(t)) -\rho_3(t)f_2(\tau(t))f_1(\tau^2(t))} =:f_3(t)>0 \] for $t\in [\tau^{n-3}(t_1),t_1^*]$. Doing the above process for $n$ terms, we obtain \eqref{e6}, which completes the proof. \end{proof} \begin{remark}\label{rmk1}\rm Under the conditions of Lemma \ref{lem1}, we can obtain a lower bound that does not use recursion to define $f_n$: \begin{equation} \label{e6b} \frac{x(\tau(t))}{x(t)} \geq \frac{1}{1-\sum_{i=1}^n \rho_i(t)\prod_{j=1}^{i-1} f_1(\tau^j(t))} =:f_n(t)\quad \forall t\in[t_1,t_1^*]\,. \end{equation} This bound is easier to compute, but is not as sharp as the one in the lemma. Let $f_1$ and \eqref{e9} be as defined as in Lemma \ref{lem1}. From \eqref{e9}, we obtain \begin{equation} \label{e10b} \begin{gathered} \frac{x(\tau^3(t))}{x(\tau^2(t))} \geq f_1(\tau^2(t)) \quad \text{for } t\in [\tau^{n-3}(t_1),t_1^*]\,,\\ \frac{x(\tau^4(t))}{x(\tau^3(t))} \geq f_1(\tau^3(t)) \quad \text{for } t\in [\tau^{n-4}(t_1),t_1^*]\,. \end{gathered} \end{equation} Then considering the first four terms in the right-hand side in \eqref{e8}, and using \eqref{e9} and \eqref{e10b} we have \[ \Big(1-\rho_1(t)-\rho_2(t)f_1(\tau(t)) -\rho_3(t)f_1(\tau(t))f_1(\tau^2(t))\Big) x(\tau(t)) \geq x(t)>0 \] for $t\in [\tau^{n-3}(t_1),t_1^*]$. Repeating this process for $n$ terms, we obtain \eqref{e6b}, which will be used in computations later. \end{remark} The functions $f_n$ defined in Lemma \ref{lem1} have the following properties: Evaluating $f_n(t)$ requires evaluating $\rho_n(t)$ which in turn requires evaluating $p(s)$ for $s\in[\tau^n(t),t]$. In general, the functions $f_n$ do not have monotonicity unless constant delay and constant coefficient. In addition, for a large index, $f_n(t)$ may not be defined, because of a division by zero in \eqref{e6}, or $f_n(t)$ may be negative. \begin{lemma}\label{lem2} Let $x(t)$ be a solution of \eqref{e1}. Assume that there exit $n\geq 1$, $t_1$, and $t_1^*$ such that $\tau(t_1^*)\geq t_1>t_0$, $\tau^{n+2}(t_1)\geq t_0$, $p(s)\geq 0$ for $s\in [\tau^{n+1}(t_1),t_1^*]$, and $x(t)> 0$ for $t\in [\tau^{n+2}(t_1),t_1^*]$. Let $\rho_i(t)$ be defined by \eqref{e2}, and $f_1(t)=1/(1-\rho_1(t))$. Then \begin{equation} \label{e12} \frac{1-\rho_1(t)}{\sum_{i=2}^n \rho_i(t)\prod_{j=2}^{i-1} f_{n-j}(\tau^j(t))} \geq \frac{x(\tau^2(t))}{x(\tau(t))} \quad \forall t\in[\tau^{-1}(t_1),t_1^*]\,, \end{equation} where $f_2,f_3,\dots$ are defined recursively by Lemma \ref{lem1}. \end{lemma} \begin{proof} Note that the conditions for Lemma \ref{lem1} are also satisfied in this lemma. Using \eqref{e8} and ignoring the term $x(t)$ which is positive, we have \[ x(\tau(t)) > \rho_1(t)x(\tau(t))+\rho_2(t)x(\tau^2(t))+\rho_3(t)x(\tau^3(t)) +\rho_4(t)x(\tau^4(t))+\dots \,. \] Using \eqref{e9b} and \eqref{e10}, we have \[ (1-\rho_1(t))x(\tau(t)) > \Big(\rho_2(t)+\rho_3(t)f_2(\tau^2(t)) +\rho_4(t)f_2(\tau^2(t))f_1(\tau^3(t)) +\dots \Big)x(\tau^2(t)) \,. \] Then \eqref{e12} follows by applying the same process $n$ times. The proof is complete. \end{proof} \begin{remark}\label{rmk2} \rm Under the assumptions of Lemma \ref{lem2}, we can obtain a bound that does not use recursion to define $f_n$: \begin{equation} \label{e12b} \frac{1-\rho_1(t)}{\sum_{i=2}^n \rho_i(t)\prod_{j=2}^{i-1} f_1(\tau^j(t))} \geq \frac{x(\tau^2(t))}{x(\tau(t))} \quad \forall t\in[\tau^{-1}(t_1),t_1^*]\,. \end{equation} Of course this bound is easier to compute, but is not as sharp as the one in the lemma. \end{remark} In the next theorems, we assume that there is time $t_k^*$ and an integer $n_k^*$ for which $f_n(t)$ may be negative or lower bound from \eqref{e6} exceeds the upper bound from \eqref{e12}. \begin{theorem}\label{thm1} Assume that there exists an increasing sequence $\{t_k\}_{k=1}^\infty$ approaching $\infty$, and a bounded sequence of positive integers $\{n_k\}_{k=1}^\infty$, such that for each $k$ there exist $t_k$ and $ t_k^*$ such that $t_k^*\geq t_k>t_0$, $\tau^{n_k+2}(t_k)\geq t_0$, and $p(s)\geq 0 $ for $s\in[\tau^{n_k+1}(t_k),t_k^*]$. If there exists a sequence of integers $\{n_k^*\}_k$ with $1\leq n_k^*\leq n_k$ for which \begin{equation} \label{e14a} {f}_{1}(t)>0,\;{f}_{2}(t)>0,\;\dots,\;{f}_{n_k^*-1}(t)>0,\;{f}_{n_k^*}(t)<0, \end{equation} where $t\in[t_k,t_k^*]$, $f_1(t)=1/(1-\rho_1(t))$ and $f_2(t),f_3(t),\dots$ are defined recursively by Lemma \ref{lem1}, then every solution of \eqref{e1} is oscillatory. \end{theorem} \begin{proof} Let $x(t)$ be a solution to \eqref{e1}. Without loss of generality, we may assume that $x(t)>0$ for $t\in[\tau^{n_k^*+2}(t_k),t_k^*]$, where $n_k^*$ satisfies (\ref{e14a}). By Lemma \ref{lem1}, we get ${f}_{n_k^*}>0$ for $t\in[t_k,t_k^*]$, which is a contradiction and completes the proof of Theorem \ref{thm1}.\end{proof} \begin{corollary}\label{cor1} Assume that $p(t)\geq p\geq 0$ on $[t_0,\;\infty)$ and $\tau(t)=t-\sigma$ for $\sigma>0$. If there exists $n\geq1$ such that \[ \beta:=\limsup_{t \to \infty}\int^{t}_{\tau(t)}p(s)ds>\frac{1} {1+\sum^{n}_{i=2}\frac{\alpha^{i-1}}{i!}\prod^{i-1}_{j=1}{f}_{n-j}}, \] where $\alpha:=\liminf_{t\to\infty}\int^t_{\tau(t)}p(s)ds=p\sigma$ and ${f}_{1},{f}_{2},\dots{f}_{n-1}$ is defined by \[{f}_{1}=\frac{1}{1-\alpha},\quad {f}_{2}=\frac{1}{1-\alpha-\frac{\alpha^{2}}{2!}{f}_{1}},\quad \dots,\quad {f}_{n-1}=\frac{1} {1- \sum^{n-1}_{i=1}\frac{\alpha^{i}}{i!}\prod^{i-1}_{j=1}{f}_{n-j-1}},\] then every solution of (\ref{e1}) oscillates. \end{corollary} \begin{remark}\label{rmk3}\rm Under the conditions of Corollary \ref{cor1}, we can obtain a lower bound that use recursion to define $f_n$ when $\alpha=1/e$. A numerical trend can be found that the lowest bound for $\beta$ is $\beta\to\frac{1}{e-1}\approx 0.5819$ since $f_n\to e$. \end{remark} \begin{theorem}\label{thm2} Assume that there exists an increasing sequence $\{t_k\}_{k=1}^\infty$ approaching $\infty$, and a bounded sequence of positive integers $\{n_k\}_{k=1}^\infty$, such that for each $k$ there exist $t_k$ and $ t_k^*$ such that $\tau(t_k^*)\geq t_k>t_0$, $\tau^{n_k+2}(t_k)\geq t_0$, and $p(s)\geq 0 $ for $s\in[\tau^{n_k+1}(t_k),t_k^*]$. If there exists a sequence of integers $\{n_k^*\}_k$ with $1\leq n_k^*\leq n_k$ for which \begin{equation} \label{e14c} \frac{1-\rho_1(t)}{\sum_{i=2}^{n_k^*} \rho_i(t)\prod_{j=2}^{i-1} f_{n-j}(\tau^j(t))} <\frac{1}{1-\sum_{i=1}^{n_k^*} \rho_i(\tau(t))\prod_{j=1}^{i-1} f_{n-j}(\tau^{j+1}(t))}, \end{equation} where $t\in[ \tau^{-1}(t_k),t_k^*]$, then every solution of \eqref{e1} is oscillatory. \end{theorem} \begin{proof} To reach a contradiction assume that a solution $x(t)$ does not have zeros on the interval $[\tau^{n+1}(t_k),t_k^*]$, and without loss of generality assume that $x(t)>0$. Then the hypotheses for both Lemmas \ref{lem1} and \ref{lem2} are satisfied. Applying \eqref{e6b} with $f_n(\tau(t))$ instead of $f_n(t)$, we obtain an inequality that combined with \eqref{e12b} contradicts the definition of $n_k^*$; Therefore, each solution of \eqref{e1} has a zero in $[\tau^{n_k+1}(t_k),t_k^*]$. This completes the proof. \end{proof} In the next corollary, we assume that there is time $t_k^*$ and an integer $n_k^*$ for which lower bound from \eqref{e6b} exceeds the upper bound from \eqref{e12b}. Note that to compare the two bounds we need $f_1(\tau(t_k^*))$, not $f_1(t_k^*)$, in \eqref{e6b}. \begin{corollary}\label{cor2} Assume that there exists an increasing sequence $\{t_k\}_{k=1}^\infty$ approaching $\infty$, and a bounded sequence of positive integers $\{n_k\}_{k=1}^\infty$, such that for each $k$ there exist $t_k$ and $ t_k^*$ such that $\tau(t_k^*)\geq t_k>t_0$, $\tau^{n_k+2}(t_k)\geq t_0$, and $p(s)\geq 0 $ for $s\in[\tau^{n_k+1}(t_k),t_k^*]$. If there exists a sequence of integers $\{n_k^*\}_k$ with $1\leq n_k^*\leq n_k$ for which \begin{equation} \label{e14d} \frac{1-\rho_1(t)}{\sum_{i=2}^{n_k^*} \rho_i(t)\prod_{j=2}^{i-1} f_1(\tau^j(t))} <\frac{1}{1-\sum_{i=1}^{n_k^*} \rho_i(\tau(t))\prod_{j=1}^{i-1} f_1(\tau^{j+1}(t))}, \end{equation} where $t\in[ \tau^{-1}(t_k),t_k^*]$, then every solution of \eqref{e1} is oscillatory. \end{corollary} \section{Examples} The following examples show equations for which Theorem \ref{thm1} or Theorem \ref{thm2} imply oscillation of all solutions, when $\limsup\int_{\tau(t)}^t p<1$ and $\liminf\int_{\tau(t)}^t p<1/e$. Thus oscillation criteria from the references cannot be applied to these examples. \begin{example} \label{examp1} \rm Consider the first-order delay differential equation with constant delay and sign-changing coefficient \begin{equation}\label{e31} x'(t)+p(t)x(t-1))=0,\quad t\geq 0, \end{equation} where \[ p(t)=\begin{cases} -\beta+2\beta t & \text{for }0\leq t< 1,\\ \beta & \text{for }1\leq t< 6,\\ \beta-2\beta (t-6) & \text{for }6\leq t< 7,\\ -\beta & \text{for }7\leq t< 8, \end{cases} \] and $p$ is periodic with $p(t+8)=p(t)$. Then $\limsup_{t\to\infty} \int_{\tau(t)}^t p(s)\,ds=\beta$ and $\liminf_{t\to\infty}\int_{\tau(t)}^t p(s)\,ds=-\beta$. So the interesting case happens when $0<\beta<1$. For $t=t_k=8k+6$ and $i=1,2,3,4$, the functions $\rho_i(t)=\beta^i/i!$ and \begin{gather*} f_1(t)=\frac{1}{1-\beta},\quad f_2(t)=\frac{1}{1-\beta-\frac{\beta^2}{2!}f_1(t-1)},\\ f_3(t)=\frac{1}{1-\beta-\frac{\beta^2}{2!}f_2(t-1) -\frac{\beta^3}{3!}f_2(t-1)f_1(t-2)},\\ f_4(t)=\frac{1}{1-\beta-\frac{\beta^2}{2!}f_3(t-1)-\frac{\beta^3}{3!}f_3(t-1)f_2(t-2) -\frac{\beta^4}{4!}f_3(t-1)f_2(t-2)f_1(t-3)}. \end{gather*} For $\beta=0.439$ and $n_k^*=4$, the software package Mathematica computes the lower bound to be \[ f_1(t)\approx 1.7825,\quad f_2(t)\approx2.5691,\quad f_3(t)\approx4.0183,\quad f_4(t)\approx-0.0038. \] Therefore by Theorem \ref{thm1} all solutions of \eqref{e31} are oscillatory. Looking for the lowest possible value of $\beta$, we consider the limit as $n_k^*\to\infty$. It is interesting to point out that the value of the lower bound of ${\beta}$ appears as close as possible to value $1/e\approx0.367879$. A numerical trend can be found in Table \ref{table1}, Then by Table \ref{table1} and Theorem \ref{thm1} every solution of \eqref{e31} has at least one zero on $[{t}_{k}-(2+n_k^*),\;{t}_{k}]$ for $k\in\mathbb{N}$, where $n_k^*$ is given by the corresponding $\beta$. So every solution of \eqref{e31} is oscillatory. \begin{table}[ht] \renewcommand{\arraystretch}{1.3} \caption{Numerical results for different lower bound of $\beta>1/e$} \label{table1} \begin{center} \begin{tabular}{|cccccc|} \hline $\beta\to 1/e$ &$f_1>0$& $f_2>0$ &\dots & $f_{n_k^{*}-1}>0$ & $f_{n_k^*}\leq0$ \\ \hline 0.37200 & 1.5924 & 1.9312 & \dots & $f_{19}\approx6.3466$ & $ f_{20}\approx-7.2066$\\ 0.36900 & 1.5848 & 1.9117 & \dots & $f_{38}\approx5.5387$ & $f_{39}\approx-56.786$\\ 0.36810 & 1.5825 & 1.9059 & \dots & $f_{89}\approx9.3745$ & $f_{90}\approx-1.8723$\\ 0.36795 & 1.5822 & 1.9049 & \dots & $f_{159}\approx37.17$ & $f_{160}\approx-0.232$\\ 0.36789 & 1.5820 & 1.9046 & \dots & $f_{413}\approx11.25$ & $f_{414}\approx-1.270$\\ \hline \end{tabular} \end{center} \end{table} \end{example} \begin{example} \label{examp2} \rm Consider the first-order delay differential equation with constant delay and variable coefficient \begin{equation}\label{e34} x'(t)+b(1+\sin (2t))x\big(t-\frac{\pi}{2}\big)=0, \quad t\geq 0\,. \end{equation} Then $\limsup_{t\to\infty} \int_{\tau(t)}^t p(s)\,ds=\frac{b}{2}(\pi+2)$ and $\liminf_{t\to\infty}\int_{\tau(t)}^t p(s)\,ds=\frac{b}{2}(\pi-2)$. So the interesting case happens when $0