\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 96, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/96\hfil Oscillation for second-order delay equations]
{Oscillation for second-order differential equations
 with delay}

\author[B. Bacul\'ikov\'a \hfil EJDE-2018/96\hfilneg]
{Blanka Bacul\'ikov\'a}

\address{Blanka Bacul\'ikov\'a \newline
Department of Mathematics,
Faculty of Electrical Engineering and Informatics,
Technical University of Ko\v{s}ice,
Letn\'a 9, 042 00~Ko\v{s}ice, Slovakia}
\email{blanka.baculikova@tuke.sk}

\thanks{Submitted March 28, 2018. Published April 24, 2018.}
\subjclass[2010]{34K11, 34C10}
\keywords{Second order differential equation; delay argument; oscillation; 
\hfill\break\indent monotonic properties}

\begin{abstract}
 We establishing monotonic properties of non-oscillatory solutions,
 and oscillation criteria for the second-order delay differential equation
 \[
 y''(t)+p(t)y(\tau(t))=0.
 \]
 The criteria obtained fulfil the gap in the oscillation theory and
 essentially improves the earlier ones. The progress is illustrated via
 Euler's differential equation. Moreover, we provide upper and lower bounds
 for the non-oscillatory solutions.
\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}

We consider the second-order delay differential equation
\begin{equation} \label{E}
y''(t)+p(t)y(\tau(t))=0,
\end{equation} 
under the following assumptions:
\begin{itemize}
\item[(H1)] $p \in C([t_0,\infty))$ and is positive;
\item[(H2)] $ \tau\in C([t_0,\infty))$ and $ \tau(t)\leq t$.
\end{itemize}

By a solution of \eqref{E} we mean a function $y$ in $C^2([t_0,\infty))$
that satisfies \eqref{E} on $[t_0,\infty)$.
We consider only those solutions
that satisfy $\sup \{|y(t)|:t \geq T\}>0$ for all $T\geq t_0$.
A solution of \eqref{E} is said to be oscillatory if
it has arbitrarily large zeros; otherwise, it is called nonoscillatory.
Equation \eqref{E} is said to be oscillatory if all its solutions are oscillatory.

There are many papers devoted to the oscillation of \eqref{E}
(see e.g. \cite{rmw}--\cite{Stu}). Various techniques have been obtained
for investigation of \eqref{E}. We mention here the pioneering work
of Sturm \cite{Stu} who introduced comparison principle to the oscillation theory.
Later Kneser \cite{Kne} contribute to the subject. Brands \cite{br} proved
 that oscillation of \eqref{E} with bounded delay is equivalent to
oscillation of ordinary differential equations.
 A new impetus to the investigation of oscillation was given by
 Mahfoud \cite{Mah} who deduce oscillation of delay equations from that
of ordinary equations.

\begin{theorem} \label{thmA}
Let $\tau'(t)>0$. If the ordinary differential equation
\[\label{E1}
y''(t)+\frac{p(\tau^{-1}(t))}{\tau'(\tau^{-1}(t))}y(t)=0
\]
is oscillatory, then so does \eqref{E}.
\end{theorem}

This comparison result permit us to extend any oscillatory criterion from
ordinary to delay differential equation.
Koplatadze et al.\ \cite{Kop} elaborated very nice technique for
investigation of \eqref{E} and presented the following criterion.

\begin{theorem} \label{thmB}
Assume that
\[
\limsup_{t\to\infty} \Big\{
 \tau(t)\int_t^\infty p(s)\,\mathrm{d} s
+ \int_{ \tau(t)}^t \tau(s)p(s)\,\mathrm{d} s
+\frac{1}{ \tau(t)}\int_{t_1}^{\tau(t)} s \tau(s) p(s)\,\mathrm{d} s
\Big\}>1.
\]
Then \eqref{E} is oscillatory.
\end{theorem}

The aim of this article is to establish new technique that improves
criteria existing for oscillation of \eqref{E}.
This fact will be illustrated via Theorems \ref{thmA}, \ref{thmB}.
 Our method is based on new monotonicity properties of possible
non-oscillatory solutions of \eqref{E}.

In this article, we assume that all functional inequalities hold eventually,
that is they are satisfied for all $t$ large enough.

\section{Preliminaries}

For non-oscillatory solutions of \eqref{E}, we restrict our attention 
to positive  solutions because if $y$ is a solution of so is $-y$.
Next we recall a well-known lemma by Kiguradze (see \cite{Ki1,KCH}) about
the structure of non-oscillatory solutions.

\begin{lemma} \label{lem1}
If $y(t)$ is a positive solution of \eqref{E}, then
\begin{equation}\label{kig}
y'(t)>0\quad \text{and}\quad y''(t)<0 ,
\end{equation}
 eventually.
\end{lemma}

As a preliminary, from \cite[Lemma 4.1]{Kop1} it follows that
the condition
\begin{equation}\label{nec}
\int_{t_0}^\infty \tau(s)p(s)\mathrm{d}{s}=\infty.
\end{equation}
is necessary for the oscillation of \eqref{E}.
So in what follows, we shall assume that \eqref{nec} holds.

\begin{lemma} \label{lem2}
If $y(t)$ is a positive solution of \eqref{E}, then
\begin{equation}\label{prop}
\frac{y(t)}{t}\downarrow 0\quad\text{and}\quad ty'(t)\leq y(t).
\end{equation}
\end{lemma}

\begin{proof}
 Assume that \eqref{E} possesses a positive solution $y(t)$.
Then \eqref{kig} is satisfied, let us say for $t\geq t_1$.
It follows from L'Hospital's rule that
$$
\lim_{t\to\infty}\frac{y(t)}{t}=\lim_{t\to\infty}{y'(t)}.
$$
We claim that \eqref{nec} implies $\lim_{t\to\infty} y'(t)=0$.
If we admit that $\lim_{t\to\infty }y'(t)=\ell>0$,
then integrating \eqref{E} yields
$$
y'(t_1)\geq \int_{t_1}^\infty p(s)y(\tau(s))\mathrm{d}{s}
\geq\int_{t_1}^\infty \tau(s)p(s)\frac{y(\tau(s))}{\tau(s)}\mathrm{d}{s}
\geq \ell \int_{t_1}^\infty \tau(s)p(s)\mathrm{d}{s}.
$$
This contradicts to \eqref{nec} and we see that $\lim_{t\to\infty }y'(t)=0$,
which implies
$$
y(t)=y(t_1)+\int_{t_1}^t y'(s)\mathrm{d}{s}\geq y(t_1)-t_1y'(t)+ty'(t)\geq ty'(t).
$$
Consequently
$$
\left(\frac{y(t)}{t}\right)'=\frac{ty'(t)-y(t)}{t^2}\leq0.
$$
The proof is complete.
\end{proof}

We recall the following comparison result, which is a particular
case of \cite[Theorem 2]{KN}.

\begin{lemma}\label{ccc}
Assume that
$a(t)\geq b(t)\geq 0$.
If the differential inequality
$$
y''(t)+a(t)y(t) \leq0
$$
has a positive solution, then the equation
$$
y''(t)+b(t)y(t) =0
$$
has a positive solution.
\end{lemma}

\begin{theorem}\label{first}
Assume that there is a constant $a_0$ such that for $t\geq t_0$
 \begin{equation}\label{stu}
 t\tau(t)p(t)\geq a_0> \frac{1}{4}\,.
 \end{equation}
Then \eqref{E} is oscillatory.
\end{theorem}

\begin{proof}
 On the contrary, assume that \eqref{E} possesses an eventually positive
solution $y(t)$. Taking the monotonicity of $y(t)/t$ into account,
we see that $y(t)$ is also solution of the inequality
 \begin{equation} \label{E2}
 y''(t)+\frac{\tau(t)}{t}p(t)y(t)\leq 0.
 \end{equation}
Lemma \ref{ccc} applied to \eqref{E2} and the Euler differential equation
 \begin{equation} \label{E_E}
 y''(t)+\frac{a_0}{t^2}y(t)=0,
 \end{equation}
guarantees that \eqref{E_E} has a positive solution.
This is a contradiction since \eqref{E_E} is oscillatory for $a_0>1/4$.
\end{proof}

In our next considerations we improve \eqref{stu}. In what follows we
shall assume that there exists a constant $a_0$ such that for $t\geq t_0$
\begin{equation}\label{sturm}
t\tau(t)p(t)\geq a_0>0 \quad\text{and}\quad a_0\leq \frac{1}{4}.
\end{equation}
We denote
\begin{equation}\label{beta}
\beta=\frac{1+\sqrt{1-4a_0}}{2}.
\end{equation}

\section{Main results}

In this section we derive new properties of non-oscillatory solutions
of \eqref{E} that will be used for establishing new oscillatory criteria.

\begin{lemma}\label{lem3}
 Assume that $y(t)$ is a positive solution of \eqref{E}.
 Then for any $\varepsilon>0$, the function
 $y(t)/t^{\beta+\varepsilon}$ is decreasing.
\end{lemma}

\begin{proof}
 Assume that $y(t)>0$ is a solution of \eqref{E}. Then \eqref{kig}
holds for $t\geq t_1$. Using the monotonicity of $\frac{y(t)}{t}$ into account,
it is easy to verify that
 \begin{equation}\label{sec}
 \begin{split}
 \Big(t^{2\beta}\Big(\frac{y(t)}{t^\beta}\Big)'\Big)'
&= y''(t)t^\beta-\beta(\beta-1)t^{\beta-2}y(t)\\
&= -t^\beta p(t)y(\tau(t))-\beta(\beta-1)t^{\beta-2}y(t)\\
&\leq t^{\beta-2}y(t)\left(-t\tau(t)p(t)-\beta(\beta-1)\right)\\
&\leq t^{\beta-2}y(t)\big(-a_0-\beta(\beta-1)\big)=0,
 \end{split}
\end{equation}
for $t\geq t_1$. Therefore,
$t^{2\beta}\left(\frac{y(t)}{t^\beta}\right)'$ is decreasing.
Denote
\begin{gather*}
\bar{ \beta}= \beta+\varepsilon \quad \text{for $\varepsilon$ small enough},\\
\delta=\varepsilon(2 \beta-1)+\varepsilon^2.
\end{gather*}
Since $- \beta( \beta-1)=a_0$, it is easy to verify, that
$-\bar{ \beta}(\bar{ \beta}-1)=a_0-\delta$.
Then
\begin{equation}\label{sec1}
\Big(t^{2\bar{ \beta}}\Big(\frac{y(t)}{t^{\bar{ \beta}}}\Big)'\Big)'
\leq t^{\bar{ \beta}-2}y(t)\left(-a_0-\bar{ \beta}(\bar{ \beta}-1)\right)
= - t^{\bar{ \beta}-2}y(t)\delta<0.
\end{equation}
Since $\big(t^{2\bar{ \beta}}\big(\frac{y(t)}{ t^{\bar{ \beta}}}\big)'\big)'<0$,
then $t^{2\bar{ \beta}}\big(\frac{y(t)}{ t^{\bar{ \beta}}}\big)'$ is decreasing
and so either
$$
\Big(\frac{y(t)}{ t^{\bar{ \beta}}}\Big)'>0\quad \text{or}\quad
\Big(\frac{y(t)}{ t^{\bar{ \beta}}}\Big)'<0,
$$
eventually.

If we admit that $\big(\frac{y(t)}{ t^{\bar{ \beta}}}\big)'>0$, then
integrating inequality \eqref{sec1} from $t_1$ to $\infty$, we have
$$
t_1^{2\bar{ \beta}}\Big(\frac{y(x)}{ x^{\bar{ \beta}}}\Big)'_{x=t_1}
\geq \delta \int_{t_1}^{\infty}s^{2{\bar{ \beta}-2}}\frac{y(s)}{ s^{\bar{ \beta}}}\,
\mathrm{d}{s}
\geq \delta \frac{y(t_1)}{ t_1^{\bar{ \beta}}}
\int_{t_1}^{\infty}s^{2{\bar{ \beta}-2}}\,\mathrm{d}{s}=\infty.
$$
It is a contradiction and we conclude, that
$\big(\frac{y(t)}{ t^{\bar{ \beta}}}\big)'>0$ and so
 $y(t)/t^{\beta+\varepsilon}$ is decreasing.
\end{proof}

\begin{lemma}\label{lem4}
Assume that there are constants ${a_1}$ and $\varepsilon$ such that for
$t\geq t_0$,
\begin{equation}\label{55}
t^2p(t)\Big(\frac{\tau(t)}{t}\Big)^{\beta+\varepsilon}\geq {a_1}.
\end{equation}
If $a_1>\frac{1}{4}$, then \eqref{E} is oscillatory.
If $a_1\leq\frac{1}{4}$, then for any positive solution $y(t)$ of \eqref{E}
$$
\frac{y(t)}{ t^{\beta}}
$$ is decreasing.
\end{lemma}

\begin{proof}
Assume that $y(t)>0$ is a solution of \eqref{E}.
The monotonicity of $\frac{y(t)}{ t^{{ \beta+\varepsilon}}}$ implies that
$y(t)$ is a positive solution of the inequality
\begin{equation}\label{etau}
 y''(t)+\Big(\frac{\tau(t)}{t}\Big)^{{ \beta+\varepsilon}}p(t)y(t)\leq 0.
\end{equation}
 Lemma \ref{ccc} implies that the Euler equation
 $$
 y''(t)+\frac{a_1}{t^2}y(t)=0,
 $$
 has a positive solution. This contradicts the fact that the
Euler equation is oscillatory for $a_1>1/4$
and so we conclude that \eqref{E} is oscillatory.

Now, we assume, that ${a_1}\leq\frac{1}{4}$.
 Denote
 $$
 { \beta_1}=\frac{1+\sqrt{ 1-4{a_1}}}{2}.
 $$
Let us consider $\varepsilon>0$, such that ${ \beta_1}+\varepsilon\leq\beta$.
It is easy to see that
$$
-({ \beta_1}+\varepsilon)({ \beta_1}+\varepsilon-1)={a_1}-\delta_1,
$$
where $\delta_1=\varepsilon(2{ \beta_1}-1)+\varepsilon^2$.

On the other hand, the monotonicity of
$y(t)/t^{{ \beta+\varepsilon}}$ yields
 $$
 y(\tau(t))\geq\Big(\frac{\tau(t)}{t}\Big)^{{ \beta+\varepsilon}}y(t).
 $$
Thus,
 \begin{align*}
 \Big(t^{2{ \beta_1}+\varepsilon}\Big(\frac{y(t)}{ t^{{ \beta_1}
 +\varepsilon}}\Big)'\Big)'
 &= -t^{{ \beta_1}+\varepsilon} p(t)y(\tau(t))
 -({ \beta_1}+\varepsilon)({ \beta_1}+\varepsilon-1)t^{{ \beta_1}+\varepsilon-2}y(t)\\
 &\leq t^{{ \beta_1}+\varepsilon-2}y(t)
\Big(-t^2p(t)\Big(\frac{\tau(t)}{t}\Big)^{{ \beta+\varepsilon}}
-({ \beta_1}+\varepsilon)({ \beta_1}+\varepsilon-1)\Big)\\
 &\leq - t^{{ \beta_1}+\varepsilon-2}y(t)\delta_1.
 \end{align*}
Proceeding similarly as in proof of Lemma \ref{lem3}, we obtain that
$\frac{y(t)}{ t^{{ \beta_1}+\varepsilon}}$ is decreasing.
 Since
 $$
 { \beta_1}+\varepsilon\leq\beta,
 $$
 we can conclude that $\frac{y(t)}{ t^{\beta}}$ is decreasing too.
The proof is complete.
\end{proof}

Now we are ready to provide the oscillatory criterion that improves
Theorem~\ref{first}.

\begin{theorem}\label{thm2}
Assume that there is a constant $a_2$ such that for $t\geq t_0$,
\begin{equation}\label{theorem}
t^{2- \beta}(\tau(t))^{ \beta}p(t)\geq a_2>\frac{1}{4},
\end{equation}
then \eqref{E}
 is oscillatory.
 \end{theorem}

\begin{proof}
Assume to the contrary that \eqref{E} has a positive solution $y(t)$.
The monotonicity of $\frac{y(t)}{ t^{\beta}}$ implies that $y(t)$ is a solution
of the differential inequality
\[\label{E3}
 y''(t)+\Big(\frac{\tau(t)}{t}\Big)^{ \beta}p(t)y(t)\leq 0.
\]
Lemma \ref{ccc} implies that the Euler equation
$$
y''(t)+\frac{a_2}{t^2}y(t)=0,
$$
has a positive solution. This contradicts to fact that considered Euler
equation is oscillatory for $a_2>1/4$. The proof is complete.
\end{proof}

\begin{remark} \rm
 In contrast to results presented in \cite{Opl} our oscillatory criterion is
easily verifiable and does not require any auxiliary constants and functions.
Unlike to \cite{Opl} our results will be supported by illustrative example.
\end{remark}

We illustrate the novelty and progress of our oscillation criterion via
its application to Euler differential equations with a delay argument:
\begin{equation}\label{EED}
y''(t)+\frac{a}{t^2}y(\lambda t)=0, \quad \lambda\in(0,1).
\end{equation}

\begin{corollary} \label{coro1}
 If
 \begin{equation}\label{C1}
 \lambda^{\beta}a>\frac{1}{4},
 \end{equation}
 then \eqref{EED} is oscillatory.
\end{corollary}

\begin{remark} \rm
By Theorem \ref{thmA}, the oscillation of \eqref{EED} follows from the oscillation of
\begin{equation}\label{R1}
y''(t)+\frac{a\lambda}{t^2}y(t)=0,
\end{equation}
which leads to the condition
 \begin{equation}\label{sC2}
 \lambda a>\frac{1}{4}.
 \end{equation}
 What is more, \cite[Corollaries 7.5 and 7.6 , and Theorem 7.9]{rg}
guarantee the oscillation of \eqref{EED} if \eqref{sC2} holds.
 Evidently criterion \eqref{C1} provides better result.
 \end{remark}

Our next considerations are intended to essentially improve Theorem \ref{thmB}.
For this reason we need the monotonicity which is opposite to that in
Lemma \ref{lem4}.

 \begin{lemma}\label{lem5}
Let \eqref{sturm} hold and $ \alpha=\frac{1-\sqrt{1-4a_0}}{2}$.
Assume that $y(t)$ is a positive solution of \eqref{E}.
 Then $y(t)/t^{\alpha}$ is increasing.
 \end{lemma}

 \begin{proof}
 Assume that $y(t)$ is a positive solution of \eqref{E}. Then  \eqref{kig} 
is satisfied for $t\geq t_1$.
 Taking the monotonicity of $\frac{y(t)}{t}$ into account, it is easy 
to verify that
 \begin{equation}\label{le5}
 \begin{split}
 \Big(t^{2\alpha}\Big(\frac{y(t)}{t^\alpha}\Big)'\Big)'
&= y''(t)t^\alpha-\alpha(\alpha-1)t^{\alpha-2}y(t)\\
&= -t^\alpha p(t)y(\tau(t))-\alpha(\alpha-1)t^{\alpha-2}y(t)\\
&\leq  t^{\alpha-2}y(t)\left(-t\tau(t)p(t)-\alpha(\alpha-1)\right)\leq 0.
\end{split}
\end{equation}
 Therefore $t^{2\alpha}\left(\frac{y(t)}{t^\alpha}\right)'$ is decreasing. 
If we admit that $t^{2\alpha}\left(\frac{y(t)}{t^\alpha}\right)'<0$ 
for $t\geq t_2\geq t_{1}$, then there exists constant $k>0$
such that
 $$
 t^{2\alpha}\Big(\frac{y(t)}{t^\alpha}\Big)'<-k<0
 $$
for $t>t_2$.
Integrating the last inequality form $t_2$ to $t$, we have
 $$
 \frac{y(t)}{t^{\alpha}}<\frac{y(t_2)}{t_{2}^{\alpha}}
-k\int_{t_2}^{t}s^{-2\alpha}\,\mathrm{d}{s}\to -\infty \quad
\text{for } t\to \infty.
 $$
 This is a contradiction and we conclude that 
$t^{2\alpha}\big(\frac{y(t)}{t^\alpha}\big)'>0$. The proof is complete.
 \end{proof}

Lemmas \ref{lem4} and \ref{lem5} provide upper and lower bound for possible
 non-oscillatory solutions of \eqref{E}.

 \begin{theorem}\label{Thbound}
 Let \eqref{sturm} hold, $ \alpha=\frac{1-\sqrt{1-4a_0}}{2}$ and 
$ \beta=\frac{1+\sqrt{1-4a_0}}{2}$.
 Then every positive solution $y(t)$ of \eqref{E} satisfies
 $$
 c_1 t^\alpha\leq y(t) \leq c_2 t^\beta,
 $$
 $c_1$, $c_2$ are constants.
 \end{theorem}

\begin{proof}
By Lemma \ref{lem5}, the function $y(t)/t^{\alpha}$ is increasing and so for 
all $t\geq t_1$,
 $$
 \frac{y(t)}{t^{\alpha}}\geq \frac{y(t_1)}{t_1^{\alpha}}=c_1.
 $$
 The second part of the theorem can be proved similarly.
 \end{proof}

 Now, we present new oscillatory results using both monotonic properties 
of non-oscillatory solutions of \eqref{E} presented in
 Lemma \ref{lem4} and Lemma \ref{lem5}.

\begin{theorem}\label{thm3}
 Let \eqref{sturm} hold, and assume that
\begin{equation}\label{T3}
\begin{split}
&\limsup_{t\to\infty} \Big\{ \tau^{-\beta}(t)\int_{t_1}^{\tau(t)}s\tau^{\beta}(s)p(s) 
\,\mathrm{d}{s}\\
&+\tau^{1-\beta}(t)\int_{\tau(t)}^{t}\tau^{\beta}(s)p(s)\,\mathrm{d}{s}
 +\tau^{1-\alpha}(t)\int_{t}^{\infty}\tau^{\alpha}(s)p(s)\,\mathrm{d}{s}\Big\}>1.
 \end{split}
 \end{equation}
 Then \eqref{E} is oscillatory.
\end{theorem}

\begin{proof}
 On the contrary, assume that \eqref{E} possesses a positive solution $y(t)$. 
Then \eqref{kig} holds for $t\geq t_1$.
 Integrating \eqref{E} twice, we get
 \begin{equation}\label{1.1}\begin{split}
 y(t)
&\geq y(t_{1})+ \int_{t_1}^{t}\int_{u}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}\,\mathrm{d}{u}\\
&=y(t_{1})+\int_{t_1}^{t}\int_{u}^{t}p(s)y(\tau(s))\,\mathrm{d}{s}\,\mathrm{d}{u}
 +\int_{t_1}^{t}\int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}\,\mathrm{d}{u}\\
&=y(t_{1})+\int_{t_1}^{t}(s-t_{1})p(s)y(\tau(s))\,\mathrm{d}{s}
 +(t-t_{1})\int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}\\
&=y(t_{1})-t_{1}\int_{t_{1}}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}
 +\int_{t_1}^{t}sp(s)y(\tau(s))\,\mathrm{d}{s} \\
&\quad  +t\int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}.
 \end{split}
 \end{equation}
 On the other hand, an integration
 of \eqref{E} yields
 $$
 y'(t)\geq \int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}
 $$
 which in view of \eqref{prop} implies
 $$
 y(t_{1})>t_{1}\int_{t_{1}}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}.
 $$
 Employing the last inequality in \eqref{1.1}, we see that
 \begin{equation}\label{1.2}\begin{split}
 y(t)\geq \int_{t_1}^{t}sp(s)y(\tau(s))\,\mathrm{d}{s}
+t\int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}.
 \end{split}
 \end{equation}
 Therefore,
 \begin{align*} % \begin{split}
 y(\tau(t))
&\geq \int_{t_1}^{\tau(t)}sp(s)y(\tau(s))\,\mathrm{d}{s}
 +\tau(t)\int_{\tau(t)}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}\\
 &=\int_{t_1}^{\tau(t)}sp(s)y(\tau(s))\,\mathrm{d}{s}
 +\tau(t)\int_{\tau(t)}^{t}p(s)y(\tau(s))\,\mathrm{d}{s} \\
&\quad +\tau(t)\int_{t}^{\infty}p(s)y(\tau(s))\,\mathrm{d}{s}.
 \end{align*}
 Using that $y(t)/t^{\beta}$ is decreasing and $\frac{y(t)}{t^{\alpha}}$ 
is increasing, we have
 \begin{align*}
 1&\geq\tau^{-\beta}(t)\int_{t_1}^{\tau(t)}s\tau^{\beta}(s)p(s) \,\mathrm{d}{s}
 +\tau^{1-\beta}(t)\int_{\tau(t)}^{t}\tau^{\beta}(s)p(s)\,\mathrm{d}{s}\\
 &\quad +\tau^{1-\alpha}(t)\int_{t}^{\infty}\tau^{\alpha}(s)p(s)\,\mathrm{d}{s}.
 \end{align*}
 Taking limit superior as $t\to \infty$ on both sides of the previous 
inequality, we are led to contradiction with assumptions of the theorem. 
The proof is complete.
\end{proof}

\begin{corollary}\label{coro2}
 Let \eqref{sturm} hold, and assume that
 \begin{align*}
 &\limsup_{t\to\infty} \Big\{ t^{-\beta}\int_{t_1}^{\lambda t}s^{1+\beta}p(s) 
 \,\mathrm{d}{s} \\
&+\lambda t^{1-\beta}\int_{\lambda t}^{t}s^{\beta}p(s)\,\mathrm{d}{s}
 +t^{1-\alpha}\lambda \int_{t}^{\infty}s^{\alpha}p(s)\,\mathrm{d}{s}\Big\}>1.
 \end{align*}
Then 
\begin{equation} \label{ED}
 y''(t)+p(t)y(\lambda t)=0, \quad \lambda\in(0,1),
\end{equation}
 is oscillatory.
\end{corollary}

\begin{proof}
Proceeding as in proof of Theorem \\ref{thm3} for $\tau (t)=\lambda t$ with 
$\lambda\in(0,1)$, we obtain the result.
\end{proof}

Following result is simple consequence of Corollary \ref{coro2} for \eqref{EED}

\begin{corollary} \label{coro3}
 If
 \begin{equation}\label{C3}
 \Big\{\frac{\lambda^{\beta}}{\beta}+\frac{\lambda^{\beta}
-\lambda}{1-\beta}+\frac{\lambda}{\beta}\Big\}>1,
 \end{equation}
 then is \eqref{EED} oscillatory.
\end{corollary}

\begin{remark} \rm
 If we employ the additional condition $\tau'(t)>0$, it is easy to see that 
each term of \eqref{T3} is greater  than the corresponding term of the 
criterion presented in Theorem \ref{thmB}.
 Consequently Theorem \ref{thm3} essentially improves the result of \cite{Kop}.
\end{remark}

We illustrate  the results obtained with example.

\begin{example}\label{ex1} \rm
We consider the Euler delay equation
 \[ %\label{EED}
 y''(t)+\frac{a}{t^2}y(\lambda t)=0, \quad \lambda\in(0,1).
 \] 
For $\lambda=0,2$ and $a=1,25$ criterion \eqref{C3} gives $2,2361>1$.
On the other hand criterion \cite[(2.5)]{Kop} gives $0,9024 \ngtr 1$.
For $\lambda=0,8$ and $a=0,3125$ our criterion holds $(1,1180 > 1)$
and Koplatadze et al.\ \cite{Kop} fails since $0,5558\ngtr 1$.
So our criterion essentially improves the known ones.
\end{example}


\subsection*{Acknowledgements}
This research was supported by S.G.A. Kega 019-025TUKE-4/2017.


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