\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 59, pp. 1--30.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2004/59\hfil Damped second order linear differential equation]
{Damped second order linear differential equation with deviating
arguments: Sharp results in oscillation properties} 

\author[Leonid Berezansky \& Yury Domshlak\hfil EJDE-2004/59\hfilneg]
{Leonid Berezansky \& Yury Domshlak}  % in alphabetical order

\address{Leonid Berezansky \hfill\break
Department of Mathematics,
Ben-Gurion University of the Negev,
Beer-Sheva 84105, Israel}
\email{brznsky@cs.bgu.ac.il}

\address{Yury Domshlak\hfill\break
Department of Mathematics,
Ben-Gurion University of the Negev,
Beer-Sheva 84105, Israel}
\email{domshlak@cs.bgu.ac.il}

\date{}
\thanks{Submitted March 23, 2004. Published April 19, 2004.}
\subjclass[2000]{34K11}
\keywords{Linear differential equation with deviating arguments,
        second order,  \hfill\break\indent
	damping term, oscillation, Sturmian comparison method}


\begin{abstract}
This article presents a new approach for investigating the oscillation
properties of second order linear differential equations with a damped term
containing a deviating argument
$$
x''(t)-[P(t)x(r(t))]'+Q(t)x(l(t))=0,\quad r(t)\leq t.
$$
To study this equation, a specially adapted  version of Sturmian Comparison
Method is developed and the following results are obtained:

\noindent (a) A comprehensive description of all critical (threshold) states
with respect to its oscillation properties
for a linear autonomous delay differential equation
$$
y''(t)-py'(t-\tau)+qy(t-\sigma)=0, \quad \tau>0,\;\infty<\sigma<\infty.
$$
(b) Two versions of Sturm-Like Comparison Theorems.
Based on these Theorems, sharp conditions under which all solutions are
oscillatory for specific realizations of $P(t), r(t)$ and $l(t)$ are obtained.
These conditions are formulated as the unimprovable analogues  of the classical
Knezer Theorem which is well-known for ordinary differential equations
($P(t)=0$, $l(t)=t$).

\noindent (c) Upper bounds for intervals, where
any solution  has at least one zero.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

It is well-known that many results for second order
linear ordinary differential equations
were obtained for the equation
\begin{equation} \label{1}
y''(t)+a(t)y(t)=0,
\end{equation}
but not for the equation
\begin{equation}\label{2}
x''(t)-(P(t)x(t)))'+ Q(t)x(t)=0
\end{equation}
with damping term $(P(t)x(t)))'$. The reason is the following: it is easy
to transform \eqref{2} into \eqref{1} by the substitution
\begin{equation} \label{3}
x(t)=y(t)\exp\big\{\frac{1}{2}\int P(t)dt\big\},
\end{equation}
where $a(t):=Q(t)-\frac{1}{4}P^2(t)-\frac{1}{2}P'(t)$.
After this transformation
the solutions of \eqref{2} and \eqref{1} have the same zeros.

Consider now a second order linear delay differential
equation without delay in the damping part
\begin{equation}
\label{4}
(lx)(t):=x''(t)-(P(t)x(t)))'+ Q(t)x(l(t))=0.
\end{equation}
The substitution \eqref{3}
transforms \eqref{4} into the following equation
\begin{equation} \label{5}
y''(t)+a(t)y(t)+b(t)y(l(t))=0,
\end{equation}
where
$$
a(t)=-\frac{1}{4}P^2(t)-\frac{1}{2}P'(t),~ b(t)=Q(t)
\exp\big\{-\frac{1}{2}\int_{l(t)}^t P(s)ds\big\}.
$$
Hence every result on oscillation properties of \eqref{5}
without a damping term implies the corresponding result for
\eqref{4}.

The situation is dramatically changed for an equation with delay
in the damping term:
\begin{equation}
\label{6}
(lx)(t):=x''(t)-(P(t)x(r(t)))'+ Q(t)x(l(t))=0.
\end{equation}
There is no good substitution
which transforms \eqref{6} into an equation without a damping term.
Hence we have to investigate \eqref{6} itself.

It is important to note  that the delay in the damping term significantly
changes oscillating properties of this equation.
Consider, for example, an autonomous equation
\begin{equation} \label{7}
x''(t)-px'(t)+qx(t-\sigma)=0, \quad-\infty<\sigma<\infty,\; t\geq t_0.
\end{equation}
It is well known that for every $-\infty<p<\infty$ and $q>\frac{p^2}{4}$
all solutions of \eqref{7} are oscillatory.

On the other hand, an equation with a delay in the damping term
\begin{equation}
\label{8}
x''(t)-px'(t-\tau)+qx(t-\sigma)=0, \quad\tau>0,\; \sigma\leq \tau,\; p<0
\end{equation}
has a nonoscillatory solution for any $q$.

Oscillation properties of damping equations  were studied
in \cite{g1,g2,g3,g4,g5} in which  very general nonlinear
differential equations with delay were considered.
Unfortunately, in \cite{g1,g2,g3,g4,g5} there are
no new results on \eqref{6} and even for \eqref{4}.
The purpose of the present paper is to obtain such results.

Our main tool is Sturmian Comparison Method (SCM) which was
developed in \cite{d1} by one of the authors of this paper for various classes
of differential and difference equations.
For delay differential equations of the first and the second order
this method was described in details in \cite[Chap.4]{d1}.
The further development of SCM for various kinds of differential equations
one can find in the papers \cite{d2,d3,d4,d5}.

We will explain here the main idea of this method.
One of the most important results in the
qualitative theory of ordinary differential equations is the classical

\begin{theorem}[Sturmian Comparison Theorem] \label{thmA}
Suppose there  exists a positive solution of the inequality
\begin{equation} \label{9}
\tilde{l}y:=y''(t)+a(t)y(t)\geq 0, \quad t\in(t_1,t_2),
\end{equation}
such that $y(t_1)=y(t_2)=0$ and $b(t)\geq a(t)$,  $t\in(t_1,t_2)$.
Then the inequality
\begin{equation} \label{10}
{l}x:=x''(t)+b(t)x(t)\leq 0
\end{equation}
has no positive solutions on $(t_1,t_2)$.
\end{theorem}

The above theorem implies the following result.

\begin{theorem}[Sturmian Oscillation Comparison Theorem] \label{thmB}
Suppose  $b(t)\geq a(t)$, $t\in (t_0, \infty)$, and all solutions
of the equation $\tilde{l}y=0$ are oscillatory. Then all solutions of
the equation $lx=0$ are also oscillatory.
\end{theorem}

There are several investigations devoted to extensions
of  Theorem \ref{thmB} to various classes of functional differential equations,
including nonlinear differential equations with delays.
We will not follow this direction, since SCM is concerned with the
generalization just of Theorem \ref{thmA}  but not of Theorem \ref{thmB}  for various
classes of
functional differential equations.

It is worth to note that Theorem \ref{thmA}  is formulated for a finite interval
$(t_1,t_2)$, but not for a semiaxis. Thus it  allows not only to obtain
conditions  of oscillation of all solutions but also to estimate
the length of a sign-preservation interval.

The most important advantage of SCM is  the fact, that for the application of
this method it is sufficient to construct only one solution of
\eqref{9} (``Testing inequality"), satisfying some required properties.
Thus to the best of our knowledge this approach is  the unique
constructive method in the oscillation theory of functional differential equations.

We present here a version of SCM which is
specially adapted to second order delay damped differential equation \eqref{6}.
However first  we consider oscillation properties of \eqref{4}
without delay in the damping term.
To this end we do not need to develop a new theory.

\section{Oscillation Properties of \eqref{4}}

Consider \eqref{4} with monotone increasing delay
argument $l(t)<t$ and denote by $k(t)$ the inverse function of $l(t)$.
A classification of oscillation properties of \eqref{4}
is based on the following statement which was published
long time ago.

\begin{theorem}[{\cite[Lemma 4.4 , Theorem \ref{thm4}.4]{d1}}] \label{thmC}
Let be $l(t_2)>t_1$, $\varphi(t)>0$, $z(t)\geq 0$,
$\varphi, z\in {\mathbf{C}}^2[l(t_1),k(t_2)]$,
\begin{equation}\label{11}
\int_{t_1}^{t_2} \varphi(s)ds= \pi,\quad
0<\int_{l(t)}^{t} \varphi(s)ds <\frac{\pi}{2}, \quad t\in[t_1, k(t_2)].
\end{equation}
If
\begin{equation}\label{12}
\begin{aligned}
a(t)\geq \tilde{a}(t)&:=\varphi^2(t)+\frac{\varphi''(t)}{2\varphi(t)}-
\frac{3}{4} \left(\frac{\varphi'(t)}{\varphi(t)}\right)^2+z'(t)-z^2(t)\\
&\quad - z(t)\frac{\varphi'(t)}{\varphi(t)}-2\varphi(t)z(t)\cot \int_{t}^{k(t)}
 \varphi(s)ds
\end{aligned}\end{equation}
and
\begin{equation} \label{13}
b(t)\geq \tilde{b}(t):=2l'(t)z[l(t)]\sqrt{\varphi(t)\varphi[l(t)]}
\csc\int_{l(t)}^t \varphi(s)ds\exp\big\{\int_{l(t)}^{t} z(s)ds\big\},
\end{equation}
then \eqref{5} has no positive solution on $[l(t_1),t_2]$ .
\end{theorem}


\begin{corollary} \label{coroC1}
Suppose
$$
\varphi(t)>0; \int^{\infty} \varphi (t)dt=+\infty, \quad
\int_{l(t)}^t \varphi (s)ds<\frac{\pi}{2};\quad z(t)\geq 0
$$
and  \eqref{12}-\eqref{13} hold on $(t_0,\infty)$. Then all
solutions of \eqref{5} are oscillatory and, moreover, every solution of
\eqref{5} has at least one zero on any interval $(l(t_1),k(t_2))$ when
$l(t_2)>t_1$ and $\int_{t_1}^{t_2} \varphi(t)dt =\pi$.
\end{corollary}

As was mentioned in Introduction, the substitution \eqref{3} in \eqref{4}
 does not change the oscillation properties of \eqref{4}
and transforms it into \eqref{5}, where
$$
a(t):=-\frac{1}{4}P^2(t)-\frac{1}{2}P'(t),\quad
b(t):=Q(t)\exp\big\{-\frac{1}{2}\int_{l(t)}^t P(s)ds\big\}.
$$
Thus \eqref{11}--\eqref{13} represent easily verified  conditions
formulated in terms of \eqref{4} only.

One can illustrate the quality of these conditions by the following example.

\begin{example} \label{ex1} \rm
Let in \eqref{4}
$P(t):=\frac{p}{t}$, $p=\mathrm{const}$, $l(t):=\frac{t}{\mu}$, $\mu>1$
and so $ k(t)=\mu t$.
Let in Theorem \ref{thmC}
$\varphi(t):=\frac{\nu}{t}$, $\nu>0$, $z(t):=\frac{s}{t}$, $s=\mathrm{const}>0$.
Then
\begin{gather*}
t^2\tilde{a}(t):=\frac{1}{4}+\nu^2-s^2-2\nu s\cot(\nu \ln \mu), \\
t^2\tilde{b}(t):= 2\nu s\mu^{s+\frac{1}{2}}\csc(\nu \ln \mu), \\
\exp\big\{-\frac{1}{2}\int_{l(t)}^tP(s)ds\big\}=\mu^{-\frac{p}{2}};\quad
t^2a(t)=-\frac{p^2}{4}+\frac{p}{2}.
\end{gather*}
\end{example}

Conditions \eqref{12}--\eqref{13} take a form
\begin{equation} \label{14}
\begin{gathered}
-\frac{p^2}{4}+\frac{p}{2}\geq \frac{1}{4}+\nu^2-s^2-2\nu s\cot(\nu \ln \mu),\\
t^2 Q(t)\mu^{-\frac{p}{2}}\geq 2\nu s\mu^{s+\frac{1}{2}}\csc(\nu \ln \mu).
\end{gathered}
\end{equation}
Denote by $s_{\nu}$ the positive root of a quadratic equation
\begin{equation}
\label{15}
s^2+2\nu s\cot(\nu \ln \mu)-\frac{1}{4}(p-1)^2-\nu^2=0,
\end{equation}
which exists for all $\mu>1$ and $\nu>0$ is sufficiently small.
Then \eqref{14}$_1$ and \eqref{14}$_2$ turn into an  equality
and into the following inequality
\begin{equation} \label{16}
t^2 Q(t)\geq 2\nu s_{\nu} \frac{\mu^{s+\frac{1+p}{2}}}{\sin(\nu \ln \mu)},
\quad \forall t
\end{equation}
respectively.
Suppose
$$
s_0:=\lim_{\nu\to 0} s_{\nu} =\frac{(p-1)^2\ln
\mu}{2\sqrt{4+(p-1)^2\ln^2\mu}+4}.
$$
Let  $t_1=T$ in Theorem \ref{thmC} be  sufficiently large. By \eqref{11}
 we have
$$ \int_T^{t_2}\varphi(s)ds=\pi \Longleftrightarrow t_2=T\exp{\frac{\pi}{\nu}}.
$$
Then $(l(t_1),t_2)=(\mu T,T\exp{\frac{\pi}{\nu}})$,
 and we have the following statement.

\begin{corollary} \label{coroC2}
Let
\begin{equation} \label{17}
\liminf_{t\to \infty} t^2 Q(t)= C>\frac{2s_0 \mu^{\frac{1+p}{2}+s_0}}{\ln
\mu}:=B(p).
\end{equation}
Then all solutions of the equation
\begin{equation}
\label{18}
x''(t)-\left[\frac{p}{t}x(t)\right]'+Q(t)x\left(\frac{t}{\mu}\right)=0
\end{equation}
are oscillatory. Moreover, there exists $\nu>0$ such that every
solution of \eqref{18}
has at least one zero on any  interval $(\frac{T}{\mu},Te^{\frac{\pi}{\nu}})$
for sufficiently large $T$.
\end{corollary}

\noindent{\bf Remark.}\;
Condition \eqref{17}  is the best possible for every $p$ in the following sense.
By direct calculations  one can check, that \eqref{18}
with $Q(t):=\frac{B(p)}{t^2}$ has a nonoscillatory solution
$x(t)=t^{s_0+\frac{1+p}{2}}$.
Let us notice, that for $p=0$ (i.e. for the equation without a damping term)
this fact was mentioned in \cite[p.177]{d1}.

\section{Critical states of the second order
 damped autonomous equation}

Instead of \eqref{8}  without loss of generality we  will
consider here the equation
\begin{equation}
\label{19}
x''(t)-px'(t-1)+qx(t-\sigma)=0,\quad - \infty<\sigma<\infty, ~t\geq 0,
\end{equation}
since one can transform \eqref{8} into \eqref{19}
by rescaling. \smallskip

\noindent {\bf Definition.} %1
We will say that \eqref{19} is {\bf in a critical state (CS)} with respect to
its  oscillation properties if there exists an eventually
positive solution $x(t)>0, t\geq t_0$ of \eqref{19}, while an equation
$$
z''(t)-pz'(t-1)+(q+\epsilon)z(t-\sigma)=0
$$
has no such solution for every $\epsilon>0$. In this case the pair
of numbers $\{p;q\}$ is said to be {\bf a critical pair}.

It is well known that for a wide class of autonomous linear differential
equations with deviating arguments, in particular, for \eqref{19}, the
following statement holds:
\begin{quote}
All solutions of the equation are oscillatory if and only if its characteristic
quasipolynomial does not change its sign for every $\lambda\in(-\infty,\infty)$.
\end{quote}

In particular,  all solutions of \eqref{19}
are oscillatory if and only if its characteristic quasipolynomial is
eventually positive:
$$
F_{p,q} (\lambda):=F(\lambda)=\lambda^2-p\lambda e^{-\lambda}+qe^{-\lambda\sigma}>0,
\quad\lambda\in(-\infty,\infty).
$$
On the other hand,  \eqref{19} is in  CS if and only if the following conditions
hold
\begin{equation} \label{20}
\begin{gathered}
F(\lambda)\geq 0 \quad  \forall\lambda\in(-\infty,\infty)\\
\exists \bar{\lambda}:F(\bar{\lambda})=0.
\end{gathered}
\end{equation}
Indeed, \eqref{20}$_2$ implies that \eqref{19} has a solution
$x(t)=e^{\bar{\lambda}t}$,
and \eqref{20}$_1$ provides that $F_{p,q+\epsilon}(\lambda)>0$,
for all $\lambda$.

Hence  CS $\{p;q\}$ implies that $\bar{\lambda}$ is a solution of the system
\begin{equation}\label{21}
\begin{gathered}
F(\lambda)=0\\ F'(\lambda)=0.
\end{gathered}
\end{equation}
Note that the existence of a solution of \eqref{21} for some pair $\{p,q\}$ does
not imply that this pair is critical.

Rewrite \eqref{21} in the form
\begin{equation} \label{22}
\begin{gathered}
pe^{-\lambda}\lambda-qe^{-\lambda\sigma}-\lambda^2=0\\
pe^{-\lambda}(1-\lambda)+qe^{-\lambda\sigma}\sigma-2\lambda=0.
\end{gathered}
\end{equation}
Then the pair $\{p,q\}$ (which may or may not be critical)
can be presented in the parametric form
\begin{equation} \label{23}
\begin{gathered}
p=\frac{\lambda(2+\lambda\sigma)}{1+\lambda(\sigma-1)}e^{\lambda}\\
q=\frac{\lambda^2(1+\lambda)}{1+\lambda(\sigma-1)}e^{\lambda\sigma}
\end{gathered}
\end{equation}
where $-\infty<\lambda<\infty$.
The pair $\{p,q\}$ belongs to the quadrant $\{p\geq 0, q\geq 0\}$ if
\begin{equation} \label{24}
\begin{gathered}
0\leq \lambda <\frac{1}{1-\sigma}, \quad\mbox{when }  -\infty<\sigma<1,\\
0<\lambda<\infty,\quad  \mbox{when } \sigma\geq 1.
\end{gathered}
\end{equation}
The pair $\{p,q\}$ represented by  \eqref{23} cannot be in CS if
\begin{equation} \label{25}
\lim_{\lambda\to -\infty} F(\lambda)=-\infty; \quad
\lim_{\lambda\to \infty} F(\lambda)=\infty.
\end{equation}
A pair $\{p<0,q>0\}$ cannot be in CS. For $-\infty<\sigma\leq 1$ this fact  follows
from \eqref{25}. For $\sigma>1$ this fact follows from the boundedness of
$F(\lambda)$ from below.

Suppose  \eqref{21} is solvable for two pairs
$\{p,q_1\}$ and $\{p,q_2\}$, where $q_1<q_2$. Then the first pair is not
a CS, but the second one may be CS.

Let us give without additional explanations the {\em full description} of
CS for \eqref{19}:

\noindent \textbf{(1) $-\infty<\sigma<1$:}
A pair  $\{p,q\}$ is a critical one if $\lambda\in[0,\frac{1}{1-\sigma}) $
in \eqref{23}. In this case
\begin{equation} \label{26}
q=q(p)\approx \begin{cases}
\frac{p^2}{4}, & 0<p\ll 1\\
\frac{1}{(1-\sigma)e}p+\frac{3-2\sigma}{(1-\sigma)^2}e^{\frac{\sigma}{1-\sigma}}
+o(1),& p\to +\infty.
\end{cases}
\end{equation}
The last expression is the right asymptote for $q=q(p)$.

\noindent\textbf{(2) $\sigma=1$:}
A pair  $\{p,q\}$ is a critical one if $\lambda\in[0,\infty)$ in \eqref{23}.
In this case \eqref{26}$_1$ holds and also
$q(p)\approx p\ln p$ as $p\to\infty$ (i.e. $\lambda\to\infty$).

\noindent\textbf{(3) $1<\sigma<2$:}
A pair  $\{p,q\}$ is a critical one if $\lambda\in
(-\infty,-\frac{1}{\sigma-1})\bigcup [0,\infty)$ in \eqref{23}. In this case
\eqref{26}$_2$ holds  for $p\to -\infty$ (i.e. $\lambda\to -
\frac{1}{\sigma-1}-0)$.
This is the left asymptote, \eqref{26}$_1$ holds and, in addition,
\begin{equation} \label{27}
q=q(p)\approx \begin{cases}
|p|^{\sigma} |\ln|p||^{2-\sigma}, & p\to 0-0\\
p^{\sigma}(\ln{p})^{2-\sigma}, & p\to +\infty \;(\mbox{i.e. } \lambda \to +\infty).
\end{cases}
\end{equation}

\noindent\textbf{(4) $\sigma=2$:}
A pair  $\{p,q\}$ is a critical one if $\lambda\in(\infty,\infty)$ and
$q(p)=\frac{p^2}{4}$.

\noindent\textbf{(5) $\sigma>2$:}
A pair  $\{p,q\}$ is a critical one if $\lambda\in (-\frac{1}{\sigma-1},\infty)$
in \eqref{23}. In this case $q=q(p)$ has the same left asymptote as in the
case (3), and
$$
q(p)\approx \begin{cases}
\frac{p^2}{4},& p\to 0 \;(\mbox{i.e.}\lambda\to 0)\\
p^{\sigma}(\ln{p})^{2-\sigma},& p\to +\infty.
\end{cases}
$$

\section{The Sturm-like comparison theorems}
Consider the inequality
\begin{equation} \label{28}
(lx)(t):= x''(t)-[P(t)x(r(t))]'+Q(t)x(l(t))\leq 0,\quad t\in(a,b),
\end{equation}
where $r(t), l(t)$ are monotone increasing continuous differentiable functions,
$$
\lim_{t\to\infty} r(t)=\lim_{t\to\infty} l(t)=\infty, \quad r(t)\leq t.
$$
(We do not assume that  $l(t)\leq t$).
Denote
$$
e_{\rm ext}:=\{t\bar{\in}(a,b):l(t)\in (a,b)\},\quad
e_{\rm int}:=\{t\in(a,b):l(t)\bar{\in} (a,b)\}.
$$
In particular, if $l(t)\leq t$, then $e_{\rm ext}=(b,k(b))$,
$e_{\rm int}=(a,k(a))$.
If $l(t)\geq t$, then $e_{\rm ext}=(k(a),a)$, $e_{\rm int}=(k(b),b)$.
Here $k(t)$ and $q(t)$ are the inverse functions
for $l(t)$ and $r(t)$, respectively.
Consider also the inequality
\begin{equation}
\label{29}
(\tilde{l}y)(t):=y''(t)+q'(t)P[q(t)]y'(q(t))+k'(t)\tilde{Q}[k(t)]y[k(t)]\geq 0,
\quad t\in (a,b).
\end{equation}
For arbitrary continuous on $(a,q(b))\bigcup e_{\rm ext}$ and possesing all
needed derivatives function $y(t)$, we have the following equalities:
\begin{gather*}
\int_a^b x''(t)y(t)dt=x'(b)y(b)-x'(a)y(a)-x(b)y'(b)+x(a)y'(b)
+\int_a^b y''(t)x(t)dt, \\
\begin{aligned}
\int_a^b Q(t)x(l(t))y(t)dt
&=\int_{(a,b)\setminus e_{\rm int}}[Q(t)-\tilde{Q}(t)]x(l(t))y(t)dt
+\int_{e_{\rm int}} Q(t)x(l(t))y(t)dt\\
&\quad -\int_{e_{\rm ext}}\tilde{Q}(t)x(l(t))y(t)dt+\int_a^b k'(t)
\tilde{Q}(k(t))y(k(t))x(t)dt,
\end{aligned}\\
\begin{aligned}
-\int_a^b \left[P(t)x(r(t))\right]'y(t)dt
&=-P(b)y(b)x(r(b))+P(a)y(a)x(r(a))\\
&\quad +\int_a^{q(a)}P(t)x(r(t))y'(t)dt
-\int_b^{q(b)} P(t)x(r(t))y'(t)dt\\
&\quad +\int_a^b q'(t)P(q(t))y'(q(t))x(t)dt.
\end{aligned}
\end{gather*}
These equalities imply the  Main  Identity-1:
\begin{equation} \label{30}
\begin{aligned}
\int_a^b (lx)(t)y(t)dt
&=\int_a^b (\tilde{l}y)(t)x(t)dt+
\left[x'(b)-P(b)x(r(b))\right]y(b)\\
&\quad -\left[x'(a)-P(a)x(r(a))\right]y(a)+x(a)y'(a)-x(b)y'(b)\\
&\quad +\int_a^{q(a)} P(t)x(r(t))y'(t)dt-\int_b^{q(b)} P(t)x(r(t))y'(t)dt\\
&\quad +\int_{(a,b)\setminus e_{\rm int}}[Q(t)-\tilde{Q}(t)]x(l(t))y(t)dt
+\int_{e_{\rm int}} Q(t)x(l(t))y(t)dt\\
&\quad -\int_a^b \tilde{Q}(t)x(l(t))y(t)dt.
\end{aligned}
\end{equation}
This Identity is a basis of the first Sturm-like Comparison Theorem for
Damped equation.


\begin{theorem} \label{thm1}
Let be $b> \max\{q(a);k(a)\}$ and assume that:
\begin{enumerate}
\item $P(t)\geq 0$, $t\in (a,q(a))\bigcup (b,q(b))$.

\item $Q(t)\geq 0$, $t\in e_{\rm int}$.

\item $\tilde{Q}(t)\geq 0$, $t\in e_{\rm ext}$.

\item
\begin{equation} \label{31}
 Q(t)\geq\tilde{Q}(t), t\in {(a,b)\setminus e_{\rm int}}.
\end{equation}

\item Inequality \eqref{29} has a solution $y(t)$ such that
\begin{gather} \label{32}
y(a)=y(b)=0;\quad  y(t)>0,\; t\in (a,b);\quad  y(t)\leq 0,\; t\in e_{\rm ext};\\
\label{33}
y'(t)\geq 0, t\in (a,q(a)); y'(t)\leq 0, t\in (b,q(b)).
\end{gather}

\item At least one of the inequalities \eqref{31}-\eqref{33} is
strong on some subinterval.
\end{enumerate}
Then  there is no  positive solution of \eqref{28}  on
$(r(a),b)\bigcup e_{\rm ext}$.
\end{theorem}

\begin{proof}
Suppose \eqref{28} has  a positive solution  on
$(r(a),b)\bigcup e_{\rm ext}$. Then the left hand-side of \eqref{30}
is zero but all terms of the right hand-side are nonnegative and at least
one is positive. We have a contradiction.
\end{proof}

\noindent{\bf Remark.}
For the case $P(t)\equiv 0$ (i.e. without the damping term) this Theorem was
published in \cite[Theorem \ref{thm4}.4, p.112]{d1}. \smallskip

Now, we state one more version of Sturm-like Comparison Theorem
for \eqref{28}, which is based on another Main Identity-2
which differs from \eqref{30}.
We will employ the preceding notation and the following:
\begin{equation} \label{34}
\begin{gathered}
\bar{e}_{\rm ext}:=\{ t\bar{\in}(a,b):l[q(t)]\in (a,b)\},\\
\bar{e}_{\rm int}:=\{ t\in(a,b):l[q(t)]\bar{\in} (a,b)\},\\
 lq(t):=l[q(t)].
\end{gathered}\end{equation}
The last notion will be applied to some other double superpositions.

In particular case $l(t)\leq r(t)\leq t$ we have
\begin{equation} \label{39}
\bar{e}_{\rm int}=(a,rk(a)),\quad  \bar{e}_{\rm ext}=(b,rk(b)).
\end{equation}
If $l(t)\geq r(t)$, then
\begin{equation}
\label{40}
\bar{e}_{\rm int}=(rk(b),b),\quad  \bar{e}_{\rm ext}=(rk(a),a).
\end{equation}
An important  particular case for us is
$r(t):=t-1$, $l(t):=t-\sigma$, $q(t)=t+1$, $k(t)=t+\sigma$,
$-\infty<\sigma<\infty$
and
\begin{gather*}
\bar{e}_{\rm int}=(a, a+\sigma-1), \quad \bar{e}_{\rm ext}=(b, b+\sigma-1)\quad
\mbox{for } \sigma\geq 1, \\
\bar{e}_{\rm int}=(b+\sigma-1,b),\quad  \bar{e}_{\rm ext}=(a+\sigma-1,a)\quad
\mbox{for }-\infty<\sigma\leq 1.
\end{gather*}
We introduce an operator $\tilde{\tilde{l}}y$ and the corresponding inequality
\begin{equation}
\label{35}
(\tilde{\tilde{l}}y)(t):= [r'(t)y'(r(t))]'+P(q(t))y'(t)+k'(t)\tilde{Q}(k(t))y(rk(t))
\geq 0,\quad t\in(a,b).
\end{equation}
Let $y(t)$ be an arbitrary smooth function, such that $y(a)=y(b)=0$.
Then
\begin{equation} \label{36}
\begin{aligned}
&\int_{q(a)}^{q(b)} x''(t)y(r(t))dt\\
& = \int_a^b \left[r'(t)y'(r(t))\right]'x(t)dt+
y'(a)\frac{x(q(a))}{q'(a)}-y'(b)\frac{x(q(b))}{q'(b)}\\
& -\int_{r(a)}^a q'(t)\left[\frac{y'(t)}{q'(t)}\right]' x(q(t))dt+
\int_{r(b)}^b q'(t)\left[\frac{y'(t)}{q'(t)}\right]' x(q(t))dt.
\end{aligned}
\end{equation}
\begin{equation} \label{37}
-\int_{q(a)}^{q(b)}[P(t)x(r(t))]'y(r(t))dt=
 \int_a^b P(q(t))y'(t)x(t)dt.
\end{equation}
\begin{equation} \label{38}
\begin{aligned}
&\int_{q(a)}^{q(b)} Q(t)x(l(t))y(r(t))dt\\
&=\int_a^b k'(t)\tilde{Q}(k(t))y(rk(t))x(t)dt\\
&\quad +\int_{(a,b)\setminus\bar{e}_{\rm int}}q'(t)\left\{Q(q(t))
-\tilde{Q}(q(t))\right\}x(lq(t))y(t)dt\\
&\quad +\int_{\bar{e}_{\rm int}} q'(t) Q(q(t))x(lq(t))y(t)dt
-\int_{\bar{e}_{\rm ext}} q'(t) \tilde{Q}(q(t))x(lq(t))y(t)dt.
\end{aligned}
\end{equation}
Equalities \eqref{36}-\eqref{38}
imply  Main Identity-2:
\begin{equation}\label{41}
\begin{aligned}
&\int_{q(a)}^{q(b)}(lx)(t)y(r(t))dt\\
&= \int_a^b (\tilde{\tilde{l}}y)(t)x(t)dt+y'(a)\frac{x(q(a))}{q'(a)}\\
&\quad -y'(b)\frac{x(q(b))}{q'(b)}-\int_{r(a)}^a q'(t)\left[\frac{y'(t)}{q'(t)}\right]' x(q(t))dt+
\int_{r(b)}^b q'(t)\left[\frac{y'(t)}{q'(t)}\right]' x(q(t))dt\\
&\quad +\int_{(a,b)\setminus\bar{e}_{\rm int}}q'(t)\left\{Q(q(t))
-\tilde{Q}(q(t))\right\}x(lq(t))y(t)dt\\
&\quad+\int_{\bar{e}_{\rm int}} q'(t) Q(q(t))x(lq(t))y(t)dt
-\int_{\bar{e}_{\rm ext}} q'(t) \tilde{Q}(q(t))x(lq(t))y(t)dt.
\end{aligned}
\end{equation}
Now we can formulate the  second version of a Sturm-like Comparison Theorem for
\eqref{28}

\begin{theorem} \label{thm2}
Let $b>\max\{q(a),k(a)\}$ and the following conditions hold:
\begin{enumerate}
\item \begin{equation} \label{42}
Q(q(t))\geq 0, t\in \bar{e}_{\rm int}, \quad  \tilde{Q}(q(t))\geq 0, \quad
t\in \bar{e}_{\rm ext}.
\end{equation}
\item
\begin{equation} \label{43}
Q(q(t))\geq \tilde{Q}(q(t)), t\in (a,b)\setminus\bar{e}_{\rm int}.
\end{equation}
\item  Inequality \eqref{35} has a solution $y(t)$ such that
\begin{gather} \label{44}
y(a)=y(b)=0;\quad  y(t)>0,\; t\in(a,b);\quad   y(t)\leq 0,\; t\in \bar{e}_{\rm ext},
\\
\label{45}
\big[\frac{y'(t)}{q'(t)}\big]'\leq 0, t\in(r(a),a);\quad
\big[\frac{y'(t)}{q'(t)}\big]'\geq 0, \quad t\in(r(b),b).
\end{gather}

\item At least one of the inequalities \eqref{42}-\eqref{45} is strong on
one of the subintervals.
\end{enumerate}
Then \eqref{28} has no positive solution $x(t)$ on
$ (r(a),b)\bigcup \bar{e}_{\rm ext}$.
\end{theorem}

\noindent{\bf Remark.}\;
This Theorem, in contrast to  Theorem \ref{thm1} ,  
does not impose any restriction on the
sign of $P(t)$.


\section{Testing Inequalities}

In both Sturm-like Comparison Theorems there exists  a requirement that an
``associated inequality" has a solution satisfying several conditions.
In this chapter we will  construct the such inequalities
(\eqref{29} and \eqref{35} in an explicit form.
Later we will employ these inequalities as testing
inequalities in applications of Sturm-like Comparison Theorems.

 We will look for the solution of \eqref{29} in the form
\begin{equation} \label{46}
y(t)=\exp\big\{\int_a^t z(s)ds\big\}\sin\int_a^t \varphi(s)ds,\quad t\in(a,b).
\end{equation}
Here
\begin{equation} \label{47}
\varphi(t)>0, \quad \int_a^b \varphi(s)ds=\pi, \quad
\int_a^{p(a)} \varphi(s)ds<\frac{\pi}{2}, \quad
\int_b^{p(b)} \varphi(s)ds<\frac{\pi}{2},
\end{equation}
where $p(t)=\max\{q(t),k(t)\}$.
Then
\begin{equation} \label{48}
\begin{aligned}
y'(t)&=\exp\big\{\int_a^t z(s)ds\big\}\big\{z(t)\sin\int_a^t \varphi(s)ds
+\varphi(t)\cos\int_a^t \varphi(s)ds\big\},\\
 y'(q(t))&=\exp\big\{\int_a^t z(s)ds\big\}
 \exp\big\{\int_t^{q(t)} z(s)ds\big\}\\
&\quad \times\big\{\big[z(q(t))\cos\int_t^{q(t)} \varphi(s)ds-\varphi(q(t))
\sin\int_t^{q(t)} \varphi(s)ds\big]\sin\int_a^t\varphi(s)ds \\
&\quad +\big[z(q(t))\sin\int_t^{q(t)} \varphi(s)ds+\varphi(q(t))
\cos\int_t^{q(t)} \varphi(s)ds\big]\cos\int_a^t\varphi(s)ds\big\},\\
y''(t)&= \exp\big\{\int_a^t z(s)ds\big\}\big\{A(t)\sin\int_a^t\varphi(s)ds
+B(t)\cos\int_a^t\varphi(s)ds\big\},
\end{aligned}
\end{equation}
where
\[
A(t):=z^2(t)+z'(t)-\varphi^2(t),\quad B(t):=\varphi'(t)+2\varphi'(t)z(t),
\]
\begin{align*}
y(k(t))&=\exp\big\{\int_a^t z(s)ds\big\}\exp\big\{\int_t^{k(t)} z(s)ds\big\}\\
&\quad \times\big\{\cos\int_t^{k(t)}\varphi(s)ds\sin\int_a^t\varphi(s)ds
+\sin\int_t^{k(t)}\varphi(s)ds\cos\int_a^t\varphi(s)ds\big\}.
\end{align*}
Hence on the interval $(a,b)$ we have
$$
(\tilde{l}y)(t)\geq 0\Longleftrightarrow S(t)\sin\int_a^t\varphi(s)ds
+R(t)\cos\int_a^t\varphi(s)ds\geq 0
$$
if and only if
\begin{equation} \label{49}
\begin{aligned}
S(t)&:=A(t)+q'(t)P(q(t))\exp\big\{\int_t^{q(t)} z(s)ds\big\}\\
&\quad \times\big[z(q(t))\cos\int_t^{q(t)} \varphi(s)ds-\varphi(q(t))
\sin\int_t^{q(t)} \varphi(s)ds\big]\\
&\quad +k'(t)\tilde{Q}(k(t))\exp\big\{\int_t^{k(t)} z(s)ds\big\}
\cos\int_t^{k(t)} \varphi(s)ds\geq 0,
\\
R(t)&:= B(t)+q'(t)P(q(t))\exp\left\{\int_t^{q(t)} z(s)ds\right\}\\
&\quad \times\big[z(q(t))\sin\int_t^{q(t)} \varphi(s)ds+\varphi(q(t))
\cos\int_t^{q(t)} \varphi(s)ds\big]\\
&\quad +k'(t)\tilde{Q}(k(t))\exp\big\{\int_t^{k(t)} z(s)ds\big\}
\sin\int_t^{k(t)} \varphi(s)ds=0.
\end{aligned}
\end{equation}
If $\sin\int_t^{k(t)} \varphi(s)ds$ is a positive or a negative function, then
\eqref{49} can be  transformed  into one of two
equivalent systems on $(a,b)$:
\begin{equation} \label{50}
\tilde{Q}(k(t))\geq M(k(t)),\quad R(t)=0
\end{equation}
or
\begin{equation} \label{51}
Z(t)\geq 0,\quad R(t)=0
\end{equation}
where
\begin{align*}
Z(t)&:=A(t)\sin\int_t^{k(t)} \varphi(s)ds-B(t)\cos\int_t^{k(t)} \varphi(s)ds\\
&\quad +q'(t)P(q(t))\exp\big\{\int_t^{q(t)} z(s)ds\big\}\\
&\quad\times \big[z(q(t))\sin\int_{q(t)}^{k(t)} \varphi(s)ds
-\varphi(q(t))\cos\int_{q(t)}^{k(t)} \varphi(s)ds\big],
\end{align*}
\begin{align*}
k'(t)M(k(t))&:= -\exp\big\{-\int_t^{k(t)} z(s)ds\big\}
\Big\{A(t)\cos\int_t^{k(t)} \varphi(s)ds\\
&\quad +B(t)\sin\int_t^{k(t)} \varphi(s)ds+
 +q'(t)P(q(t))\exp\big\{\int_t^{q(t)} z(s)ds\big\}\\
&\times\big[z(q(t))\cos\int_{q(t)}^{k(t)} \varphi(s)ds+\varphi(q(t))
\sin\int_{q(t)}^{k(t)} \varphi(s)ds\big]\Big\}.
\end{align*}
Thus if $\tilde{Q}(t)$ and $z(t)$ satisfy \eqref{49} (or
\eqref{50}, or \eqref{51}, then \eqref{29}
has a solution $y(t)$ of the form \eqref{46} for which \eqref{32}-\eqref{33} hold.

In case $l(t)\equiv t$, \eqref{49} changes its character because in this version
\eqref{49}$_2$ does not contain $\tilde{Q}(t)$ and turns into the equation
in $z(t)$ only:
\begin{equation} \label{52}
\begin{aligned}
R(t)&:=B(t)+q'(t)P[q(t)]\exp\int_t^{q(t)}z(s)ds \\
&\times\big[z(q(t))\sin \int_t^{q(t)}\varphi(s)ds+\varphi(q(t))
\cos\int_t^{q(t)}\varphi(s)ds\big]=0.
\end{aligned}
\end{equation}

For \eqref{35}, let us repeat all  calculations done for \eqref{29}.
We will look for  a solution of \eqref{35} of the same form \eqref{46},
where  $\varphi(t)$ and $z(t)$ satisfy the same conditions \eqref{47}.
After some  calculations we have the following  system  on $(a,b)$:
\begin{equation} \label{53}
\begin{aligned}
\tilde{S}(t)&:= r''(t)
\big[z(r(t))\cos\int_{r(t)}^t \varphi(s)ds +\varphi(r(t))\sin \int_{r(t)}^t
\varphi(s)ds\big]\\
&\quad +[r'(t)]^2\big[A(r(t))\cos\int_{r(t)}^t \varphi(s)ds+B(r(t))\sin
\int_{r(t)}^t\varphi(s)ds\big]\\
&\quad +P(q(t))z(t)\exp\big\{\int_{r(t)}^tz(s)ds\big\}\\
&\quad +k'(t)\tilde{Q}(k(t))\exp\big\{\int_{r(t)}^{rk(t)}z(s)ds\big\}
\cos \int_t^{rk(t)}\varphi(s)ds\geq 0,
\end{aligned}
\end{equation}
\begin{align*}
\tilde{R}(t)&:=r''(t)\big[-z(r(t))\sin\int_{r(t)}^t \varphi(s)ds
+\varphi(r(t))\cos \int_{r(t)}^t \varphi(s)ds\big]\\
&\quad +[r'(t)]^2\big[-A(r(t))\sin\int_{r(t)}^t \varphi(s)ds+B(r(t))\cos
 \int_{r(t)}^t\varphi(s)ds\big]\\
&\quad +P(q(t))\varphi(t)\exp\big\{\int_{r(t)}^tz(s)ds\big\}\\
&\quad +k'(t)\tilde{Q}(k(t))\exp\big\{\int_{r(t)}^{rk(t)}z(s)ds\big\}
\sin \int_t^{rk(t)}\varphi(s)ds= 0.
\end{align*}
Instead of \eqref{53} one of the following two equivalent systems can be
applied
\begin{equation} \label{54}
\tilde{Q}(t)\geq\tilde{N}(t), \quad \tilde{R}(t)=0,
\end{equation}
or
\begin{equation} \label{55}
\tilde{Z}(t)\geq 0,\quad \tilde{R}(t)=0,
\end{equation}
where
\begin{align*}
 &k'(t)\tilde{N}(k(t))\cos\int_t^{rk(t)} \varphi(s)ds\\
 &:=\exp\big\{-\int_{r(t)}^{rk(t)}z(s)ds\big\}
\Big\{-P(q(t))z(t)\exp\int_{r(t)}^tz(s)ds \\
&\quad -r''(t)\big[z(r(t))\cos\int_{r(t)}^t \varphi(s)ds +\varphi(r(t))
\sin \int_{r(t)}^t \varphi(s)ds\big]\\
&\quad -[r'(t)]^2 \big[A(r(t))\cos\int_{r(t)}^t \varphi(s)ds+B(r(t))\sin
\int_{r(t)}^t\varphi(s)ds\big]\Big\},
\end{align*}
\begin{align*}
\tilde{Z}(t)&:=r''(t)\big[ z(r(t))\sin \int_{r(t)}^{rk(t)}\varphi(s)ds
-\varphi(r(t))\cos\int_{r(t)}^{rk(t)}\varphi(s)ds\big]\\
&\quad +(r'(t))^2\big[A(r(t))\sin \int_{r(t)}^{rk(t)}\varphi(s)ds
-B(r(t))\cos\int_{r(t)}^{rk(t)}\varphi(s)ds\big]\\
&\quad + P(q(t))\big[z(t)\sin \int_t^{rk(t)}\varphi(s)ds
-\varphi(t)\cos\int_t^{rk(t)}\varphi(s)ds\big]\exp\int_{r(t)}^tz(s)ds.
\end{align*}

\section{Applications of Sturm-like Comparison Theorems
to Investigation of some Typical Damped Equations}

Consider the equation
\begin{equation} \label{56}
x''(t)-px'(t-1)+Q(t)x(t-\sigma)=0,\quad -\infty<\sigma<\infty,\; p>0.
\end{equation}
The equation
\begin{equation}
\label{57}
\frac{s(2+s\sigma)}{1+s(\sigma-1)}e^{s}=p
\end{equation}
has (a unique) root $s>0$. This root $s\in (0, \frac{1}{1-\sigma})$ for
$-\infty<\sigma<1$ and $s\in (0,\infty)$ for $\sigma\geq 1$.
The number $q:=\frac {s^2(s+1)}{1+s(\sigma-1)}e^{s\sigma}$ together
with $p$ forms a critical pair $\{p,q\}$ for the autonomous equation
\begin{equation} \label{58}
z''(t)-pz(t-1)+qz(t-\sigma)=0.
\end{equation}

\begin{theorem}  \label{thm3}
If
\begin{equation} \label{59}
\liminf_{t\to\infty}Q(t)= C>q,
\end{equation}
then all solutions of \eqref{56} are oscillatory and there exists $\nu >0$
such that every solution has at least one zero on any interval
$(T-1,T+\frac{\pi}{\nu})$ for sufficiently large $T$.
\end{theorem}

\noindent {\bf Remark.}\;
The constant $q$ in \eqref{59} is the best possible one. Indeed,
\eqref{56} has a nonoscillatory solution $z(t)=e^{st}$ for $Q(t)\equiv q$.

\begin{proof} The proof is based on Theorem \ref{thm2}.
Let in \eqref{49} $\varphi(t):=\nu$; $z(t)=$const, which will
be chosen later. Then \eqref{49} has the form
\begin{equation} \label{60}
\begin{gathered}
\tilde{Q}(t)e^{z(\sigma-1)}\cos\nu(\sigma-1)\geq
-e^{-z}\left[(z^2-\nu^2)\cos\nu+2\nu z\sin\nu\right]-pz\\
\tilde{Q}(t)e^{z(\sigma-1)}\sin\nu(\sigma-1)=
-e^{-z}\left[-(z^2-\nu^2)\sin\nu+2\nu z\cos\nu\right]-p\nu
\end{gathered}
\end{equation}
We will choose $z$ and $\tilde{Q}$ as a solution of a system, which is stronger
than \eqref{60}, with equality in \eqref{60}$_1$ instead of the inequality.
Then this new system  is equivalent to one of the  following two systems:
\begin{equation} \label{61}
\begin{gathered}
F(\nu,z):=pe^z-(z^2-\nu^2)\frac{\sin\nu}{\nu}+2z\cos\nu=0\\
\tilde{Q}=-pz-e^{-z}\left[(z^2-\nu^2)\cos\nu+2\nu z\sin\nu\right]
\end{gathered},
\end{equation}
for $\sigma=1$, and
\begin{equation} \label{62}
\begin{gathered}
\Phi(\nu,z):=(z^2-\nu^2)\frac{\sin\nu\sigma}{\nu}-2z\cos\nu\sigma+pe^z
\big[z\frac{\sin\nu(\sigma-1)}{\nu}-cos\nu(\sigma-1)\big]=0\\
\tilde{Q}=\frac{e^{-\sigma z}}{\cos\nu(\sigma-1)}\left\{
-(z^2-{\nu}^2)\cos\nu-2z\nu\sin\nu-pe^z z\right\}
\end{gathered}
\end{equation}
for $\sigma\neq 1$.

Evidently, a possible solution $\{z, \tilde{Q} \}$ of \eqref{61} or \eqref{62}
does not depend on $t$.
Consider first \eqref{61}:
\begin{gather*}
F_z'(\nu,z):=pe^z-2z\frac{\sin\nu}{\nu}+2\cos\nu, \\
F(0,-s)=pe^{-s} -s^2-2s=\frac{s(2+s)e^s}{1} e^{-s}-s^2-2s=0\\
F_z'(0,-s)=s(2+s)+2s+2=s^2+4s+2\neq 0.
\end{gather*}
Then for sufficiently small $\nu$ there exists the implicit function $z=z(\nu)$
such that $z(0)=-s$ and $F(\nu,z(\nu))=0$.
The number $z(\nu)$, which is constructed by this way, is taken  as the function
$z(t)$.
 From \eqref{61}$_2$ we obtain
\begin{gather}
\tilde{Q}=\tilde{Q}_{\nu} =-pz_{\nu}-e^{-z_{\nu}}\left[(z_{\nu}^2-\nu^2)\cos\nu+
2\nu z_{\nu}\sin\nu\right], \nonumber\\
\lim_{\nu\to 0}\tilde{Q}_{\nu}=s^2(2+s)e^s -e^s s^2= s^2(s+1)e^s=q.\label{63}
\end{gather}
Let us do the same calculations with \eqref{62}:
$$
\Phi(0,-s)=s^2\sigma+2s+\frac{s^2\sigma+2s}{1+s(\sigma-1)}e^se^{-s}
(-s(\sigma-1)-1)=0,
$$
\begin{align*}
&\Phi_z'(\nu,z)\\
&=2z\frac{\sin\nu\sigma}{\nu}-2\cos\nu\sigma+pe^z
\big[z\frac{\sin\nu(\sigma-1)}{\nu}-\cos\nu(\sigma-1)+\frac{\sin\nu(\sigma-1)}
{\nu}\big]
\end{align*}
\begin{align*}
\Phi_z'(0,-s)
&=-2s\sigma-2+pe^{-s}[-s(\sigma-1)-1+\sigma-1]\\
&=-2s\sigma-2+\frac{s^2\sigma+2s}{1+s(\sigma-1)}[-s(\sigma-1)+\sigma-2]\\
&=-\frac{\sigma^2(1+s)s^2-\sigma
s(s^2-2s-2)-2(s^2-s-1)}{1+s(\sigma-1)}\\
&:=-\frac{h(s,\sigma)}{1+s(\sigma-1)},
\end{align*}
We will prove that $h(s,\sigma)>0$ in two domains:

\noindent\textbf{(1)} Let $(\sigma,s)\in\{\sigma>1;s>0\}$. Then
$$
h(s,\sigma)=s^3(\sigma-1)\sigma+s^2(\sigma^2+2\sigma-2)+s(2\sigma+2)+2
>s^2+4s+2>0.
$$
\textbf{(2)} Let $(\sigma,s)\in\{-\infty<\sigma<1; 0<s<\frac{1}{1-\sigma}\}$.
Then $h(s,\sigma)>0$ on the boundary of this domain. Indeed,
$$
h\big(s,\frac{s-1}{s}\big)=s+1>0;h(s,1)=s^2+4s+2>0;h(0,\sigma)=2>0.
$$
The domain  $(\sigma,s)$ in case $2^o$ does not contains stationary points of
$h(s,\sigma)$ since there are no solutions of the system
\begin{gather*}
h'_s=0\\h'_{\sigma}=0
\end{gather*}
in this domain. Indeed,
$$
h'_{\sigma}=0\Leftrightarrow \sigma=\frac{s^2-2s-2}{2s(1+s)}\Leftrightarrow
\frac{1}{1-\sigma}-s=-\frac{s^2(s+2)}{s^2+4s+2}<0;
$$
i.e. the point $(\sigma_0,s_0)$ belongs to outside of the domain for every $s$.
Thus $\Phi_z'(0,-s)\neq 0$ for $\forall p>0, \sigma\neq 1$
and we can choose $z=z_{\nu}$ as a solution of \eqref{62}$_1$.
In addition, from \eqref{62}$_2$ we have
\begin{equation}
\label{64}
\lim_{\nu\to 0}\tilde{Q}_{\nu}=e^{\sigma s} \big\{
-s^2+s\frac{s(2+\sigma s)}{1+s(\sigma-1)}e^se^{-s}\big\}
=\frac{s^2(1+s)}{1+s(\sigma-1)}e^{\sigma s}=q.
\end{equation}
Conditions \eqref{64} and \eqref{59} imply that for sufficiently small $\nu$
we have $\tilde{Q}<C$. Then by \eqref{59} we have $Q\geq \tilde{Q}$
which is condition 2 in Theorem \ref{thm2}. One needs to check only \eqref{45}:
For $t\in(a-1,a)$ we have
\begin{gather*}
y''(t)\leq 0\Longleftrightarrow A(t)\sin\nu(t-a)+B(t)\cos\nu(t-a)\leq 0 \\
\Longleftrightarrow -A(t)\tan\nu(a-t)+B(t)\leq 0\Longleftrightarrow
-(z_{\nu}^2-\nu^2)\frac{\tan\nu(a-t)}{\nu}+2z_{\nu}\leq 0 \\
\Longleftarrow s^2\frac{\tan\nu(a-t)}{\nu}+2s>0
\Longleftarrow s>0.
\end{gather*}
 From the condition $\int_a^b\varphi(t) dt=\pi$ we have $b=a+\frac{\pi}{\nu} $.
Then for $t\in(b-1,b)$ we have
\begin{gather*}
y''(t)\geq 0\Longleftrightarrow A(t)\sin(\nu(t-b)+\pi)+
 B(t)\cos(\nu(t-b)+\pi)\geq 0 \\
\Longleftrightarrow
A(t)\sin\nu(t-b)+B(t)\cos\nu(t-b)\leq 0.
\end{gather*}
The end of the calculations is similar to the previous case.
All conditions of Theorem \ref{thm2} hold. Then Theorem \ref{thm3} is proven.
\end{proof}

We will strengthen Theorem \ref{thm3}  by formulating the following
result which will be called Knezer-like Theorem for \eqref{56}.

Let be $p>0$, $s>0$, $q>0$  denoted as in the beginning of section 6.
Denote also the constant
\begin{equation} \label{65}
K(p,\sigma):=\frac{\sigma^2s^2(1+s)+\sigma s(-s^2+2s+2)-2s^2+2s+2}
{8[1+s(\sigma-1)]}e^{s\sigma}.
\end{equation}

\begin{theorem} \label{thm4}
If
\begin{equation} \label{66}
\liminf_{t\to\infty}\left\{[Q(t)-q]t^2\right\}=C>K(p,\sigma),
\end{equation}
then all solutions of \eqref{56} are oscillatory
and there exists $\nu>0$ such that every solution has at least one zero on any
interval $(T-1,T\exp\frac{\pi}{\nu})$ for sufficiently large $T$.
\end{theorem}

\begin{proof} The proof is based on Theorem \ref{thm2}.
Suppose first  $\sigma\neq 1$ and set in \eqref{49},
$$
\varphi(t):=\frac{\nu}{t},\quad  z(t):= -s+\frac{1}{2t}+\frac{\beta}{t^2},
$$
where $\beta$ will be chosen later. Substitute
$k(t)=t+\sigma,~ q(t)=t+1,~ rk(t)=t+\sigma-1$ in \eqref{48}-\eqref{49}.
Write out several asymptotic, where we denote $f(t)\cong g(t)$
instead of $f(t)=g(t)+o(\frac{1}{t^2})$:
\begin{gather*}
\frac{t}{\nu}\varphi(t+1)\cong 1-\frac{1}{t}+\frac{1}{t^2};\quad
\cos\int_t^{t+\sigma}\varphi(s)ds\cong 1-\frac{\nu^2\sigma^2}{2t^2},\\
\cos\int_{t+1}^{t+\sigma}\varphi(s)ds\cong 1-\frac{\nu^2 (\sigma-1)^2}{2t^2};\quad
\frac{t}{\nu\sigma}\sin \int_t^{t+\sigma}\varphi(s)ds\cong 1-\frac{\sigma}{2t}+
\frac{\sigma^2(2-\nu^2)}{6t^2};\\
\frac{t}{\nu(\sigma-1)}\sin \int_{t+1}^{t+\sigma}\varphi(s)ds\cong 1-\frac{\sigma-1}{2t}
+\frac{(2-\nu^2)(\sigma-1)^2+3\sigma+3}{6t^2};\\
\frac{t}{\nu\sigma}\sin \int_{t-1}^{t+\sigma-1}\varphi(s)ds\cong
1+\frac{2-\sigma}{2t}+\frac{2\sigma^2-6\sigma+6-\sigma^2\nu^2}{6t^2};\\
\frac{t}{\nu}\sin \int_{t-1}^{t}\varphi(s)ds\cong 1+\frac{1}{2t}+
\frac{2-\nu^2}{6t^2};\\
\frac{\nu\sigma}{t}\csc \int_t^{t+\sigma}\varphi(s)ds\cong 1+\frac{\sigma}{2t}
+\frac{\sigma^2(2\nu^2-1)}{12t^2};\\
z(t+1)\cong-s+\frac{1}{2t}+\frac{2\beta-1}{2t^2};\quad
\int_t^{t+\sigma}z(s)ds\cong-s\sigma+\frac{\sigma}{2t}
+\frac{4\sigma\beta-\sigma^2}{4t^2},\\
\exp \int_t^{t+\sigma}z(s)ds\cong e^{-s\sigma}\left[1+\frac{\sigma}{2t}+
\frac{8\sigma\beta-\sigma^2}{8t^2}\right];\\
\exp\big\{- \int_t^{t+\sigma}z(s)ds\big\}\cong e^{s\sigma}
\left[1-\frac{\sigma}{2t}+
\frac{3\sigma^2-8\sigma\beta}{8t^2}\right];\\
A(t):=z^2(t)+z'(t)-\varphi^2(t)\cong s^2-\frac{s}{t}
-\frac{1+4s+4\nu^2+8\beta s}{4t^2};\\
\frac{t}{\nu}B(t)\cong-2s-\frac{2s}{t}+\frac{2\beta-2s}{t^2}.
\end{gather*}
Using these asymptotic results, rewrite \eqref{49}$_1$ as follows
\begin{align*}
&\tilde{Q}(t+\sigma)\geq \tilde{N}(t+\sigma)\\
&:=\frac{\exp\big\{-\int_t^{t+\sigma-1} z(s)ds
\big\}}{\cos\int_t^{t+\sigma-1}\varphi(s)ds}
\Big\{-pz(t)-\exp\big\{-\int_{t-1}^tz(s)ds \big\}\\
&\quad \times\big[A(t-1)\cos\int_{t-1}^t\varphi(s)ds+B(t-1)
\sin\int_{t-1}^t\varphi(s)ds\big]\Big\} =\dots\\
&=e^{s\sigma}\Big(1-\frac{\sigma-1}{2t}+\frac{3(\sigma-1)^2-8\beta(\sigma-1)
+4(\sigma-1)^2\nu^2}{8t^2}\Big)
\left\{D_0+D_1\frac{1}{t}+D_2\frac{1}{t^2}\right\},
\end{align*}
where
\begin{gather*}
D_0:=\frac{2s^2+\sigma s^3}{1+s(\sigma-1)}-s^2=\frac{s^2(1+s)}{1+s(\sigma-1)}
=qe^{-s\sigma},\\
D_1= -\frac{2s+\sigma s^2}{2(1+s(\sigma-1))}+\frac{s^2}{2}+s=
\frac{\sigma-1}{2}qe^{-s\sigma}, \\
\begin{aligned}
D_2&=-\beta\frac{2s+\sigma s^2}{1+s(\sigma-1)}+\frac{s^2+8\beta s^2}{8}-\frac{s}{2}
+\frac{1}{4}-\frac{2s-4\beta s+\nu^2(s^2+4s+2)}{2}\\
&=\beta\frac{(\sigma-1)s^2(1+s)}{1+s(\sigma-1)}
+\frac{s^2+4s+2}{8}+\nu^2\frac{s^2+4s+2}{2}.
\end{aligned}
\end{gather*}
After  these calculations we have
\begin{equation} \label{67}
\tilde{Q}(t+\sigma)\geq \bar{N}(t+\sigma)\cong q+\frac{K(p,\sigma)
+\nu^2L(\sigma,s)}{t^2},
\end{equation}
in which the   explicit form of $L(\sigma,s)$ is unessential. Note also that \eqref{67}
 does not contain parameter $\beta$.
Equality $\eqref{49}_2)$ turns into
\begin{equation}\label{68}
\begin{aligned}
\tilde{Q}(t+\sigma)
&= \frac{\exp\big\{-\int_t^{t+\sigma-1}z(s)ds\big\}}
{\sin\int_t^{t+\sigma-1}\varphi(s)ds}\Big\{-p\frac{\nu}{t}+
\exp\big\{-\int_{t-1}^{t}z(s)ds\big\}\\
&\quad \times \big[A(t-1)\sin\int_{t-1}^{t}\varphi(s)ds-
-B(t-1)\cos\int_{t-1}^{t}\varphi(s)ds\big]\Big\}\\
&\cong q+\frac{1}{t^2}\big\{L_1(s,\sigma)\nu^2-\beta R(s,\sigma)
+\tilde{K}(p,\sigma)\big\},
\end{aligned}
\end{equation}
where $R(s,\sigma):=\frac{1}{\sigma-1} [q(\sigma-1)^2+4e^{s\sigma}]$.
The explicit form of $L_1$
and $\tilde{K}$ is unessential.

Evidently  $R(s,\sigma)\neq 0, ~\forall s>0,~\sigma\neq 1$, hence
there exists $\beta$ such that
\begin{equation}
\label{69}
K(p,\sigma)<L_1-\beta R+\tilde{K}=C_1<C
\end{equation}
and choose $\nu>0$ small, such that
\begin{equation}
\label{70}
K(p,\sigma)+\nu^2L(s,\sigma)<C_1<C,
\end{equation}
which is possible due to \eqref{66}.
This inequality  implies (2) in  Theorem \ref{thm2}  for
sufficiently large $t$.
We need now to check \eqref{45} only.
For $\eqref{45}_1)$  we have for $t\in(a-1,a)$:
\begin{gather*}
y''(t)\leq 0\Leftrightarrow s^2\sin\big(\nu\ln\frac{t}{a}\big)
-\frac{2\nu s}{t}\cos\big(\nu\ln\frac{t}{a}\big)< 0\Leftrightarrow \\
\Leftrightarrow -s^2\tan\big(\nu\ln\frac{a}{t}\big)
-\frac{2\nu s}{t}< 0\Leftrightarrow g(t):=s^2\tan\big(\nu\ln\frac{a}{t}\big)
+\frac{2\nu s}{t}>0.
\end{gather*}
Since
$$
g'(t)=-s^2\frac{\nu}{t\cos^2(\nu\ln\frac{a}{t})}-\frac{2\nu s}{t}<0,
$$
it follows that $\inf_{t\in(a-1,a)}g(t)=g(a)=\frac{2\nu s}{a}>0$, for all $s>0$.

For \eqref{45}$_2$ due to $\int_a^b\varphi(s)ds=\pi$ we have for $t\in(b-1,b)$:
\begin{gather*}
y''(t)>0\Leftrightarrow s^2\sin\big(\pi+\nu\ln\frac{t}{a}\big)
-\frac{2\nu s}{t}\cos\big(\pi+\nu\ln\frac{t}{a}\big)>0\Leftrightarrow \\
s^2\sin\big(\nu\ln\frac{t}{a}\big)
-\frac{2\nu s}{t}\cos\big(\nu\ln\frac{t}{a}\big)<0,
\end{gather*}
etc. (see the above calculations).
For the case $\sigma\neq 1$ Theorem \ref{thm4}  is proven.

Now suppose   $\sigma=1$. In this case we have
$$
k(t)=q(t)=t+1,\quad  rk(t)=t, \quad q=s^2(s+1)e^{s\sigma},
$$
where $s\geq 0$ is the unique root of the equation $s(2+s)e^s=p$.
System \eqref{49} has now the  form
\begin{equation} \label{71}
\begin{gathered}
\begin{aligned}
\tilde{Q}(t+2)\geq M(t)
&:=-pz(t+1)-\exp\big\{-\int_t^{t+1}z(s)ds\big\} \\
&\quad \times\big[A(t)\cos\int_t^{t+1}\varphi(s)ds+B(t)\sin\int_t^{t+1}\varphi(s)ds\big]
\end{aligned}\\
\begin{aligned}
\tilde{R}(t)&:=-A(t)\sin\int_t^{t+1}\varphi(s)ds+B(t)\cos\int_t^{t+1}\varphi(s)ds\\
&\quad +p\varphi(t+1)\exp\big\{\int_t^{t+1}z(s)ds\big\}=0.
\end{aligned}
\end{gathered}
\end{equation}
Rewrite \eqref{71}$_2$ after the substitution $\varphi(t):=\frac{\nu}{t}$ and
expressions $A(t)$ and $B(t)$ in the form
\begin{equation}\label{72}
z'(t)+z^2(t)-\frac{\nu^2}{t^2}-\frac{p\nu}{t\sin[\nu\ln(1+\frac{1}{t})]}
\exp\big\{\int_t^{t+1}z(s)ds\big\}-\frac{\nu^2[2z(t)-\frac{1}{t}]}
{t^2\tan[\nu\ln(1+\frac{1}{t})]}=0.
\end{equation}
This equality is an equation with respect to $z(t)$ only in $(T,\infty)$ for
sufficiently  large $T$.

All coefficients in \eqref{72} are analytic functions in a neighborhood
of the infinity. Therefore there exists a  solution $z(t)$ of \eqref{72} which
has the following expansion in some neighborhood of the infinity :
$$
z(t)=\sum_{n=0}^{\infty}\frac{a_n}{t^n},\quad  t\in(T,\infty).
$$
We will look for a few  terms of this expansion:
$$
z(t)=a_0+\frac{a_1}{t}+\frac{a_2}{t^2}+o(t^{-2}).
$$

We have
\begin{align*}
A(t)&:=z^2(t)+z'(t)-\varphi^2(t)=\big[a_0+\frac{a_1}{t}+\frac{a_2}{t^2}\big]^2
-\frac{a_1}{t^2}-\frac{\nu^2}{t^2}+o(t^{-2})\\
&= a_0^2+\frac{a_0a_1}{t}+[a_1^2+2a_0a_2-a_1-\nu^2]\frac{1}{t^2}+o(t^{-2}),\\
B(t)&:=\varphi'(t)+2\varphi(t)z(t)=-\frac{\nu}{t^2}+\frac{2\nu}{t}
\big[a_0+\frac{a_1}{t}+\frac{a_2}{t^2}+o(t^{-2})\big]\\
&=\frac{\nu}{t}\big[2a_0+\frac{2a_1-1}{t}+\frac{2a_2}{t^2}+o(t^{-2})\big].
\end{align*}
Then
\begin{align*}
\int_t^{t+1} z(s)ds&=\int_t^{t+1}\big[a_0+\frac{a_1}{s}+\frac{a_2}{s^2}+o(s^{-2})\big]ds\\
&=a_0+a_1\ln(1+\frac{1}{t})+\frac{a_2}{t(t+1)}+o(t^{-2})\\
&=a_0+a_1\big(\frac{1}{t}-\frac{1}{2t^2}\big)+\frac{a_2}{t^2}+o(t^{-2})\\
&=a_0+\frac{a_1}{t}-\frac{a_1-2a_2}{2t^2}+o(t^{-2}),
\end{align*}
\begin{align*}
\exp\big\{\int_t^{t+1}z(s)ds\big\}
&=e^{a_0}\Big\{1+\big(\frac{a_1}{t}-\frac{a_1-2a_2}{2t^2}\big)
+\frac{1}{2!}\big(\frac{a_1}{t}-\frac{a_1-2a_2}{2t^2}\big)^2
+o(t^{-2})\Big\}\\
&=e^{a_0}\big\{1+\frac{a_1}{t}+\frac{a_1^2-a_1+2a_2}{2t^2}+o(t^{-2})\big\}.
\end{align*}
Substituting these expressions and the previous asymptotic in
$\eqref{72}$ and taking into account that
$p=s(2+s)e^s$,  $q=s^2(1+s)e^s$, we obtain
\begin{equation} \label{73}
\begin{aligned}
-\big[a_0^2+\frac{2a_0a_1}{t}+(a_1^2+2a_0a_1-a_1-\nu^2)\frac{1}{t^2}+o(t^{-2}\big]
\frac{\nu}{t}\big[1-\frac{1}{2t}+\frac{2-\nu^2}{6t^2}+o(t^{-2}\big]\\
+\frac{\nu}{t}\big[2a_0+\frac{2a_1-1}{t}+\frac{2a_2}{t^2}+o(t^{-2}\big]
\big[1-\frac{\nu^2}{2t^2}+o(t^{-2}\big]\\
+s(2+s)e^s\frac{\nu}{t}\big[1-\frac{1}{t}+\frac{1}{t^2}+o(t^{-2}\big]e^{a_0}
\big[1+\frac{a_1}{t}+\frac{a_1^2-a_1+2a_2}{2t^2}+o(t^{-2}\big]=0.
\end{aligned}
\end{equation}
Now we equate to zero all three coefficients of the left-hand side of the last
equation.
The first one has the form
\begin{equation}\label{74}
-a_0^2+2a_0+s(2+s)e^{s+a_0}=0,
\end{equation}
so $a_0=-s$ is the unique root of this equation.
Equating to zero the coefficient of $\frac{1}{t}$ we have:
$$
\frac{s^2}{2}+2sa_1+2a_1-1+s(2+s)(a_1-1)=0,
$$
then
\begin{equation} \label{75}
a_1=\frac{1}{2}.
\end{equation}
We will see further that the coefficient $a_2$ is not applied,
so we will not look for this coefficient.

Thus, substituting $z(t)=-s+\frac{1}{2t}+\frac{a_2}{t^2}+o(t^{-2})$
into $\eqref{50}_1)$ we have
\begin{align*}
M(t)&\cong -2s(2+s)e^s\big[-s+\frac{1}{2t}+\frac{2a_2-1}{2t^2}\big]\\
&\quad -e^s\big[1-\frac{1}{2t}-\frac{3-8a_2}{8t^2}\big]
\Big\{\big[s^2-\frac{s}{t}-\frac{1+8sa_2+4\nu^2}{4t^2}\big]
\big(1-\frac{\nu^2}{2t^2}\big)\\
&\quad +\frac{\nu^2}{t^2}\big[-2s+\frac{a_2}{t^2}\big]
\big[1-\frac{1}{2t}+\frac{2-\nu^2}{6t^2}\big]\Big\}\\
&\cong s^2(1+s)e^s+\frac{1}{t^2}\big[\frac{e^s}{8}(s^2+4s+2)+\nu^2L_2(s)\big]\\
&\cong q+\frac{1}{t^2}\big[K(p,1)+\nu^2L_2(s)+o(t^{-2})\big].
\end{align*}
Note that the latter expression does not contain the coefficient $a_2$.
We can choose $\nu$ so small that
the inequality $K(p,1)+\nu L_2(s)<C$ holds together with \eqref{60}, which  was
necessary to prove.
For the case $\sigma=1$ Theorem \ref{thm4}  is proven too.
\end{proof}

\noindent {\bf Remark.}\; Condition \eqref{66} is  unimprovable.
Indeed, the function $x_0(t):=\sqrt{t}e^{st}$ is a nonoscillatory solution of
\eqref{56}, where
\begin{equation} \label{76}
Q(t+\sigma):=Q_0(t+\sigma)=\frac{px_0'(t+\sigma-1)-x_0^{''}(t+\sigma)}{x_0(t)}.
\end{equation}
We have $x_0'(t)=\sqrt{t}e^{st}(s+\frac{1}{2t})$,
\begin{align*}
&x_0'(t+\sigma-1)\\
&=\sqrt{t}e^{st}e^{s(\sigma-1)}\Big\{\frac{1}{2t}\big[1-\frac{\sigma-1}{2t}
+o(t^{-1})\big]+
s\big[1-\frac{\sigma-1}{2t}-\frac{(\sigma-1)^2}{8t^2}+o(t^{-2}\big]\Big\},
\end{align*}
$$
x_0''(t)=\sqrt{t}e^{st}\big[s^2+\frac{s}{t}-\frac{1}{4t^2}+o(t^{-2})\big],
$$
\begin{align*}
&x_0''(t+\sigma)\\
&=\sqrt{t}e^{st}e^{s\sigma}\big\{-\frac{1}{4t^2}[1+o(1)]
+\frac{s}{t}\big[1-\frac{\sigma}{2t}+o(t^{-1})\big]
+s^2\big[1+\frac{\sigma}{2t}-\frac{\sigma^2}{8t^2}+o(t^{-2})\big]\big\}.
\end{align*}
Substituting these expressions into \eqref{76} we obtain (after calculations)
$$
Q_0(t)=q+\frac{K(p,\sigma)}{t^2}+o(t^{-2}).
$$
Thus it is impossible to improve
the constant $K(p,\sigma)$ in \eqref{66} for any $s>0$ and any $\sigma$.
\smallskip

Consider an application of Theorem \ref{thm1} and Theorem \ref{thm2} to the
following typical equation
\begin{equation} \label{77}
lx:=x''(t)-\big[\frac{p}{t-1}x(t-1)\big]'+Q(t)x(t-\sigma)=0,\quad \sigma\neq 0.
\end{equation}
Here $P(t):=\frac{p}{t-1}$, $l(t)=t-\sigma$, $r(t)=t-1$, $k(t)=t+\sigma$, and
$q(t)=t+1$.

\noindent{\bf Case $p>0$:} In this case we will apply Theorem \ref{thm1}.
Rewrite \eqref{49} as follows:
\begin{equation} \label{78}
\begin{aligned}
\tilde{Q}(t+\sigma)
&=-\frac{\exp\big\{-\int_t^{t+\sigma}z(s)ds\big\}}
{\sin \int_t^{t+\sigma}\varphi(s)ds}
\Big\{B(t)+\frac{p}{t} \exp\int_t^{t+1}z(s)ds\\
&\quad \times\Big[z(t+1)\sin \int_t^{t+1}\varphi(s)ds
+\varphi(t+1)\cos\int_t^{t+1}\varphi(s)ds\Big]\Big\},\\
\tilde{Q}(t+\sigma)&\geq M(t+\sigma)\\
&:=-\exp\big\{-\int_t^{t+\sigma}z(s)ds\big\}
\Big\{A(t)\cos\int_t^{t+\sigma}\varphi(s)ds\\
&\quad +B(t)\sin\int_t^{t+\sigma}\varphi(s)ds+
\frac{p}{t}\exp\int_t^{t+1}z(s)ds \\
&\quad \times\Big[z(t+1)\cos\int_{t+1}^{t+\sigma}\varphi(s)ds
+\varphi(t+1)\sin\int_{t+1}^{t+\sigma}\varphi(s)ds\Big]\Big\}
\end{aligned}
\end{equation}
Suppose $\varphi(t):=\frac{\nu}{t}$, $z(t):=\frac{\alpha}{t}+\frac{\beta}{t^2}$,
where $\alpha$ and $\beta$ will be chosen later.
Let us first check (5) of Theorem \ref{thm1}. For $t\in[a,a+1]$ we have
\begin{equation} \label{79}
\begin{gathered}
y'(t)\geq 0\Leftrightarrow z(t)\sin\big(\nu\ln\frac{t}{a}\big)+\frac{\nu}{t}
\cos\big(\nu\ln\frac{t}{a}\big)\geq 0\Leftrightarrow \\
\Leftrightarrow \alpha\tan\big(\nu\ln\frac{t}{a}\big)+\nu\geq 0 \Leftrightarrow
 \alpha\ln\frac{t}{a}\frac{\tan\left(\nu\ln\frac{t}{a}\right)}{\nu\ln\frac{t}{a}}+1\geq 0.
\end{gathered}
\end{equation}
Since
$\lim_{\nu\to 0} \frac{\tan(\nu\ln\frac{t}{a})}{\nu\ln\frac{t}{a}}=1$,
Inequality \eqref{33} is fulfilled if for sufficiently small $\nu$ we
have $\alpha\ln\frac{t}{a}+1>0,~t\in[a,a+1]$. The last one is true if
$$
\alpha\ln\frac{a+1}{a}+1>0\Leftrightarrow \frac{\alpha}{a}+1>0,
$$
that is {\em for every $\alpha$,} if $a$ is  sufficiently large.
Since $\int_a^b \frac{\varphi(s)}{s}ds=\pi$ we will obtain
similarly that $y'(t)\leq 0,\quad t\in[b,b+1]$.

Now, we return to \eqref{78}$_2$:
\begin{gather*}
A(t)=z^2+z'-\varphi^2=\frac{\alpha^2-\alpha-\nu^2}{t^2}+o(t^{-2});\\
B(t)=\frac{\nu(2\alpha-1)}{t^2}+\frac{2\nu\beta}{t^3}+o(t^{-3});
\end{gather*}
\begin{equation} \label{80}
\begin{gathered}
M(t+\sigma)=-\big\{\frac{\alpha^2-\alpha-\nu^2}{t^2}+\frac{p}{t}
\big[\frac{\alpha}{t}+o(t^{-1})\big]\big\}\\
\Rightarrow t^2M(t+\sigma)=\alpha(1-p)-\alpha^2+\nu^2+o(1).
\end{gathered}
\end{equation}
Let $\alpha=\alpha_0=\frac{1-p}{2}.$
Then
$$
\sup_{\alpha}[\alpha(1-p)-\alpha^2]=\alpha_0(1-p)-\alpha_0^2=\frac{(1-p)^2}{4}.
$$
Hence
$$
t^2M(t+\sigma)=\frac{(1-p)^2}{4}+\nu^2+o(1).
$$
 From \eqref{78}$_1$ we have
\begin{align*}
\tilde{Q}(t+\sigma)
&\cong -\frac{1-\frac{\alpha\sigma}{t}}{\frac{\nu\sigma}{t}}
\Big\{\frac{\nu(2\alpha-1)}{t^2}+\frac{2\nu\beta}{t^3}\\
&\quad +\frac{p}{t}\big(1+\frac{\alpha}{t}\big)\Big[\big(\frac{\alpha}{t+1}
+\frac{\beta}{(t+1)^2}\big)\frac{\nu}{t}+\frac{\nu}{t}\big(1-\frac{\nu^2}{2t^2}
\big)\Big]\Big\}\\
&\quad \cong -\frac{t}{\sigma}\big\{ \frac{2\alpha-1+p}{t^2}
+\frac{2\beta+2p\alpha}{t^3} +o(t^{-3})\big\}.
\end{align*}
Since $\alpha=\alpha_0=(1-p)/2$,
\begin{equation}\label{81}
t^2\tilde{Q}(t+\sigma)=-\frac{2}{\sigma}\beta-\frac{p(1-p)}{\sigma}+o(1).
\end{equation}
If from the beginning we require in \eqref{66} that
\begin{equation} \label{82}
\liminf_{t\to\infty} t^2Q(t):= C>\frac{(1+p)^2}{4},
\end{equation}
and we choose $\nu$ small such that
\begin{equation} \label{83}
\frac{(1+p)^2}{4}+\nu^2<C
\end{equation}
and $\beta$ such that
$$
\frac{(1+p)^2}{4}+\nu^2<-\frac{2}{\sigma}\beta-\frac{p(1-p)}{\sigma}<C,
$$
then $M(t)<\tilde{Q}(t)<Q(t)$,
and the choice of  a function $\tilde{Q}(t)$ with required properties is
completed.
\smallskip

\noindent {\bf Case $p<0$:}
Since an application of Theorem \ref{thm1} is impossible in this case, we
employ Theorem \ref{thm2}.
First consider \eqref{45} and rewrite \eqref{45}$_1$ for
$t\in[a-1,a]$ as
\begin{gather*}
y''(t)\leq 0\Leftrightarrow A(t)\sin(\nu\ln\frac{t}{a})+
B(t)\cos\big(\nu\ln\frac{t}{a}\big)\leq 0  \\
\Leftrightarrow -A(t)\sin(\nu\ln\frac{a}{t})+
B(t)\cos(\nu\ln\frac{a}{t})\leq 0 \\
\Leftrightarrow -\frac{\alpha^2-\alpha-\nu^2}{t^2}\tan(\nu\ln\frac{a}{t})
+\frac{\nu(2\alpha-1)}{t^2}+\frac{2\nu\beta}{t^3}<0\\
\Leftarrow (-\alpha^2+\alpha+\nu^2)\tan(\nu\ln\frac{a}{t})
+\nu(2\alpha-1)<0.
\end{gather*}
Evidently, the condition $2\alpha-1<0$ is necessary (put $t=a$). However it is
also sufficient, since $\tan(\nu\ln\frac{a}{t})\cong \nu/a$
and it becomes small for sufficiently large $a$.
Thus, let  $\alpha<1/2$. Since $\int_a^b \frac{\nu}{s}ds
=\nu\ln\frac{b}{a}=\pi$. Then
\eqref{45}$_2$ can be rewritten for $t\in[b-1,b]$ in the form
\begin{gather*}
A(t)\sin\int_a^t\varphi(s)ds+B(t)\cos\int_a^t\varphi(s)ds\geq 0 \\
\Leftrightarrow A(t)\sin\big(\int_a^b\varphi(s)ds+\int_b^t\varphi(s)ds\big)
+B(t)\cos\big(\int_a^b\varphi(s)ds+\int_b^t\varphi(s)ds\big)\geq 0\\
\Leftrightarrow -A(t)\sin \int_b^t\varphi(s)ds-B(t)\cos\int_b^t\varphi(s)ds\geq 0
\end{gather*}
which is proven above. Then for $\alpha<1/2$ Condition \eqref{45}
 is fulfilled.
Condition (2) of Theorem \ref{thm2}  holds  if
\begin{equation} \label{84}
t^2\tilde{M}(t+\sigma)=\alpha(1-p)-\alpha^2+\nu^2+o(1),
\end{equation}
but here we can not assume  $\alpha=\alpha_0=\frac{1-p}{2}$
since $p<0$, $\alpha_0>1/2$, and hence  \eqref{74} is not satisfied.

Let $\alpha=\frac{1}{2}$ in \eqref{84}, that is
\begin{equation} \label{85}
t^2\tilde{M}(t+\sigma)=\frac{1-2p}{4}+\nu^2+o(1).
\end{equation}
Then  \eqref{78} turns into
\begin{align*}
&\tilde{Q}(t+\sigma)\exp\big\{\int_t^{t+\sigma-1} z(s)ds\big\}\sin
\big(\nu\ln\frac{t+\sigma-1}{t}\big)+\frac{p}{t}\frac{\nu}{t}\\
&+\exp\big\{-\int_{t-1}^t z(s)ds\big\}\Big\{\big[ -\frac{\alpha^2-\alpha-\nu^2}{t^2}
+ \frac{(2\alpha-1)\beta}{t^3}\big]\sin\big(\nu\ln\frac{t}{t-1}\big)\\
&+\big[\frac{\nu(2\alpha-1)}{t^2}+\frac{2\nu\beta}{t^3}\big]
\cos(\nu\ln\frac{t}{t-1})\Big\}+o(t^{-3})=0,
\end{align*}
or
$$
\tilde{Q}(t+\sigma)=\frac{1}{\frac{\nu(\sigma-1)}{t}}
\big\{\frac{\nu(2\alpha-1-p)}{t^2}+o(t^{-2})\big\}.
$$
Since $\alpha=1/2$,
\begin{equation} \label{86}
t\tilde{Q}(t+\sigma)=-\frac{p}{\sigma-1} +o(t^{-1}).
\end{equation}
We have $p<0$, so we should require $\sigma>1$ to satisfy
Condition (1) of Theorem \ref{thm2}.
Evidently, \eqref{85}--\eqref{86} imply $\tilde{Q}(t)>\tilde{M}(t)$,
$t\gg t_0$, that is \eqref{78}$_2$ is satisfied.

Now we can formulate the final result for \eqref{77}.

\begin{theorem} \label{thm5}
Suppose one of the two following conditions holds:
\begin{itemize}
\item[(a)] $p\geq0$, $\sigma\neq 0$,
\begin{equation} \label{87}
\liminf_{t\to +\infty}[t^2 Q(t)]=C_1>\frac{(1+p)^2}{4}
\end{equation}

\item[(b)] $p<0$, $\sigma> 1$,
\begin{equation} \label{88}
\liminf_{t\to +\infty}[t Q(t)]=C_2>\frac{|p|}{\sigma-1}.
\end{equation}
\end{itemize}
Then all solutions of \eqref{77} are oscillatory and there exists $\nu>0$ such
that all solutions have at least one zero on any interval
$(T-1,T\exp\frac{\pi}{\nu})$ for sufficiently  large $T$.
\end{theorem}

\noindent{\bf Remarks:} 
\quad
(1)\; Equation \eqref{77} for  case $p>0$, $\sigma\neq 0$ has a nonoscillatory
solution $x_0(t)=t^{\frac{p+1}{2}}$ and
$$
Q(t)=Q_0(t):=\frac{\big[\frac{p}{t-1}x_0(t-1)\big]'-x_0''(t)}{x_0(t-\sigma)}=
=\frac{1}{t^2}\frac{p(\gamma-1)(1-\frac{1}{t})^{\gamma-2}-\gamma(\gamma-1)}
{(1-\frac{\sigma}{t})^{\gamma}}.
$$
It is easy to see that
$$
\lim_{t\to\infty} t^2Q_0(t)=\frac{(1-p)^2}{4}.
$$
Thus \eqref{87} which is apparently exact in order ($t^{-2}$),
but is not exact in constant.
Condition \eqref{88}, apparently, is not exact, both in order and in constant.

\noindent(2)\; The condition $\sigma>1$ in the case $p<0$ is not accidental.
We remember that an autonomous \eqref{19} with $p<0$, $\tau=1$, $\sigma\leq 1$
cannot be oscillatory for any $q$, but for $\sigma>1$ there exist such numbers $q$.
\smallskip

%{\bf 6.4}
We will investigate here oscillation properties of
another equation, which contains unbounded delay  in damped term.
Consider the equation
\begin{equation} \label{89}
x''(t)-\big[\frac{p}{t}x\big(\frac{t}{\mu}\big)\big]'+Q(t)x(t)=0,
\quad p>0,\; \mu>1.
\end{equation}
Let be $\varphi(t):=\nu/t$, $\nu>0$. Then
$$
A(t):=z^2(t)+z'(t)-\frac{\nu^2}{t^2};\quad
B(t):=\frac{\nu}{t}\big[2z(t)-\frac{1}{t}\big].
$$
Substituting these equalities in \eqref{49} we have
\begin{align*}
S(t)&:=z^2(t)-z'(t)-\frac{\nu^2}{t^2}
+\frac{p}{t}\exp\big\{\int_t^{\mu t}z(s)ds\big\} \\
&\quad\times \big[z(\mu t)\cos(\nu\ln\mu)-\frac{\nu}{\mu t}\sin(\nu\ln\mu)\big]
+\tilde{Q}(t)\geq 0\\
R(t)&:= \frac{\nu}{t}\big[2z(t)-\frac{1}{t}\big]+\frac{p}{t}
\exp\big\{\int_t^{\mu t}z(s)ds\big\} \\
&\quad \times\big[z(\mu t)\sin(\nu\ln\mu)
+\frac{\nu}{\mu t}\cos(\nu\ln\mu)\big]=0
\end{align*}
or
\begin{equation}\label{90}
\begin{aligned}
\tilde{Q}(t)\geq M(t)&:=-z'(t)-z^2(t)+\frac{\nu^2}{t^2}
-\frac{p}{t}\exp\big\{\int_t^{\mu t}z(s)ds\big\} \\
&\quad\times \big[z(\mu t)\cos(\nu\ln\mu)-\frac{\nu}{\mu t}\sin(\nu\ln\mu)\big]\\
N[z](t)&:=2tz(t)-1+p\exp\big\{\int_t^{\mu t}z(s)ds\big\}\\
&\quad\times \big[\mu tz(\mu t)\frac{1}{\mu\nu}\sin(\nu\ln\mu)
+\frac{1}{\mu}\cos(\nu\ln\mu)\big]=0.
\end{aligned}
\end{equation}
Equation \eqref{90}$_2$ has an analytic solution $z(t)$ in a neighborhood of
infinity, such that $z(\infty)=0$.

Let us find the first term of an expansion $z(t)=\frac{\alpha}{t}+\dots$:
\begin{equation} \label{91}
\begin{gathered}
2\alpha-1 +p\mu^{\alpha}\big[\alpha\frac{1}{\mu\nu}\sin(\nu\ln\mu)+\frac{1}{\mu}
\cos(\nu\ln\mu)\big]=0\\
\Leftrightarrow
F(\alpha;\nu):=2\alpha-1+p\mu^{\alpha-1}\big[\alpha\ln\mu\frac{\sin(\nu\ln\mu)}{\nu\ln\mu}
+\cos(\nu\ln\mu)\big]=0.
\end{gathered}
\end{equation}
Consider also the limit equation of \eqref{91}, where $\nu\to 0$:
\begin{equation} \label{92}
F(\alpha;0):=2\alpha-1+p\mu^{\alpha-1}(\alpha\ln\mu+1)=0.
\end{equation}
One can show  that \eqref{92} has the unique root $\alpha_0$
(we omit calculations) and, moreover, $\alpha_0\in[0,\frac{1}{2})$
for $ 0<p\leq \mu$
and $\alpha_0\in(-\frac{1}{\ln\mu},0)$ for $p>\mu$.

On the other hand
$$
F_{\alpha}'(\alpha_0,0)=2p\mu^{\alpha_0-1}\ln\mu (2+\ln\mu)>0.
$$
Then \eqref{91} has a solution $\alpha_{\nu}$ for $0<\nu<\nu_0$.
That is $z(t)=\frac{\alpha_{\nu}}{t}+o(t^{-1})$ is a solution of \eqref{90}$_2$.
Thus
$$
M(t)t^2=\alpha_{\nu}-\alpha_{\nu}^2+\nu^2-p\mu^{\alpha_{\nu}-1}
[\alpha_{\nu}\cos(\nu\ln\mu)-\nu\sin(\nu\ln\mu)]+o(t^{-2}).
$$
Since
$$
\alpha_{\nu}-\alpha_{\nu}^2+\nu^2-p\mu^{\alpha_{\nu}-1}
[\alpha_{\nu}\cos(\nu\ln\mu)-\nu\sin(\nu\ln\mu)]
=\alpha_0^2\left[1+p\mu^{\alpha_0-1}\ln\mu\right]+o(\nu^{2}),
$$
 the equality
\begin{equation} \label{93}
\liminf_{t\to\infty}
t^2Q(t)=C>\alpha_0^2\left[1+p\mu^{\alpha_0-1}\ln\mu\right]
:=K(p;\mu)
\end{equation}
implies  $Q(t)>\tilde{Q}(t)$ for $t>T$.
Then conditions (2)--(4) of Theorem \ref{thm1} hold.
Since $p>0$,  Condition (1) also holds.

We check  now  \eqref{33} of (5). For $t\in (a,\mu a)$ we have
\begin{equation} \label{94}
\begin{gathered}
y'(t)\geq 0\Leftrightarrow
\frac{\alpha_0}{t}\sin\big(\nu\ln\frac{t}{a}\big)+\frac{\nu}{t}
\cos\big(\nu\ln\frac{t}{a}\big)\geq 0\\
\Leftrightarrow
\alpha_0\tan\big(\nu\ln\frac{t}{a}\big)+\nu\geq 0\Leftrightarrow
\alpha_0\ln\mu\frac{\tan\left(\nu\ln\frac{t}{a}\right)}{\nu\ln\frac{t}{a}}
+1\geq 0.
\end{gathered}
\end{equation}
Since $\lim_{\nu\to 0}\frac{\tan\left(\nu\ln\frac{t}{a}\right)}{\nu\ln\frac{t}{a}}
=1$ uniformly on $t\in(a,\mu a)$ then an inequality $\alpha_0\ln\mu+1>0$
implies \eqref{94} for sufficiently small $\nu>0$.
Since
$$
\int_a^b \varphi(s)ds =\pi\Leftrightarrow b=ae^{\frac{\pi}{\nu}}
$$
it follows that $y'(t)\leq 0$, $t\in(b,\mu b)$, which is the second part
of \eqref{33}.
Thus we have the following statement which is the exact analogue of the
classical Knezerian Theorem:

\begin{theorem} \label{thm6}
Assume \eqref{93}.
Then all  solutions  of \eqref{89} are oscillatory, and, moreover,
there exists $\nu>0$ such that  any solution
has at least one zero on any interval
$\big(\frac{T}{\mu},Te^{\frac{\pi}{\nu}}\big)$ for sufficiently large $T$.
\end{theorem}

\noindent{\bf Remark.}\;
The constant $K(p,\mu)$ is an unimprovable in \eqref{93}. Indeed, by
direct calculations
one can check that $x_0(t)=t^{1-\alpha_0}$ is a non-oscillatory solution of
\eqref{89} with $Q_0(t)=Kt^{-2}$.
\smallskip

%{\bf 6.5}
We will investigate in detail a particular case of \eqref{89}
with $p=\mu$, i.e. the following equation
\begin{equation} \label{95}
x''(t)-\big[\frac{\mu}{t}x\big(\frac{t}{\mu}\big)\big]'+Q(t)x(t)=0,\quad \mu>1.
\end{equation}
If $p=\mu$ (and only in this case) then \eqref{92} has the root $\alpha_0=0$.
And so, in\eqref{93} $K(\mu,\mu)=0$. It means that there is no
information about the main term
 of  Knezerian  Minorant for $Q(t)$ in \eqref{95}.

Let $\varphi(t):=\frac{\nu}{t\ln t}$ in \eqref{49}.
Hence
\begin{gather*}
\varphi'(t)=-\frac{\nu(1+\ln t)}{t^2\ln^2 t};~
\int_t^{\mu t}\varphi(s)ds=
\frac{\nu\ln \mu}{\ln t}\big[1-\frac{\ln\mu}{2\ln t}+o\big(\frac{1}{\ln t}\big)
\big], \\
A(t):=z^2(t)+z'(t)-\frac{\nu^2}{t^2\ln^2t};\quad
B(t):=\frac{\nu}{t\ln t}\big[2z(t)-\frac{1}{t}-\frac{1}{t\ln t}\big].
\end{gather*}
Substituting all these equalities in \eqref{49}, with $p=\mu$, we obtain:
\begin{equation} \label{96}
\begin{aligned}
z^2(t)+z'(t)-\frac{\nu^2}{t^2\ln^2t}+\frac{\mu}{t}\exp\big\{\int_t^{\mu
t}z(s)ds\big\}\\
\times\big[z(\mu t)\cos\int_t^{\mu t}\varphi(s)ds-\frac{\nu}{\mu t\ln(\mu t)}\sin
\int_t^{\mu t}\varphi(s)ds\big]+\tilde{Q}(t)\geq 0\,,\\
\frac{\nu}{t\ln t}\big[2z(t)-\frac{1}{t}-\frac{1}{t\ln t}\big]
\frac{\mu}{t}\exp\int_t^{\mu t}z(s)ds \\
\times\big[z(\mu t)\sin\int_t^{\mu t}\varphi(s)ds+\frac{\nu}{\mu t\ln(\mu t)}\cos
\int_t^{\mu
t}\varphi(s)ds\big]=0.
\end{aligned}
\end{equation}
We will look for the main part of the solution $z(t)$ of \eqref{96}$_2$
in a neighborhood of the infinity, that is
a solution $z(t)$ of the form
$$
tz(t)=\sum_{n=1}^{\infty} \frac{a_n}{(\ln t)^n}.
$$
Let us find the number $\beta$ in
$z(t)=\frac{\beta}{t\ln t}+o(\frac{1}{t\ln t})$.
Since
$$
\exp\big\{\int_t^{\mu t}z(s)ds\big\}
=1+\frac{\beta\ln \mu}{\ln t}+o\big(\frac{1}{\ln t}\big),
$$
Equation \eqref{96}$_2$ has the form
\begin{equation} \label{97}
\begin{aligned}
\frac{\nu}{t\ln t}\big(\frac{2\beta-1}{t\ln t}-\frac{1}{t}\big)
+\frac{\mu}{t}\big(1+\frac{\beta\ln\mu}{\ln t}\big)\\
\times \Big[\frac{\beta}{\mu t\ln(\mu t)}\sin\big(\frac{\nu\ln\mu}{\ln t}\big)
+\frac{\nu}{\mu t\ln(\mu t)}\cos \big(\frac{\nu\ln\mu}{\ln t}\big)+
o\big(\frac{1}{\ln t}\big)\Big]=0.
\end{aligned}
\end{equation}
Applying the equality
$$
\frac{1}{\ln(\mu t)}
=\frac{1}{\ln t}\big(1-\frac{\ln\mu}{\ln t}+o(\frac{1}{\ln t})\big),
$$
and multiplying all terms of \eqref{96}$_2$  by $t^2\ln t$,
write its main terms:
$$
\frac{2\beta-1}{\ln t}-1+\big(1-\frac{\ln \mu}{t}\big)
\big(1+\frac{\beta\ln\mu}{\ln t}\big)
\big(1+\frac{\beta\ln\mu}{\ln t}\big)+o\big(\frac{1}{\ln t}\big)=0
$$
or
\begin{equation} \label{98}
\frac{2\beta-1}{\ln t}-1+1+\frac{(2\beta-1)\ln\mu}{\ln t}+
o\big(\frac{1}{\ln t}\big)=0\Rightarrow \beta=\frac{1}{2}.
\end{equation}
Thus, $z(t)=\frac{1}{2t\ln t}+o(\frac{1}{t\ln t})$
and we can write \eqref{96}$_1$ as
\begin{align*}
&\tilde{Q}(t)+\frac{1}{4t^2\ln^2 t}-\frac{1+\ln t}{2t^2\ln^2 t}
-\frac{\nu^2}{2t^2\ln^2 t}+
\frac{\mu}{t}\big[1+\frac{\ln\mu}{2\ln t}+o\big(\frac{1}{\ln t}\big)\big]\\
&\big\{\frac{1}{2\mu t\ln(\mu t)}\cos\frac{\nu\ln\mu}{\ln t}
-\frac{\nu}{\mu t\ln(\mu t)}\frac{\nu\ln\mu}{\ln t}+o\big(\frac{1}{\ln
t}\big)\big\}\geq 0.
\end{align*}
After some calculations we obtain
$$
t^2\ln^2t\tilde{Q}(t)\geq \frac{1+\ln \mu}{4}+o(1).
$$
Since
$$
\int_a^b\frac{\nu}{s\ln s} ds=\pi\Leftrightarrow \nu\ln\left[
\frac{\ln b}{\ln a}\right]=\pi\Leftrightarrow b=ae^{\exp\frac{\pi}{\nu}},
$$
we can formulate the following result.

\begin{theorem} \label{thm7}
Suppose in \eqref{95}
\begin{equation}
\label{99}
\liminf_{t\to\infty} [t^2\ln^2t~Q(t)]=C>\frac{1+\ln \mu}{4}.
\end{equation}
Then all its solutions  are oscillatory.
Moreover,  there exists $\nu>0$ such that every
solution has at least one zero in any interval
$(\frac{T}{\mu}, T^{\exp\frac{\pi}{\nu}})$ for sufficiently large
$T$.
\end{theorem}

This result is the best possible one in the following sense.
The function\\
 $x_0(t):=t\sqrt{\ln t}$ is a nonoscillatory solution of \eqref{95}
with
$$
Q(t)=Q_0(t):=\frac{1}{x_0(t)}\big\{\big[\frac{\mu}{t}
x_0\big(\frac{t}{\mu}\big)\big]'-x_0''(t)\big\}.
$$
On the other hand, we can show that
\begin{equation}
\label{100}
\lim_{t\to\infty} \left\{t^2\ln^2t~Q_0(t)\right\}=\frac{1+\ln\mu}{4}.
\end{equation}

\noindent{\bf Remark.}\;
This phenomenon is well-known for ordinary
differential \eqref{2}. Indeed, substituting \eqref{3}, where
$P(t):=\frac{1}{t}$, we transform
 \eqref{2} into \eqref{1} with $a(t)=Q(t)+\frac{1}{4t^2}$. Then by
generalized Kneser Theorem  \cite[Theorem 7.1, Exercise 1.2, Ch.11]{h1}
 the following statement holds
\begin{quote} If
\begin{equation}
\label{101}
\liminf_{t\to\infty} \left\{t^2\ln^2 t~Q(t)\right\} =C>\frac{1}{4},
\end{equation}
then all solutions of the equation
\begin{equation}
\label{102}
x''(t)+\big[\frac{x(t)}{t}\big]'+Q(t)x(t)=0
\end{equation}
are oscillatory.
\end{quote}

For differential equations with deviating arguments such phenomenon is described
for the first time here. Note, by the way, that \eqref{99} turns into
\eqref{101} when $\mu=1$.

%{\bf 6.6}
Now, we apply Theorem \ref{thm1} and Theorem \ref{thm2} to the investigation
of oscillation properties of \eqref{6} with {\em strong} deviating
arguments $r(t)$ or $l(t)$. We apply this term when not only $t-r(t)$ is unbounded
but $\lim_{t\to\infty}\frac{r(t)}{t}=0$ holds as well. For example, the
delay
$$
t-r(t)=t-\frac{t}{\mu}=\frac{\mu-1}{\mu}t,\quad \mu>1
$$
which was considered in the previous paragraph, is unbounded, but not strong in
our sense.
A typical example of the strong deviating argument is
$r(t)=t^{1/\mu}$, $\mu>1$. Investigation of such equations
usually meets difficulties. Below we will apply our method to
a typical equation with a strong delay.
Consider the equation
\begin{equation} \label{103}
x''(t)-\big[\frac{p}{t^{\beta}}x(t^{\frac{1}{\mu}})\big]'+Q(t)x(t^{\frac{1}{\mu}})=0,
\quad \mu>1;\; p>0,\; \beta\geq \frac{1}{\mu},
\end{equation}
and let us apply Theorem \ref{thm1}. Choose $z(t)$ and $\tilde{Q}(t)$ on
$(T,\infty)$ for $T$ sufficiently large
such that \eqref{51} is satisfied. We will seek $z(t)$ in the form
\begin{equation}
\label{104}
z(t)=\frac{s}{t\ln t}+\frac{\gamma}{t\ln^2 t},
\end{equation}
where parameters $s$ and $\gamma$ will be chosen later, and $\varphi(t)$ is
define as $\varphi(t)=\nu/(t\ln t)$.
Then
\begin{gather*}
A(t)=\frac{1}{t^2\ln t}\big[-s+\frac{s^2-s-\nu^2-\gamma}{\ln t}
+o\big(\frac{1}{\ln t}\big)\big],\\
B(t)=\frac{\nu}{t^2\ln t}\big[-1+\frac{2s-1}{\ln t}+o\big(\frac{1}{\ln
t}\big)\big],\\
r(t)=l(t)=t^{1/\mu}, \quad q(t)=k(t)=t^{\mu},
\quad \int_t^{t^{\mu}}\varphi(s)ds=\nu\ln \mu, \\
\exp \int_t^{t^{\mu}}z(s)ds =\mu^s\big[ 1+\frac{\gamma(\mu-1)}{\mu\ln t}
+o\big(\frac{1}{\ln t}\big)\big].
\end{gather*}
Substituting all these expressions in \eqref{51} we obtain
\begin{equation} \label{105}
\begin{gathered}
Z(t)\geq 0\Leftrightarrow \big[-s+\frac{s^2-s-\nu^2-\gamma}{\ln t}
+o\big(\frac{1}{\ln t}\big)\big]\sin(\nu\ln\mu)\\
-\big[\nu-1+\frac{2s-1}{\ln t}+o\big(\frac{1}{\ln t}\big)\big]
\nu\cos(\nu\ln\mu)- \frac{p\nu\mu^s}{t^{\beta\mu-1}}
\big[1+\frac{\gamma(\mu-1)}{\mu\ln t}+o\big(\frac{1}{\ln
t}\big)\big]\geq 0.
\end{gathered}
\end{equation}
Consider two cases: $\beta=\frac{1}{\mu}$ and $\beta>1/\mu$.

\noindent{\bf Case $\beta=\frac{1}{\mu}$:}\;
Equation \eqref{105} turns into
\begin{equation}\label{106}
\begin{aligned}
\big[ -s\sin(\nu\ln\mu)+\nu \cos(\nu\ln\mu)-\nu p\mu^s\big]\\
+\frac{\nu}{\ln t}\big[s^2-s-\gamma-\nu^2)
\frac{\sin(\nu\ln\mu)}{\nu}-p\mu^{s-1}(\mu-1)\gamma\big]+o\big(\frac{1}{\ln t}\big)
\geq 0.
\end{aligned}
\end{equation}
Put $s=s_{\nu}$ for sufficiently small $\nu>0$, where $s_{\nu}$ is the root of
the equation
\begin{equation} \label{107}
F(\nu,s):=\frac{\sin(\nu\ln\mu)}{\nu\ln\mu}s\ln\mu-\cos(\nu\ln\mu)+p\nu^s=0.
\end{equation}
By the Implicit Function Theorem such a root exists since there exists the unique
root $s=s_0<\frac{1}{\ln \mu}$ of the equation
\begin{equation} \label{108}
F(0,s):=s\ln \mu-1+p\nu^s=0,
\end{equation}
and $F_s'(0,s_0)=\ln\mu(1+p\mu^{s_0})\neq 0$.  The coefficient $\gamma$ in the second
bracket in \eqref{105} is negative, hence by choosing $\gamma<\gamma_0$
one can carry out \eqref{105}. The strict value of $\gamma_0$ is not important.
Thus assume
\begin{equation}
\label{110}
z(t):=\frac{s_{\nu}}{t\ln t}+\frac{\gamma}{t\ln^2 t}
\end{equation}
and $\nu>0$ is sufficiently small. Substituting \eqref{110} in $\eqref{51}_2)$,
we obtain
\begin{align*}
\mu t^{\mu-1}\tilde{Q}(t^{\mu})
&=[\sin(\nu\ln\mu)]^{-1} \Big\{\exp\big\{-\int_t^{t^{\mu}}
z(s)ds\big\}B(t)\\
&\quad +\mu t^{\mu-1}\frac{p}{t}\big[z(t^{\mu})\sin(\nu\ln\mu)
+\frac{\nu}{t^{\mu}\mu\ln t}\cos(\nu\ln\mu)\big]\Big\},
\end{align*}
hence
$$
\mu t^{\mu+1}\ln t\tilde{Q}(t^{\mu})=D(p)+\epsilon(\nu,t),
$$
where
$\lim_{\nu\to 0,t\to \infty}\epsilon(\nu,t)=0$
and
\begin{equation} \label{111}
D(p):=\frac{1}{\ln\mu}\left\{\mu^{-s_0}-p(s_0\ln\mu
+1)\right\}=\frac{s_0^2\ln\mu}{\mu^{s_0}}.
\end{equation}
Proceeding to a new time by the substitution of $t$ instead of
$t^{\frac{1}{\mu}}$ we obtain
\begin{equation} \label{113}
t^{\frac{\mu+1}{\mu}} \ln t \tilde{Q}(t)=D(p)+\epsilon(\nu,t^{\frac{1}{\mu}}).
\end{equation}
By the way, note that
\begin{equation}\label{114}
\lim_{p\to 0} D(p)=\frac{1}{e\ln\mu}.
\end{equation}
Hence we have the following statement.

\begin{theorem} \label{thm8}
Let be  $\beta=\frac{1}{\mu}$, $p>0$, $\mu>1$ in \eqref{103} and
\begin{equation}
\label{115}
\liminf_{t\to\infty}\big\{t^{\frac{\mu+1}{\mu}} \ln t \tilde{Q}(t)\big\}=C>D(p).
\end{equation}
Then all  solutions of \eqref{103} are oscillatory and there
exists $\nu >0$ such that
every solution has at least one zero in any interval
$(T^{1/\mu},T^{\exp(\pi/\nu)} )$ for sufficiently large $T$.
\end{theorem}

\begin{proof} It follows from Theorem \ref{thm1} and the fact, that a pair
$\{\tilde{Q}(t);z(t)\}$
constructed above satisfies \eqref{51}. The interval
$(T^{\frac{1}{\mu}},T^{\exp(\pi/\nu)})$ coincides with the interval
$(a,b)$ in Theorem \ref{thm1}. Indeed
$$
\int_T^b\varphi(t)dt=\pi\Leftrightarrow\nu\int_T^b\frac{dt}{t\ln t}=\pi
\Rightarrow b=T^{\exp(\pi/\nu)}.
$$
\end{proof}

\noindent {\bf Remark.}\;
Condition \eqref{115} is unimprovable in the same sense:
there exists \eqref{103} such that
\begin{equation}
\label{116}
\lim_{t\to\infty}t^{\frac{\mu+1}{\mu}} \ln t \tilde{Q}(t)=D(p),
\end{equation}
but, nevertheless, this equation has a nonoscillatory solution.

Indeed, let $x_0(t):=t(\ln t)^{-s_0}$, where $s_0$ is defined in
\eqref{108}.
Then $x_0(t)$ is a nonoscillatory solution of \eqref{103}, where
$$
Q(t)=Q_0(t):=\frac{1}{x_0(t^{\frac{1}{\mu}})}
\big\{\big[\frac{p}{t^{\frac{1}{\mu}}}x_0(t^{\frac{1}{\mu}})\big]'
-x_0''(t^{\frac{1}{\mu}})\big\}.
$$
After some calculations we get
$$
Q(t)=\big\{t^{\frac{\mu+1}{\mu}}\ln t\big\}^{-1}
\big\{D(p)-\frac{s_0(s_0+1)}{\mu^{s_0}\ln t}\big\},
$$
hence \eqref{115} holds.

\noindent{\bf Case $\beta>1/\mu$:}\;
One can write \eqref{105} as
\begin{equation} \label{117}
[-s\sin(\nu\ln\mu)+\nu\cos(\nu\ln\mu)]+\frac{1}{\ln t}(s^2-s-\gamma-\nu^2)
\sin(\nu\ln\mu)\geq 0.
\end{equation}
After calculations as in the previous case with 
$s_{\nu}=(\nu\cos(\nu\ln\mu))/(\sin(\nu\ln\mu)$,
we obtain
\begin{equation}
\label{118}
t^{\frac{\mu+1}{\mu}} \ln t \tilde{Q}(t)=\frac{1}{e\ln \mu}+\epsilon
(\nu,t^{\frac{1}{\mu}}).
\end{equation}
Thus we have the statement.

\begin{theorem} \label{thm9}
Let $\beta>1/\mu$, $p>0$ in \eqref{103}.
Then the condition
$$
\liminf_{t\to\infty}\big\{t^{\frac{\mu+1}{\mu}} \ln t Q(t)\big\}
=C>\frac{1}{e\ln \mu}
$$
implies the assertion of Theorem \ref{thm8}.
\end{theorem}

\vspace{3mm}

%{\bf 6.7}
Now we show that it is possible to extend
previous results to nonlinear damped  equations.
Here we do not intend to cover the widest class of such equations,
but to  present the sketch of the main idea only.
Consider the nonlinear equation
\begin{equation}
\label{119}
z''(t)-[P(t)z(r(t))]'+F(t,z(l(t)))=0, \quad t\in (a,b),
\end{equation}
where $F(t,u)$ satisfies the following ``one-sided" estimation:
\begin{equation} \label{120}
vF(t,v)\geq \tilde{Q}(t)v^2, \;\forall t\in (a,b)\setminus e_{\rm int},
\quad vF(t,v)\geq 0,\; t\in e_{\rm int}, \;\forall
v\in(-\infty,\infty).
\end{equation}

\begin{theorem} \label{thm10}
Suppose for $P(t), r(t), l(t), \tilde{Q}(t)$
Conditions (1), (3), (5), (6) of Theorem \ref{thm1} and
\eqref{120} hold.
Then \eqref{119} has no solution which is positive on
$(r(a),b)\bigcup e_{\rm ext}$.
\end{theorem}

\begin{proof}
Suppose by contradiction that \eqref{119} has a solution
$z_0(t)>0$, for $t\in (r(a),b)\cup e_{\rm ext}$.
 Then the linear equation
\begin{equation}
\label{121}
(lx)(t):=x''(t)-[P(t)x(r(t))]'+\frac{F(t,
z_0(l(t))}{z_0(l(t))}x(l(t))=0,~t\in (a,b)
\end{equation}
also has a positive solution $x(t):=z_0(t)$ on $t\in (r(a),b)\cup e_{\rm ext}$.
Equation \eqref{121} has the form of \eqref{28} with
$Q(t):=\frac{F(t,x_0(l(t))}{x_0(l(t))}$.
Equalities \eqref{120} and \eqref{121} imply that conditions (2) and (4)
of Theorem \ref{thm1} hold. Then the statement of Theorem \ref{thm10} is
a consequence of Theorem \ref{thm1}.
\end{proof}

\begin{corollary} \label{coro10.1}
Let
$$
\varphi(t)>0,\quad  \int^{\infty} \varphi(t)dt =\infty
$$
and let $\{\tilde{Q}(t);z(t)\}$ be a solution of \eqref{50}.
Suppose \eqref{120} holds on $(T,\infty)$ for sufficiently large $T$ and
$\tilde{Q}(t)\geq 0$ on $(T,\infty)$.
Then all solutions of \eqref{119} are oscillatory. If ,in addition,
\eqref{47} holds, then every solution of \eqref{119} has at least one zero on any
interval $(r(a),b)\bigcup e_{\rm ext}$.
\end{corollary}

\subsection*{Acknowledgement}
The first author was partially supported by
Israel Ministry of Absorbtion.


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\end{document}