\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 23, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/23\hfil Oscillation of solutions]
{Oscillation of solutions for odd-order
 neutral functional differential equations}

\author[T. Candan\hfil EJDE-2010/23\hfilneg]
{Tuncay Candan}  % in alphabetical order

\address{Tuncay Candan \newline
Department of Mathematics,
Faculty of Art and Science \\
Ni\u{g}de University, Ni\u{g}de, 51200, Turkey}
\email{tcandan@nigde.edu.tr}

\thanks{Submitted December 9, 2009. Published February 4, 2010.}
\subjclass[2000]{34K11, 34K40} 
\keywords{Neutral differential equations; oscillation of solutions; \hfill\break\indent
distributed deviating arguments}

\begin{abstract}
 In this article, we establish oscillation criteria
 for all solutions to the neutral differential equations
 \[
 [x(t)\pm ax(t\pm h)\pm bx(t\pm g)]^{(n)}
 =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
 \]
 where $n$ is odd, $h$, $g$, $a$ and $b$ are nonnegative
 constants. We consider 10 of the 16 possible combinations
 of $\pm$ signs, and give some examples to illustrate our results.
\end{abstract}

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

\section{Introduction}

In this article, we study the oscillatory behavior of solutions to
to  $n$-order mixed neutral functional
differential equations with distributed deviating arguments
\begin{equation} \label{e:1}
[x(t)\pm ax(t\pm h)\pm bx(t\pm g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation}
where $n$ is odd, $h$, $g$, $a$ and $b$ are nonnegative
constants, $p$ and $q$ are positive constants,  and $0<c<d$.
We consider 10 of the 16 possible combinations of $\pm$ signs.
The equations
\begin{equation*}
\frac{d^2}{dt^2}(x(t)\pm x[t-\tau]\pm
x[t+\sigma])+q x[t-\alpha]+px[t+\beta]=0
\end{equation*}
are encountered in the study of vibrating masses attached to
an elastic bar \cite{hale2}, and were studied by Grace and Lalli
\cite{gra4}. Later Grace extended their results to
$n$-order equations with  $n$ odd in \cite{gra5}, and
with $n$ even in \cite{gra6}.
 Moreover, Grace \cite{gra7} remarked that the
results  for the $n$-order equations
\begin{equation*}
\frac{d^n}{dt^n}(x(t)+c x[t-h]+C x[t+H])+q x[t-g]+Q x[t+G]=0
\end{equation*}
are extendable to the equations
\begin{align*}
&\Big(x(t)+\sum_{i=1}^{n_1}c_i
x(t-h_i)+\sum_{j=1}^{n_2}C_j x(t+H_j)\big)^{(n)}\\
&\pm \Big(\sum_{k=1}^{n_3}q_k x(t-g_k)+\sum_{m=1}^{n_4}Q_m
x(t+G_m)\Big)
=0.
\end{align*}

 In recent years, Candan  \cite{Tun2}, and Candan and Dahiya \cite{Tun1}
obtained some results for distributed delays, which
motivate us to study \eqref{e:1}. For books related to this topic, we
refer the reader to \cite{Bai,hale2,Lad}.

 A function $x$ is said to be a solution of \eqref{e:1} if
$x(t)\pm ax(t \pm h) \pm bx(t \pm g)$ is $n$ times continuous
differentiable and $x(t)$ satisfies \eqref{e:1} for $t\ge t_0$.

A nontrivial solution of \eqref{e:1},
for all large $t$, is called oscillatory if it
has no largest zero. Otherwise, a solution is called non-oscillatory.

The purpose of this paper is to provide sufficient conditions,
only on the coefficients and on limits of the integrals, to guarantee
that \eqref{e:1} is oscillatory.

\section{Main Results}

 The following lemmas will be used in our proofs.

\begin{lemma}[\cite{Ladas}]\label{lem1}
Suppose that $a$ and $h$ are positive constants and
$a^{1/n}\big(\frac{h}{n}\big)e>1$.
Then
\begin{itemize}
\item[(i)] the inequality
\[
x^{(n)}(t)-ax(t+h)\ge 0
\]
has no eventually positive solutions when $n$ is odd;
\item[(ii)] the inequality
\[
x^{(n)}(t)+ax(t-h)\le 0
\]
has no eventually positive solutions when $n$ is odd.
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{Kig}] \label{lem2}
Let $x(t)$ be a function such that it and each of its derivative
up to order $(n-1)$ inclusive are absolutely continuous and of
constant sign in an interval $(t_0,\infty)$.
If $x^{(n)}(t)$ is of constant sign and not identically zero
on any interval of the form $[t_1,\infty)$ for some $t_1\ge t_0$,
then there exist a $t_x\ge t_0$ and  an integer $m$, $0\le m\le n$
with $n+m$ even for $x^{(n)}(t)\ge 0$,  or $n+m$ odd
for $x^{(n)}(t)\le 0$, and such that for every $t\ge t_x$,
\[
m>0\quad \text{implies}\quad x^{(k)}(t)> 0, \quad k=0,1,\ldots,
m-1
\]
and
\[
m\le n-1\quad \text{implies}\quad (-1)^{m+k}x^{(k)}(t)> 0, \quad
k=m,m+1,\ldots, n-1.
\]
\end{lemma}

\begin{theorem}\label{thm1}
Suppose that $b>0$, either
\begin{equation} \label{e:2}
\Big(\frac{p(d-c)}{b}\Big)^{1/n}\Big(\frac{g+c}{n}\Big)e>1,
\end{equation}
or
\begin{equation} \label{e:3}
\Big(\frac{(p+q)(d-c)}{b}\Big)^{1/n}\Big(\frac{g-d}{n}\Big)e>1,\quad
g>d,
\end{equation}
and
\begin{equation} \label{e:4}
\Big(\frac{q(d-c)}{1+a}\Big)^{1/n}\big(\frac{c}{n}\big)e>1.
\end{equation}
Then
\begin{equation} \label{e:thm1}
[x(t)+ax(t-h)-bx(t+g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation}
is oscillatory.
\end{theorem}

\begin{proof}
Let $x(t)$ be  a non-oscillatory solution of \eqref{e:thm1}. We may
assume that $x(t)$ is eventually positive; that is, there exists a
$t_0\ge 0$ such that $x(t)>0$ for $t\ge t_0$. If $x(t)$ is an
eventually  negative  solution, the proof follows the same arguments.
Let
\[
z(t)=x(t)+ax(t-h)-bx(t+g),\quad t\ge t_0+h.
\]
 From \eqref{e:thm1}, we have
\begin{equation} \label{e:5}
z^{(n)}(t)=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi
\end{equation}
for $t\ge t_1\ge t_0+h$, which implies that $z^{(n)}(t)> 0$. Then
$z^{(i)}(t)$, $i=0,1,\ldots,n$ are of constant sign on
$[t_1,\infty)$. We have two possible cases to consider:
 $z(t)<0$ for $t\ge t_1 $, and $z(t)>0$ for $t\ge t_1$.

\textbf{Case 1: $z(t)<0$ for $t\ge t_1$.}
Let $v(t)=-z(t)$. Then  from \eqref{e:5}, we obtain
\begin{equation} \label{e:6}
v^{(n)}(t)+p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi=0.
\end{equation}
On the other hand, since
\begin{equation*}
0<v(t)=-z(t)=-x(t)-ax(t-h)+bx(t+g)
\le bx(t+g)\quad \text{for } t\ge t_1,
\end{equation*}
 there is a $t_2\ge t_1$ such that
\begin{equation} \label{e:7}
x(t)\ge\frac {v(t-g)}{b}\quad \text{for } t\ge t_2.
\end{equation}
In view of \eqref{e:7} it follows from \eqref{e:6} that
\begin{equation} \label{e:8}
v^{(n)}(t)+\frac{p}{b}\int_c^d v(t-g-\xi)d\xi
+\frac{q}{b}\int_c^d v(t-g+\xi)d\xi\le
0\quad \text{for } t\ge t_3>t_2.
\end{equation}
It is clear that from  either  \eqref{e:6} or \eqref{e:8},
$v^{(n)}(t)<0$ for $t\ge t_3$. Therefore, by Lemma~\ref{lem2}
 $v^{(n-1)}(t)>0$ for $t\ge t_3$. Now, we want to show that
$v'(t)<0$ for  $t\ge t_3$. Suppose on the contrary $v'(t)>0$ for
$t\ge t_3$, then there exists a constant $k>0$ and $t_4\ge t_3$
such that
\[
v(t-g-\xi)\ge k,\quad \quad v(t-g+\xi)\ge k
\]
for $t\ge t_4$ and $\xi \in [c,d]$. Thus,
\[
v^{(n)}(t)\le -\frac{k(p+q)(d-c)}{b}\quad \text{for } t\ge t_4
\]
and
\[
v^{(n-1)}(t)\le v^{(n-1)}(t_4)-\frac{k(p+q)(d-c)(t-t_4)}{b}\to
-\infty\quad \text{as } t\to\infty,
\]
 which is a contradiction. Thus, $v'(t)<0$ and  therefore  $(-1)^iv^{(i)}(t)>0$ for $t\ge t_4$ and $i=0,1,\ldots,n$.
Then from \eqref{e:8}, we have
\begin{equation} \label{e:9}
v^{(n)}(t)+\frac{p(d-c)}{b}v(t-(g+c))\le 0,
\end{equation}
and
\begin{equation} \label{e:10}
v^{(n)}(t)+\frac{(p+q)(d-c)}{b}v(t-(g-d))\le 0,\quad t\ge t_4.
\end{equation}
Thus, from Lemma~\ref{lem1} $(ii)$ and condition \eqref{e:2},
\eqref{e:9} has no eventually positive solutions or from
Lemma~\ref{lem1} (ii) and condition \eqref{e:3}, \eqref{e:10} has
no eventually positive solutions, which is a contradiction.


\textbf{Case 2: $z(t)>0$ for $t\ge t_1$.} Let
\[
w(t)=z(t)+az(t-h)-bz(t+g),\quad t\ge t_1+h.
\]
Thus, one can show that
\begin{equation} \label{e:11}
w^{(n)}(t)=p\int_c^d z(t-\xi)d\xi+q\int_c^d z(t+\xi)d\xi,
\end{equation}
then
\begin{equation} \label{e:12}
[w(t)+aw(t-h)-bw(t+g)]^{(n)}=p\int_c^d w(t-\xi)d\xi
+q\int_c^d w(t+\xi)d\xi.
\end{equation}
Since $n$ is odd, by Lemma~\ref{lem2} $z'(t)>0$ for $t\ge t^*_2\ge
t_1+h$. From equation \eqref{e:11}, $w^{(n)}(t)>0$ and
$w^{(n+1)}(t)>0$ for $t\ge t^*_3\ge t^*_2$. Therefore,
$w^{(i)}(t)>0$ for $i=0,1\ldots,n+1$ and $t\ge t^*_3$. Using this
results and \eqref{e:12} we obtain
\[
(1+a)w^{(n)}(t)\ge p\int_c^d w(t-\xi)d\xi+q\int_c^d w(t+\xi)d\xi
\ge q\int_c^d w(t+\xi)d\xi
\]
and then
\[
w^{(n)}(t)\ge\frac{q(d-c)}{1+a} w(t+c),\quad t\ge t^*_3.
\]
This last equation does not have a positive solution by
Lemma~\ref{lem1} (i) and condition \eqref{e:4}. Therefore, it is
a contradiction, and the proof is complete.
\end{proof}

\begin{example} \label{exa1} \rm
 Consider the  neutral differential equation
\[
[x(t)+x(t-\pi)-x(t+\frac{9\pi}{2})]'''
=\frac{1}{2}\int_{\pi/2}^{3\pi}x(t-\xi)d\xi
+\frac{1}{2}\int_{\pi/2}^{3\pi}x(t+\xi)d\xi,
\]
so that $n=3$, $a=b=1$, $c=\frac{\pi}{2}$, $d=3\pi$, $p=q=\frac{1}{2}$,
$h=\pi$, $g=\frac{9\pi}{2}$.
 One can  verify that the conditions of Theorem~\ref{thm1} are
satisfied. We shall note that $x(t)=\cos t$ is a solution of this
problem.
\end{example}

\begin{theorem}\label{thm2}
Suppose $c>h$, $c>g$, $a>0$,
\begin{gather} \label{e:13}
\Big(\frac{p(d-c)}{a}\Big)^{1/n}\Big(\frac{c-h}{n}\Big)e>1,
\\ \label{e:14}
\Big(\frac{q(d-c)}{1+b}\Big)^{1/n}\Big(\frac{c-g}{n}\Big)e>1.
\end{gather}
Then
\begin{equation} \label{e:thm2}
[x(t)- ax(t-h)+ bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation}
is oscillatory.
\end{theorem}

\begin{proof}
Let $x(t)$ be  a non-oscillatory solution of \eqref{e:thm2}. Without
loss of generality we may  assume that $x(t)$ is eventually
positive; that is, there exists a $t_0\ge 0$ such that $x(t)>0$
for $t\ge t_0$. If $x(t)$ is eventually negative solution,
the proof follows the same arguments. Let
\[
z(t)=x(t)-ax(t-h)+bx(t+g),\quad t\ge t_0+h.
\]
As in the proof of the Theorem~\ref{thm1} the function $z^{(i)}(t)$
are of constant sign for $t\ge t_1\ge t_0+h$ and $i=0,1,\ldots,n$,
hence we have two possible cases  to consider for $z(t)$:
$z(t)<0$ for $t\ge t_1 $,  and $z(t)>0$ for $t\ge t_1$.

\textbf{Case 1: $z(t)<0$ for $t\ge t_1$.} Let
$v(t)=-z(t)$. Then we  obtain
\begin{equation} \label{e:15}
v^{(n)}(t)+p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi=0.
\end{equation}
On the other hand, since
\[
0<v(t)=-z(t)=-x(t)+ax(t-h)-bx(t+g)
\le ax(t-h)\quad \text{for } t\ge t_1,
\]
there is a $t_2\ge t_1$ such that
\begin{equation} \label{e:16}
x(t)\ge\frac {v(t+h)}{a}\quad \text{for } t\ge t_2.
\end{equation}
In view of \eqref{e:16} it follows from \eqref{e:15} that
\begin{equation} \label{e:17}
v^{(n)}(t)+\frac{p}{a}\int_c^d v(t+h-\xi)d\xi
+\frac{q}{a}\int_c^d v(t+h+\xi)d\xi\le
0\quad \text{for } t\ge t_3\ge t_2.
\end{equation}
As in the proof of the Theorem~\ref{thm1} (case 1) we show that
$(-1)^iv^{(i)}(t)>0$ for $t\ge t_4\ge t_3$ and $i=0,1,\ldots,n$,
and using this in \eqref{e:17} we see that
\begin{equation} \label{e:18}
v^{(n)}(t)+\frac{p(d-c)}{a}v(t-(c-h))\le 0\quad \text{for } t\ge
t_4.
\end{equation}
Thus, from Lemma~\ref{lem1} (ii) and condition \eqref{e:13},
\eqref{e:18} has no eventually positive solutions, which is a
contradiction.

\textbf{Case 2: $z(t)>0$ for $t\ge t_1$.} Let
\[
w(t)=z(t)-az(t-h)+bz(t+g)\,.
\]
Then one sees that
\begin{gather*}
w^{(n)}(t)=p\int_c^d z(t-\xi)d\xi+q\int_c^d z(t+\xi)d\xi,
\\
[w(t)-aw(t-h)+bw(t+g)]^{(n)}=p\int_c^d w(t-\xi)d\xi
+q\int_c^d w(t+\xi)d\xi.
\end{gather*}
As in the proof of the Theorem~\ref{thm1} (case 2), we have
$w^{(i)}(t)>0$ for $t\ge t^*_2\ge t_1$ and $i=0,1,\ldots,n+1$.
Then, we obtain
\[
(1+b)w^{(n)}(t+g)
\ge p\int_c^d w(t-\xi)d\xi+q\int_c^d w(t+\xi)d\xi
\ge q\int_c^d w(t+\xi)d\xi.
\]
Since $w'(t)>0$ for $t\ge t^*_2$,
\[
w^{(n)}(t)\ge\frac{q(d-c)}{1+b} w(t+(c-g)).
\]
The above equation does not have a positive solution by
Lemma~\ref{lem1} (i) and condition \eqref{e:14}. Thus, the proof
is complete.
\end{proof}

\begin{example} \label{exa2} \rm
Consider the  neutral differential equation
\[
[x(t)-x(t-\pi)+2x(t+\pi)]^{(5)}=\int_{2\pi}^{4\pi}x(t-\xi)d\xi+\frac{1}{2}\int_{2\pi}^{4\pi}x(t+\xi)d\xi,
\]
so that $n=5$, $a=1$, $b=2$, $c=2\pi$, $d=4\pi$, $p=1$,
$q=\frac{1}{2}$, $g=h=\pi$.
 One can  check that the conditions of Theorem~\ref{thm2} are
satisfied. By direct substitution it is easy to see that
$x(t)=t\cos t$ is a solution of this problem.
\end{example}


\begin{example} \label{exa3} \rm
Consider the  neutral differential equation
\[
[x(t)-x(t-\pi)+2x(t+\pi)]^{(9)}
=\frac{3}{4}\int_{6\pi}^{8\pi}x(t-\xi)d\xi
+\frac{3}{4}\int_{6\pi}^{8\pi}x(t+\xi)d\xi.
\]
 We see that $n=9$, $a=1$, $b=2$, $c=6\pi$, $d=8\pi$,
$p=q=\frac{3}{4}$, $g=h=\pi$.
 One can verify that the conditions of Theorem~\ref{thm2} are
satisfied. It is easy to show that $x(t)=t\sin t$ is a solution of
this problem.
\end{example}

Since the proofs of the following two theorems are similar to that
of Theorems \ref{thm1} and \ref{thm2}, they are omitted.


\begin{theorem}\label{thm3}
Suppose that $c>g$, $b>0$,
 \eqref{e:4} holds, and
\[
\Big(\frac{p(d-c)}{b}\Big)^{1/n}\Big(\frac{c-g}{n}\Big)e>1\,.
\]
 Then
\begin{equation*} 
[x(t)+ ax(t- h)- bx(t- g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}

\begin{theorem}\label{thm4}
Suppose that $c>h$, $b>0$, \eqref{e:2} or \eqref{e:3} hold,  and
\[
\Big(\frac{q(d-c)}{1+a}\Big)^{1/n}\Big(\frac{c-h}{n}\Big)e>1\,.
\]
 Then
\begin{equation*} \label{e:thm4}
[x(t)+ ax(t+ h)- bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}

\begin{theorem}\label{thm5}
Suppose  $c>g$,  and
\begin{equation} \label{e:19}
\Big(\frac{q(d-c)}{1+a+b}\Big)^{1/n}\Big(\frac{c-g}{n}\Big)e>1.
\end{equation}
Then
\begin{equation} \label{e:thm5}
[x(t)+ ax(t- h)+ bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation}
is oscillatory.
\end{theorem}

\begin{proof}
 Suppose there exist a nonoscillatory solution $x(t)$ of
\eqref{e:thm5}. Without loss of generality we may say that $x(t)>0$
for $t\ge t_0$. Let
\[
z(t)=x(t)+ax(t-h)+bx(t+g),\quad t\ge t_0+h.
\]
Clearly $z(t)>0$ for $t\ge t_0+h$. Thus, using \eqref{e:thm5}, we
get
\[
z^{(n)}(t)=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi
\]
for $t\ge t_1$ for some $t_1\ge t_0+h$. Therefore, we conclude
that $z^{(i)}(t)$, $i=0,1,\ldots,n$ are of constant sign, by
Lemma~\ref{lem2} $z(t)>0$ and $z'(t)>0$ on $[t_1,\infty)$. Let
\[
w(t)=z(t)+az(t-h)+bz(t+g),
\]
 then we show that
\begin{equation} \label{e:20}
w^{(n)}(t)=p\int_c^d z(t-\xi)d\xi+q\int_c^d z(t+\xi)d\xi
\end{equation}
and then
\begin{equation} \label{e:21}
[w(t)+aw(t-h)+bw(t+g)]^{(n)}
=p\int_c^d w(t-\xi)d\xi+q\int_c^d w(t+\xi)d\xi.
\end{equation}
Since $z(t)>0$ and $z'(t)>0$ are eventually increasing, from
\eqref{e:20} we see that $w^{(n)}(t)>0$ and $w^{(n+1)}(t)>0$ for
$t\ge t_2\ge t_1$. As a result of this $w^{(i)}(t)>0$ for
$i=0,1,\ldots,n+1$ and $t\ge t_2$. Thus from \eqref{e:21}, we have
\[
(1+a+b)w^{(n)}(t+g)\ge q \int_c^d w(t+\xi)d\xi,
\]
and then using the eventually increasing nature of  $w(t)$,  we obtain
\[
w^{(n)}(t+g)\ge \frac{q(d-c)}{1+a+b}w(t+c)
\]
or
\begin{equation} \label{e:22}
w^{(n)}(t)\ge\frac{q(d-c)}{1+a+b} w(t+(c-g),\quad t\ge t_3\ge t_2.
\end{equation}
 In view of Lemma~\ref{lem1}$(i)$  and \eqref{e:19}, the
inequality \eqref{e:22} has no eventually positive solutions,
which leads to a  contradiction. Thus, the proof is complete.
\end{proof}

\begin{example} \label{exa4} \rm
Consider the neutral differential equation
\[
[x(t)+x(t-\pi)+x(t+\frac{3\pi}{2})]'''
=\frac{1}{4}\int_{5\pi/2}^{7\pi/2}x(t-\xi)d\xi
+\frac{1}{4}\int_{5\pi/2}^{7\pi/2}x(t+\xi)d\xi,
\]
so that $n=3$, $a=b=1$, $c=\frac{5\pi}{2}$, $d=\frac{7\pi}{2}$,
$p=q=\frac{1}{4}$, $h=\pi$, $g=\frac{3\pi}{2}$.
 One can see that the conditions of Theorem~\ref{thm5} are
satisfied. In fact $x(t)=\sin t+\cos t$ is an oscillatory solution
of this problem.
\end{example}

The proofs of the following two theorems are similar to that of
Theorem~\ref{thm5} and  therefore omitted.

\begin{theorem}\label{thm6}
Suppose that $c>g>h$, and \eqref{e:19} holds. Then the equation
\begin{equation*}
[x(t)+ ax(t+ h)+ bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}

\begin{theorem}\label{thm7}
Suppose that
\[
\Big(\frac{q(d-c)}{1+a+b}\Big)^{1/n}\big(\frac{c}{n}\big)e>1.
\]
Then
\begin{equation*} 
[x(t)+ ax(t- h)+ bx(t- g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}

\begin{theorem}\label{thm8}
Suppose $a>0$, $c>h$,
\begin{gather} \label{e:23}
\Big(\frac{p(d-c)}{a+b}\Big)^{1/n}\Big(\frac{c-h}{n}\Big)e>1, \\
\label{e:24} \Big(q(d-c)\Big)^{1/n}\big(\frac{c}{n}\big)e>1.
\end{gather}
Then
\begin{equation} \label{e:thm8}
[x(t)- ax(t- h)- bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation}
is oscillatory.
\end{theorem}

\begin{proof}
Suppose that $x(t)$ is a non-oscillatory solution of \eqref{e:thm8}.
We may  assume that $x(t)$ is eventually positive, say $x(t)>0$
for $t\ge t_0$. Let
\begin{equation} \label{e:25}
z(t)=x(t)-ax(t-h)-bx(t+g),\quad t\ge t_0+h.
\end{equation}
 From \eqref{e:thm8}, we have
\begin{equation} \label{e:26}
z^{(n)}(t)=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi
\end{equation}
for $t\ge t_1$ for some $t_1\ge t_0+h$, implies that $z^{(i)}(t)$,
$i=0,1,\ldots,n$ are of constant sign on $[t_1,\infty)$. We have
two cases:  $z(t)>0$ for $t\ge t_1 $, and $z(t)<0$ for
$t\ge t_1$.

\textbf{Case 1: $z(t)>0$ for $t\ge t_1$.} From \eqref{e:25},
\begin{equation} \label{e:27}
x(t)\ge z(t).
\end{equation}
In view of  \eqref{e:26} and \eqref{e:27},  we have
\[
z^{(n)}(t)\ge q\int_c^d z(t+\xi)d\xi \quad \text{for }
t\ge t_1.
\]
As in the proof of Theorem~\ref{thm1}, $z'(t)$ is eventually
positive. Thus
\[
z^{(n)}(t)\ge q(d-c)z(t+c),
\]
which contradicts to Lemma~\ref{lem1} (i) and condition
\eqref{e:24}.

\textbf{Case 2: $z(t)<0$ for $t\ge t_1$.} Let
\[
0<v(t)=-z(t)=-x(t)+ax(t-h)+bx(t+g),
\]
then
\[
v^{(n)}(t)+p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi=0.
\]
Set
\[
w(t)=-v(t)+av(t-h)+bv(t+g)\,.
\]
Then
\begin{equation} \label{e:28}
w^{(n)}(t)+p\int_c^d v(t-\xi)d\xi+q\int_c^d v(t+\xi)d\xi=0
\end{equation}
and since the function satisfies \eqref{e:thm8}, we obtain
\[
[-w(t)+aw(t-h)+bw(t+g)]^{(n)}+p\int_c^d w(t-\xi)d\xi
+q\int_c^d w(t+\xi)d\xi=0.
\]
If $w(t)<0$ for $t\ge t_1$, we can handle as in case 1. Now
suppose $w(t)>0$ for $t\ge t_1$. On the other hand, $v'(t)<0$ for
$t\ge t_2\ge t_1$, otherwise from \eqref{e:28} we see that
$w^{(n)}(t)<0$ and $w^{(n+1)}(t)<0$ for $t\ge t_2$ which is a
contradiction. As a result of this,
\[
(-1)^i w^{(i)}(t)>0\quad \text{for } i=0,1,\ldots,n+1\quad
\text{and}\quad t\ge t_2,
\]
and then
\begin{gather*}
(a+b)w^{(n)}(t-h)+p\int_c^d w(t-\xi)d\xi\le 0,\\
w^{(n)}(t)+\frac{p(d-c)}{a+b} w(t-(c-h))\le 0,
\end{gather*}
which leads to a contradiction by condition \eqref{e:23} and
Lemma~\ref{lem1} $(ii)$. This completes the proof.
\end{proof}

\begin{example} \label{exa5} \rm
Consider the  equation
\[
[x(t)-\frac{3}{2}x(t-\frac{3\pi}{2})-\frac{4}{3}x(t+2\pi)]'''
=\frac{7}{12}\int_{2\pi}^{7\pi/2}x(t-\xi)d\xi
+\frac{11}{12}\int_{2\pi}^{7\pi/2}x(t+\xi)d\xi.
\]
We see that $n=3$, $a=\frac{3}{2}$, $b=\frac{4}{3}$, $c=2\pi$,
$d=\frac{7\pi}{2}$, $p=\frac{7}{12}$, $q=\frac{11}{12}$,
$h=\frac{3\pi}{2}$, $g=2\pi$.
Clearly  the conditions of Theorem~\ref{thm8} are
satisfied. In fact, $x(t)=\sin t$ is a solution of this problem.
\end{example}


The proofs of the following two theorems are similar to that of
Theorem~\ref{thm8}, hence the proofs are omitted.

\begin{theorem}\label{thm9}
Suppose $a>0$, $h>g$, and \eqref{e:23} and \eqref{e:24} hold. Then
\begin{equation*} 
[x(t)- ax(t- h)- bx(t- g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}

\begin{theorem}\label{thm10}
Suppose $b>0$, $h>g$, $\lambda=\mu=-1$, $\alpha=\beta=1$. In
addition, if \eqref{e:24} and
\[
\Big(\frac{p(d-c)}{a+b}\Big)^{1/n}\Big(\frac{c+g}{n}\Big)e>1,
\]
Then 
\begin{equation*} 
[x(t)- ax(t+ h)- bx(t+ g)]^{(n)}
=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,
\end{equation*}
is oscillatory.
\end{theorem}


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\end{document}