\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 $$\label{e:1} [x(t)\pm ax(t\pm h)\pm bx(t\pm g)]^{(n)} =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,$$ where $n$ is odd, $h$, $g$, $a$ and $b$ are nonnegative constants, $p$ and $q$ are positive constants, and $01$. 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 $$\label{e:2} \Big(\frac{p(d-c)}{b}\Big)^{1/n}\Big(\frac{g+c}{n}\Big)e>1,$$ or $$\label{e:3} \Big(\frac{(p+q)(d-c)}{b}\Big)^{1/n}\Big(\frac{g-d}{n}\Big)e>1,\quad g>d,$$ and $$\label{e:4} \Big(\frac{q(d-c)}{1+a}\Big)^{1/n}\big(\frac{c}{n}\big)e>1.$$ Then $$\label{e:thm1} [x(t)+ax(t-h)-bx(t+g)]^{(n)} =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,$$ 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 $$\label{e:5} z^{(n)}(t)=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi$$ for $t\ge t_1\ge t_0+h$, 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 $$\label{e:6} v^{(n)}(t)+p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi=0.$$ On the other hand, since \begin{equation*} 0t_2. 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 $$\label{e:9} v^{(n)}(t)+\frac{p(d-c)}{b}v(t-(g+c))\le 0,$$ and $$\label{e:10} v^{(n)}(t)+\frac{(p+q)(d-c)}{b}v(t-(g-d))\le 0,\quad t\ge t_4.$$ 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 $$\label{e:11} w^{(n)}(t)=p\int_c^d z(t-\xi)d\xi+q\int_c^d z(t+\xi)d\xi,$$ then $$\label{e:12} [w(t)+aw(t-h)-bw(t+g)]^{(n)}=p\int_c^d w(t-\xi)d\xi +q\int_c^d w(t+\xi)d\xi.$$ 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 $$\label{e:thm2} [x(t)- ax(t-h)+ bx(t+ g)]^{(n)} =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,$$ 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 $$\label{e:15} v^{(n)}(t)+p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi=0.$$ On the other hand, since $00 for t\ge t_4\ge t_3 and i=0,1,\ldots,n, and using this in \eqref{e:17} we see that $$\label{e:18} v^{(n)}(t)+\frac{p(d-c)}{a}v(t-(c-h))\le 0\quad \text{for } t\ge t_4.$$ 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 $$\label{e:19} \Big(\frac{q(d-c)}{1+a+b}\Big)^{1/n}\Big(\frac{c-g}{n}\Big)e>1.$$ Then $$\label{e:thm5} [x(t)+ ax(t- h)+ bx(t+ g)]^{(n)} =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,$$ 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 $$\label{e:20} w^{(n)}(t)=p\int_c^d z(t-\xi)d\xi+q\int_c^d z(t+\xi)d\xi$$ and then $$\label{e:21} [w(t)+aw(t-h)+bw(t+g)]^{(n)} =p\int_c^d w(t-\xi)d\xi+q\int_c^d w(t+\xi)d\xi.$$ 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 $$\label{e:22} w^{(n)}(t)\ge\frac{q(d-c)}{1+a+b} w(t+(c-g),\quad t\ge t_3\ge t_2.$$ 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 $$\label{e:thm8} [x(t)- ax(t- h)- bx(t+ g)]^{(n)} =p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi,$$ 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 $$\label{e:25} z(t)=x(t)-ax(t-h)-bx(t+g),\quad t\ge t_0+h.$$ From \eqref{e:thm8}, we have $$\label{e:26} z^{(n)}(t)=p\int_c^d x(t-\xi)d\xi+q\int_c^d x(t+\xi)d\xi$$ for $t\ge t_1$ for some $t_1\ge t_0+h$, 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}, $$\label{e:27} x(t)\ge z(t).$$ 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 $00 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} \begin{thebibliography}{00} \bibitem{Bai} D. D. Bainov and D. P. Mishev; Oscillation Theory for Neutral Differential Equations with Delay, {\emph Adam Hilger, Bristol}, 1991. \bibitem{Tun2} T. 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