\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 177, pp. 1--8.\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-2010/177\hfil Li\'enard type equation] {Li\'enard type p-Laplacian neutral Rayleigh equation with a deviating argument} \author[A. Anane, O. Chakrone, L. Moutaouekkil\hfil EJDE-2010/177\hfilneg] {Aomar Anane, Omar Chakrone, Loubna Moutaouekkil} % in alphabetical order \address{Universit\'e Mohamed I, Facult\'e des Sciences, D\'epartement de Math\'ematiques et Informatique, Oujda, Maroc} \email[Aomar Anane]{anane@sciences.univ-oujda.ac.ma} \email[Omar Chakrone]{chakrone@yahoo.fr} \email[Loubna Moutaouekkil]{loubna\_anits@yahoo.fr} \thanks{Submitted September 15, 2010. Published December 22, 2010.} \subjclass[2000]{34C25, 34B15} \keywords{Periodic solution; neutral Rayleigh equation; Li\'enard equation; \hfill\break\indent Deviating argument; p-Laplacian; Man\'asevich-Mawhin continuation} \begin{abstract} Based on Man\'asevich-Mawhin continuation theorem, we prove the existence of periodic solutions for Li\'enard type $p$-Laplacian neutral Rayleigh equations with a deviating argument, $$(\phi_p(x(t)-c x(t-\sigma))')'+f(x(t))x'(t)+ g(t,x(t-\tau(t)))=e(t).$$ An example is provided to illustrate our results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} The existence of periodic solutions for Li\'enard type $p$-Laplacian equation with a deviating argument $$\label{e1.1} (\phi_p(x'(t)))'+f(x(t))x'(t)+ g(t,x(t-\tau(t)))=e(t)$$ has been studied using the coincidence degree theory \cite{l1}. Zhu and Lu \cite{z2}, studied the existence of periodic solution for $p$-Laplacian neutral functional differential equation with a deviating argument when $p>2$ $$\label{e1.2} (\phi_p(x(t)-c x(t-\sigma))')'+g(t,x(t-\tau(t)))=e(t).$$ They obtained some results by transforming \eqref{e1.2} into a two-dimensional system to which Mawhin's continuation theorem was applied. Peng \cite{p1} discussed the existence of periodic solution for $p$-Laplacian neutral Rayleigh equation with a deviating argument $$\label{e1.3} (\phi_p(x(t)-c x(t-\sigma))')'+f(x'(t))+g(t,x(t-\tau(t)))=e(t)$$ and obtained the existence of periodic solutions under the assumption $f(0)=0$ and $\int_0^T e(t)dt=0$. Throughout this paper, $20$ is fixed. $g$ is continuous function defined on $\mathbb{R}^2$ and $T$-periodic in the first argument, $c$ and $\sigma$ are constants such that $|c|\neq1$. \section{Preliminaries} Let $\mathcal{C}_T=\{x\in\mathcal{C}(\mathbb{R},\mathbb{R}): x(t+T)=x(t)\}$ and $\mathcal{C}_T^{1}=\{x\in\mathcal{C}^{1}(\mathbb{R},\mathbb{R}): x(t+T)=x(t)\}$. $\mathcal{C}_T$ is a Banach space endowed with the norm $\|x\|_{\infty}=\max|x(t)|_{t\in[0,T]}$. $\mathcal{C}_T^{1}$ is a Banach space endowed with the norm $\|x\|=\max\{\|x\|_{\infty},\|x'\|_{\infty}\}$. In what follows, we will use $\|.\|_p$ to denote the $L^{P}$-norm. We also define a linear operator $A:\mathcal{C}_T\to \mathcal{C}_T$, $$(Ax)(t)=x(t)-cx(t-\sigma).$$ \begin{lemma}[\cite{l2,z1}] \label{lem2.1} If $|c|\neq1$, then $A$ has continuous bounded inverse on $\mathcal{C}_T$, and \begin{itemize} \item[(1)] $\|A^{-1}x\|_\infty\leq\frac{ \|x\|_\infty}{ |1-|c||}$, for all $x\in\mathcal{C}_T$; \item[(2)] $(A^{-1}x)(t)= \begin{cases} \sum_{j\geq0}c^{j}x(t-j\sigma), &|c|<1\\ -\sum_{j\geq1}c^{-j}x(t+j\sigma), &|c|>1. \end{cases} %E4$ \item[(3)] $$\int_0^T|(A^{-1}x)(t)|dt\leq\frac{1}{|1-|c||}\int_0^T|x(t)|dt, \quad \forall x\in\mathcal{C}_T.$$ \end{itemize} \end{lemma} \begin{lemma}[\cite{p1}]\label{lem2.2} If $|c|\neq1$ and $p>1$, then $$\label{e2.1} \int_0^T|(A^{-1}x)(t)|^pdt\leq\frac{ 1}{ |1-|c||^p}\int_0^T|x(t)|^pdt,\quad \forall x\in\mathcal{C}_T.$$ \end{lemma} For the $T$-periodic boundary value problem $$\label{e2.2} (\phi_p(x'(t)))'=\widetilde{f}(t,x,x'),\quad x(0)=x(T),\quad x'(0)=x'(T),$$ where $\widetilde{f}\in\mathcal{C}(\mathbb{R}^{3},\mathbb{R})$, we have the following result. \begin{lemma}[\cite{m1}] \label{lem2.3} Let $\Omega$ be an open bounded set in $\mathcal{C}_T^{1}$, and let the following conditions hold: \begin{itemize} \item [(i)] For each $\lambda\in(0,1)$, the problem $$(\phi_p(x'(t)))'=\lambda\widetilde{f}(t,x,x'),\quad x(0)=x(T),\quad x'(0)=x'(T)$$ has no solution on $\partial\Omega$. \item [(ii)] The equation $$F(a)=\frac{1}{T}\int^T_0\widetilde{f}(t,a,0)dt=0$$ has no solution on $\partial\Omega\cap\mathbb{R}$. \item [(iii)] The Brouwer degree of $F$, $\deg(F,\Omega\cap\mathbb{R},0)\neq0$. \end{itemize} Then the $T$-periodic boundary value problem \eqref{e2.2} has at least one periodic solution on $\overline{\Omega}$. \end{lemma} \section{Main results} \begin{theorem}\label{thm3.1} Suppose that $p>2$ and there exist constants $r_1\geq0$, $r_2\geq 0$, $d>0$ and $k>0$ such that \begin{itemize} \item [(A1)] $|f(x)|\leq k+r_1|x|^{p-2}$ for $x\in\mathbb{R}$; \item [(A2)] $x[g(t,x)-e(t)]<0$ for $|x|>d$ and $t\in\mathbb{R}$; \item [(A3)] $\lim_{x\to -\infty}\frac{ |g(t,x)-e(t)|}{ |x|^{p-1}}=r_2$. \end{itemize} Then \eqref{e1.4} has at least one $T$-periodic solution if $$\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p.$$ \end{theorem} \begin{proof} Consider the homotopic equation of \eqref{e1.4} as follows: $$\label{e3.1} (\phi_p(x(t)-c x(t-\sigma))')'+\lambda f(x(t))x'(t)+\lambda g(t,x(t-\tau(t)))=\lambda e(t),\quad \lambda\in(0,1).$$ We claim that the set of all possible periodic solution of \eqref{e3.1} are bounded in $\mathcal{C}_T^{1}$. Let $x(t)\in \mathcal{C}_T^{1}$ be an arbitrary solution of \eqref{e3.1} with period $T$. By integrating two sides of \eqref{e3.1} over $[0,T]$, and noticing that $x'(0)=x'(T)$, we have $$\label{e3.2} \int _0^T[g(t,x(t-\tau(t)))- e(t)]dt=0.$$ By the integral mean value theorem, there is a constant $\xi\in[0,T]$ such that $g(\xi,x(\xi-\tau(\xi)))- e(\xi)=0$. So from assumption (A2), we can get $|x(\xi-\tau(\xi))|\leq d$. Let $\xi-\tau(\xi)=mT+\overline{\xi}$, where $\overline{\xi}\in[0,T]$, and $m$ is an integer. Then, we have $$|x(t)|=|x(\overline{\xi})+\int_{\overline{\xi}}^{t}x'(s)ds|\leq d+\int_{\overline{\xi}}^{t}|x'(s)|ds,\quad t\in[\overline{\xi},\overline{\xi}+T],$$ and $$|x(t)|=|x(t-T)|=|x(\overline{\xi})-\int_{t-T}^{\overline{\xi}}x'(s)ds|\leq d+\int_{t-T}^{\overline{\xi}}|x'(s)|ds,\quad t\in[\overline{\xi},\overline{\xi}+T].$$ Combining the above two inequalities, we obtain \label{e3.3} \begin{aligned} \|x\|_{\infty} &=\max_{t\in[0,T]}|x(t)| =\max_{t\in[\overline{\xi},\overline{\xi}+T]}|x(t)|\\ &\leq\max_{t\in[\overline{\xi},\overline{\xi}+T]} \Big\{d+\frac{1}{2}\Big(\int_{\overline{\xi}}^{t} |x'(s)|ds+\int_{t-T}^{\overline{\xi}}|x'(s)|ds\Big)\Big\}\\ &\leq d+\frac{1}{2}\int_0^T|x'(s)|ds. \end{aligned} In view of $\frac{1}{ 2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p$, there exist a constant $\varepsilon>0$ such that $$\frac{1}{ 2^{p-1}}(1+|c|)T^{p-1}(r_1+T(r_2+\varepsilon))<|1-|c||^p.$$ From assumption (A3), there exist a constant $\rho>d$ such that $$\label{e3.4} |g(t,x(t-\tau(t)))- e(t)|dt\leq(r_2+\varepsilon)|x|^{p-1}\;\;\;for\; t\in\mathbb{R}\;\;and\; \;x<-\rho.$$ Denote $E_1=\{t\in[0,T],x(t-\tau(t))\leq-\rho\}$, $E_2=\{t\in[0,T],|x(t-\tau(t))|<\rho\}$, $E_3=\{t\in[0,T],x(t-\tau(t))\geq\rho\}$. By \eqref{e3.2}, it is easy to see that $$\label{e3.5} \Big(\int_{E_1}+\int_{E_2}+\int_{E_3}\Big)[g(t,x(t-\tau(t)))- e(t)]dt=0.$$ Hence \label{e3.6} \begin{aligned} \int_{E_3}|g(t,x(t-\tau(t)))- e(t)|dt &=-\int_{E_3}[g(t,x(t-\tau(t)))- e(t)]dt\\& =\Big(\int_{E_1}+\int_{E_2}\Big)[g(t,x(t-\tau(t)))-e(t)]dt\\ & \leq \Big(\int_{E_1}+\int_{E_2}\Big)|g(t,x(t-\tau(t)))- e(t)|dt. \end{aligned} Therefore, by \eqref{e3.4} and \eqref{e3.6}, we obtain \label{e3.7} \begin{aligned} \int_0^T|g(t,x(t-\tau(t)))-e(t)|dt &=\Big(\int_{E_1}+\int_{E_2}+\int_{E_3}\Big)|g(t,x(t-\tau(t)))- e(t)|dt\\ &\leq 2\Big(\int_{E_1}+\int_{E_2}\Big)|g(t,x(t-\tau(t)))- e(t)|dt\\ &\leq 2 \int_{E_1}(r_2+\varepsilon)|x(t-\tau(t))|^{p-1}dt +2\widetilde{g}_{\rho}T\\ & \leq 2 (r_2+\varepsilon)T \|x\|_{\infty}^{p-1}+2\widetilde{g}_{\rho}T. \end{aligned} Where $\widetilde{g}_{\rho}=\max_{t\in E_2}|g(t,x(t-\tau(t)))- e(t)|$. Multiplying both sides of \eqref{e3.1} by $(Ax)(t)=x(t)-cx(t-\sigma)$ and integrating them over $[0,T]$, we have \label{e3.8} \begin{aligned} \|Ax'\|_p^p & =\lambda\int_0^T (Ax)(t)\left[f(x(t))x'(t)+ g(t,x(t-\tau(t)))- e(t)\right]dt\\ & \leq (1+|c|)\|x\|_{\infty}\int_0^T\left [|f(x(t))x'(t)| + |g(t,x(t-\tau(t)))- e(t)|\right]dt. \end{aligned} From assumption (A1), we obtain. $$\label{e3.9} \int_0^T|f(x(t))x'(t)|dt \leq k\int_0^T|x'(t)|dt+ r_1 \int_0^T|x'(t)||x(t)|^{p-2}dt.$$ Using H\"older inequality, and substituting \eqref{e3.3} into \eqref{e3.9}, we obtain $$\label{e3.10} \int_0^T|f(x(t))x'(t)|dt \leq kT^{1/q}\|x'\|_p+r_1T^{1/q}\|x'\|_p \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-2}.$$ From \eqref{e3.3} and \eqref{e3.7}, we have $$\label{e3.11} \int_0^T|g(t,x(t-\tau(t)))- e(t)|dt \leq 2\;\widetilde{g}_{\rho}\; T + 2(r_2+\varepsilon)T \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1}.$$ Substituting \eqref{e3.10}, \eqref{e3.11} and \eqref{e3.3} into \eqref{e3.8}, we obtain \label{e3.12} \begin{aligned} & \|Ax'\|_p^p\\ &\leq (1+|c|)\Big[ kT^{1/q}\|x'\|_p \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)\\ &\quad +\Big(d+\frac{1}{2} \int_0^T|x'(t)|dt\Big)^{p-1}r_1T^{1/q}\|x'\|_p\\ &\quad +2(r_2+\varepsilon)T \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^p +2\widetilde{g}_{\rho} T \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)\Big]. \end{aligned} Case(1). If $\int_0^T|x'(t)|dt=0$, from \eqref{e3.3}, we have $\|x\|_{\infty}0$, then $$\label{e3.13} \Big(d+\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1} =\Big(\frac{1}{2}\int_0^T|x'(t)|dt\Big)^{p-1} \Big(1+\frac{2d}{\int_0^T|x'(t)|dt}\Big)^{p-1}.$$ By elementary analysis, there is a constant $\delta>0$ such that $$\label{e3.14} (1+u)^{p-1}\leq1+pu,\quad \forall u \in [0,\delta].$$ If $2d / \int_0^T|x'(t)|dt>\delta$, then $\int_0^T|x'(t)|dt<2d/\delta$, so from \eqref{e3.3}, we have $\|x\|_{\infty}0$ such that $$\label{e3.16} (1+u)^p\leq1+(1+p)u,\quad \forall u \in [0,\delta']$$ If $2d/ \int_0^T|x'(t)|dt>\delta'$, then $\int_0^T|x'(t)|dt<2d/ \delta'$, so from \eqref{e3.3}, we have $\|x\|_{\infty}2$ and $\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p$, there exists a constant $R_3>0$ such that $$\label{e3.20} \|x'\|_p\leq R_3.$$ Which together with \eqref{e3.3} implies that there is a positive number $R_4$ such that $$\label{e3.21} \|x\|_{\infty}\leq R_4.$$ From \eqref{e3.1}, we have \label{e3.22} \begin{aligned} &\int_0^T|(\phi_p(A x')(t))'|dt\\ &\leq \int_0^T[|f(x(t))x'(t)|+|g(t,x(t-\tau(t)))+|e(t)|]dt\\&\leq k T ^{1/q}\|x'\|_p+\int_0^Tr_1|x|^{p-2}|x'(t)|+Tg_{R_4}+\int_0^T|e(t)|dt\\&\leq k T ^{1/q}\|x'\|_p+r_1\|x\|_{\infty}^{p-2}T ^{1/q}\|x'\|_p+Tg_{R_4}+\int_0^T|e(t)|dt\\&\leq kT ^{1/q}R_3+r_1R_4^{p-2}T ^{1/q}R_3+Tg_{R_4}+\int_0^T|e(t)|dt=R_5, \end{aligned} where $g_{R_4}=\max_{|x|\leq R_4, t\in[0,T]}|g(t,x(t-\tau(t)))|$. As $(Ax)(0)=(Ax)(T)$, there exists $t_0\in ]0,T[$ such that $(Ax')(t_0)=0$, while $\phi_p(0)=0$ we see $\phi_p(A x')(t_0)=0$. Thus, for any $t\in [0,T]$, we have $$|\phi_p(A x')(t))|=|\int_{t_0}^{t}\phi_p(A x')(s))ds|\leq \int_0^T|(\phi_p(A x')(s))'|dt\leq R_5.$$ From which, it follows that $$\label{e3.23} \|Ax'\|_{\infty}\leq R_5^{q-1}.$$ From Lemma \ref{lem2.1}, we derive $$\label{e3.24} \|x'\|_{\infty}=\|A^{-1}Ax'\|_{\infty} \leq\frac{\|Ax'\|_{\infty}}{|1-|c||} \leq\frac{R_5^{q-1}}{|1-|c||}=R_6.$$ Now, let $y(t)=(Ax)(t)$, we can see that \eqref{e3.1} is equivalent to the equation $$\label{e3.25} (\phi_p(y'(t)))'+\lambda f((A^{-1}y)(t))(A^{-1}y')(t)+\lambda g(t,(A^{-1}y)(t-\tau(t)))=\lambda e(t).$$ So, if $y$ is an periodic solution of \eqref{e3.25}, then $x=A^{-1}y$ is $T$-periodic solution of \eqref{e3.1}. Let $R_7=2(1+|c|)\max\{R_4,R_6,d\}$, $\Omega=\{y\in \mathcal{C}_T^{1}:\|y\|R_4$, from \eqref{e3.21} which is a contradiction. Similarly, $\|y'\|_{\infty}=R_7$ is also impossible. If $y\in\partial\Omega\cap\mathbb{R}$, then $y$ is a constant and $|y|=R_7$, $x=A^{-1}y=\frac{y}{1-c}$, $|x|\geq2\max\{R_4,R_6,d\}$. Let $$F(y)=\frac{1}{T}\int_0^T[e(t)-f((A^{-1}y)(t))(A^{-1}y')(t) -g(t,(A^{-1}y)(t-\tau(t)))].$$ Then $F(y)=\frac{1}{T}\int_0^T[e(t)-g(t,\frac{y}{1-c})]dt$ for $y\in\partial\Omega\cap\mathbb{R}$. From (A2), we know that $F(y)\neq0$ on $\partial\Omega\cap\mathbb{R}$, so condition (ii) in Lemma \ref{lem2.3} is satisfied. Define $$H(y,\mu)=\mu(A^{-1}y)+(1-\mu)F(y),$$ $y\in\partial\Omega\cap\mathbb{R}$, $\mu\in[0,1]$. Then $$(-A^{-1}y)H(y,\mu)=-\mu(A^{-1}y)^2-(1-\mu)(A^{-1}y) \frac{1}{T}\int_0^T[e(t)-g(t,(A^{-1}y)(t-\tau(t)))]dt.$$ From (A2) we obtain $(A^{-1}y)H(y,\mu)>0$. Thus $H(y,\mu)$ is a homotopic transformation and $\deg[F,\Omega\cap\mathbb{R},0]=\deg[A^{-1}y,\Omega\cap\mathbb{R},0] \neq 0$. So, for \eqref{e3.25}, all of conditions of Lemma \ref{lem2.3} are satisfied. Applying Lemma \ref{lem2.3}, we conclude that $$\label{e3.26} (\phi_p(y'(t)))'+ f((A^{-1}y)(t))(A^{-1}y')(t)+ g(t,(A^{-1}y)(t-\tau(t)))= e(t)$$ has at least one $T$-periodic solution $\overline{y}$. Therefore, $\overline{x}=A^{-1}\overline{y}$ is an $T$-periodic solution of \eqref{e1.4}. \end{proof} Similarly, we can prove the following Theorem. \begin{theorem}\label{thm3.2} Suppose that $p>2$ and that there exist constants $r_1\geq0$, $r_2\geq0$, $d>0$ and $k>0$ such that \begin{itemize} \item [(A1)] $|f(x)|\leq k+r_1|x|^{p-2}$ for $x\in\mathbb{R}$; \item [(A2)] $x[g(t,x)-e(t)]<0$ for $|x|>d$ and $t\in\mathbb{R}$; \item [(A3)] $\lim_{x\to +\infty} \frac{|g(t,x)-e(t)|}{|x|^{p-1}}=r_2$. \end{itemize} then \eqref{e1.4} has at least one $T$-periodic solution if $$\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2)<|1-|c||^p.$$ \end{theorem} \section{Example} In this section, we illustrate Theorem \ref{thm3.1} with the following example. Consider the equation $$\label{e4.1} (\phi_3(x(t)-5 x(t-\pi))')'+f(x(t))x'(t)+ g(t,x(t-\sin(t)))=e^{\cos^2t},$$ where $p=3$, $c=5$, $\sigma=4$, $\mathrm{T}=2\pi$, $\tau(t)=\sin t$, $e(t)=e^{\cos^2t}$, $f(x)=2+\frac{\sqrt{|x|}}{\pi^2}$, $$g(t,x) = \begin{cases} -xe^{\sin^2t}, & x\geq 0 \\ \frac{ x^2}{18\pi^2}, & x<0 . \end{cases}$$ Let $d=3\pi\sqrt{2e}$, $r_1=\frac{1}{\pi^2}$, $r_2=\frac{1}{18\pi^2}$, $k=4+\frac{\max_{|x|\leq1}|f(x)|}{\pi^2}$. We can easily check the condition (A1), (A2) and (A3) of Theorem \ref{thm3.1} hold. Furthermore, $$\frac{1}{2^{p-1}}(1+|c|)T^{p-1}(r_1+Tr_2) =6+\frac{2\pi}{3}<|1-|c||^p=64.$$ By Theorem \ref{thm3.1}, \eqref{e4.1} has at least one $2\pi$-periodic solution. \begin{thebibliography}{0} \bibitem{l1} B. Liu; \emph{Periodic solutions for Li\'enard type $p$-Laplacian equation with a deviating arguments}, Journal of Computational and Applied Mathematics 214(2008), 13-18. \bibitem{l2} S. P. Lu, J. Ren, W. 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