\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 140, 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 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/140\hfil Periodic solutions] {Periodic solutions for $p$-Laplacian functional differential equations with two deviating arguments} \author[C. Song, X. Gao \hfil EJDE-2011/140\hfilneg] {Changxiu Song, Xuejun Gao} \address{Changxiu Song \newline School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, China} \email{scx168@sohu.com} \address{Xuejun Gao \newline School of Applied Mathematics, Guangdong University of Technology, Guangzhou 510006, China} \email{gaoxxj@163.com} \thanks{Submitted March 5, 2011. Published October 27, 2011.} \thanks{Supported by grants 10871052 and 109010600 NNSF of China, and by grant \hfill\break\indent 10151009001000032 from NSF of Guangdong} \subjclass[2000]{34B15} \keywords{$p$-Laplacian operator; periodic solutions; coincidence degree; \hfill\break\indent deviating arguments} \begin{abstract} Using the theory of coincidence degree, we prove the existence of periodic solutions for the $p$-Laplacian functional differential equations with deviating arguments. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In recent years, the existence of periodic solutions for the Duffing equation, Rayleigh equation and Li\'enard equation has received a lot of attention; see \cite{c1,d1,h1,l1,l2,l3,m1}. For example, Liu \cite{l1} studied periodic solutions for the $p$-Laplacian Li\'enard equation with a deviating argument. Using Mawhin's continuation theorem, some results on the existence of periodic solution are obtained. But the $p$-Laplacian Li\'enard equation with two deviating arguments has been studied far less often. In this article, we study the existence of periodic solutions for the following Li\'enard equation with two deviating arguments: $$(\phi_p(x'(t)))'+f(x(t))x'(t)+g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))=e(t),\label{e1.1}$$ where $f, \tau_1, \tau_2, e\in C(\mathbb{R},\mathbb{R})$; $g_1, g_2\in C(\mathbb{R}^2,\mathbb{R})$; $\tau_1(t), \tau_2(t), g_1(t,x), g_2(t,x), e(t)$ are periodic functions with period $T$; $\phi_p(\cdot )$ is the $p$-Laplacian operator, $10$ is a constant, $C_T=\{x\in C(\mathbb{R},\mathbb{R}): x(t+T)\equiv x(t)\}$ with the norm $\|x\|_\infty=\max_{t\in [0,T]}|x(t)|, X=Z=\{z=(x,y)\in C(\mathbb{R},\mathbb{R}^2): z(t)\equiv z(x+T)\}$ with the norm $\|z\|=\max\{\|x\|_\infty,\|y\|_\infty\}$. Clearly, $X$ and $Z$ are Banach spaces. Also let $L:\operatorname{Dom} L\subset X\to Z$ be defined by $(Lz)(t)=z'(t)=\begin{pmatrix}x'(t)\\ y'(t)\end{pmatrix},$ and $N:X\to Z$ defined by $(Nz)(t)=\begin{pmatrix} \phi_q(y(t))\\ -f(x(t))x'(t)-g_1(t,x(t-\tau_1(t)))-g_2(t,x(t-\tau_2(t)))+e(t) \end{pmatrix}$ It is easy to see that $\ker L=\mathbb{R}^2$, $\operatorname{Im}L =\{z\in Z:\int_0^Tz(s)ds=0\}$. So $L$ is a Fredholm operator with index zero. Let $P:X\to \ker L$ and $Q:Z\to \operatorname{Im} Q$ be defined by $Pu=\frac{1}{T}\int_0^Tu(s)ds,\quad u\in X;\quad Qv=\frac{1}{T}\int_0^Tv(s)ds,\quad v\in Z,$ and let $K_p$ denote the inverse of $L|_{\ker P\cap \operatorname{Dom}L}$. Obviously, $\ker L=\operatorname{Im}Q=\mathbb{R}^2$ and $$(K_pz)(t)=\int_0^tz(s)ds-\frac{1}{T}\int_0^T\int_0^tz(s)\,ds\,dt. \label{e2.2}$$ From this equality, one can easily see that $N$ is $L$-compact on $\overline{\Omega}$, where $\Omega$ is an open bounded subset of $X$. \begin{theorem} \label{thm2.1} Suppose that there exist constants $d>0$ $r_1\geq 0$ and $r_2\geq 0$ such that \begin{itemize} \item[(H1)] $g_1(t,u)+g_2(t,v)-e(t)>0$ for all $t\in \mathbb{R}$, $|\max\{u,v\}|>d$; \item[(H2)] $\lim_{x\to -\infty}\sup_{t\in [0,T]}\frac{|g_1(t,x)|}{|x|^{p-1}}\leq r_1$; $\lim_{x\to -\infty}\sup_{t\in [0,T]}\frac{|g_2(t,x))|}{|x|^{p-1}}\leq r_2$. \end{itemize} Then \eqref{e1.1} has at least one $T$-periodic solution, if $4(r_1+r_2)T (T/2)^{p/q}<1$. \end{theorem} \begin{proof} Consider the parametric equation $$(Lz)(t)=\lambda (Nz)(t),\quad \lambda\in (0,1). \label{e2.3}$$ Let $z(t)=\begin{pmatrix}x(t)\\ y(t)\end{pmatrix}$ be a possible $T$-periodic solution of \eqref{e2.3} for some $\lambda\in (0,1)$. One can see $x=x(t)$ is a $T$-periodic solution of the equation $$(\phi_p(x'(t)))'+\lambda^{p-1}f(x(t))x'(t)+\lambda^pg_1 (t,x(t-\tau_1(t)))+\lambda^pg_2(t,x(t-\tau_2(t))) =\lambda^pe(t).\label{e2.4}$$ Integrating both sides of \eqref{e2.4} over $[0,T]$, we have $$\int_0^T [g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt=0, \label{e2.5}$$ which implies that there exists $\eta\in [0,T]$ such that $g_1(\eta,x(\eta-\tau_1(\eta)))+g_2(\eta,x(\eta-\tau_2(\eta))) -e(\eta)=0.$ From assumption (H1), we know that there exists $\xi\in \mathbb{R}$ such that $|x(\xi)|\leq d$. Let $\xi=kT+t_0$, where $t_0\in [0,T]$ and $k$ is an integer. Let $\chi (t)=x(t+t_0)-x(t_0)$. Then $\chi (0)=\chi (T)=0$ and $\chi\in W^{1,p}([0,T],\mathbb{R})$. By Lemma \ref{lem1.1}, we have $\|x\|_\infty\leq \|\chi\|\infty+d\leq (\frac{T}{2})^{1/q}\|\chi'\|_p+d=(\frac{T}{2})^{1/q}\|x'\|_p+d.$ On the other hand, multiplying the two sides of \eqref{e2.4} by $x(t)$ and integrating them over $[0,T]$, we obtain $-\|x'\|_p^p=-\lambda^p\int_o^Tx(t)[g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))-e(t)]dt;$ i.e., $$\|x'\|_p^p\leq \|x\|_\infty \int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt. \label{e2.6}$$ From assumption (H2), there exists a constant $\rho>0$ such that $$|g_1(t,x)|\leq r_1|x|^{p-1},\quad |g_2(t,x)|\leq r_2|x|^{p-1},\quad \forall t\in \mathbb{R},\; x<-\rho. \label{e2.7}$$ Let \begin{gather*} E_1=\{t\in[0,T]:\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}<-\rho\},\\ E_2=\{t\in[0,T]:\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}>\rho\},\\ E_3=\{t\in [0,T]:|\max\{x(t-\tau_1(t)),x(t-\tau_2(t))\}|\leq\rho\},\\ E_4=\{t\in [0,T]:-\rho\leq x(t-\tau_1(t))\leq \rho, -\rho\leq x(t-\tau_2(t))\leq \rho\},\\ E_5=\{t\in[0,T]:x(t-\tau_1(t))<- \rho,-\rho\leq x(t-\tau_2(t)) \leq \rho\}, \\ E_6=\{t\in [0,T]:-\rho\leq x(t-\tau_1(t))\leq \rho, x(t-\tau_2(t))<- \rho\}. \end{gather*} By \eqref{e2.4} it is easy to see that $\int_0^T[g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt=0.$ Hence \begin{align*} &\int_{E_2}|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\ &=\int_{E_2}[g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)]dt\\ &= -\Big(\int_{E_1}+\int_{E_3}\Big)[g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))-e(t)]dt\\ &\leq \Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))-e(t)|dt. \end{align*} % {e2.8} From the above inequality and \eqref{e2.7} we obtain \begin{align*} &\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\ &\leq 2\Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))-e(t)|dt\\ &\leq 2\Big(\int_{E_1}+\int_{E_3}\Big)|g_1(t,x(t-\tau_1(t)))|dt +2\Big(\int_{E_1}+\int_{E_3}\Big)|g_2(t,x(t-\tau_2(t)))|dt\\ &\quad +2\int_0^T|e(t)|dt \\ &\leq 2r_1\int_{E_1}|x(t-\tau_1(t)|^{p-1}dt +2r_2\int_{E_1}|x(t-\tau_2(t)|^{p-1}dt\\ &\quad +2\int_{E_3}(|g_1(t,x(t-\tau_1(t)))| +|g_2(t,x(t-\tau_2(t)))|)dt+2\int_0^T|e(t)|dt\\ &\leq 2(r_1+r_2)T\|x\|_\infty^{p-1} +2\Big(\int_{E_4}+\int_{E_5}+\int_{E_6}\Big) \Big(|g_1(t,x(t-\tau_1(t)))|\\ &\quad +|g_2(t,x(t-\tau_2(t)))|\Big)dt +2\int_0^T|e(t)|dt\\ &\leq 2(r_1+r_2)T\|x\|_\infty^{p-1}+2T(g_{1\rho}+g_{2\rho}) +2r_1\int_{E_5}|x(t-\tau_1(t)|^{p-1}dt\\ &\quad+2Tg_{2\rho}+2r_2\int_{E_6}|x(t-\tau_2(t)|^{p-1}dt +2Tg_{1\rho}+2\int_0^T|e(t)|dt\\ &\leq 4(r_1+r_2)T\|x\|_\infty^{p-1} +4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt, \end{align*} % {e2.9} where $g_{1\rho}=\max_{t\in [0,T], |x|\leq \rho}|g_1(t,x(t-\tau_1(t)))|, \quad g_{2\rho}=\max_{t\in [0,T],|x|\leq \rho}|g_2(t,x(t-\tau_2(t)))|.$ From \eqref{e2.6} and the above inequality, we have \begin{align*} \|x'\|_p^p &\leq \|x\|_\infty \int_0^T|g_1(t,x(t-\tau_1(t))) +g_2(t,x(t-\tau_2(t)))-e(t)|dt\\ &\leq \|x\|_\infty[4(r_1+r_2)T\|x\|_\infty^{p-1} +4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt]\\ &=4(r_1+r_2)T \Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big)^p +[4T(g_{1\rho}+g_{2\rho})\\ &\quad +2\int_0^T|e(t)|dt]\Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big). \end{align*} Case (1): $\|x'(t)\|=0$, from \eqref{e2.6} we see $\|x\|_\infty\leq d$. Case (2): $\|x'(t)\|>0$, then we know that $[(\frac{T}{2})^{1/q}\|x'\|_p+d]^p =(\frac{T}{2})^{p/q}\|x'\|^p_p[1+(\frac{T}{2})^{-1/q} \frac{d}{\|x'(t)\|_p}]^p.$ From mathematical analysis, there is a constant $\delta>0$ such that $$(1+x)^p<1+(1+p)x, \quad \forall x\in [0,\delta].\label{e2.10}$$ If $(\frac{T}{2})^{-1/q}\frac{d}{\|x'(t)\|_p}>\delta$, then we have $\|x'\|_p<(\frac{T}{2})^{-1/q}\frac{d}{\delta}$. If $(\frac{T}{2})^{-1/q}\frac{d}{\|x'(t)\|_p}\leq\delta$, by \eqref{e2.10} we know that $$[(\frac{T}{2})^{1/q}\|x'\|_p+d]^p\leq (\frac{T}{2})^{p/q}\|x'(t)\|_p^p+(p+1)(\frac{T}{2}) ^{p-1)/q}d\|x'(t)\|^{p-1}_p.\label{e2.11}$$ By \eqref{e2.11}, we obtain \begin{align*} \|x'\|_p^p &\leq 4(r_1+r_2)T (\frac{T}{2})^{p/q}\|x'(t)\|_p^p+(p+1)(\frac{T}{2})^{p-1)/q} d\|x'(t)\|^{p-1}_p\\ &\quad +(4T(g_{1\rho}+g_{2\rho})+2\int_0^T|e(t)|dt) \Big((\frac{T}{2})^{1/q}\|x'\|_p+d\Big). \end{align*} As $p>1$, $4(r_1+r_2)T (\frac{T}{2})^{p/q}<1$, there exists a constant $R_2>0$ such that $\|x'\|_p\leq R_2$. Let $R_1=\max\{(\frac{T}{2})^{-1/q}\frac{d}{\delta},R_2\}$. Then we have $\|x\|_\infty\leq (\frac{T}{2})^{1/q}R_1:=R_0$. By the second equation of \eqref{e2.1} we obtain $y'(t)=-f(x(t))x'(t)-\lambda g_1(t,x(t-\tau_1(t)))-\lambda g_2(t,x(t-\tau_2(t)))+\lambda e(t).$ Hence \begin{align*} \int_0^T |y'(t)|dt &\leq f_{R_0}\int_0^T|x'(t)|dt+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt\\ &\leq f_{R_0}T^{1/q}\|x'\|_p+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt\\ &\leq f_{R_0}T^{1/q}R_1+Tg_{1R_0}+Tg_{2R_0}+\int_0^T|e(t)|dt:=R_3, \end{align*} where $f_{R_0}=\max_{|s|\leq R_0}|f(s)|,\quad g_{1R_0}=\max_{t\in [0,T],s\leq R_0}|g_1(t,s)|,\quad g_{2R_0}=\max_{t\in [0,T],s\leq R_0}|g_2(t,s)|.$ By the first equation of \eqref{e2.1} we have $\int_0^T\phi_q(y(t))dt=0$, which implies there exists a constant $t_1\in [0,T]$ such that $y(t_1)=0$. So $|y(t)|=|\int_{t_1}^ty'(s)ds|\leq \int_0^T|y'(s)|ds\leq R_3,$ and $\|y\|_\infty\leq R_3$. Let $R_4> \max\{R_0,R_3\}$, $\Omega=\{z\in Z: \|z\|d$. But when $|x|=R_4$, we know that $-g_1(t,x)-g_2(t,x)+e(t)<0$, which yields a contradiction. So conditions (1) and (2) of Lemma \ref{lem1.2} is satisfied. Define the isomorphism $J:\operatorname{Im} Q\to \ker L$ as follows: $J(x,y)^T=(-y,x)^T.$ Let $H(\mu, z)=\mu x+(1-\mu)JQNz, (\mu,z)\in [0,1]\times \Omega$, then we have $H(\mu,z)=\begin{pmatrix} \mu x+(1-\mu)\frac{1}{T}\int_0^T[g_1(t,x)+g_2(t,x)-e(t)]dt\\ \mu y+(1-\mu)\phi_q (y)\end{pmatrix},$ where $(\mu,z)\in [0,1]\times (\ker L\cap \partial\Omega)$. It is obvious that $H(\mu,z)=\mu x+(1-\mu) JQNz\neq 0$ for $(\mu,z)\in [0,1]\times (\ker L\cap \partial\Omega)$. Hence $\deg \{JQN,\Omega\cap \ker L,0\}=\deg \{I,\Omega\cap \ker L,0\} =1\neq 0.$ So the condition (3) of Lemma \ref{lem1.2} is satisfied. By applying Lemma \ref{lem1.2}, we conclude that equation $Lz=Nz$ has a solution $z(t)=(x(t),y(t))^T$; i.e., \eqref{e1.1} has a $T$-periodic solution $x(t)$. \end{proof} \begin{theorem} \label{thm2.2} Suppose that there exist constants $d>0$, $r_1\geq 0$, and $r_2\geq 0$ such that {\rm (H1)} holds and \begin{itemize} \item[(H2*)] $\lim_{x\to +\infty}\sup_{t\in [0,T]}\frac{|g_1(t,x)|}{|x|^{p-1}}\leq r_1$; $\lim_{x\to +\infty}\sup_{t\in [0,T]}\frac{|g_2(t,x))|}{|x|^{p-1}}\leq r_2$. \end{itemize} Then \eqref{e1.1} has at least one $T$-periodic solution, if $4(r_1+r_2)T (\frac{T}{2})^{p/q}<1$. \end{theorem} \begin{theorem} \label{thm2.3} Suppose that $p> 2$ and there exist constants $d>0$, $b_1\geq 0$, $b_2\geq 0$ such that {\rm (H1)} holds and \begin{itemize} \item[(H3)] $|g_i(t,u)-g_i(t,v)|\leq b_i|u-v|$ for all $t, u, v\in \mathbb{R}$, $i=1,2$. \end{itemize} Then \eqref{e1.1} has at least one $T$-periodic solution. \end{theorem} \begin{proof} By the proof of Theorem \ref{thm2.1}, we have $\|x\|_\infty\leq \|\chi\|_\infty+d\leq (\frac{T}{2})^{1/q}\|\chi'\|_p+d=(\frac{T}{2})^{1/q}\|x'\|_p+d,$ and $\|x'\|_p^p\leq \|x\|_\infty \int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt.$ From assumption (H3), we have \begin{align*} &\int_0^T|g_1(t,x(t-\tau_1(t)))+g_2(t,x(t-\tau_2(t)))-e(t)|dt\\ &\leq \int_0^T|g_1(t,x(t-\tau_1(t)))-g_1(t,0)|dt +\int_0^T|g_1(t,0)|dt\\ &\quad +|g_2(t,x(t-\tau_2(t)))-g_1(t,0)|dt+\int_0^T|g_2(t,0)|dt+\int_0^T|e(t)|dt\\ &\leq b_1\int_0^T|x(t-\tau_1(t))|dt+\int_0^T|g_1(t,0)|dt+b_2\int_0^T|x(t-\tau_2(t))|dt\\ &\quad +\int_0^T|g_2(t,0)|dt+\int_0^T|e(t)|dt\\ &\leq T(b_1+b_2)\|x\|_\infty+Tb \end{align*} where $b=\max\{|g_1(t,0)|+|g_2(t,0)|+|e(t)|\}$. Thus, \begin{align*} \|x'\|_p^p &\leq \|x\|_\infty[T(b_1+b_2)\|x\|_\infty+Tb]\\ & \leq T(b_1+b_2)[(\frac{T}{2})^{1/q}\|x'\|_p+d]^2 +Tb(\frac{T}{2})^{1/q}\|x'\|_p+Tbd. \end{align*} As $p>2$, there exists a constant $R_2>0$ such that $\|x'\|_p\leq R_2$. The rest of the proof is same to Theorem \ref{thm2.1} and is omitted. \end{proof} \begin{corollary} \label{coro2.1} Suppose that $p=2$ and conditions {\rm (H1), (H3)} hold. Then \eqref{e1.1} has at least one $T$-periodic solution, if $T(b_1+b_2)(\frac{T}{2})^\frac{2}{q}<1$. \end{corollary} \begin{remark} \label{rmk2.1} \rm If condition (H1) is replaced by \begin{itemize} \item[(H1*)] $g_1(t,u)+g_2(t,v)-e(t)<0$ for all $t\in \mathbb{R}$, $|\max\{u,v\}|>d$, \end{itemize} then the results in this article still hold. \end{remark} \begin{example} \label{exmp2.1} \rm Consider the equation $$(\phi_3(x'(t)))'+e^{x(t)}x'(t)+g_1(t,x(t-\sin t))+g_2(t,x(t-\cos t))=\frac{1}{\pi}\sin t, \label{e2.12}$$ where $p=3$, $T=2\pi$, $\tau_1(t)=\sin t$, $\tau_2(t)=\cos t$, \begin{gather*} g_1(t,x)=\begin{cases} e^{\sin^2t}x^3+\frac{1}{\pi}\sin t, & x\geq 0,\\[3PT] \frac{x^2}{18e\pi^3}e^{\sin^2t}+\frac{1}{\pi}\sin t,& x<0, \end{cases} \\ g_2(t,x)=\begin{cases} e^{\cos^2t}x^3, & x\geq 0,\\[3PT] \frac{x^2}{18e\pi^3}e^{\cos^2t},& x<0. \end{cases} \end{gather*} By \eqref{e2.12}, we can get $d=1/10$ (Actually, $d$ can be an arbitrarily small positive), $r_1=r_2=1/(18\pi^3)$, $4(r_1+r_2)T (T/2)^{p/q}<1$ and check that (H1)--(H2) hold. 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