\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 66, pp. 1--19.\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-2011/66\hfil Positive periodic solutions] {Positive periodic solutions for third-order nonlinear differential equations} \author[J. Ren, S. Siegmund, Y. Chen\hfil EJDE-2011/66\hfilneg] {Jingli Ren, Stefan Siegmund, Yueli Chen} % in alphabetical order \address{Jingli Ren \newline Department of Mathematics, Zhengzhou University, Zhengzhou 450001, China.\newline Department of Mathematics, Dresden University of Technology, Dresden 01062, Germany} \email{renjl@zzu.edu.cn} \address{Stefan Siegmund \newline Department of Mathematics, Dresden University of Technology, Dresden 01062, Germany} \email{stefan.siegmund@tu-dresden.de} \address{Yueli Chen \newline Department of Mathematics, Zhengzhou University, Zhengzhou 450001, China} \thanks{Submitted April 19, 2011. Published May 18, 2011.} \thanks{Supported by AvH Foundation of Germany and by grant 10971202 from NSFC of China} \subjclass[2000]{34B18, 34B27, 35B10} \keywords{Green's function; positive periodic solution; \hfill\break\indent third-order differential equation} \begin{abstract} For several classes of third-order constant coefficient linear differential equations we obtain existence and uniqueness of periodic solutions utilizing explicit Green's functions. We discuss an iteration method for constant coefficient nonlinear differential equations and provide new conditions for the existence of periodic positive solutions for third-order time-varying nonlinear and neutral differential equations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Let $X=C[0,\omega]$ with norm $\|x\|=\max_{t\in[0,\omega]}|x(t)|$. We denote $C^{+}_{\omega}=\{u(t)\in X,~u(t)>0,~u(t+\omega)=u(t)\}$, and $C^{-}_{\omega}=\{u(t)\in X,~u(t)<0,~u(t+\omega)=u(t)\}$. We study the existence of positive periodic solutions for certain classes of third-order differential equations. Third-order differential equations arise in a variety of areas in agriculture, biology, economics and physics \cite{3,4,5,6,8} and attract a lot of attention of many researchers \cite{zengjidu,konglingbin,chujifeng,fengyuqiang,fengyuqiang1,liyongxiang,yuhengxue,guolijun,alex,sunyigang,muk,cabada,Hakl} and the reference therein. In the study of higher-order (in particular third-order) differential equations, the naive idea to translate the equation into a first order system of differential equations by defining $x_1=x,~ x_2=x',~x_3=x'', \dots$ (see \cite{lusiping,linxiaojie,panghuihui,liuyj}), works well for showing existence of periodic solutions, however, it does not obviously lead to existence proofs for positive periodic solutions, since the condition $x = x_1\geq 0$ of positivity for the higher order equation is different from the natural positivity condition $(x_1, x_2, \dots) \geq 0$ for the corresponding system. Another approach which is frequently used is to transform the third-order equation into a corresponding integral equation and to establish the existence of positive periodic solutions based on a fixed point theorem in cones. Following this path one needs an explicit representation of the Green's function for corresponding ordinary equation, see \cite{agarwal86,anderson03}. In \cite{agarwal86}, R. Agarwal gave the explicit Green's function for the $n$th-order and $2m$th-order differential equations. In addition, Anderson studied the Green's function for the third-order boundary value problem in \cite{anderson03}, \begin{gather*} u'''(t)=0,\quad t_1\leq t\leq t_3\\ u(t_1)=u'(t_2)=0,\quad \gamma u(t_3)+\delta u''(t_3)=0 \end{gather*} The singular nonlinear third-order periodic boundary value problem $${\label{111}} \begin{gathered} u'''(t)+\rho^3u(t)=f(t,u(t)),\quad t\in[0,2\pi],\\ u^{(i)}(0)=u^{(i)}(2\pi),\quad i=0,1,2, \end{gathered}$$ has been investigated in \cite{konglingbin,sunyigang,chujifeng}, where $\rho\in(0,1/\sqrt{3})$ is a constant, $f:(0,2\pi)\times(0,+\infty)\to\mathbb{R}^{+}$. By employing the Green's function for the equation \begin{gather*} u''-\rho u'+\rho^2u=0,\\ u(0)=u(2\pi),\quad u'(0)=u'(2\pi), \end{gather*} the existence and multiplicity of positive solutions of \eqref{111} were established. However, the direct Green's function of \eqref{111} was not constructed. Motivated by these excellent works, we give the explicit forms of the Green's functions for several differential third-order equations with the $\omega$-periodic boundary value conditions and then provide sufficient conditions for the existence of positive periodic solutions. This article is divided into six parts. In order to get the main result, we first consider the above four types of third-order constant coefficient linear differential equations and present their Green's functions and properties for those equations in Section 2. In Section 3, by applying the Banach fixed-point theorem and the results of Section 2, we obtain existence and uniqueness of solutions and an iteration method for the following constant coefficient nonlinear differential equations $$\label{eq1} u'''-\rho^3u=f(t, u),$$ where $f\in C([0,\omega]\times\mathbb{R}, \mathbb{R})$. In Section 4, we study third-order time-varying nonlinear differential equations, \begin{gather}\label{eq1.01} u'''-a(t)u=f(t,u), \\ \label{eq1.02} u'''+a(t)u=f(t,u), \end{gather} $a>0$, $f\in C([0,\omega]\times [0,\infty), [0,\infty))$. We provide sufficient conditions for the existence of positive solutions for linear versions of equations \eqref{eq1.01} and \eqref{eq1.02}. In Section 5, we go one step further and discuss a more general third-order nonlinear differential equation $$\label{eq1.03} y'''+p(t)y''+q(t)y'+c(t)y=g(t, y).$$ Here $p, q, c \in C(\mathbb{R}, \mathbb{R})$, $g \in (\mathbb{R}\times[0, \infty), [0, \infty))$, and $g(t, y)>0$ for $y>0$; $p, q, c, g$ are $\omega$-periodic functions in $t$ for some period $\omega>0$. In Section 6 we study a neutral functional differential equation $$\label{eq1.1} (x(t)-cx(t-\tau(t)))'''+a(t)x(t)=f(t,x(t-\tau(t))),$$ and present an existence result for positive periodic solutions for this equation with an $\omega$-periodic function $\tau\in C(\mathbb{R}, \mathbb{R})$, constants $\omega$, $c$ with $|c| < 1$, $a\in C_{\omega}^{+}$, $f\in C(\mathbb{R}\times [0,\infty), [0,\infty))$ and $f(t,x)$ is $\omega$-periodic in $t$. \section{Green's Functions} \begin{theorem}\label{thm2.01} For $\rho>0$ and $h\in X$, the equation $$\label{eq2.01} \begin{gathered} u'''-\rho^3u=h(t),\\ u(0)=u(\omega),\quad u'(0)=u'(\omega),\quad u''(0)=u''(\omega) \end{gathered}$$ has a unique $\omega$-periodic solution which is of the form $$\label{eq2.02} u(t)=\int^{\omega}_0G_1(t, s)(-h(s))\mathrm{d}s,$$ where $$\label{eq2.03} G_1(t,s)=\begin{cases} \frac{2\exp(\frac{1}{2}\rho(s-t)) [\sin(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6}) -\exp(-\frac{1}{2}\rho\omega)\sin(\frac{\sqrt{3}}{2}\rho(t-s-\omega) +\frac{\pi}{6})]}{3\rho^2(1+\exp(-\rho\omega) -2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}}{2}\rho\omega))} \\ +\frac{\exp(\rho(t-s))}{3\rho^2(\exp(\rho\omega)-1)}, \quad \text{if }0\leq s\leq t\leq \omega \\[3pt] \frac{2\exp(\frac{1}{2}\rho(s-t-\omega))[\sin(\frac{\sqrt{3}}{2} \rho(t-s+\omega)+\frac{\pi}{6})-\exp(-\frac{1}{2}\rho\omega) \sin(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6})]} {3\rho^2(1+\exp(-\rho\omega)-2\exp(-\frac{\rho\omega}{2}) \cos(\frac{\sqrt{3}}{2}\rho\omega))} \\ +\frac{\exp(\rho(t+\omega-s))}{3\rho^2(\exp(\rho\omega)-1)}, \quad \text{if }0\leq t\leq s\leq \omega \end{cases}$$ \end{theorem} \begin{proof} It is easy to check that the associated homogeneous equation of \eqref{eq2.01} has the solution $v(t)=c_1\exp(\rho t)+\exp(-\frac{\rho t}{2})(c_2 \cos\frac{\sqrt{3}\rho}{2}t+c_3\sin\frac{\sqrt{3}\rho}{2}t)$. The only periodic solution of the associated homogeneous equation of \eqref{eq2.01} is the trivial solution; i.e., $c_1, c_2, c_3 = 0$. This follows by assuming that $v(t)$ is periodic; we immediately get that $c_1=0$ and by assuming that $c_2^2 + c_3^2 > 0$ and choosing $\varphi$ such that $\sin \varphi = \frac{c_2}{\sqrt{c_2^2 + c_3^2}}$, $\cos \varphi = \frac{c_3}{\sqrt{c_2^2 + c_3^2}}$, we obtain \begin{align*} \frac{v(t)}{\sqrt{c_2^2 + c_3^2}} &= \exp\big(-\frac{\rho t}{2}\big)\big(\sin \varphi \cos\frac{\sqrt{3}\rho}{2}t+ \cos \varphi \sin\frac{\sqrt{3}\rho}{2}t\big)\\ &= \exp\big(-\frac{\rho t}{2}\big) \sin \big(\varphi + \frac{\sqrt{3}\rho}{2}t \big) \end{align*} which for $t \to \infty$ contradicts periodicity of $v$, proving that $c_2 = c_3 = 0$. Applying the method of variation of parameters, we obtain \begin{gather*} c_1'(t)=\frac{\exp(-\rho t)}{3\rho^2}h(t), \quad c_2'(t)=\frac{\frac{\sqrt{3}}{3}\sin\frac{\sqrt{3}\rho t}{2}-\frac{1}{3}\cos\frac{\sqrt{3}\rho t}{2}}{\rho^2}\exp(\frac{\rho t}{2})h(t),\\ c_3'(t)=\frac{-\frac{1}{3}\sin\frac{\sqrt{3}\rho t}{2}-\frac{\sqrt{3}}{3}\cos\frac{\sqrt{3}\rho t}{2}}{\rho^2}\exp(\frac{\rho t}{2})h(t), \end{gather*} and then \begin{gather*} c_1(t)=c_1(0)+\int^{t}_0\frac{\exp(-\rho s)}{3\rho^2}h(s) \,\mathrm{d}s,\\ c_2(t)=c_2(0)+\int^{t}_0\frac{\frac{\sqrt{3}}{3} \sin\frac{\sqrt{3}\rho s}{2}-\frac{1}{3} \cos\frac{\sqrt{3}\rho s}{2}}{\rho^2} \exp(\frac{\rho s}{2})h(s)\,\mathrm{d}s,\\ c_3(t)=c_3(0)+\int^{t}_0\frac{-\frac{1}{3} \sin\frac{\sqrt{3}\rho s}{2}-\frac{\sqrt{3}}{3} \cos\frac{\sqrt{3}\rho s}{2}}{\rho^2} \exp(\frac{\rho s}{2})h(s)\,\mathrm{d}s. \\ \begin{split} u(t)&=c_1(t)\exp(\rho t)+\exp(-\frac{\rho t}{2})(c_2(t)\cos\frac{\sqrt{3}\rho}{2}t+c_3(t)\sin\frac{\sqrt{3}\rho}{2}t)\\ &=c_1(0)\exp(\rho t)+c_2\exp(-\frac{\rho t}{2})\cos(\frac{\sqrt{3}}{2}\rho t)+c_3(0)\exp(-\frac{\rho t}{2})\sin(\frac{\sqrt{3}}{2}\rho t)\\ &\quad+\int_0^{t}\frac{\exp(\rho(t-s))}{3\rho^2}h(s)\, \mathrm{d}s+\int_0^{t}\frac{\sin(\frac{\sqrt{3}}{2}\rho(s-t) -\frac{\pi}{6})}{6\rho^2}\exp(\frac{\rho}{2}(s-t))h(s)\,\mathrm{d}s \end{split} \end{gather*} Noting that $u(0)=u(\omega)$, $u'(0)=u'(\omega)$, $u''(0)=u''(\omega)$, we obtain \begin{gather*} c_1(0)=\int_0^{\omega}\frac{\exp(\rho(\omega-s))} {3\rho^2(1-\exp(\rho\omega))}h(s)\,\mathrm{d}s,\\ c_2(0)=\int_0^{\omega}\frac{2\exp(\frac{\rho(s-\omega)}{2}) [\exp(-\frac{\rho\omega}{2})\sin(\frac{\pi}{6} -\frac{\sqrt{3}\rho s}{2})-\sin(\frac{\pi}{6} -\frac{\sqrt{3}\rho(s-\omega)}{2})]}{3\rho^2 (\exp(-\rho\omega)-2\exp(-\frac{\rho\omega}{2}) \cos\frac{\sqrt{3}\rho\omega}{2}+1)}h(s)\,\mathrm{d}s,\\ c_3(0)=\int_0^{\omega}\frac{2\exp(\frac{\rho(s-\omega)}{2}) [\exp(-\frac{\rho\omega}{2})\cos(\frac{\pi}{6} -\frac{\sqrt{3}\rho s}{2})-\cos(\frac{\pi}{6} -\frac{\sqrt{3}\rho(s-\omega)}{2})]}{3\rho^2 (\exp(-\rho\omega)-2\exp(-\frac{\rho\omega}{2}) \cos\frac{\sqrt{3}\rho\omega}{2}+1)}h(s)\,\mathrm{d}s, \end{gather*} Therefore, \begin{align*} u(t) &=c_1(t)\exp(\rho t)+\exp(-\frac{\rho t}{2})(c_2(t) \cos\frac{\sqrt{3}\rho}{2}t+c_3(t) \sin\frac{\sqrt{3}\rho}{2}t)\\ &=\int_0^{t}\Bigl\{2\exp(\frac{1}{2}\rho(s-t)) \big[\sin(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6}) -\exp(-\frac{1}{2}\rho\omega) \sin(\frac{\sqrt{3}}{2}\rho(t-s\\ &\quad -\omega) +\frac{\pi}{6})\big] \big/ \big[3\rho^2(1+\exp(-\rho\omega) -2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}}{2}\rho\omega))\big] \\ &\quad +\frac{\exp(\rho(t-s))}{3\rho^2(1-\exp(\rho\omega))} \Bigr\}h(s)\,\mathrm{d}s\\ &\quad +\int_{t}^{\omega}\Bigl\{2\exp(\frac{1}{2} \rho(s-t-\omega))[\sin(\frac{\sqrt{3}}{2}\rho(t-s+\omega) +\frac{\pi}{6})-\exp(-\frac{1}{2}\rho\omega)\\ &\quad\times \sin(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6})\big] \big/ \big[3\rho^2 (1+\exp(-\rho\omega)-2\exp(-\frac{\rho\omega}{2}) \cos(\frac{\sqrt{3}}{2}\rho\omega))\big] \\ &\quad +\frac{\exp(\rho(t+\omega-s))}{3\rho^2 (1-\exp(\rho\omega))}\Bigr\}h(s)\,\mathrm{d}s\\ &=\int_0^{\omega}G_1(t,s)h(s)\,\mathrm{d}s \end{align*} where $G_1(t, s)$ is defined as in \eqref{eq2.03}. By direct calculation, we obtain the solution $u$ satisfies the periodic boundary value condition of the problem \eqref{eq2.01}. \end{proof} Similarly, we have the following result. \begin{theorem}{\label{thm2.02}} For $\rho>0$ and $h\in X$ the equation $${\label{eq2.04}} \begin{gathered} u'''+\rho^3u=h(t),\\ u(0)=u(\omega),\quad u'(0)=u'(\omega),\quad u''(0)=u''(\omega) \end{gathered}$$ has a unique $\omega$-periodic solution $${\label{eq2.05}} u(t)=\int^{\omega}_0G_2(t, s)h(s)\mathrm{d}s,$$ where $${\label{eq2.06}} G_2(t,s)=\begin{cases} \frac{2\exp(\frac{1}{2}\rho(t-s))[\sin(\frac{\sqrt{3}}{2} \rho(t-s)-\frac{\pi}{6})-\exp(\frac{1}{2} \rho\omega)\sin(\frac{\sqrt{3}}{2}\rho(t-s-\omega)-\frac{\pi}{6})]} {3\rho^2(1+\exp(\rho\omega)-2\exp(\frac{1}{2}\rho\omega) \cos(\frac{\sqrt{3}}{2}\rho\omega))}\\ +\frac{\exp(\rho(s-t))}{3\rho^2(1-\exp(-\rho\omega))}, \quad \text{if }0\leq s\leq t\leq \omega\\[3pt] \frac{2\exp(\frac{1}{2}\rho(t+\omega-s)) [\sin(\frac{\sqrt{3}}{2}\rho(t+\omega-s)-\frac{\pi}{6}) -\exp(\frac{1}{2}\rho\omega)\sin(\frac{\sqrt{3}}{2}\rho(t-s) -\frac{\pi}{6})]}{3\rho^2(1+\exp(\rho\omega) -2\exp(\frac{1}{2}\rho\omega)\cos(\frac{\sqrt{3}}{2}\rho\omega))}\\ +\frac{\exp(\rho(s-t-\omega))}{3\rho^2(1-\exp(-\rho\omega))}, \quad \text{if }0\leq t\leq s\leq \omega \end{cases}$$ \end{theorem} \begin{theorem}\label{thm2.03} For $\rho>0$ and $h\in X$ the equation $${\label{eq2.07}} \begin{gathered} u'''-3\rho u''+3\rho^2u'-\rho^3u=h(t),\\ u(0)=u(\omega),\quad u'(0)=u'(\omega),\quad u''(0)=u''(\omega) \end{gathered}$$ has a unique $\omega$-periodic solution $${\label{eq2.08}} u(t)=\int^{\omega}_0G_3(t, s)(-h(s))\mathrm{d}s,$$ where $${\label{eq2.09}} G_3(t,s)=\begin{cases} \frac{[(s-t)\exp(\rho\omega)+\omega-s+t]^2+\omega^2 \exp(\rho\omega)}{2(\exp(\rho\omega)-1)^3}\exp(\rho(t+\omega-s)), & 0\leq t\leq s\leq \omega\\ \frac{[(s-t+\omega)\exp(\rho\omega)-s+t]^2 +\omega^2\exp(\rho\omega)}{2(\exp(\rho\omega)-1)^3} \exp(\rho(t-s)), & 0\leq s\leq t\leq \omega \end{cases}$$ \end{theorem} \begin{proof} In this case, the associated homogeneous equation of \eqref{eq2.07} has solutions $$u(t)=c_1\exp(\rho t)+c_2t\exp(\rho t)+c_3t^2\exp(\rho t).$$ Analogously, by applying the method of variation of parameters we obtain \begin{gather*} c'_1(t)=\frac{h(t)t^2}{2\exp(\rho t)},\quad c'_2(t)=\frac{-h(t)t}{\exp(\rho t)},\quad c'_3(t)=\frac{h(t)}{2\exp(\rho t)}, \\ c_1(t)=c_1(0)+\int_0^{t}\frac{h(s)s^2}{2\exp(\rho s)} \,\mathrm{d}s,\quad c_2(t)=c_2(0)+\int_0^{t}\frac{-h(s)s}{\exp(\rho s)}\,\mathrm{d}s,\\ c_3(t)=c_3(0)+\int_0^{t}\frac{h(s)}{2\exp(\rho s)}\,\mathrm{d}s. \end{gather*} Noting that $u(0)=u(\omega)$, $u'(0)=u'(\omega)$, $u''(0)=u''(\omega)$, we obtain \begin{gather*} c_1(0)=\int_0^{\omega}\frac{h(s)\exp(\rho(\omega-s)) [(s\exp(\rho\omega)+\omega-s)^2+\omega^2\exp(\rho \omega)]} {2(1-\exp(\rho\omega))^3}\,\mathrm{d}s,\\ c_2(0)=\int_0^{\omega}\frac{h(s)\exp(\rho(\omega-s)) (\omega-s+s\exp(\rho\omega))}{(1-\exp(\rho\omega))^2}\,\mathrm{d}s,\\ c_3(0)=\int_0^{\omega}\frac{h(s)\exp(\rho(\omega-s))}{2 (1-\exp(\rho\omega))}\,\mathrm{d}s. \end{gather*} Therefore, \begin{align*} u(t) &=c_1\exp(\rho t)+c_2t\exp(\rho t)+c_3t^2\exp(\rho t)\\ &=\int_0^{t}\frac{[(s-t+\omega)\exp(\rho\omega)-s+t]^2+\omega^2\exp(\rho\omega)}{2(\exp(\rho\omega)-1)^3}\exp(\rho(t-s))(-h(s))\,\mathrm{d}s\\ &\quad +\int_{t}^{\omega}\frac{[(s-t)\exp(\rho\omega)+\omega-s+t]^2 +\omega^2\exp(\rho\omega)}{2(\exp(\rho\omega)-1)^3}\\ &\quad \times \exp(\rho(t+\omega-s))(-h(s))\,\mathrm{d}s\\ &=\int_0^{\omega}G_3(t,s)(-h(s))\,\mathrm{d}s \end{align*} Where $G_3(t,s)$ is as defined in \eqref{eq2.09}. \end{proof} A dual version of Theorem \ref{thm2.03} which can be proved similarly. \begin{theorem} \label{thm2.04} For $\rho>0$ and $h\in X$ the equation $${\label{eq2.010}} \begin{gathered} u'''+3\rho u''+3\rho^2u'+\rho^3u=h(t),\\ u(0)=u(\omega),\quad u'(0)=u'(\omega),\quad u''(0)=u''(\omega) \end{gathered}$$ has a unique $\omega$-periodic solution $${\label{eq2.011}} u(t)=\int^{\omega}_0G_4(t, s)h(s)\mathrm{d}s,$$ where $${\label{eq2.012}} G_4(t,s)=\begin{cases} \frac{[(s-t)\exp(-\rho\omega)+\omega-s+t]^2+\omega^2 \exp(-\rho\omega)}{2(1-\exp(-\rho\omega))^3} \exp(-\rho(t+\omega-s)),\\ \quad\text{if } 0\leq t\leq s\leq \omega\\[3pt] \frac{[(s-t+\omega)\exp(-\rho\omega)-s+t]^2+\omega^2 \exp(-\rho\omega)}{2(1-\exp(-\rho\omega))^3}\exp(-\rho(t-s)),\\ \quad \text{if }0\leq s\leq t\leq \omega \end{cases}$$ \end{theorem} Now we present the properties of the Green's functions for \eqref{eq2.01}, \eqref{eq2.04}, \eqref{eq2.07}, \eqref{eq2.010}. For convenience we use the abbreviations \begin{gather*} A_1=\frac{1}{3\rho^2(\exp(\rho\omega)-1)},\quad B_1=\frac{3+2\exp(-\frac{\rho\omega}{2})}{3\rho^2 (1-\exp(-\frac{\rho\omega}{2}))^2},\\ A_2=\frac{\omega^2(1+\exp(\rho\omega))}{2(\exp(\rho\omega)-1)^3}, \quad B_2=\frac{\omega^2\exp(2\rho\omega)(1+\exp(\rho\omega))} {2(\exp(\rho\omega)-1)^3}. \end{gather*} \begin{theorem}{\label{thm2.05}} $\int^{\omega}_0G_1(t, s)\mathrm{d}s=1/\rho^3$ and if $\sqrt{3}\rho\omega<4\pi/3$ holds, then $00$ for $s\in[0, t]$ and $H^{*}_1(t,s)>0$ for $s\in [t,\omega]$ and $\exp(-\rho\omega)+1-2\exp(-\frac{\rho\omega}{2})\cos\frac{\sqrt{3}\rho\omega}{2}>[1-\exp(-\frac{\rho\omega}{2})]^2>0$. For convenience, we denote $\theta=\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6}$, \begin{align*} g_1(t,s)&= \sin\big(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6}\big) -\exp\big(-\frac{\rho\omega}{2}\big) \sin\big(\frac{\sqrt{3}}{2}\rho(t-s-\omega)+\frac{\pi}{6}\big)\\ &= \sin(\theta)-\exp(-\frac{\rho\omega}{2}) \sin(\theta-\frac{\sqrt{3}}{2}\rho\omega), \end{align*} \begin{align*} g^{*}_1(t,s) &= \sin\big(\frac{\sqrt{3}}{2}\rho(t-s+\omega)+\frac{\pi}{6}\big) -\exp(-\frac{\rho\omega}{2}) \sin\big(\frac{\sqrt{3}}{2}\rho(t-s)+\frac{\pi}{6}\big)\\ &= \sin(\theta+\frac{\sqrt{3}}{2}\rho\omega) -\exp(-\frac{\rho\omega}{2})\sin\theta. \end{align*} If $g_1(t,s)>0$ and $g^{*}_1(t,s)>0$, then obviously $H_2(t,s)>0$, $H^{*}_2(t,s)>0$ and $G_1(t,s)>0$. For $0\leq s\leq t\leq \omega$, since $\sqrt{3}\rho\omega<4\pi/3$, we have \begin{gather*} \frac{\pi}{6}\leq\theta\leq\frac{\sqrt{3}}{2}\rho\omega +\frac{\pi}{6}<\frac{5\pi}{6},\\ -\frac{\pi}{2}<\frac{\pi}{6}-\frac{\sqrt{3}}{2}\rho\omega\leq\theta -\frac{\sqrt{3}}{2}\rho\omega\leq\frac{\pi}{6}. \end{gather*} (i) For $-\frac{\pi}{2}<\theta-\frac{\sqrt{3}}{2}\rho\omega\leq 0$, then $\sin\theta>0$, $\sin(\theta-\frac{\sqrt{3}}{2}\rho\omega)<0$, we obtain $g_1(t,s)>0$ (ii) For $0<\theta-\frac{\sqrt{3}}{2}\rho\omega\leq \frac{\pi}{6}$, we have $\sin\theta>0$, $\sin(\theta-\frac{\sqrt{3}}{2}\rho\omega)>0$, and $$0<\frac{\sqrt{3}}{4}\rho\omega \leq \theta-\frac{\sqrt{3}}{4}\rho\omega \leq\frac{\pi}{6}+\frac{\sqrt{3}}{4}\rho\omega<\frac{\pi}{2}.$$ \begin{align*} g_1(t,s)&= \sin(\theta)-\exp(-\frac{\rho\omega}{2}) \sin(\theta-\frac{\sqrt{3}}{2}\rho\omega)\\ &\geq \sin\theta-\sin(\theta-\frac{\sqrt{3}}{2}\rho\omega)\\ &= \sin(\theta-\frac{\sqrt{3}}{4}\rho\omega +\frac{\sqrt{3}}{4}\rho\omega)-\sin(\theta-\frac{\sqrt{3}}{4} \rho\omega-\frac{\sqrt{3}}{4}\rho\omega)\\ &= 2\cos(\theta-\frac{\sqrt{3}}{4}\rho\omega) \sin(\frac{\sqrt{3}}{4}\rho\omega)>0 \end{align*} For $0\leq t\leq s\leq \omega$, \begin{gather*} -\frac{\pi}{2}<-\frac{\sqrt{3}}{2}\rho\omega+\frac{\pi}{6} \leq\theta\leq \frac{\pi}{6},\\ \frac{\pi}{6}\leq \theta+\frac{\sqrt{3}}{2}\rho\omega \leq \frac{\pi}{6}+\frac{\sqrt{3}}{2}\rho\omega<\frac{5}{6}\pi. \end{gather*} (i) For $-\frac{\pi}{2}<\theta\leq0$, we have $\sin\theta<0$, $\sin(\theta+\frac{\sqrt{3}}{2}\rho\omega)>0$, then $g^{*}_1(t,s)>0$. (ii)For $0<\theta\leq\frac{\pi}{6}$, we have $\sin\theta>0$, $\sin(\theta+\frac{\sqrt{3}}{2}\rho\omega)>0$, and $$0<\theta+\frac{\sqrt{3}}{4}\rho\omega<\frac{\pi}{2}$$ \begin{align*} g^{*}_1(t,s) &= \sin(\theta+\frac{\sqrt{3}}{2}\rho\omega) -\exp(-\frac{\rho\omega}{2})\sin\theta\\ &\geq \sin(\theta+\frac{\sqrt{3}}{2}\rho\omega)-\sin\theta\\ &= \sin(\theta+\frac{\sqrt{3}}{4}\rho\omega +\frac{\sqrt{3}}{4}\rho\omega)-\sin(\theta+\frac{\sqrt{3}}{4} \rho\omega-\frac{\sqrt{3}}{4}\rho\omega)\\ &= 2\cos(\theta+\frac{\sqrt{3}}{4}\rho\omega) \sin(\frac{\sqrt{3}}{4}\rho\omega) >0 \end{align*} If $\sqrt{3}\rho\omega<4\pi/$, we obtain $g_1(t,s)>0$ and $g^{*}_1(t,s)>0$, proving that $G(t,s)>0$ for all $t\in[0,\omega]$ and $s\in[0,\omega]$. Next we compute a lower and an upper bound for $G_1(t, s)$ for $s\in[0, \omega]$. We have \begin{align*} A_1&=\frac{1}{3\rho^2(\exp(\rho\omega)-1)}\leq\frac{\exp(\rho (t+\omega-s)}{3\rho^2(\exp(\rho\omega)-1)}0, \quad 0\leq t\leq s\leq\omega, \\ \frac{\partial H^{*}_3(t, s)}{\partial s}=2[(s-t+\omega)\exp(\rho\omega)-s+t](\exp(\rho\omega)-1)>0,\quad 0\leq s\leq t\leq \omega, \end{gather*} the functions $s \mapsto H_3(t,s)$ and $s \mapsto H^{*}_3(t,s)$ are increasing. Recalling that \begin{gather*} H_3(t,t)=\omega^2+\omega^2\exp(\rho\omega) =\omega^2(1+\exp(\rho\omega)),\\ H_3(t, \omega)=[(\omega-t)\exp(\rho\omega)+t]^2 +\omega^2\exp(\rho\omega)\leq \omega^2\exp(2\rho\omega)+\omega^2\exp(\rho\omega),\\ \omega^2+\omega^2\exp(\rho\omega)\leq H^{*}_3(t,0) =[(\omega-t)\exp(\rho\omega)+t]^2+\omega^2\exp(\rho\omega),\\ H^{*}(t,t)=\omega^2\exp(2\rho\omega)+\omega^2\exp(\rho\omega). \end{gather*} and \begin{gather*} 00$for$u>0$. We introduce the following abbreviations \begin{gather*} a^{*} = \max\{a(t): t\in [0,\omega]\},\quad a_{*} =\min\{a(t): t\in [0,\omega]\},\quad rho=\sqrt[3]{a^{*}},\\ \overline{f}_0=\lim_{x\to0^{+}}\sup_{t\in[0, \omega]}\frac{f(t,x)}{x},\quad \underline{f}_{\infty}=\lim_{x\to\infty}\inf_{t\in[0, \omega]}\frac{f(t, x)}{x}. \end{gather*} % Let$X$be defined as in the beginning of Section 1. Moreover, define a cone$K_0$in$X$by$K_0=\{x\in X: x(t)>\theta\|x\|\}$, where$0<\theta=\frac{a_{*}A_1}{a^{*}B_1}<1$and for$r>0$define$K_{0r}=\{x\in K_0: \|x\|0$for$-h(t)>0$and$\|B_1\|\leq a^{*}-a_{*}$. By Theorem \ref{thm2.01}, the solution of \eqref{eq4.03} can be written in the form $${\label{eq4.05}} u(t)=(T_1h)(t)+(T_1B_1u)(t).$$ And for$\|T_1B_1\|\leq\|T_1\|\|B_1\|\leq\frac{1}{a^{*}}(a^{*}-a_{*})<1$, we have $${\label{eq4.06}} u(t)=(I-T_1B_1)^{-1}(T_1h)(t).$$ Define an operator$P_1: X\to X$by $${\label{eq4.07}} (P_1h)(t)=(I-T_1B_1)^{-1}(T_1h)(t).$$ Obviously, for any$h\in C_{\omega}^{-}$,$u(t)=(P_1h)(t)$is the unique positive solution of \eqref{eq4.03}. \begin{lemma} \label{lem3.1} If$\sqrt{3}\rho\omega<4\pi/3$holds, then$P_1$is completely continuous and $${\label{eq3.6}} (T_1h)(t)\leq (P_1h)(t)\leq\frac{a^{*}}{a_{*}}\|(T_1h)(t)\|,\quad \text{for all } h\in C_{\omega}^{-}.$$ \end{lemma} \begin{proof} Since$\|T_1B_1\|\leq\|T_1\|\|B_1\|\leq1-\frac{a_{*}}{a^{*}}<1$, by a Neumann expansion of$P_1$, we have $${\label{eq3.7}} \begin{split} P_1&=(I-T_1B_1)^{-1}T_1 \\ &= (I+T_1B_1+(T_1B_1)^2+\dots+(T_1B_1)^{n}+\dots)T_1\\ &=T_1+T_1B_1T_1+(T_1B_1)^2T_1+\dots+(T_1B_1)^{n}T_1+\dots \end{split}$$ By \eqref{eq3.7}, and recalling that$\|T_1B_1\|\leq1-\frac{a_{*}}{a^{*}}$and$(T_1h)(t)>0$, we obtain $${\label{eqp1}} (T_1h)(t)\leq(P_1h)(t)\leq\frac{a^{*}}{a_{*}}\|(T_1h)(t)\|, \quad h\in C^{+}_{\omega}.$$ From \cite{Guo da jun}, we obtain that$T_1$is completely continuous and$P_1$is also completely continuous. \end{proof} Define an operator$Q_1: X\to X$by $${\label{eq4.010}} Q_1u(t)=P_1(-f(t, u)).$$ From the continuity of$P_1$, it is easy to verify that$Q_1$is completely continuous in$X$. Comparing \eqref{eq4.01} with \eqref{eq4.03}, we see that the existence of solutions for equation \eqref{eq4.01} is equivalent to the existence of fixed-points for the equation$u = Q_1u$. \begin{lemma} \label{lem3.02}$Q_1(K_0)\subset K_0$. \end{lemma} \begin{proof} It is easy to verify that$Q_1u(t+\omega)=Q_1u(t)$. For$u\in K_0, we have from Lemma \ref{lem3.1} that \begin{align*} Q_1u(t)&=P_1(-f(t, u))\geq T_1(-f(t, u))\\ &=\int^{\omega}_0G_1(t,s)f(s, u(s))\mathrm{d}s > A_1\int^{\omega}_0f(s, u(s))\mathrm{d}s, \end{align*} on the other hand \begin{align*} Q_1u(t)&=P_1(-f(t, u))\leq\frac{a^{*}}{a_{*}}\|T_1(-f(t,u))\|\\ &=\frac{a^{*}}{a_{*}}\max_{t\in[0, \omega]}\int^{\omega}_0G_1(t,s)f(s,u(s))\mathrm{d}s \leq\frac{a^{*}B_1}{a_{*}}\int^{\omega}_0f(s, u(s))\mathrm{d}s. \end{align*} Therefore, $$Q_1u(t)>\frac{a_{*}A_1}{a^{*}B_1}\|Q_1u\|=\theta\|Q_1u\|;$$ i.e.,Q_1(K_0)\subset K_0$. \end{proof} Next, we study the following equation corresponding to \eqref{eq4.02}. $${\label{eq4.011}} u'''+a(t)u=h(t),\quad h\in C_{\omega}^{+}$$ Define operators$T_2, B_2: X\to X$by $${\label{eq4.012}} (T_2h)(t)=\int^{\omega}_0G_2(t,s)h(s)\mathrm{d}s,\quad (B_2u)(t)=(a^{*}-a(t))u(t).$$ If$\sqrt{3}\rho\omega<4\pi/3$holds,$T_2, B_2$are completely continuous,$(T_2h)(t)>0$for$h(t)>0$and$\|B_2\|\leq a^{*}-a_{*}$. By Theorem \ref{thm2.02}, the solution of \eqref{eq4.011} can be written in the form $${\label{eq4.013}} u(t)=(T_2h)(t)+(T_2B_2u)(t).$$ And for$\|T_2B_2\|\leq\|T_2\|\|B_2\|\leq\frac{1}{a^{*}}(a^{*}-a_{*})<1$, we have $${\label{eq4.014}} u(t)=(I-T_2B_2)^{-1}(T_2h)(t).$$ Define an operator$P_2: X\to X$by $${\label{eq4.015}} (P_2h)(t)=(I-T_2B_2)^{-1}(T_2h)(t).$$ Obviously, for any$h\in C_{\omega}^{+}$,$u(t)=(P_2h)(t)$is the unique positive solution of \eqref{eq4.011}. Similar to Lemma \ref{lem3.1}, we can prove the following result. \begin{lemma} \label{lem4.03} If$\sqrt{3}\rho\omega<4\pi/3$holds, then$P_2$is completely continuous and $${\label{eq4.016}} (T_2h)(t)\leq (P_2h)(t)\leq\frac{a^{*}}{a_{*}}(T_2h)(t),\quad \text{for all } h\in C_{\omega}^{+}.$$ \end{lemma} Define an operator$Q_2: X\to X$by $${\label{eq4.018}} Q_2u(t)=P_2(f(t, u)).$$ Clearly,$Q_2$is completely continuous in$X$. Comparing \eqref{eq4.02} with \eqref{eq4.011}, it is clear that the existence of solutions for equation \eqref{eq4.02} is equivalent to the existence of fixed-points for the equation$u = Q_2 u$. \begin{lemma} \label{lem4.04}$Q_2(K_0)\subset K_0$. \end{lemma} \begin{proof} From the definition of$Q_2$, it is easy to verify that$Q_2u(t+\omega)=Q_2u(t)$. For$u\in K_0, we have from Lemma \ref{lem4.03} that $$Q_2u(t)=P_2(f(t, u))\geq T_2(f(t, u))=\int^{\omega}_0G_2(t,s)f(s, u(s))\mathrm{d}s> A_1\int^{\omega}_0 \! f(s, u(s))\mathrm{d}s,$$ on the other hand \begin{align*} Q_2u(t)&=P_2(f(t, u))\leq\frac{a^{*}}{a_{*}}\|T_2f(t,u)\|\\ &=\frac{a^{*}}{a_{*}}\max_{t\in[0, \omega]}\int^{\omega}_0G_2(t,s)f(s, u(s))\mathrm{d}s\leq\frac{a^{*}B_1}{a_{*}}\int^{\omega}_0f(s, u(s))\mathrm{d}s. \end{align*} Therefore, $$Q_2u(t)>\frac{a_{*}A_1}{a^{*}B_1}\|Q_2u\|=\theta\|Q_2u\|;$$ i.e.,Q_2(K_0)\subset K_0$. \end{proof} \begin{lemma}[\cite{guolaks}] \label{lem4.05} Let$E$be a Banach space and$K$a cone in$E$. For$r>0$, define$K_{r}=\{u\in K:\|u\|0$satisfies $$\frac{a^{*}B_1}{a_{*}}\varepsilon\omega<1.$$ By recalling the proof of Lemma \ref{lem3.1}, we obtain that $$\|Q_1u\|\leq\frac{a^{*}B_1}{a_{*}}\int^{\omega}_0f(s, u(s))\mathrm{d}s \leq\frac{a^{*}B_1}{a_{*}}\varepsilon\int^{\omega}_0u(s)\mathrm{d}s\\ \leq\frac{a^{*}B_1}{a_{*}}\varepsilon\omega\|u\|<\|u\|$$ for$u\in\partial K_{0r_1}$,$t\in[0, \omega]$. On the other hand, if$\underline{f}_{\infty}=\infty$, there exists a constant$\tilde{H}>r_1$such that$f(t, u)\geq\eta u$for$u\geq \tilde{H}, t\in[0, \omega]$, where the constant$\eta>0$satisfies$A_1\eta\omega\theta>1$, and again from the proof of Lemma \ref{lem3.1}, one can easily see that $$Q_1u> A_1\int^{\omega}_0f(s, u(s))\mathrm{d}s> A_1\eta\int^{\omega}_0u(s)\mathrm{d}s> A_1\eta\omega\theta\|u\|>\|u\|$$ for$u\in\partial K_{0\tilde{H}}$,$t\in[0, \omega]$. % By Lemma \ref{lem4.05}, we know that$i(Q_1, K_{0r_1}, K_0)=1, i(Q_1, K_{0\tilde{H}}, K_0)=0$, i.e.$i(Q_1, K_{0\tilde{H}}\backslash \bar{K}_{0r_1}, K_0)=-1$, and$Q_1$has a fixed point in$K_{0\tilde{H}}\backslash\bar{K}_{0r_1}$. Consequently, \eqref{eq4.01} has a positive$\omega$-periodic solution for$r_10$for$u>0$. \end{proof} By Theorem \ref{thm5.01}, under the assumption that$\frac{1}{3}p^2(t)+p'(t)=q(t)$, we know that if$u$is a positive solution for \eqref{eq5.01}, then$y=u\exp(-\int\frac{p(t)}{3}\mathrm{d}t)$is also a positive solution for \eqref{eq1.03}. Next we discuss Eq \eqref{eq5.01} in two cases \emph{(i)}$b\in C([0,\omega], (0,\infty))$and \emph{(ii)}$b\in C([0,\omega], (-\infty, 0))$. \emph{Case (i):}$b \in C([0,\omega], (-\infty, 0))$. In this case, \eqref{eq5.01} is equivalent to $${\label{eq5.04}} u'''-a(t)u=f(t, u),$$ with$a(t)=-b(t)$, clearly$a\in C([0,\omega], (0, \infty))$, and$a^{*} = \max\{-b(t): t\in [0,\omega]\}$,$a_{*} =\min\{-b(t): t\in [0,\omega]\}$. \emph{Case (ii):}$b\in C([0,\omega], (0, \infty))$. In this case, \eqref{eq5.01} is equivalent to $${\label{eq5.05}} u'''+a(t)u=f(t,u)$$ here$a(t)=b(t)$. Finally, by recalling the proofs for the existence of positive solutions of \eqref{eq5.04} and \eqref{eq5.05} in Section 3 and by applying Theorem \ref{thm5.01} we obtain \begin{theorem} \label{thm5.02} If$\frac{1}{3}p^2(t)+p'(t)=q(t)$and$\sqrt{3}\rho\omega<4\pi/3$hold,$\overline{f}_0=0, \underline{f}_{\infty}=\infty$, then \eqref{eq1.03} has at least one positive solution. \end{theorem} We illustrate our results with an example. \begin{example} \label{exmp5.01} \rm Consider the third-order differential equation $${\label{eq5.06}} \begin{split} &y'''+\sin ty''+ \Big(\frac{1}{3}\sin^2t+\cos t\Big)y'\\ &+ \Big(-\frac{1}{1000}\exp(\sin t)-\frac{1}{3}\sin t+\frac{1}{27}\sin^3t+\frac{1}{3}\sin t\cos t\Big)y\\ &=\frac{1}{1000}\exp(\sin t)y^2. \end{split}$$ Comparing with \eqref{eq1.03}, we are lead to the definitions \begin{gather*} p(t)=\sin t, \quad q(t)=\frac{1}{3}\sin^2t+\cos t,\\ c(t)=-\frac{1}{1000}\exp(\sin t)-\frac{1}{3}\sin t +\frac{1}{27}\sin^3t-\frac{1}{3}\sin t\cos t, \\ g(t, y)=\frac{1}{1000}\exp(\sin t)y^2. \end{gather*} It is easy to see that$\frac{1}{3}p^2(t)+p'(t)=q(t)$. Then by Theorem \ref{thm5.01}, we can transform \eqref{eq5.06} into $$\label{eq5.07} u'''+b(t)u=f(t,u),$$ where$b(t)=-\frac{1}{1000}\exp(\sin t)$and$f(t,u)=\frac{1}{1000}u^2\exp\big(\sin t+\frac{\cos t}{3}\big)$. Since$\sqrt{3}\rho\omega= 1.5188 <4\pi/3$, and noticing that$\overline{f}_0=0, \underline{f}_{\infty}=\infty$, we know from Theorem \ref{thm5.02} that \eqref{eq5.07} has a positive solution$u$, and then \eqref{eq5.06} has a positive solution$y=u\exp(\frac{\cos s}{3})$. \end{example} \section{Positive Periodic Solution for \eqref{eq1.1}} Equation \eqref{eq1.1} can be rewritten as $${\label{eq3.8}} (x(t)-cx(t-\tau(t)))'''+a(t)(x(t)-cx(t-\tau(t)))=f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t)).$$ With$y(t)=x(t)-cx(t-\tau(t))$Equation \eqref{eq3.8} can be transformed into $${\label{eq3.9}} y'''+a(t)y(t)=f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t))$$ % Define$P:X\to X$by $${\label{eq3.5}} (Ph)(t)=(I-TB)^{-1}(Th)(t),$$ where$T, B$are defined as$T_2, B_2$in Section 3. Define operators$Q,S:X\to X$by $${\label{eq3.10}} (Qx)(t)=P(f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t))),\quad (Sx)(t)=cx(t-\tau(t)).$$ % From \eqref{eq3.9} and \eqref{eq3.10} and the results of Section 3, we know that the existence of periodic solutions for \eqref{eq1.1} is equivalent to the existence of solutions for the operator equation $${\label{eq3.11}} Qx+Sx=x \quad \text{in }X.$$ \begin{lemma}[\cite{BL You}] \label{lem3.4} Let$X$be a Banach space, assume$K$is a bounded closed convex subset of$X$and$Q,S:K\to X$satisfy the following assumptions: \begin{itemize} \item[(i)]$Qx+Sy\in K,\forall x,y\in K$, \item[(ii)]$S$is a contractive operator, \item[(iii)]$Q$is a completely continuous operator in$K$. \end{itemize} Then$Q+S$has a fixed point in$K$. \end{lemma} \begin{theorem} \label{thm3.1} If$\sqrt{3}\rho\omega<4\pi/3$holds,$c\in(0,1)$, and$ca_{*}\leq f(t,x)-ca(t)x\leq a^{*}$for all$t\in[0,\omega]$and for all$x\in[\frac{ca_{*}}{(1-c)a^{*}},\frac{a^{*}}{(1-c)a_{*}}]$, then \eqref{eq1.1} has at least one positive$\omega$-periodic solution$x(t)$with$0<\frac{ca_{*}}{(1-c)a^{*}}\leq x(t)\leq\frac{a^{*}}{(1-c)a_{*}}$. \end{theorem} \begin{proof} Define$K_1=\{x\in X:x\in[\frac{ca_{*}}{(1-c)a^{*}},\frac{a^{*}}{(1-c)a_{*}}]\}$. Obviously,$K_1$is a bounded closed convex set in$X$. Since$P$is completely continuous, so is$Q$. Besides, it is easy to see that$S$is contractive if$|c| < 1$. Now we prove that$Qx+Sy\in K_1$for all$x,y\in K_1$. By Lemma \ref{lem4.03}, we obtain $${\label{eq3.12}} \begin{split} &Qx(t)+Sy(t)\\ &= P(f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t)))+cy(t-\tau(t))\\ &\leq \frac{a^{*}}{a_{*}}\|T(f(t,x(t-\tau(t))) -ca(t)x(t-\tau(t)))\|+cy(t-\tau(t))\\ &\leq \frac{a^{*}}{a_{*}}\max_{t\in[0,\omega]} \int^{t+\omega}_{t}G_2(t,s)(f(s,x(s-\tau(s))) -ca(s)x(s-\tau(s)))ds+cy(t-\tau(t))\\ &\leq \frac{a^{*}}{a_{*}}\max_{t\in[0,\omega]} \int^{t+\omega}_{t}G_2(t,s)a^{*}ds+c\frac{a^{*}}{(1-c)a_{*}}\\ &= \frac{a^{*}}{a_{*}}a^{*}\frac{1}{a^{*}} +\frac{ca^{*}}{(1-c)a_{*}} = \frac{a^{*}}{(1-c)a_{*}} \end{split}$$ On the other hand, $${\label{eq3.13}} \begin{split} &Qx(t)+Sy(t)\\ &=P(f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t)))+cy(t-\tau(t))\\ &\geq T(f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t))+cy(t-\tau(t))\\ &\geq\int^{t+\omega}_{t}G_2(t,s)(f(s,x(s-\tau(s))) -ca(s)x(s-\tau(s)))ds+cy(t-\tau(t))\\ &\geq\frac{1}{a^{*}}ca_{*}+\frac{c^2a_{*}}{(1-c)a^{*}} =\frac{ca_{*}}{(1-c)a^{*}} \end{split}$$ Combining \eqref{eq3.12} and \eqref{eq3.13}, we obtain$Qx+Sy\in K_1$, for all$x,y\in K_1$. By Lemma \ref{lem3.4} we obtain that$Q+S$has a fixed point$x\in K_1$; i.e., \eqref{eq1.1} has a positive$\omega$-periodic solution$x(t)$with$0<\frac{ca_{*}}{(1-c)a^{*}}\leq x(t)\leq \frac{a^{*}}{(1-c)a_{*}}. \end{proof} \begin{example} \label{exmp3.1} \rm Consider the equation \begin{align*} &\big(x(t)-\frac{1}{2}x(t-\cos^2t)\big)''' +\frac{1}{1000}(1-\frac{1}{2}\sin^2t)x(t)\\ &=\frac{1}{1000}(1-\frac{3}{4}\sin^2t) +\frac{1}{2000}(1-\frac{1}{2}\sin^2t)x(t-\cos^2t), \end{align*} herec=\frac{1}{2}$,$a(t)=\frac{1}{1000}(1-\frac{1}{2}\sin^2t)$and$\tau(t)=\cos^2t$. Obviously$a\in C(\mathbb{R},(0,\infty))$is a$\pi$-periodic function with$a^{*}=\frac{1}{1000}$,$a_{*}=\frac{1}{2000}$, and then$\rho=\frac{1}{10}$. Noticing that$\frac{\sqrt{3} \pi}{10}<4\pi/3$holds. Moreover, it is easy to see that$\frac{1}{4000}\leq f(t,x)-ca(t)x=\frac{1}{1000}(1-\frac{3}{4}\sin^2t)\leq\frac{1}{1000}$. Then by Theorem \ref{thm3.1}, we know the equation has at least one positive solution$x$with$\frac{1}{2}\leq x(t)\leq 4$. \end{example} \begin{theorem} \label{thm3.2} If$\sqrt{3}\rho\omega<4\pi/3$holds,$c=0$, and$00$, it is easy to see from \eqref{eq3.11} and \eqref{eq3.13}, that$x(t)>0$; i.e., \eqref{eq1.1} has at least one positive$\omega$-periodic solution$x(t)$with$0-ca^{*}\frac{1}{a^{*}}+c = 0. \end{split} Combining \eqref{eq3.14} and \eqref{eq3.15}, we obtain $Qx+Sy\in K_3$ for any $x,y\in K_3$. By Lemma \ref{lem3.4}, we obtain that \eqref{eq1.1} has at least one nonnegative $\omega$-periodic solution $x(t)$ with $0\leq x(t)\leq1$. Since $F(x)>-ca^{*}$, by \eqref{eq3.15}, we obtain $x(t)>0$. So \eqref{eq1.1} has at least one positive $\omega$-periodic solution $x$ with \$0