\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
\begin{equation}{\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}
\end{equation}
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
\begin{equation}\label{eq1}
u'''-\rho^3u=f(t, u),
\end{equation}
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
\begin{equation}\label{eq1.03}
y'''+p(t)y''+q(t)y'+c(t)y=g(t, y).
\end{equation}
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
\begin{equation} \label{eq1.1}
(x(t)-cx(t-\tau(t)))'''+a(t)x(t)=f(t,x(t-\tau(t))),
\end{equation}
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
\begin{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}
\end{equation}
has a unique $\omega$-periodic solution which is of the form
\begin{equation} \label{eq2.02}
u(t)=\int^{\omega}_0G_1(t, s)(-h(s))\mathrm{d}s,
\end{equation}
 where
\begin{equation} \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{equation}
\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
\begin{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}
\end{equation}
has a unique $\omega$-periodic solution
\begin{equation}{\label{eq2.05}}
u(t)=\int^{\omega}_0G_2(t, s)h(s)\mathrm{d}s,
\end{equation}
where
\begin{equation}{\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{equation}
\end{theorem}

\begin{theorem}\label{thm2.03}
 For $\rho>0$ and $h\in X$ the equation
\begin{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}
\end{equation}
has a unique $\omega$-periodic solution
\begin{equation}{\label{eq2.08}}
u(t)=\int^{\omega}_0G_3(t, s)(-h(s))\mathrm{d}s,
\end{equation}
where
\begin{equation}{\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{equation}
\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
\begin{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}
\end{equation}
has a unique $\omega$-periodic solution
\begin{equation}{\label{eq2.011}}
u(t)=\int^{\omega}_0G_4(t, s)h(s)\mathrm{d}s,
\end{equation}
where
\begin{equation}{\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{equation}
\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 $0<A_1<
G_1(t, s)\leq B_1$ for all $t\in[0,\omega]$ and $s\in[0,\omega]$.
\end{theorem}

\begin{proof} Let
\begin{gather*}
H_1(t, s)=\frac{\exp(\rho(t-s))}{3\rho^2[\exp(\rho\omega)-1]},\quad
^{*}_1(t,s)=\frac{\exp(\rho(t+\omega-s))}{3\rho^2
[\exp(\rho\omega)-1]},\\
\begin{aligned}
H_2(t,s)
&=\Big(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/
 \Big(3\rho^2(1+\exp(-\rho\omega)
 -2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}}{2}\rho\omega))\Big),
\end{aligned}\\
\begin{aligned}
&H^{*}_2(t,s)\\
&=\Big(2\exp(\frac{1}{2}\rho(s-t-\omega))\big[\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).
\end{aligned}
\end{gather*}
A direct computation shows that $\int^{\omega}_0G_1(t,
s)\mathrm{d}s=1/\rho^3$. It is easy to see that $H_1(t,
s)>0$ 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)}<G_1(t,s)\\
&\leq\frac{\exp(\rho(t+\omega-s))}{3\rho^2[\exp(\rho\omega)-1]}
+\frac{\exp(\frac{\rho(s-t-\omega)}{2})
 [2+2\exp(-\frac{\rho\omega}{2})]}{3\rho^2[\exp(-\rho\omega)
 +1-2\exp(-\frac{\rho\omega}{2})\cos\frac{\sqrt{3}\rho\omega}{2}]}\\
&\leq\frac{\exp(\rho\omega)}{3\rho^2[\exp(\rho\omega)-1]}
 +\frac{2+2\exp(-\frac{\rho\omega}{2})}{3\rho^2[\exp(-\rho\omega)
 +1-2\exp(-\frac{\rho\omega}{2})\cos\frac{\sqrt{3}\rho\omega}{2}]}\\
&\leq\frac{1}{3\rho^2[1-\exp(-\rho\omega)]}
 +\frac{2+2\exp(-\frac{\rho\omega}{2})}{3\rho^2
 [1-\exp(\frac{-\rho\omega}{2})]^2}\\
&\leq\frac{3+2\exp(-\frac{\rho\omega}{2})}{3\rho^2
[1-\exp(\frac{-\rho\omega}{2})]^2}=B_1
\end{align*}
 and the proof is complete.
\end{proof}

Similarly, the following dual theorem can be proved.

\begin{theorem} \label{thm2.06}
$\int^{\omega}_0G_2(t, s)\mathrm{d}s=1/\rho^3$ and if
$\sqrt{3}\rho\omega<4\pi/3$ holds, then $0<A_1< G_2(t,
s)\leq B_1$ for all $t\in[0,\omega]$ and $s\in[0,\omega]$.
\end{theorem}

The next theorem provides bounds for $G_3$.

\begin{theorem} \label{thm2.07}
$\int^{\omega}_0G_3(t, s)\mathrm{d}s=1/\rho^3$ and
$0<A_2\leq G_3(t, s)\leq B_2$ for all $t\in[0,\omega]$ and
$s\in[0,\omega]$.
\end{theorem}

\begin{proof}
 A direct computation shows that
$\int^{\omega}_0G_3(t,s)\mathrm{d}s=1/\rho^3$.
Next we compute bounds for $G_3(t, s)$. For convenience rewrite
\[
G_3(t,s) = \begin{cases}
\frac{\exp(\rho(\omega-s+t))}{2(\exp(\rho
\omega) - 1)^3} H_3(t,s),& 0\leq t\leq s\leq \omega\\
\frac{\exp(\rho(t-s))}{2(\exp(\rho \omega) - 1)^3}
H^{*}_3(t,s), & 0\leq s\leq t\leq \omega
\end{cases}
\]
with
\begin{gather*}
H_3(t,s)=[(s-t)(\exp(\rho\omega)-1)
+\omega]^2+\omega^2\exp(\rho\omega),\quad 0\leq t\leq s\leq \omega
\\
H^{*}_3(t,s)=[(s-t+\omega)\exp(\rho\omega)-s+t]^2
+\omega^2\exp(\rho\omega),\quad 0\leq s\leq t\leq\omega
\end{gather*}
Since
\begin{gather*}
\frac{\partial H_3(t, s)}{\partial
s}=2[(s-t)(\exp(\rho\omega)-1)+\omega](\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*}
0<H_3(t, t)\leq H_3(t,s)<H_3(t,
\omega)\leq\omega^2\exp(2\rho\omega)+\omega^2\exp(\rho\omega),\\
0<\omega^2+\omega^2\exp(\rho\omega)\leq H^{*}_3(t,0)\leq
H^{*}_3(t,s)\leq H^{*}(t,t),
\end{gather*}
Using these inequalities we obtain
$0<A_2\leq G_3(t,s)\leq B_2$ and the proof is complete.
\end{proof}

Similarly, we obtain the following analog theorem.

\begin{theorem} \label{thm2.08}
$\int^{\omega}_0G_4(t, s)\mathrm{d}s=1/\rho^3$ and
$0<A_2\leq G_4(t, s)\leq B_2$ for all $t\in[0,\omega]$
and $s\in[0,\omega]$.
\end{theorem}


\section{Periodic Solution of \eqref{eq1}}

For reference we briefly recall Banach's fixed point theorem
and related error estimates.

\begin{lemma}[\cite{zeidler}] \label{lem31}
Let $M$ be a closed nonempty set in the Banach space $X$ and
$A: M\to M$ a $k$-contractive operator; i.e.,
there exists $k$, $0\leq k<1$, with
\begin{equation} \label{eq15}
 \| Au-Av\|\leq k \| u-v \| \quad \text{for all  }
u, v\in M.
\end{equation}
Consider the operator equation
\begin{equation}{\label{eq13}}
u=Au,\quad u\in  M,\end{equation}
and for any $u_0 \in M$ the iteration
\begin{equation}{\label{eq14}}
u_{n+1}=Au_n,\quad  n=0, 1, \dots
\end{equation}
Then the following statements hold:
\begin{itemize}
\item[(i)] \emph{Existence and uniqueness:}
there exists a unique $u^*$ which solves \eqref{eq13}; i.e.,
 $Au^* = u^*$.

\item[(ii)] \emph{Convergence of the iteration method:}
For all $u_0\in M$ one has $\lim_{n \to \infty} u_n = u^*$.

\item[(iii)] \emph{Error estimates:} For all $n=0,1, \dots$,
one has the so-called a priori error estimate
\begin{equation}{\label{eq16}}
\| u_n-u\| \leq k^n(1-k)^{-1}\|u_1-u_0\|,
\end{equation}
and the so-called a posteriori error estimate
\begin{equation}{\label{eq17}}
 \| u_{n+1}-u\| \leq k(1-k)^{-1}\|u_{n+1}-u_n\|.
\end{equation}

\item[(iv)] \emph{Rate of convergence:} For all $n=0, 1, \dots$,
 one has $\| u_{n+1}-u\| \leq k\|u_{n}-u\|$.
\end{itemize}
\end{lemma}

 Now we consider \eqref{eq1}. Let $X$ be defined as in
Section 1. Define an operator $D: X\to X$ by
$$
D u(t)=\int^{\omega}_0 G_1(t, s)f(s, u(s))\mathrm{d}s.
$$
By Theorem \ref{thm2.01}, we know that the periodic solution problem
of \eqref{eq1} is equal to the fixed point problem $u=D u$.
For any $u_0 \in X$ define the sequence $(u_n)$ by
\begin{equation}{\label{eq20}}
u_{n+1}(t)=\int^{\omega}_0G_1(t, s)f(s, u_{n}(s))\mathrm{d}s,
\quad n=0,1,\dots.
\end{equation}
We introduce the abbreviation:
\begin{equation}{\label{eq33}}
\vartheta=\frac{2}{3\rho\sqrt{1+\exp(-\rho\omega)
-2\exp(-\frac{1}{2}\rho\omega)\cos(\frac{\sqrt{3}}{2}\rho\omega)}}
+\frac{\exp(\rho\omega)}{3\rho(\exp(\rho\omega)-1)}.
\end{equation}

\begin{theorem} \label{thm31}
Assume that the partial derivative
$f_u\in C([0,\omega]\times\mathbb{R},
\mathbb{R})$ and there is a number $l$ such that
$| f_u(t,u)| \leq l$ for all $t\in[0,\omega]$, $u\in\mathbb{R}$.
Then if $l\vartheta\omega<\rho$, the following statements hold:
\begin{itemize}
\item[(i)] the original problem \eqref{eq1} has a unique solution $u\in X$;

\item[(ii)] the sequence $(u_n)$ constructed by \eqref{eq20} converges to $u$ in $X$;

\item[(iii)] for all $n=0, 1, 2, \dots$, we obtain for
$k:=\frac{l\vartheta\omega}{\rho}$ the following error estimates
$$
\|u_n-u\|\leq k^n(1-k)^{-1}\|u_1-u_0\|, \quad
 \|u_{n+1}-u\|\leq k^n(1-k)^{-1}\| u_{n+1}-u_n\|.
$$
\end{itemize}
\end{theorem}

\begin{proof}
By calculation, we obtain:
For $0\leq t\leq s\leq\omega$,
\begin{align*}
&|G_1(t,s)|\\
&\leq\big|\frac{\exp(\rho(t+\omega-s))}{3\rho^2
[\exp(\rho\omega)-1]}\big|\\
&\quad +\Big|\frac{\exp(\frac{\rho(s-t-\omega)}{2})
 [2\sin(\frac{\pi}{6}-\frac{\sqrt{3}\rho}{2}(s-t-\omega))
 -2\exp(-\frac{\rho\omega}{2})\sin(\frac{\pi}{6}
 -\frac{\sqrt{3}\rho}{2}(s-t))]}{3\rho^2[\exp(-\rho\omega)
 +1-2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}\rho\omega}{2})]}\Big|\\
&\leq\big|\frac{\exp(\rho\omega)}{3\rho^2
 [\exp(\rho\omega)-1]}\big|+\Big|\frac{2\exp(\frac{\rho(s-t
 -\omega)}{2})\sin(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\rho(s-t)
 +\varphi_1)}{3\rho^2\sqrt{\exp(-\rho\omega)
 +1-2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}\rho\omega}{2})}}
 \Big|\\
&\leq\frac{\exp(\rho\omega)}{3\rho^2[\exp(\rho\omega)-1]}
 +\frac{2}{3\rho^2\sqrt{\exp(-\rho\omega)
 +1-2\exp(-\frac{\rho\omega}{2})\cos(\frac{\sqrt{3}\rho\omega}{2})}}
= \frac{\vartheta}{\rho}.
\end{align*}
Similarly, we can obtain:
For $0\leq s\leq t\leq \omega$,
$$
|G_1(t,s)|\leq \frac{\vartheta}{\rho}.
$$
So for all $t\in[0,\omega]$ and $s\in[0,\omega]$, we have
$|G_1(t,s)|\leq \frac{\vartheta}{\rho}$.

 On the other hand, for any $t\in[0,\omega]$, and $u, v\in
\mathbb{R}$, by the mean value theorem, there exists a $w\in
\mathbb{R}$ such that
$$
| f(s, u)-f(s, v)|  \leq |  f_u(s, w)|  |  u-v|  \leq l |  u-v| .
$$
Therefore,
\begin{align*}
|Du-Dv|
&=  \Big|\int^{\omega}_0G_1(t, s)[f(s, u(s))-f(s, v(s))]
 \mathrm{d}s\Big|  \\
& \leq \int^{\omega}_0|  G_1(t, s)|  |  f(s, u(s))-f(s, v(s))|
  \mathrm{d}s\\
& \leq  l|  u-v|\int^{\omega}_0|  G_1(t, s)| \mathrm{d}s
\leq \frac{l\vartheta\omega}{\rho}|u-v| ;
\end{align*}
i.e., $\| Du-Dv\|\leq  k\| u-v\|$ for all $u, x\in X$.
Let $M:=X$, the assertions follow directly from Lemma \ref{lem31}.
\end{proof}

Similarly, we can obtain the corresponding results for the equations
\begin{gather*} %{\label{eq2}}
u'''+\rho^3u=f(t, u), \\
%\label{eq3}
u'''-3\rho u''+3\rho^2u'-\rho^3u=f(t, u), \\
% \label{eq4}
u'''+3\rho u''+3\rho^2u'+\rho^3u=f(t, u),
\end{gather*}
where $f\in C([0,\omega]\times\mathbb{R}, (0,+\infty))$.

\section{Positive  Solutions of \eqref{eq1.01}, \eqref{eq1.02}}

In this section, we consider the existence of  positive periodic
solutions for the third-order nonlinear differential equations with
$\omega$-periodic boundary value condition
\begin{equation}{\label{eq4.01}}
u'''-a(t)u=f(t,u),
\end{equation}
and
\begin{equation}{\label{eq4.02}}
u'''+a(t)u=f(t,u),
\end{equation} where $a\in C([0,\omega], (0,
\infty))$, and $f\in C([0,\omega]\times[0, \infty), [0, \infty))$,
and $f(t, u)>0$ 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
$K_{0r}=\{x\in K_0: \|x\|<r\}$ and $\partial K_{0r}=\{x\in K_0:
\|x\|=r\}$.

First, we study the following equation corresponding to
\eqref{eq4.01}
\begin{equation}{\label{eq4.03}}
u'''-a(t)u=h(t),\quad h\in C^{-}_{\omega}
\end{equation}
Define operators $T_1, B_1: X\to X$ by
\begin{equation}{\label{eq4.04}}
(T_1h)(t)=\int^{\omega}_0G_1(t,s)(-h(s))\mathrm{d}s,\quad
(B_1u)(t)=(a(t)-a^{*})u(t).
\end{equation}
Clearly, if  $\sqrt{3}\rho\omega<4\pi/3$ holds, then
$T_1, B_1$ are completely continuous, $(T_1h)(t)>0$ 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
\begin{equation}{\label{eq4.05}}
u(t)=(T_1h)(t)+(T_1B_1u)(t).
\end{equation}
And for
$\|T_1B_1\|\leq\|T_1\|\|B_1\|\leq\frac{1}{a^{*}}(a^{*}-a_{*})<1$,
we have
\begin{equation}{\label{eq4.06}}
u(t)=(I-T_1B_1)^{-1}(T_1h)(t).
\end{equation}
Define an operator $P_1: X\to X$ by
\begin{equation}{\label{eq4.07}}
(P_1h)(t)=(I-T_1B_1)^{-1}(T_1h)(t).
\end{equation}
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
\begin{equation}{\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{equation}
\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
\begin{equation}{\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}
\end{equation}
 By \eqref{eq3.7}, and recalling that
$\|T_1B_1\|\leq1-\frac{a_{*}}{a^{*}}$
and $(T_1h)(t)>0$, we obtain
\begin{equation}{\label{eqp1}}
(T_1h)(t)\leq(P_1h)(t)\leq\frac{a^{*}}{a_{*}}\|(T_1h)(t)\|,
\quad h\in C^{+}_{\omega}.
\end{equation}
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
\begin{equation}{\label{eq4.010}}
Q_1u(t)=P_1(-f(t, u)).
\end{equation}
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}.
\begin{equation}{\label{eq4.011}}
 u'''+a(t)u=h(t),\quad h\in C_{\omega}^{+}
\end{equation}
Define operators $T_2, B_2: X\to X$ by
\begin{equation}{\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).
\end{equation}
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
\begin{equation}{\label{eq4.013}}
u(t)=(T_2h)(t)+(T_2B_2u)(t).
\end{equation}
And for
$\|T_2B_2\|\leq\|T_2\|\|B_2\|\leq\frac{1}{a^{*}}(a^{*}-a_{*})<1$,
we have
\begin{equation}{\label{eq4.014}}
u(t)=(I-T_2B_2)^{-1}(T_2h)(t).
\end{equation}
Define an operator $P_2: X\to X$ by
\begin{equation}{\label{eq4.015}}
(P_2h)(t)=(I-T_2B_2)^{-1}(T_2h)(t).
\end{equation}
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
\begin{equation}{\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{equation}
\end{lemma}

Define an operator $Q_2: X\to X$ by
\begin{equation}{\label{eq4.018}}
Q_2u(t)=P_2(f(t, u)).
\end{equation}
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
$K_{r}=\{u\in K:\|u\|<r\}$. Assume that $T:\bar{K}_{r}\to
K$ is completely continuous operator such that $Tx\neq x$
for $x\in\partial K_{r}=\{u\in K:\|u\|=r\}$, so
\begin{itemize}
\item[(i)] if $\|Tx\|\geq\|x\|$ for $x\in\partial K_{r}$, then
$i(T,K_{r},K)=0$;

\item[(ii)] if $\|Tx\|\leq\|x\|$ for $x\in\partial K_{r}$, then
$i(T,K_{r},K)=1$.
\end{itemize}
\end{lemma}

\begin{theorem} \label{thm4.01}
If $\sqrt{3}\rho\omega<4\pi/3$ holds, and
$\overline{f}_0=0, \underline{f}_{\infty}=\infty$, then
\eqref{eq4.01} has at least one positive  solution.
\end{theorem}

\begin{proof}
If $\overline{f}_0=0$, we can choose $0<r_1<1$, such
that $f(t, u)\leq\varepsilon u$ for $0\leq u\leq r_1, t\in[0,
\omega]$, where the constant $\varepsilon>0$ 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 $r_1<u<\tilde{H}$.
\end{proof}

By Lemmas \ref{lem4.03} and  \ref{lem4.04}, we obtain the following
result.

\begin{theorem} \label{thm4.02}
If $\sqrt{3}\rho\omega<4\pi/3$ holds, and $\overline{f}_0=0$,
$\underline{f}_{\infty}=\infty$, then
 \eqref{eq4.02} has at least one
positive  solution.
\end{theorem}

\section{Positive  Solutions for \eqref{eq1.03}}

\begin{theorem} \label{thm5.01}
If $\frac{1}{3}p^2(t)+p'(t)=q(t)$, then  \eqref{eq1.03}
can be transformed into
\begin{equation}{\label{eq5.01}}
u'''(t)+b(t)u(t)=f(t, u),
\end{equation}
where
\begin{equation}{\label{eq5.02}}
\begin{gathered}
b(t)=-\frac{1}{3}p''(t)+\frac{2}{27}p^3(t)-\frac{1}{3}p(t)q(t)+c(t),\\
f(t,u)=\frac{g(t, u\exp(-\int\frac{p(t)}{3}\mathrm{d}t))}
{\exp(-\int\frac{p(t)}{3}\mathrm{d}t)},
\end{gathered}
\end{equation}
$b\in C([0,\omega], \mathbb{R})$;
$f\in C([0,\omega]\times[0, \infty), [0, \infty))$.
\end{theorem}

\begin{proof}
Let $y=ux$, then
\begin{equation}{\label{eq5.03}}
y'=u'x+ux',\quad
y''=u''x+2u'x'+ux'',\quad
y'''=u'''x+3u''x'+3u'x''+ux'''.
\end{equation}
Substituting \eqref{eq5.03} into  \eqref{eq1.03} yields
\begin{align*}
&u'''x+[3x'+p(t)x]u''+[3x''+2p(t)x'+q(t)x]u'+[x'''+p(t)x''
+q(t)x'+c(t)x]u\\
&=g(t, ux),
\end{align*}
or equivalently for $x \neq 0$,
\begin{align*}
&u'''+\frac{3x'+p(t)x}{x}u''+\frac{3x''+2p(t)x'+q(t)x}{x}u'
+\frac{x'''+p(t)x''+q(t)x'+c(t)x}{x}u\\
&=\frac{g(t, ux)}{x}.
\end{align*}
It is easy to verify that $3x'+p(t)x=0$ and $3x''+2p(t)x'+q(t)x=0$
if $x=\exp(-\int\frac{p(t)}{3}\mathrm{d}t)$ and
$\frac{1}{3}p^2(t)+p'(t)=q(t)$. Hence \eqref{eq1.03} can be
transformed to
$$
u'''(t)+b(t)u(t)=f(t, u),
$$
where $b(t), f(t, u)$ are given in \eqref{eq5.02}. It is easy to see
that $b\in C([0,\omega], \mathbb{R})$; $f\in C([0,\omega]\times[0,
\infty), [0, \infty))$, $f(t, u)>0$ 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
\begin{equation}{\label{eq5.04}}
u'''-a(t)u=f(t, u),
\end{equation}
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
\begin{equation}{\label{eq5.05}}
u'''+a(t)u=f(t,u)
\end{equation}
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
\begin{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}
\end{equation}

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
\begin{equation} \label{eq5.07}
u'''+b(t)u=f(t,u),
\end{equation}
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
\begin{equation}{\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)).
\end{equation}
With $y(t)=x(t)-cx(t-\tau(t))$ Equation \eqref{eq3.8} can be
transformed into
\begin{equation}{\label{eq3.9}}
y'''+a(t)y(t)=f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t))
\end{equation}
%
Define $P:X\to X$ by
\begin{equation}{\label{eq3.5}}
(Ph)(t)=(I-TB)^{-1}(Th)(t),
\end{equation}
where $T, B$ are defined as $T_2, B_2$ in Section 3. Define
operators $Q,S:X\to X$ by
\begin{equation}{\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)).
\end{equation}
%
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
\begin{equation}{\label{eq3.11}}
Qx+Sx=x  \quad \text{in }X.
\end{equation}

\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
\begin{equation}{\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}
\end{equation}
On the other hand,
\begin{equation}{\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}
\end{equation}
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*}
here $c=\frac{1}{2}$, $a(t)=\frac{1}{1000}(1-\frac{1}{2}\sin^2t)$
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
$0<f(t,x(t-\tau(t)))\leq a^{*}$ for all $t\in[0,\omega]$ and for all
$x\in[0,a^{*}/a_{*}]$, then \eqref{eq1.1} has at least one
positive $\omega$-periodic solution $x$ with $0<x(t)\leq a^{*}/a_{*}$.
\end{theorem}

\begin{proof}
 By \eqref{eq3.10}, $S=0$. We define
$K_2=\{x\in X:x\in[0,\frac{a^{*}}{a_{*}}]\}$.
Similarly as in the proof of Theorem \ref{thm3.1} we obtain
that \eqref{eq1.1} has at least one nonnegative $\omega$-periodic
solution $x(t)$ with $0\leq x(t)\leq \frac{a^{*}}{a_{*}}$. Since
$F(x)>0$, 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
$0<x(t)\leq\frac{a^{*}}{a_{*}}$.
\end{proof}

\begin{example} \label{exmp3.2} \rm
Consider the equation
$$x'''(t)+\frac{1}{1000}\big(1-\frac{1}{2}\sin^2t\big)x(t)=\frac{1}{1000}\big(1-\frac{1}{3}\sin^2t\big)-\frac{x(t-\cos^2t)}{4000}.$$
 as in Example \ref{exmp3.1}, we obtain
$\frac{\sqrt{3}\pi}{10}<4\pi/3$ holds. Moreover,
$0<\frac{1}{6000}\leq
f(t,x(t-\tau(t)))=\frac{1}{1000}(1-\frac{1}{3}\sin^2t)-\frac{x(t-\cos^2t)}{4000}\leq\frac{1}{1000}$
for all $t\in[0,\pi]$ and $x\in[0,2]$. All assumptions of Theorem
\ref{thm3.2} are satisfied and hence the equation has at least one
positive solution $x(t)$ with $0<x(t)\leq2$.
\end{example}

\begin{theorem} \label{thm3.3}
If $\sqrt{3}\rho\omega<4\pi/3$
holds, $c\in(-\frac{a_{*}}{a^{*}},0)$, and
$-ca^{*}<f(t,x)-ca(t)x\leq a_{*}$ for all $t\in[0,\omega]$ and
$x\in[0,1]$, then \eqref{eq1.1} has at least one positive
$\omega$-periodic solution $x$ with $0<x(t)\leq1$.
\end{theorem}

\begin{proof}
As in the proof of Theorem \ref{thm3.1}, define
$K_3=\{x\in X:x\in[0,1]\}$ and then $K_3$ is a bounded closed
convex set in $X$. $Q$ is a completely continuous, and $S$
is contractive since $|c| < 1$.
Now we prove $Qx+Sy\in K_3$ for all $x,y\in K_3$.
By Lemma \ref{lem4.03}, we obtain
\begin{align}
&(Qx)(t)+(Sy)(t) \nonumber \\
&= P(f(t,x(t-\tau(t)))-ca(t)x(t-\tau(t)))+cy(t-\tau(t)) \nonumber\\
&\leq \frac{a^{*}}{a_{*}}\|T(f(t,x(t-\tau(t)))
 -ca(t)x(t-\tau(t)))\|+cy(t-\tau(t)) \nonumber\\
&\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)) \nonumber\\
&\leq \frac{a^{*}}{a_{*}}\int^{t+\omega}_{t}G_2(t,s)a_{*}ds = 1.
\label{eq3.14}
\end{align}
On the other hand, by Theorem \ref{thm2.06} and Lemma \ref{lem3.1}
\begin{equation}\label{eq3.15}
\begin{split}
&(Qx)(t)+(Sy)(t)\\
&= P(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))\\
&>-ca^{*}\frac{1}{a^{*}}+c = 0.
\end{split}
\end{equation}
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 $0<x(t)\leq1$.
\end{proof}

\begin{example} \label{exmp3.3} \rm
Consider the equation
\begin{align*}
&\big(x(t)+\frac{1}{4}x(t-\cos^2t)\big)'''
+\frac{1}{1000}\big(1-\frac{1}{3}\sin^2t\big)x(t)\\
&=\big(1-\frac{1}{3}\sin^2t\big)
\big[\frac{1}{1500}-\frac{1}{4000}x(t-\cos^2t)\big].
\end{align*}
As in Example \ref{thm3.1}, we can verify that all the assumptions of
Theorem \ref{thm3.3} hold, then the equation has at least
one positive solution $x$ with $0<x(t)\leq1$.
\end{example}

\begin{remark} \label{rmk6.1} \rm
In a similar way, we can discuss the third order neutral differential
equation
$$
(x(t)-cx(t-\tau(t)))'''-a(t)x(t)=f(t,x(t-\tau(t))),
$$
with $a\in C(\mathbb{R},(0,\infty))$, $\tau\in C(\mathbb{R},
\mathbb{R})$, $f\in C(\mathbb{R}\times [0,\infty),  [0,\infty))$,
$a(t), \tau(t)$ are $\omega$-periodic functions, $f(t,x)$ is
$\omega$-periodic in $t$, and $\omega$, $c$ are constants with
$|c| < 1$.
\end{remark}

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