\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 25, 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  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/25\hfil
 On second order periodic boundary-value problems]
{On second order periodic boundary-value problems
with upper and lower solutions in the reversed order}
\author[H. Gao, S. Weng, D. Jiang, X. Hou\hfil EJDE-2006/25\hfilneg]
{Haiyin Gao, Shiyou Weng, Daqing Jiang, Xuezhang Hou} 

\address{Haiyin Gao  \newline
School of Mathematics and Statistics, Northeast Normal University,
 Changchun 130024, China}
\email{gaohaiyinhealthy@yahoo.com.cn}

\address{Shiyou Weng \newline
Applied Science College, Changchun University, Jilin 130022,
China} \email{wengshiyou2001@yahoo.com.cn}

\address{Daqing Jiang \newline
School of Mathematics and Statistics, Northeast Normal University,
 Changchun 130024, China}
\email{daqingjiang@vip.163.com}

\address{Xuezhang Hou \newline
Mathematics Department, Towson University,
 Baltimore, MD 21252, USA}
\email{xhou@towson.edu}


\date{}
\thanks{Submitted November 7, 2005. Published February 28, 2006.}
\thanks{Supported by grant 10171010 from the National Natural
Science Foundation of China}
\subjclass[2000]{34B15, 34B16}
\keywords{Periodic boundary value; existence; upper and lower solutions}

\begin{abstract}
 In this paper, we study the differential equation with the periodic
 boundary value
 \begin{gather*}
 u''(t)=f(t, u(t), u'(t)),\quad t\in [0,  2\pi]\\
 u(0)=u(2\pi), \quad u'(0)=u'(2\pi).
 \end{gather*}
 The existence of solutions to the periodic boundary problem above
 with appropriate conditions is proved by using an upper and lower
 solution method.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

 \section{Introduction and Main Results}

 In this paper,  we study the second-order periodic
boundary-value problem
\begin{equation}
\begin{gathered}
u''(t)=f(t, u(t), u'(t)),\quad t\in [0,  2\pi]\\
u(0)=u(2\pi), \quad u'(0)=u'(2\pi),
\end{gathered} \label{e1.1}
\end{equation}
where $f(t, u, v)$ is a Caratheodory function.
A function $f:  [0, 2\pi]\times \mathbb{R}^2\to \mathbb{R}$ is said to be a
Carathrodary function  if it possess the following three properties:
\begin{itemize}
\item[(i)] For all $(u,  v)\in \mathbb{R}^2$, the mapping $t\to f(t, u, v)$
is measurable on $[0,  2\pi]$.

\item[(ii)] For almost all $t\in [0,  2\pi]$,  the mapping $(u,
v)\to f(t, u, v)$ is continuous on $\mathbb{R}^2$.

\item[(iii)] For any given $N>0$,  there exists $g_N(t)$,  a Lebesgue
integrable function defined on $[0,  2\pi]$, such that
$$
|f(t, u, v)|\leq g_N(t)\quad \hbox{for a. e. }t\in [0, 2\pi],
$$
whenever $|u|, |v|\leq N$.
\end{itemize}

To develop upper and lower solutions method,  we need the concepts
of upper and lower solutions. We say that
$\beta\in W^{2, 1}[0,2\pi]$ is an upper solution to \eqref{e1.1},
if it satisfies
\begin{equation}
\begin{gathered}
\beta''(t)\leq f(t, \beta(t), \beta'(t)),\quad t\in [0,  2\pi]\\
\beta(0)=\beta(2\pi),\quad \beta'(0)\leq \beta'(2\pi).
\end{gathered}\label{e1.2}
\end{equation}
Similarly,  a function $\alpha\in W^{2, 1}[0,  2\pi]$ is said to
be a lower solution to \eqref{e1.1}, if it satisfies
\begin{equation}
\begin{gathered}
\alpha''(t)\geq f(t, \alpha(t), \alpha'(t)),\quad t\in [0,  2\pi]\\
\alpha(0)=\alpha(2\pi),\quad \alpha'(0)\geq \alpha'(2\pi).
\end{gathered} \label{e1.3}
\end{equation}
We call a function $u\in W^{2, 1}[0,  2\pi]$ a solution to
 \eqref{e1.1}, if it is an upper and a lower solution
 to \eqref{e1.1}.

Under the classical assumption that $\alpha(t)\leq \beta(t)$, a
number of authors have studied the existence of the methods of
lower and upper solutions or the monotone iterative technique
\cite{c1,c3,c4,g1,l1,j1,n2,n3,s1,w1}.
 Only a few have study the case where $\alpha(t), \beta(t)$ satisfy the
opposite ordering condition $\beta(t)\leq\alpha(t)$; see
\cite{c1,c2,j2,n1,o2,r1,r2,w2}
  Wang \cite{w2} has investigated a special case of \eqref{e1.1}
where $f(t, u,v)=-kv + F(t, u)$ and $F(t, u)$ is increasing with
respect to $u$,
in the presence of a lower solution $\alpha(t)$ and an upper
solution $\beta(t)$ with $\beta(t)\leq \alpha(t)$.
Rachunkova \cite{r2} has recently proved that \eqref{e1.1} has at
least one solution $u(t)$ under the case $\beta(t)\leq\alpha(t)$.
However,  the proof of the result in \cite{r2} is not constructive and
is not able to guarantee that $u(t)$ satisfies
$\beta(t)\leq u(t)\leq \alpha(t)$. Recently,  Jiang, Fan and Wan \cite{j2}
have studied \eqref{e1.1} by means of  a monotone iterative technique
in the presence of a lower solution $\alpha(t)$ and an upper solution
$\beta(t)$ with $\beta(t)\leq \alpha(t)$. To develop a monotone
method,  the following hypotheses are needed in \cite{j2}.
\begin{itemize}
\item[(A1)] For any given $\beta,  \alpha\in C[0,  2\pi]$ with
$\beta(t)\leq \alpha(t)$ on $[0,  2\pi]$,  there exist $0<A\leq B$
such that
$$
A(v_2-v_1)\leq f(t, u, v_2)-f(t, u, v_1)\leq B(v_2-v_1)
$$
or
$$
-B(v_2-v_1)\leq f(t, u, v_2)-f(t, u, v_1)\leq -A(v_2-v_1)
$$
for a.e. $t\in [0, 2\pi]$ whenever $\beta(t)\leq u\leq\alpha(t),
v_1,  v_2\in \mathbb{R}$, and $v_1 \leq v_2$.

\item[(A2)] Inequality
$$
f(t, u_2, v)-f(t, u_1, v)\geq - \frac{A^2}{4} (u_2-u_1)
$$
holds for a.e. $t\in [0,  2\pi]$, whenever $\beta(t)\leq u_1\leq
u_2\leq \alpha(t), v\in \mathbb{R}$.
\end{itemize}

The purpose of this paper is to prove the existence
of solutions to \eqref{e1.1} under the assumption that there exist
a lower solution $\alpha(t)$ and an upper solution $\beta(t)$ of
\eqref{e1.1} with $\beta(t)\leq\alpha(t)$ and $f(t,u,v)$ only satisfies
one side Lipschitz condition. We use the upper and lower
solutions method  and prove that the solution $u(t)$ of \eqref{e1.1}
satisfies $\beta(t)\leq u(t)\leq\alpha(t)$. Our result extends and
complements those in \cite{w2,r2,j2}.

 To develop upper and lower solutions  method,  we  need
one of the following hypotheses
\begin{itemize}
\item[(H1)] For any given $\beta,  \alpha\in C[0,  2\pi]$ with
$\beta(t)\leq \alpha(t)$ on $[0,  2\pi]$,  there exist $A>0$ and
$B>0$ such that $B^2\geq 4A$ and
\begin{equation}
f(t, u_2, v_2)-f(t, u_1, v_1)\geq -A(u_2-u_1)+B(v_2-v_1)
\label{e1.4}
\end{equation}
 for a.e. $t\in [0, 2\pi]$ whenever $\beta(t)\leq u_1\leq
u_2\leq\alpha(t)$, $v_1,  v_2\in \mathbb{R}$, and $v_1 \leq v_2$.

\item[(H1')] For any given $\beta,  \alpha\in C[0,  2\pi]$
with $\beta(t)\leq \alpha(t)$ on $[0,  2\pi]$,  there exist $A>0$
and $B>0$ such that $B^2\geq 4A$ and
\begin{equation}
f(t, u_2, v_2)-f(t, u_1, v_1)\geq -A(u_2-u_1)+B(v_1-v_2)
\label{e1.4b}
\end{equation}
 for a.e. $t\in [0, 2\pi]$ whenever $\beta(t)\leq u_1\leq
u_2\leq\alpha(t)$, $v_1,  v_2\in \mathbb{R}$, and $v_1 \geq v_2$.
\end{itemize}
%\begin{remark} \label{rmk1} \rm
We remark that condition (H1') is  equivalent to
\begin{itemize}
\item[(a1)] For any given $\beta,  \alpha\in C[0,  2\pi]$ with
$\beta(t)\leq \alpha(t)$ on $[0,  2\pi]$,  there exists $B>0$ such
that
$$
 f(t, u, v_1)-f(t, u, v_2)\leq  -B(v_1-v_2)
$$
for a.e. $t\in [0, 2\pi]$ whenever $\beta(t)\leq u\leq\alpha(t)$,
$v_1,  v_2\in \mathbb{R}$, and $v_1 \geq v_2$.

\item[(a2)] There exists $A>0$ such that $B^2\geq 4A$ and
$$
f(t, u_2, v)-f(t, u_1, v)\geq - A (u_2-u_1)
$$
holds for a.e. $t\in [0,  2\pi]$, whenever $\beta(t)\leq u_1\leq
u_2\leq \alpha(t)$, $v\in \mathbb{R}$.
\end{itemize}
Also we remark that  (H1) or (H1')  weaker than (A1)-(A2) in \cite{j2}.
\smallskip


Let $m<0$ and $M<0$ be
two real roots to the equation $x^2+Bx+A=0$, then
$$
m+M=-B,\quad m M=A.
$$
Let $m_0>0$ and $M_0>0$ are two roots to the equation
$x^2-Bx+A=0$, then
$$
m_0+M_0=B,\quad m_0 M_0=A.
$$
Let
\begin{equation}
A(t):=\alpha'(t)+m\alpha (t),\quad
B(t):=\beta'(t)+m\beta(t), \label{e1.5}
\end{equation}
and
\begin{equation}
A_0(t):=\alpha'(t)+m_0\alpha (t),\quad
B_0(t):=\beta'(t)+m_0\beta(t).\label{e1.5b}
\end{equation}
The main results of this paper are stated as follows.

\begin{theorem} \label{thm1}
Suppose that there exists a lower solution $\alpha(t)$ and an
upper solution $\beta(t)$ of \eqref{e1.1} such that
$\beta(t)\leq\alpha(t)$ on $[0,2\pi]$,  and $f(t, u, v)$
is a Caratheodory function satisfying
the hypothesis (H1).  Then $A(t) \leq B(t)$ on $[0,2\pi]$ and
\eqref{e1.1}
 has a solution $u\in W^{2,1}[0,2\pi]$ such that
$$
\beta(t)\leq u(t)\leq \alpha(t),\quad
A(t)\leq u'(t)+mu(t)\leq B(t).
$$
\end{theorem}

\begin{theorem} \label{thm2}
Suppose that there exists a lower solution $\alpha(t)$ and an
upper solution $\beta(t)$ of \eqref{e1.1} such that
$\beta(t)\leq\alpha(t)$ on $[0,2\pi]$,  and $f(t, u, v)$
is a Caratheodory function satisfying
the hypothesis (H1').  Then $B_0(t) \leq A_0(t)$ on $[0,2\pi]$
and \eqref{e1.1}  has a solution $u\in W^{2,1}[0,2\pi]$ such that
$$
\beta(t)\leq u(t)\leq \alpha(t),\quad
B_0(t)\leq u'(t)+m_0u(t)\leq A_0(t).
$$
\end{theorem}


\section{Proof of Theorems \ref{thm1} and \ref{thm2}}

To prove the validity of upper and lower solutions
 method,  we use the following
maximum-minimum principle, see \cite{j2}.

\begin{lemma} \label{lem1}
Let $y\in W^{1, 1}[0,2\pi]$, and satisfy
\begin{gather*}
y'(t)+ Ly(t)\geq 0 \quad \hbox{for a. e. } t\in [0, 2\pi],\\
y(0)\geq y(2\pi),
\end{gather*}
where $|L|> 0$.  Then $Ly(t)\geq 0$ on $[0,  2\pi]$, i.e., when
$L>0$ the minimum of $y(t)$ is nonnegative;  when $L<0$ the
maximum of $y(t)$ is nonpositive.
\end{lemma}

\begin{lemma} \label{lem2}
Suppose that there exists a lower solution $\alpha(t)$ and an upper
solution $\beta(t)$ of \eqref{e1.1} such that
$\beta(t)\leq\alpha(t)$ on $[0,2\pi]$,  and $f(t, u, v)$ is a
 Caratheodory function satisfying the hypothesis (H1).
Then $A(t)\leq B(t)$ on $[0,2\pi]$.
\end{lemma}

\begin{proof}
It follows from \eqref{e1.2} and \eqref{e1.3} that
\begin{gather*}
A'(t)+MA(t)\geq f(t,\alpha
(t),A(t)-m\alpha(t))+(m+M)A(t)-m^2\alpha (t),\quad t\in[0, 2\pi]\\
A(0)\geq A(2\pi),
\end{gather*}
and
\begin{gather*}
B'(t)+MB(t)\leq f(t,\beta
(t),B(t)-m\beta(t))+(m+M)B(t)-m^2\beta (t),\quad t\in[0, 2\pi]\\
B(0)\leq B(2\pi).
\end{gather*}
Let $y(t)=A(t)-B(t)$, then $y(0)\geq y(2\pi)$.
Assume that $y(t)>0$ for some $t\in[0,2\pi]$. Indeed, if $y(t)>0$
on $[0,2\pi]$, then by (H1) we have
\begin{align*}
y'(t)+My(t)&\geq f(t,\alpha (t),A(t)-m\alpha(t))-f(t,\beta
(t),B(t)-m\beta(t))\\
&\quad +(m+M)y(t)-m^2(\alpha(t)-\beta(t))\\
&\geq -(A+Bm+m^2)(\alpha(t)-\beta(t))+(B+m+M)y(t)\\
&= 0,\quad t\in[0, 2\pi],
\end{align*}
then by Lemma \ref{lem1}, we have $y(t)\leq 0$ on $[0,2\pi]$, which is a
contradiction.

If $y(0)\leq 0$ (then $y(2\pi)\leq y(0)\leq 0$), and hence there
exists $s$ $\in (0,2\pi)$ with $y(s)>0$, then there would be
$0\leq a<s<b\leq 2\pi$ such that $y(t)>0$ in $(a,b)$ with
$y(a)=y(b)=0$.
 By \eqref{e1.2} and \eqref{e1.3}, we have
$$
y'(t)+My(t)\geq 0,\quad t\in [a,b],\quad y(a)=y(b)=0.
$$
This leads to $y'(t)\geq -My(t)>0$ on $[a,b]$, which is
again a contradiction.

If $y(0)>0$, then there exists $a \in (0,2\pi)$ such that $y(t)>0$
on $[0,a)$ with $y(a)=0$. So we have $y'(t)+My(t)\geq 0$ on
$[0,a)$, hence $y'(t)>0$  in $[0, a)$, which implies that
$y(0)<y(a)=0$, this is also a contradiction.
The proof of Lemma \ref{lem2} is completed.
\end{proof}

 Similarly,  we have the following result.

\begin{lemma} \label{lem3} Suppose that there
exists a lower solution $\alpha(t)$ and an upper solution
$\beta(t)$ of \eqref{e1.1} such that $\beta(t)\leq\alpha(t)$ on
$[0,2\pi]$,  and $f(t, u, v)$ is a Caratheodory function satisfying
the hypothesis (H1'). Then $B_0(t)\leq A_0(t)$ on $[0,2\pi]$.
\end{lemma}

\begin{proof} It follows from \eqref{e1.2} and
\eqref{e1.3} that for $t\in[0, 2\pi]$,
\begin{gather*}
A_0'(t)+M_0A_0(t)\geq f(t,\alpha
(t),A_0(t)-m_0\alpha(t))+(m_0+M_0)A_0(t)-m_0^2\alpha (t),\\
A_0(0)\geq A_0(2\pi)
\end{gather*}
and for $t\in[0, 2\pi]$
\begin{gather*}
B_0'(t)+MB(t)\leq f(t,\beta
(t),B_0(t)-m_0\beta(t))+(m_0+M_0)B_0(t)-m_0^2\beta (t),
\\
B_0(0)\leq B_0(2\pi).
\end{gather*}
Let $y(t)=A_0(t)-B_0(t)$, then $y(0)\geq y(2\pi)$.
Assume that $y(t)<0$ for some $t\in[0,2\pi]$. Indeed, if $y(t)<0$
on $[0,2\pi]$, then by (H1') we have
\begin{align*}
y'(t)+M_0y(t)&\geq f(t,\alpha (t),A_0(t)-m_0\alpha(t))-f(t,\beta
(t),B_0(t)-m_0\beta(t))\\
&\quad +(m_0+M_0)y(t)-m_0^2(\alpha(t)-\beta(t))\\
&\geq -(A-Bm_0+m_0^2)(\alpha(t)-\beta(t))+(-B+m_0+M_0)y(t)\\
&= 0,\quad t\in[0, 2\pi],
\end{align*}
then by Lemma \ref{lem1}, we have $y(t)\geq 0$ on $[0,2\pi]$, which is a
contradiction.

If $y(2\pi)\geq 0$ (then $y(0)\geq y(2\pi)\geq 0$), and hence
there exists $s$ $\in (0,2\pi)$ with $y(s)<0$, then there would be
$0\leq a<s<b\leq 2\pi$ such that $y(t)<0$ in $(a,b)$ with
$y(a)=y(b)=0$.  By \eqref{e1.2} and \eqref{e1.3}, we have
$$
y'(t)+M_0y(t)\geq 0, \quad t\in [a,b],\quad y(a)=y(b)=0.
$$
This leads to $y'(t)\geq -M_0y(t)>0$ on $[a,b]$, which is again a
contradiction.

If $y(2\pi)<0$, then there exists $a \in (0,2\pi)$ such that
$y(t)<0$ on $(a, 2\pi]$ with $y(a)=0$.
 So we have $y'(t)+M_0y(t)\geq 0$ on $(a, 2\pi]$, hence $y'(t)>0$
in $(a, 2\pi]$, which implies that $y(2\pi)>y(a)=0$, this is also a
contradiction.
The proof of Lemma \ref{lem3} is complete.
\end{proof}

 In the following arguments, we only  give the proof of
Theorem \ref{thm1},  since the proof of Theorem \ref{thm2}
  can be treated in a similar way.

Let
$$
p(t,x)=\begin{cases}
A(t), &  x<A(t),\\
x, & A(t)\leq x\leq B(t),\\
B(t), & x>B(t).
\end{cases}
$$
It is interesting to give an introduction to Lemma \ref{lem4} and a
reference where it can be found.

\begin{lemma} \label{lem4}
 If $m>0$, then for any $q(t)\in L^1[0,2\pi]$, the problem
\begin{gather*}
u'(t)+mu(t)=q(t),\quad \mbox{for  a.e. }t\in[0,2\pi]\\
u(0)=u(2\pi),
\end{gather*}
has a unique solution $u\in W^{1,1}[0,2\pi]$, and
$$
u(t)=L^{-1}q(t)=\int_0^{2\pi}G_m(t,s)q(s)ds,
$$
where
$$
G_m(t, s): =\begin{cases}
\frac{e^{m(2\pi+s-t)}}{e^{2m\pi}-1}, & 0\leq s \leq t \leq 2\pi,\\
 \frac{e^{m(s-t)}}{e^{2m\pi}-1}, & 0\leq t \leq s \leq 2\pi.
\end{cases}
$$
\end{lemma}

By Lemma \ref{lem1}, we have
$$
\alpha (t)=L^{-1}A(t),\quad \beta (t)=L^{-1}B(t),\quad
\beta(t)\leq L^{-1}p(t,x)\leq \alpha(t).
$$
Now we consider the  modified problem
\begin{equation}
\begin{gathered}
x'(t)+Mx(t)=f\big(t,L^{-1}p(t,x(t)),(I-mL^{-1})p(t,x(t))\big)\\
+(m+M)p(t,x(t))-m^2L^{-1}p(t,x(t)),
x(0)=x(2\pi).
\end{gathered} \label{e2.1}
\end{equation}
For each $x\in C[0,2\pi]$, we define the mapping
$\Phi:C[0,2\pi]\to C[0,2\pi]$,
\begin{equation}
(\Phi x)(t)=\int_0^{2\pi}G_M(t,s)(Fx)(s)ds,\label{e2.2}
\end{equation}
where
\begin{align*}
(Fx)(t)&:=f\big(t,L^{-1}p(t,x(t)),(I-mL^{-1})p(t,x(t))\Big)\\
&\quad +(m+M)p(t,x(t))-m^2L^{-1}p(t,x(t)).
\end{align*}
Since $p(t,x(t))$ and $L^{-1}p(t,x(t))$ are bounded and
$f(t,u,v)$ is  a Caratheodary function,
  there exists $g(t)$,  a Lebesgue integrable function defined
on $[0,2\pi]$ such that
$$
|(Fx)(t)|\leq g(t)\quad \hbox{for a. e. }t\in [0, 2\pi].
$$
Thus $(\Phi x)(t)$ is also bounded.

We can easily prove that $\Phi:C[0,2\pi]\to C[0,2\pi]$ is
completely continuous. Then Leray-Schauder fixed point Theorem
assures that $\Phi$ has a fixed point $x\in C[0,2\pi]$ and
\begin{equation}
x(t)=\int_0^{2\pi}G_M(t,s)(Fx)(s)ds,\label{e2.3}
\end{equation}
thus the modified problem \eqref{e2.1} has one solution
$x\in W^{1,1}[0,2\pi]$.


 \begin{lemma} \label{lem5}
Suppose that (H1) holds. Assume that $\alpha (t)$ and $\beta (t)$
are lower and upper solutions to $\eqref{e1.1}$ and
$\beta (t) \leq \alpha (t)$ on $[0,2\pi]$. Let $x\in W^{1,1}[0,2\pi]$
be a solution to \eqref{e2.1},
then $A(t)\leq x(t)\leq B(t)$ on $[0,2\pi]$.
\end{lemma}

\begin{remark} \label{rmk2} \rm
Lemma \ref{lem4} implies
$u(t)=L^{-1}x(t)=\int_0^{2\pi}G_m(t,s)x(s)ds$ is a solution to
\eqref{e1.1}, since $u'(t)+mu(t)=x(t),u(0)=u(2\pi)$ and
$A(t)\leq x(t)\leq B(t)$.
\end{remark}

\begin{proof}[Proof of Lemma \ref{lem5}]
Since $\alpha (t)=L^{-1}A(t),\beta (t)=L^{-1}B(t)$,
\begin{gather*}
B'(t)+MB(t)\leq
f(t,L^{-1}B(t),(I-mL^{-1})B(t))-m^2L^{-1}B(t)+(m+M)B(t),\\
B(0)\leq B(2\pi)
\end {gather*}
 and
\begin{gather*}
A'(t)+MA(t)\geq
f(t,L^{-1}A(t),(I-mL^{-1})A(t))-m^2L^{-1}A(t)+(m+M)A(t),\\
A(0)\geq A(2\pi).
\end {gather*}
We shall  prove only that $x(t)\leq B(t)$ on $[0, 2\pi]$, because
$A(t)\leq x(t)$ can be proved by a similar manner. Let
$y(t)=x(t)-B(t)$, then
$$y(0)\geq y(2\pi).$$
Assume that $y(t)>0$ for some $t\in[0,2\pi]$. Indeed, if $y(t)>0$
on $[0,2\pi]$, we have
\begin{align*}
x'(t)+Mx(t)&=
f(t,L^{-1}B(t),(I-mL^{-1})B(t))-m^2L^{-1}B(t)+(m+M)B(t)\\
&\geq B'(t)+MB(t),
\end{align*}
i.e., $y'(t)+My(t)\geq 0$ on $[0,2\pi]$. Lemma \ref{lem1} implies
$y(t)\leq 0$
 on $[0,2\pi]$, which is  a contradiction.
Therefore, there would be a point $s\in [0,2\pi]$ with $y(s)\leq 0$.

 If $y(0)\leq 0$ (then $y(2\pi)\leq y(0)\leq 0$), and hence there
 exist $0\leq a<s<b\leq 2\pi$ such that $y(t)>0$ in $(a,b)$ with
 $y(a)=y(b)=0$. Then $p(t,x(t))=B(t)$ on $[a,b]$ and
\begin{align*}
 &y'(t)+M y(t)\\
 &\geq
 f(t,L^{-1}p(t,x(t)),B(t)-mL^{-1}p(t,x(t)))+(m+M)B(t)-m^2L^{-1}p(t,x(t))\\
 &\quad - [f(t,L^{-1}B(t),B(t)-mL^{-1}B(t))+(m+M)B(t)-m^2L^{-1}B(t)]\\
 &\geq (-A-Bm-m^2)(L^{-1}p(t,x(t))-L^{-1}B(t))\\
 &= 0,\quad t\in (a,b).
 \end{align*}
This leads to $y'(t)\geq -My(t)>0$ on $(a,b)$, which is again a
contradiction.

If $y(0)>0$, there exists $a\in(0,2\pi)$ such that $y(t)>0$ on
$[0, a)$ with $y(a)=0$. So we have $y'(t)+My(t)\geq 0$, hence
$y'(t)>0$  in $[0, a)$, which implies that $y(0)<y(a)=0$, this
is also a contradiction.
The proof is complete.
\end{proof}

By Remark \ref{rmk2},
we have obtained the results of Theorem \ref{thm1}.

\section{Example}

In this section, we consider the periodic boundary-value problem
\begin{equation}
\begin{gathered}
u''(t) + k u'(t) = F(t, u),\quad t\in [0,  2\pi]\\
u(0)=u(2\pi), u'(0)=u'(2\pi),
\end{gathered} \label{e3.1}
\end{equation}
where $F(t, u)$ is a Caratheodory function,  $k >0$ or $k<0$.

      We say that $\beta\in W^{2, 1}[0, 2\pi]$ is an upper
solution to the problem \eqref{e3.1},  if it satisfies
\begin{equation}
\begin{gathered}
\beta''(t) + k \beta'(t) \leq F(t, \beta(t)),\quad t\in [0,  2\pi]\\
\beta(0)=\beta(2\pi), \beta'(0)\leq\beta'(2\pi).
\end{gathered}\label{e3.2}
\end{equation}
Similarly,  a function $\alpha\in W^{2, 1}[0,  2\pi]$ is said to
be a lower solution to \eqref{e3.1}, if it satisfies
\begin{equation}
\begin{gathered}
\alpha''(t) + k \alpha'(t)  \geq F(t, \alpha(t)),\quad t\in [0,  2\pi]\\
\alpha(0)=\alpha(2\pi), \alpha'(0)\geq\alpha'(2\pi).
\end{gathered} \label{e3.3}
\end{equation}
To develop the upper and lower solutions method,  we also need the
following hypothesis:
\begin{itemize}
\item[(H)] For any given $\beta,  \alpha\in C[0,  2\pi]$ with
$\beta(t)\leq \alpha(t)$ on $[0,  2\pi]$, the inequality
$$
F(t, u_2)-F(t, u_1) \geq - \frac{k^2}{4} (u_2-u_1)
$$
holds for a.e. $t\in [0,  2\pi]$, whenever $\beta(t)\leq u_1\leq
u_2\leq \alpha(t)$.
\end{itemize}


Let $ A=k^2/4$, $B=|k|$,    then (H1)  holds when $k<0$,
and (H1')  holds when $k>0$. Hence the conclusions of Theorem \ref{thm1}
hold when $k<0$, thus
$\alpha'(t)+\frac{k}{2}\alpha(t)\leq \beta'(t)+\frac{k}{2}\beta(t)$
and problem \eqref{e3.1} has one solution
$u\in W^{2,1}[0,2\pi]$ such that
$$
\beta(t)\leq u(t)\leq \alpha (t), \alpha'(t)
+\frac{k}{2}\alpha(t)\leq u'(t)+\frac{k}{2}u(t)
\leq \beta'(t)+\frac{k}{2}\beta(t).
$$
The conclusions of Theorem \ref{thm2} hold when $k>0$, thus
$\alpha'(t)+\frac{k}{2}\alpha(t)\geq \beta'(t)+\frac{k}{2}\beta(t)$
 and problem \eqref{e3.1} has one solution
$u\in W^{2,1}[0,2\pi]$ such that
$$
\beta(t)\leq u(t)\leq \alpha (t),~\beta'(t)+\frac{k}{2}\beta(t)
\leq u'(t)+\frac{k}{2}u(t)
\leq \alpha'(t)+\frac{k}{2}\alpha(t).
$$
In \cite{j2,w2}, the authors obtained one solution
$u\in W^{2,1}[0,2\pi]$ of \eqref{e3.1} such that
$\beta(t)\leq u(t)\leq \alpha (t)$.
Here we have improved the results of \cite{j2,w2}.

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\end{document}