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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 116, pp. 1--10.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/116\hfil Upper and lower solutions]
{The Method of upper and lower solutions for
 second-order non-homogeneous two-point boundary-value problem}

\author[M. Jia,  X. Liu\hfil EJDE-2007/116\hfilneg]
{Mei Jia,  Xiping Liu}  % in alphabetical order

\address{Mei Jia \newline
College of Science, University of Shanghai for Science
and Technology, Shanghai 200093,  China}
\email{jiamei-usst@163.com}

\address{Xiping Liu \newline
College of Science, University of Shanghai for Science and
Technology, Shanghai 200093,  China} 
\email{xipingliu@163.com}

\thanks{Submitted June 7, 2007. Published August 30, 2007.}
\thanks{Supported by grant 05EZ52 from the Foundation of
 Educational Department of Shanghai}
\subjclass[2000]{34B15, 34B27}
\keywords{Upper and lower solutions; cone; monotone iterative method}

\begin{abstract}
  This paper studies the existence and uniqueness of
  solutions for a type of second-order two-point boundary-value
  problem depending on the first-order derivative through a non-linear
  term. By constructing a special cone and using the upper and lower
  solutions method, we obtain the sufficient conditions of the
  existence and uniqueness of solutions, and a monotone
  iterative sequence  solving the boundary-value problem.
  An error estimate formula is also given under the condition of
  a unique solution.
\end{abstract}

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

\section{Introduction}

 In this paper, we study the existence and uniqueness of
solutions to the second-order non-homogeneous two-point
boundary-value problem
\begin{equation} \label{e1.1}
\begin{gathered}
x''(t)+f(t,x(t),x'(t))=0,\quad t\in(0,1),\\
 x'(0)=a,\quad x(1)=b,
\end{gathered}
\end{equation}
where $f\in C([0,1]\times \mathbb{R}^2,\mathbb{R})$, and $a$,
$b\in \mathbb{R}$.

It is well known that the upper and lower solutions method is an
important tool in studying boundary-value problem of ordinary
differential equation. Recently, there are numerous results of the
problem by means of the method (see the references in this article).
We notice that most of these papers study the existence and
uniqueness of solutions of the boundary-value problem with
nonlinear term $f(t,u)$. The nonlinear term $f$, however, usually
satisfies Nagumo condition when the $f$ depends on the first order
derivative (see for example \cite{b1,d1,j1,j2}), which weakens the role of
the first order derivative term.

In this paper, the nonlinear term $f$ depends on the first order
derivative and does not need to satisfy the Nagumo condition. By
constructing a special cone and using the upper and lower
solutions method, we obtain the sufficient conditions of the
existence and uniqueness of solutions, as well as the monotone
iterative sequence which is used to solve the boundary-value
problem. The error estimate formula is also given under the
condition of unique solution. And the method we adopt is new and
so are the conclusions we obtain.

\section{Preliminaries}

 Throughout this paper, we assume that $N$ satisfies the hypothesis
\begin{itemize}
\item[(H1)]  $0<N<\frac{\pi}{2}$.
\end{itemize}
To investigate the boundary-value problem \eqref{e1.1}, we
consider the boundary-value problem
\begin{equation} \label{e2.1}
\begin{gathered}
-x''(t)-N^2x(t)=h(t),\quad t\in(0,1),\\
 x'(0)=0,\quad  x(1)=0,
\end{gathered}
\end{equation}
where $h\in C[0,1]$.

\begin{lemma} \label{lem2.1}
 The Green's function of the boundary-value problem
\begin{equation} \label{e2.2}
\begin{gathered}
-x''(t)-N^2x(t)=0,\quad t\in(0,1),\\
x'(0)=0,\quad x(1)=0,
\end{gathered}
\end{equation}
is
\begin{equation} \label{e2.3}
G (t,s)=\frac{1}{N\cos N} \begin{cases}
\cos(Nt)\sin(N(1-s)), & 0\leq t\leq s\leq 1,\\
\cos(N s)\sin(N(1-t)), & 0\leq s\leq t\leq 1.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof} We look for a Green's function of the form
\[
G (t,s)=\begin{cases}
A\cos(N t)+B\sin(N t),\quad 0\leq t\leq s\leq 1,\\
C\cos(N t)+D\sin(N t),\quad 0\leq s\leq t\leq 1.
\end{cases}
\]
By the definition and properties of the Green's function and the
boundary  conditions, it is easy to obtain that
\begin{gather*}
A=\frac{\sin(N(1-s))}{N\cos N}, \quad B=0, \\
C=\frac{\sin N\cos(Ns)}{N\cos N}, \quad D=\frac{-\cos(Ns)}{N}.
\end{gather*}
Hence, the Green's function is as stated in the Lemma.
\end{proof}

It is easy to show that the following lemma holds by means of
calculations.

\begin{lemma} \label{lem2.2}
 If (H1) holds. Then: (1)
$$
\frac{\partial G(t,s)}{\partial t}=-\frac{1}{\cos N}
\begin{cases}
\sin(Nt)\sin(N(1-s)), & 0\leq t< s\leq 1,\\
\cos(N s)\cos(N(1-t)), & 0\leq s< t\leq 1,
\end{cases}
$$
$G(t,s)\geq 0$ and $\frac{\partial G(t,s)}{\partial t}\leq 0$ for
all $t,s\in [0,1]$; (2)
\begin{gather*}
\int^1_0G(t,s)\,\mathrm{d}s= \frac{\cos(Nt)-\cos N}{N^2\cos N}\quad
\text{for all }t\in [0,1] ,\\
 \max_{t\in [0,1]}\int^1_0G(t,s)\,\mathrm{d}s= \frac{1-\cos N}{N^2\cos N};
\end{gather*}
(3)
\begin{gather*}
 \int^1_0\Big(-\frac{\partial G(t,s)} {\partial
t}\Big)\,\mathrm{d}s=\frac{\sin(Nt)}{N\cos N}\quad
\text{for all } t\in [0,1],\\
 \max_{t\in [0,1]}\int^1_0\Big(-\frac{\partial G(t,s)} {\partial
t}\Big)\,\mathrm{d}s= \frac{\sin N}{N\cos N}
\end{gather*}
\end{lemma}

\begin{lemma} \label{lem2.3}
Suppose $h\in C[0,1]$, $\overline{a}$ and $\overline{b}\in
\mathbb{R}$ , then the unique solution of the boundary-value
problem
\begin{equation} \label{e2.4}
\begin{gathered}
-x''(t)-N^2x(t)=h(t),\quad t\in(0,1),\\
x'(0)=\overline{a},\quad x(1)=\overline{b},
\end{gathered}
\end{equation}
is
$$
 x(t)=x_0(t)+\int_0^1G (t,s)h(s)\,\mathrm{d}s
$$
where
\begin{equation} \label{e2.5}
x_0(t)=\frac{1}{N\cos N}[\overline{b}N\cos(N
t)-\overline{a}\sin(N(1-t))]
\end{equation}
\end{lemma}

\begin{proof}
Note that the equation $-x''(t)-N^2x(t)=0$, has solutions of the form
$$
x(t)=c_1\cos(N t)+c_2\sin(N t)\,.
$$
 Using the boundary condition in \eqref{e2.4}, we
obtain that the unique solution of the boundary-value problem
 $$
\begin{gathered}
-x''(t)-N^2x(t)=0,\quad t\in(0,1),\\
x'(0)=\overline{a},\quad x(1)=\overline{b},
\end{gathered}
$$
is
$$
x_0(t)=\frac{1}{N\cos N}[\overline{b}N\cos(N
t)-\overline{a}\sin(N(1-t))]
$$
Since $G(t,s)$ is the Green's function of the
boundary-value problem \eqref{e2.2}. Then the unique solution of the
boundary-value problem \eqref{e2.4} is
$$
x(t)=x_0(t)+\int_0^1G (t,s)h(s)\,\mathrm{d}s
$$
\end{proof}

 From the hypothesis (H1) and the definition of $x_0(t)$,
  it is easy to see that the following lemma holds.

\begin{lemma} \label{lem2.4}
 If $\overline{b}\geq 0$, $\overline{a}=0$
and (H1) hold, then $x_0(t)\geq 0$ for all $t\in [0,1]$, where
$x_0(t)$ is defined by \eqref{e2.5}.
\end{lemma}

In the following, we establish a comparison principle.

\begin{lemma} \label{lem2.5}
Suppose that $x\in C^2[0,1]$ satisfies
\begin{gather*}
-x''(t)-N^2x(t)\geq 0,\quad t\in(0,1),\\
x'(0)=0,\quad x(1)\geq 0.
\end{gather*}
Then $x(t)\geq 0$ for all $t\in [0,1]$.
\end{lemma}

\begin{proof}  Let
$$
h(t)=-x''(t)-N^2x(t),\quad
\overline{a}=0,\quad x(1)=\overline{b}\,.
$$
Then
$h(t)\geq 0$ for all $t\in [0,1]$ and $\overline{b}\geq 0$.
Consider the boundary-value problem
\begin{equation} \label{e2.6}
\begin{gathered}
-u''(t)-N^2u(t)=h(t),\quad t\in(0,1),\\
u'(0)=0,\quad u(1)=\overline{b},
\end{gathered}
\end{equation}
By Lemma \ref{lem2.3}, this boundary-value problem has the unique
solution
\[
u(t)=x_0(t)+\int_0^1G (t,s)h(s)\,\mathrm{d}s
\]
Since $x_0(t)\geq 0$ by  Lemma \ref{lem2.4} and $G(t,s)\geq 0$ by
Lemma \ref{lem2.2} for all $t, s\in [0,1]$, we have $u(t)\geq 0$ for all $t\in
[0,1]$. It follows from the definition of $h$ that $x$ is a
solution of the boundary-value problem \eqref{e2.6}. Hence, $u=x$ which
gives $x(t)\geq 0$ for all $t\in [0,1]$.
\end{proof}

\begin{lemma}[{\cite[Lemma 1.1.2]{g1}}] \label{lem2.6}
Let $E$ be partially ordered Banach space, $\{x_n\}\subset E$ is
monotone sequence and relatively compact set, then $\{x_n\}$ is
convergent.
\end{lemma}

\begin{lemma}[{\cite[Lemma 1.1.2]{g1}}] \label{lem2.7}
Let $E$ be partially ordering Banach space, $x_n\preceq y_n$,
$(n=1,2,3\dots)$, if $x_n\to x^*$, $y_n\to y^*$,
we have $x^*\preceq y^*$.
\end{lemma}

Here the symbol  $\preceq$ denotes the partially order in the
Banach space $E$.

\noindent\textbf{Definition.} \; %2.1
A function $\varphi_0\in C^2([0,1])$ is said to be a lower solution of
boundary-value problem \eqref{e1.1}, if
\begin{gather*}
-{\varphi_0''}(t)\leq f(t,\varphi_0(t),\varphi'_0(t)),\\
\varphi'_0(0)=a,\quad
\varphi_0(1)\leq b .
\end{gather*}
A function $\psi_0\in C^2([0,1])$ is said to be an upper
solution of the boundary-value problem \eqref{e1.1}, if
\begin{gather*}
-{\psi_0''}(t)\geq f(t,\psi_0(t),\psi'_0(t)),\\
\psi'_0(0)=a,\quad
\psi_0(1)\geq b.
\end{gather*}

 \section{Existence of solutions of the boundary-value problem}

 Let $E=C^1[0,1]$ with
$\|x\|=\max\{|x|_\infty,|x'|_\infty \}$, where
$|x|_\infty=\max\limits_{t\in [0,1]} |x(t)|$. Let
\[% \begin{align*}
P=\{x\in E: x(t)\geq 0 \text{ for all }t\in [0,1],
    x' \text{ is decreasing and } x'(0)\leq 0 \}\,.
\]% \end{align*}
Then $P$ is a cone in $E$ and $E$ is a partially ordered Banach space.

Obviously, for any $x\preceq y\in E$ if only and if $y-x \in P$,
namely, $x(t)\leq y(t)$ for all $t\in [0,1]$, $y'-x'$ is monotone
decreasing and $ y'(0)-x'(0)\leq 0$. Then $y'(t)-x'(t)\leq
y'(0)-x'(0)\leq 0$ for all $t\in [0,1]$. Therefore
\begin{equation} \label{e3.1}
x\preceq y\in E\;\Rightarrow\; x(t)\leq
y(t)\text{ and } y'(t)-x'(t)\leq 0
\text{ for all } t\in [0,1].
\end{equation}
For any $\alpha\preceq\beta\in E$, denote $D_0=[\alpha,
\beta]=\{x\in E:\alpha\preceq x\preceq\beta\}$. It is easy to
see that $D_0$ is a bounded set.

\begin{theorem} \label{thm3.1}
Suppose (H1) holds, and there exist a
upper solution $\psi_0$ and a lower solution $\varphi_0$ of
boundary-value problem \eqref{e1.1} such that
$\varphi_0\preceq\psi_0$ and $f$ satisfies:
\begin{itemize}
\item[(H2)] $f(t,u_2,v)-f(t,u_1, v)\geq N^2(u_2-u_1)$
  for all $t\in [0,1]$, $\psi'_0(t)\leq v \leq\varphi'_0(t)$ and
  $\varphi_0(t)\leq u_1\leq   u_2\leq\psi_0(t)$;

\item[(H3)] $f(t,u,v_2)-f(t,u,v_1)\leq 0$ for all $t\in
[0,1]$, $\varphi_0(t)\leq u \leq\psi_0(t)$ and $\psi'_0(t)\leq
v_1\leq v_2\leq\varphi'_0(t)$.
\end{itemize}
Then the boundary-value problem \eqref{e1.1} has a minimal
solution $\varphi^*$ and a maximal solution $\psi^*$ on the ordered
interval $[\varphi_0,\psi_0]$.
Moreover, the iterative sequences
\begin{gather*}
\varphi_n(t)=\overline{x}_0(t)+\int_0^1G (t,s)(f(s,
\varphi_{n-1}(s),\varphi'_{n-1}(s))
-N^2\varphi_{n-1}(s))\,\mathrm{d}s,
\\
\psi_n(t)=\overline{x}_0(t)+\int_0^1G (t,s)(f(s,\psi_{n-1}(s),
\psi'_{n-1}(s)) -N^2\psi_{n-1}(s))\,\mathrm{d}s
\end{gather*}
converge uniformly on $[0,1]$ to $\varphi^*$ and $\psi^*$
respectively. Here
$$
\overline{x}_0(t)=\frac{1}{N\cos N}[bN\cos(N
t)-a\sin(N(1-t))].
$$
\end{theorem}

\begin{proof}
 It is easy to see that $x=\varphi_0=\psi_0$ is
the solution of the boundary-value problem \eqref{e1.1} if
$\varphi_0\equiv\psi_0$. Next we consider
$\varphi_0\not\equiv\psi_0$.
 Denote $D=[\varphi_0,\psi_0]$.
For any $h\in D$, we consider the boundary-value problem
\begin{equation} \label{e3.2}
\begin{gathered}
-x''(t)-N^2x(t)=f(t,h(t),h'(t))-N^2h(t),\\
x'(0)=a,\quad x(1)=b.
\end{gathered}
\end{equation}
By Lemma \ref{lem2.3}, the unique solution of the above boundary-value problem
is
\begin{equation} \label{e3.3}
x(t)=\overline{x}_0(t)+\int_0^1G (t,s)(f(s,h(s),h'(s))
 -N^2h(s))\,\mathrm{d}s:=(Qh)(t)
\end{equation}
where
$$
\overline{x}_0(t)=\frac{1}{N\cos N}[bN\cos(N t)-a\sin(N(1-t))]
$$
It is clear that $x$ is a solution of boundary-value problem
\eqref{e1.1} if and only if $x$ is a fixed point of $Q$.

Let $F:D\to C([0,1])$, $(Fh)(t)=f(t,h(t),h'(t))-N^2h(t)$.
 Then $F$ is a continuous and bounded operator.

 Define $T:C([0,1])\to C^1([0,1])$,
$(Th)(t)=\overline{x}_0(t)+ \int_0^1G (t,s)h(s)
\,\mathrm{d}s$.
It is obvious that $T$ is a linear completely continuous operator.

Denote $Q=T\circ F$, so $Q:D\to C^1([0,1])$ is continuous
and relatively compact, $Q(D)$ is a relatively compact set.

\noindent (1) We prove $Q$ is an increasing operator.
For any $h_1$, $h_2\in D$ and
$h_1\preceq h_2$, by \eqref{e3.1}, we have
\[
\varphi_0(t)\leq h_1(t)\leq h_2(t)\leq
\psi_0(t)\quad\text{and}\quad \psi'_0(t)\leq
h'_2(t)\leq h'_1(t)\leq\varphi'_0(t)
\]
for all $t\in [0,1]$.
By (H2) and (H3),
\begin{align*}
&[f(t,h_2(t),h'_2(t))-N^2h_2(t)]-[f(t,h_1(t),h'_1(t))-N^2h_1(t)]\\
&=[f(t,h_2(t),h'_2(t))-f(t,h_1(t),h'_1(t)))]-N^2(h_2(t)-h_1(t))\\
&=[f(t,h_2(t),h'_2(t))-f(t,h_1(t),h'_2(t)))]+[f(t,h_1(t),h'_2(t))
-f(t,h_1(t),h'_1(t)))]\\
&\quad-N^2(h_2(t)-h_1(t))\\
&\geq N^2(h_2(t)-h_1(t))-N^2(h_2(t)-h_1(t))
 \geq 0
\end{align*}
 Therefore, $(Qh_1)(t)\leq (Qh_2)(t)$ by Lemma \ref{lem2.2}(1) for all $t\in [0,1]$.
Also for all $t\in [0,1]$,
\begin{align*}
(Qh_2)''(t)-(Qh_1)''(t)
&=-[N^2(Qh_2)(t)+f(t,h_2(t),h'_2(t))-N^2h_2(t)]\\
&\quad +[N^2(Qh_1)(t)+f(t,h_1(t),h'_1(t))-N^2h_1(t)]\\
&=N^2[(Qh_1)(t)-(Qh_2)(t)]  -[f(t,h_2(t),h'_2(t))-N^2h_2(t)]\\
&\quad +[f(t,h_1(t),h'_1(t))-N^2h_1(t)]\\
&\leq 0
\end{align*}
Hence, $(Qh_2)'(t)-(Qh_1)'(t)$ is monotonically decreasing
for all $t\in [0,1]$.

Obviously, $(Qh_2)'(0)-(Qh_1)'(0)=a-a=0$.
So $(Qh_2)-(Qh_1)\in P$, namely $Qh_1\preceq Qh_2$.
We get $Q:D\to C^1([0,1])$ is an increasing operator.

\noindent (2) We prove $Q\psi_0\preceq \psi_0$, $\varphi_0\preceq
Q\varphi_0$.
Denote $\psi_1=Q\psi_0$, since $\psi_0$ is the upper solution of
the boundary-value problem \eqref{e1.1}. Then
\begin{gather*}
-{\psi_0''}(t)\geq f(t,\psi_0(t),\psi'_0(t)),\\
\psi'_0(0)=a,\quad
\psi_0(1)\geq b.
\end{gather*}
Let $\psi=\psi_0-\psi_1$, the definition of $Q$ yields
\begin{align*}
&-\psi''(t)-N^2\psi(t)\\
&=-(\psi_0(t)-\psi_1(t))''-N^2(\psi_0(t)-\psi_1(t))\\
&=(-\psi_0''(t)-N^2\psi_0(t))-(-\psi_1''(t)-N^2\psi_1(t))\\
&\geq (f(t,\psi_0(t),\psi'_0(t))-N^2\psi_0(t))-
(f(t,\psi_0(t),\psi'_0(t))-N^2\psi_0(t))
=0
\end{align*}
for all $t\in [0,1]$ and
\begin{gather*}
\psi'(0)=\psi'_0(0)-\psi'_1(0)=a-a=0,\\
\psi(1)=\psi_0(1)-\psi_1(1)\geq b-b=0.
\end{gather*}
By Lemma \ref{lem2.5}, we have $\psi(t)\geq 0$ on $[0,1]$. That is
$(Q\psi_0)(t)\leq\psi_0(t)$ for all $t\in [0,1]$.
 Moreover
\begin{align*}
&\psi_0''(t)-(Q\psi_0)''(t)\\
&\leq -f(t,\psi_0(t),\psi'_0(t))+
[N^2(Q\psi_0)(t)+f(t,\psi_0(t),\psi'_0(t))-N^2\psi_0(t)]\\
&=N^2[(Q\psi_0)(t)-\psi_0(t)]
\leq 0
\end{align*}
 for all $t\in[0,1]$. Hence, $\psi_0'(t)-(Q\psi_0)'(t)$ is monotone
decreasing  for all $t\in [0,1]$.

Obviously, $\psi_0'(0)-(Q\psi_0)'(0)=a-a=0$.
Therefore, we get $Q\psi_0\preceq\psi_0$.
Similarly, we can prove that $\varphi_0\preceq Q\varphi_0$. So
$\varphi_0\preceq\varphi_1\preceq\psi_1\preceq\psi_0$.

\noindent (3) We prove the existence of the minimal solution and the maximal
solution of boundary-value problem \eqref{e1.1}.
We can repeat step (2) and construct an iterative sequence
\begin{gather*}
\psi_n=Q\psi_{n-1}
=\overline{x}_0(t)+\int_0^1G (t,s)(f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -N^2\psi_{n-1}(s))\,\mathrm{d}s,
\\
 \varphi_n=Q\varphi_{n-1}
=\overline{x}_0(t)+\int_0^1G (t,s)(f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s))
-N^2\varphi_{n-1}(s))\,\mathrm{d}s
\end{gather*}
for $n=1,2,\dots$. We obtain
\[
\varphi_0\preceq\varphi_1\preceq\varphi_2\preceq \dots\leq
\varphi_n\preceq\dots\preceq\psi_n\preceq\dots\preceq\psi_2
\preceq\psi_1\preceq\psi_0
\]
By $\{\psi_n\}$, $\{\varphi_n\}\subset Q(D)$ and Lemma \ref{lem2.6} we can
show that there exist $\varphi^*$, $\psi^*\in D$ such that
$\psi_n\to\psi^*$, $\varphi_n\to\varphi^*$
$(n\to \infty)$. By the continuity of $Q$, we have
$\varphi^*=Q\varphi^*$ and $\psi^*=Q\psi^*$ as $n\to
\infty$. So $\varphi^*$ and $\psi^*$ are the fixed points of $Q$.

In the following, we prove that $\varphi^*$, $\psi^*$ are the
minimal solution and the maximal solution of boundary-value
problem \eqref{e1.1}, respectively.

Assume $z\in D=[\varphi_0, \psi_0]$
 is a fixed point of $Q$, then $\varphi_0\preceq z\preceq\psi_0$.
 As $Q$ is an increasing operator, we get $Q\varphi_0\preceq Qz\preceq
 Q\psi_0$, that is $\varphi_1\preceq z\preceq\psi_1$.
 In a similar way we have $Q\varphi_1\preceq Qz\preceq
 Q\psi_1$, that is $\varphi_2\preceq z\preceq\psi_2$.
 Repeat it, and we have $\varphi_n\preceq z\preceq\psi_n$ for $n=3,4,\dots$.

By Lemma \ref{lem2.7}, we can obtain $\varphi^*\preceq z\preceq\psi^*$.
Namely, $\varphi^*$, $\psi^*$ are  the minimal fixed point and the
maximal point of $Q$, respectively.
Therefore, $\varphi^*$, $\psi^*$ are the minimal solution and
maximal solution of boundary-value problem \eqref{e1.1} in the
ordered interval  $[\varphi_0, \psi_0]$, respectively.
\end{proof}

\section{Uniqueness of solutions of the boundary-value problem}

It is easy to show that the following lemma holds.

\begin{lemma} \label{lem4.1}
If (H1) holds, then $\sin N>\frac{1-\cos N}{N}$.
\end{lemma}

\begin{theorem} \label{thm4.1}
 Suppose that the hypotheses of Theorem \ref{thm3.1} hold, and
\begin{itemize}
\item[(H4)] There exists a constant $M_1$ with $0<M_1<N\cot N$,
such that
$$
f(t,u,v_1)-f(t,u,v_2)\leq M_1(v_2-v_1)
$$
for all $t\in [0,1]$,
$\varphi_0(t)\leq u \leq\psi_0(t)$ and $\psi'_0(t)\leq v_1\leq
v_2\leq\varphi'_0(t)$;

\item[(H5)] There exists a constant $M_2$ with
$N^2<M_2<N^2+N\cot N-M_1$, such that $$f(t,u_2,v)-f(t,u_1,v)\leq
M_2(u_2-u_1)$$ for all $t\in [0,1]$, $\psi'_0(t)\leq
v\leq\varphi'_0(t)$ and $\varphi_0(t)\leq u_1\leq
  u_2\leq\psi_0(t)$.

\end{itemize}
Then the boundary-value problem \eqref{e1.1} has a unique solution
$x^*$ on $[\varphi_0, \psi_0]$ and for any
$x_0\in [\varphi_0, \psi_0]$, iterative sequence
\[
x_n(t)=\overline{x}_0(t)+ \int_0^1
G (t,s)(f(s,x_{n-1}(s),x'_{n-1}(s))-N^2x_{n-1}(s))\,\mathrm{d}s,\quad
n=1,2,\dots
\]
converge uniformly to $x^*$ on [0,1], and its error estimate formula is
\[
\|x_n-x^*\|\leq 2\Big(\frac{M_1+M_2-N^2}{N\cot N}\Big)^n
\|\psi_0-\varphi_0\|,\quad n=1,2,\dots.
\]
\end{theorem}

\begin{proof}
 Let $\varphi_n$ and $\psi_n$ be defined as in Theorem \ref{thm3.1}.
 According to (H4), (H5), Lemma \ref{lem2.2} and the assumptions
of Theorem \ref{thm4.1} we have
\begin{align*}
0&\leq\psi_n(t)-\varphi_n(t)\\
&=(Q\psi_{n-1})(t)-(Q\varphi_{n-1})(t)\\
&= \int_0^1G(t,s)(f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -N^2\psi_{n-1}(s))\,\mathrm{d}s\\
&\quad - \int_0^1G(t,s)(f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s))
 -N^2\varphi_{n-1}(s))\,\mathrm{d}s\\
&= \int_0^1G(t,s)\big[f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s))\\
&\quad +N^2(\varphi_{n-1}(s)-\psi_{n-1}(s))\big]\,\mathrm{d}s \\
&= \int_0^1G(t,s)[(f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -f(s,\psi_{n-1}(s),\varphi'_{n-1}(s)))\\
&\quad +(f(s,\psi_{n-1}(s),\varphi'_{n-1}(s))
 -f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s)))\\
&\quad +N^2(\varphi_{n-1}(s)-\psi_{n-1}(s))]\,\mathrm{d}s \\
&\leq \int_0^1G(t,s)[M_1(\varphi'_{n-1}(s))-\psi'_{n-1}(s))
 +M_2(\psi_{n-1}(s))-\varphi_{n-1}(s))\\
&\quad +N^2(\varphi_{n-1}(s)
 -\psi_{n-1}(s)]\,\mathrm{d}s\\
&= \int_0^1G(t,s)[M_1(\varphi'_{n-1}(s))-\psi'_{n-1}(s))
 +(M_2-N^2)(\psi_{n-1}(s))-\varphi_{n-1}(s))]\,\mathrm{d}s\\
&\leq(M_1+M_2-N^2) \int_0^1G(t,s)\|\psi_{n-1}
 -\varphi_{n-1}\| \,\mathrm{d}s\\
&\leq(M_1+M_2-N^2)\|\psi_{n-1}-\varphi_{n-1}\|
\frac{1-\cos N}{N^2\cos N}\,.
\end{align*}
Similarly
\begin{align*}
0&\leq \varphi'_n(t)-\psi'_n(t)\\
&=(Q\varphi_{n-1})'(t)-(Q\psi_{n-1})'(t)\\
&=- \int_0^1\frac{\partial
 G(t,s)}{\partial t }(f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -N^2\psi_{n-1}(s))\,\mathrm{d}s\\
&\quad + \int_0^1\frac{\partial
 G(t,s)}{\partial t }(f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s))
 -N^2\varphi_{n-1}(s))\,\mathrm{d}s\\
&= \int_0^1\Big(-\frac{\partial
 G(t,s)}{\partial t }\Big)[f(s,\psi_{n-1}(s),\psi'_{n-1}(s))
 -f(s,\varphi_{n-1}(s),\varphi'_{n-1}(s)) \\
&\quad +N^2(\varphi_{n-1}(s)-\psi_{n-1}(s))]\,\mathrm{d}s\\
&\leq (M_1+M_2-N^2) \int_0^1\Big(-\frac{\partial
 G(t,s)}{\partial t }\Big)\|\psi_{n-1}-\varphi_{n-1}\|
\,\mathrm{d}s\\
&\leq (M_1+M_2-N^2)\|\psi_{n-1}-\varphi_{n-1}\|\frac{\sin N}{N\cos N}\,.
\end{align*}
By Lemma \ref{lem4.1},
\begin{align*}
\|\psi_{n}-\varphi_{n}\|
&\leq\max\Big\{\frac{\sin
N}{N\cos N},\frac{1-\cos N}{N^2\cos
N}\Big\}(M_1+M_2-N^2)\|\psi_{n-1}-\varphi_{n-1}\| \\
&=\frac{1}{N\cot N}(M_1+M_2-N^2)\|\psi_{n-1}-\varphi_{n-1}\|
\end{align*}
Using the inequality repeatedly, we have
$$
\|\psi_{n}-\varphi_{n}\|\leq\Big(\frac{M_1+M_2-N^2}{N\cot
N}\Big)^n\|\psi_0-\varphi_0\|
$$
Noting that $0<\frac{M_1+M_2-N^2}{N\cot N}<1 $, we have
\[
\|\psi_n-\varphi_n\|\to 0,\quad\text{as }n\to \infty.
\]
Since $\psi_n\to\psi^*$, $\varphi_n\to \varphi^*$,
there exists the unique $x^*\in
\bigcap\limits_{n=1}^\infty[\varphi_n,\psi_n]$ such that
$\psi_n\to x^*$, $\varphi_n\to x^*$,
$(n\to \infty)$. So by Lemma \ref{lem2.7},
\[
\varphi_n\preceq x^*\preceq\psi_n,\quad x^*\in D.
\]
The monotonicity of $Q$ implies
\[
\varphi_{n+1}=Q\varphi_n\preceq Qx^*\preceq Q\psi_n=\psi_{n+1}.
\]
Let $n\to \infty$ we can show that $x^*\preceq Qx^*\preceq
x^*$. So $x^*=Qx^*$.
 Consequently, $x^*$ is the unique solution of boundary-value
problem \eqref{e1.1}.
For any $x_0\in [\varphi_0,\psi_0]$, we have
\[
\|x_n-x^*\|\leq \|x_n-\varphi_n\|+\|\varphi_n-x^*\|
\leq 2\|\psi_n-\varphi_n\|
\leq 2\Big(\frac{M_1+M_2-N^2}{N\cot N}\Big)^n\|\psi_0-\varphi_0\|
\]
where
\[
x_n(t)=\overline{x}_0(t)+ \int_0^1
G (t,s)(f(s,x_{n-1}(s),x'_{n-1}(s))-N^2x_{n-1}(s))\,\mathrm{d}s,\quad
n=1,2,\dots
\]
for all $t\in[0,1]$.
\end{proof}

\section{Illustration}

In this section, we give an example about the theoretical results.
 Let $a=0$, $b=1$, $N=1$,
$f(t,u,v)=1+(1+\frac{1}{8}t^2)u-\frac{1}{8}v$. Then $f\in
C([0,1]\times \mathbb{R}^2,\mathbb{R})$, $a$, $b\in \mathbb{R}$
and $N$ satisfies the hypothesis (H1).
Consider the boundary-value problem
\begin{equation} \label{e5.1}
\begin{gathered}
x''(t)+1+(1+\frac{1}{8}t^2)x(t)-\frac{1}{8}x'(t)=0,\quad t\in(0,1)\\
x'(0)=0,\quad x(1)=1\,.
\end{gathered}
\end{equation}
Let $\varphi_0(t)=\int_0^1G(t,s)\,\mathrm{d}s$,
$\psi_0(t)=4\int_0^1G(t,s)\,\mathrm{d}s+1$ for all $t\in[0,1]$,
where
\begin{equation} \label{e5.2}
G (t,s)=\frac{1}{\cos 1}
\begin{cases}
\cos t\sin(1-s), & 0\leq t\leq s\leq 1,\\
\cos s\sin (1-t), & 0\leq s\leq t\leq 1 .
\end{cases}
\end{equation}
So $\psi'_0(t)=4\varphi_0'(t)$ and $\psi''_0(t)=4\varphi''_0(t)$
for all $t\in[0,1]$. By Lemma \ref{lem2.2}, we have
\begin{gather*}
 \max_{t\in[0,1]}\int_0^1G(t,s)\,\mathrm{d}s
 =\frac{1}{\cos 1}-1\approx 0.8508, \\
 \max_{t\in[0,1]}\int_0^1\big(-\frac{\partial
G(t,s)}{\partial t }\big)\,\mathrm{d}s=\tan 1\approx 1.5574,
\end{gather*}
and $\varphi_0'(t)\leq 0$ for all $t\in[0,1]$.
By Lemma \ref{lem2.1}, Lemma \ref{lem2.2} and Lemma \ref{lem2.3}, we can obtain
\begin{gather*}
-\varphi_0''(t)
=\varphi_0(t)+1\leq 1+(1+\frac{1}{8}t^2)\varphi_0(t)-\frac{1}{8}\varphi_0'(t)
=f(t, \varphi_0(t), \varphi_0'(t)),\quad t\in(0,1)\\
\varphi'_0(0)=0,\quad \varphi_0(1)=0<1
\end{gather*}
and
\begin{gather*}
-\psi_0''(t)
=\psi_0(t)+3\geq 1+(1+\frac{1}{8}t^2)\psi_0(t)-\frac{1}{8}\psi_0'(t)
=f(t,\psi_0(t),\psi_0'(t)),\quad t\in(0,1)\\
\psi'_0(0)=0,\quad \psi_0(1)=1
\end{gather*}
Hence, $\varphi_0$, $\psi_0$ are the lower solution and the upper
solution of the boundary-value problem \eqref{e5.1}, respectively, and
$\varphi_0\preceq\psi_0$.

Let $M_1=\frac{1}{8}$, $M_2=\frac{9}{8}$. Then $0<M_1<\cot 1$ and
$1<M_2<1+\cot 1-M_1$.
Therefore, the boundary-value problem \eqref{e5.1} satisfies the
conditions of Theorem \ref{thm4.1}. Then the boundary-value problem \eqref{e5.1}
has the unique solution $x^*$ on $[\varphi_0, \psi_0]$ and for any
$x_0\in [\varphi_0, \psi_0]$, iterative sequence
\[
x_n(t)=\overline{x}_0(t)+
\int_0^1G(t,s)(f(s,x_{n-1}(s),x'_{n-1}(s))
-x_{n-1}(s))\,\mathrm{d}s,\quad n=1,2,\dots
\]
converge uniformly to $x^*$ on $[0,1]$, and its error estimate formula is
\[
\|x_n-x^*\|\leq 2\big(\frac{\tan 1}{4}\big)^n\|\psi_0-\varphi_0\|
,\quad n=1,2,\dots.
\]


\subsection*{Acknowledgements}
We are grateful to the anonymous referees for their
valuable comments and suggestions.

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\end{document}