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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 219, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/219 \hfil  Monotone iterative method ]
{Monotone iterative method for obtaining positive solutions 
of integral boundary-value problems with $\phi$-Laplacian operator}

\author[Y. Ding \hfil EJDE-2012/219\hfilneg]
{Yonghong Ding}  % in alphabetical order

\address{Yonghong Ding \newline
 Department of Mathematics, Tianshui Normal University,
 Tianshui 741000, China}
\email{dyh198510@126.com}

\thanks{Submitted October 11, 2012. Published November 29, 2012.}
\subjclass[2000]{34B15, 34B18}
\keywords{$\phi$-Laplacian; monotone iterative; cone; positive solutions}

\begin{abstract}
 This article concerns the existence, multiplicity of positive solutions
 for the integral boundary-value problem with $\phi$-Laplacian,
 \begin{gather*}
 \big(\phi(u'(t))\big)'+f(t,u(t),u'(t))=0,\quad t\in[0,1],\\
 u(0)=\int_0^1 u(r)g(r)\,\mathrm{d}r,\quad u(1)=\int_0^1u(r)h(r)\,\mathrm{d}r,
 \end{gather*}
 where $\phi$ is an odd, increasing homeomorphism from $\mathbb{R}$
 to $\mathbb{R}$. Using a monotone iterative technique,
 we obtain the existence of positive solutions for this problem,
 and present iterative schemes for approximating the solutions.
\end{abstract}

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

\section{Introduction}

Consider the integral boundary-value problem
\begin{equation}
\begin{gathered}
    \big(\phi(u'(t))\big)'+f(t,u(t),u'(t))=0,\quad t\in[0,1],\\
      u(0)=\int_0^1u(r)g(r)\,\mathrm{d}r,\quad
u(1)=\int_0^1u(r)h(r)\,\mathrm{d}r,
    \end{gathered} \label{e1}
\end{equation}
where $\phi,f,g$ and $h$ satisfy
\begin{itemize}

\item[(H1)] $\phi$ is an odd, increasing homeomorphism from $\mathbb{R}$
onto $\mathbb{R}$ and there exist two increasing homeomorphisms
$\psi_1$ and $\psi_2$ of $(0, \infty)$ onto $(0, \infty)$ such that
$$
\psi_1(u)\phi(v) \leq \phi(uv) \leq \psi_2(u)\phi(v), u, v > 0.
$$
Moreover,
$\phi, \phi^{-1}\in C^1(R)$, where
$\phi^{-1}$ denotes the inverse of $\phi$.

\item[(H2)] $f:[0, 1]\times [0, +\infty)\times
(-\infty, +\infty)\to (0, +\infty)$ is continuous.
$g, h\in L^1[0, 1]$ are nonnegative, and
$0<\int_0^1g(t)\,\mathrm{d}t<1, 0<\int_0^1h(t)\,\mathrm{d}t<1$.
\end{itemize}
The assumption (H1) on the function $\phi$ was first introduced by
Wang [1, 2]. It covers two special cases:  $\phi(u)=u$ and
$\phi(u)=|u|^{p-2}u, p>1$. Many authors have studied the positive solutions
of two-point and multi-point boundary value problems when $\phi(u)=u$ or
$\phi(u)=|u|^{p-2}u, p>1$.
For details and references see \cite{a1,f1,l1,s1,w3,w4}.

In a recent paper \cite{d1}, the author proved the existence and
multiplicity of positive solutions of \eqref{e1} by
applying  the Krasnoselskii fixed point theorem and Avery-Peterson
fixed point theorem. However there is an interesting question which is
showing how to find these solutions, since they exist.
Motivated by the question and all the works above, in this article,
by applying classical monotone iterative techniques,  we not only obtain
the existence of positive solutions of \eqref{e1}, but also give iterative
schemes for approximating the solutions. It is worth stating that we will
not require the existence of lower and upper solutions, and
the first term of iterative scheme is a constant function or a simple function.
 Therefore, the iterative scheme is significant and feasible.

\section{Preliminaries}

 The basic space used in this article is the real Banach space
 $C^1[0, 1]$ with norm
 $\|u\|_1=\max\{\|u\|_{c}, \|u'\|_{c}\}$, where
$\|u\|_{c}=\max_{0\leq t\leq  1}|u(t)|$.
Let
\[
P=\big\{u\in C^1[0, 1]: u(t)\geq 0, u \text{ is concave on $[0, 1]$}\big\}.
\]
It is obvious that $P$ is a cone in $C^1[0, 1]$.

For any $x\in C^1[0, 1]$, $x(t)\geq 0$, $t\in [0,  1]$, we suppose that $u$ is
 a solution of the boundary-value problem
\begin{equation}
\begin{gathered}
    \big(\phi(u'(t))\big)'+f(t,x(t),x'(t))=0, \quad t\in[0,1],\\
  u(0)=\int_0^1u(r)g(r)\,\mathrm{d}r, \quad
  u(1)=\int_0^1u(r)h(r)\,\mathrm{d}r.
    \end{gathered} \label{e2}
\end{equation}
By integrating  \eqref{e2} on $[0, t]$, we have
$$
\phi(u'(t))=A_x-\int_0^t f(s,x(s),x'(s))\,\mathrm{d}s,
 $$
then
\begin{equation}
u'(t)=\phi^{-1}\Big(A_x-\int_0^t f(s,x(s),x'(s))\,\mathrm{d}s\Big).
 \label{e3}
\end{equation}
Thus
\begin{equation}
u(t)=u(0)+\int_0^t\phi^{-1}\Big(A_x-\int_0^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s
\label{e4}
\end{equation}
or
\begin{equation}
u(t)=u(1)-\int_t^1\phi^{-1}\Big(A_x-\int_0^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s.
\label{e5}
\end{equation}
According to the boundary condition, we have
$$
u(0)=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(A_x-\int_0^s f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,
\mathrm{d}s\,\mathrm{d}r
$$
and
$$
u(1)=-\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\int_0^1h(r)\int_r^1\phi^{-1}
\Big(A_x-\int_0^sf(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)
\,\mathrm{d}s\,\mathrm{d}r,
$$
where $A_x$ satisfies the equation
\begin{equation}
\begin{aligned}
H_x(c)
&=\frac{1-\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1g(r)\,\mathrm{d}r}
 \int_0^1g(r)\int_0^r\phi^{-1}\Big(c-\int_0^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\Big(1-\int_0^1h(r)\,\mathrm{d}r\Big)\int_0^1\phi^{-1}\Big(c-\int_0^s
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\\
&\quad +\int_0^1h(r)\int_r^1\phi^{-1}\Big(c-\int_0^sf(\tau,x(\tau),x'(\tau))
 \,\mathrm{d}\tau\Big) \,\mathrm{d}s\,\mathrm{d}r=0.
\end{aligned}
\label{e6}
\end{equation}

\begin{lemma} \label{lem2.1}
Assume {\rm (H1)} and {\rm (H2)} hold, for any $x\in C^1[0, 1]$
 with $x(t)\geq 0, t\in [0, 1]$, there exists a
unique $A_x\in (-\infty, +\infty)$ satisfying \eqref{e6}.
Moreover, there is a unique $\delta_x\in (0,1)$ such that
$$
A_x=\int_0^{\delta_x} f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau.
$$
\end{lemma}

\begin{proof} 
From the expression of $H_x(c)$ it is easy to see that 
$H_x: \mathbb{R}\to\mathbb{R}$  is continuous and strictly increasing, and
 $$
H_x(0)<0,  \quad H_x\Big(\int_0^1 f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)>0.
 $$
 Hence there exists a unique 
$A_x\in (0,  \int_0^1 f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau)
\subset(-\infty,  +\infty)$ satisfying \eqref{e6}. Let
 $$
 F(t)=\int_0^t f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau,
 $$
 then $ F(t)$ is continuous and strictly increasing on $[0, 1]$, and 
$F(0)=0$, $F(1)=\int_0^1 f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau$. So
 $$
 0=F(0)<A_x<F(1)=\int_0^1 f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau.
 $$
 Therefore, there exists a unique $\delta_x\in (0,  1)$ such that
$$
A_x=\int_0^{\delta_x} f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau.
$$
The proof is complete.
\end{proof}

 By \eqref{e4}, \eqref{e5} and  Lemma \ref{lem2.1}, if $u$ is a solution 
of \eqref{e2}, then $u(t)$ can be expressed as
\begin{align*}
u(t)&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)
 \int_0^r\phi^{-1}\Big(\int_s^{\delta_x}
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
&\quad +\int_0^t\phi^{-1}\Big(\int_s^{\delta_x}
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s
\end{align*}
or 
\begin{align*}
u(t)&=\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}
 \int_0^1 h(r)\int_r^1\phi^{-1} \Big(\int_{\delta_x}^sf(\tau,x(\tau),
 x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s \,\mathrm{d}r\\
&\quad +\int_t^1\phi^{-1}\Big(\int_{\delta_x}^sf(\tau,x(\tau),x'(\tau))
 \,\mathrm{d}\tau\Big)\,\mathrm{d}s.
\end{align*}

\begin{lemma} \label{lem2.2}
 Suppose {\rm (H1)} and {\rm (H2)} hold. 
If $u(t)$ is a solution of \eqref{e2}, then
\begin{itemize}
\item[(i)] $u(t)$ is concave and $u(t)\geq 0$ on $[0, 1]$;

\item[(ii)] there exists a unique $t_0\in (0,  1)$ such that
$u(t_0)=\max_{0\leq t\leq  1}u(t)$ and $u'(t_0)=0$;

\item[(iii)] $\delta_x=t_0$.
\end{itemize}
\end{lemma}

\begin{proof} (i) Let $u(t)$ is a solution of \eqref{e2}. Then
$$
(\phi(u'(t)))'=-f(t,x(t),x'(t))<0, \quad t\in[0,1].
$$
Therefore, $\phi(u'(t))$ is strictly decreasing. 
It follows that $u'(t)$ is also strictly decreasing.
 Thus, $u(t)$ is  strictly concave on $[0, 1]$. Without loss of generality, 
we assume that $u(0)=\min\{u(0), u(1)\}$. By the concavity of $u$, we know
 that $u(t)\geq u(0)$, $t\in [0, 1]$. So we obtain
 $$
u(0)=\int_0^1u(t)g(t)\,\mathrm{d}t\geq
 u(0)\int_0^1g(t)\,\mathrm{d}t.
$$
By
 $0<\int_0^1g(t)\,\mathrm{d}t<1$, it is obvious that $u(0)\geq 0$.
 Hence, $u(t)\geq 0$,  $t\in [0, 1]$.

 (ii) Since $u(t)$ is  strictly concave on $[0, 1]$, there exist a unique 
$t_0\in [0, 1]$ such that $u(t_0)=\max_{0\leq t\leq  1}u(t)$.
By the boundary conditions and $u(t)\geq 0$, we know that $t_0\neq 0$ or 1,
that is, $t_0\in (0, ~1)$ such that $u(t_0)=\max_{0\leq t\leq 1}u(t)$.

 (iii) By \eqref{e3} and Lemma \ref{lem2.1}, it is easy to see that
 $$
u'(t)=\phi^{-1}\Big(\int_t^{\delta_x} f(s,x(s),x'(s))\,\mathrm{d}s\Big),
 $$
therefore, $u'(\delta_x)=0$, this implies $\delta_x=t_0$.
The proof is complete.
\end{proof}

Now we define an operator $T$ by
\begin{equation}
Tx(t)=\begin{cases}
\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{\delta_x}
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
+\int_0^t\phi^{-1}\Big(\int_s^{\delta_x} f(\tau,x(\tau),x'(\tau))
 \,\mathrm{d}\tau\Big)\,\mathrm{d}s, & 0\leq t\leq {\delta_x},
\\
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\int_0^1h(r)\int_r^1\phi^{-1}
 \Big(\int_{\delta_x}^s f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)
 \,\mathrm{d}s\,\mathrm{d}r
\\
+\int_t^1\phi^{-1}\Big(\int_{\delta_x}^sf(\tau,x(\tau),x'(\tau))\,
\mathrm{d}\tau\Big)\,\mathrm{d}s,
& {\delta_x}\leq t\leq 1 .
\end{cases} \label{e7}
\end{equation}
It is easy to prove that each fixed point of $T$ is a solution for
\eqref{e1}.

\begin{lemma} \label{lem2.3}
the operator $T:P\to P$ is completely continuous.
\end{lemma}

\begin{proof}  Let $x\in P$, then from the definition
 of $T$, we have
\begin{equation}
(Tx)'(t)=\begin{cases}
\phi^{-1}\Big(\int_t^{\delta_x} f(\tau,x(\tau),x'(\tau))\,
 \mathrm{d}\tau\Big)\geq 0, & 0\leq t\leq \delta_x,\\
-\phi^{-1}\Big(\int_{\delta_x}^t
f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau\Big)\leq 0,
& \delta_x\leq t\leq 1 .
\end{cases} \label{e8}
\end{equation}
So $(Tx)'(t)$ is monotone decreasing continuous and
$(Tx)'(\delta_x)=0$. Hence, $(Tx)(t)$ is nonnegative and concave on
$[0, 1]$. This shows that $T(P)\subset P$.
Next, we prove $T$ is compact on $C^1[0, 1]$.

Let $D$ be a bounded subset of $P$ and $m>0$ is a constant such that
$$
\int_0^1f(\tau,x(\tau),x'(\tau))\,\mathrm{d}\tau< m, \quad
\forall ~x\in D.
$$ 
From the definition of $T$, for any $x\in D$, we obtain
$$
 |Tx(t)|<\begin{cases}
\frac{\phi^{-1}(m)}{1-\int_0^1g(r)\,\mathrm{d}r}, & 0\leq t\leq\delta_x,\\
\frac{\phi^{-1}(m)}{1-\int_0^1h(r)\,\mathrm{d}r}, & \delta_x\leq t\leq 1,
\end{cases}
$$
$$
|(Tx)'(t)|<\phi^{-1}(m), \quad 0\leq t\leq 1.
$$
Hence, $TD$ is uniformly bounded and equicontinuous. So we have $TD$
is compact on $C[0, 1]$. From \eqref{e8} we know that for all
$\varepsilon>0$ there exists $\kappa>0$, such that when
$|t_1-t_2|<\kappa$, we have
$$
|\phi(Tx)'(t_1)-\phi(Tx)'(t_2)|<\varepsilon.
 $$
So $\phi(TD)'$ is compact on $C[0,  1]$, it follows that $(TD)'$ is
 compact on $C[0, 1]$. Therefore, $TD$ is compact on $C^1[0, 1]$.
Thus, $T:P\to P$ is completely continuous.
The proof is complete.
\end{proof}

\section{Existence of positive solutions}

 For convenience, we denote
$$
A=\max\Big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}, 
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\Big\}.
$$
Our result is as follows.

 \begin{theorem} \label{thm3.1}
 Assume {\rm (H1)} and {\rm (H2)} hold. If there exists $a>0$ such that
\begin{itemize}
\item[(i)] $f(t,x_1,y_1)\leq f(t,x_2,y_2)$ for any
 $0\leq t\leq 1$, $0\leq x_1\leq x_2\leq a$,
$0\leq|y_1|\leq| y_2|\leq a$;

\item[(ii)] $\max_{0\leq t\leq 1}f(t,a,a)\leq\phi(\frac{a}{A})$.
\end{itemize}
Then \eqref{e1} has at least two positive, concave solutions
 $w^{\ast}$ and $v^{\ast}$ satisfying
\begin{gather*}
0<w^{\ast}\leq a, \quad 0<|(w^{\ast})'|\leq a,\\
\lim_{n\to\infty}w_n=\lim_{n\to\infty}T^{n}w_0=w^{\ast},\\
\lim_{n\to\infty}(w_n)'=\lim_{n\to\infty}(T^{n}w_0)'=(w^{\ast})',
\end{gather*}
where
$$
w_0(t)=a\frac{\min\big\{\frac{\int_0^1g(r)\,
\mathrm{d}r}{1-\int_0^1g(r)\,\mathrm{d}r} + t, 
\frac{\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1h(r)\,\mathrm{d}r} + 1-t
\big\}}{\max\big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}, 
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\big\}}, \quad
0\leq t\leq1,
$$
and
\begin{gather*}
0<v^{\ast}\leq a, \quad 0<|(v^{\ast})'|\leq a, \\
\lim_{n\to\infty}v_n=\lim_{n\to\infty}T^{n}v_0=v^{\ast},\\
\lim_{n\to\infty}(v_n)'=\lim_{n\to\infty}(T^{n}v_0)'=(v^{\ast})',
\end{gather*}
where $v_0(t)=0$, $0\leq t\leq1$.
\end{theorem}

 \begin{proof}
Let $P_{a}=\{u\in P|~\|u\|_1<a\}$, 
$\overline{P}_{a}=\{u\in P|~\|u\|_1\leq a\}$. Next, we show that 
$T(\overline{P}_{a})\subset\overline{P}_{a}$.
If $u\in\overline{P}_{a}$, then $\|u\|_1\leq a$. Hence,
$$
0\leq u(t)\leq\|u\|_{c}\leq a, \quad 0\leq |u'(t)|\leq\|u'\|_{c}\leq a.
$$
From (i) and (ii), we have that
\begin{equation}
0<f(t,u(t),u'(t))\leq f(t,a,a)\leq\max_{0\leq t\leq
 1}f(t,a,a)\leq\phi(\frac{a}{A}).\label{e9}
\end{equation}
By this inequality and the definition of $T$, we have
\begin{align*}
&\|Tu\|_{c}=(Tu)(\delta_{u})\\
&=\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)
\int_0^r\phi^{-1}\Big(\int_s^{\delta_{u}}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
&\quad +\int_0^{\delta_{u}}\phi^{-1}\Big(\int_s^{\delta_{u}}
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s
\\
&=\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\int_0^1h(r)\int_r^1\phi^{-1}
\Big(\int_{\delta_{u}}^s f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)
 \,\mathrm{d}s\,\mathrm{d}r
\\
&\quad +\int_{\delta_{u}}^1\phi^{-1}
 \Big(\int_{\delta_{u}}^sf(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)
 \,\mathrm{d}s
\\
&\leq\max\Big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)
\int_0^1\phi^{-1}\Big(\int_0^1
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
&\quad +\int_0^1\phi^{-1}\Big(\int_0^1
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s,
\\
&\quad \frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\int_0^1h(r)\int_0^1\phi^{-1}
\Big(\int_0^1
f(\tau,u(\tau),u'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
&\quad +\int_0^1\phi^{-1}\Big(\int_0^1f(\tau,u(\tau),u'(\tau))\,
 \mathrm{d}\tau\Big)\,\mathrm{d}s\big\}
\\
&\leq\frac{a}{A}\max\big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r},
 \frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\Big\}\\
&=a.
\end{align*}
Furthermore, by \eqref{e8} it is easy to verify that
$$
\|(Tu)'\|_{c}\leq\frac{a}{A}<a.
$$
Hence, $\|Tu\|_1\leq a$. So we have
$T(\overline{P}_{a})\subset\overline{P}_{a}$.
Let
$$
w_0(t)=a\frac{\min\big\{\frac{\int_0^1g(r)\,\mathrm{d}r}{1-\int_0^1g(r)\,\mathrm{d}r} + t, ~~\frac{\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1h(r)\,\mathrm{d}r} + 1-t
\big\}}{\max\big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r},
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\big\}},
\quad 0\leq t\leq1.
$$
Now we define a sequence $\{w_n\}$ by the iterative scheme
\begin{equation}
w_{n+1}=Tw_n=T^{n}w_0,  \quad n=0,1,2,\dots \label{e10}
\end{equation}
Since $T(\overline{P}_{a})\subset\overline{P}_{a}$ and
$w_0(t)\in\overline{P}_{a}$, we have
$w_n\in\overline{P}_{a}, n=0,1,2,\dots $.
From Lemma \ref{lem2.3}, $T$ is compact, we assert that
$\{w_n\}_{n=1}^{\infty}$ has a convergent subsequence
$\{w_{n_k}\}_{k=1}^{\infty}$ and there exists $w^{\ast}\in\overline{P}_{a}$,
such that $w_{n_k}\to w^{\ast}$.

On the other hand, since
\begin{align*} 
w_1(t)&=Tw_0(t)\\
&=\begin{cases}
\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}\int_0^1g(r)\int_0^r\phi^{-1}
\Big(\int_s^{\delta_{w_0}}
f(\tau,w_0(\tau),w_0'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
+\int_0^t\phi^{-1}\Big(\int_s^{\delta_{w_0}}
 f(\tau,w_0(\tau),w_0'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s, \\
 \quad\text{if } 0\leq t\leq {\delta_{w_0}},\\[3pt]
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\int_0^1h(r)\int_r^1\phi^{-1}
\Big(\int_{\delta_{w_0}}^s
f(\tau,w_0(\tau),w_0'(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
+\int_t^1\phi^{-1}\Big(\int_{\delta_{w_0}}^sf(\tau,w_0(\tau),w_0'(\tau))\,
\mathrm{d}\tau\Big)\,\mathrm{d}s,\\
 \quad\text{if } {\delta_{w_0}}\leq t\leq 1
\end{cases} 
\\
&\leq\min\Big\{\frac{a}{A}\frac{\int_0^1g(r)\,
\mathrm{d}r}{1-\int_0^1g(r)\,\mathrm{d}r}+\frac{a}{A}t,
\frac{a}{A}\frac{\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1h(r)\,\mathrm{d}r}
+\frac{a}{A}(1-t)\Big\}
\\
&=a\frac{\min\big\{\frac{\int_0^1g(r)\,\mathrm{d}r}{1-\int_0^1g(r)
 \,\mathrm{d}r}+t, \frac{\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1h(r)
 \,\mathrm{d}r}+1-t\big\}}
{\max\big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r}, 
\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\big\}}
\\
&=w_0(t), \quad 0\leq t\leq 1,
\end{align*}
and
\begin{align*}
|w_1'(t)|&=|(Tw_0)'(t)|\\
&=\begin{cases}
\phi^{-1}\Big(\int_t^{\delta_{w_0}} f(\tau,w_0(\tau),w_0'(\tau))
 \,\mathrm{d}\tau\Big),  & 0\leq t\leq \delta_{w_0},\\
\phi^{-1}\Big(\int_{\delta_{w_0}}^t
f(\tau,w_0(\tau),w_0'(\tau))\,\mathrm{d}\tau\Big),
 & \delta_{w_0}\leq t\leq 1
\end{cases} \\
&\leq\phi^{-1}\Big(\int_0^1 f(\tau,w_0(\tau),w_0'(\tau))\,\mathrm{d}\tau\Big)
\\
&\leq\frac{a}{A}=|w_0'(t)|, \quad 0\leq t\leq 1,
\end{align*}
we have
$$
w_1(t)\leq w_0(t), \quad |w_1'(t)|\leq |w_0'(t)|, \quad
0\leq t\leq 1.
$$
By (i) and \eqref{e7}, we easily see that $T$ is increasing, it follows that
\begin{gather*}
w_2(t)=Tw_1(t)\leq Tw_0(t)=w_1(t), \quad 0\leq t\leq 1, \\
|w_2'(t)|= |(Tw_1)'(t)|\leq |(Tw_0)'(t)| =|w_1'(t)|, \quad 0\leq t\leq 1.
\end{gather*}
Moreover, we have
$$
w_{n+1}(t)\leq w_n(t), ~|w_{n+1}'(t)|\leq |w_n'(t)|, \quad 0\leq t\leq 1,
\quad n=0,1,2,\dots
$$
Therefore, $w_n\to w^{\ast}$. Let $n\to \infty$ in \eqref{e10} to obtain 
$Tw^{\ast}=w^{\ast}$.

 Next, Let $v_0(t)=0$, $0\leq t\leq 1$, then $v_0(t)\in\overline{P}_{a}$.
Let $v_1=Tv_0$, then  $v_1\in\overline{P}_{a}$. We denote
\begin{equation}
v_{n+1}=Tv_n=T^{n}v_0, \quad n=0,1,2,\dots \label{e11}
\end{equation}
Similar to $\{w_n\}_{n=1}^{\infty}$, we assert that
 $\{v_n\}_{n=1}^{\infty}$ has a convergent subsequence
$\{v_{n_k}\}_{k=1}^{\infty}$ and there exists $v^{\ast}\in\overline{P}_{a}$,
 such that $v_{n_k}\to v^{\ast}$.
 Since
\begin{gather*}
v_1(t)=(Tv_0)(t)= (T0)(t)\geq 0, \quad 0\leq t\leq 1, \\
|v_1'(t)|= |(Tv_0)'(t)|= |(T0)'(t)| \geq 0, \quad 0\leq t\leq 1,
\end{gather*}
we have
\begin{gather*}
v_2(t)=(Tv_1)(t)\geq (T0)(t)=v_1(t), \quad 0\leq t\leq 1,\\
|v_2'(t)|= |(Tv_1)'(t)|\leq |(T0)'(t)| =|v_1'(t)|, \quad 0\leq t\leq 1.
\end{gather*}
Moreover, we have
$$
v_{n+1}(t)\leq v_n(t), \quad |v_{n+1}'(t)|\leq |v_n'(t)|, \quad
0\leq t\leq 1, \quad n=0,1,2,\dots
$$
Thus $v_n\to v^{\ast}$ and $Tv^{\ast}=v^{\ast}$.

It is well know that the fixed point of $T$ is the solution of \eqref{e1}. 
Therefore, $w^{\ast}$ and $v^{\ast}$
are positive, concave solutions of \eqref{e1}.
The proof is complete.
\end{proof}


\begin{remark} \label{rmk1}\rm
 we can see easily that $w^{\ast}$ and $v^{\ast}$ are the maximal and 
minimal solutions of  \eqref{e1}. If $w^{\ast}\equiv v^{\ast}$, then 
\eqref{e1} has a unique positive solution in $\overline{P}_{a}$.
\end{remark}

 From Theorem \ref{thm3.1} we immediately obtain the following result.

\begin{corollary} \label{coro2.1} 
 Assume {\rm (H1), (H2)} and Theorem \ref{thm3.1}(i) hold. 
If there exists $a>0$ such that
\begin{itemize}
\item[(iii)] $\lim_{\ell\to\infty}\max_{0\leq t\leq  1}f(t,\ell,a)
\leq\phi(\frac{1}{A})$.
\end{itemize}
Then \eqref{e1} has at least two positive, concave solutions 
$w^{\ast}$ and $v^{\ast}$ such that the conclusion of
 Theorem \ref{thm3.1} hold.
\end{corollary}

\begin{corollary} \label{coro2.2} Assume {\rm (H1), (H2)} and 
Theorem \ref{thm3.1}(i) hold. If there exists $0<a_1<a_2<\dots <a_n$ 
such that
\begin{itemize}
\item[(iv)] $\max_{0\leq t\leq 1}f(t,a_k,a_k)\leq\phi(\frac{a_k}{A})$, 
$k=1,2,\dots ,n$.
\end{itemize}
Then \eqref{e1} has at least $2n$ positive, concave solutions 
$w^{\ast}_k$ and $v^{\ast}_k$ satisfying
\begin{gather*}
0<w^{\ast}_k\leq a_k, ~~0<|(w^{\ast}_k)'|\leq a_k,\\
\lim_{n\to\infty}w_{k_n}=\lim_{n\to\infty}T^{n}w_{k_0}=w^{\ast}_k,\\
\lim_{n\to\infty}(w_{k_n})'=\lim_{n\to\infty}(T^{n}w_{k_0})'=(w^{\ast}_k)',
\end{gather*}
where
$$
w_{k_0}(t)=a_k\frac{\min\big\{\frac{\int_0^1g(r)\,
\mathrm{d}r}{1-\int_0^1g(r)\,\mathrm{d}r} + t, 
\frac{\int_0^1h(r)\,\mathrm{d}r}{1-\int_0^1h(r)\,\mathrm{d}r} + 1-t
\big\}}{\max\big\{\frac{1}{1-\int_0^1g(r)\,\mathrm{d}r},
 ~\frac{1}{1-\int_0^1h(r)\,\mathrm{d}r}\big\}}, \quad
0\leq t\leq1,
$$
and
\begin{gather*}
0<v^{\ast}_k\leq a_k, \quad 0<|(v^{\ast}_k)'|\leq a_k, \\
\lim_{n\to\infty}v_{k_n}=\lim_{n\to\infty}T^{n}v_{k_0}=v^{\ast}_k,\\
\lim_{n\to\infty}(v_{k_n})'=\lim_{n\to\infty}(T^{n}v_{k_0})'=(v^{\ast}_k)',
\end{gather*}
where $v_{k_0}(t)=0, 0\leq t\leq1$.
\end{corollary}

\section{Example} 

In this section, we give a example as an application of the main results.

Let $\phi(u)=|u|u, \quad g(t)=h(t)=1/2$. 
We consider the boundary-value problem
\begin{equation} 
\begin{gathered}
    \big(\phi(u')\big)'+f(t,u(t),u'(t))=0, \quad t\in[0,1],\\
 u(0)=\frac{1}{2}\int_0^1u(t)\,\mathrm{d}t,\quad
 u(1)=\frac{1}{2}\int_0^1u(t)\,\mathrm{d}t,
    \end{gathered} \label{e12}
\end{equation}
where
$$
f(t,u,v)=-t^{2}+t+\frac{1}{8}u+\frac{1}{16}v^{2}+2, \quad
(t,u,v)\in [0, 1]\times[0, \infty)\times(-\infty, \infty).
$$
Let $\psi_1(u)=\psi_2(u)=u^{2}$, $u>0$. Choosing $a=4$. By
calculations we obtain $A=2$. It is easy to verify that $f(t,u,v)$ satisfies
\begin{itemize}
\item[(1)] $f(t,x_1,y_1)\leq f(t,x_2,y_2)$ for any
$0\leq t\leq 1, 0\leq x_1\leq x_2\leq 4, 0\leq|y_1|\leq| y_2|\leq 4$;

\item[(2)] $\max_{0\leq t\leq  1}f(t,a,a)=f(\frac{1}{2},4,4)
\leq\phi(\frac{a}{A})=4$.
\end{itemize}
Hence, by Theorem \ref{thm3.1}, \eqref{e12} has two positive solutions 
$w^{\ast}$ and $v^{\ast}$. For $n=0,1,2,\dots $,
the two iterative schemes are:
\begin{gather*}
w_0(t)=\begin{cases}
2+2t, & 0\leq t\leq \frac{1}{2},\\
4-2t, & \frac{1}{2}\leq t\leq 1,
\end{cases}
\\
 w_{n+1}(t)=\begin{cases}
\int_0^1\int_0^r\phi^{-1}\Big(\int_s^{\delta_n}
(-\tau^{2}+\tau+\frac{1}{8}w_n(\tau)+\frac{1}{16}w_n'^{2}(\tau))
 \,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
+\int_0^t\phi^{-1}\Big(\int_s^{\delta_n} (-\tau^{2}+\tau+\frac{1}{8}w_n(\tau)
 +\frac{1}{16}w_n'^{2}(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s,
 & 0\leq t\leq \delta_n,\\
\int_0^1\int_r^1\phi^{-1}\Big(\int_{\delta_n}^s
(-\tau^{2}+\tau+\frac{1}{8}w_n(\tau)+\frac{1}{16}w_n'^{2}(\tau))
 \,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
+\int_t^1\phi^{-1}\Big(\int_{\delta_n}^s(-\tau^{2}+\tau+\frac{1}{8}w_n(\tau)
+\frac{1}{16}w_n'^{2}(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s,
& \delta_n\leq t\leq 1.
\end{cases} 
\end{gather*}
\begin{gather*}
v_0(t)=0,
\\
v_{n+1}(t)=\begin{cases}
\int_0^1\int_0^r\phi^{-1}\Big(\int_s^{\delta_n}
(-\tau^{2}+\tau+\frac{1}{8}v_n(\tau)+\frac{1}{16}v_n'^{2}(\tau))
 \,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r
\\
+\int_0^t\phi^{-1}\Big(\int_s^{\delta_n} (-\tau^{2}+\tau
+\frac{1}{8}v_n(\tau)+\frac{1}{16}v_n'^{2}(\tau))\,\mathrm{d}\tau\Big)
\,\mathrm{d}s, 
 & 0\leq t\leq \delta_n,
\\
\int_0^1\int_r^1\phi^{-1}\Big(\int_{\delta_n}^s
(-\tau^{2}+\tau+\frac{1}{8}v_n(\tau)
+\frac{1}{16}v_n'^{2}(\tau))\,\mathrm{d}\tau\Big)\,\mathrm{d}s\,\mathrm{d}r\\
+\int_t^1\phi^{-1}\Big(\int_{\delta_n}^s(-\tau^{2}+\tau
+\frac{1}{8}v_n(\tau)+\frac{1}{16}v_n'^{2}(\tau))\,
\mathrm{d}\tau\Big)\,\mathrm{d}s,
 &\delta_n\leq t\leq 1 .
\end{cases} 
\end{gather*}


\begin{thebibliography}{9}

\bibitem{a1}  R. Avery, J. Henderson;
\emph{Existence of three positive pseudo-symmetric solutions
for a one-dimensional $p$-Laplacian}, J. Math. Anal. Appl. 277~(2003)~395-404.

\bibitem{d1}  Y. Ding;
\emph{Positive solutions for integral boundary value problem with
$\phi$-Laplacian operator}, Boundary Value Problem. Volume(2011),
Article ID 827510.

\bibitem{f1} H. Feng, W. Ge, M. Jiang;
\emph{Multiple positive solutions for m-point boundary value
problems with a one-dimensional $p$-Laplacian}, Nonlinear Anal. 68
(2008) 2269-2279.

\bibitem{l1}  B. Liu;
\emph{Positive solutions of three-point boundary value
problems for one-dimensional $p$-Laplacian with infinitely many
singularities},  Appl. Math. Lett.  17 (2004) 655-661.

\bibitem{s1}  B. Sun, W. Ge;
\emph{Existence and iteration of positive solutions for some
$p$-Laplacian boundary value problems}, Nonlinear Anal. 67
(2007) 1820-1830.

\bibitem{w1}  H. Wang;
\emph{On the number of positive solutions of nonlinear systems}, 
J. Math. Anal. Appl. 281(2003) 287-306.

\bibitem{w2}  H. Wang;
\emph{On the structure of positive radial solutions for quasilinear 
equations in annular domain}, Adv. Differential Equations. 8 (2003) 111-128.

\bibitem{w3}  J. Wang;
\emph{The existence of positive solutions for the one-dimensional $p$-Laplacian}, 
Proc. Amer. Math. Soc. 125 (1997) 2275-2283.

\bibitem{w4} Z. Wang, J. Zhang;
\emph{Positive solutions for one-dimensional $p$-Laplacian boundary value
problems with dependence on the first order derivative}, J. Math.
Anal. Appl. 314 (2006) 618-630.

\end{thebibliography}


\end{document}