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\markboth{\hfil Minimal and maximal solutions  \hfil EJDE--2003/21}
{EJDE--2003/21\hfil Myron K. Grammatikopoulos \& Petio S. Kelevedjiev \hfil}

\begin{document}

\title{\vspace{-1in}%
\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2003}(2003), No. 21, pp. 1--14. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)} 
\vspace{\bigskipamount} \\
%
 Minimal and maximal solutions for two-point boundary-value problems 
%
\thanks{\emph{Mathematics Subject Classifications:} 34B15. \hfil\break
\indent
{\em Key words:} Boundary-value problems, minimal and maximal solutions, 
\hfil\break\indent
monotone method, barrier strips. \hfil\break\indent
\copyright 2003 Southwest Texas State University. 
\hfil\break\indent
Submitted December 12, 2002. Published February 28, 2003.} }
\date{}
\author{Myron K. Grammatikopoulos \& Petio S. Kelevedjiev}
\maketitle

\begin{abstract}
 In this article we consider a boundary-value  problem for the 
 equation  ${f(t,x,x',x'')=0}$ with mixed boundary  conditions.
 Assuming the existence of suitable barrier strips, and using 
 the monotone iterative method, we obtain the minimal and 
 maximal solutions.
\end{abstract}

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

\section{Introduction}

We apply the monotone iterative method to obtain minimal and maximal
solutions to the nonlinear boundary-value problem (BVP) 
\begin{equation}
\begin{gathered} f(t,x,x',x'')=0,\quad 0\le a\le t\le b,\\ x(a)=A,\quad
x'(b)=B, \end{gathered}  \tag{1.1}
\end{equation}
where the scalar function $f(t,x,p,q)$ is continuous and has continuous
first derivatives on suitable subsets of $[a,b]\times \mathbb{R}^{3}$. For
results, which guarantee the existence of $C^{2}[a,b]$-solutions to BVPs for
the equation $x''=f(t,x,x',x'')-y(t)$ with
various linear boundary conditions, see \cite{f1,f2,m1,m2,p1,p2,p3}.
Concerning the uniqueness results, we refer to \cite{p1}. A result,
concerning the existence and uniqueness of $C^{2}[a,b]$-solutions to the BVP
for the equation $x''=f(t,x,x',x'')$, with
general linear boundary conditions, can be found in \cite{t2}. The results
of \cite{m3} guarantee the existence of $W^{2,\infty }[a,b]$-solutions or of 
$C^{2}[a,b]$-solutions to the Dirichlet BVP for the equation $%
f(t,x,x',x'')=0$, where the function $f(t,x,p,q)$is
defined on $[a,b]\times \mathbb{R}^{n}\times \mathbb{R}^{n}\times Y$, and $Y$ is a
non-empty closed connected or locally connected subset of $\mathbb{R}^{n}$.
Finally, the $C^{2}[a,b]$-solvability of BVPs for the equation ${%
f(t,x,x',x'')=0}$ with fully nonlinear boundary
conditions is studied in \cite{k2}.

Note that, in the literature, the monotone iterative method is applied on
BVPs for equations of the forms $x''=f(t,x,x')$ and $%
(\phi (x'))'=f(t,x,x')$ with various boundary
conditions (see, for example, \cite{c1,c2,c3,d1,h1,j1,k1,k3,l2,m4,t1,w1}).
The sequences of iterates, considered in \cite{c1,c2,c3,d1,j1,k3,w1},
converge to the extremal solutions, while the sequences of iterates,
considered in\cite{h1,l2,m3}, converge to the unique solution. The first
elements $u_{0}(t)$ and $v_{0}(t)$ of such sequences of iterates usually are
lower and upper solutions respectively of the problems under consideration
(see, for example, \cite{c1,c2,c3,d1,j1,k3,w1}. To derive the needed
monotone iterates, the authors of \cite{c1,c2,c3,d1,j1,k3,l2,w1} use
suitable growth conditions. For more applications of the monotone iterative
method, see cite{b1,l1,l3,s1,y1}.

In this article, following cite{k1}, we obtain the extremal solutions to
(1.1) under assumption of the existence of suitable barrier strips (see
Remarks 2.1 and 2.2 below), which immediately imply the first iterates $%
u_{0}(t)$ and $v_{0}(t)$. A version of \cite[Theorem 5.1]{k2} implies the
existence of the next iterates, and a suitable comparison result guarantees
the monotone properties for the sequences of iterates. Finally, the
Arzela-Askoli's theorem ensures the existence of the extremal solutions of
the problem (1.1) as limits of the sequences of iterates.

\section{Basic hypotheses}

The following four hypotheses will be a tool for obtaining our results.

\begin{itemize}
\item[(H1)]  There are constants $K>0$, $F,F_{1},L,L_{1}$ such that 
\[
Fa\leq A\leq La,\quad F_{1}<F\leq B\leq L<L_{1}\,.
\]
For the set $T:=\{(t,x):a\leq t\leq b,Ft\leq x\leq Lt\}$, we assume that 
\[
f(t,x,p,q)+Kq\geq 0
\]
on $\bigl\{(t,x,p,q):(t,x)\in T,\;p\in \lbrack L,L_{1}],\;q\in (-\infty ,0)%
\bigr\}$, and 
\[
f(t,x,p,q)+Kq\leq 0
\]
on $\bigl\{(t,x,p,q):(t,x)\in T,\;p\in \lbrack F_{1},F],\;q\in (0,\infty )%
\bigr\}$.
\end{itemize}

\begin{remark} \label{rmk2.1} \rm
Set $\Phi _{1}(t,x,p,q)\equiv $
$f(t,x,p,q)+Kq$. Then, the strip $\Delta _{1}=[a,b]\times [
L,L_{1}], $ on which $\Phi _{1}(t,x,p,q)\geq 0$, and the strip
$\Delta_{2}=[a,b]\times [ F_{1},F]$, on which $\Phi _{1}(t,x,p,q)\leq 0$,
are such that the graph of the function $x'(t),t\in [ a,b]$,
does not cross $\Delta _{1}$ and\ $\Delta _{2}$, and is located between
them. For this reason $\Delta _{1}$ and $\Delta _{2}$ are called
barrier strips for $x'(t)$, $t\in [ a,b]$.
\end{remark}

\begin{itemize}
\item[(H2)]  There are constants $G_{i}^{-}$, $G_{i}^{+}$, $H_{i}^{-}$, $%
H_{i}^{+},\;i=1,2$, such that 
\begin{gather*}
G_{2}^{+}>G_{1}^{+}\geq 2C,\quad G_{2}^{-}>G_{1}^{-}\geq 2C, \\
H_{2}^{+}<H_{1}^{+}\leq -2C,\quad H_{2}^{-}<H_{1}^{-}\leq -2C,
\end{gather*}
where $C=\max \{|L|,|F|\}/(b-a)$, $f(t,x,p,q)$ and $f_{q}(t,x,p,q)$ are
continuous and $f_{q}(t,x,p,q)<0$ for 
\[
(t,x,p,q)\in[a,b]\times[m_1-\varepsilon,M_1+\varepsilon] \times[%
F-\varepsilon,L+\varepsilon]\times[m_2-\varepsilon , M_2+\varepsilon], 
\]
where $m_1=\min\{Fa,Fb\}$ $M_1=\max\{La,Lb\}$, $m_2=\min\bigl\{%
H_{2}^{+},H_{2}^{-}\bigr\}$, $M_2=\max \bigl\{G_{2}^{+},G_{2}^{-}\bigr\}$,
and $\varepsilon>0$ is fixed and such that 
\begin{equation}
H_{1}^{+}>H_{2}^{+}+\varepsilon, \quad H_{1}^{-}>H_{2}^{-}+\varepsilon,\quad
G_{2}^{+}>G_{1}^{+}+\varepsilon,\quad G_{2}^{-}>G_{1}^{-}+\varepsilon. 
\tag{2.1}
\end{equation}
$f_{t}(t,x,p,q)$, $f_{x}(t,x,p,q)$ and $f_{p}(t,x,p,q)$ are continuous for $%
(t,x,p,q)$ in $[a,b]\times[m_1,M_1]\times[F,L]\times[m_2,M_2]$; 
\[
f_{t}(t,x,p,q)+\ f_{x}(t,x,p,q)p+ f_{p}(t,x,p,q)q\geq 0 
\]
for $(t,x,p,q)$ in $[a,b]\times[m_1,M_1]\times[F,L]\times\big([
H_{2}^{+},H_{1}^{+}]\cup [ G_{1}^{+},G_{2}^{+}]\big)$, and 
\[
f_{t}(t,x,p,q)+\ f_{x}(t,x,p,q)p+\ f_{p}(t,x,p,q)q\leq 0 
\]
for $(t,x,p,q)$ in $[a,b]\times[m_1,M_1]\times[F,L]\times\bigl([
H_{2}^{-},H_{1}^{-}]\cup [ G_{1}^{-},G_{2}^{-}]\bigr)$, where $F$ and $L$
are the constants of {H1}.
\end{itemize}

\begin{remark} \label{rmk2.2} \rm
Set
$\Phi _{2}(t,x,p,q)\equiv f_{t}(t,x,p,q)+f_{x}(t,x,p,q)p+f_{p}(t,x,p,q)q$.
Then, the pair of strips
$\Omega _{1}=[a,b]\times ([H_{2}^{+},H_{1}^{+}]\cup [G_{1}^{+},G_{2}^{+}])$,
where $\Phi _{2}(t,x,p,q)\geq 0$, and the pair
of strips $\Omega _{2}=[a,b]\times ([H_{2}^{-},H_{1}^{-}]\cup [
G_{1}^{-},G_{2}^{-}])$, where $\Phi _{2}(t,x,p,q)\leq 0$,
 are such that the graph of the function $x''(t),t\in [ a,b]$, can
not cross the outer strips, of the four such ones, defined by
$\Omega _{1}$ and $\Omega _{2}$. For this reason the outer strips of
$\Omega _{1}$ and $\Omega _{2}$ are called barrier strips for
$x''(t),t\in [ a,b]$.
\end{remark}

\begin{itemize}
\item[(H3)]  For $m_{3}=\min \{H_{1}^{+},H_{1}^{-}\} $ and $M_{3}=\max
\{G_{1}^{+},G_{1}^{-}\}$ 
\[
h(\lambda ,t,x,p,m_{3}-\varepsilon)h(\lambda ,t,x,p,M_{3}+\varepsilon) \leq
0 
\]
for $(\lambda,t,x,p)$ in $[0,1]\times[a,b]\times[ m_1-\varepsilon,M_1+%
\varepsilon]\times[F-\varepsilon,L+\varepsilon]$, where $h(\lambda
,t,x,p,q)=(\lambda -1)Kq+\lambda f(t,x,p,q)$, $F,L,K$ are the constants of {%
H1}, and $H_{1}^{+}$, $H_{1}^{-}$, $G_{1}^{+}$, $G_{1}^{-}$, $C$, $m_1$, $M_1
$, and $\varepsilon $ are as in {H2}.

\item[(H4)]  For $(t,x,p,q)$ in $T\times[F,L]\times[\min%
\{H_{1}^{+},H_{1}^{-}\},\max \{G_{1}^{+},G_{1}^{-}\}]$, $f_{x}(t,x,p,q)\ge 0$%
,  where the trapezoid $T$ and the constants $F$ and $L$ are as in H1, and $%
H_{1}^{+}$, $H_{1}^{-}$, $G_{1}^{+}$ and $G_{1}^{-}$ are the constants in {H2%
}, and $m_3$ and $M_3$ are as in {H3}.
\end{itemize}

\section{Main result}

For a function $y(t)\in C[a,b]$ bounded on $[a,b]$, we define a mapping 
\[
\mathcal{A}y=x, 
\]
where $x(t)\in C^{2}[a,b]$ is a solution to the BVP 
\begin{equation}
\begin{gathered} f(t,y(t),x',x'')=0,\quad t\in[a,b],\\ x(a)=A,\quad x'(b)=B.
\end{gathered}  \tag{3.1}
\end{equation}
We will show that under the hypotheses H1, H2, and H3, the map $\mathcal{A}$
is uniquely determined. For this reason, we consider two sequences $\{u_{n}\}
$ and $\{v_{n}\}$, $n=0,1,\dots$, defined by 
\[
u_{n+1}=\mathcal{A}u_{n}\quad \mbox{and}\quad v_{n+1}=\mathcal{A}v_{n}, 
\]
where $u_{0}=Ft$, $v_{0}=Lt$, $t\in [ a,b]$, and $F$ and $L$ are as in H1.
Now we formulate our main result.

\begin{theorem} \label{thm3.1}
Under hypotheses H1--H4, there are sequences $\{u_{n}\}$ and
$\{v_{n}\}$, $n=0,1,\dots$, such that for $n\to +\infty $:
$u_{n}\to u^{m}$, $v_{n}\to v^{M}$ and
\[
u_{0}\leq u_{1}\leq \dots\leq u_{n}\leq \dots\leq u^{m}\leq x\leq v^{M}
\leq \dots \leq v_{n}\leq \dots\leq v_{1}\leq v_{0},
\]
where $u^{m}(t)$ and $v^{M}(t)$ are the minimal and maximal
solutions of the BVP (1.1) respectively, and $x(t)\in C^{2}[a,b]$
is a solution of (1.1).
\end{theorem}

The proof of this theorem can be found at the end of this article and is
based on the auxiliary results, which we present in the next section.

\section{ Auxiliary results}

%sec. 4

We begin this section with an existence result, which is a modification of 
\cite[Theorem 6.1, Chapter II]{g1}. Namely, we consider the family of BVPs 
\begin{equation}
\begin{gathered} Kx''=\lambda\big(Kx''+f(t,y(t),x',x'')\big),\quad
t\in[a,b],\\ x(a)=A,\quad x'(b)=B, \end{gathered}  \label{4.1l}
\end{equation}
where $\lambda \in [ 0,1]$ and $K>0$.

\begin{lemma} \label{lm4.1}
Assume that there are constants $Q_{i}$,
$i=0,1,\dots,5$, independent of $\lambda $ such that
\begin{itemize}
\item[(i)]  For each solution $x(t)\in C^{2}[a,b]$ of
\eqref{4.1l}  it holds
\[
Q_{0}<x(t)<Q_{1},\;Q_{2}<x'(t)<Q_{3},\;Q_{4}<x''(t)<Q_{5},\quad t\in [ a,b].
\]
\end{itemize}
Also assume that:
\begin{itemize}
\item[(ii)]  $f(t,x,p,q)$ and $f_{q}(t,x,p,q)$ are continuous,
and $f_{q}(t,x,p,q)<0$ for all
$(t,x,p,q)$ in $[a,b]\times[Q_{0},Q_{1}]\times[Q_{2},Q_{3}]
\times[Q_{4},Q_{5}]$

\item[(iii)] $h(\lambda ,t,x,p,Q_{4})h(\lambda ,t,x,p,Q_{5})\leq
0$ for $(\lambda ,t,x,p)$ in
$\Lambda :=[0,1]\times[a,b]\times[Q_{0},Q_{1}]\times[Q_{2},Q_{3}]$,
where $h(\lambda ,t,x,p,q)=(\lambda -1)Kq+\lambda f(t,x,p,q)$.
\end{itemize}
Then the BVP (3.1) has a $C^{2}[a,b]$-solution for each
$y(t)\in C[a,b]\; $ such that $Q_{0}<y(t)<Q_{1}$, $t\in [a,b]$.
\end{lemma}

\paragraph{Proof}

In view of (ii) and (iii), we conclude that there is a unique function $%
G(\lambda ,t,x,p)$ which is continuous on $\Lambda $ and such that 
\[
q=G(\lambda ,t,x,p)\quad\mbox{for}\quad (\lambda ,t,x,p)\in \Lambda 
\]
is equivalent to the equation 
\[
h(\lambda ,t,x,p,q)=0\quad\mbox{on}\quad\Lambda \times [Q_{4},Q_{5}]. 
\]
Note that $h(0,t,x,p,0)=0$ yields 
\begin{equation}
G(0,t,x,p)=0\quad \mbox{for}\quad (t,x,p)\in[a,b]\times[Q_{0},Q_{1}]\times[%
Q_{2},Q_{3}].  \tag{4.2}
\end{equation}
Thus, the family \eqref{4.1l} is equivalent to the family of BVPs 
\begin{equation}
\begin{gathered} x''=G(\lambda ,t,y(t),x'),\quad t\in[a,b],\\ x(a)=A,\quad
x'(b)=B, \end{gathered}  \tag{4.3}
\end{equation}
where $\lambda \in [ 0,1]$. Now, define the set 
\[
U=\bigl\{x(t)\in C^{2}[a,b]:\,x(t)\in (Q_{0},Q_{1}),\,x'(t)\in
(Q_{2},Q_{3}),\,x''(t)\in (Q_{4},Q_{5})\bigr\}, 
\]
which is an open subset of the convex set $C_{Q}^{2}[a,b]$ of the Banach
space $C^{2}[a,b]$ and consider the map $\mathbf{N}:C_{Q}^{2}[a,b]\to C[a,b]$%
, defined by 
\[
\mathbf{N} x=x'', 
\]
where ${C_{Q}^{2}[a,b]}=\{x\in {C^{2}[a,b]}:x(a)=A,x'(b)=B\}$. It is
easy to see that the map $\mathbf{S}:C_{Q_{0}}^{2}[a,b]\to C[a,b]$, defined
by 
\[
\mathbf{S} x=x''\, 
\]
with $C_{Q_{0}}^{2}[a,b]=\{x\in C^{2}[a,b]:\,x(a)=0,\; x'(b)=0\}$,
is one-to-one and the problem $\mathbf{S}x=0,\,x(a)=A,\,x'(b)=B$,
has a unique solution $l$. Then $\mathbf{N}^{-1}:C[a,b]\to C_{Q}^{2}[a,b]$
exists, is continuous, and moreover 
\[
\mathbf{N}^{-1}s=\mathbf{S}^{-1}s+l. 
\]
Let $\mathbf{H}_{\lambda }:\overline{U}\to C_{Q}^{2}[a,b]$ be defined by 
\[
\mathbf{H}_{\lambda }x=\mathbf{N}^{-1}\mathbf{G}_{\lambda }\mathbf{j}%
(x),\quad \lambda \in [ 0,1], 
\]
where $\mathbf{j}:C_{Q}^{2}[a,b]\to C^{1}[a,b]$ is defined by $\mathbf{j}x=x$%
, $\mathbf{G_{\lambda }}:C^{1}[a,b]\to C[a,b]$ is defined by 
\[
\left( \mathbf{G_{\lambda }}x\right) (t) =G\bigl(\lambda
,t,y(t),x'(t)\bigr),\quad \lambda \in [ 0,1]\,. 
\]
Clearly, $\mathbf{H}_{\lambda }$ is a compact homotopy, because $\mathbf{j}$
is a completely continuous embeding, and $\mathbf{G_{\lambda }}$ and $%
\mathbf{N}^{-1}$ are continuous. Moreover, $\mathbf{H}_{\lambda }x=x$
implies 
\[
x=\mathbf{N}^{-1}\mathbf{G}_{\lambda }\mathbf{j}(x). 
\]
Hence, by the definition of $\mathbf{N}^{-1}$, we have 
\[
x=\mathbf{S}^{-1}\mathbf{G}_{\lambda }\mathbf{j}(x)+l. 
\]
Finally, since $\mathbf{S}l=0$, it follows that 
\[
\mathbf{S}x=\mathbf{G}_{\lambda }\mathbf{j}(x). 
\]
Thus, the fixed points of $\mathbf{H}_{\lambda }$ are solutions to (4.3) and
obviously$\mathbf{H}_{\lambda }$ has no fixed points on $\partial U. $In
view of (4.2), the map $\mathbf{H}_{0}$, which has the form $\mathbf{H}%
_{0}x=l$, is constant. Moreover, $l$, as the unique solution of (4.1)$_{0}$,
belongs to the set $U$. Hence, by \cite[Theorem 2.2]{g1}, the map $\mathbf{H}%
_{0}$ is essential. The topological transversality theorem of \cite{g1}
implies that ${\mathbf{H}_{1}}$ is also essential, i.e. for $\lambda =1$
(4.3) has a solution. Moreover, for $\lambda =1$ (4.3) coincides with (3.1).
Therefore, the problem (3.1) has a solution. The proof of the lemma is
complete. \hfill$\diamondsuit $

To obtain our next auxiliary results, we introduce the following two sets 
\begin{gather*}
V=\{y(t)\in C[a,b]: Ft\leq y(t)\leq Lt,\,t\in [ a,b]\}, \\
V_{1}=\{y(t)\in C^{1}[a,b]:Ft\leq y(t)\leq Lt,\,F\leq y'(t)\leq
L,\,t\in [ a,b]\},
\end{gather*}
where the constants $L$ and $F$ are as in H1. Then we formulate the
following results.

\begin{lemma} \label{lm4.2}
Let H1 hold and $x(t)\in
C^{2}[a,b]$ be a solution to \eqref{4.1l} with $y(t)\in V$.
Then the following statements hold:
\begin{itemize}
\item[(i)] If there is an interval $T_{1}\subseteq [ a,b]$
such that
\[
L\leq x'(t)\leq L_{1}\quad \mbox{for}\quad t\in T_{1},\tag{4.4}
\]
then $x''(t)\geq 0$ for $t\in T_{1}$.

\item[(ii)] If there is an interval $T_{2}\subseteq [a,b]$
such that $F_{1}\leq x'(t)\leq F$  for $t\in T_{2}$,
then $x''(t)\leq 0$ for $t\in T_{2}$.
\end{itemize}
\end{lemma}

\paragraph{Proof}

Since the proofs of (i) and (ii) are similar, it is sufficient to show that
(4.4) implies $x''(t)\geq 0$ for $t\in T_{1}$. Indeed, the
assertion is true for $\lambda =0$. Now, let $\lambda \in (0,1]$ and assume
that there is a $t_{0}\in T_{1}$ such that $x''(t_{0})<0$. Then 
\[
0>Kx''(t_{0})=\lambda \left[ Kx^{\prime%
\prime}(t_{0})+f(t_{0},x(t_{0}),x'(t_{0}),x''(t_{0}))%
\right] \geq 0. 
\]
This contradiction proves the assertion. \hfill$\diamondsuit$

\begin{lemma} \label{lm4.3}
Let H1 hold, and $x(t)\in C^{2}[a,b]$ be a solution to \eqref{4.1l}
with $y(t)\in V$. Then
\[
Ft\leq x(t)\leq Lt,\quad F\leq x'(t)\leq L\quad \mbox{for}\quad
t\in [ a,b].
\]
\end{lemma}

\paragraph{Proof}

Consider the sets 
\[
Y_{0}=\left\{ t\in [ a,b]:L<x'(t)\leq L_{1}\right\} \quad \mbox{and}%
\quad Y_{1}=\left\{ t\in [ a,b]:F_{1}\leq x'(t)<F\right\} 
\]
and suppose that they are not empty. Then, using the continuity of $%
x'(t)$ and the inequality $F\leq x'(b)\leq L$, we easily
conclude that there are closed intervals $[ t_{0},\tau _{0}]\subseteq Y_{0}$
and $[t_{1},\tau _{1}]\subseteq Y_{1}$ such that 
\begin{equation}
x'(t_{0})>x'(\tau _{0})\quad \mbox{and}\quad
x'(t_{1})<x'(\tau _{1}).  \tag{4.5}
\end{equation}
On the other hand, by Lemma \ref{lm4.2}, we have 
\[
x''(t)\geq 0\quad\mbox{for}\quad t\in [ t_{0},\tau _{0}]\quad %
\mbox{and}\quad x''(t)\leq 0\quad \mbox{for }t\in [ t_{1},\tau
_{1}] 
\]
and therefore, we have 
\[
x'(t_{0})\leq x'(\tau _{0})\quad \mbox{and} \quad
x'(t_{1})\geq x'(\tau _{1}). 
\]
But this contradicts (4.5). The obtained contradiction shows that $Y_{0}$
and $Y_{1}$are empty, and so we see that 
\[
F\leq x'(t)\leq L\quad \mbox{for}\quad t\in [ a,b]. 
\]
Integrating this expression from $a$ to $t$ and using the fact that $Fa\leq
A\leq La$, we get 
\[
Ft\leq x(t)\leq Lt,\quad t\in [ a,b] 
\]
which concludes the proof.\hfill $\diamondsuit $

\begin{remark} \label{rmk4.1} \rm
Let $x(t)\in C^{2}[a,b]$ be a solution to
(1.1). Then, in view of Lemma \ref{lm4.3},  if $F=L$, it follows
that$x'(t)=B$, $t\in [ a,b]$. Now, using $Fa\leq A\leq La$, we see that
$x(t)=Bt$, $t\in [ a,b]$, is the unique  $C^{2}[a,b]$-solution to the
problem (1.1).
\end{remark}

\begin{lemma} \label{lm4.4}
Let H1 and  H2 hold, and
$x(t)\in C^{2}[a,b]$ be a solution to \eqref{4.1l} with
$y(t)\in V_{1}$. Then
\[
m_3\leq x''(t)\leq M_3,\quad t\in [ a,b],
\]
and there is a constant $D$ independent of $\lambda $ such that
\[
|x''' (t)|\leq D\quad \mbox{for } t\in [ a,b].
\]
\end{lemma}

\paragraph{Proof}

By the mean value theorem, there is a $\xi \in (a,b)$ such that $%
x''(\xi )=[x'(b)-x'(a)]/(b-a)$. Since Lemma \ref
{lm4.3} implies 
\[
F\leq x'(t)\leq L\quad \mbox{for}\quad t\in [ a,b],\tag{4.6} 
\]
we see that 
\[
x''(\xi )\leq 2C\leq G_{1}^{+},\tag{4.7} 
\]
where $C=\max \{|L|,|F|\}/(b-a)$. Now suppose that the set 
\[
Y=\left\{ t\in [ a,\xi ]:G_{1}^{+}<x''(t)\leq G_{2}^{+}\right\} 
\]
is not empty. The continuity of $x''(t)$ and (4.7) imply that
there is a closed interval $[ t_{0},\tau _{0}]\subseteq Y$ such that 
\[
x''(t_{0})>x''(\tau _{0}).\tag{4.8} 
\]
Since (4.6) holds for $t\in [ t_{0},\tau _{0}]$ and 
\[
\begin{gathered}
G_{1}^{+}<x''(t)\leq G_{2}^{+}\;\;\mbox{for }t\in [t_{0},\tau _{0}], \\
m_1\le Ft\le y(t)\le Lt\le M_1\quad\mbox{for }t\in[t_0, \tau_0 ],\\
 F\le y'(t)\le L\quad\mbox{for } t\in[t_0,\tau_0 ],
\end{gathered} \tag{4.9} 
\]
in view of H2, we have 
\[
\Psi _{1}(t)\equiv f_{q}\bigl(t,y(t),x'(t),x''(t)\bigr)%
<0,\quad t\in [ t_{0},\tau _{0}], 
\]
and for $t\in [ t_{0},\tau _{0}]$, 
\begin{align*}
\Psi _{2}(t)\equiv & f_{t}\bigl(t,y(t),x'(t),x''(t)%
\bigr) +f_{x}\bigl(t,y(t),x'(t),x''(t)\bigr)%
y'(t) \\
&+f_{p}\Bigl(t,y(t),x'(t),x''(t)\bigr)%
x''(t)\geq 0.
\end{align*}
Thus, using the last two inequalities and the continuity of $%
f_{t},\,f_{x},\,f_{p}$ and $f_{q}$ on $[t_{0},\tau _{0}]$, we conclude that $%
x^{\prime\prime\prime}$ is continuous on $[t_{0},\tau _{0}]$ and 
\[
x^{\prime\prime\prime}(t)=\lambda \Psi _{2}\ (t)/[K(1-\lambda )-\lambda \Psi
_{1}(t)] \geq 0\quad \mbox{for } t\in [ t_{0},\tau _{0}]. \tag{4.10} 
\]
Consequently, $x''(t_{0})\leq x''(\tau _{0})$,
which contradicts (4.8). Thus, 
\[
x''(t)\leq G_{1}^{+}\quad \mbox{for } t\in [ a,\xi]. 
\]
The inequality $H_{1}^{-}\leq x''(t)$, $t\in [ a,\xi ]$ can be
obtained in the same manner. Similarly, it is easy to show that 
\[
H_{1}^{+}\leq x''(t)\leq G_{1}^{-},\quad t\in [ \xi ,b]. 
\]
Finally, using (4.6), (4.9), the fact that $x''$is bounded on $%
[a,b]$ and the continuity of the partial derivatives of $f(t,x,p,q)\;\;%
\mbox{on}$ the set $[a,b]\times[m_1,M_1]\times[F,L]\times[m_3,M_3]$,  from
(4.10) it follows that there is a constant $D$ independent of $\lambda $
such that 
\[
|x^{\prime\prime\prime}(t)|\leq D\quad \mbox{for } t\in [ a,b]. 
\]
The proof of the lemma is complete. \hfill$\diamondsuit $

\begin{lemma} \label{lm4.5}
Suppose that H1, H2 and H3 hold.
Then the BVP (3.1) has a $C^{2}[a,b]$-solution, if $y(t)\in V_{1}$.
\end{lemma}

\paragraph{Proof}

Let $x(t)\in C^{2}[a,b]$ be a solution to (4.1)$_{\lambda }$. Then, by Lemma 
\ref{lm4.3}, we have 
\begin{gather*}
F-\varepsilon <x'(t)<L+\varepsilon \quad \mbox{for } t\in [ a,b] \\
m_1-\varepsilon <x(t)<M_1+\varepsilon \quad \mbox{for } t\in [a,b],
\end{gather*}
while, by Lemma \ref{lm4.4}, 
\[
m_3-\varepsilon <x''(t)<M_3+\varepsilon \quad \mbox{for } t\in
[ a,b], 
\]
where $\varepsilon >0$ is as in H2. Thus, the condition (i) of Lemma \ref
{lm4.1} holds for $Q_{0}=m_1-\varepsilon $, $Q_{1}=M_1+\varepsilon $, $%
Q_{2}=F-\varepsilon$, $Q_{3}=L+\varepsilon $, $Q_{4}=m_3-\varepsilon $ and $%
Q_{5}=M_3+\varepsilon$. Moreover, from (2.1) and H3 it follows that the
conditions (ii) and (iii) of Lemma \ref{lm4.1} are satisfied. Also, 
\[
m_1-\varepsilon <y(t)<M_1+\varepsilon\quad\mbox{for }  t\in [a,b]. 
\]
So, we can apply Lemma \ref{lm4.1} to conclude that the problem (3.1) has a
solution in $C^{2}[a,b]$. The proof of the lemma is complete. \hfill$%
\diamondsuit $

We need the following two lemmas which are adopted from \cite{p4}.

\begin{lemma}[{\cite[Chapter I, Theorem 1]{p4}}] \label{lm4.6}
Suppose $\phi(t)$ satisfies the differential inequality
\[
\phi ''+g(t)\phi '\geq 0\quad \mbox{for } a<t<b,
\tag{4.11}
\]
with $g(t)$ a bounded function. If $\phi (t)\leq M$ in $(a,b)$
and if the maximum $M$ of $\phi $ is attained at an interior point
$c$ of $(a,b)$, then $\phi \equiv M$.
\end{lemma}

\begin{lemma}[{\cite[Chapter I, Theorem 2]{p4}}] \label{lm4.7}
Suppose $\phi (t)$ is a nonconstant function which satisfies the
inequality (4.11) and has one-sided derivatives at $a$ and $b$,
and suppose $g$ is bounded on every
closed subinterval of $(a,b)$. If the maximum of $\phi $ occurs at
$t=a$ and $g$ is bounded below at $t=a$, then $\phi '(a)<0$.
If the maximum occurs at $t=b$ and $g$ is bounded above
at $t=b$, then $\phi '(b)>0$.
\end{lemma}

\begin{lemma} \label{lm4.8}
Suppose that $\phi \in C^{2}(a,b)\cap C^{1}[a,b]$ satisfies the inequality
\[
\phi ''(t)+g(t)\phi '(t)\geq 0\quad \mbox{for }t\in (a,b),
\]
where $g(t)$ is bounded on $(a,b)$. If $\phi (a)\leq 0$ and
\[
\phi '(b)\leq 0,\tag{4.12}
\]
then
\[
\phi (t)\leq 0\quad\mbox{for }t\in [ a,b].\tag{4.13}
\]
\end{lemma}

\paragraph{Proof}

First, assume that $\phi (t)$achieves its maximum at $t_{0}\in (a,b)$. By
Lemma \ref{lm4.6}, for $t\in[a,b]$ we obtain $\phi (t)\equiv \phi
(t_{0})=\phi (a)\le 0$ and so (4.13) holds.

Next, suppose that $\phi (t)$ achieves its maximum at the ends of the
interval $[a,b]$. If we assume $\phi (t)\leq \phi(b)$, $t\in [ a,b]$, the
application of Lemma \ref{lm4.7} shows that $\phi '(b)>0$, which
contradicts (4.12). Thus, by our assumtions, $\phi (t)\leq \phi (a)\leq 0$, $%
t\in [ a,b]$, and so (4.13) follows. The proof is complete. \hfill$%
\diamondsuit$

In the last two lemmas we use the map $\mathcal{A}$ defined in the section 3.

\begin{lemma} \label{lm4.9}
Under assumptions  H1, H2, and H3, for
any $y\in V_{1}$, the image $x$ by the map $\mathcal{A}$ exists and it is
unique.
\end{lemma}

\paragraph{Proof}

The existence of the image of $x$ follows from Lemma \ref{lm4.5}. In order
to see that $x$ is unique, fix $y$ and assume that$z$is an other image of $y$
by $\mathcal{A}$ and consider the function $\phi (t)=x(t)-z(t)$, $t\in [ a,b]
$. Then, it is evident that 
\[
f\bigl(t,y(t),x'(t),x''(t)\bigr)-f\bigl(%
t,y(t),z'(t),z''(t)\bigr)=0,\quad t\in [ a,b]. 
\]
Next, we construct the equality 
\begin{align*}
f\bigl(t,y(t),x'(t),x''(t)\bigr)-f\bigl(%
t,y(t),z'(t),x''(t)\bigr)& \\
+f\bigl(t,y(t),z'(t),x''(t)\bigr)-f\bigl(%
t,y(t),z'(t),z''(t)\bigr)&=0,
\end{align*}
which can be rewritten in the form $I_{1}(t)\phi '(t)+\ I_{2}(t)\phi
''(t)=0$, where 
\begin{gather*}
I_{1}(t) =\int_{0}^{1} f_{p}\bigl(t,y(t),z'(t)+\theta
(x'(t)-z'(t)),x''(t)\bigr)d\theta , \\
I_{2}(t) =\int_{0}^{1} f_{q}\bigl(t,y(t),z'(t),z''(t)
+\theta (x''(t)-z''(t))\bigr)d\theta .
\end{gather*}
Hence, the function $\phi (t)$ is a solution to the BVP 
\begin{gather*}
\phi ''(t)+ \frac{I_{1}(t)}{I_{2}(t)}\phi '(t)=0,\quad
t\in [ a,b], \\
\phi (a)=0,\quad \phi '(b)=0.
\end{gather*}
Moreover, it is easy to conclude that $\phi (t)\equiv 0$, $t\in [ a,b]$, is
the unique solution of the above BVP. Consequently, $x(t)\equiv z(t)$, $t\in
[ a,b]$. The proof of the lemma is complete. \hfill $\diamondsuit $

\begin{lemma} \label{lm4.10}
Under the hypotheses  H1--H4,  if
$y_{1}(t),y_{2}(t)\in V_{1}$ are such that ${y_{1}(t)\leq y_{2}(t)}$ for
$t\in [ a,b]$, then
\[
x_{1}(t)\leq x_{2}(t)\;\;\mbox{for}\quad t\in [ a,b],
\]
where $x_{i}=\mathcal{A}y_{i},i=1,2$.
\end{lemma}

\paragraph{Proof}

Observe that, by Lemma \ref{lm4.3}, we have $F\leq x_{1}'(t)\leq L$, 
$t\in [ a,b]$, and, by Lemma \ref{lm4.4}, 
\[
m_3\leq {x_1}''(t)\,\leq M_3,\quad t\in [ a,b]. 
\]
Moreover, 
\[
Ft\leq y_{1}(t)\leq y_{2}(t)\leq Lt,\quad t\in [ a,b]. 
\]
Thus, from $f_{x}(t,x,p,q)\geq 0$ for $(t,x,p,q)$ in $T\times [ F,L]\times[%
m_3,M_3]$ it follows that 
\[
0=f\bigl(t,y_{1}(t),x_{1}'(t),x_{1}''(t)\bigr)\leq f%
\bigl(t,y_{2}(t),x_{1}'(t),x_{1}''(t)\bigr),\quad t\in
[ a,b]. 
\]
Hence, for $t\in [ a,b]$ we have 
\[
f\bigl(t,y_{2}(t),x_{2}'(t),x_{2}''(t)\bigr)
-f\bigl(t,y_{2}(t),x_{1}'(t),x_{1}''(t)\bigr)\leq 0 
\]
and then, as in Lemma \ref{lm4.9}, we construct the inequality 
\begin{align*}
f\bigl(t,y_{2}(t),x_{1}'(t),x_{1}''(t)\bigr) -f\bigl(%
t,y_{2}(t),x_{2}'(t),x_{1}''(t)\bigr)& \\
+f\bigl(t,y_{2}(t),x_{2}'(t),x_{1}''(t)\bigr) -f\bigl(%
t,y_{2}(t),x_{2}'(t),x_{2}''(t)\bigr)&\geq 0
\end{align*}
from which for $\phi (t)=x_{1}(t)-x_{2}(t)$, $t\in [ a,b]$, we find 
\[
\phi ''(t)+\frac{\ J_{1}(t)}{\ J_{2}(t)}\phi '(t)\geq
0, \quad t\in [ a,b], 
\]
where 
\begin{gather*}
J_{1}(t) =\int_{0}^{1}f_{p}\bigl(t,y_{2}(t),x_{2}'(t) +\theta
(x_{1}'(t)-x_{2}'(t)),x_{1}''(t)\bigr)d\theta ,
\\
\ J_{2}(t) =\int_{0}^{1}f_{q}\bigl(t,y_{2}(t),x_{2}'(t),x_{2}^{%
\prime\prime}(t) +\theta (x_{1}''(t)-x_{2}''(t))%
\bigr)d\theta .
\end{gather*}
Furthermore, $\phi (a)=0$, $\phi '(b)=0$. Finally, applying Lemma 
\ref{lm4.8}, we see that $\phi (t)\leq 0$ for $t\in [ a,b]$, which completes
the proof. \hfill $\diamondsuit $

\section{Proof of Theorem \ref{thm3.1}}

%sec. 5
Consider the sequences $\{u_{n}\,\}$ and $\{v_{n}\,\}$, defined by 
\[
u_{n+1}=\mathcal{A}u_{n}\quad \mbox{and}\quad v_{n+1}=\mathcal{A}
v_{n},\quad n=0,1,\dots 
\]
In view of Lemma \ref{lm4.5}, from Lemma \ref{lm4.3} it follows that $%
Ft=u_{0}\leq u_{1}$ and $v_{1}\leq v_{0}=Lt$. Moreover, Lemma \ref{lm4.10}
and induction arguments imply that 
\[
u_{n-1}\leq u_{n},\quad v_{n}\leq v_{n-1},\quad n=1,2,\dots 
\]
On the other hand, since $u_{0}\leq v_{0}$, by Lemma \ref{lm4.10} and
induction arguments, we conclude that $u_{n}\leq v_{n}$, $n=0,1,\dots$.
>From this observation it follows that 
\[
u_{0}\leq u_{n}\leq v_{0},\;\;n=0,1,\dots. 
\]
Therefore, $\{u_{n}\}$ is uniformly bounded. Furthermore, since, by Lemma 
\ref{lm4.3}, $\{u_{n}'\;\}$ is uniformly bounded, we see that $%
\{u_{n}\;\} $ is equicontinuous. Finally, since, by Lemma \ref{lm4.4}, $%
\{u_{n}^{\prime\prime\prime}\}$ is uniformly bounded, it follows that the
sequence $\{u_{n}''\}$ is uniformly bounded and equicontinuous.
Thus, we can apply the Arzela-Ascoli theorem to conclude that there are a
subsequence $\{u_{n_{i}}\}$ and a function $u\in C^{2}[a,b]$ such that $%
\{u_{n_{i}}\}$, $\{u_{n_{i}}'\}$ and $\{u_{n_{i}}''\}$
are uniformly convergent on $[a,b]$ to $u,u'$ and $u''$
respectively. Now, using the fact that $u_{n_{i}}=\mathcal{A}u_{n_{i-1}}$can
be rewritten equivalently in the form 
\begin{align*}
u_{n_{i}}(t)&=\frac{1}{K}\int_{a}^{t}\Big( \int_{b}^{r}\bigl(%
Ku_{n_{i}}''(s)+f(s,u_{n_{i-1}}(s),u_{n_{i}}'(s),
u_{n_{i}}''(s))\bigr)ds\Big) dr \\
&\quad+B(t-a)+A,
\end{align*}
letting $i\to +\infty $, we obtain 
\[
\ u(t)=\frac{1}{K}\int_{a}^{t}\Big(
\int_{b}^{r}\bigl(Ku''(s)+f(s,u(s),u'(s),u^{\prime%
\prime}(s))\bigr)ds\Big) dr +B(t-a)+A, 
\]
from which it follows that $u(t)$ is a solution to the BVP (1.1).

Remark that, if $x(t)$ is a solution of (1.1), then, by Lemma \ref{lm4.3},
we have 
$u_{0}(t)\leq x(t)$ for $t\in [ a,b]$. 
Applying Lemma \ref{lm4.10} (it is possible, because $x=\mathcal{A}x$), by
induction we obtain 
\[
u_{n}(t)\leq x(t),\quad t\in [ a,b],\quad n=0,1,\dots, 
\]
and then $u(t)\leq x(t)$, $t\in [ a,b]$, which holds for each solution $%
x(t)\in C^{2}[a,b]$ of the problem (1.1). Consequently, it follows that 
\[
u(t)\equiv u^{m}(t),\quad t\in [ a,b]. 
\]
By similar arguments, we conclude that $\lim v_{n}=v^{M}(t)$, $t\in [ a,b]$.
Thus, the proof is complete. \hfill$\diamondsuit $

\paragraph{Acknowledgement}

P. S. Kelevedjiev would like to thank the Ministry of National Economy of
Helenic Republic for providing the NATO Science Fellowship (No. DOO 850/02)
and the University of Ioannina for its hospitality.

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\noindent\textsc{Myron K. Grammatikopoulos}\newline
Department of Mathematics, University of Ioannina \newline
451 10 Ioannina, Hellas, Greece\newline
e-mail: mgrammat@cc.uoi.gr \smallskip

\noindent\textsc{Petio S. Kelevedjiev} \newline
Department of Mathematics, Technical University of Sliven \newline
8800 Sliven, Bulgaria \newline
e-mail: keleved@mailcity.com

\end{document}