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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 127, 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-2007/127\hfil Existence and localization of solutions]
{Existence and localization of solutions
 for fourth-order boundary-value problems}

\author[R. Engui\c{c}a, L. Sanchez\hfil EJDE-2007/127\hfilneg]
{Ricardo Engui\c{c}a, Lu\'is Sanchez}  % in alphabetical order

\address{Ricardo Engui\c{c}a \newline
Area Cient\'ifica de Matem\'atica,
Instituto Superior de Engenharia de Lisboa \\
Rua Conselheiro Em\'idio Navarro, 1 - 1950-062 Lisboa, Portugal}
\email{rroque@dec.isel.ipl.pt}

\address{Lu\'is Sanchez \newline
Faculdade de Ci\^encias da Universidade de Lisboa \\
Avenida Professor Gama Pinto 2, 1649-003 Lisboa, Portugal}
\email{sanchez@ptmat.fc.ul.pt}

\thanks{Submitted July 20, 2007. Published September 28, 2007.}
\subjclass[2000]{34B15, 34C25}
\keywords{Beam equation; fourth order boundary value problem; \hfill\break\indent
 maximum principles; lower and upper solutions; reversed order}

\begin{abstract}
 In this paper, we study the existence of solutions for the
 differential equation
 $$
 u^{(4)}(t)=f\big(t,u(t),u''(t)\big),
 $$
 where $f$ satisfies one-sided Lipschitz conditions with respect
 to $u$ and $u''$, with periodic conditions or boundary conditions
 from ``simply supported'' beam theory.
 We assume the existence of lower and upper solutions
 (well-ordered and in some cases reversely ordered) and we make
 use of a fourth-order linear differential operator factorization.
\end{abstract}

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

\section{Introduction}
 It is well known that fourth order boundary value problems are
motivated by the theory of beam deflection. However, the
corresponding boundary value problems are also mathematically
interesting as they are usually not so simple as similar 2nd order
problems, and well known results for the second order case do not
necessarily hold for the corresponding fourth order problems
without some strengthening of assumptions. An example of this is
the theorem that states that if a given boundary-value problem has
well ordered lower and upper solutions, then it has a solution
lying between those two functions: Roughly, this is true in the
second-order case, but one needs to add monotonicity assumptions
to obtain some true statement in the fourth order case. In this
note we propose a step in the direction of establishing this type
of results.

Recently, several authors have studied the existence and
multiplicity of solutions to the equation
\begin{equation}\label{eqd1}
 u^{(4)}(t)=f\big(t,u(t),u''(t)\big), \quad t\in [0,2\pi]
\end{equation}
with various boundary conditions. Here we are interested in the
periodic boundary conditions
 \[
 u(0)=u(2\pi),\quad u'(0)=u'(2\pi),\quad u''(0)=u''(2\pi),\quad
 u'''(0)=u'''(2\pi)
 \]
and the ``simply supported'' boundary conditions
 \[
 u(0)=u(\pi)=u''(0)=u''(\pi)=0.
 \]

The periodic problem with $f$ independent of $u''$ was
studied by Cabada \cite{C94}, via maximum principles and the
monotone method. Jiang, Gao and Wan \cite{dJwGaW02} obtained
results for the full nonlinear problem using the monotone method.
Allowing a linear dependence of $f$ on $u''$, Li \cite{L03} and
Liu and Li \cite{LiuLi05} have obtained existence results with
fixed point theory.

For the ``simply supported'' case, Bai and Wang \cite{BW02} have
obtained existence and multiplicity results without dependence on
$u''$. With linear dependence on $u''$, we can find results of
existence in Li \cite{Li} and existence and multiplicity in Yao
\cite{Y04}. Cabada, Cid and Sanchez \cite{CCS06}
 obtained results for the problem without dependence on $u''$,
using upper and lower solutions in reversed order. The superlinear
case has been studied by Rynne \cite{R02} using a bifurcation
technique.

Our paper deals with fourth-order boundary-value problems in a way
that Gao, Weng, Jiang and Hou \cite{GWJH06} did for second order.
We consider equation \eqref{eqd1} with periodic as well as
``simply supported'' boundary conditions, and prove existence
results (considering $f$ one-sided Lipschitz in both variables $u$
and $u''$) if there exist lower and upper solutions (ordered or in
reversed order for the periodic case, and ordered in the ``simply
supported'' case). Habets and Sanchez \cite{HS06} have obtained
results similar to those we obtain here, using Lipschitz
conditions. The main difference is that in our case only
localization is obtainable, no iterative method is available.

In section 2 we present some maximum principles, in sections 3 and
4 we prove existence results for the periodic case, respectively,
for ordered and reversely ordered upper and lower solutions. In
section 5 we deal with the ``simply supported'' case with ordered
upper and lower solutions.

\section{Auxiliary results}

In this section we state some simple maximum principles that will
be useful in following sections.

\begin{lemma} [Maximum Principle 1]\label{mp1}
 Let $L<0$, $p,q,r\in \mathbb{R}$ with $p<r<q$ and
$y\in C[p,q]\cap W^{2,1}(p,r)\cap W^{2,1}(r,q)$ such that
\begin{gather*}
 y''(t)+Ly(t)=f(t)\ge 0 \quad\text{a.e. } (p,q),\\
 y(p)=y(q),\quad y'(p)\ge y'(q),\quad y'(r^+)\ge y'(r^-).
\end{gather*}
Then $y(t)\le0$ for all $t\in[0,2\pi]$. Moreover, if
$y''(t)+Ly(t)\not \equiv 0$, then $y(t)<0$ for all
$t\in\,]0,2\pi[$.
 \end{lemma}

\begin{proof}
Suppose that $y(t)>0$ for all $t\in ]p,q[$. Then we would have the
contradiction
 \[
 0\ge y'(r^-)-y'(r^+)+y'(q)-y'(p)=\int_p^r f(t)-Ly(t)\,dt
+\int_r^q f(t)-Ly(t)\,dt>0.
 \]
If $y(p)>0$, then $y(q)>0$ and therefore there exist two intervals
$[p,p_1]$ and $[q_1,q]$ where $y>0$, $y(p_1)=y(q_1)=0$,
$y'(p_1)\le0$ and $y'(q_1)\ge 0$. If $r$ belongs to one of the
intervals, then we would have the contradiction
\begin{align*}
0&\ge y'(r^-)-y'(r^+)+y'(q)-y'(p)+y'(p_1)-y'(q_1)\\
&=\int_p^{p_1} f(t)-Ly(t)\,dt+\int_{q_1}^q f(t)-Ly(t)\,dt>0,
\end{align*}
 otherwise, the contradiction is the same, without the terms involving $r$.

If $y(p)<0$ then the exists an interval $]p_1,q_1[$ where $y>0$
and $y(p_1)=y(q_1)=0$, and we can apply the arguments used in the
first case.
 \end{proof}


\begin{lemma}[Maximum Principle 2]\label{mp2}
 Let $0<L<\frac{1}{4}$, $p<r<q$ with $q-p\le2\pi$ and $y\in C[p,q]\cap W^{2,1}(p,r)\cap W^{2,1}(r,q)$ such that
 \[
 y''(t)+Ly(t)=f(t)\ge 0\;\,a.e.\; (p,q),\quad y(p)=y(q),\;\;y'(p)\ge y'(q),\;\;y'(r^+)\ge y'(r^-).
 \]
Then $y(t)\ge0$ for all $t\in[0,2\pi]$. Moreover, if
$y''(t)+Ly(t)\not \equiv 0$, then $y(t)>0$ for all
$t\in\,]0,2\pi[$.
\end{lemma}

\begin{proof}
It follows easily combining the arguments used in the proof of the
previous lemma and the proof of \cite[Proposition 2.3]{HS06}.
 \end{proof}

\begin{lemma}[Maximum Principle 3]\label{mp3}
Let $L<1$ and $y\in W^{2,1}(0,\pi)$ such that
 \[
y''(t)+Ly(t)\ge 0 \quad\text{a.e. } (0,\pi),\quad y(0)\le0,\quad
y(\pi)\le0.
 \]
Then $y(t)\le0$ for all $t\in[0,\pi]$. Moreover, if
$y''(t)+Ly(t)\not \equiv 0$, then $y(t)<0$ for all
$t\in\,]0,\pi[$.
\end{lemma}

\begin{lemma}[Maximum Principle 4]\label{mp4}
Let $L<1$, $M\in\mathbb{R}$ and $y\in W^{2,1}(0,\pi)$ such that
 \[
 y''(t)+Ly^+(t)-My^-(t)\ge 0 \;\,a.e.\; (0,\pi),\quad y(0)\le0,\;\;y(\pi)\le0,
 \]
where $y^+,y^-$ are respectively the positive and negative parts
of $y$. Then $y(t)\le0$ for all $t\in[0,\pi]$.
 \end{lemma}

 \section{Periodic problem with ordered upper and lower solutions}

Let us consider the fourth-order equation
\begin{equation}\label{eq}
 u^{(4)}(t)=f\big(t,u(t),u''(t)\big), \quad t\in [0,2\pi]
\end{equation}
 where $f$ is a $L^1$-Carath\'eodory function, with periodic boundary
conditions
 \begin{equation}\label{bcper}
 u(0)=u(2\pi),\quad u'(0)=u'(2\pi),\quad u''(0)=u''(2\pi),\quad
 u'''(0)=u'''(2\pi).
 \end{equation}
We say that $\alpha\in W^{4,1}(0,2\pi)$ is a lower solution of the
boundary-value problem  \eqref{eq}--\eqref{bcper} if
\begin{gather*}
 \alpha^{(4)}(t)\le f\left(t,\alpha(t),\alpha''(t)\right), \quad t\in [0,2\pi]
\\
 \alpha(0)=\alpha(2\pi),\quad \alpha'(0)=\alpha'(2\pi),\quad
 \alpha''(0)=\alpha''(2\pi),\quad \alpha'''(0)\le \alpha'''(2\pi).
\end{gather*}
A function $\beta \in W^{4,1}(0,2\pi)$ satisfying the reversed
inequalities is called an upper solution.

Let $\alpha,\, \beta$ be respectively a lower and an upper
solution of \eqref{eq}--\eqref{bcper}, such that $\alpha(t)\le
\beta(t)$ for all $t\in[0,2\pi]$.

In the following, we assume the hypothesis
 \begin{itemize}
 \item[(H1)] There exist constants $C,D >0$ with $D^2>4C$, such that
 \begin{equation}
 f\left(t,u_2,v_2\right)-f\left(t,u_1,v_1\right)\ge
 -C\left(u_2-u_1\right)+D\left(v_2-v_1\right)
 \end{equation}
 for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$, $v_1\le v_2$.
 \end{itemize}

 \begin{remark} \label{rmk3.1} \rm
 If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H1)
is equivalent to
 $\frac{\partial f}{\partial u}\ge -C$
 and $\frac{\partial f}{\partial v}\ge D$.
 \end{remark}

 Let $m,M<0$ be the two roots of the equation $x^2+Dx+C=0$
(note that $C=mM,\,D=-(m+M)$). Setting
$a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$,
 we have the following result.

 \begin{proposition}\label{ab1}
If $f$ is a $L^1$-Carath\'eodory function satisfying (H1) for
$\alpha(t),\beta(t)$ lower and upper solutions such that
$\alpha(t)\le\beta(t)$ for all $t\in[0,2\pi]$, then $b(t)\le
a(t)$.
 \end{proposition}

 \begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)=y(2\pi)$ and $y'(0)\ge
y'(2\pi)$. Suppose towards a contradiction that there exists
$t_0\in [0,2\pi]$ such that $y(t_0)>0$.

 If $y(t)>0$ for all $t\in[0,2\pi]$, we have (noting that
$b(t)-m\beta(t)\ge a(t)-m\alpha(t)$ and $m^2+mD+C=0$)
 \begin{align*}
 y''(t)+My(t)=&b''(t)-a''(t)+Mb(t)-Ma(t)\\
=&\beta^{(4)}(t)+(m+M)\beta''(t)+(m+M)m\beta(t)-m^2\beta(t)-\\
 &-(\alpha^{(4)}(t)+(m+M)\alpha''(t)+(m+M)m\alpha(t)-m^2\alpha(t))\\
\ge&f(t,\beta(t),b(t)-m\beta(t))-f(t,\alpha(t),a(t)-m\alpha(t))\\
 &+(m+M)y(t)-m^2(\beta(t)-\alpha(t))\ge0,
 \end{align*}
 and this is a contradiction, since by the Maximum Principle \ref{mp1}
 we would have $y(t)\le0$.

Otherwise, considering if necessary the periodic extension of
$y(t)$, there exists an interval $[p,q]$ with $q-p<2\pi$ such that
$y(p)=y(q)=0$, $y'(p)\ge 0\ge y'(q)$, and $y(t)>0$ for
$t\in\,]p,q[$, possibly with $p<0<q$, in which case $y'(0^+)\ge
y'(0^-)$. Applying Maximum Principle \ref{mp1} again, we reach a
contradiction.
 \end{proof}

 Let
 \[
 p(t,x)=  \begin{cases}
 b(t), & x<b(t)\\
 x, & b(t)\le x\le a(t)\\
 a(t), & x>a(t).
 \end{cases}
 \]
Consider the boundary-value problem
 \[
 u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=u(2\pi),\quad u'(0)=u'(2\pi),
 \]
with $q\in L^1[0,1]$. Since $m<0$, the operator $L_m$ is
invertible so that we can write its unique solution $u$ as
$u=L_m^{-1}q$, and by the Maximum principle \ref{mp1} we know that
if $q(t)\ge0$ then $u(t)\le0$. Since $\alpha(t)=L_m^{-1}a(t)$,
$\beta(t)=L_m^{-1}b(t)$ and $b(t)\le p(t,x(t))\le a(t)$, we have
 \[
 \alpha(t)\le L_m^{-1}p(t,x(t))\le \beta(t).
 \]
Let us consider the modified problem
\begin{equation} \label{modif}
\begin{gathered}
 \begin{aligned}
 x''(t)+Mx(t)=(Fx)(t)
&\equiv f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t))-mL_m^{-1}p(t,x(t))\right)\\
&\quad +(m+M)p(t,x(t))-m^2L_m^{-1}p(t,x(t)),
\end{aligned} \\
 x(0)=x(2\pi),\quad  x'(0)=x'(2\pi).
 \end{gathered}
\end{equation}
 Considering the operator $\Phi:C[0,2\pi]\to C[0,2\pi]$ with
$\Phi x=L_M^{-1}(Fx)$, since $p(t,x(t))$ and $L_m^{-1}p(t,x(t))$
are bounded and $f$ is a Carath\'eodory function, there exists a
$L^1[0,2\pi]$ function $g(t)$ such that $|(Fx)(t)|\le g(t)$ for
a.e. $t\in[0,2\pi]$. Therefore, applying Leray-Schauder's fixed
point Theorem, we can conclude that $\Phi$ has a fixed point
$x(t)$ which is a solution of the modified problem \eqref{modif}.

 \begin{proposition}\label{prop1}
Let $x(t)$ be a solution of the modified problem \eqref{modif}.
Assuming (H1), for given lower and upper solutions $\alpha(t)$ and
$\beta(t)$, with $\alpha\le \beta$ for all $t\in[0,2\pi]$, we have
 \[
b(t)\le x(t)\le a(t).
 \]
 \end{proposition}

 \begin{proof}
We will only prove that $x(t)\le a(t)$, since the other inequality
can be obtained with similar arguments.
 We have
\begin{gather*}
 a''(t)+Ma(t)\le f\left(t,L_m^{-1}a(t),a(t)-mL_m^{-1}a(t)\right)+(m+M)a(t)
 -m^2L_m^{-1}a(t), \\
 a(0)=a(2\pi),\quad a'(0)\le a'(2\pi).
\end{gather*}
Setting $y(t)=x(t)-a(t)$, we have $y(0)=y(2\pi)$, $y'(0)\ge
y'(2\pi)$.  Suppose towards a contradiction that there exists
$t_0\in [0,2\pi]$ such that  $y(t_0)>0$.

If $y(t)>0$ for all $t$, we have $x(t)>a(t)$ and, therefore,
$p(t,x(t))=a(t)$, so
 \begin{align*}
 x''(t)+Mx(t)=& f\left(t,L_m^{-1}a(t),a(t)-mL_m^{-1}a(t)\right)+(m+M)a(t)
 -m^2L_m^{-1}a(t) \\
 \ge & a''(t)+Ma(t),
 \end{align*}
which is a contradiction, since by the Maximum Principle \ref{mp1}
we would have $y(t)\le0$.

Otherwise, considering if necessary the periodic extension of
$y(t)$, there exists an interval $[p,q]$ with $q-p<2\pi$ such that
$y(p)=y(q)=0$, $y'(p)\ge 0 \ge y'(q)$, $y(t)>0$ for $t\in\,]p,q[$,
possibly with $p<0<q$, in which case $y'(0^+)\ge y'(0^-)$. Then,
recalling that $L_m^{-1}p(t,x(t))\ge L_m^{-1}a(t)$,
 \begin{align*}
&y''(t)+My(t)\\
&=x''(t)+Mx(t)-a''(t)-Ma(t) \\
&\ge f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t))-mL_m^{-1}p(t,x(t))\right)
 -f(t,L_m^{-1}a(t),a(t)-mL_m^{-1}a(t)) \\
&\quad +(m+M)y(t)-m^2(L_m^{-1}p(t,x(t))-\alpha(t))\ge0,
 \end{align*}
so, we have again a contradiction by the Maximum Principle
\ref{mp1}.
 \end{proof}

\begin{theorem}\label{teo1}
Assuming (H1), for given lower and upper solutions $\alpha$ and
$\beta$, with $\alpha\le \beta$, the boundary value problem
\eqref{eq}--\eqref{bcper} has a solution $u(t)\in W^{4,1}(0,2\pi)$
such that $\alpha\le u\le\beta$.
 \end{theorem}

 \begin{proof}
Let $u(t)=L_m^{-1}x(t)$, where $x(t)$ is a solution of the
modified problem \eqref{modif}. Since $x(t)=u''(t)+mu(t)$,
 we have $x''(t)+Mx(t)=u^{(4)}(t)-Du''(t)+Cu(t)$.
 On the other hand,
 \begin{align*}
 x''(t)+Mx(t)=&f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t))-mL_m^{-1}p(t,x(t))\right)\\
 &+(m+M)p(t,x(t))-m^2L_m^{-1}p(t,x(t)) \\
 =&f\big(t,u(t),u''(t)\big)-Du''(t)+Cu(t),
 \end{align*}
so $u(t)$ satisfies \eqref{eq}--\eqref{bcper}. Moreover, from
$$
\beta''+m\beta\le u''+mu\le\alpha''+m\alpha,
$$
taking into account the boundary conditions and Maximum Principle
\ref{mp1}, we obtain $\alpha \le u\le\beta$.
\end{proof}

We can reach a similar conclusion, assuming the following
hypothesis
 \begin{itemize}
\item[(H1')]there exist constants $C,D >0$ with $D<4C+1/4$ and $D^2>4C$,
such that
 \begin{equation}
 f\left(t,u_2,v_1\right)-f\left(t,u_1,v_2\right)\ge
 -C\left(u_2-u_1\right)-D\left(v_1-v_2\right)
 \end{equation}
 for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$, $v_1\le v_2$.
\end{itemize}

\begin{remark} \label{rmk3.5} \rm
 If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H1')
is equivalent to
 $\frac{\partial f}{\partial u}\ge -C$
 and $\frac{\partial f}{\partial v}\le -D$.
 \end{remark}

Let $0<m,M<\frac{1}{4}$ be the two roots of the equation
$x^2-Dx+C=0$ (note that $C=mM$, $D=m+M)$). Defining $a(t)$ and
$b(t)$ as above, we have the following result.

 \begin{proposition}\label{ab2}
If $f$ is a $L^1$-Carath\'eodory function satisfying (H1') for
$\alpha(t),\beta(t)$ lower and upper solutions such that
$\alpha(t)\le\beta(t)$, then $b(t)\ge a(t)$.
 \end{proposition}

\begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)=y(2\pi)$ and $y'(0)\ge
y'(2\pi)$. Suppose towards a contradiction that there exists
$t_0\in [0,2\pi]$ such that $y(t_0)<0$. We can reach a
contradiction with similar arguments from the ones used in
Proposition \ref{ab1}, using instead the Maximum principle
\ref{mp2}.
\end{proof}

 Using the same arguments as in the previous case, we prove the
following result.

 \begin{theorem}
Assuming (H1'), for given lower and upper solutions $\alpha$ and
$\beta$, with $\alpha\le \beta$, the boundary value problem
\eqref{eq}--\eqref{bcper} has a solution $u(t)\in W^{4,1}(0,2\pi)$
such that $\alpha\le u\le\beta$.
 \end{theorem}

\section{Periodic problem with upper and lower solutions in reversed order}

In this section we prove similar results from the ones above, but
with lower and upper solutions in reversed order, that is
$\beta(t) \le \alpha(t)$, for all $t\in[0,2\pi]$.

 \begin{itemize}
 \item[(H2)] There exist constants $C,D$ with $C<0$ and $D>-4C-1/4$,
 such that
 \begin{equation}
 f\left(t,u_1,v_2\right)-f\left(t,u_2,v_1\right)\ge
 -C\left(u_1-u_2\right)+D\left(v_2-v_1\right)
 \end{equation}
 for a.e. $t\in[0,2\pi]$, $\beta(t)\le u_1\le u_2\le\alpha(t)$, $v_1\le v_2$.
 \end{itemize}



\begin{remark} \label{rmk4.1} \rm
If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in
(H2) is equivalent to $\frac{\partial f}{\partial u}\le -C$
 and $\frac{\partial f}{\partial v}\ge D$.
 \end{remark}

Let $M<0$ and $0<m<\frac{1}{4}$ be the two roots of the equation
$x^2+Dx+C=0$ (note that $C=mM,\,D=-(m+M)$). Defining
$a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$,
 we have the following result.

 \begin{proposition}\label{ab3}
If $f$ is a $L^1$-Carath\'eodory function satisfying (H2) for
$\alpha(t),\beta(t)$ lower and upper solutions such that
$\beta(t)\le\alpha(t)$, then $b(t)\le a(t)$.
 \end{proposition}

 \begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)=y(2\pi)$ and $y'(0)\ge
y'(2\pi)$. Suppose towards a contradiction that there exists
$t_0\in [0,2\pi]$ such that $y(t_0)>0$.

If $y(t)>0$ for all $t$, we have (noting that $b(t)-m\beta(t)\ge
a(t)-m\alpha(t)$)
 \begin{align*}
 y''(t)+My(t)\ge&f(t,\beta(t),b(t)-m\beta(t))-f(t,\alpha(t),a(t)-m\alpha(t))\\
 &+(m+M)y(t)-m^2(\beta(t)-\alpha(t))\ge0,
 \end{align*}
and this is a contradiction, since by the Maximum Principle
\ref{mp1} we would have $y(t)\le0$.

Otherwise, considering if necessary the periodic extension of
$y(t)$, there exists an interval $[p,q]$ with $q-p<2\pi$ such that
$y(p)=y(q)=0$, $y'(p)\ge 0 \ge y'(q)$, $y(t)>0$ for $t\in\,]p,q[$,
possibly with $p<0<q$, in which case $y'(0^+)\ge y'(0^-)$.
Applying Maximum Principle \ref{mp1} again, we reach a
contradiction.
\end{proof}

 Let
 \[
 p(t,x)= \begin{cases}
 b(t), & x<b(t)\\
 x, & b(t)\le x\le a(t)\\
 a(t), & x>a(t).
 \end{cases}
 \]
Consider the boundary-value problem
 \[
 u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=u(2\pi),\quad u'(0)=u'(2\pi),
 \]
with $q\in L^1[0,1]$. Since $m<1$ , the operator $L_m$ is
invertible so that we can write its unique solution $u$ as
$u=L_m^{-1}q$, and by the Maximum principle \ref{mp2} we know that
if $q(t)\ge0$, then $u(t)\ge0$. Since $\alpha(t)=L_m^{-1}a(t)$,
$\beta(t)=L_m^{-1}b(t)$ and $b(t)\le p(t,x(t))\le a(t)$, we have
 \[
\beta(t)\le L_m^{-1}p(t,x(t))\le \alpha(t).
 \]
Let us consider the modified problem
\begin{equation} \label{modif2}
\begin{gathered}
\begin{aligned}
 x''(t)+Mx(t)=(Fx)(t)&\equiv f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t))
-mL_m^{-1}p(t,x(t))\right)\\
 &\quad +(m+M)p(t,x(t))-m^2L_m^{-1}p(t,x(t)),
\end{aligned}\\
x(0)=x(2\pi),\quad  x'(0)=x'(2\pi).
\end{gathered}
\end{equation}
Considering the operator $\Phi:C[0,2\pi]\to C[0,2\pi]$ with
$\Phi x=L_M^{-1}(Fx)$ since $p(t,x(t))$ and $L_m^{-1}p(t,x(t))$
are bounded and $f$ is a Carathéodory function, there exists a
$L^1[0,2\pi]$ function $g(t)$ such that $|(Fx)(t)|\le g(t)$
for a.e. $t\in[0,2\pi]$. Therefore, applying
Leray-Schauder's fixed point Theorem, we can conclude that $\Phi$
has a fixed point $x(t)$ which is a solution of the modified
problem \eqref{modif2}.

 \begin{proposition} \label{prop4.3}
Let $x(t)$ be a solution of the modified problem \eqref{modif2}.
Assuming (H2), for given lower and upper solutions $\alpha$ and
$\beta$, with $\alpha\le \beta$, we have
\[
 b(t)\le x(t)\le a(t).
\]
\end{proposition}

The proof of the above proposition is similar to the one of
proposition \ref{prop1}.

\begin{theorem} \label{thm4.4}
Assuming (H2), for given lower and upper solutions $\alpha$ and
$\beta$, with  $\beta\le \alpha$, the boundary value problem
\eqref{eq}--\eqref{bcper} has a solution  $u(t)\in
W^{4,1}(0,2\pi)$ such that $\beta\le u\le\alpha$.
 \end{theorem}

The proof of the above theorem is similar to the one of theorem
\ref{teo1}.

 We can reach a similar conclusion, assuming the following hypothesis
 \begin{itemize}
 \item[(H2')]there exist constants $C,D$ with $C<0$ and $D<4C+1/4$ such that
 \begin{equation}
 f\left(t,u_1,v_1\right)-f\left(t,u_2,v_2\right)\ge
 -C\left(u_1-u_2\right)-D\left(v_1-v_2\right)
 \end{equation}
 for a.e. $t\in[0,2\pi]$, $\beta(t)\le u_1\le u_2\le\alpha(t)$, $v_1\le v_2$.
 \end{itemize}


\begin{remark} \label{rmk4.5} \rm
If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in
(H2') is equivalent to $\frac{\partial f}{\partial u}\le -C$ and
$\frac{\partial f}{\partial v}\le -D$.
\end{remark}

Let $m<0$ and $0<M<\frac{1}{4}$ be the two roots of the equation
$x^2-Dx+C=0$ (note that $C=mM,\,D=m+M)$). Defining $a(t)$ and
$b(t)$ as above, we have the following result.

 \begin{proposition}\label{ab4}
 If $f$ is a $L^1$-Carath\'eodory function satisfying (H2')
for $\alpha(t),\beta(t)$ lower and upper solutions such that
$\beta(t)\le\alpha(t)$, then $b(t)\ge a(t)$.
 \end{proposition}

\begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)=y(2\pi)$ and $y'(0)\ge
y'(2\pi)$. Suppose towards a contradiction that there exists
$t_0\in [0,2\pi]$ such that $y(t_0)<0$. We can reach a
contradiction with similar arguments from the ones used in
Proposition \ref{ab3}, using instead the Maximum principle
\ref{mp2}.
 \end{proof}

 Using the same arguments as in the previous case, we prove the
following result.

\begin{theorem} \label{thm4.7}
Assuming (H2'), for given lower and upper solutions $\alpha$ and
$\beta$, with $\beta\le \alpha$, the boundary value problem
\eqref{eq}--\eqref{bcper} has a solution $u(t)\in W^{4,1}(0,2\pi)$
such that $\beta\le u\le\alpha$.
 \end{theorem}

 \section{``Simply supported'' problem with ordered upper and lower solutions}

Let us now consider the fourth-order equation
 \begin{equation}\label{eq2}
 u^{(4)}(t)=f\big(t,u(t),u''(t)\big), \quad t\in [0,\pi]
 \end{equation}
 where $f$ is a $L^1$-Carath\'eodory function, with the boundary conditions
 \begin{equation}\label{bcss}
 u(0)=u(\pi)=u''(0)=u''(\pi)=0.
 \end{equation}
We say that $\alpha\in W^{4,1}(0,\pi)$ is a lower solution of the
boundary value problem
 \eqref{eq2}--\eqref{bcss} if
\begin{gather*}
 \alpha^{(4)}(t)\le f\left(t,\alpha(t),\alpha''(t)\right), \quad t\in [0,\pi]
\\
 \alpha(0)\le0,\quad \alpha(\pi)\le 0,\quad
 \alpha''(0)\ge0,\quad \alpha''(\pi)\ge0.
\end{gather*}
A function $\beta \in W^{4,1}(0,\pi)$ satisfying the reversed
inequalities is called an upper solution.

Let $\alpha,\, \beta$ be respectively a lower and an upper
solution of  \eqref{eq}--\eqref{bcper}, such that $\alpha(t)\le
\beta(t)$ for all $t\in[0,\pi]$.

In the following, we assume the hypothesis
 \begin{itemize}
 \item[(H3)] There exist constants $C,D$ such that $C<0$ or $D>0$,
with $D>-C-1$, $D^2>4C$ , and
 \begin{equation}
 f\left(t,u_2,v_2\right)-f\left(t,u_1,v_1\right)\ge
 -C\left(u_2-u_1\right)+D\left(v_2-v_1\right)
 \end{equation}
 for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$, $v_1\le v_2$.
 \end{itemize}

 \begin{remark} \label{rmk5.1} \rm
If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in
(H3) is equivalent to $\frac{\partial f}{\partial u}\ge -C$ and
$\frac{\partial f}{\partial v}\ge D$.
 \end{remark}

Let $m<0$ and $M<1$ be the two roots of the equation $x^2+Dx+C=0$
(note that $C=mM,\,D=-(m+M)$). Defining
$a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$,
 we have the following result.

 \begin{proposition}\label{ab5}
If $f$ is a $L^1$-Carath\'eodory function satisfying (H3) for
$\alpha(t),\beta(t)$ lower and upper solutions such that
$\alpha(t)\le\beta(t)$, then $b(t)\le a(t)$.
 \end{proposition}

 \begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)\le0$ and $y(\pi)\le0$.
Suppose towards a contradiction that there exists $t_0\in [0,\pi]$
such that $y(t_0)>0$.  The result follows using similar arguments
to the ones of Proposition \ref{ab1} (using Maximum principle
\ref{mp3} instead).
\end{proof}

 Let
 \[
 p(t,x)= \begin{cases}
 b(t), & x<b(t)\\
 x, & b(t)\le x\le a(t)\\
 a(t), & x>a(t).
 \end{cases}
 \]
Consider the boundary value problem
 \[
 u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=0,\;\quad u(\pi)=0,
 \]
with $q\in L^1[0,1]$. Since $m<0$, the operator $L_m$ is
invertible so that we can write its unique solution $u$ as
 $u=L_m^{-1}q$.

Let us define $\tilde{a}(t)$ such that $\tilde{a}''+m\tilde{a}=0$,
$\tilde{a}(0)=\alpha(0)$, $\tilde{a}(\pi)=\alpha(\pi)$, and
$\tilde{b}(t)$ such that $\tilde{b}''+m\tilde{b}=0$,
$\tilde{b}(0)=\beta(0)$, $\tilde{b}(\pi)=\beta(\pi)$. It is
obvious that $\tilde{a}(t)\le0$ and $\tilde{b}(t)\ge0$ for all
$t\in [0,\pi]$.

We have $\alpha(t)=L_m^{-1}a(t)+\tilde{a}(t)$ and
$\beta(t)=L_m^{-1}b(t)+\tilde{b}(t)$, then,
 by the Maximum principle \ref{mp3},
 \[
 \alpha(t)\le L_m^{-1}p(t,x(t))\le \beta(t).
 \]
Proceeding in a similar way as in the previous cases, we can reach
an analogue conclusion:

\begin{theorem}
Assuming (H3), for given lower and upper solutions $\alpha$ and
$\beta$, with $\alpha\le \beta$, the boundary value problem
\eqref{eq2}--\eqref{bcss} has a solution $u(t)\in W^{4,1}(0,\pi)$
such that $\alpha\le u\le\beta$.
\end{theorem}

Let us now consider a hypothesis somehow different from the ones
considered above. Suppose that
 \begin{itemize}
 \item[(H4)] There exist constants $C,D$ with $C>0$, $0<D<1$, and
 \begin{equation}
 f\left(t,u_2,v_2\right)-f\left(t,u_1,v_1\right)\ge
 C\left(u_2-u_1\right)-D\left|v_2-v_1\right|
 \end{equation}
for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$,
$v_1, v_2\in \mathbb{R}$.
 \end{itemize}

\begin{remark} \label{rmk5.4} \rm
If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in
(H4) is equivalent to $\frac{\partial f}{\partial u}\ge C$ and
$|\frac{\partial f}{\partial v}|\le D$.
 \end{remark}

Let $m<0$ be such that $C+Dm-m^2>0$ and $D-m<1$. Defining
$a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$, we
have the following result.

\begin{proposition}\label{ab6}
If $f$ is a $L^1$-Carath\'eodory function satisfying (H4) for
$\alpha(t),\beta(t)$ lower and upper solutions such that
$\alpha(t)\le\beta(t)$, then $b(t)\le a(t)$.
 \end{proposition}

 \begin{proof}
Setting $y(t)=b(t)-a(t)$, we have $y(0)\le0$, $y(\pi)\le0$ and
 \begin{align*}
 y''(t)=&\beta^{(4)}(t)-\alpha^{(4)}(t)+m(\beta''(t)-\alpha''(t))
 +m^2(\beta(t)-\alpha(t))-m^2(\beta(t)-\alpha(t))\\
\ge&f(t,\beta(t),b(t)-m\beta(t))-f(t,\alpha(t),a(t)-m\alpha(t))+my(t)
 -m^2(\beta(t)-\alpha(t)) \\
\ge&C(\beta(t)-\alpha(t))-D\left|y(t)-m(\beta(t)-\alpha(t))\right|
 +my(t)-m^2(\beta(t)-\alpha(t)) \\
\ge&(C+Dm-m^2)(\beta(t)-\alpha(t))-D\left|y(t)\right|+my(t).
 \end{align*}
To apply Maximum principle \ref{mp4}, we rewrite the previous
inequality as
 \[
 y''(t)+(D-m)y^+(t)+(D+m)y^-(t)\ge0
 \]
and conclude that $y(t)\le 0$.
\end{proof}


Proceeding in a similar way as above, we can reach an analogue
conclusion:

 \begin{theorem} \label{thm5.6}
Assuming (H4), for given lower and upper solutions $\alpha$ and
$\beta$, with $\alpha\le \beta$, the boundary value problem
\eqref{eq2}--\eqref{bcss} has a solution $u(t)\in W^{4,1}(0,\pi)$
such that $\alpha\le u\le\beta$.
\end{theorem}


 \subsection*{Acknowledgments}
 This work was supported by grant POCTI/Mat/57258/2004
from Funda\c{c}\~ao para a Ci\^encia e Tecnologia,  and
by grant POCTI-ISFL-1-209 from Centro de Matem\'atica e
Aplica\c{c}\~{o}es Fundamentais. The authors would like to thank
the anonymous referee for his/her  careful reading of the
original manuscript.

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