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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 78, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/78\hfil Positive solutions for beam equation]
{Positive solutions for the beam equation under certain boundary conditions}
\author[B. Yang\hfil EJDE-2005/78\hfilneg]
{Bo Yang}

\address{Bo Yang \hfill\break
Department of Mathematics, Kennesaw State University,
 Kennesaw, GA 30144, USA}
\email{byang@kennesaw.edu}

\date{}
\thanks{Submitted April 29, 2005. Published July 8, 2005.}
\subjclass[2000]{34B18}
\keywords{Fixed point theorem; cone; nonlinear boundary-value problem; 
\hfill\break\indent  positive solution}

\begin{abstract}
 We consider a boundary-value problem for the beam equation, in
 which the boundary conditions mean that the beam is embedded at
 one end and fastened with a sliding clamp at the other end. Some
 priori estimates to the positive solutions for the boundary-value
 problem are obtained. Sufficient conditions for the existence and
 nonexistence of positive solutions for the boundary-value problem
 are established.
\end{abstract}

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

\section{Introduction}

In this paper, we consider the fourth order beam equation
\begin{equation}
u''''(t) =  g(t)f(u(t)) , \quad 0 \le t \le 1,
\label{ee}
\end{equation}
together with boundary conditions
\begin{equation}
u(0) = u'(0) = u'(1) = u'''(1) = 0.
\label{bb}
\end{equation}
Throughout this paper, we assume that
\begin{itemize}
\item[(H1)] $f: [0,\infty) \to [0,\infty) $ is continuous

\item[(H2)] $ g: [0,1] \to [0,\infty) $ is a continuous function such that
$  \int_0^1 g(t)\,dt > 0$.
\end{itemize}
Equation \eqref{ee} and the boundary conditions \eqref{bb} arise from the
study of elasticity and have definite physical meanings.
Equation \eqref{ee} describes the deflection or deformation of an elastic
beam under a certain force.
The boundary conditions \eqref{bb} mean that the beam is embedded at
the end $ t=0 $, and fastened with a sliding clamp at the end $ t=1 $.

In 1989, Gupta \cite{Gu1} considered the boundary-value problem
\begin{gather}
u''''(t)+f(t)u(t)=e(t),\quad 0<t<\pi, \label{ee03} \\
u'(0) = u'''(0) = u'''(\pi) = u'(\pi) = 0,
\label{bb03}
\end{gather}
where \eqref{bb03} means that the beam is fastened with sliding clamps at
both ends $ t=0 $ and $ t=\pi$.
In 2004, Kosmatov \cite{Ko} considered \eqref{ee} together with the
boundary conditions
\begin{equation}
u(0) = u'(0) = u'(1) = u(1) = 0,
\label{bb02}
\end{equation}
and obtained sufficient conditions for existence of infinitely many
solutions to the problem \eqref{ee}-\eqref{bb02}.
Note that the boundary conditions \eqref{bb02} mean that the beam
is embedded at both ends $ t=0 $ and $ t=1$.


In fact, \eqref{ee} has been studied by many authors under
various boundary conditions and by different approaches. For some
other results on boundary-value problems of the beam equation, we
refer the reader to the papers of Agarwal \cite{A}, Bai and Wang
\cite{BW}, Davis and Henderson \cite{DH}, Dalmasso \cite{Dal},
Dunninger \cite{Du}, Elgindi and Guan \cite{EG}, Eloe, Henderson,
and Kosmatov \cite{EHK}, Graef and B.\,Yang \cite{GY}, Gupta
\cite{Gu},
Ma \cite{Ma, M}, Ma and Wang \cite{MW}, B.\,Yang \cite{BY},
and Y. Yang \cite{Y}.

In this paper, we will study the positive solutions of the problem
\eqref{ee}-\eqref{bb}. By positive solution, we mean a solution $
u(t) $ such that $ u(t)>0 $ for $ t\in (0,1) $. A beam can have
different shapes under different boundary constraints. One of the
purposes of this paper is to make some estimates to the shape of
the beam under boundary conditions \eqref{bb}.

This paper is organized as follows. In Section 2, we give the
Green's function for the problem  \eqref{ee}-\eqref{bb}, state the
Krasnosel'skii's fixed point theorem, and fix some notations. In
Section 3, we present some priori estimates to positive solutions
to the problem \eqref{ee}-\eqref{bb}. In Sections 4 and 5, we
establish some existence and nonexistence results for positive
solutions to the problem \eqref{ee}-\eqref{bb}.

\section{Preliminaries}
The Green's function $ G:[0,1]\times[0,1]\to [0,\infty) $ for the
problem \eqref{ee}-\eqref{bb} is
$$
G(t,s) =\begin{cases}  \frac{1}{12} t^2 ( 6s - 3s^2 - 2t), &
\mbox{if } 0\le t\le s\le 1, \\[3pt]
\frac{1}{12} s^2 ( 6t -3t^2  - 2s), &
 \mbox{if } 0\le s\le t\le 1.
\end{cases}
$$
Then problem \eqref{ee}-\eqref{bb} is equivalent to the integral
equation
\begin{equation}
u(t) = \int_0^1 G(t,s) g(s) f(u(s))\,ds,\quad 0\leq t\leq 1.
\label{int}
\end{equation}
It is easy to verify that $G$ is a continuous function, and
$G(t,s)>0$ if $t,s\in(0,1)$. We will need the following fixed
point theorem, which is due to Krasnosel'skii \cite{K}, to prove
some of our results.

\begin{theorem} \label{thmK}
Let $ (X,\|\cdot\|) $ be  Banach space over the reals, and let
$P\subset X $ be a cone in $ X $. Let $ H_1 $ and $ H_2 $ be real
numbers such that $ H_2>H_1>0 $, and let
$$
\Omega_i = \{ v\in X\ |\ \|v\| < H_i\}, \quad i=1,2.
$$
If $L:P\cap(\overline{\Omega_2}-\Omega_1 )\to P$ is a completely
continuous operator such that, either
\begin{itemize}
\item[(K1)] $ \|Lv\|\leq\|v\| $ if $ v\in P\cap\partial \Omega_1$,
and $   \|Lv\|\geq\|v\| $ if $ v\in P\cap\partial \Omega_2$, or

\item[(K2)] $ \|Lv\|\geq\|v\| $ if $ v\in P\cap\partial \Omega_1 $,
and $   \|Lv\|\leq\|v\| $ if $ v \in P\cap\partial \Omega_2 $.
\end{itemize}
Then $ L $ has a fixed point in $
P\cap(\overline{\Omega_2}-\Omega_1) $.
\end{theorem}

For the rest of this paper, we let $ X = C[0,1] $ be with norm
$$
\|v\|=\max_{t\in [0,1]} |v(t)|,\quad \forall v\in X.
$$
Clearly $X$ is a Banach space. We define
$Y=\{v\in X\,|\,v(t)\ge 0\mbox{ for }0\le t\le 1\}$, and define
the operator $ T : Y \to X $ by
\begin{equation}
    (Tu)(t) = \int_0^1 G(t,s)g(s)f(u(s))\,ds,\quad 0\leq t\leq 1.
\label{op}
\end{equation}
It is clear that if (H1) and (H2) hold, then $ T: Y\to Y $ is a
completely continuous operator. We also define the constants
\begin{gather*}
F_0 = \limsup_{x\to 0^+}\frac{f(x)}{x},     \quad
f_0=\liminf_{x\to 0^+}\frac{f(x)}{x}, \\
F_{\infty} = \limsup_{x\to +\infty}\frac{f(x)}{x},     \quad
f_{\infty}=\liminf_{x\to +\infty}\frac{f(x)}{x}.
\end{gather*}
These constants, which are associated with the function $f$, will be used
in Sections 4 and 5.

\section{Estimates for Positive Solutions}

In this section, we shall give some  estimates for positive
solutions of the problem \eqref{ee}-\eqref{bb}. To this purpose,
we define the functions $ a:[0,1]\to [0,1] $,  $ b:[0,1]\to [0,1]
$, and $ c:[0,1]\to [0,1] $ by
$$
a(t) = 3t^2 - 2t^3, \quad
b(t) = 2t-t^2 ,\quad
c(t) = 4t^2 - 4t^3 + t^4.
$$
It is easy to see that $ b(t)\ge c(t)\ge a(t)\ge t^2 $ for $0\le
t\le 1$.

\begin{lemma} \label{lem3.1}
If $ u \in C^4[0,1] $ satisfies the boundary conditions \eqref{bb},
and
\begin{equation}
u''''(t)\ge 0\quad\mbox{for}\quad 0\le t\le 1, \label{2101}
\end{equation}
then
\begin{equation}
u'''(t)\le 0,\quad u'(t)\ge 0, \quad u(t)\ge 0 \quad\mbox{for}\quad 0\le t\le 1.
\label{2102}
\end{equation}
\end{lemma}

\begin{proof}  Note that \eqref{2101} implies that $u'''$ is nondecreasing.
Since $ u'''(1)=0$, we have $u'''(t)\le 0$ on $ [0,1]$, which means that
$u'$ is concave downward on $[0,1]$.
Since $u'(0)=u'(1)=0$, we have $u'(t)\ge 0 $ on $[0,1]$. Since $u(0)=0$,
we have $u(t)\ge 0 $ on $[0,1]$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem3.2}
 If $ u \in C^4[0,1] $ satisfies \eqref{bb} and \eqref{2101}, then
\begin{equation}
u(t)\ge a(t)u(1)  \quad \mbox{ for }\quad 0\le t\le 1.
\label{2201}
\end{equation}
\end{lemma}

\begin{proof}
 If $ u \in C^4[0,1] $ satisfies \eqref{bb} and \eqref{2101}, then
$ u(0)=0 $, $ u(1)\ge 0 $, and $ u'(t) \ge 0 $ for $ 0\le t\le 1
$. If $ u(1)=0 $, then $ u(t)\equiv 0 $, and it is easy to see
that \eqref{2201} is true in this case.

Now we prove \eqref{2201} when $ u(1)>0 $. Without loss of
generality, we assume that $ u(1)=1 $. If we define
$$
h(t) = u(t) -a(t)u(1) = u(t) - (3t^2-2t^3),\quad 0\le t\le 1,
$$
then
\begin{equation*}
\begin{gathered}
h'(t)= u'(t)-(6t-6t^2), \quad h''(t) = u''(t)-(6-12t), \\
h'''(t) = u'''(t)+12,
\end{gathered} 
\end{equation*}
%
\begin{equation}
h''''(t) = u''''(t)\geq 0,\quad 0\le t \le 1.
\label{2202}
\end{equation}
To prove the lemma, it suffices to show that $ h(t)\ge 0 $ on $ [0,1] $.
It is easy to see that $ h(0) = h(1) = 0$.
By mean value theorem, there exists $ r_1\in (0,1) $ such that
$ h'(r_1)=0 $. It is also easy to see that $ h'(0) = h'(1) = 0$.
Since $ h'(0) = h'(r_1) = h'(1) = 0 $, there exist $ r_2\in (0,r_1)$
and $ t_2\in (r_1,1) $ such that $ h''(r_2) = h''(t_2) = 0 $.

Note that \eqref{2202} implies that $ h''$ is concave upward.
Since $ h''(r_2) = h''(t_2) = 0 $, we have
$$
h''(t)\geq 0 \ \ {\rm on}\ \  (0,r_2),\ \
h''(t)\leq 0 \ \ {\rm on}\ \  (r_2,t_2),\ \
{\rm and}\ \
h''(t)\geq 0 \ \ {\rm on}\ \  (t_2,1).
$$
These inequalities, together with the fact that $ h'(0)=h'(r_1) = h'(1)=0 $,
imply that
$$
h'(t)\geq 0 \text{ on } (0,r_1),\quad
h'(t)\leq 0 \text{ on }  (r_1,1).
$$
Since $ h(0)=h(1)=0 $, we have
$h(t)\geq 0$ on $(0,1)$.
The proof is complete.
\end{proof}


\begin{lemma} \label{lem3.3}
 If $ u \in C^4[0,1] $ satisfies \eqref{bb} and \eqref{2101}, then
\begin{equation}
u(t) \leq u(1)b(t) \quad \mbox{ for }\quad t\in [0,1].
\label{2301}
\end{equation}
\end{lemma}

\begin{proof} Without loss of  generality, we assume that $ u(1)=1$.
If we define
$$
h(t)= b(t) u(1) -u(t) = 2t -t^2 - u(t),\quad 0\le t\le 1,
$$
then
\begin{equation*}
h'(t)= 2-2t - u'(t),\quad h''(t) = -2 - u''(t),
\end{equation*}
\begin{equation}
h'''(t) = - u'''(t),\quad 0\le t \le 1.
\label{2302}
\end{equation}
It is easy to see that $ h(0) =  h(1) = h'(1) = 0$.
By mean value theorem, because $ h(0) = h(1) = 0 $, there exists
$ r_1\in (0,1) $ such that $ h'(r_1)=0 $.
We see from \eqref{2302} and \eqref{2102} that $h'''(t)\ge 0 $ on $[0,1]$,
 which implies that $ h'$ is concave upward on $ [0,1]$.
Since $ h'(r_1) = h'(1) = 0 $, we have
$$
h'(t)\geq 0 \text{ on }  (0,r_1),\quad
h'(t)\leq 0 \text{ on } (r_1,1).
$$
Since $ h(0) = h(1) = 0 $, we have $h(t)\ge 0$ on $(0,1)$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem3.4}
If $ u \in C^4[0,1] $ satisfies \eqref{bb} and \eqref{2101}, and
$ u''''(t) $ is nondecreasing on $ [0,1] $, then
\begin{equation}
u(t) \leq u(1)c(t) \quad \mbox{ for }\quad t\in [0,1].
\label{2401}
\end{equation}
\end{lemma}

\begin{proof} Without loss of generality, we assume that $ u(1)=1 $.
If we define
$$
h(t) = c(t)u(1)-u(t) =  4t^2-4t^3+t^4 - u(t),\quad 0\le t\le 1,
$$
then
\begin{equation*}
\begin{gathered}
h'(t) = 8t -12t^2 +4t^3 - u'(t), \quad
h''(t) = 8-24t +12t^2 - u''(t), \\
h'''(t) = -24 +24 t - u'''(t),
\end{gathered}
\end{equation*}
\begin{equation}
h''''(t) = 24 - u''''(t),\quad 0\le t \le 1.
\label{2402}
\end{equation}
It is easily seen that $ h(0) = h(1) =  h'(0) = h'(1)=0 $.
By the mean value theorem, because $ h(0) = h(1) = 0 $, there exists
$ r_1 \in (0,1) $ such that $ h'(r_1)=0 $.
Since $ h'(0) = h'(r_1) = h'(1) = 0 $, there exist $ r_2 \in (0,r_1) $
and $ t_2\in(r_1,1) $ such that $ h''(r_2) = h''(t_2) = 0 $.
As a consequence, there exists $ r_3\in(r_2,t_2) $ such that $ h'''(r_3)=0 $.

Note that $ u'''' $ is nondecreasing by assumption. It follows
from \eqref{2402} that $ h'''(t) $ is concave downward.
It is easy to see that $ h'''(1)=0 $.
Because $ h'''(r_3) = h'''(1) = 0 $, we have
$$
h'''(t) \le 0 \mbox{ on } (0,r_3),\quad \mbox{and}\quad
h'''(t) \ge 0 \mbox{ on } (r_3,1).
$$
Since $ h''(r_2) = h''(t_2) = 0 $, we have
$$
h''(t) \ge 0 \mbox{ on } (0,r_2), \quad
h''(t) \le 0 \mbox{ on } (r_2,t_2), \quad
h''(t) \ge 0 \mbox{ on } (t_2,1).
$$
Because $ h'(0) = h'(r_1) = h'(1) = 0 $, we have
$$
h'(t) \ge 0 \mbox{ for } t\in (0,r_1), \quad
h'(t) \le 0 \mbox{ for } t\in (r_1,1).
$$
Hence, $h(t) \ge 0 $ for $ 0\le t\le 1$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.5}
Suppose that (H1) and (H2) hold. If $ u(t) $ is a nonnegative solution
to the problem \eqref{ee}-\eqref{bb}, then $ u(t) $ satisfies
\eqref{2102}, \eqref{2201}, and \eqref{2301}.
\end{theorem}

\begin{proof} If $ u(t) $ is a nonnegative solution to the problem
\eqref{ee}-\eqref{bb}, then $ u(t) $ satisfies the boundary conditions
\eqref{bb}, and
$$
u''''(t)=g(t)f(u(t))\ge 0,\quad 0\le t\le 1.
$$
Now Theorem \ref{thm3.5} follows directly from
Lemmas \ref{lem3.1}, \ref{lem3.2}, and \ref{lem3.3}.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.6}
 Suppose that (H1), (H2), and the following condition hold.
\begin{itemize}
\item[(H3)] Both $ f $ and $ g $ are nondecreasing functions.
\end{itemize}
If $ u(t) $ is a nonnegative solution to the problem \eqref{ee}-\eqref{bb},
then $ u(t) $ satisfies \eqref{2102}, \eqref{2201}, and \eqref{2401}.
\end{theorem}

\begin{proof} By Theorem \ref{thm3.5}, $ u(t) $ satisfies \eqref{2102} and \eqref{2201}.
Therefore $ u(t) $ is nondecreasing on $[0,1] $. It is obvious that
$ u''''(t) = g(t)f(u(t)) \ge 0 $.
By (H3), we have that $ u''''(t) = g(t)f(u(t)) $ is nondecreasing
on the interval $ [0,1] $. It follows directly from Lemma \ref{lem3.4} that
$ u(t) $ satisfies \eqref{2401}. The proof is complete.
\end{proof}

\section{Existence and Nonexistence Results}

First, we define some important constants:
$$
A = \int_0^1 G(1,s)g(s)a(s)\,ds,
    \quad
B = \int_0^1 G(1,s)g(s)b(s)\,ds.
$$
We also define
\begin{align*}
P = \big\{& v\in X :  v(1)\geq 0,\ v(t)\mbox{ is nondecreasing on }[0,1],\\
     & a(t)v(1)\leq v(t)\leq b(t) v(1) \mbox{ on } [0,1]\big\}.
\end{align*}
Clearly $ P $ is a positive cone in $ X $. It is obvious that if $u\in P$,
then $u(1)=\|u\|$. We see from Theorem \ref{thm3.5} that if $ u(t) $ is a
nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then $ u\in P $.
In a similar fashion to Theorem \ref{thm3.5}, we can show that $T(P)\subset P$.
To find a positive solution to the problem \eqref{ee}-\eqref{bb}, we need
only to find a fixed point $ u $ of $ T $ such that $ u\in P $ and
$ u(1)=\|u\|>0 $.

The next two theorems provide sufficient conditions for the existence
of at least one positive solution for the problem \eqref{ee}-\eqref{bb}.


\begin{theorem} \label{thm4.1}
Suppose that (H1) and (H2) hold. If $ B F_0 < 1 < A f_\infty $, then
 problem \eqref{ee}-\eqref{bb} has at least one positive solution.
\end{theorem}

\begin{proof} First, we choose $ \varepsilon >0 $
such that $ (F_0+\varepsilon)  B\leq 1 $. By the definition of $
F_0 $, there exists $ H_1 >0 $ such that
$ f(x)\leq (F_0+\varepsilon)x$ for $ 0 < x\leq H_1 $. Now for each
$ u\in P $ with $ \|u\|=H_1 $, we have
\begin{align*}
(Tu)(1) & =   \int_0^1 G(1,s) g(s) f(u(s))\,ds\\
&\leq  \int_0^1 G(1,s) g(s) (F_0+\varepsilon)u(s)\,ds\\
& \leq (F_0+\varepsilon)\|u\|  \int_0^1 G(1,s) g(s) b(s)\,ds\\
& = (F_0+\varepsilon)\|u\|  B
\leq  \|u\|,
\end{align*}
which means $ \|Tu\|\leq \|u\| $. Thus, if we let
$ \Omega_1=\{ u\in X\ |\ \|u\|<H_1\}$,
then
$$
\|Tu\|\leq \|u\| \ \ {\rm for}\ \ u\in P\cap\partial\Omega_1.
$$
To construct $ \Omega_2 $, we choose $ \delta >0 $ and $ \tau \in (0,1/4) $
such that
$$
\int_\tau^{1} G(1,s) g(s) a(s)\,ds \cdot(f_\infty-\delta) \ge 1.
$$
There exists $ H_3 >2H_1 $ such that
$f(x)\geq (f_\infty-\delta)x \ \mbox{for}\ x\geq H_3$.
Let $ H_2=H_3/{\tau^2} $.
If $ u\in P $ such that $ \|u\| = H_2 $, then for each $ t\in[\tau,1] $, we have
$$
u(t)\ge H_2 a(t)\ge H_2 t^2 \ge H_2 \tau^2 \ge H_3 .
$$
Therefore, for each $ u\in P $ with $ \|u\| = H_2 $, we have
\begin{align*}
(Tu)(1)
& =  \int_0^{1} G(1,s) g(s) f(u(s))\,ds\\
&\geq  \int_\tau^{1} G(1,s) g(s) f(u(s))\,ds\\
& \geq  \int_\tau^{1} G(1,s) g(s) (f_\infty-\delta)u(s)\,ds\\
& \geq  \int_\tau^{1} G(1,s) g(s) a(s)\, ds\cdot (f_\infty-\delta)\|u\|
\geq     \|u\|,
\end{align*}
which means $ \|Tu\|\geq \|u\| $. Thus, if we let
$ \Omega_2=\{u\in X\ |\ \|u\|<H_2\}$, then
$ \overline{\Omega_1}\subset\Omega_2 $, and
$$
\|Tu\|\geq \|u\| \quad \text{for}\quad u\in P\cap\partial\Omega_2.
$$
Now that the condition (K1) of Theorem \ref{thmK} is satisfied, there
exists a fixed point of $ T $ in $ P\cap(\overline{\Omega_2}-\Omega_1) $.
The proof is now complete.
\end{proof}

\begin{theorem} \label{thm4.2}
 Suppose that (H1) and (H2) hold. If $ B F_{\infty} < 1 < A f_0 $, then
 the problem \eqref{ee}-\eqref{bb} has at least one positive solution.
\end{theorem}

The proof of Theorem \ref{thm4.2} is very similar to that of
Theorem \ref{thm4.1} and
therefore omitted. The next two theorems provide sufficient conditions
for the nonexistence of positive solutions to the problem \eqref{ee}-\eqref{bb}.

\begin{theorem} \label{thm4.3} Suppose (H1) and (H2)
hold. If $B f(x)<x $ for all $ x>0 $, then the problem
\eqref{ee}-\eqref{bb} has no positive solutions.
\end{theorem}

\begin{proof} Assume the contrary that $ u(t) $ is a
positive solution of the problem \eqref{ee}-\eqref{bb}. Then
$u\in P $, $u(t)>0 $ for $ 0<t\le 1 $, and
\begin{align*}
u(1) & =     \int_0^1 G(1,s) g(s)f(u(s)) \, ds\\
&<    B^{-1} \int_0^1 G(1,s) g(s) u(s) \, ds \\
& \leq B^{-1}\int_{0}^{1} G(1,s) g(s) b(s) \, ds\cdot u(1)\\
&=    B^{-1}Bu(1)
=    u(1),
\end{align*}
which is a contradiction. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4.4}
Suppose (H1) and (H2) hold. If $A f(x) > x $ for all $x>0$, then the
problem \eqref{ee}-\eqref{bb} has no positive solutions.
\end{theorem}

\section{More Existence and Nonexistence Results}

In this section, we define a new constant
$$
C = \int_0^1 G(1,s)g(s)c(s)\,ds,
$$
and define the positive cone $ Q $ of $ X $ by
\begin{align*}
Q = \big\{& v\in X : v(1)\geq 0,\ v(t)\mbox{ is nondecreasing on } [0,1],    \\
    & a(t)v(1)\leq v(t)\leq c(t) v(1)\mbox{ on } [0,1]\big\}.
\end{align*}
It is obvious that if $u\in Q$, then $u(1)=\|u\|$. We see from
Theorem \ref{thm3.6}
that if (H1), (H2), and (H3) hold, and $ u(t) $ is a nonnegative
solution to the problem \eqref{ee}-\eqref{bb}, then $ u\in Q $. In a similar
fashion to Theorem \ref{thm3.6}, we can show that if (H1), (H2), and (H3) hold,
then $T(Q)\subset Q$.

\begin{theorem} \label{thm5.1} Suppose that (H1), (H2), and (H3) hold.
If either $ C F_0 < 1 < A f_\infty $ or $ C F_{\infty} < 1 < A f_0 $,
then  problem \eqref{ee}-\eqref{bb} has at least one positive solution.
\end{theorem}

The proof of the above theorem is omitted, because it is very similar to
that of Theorem \ref{thm4.1}. The only difference is that we use the positive
cone $ Q $, instead of $ P $, in the proof of Theorem \ref{thm5.1}.

\begin{theorem} \label{thm5.2}
Suppose (H1), (H2), and (H3) hold. If $ C f(x)<x $ for all $ x >0$,
then  problem \eqref{ee}-\eqref{bb} has no positive solutions.
\end{theorem}

The proof of the above theorem is quite similar to that of
Theorem \ref{thm4.3}
and therefore omitted.

\begin{example} \label{ex5.3} \rm
Consider the beam equation
\begin{equation}
u''''(t) =  \lambda (t + 2t^2) u(t)(1+2u(t))/(1+u(t)),\quad 0\le t\le 1,
\label{eee}
\end{equation}
where $\lambda>0$ is a parameter, together with the boundary
conditions \eqref{bb}. In this example, $g(t)=t + 2t^2$ and
$f(u)=\lambda u(1+2u)/(1+u)$. It is easy to see that
$ f_0 = F_0 =\lambda $, $ f_\infty = F_\infty = 2\lambda $, and
$$
\lambda x < f(x) < 2\lambda x \quad \mbox{for}\quad x > 0.
$$
Calculations indicate that $ A = 47/756$, $B = 43/630$, and $C =
1937/30240$. By Theorem \ref{thm4.1}, if
$$ 8.04 \approx1/(2A) < \lambda < 1/B\approx 14.56 , $$
then the problem \eqref{eee}-\eqref{bb} has at least one positive
solution. From Theorems \ref{thm4.3} and \ref{thm4.4} we see that if
$$
\lambda \le 1/(2B) \approx 7.326
\quad \mbox{or}\quad
\lambda \ge 1/A\approx 16.085,
$$
then the problem \eqref{eee}-\eqref{bb} has no positive solutions.

Note that $ g(t) $ is increasing on $[0,1]$, and $ f(u) $ is
increasing on $[0,+\infty)$. Therefore Theorems \ref{thm5.1} and
\ref{thm5.2} apply.
From Theorem \ref{thm5.1} we see that if
$$
8.04 \approx1/(2A) < \lambda < 1/C\approx 15.612 ,
$$
then the problem \eqref{eee}-\eqref{bb} has at least one positive
solution. From Theorem \ref{thm5.2} we see that if
$$
\lambda \le 1/(2C) \approx 7.806,
$$
then the problem \eqref{eee}-\eqref{bb} has no positive solutions.
This example shows that our existence and nonexistence results are
quite sharp indeed.
\end{example}

\subsection*{Acknowledgment}
The author is grateful to the anonymous referee for his/her valuable
comments and suggestions.

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\end{document}