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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 03, 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  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/03\hfil Positive solutions]
{Positive solutions for singular nonlinear beam equation}

\author[ R. Song, H. L\"u\hfil EJDE-2007/03\hfilneg]
{Ruhao Song, Haishen L\"u}  

\address{Department of Applied Mathematics, Hohai University,
Nanjing 210098, China}
\email[Ruhao Song]{songruhao@163.com}
\email[Haishen L\"u]{haishen2001@yahoo.com.cn}

\thanks{Submitted November 13, 2006. Published January 2, 2007.}
\thanks{Supported by grant 10301033 from NNSF of China}
\subjclass[2000]{34B15, 34B16}
\keywords{p-laplacian; singular; fixed point theorem}

\begin{abstract}
 In this paper, we study the existence of solutions for the
 singular p-Laplacian equation
 \begin{gather*}
 \big(|u''|^{p-2}u''\big)''-f(t,u)=0,\quad t\in (0,1) \\
 u(0)=u(1)=0, \\
 u''(0)=u''(1)=0,
 \end{gather*}
 where $f(t,u)$ is singular at $t=0,1$ and at $u=0$. We prove the
 existence of  at least one solution.
\end{abstract}

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


\section{Introduction}

In this paper, we establish the existence of solutions to the
singular boundary-value problem
\begin{equation}
\begin{gathered}
\big(|u''|^{p-2}u''\big)''-f(t,u)=0,\quad t\in (0,1) \\
u(0)=u(1)=0, \\
u''(0)=u''(1)=0,
\end{gathered} \label{e1.1}
\end{equation}
where $p>1$ and $f(t,u)$ has singularity at $t=0,1$ and at $u=0$. For
convenience, we denote $\varphi _{p}(s)=|s|^{p-2}s$, for $p>1$.

Equation \eqref{e1.1} occurs in the following models of beams
\cite{b1}:
Beams with small deformations (also called geometric linearity);
beams of a material which satisfies a nonlinear power-like
stress-strain
law; beams with two-sided links (for example, springs) which satisfy a
nonlinear power-like elasticity law.
The best known setting is the boundary-value problem, for $p=2$,
\begin{equation*}
u^{(4)}-f(t,u(t))=0,\quad t\in (0,1)\,.
\end{equation*}
This model describes deformations of an elastic
beams with the boundary conditions  reflecting both ends
simply supported, also for one end simply supported and the other end
clamped by sliding clamps. Vanishing moments and shear
forces at the tail ends are frequently included in the boundary
conditions;  see for example Gupta \cite{g2} and its references. One
derivation of lines is used in the description over regions of
certain partial differential equations describing the deflection
of an elastic beam.



 Agarwal et al. \cite{a2,a3} consider the  boundary-value problem
\begin{gather*}
(-1)^{n}x^{(2n)}(t)=\mu f(t,x(t),\dots ,x^{(2n-2) }(t) ),
\\
x^{(2j)}(0)=x^{(2j)}(T)=0,\ 0\leq j\leq n-1
\end{gather*}
under the  critical condition:
\begin{itemize}
\item[(A)]  For a.e. $t\in [ 0,T] $ and for each
$(x_{0},\dots ,x_{2n-2})\in D$ (defined in \cite{a2})
\begin{equation*}
f(t,x_{0},\dots ,x_{2n-2})\leq \phi (t)
+\sum_{j=0}^{2n-2}q_{j}(t)\omega _{j}(|x_{j}|
)+\sum_{j=0}^{2n-2}h_{j}(t)|x_{j}|^{\alpha
_{j}}
\end{equation*}
where $\phi $, $h_{j}\in L^{1}(0,T)$ and $q_{j}\in L^{\infty
}(0,T)$ are nonnegative, $\omega _{j}:( 0,\infty ) \to
(0,\infty )$ are nonincreasing, $\alpha _{j}\in (0,1)$ and
\begin{equation*}
\int_{0}^{T}\omega _{j}(s)ds<\infty ,\quad \omega _{j}( uv)\leq
\Lambda \omega _{j}(u)\omega _{j}( v)
\end{equation*}
for $0\leq j\leq 2n-2$ and $u,v\in (0,\infty )$ with a positive
constant $\Lambda $.
\end{itemize}

Closely related to the results of this paper is the recent work
by Agarwal, L\"{u} and O'Regan \cite{a1}. There the authors consider
 positive solutions for the boundary-value problem
\begin{equation*}
\big(|u''|^{p-2}u''\big)''-\lambda q( t) f(u(t))=0,
\end{equation*}
where the nonlinearity $f$ is nonsingular. In this paper consider
nonlinearity $f$ may be singular. We point out a sufficient
condition for problem \eqref{e1.1} has a positive solution, but it
doesn't satisfies the condition (A), for example
\begin{equation*}
f(t,u)=\frac{t^{\alpha }(1+t)^{\alpha }}{u^{\beta } }
\end{equation*}
where $\alpha +1>\beta >0$.

Singular nonlinear two point boundary-value problems arise
naturally in applications and usually, only positive solutions are
meaningful. By a positive solution of \eqref{e1.1}, we mean a function
$u\in C^{(2)}[0,1]$ with $\varphi _{p}(u'')\in C^{(2) }(0,1) $
satisfying \eqref{e1.1}.

We next give definitions and some properties of cones in Banach spaces.
After
that, we state a fixed point theorem  for operators that are
decreasing with respect to a cone \cite{e1,g1}.

Let $B$ be a Banach space, and $K$ a closed, nonempty subset of
$B$. $K$ is a cone provided
(i) $\alpha u+\beta v\in K$, for all
$u,v\in K$ and all $\alpha ,\beta \geq 0$ and
(ii) $u,-u\in K$ imply $u=0$.

Given a cone $K$, a partial order, $\leq $, is
induced on $B$ by $x\leq y$, for $x,y\in B$ if $y-x\in K$. (For
clarity, we may sometimes write $x\leq y$ (wrt $K$).
If $x,y\in B $
with $x\leq y$, let $\langle x,y\rangle $ denote the closed order
interval between x and y given by,
$\langle x,y\rangle =\{z\in B|x\leq z\leq y\} $.
A cone $K$ is normal in $B$ provided, there exists $\delta >0$,
such that $\Vert e_{1}+e_{2}\Vert \geq \delta $, for all
$e_{1},e_{2}\in K$, with $\|e_{1}\|=\|e_{2}\|=1$.

The following fixed point theorem can be found in \cite{e1,g1}.

\begin{theorem} \label{thm1.1}
 Let $B$ be a Banach space, $K$ a
normal cone in $B$, $E\subseteq K$ such that, if $x,y\in E$ with
$x\leq y$, then $\langle x,y\rangle \subseteq E$, and
let $T:E\to K$ be a continuous mapping that is decreasing
with respect to $K$, and which is compact on any closed order
interval contained in $E$. Suppose there exists $x_{0}\in E$ such
that $T^{2}(x_{0})=T(Tx_{0})$ is defined, and furthermore,
$Tx_{0}$, $T^{2}x_{0}$ are order comparable to $x_{0}$. If, either
\begin{itemize}
\item[(I)] $Tx_{0}\leq x_{0}$ and $T^{2}x_{0}\leq x_{0}$, or
$x_{0}\leq Tx_{0}$ and $x_{0}\leq T^{2}x_{0}$, or

\item[(II)] The complete sequence of iterates
$\{T^{n}x_{0}\}_{n=0}^{\infty }$ is defined, and there exists
$y_{0}\in E$ such that $Ty_{0}\in E$ and $y_{0}\leq T^{n}x_{0}$,
for all $n\geq 0$, then $T$ has a fixed point in $E$.
\end{itemize}
\end{theorem}

\section{Main Theorem}

\begin{theorem} \label{thm2.1}
 Assume the following conditions hold:
\begin{itemize}
\item[(a)] $f(t,u):(0,1)\times (0,\infty )\to (0,\infty )$ is
continuous,

\item[(b)]  $f(t.u)$ is decreasing in $u$, for each fixed $t\in (
0,1)$,

\item[(c)] $\int_{0}^{1}f(t,u)dt<\infty $, for each fixed $u$,

\item[(d)] $\lim_{u\to 0^{+}}f(t,u)=\infty $ uniformly on compact
subsets
of $(0,1)$,

\item[(e)] $\lim_{u\to \infty }f(t,u)=0$ uniformly on compact
subsets of $(0,1)$.

\item[(f)] for each $\tau >0$, $0<\int_{0}^{1}f(t,g_{\tau }(t) )
dt<\infty $, where $g_{\tau }(x)=\tau g(x)$ and
\begin{equation*}
g(t)=\begin{cases}
t,& 0\leq t\leq \frac{1}{2}, \\
(1-t), &\frac{1}{2}\leq t\leq 1.
\end{cases}
\end{equation*}
\end{itemize}
Then the boundary-value problem \eqref{e1.1} has a positive solution
$u\in C^{(2)}[0,1]$ with $\varphi _{p}(u'')\in C^{(2) }(0,1) $.
\end{theorem}


Before the proof of Theorem \ref{thm2.1}, We give some Lemmas which we
will uses
in its proof.

\begin{lemma} \label{lem2.1}
 If $u\in C^{(2)}[0,1]$, $\varphi_{p}(u'')\in C^{(2)}(0,1)$ such that
$(|u''|^{p-2}u'')''>0$ on $(0,1)$, and
$u(0)=u(1)=u''(0)=u''(1)=0$, then
\begin{equation}
u(t)\geq \frac{1}{4}\max_{0\leq t\leq 1}|u(t)|,
\quad \frac{1 }{4}\leq t\leq \frac{3}{4}.\label{e2.1}
\end{equation}
\end{lemma}
The proof of the above lemma is easy; so we omit it.

Recall that the Green function for the problem
\begin{gather*}
u''(t)=0,\quad 0\leq t\leq 1, \\
u(0)=u(1)=0
\end{gather*}
is defined as
\begin{equation}
G(t,s)=\begin{cases}
(1-s)t,& 0\leq t\leq s\leq 1, \\
(1-t)s, &0\leq s\leq t\leq 1.
\end{cases}\label{e2.2}
\end{equation}
A direct calculation shows that
\begin{equation*}
G(t,s)\leq G(s,s)\quad \hbox{for }(t,s)\in [0,1]\times [0,1],
\end{equation*}
\begin{gather}
G(t,s)\leq \frac{1}{4}\quad \text{for }(t,s)\in [0,1]
\times [0,1],\label{e2.3} \\
G(t,s)\geq \frac{1}{4}G(s,s)\geq \frac{3}{64}\quad
 \text{for }(t,s)\in [\frac{1}{4},\frac{3}{4}]
\times [\frac{1}{4},\frac{3}{4}]\label{e2.4}
\end{gather}


\begin{lemma} \label{lem2.2}
 If $u\in C^{(2)}[0,1]$, $\varphi_{p}(u'')\in C^{(2)}(0,1)$ such that
$(|u''|^{p-2}u'')''>0$ on $(0,1)$, and
$u(0)=u(1)=u''(0)=u''(1)=0$, then $u(t)\geq 0$ on $[0,1]$.
\end{lemma}

\begin{proof}
Let $v=u''$. Then
\begin{gather*}
(\varphi _{p}(v))''(t)>0, \\
v(0)=v(1)=0.
\end{gather*}
This implies
\begin{gather*}
(\varphi _{p}(v))''(t) >0, \\
(\varphi _{p}(v))(0)=(\varphi _{p}( v))(1)=0.
\end{gather*}
By the convexity of $\varphi _{p}(v)$, we  obtain
$( \varphi_{p}(v))(t)\leq 0$, for $0\leq t\leq 1$. So $v(t)\leq 0$,
i.e.
\begin{gather*}
u''\leq 0\quad \hbox{on }[ 0,1], \\
u(0)=u(1)=0.
\end{gather*}
Then by the concavity of $u$, we have $u\geq 0$ on $[0,1]$.
\end{proof}

It follows from Lemma \ref{lem2.2} and  Rolle's theorem, that
$u( t)$ has an extreme point, say at $t_{0}\in [ 0,1]$. Then
we define a piecewise polynomial function,
\begin{equation}
p(t)=\begin{cases}
\frac{|u|_{\infty }}{t_{0}}t,& 0\leq t\leq t_{0}, \\[3pt]
\frac{|u|_{\infty }}{1-t_{0}}(1-t),& t_{0}\leq t\leq 1,
\end{cases} \label{e2.5}
\end{equation}
where $|u|_{\infty }=\sup_{0\leq t\leq 1}|u(t)|=u(t_{0})$. Then we have
the
following Lemma.

\begin{lemma} \label{lem2.3}
 Assume $u\in C^{(2)}[0,1]$. Let $\varphi _{p}(u'')$ be a function in
$C^{(2)}(0,1)$ such that $(\varphi_{p}(u''(t)))''>0$, $0<t<1$, and
$u(0)=u(1)=u''(0)=u''( 1) =0$.
Then $u(t)\geq p(t)$, on $[0,1]$, where $p(t)$ is defined by
\eqref{e2.5}.
\end{lemma}

\begin{lemma} \label{lem2.4}
Assume $u\in C^{(2)}[0,1]$. Let $\varphi _{p}(u'')$ be a function
in $C^{(2)}(0,1)$ be such that
$(|u''|^{p-2}u'')''>0$ on $( 0,1)$ and
$u(0) =u(1)=u''(0)=u''( 1)=0$. Then, there exists
$\tau>0$ such that $u(t)\geq g_{\tau }(t)$ on
$[0,1]$.
\end{lemma}
The proof of the above lemma is easy; so we omit it.
Our next work is applying Theorem \ref{thm1.1} to a sequence of
operators that are
decreasing with respect to a cone. The obtained fixed points provide a
sequence of iterates which converges to a solution of \eqref{e1.1}.
Positivity of solutions and Lemmas \ref{lem2.1}--\ref{lem2.3} are
fundamental in
this construction.

Let $B$ be the Banach space $C[0,1]$  with the norm $\|u\|=|u|_{\infty
}$.
Let
\begin{equation*}
K=\{u\in B : u(t)\geq 0, \hbox{ on }[0,1]\},
\end{equation*}
which is a normal cone in $B$. Let $D\subseteq K$ be defined by
\begin{equation*}
D=\{\varphi \in B : \hbox{there exist $\tau (\varphi )>0$
such that $g_{\tau }(t)\leq \varphi (t)$  on $[0,1]$}\}.
\end{equation*}
Define $T:D\to K$ by
\begin{equation*}
T\varphi (t)=\int_{0}^{1}G(t,x)\varphi
_{p}^{-1}(\int_{0}^{1}G(x,s)f(s,\varphi (s))ds)dx,\quad
 0\leq t\leq 1.
\end{equation*}

If $\varphi (t)>0$ for $t\in [0,1]$, by  assumption
(a), we know $f(t,\varphi (t))>0$. If $\varphi (t)\in D$, then
\begin{equation*}
(\varphi _{p}((T\varphi )''(t)))''=f(t,\varphi (t) )>0.
\end{equation*}
Note that $T\varphi (t)$ satisfies the boundary condition of
\eqref{e1.1}.
Lemma \ref{lem2.4} yields that $T\varphi (t)\in D$. So
$T:D\to D$.
Moreover, if $\varphi (t)$ is a positive solution of \eqref{e1.1}, then
by Lemma \ref{lem2.4} $\varphi (t)\in D$ and $T\varphi ( t)=\varphi
(t)$.
Next we prove that all of the solutions of \eqref{e1.1} which belong to
$D$ have a priori bounds.


\begin{lemma} \label{lem2.5}
Assume that conditions (a)-(f) are satisfied. Then there exist an
$R>0$ such that $\|\varphi \|=|\varphi |_{\infty }\leq R$, for all
solutions,
$\varphi $, of \eqref{e1.1} that belong to $D$.
\end{lemma}

\begin{proof}
Suppose that the conclusion is false.
Then there exists a sequence, $\{\varphi _{n}\}\subset D$,
of solutions of \eqref{e1.1} such that
$\lim_{n\to \infty}|\varphi _{n}|=\infty $. Without out loss of
generality, we may assume that, for each $n\geq 1$,
\begin{equation}
|\varphi _{n}|_{\infty }\leq |\varphi
_{n+1}|_{\infty }.\label{e2.6}
\end{equation}
For each $n\geq 1$, let $t_{n}\in (0,1)$ be the unique point such
that
\[
0<\varphi _{n}(t_{n})=|\varphi _{n}|_{\infty }.
\]
Then we have $\varphi _{n}(t_{n})\geq \varphi _{n-1}( t_{n-1})\geq
\dots \geq \varphi _{1}(t_{1})$. Let
$\tau =\frac{1}{4}\varphi_{1}(t_{1})$ and
\begin{equation}
g_{\tau }(t)=\begin{cases}
\tau t & \text{for }t\in [0,\frac{1}{2}], \\
\tau (1-t) &\text{for }t\in [\frac{1}{2},1].
\end{cases} \label{e2.7}
\end{equation}
By the inequality \eqref{e2.1}, we obtain
\[
\varphi _{n}(t)\geq \frac{|\varphi _{n}|
_{\infty }}{4}=\frac{\varphi _{n}(t_{n})}{4}
\geq \frac{1}{4}\varphi _{1}(t_{1})\geq g_{\tau }( t)\quad
\text{for }t\in [\frac{1}{4},\frac{3}{4}].
\]
Next, we claim that
\begin{equation*}
\varphi _{n}(t)\geq g_{\tau }(t)\quad \text{for }t\in [
0,\frac{1}{4}].
\end{equation*}
Let $p_{n}$ be the corresponding piecewise polynomial defined by
\eqref{e2.5} relative to $\varphi _{n}$ and $t_{n}$. There two case for
$t_{n}$

\noindent Case 1. $t_{n}\geq \frac{1}{4}$. Then, for $0\leq t\leq
\frac{1}{4}$,
\[
p_{n}(t)=\frac{|\varphi _{n}|_{\infty }}{
t_{n}}t\geq |\varphi _{n}|_{\infty }t
\geq \frac{|\varphi _{1}|_{\infty}}{4}t\geq g_{\tau }(t).
\]

\noindent Case 2. $t_{n}<\frac{1}{4}$. Then, for $0\leq t\leq t_{n}$,
as the
proof of Case 1, we have
\begin{equation*}
p_{n}(t)\geq g_{\tau }(t)\quad \text{for }t\in [0,t_{n}].
\end{equation*}
On the other hand, on $[t_{n},\frac{1}{4}]$,
\[
p_{n}(t)=\frac{|\varphi _{n}|_{\infty }}{1-t_{n}}(1-t)
\geq |\varphi _{n}|_{\infty }(1-t)
\geq |\varphi _{1}|_{\infty }t\geq g_{\tau}( t).
\]
Thus, again for $0\leq t\leq \frac{1}{4}$,
\begin{equation*}
p_{n}(t)\geq g_{\tau }(t).
\end{equation*}
Using  analogous methods, we have
$p_{n}(t)\geq g_{\tau }(t)$ for $t\in [\frac{3}{4},1]$.
In conclusion,
\begin{equation*}
p_{n}(t)\geq g_{\tau }(t)\quad \text{for }t\in [0,1]
\end{equation*}
which implies
\begin{equation}
\varphi _{n}(t)\geq g_{\tau }(t)\quad
\text{for $t\in [0,1]$ and $n\geq 1$.} \label{e2.8}
\end{equation}
Assumptions (b) and (f) yield, for $0\leq t\leq 1$ and all
$n\geq 1$,
\begin{align*}
\varphi _{n}(t)&=(T\varphi _{n})(t)\\
&=\int_{0}^{1}G(t,x)\varphi _{p}^{-1}\Big( \int_{0}^{1}G( x,s)
f(s,\varphi _{n}(s))
ds\Big)dx \\
&\leq \int_{0}^{1}\frac{1}{4}\varphi
_{p}^{-1}\Big(\int_{0}^{1}\frac{1}{4}
f(s,\varphi _{n}(s))ds\Big)dx \\
&\leq \int_{0}^{1}\frac{1}{4}\varphi
_{p}^{-1}\Big(\int_{0}^{1}\frac{1}{4}
f(s,g_{\tau }(s))ds\Big)dx
=N
\end{align*}
for some $0<N<\infty $. In particular,
$|\varphi _{n}|_{\infty }\leq N$  for all
$n\geq 1$ which contradicts
$\lim_{n\to \infty }|\varphi _{n}|_{\infty }=\infty $.
The proof is complete.
\end{proof}

Our next step in obtaining solutions of \eqref{e1.1} is to construct
a sequence of nonsingular perturbations of $f$.
 For each $n\geq 1$, define $\psi_{n}:[0,1]\to [0,1]$ by
\begin{equation*}
\psi _{n}(t)=\int_{0}^{1}G(t,x)\varphi
_{p}^{-1}\Big(\int_{0}^{1}G(x,s)f(s,n)ds\Big)dx.
\end{equation*}
Because $\varphi _{p}^{-1}$ is increasing and conditions (a)--(g),
for $n\geq 1$,
\begin{equation*}
0<\psi _{n+1}(t)\leq \psi _{n}(t)\quad \text{for } t\in (0,1),
\end{equation*}
and
\begin{equation}
\lim_{n\to \infty }\psi _{n}(t)=0\quad\hbox{uniformly on } \
[0,1].\label{e2.9}
\end{equation}
Now define a sequence of functions
$f_{n}:(0,1)\times [0,\infty )\to (0,\infty )$, $n\geq 1$, by
\begin{equation}
f_{n}(t,u)=f(t,\max \{u,\psi _{n}(t)\}).\label{e2.10}
\end{equation}
Then, for each $n\geq 1$, $f_{n}$ is continuous, nonsingular and
satisfies (b). Furthermore, for $n\geq 1$,

\begin{equation*}
f_{n}(t,u)\leq f(t,u) \quad \hbox{on } ( 0,1)\times (0,\infty ),
\end{equation*}
\begin{equation}
f_{n}(t,u)\leq f(t,\psi _{n})\quad \hbox{on } (0,1)\times
(0,\infty )\label{e2.11}
\end{equation}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
We begin by defining a sequence of operators $T_{n}:K\to K$, $n\geq 1$
by
\begin{equation*}
T_{n}\varphi (t)=\int_{0}^{1}G(t,x)\varphi
_{p}^{-1}\Big(\int_{0}^{1}G(x,s)f_{n}(s,\varphi (s))ds\Big)dx.
\end{equation*}
Note that, for $n\geq 1$ and $\varphi \in K$, we have
\begin{gather*}
(\varphi _{p}((T_{n}\varphi )''))''
=f_{n}(t,\varphi (t))>0 \quad \text{for }t\in (0,1), \\
T_{n}\varphi (0)=T_{n}\varphi (1)=0, \\
(T_{n}\varphi )''(0)=(T_{n}\varphi )''(1)=0.
\end{gather*}
and $T_{n}\varphi >0$ on $(0,1)$. In particular,
$T_{n}\varphi \in D$. Since each $f_{n}$ satisfies $( b)$,
it follows that if $\varphi _{1},\varphi _{2}\in K$ with
$\varphi _{1}\leq \varphi_{2}$, then for $n\geq 1$,
 $T_{n}\varphi _{2}\leq T_{n}\varphi_{1}$; that is, each
$T_{n}$ is decreasing with respect to $K$. It
is also clear that $0\leq T_{n}(0)$ and $0\leq T_{n}^{2}(0) $, for
each $n$.

By Theorem \ref{thm1.1}, for each $n$, there exists a $\varphi _{n}\in
K$,
satisfies $T_{n}\varphi _{n}=\varphi _{n}$, and $\varphi _{n}$
satisfies the boundary condition of \eqref{e1.1}.

In addition, by \eqref{e2.11} we have $T_{n}\varphi \leq T\Psi _{n}$,
for
each $\varphi \in K$ and $n\geq 1$. Thus
\begin{equation}
\varphi _{n}=T_{n}\varphi _{n}\leq T\Psi _{n},\;n\geq 1.\label{e2.12}
\end{equation}

By essentially the same argument as in Lemma \ref{lem2.5}, there exist
an
$R>0$, such that, for each $n\geq 1$
\begin{equation}
\varphi _{n}\leq R\label{e2.13}
\end{equation}
Our next claim is that there exist a $\kappa >0$ such that
$\kappa \leq |\varphi _{n}|_{\infty }$ for all
$n\geq 1$. We assume this claim to be false. Then, by passing to a
subsequence and relabelling, we assume with no loss of generality
that
\begin{equation}
\lim_{n\to \infty }\varphi _{n}(t)=0,\; \hbox{uniformly on
}[0,1].\label{e2.14}
\end{equation}
By condition (d), there exists a $\delta >0$ such that, for
$t\in [\frac{1}{4},\frac{3}{4}]$ and $0<u<\delta $,
$f(t,u)>1$.
By \eqref{e2.14}, there exist an $n_{0}\geq 1$ such that for $n\geq
n_{0}$,
\begin{equation*}
0<\varphi _{n}(t)<\frac{\delta }{2}\quad \hbox{for }\ t\in (0,1).
\end{equation*}
Also from \eqref{e2.9}, there exist an $n_{1}\geq n_{0}$ such that, for
$n\geq n_{1}$,
\begin{equation*}
0<\psi _{n}(t)<\frac{\delta }{2},\;\text{for }t\in ( 0,1).
\end{equation*}
Thus for $n\geq n_{1}$ and $\frac{1}{4}\leq t\leq \frac{3}{4}$,
\begin{align*}
\varphi _{n}(t)&= T_{n}\varphi _{n}(t)\\
&= \int_{0}^{1}G(t,x)\varphi _{p}^{-1}\Big( \int_{0}^{1}G( x,s)
f_{n}(s,\varphi _{n}(s)
)ds\Big)dx \\
&\geq \int_{\frac{1}{4}}^{3/4}G(t,x)\varphi
_{p}^{-1}\Big(\int_{\frac{1}{4}}^{3/4}G(x,s)
f_{n}(s,\varphi _{n}(s))ds\Big)dx \\
&\geq \frac{1}{2}\times \frac{3}{64}\varphi _{p}^{-1}\Big(
\int_{1/4}^{3/4}\frac{3}{64}f(s,\max \{
\varphi _{n}(s),\psi _{n}(s)\} )ds\Big)\\
&\geq \frac{1}{2}\times \frac{3}{64}\varphi
_{p}^{-1}\Big(\int_{1/4}^{3/4}\frac{3}{64}f(s,\frac{\delta
}{2})ds\Big)\\
&\geq \kappa >0.
\end{align*}
This contradicts the uniform limit \eqref{e2.14}. Our claim is
verified.
That is
there exists a $\kappa >0$ such that
\begin{equation*}
\kappa \leq |\varphi _{n}|_{\infty }\leq R \quad\hbox{for all }n
\end{equation*}
Applying Lemma \ref{lem2.1},
\[
\varphi _{n}(t)\geq \frac{1}{4}|\varphi_{n}|_{\infty }
\geq \frac{\kappa }{4},\quad t\in [\frac{1}{4},\frac{3}{4}],\;
n\geq 1.
\]
Let $\tau =\kappa/4$. Using a mimic methods in the proof
of Lemma \ref{lem2.5}, we have
\begin{equation*}
g_{\tau }(t)\leq \varphi _{n}(t)\quad \hbox{on } [
0,1],\hbox{ for } n\geq 1
\end{equation*}
By \eqref{e2.13}, we now have
\begin{equation*}
g_{\tau }(t)\leq \varphi _{n}(t)\leq R\quad \hbox{for all }n\geq 1;
\end{equation*}
that is, the sequence $\{\varphi _{n}(t)\}$ belongs to the closed
order interval $\langle g_{\tau },R\rangle \subset D$.

When restricted to this closed order interval, $T$ is a compact
mapping, and so, there is a subsequence of $\{T\varphi _{n}(t)\}$
which converges to some $\varphi ^{\ast }\in K$. We relabel the
subsequence as the original sequence so that
\begin{equation}
\lim_{n\to \infty }\|T\varphi _{n}-\varphi ^{\ast
}\|=0.\label{e2.15}
\end{equation}
The final part of the proof is to establish that
\begin{equation*}
\lim_{n\to \infty }\|T\varphi _{n}-\varphi _{n}\|
=0.
\end{equation*}
Let $C=\frac{1}{4}\int_{0}^{1}f(s,g_{\tau }(s))ds$. Then
\begin{equation*}
\int_{0}^{1}G(x,s)f(s,\varphi _{n}(s))ds\leq C\quad
 \text{for all }n\geq 1.
\end{equation*}
By the uniformly continuous of $\varphi _{p}^{-1}$ on
 $[0,C]$, let $\varepsilon >0$ be given, there exists
$\delta >0$, such that if $s_{1},s_{2}\in [0,C]$ and
$|s_{1}-s_{2}|<\delta $, we have
\begin{equation*}
|\varphi _{p}^{-1}(s_{1})-\varphi _{p}^{-1}(
s_{2})|<\varepsilon .
\end{equation*}
By the integrability condition $(f)$, for above $\delta $, there
exists $0<\delta _{1}<1$, such that
\begin{equation}
\int_{0}^{\delta _{1}}f(s,g_{\tau }(s))ds+\int_{1-\delta
_{1}}^{1}f(s,g_{\tau }(s))ds\leq \delta .\label{e2.16}
\end{equation}
Further, by \eqref{e2.9} there exists an $n_{0}$ such that, for $n\geq
n_{0}$,
\begin{equation*}
\psi _{n}(t)\leq g_{\tau }(t)\leq \varphi _{n}( t) \quad
\hbox{on }[\delta _{1},1-\delta _{1}].
\end{equation*}
from the definition of \eqref{e2.10}, we know
\begin{equation*}
f_{n}(s,\varphi _{n}(s))=f(s,\varphi _{n}( s)),\;\hbox{for }s\in
[\delta _{1},1-\delta _{1}]\;\hbox{and }n\geq n_{0}.
\end{equation*}
Thus, for $t\in [0,1]$ and $n\geq n_{0}$, by \eqref{e2.16},
\begin{align*}
&\int_{0}^{1}G(x,s)f(s,\varphi _{n}(s))
ds-\int_{0}^{1}G(x,s)f_{n}(s,\varphi _{n}(
s))ds \\
&=\int_{0}^{\delta _{1}}G(x,s)f(s,\varphi _{n}( s))
ds+\int_{1-\delta _{1}}^{1}G(x,s)f(
s,\varphi _{n}(s))ds \\
&\quad -(\int_{0}^{\delta _{1}}G(x,s)f_{n}(s,\varphi _{n}( s) )
ds+\int_{1-\delta _{1}}^{1}G(x,s)
f_{n}(s,\varphi _{n}(s))ds)\\
&\leq \int_{0}^{\delta _{1}}f(s,g_{\tau }(s))
ds+\int_{1-\delta _{1}}^{1}f(s,g_{\tau }(s))ds
\leq \delta .
\end{align*}
 So
\begin{equation*}
\Big|\varphi _{p}^{-1}\Big(\int_{0}^{1}G(x,s)f( s,\varphi
_{n}(s))ds\Big)
-\varphi _{p}^{-1}\Big( \int_{0}^{1}G(x,s) f_{n}(s,\varphi
(s))ds\Big))\Big|\leq \varepsilon .
\end{equation*}
Then for $n\geq n_{0}$, we have
\begin{align*}
|T\varphi _{n}(t)-\varphi _{n}(t)|
&=\Big|\int_{0}^{1}G(t,x)\varphi _{p}^{-1}\Big(
\int_{0}^{1}G(x,s)f(s,\varphi _{n}(s))
ds\Big)dx \\
&\quad -\int_{0}^{1}G(t,x)\varphi _{p}^{-1}\Big(
\int_{0}^{1}G(x,s)f_{n}(s,\varphi _{n}(s)
)ds\Big)dx\Big|\\
&= \int_{0}^{1}G(t,x)\Big|\varphi _{p}^{-1}\Big(
\int_{0}^{1}G(x,s)f(s,\varphi _{n}(s))ds\Big)\\
&\quad -\varphi_{p}^{-1}\Big(\int_{0}^{1}G(x,s)
f_{n}(s,\varphi _{n}(s))ds\Big)\Big|dx \\
&\leq \frac{1}{4}\varepsilon <\varepsilon
\end{align*}
In particular,
\begin{equation*}
\lim_{n\to \infty }\|T\varphi _{n}(t) -\varphi_{n}(t)\|=0.
\end{equation*}
Then in conjunction with \eqref{e2.15} we can easily obtain
\begin{equation*}
\lim_{n\to \infty }\|\varphi _{n}-\varphi ^{\ast}\|=0,
\end{equation*}
and this implies
$\varphi ^{\ast }\in \langle g_{\tau },K\rangle \subset D$
and
\begin{equation*}
\varphi ^{\ast }=\lim_{n\to \infty }T\varphi _{n}=T\big(
\lim_{n\to \infty }\varphi _{n}\big)=T\varphi ^{\ast },
\end{equation*}
which is sufficient for the conclusion of the Theorem \ref{thm2.1}.
\end{proof}

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\end{document}