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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 39, 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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/39\hfil Positive solutions]
{Positive solutions of singular fourth-order  boundary-value problems}
\author[Y. Cui, Y. Zou\hfil EJDE-2006/39\hfilneg]
{Yujun Cui, Yumei Zou}  % in alphabetical order

\address{Department of Applied Mathematics,
 Shandong  University of Science
and technology, Qingdao, 266510, China}
\email[Y. Cui]{cyj720201@163.com}

\date{}
\thanks{Submitted September 6, 2005. Published March 21, 2006}
\thanks{Supported by grant 10371066 from the National Science Foundation
of China}
\subjclass[2000]{34A34, 34B15, 45G15}
\keywords{Singular boundary value problem; fixed point theorem;
\hfill\break\indent positive solution}

\begin{abstract}
 In this paper, we present necessary and sufficient conditions
 for the existence of  positive $C^3[0,1]\cap C^4(0,1)$ solutions
 for the singular boundary-value problem
 \begin{gather*}
 x''''(t)=p(t)f(x(t)),\quad t\in(0,1);\\
 x(0)=x(1)=x'(0)=x'(1)=0,
 \end {gather*}
 where  $f(x)$ is either superlinear or sublinear,
 $p:(0,1)\to [0,+\infty)$ may be singular at both ends $t=0$
 and $t=1$.
 For this goal, we  use  fixed-point index results.
\end{abstract}

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

\section{Introduction}

In this paper, we consider the  fourth order differential
equation
\begin{gather}
x''''(t)=p(t)f(x(t)),\quad t\in(0,1);\label{e1.1}\\
x(0)=x(1)=x'(0)=x'(1)=0.\label{e1.2}
\end{gather}
where  $f(x)$ is either superlinear or sublinear,
$p:(0,1)\to [0,+\infty)$ may be singular at both ends $t=0$
and $t=1$.

Recently, the existence and multiplicity of positive solutions of
\eqref{e1.1}-\eqref{e1.2} in the non-singular case has been
extensively studied in the literature; see \cite{m2,k1,y1} and
references therein. However for singular fourth order boundary-value
problems, the research has proceeded very slowly. Ma and Tisdell
\cite{m1} studied the  singular sublinear fourth order
boundary value problems
\begin{gather}
x''''(t)=p(t)x^{\lambda}(t),\quad t\in(0,1);\label{e1.3} \\
x(0)=x(1)=x'(0)=x'(1)=0.\label{e1.4}
\end{gather}
where $\lambda\in (0,1)$ is given, and
$p:(0,1)\to [0,\infty)$ may be singular at both ends $t=0$ and $t=1$.
Base upon the method  of  lower and upper solutions,
Ma and Tisdell showed that \eqref{e1.3}-\eqref{e1.4} has a  positive
solution in $C^2[0,1]\cap C^4(0,1)$ if and only if
$$
0<\int_0^1t^{1+2\lambda}(1-t)^{1+2\lambda}p(t)dt<+\infty.
$$
Moreover, this positive solution is in $C^3[0,1]\cap C^4(0,1)$ if
and only if
$$
0<\int_0^1t^{2\lambda}(1-t)^{2\lambda}p(t)dt<+\infty.
$$
But necessary and sufficient conditions for the existence of
positive solution of superlinear BVPs \eqref{e1.3}-\eqref{e1.4}
still remain unknown. In this paper, by using the  fixed point
index, we give some  necessary and sufficient conditions for the
existence of $C^3[0,1]\cap C^4(0,1)$ positive solutions to the
singular boundary value problem \eqref{e1.1}-\eqref{e1.2}.

In our discussion, by a $C^k[0,1]$ solution $(k=2,3)$ of
\eqref{e1.1}-\eqref{e1.2} we mean a function
$y(t)\in C^k[0,1]\cap C^4(0,1)$ which satisfies \eqref{e1.2}
and \eqref{e1.1} on (0,1).
We call a solution $y(t)$ is a positive solution if $y(t)>0$ for
$t\in (0,1)$.


This paper is organized as follows. Section 2 gives some preliminary
lemmas corresponding to \eqref{e1.1}-\eqref{e1.2}. Section 3 is
devoted to the the existence of $C^3[0,1]\cap C^4(0,1)$ positive
solutions for \eqref{e1.1}-\eqref{e1.2}. At the end of this
section we state some lemmas of the fixed point theory, which will
be used in Section 3.

Let $E$ be a Banach space, $P$ a cone in $E$, $\Omega$ a bounded
open set in $E$.

\begin{lemma}[\cite{g1}] \label{lem1.1}
Let $\theta\in \Omega$, $A:\overline{\Omega}\cap P\to P$ be
completely continuous. Suppose that there exists
$u_{0}\in P\backslash\{\theta\}$ such that
$$
u-Au\neq\mu u_{0},\ \forall\ u\in\partial\Omega\cap P,\
\mu\geq0,
$$
then the fixed point index $i(A,\ \Omega\cap P,\ P)=0$.
\end{lemma}

\begin{lemma}[\cite{g1}] \label{lem1.2}
Let $\theta\in \Omega$, $A:\overline{\Omega}\cap P\to P$ be
completely continuous. Suppose that
$$
Au\neq\mu u,\quad \forall\ u\in\partial\Omega\cap P,\quad
\mu\geq1,
$$
then the fixed point index $i(A,\ \Omega\cap P,\ P)$ is equal to 1.
\end{lemma}

\section{Preliminaries}

 We give some notations, which will be used below.
Let $C[0,1]$, $C^k[0,1]$ and $L^1[0,1]$
 be the classical Banach spaces with their usual norms $\|\cdot\|$,
$\|\cdot\|_{C^k}$ and $\|\cdot\|_{L^1}$, respectively. Let $AC[0,1]$
be the space of all absolutely continuous functions on [0,1]. Let
$$
AC^k[0,1]=\{u\in C^k[0,1]: u^{(k)}\in AC[0,1]\}.
$$
Clearly $AC^0[0,1]=AC[0,1]$. Let $I$ be an interval of $R$. We
denote by  $L^1_{\rm loc}I$ the spaces of functions
defined by
$$
L^1_{\rm loc}I=\{u:I\to R: u|_{[c,d]}\in  L^1[c,d]\ \
\text{for\ every\ compact\ interval}\ [c,d]\subset I\}.
$$
For $n,m\in N$, we denote by $X[n,m]$ the Banach space
$$
X[n,m]=\{\varphi\in L^1_{\rm loc}(0,1)\Big|\
\int_0^1t^n(1-t)^m|\varphi(t)|dt<+\infty\},
$$
equipped with the norm
$$
\|\varphi\|_{X[n,m]}=\int_0^1t^n(1-t)^m|\varphi(t)|dt.
$$
Now let $G(t,s)$ be the Green's function of the linear problem
\begin{gather*}
x''''(t)=0,\quad t\in(0,1);\\
x(0)=x(1)=x'(0)=x'(1)=0,
\end{gather*}
which can be explicitly given by
$$
G(t,s)=\frac16\begin{cases}
t^2(1-s)^2[(s-t)+2(1-t)s],&  0\leq t\leq s\leq 1,\\
s^2(1-t)^2[(t-s)+2(1-s)t], & 0\leq s\leq t\leq 1.
\end{cases}
$$
It is clear that for all $t,s\in [0,1]$,
\begin{equation}
\frac13t^2(1-t)^2s^2(1-s)^2\leq G(t,s)\leq \frac12t^2(1-t)^2,\
 G(t,s)\leq \frac12s^2(1-s)^2.\label{e2.1}
\end{equation}
Suppose that $\varphi\in X[2,2]$. We denote
$$
T(\varphi)(t)=\int^1_0 G(t,s)\varphi(s)ds,
$$
i.e.
\begin{align*}
T(\varphi)(t)&= \frac16\int_0^t s^2(1-t)^2[(t-s)+2(1-s)t]\varphi(s)ds\\
&\quad +\frac16\int_t^1
t^2(1-s)^2[(s-t)+2(1-t)s]\varphi(s)ds.
\end{align*}


\begin{lemma}[\cite{m1}] \label{lem2.1}
Let $\varphi\in X[2,2]$. Then $T(\varphi)(t)$, $[T(\varphi)]'(t)$,
 $[T(\varphi)]''(t)$,
 $[T(\varphi)]'''(t)$ are  $AC_{\rm loc}(0,1)\cap C^1(0,1)$, and
$$
[T(\varphi)]''''(t)=\varphi(t),\quad  \text{a.e. } t\in (0,1).
$$
\end{lemma}

\begin{lemma}[\cite{m1}] \label{lem2.2}
Let $\varphi\in X[2,2]$. Then
$$
T(\varphi)(0)=T(\varphi)(1)=T(\varphi)'(0)=T(\varphi)'(1)=0.
$$
\end{lemma}

\begin{lemma}[\cite{m1}] \label{lem2.3}
Let $\varphi\in L^1(0,1)$. Then $[T(\varphi)](t)\in AC^3[0,1]$.
\end{lemma}


\section{Main Result}

We shall assume the following conditions:
\begin{itemize}
\item[(H1)] $f:[0,\infty)\to [0,\infty)$ is continuous and
nondecreasing in $x$,  $f(x)>0$ on $(0,\infty)$, and
there exists $\lambda>1$ such that
\begin{equation}
f(cx)\leq c^{\lambda}f(x),\ \ \forall\ c\geq 1,\ x\in
[0,+\infty).\label{e3.1}
\end{equation}

\item[(H2)] $p:(0,1)\to[0,\infty)$ is continuous,
$ \int_0^1s^2(1-s)^2p(s)ds<+\infty$,
 and there exists $\theta\in (0,1/2)$ such that
$$
0<\int_\theta^{1-\theta}s^2(1-s)^2p(s)ds.
$$

\item[(H3)] $0\leq \limsup_{x\to 0+}\frac{f(x)}{x}<M_1,
m_1<\liminf_{x\to +\infty}\frac{f(x)}{x}\leq
+\infty$, where
\begin{gather*}
M_1=(\max_{t\in [0,1]}\int_0^1G(t,s)p(s)ds)^{-1},
\\
m_1=(\min_{t\in [\theta,1-\theta]}\int_\theta^{1-\theta}G(t,s)p(s)ds)^{-1}.
\end{gather*}
\end{itemize}

\begin{theorem} \label{thm3.1}
Under assumptions {\rm (H1)-(H3)},
 a necessary and sufficient condition for
\eqref{e1.1}-\eqref{e1.2} to have a positive solution in
$C^3[0,1]\cap C^4(0,1)$ is that
\begin{equation}
\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty.\label{e3.2}
\end{equation}
\end{theorem}

\begin{remark} \label{rmk3.1} \rm
Inequality \eqref{e3.1} implies
\begin{equation}
f(cx)\geq c^{\lambda}f(x),\quad \forall\ c\in (0,1),\ x\in
[0,+\infty).\label{e3.3}
\end{equation}
Conversely, \eqref{e3.3} implies \eqref{e3.1}.
\end{remark}

\begin{remark} \label{rmk3.2} \rm
(H2) is equivalent to
\begin{itemize}
\item[(H2')]  $p\in C((0,1),[0,+\infty))\cap X[2,2]$, and there exists
$t_0\in (0,1)$ with $p(t_0)>0$.
\end{itemize}
\end{remark}


\begin{proof}[Proof of Theorem \ref{thm3.1}]
  Necessity. Let $x\in C^2[0,1]\cap C^4(0,1)$ be a
positive solution of \eqref{e1.1}  and \eqref{e1.2}. Then by the
fact
\begin{align*}
x''(t)&= \frac16\int_0^t\{2s^2[(t-s)+2(1-s)t]-4s^2(1-t)[1+2(1-s)]\}
p(s)f(x(s))ds\\
&\quad +\frac16\int_t^1\{2(1-s)^2[(s-t)+2(1-t)s]+4t(1-s)^2[-1-2s]\}
p(s)f(x(s))ds.
\end{align*}
we have that
\begin{gather*}
x''(0)=\int_0^1(1-s)^2sp(s)f(x(s))ds>0, \\
x''(1)=\int_0^1s^2(1-s)p(s)f(x(s))ds>0.
\end{gather*}
and accordingly, there exist $I_1,I_2\in(0,+\infty)$ such that
$$
I_1t^2(1-t)^2\leq x(t)\leq I_2t^2(1-t)^2,\quad t\in [0,1].
$$
Let $c_1\geq \max\{1,1/I_1\}$, then
$$
t^2(1-t)^2\leq c_1x(t),\quad  t\in [0,1].
$$
So by (H1),
\begin{align*}
\int_0^1p(s)f(s^2(1-s)^2)ds
&\leq \int_0^1p(s)f(c_1x(s))ds\\
&\leq c_1^{\lambda}\int_0^1p(s)f(x(s))ds\\
&= c_1^{\lambda}\int_0^1x''''(s)ds \\
&\leq c_1^{\lambda}[x'''(1)-x'''(0)]<\infty.
\end{align*}
On the other hand, if $c_2\leq \min\{1/2,1/I_2\}$, then
$$
t^2(1-t)^2\geq c_2x(t),\ \ \ t\in [0,1].$$
So by (H1) and \eqref{e3.3},
$$
  \int_0^1p(s)f(s^2(1-s)^2)ds\geq
\int_0^1p(s)f(c_2x(s))ds\geq c_2^{\lambda}\int_0^1p(s)f(x(s))ds\geq
0
$$
Notice that $\int_0^1p(s)f(x(s))ds>0$, for otherwise
$p(s)f(x(s))\equiv0$ on (0,1). In this case
\eqref{e1.1}-\eqref{e1.2} has only trivial solution $x\equiv 0$.
This contradicts the assumption that $x$ is a positive solution.
Thus \eqref{e3.2} holds.
\smallskip


Sufficiency. Suppose that \eqref{e3.2} holds. we define a set
$P\subset C[0,1]$ by
\begin{align*}
P=\big\{&x\in C[0,1]: \exists c_x>0,\ 0\leq x(t)\leq c_xt^2(1-t)^2,\\
& x(t)\geq \frac23t^2(1-t)^2\|x\|,\ t\in [0,1]\big\}.
\end{align*}
By its definition, it is easy to verify that $P$ is a cone.
We define $T:P\to C[0,1]$ by
$$
T(x)(t)=\int^1_0 G(t,s)p(s)f(x(s))ds,\quad t\in[0,1],\ x\in P.
$$
In the following, we  prove that
$T:P\to P$ is completely continuous.

\noindent 1. We first show that $T:P\to P$ is well defined.
For $x\in P$, there exist $c_x\geq 1$ such that
$0\leq x(t)\leq c_xt^2(1-t)^2$ and for $t\in [0,1]$, by \eqref{e2.1},
we get
$$
(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds \leq \frac12
c_x^{\lambda}t^2(1-t)^2\int_0^1p(s)f(s^2(1-s)^2)ds.
$$
This implies that $p(t)f(x(t))\in L^1[0,1]$, by Lemma \ref{lem2.3},
we have $Tx\in C[0,1]$. Let $c_{Tx}=\frac12
c_x^{\lambda}\int_0^1p(s)f(s^2(1-s)^2)ds$. By (3.2), we know
$c_{Tx}>0$, so
$$
(Tx)(t)\leq c_{Tx}t^2(1-t)^2,\ \ t\in [0,1].
$$
In addition, for $t\in [0,1]$, by \eqref{e2.1}, we get
\begin{equation}
(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds\geq
\frac13t^2(1-t)^2\int_0^1s^2(1-s)^2p(s)f(x(s))ds,\label{e3.4}
\end{equation}
and
$$
(Tx)(t)=\int_0^1G(t,s)p(s)f(x(s))ds\leq
\frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds.
$$
Hence
$$
\|Tx\|\leq \frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds.
$$
Combining the above  with \eqref{e3.4}, we have
$$
(Tx)(t)\geq \frac13t^2(1-t)^2\int_0^1s^2(1-s)^2p(s)f(x(s))ds\geq
\frac23t^2(1-t)^2\|Tx\|,
$$
i.e., $T(P)\subset P$.
\smallskip


\noindent 2.  We  show that $T:P\to P$ is compact.
Let $D\subset P$ be bounded, i.e., $\|x\|\leq M$ for all $x\in D$
and some $M>0$. It is clear that if $x\in P$ satisfies $x\in D$, by
(H2) we have
$$
|(Tx)(t)|\leq \frac12\int_0^1s^2(1-s)^2p(s)f(x(s))ds\leq
\frac12\int_0^1s^2(1-s)^2p(s)f(M)ds.
$$
So $T(D)$ is uniformly bounded.

Next we prove that $\|(Tx)'\|\leq N$ for all $x\in D$ and some
$N>0$. In fact, for  $x\in D$. By Lemma \ref{lem2.3}, we know $Tx\in
C^2[0,1]$ and
\begin{align*}
& |(Tx)'(t)|\\
&= \Big|
\frac16\int_0^t\{-2s^2(1-t)[(t-s)+2(1-s)t]+s^2(1-t)^2[1+2(1-s)]\}p(s)f(x(s))ds\\
&\quad
+\frac16\int_t^1\{2t(1-s)^2[(s-t)+2(1-t)s]+t^2(1-s)^2[-1-2s]\}p(s)f(x(s))ds\Big|\\
& \leq \frac16\int_0^t\{2s^2(1-s)[(1-s)+2(1-s)]+s^2(1-s)^2[1+2(1-s)]\}p(s)f(M)ds\\
&\quad +\frac16\int_t^1\{2t(1-s)^2[s+2s]+s^2(1-s)^2[1+2s]\}p(s)f(M)ds\\
& \leq \frac96\int_0^ts^2(1-s)^2p(s)f(M)ds
\quad +\frac96\int_t^1s^2(1-s)^2p(s)f(M)ds\\
&= \frac32\int_0^1s^2(1-s)^2p(s)f(M)ds=N.
\end{align*}
This means that $T(D)$ is equicontinuous.
From the Ascoli-Arzela theorem, $T(D)$ is relatively compact. This
completes the proof that $T$ is compact.
\smallskip

\noindent 3. We prove $T:P\to P$ is continuous.
Assume that $x_n,x\in P$ and $x_n\to x$. Then there exists
$M>0$ such that $\|x\|\leq M,\ \|x_n\|\leq M$ for every $n>0$. Since
$f(x)$ is continuous, we have
$$
|f(x_n(s))-f(x(s))|\to 0,\quad \text{as } n\to \infty,\quad
 \forall\ s\in [0,1],
 $$
and
$$
|f(x_n(s))-f(x(s))|\leq 2f(M),\quad \forall\ t\in [0,1],\ (n=1,2,3\dots).
 $$
Consequently,  for all $t\in [0,1]$,
\begin{equation}
\|(Tx_n)(t)-(Tx)(t)\|\leq
\int_0^1s^2(1-s)^2p(s)|f(x_n(s))-f(x(s))|ds\to 0. \label{e3.5}
\end{equation}
We now show
\begin{equation} \|Tx_n-Tx\|\to 0\quad \text{as } n\to
\infty).\label{e3.6}
\end{equation}
If \eqref{e3.6} is not true, then there exist a positive
number $\varepsilon>0$ and a sequence $\{x_{n_i}\}\subset\{x_n\}$
such that
\begin{equation}
\|Tx_{n_i}-Tx\|\geq\varepsilon,\quad (i=1,2,3\ldots).\label{e3.7}
\end{equation}
Since $\{x_n\}$ is bounded, $\{Tx_n\}$ is relatively compact and
there is a subsequence of $\{Tx_{n_i}\}$ which converges in $C[0,1]$
to some $y\in C[0,1]$. Without loss of generality, we may assume
that $\{Tx_{n_i}\}$ itself  converges to $y$:
\begin{equation}
\|Tx_{n_i}-y\|\to 0,\quad  \text{as } i\to
\infty.\label{e3.8}
\end{equation}
By virtue of \eqref{e3.5} and \eqref{e3.8}, we have $y=Tx$, and so,
\eqref{e3.8} contradicts \eqref{e3.7}. Hence, \eqref{e3.6} holds,
and the continuity of $T$ is proved.
To sum up, we have proved $T:P\to P$ is  completely
continuous.

For all $x\in P$, from the above proof, we know $Tx\in P$, By
Lemma \ref{lem2.1} and Lemma \ref{lem2.2}, the fixed point of the
equation
$$
Tx=x,\quad  x\in P.
$$
is the solution of  \eqref{e1.1}-\eqref{e1.2}. Next we will look
for the fixed point.

By the first part of (H3), there exist $1>r>0$, $\varepsilon>0$
such that $0<u<r$ implies $f(x)/x \leq (M_1-\varepsilon)$.
Therefore, we have
$$
f(x)\leq (M_1-\varepsilon)x\leq (M_1-\varepsilon)r,\ \ 0<x\leq r.
$$
Set $B_r=\{x\in C[0,1]: \|x\|<r\}$. For $\forall\ x\in\partial
B_r\cap P$, we have
\begin{align*}
\|Tx\|&= \max_{t\in
[0,1]}\int_0^1G(t,s)p(s)f(x(s))ds
\leq (M_1-\varepsilon)r\max_{t\in [0,1]}\int_0^1G(t,s)p(s)ds\\
&\leq   r-\varepsilon r\max_{t\in
[0,1]}\int_0^1G(t,s)p(s)ds<r.
\end{align*}
Then for $x\in \partial B_r\cap P$ and $\mu \geq 1$, we have
$$
Tx\neq \mu x.
$$
In  not, there exist $x_0\in \partial B_r\cap P$ and
$\mu_0\geq 1$ such that $Tx_0=\mu_0x_0$, then $\|Tx_0\|\geq
\|x_0\|$, which is a contradiction. According to Lemma \ref{lem1.2},
we have
\begin{equation}
i(T,\ B_r\cap P,\ P)=1.\label{e3.9}
\end{equation}
By the second part of (H3), $m_1<\liminf_{x\to
+\infty}\frac{f(x)}{x}\leq +\infty$, there exist $R_1>\max\{\theta
r,1\}$, $\varepsilon_1>0$ such that
$$
f(x)\geq (m_1+\varepsilon_1)x,\quad  x\geq R_1.
$$
Let $R_2>\frac{3R_1}{2\theta^2(1-\theta)^2}$, and
$B_{R_2}=\{x\in C[0,1]: \|x\|<R_2\}$, then
$$
\min_{t\in [\theta,1-\theta]}x(t)\geq \min_{t\in
[\theta,1-\theta]}\frac{2}{3}t^2(1-t)^2\|x\|\geq R_1,\quad
 \forall\ x\in \partial B_{R_2}\cap P.
$$
We now prove that
$$
x-Tx\neq\mu t^2(1-t)^2,\quad  \text{for } \mu\geq 0 \text{ and }
 x\in \partial B_{R_2}\cap P.
$$
If not, then there are $\mu_1\geq 0$ and $x_1\in \partial
B_{R_2}\cap P$ such that $x_1-Tx_1=\mu_1 t^2(1-t)^2$. So $\mu_1>0$,
otherwise there is a fixed point in $\partial B_{R_2}\cap P$ and
this would complete the proof. Let
$\eta=\min_{t\in[\theta,1-\theta]}x_1(t)$. Then if
$t\in[\theta,1-\theta]$, we have
\begin{align*}
x_1(t)&= \int_0^1G(t,s)p(s)f(x_1(s))ds+\mu_1t^2(1-t)^2\\
&\geq \int_\theta^{1-\theta}G(t,s)p(s)f(x_1(s))ds+\mu_1t^2(1-t)^2\\
&\geq (m_1+\varepsilon_1)\int_\theta^{1-\theta}G(t,s)p(s)x_1(s)ds
+\mu_1t^2(1-t)^2\\
&\geq \eta(m_1+\varepsilon_1)\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_1t^2(1-t)^2\\
&\geq \eta+\eta\varepsilon_1\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_1t^2(1-t)^2.
\end{align*}
Therefore,
$$
x_1(t)>\eta,\quad t\in [\theta,1-\theta],
$$
which is a contradiction. According to Lemma \ref{lem1.1}, we get
\begin{equation}
i(T, B_{R_2}\cap P,\ P)=0.\label{e3.10}
\end{equation}
By \eqref{e3.9} and \eqref{e3.10}, we have
$$
i(T,\ (B_{R_2}\overline{\backslash{B_r}})\cap P,\ P) =i(T,\
B_{R_2}\cap P,\ P)-i(T,\ B_r\cap P,\ P)=-1.
$$
Then $T$ has at least a fixed point $x^*$ in
$(B_{R_2}\overline{\backslash{B_r}})\cap P$ satisfying $0<r\leq
\|x^*\|\leq R_2$. Since $x^*\in P$, there exists $r_{x^*}>1$ such
that $x^*\leq r_{x^*}t^2(1-t)^2$, then
\begin{align*}
\int_0^1p(s)f(x^*(s))ds
&\leq \int_0^1p(s)f(r_{x^*}s^2(1-s)^2)ds\\
&\leq r_{x^*}^{\lambda}\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty,
\end{align*}
that is $p(t)f(x^*(t))\in L^1(0,1)$, then by Lemma \ref{lem2.3}, we
have $x^*\in AC^3[0,1]$,
so $x^*$ is a $C^3[0,1]\cap C^4(0,1)$ positive solution of
\eqref{e1.1}-\eqref{e1.2}.
This completes the proof of sufficiency.
\end{proof}

\begin{corollary} \label{coro1}
Let $p$ be as above, $0<\int_0^1s^2(1-s)^2p(s)ds<+\infty$, and
$\lambda>1$. Then  BVP
\eqref{e1.3}-\eqref{e1.4} has at least a positive solution in
 $C^3[0,1]\cap C^4(0,1)$
 \end{corollary}

\begin{proof} The hypotheses on the function $p(s)$ implies
$0<\int_0^1(s(1-s))^{2\lambda}p(s)ds<+\infty$  for $\lambda>1$. The
result now follows from Theorem \ref{thm3.1}.
\end{proof}

\begin{theorem} \label{thm3.2}
Assume that (H1) and (H2) are satisfied. If
$$
\lim_{x\to 0+}\frac{f(x)}{x}=0, \quad
\lim_{x\to +\infty}\frac{f(x)}{x}+\infty,
$$
Then a necessary and sufficient condition for
\eqref{e1.1}-\eqref{e1.2} to have a positive solution in
$C^3[0,1]\cap C^4(0,1)$ is that
$$
\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty.
$$
\end{theorem}


\begin{proof} Clearly (H1)-(H3) hold, and result follows
from Theorem  \ref{thm3.1}. We omit the detail.
\end{proof}

Next, we shall study \eqref{e1.1}-\eqref{e1.2} in the sublinear
case. We assume:
\begin{itemize}
\item[(H1')]  $f:[0,\infty)\to [0,\infty)$ is continuous and
nondecreasing in $x$,
 $f(x)>0$ on $(0,\infty)$, and
there exists $0<\lambda_1<1$ such that
$$
f(cx)\geq c^{\lambda_1}f(x),\quad  \forall\ c\in(0,1),\ x\in [0,+\infty).
$$
\item[(H3')] $0\leq \limsup_{x\to
+\infty}\frac{f(x)}{x}<M_1, m_1<\liminf_{x\to
0+}\frac{f(x)}{x}\leq +\infty$, where
\begin{gather*}
M_1=(\max_{t\in [0,1]}\int_0^1G(t,s)p(s)ds)^{-1}, \\
m_1=(\min_{t\in [\theta,1-\theta]}\int_\theta^{1-\theta}G(t,s)p(s)ds)^{-1}.
\end{gather*}
\end{itemize}

\begin{theorem} \label{thm3.2b}
Assume {\rm (H1'), (H2)}, and {\rm (H3')}.
Then a necessary and sufficient condition for
\eqref{e1.1}-\eqref{e1.2} to have a positive solution in
$C^3[0,1]\cap C^4(0,1)$ is that
\begin{equation}
\int_0^1p(s)f(s^2(1-s)^2)ds<+\infty.\label{e3.11}
\end{equation}
\end{theorem}

\begin{proof} By (H1'), we have
$f(cx)\leq c^{\lambda_1}f(x)$, $c\geq 1,\ x\in [0,+\infty)$.
 The proof of necessity is almost the same as that in Theorem \ref{thm3.1}.

 We will show the roof of the sufficiency.
 We base the proof on the argument in Theorem
\ref{thm3.1} and need only show completely continuous operator
$T:P\to P$ has a fixed point.

 By the first part of (H3'), there are $R_3>1$, $\varepsilon_3>0$ such
that $x\geq R_3$ implies $f(x) \leq (M_1-\varepsilon_3)x$.
 Let $M=\max\{f(x): 0\leq x\leq R_3\}$, then
$$
f(x)\leq (M_1-\varepsilon_3)x+M,\quad x\in [0,+\infty).
$$
Choose $R_4>\max\{M\varepsilon_3^{-1},1\}$.
Let $B_{R_4}=\{x\in C[0,1]: \|x\|<R_4\}$. Then for all
$x\in\partial B_{R_4}\cap P$, we have
\begin{align*}
\|Tx\|
&= \max_{t\in [0,1]}\int_0^1G(t,s)p(s)f(x(s))ds\\
&\leq (M+(M_1-\varepsilon_3)\|x\|)\max_{t\in [0,1]}\int_0^1G(t,s)p(s)ds\\
&\leq   M_1R_4\max_{t\in
[0,1]}\int_0^1G(t,s)p(s)ds
+(M-\varepsilon_3R_4)\frac12\int_0^1s^2(1-s)^2p(s)ds \\
&= R_4+(M-\varepsilon_3R_4)\frac12\int_0^1s^2(1-s)^2p(s)ds\\
&<R_4=\|x\|.
\end{align*}
So it is easy to know that $ Tx\neq \mu x$ for
$x\in \partial B_{R_4}\cap P$ and $\mu\geq 1$. According to
Lemma \ref{lem1.2}, we have
\begin{equation}
 i(T,\ B_{R_4}\cap P,\ P)=1.\label{e3.12}
\end{equation}
By the second part of (H3'), $m_1<\liminf_{x\to
+\infty}\frac{f(x)}{x}\leq +\infty$, there exist $0<r_1<1$,
$\varepsilon_5>0$ such that $0<x<r_1$  implies
$$
\frac{f(x)}{x}\geq (m_1+\varepsilon_5)x.
$$
Let $B_{r_1}=\{x\in C[0,1]: \|x\|<r_1\}$. We now prove that
$$
x-Tx\neq\mu t^2(1-t)^2,\ \text{for}\ \mu\geq 0\ \text{and}\
 x\in \partial B_{R_1}\cap P.
$$
If not, there are $\mu_2\geq $ and $x_2\in \partial
B_{r_1}\cap P$ such that $x_2-Tx_2=\mu_2 t^2(1-t)^2$. So $\mu_2>0$,
otherwise there is a fixed point in $\partial B_{r_1}\cap P$ and
this would complete the proof. Let
$\eta=\min_{t\in[\theta,1-\theta]}x_2(t)$. Then if
$t\in[\theta,1-\theta]$, we have
\begin{align*}
x_2(t)&= \int_0^1G(t,s)p(s)f(x_2(s))ds+\mu_2t^2(1-t)^2\\
&\geq \int_\theta^{1-\theta}G(t,s)p(s)f(x_2(s))ds+\mu_2t^2(1-t)^2\\
&\geq (m_1+\varepsilon_5)\int_\theta^{1-\theta}G(t,s)p(s)x_2(s)ds
+\mu_2t^2(1-t)^2\\
&\geq \eta(m_1+\varepsilon_5)\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_2t^2(1-t)^2\\
&\geq \eta+\eta\varepsilon_5\int_\theta^{1-\theta}G(t,s)p(s)ds+\mu_2t^2(1-t)^2.
\end{align*}
Therefore,
$$
x_2(t)>\eta,\quad t\in [\theta,1-\theta].
$$
which is a contradiction. According to Lemma 1.1, we get
\begin{equation}
i(T,\ B_{r_1}\cap P,\ P)=0.\label{e3.13}
\end{equation}
By \eqref{e3.12} and \eqref{e3.13}, we have
$$
i(T,\ (B_{R_4}\overline{\backslash{B_{r_1}}})\cap P,\ P) =i(T,\
B_{R_4}\cap P,\ P)-i(T,\ B_{r_1}\cap P,\ P)=1.
$$
Then $T$ has at least a fixed point $x^*$ in
$(B_{R_4}\overline{\backslash{B_{r_1}}})\cap P$, satisfying
$0<r_1\leq \|x^*\|\leq R_4$, and $x^*$ is also a $C^3[0,1]\cap
C^4(0,1)$ positive solution of \eqref{e1.1}-\eqref{e1.2}. This
completes the proof.
\end{proof}

\begin{corollary} \label{coro2}
Let $p$ be as above,
$0<\int_0^1s^2(1-s)^2p(s)ds<+\infty$, and
 $0<\lambda<1$. Then   a necessary
 and sufficient condition for \eqref{e1.3}-\eqref{e1.4} to have a
positive solution in
 $C^3[0,1]\cap C^4(0,1)$ is that
$$
0<\int_0^1(s(1-s))^{2\lambda}p(s)ds<+\infty.
$$
\end{corollary}

\begin{example} \label{ex1} \rm
The singular boundary-value problem
\begin{gather*}
x''''(t)=t^{-5/2}(1-t)^{-4/3}x^{\lambda},\quad
 t\in(0,1),\; \lambda>1,\\
x(0)=x(1)=x'(0)=x'(1)=0,
\end{gather*}
has a solution $x\in C^3[0,1]\cap C^4(0,1)$ with $x(t)>0$ on
$(0,1)$.
To see this, we will apply Theorem \ref{thm3.1} with
$p(t)=t^{-5/2}(1-t)^{-4/3}$, $f(x)=x^{\lambda}$
$(\lambda>1)$.
Clearly (H1) holds. Note that
$$
\int_0^1p(s)s^2(1-s)^2ds=\int_0^1s^{-1/2}(1-s)^{2/3}ds\leq 2.
$$
Consequently (H2) holds (with $\theta=1/4$). Also note that
(H3) holds since
$$
\lim_{x\to 0+}\frac{f(x)}{x}=0, \quad
\lim_{x\to +\infty}\frac{f(x)}{x}=+\infty.
$$
Finally note that
$\int_0^1p(s)f(s^2(1-s)^2)ds=\int_0^1p(s)(s(1-s))^{2\lambda}ds<+\infty$.
The result now follows from Theorem \ref{thm3.1}.
\end{example}

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\end{document}