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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 105, 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 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/105\hfil Existence of positive solutions]
{Existence of positive solutions for singular eigenvalue problems}

\author[M. Feng, W. Ge\hfil EJDE-2006/105\hfilneg]
{Meiqiang Feng, Weigao Ge}  % in alphabetical order

\address{Meiqiang Feng\newline
Department of Applied Mathematics, Beijing Institute of Technology,
Beijing 100081, China\newline
Department of Fundamental Sciences, Beijing Information Technology
Institute, Beijing 100101, China}
\email{meiqiangfeng@sina.com}

\address{Weigao Ge \newline
Department of Applied Mathematics, Beijing Institute of Technology,
Beijing 100081, China}
\email{gew@bit.edu.cn}

\date{}
\thanks{Submitted May 10, 2006. Published September 8, 2006.}
\thanks{Supported by grants 10371006 from the National Nature Science
Foundation of China, and \hfill\break\indent
20050007011 from the Doctoral Program
Foundation of Education Ministry of China}
\subjclass[2000]{34B15}
\keywords{Positive solution; nonexistence; existence;
 complete continuity; \hfill\break\indent singularity}

\begin{abstract}
 In this paper, we discuss  the existence, nonexistence,
 and multiplicity of positive solutions  for a class of singular
 eigenvalue problems. Some of our theorems are new, while others
 extend earlier results obtained by Zhang and Kong \cite{z1}.
 The interesting point is that the authors obtain the relation
 between the existence of solutions and the parameter $\lambda$.
 The arguments are based on the fixed point index theory and
 the upper and lower solutions method.
\end{abstract}

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

\section{Introduction}

 The deformations of an elastic beam are described by a fourth-order
 differential equation
$$
u^{(4)}=f(t,u,u'').
$$
 Most of the available literature on fourth-order boundary value problems,
for example \cite{a1,a2,g1,g3,j1,o1,w1}, discusses the case when $f$ is either
continuous or a Caratheodory function
  and is concerned with the existence and uniqueness
  of positive solutions for boundary value problems for the above
  differential equation.
 However, only a small number of papers have discussed fourth-order singular
  eigenvalue problems; see  for example \cite{y1,z3}.

 In this paper, we study
  the  fourth-order singular differential equation
\begin{equation}
  u^{(4)}(t)=\lambda g(t)f(u(t)),\quad  0< t <1 , \label{e1.1}
\end{equation}
  subject to one of the following boundary conditions:
\begin{gather}
u(0)=u(1)=u''(0)=u''(1)=0,  \label{e1.2}\\
u(0)=u'(1)=u''(0)=u'''(1)=0,  \label{e1.3}
\end{gather}
where  $\lambda > 0 $. The following assumptions will stand throughout
 this paper:
\begin{itemize}
\item[(H1)] $f\in C([0,+ \infty ),  (0,+ \infty ))$ and is
nondecreasing on $[0,+\infty)$. Furthermore,
there exist $\bar \delta>0,m\geq 2$ such that
$f(u)>\bar \delta u^{m},u\in[0,+\infty)$;

\item[(H2)] $g \in C((0,1),  (0,+ \infty ))$ and
$0<\int_{0}^{1}s(1-s)g(s)ds<+\infty$.

\end{itemize}
 It is the purpose of this paper to obtain the existence and the
nonexistence of positive solutions, and
multiplicity results for the eigenvalue problems (EP)
\eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3} by employing
new technique (different from the one used in \cite{z1}).
Very few papers discuss the connection between the existence of
solutions and the parameter $\lambda$. The work done by others \cite{z3}
 does not cover the general case given in \eqref{e1.1}-\eqref{e1.2} and
\eqref{e1.1}-\eqref{e1.3}.

 In this paper, we use mainly the following fixed point index theory to
obtain multiplicity results for
\eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3}.

\begin{lemma}[\cite{g2}] \label{lem1.1}
Let $P$ be a cone in a real Banach space $E$ and
 $\Omega$ be a bounded open subset of $E$ with $\theta \in \Omega$.
Suppose $A:P \cap \bar \Omega \to P $ is a  completely
continuous operator, that satisfies
$$
Ax=\mu x,\; x \in P \cap \partial\Omega \Longrightarrow \mu<1.
$$
Then $ i(A,P \cap \Omega ,P)=1$.
\end{lemma}

\begin{lemma}[\cite{g2}] \label{lem1.2}
 Suppose $A:P \cap \bar \Omega \to P $ is a completely
continuous operator, and satisfies:
\begin{enumerate}
\item $\inf_{x \in P \cap \partial\Omega}\|Ax\|>0$;
\item $Ax=\mu x,x \in P \cap \partial\Omega \Longrightarrow \mu\not\in(0,1].$\\
Then $i(A,P \cap \Omega ,P)=0$.
\end{enumerate}
\end{lemma}

In Section 2, we provide some necessary background. In particular, we state
 some properties of the Green's
function associated with   \eqref{e1.1}-\eqref{e1.2} and some
Lemmas. In Section 3, we present our main result and discuss an
example.

\section{Preliminaries}

 For the convenience of the reader, we present here the necessary
 definitions and Lemmas.
  Let $E=C[0,1]$ be a real Banach space with the norm
$\|u\|=\max_{0 \leq t\leq 1}|u(t)|$. Let
$S=\{\lambda >0\text{ such that \eqref{e1.1} has at least one solution}\}$
 and
$P=\{u\in E : u(t)\geq 0,t\in [0,1]\}$. It is clear that $P$ is a cone
of $E$.

 We  deal first with  \eqref{e1.1}-\eqref{e1.2}.
 Define
 $$
G_{1}(t,\xi)=\begin{cases}
   t(1-\xi), & 0\leq t \leq \xi \leq 1,\\
   \xi(1-t), & 0\leq \xi \leq t \leq 1.
\end{cases}
$$
   \begin{align*}
G(t,s)&=\int_{0}^{1}G_{1}(t,\xi)G_{1}(\xi,s)d \xi  \\
 &=\begin{cases}
   t(1-s)\frac{2s-s^2-t^2}{6}, \quad 0\leq t \leq s \leq 1,\\
   s(1-t)\frac{2t-t^2-s^2}{6}, \quad 0\leq s \leq t \leq 1.
  \end{cases}
\end{align*}
 It is easy to prove that
 $G_{1}(t,s)$ and $G(t,s)$ have the following properties.

\begin{proposition} \label{prop2.1}
 For all  $ t,s\in[0,1]$, we have
\begin{equation}
\begin{gathered}
 G_{1}(t,s)>0,\quad\text{for }(t,s)\in (0,1)\times(0,1); \\
 G_{1}(t,s)\leq G_{1}(s,s)=s(1-s),\quad \text{for }  0\leq t, s \leq 1;\\
 0\leq G_{1}(t,s)\leq \frac{1}{4},\quad\text{for }  0\leq t, s \leq 1;\\
 G(t,s)\leq  \frac{1}{6}G_{1}(s,s)=\frac{1}{6}s(1-s),\quad
\text{for } 0\leq t, s \leq 1.
\end{gathered} \label{e2.1}
\end{equation}
\end{proposition}

\begin{proposition} \label{prop2.2}
For all $t \in [\theta,1-\theta]$, we have
\begin{equation}
G_{1}(t,s)\geq \theta G_{1}(s,s),\quad
\theta \in (0,\frac{1}{2}),\; s\in [0,1].  \label{e2.2}
\end{equation}
 In fact
$$
\frac{G_{1}(t,s)}{G_{1}(s,s)}=  \begin{cases}
               \frac{t}{s}, & 0\leq t \leq s \leq 1,\\
              \frac{1-t}{1-s},& 1\geq t \geq s \geq 0.
  \end{cases}
\geq \begin{cases}
t\geq \theta, & t\leq s,\\
1-t\geq \theta, &t\geq s.
\end{cases}
$$
Therefore, for all $t \in [\theta,1-\theta]$, we have
$$
G_{1}(t,s)\geq \theta G_{1}(s,s),\quad \theta \in (0,\frac{1}{2}),\;
s \in [0,1].
$$
\end{proposition}

\begin{definition} \label{def2.1} \rm
 Let $\alpha(t)\in C^2[0,1] \cap C^4(0,1)$.
We say that $\alpha$ is a lower solution of
\eqref{e1.1}-\eqref{e1.2}  if it satisfies
\begin{gather*}
     \alpha^{(4)}(t)\leq \lambda g(t)f(u(t)), \quad  0< t <1 ,   \\
\alpha(0)\leq 0, \alpha(1)\leq 0, \alpha''(0)\geq 0 ,\quad \alpha''(1)\geq 0.
\end{gather*}
\end{definition}

\begin{definition} \label{def2.2}\rm
Let $\beta(t)\in C^2[0,1] \cap C^4(0,1)$. We
say that $\beta$ is an upper solution of
\eqref{e1.1}-\eqref{e1.2}  if it satisfies
\begin{gather*}
     \beta^{(4)}(t)\geq \lambda g(t)f(u(t)), \quad  0< t <1 ,   \\
     \beta(0)\geq 0, \beta(1)\geq 0, \beta''(0)\leq 0 ,\quad
\beta''(1)\leq 0.
\end{gather*}
\end{definition}

First, we consider the following eigenvalue problem
\begin{equation}
\begin{gathered}
     u^{(4)}(t)=\lambda g(t)f(u(t)), \quad  0< t <1 ,   \\
    u(0)=u(1)=u''(0)=u''(1)=h\geq 0.
\end{gathered} \label{e2.3}
\end{equation}
 Define $T_{\lambda}^h: E \to E $ by
\begin{equation}
T_{\lambda}^h u(t) =h+\int_{0}^{1}G(t,s)\lambda
g(s)f(u(s))ds-\int_{0}^{1}G_{1}(t,s)hds. \label{e2.4}
\end{equation}
 From \eqref{e2.4}, it is easy to obtain the following lemma,
which is proved by a direct computation.

\begin{lemma} \label{lem2.1}
 Suppose that (H1) and (H2) are satisfied.
 Then \eqref{e1.1}-\eqref{e1.2} has a solution $u$ if and only if $u$ is
a fixed point of  $T_{\lambda}^0 $.
\end{lemma}

To prove the following results we define the cone
\begin{equation}
Q=\{u\in C[0,1]|u(t)\geq 0,\min_{\theta\leq t \leq 1-
\theta}u(t) \geq M_{\theta}\|u\|\} \label{e2.5}
\end{equation}
where $\|u\|=\max_{t\in [0,1]}|u(t)|$,
$M_{\theta}=\theta^2(1-6\theta^2+4\theta^3)$,
$\theta \in (0,\frac{1}{2})$.
 It is clear that $Q\subset P$.

\begin{lemma} \label{lem2.2}
 Suppose that (H1) and (H2) are
satisfied. Then $T_{\lambda}^{0}Q\subset Q$
 is completely continuous and nondecreasing.
\end{lemma}

\begin{proof} For any $u\in P$, by \eqref{e2.1} and \eqref{e2.4},  we have
\begin{align*}
T_{\lambda}^0 u(t)&=\int_{0}^{1}G(t,s)\lambda g(s)f(u(s))ds \\
&\leq\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds.
\end{align*}
Therefore,
\[
\|T_{\lambda}^0 u\|\leq\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds.
\]
On the other hand, by \eqref{e2.2}, for any $\theta\leq t\leq 1-\theta$,
we have
\begin{equation}
G(t,s)=\int_{0}^{1}G_{1}(t,\xi)G_{1}(\xi,s)d \xi \geq
M_{\theta}\frac{1}{6}s(1-s).\label{e2.6}
\end{equation}
Therefore,
\begin{align*}
\min_{\theta\leq t\leq 1-\theta}T_{\lambda}^0 u(t)
&=\min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)\lambda g(s)f(u(s))ds\\
& \geq M_{\theta}\frac{1}{6}\int_{0}^{1}\lambda s(1-s)g(s)f(u(s))ds \\
& \geq M_{\theta}\|T_{\lambda}^0 u\|.
\end{align*}
 Hence $T_{\lambda}^0 P\subset Q$ and then
$T_{\lambda}^0 Q\subset  Q $ by $Q\subset P$.
 By similar arguments in \cite{a2,o1,z1,z3},
 $T_{\lambda}^0: Q\to Q$ is completely continuous.
 Since f is increasing
 on $[0,+ \infty)$, it is easy to obtain that $T_{\lambda}^{0}$ is
nondecreasing on $[0,+ \infty)$.
\end{proof}

\begin{remark} \label{rmk2.1}\rm
 Reasoning as in the proofs of  Lemmas \ref{lem2.1} and \ref{lem2.2},
we conclude that
$ T_{\lambda}^{h}:Q\to Q$  is completely continuous and that $u(t)$
is a solution of \eqref{e2.3} if and only if $u(t)$
 is a fixed point of $ T_{\lambda}^{h}$.
\end{remark}

\begin{lemma} \label{lem2.3}
Suppose that $\lambda \in S,S_{1}=(\lambda,+\infty)\cap S\not\equiv \emptyset$.
Then there exists $R(\lambda )>0$, such that
$\|u_{\lambda'}\|\leq R(\lambda )$, where $\lambda'\in S_{1}$,
and $u_{\lambda'}\in Q$ is a solution of
\eqref{e1.1}-\eqref{e1.2} with $\lambda'$ instead of $\lambda$.
\end{lemma}

\begin{proof}
For any $\lambda'\in S$, let $u_{\lambda'}$ be a
solution of \eqref{e1.1}-\eqref{e1.2} with $\lambda'$ instead of $\lambda$.
Then
  $$
u_{\lambda'}(t)= T_{\lambda'}^{0}u_{\lambda'}(t)
  = \int_{0}^{1}G(t,s)\lambda'g(s)f(u_{\lambda'}(s))ds.
$$
Let $R(\lambda)
=\max\{[\frac{1}{6}\lambda'M_{\theta}^{m+1}\bar\delta
\int_{\theta}^{1-\theta}G_{1}(s,s) g(s)ds]^{-1},1\}$.
Next we shall prove that $\|u_{\lambda'}\|\leq R(\lambda )$.
Indeed, if $\|u_{\lambda'}\|< 1$, the result is easily
obtained. On the other hand,
  if $\|u_{\lambda'}\|\geq 1$, then we have by (H1) and \eqref{e2.6},
 \begin{align*}
\frac{1}{\|u_{\lambda'}\|}
&\geq \frac{\min_{\theta\leq t\leq 1-\theta}u_{\lambda'}(t)}
  {\|u_{\lambda'}\|^{2}}\\
& = \frac{1}{\|u_{\lambda'}\|^{2}}\min_{\theta\leq t\leq 1-\theta}
  \int_{0}^{1}G(t,s)\lambda' g(s)f(u_{\lambda'}(s))ds\\
& \geq \frac{1}{\|u_{\lambda'}\|^{2}}M_{\theta}
  \int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)\lambda' g(s)
  \bar\delta u_{\lambda'}(s)^{m}ds\\
& \geq \frac{1}{\|u_{\lambda'}\|^{2}}M_{\theta}^{m+1}\int_{\theta}^{1-\theta}
  \frac{1}{6}G_{1}(s,s)\lambda' g(s)\bar\delta \|u_{\lambda'}\|^{m}ds\\
& \geq \frac{1}{6}\lambda'
M_{\theta}^{m+1}\bar\delta\int_{\theta}^{1-\theta}G_{1}(s,s)g(s)ds.
\end{align*}
Therefore, $\|u_{\lambda'}\|\leq R(\lambda)$ and the conclusion of
Lemma \ref{lem2.3} follows.
\end{proof}

\begin{lemma}[\cite{d1}] \label{lem2.4}
Suppose that $f:[0,+\infty) \to (0,+\infty)$ is continuous and increasing.
 If $s,s_{0}$ and $M$ are such that
  $0<s<s_{0},M>0$, then there exist $\bar s\in(s,s_{0}),h_{0}\in (0,1)$
such that
 $$
sf(u+h)<\bar s f(u),u\in[0,M],h\in(0,h_{0}).
$$
\end{lemma}

\section{Main results}

 In this section, we apply Lemmas \ref{lem1.1} and \ref{lem1.2}
to establish nonexistence and existence of positive solutions, as
well as multiplicity results
 for \eqref{e1.1}-\eqref{e1.2} and \eqref{e1.1}-\eqref{e1.3}.
 Our approach depends on
the upper and lower solutions method and the fixed point index
theory.
We deal with  \eqref{e1.1}-\eqref{e1.2} first.

\begin{theorem} \label{thm3.1}
Let (H1) and (H2) be satisfied.
 Then there exists $0<\lambda^{*}<+\infty$ such that
\begin{enumerate}
\item  EP \eqref{e1.1}-\eqref{e1.2} has no solution for $\lambda >\lambda^{*}$;
\item  EP \eqref{e1.1}-\eqref{e1.2} has at least one positive solution for
 $\lambda =\lambda^{*}$;
\item  EP \eqref{e1.1}-\eqref{e1.2} has  at least two positive solutions
for $0<\lambda <\lambda^{*}$.
\end{enumerate}
\end{theorem}

\begin{proof} First, we prove that the conclusion (1) of Theorem \ref{thm3.1}
 holds.  If $\beta(t)$ is a solution of the  boundary-value problem
\begin{equation}
\begin{gathered}
     u^{(4)}(t)=g(t) \quad 0< t <1 ,   \\
    u(0)=u(1)=u''(0)=u''(1)=0,
\end{gathered}  \label{e3.1}
\end{equation}
then,  by Lemma \ref{lem2.1}, we have
$\beta(t)=\int_{0}^{1}G(t,s)g(s)ds$.
 Let $\beta_{0}=\max_{t\in [0,1]}\beta(t)$;
 therefore, by (H1) and \eqref{e2.4},
$$
T_{\lambda}^{0}\beta(t)\leq T_{\lambda}^{0}\beta_{0}
=\int_{0}^{1}G(t,s)\lambda g(s)f(\beta_{0})ds<\beta(t),\quad \forall
0<\lambda<\frac{1}{f(\beta_{0})}.
$$
This implies that $\beta(t)$ is an upper solution of $T_{\lambda}^{0}$.
On the other hand, let $\alpha(t)\equiv 0,t\in [0,1]$, then $\alpha(t)$
is a lower solution of $T_{\lambda}^{0}$, and $\alpha(t)<\beta(t),t\in[0,1]$.
Clearly $T_{\lambda}^{0}$ is completely continuous on
$[\alpha,\beta]$. Therefore,
 $T_{\lambda}^{0}$ has a fixed point $u_{\lambda}\in [\alpha,\beta]$,
and therefore $u_{\lambda}$ is a solution of
  \eqref{e1.1}-\eqref{e1.2}. Hence, for any
$0<\lambda<\frac{1}{f(\beta_{0})}$, we have $\lambda \in S$,
which implies that  $S\not=\emptyset$.

 On the other hand, if $\lambda_{1}\in S $, then we must have
$(0,\lambda_{1})\subset S$.
In fact, let $u_{\lambda_{1}}$ be a solution of
\eqref{e1.1}-\eqref{e1.2}. Then,  by Lemma \ref{lem2.1}, we have
$$
u_{\lambda_{1}}(t)=T_{\lambda_{1}}^{0}u_{\lambda_{1}}(t),t\in [0,1].
$$
Therefore, for any $\lambda\in (0,\lambda_{1})$,  by \eqref{e2.4}, we have
\begin{align*}
T_{\lambda}^{0}u_{\lambda_{1}}(t)
&=\int_{0}^{1}G(t,s)\lambda g(s) f(u_{\lambda_{1}}(s)ds \\
& \leq \int_{0}^{1}G(t,s)\lambda_{1} g(s) f(u_{\lambda_{1}}(s)ds \\
& =T_{\lambda_{1}}^{0}u_{\lambda_{1}}(t)\\
& =u_{\lambda_{1}}(t),
\end{align*}
which implies that $u_{\lambda_{1}}$ is an upper solution of
$T_{\lambda}^{0}$. Combining this with the fact that
 $\alpha(t)\equiv 0$  ($t\in [0,1]$)  is a lower solution of
$T_{\lambda}^{0}$,  then, by Lemma \ref{lem2.1},  EP
\eqref{e1.1}-\eqref{e1.2} has a solution, therefore $\lambda \in
S$. Thus we have $(0,\lambda_{1})\subset S$.

  Let $\lambda^{*}=\sup S$, now we prove that $\lambda^{*}<+\infty$.
If this is not true, then we must have $N\subset S$, where $N$ denotes
natural number numbers. Therefore, for any $n\in N$, by Lemma \ref{lem2.1},
there exists $u_{n}\in Q $ satisfying
$$
u_{n}=T_{n}^{0}u_{n}
=\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds.
$$
Let $K=[\frac{\bar\delta M_{\theta}^{m+1}}{6}
\int_{\theta}^{1-\theta}G_{1}(s,s) g(s) ds]^{-1}$. Suppose
 $\|u_{n}\|\geq 1$. Then we have
\begin{align*}
1&\geq\frac{1}{\|u_{n}\|}\\
&\geq\frac{\min_{\theta\leq t\leq 1-\theta}u_{n}(t)}{\|u_{n}\|^{2}}\\
&= \frac{1}{\|u_{n}\|^2}
\min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds\\
&\geq \frac{1}{\|u_{n}\|^2}M_{\theta}
\int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)\bar\delta u_{n}(s)^m ds\\
&\geq \frac{1}{\|u_{n}\|^2}M_{\theta}^{m+1}
\int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)\bar\delta \|u_{n}\|^m ds\\
&\geq \frac{1}{6}n M_{\theta}^{m+1}\bar\delta
\int_{\theta}^{1-\theta}G_{1}(s,s) g(s) ds.
\end{align*}
If $\|u_{n}\|\leq 1$, then
\begin{align*}
1\geq\|u_{n}\|
&\geq\min_{\theta\leq t\leq 1-\theta}\int_{0}^{1}G(t,s)n g(s) f(u_{n}(s)ds \\
&\geq M_{\theta}
\int_{\theta}^{1-\theta}\frac{1}{6}G_{1}(s,s)n g(s)f(0)ds.
\end{align*}
Hence $n\leq \{K,(M_{\theta} \int_{\theta}^{1-\theta}\frac{1}{6}
G_{1}(s,s)g(s)f(0)ds)^{-1}\}$,
this contradicts the fact that $N$ is unbounded; therefore
$\lambda^{*}<+\infty$,
 and  the proof of the conclusion (1) is complete.

Secondly, we verify the conclusion (2) of Theorem \ref{thm3.1}.
Let $\{\lambda_{n}\}\subset [\frac{\lambda^{*}}{2},
\lambda^{*}),\lambda_{n}\to \lambda^{*}(n\to
\infty)$, $\{\lambda_{n}\}$ be an increasing sequence. Suppose
$u_{n}$ is solution of \eqref{e1.1} with $\lambda_n$ instead of $\lambda$.
 By Lemma \ref{lem2.3}, there
exists $R(\frac{\lambda^{*}}{2})>0$ such that
$\|u_{n}\|\leq R(\frac{\lambda^{*}}{2})$, $n=1,2,\cdots$. Hence $u_{n}$ is a
bounded set. It is clear that $\{u_{n}\}$ is an equicontinuous set
of $C[0,1]$.
 Therefore, by the Ascoli-Arzela theorem, it follows that
 $\{u_{n}\}$ is compact set, and therefore $\{u_{n}\}$  has a convergent
subsequence.
 Without loss of generality, we
 suppose that $u_{n}$ is convergent: $u_{n}\to u^{*}(n\to +\infty)$.
 Since $u_{n}=T_{\lambda_{n}}^{0}u_{n}$,
 by control convergence theorem ($f$ is bounded),
we have $u^{*}=T_{\lambda^{*}}^{0}u^{*}$. Therefore, by Lemma \ref{lem2.1},
   $u^{*}$ is a solution of \eqref{e1.1}-\eqref{e1.2} with
$\lambda^*$ instead of $\lambda$.
 Hence the conclusion (2) of Theorem \ref{thm3.1} holds.

 Finally, we prove the conclusion (3) of Theorem \ref{thm3.1}.
 Let $\alpha(t)\equiv h$   $(t\in[0,1])$. Then for any
$\lambda \in (0,\lambda^{*})$,
 $\alpha(t)$ is a lower solution of  \eqref{e2.3}.
On the other hand, by Lemma \ref{lem2.3}, there exists $R(\lambda)>0$ such that
$\|u_{\lambda'}\|\leq R(\lambda)$, $ \lambda'\in [\lambda, \lambda^{*}]$,
where $u_{\lambda'}$ is a solution of \eqref{e1.1} with $\lambda'$ instead of
$\lambda$.
Also by Lemma \ref{lem2.4}, there exist
$\bar \lambda \in [\lambda, \lambda^{*}],h_{0}\in(0,1)$ satisfying
 $$
\lambda f(u+h)<\bar \lambda f(u),u\in[0,R(\lambda)],h\in (0,h_{0}).
$$
 Let $u_{\bar \lambda}$ be a solution of  \eqref{e1.1}-\eqref{e1.2}
with $\bar \lambda$,  and
 $\bar u_{\lambda}(t)=u_{\bar \lambda}+h$, $h\in (0,h_{0})$. Then
\begin{align*}
\bar u_{\lambda}(t)
&=u_{\bar \lambda}+h\\
& =\int_{0}^{1}G(t,s) \bar \lambda g(s) f(u_{\bar \lambda}(s))ds+h\\
& \geq h+\int_{0}^{1}G(t,s) \lambda g(s) f(u_{\bar \lambda}(s)+h)ds-\int_{0}^{1}G_{1}(t,s)hds\\
& = T_{\lambda}^{h}\bar u_{\lambda}(t).
\end{align*}
Combining this with $\bar u_{\lambda}(0)=\bar u_{\lambda}(1)\geq h$,
$\bar u_{\lambda}''(0)=0\leq h$,
$\bar u_{\lambda}''(1)=0\leq h$, we have the $\bar u_{\lambda}(t)$
is an upper solution of \eqref{e2.3}. Therefore
\eqref{e2.3} has solution.
Let  $v_{\lambda}(t)$ be a solution of \eqref{e2.3}. Let
 $\Omega=\{u\in Q |u(t)<v_{\lambda}(t),t\in [0,1]\}$.
It is clear that $\Omega \subset Q$ is a bounded  open set. If
$u\in \partial \Omega$, then there exists $t_{0}\in[0,1]$, such
that $ u(t_{0})=v_{\lambda}(t_{0})$. Therefore, for any
$\mu \geq 1$, $h\in (0,h_{0})$, $u\in \partial \Omega $, we have
\begin{align*}
T_{\lambda}^{0}u(t_{0})
&< h + T_{\lambda}^{0}u(t_{0})- \int_{0}^{1}G_{1}(t,s)hds\\
& \leq h + T_{\lambda}^{0}v_{\lambda}(t_{0})-\int_{0}^{1}G_{1}(t,s)hds\\
& =T_{\lambda}^{h}v_{\lambda}(t_{0})\\
& =v_{\lambda}(t_{0})\\
& =u(t_{0})\\
& \leq \mu u(t_{0}).
\end{align*}
Hence for any $\mu \geq 1$, we have
$T_{\lambda}^{0}u\neq \mu u$, $u\in \partial \Omega$.
Therefore,  by Lemma \ref{lem1.1},
\begin{equation}
i(T_{\lambda}^{0},\Omega,Q)=1. \label{e3.2}
\end{equation}

It remains to prove that the conditions of Lemma \ref{lem1.2} are satisfied
Firstly, we check the condition (1) of Lemma \ref{lem1.2} is fulfilled.
In fact, for any $u\in Q$, we have by (H1) and \eqref{e2.5},
\begin{equation}
\begin{aligned}
T_{\lambda}^{0}u(\frac{1}{2})
&=\int_{0}^{1}G(\frac{1}{2},s) \lambda g(s) f(u(s))ds\\
& \geq \int_{\theta}^{1-\theta}G(\frac{1}{2},s)
  \lambda g(s)\bar \delta M_{\theta}^m\|u\|^m ds\\
& =\|u\|^m\int_{\theta}^{1-\theta}G(\frac{1}{2},s) \lambda g(s)\bar
  \delta M_{\theta}^m ds\\
& =\|u\|^{m-1}\int_{\theta}^{1-\theta}G(\frac{1}{2},s) \lambda g(s)\bar
  \delta M_{\theta}^m ds\|u\|
\end{aligned}
\label{e3.3}
\end{equation}
Choose $\bar R>0$ such that $\bar
R^{m-1}\int_{\theta}^{1-\theta}G(\frac{1}{2},s) \lambda g(s)\bar
\delta M_{\theta}^mds>1$. Therefore, for any $R>\bar R$ and
$B_{R}\subset Q$, by \eqref{e3.3},
\begin{equation}
\|T_{\lambda}^{0}u\|>\|u\| >0,u\in \partial B_{R}, \label{e3.4}
\end{equation}
 where $B_{R}=\{u\in Q| \|u\|<R\}$.
 Hence the condition (1) of Lemma \ref{lem1.2} is fulfilled.

 Now we prove that the condition (2) of Lemma \ref{lem1.2} is satisfied.
 In fact, if the condition (2) of Lemma \ref{lem1.2} does not hold, then there exist
 $u_{1}\in Q\cap\partial B_{R},0<\mu_{1}\leq 1$, such that
  $T_{\lambda}^{0}u_{1}=\mu_{1}u_{1}$.
  Therefore, $\|T_{\lambda}^{0}u_{1}\|\leq \|u_{1}\|$. This conflicts
with \eqref{e3.4}.
 Hence the condition (2) of Lemma \ref{lem1.2} is satisfied. Therefore
by Lemma \ref{lem1.2},
 we  have
\begin{equation}
i(T_{\lambda}^{0}, B_{R},Q)=0. \label{e3.5}
\end{equation}
  Consequently, by the additivity of the fixed point index, we get
 $$
0=i(T_{\lambda}^{0},B_{R},Q)=i(T_{\lambda}^{0},\Omega,Q)
+i(T_{\lambda}^{0},B_{R}\setminus\bar \Omega,Q).
$$
 Since $i(T_{\lambda}^{0},\Omega,Q)=1$,
$i(T_{\lambda}^{0},B_{R}\setminus\bar \Omega,Q)=-1$.
 Therefore, by the solution property of the fixed point
  index, there is a fixed point of $T_{\lambda}^{0}$ in $\Omega$ and
   a fixed point of $T_{\lambda}^{0}$ in $B_{R}\setminus \bar \Omega$, respectively.
 Therefore by Lemma \ref{lem2.1}, EP \eqref{e1.1}-\eqref{e1.2}  has at least
two solutions.
 Furthermore,
 \eqref{e1.1}-\eqref{e1.2}  has at least two positive solutions by (H1) and (H2).
  The proof of Theorem \ref{thm3.1} is complete.
\end{proof}

  Now we study  \eqref{e1.1}-\eqref{e1.3}. The method is  similar to the
method above.  Define
$$
\hat{G}(t,s)=\min\{t,s\}=
\begin{cases}
    t, & t \leq s, \;      0\leq t\leq s \leq 1, \\
    s,  &   s \leq t, \;     0\leq s\leq t \leq 1,
    \end {cases}
$$
\begin{align*}
  \tilde G(t,s)
   &=\int_{0}^{1}\hat{G}(t,r)\hat{G}(r,s)dr\\
   &=   \begin{cases}
    \frac{s^3}{3}+\frac{s(t^2-s^2)}{2}+st(1-t),  & 0\leq s\leq t \leq 1,  \\
    \frac{t^3}{3}+\frac{t(s^2-t^2)}{2}+ts(1-s),  & 0\leq t\leq s \leq 1.
 \end{cases}
 \end{align*}
It is easy to prove that $\hat{G}(t,s)$ and $\tilde G(t,s) $
 have the following properties.

\begin{proposition} \label{prop3.1}
For all  $ t,s\in[0,1]$, $\alpha \in(0,\frac{1}{2})$ we have
\begin{gather*}
\hat{G}(t,s)>0, \quad  t,s \in (0,1),\\
\hat{G}(t,s) \leq \hat{G}(s,s) = s,  \quad  t,s \in [0,1],\\
\hat{G}(t,s) \geq \alpha \hat{G}(s,s) ,\quad  t \in[\alpha,1-\alpha],\;
 s \in [0,1],\\
\tilde G(t,s) \leq \frac{1}{2}s, \quad  t,s \in [0,1];\\
\tilde G(t,s) \geq \frac{1}{2}M_{\alpha}s, \quad
 t \in [\alpha,1-\alpha],\;  s \in [0,1]
\end{gather*}
 where $M_{\alpha}= \alpha^{2}(1-2\alpha)$.
\end{proposition}

Define the cone
$$
\hat Q=\{u\in C[0,1]|u(t)\geq 0,\min_{\alpha\leq t \leq 1-
\alpha}u(t) \geq M_{\alpha} \|u\|\}
$$
and let
\begin{itemize}
\item[(H3)] $g \in C((0,1),  (0,+ \infty ))$ and
    $0<\int_{0}^{1}s g(s)ds<+\infty$
\end{itemize}

\begin{theorem} \label{thm3.2}
Let (H1) and (H3) be satisfied.
 Then there exists $0<\lambda^{*}<+\infty$ such that:
\begin{enumerate}
\item  EP \eqref{e1.1}-\eqref{e1.3} has no solution for $\lambda >\lambda^{*}$;
\item  EP \eqref{e1.1}-\eqref{e1.3} has at least one positive solution
for $\lambda =\lambda^{*}$;
\item  EP \eqref{e1.1}-\eqref{e1.3} has  at least two positive solutions
for $0<\lambda <\lambda^{*}$.
\end{enumerate}
\end{theorem}

As an example we consider the eigenvalue problem
\begin{equation}
\begin{gathered}
    u^{(4)}(t)=\lambda \frac{1}{t(1-t)}2^{2u},\quad  0<t<1,\\
    u(0)=u(1)=u''(0)=u''(1)=0.
\end{gathered} \label{e3.6}
\end{equation}
It is clear that  \eqref{e3.6} is not covered by the results
in  \cite{a1,a2,g1,g3,j1,o1,w1,y1,z1,z2,z3}.

 Let $g(t)=\frac{1}{t(1-t)}$, $f(u)=2^{2u}$. It is obvious that $g(t)$
is singular at both $t=0$ and at $t=1$.
However, hypothesis (H2) is satisfied.
In addition, for $\bar\delta = 1 >0, m=2 $,
 we have   $f(u)=2^{2u}=\bar\delta 2^{2u}>u^{2}=u^{m}>0$.
So that (H1) is satisfied.


\subsection*{Acknowledgments}
The author is thankful to the referee for his/her valuable suggestions
regarding the original manuscript.

\begin{thebibliography}{00}

\bibitem{a1} R. P. Agarwal, \emph{Some new results on two point boundary value
problems  for higher order differential equations},
Funkcial. Ekvac. 29(1986), 197-212.

\bibitem{a2} A. R. Aftabizadeh, \emph{Existence and uniqueness theorems for
fourth-order boundary value problem}, J. Math. Anal. Appl., 116(1986), 415-426.

\bibitem{c1} Y. S. Choi, \emph{A singular boundary value problem arising from
near-ignition analysis  of flame structure}, Diff. Intergral Eqns, 4(1991),
891-895.

\bibitem{d1} R. Dalmasso, \emph{Positive solutions of singular boundary value
problems}. Nonlinear  Analysis,1996, 27(6), 645-652.

\bibitem{g1} A. Granas, R. B. Guenther, and J. W. Lee;
\emph{Nonlinear boundary value problems for
 some classes of ordinary differential equations},
Rocky Mountain J. Math. 10(1980), 35-58.

\bibitem{g2} D. Guo, V. Lakshmikantham, X. Z. Liu;
\emph{Nonlinear integral equations in abstract
  cones, New York}; Academic Press, 1988.

\bibitem{g3} C. P. Gupta, \emph{Existence and uniqueness theorems
for the bending of an elastic beam  equation}, Appl.Anal., 26(1988), 289-304.

\bibitem{j1} L. K. Jackson, \emph{Existence and uniqueness of solutions of
boundary  value problems  for Lipschitz equations},
 J. Differential Equations 32(1979), 76-90.

\bibitem{o1} D. O'Regan,  \emph{Solvability of some fourth (and higher)
order singular boundary value problems}, J. Math. Anal. Appl., 161(1991),
78-116.

\bibitem{w1} Z. Wei, \emph{Positive solutions of singular boundary value
problems of fourth order differential equations} (in  Chinese), Acta
Mathematica Sinica, 1999, 42(4), 715-722.

\bibitem{y1} Q. Yao, Z. Bai, \emph{The existence of positive solutions of
boundary value  problem $u^{(4)} (t)-\lambda h(t)f(u(t)) =
0$} (in chinese), Chinese Annals of Mathematics (A), 1999, 20(5),
575-578.

\bibitem{z1} B. Zhang , L. Kong, \emph{Positive solutions of fourth order
singular boundary value problems}, Nonlinear Studies, 7(2000), 70-77.

\bibitem{z2} B. Zhang , L. Kong, \emph{Existence of positive solutions
of fourth order singular boundary value problems},
Chinese Annals of Mathematics, 22A: 4(2001), 397-402.

\bibitem{z3} Y. Zhou, \emph{Positive solutions of fourth order nonlinear
eigenvalue problems} (in chinese),  J. Sys. Sci. \&  Math. Scis,
2004, 24(4): 433-442.

\end{thebibliography}
\end{document}