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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 147, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2013/147\hfil Positive solutions]
{Positive solutions for a third-order three-point boundary-value problem}

\author[F. J. Torres \hfil EJDE-2013/147\hfilneg]
{Francisco J. Torres}  % in alphabetical order

\address{Francisco J. Torres \newline
Departmento de Matem\'atica, Universidad de Atacama,
Avenida Copayapu 485, Copiap\'o, Chile}
\email{ftorres@mat.uda.cl} 

\thanks{Submitted March 29, 2013. Published June 27, 2013.}
\thanks{Partially supported by DIUDA grant 221181, Universidad de Atacama.}
\subjclass[2000]{34B18, 37C25}
\keywords{Cone; fixed point index; fixed point theorem; positive solution}

\begin{abstract}
 In this article, we study the existence of positive solutions to a
 nonlinear third-order three point boundary value problem. The main
 tools are Krasnosel'skii fixed point theorem on cones, and the
 fixed point index theory.
\end{abstract}

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

\section{Introduction}

In this article, we are interested in the existence of single and
multiple positive solutions to nonlinear third-order three-point
boundary-value problem
\begin{gather}\label{ecua}
  u'''(t)+a(t)f(t,u(t))=0 ,\quad 0<t<1,\\
\label{bord}
 u(0)=0, \quad u'(0)=u'(1)=\alpha u(\eta)\,,
\end{gather}
where $ \eta \in (0,1)$, $\alpha \in [0,\frac{1}{\eta})$.
We assume the following conditions hold in this article:
\begin{itemize}
\item[(H1)] $f\in C([0,1]\times[0,\infty), [0,\infty))$.
\item[(H2)] $a\in L^{1}[0,1]$ is nonnegative and $a(t)\not\equiv 0$ on any subinterval of $[0,1]$.
\end{itemize}

Third-order differential equation arise in a variety of different
areas of applied mathematics and physics, as the deflection of a
curved beam having a constant or varying cross section, three layer
beam, electromagnetic waves o gravity driven flows and so on. Li in
\cite{Li}  by using Krasnosel'skii fixed point theorem on cone
establish various results on the existence of positive solutions.
Sun \cite{Sun} use the Krasnosel'skii fixed point theorem and
Schauder's fixed point theorem to obtain existence and nonexistence
of positive solutions. In \cite{Liu} Liu et al obtain results for
the existence of at least one, two, three and infinitely many
monotone positive solutions by using Krasnosel'skii and
Leggett-Williams fixed point theorem. In \cite{Luan} Luan et al
obtain existence results under conditions that the nonlinear term
satisfies Carath\'eodory condition, semipositone and lower unbounded
by using the fixed point index theory. In \cite{Bai}, Bai the
nonlinear term depends on $u$, $u'$ and $u''$, prove the existence
of at least one solution with the use of lower and upper solutions
methods and Schauder fixed point theorem. Motivated by the above
works, we obtain some sufficient conditions for the existence of at
least one and two positive solutions for \eqref{ecua} and
\eqref{bord}.
The organization of this article is as follows. In section 2, we
present some necessary definitions and preliminary results that will
be used to prove our results. In section 3, we discuss the existence
of at least one positive solution for \eqref{ecua} and \eqref{bord}.
In section 4, we discuss the existence of multiple positive
solutions for \eqref{ecua} and \eqref{bord}. Finally, we give some
examples to illustrate our results in section 5.



\section{Preliminaries}

\begin{definition} \label{def2.1}\rm
Let $E$ be a real Banach space. A nonempty closed convex set
$K\subset E$ is called cone if
 \begin{enumerate}
 \item if $x\in K$ and $\lambda >0$, then $\lambda x \in K$
 \item it $x\in K$ and $-x\in K$, then $x=0$.
 \end{enumerate}
\end{definition}

\begin{definition} \label{def2.2} \rm
An operator is called completely continuous if it is continuous and
maps bounded sets into precompact sets.
\end{definition}

\begin{remark}\rm \label{rmk1}
By the positive solution of \eqref{ecua}, \eqref{bord} we understand
a function $u(t)$ wich is positive on $[0,1]$ and satisfies the
differential equation \eqref{ecua} and the boundary conditions
\eqref{bord}.
\end{remark}

We shall consider the Banach space $E=C[0,1]$ equipped with standard
norm
\[
\|u\|=\max_{0 \leq t\leq 1} |u(t)|\,.
\]
The proof of existence of solution is based on an applications of
the following theorems.

\begin{theorem}[\cite{Deimling,Guo}] \label{teor1}
 Let $E$ be a Banach space and let $K\subseteq E$ be a cone.
Assume $\Omega _{1}$ and $\Omega_{2}$ are open subsets of $E$ with
$0\in \Omega_{1}\subseteq \overline{\Omega _{1}}\subseteq\Omega_{2}$
and let
\[
T:K\cap (\overline{\Omega_{2}}\backslash\Omega_{1})\to K
\]
be  completely continuous such that
\begin{itemize}
\item[(i)] $\|Tu\|\leq \|u\|$ if $u\in K\cap \partial \Omega_{1}$ and
$\|Tu\|\geq \|u\|$ if $u\in K\cap \partial \Omega_{2}$;
or
\item[(ii)] $\|Tu\|\geq \|u\|$ if $u\in K\cap \partial \Omega_{1}$ and
$\|Tu\|\leq\|u\|$ if $u\in K\cap \partial \Omega_{2}$\,.
\end{itemize}
Then $T$ has a fixed point in
$K\cap (\overline{\Omega_{2}}\backslash\Omega_{1})$
\end{theorem}

\begin{theorem}[\cite{Deimling,Guo}]\label{teor2}
Let $E$ be a Banach space and $K$ be a cone of $E$. For $r>0$,
define $K_{r}=\{u\in K :\|u\|\leq r\}$ and assume that
$T:K_{r}\to K$ is a completely continuous operator such that
$Tu\neq u$ for $u\in \partial K_{r}$
\begin{enumerate}
\item If $\|Tu\|\leq \|u\|$ for all $u\in\partial K_{r}$, then
$i(T,K_{r},K)=1$
\item If $\|Tu\|\geq \|u\|$ for all $u\in\partial K_{r}$, then
$i(T,K_{r},K)=0$.
\end{enumerate}
\end{theorem}

Consider the three-point boundary-value problem
\begin{gather}\label{ecua2}
u'''+h(t)=0 ,\quad 0<t<1,\\
\label{bord2}
u(0)=0, \quad u'(0)=u'(1)=\alpha u(\eta)\,,
\end{gather}
where $\eta \in (0,1)$, $\alpha \in [0,1/\eta)$.

\begin{lemma}\label{lem1.5}
Let $\alpha \eta \neq 1$, $h\in L^{1}[0,1]$. Then the three-point
boundary-value problem \eqref{ecua2} and \eqref{bord2} has a unique
solution
\begin{equation*}
u(t)=\int_0^{1} G(t,s)h(s)ds,
\end{equation*}
where $G(t,s)=g(t,s)+\frac{\alpha t}{1-\alpha \eta}g(\eta,s)$, and
\begin{equation}\label{green}
g(x,y) = \begin{cases} \frac{1}{2}(2x-x^{2}-y)y  &  0\leq y\leq x \leq1 \\
\frac{1}{2} x^{2}(1-y) &  0\leq x \leq y \leq 1\,.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof}
From \eqref{ecua2},
$u'''=-h(t)$.
Applying the method of variation of parameter, we obtain
\begin{equation}\label{var}
u(t)=-\frac{1}{2}\int_0^t(t-s)^{2}h(s)ds + At^{2}+Bt+C,
\end{equation}
where $A,B,C \in \mathbb{R}$.
From \eqref{bord2}, $C=0$.
Since $u'(0)=u'(1)$,
\[
B=-\int_0^{1}(1-s)h(s)ds+2A+B.
\]
Therefore,
\[
A=\frac{1}{2}\int_0^{1}(1-s)h(s)ds\,.
\]
Since $u'(0)=\alpha u(\eta)$, we obtain:
\begin{gather*}
B=-\frac{\alpha}{2}\int_0^{\eta}(\eta-s)^{2}h(s)ds+\frac{\alpha
\eta^{2}}{2}\int_0^{1}(1-s)h(s)ds +B\alpha \eta,
\\
(1-\alpha\eta)B=-\frac{\alpha}{2}\int_0^{\eta}(\eta-s)^{2}h(s)ds
+\frac{\alpha\eta^{2}}{2}\int_0^{1}(1-s)h(s)ds,
\\
B=-\frac{\alpha}{2(1-\alpha\eta)}\int_0^{\eta}(\eta-s)^{2}h(s)ds
+\frac{\alpha\eta^{2}}{2(1-\alpha\eta)}\int_0^{1}(1-s)h(s)ds\,.
\end{gather*}
Replacing these expressions in \eqref{var},
\begin{align*}
u(t)&=-\frac{1}{2}\int_0^t(t-s)^{2}h(s)ds+\frac{t^{2}}{2}\int_0^{1}(1-s)h(s)ds
 -\frac{\alpha t}{2(1-\alpha\eta)}\int_0^{\eta}(\eta-s)^{2}h(s)ds\\
&\quad +\frac{\alpha t\eta^{2}}{2(1-\alpha\eta)}\int_0^{1}(1-s)h(s)ds \\
&=-\frac{1}{2}\int_0^t(t-s)^{2}h(s)ds+\frac{1}{2}\int_0^tt^{2}(1-s)h(s)ds
+\frac{1}{2}\int_{t}^{1}t^{2}(1-s)h(s)ds\\ 
& \quad -\frac{\alpha t}{2(1-\alpha\eta)}
 \int_0^{\eta}(\eta-s)^{2}h(s)ds+\frac{\alpha t\eta^{2}}{2(1-\alpha\eta)} 
 \int_0^{1}(1-s)h(s)ds \\
&=\frac{1}{2}\Big[\int_0^t(2t-t^{2}-s)sh(s)ds
 +\int_{t}^{1}t^{2}(1-s)h(s)ds \Big] \\
& \quad +\frac{\alpha t}{1-\alpha\eta}\frac{1}{2}
 \Big[\int_0^{1}\eta^{2}(1-s)h(s)ds
 -\int_0^{\eta}(\eta-s)^{2}h(s)ds \Big] \\
&= \int_0^{1}g(s,t)h(s)ds +\frac{\alpha t}{1-\alpha\eta}
 \int_0^{1}g(\eta,s)h(s)ds \\
&= \int_0^{1}G(t,s)h(s)ds\,.
\end{align*}
\end{proof}

\begin{lemma} \label{lem1.6}
Let  $\sigma\in (0,1]$ be fixed. Then
\[
 \frac{1}{2}\gamma s(1-s)\leq g(t,s)\leq \frac{1}{2}s(1-s), \quad
 \forall (t,s)\in [\sigma,1]\times  [0,1],
\]
where $\gamma=\sigma^2$.
\end{lemma}

\begin{proof}
 If $s\leq t$, from \eqref{green},
\begin{align*}
 g(t,s)
&=\frac{1}{2}(2t-t^2-s)s   \\ 
&= \frac{1}{2}(-(t^2-2t)-s)s   \\ 
&= \frac{1}{2}(-[(t-1)^2-1]-s)s   \\ 
&=  \frac{1}{2}[1-(1-t)^2-s]s   \\ 
&=  \frac{1}{2}[(1-s)-(1-t)^2]s  \,.
\end{align*}
Then
\[
g(t,s)\leq \frac{1}{2}s(1-s)\,.
\]
On the other hand,
\begin{align*}
 g(t,s)
&= \frac{1}{2}(2t-t^2-s)s  \\ 
&=  \frac{1}{2}ts(1-s)+\frac{1}{2}[(1-t)(t-s)s]  \\ 
&\geq \frac{1}{2}t s(1-s) \\ 
&\geq \frac{1}{2}t^{2} s(1-s) \,.
\end{align*}
If $t\leq s$, from \eqref{green},
\begin{align*}
\frac{1}{2}t^2(1-s)s
&\leq g(t,s)  \\ 
&= \frac{1}{2}t^2(1-s) \\ 
&\leq  \frac{1}{2}s^{2}(1-s) \\ 
&\leq  \frac{1}{2}s(1-s)\,. 
\end{align*}
Therefore
\begin{equation}\label{not2}
\frac{1}{2}t^2(1-s)s\leq g(t,s)\leq\frac{1}{2}(1-s)s \quad\forall
(t,s)\in [0,1]\times [0,1]\,.
\end{equation}
For $t \in [\sigma,1]$, we have
\[
\frac{1}{2}\sigma^2(1-s)s\leq g(t,s)\leq\frac{1}{2}(1-s)s \quad
\forall (t,s)\in [\sigma,1]\times [0,1] \,.
\]
\end{proof}

\begin{remark} \label{rmk2} \rm
For $t=1$ in \eqref{not2}, we have
\begin{equation} \label{not1}
\frac{1}{2}(1-s)s = g(1,s)\,.
\end{equation}
\end{remark}

\begin{lemma}\label{lem1.7}
 Let $h(t)\in C^+[0,1]$. The unique solution $u(t)$ of \eqref{ecua2},
\eqref{bord2} is nonnegative and satisfies
\[
\min_{\sigma\leq t\leq  1} u(t)\geq \gamma \|u\|\,.
\]
\end{lemma}

\begin{proof} 
From Lemma \ref{lem1.5} and Lemma \ref{lem1.6}, $u(t)$ is nonnegative.
For $t\in [0,1]$, from Lemma \ref{lem1.5} and Lemma \ref{lem1.6}, we
have that
\begin{align*}
u(t)
&= \int_0^{1}g(t,s)h(s)ds+\frac{\alpha t}{1-\alpha \eta}
 \int_0^{1}g(\eta,s)h(s)ds  \\
&\leq \frac{1}{2}\int_0^{1}s(1-s)h(s)ds+\frac{\alpha}{1-\alpha
\eta}\int_0^{1}g(\eta,s)h(s)ds\,.
\end{align*}
Then
\begin{equation}
\|u\|\leq
\frac{1}{2}\int_0^{1}s(1-s)h(s)ds+\frac{\alpha}{1-\alpha
\eta}\int_0^{1}g(\eta,s)h(s)ds\,.
\end{equation}
On the other hand,  Lemma \ref{lem1.6} imply that, for any $t\in
[\sigma,1]$,
\begin{align*}
u(t)&= \int_0^{1}g(t,s)h(s)ds+\frac{\alpha t}{1-\alpha \eta}\int_0^{1}g(\eta,s)h(s)ds  \\
&\geq \frac{1}{2}\gamma\int_0^{1}s(1-s)h(s)ds
 +\frac{\alpha t^2}{1-\alpha \eta}\int_0^{1}g(\eta,s)h(s)ds   \\
&\geq \frac{1}{2}\gamma\int_0^{1}s(1-s)h(s)ds+\frac{\alpha
\sigma^{2}}{1-\alpha \eta}\int_0^{1}g(\eta,s)h(s)ds  \\
&= \gamma\Big[\frac{1}{2}\int_0^{1}s(1-s)h(s)ds
+\frac{\alpha }{1-\alpha \eta}\int_0^{1}g(\eta,s)h(s)ds \Big]  \\
&\geq  \gamma \|u\|\,.
\end{align*}
Therefore
\[
\min_{\sigma\leq t\leq  1} u(t)\geq \gamma \|u\|\,.
\]
\end{proof}

 We introduce the notation
\[
f_{a}:=\liminf_{u\to a}\underset{0\leq t\leq
 1}{\min}\frac{f(t,u)}{u}, \quad 
f^{b}:=\limsup_{u\to b}\underset{0\leq t\leq
 1}{\max}\frac{f(t,u)}{u}, 
\]
where $a,b=0^{+},\infty$,
\begin{gather*}
N=\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)
ds+\frac{\alpha\gamma}{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds,\\
M=\int_0^{1}\frac{1}{2}s(1-s)a(s)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds\,.
\end{gather*}
Define the cone 
\[
K=\{u\in C[0,1]: u(t)\geq 0, \, \min_{\sigma\leq t\leq 1} u(t)\geq \gamma \|u\|\}
\]
and the operator $T:K \to E$ by
\begin{equation}\label{oper}
Tu(t)=\int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha
t}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
\end{equation}

\begin{remark} \label{rmk3} \rm
By Lemma \ref{lem1.5},  problem \eqref{ecua}, \eqref{bord}
has a positive solution $u(t)$ if and only if $u$ is a fixed point
of $T$.
\end{remark}

\begin{lemma}
The operator defined in \eqref{oper}, is completely continuous and
satisfies $T(K)\subseteq K$.
\end{lemma}

\begin{proof} 
By Lemma \ref{lem1.7}, $T(K)\subseteq K$. $T$ is
completely continuous by an application of Arzela-Ascoli theorem.
\end{proof}


In what follow, we will use the following conditions
\begin{itemize}
\item[(a)] $f^{0}=0$ and $f_{\infty}=\infty$;  

\item[(b)] $f_0=\infty$ and $f^{\infty}={0}$;  

\item[(c)] $f_0=\infty$ and $f_{\infty}=\infty$;  

\item[(d)] $f^{0}=0$ and $f^{\infty}=0$;

\item[(e)] $0\leq f^{0}<R$ and $r<f_{\infty}\leq \infty $; 

\item[(f)]  $r<f_0\leq\infty$ and $0\leq f^{\infty}<R$;

\item[(g)] $\exists{\rho}>0$ : $f(t,u)<R \rho$, $0<u\leq \rho$, $t\in [0,1]$;

\item[(h)] $\exists{\rho}>0$ : $f(t,u)>r \rho$, 
$\rho<u\leq \frac{\rho}{\gamma}$, $t\in [\sigma,1]$. 
\end{itemize}

\begin{remark} \label{rmk4} \rm
We note that (a) corresponds to the superlinear case
and (b) corresponds to the sublinear case.
In conditions (e) and (f), $r=N^{-1}$ and
$R=M^{-1}$. It is obvious that $r>R>0$.
\end{remark}

\section{Existence of Positive Solutions}\label{sec.2}

\begin{theorem}\label{teor7}
Assume that the conditions on $a$, $f$ and {\rm(a)} hold. Then
\eqref{ecua}, \eqref{bord} has at least one positive solution.
\end{theorem}

\begin{proof} 
Since $f^{0}=0$, $\exists \ H_{1}>0$ such that 
$f(t,u)\leq \varepsilon u$, for all $t\in[0,1]$, $0<u \leq H_{1}$,
 where $\varepsilon>0$. Then for $u\in K\cap
\partial \Omega_{1}$, with $\Omega_{1}=\{u\in X:\|u\|<H_{1}\}$, we
have
\begin{align*}
Tu(t)&= \int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha
t}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
 \\ 
&\leq \int_0^{1}\frac{1}{2}s(1-s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds  \\
&\leq  \int_0^{1}\frac{1}{2}s(1-s)a(s)\varepsilon
uds+\frac{\alpha }{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)\varepsilon u ds  \\
&\leq
\varepsilon\Big[\int_0^{1}\frac{1}{2}s(1-s)a(s)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds\Big]\|u\|\,.
\end{align*}
If  $\varepsilon M \leq 1$, then
$Tu(t)\leq \|u\|$.
Therefore,
\[
\|Tu\|\leq \|u\|\,.
\]
On the other hand, since $f_{\infty}=\infty$, there exists 
$\bar{H_{2}}>0$ such that $f(t,u)\geq\delta u$, for all 
$t\in [\sigma,1]$ with $\bar{H_{2}}\leq u$ and $\delta>0$. 
Then for $u\in K\cap\partial\Omega_{2}$, where 
$\Omega_{2}=\{u\in X:\|u\|<H_{2}\}$
with $H_{2}=\max\{2H_{1},\frac{\bar{H_{2}}}{\gamma}\}$. 
Then $u\in K\cap \partial \Omega_{2}$ implies that 
$\min_{\sigma\leq t\leq 1} u(t)\geq \gamma \|u\|=\gamma H_{2}> \bar{H_{2}}$.
So, by \eqref{not1},  we obtain
\begin{align*}
(Tu)(1)
&=  \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
 \\
&\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)f(s,u)ds
 \\
&\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)\delta
u(s)ds+\frac{\alpha }{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)\delta u(s)ds
  \\
&\geq \delta\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)
ds+\frac{\alpha\gamma}{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\|\,.
\end{align*}
If  $\delta N \geq 1$, then
\begin{equation}
Tu(1)\geq\|u\|
\end{equation}
which implies that
\begin{equation}
\|Tu\|\geq \|u\|\,.
\end{equation}
Therefore, by Theorem \ref{teor1},  the operator $T$ has
at least one fixed point, which is a positive solution of
\eqref{ecua}, \eqref{bord}.
\end{proof}

\begin{theorem}\label{teor8}
Assume that the conditions on $a$, $f$ and {\rm (b)} hold. Then
\eqref{ecua}, \eqref{bord} has at least one positive solution.
\end{theorem}

\begin{proof} 
Since $f_0=\infty$, there exists $H_{1}>0$ such that 
$f(t,u)\geq \xi u$, for all $t\in [\sigma,1]$, $0<u\leq H_{1}$ 
where $\xi>0$; thus,
for $u\in K\cap
\partial \Omega_{1}$, with 
$\Omega_{1}=\{u\in X:\|u\|<H_{1}\}$, by \eqref{not1}, we
have
\begin{align*}
(Tu)(1)&=  \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
 \\
&\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)f(s,u)ds
 \\
&\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)\delta
u(s)ds+\frac{\alpha }{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)\delta u(s)ds
  \\
&\geq \xi\left[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)
ds+\frac{\alpha\gamma}{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds \right]\|u\|\,.
\end{align*}
If $\xi N\geq 1$, then
$Tu(1)\geq\|u\|$.
Therefore
\[
\|Tu\|\geq \|u\|\,.
\]
On the other hand, since $f^{\infty}=0$,  there exists $\bar{H_{2}}>0$
such that $f(t,u)\leq\lambda u$, for all $t\in [0,1]$ with
$\bar{H_{2}}\leq u$ and $\lambda>0$. 

We consider two cases:

\noindent\textbf{Case 1.}
 Suppose $f$ is bounded.
Let $L$ such that $f(t,u)\leq L$ and  
$\Omega_{2}=\{u\in X:\|u\|<H_{2}\}$ where 
$H_{2}=\max\{2H_{1},LM\}$. If $u\in K\cap\partial\Omega_{2}$, 
then by Lemma \ref{lem1.6}, we have
\begin{align*}
Tu(t)&= \int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha
t}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds \\ 
&\leq
\int_0^{1}\frac{1}{2}s(1-s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds  \\
&\leq  \int_0^{1}\frac{1}{2}s(1-s)a(s)L ds+\frac{\alpha
}{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)L ds  \\
&\leq
L\Big[\int_0^{1}\frac{1}{2}s(1-s)a(s)+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds \Big]  \\
&\leq  H_{2}=\|u\|
\end{align*}
and consequently,
$\|Tu\|\leq\|u\|$.


\noindent\textbf{Case 2.} Suppose $f$ is unbounded, then
  from (H1)  there
is  $H_{2}>\max\{2H_{1},\bar{H_{2}}\}$ such that 
$f(t,u)\leq f(t,H_{2})$ with $0<u\leq H_{2}$ and let 
$\Omega_{2}=\{u\in X:\|u\|<H_{2}\}$.
 If $u\in  K \cap \partial \Omega_{2}$ and $\lambda
M\leq 1$,  we have
\begin{align*}
Tu(t)&= \int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha
t}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
 \\ &\leq
\int_0^{1}\frac{1}{2}s(1-s)a(s)f(s,H_{2})ds+\frac{\alpha
}{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)f(s,H_{2})ds  \\
&\leq  \int_0^{1}\frac{1}{2}s(1-s)a(s)\lambda  H_{2}
ds+\frac{\alpha }{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)\lambda H_{2} ds  \\
&\leq
\lambda\left[\int_0^{1}\frac{1}{2}s(1-s)a(s)+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds\right]H_{2}  \\
&\leq  H_{2}=\|u\|\,.
\end{align*}
Thus,
$\|Tu\|\leq\|u\|$.

Therefore by Theorem \ref{teor1},  the operator $T$ has
at least one fixed point, which is a positive solution of
\eqref{ecua}, \eqref{bord}.
\end{proof}

\begin{theorem}\label{teor9}
Assume that the conditions on $a$, $f$ and {\rm (e)} hold. Then
\eqref{ecua}, \eqref{bord} has at least one positive solution.
\end{theorem}

\begin{proof} 
Since $0\leq f^{0}<R$, there exists $H_{1}>0$ and
$0<\varepsilon_{1}<R$ such that $f(t,u)\leq(R-\varepsilon_{1})u$,
$0\leq t\leq 1$, $0< u \leq H_{1}$. Let 
$\Omega_{1}=\{u\in X:\|u\|<H_{1}\}$. So for any 
$u\in K\cap \partial \Omega_{1}$,
\begin{align*}
Tu(t)&= \int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha t
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
\\ 
&\leq \int_0^{1}\frac{1}{2}s(1-s)a(s)(R-\varepsilon_{1})u
ds+\frac{\alpha }{1-\alpha
\eta}\int_0^{1}g(\eta,s)a(s)(R-\varepsilon_{1})u ds  \\
&\leq (R-\varepsilon_{1})\left[\int_0^{1}\frac{1}{2}s(1-s)a(s)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds\right]\|u\|  \\
&=  (R-\varepsilon_{1})M\|u\|   < \|u\|\,.
\end{align*}
Thus
$\|Tu\|< \|u\|$.


Since $r<f_{\infty}\leq \infty$, there exist $\bar{H_{2}}>0$ and
$\varepsilon_{2}> 0$ such that $f(t,u)\geq(r+\varepsilon_{2})u$ for
$u\geq \bar{H_{2}}$ and $\sigma\leq t \leq 1$. Let
$H_{2}=\max\{2H_{1},\frac{\bar{H_{2}}}{\gamma}\}$ and
$\Omega_{2}=\{u\in X:\|u\|<H_{2}\}$. Then $u\in K\cap
\partial \Omega_{2}$ implies 
$\min_{\sigma\leq t\leq 1} u(t)\geq \gamma \|u\|=\gamma H_{2}> \bar{H_{2}}$.
So, by \eqref{not1} we obtain
\begin{align*}
Tu(1)
&= \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
\\ 
&\geq \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)(r+\varepsilon_{2})u
ds+\frac{\alpha }{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)(r+\varepsilon_{2})uds  
\\
&\geq (r+\varepsilon_{2})\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\alpha\gamma
}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\|  
\\
&=  (r+\varepsilon_{2})N\|u\|  > \|u\|\,.
\end{align*}
Thus,
$\|Tu\| > \|u\|$.

Therefore, by Theorem \ref{teor1}, the operator $T$ has
at least one fixed point, which is a positive solution of
\eqref{ecua}, \eqref{bord}.
\end{proof}

\begin{theorem}\label{teor10}
Assume that the conditions on $a$, $f$ and {\rm (f)} hold. Then
\eqref{ecua}, \eqref{bord} has at least one positive solution.
\end{theorem}

\section{Multiplicity results}

\begin{theorem}\label{teor11}
Assume that the conditions on  $a$, $f$, {\rm (c)} and
{\rm (g)} hold. Then \eqref{ecua}, \eqref{bord} has at least two
positive solutions.
\end{theorem}

\begin{proof} Since $f_0=\infty$, $\exists H_{1}>0$ where $0<H_{1}<\rho $ 
such that $f(t,u)>ru$ with
$0<u\leq H_{1}$ and $t\in [\sigma,1]$. Let $\Omega_{1}=\{u\in
X:\|u\|<H_{1}\}$. Then for any $u\in K\cap \partial \Omega_{1}$,
\begin{align*}
Tu(1)&= \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
\\ 
&> \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)ru ds+\frac{\alpha
}{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)ru ds  \\
&> r\left[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\gamma\alpha
}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\right]\|u\|  \\
&=  r N\|u\|=\|u\|
\end{align*}
Thus,
$\|Tu\|>\|u\|$.
Therefore, by Theorem \ref{teor2}
\[
i(T,K_{H_{1}},K)=0\,.
\]
Since $f_{\infty}=\infty$, there exists $\bar{H_{2}}>\rho$ such that
$f(t,u)>ru$ with $u\geq \bar{H_{2}}$, $t\in [\sigma,1]$. Let
$H_{2}=\frac{\bar{H_{2}}}{\gamma}$ and $\Omega_{2}=\{u\in
X:\|u\|<H_{2}\}$. Then for $u\in K\cap \partial \Omega_{2}$, we have
$\underset{\sigma\leq t\leq  1}{\min}u(t)\geq \gamma
\|u\|=\gamma H_{2}= \bar{H_{2}}$. Hence,
\begin{align*}
Tu(1)
&= \int_0^{1}g(1,s)a(s)f(s,u)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds
\\ 
&> \int_{\sigma}^{1}\frac{1}{2}s(1-s)a(s)ru ds+\frac{\alpha
}{1-\alpha
\eta}\int_{\sigma}^{1}g(\eta,s)a(s)ru ds  \\
&> r\Big[\int_{\sigma}^{1}\frac{\gamma}{2}s(1-s)a(s)ds+\frac{\gamma\alpha
}{1-\alpha \eta}\int_{\sigma}^{1}g(\eta,s)a(s)ds\Big]\|u\|  \\
&=  r N\|u\|=\|u\|\,.
\end{align*}
Thus,
$\|Tu\|>\|u\|$.
Therefore, by Theorem \ref{teor2}
\[
i(T,K_{H_{2}},K)=0\,.
\]

On the other hand, let $\Omega_{3}=\{u\in X:\|u\|<\rho\}$. For any
$u\in K\cap \partial\Omega_{3}$, we get from (e) that
$f(t,u)<R \rho $ for $0 \leq t \leq 1$, then
\begin{align*}
Tu(t)&= \int_0^{1}g(t,s)a(s)f(s,u)ds+\frac{\alpha
t}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)f(s,u)ds \\ 
&< \int_0^{1}\frac{1}{2}s(1-s)a(s)R \rho\,ds
+\frac{\alpha }{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)R \rho\,ds  \\
&< R\Big[\int_0^{1}\frac{1}{2}s(1-s)a(s)ds+\frac{\alpha
}{1-\alpha \eta}\int_0^{1}g(\eta,s)a(s)ds\Big]\rho  \\
&< RM\rho  \leq  \|u\|\,.
\end{align*}
Thus,
$\|Tu\|<\|u\|$.
Therefore, by Theorem \ref{teor2},
\[
i(T,K_{\rho},K)=1\,.
\]
Hence,
\begin{gather*}
i(T,K_{H_{2}}\setminus
\overline{K_{\rho}},K)=i(T,K_{H_{2}},K)-i(T,K_{\rho},K)=0-1=-1
\\
i(T,K_{\rho}\setminus
\overline{K_{H_{1}}},K)=i(T,K_{\rho},K)-i(T,K_{H_{1}},K)=1-0=1
\end{gather*}
Therefore, there exist at least two positive solutions 
$u_{1}\in K\cap(\bar{\Omega_{3}}\backslash\Omega_{1})$ and 
$u_{2}\in K\cap(\bar{\Omega_{2}}\backslash\Omega_{3})$ of
\eqref{ecua},\eqref{bord} in $K$, such that
\begin{equation}
0<\|u_{1}\|<\rho<\|u_{2}\|\,.
\end{equation}
\end{proof}

\begin{theorem}\label{teor12}
Assume that the conditions on  $a$, $f$, {\rm (d)} and
{\rm (h)} hold. Then \eqref{ecua}, \eqref{bord} has at least two
positive solutions.
\end{theorem}

\section{Examples}

\begin{example}[Superlinear and Sublinear Case]\rm
\begin{itemize}
\item[(a)] If $f(t,u)=u^{\alpha}$, $\alpha >1$, the
conclusions of Theorem \ref{teor7}, hold.
\item[(b)] If $f(t,u)=1+u^{\alpha}$, $\alpha \in(0,1)$ the
conclusions of Theorem \ref{teor8}, hold. 
\end{itemize}
\end{example}

\begin{example} \rm
Let $f(t,u)=\lambda t \ln{(1+u)}+u^2$, fix $\lambda >0$,
sufficiently small. Clearly $f^{0}=\lambda$ and $f_{\infty}=\infty$.
By Theorem \ref{teor9}, \eqref{ecua} and \eqref{bord} have at least
one positive solution.
\end{example}

\begin{example} \rm
Let $f(t,u)=u^2e^{-u}+ \mu \sin{u}$, fix $\mu > 0$ sufficiently
large. Then $f_0=\mu$ and $f^{\infty}=0$. By Theorem \ref{teor10},
\eqref{ecua} and \eqref{bord} have at least one positive solution.
\end{example}

\begin{example} \rm
Consider the boundary-value problem
\begin{gather} \label{ecuaex}
  u'''(t)+u^{b}+u^{c}=0 ,\quad 0<t<1, \\
\label{bordex}
 u(0)=0, \quad u'(0)=u'(1)=\frac{1}{4} u(\frac{1}{2}),
\end{gather}
where $f(t,u)=f(u)=u^{b}+u^{c}$, $a(t)=1$, $b\in(0,1)$ and $c>1$.
Then $f_0=\infty$ and $f_{\infty}=\infty$. By a simple calculation,
$M=2/21$ then $R=21/2$. 

On the other hand, we could choose $ \rho =1$, then 
$f(t,u)\leq 2< \frac{21}{2}1=R \rho$ for 
$(t,u)\in[0,1]\times[0,\rho]$. By Theorem
\ref{teor11}, \eqref{ecuaex} and \eqref{bordex} have at least two
positive solutions.
\end{example}

\begin{thebibliography}{99}

\bibitem{Bai} Z. Bai;
 Existence of Solutions for Some Third-order Boundary Value Problems, 
\emph{Electron. J. Differential Equations,} 25 (2008), 1--6.

\bibitem{Deimling} K. Deimling;
 \emph{Nonlinear Functional Analysis,} Springer, Berlin, 1985.

\bibitem{Guo}D. Guo and V. Lakshmikantham;
 \emph{Nonlinear Problems in Abstract Cones,} Academic Press, San Diego, 1988.

\bibitem{Li} S. Li;
 Positive Solutions of Nonlinear Singular Third-order Two-point Boundary 
Value Problems, \emph{J. Math. Anal. Appl.,} 323 (2006), 413--425.

\bibitem{Liu} Z. Liu, L. Debnath, S. Kang;
 Existence of Positive Solutions to a Third-order Two-point Generaliza 
Right Focal Boundary Value Problems, 
\emph{Computer and Mathematics with Applications,} \textbf{55} (2008), 356--367.

\bibitem{Luan} S. Luan, H Su, Q Sun;
 Positive Solutions to Third-order Two-point Semipositone Boundary 
Value Problems, \emph{Int. J. Math. Anal.,} 3 Vol 3 (2009), 99--108.

\bibitem{Sun} Y. Sun;
 Positive Solutions for Third-Order Three-Point Nonhomogeneous Boundary
 Value Problem, \emph{Appl. Math. Letters,} 22 (2009), 45--51.

\end{thebibliography}

\end{document}