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

\AtBeginDocument{{\noindent\small
Eighth Mississippi State - UAB Conference on Differential Equations and
Computational Simulations.
{\em Electronic Journal of Differential Equations},
Conf. 19 (2010),  pp. 151--159.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document} \setcounter{page}{151}
\title[\hfilneg EJDE-2010/Conf/19/\hfil Third order three-point BVP]
{Positive solutions to a nonlinear third order three-point boundary
value problem}

\author[J. R. Graef, L. Kong, B. Yang\hfil EJDE/Conf/19 \hfilneg]
{John R. Graef, Lingju Kong, Bo Yang}  % in alphabetical order

\address{John R. Graef \newline
Department of Mathematics,
University of Tennessee at Chattanooga, Chattanooga, TN 37403, USA}
\email{John-Graef@utc.edu}

\address{Lingju Kong \newline
Department of Mathematics,
University of Tennessee at Chattanooga, Chattanooga, TN 37403, USA}
\email{Lingju-Kong@utc.edu}

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

\thanks{Published September 25, 2010.}
\subjclass[2000]{34B15}
\keywords{Boundary value problems; existence of positive
solutions; \hfill\break\indent
nonexistence of positive solutions; nonlinear
equations; third order problem; \hfill\break\indent
three point boundary conditions}

\begin{abstract}
 We consider a third order three point boundary value problem.
 Some upper and lower estimates for positive solutions of the
 problem are proved. Sufficient conditions for the existence
 and nonexistence of positive solutions for the problem are
 obtained. An example is included to illustrate the results.
\end{abstract}

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

\section{Introduction}

Recently, second and higher order multi-point boundary value
problems have attracted a lot of attention. In 2004, Henderson
\cite{H} considered the second order three point boundary value
problem
\begin{equation} \label{ee00}
\begin{gathered}
u''(t)+f(u(t))=0,\quad 0\le t\le 1, \\
u(0)=u(p)-u(1)=0.
\end{gathered}
\end{equation}
In 2006, Graef and Yang \cite{GY-2} studied the third order
nonlocal boundary value problem
\begin{gather}
u'''(t) = g(t)f(u(t)), \quad 0\le t\le 1, \label{ee1} \\
u(0) = u'(p) = u''(1) = 0. \label{ee2}
\end{gather}
For some other results on third order boundary value problems
we refer the reader to the papers \cite{AD,GY,GSZ,HT,M,W}.
%
Motivated by these works, in this paper we consider the third
order three point nonlinear boundary value problem
\begin{gather}
u'''(t) = g(t)f(u(t)), \quad 0\le t\le 1, \label{bhu}\\
u(0)= u(p) -u(1) = u''(1) = 0. \label{rfv}
\end{gather}
%
To our knowledge, the problem \eqref{bhu}--\eqref{rfv} has not
been considered before. The boundary conditions \eqref{rfv}
are closely related to some other boundary conditions.
Firstly, \eqref{rfv} contains \eqref{ee00} as a part.
We also note that $u(p)-u(1)=0$ implies that there exists
$\beta\in(p,1)$ such that $u'(\beta)=0$, and therefore the
boundary conditions \eqref{rfv} imply
\[
u(0)=u'(\beta)=u''(1)=0.
\]
Hence, boundary conditions \eqref{rfv} are closely related
to the conditions \eqref{ee2}. If we let $p\to 1^-$,
then \eqref{rfv} ``tends to''
\begin{equation}
u(0) = u'(1) = u''(1) = 0, \label{rfvvv}
\end{equation}
which are often referred to as the (1,2) focal boundary conditions.

In this paper, we are interested in the existence and nonexistence
of positive solutions of the problem \eqref{bhu}--\eqref{rfv}.
%
By a {\it positive solution}, we mean a solution $u(t)$ to
the boundary value problem such that $u(t)>0$ for $0<t<1$.
%
Throughout the paper, we assume that
\begin{itemize}
\item[(H1)] The functions $f:[0,\infty)\to [0,\infty) $ and
 $g:[0,1]\to [0,\infty)$ are continuous, and
 $g(t)\not\equiv 0$ on $[0,1]$;
\item[(H2)] The constant $p$ satisfies $1/2<p<1$.
\end{itemize}

We will use the following fixed point theorem, which is due
to Krasnosel'skii, to prove the existence results.

\begin{theorem}[\cite{K}] \label{tt-K}
Let $X$ be a Banach space over the reals, and let $P\subset X$ be
a cone in $X$.
Let $\le$ be the partial order on $X$ determined by $P$.
Assume that $\Omega_1$ and $\Omega_2$ are bounded open subsets
of $X$ with $0\in\Omega_1$ and $\overline{\Omega_1}\subset \Omega_2$.
Let
\[
L:P\cap( \overline{\Omega_2}-\Omega_1 )\to P
\]
be a completely continuous operator such that, either
\begin{itemize}
\item[(K1)] $Lu\not\ge u$ if $u\in P\cap\partial \Omega_1$,
    and $Lu\not\le u$ if $u\in P\cap\partial \Omega_2$;
    or
\item[(K2)] $Lu\not\le u$ if $u\in P\cap\partial \Omega_1$,
    and $Lu\not\ge u$ if $u \in P\cap\partial \Omega_2$.
\end{itemize}
Then $L$ has a fixed point in
$P\cap(\overline{\Omega}_2-\Omega_1) $.
\end{theorem}

Before the Krasnosel'skii fixed point theorem can be used to
obtain any existence result, we need to find some nice estimates
to positive solutions to the problem \eqref{bhu}--\eqref{rfv} first.
These {\it a priori} estimates are essential to a successful
application of the Krasnosel'skii fixed point theorem.
It is based on these estimates that we can define an appropriate
cone on which Theorem \ref{tt-K} can be applied. Better
estimates will result in sharper existence and nonexistence conditions.

We now fix some notation. Throughout we let $X=C[0,1]$
with the supremum norm
$$
    \|v\|=\max_{t\in [0,1]} |v(t)| \quad \text{for all }  v\in X.
$$
Clearly, $X$ is a Banach space. Also, we define the constants
\begin{gather*}
    F_0=\limsup_{x\to 0^+}\frac{f(x)}{x},\quad
    f_0=\liminf_{x\to 0^+}\frac{f(x)}{x}, \\
    F_{\infty}=\limsup_{x\to + \infty} \frac{f(x)}{x},\quad
    f_{\infty}=\liminf_{x\to + \infty} \frac{f(x)}{x}.
\end{gather*}
These constants will be used later in the statements of our
existence theorems.

This paper is organized as follows. In Section 2, we obtain
some {\it a priori} estimates to positive solutions to the
problem \eqref{bhu}--\eqref{rfv}. In Section 3, we define a
positive cone of the Banach space $X$ using the estimates
obtained in Section 2, and apply Theorem \ref{tt-K} to establish
some existence results for positive solutions of the
problem \eqref{bhu}--\eqref{rfv}. In Section 4, we present
some nonexistence results. An example is given at the end of
the paper to illustrate our results.

\section{Green Function and Estimates of Positive Solutions}

In this section, we give the Green function for the
problem \eqref{bhu}--\eqref{rfv} and prove some estimates
for positive solutions of the problem.

We need the indicator function $\chi$ to write the expression
for the Green's function for the problem \eqref{bhu}--\eqref{rfv}.
Recall that if $[a,b]\subset R:=(-\infty,+\infty)$ is a closed
interval, then the indicator function $\chi$ of $[a,b]$ is given by
\[
\chi_{[a,b]}(t)=
\begin{cases}
1, & \text{if }t\in [a,b],     \\
0, & \text{if }t\not\in [a,b].
\end{cases}
\]
%
Now we define the function $G:[0,1]\times[0,1]\to [0,\infty)$ by
\begin{equation}
\begin{aligned}
G(t,s)&=
\frac{t(1+p)-t^2}{2}-\frac{t(1-s)^2}{2(1-p)}
+ \frac{t(p-s)^2}{2(1-p)}\chi_{[0,p]}(s)\\
&\quad + \frac{(t-s)^2}{2}\chi_{[0,t]}(s).
\end{aligned}
\end{equation}
Then $G(t,s)$ is the Green function associated with the
problem \eqref{bhu}--\eqref{rfv}.
Moreover, the problem \eqref{bhu}--\eqref{rfv} is equivalent
to the integral equation
\begin{equation} \label{ee3}
u(t)=\int_0^1 G(t,s)g(s)f(u(s))ds, \quad 0\le t\le 1.
\end{equation}
It can be shown that $G(t,s)\ge 0$ if $(t,s)\in[0,1]^2$.

The following result is based on Lemmas 2.1 and 2.2 of Graef
and Yang \cite{GY-2}.

\begin{lemma} \label{th2.1}
If $u'''(t)\ge 0$ for $0\le t\le 1$, and $u(0)=u'(\beta)=u''(1)=0$,
where $\beta\in(1/2,1)$ is a constant, then $u(t)\ge 0$ for
$0\le t\le 1$, and
\[
\frac{2\beta t-t^2}{\beta^2}u(\beta)\le u(t)\le u(\beta)\quad
\text{for } 0\le t\le 1.
\]
\end{lemma}

Throughout this paper we let
\[
a(t)=\begin{cases}
2t-t^2, &\text{if } t\le\frac{2p}{1+p},\\
\frac{2pt-t^2}{p^2}, &\text{if } t\ge \frac{2p}{1+p}.
\end{cases}
\]
It can be shown that
\[
a(t)\geq \min\{t,1-t\},\quad 0\le t\le 1.
\]
The proof of the last inequality is straightforward and
therefore is omitted.


\begin{lemma} \label{lem2.2}
If $u\in C^3[0,1]$ is such that
\begin{equation} \label{nnm}
u'''(t)\ge 0,\quad 0\le t\le 1,
\end{equation}
and
\[
u(0)=u(p)-u(1)=u''(1)=0,
\]
then $u(t)\ge a(t)\|u\|$ for $0\le t\le 1$.
\end{lemma}

\begin{proof}
Since $u(p)=u(1)$, there exists $\beta\in(p,1)\subset (1/2, 1)$
such that $u'(\beta)=0$.
Since $u'''(t)\ge 0$ for $0\le t\le 1$ and
$u(0)=u'(\beta)=u''(1)=0$, by Lemma \ref{th2.1} we have
\[
\frac{2\beta t-t^2}{\beta^2}u(\beta)\le u(t)
\le u(\beta)=\|u\|,\quad 0\le t\le 1.
\]
If $0\le t\le 2p/(1+p)$, then
\begin{align*}
u(t)-a(t)\|u\|
&\geq  u(\beta)\left[\frac{2\beta t-t^2}{\beta^2}-(2t-t^2)\right]
\\
&=  \frac{u(\beta)}{\beta^2}t(1-\beta)[2\beta -t(1+\beta)]
\\
&\geq  \frac{u(\beta)}{\beta^2}t(1-\beta)
\left[2\beta -\frac{2p}{1+p}(1+\beta)\right]
\\
&=  \frac{2tu(\beta)(1-\beta)}{\beta^2(1+p)}(\beta-p)
\ge 0.
\end{align*}
If $2p/(1+p)\le t\le 1$, then
\begin{align*}
u(t)-a(t)\|u\|
&\geq  u(\beta)\left[\frac{2\beta t-t^2}{\beta^2}
-\frac{2pt-t^2}{p^2}\right]
\\
&=  \frac{t u(\beta)(\beta -p)}{\beta^2 p^2}[(p+\beta)t-2p\beta]
\\
&=  \frac{t u(\beta)(\beta -p)}{\beta^2 p^2}( pt +\beta(t-2p))
\\
&\geq \frac{t u(\beta)(\beta -p)}{\beta^2 p^2}
\left[\frac{2p^2}{1 + p} + \beta \left(\frac{2p}{1+p} - 2p\right)\right]
\\
&=  \frac{t u(\beta)(\beta -p)}{\beta^2 p^2}\frac{2p^2}{1+p}(1-\beta)
\ge  0.
\end{align*}
Thus, we have proved that $u(t)\ge a(t)\|u\|$ on $[0,1]$.
\end{proof}


The following lemma is immediate.

\begin{lemma} \label{lem2.3}
If $u\in C^3[0,1]$ is such that
$u'''(t)\ge 0$, $0\le t\le 1$, and
\[
u(0)=u(p)-u(1)=u''(1)=0,
\]
then
\[
u(2p/(1+p))\ge \frac{4p}{(1+p)^2}\|u\|,
\]
or equivalently,
\[
\|u\|\le \frac{(1+p)^2}{4p}u(2p/(1+p)).
\]
\end{lemma}

We can summarize our findings in the following theorem.

\begin{theorem} \label{uhj}
Suppose that {\rm (H1)} and {\rm (H2)} hold.
If $u\in C^3[0,1]$ satisfies \eqref{nnm} and the boundary conditions
 \eqref{rfv}, then $u(t)\ge a(t)\|u\|$ on $[0,1]$.
 In particular, if $u\in C^3[0,1]$ is a nonnegative solution to
the boundary value problem \eqref{bhu}--\eqref{rfv},
then $u(t)\ge a(t)\|u\|$ on $[0,1]$.
\end{theorem}

Now we define
$$
P=
\{v\in X: a(t)\|v\|\le v(t) \text{ \ on \ }\ [0,1]\}.
$$
Clearly, $P$ is a positive cone of the Banach space $X$.
Define an operator $T:P\to X$ by
$$
Tu(t)= \int_0^1 G(t,s)g(s)f(u(s))ds,\quad
 0\leq t\leq 1, \text{ for all }  u\in X.
$$
It is well known that $T:P\to X$ is a completely continuous operator.
%
And by the same arguments as those used to prove Theorem \ref{uhj},
we can show that $T(P)\subset P$. We also note that if $v\in P$, then
\[
\|v\|\le\frac{(1+p)^2}{4p}v(2p/(1+p)).
\]

Now the integral equation \eqref{ee3} is equivalent to the equality
$$
Tu=u,\quad u\in P,
$$
so in order to solve the problem \eqref{bhu}--\eqref{rfv}, we
only need to find a fixed point of $T$ in $P$.

\section{Existence of Positive Solutions}

We begin by defining the constants
\[
    A = \int_0^1 G(2p/(1+p),s) g(s) a(s)\,ds \quad \text{and} \quad
    B = \int_0^1 G(2p/(1+p),s) g(s) \,ds.
\]
Our first existence result is the following.

\begin{theorem} \label{tt6}
If
\[
BF_0 \frac{(1+p)^2}{4p}<1< A f_{\infty},
\]
then the problem \eqref{bhu}--\eqref{rfv} has at least one
positive solution.
\end{theorem}

\begin{proof}
Choose $\epsilon>0$ such that $(F_0+\epsilon)B(1+p)^2/4p <1$.
Then there exists $H_1 >0 $ such that
$$
f(x)\leq (F_0+\epsilon)x \ \ {\rm for}\ \ 0<x\leq H_1.
$$
For each $u\in P$ with $\|u\|=H_1$, we have
%
\begin{align*}
(Tu)(2p/(1+p))
& =
\int_0^1 G (2p/(1+p),s) g(s) f(u(s))\,ds
\\
&\le
\int_0^1 G(2p/(1+p),s) g(s) (F_0+\epsilon)u(s)\,ds
\\
& \le
(F_0+\epsilon)\|u\| \int_0^1 G(2p/(1+p),s) g(s)ds
\\
& =
(F_0+\epsilon)\|u\| B
\\
&\le
B(F_0+\epsilon)\frac{(1+p)^2}{4p}u(2p/(1+p))
\\
&< u(2p/(1+p)),
\end{align*}
which means $ Tu\not\ge u$. If we let
$\Omega_1=\{ u\in X\,|\ \|u\|<H_1\}$, then
$$
Tu\not\ge u,\quad\text{for  any } u\in P\cap\partial\Omega_1.
$$

To construct $\Omega_2$, we first choose $c\in (0,1/4)$ and
$\delta>0$ such that
$$
(f_\infty-\delta)\int_c^{1-c} G(2p/(1+p),s) g(s) a(s)\,ds >1.
$$
Now, there exists $ H_3 >0 $ such that
$f(x)\geq (f_\infty-\delta)x$ for $x\geq H_3$.
Let $H_2=H_1 + {H_3}/{c}$.
If $u\in P$ with $\|u\| = H_2$, then for $c\le t\le 1-c$, we have
$$
u(t)\geq\min\{t,1-t\}\|u\| \ge c H_2\ge H_3.
$$
So, if $u\in P$ with $\|u\| = H_2$, then
%
\begin{align*}
(Tu)(2p/(1+p))
& \geq
 \int_c^{1-c} G(2p/(1+p),s) g(s) f(u(s))\,ds
\\
& \geq
\int_c^{1-c} G(2p/(1+p),s) g(s) (f_\infty-\delta)u(s)ds
\\
& \ge
(f_\infty-\delta) \|u\|  \int_c^{1-c} G(2p/(1+p),s) g(s) a(s)\,ds
\\
& >
\|u\|
\\
&\ge
u(2p/(1+p)),
\end{align*}
%
which means $Tu\not\le u$. So, if we let
$\Omega_2=\{ u\in X: \|u\|<H_2\}$, then $\Omega_1\subset \Omega_2$, and
$$
Tu\not\le u,\quad\text{for any } u\in P\cap\partial\Omega_2.
$$
Therefore, condition (K1) of Theorem \ref{tt-K} is satisfied,
and so there exists a fixed point of $T$ in $P$.
This completes the proof of the theorem.
\end{proof}

Our next theorem is a companion result to the one above.

\begin{theorem} \label{tt7}
If
\[
B F_{\infty}\frac{(1+p)^2}{4p}< 1 < A f_0,\]
then the problem \eqref{bhu}--\eqref{rfv} has at least one
positive solution.
\end{theorem}

\begin{proof}
We first choose $\epsilon>0$ such that
$$
A(f_0-\epsilon)>1.
$$
There exists $H_1 >0$ such that $f(x)\ge (f_0-\epsilon)x$ for
$x\geq H_1$.
If $u\in P$ with $\|u\| = H_1$, then
%
\begin{align*}
(Tu)(2p/(1+p))
& \ge
 \int_0^1 G(2p/(1+p),s) g(s) f(u(s))\,ds
\\
& \ge
\int_0^1 G(2p/(1+p),s) g(s) (f_0-\epsilon)u(s)ds
\\
& \ge
(f_0-\epsilon) \|u\| \int_0^1 G(2p/(1+p),s) g(s) a(s)\,ds
\\
& >
\|u\|
\\
&\ge
u(2p/(1+p)),
\end{align*}
%
which means $Tu\not\le u$. So, if we let
$\Omega_1=\{ u\in X\,|\ \|u\|<H_1\}$, then
$$
Tu\not\le u,\quad \text{for  any } u\in P\cap\partial\Omega_1.
$$

To construct $\Omega_2$, we choose $\delta\in (0,1)$ such that
\[
((F_{\infty} + \delta) B + \delta)\frac{(1+p)^2}{4p}  < 1.
\]
There exists $H_3>0$ such that $f(x)\le (F_{\infty}+\delta) x$
for $x\geq H_3$.
If we let $M = \max_{0\leq x\leq H_3}f(x)$, then
$f(x)\le M + (F_{\infty}+\delta)x$ \text{for} $x\ge 0$. Let
\[
K = M \int_0^1 G(2p/(1+p),s)g(s)ds+1,
\]
and let $H_2 = H_1+K \big(\frac{4p}{(1+p)^2}-(F_{\infty}+\delta)B
\big)^{-1}$.
%
Now for each $u\in P$ with $\|u\| = H_2$, we have
\begin{align*}
(Tu)(2p/(1+p))
& =
  \int_0^1 G(2p/(1+p),s) g(s) f(u(s))\,ds
\\
& \le
  \int_0^1 G(2p/(1+p),s) g(s) (M+(F_{\infty}+\delta)u(s))\,ds
\\
& <
   K + (F_{\infty}+\delta)\int_0^{1} G(2p/(1+p),s) g(s) u(s)\,ds
\\
& \le
   K + (F_{\infty}+\delta)\|u\| \int_0^{1} G(2p/(1+p),s) g(s) \,ds
\\
&\le
   K + (F_{\infty}+\delta)B\|u\|
\\
& \le
\Big(\frac{4p}{(1+p)^2} - (F_{\infty}+\delta)B\Big) H_2
+ (F_{\infty}+\delta)B H_2
\\
&=
\frac{4p}{(1+p)^2} \|u\|
\\
&\le
u(2p/(1+p)),
\end{align*}
%
which means $ Tu\not\ge u $.
%
So, if we let $\Omega_2=\{ u\in X\ |\ \|u\|<H_2\}$, then
$$
Tu\not\ge u,\quad\text{for  any } u\in P\cap\partial\Omega_2.
$$
By Theorem \ref{tt-K}, $T$ has a fixed point in
 $P\cap(\overline{\Omega}_2-\Omega_1)$. Therefore,
problem \eqref{bhu}--\eqref{rfv} has at least one positive solution,
and this completes the proof of the theorem.
\end{proof}

\section{Nonexistence Results and Example}

In this section, we give some sufficient conditions for
the nonexistence of positive solutions.

\begin{theorem} \label{tt8}
Suppose that {\rm (H1)} and {\rm (H2)} hold.
If $\frac{(1+p)^2}{4p}Bf(x)<x$ for all $x\in(0,+\infty)$,
then the problem \eqref{bhu}--\eqref{rfv} has no positive solutions.
\end{theorem}

\begin{proof}
Assume to the contrary that $u(t)$ is a positive solution of
 problem \eqref{bhu}--\eqref{rfv}.
Then $u\in P$, $u(t)>0$ for $0<t<1$, and
\begin{align*}
%
u(2p/(1+p))
& =
\int_0^1 G(2p/(1+p),s) g(s) f(u(s))\,ds
\\
& <
\frac{4p}{(1+p)^2} B^{-1} \int_0^1 G(2p/(1+p),s) g(s) u(s)\,ds
\\
& \le
\frac{4p}{(1+p)^2} B^{-1} \|u\|  \int_{0}^{1} G(2p/(1+p),s) g(s) ds
\\
& \leq
\frac{4p}{(1+p)^2} \|u\|
\\
&\le
u(2p/(1+p),
\end{align*}
%
which is a contradiction.
\end{proof}

In a similar fashion, we can prove the next theorem.

\begin{theorem} \label{tt9}
Suppose that {\rm (H1)} and {\rm (H2)} hold. If $A f(x)>x$ for
all $x\in(0,+\infty)$, then the problem \eqref{bhu}--\eqref{rfv}
has no positive solutions.
\end{theorem}

We conclude the paper with an example.

\begin{example} \rm
Consider the third-order boundary-value problem
\begin{gather}
u'''(t)=g(t)f(u(t)), \quad 0<t<1, \label{rfv2} \\
u(0)=u(3/4)-u(1)=u''(1)=0, \label{rfv3}
\end{gather}
where
\begin{gather*}
g(t)=(1+t)/10,\quad 0\le t\le 1, \\
f(u)=\lambda u\frac{1+3u}{1+u}, \quad u\ge 0.
\end{gather*}
Here $\lambda>0$ is a parameter.
We easily see that $F_0=f_0=\lambda$ and
$F_{\infty}=f_{\infty}=3\lambda$.
Calculations indicate that
$$
A=\frac{5268393409}{216850636800},\quad
B=\frac{33611}{1229312}.
$$
 From Theorem \ref{tt6} we see that if
$$
13.7203 \approx\frac{ 1}{ 3A } < \lambda
< \frac{48}{49B}\approx 35.8282,
$$
then problem \eqref{rfv2}--\eqref{rfv3} has at least one
positive solution.
 From Theorems \ref{tt8} and \ref{tt9}, we see that if
$$
\lambda<\frac{ 16 }{49B}\approx 11.9427 \quad\text{or}\quad \lambda>\frac{1}{A}\approx 41.1607,
$$
then  problem \eqref{rfv2}--\eqref{rfv3} has no positive solutions.

This example shows that our existence and nonexistence
conditions work very well.
\end{example}

\begin{thebibliography}{0}

\bibitem{AD}
D. R. Anderson and J. M. Davis;
\emph{Multiple solutions and eigenvalues for third-order right
focal boundary value problems},
J. Math. Anal. Appl. 267 (2002), 135--157.

\bibitem{GY} J. R. Graef and B. Yang
\emph{Positive solutions for a third order nonlocal boundary value
problem}, Discrete Contin. Dyn. Syst., Ser. S 1 (2008), 89-97.

\bibitem{GY-2} {J. R. Graef} and B. Yang
\emph{Positive solutions of a third order eigenvalue problem},
{Dynam. Systems Appl.} 15 (2006), 97--110.

\bibitem{GSZ} Li-Jun Guo, Jian-Ping Sun, Ya-Hong Zhao
\emph{Existence of positive solutions for nonlinear third-order
three-point boundary value problems}, Nonlinear Anal. 68 (2008),
3151--3158.

\bibitem{H} J. Henderson
\emph{Double solutions of three point boundary value problem for
second order differential equations},
Electronic J. Differential Equations 2004 (2004), No. 115, 1--7.

\bibitem{HT} J. Henderson and C. C. Tisdell
\emph{Five-point boundary value problems for third-order differential
equations by solution matching}, Math. Comput. Modelling 42 (2005), 133--137.

\bibitem{K} M. A. Krasnosel'skii
\emph{Positive Solutions of Operator Equations},
 Noordhoff, Groningen, 1964.

\bibitem{M} R. Ma
\emph{Multiplicity results for a third order boundary value problem
 at resonance}, Nonlinear Anal. 32 (1998), 493--499.

\bibitem{W}  P. J. Y. Wong
\emph{Triple fixed-sign solutions for a system of third-order
generalized right focal boundary value problems},
Differential \& Difference Equations and Applications, 1139--1148,
Hindawi Publ. Corp., New York, 2006.

\end{thebibliography}

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