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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 118, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/118\hfil Multiple positive solutions]
{Multiple positive solutions for a third-order three-point BVP with
 sign-changing Green's function}

\author[J.-P. Sun, J. Zhao\hfil EJDE-2012/118\hfilneg]
{Jian-Ping Sun, Juan Zhao}  % in alphabetical order

\address{Department of Applied Mathematics, Lanzhou University of Technology,
Lanzhou, Gansu 730050, China}
\email[Jian-Ping]{jpsun@lut.cn}
\email[Juan Zhao]{z\_1111z@163.com}


\thanks{Submitted May 29, 2012. Published July 14, 2012.}
\subjclass[2000]{34B10, 34B18}
\keywords{Third-order three-point boundary-value problem; \hfill\break\indent
sign-changing Green's function; positive solution}

\begin{abstract}
 This article concerns the third-order three-point
 boundary-value problem
 \begin{gather*}
 u'''(t)=f(t,u(t)),\quad t\in [0,1], \\
 u'(0)=u(1)=u''(\eta)=0.
 \end{gather*}
 Although the corresponding Green's function
 is sign-changing, we still obtain the existence of at least
 $2m-1$ positive solutions for arbitrary positive integer
 $m$ under suitable conditions on $f$.
\end{abstract}

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


\section{Introduction}

Third-order differential equations arise from a variety of 
areas of applied mathematics and physics, e.g., in the deflection of
a curved beam having a constant or varying cross section, a
three-layer beam, electromagnetic waves or gravity driven flows and
so on \cite{4}.

Recently, the existence of single or multiple positive solutions to
some third-order three-point boundary-value problems (BVPs for
short) has received much attention from many authors. For example,
in 1998, by using the Leggett-Williams fixed point theorem, Anderson
\cite{0} proved the existence of at least three positive solutions
to the problem
\begin{gather*}
-x'''(t)+f(x(t))=0,\quad t\in [0,1],\\
x(0)=x'(t_2)=x''(1)=0,
\end{gather*}
where $t_2\in [\frac{1}{2},1)$. In 2003, Anderson
\cite{1.1} obtained some existence results of positive solutions for
the problem
\begin{gather*}
x'''(t)=f(t,x(t)),\quad t_1\leq t\leq t_3,\\
x(t_1)=x'(t_2)=0,\quad \gamma x(t_3)+\delta x''(t_3)=0.
\end{gather*}
The main tools used were the Guo-Krasnosel'skii and Leggett-Williams
fixed point theorems. In 2005, Sun \cite{9} studied the existence of
single and multiple positive solutions for the singular BVP
\begin{gather*}
u'''(t)-\lambda a(t)F(t,u(t))=0,\quad t\in (0,1),\\
u(0)=u'(\eta)=u''(1)=0,
\end{gather*}
where $\eta\in [\frac{1}{2},1)$, $\lambda$ was a positive
parameter and $a(t)$ was a nonnegative continuous function defined
on $(0, 1)$. His main tool was the Guo-Krasnosel'skii fixed point
theorem. In 2008, by using the Guo-Krasnosel'skii fixed point
theorem, Guo, Sun and Zhao \cite{3} obtained the existence of at
least one positive solution for the problem
\begin{gather*}
u'''(t)+h(t)f(u(t))=0,\quad t\in (0,1),\\
u(0)=u'(0)=0,\quad u'(1)=\alpha u'(\eta ),
\end{gather*}
where $0<\eta <1$ and $1<\alpha <1/\eta $. For more results
concerning the existence of positive solutions to third-order
three-point BVPs, one can refer to \cite{3.1,3.2,13,12,10,11}.

It is necessary to point out that all the above-mentioned works are
achieved when the corresponding Green's functions are positive,
which is a very important condition. A natural question is that
whether we can obtain the existence of positive solutions to some
third-order three-point BVPs when the corresponding Green's
functions are sign-changing. It is worth mentioning that Palamides
and Smyrlis \cite{7} discussed the existence of at least one
positive solution to the singular third-order three-point BVP with
an indefinitely signed Green's function
\begin{gather*}
u'''(t)=a(t)f(t,u(t)),\quad t\in (0,1),\\
u(0)=u(1)=u''(\eta)=0,\quad \eta\in (\frac{17}{24},1).
\end{gather*}
Their technique was a combination of the Guo-Krasnosel'skii fixed
point theorem and properties of the corresponding vector field. The
following equality
\begin{equation}
\max_{t\in[0,1]}\int_0^1G(t,s)a(s)f(s,u(s))ds
=\int_0^1\max_{t\in[0,1]} G( t,s)a(s)f(s,u(s))ds \label{0}
\end{equation}
played an important role in the process of their proof.
Unfortunately, the equality \eqref{0} is not right. For a
counterexample, one can refer to our paper \cite{16}.

 Motivated greatly by the above-mentioned works, in this
paper we study the following third-order three-point BVP
\begin{equation}
\begin{gathered}
u'''(t)=f(t,u(t)),\quad t\in [0,1], \\
u'(0)=u(1)=u''(\eta)=0,
\end{gathered}\label{1.1}
\end{equation}
where $f\in C([0,1]\times[0,+\infty),\ [0,+\infty))$ and 
$\eta\in(\frac{1}{2},1)$. Although the corresponding Green's function
is sign-changing, we still obtain the existence of at least
$2m-1$ positive solutions for arbitrary positive integer
$m$ under suitable conditions on $f$.

 In the remainder of this section, we state some fundamental
concepts and the Leggett-Williams fixed point theorem \cite{2}.

 Let $E$ be a real Banach space with cone $P$. A map $\sigma:
P \to (-\infty,+\infty)$ is said to be a concave functional
if
\[
\sigma(tx+(1-t)y)\geq t\sigma(x)+(1-t)\sigma(y)
\]
for all $x,y\in P$ and $t\in [0,1]$. Let $a$ and $b$ be two numbers
with $0<a<b$ and $\sigma$ be a nonnegative continuous concave
functional on $P$. We define the following convex sets
\begin{gather*}
P_a=\{x\in P:\| x\|<a\}, \\
P(\sigma,a,b)=\{x\in P:a\leq \sigma(x),\ \| x\|\leq b\}.
\end{gather*}

\begin{theorem}[Leggett-Williams fixed point theorem]
Let $A: \overline{P_c}\to \overline{P_c}$ be completely
continuous and $\sigma$ be a nonnegative continuous concave
functional on $P$ such that $\sigma(x) \leq \| x\|$ for
all $x \in \overline{P_c}$.
Suppose that there exist $0<d<a<b \leq c$ such that
\begin{itemize}
\item[(i)] $\{x \in P(\sigma,a,b):\sigma(x)>a\}\neq \emptyset$ and 
 $\sigma(Ax)>a$ for $x\in P(\sigma,a,b)$;
\item[(ii)] $\| Ax\|<d$ for $\| x\| \leq d$;
\item[(iii)] $\sigma(Ax)>a$ for $x\in P(\sigma,a,c)$ with $\| Ax\|>b$.
\end{itemize}
Then $A$ has at least three fixed points $x_1,x_2,x_{3}$ in
$\overline{P_c}$ satisfying
\[
\| x_1\|<d,\ a<\sigma(x_2),\ \| x_{3}\|>d,\ \sigma(x_{3})<a.
\]
\end{theorem}

\section{Preliminaries}

\noindent

In this article, we  assume that Banach space
$E=C[0,1]$ is equipped with the norm $\| u\|=
\max_{t\in [0,1]}| u(t)|$.

For any $y\in E$, we consider the BVP
\begin{equation}
\begin{gathered}
u'''(t)=y(t),\quad t\in [0,1], \\
u'(0)=u(1)=u''(\eta)=0.
\end{gathered}{\label{4}}
\end{equation}
After a simple computation, we obtain the following expression of Green's 
function $G(t,s)$ of the BVP \eqref{4}: for $s\geq\eta$,
\[
G(t,s)=\begin{cases}
-\frac{(1-s)^{2}}{2},&  0\leq t\leq s\leq 1, \\
\frac{t^{2}-2st+2s-1}{2},& 0\leq s\leq t\leq 1
\end{cases}
\]
and for $s<\eta$,
\[
G(t,s)=\begin{cases}
\frac{-t^{2}-s^{2}+2s}{2}, & 0\leq t\leq s\leq 1, \\
-st+s, & 0\leq s\leq t\leq 1.
\end{cases}
\]
Obviously,
$G(t,s)\geq 0$  for $0\leq s< \eta$,  and 
$G(t,s)\leq 0$ for $\eta\leq s\leq 1$.
Moreover, for $s\geq\eta$,
\[
\max\{G(t,s):t\in[0,1]\}=G(1,s)=0
\]
and for $s<\eta$,
\[
\max\{G(t,s):t\in[0,1]\}=G(0,s)=-\frac{s^{2}}{2}+s.
\]


To obtain the existence of positive solutions for
\eqref{1.1}, we need to construct a suitable cone in $E$. 
Let $u$ be a solution of  \eqref{1.1}. Then it is easy to verify that
$u(t)\geq 0$  for $t\in [0,1]$ provided that $u'(1)\leq 0$.
In fact, since $f$ is nonnegative, we know that 
$u'''(t)\geq 0$ for $t\in [0,1]$, which together with $u''(\eta)=0$
implies that
\begin{equation}
u''(t)\leq 0 \text{ for  }t\in [0,\eta]\quad \text{and}\quad
u''(t)\geq 0 \text{ for  }t\in [\eta,1].{\label{1.001}}
\end{equation}
In view of \eqref{1.001} and $u'(0)=0$, we have
\begin{equation}
u'(t)\leq 0 \text{ for  }t\in [0,\eta]\quad \text{and}\quad
u'(t)\leq u'(1) \text{ for  }t\in [\eta,1]. {\label{2}}
\end{equation}
If $u'(1)\leq 0$, then it follows from \eqref{2} that 
$u'(t)\leq 0$  for $t\in [0,1]$, which together with  $u(1)=0$ implies
that $u(t)\geq 0$ for $t\in [0,1]$. Therefore, we define a
cone in $E$ as follows:
\[
\hat{P}=\{y\in E: y(t) \text{ is nonnegative and decreasing
on }[0,1]\}.
\]

\begin{lemma}[\cite{16}] \label{lem2.1}
 Let $y\in \hat{P}$  and $u(t)=\int_0^1G(t,s)y(s)ds$, $t\in [0,1]$. 
Then $u\in \hat{P}$ and $u$ is the unique solution of 
\eqref{4}. Moreover, $u$ satisfies
\[
\min_{t\in[1-\theta,\theta]}u(t)\geq\theta^{*}\| u\|,
\]
where $\theta\in (\frac{1}{2},\eta)$ and
$\theta^{*}=(\eta-\theta)/\eta$.
\end{lemma}

\section{Main results}

In the remainder of this paper, we  assume that $f:[0,1]\times
[0,+\infty)\to [0,+\infty)$
is continuous and satisfies the following two conditions:
\begin{itemize}
\item[(D1)] For each $x\in [0,+\infty)$, the mapping $t\mapsto f(t,x)$ is decreasing;

\item[(D2)] For each $t\in [0,1]$, the mapping $x\mapsto f(t,x)$ is increasing.
\end{itemize}
Let
\[
P=\{u\in
\hat{P}:\min_{t\in[1-\theta,\theta]}u(t)\geq\theta^{*}\|
u\|\}.
\]
Then it is easy to check that $P$ is a cone in $E$. Now, we define an
operator $A$ on $P$ by
\[
(Au)(t)=\int_0^1 G(t,s)f(s,u(s))ds,\quad t\in [0,1].
\]
Obviously, if $u$ is a fixed point of $A$ in $P$, then $u$ is a
nonnegative solution of  \eqref{1.1}.
For convenience, we denote
\[
H_1=\int_0^\eta\big(-\frac{s^{2}}{2}+s\big)ds, \quad 
H_2=\min_{t\in[1-\theta,\theta]}\int_{1-\theta}^\theta G(t,s)ds.
\]

\begin{theorem} \label{thm3.1}
Assume that there exist numbers $d,a$ and $c$
with $0<d<a<\frac{a}{\theta^{*}}\leq c$ such that
\begin{gather}
f(t,u)<\frac{d}{H_1},\quad t\in [0,\eta],\; u\in[0,d],\label{3.1}\\
f(t,u)>\frac{a}{H_2},\quad t\in [1-\theta,\theta],\;
u\in[a,\frac{a}{\theta^{*}}],\label{3.2} \\
f(t,u)<\frac{c}{H_1},\quad t\in [0,\eta],\; u\in[0,c]. \label{3.3}
\end{gather}
Then  \eqref{1.1} has at least three positive solutions $u$,
$v$ and $w$ satisfying
\[
\|u\|<d,\quad a<\min_{t\in[1-\theta,\theta]}v(t),\quad
d<\|w\|,\quad \min_{t\in[1-\theta,\theta]}w(t)<a.
\]
\end{theorem}

\begin{proof}
For $u\in P$, we define
\[
\sigma(u)=\min_{t\in[1-\theta,\theta]}u(t).
\]
It is easy to check that $\sigma$ is a nonnegative continuous concave 
functional on $P$ with $\sigma(u) \leq \|u\|$ for $u \in P$ and that $A:P\to P$ 
is completely continuous.

 We first assert that if there exists a positive number $r$
such that $f(t,u) < \frac{r}{H_1}$ for $t\in [0,\eta]$ and 
$u \in [0,r]$, then $A :\overline{P_{r}}\to P_{r}$.
Indeed, if $u\in \overline{P_{r}}$, then
\begin{align*}
\| Au\|
&= \max_{t\in[0,1]}\int_0^1G(t,s)f(s,u(s))ds\\
&\leq \int_0^1\max_{t\in[0,1]}G(t,s)f(s,u(s))ds\\
&= \int_0^\eta\max_{t\in[0,1]}G(t,s)f(s,u(s))ds
 +\int_\eta^1\max_{t\in[0,1]}G(t,s)f(s,u(s))ds\\
&= \int_0^\eta(-\frac{s^{2}}{2}+s)f(s,u(s))ds\\
&< \frac{r}{H_1}\int_0^\eta(-\frac{s^{2}}{2}+s)ds
= r;
\end{align*}
that is, $Au \in P_{r}$.

Hence, we have shown that if \eqref{3.1} and \eqref{3.3}
hold, then $A$ maps $\overline{P_{d}}$ into $P_{d}$ and $\overline{P_c}$ into $_c$.

 Next, we assert that 
$\{u \in P(\sigma,a,\frac{a}{\theta^{*}}): \sigma(u) > a\}\neq \emptyset$
and $\sigma(Au) > a$ for all $u \in P(\sigma, a, \frac{a}{\theta^{*}})$.
In fact, the constant function
$\frac{a+\frac{a}{\theta^{*}}}{2}$ belongs to 
$\{u \in P(\sigma,a,\frac{a}{\theta^{*}}): \sigma(u) > a\}$.

On the one hand, for $u\in P(\sigma,a,\frac{a}{\theta^{*}})$, we
have
\begin{equation}
a\leq\sigma(u)=\min_{t\in[1-\theta,\theta]}u(t)\leq u(t)
\leq \|u\|\leq \frac{a}{\theta^{*}} \label{3.35}
\end{equation}
for all $t\in [1-\theta,\theta]$.

Also,  for any $u\in P$ and $t\in [1-\theta,\theta]$, we
have
\begin{align*}
&\int_0^{1-\theta} G(t,s)f(s,u(s))ds+\int_\theta^\eta G(t,s)f(s,u(s))ds
 +\int_\eta^1 G(t,s)f(s,u(s))ds\\
&\geq \int_0^{1-\theta} (1-t)s f(s,u(s))ds
 -\int_\eta^1 \frac{(1-s)^{2}}{2}f(s,u(s))ds\\
&\geq  f(\eta,u(\eta))[\int_0^{1-\theta} (1-t)s ds
 -\int_\eta^1 \frac{(1-s)^{2}}{2}ds]\\
&\geq  f(\eta,u(\eta))[\int_0^{1-\theta} (1-t)s ds
 -\int_\theta^1 \frac{(1-s)^{2}}{2}ds]\\
&=  f(\eta,u(\eta))[\frac{(1-t)(1-\theta)^{2}}{2}-\frac{(1-\theta)^{3}}{6}]\\
&\geq  f(\eta,u(\eta))[\frac{(1-\theta)(1-\theta)^{2}}{2}
 -\frac{(1-\theta)^{3}}{6}]\\
&=  f(\eta,u(\eta))\frac{(1-\theta)^{3}}{3}
\geq  0,
\end{align*}
which together with \eqref{3.2} and \eqref{3.35} implies 
\begin{align*}
\sigma(Au)
&= \min_{t\in[1-\theta,\theta]}\int_0^1 G(t,s)f(s,u(s))ds\\
&\geq \min_{t\in[1-\theta,\theta]}\int_{1-\theta}^\theta G(t,s)f(s,u(s))ds\\
&> \frac{a}{H_2}\min_{t\in[1-\theta,\theta]}\int_{1-\theta}^\theta G(t,s)ds
=  a
\end{align*}
for $u\in P(\sigma,a,\frac{a}{\theta^{*}})$.

 Finally, we verify that if $u \in P(\sigma, a, c)$ and
$\| Au\| > a/\theta^{*}$ , then $\sigma(Au) > a$.
To see this, we suppose that $u \in P(\sigma, a, c)$ and
 $\| Au\| > a/\theta^{*}$. Then it follows from $Au \in P$
that
\[
\sigma(Au)= \min_{t\in[1-\theta,\theta]}(Au)(t)
\geq  \theta^{*}\| Au\| > a.
\]

 To sum up, all the hypotheses of the Leggett-Williams fixed
point theorem are satisfied. Therefore, $A$ has at least three fixed
points; that is,  \eqref{1.1} has at least three positive
solutions $u, v$ and $w$ satisfying
\[
\|u\|<d,\quad a<\min_{t\in[1-\theta,\theta]}v(t),\quad
d<\|w\|,\quad  \min_{t\in[1-\theta,\theta]}w(t)<a.
\]
\end{proof}

\begin{theorem}\label{thm3.2}
 Let $m$ be an arbitrary positive integer. Assume that there exist
numbers $d_i$ $(1\leq i\leq m)$ and $a_j$ $(1\leq j \leq m-1)$
with $0<d_1<a_1<\frac{a_1}{\theta^{*}} <d_2<a_2<
\frac{a_2}{\theta^{*}}<\dots<d_{m-1}<a_{m-1}<\frac{a_{m-1}}{\theta^{*}}<d_{m}$
such that
\begin{gather}
f(t,u)<\frac{d_i}{H_1},\quad t\in [0,\eta],\; u\in[0,d_i],\; 1\leq
i\leq m,\label{3.4} \\
f(t,u)>\frac{a_j}{H_2},\quad t\in [1-\theta,\theta],\;
u\in[a_j,\frac{a_j}{\theta^{*}}],\; 1\leq j \leq
m-1.\label{3.5}
\end{gather}
Then  \eqref{1.1} has at least $2m-1$ positive solutions in
$\overline{P_{d_{m}}}$.
\end{theorem}

\begin{proof}
We use induction on $m$. First, for $m = 1$, we know from
\eqref{3.4} that $A :\overline{P_{d_1}}\to P_{d_1}$.
Then it follows from Schauder fixed point theorem that 
\eqref{1.1} has at least one positive solution in
$\overline{P_{d_1}}$.

Next, we assume that this conclusion holds for $m = k$. To
show that this conclusion also holds for $m=k+1$, we suppose that
there exist numbers $d_i$  $(1\leq i\leq k+1)$ and $a_j$
$(1\leq j \leq k)$ with
 $0<d_1<a_1<\frac{a_1}{\theta^{*}} <d_2<a_2<
\frac{a_2}{\theta^{*}}<\dots<d_k<a_k<\frac{a_k}{\theta^{*}}<d_{k+1}$
such that
\begin{gather}
f(t,u)<\frac{d_i}{H_1},\quad t\in [0,\eta],\; u\in[0,d_i],\; 1\leq
i\leq k+1,\label{3.6}
\\
f(t,u)>\frac{a_j}{H_2},\quad t\in [1-\theta,\theta],\;
u\in[a_j,\frac{a_j}{\theta^{*}}],\; 1\leq j \leq k.\label{3.7}
\end{gather}
By assumption,  \eqref{1.1} has at least $2k-1$ positive
solutions $u_i\ (i = 1,2,\dots,2k-1)$ in $\overline{P_{d_k}}$.
At the same time, it follows from Theorem \ref{thm3.1}, 
\eqref{3.6} and \eqref{3.7} that  \eqref{1.1} has at least three positive
solutions $u, v$ and $w$ in $\overline{P_{d_{k+1}}}$
 such that
\[
\|u\|<d_k,\quad a_k<\min_{t\in[1-\theta,\theta]}v(t),\quad
d_k<\|w\|,\quad \min_{t\in[1-\theta,\theta]}w(t)<a_k.
\]
Obviously, $v$ and $w$ are different from $u_i$\
$(i =1,2,\dots,2k-1)$. Therefore,  \eqref{1.1} has at least
$2k+1$ positive solutions in $\overline{P_{d_{k+1}}}$, which shows
that this conclusion also holds for $m=k+1$.
\end{proof}

\begin{example} \label{examp1} \rm 
We consider the BVP
\begin{equation}
\begin{gathered}
u'''(t)=f(t,u(t)),\quad t\in [0,1], \\
u'(0)=u(1)=u''(\frac{2}{3})=0,
\end{gathered}\label{example}
\end{equation}
where
\[
f(t,u)=\begin{cases}
(1-t)(u+1)^{2}, &(t,u)\in[0,1]\times[0,1],\\
(1-t)[122(u-1)+4], &(t,u)\in[0,1]\times[1,2],\\
14(1-t)(u+1)^{2}, & (t,u)\in[0,1]\times[2,20],\\
6174(1-t), & (t,u)\in[0,1]\times[20,+\infty).
\end{cases}
\]
Let $\theta=3/5$. Then $\theta^{*}=1/10$. A simple
calculation shows that
$H_1=14/81$ and $H_2=1/25$.
If we choose
$d=1$, $a=2$, $c=1068$,
then all the conditions of Theorem \ref{thm3.1} are satisfied. Therefore, it
follows from Theorem \ref{thm3.1} that  \eqref{example} has at least three
positive solutions.
\end{example}

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\end{document}