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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 127, 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  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2005/127\hfil Eigenvalues and symmetric positive solutions]
{Eigenvalues and symmetric positive solutions for a
 three-point boundary-value problem}

\author[Y. Sun\hfil EJDE-2005/127\hfilneg]
{Yongping Sun}

\address{Yongping Sun \hfill\break
Department of Fundamental Courses,
Hangzhou Radio  \& TV University,
Hangzhou,  Zhejiang 310012, China}
\email{sunyongping@126.com}

\date{}
\thanks{Submitted September 15, 2005. Published November 23, 2005.}
\thanks{Supported by grants 10471075 from NSFC, and 20051897
 from the Zhejiang Provincial \hfill\break\indent
  Department of Education of China}
\subjclass[2000]{34B10, 34B15} 
\keywords{Symmetric positive 
solution; three-point boundary-value problem; \hfill\break\indent
Schauder fixed point theorem; eigenvalue}


\begin{abstract}
 In this paper, we consider the
 second-order three-point boundary-value problem
 \begin{gather*}
 u''(t)+f(t,u,u',u'')=0,\quad  0\leq t\leq 1,\\
 u(0)=u(1)=\alpha u(\eta).
 \end{gather*}
 Under suitable conditions and using Schauder fixed point
 theorem, we prove the existence of at least one symmetric positive
 solution. We also study the existence of positive
 eigenvalues for this problem. We emphasis the
 highest-order derivative occurs nonlinearly in our problem.
\end{abstract}

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

\section{Introduction}

 In this paper, we discuss the existence of symmetric positive
solution for the  second-order three-point boundary-value problem (BVP)
\begin{gather}
u''(t)+f(t,u,u',u'')=0,\quad  0\leq t\leq 1,  \label{e1.1}\\
u(0)=u(1)=\alpha u(\eta), \label{e1.2}
\end{gather}
where $\alpha\in (0,1)$,
$\eta\in (0,\frac{1}{2}]$;
$f:[0,1]\times [0,\infty)\times(-\infty,\infty)
\times(-\infty,0]\to[0,\infty)$ is
continuous, $ f(t,0,0,0)\not\equiv 0$, $t\in [0,1]$,
$f(\cdot,u,v,w)$ is symmetric on $[0,1]$ for all
$(u,v,w)\in[0,\infty)\times(-\infty,\infty)\times(-\infty,0]$, in
which the second-order derivative may occur nonlinearly. We also
establish new result for second order differential equations of
the form
\[
u''(t)+\lambda f(t,u,u',u'')=0,\quad  0\leq t\leq 1,
\]
subject to condition \eqref{e1.2}.

The three-point boundary-value problems for ordinary differential
equations arise in a variety of different areas of applied
mathematics and physics. In the past few years, there has been much
attention focused on questions of positive solutions of three-point
boundary-value problems for nonlinear ordinary differential
equations. The main approach is reformulate the problem to an
operator equation of the form $u=Au$, where $A$ is a suitable
operator and then applying a fixed point theorem, such as
Krasnoselskii's fixed point theorem or Leggett-Williams fixed point
theorem and so on; see for example \cite{f1}-\cite{w1} and
references therein. But for the existence of symmetric positive
solutions has received very little attention in the literature. Thus
the aim of the present paper is to establish a simple criterion for
the existence of positive solution to the BVP
\eqref{e1.1}-\eqref{e1.2} by Schauder fixed point theorem. The
emphasis in this work is the highest-order derivative occurs
nonlinearly in our problem and we study the existence of the
symmetric positive solution.

The organization of this paper is as follows. In Section 2, we
present some preliminary results that will be used to prove our
main results. In Section 3, we discuss the existence of symmetric
positive solution for the BVP \eqref{e1.1}-\eqref{e1.2}, the existence of
positive eigenvalues for this problem and we give two examples to
illustrate our results.

\section{Preliminaries}

Consider the three-point boundary-value problem
\begin{gather}
 u''+h(t)=0,\quad 0\leq t\leq 1,  \label{e2.1}\\
 u(0)=u(1)=\alpha u(\eta), \label{e2.2}
 \end{gather}
where $\eta\in(0,1)$.

\begin{lemma}\label{lem2.1}
Let $\alpha\neq 1$, $h\in C[0,1]$. Then the three-point
BVP \eqref{e2.1}-\eqref{e2.2} has a unique solution
\begin{equation}
u(t)=\int_0^1G(t,s)h(s)ds+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds,
\label{e2.3}
\end{equation}
where
\begin{equation}
G(x,y)= \begin{cases}
x(1-y), &0\leq x\leq y\leq 1,\\
y(1-x), &0\leq y\leq x\leq 1.
\end{cases} \label{e2.4}
\end{equation}
\end{lemma}

\begin{proof} From \eqref{e2.1} we have
$u''(t)=-h(t)$.
For $t\in [0,1]$, integrating from $0$ to $t$, we get
$$
u'(t)=-\int_0^th(s)ds+B.
$$
For $t\in [0,1]$, integrate from $0$ to $t$ yields
$$
u(t)=-\int_0^t\Big(\int_0^xh(s)ds\Big)dx+Bt+A,
$$
i.e.
$$ u(t)=-\int_0^t(t-s)h(s)ds+Bt+A. $$
So,
\begin{gather*}
u(0)=A,\\
 u(1)=-\int_0^1(1-s)h(s)ds+B+A,\\
u(\eta)=-\int_0^{\eta}(\eta-s)h(s)ds+\eta B+A.
\end{gather*}
Combining this with \eqref{e2.2}, we have
\begin{gather*}
B=\int_0^1(1-s)h(s)ds,\\
A=\frac{\alpha\eta}{1-\alpha}\int_0^1(1-s)h(s)ds
-\frac{\alpha}{1-\alpha}\int_0^{\eta}(\eta-s)h(s)ds.
\end{gather*}
Therefore, the three-point BVP \eqref{e2.1}-\eqref{e2.2} has a unique
solution
\begin{align*}
u(t)=&-\int_0^t(t-s)h(s)ds+t\int_0^1(1-s)h(s)ds\\
     &+\frac{\alpha\eta}{1-\alpha}\int_0^1(1-s)h(s)ds-\frac{\alpha}{1-\alpha}\int_0^\eta(\eta-s)h(s)ds\\
    =&\int_0^1G(t,s)h(s)ds+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds\,.
\end{align*}
This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.1} \rm
For any $x,y\in[0,1]$, we have
$$ G(x,y)\leq G(x,x)\ \hbox{and}\ G(x,y)=G(1-x,1-y).$$
\end{remark}

 From \eqref{e2.3} we obtain the following lemma.

\begin{lemma}\label{lem2.2}
Let $\alpha\in(0,1)$, $h\in C^+[0,1]$. Then the unique solution $u(t)$ of the
BVP \eqref{e2.1}-\eqref{e2.2} is nonnegative on $[0,1]$, and if
$h(t)\not\equiv 0$, then $u(t)>0$, for all $t\in [0,1]$.
\end{lemma}

\begin{proof} Let $y\in C^+[0,1]$. From the fact that
$u''(t)=-y(t)\leq 0$, for all $t\in [0,1]$, we know that the
graph of $u(t)$ is concave on $[0,1]$. From \eqref{e2.2} and \eqref{e2.3} we
have that
\begin{align*}
u(1)=u(0)=& \frac{\alpha\eta}{1-\alpha}\int_0^1(1-s)y(s)ds
 -\frac{\alpha}{1-\alpha}\int_0^{\eta}(\eta-s)y(s)ds\\
=&\frac{\alpha\eta}{1-\alpha}\int_\eta^1(1-s)y(s)ds
 +\frac{\alpha}{1-\alpha}\int_0^{\eta}[\eta(1-s)-(\eta-s)]y(s)ds\\
=&\frac{\alpha\eta}{1-\alpha}\int_\eta^1(1-s)y(s)ds
 +\frac{\alpha(1-\eta)}{1-\alpha}\int_0^{\eta}sy(s)ds\geq 0.
\end{align*}
It follows that $u(t)\geq 0$, for all $t\in [0,1]$, and if
$h(t)\not\equiv 0$, then $u(t)>0$, for all  $t\in [0,1]$.
\end{proof}

\begin{lemma}\label{lem2.3}
Let $\alpha,\eta\in(0,1)$, $h(t)$ be symmetric on $[0,1]$.
Then the unique solution $u(t)$ of the BVP \eqref{e2.1}-\eqref{e2.2} is
symmetric on $[0,1]$.
\end{lemma}

\begin{proof} From \eqref{e2.3}, we have
$$
u(t)=\int_0^1G(t,s)h(s)ds+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds.
$$
Therefore,
\begin{align*}
u(1-t)=&\int_0^1G(1-t,s)h(s)ds+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds\\
=&\int_1^0G(1-t,1-s)h(1-s)d(1-s)+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds\\
=&\int_0^1G(t,s)h(s)ds+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)h(s)ds\\
=& u(t).
\end{align*}
i.e., $u(t)$ is symmetric on $[0,1]$.
\end{proof}

Define an integral operator $T: E\to E$ by
\begin{equation}
T u(t)=\int_0^1G(t,s)f(s,u,u',u'')ds
+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)f(s,u,u',u'')ds.
\end{equation}
for $t\in [0,1]$.
 It is easy to see that \eqref{e1.1}-\eqref{e1.2} has a solution
 $u=u(t)$ if and only
if $u$ is a fixed point of the operator $T$.
The main tool in our approach is the following Schauder fixed point
theorem (See \cite{d1}).

\begin{theorem}\label{thm2.1}
 Let $E$ be Banach space and $B\subset E$ be a bounded closed convex
subset, $T:E\to E$ be a completely
continuous operator such that
$T(B)\subset B$.
Then $T$ has a fixed point in $B$.
\end{theorem}

\section{Main results}

 In this section, we study the
existence of positive solution for the BVP \eqref{e1.1}-\eqref{e1.2}. We obtain
the following existence results.

\begin{theorem}\label{thm3.1}
Assume that $\alpha,\eta\in(0,1),
f:[0,1]\times[0,\infty)\times(-\infty,\infty)\times(-\infty,0]\to[0,\infty)$
is continuous, $ f(t,0,0,0)\not\equiv 0$,
$t\in [0,1]$, $f(\cdot,u,v,w)$ is symmetric on $[0,1]$ for all
$(u,v,w)\in[0,\infty)\times(-\infty,\infty)\times(-\infty,0]$. If
there exist constant $M$ such that
$$
\max\Big\{1,\frac{1-\alpha+4\alpha\eta(1-\eta)}{4(1-\alpha)}\Big\}L\leq 2M,
$$
where
$$
L=\max\{f(t,u,v,w): 0\leq t\leq 1,\, 0\leq u\leq M,\, -M\leq
v\leq M,\, -2M\leq w\leq 0\}.
$$
Then $\eqref{e1.1}-\eqref{e1.2}$ has at least one symmetric
positive solution.
\end{theorem}

\begin{proof} Let $E=C^2[0,1]$ be a Banach space with
norm $\|u\|=\max\{|u|_0,|u'|_0,|u''|_0\}$, where
$|u|_0=\max_{0\leq t\leq 1} |u(t)|$,
$
B=\{u(t):u(t)\in C^2[0,1] \hbox{ is symmetric on }[0,1],
\, 0\leq u(t)\leq M,\, -M\leq u'(t)\leq M,\, -2M\leq u''(t)\leq
0\}$,
then $B$ is a bounded closed convex subset of $E$. Now we
prove
\begin{equation}
T(B)\subset B. \label{e3.1}
\end{equation}
In fact, for any $u\in B$, from \eqref{e2.3}, \eqref{e2.4} and
Remark 2.2 we have
\begin{align*}
Tu(t)=&\int_0^1G(t,s)f(s,u,u',u'')ds
+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)f(s,u,u',u'')ds\\
\leq&\Big(\int_0^1G(t,s)ds
+\frac{\alpha}{1-\alpha}\int_0^1G(\eta,s)ds\Big)L\\
\leq&\left(\int_0^1t(1-t)ds
+\frac{\alpha}{1-\alpha}\int_0^\eta (1-\eta)sds+\frac{\alpha}{1-\alpha}\int_\eta^1\eta(1-s)ds\right)L\\
=&\Big(\frac{1}{2}t(1-t)+\frac{\alpha\eta(1-\eta)}{2(1-\alpha)}\Big) L\\
\leq&\frac{1-\alpha+4\alpha\eta(1-\eta)}{8(1-\alpha)}L\leq M,
\end{align*}
which implies
\begin{equation}
0\leq Tu(t)\leq M,\quad \forall\ t\in [0,1].  \label{e3.2}
\end{equation}
In addition, from
\[
(Tu)'(t)=-\int_0^tsf(s,u,u',u'')ds +\int_t^1(1-s)f(s,u,u',u'')ds
\]
we have
\begin{align*}
 (Tu)'(t)&\leq \int_0^1(1-s)f(s,u,u',u'')ds\\
&\leq L\int_0^1(1-s)ds\\
&=\frac{1}{2}L\leq M
\end{align*}
and
\begin{align*}
(Tu)'(t)&\geq -\int_0^tsf(s,u,u',u'')ds\\
&\geq -\int_0^1sf(s,u,u',u'')ds\\
&\geq-\frac{1}{2}L\geq-M
\end{align*}
 which implies
\begin{equation}
-M\leq (Tu)'(t)\leq M,\quad \forall\ t\in [0,1]. \label{e3.3}
\end{equation}
Finally, from
\[
(Tu)''(t)=-f(t,u,u',u''),\quad \forall\ t\in [0,1].
\]
 we know
\begin{equation}
0\geq (Tu)''(t)\geq -L\geq -2M,\quad \forall\ t\in [0,1].
\label{e3.4}
\end{equation}
Therefore, from
\eqref{e3.2}, \eqref{e3.3}, \eqref{e3.4}  and Lemma \ref{lem2.1} we have
$Tu\in B$ which implies \eqref{e3.1}.
Thus, by Schauder fixed theorem, $T$ has a fixed point
$u^\ast\in B$. From $f(t,0,0,0)\not\equiv 0$, $t\in [0,1]$.
We know $u^\ast$ is a symmetric positive solution of
 \eqref{e1.1}-\eqref{e1.2}.
\end{proof}

\begin{example} \label{ex3.1} \rm
 Consider the three-point boundary-value problem
\begin{gather}
u''(t)+4t(1-t)(1+\sqrt{u})+\frac{1}{8}\min\{t,1-t\}(u')^2+\sqrt{-2u''}=0,\quad
 0\leq t\leq 1, \label{e3.5}\\
 u(0)=u(1)=\frac{1}{2}u(\frac{1}{4}). \label{e3.6}
\end{gather}
Set $f(t,u,v,w)=4t(1-t)(1+\sqrt{u})+\frac{1}{8}\min\{t,1-t\}v^2+\sqrt{-2w}$,
$M=16$. A direct computation shows $L=29$ and
$\max\{1,\frac{1-\alpha+4\alpha\eta(1-\eta)}{4(1-\alpha)}\}L=29<
32= 2M$, by Theorem \ref{thm3.1},
the BVP \eqref{e3.5}-\eqref{e3.6} has at least one symmetric positive solution.
\end{example}

As an application of Theorem \ref{thm3.1}, we consider the eigenvalue
problem
\begin{gather}
 u''(t)+\lambda f(t,u,u',u'')=0,\quad  0\leq t\leq 1, \label{e3.7}\\
 u(0)=u(1)=\alpha u(\eta), \label{e3.8}
\end{gather}
We have the following existence result.

\begin{theorem}\label{thm3.2}
Assume that $\alpha,\, \eta\in(0,1)$ are
constants, $\lambda>0$ is parameter,
$f:[0,1]\times[0,\infty)\times(-\infty,\infty)\times(-\infty,0]\to[0,\infty)$
is continuous, $ f(t,0,0,0)\not\equiv 0$, $t\in [0,1]$,
$f(\cdot,u,v,w)$ is symmetric on $[0,1]$ for all
$(u,v,w)\in[0,\infty)\times(-\infty,\infty)\times(-\infty,0] $. Then
for each
$$
\lambda\in
\Big(0,\frac{2M}{\max\{1,\frac{1-\alpha+4\alpha\eta(1-\eta)}{4(1-\alpha)}\}L}
\Big],
$$
where $M>0$ and $L=\max\{f(t,u,v,w)|\ 0\leq t\leq 1,\ 0\leq u\leq
M,\ -M\leq v\leq M,\ -2M\leq w\leq 0\}$, the eigenvalue problem
\eqref{e3.7}-\eqref{e3.8} has at least one symmetric positive solution.
\end{theorem}

\begin{example} \label{ex3.2}\rm
 Consider the eigenvalue problem
\begin{gather}
u''(t)+\lambda f(t,u,u',u'')=0,\quad  0\leq t\leq 1, \label{e3.9}\\
u(0)=u(1)=\frac{15}{16}u(\frac{1}{2}), \label{e3.10}
\end{gather}
where
$f(t,u,v,w)=1+\frac{4}{13}t(1-t)(1+3\sqrt{u})+\frac{1}{4}t(1-t)|v|
-\frac{1}{16}\min\{t,1-t\}w$. Set $ M=16$. A direct computation
shows $L=4$ and
$\max\{1,\frac{1-\alpha+4\alpha\eta(1-\eta)}{4(1-\alpha)}\}L=16$, by
Theorem \ref{thm3.2}, the eigenvalue
problem \eqref{e3.9}-\eqref{e3.10} has at least one symmetric
positive solution.
\end{example}

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\end{document}