\documentclass[reqno]{amsart}
\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2004(2004), No. 06, pp. 1--8.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2004/06\hfil Triple positive solutions]
{Triple positive solutions for a class of two-point boundary-value problems}
\author[Z. Bai, Y. Wang, \& W. Ge\hfil EJDE-2004/06\hfilneg]
{Zhanbing Bai, Yifu Wang, \& Weigao Ge} % in alphabetical order
\address{Zhanbing Bai\hfill\break
Department of Applied Mathematics,
Beijing Institute of Technology, Beijing 100081, China
\\
Department of Applied
Mathematics, University of Petroleum, Dongying 257061, China}
\email{baizhanbing@263.net}
\address{Yifu Wang \hfill\break
Department of Applied Mathematics,
Beijing Institute of Technology, Beijing 100081, China}
\email{yifu-wang@163.com}
\address{Weigao Ge \hfill\break
Department of Applied Mathematics,
Beijing Institute of Technology,
Beijing 100081, China}
\email{gew@bit.edu.cn}
\date{}
\thanks{Submitted November 25, 2003. Published January 2, 2004.}
\thanks{Supported by grants 10371006 from the National Nature Science
Foundation of China, \hfill\break\indent
and 1999000722 from the Doctoral Program Foundation of Education Ministry of China.}
\subjclass[2000]{34B15}
\keywords{Triple positive solutions, boundary-value problem, \hfill\break\indent
fixed-point theorem}
\begin{abstract}
We obtain sufficient conditions for the existence of
at least three positive solutions for the equation
$ x''(t) + q(t)f(t, x(t), x'(t)) = 0 $ subject to some boundary
conditions. This is an application of a new fixed-point theorem
introduced by Avery and Peterson \cite{AvP}.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks
\section{Introduction}
Recently, the existence and multiplicity of positive solutions for
nonlinear ordinary differential equations and difference equations
have been studied extensively. To identify a few, we refer the
reader to \cite{Ag,Av,AvA,AvH,AvH2,AvP,BW,Guo,HeW,LW,Kr,Li,LL}.
The main tools used in above works are fixed-point theorems.
Fixed-point theorems and their applications to nonlinear problems
have a long history, some of which is documented in Zeidler's book
\cite{Ze}, and the recent book by Agarwal, O'Regan and Wong
\cite{Ag} contains an excellent summary of the current results and
applications.
An interest in triple solutions evolved from the Leggett-Williams
multiple fixed-point theorem \cite{LW}. And lately, two triple
fixed-point theorems due to Avery \cite{Av} and Avery and Peterson
\cite{AvP} have been applied to obtain triple solutions of certain
boundary-value problems for ordinary differential equations as
well as for their discrete analogues.
Avery and Peterson \cite{AvP}, generalize the fixed-point
theorem of Leggett-Williams by using theory of fixed-point index
and Dugundji extension theorem. An application of the theorem be
given to prove the existence of three positive solutions to the
following second-order discrete boundary-value problem
\begin{gather*}
\Delta^2x(k-1) + f(x(k)) = 0, \quad \mbox{for all } k \in [a+1, b+1], \\
x(a)=x(b+2)=0,
\end{gather*}
where $f: \mathbb{R} \to \mathbb{R}$ is continuous and nonnegative
for $x \ge 0$.
In this paper, we concentrate in getting three positive
solutions for the second-order differential equation
\begin{equation}\label{11}
x''(t) + q(t)f(t, x(t), x'(t)) = 0, \quad 0b\}
\neq \empty$ and $\alpha(Tx) >b $ for $x \in P(\gamma, \theta, \alpha, b, c,
d)$;
\item [(S2)] $\alpha(Tx) >b$ for $x \in P(\gamma, \alpha, b,
d)$ with $\theta(Tx) >c$;
\item [(S3)] $0 \not\in R(\gamma, \psi, a, d)$ and $\psi(Tx) From the fact $x''(t)= -f(t, x, x') \le 0$, we know
that $x$ is concave on $[0, 1]$. So, define the cone $P \subset X
$ by
$$
P= \{ x \in X : x(t) \geq 0, x(0)=x(1)=0, x \text{ is concave on }
[0, 1]\} \subset X .
$$
Let the nonnegative continuous concave functional $\alpha$, the
nonnegative continuous convex functional $\theta, \gamma$, and the
nonnegative continuous functional $\psi$ be defined
on the cone $P$ by
$$\gamma(x) = \max_{0 \leq t \leq 1}|x'(t)|,
\quad \psi(x) = \theta(x) = \max_{0 \leq t \leq 1}|x(t)|,
\quad \alpha(x)=\min_{\frac{1}{4} \le t\le \frac{3}{4} }|x(t)|.
$$
\begin{lemma}\label{lm31}
If $x \in P$, then
$\max_{0 \leq t \leq 1}|x(t)| \le \frac{1}{2}\max_{0 \leq t \leq 1}|x'(t)|$.
\end{lemma}
\begin{proof} To the contrary, suppose that there exist $t_0 \in (0,1)$ such that
$|x(t_0)| > \frac{1}{2}\max_{0 \leq t \leq 1}|x'(t)|=:A$.
Then by the mid-value theorem there exist
$t_1\in (0, t_0)$, $t_2 \in (t_0, 1)$ such that
$$
x'(t_1) = \frac{x(t_0)-x(0)}{t_0}=\frac{x(t_0)}{t_0}, \quad
x'(t_2) = \frac{x(1)-x(t_0)}{1-t_0}=\frac{-x(t_0)}{1-t_0}.
$$
Thus, $\max_{0 \leq t \leq 1}|x'(t)| \ge \max\left\{|x'(t_1)|,
|x'(t_2)|\right\} >2A$, it is a contradiction. The proof is
complete.
\end{proof}
By Lemma \ref{lm31} and their definitions, and the
concavity of $x$, the functionals defined above satisfy:
\begin{equation}\label{31}
\frac{1}{4} \theta(x) \le \alpha(x) \le \theta(x) = \psi(x),\quad
\|x\| = \max\{\theta(x), \gamma(x)\} = \gamma(x),
\end{equation}
for all $x \in \overline{P(\gamma, d)} \subset P$.
Therefore, Condition \eqref{21} is satisfied.
Denote by $G(t, s) $ the Green's function for boundary-value problem
\begin{gather*}
-x''(t) =0, \quad 0 \frac{b}{\delta}$, for
$(t, u, v) \in [1/4, 3/4] \times [b, 4b] \times [-d, d]$;
\item[(A3)] $ f(t, u, v) < \frac{a}{N}$, for
$(t, u, v) \in [0, 1] \times [0, a] \times [-d, d]$.
\end{itemize}
\begin{theorem} \label{th31}
Under assumptions (A1)--(A3), the boundary-value problem \eqref{11}-\eqref{12}
has at least three positive solutions $x_1$, $x_2$, and $x_3$ satisfying
\begin{equation} \label{x123}
\begin{gathered}
\max_{0 \leq t \leq 1}|x_i'(t)| \le d, \quad \mbox{for } i=1, 2, 3;\\
b < \min_{\frac{1}{4} \le t\le \frac{3}{4} }|x_1(t)|; \\
a< \max_{0 \leq t \leq 1}|x_2(t)|, \quad \mbox{with }
\min_{\frac{1}{4} \le t\le \frac{3}{4} }|x_2(t)| **b$, and so $\{x \in P(\gamma, \theta, \alpha, b, 4b, d) \mid \alpha(x)>b\} \neq \emptyset
$. Hence, if $x \in P(\gamma, \theta, \alpha, b, 4b, d)$, then $b
\le x(t) \le 4b, |x'(t)| \le d $ for $1/4 \le t \le 3/4$. From
assumption (A2), we have $f(t, x(t), x'(t)) \ge \frac{b}{\delta}$
for $1/4 \le t \le 3/4$, and by the conditions of $\alpha$ and the
cone $P$, we have to distinguish two cases, (i) $\alpha(Tx)=
(Tx)(1/4)$ and (ii) $\alpha(Tx)= (Tx)(3/4)$.
In case (i), we have
\[
\alpha(Tx)=(Tx)(\frac{1}{4})
= \int_0^1G(\frac{1}{4}, s)q(s)f(s, x(s), x'(s))
ds > \frac{b}{\delta}\cdot \int_{1/4}^{3/4} G(\frac{1}{4}, s)q(s)ds \ge b\,.
\]
In case (ii), we have
\[
\alpha(Tx)=(Tx)(\frac{3}{4})
= \int_0^1G(\frac{3}{4}, s)f(s, x(s), x'(s))q(s) ds
> \frac{b}{\delta}\cdot \int_{1/4}^{3/4} G(\frac{3}{4}, s)q(s)ds \ge b;
\]
i.e.,
\begin{equation*}
\alpha(Tx) >b, \text{ for all } x \in P(\gamma, \theta, \alpha, b,
4b, d).
\end{equation*}
This show that condition (S1) of Theorem \ref{th21} is satisfied.
Secondly, with \eqref{31} and $b \le \frac{d}{8}$, we have
$$
\alpha(Tx) \ge \frac{1}{4}\theta(Tx) > \frac{4b}{4}=b,
$$
for all $x \in P(\gamma, \alpha, b, d)$ with $\theta(Tx) >4b$.
Thus, condition (S2) of Theorem \ref{th21} is satisfied.
We finally show that (S3) of Theorem \ref{th21} also holds.
Clearly, as $\psi(0)=0 b/\delta_1$, for
$(t, u, v) \in [1/2, 1] \times [b, 2b] \times [-d, d]$
\item[(A6)] $ f(t, u, v) < \frac{a}{N_1}$, for
$(t, u, v) \in [0, 1] \times [0, a] \times [-d, d]$.
\end{itemize}
\begin{theorem} \label{th32}
Under assumption (A4)--(A6), the boundary-value problem \eqref{11}-\eqref{13}
has at least three positive solutions $x_1$, $x_2$, and $x_3$ satisfying
\begin{gather*}
\max_{0 \leq t \leq 1}|x_i'(t)| \le d, \quad \mbox{for } i=1, 2, 3;\\
b < \min_{\frac{1}{2} \le t\le 1 }|x_1(t)|;\\
a< \max_{0 \leq t \leq 1}|x_2(t)|, \quad \mbox{with }
\min_{\frac{1}{2} \le t\le 1 }|x_2(t)| ****\frac{b}{\delta}=32,\; \mbox{ for } 1/4 \le t \le 3/4, 2 \le u \le 8, -3000 \le v\le 3000; \\
f(t, u, v) & <\frac{d}{M}=6000,\; \mbox{ for } 0 \le t \le 1, 0 \le u \le 1500, -3000 \le v\le 3000.
\end{align*}
Then all assumptions of Theorem \ref{th31} hold. Thus, with
Theorem \ref{th31}, Problem \eqref{33} has at least
three positive solutions $x_1, x_2, x_3 $ such that
\begin{gather*}
\max_{0 \leq t \leq 1}|x_i'(t)| \le 3000, \quad \mbox{for } i=1, 2, 3;\\
2 < \min_{\frac{1}{4} \le t\le \frac{3}{4} }|x_1(t)|; \\
1< \max_{0 \leq t \leq 1}|x_2(t)|, \quad \mbox{with }
\min_{\frac{1}{4} \le t\le \frac{3}{4} }|x_2(t)| <2; \\
\max_{0 \leq t \leq 1}|x_3(t)| <1\,.
\end{gather*}
\begin{remark} \rm
The early results, see \cite{Ag, Av, AvA, AvH2, AvP, LW}, for example, are not
applicable to the above problem. In conclusion, we see that the nonlinear term is
involved in first derivative explicitly.
\end{remark}
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\end{document}
**