\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 0<t<1
\end{equation}
subject to one of the following two pairs of boundary conditions:
\begin{gather}\label{12}
x(0)=0=x(1), \\
\label{13}
x(0)=0=x'(1).
\end{gather}
We are concerned with positive solutions to the above problem, i.e.,
$x(t) \ge 0$ on $[0, 1]$.
In this article, it is assumed that:
\begin{itemize}
\item[(C1)]  $f \in C([0, 1] \times [0, \infty ) \times \mathbb{R}, [0, \infty))$;

\item[(C2)]  $q(t)$ is nonnegative measurable function defined in
$(0,1)$, and $q(t)$ does not identically vanish on any subinterval of
$(0, 1)$. Furthermore, $q(t)$ satisfies $0 <\int_0^1t(1-t)q(t)dt<\infty$.
\end{itemize}
Our main results will depend on an application of a fixed-point
theorem due to Avery and Peterson which deals with fixed points of
a cone-preserving operator defined on an ordered Banach space. The
emphasis here is the nonlinear term be involved explicitly with
the first-order derivative. To the best of the authors knowledge,
there are no results for triple positive solutions by using the
Leggett-Williams fixed-point theorem or its generalizations.

\section{Background materials and definitions}

For the convenience of the reader, we present here the necessary
definitions from cone theory in Banach spaces; these definitions
can be found in  recent literature.

\begin{definition} \rm
Let $E$ be a real Banach space over $\mathbb{R}$. A nonempty
convex closed set $P \subset E$ is said to be a cone provided that
\begin{itemize}
  \item [(i)] $au  \in P$ for all $u\in P$ and all $a \ge
  0$ and
  \item [(ii)] $u, -u \in P$ implies $u=0$.
\end{itemize}
\end{definition}
Note that every cone $P\subset E$ induces an ordering in $E$ given by
$x \le y$ if $y-x \in P$.

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

\begin{definition} \rm
The map $\alpha$ is said to be a nonnegative continuous concave
functional on a cone $P$ of a real Banach space $E$ provided that
$\alpha: P \to [0, \infty)$ is continuous and
$$\alpha(tx +(1-t)y) \ge t \alpha(x) + (1-t)\alpha(y)$$
for all $x, y \in P$ and $0 \le t \le 1$. Similarly, we say the
map $\beta$ is a nonnegative continuous convex functional on a
cone $P$ of a real Banach space $E$ provided that $\beta: P \to
[0, \infty)$ is continuous and
$$\beta(tx +(1-t)y) \le t \beta(x) + (1-t)\beta(y)$$
for all $x, y \in P$ and $0 \le t \le 1$.
\end{definition}

Let $\gamma$ and $\theta$ be nonnegative continuous convex
functionals on $P$, $\alpha$ be a nonnegative continuous concave
functional on $P$, and $\psi$ be a nonnegative continuous
functional on $P$. Then for positive real numbers $a, b, c$, and
$d$, we define the following convex sets:
\begin{gather*}
P(\gamma, d)  = \{x \in P \mid \gamma(x) <d \},\\
P(\gamma, \alpha, b, d)  = \{x \in P \mid b \le \alpha(x), \gamma(x) \le d \},\\
P(\gamma, \theta, \alpha, b, c, d)  = \{x \in P \mid b \le
\alpha(x), \theta(x) \le c, \gamma(x) \le d \},
\end{gather*}
and a closed set
$$
R(\gamma, \psi, a, d)  = \{x \in P \mid a \le \psi(x), \gamma(x) \le d\}.
$$

The following fixed-point theorem due to Avery and Peterson is
fundamental in the proofs of our main results.

\begin{theorem}[\cite{AvP}] \label{th21}
Let $P$ be a cone in a real Banach space $E$. Let $\gamma$ and
$\theta$ be nonnegative continuous convex functionals on $P$,
$\alpha$ be a nonnegative continuous concave functional on $P$,
and $\psi$ be a nonnegative continuous functional on $P$
satisfying $\psi(\lambda x) \le \lambda \psi(x)$ for $0 \le
\lambda \le 1$, such that for some positive numbers $M$ and $d$,
\begin{equation}\label{21}
    \alpha(x) \le \psi(x) \;\; \mbox{ and }\;\; \|x\| \le M \gamma(x),
\end{equation}
for all $x \in \overline{P(\gamma, d)}$. Suppose
$T: \overline{P(\gamma, d)} \to \overline{P(\gamma, d)}$
is completely continuous and there exist positive numbers $a, b$,
and $c$ with $a<b$ such that
\begin{itemize}
    \item [(S1)] $\{x \in P(\gamma, \theta, \alpha, b, c, d) \mid \alpha(x) >b\}
    \neq \empty$ 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 $\psi(Tx) <a$
    for $ x \in R(\gamma, \psi, a, d) $ with $\psi(x) =a$.
\end{itemize}
Then $T$ has at least three fixed points $x_1, x_2, x_3 \in
\overline{P(\gamma, d)}$, such that
\begin{gather*}
\gamma(x_i) \le d \quad\mbox{for }i=1, 2, 3;\\
b < \alpha(x_1);\\
a< \psi(x_2)\quad\mbox{with }\alpha(x_2) <b;\\
\psi(x_3) <a\,.
\end{gather*}
\end{theorem}

\section{Existence of triple positive solutions}

In this section, we impose growth conditions on $f$ which allow us
to apply Theorem \ref{th21} to establish the existence of triple
positive solutions of Problem \eqref{11}-\eqref{12}, and
\eqref{11}-\eqref{13}.

We first deal with the boundary-value problem
\eqref{11}-\eqref{12}. Let $X= C^1[0, 1]$  be endowed with the
ordering $x \le y$ if $x(t) \le y(t)$ for all $t \in [0, 1]$, and
the maximum norm,
$$
\Vert x \Vert = \max \Big\{ \max_{0 \leq t \leq 1}|x(t)|,\;
\max_{0 \leq t \leq 1}|x'(t)| \Big\}.
$$
>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<t<1, \\
 x(0) =x(1)= 0.
\end{gather*}
then $G(t, s) \geq 0$ for $0 \leq t, s \leq 1$ and
$$
G(t, s) =  \begin{cases}
     t(1-s) &\mbox{if } 0 \leq t \leq s \leq 1,\\
     s(1-t) &\mbox{if } 0 \leq s \leq t \leq 1.
  \end{cases}
$$
Let
\begin{gather*}
\delta = \min\Big\{\int_{1/4}^{3/4} G(1/4, s)q(s)
ds, \int_{1/4}^{3/4} G(3/4, s)q(s)ds\Big\}, \\
M = \max \Big\{ \int_0^1 (1-s)q(s) ds, \int_0^1 s q(s)ds \Big\}, \\
N = \max_{0 \leq t \leq 1}\int_0^1G(t, s)q(s)ds.
\end{gather*}

To present our main result, we assume there exist constants $0<a<b\le d/8$ such that

\begin{itemize}
 \item[(A1)] $ f(t, u, v) \le d/M$,
for $(t, u, v) \in [0, 1] \times [0,d/] \times [-d, d]$

 \item[(A2)] $f(t, u, v) > \frac{b}{\delta}$, 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; \\
\max_{0 \leq t \leq 1}|x_3(t)| <a.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
Problem \eqref{11}-\eqref{12} has a solution $x=x(t)$ if and only
if $x$ solves the operator equation
$$
x(t) = Tx(t) := \int_0^1G(t, s)q(s)f(s, x(s), x'(s)) ds.
$$
It is well know that this operator, $T: P\to P$, is completely
continuous.
We now show that all the conditions of Theorem \ref{th21} are
satisfied.

If $x \in \overline {P(\gamma, d)}$, then  $\gamma(x) =
\max_{0 \leq t \leq 1}|x'(t)| \le d $. With Lemma
\ref{lm31} and $\max_{0 \leq t \leq 1}|x(t)| \le \frac{d}{2}$,
then assumption (A1) implies $f(t, x(t), x'(t)) \le \frac{d}{M}$.
On the other hand, for $x \in P$, there is $Tx \in P$, then $Tx$
is concave on $[0, 1]$, and $\max_{t \in [0, 1]}|(Tx)'(t)| = \max
\{|(Tx)'(0)|,$ $ |(Tx)'(1)| \}$, so
\begin{align*}
  \gamma(Tx) &= \max_{t \in [0, 1]}|(Tx)'(t)| \\
&=\max_{t \in [0, 1]} \Big| -\int_0^t s q(s)f(s, x(s), x'(s))ds
+ \int_t^1 (1-s) q(s)f(s, x(s), x'(s))ds \Big|  \\
&=\max \Big\{ \int_0^1 (1-s) q(s)f(s, x(s), x'(s))ds,
\int_0^1 s q(s)f(s, x(s), x'(s))ds \Big\}  \\
&\leq \frac{d}{M} \cdot \max \Big\{ \int_0^1 (1-s)q(s) ds,
\int_0^1 s q(s)ds \Big\}  \\
&= \frac{d}{M} \cdot M=d.
\end{align*}
Hence, $T: \overline {P(\gamma, d)} \to \overline {P(\gamma, d)}$.

To check condition (S1) of Theorem \ref{th21}, we choose $x(t) =
4b, \;0 \le t \le 1 $. It is easy to see that $x(t)= 4b \in
P(\gamma, \theta, \alpha, b, 4b, d)$ and $\alpha(x)=\alpha(4b)
>b$, 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 $\psi(0)=0<a$, there holds that $0 \not\in R(\gamma,
\psi, a, d)$. Suppose that $ x \in R(\gamma, \psi, a, d)$ with
$\psi(x)= a$. Then, by the assumption (A3),
\begin{align*}
    \psi(Tx) &=\max_{0 \leq t \leq 1}|(Tx)(t)| \\
 &=\max_{0 \leq t \leq 1}\int_0^1G(t, s)q(s)f(s, x(s),
 x'(s))ds \\
 &< \frac{a}{N} \cdot \max_{0 \leq t \leq 1}\int_0^1G(t,
 s)q(s)ds=a.
\end{align*}
So, Condition (S3) of Theorem \ref{th21} is satisfied.
Therefore, an application of Theorem \ref{th21} imply the boundary-value problem
 \eqref{11}-\eqref{12} has at least three positive
solutions $x_1, x_2$, and $x_3$ satisfying \eqref{x123}.
The proof is complete.
\end{proof}

\begin{remark}
To apply Theorem \ref{th21}, we only need $T: \overline {P(\gamma,
d)} \to \overline {P(\gamma, d)}$, therefore, condition (C1) can
be substituted with a weaker condition
\begin{itemize}
\item[(C1)'] $f \in C([0, 1]\times [0, d/2] \times [-d, d], [0,\infty))$
\end{itemize}
\end{remark}

Now we deal with Problem \eqref{11}-\eqref{13}. The method is just
similar to what we have done above. Moreover, the solutions of
Problem \eqref{11}-\eqref{13} are monotone increasing, which leads
to the situation more simple. Define the cone $P_1 \subset X $ by
$$
P_1= \{ x \in X \mid x(t) \geq 0, x(0)=x'(1)=0, x
\text{ is concave and increasing on } [0, 1]\}.
$$
Let the nonnegative continuous concave functional $\alpha_1$, the
nonnegative continuous convex functional $\theta_1, \gamma_1$, and
the nonnegative continuous functional $\psi_1$ be defined
 on the cone $P_1$ by
\begin{gather*}
\gamma_1(x) = \max_{t \in [0, 1]}|x'(t)| = x'(0), \quad
\psi_1(x) = \theta_1(x) = \max_{t \in [0, 1]}|x(t)|  = x(1),\\
\alpha_1(x)=\min_{t \in [\frac{1}{2}, 1]}|x(t)| = x(\frac{1}{2}),
\quad \mbox{for } x \in P_1\,.
\end{gather*}

\begin{lemma}\label{lm32} If $x \in P_1$, then
$x(1) \le x'(0)$.
\end{lemma}

With Lemma \ref{lm32}, their definition, and the
concavity of $x$, the functionals defined above satisfy
\begin{equation}\label{32}
\frac{1}{2} \theta_1(x) \le \alpha_1(x) \le \theta_1(x) =
\psi_1(x),\quad \|x\| = \max\{\theta_1(x), \gamma_1(x)\} \le
\gamma_1(x),
\end{equation}
for all $x \in \overline {P_1(\gamma, d)} \subset P_1$.

Denote by $G_1(t, s) $ is Green's function for boundary-value problem
\begin{gather*}
  -x''(t) =0, \quad 0<t<1, \\
 x(0) =x'(1)= 0.
\end{gather*}
Then $G_1(t, s) \geq 0$ for $0 \leq t, s \leq 1$ and
$$
G_1(t, s) =  \begin{cases}
     t &\mbox{if } 0 \leq t \leq s \leq 1, \\
     s &\mbox{if } 0 \leq s \leq t \leq 1.
  \end{cases}
$$
 Let
\begin{gather*}
\delta_1 = \int_{\frac{1}{2}}^1G(1/2, s)q(s) ds =
\frac{1}{2}\int_{\frac{1}{2}}^1q(s)ds , \\
M_1 =  \int_0^1 (1-s) q(s)ds , \\
N_1 = \int_0^1 s q(s)ds.
\end{gather*}

Suppose there exist constants $0<a<b\le d/2$ such that
\begin{itemize}
\item[(A4)] $ f(t, u, v) \le d/M_1$, for
  $(t, u, v) \in [0, 1] \times [0, d] \times [-d, d]$
\item[(A5)] $f(t, u, v) > b/\delta_1$, 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)| <b;\\
\max_{0 \leq t \leq 1}|x_3(t)| <a\,.
\end{gather*}
\end{theorem}


\subsection*{Example} Consider the boundary-value problem
\begin{equation} \label{33}
\begin{gathered}
  x''(t) +f(t, x(t), x'(t))= 0, \ \ 0<t<1,  \\
  x(0)=x(1)=0,
\end{gathered}
\end{equation}
where
$$f(t, u, v)=
  \begin{cases}
    e^t+\frac{9}{2}u^3+(\frac{v}{3000})^3 & \text{for } u\le 8, \\
    e^t+\frac{9}{2}(9-u)u^3+(\frac{v}{3000})^3 & \text{for } 8 \le u\le 9, \\
     e^t+\frac{9}{2}(u-9)u^3+(\frac{v}{3000})^3 & \text{for } 9 \le u\le 10, \\
     e^t+4500+(\frac{v}{3000})^3 & \text{for } u \ge 10.
  \end{cases}
  $$
Choose $a=1, b=2, d=3000$, we note $\delta =1/16, M=1/2, N=1/8$.
Consequently, $f(t, u, v) $ satisfy
\begin{align*}
  f(t, u, v) & <\frac{a}{N}=8,\;  \mbox{ for } 0 \le t \le 1, 0 \le u \le 1, -3000 \le v\le 3000; \\
  f(t, u, v) & >\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}