\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 17, pp. 1--14.\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-2005/17\hfil Three-point problems on time scales]
{Positive solutions to a generalized second-order three-point
boundary-value problem on time scales}
\author[H. Luo, Q. Ma\hfil EJDE-2005/17\hfilneg]
{Hua Luo, Qiaozhen Ma}  % in alphabetical order

\address{Hua Luo\hfill\break
College of Mathematics and Information Science, Northwest Normal
University, \hfill\break
Lanzhou 730070, Gansu, China}
\email{luohua@nwnu.edu.cn}

\address{Qiaozhen Ma \hfill\break
College of Mathematics and Information Science, Northwest Normal
University, \hfill\break
Lanzhou 730070, Gansu, China}
\email{maqzh@nwnu.edu.cn}


\date{}
\thanks{Submitted September 1, 2004. Published February 1, 2005.}
\subjclass[2000]{34B18, 39A10} 
\keywords{Time scales; three-point boundary value problems; cone; 
fixed points;\hfill\break\indent
positive solutions }

\begin{abstract}
 Let $\mathbb{T}$ be a time scale with $0,T \in \mathbb{T}$. We
 investigate the existence and multiplicity of positive solutions
 to the nonlinear second-order three-point boundary-value problem
 \begin{gather*}
 u^{\Delta\nabla}(t)+a(t)f(u(t))=0,\quad t\in[0, T]\subset \mathbb{T},\\
 u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta)
 \end{gather*}
 on time scales $\mathbb{T}$, where $0<\eta<T$,
 $0<\alpha<\frac{T}{\eta}$, $0<\beta<\frac{T-\alpha\eta}{T-\eta}$
 are given constants.
\end{abstract}

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

\section{Introduction}

In recent years, many authors have begun to pay attention to the
study of boundary-value problems on time scales. Here two-point
boundary-value problems have been extensively studied; see
\cite{atici,avery,agarw,erbep, maluo} and the references therein.
However, the research for three-point boundary-value problems is
still a fairly new subject, even though it is  growing rapidly;
see \cite{Ande1,Ande2,Kaufm,Ande3}.

In 2002, inspired by the study of the existence of positive
solutions in \cite{MaRy1} for the three-point boundary-value
problem of differential equations, Anderson \cite{Ande3}
considered the following three-point boundary-value problem on a
time scale $\mathbb{T}$,
\begin{gather}
u^{\Delta\nabla}(t)+a(t)f(u(t))=0,\quad t\in[0, T]\subset
\mathbb{T},\label{e1.1} \\
u(0)=0,\quad u(T)=\alpha u(\eta). \label{e1.2}
\end{gather}
He investigated the existence of at least one positive solution
and of at least three positive solutions for the problem
\eqref{e1.1}-\eqref{e1.2} by using Guo-Krasnoselskii's fixed-point theorem and
Leggett-Williams fixed-point theorem, respectively.

In this paper, we extend Anderson's results to the more general
boundary-value problem on time scale $\mathbb{T}$,
\begin{gather}
u^{\Delta\nabla}(t)+a(t)f(u(t))=0,\quad t\in[0, T]\subset
\mathbb{T}, \label{e1.3} \\
u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta), \label{e1.4}
\end{gather}
where  $\alpha>0$,\ $\beta\ge0$,\ $\eta\in (0,
T)\subset\mathbb{T}$ are given constants. Clearly if $\beta=0$,
then \eqref{e1.4} reduces to \eqref{e1.2}.  We also point out that
when $\mathbb{T}=\mathbb{R},\ \beta=0$, \eqref{e1.3}-\eqref{e1.4}
becomes a boundary-value problem of differential equations and
just is the problem considered in \cite{MaRy1}; when
$\mathbb{T}=\mathbb{Z},\ \beta=0$, \eqref{e1.3}-\eqref{e1.4}
becomes a boundary-value problem of difference equations and just
is the problem considered in \cite{MaRy2}. We will use
Guo-Krasnoselskii's fixed-point theorem and Leggett-Williams
fixed-point theorem to investigate the existence and multiplicity
of positive solutions for the problem \eqref{e1.3}-\eqref{e1.4}.
Our main results extend the main results of Ma\cite{MaRy1},
Anderson\cite{Ande3}, Ma and Raffoul\cite{MaRy2}.

\vskip 2mm

The rest of the paper is arranged as follows:  we state some basic
time-scale definitions and prove several preliminary results in
Section 2. Section 3 is devoted to the existence of a positive
solution of \eqref{e1.3}-\eqref{e1.4}, the main tool being the
Guo-Krasnoselskii's fixed-point theorem. Next in Section 4, we
give a multiplicity result by using the Leggett-Williams
fixed-point theorem. Finally we give two examples to illustrate
our results in Section 5.


\section{Preliminaries}

For convenience, we list here the following definitions
which are needed later.

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset
of real numbers $\mathbb{R}$. The operators $\sigma$ and $\rho$
from $\mathbb{T}$ to $\mathbb{T}$, defined by \cite{Hilge},
\begin{gather*}
\sigma(t)=\inf\{\tau\in\mathbb{T}:\tau>t\}\in\mathbb{T}, \\
\rho(t)=\sup\{\tau\in\mathbb{T}:\tau<t\}\in\mathbb{T}
\end{gather*}
are called the {\it forward jump operator} and  the {\it backward
jump operator}, respectively.  In this definition
$$
\inf\emptyset:=\sup\mathbb{T}, \quad  \sup\emptyset:=\inf\mathbb{T}.
$$
 The point $t\in\mathbb{T}$ is
{\it left-dense, left-scattered, right-dense, right-scattered} if
$\rho (t)=t$, $\rho(t)<t$, $\sigma(t)=t$, $\sigma(t)>t$,
respectively.


Let $f:\mathbb{T}\to \mathbb{R}$ and $t\in \mathbb{T}$ (assume
$t$ is not left-scattered if $t=\sup\mathbb{T}$), then the {\it
delta derivative of $f$ at the point $t$} is defined to be the
number $f^{\Delta}(t)$(provided it exists) with the property that
for each $\epsilon>0$ there is a neighborhood $U$ of $t$ such that
$$
|f(\sigma(t))-f(s)-f^{\Delta}(t)(\sigma(t)-s) |\le | \sigma(t)-s|,
\quad \text{for all } s\in U.
$$
Similarly, for $t\in \mathbb{T}$ (assume $t$ is not
right-scattered if $t=\inf\mathbb{T}$),  the {\it nabla derivative
of $f$ at the point $t$} is defined in \cite{atici} to be the
number $f^{\nabla}(t)$(provided it exists) with the property that
for each $\epsilon >0$ there is a neighborhood $U$ of $t$ such
that
$$
|f(\rho(t))-f(s)-f^{\nabla}(t)(\rho(t)-s) |\le | \rho(t)-s |,
\quad \text{for all } s\in U.
$$
A function $f$ is {\it left-dense continuous}\ (i.e.
{ld-continuous}), if $f$ is continuous at each left-dense point in
$\mathbb{T}$ and its right-sided limit exists at each right-dense
point in $\mathbb{T}$. It is well-known\cite{Bohne} that if $f$ is
ld-continuous, then there is a function $F(t)$ such that
$F^{\nabla}(t)=f(t)$. In this case, it is defined that
$$ \int^b_a f(t)\nabla t=F(b)-F(a).
$$
For the rest of this article, $\mathbb{T}$ denotes a time scale with
$0, T\in\mathbb{T}$. Also we denote the set of left-dense continuous
functions from $[0,  T]\subset\mathbb{T}$ to $E\subset\mathbb{R}$
by $C_{ld}([0, T],\ E)$, which is a Banach space with the maximum
norm $\|u\|=\max_{t\in[0, T]}|u(t)|$. We now state and
prove several lemmas before stating our main results.


\begin{lemma}\label{lemm1}
Let $\beta\neq\frac{T-\alpha\eta}{T-\eta}$. Then for $y\in
C_{ld}([0, T],\ \mathbb{R})$, the problem
\begin{gather}
u^{\Delta\nabla}(t)+y(t)=0, \quad t\in[0, T]\subset\mathbb{T},
\label{e2.1} \\
u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta) \label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
\begin{aligned}
 u(t)&=-\int_0^t (t-s)y(s)\nabla
s+\frac{(\beta-\alpha)t-\beta T}
{(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s \\
&\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
\int_0^T (T-s)y(s)\nabla s.
\end{aligned} \label{e2.3}
\end{equation}
\end{lemma}

\begin{proof}
 From \eqref{e2.1}, we have
$$ u(t)=u(0)+u^\Delta (0)t-\int_0^t (t-s)y(s)\nabla s\ :=A+Bt-
\int_0^t (t-s)y(s)\nabla s.
$$
Since
\begin{align*}
u(0)=A;\\
u(\eta)=A+B\eta-\int_0^\eta (\eta-s)y(s)\nabla s;\\
u(T)=A+BT-\int_0^T (T-s)y(s)\nabla s,
\end{align*}
 by \eqref{e2.2} from $u(0)=\beta u(\eta)$,  we have
$$
(1-\beta)A-B\beta\eta=-\beta\int_0^\eta (\eta-s)y(s)\nabla s;
$$
 from $u(T)=\alpha u(\eta)$, we have
$$
(1-\alpha)A+B(T-\alpha\eta)=\int_0^T (T-s)y(s)\nabla s-\alpha
\int_0^\eta (\eta-s)y(s)\nabla s.
$$
Therefore,
\begin{align*}
A&=\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T
(T-s)y(s)\nabla s \\
&\quad -\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
\int_0^\eta (\eta-s)y(s)\nabla s;\\
B&=\frac{1-\beta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T
(T-s)y(s)\nabla s\\
&\quad -\frac{\alpha-\beta} {(T-\alpha\eta)-
\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s\,,
\end{align*}
 from which it follows that
\begin{align*}
u(t)=&\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T
       (T-s)y(s)\nabla s\\
&\quad -\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
       \int_0^\eta (\eta-s)y(s)\nabla s\\
     &\quad +\frac{(1-\beta)t}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T
       (T-s)y(s)\nabla s\\
&\quad -\frac{(\alpha-\beta)t} {(T-\alpha\eta)-
        \beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s
     -\int_0^t (t-s)y(s)\nabla s\\
&=-\int_0^t (t-s)y(s)\nabla s+\frac{(\beta-\alpha)t-\beta T}
      {(T-\alpha\eta)-\beta(T-\eta)} \int_0^\eta (\eta-s)y(s)\nabla s \\
     &\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
      \int_0^T (T-s)y(s)\nabla s.
\end{align*}
The function $u$ presented above is a solution to the
problem \eqref{e2.1}-\eqref{e2.2}, and the uniqueness of $u$ is obvious.
\end{proof}

\begin{lemma}\label{lemm2}
Let $0<\alpha<\frac{T}{\eta}$,\ $0\le
\beta<\frac{T-\alpha\eta}{T-\eta}$.  If $y\in C_{ld}([0, T],\
[0, \infty))$, then the unique solution $u$ of the problem
\eqref{e2.1}-\eqref{e2.2} satisfies
$$ u(t)\ge 0,\quad t\in[0, T]\subset\mathbb{T}.$$
\end{lemma}

\begin{proof}
 It is known that the graph of $u$ is concave down on
 $[0, T]$ from $u^{\Delta\nabla}(t)=-y(t)\le 0$, so
 $$
 \frac{u(\eta)-u(0)}{\eta}\ge\frac{u(T)-u(0)}{T}.
$$
Combining this with \eqref{e2.2}, we have
$$
\frac{1-\beta}{\eta} u(\eta)\ge\frac{\alpha-\beta}{T} u(\eta).
$$
If $u(0)<0$, then $u(\eta)<0$. It implies that
$\beta\ge\frac{T-\alpha\eta}{T-\eta}$, a contradiction to
$\beta<\frac{T-\alpha\eta}{T-\eta}$.

If $u(T)<0$, then $u(\eta)<0$, and the same contradiction emerges.
Thus, it is true that $u(0)\ge 0$,\ $u(T)\ge 0$, together with the
concavity of $u$, we have
$$u(t)\ge 0,\quad t\in[0, T]\subset\mathbb{T}.$$
as required.
\end{proof}


\begin{lemma}\label{lemm3}
Let $\alpha\eta\neq T$, $\beta>\max\{\frac{T-\alpha\eta}{T-\eta},0\}$.
If $y\in C_{ld}([0, T], [0, \infty))$, then problem \eqref{e2.1}-\eqref{e2.2} has
no nonnegative solutions.
\end{lemma}

\begin{proof}
Suppose that problem \eqref{e2.1}-\eqref{e2.2} has a nonnegative
solution $u$ satisfying $u(t)\ge 0, t\in[0, T]$ and there is a
$t_0\in(0, T)$ such that $u(t_0)>0$.

 If $u(T)>0$, then $u(\eta)>0$. It implies
$$
u(0)=\beta u(\eta)>\frac{T-\alpha\eta}{T-\eta} u(\eta)=\frac
 {Tu(\eta)-\eta u(T)}{T-\eta},
$$
that is
$$
\frac{u(T)-u(0)}{T}>\frac{u(\eta)-u(0)}{\eta},
$$
which is a contradiction to the concavity of $u$.

If $u(T)=0$, then $u(\eta)=0$. When $t_0\in (0, \eta)$, we get
$u(t_0)>u(\eta)=u(T)$, a violation of the concavity of $u$. When
$t_0\in (\eta, T)$, we get $u(0)=\beta u(\eta)=0=u(\eta)<u(t_0)$,
another violation of the concavity of $u$. Therefore, no
nonnegative solutions exist.
\end{proof}

\begin{remark} \label{rmk2.4} \rm
When $\beta=0$, the result similar to Lemma 2.3 has been obtained
in Lemma 5 of \cite{Ande3} for $\alpha\eta>T$.
\end{remark}

\begin{lemma}\label{lemm4}
Let $0<\alpha<\frac{T}{\eta}$, $0<\beta<\frac{T-\alpha\eta}{T-\eta}$.
If $y\in C_{ld}([0, T],[0, \infty))$, then the unique solution to the problem
\eqref{e2.1}-\eqref{e2.2} satisfies
\begin{equation}
\min_{t\in [0, T]} u(t)\ge \gamma \|u\|,  \label{e2.4}
\end{equation}
where
\begin{equation}
\gamma:=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\;
\frac{\alpha\eta}{T},\; \frac{\beta(T-\eta)}{T},\;
\frac{\beta\eta}{T}\Big\}.  \label{e2.5}
\end{equation}
\end{lemma}

\begin{proof}
It is known that the graph of $u$ is concave down on
 $[0, T]$ from $u^{\Delta\nabla}(t)=-y(t)\le 0$. We divide the
 proof into two cases.


\noindent Case 1.\; $0<\alpha<1$, then
$\frac{T-\alpha\eta}{T-\eta}>\alpha$.
For $u(0)=\beta u(\eta)=\frac \beta\alpha u(T)$, it may develop in
the following two possible directions.
\\
(i)\; $0<\alpha\le\beta$. It implies that $u(0)\ge u(T)$, so
$$\min_{t\in[0, T]} u(t)=u(T).$$
Assume $\|u\|=u(t_1)$,\ $t_1\in [0, T)$, then either $0\le
t_1\le\eta<\rho(T)$, or $0<\eta<t_1<T$.
\\
If $0\le t_1\le\eta<\rho(T)$, then
\begin{align*}
u(t_1)&\le u(T)+\frac{u(T)-u(\eta)}{T-\eta} (t_1-T)\\
      &\le u(T)+\frac{u(T)-u(\eta)}{T-\eta} (0-T)\\
      &=\frac{Tu(\eta)-\eta u(T)}{T-\eta}\\
      &=\frac{T-\alpha\eta}{\alpha(T-\eta)} u(T),
\end{align*}
from which it follows that $\min_{t\in[0, T]}
u(t)\ge\frac{\alpha(T-\eta)}{T-\alpha\eta} \|u\|$.
\\
If $0<\eta<t_1<T$, from
$$
\frac{u(\eta)}{\eta}\ge\frac{u(t_1)}{t_1}\ge\frac{u(t_1)}{T},
$$
together with $u(T)=\alpha u(\eta)$, we have
$$
u(T)>\frac{\alpha\eta}{T} u(t_1),
$$
so that, $\min_{t\in[0, T]} u(t)\ge\frac{\alpha\eta}{T}
\|u\|$.


\noindent(ii)\; $0<\beta<\alpha$. It implies that $u(0)\le u(T)$, so
$$ \min_{t\in[0, T]} u(t)=u(0). $$
Assume $\|u\|=u(t_2)$, $t_2\in (0, T]$, then either $0<t_2<\eta<\rho(T)$,
or $0<\eta\le t_2\le T$.
\\
If $0< t_2<\eta<\rho(T)$, from
$$\frac{u(\eta)}{T-\eta}\ge\frac{u(t_2)}{T-t_2}\ge\frac{u(t_2)}{T},
$$
together with $u(0)=\beta u(\eta)$, we have
$$u(0)\ge\frac{\beta(T-\eta)}{T} u(t_2),
$$
hence, $\min_{t\in[0, T]} u(t)\ge\frac{\beta(T-\eta)}{T}\|u\|$.
\\
If $0<\eta\le t_2\le T$, from
$$\frac{u(t_2)}{T}\le\frac{u(t_2)}{t_2}\le \frac{u(\eta)}{\eta},
$$
together with $u(0)=\beta u(\eta)$, we have
$$u(0)\ge\frac{\beta\eta}{T} u(t_2),
$$
so that, $\min_{t\in[0, T]} u(t)\ge\frac{\beta\eta}{T} \|u\|$.

\noindent Case 2.\; $\frac T\eta>\alpha\ge 1$, then
$\frac{T-\alpha\eta}{T-\eta}\le\alpha$.
In this case,  $\beta<\alpha$ is true.  It implies that
$u(0)\le u(T)$. So,
$$ \min_{t\in [0, T]} u(t)=u(0). $$
Assume $\|u\|=u(t_2),\ t_2\in(0, T]$ again. Since $\alpha\ge 1$,
it is known that $u(\eta)\le u(T)$, together with the concavity of
$u$, we have $0<\eta\le t_2\le T$. Similar to the above
discussion,
$$
\min_{t\in [0, T]} u(t)\ge\frac{\beta\eta}{T} \|u\|.
$$
Summing up, we have
$$ \min_{t\in [0, T]} u(t)\ge\gamma \|u\|, $$
where
$$
0<\gamma=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\;
\frac{\alpha\eta}{T},\; \frac{\beta(T-\eta)}{T},\;
\frac{\beta\eta}{T}\Big\}<1.
$$
This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.6} \rm
If $\beta=0$,  Anderson obtained the inequality in
\cite[Lemma 7]{Ande3} that is
$$ \min_{t\in [\eta, T]} u(t)\ge r\|u\|, $$
where
$$r\,:=\min\Big\{\frac{\alpha(T-\eta)}{T-\alpha\eta},\
\frac{\alpha\eta}{T},\ \frac{\eta}{T}\Big\}.$$
\end{remark}

The following two theorems, Theorem \ref{thm1}
(Guo-Krasnoselskii's fixed-point theorem)and Theorem \ref{thm2}
(Leggett-Williams fixed-point theorem), will play an important
role in the proof of our main results.

\begin{theorem}[\cite{GuoDj}]\label{thm1}
Let $E$ be a Banach space, and let $K\subset E$ be a cone. Assume
$\Omega_1,\Omega_2$ are open bounded subsets of $E$ with $0\in
\Omega_1, \ \overline\Omega_1\subset\Omega_2$, and let
$$
A:K\cap(\overline \Omega_2 \setminus \Omega_1)\longrightarrow K
$$
be a completely continuous operator such that either
\begin{enumerate}
\item[(i)]
 $\|Au\| \le \|u\|, \ \  u\in K\cap
\partial \Omega_1$, and $\|Au\| \ge\|u\|, \ \ u\in K\cap \partial \Omega_2$; or
\item[(ii)] $\|Au\| \ge \|u\|, \ \ u\in K\cap
\partial \Omega_1$, and $\|Au\| \le \|u\|, \ \ u\in K\cap \partial \Omega_2$
\end{enumerate}
hold. Then $A$ has a fixed point in $K\cap (\overline
\Omega_2 \setminus \Omega_1)$.
\end{theorem}

\begin{theorem}[\cite{Legge}] \label{thm2}
 Let $P$ be a cone in the real Banach space $E$. Set
\begin{gather}
P_c:=\{x\in P:\|x\|<c\}\,,\label{e2.6} \\
P(\psi, a, b):=\{x\in P : a\le\psi(x), \|x\|\le b\}.\label{e2.7}
\end{gather}
Suppose $A:\overline P_c\to \overline P_c$ be a completely
continuous operator and $\psi$ be a nonnegative continuous concave
functional on $P$ with $\psi(x)\le\|x\|$ for all $x\in\overline
P_c$. If there exists $0<a<b<d\le c$ such that the following
conditions hold,
\begin{enumerate}
\item[(i)]$\{x\in P(\psi, b, d) : \psi(x)>b\}\neq\emptyset$
and $\psi(Ax)>b$ for all $x\in P(\psi, b, d)$;
\item[(ii)]$\|Ax\|<a$ for $\|x\|\le a$;
\item[(iii)]$\psi(Ax)>b$
for $x\in P(\psi, b, c)$ with $\|Ax\|>d$.
\end{enumerate}
Then $A$ has at least three fixed points $x_1, x_2$ and
$x_3$ in $\overline P_c $ satisfying
$$
\|x_1\|<a,\quad\psi(x_2)>b,\quad a<\|x_3\|\quad\text{with }
\psi(x_3)<b.
$$
\end{theorem}

\section{ Existence of Positive Solutions}

 We assume the following hypotheses:
\begin{enumerate}
\item[(A1)] $f\in C([0,\infty), [0,\infty))$;
\item[(A2)] $a\in C_{ld}([0,T], [0,\infty))$ and there exists $t_0\in (0,T)$,
such that $a(t_0)>0$.
\end{enumerate}
Define
$$
f_0=\lim_{u\to 0^+}\frac{f(u)}{u},\quad f_{\infty}=\lim
_{u\to\infty}\frac{f(u)}{u}.$$

 For the boundary-value problem \eqref{e1.3}-\eqref{e1.4}, we establish the
 following existence theorem by using Theorem \ref{thm1}
(Guo-Krasnoselskii's fixed-point theorem).

\begin{theorem}\label{onesolution}
Assume (A1), (A2) hold, and $0<\alpha<\frac T\eta$,
$0<\beta<\frac{T-\alpha\eta}{T-\eta}$.  If either
\begin{enumerate}
\item[(C1)]$f_0=0$ and $f_\infty=\infty$ ({\it $f$ is
superlinear}), or
\item[(C2)]$f_0=\infty$ and  $f_\infty=0$ ({\it
$f$ is sublinear}),
\end{enumerate}
\noindent then  problem \eqref{e1.3}-\eqref{e1.4} has at least one positive
solution.
\end{theorem}

\begin{proof}
It is known that $0<\alpha<\frac T\eta$,
$0<\beta<\frac{T-\alpha\eta}{T-\eta}$. From Lemma \ref{lemm1}, $u$
is a solution to the boundary-value problem
\eqref{e1.3}-\eqref{e1.4} if and only if $u$ is a fixed point of
operator $A$, where $A$ is defined by
\begin{equation}
\begin{aligned}
&Au(t)\\
&=-\int^t_0 (t-s)a(s)f(u(s))\nabla s
                  +\frac{(\beta-\alpha)t-\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
                   \int^{\eta}_0 (\eta-s)a(s)f(u(s))\nabla s\\
&\quad+\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
                 \int^T_0 (T-s)a(s)f(u(s))\nabla s.
\end{aligned} \label{e3.1}
\end{equation}
Denote
$$
K=\{u\in C_{ld}([0,T], \mathbb{R}) :  u \ge 0, \min_{t\in [0, T]} u(t)
\ge\gamma \|u\|\},
$$
where $\gamma$ is defined in \eqref{e2.5}.

It is obvious that $K$ is a cone in  $C_{ld}([0,T], \mathbb{R})$.
Moreover, from (A1), (A2), Lemma \ref{lemm2} and Lemma
\ref{lemm4}, $AK\subset K$. It is also easy to check that
$A: K\to K$ is completely continuous.

\subsection*{Superlinear case}  $f_0=0$ and $f_\infty =\infty$.
Since $f_0=0$, we may choose $H_1>0$ so that $f(u) \le \epsilon
u$, for $0<u\le H_1$, where $\epsilon > 0$ satisfies
$$
  \epsilon\frac {T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
        \int^T_0(T-s)a(s)\nabla s\le 1.
$$
Thus, if we let
$$
\Omega_1=\{u\in C_{ld}([0,T],\ \mathbb{R})\ : \  \|u\| <H_1 \},
$$
then for $u\in K\cap \partial \Omega_1$, we get
\begin{align*}
Au(t) &\le \frac{(\beta-\alpha)t-\beta
             T}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^\eta_0(\eta-s)a(s)f(u(s))\nabla s\\
&\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
& \le \frac{\beta t}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^\eta_0(\eta-s)a(s)f(u(s))\nabla s\\
&\quad +\frac{t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
& \le \frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^\eta_0(\eta-s)a(s)f(u(s))\nabla s\\
&\quad +\frac{T+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
& \le \frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
& \le\epsilon\|u\|\frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)\nabla s
\le\|u\|.
\end{align*}
Thus  $\|Au\|\le\|u\|$,\ $u\in K\cap\partial \Omega_1$.

Further, since $f_\infty = \infty$, there exists $\hat H_2 >0$
such that $f(u)\ge \rho u,\ \text{for}\ u \ge \hat H_2$, where
$\rho >0 $ is chosen so that
$$
\rho\gamma\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)}
\int^T_0 sa(s) \nabla s \ge 1.
$$
 Let $H_2 =\max \{2H_1,\; \frac{\hat H_2}\gamma \}$ and
$$
\Omega_2=\{u\in C_{ld}([0, T], \mathbb{R}) : \|u\|<H_2\}.
$$
Then $u\in K\cap \partial\Omega_2$ implies
$$
\min_{t\in [0, T]}u(t) \ge \gamma\|u\|=\gamma H_2\ge\hat H_2,
$$
and so
\begin{align*}
&Au(\eta)\\
&=-\int^\eta_0(\eta-s)a(s)f(u(s))\nabla s
            +\frac{\beta\eta-\alpha\eta-\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^\eta_0(\eta-s)a(s)f(u(s))\nabla s\\
&\quad +\frac{\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
&=\frac{-T}{(T-\alpha\eta)-\beta(T-\eta)}\int^\eta_0(\eta-s)a(s)f(u(s))\nabla s\\
&\quad +\frac{\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
&\ge \frac{1}{(T-\alpha\eta)-\beta(T-\eta)}\int^T_0[-T(\eta-s)+\eta(T-s)]
       a(s)f(u(s))\nabla s\\
&=\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0 sa(s)f(u(s))\nabla s\\
&\ge \gamma\rho\|u\|\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0 sa(s)\nabla s
 \ge \|u\|.
\end{align*}
Hence, $\|Au\|\ge\|u\|$, $u\in K\cap\partial \Omega_2$.
By the first part of Theorem \ref{thm1},  $A$ has a fixed point in
$K\cap (\overline \Omega_2 \setminus \Omega_1)$, such that
$H_1\le\|u\|\le H_2$.  This completes the superlinear part of the
theorem.

\subsection*{Sublinear case} $f_0=\infty$ and $f_\infty =0$.
Since $f_0=\infty$, choose $ H_3 >0$ such that $f(u)\ge M u $ for
$0<u\le H_3$, where $M>0$ satisfies
$$
M\gamma\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T sa(s)\nabla s\ge 1.
$$
Let
$$
\Omega_3=\{u\in C_{ld}([0,T], \mathbb{R}) :   \|u\| <H_3 \},
$$
then for $u\in K\cap \partial \Omega_3$, we get
\begin{align*}
 Ay(\eta)\ge &\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)}
              \int^T_0 sa(s)f(u(s))\nabla s\\
        \ge &M\gamma \|u\|\frac{T-\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0 sa(s)\nabla s
        \ge \|u\|.
\end{align*}
Thus, $\|Au\|\ge\|u\|$,\ $u\in K\cap\partial \Omega_3$.
Now, since $f_\infty=0$, there exists $\hat H_4 >0$ so that
$f(u) \le \lambda u$ for $u\ge \hat H_4$, where $\lambda>0$
satisfies
$$
 \lambda\frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
 \int_0^T (T-s)a(s)\nabla s\le 1.
$$
Choose $H_4 =\max \{2H_3,\ \frac {\hat H_4}\gamma \}$. Let
$$
\Omega_4=\{u\in C_{ld}([0, T],\ \mathbb{R})  : \|u\|<H_4\},
$$
then $u\in K\cap \partial\Omega_4$ implies
$$
\min_{t\in [0, T]}u(t) \ge \gamma\|u\|=\gamma H_4\ge\hat
H_4\,.
$$
Therefore,
\begin{align*}
Au(t) & \le \frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)f(u(s))\nabla s\\
        & \le \lambda\|u\|\frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)a(s)\nabla s
        \le\|u\|.
\end{align*}
Thus $\|Au\|\le\|u\|$,\ $u\in K\cap\partial\Omega_4$.

By the second part of Theorem \ref{thm1}, $A$ has a fixed point
$u$ in $K\cap (\overline \Omega_4 \setminus \Omega_3)$, such that
$H_3\le\|u\|\le H_4$.  This completes the sublinear part of the
theorem.
Therefore, the problem \eqref{e1.3}-\eqref{e1.4} has at least one positive
solution. It finishes the proof of Theorem \ref{onesolution}.
\end{proof}

\section{Multiplicity of Positive Solutions}

In this section, we discuss the multiplicity of positive solutions
for the general boundary-value problem
\begin{gather}
u^{\Delta\nabla}(t)+f(t, u(t))=0,\quad t\in[0, T]\subset
\mathbb{T},\label{e4.1}\\
u(0)=\beta u(\eta),\quad u(T)=\alpha u(\eta),\label{e4.2}
\end{gather}
where $\eta\in (0,\ \rho(T))\subset\mathbb{T}$, $0<\alpha<\frac
T\eta$, $0<\beta<\frac{T-\alpha\eta}{T-\eta}$ are given
constants.

To state the next theorem we assume
\begin{itemize}
\item[(A3)]  $f\in C_{ld}([0, T]\times [0, \infty),[0, \infty))$.
\end{itemize}
Define constants
\begin{gather}
m:=\Big(\frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
\int_0^T (T-s)\nabla s\Big)^{-1},  \label{e4.3}\\
\begin{aligned}
\delta :=\min\Big\{&\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
\int_\eta^T (T-s)\nabla s,\\
& \frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T
(T-s)\nabla s\Big\}
\end{aligned} \label{e4.4}
\end{gather}
Note that $\delta>0$ from $0<\eta<\rho(T)$, $0<\alpha<\frac
T\eta$, $0<\beta<\frac{T-\alpha\eta}{T-\eta}$. Using Theorem
\ref{thm2}(the Leggett-Williams fixed-point theorem), we
established the following existence theorem for the boundary-value
problem \eqref{e4.1}-\eqref{e4.2}.

 \begin{theorem}\label{threesolution}
 Assume (A3) holds, and
$0<\alpha<\frac T\eta$, $0<\beta<\frac{T-\alpha\eta}{T-\eta}$.
Suppose there exists constants $0<a<b<b/\gamma\le c$ such that
\begin{enumerate}
\item[(D1)] $f(t, u)<ma$ for  $t\in[0, T]$, $u\in[0, a]$;
\item[(D2)] $f(t, u)\ge\frac b\delta$  for $t\in[\eta, T]$,
$u\in[b, \frac b\gamma]$;
\item[(D3)] $f(t, u)\le mc$ for
$t\in[0, T]$, $u\in[0,c]$,
\end{enumerate}
where $\gamma, m, \delta$ are as defined in \eqref{e2.5},
\eqref{e4.3} and \eqref{e4.4}, respectively. Then the
boundary-value problem \eqref{e4.1}-\eqref{e4.2} has at least
three positive solutions $u_1, u_2$ and $u_3$ satisfying
$$
\|u_1\|<a,\quad\min_{t\in[0, T]}(u_2)(t)>b,\quad a<\|u_3\|\quad\text{with }
\min_{t\in[0, T]}(u_3)(t)<b.
$$
\end{theorem}

\begin{proof}
It is known that $0<\alpha<\frac T\eta$,
$0<\beta<\frac{T-\alpha\eta}{T-\eta}$.
Define the cone $P\subset C_{ld}([0, T], \mathbb{R})$ by
\begin{equation}
P=\{u\in C_{ld}([0, T], \mathbb{R}) : u \text{ concave down and }
u(t)\ge 0 \text{ on } [0, T]\}.  \label{e4.5}
\end{equation}
Let $\psi:P\to [0,\infty)$ be defined by
\begin{equation}
\psi(u)=\min_{t\in[0, T]} u(t),\quad u\in P. \label{e4.6}
\end{equation}
then $\psi$ is a nonnegative continuous concave functional and
$\psi(u)\le\|u\|, u\in P$.

Define the operator $A: P\to C_{ld}([0, T], \mathbb{R})$ by
\begin{equation}
\begin{aligned} Au(t)&=-\int^t_0 (t-s)f(s, u(s))\nabla s
                  +\frac{(\beta-\alpha)t-\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
                   \int^{\eta}_0 (\eta-s)f(s, u(s))\nabla s\\
               &\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
                 \int^T_0 (T-s)f(s, u(s))\nabla s.
\end{aligned} \label{e4.7}
\end{equation}
Then the fixed points of $A$ just are the solutions of the
boundary-value problem \eqref{e4.1}-\eqref{e4.2} from Lemma
\ref{lemm1}. Since $(Au)^{\Delta\nabla}(t)=-f(t,  u(t))$ for $t\in
(0, T)$, together with (A3) and Lemma \ref{lemm2}, we see that
$Au(t)\ge 0,\ t\in [0, T]$ and $(Au)^{\Delta\nabla}(t)\le 0,\
t\in(0, T)$. Thus $A:P\to P$. Moreover, $A$ is completely
continuous.

We now verify that all of the conditions of Theorem \ref{thm2} are
satisfied.
Since
$$
\psi(u)=\min_{t\in[0, T]} u(t),\quad u\in P.
$$
we have $\psi(u)\le\|u\|$. Now we show $A:\overline {P_c}\to
\overline {P_c}$, where $P_c$ is given in \eqref{e2.6}.
If $u\in\overline {P_c}$, then $0\le u\le c$, together with (D3),
we find $\forall\ t\in[0, T]$,
\begin{align*}
Au(t)&\le \frac{(\beta-\alpha)t-\beta
             T}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^\eta_0(\eta-s)f(s, u(s))\nabla s\\
&\quad +\frac{(1-\beta)t+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)f(s, u(s))\nabla s\\
&\le  \frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)f(s, u(s))\nabla s\\
&\le   m c \frac{T+\beta(T+\eta)}{(T-\alpha\eta)-\beta(T-\eta)}
             \int^T_0(T-s)\nabla s
        = c.
\end{align*}
Thus, $A:\overline {P_c}\to \overline {P_c}$.

By (D1) and the argument above, we can get that
$A:\overline{P_a}\to P_a$. So, $\|Au\|<a$ for $\|u\|\le a$, the condition (ii)
of Theorem \ref{thm2} holds.

Consider the condition (i) of Theorem \ref{thm2} now. Since
$\psi(b/\gamma)=b/\gamma>b$, let $d=b/\gamma$, then $\{u\in
P(\psi, b, d):\psi(u)>b\}\neq\emptyset$. For $u\in
P(\psi, b, d)$, we have $b\le u(t)\le b/\gamma,\ t\in[0, T]$.
Combining with (D2), we get
$$
f(t, u)\ge\frac b\delta,\quad t\in[\eta, T].
$$
Since $u\in P(\psi, b, d)$, then there are two cases that either
$\psi(Au)(t)=Au(0)$, or $\psi(Au)(t)=Au(T)$. As the former holds,
we have
\begin{align*}
\psi(Au)(t)
&=\frac{-\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\
&\quad +\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)f(s, u(s))\nabla s\\
&=\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T T f(s, u(s))\nabla s\\
&\quad+\frac{\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_0^\eta s f(s, u(s))\nabla s\\
&\quad-\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T s f(s, u(s))\nabla s\\
&>\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T T f(s, u(s))\nabla
           s\\
&\quad-\frac{\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_\eta^T s f(s, u(s))\nabla s\\
&\ge \frac{b\beta\eta}{\delta[(T-\alpha\eta)-\beta(T-\eta)]}
           \int_\eta^T (T-s)\nabla s
           \ge b.
\end{align*}
As the later holds, we have
\begin{align*}
&\psi(Au)(t)\\
&=-\int_0^T (T-s)f(s, u(s))\nabla
              s+\frac{(\beta-\alpha)T-\beta T}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\
&\quad +\frac{(1-\beta)T+\beta\eta}{(T-\alpha\eta)-\beta(T-\eta)} \int_0^T (T-s)f(s, u(s))\nabla s\\
&=\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_0^T (T-s)f(s, u(s))\nabla s\\
&\quad -\frac{\alpha T}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_0^\eta (\eta-s)f(s, u(s))\nabla s\\
&=\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_\eta^T T f(s, u(s))\nabla s\\
&\quad -\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}\int_0^T s f(s, u(s))\nabla s\\
&\quad +\frac{\alpha T}{(T-\alpha\eta)-\beta(T-\eta)}
               \int^\eta_0 s f(s, u(s))\nabla s\\
&> \frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_\eta^T T f(s, u(s))\nabla s\\
&\quad -\frac{\alpha\eta}{(T-\alpha\eta)-\beta(T-\eta)}
               \int_\eta^T s f(s, u(s))\nabla s\\
&\ge \frac{b\alpha\eta}{\delta[(T-\alpha\eta)-\beta(T-\eta)]}
           \int_\eta^T (T-s)\nabla s
           \ge b.
\end{align*}
So, $\psi(Au)>b,\ u\in P(\psi, b,  b/\gamma)$, as required.

For the condition (iii) of the Theorem \ref{thm2}, we can verify
it easily under our assumptions using Lemma \ref{lemm4}. Here
$$
\psi(Au)=\min_{t\in [0, T]} Au(t) \ge \gamma\|Au\|
        > \gamma \frac b\gamma = b
$$
as long as $u\in P(\psi, b, c)$ with $\|Au\|>b/\gamma$.

Since all conditions of Theorem \ref{thm2} are satisfied. We say
the problem \eqref{e4.1}-\eqref{e4.2} has at least three positive solutions
$u_1$,\ $u_2$,\ $u_3$ with
$$
\|u_1\|<a,\quad\psi(u_2)>b,\quad a<\|u_3\|\quad\text{with }\psi(u_3)<b.
$$
\end{proof}

\section{Examples}

\begin{example} \label{exa5.1}
 Let $\mathbb{T}=[0,1]\cup [2,3]$.
Considering the boundary-value problem on $\mathbb{T}$
\begin{gather}
u^{\Delta\nabla}(t)+tu^p=0,\quad t\in[0, 3]\subset \mathbb{T},
\label{e5.1}\\
u(0)=\frac12 u(2),\quad u(3)=u(2), \label{e5.2}
\end{gather}
where $p\neq 1$. When taking $T=3$, $\eta=2$, $\alpha=1$,
$\beta=\frac 12$, and
$$ a(t)=t,\quad t\in[0,3]\subset\mathbb{T};\quad f(u)=u^p,\quad
u\in[0,\infty),
$$
we prove the solvability of problem \eqref{e5.1}-\eqref{e5.2} by means of
Theorem \ref{onesolution}.
It is clear that $a(\cdot)$ and $f(\cdot)$ satisfy $(A1)$ and
$(A2)$. We can also show that
$$
0<\alpha\eta=2<3=T,\quad 0<\beta(T-\eta)=\frac 12<T-\alpha\eta=1.
$$
Now we consider the existence of positive solutions of the problem
\eqref{e5.1}-\eqref{e5.2} in two cases.


\noindent Case 1: $p>1$. In this case,
$$
\lim_{u\to 0^+} \frac{f(u)}{u}=\lim_{u\to 0^+}u^{p-1}=0, \quad
\lim_{u\to\infty} \frac{f(u)}{u}=\lim_{u\to
\infty}u^{p-1}=\infty,
$$
and (C1) of Theorem \ref{onesolution} holds. So the problem
\eqref{e5.1}-\eqref{e5.2} has at least one positive solution by Theorem
\ref{onesolution}.

\noindent Case 2.  $p<1$. In this case,
$$
\lim_{u\to 0^+} \frac{f(u)}{u}=\lim_{u\to 0^+}\frac
1{u^{1-p}}=\infty,
\quad \lim_{u\to\infty}
\frac{f(u)}{u}=\lim_{u\to \infty}\frac 1{u^{1-p}}=0,
$$
and (C2) of Theorem \ref{onesolution} holds. So the problem
\eqref{e5.1}-\eqref{e5.2} has at least one positive solution by Theorem
\ref{onesolution}.
Therefore, the boundary-value problem \eqref{e5.1}-\eqref{e5.2} has at least one
positive solution when $p\neq 1$.
\end{example}


\begin{example} \label{exa5.2}
 Let $\mathbb{T}=\{0\}\cup\{1/2^n:n\in\mathbb{N}_0\}$.
Considering the boundary-value problem on $\mathbb{T}$
\begin{gather}
u^{\Delta\nabla}(t)+\frac{2005u^3}{u^3+5000}=0,\quad
t\in[0, 1]\subset \mathbb{T}, \label{e5.3}\\
u(0)=\frac13 u(\frac1{16}),\quad u(1)=8u(\frac1{16}), \label{e5.4}
\end{gather}
When taking $T=1$, $\eta=1/16$, $\alpha=8$, $\beta=1/3$, and
$$
f(t,u)=f(u)=\frac{2005u^3}{u^3+5000},\quad u\ge 0,
$$
we prove the solvability of the problem \eqref{e5.1}-\eqref{e5.2} by means of
Theorem \ref{threesolution}.
It is clear that $f(\cdot)$ is continuous and increasing on
$[0,\infty)$. We can also seen that
$$
0<\alpha\eta=\frac 12<1=T,\quad 0<\beta(T-\eta)=\frac
5{16}<T-\alpha\eta=\frac 12.
$$
Now we check that (D1), (D2) and (D3) of Theorem
\ref{threesolution} are satisfied.
By \eqref{e2.5}, \eqref{e4.3} and \eqref{e4.4}, we get
$\gamma=1/48$, $m=27/65$, $\delta=35/1152$. Let $c=5000$, we have
$$
f(u)\leq 2005<mc\approx 2076.92, \quad u\in[0,c]
$$
from $\lim_{u\to\infty}f(u)=2005$, so that (D3) is met. Note that
$f(10)\approx334.17$, when we set $b=10$,
$$
f(u)\geq \frac b\delta\approx 329.14, \quad u\in[b,48b]
$$
holds. It means that (D2) are satisfied.
To verify (D1), as $f(\frac 15)\approx0.0032$, we take $a=1/5$, then
$$
f(u)<ma\approx 0.083, \quad u\in[0,a],
$$
and (D1) holds. Summing up,  there exists constants $a=1/5$,
$b=10$, $c=5000$ satisfying
$$
0<a<b<\frac b\gamma\leq c
$$
such that (D1), (D2) and (D3) of Theorem \ref{threesolution} hold.
So the boundary-value problem \eqref{e5.3}-\eqref{e5.4} has at least three
positive solutions $u_1, u_2$ and $u_3$ satisfying
$$
\|u_1\|<\frac 15,\quad\min_{t\in[0, T]}(u_2)(t)>10,\quad
\frac 15<\|u_3\|\quad\text{with}\quad
\min_{t\in[0, T]}(u_3)(t)<10.
$$
\end{example}

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\end{document}