\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 75, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/75\hfil Three positive solutions]
{Three positive solutions for $m$-point boundary-value problems
 with one-dimensional $p$-Laplacian}

\author[D. Bai, H. Feng\hfil EJDE-2011/75\hfilneg]
{Donglong Bai, Hanying Feng}  % in alphabetical order

\address{Donglong Bai \newline
Department of Mathematics, 
Shijiazhuang Mechanical Engineering College,  
Shijiazhuang 050003, Hebei, China}
\email{baidonglong@yeal.net}

\address{Hanying Feng \newline
Department of Mathematics, 
Shijiazhuang Mechanical Engineering College,  
Shijiazhuang 050003, Hebei, China}
\email{fhanying@yahoo.com.cn}


\thanks{Submitted April 5, 2011. Published June 16, 2011.}
\thanks{Supported by grants 10971045 from the NNSF of
China, and A2009001426 from the \hfill\break\indent HEBNSF of
China} \subjclass[2000]{34B10, 34B15, 34B18} 
\keywords{Multipoint boundary value problem; positive solution; \hfill\break\indent
Avery-Peterson fixed point theorem; one-dimensional $p$-Laplacian}

\begin{abstract}
 In this article, we study the multipoint boundary  value
 problem  for the one-dimensional $p$-Laplacian
 $$
 (\phi_p(u'))'+ q(t)f(t,u(t),u'(t))=0,\quad t\in (0,1),
 $$
 subject to the boundary  conditions
 $$
 u(0)=\sum_{i=1}^{m-2} a_iu(\xi_i),\quad  u'(1)=\beta u'(0).
 $$
 Using a fixed point theorem due to Avery and Peterson, we provide
 sufficient conditions for the existence of at least three positive
 solutions to the above boundary value problem. The interesting point
 is that the nonlinear term $f$ involves  the first
 derivative of the unknown function.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

In this article, we study  the  existence of multiple positive
solutions to the  boundary  value problem
 (BVP for short) for the one-dimensional $p$-Laplacian
\begin{gather}\label{e1.1}
(\phi_p(u'))'+ q(t)f(t,u(t),u'(t))=0, \quad t\in (0,1),\\
\label{e1.2}
 u(0)=\sum_{i=1}^{m-2}\ a_iu(\xi_i),\quad
 u'(1)=\beta u'(0),
\end{gather}
where $\phi_p(s)=|s|^{p-2}s$, $p>1$,
$\xi_i\in(0,1)$ with $0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$.
use the following assumptions:
\begin{itemize}
\item[(H1)] $ a_i\in [0,1) $ satisfies
$\sum_{i=1}^{m-2} a_i<1$, $\beta\in(0,1)$;

\item[(H2)] $f\in C([0,1]\times [0,+\infty)\times\mathbb{R},
[0,+\infty))$;

\item[(H3)] $q\in L^1[0,1]$ is nonnegative on $(0,1)$ and
$q$ is not identically zero on any subinterval of $(0,1)$.
Furthermore, $q$ satisfies
$0<\int_0^1q(t)dt<\infty$.
\end{itemize}

Multipoint boundary value problems of ordinary differential
equations arise in a variety of  areas of applied mathematics and
physics. For example, the vibrations of a guy wire of a uniform
cross-section and composed of N parts of different densities can
be set up as a multipoint boundary value problem (see \cite{mo}).
The study of multipoint boundary value problems for linear
second-order ordinary differential equations was initiated by
Il'in and Moiseev \cite{il}. Since then there has been much
current attention focused on the study of nonlinear multipoint
boundary value problems, see
\cite{ag,bz,fh1,fh2,fh3,fm,gu1,li,lu,md,su,wf,zh}. However, to the
best knowledge of the authors, no work has been done for
\eqref{e1.1}, \eqref{e1.2}. The aim of this paper is to fill this
gap in the relevant literature.

Karakostas \cite{ka} proved the existence of positive
solutions for the multipoint boundary-value problem
$$
x''(t)-sign(1-\alpha)q(t)f(x,x')x'=0, \  t\in (0,1),
$$
with one of the following sets of boundary conditions:
$$
x(0)=0, \quad x'(1)=\alpha x'(0),
$$
or
$$
x(1)=0, \quad x'(1)=\alpha x'(0),
$$
where $\alpha>0$, $\alpha\neq1$. By using indices of convergence
of the nonlinearities at $0$ and at $+\infty$, the author provide
a priori upper and lower bounds for the slope of the solutions.

Ma \cite{ma} proved  the existence of positive
solutions for the multipoint boundary-value problem
\begin{gather*}
x''(t)-q(t)f(x,x')x'=0, \quad  t\in (0,1),\\
x(0)=\sum_{i=1}^{n-2} b_ix(\xi_i),\quad
x'(1)=\alpha x'(0),
\end{gather*}
where $\xi_i\in(0, 1)$ with
$0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$, $b_i\in[0,1)$, $\alpha>1$.
They provided sufficient conditions for the existence of multiple
positive solutions to the above BVP by applying the fixed point
theorem in cones.


Motivated by these results, our purpose of this paper is to show the
existence of at least three positive solutions to multipoint BVP
\eqref{e1.1} and \eqref{e1.2}.


\section{Preliminaries}

For the convenience of readers, we provide some background material
from the theory of cones in Banach spaces. We also state in
this section the Avery-Peterson's fixed point theorem.

\begin{definition} \label{def2.1} \rm
 Let $E$ be a real Banach space over $\mathbb{R}$. A
nonempty closed set $P \subset E$ is said to be a cone provide
that
\begin{itemize}
\item[(i)] $au+bv\in P$ for all $u, v\in P$ and all $a\geq0, b\geq0$;
and
\item[(ii)] $u, -u \in P$ implies $u=0$.
\end{itemize}
Every cone $P\subset E$
induces an ordering in $E$ given by  $x\leq y$
  if and only if  $y-x\in P$.
\end{definition}

\begin{definition} \label{def2.2} \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
$$
\alpha(tx+(1-t)y)\geq t\alpha(x)+(1-t)\alpha(y)
$$
for all $x,y\in P$ and $0 \leq t \leq 1$. Similarly, we say the
map $\gamma$ is a nonnegative continuous convex functional on
a cone $P$ of a real
Banach space $E$ provided that $\gamma: P\to [0, \infty)$ is
continuous and
$$
\gamma(tx+(1-t)y)\leq t\gamma(x)+(1-t)\gamma(y)
$$
for all $x,y\in P$ and $0 \leq t\leq1$.
\end{definition}

Let $\gamma$ and $\theta$ be nonnegative continuous convex
functionals on a cone $P$, $\alpha$ be a nonnegative continuous
concave functional on a cone $P$, and $\psi$ a nonnegative
continuous functional on a cone $P$. Then for positive real numbers
$a, b, c, d$, we define the following convex sets:
\begin{gather*}
P(\gamma, d)=\{u\in P: \gamma(u)<d\},\\
P(\gamma, \alpha, b, d)=\{u \in P: b\leq \alpha(u),\gamma(u)\leq
d\},\\
P(\gamma, \theta,\alpha, b, c, d)=\{u\in P: b\leq \alpha(u),
\theta(u)\leq c, \gamma(u)\leq d\}, \\
R(\gamma, \psi, a, d)=\{u\in P: a\leq \psi(u), \gamma(u)\leq d\}.
\end{gather*}

To prove our results, we need the following fixed point theorem due
to Avery and Peterson \cite{av}.

\begin{theorem} \label{thm2.1}
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 $\psi(\lambda u)\leq \lambda \psi(u)$ for
$0\leq\lambda\leq 1$, such that for some positive numbers
$\overline{M}$ and $d$,
\begin{equation}\label{e2.1}
\alpha(u)\leq \psi(u)\quad\text{and}\quad
\|u\|\leq \overline{M}\gamma(u)
\end{equation}
for all $u\in \overline{P(\gamma, d)}$. 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)] the set
$\{u\in P(\gamma, \theta,\alpha, b, c, d):\alpha(u)>b\} \neq \phi$
and $\alpha(Tu)>b$ for all $u$ in $P(\gamma, \theta,\alpha, b, c, d)$;
\item[(S2)] $\alpha(Tu)>b$ for $u\in P(\gamma, \alpha, b, d)$ with
$\theta(Tu)>c$;

\item[(S3)]  $0 \not\in R(\gamma, \psi, a, d)$ and $\psi(Tu)<a$ for
$u\in R(\gamma, \psi, a, d)$ with $\psi(u)=a$.
\end{itemize}
Then $T$ has at least three fixed points
$u_1, u_2, u_3\in\overline{P(\gamma, d)}$, such that
$\gamma(u_i)\leq d$ for $i=1,2,3$, $b<\alpha(u_1)$,
$a<\psi (u_2)$, with $\alpha(u_2)<b$,
$\psi (u_3)<a$.
\end{theorem}

\section{Related lemmas}

Let the Banach space $E=C^1[0,1]$ be endowed with  the norm
$$
\|u\| =\max\big\{\max_{ t\in [0,1]} |u(t)|,
\max_{t\in [0,1]}|u'(t)|\big\}.
$$
Define the cone $P\subset E$ by
$P=\{u\in E: u(t)\geq 0,\; u(0)=\sum_{i=1}^{m-2}
a_iu(\xi_i),\; u'(1)=\beta u'(0), \; u\text{ is concave on }
[0,1]\}\subset E$.

It follows from (H3) that there exists a natural number
$ k > \max\{\frac{1}{\xi_1},\frac{1}{1-\xi_{m-2}} \}$
such that
$0<\int_{1/k}^{1-(1/k)}q(t)dt<\infty$.

Let the nonnegative continuous concave functional $\alpha$, the
nonnegative continuous convex functionals $\theta, \gamma$, and the
nonnegative continuous functional $\psi$ be defined on the cone $P$
by
\begin{gather*}
\gamma(u)=\max _{0\leq t\leq 1}|u'(t)|,\quad
\psi(u)=\theta(u)=\max _{0\leq t\leq 1}|u(t)|,\\
\alpha(u)=\min_{1/k\leq t\leq 1-(1/k) } |u(t)|
\quad\text{for } u\in P.
\end{gather*}

\begin{lemma} \label{lem3.1}
   Assume  that {\rm (H1)--(H3)} hold.  Then, for
any $x\in C^{+}[0,1]=:\{x\in C^1[0,1]:x(t)\geq 0\}$,
\begin{gather}\label{e3.1}
(\phi_p(u'))'+ q(t)f(t,x(t),x'(t))=0,\quad t \in (0,1), \\
\label{e3.2}
u(0)=\sum_{i=1}^{m-2}\ a_iu(\xi_i),\ u'(1)=\beta u'(0),
\end{gather}
has the unique solution
\begin{equation} \label{e3.3}
\begin{aligned}
u(t)&=\int_0^{t}\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{1-
\sum_{i=1}^{m-2}a_i}\sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau \\
&\quad + \frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
For an $x\in C^{+}[0,1]$, suppose $u$ is a solution
of  \eqref{e3.1}, \eqref{e3.2}. By integration of \eqref{e3.1},
it follows that
\begin{gather*}
u'(t)=\phi^{-1}_p\Big(\phi_p(u'(0))
 -\int_0^{t}q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big),\\
u(t)=u(0)+\int_0^{t}\phi^{-1}_p\Big(\phi_p(u'(0))
-\int_0^{s}q(\tau) f(\tau,x(\tau),x'(\tau))d\tau\Big)ds.
\end{gather*}
Using the boundary condition \eqref{e3.2}, we can easily show that
\begin{align*}
u(t)
&=\int_0^{t}\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
 \sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,x(\tau),x'(\tau))d\tau\Big)ds.
\end{align*}
One the other hand, it is easy to verify that if $u$ is the solution
of \eqref{e3.3}, then $u$ is a solution of \eqref{e3.1} and
 \eqref{e3.2}.
\end{proof}

\begin{lemma} \label{lem3.2}
Assume  that {\rm (H1)--(H3)} hold. If $x\in C^{+}[0,1]$,
then the unique solution $u(t)$ of  \eqref{e3.1} and \eqref{e3.2}
is concave and $u(t)\geq 0$, $u'(t)\geq 0$, $t\in[0,1]$.
\end{lemma}

\begin{proof}
From the fact that
$(\phi_p(u'))'(t)=-q(t)f(t,x(t),x'(t))\leq 0$, we have
$\phi_p(u'(t))$ is nonincreasing. It follows that $u'(t)$ is also
nonincreasing. Thus, we know that the graph of $u(t)$ is concave
down on $(0,1)$. Then the concavity of $u$ together with boundary
condition $u'(1)=\beta u'(0)$ implies that $u'(t)\geq0$ for
$t\in[0,1]$.

From $u'(t)\geq0$, we know that
$u(\xi_i)\geq u(0)$, for $i=1,2,\dots,m-2$.
This implies
$$
u(0)=\sum _{i=1}^{m-2}a_iu(\xi_i)\geq\sum
_{i=1}^{m-2}a_iu(0).
$$
By $1-\sum _{i=1}^{m-2}a_i>0$, it is obvious that
$u(0)\geq0$. Hence, we know
that $u(t)\geq0$, $t\in[0,1]$.
\end{proof}

\begin{lemma} \label{lem3.3}
If $u\in P$, then $\max _{0\leq t\leq 1}|u(t)|
\leq\overline{M}\max _{0\leq t\leq 1}|u'(t)|$, where
$\overline{M}=1+\frac{\sum _{i=1}^{m-2}
a_i\xi_i}{1-\sum _{i=1}^{m-2}a_i}$.
\end{lemma}

\begin{proof}
 For $u\in P$, by the concavity of $u$ and that
$u'(t)\geq0$, one arrives at
$$
u(1)-u(0)\leq u'(0)=\max _{0\leq t\leq 1}|u'(t)|.
$$
On the other hand,
\begin{align*}
(1-\sum_{i=1}^{m-2}a_i)u(0)
&=u(0)-\sum_{i=1}^{m-2}a_iu(0)
 =\sum_{i=1}^{m-2}a_iu(\xi_i)-\sum_{i=1}^{m-2}a_iu(0)\\
&=\sum_{i=1}^{m-2}a_i(u(\xi_i)-u(0))
 =\sum_{i=1}^{m-2}a_i\xi_iu'(\eta_i),
\end{align*}
where $\eta_i\in(0, \xi_i)$. So
$$
u(0)= \frac{\sum_{i=1}^{m-2}a_i\xi_iu'(\eta_i)}{1-\sum
_{i=1}^{m-2}a_i}
\leq\frac{\sum_{i=1}^{m-2}a_i\xi_i}{1-\sum
_{i=1}^{m-2}a_i}\max _{0\leq t\leq 1}|u'(t)|.
$$
Thus one has
$$
\max _{0\leq t\leq
1}|u(t)|=u(1)\leq\Big(1+\frac{\sum
_{i=1}^{m-2}a_i\xi_i}{1-\sum
_{i=1}^{m-2}a_i}\Big)u'(0)=\overline{M}\max
_{0\leq t\leq 1}|u'(t)|.
$$
With Lemma \ref{lem3.3} and the concavity of $u$, for all $u\in P$,
we obtain
\begin{equation}\label{e3.4}
\frac{1}{k}\theta(u)\leq\alpha(u)\leq\theta(u)
=\psi(u), \quad
\|u\|=\max\{\theta(u),\gamma(u)\}\leq\overline{M}\gamma(u).
\end{equation}
\end{proof}

\begin{lemma} \label{lem3.4}
 Define an operator $T: P\to P$,
\begin{equation} \label{e3.5}
\begin{aligned}
(Tu)(t)
&=\int_0^{t}\phi^{-1}_p\Big(\int_s^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau {l}\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
 \sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
 \phi^{-1}_p\Big(\int_s^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau \\
&\quad + \frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds.
\end{aligned}
\end{equation}
Then $T:\ P\to P$ is completely continuous.
\end{lemma}

\begin{proof}
 According to the definition of $T$ and Lemma \ref{lem3.2}, it
is easy to show that $T(P)\subset P$. By similar arguments in
\cite{ka, wf}, $T : P\to P$ is completely continuous.
\end{proof}

\section{Existence of positive solutions}

We are now ready to apply the Avery-Peterson's fixed point theorem
to the operator $T$ to give sufficient conditions for the existence
of at least three positive solutions to \eqref{e1.1}, \eqref{e1.2}.
For convenience we introduce following notation. Let
\begin{gather*}
 K=\frac{4k^{2}(1-
\sum_{i=1}^{m-2}a_i)}{(k\beta+\beta+3k-1)
(1- \sum_{i=1}^{m-2}a_i)+(k^{2}+k) \sum_{i=1}^{m-2}a_i
[2\xi_i-(1-\beta)\xi_i^{2}]},
\\
 L=\phi_p^{-1}\Big(\frac{1}{1-\phi_p(\beta)}\int_0^1q(\tau)d\tau\Big),
\\
\begin{split}
M&=\int_{1/k}^{1-(1/k)}\phi_p^{-1}
\Big(\int_s^{1-(1/k)}q(\tau) d\tau
+\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_{1/k}
^{1-(1/k)}q(\tau)d\tau\Big)ds \\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \sum_{i=1}^{m-2}a_i
\int_{1/k}^{\xi_i}\phi_p^{-1}
\Big(\int_s^{1-(1/k)}q(\tau) d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}
\int_{1/k}^{1-(1/k)}q(\tau) d\tau\Big)ds,
\end{split}\\
\begin{split}
N&=\int_0^1\phi_p^{-1}\Big(\int_s^1q(\tau)
d\tau+\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1q(\tau)
d\tau\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i} \sum_{i=1}^{m-2}a_i
\int_0^{\xi_i}\phi_p^{-1}\Big(\int_s^1q(\tau)
d\tau+\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1q(\tau)
d\tau\Big)ds.
\end{split}
\end{gather*}

\begin{theorem} \label{thm4.1}
Assume that {\rm (H1)--(H3)} hold. Let
$0<a<b\leq \min\{\frac{1}{K}, \frac{\overline{M}}{k}\}d$,
and suppose that $f$ satisfies the following conditions:
\begin{itemize}
\item[(A1)] $f(t,u,v)\leq \phi_p(d/L)$ for $(t,u,v)\in [0,1]
\times [0,\overline{M}d]\times [0,d]$;

\item[(A2)] $f(t,u,v)\geq\phi_p(kb/M)$ for $(t,u,v)\in
[1/k,1-1/k]\times [b,kb]\times [0,d]$;

\item[(A3)] $f(t,u,v)<\phi_p(a/N)$  for $(t,u,v)\in [0,1]
\times [0,a]\times [0,d]$.
\end{itemize}
Then boundary-value problem \eqref{e1.1}, \eqref{e1.2} has at least three positive
solutions $u_1,u_2,u_3$  such that
$\max _{0\leq t \leq 1}|u'_i(t)|\leq d$, for $i=1,2, 3$, and
$$
b<\min_{1/k\leq t\leq 1-(1/k) }|u_1(t)|,\quad
\max _{0\leq t\leq 1}|u_1(t)|\leq \overline{M}d,\quad
a<\max _{0\leq t\leq 1}|u_2(t)|<k b,
$$
with
$$
\min_{1/k \leq t\leq 1-(1/k) }|u_2(t)|<b,\quad
\max _{0\leq t\leq 1}|u_3(t)|<a.
$$
\end{theorem}


\begin{proof}
Recall that  \eqref{e1.1}, \eqref{e1.2} has a solution
$u=u(t)$ if and only if $u$ solves the operator equation $u=Tu$.
Thus we set out to verify that the operator $T$ satisfies
Avery-Peterson's fixed point theorem which will prove the existence
of three fixed points of $T$. The proof is divided into four
steps.

(1) We will show that (A1) implies
\begin{equation}\label{e4.1}
 T:\overline{P(\gamma, d)}
\to\overline{P(\gamma, d)}.
\end{equation}
In fact, for $u\in \overline{P(\gamma, d)}$, there is
$\gamma(u)=\max _{0\leq t \leq 1}|u'(t)|\leq d$.
With Lemma \ref{lem3.3}, we have $\max _{0\leq t \leq 1}|u(t)|\leq
{\overline{M}}d$.
Then condition (A1) implies $f(t,u,v)\leq \phi_p(d/L)$.
On the other hand, one has $Tu\in P$ for $u\in P$,
then $Tu$ is concave and
$\max _{0\leq t \leq 1}|(Tu)'(t)|=(Tu)'(0)$, so
\begin{align*}
\gamma(Tu)&=\max _{0\leq t \leq 1}|(Tu)'(t)|=(Tu)'(0)\\
&=\phi^{-1}_p\Big(\int_0^1 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau
+\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)
\\
&=\phi^{-1}_p\Big(\frac{1}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)\\
&\leq\phi^{-1}_p\Big(\frac{1}{1-\phi_p(\beta)}\int_0^1
q(\tau)d\tau\phi_p(d/L)\Big)\\
& = \frac{d}{L} \phi^{-1}_p
\Big(\frac{1}{1-\phi_p(\beta)}\int_0^1q(\tau)d\tau
\Big)= \frac{d}{L}L=d.
\end{align*}
Thus, \eqref{e4.1} holds.

(2) We show that the condition (S1) in Theorem \ref{thm2.1} holds.
We take
$$
u_0(t)=-\frac{2k^{2}b(1-
\sum_{i=1}^{m-2}a_i)[(1-\beta)t^{2}-2t]+2k^{2}b\sum_{i=1}^{m-2}a_i[(1-\beta)\xi_i^{2}-2\xi_i]}{(k\beta+\beta+3k-1)
(1- \sum_{i=1}^{m-2}a_i)+(k^{2}+k)
\sum_{i=1}^{m-2}a_i[2\xi_i-(1-\beta)\xi_i^{2}]},
$$
for $t\in[0,1]$.
It is easy to see that $u_0(0)=\sum_{i=1}^{m-2}a_iu_0(\xi_i)$,
$u_0'(1)=\beta u_0'(0)$, $u_0(t)\geq0$ and
is concave on $[0,1]$ so that $u_0(t)\in P$. At the same time,
\begin{align*}
\alpha(u_0)
&=\min_{1/k\leq t\leq 1-(1/k)} |u_0|=u_0(\frac{1}{k})\\
&=\frac{2b\Big((1-
\sum_{i=1}^{m-2}a_i)(2k+\beta-1)+k^{2}
\sum_{i=1}^{m-2}a_i[2\xi_i-(1-\beta)\xi_i^{2}]\Big)}
{(k\beta+\beta+3k-1) (1- \sum_{i=1}^{m-2}a_i)+(k^{2}+k)
\sum_{i=1}^{m-2}a_i[2\xi_i-(1-\beta)\xi_i^{2}]}\\
&>b,
\end{align*}
\begin{align*}
\theta(u_0)
&=\max _{0\leq t\leq 1}|u_0(t)|=u_0(1)\\
&=\frac{2k^{2}b\Big((1-
\sum_{i=1}^{m-2}a_i)(\beta+1)+\sum_{i=1}^{m-2}
a_i[2\xi_i-(1-\beta)\xi_i^{2}]\Big)}{(k\beta+\beta+3k-1)
(1- \sum_{i=1}^{m-2}a_i)+(k^{2}+k)
\sum_{i=1}^{m-2}a_i[2\xi_i-(1-\beta)\xi_i^{2}]}\\
&<kb,
\end{align*}
\begin{align*}
\gamma(u_0)
&=\max _{0\leq t\leq 1}|u'(t)|=u_0'(0)\\
&=\frac{4k^{2}b(1-
\sum_{i=1}^{m-2}a_i)}{(k\beta+\beta+3k-1) (1-
\sum_{i=1}^{m-2}a_i)+(k^{2}+k)
\sum_{i=1}^{m-2}a_i[2\xi_i-(1-\beta)\xi_i^{2}]}\\
&=Kb\leq d.
\end{align*}
So $u_0(t)\in P(\gamma, \theta,\alpha, b, kb, d)$ and
$\{u\in P(\gamma, \theta,\alpha, b, kb, d)\mid \alpha(u)>b\} \neq \phi$.
Thus for $ u\in P(\gamma, \theta,\alpha, b, kb, d)$, there is
$b\leq u(t)\leq kb$, $0\leq u'(t)\leq d$, for $1/k\leq t\leq 1-1/k$.
Hence by condition (A2) of this theorem, one has
$f(t,u,u'(t))\geq\phi_p(kb/M)$, for $t\in [1/k,1-1/k]$. Noting
$(Tu)(1)\geq0$ from Lemma \ref{lem3.2} and combining the conditions on
$\alpha$ and $P$, one arrives at
\begin{align*}
\alpha(Tu)
&=\min _{1/k \leq t \leq 1-(1/k)}|(Tu)(t)|\geq
 \frac{1}{k}\max_{0\leq t \leq 1}|(Tu)(t)|= \frac{1}{k}(Tu)(1)\\
&=\frac{1}{k}\int_0^1\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&\quad+ \frac{1}{k(1- \sum_{i=1}^{m-2}a_i)}
\sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
\phi^{-1}_p\Big(\int_s^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\\
&\quad + \frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds \\
&>\frac{1}{k}\int_{1/k}^{1-(1/k)}\phi^{-1}_p
 \Big(\int_s^{1-(1/k)}
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\\
&\quad+ \frac{\phi_p(\beta)}{1-\phi_p(\beta)}
 \int_{1/k}^{1-(1/k)}
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{k(1-\sum_{i=1}^{m-2}a_i)}
 \sum_{i=1}^{m-2}a_i\int_{1/k}^{\xi_i}
 \phi^{-1}_p\Big(\int_s^{1-(1/k)}
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau \\
&\quad+\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_{1/k}^{1-(1/k)}
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&\geq \frac{1}{k}\int_{1/k}^{1-(1/k)}\phi^{-1}_p
 \Big(\int_s^{1-(1/k)} q(\tau)d\tau\phi_p(\frac{kb}{M})\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_{1/k}^{1-(1/k)}
 q(\tau)d\tau\phi_p(\frac{kb}{M})\Big)ds\\
&\quad +\frac{1}{k(1-\sum_{i=1}^{m-2}a_i)}
 \sum_{i=1}^{m-2}a_i\int_{1/k}^{\xi_i}
 \phi^{-1}_p\Big(\int_s^{1-(1/k)}q(\tau)d\tau\phi_p(\frac{kb}{M})\\
&\quad + \frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_{1/k}^{1-(1/k)}
 q(\tau)d\tau\phi_p(\frac{kb}{M})\Big)ds \\
&= \frac{1}{k}\frac{kb}{M}M =b.
\end{align*}
Therefore,
$\alpha(Tu)>b$, for all $u\in P(\gamma, \theta,\alpha, b, kb, d)$.
Consequently, condition (S1) in Theorem \ref{thm2.1} holds.

(3) We now prove (S2) in Theorem \ref{thm2.1} holds.
With \eqref{e3.4} we have
$$
\alpha(Tu)\geq \frac{1}{k}\theta(Tu)>\frac{1}{k}kb=b,
$$
for $u\in P(\gamma,\alpha,b,d)$ with $\theta(Tu)>kb$. Hence, condition
(S2) in Theorem \ref{thm2.1} is satisfied.

(4) Finally, we show that (S3) in Theorem \ref{thm2.1} is satisfied.
Since $\psi(0)=0<a$, so $0\not\in R(\gamma,\psi,a,d)$. Suppose that
$u\in R(\gamma,\psi,a,d)$ with $\psi(u)=a$. Then by the condition
(A3) of this theorem,
\begin{align*}
\psi(Tu)
&=\max _{0\leq t\leq1}|(Tu)(t)|=(Tu)(1)\\
&=\int_0^1\phi^{-1}_p\Big(\int_s^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
\sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
\phi^{-1}_p\Big(\int_s^1 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau \\
&\quad + \frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
 q(\tau)f(\tau,u(\tau),u'(\tau))d\tau\Big)ds\\
&< \int_0^1\phi^{-1}_p\Big(\int_s^1 q(\tau)d\tau\phi_p(a/N)
 +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
 q(\tau)d\tau\phi_p(a/N)\Big)ds\\
&\quad +\frac{1}{1-\sum_{i=1}^{m-2}a_i}
 \sum_{i=1}^{m-2}a_i\int_0^{\xi_i}
 \phi^{-1}_p\Big(\int_s^1 q(\tau)d\tau\phi_p(a/N) \\
&\quad +\frac{\phi_p(\beta)}{1-\phi_p(\beta)}\int_0^1
 q(\tau)d\tau\phi_p(a/N)\Big)ds \\
&= \frac{a}{N}N=a.
\end{align*}
Thus, condition (S3) in Theorem \ref{thm2.1} holds.

Then Theorem \ref{thm2.1} implies that \eqref{e1.1}, \eqref{e1.2} has
at least three positive solutions satisfying the statement in
Theorem \ref{thm4.1}. The proof is complete.
\end{proof}

\section{Example}

Let $p=3$, $q(t)=1$ in \eqref{e1.1} and $ m=4$,
$\beta=1/2$, $\xi_1=1/3$, $\xi_2=2.3$,
$a_1=1/2$, $a_2=1/4$ in \eqref{e1.2}. Now we consider the
boundary-value problem
\begin{gather}\label{e5.1}
(|u'(t)|u'(t))'+ f(t,u(t),u'(t))=0, \quad t\in (0,1), \\
\label{e5.2}
u(0)=\frac{1}{2}u(\frac{1}{3})+\frac{1}{4}u(\frac{2}{3}),\quad
u'(1)=\frac{1}{2}u'(0),
\end{gather}
where
$$
f(t,u,v)=\begin{cases}
\frac{t}{20}+1.8\times10^{4}\cdot u^{40}
 +\frac{1}{100}(\frac{v}{3\times10^{21}})^{4}+\frac{1}{1000},
& u\leq6; \\
\frac{t}{20}+1.8\times10^{4}\cdot6^{40}
+\frac{1}{100}(\frac{v}{3\times10^{21}})^{4}+\frac{1}{1000},
& u>6.
\end{cases}
$$
Choose $a=2/3$, $b=1$, $k=6$, $d=3\times 10^{21}$, we note
that $K=4/5$, $L=2\sqrt{3}/3$, $\overline{M}=7/3$,
$M\doteq 1.185$, $N\doteq 2.281$.
Consequently, $f(t,u,v)$ satisfies
\begin{itemize}
\item[(1)] $ f(t,u,v)<2.407\times
10^{35}<\phi_3(d/L)\doteq 6.75\times 10^{42}$
for $(t,u,v)\in [0,1] \times [0,7\times 10^{21}]\times
[0,3\times 10^{21}]$;

\item[(2)] $ f(t,u,v)>1.8\times
10^{4}>\phi_3(kb/M)\doteq25.637$
for $(t,u,v)\in
[1/6,5/6]\times [1,6]\times [0,3\times 10^{21}]$;

\item[(3)] $ f(t,u,v)<0.063<\phi_3(a/N)\doteq 0.085$
for $(t,u,v)\in [0,1] \times [0,2/3]\times
[0,3\times 10^{21}]$.

\end{itemize}
Then all conditions of Theorem \ref{thm4.1} hold. Therefore,
\eqref{e5.1}, \eqref{e5.2} has at least three positive solutions
$u_1, u_2, u_3$ satisfying
\begin{gather*}
\max _{0\leq t \leq 1}|u'_i(t)|\leq 3\times 10^{21}\quad\text{for }
i=1,2,3, \\
1<\min_{1/k\leq t\leq 1-(1/k) }|u_1(t)|,\quad
\max _{0\leq t\leq 1}|u_1(t)|\leq 7\times 10^{21},\quad
\frac{2}{3}<\max _{0\leq t\leq 1}|u_2(t)|<6,
\end{gather*}
with
$\min_{1/k\leq t\leq 1-(1/k)}|u_2(t)|<1$,
$\max _{0\leq t\leq 1}|u_3(t)|< 2/3$.

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\end{document}
