\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 83, 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-2011/83\hfil Positive solutions]
{Positive solutions for a nonlinear $n$-th
order $m$-point boundary-value problem}

\author[J. Zhang, Y. Guo, Y. Ji\hfil EJDE-2011/83\hfilneg]
{Jiehua Zhang, Yanping Guo, Yude Ji}

\address{Jiehua Zhang \newline
College of Sunshine, Fuzhou University,
Fuzhou 350015,  China}
\email{jiehuahappy@163.com}

\address{Yanping Guo \newline
College of Sciences, Hebei University
of Science and Technology, Shijiazhuang 050018, China}
\email{guoyanping65@sohu.com}

\address{Yude Ji \newline
College of Sciences, Hebei University
of Science and Technology, Shijiazhuang 050018, China}
\email{jiyude-1980@163.com}


\thanks{Submitted March 12, 2010. Published June 24, 2011.}
\thanks{Supported by grants: 10971045 the Natural Science 
Foundation of China, and \hfill\break\indent 
A2009000664 from the Natural Science Foundation of Hebei Province}

\subjclass[2000]{39A10}
\keywords{Boundary value problem; positive solution; fixed
point theorem; \hfill\break\indent Green's function}

\begin{abstract}
 Using the Leggett-Williams fixed point theorem in cones,
 we prove the existence of at least three
 positive solutions to the nonlinear $n$-th order $m$-point
 boundary-value problem
 \begin{gather*}
 \Delta^{n}u(k)+a(k)f(k,u)=0, \quad  k\in \{0,N\},\\
  u(0)=0,\; \Delta u(0)=0, \dots, \Delta^{n-2}u(0)=0,\quad
  u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i).
 \end{gather*}
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

 Multi-point boundary value problems arise in a variety
of areas of applied mathematics and physics. The
solvability of two-point difference and multi-point differential
boundary value problems has been studied extensively in the
literature in recent years; see \cite{e1,e2,e3,e4,e5,e6,g1,h1,i1,m1}
and their references.
Guo \cite{g1} used Leggett-Williams fixed point theorem to
obtain the existence of at least three positive solutions for the
second-order $m$-point boundary value problem
\begin{gather*}
 u''(t)+f(t,u)=0,\quad 0\leq t\leq 1,\\
 u(0)=0,\quad  u(1)-\sum_{i=1}^{m-2}k_iu(\xi_i)=0,
\end{gather*}
where
$k_i>0$ $(i=1,2,\dots,m-2)$, $0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$,
$0<\sum_{i=1}^{m-2}k_i\xi_i<1$ are given, and
$f:[0,1]\times[0,\infty)\to [0,\infty)$ is continuous.

Recently,  Eloe and  Ahmad \cite{e7} discussed the existence of at
least one positive solution for the nonlinear $n$-th order
three-point boundary value problem
\begin{gather*}
u^{(n)}(t)+a(t)f(u)=0, \quad  t\in(0,1),\\
u(0)=0,\; u'(0)=\dots=u^{(n-2)}(0)=0,\quad u(1)=\alpha u(\eta),
\end{gather*}
where $n\geq 2,0<\eta<1,0<\alpha\eta^{n-1}<1,f(t)\in
C([0,1],[0,\infty))$ is either superlinear or sublinear. The method
they used is the Krasnoselskii's fixed point theorem in cones.

Motivated by the results \cite{e7,l1}, in this paper, we investigate the
existence of positive solutions for the following nonlinear $n$-th
order $m$-point boundary value problem
\begin{gather} \label{e1}
\Delta^{n}u(k)+a(k)f(k,u)=0, \quad  k\in \{0,N\},\\
u(0)=0,\quad \Delta u(0)=0, \dots, \Delta^{n-2}u(0)=0,\quad
u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i), \label{e2}
\end{gather}
where $n\geq2$, $\alpha_i\geq0$ for $i=1,2,\dots,m-3$, and
$\alpha_{m-2}>0$, $\xi_i$ is an integer, satisfying
$n=\xi_{0}\leq\xi_1<\xi_2<\dots<\xi_{m-2}<\xi_{m-1}=N+n$,
\[
0<\sum_{i=1}^{m-2}\alpha_i(\sum_{j=1}^{n-1}\prod_{l=1}^{j}
(\xi_i-n+l)+1) <\sum_{j=1}^{n-1}\prod_{l=1}^{j}(N+l)+1.
\]
We denote
$\{i,j\}=\{ k\in\mathbb{N}:i\leq k\leq j\}$ and assume that:
\begin{itemize}
\item[(A1)]  $f:\{0,N\}\times[0,\infty)\to [0,\infty)$ is continuous;
\item[(A2)] $a(k)\geq 0$, for $k\in \{0,N\}$ and there exists
$k_{0}\in\{\xi_{m-2},N\}$ such that $a(k_{0})>0$.
\end{itemize}
This article is organized as follows.
In Section 2, we present some
preliminaries that will be used to prove our main results.
In Section 3, using the Leggett-Williams fixed point theorem,
we show that \eqref{e1}--\eqref{e2}  has at least three
positive solutions.

\section{Preliminaries}

 In this section, we present some notation and lemmas, which
are fundamental in the proof of our main results.

Let $E$ be a Banach space over $\mathbb{R}$. A
nonempty convex closed set $K\subset E$ is said to be a cone
provided that
\begin{itemize}
\item[(i)] $au\in K$ for all $u\in K$ and all $a\geq 0$;
\item[(ii)] $u,-u\in K$ implies $u=0$.
\end{itemize}

A map $\alpha$ is said to be a
nonnegative continuous concave functional on $K$ provided that
$\alpha:K\to [0,\infty) $ is continuous and
$$
\alpha(tx+(1-t)y)\geq t\alpha(x)+(1-t)\alpha(y)
$$
for all $ x,y\in K $ and $0\leq  t\leq 1$.
Similarly, we say a map $\beta$ is a nonnegative continuous
convex functional on $K$
provided that $\beta:K\to [0,\infty) $ is continuous and
$$
\beta(tx+(1-t)y)\leq t\beta(x)+(1-t)\beta(y)
$$
for all $ x,y\in K $ and $0\leq  t\leq 1$.

Let $\alpha$ be a nonnegative continuous concave functional on $K$.
Then, for nonnegative real numbers $0<b<d$ and $c$, we define the
convex sets
\begin{gather*}
P_{c}=\{x\in K|\|x\|<c\},\\
P(\alpha,b,d)=\{x\in K|b\leq\alpha(x),\|x\|\leq d\}.
\end{gather*}

\begin{theorem}[Leggett-Williams fixed point theorem]  \label{thm2.1}
Let $A:\overline{P_{c}}\to \overline{P_{c}}$ be a completely
continuous operator and  let $\alpha$ be a nonnegative continuous
concave functional on $K$ such that $\alpha(x)\leq\|x\|$ for all
$x\in\overline{P_{c}}$. Suppose there exist $0<a<b<d\leq c$ such
that
\begin{itemize}
\item[(C1)] $\{x\in P(\alpha,b,d)|\alpha (x)>b\}\neq \emptyset$,
and $\alpha (Ax)>b$ for $x\in P(\alpha,b,d)$,
\item[(C2)] $ \|Ax\|<a$ for $\|x\|\leq a$, and
\item[(C3)] $\alpha (Ax)>b$ for $x\in P(\alpha,b,c)$, with
$\|Ax\|>d$.
\end{itemize}
Then $A$ has at least three fixed point $x_1,x_2$ and $x_3$
such that
$\|x_1\|<a$, $b<\alpha (x_2)$ and $\|x_3\|>a$ with
$\alpha(x_3)<b$.
\end{theorem}

\begin{lemma}[\cite{m1}] \label{lem2.1}
Assume that $u$ satisfies the difference
inequality $\Delta^{n}u(k)\leq0$, $k\in\{0,N\}$, and the homogeneous
boundary conditions, $u(0)=\dots=u(n-2)=0$, $u(N+n)=0$. Then,
$u(k)\geq0$, $k\in\{0,N+n\}$.
\end{lemma}

For a finite or infinite sequence $u(0),u(1),\dots$, the value
$k=0$ is a node for the sequence if $u(0)=0$, and a value $k>0$
is a node for $u$ if $u(k)=0$ or $u(k-1)u(k)<0$.
The following lemma, obtained in \cite{m1},
is a discrete analogue of Rolle's Theorem.

\begin{lemma} \label{lem2.2}
Suppose that the finite sequence
$u(0),\dots,u(j)$ has $N_{j}$ nodes and the sequence
$\Delta u(0),\dots,\Delta u(j-1)$ has $M_{j}$ nodes.
Then, $M_{j}\geq N_{j}-1$.
\end{lemma}

\begin{theorem} \label{thm2.2}
Assume $n\leq\xi_1<\xi_2<\dots<\xi_{m-2}<N+n$,
$$
0<\sum_{i=1}^{m-2}\alpha_i(\sum_{j=1}^{n-1}\prod_{l=1}^{j}
(\xi_i-n+l)+1) <\sum_{j=1}^{n-1}\prod_{l=1}^{j}(N+l)+1,
$$
and $y(k)\geq0$, $k\in\{0,N\}$. Then, the difference equation
\begin{equation} \label{e3}
\Delta^{n}u(k)+y(k)=0,\quad k\in \{0,N\},
\end{equation}
coupled with the boundary conditions \eqref{e2}, has a unique solution
\begin{equation} \label{e4}
u(k)=\begin{cases}
0, & \text{for }  k\in \{0,n-2\},\\[3pt]
\frac{\delta}{M(n-1)!}, &\text{for }   k=n-1,\\[3pt]
-\frac{1}{(n-1)!}\sum_{s=0}^{k-n}y(s)\prod_{j=1}^{n-1}(k-n+j-s)\\
+\frac{\delta}{M(n-1)!}\sigma,
&\text{for }  k\in\{n,N+n\},
\end{cases}
\end{equation}
where
\begin{gather*}
M=\big(\sum_{j=1}^{n-1}\prod_{l=1}^{j}(N+l)+1\big)-
\sum_{i=1}^{m-2}\alpha_i\big(\sum_{j=1}^{n-1}
\prod_{l=1}^{j}(\xi_i-n+l)+1\big),
\\
\delta=\sum_{s=0}^{N}y(s)\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\sum_{s=0}^{\xi_i-n}y(s)
 \prod_{j=1}^{n-1}(\xi_i-n+j-s),
\\
\sigma=\sum_{j=1}^{n-1}\prod_{l=1}^{j}(k-n+l)+1.
\end{gather*}
\end{theorem}

\begin{proof}
Let $\Delta^{n-1}u(0)=A$, since $u(0)=0,\ \Delta
u(0)=0, \dots, \Delta^{n-2}u(0)=0$, it follows that
$\Delta^{n-z}u(z-1)=A$, for $z\in\{1,n-1\}$, $u(0)=\dots=u(n-2)=0$,
$u(n-1)=A$.

Summing \eqref{e3} from $0$ to $k-1$, one gets
$\Delta^{n-1}u(k)=-\sum_{s=0}^{k-1}y(s)+A$. Again summing the
equality above, from $1$ to $k-1$, it follows that
\[
\Delta^{n-2}u(k)=-\sum_{s_1=0}^{k-2}\sum_{s=0}^{s_1}y(s)+(k-1)A+A.
\]
Repeat the summing in this way in proper order, we get
\[
u(k)=-\sum_{s_{n-1}=0}^{k-n}\dots\sum_{s=0}^{s_1}y(s)+A\sigma.
\]
It can be expressed that
\begin{align*}
\sum_{s_1=0}^{k-2}\sum_{s=0}^{s_1}y(s)
&= \sum_{s=0}^{0}y(s)+\sum_{s=0}^{1}y(s)+\dots+\sum_{s=0}^{s_2}y(s)\\
&=  (s_2+1)y(0)+s_2y\eqref{e1}+\dots+y(s_2)\\
&=  \sum_{s=0}^{s_2}(s_2+1-s)y(s),
\end{align*}
by repeating this process coupled with the mathematical induction, we
have
\[
\sum_{s_{n-1}=0}^{k-n}\dots\sum_{s=0}^{s_1}y(s)=
\frac{1}{(n-1)!}\sum_{s=0}^{k-n}y(s)\prod_{j=1}^{n-1}(k-n+j-s).
\]
From $u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)$, we have
$A=\delta/(M(n-1)!)$.
Hence, \eqref{e4} is the unique solution.
\end{proof}

\begin{theorem} \label{thm2.3}  Assume that
$n\leq\xi_1<\xi_2<\dots<\xi_{m-2}<N+n$ and that
$0<\sum_{i=1}^{m-2}\alpha_i(\sum_{j=1}^{n-1}\prod_{l=1}^{j}
(\xi_i-n+l)+1)
<\sum_{j=1}^{n-1}\prod_{l=1}^{j}(N+l)+1$. Then, the
Green's function for the boundary value problem
\begin{gather*}
-\Delta^{n}u(k)=0,\quad k\in \{0,N\}, \\
u(0)=0,\quad \Delta u(0)=0, \dots, \Delta^{n-2}u(0)=0,\quad
u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i),
\end{gather*}
is given by
\[
G(k,s)=\begin{cases} 0, & \text{for }  k\in \{0,n-2\},\\
\frac{h(\xi_{r-1},\xi_{r};s)}{(n-1)!}, & \text{for }  k=n-1,\\
\frac{-\prod_{j=1}^{n-1}(k-n+j-s)+h(\xi_{r-1},
\xi_{r};s)\sigma}{(n-1)!},
&\text{for }  0\leq s\leq k-n\leq N,\\
\frac{h(\xi_{r-1},\xi_{r};s)\sigma}{(n-1)!}, & \text{for }
0<k-n+1\leq s\leq N,
\end{cases}
\]
where
\[
h(\xi_{r-1},\xi_{r};s)
=\begin{cases}
\frac{\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)}{M},\\
\quad \text{for } 0\leq s\leq\xi_1-n,\\[3pt]
\frac{\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=r}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)}{M},\\
\quad\text{for } s\in\{\xi_{r-1}-n+1,\xi_{r}-n\}, r\in\{2,m-1\}.
\end{cases}
\]
\end{theorem}

\begin{proof} Make the assumption that
$\sum_{i=m_1}^{m_2}f(i)=0$ for $m_2<m_1$.
For $n\leq k\leq \xi_1$, the unique solution of \eqref{e3} \eqref{e2}
can be expressed as
\begin{align*}
u(k)&= \frac{1}{M(n-1)!}\{\sum_{s=0}^{k-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{s=k-n+1}^{\xi_1-n}\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma y(s)\\
&\quad +\sum_{r=2}^{m-1}\sum_{s=\xi_{r-1}-n+1}^{\xi_{r}-n}
 \big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=r}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma y(s)\}
\end{align*}
If $\xi_{t-1}+1\leq k\leq \xi_{t}$, $2\leq t\leq m-2$, the unique
solution of \eqref{e3} \eqref{e2} can be expressed as
\begin{align*}
u(k)
&= \frac{1}{M(n-1)!}\{\sum_{s=0}^{\xi_1-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{r=2}^{t-1}\sum_{s=\xi_{r-1}-n+1}^{\xi_{r}-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=r}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{s=\xi_{t-1}-n+1}^{k-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=t}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{s=k-n+1}^{\xi_{t}-n}(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=t}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma y(s)\\
&\quad +\sum_{r=t+1}^{m-1}\sum_{s=\xi_{r-1}-n+1}^{\xi_{r}-n}(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=r}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma
y(s)\}.
\end{align*}
For $\xi_{m-2}+1\leq k\leq N+n$, the unique solution
of \eqref{e3} \eqref{e2} can
be expressed as
\begin{align*}
u(k)&= \frac{1}{M(n-1)!}\{\sum_{s=0}^{\xi_1-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{r=2}^{m-2}\sum_{s=\xi_{r-1}-n+1}^{\xi_{r}-n}[-M\prod_{j=1}^{n-1}(k-n+j-s)\\
&\quad +\big(\prod_{j=1}^{n-1}(N+j-s)-
\sum_{i=r}^{m-2}\alpha_i\prod_{j=1}^{n-1}(\xi_i-n+j-s)\big)\sigma]y(s)\\
&\quad +\sum_{s=\xi_{m-2}-n+1}^{k-n}\big(-M\prod_{j=1}^{n-1}(k-n+j-s)
+\sigma\prod_{j=1}^{n-1}(N+j-s)\big)y(s)\\
&\quad +\sum_{s=k-n+1}^{N}\big(\prod_{j=1}^{n-1}(N+j-s)\big)\sigma
y(s).
\end{align*}

Therefore, the unique solution of \eqref{e3} \eqref{e2} is
$u(k)=\sum_{s=0}^{N}G(k,s)y(s)$. By the method which Eloe has
recently used to obtain the sign of Green's function and related
inequalities in \cite{e6}, it can be verified directly
that $G(k,s)\geq0$ on $\{0,N+n\}\times\{0,N\}$.
So, $u(k)\geq0$, $k\in\{0,N+n\}$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm2.4}
Assume that
$n\leq\xi_1<\xi_2<\dots<\xi_{m-2}<N+n$, and that
$0<\sum_{i=1}^{m-2}\alpha_i(\sum_{j=1}^{n-1}\prod_{l=1}^{j}(\xi_i-n+l)+1)
<\sum_{j=1}^{n-1}\prod_{l=1}^{j}(N+l)+1$. If $u$
satisfies $\Delta^{n}u(k)\leq0$, $k\in\{0,N\}$, with the nonlocal
conditions \eqref{e2}, then
\begin{equation} \label{e5}
\min_{k\in \{\xi_{m-2},N+n\}}u(k)\geq\gamma\|u\|,
\end{equation}
where
\begin{align*}
\gamma=\min\Big\{&\frac{\alpha_{m-2}(N+n-\xi_{m-2})}{N+n-\alpha_{m-2}
\xi_{m-2}},\;
\frac{\alpha_{m-2}\prod_{i=0}^{n-2}(\xi_{m-2}-i)}
{\prod_{i=0}^{n-2}(N+n-i)},\;
\frac{\alpha_1\prod_{i=0}^{n-2}(\xi_1-i)}{\prod_{i=0}^{n-2}(N+n-i)},\\
&\frac{\prod_{i=0}^{n-2}(\xi_{m-2}-i)}{\prod_{i=0}^{n-2}(N+n-i)}\Big\}.
\end{align*}
\end{theorem}

\begin{proof} We will show the details in the case that $u$
satisfies the strict difference inequality $\Delta^{n}u(k)<0$,
$k\in\{0,N\}$. Once \eqref{e5} is obtained for functions satisfying the
strict inequality, one assumes that $u$ satisfies the difference
inequality and sets
\begin{align*}
u(\epsilon,k)
&=u(k)+\epsilon\big(\prod_{j=0}^{n-2}(k-j)\big)\\
&\quad\times \Big(\frac{(N+n)\prod_{j=0}^{n-2}(N+n-j)-
\sum_{i=1}^{m-2}\alpha_i\xi_i\prod_{j=0}^{n-2}(\xi_i-j)}
{\prod_{j=0}^{n-2}(N+n-j)-
\sum_{i=1}^{m-2}\alpha_i\prod_{j=0}^{n-2}(\xi_i-j)}-k\Big).
\end{align*}
Then for each $\epsilon>0$, $u(\epsilon,k)$ satisfies the strict
difference inequality and the nonlocal conditions \eqref{e2}.
Thus, \eqref{e5} holds for each $\epsilon>0$ and by limiting,
it holds for $\epsilon=0$.

Under the assumption $\Delta^{n}u(k)<0$, $k\in\{0,N\}$, we have to
distinguish two cases.

Case (i): $0<\sum_{i=1}^{m-2} \alpha_i<1$. Suppose
$u(\xi_{r})=\max_{i\in \{1,m-2\}}u(\xi_i)$, then
$u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)\leq
\sum_{i=1}^{m-2}\alpha_iu(\xi_{r})<u(\xi_{r})$. It follows
by repeated applications of Lemma \ref{lem2.2} that for each $j\in\{1,n-1\}$,
$\Delta^{j}u$ has precisely one node, $k_{j}\in\{n-1-j,N+n-j\}$ and
$k_{j+1}<k_{j}$, $j\in\{1,n-2\}$. Assume that
$\|u\|=u(\overline{k})$, if $\Delta u$ vanishes and $\|u\|$ is
attained at more than one point, choose $\overline{k}$ to be the
largest value producing $\|u\|$, then that node occurs at
$k_1=\overline{k}-1$. Otherwise, $k_1=\overline{k}$. Moreover,
with the strict difference inequality $\Delta^{n}u(k)<0$,
$k\in\{0,N\}$, we know that $u$ is increasing on
$\{n-2,\overline{k}\}$ and decreasing, concave down on
$\{\overline{k},N+n\}$. And, if $k\neq k_{j}$,
$k\in\{n-1-j,N+n-j\}$, $\Delta^{j}u$ does not have a node at $k$.
So, it is easy to see that
$\min_{k\in\{\xi_{m-2},N+n\}}u(k)=u(N+n)$.

First assume that $\overline{k}\leq \xi_{m-2}<N+n$. Since
$u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)\geq
\alpha_{m-2}u(\xi_{m-2})$, and by the decreasing, negative concavity
nature of $u$, we have
\begin{align*}
u(\overline{k}) &\leq
u(N+n)+\frac{u(N+n)-u(\xi_{m-2})}{N+n-\xi_{m-2}}
\big(\overline{k}-(N+n)\big)\\
&\leq  u(N+n)+
\big(\frac{1}{\alpha_{m-2}}u(N+n)-u(N+n)\big)
\frac{N+n}{N+n-\xi_{m-2}}\\
&=  \frac{N+n-\alpha_{m-2}
\xi_{m-2}}{\alpha_{m-2}(N+n-\xi_{m-2})}u(N+n);
\end{align*}
i.e.,
\[
\min_{k\in\{\xi_{m-2},N+n\}}u(k)
\geq\frac{\alpha_{m-2}(N+n-\xi_{m-2})}{N+n-\alpha_{m-2}
\xi_{m-2}}\|u\|.
\]

Second, if $ \xi_{m-2}<\overline{k}<N+n$, let
\[
h(k)=u(k)-\frac{\|u\|\prod_{i=0}^{n-2}(k-i)}{\prod_{i=0}^{n-2}
(\overline{k}-i)}, \quad k\in \{0,\overline{k}\}.
\]
 We can prove directly that
$\Delta^{n}h(k)<0$, $k\in\{0,\overline{k}-n\}$,
$h(0)=\dots=h(n-2)=0$, $h(\overline{k})=0$. Apply Lemma \ref{lem2.1}, it
follows that $h(k)\geq0$; i.e.,
\[
u(k)\geq\frac{\|u\|\prod_{i=0}^{n-2}(k-i)}{\prod_{i=0}^{n-2}
(\overline{k}-i)},
\quad k\in\{0,\overline{k}\}.
\]
 So, in particular,
\begin{equation} \label{e6}
u(\xi_{m-2})\geq\frac{\|u\|\prod_{i=0}^{n-2}
(\xi_{m-2}-i)}{\prod_{i=0}^{n-2}(\overline{k}-i)}
>\frac{\|u\|\prod_{i=0}^{n-2}(\xi_{m-2}-i)}{\prod_{i=0}^{n-2}(N+n-i)},
\end{equation}
which implies
\[
u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)\geq
\alpha_{m-2}u(\xi_{m-2})\geq
\frac{\alpha_{m-2}\prod_{i=0}^{n-2}
(\xi_{m-2}-i)}{\prod_{i=0}^{n-2}(N+n-i)}\|u\|.
\]

Case (ii): $\sum_{i=1}^{m-2} \alpha_i\geq1$. Again, using
the argument given in the first case, we obtain the similar nature
of $u$.

Firstly, suppose $u(\xi_{m-2})>u(N+n)$, then
$\min_{k\in\{\xi_{m-2},N+n\}}u(k)=u(N+n)$, which implies
$\xi_1<\overline{k}<N+n$. In fact, if
$n-2<\overline{k}\leq\xi_1$, then $u(\xi_1)\geq
u(\xi_2)\geq\dots\geq u(\xi_{m-2})>u(N+n)$, and
\[
u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)
>\sum_{i=1}^{m-2}\alpha_iu(N+n)\geq u(N+n).
\]
Which is a contradiction. Thus \eqref{e6} is readily modified to
obtain
\[
u(\xi_1)\geq \frac{\|u\|\prod_{i=0}^{n-2}(\xi_1-i)}{\prod_{i=0}^{n-2}
(N+n-i)},
\]
which implies
$$
u(N+n)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i)\geq
\alpha_1u(\xi_1)\geq
\frac{\alpha_1\prod_{i=0}^{n-2}(\xi_1-i)}{\prod_{i=0}^{n-2}
(N+n-i)}\|u\|.
$$
Secondly, if $u(\xi_{m-2})\leq u(N+n)$, then
$\min_{k\in\{\xi_{m-2},N+n\}}u(k)=u(\xi_{m-2})$; thus,
$\xi_{m-2}\leq\overline{k}\leq N+n$. Hence, we have \eqref{e6}.
The proof is complete.
\end{proof}

\section{Main results}

In this section, we will impose suitable growth conditions on $f$,
which enable us to apply Theorem \ref{thm2.1} to obtain three positive
solutions for \eqref{e1}) \eqref{e2}.

Let $E=\big\{u:\{0, N+n\}\to\mathbb{R}\big\}$, and choose the
cone $K\subset E$,
\[
K=\big\{u\in E:u(k)\geq 0,\; k\in
\{0,N+n\},\text{ and }\min_{k\in
\{\xi_{m-2},N+n\}}u(k)\geq\gamma\|u\|\big\}.
\]
Define an operator $A$ by
\[
Au(k)=\sum_{s=0}^{N}G(k,s)a(s)f\big(s,u(s)\big).
\]
Obviously, $u$ is a solution of \eqref{e1} \eqref{e2}
if and only if $u$ is a fixed point of operator $A$.

Finally, we define the nonnegative continuous concave functional
$\alpha$ on $K$ by
$$
\alpha(u)=\min_{k\in\{\xi_{m-2},N+n\}}u(k).
$$
Note that, for each $u\in K$, $\alpha(u)\leq\|u\|$.

For of convenience, we denote
\[
\lambda_1=\max_{k\in\{0,N+n\}}\sum_{s=0}^{N}G(k,s)a(s),\quad
\lambda_2=\min_{k\in\{\xi_{m-2},N+n\}}\sum_{s=\xi_{m-2}}^{N}G(k,s)a(s).
\]
Then $0<\lambda_2<\lambda_1$. To present our main result, we
assume there exist constants
$0<a<b<\min\{\gamma,\frac{\lambda_2}{\lambda_1}\}c$ such that
\begin{itemize}
\item[(H1)] $f(k,u)\leq c/\lambda_1$, for $(k,u)\in
\{0,N+n\}\times [0,c]$;

\item[(H2)]  $f(k,u)<a/\lambda_1$, for $(k,u)\in
\{0,N+n\}\times [0,a]$;

\item[(H3)] $f(k,u)>b/\lambda_2$, for $(k,u)\in
\{\xi_{m-2},N+n\}\times [b,b/\gamma]$.
\end{itemize}

\begin{theorem} \label{thm3.1}
Under assumptions {\rm (H1)--(H3)},
the boundary value problem \eqref{e1} \eqref{e2} has at least
three positive solutions $u_1$, $u_2$ and $u_3$ satisfying
\begin{equation} \label{e7}
\|u_1\|<a,\quad
b<\min_{k\in \{\xi_{m-2},N+n\}}u_2(k),\quad
\|u_3\|>a, \quad
\min_{k\in \{\xi_{m-2},N+n\}}u_3(k)<b.
\end{equation}
\end{theorem}

\begin{proof}
First, We note that  $A:\overline{P_{c}}\to \overline{P_{c}}$
is completely continuous. If $u\in\overline{P_{c}}$,
then $\|u\|\leq c$, and by condition (H1), we have
\[
\|Au\|
=\max_{k\in\{0,N+n\}}\sum_{s=0}^{N}G(k,s)a(s)f\big(s,u(s)\big)\\
\leq\frac{c}{\lambda_1}\max_{k\in\{0,N+n\}}\sum_{s=0}^{N}G(k,s)a(s)=c.
\]
Hence, $A:\overline{P_{c}}\to \overline{P_{c}}$. Standard
applications of Arzela-Ascoli theorem imply that $A$ is completely
continuous.
In an analogous argument, the condition (H2) implies
the condition (C2) of Theorem \ref{thm2.1}.

We now show that condition (C1) of Theorem \ref{thm2.1} is satisfied.
Obviously,
$$
\{u\in P(\alpha,b,\frac{b}{\gamma}):\alpha (u)>b\}\neq \emptyset.
$$
If $u\in P(\alpha,b,\frac{b}{\gamma})$, then
$b\leq u(k)\leq \frac{b}{\gamma}$, for $k\in\{\xi_{m-2},N+n\}$.
 By condition (H3), we obtain
\begin{align*}
\alpha(Au)&= \min_{k\in
\{\xi_{m-2},N+n\}}\sum_{s=0}^{N}G(k,s)a(s)f\big(s,u(s)\big)\\
&\geq \min_{k\in
\{\xi_{m-2},N+n\}}\sum_{s=\xi_{m-2}}^{N}G(k,s)a(s)f\big(s,u(s)\big)\\
&> \frac{b}{\lambda_2}\min_{k\in
\{\xi_{m-2},N+n\}}\sum_{s=\xi_{m-2}}^{N}G(k,s)a(s)=b.
\end{align*}
Therefore, condition (C1) of Theorem \ref{thm2.1} is satisfied.

Finally, we show that condition (C3) of Theorem \ref{thm2.1} also holds.
If $u\in P(\alpha,b,c)$ and $\|Au\|>\frac{b}{\gamma}$, then
$$
\alpha(Au)=\min_{k\in \{\xi_{m-2},N+n\}}Au(k)\geq\gamma\|Au\|>b.
$$
So, condition (C3) of Theorem \ref{thm2.1} is satisfied.

Applying Theorem \ref{thm2.1}, we know that the boundary value problem
\eqref{e1} \eqref{e2} has at least three positive
solutions $u_1$, $u_2$ and $u_3$
satisfying \eqref{e7}. The proof is complete.
\end{proof}

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\end{document}