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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 87, 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  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/87\hfil Positive solutions]
{Positive solutions to nonlinear second-order three-point boundary-value
problems for difference equation with change of sign}

\author[C. Wang, X. Han, C. Li\hfil EJDE-2008/87\hfilneg]
{Chunli Wang, Xiaoshuang Han, Chunhong Li}  

\address{Chunli Wang \newline
Institute of Information Technology \\
University of Electronic Technology\\
Guilin, Guangxi 541004, China \newline
Department of  Mathematics \\
Huaiyin Teachers College\\
Huaian, Jiangsu 223001,  China}
\email{wangchunliwcl821222@sina.com}

\address{Xiaoshuang Han \newline
Yanbian University of Science and Technology\\
Yanji, Jilin 133000, China}
\email{petty\_hxs@hotmail.com}

\address{Chunhong Li \newline
Department of Mathematics\\
Yanbian University \\
Yanji, Jilin 133000, China \newline
Department of  Mathematics \\
Huaiyin Teachers College\\
Huaian, Jiangsu 223001,  China}
\email{abbccc2007@163.com}

\thanks{Submitted March 6, 2008. Published June 11, 2008.}
\subjclass[2000]{39A05, 39A10}
\keywords{Boundary value problem; positive solution;
difference equation; \hfill\break\indent
fixed point; changing sign coefficients}

\begin{abstract}
 In this paper we investigate the existence of positive solution to
 the  discrete second-order three-point boundary-value problem
 \begin{gather*}
 \Delta^2 x_{k-1}+ h(k) f(x_k)=0, \quad  k \in [1, n], \\
 x_0 =0,  \quad  a x_l = x_{n+1},
 \end{gather*}
 where  $n \in [2, \infty)$, $l \in [1, n]$, $0 < a < 1$,
 $(1-a)l \geq 2$, $(1+a)l\leq n+1$, $f \in C(\mathbb{R}^+,\mathbb{R}^+)$ and $h(t)$
 is a function that may change sign on $[1, n]$.
 Using the fixed-point index theory, we prove the existence of
 positive solution for the above boundary-value problem.
\end{abstract}

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

\section{Introduction}

Recently, some authors considered the existence of positive
solutions to discrete boundary-value problems and obtained some
existence results; see for example,\cite{a1,a2,c1,m1,y1}.
Motivated by the papers \cite{l1,z1},  we
consider the existence of positive solution for the nonlinear
discrete three-point boundary-value problem
\begin{equation}
\begin{gathered}
\Delta^2 x_{k-1} + h(k) f(x_k)=0, \quad  k \in [1, n], \\
x_0=0, \quad  a x_l = x_{n+1},
\end{gathered} \label{e1.1}
\end{equation}
where $n \in \{2, 3, \dots\}$, $l \in [1, n] = \{1, 2, \dots ,
n\}$, $0 < a < 1$,  $(1-a)l \geq 2$, $(1+a)l\leq n+1$, $f \in
C(\mathbb{R}^+,\mathbb{R}^+)$ and $h \in C(\mathbb{R}^+, \mathbb{R})$.

When $h(k) \equiv 1$, Equation  \eqref{e1.1} reduces to the nonlinear
discrete three-point boundary-value problem studied by Zhang and Medina
\cite{z1}. Using the same approach as in \cite{z1},
we obtain the  the existence of positive solutions to
\eqref{e1.1} when $h \in C(\mathbb{R}^+, \mathbb{R}^+)$.
To the author's knowledge, no one has studied the existence of
positive solution for \eqref{e1.1} when $h$ is allowed to change sign
on $[1,n]$. Hence, the aim of the present paper is to establish
simple criteria for the existence of at least
one positive solution of the \eqref{e1.1}. Our main tool is the fixed-point
index theory  \cite{g1}.

\begin{theorem}[\cite{g1}]\label{thm1.1}
 Suppose $E$ is a real Banach space, $K \subset E$ is a cone,
and $\Omega_r = \{u \in K :\|u\| \leq r\}$. Let the operator
$T : \Omega_r \to K$ be completely continuous and satisfy $T x \neq  x$,
for all $x \in \partial \Omega_r$. Then
\begin{itemize}
\item[(i)]  If $\|T x\| \leq \|x\|$, for all $x \in \partial \Omega_r$,
then $i(T, \Omega_r, K) = 1$;

\item[(ii)]  If $\|T x\| \geq \|x\|$, for all $x \in \partial
\Omega_r$,  then $i(T, \Omega_r, K) = 0$.
\end{itemize}
\end{theorem}

In this paper, by a positive solution $x$ of  \eqref{e1.1}, we
mean a solution  of the \eqref{e1.1} satisfying
$x_k > 0$, $k \in [1, n+1]$.

We will use the following notation:
$\mathbb{Z}=\{0,\pm 1,\pm 2,\dots\}$;  $\mathbb{N}=\{0,1,2,\dots\}$;
$[m,n]=\{m,m+1,m+2,\dots,n\} \subset \mathbb{Z}$;
$[x]$ is the integer value function;
$\Delta y_k = y_{k+1}-y_k$,
$ \Delta^n y_k = \Delta(\Delta^{n-1}y_k)$,
$n \geq 2$, $k\in \mathbb{N}$.

Moreover, we shall use the following assumptions:
\begin{itemize}
\item[(H1)]  $f \in C(\mathbb{R}^+,\mathbb{R}^+)$ is continuous
and nondecreasing.

\item[(H2)]  $h:[1,n] \to (-\infty,+\infty)$  such that
$h(k) \geq 0$, $k \in [1, l];  h(k) \leq 0$, $k \in [l, n]$.
Moreover, $h(k)$ does not vanish identically on any subinterval
of $[1,n]$.


\item[((H3)] There exist nonnegative constants in the extended reals,
$f_0$, $f_\infty$, such that
$$
f_0 = \lim _{u\to 0^+}\frac{f(u)}{u}, \quad
f_\infty  = \lim _{u\to +\infty}\frac{f(u)}{u}.
$$
\end{itemize}

This paper is organized  as follows:
In section 2, preliminary lemmas are given. In section 3,
we prove the existence of positive solutions for \eqref{e1.1}.
In section an example is provided.

\section{Preliminaries}

In this section, we give some lemmas that will be used to prove our
main results.

\begin{lemma}[\cite{z1}] \label{lem2.1}
Let $a l \neq n+1$. For $\{y_k\}_{k=0}^{n+1}$, the  problem
\begin{equation}
\begin{gathered}
\Delta^2 x_{k-1}+ y_k=0, \quad  k \in [1,n], \\
x_0 =0, \quad a x_l = x_{n+1}\,.
\end{gathered}\label{e2.1}
\end{equation}
has a unique solution
$$
x_k = \frac{k}{n+1-al} \Big(\sum_{i=0}^n
\sum_{j=0}^i y_j - a
\sum_{i=0}^{l-1}\sum_{j=0}^i y_j \Big) -
\sum_{i=0}^{k-1}\sum_{j=0}^i y_j, \quad k \in [0,n+1].
$$
\end{lemma}

Using the above lemma, for $0 < a < 1$,  it is easy to prove the
following result.

\begin{lemma}\label{lem2.2}
The Green's function for \eqref{e2.1} is
\begin{equation}
 G(k, j)=
\begin{cases}
\dfrac{((a-1)k + n+1-al)j}{n+1-al}, &  j < k, \; j < l,\\[3pt]
\dfrac{((a-1)j + n+1 - al)k}{n+1- a l}, & k \leq j < l,\\[3pt]
\dfrac{a l(k-j)+(n+1-k)j}{n+1 - a l}, &  l < j \leq k,\\[3pt]
\dfrac{(n+1-j)k}{n+1 - a l}, &  j \geq k, \; j \geq l,
\end{cases} \label{e2.2}
\end{equation}

\end{lemma}

We want to point that this Green's function is new.
\begin{remark} \label{rmk2.1} \rm
Note that $G(k, j) \geq 0$ for $(k, j) \in [0,n+1] \times [1, n]$.
\end{remark}

\begin{lemma}\label{lem2.3}
Let $(1-a)l \geq 2$.  For all $j_1 \in [\tau, l]$ and $j_2 \in [l, n]$, we have
\begin{equation}
G(k, j_1) \geq M G(k, j_2), \quad k \in [0, n+1],\label{e2.3}
\end{equation}
 where $\tau \in [[al] + 1, l - 1]$ and $M = a^2 l/n$.
\end{lemma}

\begin{proof} It is easy to check that $(1-a)l \geq 2$
implies $[[al] + 1, l - 1]$. We divide the proof into two cases.

\noindent\textbf{Case 1:}  $k \leq l$.  By \eqref{e2.2},
\begin{align*}
\frac{G(k, j_1)}{G(k, j_2)}
&= \begin{cases}
\dfrac{[(a-1)k+n+1-al]j_1}{(n+1-j_2)k}, &  j_1 \leq k, \\[3pt]
\dfrac{[(a-1)j_1+n+1-al]k}{(n+1-j_2)k}, &  k < j_1
\end{cases}
\\
&\geq \begin{cases}
\dfrac{(n+1-l)\tau}{(n+1-l)k}, &  j_1 \leq k, \\[3pt]
\dfrac{(n+1-l)k}{(n+1-l)k}, &  k < j_1
\end{cases}
\\
&\geq \min \{\frac{\tau}{l}, 1\} \\
&= \frac{\tau}{l} \\
&\geq a > M.
\end{align*}

\noindent\textbf{Case 2:} $k \geq l$.
 For $j_2 \in [l, n]$ and $(1+a)l\leq n+1$, it is
easy to check that
\begin{equation}
\begin{aligned}
(n+1-k-al)j_2 + alk
&=(n+1-al)j_2+(al-j_2)k \\
&\leq(n+1-al)j_2+(al-j_2)l \\
&=(n+1-al-l)j_2+al^2 \\
&\leq(n+1-al-l)n+al^2\\
&= (n - al)(n+1-l) + al,
\end{aligned} \label{e2.4}
\end{equation}
 and
\begin{equation}
(a-1)k+n+1-al \geq (a-1)(n+1) + n+1-al = a(n+1-l), \label{e2.5}
\end{equation}
  From (\ref{e2.2}), (\ref{e2.4}) and (\ref{e2.5}), we
obtain
\begin{align*}
\frac{G(k, j_1)}{G(k, j_2)}
&= \begin{cases}
\dfrac{[(a-1)k+n+1-al]j_1}{(n+1-k-al)j_2 + alk}, &  j_2 \leq k, \\[3pt]
\dfrac{[(a-1)k+n+1-al]j_1}{(n+1-j_2)k}, &  k < j_2
\end{cases} \\
&\geq \begin{cases}
\dfrac{a(n+1-l)\tau}{(n-al)(n+1-l) + al}, &  j_2 \leq k, \\
\dfrac{a(n+1-l) \tau}{(n+1-l)n}, &  k < j_2
\end{cases} \\
&\geq \min \big\{ \frac{a^2 l(n+1-l)
}{(n-al)(n+1-l) + al}, \frac{a^2 l}{n} \big\} \\
&= \frac{a^2 l}{n} = M.
\end{align*}
 Hence, $G(t, j_1) \geq  M G(t, j_2)$ holds.
\end{proof}

Let  $C[0,n+1]$ be the Banach space with the norm $\|x\|=\sup_{k \in
[0,n+1]}|x_k|$. Denote
\begin{gather*}
C_0^+[0,n+1]=\{x \in C[0,n+1]:  \min_{k \in [0,n+1]} x_k
\geq 0 \text{ and } x_0=0,\;  x_{n+1} = a x_l\},
\\
P=\{x \in C_0^+[0,n+1]: x_k \text{ is  concave  on $[0,l]$
 and  convex  on  $[l,n+1]$}\}.
\end{gather*}
 It is obvious that $P$ is a cone in $C[0,n+1]$.

\begin{lemma}\label{lem2.4}
 If $x \in P$, then
$$
x_k \geq  A(k) x_l, \quad k \in [0,l];\quad
x_k \leq  A(k)  x_l, \quad k \in [l,n+1].
$$
 where
$$
 A(k) = \begin{cases}
k/l, &   k \in [0, l],\\
\frac{n+1-al+(a-1)k}{n+1-l}, &  k \in [l+1, n+1],
\end{cases}
$$
\end{lemma}

\begin{proof}  Since $x \in P$, we have  $x_k$ is concave
on $[0,l]$, convex on $[l,n+1]$, $x_0=0$, and $x_{n+1} = a x_l$.
Thus,
$$
x_k \geq x_0 +\frac{x_l-x_0}{l}k=\frac{k}{l} x_l, \quad
\text{for }  k \in[0,l],
$$
 and
$$
x_k \leq x_{n+1} +
\frac{x_{n+1}-x_l}{n+1-l}(k-1-n) =\frac{n+1-al+(a-1)k}{n+1-l} x_l,
\quad\text{for }   k \in [l,n+1].
$$
  Hence, we have
$$
x_k \geq A(k) x_l, \quad k \in[0,l]; \quad
x_k \leq A(k) x_l, \quad  k \in [l,n+1].
$$
\end{proof}

\begin{lemma}\label{lem2.5}
Assume that $(1-a)l \geq 2$. Let $x \in P$, then
$$
 x_k  \geq  \mu \|x\|, \quad  k  \in [[al]+1, \tau],
$$
 where $\mu = \min \{a,1-\frac{\tau}{l}\}$,
$\tau \in [[al]+1, l-1]$.
\end{lemma}

 \begin{proof}  Let $x \in P$, then $x$ is concave on
$[0,l]$, and convex on $[l,n+1]$. Since $0 < a < 1$,
$x_{n+1} = a x_l < x_l$, then
$ \|x\| = \sup_{k \in [0,n+1]}|x_k| = \sup_{k \in[0,l]}|x_k|$.
Set $r = \inf\{r \in [0,l]: \sup_{k \in [0,l]}x_k = x_r\}$.
We now consider the following two cases:

\noindent\textbf{Case (i):}  $k \in [0, r]$.
By the concavity of $x_k$, we have
$$
x_k \geq x_0 + \frac{x_r - x_0}{r}k =
\frac{k}{r} x_r \geq \frac{k}{l} x_r = \frac{k}{l}\|x\|.
$$

\noindent\textbf{Case(ii):}  $k \in [r,l]$.
Similarly, we obtain
\begin{align*}
x_k  &\geq x_r + \frac{x_r -x_l}{r-l}(k-r)\\
&= \frac{l-k}{l-r} x_r + \frac{k-r}{l-r} x_l\\
& \geq \frac{l-k}{l-r} x_r \geq \frac{l-k}{l} x_r \\
&= \big[1- \frac{k}{l}\big] \|x\|.
\end{align*}
 Thus, we have
$$
x_k \geq \min \big\{\frac{k}{l},
1-\frac{k}{l}\big\}\|x\|, \quad  k \in [0,l],
$$
which yields
$$
\min _{k \in [[al]+1, \tau]} x_k \geq \min
\{a, 1-\frac{\tau}{l}\} \|x\| = \mu\|x\|.
$$
The proof is completed.
\end{proof}

\begin{lemma}\label{lem2.6}
Assume that $(1-a)l \geq 2$. If conditions
{\rm (H1), (H2), (H4)} hold $\forall  k \in
[0,n-l]$, then there exists a constant $\tau \in [[al]+1, l-2]$  such
that
\begin{equation}
{ B(k) = h^+(l- [\delta k]) - \frac{1}{M} h^-(l+k) \geq
0 }, \label{e2.6}
\end{equation}
where $h^+(k) = \max\{h(k),0\}, h^-(k) = -
\min\{h(k),0\}$, $\delta = \frac{l-\tau-1}{n-l+1}$, and
$M = a^2 l/n$. Then for all $q \in [0,\infty)$, we have
\begin{equation}
{\sum _{j=\tau+1}^n G(k,j) h(j) f(q A(j)) \geq 0. } \label{e2.7}
\end{equation}
\end{lemma}

\begin{proof}
By the definition of $A(k)$, it is easy to check that
\begin{equation}
{ A(l-[\delta r]) = \frac{1}{l}
\Big(l-\big[\frac{l-\tau-1}{n-l+1}r\big]\Big)
=1- \frac{1}{l} \big[\frac{l-\tau-1}{n-l+1}r\big], \quad r \in [0, n-l+1], }
\label{e2.8}
\end{equation}
  and
\begin{equation}
{A(l+r)=1-\frac{r}{n-l+1}(1-a), \quad r \in [0, n-l]. }
\label{e2.9}
\end{equation}
Set $j=l- [\delta r]$, $r \in [0,n-l+1]$
($\delta$ as in (H4)).  For all $q \in [0,\infty)$, by view of Lemma
\ref{lem2.3}, Remark \ref{rmk2.1},  (\ref{e2.6}), (\ref{e2.8}), (\ref{e2.9}),
(H4), and that $f$ is nondecreasing, we have
\begin{equation}
\begin{aligned}
&\sum_{j=\tau+1}^l G(k,j) h^+(j) f(q A(j))\\
&=\sum_{r=0}^{n-l+1} G(k, l-[\delta r]) h^+(l-
[\delta r]) f(q A(l-[\delta r])) \\
& = \sum _{r=0}^{n-l+1} G(k,
l-[\delta r]) h^+(l-[\delta r]) f \Big(q \big(1- \frac{1}{l}
\big[\frac{l-\tau-1}{n-l+1}r\big] \big)\Big) \\
&{\geq \sum _{r=0}^{n-l+1} G(k,
l-[\delta r]) h^+(l-[\delta r]) f \Big(q
\big(1-\frac{r}{n-l+1}\big(1-\frac{\tau+1}{l}\big)\big)\Big)}\\
& \geq M \sum _{r=0}^{n-l} G(k,
l+r) h^+(l - [\delta r])f \Big(q
\big(1-\frac{r}{n-l+1}\big(1-\frac{\tau+1}{l}\big)\big)\Big)\\
&\geq \sum _{r=0}^{n-l} G(k, l+r) h^-(l+r)
f \Big(q \big(1-\frac{r}{n-l+1}(1-a)\big)\Big).
\end{aligned}\label{e2.10}
\end{equation}
 Again, setting $j=l+r$, $r \in[0,n-l]$, for
$q \in [0,\infty)$, we obtain
\begin{equation}
{ \sum _{j=l+1}^n G(k,j)h^-(j)f(q A(j)) = \sum
_{r=1}^{n-l} G(k, l+r) h^-(l+r)
  f \Big(q \big(1-\frac{r}{n-l+1}(1-a)\big)\Big). }
\label{e2.11}
\end{equation}
Thus, by (\ref{e2.10}) and (\ref{e2.11}), we get
\begin{align*}
&\sum _{j=\tau+1}^n G(k,j) h(j) f(q A(j))\\
& = \sum _{j=\tau+1}^l G(k,j) h^+(j) f(q A(j))
- \sum _{j=l+1}^n G(k,j) h^-(j) f(q A(j)) \geq 0.
\end{align*}
 The proof is completed.
\end{proof}

We define the operator $T : C[0, n+1] \to C[0, n+1]$ by
\begin{equation}
{(Tx)_k = \sum _{j=1}^n G(k,j) h(j) f(x_j),
\quad  (k,j) \in [0,n+1] \times [1,n]. } \label{e2.12}
\end{equation}
 where $G(k, j)$ as in (\ref{e2.2}).  From Lemma
\ref{lem2.3}, we easily know that $x(t)$ is a solution of the \eqref{e1.1} if and only if $x(t)$ is a fixed point of the operator
$T$.

\begin{lemma}\label{lem2.7}
Let $(1-a)l \geq 2$. Assume that conditions {\rm (H1), (H2),(H4)}
 are satisfied. Then $T$ maps $P$ into $P$.
\end{lemma}

\begin{proof}   For $x \in P$, by Lemmas \ref{lem2.4},
\ref{lem2.6},  and $f$ is nondecreasing, we have
\begin{equation}
\begin{aligned}
&\sum _{j=\tau+1}^n G(k,j) h(j)f(x_j) \\
&= \sum _{j=\tau}^l G(k,j) h^+(j) f(x_j)
- \sum_{j=l+1}^n G(k,j) h^-(j) f(x_j) \\
& \geq \sum _{j=\tau+1}^l G(k,j) h^+(j) f(A(j) x_l)
 - \sum _{j=l+1}^n G(k,j) h^-(j)f(A(j)x_l) \\
&=  \sum _{j=\tau+1}^n G(k,j) h(j)
f(x_l A(j)) \geq 0,
\end{aligned} \label{e2.13}
\end{equation}
which implies
\begin{align*}
(Tx)_k &= \sum _{j=1}^n G(k,j) h(j) f(x_j)\\
& = \sum _{j=1}^\tau G(k,j)
h^+(j) f(x_j)+ \sum _{j=\tau+1}^n G(k,j) h(j) f(x_j)\\
&\geq  \sum _{j=1}^\tau G(k,j)
h^+(j) f(x_j)\geq 0,
\end{align*}
again $(Tx)_0=0$, $(Tx)_{n+1} = a(Tx)_l$, it follows that
$T: P \to C_0^+[0,n+1]$. On the other hand,
\begin{gather*}
\Delta^2(Tx)_k = - h^+(j) f(x_j) \leq 0, \quad j \in [0,l],\\
\Delta^2(Tx)_k = h^-(j) f(x_j) \geq 0, \quad j \in [l,n+1].
\end{gather*}
 Thus, $T$ maps $P$ into $P$.
\end{proof}

\begin{lemma} \label{lem2.8}
 Let $(1-a)l \geq 2$. Assume that conditions
{\rm (H1), (H2), (H4)} are satisfied. If $z \in P$ is a fixed point
of $T$ and $\|z\|>0$, then $z$ is a positive
solution of the \eqref{e1.1}.
\end{lemma}

\begin{proof}
At first, we claim that $z_l>0$. Otherwise,
$z_l=0$ implies $z_{n+1} = a z_l = 0$. By the convexity and
the nonnegativity of $z$ on $[l,n+1]$, we have
$$
z_k \equiv 0, \quad k \in [l,n+1],
$$
this implies $\Delta z_l = z_{l+1}-z_l=0$. Since $z = Tz$,
we have  $\Delta^2 z_k = - h^+(k) f(z_k) \leq 0$,  $k \in [0,l]$.
Then
$$
\Delta z_k \geq \Delta z_l = 0, \quad  k \in [0,l-1].
$$
  Thus, $z_k \leq z_l=0$,  $k \in [0,l]$.
  By the nonnegativity of $z$,  we get
$$
z_k \equiv 0, \quad  k \in [0,l],
$$
which yields a contradiction with $\|z\|>0$.
\end{proof}

Next, in view of Lemma \ref{lem2.1}, for $z \in P$, we have
\begin{equation}
 z_k  \geq \frac{k}{l} z_l > 0, \quad k \in [1, l].
\label{e2.14}
\end{equation}
Note that $h(k)$ does not vanish identically on any
subinterval of $k \in [1,l]$, for any $k \in [1, n]$. By
(\ref{e2.13}) we have
\begin{align*}
z_k &= (Tz)_k  \\
&= \sum_{j=1}^n G(k,j) h(j) f(z_j) \\
&= \sum _{j=1}^\tau   G(k,j) h^+(j) f(z_j)
 + \sum_{j=\tau}^n G(k,j) h(j) f(z_j) \\
&\geq \sum _{j=1}^{\tau} G(k,j)
h^+(j) f(z_j)>0.
\end{align*}
Thus, we assert that $z$ is a positive solution of \eqref{e1.1}.

\section {Existence of solutions}

For convenience, we set
\begin{gather*}
 M = \Big(\mu \max _{k \in [0,n+1]} \sum
_{j=[al]+1}^\tau G(k,j)h^+(j)\Big)^{-1},\\
m = \Big(\max _{k \in [0,n+1]} \sum _{j=1}^l G(k,j)
h^+(j)\Big)^{-1}
\end{gather*}
where  $\mu$ as in Lemma \ref{lem2.5}.

\begin{theorem}\label{thm3.1}
Let $(1-a)l \geq 2$. Assume
that conditions {\rm (H1)--(H4)} are satisfied.
If {\rm (H5)}, $0 \leq f_0 < m$, and $M < f_{\infty} \leq + \infty$
hold, then  \eqref{e1.1} has at least one positive
solution.
\end{theorem}

\begin{proof}  By Lemma \ref{lem2.7},  $T: P \to P$.
Moreover, it is easy to check by Arzela-Ascoli theorem that $T$ is
completely continuous. By (H5), we have $f_0 <  m$. There exist
$\rho_1> 0$ and $\varepsilon_1 > 0$ such that
\begin{equation}
 f(u) \leq (m - \varepsilon_1) u, \quad for \quad  0 <
u \leq \rho_1. \label{e3.1}
\end{equation}
Let $\Omega_1 = \{x \in P: \|x\| < \rho_1\}$. For $x \in
\partial\Omega_1$, by (\ref{e3.1}), we have
\begin{align*}
(Tx)_k &= \sum _{j=1}^n G(k,j) h(j) f(x_j) \\
&= \sum _{j=1}^l G(k,j) h^+(j)
f(x_j) - \sum_{k=l+1}^n G(k,j) h^-(j) f(x_j) \\
& \leq \sum _{j=1}^l G(k,j) h^+(j) f(x_j)\\
&\leq  \rho_1 (m - \varepsilon_1) \max
_{k \in [0, n+1]} \sum_{j=1}^l G(k,j) h^+(j) \\
&= \rho_1 (m-\varepsilon_1) \frac{1}{m} \\
&< \rho_1 = \|x\|,
\end{align*}
which yields
$\|Tx\| < \|x\|$ for $x \in \partial \Omega_1$.
 Then by Theorem \ref{thm3.1}, we have
\begin{equation}
i(T, \Omega_1, P) = 1. \label{e3.2}
\end{equation}

 On the other hand,  from  (H5), we have $f_{\infty} > M$.
 There exist $\rho_2 > \rho_1 > 0$ and $\varepsilon_2$ such that
\begin{equation}
 f(u) \geq  (M + \varepsilon_2) u, \quad  {\rm for}
\quad u \geq \mu \rho_2.   \label{e3.3}
\end{equation}
Set $\Omega_2 = \{x \in P: \|x\| < \rho_2\}$. For any
$x \in \partial\Omega_2$, from Lemma \ref{lem2.5}, we have
$x_k \geq \mu \|x\| = \mu \rho_2$, for $k \in [[al]+1, \tau]$. Then from
(\ref{e2.7}) and (\ref{e3.3}), we obtain
\begin{align*}
 \|Tx\| &= \max _{k \in [0,n+1]}
\Big[\sum_{j=1}^{\tau} G(k,j) h^+(j)f(x_j) +
\sum_{j=\tau+1}^n G(k,j) h(j)f(x_j)\Big] \\
&\geq \max _{k \in [0,n+1]} \sum_{j=1}^\tau G(k,j) h^+(j)f(x_j)\\
&\geq \max_{k \in [0,n+1]}\sum_{j=[al]+1}^\tau  G(k,j) h^+(j)f(x_j) \\
&\geq \mu (M + \varepsilon_2) \rho_2 \max_{k\in[0,n+1]}
\sum_{j=[al]+1}^\tau G(k,j) h^+(j) \\
&=  \frac{1}{M} (M + \varepsilon_2) \rho_2 > \rho_2 = \|x\|;
\end{align*}
this is,
$\|Tx\| > \|x\|$,   for $ x \in \partial \Omega_2$.
 Then, by Theorem \ref{thm3.1},
\begin{equation}
i(T, \Omega_2, P) = 0.  \label{e3.4}
\end{equation}
Therefore, by (\ref{e3.2}), (\ref{e3.4}), and  $\rho_1 < \rho_2$, we
have
$$
i(T, \Omega_2 \setminus \bar{\Omega}_1, P)= -1.
$$
 Then operator $T$ has a fixed point in
$\Omega_2 \setminus \bar{\Omega}_1$.  So, \eqref{e1.1} has at
least one positive solution.
\end{proof}

\begin{theorem}\label{thm3.2}
 Let $(1-a)l \geq 2$. Assume that  {\rm (H1)--(H4)} are satisfied.
 If {\rm (H6)}, $M < f_0 \leq + \infty$, and $0 \leq f_{\infty} < m$
 hold, then  \eqref{e1.1} has at least one positive solution.
\end{theorem}

\begin{proof}  At first, by  (H6), we get $f_0 > M$,
there exist $\rho_3$ and $\varepsilon_3$ such that
\begin{equation}
 f(u) \geq (M + \varepsilon_3)u,\quad \text{for }  0 < u
< \rho_3.  \label{e3.5}
\end{equation}
  Set $\Omega_3 = \{x \in P : \|x\| < \rho_3\}$.
For any $x \in \partial \Omega_3$, by Lemma \ref{lem2.5}, we get
$x_k \geq \mu \|x\| = \mu \rho_3$, for $k \in [[al]+1, \tau]$, then by
(\ref{e2.7}) and (\ref{e3.5}), we have
\begin{align*}
 \|Tx\|& = \max_{k \in [0,n+1]}
\Big[\sum_{k=1}^\tau
 G(k,j) h^+(j) f(x_j) + \sum_{j=\tau+1}^n G(k,j) h(j) f(x_j)\Big] \\
& \geq \max_{k \in [0,n+1]}
\sum_{j=1}^\tau  G(k,j) h^+(j) f(x_j) \\
&\geq \max_{k \in [0,n+1]} \sum_{j=[al]+1}^\tau G(k,j) h^+(j) f(x_j) \\
&\geq  \mu (M+\varepsilon_3) \rho_3 \max_{k \in [0,n+1]}
\sum_{j=[al]+1}^\tau G(k,j) h^+(j) \\
&= \frac{1}{M} (M+\varepsilon_3) \rho_3 > \rho_3 = \|x\|;
\end{align*}
 this is,
$\|Tx\| > \|x\|$,  for $x \in \partial \Omega_3$. Thus,  by
Theorem \ref{thm1.1},
\begin{equation}
i(T, \Omega_3, P) = 0.  \label{e3.6}
\end{equation}
Next, from (H6), we have $f_{\infty} < m$. There exist
$\rho_4> 0$ and $0 < \varepsilon_4 < \rho_4$ such that
 $$
f(u) \leq (m-\varepsilon_4)u, \quad \text{for } u \geq \rho_4.
$$
Set $L = \max_{0 \leq u \leq \rho_4} f(u)$. Then
\begin{equation}
f(u) \leq L + (m-\varepsilon_4)u, \quad \text{for }
\quad u \geq 0. \label{e3.7}
\end{equation}
Choose $\rho_5 > \max \{\rho_4, L/\varepsilon_4\}$. Let
$\Omega_4 = \{x \in P : \|x\| < \rho_5\}$. Then for
 $x \in \partial \Omega_4$,  by (\ref{e2.7}) and (\ref{e3.7}),
we have
\begin{align*}
(Tx)_k  &= \sum _{j=1}^n G(k,j) h(j) f(x_j)\\
&= \sum _{j=1}^l G(k,j) h^+(j) f(x_j) - \sum_{j=l+1}^n G(k,j) h^-(j)f(x_j) \\
&\leq  \sum _{j=1}^l G(k,j)  h^+(j) f(x_j) \\
&\leq (L + (m-\varepsilon_4) \rho_5) \frac{1}{m} \\
&=  \rho_5 -(\varepsilon_4 \rho_5 -L)\frac{1}{m}\\
&< \rho_5 = \|x\|.
\end{align*}
 Thus,  by Theorem \ref{thm3.1},
\begin{equation}
i(T, \Omega_4, P) = 1.
 \label{e3.8}
\end{equation}
Therefore, by (\ref{e3.6}), (\ref{e3.8}), and  $\rho_3 <
\rho_5$, we have
$$
i(T, \Omega_4 \setminus \bar{\Omega}_3, P)= 1.
$$
 Then operator $T$ has a fixed point in $\Omega_4 \setminus
\bar{\Omega}_3$.  So, \eqref{e1.1} has at least one positive
solution.
\end{proof}

\section{Example}

In this section, we illustrates our main results.
Consider the boundary-value problem
\begin{equation}
\begin{gathered}
 \Delta^2 x_{k-1}+ h(k) x_k^{\alpha} = 0, \quad  k \in [1,11], \\
 x_0 =0, \quad  \frac{1}{3} x_6 = x_{12},
\end{gathered}  \label{e4.1}
\end{equation}
where $0 < \alpha < 1$, and
\[
 h(k)= \begin{cases}
 3(k-8)^2, &   k \in [1, 8],\\
\frac{8}{99}(8-k)^3,  &  k \in [8, 11].\\
\end{cases}
\]
Let $n=11$,  $l=8$, $a = 1/3$, then we have $M =8/99$.
Now taking $\tau=3$, then  $\tau \in [[al] + 1, l -2] = [3, 6]$,
 $\delta = 1$, and for all $k \in [0, n - l]=[0, 3]$,
we have
$$
B(k) = h^+(l - [\delta k]) - \frac{1}{M} h^-(l +
k) = k^2(3-k) \geq 0.
$$
 Hence, Condition (H2) and (H4) hold.
Set $f(u) = u^{\alpha}$, it is easy to see that
$$
f_0 = \infty, \quad f_{\infty} = 0,
$$
 that is, (H6) holds. Thus, by Theorem \ref{thm3.2},
\eqref{e4.1} has at least one positive solution.

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\end{document}