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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 70, pp. 1--11.\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-2008/70\hfil Positive solutions]
{Positive solutions for a high-order multi-point boundary-value
problem in Banach spaces}

\author[W. Jiang \hfil EJDE-2008/70\hfilneg]
{Weihua Jiang}

\address{Weihua Jiang \newline
 College of Mathematics and Science of Information,
 Hebei Normal University \\
 Shijiazhuang, 050016, Hebei,  China\newline
 College of Sciences, Hebei University of Science and Technology \\
 Shijiazhuang, 050018, Hebei,  China}
\email{weihuajiang@hebust.edu.cn}

\thanks{Submitted March 30, 2008. Published May 13, 2008.}
\thanks{Supported by grants 10701032 from the Natural Science Foundation
 of China,  and \hfill\break\indent
07217169 from the Science and Technology Key Project of Hebei province}
\subjclass[2000]{34B15}
\keywords{Banach space; positive solution; strict set contraction; 
\hfill\break\indent boundary-value problem}

\begin{abstract}
 Using the fixed point theory of strict set
 contractions, we study the existence of at least one, two, and
 multiple positive solutions for higher order multiple point
 boundary-value problems in Banach spaces. Our result extends
 some of the existing results.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}



\section{Introduction}

In the previous 30 years, the theory of ordinary differential
equations in Banach spaces has become a new important branch (see,
for example, \cite{d1,g1,g2,l1} and references cited therein).
In 1988, Guo and Lakshmikantham \cite{g4}
 discussed the existence of multiple solutions
for two-point boundary-value problem of ordinary differential
equations in Banach spaces. Since then, nonlinear second-order
multi-point boundary-value problems in Banach spaces have been
studied by several authors (see, for example, \cite{f1,l2,l3}
 and references
cited therein). On the other hand, recently, high-order multi-point
boundary-value problems for scalar ordinary differential equations
have received a great deal of attention in the literature (see,
for instance, \cite{e1,g3,j2}
 and references cited therein). However, to the
best of our knowledge, no one has  considered the existence of
multiple positive solutions (at least three or more) for high-order
multi-point boundary-value problems in Banach spaces. We will fill
this gap in the literature. In this paper, we shall discuss the
existence of  at least one, two, and
multiple positive solutions for the $n$th-order $m$-point boundary-value
problem value problem
\begin{gather}
y^{(n)}(t)+f(t,y)=\theta,\quad 0<t<1, \label{e1.1}\\
y(0)=y'(0)=\dots=y^{(n-2)}(0)=\theta,\quad
y(1)=\sum_{i=1}^{m-2}k_{i}y(\xi_{i})  \label{e1.2}
\end{gather}
in a real Banach space $E$, where $n\geq 2$, $\theta$ is the zero
element of $E$, $0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$,
 $k_i>0$, $i=1,2,\dots,m-2$. In the scalar case, the existence of positive
solutions to \eqref{e1.1}-\eqref{e1.2} had been solved in \cite{e1,g3};
So our result extends those results, to some degree.
  The key tool in our approach is the following fixed point theorem of
 strict-set-contractions.

\begin{theorem}[\cite{c1,p1}] \label{thm1.1}
Let $K$ be a cone of the real Banach
 space $X$ and $K_{r,R}=\{x\in K :r\leq \|x\|\leq R\}$ with $R>r>0$.
 Suppose that $A:K_{r,R}\to K$ is a strict set contraction
 such that one of the following two conditions is satisfied
\begin{itemize}
\item[(i)] $Ax\not\leq x$, for all $x\in K$,
$\|x\|=r$ and $Ax\not\geq x$, for all $x\in  K$, $\|x\|=R$.

\item[(ii)] $Ax\not\geq x$, for all $x\in K$, $\|x\|=r$ and
$Ax\not\leq x$, for all $x\in K$, $\|x\|=R$.
\end{itemize}
Then $A$ has at least one fixed point $x\in K$ satisfying
$r<\|x\|<R$.
\end{theorem}

Let the real Banach space $E$ with norm $\|\cdot\|$ be partially
ordered by a normal cone $P$ of $E$; i.e., $x\leq y$ if and only if
$y-x\in P$, and $P^*$ denotes the dual cone of $P$. Denote the
normal constant of $P$ by $N$; i.e., $\theta\leq x\leq y$ implies
$\|x\|\leq N\|y\|$. Take $I=[0,1]$. For any $x\in C[I,E]$,
evidently, $(C[I,E],\|\cdot\|_c)$ is a Banach space with
$\|x\|_c=\max_{t\in I}\|x(t)\|$, and $Q=\{x\in C[I,E]:x(t)\geq
\theta \text{ for }t\in I\}$ is a cone of the Banach space $C[I,E]$.
A function $x\in C^n[I,E]$ is called a positive solution of the
boundary-value problem \eqref{e1.1}-\eqref{e1.2} if it satisfies
\eqref{e1.1}-\eqref{e1.2} and $x\in Q,~x(t)\not\equiv \theta$.

For a bounded set $S$ in a Banach space, we denote $\alpha(S)$ the
Kuratowski measure of non-compactness. In this paper, we denote
$\alpha(\cdot)$ the Kuratowski measure of non-compactness of a
bounded set in $E$ and $C[I,E]$. The closed balls in spaces $E$ and
$C[I,E]$ are denoted by $T_r=\{x\in E:\|x\|\leq r\}(r>0)$ and
$B_r=\{y\in C[I,E]:\|y\|_c\leq r\}(r>0)$, respectively.

For convenience, we set
$$
a_0=\sum_{i=1}^{m-2}k_{i}\xi_i^{n-1}, \quad
a_1=\sum_{i=1}^{m-2}k_{i}\xi_i^{n-1}(1-\xi_{m-2})^n.
$$
In this paper, we assume the following conditions.
\begin{itemize}
\item[(H1)] $n\geq 2$, $k_i>0$, $i=1,2,\dots,m-2$,
$0<\xi_1<\xi_2<\dots<\xi_{m-2}<1$, $0<a_0<1$.

\item[(H2)] $P$ is a normal cone of $E$ and $N$ is the normal constant;
$f:I\times P\to P$, $f(t,\theta)=\theta$, for all $t\in I$; for
any $r>0$, $f(t,x)$ is uniformly continuous and bounded on $I\times
(P\cap T_r)$ and there exists a constant $L_r$ with $0\leq
L_r<\frac{(n-1)!(1-a_0)}{4}$ such that
$$
\alpha(f(I\times D))\leq L_r\alpha(D),\quad \forall D\subset P\cap T_r.
$$
\end{itemize}

\section{Preliminary lemmas}

\begin{lemma} \label{lem2.1}
Suppose $a_0\neq 1$, then for $h(t)\in C[I,E]$,
the problem
\begin{gather}
y^{(n)}(t)+h(t)=\theta,\quad 0<t<1, \label{e2.1}\\
y(0)=y'(0)=\dots=y^{(n-2)}(0)=\theta,\quad
y(1)=\sum_{i=1}^{m-2}k_{i}y(\xi_{i}) \label{e2.2}
\end{gather}
has a unique solution
\begin{equation} \label{e2.3}
\begin{aligned}
y(t)=&-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}h(s)ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}h(s)ds}\\
&\quad -\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}\int_0^{\xi_i}
 \frac{(\xi_i-s)^{n-1}}{(n-1)!}h(s)ds.
\end{aligned}
\end{equation}
\end{lemma}

 The proof of the above lemma is easy, so we omit it.

\begin{lemma} \label{lem2.2}
Let {\rm (H1)} hold. If $h\in Q$, then the unique
solution $y$ of  \eqref{e2.1}-\eqref{e2.2} satisfies
$y(t)\geq \theta$, $t\in I$, that is $y\in Q$.
\end{lemma}

\begin{proof} By \eqref{e2.3}, we get
\begin{align*}
y(t)&=-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}h(s)ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}h(s)ds}\\
&\quad-\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}
\int_0^{\xi_i}\frac{(\xi_i-s)^{n-1}}{(n-1)!}h(s)ds\\
&\geq\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}\xi_i^{n-1}
\int_{\xi_i}^1\frac{(1-s)^{n-1}}{(n-1)!}h(s)ds\geq \theta.
\end{align*}
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.3}
Assume {\rm (H1)}. If $h\in Q$, then the unique solution $y$ of
\eqref{e2.1}-\eqref{e2.2} satisfies
$$
y(t)\geq \gamma y(s),\quad \forall t\in[\xi_{m-2},1],\; s\in I,
$$
where
$$
\gamma=\min\{\frac{k_{m-2}(1-\xi_{m-2})}{1-k_{m-2}\xi_{m-2}},\,
k_{m-2}\xi_{m-2}^{n-1},\, k_1\xi_1^{n-1},\, \xi_{m-2}^{n-1}\}.
$$
\end{lemma}

\begin{proof}
For any $\varphi\in P^*$, we have
$\varphi(h(t))\geq 0,~t\in I$. It follows from
\begin{align*}
\varphi(y(t))
&=-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}\varphi(h(s))ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}\varphi(h(s))ds}\\
&\quad -\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}
\int_0^{\xi_i}\frac{(\xi_i-s)^{n-1}}{(n-1)!}\varphi(h(s))ds
\end{align*}
and \cite[Lemma 3.2]{g3} that
$$
\varphi(y(t))\geq \gamma\varphi(y(s)),\quad \forall t\in[\xi_{m-2},1],\;
s\in I.
$$
 So, we have
$$
\varphi(y(t)-\gamma y(s))\geq 0,\quad \forall t\in[\xi_{m-2},1],\; s\in I.
$$
Since $\varphi\in P^*$ is arbitrary, we get
$$
y(t)-\gamma y(s)\geq \theta,\quad \forall t\in[\xi_{m-2},1],\;
s\in I.
$$
The proof is complete.
\end{proof}

Define an operator $A:Q\to C[I,E]$ as follows
\begin{equation} \label{e2.4}
\begin{aligned}
A(y(t))&:=-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,y(s))ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}f(s,y(s))ds}\\
&\quad -\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}
 \int_0^{\xi_i}\frac{(\xi_i-s)^{n-1}}{(n-1)!}f(s,y(s))ds.
\end{aligned}
\end{equation}
By Lemmas  2.1 and  2.2, we get that $A:Q\to C^n[I,E]\cap Q$, and
$y(t)$ is a positive solution of  \eqref{e1.1}-\eqref{e1.2} if
and only if $y(t)\in C^n[I,E]\cap Q$ and $y(t)\not\equiv \theta$ is
a fixed point of the operator $A$.


\begin{lemma} \label{lem2.4}
Suppose {\rm (H1)--(H2)} hold. Then, for any
$r>0$, the operator $A$ is a strict set contraction on $Q\cap
B_r$.
\end{lemma}

\begin{proof}
Since $f(t,x)$ is uniformly continuous and bounded on
$I\times (P\cap T_r)$, we see from \eqref{e2.4} that $A$ is continuous and
bounded on $Q\cap B_r$.  For any $S\subset Q\cap B_r$, by \eqref{e2.4}, we
can easily show that the functions $A(S)=\{Ay|y\in S\}$ are uniformly
bounded and equicontinuous. By \cite{l1}, we have
\begin{equation}
\alpha(A(S))=\sup_{t\in I}\alpha(A(S(t))),\label{e2.5}
\end{equation}
where $A(S(t))=\{Ay(t):y\in S,\,t\in I~\text{is fixed}\}$.
 For any
$y\in C[I,E],\, g\in C[I,I]$, by
$\int_0^tg(s)y(s)ds\in \overline{co}(\{g(t)y(t)|t\in
I\}\cup\{\theta\})\subset\overline{co}(\{y(t)|t\in
I\}\cup\{\theta\})$, we get
\begin{align*}
\alpha(A(S(t)))
&=\alpha(\{-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,y(s))ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}f(s,y(s))ds}\\
&\quad -\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}
 \int_0^{\xi_i}\frac{(\xi_i-s)^{n-1}}{(n-1)!} f(s,y(s))ds|y\in S\})\\
&\leq\frac{1}{(n-1)!}\alpha(\overline{co}(\{f(s,y(s))|s\in
I,~y\in S\}\cup\{\theta\}))\\
&\quad +\frac{1}{(1-a_0)(n-1)!}\alpha(\overline{co}(\{f(s,y(s))|s\in
I,~y\in S\}\cup\{\theta\}))\\
&\quad +\frac{a_0}{(1-a_0)(n-1)!}\alpha(\overline{co}(\{f(s,y(s))|s\in
I,~y\in S\}\cup\{\theta\}))\\
&=\frac{2}{(1-a_0)(n-1)!}\alpha(\{f(s,y(s))|s\in
I,~y\in S\})\\
&\leq\frac{2}{(1-a_0)(n-1)!}\alpha(f(I\times B)),
\end{align*}
where $B=\{y(s):s\in I,~y\in S\}\subset P\cap T_r$.
By (H2), we get
\begin{equation}
\alpha(A(S(t)))\leq\frac{2}{(1-a_0)(n-1)!}L_r\alpha(B).\label{e2.6}
\end{equation}
For each $\varepsilon>0$, there exists a partition
$S=\bigcup_{j=1}^{l}S_j$ such that
\begin{equation}
\mathop{\rm diam}(S_j)<\alpha(S)+\frac{\varepsilon}{3},
\quad j=1,2,\dots,l.\label{e2.7}
\end{equation}
Now, choose $y_j\in S_j$, $j=1,2,\dots,l$ and a partition
$0=t_0<t_1<\dots<t_k=1$ such that
\begin{equation}
\|y_j(t)-y_j(\overline{t})\|<\frac{\varepsilon}{3},\quad
\forall t,\; \overline{t}\in[t_{i-1},t_i],\;
j=1,2,\dots,l,\; i=1,2,\dots,k.\label{e2.8}
\end{equation}
Obviously,
$B=\cup_{j=1}^{l}\cup_{i=1}^{k}B_{ij}$, where
$B_{ij}=\{y(t):y\in S_j,t\in[t_{i-1},~t_i]\}$. For any
$y(t),\;\overline{y}(\overline{t})\in B_{ij}$,
by \eqref{e2.7} and \eqref{e2.8},
we obtain
\begin{align*}
\|y(t)-\overline{y}(\overline{t})\|
&\leq\|y(t)-y_j(t)\|+\|y_j(t)-y_j(\overline{t})\|
 +\|y_j(\overline{t})-\overline{y}(\overline{t})\|\\
&\leq\|y-y_j\|_c+\frac{\varepsilon}{3}+\|y_j-\overline{y}\|_c\\
&\leq2\mathop{\rm diam}(S_j)+\frac{\varepsilon}{3}<2\alpha(S)+\varepsilon,
\end{align*}
which implies $\mathop{\rm diam}(B_{ij})\leq 2\alpha(S)+\varepsilon$,
and so, $\alpha(B)\leq 2\alpha(S)+\varepsilon$. Since $\varepsilon$ is
arbitrary, we get
\begin{equation}
\alpha(B)\leq 2\alpha(S).\label{e2.9}
\end{equation}
It follows from \eqref{e2.5}, \eqref{e2.6} and \eqref{e2.9} that
$$
\alpha(A(S))\leq \frac{4}{(n-1)!(1-a_0)}L_r\alpha(S),\quad
\forall S\subset Q\cap B_r.
$$
 By (H2), we get that $A$ is a strict set contraction on $Q\cap B_r$.
\end{proof}

\section{Main results}

 Let $K=\{y\in Q:y(t)\geq \gamma y(s),\;\forall
t\in[\xi_{m-2},1],\; s\in I\}$. Clearly, $K\subset Q$ is a cone of
$C[I,E]$. By Lemma 2.2 and Lemma 2.3, we get $AQ\subset K$. So,
$AK\subset K$.

For convenience, for any $x\in P$ and $\varphi\in P^*$, we set
\begin{gather*}
f^{0}=\limsup_{\|x\|\to 0}\sup_{t\in I}\frac{\|f(t,x)\|}{\|x\|},\quad
f^{\infty}=\limsup_{\|x\|\to\infty}\sup_{t\in I}\frac{\|f(t,x)\|}{\|x\|},\\
f^\varphi_{0}=\liminf_{\|x\|\to 0}\inf_{t\in I}
 \frac{\varphi(f(t,x))}{\varphi(x)},\quad
f^\varphi_{\infty}=\liminf_{\|x\|\to \infty}\inf_{t\in I}
 \frac{\varphi(f(t,x))}{\varphi(x)}\,.
\end{gather*}
Then we list the following assumptions:
\begin{itemize}
\item[(H3)] There exists $\varphi\in P^*$ with $\varphi(x)>0$, for all
$x>\theta$ such that $\frac{n!(1-a_0)}{\gamma a_1}<f^\varphi_{0}\leq\infty$.

\item[(H4)] There exists $\varphi\in P^*$ with $\varphi(x)>0$, for all
$x>\theta$ such that
$\frac{n!(1-a_0)}{\gamma a_1}<f^\varphi_{\infty}\leq\infty$.

\item[(H5)] $0\leq f^{0}<\frac{n!(1-a_0)}{N}$.

\item[(H6)] $ 0\leq f^{\infty}<\frac{n!(1-a_0)}{N}$.

\item[(H7)]  There exists $r_0>0$ such that
$\sup_{t\in I,\, x\in P\cap T_{r_0}} \|f(t,x)\|<\frac{n!(1-a_0)}{N}r_0$.

\item[(H8)] There exist $R_0>0$ and $\varphi\in P^*$ with $\varphi(x)>0$
for any $x>\theta$ such that
$$
\inf_{t\in[\xi_{m-2},1],\, x\in P,\,\gamma R_0/N\leq\|x\|\leq R_0}
\frac{\varphi(f(t,x))}{\varphi(x)}> \frac{n!(1-a_0)}{\gamma a_1}.
$$
\end{itemize}

\begin{theorem} \label{thm3.1}
Suppose {\rm (H1)--(H2)} hold. In addition suppose {\rm (H4)} and
{\rm (H5)} or {\rm (H3)} and {\rm (H6)} are satisfied.
Then  \eqref{e1.1}-\eqref{e1.2} has at least one positive solution.
\end{theorem}

\begin{proof} (i) Suppose (H4) and (H5) hold.
By (H4),  there exist constants
\begin{equation}
M>\frac{n!(1-a_0)}{\gamma a_1}\label{e3.1}
\end{equation}
and $r_1>0$ such that
\begin{equation}
\varphi(f(t,x))\geq M\varphi(x),\quad \forall t\in I, x\in P,~\|x\|>r_1.
\label{e3.2}
\end{equation}
For  $R>\frac{N}{\gamma}r_1$, we will show that
\begin{equation}
Ay\not\leq y,~\forall y\in K,~\|y\|_c=R.\label{e3.3}
\end{equation}
In fact, if not, there exists $y_0\in K$, $\|y_0\|_c=R$ such that
$Ay_0\leq y_0$. By
\begin{equation}
y_0(t)\geq \gamma y_0(s)\geq \theta,\quad
\forall t\in [\xi_{m-2},1],\; s\in I,\label{e3.4}
\end{equation}
we have
\begin{equation}
\|y_0(t)\|\geq \frac{\gamma}{N}\|y_0\|_c>r_1,\quad
\forall t\in [\xi_{m-2},1].\label{e3.5}
\end{equation}
By \eqref{e2.4}, for any $t\in I$, we have
\begin{align*}
A(y_0(t))
&=-{\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds
+\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds}\\
&\quad -\frac{t^{n-1}}{1-a_0}\sum_{i=1}^{m-2}k_{i}
\int_0^{\xi_i}\frac{(\xi_i-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds.\\
&\geq\frac{a_0t^{n-1}}{1-a_0}
\int_{\xi_{m-2}}^{1}\frac{(1-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds.
\end{align*}
This, together with \eqref{e3.2}, \eqref{e3.4} and \eqref{e3.5}, implies
\begin{align*}
\varphi(Ay_0(1))
&\geq\frac{a_0}{1-a_0} \int_{\xi_{m-2}}^{1}\frac{(1-s)^{n-1}}{(n-1)!}M\gamma
\varphi(y_0(1))ds\\
&=\frac{a_1}{n!(1-a_0)}M\gamma \varphi(y_0(1)).
\end{align*}
Considering $Ay_0\leq y_0$, we get
\begin{equation}
\varphi(y_0(1))\geq\frac{\gamma a_1}{n!(1-a_0)}M\varphi(y_0(1)).\label{e3.6}
\end{equation}
It is easy to see that $\varphi(y_0(1))>0$ (In fact, if
$\varphi(y_0(1))=0$, by \eqref{e3.4}, we get
$\varphi(y_0(1))\geq \gamma\varphi(y_0(s))\geq 0$, for all $s\in I$.
So, we have $\varphi(y_0(s))\equiv 0$, for all $s\in I$, that is
$y_0(s)\equiv\theta$. This is a contradiction with $\|y_0\|_c=R$).
So, \eqref{e3.6} contradicts with \eqref{e3.1}. Therefore,
\eqref{e3.3} is true.

On the other hand, by (H5) and $f(t,\theta)=\theta$, we get that
there exist constants $\varepsilon\in(0,n!(1-a_0)/N)$ and $0<r_2<R$
such that
\begin{equation}
\|f(t,x)\|\leq\varepsilon\|x\|,\quad \forall t\in I,\;
x\in P,\; \|x\|<r_2.\label{e3.7}
\end{equation}
For any $0<r<r_2$, we now prove
\begin{equation}
Ay\not\geq y,~\forall y\in K,~\|y\|_c=r.\label{e3.8}
\end{equation}
In fact, if not, there exists $y_0\in K,~\|y_0\|_c=r$ such that
$Ay_0\geq y_0$. Since \eqref{e2.4} implies
\begin{equation}
Ay_0(t)\leq
\frac{t^{n-1}}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds,
\quad \forall t\in I,\label{e3.9}
\end{equation}
we have
$$
\theta\leq y_0(t)\leq\frac{t^{n-1}}{1-a_0}
\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}f(s,y_0(s))ds,\quad
\forall t\in I.
$$
This, together with \eqref{e3.7}, implies
$$
\|y_0(t)\|\leq\frac{N\varepsilon}{1-a_0}
\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}\|y_0(s)\|ds
\leq\frac{N\varepsilon\|y_0\|_c}{n!(1-a_0)},
\quad \forall t\in I.
$$
Therefore, we get $\varepsilon\geq n!(1-a_0)/N$. This is a
contradiction. So, \eqref{e3.8} is true.

By \eqref{e3.3}, \eqref{e3.8}, Lemma 2.4 and Theorem 1.1, we get
that the operator $A$ has at least one fixed point $y\in K$ satisfying
$r<\|y\|_c<R$.
\smallskip

\noindent(ii) Suppose (H3) and (H6) hold.
By (H3), in the same way as establishing \eqref{e3.3} we can assert
 that there exists $r_2>0$ such that for any $0<r<r_2$,
\begin{equation}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=r.\label{e3.10}
\end{equation}
On the other hand, by (H6), we get that there exist constants
$0<\varepsilon<n!(1-a_0)/N$ and $r_1>0$ such that
$$
\|f(t,x)\|\leq\varepsilon\|x\|,\quad \forall t\in I,\;
x\in P,\; \|x\|>r_1.
$$
By (H2), we get
$$
\sup_{t\in I,\, x\in P\cap T_{r_1}}\|f(t,x)\|:=b<\infty.
$$
So, we have
\begin{equation}
\|f(t,x)\|\leq\varepsilon\|x\|+b,\quad \forall t\in I,\;
x\in P.\label{e3.11}
\end{equation}
Take $R>\max\{r_2,\,\frac{Nb}{n!(1-a_0)-N\varepsilon}\}$. We will
prove that
\begin{equation}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=R.\label{e3.12}
\end{equation}
In fact, if there exists $y_0\in K$, $\|y_0\|_c=R$ such that
$Ay_0\geq y_0$. Then, by \eqref{e3.9} and \eqref{e3.11}, we get
$$
\|y_0(t)\|\leq\frac{Nt^{n-1}}{1-a_0}
\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}(\varepsilon\|y_0(s)\|+b)ds
\leq\frac{N}{n!(1-a_0)}(\varepsilon\|y_0\|_c+b),~~\forall t\in I.
$$
So, we have
$$
\|y_0\|_c\leq\frac{Nb}{n!(1-a_0)-N\varepsilon}<R.
$$
A contradiction. Therefore, \eqref{e3.12} holds.

By \eqref{e3.10}, \eqref{e3.12}, Lemma 2.4 and Theorem 1.1,
 the operator $A$ has at least one fixed point $y\in K$ satisfying
$r<\|y\|_c<R$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.2.} Suppose {\rm (H1)} and {\rm (H2)} hold.
In addition suppose that one of the following conditions is
satisfied
\begin{itemize}
\item[(i)] {\rm (H3), (H4), (H7)} hold.

\item[(ii)] {\rm (H5), (H6), (H8)} hold.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has at least two positive solutions.
\end{theorem}

\begin{proof} (i) By (H3), (H4) and the proof of
Theorem 3.1,  there exist $r, R$ with $0<r<r_0<R$ such
that
\begin{equation}
\begin{gathered}
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=r.\\
Ay\not\leq y,\quad \forall y\in K,\; \|y\|_c=R.
\end{gathered}\label{e3.13}
\end{equation}
Now, we will prove
\begin{equation}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=r_0.\label{e3.14}
\end{equation}
In fact, if there exists $y_0\in K$, $\|y_0\|_c=r_0$ such that
$Ay_0\geq y_0$. By \eqref{e3.9} and (H7), we get
$$
\|y_0\|_c<\frac{N}{1-a_0}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}\cdot
\frac{n!(1-a_0)}{N}r_0ds=r_0.
$$
A contradiction. So, \eqref{e3.14} is true. By \eqref{e3.12}-\eqref{e3.14},
 Lemma 2.4 and Theorem 1.1, we get that the operator $A$ has at least two
fixed points $y_1$, $y_2\in K$ satisfying
$$
r<\|y_1\|_c<r_0<\|y_2\|_c<R.
$$

\noindent (ii) By (H5), (H6) and the proof of Theorem 3.1,
 there exist $r, R$ with $0<r<R_0<R$ such that
\begin{gather}
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=r.\label{e3.15}\\
Ay\not\geq y,\quad \forall y\in K,\; \|y\|_c=R.\label{e3.16}
\end{gather}
On the other hand, by (H8) and the methods used in the proof
of \eqref{e3.3}, we can prove that
\begin{equation}
Ay\not\leq y,\quad \forall y\in K,\quad \|y\|_c=R_0.\label{e3.17}
\end{equation}
By \eqref{e3.15}-\eqref{e3.17}, Lemma 2.4 and Theorem 1.1, the
operator $A$ has at least two fixed points $y_1, y_2\in K$
satisfying
$$
r<\|y_1\|_c<R_0<\|y_2\|_c<R.$$
The proof is complete.
\end{proof}

Similar to the proofs of Theorem 3.1 and Theorem 3.2, we can easily
get the following corollaries.

\begin{corollary} \label{coro3.1}
 Suppose {\rm (H1), (H2)} hold. In addition suppose that one of the
following conditions is satisfied
\begin{itemize}
\item[(i)] {\rm (H4), (H5), (H7), (H8)} hold with $R_0<r_0$.

\item[(ii)] {\rm (H3), (H6), (H7),  (H8)} hold with $r_0<R_0$.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has at least three positive solutions.
\end{corollary}

\begin{corollary} \label{coro3.2}
 Suppose {\rm (H1), (H2)} hold. In addition suppose that one of the
following conditions is satisfied
\begin{itemize}
\item[(i)] {\rm (H5)--(H7)} hold, and there exist
$R_i>0$, $\varphi_i\in P^*$ with $\varphi_i(x)>0$ for $x>\theta$,
$i=1,2$ such that
$$
\inf_{t\in[\xi_{m-2},1],\, x\in P,\, \gamma R_i/N\leq\|x\|\leq R_i}
\frac{\varphi_i(f(t,x))}{\varphi_i(x)}>
\frac{n!(1-a_0)}{\gamma a_1},\quad i=1,2,
$$
where $R_1<r_0<R_2$.

\item[(ii)] {\rm (H3), (H4), (H8)} hold, and there exist $r_1, r_2>0$
such that
$$
\sup_{t\in I,\, x\in P\cap T_{r_i}}\|f(t,~x)\|<\frac{n!(1-a_0)}{N}r_i,
\quad i=1,2,
$$
where $r_1<R_0<r_2$.
\end{itemize}
Then  \eqref{e1.1}-\eqref{e1.2} has at least four positive solutions.
\end{corollary}

We can easily obtain  the existence of multiple positive
solutions for  \eqref{e1.1}-\eqref{e1.2}.

\section{Examples}

In this section, we give some examples to illustrate our results.

\begin{example} \label{exam4.1}
Consider the boundary value problem
\begin{gather}
y_i^{'''}(t)+f_i(t,y_1,y_2,\ldots,y_l)=0,\quad 0<t<1, \label{e4.1}\\
y_i(0)=y_i'(0)=0,\quad y_i(1)=y_i(\frac{1}{2}),\quad i=1,2,\ldots,l,
\label{e4.2}
\end{gather}
where
$f_i(t,y_1,y_2,\ldots,y_l)=y_{i+1}^{\frac{2}{3}}+e^{-t}y_{i+2}^{\frac{3}{2}},$
$i=1,2,\ldots,l-2,$
$f_{l-1}(t,y_1,y_2,\ldots,y_l)=y_{l}^{\frac{2}{3}}+e^{-t}y_{1}^{\frac{3}{2}},$
$f_{l}(t,y_1,y_2,\ldots,y_l)=y_{1}^{\frac{2}{3}}+e^{-t}y_{2}^{\frac{3}{2}}.$
\end{example}
{\bf Conclusion.} The problem (4.1)-(4.2) has at least two positive
solutions.

\begin{proof} Let $E=R^l=\{y=(y_1,y_2,\ldots,y_l)|y_i\in
R,~i=1,2,\ldots,l\}$ with the norm $\|y\|=\max\limits_{1\leq i\leq
l}|y_i|$, and $P=\{y=(y_1,y_2,\ldots,y_l)|y_i\geq
0,~i=1,2,\ldots,l\}$. Then $P$ is a normal cone in $E$ and the
normal constant $N=1$. Corresponding to (1.1)-(1.2), we get
$n=m=3,~k_1=1,~\xi_1=\frac{1}{2},~a_0=\frac{1}{4},~a_1=\frac{1}{32},~\gamma=\frac{1}{4}.$
Obviously, {\rm (H1)} is satisfied. Set $\theta=(0,0,\ldots,0)$ and
$f=(f_1,f_2,\ldots,f_l).$ Then $f:I\times P\rightarrow P$ is
continuous and $f(t,\theta)=\theta,$ for all $t\in I$. It is clear
that $\alpha(f(D))=0$ for any $D\subset P\cap T_r$. So, {\rm (H2)}
holds. It is easy to see that $P^*=P$. So, we choose
$\varphi=(1,1,\ldots,1),$ and then
$$
\frac{\varphi(f(t,y))}{\varphi(y)}=\frac{\sum\limits_{i=1}^lf_i(t,y_1,y_2,\ldots,y_l)}
{\sum\limits_{i=1}^ly_i}.
$$

We now prove that the conditions {\rm (H3)} and {\rm (H4)} are
satisfied.

For any $y\in P,$ $y\neq \theta$, we can easily get $\varphi(y)>0$
and
$$
\frac{\varphi(f(t,y))}{\varphi(y)}=\frac{\sum\limits_{i=1}^ly_i^{\frac{2}{3}}
+e^{-t}\sum\limits_{i=1}^ly_i^{\frac{3}{2}}}{\sum\limits_{i=1}^ly_i}
\geq\frac{\max\limits_{1\leq i\leq
l}y_i^{\frac{2}{3}}}{n\max\limits_{1\leq i\leq l}y_i}
=\frac{1}{n}\frac{1}{\max\limits_{1\leq i\leq
l}y_i^{\frac{1}{3}}}\rightarrow \infty,~~(\|y\|\rightarrow 0).
$$
and
$$
\frac{\varphi(f(t,y))}{\varphi(y)}=\frac{\sum\limits_{i=1}^ly_i^{\frac{2}{3}}
+e^{-t}\sum\limits_{i=1}^ly_i^{\frac{3}{2}}}{\sum\limits_{i=1}^ly_i}
\geq\frac{e^{-1}\max\limits_{1\leq i\leq
l}y_i^{\frac{3}{2}}}{n\max\limits_{1\leq i\leq l}y_i}
=\frac{1}{ne}\max\limits_{1\leq i\leq l}y_i^{\frac{1}{2}}\rightarrow
\infty,~~(\|y\|\rightarrow \infty).
$$

So, {\rm (H3)} and {\rm (H4)} hold. Finally, we will show {\rm (H7)}
is satisfied.

Since $\frac{n!(1-a_0)}{N}r_0=4.5r_0$, taking $r_0=1$, we get
$$
\sup_{t\in I,\, y\in P\cap T_{r_0}} \|f(t,y)\|\leq
\max\limits_{1\leq i\leq l}y_i^{\frac{2}{3}}+\max\limits_{1\leq
i\leq l}y_i^{\frac{3}{2}}\leq 2.
$$
Therefore, {\rm (H7)} holds. By Theorem 3.2 (i), we get that the
problem (4.1)-(4.2) has at least two positive solutions.
\end{proof}

\begin{example} \label{exam4.2}
The boundary value problem
\begin{gather}
y_i^{'''}(t)+e^{at}\sin^2(\frac{\pi}{2}y_i)=0,\quad 0<t<1, \label{e4.3}\\
y_i(0)=y_i'(0)=0,\quad y_i(1)=y_i(\frac{1}{2}),\quad i=1,2,\ldots,l.
\label{e4.4}
\end{gather}
has at least two positive solutions, where
$a>2\ln\frac{576l}{\sin^2\frac{\pi}{8}}$.
\end{example}

\begin{proof} Let $E,~\|\cdot\|,~P,~\theta,~\varphi$ be the same as
in Example 4.1. Take $f=(e^{at}\sin^2(\frac{\pi}{2}y_1),$
$e^{at}\sin^2(\frac{\pi}{2}y_2),\ldots,e^{at}\sin^2(\frac{\pi}{2}y_l)).$
Then $f:I\times P\rightarrow P$ is continuous and
$f(t,\theta)=\theta,$ for all $t\in I$. Similar to the proof of
Example 4.1, we get that {\rm (H1)} and {\rm (H2)} are satisfied.
Now, we prove that {\rm (H5)} and {\rm (H6)} are satisfied. Because
$$
\frac{\|f(t,y)\|}{\|y\|}=\frac{\max\limits_{1\leq i\leq
l}e^{at}\sin^2(\frac{\pi}{2}y_i)}{\max\limits_{1\leq i\leq l}y_i}
\leq e^{a}\frac{\max\limits_{1\leq i\leq
l}\sin^2(\frac{\pi}{2}y_i)}{\max\limits_{1\leq i\leq
l}y_i}\rightarrow 0,\quad (\|y\|\rightarrow 0),
$$
and
$$
\frac{\|f(t,y)\|}{\|y\|}=\frac{\max\limits_{1\leq i\leq
l}e^{at}\sin^2(\frac{\pi}{2}y_i)}{\max\limits_{1\leq i\leq l}y_i}
\leq e^{a}\frac{\max\limits_{1\leq i\leq
l}\sin^2(\frac{\pi}{2}y_i)}{\max\limits_{1\leq i\leq
l}y_i}\rightarrow 0,\quad (\|y\|\rightarrow \infty),
$$
 {\rm (H5)} and {\rm (H6)} hold. Now, we prove that {\rm (H8)} is
satisfied. Since $\frac{n!(1-a_0)}{\gamma a_1}=576$ (where,
$n,~a_0,~a_1$ and $\gamma$ are the same as in Example 4.1), taking
$R_0=1$, for $t\in [\frac{1}{2},1],$ $y\in P$, $\frac{1}{4}
\leq\|y\|\leq 1$, we have
$$
\frac{\varphi(f(t,y))}{\varphi(y)}=\frac{\sum\limits_{i=1}^{l}e^{at}\sin^2(\frac{\pi}{2}y_i)}
{\sum\limits_{i=1}^{l}y_i} \geq
e^{\frac{a}{2}}\frac{\sin^2(\frac{\pi}{8})}{l}>576.
$$
So, {\rm (H8)} holds. By Theorem 3.2 (ii), we get that the problem
(4.3)-(4.4) has at least two positive solutions.
\end{proof}
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