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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 163, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/163\hfil Existence of positive solutions]
{Existence of positive solutions for boundary-value problems
with integral boundary conditions and sign changing
nonlinearities}

\author[S. Liu, M. Jia,  Y. Tian \hfil EJDE-2010/163\hfilneg]
{Shu Liu, Mei Jia, Yingqiang Tian}

\address{College of Science,
Shanghai University for Science and Technology, Shanghai 200093,
China}
 \email[Shu Liu]{liu2008shu@126.com}
 \email[Mei Jia]{jiamei-usst@163.com }
 \email[Yingqiang Tian]{tian0013@163.com}


\thanks{Submitted June 15, 2010. Published November 12, 2010.}
\thanks{Supported by grant 10ZZ93 from the Innovation Program of
Shanghai Municipal \hfill\break\indent Education Commission}
\subjclass[2000]{34B15, 34B18}
\keywords{Boundary value problems;  $p$-Laplacian;
 integral boundary conditions; \hfill\break\indent
fixed point theorem in double cones;  positive solutions}

\begin{abstract}
 In this article, we show the existence of positive
 solutions  for boundary value problems with integral boundary
 conditions  and sign changing nonlinearities. By using a fixed point
 theorem in double cones, we obtain sufficient conditions for the
 existence of two positive solutions.
\end{abstract}

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

\section{Introduction}


The theory of boundary value problems with integral boundary
conditions for ordinary differential equations arises in different
areas of applied mathematics and physics. For example, heat
conduction, chemical engineering, underground water flow,
thermo-elasticity, and plasma physics can all be reduced to nonlocal
problems with integral boundary conditions. For boundary value
problems with integral boundary conditions and comments on their
importance, we refer the reader to \cite{g1,k1,l1}
and the references therein.
For more information about the general theory of integral equations
and their relation with boundary value problems we refer to
\cite{a1,c2}.

Moreover, boundary value problems with integral boundary conditions
constitute a very interesting and important class of problems. They
include two, three, multi-point and nonlocal boundary value problems
as special cases. The existence and multiplicity of positive
solutions for such problems have received a great deal of attention.
To identify a few, we refer the reader to \cite{b1,g3,m1,w1,w2,x1}
and references therein. But the corresponding theory for
boundary-value problem
with integral boundary conditions and sign changing nonlinearities
of one-dimensional $p$-Laplacian is not investigated till now.
By using the fixed point theorem in double cones, Guo in \cite{g2}
discussed the existence of positive solutions for second-order
three-point boundary-value problem
\begin{gather*}
 x''+f(t,x)=0, \quad 0\leq t\leq 1,\\
 x(0)-\beta x'(0)=0,\quad x(1)=\alpha x(\eta),
\end{gather*}
where $f$ is allowed to change sign. Sufficient conditions are
obtained by imposing growth conditions on $f$ which ensure the
existence of at least two positive solutions for the above boundary
value problems. Meanwhile, they proved a fixed  point theorem in
double cones which generalize the fixed point theorem in a cones to
some degree. 
By using a theorem similar to the one in \cite{g2}, Cheung
\cite{c1} proved the existence of two positive solutions for the
problem
$$
      (\Phi_p(u'))'+h(t)f(t,u)=0, \quad 0<t<1,
$$
with each of the following two sets of boundary conditions
\begin{gather*}
      u'(0)=0, \quad u(1)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i);\\
u(0)=\sum_{i=1}^{m-2}\alpha_iu(\xi_i), \quad u'(1)=0,
\end{gather*}
where $h:[0,1]\to \mathbb{R}^+$ and
$f:[0,1]\times [0,\infty)\to \mathbb{R}$ are continuous functions.


Recently, by using the classical fixed-point index theorem for
compact maps, Feng \cite{f1} established  sufficient conditions for the
existence of multiple positive solutions for the second-order
impulsive differential equations, with $p$-Laplacian and integral
boundary conditions,
\begin{gather*}
      -(\phi_p(u'(t)))'=f(t,u(t)), \quad  t\neq t_k, \; t\in (0,1)\\
      -\Delta u|_{t=t_k}=I_k(u(t_k)), \quad  t=1,2,\dots ,n,\\
      u'(0)=0, \quad u(1)=\int_0^1 g(t)u(t) dt\,.
\end{gather*}

Motivated by the above, we consider the existence of positive
solutions for the boundary-value problem, with integral boundary
conditions and sign changing nonlinearities of one-dimensional
$p$-Laplacian,
\begin{equation} \label{e1.1}
   \begin{gathered}
      (\varphi_p(u'))'+f(t,u)=0, \quad 0\leq t\leq 1, \\
      au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i), \\
      u(1)=\int_0^1 g(s)u(s) ds,
   \end{gathered}
\end{equation}
where $a, b\in[0,+\infty)$,
$a_i\in(0,+\infty)$, $i=1,2,\dots ,m-2$,
$0<\xi_1<\dots <\xi_{m-2}<1$, $m\geq 3$,
$\varphi_p(u)=|u|^{p-2}u,p>1$,
$\varphi_q=(\varphi_p)^{-1}$,
and $\frac{1}{q}+\frac{1}{p}=1$.

We will assume the following conditions:
\begin{itemize}

\item[(H1)] $f:[0,1] \times [0,+\infty)\to \mathbb{R}$ is
continuous;

\item[(H2)] $f(t,0)\geq 0$ $(f\not\equiv 0)$ for $t\in [0,1]$;

\item[(H3)] $a>\sum_{i=1}^{m-2}a_i$; $g$ is non-negative,
integrable,  and $\sigma:=\int_0^1 g(t) dt \in [0,1)$.
\end{itemize}

\section{Preliminaries}

For a cone $K$ in a Banach space $X$ with norm $\| \cdot \|$
and a constant $r>0$, let $K_r=\{ x\in K:\| x \| <r \}$,
$\partial K_r=\{x\in K:\| x \| =r\}$. Suppose
$\alpha :K\to[0,+\infty)$ is a continuously increasing functional,
i.e., $ \alpha$ is continuous and $\alpha(\lambda x)\leq \lambda
\alpha (x)$ for $\lambda \in (0,1)$. Denote
$K(b)=\{x\in K:\alpha (x)<b\}$,
$\partial K(b)=\{x\in K:\alpha (x)=b\}$
 and $K_a(b)=\{x\in K:a<\| x \| , \alpha (x)<b\}$.


\begin{lemma}[\cite{g2}] \label{lem2.1}
Let $X$ be a real Banach space with
norm $\| \cdot \|$  and $K, K' \subset X$ two cones with
$K'\subset K$. Suppose $T: K\to K$ and $T^*: K'\to K'$ are two
completely continuous operators and $\alpha : K'\to
\mathbb{R}^+$ a continuously increasing functional satisfying
$\alpha(x)\leq \| x \| \leq M\alpha(x)$ for all $x$ in $K'$, where
$M\geq 1$ is a constant.
If there are constants $b>a>0$ such that
\begin{itemize}
\item[(C1)] $\| Tx\| <a$, for $x\in \partial K_a$;
\item[(C2)] $\| T^*x\| <a$, for $x\in \partial K'_a$ and
$\alpha(T^*x)>b$ for $x\in \partial K'(b)$;
\item[(C3)] $Tx=T^*x$, for $x\in \partial K'_a(b)\cap\{u:T^*u=u\}$,
\end{itemize}
then $T$ has at least two fixed points $y_1$ and $y_2$ in $K$ such that
$$
0\leq \| y_1\| <a< \| y_2\|, \quad \alpha(y_2)<b.
$$
\end{lemma}

We also denote $C^+[0,1]=\{u\in C[0,1]:u(t)\geq 0, t\in [0,1]\}$.


\begin{lemma} \label{lem2.2}
For any $h\in C^+[0,1]$,  the boundary-value problem
\begin{equation} \label{e2.1}
   \begin{gathered}
      (\varphi_p(u'))'+h(t)=0, \quad 0\leq t\leq 1, \\
      au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i), \\
      u(1)=\int_0^1 g(s)u(s) ds,
   \end{gathered}
\end{equation}
has a unique solution of the form
\begin{equation} \label{e2.2}
u(t)=\frac{b\varphi_q(A_h)+\sum_{i=1}^{m-2}a_{i}
\int_0^{\xi_i}\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}
+\int_0^t\varphi_q(A_h -\int_0^sh(\tau)d\tau)ds,
\end{equation}
or
\begin{equation} \label{e2.3}
u(t)=\frac{\int_0^1g(s)(\int_s^1\varphi_q
(\int_0^rh(\tau)d\tau-A_h)dr)ds}
{1-\int_0^1g(s)ds}
+\int_t^1\varphi_q(\int_0^sh(\tau)d\tau-A_h)ds,
\end{equation}
where $A_h$ satisfies
\begin{equation} \label{e2.4}
\begin{aligned}
 b\varphi_q(A_h)
  & =\frac{(a-\sum_{i=1}^{m-2}a_{i})\int_0^1g(s)
  (\int_s^1\varphi_q(\int_0^rh(\tau)d\tau-A_h)dr)ds}{1-\int_0^1g(s)ds}\\
 &\quad  +a\int_0^1\varphi_q(\int_0^sh(\tau)d\tau-A_h)ds
   -\sum_{i=1}^{m-2}a_{i}\int_{\xi_i}^1\varphi_q(\int_0^sh(\tau)d\tau-A_h)ds.
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
 If $u$ is a solution of \eqref{e2.1},  it is
easy to see that
$$
\varphi_p(u'(t))-\varphi_p(u'(0))=-\int_0^th(s)ds.
$$
Let $A_h=\varphi_p(u'(0))$. Integrating, we can have
\begin{equation}
u'(t)=\varphi_q(A_h-\int_0^th(s)ds).
\label{e2.5}
\end{equation}
Integrating  from $t$ to $1$, we have
$$
u(t)=u(1)-\int_t^1\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds.
$$
Substituting $u(1)=\int_0^1 g(s)u(s)ds$, we obtain
$$
u(t)=\int_0^1g(s)u(s)ds-\int_t^1
\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds.
$$
Using the above equation and $u(1)=\int_0^1 g(s)u(s) ds$, we have
$$
\int_0^1g(s)u(s)ds\int_0^1g(s)ds
-\int_0^1g(s)(\int_s^1\varphi_q(A_h-\int_0^rh(\tau)d\tau)dr)ds
=\int_0^1g(s)u(s)ds,
$$
i.e.,
$$
\int_0^1g(s)u(s)ds=\frac{\int_0^1g(s)
(\int_s^1\varphi_q(\int_0^rh(\tau)d\tau-A_h)dr)ds}
{1-\int_0^1g(s)ds}.
$$
So
\begin{equation}
u(t)=\frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_0^rh(\tau)d\tau-A_h)dr)ds}
{1-\int_0^1g(s)ds}+\int_t^1\varphi_q(\int_0^sh(\tau)d\tau-A_h)ds.
\label{e2.6}
\end{equation}
Integrating \eqref{e2.5} from $0$ to $t$, we obtain
$$
u(t)=u(0)+\int_0^t\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds.
$$
Using the boundary condition
$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$, we have
$$
u(t)=\frac{b\varphi_q(A_h)+\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}+\int_0^t\varphi_q(A_h-\int_0^sh(\tau)d\tau)ds,
$$
where $A_h$ can be easily obtained from the boundary condition
$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$ and \eqref{e2.6}.
\end{proof}



\begin{lemma} \label{lem2.3}
 Assume that {\rm (H3)} holds. Then for
each $h\in C^+[0,1]$, there exists a unique
$A_h\in (0,\int_0^1h(\tau)d\tau)$ satisfying \eqref{e2.2} and
\eqref{e2.3}. Moreover,
there is a unique $\sigma_h\in(0,1)$ such that
$A_h=\int_0^{\sigma_h}h(\tau)d\tau$.
\end{lemma}

\begin{proof}  For  $h\in C^+[0,1]$, define
\begin{align*}
    H_h(\eta)&=
b\varphi_q(\eta)-\frac{(a-\sum_{i=1}^{m-2}a_{i})\int_0^1g(s)(\int_s^1\varphi_q(\int_0^rh(\tau)d\tau-\eta)dr)ds}
{1-\int_0^1g(s)ds}\\
&\quad-a\int_0^1\varphi_q(\int_0^sh(\tau)d\tau-\eta)ds
+\sum_{i=1}^{m-2}a_{i}\int_{\xi_i}^1\varphi_q(\int_0^sh(\tau)d\tau-\eta)ds,
\end{align*}
where $\eta\in(-\infty,+\infty)$. Hence
\begin{align*}
    H_h(\eta)
& = (\sum_{i=1}^{m-2}a_{i}-a)\int_0^1\varphi_q
 (\int_0^sh(\tau)d\tau-\eta)ds
  -\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q
  (\int_0^sh(\tau)d\tau-\eta)ds\\
  &\quad +b\varphi_q(\eta)-\frac{(a-\sum_{i=1}^{m-2}a_{i})
  \int_0^1g(s)(\int_s^1\varphi_q(\int_0^rh(\tau)d\tau-\eta)dr)ds}
{1-\int_0^1g(s)ds}.
\end{align*}

By the properties of the function $\varphi_q(\cdot)$, we know that
$H_h(\eta)$ is strictly increasing function on $(-\infty,+\infty)$.
We also easily know that $H_h(0)<0$, $H_h(\int_0^1h(\tau)d\tau)>0$,
so we can prove the lemma by the continuity of $H_h$. Namely, there
is a unique $A_h\in (0,\int_0^1h(\tau)d\tau)$ satisfying \eqref{e2.2}
and \eqref{e2.3}. Meanwhile, we could have that there is unique
$\sigma_h\in(0,1)$ such that $A_h=\int_0^{\sigma_h}h(\tau)d\tau$ by
the continuity of $\int_0^th(\tau)d\tau$ on $[0,1]$.
\end{proof}

\begin{lemma} \label{lem2.4}
 Assume that {\rm (H3)} holds. If $h\in C^+[0,1]$, then the solution
of boundary value problem \eqref{e2.1} has
the following properties:
\begin{itemize}
\item[(i)]  $u(t)$ is a concave function;
\item[(ii)] $u(t)\geq 0$, $t\in [0,1]$;
\item[(iii)] $u(\sigma_h)=\max_{0\leq t\leq 1}u(t)$, where
$\sigma_h$ is defined in lemma \ref{lem2.3}.
\end{itemize}
\end{lemma}

\begin{proof}
Suppose that $u(t)$ is the solution of \eqref{e2.1}.

(i) From the fact that $(\varphi_p(u'))'(t)=-h(t)\leq 0$  we know
that $\varphi_p(u')$ is non-increasing. It follows that $u'(t)$ is
also non-increasing. Thus, $u(t)$ is concave down on $[0, 1]$.

(ii) From (i), we know that the minimum of $u(t)$ is obtained at 0
or 1. So we only have to prove that $u(0)\geq 0$ and also
$u(1)\geq 0$.

Firstly, we prove that $u(0)\geq 0$. If $u(0)<0$ and $u'(0)\geq 0$,
by the condition $au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$,
we have
$$
au(0)-\sum_{i=1}^{m-2}a_iu(\xi_i)=bu'(0)\geq 0,
$$
so
\begin{equation}
au(0)\geq \sum_{i=1}^{m-2}a_iu(\xi_i).
\label{e2.7}
\end{equation}
 From $a>\sum_{i=1}^{m-2}a_i>0$, we have
\begin{equation}
au(0)< \sum_{i=1}^{m-2}a_iu(0).
\label{e2.8}
\end{equation}
 From \eqref{e2.7} and \eqref{e2.8}, we obtain
$$
\sum_{i=1}^{m-2}a_iu(0)> \sum_{i=1}^{m-2}a_iu(\xi_i).
$$
Hence, there is a $j$, $1\leq j\leq {m-2}$ such that
$0>u(0)>u(\xi_j)$.
 Then we  obtain that $u(1)=\min_{t\in[0,1]}u(t)<0$
and that there is $\eta_1\leq \eta_2\in [0,1]$ such that
$u(t)\leq 0$ when $t\in [0,\eta_1]\cup[\eta_2,1]$ and
$u(t)\geq 0$ when $t\in [\eta_1,\eta_2]$.

 From the boundary condition $u(1)=\int_0^1 g(s)u(s) ds$, we have
\begin{equation} \label{e2.9}
\begin{aligned}
|u(1)|
& =-u(1)=-\int_0^{\eta_1}g(s)u(s)ds
 -\int_{\eta_1}^{\eta_2}g(s)u(s) ds-\int_{\eta_2}^1g(s)u(s) ds\\
& \leq \int_0^{\eta_1}g(s)|u(s)| ds
 +\int_{\eta_1}^{\eta_2}g(s)|u(s)|ds
 +\int_{\eta_2}^1g(s)|u(s)|ds \\
&=|u(1)|\int_0^1g(s) ds.
\end{aligned}
\end{equation}
We  have$\int_0^1g(s) ds \geq 1$ which is contradiction to
with (H3), so $u(0)\geq 0$.

If $u(0)<0$ and $u'(0)<0$, we can prove that $u(0)\geq 0$ by using
the same way as in \eqref{e2.9}. Similarly, we can prove
$u(1)\geq 0$.
Therefore, we obtain that  $u(t)\geq 0$ for $t\in [0,1]$.

(iii) From the boundary conditions, we can easily show that the
maximum of $u(t)$ can not be at 0 or 1. If $u(0)$ is the maximum, then
$u'(0)\leq 0$ which is contradicted with boundary condition
$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$. If $u(1)$ is the
maximum, it is contradicts the boundary condition
$u(1)=\int_0^1 g(s)u(s) ds$.
\end{proof}

 From  Lemmas \ref{lem2.2} and  \ref{lem2.3}, we  have
$u'(t)=\varphi_q(\int_t^{\sigma_h}h(\tau)d\tau)$. By the concavity
of $u(t)$, we know $u(\sigma_h)=\max_{0\leq t\leq 1}u(t)$.

\begin{lemma} \label{lem2.5}
 If $u\in C[0,1]$ and $u(t)\geq 0$ is
a concave function, then for any $\delta\in(0,\frac{1}{2})$, we have
$$
\min_{\delta\leq t\leq {1-\delta}}u(t)\geq \delta \| u\|.
$$
\end{lemma}

\begin{proof}
Let $u(\sigma)=\|u\|$. We discuss  three cases:

(i) If $\sigma<\delta$, then
$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(1-\delta)$.
By the concavity of $u(t)$, we  have
$$
\frac{u(\sigma)-u(1)}{1-\sigma}\leq \frac{u(1-\delta)-u(1)}{\delta};
$$
i.e.,
$$
\frac{u(1-\delta)}{\delta}\geq
\frac{u(\sigma)}{1-\sigma}+(\frac{1}{\delta}-\frac{1}{1-\sigma})u(1).
$$
For $\sigma< \delta<{1-\delta}$, we have ${1-\sigma}>\delta$.
So
$$
u(1-\delta)\geq \frac{\delta}{1-\sigma}u(\sigma)\geq {\delta
u(\sigma)}.
$$

(ii) If $\sigma>{1-\delta}$, then $\min_{\delta\leq t\leq
{1-\delta}}u(t)=u(\delta)$ and $\sigma >\delta$. By the concavity of
$u(t)$, we  have
$$
\frac{u(\sigma)-u(0)}{\sigma}\leq \frac{u(\delta)-u(0)}{\delta};
$$
i.e.,
$$
\frac{u(\delta)}{\delta}\geq
\frac{u(\sigma)}{\sigma}+(\frac{1}{\delta}-\frac{1}{\sigma})u(0).
$$
So
$$
u(\delta)\geq \frac{\delta}{\sigma}u(\sigma)\geq {\delta u(\sigma)}.
$$

(iii) For $\delta\leq\sigma\leq{1-\delta}$, if
$\min_{\delta\leq t\leq {1-\delta}}u(t)=u(1-\delta)$, the
proof is the same as (i); if $\min_{\delta\leq t\leq
{1-\delta}}u(t)=u(\delta)$, the proof is the same as (ii).

From  the three cases above, the proof is complete.
\end{proof}

Let $X=C[0,1]$ and $\|u\|=\max_{0\leq t\leq 1}|u(t)|$, denote
$K=\{u\in X:u(t)\geq 0, t\in [0,1]\}$ and $K'=\{u\in K: u(t)$ is a
concave function on $[0,1]\}$. Obviously, $K,K'\subset X$ are two
cones of $X$ with $K'\subset K$. For $u\in K$, from
Lemma \ref{lem2.5} we
know  $\min_{t\in[\frac{1}{k},1-\frac{1}{k}]}u(t)\geq
{\frac{1}{k}\|u\|}$ with
$k>\max\{\frac{1}{\xi_1},\frac{2}{1-\xi_{m-2}}\}$, and we can show
$1-\frac{1}{k}>\xi_1$.

Denote
$$
\alpha(u)=\min_{t\in[\frac{1}{k},1-\frac{1}{k}]}u(t),\quad
\text{for } u\in K'.
$$
We have $\alpha(u)\leq \|u\| \leq {k\alpha(u)}$.
Define $T:K\to K$ by
$$
 Tu(t)=\begin{cases}
\Big[\frac{b\varphi_q(\int_0^{\sigma_u}f(\tau,u(\tau))d\tau)
 +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(\int_s^{\sigma_u}
 f(\tau,u(\tau))d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}} \\
+\int_0^t\varphi_q(\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds\Big]^+,
& 0\leq t\leq \sigma_u,  \\[4pt]
\Big[\frac{\int_0^1g(s)(\int_s^1\varphi_q
 (\int_{\sigma_u}^rf(\tau,u(\tau))d\tau)dr)ds}{1-\int_0^1g(s)ds}\\
+\int_t^1\varphi_q(\int_{\sigma_u}^sf(\tau,u(\tau))d\tau)ds\Big]^+,
& \sigma_u\leq t\leq 1,
\end{cases}
$$
where $B^+=\max\{B,0\}$ and $\sigma_u$ is defined in
Lemma \ref{lem2.3}.

Define $A:K\to X$ by
$$
Au(t)=\begin{cases}
\frac{b\varphi_q(\int_0^{\sigma_u}f(\tau,u(\tau))d\tau)
 +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q
 (\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}\\
+\int_0^t\varphi_q(\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds,
 & 0\leq t\leq \sigma_u, \\[4pt]
\frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^r
 f(\tau,u(\tau))d\tau)dr)ds} {1-\int_0^1g(s)ds}\\
+\int_t^1\varphi_q(\int_{\sigma_u}^s
 f(\tau,u(\tau))d\tau)ds,&
\sigma_u\leq t\leq 1.
\end{cases}
$$

For $u\in X$, denote $\theta : X\to K$ while
$(\theta u)(t)=\max \{u(t),0\}$, then
$T=\theta \circ A$. For $u\in K'$,
define $T^*:K'\to K'$ by
$$
T^*u(t)=\begin{cases}
\frac{b\varphi_q(\int_0^{\sigma_u}f^+(\tau,u(\tau))d\tau)
   +\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(\int_s^{\sigma_u}
f^+(\tau,u(\tau))d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}} \\
+\int_0^t\varphi_q(\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds,
& 0\leq t\leq \sigma_u, \\[4pt]
\frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^r
 f^+(\tau,u(\tau))d\tau)dr)ds} {1-\int_0^1g(s)ds}\\
+\int_t^1\varphi_q(\int_{\sigma_u}^s
 f^+(\tau,u(\tau))d\tau)ds,&
\sigma_u\leq t\leq 1.
\end{cases}
$$


\begin{lemma} \label{lem2.6}
 $T^*:K'\to K'$ is completely continuous.
\end{lemma}

\begin{proof}
Obviously, we know $T^*(u)\geq 0$. By
$[\varphi_p((T^*u)'(t))]'=-f^+(t,u(t))\leq 0$, it is easy to show
$\varphi_p((T^*u)'(t))$ is non-increasing. Then $(T^*u)'(t)$ is also
monotone non-increasing, i.e., $(T^*u)(t)$ is concave function. Thus
$T^*:K'\to K'$. We can have $T^*$ is completely continuous
from Ascoli-Arzela theorem and concavity of $f^+$.
\end{proof}

 From Lemma \ref{lem2.2}, we have the following lemma.


\begin{lemma} \label{lem2.7}
A function $u(t)$ is a solution of boundary
value problem \eqref{e1.1} if and only if  $u(t)$ is a
fixed point of the operator $A$.
\end{lemma}


\begin{lemma}[\cite{g2}] \label{lem2.8}
If $A:K\to X$ is completely continuous, then
$T=\theta \circ A :K\to K$ is also completely continuous.
\end{lemma}


\begin{lemma} \label{lem2.9}
If $u$ is a fixed point of operator $T$, then $u$ is also a
fixed point of operator $A$.
\end{lemma}

\begin{proof}
Let $u$ be a fixed point of operator of $T$. Obviously,
if $Au(t)\geq 0$, $t\in [0,1]$, then $u(t)$ is a fixed point of
operator $A$.

We prove that if $Tu(t)=u(t)$, then $Au(t)\geq 0$ for $t\in[0,1]$.

Suppose this is not true, then there is a $t_0\in (0,1)$ such that
$Au(t_0)<0=u(t_0)$. Let $(t_1,t_2)$ be the maximal interval which
contain $t_0$ and such that $Au(t)<0$, $t\in (t_1,t_2)$. It follows
$[t_1,t_2]\neq [0,1]$ from (H2).

If $t_2<1$, we have $u(t)=0$ for $t\in [t_1,t_2]$, $Au(t)<0$ for
$t\in (t_1,t_2)$ and $Au(t_2)=0$. Thus $(Au)'(t_2)\geq 0$. From
(H2), we know $[\varphi_p((Au)'(t))]'=-f(t,0)\leq 0$, so $t_1=0$.

On the other hand, from the fact that $Au(t)<0$ for $t\in [0,t_2)$
and $Au(t_2)=0, (Au)'(t_2)\geq 0$ and $(Au)''(t)\leq 0$, we
can obtain $(Au)'(0)>0$. Without loss of generality, we suppose that
there exists $i=k$ such that $u(\xi_i)\leq 0$ for $1\leq i\leq k$
and $u(\xi_i)>0$ for $k<i\leq {m-2}$. Since $(Au)'(0)>0$,
$a>\sum_{i=1}^{m-2}a_i$ and
$aAu(0)-bAu'(0)=\sum_{i=1}^{m-2}a_iAu(\xi_i)<0$, we have
\begin{align*}
    |aAu(0)-bAu'(0)|
&>a|Au(0)|>\sum_{i=1}^ka_i|Au(0)|+\sum_{i=k+1}^{m-2}a_i|Au(0)|\\
  &>\sum_{i=1}^ka_i|Au(\xi_i)|-\sum_{i=k+1}^{m-2}a_iAu(\xi_i)\\
  & =-(\sum_{i=1}^{m-2}a_iAu(\xi_i))\\
  &=|\sum_{i=1}^{m-2}a_iAu(\xi_i)|,
\end{align*}
a contradiction. So $t_2=1$.

If $t_1>0$, then we have $Au(t)=0$ for $t\in [t_1,t_2]$, $Au(t)<0$
for $t\in (t_1,1)$ and $Au(t_1)=0$. Thus $(Au)'(t_1)\leq 0$. We have
$[\varphi_p((Au)'(t))]'=-f(t,0)\leq 0$ by (H2). This implies
$Au(t)<0$ for $t\in (t_1,1]$ and
$Au(1)=\min_{t\in [t_1,1]}Au(t)$.

We can prove that $Au(t)\geq 0$ for $t\in [0,t_1]$. If there exists
a $t_3\not\in [t_1,1]$ such that $Au(t_3)<0$ and there is a maximal
interval $[t_4,t_5]$ which contains $t_3$ such that $Au(t)<0$ for
$t\in (t_4,t_5)$. Obviously $[t_4,t_5)\cap [t_1,1]=\emptyset$, so
$1\not\in(t_4,t_5)$; i.e., $t_5<1$, this is a contradiction with the
above discussion. Thus we can show $Au(t)\geq 0$ for $t\in [0,t_1]$.

For $Au(1)<0$, we have
$$
Au(1)=\int_0^1g(s)Au(s)ds.
$$
Then
\begin{align*}
    |Au(1)|
&=-Au(1)=-\int_0^{t_1}g(s)Au(s)ds-\int_{t_1}^1g(s)Au(s)ds\\
  &\leq\int_0^{t_1}g(s)|Au(1)|ds+\int_{t_1}^1g(s)|Au(1)|ds\\
  &=|Au(1)|\int_0^1g(s)ds.
\end{align*}
So $|Au(1)|\leq |Au(1)|\int_0^1g(s)ds$ which is a contradiction with
$(H_3)$. Thus $t_1=0$.

The above  also  contradicts $[t_1,t_2]\neq [0,1]$. Thus the
proof is complete.
\end{proof}

\section{Main result}
Denote
$$
 t^*=\frac{\xi_{m-2}+1}{2}, \quad
M=\frac{1}{q}(\frac{1-\xi_{m-2}}{2}-\frac{1}{k})^q ,\quad
N=\frac{1}{\max\big\{\frac{a+b}{a-\sum_{i=1}^{m-2}a_i},
\frac{1}{1-\int_0^1g(s)ds}\big\}},
$$
\begin{align*}
N_0=\min\Big\{&\frac{b+a_1\xi_1}{(b+a\xi_1)q}(\xi_1-\frac{1}{k})^q,\;
\frac{1}{q}\int_{\frac{1}{k}}^{\xi_1}g(s)((\xi_1-\frac{1}{k})^q
 -(\xi_1-s)^q)ds,\\
&\frac{b+a_1\xi_1}{(b+a\xi_1)q}(1-\frac{1}{k}-\xi_1)^q,\;
\frac{1}{q}\int_{\xi_1}^{1-\frac{1}{k}}g(s)((1-\xi_1-\frac{1}{k})^q
-(s-\xi_1)^q)ds\Big\}.
\end{align*}
It is easy to see that $N_0>0$ and $\frac{1}{k}<t^*<1-\frac{1}{k}$.

\begin{theorem} \label{thm3.1}
Suppose {\rm (H1)--(H3)} hold, that there exist nonnegative
constants $c_1, c_2, c_3$
such that
$$
0<c_1<\max\{\frac{N_0}{N},1\}c_2<\frac{1}{k}c_3<c_3
$$
 and that $f$ satisfies the following growth conditions:
\begin{itemize}
\item[(H4)]  $f(t,u)\geq 0$ for  $(t,u)\in [0,1]\times [c_1,c_3]$;

\item[(H5)]  $f(t,u)<\varphi_p(\frac{c_2}{N})$ for
 $(t,u)\in [0,1]\times [0,c_2]$;

\item[(H6)] $f(t,u)\geq \varphi_p(\frac{c_3}{M})$ for
$(t,u)\in [\frac{1}{k},1-\frac{1}{k}]\times [\frac{1}{k}c_3,c_3]$;

\item[(H7)]  $f(t,u)\geq \varphi_p(\frac{c_1}{N_0})$ for
$(t,u)\in[\frac{1}{k},1-\frac{1}{k}]\times [\frac{1}{k}c_2,c_3]$.

\end{itemize}
Then  \eqref{e1.1} has at least two positive
solutions $u_1$ and $u_2$ such that
$$
0<\|u_1 \|<c_2\leq \|u_2 \|, \quad  \alpha(u_2)<\frac{1}{k}c_3.
$$
\end{theorem}

\begin{proof}
For any $u\in \partial K_{c_2}$, from $(H_5)$ we have
$$
\| Tu \|=\max_{0\leq t \leq 1}|Tu(t)|=Tu(\overline{t}).
$$
If $\overline{t}<\sigma _u$, we have
\begin{align*}
Tu(\overline{t})
  & =\Big[\frac{b\varphi_q(\int_0^{\sigma_u}f(\tau,u(\tau))d\tau)
   +\sum _{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q
 (\int_s^{\sigma_u}f(\tau,u(\tau))d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}\\
&\quad +\int_0^{\overline{t}}\varphi_q(\int_s^{\sigma_u}
 f(\tau,u(\tau))d\tau)ds\Big]^+ \\
& < \frac{b\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)
   +\sum_{i=1}^{m-2}a_{i}\int_0^1\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}\\
&\quad +\int_0^1\varphi_q(\int_0^1\varphi_p (\frac{c_2}{N})d\tau)ds \\
& =\frac{c_2}{N}\frac{a+b}{a-\sum_{i=1}^{m-2}a_{i}}
  \leq c_2,
\end{align*}
and if $\overline{t}>\sigma _u$, we have
\begin{align*}
Tu(\overline{t})
  & =\Big[\frac{\int_0^1g(s)(\int_s^1\varphi_q(\int_{\sigma_u}^rf(\tau,u(\tau))d\tau)dr)ds}
{1-\int_0^1g(s)ds}+\int_{\overline{t}}^1\varphi_q(\int_{\sigma_u}^sf(\tau,u(\tau))d\tau)ds\Big]^+ \\
  & <\frac{\int_0^1g(s)(\int_0^1\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)dr)ds}{1-\int_0^1g(s)ds}+\int_0^1\varphi_q(\int_0^1\varphi_p(\frac{c_2}{N})d\tau)ds \\
  & =\frac{c_2}{N}\frac{1}{1-\int_0^1g(s)ds}
   \leq c_2.
\end{align*}
 From the above, we have $ \| Tu \| < c_2$, So (C1) of
Lemma \ref{lem2.1} is satisfied.

For $u\in \partial K'_{c_2}$; i.e., $u\in K'$ and $\|u\|=c_2$. From
Lemma \ref{lem2.4} and (H5), we can prove that $\| T^*u \| <c_2$ using
the same way as the $Tu$ above.

For $u\in \partial K'(\frac{1}{k}c_3)$; i.e., $u\in K'$ and
$\alpha(u)=\frac{1}{k}c_3$. So $\frac{1}{k}c_3\leq u(t)\leq c_3$ for
$\frac{1}{k}\leq t\leq {1-\frac{1}{k}}$.
From Lemmas \ref{lem2.6}, \ref{lem2.7}
and (H6), we have
\begin{align*}
    \alpha(T^*u)
&=\min_{t\in [\frac{1}{k},1-\frac{1}{k}]}T^*u(t)\geq\frac{1}{k}\| T^*u\|=\frac{1}{k}T^*u(\sigma_u)\\
  &=\frac{1}{k}\Big(\frac{b\varphi_q(\int_0^{\sigma_u}f^+(\tau,u(\tau))d\tau)+\sum_{i=1}^{m-2}a_{i}\int_0^{\xi_i}\varphi_q(\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds}
{a-\sum_{i=1}^{m-2}a_{i}}\\
&\quad +\int_0^{\sigma_u}\varphi_q
  (\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds\Big)\\
&=\frac{1}{k}\Big(\frac{\int_0^1g(s)(\int_s^1\varphi_q
(\int_{\sigma_u}^rf^+(\tau,u(\tau))d\tau)dr)ds}{1-\int_0^1g(s)ds}\\
&\quad +\int_{\sigma_u}^1\varphi_q(\int_{\sigma_u}^sf^+
(\tau,u(\tau))d\tau)ds\Big)\\
&\geq\frac{1}{k}\min\Big\{\int_0^{t^*}\varphi_q(\int_s^{t^*}f^+(\tau,u(\tau))d\tau)ds,\int_{t^*}^1\varphi_q(\int_{t^*}^sf^+(\tau,u(\tau))d\tau)ds\Big\}\\
&\geq \frac{1}{k} \min\Big\{\int_\frac{1}{k}^{t^*}\varphi_q(\int_s^{t^*}\varphi_p(\frac{c_3}{M})d\tau)ds,\int_{t^*}^{1-\frac{1}{k}}\varphi_q(\int_{t^*}^s\varphi_p(\frac{c_3}{M})d\tau)ds\Big\}\\
&\geq\frac{c_3}{kM}\min\frac{1}{q}\{(\frac{1+\xi_{m-2}}{2}-\frac{1}{k})^q,(\frac{1-\xi_{m-2}}{2}-\frac{1}{k})^q\}\\
&=\frac{1}{k}c_3.
\end{align*}
Finally, we show that (C3) of Lemma \ref{lem2.1} is also satisfied.

Let $u\in \partial K'_{c_2}(\frac{1}{k}c_3)\cap\{u:u=T^*u\}$, then
$u=T^*u$, $u\in K'$, $\|u\|>c_2$ and $\alpha(u)\leq \frac{1}{k}c_3$.

We know $\|u\|\leq k\alpha(u)\leq c_3$ and $\alpha(u)\geq
\frac{1}{k}\|u\|\ >\frac{1}{k}c_2$ by Lemma \ref{lem2.5}. By
$T^*:K'\to K'$ and the definition of $T^*$, we can get
$$
\min_{0\leq t \leq 1}u(t)=\min\{u(0),u(1)\}\geq 0.
$$
From the boundary condition
$au(0)-bu'(0)=\sum_{i=1}^{m-2}a_iu(\xi_i)$, we have
\begin{align*}
au(0)&=bu'(0)+\sum_{i=1}^{m-2}a_iu(\xi_i)\\
&\geq bu'(0)+a_1u(\xi_1)\\
&\geq bu'(\xi)+a_1u(\xi_1)\\
&=\frac{b}{\xi_1}(u(\xi_1)-u(0))+a_1u(\xi_1),
\end{align*}
where $\xi \in (0,\xi_1)$. So we can obtain
$$
u(0)\geq \frac{b+a_1\xi_1}{b+a\xi_1}u(\xi_1).
$$
 From the definition $T^*$ and the use of condition (H7), we have
\begin{align*}
    u(\xi_1)
  &>\int_0^{\xi_1}\varphi_q(\int_s^{\sigma_u}f^+(\tau,u(\tau))d\tau)ds > \int_{\frac{1}{k}}^{\xi_1}\varphi_q(\int_s^{\xi_1}f^+(\tau,u(\tau))d\tau)ds\\
  &>\frac{c_1}{qN_0}(\xi_1-\frac{1}{k})^q,
  \quad\text{if }  \sigma_u>\xi_1,
\end{align*}
and
\begin{align*}
    u(\xi_1)
  &>\int_{\xi_1}^{1-\frac{1}{k}}\varphi_q(\int_{\sigma_u}^sf^+(\tau,u(\tau))d\tau)ds >\int_{\xi_1}^{1-\frac{1}{k}}\varphi_q(\int_{\xi_1}^sf^+(\tau,u(\tau))d\tau)ds\\
  &>\frac{c_1}{qN_0}(1-\frac{1}{k}-\xi_1)^q,
  \quad\text{if }  \sigma_u\leq \xi_1.
\end{align*}
Hence,
$$
u(0)\geq\frac{b+a_1\xi_1}{b+a_1\xi}\cdot\frac{c_1}{qN_0}\min\{(\xi_1-\frac{1}{k})^q,(1-\frac{1}{k}-\xi_1)^q\}\geq
c_1.
$$
 From the boundary condition $u(1)=\int_0^1 g(s)u(s) ds$, we have
\begin{align*}
    u(1)
&>\int_0^{\xi_1}g(s)u(s)ds>\int_{\frac{1}{k}}^{\xi_1}g(s)(\int_{\frac{1}{k}}^{\xi_1}\varphi_q(\int_r^{\xi_1}f^+(\tau,u(\tau))d\tau)dr)ds\\
  &>\frac{c_1}{qN_0}(\int_{\frac{1}{k}}^{\xi_1}g(s)((\xi_1-\frac{1}{k})^q-(\xi_1-s)^q)ds),
  \quad\text{if } \sigma_u>\xi_1,
\end{align*}
and
\begin{align*}
    u(1)
&>\int_{\xi_1}^{1-\frac{1}{k}}g(s)u(s)ds>\int_{\xi_1}^{1-\frac{1}{k}}g(s)(\int_s^{1-\frac{1}{k}}\varphi_q(\int_{\xi_1}^rf^+(\tau,u(\tau))d\tau)dr)ds\\
&>\frac{c_1}{qN_0}(\int_{\xi_1}^{1-\frac{1}{k}}g(s)((1-\xi_1-\frac{1}{k})^q-(s-\xi_1)^q)ds),
  \quad \text{if }\sigma_u\leq \xi_1.
\end{align*}
So, $u(1)\geq c_1$. Therefore, for $u\in \partial
K'_{c_2}(\frac{1}{k}c_3)\cap\{u:u=T^*u\}$, we have
$$
c_1\leq u(t)\leq \|u\| \leq k\alpha(u)=c_3.
$$
It follows $f(t,u(t))\geq 0$, $t\in [0,1]$ from (H4). Thus,
$T^*u=Tu$. So the condition of Lemma \ref{lem2.1} is satisfied.
Then by Lemma \ref{lem2.7}, we know that operator $T$ has two fixed
points $u_1$ and $u_2$ satisfying
$$
0<\|u_1 \|<c_2\leq \|u_2 \|, \quad \alpha(u_2)<\frac{1}{k}c_3.
$$
The proof is complete.
\end{proof}


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\end{document}