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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 66, 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 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/66\hfil Multiple nonnegative solutions]
{Multiple nonnegative solutions for second-order boundary-value
problems with sign-changing nonlinearities}

\author[S. Xi,  M. Jia, H. Ji \hfil EJDE-2009/66\hfilneg]
{Shouliang Xi,  Mei Jia, Huipeng Ji}  % in alphabetical order

\address{College of Science, University of Shanghai for Science and
Technology, Shanghai, 200093, China}
\email[S.Xi]{xishouliang@163.com}
\email[M.Jia]{jiamei-usst@163.com}
\email[H.Ji]{jihuipeng1983@163.com}

\thanks{Submitted December 8, 2008. Published May 14, 2009.}
\thanks{Supported by grant 05EZ53 from the
Foundation of Educational Commission of Shanghai.}
\subjclass[2000]{34B10, 34B18}
\keywords{Nonnegative solutions; fixed-point theorem in double cones;
\hfill\break\indent integral kernel; integral boundary conditions}

\begin{abstract}
 In this paper, we study the existence of multiple nonnegative
 solutions for second-order boundary-value problems of differential
 equations with sign-changing nonlinearities.
 Our main tools are the fixed-point theorem in double cones and
 the Leggett-Williams fixed point theorem. We present also the
 integral kernel associated with the boundary-value problem.
\end{abstract}

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

\section{Introduction}

Boundary-value problems with nonnegative solutions
describe many phenomena in the applied science, and they are widely
used in  fields, such as chemistry, biological, etc.;
see for example \cite{c2,g1,j1,j2,j3,k1}. Problems with
integral boundary conditions have been applied in heat
conduction, chemical engineering, underground water flow-elasticity,
etc. The existence of nonnegative solutions to these problems
have received a lot of attention; see \cite{f1,k1,k2,k3,k4,k5} and
reference therein.

Recently, by constructing a special cone and using the fixed point
index theory, Liu and Yan \cite{k2} proved the existence
of multiple solutions to the singular boundary-value problem
\begin{gather*}
  (p(t)x'(t))'+\lambda f(t,x(t),y(t))=0\\
  (p(t)y'(t))'+\lambda g(t,x(t),y(t))=0\\
  \alpha x(0)-\beta x'(0)=\gamma x(1)+\delta x'(1)=0\\
  \alpha y(0)-\beta y'(0)=\gamma y(1)+\delta y'(1)=0,
\end{gather*}
where the parameter $\lambda$ in $\mathbb{R}^+$,
$p\in C([0,1],\mathbb{R}^+)$,
$\alpha, \beta, \gamma, \delta \geq 0$,
$\beta\gamma+\alpha\delta+\alpha\gamma>0$,
$f\in C((0,1) \times \mathbb{R}^+ \times \mathbb{R}, \mathbb{R}^+)$,
$g\in C((0,1) \times \mathbb{R}^+ \times \mathbb{R}, \mathbb{R})$,
but $g$ must be controlled by $f$.

By using fixed point index theory in a cone, Yang \cite{k3} studied
the existence of positive solutions to a system of second-order
nonlocal boundary value problems
\begin{gather*}
  -u''=f(t,u,v)\\
  -v''=g(t,u,v)\\
  u(0)=v(0)=0\\
  u(1)=H_1 \Big(\int_0^1 u(\tau)d\alpha (\tau)\Big)\\
  v(1)=H_2 \Big(\int_0^1 v(\tau)d\beta (\tau)\Big),
\end{gather*}
where $\alpha$ and $\beta$ are increasing nonconstant functions
defined on $[0, 1]$ with $\alpha (0)=0=\beta (0)$ and $f, g\in
C((0,1) \times \mathbb{R}^+ \times \mathbb{R}^+, \mathbb{R}^+)$,
$H_i\in C(\mathbb{R}^+, \mathbb{R}^+)$.

By using fixed point theory in a cone, Feng \cite{k5} studied
positive solutions for the boundary-value problem, with
integral boundary conditions in Banach spaces,
\begin{equation*}
  x''+f(t,x)=0\\
\end{equation*}
with
\begin{equation*}
  x(0)=\int_0^1 g(t)x(t)dt, \quad x(1)=0\\
\end{equation*}
or
\begin{equation*}
  x(0)=0, x(1)=\int_0^1 g(t)x(t)dt,
\end{equation*}
where $f\in C([0,1]\times P, P), g\in L^1[0,1]$, and $P$ is a cone
of $E$. All of these, we can find the nonlinear term $f$ is
nonnegative.

 In this paper, by using the fixed point theorem in double cones and
the Leggett-Williams fixed point theorem, we study the existence of
multiple nonnegative solutions to the \ boundary value
problem
\begin{equation} \label{e1.1}
\begin{gathered}
  u_1''(t)+f_1(t,u_1(t),u_2(t))=0\\
  u_2''(t)+f_2(t,u_1(t),u_2(t))=0\\
  u_1(0)=u_2(0)=0\\
  u_1(1)=\int_0^1 g_1(s)u_1(s)ds,u_2(1)=\int_0^1 g_2(s)u_2(s)ds,
\end{gathered}
\end{equation}
where $f_1, f_2\in C((0,1) \times \mathbb{R}^+ \times \mathbb{R}^+,
\mathbb{R})$, and
$g_1, g_2$ are nonnegative functions in $L^1 [0,1]$.

In this paper we assume that the following conditions:
\begin{itemize}
\item[(H1)] $f_i\in C((0,1) \times \mathbb{R}^+ \times \mathbb{R}^+,
\mathbb{R})$, $g_i\in L^1 [0,1]$ is nonnegative, $i=1,2$;
\item[(H2)] $1-\int_0^1sg_i(s)ds>0$;
\item[(H3)] $f_1(t,0,u_2(t))\geq 0(\not\equiv0)$,
$f_2(t,u_1(t),0)\geq 0(\not\equiv0)$, $t\in[0,1]$.
\end{itemize}

\section{Preliminaries}

 Let $X$ be a Banach space  with norm $\|\cdot\|$ and
$K\subset X$ be a cone. For a constant $r>0$, denote $K_r=\{x\in K:
\|x\|<r\}$, $\partial K_r=\{x\in K: \|x\|=r\}$. Suppose $\alpha:
K\to \mathbb{R}^{+}$ is a continuously increasing functional; i.e.
$\alpha$ is continuous and $\alpha(\lambda x)\leq \alpha(x)$ for
$\lambda\in(0,1)$. Let
$$
K(b)=\{x\in K: \alpha(x)<b\},\partial K(b)=\{x\in K: \alpha(x)=b\}.
$$
and $K_{a}(b)=\{x\in K: \|x\|>a, \alpha(x)<b\}$.
$\phi: K\to \mathbb{R}^{+}$ is a continuously concave functional.
Denote
$$
  K(\phi ,a,b)=\{x\in K: \phi(x)\geq a, \|x\|\leq b\}.
$$

We will use the following two theorem.

\begin{theorem}[\cite{c1}]  \label{thm2.1}
 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 $\alpha: K'\to \mathbb{R}^+$ is
 a continuously increasing functional satisfying
 $\alpha(x)\leq \|x\|\leq M  \alpha(x)$ for all $x\in K'$,
where $M\geq1$ 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 $x\in K'_{a}(b)\cap\{u:T^*u=u\}$.

\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{theorem}

\begin{theorem}[Leggett-Williams fixed point theorem \cite{k6}]
 \label{thm2.2}
 Let $A :\overline{K_c}\to \overline{K_c}$ be completely continuous and
$\phi $ be a nonnegative continuous concave functional on $K$ such
that $\phi (x)\leq \| x\| $ for all $x\in \overline{K_c}$. Suppose
that there exist $0<d<a<b\leq c$ such that
\begin{itemize}
\item[(C4)] $\{ x\in K(\phi ,a,b):\phi (x)>a\} \neq
\emptyset $ and $\phi (Ax)>a$ for $x\in K(\phi ,a,b)$;

\item[(C5)]  $\| Ax\| <d$ for $\|x\| \leq d$;

\item[(C6)] $\phi (Ax)>a $ for $x\in K(\phi ,a,c)$ with $\| Ax\| >b$.
\end{itemize}
Then $A$ has at least three fixed points $x_1$, $x_2$, $x_3$ in
$\overline{K_c}$ satisfying
\[
\| x_1\| <d, \quad a<\phi (x_2),\quad \|x_3\| >d, \quad
\phi (x_3)<a.
\]
\end{theorem}

\begin{lemma}\label{lem2.3}
Assume that {\rm (H2)} holds. Then for any $y_i\in C[0,1]$, the boundary
value problem
\begin{gather}
  u_i''(t)+y_i(t)=0\label{e2.1}\\
  u_i(0)=0, u_i(1)=\int_0^1 g_i(s)u_i(s)ds,\label{e2.2}
\end{gather}
has a unique solution
\begin{equation}
  u_i(t)=\int_0^1 H_i(t,s)y_i(s)ds, \quad i=1,2,\label{e2.3}
\end{equation}
where
\begin{equation*}
  H_i(t,s)=G(t,s)+\frac{t\int_0^1 g_i(r)G(r,s)dr}{1-\int_0^1 sg_i(s)ds},
  \quad i=1,2,
\end{equation*}
\begin{equation*}
 G(t,s)=\begin{cases}
t(1-s), &\text{if } 0\leq t\leq s\leq 1,\\
s(1-t), &\text{if } 0\leq s\leq t\leq 1,
\end{cases}
\end{equation*}
\end{lemma}
The proof is similar to  \cite[Lemma 2.1]{k5}, and is
omitted.

\begin{lemma} \label{lem2.4}
Assume that {\rm (H2)} holds. Let $\delta \in(0,\frac{1}{2})$, then for
all $t\in [\delta, 1-\delta], \sigma ,s\in [0,1]$, we have
\begin{equation}
 H_i(\sigma,s) \geq 0,\quad
 H_i(t,s) \geq \delta H_i(\sigma,s). \label{e2.4}
\end{equation}
\end{lemma}

\begin{proof} It is clear that $H_i(\sigma,s) \geq 0$, From
the properties of $G(t,s)$, we obtain
\begin{equation*}
  G(t,s) \geq \delta G(\sigma,s), \quad
t \in [\delta, 1-\delta],\; \sigma, s \in [0,1],
\end{equation*}
then
\begin{align*}
H_i(t,s)&=G(t,s)+\frac{t\int_0^1 g_i(r)G(r,s)dr}{1-\int_0^1 sg_i
(s)ds}\\
&\geq\delta G(\sigma,s)+\frac{\delta \int_0^1 g_i
(r)G(r,s)dr}{1-\int_0^1 sg_i(s)ds}\\
&\geq\delta G(\sigma,s)+\frac{\delta \sigma \int_0^1 g_i
(r)G(r,s)dr}{1-\int_0^1 sg_i(s)ds}
=\delta H_i(\sigma,s)
\end{align*}
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
Assume that {\rm (H2)} holds. If $y_i\in C[0,1], y_i\geq 0$, then the
unique solution $u_i(t)$ of the boundary-value problem
\eqref{e2.1}-\eqref{e2.2} satisfies $u_i(t)\geq 0$ and
$\min_{t\in[\delta, 1-\delta]}u_i(t)\geq \delta \|u_i\|, i=1, 2$.
\end{lemma}

\begin{proof}
 It is clear that $u_i(t)\geq 0$, for all $t\in [0,1]$, $i=1, 2$.
 In fact, from \eqref{e2.3} and \eqref{e2.4}, for any $t\in
[\delta,1-\delta], s, \sigma\in [0,1], i=1, 2$, we have
\[
u_i(t)=\int_0^1 H_i(t,s)y_i(s)ds
\geq \int_0^1 \delta H_i(\sigma,s)y_i(s)ds
=\delta u_i(\sigma).
\]
Hence,
\begin{equation*}
  u_i(t)\geq \delta \max_{0\leq \sigma\leq 1}|u_i(\sigma)|=\delta \|u_i\|,
\end{equation*}
and
$  \min_{\delta\leq t\leq 1-\delta}u_i(t)\geq \delta \|u_i\|$.
The proof is complete.
\end{proof}


Let $X=C[0,1] \times C[0,1]$ with the norm
$\|(u_1,u_2)\|:=\|u_1\|+\|u_2\|$,
$K=\{(u_1,u_2)\in X:u_i\geq 0,i=1,2\}$ and
$K '=\{(u_1,u_2)\in K$: $u_i(t)$ is concave in
$[0,1], \min_{t\in[\delta, 1-\delta]}u_i(t)\geq \delta \|u_i\|, i=1,
2 \}$.

Clearly, $K,K'\subset X$  are cones with $K'\subset K$.
 Let $T_i:K\to C[0,1], i=1,2$ be defined
 by
\begin{gather*}
  T_1(u_1,u_2)(t)=\Big(\int_0^1 H_1(t,s)f_1(s,u_1(s),u_2(s))ds\Big)^{+},
\quad t\in[0,1],\\
  T_2(u_1,u_2)(t)=\Big(\int_0^1 H_2(t,s)f_2(s,u_1(s),u_2(s))ds\Big)^{+},\quad
  t\in[0,1],
\end{gather*}
where $(B)^{+}=\max\{B, 0\}$.
Let
\begin{gather*}
  T(u_1,u_2)(t)=(T_1(u_1,u_2)(t), T_2(u_1,u_2)(t)),\\
  A_1(u_1,u_2)(t)=\int_0^1 H_1(t,s)f_1(s,u_1(s),u_2(s))ds,\quad t\in[0,1],\\
  A_2(u_1,u_2)(t)=\int_0^1 H_2(t,s)f_2(s,u_1(s),u_2(s))ds,\quad
  t\in[0,1], \\
  A(u_1 ,u_2 )(t)=(A_1(u_1 ,u_2 )(t), A_2(u_1 ,u_2 )(t)).
\end{gather*}
 For $(u_1 ,u_2 )\in X$, define $\theta: X\to K$ by
$$
(\theta (u_1,u_2 ))(t)=(\max\{u_1(t), 0\}, \max\{u_2(t), 0\}),
$$
then $T=\theta \circ A$.

 Let $T^*_i: K'\to C[0,1], i=1, 2$ be defined by
\begin{equation}
\begin{gathered}
  T_1^{*}(u_1 ,u_2 )(t)=\int_{0}^{1}H_1(t,s)f_1^{+}(s,u_1(s),u_2(s))ds,\quad t\in[0,1],\\
  T_2^{*}(u_1 ,u_2 )(t)=\int_{0}^{1}H_2(t,s)f_2^{+}(s,u_1(s),u_2(s))ds,\quad
  t\in[0,1],
\end{gathered}\label{e2.7}
\end{equation}
and
$$
  T^{*}(u_1 ,u_2 )(t)=(T_1^{*}(u_1 ,u_2 )(t), T_2^{*}(u_1 ,u_2 )(t)).
$$
Define $\alpha:K'\to R^+$ by
$$
  \alpha(u_1 ,u_2 )=\min_{\delta\leq t\leq 1-\delta}u_1(t)
+\min_{\delta\leq t\leq
  1-\delta}u_2(t).
$$
 It is clear that $\alpha$ is a continuous increasing functional
and $\alpha(u_1 ,u_2 )\leq \|(u_1,u_2)\|$.
 For $u\in K'$, we have
\[
\alpha(u_1 ,u_2 )=\min_{\delta\leq t\leq
1-\delta}u_1(t)+\min_{\delta\leq t\leq 1-\delta}u_2(t)
\geq \delta \|u_1\|+\delta \|u_2\|
=\delta \|(u_1,u_2)\|.
\]
Therefore,
$$
  \alpha(u_1 ,u_2 )\leq \|(u_1,u_2)\|\leq \frac{1}{\delta}\alpha(u_1 ,u_2).
$$

\begin{lemma}\label{lem2.6}
Suppose $A:K\to X$ is completely continuous. Then
$\theta \circ A:K\to K$ is also a completely continuous operator.
\end{lemma}

\begin{proof}
 The complete continuity of $A$ implies that
$A$ is continuous and maps each bounded subset of  $K$ to a
relatively compact set of $X$. Let $D\subset K$ be a bounded set,
for any $\epsilon>0$, there exist $P_i (x_i ,y_i )\in X,i=1,2,\dots,
m$, such that
$$
A D\subset\cup_{i=1}^{m}B(P_i,\epsilon),
$$
where $B(P_i,\epsilon):=\{(u_1,u_2)\in K:
\|u_1-x_i\|+\|u_2-y_i\|<\epsilon\}$. Then for any
$Q^{*}(x^{*}_Q , y^{*}_Q)\in (\theta \circ A)(D)$, there exists
$Q(x_Q ,y_Q )\in AD$, such that
$$
  (x^{*}_Q , y^{*}_Q )=(\max\{x_Q ,0\},\max\{y_Q ,0\}).
$$
We choose a $P_i \in \{P_1, P_2,\dots,P_m\}$, such that
$$
\|x_Q-x_i\|+\|y_Q-y_i\|<\epsilon.
$$
Since
$$
  \|x^{*}_Q-x^{*}_i\|+\|y^{*}_Q-y^{*}_i\|\leq \|x_Q-x_i\|+\|y_Q-y_i\|<\epsilon,
$$
we have
$ Q^{*}(x^{*}_Q , y^{*}_Q )\in B(P^{*}_i,\epsilon)$,
and so $(\theta \circ A)(D)$ is relatively compact.

For each $\epsilon>0$, there exists $\eta>0$, such that
$ \|A(x_1,y_1)-A(x_2,y_2)\|<\epsilon$, for
$\|x_1-x_2\|+\|y_1-y_2\|<\eta$.
Since
\begin{align*}
&\|(\theta \circ A)(x_1,y_1)-(\theta \circ A)(x_2,y_2)\|\\
&=\|\Big(\max\{A_1(x_1,y_1),0\}-\max\{A_1(x_2,y_2),0\}, \\
&\quad \max\{A_2(x_1,y_1),0\}-\max\{A_2(x_2,y_2),0\}\Big)\|\\
&\leq\|A(x_1,y_1)-A(x_2,y_2)\|<\epsilon.
\end{align*}
We have
    $\|(\theta \circ A)(x_1,y_1)-(\theta \circ A)(x_2,y_2)\|<\epsilon$,
for $\|x_1-x_2\|+\|y_1-y_2\|<\eta.$

Hence, $\theta \circ A$ is continuous in $K$ and $\theta \circ A$ is
completely continuous.
The proof is complete.
\end{proof}

 Since $f_i$ is continuous, it is clear that $A:K\to X$ and
$T^{*}:K '\to X$ are completely continuous. From
Lemmas \ref{lem2.6} and  \ref{lem2.5}, we have $T:K\to K$ and $T^*:K'\to K'$ are completely
continuous.

\begin{lemma}\label{lem2.7}
If $(u_1,u_2)$ is a fixed point of $T$, then $(u_1,u_2)$ is a fixed
point of $A$.
\end{lemma}

\begin{proof}
 Suppose $(u_1,u_2)$ is a fixed point of $T$,
obviously, we just need to prove that $A_i (u_1,u_2)(t)\geq 0$,
$i=1,2$, for $t\in [0,1]$.

If there exist $t_0\in (0,1)$ and an $i$ ($i=1,2$) such that
$u_i(t_0)=T_i (u_1,u_2)(t_0)=0$ but $A_i(u_1,u_2)(t_0)<0$. Without
loss of generalization, let $i=1$ and  $(t_1,t_2)$ be the maximal
interval and contains $t_0$ such that $A_1(u_1,u_2)(t)<0$ for all
$t\in(t_1,t_2)$. Obviously, $(t_1,t_2)\neq(0,1)$. Or else,
$T_1(u_1,u_2)(t)=u_1(t)=0$, for all $t\in [0,1]$. This is in
contradiction with (H3).

\textbf{Case i:} If $t_2<1$, then $A_1(u_1,u_2)(t_2)=0$. Thus,
$A_1'(u_1,u_2)(t_2)\geq 0$, We obtain
$$
  A_1''(u_1,u_2)(t)=-f_1(t,0,u_2)\leq 0, \quad\text{for }  t\in (t_1,t_2).
$$
So
$$
  A_1'(u_1,u_2)(t)\geq 0, \quad\text{for } t\in [t_1,t_2]
$$
We obtain $t_1=0$, and $A_1'(u_1,u_2)(0)\geq 0$, $A_1(u_1,u_2)(0)<0$.
This is in contradiction with $A_1(u_1,u_2)(0)=0$.

\textbf{Case ii:} If $t_1>0$, we have $A_1(u_1,u_2)(t_1)=0$. Thus
$A_1'(u_1,u_2)(t_1)\leq 0$. We obtain
$$
  A_1''(u_1,u_2)(t)=-f_1(t,0,u_2)\leq 0, \quad\text{for }  t\in (t_1,t_2).
$$
So
$$
   A_1'(u_1,u_2)(t)<0, \quad\text{for } t\in [t_1,t_2].
$$
We obtain $t_2=1, A_1'(u_1,u_2)(1)\leq 0$.

On the other hand, $A_1(u_1,u_2)(t)<0$, for
$t\in (t_1,t_2), A_1'(u_1,u_2)(1)\leq 0$ imply
$ A_1(u_1,u_2)(1)<0$. By (H1),
$A_1(u_1,u_2)(1)=\int_0^1 g_1(s)u_1(s)ds\geq 0$. This is a
contradiction.
The proof is complete.
\end{proof}

\section{Main result}

Denote
$$
  M_i=\max_{t\in[0,1]}\int_{0}^{1}H_i(t,s)ds,\quad
  m_i=\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}H_i(t,s)ds,
  i=1, 2
$$

\begin{theorem}\label{thm3.1}
Suppose that condition {\rm (H1)--(H3)} hold. Assume that there exist
positive numbers $\delta, a, b, \lambda_i, \mu_i$, $i=1, 2$, such that
$\delta\in (0,\frac{1}{2})$, $0<a<\delta b<b$,
$\lambda_1+\lambda_2\leq 1$,  $\mu_1+\mu_2>1$, and satisfy
\begin{itemize}
\item[(H4)] $f_i(t,u_1,u_2)\geq0$, for $t\in[0,1], u_1+u_2\in[0,b]$;

\item[(H5)] $f_i(t,u_1,u_2)<\frac{\lambda_ia}{M_i}$, for
$t\in [0,1], u_1+u_2\in[0,a]$;

\item[(H6)] $f_i(t,u_1,u_2)\geq\frac{\mu_i\delta b}{m_i}$, for
$t\in[\delta,1-\delta], u_1+u_2\in[\delta b, b]$.

\end{itemize}
Then, \eqref{e1.1} has at least two nonnegative
solutions $(u_1,u_2)$ and $(u_1^{*},u_2^{*})$ such that $0\leq
\|(u_1,u_2)\|<a<\|(u_1^{*},u_2^{*})\|$,
$\alpha(u_1^{*},u_2^{*})<\delta b$.
\end{theorem}

\begin{proof}
 For all $(u_1,u_2)\in \partial K_a$, from
(H5) we have
\begin{align*}
\|T_i(u_1,u_2)\|
&=\max_{t\in[0,1]}(\int_0^1H_i(t,s)f_i(s,u_1(s),u_2(s))ds)^{+}\\
&=\max_{t\in[0,1]}\max\{\int_0^1H_i(t,s)f_i(s,u_1(s),u_2(s))ds,0\}\\
&<\frac{\lambda_ia}{M_i}\max_{t\in[0,1]}\int_0^1H_i(t,s)ds=\lambda_ia.
\end{align*}
Therefore,
\[
\|T(u_1,u_2)\|=\|T_1(u_1,u_2)\|+\|T_2(u_1,u_2)\|
<\lambda_1a+\lambda_2a\leq a.
\]
So (C1) of Theorem \ref{thm2.1} is satisfied.

 For $(u_1,u_2)\in \partial K'_a$, from (H5), we have
\begin{align*}
\|T^*_i(u_1,u_2)\|
&=\max_{t\in[0,1]}\int_0^1H_i(t,s)f^{+}_i(s,u_1(s),u_2(s))ds\\
&<\frac{\lambda_ia}{M_i}\max_{t\in[0,1]}\int_0^1H_i(t,s)ds=\lambda_ia.
\end{align*}
We also obtain
\[
\|T^*_i(u_1,u_2)\|=\|T^*_1(u_1,u_2)\|+\|T^*_2(u_1,u_2)\|
<\lambda_1a+\lambda_2a\leq a.
\]
For $(u_1,u_2)\in \partial K'(\delta b)$, i.e.,
$\alpha(u_1,u_2)=\delta b$, For $t\in [\delta,1-\delta]$, by
Lemma \ref{lem2.5}, we have
$\delta b\leq u_1(t)+u_2(t)\leq b$.
 From (H6), we obtain
\begin{align*}
\alpha(T^*(u_1,u_2))&=\min_{t\in[\delta,1-\delta]}\int_0^1H_1(t,s)f^{+}_1(s,u_1(s),u_2(s))ds\\
&\quad +\min_{t\in[\delta,1-\delta]}\int_0^1H_2(t,s)f^{+}_2(s,u_1(s),u_2(s))ds\\
&\geq \min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}
H_1(t,s)f^{+}_1(s,u_1(s),u_2(s))ds\\
&\quad +\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}
H_2(t,s)f^{+}_2(s,u_1(s),u_2(s))ds\\
&\geq \frac{\mu_1\delta
b}{m_1}\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}
H_1(t,s)ds+\frac{\mu_2\delta
b}{m_2}\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta} H_2(t,s)ds\\
&=\mu_1\delta b+\mu_2\delta b>\delta b.
\end{align*}
 Therefore (C2) of Theorem \ref{thm2.1} is satisfied.

 Finally, we show that (C3) of Theorem \ref{thm2.1} is  satisfied.
 Let $(u_1,u_2)\in K'_a(\delta b)\cap\{(u_1,u_2):
T^*(u_1,u_2)=(u_1,u_2)\}$, we have
$$
  \alpha(u_1,u_2)<\delta b, \|(u_1,u_2)\|>a.
$$
 From Lemma \ref{lem2.5}, we know that
\begin{gather*}
  \|(u_1,u_2)\|\leq \frac{1}{\delta}\alpha(u_1,u_2)<b,\\
  0\leq u_1(t)+u_2(t)<b.
\end{gather*}
 From (H4), we  obtain
$$
  f^{+}_i(s,u_1(s),u_2(s))=f_i(s,u_1(s),u_2(s)).
$$
This implies that $T(u_1,u_2)=T^*(u_1,u_2)$ for
\[
(u_1,u_2)\in K'_a(\delta b)\cap\{(u_1,u_2): T^*(u_1,u_2)=(u_1,u_2)\}.
\]
 By Theorem \ref{thm2.1} and Lemma \ref{lem2.7}, we know that
\eqref{e1.1} has at least two nonnegative solutions
$(u_1,u_2)$ and $(u_1^*,u_2^*)$ such that
$$
  0\leq \|(u_1,u_2)\|<a<\|(u_1^*,u_2^*)\|, \alpha(u_1^*,u_2^*)<b.
$$
The proof is complete.
\end{proof}

Define $\phi:K\to R^+$ by
$$
  \phi(u_1 ,u_2 )=\min_{\delta\leq t\leq 1-\delta}u_1(t)
+\min_{\delta\leq t\leq 1-\delta}u_2(t)
$$

\begin{theorem}\label{thm3.2}
Suppose that condition {\rm (H1)--(H3)} hold. There exist
 $\delta\in (0,\frac{1}{2})$, $a, b, \lambda_i, \mu_i>0$, $i=1, 2$,
such that $0<a<\delta b<b$, $\lambda_1+\lambda_2\leq 1, \mu_1+\mu_2>1$, and
{\rm (H5), (H6)} hold, and satisfy
\begin{itemize}
\item[(H7)] $f_i(t,u_1,u_2)\geq 0$, for $t\in[0,1]$,
$u_1+u_2\in[\delta b,b]$.
\item[(H8)] $f_i(t,u_1,u_2)\leq \frac{\lambda_ib}{M_i}$, for
$t\in [0,1], u_1+u_2\in[0,b]$.
\end{itemize}
Then,  \eqref{e1.1} has at least three nonnegative
solutions $(u_1,u_2)$, $(u_1^*,u_2^*)$, $(u_1^{**},u_2^{**})$, such
that $0\leq \|(u_1,u_2)\|<a<\|(u_1^*,u_2^*)\|, \phi(u_1^*,u_2^*)<b,
\phi(u_1^{**},u_2^{**})\geq b$.
\end{theorem}

\begin{proof}
 Firstly, we prove $T:\overline{K_b} \to \overline{K_b}$ is a completely
continuous operator.
 From (H8), for $i=1,2$, we obtain
\begin{align*}
\|T_i(u_1,u_2)\|
&=\max_{t\in[0,1]}\Big(\int_0^1H_i(t,s)f_i(s,u_1(s),u_2(s))ds\Big)^{+}\\
&=\max_{t\in[0,1]}\max\big\{\int_0^1H_i(t,s)f_i(s,u_1(s),u_2(s))ds,0\big\}\\
&<\frac{\lambda_ib}{M_i}\max_{t\in[0,1]}\int_0^1H_i(t,s)ds=\lambda_ib.
\end{align*}
Therefore,
\[
\|T(u_1,u_2)\|=\|T_1(u_1,u_2)\|+\|T_2(u_1,u_2)\|
<\lambda_1b+\lambda_2b\leq b.
\]
 From Lemma \ref{lem2.6}, we know $T:\overline{K_b} \to \overline{K_b}$ is a
completely continuous operator.
 For the operator $T$ and any $u_1+u_2\in [0,a]$, from (H5) and
Theorem \ref{thm3.1}, we know (C5) of Theorem \ref{thm2.2} is satisfied.

 Next, we show that (C4) of Theorem \ref{thm2.2} is satisfied.
Clearly,
$$
\{(u_1,u_2)\in K(\phi, \delta b, b):\phi(u_1,u_2)>\delta b\}\neq
\emptyset.
$$
Assume $(u_1,u_2)\in K(\phi, \delta b, b)$, for any
$t\in [\delta,1-\delta]$, we have $\delta b\leq u_1+u_2\leq b$. From
(H6) and (H7) we obtain
\begin{align*}
\phi(T(u_1,u_2))
&=\min_{t\in[\delta,1-\delta]}
 \Big(\int_0^1H_1(t,s)f_1(s,u_1(s),u_2(s))ds\Big)^{+}\\
&\quad +\min_{t\in[\delta,1-\delta]}
 \Big(\int_0^1H_2(t,s)f_2(s,u_1(s),u_2(s))ds\Big)^{+}\\
&\geq \min_{t\in[\delta,1-\delta]}\int_{0}^{1}
 H_1(t,s)f_1(s,u_1(s),u_2(s))ds\\
&\quad +\min_{t\in[\delta,1-\delta]}\int_{0}^{1}
 H_2(t,s)f_2(s,u_1(s),u_2(s))ds\\
&\geq \min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}H_1(t,s)f_1(s,u_1(s),u_2(s))ds\\
&\quad +\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}H_2(t,s)f_2(s,u_1(s),u_2(s))ds\\
&\geq \frac{\mu_1\delta
b}{m_1}\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta}
H_1(t,s)ds+\frac{\mu_2\delta
b}{m_2}\min_{t\in[\delta,1-\delta]}\int_{\delta}^{1-\delta} H_2(t,s)ds\\
&=\mu_1\delta b+\mu_2\delta b>\delta b.
\end{align*}
 Finally, for
$(u_1,u_2)\in K(\phi, \delta b, b)$ and $\|T(u_1,u_2)\|>b$, it is
easy to prove that
\[
\phi(T(u_1,u_2))\geq \delta \|T(u_1,u_2)\|>\delta b.
\]
Then (C6) of Theorem \ref{thm2.2} is satisfied.
Therefore from Theorem \ref{thm2.2} and Lemma \ref{lem2.7} we know that
\eqref{e1.1} has at least three nonnegative solutions
$(u_1,u_2)$, $(u_1^*,u_2^*)$, $(u_1^{**},u_2^{**})$, such that
\[
0\leq \|(u_1,u_2)\|<a<\|(u_1^*,u_2^*)\|,\quad
\alpha(u_1^*,u_2^*)<b, \quad
\alpha(u_1^{**},u_2^{**})\geq b.
\]
The proof is complete.\end{proof}

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\end{document}