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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 13, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/13\hfil Positive solutions]
{Positive solutions for a second-order system with
integral boundary conditions}

\author[W. Song, W. Gao\hfil EJDE-2011/13\hfilneg]
{Wenjing Song, Wenjie Gao}

\address{Wenjing Song \newline
Institute of Mathematics, Jilin University, Changchun 130012, China.
Institute of Applied Mathematics, Jilin University of Finance and Economics, 
Changchun 130017, China}
\email{swj-78@163.com}

\address{Wenjie Gao \newline
Institute of Mathematics, Jilin University, Changchun 130012,
China}
\email{wjgao@jlu.edu.cn}

\thanks{Submitted November 29, 2010. Published January 26, 2011.}
\thanks{Supported by grant 10771085 from NSFC,  by
Key Lab of Symbolic Computation and \hfill\break\indent
Knowledge Engineering of Ministry of Education, and by the
985 program of Jilin University}
\subjclass[2000]{34B15, 34B27}
\keywords{Positive solution; integral boundary condition;
 fixed point theorem}

\begin{abstract}
 This article concerns the existence of
 positive solutions to a second-order system with integral boundary
 conditions. By applying Krasnoselskii fixed point
 theorem, we show the existence of solutions
 under certain conditions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}

\section{Introduction}

In this article, we investigate the existence of positive solutions to
the following system of second order ordinary differential equations
with integral boundary conditions:
\begin{equation}\label{e1.1}
\begin{gathered}
x''(t)=-f(t,x(t),y(t)),\quad
(t,x,y)\in(0,1)\times[0,+\infty)\times[0,+\infty),\\
y''(t)=-g(t,x(t),y(t)),\quad (t,x,y)\in(0,1)\times[0,+\infty)\times[0,+\infty),\\
x(0)-ax'(0)=\int_0^1 \varphi_0(s)y(s)ds,\quad
x(1)+bx'(1)=\int_0^1 \varphi_1(s)y(s)ds,\\
y(0)-ay'(0)=\int_0^1 \psi_0(s)x(s)ds,\quad
y(1)+by'(1)=\int_0^1 \psi_1(s)x(s)ds,
\end{gathered}
\end{equation}
where $f,g\in C([0,1]\times[0,+\infty)\times[0,+\infty),[0,+\infty))$,
$\varphi_0,\varphi_1,\psi_0,\psi_1\in C([0,1],[0,+\infty))$,
$a$ and $b$ are positive real parameters.

Boundary value problems with positive solutions describe many
phenomena in the applied sciences found in the theory of nonlinear
diffusion generated by nonlinear sources, thermal ignition of gases,
and concentration in chemical or biological problems. Readers may
refer to \cite{b2,c1,g1} for details. In the past few years, much effort has
been devoted to the study of the existence of positive solutions to
ordinary differential equations or systems with different kinds of
boundary conditions, see \cite{a1,l1,l2,l3,m1}.

On the other hand, problems with integral boundary conditions arise
naturally in thermal conduction problems \cite{c2},
semiconductor problems \cite{i1} and hydrodynamic problems
\cite{c3}. Many
authors have investigated scalar problems with integral boundary
conditions; see for instance \cite{b1,k1,y2,z1}.
Particularly, in \cite{b1},  Boucherif discussed the following
boundary value problem with
integral boundary condition:
\begin{equation} \label{e1.2}
\begin{gathered}
y''(t)=f(t,y(t)),\quad  0<t<1,\\
y(0)-ay'(0)=\int_0^1 g_0(s)y(s)ds, \\
y(1)-by'(1)=\int_0^1 g_1(s)y(s)ds.
\end{gathered}
\end{equation}
He obtained the existence of positive solutions of Problem \eqref{e1.2} by
applying Krasnoselskii fixed point theorem in a cone.

However, to the best of our knowledge, there seem to be quite few
works on ordinary differential systems of second order with integral
boundary conditions. In 2005, Yang \cite{y1} studied the following system
with integral boundary condition:
\begin{equation} \label{e1.3}
\begin{gathered}
-u''(t)=f(t,u,v),\qquad \qquad -v''(t)=g(t,u,v), \\
u(1)=H_1(\int_0^1u(\tau)d\alpha(\tau)),\quad v(1)
=H_2(\int_0^1v(\tau)d\beta(\tau)),\\
u(0)=v(0)=0,
\end{gathered}
\end{equation}
where $\alpha$ and $\beta$ are increasing nonconstant functions
defined on $[0, 1]$ with $\alpha(0)=0=\beta(0)$; $f\in C([0,
1]\times R^+ \times R^+ ,R^+)$ and $g\in C([0, 1]\times R^+ \times
R^+ ,R^+)$; and $H_i\in C(R^+ ,R^+ )(i =1,2)$. Here
$\int_0^1u(\tau)d\alpha(\tau)$ and $\int_0^1v(\tau)d\beta(\tau)$
denote the Riemann-Stieltjes integrals. By using the fixed point
index theory in a cone and a priori estimates as the main tools in
the proofs, he proved the existence of positive solutions to \eqref{e1.3}.


Motivated by the works mentioned above, we intend to study the
existence of positive solutions of Problem \eqref{e1.1}. Compared with the
scalar case \eqref{e1.2} and Problem \eqref{e1.3}, the characteristic of \eqref{e1.1} is
that the two exponents $x$ and $y$ are coupled not only in the
equations, but also on the boundary conditions, which make the
resolvent kernel $R(t,s)$ in our system much more complicated. This
in turn brings substantial difficulties in proving the complete
continuity of the operator $T$ (see Section 2 for its definition).
Therefore our results cannot be routinely deduced from the ones of
\eqref{e1.2} and \eqref{e1.3} in the above literature. The outline of this paper
is as follows. We present some preliminaries in Section 2 and the
main results are proved in Section 3. In Section 4 we will give two
examples to illustrate our results.

 \section{Preliminaries}

 In this section, we present some  propositions and lemmas that will be
 used in the proof of our main results.

 We shall denote by  $C[0,1]$ the Banach Space consisting of all continuous
functions on $[0,1]$ equipped with the
 standard norm
 $$
\| u \|=\max_{0\leq t\leq 1}| u(t)|,
$$
and equip the Banach space $C[0,1]\times C[0,1]$ with the
standard norm
 $$
\| (u,v) \|=\| u \|+\| v \|=\max_{0\leq t\leq 1}| u(t)|
+\max_{0\leq t\leq 1} | v(t)|.
$$
We will use the following assumptions:
\begin{itemize}
\item[((H0)] $f,g\in  C([0,1]\times[0,+\infty)\times[0,+\infty),
[0,+\infty))$, $\varphi_{i}, \psi_{i}\in
 C([0,1],[0,+\infty))$, $i=1,2$, $a$ and $b$ are positive real
 parameters.

\item[(H1)] $\varphi_0, \varphi_1$ are continuous, positive and
 the auxiliary function
 $$
\Phi(t,s)=\frac{1}{1+a+b}[(1+b-t)\varphi_0(s)
+(a+t)\varphi_1(s)],\quad t,s\in[0,1],
$$
 satisfies
 $$
0\leq m_{\Phi}:=\min\{\Phi(t,s):t,s\in[0,1]\}
\leq M_{\Phi}:=\max\{\Phi(t,s):t,s\in  [0,1]\}<1.
$$

\item[(H2)] $\psi_0,\psi_1$ are continuous, positive functions
on $[0,1]$ and
the auxiliary function
 $$
\Psi(t,s)=\frac{1}{1+a+b}[(1+b-t)\psi_0(s)+(a+t)\psi_1(s)],\quad
 t,s\in[0,1],
$$
 satisfies
 $$
0\leq m_{\Psi}:=\min\{\Psi(t,s):t,s\in[0,1]\}
\leq M_{\Psi}:=\max\{\Psi(t,s):t,s\in  [0,1]\}<1.
$$
\end{itemize}

Evidently, $(x,y)\in C^2(0,1)\times C^2(0,1)$ is a solution of
Problem \eqref{e1.1} if and only if $(x,y)\in C[0,1]\times C[0,1]$ is a
solution to the system of integral equations
\begin{equation}\label{e2.1}
\begin{aligned}
x(t)&=\int _0^1  G(t,s)f(s,x(s),y(s))ds
+\frac{1+b-t}{1+a+b}\int_0^1 \varphi_0(s)y(s)ds\\
&\quad +\frac{a+t}{1+a+b}\int_0^1 \varphi_1(s)y(s)ds, \quad t\in[0,1],
\\
y(t)&=\int _0^1  G(t,s)g(s,x(s),y(s))ds
+\frac{1+b-t}{1+a+b}\int_0^1 \psi_0(s)x(s)ds\\
&\quad +\frac{a+t}{1+a+b}\int_0^1 \psi_1(s)x(s)ds,\quad t\in[0,1],
\end{aligned}
\end{equation}
where for $(t,s)\in [0,1]\times [0,1]$,
\[
G(t,s)=
\begin{cases}
k_1(t)k_2(s), & 0\leq t\leq s,\\
k_1(s)k_2(t), & 0\leq s\leq t\,,
\end{cases}
\]
where
$$
k_1(t)=a+t,\quad  k_2(t)=\frac{1+b-t}{1+a+b}.
$$

It is clear that $k_1(t)>0$ and $k_2(t)>0$ for all $t\in[0,1]$,
and $G(t,s)>0$ for all $(t,s)\in[0,1]\times[0,1]$. Moreover, we have
the following propositions:

\begin{proposition} \label{prop2.1}
 There exists a positive continuous function
$\gamma:[0,1]\to\mathbb{R}$ such that
$G(t,s)\geq \gamma(t)G(s,s)$  for all $t,s\in[0,1]$. Moreover,
$\gamma_0:=\min\{\gamma(t):t\in[0,1]\}>0$.
\end{proposition}

 The proof of the above proposition is similar to \cite[Lemma 2]{b1},
and we omit it here.

\begin{proposition} \label{prop2.2}
Under assumption {\rm (H0)}, for all
$t,s\in[0,1]$, we have $G(t,s)\leq G(s,s)$.
\end{proposition}

The of the above proposition follows standard argument, it is
omitted here.

Let us denote two operators $A, B:=C[0,1]\times
C[0,1]\to C[0,1]$ as follows:
\begin{gather*}
A(x,y)(t)=\int_0^1 G(t,s)f(s,x(s),y(s))ds+\int_0^1 \Phi(t,s)y(s)ds,\\
B(x,y)(t)=\int_0^1 G(t,s)g(s,x(s),y(s))ds+\int_0^1 \Psi(t,s)x(s)ds.\\
\end{gather*}
Then we  define an operator $T:C[0,1]\times C[0,1]\to
C[0,1]\times C[0,1]$ as
\begin{equation} \label{e2.2}
Tz(t)=\int_0^1 H(t,s)F(s,x(s),y(s))ds
+\int_0^1 K(t,s)z(s)ds
=\begin{pmatrix}
A(x,y)(t)\\
B(x,y)(t)
\end{pmatrix},
\end{equation}
where
\begin{gather*}
z(t)=\begin{pmatrix}
x(t)\\
y(t)
\end{pmatrix}, \; (x,y)\in C[0,1]\times C[0,1],\quad
H(t,s)=\begin{pmatrix}
G(t,s)&0\\
0&G(t,s)
\end{pmatrix},
\\
F(s,x(s),y(s))=\begin{pmatrix}
f(s,x(s),y(s))\\
g(s,x(s),y(s))
\end{pmatrix} \quad
 K(t,s)=\begin{pmatrix}
0&\Phi(t,s)\\
\Psi(t,s)&0
\end{pmatrix}.
\end{gather*}

It is clear that the existence of a positive solution for
\eqref{e2.1} is equivalent to the existence of a nontrivial
fixed point of $T$ in $C[0,1]\times C[0,1]$.
To obtain a positive solution of  \eqref{e1.1}, we need the following
lemma.

\begin{lemma} \label{lem2.1}
Assume {\rm (H0)--(H2)} hold. Then
$T:C[0,1]\times C[0,1]\to C[0,1]\times C[0,1]$ is a
completely continuous operator.
\end{lemma}

\begin{proof}
Firstly, we prove that $T$ is a compact operator.
That is, for any bounded subset $D\subset C[0,1]\times C[0,1]$, we
show that $T(D)$ is relatively compact in $C[0,1]\times C[0,1]$.
Since $D\subset C[0,1]\times C[0,1]$ is a bounded subset, there
exists a constant $\overline{M}>0$ such that
$\| z\|=\| x\|+\| y\|\leq
\overline{M}$ for any $z\in D$.

By applying (H0), (H1) and
Proposition 2.2, we obtain
\begin{align*}
\| A(x,y)\|
&=\max_{0\leq t\leq 1}\big|\int_0^1 G(t,s)f(s,x(s),y(s))ds
+\int_0^1 \Phi(t,s)y(s)ds\big|\\
&\leq L\int_0^1 | G(s,s)| ds+M_{\Phi}\overline{M}<+\infty.
\end{align*}
Here $L=\max\{f(t,x,y):0\leq t\leq1,| x|\leq \overline{M},
| y|\leq \overline{M}\}+\max\{g(t,x,y):0\leq t\leq1,|
x|\leq \overline{M},| y|\leq \overline{M}\}$.
Similarly, we can obtain
$$
\| B(x,y)\|\leq L\int_0^1 | G(s,s)|
ds+M_{\Psi}\overline{M}<+\infty.
$$
Then from the definition of the
norm of the product space $C[0,1]\times C[0,1]$, we have
\begin{align*}
\| T(z)\|&=\| A(x,y)\|+\|B(x,y)\| \\
&\leq 2L\int_0^1 | G(s,s)| ds+(M_{\Phi}+M_{\Psi})\overline{M}
<+\infty.
\end{align*}
Therefore, $T(D)$ is
uniformly bounded with the norm of $C[0,1]\times C[0,1]$.
Moreover, for any $t\in(0,1)$, we have
\begin{align*}
&|\frac{d}{dt} A(x,y)(t)|\\
&=\Big| \Big(\int_0^{t}G(t,s)f(s,x(s),y(s))ds+
\int_{t}^1 G(t,s)f(s,x(s),y(s))ds\Big)' \\
&\quad -\frac{1}{1+a+b}\int_0^1 \varphi_0(s)y(s)ds
+\frac{1}{1+a+b}\int_0^1 \varphi_1(s)y(s)ds\Big|\\
& =\Big|
\frac{1}{1+a+b}\Big[-\int_0^1 sf(s,x(s),y(s))ds
-a\int_0^{t}f(s,x(s),y(s))ds \\
&\quad+(b+1)\int_{t}^1 f(s,x(s),y(s))ds
-\int_0^1 \varphi_0(s)y(s)ds+\int_0^1 \varphi_1(s)y(s)ds\Big]\Big|\\
&\leq\frac{1}{1+a+b}[(2+a+b)L+2K\overline{M}]<+\infty ,
\end{align*}
where
\begin{align*}
K&=\max\{\varphi_0(t):0\leq t\leq1\}+\max\{\varphi_1(t):0\leq t\leq1\}\\
&\quad + \max\{\psi_0(t):0\leq t\leq1\}
+\max\{\psi_1(t):0\leq t\leq1\}.
\end{align*}
Thus, it is easy to prove that $A(D)$ is equicontinuous.
 This together with the
Arzel\'a-Ascoli theorem guarantees that $A(D)$ is relatively
compact in $C[0,1]$.

Similarly, we can prove that $B(D)$ is relatively compact in
$C[0,1]$. Therefore, $T(D)$ is relatively compact in $C[0,1]\times
C[0,1]$. On the other hand, according to the definition of $T$, it
is easily seen that $T$ is continuous. We obtain that $T$ is
completely continuous. The proof is complete.
\end{proof}

We shall discuss the existence of a positive solution of
\eqref{e1.1} by using the following fixed point theorem of
cone expansion and compression.

\begin{lemma}[{\cite[Theorem 4]{b1}}] \label{lem2.2}
Let $E$ be a Banach space and
$K\subset E$ be a cone. Suppose $\Omega_1$ and $\Omega_2$ are
two bounded open sets in Banach space $E$ such that
$\theta\in\Omega_1$, $\overline{\Omega_1}\subset \Omega_2$ and
suppose that the operator
$T:K\cap(\overline{\Omega_2}\setminus\Omega_1)\to K$
is completely continuous such that
\begin{itemize}
\item[(A1)] $\| Tx\|\leq \| x\|$ for all
$x\in K\cap\partial\Omega_1$ and
$\| Tx\|\geq \| x\|$ for all $x\in K\cap\partial\Omega_2$ or

\item[(A2)] $ \| Tx\|\geq \| x\|$ for all
$x\in K\cap\partial\Omega_1$ and
$\| Tx\|\leq \| x\|$ for all $x\in K\cap\partial\Omega_2$.
\end{itemize}
Then $T$ has a fixed point in
$K\cap(\overline{\Omega_2}\setminus\Omega_1)$.
\end{lemma}

To use Lemma \ref{lem2.2}, let $E=C[0,1]\times C[0,1]$,
$$
P=\{u\in C[0,1],\ u(t)\geq 0,\ t\in[0,1]\},
$$
and
$$
P_0=\{(u,v)\in P\times P,\min_{0\leq t\leq 1}((u,v))= \min_{0\leq t\leq 1}(u(t)+v(t))\geq\frac{1-M}{1-m}\gamma_0\|
(u,v)\|\},
$$
where
$$
M=\max\{M_{\Phi},M_{\Psi}\},\quad
m=\min\{m_{\Phi},m_{\Psi}\}.
$$
It is easy to see that $P_0$ is a cone in $E$.

\begin{lemma} \label{lem2.3}
Under Assumptions {\rm (H0)--(H2)}, the operator
$T:P_0\to P_0$ is a completely continuous.
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.1}, we only need to prove that
$T(P_0)\subset P_0$.
Define an operator $N:C[0,1]\times C[0,1]\to
C[0,1]\times C[0,1]$ by
$N(z)(t)=\int_0^1 K(t,s)z(s)ds$. Then
$N(P\times P)\subset P\times P$.
Noting that
\begin{align*}
 \| Nz(t)\|
&=\max_{0\leq t\leq 1}|\int_0^1 \Phi(t,s)y(s)ds|+
\max_{0\leq t\leq 1}|\int_0^1 \Psi(t,s)x(s)ds|\\
&\leq \max\{M_{\Phi},M_{\Psi}\}\| z\|,
\end{align*}
one has $\| N \|\leq \max\{M_{\Phi},M_{\Psi}\}<1$.
Then $I-N$ is invertible.
Similarly to \cite[Lemma 3]{b1}, we have
$$
Tz(t)=\int_0^1 H(t,s)F(s,x(s),y(s))ds
+\int_0^1 R(t,s)\int_0^1 H(s,\tau)F(\tau,x(\tau),y(\tau))d\tau
ds,
$$
where $R(t,s)$ is the resolvent kernel by
$R(t,s)=\sum_{j=1}^{\infty}K_{j}(t,s)$ and
$K_{j}(t,s)=\int_0^1 K (t,\tau)K _{j-1}(\tau,s)d\tau$,
$j=2,3,\dots$, and $K _1(t,s)=K(t,s)$.
Let
\[
R(t,s)=\begin{pmatrix}
R_1(t,s) & R_2(t,s)\\
R_3(t,s) & R_4(t,s)
\end{pmatrix}.
\]
It can be easily verified that
\begin{gather*}
\frac{m^2}{1-m^2}\leq R_1(t,s),\;
R_4(t,s)\leq\frac{M^2}{1-M^2},\\
\frac{m}{1-m^2}\leq R_2(t,s),\;
R_3(t,s)\leq\frac{M}{1-M^2}\,.
\end{gather*}
From Propositions 2.1, 2.2 and (H1), we find that
\begin{equation} \label{e2.3}
\begin{aligned}
\|Tz(t)\|
&\leq \frac{1}{1-M}[\int_0^1 G(\tau,\tau)f(\tau,x(\tau),y(\tau))d\tau\\
&\quad + \int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau],
\end{aligned}
\end{equation}
\begin{equation} \label{e2.4}
\begin{aligned}
\min_{0\leq t\leq 1}Tz(t)
&\geq \frac{\gamma_0}{1-m}[\int_0^1 G(\tau,\tau)f(\tau,x(\tau),y(\tau))
d\tau\\
&\quad +\int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau].
\end{aligned}
\end{equation}
By \eqref{e2.3} and \eqref{e2.4}, we have
$$
\min_{0\leq t\leq1}Tz(t)\geq \frac{1-M}{1-m}\gamma_0\|
Tz(t)\|.
$$
Therefore, $T(P_0)\subset P_0$.
\end{proof}

 \section{Main results}

In this section, we show the existence of positive solutions to
 \eqref{e1.1}. Firstly, we introduce some notation.
\begin{gather*}
f_{\beta}=\liminf_{| x| +|
y|\to\beta}\min_{0\leq t\leq 1}\frac{f(t,x,y)}{| x| +| y|},\quad
f^{\beta}=\limsup_{| x| +|
y|\to\beta}\max_{0\leq t\leq 1}\frac{f(t,x,y)}{| x| +| y|},\\
g_{\beta}=\liminf_{| x| +|
y|\to\beta}\min_{0\leq t\leq 1}\frac{g(t,x,y)}{| x| +| y|},\quad
g^{\beta}=\limsup_{| x| +|y|\to\beta}\max_{0\leq t\leq
1}\frac{g(t,x,y)}{| x| +| y|},
\end{gather*}
where $\beta=0$ or $\infty$.

\begin{theorem} \label{thm3.1}
Assume that {\rm (H0)--(H2)} hold. If
$$
f^{0},g^{0}<\frac{1-M}{2\int_0^1 G(s,s)ds}\quad\text{and}\quad
f_{\infty},g_{\infty}>\frac{(1-m)^2}{2\gamma_0^2(1-M)
\int_0^1 G(s,s)ds},
$$
then Problem \eqref{e1.1} has at least one positive solution.
\end{theorem}

\begin{proof}
 Since $f^{0},g^{0}<\frac{1-M}{2\int_0^1 G(s,s)ds}$, there exists an
$r>0$, such that
$f(t,x,y)\leq (f^{0}+\varepsilon_1)(| x|+| y |)$, and
$g(t,x,y)\leq (g^{0}+\varepsilon_1)(|x|+| y |)$ for
$t\in[0,1]$, $| x|+| y |\leq r$,
where $\varepsilon_1$ satisfies
$f^{0}+\varepsilon_1\leq\frac{1-M}{2\int_0^1 G(s,s)ds}$ and
$g^{0}+\varepsilon_1\leq\frac{1-M}{2\int_0^1 G(s,s)ds}$.

Let $\Omega_1=\{z=(x,y)\in P\times P,\ \| z\|<r\}$.
For any $z=(x,y)\in \partial \Omega_1\cap P_0$, we have
\begin{equation} \label{e3.1}
\begin{split}
\| Tz\|&\leq
\frac{1}{1-M}\int_0^1 G(s,s)ds\cdot
(f^{0}+\varepsilon_1+g^{0}+\varepsilon_1)\cdot\|(x,y)\|\\
&\leq \| z\|.
\end{split}
\end{equation}
On the other hand, since
$f_{\infty},g_{\infty}>\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}
$, there exists an $R>r>0$, such that
$f(t,x,y)\geq (f_{\infty}-\varepsilon _2)(| x|+| y|)$ and
$g(t,x,y)\geq (g_{\infty}-\varepsilon _2)(| x|+| y|)$
for $t\in [0,1],\ | x|+| y|\geq R$, where
$\varepsilon_2$ satisfies $f_{\infty}-\varepsilon _2\geq
\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$ and
$g_{\infty}-\varepsilon _2\geq
\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$.

Let $\Omega_2=\{z=(x,y)\in P\times P,\ \|
z\|<R_1\}$, where $R_1=\frac{1-m}{(1-M)\gamma_0}R$.
For any $z=(x,y)\in \partial \Omega_2\cap P_0$, we have
\begin{equation} \label{e3.2}
\begin{split}
\| Tz(t)\|\geq\min _{0\leq t\leq 1}Tz(t)
&\geq \frac{\gamma_0^2(1-M)}{(1-m)^2}\int_0^1 G(s,s)ds\cdot
(f_{\infty}-\varepsilon _2+g_{\infty}-\varepsilon _2)\cdot
\| z\|\\
&\geq\| z\|.
\end{split}
\end{equation}
Applying Lemma \ref{lem2.2} to \eqref{e3.1} and \eqref{e3.2} yields
that $T$ has a fixed point
$z^{*}\in P_0\cap (\overline{\Omega_2\setminus}\Omega_1)$
and hence $z^{*}$ is a positive solution of \eqref{e1.1}.
\end{proof}

From the proof of Theorem 3.1, we can also obtain the
following result.

\begin{theorem} \label{thm3.2}
Assume that {\rm (H0)--(H2)} hold. If
$f^{\infty},g^{\infty}<\frac{1-M}{2\int_0^1 G(s,s)ds}$ and
$f_0,g_0>\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}
$, then  \eqref{e1.1} has at least one positive solution.
\end{theorem}

Next we discuss the multiplicity of positive solutions for Problem
\eqref{e1.1}. We obtain the following results.

\begin{theorem} \label{thm3.3}
Assume that {\rm (H0)--(H2)} hold, and
\begin{itemize}
\item[(i)]
$f_0>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$
and
$g_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$;

\item[(ii)] There exists an $l>0$ such that $\max_{0\leq
t\leq1,(x,y)\in\partial\Omega_1}f(t,x,y)<\frac{1-M}{2\int_0^1 G(s,s)ds}
l$ and $\max_{0\leq
t\leq1,(x,y)\in\partial\Omega_1}g(t,x,y)
<\frac{1-M}{2\int_0^1 G(s,s)ds}l$, where
$\Omega_1:=\{z=(x,y)\in P\times P,\|z\|<l\}$.
\end{itemize}
Then Problem \eqref{e1.1} has at least two positive
solutions.
\end{theorem}

\begin{proof}
Since $f_0>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$,
we can choose $\varepsilon_3>0$ such that
$f_0-\varepsilon_3\geq
\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, and also
there exists an $0<l_1<l$, such that $f(t,x,y)\geq
(f_0-\varepsilon_3)(| x|+| y|)$ for $t\in[0,1],
| x|+| y|\leq l_1$. Let
$\Omega_{l_1}:=\{z=(x,y)\in
P\times P, \| z\|<l_1\}$, For any
$z=(x,y)\in\partial \Omega_{l_1}\cap P_0$, we have
\begin{equation} \label{e3.3}
\begin{split}
\| Tz\|& \geq\frac{\gamma_0}{1-m}\int_0
^1 G(s,s)ds\cdot(f_0-\varepsilon_3)(\frac{1-M}{1-m}\gamma_0\|
z\|) \\
&\geq\| z\|.
\end{split}
\end{equation}

Again, by using $g_{\infty}\geq
\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, we can
choose $\varepsilon_4>0$, such that
$g_{\infty}-\varepsilon_4\geq
\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$, and also
there exists an $l_2>l$, such that $g(t,x,y)\geq
(g_{\infty}-\varepsilon_4)(| x|+| y|)$ for $t\in[0,1],
| x|+| y|\geq l_2$. Let
$\Omega_{\widetilde{l}_2}:=\{z=(x,y)\in P\times P, \|
z\|<\widetilde{l}_2\}$, where
$\widetilde{l}_2=\frac{1-m}{(1-M)\gamma_0}l_2$. For any
$z=(x,y)\in\partial \Omega_{\widetilde{l}_2}\cap P_0$, we have
\begin{equation} \label{e3.4}
\begin{split}
\| Tz\|&\geq \frac{\gamma_0}{1-m}\int_0
^1 G(s,s)ds\cdot(g_{\infty}-\varepsilon_4)(\frac{1-M}{1-m}\gamma_0\|
z\|) \\
&\geq\| z\|.
\end{split}
\end{equation}

By (ii), for $z=(x,y)\in\partial\Omega_1\cap P_0$, we have
\begin{equation} \label{e3.5}
\begin{split}
\|Tz(t)\|&\leq\frac{1}{1-M}[\int_0^1 G(\tau,\tau)
f(\tau,x(\tau),y(\tau))d\tau+
\int_0^1 G(\tau,\tau)g(\tau,x(\tau),y(\tau))d\tau] \\
&< \frac{1}{1-M} \int_0^1 G(s,s)ds\cdot\frac{1-M}{\int
_0^1 G(s,s)ds}l=l.
\end{split}
\end{equation}
Therefore, from \eqref{e3.3}, \eqref{e3.5} and Lemma \ref{lem2.2},
it follows that \eqref{e1.1} has at least one positive
solution $z_1\in P_0$ with $l_1\leq \| z_1\|<l$.
Similarly, from \eqref{e3.4}, \eqref{e3.5} and Lemma \ref{lem2.2},
it follows that \eqref{e1.1} has at least one
positive solution $z_2\in P_0$ with
$l< \| z_2\|\leq \widetilde{l}_2$. Therefore,  \eqref{e1.1} has at
least two positive solutions. The proof is complete.
\end{proof}

Similarly, we have the following results.

\begin{theorem} \label{thm3.4}
Assume  {\rm (H0)--(H2)}, and
\begin{itemize}
\item[(i)] $f_0,f_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)
\int_0^1 G(s,s)ds}$
or
$g_0,f_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$
or
$g_0,g_{\infty}>\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}$;

\item[(ii)] There exists an $l>0$ such that
$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}f(t,x,y)
<\frac{1-M}{2\int_0^1 G(s,s)ds} l$ and
$\max_{0\leq t\leq1,(x,y)\in\partial\Omega_1}g(t,x,y)
<\frac{1-M}{2\int_0^1 G(s,s)ds} l$, where
$\Omega_1:=\{z=(x,y)\in P\times P,\|z\|<l\}$.
\end{itemize}
Then Problem \eqref{e1.1} has at least two positive
solutions.
\end{theorem}

 \section{Examples}

\begin{example} \label{exa4.1} \rm
Set $f(t,x,y)=\sqrt{\frac{1+t}{8}}(x^2+y^2)$,
$g(t,x,y)=\sqrt{1-\frac{t}{4}}[(x^2+y^2)^2+(x^2+y^2)e^{-(x^2+y^2)}]$,
$a=1$, $b=1$, $\varphi_{i}=\psi_{i}=1/3$, $i=0,1$.
Then
$\int_0^1 G(s,s)ds=13/6$, $\gamma_0=1/6$,
$M=m=1/3$,
$$
\frac{1-M}{2\int_0^1G(s,s)ds}=\frac{2}{13},\quad
\frac{(1-m)^2}{2\gamma_0^2(1-M)\int_0^1 G(s,s)ds}=\frac{72}{13}.
$$
Then conditions of Theorem 3.1 are satisfied. We obtain that Problem
\eqref{e1.1} has at least one positive solution.
\end{example}

\begin{example} \label{exa4.2} \rm
Set
$f(t,x,y)=\sqrt{\frac{1+t}{32}}\cdot\sqrt[3]{x^2+y^2}$,
$g(t,x,y)=\frac{2-t}{208}(x^2+y^2)(1+e^{-(x^2+y^2)})$,
$a=1$, $b=1$, $l=6$, $\varphi_{i}=\psi_{i}=1/3$, $i=0,1$.
Then
$\int_0^1 G(s,s)ds=13/6$, $\gamma_0=1/6$,
$M=m=1/3$,
$$
\frac{1-M}{2\int_0^1 G(s,s)ds}=\frac{2}{13},\quad
\frac{(1-m)^2}{\gamma_0^2(1-M)\int_0^1 G(s,s)ds}=\frac{144}{13}.
$$
Then the conditions of Theorem 3.3 are satisfied.
We obtain that Problem \eqref{e1.1} has at least two positive
solution.
\end{example}

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\end{document}