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

\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations},
 Vol. 2008(2008), No. 74, pp. 1--14.\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-2008/74\hfil Positive solutions]
{Positive solutions for systems of nonlinear
singular differential equations}

\author[Y. Yuan, C. Zhao, Y. Liu\hfil EJDE-2008/74\hfilneg]
{Yanyan Yuan, Chenglong Zhao, Yansheng Liu}  % not in alphabetical order

\address{Yanyan Yuan\newline
Institute of  Systems Science, Chinese Academy of Sciences,
Beijing 100190, China}
\email{yyyuan@amss.ac.cn}

\address{Chenglong Zhao \newline
Department of Mathematics and Physics,
Jinan Engineering Vocational Technical College,
Jinan 250014,  China}
\email{chlzhao666@126.com}

\address{Yansheng Liu \newline
Department of Mathematics, Shandong Normal University, Jinan
250014,  China}
\email{ysliu6668@sohu.com}

\thanks{Submitted January 16, 2008. Published May 21, 2008.}
\thanks{Supported by grants 10571111 from  the NNSF of China, and Y2006A22
 from China  \hfill\break\indent Scholarship
 Council and  Natural Science Foundation of Shandong Province}
\subjclass[2000]{34B16} 
\keywords{Singular boundary value problem; cone; positive Solution}

\begin{abstract}
 By constructing a special cone and using the fixed
 point theorem of cone expansion and compression, this paper
 shows the existence of positive solutions for two-point
 boundary-value problems  of  nonlinear singular
 differential systems. To illustrate the applications of our
 main results, some examples are given.
\end{abstract}

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

\section{Introduction}

 Recently, singular boundary value problems (SBVP for short) have been
studied extensively (see \cite{g1,l2,l3,o1,s1,w1,w2,z1,z2}
 and references therein).
Under the superlinear effect, Wei and Zhang \cite{w2}
obtained  necessary and sufficient conditions for the existence of
$C^2[0,1]$ and $C^3[0,1]$ positive
solutions for fourth-order singular boundary value problems by using the
fixed point theorem of cone
expansion and compression. Under the sublinear effect, Wei \cite{w1}
obtained  necessary and sufficient conditions for the existence of positive
solutions  for fourth-order singular boundary value problems by using
the upper and lower solution method and the maximal principal. However, in
this paper, we will investigate the existence of positive
solutions of second and fourth order singular boundary value
 problems  of nonlinear singular differential
systems.  We obtain  necessary and sufficient conditions for the
existence of $C^2[0,1]\times C[0,1]$ and $C^3[0,1]\times C^1[0,1]$
positive solutions for the coupled systems. Two examples are given
to show the applications of our results.

In this article, we investigate the  boundary-value problem
\begin{equation}
\begin{gathered}
  u^{(4)}=f(t,u,v);\\
 -v''=g(t,u,v);\\
  u(0)=u(1)=u''(0)=u''(1)=0;\\
  v(0)=v(1)=0,
\end{gathered} \label{e1.1}
\end{equation}
where $t\in (0,1),  f, g\in C[(0,1)\times [0,\infty)\times
[0,\infty),[0,\infty)]$; that is, $f, g$ may be singular at
$t=0$ and $t=1$.

Let
\begin{align*}
 E=\big\{&(u,v)\in C^2[0,1]\times C[0,1]:
u(0)=u(1)=u''(0)=u''(1)=0, \\
& v(0)=v(1)=0\big\} ,
\end{align*}
with the norm $\|(u,v)\|=\|u\|_2+\|v\|_0$, where
$ \|u\|_2=\max_{t\in J}|u''(t)|$,
$ \|v\|_0=\max_{t \in J}|v(t)|$, $J=[0,1]$. Then
$(E,\|\cdot\|)$ is a Banach space. In this paper, $E$ will be the
basic space to study \eqref{e1.1}.
Define
\begin{align*}
 P=\Big\{&(u,v)\in E:v(t)\geq t(1-t)v(s),\,
 u(t)\geq t(1-t)u(s),\\
& u(t)\geq -t(1-t)u''(s)/30,\,
 u''(t)\leq t(1-t) u''(s)\leq 0 \text{ and}\\
&u(t), v(t) \text{ are nonnegative concave functions, for all
$t,s \in J$}\Big\}.
\end{align*}
 It is easy to see that $P$ is a cone of $E$.

A pair $(u,v)$ is said to be a $C^2[0,1]\times C[0,1]$ positive solution
 of \eqref{e1.1} if  $u\in C^2[0,1]\cap C^{(4)}(0,1)$,
$ v\in C[0,1]\cap C^2(0,1)$ satisfy \eqref{e1.1} and
$u(t)>0$, $u''(t)\leq0$, $v(t)>0 $ for $ t\in(0,1)$.
In addition, if $(u,v)$ is a $C^2[0,1]\times C[0,1]$ positive solution
of \eqref{e1.1} and both $u'''(0^+)$, $u'''(1^-)$, $v'(0^+)$ and
$v'(1^-)$ exist, then $(u,v)$ is said to be a $C^3[0,1]\times C^1[0,1]$
positive solution of \eqref{e1.1}.

Now, we state a lemma which will be used in Section 2.

\begin{lemma}[\cite{g1}] \label{lem1.1}
Let $P$ be a cone of real Banach space $E$,
$\Omega_1$, $\Omega_2$ be bounded open sets of $E$, and $\theta$
be in $\overline{\Omega}_1\subset\Omega_2$. Suppose that
$A:P\cap(\overline{\Omega}_2\setminus\Omega_1)\to P$ is
completely continuous such that one of the following two conditions
is satisfied:
\begin{itemize}
\item[(i)] $\|Ax\|\leq\|x\|$ for $x\in P\cap\partial\Omega_1$;
$\|Ax\|\geq\|x\|$ for $x\in P\cap\partial\Omega_2$.

\item[(ii)] $\|Ax\|\leq\|x\|$ for $x\in P\cap\partial\Omega_2$;
$\|Ax\|\geq\|x\|$ for $x\in P\cap\partial\Omega_1$.

\end{itemize}
Then, $A$ has a fixed point in
$P\cap(\overline{\Omega}_2\setminus\Omega_1)$.
\end{lemma}

\section{Main Results}

 Let us list some conditions to be used later.
\begin{itemize}
\item[(H1)] $  g\in C[(0,1)\times [0,\infty)\times
 [0,\infty),[0,\infty)]$ and satisfy
 $$
\int_0^1 t(1-t)f(t,t(1-t),1)dt<\infty,\quad
 \int_0^1 t(1-t)g(t,t(1-t),1)dt<\infty .
$$

\item[(H2)]
$$
\int_0^1 f(t,t(1-t),t(1-t))dt<\infty ,\quad
   \int_0^1 g(t,t(1-t),t(1-t))dt<\infty.
$$

\item[(H3)] $f$ is quasi-homogeneous with respect to the last two
 variables, that is, there are constants
$\lambda_1, \mu_1, \alpha _1, \beta_1, N_1, M_1,   N_2, M_2$
 with $0\leq\lambda_1\leq\mu_1<+\infty$,
$0\leq \alpha _1\leq \beta_1\leq1$,
$\mu_1+\beta_1<1$, $ 0<  N_1\leq1\leq  M_1$,
$0<  N_2\leq1\leq  M_2$ such that for
all $0<t<1$, $u\geq 0$, $v\geq0$ satisfying
\begin{itemize}
\item[(a)] $c^{ \mu_1}f(t,u,v)\leq f(t,cu,v)
\leq c^{ \lambda_1 }f(t,u,v)$, $0<c\leq N_1$;\\
$c^{ \lambda_1 }f(t,u,v)\leq f(t,cu,v)\leq c^{\mu_1  }f(t,u,v)$,
$ c\geq M_1$;

\item[(b)] $c^{ \beta_1 }f(t,u,v)\leq f(t,u,cv)\leq
c^{\alpha _1 }f(t,u,v)$, $0<c\leq N_2$;\\
$c^{\alpha _1 }f(t,u,v)\leq f(t,u,cv)\leq c^{\beta_1  }f(t,u,v)$,
$ c\geq M_2$.
\end{itemize}

\item[(H4)] $g$ is quasi-homogeneous with respect to the last two
 variables; that is, there are constants $\lambda_2, \mu_2, \alpha _2,
\beta_2, N_3, M_3, N_4, M_4$
 with $ 0\leq\lambda_2\leq\mu_2<+\infty$,
$0\leq \alpha _2\leq \beta_2\leq1$, $\mu_2+\beta_2<1$,
$ 0<  N_3\leq1\leq  M_3$, $0<  N_4\leq 1\leq  M_4$ such that for
 all $0<t<1$, $u\geq0$, $v\geq0$ satisfying
\begin{itemize}
\item[(a)] $c^{ \mu_2}g(t,u,v)\leq g(t,cu,v)\leq c^{ \lambda_2 }g(t,u,v)$,
  $0<c\leq N_3$;\\
   $c^{ \lambda_2 }g(t,u,v)\leq g(t,cu,v)\leq c^{\mu_2  }g(t,u,v)$,
  $c\geq M_3$;

  \item[(b)] $c^{ \beta_2 }g(t,u,v)\leq g(t,u,cv)\leq c^{\alpha _2 }g(t,u,v)$,
     $0<c\leq N_4$;\\
     $c^{\alpha _2}g(t,u,v)\leq g(t,u,cv)\leq c^{\beta_2  }g(t,u,v)$,
     $c\geq M_4$.
  \end{itemize}
\item[(H5)] There exist $0<\gamma_i<1, k_i\geq0(i=1,2)$ such that
 $$
f(t,u,v)\geq k_1(u+v)^{\gamma_1},\quad
 g(t,u,v)\geq   k_2(u+v)^{\gamma_2}
$$
for any $t\in J, \ (u,v)\in P$.

\end{itemize}
The main results of this paper are as follows.

\begin{theorem} \label{thm2.1}
 Suppose {\rm (H3)-(H5)} hold. Then \eqref{e1.1} has
a $C^2[0,1]\times C[0,1]$ positive solution $(u,v)$, if and only if
{\rm (H1)} holds.
\end{theorem}

\begin{theorem} \label{thm2.2}
 Suppose {\rm (H3)-(H5)} hold. Then \eqref{e1.1}
has a $C^3[0,1]\times C^1[0,1]$ positive solution $(u,v)$, if and
only if {\rm (H2)} holds.
\end{theorem}

 To prove Theorems  \ref{thm2.1} and  \ref{thm2.2}, we need some
 preliminary  lemmas.

\begin{lemma} \label{lem2.1}
The functions $u\in C^2[0,1]\cap C^{(4)}(0,1), v\in
C[0,1]\cap C^2(0,1)$ form a solution to \eqref{e1.1} if and only if
$(u,v)$ is a fixed point of the integral operator
$A(u,v)=(A_1(u,v),A_2(u,v))$ in $C^2[0,1]\times C[0,1]$,
where
\begin{equation}
\begin{gathered}
A_1(u,v)(t)=\int_0^1 G(t,s)\int_0^1 G(s,\tau)f(\tau,u(\tau),v(\tau))
d\tau ds, \\
A_2(u,v)(t)=\int_0^1 G(t,s)g(s,u(s),v(s))ds,\\
G(t,s)=\begin{cases}
  t(1-s), & 0\leq t\leq s\leq1; \\
  s(1-t), & 0\leq s\leq t\leq 1,
\end{cases}
\end{gathered}\label{e2.1}
\end{equation}
\end{lemma}
The proof of the above lemma is obvious; we omit it.

\begin{lemma} \label{lem2.2}
Assume that {\rm (H1), (H3), (H4)} hold.
Then $A:P\to P$ is a completely continuous operator.
\end{lemma}

\begin{proof}
First of all, we show that $A(P)\subset P$. Note that
$G(t,s)\geq t(1-t)G(s,\tau)$ for all $\tau, s\in J$.  Then
 \begin{align*}
A_1(u,v)(t)
&=\int_0^1 G(t,\tau)\int_0^1 G(\tau ,\xi)f(\xi,u(\xi),v(\xi))d\xi d\tau \\
&\geq t(1-t)\int_0^1 G(s,\tau)\int_0^1 G(\tau ,\xi)f(\xi,u(\xi),v(\xi))d\xi d\tau \\
 &= t(1-t)A_1(u,v)(s), \quad \forall t,s\in J, \; (u,v)\in P.
\end{align*}
\begin{align*}
( A_1(u,v))''(t) &=-\int_0^1 G(t,\tau)f(\tau,u(\tau),v(\tau))d\tau \\
& \leq -t(1-t)\int_0^1 G(s,\tau)f(\tau,u(\tau),v(\tau))d\tau \\
&= t(1-t)( A_1(u,v))''(s) \leq0 , \quad \forall t,s \in J,\; (u,v)\in P.
\end{align*}
 \begin{align*}
A_2(u,v)(t)&= \int_0^1 G(t,s)g(s, u(s),v(s))ds \\
&\geq  t(1-t)\int_0^1 G(s,\tau)g(\tau, u(\tau),v(\tau))d\tau \\
&= t(1-t)A_2(u,v)(s),  \quad \forall t,s \in J,\; (u,v)\in P.
\end{align*}
$$
(A_2(u,v))''(t)=-g(t, u(t),v(t))\leq0, \quad \forall t \in J,\;
(u,v)\in P.
$$
It is easy to see that $G(t,s)\geq s(1-s)t(1-t)$. Then
\begin{align*}
A_1(u,v)(t)
&=-\int_0^1 G(t,\tau)( A_1(u,v))''(\tau) d\tau\\
&\geq -\int_0^1 G(t,\tau)\tau(1-\tau)( A_1(u,v))''(s) d\tau\\
&\geq -t(1-t) \int_0^1 \tau^2(1-\tau)^2( A_1(u,v))''(s) d\tau \\
&= -\frac{1}{30}t(1-t)( A_1(u,v))''(s) , \quad \forall t,s \in J, \;
(u,v)\in P.
\end{align*}
Therefore $A(P)\subset P$.

 Next we show that $A$ is bounded. Suppose $V$ is an any bounded set of $P$,
then there exists a $M>0$ such that $\|(u,v)\| \leq M$ for any
$ (u,v)\in V$. It follows from $u(t)=\int_0^1 G(t,s)(-u''(s))ds$
that $u(t)\leq \frac{1}{2}t(1-t)\|u\|_2$.
On the other hand, it follows from $u(t)\geq t(1-t)u(s)$, for all
$t,s\in J$ that $u(t)\geq t(1-t)\|u\|_0$. Hence
\begin{equation}
t(1-t)\|u\|_0\leq u(t)\leq \frac{1}{2}t(1-t)\|u\|_2, \quad t\in J.\label{e2.2}
\end{equation}
 Choose positive numbers $c_1\geq \max \{M_1,\frac{M}{2N_1}\}$ and
$ c_2\geq \max\{M_2, \frac{M}{N_2} \}$. For any $(u,v)\in V$, $t\in J$,
we can get
\begin{align*}
|( A_1(u,v))''(t)|
&= \int_0^1 G(t,s)f(s,u(s),v(s))ds \\
&\leq \int_0^1 s(1-s) f(s,c_1\frac {u(s)}{c_1s(1-s)}s(1-s),
 c_2\frac{v(s)}{c_2})ds\\
&\leq c_1^{\mu_1}c_2^{\beta_1}\int_0^1 (\frac{u(s)}{c_1s(1-s)})^{\lambda_1}
(\frac{v(s)}{c_2})^{\alpha_1}s(1-s) f(s,s(1-s),1)ds\\
&\leq  c_1^{\mu_1}(\frac{1}{2c_1})^{\lambda_1}{\|u\|_2}^{\lambda_1}
 c_2^{\beta_1}\|v\|_0^ {\alpha_1}c_2^{-\alpha_1}\int_0^1 s(1-s)
 f(s,s(1-s),1)ds\\
&\leq 2^{-\lambda_1} c_1^{\mu_1-\lambda_1}c_2^{\beta_1-\alpha_1}
 M^{\lambda_1+\alpha_1}\int_0^1 s(1-s) f(s,s(1-s),1)ds<+\infty.
\end{align*}
Let $c_3\geq\max\{M_3,\frac{M}{2N_3}\}$ and
$ c_4\geq \max\{M_4,\frac{M}{N_4} \}$. For any
$(u,v)\in V$, $t\in J$, we have
\begin{align*}|A_2(u,v)(t)|
&= \int_0^1 G(t,s)g(s,u(s),v(s))ds \\
&\leq \int_0^1 s(1-s)g(s, c_3\frac {u(s)}{c_3s(1-s)}s(1-s),
  c_4\frac{v(s)}{c_4})ds\\
&\leq 2^{-\lambda_2} c_3^{\mu_2-\lambda_2}c_4^{\beta_2-\alpha_2}
  M^{\lambda_2+\alpha_2}\int_0^1 s(1-s) g(s,s(1-s),1)ds<+\infty.
\end{align*}
Consequently,  $A$ is  bounded on $P$.

Thirdly, we show that $AV$ is equicontinuous for arbitrary
 bounded set $V \subset P $. Choose positive
 numbers $c_1\geq\max \{M_1,\frac{M}{2N_1}\}$,
$c_2\geq \max\{M_2, \frac{M}{N_2} \}$, it follows from
$$
A_1(u,v)(t)=\int_0^1 G(t,s)\int_0^1 G(s,\tau)f(\tau,u(\tau),v(\tau)) d\tau ds
$$
that
\begin{align*}
( A_1(u,v))''(t)
&= -\int_0^1 G(t,s)f(s,u(s),v(s))ds\\
&= -\int_{0}^{t}s(1-t)f(s,u(s),v(s))ds-\int_{t}^{1} t(1-s) f(s,u(s),v(s))ds ,
\end{align*}
\begin{align*}
( A_1(u,v))'''(t)
&= \int_{0}^{t}sf(s,u(s),v(s))ds- \int_{t}^{1}(1-s)f(s,u(s),v(s))ds\\
&\leq \int_{0}^{t} sf(s,u(s),v(s))ds+\int_{t}^{1}(1-s)f(s,u(s),v(s))ds\\
&\leq c_0\Big(\int_{0}^{t}sf(s,s(1-s),1)ds+
\int_{t}^{1}(1-s)f(s,s(1-s),1)ds\Big),
\end{align*}
where
$c_0=2^{-\lambda_1}c_1^{\mu_1-\lambda_1}c_2^{\beta_1-\alpha_1}M^{\lambda_1+\alpha_1}$.
Assume
$$
H(t)= c_0\Big(\int_{0}^{t}sf(s,s(1-s),1)ds+
\int_{t}^{1}(1-s)f(s,s(1-s),1)ds\Big).
$$
So we can obtain
\begin{equation}
\begin{aligned}
\int_0^1 H(t)dt
&= c_0\Big(\int_0^1 dt \int_{0}^{t}sf(s,s(1-s),1)ds\\
&\quad +\int_0^1 dt\int_{t}^{1}(1-s)f(s,s(1-s),1)ds\Big) \\
&= 2c_0\int_0^1 s(1-s)f(s,s(1-s),1)ds<+\infty.
\end{aligned} \label{e2.3}
\end{equation}
Thus for any given $t_1,t_2\in J$ with $t_1\leq t_2$ and all
$(u,v)\in V$, we obtain
\[
\| (A_1(u,v))''(t_2)-(A_1(u,v))''(t_1)\|
=|\int_{t_1}^{t_2}A_1(u,v))'''(t) dt|\leq
\int_{t_1}^{t_2}H(t)dt. %\label{e2.4}
\]
 From this inequality, \eqref{e2.3} and the
absolute continuity of integral, it follows that $A_1V$ is
equicontinuous on $J$.

On the other hand
\begin{align*}|
(A_2(u,v))'(t)|
&=-\int_{0}^{t}s g(s,u(s),v(s))ds+ \int_{t}^{1}(1-s)g(s,u(s),v(s))ds\\
&\leq \int_{0}^{t} s g(s,u(s),v(s))d s+ \int_{t}^{1}(1-s)g(s,u(s),v(s))ds
\end{align*}
Let $G(t)= \int_{0}^{t} s g(s,u(s),v(s))d s+
\int_{t}^{1}(1-s)g(s,u(s),v(s))ds, c_3\geq\max \{M_3,\frac{M}{2N_3}\}$
and $ c_4\geq \max\{M_4, \frac{M}{N_4}\}$,
then
\begin{equation}
\begin{aligned}
\int_0^1  G(t)dt
&= 2\int_0^1 s(1-s) g(s,u(s),v(s))d s \\
&\leq  2^{1-\lambda_2} c_3^{\mu_2-\lambda_2}c_4^{\beta_2-\alpha_2}
M^{\lambda_2+\alpha_2}\int_0^1 s(1-s) g(s,s(1-s),1)ds<+\infty.
\end{aligned} \label{e2.5}
\end{equation}
 Thus for any given $t_1,t_2\in J$ with $t_1\leq t_2$ and all $(u,v)\in V$,
 we obtain
\[
 \| A_2(u,v)(t_2)-A_2(u,v)(t_1)\|
=\big|\int_{t_1}^{t_2}(A_2(u,v))'(t) dt\big|\leq
\int_{t_1}^{t_2}G(t)dt. %\label{e2.6}
\]
From this inequality, \eqref{e2.5}, and the absolute continuity of
integral, it follows that $A_2V$ is equicontinuous on $J$. Therefore
$AV$ are equicontinuous on $J$. It follows from the
 Ascoli-Arzela theorem that $A_1V$ and $A_2V$ is relatively compact.

Finally, we show that $A:P\to P$ is a continuous operator.
Notice that $A$ is continuous on $C^2[0,1]\times C[0,1]$ if and only
if $A_1$ is continuous on $C^2[0,1]$ and $A_2$ is continuous on
$C[0,1]$.

Suppose $\{(u_n,v_n)\}\subset P, (u,v) \in P$ and $\|u_n-u\|_2\to
0$, $\| v_n-v\|_0\to 0$ as $n\to\infty$. It follows from
\eqref{e2.2} that $\|u\|_0\leq\frac{1}{2}\|u\|_2$. So we can get
$\|u_n-u\|_0\to0(n\to\infty)$ from $\|u_n-u\|_2\to0(n\to\infty)$.
Then $u_n(t)\to u(t)$ and $v_n(t)\to v(t)$ as $n\to\infty$ uniformly
with respect to $t\in J$.
 Therefore
\begin{align*}
&| (A_1(u_n,v_n))''(t)-  (A_1(u,v))''(t)|\\
&= \big|\int_0^1 G(t,s)f(s,u_n(s),v_n(s))ds-\int_0^1 G(t,s)f(s,u(s),v(s))ds\big| \\
&\leq \int_0^1  G(t,s)\big|f(s,u_n(s),v_n(s))- f(s,u(s),v(s))\big|ds.
\end{align*}

 From (H1), (H3) and the Lebesgue dominated convergence
theorem, it follows that
$$
 |( A_1(u_n,v_n))''(t)- ( A_1(u,v))''(t)|\to0 \quad \text{as }
n\to\infty.
$$
Hence one can conclude that
$$
\|A_1(u_n,v_n)-  A_1(u,v)\|_2\to0 \quad \text{as } n\to\infty.
$$
In fact, if this is not true, then there exist $\epsilon_0$ and
$\{u_{n_i}\}\subset\{u_n\}, \{v_{n_i}\}\subset\{v_n\}$ such that $\|
A_1(u_{n_i},v_{n_i})-  A_1(u,v)\|_2 \geq\epsilon_0$ ($i=1,2\dots$).
Since $ \{ A_1(u_n,v_n)\}$ is relatively compact, there exists a
sequence of  $ \{ A_1(u_n,v_n)\}$ which convergence in $C^2[0,1]$ to
some $y$. Not loss of generality, we may assume that
 $\{A_1(u_{n_i},v_{n_i})\}$ itself converge to $y$, then
 $y=A_1(u,v)$.  This is a contradiction. Consequently $A_1$ is
 continuous. In the same way, we can get $A_2$ is continuous, too.
This completes the proof.
\end{proof}

  \begin{lemma}[\cite{l1}] \label{lem2.3}
 Suppose $(u,v)\in P$ and $\mu \in (0,  \frac{1}{2})$.
Then $u(t)+v(t)\geq\mu(1-\mu)(\|u\|_0+\|v\|_0), t\in [\mu, 1-\mu]$.
\end{lemma}

The  proof  of the above lemma is  obvious; we omit it.

In the following we prove Theorem \ref{thm2.1}.

\begin{proof}
\textbf{Sufficiency.}
  From Lemma \ref{lem2.3}, we can choose $\mu=\frac{1}{4}$.
Then   $u(t)+v(t)\geq\frac{3}{16}(\|u\|_0+\|v\|_0),  t\in [\frac{1}{4},
  \frac{3}{4}]$. First of all, we
 prove
\begin{equation}
\|A(u,v)\|\geq\|(u,v)\|, \quad \forall (u,v)\in \partial P_r,\label{e2.7}
\end{equation}
where $P_r=\big\{\|(u,v)\|<r\big\}$,
\begin{align*}
r \leq\min\Big\{&2N_1, N_2, 2N_3, N_4,
{\big(\frac {k_1}{2{(160)}^{\gamma_1}}
\int_{1/4}^{3/4}[s(1-s)]^{1+\gamma_1}ds\big)}
^\frac{1}{1-\gamma_1},\\
&{\big(\frac {k_2}{2{(160)}^{\gamma_2}}
\int_{1/4}^{3/4}[s(1-s)]^{1+\gamma_2}ds
\big)}^\frac{1}{1-\gamma_2} \Big\}.
\end{align*}
 From the definition of $P$, we know that
$\|u\|_0\geq \frac{1}{30}t(1-t)\|u\|_2$ for any
$(u,v)\in P, t\in J$.
By condition  (H5) and Lemma \ref{lem2.3}, we obtain
\begin{align*}
-(A_1(u,v))''(t)&= \int_0^1 G(t,s)f(s,u(s),v(s))ds\\
 &\geq \frac{1}{4}\int_{1/4}^{3/4}s(1-s)f(s,u(s),v(s))ds\\
 &\geq \frac{1}{4}\int_{1/4}^{3/4}k_1s(1-s)(u(s)
   +v(s))^{\gamma_1}ds \\
 &\geq \frac{k_1}{4}\int_{1/4}^{3/4}s(1-s)(\frac{3}{16}
   (\|u\|_0+\|v\|_0))^{\gamma_1} ds \\
 &\geq \frac{k_1}{4(160)^{\gamma_1}}r^{\gamma_1}
  \int_{1/4}^{3/4}[s(1-s)]^{1+\gamma_1}ds\\
 &\geq \frac{r}{2}=\frac {\|(u,v)\|}{2}, \quad
   \forall t\in J, (u,v)\in \partial P_r.
\end{align*}
Consequently
$$
\|A_1(u,v)\|_2\geq \frac {\|(u,v)\|}{2}, \quad
\forall(u,v)\in \partial P_r.
$$
 For any $t\in J$, $(u,v)\in \partial P_r$, by virtue of (H5)
and Lemma \ref{lem2.3}, one can see
\begin{align*}
 A_2(u,v)(t)
&=\int_0^1 G(t,s)g(s,u(s),v(s))ds \\
&\geq \frac{1}{4}\int_{1/4}^{3/4}s(1-s))g(s,u(s),v(s))ds\\
&\geq  \frac{k_2}{4}\int_{1/4}^{3/4}s(1-s)
  (\frac{3}{16}(\|u\|_0+\|v\|_0))^{\gamma_2}ds \\
&\geq \frac{k_2}{4(160)^{\gamma_2}}r^{\gamma_2}
  \int_{1/4}^{3/4}[s(1-s)]^{1+\gamma_2}ds\\
&\geq \frac{r}{2}=\frac {\|(u,v)\|}{2}, \quad
  \forall t\in J, (u,v)\in  \partial P_r.
\end{align*}
 Therefore,  $\|A_2(u,v)\|_0\geq \frac {\|(u,v)\|}{2}$,
for all $(u,v)\in \partial P_r$. Consequently,  \eqref{e2.7} holds.

Next we claim that
\begin{equation}
\|A(u,v)\|\leq\|(u,v)\|, \forall (u,v)\in \partial P_R,  \label{e2.8}
\end{equation}
where
\begin{align*}
R\geq\max\Big\{&2N_1M_1,\; M_2N_2,\; 2N_3M_3,\; M_4N_4,\\
&\big(2^{1-\mu_1}N_1^{\lambda_1-\mu_1}N_2^{\alpha_1
 -\beta_1}\int_0^1 s(1-s)f(s,s(1-s),1)ds\big)^\frac{1}{1-(\mu_1+\beta_1)},\\
&\big(2^{1-\mu_2}N_3^{\lambda_2-\mu_2}N_4^{\alpha_2-\beta_2}\int_0^1
s(1-s)g(s,s(1-s),1)ds\big)^\frac{1}{1-(\mu_2+\beta_2)}\Big\},
\end{align*}
$P_R=\big\{\|(u,v)\|<R\big\}$.

Let $c_1=\frac{R}{2N_1}$ and $c_2=\frac{N_2}{R}$. Then for any
$ (u,v)\in \partial P_R$, by virtue of (H3), we have
 \begin{align*}
 -(A_1(u,v))''(t)
&= \int_0^1 G(t,s)f(s,u(s),v(s))ds\\
&\leq\int_0^1 s(1-s)f(s,u(s),v(s))ds\\
&=\int_0^1 s(1-s)f(s,c_1\frac{u(s)}{c_1s(1-s)}s(1-s),\frac{c_2v(s)}{c_2})ds\\
&\leq \int_0^1 s(1-s) c_1^{\mu_1}{(\frac{u(s)}{c_1s(1-s)})}
  ^{\lambda_1}{(\frac{1}{c_2})}^{\beta_1}{(c_2v(s))}^{\alpha_1}f(s,s(1-s),1)ds\\
&\leq 2^{-\lambda_1}c_1^{\mu_1-\lambda_1}c_2^{\alpha_1-\beta_1}
  R^{\lambda_1+\alpha_1}\int_0^1 s(1-s)f(s,s(1-s),1)ds\\
&= 2^{-\mu_1}N_1^{\lambda_1-\mu_1}N_2^{\alpha_1-\beta_1}
  R^{\mu_1+\beta_1}\int_0^1 s(1-s)f(s,s(1-s),1)ds\\
&\leq  \frac{R}{2}=\frac {\|(u,v)\|}{2}.
\end{align*}
Therefore, for any $(u,v)\in \partial P_R$, we have
 $$\|A_1(u,v)\|_2\leq \displaystyle\frac{\|(u,v)\|}{2}.$$

 For any $(u,v)\in \partial P_R, t\in J$, by virtue of
 (H4), one can also see
\begin{align*}
A_2(u,v)(t)
&= \int_0^1 G(t,s)g(s,u(s),v(s))ds\\
&\leq \int_0^1 s(1-s)g(s,u(s),v(s))ds\\
&\leq 2^{-\mu_2}N_3^{\lambda_2-\mu_2}N_4^{\alpha_2-\beta_2}R^{\mu_2+\beta_2}
 \int_0^1 s(1-s)g(s,s(1-s),1)ds\\
&\leq \frac{R}{2}=\frac{\|(u,v)\|}{2}.
\end{align*}
Then for any $(u,v)\in \partial P_R$, we have $$\|A_2(u,v)\|_0\leq
\displaystyle\frac{\|(u,v)\|}{2}.$$

 Consequently, \eqref{e2.8} holds.
By Lemma \ref{lem1.1} and Lemma \ref{lem2.2}, we obtain that $A$ has a fixed point
$(u,v)$ in $\overline{P_R}\setminus P_r$ and satisfies $u''(t)<0$,
$u(t)>0$, $v(t)>0$, for all $t\in (0,1)$.
\smallskip

\noindent\textbf{Necessity.}
Let $u\in C^2[0,1]\cap C^{(4)}(0,1), v\in
C[0,1]\cap C^2(0,1)$ be a positive solution of \eqref{e1.1}. It
follows from $u(0)=u(1)=0$ and $ u''(t)\leq0$ for $t\in J$ that
there exist $0<m_1<1<m_2$ such that $m_1t(1-t)\leq u(t)\leq
m_2t(1-t)$. In the same way, there also exist $0<n_1<1<n_2$ such
that $n_1t(1-t)\leq v(t)\leq n_2t(1-t)$. There exists $t_0\in (0,1)$
such that $v'(t_0)=0$. This together with $v''(t)\leq 0$ for $t\in
(0,1)$ yields that $v'(t)\geq0$ as $t\in(0,t_0)$ and $v'(t)\leq0$ as
$ t\in(t_0,1)$. Choose positive numbers
$c_3\leq\min\{N_3,\frac{1}{m_2M_3}\}, c_4\leq\min\{N_4,
\frac{1}{M_4n_2}\}$. Then
\begin{align*}
g(t,t(1-t),1)
&= g(t, c_3\frac{t(1-t)}{c_3u(t)}u(t), c_4\frac{1}{c_4v(t)}v(t))\\
&\leq c_3^{\lambda_2}{(\frac{t(1-t)}{c_3u(t)})}^{\mu_2}
  c_4^{\alpha_2}{(\frac{1}{c_4v(t)})}^{\beta_2}g(t,u(t),v(t))\\
&\leq c_3^{\lambda_2}{(\frac{1}{c_3m_1})}^{\mu_2}
  c_4^{\alpha_2-\beta_2}{(\frac{1}{v(t)})}^{\beta_2}g(t,u(t),v(t))\\
&= c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}{(v(t))}^{-\beta_2}
 g(t,u(t),v(t)).
\end{align*}
Namely,
\begin{equation}
{(v(t))}^{\beta_2}g(t,t(1-t),1)\leq
c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}g(t,u(t),v(t)).
\label{e2.9}
\end{equation}
Hence, integrate \eqref{e2.9} from $t_0$ to $t$ to obtain
\begin{align*}
\int_{t_0}^{t}{(v(s))}^{\beta_2}g(s,s(1-s),1)ds
&\leq c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}
 \int_{t_0}^{t}g(s,u(s),v(s))ds\\
&= c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}v'(t).
\end{align*}
Since $v(t)$ is decreasing on $[t_0,1]$, we get
$$
{(v(t))}^{\beta_2} \int_{t_0}^{t}g(s,s(1-s),1)ds\leq
-c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}v'(t);
$$
namely,
\begin{equation}
\int_{t_0}^{t}g(s,s(1-s),1)ds\leq
-c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}
\frac{v'(t)}{{(v(t))}^{\beta_2}}.\label{e2.10}
\end{equation}
Note that $\beta_2<1$, then integrate \eqref{e2.10} from $t_0$ to $1$ to have
$$
\int_{t_0}^{1}dt\int_{t_0}^{t}g(s,s(1-s),1)ds\leq
-c_3^{\lambda_2-\mu_2}c_4^{\alpha_2-\beta_2}m_1^{-\mu_2}\int_{t_0}^{1}
\frac{v'(t)}{{(v(t))}^{\beta_2}}dt.
$$
Therefore,
$$
\int_{t_0}^{1}(1-s)g(s,s(1-s),1)ds\leq
c_3^{\lambda_2-\mu_2}c_4^{\alpha_2
-\beta_2}m_1^{-\mu_2}{(1-\beta_2)}^{-1}{(v(t_0))}^{1-\beta_2}<\infty.
$$
On the other hand, we can also prove
$$
\int_{0}^{t_0}sg(s,s(1-s),1)ds<\infty.
$$
Thus
$$
\int_0^1 s(1-s)g(s,s(1-s),1)ds<\infty.
$$

Next, we prove that
$$
\int_0^1 s(1-s)f(s,s(1-s),1)ds<\infty.
$$
Let $c_1\leq\min\{N_1,\frac{1}{m_2M_1}\}, c_2\leq\min\{N_2,
\frac{1}{M_2n_2}\}$, then
\begin{align*}
f(t,t(1-t),1)
&= f(t, c_1\frac{t(1-t)}{c_1u(t)}u(t), c_2\frac{1}{c_2v(t)}v(t))\\
&\leq c_1^{\lambda_1}{(\frac{t(1-t)}{c_1u(t)})}^{\mu_1}
  c_2^{\alpha_1}{(\frac{1}{c_2v(t)})}^{\beta_1}f(t,u(t),v(t))\\
&\leq c_1^{\lambda_1}{(\frac{1}{c_1m_1})}^{\mu_1}
  c_2^{\alpha_1-\beta_1}{(\frac{1}{v(t)})}^{\beta_1}f(t,u(t),v(t))\\
&= c_1^{\lambda_1-\mu_1}c_2^{\alpha_1-\beta_1}m_1^{-\mu_1}{(v(t))}^{-\beta_1}
  f(t,u(t),v(t)).
\end{align*}
There exists $t_0\in (0,1)$ such that $u'''(t_0)=0$ from
$u''(0)=u''(1)=0$. This together with $u^{(4)}\geq0$ for
$t_0\in (0,1)$ yields that $u'''(t)\geq0$ as $t\in(0,t_0)$ and
$u'''(t)\leq0$ as $t\in(t_0,1)$. Integrate
$u^{(4)}(t)=f(t,u(t),v(t))$ from $t$ to $t_0$, we can get
$$
-u^{(3)}(t)=\int_{t}^{t_0}f(s,u(s),v(s))ds, t\in (0,t_0).
$$
However,
 \begin{align*}
\int_{0}^{t_0}t f(t,t(1-t),1)dt
&= \int_{0}^{t_0}dt\int_{t}^{t_0}f(s,s(1-s),1)ds\\
&\leq \int_{0}^{t_0}\int_{t}^{t_0}c_1^{\lambda_1-\mu_1}
 c_2^{\alpha_1-\beta_1}m_1^{-\mu_1}{(v(s))}^{-\beta_1}f(s,u(s),v(s))ds.
\end{align*}
On the other hand
\begin{align*}
v(t)&= \int_0^1 G(t,s)g(s,u(s),v(s))ds\\
& \geq \frac{1}{4}\int_{1/4}^{3/4}
  s(1-s)g(s,s(1-s)c_5 \frac{u(s)}{c_5s(1-s)},c_6\frac{v(s)}{c_6})ds.
\end{align*}
 Let $c_5\geq\max \{M_1,\frac{m_2}{N_1}\}$,
$c_6\geq\max\{M_2,\frac{\|v\|_0}{N_2}\}$. Then
$$
v(t)\geq\frac{1}{4}c_5^{\lambda_2-\mu_2}c_6^{\alpha_2
-\beta_2}{m_1}^{\mu_2}\int_{1/4}^{3/4}s(1-s){((v(s))}^{\beta_2}
g(s,s(1-s),1)ds.
$$
However, one can see that $v(t)\geq\frac{1}{16}\|v\|_0$ as
$t\in[\frac{1}{4},\frac{3}{4}]$.
 Hence, for  $t\in J$,
$$
v(t)\geq 2^{-(2+4\beta_2)}c_5^{\lambda_2-\mu_2}
 c_6^{\alpha_2-\beta_2}{m_1}^{\mu_2}\|v\|_0^{\beta_2}
 \int_{1/4}^{3/4}s(1-s)g(s,s(1-s),1)ds.
$$
Let
\begin{align*}
k_0&=c_1^{\lambda_1-\mu_1}c_2^{\alpha_1-\beta_1}{m_1}^{-\mu_1}
\Big(2^{-(2+4\beta_2)}c_5^{\alpha_2-\mu_2}
 c_6^{\alpha_2-\beta_2}{m_1}^{\mu_2}\|v\|_0^{\beta_2} \\
&\quad\times \int_{1/4}^{3/4}s(1-s)g(s,s(1-s),1)ds\Big)^{-\beta_1}.
\end{align*}
Thus
$$
\int_{0}^{t_0}t f(t,t(1-t),1)dt\leq
k_0\int_{0}^{t_0}dt\int_{t}^{t_0}f(s,u(s),v(s))ds=k_0(-u''(t_0))<\infty.
$$
In the same way, we can also prove
$$
\int_{t_0}^{1}(1-t)f(t,t(1-t),1)dt<+\infty.
$$
Hence
$$
\int_0^1 t(1-t)f(t,t(1-t),1)dt<+\infty.
$$
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm2.2}]
\textbf{Sufficiency.}
 First of all, we prove that
$$
\int_0^1 f(t,t(1-t),t(1-t))dt<+\infty
$$
implies
$$
\int_0^1 t(1-t)f(t,t(1-t),1)dt<+\infty.
$$
Choose positive number $c\geq\max\{M_2,\frac{1}{4N_2}\}$. Then
\begin{align*}
f(t,t(1-t),t(1-t))
&= f(t,t(1-t),c \frac{t(1-t)}{c})\\
&\geq c^{\alpha_1-\beta_1}{(t(1-t))}^{\beta_1}
f(t,t(1-t),1)\\
&\geq c^{\alpha_1-\beta_1}t(1-t)f(t,t(1-t),1).
\end{align*}
Consequently, we can get
$$
\int_0^1 f(t,t(1-t),t(1-t))dt\geq
c^{\alpha_1-\beta_1}\int_0^1 t(1-t)f(t,t(1-t),1)dt,
$$
namely, $$\int_0^1t(1-t)f(t,t(1-t),1)dt<+\infty.$$

On the other hand, we can also prove that
$$
\int_0^1 g(t,t(1-t),t(1-t))dt<+\infty.
$$
This implies
$$
\int_0^1 t(1-t)g(t,t(1-t),1)dt<+\infty.
$$
 From above inequalities, we know that \eqref{e1.1} exists a
$C^2[0,1]\times C[0,1]$ positive solution $(u,v)$. Therefore it
suffices to show that $u'''(0^+), u'''(1^-), v'(0^+)$ and
$v'(1^-)$ exist. The same reason as the proof of Theorem \ref{thm2.1} of
necessity asserts that there exist $0<m_1<1<m_2$ and $ 0<n_1<1<n_2$
satisfying $m_1t(1-t)\leq u(t)\leq m_2t(1-t)$ and
$n_1t(1-t)\leq v(t)\leq n_2t(1-t), t\in J$.

 Let $c_1\geq\max\{M_1,\frac {m_2}{N_1}\}$ and
$c_2\geq\max\{M_2,\frac {n_2}{N_2}\}$, then we have
\begin{align*}
\int_0^1 |u^{(4)}(t)|dt
&= \int_0^1 f(t,u(t),v(t))dt\\
&= \int_0^1  f(t, c_1\frac {u(t)}{c_1t(1-t)}t(1-t),
  c_2\frac {v(t)}{c_2t(1-t)}t(1-t))dt\\
&\leq c_1^{\mu_1}{(\frac{u(t)}{c_1t(1-t)})}^{\lambda_1}
  c_2^{\beta_1}{(\frac{v(t)}{c_2t(1-t)})}^{\alpha_1}
  \int_0^1 f(t,t(1-t),t(1-t))dt\\
&\leq c_1^{\mu_1-\lambda_1}c_2^{\beta_1
 -\alpha_1}{m_2}^{\lambda_1}{n_2}^{\alpha_1}\int_0^1
 f(t,t(1-t),t(1-t))dt<+\infty.
\end{align*}
This guarantees $u'''(0^+)$ and $u'''(1^-)$ exist.

On the other hand,
 let  $c_3\geq\max\{M_3,\frac {m_2}{N_3}\}$ and
$c_4\geq\max\{M_4,\frac {n_2}{N_4}\}$, then
\begin{align*}
\int_0^1 |-v''(t)|dt
&= \int_0^1 g(t,u(t),v(t))dt\\
&= \int_0^1 g(t,c_3\frac {u(t)}{c_3t(1-t)}t(1-t),
 c_4\frac {v(t)}{c_4t(1-t)}t(1-t))dt\\
&\leq c_3^{\mu_2-\lambda_2}c_4^{\beta_2-\alpha_2}{m_2}^{\lambda_2}{n_2}^{\alpha_2}
 \int_0^1 g(t,t(1-t),t(1-t))dt<+\infty.
\end{align*}
This means that $v'(0^+)$ and $v'(1^-)$ exist.
\smallskip

\noindent\textbf{Necessity.}
 Let $(u,v)$ be a $C^3[0,1]\times C^1[0,1]$ positive
solution of \eqref{e1.1}. The same reason as the beginning of the proof
of sufficiency asserts that there exist $0<m_1<1<m_2$ and
$0<n_1<1<n_2$ satisfying $m_1t(1-t)\leq u(t)\leq m_2t(1-t)$ and
$n_1t(1-t)\leq v(t)\leq n_2t(1-t), t\in J$. Suppose
$c_1\leq\min\{N_1,\frac{1}{M_1m_2}\}$,
$c_2\leq\min\{N_2,\frac{1}{M_2n_2}\}$,
$c_3\leq\min\{N_3,\frac{1}{M_3m_2}\}$ and
$c_4\leq\min\{N_4,\frac{1}{M_4n_2}\}$.
Then we have
\begin{align*}
f(t,t(1-t),t(1-t))
&= f(t, c_1\frac {t(1-t)}{c_1u(t)}u(t), c_2\frac {t(1-t)}{c_2v(t)}v(t))\\
&\leq {c_1}^{\lambda_1}{(\frac {t(1-t}{c_1u(t)})}^{\mu_1}
  {c_2}^{\alpha_1}{(\frac {t(1-t}{c_1v(t)})}^{\beta_1}f(t,u(t),v(t))\\
&\leq {c_1}^{\lambda_1-\mu_1}{c_2}^{\alpha_1-\beta_1}{m_1}^{-\mu_1}
 {n_1}^{-\beta_1}f(t,u(t),v(t)).
\end{align*}
 Consequently,
$$
\int_0^1 f(t,t(1-t),t(1-t))dt\leq{c_1}^{\lambda_1
-\mu_1}{c_2}^{\alpha_1-\beta_1}{m_1}^{-\mu_1}{n_1}^{-\beta_1}
[ u'''(1^-)-u'''(0^+)]<+\infty.
$$
On the other hand, we can also prove
$$
\int_0^1 g(t,t(1-t),t(1-t))dt\leq{c_3}^{\lambda_2-\mu_2}{c_4}^{\alpha_2
 -\beta_2}{m_1}^{-\mu_2}{n_1}^{-\beta_2}
[ v'(1^-)-v'(0^+)]<+\infty.
$$
Therefore, our conclusion follows.
\end{proof}

In the following we give some examples to illustrate the theorems
obtained in Section 2.

\begin{example} \label{exa1} \rm
 Consider \eqref{e1.1} with
$$
f(t,u,v)=p(t)u^{10}v^{1/3}+(u+v)^{1/3},\quad
g(t,u,v)=a(t)u^{2}v^{2/5}+(u+v)^{2/5},
$$
where $p, a\in C[(0,1),R^+]$ and
\begin{gather*}
\int_0^1 [p(t)(t(1-t))^{11}+(t(1-t))(t(1-t)+1)^{1/3}]dt<+\infty,\\
\int_0^1 [a(t)(t(1-t))^{3}+(t(1-t))(t(1-t)+1)^{2/5}]dt<+\infty.
\end{gather*}
It is obvious that $f, g$ satisfy  (H3)-(H5). So it is easy to
see, by Theorem \ref{thm2.1}, that \eqref{e1.1} has a $C^2[0,1]\times C[0,1]$
positive solution.
\end{example}

\begin{example} \label{exa2}\rm
 In \eqref{e1.1}, let
$$
f(t,u,v)=q(t)u^{3}v^{1/4}+(u+v)^{1/4},\quad
g(t,u,v)=e(t)u^{3}v^{2/3}+(u+v)^{2/3},
$$
where $q, e\in C[(0,1),R^+]$ and
\begin{gather*}
\int_0^1 [q(t){(t(1-t))}^{13/4}+(2t(1-t))^{1/4}]dt<+\infty,\\
\int_0^1 [e(t){(t(1-t))}^{11/3}+(2t(1-t))^{2/3}]dt<+\infty.
\end{gather*}
It is obvious that $f, g$ satisfy (H3)-(H5). So it is easy to
see, by Theorem \ref{thm2.2}, that \eqref{e1.1} has a $C^3[0,1]\times C^1[0,1]$
positive solution.
\end{example}


\subsection*{Acknowledgements} The authors are grateful to the
anonymous referees for their helpful comments.


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\end{thebibliography}

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