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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 19, 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 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/19\hfil Existence of positive solutions]
{Existence of positive solutions for systems of bending elastic
beam equations}

\author[P. Kang, Z. Wei \hfil EJDE-2012/19\hfilneg]
{Ping Kang, Zhongli Wei}

\address{Ping Kang \newline
Department of Mathematics, Tianjin Polytechnic University,
Tianjin, 300160, China}
\email{jnkp1980@163.com, Tel: 86-22-24493329}

\address{Zhongli Wei \newline
School of  Mathematics, Shandong University, Jinan, Shandong,
250100,  China\newline
Department of Mathematics, Shandong   Jianzhu University,
 Jinan,  Shandong, 250101,  China}
\email{jnwzl@yahoo.com.cn}

\thanks{Submitted September 5, 2011. Published January 31, 2012.}
\thanks{Supported  by grants 10971046 from the NNSF-China,
and ZR2009AM004 the  NSF \hfill\break\indent of Shandong Province}
\subjclass[2000]{34B15, 39A10}
\keywords{Positive solutions; fixed points; cone}

\begin{abstract}
 This article discusses the existence of positive solutions
 for systems of bending elastic beam equations. In mechanics,
 the problem describes the deformations of two elastic beams
 in equilibrium state, whose two ends are simply supported.
\end{abstract}

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

\section{Introduction}

The deformations of two elastic beams in equilibrium state, whose
two ends are simply supported, can be described by the
systems of bending elastic beam equations:
\begin{equation}
\begin{gathered}
u^{(4)}(t)=f_1(t,u(t),v(t),u''(t),v''(t)),
\quad 0<t<1,\\
v^{(4)}(t)= f_2(t,u(t),v(t),u''(t),v''(t)),\quad 0<t<1,\\
u(0)=u(1)=u''(0)=u''(1)=0,\\
v(0)=v(1)=v''(0)=v''(1)=0,
\end{gathered} \label{e1.1}
\end{equation}
where $f_1,f_2: I\times \mathbb{R}^{+}\times
\mathbb{R}^{+}\times \mathbb{R}^{-}\times \mathbb{R}^{-}\to
\mathbb{R}^{+}$ are continuous functions, and the $u'',v''$ in
$f_1$ and $f_2$ are the bending moment terms which represent
bending effect, $I=[0,1],\mathbb{R}^{+}=[0,+\infty)$,
$\mathbb{R}^{-}=(-\infty,0]$.

In recent years, due to its importance in physics, some authors
(see \cite{k1,l1,w1,w2})
have studied the existence of solutions to the equation
\begin{equation}
\begin{gathered}
u^{(4)}(t)=f(t,u(t),u''(t)),\quad  0<t<1,\\
u(0)=u(1)=u''(0)=u''(1)=0.
\end{gathered}
\label{e1.2}
\end{equation}
 Naturally, further study in this specific field is on
the system of fourth-order ordinary differential equations. However,
to our knowledge, results for systems of fourth-order ordinary
differential equations are rarely seen (see \cite{l2,x1}). For example, In
\cite{l2}, by applying the fixed-point theorem of cone expansion and
compression type due to Krasnosel'skii, the authors show the
existence of single and multiple positive solutions of the singular
boundary value problems for systems of nonlinear fourth-order
differential equations of the form
\begin{gather}
  u^{(4)}(t)= a_1(t)f_1(t,u(t),v(t),u''(t),v''(t))
+b_1(t)g_1(t,u(t),v(t),u''(t),v''(t)),
\nonumber \\
v^{(4)}(t)= a_2(t)f_2(t,u(t),v(t),u''(t),v''(t))
+b_2(t)g_2(t,u(t),v(t),u''(t),v''(t)), \nonumber\\
0<t<1,  \nonumber 
\\
u(0)=u(1)=v(0)=v(1)=0, \label{e1.3} \\
\alpha_1 u''(0)-\beta_1 u'''(0)=0,\quad
\gamma_1 u''(1)+\delta_1 u'''(1)=0, \nonumber \\
\alpha_2 v''(0)-\beta_2 v'''(0)=0,\quad
\gamma_2 v''(1)+\delta_2 v'''(1)=0, \nonumber
\end{gather}
where $f_i, g_i$ satisfied some weaker conditions and are
continuous; $a_i(t)$ and $b_i(t)$ are allowed to be singular at
$t=0$ or $t=1, i=1,2$.

In the above articles, it is always supposed that the
nonlinear terms satisfy the superlinear and sublinear conditions, or
some weaker conditions which are similar to them (see \cite{l2,x1}).
Therefore, the purpose of this paper is to improve these results. We
shall employ the theory of the fixed point index in cones to present
some precise conditions on $f_1$ and $f_2$ guaranteeing the
existence of positive solutions of the system \eqref{e1.1}.

Moreover, in this paper, we study the existence of positive
solutions for system \eqref{e1.1} in the case that the nonlinear
terms have the different features.
However, it is difficult to directly
construct proper open sets in a single cone in product space.
Therefore, we will construct a cone $K_1\times K_2$ which is the
Cartesian product of two cones in space $C^2[0,1]$ and choose the
proper open sets $O=O_1\times O_2\subset K_1\times K_2$. Applying
the product formula for the fixed point index on product cone and
the fixed point index theory, we obtain the existence of positive
solutions for system \eqref{e1.1}.

This paper is organized as follows. In Section 2, we present some
preliminaries and main result. In Section 3, we present some basic
lemmas that will be used to prove our main result. In Section 4, we
will prove the main result in Section 2.


\section{Preliminaries and main result}

In this Section, we will give some useful preliminary results and
change the system \eqref{e1.1} into the fixed point problem
in a cone which is the Cartesian product of two cones.

We shall consider the Banach space $C^2[0,1]$ equipped with the
 norm
$$
\|u\|_2=\|u\|+\|u''\|=\max_{0\leq t\leq1}|u(t)|
+\max_{0\leq t\leq1}|u''(t)|,
$$
and the Banach space $C^2[0,1]\times C^2[0,1]$ equipped with the
 norm
$$
\|(u,v)\|_2=\|u\|_2+\|v\|_2.
$$
Let $G(t,s)$ be the Green function to the linear boundary value
problem
$$
-u''=0,\quad u(0)=u(1)=0.
$$
which is explicitly expressed by
\begin{equation}
G(t,s)=\begin{cases} t(1-s), & 0\leq t\leq s\leq 1,\\
s(1-t), &0\leq s\leq t\leq 1.
\end{cases} \label{e2.1}
\end{equation}
It is clear that
\begin{equation}
\begin{gathered}
G(t,s)>0,\quad 0<t,s<1,\\
G(t,t)G(s,s)\leq G(t,s)\leq G(s,s)=s(1-s),\quad t, s\in I.
\end{gathered}
\label{e2.2}
\end{equation}
For convenience, we now introduce the notation, for $r>0$,
\[
K_r=\{u\in K: \|u\|_2<r \},\quad
\partial K_r=\{u\in K:\|u\|_2=r\},\\
\overline{K_r}=\{u\in K:\|u\|_2\leq r \},
\]
and
\begin{align*}
K=\big\{&u\in C^2[0,1]: u(t)\geq 0,u''(t)\leq 0,\;
u(t)\geq q(t)\|u\|,\\
&u''(t)\leq -q(t) \|u''\|,\; t\in I\big\},
\end{align*}
where $q(t)=t(1-t)$.
It is easy to prove that $K$ is a cone in
$C^2[0,1]\times C^2[0,1]$.

Let us list the following assumptions:
\begin{itemize}
\item[(H1)] $f_1,f_2: I\times \mathbb{R}^{+}\times
\mathbb{R}^{+}\times \mathbb{R}^{-}\times \mathbb{R}^{-}\to
\mathbb{R}^{+}$ are continuous functions;

\item[(H2)] there exist $h_1\in C(I\times \mathbb{R}^{+}\times
\mathbb{R}^{-},\mathbb{R}^{+})$, such that
$$
f_1(t,x,y,r,s)\geq h_1(t,x,r),\quad \forall  t\in I,\; x, y\in
\mathbb{R}^{+},\; r, s\in \mathbb{R}^{-},
$$
where
$$
\liminf_{|x|+|r|\to +\infty} \min_{t\in
I}\frac{h_1(t,x,r)}{|x|+|r|}>\frac{\pi^4}{1+\pi^2};
$$

\item[(H3)] there exist $h_2\in C(I\times
\mathbb{R}^{+}\times \mathbb{R}^{-},\mathbb{R}^{+})$, such that
 $$
f_2(t,x,y,r,s)\leq h_2(t,y,s),\quad \forall
t\in I,x, y\in \mathbb{R}^{+},\; r, s\in \mathbb{R}^{-},
$$
where
$$
\limsup_{|y|+|s|\to +\infty} \max_{t\in
I}\frac{h_2(t,y,s)}{|y|+|s|}<\frac{\pi^4}{1+\pi^2};
$$


\item[(H4)] there exist $\alpha_1$, $\beta_1\geq 0$,
with $\frac{\alpha_1}{\pi^4}+\frac{\beta_1}{\pi^2}<1$, and $r_0>0$,
such that
 $$
f_1(t,x,y,r,s)\leq \alpha_1 x- \beta_1 r,\quad
 \forall  t\in I,\;x\in[0,r_0],\; r\in [-r_0,0],\;
 y\in \mathbb{R}^{+},\; s\in \mathbb{R}^{-};
$$

\item[(H5)] there exist $\alpha_2>0$, $\beta_2\geq0$,
$\frac{\alpha_2}{\pi^4}+\frac{\beta_2}{\pi^2}>1$, and $r_0^*>0$,
such that
$$
f_2(t,x,y,r,s)\geq \alpha_2 y- \beta_2 s,\quad \forall  t\in I,\;
y\in[0,r_0^*],\; s\in [-r_0^*,0],\; x\in \mathbb{R}^{+},\;
 r\in \mathbb{R}^{-}.
$$
\end{itemize}

We obtain the following results concerned with positive solutions
for system \eqref{e1.1}.

\begin{theorem} \label{thm2.1}
Assume that {\rm (H1)--(H5)} hold. Then \eqref{e1.1} has at least
one positive solution.
\end{theorem}

 It is easy to see that  conditions
(H4) and (H5)  are weaker than the
superlinear and sublinear conditions.

\section{Basic lemmas}

For $\lambda \in I$, $u, v\in C^2[0,1]$, we define two operators
 $A_\lambda, B_\lambda :C^2[0,1]\times C^2[0,1]\to C^2[0,1]$ by
\begin{equation}
\begin{gathered}
\begin{aligned}
 A_{\lambda}(u,v)(t)
&=\int_0^1\int_0^1  G(t,s)G(s,\tau)[\lambda
f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s,
\end{aligned}
\\
\begin{aligned}
 B_{\lambda}(u,v)(t)
&=\int_0^1\int_0^1 G(t,s)G(s,\tau)[\lambda
f_2(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\\
&\quad +(1-\lambda)h_2(\tau,v(\tau),v''(\tau))]\mathrm{d}\tau
\mathrm{d}s.
\end{aligned}
\end{gathered}\label{e3.1}
\end{equation}
Then we  define an operator $T_{\lambda}:C^2[0,1]\times C^2[0,1]\to
C^2[0,1]\times C^2[0,1]$ by
\begin{equation}
T_{\lambda}(u,v)=(A_{\lambda}(u,v),B_{\lambda}(u,v)),\ (u,v)\in
C^2[0,1]\times C^2[0,1].\label{e3.2}
\end{equation}

\begin{lemma} \label{lem3.1}
 Assume that {\rm (H1)} holds. Then
\begin{itemize}
\item[(1)] $T_{\lambda}: C^2[0,1]\times C^2[0,1]\to C^2[0,1]\times
C^2[0,1]$ is completely continuous.

\item[(2)] $T_{\lambda}: K\times K\to K\times K$ is completely
continuous.

\item[(3)]  If  $(u,v)\in K\times K$ is a nontrivial fixed point of
$T_1$, then  $(u,v)$ is a positive solution of system \eqref{e1.1}.
\end{itemize}
\end{lemma}

\begin{proof} (1)
 The proof is similar to that of \cite[Lemma 2.1]{l2},
and we omit it.
(2) By (1), we only need to prove that
operator $T_{\lambda}: K\times K\to K\times K$. In fact, for
any $(u,v)\in K\times K$, it follows from \eqref{e3.1} that
\begin{equation} \begin{gathered}
A_{\lambda}(u,v)(t)\geq 0,\quad A_{\lambda}(u,v)''(t)\leq 0,\quad
 t\in [0,1],\\
\begin{aligned}
 \|A_{\lambda}(u,v)\|
&\leq \int^1_0\int^1_0  s(1-s)G(s,\tau)[\lambda
f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\\
 &\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s,
\end{aligned} \\
\begin{aligned}
\|A_{\lambda}''(u,v)\|
&\leq \int^1_0G(s,s)[\lambda f_1(s,u(s),v(s),u''(s),v''(s))\\
&\quad +(1-\lambda)h_1(s,u(s),u''(s))]
\mathrm{d}s.
\end{aligned}
\end{gathered}\label{e3.3}
\end{equation}
On the other hand, for any $(u,v)\in K\times K$ and any
$0\leq t\leq1$, It follows from \eqref{e2.2}, \eqref{e3.1}, and
\eqref{e3.3} that
\begin{align*}
A_{\lambda}(u,v)(t)
&=  \int_0^1\int_0^1 G(t,s)G(s,\tau)[\lambda
f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s\\
&\geq q(t)\int_0^1\int_0^1s(1-s)G(s,\tau) [\lambda
f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s\\
&  \geq q(t)\|A_{\lambda}(u,v)\|,
\end{align*}
and
\begin{align*}
A_{\lambda}''(u,v)(t)
&= -\int_0^1G(t,s)[\lambda f_1(s,u(s),v(s),u''(s),v''(s))\\
&\quad +(1-\lambda)h_1(s,u(s),u''(s))]
\mathrm{d}s\\
&\leq -q(t) \int_0^1G(s,s)[\lambda
f_1(s,u(s),v(s),u''(s),v''(s))\\
&\quad +(1-\lambda)h_1(s,u(s),u''(s))]\mathrm{d}s\\
& \leq -q(t) \|A_{\lambda}''(u,v)\|.
\end{align*}
In a similar way, it follows that
$$
B_{\lambda}(u,v)(t)\geq 0,\quad
B_{\lambda}(u,v)''(t)\leq 0,\quad t\in I,
$$
and
$$
B_{\lambda}(u,v)(t)\geq q(t)\|B_{\lambda}(u,v)\|,\quad
B_{\lambda}''(u,v)(t)\leq -q(t) \|B_{\lambda}''(u,v)\|,\quad
\forall\ t\in I.
$$
From the above, we assert that
$T_{\lambda}(u,v)=(A_{\lambda}(u,v),B_{\lambda}(u,v))\in K\times K$;
that is, $T_{\lambda}:K\times K\to K\times K$.

(3) Let $(u,v)\in K\times K$ is a fixed point of $T_1$, Then
\begin{align*}
u(t)&=A_1(u,v)(t)\\
&=\int_0^1\Big[\int_0^1 G(t,s)G(s,\tau)
 f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau))\mathrm{d}\tau\Big]
\mathrm{d}s,\quad t\in I,\\
v(t)&=B_1(u,v)(t)\\
&=\int_0^1\Big[\int_0^1 G(t,s)G(s,\tau)f_2(\tau,u(\tau),
v(\tau),u''(\tau),v''(\tau))\mathrm{d}\tau\Big]\mathrm{d}s,\quad
t\in I.
\end{align*}
After direct computations, we obtain
\begin{gather*}
u''(t) =-\int_0^1G(t,s)f_1(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s,\\
\begin{aligned}
u'''(t) &= \int_0^t sf_1(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s\\
&\quad -\int_t^1(1-s)f_1(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s,
\end{aligned}\\
u^{(4)}(t) =f_1(t,u(t),v(t),u''(t),v''(t)),\\
v''(t)=-\int_0^1G(t,s)f_2(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s,\\
\begin{aligned}
v'''(t) &=\int_0^t
sf_2(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s\\
&\quad -\int_t^1(1-s)f_2(s,u(s),v(s),u''(s),v''(s))\mathrm{d}s,
\end{aligned} \\
v^{(4)}(t)=f_2(t,u(t),v(t),u''(t),v''(t)).
\end{gather*}
Moreover, since $G(0,s)=G(1,s)=0$, we see that
$u(0)=u(1)=u''(0)=u''(1)=v(0)=v(1)=v''(0)=v''(1)=0$.

Therefore, $(u,v)$ is a solution of  \eqref{e1.1}. Moreover, since
the graphs of $u\in K$ and $v\in K$ are concave down on $I$, we
assert that $(u,v)$ is a positive solution of system \eqref{e1.1}.
This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.1}\rm
Denoting
$T(\lambda,u,v)(t)=T_{\lambda}(u,v)(t)$, we see
that $\overline{T(\lambda\times K\times K)}$ is a compact set by the
Arzela-Ascoli theorem.
\end{remark}

\begin{lemma}[\cite{g1,z1}] \label{lem3.2}
Let $E$ be a real Banach space and let $P$ be a closed convex cone
in $E$. $\Omega$ be a bounded open set of $E$, $\theta \in \Omega$,
$A:P\cap \overline{\Omega}\to P$ be completely continuous. Then
the following conclusions are valid:
\begin{itemize}
\item[(i)] if $\mu Au\neq u$ for every $u\in P\cap\partial \Omega$
and $\mu\in (0,1]$, then  $i(A,P\cap \Omega,P)=1$;

\item[(ii)] if mapping $A$ satisfies the following two conditions:\\
(a) $\inf_{u\in P\cap\partial \Omega}\|Au\|>0$,\\
(b) $\mu Au\neq u$ for every $u\in P\cap\partial \Omega$ and
$\mu\geq 1$,
then $i(A,P\cap \Omega,P)=0$.
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{c1}] \label{lem3.3}
 Let $E$ be a Banach space and let $K_i\subset E(i=1,2)$ be a closed
convex cone in $E$. For $r_i>0$ $(i=1,2)$, denote
$K_{r_i}=\{u\in K_i: \|u\|<r_i\}$,
$\partial K_{r_i}=\{u\in K_i: \|u\|=r_i\}$. Let $A_i:K_i\to K_i$ be
completely continuous. If $A_iu_i\neq u_i$, for all $u\in \partial
K_{r_i}$, then
$$
i(A,K_{r_1}\times K_{r_2}, K_1\times K_2)=i(A_1,K_{r_1},K_1)
\times i(A_2,K_{r_2},K_2),
$$
where $A(u,v)=(A_1u,A_2v)$, for all $(u,v)\in K_1\times K_2$.
\end{lemma}


\section{Proof of main result}

We separate the proof of Theorem \ref{thm2.1}
into five steps.

\textbf{Step 1.} For each $r_1\in (0, r_0)$, we will prove that
\begin{equation}
\mu A_{\lambda}(u,v)\neq u , \quad \forall \mu\in (0,1],\;
 (u,v)\in \partial K_{r_1}\times K.\label{e4.1}
\end{equation}
In fact, if there exist $\mu_0 \in (0,1]$ and
$(u_0,v_0)\in \partial K_{r_1}\times K$, such that
$\mu_0 A_{\lambda}(u_0,v_0)= u_0$,
then $u_0(t)$ satisfies the differential equation
\begin{gather*}
u_0^{(4)}(t)=\mu_0[\lambda f_1(t,u_0(t),v_0(t),u_0''(t),v_0''(t))
+(1-\lambda)h_1(t,u_0(t),u_0''(t))], \\
u_0(0)=u_0(1)=u_0''(0)=u_0''(1)=0.
\end{gather*}
Since $0\leq u_0(t),-u_0''(t)\leq \|u_0\|_2=r_1<r_0$, from
(H2) and (H4), we obtain
\begin{align*}
u_0^{(4)}(t)
&\leq \lambda f_1(t,u_0(t),v_0(t),u_0''(t),v_0''(t))
  +(1-\lambda)h_1(t,u_0(t),u_0''(t))
\\  &\leq \alpha_1 u_0(t)- \beta_1 u_0''(t),
\end{align*}
Multiplying  both sides of this
inequality by $\sin(\pi t)$ and integrating on $I$, then using
integrating by parts, we obtain
\begin{equation}
\pi^4\int_0^1u_0(t)\sin(\pi t)\mathrm{d}t
\leq (\alpha_1+\beta_1\pi^2)\int_0^1u_0(t)\sin(\pi t)\mathrm{d}t.
\label{e4.2}
\end{equation}
By \cite[Lemma 1]{l1},
\begin{equation}
\frac{\pi^3+\pi^5}{4}\int_0^1u_0(t)\sin(\pi t)\mathrm{d}t\geq
\|u_0\|+\|u_0''\|=\|u_0\|_2=r_1>0.\label{e4.3}
\end{equation}
Hence, $\int_0^1u_0(t)\sin(\pi t)\mathrm{d}t>0$.
From \eqref{e4.2} and \eqref{e4.3},
 we obtain that $\pi^4\leq (\alpha_1+\beta_1\pi^2)$, which is a
contradiction.

\textbf{Step 2.} From  (H2), there exist
$\epsilon>0$, $m>0$, $C>0$, such that
\begin{equation}
h_1(t,u,u'')\geq (\frac{\pi^4}{1+\pi^2}+\epsilon)(|u|+|u''|),\quad
\forall  t\in I,\; |u|+|u''|\geq m,\label{e4.4}
\end{equation}
and
\begin{equation}
h_1(t,u,u'')\geq
(\frac{\pi^4}{1+\pi^2}+\epsilon)(|u|+|u''|)-C,\quad \forall  t\in I,
 \; u\in \mathbb{R}^{+}.\label{e4.5}
\end{equation}
We will prove that there exist $R_1>r_1$, such that
\begin{equation}
\mu A_{\lambda}(u,v)\neq u ,\quad
\inf_{u\in \partial K_{R_1}}\|A_{\lambda}(u,v)\|_2>0, \quad
 \forall \mu\geq 1,\; (u,v)\in \partial K_{R_1}\times K.
\label{e4.6}
\end{equation}
In fact, if there exist $\mu_1 \geq 1$ and $(u_1,v_1)\in \partial
K_{R_1}\times K$, such that  $\mu_1 A_{\lambda}(u_1,v_1)= u_1$, then
$u_1(t)$ satisfies the differential equation
\begin{gather*}
u_1^{(4)}(t)=\mu_1[\lambda f_1(t,u_1(t),v_1(t),u_1''(t),v_1''(t))
+(1-\lambda)h_1(t,u_1(t),u_1''(t))] \\
u_1(0)=u_1(1)=u_1''(0)=u_1''(1)=0.
\end{gather*}
In combination with \eqref{e4.5} and the condition (H2),
we obtain that
\begin{align*}
u_1^{(4)}(t)&\geq \lambda f_1(t,u_1(t),v_1(t),u_1''(t),v_1''(t))
 +(1-\lambda)h_1(t,u_1(t),u_1''(t))\\
&=\lambda (f_1(t,u_1(t),v_1(t),u_1''(t),v_1''(t))
 -h_1(t,u_1(t),u_1''(t)))+h_1(t,u_1(t),u_1''(t))\\
&\geq h_1(t,u_1(t),u_1''(t))\\
&\geq(\frac{\pi^4}{1+\pi^2}+\epsilon)(u_1-u_1'')-C,\quad
 \forall  t\in I.
\end{align*}
Multiplying the both sides of this inequality by
$\sin(\pi t)$ and integrating on $I$, then using integrating by parts,
we obtain
$$
\pi^4\int_0^1u_1(t)\sin(\pi t)\mathrm{d}t
\geq(\frac{\pi^4}{1+\pi^2}+\epsilon)(1+\pi^2)\int_0^1u_1(t)
\sin(\pi t)\mathrm{d}t -\frac{2C}{\pi}.
$$
Hence
$$ \int_0^1u_1(t)\sin(\pi t)\mathrm{d}t
\leq \frac{1}{(1+\pi^2)\epsilon}\frac{2C}{\pi}.
$$
In combination with \eqref{e4.3}, we obtain
$$
\|u_1\|_2\leq \frac{\pi^3+\pi^5}{4(1+\pi^2)\epsilon}\frac{2C}{\pi}
=\frac{\pi^2C}{2\epsilon}:=R^*.
$$
So, as $R_1>R^{*}$, we have $\mu A_{\lambda}(u,v)\neq u $,
 for all $(u,v)\in \partial K_{R_1}\times K$ and $\mu\geq 1$.
In addition, if $R_1>\frac{5C}{\epsilon}$,
by \eqref{e4.5}, we know that for all
$(u,v)\in \partial K_{R_1}\times K$,
\begin{align*} 
&A_{\lambda}(u,v)(\frac{1}{2})\\
&= \int_0^1\int_0^1 G(\frac{1}{2},s)G(s,\tau)[\lambda
f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau)) \\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s\\
&\geq \frac{1}{4}\int_0^1\int_0^1G(s,s)G(s,\tau)
[\lambda f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau)) \\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s\\
&\geq \frac{1}{4}\int_0^1G(s,s)\int_0^1q(s)G(\tau,\tau)
[\lambda f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau)) \\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\mathrm{d}s\\
&= \frac{1}{4}\int_0^1G(s,s)q(s)\mathrm{d}s\int_0^1G(\tau,\tau)
[\lambda f_1(\tau,u(\tau),v(\tau),u''(\tau),v''(\tau)) \\
&\quad +(1-\lambda)h_1(\tau,u(\tau),u''(\tau))]\mathrm{d}\tau
\\
&\geq \frac{1}{120}\int_{0}^{1}G(\tau,\tau)[(\frac{\pi^4}{1+\pi^2}
+\epsilon)(u(\tau)-u''(\tau))]\mathrm{d}\tau
-\frac{1}{120}\int_{0}^{1}G(\tau,\tau)\mathrm{d}\tau\cdot C\\
&\geq \frac{1}{120}\Big[\epsilon\int_{0}^{1}G(\tau,\tau)
q(\tau)(\|u\|+\|u''\|)\mathrm{d}\tau\Big]-\frac{1}{720} C\\
&= \frac{1}{120}\frac{\epsilon}{30}\|u\|_2-\frac{1}{720} C
>0,
\end{align*}
which follows that
$\inf_{u\in \partial K_{R_1}}\|A_{\lambda}(u,v)\|_2>0$. Hence,
we choose
\begin{equation}
R_1>\max\{R^*, \frac{5C}{\epsilon},r_1\}. \label{e4.7}
\end{equation}


\textbf{Step 3.} For each $r_2\in (0, r_0^*)$, we will prove that
\begin{equation}
\mu B_{\lambda}(u,v)\neq v,\quad
 \inf_{v\in \partial K_{r_2}}\|B_{\lambda}(u,v)\|_2>0, \quad
 \forall \mu\geq 1,\; (u,v)\in K\times \partial K_{r_2}.\label{e4.8}
\end{equation}
From (H3) and (H5),
\begin{equation}
\begin{gathered}
\lambda f_2(t,u,v,u'',v'')+(1-\lambda)h_2(t,v,v'')
\geq \alpha_2v-\beta_2v'',\\\
 \forall  t\in I,\; v\in[0,r_2],\;
v''\in [-r_2,0],\; u\in \mathbb{R}^{+},\;
u''\in \mathbb{R}^{-}.
\end{gathered}
\label{e4.9}
\end{equation}
By \eqref{e4.9} and a proof
similar to Step 1 and 2, we  deduce that \eqref{e4.8} holds.


\textbf{Step 4.} We will prove that
\begin{equation}
\mu B_{\lambda}(u,v)\neq v, \quad \forall \mu\in (0,1],\;
(u,v)\in K\times \partial K_{R_2}.\label{e4.10}
\end{equation}
From (H3),we know that there exist $\epsilon>0$, $m>0$, $C>0$,
such that
\begin{gather*}
\lambda f_2(t,u,v,u'',v'')+(1-\lambda)h_2(t,v,v'')
\leq (\frac{\pi^4}{1+\pi^2}-\epsilon)( |v|+|v''|),\\
 \forall  t\in I,\;  |v|+|v''|\geq m,\; u\in \mathbb{R}^{+},
\; u''\in \mathbb{R}^{-};\\
\lambda f_2(t,u,v,u'',v'')+(1-\lambda)h_2(t,v,v'')
\leq (\frac{\pi^4}{1+\pi^2}-\epsilon)( |v|+|v''|)+C,\\
 \forall t\in I,\;  u\in \mathbb{R}^{+},\;
v\in \mathbb{R}^{+},\; u''\in \mathbb{R}^{-},\;
 v''\in \mathbb{R}^{-}.
\end{gather*} %\label{e4.11}
Then the proof similar to Step 2.
If we choose $R_2>\max\{R^*,r_2\}$, we deduce that
\eqref{e4.10} holds.

\textbf{Step 5.} We choose an open set
 $D=(K_{R_1}\backslash \overline{K_{r_1}})\times (K_{R_2}\backslash
\overline{K_{r_2}})$.
By \eqref{e4.1}, \eqref{e4.6}, \eqref{e4.8}, and  \eqref{e4.10},
it is easy to verify that
$\{T_{\lambda}\}_{\lambda\in I}$ satisfy the sufficient conditions
for the homotopy invariance of fixed point index on $\partial D$; on
the other hand, in combination with the classical fixed point index
results (see Lemma \ref{lem3.2}), we have
\begin{gather*}
i(A_{0},K_{r_1},K)= i(B_{0},K_{R_2},K)=1,\\
i(A_{0},K_{R_1},K)=i(B_{0},K_{r_2},K)=0.
\end{gather*}
Applying the homotopy invariance of fixed point index and the
product formula for the fixed point index (see Lemma \ref{lem3.3}),
we obtain
\begin{align*}
&i(T_1,D, K\times K)=i(T_{0},D, K\times K)\\
&=i(A_{0},K_{R_1}\backslash \overline{K_{r_1}},K)
  \times i(B_{0},K_{R_2}\backslash \overline{K_{r_2}},K)\\
&=[i(A_{0},K_{R_1},K)-i(A_{0},K_{r_1},K)]\times
  [i(B_{0},K_{R_2},K)-i(B_{0}, K_{r_2},K)]
=-1.
\end{align*}
Thus, $T_1$  has at least a fixed
point$(u^{*},v^{*})\in (K_{R_1}\backslash \overline{K_{r_1}})\times
(K_{R_2}\backslash \overline{K_{r_2}})$.  Hence, by
Lemma \ref{lem3.1}, system \eqref{e1.1} has at least one
positive solution $(u^{*},v^{*})$. The
proof of Theorem \ref{thm2.1} is complete.

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\end{document}