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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 163, 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 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/163\hfil Three positive solutions]
{Three positive solutions for a  system of singular generalized
Lidstone problems}

\author[J. Xu, Z. Yang\hfil EJDE-2009/163\hfilneg]
{Jiafa Xu, Zhilin Yang}  % in alphabetical order

\address{Jiafa Xu \newline
Department of Mathematics, Qingdao Technological University, No 11,
Fushun Road, Qingdao, China}
\email{xujiafa292@sina.com}

\address{Zhilin Yang \newline
Department of Mathematics, Qingdao Technological University, No 11,
Fushun Road, Qingdao, China}
\email{zhilinyang@sina.com}

\thanks{Submitted October 7, 2009. Published December 21, 2009.}
\thanks{Supported by grants 10871116 and 10971179 from
the NNSF of China}
\subjclass[2000]{34A34, 34B18, 45G15, 47H10}
\keywords{Singular generalized Lidstone problem; positive solution;
cone; \hfill\break\indent concave functional}

\begin{abstract}
 In this article, we show the existence of at least three positive
 solutions for the system of singular generalized Lidstone boundary
 value problems
 \begin{gather*}
 (-1)^m x^{(2m)}=a(t)f_1(t,x,-x'',\dots,(-1)^{m-1}x^{(2m-2)},y,-y'',\\
  \dots,(-1)^{n-1}y^{(2n-2)}),  \\
 (-1)^n y^{(2n)}=b(t)f_2(t,x,-x'',\dots,(-1)^{m-1}x^{(2m-2)},y,-y'',\\
  \dots,(-1)^{n-1}y^{(2n-2)}), \\
 a_1 x^{(2i)}(0)-b_1 x^{(2i+1)}(0)=c_1x^{(2i)}(1)+d_1 x^{(2i+1)}(1)=0,\\
 a_2y^{(2j)}(0)-b_2y^{(2j+1)}(0)=c_2y^{(2j)}(1)+d_2y^{(2j+1)}(1)=0.
 \end{gather*}
 The proofs of our main results are based on the Leggett-Williams
 fixed point theorem. Also, we give an example to illustrate our
 results.
\end{abstract}

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

\section{Introduction}

In this article, we study the existence positive
solutions  for the following  system of singular generalized
Lidstone boundary value problems
\begin{equation}\label{eq1}
\begin{gathered}
(-1)^m x^{(2m)}=a(t)f_1(t,x,-x'',\dots,(-1)^{m-1}x^{(2m-2)},y,-y'',
 \dots,(-1)^{n-1}y^{(2n-2)}),\\
(-1)^n y^{(2n)}=b(t)f_2(t,x,-x'',\dots,(-1)^{m-1}x^{(2m-2)},y,-y'',
 \dots,(-1)^{n-1}y^{(2n-2)}),\\
a_1 x^{(2i)}(0)-b_1 x^{(2i+1)}(0)=c_1x^{(2i)}(1)+d_1 x^{(2i+1)}(1)=0
\quad  (i=0,1,\dots,m-1),\\
a_2y^{(2j)}(0)-b_2y^{(2j+1)}(0)=c_2y^{(2j)}(1)+d_2y^{(2j+1)}(1)=0\quad
(j=0,1,\dots,n-1).
\end{gathered}
\end{equation}
where $m, n\geq 1$, $a(t),b(t) \in C((0, 1), [0,+\infty))$, $a(t)$
and $b(t)$ are allowed to be singular at $t = 0$ and/or $t = 1$;
$f_i\in C([0,1]\times \mathbb{R}^{m+n}_+,\mathbb{R}_+)(\mathbb{R}_+:=[0,+\infty))$;
$a_i,b_i,c_i,d_i\in \mathbb{R}_+$ with
$\rho_i := a_ic_i+a_id_i+b_ic_i> 0$, $i=1,2$.

Singular boundary value problems for ordinary differential equation
describe many phenomena in applied mathematics and physical
sciences, which can be found in the theory of nonlinear diffusion
generated by nonlinear sources and in the thermal ignition of gases,
see \cite{Anderson,Lin}. Very recently, increasing attention is paid
to question of positive solutions for  systems of second-order or
higher order singular differential equations, see for example
\cite{Kang,Li,Liu_1,Liu} and references therein.

In \cite{Kang}, by using Leggett-Williams fixed point theorem
\cite{Leggett}, Kang et al obtained at least three positive
solutions to the following singular nonlocal boundary value problems
for systems of nonlinear second-order ordinary differential
equations.
\begin{equation}\label{kang}
\begin{gathered}
u''(t)+a_1(t)f_1(t,u(t),v(t))=0, \quad  0<t<1,\\
v''(t)+a_2(t)f_2(t,u(t),v(t))=0,\quad  0<t<1,\\
\alpha_1u(0)-\beta_1u'(0)=0, \quad
\gamma_1u(1)+\delta_1u'(1)=g_1(\int_0^1 u(s)d\phi_1(s),\int_0^1
v(s)d\phi_1(s)),\\
\alpha_2v(0)-\beta_2v'(0)=0, \quad
\gamma_2v(1)+\delta_2v'(1)=g_2(\int_0^1 u(s)d\phi_2(s),\int_0^1
v(s)d\phi_2(s)),
\end{gathered}
\end{equation}
where $a_i \in C((0, 1), \mathbb{R}_+)$ is allowed to be singular at
$t = 0$ or $t = 1$;  $f_i \in C( [0, 1]\times \mathbb{R}_+\times
\mathbb{R}_+,\mathbb{R}_+)$ and
$g_i  \in C( [0, 1]\times \mathbb{R}_+\times \mathbb R_+,\mathbb{R}_+)$
are such that  $a_1(t)f_1(t, 0, 0)$ or $a_2(t)f_2(t, 0, 0)$ does not
vanish identically on any
subinterval of $(0, 1)$;  $\alpha_i \geq 0$, $\beta_i \geq  0$,
$\gamma_i \geq 0$, $\delta_i\geq 0$, and $\rho_i = \alpha_i\gamma_i +
\alpha_i \delta_i + \beta_i\gamma_i
> 0$; $\int_0^1 u(s)d\phi_1(s)$ and $\int_0^1
v(s)d\phi_1(s)$ denote the Riemann-Stieltjes integrals, $i = 1, 2$.

Motivated by \cite{Kang}, we deal with the  system of singular
generalized Lidstone problems \eqref{eq1}. To overcome the
difficulties of \eqref{eq1} resulting from the derivatives of even
orders, as in \cite{Yang}, we use the method of order reduction to
transform \eqref{eq1} into an equivalent system of integro-integral
equations, then prove the existence of positive solutions for the
resulting system, thereby establishing that of positive solutions
for \eqref{eq1}(see the main result in Section 3). The features of
this paper mainly include the following aspects. Firstly, our study
is on systems of singular generalized Lidstone problems. Secondly,
$a$ and $b$ are allowed to be singular at $t = 0$ and/or $t=1$.
Finally, the system contains two equations, which can be of
different orders. Thus the results presented here are different from
those in \cite{Kang,Li,Liu_1,Liu}.

The remaining of this paper is organized as follows:  Section 2
gives some preliminary results.  The main result is stated and
proved in Section 3, then followed by an example to illustrate the
validity of our main result.

\section{Preliminaries}

Given a cone $K$ in a real Banach space $E$, a map $\alpha$ is said
to be a nonnegative continuous concave functional on $K$ provided
that $\alpha : K \to [0,+\infty)$ is continuous and
\[
\alpha(tx+(1-t)y)\geq t\alpha(x)+(1-t)\alpha(y)
\]
for all $x, y \in K$ and $0 \leq t \leq 1$.

Let $0 < a < b$  be given and let $\alpha$ be a nonnegative
continuous concave functional on $K$. Define the convex sets $P_r$
and $P(\alpha, a, b)$ by
\begin{gather*}
P_r:=\{x\in K:\|x\|<r\}, \\
P(\alpha, a, b):=\{x\in K: a\leq \alpha(x), \|x\|\leq b\}.
\end{gather*}
To prove our main result, we need the following Leggett-Williams
fixed point theorem.

\begin{lemma}[see \cite{Leggett}] \label{lem1}
 Let $T : \overline{P_c}\to \overline{P_c}$ be a completely
continuous operator and let $\alpha$ be a nonnegative continuous
concave functional on $K$ such that $\alpha(x) \leq \|x\|$ for all
$x \in \overline{P_c}$. Suppose there exist $0 < a < b < d \leq c$
such that
\begin{itemize}
\item[(C1)]  $\{x\in P(\alpha,b,d):\alpha(x)>b\}\neq \emptyset$, and
$\alpha(Tx)>b$ for $x\in P(\alpha,b,d)$,

\item[(C2)]  $\|Tx\|<a$ for $\|x\|\leq a$, and

\item[(C3)] $\alpha(Tx)>b$ for $x\in P(\alpha,b,c)$ with $\|Tx\|>d$.
\end{itemize}
Then $T$ has at least three fixed points $x_1,x_2$, and $x_3$ such
that
$\|x_1\|<a$, $b<\alpha(x_2)$ and $\|x_3\|>a$
with $\alpha(x_3)<b$.
\end{lemma}

For  fixed nonnegative constants $a,b,c,d$ with  $\rho
:=ac+ad+bc>0$, we let
\begin{equation}\label{k}
k(t,s):=\frac{1}{\rho}
\begin{cases}
(b+as)(c+d-ct), & 0\leq s\leq t\leq 1,\\
(b+at)(c+d-cs), & 0\leq t\leq s\leq 1,
\end{cases}
\end{equation}
 By definition, $k\in C([0,1]\times [0,1],\mathbb{R}_+)$ has the following
 properties:
\begin{itemize}
\item[(i)] $k(t,s)\leq k(s,s)$ for all $t,s\in [0,1]$.

\item[(ii)] For any  $\theta\in (0,1/2)$, there exists
$\gamma\in (0,\min\{\frac{\theta a+ b}{a+b},\frac{\theta c+ d}{c+d}\} ]$
such that
 \[
 k(t,s)\geq \gamma k(s,s),\forall  t\in[\theta,1-\theta],s\in
[0,1].
\]
\end{itemize}

\begin{lemma} \label{lem2}
 Let $f\in C[0,1],h(t)\in C(0,1)$ and $\int_0^1 k(s,s)h(s)ds<+\infty$. The
  boundary value problem
\begin{gather*}
-u''=h(t)f(t),\\
au(0)-bu'(0)=0,\\
cu(1)+du'(1)=0,
\end{gather*}
has a unique solution
\[
u(t)=\int_0^1 k(t,s)h(s)f(s)ds,
\]
where $k(t,s)$ is given by \eqref{k}.
\end{lemma}

Let
\begin{gather}\label{k1}
k_1(t,s):=\frac{1}{\rho_1}
\begin{cases}
(b_1+a_1s)(c_1+d_1-c_1t), & 0\leq s\leq t\leq 1,\\
(b_1+a_1t)(c_1+d_1-c_1s), & 0\leq t\leq s\leq 1,
\end{cases}\\
\label{g1}
g_1(t,s):=\frac{1}{\rho_2}
\begin{cases}
(b_2+a_2s)(c_2+d_2-c_2t), & 0\leq s\leq t\leq 1,\\
(b_2+a_2t)(c_2+d_2-c_2s), & 0\leq t\leq s\leq 1.
\end{cases}
\end{gather}
 For $i,j=2,\dots$, define
\begin{equation}
k_i(t,s):=\int_0^1 k_1(t,\tau)k_{i-1}(\tau,s)d\tau,\quad
g_j(t,s):=\int_0^1 g_1(t,\tau)g_{j-1}(\tau,s)d\tau
\end{equation}
and the operators $A_i: C[0,1]\to C[0,1]$ and $B_j: C[0,1]\to
C[0,1]$ by
\begin{gather*}
(A_iu)(t):=\int_0^1 k_i(t,s)u(s)ds,\quad  i=1,2,\dots,m,\\
(B_jv)(t):=\int_0^1 g_j(t,s)v(s)ds,\quad  j=1,2,\dots,n.
\end{gather*}
 For $1\leq i\leq m-1$ and $1\leq j \leq n-1$,
let
\begin{gather*}
\xi_i:=\min_{\theta\leq t\leq 1-\theta}\int_0^1 k_i(t,s)ds,\quad
\eta_i:=\max_{0\leq t\leq 1}\int_0^1 k_i(t,s)ds,\\
\mu_j:=\min_{\theta\leq t\leq 1-\theta}\int_0^1 g_j(t,s)ds,\quad
\nu_j:=\max_{0\leq t\leq 1}\int_0^1 g_j(t,s)ds,\\
\xi:=\min_{1\le i\le m-1,1\le j\le n-1}\{\xi_i,\mu_j\},\quad
\eta:=\max_{1\le i\le m-1,1\le j\le n-1}\{\eta_i,\nu_j\}.
\end{gather*}

\section{Main result}

Let $ u(t)=(-1)^{m-1}x^{(2m-2)}$ and $v(t)=(-1)^{n-1}y^{(2n-2)}$. It is
easy to see that \eqref{eq1} is equivalent to the  system
of integro-ordinary differential equations
\begin{align*}
 -u''(t)&= a(t)f_1\Big(t,\int_0^1
k_{m-1}(t,s)u(s)ds,\dots,\int_0^1 k_1(t,s)u(s)ds,u(t),\\
&\quad \int_0^1 g_{n-1}(t,s)v(s)ds,\dots,\int_0^1
g_1(t,s)v(s)ds,v(t)\Big),\\
-v''(t)&= b(t)f_2\Big(t,\int_0^1
k_{m-1}(t,s)u(s)ds,\dots,\int_0^1 k_1(t,s)u(s)ds,u(t),\\
&\quad \int_0^1 g_{n-1}(t,s)v(s)ds,\dots,\int_0^1
g_1(t,s)v(s)ds,v(t)\Big),\\
\end{align*}
subject to the boundary conditions
\[
a_1u(0)-b_1u'(0)= c_1u(1)+d_1u'(1)=0 ,\quad
a_2v(0)-b_2v'(0)= c_2v(1)+d_2v'(1)=0 .
\]
 Furthermore, the above system is
equivalent to the system
\begin{equation}\label{sys}
\begin{aligned}
 u(t)&= \int_0^1 k_1(t,s)a(s)f_1\Big(s,\int_0^1
k_{m-1}(s,\tau)u(\tau)d\tau,\dots,\int_0^1
k_1(s,\tau)u(\tau)d\tau,u(s),\\
&\quad  \int_0^1g_{n-1}(s,\tau)v(\tau)d\tau,\dots,\int_0^1
g_1(s,\tau)v(\tau)d\tau,v(s)\Big)ds,\\
v(t)&= \int_0^1 g_1(t,s)b(s)f_2\Big(s,\int_0^1
k_{m-1}(s,\tau)u(\tau)d\tau,\dots,\int_0^1
k_1(s,\tau)u(\tau)d\tau,u(s),\\
\quad & \int_0^1 g_{n-1}(s,\tau)v(\tau)d\tau,\dots,\int_0^1
g_1(s,\tau)v(\tau)d\tau,v(s)\Big)ds.
\end{aligned}
\end{equation}
Let $E := C([0, 1],\mathbb{R})\times C([0, 1],\mathbb{R})$ endowed
with the norm $\|(u, v)\|:= \|u\|+\|v\|$, where
$\|u\|:= \max_{0\leq t\leq 1} |u(t)|$, and define the cone
$K\subset E$ by
\[
K := \Big\{(u, v)\in E : u(t) \geq 0, v(t) \geq 0, t \in [0, 1], \;
 \min_{\theta \leq t\leq 1-\theta} (u(t) + v(t)) \geq \gamma
\|(u, v)\| \Big\}.
\]
Clearly, $(E,\|\cdot\|)$ is a real Banach
space and $P$ is a cone on $E$.
Define the operator $T:K\to  K$ by
\[
 T(u,v)(t):=(T_1(u,v)(t),T_2(u,v)(t)),
\]
 where
\begin{align*}
 T_1(u,v)(t)
& := \int_0^1 k_1(t,s)a(s)f_1\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s),(B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds,
\\
 T_2(u,v)(t)
& := \int_0^1 g_1(t,s)b(s)f_2\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s),(B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds.
\end{align*}
 Now $f\in C([0,1]\times \mathbb{R}^{m+n}_+,\mathbb{R}_+)$ and $g\in
C([0,1]\times \mathbb{R}^{m+n}_+,\mathbb{R}_+)$  imply that $T: K\to
K$ is a completely continuous operator. In our setting,  the
existence of positive solutions for \eqref{sys} is equivalent to
that of positive fixed points of $T$.

\begin{lemma}
The operator $T$ maps $K$ into $K$.
\end{lemma}

\begin{proof}
If $(u,v) \in K$, then
\begin{align*}
& (T_1(u,v)(t))''\\
&=-a(t)f_1\Big(t,(A_{m-1}u)(t),\dots,(A_1u)(t),u(t),
(B_{n-1}v)(t),\dots,(B_1v)(t),v(t)\Big)\\
& \leq 0,
\\
&(T_2(u,v)(t))''\\
 &=- b(t)f_2\Big(t,(A_{m-1}u)(t),\dots,(A_1u)(t),u(t),
(B_{n-1}v)(t),\dots,(B_1v)(t),v(t)\Big)\\
& \leq 0.
\end{align*}
So $T_1(u,v)(t)$ and $T_2(u,v)(t)$ are concave on [0,1]. If $(u, v)
\in K$, then from the  properties of $k_1(t, s)$ and $g_1(t, s)$, we
have
\begin{align*}
 \|T_1(u,v)(t)\|
& \leq \int_0^1
k_1(s,s) a(s)f_1\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s), (B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds,
\end{align*}
and
\begin{align*}
 \|T_2(u,v)(t)\|
& \leq \int_0^1 g_1(s,s)b(s)f_2\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s), (B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds.
\end{align*}
On the other hand,
\begin{align*}
 \min_{\theta\leq t\leq 1-\theta}T_1(u,v)(t)
& \geq \gamma \int_0^1
k_1(s,s)a(s) f_1\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),u(s),\\
&\quad (B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds\\ & \geq
\gamma\|T_1(u,v)(t)\|,
\end{align*}
and
\begin{align*}
 \min_{\theta\leq t\leq 1-\theta}T_2(u,v)(t)
& \geq \gamma \int_0^1 g_1(s,s)b(s) f_2\Big(s,(A_{m-1}u)(s),
\dots,(A_1u)(s),u(s),\\
&\quad (B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds\\
& \geq \gamma\|T_2(u,v)(t)\|.
\end{align*}
Combining the preceding inequalities, we arrive at
\begin{align*}
  \min_{\theta\leq t\leq 1-\theta} (T_1(u,v)(t)+T_2(u,v)(t))
& \geq \min_{\theta\leq t\leq 1-\theta}(T_1(u,v)(t))
 +\min_{\theta\leq t\leq 1-\theta}(T_2(u,v)(t))\\
& \geq \gamma (\|T_1(u,v)(t)\|+\|T_2(u,v)(t)\|).
\end{align*}
This completes the proof.
\end{proof}

In this paper, we use the following assumptions:
\begin{itemize}
\item[((H1)] $a(t)$ and $b(t)$ do not vanish identically on
any subinterval of $(0, 1)$, and there exists $t_0 \in (0, 1)$ such
that $a(t_0) > 0$, $b(t_0) > 0$ and
$0 < \int_0^1  k_1(s, s)a(s)ds < +\infty$,
$0 < \int_0^1  g_1(s, s)b(s)ds < +\infty$.

\end{itemize}
Finally, we define the nonnegative continuous concave functional
\[
\alpha(u,v):=\min_{\theta\leq t\leq 1-\theta}(u(t)+v(t)).
\]
We observe here that, for each $(u, v) \in K$,
$\alpha(u, v) \leq \|(u, v)\|$.
Let
\begin{gather*}
\widetilde{\xi}_1:=\min_{\theta\leq t\leq 1-\theta}
 \int_0^1 k_1(t,s)a(s)ds,\quad
\widetilde{\eta}_1:=\max_{0\leq t\leq 1}\int_0^1 k_1(t,s)a(s)ds,\\
\widetilde{\mu}_1:=\min_{\theta\leq t\leq 1-\theta}\int_0^1 g_1(t,s)b(s)ds,
\quad
\widetilde{\nu}_1:=\max_{0\leq t\leq 1}\int_0^1 g_1(t,s)b(s)ds\,.
\end{gather*}

\begin{theorem}\label{thm1}
Let $1\le i\le m-1$, and $1\le j\le n-1$.Assume there exist nonnegative
numbers $a, b, c  $  such that
$0 < a < b \leq \min\{\gamma, \widetilde{\xi}_{1}/\widetilde{\eta}_{1},
\xi_i/\eta_i, \widetilde{\mu}_1/\widetilde{\nu}_1,\mu_j/\nu_j\}c$, and
$f (t, x_1,\dots,x_{m-1},x_m,y_{1}, \dots,y_{n-1},y_n)$, $g(t,
x_1,\dots,x_{m-1},x_m,y_{1}, \dots,y_{n-1},y_n)$ satisfy the
following growth conditions:
\begin{itemize}
\item[(H2)] $f_1 (t, x_1,\dots,x_{m-1},x_m,y_{1},
\dots,y_{n-1},y_n)\leq c/2\widetilde{\eta}_1$,\\
$f_2 (t, x_1,\dots,x_{m-1},x_m,y_{1}, \dots,y_{n-1},y_n)\leq
c/2\widetilde{\nu}_1$, for all $t\in [0,1]$,
$x_i+y_j\in [0,\eta c]$, $x_m+y_n \in[0,c]$.

\item[(H3)] $f_1 (t, x_1,\dots,x_{m-1},x_m,y_{1},
\dots,y_{n-1},y_n)< a/2\widetilde{\eta}_1$,\\
 $f_2 (t, x_1,\dots,x_{m-1},x_m,y_{1}, \dots,y_{n-1},y_n)<
a/2\widetilde{\nu}_1$,
for all $t\in [0,1]$, $x_i+y_j\in [0,\eta a]$,
$x_m+y_n \in[0,a]$.

\item[(H4)] $f_1 (t, x_1,\dots,x_{m-1},x_m,y_{1},
\dots,y_{n-1},y_n)\geq b/2\widetilde{\xi}_1$,\\
$f_2 (t, x_1,\dots,x_{m-1},x_m,y_{1}, \dots,y_{n-1},y_n)\geq
b/2\widetilde{\mu}_1$, for all $t\in [\theta,1-\theta],
x_i+y_j\in [\xi b,\eta b/\gamma]$, $x_m+y_n \in[b,b/\gamma]$.

\end{itemize}
Then  \eqref{eq1} has at least three positive fixed
points $(u_1, v_1)$, $(u_2, v_2)$, $(u_3, v_3)$ such that
$\|(u_1, v_1)\| < a$,
$b < \min_{\theta\leq t\leq 1-\theta} (u_2(t) + v_2(t))$,
 and $\|(u_3, v_3)\|> a$,  with
$\min_{\theta\leq t\leq 1-\theta} (u_3(t) + v_3(t)) < b$.
\end{theorem}

\begin{proof}
We note first that $T : \overline{P_c} \to \overline{P_c}$ is a
completely continuous operator. If $(u, v)\in \overline{P_c} $, then
$\|(u, v)\| \leq c $. For $1 \leq i \leq m-1$ and $1 \leq j \leq
n-1$,
\[
(A_iu)(t)+(B_jv)(t)=\int_0^1 k_i(t,s)u(s)ds
+\int_0^1 g_j(t,s)v(s)ds\leq \eta_i\|u\|+\nu_j\|v\|\leq \eta c.
\]
From (H2), we have
\begin{align*}
  \|T(u, v)(t)\|
&=\max_{0\leq t\leq 1} |T_1(u, v)(t)| +\max _{0\leq t\leq 1}  |T_2(u,
v)(t)|\\
& =\max _{0\leq t\leq 1} \int_0^1
k_1(t,s)a(s)f_1\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s),
(B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds\\
&\quad + \max _{0\leq t\leq 1}\int_0^1
g_1(t,s)b(s)f_2\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s),
(B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds\\
& \leq c/2\widetilde{\eta}_1 \max _{0\leq t\leq 1}
\int_0^1k_1(t,s)a(s)ds+c/2\widetilde{\nu}_1 \max _{0\leq t\leq 1}
\int_0^1g_1(t,s)b(s)ds\leq c.
\end{align*}
Therefore, $T : \overline{P_c} \to \overline{P_c}$. In an analogous
argument, one can verify that (H3) implies condition (C2) of
Lemma \ref{lem1}.
Clearly, $\{(u,v)\in P(\alpha,b,b/\gamma):\alpha(u,v)>b\}\neq
\emptyset$. If $(u, v) \in P(\alpha, b, b/\gamma )$, then
$b \leq u(t) + v(t) \leq b/\gamma $, $t \in [\theta, 1 - \theta]$.

For $1 \leq i \leq m-1$ and $1 \leq j \leq n-1$, we have
\begin{gather*}
(A_iu)(t)+(B_jv)(t)=\int_0^1 k_i(t,s)u(s)ds
+\int_0^1 g_j(t,s)v(s)ds\geq \xi b,\\
(A_iu)(t)+(B_jv)(t)=\int_0^1 k_i(t,s)u(s)ds
+\int_0^1 g_j(t,s)v(s)ds\leq \eta b/\gamma.
\end{gather*}
By (H4),
\begin{align*}
 \alpha(T(u,v)(t))
&=\min_{\theta\leq t\leq 1-\theta}(T_1(u,v)(t)+T_2(u,v)(t))\\
& \geq\min_{\theta\leq t\leq 1-\theta}T_1(u,v)(t)
 +\min_{\theta\leq t\leq 1-\theta}T_2(u,v)(t)\\
& \geq \min_{\theta\leq t\leq 1-\theta}
 \int_0^1 k_1(t,s)a(s)
 f_1\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s), (B_{n-1}v)(s),\dots,(A_1v)(s),v(s)\Big)ds\\
&\quad + \min_{\theta\leq t\leq 1-\theta} \int_0^1 g_1(t,s)
 b(s)f_2\Big(s,(A_{m-1}u)(s),\dots,(A_1u)(s),\\
&\quad u(s), (B_{n-1}v)(s),\dots,(B_1v)(s),v(s)\Big)ds\\
& \geq b/2\widetilde{\xi}_1\min_{\theta\leq t\leq 1-\theta}
 \int_0^1 k_1(t,s)a(s)ds+b/2\widetilde{\mu}_1
 \min_{\theta\leq t\leq 1-\theta}
 \int_0^1 g_1(t,s)b(s)ds\\
&=b.
\end{align*}
Therefore, condition (C1) of Lemma \ref{lem1} is satisfied.

Finally, we show that  (C3) is also satisfied.
If $(u,v)\in P(\alpha,b,c)\text{ and }\|T(u,v)\|> b/\gamma $, then
\[
\alpha(T(u, v)(t)) = \min_{\theta\leq t\leq 1-\theta} (T_1(u, v)(t)
+ T_2(u, v)(t)) \geq \gamma \|T(u, v)\| > b.
\]
Therefore,  (C3) of Lemma \ref{lem1} is also satisfied.  By
Lemma \ref{lem1},  the system \eqref{eq1} has at least three positive fixed
points $(u_1, v_1), (u_2, v_2), (u_3, v_3)$ such that
$\|(u_1,v_1)\| < a$,
$b < \min_{\theta\leq t\leq 1-\theta} (u_2(t) + v_2(t))$,
 and $\|(u_3, v_3)\|> a$, with
$\min_{\theta\leq t\leq 1-\theta} (u_3(t) + v_3(t)) < b$.
\end{proof}

\subsection*{An example}
   Consider a system of
nonlinear second-order and fourth-order ordinary differential
equations (with $m=1, n=2$):
\begin{equation}\label{system}
\begin{gathered}
-x''=a(t)f_1(t,x,y,-y'')=0,\\
y^{(4)}=b(t)f_2(t,x,y,-y'')=0,\\
 x(0)-x'(0)=x(1)+ x'(1)=0,\\
y(0)-y'(0)=y(1)+y'(1)=0,\\
y''(0)-y'''(0)=y''(1)+y'''(1)=0.
\end{gathered}
\end{equation}
Let $ u:=x$ and $v:=-y''$, then the problem \eqref{system} is
equivalent to the following system of nonlinear  integral equations
\begin{equation}\label{equivalentsystem}
\begin{gathered}
u(t)=\int_0^1 k_1(t,s)a(s)f_1(s,u(s),\int_0^1
g_1(s,\tau)v(\tau)d\tau,v(s)),\\
 v(t)=\int_0^1 g_1(t,s)b(s)f_2(s,u(s),\int_0^1
g_1(s,\tau)v(\tau)d\tau,v(s)),
\end{gathered}
\end{equation}
where
\[
k_1(t,s)=g_1(t,s):=\frac{1}{3}\begin{cases}
(1+s)(2-t), 0\le s\le t\le 1,\\
(1+t)(2-s), 0\le t\le s\le 1.
\end{cases}
\]
We choose $a(t):=(1-t)^{-1/2}$,
$b(t):=t^{-1/2}$,
$\theta:=1/4$,
$f_1(t,x_1,x_2,x_3)$ equals
\[
\begin{cases}
\frac{t}{100}+\frac{1}{10}(x_1+x_3)^2,
&  t\in [0,1],\;  0\leq x_1+x_3\leq 2,\; x_2\ge 0,\\
\frac{t}{100}+6[(x_1+x_3)^2-2(x_1+x_3)]+\frac{2}{5},
&  t\in [0,1],\;  2< x_1+x_3< 4,\; x_2\ge 0,\\
\frac{t}{100}+20\log_2(x_1+x_3)+2(x_1+x_3)+\frac{2}{5},
&  t\in [0,1],\; 4\leq x_1+x_3\leq 16,\; x_2\ge 0,\\
\frac{t}{100}+\frac{1}{4}(x_1+x_3)+\frac{542}{5},
&  t\in [0,1],\; x_1+x_3>16,x_2\ge 0.
\end{cases}
\]
and
$f_2(t,x_1,x_2,x_3)$ equals
\[
\begin{cases}
\frac{t}{100}+\frac{1}{10}(x_1+x_3)^2,
&   t\in [0,1],\; 0\leq x_1+x_3\leq 2,\; x_2\ge 0,\\
\frac{t}{100}+\frac{13}{8}[(x_1+x_3)^2-2(x_1+x_3)]+\frac{2}{5},
&   t\in [0,1],\;  2< x_1+x_3< 4,\; x_2\ge 0,\\
\frac{t}{100}+\frac{5}{2}\log_2(x_1+x_3)+2(x_1+x_3)+\frac{2}{5},
&   t\in [0,1],\;  4\leq x_1+x_3\leq 16,\; x_2\ge 0,\\
\frac{t}{100}+\frac{1}{4}(x_1+x_3)+\frac{192}{5},
&   t\in [0,1],\; x_1+x_3>16,x_2\ge 0.
\end{cases}
\]
Then by direct calculation, we obtain
\[
\xi\approx 0.4120,\quad \eta=0.5,\quad
\widetilde{\xi}_1 \approx 0.0417,\quad
\widetilde{\mu}_1 \approx 0.9273,\quad
\widetilde{\eta}_1\approx 0.8889,\quad
\widetilde{\nu}_1\approx 1.1111 .
\]
It is easy to check  that (H1)
holds. Choose $\gamma=\frac{1}{4}$, $a=1$, $b=4$, $c=800$. Also, it
is easy to verify  that $f_1$ and $f_2$ satisfy  conditions
(H2)-(H4). So the system \eqref{system} has at least three positive
solutions $(u_1, v_1), (u_2, v_2), (u_3, v_3)$ such that
$\|(u_1, v_1)\| < 1$,
$4 < \min_{ 1/ 4\leq t\leq 3 /4 }(u_2(t) + v_2(t))$, and
$\|(u_3, v_3)\|> 1$, with
$\min_{1/ 4\leq t\leq 3 /4 } (u_3(t) + v_3(t)) < 4$.

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\end{document}