\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 131, pp. 1--11.\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-2011/131\hfil Existence of solutions]
{Existence of solutions for a coupled quasilinear system on time scales}

\author[M. R. Sidi Ammi, A. B. Malinowska\hfil EJDE-2011/131\hfilneg]
{Moulay Rchid Sidi Ammi, Agnieszka B. Malinowska} % in alphabetical order

\address{Moulay Rchid Sidi Ammi  \newline
Faculty of Sciences and Techniques, Moulay Ismail University,
AMNEA Group, Errachidia, Morocco}
\email{sidiammi@ua.pt}

\address{Agnieszka B. Malinowska \newline
Faculty of Computer Science,
Bia{\l}ystok University of Technology,
15-351 Bialystok, Poland}
\email{abmalinowska@ua.pt}

\thanks{Submitted March 1, 2011. Published October 11, 2011.}
\subjclass[2000]{37C25, 34B18, 47H10}
\keywords{Time scales; fixed point theorem; positive
solution; cone}

\begin{abstract}
 In this work we study a  quasilinear system of equations
 on time scales. Using the Krasnosel´skii fixed-point theorem,
 sufficient conditions are given for the existence of positive
 solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

In this paper we consider the existence of positive
solutions for the system of dynamic equations
\begin{equation} \label{eq1}
\begin{gathered}
 -\Big( u^\Delta(t)\Big)^\Delta= \lambda f(v(t)) ,\quad
 \forall t \in [0, T]_{\mathbb{T}},\\
 -\Big( v^\Delta(t)\Big)^\Delta= \lambda g(u(t)) ,\quad
 \forall t \in [0, T]_{\mathbb{T}},
  \end{gathered}
\end{equation}
satisfying the boundary conditions
\begin{equation} \label{eq2}
\begin{gathered}
u^\Delta (0) - \beta u^\Delta (\eta)= 0, \quad u(T) - \beta
u(\eta)=0, \\
v^\Delta (0) - \beta v^\Delta (\eta)= 0, \quad v(T) - \beta
v(\eta)=0,
\end{gathered}
\end{equation}
where $\eta \in [0, T]_{\mathbb{T}}, \beta \in \mathbb{R}$,
such that $ 0< \beta <1$. We are seeking for a pair $(u, v)$
of solutions for the system \eqref{eq1}-\eqref{eq2}.
Our general  assumptions are:
\begin{itemize}

\item[(H1)] the functions $f, g$ belong to $C([0, \infty), [0, \infty))$;

\item[(H2)] the following limits exist as real numbers:
\begin{gather*}
f_0:=\lim_{x\to 0^{+}}f(x)/x,\quad
g_0:=\lim_{x\to 0^{+}}g(x)/x,  \\
f_\infty:=\lim_{x\to \infty}f(x)/x, \quad
g_\infty:=\lim_{x\to \infty}g(x)/x, \quad
f_{\infty}g_{\infty} \neq 0.
\end{gather*}
\end{itemize}

The theory of dynamic equations on time scales (more
generally, on measure chains) was introduced in 1988 by Stefan Hilger
in his PhD thesis (see \cite{h1,h2}). The theory presents a
structure where, once a result is established for a general time
scale, then special cases can be obtained by taking the particular
time scale. If $\mathbb{T}=\mathbb{R}$, then we have the result
for differential equations. Choosing $\mathbb{T}=\mathbb{Z}$
we immediately get the result for difference equations.
A great deal of work has been
done since 1988, unifying and extending the theories of
differential and difference equations, and many results are now
available in the general setting of time scales, see
\cite{a1,a2,a3,abra} and references therein. In this paper we
prove existence of positive solutions to the problem
\eqref{eq1}-\eqref{eq2} on a general time scale $\mathbb{T}$.

The outline of this paper is as follows. In section~\ref{sec2}, we
give some  preliminary results with respect to the calculus on
time scales. For more details see for example \cite{bp2, bp3}.
Section~\ref{sec3} is devoted to the existence of positive solutions
using fixed-point theory. We are concerned with determining values
of $\lambda$ (eigenvalues) for which  there exist positive solutions
of \eqref{eq1}--\eqref{eq2}. In \cite{kam}, a Green function plays a fundamental
role to define an appropriate operator on a suitable cone and to
prove existence of solutions for a system of dynamic equations.
Differently, here we will use the well known Guo-Krasnoselskii
fixed point theorem without introducing the Green function.

\section{Preliminary results on time scales}\label{sec2}

 Now, we introduce some basic concepts of time scales,
preliminaries and
lemmas that will be needed later. For a deep details, the reader
can see \cite{a1,a2,a3,abra,bp1,bp2} and references therein. A
time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of
real numbers $ \mathbb{R}$. The operators $\sigma$ and $\rho$ from
$\mathbb{T}$ to $\mathbb{T}$ which are defined in \cite{h1,h2},
$$
\sigma(t)=\inf\{\tau\in\mathbb{T}:\tau> t\}\in\mathbb{T}, \quad
\rho(t)=\sup\{\tau\in\mathbb{T}:\tau< t\}\in\mathbb{T}
$$
are called the  forward jump operator and  the backward jump operator,
respectively.

 The point $t\in\mathbb{T}$ is
left-dense, left-scattered, right-dense, right-scattered if
$\rho (t)=t$, $\rho(t)<t$, $\sigma(t)=t$, $\sigma(t)>t$,
respectively. If $\mathbb{T}$ has a right scattered minimum $m$,
define $\mathbb{T}_{k}=\mathbb{T}-\{m\}$; otherwise set
$\mathbb{T}_{k}=\mathbb{T}$. If $\mathbb{T}$ has a left scattered
maximum $M$, define $\mathbb{T}^{k}=\mathbb{T}-\{M\}$; otherwise
set $\mathbb{T}^{k}=\mathbb{T}$.

Let $f:\mathbb{T}\to  \mathbb{R}$ and $t\in \mathbb{T}^{k}$
(assume $t$ is not left-scattered if $t=\sup\mathbb{T}$), we define
$f^\Delta(t)$ to be the number (provided it exists) such that
for all $\epsilon>0$ there is a neighborhood $U$ of $t$ such
that
$$
|f(\sigma(t))-f(s)-f^{\Delta}(t)(\sigma(t)-s) |\le | \sigma(t)-s|,
\quad \mbox{for all } s\in U.
$$
We call $f^\Delta(t)$ the delta derivative of $f$ at $t$.
If $\mathbb{T}=\mathbb{R}$, then
$f^\Delta$ coincides with the usual derivative $f'$.
If $\mathbb{T}=\mathbb{Z}$, then
$f^\Delta=\Delta f:=f(t+1)-f(t)$ is the forward difference.

Similarly, for $t\in \mathbb{T}$ (assume
$t$ is not right-scattered if $t=\inf\mathbb{T}$),  the nabla
derivative of $f$ at the point $t$ is defined in \cite{a4} to be
the number $f^{\nabla}(t)$ (provided it exists) with the property
that for all $\epsilon >0$ there is a neighborhood $U$ of $t$
such that
$$
|f(\rho(t))-f(s)-f^{\nabla}(t)(\rho(t)-s) |\le | \rho(t)-s |,
\quad \mbox{for all } s\in U.
$$
If $\mathbb{T}=\mathbb{R}$, then $f^\Delta(t)=f^\nabla(t)=f'(t)$.
If $\mathbb{T}=Z$, then $f^\nabla(t)=\nabla f:=f(t)-f(t-1)$ is
the backward difference operator.

We say that a function $f$ is left-dense continuous (i.e.,
$ld$-continuous), provided $f$ is continuous at each left-dense
point in $\mathbb{T}$ and its right-sided limit exists at each
right-dense point in $\mathbb{T}$. It is well-known that if $f$ is
$ld$-continuous and if $F^{\nabla}(t)=f(t)$, then we can define the
nabla integral by
$$
\int^b_a f(t)\nabla t=F(b)-F(a)
$$
for all $a,b\in \mathbb{T}$.
A function $f:\mathbb{T}\to\mathbb{R}$ is called
$rd$-continuous if it is continuous at right-dense points and
if its left-sided limit exists at left-dense points.
If $f$ is $rd$-continuous and $F^{\Delta}(t)=f(t)$,
then we define the delta integral by
$$
\int^b_a f(t)\Delta t=F(b)-F(a)
$$
for all $a,b\in \mathbb{T}$.

 From now on, $\mathbb{T}$ is a closed subset of
$\mathbb{R}$ such that $0\in\mathbb{T}_k$, $T\in\mathbb{T}^k$. Let
$\mathbb{E}= \mathbb{C}_{ld}([0, T], \mathbb{R})$ which is a
Banach space of $ld$-continuous functions with the maximum norm
$\|u\|= \max_{[0, T]}|u(t)| $.

The main tool in this paper is an application of the
Guo-Krasnosel’skii fixed point theorem for operators leaving a
Banach space cone invariant.

\begin{theorem}[Krasnosel'skii \cite{kra}] \label{th21}
Let $\mathcal{B}$ be a Banach space, and let $\mathcal{P}\subset
\mathcal{B}$ be a cone in $\mathcal{B}$. Assume that $\Omega_1$
and $\Omega_2$ are open subsets of $\mathcal{B}$ with
$0\in\Omega_1\subset\overline{\Omega}_1\subset\Omega_2$, and
let
\begin{equation*}
G:\mathcal{P}\cap(\overline{\Omega}_2 \backslash \Omega_1)\to
\mathcal{P}
\end{equation*}
be a completely continuous operator such that either
\begin{itemize}
\item[(i)] $\|Gu\|\leq\|u\|$,
$u\in\mathcal{P}\cap\partial\Omega_1$, and  $\|Gu\|\geq\|u\|$,
$u\in \mathcal{P}\cap\partial\Omega_2$; or \item [(ii)]
$\|Gu\|\geq\|u\|$, $u\in \mathcal{P}\cap\partial\Omega_1$, and
$\| Gu\|\leq\|u\|$, $u\in \mathcal{P}\cap\partial\Omega_2$.
\end{itemize}
  Then, $G$ has a fixed
point in $\mathcal{P}\cap(\overline{\Omega}_2 \backslash
\Omega_1)$.
\end{theorem}

\section{ Main Results}\label{sec3}

By a positive solution of the eigenvalue problem
\eqref{eq1}-\eqref{eq2}, we understand a pair of  function $(u(t),
v(t))$ which is positive on $[0, T]_{\mathbb{T}}$, and satisfies
\eqref{eq1}-\eqref{eq2}.

\begin{lemma} \label{lm1}
Assume that hypotheses {\rm (H1)--(H2)} are satisfied. Then a pair
of functions $(u(t),
v(t))$ is a solution of \eqref{eq1}-\eqref{eq2} if and only if
$u(t), v(t) \in \mathbb{E}$ and $(u(t), v(t))$ satisfies the
system integral equations:
\begin{gather}\label{eq3}
u(t)= u(0) + \int_{0}^{t} \Big( A_1 -\lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s, \\
\label{eq4}
v(t)= v(0) + \int_{0}^{t} \Big( A_2 -\lambda \int_{0}^{s}
g(u(r)) \Delta r \Big) \Delta s,
\end{gather}
where
\begin{gather*}
u(0)= \frac{1}{1-\beta} \Big( \beta \int_{0}^{\eta} h_1(s)
\Delta s - \int_{0}^{T} h_1(s) \Delta s \Big), \\
v(0)= \frac{1}{1-\beta} \Big( \beta \int_{0}^{\eta} h_2(s)
\Delta s - \int_{0}^{T} h_2(s) \Delta
s \Big), \\
- h_1(s)= \int_{0}^{s}\lambda f(v(r)) \Delta r -A_1,\\
A_1= u^\Delta(0)= -\frac{\lambda
\beta}{1-\beta}\int_{0}^{\eta} f(v(r))
\Delta r, \\
- h_2(s)= \int_{0}^{s}\lambda g(u(r)) \Delta r -A_2,\\
A_2= u^\Delta(0)= -\frac{\lambda
\beta}{1-\beta}\int_{0}^{\eta} g(u(r)) \Delta r.
\end{gather*}
\end{lemma}

\begin{proof}
 Necessity. Integrating  \eqref{eq1} we have
$$
u^\Delta(s)= u^\Delta(0) - \int_{0}^{s}\lambda f(v(r))
\Delta r.
$$
On the other hand, by the boundary condition \eqref{eq2} we have
$$
u^\Delta(0) = \beta u^\Delta(\eta)= \beta \Big(
u^\Delta(0) - \int_{0}^{\eta}\lambda f(v(r)) \Delta r
\Big).
$$
Then
$$
A_1= u^\Delta(0) = \frac{-\lambda \beta}{1- \beta}
\int_{0}^{\eta}f(v(r)) \Delta r.
$$
It follows that
\begin{equation*}
u^\Delta(s)=  -\lambda \int_{0}^{s} f(v(r)) \Delta r +A_1.
\end{equation*}
Integrating the above equation we obtain
\begin{equation} \label{equa3}
u(t)= u(0) + \int_{0}^{t}\Big( A_1- \lambda \int_{0}^{s}
f(v(r)) \Delta r \Big)\Delta s.
\end{equation}
Moreover, by \eqref{equa3} and the boundary condition \eqref{eq2},
we have
\begin{align*}
u(0)&= u(T) - \int_{0}^{T}\Big( A_1- \lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s\\
&= \beta u(\eta) - \int_{0}^{T}\Big( A_1- \lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s \\
&= \beta \Big( u(0)+ \int_{0}^{\eta}\Big( A_1- \lambda
\int_{0}^{s} f(v(r)) \Delta r \Big) \Delta s \Big)
\\
&\quad - \int_{0}^{T} \Big( A_1- \lambda \int_{0}^{s} f(v(r)) \Delta
r \Big) \Delta s.
\end{align*}
Then
\begin{align*}
u(0)&= \frac{1}{1-\beta} \Big( \beta \int_{0}^{\eta} \Big(
A_1- \lambda \int_{0}^{s} f(v(r)) \Delta r \Big) \Delta s \\
&\quad - \int_{0}^{T} \Big( A_1- \lambda \int_{0}^{s} f(v(r))
\Delta r \Big) \Delta s \Big).
\end{align*}
Then we have \eqref{eq3}. Similar arguments need to be done  to
prove \eqref{eq4}.

Sufficiency. Simple calculations by taking the delta derivative of
$u(t)$ and $v(t)$ lead to the result.
\end{proof}


\begin{lemma}\label{lm32}
Suppose that {\rm (H1)--(H2)}  hold, then a solution $(u, v)$ of
\eqref{eq1}-\eqref{eq2} satisfies $u(t) \geq 0$ and $v(t) \geq 0$,
for $t \in [0, T]_{\mathbb{T}}$.
\end{lemma}

\begin{proof}
 Since $0 < \beta < 1$, we have
\begin{align*}
&u(0)\\
&= \frac{\beta}{1-\beta}  \int_{0}^{\eta} \Big( A_1-
\lambda \int_{0}^{s} f(v(r)) \Delta r \Big) \Delta s
- \frac{1}{1-\beta} \int_{0}^{T} \Big( A_1- \lambda \int_{0}^{s}
 f(v(r)) \Delta r \Big) \Delta s \\
& \geq \frac{\beta}{1-\beta} \Big\{ \int_{0}^{\eta} \Big(
A_1- \lambda \int_{0}^{s} f(v(r)) \Delta r \Big) \Delta
s  -  \int_{0}^{T} \Big( A_1- \lambda \int_{0}^{s} f(v(r))
\Delta r \Big) \Delta s
 \Big\}\\
&\geq 0,
\end{align*}
and
 \begin{align*}
u(T) &=  u(0) + \int_{0}^{T} h_1(s) \Delta s \\
& =  \frac{\beta}{1-\beta} \int_{0}^{\eta} h_1(s) \Delta s -
\frac{1}{1-\beta} \int_{0}^{T} h_1(s) \Delta s +
 \int_{0}^{T} h_1(s) \Delta s \\
& \geq \frac{\beta}{1-\beta} \int_{0}^{\eta} h_1(s) \Delta s
- \frac{\beta}{1-\beta} \int_{0}^{T} h_1(s) \Delta s \\
 &  \geq \frac{\beta}{1-\beta}   \Big(   \int_{0}^{\eta} h_1(s) \Delta s
-  \int_{0}^{T} h_1(s) \Delta s  \Big) \\
& \geq  \frac{-\beta}{1-\beta} \int_{\eta}^{T} h_1(s) \Delta s \\
& \geq \frac{\beta}{1-\beta} \int_{\eta}^{T} (-h_1(s)) \Delta
s \geq 0.
\end{align*}

If $t \in [0, T]_{\mathbb{T}}$, then
\begin{align*}
u(t)&= u(0) + \int_{0}^{t}\Big( A_1- \lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s\\
& \geq  u(0) +  \int_{0}^{T}\Big( A_1- \lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s \\
&= u(T)  \geq 0 \,.
\end{align*}

Arguing exactly as above, we have $v(t) \geq 0$. The proof
is now complete.
\end{proof}

\begin{lemma}\label{lm33}
Suppose that {\rm (H1)-(H2)} are satisfied, then
$$
u(T)= \inf_{t \in \mathbb{T}} u(t) \geq \rho u(0)= \rho \|u\|,
$$
where $ 1\geq  \rho = \beta \frac{T-\eta}{T- \beta \eta} \geq 0$.
\end{lemma}

\begin{proof} Since $ u^\Delta(0) \leq 0$  we have
$u^\Delta(s)= u^\Delta(0) - \lambda \int_{0}^{s} f(v(r))
\Delta r \leq 0$, then $u^\Delta \leq 0$, which implies  that
$u$ is a non-increasing function on $[0, T]_{\mathbb{T}}$.
Moreover, we have $\|u\|=u(0), \, \inf_{t \in (0,
T)_{\mathbb{T}}}u(t) = u(T)$. Hence,
 from the concavity of $u(t)$ that each point on chord
between $(0, u(0))$ and $(T, u(T))$ is below the graph of $u(t)$,
we have
$$
u(T) \geq u(0)+T \frac{u(T)-u(\eta)}{T-\eta}.
$$
On other terms
 $$
Tu(\eta) - \eta u(T) \geq (T- \eta)u(0).
 $$
Using the boundary condition \eqref{eq2}, it follows
 $$
 (\frac{T}{\beta}- \eta) u(T) \geq (T- \eta)u(0).
 $$
Then
$$
 u(T) \geq \beta \frac{T-\eta}{T- \beta \eta} u(0).
$$
\end{proof}

For our construction of an operator $G$, define a cone
 $\mathbb{K}\subset\mathbb{E}$
by
\begin{equation}\label{e36}
\mathbb{K}=\big\{u\in \mathbb{E}:  u(t)\geq 0
\text{ on $\mathbb{T}$ and } u(t)\geq \rho \| u\|, \text{ for }
t\in \mathbb{T}\big\}.
\end{equation}
Also define the positive numbers
\begin{gather*}
\lambda_1 = \max \Big( \Big(  \frac{\beta(T-\eta)
\eta}{(1-\beta)}\sqrt{\rho
(f_{\infty}-\varepsilon)(g_{\infty}-\varepsilon)} \Big)^{-1},
 \Big( \frac{\beta}{1-\beta}(g_{\infty} - \epsilon) \eta
 \Big)^{-1} \Big),
\\
\lambda_2 = \min \Big( \Big(\frac{2\sqrt{2}
T^{2}}{(1-\beta)^{2}}
\sqrt{(f_{0}+\varepsilon)(g_{0}+\varepsilon)} \Big)^{-1},
\Big( \frac{2T^{2}}{(1-\beta)^{2}}(g_{0}+\epsilon ) \Big)^{-1}
\Big).
\end{gather*}


\begin{theorem}\label{t31}
Assume that conditions {\rm (h1)--(H2)} are satisfied.
Then, for each $\lambda$ satisfying
$\lambda_1 < \lambda < \lambda_2$,
there exists a pair $(u, v)$ satisfying \eqref{eq1}-\eqref{eq2}
such that $u(t) \geq 0$ and $v(t) \geq  0$ on $[0, T]_{\mathbb{T}}$.
\end{theorem}

\begin{proof}
The main idea of the proof is to use Theorem \ref{th21}.
First observe that we seek suitable fixed points in the cone
$\mathbb{K}$ \ref{lm32} of the  integral operator
\begin{equation}\label{eq6}
Gu(t)= u(0) + \int_{0}^{t} \Big( A_1 -\lambda \int_{0}^{s}
f(v(r)) \Delta r \Big) \Delta s,
\end{equation}
where $v$ is a function of $u$ given by \eqref{eq4}.

\begin{lemma}
Let $G$ defined by \eqref{eq6}, then
\begin{itemize}
\item[(i)] $G(\mathbb{K}) \subseteq \mathbb{K}.$
\item[(ii)] $G: \mathbb{K} \to \mathbb{K}$ is completely continuous.
\end{itemize}
\end{lemma}

Notice that by Lemma~\ref{lm32} and  Lemma~\ref{lm33}, (i)
is satisfied. On the other hand standard arguments show that $G$
is completely continuous.

Since $\beta <1$ we have
\begin{equation}\label{eq7}
\begin{aligned}
Gu(t)& \leq u(0)  = \frac{1}{1-\beta} \Big( -\beta
\int_{0}^{\eta} h_1(s) \Delta s + \int_{0}^{T} h_1(s)
\Delta
s \Big) \\
& \leq \frac{2}{1-\beta} \int_{0}^{T}| h_1(s)| \Delta s.
\end{aligned}
\end{equation}
 Now, from the definitions of $f_0$ and $g_0$, there exists
$H_1>0$ such that
\begin{equation}\label{cond1}
f(x)\leq(f_0+\epsilon)x,~~g(x)\leq(g_0+\epsilon)x,\quad  \mbox{
for all } x  \mbox{ such }  0<x\leq H_1.
\end{equation}
Let $u \in \mathbb{K}$ with $\|u\|= H_1$. We have
\begin{equation}\label{eq9}
\begin{aligned}
|v(r)|
& = \big |v(0) + \int_{0}^{r} h_2(s) \Delta s \big |\\
& \leq |v(0)| + \int_{0}^{r} |h_2(s)| \Delta s\\
& \leq |v(0)| + \int_{0}^{\eta} |h_2(s)| \Delta s.
\end{aligned}
\end{equation}
Furthermore,
\begin{equation}\label{eq10}
\begin{aligned}
|h_2(s)|
& = \big |A_2- \lambda \int_{0}^{s}g(u(r)) \Delta r \big |\\
& \leq |A_2| + \lambda \int_{0}^{s}|g(u(r))| \Delta r\\
& \leq |A_2| + \lambda  (g_{0}+\epsilon) |s| \|u\|\\
& \leq \frac{\lambda \beta}{1-\beta} \int_{0}^{\eta}|g(u(r))|
\Delta r + \lambda  (g_{0}+\epsilon) |s| \|u\|\\
& \leq \frac{\lambda \beta}{1-\beta}  (g_{0}+\epsilon)
\eta \|u\| + \lambda  (g_{0}+\epsilon) |s| \|u\|\\
& \leq \frac{\lambda T}{1-\beta} (g_{0}+\epsilon) \|u\|.
\end{aligned}
\end{equation}
Then
\begin{equation}\label{eq11}
\int_{0}^{T} |h_2(s)| \Delta s  \leq \frac{\lambda
T^{2}}{1-\beta}(g_{0}+ \epsilon) \|u\|.
\end{equation}
Then using \eqref{eq11}, we have
\begin{equation}\label{eq12}
\begin{aligned}
|v(0)| & = |\frac{\beta}{1-\beta}  \int_{0}^{\eta} h_2(s)
\Delta s - \frac{1}{1-\beta} \int_{0}^{T} h_2(s) \Delta s| \\
&  \leq \frac{\beta}{1-\beta}  \int_{0}^{\eta} |h_2(s)|
\Delta s +
\frac{1}{1-\beta} \int_{0}^{T} |h_2(s)| \Delta s  \\
& \leq \frac{1+ \beta}{1-\beta}  \int_{0}^{T} |h_2(s)| \Delta
s \\
& \leq \lambda T^{2} \frac{1+ \beta}{(1-\beta)^{2}} (g_{0}+
\epsilon) \|u\|.
\end{aligned}
\end{equation}
Then from \eqref{eq9}-\eqref{eq12}, we obtain
\begin{equation}\label{eq13}
\begin{aligned}
|v(r)| & \leq \lambda T^{2} \frac{1+ \beta}{(1-\beta)^{2}} (g_{0}+
\epsilon) \|u\| + \frac{\lambda T^{2}}{(1-\beta)} (g_{0}+
\epsilon)
\|u\| \\
& \leq \frac{2\lambda T^{2}}{(1-\beta)^{2}} (g_{0}+ \epsilon)
\|u\|.
\end{aligned}
\end{equation}
The choice of $\lambda_2$ it yields
\begin{equation}\label{eq12bis}
\|v\|  \leq \frac{2\lambda T^{2}}{(1-\beta)^{2}} (g_{0}+ \epsilon)
 \|u\|
 \leq \frac{2\lambda_2 T^{2}}{(1-\beta)^{2}} (g_{0}+ \epsilon) \|u\|
 \leq  \|u\| = H_1.
\end{equation}
Then by using \eqref{cond1} and \eqref{eq12bis} we obtain
\begin{equation}\label{eq8}
|A_1| \leq \frac{\lambda \beta}{1-\beta}
\int_{0}^{\eta}|f(v(r))| \Delta r  \leq \frac{\lambda
\beta}{1-\beta} (f_{0} + \epsilon) \int_{0}^{\eta}|v(r)| \Delta r.
\end{equation}
It follows from \eqref{eq13}-\eqref{eq8} that
\begin{equation}\label{eq14}
\begin{aligned}
|A_1| &  \leq \frac{\lambda \beta}{1-\beta} (f_{0}+ \epsilon)
\eta \frac{2 \lambda T^{2}}{(1-\beta)^{2}} (g_{0}+ \epsilon) \|u\|\\
&= \frac{2\lambda^{2} \beta T^{2} \eta}{(1-\beta)^{3}}(f_{0}+
\epsilon)(g_{0}+ \epsilon)  \|u\|.
\end{aligned}
\end{equation}
Since $h_1(s) = A_1- \lambda \int_{0}^{s} f(v(r)) \Delta r$,
it follows that
$$
|h_1(s)| \leq |A_1| + \lambda \int_{0}^{s} |f(v(r))| \Delta r.
$$
We obtain from \eqref{cond1} and \eqref{eq13}, that
\begin{equation}\label{eq15}
\begin{aligned}
 \int_{0}^{s} |f(v(r))| \Delta r & \leq \int_{0}^{s}
(f_{0} + \epsilon)  \|v\| \Delta r \leq  T (f_{0} +
\epsilon) \|v\| \\
&  \leq \frac{2 \lambda T^{3}}{(1-\beta)^{2}}  (f_{0}+ \epsilon)
(g_{0}+ \epsilon) \|u\|.
\end{aligned}
\end{equation}
It follows from \eqref{eq14}-\eqref{eq15} that
\begin{equation}\label{eq156}
\begin{aligned}
|h_1(s)|
& \leq   \Big( \frac{2 \lambda^{2} \beta
T^{2}\eta}{(1- \beta)^{3}} (f_{0}+ \epsilon) (g_{0}+ \epsilon)
\|u\| + \frac{2 \lambda^{2} T^{3}}{(1-
\beta)^{2}} (f_{0}+ \epsilon) (g_{0}+ \epsilon) \|u\| \Big) \\
& \leq     \frac{4 \lambda^{2}T^{3}}{(1- \beta)^{3}} (f_{0}+
\epsilon) (g_{0}+ \epsilon) \|u\|.
\end{aligned}
\end{equation}
Combining inequalities \eqref{eq13}-\eqref{eq15}, we have,
from \eqref{eq7} and the choice of $\lambda_2$,
\begin{align*}%\label{eq17}
\|Gu\| & \leq \frac{2}{1- \beta} \int_{0}^{T} |h_1(s)| \Delta s
\\
& \leq   \frac{8 \lambda^{2}T^{4}}{(1- \beta)^{4}}
(f_{0}+ \epsilon) (g_{0}+ \epsilon) \|u\| \\
& \leq  \lambda^{2}_2 \frac{8 T^{4}}{(1- \beta)^{4}} (f_{0}+
\epsilon) (g_{0}+ \epsilon) \|u\| = \|u\|.
\end{align*}
So $\|Gu\|  \leq \|u\| $. If we set
$\Omega_1 = \{ u \in \mathbb{E} : \|u\| < H_1 \}$, then
$$
\|Gu\|  \leq \|u\| \quad \mbox{for } u \in \mathbb{K}
\cap \partial \Omega_1.
$$
Now, from the definitions of $f_{\infty}$ and
$g_{\infty}$, there exists $\overline{H}_2>0$ such that
\begin{equation}\label{cond2}
f(x) \geq (f_{\infty}-\varepsilon)x, \quad
g(x) \geq (g_{\infty}-\varepsilon)x, \quad \mbox{for }
x \geq \overline{H}_2>0.
\end{equation}
Let $H_2= \max \{ 2H_1, \frac{\overline{H}_2}{\rho}\} $. Let
$u \in \mathbb{K}$ with $\|u\| = H_2, \min u(t) \geq \rho \|u\|
\geq \overline{H}_2$.

On the other hand,
\begin{equation}\label{eq18bis}
\begin{aligned}
Gu(t) &=  u(0) + \int_{0}^{t} h_1(s) \Delta s \\
& =  \frac{\beta}{1-\beta} \int_{0}^{\eta} h_1(s) \Delta s
- \frac{1}{1-\beta} \int_{0}^{T} h_1(s) \Delta s +
 \int_{0}^{t} h_1(s) \Delta s \\
& \geq \frac{\beta}{1-\beta} \int_{0}^{\eta} h_1(s) \Delta s
- \frac{\beta}{1-\beta} \int_{0}^{T} h_1(s) \Delta s \\
 &  \geq \frac{\beta}{1-\beta}   \Big(   \int_{0}^{\eta} h_1(s) \Delta s
-  \int_{0}^{T} h_1(s) \Delta s  \Big) \\
& \geq  \frac{-\beta}{1-\beta} \int_{\eta}^{T} h_1(s) \Delta s \\
& \geq \frac{\beta}{1-\beta} \int_{\eta}^{T} (-h_1(s)) \Delta s.
\end{aligned}
\end{equation}
Furthermore, from \eqref{eq4} and as in \eqref{eq18bis} we have
$$
v(t) \geq \frac{\beta}{1-\beta} \int_{\eta}^{T} (-h_2(s)) \Delta s.
$$
Since $A_2 \leq 0$, we have from \eqref{cond2}
\begin{align*}
-h_2(s) & \geq \lambda \int_{0}^{s} g(u(r)) \Delta r -A_2\\
& \geq  \lambda \int_{0}^{s} g(u(r)) \Delta r \\
& \geq \lambda(g_{\infty}- \epsilon) \int_{0}^{s}  u(r) \Delta r \\
& \geq \lambda (g_{\infty}- \epsilon) \int_{0}^{s} \rho \|u\| \Delta r \\
& \geq \lambda (g_{\infty}- \epsilon) \rho \|u\| s.
\end{align*}
Then by the choice of $\lambda_1$,
\begin{equation}\label{eq22bis}
\begin{aligned}
 v(t)  &  \geq \frac{\lambda \beta}{1-\beta}  (g_{\infty}- \epsilon) \rho \|u\|\int_{\eta}^{T} s  \Delta s \\
 & \geq  \lambda \frac{\beta}{1-\beta} (g_{\infty}- \epsilon) \rho \|u\|  \eta \\
 & \geq  \lambda_1 \frac{\beta}{1-\beta} (g_{\infty}- \epsilon) \rho   \eta \|u\| \\
 & \geq  \rho \|u\|
  \geq \overline{H_2}.
\end{aligned}
\end{equation}

Since $A_1 \leq 0$, then  by \eqref{eq22bis} we have
\begin{align*} %\label{eq18}
- h_1(s)  = \lambda \int_{0}^{s} f(v(r)) \Delta r - A_1
 \geq \lambda \int_{0}^{s} f(v(r)) \Delta r
 \geq  \lambda (f_{\infty} -\epsilon) \int_{0}^{\eta} v(r)
\Delta r.
\end{align*}
We also have
\[ %\label{eq19}
v(r)  = v(0) + \int_{0}^{s} h_2(s) \Delta s
 \geq   \frac{\beta}{1- \beta} \int_{\eta}^{T}(-h_2(s)) \Delta s.
\]
Applying  Lemma~\ref{lm33}, we have
\begin{align*} %\label{eq20}
 -h_2(s)&= \lambda \int_{0}^{s} g(u(r)) \Delta r - A_2\\
& \geq \lambda \int_{0}^{s} g(u(r)) \Delta r\\
& \geq \lambda \int_{0}^{\eta} g(u(r)) \Delta r\\
& \geq \lambda  (g_{\infty} -\epsilon)\int_{0}^{\eta} u(r) \Delta r\\
& \geq \lambda (g_{\infty}-\epsilon) \rho \eta \|u\|.
\end{align*}
Consequently,
\begin{equation*} %\label{eq21}
v(r) \geq  \lambda \frac{\beta}{1-\beta} (T- \eta)
(g_{\infty}-\epsilon) \rho \eta \|u\|.
\end{equation*}
Then
\begin{equation*} % \label{eq22}
-h_1(s) \geq \lambda^{2} (T - \eta) \frac{\beta  \rho
\eta^{2}}{1-\beta} (f_{\infty}-\epsilon) (g_{\infty}-\epsilon)
\|u\|.
\end{equation*}
Finally, for $\lambda \geq \lambda_1$, we obtain
\begin{align*} %\label{eq23}
Gu(t) & = \frac{\beta}{1-\beta} \int_{\eta}^{T} (-h_1(s))
\Delta s \\
& \geq  \lambda^{2} \frac{\beta^{2} \eta^{2}(T-
\eta)^{2}}{(1-\beta)^{2}} \rho (f_{\infty}-\epsilon)
(g_{\infty}-\epsilon) \|u\| \\
& \geq \lambda_1^{2} \frac{\beta^{2} \eta^{2}(T-
\eta)^{2}}{(1-\beta)^{2}} \rho (f_{\infty}-\epsilon)
(g_{\infty}-\epsilon) \|u\| \\
& \geq \|u\|= H_2.
\end{align*}
Hence, $ \|Gu\| \geq  \|u\|$. So, if we set
$\Omega_2 = \big\{u \in \mathbb{E}: \|u\| < H_2 \big\}$, then
$$
\|Gu\| \geq  \|u\|, \quad \mbox{for } u \in \mathbb{K} \bigcap
\partial \Omega_2.
$$
Applying Theorem~\ref{th21}, we obtain that $G$ has a fixed point
$u \in \mathbb{K} \bigcap (\overline{\Omega_2} / \Omega_1)$.
 With $v$ being defined by
$$
v(t)= v(0) + \int_{0}^{t} \Big( A_2 -\lambda \int_{0}^{s}
g(u(r)) \Delta r \Big) \Delta s,
$$
the pair $(u, v)$ is a  solution of \eqref{eq1}-\eqref{eq2}
for the given $\lambda$. The proof is now complete.
\end{proof}

  We point out that, under technical calculus,
results of this paper can be extended to the following
$p$-Laplacian case with $2\leq p \leq +\infty$:
\begin{equation} \label{eqp}
\begin{gathered}
 -\Big(\phi_{p}( u^\Delta(t))\Big)^\Delta= \lambda f(v(t)) ,
\quad \forall t \in [0, T]_{\mathbb{T}},\\
 -\Big(\phi_{p}( v^\Delta(t))\Big)^\Delta= \lambda g(u(t)) , \quad
\forall t \in [0, T]_{\mathbb{T}},
  \end{gathered}
\end{equation}
where $\phi_{p}(\xi)= | \xi |^{p-2}\xi$.
If $p=2$ we obtain problem \eqref{eq1}.

\subsection*{Acknowledgements}
This work was supported by the project
\emph{New Explorations in Control Theory Through Advanced Research}
(NECTAR) cofinanced by \emph{Funda\c{c}\~{a}o para a Ci\^{e}ncia e
a Tecnologia} (FCT), Portugal,
and the \emph{Centre National de la Recherche Scientifique et Technique} (CNRST), Morocco.
A.B. Malinowska is a senior researcher at the
\emph{Center for Research and Development in Mathematics and
Applications} (CIDMA) at the University of Aveiro, Portugal,
under the support of BUT grant S/WI/02/2011.


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\end{document}
