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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 127, 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-2009/127\hfil Positive solutions]
{Positive solutions for a system of nonlinear
 boundary-value problems on time scales}

\author[A. K. Rao \hfil EJDE-2009/127\hfilneg]
{A. Kameswara Rao}

\address{A. Kameswara Rao \newline
Department of Applied Mathematics, Andhra University,
Visakhapatnam 530 003, India}
 \email{kamesh\_1724@yahoo.com}


\thanks{Submitted July 6, 2009. Published October 4, 2009.}
\subjclass[2000]{39A10, 34B15, 34A40}
\keywords{Dynamic equations; eigenvalue intervals; positive
solution; cone}

\begin{abstract}
 We determine the values of a parameter $\lambda$ for which there
 exist positive solutions to the system of dynamic equations
 \begin{gather*}
 u^{\Delta \Delta}(t)+\lambda p(t)f(v(\sigma(t)))=0,\quad t\in[a, b]_\mathbb{T}, \\
 v^{\Delta \Delta}(t)+\lambda q(t)g(u(\sigma(t)))=0,
 \quad t\in[a, b]_\mathbb{T},
 \end{gather*}
 with the  boundary conditions,
 $\alpha u(a)-\beta u^{\Delta}(a)=0$,
 $\gamma u(\sigma^2(b))+\delta u^{\Delta}(\sigma(b))=0$,
 $\alpha v(a)-\beta v^{\Delta}(a)=0$,
 $\gamma v(\sigma^2(b))+\delta v^{\Delta}(\sigma(b))=0$, where
 $\mathbb{T}$ is a time scale. To this end we apply a Guo-Krasnosel'skii
 fixed point theorem.
\end{abstract}

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

\section{Introduction}

Let $\mathbb{T}$ be a time scale with $a, \sigma^2(b)\in \mathbb{T}$.
Given an interval $J$ of $\mathbb{R}$, we will use the interval notation
\begin{equation}\label{e11}
J_\mathbb{T}=J\cap\mathbb{T}.
\end{equation}
 We are concerned with determining values of $\lambda$
(eigenvalues) for which there exist positive solutions for the
system of dynamic equations
\begin{equation}\label{e12}
\begin{gathered}
u^{\Delta \Delta}(t)+\lambda p(t)f(v(\sigma(t)))=0, \quad t\in[a,b]_\mathbb{T},\\
v^{\Delta \Delta}(t)+\lambda q(t)g(u(\sigma(t)))=0, \quad t\in[a,b]_\mathbb{T},
 \end{gathered}
\end{equation}
satisfying the boundary conditions
\begin{equation}\label{e13}
\begin{gathered}
\alpha u(a)-\beta u^{\Delta}(a)=0, \quad
 \gamma u(\sigma^2(b))+\delta u^{\Delta}(\sigma(b))=0,\\
\alpha v(a)-\beta v^{\Delta}(a) =0,\quad
\gamma v(\sigma^2(b))+\delta v^{\Delta}(\sigma(b))=0\,.
 \end{gathered}
\end{equation}
We will use the following assumptions:
\begin{itemize}
\item[(A1)] $f, g\in C([0, \infty), [0, \infty))$;
\item[(A2)] $p, q\in C([a, \sigma(b)]_\mathbb{T}, [0, \infty))$,
 and each function does not vanish
identically on any closed subinterval of $[a,\sigma(b)]_\mathbb{T}$;
\item[(A3)] the following limits exist as real numbers:\\
$f_0:=\lim_{x\to 0^{+}}f(x)/x$,
$g_0:=\lim_{x\to 0^{+}}g(x)/x$,\\
$f_\infty:=\lim_{x\to \infty}f(x)/x$, and
$g_\infty:=\lim_{x\to \infty}g(x)/x$
\end{itemize}

There is an ongoing flurry of research activities devoted to
positive solutions of dynamic equations on time scales. This work
entails an extension of the paper by Chyan and Henderson \cite{c2}
to eigenvalue problem for system of nonlinear boundary value
problems on time scales. Also, in that light, this paper is
closely related to the works of Li and Sun \cite{l1,s1}.

On a larger scale, there has been a great deal of study focused on
positive solutions of boundary value problems for ordinary
differential equations. Interest in such solutions is high from a
theoretical sense \cite{e2,g2,h2,k1,k2} and as applications
for which only positive solutions are meaningful \cite{a2,g1,h5,w2}.
These considerations are caste primarily for scalar problems,
but good attention has been given to boundary value problems for
systems of differential equations \cite{h3,h4,i1,i2}.

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 \cite{e2}. A Green function plays a
fundamental role in defining an appropriate operator on a suitable
cone.

 \section{Green's Function and Bounds}

In this section, we state the well-known Guo-Krasnosel'skii fixed
point theorem which we will apply to a completely continuous
operator whose kernel, $G(t, s)$ is the Green's function for
 \begin{equation}\label{e21}
 \begin{gathered}
-y^{\Delta \Delta}=0,\\
\alpha u(a)-\beta u^{\Delta}(a)=0, \quad
\gamma u(\sigma^2(b))+\delta u^{\Delta}(\sigma(b))=0
\end{gathered}
\end{equation}
is given by
 \begin{equation}\label{e22}
 G(t,s)= \frac{1}{d}
\begin{cases}
\{\alpha(t-a)+\beta\}\{\gamma(\sigma^2(b)-\sigma(s))+\delta\}:
&  a\leq t\leq s\leq\sigma^2(b)\\
\{\alpha(\sigma(s)-a)+\beta\}\{\gamma(\sigma^2(b)-t)+\delta\}: &
a\leq \sigma(s)\leq t\leq \sigma^2(b)
\end{cases}
\end{equation}
where $\alpha, \beta, \gamma, \delta\geq 0$ and
$$
d:=\gamma \beta+\alpha \delta+\alpha \gamma(\sigma^2(b)-a)>0.
$$
One can easily check that
\begin{equation}\label{e23}
G(t, s)>0, \quad
(t, s)\in (a, \sigma^2(b))_\mathbb{T}\times(a, \sigma(b))_\mathbb{T}
\end{equation}
and
\begin{equation}\label{e24}
G(t,s)\leq G(\sigma(s),s)=\frac{[\alpha(\sigma(s)-a)+\beta]
[\gamma(\sigma^2(b)-\sigma(s))+\delta]}{d}
\end{equation}
for
$t\in [a, \sigma^2(b)]_\mathbb{T}$, $s\in [a, \sigma(b)]_\mathbb{T}$.
Let $I=\big[\frac{3a+\sigma^2(b)}{4}, \frac{a+3\sigma^2(b)}{4}\big]_\mathbb{T}$. Then
\begin{equation}\label{e25}
G(t, s)\geq kG(\sigma(s), s)
=k\frac{[\alpha(\sigma(s)-a)+\beta][\gamma(\sigma^2(b)-\sigma(s))
+\delta]}{d}
\end{equation}
for $t\in I$, $s\in [a, \sigma(b)]_\mathbb{T}$, where
\begin{equation}\label{e26}
k=\min\Bigg\{\frac{\gamma(\sigma^2(b)-a)+4\delta}{4(\gamma(\sigma^2(b)-a)
+\delta)}, \;
\frac{\alpha(\sigma^2(b)-a)+4\beta}{4(\alpha(\sigma^2(b)-a)+\beta)}\Bigg\}.
\end{equation}
We note that a pair $(u(t), v(t))$ is a solution of the eigenvalue
problem \eqref{e12}, \eqref{e13} if and only if
\begin{equation}\label{e27}
\begin{gathered}
u(t)=\lambda\int_{a}^{\sigma(b)} G(t,s)p(s)f
\Big(\lambda\int_{a}^{\sigma(b)} G(\sigma(s),r)q(r)g(u(\sigma(r)))\Delta
r\Big)\Delta s,
a\leq t \leq\sigma^2 (b),\\
v(t)=\lambda \int_{a}^{\sigma(b)}G(t,s)q(s)g(u(\sigma(s)))\Delta s,
\quad a\leq t\leq \sigma^2 (b).
\end{gathered}
\end{equation}

 Values of $\lambda$ for which there are positive solutions
(positive with respect to a cone) of \eqref{e12}, \eqref{e13} will
be determined via applications of the following fixed point theorem
\cite{k1}.

\begin{theorem}[Krasnosel'skii]  \label{t21}
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}\label{e28}
T:\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)] $\|Tu\|\leq\|u\|$, $u\in\mathcal{P}\cap\partial\Omega_{1}$,
and  $\|Tu\|\geq\|u\|$, $u\in \mathcal{P}\cap\partial\Omega_{2}$;
or
\item [(ii)]
$\|Tu\|\geq\|u\|$, $u\in \mathcal{P}\cap\partial\Omega_{1}$, and
$\| Tu\|\leq\|u\|$, $u\in \mathcal{P}\cap\partial\Omega_{2}$.
\end{itemize}
  Then, $T$ has a fixed
point in $\mathcal{P}\cap(\overline{\Omega}_{2} \backslash \Omega_{1})$.
\end{theorem}

\section{Positive Solutions in a Cone}

In this section, we apply Theorem \ref{t21} to obtain solutions in a
cone (i.e., positive solutions) of \eqref{e12}, \eqref{e13}. Assume
throughout that $[a, \sigma^2(b)]_\mathbb{T}$ is such that
\begin{equation}\label{e31}
\begin{gathered}
\xi =\min \Big\{ t\in \mathbb{T}: t\geq\frac{3a+\sigma^2(b)}{4}\Big\},\\
\omega =\max\Big\{t\in \mathbb{T}: t\leq
\frac{a+3\sigma^2(b)}{4}\Big\};
\end{gathered}
\end{equation}
both exist and satisfy
\begin{equation}\label{e32}
\frac{3a+\sigma^2(b)}{4}\leq \xi < \omega \leq
\frac{a+3\sigma^2(b)}{4}.
\end{equation}
Next, let $\tau \in [\xi, \omega]_\mathbb{T}$ be defied by
\begin{equation}\label{e33}
\int_{\xi}^{\omega}G(\tau, s)p(s)\Delta s=\max_{t\in
[\xi,\omega]_\mathbb{T}}\int_{\xi}^{\omega}G(t,s)p(s)\Delta s.
\end{equation}
Finally, we define
\begin{gather}\label{e34}
l=\min_{s\in[a,\sigma(b)]_\mathbb{T}}
\frac{G(\sigma(\omega),s)}{G(\sigma(s), s)}, \\
\gamma=\min\{k,l\}.  \label{e35}
\end{gather}
For our construction, let
$\mathcal{B}=\{x:[a, \sigma^2(b)]_\mathbb{T}\to \mathbb{R}\}$ with
supremum norm $\|x\|=\sup \{| x(t)|:t\in [a,\sigma^2(b)]_\mathbb{T}\}$
and define a cone
$\mathcal{P}\subset\mathcal{B}$ by
\begin{equation}\label{e36}
\mathcal{P}=\Big\{x\in \mathcal{B}| x(t)\geq 0~ {\rm on}~
[a,\sigma^2(b)]_\mathbb{T}, \quad\text{and}\quad x(t)\geq \gamma \|
x\|, ~{\rm for} ~t\in [\xi, \omega]_\mathbb{T}\Big\}.
\end{equation}
For our first result, define positive numbers $L_1$ and $L_2$, by
% \label{e37}
\begin{gather*}
L_1:=\max\Big\{\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)p(s)\Delta
sf_{\infty}\Big]^{-1},
\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)q(s)\Delta s
 g_{\infty}\Big]^{-1}\Big\},\\
L_2:=\min\Big\{\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)p(s)\Delta s
f_{0}\Big]^{-1},
\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)q(s)\Delta s g_{0}\Big]^{-1}\Big\}.
\end{gather*}

\begin{theorem}\label{t31}
Assume that conditions {\rm (A1)--(A3)} are satisfied. Then, for each
$\lambda$ satisfying
\begin{equation}\label{e38}
L_1<\lambda< L_2,
\end{equation}
there exists a pair $(u, v)$ satisfying \eqref{e12}, \eqref{e13}
such that $u(x)>0$ and $v(x)>0$ on $(a,\sigma^2(b))_\mathbb{T}$.
\end{theorem}

\begin{proof}
Let $\lambda$ be as in \eqref{e38}. And let $\epsilon >0$ be chosen
such that %\label{e39}
\[
\max\Big\{\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)p(s)\Delta s
(f_{\infty}-\epsilon)\Big]^{-1},
\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)q(s)\Delta s
(g_{\infty}-\epsilon)\Big]^{-1}\Big\}\leq \lambda
\]
\begin{align*}
\lambda \leq\min\Big\{&\Big[\int_{a}^{\sigma(b)}G(\sigma(s),
s)p(s)\Delta s (f_{0}+\epsilon)\Big]^{-1},\\
&\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)q(s)\Delta s
(g_{0}+\epsilon)\Big]^{-1}\Big\}.
\end{align*}
Define an integral operator $T:\mathcal{P}\to \mathcal{B}$
by
\begin{equation}\label{e310}
Tu(t)=\lambda\int_{a}^{\sigma(b)}G(t,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s.
\end{equation}
By the remarks in Section 2, we seek suitable fixed points of $T$ in
the cone $\mathcal{P}$.

 Notice from (A1), (A2), and \eqref{e23} that, for $u\in \mathcal{P}$,
$Tu(t)\geq 0$ on $[a,\sigma^2(b)]_\mathbb{T}$. Also, for $u\in \mathcal{P}$,
we have from \eqref{e24} that
\begin{equation}\label{e311}
\begin{aligned}
Tu(t)&:=\lambda\int_{a}^{\sigma(b)}G(t,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\leq \lambda\int_{a}^{\sigma(b)}G(\sigma(s),s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s
\end{aligned}
\end{equation}
 so that
 \begin{equation}\label{e312}
\|Tu\|\leq\lambda\int_{a}^{\sigma(b)}G(\sigma(s),s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s), r)q(r)g(u(\sigma(r)))\Delta
r\Big)\Delta s.
\end{equation}
Next, if $u\in \mathcal{P}$, we have from \eqref{e25}, \eqref{e35},
and \eqref{e310} that
\begin{equation}\label{e313}
\begin{aligned}
&\min _{t\in [\xi, \omega]_\mathbb{T}}Tu(t)\\
&=\min _{t\in [\xi,
\omega]_\mathbb{T}}\lambda\int_{a}^{\sigma(b)}G(t,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\geq \lambda\gamma\int_{a}^{\sigma(b)}G(\sigma(s),s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\geq \gamma \|Tu\|.
\end{aligned}
\end{equation}
Consequently, $T:\mathcal{P}\to\mathcal{P}$. In addition,
standard arguments shows that $T$ is completely continuous.

 Now, from the definitions of $f_0$ and $g_0$, there exists
$H_1>0$ such that
\[ %\label{e314}
f(x)\leq(f_0+\epsilon)x,~~g(x)\leq(g_0+\epsilon)x,\quad 0<x\leq H_1.
\]
 Let $u\in\mathcal{P}$ with $\|u \|=H_1$. We
first have from \eqref{e24} and choice of $\epsilon$, for
$a\leq s\leq \sigma(b)$, that
%\label{e315}
\begin{align*}
\lambda\int_{a}^{\sigma(b)}G(\sigma(s), r)q(r)g(u(\sigma(r)))\Delta r
&\leq \lambda
\int_{a}^{\sigma(b)}G(\sigma(r), r)q(r)g(u(\sigma(r)))\Delta r\\
&\leq \lambda
\int_{a}^{\sigma(b)}G(\sigma(r), r)q(r)(g_0+\epsilon)u(r)\Delta r\\
& \leq\lambda\int_{a}^{\sigma(b)}G(\sigma(r), r)q(r)
\Delta r(g_0+\epsilon)\|u\|\\
&\leq\|u\|=H_1.
\end{align*}
As a consequence, we next have from \eqref{e24} and choice of
$\epsilon$, for $a\leq t\leq \sigma^2(b)$, that
 % \label{e316}
\begin{align*}
Tu(t)&=\lambda\int_{a}^{\sigma(b)}G(t,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\leq\lambda\int_{a}^{\sigma(b)}G(\sigma(s),s)p(s)
(f_0+\epsilon)\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r \Delta s\\
&\leq\lambda \int_a^{\sigma(b)} G(\sigma(s), s)p(s)(f_0+\epsilon)H_1 \Delta s\\
&\leq H_1=\|u\|.
\end{align*}
So, $\|Tu \|\leq \|u \|$. If we set%\label{e317}
$\Omega_1=\{x\in\mathcal{B}|~~\|x\|<H_1\}$,
then
\begin{equation}\label{e318}
\|Tu \|\leq \|u \|, \quad\text{for } u\in \mathcal{P}\cap \partial
 \Omega_1.
\end{equation}
 Next, from the definitions of $f_\infty$ and $g_\infty$, there
exists $\overline{H}_2>0$ such that
\begin{equation}\label{e319}
f(x)\geq(f_\infty-\epsilon)x,\quad
g(x)\geq(g_\infty-\epsilon)x,\quad x\geq\overline{H}_2.
\end{equation}
Let %\label{e320}
$H_2=\max\{2H_1, \overline{H}_2/\gamma\}$.
Let $u\in \mathcal{P}$ and $\|u\|=H_2$. Then,
\begin{equation}\label{e321}
\min_{t\in[\xi,\omega]_\mathbb{T}}u(t)\geq\gamma\|
u\|\geq\overline{H}_2.
\end{equation}
Consequently, from \eqref{e25} and choice of $\epsilon$, for $a\leq
s\leq\sigma(b)$, we have that
\begin{equation}\label{e322}
\begin{aligned}
\lambda\int_{a}^{\sigma(b)}G(\sigma(s), r)q(r)g(u(\sigma(r)))\Delta r
&\geq \lambda
\int_{\xi}^{\omega}G(\sigma(s), r)q(r)g(u(\sigma(r)))\Delta r\\
&\geq\lambda\int_{\xi}^{\omega}G(\tau, r)q(r)g(u(\sigma(r)))\Delta r\\
&\geq \lambda
\int_{\xi}^{\omega}G(\tau, r)q(r)(g_\infty-\epsilon)u(r)\Delta r\\
&\geq\gamma\lambda\int_{\xi}^{\omega}G(\tau, r)q(r)(g_\infty-\epsilon)\Delta r\|u\|\\
&\geq\|u\|=H_2.
\end{aligned}
\end{equation}
And so, we have from \eqref{e25} and choice of $\epsilon$ that
 % \label{e323}
\begin{align*}
Tu(\tau)&=\lambda\int_{a}^{\sigma(b)}G(\tau,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\geq\lambda\int_{a}^{\sigma(b)}G(\tau,s)p(s)
(f_\infty-\epsilon)\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r \Delta s\\
&\geq \lambda\int_a^{\sigma(b)} G(\tau, s)p(s)(f_\infty-\epsilon)H_2 \Delta s\\
&\geq \gamma H_2>H_2=\|u\|.
\end{align*}
Hence, $\|Tu \|\geq \|u \|$. So if we
set %\label{e324}
$\Omega_2=\{x\in\mathcal{B}:\|x\|<H_2\}$, then
\begin{equation}\label{e325}
\|Tu \|\geq \|u \|,\quad\text{for } u\in \mathcal{P}\cap \partial \Omega_2.
\end{equation}
 Applying Theorem \ref{t21} to \eqref{e318} and \eqref{e325}, we
obtain that $T$ has a fixed point
$u\in \mathcal{P}\cap(\overline{\Omega}_{2} \backslash \Omega_{1})$. As
such, and with $v$ being defined by
\begin{equation}\label{e326}
v(t)=\lambda\int_a^{\sigma(b)}G(t,s)q(s)g(u(\sigma(s)))\Delta s,
\end{equation}
the pair $(u, v)$ is a desired solution of \eqref{e12}, \eqref{e13}
for the given $\lambda$. The proof is complete.
\end{proof}

Prior to our next result, we introduce another hypothesis.
\begin{itemize}
\item[(A4)] $g(0)=0$, and $f$ is an increasing function.
\end{itemize}
We now define positive numbers $L_3$ and $L_4$ by
 %\label{e327}
\begin{gather*}
L_3:=\max\Big\{\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)p(s)\Delta s
f_{0}\Big]^{-1},
\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)q(s)\Delta s g_{0}\Big]^{-1}\Big\},
\\
L_4:=\min\Big\{\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)p(s)\Delta s
f_{\infty}\Big]^{-1},
\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)q(s)\Delta s g_{\infty}\Big]^{-1}
\Big\}.
\end{gather*}

\begin{theorem}\label{t32}
Assume that conditions {\rm (A1)--(A4)} are satisfied. Then, for each
$\lambda$ satisfying
\begin{equation}\label{e328}
L_3<\lambda< L_4,
\end{equation}
there exists a pair $(u, v)$ satisfying \eqref{e12}, \eqref{e13}
such that $u(x)>0$ and $v(x)>0$ on $(a, \sigma^2(b))_\mathbb{T}$.
\end{theorem}

\begin{proof}
Let $\lambda$ be as in \eqref{e328}. And let $\epsilon >0$ be chosen
such that
\[ %\label{e329}
\max\Big\{\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)p(s)\Delta s
(f_{0}-\epsilon)\Big]^{-1},
\Big[\gamma\int_{\xi}^{\omega}G(\tau, s)q(s)\Delta s (g_{0}-\epsilon)
\Big]^{-1}\Big\}\leq\lambda,
\]
\begin{align*}
\lambda\leq\min\Big\{&\Big[\int_{a}^{\sigma(b)}G(\sigma(s),
s)p(s)\Delta s (f_{\infty}+\epsilon)\Big]^{-1},\\
&\Big[\int_{a}^{\sigma(b)}G(\sigma(s), s)q(s)\Delta s
(g_{\infty}+\epsilon)\Big]^{-1}\Big\}.
\end{align*}

 Let $T$ be the cone preserving, completely continuous operator
that was defined by \eqref{e310}.
 From the definitions of $f_0$ and $g_0$, there exists $H_1>0$ such
that
\begin{equation}\label{e330}
f(x)\geq(f_0-\epsilon)x,\quad
g(x)\geq(g_0-\epsilon)x,\quad 0<x\leq H_1
\end{equation}
Now, $g(0)=0$, and so there exists $0<H_2<H_1$ such that
\begin{equation}\label{e331}
\lambda g(x)\leq \frac{H_1}{\int_a^{\sigma(b)}G(\sigma(s),s)q(s)\Delta s},
\quad 0\leq x\leq H_2.
\end{equation}
Choose $u\in\mathcal{P}$ with $\|u\|=H_2$. Then, for
$a\leq s\leq \sigma(b)$, we have
\begin{equation}\label{e332}
\lambda\int_a^{\sigma(b)}G(\sigma(s), r)q(r)g(u(\sigma(r))\Delta
r\leq \frac{\int_a^{\sigma(b)}G(\sigma(s), r)q(r)H_1 \Delta
r}{\int_a^{\sigma(b)}G(\sigma(s), s)q(s)\Delta s}\leq H_1.
\end{equation}
Then
\begin{equation}\label{e333}
\begin{aligned}
Tu(\tau)&=\lambda\int_{a}^{\sigma(b)}G(\tau,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r\Big)\Delta s\\
&\geq\lambda\int_{\xi}^{\omega}G(\tau,s)p(s)
(f_0-\epsilon)\lambda\int_{a}^{\sigma(b)}G(\sigma(s),
r)q(r)g(u(\sigma(r)))\Delta r \Delta s\\
&\geq \lambda\int_\xi^\omega G(\tau, s)p(s)(f_0-\epsilon)\lambda\int_\xi^\omega G(\tau, r)q(r)g(u(\sigma(r)))\Delta r \Delta s\\
&\geq \lambda\int_\xi^\omega G(\tau, s)p(s)(f_0-\epsilon)\lambda \gamma\int_\xi^\omega G(\tau, r)q(r)(g_0-\epsilon)\|u\|\Delta r \Delta s\\
&\geq \lambda\int_\xi^\omega G(\tau, s)p(s)(f_0-\epsilon)
\|u\|\Delta s\\
&\geq \lambda\gamma\int_\xi^\omega G(\tau, s)p(s)(f_0-\epsilon)
\|u\|\Delta s \geq\|u\|.
\end{aligned}
\end{equation}
So, $\|Tu \|\geq \|u \|$. If we put %\label{e334}
$\Omega_1=\{x\in\mathcal{B}|~~\|x\|<H_2\}$,
then
\begin{equation}\label{e335}
\|Tu \|\geq \|u \|, \quad\text{for } u\in \mathcal{P}\cap
\partial \Omega_1.
\end{equation}
 Next, by definitions of $f_\infty$ and $g_\infty$, there exists
$\overline{H}_1$ such that
\begin{equation}\label{e336}
f(x)\leq(f_\infty-\epsilon)x,\quad
g(x)\leq(g_\infty-\epsilon)x,\quad x\geq
\overline{H}_1
\end{equation}
There are two cases:
(i) $g$ is bounded, and (ii) $g$ is unbounded.

 For case (i), suppose $N>0$ is such that $g(x)\leq N$ for all
$0<x<\infty$. Then, for $a\leq s \leq \sigma(b)$ and
$u\in\mathcal{P}$,
\begin{equation}\label{e337}
\lambda \int_a^{\sigma(b)}G(\sigma(s), r)q(r)g(u(\sigma(r)))\Delta
r\leq N\lambda\int_a^{\sigma(b)}G(\sigma(r),r)q(r)\Delta r.
\end{equation}
Let
\begin{equation}\label{e338}
M=\max\Big\{f(x)|0\leq x\leq N\lambda
\int_a^{\sigma(b)}G(\sigma(r),r)q(r)\Delta r\Big\},
\end{equation}
and let
\begin{equation}\label{e339}
H_3>\max\Big\{2H_2, M\lambda\int_a^{\sigma(b)}G(\sigma(s),s)p(s)\Delta
s\Big\}.
\end{equation}
Then, for $u\in \mathcal{P}$ with $\|u \|=H_3$,
\begin{equation}\label{e340}
Tu(t)\leq\lambda\int_a^{\sigma(b)}G(\sigma(s), s)p(s)M \Delta s
\leq H_3=\|u \|
\end{equation}
so that $\|Tu \|\leq \|u \|$.
If %\label{e341}
$\Omega_2=\{x\in\mathcal{B}|~~\|x\|<H_3\}$,
then
\begin{equation}\label{e342}
\|Tu \|\leq \|u \|, \quad\text{for } u\in
\mathcal{P}\cap \partial \Omega_2.
\end{equation}
 For case (ii), there exists $H_3>\max\{2H_2, \overline{H}_1\}$
such that $g(x)\leq g(H_3)$, for $0<x\leq H_3$. Similarly, there
exists $H_4>\max\{H_3, \lambda \int_a^{\sigma(b)}G(\sigma(r),
r)q(r)g(H_3) \Delta r\}$ such that $f(x)\leq f(H_4)$, for $0<x\leq
H_4$. Choosing $u \in \mathcal{P}$ with $\|u\|=H_4$,
we have by $(A4)$ that
\begin{equation}\label{e343}
\begin{aligned}
Tu(t)&\leq\lambda\int_{a}^{\sigma(b)}G(t,s)p(s)
f\Big(\lambda\int_{a}^{\sigma(b)}G(\sigma(r),
r)q(r)g(H_3)\Delta r\Big)\Delta s\\
&\leq\lambda\int_{a}^{\sigma(b)}G(t,s)p(s) f(H_4) \Delta s\\
&\leq \lambda\int_a^{\sigma(b)} G(\sigma(s), s)p(s) \Delta s
(f_\infty+\epsilon)H_4\\
&\leq  H_4=\|u\|,
\end{aligned}
\end{equation}
and so $\|Tu \|\leq \|u \|$. For this
case, if we let %\label{e344}
$\Omega_2=\{x\in\mathcal{B}:\|x\|<H_4\}$,
then
\[ % \label{e345}
\|Tu \|\leq \|u \|, \quad\text{for } u\in\mathcal{P}\cap \partial
\Omega_2.
\]
In either case, application of part (ii) of Theorem
\ref{t21} yields a fixed point $u$ of $T$  belonging to
$\mathcal{P}\cap(\overline{\Omega}_{2} \backslash \Omega_{1})$, which
in turn yields a pair $(u, v)$ satisfying \eqref{e12}, \eqref{e13}
for the chosen value of $\lambda$. The proof is complete.
\end{proof}

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