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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 114, pp. 1--10.\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-2008/114\hfil Existence of positive solutions]
{Existence of positive solutions for semipositone
 dynamic system on time scales}

\author[N. N. Shao, Y. W. Zhang\hfil EJDE-2008/114\hfilneg]
{Na-Na Shao, You-Wei Zhang}  % in alphabetical order

\address{Na-Na Shao \newline
 School of Mathematics and Statistics, Lanzhou University,
 Lanzhou, 730000, Gansu,   China}
\email{shaonn06@lzu.cn}

\address{You-Wei Zhang \newline
 Department of Mathematics, Hexi University,
 Zhangye, 734000, Gansu, China}
\email{zhangyw05@lzu.cn}


\thanks{Submitted April 8, 2008. Published August 20, 2008.}
\thanks{Supported by grants 10726049 from Tianyuan Youth,
and Lzu05003 from Fundamental \hfill\break\indent
Research Fund for Physics and
Mathematics of Lanzhou University}
\subjclass[2000]{34B15, 39A10}
\keywords{Positive solution; semipositone dynamic system;
  cone; \hfill\break\indent fixed point index; time scales}

\begin{abstract}
 In this paper, we study the following  semipositone
 dynamic system on time scales
 \begin{gather*}
 -x^{\Delta\Delta}(t)=f(t,y)+p(t),  \quad  t\in(0,T)_{\mathbb{T}},\\
 -y^{\Delta\Delta}(t)=g(t,x),  \quad  t\in(0,T)_{\mathbb{T}},\\
 x(0)=x(\sigma^{2}(T))=0, \\
 \alpha{y(0)}-\beta{y^{\Delta}{(0)}}=
 \gamma{y(\sigma(T))}+\delta{y^{\Delta}(\sigma(T))}=0.
 \end{gather*}
 Using fixed point index theory, we show the existence of at
 least one positive solution. The interesting point is the
 that nonlinear term is allowed to change sign and may tend to
 negative infinity.

\end{abstract}

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

\section{Introduction}

The theory of dynamic equations on time scales is
undergoing rapid development. This is not only because
it can provide a unifying structure for the study of differential
equations in the continuous case and the study of finite difference
equations in the discrete case, but also because the study of time
scales has led to many important applications, e.g., in the study of
insect population models, neural networks, biology, heat transfer,
stock market, crop harvesting and epidemic models
\cite{1, BP1, BP2, jam, KLS}.


Let $\mathbb{T}$ be a time scale
(arbitrary nonempty closed subset of the real numbers $\mathbb{R}$). For
each internal $I$ of $\mathbb{R}$, we denote by
$I_{\mathbb{T}}=I\bigcap{\mathbb{T}}$.  In this paper, we are
interested in the nonlinear dynamic system on a time scale
$\mathbb{T}$,
\begin{equation}\label{1.1}
\begin{gathered}
-x^{\Delta\Delta}(t)=f(t,y)+p(t),  \quad  t\in(0,T)_{\mathbb{T}},\\
-y^{\Delta\Delta}(t)=g(t,x),  \quad  t\in(0,T)_{\mathbb{T}},\\
x(0)=x(\sigma^{2}(T))=0, \\
\alpha{y(0)}-\beta{y^{\Delta}(0)}=\gamma{y(\sigma(T))}
+\delta{y^{\Delta}(\sigma(T))}=0,
\end{gathered}
\end{equation}
 where $p:(0,T)\to \mathbb{R}$ is Lebesgue integrable and
$f,g\in C((0,T)_{\mathbb{T}}\times \mathbb{R} ^+, \mathbb{R} ^+)$,
$0,T\in\mathbb{T}$, $\alpha,\beta,\gamma,\delta>0 $ are constants
such that $\rho=\alpha\delta+\beta\gamma+\alpha\gamma\sigma(T)>0$.


  Recently, there is much attention paid to question
of positive solutions of boundary value problems on time scales, see
\cite{BP2, 4, EM,  7, SunLi, 9} and the references therein. However,
to the best of our knowledge, very few works can be found
 for the problem when the nonlinearity can change sign.

Motivated by the main ideas in \cite{10}, the purpose of this paper
is to study the existence at least one positive solution for the
semipositone differential system \eqref{1.1}. Different from the
previous papers, in this paper the nonlinearity term $p(t)$ is
allowed to be negative and may tend to infinity. Our
results are new even for the special case of difference equations.

The rest of the paper is organized as follows. In section 2, we
provide some lemmas which are useful later. In Section 3, we give
the main result of the paper and an example is presented to
demonstrate the application of our main results.

\section{Preliminaries}

Let $\mathbb{X}=\{{x|x:[0,\sigma^{2}(T)]_{\mathbb{T}}}\to\mathbb{R}\text{
is continuous}\}$ be a Banach space endowed with the norm
$\|u\|=\max_{t\in[0,\sigma^{2}(T)]_{\mathbb{T}}}|u(t)|$.

Define
$P=\{x\in{\mathbb{X}}:x(t)\geq0,t\in[0,\sigma^2(T)]_{\mathbb{T}}\}$
and $K=\{x\in{P}:x(t)\geq{q(t)}\|x\|,t\in[0,\sigma^{2}(T)]\}$, where
$q(t)=\frac{t(\sigma^{2}(T)-t)}{(\sigma^{2}(T))^{2}}$, it is easy to
see that $P$ and $K$ are cones of $\mathbb{X}$ and $K\subset{P}$. To
obtain solutions of the system \eqref{1.1}, we first denote the
Green's functions of the following boundary value problems:
\begin{gather*}
-x^{\Delta\Delta}(t)=0,  \quad   t\in(0,T)_{\mathbb{T}},\\
x(0)=x(\sigma^{2}(T))=0,
\end{gather*}
and
\begin{gather*}
-y^{\Delta\Delta}(t)=0,  \quad   t\in(0,T)_{\mathbb{T}},\\
\alpha{y(0)}-\beta{y^{\Delta}(0)}
=\gamma{y(\sigma(T))}+\delta{y^{\Delta}(\sigma(T))}=0,
\end{gather*}
by $G(t,s)$ and $H(t,s)$ respectively. From \cite{4,EM}, we know
that
\begin{gather}\label{2.1}
G(t,s)=\frac{1}{\sigma^{2}(T)}\begin{cases}
t(\sigma^{2}(T)-\sigma(s)),  &  t\leq{s},\\
\sigma(s)(\sigma^{2}(T)-t),  &  t\geq\sigma(s);
\end{cases}\\
\label{2.2}
H(t,s)=\frac{1}{\rho} \begin{cases}
(\beta+\alpha{t})(\gamma(\sigma(T)-\sigma(s))+\delta),   &  t\leq{s},\\
(\beta+\alpha\sigma(s))(\gamma(\sigma(T)-t)+\delta),   &  t\geq\sigma(s),
\end{cases}
\end{gather}
 Obviously,
\begin{equation} \label{2.3}
0\leq{G(t,s)}\leq{\frac{t(\sigma^{2}(T)-t)}{\sigma^{2}(T)}},\quad
0\leq{H(t,s)}\leq{H(\sigma(s),s)}.
\end{equation}

For the sake of convenience, we state the following hypotheses:
\begin{itemize}
\item[(C1)]
$f:(0,T)_{\mathbb{T}}\times[0,+\infty)\to[0,+\infty)$ is
continuous and for any $t\in(0,T)_{\mathbb{T}}$,
$y\in[0,+\infty),f(t,y)$ is nondecreasing on $y$ and satisfying
$f(t,y)\leq{p^*(t)h(y)}$, where
$p^*:(0,T)_{\mathbb{T}}\to[0,+\infty)$ and
$h:(0,+\infty)\to[0,+\infty)$ are continuous,
$\lim_{y\to+\infty}\frac{f(t,y)}{y}=+\infty$ uniformly for
$t$ on any close subinterval of $(0,T)_{\mathbb{T}}$.

\item[(C2)] $g:(0,T)_{\mathbb{T}}\times[0,+\infty)\to[0,+\infty)$
is continuous, there exist constants
$\lambda_{1}\geq\lambda_{2}\geq{1}$ such that for any
$t\in[0,T]_{\mathbb{T}},x\in[0,+\infty)$,
\begin{equation}\label{2.4}
c^{\lambda_{1}}g(t,x)\leq{g(t,cx)}\leq{c^{\lambda_{2}}g(t,x)},\quad
\forall0\leq{c}\leq1,
\end{equation} with
$0<\int_{0}^{\sigma(T)}H(\sigma(t),t)g(t,1)\Delta{t}<+\infty$.

\item[(C3)]$p:(0,T)_\mathbb{T}\to(-\infty,+\infty)$
 is Lebesgue integrable such that $\int_0^{\sigma{(T)}}p_{-}(t)\Delta{t}>0$
and \[0<\int_{0}^{\sigma(T)}G(\sigma(t),t)[p^*(t)+p_{+}(t)]
\Delta{t}<\frac{\sigma^2(T)\int_0^{\sigma{(t)}}p_{-}(t)
\Delta{t}}{2(\max_{0\leq \tau\leq {R}}h(\tau)+1)},\]where
$p_{+}(t)=\max\{p(t),0\}, p_{-}(t)=\max\{0,-p(t)\}$,
and
\[
R=\Big(\int_0^{\sigma(T)}P_{-}(t)\Delta{t}+1\Big)^{\lambda_{1}}
\int_{0}^{\sigma(T)} H(\sigma(t),t)g(t,1)\Delta{t}.
\]
\end{itemize}
From $(C2)$,  as in [10, Remark2.2, Lemma2.2] we  obtain the
following result.

\begin{remark}\label{rem2.1} \rm
If $(C2)$ is satisfied,
then for $t\in[0,T]_{\mathbb{T}}$, $g(t,x)$ is increasing on $x$ and
for $(t,x)\in[0,T]_{\mathbb{T}}\times[0,+\infty)$, $c\in[1,+\infty),
\lambda_{1}\geq\lambda_{2}>1$,
\begin{equation}\label{2.5}
c^{\lambda_{2}}g(t,x)\leq{g(t,cx)}\leq{c^{\lambda_{1}}g(t,x)},
\quad
\lim_{x\to{+\infty}}\min_{t\in[0,T]_{\mathbb{T}}}{\frac{g(t,x)}{x}}={+\infty}
\end{equation}
\end{remark}

For convenience, we define a functions
\[
[x(t)]^{*}= \begin{cases}
x(t), &  x(t)\geq0,\\
0,    &    x(t)<0,
\end{cases}
\]
and
\[
\omega(t)=\int_0^{\sigma(T)}G(t,s)p_{-}(s)\Delta{s},\quad
t\in[0,\sigma^{2}(T)]_{\mathbb{T}}.
\]
By the definition of the function $\omega(t)$ and  $G(t,s)\geq0$, we
have
\[
0\leq \omega(t)=\int_{0}^{\sigma(T)}G(t,s)p_{-}(s)\Delta{s}\leq
\frac{t(\sigma^{2}(t)-t)}{\sigma^{2}({T})}
 \int_{0}^{\sigma(T)}p_{-}(s)\Delta{s}
<\infty,
\]
thus $\omega(t)\in{P}$, $-\omega^{\Delta\Delta}(t)=p_{-}(t)$  and
$\omega(0)=\omega(\sigma^{2}(T))=0$.

 Next, we consider the approximate system
\begin{equation}\label{2.6}
\begin{gathered}
-x^{\Delta\Delta}(t)=f(t,y)+p_{+}(t),  \quad t\in(0,T)_{\mathbb{T}},\\
-y^{\Delta\Delta}(t)=g(t,[x(t)-\omega(t)]^{*}),  \quad t\in(0,T)_{\mathbb{T}},\\
x(0)=x(\sigma^{2}(T))=0, \\
\alpha{y(0)}-\beta{y^{\Delta}(0)}=\gamma{y(\sigma(T))}
+\delta{y^{\Delta}(\sigma(T))}=0.
\end{gathered}
\end{equation}
It is well known that $(x,y)$ is a solution of system \eqref{2.6}
if and only if $(x,y)$ is the solution of the  nonlinear
integral equation system
\begin{equation}\label{2.7}
\begin{gathered}
x(t)=\int_{0}^{\sigma(T)}G(t,s)[f(s,y(s))+p_{+}(s)]\Delta{s}, \quad
 t\in[0,\sigma^{2}(T)]_{\mathbb{{T}}},\\
y(t)=\int_{0}^{\sigma(T)}H(t,s)g(s,[x(s)-\omega(s)]^{*})\Delta{s}.
\end{gathered}
\end{equation}
Obviously, the integral system \eqref{2.7} is equivalent to the
 nonlinear integral equation
\begin{equation}\label{2.8}
x(t)=\int_{0}^{\sigma(T)} G(t,s)
\Big[f\Big(s,\int_{0}^{\sigma(T)}H(s,\tau)g(\tau,[x(\tau)
-\omega(\tau)]^{*})
\Delta{\tau}\Big)+p_{+}(s)\Big]\Delta{s},
\end{equation}
$t\in[0,\sigma^{2}(T)]_{\mathbb{T}}$.

Now, we define the operator $F:P\to{\mathbb{X}}$ by
\begin{equation}\label{2.9}
(Fx)(t)=\int_{0}^{\sigma(T)}G(t,s)
\Big[f\Big(s,\int_{0}^{\sigma(T)}H(s,\tau)
g(\tau,[x(\tau)-\omega(\tau)]^{*})\Delta{\tau}\Big)+p_{+}(s)\Big]\Delta{s}.
\end{equation}
Then the existence of solutions to system \eqref{2.7} is equivalent
to the existence of solution for nonlinear integral equation
\eqref{2.8}. Therefore, if $x$ is a fixed point of the operator $F$
in $\mathbb{X}$, then the system \eqref{2.7} has one solution
$(u,v)$ which can be written as
\begin{equation}\label{2.10}
\begin{gathered}
u(t)=x(t),\\
v(t)=\int_{0}^{\sigma(T)}H(t,s)g(s,[x(s)-\omega(s)]^{*})\Delta{s},
\quad  t\in[0,\sigma^{2}(T)]_{\mathbb{T}}.
\end{gathered}
\end{equation}

\begin{lemma}\label{lem2.2}
 If $(u,v)$ with $u(t)\geq\omega(t)$
for $ t\in[0,\sigma^{2}(T)]_{\mathbb{T}}$ is a positive solution of
system \eqref{2.7}, then $(u-\omega,v)$ is a positive solution of
semipositone dynamical system \eqref{1.1}.
\end{lemma}

\begin{proof}
Suppose that $(u,v)$ is a positive solution of \eqref{2.7} with
$u(t)\geq\omega(t)$ for $t\in[0,\sigma^{2}(T)]_{\mathbb{T}}$, then
from \eqref{2.7} and the definition of $[u(t)]^{*}$, we have
\begin{equation}\label{2.11}
\begin{gathered}
-u^{\Delta\Delta}(t)=f(t,v(t))+p_{+}(t),  \quad  t\in(0,T)_{\mathbb{T}},\\
-v^{\Delta\Delta}(t)=g(t,[u(t)-\omega(t)]^*),  \quad t\in(0,T)_{\mathbb{T}},\\
u(0)=u(\sigma^{2}(T))=0, \\
\alpha{v(0)}-\beta{v^{\Delta}(0)}=\gamma{v(\sigma(T))}
+\delta{v^{\Delta}(\sigma(T))}=0.
\end{gathered}
\end{equation}
Set $u_{1}=u-\omega$, then
$u_{1}^{{\Delta}{\Delta}}=u^{{\Delta}{\Delta}}-\omega^{{\Delta}{\Delta}}$
which implies
\begin{equation}
\begin{gathered}\label{2.12}
-u_{1}^{\Delta\Delta}(t)=f(t,v(t))+p_+(t)-p_-(t),  \quad   t\in(0,T)_{\mathbb{T}},\\
-v^{\Delta\Delta}(t)=g(t,u_{1}(t)),  \quad  t\in(0,T)_{\mathbb{T}},\\
u_{1}(0)=u_{1}(\sigma^{2}(T))=0, \\
\alpha{v(0)}-\beta{v^{\Delta}(0)}
=\gamma{v(\sigma(T))}+\delta{v^{\Delta}(\sigma(T))}=0,.
\end{gathered}
\end{equation}
 From (C3) we know $p_+(t)-p_-(t)=p(t)$, then $(u_{1},v)$ is a
positive solution of  \eqref{1.1}.
\end{proof}

\begin{lemma}\label{lem2.4}
If {\rm(C1)--(C3)} are satisfied, then $F:K\to{K}$ is completely
continuous.
\end{lemma}
\begin{proof}
For any $x\in{K}$, let $u(t)=(Fx)(t)$. By the definition of the
operator $F$, we have $u(t)\geq0$, and $u(0)=u(\sigma^{2}(T))=0$.
Hence, there exists a $t_{0}\in[0,\sigma^{2}(T)]_\mathbb{T}$ such
that $\|u\|=u(t_{0})$. Since
\begin{equation} \label{2.13}
\frac{G(t,s)}{G(t_{0},s)}= \begin{cases}
\frac{t}{t_{0}},      & t,t_{0}\leq{s},\\[3pt]
\frac{t(\sigma^{2}(T)-\sigma(s))}{\sigma(s)(\sigma^{2}(T)-t_{0})},
  & t\leq{s}<t_{0},\\[3pt]
\frac{\sigma(s)(\sigma^{2}(T)-t)}{t_{0}(\sigma^{2}(T)-\sigma(s))},
 &  t_{0}\leq{s}<t,\\[3pt]
\frac{\sigma^{2}(T)-t}{\sigma^{2}(T)-t_{0}},& t,t_{0}\geq{s},
\end{cases}
\end{equation}
we have
\[
\frac{G(t,s)}{G(t_{0},s)}
\geq{\frac{t(\sigma^{2}(T)-t)}{(\sigma^{2}(T))^{2}}}=q(t).
\]
Thus
\begin{align*}
&(Fx)(t)\\
&=\int_{0}^{\sigma(T)}G(t,s)
 \Big[f\Big(s,\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,[x(\tau)-\omega(\tau)]^{*})\Delta\tau\Big)+p_+(s)\Big]\Delta{s}\\
&\geq {q(t)}\int_{0}^{\sigma(T)}G(t_{0},s)
 \Big[f\Big(s,\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,[x(\tau)-\omega(\tau)]^{*})\Delta\tau\Big)+p_+(s)\Big]\Delta{s}\\
&\geq {q(t)}(Fx)(t_{0})={q(t)}\|Fx\|,
\end{align*}
and $F(K)\subseteq{K}$. By standard argument, we can easily obtain
that the operator $F:K\to K$ is completely continuous. The
proof is complete.
\end{proof}

For our arguments, the following fixed point index theory \cite{5} is
crucial.

\begin{lemma} \label{th1.1}
Let $\mathbb{X}$ be a Banach space and $K$ be a cone
in $\mathbb{X}$. Assume that $\Omega$ is a bounded open subset of
$\mathbb{X}$ with $\theta\in\Omega$ and let
$\Phi:K\cap\overline{\Omega}\to{K}$ be a completely
continuous operator.
\begin{itemize}
\item[(i)] If $\Phi{z}\neq\lambda{z}$
for all ${z}\in{K}\bigcap{\partial\Omega}$, $\lambda\geq1$, then
$i(\Phi, K\bigcap\Omega,K)=1$;

\item[(ii)] if $\Phi{z}\not\leq {z}$ for all ${z}\in{K\bigcap\partial\Omega}$,
then $i(\Phi, K\bigcap\Omega, K)=0$.
\end{itemize}
\end{lemma}

\section{Main Results}

Before presenting the main result, we give two lemmas which are
important in establishing the existence of one positive solutions to
the problem \eqref{1.1}.

\begin{lemma}\label{lem2.5}
Assume {\rm (C1)--(C3)} hold, if we let
$r=\sigma^{2}(T)\int_0^{\sigma(T)}p_-(t)\Delta{t}$,
$\Omega_{r}=\{x\in\mathbb{X}:\|x\|<{r}\}$, then $i(F,K\cap
\Omega_{r},K)=1$.
\end{lemma}

\begin{proof}
Suppose that there exist $\lambda_{0}\geq1$ and
$x_{0}\in{K}\cap\partial\Omega_{r}$ such that
$Fx_{0}=\lambda_{0}x_{0}$. By
$0\leq{G(t,s)}\leq t(\sigma^2(T)-t)/\sigma^2(T)$, we have
\begin{equation}\label{2.16}
0<\omega(t)=\int_{0}^{\sigma(T)}G(t,s)p_-(s)\Delta{s}\leq
\frac{t(\sigma^{2}(T)-t)}{\sigma^{2}(T)}
\int_{0}^{\sigma(T)}p_-(s)\Delta{s}<+\infty,
\end{equation}
for $t\in[0,\sigma^{2}(T)]_{\mathbb{T}}$,
which together with
$x_{0}(t)\geq{q(t)\|x_{0}\|=rq(t)}$, $t\in[0,\sigma^{2}(T)]_{\mathbb{T}}$,
yield
\begin{equation}\label{2.17}
x_{0}(t)-\omega(t)\geq{rq(t)-\frac{t(\sigma^{2}(T)-t)}{\sigma^{2}(T)}}
\int_{0}^{\sigma(T)}p_-(s)\Delta{s}=0,
\quad t\in[0,\sigma^{2}(T)]_{\mathbb{T}}.
\end{equation}
Thus, from $x_{0}=\frac{1}{\lambda_{0}}Fx_{0}$ and (\ref{2.17}), we
obtain
\begin{equation}
\begin{gathered}\label{2.18}
-x_{0}^{\Delta\Delta}(t)=\frac{1}{\lambda_{0}}
 \Big[f\Big(t,\int_{0}^{\sigma(T)}
H(t,\tau)g(\tau,x_0(\tau)-\omega(\tau))\Delta{\tau}\Big)+p_+(t)\Big],
\quad  t\in[0,T]_{\mathbb{T}},\\
x_{0}(0)=x_{0}(\sigma^{2}(T))=0,
\end{gathered}
\end{equation}
which shows that there exists a
$t_{0}\in[0,\sigma^{2}(T)]_{\mathbb{T}}$ such that
$x_{0}(t_{0})=\|x_{0}\|=r$, $x_{0}^{\Delta}(t_{0})\leq0$. Then
\begin{equation}\label{2.19}
0\leq{x_{0}(t)-\omega(t)}\leq{x_{0}(t)}\leq\|x_{0}\|=r<r+1.
\end{equation}
Let $t\in[0,t_{0}]_{\mathbb{T}}$, integrating the equation in
(\ref{2.18}) from $t$ to $t_{0}$, we have
\begin{align*}
x_{0}^{\Delta}(t)-x_{0}^{\Delta}(t_{0})
&= \int_{t}^{t_{0}}-x_{0}^{\Delta\Delta}(s)\Delta{s}\\
&= \int_{t}^{t_{0}}\frac{1}{\lambda_{0}}
 \Big[f\Big(s,\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,x_0(\tau)-\omega(\tau))\Delta{\tau}\Big)+p_+(s)\Big]\Delta{s}\\
&\leq \int_{t}^{t_{0}}\Big[f\Big(s,\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,x_0(\tau)-\omega(\tau))\Delta{\tau}\Big)+p_+(s)\Big]\Delta{s}\\
&\leq \int_{t}^{t_{0}}\Big[p^*(s)h\Big(\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,x_0(\tau)-\omega(\tau))\Delta{\tau}\Big)+p_+(s)\Big]\Delta{s}.
\end{align*}
Since $g(t,x)$ is increasing with respect  to $x$,
\begin{align*}
\int_{0}^{\sigma(T)}H(s,\tau)g(\tau,x_0(\tau)-\omega(\tau))\Delta\tau
&\leq \int_{0}^{\sigma(T)}
H(\sigma(\tau),\tau)g(\tau,x_{0}(\tau)-\omega(\tau))\Delta\tau\\
&\leq \int_{0}^{\sigma(T)}H(\sigma(\tau),\tau)g(\tau,r+1)\Delta\tau\\
&\leq (r+1)^{\lambda_{1}}\int_{0}^{\sigma(T)}H(\sigma(\tau),\tau)g(\tau,1)
 \Delta\tau.
\end{align*}
So
\begin{equation}\label{2.20}
x_{0}^{\Delta}(t)\leq \big(\max_{0\leq\tau\leq{R}}h(\tau)+1\big)
\int_{t}^{t_{0}}(p^*(s)+p_+(s))\Delta{s},
\end{equation}
where
$R=(r+1)^{\lambda_{1}}\int_{0}^{\sigma(T)}H(\sigma(\tau),\tau)g(\tau,1)
\Delta\tau$.
Integrating (\ref{2.20}) from $0$ to $t_{0}$, we have
\begin{align*}
r&= \int_{0}^{t_{0}}x_{0}^{\Delta}(s)\Delta{s}
\leq[\max_{0\leq\tau\leq{R}}h(\tau)+1]\int_{0}^{t_{0}}\int_{s}^{t_{0}}(p^*(\tau)+p_+(\tau))\Delta{\tau}\Delta{s}\\
&\leq [\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^2(T)\int_{0}^{t_{0}}(p^*(\tau)+p_+(\tau))\Delta{\tau}\\
&\leq \frac{[\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^{2}(T)}{\sigma^{2}(T)-t_{0}}\int_{0}^{t_{0}}\frac{\sigma(\tau)(\sigma^{2}(T)-\sigma(\tau))}
{\sigma^{2}(T)}(p^*(\tau)+p_+(\tau))\Delta{\tau}.\\
&\leq \frac{[\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^{2}(T)}{\sigma^{2}(T)-t_{0}}\int_{0}^{\sigma(T)}\frac{\sigma(\tau)(\sigma^{2}(T)-\sigma(\tau))}
{\sigma^{2}(T)}(p^*(\tau)+p_+(\tau))\Delta{\tau}.
\end{align*}
Consequently,
\begin{equation}\label{2.21}
\int_{0}^{\sigma(T)}G(\sigma(\tau),\tau)(p^*(\tau)+p_+(\tau))\Delta\tau\geq
\frac{r(\sigma^{2}(T)-t_{0})}{[\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^{2}(T)}.
\end{equation}
Integrate (\ref{2.20}) from $t_{0}$ to $t$, we  have the
following inequality in the same way
\begin{equation}\label{2.22}
\int_{0}^{\sigma(T)}G(\sigma(\tau),\tau)(p^*(\tau)+p_+(\tau))
 \Delta\tau\geq\frac{rt_{0}}{[\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^{2}(T)}\,.
\end{equation}
Combining with (\ref{2.21}) and (\ref{2.22}),
\begin{align*}
2\int_{0}^{\sigma(T)}G(\sigma(\tau),\tau)(p^*(\tau)+p_+(\tau))\Delta\tau
&\geq \frac{r\sigma^{2}(T)}{[\max_{0\leq\tau\leq{R}}h(\tau)+1]\sigma^{2}(T)}\\
&= \frac{\sigma^{2}(T)}{\max_{0\leq\tau\leq{R}}h(\tau)+1}
  \int_0^{\sigma(T)}p_-(\tau)\Delta\tau.
\end{align*}
As a result,
\begin{equation}\label{2.23}
\int_{0}^{\sigma(T)}G(\sigma(\tau),\tau)(p^*(\tau)+p_+(\tau))
 \Delta\tau\geq\frac{\sigma^{2}(T)}
{2[\max_{0\leq\tau\leq{R}}h(\tau)+1]}\int_0^{\sigma{(T)}}p_-(\tau)\Delta\tau,
\end{equation}
which is a contradiction with (C3). So applying Lemma
{\ref{th1.1}}, $i(F,K\cap\Omega_{r},K)=1$.
\end{proof}

\begin{lemma}\label{lem2.6}
Assume {\rm (C1)--(C3)}. Then there exist
$R^{*}>r=\sigma^2(T)\int_0^{\sigma(T)}p_-(t)\Delta{s}$, such that
$i(F,K\cap\Omega_{R^{*}},K)=0$, where
$\Omega_{R^{*}}=\{x|x\in{\mathbb{X},\|x\|<{R^{*}}}\}$.
\end{lemma}

\begin{proof}
Choose constants $0<\alpha^*\leq\beta^*$ and $L$, such that
$[\alpha^*,\beta^*]_{\mathbb{T}}\subseteq[0,T]_{\mathbb{T}}$  and
\[
L>\frac{2\sigma^{2}(T)}{\alpha^*(\sigma^{2}(T)-\beta^*)}
\Big(\min_{t\in[0,\sigma^{2}(T)]_{\mathbb{T}}}{\int_{\alpha^*}
^{\beta^*}G(t,s)\Delta{s}}\Big)^{-1}
\]
By (C1), there exists $R_{1}^{*}>2r$ such that $f(t,y)>Ly$, for
$t\in[\alpha^*,\beta^*]_{\mathbb{T}}$, $y\geq{R_{1}^{*}}$. Since
$\lim_{x\to+\infty}\min g(t,x)/x=+\infty$,  there
is $R_{2}^{*}>R_{1}^{*}$, ${t\in[\alpha^*,\beta^*]_{\mathbb{T}}}$.
When $x>R_{2}^{*}$, from Remark \ref{rem2.1} we have
\begin{equation}
\frac{g(t,x)}{x}\geq\min_{t\in[\alpha^*,\beta^*]_{\mathbb{T}}}
 \frac{g(t,x)}{x}\geq\frac{1}
{\min_{\alpha\leq{s}\leq\beta}\int_{\alpha^*}^{\beta^*}H(s,\tau)
 \Delta{\tau}}.
\end{equation}
So that
\[
g(t,x)\geq\frac{x}{\min_{\alpha^*\leq{s}
\leq\beta^*}\int_{\alpha^*}^{\beta^*}H(s,\tau)\Delta{\tau}} \quad
\text{for } t \in [\alpha,\beta]_{\mathbb{T}},\; x\geq{R_{2}^{*}}.
\]
Let $R^{*}=\frac{2R_{2}^{*}(\sigma^{2}(T))^{2}}
 {\alpha^*(\sigma^{2}(T)-\beta^*)}$,
then $R^{*}>{R_{2}^{*}}>{R_{1}^{*}}>2r$, we assert that $
Fx \not\leq {x}$, ${x}\in{K\cap\partial\Omega_{R^{*}}}$.

Suppose on the contrary that there exists
$x_{1}\in{K\cap\partial\Omega_{R^{*}}}$ such that
$Fx_{1}\leq{x_{1}}$. Then for $t\in[\alpha,\beta]_{\mathbb{T}}$, we
have
\begin{align*}
x_{1}(t)-\omega(t)
&\geq {x_{1}(t)-\frac{t(\sigma^{2}(T)-t)}{\sigma^{2}(T)}}
  \int_0^{\sigma(T)}P_-(\tau)\Delta\tau\\
&=x_{1}(t)-rq(t)\\
&\geq {x_{1}(t)-\frac{x_{1}(t)}{\|x_{1}(t)\|}r}\\
&>\frac{1}{2}x_{1}(t)\\
&\geq\frac{1}{2}q(t)\|x_{1}\|\\
&= \frac{t(\sigma^{2}(T)-t)R^{*}}{2(\sigma^{2}(T))^{2}}\\
&\geq\frac{R^{*}\alpha^*(\sigma^{2}(T)-\beta^*)}{2(\sigma^{2}(T))^{2}}
=R_{2}^{*}>0.
\end{align*}
Then
\begin{align*}
&\int_{\alpha}^{\beta}H(s,\tau)g(\tau,[x_{1}(\tau)-\omega(\tau)]^{*})
 \Delta\tau\\
&=\int_{\alpha}^{\beta}H(s,\tau)g(\tau,[x_{1}(\tau)-\omega(\tau)])\Delta\tau\\
&\geq \frac{1}{\min_{\alpha^*\leq{s}\leq\beta^*}\int_{\alpha^*}^{\beta^*}H(s,\tau)\Delta\tau}
\int_{\alpha^*}^{\beta^*}H(s,\tau)(x_{1}(\tau)-\omega(\tau))\Delta\tau\\
&\geq \frac{R^{*}\alpha^*(\sigma^{2}(T)-\beta^*)}{2(\sigma^{2}(T))^{2}}
=R_{2}^{*}\\
&>R_{1}^{*}>0,
\end{align*}
Since $f(t,y)$ is nondecreasing on $y$,
\begin{align*}
R^{*}&\geq {x_{1}(t)}\geq(Fx_{1})(t)\\
&=\int_{0}^{\sigma(T)}G(t,s) \Big[f\Big(s,\int_{0}^{\sigma(T)}
H(s,\tau)g(\tau,[x_{1}(\tau)-\omega(\tau)]^{*})\Delta\tau\Big)+p_+(s)\Big]\Delta{s}\\
&\geq \int_{\alpha}^{\beta}G(t,s)\Big[f\Big(s,\int_{0}^{\sigma(T)}H(s,\tau)g(\tau,[x_{1}
(\tau)-\omega(\tau)])\Delta\tau\Big)+p_+(s)\Big]\Delta{s}\\
&\geq \int_{\alpha}^{\beta}G(t,s)f
 \Big(s,\int_{0}^{\sigma(T)}H(s,\tau)g(\tau,[x_{1}(\tau)-\omega(\tau)])
 \Delta\tau\Big)\Delta{s}\\
&\geq \int_{\alpha}^{\beta}G(t,s)L
 \Big(\int_{0}^{\sigma(T)}H(s,\tau)g(\tau,[x_{1}(\tau)-\omega(\tau)])
 \Delta\tau\Big)\Delta{s}\\
&\geq \frac{LR^{*}\alpha^*(\sigma^{2}(T)-\beta^*)}{2(\sigma^{2}(T))^{2}}
 \int_{\alpha^*}^{\beta^*}G(t,s)\Delta{s}.
\end{align*}
So,
\[
L\leq\frac{2\sigma^{2}(T)^{2}}{\alpha^*(\sigma^{2}-\beta^*)}
\Big[\min_{t\in[0,\sigma^{2}(T)]_{\mathbb{T}}}{\int_
{\alpha^*}^{\beta^*}G(t,s)\Delta{s}}\Big]^{-1}.
\]
 This contradicts the choice of the constant $L$.
Thus from Lemma \ref{th1.1}, $i(F,K\cap\Omega_{R^{*}},K)=0$.
\end{proof}

Now we present the main result of this paper.
\begin{theorem}\label{th3.1}
Suppose that {\rm (C1)--(C3)} are satisfied, then the semipositone
dynamic system \eqref{1.1} has at least one positive solution.
\end{theorem}

\begin{proof}
By Lemmas \ref{lem2.5}, \ref{lem2.6} and the properties of the
fixed point index, we have
\[
i(F,K_{R^{*}}\backslash\overline{K_{r}},K)=-1.
\]
Thus the operator
$F$ has a fixed point $u_{0}$ in
$K_{R^{*}}\backslash\overline{K_{r}}$ with $r<\|u_{0}\|<{R^{*}}$.
Since $\|u_{0}\|>r$, we have
\begin{align*}
u_{0}(t)-\omega(t)
&\geq {q(t)\|u_{0}\|}-\int_{0}^{\sigma(T)}G(t,s)p_-(s)\Delta{s}\\
&\geq {q(t)\|u_{0}\|}-\frac{t(\sigma^{2}(T)-t)}{\sigma^{2}(T)}
  \int_0^{\sigma(T)}p_-(s)\Delta{s}\\
&\geq {q(t)({\|u_{0}\|}-r)>0}.
\end{align*}
It follows from Lemma \ref{lem2.2} that
\begin{equation}
\begin{gathered}
u(t)=u_{0}(t)-\omega(t),\\
v(t)=\int_{0}^{\sigma(T)}{H(t,s)g(s,u_{0}(s))\Delta{s}}.
\end{gathered}
\end{equation}
is the positive solution of system \eqref{1.1}. The
proof is complete.
\end{proof}

\subsection*{Example}
Let $\mathbb{T}=[0,\frac{1}{3}]\bigcup[\frac{2}{3},1]$.
We consider the  dynamic system
\begin{equation}{\label{3.4}}
\begin{gathered}
-x^{\Delta\Delta}(t)=\frac{y}{t+1}-\frac{1}{\sqrt{t}}, \quad
 t\in(0,1)_{\mathbb{T}},\\
-y^{\Delta\Delta}(t)=\frac{x^{2}}{(t-\frac{1}{2})^{2}},\quad
 t\in(0,1)_{\mathbb{T}},\\
x(0)=x(\sigma^{2}(1))=0, \\
{y(0)}-{y{(0)}}=y(\sigma(1))+y^{\Delta}(\sigma(1))=0.
\end{gathered}
\end{equation}
In fact, for $f(t,y)=\frac{y}{t+1}-3$,
$p(t)=-\frac{1}{\sqrt{t}}$,
$g(t,x)=\frac{x^{2}}{(t-\frac{1}{2})^{2}}$,
$\lambda_{1}=3$, $\lambda_{2}=\frac{3}{2}$,
all conditions of Theorem \ref{th3.1} are satisfied.
Therefore,  problem (\ref{3.4}) has at least one positive solution.

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\end{document}