\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 133, pp. 1--8.\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 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2004/133\hfil Positive solutions for semi-positone Neumann BVP]
{Positive solutions for singular semi-positone  Neumann
 boundary-value problems} 

\author[Y. P. Sun \& Y. Sun\hfil EJDE-2004/133\hfilneg]
{Yong-Ping Sun, Yan Sun}  % in alphabetical order

\address{Yong-Ping Sun \hfill\break
Department of Mathematics, Qufu Normal University \\
Qufu, Shandong 273165, China. \hfill\break
Department of Fundamental Courses, Hangzhou Radio
 \& TV University \\
Hangzhou, Zhejiang 310012, China}
\email{syp@mail.hzrtvu.edu.cn}

\address{Yan Sun \hfill\break
Department of Mathematics, Qufu Normal University \\
Qufu, Shandong 273165, China} 
\email{ysun@163169.net}

\date{}
\thanks{Submitted October 12, 2004. Published November 16, 2004.}
\thanks{Supported by  NSFC (19871075), NSFSP (Z2003A01) and EDZP (20040495)}
\subjclass[2000]{34B10, 34B15}
\keywords{Positive solution; semi-positone; fixed points; cone; \hfill\break\indent
singular Neumann boundary-value problem}


\begin{abstract}
 In  this paper, we study the singular semi-positone Neumann 
 bound\-ary-value problem
 \begin{gather*}
  -u''+m^2u=\lambda f(t,u)+g(t,u),\quad 0<t<1,\\
 u'(0)=u'(1)=0,
 \end{gather*}
 where $m$ is a positive constant.
 Under some suitable assumptions on the functions $f$ and $g$,
 for sufficiently small $\lambda$, we prove the existence of a
 positive solution. Our approach is based on the Krasnasel'skii
 fixed point theorem in cones.
\end{abstract}

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


\section{Introduction}

In  this paper, we  shall study the following singular
semi-positone Neumann boundary-value problem (NBVP)
\begin{equation}
\begin{gathered}
-u''+m^2u=\lambda f(t,u)+g(t,u),\quad 0<t<1,\\
u'(0)=u'(1)=0,
\end{gathered}\label{e1.1}
\end{equation}
where $m>0$ is a constant, $\lambda>0$ is a parameter, $f:
(0,1)\times[0,+\infty)\to [0,+\infty)$ and $g:[0,1]\times
[0,+\infty)\to (-\infty,+\infty)$ are continuous.

We say  problem \eqref{e1.1}  is singular  because  $f$ may be
singular at $t=0$ and/or $t=1$. When $g(t,u)\not\equiv0$, problem
\eqref{e1.1}  is a semi-positone problem, this situation arises
naturally in chemical reaction theory \cite{a}. In recent years,
attention has been paid to \eqref{e1.1} in the case of
$g(t,u)\equiv 0$; see, for example, \cite{rs,do,ma} and the
references therein. Attention has been paid also to the
semi-positone boundary-value problem; see, for example,
\cite{ag,mwr,x} and the references therein. As far as the authors
know,  there are no existence results for the singular
semi-positone NBVP. Recently, Xu \cite{x} studied the existence of
positive solutions for the singular semi-positone boundary-value
problem
\begin{gather*}
u''+f(t,u)+q(t)=0 ,\quad 0<t<1,\\
u(0)=u(1)=0,
\end{gather*}
where $f: (0,1)\times[0,+\infty)\to [0,+\infty)$ and
$q:(0,1)\to(-\infty,+\infty)$ are continuous.

Motivated by the papers mentioned above,  we
study the existence of positive solutions for the singular
semi-positone NBVP \eqref{e1.1}, and give an explicit interval
for $\lambda$. Our results can be regarded as an extension and
improvement of the corresponding results of \cite{do,ma}. The
paper is organized as follows. In Section 2, we present some
lemmas that will be used to prove the main result and
Krasnasel'skii fixed point theorem in cones.
In Section 3, we prove the main result of this paper.

\section{Preliminaries}

 We consider problem in the Banach space $E=C[0,1]$
equipped with the norm $\|u\|=\sup_{t\in [0,1]}|u(t)|$. Let
$G(t,s)$ be the Green's function for the Boundary-value problem
\begin{equation}\begin{gathered}
-u''+m^2u=0 ,\quad 0<t<1,\\
 u'(0)=u'(1)=0\,.
\end{gathered}\label{e2.1}
\end{equation}
Then
$$
G(t,s)=\frac{1}{\rho}\begin{cases} \varphi (s)\varphi (1-t),&
  0\leq s\leq t\leq 1, \\
  \varphi (t)\varphi (1-s),&  0\leq t\leq s\leq 1,
 \end{cases}
$$
where $\rho =\frac {1}{2}m(e^m-e^{-m})$,
$\varphi (t)=\frac {1}{2}(e^{mt}+e^{-mt})$. It is obvious that
$\varphi (t)$ is increasing on $[0,1]$, and
$$
G(t,s)\leq G(s,s),\quad  0\leq t,s\leq 1.
$$

\begin{lemma}\label{lem1}
Let $G(t,s)$ be the Green's function for the
NBVP \eqref{e2.1}.
\begin{itemize}
\item[(1)] Assume that $0<\theta <\frac{1}{2}$,
then
$$
G(t,s)\geq M_{\theta}G(s,s),\quad \theta \leq t\leq 1-\theta,\; 0\leq s\leq 1.
$$
where
$$ M_{\theta}=\frac{e^{m\theta }+e^{-m\theta}}{e^m+e^{-m}}.$$

\item[(2)]  $$
G(t,s)\geq C\varphi(t)\varphi(1-t)G(t_0,s),
\quad t,t_0, s\in [0,1],
$$ 
where $C=1/\varphi^2(1)$.
\end{itemize}\end{lemma}

\begin{proof}
(1) Let $t\in [\theta ,1-\theta ]$.
For $s\leq t$,
 $$
\frac{G(t,s)}{G(s,s)}=\frac{\varphi (1-t)}{\varphi (1-s)}
 \geq \frac{\varphi (\theta)}{\varphi (1)}=\frac{e^{m\theta}+e^{-m\theta}}
 {e^m+e^{-m}}=M_\theta.
$$
If $t\leq s$, then
 $$
\frac{G(t,s)}{G(s,s)}=\frac{\varphi (t)}{\varphi (s)}
 \geq \frac{\varphi (\theta)}{\varphi (1)}=\frac{e^{m\theta}+e^{-m\theta}}
 {e^m+e^{-m}}=M_\theta.
$$
Thus
$$
G(t,s)\geq M_{\theta}G(s,s),\quad \theta \leq t\leq
1-\theta,\; 0\leq s\leq 1.
$$
(2)\ When $t, t_0\leq s$,
\begin{align*}
\frac{G(t,s)}{G(t_0,s)}&=\frac{\varphi(t)\varphi(1-s)}{\varphi(t_0)\varphi(1-s)}
=\frac{\varphi(t)\varphi(1-t)}{\varphi(t_0)\varphi(1-t)}\\
 &\geq\frac{1}{\varphi^2(1)}\varphi(t)\varphi(1-t)=C\varphi(t)\varphi(1-t).
\end{align*}
If $t\leq s\leq t_0$,
\begin{align*}
\frac{G(t,s)}{G(t_0,s)}&=\frac{\varphi(t)\varphi(1-s)}{\varphi(s)\varphi(1-t_0)}
=\frac{\varphi(t)\varphi(1-t)}{\varphi(s)\varphi(1-t)}\cdot\frac{\varphi(1-s)}{\varphi(1-t_0)}\\
&\geq
\frac{1}{\varphi^2(1)}\varphi(t)\varphi(1-t)=C\varphi(t)\varphi(1-t).
\end{align*}
If $t_0\leq s\leq t$,
\begin{align*}
\frac{G(t,s)}{G(t_0,s)}&=\frac{\varphi(s)\varphi(1-t)}{\varphi(t_0)\varphi(1-s)}
=\frac{\varphi(t)\varphi(1-t)}{\varphi(t)\varphi(1-s)}\cdot\frac{\varphi(s)}{\varphi(t_0)}\\
&\geq
\frac{1}{\varphi^2(1)}\varphi(t)\varphi(1-t)=C\varphi(t)\varphi(1-t).
\end{align*}
For $s\leq t, t_0$,
\begin{align*}\frac{G(t,s)}{G(t_0,s)}&=\frac{\varphi(s)\varphi(1-t)}{\varphi(s)\varphi(1-t_0)}
=\frac{\varphi(t)\varphi(1-t)}{\varphi(t)\varphi(1-t_0)}\\
&\geq\frac{1}{\varphi^2(1)}\varphi(t)\varphi(1-t)=C\varphi(t)\varphi(1-t).
\end{align*}
Therefore,
\begin{equation*}G(t,s)\geq C\varphi(t)\varphi(1-t)G(t_0,s),\quad
t,\ t_0,\ s\in [0,1].
\end{equation*}
This completes the proof.\end{proof}

\begin{lemma}\label{lem2}
Let $y\in C((0,1),[0,\infty)), 0<\int_0^1y(s)ds<\infty$.
 Then the NBVP
\begin{equation}\begin{gathered}
-w''+m^2w=y(t),\quad  0<t<1, \\
w'(0)=w'(1)=0,
\end{gathered}\label{e2.2}
\end{equation}
 has a unique solution $w$ and there exists a constant
$C_y$ such that
\begin{equation}\label{e2.3}
C\|w\|\varphi(t)\varphi(1-t)\leq w(t)\leq C_y\varphi(t)\varphi(1-t),\quad 0\leq t\leq
1.\end{equation}
\end{lemma}

\begin {proof} It is obvious
that $w(t)=\int _0^1G(t,s)y(s)ds$ is the unique solution of
\eqref{e2.2}. First, let $t_0\in (0,1)$ such that
$\|w\|=w(t_0)=\int_0^1G(t_0,s)y(s)ds$. By Lemma \ref{lem1}, we have
\begin{align*}
 w(t)&=\int_0^1G(t,s)y(s)ds\\
&\geq\int_0^1C\varphi(t)\varphi(1-t)G(t_0,s)y(s)ds\\
&=C\varphi(t)\varphi(1-t)\int_0^1G(t_0,s)y(s)ds\\
&=C\varphi(t)\varphi(1-t)\|w\|,
\end{align*}
which is the first inequality of \eqref{e2.3}.
 On the other hand,
\begin{align*} w(t)&=\int_0^1G(t,s)y(s)ds\\
&=\frac{1}{\rho}\int_0^t\varphi (s)\varphi (1-t)y(s)ds+
\frac{1}{\rho}\int_t^1\varphi (t)\varphi (1-s)y(s)ds\\
&\leq\frac{1}{\rho}\varphi (1-t)\varphi (t)\int_0^ty(s)ds+
\frac{1}{\rho}\varphi (t)\varphi (1-t)\int_t^1y(s)ds\\
&=\frac{1}{\rho}\varphi (1-t)\varphi (t)\int_0^1y(s)ds.
\end{align*}
By setting
$$ C_y=\frac{1}{\rho}\int_0^1y(s)ds,
$$
then the second inequality of \eqref{e2.3} is proved.\end{proof}


\begin{remark} \label{rmk2.1} \rm
 From Lemma \ref{lem2} we know, if
$y(t)\equiv M$, then $C_y=C_M=\frac{M}{\rho}$.
\end{remark}

We make the following assumptions
\begin{itemize}
\item[(H1)] $ f(t,u)\leq
p(t)q(u)$, where $p:(0,1)\to [0,+\infty)$ and $q:[0,+\infty)\to
[0,+\infty)$ are continuous.

\item[(H2)] $|g(t,u)|\leq M$, where $M>0$ is a constant.

\item[(H3)]  $ 0<\int_0^1G(s,s)p(s)ds<+\infty$.

\item[(H4)] $\lim_{u\to+\infty}\frac{f(t,u)}{u}=+\infty$
uniformly on any compact subinterval of $(0,1)$.
\end{itemize}
Let
\begin{gather*}
C^+[0,1]=\{u\in C[0,1]:u(t)\geq 0,\ 0\leq t\leq 1\},\\
K=\{u:u\in C^+[0,1],\min_{\theta\leq t\leq 1-\theta}u(t)\geq M_{\theta}\|u\|\}.
\end{gather*}
It is obvious that $C^+[0,1]$ and $K$ are cones of $E$.
Let $v(t)$ be the solution  of the boundrary-value problem
\begin{gather*}
-v''+m^2v=M ,\  0<t<1,\\
 v'(0)=v'(1)=0.
\end{gather*}
By Lemma \ref{lem2}, $v(t)\leq
C_M\varphi(t)\varphi(1-t)=\frac{M}{\rho}\varphi(t)\varphi(1-t)$.
Set
\begin{gather*}
[y(t)]^*=\begin{cases} y(t), & y(t)\geq 0,\\
0, & y(t)<0,\end{cases} \quad 0<t<1,
\\
F(t,u)=\lambda f(t,[u-v]^*)+g(t,[u-v]^*)+M,\ 0\leq t\leq 1.
\end{gather*}
Consider the boundary-value problem
\begin{equation}\begin{gathered}
-u''+m^2u= F(t,u),\quad 0<t<1,\\
u'(0)=u'(1)=0.
\end{gathered}\label{e2.4}
\end{equation}
It is no difficulty to prove that $u=u_0-v$ is a
positive solution of \eqref{e1.1} if and only if $u_0$ is
a positive solution of \eqref{e2.4} and $u_0(t)>v(t)$,
$0<t<1$.

Define an operator $T_\lambda:C^+[0,1]\to C^+[0,1]$ by
 $$(T_\lambda u)(t)=\lambda \int_0^1G(t,s) F(s,u(s))ds,\quad u\in K.$$

\begin{lemma}\label{lem3}
Let (H1)--(H3) hold. Then $T_\lambda: K\to K$ is a completely
continuous operator.
\end{lemma}

\begin{proof} For any $u\in K$, $t\in [0,1]$, we have
$$
(T_\lambda u)(t)=\lambda \int_0^1G(t,s)F(s,u(s))ds
\leq \lambda \int_0^1G(s,s)F(s,u(s))ds.
$$
thus
 $$
 \|T_\lambda u\|\leq \lambda \int_0^1G(s,s)F(s,u(s))ds.
$$
On the other hand, by Lemma \ref{lem1},
 \begin{align*}
 \min_{\theta \leq t\leq 1-\theta}(T_\lambda u)(t)
&=\min_{\theta \leq t\leq 1-\theta}\lambda \int_0^1G(t,s)F(s,u(s))ds\\
 &\geq M_\theta \lambda \int_0^1G(s,s)F(s,u(s))ds\\
&\geq M_\theta \|T_\lambda u\|.
 \end{align*}
Therefore, $T_\lambda (K) \subset K$. For a natural number $n\geq
2$, define
$$
F_n(t,x)=\begin{cases} \min\{ F(t,x),\ F(\frac{1}{n},x)\},& 0<t\leq\frac{1}{n},\\
F(t,x),& \frac{1}{n}<t<1-\frac{1}{n},\\
\min\{ F(t,x),\ F(\frac{1}{n},x)\},& 1-\frac{1}{n}\leq t<1,
\end{cases}
$$
and
$$
 (T_nu)(t)=\lambda \int_0^1G(t,s)F_n(t,u(s))ds,\quad \forall  u\in E.
$$
It is easy to prove that $T_n$ is completely continuous. Let
$D\subset E$ be a bounded set, then there is a constant $L>0$ such
that $\|u\|\leq L$  for  all $u\in D$, hence $[u(s)-x(s)]^*\leq
u(s)\leq \|u\|\leq L.$
 We have
\begin{align*}& \Big| (T_\lambda u)(t)-(T_nu)(t)\Big|\\
 &\leq \lambda \int_0^{1/n}G(t,s)\Big|F(s,u(s))-F(\frac{1}{n},u(s))\Big|ds\\
&\quad +\lambda\int_{1-\frac{1}{n}}^1G(t,s)\Big|F(s,u(s))-F(1-\frac{1}{n},u(s))\Big|ds\\
&\leq 2\lambda\Big(\int_0^{1/n}G(t,s)[p(s)q(u(s))+M]ds+
\int_{1-\frac{1}{n}}^1G(t,s)[p(s)q(u(s))+M]ds\Big)\\
&\leq 2\lambda\max _{0\leq x\leq
L}q(x)\Big(\int_0^{1/n}G(s,s)p(s)ds+
\int_{1-\frac{1}{n}}^1G(s,s)p(s)ds\Big)\\
&\quad +2M\Big(\int_0^{1/n}G(s,s)ds+
\int_{1-\frac{1}{n}}^1G(s,s)ds\Big)\\
& \to 0(n\to \infty).
\end{align*}
Therefore, $T_n$ converge uniformly to $T_\lambda$ on any bounded
subset of $E$. This implies that $T_\lambda$ is a completely
continuous operator.
\end{proof}

The following Krasnosel'skii fixed  point  theorem  in a cone
plays an important role in proving the main result \cite{g}.

\begin{theorem}\label{thm4}
Let $E$ be Banach space and $K\subset E$ be a cone in $E$.
Suppose\ $\Omega_1$\ and\ $\Omega_2$ are open subset of $E$ with
$0 \in \Omega_1$ and $\overline{\Omega}_1
 \subset \Omega_2$. Let $\ T:K\cap (\overline{\Omega}_2\setminus
 \Omega_1)\to
 K$ be a completely continuous operator such that
\begin{itemize}
\item[(A)]  $\|Tu\|\leq \|u\|$ for all $u\in K\cap \partial
\Omega_1 $ and $ \|Tu\| \geq \|u\|$ for all $u\in K\cap
\partial \Omega_2$ or

\item[(B)]  $ \|Tu\|\leq \|u\|$ for all $u\in K\cap
\partial \Omega_2 $ and $
\|Tu\|\geq \|u\|$ for all $u\in K\cap \partial \Omega_1$.
\end{itemize}
Then $T$ has a fixed point in $K\cap (\overline{\Omega}_2\setminus
\Omega_1)$.
\end{theorem}

\section{Main Result}

In this section, we present and prove our main result.

\begin{theorem}\label{thm1}
Suppose (H1)--(H4)hold, then \eqref{e1.1} has  at least one
positive solution $u\in C(0,1)\cap C^2[0,1]$ if
$$
0<\lambda\leq \Big[\max_{0\leq t\leq r}q(\tau)\int_0^1G(s,s)p(s)ds)\Big]^{-1},
$$
where $r=\max\big\{1+2M\int_0^1G(s,s)ds,\ \frac{M}{\rho
C}\big\}$.
\end{theorem}

\begin{proof}  By  Lemma \ref{lem3}, we  know that $T_\lambda$
is a completely continuous operator. Let
$$
\Omega_1=\{u\in C[0,1]:\|u\|<r\}.
$$
For all $ u\in K\cap \partial\Omega_1,\ t\in [0,1]$,
 we have
\begin{align*} (T_\lambda u)(t)&=\lambda \int_0^1G(t,s)F(t,u(s))ds\\
&=\int_0^1G(t,s)(\lambda f(s,[u(s)-x(s)]^*)+g(s,[u(s)-x(s)]^*)+M)ds\\
&\leq\int_0^1G(t,s)\lambda p(s)q([u(s)-x(s)]^*)ds+2M\int_0^1G(t,s)ds\\
&\leq\lambda\max_{0\leq\tau \leq r}q(\tau)
\int_0^1G(s,s)p(s)ds+2M\int_0^1G(s,s)ds\\
&\leq 1+2M\int_0^1G(s,s)ds\leq r=\|u\|.
\end{align*}
This implies
\begin{equation}\label{s3.1}\|T_\lambda u\|\leq \|u\|,\quad \mbox{for }
u\in K\cap \partial\Omega_1.
\end{equation}
On the other hand, choose $N$ large enough such that
$$
\frac{1}{2}\lambda M_\theta NC\varphi^2(\theta)
\int_\theta^{1-\theta}G(s,s)ds\geq 1.
$$
By (H4), there  exists a constant $B>0$ such that
 $$
 \frac{f(s,u)}{u}>N,\quad\hbox{for }(s,u)\in [\theta,1-\theta]\times [B,\infty).
$$
Set
 $$
\Omega_2=\{u\in C[0,1]:\|u\|<R\},\quad
R=\max\Big\{2r,\frac{2M}{\rho C},\frac{2B}{C\varphi^2(\theta)}\Big\}.
$$
For  any $u\in K\cap \partial\Omega_2$, $s\in [0,1]$,
\begin{align*} u(s)-v(s)&\geq
u(s)-C_M\varphi(s)\varphi(1-s)=u(s)-\frac{M}{\rho}\varphi(s)\varphi(1-s)\\
&=u(s)-\frac{M}{\rho CR}C\|u\|\varphi(s)\varphi(1-s) \geq
u(s)-\frac{1}{2}u(s)= \frac{1}{2}u(s)\geq 0.
\end{align*}
 Thus
\begin{align*} \min_{\theta\leq s\leq 1-\theta}(u(s)-v(s))
&\geq\min_{\theta\leq s\leq 1-\theta}\frac{1}{2}u(s)\\
&\geq \min_{\theta\leq s\leq
1-\theta}\frac{C}{2}\|u\|\varphi(s)\varphi (1-s)\\
&\geq\frac{CR}{2}\varphi^2(\theta)\geq B.
\end{align*}
 Therefore, for
$t\in [\theta,1-\theta]$,
\begin{align*} (T_\lambda u)(t)&=\lambda \int_0^1G(t,s)F(s,u(s))\\
&=\int_0^1G(t,s)(\lambda
f(s,[u(s)-v(s)]^*)+g(s,[u(s)-v(s)]^*)+M)ds\\
&\geq\int_\theta^{1-\theta} G(t,s)\lambda f(s,[u(s)-v(s)]^*)ds\\
&\geq\lambda M_\theta \int_\theta^{1-\theta} G(s,s) N(u(s)-v(s))ds\\
&\geq\lambda M_\theta\int_\theta^{1-\theta} G(s,s) N\frac{u(s)}{2}ds\\
&\geq\lambda M_\theta \int_\theta^{1-\theta} G(s,s) N\frac{C}{2}\|u\|\varphi(s)\varphi(1-s)ds\\
&\geq\frac{1}{2}\lambda M_\theta NC\varphi^2(\theta)
\int_\theta^{1-\theta}G(s,s)ds\|u\|\geq\|u\|.
\end{align*}
Thus
\begin{equation}\label{s3.2}
\|T_\lambda u\|\geq\|u\|,\quad \hbox{for}\  u\in
K\cap\partial\Omega_2.\end{equation}
Applying (B) of Theorem
\ref{thm4} to (\ref{s3.1}) and (\ref{s3.2}) yields that
$T_\lambda$ has a fixed point $u_0$ with $r\leq \|u_0\|\leq R$. By
Lemma \ref{lem1} it follows that
\begin{align*}
u_0(t)&\geq C\|u_0\|\varphi(t)\varphi(1-t)\\
&= Cr\varphi(t)\varphi(1-t)\\
&=\frac{r\rho C}{M}\cdot\frac{M}{\rho}\varphi(t)\varphi(1-t)\\
&\geq\frac{M}{\rho}\varphi(t)\varphi(1-t)=v(t).
\end{align*}
 Set $u(t)=u_0(t)-v(t)$,
 then $u(t)$ is a $C[0,1]\cap C^2(0,1)$ positive solution to \eqref{e1.1}.
This completes the proof.
\end{proof}


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\end{document}