\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 13, pp. 1--6.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/13\hfil Positive solutions]
{Positive solutions for a class of singular boundary-value problems}
\author[D. D. Hai\hfil EJDE-2005/13\hfilneg]
{Dang Dinh Hai}

\address{Dang Dinh Hai \hfill\break
Department of Mathematics, Mississippi State University, 
Mississippi State, MS 39762, USA}
\email{dang@math.msstate.edu}

\date{}
\thanks{Submitted August 25, 2005. Published January 26, 2006.}
\subjclass[2000]{34B16, 34B18} 
\keywords{Positive solutions; singular; sub-super solutions}

\begin{abstract}
 This paper concerns the existence and multiplicity of positive
 solutions for Sturm-Liouville boundary-value problems. We use
 fixed point theorems and the sub-super solutions method to
 two solutions to the problem studied.
\end{abstract}

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

\textbf{Introduction}

Consider the boundary-value problem
\begin{equation} \label{e1}
\begin{gathered}
Lu=\lambda f(t,u),\quad 0<t<1 \\
\alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0,
\end{gathered}
\end{equation}
where $Lu=-(ru')'+qu$, $r,q\in C[0,1]$ with $r>0$, $q\geq 0$
on $[0,1]$, $\alpha ,\beta ,\gamma ,\delta \geq 0$ with
$\alpha \delta +\alpha \gamma +\beta \gamma >0$,
$f:(0,1)\times [ 0,\infty)\to  [ 0,\infty )$, and $\lambda $ is a
positive parameter.

The existence and nonexistence of positive solutions of problem \eqref{e1}
with $f$ possibly singular have been established by Choi \cite{Ch},
Dalmasso \cite{Dal}, Wong \cite{W}, and recently by Erbe and Mathsen
\cite{Erb}. In this
paper, we shall obtain positive solutions to \eqref{e1} under assumptions
less stringent than in \cite{Erb}. In particular, we do not need the
condition that $f(t,u)$ be nondecreasing in $u$, which is essential in
\cite{Ch,Dal,Erb,W}. Our
approach depends on fixed point theorems and sub-super solutions method.

\section{Main results}

Let $G(t,s)$ be the Green's function for \eqref{e1}. Then $u$ is a
solution of \eqref{e1} if and only if
\[
u(t)=\lambda \int_{0}^{1}G(t,s)f(s,u(s))ds.
\]
Recall that
\[
G(t,s)=\begin{cases}
c^{-1}\phi (t)\psi (s) &\text{if }t\leq s \\
c^{-1}\phi (s)\psi (t) &\text{if }t>s,
\end{cases}
\]
where $\phi $ and $\psi $ satisfy%
\begin{equation}
\begin{gathered}
L\phi =0,\quad \phi (0)=\beta ,\quad \phi '(0)=\alpha \\
L\psi =0,\quad \psi (1)=\delta ,\quad \psi '(1)=-\gamma
\end{gathered} \label{e2.1}
\end{equation}
and $c=-r(t)(\phi (t)\psi '(t)-\phi '(t)\psi (t))>0$. Note
that $\phi '>0$ on $(0,1]$, $\psi '<0$ on $[0,1)$.

We shall make the following assumptions:
\begin{itemize}
\item[(H1)] $f:(0,1)\times [ 0,\infty )\to  [ 0,\infty )$ is
continuous

\item[(H2)] For each $M>0$, there exists a continuous function
$g_{M}$ on $(0,1)$ such that
$f(t,u)\leq g_{M}(t)$  for $t\in (0,1)$, $0\leq u\leq M$,
 and
\[
\int_{0}^{1}G(s,s)g_{M}(s)ds<\infty .
\]

\item[(H3)] There exist an interval $I\subset (0,1)$ and a function
$m\in L^{1}(I)$ with
$m\geq 0$, $m\not\equiv 0$ such that for every $a>0$, there exists
$r_{a}>0$ such that
\[
f(t,u)\geq am(t)u\quad \text{for }t\in I,\;u\in (0,r_{a})
\]

\item[(H4)] There exist an interval $J\subset (0,1)$  and a positive
number $d$ such that
\[
f(t,u)\geq du\quad \text{for }t\in J,\;u\geq 0.
\]

\item[(H5)] There exist an interval $I_{1}\subset (0,1)$ and a function
$m_{1}\in L^{1}(I_{1})$ with $m_{1}\geq 0,m_{1}\not\equiv 0$ such that
for every $b>0$, there exists $R_{b}>0$ such that
\[
f(t,u)\geq bm_{1}(t)u\quad \text{for }t\in I_{1},\;u\geq R_{b}.
\]
\end{itemize}
 Our main results are stated as follows.

\begin{theorem} \label{thm1}
Let \textrm{(H1)--(H3)} hold. Then there exists $\lambda _{0}$ $>0$
such that \eqref{e1} has a positive solution for $0<\lambda <\lambda _{0}$.
If, in addition, \textrm{(H5)} holds, then \eqref{e1} has at least
two positive solutions for $0<\lambda <\lambda _{0}$.
\end{theorem}

\begin{theorem} \label{thm2}
Let \textrm{(H1)--(H4)} hold. Then there exists $\lambda ^{\ast }>0$
such that \eqref{e1} has a positive solution for
$0<\lambda <\lambda ^{\ast }$ and no positive
solution for $\lambda >\lambda ^{\ast }$.
\end{theorem}

\begin{remark} \label{rmk3} \rm
Let $f(t,u)=m(t)g(u)$, where $g:[0,\infty )\to  [ 0,\infty )$
be continuous with $\lim_{u\to  0^{+}}\frac{g(u)}{u}=\infty $,
$\lim_{u\to  \infty }\frac{g(u)}{u}=\infty $, and $m\in L^{1}(0,1)$
with $m\geq 0$, $m\not\equiv 0$. Then $f$ satisfies (H1)--(H3) and (H5) and
therefore Theorem \ref{thm1} applies. If we take $m(t)=1/\sqrt{t}$,
$g(u)=u^{p}+u^{q}+h(u)$, where $p<1\leq q$ and $h$ is a nonnegative
continuous function, then it is easily seen that $f(t,u)$ satisfies
(H1)--(H5) and Theorem \ref{thm2} applies. However, the results
in \cite{Ch,Dal,Erb,W} may not apply since $g$ may not be nondecreasing.
\end{remark}

To prove our main results, we first establish the following results.

\begin{lemma}\label{lem4}
Let $h\in L^{1}(0,1)$ be such that $h\geq 0$ and let $u$ satisfy
\begin{gather*}
Lu=h\quad \text{in }(0,1) \\
\alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0.
\end{gather*}
Then
\[
u(t)\geq |u|_{0}p(t),
\]
where
$p(t)=\min \big( \frac{\phi (t)}{|\phi |_{0}},
\frac{\psi (t)}{|\psi|_{0}}\big) $, and $\|\cdot\|_{0}$ denotes the supremum  norm.
\end{lemma}


\begin{proof} We proceed as in \cite{Dan}. It is easy to see that
\[
u(t)=\int_{0}^{1}G(t,s)h(s)ds.
\]
Let $|u|_{0}=u(t_{0})$ for some $t_{0}\in (0,1)$. We verify that
\[
\frac{G(t,s)}{G(t_{0},s)}\geq p(t).
\]
If $t,t_{0}\leq s$ then
\[
\frac{G(t,s)}{G(t_{0},s)}=\frac{\phi (t)}{\phi (t_{0})}
\geq \frac{\phi (t)}{|\phi |_{0}},
\]
and if $t_{0}\leq s\leq t$ then
\[
\frac{G(t,s)}{G(t_{0},s)}=\frac{\phi (s)\psi (t)}{\phi (t_{0})\psi (s)}\geq
\frac{\psi (t)}{\psi (s)}\geq \frac{\psi (t)}{|\psi |_{0}}
\]
since $\phi (s)\geq \phi (t_{0})$. The other two cases are treated in a
similar manner. Hence
\[
u(t)\geq p(t)u(t_{0})=|u|_{0}p(t).
\]
\end{proof}

\begin{lemma} \label{lem5}
Let \textrm{(H1)--(H3)} hold. Then for each $\lambda >0$, there exists
$c_{\lambda }>0$ such that if $u$ is a nonzero solution of \eqref{e1} then
$|u|_{0}\geq c_{\lambda }$.
Furthermore, $(c_{\lambda })$ is nondecreasing in $\lambda $.
\end{lemma}

\begin{proof} Let $p_{0}=\min_{t\in I}p(t)$, where $p$
is defined in Lemma \ref{lem4}, and
$$
K= \int_{I}G(\frac{1}{2},s)m(s)ds .
$$
By (H3), there exists $r_{\lambda }\in (0,1)$ such that
\[
\frac{f(t,u)}{u}\geq \frac{2m(t)}{\lambda p_{0}K}\quad \text{for }t\in
I,\;0<u<r_{\lambda }
\]
Define
\[
c_{\lambda }=\sup \big\{ r\in (0,1):\frac{f(t,u)}{u}\geq
 \frac{2m(t)}{\lambda p_{0}K}\text{ for }t\in I,\;0<u<r\big\} .
\]
Then $0<c_{\lambda }\leq 1$ and
\begin{equation}
\frac{f(t,u)}{u}\geq \frac{2m(t)}{\lambda p_{0}K}\quad \text{for }t\in
I,\;0<u\leq c_{\lambda }.  \label{e2.2}
\end{equation}
Clearly $(c_{\lambda })$ is nondecreasing in $\lambda $. Let $u$ be a
nonzero solution of \eqref{e1} and suppose that $|u|_{0}<c_{\lambda }$.
Using Lemma \ref{lem4} and \eqref{e2.2}, we obtain
\begin{align*}
u(t) &= \lambda \int_{0}^{1}G(t,s)f(s,u(s))ds\\
&\geq \lambda \int_{I}\frac{2m(s)}{\lambda p_{0}K}G(t,s)u(s)ds \\
&\geq 2K^{-1}|u|_{0}\int_{I}G(t,s)m(s)ds,
\end{align*}
which implies
\[
|u|_{0}\geq u(\frac{1}{2})\geq 2K^{-1}\Big( \int_{I}G(\frac{1}{2}
,s)m(s)ds)\Big) |u|_{0}=2|u|_{0},
\]
a contradiction. This completes the proof.
\end{proof}

\begin{lemma} \label{lem6}
Let \textrm{(H1), (H2), (H4)} hold. Then \eqref{e1} has no positive
solution for $\lambda $ large.
\end{lemma}

\begin{proof}
Let $u$ be a positive solution of \eqref{e1}. Using
(H4) and Lemma \ref{lem4}, we obtain
\[
u\big( \frac{1}{2}\big) =\lambda \int_{0}^{1}G(\frac{1}{2},s)f(s,u(s))ds
\geq \lambda d\int_{J}G(\frac{1}{2},s)u(s)ds\geq \lambda dC|u|_{0},
\]
where $C=\big( \min_{s\in J}p(s)\big) \big( \int_{J}G(\frac{1}{2},s)ds\big)$,
which implies $\lambda \leq (dC)^{-1}$.
\end{proof}

The next Lemma establishes the existence of a solution once a pair of
ordered sub- and supersolution are known, without assuming monotonicity of
$f(t,u)$ in $u$.

\begin{lemma} \label{lem7}
Let \textrm{(H1), (H2)} hold. Suppose that $\underline{u}$ and
$\bar{u}$ in $C[0,1]\cap C^{1}(0,1)$ are sub- and supersolutions of
\eqref{e1} respectively with $0\leq \underline{u}\leq \bar{u}$, i.e.,
\begin{gather*}
L\underline{u}(t) \leq \lambda f(t,\underline{u}) \quad \mbox{in }(0,1) \\
\alpha \underline{u}(0)-\beta \underline{u}'(0)\leq 0,\quad \gamma \underline{u}(1)
+\delta \underline{u}'(1)\leq 0
\end{gather*}
and
\begin{gather*}
L\bar{u}(t) \geq \lambda f(t,\bar{u}(t))\quad\mbox{in }(0,1) \\
\alpha \bar{u}(0)-\beta \bar{u}'(0)\geq 0,\quad
\gamma \bar{u} (1)+\delta \bar{u}'(1)\geq 0.
\end{gather*}
Then \eqref{e1} has a solution $u$ with $\underline{u}\leq u\leq \bar{u}$.
\end{lemma}

\begin{proof}
The proof is essentially given in \cite{Sc}, where nonsingular
problems were considered. For convenience, we give a proof. Without loss of
generality, we assume that $\lambda =1$. Define
\[
\bar{f}(t,v)=
\begin{cases}
f(t,\bar{u}(t))+\frac{\bar{u}(t)-v}{1+\;v^{2}} & \text{if  }v>\bar{u}(t) \\
f(t,v) & \text{if } \underline{u}(t)\leq v\leq \bar{u}(t) \\
f(t,\underline{u}(t))+\frac{\underline{u}(t)-v}{1+v^{2}} &
\text{if }v\leq \underline{u}(t).
\end{cases}
\]
For each $v\in C[0,1]$, let $u=Tv$ be the solution of
\begin{gather*}
Lu=\bar{f}(t,v),\quad 0<t<1 \\
\alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0.
\end{gather*}
Then $T:C[0,1]\to  C[0,1]$ is completely continuous. Since $T$ is
bounded, $T$ has a fixed point $u$ by the Schauder fixed point Theorem. We
verify that $\underline{u}\leq u\leq \bar{u}$. Suppose to the
contrary that there exists $t_{0}\in (0,1)$ such that
$u(t_{0})>\bar{u}(t_{0})$. Let $w=u-\bar{u}$ and $t_{1}\in [ 0,1]$
be such that $w(t_{1})=\max_{0\leq t\leq 1}w(t)>0$. If
$t_{1}\in (0,1)$ then $w'(t_{1})=0$ and $(rw')'(t_{1})\leq 0$, which
implies that $Lw(t_{1})\geq 0$. On the other hand,
\[
Lw(t_{1})=Lu(t_{1})-L\bar{u}(t_{1})\leq -\frac{w(t_{1})}{1+u^{2}(t_{1})}<0,
\]
a contradiction. Suppose that $t_{1}=0$. Then $w'(0)\leq 0$, and
since $\alpha w(0)-\beta w'(0)\leq 0$, we have a contradiction if
$\alpha >0$. If $\alpha =0$ then $\beta >0$ and therefore $w'(0)=0$.
Since $-(rw')'(t)+qw(t)\equiv Lw(t)<0$ for
small $t>0$, it follows by integrating that $(rw')(t)>0$ and so
$w'(t)>0$ for small $t>0$, a contradiction. Similarly, we reach a
contradiction if $t_{1}=1$. Hence $u\leq \bar{u}$ on $(0,1)$. The lower
inequality can be derived in a similar manner.
\end{proof}

 In view of Lemmas \ref{lem4} and \ref{lem5}, we see that $u$ is a positive solution of
\eqref{e1} if and only if $u$ is a solution of
\begin{equation}
\begin{gathered}
Lu=\lambda \tilde{f}(t,u),\quad 0<t<1 \\
\alpha u(0)-\beta u'(0)=0,\quad \gamma u(1)+\delta u'(1)=0,
\end{gathered}
\label{e1'}
\end{equation}
where $\tilde{f}(t,u(t))=f(t,\max (u(t),c_{\lambda }p(t))$, or
equivalently, $u$ is a fixed point of $A_{\lambda }$, where
\[
A_{\lambda }u(t)=\lambda \int_{0}^{1}G(t,s)\tilde{f}(s,u(s))ds
\]
Note that $A_{\lambda }:C[0,1]\to  C[0,1]$ is completely continuous
(see \cite{Dan}).

We are now in a position to prove our main result.

\begin{proof}[Proof of Theorem \ref{thm1}]
Let
\[
\lambda _{0}=\Big( \int_{0}^{1}G(s,s)g_{1}(s)ds\Big) ^{-1}
\]
and suppose that $0<\lambda <\lambda _{0}$, where $g_{1}$ is defined in
(H2). Let $u$ be a solution of
\[
u=\theta A_{\lambda }u\quad \text{for some }\theta \in [ 0,1].
\]
We claim that $|u|_{0}\neq 1$. Indeed, if $|u|_{0}=1$ then since
$c_{\lambda }|p|_{0}\leq c_{\lambda }\leq 1$,
it follows from (H2) that $\tilde{f}(s,u(s))\leq g_{1}(s)$, which implies
\[
1=|u|_{0}\leq \lambda \int_{0}^{1}G(s,s)g_{1}(s)ds<1
\]
for $\lambda <\lambda _{0}$, a contradiction, and the claim is proved.
Hence the Leray-Schauder fixed point Theorem gives the existence of a fixed
point $u$ of $A_{\lambda }$ with $|u|_{0}<1$.

Next, suppose that (H5) holds. We shall employ fixed point theorems in a
cone to show the existence of a second solution.
Let $\mathbb{K}$ be the cone of nonnegative functions in $C[0,1]$. By the
above arguments, we have
\[
u\in \mathbb{K}\text{ and }u\leq A_{\lambda }u\;\Rightarrow\;
 |u|_{0}\neq 1.
\]
Let
\[
b=2\Big( \lambda p_{1}\int_{I_{1}}G\big( \frac{1}{2},s\big)
m_{1}(s)ds\Big) ^{-1},
\]
where $p_{1}=\min_{s\in I_{1}}p(s)$. By (H5), there exists $R_{b}>p_{1}$
such that
\[
\tilde{f}(s,u)\geq bm_{1}(s)u\quad \text{for }s\in I_{1},\;u\geq R_{b}.
\]
We claim that
\[
u\in \mathbb{K}\text{ and }u\geq A_{\lambda }u\;\Rightarrow\; |u|_{0}\neq
R_{b}p_{1}^{-1}
\]
Suppose that $u\in \mathbb{K}$ and $u\geq A_{\lambda }u$. If
$|u|_{0}=R_{b}p_{1}^{-1}$  then it follows from Lemma \ref{lem4} that
\[
u(s)\geq R_{b}p_{1}^{-1}p(s)\geq R_{b}\quad \text{for }s\in I_{1}.
\]
Hence
\begin{align*}
R_{b}p_{1}^{-1} &=|u|_{0}\geq u\big( \frac{1}{2}\big) \\
&\geq \lambda
\int_{0}^{1}G\big( \frac{1}{2},s\big) \tilde{f}(s,u(s))ds \\
&\geq bR_{b}\lambda \Big( \int_{I_{1}}G\big( \frac{1}{2},s)\big)
m_{1}(s)ds\Big) =2R_{b}p_{1}^{-1},
\end{align*}
a contradiction, and the claim is proved. By Krasnoselskii's fixed point
Theorem, \cite{K}, $A_{\lambda }$ has a fixed point $\tilde{u}$ in $\mathbb{K}$
with $1<|\tilde{u}|_{0}<R_{b}p_{1}^{-1}$.
This completes the proof
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
Let $\Lambda $ be the set of all $\lambda >0$ such that \eqref{e1} has
 a positive solution and let $\lambda ^{\ast}=\sup \Lambda $.
By Theorem \ref{thm1} and Lemma \ref{lem6}, $0<\lambda ^{\ast }<\infty $.
Let $0<\lambda <\lambda ^{\ast }$. Then there exists $\lambda _{0}>0$ such
that $\lambda <\lambda _{0}$ and $\eqref{e1}_{\lambda _{0}}$ has a positive
solution $u_{\lambda _{0}}$. Then $u_{\lambda _{0}}$ satisfies
\[
u_{\lambda _{0}}(t)\geq c_{\lambda _{0}}p(t)\geq c_{\lambda }p(t),
\]
and therefore
\begin{align*}
Lu_{\lambda _{0}}(t)
&=\lambda _{0}f(t,u_{\lambda _{0}}(t))\\
&=\lambda_{0}f(t,\max (u_{\lambda _{0}}(t),c_{\lambda }p(t)) \\
&\geq \lambda f(t,\max (u_{\lambda _{0}}(t),c_{\lambda }p(t))\\
&=\lambda \tilde{f}(t,u_{\lambda _{0}}(t)),
\end{align*}
i.e., $u_{\lambda _{0}}$ is a supersolution of  \eqref{e1'}. Since $0$ is a
subsolution of \eqref{e1'}, it follows from Lemma \ref{lem7} that \eqref{e1'} has
a solution $u_{\lambda }$ with $0\leq u_{\lambda }\leq u_{\lambda _{0}}$.
Thus $u_\lambda$ is a positive solution of \eqref{e1}, completing the
proof of Theorem \ref{thm2}.
\end{proof}


\begin{thebibliography}{9}

\bibitem{Ch} Y. S. Choi, \emph{A singular boundary value problem arising from
near-ignition analysis of flame structures}, Diff. Integral Eqns.
\textbf{4} (1991), 891-895.

\bibitem{Dal} R. Dalmasso, \emph{Positive solutions of singular boundary value
problems}, Nonlinear Anal. \textbf{27} (1996), 645-652.

\bibitem{Dan} H. Dang and K. Schmitt, \emph{Existence of positive solutions for
semilinear elliptic equations in annular domains}, Diff. Integral Eqns
\textbf{7} (1994), 747-758.

\bibitem{Erb} L. H. Erbe and R. M. Mathsen,
\emph{Positive solutions for singular nonlinear boundary value problems},
Nonlinear Anal.\textbf{46} (2001), 979-986.

\bibitem{K} M. A. Krasnoselskii,
\emph{Positive solutions of operator equations},
Noordhoff, Groningen (1964).

\bibitem{Sc} K. Schmitt, \emph{Boundary value problems for quasilinear second
order elliptic equations}, Nonlinear Anal. \textbf{3} (1978), 263-309.

\bibitem{W} F. H. Wong, \emph{Existence of positive solutions of singular boundary
value problems}, Nonlinear Anal. \textbf{16} (1993), 397-406.

\end{thebibliography}

\end{document}