\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 43, pp. 1--5.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/43\hfil Positive solutions]
{Positive solutions for a class of
singular boundary-value problems}

\author[L. Xia, Z. Yao\hfil EJDE-2007/43\hfilneg]
{Li Xia, Zhengan Yao}  % in alphabetical order

\address{Li Xia \newline
Department of Mathematics, Sun Yat-Sen University, Guangzhou
510275, China} 
\email{xaleysherry@163.com}

\address{Zhengan Yao \newline
Department of Mathematics, Sun Yat-Sen University, Guangzhou
510275, China} 
\email{mcsyao@mail.sysu.edu.cn}

\thanks{Submitted September 19, 2006. Published March 15, 2007.}
\thanks{Supported by grants NNSFC-10171113 and NNSFC-10471156
 from the National Natural \hfill\break\indent
 Science Foundation of China, and
 by grant NSFGD-4009793 from the Natural Science \hfill\break\indent
 Foundation of Guang Dong.}
\subjclass[2000]{34B10, 34B16, 34B18}
\keywords{Upper and lower solution; existence; singular equation}

\begin{abstract}
 Using  regularization  and the sub-super solutions method,
 this note shows the existence of positive solutions for
 singular differential equation subject to four-point boundary conditions.
\end{abstract}

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

\section{Introduction}

  This note concerns the existence of positive solutions to
the boundary-value problem (BVP)
\begin{gather}
 y''=-\frac{\beta}{t}y'+\frac{\gamma}{y}|y'|^2-f(t,y), \quad 0 < t < 1, \label{e1}\\
y(0)=y(1)=0, \label{e2}\\
y'(0)=y'(1)=0 \label{e3},
\end{gather}
where
%$g(t,y,y')=-\frac{\beta}{t}y'+\frac{\gamma}{y}|y'|^2-f(t,y)$,
$\beta>0, \gamma>\beta+1$ are constants, and $f$ satisfies
\begin{itemize}
\item[(H1)] $f(t, y)\in C^1([0,1]\times [0,\infty), [c_0,\infty))$
for sufficiently small $c_0>0$, and $f$ is non-increasing
with respect to $y$.
\end{itemize}
Equation \eqref{e1} with the nonlinear right-hand side
independent of $y'$  has been discussed extensively in the literature;
see for example \cite{a1,h1} and the references therein.
Because of its background in applied mathematics and physics,
problem \eqref{e1} with  right-hand side depending on $y'$
has attracted the attention of many authors; see for instance
\cite{g1,t1} and their references.

Guo et al. \cite{g1} studied the existence
of positive solutions for the singular boundary-value problem with
nonlinear boundary conditions
\begin{gather*}
 y''+q(t)f(t,y,y')=0,\quad 0<t<1,\\
y(0)=0,\quad \theta (y'(1))+y(1)=0\,,
\end{gather*}
where $f(t,y,y')\geq 0$ is singular at $y=0$.
They use a nonlinear alternative of Leray-Schauder type and Urysohn's
lemma.

This work is motivated by \cite{b2} where the authors studied the
problem
\begin{gather*}
y''+ \frac{N-1}{t}y'- \frac{\gamma}{y}|y'|^2+1=0, \quad 0<t<1, \\
y(1)=0,\quad  y'(0)=0.
\end{gather*}
There $N$ is a positive integer, and the problem corresponds to
$\beta=N-1$, $f\equiv 1$ in \eqref{e1}.
Applying ordinary differential equation techniques,  they obtained a
decreasing positive solution which, subsequently, was used in
\cite{b3} to study some properties of
solutions for a class of degenerate parabolic equations (see \cite{b1}
for further information).

In this note, we study problem \eqref{e1} under boundary
conditions that are mote complicated than those in \cite{b2}.
By using a regularization method and constructing
sub- and supersolutions, we obtain an existence result.

%\noindent{\bf Definition.}
A function $y \in C^2(0,1) \cap C[0,1]$ is called a solution
for \eqref{e1} if it is positive in $(0,1)$ and satisfies
 \eqref{e1} pointwise.

The main result of this note is as follows.

\begin{theorem} \label{thm1}
 Under  assumption (H1), the boundary-value problem \eqref{e1}--\eqref{e3}
 admits at least one solution.
\end{theorem}

%\noindent{\bf Remark.}
Since we need to calculate the derivatives of $f$, we assume that
$f\in C^1([0,1]\times [0,\infty), [c_0, \infty))$. However,
if $f\in C([0,1]\times [0,\infty), [c_0, \infty))$, Theorem \ref{thm1}
remains valid.

\section{Proof of Theorem \ref{thm1}}

 Since problem \eqref{e1} is singular at point $t=0$, or
$y(t)=0$, we need to regularize it. Precisely, we discuss positive
solutions of the  regularized problem
\begin{equation} \label{e4}
 -y''-\frac{\beta}{t+\varepsilon}y'+\frac{\gamma }{|y|+\varepsilon^2}|y'|^2
-f(t,y)=0, \quad 0 < t < 1,
\end{equation}
subject to the boundary conditions \eqref{e2}, where
$\varepsilon\in (0,1]$.

Denote $Ay=-y''$ and
$$
b_\varepsilon (t,\xi,
\eta)=\frac{\beta}{t+\varepsilon}\eta-\frac{\gamma
}{|\xi|+\varepsilon^2}|\eta|^2+f(t, \xi).
$$
Note that $b_\varepsilon (\cdot, \xi, \eta)\in C^\mu[0,1]$ uniformly for
$(\xi, \eta) $ in bounded subsets of $\mathbb{R}\times \mathbb{R}$ for
some $\mu\in (0,1]$, $\partial b_\varepsilon/\partial \xi,
\partial b_\varepsilon/\partial \eta$ exist and are continuous on
$[0,1]\times \mathbb{R}^2$. Moreover, there exists some positive
constant $C$ dependent of $\varepsilon^{-1}, \sigma$ such that
\begin{equation*}
 |b_\varepsilon(t,\xi,\eta)|\leq C(1+|\eta|^2)
\end{equation*}
for every $\sigma\geq 0$ and $(t, \xi, \eta)\in
[0,1]\times[-\sigma, \sigma]\times \mathbb{R}$.

A function $y$ is called a subsolution for BVP \eqref{e4} \eqref{e2} if
$y\in C^{2+\mu}[0,1]$ and
\begin{gather*}
 Ay\leq b_\varepsilon(\cdot, y, y')\quad \mbox{in }[0,1],\\
y(0)\leq 0,\quad  y(1)\leq 0.
\end{gather*}
Supersolutions are defined by reversing the above inequality
signs. We call $y$ a solution for \eqref{e4} \eqref{e2}, if $y$ is
a subsolution and a supersolution of \eqref{e4} \eqref{e2}.

Let $v(t)=\frac{1}{2}t-\frac{1}{2}t^2$, it is easy to see that $v$
is a nonnegative solution for problem
\begin{gather*}
 -v''=1,\quad 0<t<1,\\
v(0)=v(1)=0.
\end{gather*}


\begin{lemma} \label{lem1}
 Let $\underline{y}=C_1 v^2$,
$y_{1\varepsilon}=C_2(t+\varepsilon)^2$,
$y_{2\varepsilon}=C_2(1+\varepsilon-t)^2$,
$\overline{y}_\varepsilon=\min\{y_{1\varepsilon},y_{2\varepsilon}\}$,
then \eqref{e4} \eqref{e2} admits at least one solution
$y_\varepsilon\in [\underline{y}, \overline{y}_\varepsilon]$. Here
$C_1$ and $C_2\geq 1$ are some positive constants.
\end{lemma}

\begin{proof}  By \cite[Theorem 1.1]{a2}, it suffices
 to prove $\underline{y}(\overline{y})$ is a subsolution (supersolution)
  for  \eqref{e4} \eqref{e2}. Hence we need to prove
$A\underline{y}\leq   b_\varepsilon(t, \underline{y}, \underline{y}')$,
$A y_{i \varepsilon}\geq b_\varepsilon(t, y_{i\varepsilon}, y_{i \varepsilon}')$
($i=1,2$).

 From $0\leq v(t)\leq t$ and $\underline{y}'=2C_1 vv'$,
 $\underline{y}''=2C_1 |v'|^2-2C_1v$, we have
\begin{align*}
 A\underline{y}-b_\varepsilon(t, \underline{y}, \underline{y}')
&=2C_1[v-\frac{\beta}{t+\varepsilon}vv'+|v'|^2
(2\gamma \frac{C_1 v^2}{C_1 v^2+\varepsilon^2}-1)]-f(t, \underline{y})\\
&\leq 2C_1[v+\beta|v'|+(2\gamma+1) |v'|^2]-f(t, \underline{y}).
\end{align*}
Since $f(t, \xi)\geq c_0>0$, we can choose
\begin{equation*}
 C_1\leq \min\big\{\frac{c_0}{2\max_{[0,1]}[v+\beta
 |v'|+(2\gamma+1) |v'|^2]}, 1/2\big\},
\end{equation*}
hence
\begin{equation*}
A \underline{y}\leq b_\varepsilon(t, \underline{y},
\underline{y}'),\quad 0 <t< 1.
\end{equation*}
Since $C_2(t+\varepsilon)^2\geq \varepsilon^2$, it is easy to
 calculate that
\begin{align*}
 A y_{1\varepsilon}-b_\varepsilon(t, y_{1\varepsilon}, y_{1\varepsilon}')
&=2C_2[\gamma \frac{2C_2(t+\varepsilon)^2}
{C_2(t+\varepsilon)^2+\varepsilon^2}-\beta-1]-f(t, y_{1\varepsilon}),\\
&\geq 2C_2(\gamma-\beta-1)-f(t, y_{1\varepsilon}).
\end{align*}
Choosing
$$
C_2\geq \max\{\frac{1}{2(\gamma-\beta-1)}\max_{[0,1]} f(t,
\underline{y}(t)), 1\},
$$
 we see that $y_{1\varepsilon}\geq \underline{y}$ in $[0,1]$.
It follows from (H1) that
\begin{equation*}
A y_{1\varepsilon}\geq b_\varepsilon(t, y_{1 \varepsilon},
y_{1\varepsilon}'),\quad 0 < t < 1,
\end{equation*}
as asserted. The other inequality can be proved similarly.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2}
For any $\tau\in (0,1)$, there exists a
positive constant $C_\tau$ independent of $\varepsilon$ such that
\begin{equation} \label{e5}
|y_\varepsilon'|\leq C_\tau, \quad |y_\varepsilon''|\leq C_\tau, \quad
\tau\leq t\leq 1-\tau.
\end{equation}
\end{lemma}

\begin{proof}
From Lemma \ref{lem1}, BVP \eqref{e4} \eqref{e2} admits a solution
$y_\varepsilon\in C^{2+\mu}[0,1]$ which satisfies \eqref{e4} \eqref{e2}
pointwise, hence it is also a solution of
\begin{equation*}
[(t+\varepsilon)^\beta
y_\varepsilon']'=\frac{\gamma(t+\varepsilon)^\beta
}{y_\varepsilon+\varepsilon^2}|y_\varepsilon'|^2-(t+\varepsilon)^\beta
f(t, y_\varepsilon).
\end{equation*}
Since $\gamma>0$, from (H1) and Lemma \ref{lem1} we obtain
$$
 [(t+\varepsilon)^\beta
y_\varepsilon']'\geq -(t+\varepsilon)^\beta
f(t, y_\varepsilon)
\geq -2^{\beta} \max_{[0,1]} f(t,
\underline{y}(t)):=-M.
$$
Therefore,
\begin{equation*}
[(t+\varepsilon)^\beta y_\varepsilon'+Mt]'\geq 0,\quad 0<t<1,
\end{equation*}
which implies that the function
$\varphi(t):=(t+\varepsilon)^\beta
y_\varepsilon'+Mt$ is  non-decreasing on $[0,1]$.

Since $y_\varepsilon\geq 0$ for all $t\in [0,1]$ and
$y_\varepsilon(0)=y_\varepsilon(1)=0$, we have
\begin{gather*}
y_\varepsilon'(0)=\lim_{t \to 0^+}
\frac{y_\varepsilon(t)}{t}\geq 0, \\
y_\varepsilon'(1)=\lim_{t \to 1^-}
\frac{y_\varepsilon(t)}{t-1}\leq 0.
\end{gather*}
From which, it follows that
\begin{equation*}
0\leq \varphi(0) \leq \varphi(t)\leq \varphi(1)\leq M,\quad t\in
[0,1],
\end{equation*}
which implies
\begin{equation} \label{e6}
|(t+\varepsilon)^\beta y_\varepsilon'(t)|\leq M.
\end{equation}
Hence for any $\tau\in (0,1)$ there exists a positive constant
$C_\tau$ independent of $\varepsilon$ such that
\begin{equation*}
|y_\varepsilon'|\leq C_\tau, \quad \tau\leq t\leq 1.
\end{equation*}
Multiplying \eqref{e4} by $(t+\varepsilon)^{2\beta+1}$, from
\eqref{e6} (H1)
and Lemma \ref{lem1} it follows
\begin{align*}
&|(t+\varepsilon)^{2\beta+1}y_\varepsilon''|\\
=& \big|\gamma \frac{(t+\varepsilon)}
{y_\varepsilon+\varepsilon^2}[(t+\varepsilon)^\beta
y_\varepsilon']^2-(t+\varepsilon)^{2\beta+1}f(t, y_\varepsilon)
 -(2\beta+1)(t+\varepsilon)^\beta ((t+\varepsilon)^\beta
y_\varepsilon')\big|\\
\leq& C\Big(1+\frac{t+\varepsilon}{\underline{y}+\varepsilon^2}
 +f(t, \underline{y})\Big),
\end{align*}
where $C$ is independent of $\varepsilon$.
The second conclusion follows easily from the above inequality.
\end{proof}

Now we complete the proof of Theorem \ref{thm1}.
Differentiating formally \eqref{e4}
with respect to $t$, from (H1) and Lemma \ref{lem1} we obtain
\begin{align*}
|y_\varepsilon'''|
&= \big|\frac{\beta}{t+\varepsilon}(\frac{y_\varepsilon'}{t+\varepsilon}
-y_\varepsilon'')+\gamma\frac{2(y_\varepsilon+\varepsilon^2)y_\varepsilon'
y_\varepsilon'' -y_\varepsilon'|y_\varepsilon'|^2}
{(y_\varepsilon+\varepsilon^2)^2}
-f_t'(t,y_\varepsilon)-f_{y}'(t,y_\varepsilon)y_\varepsilon'(t)\big|\\
&\leq
\frac{\beta}{t+\varepsilon}(\frac{|y_\varepsilon'|}{t+\varepsilon}
+|y_\varepsilon''|)+\gamma[\frac{2|y_\varepsilon'|
|y_\varepsilon''|}{\underline{y}+\varepsilon^2}+\frac{|y_\varepsilon'|^3
}{(\underline{y}+\varepsilon^2)^2}]\\
&\quad +\max_{t\in [0,1], y\in [a, b]} |f_t'(t,
y)|+|y_\varepsilon'|\cdot \max_{t\in [0,1], y\in [a, b]}
|f_y'(t, y)|,
\end{align*}
where $a=\min_{t\in [0,1]} \underline{y}(t)$,
$b=\max_{t\in [0,1]}\overline{y}_\varepsilon(t)|_{\varepsilon=1}$.
From \eqref{e5} one infers that for any $\tau\in (0,1)$ there exists
a positive
constant $C_\tau$ independent of $\varepsilon$ such that
\begin{equation*}
|y_\varepsilon'''|\leq C_\tau, \quad  \tau\leq t\leq 1-\tau.
\end{equation*}
This implies that
\begin{equation*}
\| y_\varepsilon \|_{C^{2,1}[\tau, 1-\tau]}\leq C_\tau.
\end{equation*}
Using Arzel\'a-Ascoli theorem and diagonal sequential
process, we obtain that there exists a subsequence
$\{y_{\varepsilon_n}\}$ of $\{y_\varepsilon\}$ and a function $y
\in C^2(0,1)$ such that
\begin{equation*}
y_{\varepsilon_n} \to y,\quad \text{uniformly  in } C^2[\tau,1-\tau],
\end{equation*}
as $\varepsilon_n \to 0$.
By Lemma \ref{lem1}, we obtain
\begin{gather*}
C_1 t^2(1-t)^2\leq y(t)\leq C_2 t^2,\quad t\in [0,1], \\
C_1 t^2(1-t)^2\leq y(t)\leq C_2 (1-t)^2,\quad t\in [0,1].
\end{gather*}
From this, it is not difficult to show that $y'(0)=y'(1)=0$ and
$y\in C[0,1]$. Clearly, $y$ solves BVP \eqref{e1}-\eqref{e3}, hence
Theorem \ref{thm1} is proved.

\subsection*{Example}
 Consider boundary-value problem
\begin{equation} \label{e7}
\begin{gathered}
y''+\frac{N-1}{t}y'-\frac{N+1}{y}|y'|^2+t^2+e^{-y}+1=0,\quad 0<t<1,\\
y(0)=y(1)=y'(0)=y'(1)=0.
\end{gathered}
\end{equation}
Let $N\geq 1$, $\beta=N-1$, $\gamma=N+1$, $f(t,y)=t^2+e^{-y}+1$,
$c_0=1$. Clearly, all assumptions of Theorem \ref{thm1} are satisfied.
Hence the problem \eqref{e7} has at least one positive solution
$y\in  C^2(0, 1)\cap C[0, 1]$. But the theorems in \cite{g1,t1} are
not applicable to this example.

\subsection*{Acknowledgement}
The authors are highly grateful for the referee's careful reading
and comments on this note.

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\end{document}