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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 260, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/260\hfil Existence of positive solutions]
{Existence of positive solutions for singular p-Laplacian Sturm-Liouville
boundary value problems}

\author[D. D. Hai \hfil EJDE-2016/260\hfilneg]
{D. D. Hai}

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

\thanks{Submitted May 20, 2016. Published September 26, 2016.}
\subjclass[2010]{34B16, 34B18}
\keywords{Singular Sturm-Liouville boundary value problem;
positive solution}

\begin{abstract}
 We prove the existence of positive solutions of the Sturm-Liouville
 boundary value problem
 \begin{gather*}
 -(r(t)\phi (u'))'=\lambda g(t)f(t,u),\quad t\in (0,1),\\
 au(0)-b\phi ^{-1}(r(0))u'(0)=0,\quad cu(1)+d\phi ^{-1}(r(1))u'(1)=0,
 \end{gather*}
 where $\phi (u')=|u'|^{p-2}u'$, $p>1$,
 $f:(0,1)\times(0,\infty )\to \mathbb{R}$ satisfies a $p$-sublinear
 condition and is allowed to be singular at $u=0$ with semipositone structure.
 Our results extend previously known results in the literature.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks



\section{Introduction}

We consider the boundary-value problem
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'=\lambda g(t)f(t,u),\quad t\in (0,1),\\
au(0)-b\phi ^{-1}(r(0))u'(0)=0,\quad
cu(1)+d\phi ^{-1}(r(1))u'(1)=0,
\end{gathered}   \label{e1.1}
\end{equation}
where $\phi (u')=|u'|^{p-2}u'$, $p>1$, $a,b,c,d$ are
nonnegative constants with $ac+ad+bc>0$, $f:(0,1)\times (0,\infty )\to
\mathbb{R}$ is allowed to be singular at $u=0$, and $\lambda $ is a
positive parameter.

When $p=2$ and $f:[0,1]\times [ 0,\infty )\to \mathbb{R}$ is
continuous, Yang and Zhou \cite{y1} prove the existence of a positive solution to
\eqref{e1.1} under the assumption
\[
\lim_{u\to \infty }\sup_{t\in [ 0,1]}\frac{f(t,u)}{u}
<\frac{\lambda _1}{\lambda }
<\lim_{u\to 0^{+}}\inf_{t\in [ 0,1]}\frac{f(t,u)}{u},
\]
where $\lambda _1>0$ denotes the first eigenvalue of $-(r(t)u')'=\lambda g(t)u$
in $(0,1)$ with Sturm-Liouville boundary conditions.
Their result allows $\lim_{u\to \infty}\sup_{t\in [ 0,1]}\frac{f(t,u)}{u}=-\infty $,
which complements previous existence results in
\cite{a1,c1,e1,h1,l1,l2,w1,y2}.

In this article, we shall extend the result in \cite{y1}
to the general case $p>1$ and also allow $f$ to be singular at $u=0$.
 We also establish the existence of a positive solution to \eqref{e1.1}
for $\lambda $ large allowing $
\lim_{u\to 0^{+}}\inf_{t\in (0,1)} f(t,u)/u^{p-1}=-\infty $
and $\lim_{u\to \infty }\inf_{t\in (0,1)}f(t,u)=0$, which does not
seem to have been considered in the literature even when $p=2$.
Note that the approach in \cite{y1} depends on the Green function
and can not apply to the
nonlinear case $p>1$ or the case when $f$ is singular at $u=0$.
 Our approach depends on a new sub- and super solutions type argument and
comparison principle.

Let $g$ satisfy condition (A2) below. Then the eigenvalue problem
$-(r(t)\phi (u'))'=\lambda g(t)\phi (u)$ in $(0,1)$
with the Sturm-Liouville boundary conditions in \eqref{e1.1}
has a positive first eigenvalue $\lambda _1$ with corresponding positive
eigenfunctions (see e.g. \cite{b1,m1}).

We shall make the following assumptions:
\begin{itemize}
\item[(A1)] $r:[0,1]\to (0,\infty )$ and
$f:(0,1)\times (0,\infty)\to \mathbb{R}\ $are continuous.

\item[(A2)] $g\in L^{1}(0,1)$ with $g\geq 0,g\not\equiv 0$ and there exists a
constant $\gamma \geq 0$ such that
\[
\int_0^{1}\frac{g(t)}{q^{\gamma }(t)}dt<\infty ,
\]
where $q(t)=\min (b+at,d+c(1-t))$.

\item[(A3)] For each $r>0$, there exists a constant $K_{r}>0$ such that
\[
|f(t,u)|\leq \frac{K_{r}}{u^{\gamma }}
\]
for $t\in (0,1),u\in (0,r]$, where $\gamma $ is defined in (A2).

\item[(A4)]  $\lim_{u\to \infty }\sup \frac{f(t,u)}{\phi (u)}<\frac{
\lambda _1}{\lambda }<$ \ $\lim_{u\to 0^{+}}\inf \frac{
f(t,u)}{\phi (u)}$, where the limits are uniform in $t\in (0,1)$.

\item[(A5)] $\lim_{u\to \infty }\sup \frac{f(t,u)}{\phi (u)}<\frac{
\lambda _1}{\lambda }$ uniformly in $t\in (0,1)$.

\item[(A6)] There exist positive constants $A,L$ such that
\[
f(t,u)\geq \frac{L}{u^{\gamma }}
\]
for $t\in (0,1)$ and $u\geq A$.
\end{itemize}

By a solution of \eqref{e1.1}, we mean a function $u\in C^{1}[0,1]$ with
$r(t)\phi (u')$ absolutely continuous on $[0,1]$ and satisfying \eqref{e1.1}.

Our main results read as follows:


\begin{theorem} \label{thm1.1}
Let {\rm (A1)--(A4)} hold. Then \eqref{e1.1} has a positive
solution $u$ with $\inf_{(0,1)}(u/q)>0$.
\end{theorem}

\begin{theorem} \label{thm1.2}
 Let {\rm (A1)--(A3), (A5), (A6)} hold. Then there
exists a constant $\lambda _0>0$  such that for
$\lambda >\lambda _0$, Equation \eqref{e1.1} has a positive solution
$u_{\lambda }$ with $\inf_{(0,1)}(u_{\lambda }/q)\to \infty $
 as $\lambda \to \infty $.
\end{theorem}

Let $\bar{\lambda}<\lambda _1$ and consider the problem
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'-\bar{\lambda}g(t)\phi (u)=\lambda
g(t)f(t,u),\quad  t\in (0,1), \\
au(0)-b\phi ^{-1}(r(0))u'(0)=0,cu(1)+d\phi ^{-1}(r(1))u'(1)=0.
\end{gathered} \label{e1.2}
\end{equation}
Then, as an immediate consequence of Theorem \ref{thm1.1}, we obtain the following
corollary.


\begin{corollary} \label{coro1.1}
Let {\rm (A1)--(A3)} hold and suppose that
\[
\lim_{u\to \infty }\sup \frac{f(t,u)}{\phi (u)}
<\frac{ \lambda _1-\bar{\lambda}}{\lambda }
< \lim_{u\to 0^{+}}\inf \frac{f(t,u)}{\phi (u)}.
\]
Then \eqref{e1.2} has a positive solution.
\end{corollary}

\begin{remark} \label{rmk1.1} \rm
When $p=2$ and $f:[0,1]\times [ 0,\infty )\to \mathbb{R}$
 is continuous, \cite[Theorem 3.1]{y1} follows from
Theorem \ref{thm1.1} with $\gamma =0$.
\end{remark}

\begin{example} \label{examp1.1} \rm
Let $g(t)\equiv 1\equiv r(t)$  and consider the BVP
\begin{equation}
\begin{gathered}
-(\phi (u'))'=\lambda f(t,u),\quad t\in (0,1), \\
u(0)=u(1)=0.
\end{gathered}  \label{e1.3}
\end{equation}
 Note that $\lambda _1=\pi _{p}^{p}$, where
\[
\pi _{p}=2(p-1)^{1/p}\int_0^{1}\frac{ds}{(1-s^{p})^{1/p}}
\]
is the first eigenvalue of $-(\phi (u'))'$ with zero
boundary conditions (see \cite{d1,d2}).

(i)  Let $f(t,u)=u^{p-1}\big( \frac{e^{t}}{u^{\gamma }}-u^{\beta
}\big) $, where $\gamma \in [ 0,1)$, and
$\beta >0$. Suppose $\lambda >\lambda _1$ if $\gamma =0$, and
$\lambda $ is any positive constant if $\gamma >0$.
Then (A1)--(A4) hold and therefore Theorem \ref{thm1.1} gives the existence
of a positive solution to \eqref{e1.3}.

(ii) Let $f(t,u)=-\frac{1}{u^{\gamma }}+\frac{1}{u^{\beta }}$, 
 where $0<\beta <\gamma <1$.  Then it is easy to see that
the assumptions of Theorem \ref{thm1.2} are satisfied and therefore \eqref{e1.3} has a
positive solution for $\lambda $ large. Note that since 
$\lim_{u\to 0^{+}}\inf_{t\in (0,1)}\frac{f(t,u)}{u^{p-1}}=-\infty $
and $\lim_{u\to \infty }\inf_{t\in (0,1)}f(t,u)=0$, 
the results in \cite{a1,c1,e1,h1,l1,l2,w1,y1,y2} do not apply here.

(iii) Let $f(t,u)=(1-u^{p-1})\cos t$. Then 
\[
\lim_{u\to \infty }\sup \frac{f(t,u)}{\phi (u)}<0\quad\text{and}\quad
\lim_{u\to 0^{+}}\inf \frac{f(t,u)}{\phi (u)}=\infty 
\]
 uniformly in $t\in (0,1)$  and so \eqref{e1.2} has a positive
solution for all $\lambda >0$, by Corollary~\ref{coro1.1}.
\end{example}

\section{Preliminaries}

We shall denote the norms in $C^{1}[0,1]$ and $L^{q}(0,1)$ by $|\cdot|_1$ and $
\|\cdot\|_{q}$ respectively. Here $|u|_1=\max (\|u\|_{\infty },\|u'\|_{\infty })$.
We first recall the following results in \cite{h1}.

\begin{lemma} \label{lemA}
 Let $h\in L^{1}(0,1)$. Then the problem
\begin{gather*}
-(r(t)\phi (u'))'=h,\quad t\in (0,1) \\
au(0)-b\phi ^{-1}(r(0))u'(0)=0,\quad
cu(1)+d\phi ^{-1}(r(1))u'(1)=0
\end{gather*}
has a unique solution $u=Sh\in C^{1}[0,1]$. 
Furthermore, $S$  is completely continuous and there exists a constant 
$m>0$  such that
\[
|u|_1\leq m\phi ^{-1}(\|h\|_1).
\]
\end{lemma}

\begin{lemma} \label{lemB}
Suppose $u\in C^{1}[0,1]$ satisfies 
\begin{gather*}
-(r(t)\phi (u'))'\geq 0,\quad t\in (0,1) \\
au(0)-b\phi ^{-1}(r(0))u'(0)\geq 0,\quad
 cu(1)+d\phi^{-1}(r(1))u'(1)\geq 0.
\end{gather*}
Then there exists a constant $m_0>0$ independent of $u$ such that
\[
u(t)\geq m_0\|u\|_{\infty }q(t)
\]
for $t\in [ 0,1]$,  where $q$  is defined by {\rm (A2)}.
\end{lemma}

\begin{remark} \label{rmk2.1} \rm
Lemma \ref{lemB} is a special case of \cite[Lemma 3.4]{h1}
when $h=0$. Note that the proof of  \cite[Lemma 3.4]{h1} is incorrect for
$1<p<2$ when $h\not\equiv 0$  since it uses the inequality
\[
|\phi ^{-1}(x)-\phi ^{-1}(y)|\leq 2\phi ^{-1}(|x-y|)\quad
\text{for all } x,y\in \mathbb{R},
\]
which is not true when  $1<p<2$.  However, when 
$h=0$,  this inequality is not needed in \cite[Proof of Lemma 3.4]{h1},
which guarantees the validity of Lemma \ref{lemB}.
\end{remark}

\begin{lemma} \label{lem2.1} 
There exists a constant $k>0$  such that $|u|\leq k|u|_1q$ in $[0,1]$
 for all $u\in C^{1}[0,1]$  satisfying the Sturm-Liouville boundary
conditions in \eqref{e1.1}.
\end{lemma}

\begin{proof}
Let $u\in C^{1}[0,1]$. Then, if $b>0$,
\[
u(t)=u(0)+\int_0^{t}u'\leq 2|u|_1\leq \frac{2}{b}|u|_1(b+at)
\]
for $t\in [ 0,1]$, while if $b=0$ then $a>0$, this implies $u(0)=0$
and $u(t)\leq |u|_1t$ for $t\in [ 0,1]$.
Hence
\begin{equation}
u(t)\leq k_0|u|_1(b+at),  \label{e2.1}
\end{equation}
for $t\in [ 0,1]$, where $k_0=2/b$ if $b>0$, and $1/a$ if $b=0$.
Similarly, using
\[
u(t)=u(1)-\int_{t}^{1}u',
\]
we obtain
\begin{equation}
u(t)\leq k_1|u|_1(d+c(1-t))  \label{e2.2}
\end{equation}
for $t\in [ 0,1]$, where $k_1=2/d$ if $d>0$, and $1/c$ if $d=0$.

Combining \eqref{e2.1} and \eqref{e2.2}, we see that $u\leq k|u|_1q$ in $(0,1)$,
 where $k=\max (k_0,k_1)$. By replacing $u$ by $-u$, we see that 
Lemma \ref{lem2.1} holds.
\end{proof}

\begin{lemma} \label{lem2.2}
Let $h_0,h_1\in L^{1}(0,1)$.
Suppose $u_0,u_1\in C^{1}[0,1]$  satisfy
\begin{gather*}
-(r(t)\phi (u_i'))'=h_i,\quad t\in (0,1), \\
au_i(0)-b\phi ^{-1}(r(0))u_i'(0)=0, \quad
cu_i(1)+d\phi^{-1}(r(1))u_i'(1)=0,
\end{gather*}
for $i=0,1$.  Then there exists a constant $M_0>0$
depending on $p,a,b,c,d$, and $C$ such that 
\begin{equation}
|u_1-u_0|_1\leq M_0\max \{\|h_1-h_0\|_1,\|h_1-h_0\|_1^{
\frac{1}{p-1}}\},  \label{e2.3}
\end{equation}
where $C>0$ is such that $\|h_i\|_1<C$ for $i=0,1$.
\end{lemma}

\begin{proof} 
By integrating, we obtain
\begin{equation}
u_i(t)=C_i+\int_0^{t}\phi ^{-1}\Big( \frac{D_i-\int_0^{s}h_i}{
r(s)}\Big) ds  \label{e2.4}
\end{equation}
for $i=0,1$, where $C_i,D_i$ are constants satisfying
\begin{gather*}
aC_i-b\phi ^{-1}(D_i)=0, \\
c\Big( C_i+\int_0^{1}\phi ^{-1}\big( \frac{D_i\ -\int_0^{s}h_i}{
r(s)}\big) ds\Big) +d\phi ^{-1}\big( D_i-\int_0^{1}h_i\big) =0.
\end{gather*}
Suppose first that $a=0$. Then $b,c>0$, $D_i=0$, and
\[
C_i=\frac{d}{c}\phi ^{-1}\Big( \int_0^{1}h_i\Big) +\int_0^{1}\phi
^{-1}\Big( \frac{\int_0^{s}h_i}{r(s)}\Big) ds,
\]
and so
\[
u_i(t)=\frac{d}{c}\phi ^{-1}\Big( \int_0^{1}h_i\Big)
+\int_{t}^{1}\phi ^{-1}\Big( \frac{\int_0^{s}h_i}{r(s)}\Big) ds.
\]
For $p\geq 2$, using the inequality
\[
|\phi ^{-1}(x)-\phi ^{-1}(y)|\leq 2\phi ^{-1}(|x-y|)\quad \text{for }x,y\in
\mathbb{R},
\]
we obtain
\begin{equation}
\max \{|u_1(t)-u_0(t)|\ ,|u_1'(t)-u_0'(t)|\}\leq
M_1\|h_1-h_0\|_1^{\frac{1}{p-1}},  \label{e2.5}
\end{equation}
for $t\in [ 0,1]$, where 
$r_0=\min_{t\in [0,1]}r(t)>0$, $M_1=2\left( d/c+\phi ^{-1}(1/r_0\right) )$.

For $1<p<2$, using the Mean Value Theorem, we obtain
\[
|\phi ^{-1}(x)-\phi ^{-1}(y)|\leq (p-1)^{-1}|x-y|(\max \{|x|,|y|\})^{\frac{
2-p}{p-1}}
\]
for $x,y\in \mathbb{R}$, which implies
\begin{equation}
\max \{|u_1(t)-u_0(t)|, |u_1'(t)-u_0'(t)|\}
\leq M_{2}\|h_1-h_0\|_1,  \label{e2.6}
\end{equation}
for $t\in [ 0,1]$, where $M_{2}=(p-1)^{-1}\big(
dc^{-1}+r_0^{-1/(p-1)}\big) C^{\frac{2-p}{p-1}}$.

Suppose next that $a>0$. Then $C_i=(b/a)\phi ^{-1}(D_i)$, and $D_i$
satisfies
\begin{equation}
c\Big( \frac{b}{a}\phi ^{-1}(D_i)+\int_0^{1}\phi ^{-1}\Big( \frac{
D_i\ -\int_0^{s}h_i}{r(s)}\Big) ds\Big) +d\phi ^{-1}
\Big(D_i-\int_0^{1}h_i\Big) =0  \label{e2.7}
\end{equation}
for $i=0,1$. Since $\phi ^{-1}$ is increasing and $\phi ^{-1}(0)=0$, it
follows from \eqref{e2.7} that $|D_i|\leq \|h_i\|_1$, and
\[
|D_1-D_0|\leq \|h_1-h_0\|_1,
\]
which, together with \eqref{e2.4}, imply
\begin{equation}
\max \{|u_1(t)-u_0(t)|, |u_1'(t)-u_0'(t)|\}
\leq M_{3}\max \{\|h_1-h_0\|_1,\|h_1-h_0\|_1^{\frac{1}{p-1}}\}
\label{e2.8}
\end{equation}
for $t\in [ 0,1]$, where $M_{3}=2(b/a+(2/r_0)^{\frac{1}{p-1}})$ if 
$p\geq 2$, and $M_{3}=(p-1)^{-1}(b/a+(2/r_0)^{1/(p-1)})C^{\frac{2-p}{p-1}}$
if $1<p<2$.
Combining \eqref{e2.5},\eqref{e2.6}, and \eqref{e2.8}, we obtain \eqref{e2.3}
 with $M_0=\max_{1\leq i\leq 3}M_i$, which completes the proof.
\end{proof}

\section{Proofs of main results}

 Let $z_1\in C^{1}[0,1]$ be the normalized positive eigenfunction
of $-(r(t)\phi (u'))'=\lambda g(t)\phi (u)$ in $(0,1)$
with Sturm-Liouville boundary conditions corresponding to $\lambda _1$
i.e. $z_1>0$ on $(0,1)$ and $\|z_1\|_{\infty }=1$. By Lemma \ref{lemB}, there
exists a constant $m_0>0$ such that $z_1\geq m_0q$ in $(0,1)$.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Since  $\lim_{z\to 0^{+}}\inf \frac{f(t,z)}{\phi (z)}>\frac{\lambda _1}{\lambda }$ 
uniformly in $t\in (0,1)$, there exists a constant $c>0$ such that
\begin{equation}
\frac{f(t,z)}{\phi (z)}>\frac{\lambda _1}{\lambda }   \label{e3.1}
\end{equation}
for $z\in (0,c]$ and $t\in (0,1)$. 
Let $Z=cz_1$ and $Z_1=Mz_1$, where $M>c$
is a large constant to be determined later. 
In view of \eqref{e3.1}, $Z$ satisfies
\begin{equation}
-(r(t)\phi (Z'))'=\lambda _1g(t)\phi (Z)\leq \lambda g(t)f(t,Z)  \label{e3.2}
\end{equation}
for $t\in (0,1)$. For $v\in C[0,1]$, let 
$\tilde{v}=\min \{\max \{v,Z\},Z_1\}$. Then 
$Z\leq \tilde{v}\leq Z_1\leq M$ in $(0,1)$ and (A3) gives
\begin{equation}
|g(t)f(t,\tilde{v})|\leq \frac{K_{M}g(t)}{\tilde{v}^{\gamma }}
\leq \frac{K_{M}g(t)}{(cz_1)^{\gamma }}
\leq \frac{K_{M}g(t)}{(cm_0)^{\gamma}q^{\gamma }(t)}  \label{e3.3}
\end{equation}
for $t\in (0,1)$. Hence $g(t)f(t,\tilde{v})\in L^{1}(0,1)$ by (A2). 
Define $Tv=u$, where $u$ is the solution of
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'=\lambda g(t)f(t,\tilde{v}),\quad t\in (0,1), \\
au(0)-b\phi ^{-1}(r(0))u'(0)=0, \quad
cu(1)+d\phi ^{-1}(r(1))u'(1)=0,
\end{gathered}  \label{e3.4}
\end{equation}
whose existence follows from \eqref{e3.3} and Lemma \ref{lemA}. 
Define $S_1v=\lambda g(t)f(t,\tilde{v})$. Using \eqref{e3.3} and 
the Lebesgue Dominated Convergence Theorem, 
we see that $S_1:C[0,1]\to L^{1}(0,1)$ is continuous
and bounded. Since $T=S\circ S_1$, where $S$ is defined in Lemma \ref{lemA},
it follows that $T:C[0,1]\to C[0,1]$ is completely continuous and
bounded. Hence, by the Schauder Fixed Point Theorem, $T$ has a fixed point 
$u$. To complete the proof, we will first show that $u\geq Z$ in $(0,1)$.
Indeed, suppose $u(t^{\ast })<Z(t^{\ast })$ for some $t^{\ast }\in (0,1)$.
Let $(t_0,t_1)\subset (0,1)$ be the largest interval containing
 $t^{\ast}$ such that $u<Z$ in $(t_0,t_1)$. Then $\tilde{u}=Z$ in 
$(t_0,t_1)$ and
\begin{equation}
au(t_0)-b\phi ^{-1}(r(t_0))u'(t_0)\geq aZ(t_0)-b\phi ^{-1}(r(t_0))Z'(t_0).  \label{e3.5}
\end{equation}
Indeed, if $t_0>0$ then $u(t_0)=Z(t_0)$ and
 $u'(t_0)\leq Z'(t_0)$, while if $t_0=0$ then we have equality in \eqref{e3.5}.
Similarly,
\begin{equation}
cu(t_1)+d\phi ^{-1}(r(t_1))u'(t_1)
\geq cZ(t_1)+d\phi ^{-1}(r(t_1))Z'(t_1).  \label{e3.6}
\end{equation}
Since
\[
-(r(t)\phi (u'))'=\lambda g(t)f(t,Z),\quad t\in (t_0,t_1),
\]
it follows from \eqref{e3.2}, \eqref{e3.5}, \eqref{e3.6}, and the comparison 
principle (see e.g. \cite[Lemma 3.2]{h1}) 
that $u\geq Z$ in $(t_0,t_1)$, a contradiction. Thus 
$u\geq Z$ in $(0,1)$ and so $\tilde{u}=\min \{u,Z_1\}$ in $(0,1)$.

Next, we show that $u\leq Z_1$ in $(0,1)$. Using (A3) and 
$\lim_{z\to \infty }\sup \frac{f(t,z)}{\phi (z)}<\frac{\lambda
_1}{\lambda }$ 
uniformly in $t\in (0,1)$, we deduce the existence of
constants $A,K_{\lambda }>0$ and $\bar{\lambda}\in (0,\lambda _1)$ such
that
\[
\lambda f(t,z)\leq \bar{\lambda}\phi (z)+\frac{K_{\lambda }}{z^{\gamma }}
\]
for $z>0$ and $t\in (0,1)$. Hence
\begin{align*}
-(r(t)\phi (u'))'
&=\lambda g(t)f(t,\tilde{u})
 \leq g(t)\Big( \bar{\lambda}\phi (\tilde{u})+\frac{K_{\lambda }}{\tilde{u}
^{\gamma }}\Big) \\
&\leq g(t)\Big( \bar{\lambda}(Mz_1)^{p-1}+\frac{K_{\lambda }}{
(cz_1)^{\gamma }}\Big) \\
&\leq \bar{\lambda}g(t)(Mz_1)^{p-1}+\frac{K_{\lambda }g(t)}{
(cm_0)^{\gamma }q^{\gamma }(t)}
\end{align*}
for $t\in (0,1)$. Let $u_{M}=u/M$. Then $u_{M}$ satisfies
\[
-(r(t)\phi (u_{M}'))'\leq \bar{\lambda}g(t)z_1{}^{p-1}+
\frac{K_{\lambda }g(t)}{(cm_0)^{\gamma }M^{p-1}q^{\gamma }(t)}
\]
for $t\in (0,1)$. Let $\bar{u}_{M}$ and $\bar{u}$ satisfy
\[
-(r(t)\phi (\bar{u}_{M}'))'=\bar{\lambda}g(t)z_1{}^{p-1}+
\frac{K_{\lambda }g(t)}{(cm_0)^{\gamma }M^{p-1}q^{\gamma }(t)}\equiv
h_{M},\quad t\in (0,1),
\]
and
\[
-(r(t)\phi (\bar{u}'))'=\bar{\lambda}g(t)z_1^{p-1}\equiv
h,\quad t\in (0,1),
\]
with Sturm-Liouville boundary conditions in \eqref{e1.1}. 
Note that $\bar{u} =(\bar{\lambda}/\lambda _1)^{\frac{1}{p-1}}z_1$. 
By the comparison principle, $u_{M}\leq \bar{u}_{M}$ in $(0,1)$.
Let $\varepsilon >0$ be such that 
$(\bar{\lambda}/\lambda _1)^{1/(p-1)}+\varepsilon <1$.
Since
\[
\|h_{M}-h\|_1=\frac{K_{\lambda }}{(cm_0)^{\gamma }M^{p-1}}\Big(
\int_0^{1}\frac{g(t)}{q^{\gamma }(t)}dt\Big) \to 0\quad\text{as }
M\to \infty ,
\]
it follows from Lemmas \ref{lem2.1} and \ref{lem2.2} that
\begin{align*}
\bar{u}_{M}-\bar{u}
&\leq k|\bar{u}_{M}-\bar{u}|_1q\leq km_0^{-1}|\bar{u}_{M}-\bar{u}|_1z_1 \\
&\leq km_0^{-1}M_0\max \{\|h_{M}-h\|_1,\|h_{M}-h\|_1^{\frac{1}{p-1}}\}z_1
 <\varepsilon z_1,
\end{align*}
provided that $M$ is large enough. Consequently,
\[
u_{M}\leq \bar{u}_{M}\leq \bar{u}+\varepsilon z_1
=\Big( (\bar{\lambda} /\lambda _1)^{1/(p-1)}+\varepsilon \Big) z_1\leq z_1\quad
\text{in } (0,1),
\]
i.e. $u\leq Mz_1=Z_1$ in $(0,1)$. Hence $Z\leq u\leq Z_1$ in $(0,1)$
i.e. $u$ is a positive solution of \eqref{e1.1}, which completes the proof.
\end{proof} 


\begin{proof}[Proof of Theorem \ref{thm1.2}]
By Theorem \ref{thm1.1}, there exists a positive solution $w$ of the problem
\begin{gather*}
-(r(t)\phi (w'))'=\frac{g(t)}{w^{\gamma }},\quad t\in (0,1), \\
aw(0)-b\phi ^{-1}(r(0))w'(0))=0, \quad
cw(1)+d\phi ^{-1}(r(1))w'(1)=0
\end{gather*}
with $w\geq \alpha q$ in $(0,1)$ for some $\alpha >0$. Let $w_0$ satisfy
\[
-(r(t)\phi (w_0'))'
=\begin{cases}
\frac{L_1g(t)}{w^{\gamma }} &\text{if } w>\frac{2AL_1^{-1/(p-1)}}{
\lambda ^{\delta }}, \\
-\frac{K_1g(t)}{w^{\gamma }} &\text{if } w\leq \frac{2AL_1^{-1/(p-1)}}{
\lambda ^{\delta }}
\end{cases}
\; \equiv h_{\lambda }\quad \text{in }(0,1),
\]
with Sturm-Liouville boundary conditions, where 
$\delta =(\gamma +p-1)^{-1},L_1=L^{\frac{p-1}{p-1+\gamma }}$ and 
$K_1=2^{\gamma}L_1^{-\gamma /(p-1)}K_{2A}$, and $K_{2A}$ is defined in (A3).
Let $w_1$ satisfy
\[
-(r(t)\phi (w_1'))'=\frac{L_1g(t)}{w^{\gamma }}
\equiv h\quad  \text{in }(0,1)
\]
with Sturm-Liouville boundary conditions.
Then $w_1=L_1^{1/(p-1)}w$ and $w_0\leq w_1$ in $(0,1)$ by the comparison principle.
Since
\[
\|h_{\lambda }-h\|_1=(L_1+K_1)\int_{w\leq \frac{2AL_1^{-1/(p-1)}}{
\lambda ^{\delta }}}\frac{g(t)}{w^{\gamma }(t)}dt\to 0\quad \text{as }
\lambda \to \infty ,
\]
it follows from Lemma \ref{lem2.2} that
\[
|w_0-w_1|_1\leq M_0\max \{\|h_{\lambda }-h\|_1,\|h_{\lambda
}-h\|_1^{\frac{1}{p-1}}\}\to 0\quad \text{as }\lambda \to \infty .
\]
Hence by Lemma \ref{lem2.1}, there exists a constant $\lambda _0>0$ such that
\begin{equation}
w_0\geq w_1-k|w_0-w_1|_1q
\geq \frac{L_1^{1/(p-1)}w}{2}\quad \text{in }(0,1)  \label{e3.7}
\end{equation}
for $\lambda >\lambda _0$. Let $Z=\lambda ^{\delta }w_0$
and $Z_1=Mz_1$ where $M>\lambda ^{\delta }km_0^{-1}|w_1|_1$ 
(so that $Z_1>Z$ in $(0,1))$. We shall verify that $Z$ satisfies
\begin{equation}
-(r(t)\phi (Z'))'\leq \lambda g(t)f(t,Z)\quad \text{in }(0,1).
\label{e3.8}
\end{equation}
Indeed,
\[
-(r(t)\phi (Z'))'=\begin{cases}
\frac{\lambda ^{\delta (p-1)}L_1g(t)}{w^{\gamma }}
&\text{if }w>\frac{2AL_1^{-1/(p-1)}}{\lambda ^{\delta }}, \\[4pt]
-\frac{\lambda ^{\delta (p-1)}K_1g(t)}{w^{\gamma }}
&\text{if }w\leq \frac{2AL_1^{-1/(p-1)}}{\lambda ^{\delta }}.
\end{cases}
\]
If $w>2AL_1^{-1/(p-1)}/\lambda ^{\delta }$ then by \eqref{e3.7},
\[
Z\geq \frac{\lambda ^{\delta }L_1^{1/(p-1)}w}{2}\geq A,
\]
from which (A6) gives
\begin{equation}
\begin{aligned}
\lambda g(t)f(t,Z)
&\geq \frac{\lambda Lg(t)}{Z^{\gamma }}
 =\frac{\lambda^{1-\gamma \delta }Lg(t)}{w_0^{\gamma }} \\
&\geq \frac{\lambda ^{1-\gamma \delta }Lg(t)}{w_1^{\gamma }}
 =\frac{\lambda^{\delta (p-1)}Lg(t)}{L_1^{\gamma /(p-1)}w^{\gamma }}\\
& =\frac{\lambda^{\delta (p-1)}L_1g(t)}{w^{\gamma }}.
\end{aligned}  \label{e3.9}
\end{equation}
On the other hand, if $w\leq \frac{2AL_1^{-1/(p-1)}}{\lambda ^{\delta }}$,
then
\[
Z\leq \lambda ^{\delta }w_1=L_1^{1/(p-1)}\lambda ^{\delta }w\leq 2A,
\]
from which (A3) and \eqref{e3.7} give
\begin{equation}
\begin{aligned}
\lambda g(t)f(t,Z)
&\geq -\frac{\lambda K_{2A}g(t)}{Z^{\gamma }}
 =-\ \frac{\lambda ^{1-\gamma \delta }K_{2A}g(t)}{w_0^{\gamma }}\\
&\geq - \frac{\lambda ^{\delta (p-1)}K_{2A}g(t)}{\big(
L_1^{1/(p-1)}/2\big) ^{\gamma }w^{\gamma }}
 =-\ \frac{\lambda ^{\delta (p-1)}K_1g(t)}{w^{\gamma }}.
\end{aligned}  \label{e3.10}
\end{equation}
Combining \eqref{e3.9} and \eqref{e3.10}, we see that \eqref{e3.8} holds. 
Let $T$ be the operator defined in the proof of Theorem \ref{thm1.1} i.e. 
for $v\in C[0,1]$, $u=Tv$ satisfies \eqref{e3.4}; i.e.,
\begin{gather*}
-(r(t)\phi (u'))'=\lambda g(t)f(t,\tilde{v}),\quad t\in (0,1), \\
au(0)-b\phi ^{-1}(r(0))u'(0))=0, \quad
cu(1)+d\phi ^{-1}(r(1))u'(1)=0,
\end{gather*}
where $\tilde{v}=\min \{\max \{v,Z\},Z_1\}$.
Then $T$ has a fixed point $u_{\lambda }$ in $C[0,1]$.
Using the same arguments as in the proof of Theorem \ref{thm1.1}, we see that
 $u_{\lambda }\geq Z$ and, for $M$ large enough $u_{\lambda }\leq Z_1$ in 
$(0,1)$; i.e., $u_{\lambda }$ is a positive
solution of \eqref{e1.1} for $\lambda >\lambda _0$ with 
$u_{\lambda }\geq \lambda ^{\delta }\big( L_1^{1/(p-1)}/2\big) w$ in 
$(0,1)$, which completes the proof.
\end{proof}


\subsection*{Acknowledgements}
The author wants to thank the anonymous referee for
pointing out some errors in the original manuscript and providing helpful
suggestions.

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\end{document}