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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 92, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/92\hfil 
Sturm-Liouville superlinear $p$-Laplacian problem]
{Positive solutions for the one-dimensional Sturm-Liouville
superlinear $p$-Laplacian problem}

\author[K. D. Chu, D. D. Hai \hfil EJDE-2018/92\hfilneg]
{Khanh Duc Chu, Dang Dinh Hai}

\address{Khanh Duc Chu \newline
Faculty of Mathematics and Statistics\\
Ton Duc Thang University\\
Ho chi Minh City, Vietnam}
\email{chuduckhanh@tdt.edu.vn}

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

\dedicatory{Communicated by Pavel Drabek}

\thanks{Submitted February 12, 2018. Published April 17, 2018.}
\subjclass[2010]{34B15, 34B18}
\keywords{p-Laplacian; superlinear; positive solutions}

\begin{abstract}
 We prove the existence of positive classical solutions for the $p$-Laplacian
 problem
 \begin{gather*}
 -(r(t)\phi (u'))'=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{gather*}
 where $\phi (s)=|s|^{p-2}s$, $p>1$, $f:(0,1)\times [ 0,\infty )\to\mathbb{R}$
 is a Carath\'{e}odory function satisfying
 \[
 \limsup_{z\to 0^{+}}  \frac{f(t,z)}{z^{p-1}}<\lambda_1
 <\liminf_{z\to \infty }\frac{f(t,z)}{z^{p-1}}
 \]
 uniformly for a.e. $t\in (0,1)$, where $\lambda _1$ denotes the principal
 eigenvalue of $-(r(t)\phi (u'))'$ with Sturm-Liouville
 boundary conditions. Our result extends a previous work by Man\'{a}sevich,
 Njoku, and Zanolin to the Sturm-Liouville boundary conditions with more
 general operator.
\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}

Consider the one-dimensional $p$-Laplacian problem
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'=f(t,u)\quad \text{a.e. on }(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 $\phi (s)=|s|^{p-2}s$, $p>1$, $a,b,c,d$ are nonnegative constants with 
$ac+ad+bc>0$, $r:[0,1]\to (0,\infty )$ and $f:(0,1)\times [0,\infty )\to \mathbb{R}$.

We are interested in positive classical solution of \eqref{e1.1}, that is,
solutions $u\in C^1[0,1]$ with $u>0$ on $(0,1),\ \phi (u')$
absolutely continuous on $[0,1]$ and satisfying \eqref{e1.1}.

Let us look at the literature on problem \eqref{e1.1} with Dirichlet boundary
conditions i.e.\ $b=d=0$. In the sublinear case, Lan, Yang, and Yang \cite{L}
proved the existence of a classical positive solution to \eqref{e1.1} when 
$r(t)\equiv 1$ and $f$ is nonnegative with
\begin{equation}
\limsup_{z\to \infty }\frac{f(t,z)}{z^{p-1}}<\lambda
_1<\liminf_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}\leq \infty
\label{e1.2}
\end{equation}
uniformly for a.e. $t\in (0,1)$, where $\lambda _1=2^p(p-1)(
\int_0^1\frac{ds}{(1-s^p)^{1/p}}) ^p$ is the principal
eigenvalue of $-(\phi (u'))'$ with zero boundary
conditions (see \cite{DE,DR}). In particular, when $p=2$ and 
$f:[0,\infty)\to [ 0,\infty )$ is continuous, \eqref{e1.2} becomes
\[
\limsup_{z\to \infty }\frac{f(z)}{z}<\pi
^{2}<\liminf_{z\to 0^{+}}\frac{f(z)}{z}\leq \infty ,
\]
which was used by Webb and Lan \cite{W} to obtain nonnegative solutions to \eqref{e1.1}
with $\phi (s)=s$. In fact, \cite{W} gave a general method with covered many
boundary conditions including nonlocal ones and included both sublinear and
superlinear types of conditions. In the superlinear case, Man\'{a}sevich,
Njoku, and Zanolin \cite{M} used time-mapping estimates to prove the existence
of a classical positive solution to \eqref{e1.1} with Dirichlet boundary conditions
when\ $r(t)\equiv 1$,
\begin{equation}
\limsup_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}<\lambda
_1<\liminf_{z\to \infty }\frac{f(t,z)}{z^{p-1}}\leq \infty
\label{e1.3}
\end{equation}
and $\liminf_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}>-\infty$
uniformly for a.e.\ $t\in (0,1)$, which improves a previous result by Kaper,
Knapp, and Kwong \cite{K} where the stronger condition
\[
\lim_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}=l\leq 0\quad\text{and}\quad
\lim_{z\to \infty }\frac{f(t,z)}{z^{p-1}}=\infty
\]
uniformly for $t\in (0,1)\ $was used. Note that when $p=2$ and $f$ is
independent of $t,\ $condition \eqref{e1.3} together with $f(0)=0$ and 
$f\geq 0$
was used in \cite{F} to show the existence of a positive solution to the PDE
problem
\[
-\Delta u=f(u)\text{ in }\Omega ,\quad  u=0\text{ on } \partial \Omega .
\]
 Wang \cite{WA} showed the existence of a positive solution to \eqref{e1.1} under
nonlinear boundary conditions that include the Sturm-Liouville one when $f$
is nonnegative and satisfies either the sublinear condition
\[
\lim_{z\to 0^{+}}\frac{f(z)}{z^{p-1}}=\infty \quad\text{and}\quad
\lim_{z\to \infty }\frac{f(z)}{z^{p-1}}=0,
\]
or the superlinear one
\[
\lim_{z\to 0^{+}}\frac{f(z)}{z^{p-1}}=0\quad\text{and}\quad
\lim_{z\to \infty }\frac{f(z)}{z^{p-1}}=\infty ,
\]
which extended a previous result by Erbe and Wang \cite{E} when $p=2$. Similar
results were established in \cite{H} for singular Sturm-Liouville boundary value
problems. Note that the conditions in \cite{E,H,WA} do not involve the principal
eigenvalue of the corresponding operator.\ Existence results in the PDE
version of \eqref{e1.1} involving the principal eigenvalue of the $p$-Laplacian
operator for $p\geq 2$ was studied in \cite{CM}. In particular, the existence of a
nontrivial nonnegative weak solution $u\in W_0^{1,p}(\Omega )$ to the
problem
\begin{gather*}
-\Delta _{p}u=f(u)\quad \text{in }\Omega , \\
u=0\quad \text{on }\partial \Omega ,
\end{gather*}
was established for $f$ satisfying $|f(z)|((1+z^{p-1})^{-1}$ bounded on 
$[0,\infty )$ and either
\[
-\infty <\lim_{z\to 0^{+}}\frac{f(z)}{z^{p-1}}<\lambda
_1<\,\lim_{z\to \infty }\frac{f(z)}{z^{p-1}}<\infty ,
\]
or
\[
-\infty <\lim_{z\to \infty }\frac{f(z)}{z^{p-1}}<\lambda
_1<\,\lim_{z\to 0^{+}}\frac{f(z)}{z^{p-1}}<\infty
\]
holds. The approach used in \cite{CM} was via the Granas fixed point index (see
\cite{DU}). In this paper, we shall extend the result in \cite{M} to include the
general Sturm-Liouville boundary conditions with more general operator 
e.g.\ allowing the case $r\not\equiv 1$. Note that the proof in \cite{M} does not
apply to this general context. Since we do not require that $f$ be
non-negative but that there exists $\eta \in L^1(0,1)$ with $\eta \geq 0$
such that $\liminf_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}\geq
-\eta (t)$ uniformly for a.e.\ $t\in (0,1)$, our result also improves a
corresponding result in \cite{Ko}. In addition, some estimates on the principal
eigenvalue $\lambda _1$ for $p>1$ are provided (see Lemma \ref{lem2.5} below). 
We refer to \cite{H1,K1,R,Y} for existence results related to \eqref{e1.1}
 under suitable sublinear or superlinear conditions. Our approach is based on a
Krasnoselskii type fixed point theorem in a Banach space.

We shall make the following assumptions:
\begin{itemize}
\item[(A1)] $r:[0,1]\to (0,\infty )$ is continuous.

\item[(A2)] $f:(0,1)\times [ 0,\infty )$ is a Carath\'{e}odory function, that
is $f(\cdot,z)$ is measurable for each $z\geq 0$ and $f(t,\cdot)$ is continuous for
a.e.\ $t\in (0,1)$.

\item[(A3)] For each $k>0$, there exists $\gamma _{k}\in L^1(0,1)$ such that
\[
|f(t,z)|\leq \gamma _{k}(t)
\]
for a.e. $t\in (0,1)$ and $z\in [ 0,k]$.

\item[(A4)] There exists $\eta \in L^1(0,1)$\ with $\eta \geq 0$ such that
\[
\liminf_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}\geq -\eta (t)
\]
uniformly for a.e.\ $t\in (0,1)$.

\item[(A5)]
\[
\limsup_{z\to 0^{+}}\frac{f(t,z)}{z^{p-1}}<\lambda
_1<\liminf_{z\to \infty }\frac{f(t,z)}{z^{p-1}}
\]
uniformly for a.e.\ $t\in (0,1)$.
\end{itemize}
 Our main result reads as follows.

\begin{theorem} \label{thm1.1} 
Let {\rm (A1)--(A5)} hold. Then \eqref{e1.1} has a
positive classical solution $u$ with 
$\inf_{t\in (0,1)}\frac{u(t)}{p(t)}>0$,  where 
$p(t)=\min (at+b,d+c(1-t))$.
\end{theorem}

In particular, when $f$ is independent of $t$, we obtain the following result.


\begin{corollary} \label{coro1.1}
 Let $r$ satisfy {\rm (A1)} and let $f:[0,\infty )\to \mathbb{R}$
 be continuous with 
\[
-\infty <\lim_{z\to 0^{+}}\frac{f(z)}{z^{p-1}}<\lambda
_1<\,\lim_{z\to \infty }\frac{f(z)}{z^{p-1}}\leq \infty .
\]
Then \eqref{e1.1} has a positive classical solution $u$  with 
$\inf_{t\in (0,1)}\frac{u(t)}{p(t)}>0$.
\end{corollary}

\section{Preliminaries} 

Let $AC^1[0,1]=\{u\in C^1[0,1]:u'$ is absolutely continuous on
$[0,1]\}$. We shall denote the norm in $L^{q}(0,1)$ and $C^1[0,1]$
by $ \|\cdot\|_q$ and $|\cdot|_{C^1}$ respectively.
 Let $\lambda _1\ $be the
principal eigenvalue of $-(r(t)\phi (u'))'$ on $(0,1)$
with Sturm-Liouville boundary conditions, and let $\phi _1\ $be the
corresponding positive, normalized eigenfunction, i.e.\
$-(r(t)|\phi_1'|^{p-2}\phi _1')'=\lambda _1\phi_1^{p-1}$
a.e. on $(0,1),\phi _1>0$ on $(0,1)$, $\|\phi _1\|_{\infty}=1$ and 
$\phi _1$ satisfies the Sturm-Liouville boundary conditions in
\eqref{e1.1} (see \cite[Theorem 3.1]{B}).
 Note that $\lambda _1>0$. We recall the
following fixed point theorem of Krasnoselskii type in a Banach space (see
Amann \cite[Theorem 12.3]{A}).


\begin{lemma} \label{lemA}
Let $E$  be a Banach space and $A:E\to E$ be a completely continuous operator. 
Suppose there exist $h\in E,h\neq 0$  and positive constants 
$r,R$  with $r\neq R$ such that
\begin{itemize}
\item[(a)] If $y\in E$ satisfies $y=\theta Ay$ for some
$\theta \in (0,1]$  then $\|y\|\neq r$,

\item[(b)] If $y\in E$ satisfies $y=Ay+\xi h$ for some 
$\xi \geq 0$ then $\|y\|\neq R$.
\end{itemize}
Then $A$ has a fixed point $y\in E$ with $\min(r,R)<\|y\|<\max (r,R)$.
\end{lemma}

\begin{lemma} \label{lem2.1}
Let $t_0,t_1,\alpha ,\beta $  be
constants with $0\leq t_0<t_1\leq 1$, and 
$h\in L^1(t_0,t_1)$. Then the problem
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'=h\quad \text{a.e. on }(t_0,t_1), \\
au(t_0)-b\phi ^{-1}(r(t_0))u'(t_0)=\alpha ,\quad
cu(t_1)+d\phi ^{-1}(r(t_1))u'(t_1)=\beta
\end{gathered}  \label{e2.1}
\end{equation}
has a unique solution $u=Th\in AC^1[t_0,t_1]$.
Furthermore $T:L^1(t_0,t_1)\to C[t_0,t_1]$ is
completely continuous. 
\end{lemma}

\begin{proof}
 By integrating, it follows that \eqref{e2.1} has a unique solution $
u\in AC^1[t_0,t_1]$ given by
\[
u(t)=C+\int_{t_0}^{t}\phi ^{-1}\Big( \frac{D-\int_{t_0}^{s}h}{r(s)}\Big) ds,
\]
where $C$ and $D$ are constants satisfying
\begin{equation}
\begin{gathered}
aC-b\phi ^{-1}(D)=\alpha , \\
c\Big( C+\int_{t_0}^{t_1}\phi ^{-1}( \frac{D-\int_{t_0}^{s}h}{
r(s)}) ds\Big) +d\phi ^{-1}\Big( D-\int_{t_0}^{t_1}h\Big)
=\beta .
\end{gathered} \label{e*}
\end{equation}
In what follows, we shall see, in particular, that $C,D$ are uniquely
determined. We shall denote by $K_i,i=0,1,2,\dots$, positive constants
independent of $u$ and $h$.
\smallskip

\noindent\textbf{Case 1: $a=0$.}
Then $b,c>0$, $D=-\phi (\alpha /b)$ and
\[
C=\frac{\beta -d\phi ^{-1}\big( D-\int_{t_0}^{t_1}h\big) }{c}
-\int_{t_0}^{t_1}\phi ^{-1}\Big( \frac{D-\int_{t_0}^{s}h}{r(s)}
\Big) ds.
\]
Using the inequality
\begin{equation}
(x+y)^{q}\leq m(x^{q}+y^{q})\text{ for }x,y\geq 0,q>0,  \label{e2.2}
\end{equation}
where $m=2^{(q-1)^{+}}$, we deduce that
$|C|\leq K_1+K_2\phi ^{-1}(\|h\|_1)$,
which implies
\[
\|u\|_{\infty }\leq K_3+K_{4}\phi ^{-1}(\|h\|_1).
\]
\smallskip

\noindent\textbf{Case 2: $a>0$.}
Then \eqref{e*}  is equivalent to $C=\frac{\alpha +b\phi ^{-1}(D)}{a}$, where $D$
is the solution of
\[
\gamma (D)\equiv \frac{cb\phi ^{-1}(D)}{a}+c\int_{t_0}^{t_1}\phi
^{-1}\Big( \frac{D-\int_{t_0}^{s}h}{r(s)}\Big) ds
+d\phi ^{-1}\Big(D-\int_{t_0}^{t_1}h\Big) 
=\beta -\frac{\alpha c}{a}.
\]
Note that $D$ is uniquely determined since $\gamma (D)$ is increasing in $D$,
$\lim_{D\to \infty }\gamma (D)=\infty $ and 
$\lim_{D\to -\infty }\gamma (D)=-\infty$.

If $c=0$ then $d>0$ and it follows that 
$|D|\leq \|h\|_1+\phi (|\beta|/d)$, while if $c>0$ then
\[
|D|\leq \|h\|_1+\|r\|_{\infty }\phi \Big( \frac{1}{c(t_1-t_0)}
| \beta -\frac{\alpha c}{a}| \Big) .
\]
Hence in both cases,
\[
|u|_{C^1[t_0,t_1]}=\|u\|_{\infty }+\|u'\|_{\infty }\leq
K_{5}+K_0\phi ^{-1}(\|h\|_1).
\]
i.e.\ $T$ maps bounded sets in $L^1(t_0,t_1)$ into bounded sets in $
C^1[t_0,t_1]$. To show that $T$ is continuous, let $\varepsilon >0,\
h_i\in L^1(t_0,t_1)$ and $u_i=Th_i,i=1,2$. We shall show that
there exists a constant $\delta >0$ depending on $\varepsilon $ and an upper
bound of $\|h_i\|_{L^1(t_0,t_1)}$, $i=1,2$, such that
\begin{equation}
\|h_1-h_2\|_{L^1(t_0,t_1)}<\delta \Longrightarrow
|u_1-u_2|_{C^1[t_0,t_1]}<\varepsilon .  \label{e2.3}
\end{equation}
Note that
\[
u_i(t)=C_i+\int_{t_0}^{t}\phi ^{-1}\Big( \frac{D_i-
\int_{t_0}^{s}h_i}{r(s)}\Big) ds,
\]
and from the above calculation we obtain
\[
|D_i|\leq \max_{i=1,2}\|h_i\|_{L^1(t_0,t_1)}+K\equiv M_0
\]
for $i=1,2$, where $K>0$ independent of $u_i$ and $h_i$. This implies
\[
| D_i-\int_{t_0}^{s}h_i| ,\ | \frac{
D_i-\int_{t_0}^{s}h_i}{r(s)}| \leq 2M_0\max
(r_0^{-1},1)\equiv M
\]
for all $s\in [ t_0,t_1],i=1,2$, where $r_0=\min_{[0,1]}r>0$.
Since $\phi ^{-1}$ is uniformly continuous on
$I=[-M,M]$, it follows from the formulas for $C_i,D_i$, and the fact that
$|D_1-D_2|\leq \|h_1-h_2\|_{L^1(t_0,t_1)}$ that there exists a
constant $\delta >0$ such that \eqref{e2.3} holds. 
This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.1}\rm
If $\alpha =\beta =0$
then Lemma \ref{lem2.1} is reduced to \cite[Lemma 3.1]{H}.
Note that in this case $K_{5}=0$  in the above proof i.e. 
$|u|_{C^1[t_0,t_1]}\leq K_0\phi ^{-1}(\|h\|_1)$  for all 
$u $ satisfying \eqref{e2.1}.
\end{remark}

\begin{lemma} \label{lem2.2} 
Let $t_0,t_1,\alpha ,\beta $ be
constants with $0\leq t_0<t_1\leq 1$, and 
$\gamma,h\in L^1(t_0,t_1)$ with $\gamma \geq 0$. Then the
problem
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'+\gamma (t)\phi (u)=h(t)\quad \text{a.e. on }(t_0,t_1), \\
au(t_0)-b\phi ^{-1}(r(t_0))u'(t_0)=\alpha ,\quad
cu(t_1)+d\phi ^{-1}(r(t_1))u'(t_1)=\beta
\end{gathered}  \label{e2.4}
\end{equation}
has a unique solution $u\equiv T_0h\in AC^1[t_0,t_1]$.
Furthermore $T_0:L^1(t_0,t_1)\to C[t_0,t_1]$ is completely continuous.
\end{lemma}


\begin{proof}
Let $E=C[t_0,t_1]$ be equipped with norm $\|u\|=\sup_{[t_0,t_{1]}}|u|$.
By Lemma \ref{lem2.1}, for each $v\in E$, the problem
\begin{align*}
-(r(t)\phi (u'))'=h(t)-\gamma (t)\phi (v)\quad \text{a.e. on }(t_0,t_1), \\
au(t_0)-b\phi ^{-1}(r(t_0))u'(t_0)=\alpha ,\quad
cu(t_1)+d\phi ^{-1}(r(t_1))u'(t_1)=\beta
\end{align*}
has a unique solution $u=Sv\in AC^1[t_0,t_1]$ and $S:E\to E$
is completely continuous. Let $u\in E$ satisfy $u=\theta Su$ for some 
$\theta \in (0,1]$. Then
\begin{equation}
\begin{gathered}
-(r(t)\phi (u'))'+\theta ^{p-1}\gamma (t)\phi (u)
=\theta ^{p-1}h(t)\quad\text{a.e. on }(t_0,t_1), \\
au(t_0)-b\phi ^{-1}(r(t_0))u'(t_0)=\theta \alpha ,\quad
cu(t_1)+d\phi ^{-1}(r(t_1))u'(t_1)=\theta \beta
\end{gathered}  \label{e2.5}
\end{equation}
By integrating \eqref{e2.5}, we obtain
\begin{equation}
\phi (u'(t))=\frac{r(t_1)\phi (u'(t_1))+\theta
^{p-1}\int_{t}^{t_1}( h-\gamma \phi (u)) ds}{r(t)}  \label{e2.6}
\end{equation}
for $t\in [ t_0,t_1]$. Multiplying the equation in \eqref{e2.5} by $u$
and integrating gives
\begin{equation}
-r(t_1)\phi (u'(t_1))u(t_1)+r(t_0)\phi (u'(t_0))u(t_0)+\int_{t_0}^{t_1}r(t)|u'|^p
\leq \int_{t_0}^{t_1}|hu|.  \label{e2.7}
\end{equation}
We shall consider two cases.
\smallskip

\noindent\textbf{Case 1. $b=0$ or $d=0$.}
Without loss of generality, we suppose $b=0$. Then 
$u(t_0)=\theta \alpha /a\equiv \theta \alpha _0$. By the mean value theorem,
\begin{equation}
\|u\|\leq |\alpha _0|+\int_{t_0}^{t_1}|u'|.  \label{e2.8}
\end{equation}
Suppose first that $d=0$. Then 
$u(t_1)=\theta \beta /c\equiv \theta \beta_0$. 
Let $\xi (t)=\theta (At+B)$, where $A,B$ are constants such that 
$\xi (t_0)=\theta \alpha _0,\xi (t_1)=\theta \beta _0$ i.e. 
$A=\frac{\beta _0-\alpha _0}{t_1-t_0},B=\frac{\alpha _0t_1-\beta _0t_0
}{t_1-t_0}$. In what follows, we shall denote by $R_i$, $i=0,1,\dots$,
positive constants independent of $u$ and $\theta$.

Multiplying the equation in \eqref{e2.5} by $(u-\xi )$ and integrating, we obtain
\begin{align*}
r_0\int_{t_0}^{t_1}|u'|^p
&\leq |A\||r\|_{\infty}\int_{t_0}^{t_1}|u'|^{p-1}+(|A|+|B|)|
\Big(\int_{t_0}^{t_1}\gamma \Big) \|u\|^{p-1} \\
&\quad +( \|u\|+A+B) \int_{t_0}^{t_1}h.
\end{align*}
This, together with \eqref{e2.8}, implies $\int_{t_0}^{t_1}|u'|^p\leq R_0$.

Suppose next that $d>0$. Then from the boundary condition at $t_1$,
we obtain $u'(t_1)=\frac{\theta \beta -cu(t_1)}{d\phi^{-1}(r(t_1))}$.
Hence if $c=0$ then $u'(t_1)=\frac{\theta
\beta }{d\phi ^{-1}(r(t_1))}\equiv \theta \beta _1$ from which \eqref{e2.6} and
\eqref{e2.8} imply 
\begin{equation}
\|u'\|\leq R_1\Big( 1+\int_{t_0}^{t_1}|u'|\Big) .
\label{e2.9}
\end{equation}
Consequently, \eqref{e2.7} gives
\[
\int_{t_0}^{t_1}r(t)|u'|^p
\leq \|r\|_{\infty }\ ( |\beta_1|^{p-1}|\|u\| +|\alpha _0\||u'\|^{p-1}) 
+\Big(\int_{t_0}^{t_1}|h|\Big) \|u\|,
\]
which, together with \eqref{e2.8} and \eqref{e2.9}, implies that 
$\int_{t_0}^{t_1}|u'|^p\leq R_2$.

If $c>0$, then
\begin{equation}
\begin{aligned}
&-r(t_1)\phi (u'(t_1))u(t_1) \\
&=r(t_1)\phi \Big( \frac{cu(t_1)-\theta \beta }
  {d\phi ^{-1}(r(t_1))}\Big) u(t_1) \\
&=r(t_1)\phi \Big( \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}
\Big) \Big( \Big( \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}
\Big) \Big( \frac{d\phi ^{-1}(r(t_1))}{c}\Big) 
 +\frac{\theta \beta }{c}\Big)\\
&\geq R_2 \Big| \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}
\Big| ^p-R_3. 
\end{aligned} \label{e2.10}
\end{equation}
By \eqref{e2.6} and \eqref{e2.8},
\begin{equation}
|\phi (u'(t_0)|
\leq \frac{1}{r_0}\Big( \|r\|_{\infty
}| \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}|
^{p-1}+\int_{t_0}^{t_1}|h|\ +( \int_{t_0}^{t_1}\gamma )
\|u\|^{p-1}\Big) .  \label{e2.11}
\end{equation}
Using \eqref{e2.8}, \eqref{e2.10} and \eqref{e2.11} together with 
$u(t_0)=\theta \alpha _0$
in \eqref{e2.7}, we deduce that $\int_{t_0}^{t_1}|u'|^p\leq R_{4}$.
Hence in either case $\int_{t_0}^{t_1}|u'|^p\leq R_{5}$,
where $R_{5}=\max (R_0,R_2,R_{4})$ and so $\|u\|\leq |\alpha
_0|+R_{5}^{1/p}$ .
\smallskip

\noindent\textbf{Case 2. $b>0,d>0$.}
Then $u'(t_0)=\frac{\alpha u(t_0)-\theta \alpha }{b\phi
^{-1}(r(t_0))}$ and $u'(t_1)=\frac{\theta \beta -cu(t_1)}{
d\phi ^{-1}(r(t_1))}$.
Hence \eqref{e2.7} and \eqref{e2.8} give
\begin{equation}
\begin{aligned}
&r(t_1)\phi \Big( \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}
\Big) u(t_1)
+r(t_0)\phi \Big( \frac{\alpha u(t_0)-\theta \alpha }{
b\phi ^{-1}(r(t_0))}\Big) u(t_0)
+\int_{t_0}^{t_1}r(t)|u'|^p \\
&\leq \Big(\int_{t_0}^{t_1}|h|\Big) \|u\|.
\end{aligned}  \label{e2.12}
\end{equation}
Since $a+c>0$, we can assume without loss of generality that $c>0$.
Then 
\begin{align*}
\|u\|
&\leq |u(t_1)|+\int_{t_0}^{t_1}|u'| \\
&\leq \frac{d\phi ^{-1}(r(t_1))}{c}| 
 \frac{cu(t_1)-\theta \beta}{d\phi ^{-1}(r(t_1))}| +\frac{|\beta |}{c}
+\int_{t_0}^{t_1}|u'|,
\end{align*}
which, together with \eqref{e2.10} and \eqref{e2.12}, imply
\[
\Big| \frac{cu(t_1)-\theta \beta }{d\phi ^{-1}(r(t_1))}\Big|
^p+\int_{t_0}^{t_1}|u'|^p\leq R_{6}.
\]
Consequently, $\|u\|<R_{8}$.
Thus, we have shown that in both cases that 
$\|u\|$ is bounded by a constant independent of $u$ and $\theta $. By the
Leray-Schauder fixed point theorem, $S$ has a fixed point $u$, which is a
solution of \eqref{e2.4} in $AC^1[t_0,t_1]$. To show uniqueness, let $u,v$
be solutions of \eqref{e2.4}. Then
\begin{equation}
-(r(t)(\phi (u')-\phi (v'))'+\gamma (t)(\phi
(u)-\phi (v))=0\quad \text{a.e. on }(t_0,t_1).  \label{e2.13}
\end{equation}
We claim that $(\phi (u'(t_0))-\phi (v'(t_0))(u(t_0-v(t_0)\geq 0$.
This is true when $b=0$ since $u(t_0)=\alpha /a=v(t_0)$ in this case. 
If $b>0$ then $u'(t_0)=\frac{au(t_0)-\alpha }{b\phi ^{-1}(r(t_0))},v'(t_0)
=\frac{av(t_0)-\alpha }{b\phi ^{-1}(r(t_0))}$, which implies
\begin{align*}
&(\phi (u'(t_0))-\phi (v'(t_0))(u(t_0)-v(t_0) \\
&=\Big( \phi \Big( \frac{au(t_0)-\alpha }{b\phi ^{-1}(r(t_0))}\Big)
-\phi \Big( \frac{av(t_0)-\alpha }{b\phi ^{-1}(r(t_0))}\Big) \Big)
(u(t_0-v(t_0)\geq 0.
\end{align*}
Similarly, $(\phi (u'(t_1))-\phi (v'(t_1))(u(t_1-v(t_1)\leq 0$. 
Hence, multiplying \eqref{e2.13} by $u-v$ and
integrating, we get
\[
\int_{t_0}^{t_1}r(t)(\phi (u')-\phi (v'))(u'-v')dt\leq 0,
\]
which implies $u'=v'$ on $(t_1,t_2)$. Hence there
exists a constant $k$ such that $u(t)=v(t)+k$ for all 
$t\in [t_1,t_2]$. The boundary conditions then give $ak=ck=0$.
 Hence $k=0$, which completes the proof.
\end{proof}

Next, we prove a comparison principle, which extends \cite[Lemma 3.2]{H}
to the case $\gamma \geq 0$, $\gamma \not\equiv 0$. 

\begin{lemma} \label{lem2.3}
 Let $\gamma,h_i\in L^1(t_0,t_1)$, $i=1,2$, with 
$\gamma \geq 0$ and $h_1\geq h_2$.  Let $u_i\in AC^1[t_0,t_1]$,
$i=1,2$ satisfy
\begin{gather*}
-(r(t)\phi (u_i'))'+\gamma (t)\phi (u_i)=h_i\quad \text{a.e. on }(t_0,t_1), \\
au_1(t_0)-b\phi ^{-1}(r(t_0))u_1'(t_0)\geq
au_2(t_0)-b\phi ^{-1}(r(t_0))u_2'(t_0), \\
cu_1(t_1)+d\phi ^{-1}(r(t_1))u_1'(t_1)\geq
cu_2(t_1)+d\phi ^{-1}(r(t_1))u_2'(t_1).
\end{gather*}
Then $u_1\geq u_2$  on $[t_0,t_1]$.
\end{lemma}

\begin{proof} 
Suppose on the contrary that there exists $\tilde{t}\in(t_0,t_1)$ such that 
$u_1(\tilde{t})<u_2(\tilde{t})$. Let $(\alpha,\beta )\subset (t_0,t_1)$ 
be the largest open interval containing $\tilde{t}\ $such that $u_1<u_2$ 
on $(\alpha ,\beta )$. Hence
\begin{equation}
(r(t)(\phi (u_1')-\phi (u_2'))'\leq 0\quad \text{a.e. on }(\alpha ,\beta ), 
 \label{e2.14}
\end{equation}
\smallskip

\noindent\textbf{Case 1. $u_1(\alpha )=u_2(\alpha )$ or $u_1(\beta)=u_2(\beta )$.}
Suppose $u_1(\alpha )=u_2(\alpha )$. 
Then $u_1'(\alpha )\leq u_2'(\alpha )$.
Hence \eqref{e2.14} implies $u_1'\leq u_2'$ on $(\alpha ,\beta )$. 
If $u_1(\beta )=u_2(\beta )$ then this gives $u_1\geq u_2$ on $(\alpha ,\beta )$, 
a contradiction. If $u_1(\beta )<u_2(\beta )$ then $\beta =t_1$ and from 
the boundary condition at $t_1$, we get $d(u_2'(t_1)-u_1'(t_1))\leq 0$.
Hence if $d>0$ we get $u_2'(t_1)\leq u_1'(t_1)$ from which \eqref{e2.14} gives 
$u_1'\geq u_2'$ on $(\alpha ,\beta )$ and so $u_1\geq u_2$ on 
$(\alpha ,\beta )$, a contradiction. On the other hand, if $d=0$ then 
$ c(u_1(t_1)-u_2(t_1))\geq 0$, which implies $u_1(t_1)\geq u_2(t_1)$, 
a contradiction. Similarly, we get a contradiction if 
$u_1(\beta )=u_2(\beta )$.
\smallskip

\noindent\textbf{Case 2. $u_1<u_2$ on $[\alpha ,\beta ]$ i.e. $\alpha =t_0$
and $\beta =t_1$.}
Suppose $\min_{[\alpha ,\beta ]}(u_1-u_2)=u_1(\tau )-u_2(\tau)<0$ for some 
$\tau \in [ \alpha ,\beta ]$. If $\tau \in (t_0,t_1)$
then $u_1'(\tau )=u_2'(\tau )$ and it follows from
\eqref{e2.14} that there exists a constant $k<0$ such that $u_1=u_2+k$ on 
$[t_0,t_1]$. Using the boundary conditions, we deduce that $ak,ck\geq 0$,
a contradiction. Suppose $\tau =t_0$. Then
\[
a(u_1(t_0)-u_2(t_0))\geq b\phi ^{-1}(r(t_0))(u_1'(t_0)-u_2'(t_0))\geq 0,
\]
which implies $a=0$. Hence $b>0$ and the boundary condition at $t_0$ imply
$u_1'(t_0)-u_2'(t_0)\leq 0$, from which \eqref{e2.14}
gives $u_1'\leq u_2'$ on $(t_0,t_1)$.
Consequently, $u_1=u_2+\tilde{k}$ on $(t_0,t_1)$ for some constant 
$\tilde{k}<0$, a contradiction. Similarly, we reach a contradiction when 
$\tau =t_1$, which completes the proof.
\end{proof}

The next result plays an important role in the proof of the main result.
When $\gamma \equiv 0$, it was obtained in \cite[Lemma 3.4]{H} but the proof
there does not apply to the case $\gamma \not\equiv 0$.

\begin{lemma} \label{lem2.4}
Let $\gamma \in L^1(0,1)$ with $\gamma \geq 0$ and let 
$u\in AC^1[0,1]$  satisfy
\begin{gather*}
-(r(t)(\phi (u'))'+\gamma (t)\phi (u)\geq 0\quad \text{a.e. on }(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 $\kappa >0$  independent of $u$
 such that for all $t\in [ 0,1]$,
\[
u(t)\geq \kappa \|u\|_{\infty }p(t).
\]
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.3}, $u\geq 0$ on $[0,1]$. Suppose $\|u\|_{\infty}=u(\tau )$ for 
some $\tau \in (0,1)$. By Lemma \ref{lem2.2}, the problem
\begin{gather*}
-(r(t)\phi (z'))'+\gamma (t)\phi (z)=0\quad \text{a.e. on }(0,\tau ), \\
az(0)-b\phi ^{-1}(r(0))z'(0)=0,\quad
z(\tau )=\|u\|_{\infty } 
\end{gather*}
has a unique solution $z\in AC^1[0,\tau ]$.
By Lemma \ref{lem2.3}, $u\geq z\geq 0$ on $[0,\tau ]$, from which the boundary condition 
on $z$ at $0$ gives $z'(0)\geq 0$. Note that
\[
z(t)=z(0)+\int_0^{t}\phi ^{-1}
\Big( \frac{r(0)\phi (z'(0))+\int_0^{s}\gamma (\xi )\phi (z)d\xi }{r(s)}\Big) ds,
\]
from which \eqref{e2.2} gives
\[
z(t)\leq z(0)+m_0\Big( z'(0)+\phi ^{-1}( \int_0^{t}\gamma
(s)\phi (z)ds) \Big) ,
\]
where $m_0>0$ is a constant independent of $u$. Hence using \eqref{e2.2} again,
it follows that
\[
\phi (z(t))\leq m_1\Big( \phi (z(0)+z'(0))+\int_0^{t}\gamma
(s)\phi (z)ds\Big)
\]
for $t\in [ 0,\tau ]$, where $m_1>0$ is a constant independent of 
$u$. By Gronwall's inequality,
\[
\phi (z(t)\leq m_1\phi \big(z(0)+z'(0)\big)e^{m_1\int_0^{t}\gamma(s)ds}
\]
for $t\in [ 0,\tau ]$. In particular when $t=\tau $, we obtain
\begin{equation}
z(0)+z'(0)\geq \kappa _0\|u\|_{\infty },  \label{e2.15}
\end{equation}
where $\kappa _0=( e^{-m_1\|\gamma \|_1}/m_1) ^{1/(p-1)}$.
Since $(r(t)\phi (z'))'=\gamma (t)\phi (z)\geq 0$ on
$(0,\tau )$, it follows that $r(t)\phi (z')\geq
r(0)\phi (z'(0))$, which implies
\[
z'(t)\geq ( r(0)/\|r\|_{\infty }) ^{1/(p-1)}z'(0).\
\]
If $b=0$ then $z(0)=0$ and \eqref{e2.15} give
\begin{equation}
z(t)=\int_0^{t}z'\geq \Big( \frac{r(0)}{\|r\|_{\infty }}\Big)
^{\frac{1}{p-1}}\kappa _0\|u\|_{\infty }t=\kappa _1(at+b)\|u\|_{\infty }
\label{e2.16}
\end{equation}
for\ $t\in [ 0,\tau ]$, where
 $\kappa _1=a^{-1}( r(0)/\|r\|_{\infty }) ^{1/(p-1)}\kappa _0$.

On the other hand, if $b>0$ then $z'(0)=\frac{a}{b\phi ^{-1}(r(0))}
z(0)$ and \eqref{e2.15} becomes $z(0)\geq \tilde{\kappa}_1\|u\|_{\infty }$, where
$\tilde{\kappa}_1=\kappa _0( 1+\frac{a}{b\phi ^{-1}(r(0))})^{-1}$. Hence
\begin{equation}
z(t)\geq z(0)\geq \tilde{\kappa}_1\|u\|_{\infty }\geq \kappa
_2(at+b)\|u\|_{\infty }  \label{e2.17}
\end{equation}
for $t\in [ 0,\tau ]$, where $\kappa _2=\tilde{\kappa}_1/(a+b)$.
Combining \eqref{e2.16} and \eqref{e2.17}, we obtain 
$z(t)\geq \kappa_3(at+b)\|u\|_{\infty }$ for $t\in [ 0,\tau ]$, where 
$\kappa _3>0$ is independent of $u,\lambda ,h$.

Next, let $w\in AC^1[\tau ,1]$ be the unique solution of
\begin{gather*}
-(r(t)\phi (w'))'+\gamma (t)\phi (w)=0\quad \text{a.e. on }(\tau ,1), \\
w(\tau )=\|u\|_{\infty },\quad cw(1)+d\phi ^{-1}(r(1))w'(1)=0.
\end{gather*}
Then $u\geq w\geq 0$ on $[\tau ,1]$ and the boundary condition on $w$ at $1$
gives $w'(1)\leq 0$. Using the integral formula
\[
w(t)=w(1)-\int_{t}^1\phi ^{-1}
\Big( \frac{r(1)\phi (w'(1))-\int_{s}^1\gamma (\xi )\phi (w)d\xi }{r(s)}\Big) ds
\]
for $t\in [ \tau ,1]$ and using similar arguments as above, we obtain
$w(t)\geq \kappa _{4}(d+c(1-t))\|u\|_{\infty }$ for $t\in [ \tau ,1]$,
where $\kappa _{4}>0$ is a constant independent of $u$. 
If $\tau =0$ then $u\geq w$ on $[0,1]$ while if $\tau =1$ then 
$u\geq z$ on $[0,1]$. 
Thus $u(t)\geq \kappa \|u\|_{\infty }p(t)$ for $t\in [ 0,1]$, where $\kappa
=\min (\kappa _3,\kappa _{4})$, which completes the proof.
\end{proof}

The next result provides some estimates on $\lambda _1$ for $p>1$.

\begin{lemma} \label{lem2.5}
 Suppose $b+d>0$  and $r\equiv 1$.
 If $d>0$ then 
\begin{equation}
\frac{\min (A_1,1)}{2^{(p-1)^{+}}}\leq \lambda _1
\leq (A_1+(m_1+2)^pe^{m_1p})(2p+1),  \label{e2.18}
\end{equation}
where $A_1=(c/d)^{p-1},m_1=(c+2d)/d$,  while if $b>0$,
then 
\begin{equation}
\frac{\min (B_1,1)}{2^{(p-1)^{+}}}\leq \lambda _1\leq
(B_1+(m_2+2)^pe^{m_2p}(2p+1),  \label{e2.19}
\end{equation}
where $B_1=(a/b)^{p-1}$, $m_2=(a+2b)/b$.
\end{lemma}

\begin{proof} 
Using the Rayleigh quotient, we obtain
\begin{equation}
\lambda _1=\inf_{u\in V}\frac{\phi (u'(0))u(0)
-\phi (u'(1))u(1)+\int_0^1|u'|^pdt}{\int_0^1|u|^pdt}  \label{e2.20}
\end{equation}
where $V=\{u\in C^1[0,1]:au(0)-bu'(0)=0,cu(1)+du'(1)=0\}$.

Suppose $d>0$. Then $u'(1)=-(c/d)u(1)$ and $\phi (u'(0)u(0)\geq 0$ 
for $u\in V$. Hence
\begin{equation}
\begin{aligned}
\lambda _1
&=\inf_{u\in V}\frac{\phi (u'(0))u(0)+A_1|u(1)|^p+\int_0^1|u'|^pdt}{
\int_0^1|u|^pdt} \\
&\geq \inf_{u\in V}\frac{A_1|u(1)|^p+
\int_0^1|u'|^pdt}{\int_0^1|u|^pdt}.
\end{aligned}  \label{e2.21}
\end{equation}
Let $u\in V$. Then
\[
|u(t)|\leq |u(1)|+\int_0^1|u'|dt,
\]
which implies
\begin{align*}
\int_0^1|u|^pdt
&\leq 2^{(p-1)^{+}}\Big(|u(1)|^p+\int_0^1|u'|^pdt\Big)\\
&\leq \frac{2^{(p-1)^{+}}}{\min (A_1,1)}\Big(A_1|u(1)|^p+\int_0^1|u'|^pdt\Big) .
\end{align*}
Consequently, \eqref{e2.21} gives 
$\lambda _1\geq \frac{\min (A_1,1)}{2^{(p-1)^{+}}}$.

Next, we choose $u(t)=t^{2}e^{m_1(1-t)}$, where $m_1=(c+2d)/d$. 
Then $u\in V$ and
\begin{gather*}
u(t)\geq t^{2}, \\
|u'(t)|=te^{m_1(1-t)}|2-m_1t|\leq (m_1+2)e^{m_1}
\end{gather*}
for $t\in [ 0,1]$. Hence
\begin{equation}
\int_0^1|u|^pdt \geq \frac{1}{2p+1},\quad
\int_0^1|u'|^pdt\leq (m_1+2)^pe^{m_1p}.   \label{e2.22}
\end{equation}
Since $u(0)=0,u(1)=1$, it follows from \eqref{e2.22} and the equality 
in \eqref{e2.21}
that
\[
\lambda _1\leq (A_1+(m_1+2)^pe^{m_1p})(2p+1)
\]
i.e.\ \eqref{e2.18} holds. Suppose next that $b>0$. Then
\begin{equation}
\begin{aligned}
\lambda _1
&=\inf_{u\in V}\frac{B_1|u(0)|^p-\phi (u'(1))u(1)
 +\int_0^1|u'|^pdt}{\int_0^1|u|^pdt} \\
&\geq \inf_{u\in V}\frac{B_1|u(0)|^p+\int_0^1|u'|^pdt}{
\int_0^1|u|^pdt}. 
\end{aligned} \label{e2.23}
\end{equation}
Using the inequality
\[
|u(t)|\leq |u(0)|+\int_0^1|u'|dt,
\]
it follows that
\[
\int_0^1|u|^pdt
\leq \frac{2^{(p-1)^{+}}}{\min (B_1,1)}\Big(
B_1|u(0)|^p+\int_0^1|u'|^pdt\Big) ,
\]
from which \eqref{e2.23} implies 
$\lambda _1\geq \frac{\min (B_1,1)}{2^{(p-1)^{+}}}$. By choosing 
$u(t)=(1-t)^{2}e^{m_2t}$, where 
$m_2=(a+2b)/b$, we see that $u\in V$ and the equality in \eqref{e2.23} gives
\[
\lambda _1\leq (B_1+(m_2+2)^pe^{m_2p})(2p+1),
\]
which establishes \eqref{e2.19}. This completes the proof.
\end{proof}

\begin{example} \label{examp2.1} \rm
It follows from \eqref{e2.18} that the principal
eigenvalue $\lambda _1$ of $-(\phi (u'))'$
 with boundary conditions $u(0)-u'(0)=0=u(1)+u'(1)$ satisfies 
\[
\frac{1}{2^{(p-1)^{+}}}\leq \lambda _1\leq (1+5^pe^{3p})(2p+1)
\]
\end{example}

\section{Proof of main results}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
In view of (A2)--(A5), there exist constants $r,r_1,\bar{\lambda}>0$ 
with $r<r_1$ and $\bar{\lambda}<\lambda _1$
such that for a.e. $t\in (0,1)$,
\begin{equation}
f(t,z)\leq \bar{\lambda}z^{p-1},\quad f(t,z)+(\eta (t)+1)z^{p-1}\geq 0
\label{e3.1}
\end{equation}
for $z\leq r$;
\[
|f(t,z)|\leq \gamma _{r_1}(t)\leq \gamma _{r_1}(t)(z/r)^{p-1}
\]
for $r<z<r_1$, and $f(t,z)>0$ for $z>r_1$ and a.e.\ $t$. Hence
\[
f(t,z)+\gamma (t)z^{p-1}\geq 0
\]
for a.e.\ $t\in (0,1)$ and all $z\geq 0$, where 
$\gamma (t)=\max (\eta (t)+1,\gamma _{r_1}(t)/r^{p-1})$.
For $v\in E=C[0,1]$, we have $f(t,|v|)+\gamma (t)|v|^{p-1}\in L^1(0,1)$ 
in view of (A3). Hence by Lemma \ref{lem2.2}, the problem
\begin{gather*}
-(r(t)\phi (u'))'+\gamma (t)\phi (u)=f(t,|v|)+\gamma
(t)|v|^{p-1}\quad \text{a.e. on }(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 $u=Av\in C^1[0,1]$. Since $A=T_0\circ S_0$,
where $S_0:C[0,1]\to L^1(0,1)$ is defined by 
$(S_0v)(t)=f(t,|v|)+\gamma (t)|v|^{p-1}$ and $T_0$ is defined in 
Lemma \ref{lem2.2} with $\alpha =\beta =0$, we see that $A:E\to E$ is completely
continuous. We shall verify that
\begin{itemize}
\item[(i)] $u=\theta Au$, $\theta \in (0,1]\Longrightarrow \|u\|_{\infty
}\neq r$.
\end{itemize}
Indeed, let $u\in E$ satisfy $u=\theta Au$ for some $\theta \in (0,1]$ and
suppose $\|u\|_{\infty }=r$. Then $u\in AC^1[0,1]$ and 
\[
-(r(t)\phi (u'))'+\gamma (t)\phi (u)=\theta ^{p-1}(
f(t,|u|)+\gamma (t)|u|^{p-1}) \geq 0\quad \text{a.e. on }(0,1),
\]
which implies $u\geq 0$ on $(0,1)$ by Lemma \ref{lem2.4}. Hence
\begin{equation}
-(r(t)\phi (u'))'=\theta ^{p-1}f(t,u)-(1-\theta
^{p-1})\gamma (t)u^{p-1}\leq \theta ^{p-1}f(t,u)
\label{e3.2}
\end{equation}
 a.e.\ on $(0,1)$.

By \cite[Lemma 2.1]{H1}, there exists a constant $k_0>0$ such that 
$|z(t)|\leq k_0|z|_{C^1}p(t)$ for all $t\in [ 0,1]$ and $z\in C^1[0,1]$
satisfying the Sturm-Liouville boundary conditions in \eqref{e1.1}.
 In particular,
$\sup_{t\in (0,1)}\frac{u(t)}{p(t)}<\infty $.
 Since
\[
-(r(t)\phi (\phi _1')')=\lambda _1\phi _1^{p-1}>0\quad
\text{a.e. on }(0,1),
\]
it follows from Lemma \ref{lem2.4} (with $\gamma \equiv 0$) that
 $\inf_{t\in (0,1)}\frac{\phi _1(t)}{p(t)}>0$.
Hence there exists a smallest positive constant $\delta _0$ such that 
$u\leq \delta _0\phi _1$ on $[0,1]$. Then it follows from \eqref{e3.1} 
and \eqref{e3.2} that
\[
-(r(t)\phi (u'))'\leq \bar{\lambda}u^{p-1}\leq \bar{\lambda
}\delta _0^{p-1}\phi _1^{p-1}\quad \text{ a.e. on }(0,1),
\]
from which the weak comparison principle (see \cite[Lemma 3.2]{H}, 
\cite[Lemma A2]{S})  gives
\[
u\leq (\bar{\lambda}\delta _0^{p-1}/\lambda _1)^{\frac{1}{p-1}}\phi _1
\]
on $[0,1]$, a contradiction with the definition of $\delta _0$. 
Thus $\|u\|_{\infty }\neq r$ i.e. (i) holds.

Next, we claim that
\begin{itemize}
\item[(ii)] There exists a constant $R>r$ such that 
$u=Au+\xi$, $\xi \geq 0$  implies $\|u\|_{\infty }\neq R$.
\end{itemize}
Let $u\in E$ satisfy $u=Au+\xi $ for some $\xi \in [ 0,\infty )$. Then
$u-\xi =Au$ and therefore
\[
-(r(t)\phi (u'))'+\gamma (t)\phi (u-\xi )=f(t,|u|)+\gamma
(t)|u|^{p-1}\quad \text{a.e. on }(0,1),
\]
which implies
\begin{equation}
-(r(t)\phi (u'))'+\gamma (t)\phi (u)\geq f(t,|u|)+\gamma
(t)|u|^{p-1}\geq 0\text{ a.e. on }(0,1).  \label{e3.3}
\end{equation}
Since $\liminf_{z\to \infty }\frac{f(t,z)}{z^{p-1}}>\lambda_1$ uniformly 
for a.e. $t\in (0,1)$, there exist positive constants 
$L,\tilde{\lambda},\lambda _0$ with 
$\tilde{\lambda}>\lambda _0>\lambda _1$ such that 
$f(t,z)\geq \tilde{\lambda}z^{p-1}$ for a.e. $t\in (0,1)$
and $z>L$.

Let $\varepsilon =(k_0l)^{-1}\Big( (\tilde{\lambda}/\lambda _1)^{\frac{1
}{p-1}}-(\lambda _0/\lambda _1)^{\frac{1}{p-1}}\Big) $, 
where $l=\sup_{t\in (0,1)}\frac{p(t)}{\phi _1(t)}\in (0,\infty )$, and let 
$\delta $ be given by \eqref{e2.3}$.
 $Choose $I=[\alpha ,\beta ]\subset [0,1] $ such that
\[
\int_{[ 0,1\backslash I}( \tilde{\lambda}+\gamma _{L}(t))
<\delta ,
\]
where $\gamma _{L}$ is defined by (A3).
 Let $R>\max ( r,\dfrac{1}{\kappa l_0},\dfrac{L}{\kappa \min_{[ \alpha ,\beta ]}p}
) $, 
where $l_0=$ $\inf_{(0,1)}\frac{p}{\phi _1}>0$ and 
$\kappa $ is defined in Lemma \ref{lem2.4}. 
We claim that $\|u\|_{\infty }\neq R$.
Indeed, suppose $\|u\|_{\infty }=R$. Then it follows from \eqref{e3.3} 
and Lemma \ref{lem2.4} that $u(t)\geq \kappa \|u\|_{\infty }p(t)$ for $t\in (0,1)$. 
In particular, \eqref{e3.3} becomes
\begin{equation}
-(r(t)\phi (u'))'\geq f(t,u)\quad \text{on }(0,1),  \label{e3.4}
\end{equation}
and
\[
u(t)\geq \kappa Rp(t)\geq \kappa R\min_{[\alpha ,\beta ]}p>L
\]
for $t\in I$. Hence $f(t,u)\geq \tilde{\lambda}u^{p-1}$ for a.e. 
$t\in I$. Let $\delta _1$ be the largest positive number such that 
$u\geq \delta _1\phi _1$ on $(0,1)$. Then $\delta _1\geq \kappa l_0R>1$ and
\[
-\Big( r(t)\phi ( \frac{u'}{\delta _1}) \Big)'
\geq \begin{cases}
\tilde{\lambda}\phi _1{}^{p-1} &\text{if }t\in I, \\
-\gamma _{L}(t) &\text{if }t\notin I.
\end{cases}
\]
Let $u_1,u_2 \in AC^1[0,1]$ satisfy
\begin{align*}
-( r(t)\phi ( u_1') ) '
&=\begin{cases}
\tilde{\lambda}\phi _1{}^{p-1}& \text{if }t\in I, \\
-\gamma _{L}(t)& \text{if }t \notin I
\end{cases} \\
& \equiv h_1\quad \text{ a.e. on }(0,1),
\end{align*}
and
\[
-(r(t)\phi (u_2'))'=\tilde{\lambda}\phi _1{}^{p-1}\
\equiv h_2\quad \text{a.e. on }(0,1).
\]
with Sturm-Liouville boundary conditions. Note that 
$u_2=(\tilde{\lambda} /\lambda _1)^{\frac{1}{p-1}}\phi _1$ and 
$u\geq \delta _1u_1$ on $(0,1)$. Since
\[
\|h_1-h_2\|_1\leq \int_{[ 0,1]\backslash I}( \tilde{\lambda}
+\gamma _{L}(t)) <\delta ,
\]
it follows from \eqref{e2.3} that $|u_1-u_2|_{C^1}<\varepsilon $. Hence
\begin{align*}
u_1 &\geq u_2-k_0\varepsilon p\geq u_2-k_0l\varepsilon \phi _1 \\
&=\Big( \tilde{\lambda}/\lambda _1\Big) ^{\frac{1}{p-1}}\phi _1
 -\Big((\tilde{\lambda}/\lambda _1)^{\frac{1}{p-1}}-(\lambda _0/\lambda _1)^{
\frac{1}{p-1}}\Big) \phi _1 \\
& =(\lambda _0/\lambda _1)^{\frac{1}{p-1}}\phi _1\quad \text{on }(0,1),
\end{align*}
and consequently, 
$u\geq \delta _1(\lambda _0/\lambda _1)^{\frac{1}{p-1}}\phi _1$ on $(0,1)$,
 a contradiction with the definition of $\delta_1$.
 Thus $\|u\|_{\infty }\neq R$, as claimed i.e. (ii) holds.

By Lemma \ref{lemA}, operator $A$ has a fixed point $u\in E$ with 
$\|u\|_{\infty }>r$, which
is a classical positive solution of \eqref{e1.1} in view of 
Lemmas \ref{lem2.2} and \ref{lem2.4}. This completes the proof.
\end{proof}

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\end{document}