\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 106, pp. 1--18.\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-2005/106\hfil Three-point boundary-value problem]
{Positive solutions of three-point boundary-value problems for p-Laplacian
singular differential equations}
\author[G. N. Galanis, A. P. Palamides\hfil EJDE-2005/106\hfilneg]
{George N. Galanis, Alex P. Palamides}

\address{George N. Galanis \hfill\break 
Naval Academy of Greece, Piraeus, 185 39, Greece}
\email{ggalanis@math.uoa.gr}

\address{Alex P. Palamides \hfill\break
Department of Communication Sciences, 
University of Peloponnese, 22100 Tripolis, Greece}
\email{palamid@uop.gr}
\date{}

\thanks{Submitted May 13, 2005. Published October 7, 2005.}
\subjclass[2000]{34B15, 34B18}
\keywords{Three-point singular boundary-value problem; p-Laplacian;
\hfill\break\indent
 positive and negative solutions; vector field; Nonlinear alternative of
Leray-Schauder}

\begin{abstract}
In this paper we prove the existence of positive solutions for the
three-point singular boundary-value problem
\begin{equation*}
-[\phi _{p}(u')]'=q(t)f(t,u(t)),\quad 0<t<1
\end{equation*}
subject to
\begin{equation*}
u(0)-g(u'(0))=0,\quad u(1)-\beta u(\eta )=0
\end{equation*}
or to
\begin{equation*}
u(0)-\alpha u(\eta )=0,\quad u(1)+g(u'(1))=0,
\end{equation*}
where $\phi _{p}$ is the $p$-Laplacian operator, $0<\eta <1$; $0<\alpha
,\beta <1$ are fixed points and $g$ is a monotone continuous function
defined on the real line $\mathbb{R}$ with $g(0)=0$ and $ug(u)\geq 0$. Our
approach is a combination of Nonlinear Alternative of Leray-Schauder with
the properties of the associated vector field at the $(u,u')$ plane.
More precisely, we show that the solutions of the above boundary-value
problem remains away from the origin for the case where the nonlinearity is
sublinear and so we avoid its singularity at $u=0$.
\end{abstract}

\maketitle

\numberwithin{equation}{section} 
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma} 
\newtheorem{remark}[theorem]{Remar}
\newtheorem{corollary}[theorem]{Corollary} 
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction}

In this note we consider the nonlinear 3-point singular problem
\begin{equation}
-[\phi _{p}(u')]'=q(t)f(t,u(t)),\quad 0<t<1  \label{1.1}
\end{equation}
subject to
\begin{equation}
u(0)-g(u'(0))=0,\quad u(1) -\beta u(\eta )=0  \label{1.2}
\end{equation}
or to
\begin{equation}
u(0)-\alpha u(\eta )=0,\quad u(1)+g(u'(1))=0  \label{1.3}
\end{equation}
where $\phi _{p}(s)=|s|^{p-2}s$, $(p>1)$ is the well known $p$-Laplacian
operator, $0<\eta <1$; $0<\alpha ,\beta <1$ are fixed points and $g$ is a
monotone continuous function defined on the real line $\mathbb{R}$ with 
$g(0)=0$ and $ug(u)\geq 0$. The inhomogeneous term in \eqref{1.1} is allowed
to be singular at $u=0$ and $q(t)$ may be singular at $t=0$ or/and $t=1$ and
finally we suppose that $q(0)>0$. Further assumptions concerning the
nonlinearity $f(t,u)$ will be clarified later.

Let $B$ be the Banach space $C[0,1]$ endowed with the norm $\Vert x\Vert
=\max_{t\in [ 0,1]}|x(t)|$. A solution $u(t)$ of \eqref{1.1} subject
to \eqref{1.2} or \eqref{1.3} means that $u(t)\in C^{1}[0,1]$, is positive
on $(0,1)$, $\phi _{p}(u'(.))\in C(0,1)\cap L^{1}[0,1]$ and
satisfies the differential equation as well as the corresponding boundary
conditions. It is well known that when $p>1$, $\phi _{p}(s)$ is strictly
increasing on $\mathbb{R}$ and so its inverse $\phi _{p}^{-1}$ exists and
further $\phi _{p}^{-1}=\phi _{q}$, where $1/p+1/q=1$.

In \cite{EW}, Erbe and Wang by using Green's functions and Krasnoselskii's
fixed point theorem in cones proved existence of a positive solution of the
boundary-value problem studied the Sturm-Liouville boundary-value problem
\begin{gather*}
x^{\prime \prime }(t)=-f(t,x(t)), \\
\alpha x(0)-\beta x'(0)=0,\quad \gamma x(1)+\delta x'(1)=0,
\end{gather*}%
where $\alpha ,\beta ,\gamma ,\delta \geq 0$ and $\rho :=\beta \gamma
+\alpha \gamma +\alpha \delta >0$, mainly under the assumptions:
\begin{gather*}
f_{0}:=\lim_{x\rightarrow 0+}\max_{0\leq t\leq 1}\frac{f(t,x)}{x}=0, \\
f_{\infty }:=\lim_{x\rightarrow +\infty }\min_{0\leq t\leq 1}
\frac{f(t,x)}{x}=+\infty
\end{gather*}%
i.e., $f$ is \emph{supelinear} at both ends points $x=0$ and $x=\infty $ or
under
\begin{gather*}
f_{0}:=\lim_{x\rightarrow 0+}\min_{0\leq t\leq 1}\frac{f(t,x)}{x}=+\infty \\
f_{\infty }:=\lim_{x\rightarrow +\infty }\max_{0\leq t\leq 1}
\frac{f(t,x)}{x}=0,
\end{gather*}%
i.e., $f$ is \emph{sublinear} at both $x=0$ and $x=\infty $.

The study of multi-point boundary-value problems was initiated by Il'in and
Moiseev in \cite{im1,im2}. Many authors since then considered nonlinear
3-point boundary-value problems (see e.g., \cite{da, fe, fw, ghm, ek, ma1,
ma2, We, w2} and the references therein). In particular, Ma in \cite{ma2}
proved the existence of a positive solution to the three-point nonlinear
boundary-value problem
\begin{gather*}
-u''(t)=q(t)f(u(t)),\quad 0<t<1, \\
u(0)=0,\quad \alpha u(\eta )=u(1),
\end{gather*}
where $0<\alpha $, $0<\eta <1$ and $\alpha \eta <1$. The results of Ma were
complemented in the works of Webb \cite{w2}, Kaufmann \cite{ek}, Kaufmann
and Kosmatov \cite{kk}, and Kaufmann and Raffoul \cite{kar}.

Among the studies on semipositone multi-point boundary-value problems, we
mention the papers by Cao and Ma \cite{cm} and Liu \cite{liu}. Cao and Ma
considered the boundary-value problem
\begin{gather*}
-u''(t)=\lambda q(t)a(t)f(u(t),u'(t)),\quad 0<t<1, \\
u(0)=0,\quad \sum_{i=1}^{m-2}\alpha _{i}u(\eta _{i})=u(1)\,.
\end{gather*}
They applied the Leray-Schauder fixed point theorem to obtain an interval of
eigenvalues for which at least one positive solution exists. Liu applied a
fixed point index method to obtain such an interval for
\begin{gather*}
-u''(t)=\lambda q(t)a(t)f(u(t)),\quad 0<t<1, \\
u'(0)=0,\quad \beta u(\eta )=u(1).
\end{gather*}

In the above papers there are no assumptions for singularity of the
nonlilearity $f$ at the point $u=0$. Zhang and Wang \cite{ZW} and recently
Liu \cite{liu1} obtained some existence results for a singular nonlinear
second order 3-point boundary-value problem, for the case when only
singularity of $q(t)$ at $t=0$ or $t=1$ is permitted. Also recently, by
using the method of fixed point index, Xu \cite{Xu} studied the problem
\begin{equation*}
-u^{\prime \prime }(t)=f(u(t)),\quad 0<t<1,\quad u(0)=0,\quad \alpha u(\eta
)=u(1),
\end{equation*}%
where $f(t,u)$ is allowed to have singularity at $u=0$. Other applications
of Krasnosel'ski\u{\i}'s fixed point theorem to semipositone problems can,
for example, be found in \cite{ag}. Further recently interesting results
have been proved in \cite{BGG}, \cite{HG} or \cite{liu1}.

Finally, Ma and Ge in \cite{MG}, proved the existence of a positive solution
of the 3-point singular boundary-value problem \eqref{1.1}-\eqref{1.2} under
the following assumptions:

\begin{itemize}
\item[(H1)] $q(t)\in C(0,1)\cap L^{1}[0,1]$ with $q(t)\geq 0$ and
nondecreasing on $(0,1)$

\item[(H2)] $f\in C([0,1]\times (0,+\infty),(0,+\infty ))$

\item[(H3)] $0\leq f(t,y)\leq f_{1}(y)+f_{2}(y)$ on $[0,1]\times (0,+\infty
) $ with $f_{1}>0$ continuous non-increasing on $(0,+\infty )$ and $%
\int_{0}^{L}f_{1}(u)du<+\infty $ for any fixed $L>0$; $f_{2}\geq 0$ and
continuous on $[0,+\infty )$

\item[(H4)] For any $K>0$ there exists $\psi _{K}(t):(0,1)\to (0,+\infty )$
such that $f(t,y)\geq \psi _{K}(t),\ t\in (0,1)$ for any $y(t)\in C[0,1]$
with $0\leq y(t)\leq K$

\item[(H5)] $\int_{0}^{\eta }\phi _{p}^{-1}(\int_{s}^{\eta }q(r)\psi
(r)dr)ds>0$, $\int_{\eta }^{1}\phi _{p}^{-1}(\int_{\eta }^{s}q(r)\psi (r)
dr)ds>0$ and for any $k_{1}>0$ and $k_{2}>0$, $\int_{0}^{%
\eta}f_{1}(k_{1}s)q(s)ds +\int_{\eta }^{1}f_{1}(k_{2}(1-s))q(s)ds<+\infty $
and mainly
\begin{equation*}
\sup_{c>0}\frac{c}{\phi _{p}^{-1}(I^{-1}[G_{0}(c) ])(\frac{p}{p-1})^{\frac{1%
}{p}}[\frac{1}{ 1-\beta }(\int_{0}^{\eta }[q(s)]^{\frac{1}{p} }ds+\int_{\eta
}^{1}[q(s)]^{\frac{1}{p}}ds) ]}>1,
\end{equation*}
where
\begin{equation*}
I(c):=\int_{0}^{c}\phi _{p}^{-1}(z)dz=\frac{p-1}{p} c^{\frac{p}{p-1}}\text{
\ \ and \ }G_{0}(c) =\int_{0}^{c}[f_{1}(u)+f_{2}(u)]du\text{\ }.
\end{equation*}
\end{itemize}

It is not difficult to prove the next useful properties of $I(c)$:
\begin{equation*}
I^{-1}(uv)\leq I^{-1}(u)I^{-1}(v) \quad\text{for }u\geq 0,v\geq 0
\end{equation*}
and whenever $c<0$, we have $I(-c)=I(c)$. Indeed, since $-c>0$,
\begin{equation*}
I(-c)=\int_{0}^{-c}\phi _{p}^{-1}(z) dz=\int_{0}^{-c}\phi _{q}(z)
dz=\int_{0}^{-c}|z|^{q-2}zdz=\int_{0}^{-c}z^{q-1}dz=\frac{(-c) ^{q}}{q}
\end{equation*}
and
\begin{equation*}
I(c) =\int_{0}^{c}|z|^{q-2}zdz=\int_{0}^{c}(-z)^{q-2}zdz=-
\int_{0}^{c}(-z)^{q-1}d(-z)=\frac{(-c)^{q}}{q}.
\end{equation*}

In this work, mainly motivated by the above mentioned paper of Ma and Ge
\cite{MG}, we combine the properties of the vector field at the face 
$(u,u')$ plane and sublinearity of $f(t,u)$ at the origin $u=0$ with
the alternative continuation principle of Leray-Shauder, proving the
existence of a positive solution for the boundary-value problem 
\eqref{1.1}-\eqref{1.3} and eliminating several of the assumptions (H1)-(H5).

\section{Preliminaries}

We now proceed with the auxiliaries. Consider the boundary-value problem
\begin{gather}
-[\phi _{p}(u')]'=q(t)f(t,u(t)),\quad 0<t<1,  \label{2.1}
\\
\ u(0)-g(u'(0))=0,\quad \beta u(\eta )=u(1)  \label{2.2}
\end{gather}
and give two concept-assumptions as follows:
\begin{equation}
f_{0}:=\lim_{u\rightarrow 0}\max_{0\leq t\leq 1}\frac{f(t,u)}{u}=+\infty
\label{A2S}
\end{equation}
i.e., $f$ is \emph{sublinear} at the end point $0$, and
\begin{equation}
\lim_{u\rightarrow 0}\frac{g^{-1}(u)}{u}=\mu \in [ 0,\frac{1}{2}).
\label{G}
\end{equation}
We further consider that :

\begin{itemize}
\item[(A1)] $q(t)\in C(0,1)\cap L^{1}[0,1]$ with $q(t)>0$ and nondecreasing
on $(0,1)$

\item[(A2)] $f\in C([0,1]\times (0,+\infty),(0,+\infty ))$

\item[(A3)] $\int_{0}^{L}\max_{0\leq t\leq 1}f(t,u)du<+\infty $, for any
fixed $L>0$.

\item[(A4)] $g\in C(\mathbb{R},\mathbb{R})$, is a nondecreasing function
with $ug(u)>0$, $u\neq 0$.
\end{itemize}

\begin{remark} \label{rmk1} \rm
Note that the differential equation \eqref{2.1} defines a vector
field whose properties will be crucial for our study. More specifically,
working at the $(u,u')$ \textit{face semi-plane} $(u>0)$,
the sign condition on $f$ (see assumption (A2)), immediately gives
(since $\phi _{p}'(u')>0$ for all $u'\in \mathbb{R}$) that
$u''<0$. Thus, any trajectory $(u(t),u'(t))$, $t\geq 0$,
emanating from the semi-line
\begin{equation*}
E:=\{(u,u'):u-g(u')=0,\quad \;u>0\}
\end{equation*}
``trends'' in a natural way, (when $u'(t)>0$) toward the positive
$u$-semi-axis and then (when $u'(t)<0$) turns toward the negative
$u'$-semi-axis. Finally, by setting a certain growth rate on
$f$ (say sublinearity) we can control the vector field, so that
all trajectories with $u(0)$ small enough satisfy the
relation
\begin{equation*}
u(1)-\beta u(\eta )\neq 0.
\end{equation*}
So, all solutions of the given boundary-value problem cannot have their
initial values arbitrary small, avoiding in this way the singular point
$u=0$ of the nonlinearity.
\end{remark}

Namely we have the next result.

\begin{lemma} \label{Le1}
Let $0<\beta <1$. If $u\in C[0,1]$ is a solution of
\eqref{2.1}-\eqref{2.2}, then $u=u(t)$ is concave.
Furthermore for every solution with $u(0)>0$, it follows that
\begin{itemize}
\item[(i)] There exists a $t_{0}\in [0,1)$ such that
$u(t_{0})=\max_{0\leq t\leq 1}\|u(t)\|=\|u\|$,

\item[(ii)] $u(t)>0$, $t\in [0,1]$ and

\item[(iii)]
$\inf_{t\in [\eta ,1]}u(t)\geq \gamma u(t_{0})=\gamma \|u\|$,
where $\gamma =\min \{ \beta \eta ,\;\frac{\beta (1-\eta )
}{1-\beta \eta }\} $.
\end{itemize}
\end{lemma}

\begin{proof}
Let $u(t)$ be a solution to \eqref{2.1}-\eqref{2.2}. Then, since $[\phi
_{p}(u')]'=-q(t)f(t,u(t))\leq 0$, $\phi _{p}(u')$
is non-increasing. Consequently $u'(t)$ is non-increasing which
implies the concavity of $u(t)$.

(i) Since $u(0)>0$, by the first condition in \eqref{2.2} and the assumption
(A4), we get $u'(0)>0$. If $u'(t)\geq 0$, $t\in [0,1]$ then $%
u(1)\geq u(\eta )>\beta u(\eta )$, a contradiction. Hence there exists 
$t_{0}>0$ such that $u(t_{0}) =\max_{0\leq t\leq 1}\|u(t)\|=\|u\|$.

(ii) If $\eta \in (0,t_{0})$, then $u(\eta )>u(0)>0$ and so $u(1)=\beta
u(\eta )>0$. If $\eta \in (t_{0},1)$, then $u(\eta )>u(1)$ and so
\begin{equation*}
0=u(1)-\beta u(\eta )<u(1)-\beta u(1)
\end{equation*}
and hence $u(1)>0$. Finally, the concavity of $u(t)$ yields $u(t)>0$, 
$t\in [0,1]$.

(iii) The proof follows the concavity of the solution. Indeed, since 
$u(1)=\beta u(\eta )<u(\eta )$, let first consider the case 
$t_{0}\leq \eta <1 $. Then,
\begin{equation*}
\min_{t\in [ \eta ,1]}u(t)=u(1).
\end{equation*}%
Furthermore, we have
\begin{equation*}
u(t_{0})\leq u(1)+\frac{u(\eta )-u(1)}{1-\eta }=u(1)\{1-\frac{1
-\frac{1}{\beta }}{1-\eta }\}=u(1)\frac{1-\beta \eta }{\beta (1-\eta )}.
\end{equation*}
Consequently,
\begin{equation*}
\min_{t\in [ \eta ,1]}u(t)\geq \frac{\beta (1-\eta )}{1-\beta \eta }
\Vert u\Vert .
\end{equation*}
Let us now assume that $\eta <t_{0}<1$. Since $u(\eta )>u(1)$, we have again
\begin{equation*}
\min_{t\in [ \eta ,1]}u(t)=u(1).
\end{equation*}%
>From the concavity of $u$, we know that
\begin{equation*}
\frac{u(\eta )}{\eta }\geq \frac{u(t_{0})}{t_{0}}.
\end{equation*}%
Combining the above and the boundary condition $u(1)=\beta u(\eta )$, we
conclude that
\begin{equation*}
\frac{u(\eta )}{\beta \eta }\geq \frac{u(t_{0})}{t_{0}}\geq u(t_{0})=\Vert
u\Vert ,
\end{equation*}
that is $\min_{t\in [ \eta ,1]}u(t)\geq \beta \eta \Vert u\Vert $.
\end{proof}

\section{Existence for the first boundary-value problem}

In this section we consider the boundary-value problem 
\eqref{2.1}-\eqref{2.2} and prove the next result.

\begin{lemma}
Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then, there exists an
$\eta _{0}>0$ such that for any $\eta \leq \eta _{0}$ any solution of
\eqref{2.1} with $u(0)=\frac{\eta }{2}$, satisfies the inequality
\begin{equation}
0<u(t)\leq \eta \leq \eta _{0},\quad t\in [0,1]. \label{2.4}
\end{equation}
\end{lemma}

\begin{proof}
By assumption \eqref{A2S} it follows that for any $K>0$\ there exists $\eta
_{0}$ such that
\begin{equation}
\min_{0\leq t\leq 1}f(t,u)>Ku,\;0<u\leq \eta _{0}.  \label{2.400}
\end{equation}%
Let $K>\max \{2\mu ^{2},\;2\frac{1+2\mu }{\min \{\gamma ,1\}}\}$. We examine
first the case $p>2$. Taking into account\ \eqref{G}, we may chose $\eta
_{0} $ small enough so that
\begin{equation}
\frac{g^{-1}(\frac{\eta _{0}}{2})}{\frac{\eta _{0}}{2}}\leq 2\mu \quad 
\text{and}\quad (p-1)\frac{[g^{-1}(\eta )]^{p-2}}{\min_{0\leq t\leq 1}q(t)}
<1,\quad \eta \in (0,\eta _{0}]  \label{2.40}
\end{equation}
(if $\lim_{u\rightarrow 0}\frac{g^{-1}(u)}{u}=0$, then we may find $\mu >0$
so that \eqref{2.40} still holds true).

Assume that the boundary-value problem \eqref{2.1}-\eqref{2.2} has a
solution $u(t)$, $t\in [0,1]$ with initial value $u(0)$ arbitrary small.
Then, we may assume that $u(0)=\eta /2$ for some $\eta \in (0,\eta _{0}]$
with
\begin{equation*}
\frac{\min_{0\leq t\leq 1}q(t)}{(p-1)[ g^{-1}(\frac{\eta }{2})]^{p-2}}\geq 1.
\end{equation*}

We demonstrate first that \eqref{2.4} holds true. If not, by Lemma 
\ref{Le1}, there exist $t^{\ast }\in (0,1]$ such that 
$\frac{\eta }{2}\leq u(t)<\eta $, $0\leq t<t^{\ast }$ and 
$u(t^{\ast })=\eta $. Then by \eqref{2.400} it
follows that
\begin{equation*}
[ \phi _{p}(u')]'=[\phi _{p}'(u^{\prime
})]u^{\prime \prime }=-q(t)f(t,u(t))\leq -Kq(t)u(t)\leq -Kq(t)\frac{\eta }{2},
\end{equation*}%
i.e.,
\begin{align*}
u^{\prime \prime }(t)& \leq -Kq(t)\frac{\eta }{2}\frac{1}{[\phi _{p}'(u')]} \\
& \leq -K\frac{\eta _{0}}{2}\frac{1}{[\phi _{p}'(u')]}%
\min_{0\leq t\leq 1}q(t) \\
& \leq -K\frac{\eta }{2}\frac{\min_{0\leq t\leq 1}q(t)}{(p-1)[g^{-1}(\frac{%
\eta }{2})]^{p-2}}\leq -K\frac{\eta }{2}
\end{align*}%
Consequently, by \eqref{2.40} and the Taylor formula, we get that for some 
$t\in [ 0,t^{\ast }]$,
\begin{align*}
\eta =u(t^{\ast })& \leq \frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2})+%
\frac{(t^{\ast })^{2}}{2}u^{\prime \prime }(t) \\
& \leq \frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2})-\frac{(t^{\ast })^{2}
}{2}K\frac{\eta }{2} \\
& \leq \frac{\eta }{2}+t^{\ast }2\mu \frac{\eta }{2}-\frac{(t^{\ast })^{2}}{2}
K\frac{\eta }{2}.
\end{align*}
Considering now the map
\begin{equation*}
\phi (t^{\ast }):=Kt^{\ast }{}^{2}-4\mu t^{\ast }+2,
\end{equation*}%
the above inequality yields $\phi (t^{\ast })\leq 0$. This is a
contradiction, since the above choice of $K>2\mu ^{2}$, yields $\phi (t)>0$
for all $t\in [ 0,1]$. As a result, noticing Lemma \ref{Le1}, we
obtain $0<u(t)\leq \eta _{0},$ $t\in [ 0,1]$.

If now $p\leq 2$, then since $\lim_{u\rightarrow 0}g^{-1}(u)=0$, we easily
get $u^{\prime \prime }(t)\leq 0$. As a result, $\frac{\eta }{2}<t^{\ast
}g^{-1}(\frac{\eta }{2})$ and thus $t^{\ast }>\frac{1}{2\mu}>1$ 
(in view of \eqref{G}), a contradiction due to the initial choice of 
$t^{\ast }\in [ 0,1]$.
\end{proof}

\begin{lemma} \label{lem3}
Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then for any $\eta \in
(0,\eta _{0})$\ ($\eta _{0}$ as above) there exists an $\alpha _{0}=\alpha
_{0}(\eta )>0$ such that for any (possible) solution
$u=u(t)$ of \eqref{2.1}-\eqref{2.2}, with $u(0)=\frac{\eta }{2}$, the
following inequality holds:
\[
u(t)\geq \alpha _{0},\quad t\in [0,1]\,.
\]
\end{lemma}

\begin{proof}
By the concavity of $u(t)$, it is obvious that
\begin{equation*}
\min_{t\in [0,1]}u(t)=\min \{ u(0),\;u(1)\} .
\end{equation*}
However, in view of Lemma \ref{Le1}, $\min_{t\in [\eta ,1]}u(t)\geq \gamma
u(t_{0})=\gamma \|u\|\geq \gamma u(0)=\gamma \frac{\eta }{2}$ and so
\begin{equation*}
u(t)\geq \min_{t\in [0,1]}u(t)\geq \frac{\eta }{2}\min \{ 1,\gamma \}
:=\alpha _{0}(\eta).
\end{equation*}
\end{proof}

\begin{proposition} \label{Pr1}
Suppose that conditions \eqref{A2S}-\eqref{G} hold. Then, there
exists an $\eta _{0}^{\ast }>0$ such that any solution of
\eqref{2.1}-\eqref{2.2} satisfies the inequality $u(0)\geq \eta _{0}^{\ast }$
and so, by the previous Lemma,
\begin{equation*}
u(t)\geq \alpha _{0}(\eta _{0}^{\ast }):=\alpha
_{0}^{\ast },\quad t\in [0,1],
\end{equation*}
$\alpha _{0}^{\ast }$ being a positive constant.
\end{proposition}

\begin{proof}
Supposing the opposite, we may find a solution $u(t)$ of 
\eqref{2.1}-\eqref{2.2}, such that $u(0)=\frac{\eta }{2}$, $\eta $
 being the same as in 
\eqref{2.4}, i.e. $\eta $ is arbitrarily small.

Then, noticing \eqref{2.400}, as above by Taylor formula we can find a $t\in
[0,1]$ such that
\begin{align*}
u(1)-\beta u(\eta ) & \leq \frac{\eta }{2} +g^{-1}(\frac{\eta }{2})
+\frac{1}{2}u''(t)-\beta u(\eta ) \\
& <\frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{1}{2}Ku(t) \\
& \leq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\min \{ \alpha
_{0},\frac{\eta }{2}\} \\
& \leq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2} \frac{\eta }{2}
\min \{ 1,\gamma \} <0,
\end{align*}
due to the choice $K>2\frac{1+2\mu }{\min \{ \gamma ,1\} }$. This
contradiction completes the proof.
\end{proof}

We give now an existence principle, which is crucial for the proof of our
results.

\begin{lemma}
\label{Le4}(Nonlinear Alternative of Leray-Shauder Type)
\cite{AOW} Let $V$ be a Banach space and $C\subset V$ a convex
set. Assume that $U$ is a relative open subset of $C$ with
$u_{0}\in U$ and $T:\bar{U}\rightarrow C$ a\ completely
continuous\ (continuous and compact) map. Then either

(I)\qquad $T$ has a fixed point, or

(II)\qquad there exists $u\in \partial U$ and $\lambda \in \left(
0,1\right) $ with $u=\lambda T\left( u\right) +\left( 1-\lambda
\right) u_{0}$.
\end{lemma}

\begin{theorem} \label{Th1}
Assume (A1)-(A4) hold and
\begin{equation}
\sup_{c>0,\;0\leq t\leq 1}\frac{c^{p}}{\int_{0}^{c}f(t,u)du}>
\frac{p}{p-1}\Big[\frac{1}{1-\beta }\int_{\eta }^{1}[q(t)
]^{1/p}dt+\int_{0}^{\eta }[q(t)]^{1/p}dt\Big]^{p}.
\label{2.70}
\end{equation}
Then the 3-point boundary-value problem \eqref{2.1}-\eqref{2.2} has at least
a positive solution.
\end{theorem}

\begin{proof}
In order to show that \eqref{2.1}-\eqref{2.2} has a solution, we consider
the boundary-value problem
\begin{equation}
\begin{gathered} -[\phi _{p}(u')]'=q(t)F(t,u(t)),\quad 0<t<1,\\
u(0)-g(u'(0))=0,\quad u(1) -\beta u(\eta )=0, \end{gathered}  \label{2.8}
\end{equation}%
where
\begin{equation*}
F(t,u(t))=
\begin{cases}
f(t,u), & u\geq \alpha _{0}^{\ast } \\
f(t,\alpha _{0}^{\ast }), & u<\alpha _{0}^{\ast }
\end{cases}
\end{equation*}
and $\alpha _{0}^{\ast }$ is given in Proposition \ref{Pr1}. Then clearly 
$F\in C([0,1]\times [ 0,+\infty ),[0,+\infty ))$. Consider further the
family of problems
\begin{equation}
\begin{gathered} -[\phi _{p}(u')]'=q(t)\lambda F(t,u(t)),\quad 0<t<1,\
0<\lambda <1 \\ 
u(0)-g(u'(0))=0,\quad u(1) -\beta u(\eta )=0. \end{gathered}
\label{2.9}
\end{equation}
If $u=u(t)$, $t\in [ 0,1]$ is a solution of \eqref{2.9}, then again by
Proposition \ref{Pr1}, $u(t)\geq \alpha _{0}^{\ast }$, $t\in [ 0,1]$.
We are going to prove the existence of another constant $A_{0}^{\ast
}>\alpha _{0}^{\ast }$ such that $u(0)\leq A_{0}^{\ast }$. Indeed, setting $%
(u(0),u'(0))=(u_{0},u_{0}')\in E$ and since
\begin{gather*}
u'(t)=\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda
\int_{0}^{t}q(s)F(s,u(s))ds], \\
u(t)=u_{0}+\int_{0}^{t}\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda
\int_{0}^{t}q(s)F(s,u(s))ds]dt,
\end{gather*}%
the initial values must be chosen so that
\begin{equation}
\begin{aligned} Q(u_{0}') & :=u(1)-\beta u(\eta ) \\ &
=g(u_{0}')+\int_{0}^{1}\phi _{p}^{-1}[\phi _{p}(u_{0}')-\lambda
\int_{0}^{t}q(s) F(s,u(s))ds]dt \\ & -\beta [g(u_{0}')+\int_{0}^{\eta }\phi
_{p}^{-1}[\phi _{p}(u_{0}')-\lambda \int_{0}^{t}q(s)F(s,u(s))ds]dt]=0.
\end{aligned}  \label{2.12}
\end{equation}%
By Proposition \ref{Pr1} and its proof, there is an $\eta >0$ such that 
$Q(g^{-1}(\frac{\eta }{2}))<0$, and moreover by the definition of $Q$,
\begin{equation*}
Q\big\{\phi _{p}^{-1}\big(\lambda \int_{0}^{t}q(s)F(s,u(s))ds\big)\big\}>0.
\end{equation*}
Hence $u_{0}'$ is upper bounded and similarly 
$u_{0}=g(u_{0}')$, i.e. $(u_{0},u_{0}')\in E_{0}\subset E$, $E_{0}$ being
a compact subset of $\mathbb{R}^{2}$.

We consider now the Banach space $B=C[0,1]$ and for any $x\in B, $ let 
$u=u(t)$ be a solution of the boundary-value problem
\begin{gather*}
-[\phi _{p}(u')]'=q(t)F(t,x(t)),\quad 0<t<1, \\
u(0)-g(u'(0))=0,\quad u(1)-\beta u(\eta )=0.
\end{gather*}
By the monotonicity of functions $g$ and $Q$ for each solution $u(t)$ of
this boundary-value problem, its initial value $(u_{0},u_{0}')$ is
uniquely determined and furthermore the map
\begin{equation*}
x\to \phi _{p}[g^{-1}(u_{0})]
\end{equation*}
is continuous (see \cite{MG}).

Consider now the operator
\begin{equation*}
T_{\lambda }x(t)=u_{0}+\int_{0}^{t}\phi _{p}^{-1}\big[\phi _{p}[g^{-1}(u_{0})
\big]-\lambda \int_{0}^{s}q(r)F(r,x(r))dr]ds,\ x\in B,
\end{equation*}%
where $u_{0}$ is the unique constant corresponding to function $x(t)$ and
satisfying \eqref{2.12}. It is easily verified that $u(t)$ is a solution to 
\eqref{2.9} if and only if $u$ is a fixed point of $T_{1}$ in $C[0,1]$.

(I) We shall prove that $T=T_{1}:B\to B$ is completely continuous. By
continuity of the map $x\to \phi _{p}[g^{-1}(u_{0})]\ $it is not difficult
to be proved that $T$ is continuous. So we must only prove that $T$ is
compact i.e. it maps every bounded subset of $B$ into a relatively compact
set. Consider the closed ball $\Sigma =\{ x\in B:\|x\|\leq R\} $. Since 
$\alpha _{0}^{\ast }\leq u(0)\leq A_{0}^{\ast }$,
\begin{align*}
\|(Tx)(t)\| &\leq A_{0}^{\ast }+\phi _{p}^{-1}[\phi_{p}(g^{-1}(A_{0}^{\ast
})) +\max_{x\in \Sigma , s\in [0,1]}F(s,x) \int_{0}^{t}q(s)ds] \\
&\leq A_{0}^{\ast}+\phi _{p}^{-1}(M_{1})
\end{align*}
and
\begin{equation*}
\|(Tx)'(t)\|=\phi _{p}^{-1}[\phi _{p}( g^{-1}(A_{0}^{\ast
}))+\int_{0}^{t}q(s)F(s,x(s))ds] \leq \phi _{p}^{-1}(M_{1}).
\end{equation*}
Hence the Arzela-Ascoli Theorem guarantees the compactness of $T$.

(II) We will show that there exists a $M>0$ such that $\Vert u\Vert \leq M$
for any solution of \eqref{2.9}. We set
\begin{equation*}
G(c)=\int_{0}^{c}\max_{0\leq t\leq 1}f(t,u)ds
\end{equation*}%
Noting \eqref{2.70}, we may indeed find a $M>0$ such that
\begin{equation*}
\frac{M}{\Big(\int_{0}^{M}\sup_{\;0\leq t\leq 1}f(t,u)du\Big)^{\frac{1}{p}}(%
\frac{p}{p-1})^{\frac{1}{p}}\Big[\frac{1}{1-\beta }\int_{\eta
}^{1}[q(t)]^{1/p}dt+\int_{0}^{\eta }[q(t)]^{1/p}dt\Big]}>1.
\end{equation*}%
Also by Proposition \ref{Pr1}, any solution of the boundary-value problem 
\eqref{2.9} is convex and there exists a point $t_{0}\in (0,1)$ such that
\begin{equation*}
u'(t)\geq 0,\;t\in [ 0,t_{0}),\ u'(t_{0})=0\quad
\text{and}\quad u'(t)\leq 0,\;t\in (t_{0},1].
\end{equation*}%
Working in the interval $[t_{0},t]\subset [ t_{0},1]$, we have
\begin{equation*}
0\leq -(\phi _{p}(u'))'=\lambda q(t)F(t,u)\leq
q(t)\max_{0\leq t\leq 1}f(t,u).
\end{equation*}
Multiplying by $-u'>0$, we get
\begin{equation*}
(\phi _{p}(u'))'\phi _{p}^{-1}(\phi _{p}(u'))\leq
q(t)\max_{0\leq t\leq 1}f(t,u),\quad t\in [ t_{0},1]
\end{equation*}
and then integrating on $[t_{0},t]$, we obtain
\begin{align*}
\int_{0}^{\phi _{p}(u'(t))}\phi _{p}^{-1}(u'(t))u^{\prime
}(t)dt& \leq q(t)\int_{u(t)}^{u(t_{0})}\max_{0\leq t\leq 1}f(t,u)du \\
& \leq q(t)\int_{0}^{u(t_{0})}\max_{0\leq t\leq 1}f(t,u)du=q(t)G(u(t_{0})),
\end{align*}
hence
\begin{equation*}
I(-\phi _{p}(u'(t)))=I(\phi _{p}(u'(t)))\leq
q(t)G(u(t_{0}))
\end{equation*}
and so
\begin{equation}
0\leq -u'(t)\leq \phi
_{p}^{-1}\{[I^{-1}(q(t))]I^{-1}(G(u(t_{0})))\},\quad t\in [ t_{0},t].
\label{2.11}
\end{equation}
If $\eta \in (t_{0},1]$, an integration over $[\eta ,1]$ yields
\begin{equation*}
u(\eta )-u(1)\leq \phi _{p}^{-1}\{I^{-1}[(G(u(t_{0})))]\}\int_{\eta
}^{1}\phi _{p}^{-1}\{I^{-1}((q(t))\}.
\end{equation*}
If $\eta \in (0,t_{0}]$, we integrate over $[t_{0},1]$ to obtain
\begin{align*}
u(t_{0})-u(1)& \leq \phi
_{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{t_{0}}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt.
\end{align*}
Then clearly it follows that
\begin{equation*}
u(\eta )-u(1)\leq u(t_{0})-u(1)\leq \phi
_{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt.
\end{equation*}
Moreover, since $u(1)=\beta u(\eta )$, we get
\begin{equation*}
u(1)\leq \frac{\beta }{1-\beta }\phi
_{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt
\end{equation*}%
and so a new integration from $t_{0}$ to $1$ of (\ref{2.11}) yields
\begin{align*}
u(t_{0})& =u(1)+\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{t_{0}}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \leq u(1)+\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\int_{0}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}[\frac{\beta }{1-\beta }
\int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt+\int_{0}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt] \\
& \leq \phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}\big[\frac{1}{1-\beta }
\int_{\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))\big]dt+\int_{0}^{\eta }\phi
_{p}^{-1}[I^{-1}(q(t))]dt] \\
& =\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}(\frac{p}{p-1})^{\frac{1}{p}}\big[
\frac{1}{1-\beta }\int_{\eta }^{1}q^{\frac{1}{p}}(t)dt+\int_{0}^{\eta }q^{
\frac{1}{p}}(t)dt\big].
\end{align*}
Consequently
\begin{equation*}
\frac{u(t_{0})}{\phi _{p}^{-1}\{I^{-1}(G(u(t_{0})))\}(\frac{p}{p-1})^{\frac{1
}{p}}\big[\frac{1}{1-\beta }\int_{\eta }^{1}q^{\frac{1}{p}
}(t)dt+\int_{0}^{\eta }q^{\frac{1}{p}}(t)dt\big]}<1
\end{equation*}%
which by the assumption \eqref{2.70} implies that $u(t_{0})<M$. Finally in
view of Lemma \ref{Le4}, we may set
\begin{equation*}
C:=\{u\in B=C[0,1]:\Vert u\Vert \leq M\}\text{ \ and }U:=\{u\in C:\Vert
u\Vert <M\}.
\end{equation*}
Then, the second part of the nonlinear Alternative of Leray-Shauder Type is
ruled out and so we conclude that there exists a fixed point of the operator
\begin{equation*}
Tx(t)=T_{1}x(t)=u_{0}+\int_{0}^{t}\phi _{p}^{-1}\Big[\phi
_{p}[g^{-1}(u_{0})]-\int_{0}^{s}q(r)F(r,x(r))dr\Big]ds.
\end{equation*}
This of course yields a solution $u=u(t)$ of \eqref{2.8} and noting
Proposition \ref{Pr1} and the definition of the modification $F,u(t)$ is
actually a solution of our original boundary-value problem 
\eqref{2.1}-\eqref{2.2}.
\end{proof}

\section{Existence for the second boundary-value problem}

In the following we will study the boundary-value problem 
\eqref{1.1}-\eqref{1.3}. For this purpose, we give the next result concerning the
boundary-value problem
\begin{gather}
-[\phi _{p}(y'(s))]'=q(s)f(s,y(s)),\quad 0<t<1,
\label{2.13} \\
y(0)-g(y'(0))=0,\quad y(1)-\alpha y(1-\eta )=0\,.  \label{2.14}
\end{gather}

\begin{theorem} \label{Th2}
Assume that (A1), (A3) and (A4) hold. Instead of
(A2) we assume
\begin{itemize}
\item[(A2$^{\ast }$)] $f\in C([0,1]\times (-\infty ,0),(-\infty ,0))$.
\end{itemize}
Also assume that
\begin{equation}
\sup_{c<0,\;0\leq t\leq 1}\frac{(-c)^{p}}{\int_{c}^{0}f(t,u)du}>
\frac{p}{p-1}\Big[\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)
]^{1/p}dt+\int_{0}^{1-\eta }[q(t)]^{1/p}dt\Big]^{p}.
\label{2.1400}
\end{equation}
Then, the 3-point boundary-value problem \eqref{2.13}-\eqref{2.14} has at
least one negative solution.
\end{theorem}

To prove the above Theorem, we give some lemmas symmetrical to the previous
case, with similar proofs which are partly omitted.

Note that the differential equation \eqref{2.13} defines also a vector
field. So if we focus on the $(y,y')$ face semi-plane $(y<0)$, then
by \eqref{2.1400}, we see that $y^{\prime \prime }>0$. Thus, any trajectory 
$(y(t),y'(t))$, $t\geq 0$, emanating from the semi-line
\begin{equation*}
E^{\ast }:=\{(y,y'):y-g(y')=0,\quad \;y<0\}
\end{equation*}%
\textquotedblleft trends\textquotedblright\ in a natural way, (when 
$y'(t)<0$) toward the negative $y$-semi-axis and then (when 
$y'(t)>0$) trends toward the positive $y'$-semi-axis. As a
result, we may control the vector field, so that $y(1)+\alpha y(\eta )=0$.

\begin{lemma}\label{Le11}
Let $0<\alpha <1$. If $y\in C[0,1]$ is a solution
of the boundary-value problem \eqref{2.13}-\eqref{2.14}, then $y$ is convex.
Furthermore for every solution with $y(0)<0$, it follows that
\begin{itemize}
\item[(i)] $y(t)<0$, $t\in [0,1]$

\item[(ii)] There exists a $t_{0}\in [0,1)$ such that
\begin{equation*}
\sup_{t\in [\eta ,1]}y(t)\leq \delta y(t_{0})=-\delta \|y\|,
\end{equation*}
where $\delta =\min \{ \alpha (1-\eta ),\;\frac{\alpha (1-\eta
)}{1-\alpha (1-\eta )}\}$

\item[(iii)] There exists a $t_{0}\in [0,1)$ such that $y(
t_{0})=-\max_{0\leq t\leq 1}y(t)=-\|y\|$
\end{itemize}
\end{lemma}

\begin{proof}
Let $y(t)$ be a solution of \eqref{2.13}-\eqref{2.14}. Then since,
\begin{equation*}
[ \phi _{p}(y')]'=-q(t)f(t,y(t))\geq 0,
\end{equation*}
$[\phi _{p}(y')]$ is nondecreasing and so is $y'(t)$, a fact
that implies the convexity of $y(t)$.

(i) Since $y(0)<0$, by the first condition in \eqref{2.14} we get 
$y'(0)<0$. If $y'(t)\leq 0$, $t\in [ 0,1]$, then $y(1)\leq
y(\eta )<\alpha y(\eta )$, a contradiction. Hence, there exists a $t_{0}>0$
such that $y(t_{0})=\min_{0\leq t\leq 1}y(t)=-\Vert y\Vert $.

(ii) If $1-\eta \in (0,t_{0})$, then $y(1-\eta )<y(0)<0$ and so $y(1)=\alpha
y(1-\eta )<0$. If $1-\eta \in (t_{0},1)$ then $y(1-\eta )<y(1)$ and so
\begin{equation*}
0=y(1)-\alpha y(1-\eta )>y(1)-\alpha y(1).
\end{equation*}
Hence, $y(1)<0$. Finally, the convexity of $y(t)$ yields $y(t)<0$, $t\in
[ 0,1]$.

(iii) The proof follows by the convexity of the solution and since it is
analogous to the given one at Lemma \ref{Le1}, we omit it.
\end{proof}

\begin{lemma} \label{lem6}
Suppose that conditions \eqref{A2S} and \eqref{G} hold. Then, there exists an
$\eta _{0}<0$ such that for any $\eta \in (\eta _{0},0)$, any solution of
\eqref{2.13} with $y(0)=\eta / 2$, satisfies the
inequality
\begin{equation*}
\eta _{0}\leq \eta \leq y(t)<0,\quad t\in [0,1].
\end{equation*}
Furthermore, there exists an $\alpha _{0}=\alpha _{0}(\eta )<0$
such that any (possible) solution $y=y(t)$ of
\eqref{2.13}-\eqref{2.14} with $y(0)=\frac{\eta }{2}$, satisfies the
inequality
\begin{equation}
y(t)\leq \alpha _{0}(\eta ),\quad t\in [0,1].  \label{2.510}
\end{equation}
\end{lemma}

\begin{proof}
By the sublinearity of the function $f(t,y)$ at the point $y=0$, for every 
$K>\max \{2\mu ^{2},2\frac{1+2\mu }{\min \{\delta ,1\}}\}$ there exists an 
$\eta _{0}<0$ such that
\begin{equation}
\max_{0\leq t\leq 1}f(t,y)<Ky,\quad \eta _{0}\leq y<0.  \label{2.140}
\end{equation}%
Consider a solution $y=y(t)$\ of \eqref{2.13} with $y(0)=\frac{\eta }{2}$,
where $\eta \in (\eta _{0},0)$ is chosen small enough so that
\begin{equation*}
\frac{\min_{0\leq t\leq 1}q(t)}{(p-1)\eta ^{p-2}}>1.
\end{equation*}%
We shall prove first that $\eta \leq y(t)<0$, $t\in [ 0,1]$. If not,
by Lemma \ref{Le11}, there exists a $t^{\ast }\in (0,1]$ such that $\eta
\leq y(t)<\frac{\eta }{2}$, $0\leq t<t^{\ast }$ and $y(t^{\ast })=\eta $.
Then, by \eqref{2.140} follows that
\begin{equation*}
[ \phi _{p}(y')]'=[\phi _{p}'(y')]y^{\prime \prime }=-q(t)f(t,y)\geq -Kq(t)y(t)\geq -Kq(t)\frac{\eta }{2},
\end{equation*}
i.e.
\begin{align*}
y^{\prime \prime }(t)& \geq -Kq(t)\frac{\eta }{2}\frac{1}{[\phi _{p}'(y')]} \\
& \geq -K\frac{\eta }{2}\frac{1}{[\phi _{p}'(y')]}
\max_{0\leq t\leq 1}q(t) \\
& \geq -K\frac{\eta }{2}\frac{\max_{0\leq t\leq 1}q(t)}{M_{3}}\geq -K
\frac{\eta }{2}>0,
\end{align*}
where, noticing the monotonicity of $\phi _{p}^{\prime
}(s)=(p-1)(-s)^{p-2}>0 $, $s<0$ and of $y'(t)$, $0\leq t<1$,
\begin{equation*}
M_{3}=\max \big\{\phi _{p}'(y')=
\begin{cases}
\phi _{p}'(g^{-1}[\eta /2]), & \mbox{if }p>2 \\
\phi _{p}'(g^{-1}[\eta ]), & \mbox{if }p\in (1,2)%
\end{cases}
\big\}>0
\end{equation*}
Consequently, by \eqref{2.40} and Taylor's formula we conclude that for some
$t\in [ 0,t^{\ast }]$,
\begin{align*}
\eta =y(t^{\ast })& =\frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2})
+\frac{(t^{\ast })^{2}}{2}y^{\prime \prime }(t) \\
& \geq \frac{\eta }{2}+t^{\ast }g^{-1}(\frac{\eta }{2})-\frac{(t^{\ast })^{2}
}{2}K\frac{\eta }{2} \\
& \geq \frac{\eta }{2}+t^{\ast }2\mu \frac{\eta }{2}-\frac{(t^{\ast })^{2}}{2
}K\frac{\eta }{2}.
\end{align*}%
Considering now the map
\begin{equation*}
\phi (t^{\ast }):=Kt^{\ast }{}^{2}+4\mu t^{\ast }+2,
\end{equation*}%
the above inequality yields $\phi (t^{\ast })\leq 0$, given that $\eta <0$.
This is a contradiction, since the above choice of $K>2\mu ^{2}$, yields 
$\phi (t)>0$ for all $t\in [ 0,1]$. Consequently, noticing Lemma 
\ref{Le11}, we obtain $\eta \leq y(t)<0$, $t\in [ 0,1]$.

We proceed now with the proof of inequality \eqref{2.510}. By the convexity
of $y(t)$, it is obvious that
\begin{equation*}
\max_{t\in [ 0,1]}y(t)=\max \{y(0),\;y(1)\}.
\end{equation*}
However, in view of the same Lemma \ref{Le11},
\begin{equation*}
\sup_{t\in [ \eta ,1]}y(t)\leq \delta y(t_{0})=-\delta \Vert y\Vert
\leq \delta y(0)=\delta \frac{\eta }{2}
\end{equation*}
and so
\begin{equation*}
y(t)\leq \sup_{t\in [ \eta ,1]}y(t)\leq \frac{\eta }{2}\min \{1,\delta
\}:=\alpha _{0}(\eta )<0.
\end{equation*}
\end{proof}

\begin{proposition}\label{Pr11}
Suppose that conditions \eqref{A2S} and \eqref{G} hold. Then
there exists an $\eta _{0}^{\ast }<0$ such that any solution of
\eqref{2.13}-\eqref{2.14} satisfies the inequality
$y(0)\leq \eta_{0}^{\ast }$ and furthermore,
\begin{equation*}
y(t)\leq \alpha _{0}(\eta _{0}^{\ast }):=\alpha_{0}^{\ast },\quad t\in [0,1],
\end{equation*}
$\alpha _{0}^{\ast }$ being a negative constant.
\end{proposition}

\begin{proof}
Supposing the opposite, we may find a solution $y(t)$ of the\ boundary-value
problem \eqref{2.13}-\eqref{2.14}, such that $y(0)=\frac{\eta }{2}$, with 
$\eta \in (\eta _{0},0),\;\eta _{0}$ being the same as in the previous Lemma.
Then, noticing \eqref{2.140}, we may find a $t\in [ 0,1]$ such that
\begin{align*}
y(1)-\alpha y(1-\eta )& \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})+\frac{1%
}{2}y^{\prime \prime }(t)-\alpha y(1-\eta ) \\
& >\frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{1}{2}Ky(t) \\
& \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\min \{\alpha
_{0}(\eta ),\frac{\eta }{2}\} \\
& \geq \frac{\eta }{2}+g^{-1}(\frac{\eta }{2})-\frac{K}{2}\frac{\eta }{2}
\min \{1,\delta \}>0,
\end{align*}%
due to the choice $K>2\frac{1+2\mu }{\min \{\delta ,1\}}$ and since 
$y(1-\eta )<0$. This contradiction completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem]
\ref{Th2}. To show that \eqref{2.13}-\eqref{2.14} has a solution, we
consider the problem
\begin{equation}
\begin{gathered} -[\phi _{p}(y')]'=q(t)F(t,y(t)),\quad 0<t<1, \\
y(0)-g(y'(0))=0,\quad y(1)-\alpha y(1-\eta )=0, \end{gathered}  \label{220}
\end{equation}
where
\begin{equation*}
F(t,y(t))=
\begin{cases}
f(t,y), & \mbox{if }y\leq \alpha _{0}^{\ast } \\
f(t,f(t,\alpha _{0}^{\ast })), & \mbox{if }y>\alpha _{0}^{\ast }.
\end{cases}
\end{equation*}
Then, clearly $F\in C([0,1]\times (-\infty ,0],(-\infty ,0))$. Consider now
the family of problems
\begin{equation}
\begin{gathered} -[\phi _{p}(y')]'=q(t)\lambda F(t,y(t)),\quad 0<t<1,\;
0<\lambda <1 \\ y(0)-g(y'(0))=0,\quad y(1)-\alpha y(1-\eta )=0.
\end{gathered}  \label{221}
\end{equation}
Let $y=y(t)$, $t\in [ 0,1]$, be a solution of \eqref{221}. By
Proposition \ref{Pr11}, there is a (fixed) $\alpha _{0}^{\ast }<0$ such that
$y(t)\leq \alpha _{0}^{\ast }$, $t\in [ 0,1]$. We are going to prove
the existence of another constant $A_{0}^{\ast }<\alpha _{0}^{\ast }$ such
that $y(0)\geq A_{0}^{\ast }$. This is the case since, setting 
$(y(0),y'(0))=(y_{0},y_{0}')\in E^{\ast }$ we obtain
\begin{equation}
\begin{gathered} y'(t)=\phi _{p}^{-1}\Big[\phi _{p}( y_{0}')-\lambda
\int_{0}^{t}q(s)F(s,y(s))ds\Big],\\ y(t)=y_{0}+\int_{0}^{t}\phi
_{p}^{-1}\Big[\phi _{p}(y_{0}') -\lambda \int_{0}^{t}q(s)F(s,y(s))ds
\Big]dt. \end{gathered}  \label{222}
\end{equation}
The initial values must be chosen so that
\begin{align*}
Q^{\ast }(y_{0}')& :=y(1)-\alpha y(1-\eta )=g(y_{0}^{\prime
})+\int_{0}^{1}\phi _{p}^{-1}\Big[\phi _{p}(y_{0}')-\lambda
\int_{0}^{t}q(s)F(s,y(s))ds\Big]dt \\
& \quad -\alpha \Big[g(y_{0}')+\int_{0}^{1-\eta }\phi _{p}^{-1}\Big[
\phi _{p}(y_{0}')-\lambda \int_{0}^{t}q(s)F(s,y(s))ds\Big]dt\Big]=0.
\end{align*}
By Proposition \ref{Pr11} and its proof, there exists an $\eta \in (\eta
_{0},0)$ such that $Q^{\ast }(g^{-1}(\frac{\eta }{2}))>0$ and moreover by
the definition of $Q^{\ast }$ and the fact that $g(y_{0}')<0$,
\begin{equation*}
Q^{\ast }\Big\{\phi _{p}^{-1}\Big(\lambda \int_{0}^{t}q(s)F(s,y(s))ds\Big)
\Big\}<0.
\end{equation*}
Hence $y_{0}'$ is lower bounded and similar $y_{0}=g(y_{0}')$, i.e. 
$(y_{0},y_{0}')\in E_{0}^{\ast }$ with $E_{0}^{\ast }$ a
compact subset of $\mathbb{R}^{2}$. Furthermore, by the monotonicity of the
functions $g$ and $Q^{\ast }$, for each solution $y(t)$ of the
boundary-value problem \eqref{221}, its initial value 
$(y_{0},y_{0}')$ is uniquely determined and continuous.

Consider now the operator
\begin{equation*}
T_{\lambda }y(t)=y_{0}+\int_{0}^{t}\phi _{p}^{-1}\Big[\phi
_{p}[g^{-1}(y_{0})]-\lambda \int_{0}^{s}q(r)F(r,y(r))dr\Big]ds,
\end{equation*}%
where $y_{0}$ is the unique constant corresponding to the function $y(t)$
satisfying (\ref{222}). It is easily verified that $y(t)$ is a solution of 
\eqref{220} if and only if $y$ is a fixed point of $T_{1}$ in $C[0,1]$.

(i) We consider the Banach space $B=C[0,1]$ and we may easily, as above,
prove that $T:=T_{1}:B\rightarrow B$ is completely continuous.

(ii) We will show that there exists a $M>0$ such that $\Vert y\Vert \leq M$
for any solution of \eqref{221}. We set
\begin{equation*}
G(c)=\int_{c}^{0}\min_{0\leq t\leq 1}f(t,y)ds>0,\quad c\leq 0.
\end{equation*}%
Noting \eqref{2.1400}, we may find a $M>0$ such that
\begin{equation*}
\frac{-M}{\big(\int_{-M}^{0}\min_{0\leq t\leq 1}f(t,y)dy\big)^{\frac{1}{p}}
(\frac{p}{p-1})^{\frac{2}{p}}\big[\frac{1}{1-\alpha }\int_{1-\eta
}^{1}[q(t)]^{1/p}dt+\int_{0}^{1-\eta }[q(t)]^{1/p}dt\big]}<-1.
\end{equation*}
Also by the previous, any solution of \eqref{221} being convex, satisfies
\begin{equation*}
y(t)\leq \alpha _{0}(\eta _{0}^{\ast }):=\alpha _{0}^{\ast }<0,\quad t\in
[ 0,1],
\end{equation*}
and further there is a $t_{0}\in (0,1)$ such that
\begin{equation*}
y'(t)\leq 0,\quad t\in [ 0,t_{0}),\quad y^{\prime
}(t_{0})=0\quad \text{and}\quad y'(t)\geq 0,\;t\in (t_{0},1].
\end{equation*}
Working in the interval $[t_{0},t]\subset [ t_{0},1]$, we have
\begin{equation*}
-(\phi _{p}(y'))'=\lambda q(t)F(t,y)\geq q(t)\min_{t\in
[ 0,1]}F(t,y).
\end{equation*}
Multiplying by $-y'<0$, integrating on $[t_{0},t]$, and given that 
$q(t)$ is non-increasing, we obtain
\begin{align*}
\int_{0}^{\phi _{p}[y'(t)]}\phi _{p}^{-1}(z)dz& \leq
-q(t)\int_{y(t_{0})}^{y(t)}\min_{0\leq t\leq 1}F(t,y)dy \\
& \leq -q(t)\int_{y(t_{0})}^{0}\min_{0\leq t\leq 1}F(t,y)dy \\
& =-q(t)G(y(t_{0}))<0.
\end{align*}
Hence,
\begin{equation*}
I(\phi _{p}^{-1}(y'(t)))\leq -q(t)G(y(t_{0}))\leq q(t)G(y(t_{0}))
\end{equation*}%
and so
\begin{equation}
0\leq y'(t)\leq \phi
_{p}^{-1}\{I^{-1}(q(t))I^{-1}[G(y(t_{0}))]\},\quad t\in [ t_{0},t].
\label{2.110}
\end{equation}
If $1-\eta \in (t_{0},1]$, an integration over $[1-\eta ,1]$ yields
\begin{equation*}
y(1)-y(1-\eta )\leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta
}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt.
\end{equation*}
If $1-\eta \in (0,t_{0}]$, we integrate over $[t_{0},1]$ to obtain
\begin{align*}
y(1)-y(t_{0})& \leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{t_{0}}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \leq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt.
\end{align*}
Since $\ y(t_{0})\leq y(1-\eta )$, it follows that
\begin{equation*}
y(1)-y(1-\eta )\leq y(1)-y(t_{0})\leq \phi
_{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt.
\end{equation*}
Moreover, since $y(1)=\alpha y(1-\eta )$, we get
\begin{equation*}
y(1)\geq \frac{\alpha }{\alpha -1}\phi
_{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{1-\eta }^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt,
\end{equation*}
and so a new integration from $t_{0}$ to $1$ of \eqref{2.110} yields
\begin{align*}
& y(t_{0}) \\
& =y(1)-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{t_{0}}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \geq y(1)-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))]\int_{0}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt \\
& \geq \phi _{p}^{-1}[I^{-1}(G(y(t_{0})))][\frac{\alpha }{\alpha -1}
\int_{1-\eta }^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt-\int_{0}^{1}\phi
_{p}^{-1}[I^{-1}(q(t))]dt] \\
& =-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))][\frac{1}{1-\alpha }\int_{1-\eta
}^{1}\phi _{p}^{-1}[I^{-1}(q(t))]dt+\int_{0}^{1-\eta }\phi
_{p}^{-1}[I^{-1}(q(t))]dt] \\
& =-\phi _{p}^{-1}[I^{-1}(G(y(t_{0})))](\frac{p}{p-1})^{\frac{1}{p}}
[\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)]^{\frac{1}{p}}(t)dt+\int_{0}^{1-\eta
}[q(t)]^{\frac{1}{p}}(t)dt].
\end{align*}
Consequently, (recall that $y(t_{0})<0$)
\begin{equation*}
\frac{y(t_{0})}{[G(y(t_{0}))]^{\frac{1}{p}}(\frac{p}{p-1})^{\frac{1}{p}}
[\frac{1}{1-\alpha }\int_{1-\eta }^{1}[q(t)]^{\frac{1}{p}}(t)dt+\int_{0}
^{1-\eta }[q(t)]^{\frac{1}{p}}(t)dt]}>-1,
\end{equation*}
which in turn, by the assumption \eqref{2.1400} and the choice of $M$,
implies $y(t_{0})>-M$. Hence we obtain $\Vert y\Vert \leq M$.

Finally, in view of Lemma \ref{Le4}, we may set
\begin{equation*}
C:=\{y\in B=C[0,1]:\Vert y\Vert \leq M\}\quad \text{and}\quad U:=\{y\in
C:\Vert y\Vert <M\}\,.
\end{equation*}
Then, the second part of the nonlinear Alternative of Leray-Shauder Type is
ruled out and so we conclude that there exists a fixed point of the operator
\begin{equation*}
Ty(t)=T_{1}y(t)=y_{0}+\int_{0}^{t}\phi _{p}^{-1}\Big[\phi
_{p}[g^{-1}(y_{0})]-\int_{0}^{s}q(r)F(r,y(r))dr\Big]ds.
\end{equation*}
This of course yields a solution $y=y(t)$ of the boundary-value problem 
\eqref{220} and noting Proposition \ref{Pr1} and the definition of the
modification $F,y(t)$ is actually a solution of our boundary-value problem 
\eqref{2.13}-\eqref{2.14}.
\end{proof}

Consider now the boundary-value problem
\begin{gather}
-[\phi _{p}(u'(s))]'=q^{\ast}(s)f^{\ast }(s,u\ (s)),\quad
0<t<1,  \label{2.35} \\
u (0)-\alpha u (\eta )=0,\quad u (1)-g(u'(1))=0 \,.  \label{2.36}
\end{gather}

\begin{theorem}
Under the  assumptions of Theorem \ref{Th1}, the boundary-value problem
\eqref{2.35}--\eqref{2.36} has at least one positive solution.
\end{theorem}

\begin{proof}
We make the transformation $u(s)=-y(1-s)$, $s\in (0,1)$, where $y=y(t)$ is a
solution of the boundary-value problem \eqref{2.13}-\eqref{2.14} (it exists
by Theorem \ref{Th2}). Then, clearly
\begin{equation*}
u'(s)=y'(1-s),\quad \phi _{p}(u'(s))=\phi
_{p}(y'(1-s))
\end{equation*}
and
\begin{align*}
-(\phi _{p}(u'(s)))'& =-(\phi _{p}(y^{\prime
}(1-s)))'=q(1-s)f(1-s,y(1-s)) \\
& =q(1-s)f(1-s,-u(s)):=-q^{\ast }(s)f^{\ast }(s,u(s)),
\end{align*}
where $f^{\ast }(s,u(s)):=f(1-s,-u(s))$ and $q^{\ast }(s):=q(1-s)$, 
$s\in (0,1)$. Consequently, the function $u=u(t)$ is a solution of the
boundary-value problem
\begin{equation*}
[ \phi _{p}(u'(t))]'=-q^{\ast }(t)f^{\ast
}(t,u(t)),\quad 0<t<1.
\end{equation*}%
Moreover, since $y=y(t)$ satisfies the boundary conditions \eqref{2.14}, we
obtain
\begin{equation*}
u(1)+g(u'(1))=0,\quad u(0)-\alpha u(\eta )=0,
\end{equation*}%
that is the function $u(s)=-y(1-s)$, $s\in (0,1)$, is actually the required
solution of \eqref{2.13}-\eqref{2.14}.
\end{proof}

\section{An example}

Consider the boundary-value problem
\begin{equation}
\begin{gathered} [\phi _{p}(u')]'=-\frac{a}{\sqrt{
(1-t)}}[u^{-\frac{1}{2}}+\sin ^{2}u^{-\frac{1}{4}}] ,\quad 0<t<1, \\
u(0)=[u'(0)]^{1/3},\quad u(1)=\frac{1}{2}y(\frac{1}{3}), \end{gathered}
\label{41}
\end{equation}
where $\phi _{p}(s)=|s|^{p-2}s$, $p=3$ and $a>0$ is a constant.

Comparing to Theorem \ref{Th1}, we have chosen $g(v)=v^{1/3}$, 
$q(t)=a/\sqrt{(1-t)}$, $\beta =1/2$ and $\eta =1/3$. It is trivial to verify that
assumptions (A1)-(A4) hold true for the system \eqref{41}. Furthermore,
since
\begin{equation*}
\frac{c^{3}}{\int_{0}^{c}[u^{-1/2}+\sin ^{2}u^{-1/4}]}\geq 
\frac{c^{3}}{c+2\sqrt{c}}
\end{equation*}%
and
\begin{equation*}
\frac{3}{3-1}\Big[\frac{1}{1-1/2}\int_{1/3}^{1}\frac{a^{1/3}}{(\sqrt{1-t}
)^{1/3}}dt+\int_{0}^{1/3}\frac{a^{1/3}}{(\sqrt{1-t})^{1/3}}dt\Big]
=\frac{18}{5}a^{1/3},
\end{equation*}
it follows that \eqref{2.70} is fulfilled for every $a>0$. Hence the
boundary-value problem \eqref{41} admits a positive solution.

\begin{remark} \rm
The results in \cite{MG} can not be applied to \eqref{41}, since the
assumption (H5) is not satisfied. Indeed,
\begin{equation*}
\int_{\eta }^{1}f_{1}(k_{2}(1-s))q(s)
ds=\int_{1/3}^{1}\frac{2a}{k_{2}^{1/2}(1-s)^{1/2}(
1-s)^{1/2}}=+\infty .
\end{equation*}
\end{remark}

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\end{document}