\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 28, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/28\hfil The barrier strip technique]
{The barrier strip technique for a boundary value problem with p-Laplacian}

\author[P. S. Kelevedjiev, S. A. Tersian \hfil EJDE-2013/28\hfilneg]
{Petio S. Kelevedjiev, Stepan A. Tersian}  % in alphabetical order

\address{Petio S. Kelevedjiev \newline
Department of Mathematics, Technical University of Sliven,
Sliven, Bulgaria}
\email{keleved@mailcity.com}

\address{Stepan A. Tersian \newline
Department of Mathematical Analysis, University of Ruse,
Ruse, Bulgaria}
\email{sterzian@uni-ru.acad.bg}

\dedicatory{Dedicated to Professor Jean Mawhin on his 70th birthday}

\thanks{Submitted June 27, 2012. Published January 28, 2013.}
\subjclass[2000]{34B15, 34B18}
\keywords{Boundary value problem; second order differential
equation; \hfill\break\indent $p$-Laplacian, sign condition}

\begin{abstract}
 We study the solvability of the boundary value problem
 $$
 (\phi_p(x'))'=f(t,x,x'),\quad x(0)=A,\;x'(1)=B,
 $$
 where $\phi_p(s)=s|s|^{p-2}$, using the barrier strip type arguments.
 We establish the existence of $C^2[0,1]$-solutions, restricting our
 considerations to $p\in(1,2]$. The existence of positive monotone
 solutions is also considered.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

In this article, we study the existence of
$C^2$-solutions to the boundary-value problem (BVP)
\begin{equation}
\begin{gathered}
(\phi_p(x'))'=f(t,x,x'),\quad t\in(0,1),\\
x(0)=A,\quad x'(1)=B,\quad B>0
\end{gathered}\label{e1.1}
\end{equation}
where $\phi_p(s)=s|s|^{p-2}$,  $p\in(1,2]$, and the scalar function
$f(t,x,y)$ is defined for $(t,x,y)\in[0,1]\times D_x\times D_y$,
$D_x, D_y\subseteq R$, and continuous on a suitable subset of its domain.

Various boundary-value problems for \eqref{e1.1} 
have been studied in the general case $p>1$, and the obtained results 
guarantee $C^1$-solutions.

 Guo and Tian \cite{g2} discussed the existence of positive solutions of the
differential equation 
$\phi _p(x'))'+q(t)f(t,x)=0$, $t\in(0,1)$, satisfying either
$x(0)=x'(1)=0$ or $x(0)=x(1)=0$, where $p>1$,
$f:[0,1]\times [0,\infty )\to [-M,\infty )$ and $q:(0,1)\to [0,\infty )$ are
continuous.

The solvability of BVPs for the equation
\[
-(\phi(x'))'=q(x'(t))f(t,x(t),x'(t)),
\]
with nonlinear functional boundary conditions, and for the equation
\[
(\phi(x'))'=f(t,x(t),x(\tau(t)),x'(t)),
\]
with homogeneous Neumann boundary conditions, has been studied in Cabada and
Pouso \cite{c1} and Liu  \cite{l1}, respectively. 
In these works $\phi:\mathbb{R}\to\mathbb{R}$ is an
increasing homeomorphism, and $f$ is a Carath\'{e}odory function; in \cite{c1},
the right side is discontinuous in the $x'$ argument.

Much attention has been paid to singular problems with $p$-Laplacian.
L\"{u} and Zhong \cite{l2} consider the BVP
\begin{equation}
\begin{gathered}
(\phi(x'))'+f(t,x(t))=0,\quad t\in(0,1), \\
x(0)=x(1)=0
\end{gathered}\label{e1.2}
\end{equation}
where $\phi_p(s)=s|s|^{p-2}$, $p>1$, and
$f:C((0,1)\times[0,\infty),[0,\infty))$ may be singular at the ends of the
interval. Similar problems, singular not only at $t=0$ and $t=1$ but also at
$x=0$, are studied in Agarwal et al \cite{a1} and Jiang et al \cite{j1}; the main
nonlinearity in \cite{j1} depends on $x'$.

Stan\v{e}k \cite{s1} showed that the equation
\begin{equation}
(\phi(x'))'+\mu f(t,x,x')=0,\quad t\in(0,T),\label{e1.3}
\end{equation}
where the parameter $\mu$ is positive, has a solution satisfying  boundary
conditions of the form \eqref{e1.2}.
Here $\phi\in C(\mathbb{R})$ is an odd and increasing
function, and $f\in C([0,T]\times (0,\infty)\times (\mathbb{R}\setminus\{0\}))$ is
singular at $x=0$. Stan\v{e}k \cite{s2}
 study the solvability of a BVP for \eqref{e1.3} (in
the case $\mu<0$) with the boundary conditions
\[
x(0)-\alpha x'(0)=A,\quad
x(T)+\beta x'(0)+\gamma x'(T)=A,\quad
\alpha, A>0,\;\beta,\gamma\geq 0\,.
\]
 Now $\phi:\mathbb{R}\to\mathbb{R}$ is an increasing and odd homeomorphism, and
 $f(t,x,y)$ satisfies the Carath\'{e}odory conditions on
$[0,T]\times D$, $D=(0,(1+\beta/\alpha)A]\times(\mathbb{R}\setminus\{0\})$,
is singular at $x=0$ and may be singular at $y=0$.

Note that in the most of the cited papers, the obtained results
guarantee positive solutions. As a rule, they are established under the
assumption that the considered problems admit lower and upper solutions or
that growth type conditions are satisfied.

To prove our existence result, we use the Topological transversality theorem
\cite{g1}.  For its application, the needed a priori bounds follow from the
assumption:

\begin{itemize}
\item[(R1)]  There are constants $L_i,F_i,i=1,2$, and a sufficiently
small $\sigma >0$ such that
\begin{gather*}
F_1>0,\quad L_{2}-\sigma \geq L_1\geq B\geq F_1\geq F_{2}+\sigma,\\
[A-\sigma ,\quad L+\sigma ]\subseteq D_{x},\quad [F_{2},L_{2}]\subseteq D_{y},
\end{gather*}
where $L=L_1+A$,
\begin{gather}
f(t,x,y)\in C\big([0,1]\times [A-\sigma ,L+\sigma ]\times [F_1-\sigma
,L_1+\sigma]\big), \nonumber\\
f(t,x,y)\geq 0\quad \text{for }(t,x,y)\in [0,1]\times D_{x}\times
[L_1,L_{2}], nonumber\\
f(t,x,y)\leq 0\quad \text{for }(t,x,y)\in [0,1]\times D_{A}\times
[F_{2},F_1] \label{e1.4}
\end{gather}
where $D_{A}=(-\infty ,L]\cap D_{x}$.
\end{itemize}

Let us recall that the strips $[0,1]\times [L_1,L_2]$ and
$[0,1]\times [F_2,F_1]$ are called barrier strips since they keep the
values of $x'$ between themselves. 

\section{Fixed point theorem}

 The proofs of the following theorems
can be found in Granas et al \cite{g1}. To state them, we need standard
topological notions.

Let $Y$ be a convex subset of a Banach space $E$ and $U\subset Y$ be open in
$Y$. Let $L_{\partial U}(\overline{U},Y)$ be the set of compact maps
from $\overline{U}$ to $Y$ which are fixed point free on ${\partial U}$;
here, as usual, $\overline{U}$ and ${\partial U}$ are the closure of $U$ and
boundary of $U$ in $Y$.

A map $F$ in $L_{\partial U}(\overline{U},Y)$ is essential if every
map $G$ in $L_{\partial U}(\overline{U},Y)$ such that $G/\partial
U=F/\partial U$ has a fixed point in $U$. It is clear, in particular, every
essential map has a fixed point in $U$.

\begin{theorem}[Topological transversality theorem] \label{thm2.1} 
 Let $Y$ be a convex subset of a Banach space $E$ and $U\subset Y$ be open. 
Suppose:
\begin{itemize}
\item[(i)]  $F,G:\overline{U}\to Y$ are compact maps.
\item[(ii)]  $G\in L_{\partial U}(\overline{U},Y)$ is essential.
\item[(iii)]  $H(x,\lambda), \lambda\in[0,1]$, is a compact
homotopy joining $F$ and $G$; i.e., $ H(x,1)=F(x)$ and $ H(x,0)=G(x)$.
\item[(iv)] $H(x,\lambda), \lambda\in[0,1]$, is fixed point
free on $\partial U$.
\end{itemize}
Then $ H(x,\lambda)$, $\lambda\in[0,1]$, has at least one fixed point in 
$U$ and in particular there is a $x_0\in U$ such that $x_0=F(x_0)$.
\end{theorem}

\begin{theorem} \label{thm2.2} 
Let $l\in U$ be fixed and $F\in L_{\partial U}(\overline{U},Y)$ be the 
constant map $F(x)=l$ for $x\in\overline{U}$. Then $F$ is essential.
\end{theorem}

\section{Auxiliary results}

 For $\lambda\in[0,1]$ consider the family of BVPs
\begin{equation}
\begin{gathered}
(\phi_p(x'))'=\lambda f(t,x,x'),\quad t\in(0,1),\\
 x(0)=A,\quad x'(1)=B,
\end{gathered}\label{e3.1}
\end{equation}
where $f:[0,1]\times D_x\times D_y\to \mathbb{R}$,
 $D_x, D_y\subseteq {\mathbb{R}}$.
Since
\[
\phi_p(s)=s|s|^{p-2}=\begin{cases}  s^{p-1}, & s\geq 0\\
-(-s)^{p-1}, & s<0,
\end{cases}
\]
we obtain
\[
\phi'_p(s)=(p-1)|s|^{p-2}
=\begin{cases}
(p-1)s^{p-2}, &s\geq 0\\
(p-1)(-s)^{p-2}, &s<0
\end{cases}
\]
and $(\phi_p(x'(t)))'=(p-1)|x'(t)|^{p-2}x''(t)$, if $x''(t)$ exists.
So, we can write \eqref{e3.1} in the form
\begin{equation}
\begin{gathered}
(p-1)|x'(t)|^{p-2}x''(t)=\lambda f(t,x,x'),\;t\in(0,1),\\
 x(0)=A,\quad x'(1)=B,
\end{gathered}\label{e3.1'}
\end{equation}

Our first auxiliary result gives a priori bounds for the 
$C^2[0,1]$-solutions of the family \eqref{e3.1} 
(as well as of \eqref{e3.1'}).



\begin{lemma} \label{lem3.1} 
Let {\rm (R1)} hold and $x\in C^2[0,1]$ be a solution to
family \eqref{e3.1} for each fixed $p\in(1,2]$. Then
$$
A\leq x(t)\leq L,\;F_1\leq x'(t)\leq L_1, \quad
m_p\leq x''(t)\leq M_p\quad \text{for }t\in[0,1],
$$
where $m_p=m(p-1)^{-1}L_1^{2-p}$, $M_p=M(p-1)^{-1}L_1^{2-p}$, 
$m=\min\{f(t,x,y):(t,x,y)\in[0,1]\times[A,L]\times[F_1,L_1]\}$ and 
$M=\max\{f(t,x,y):(t,x,y)\in[0,1]\times[A,L]\times[F_1,L_1]\}$.
\end{lemma}

\begin{proof} Let us assume on the contrary that
\begin{equation}
x'(t)\leq L_1\quad \text{for }t\in[0,1]\label{e3.2}
\end{equation}
is not true. Then $x'(1)=B\leq L_1$ and $x'\in C[0,1]$ imply
that the set
\[
S_+=\{t\in[0,1]:L_1<x'(t)\leq L_2\}
\]
is not empty. Then, there exists an interval $[\alpha,\beta]\subset S_+$
 with the property
\begin{equation}
x'(\alpha)>x'(\beta).\label{e3.3}
\end{equation}
This inequality and the continuity of $x'(t)$ guarantee the existence
of a $\gamma\in[\alpha,\beta]$ such that
\[
x''(\gamma)<0.
\]
On the other hand, as $x(t)$ is a $C^2[0,1]$-solution of \eqref{e3.1},
 we have
\[
(\gamma,x(\gamma),x'(\gamma))\in[0,1]\times D_x\times D_y.
\]
More precisely, $(\gamma,x(\gamma),x'(\gamma))\in S_+\times
D_x\times (L_1,L_2]$, which allows to use (R1) to obtain
\[
0>(p-1)|x'(\gamma)|^{p-2}x''(\gamma)=\lambda
f(\gamma,x(\gamma),x'(\gamma))\geq0,
\]
a contradiction. Thus \eqref{e3.2} is true.

Now, by the mean value theorem, for each $t\in(0,1]$ there exists 
$\xi\in(0,t)$ such that $x(t)-x(0)=x'(\xi)t$, which yields
\[
x(t)\leq L\quad\text{for }t\in[0,1].
\]
Next, suppose that the set
\[
S_-=\{t\in[0,1]:F_2\leq x'(t)<F_1\}
\]
is not empty. Following the reasoning giving \eqref{e3.2} and 
using \eqref{e1.4}, we reach again a contradiction from which we 
conclude that
\begin{gather*}
0<F_1\leq x'(t)\quad \text{for }t\in[0,1],\\
A\leq x(t)\quad \text{for }t\in[0,1].
\end{gather*}
To establish the bounds for $x''(t)$, we observe that from the assumptions
$$
f(t,x,y)\geq 0\quad \text{for }(t,x,y)\in [0,1]\times D_{x}\times
[L_1,L_{2}]
$$
and $[A-\sigma ,L+\sigma ]\subseteq D_{x}$ we have, in particular,
$$
f(t,x,L_1)\geq0\quad \text{for }t\in[0,1]\times[A,L],
$$
which implies $M\geq 0$. Similarly, \eqref{e1.4} implies
$$
f(t,x,F_1)\leq0\quad\text{for }t\in[0,1]\times[A,L],
$$
from where it follows $m\leq0$.

Further, using $0<x'(t)\leq L_1,\;t\in[0,1]$, and $-(p-2)\geq0$, we get
$$
0<(x'(t))^{-(p-2)}\leq L_1^{-(p-2)},\quad t\in[0,1].
$$
Then
$$
0<{\frac{1}{(p-1)(x'(t))^{p-2}}}\leq{\frac{1}{(p-1)L_1^{p-2}}}.
$$
Now, multiplying both sides of this inequality by $\lambda M\geq 0$ and 
$\lambda m\leq0$, we obtain respectively
$$
{{\lambda M}\over{(p-1)(x'(t))^{p-2}}}
\leq{{\lambda M}\over{(p-1)L_1^{p-2}}}
\leq{{M}\over{(p-1)L_1^{p-2}}}= M_p,\;t\in[0,1],
$$
and
$$
{{\lambda m}\over{(p-1)(x'(t))^{p-2}}}
\geq{{\lambda m}\over{(p-1)L_1^{p-2}}}\geq{{m}\over{(p-1)L_1^{p-2}}}
= m_p,\quad t\in[0,1].
$$
On the other hand, by the obtained a priori bounds for $x(t)$ and $x'(t)$, 
for each $t\in[0,1]$ we have $x(t)\in[A,L]$ and $x'(t)\in[F_1,L_1]$. 
Consequently,
$$
m\leq f(t,x(t),x'(t))\leq M\quad \text{for }t\in[0,1]
$$
and multiplying by 
${\lambda(p-1)^{-1}|x'(t)|^{2-p}=\lambda(p-1)^{-1}(x'(t))^{2-p}\geq0}$, 
$\lambda,t\in[0,1]$, we reach
$$
{{\lambda m}\over{(p-1)|x'(t)|^{p-2}}}
\leq{{\lambda f(t,x(t),x'(t))}\over{(p-1)|x'(t)|^{p-2}}}
\leq{{\lambda M}\over{(p-1)|x'(t)|^{p-2}}}
$$
for $\lambda,t\in[0,1]$ which combined with the obtained above yields 
the bounds for $x''(t)$. 
\end{proof}

Now, we introduce the set $C_{BC}^2[0,1]=\{x\in
C^2[0,1]:x(0)=A,x'(1)=B\}$ and the operator
$V:C_{BC}^2[0,1]\to C[0,1]$,
defined by $Vx=(\phi_p(x'))'=(p-1)|x'(t)|^{p-2}x''(t)$, and
the operator $W:C[0,1]\to C_{BC}^2[0,1]$,
defined by
\[
(Wy)(t)=A+\int_0^t{\phi_q\bigl(\int_1^s{y(v)dv+\phi_p(B)\bigr)}}ds,
\]
where $\phi_q(s)=s|s|^{q-2}$, with $p^{-1}+q^{-1}=1,p\in(1,2]$, 
is the inverse of the function $\phi_p(s)$.

The following lemmas give some useful properties of $W$. 

\begin{lemma} \label{lem3.2} 
The operator $W$ is well defined for each $p\in(1,2]$.
\end{lemma}

\begin{proof}
 It is clear that for each $y\in C[0,1]$ the functions
$$
h(t)=\int_1^{t}y(v)dv+\phi _p(B)
$$
and $h'(t)=y(t)$ are continuous for $t\in [0,1]$. Then
$$
(\phi_q(h(t)))'=(q-1)|h(t)|^{q-2}h'(t)
$$
is also continuous for $t\in [0,1]$ since $q-2={\frac{{2-p}}{{p-1}}}\geq0$ 
for $p\in(1,2]$.
Thus, $(Wy)''(t)=(\phi_q(h(t)))'$ is in $C[0,1]$. 
Finally, it is easy to check that
$$
(Wy)(0)=A, \quad (Wy)'(1)=B,
$$
which means $(Wy)(t)\in C_{BC}^2[0,1]$.
\end{proof}  

\begin{lemma} \label{lem3.3} 
The operator $W$ is continuous for each $p\in(1,2]$.
\end{lemma}

\begin{proof}
 Let $y_{n},y\in C[0,1],\,n\in N$, be such that
 $\|y_{n}-y\|\to 0$ as $n\to \infty$.
According to Lemma \ref{lem3.2}, $Wy_{n},Wy\in C_{BC}^2[0,1]$.
We have to show that
$$
\|(Wy_{n})(t)-(Wy)(t)\|_{C_{BC}^2[0,1]}
=\|(Wy_{n})(t)-(Wy)(t)\|_{C^2[0,1]}\to 0;
$$
i.e.,
\begin{equation}
\begin{aligned}
&\lim_{n\to\infty}\|(Wy_{n})^{(i)}(t)-(Wy)^{(i)}(t)\|_{C[0,1]}\\
&=\lim_{n\to\infty}\max_{t\in [0,1]}|(Wy_{n})^{(i)}(t)-(Wy)^{(i)}(t)|=0,
\quad i=0,1,2.
\end{aligned}\label{e3.3b}
\end{equation}
In other words, we have to show that the sequences $\{(Wy_{n})^{(i)}\}$
converge uniformly on $[0,1]$ to $(Wy)^{(i)}$, $i=0,1,2$, respectively.
To this end, we see firstly that
\begin{align*}
&\lim_{n\to\infty}\max_{t\in [0,1]}|(Wy_{n})''(t)-(Wy)''(t)|\\
&=\lim_{n\to\infty}\max_{t\in [0,1]}\Bigl|{{d}\over{dt}}\phi_{q}
 \Bigl(\int_1^{t}y_{n}(v)dv+\phi_p(B)\Bigr)
 -{{d}\over{dt}}\phi_{q}\Bigl(\int_1^{t}y(v)dv+\phi_p(B)\Bigr)\Bigl| \\
&= (q-1)\lim_{n\to\infty}\max_{t\in [0,1]}\Bigl|\Bigl|\int_1^{t}y_n(v)dv
 +\phi_p(B)\Bigr|^{q-2}y_n(t)
 -\Bigr|\int_1^{t}y(v)dv+\phi_p(B)\Bigr|^{q-2}y(t)\Bigr|\\
&=0
\end{align*}
and so $\{(Wy_{n})''\}, n\in N$, converges uniformly on $[0,1]$ to $(Wy)''$.
By the continuity and the uniform convergence of the functions
$\{(Wy_{n})''\}, n\in N$, it follows that the sequence
$\{\int_0^t (Wy_{n})''(v)dv\}, n\in N$, converges uniformly to
$\int_0^t (Wy)''(v)dv$ on $[0,1]$. Then, the sequence $\{(Wy_{n})'\}, n\in N$,
converges uniformly to $(Wy)'$ on $[0,1]$ which means that $\{Wy_{n}\}, n\in N$,
converges uniformly to $Wy$ on $[0,1]$ and so \eqref{e3.3b} holds.
\end{proof}

\begin{lemma} \label{lem3.4} 
The operator $W$ is the inverse operator of $V$.
\end{lemma}

\begin{proof}
 It is clear, each function $x\in C_{BC}^2[0,1]$ has a unique
image $Vx\in C[0,1]$. Also, each function $y\in C[0,1]$ has a unique inverse
image $x\in C_{BC}^2[0,1]$ of the form
\[
x(t)=A+\int_{0}^{t} \phi _{q}\Bigl(\int_1^{s} y(v)dv+\phi _p(B)\Bigr)ds,
\]
which is the solution of the BVP
\begin{gather*}
(\phi _p(x'))'=y(t),\quad t\in (0,1), \\
x(0)=A,\quad x'(1)=B.
\end{gather*}
So, the operator $V$ is one-to-one. Further, to show that $W$ is an invertible
map, let $Vx=y$; i.e., $(\phi _p(x'))'=y$. Then
\begin{align*}
W(Vx) &= Wy=A+\int_{0}^{t}{\phi _{q}\bigl(\int_1^{s}{y(v)dv+\phi _p(B)
\bigr)}}ds \\
&= A+\int_{0}^{t}{\phi _{q}\bigl(\int_1^{s}{(\phi _p(x'(v)))'dv+\phi _p(B)\bigr)}}ds \\
&= A+\int_{0}^{t}{\phi _{q}\bigl(\phi _p(x'(s))-\phi
_p(x'(1))+\phi _p(B)\bigr)}ds \\
&= A+\int_{0}^{t}{\phi _{q}\bigl(\phi _p(x'(s))\bigr)}ds \\
&= A+\int_{0}^{t}{x'(s)}ds=A+x(t)-x(0)=x(t).
\end{align*}
\end{proof}

\section{Main result} 

We state our existence result as follows.  

\begin{theorem} \label{thm4.1}
 Let {\rm (R1)} hold. Then for each $p\in(1,2]$, BVP \eqref{e1.1} 
has at least one solution in $C^2[0,1]$.
\end{theorem}

\begin{proof}
 We will prove the assertion for an arbitrary fixed $p\in(1,2]$.
 At first, preparing the application of Theorem \ref{thm2.1}, we introduce
the set
\begin{align*}
U=\big\{&x\in C_{BC}^2[0,1]:A-\sigma<x<L+\sigma, F_1-\sigma<x'<L_1+\sigma,\\
   &m_p-\sigma<x''<M_p+\sigma\big\}.
\end{align*}
According to Lemma \ref{lem3.1}, all $C^2[0,1]$-solutions of family \eqref{e3.1} (or
\eqref{e3.1'}) are interior points of $U$. Next, consider the maps
\[
j:C_{BC}^2[0,1]\to C^1[0,1],\quad \text{defined by }jx=x,
\]
and
\[
\Phi:C^1[0,1]\to C[0,1],\quad \text{defined by }(\Phi x)(t)=f(t,x(t),x'(t))
\]
for $t\in[0,1]$ and $x(t)\in j(\overline{U})$.

Now, using the map $W$, introduce the homotopy
\[
H_\lambda:\overline{U}\times[0,1]\to C^2_{BC}[0,1],
\]
defined by $H(x,\lambda )\equiv H_{\lambda }(x)\equiv \lambda W\Phi
j(x)+(1-\lambda)l$, where $l=Bt+A$ is the unique solution of
\begin{gather*}
(\phi_p(x'))'=0,\;t\in(0,1), \\
x(0)=A,\;x'(1)=B.
\end{gather*}

Since the map $j$ is completely continuous, $\Phi$ is continuous
because $f$ is continuous, and $W$ is continuous by Lemma \ref{lem3.2}, the
homotopy is compact. For its fixed points we have
\[
\lambda W\Phi j(x)+(1-\lambda)l=x
\]
and
\[
Vx=\lambda\Phi j(x).
\]
The last means that the fixed points of $H_\lambda$ coincide with the
 $ C^2[0,1]$-solutions of \eqref{e3.1} which are not in $\partial U$,
 by Lemma \ref{lem3.1}.
Besides, $H_0(x)$ maps each function $x\in\overline{U}$ in $l$; i.e., it is a
constant map and so is essential, by Theorem \ref{thm2.2}.

So, we can apply Theorem \ref{thm2.1}. It infers that the map $H_1(x)$ has a
fixed point in $U$. It is easy to see that it is a
$C^2[0,1]$-solution of the BVPs of families \eqref{e3.1} and
\eqref{e3.1'} obtained for $\lambda=1$ and, what is the same, of
\eqref{e1.1}. 
\end{proof}

The next result guarantees important  properties of the established solutions.

\begin{theorem} \label{thm4.2} 
Let $A>0$ $(A=0)$ and {\rm (R1)} hold. Then for each $p\in(1,2]$
BVP \eqref{e1.1} has at least one positive (nonnegative) increasing solution 
in $C^2[0,1]$.
\end{theorem}

\begin{proof} 
By Theorem \ref{thm4.1}, BVP \eqref{e1.1} has a solution $x(t)\in C^2[0,1]$ for each 
$p\in(1,2]$ and by Lemma \ref{lem3.1} it is such that
\[
x(t)\geq A\quad\text{and}\quad x'(t)\geq F_1>0\quad \text{for }t\in[0,1],
\]
from where the assertion follows immediately. 
\end{proof}

\begin{example} \label{examp4.1} \rm
 Consider the BVP
\begin{gather*}
(\phi_p(x'))'=P_n(x'),\quad t\in(0,1), \\
x(0)=0,\quad x'(1)=B,\quad B>0,
\end{gather*}
where $p\in(1,2]$, and the polynomial $P_n(y),n\geq2$, has two simple 
zeros $p_1$ and $p_2$ such that $p_2>B>p_1>0$.

Clearly, there is a sufficiently small $\delta>0$ such that
\[
p_2-\delta>B>p_1+\delta,\quad p_1-\delta>0
\]
and $P_n(y)\neq0$ for $t\in(p_1-\delta,p_1)\cup(p_1,p_1+\delta)
\cup(p_2- \delta,p_2)\cup(p_2,p_2+\delta)$.

So, in the case $P_n(y)<0$ for $t\in(p_1-\delta,p_1)$ and $P_n(y)>0$ for 
$t\in(p_2,p_2+\delta)$, we can choose $F_2=p_1-\delta$,
$F_1=p_1$, $L_1=p_2$ and $L_2=p_2+\delta$ to see that (R1) holds 
and so the considered problem has a nonnegative increasing solution 
in $C^2[0,1]$, by Theorem \ref{thm4.2}.

The same conclusion follows similarly in the rest three cases for
the sign of $P_n(y)$ near $p_1$ and $p_2$. 
\end{example}

\subsection*{Acknowledgements}
 The second author would like to thank to the
Department of Mathematics and Theoretical Informatics at Technical
University of Kosice, Slovakia, where this paper was prepared during
his visit by the SAIA Fellowship programme.

The authors thank to anonymous referee for careful reading 
of the manuscript and comments.

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\end{document}
