\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 51, 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 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/51\hfil Positive solutions]
{Positive solutions for second-order boundary-value problems with
$\phi$-Laplacian}

\author[D.-R. Herlea \hfil EJDE-2016/51\hfilneg]
{Diana-Raluca Herlea}

\address{Diana-Raluca Herlea \newline
Babe\c{s}-Bolyai University,
Department of Mathematics,
400084 Cluj, Romania}
\email{dherlea@math.ubbcluj.ro}


\thanks{Submitted February 3, 2016. Published February 18, 2016.}
\subjclass[2010]{34B18, 47H10}
\keywords{Positive solution; $\phi$-Laplacian, boundary value problem;
\hfill\break\indent  Krasnosel'ski\u{\i} fixed point theorem; weak Harnack inequality}

\begin{abstract}
 This article concerns the existence, localization and multiplicity 
 of positive solutions for the boundary-value problem
 \begin{gather*}
 \big( \phi( u') \big) '+f(t,u) =0,  \\
  u(0) - a u'(0) = u'(1)= 0,
 \end{gather*}
 where $f:[0,1]\times \mathbb{R}_{+}\to \mathbb{R}_{+}$ is a continuous
 function and $\phi :\mathbb{R}\to (-b,b)$ is an increasing homeomorphism
 with $\phi (0)=0$. We obtain existence, localization and multiplicity
 results of positive solutions using Krasnosel'ski\u{\i} fixed point
 theorem in cones, and a weak Harnack type inequality. Concerning systems,
 the localization is established by the vector version of
 Krasnosel'ski\u{\i} theorem, where the compression-expansion conditions
 are expressed on components.
\end{abstract}

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

\section{Introduction}

The aim of this article is to present new results regarding the existence,
localization and multiplicity of positive solutions for the problem
\begin{equation}
\begin{gathered}
(\phi (u') ) '+f(t,u) =0,\quad 0<t<1, \\
u(0) - a u'(0) = 0, \quad u'(1) =0,
\end{gathered} \label{eq1}
\end{equation}
 where $a>0$, $\phi $ is a homeomorphism from $\mathbb{R}$ to $(-b,b)$
and $0<b \leq \infty$.

According   \cite{be}-\cite{Chen} and \cite{Maw}, there are two remarkable
 models in this context:
\begin{itemize}
\item[(1)] The  $p$-Laplacian operator, where $b= \infty$ and
$\phi(u)=| u| ^{p-2} u$, with $p>1$.

\item[(2)] The curvature operator, where $b< \infty$ and
\[
\phi(u)=\frac{u}{\sqrt{1+u^{2}}}.
\]
\end{itemize}
Problem \eqref{eq1} can be considered as a particular,  $n=1$ of the
corresponding problem for an $n$-dimensional system
\begin{equation}
\begin{gathered}
(\phi _i(u_i') ) '+f_i(t,u_1,u_2,\dots,u_{n}) =0,\quad 0<t<1, \\
u_i(0) - a_i u_i'(0) = 0,
u_i'(1) =0\quad (i=1,2,\dots,n),
\end{gathered} \label{ps}
\end{equation}
where $a_i>0$.

First we shall concentrate on the problem \eqref{eq1} for a single equation,
and then we shall extend the results to the system \eqref{ps}.
The study of  $\phi$-Laplacian equations is a classical topic that has
attracted the attention of many experts because of its  applications
(see for example \cite{Ag}). These equations, with different boundary
conditions have been studied in a large number of papers using
fixed point methods, degree theory, upper and lower solution techniques
 and variational methods. Robin boundary conditions
\[
\alpha_1 u(0)-\beta_1 u'(0) = 0 = \alpha_2 u(1)+\beta_2 u'(1),
\]
are commonly used in solving Sturm-Liouville problems which appear
in many contexts in science and engineering. These problems have been
considered in the literature by many authors in order to search the existence
of positive solutions (see \cite{Erbe,Ge}). Some of them worked with
special cases. For example \cite{Sm1,Ge2,her} studied the case
$\beta_1=\beta_2=0$ and $\alpha_1=\alpha_2=1$, while \cite{hp} discussed
the case $\alpha_1=\beta_2=0$, $\alpha_2=1$ and $\beta_1=-1$.

In this article, we considered the case $\alpha_1=\beta_2=1$, $\beta_1=a$,
with $a>0$ and $\alpha_2=0$;  we are interested not only on the existence
of positive solutions to  \eqref{eq1} and \eqref{ps}, but also on their
localization and multiplicity. We shall achive this by using a technique
based on Krasnosel'ski\u{\i}'s fixed point theorem in cones \cite{Krs}.
This result has been extensively employed in the related literature
(see for instance \cite{hd}-\cite{hp}, \cite{Prp,Tor}).

\begin{theorem}[Krasnosel'ski\u{\i}]
Let $(X, | \cdot |)$ be a normed linear space; $K \subset X$ a cone;
$r, R \in \mathbb{R}_{+}$, $0 < r < R$, $K_{r,R} = \{u \in K : r \leq |u| \leq R\}$,
and let  $N: K_{r,R} \to K$ be a compact map. Assume that one of the
following conditions is satisfied:
\begin{itemize}
\item[(a)] $N(u)\nless u$ if $|u|=r$, and $N(u)\ngtr u$ if $|u|=R$;

\item[(b)] $N(u)\ngtr u$ if $|u|=r$, and $N(u)\nless u$ if $|u|=R$.

\end{itemize}
Then $N$ has a fixed point $u$ in $K$ with $r \leq |u| \leq R$.
\end{theorem}

Here for two elements $u,v\in X$, the strict ordering $u<v$ means $v-u\in
K\setminus \{ 0\}$.

The technique based on the application of Krasnosel'ski\u{\i} theorem
for completely continuous operators on a Banach space, requires the
construction of a suitable cone of positive functions. To this end,
in the case of most boundary value problems, the corresponding Green
functions and their properties play an important role. Alternatively,
for many problems for which Green functions are not
known, one can use weak Harnack type inequalities associated to the
differential operators and the boundary conditions, as shown in \cite{Prc}
and \cite{Prp}. In our case, such an inequality will arise as a consequence
of the concavity of the positive solutions.

In the case of systems, we shall allow the homeomorphisms $\phi _i$ have
different ranges and we shall be interested to localize each
component of a solution $u=(u_1,u_2,\dots,u_{n})$ individually. In this
respect we shall use the following vector version of Krasnosel'ski\u{\i}
theorem given in \cite{Pre} (see also \cite{Pre2}).

\begin{theorem}
Let $(X,|\cdot|)$ be a normed linear space;
$K_1,K_2,\dots,K_{n}\subset X$ cones;
$K:=K_1\times K_2\times \dots\times K_{n}$; $r,R\in \mathbb{R}
_{+}^n$, $r=(r_1,r_2,\dots,r_{n})$, $R=(R_1,R_2,\dots,R_{n})$ with
$0<r_i<R_i$ for all $i$, $K_{r,R}=\{u\in K:r_i\leq |u_i|\leq R_i$,
$i=1,2,\dots,n\}$ and let $N:K_{r,R}\to K$, $N=(N_1,N_2,\dots,N_{n})$
be a compact map. Assume that for each $i=1,2,\dots,n$, one of the following
conditions are satisfied in $K_{r,R}$:
\begin{itemize}
\item[(a)] $N_i(u)\nless u_i$ if $|u_i|=r_i$, and $N_i(u)\ngtr
u_i$ if $|u_i|=R_i$;

\item[(b)] $N_i(u)\ngtr u_i$ if $|u_i|=r_i$, and $N_i(u)\nless
u_i$ if $|u_i|=R_i.\vspace{0.1cm}$

\end{itemize}
 Then $N$ has a fixed point $u=(u_1,u_2,\dots,u_{n})$ in $K$ with
$r_i\leq |u_i|\leq R_i$ for $i=1,2,\dots,n$.
\end{theorem}

Note that in the previous theorem, the same symbol $<$ is used to denote the
strict ordering induced by any of the cones $K_1,K_2,\dots,K_{n}$.

It deserves to be underlined the fact that asking the compression condition
(a) to be satisfied by some indices $i$, and the expansion condition (b) by
the others, it is allowed that the system nonlinearities behave differently
one from the other.

\section{Positive solutions of $\phi $-Laplace equations}

In this section, we seek for positive solutions for  \eqref{eq1} and prove
some existence, localization and multiplicity results. For this,
we make the following assumptions:
\begin{itemize}
\item[(A1)] $\phi :\mathbb{R}\to (-b,b)$, $0<b\leq \infty $ is an increasing
homeomorphism such that $\phi (0)=0$.

\item[(A2)] $f:[0,1]\times \mathbb{R}_{+}\to \mathbb{R}_{+}$ is continuous,
$f(t,.)$ is nondecreasing on $\mathbb{R}_{+}$ for each $t\in \lbrack 0,1]$.
\end{itemize}
First we obtain the equivalent integral equation for  problem ~\eqref{eq1}
of positive solutions. Integration of the differential equation
\eqref{eq1} gives
\[
-\phi (u'(t))=- \phi(u'(0))+ \int_0^{t}f(s,u(s))\,ds .
\]
Then
\begin{equation}
u'(t)=\phi ^{-1}\Big(\phi(u'(0))-\int_0^{t}f(s,u(s))\,ds\Big) . \label{ei0}
\end{equation}
Integrating from $0$ to $t$ we obtain
\begin{equation}
u(t)=u(0)+ \int_0^{t}\phi ^{-1}\Big(\phi(u'(0)) -\int_0^{\tau }f(s,u(s))\,ds
\Big)\,d\tau .  \label{ei}
\end{equation}
If we denote $x:=u(0)$, substituting it into \eqref{ei} and taking into account
 the first boundary condition, we have
\begin{equation}
u(t)=x+ \int_0^{t}\phi ^{-1}\Big(\phi(\frac{x}{a})
-\int_0^{\tau }f(s,u(s))\,ds\Big)\,d\tau .  \label{ei2}
\end{equation}
For $t=1$, \eqref{ei0} gives
\[
\phi ^{-1}\Big(\phi(\frac{x}{a}) -\int_0^1f(s,u(s))\,ds\Big) =0,
\]
whence
\begin{equation}
x=a \phi ^{-1}\Big(\int_0^1f(s,u(s))\,ds\Big).
\label{eqx}
\end{equation}
Next, we may define the integral operator
$N:C([0,1];\mathbb{R}_{+})\to C([0,1];\mathbb{R}_{+})$ by
\begin{equation}
N(u)(t)=a \phi ^{-1}\Big(\int_0^1f(s,u(s))\,ds\Big)
+ \int_0^{t}\phi ^{-1}\Big(\int_{\tau}^1f(s,u(s))\,ds\Big)\,d\tau ,  \label{op}
\end{equation}
and thus, finding positive solutions to \eqref{eq1} is equivalent to the
fixed point problem for the operator $N$ on $C([0,1];\mathbb{R}_{+})$. Note
that by standard arguments, $N$ is completely continuous.
Let $| \cdot| _{\infty }$ denotes the max norm on
$C[ 0,1] $.

To apply Krasnosel'ski\u{\i}'s fixed point theorem in cones we need a
weak Harnack type inequality for the differential operator
$Lu:=-(\phi(u'))'$ subjected to the boundary conditions.

\begin{lemma} \label{lem1}
For each $c \in (0,1)$, and any $u\in C^1[0,1]\cap C([ 0,1];\mathbb{R}_{+}) $
 with $u(0)-au'(0) =u'(1)=0$, $\phi \circ u'\in W^{1,1}(0,1)$ and
$(\phi(u'))'\leq 0$ a.e. on $[0,1]$, one has
\begin{equation}
u(t)\geq \gamma(t)|u|_{\infty },\quad \text{for all }t\in \lbrack 0,1],  \label{eq2}
\end{equation}
where
\[
\gamma(t) =   \begin{cases}
   \frac{a+c}{a+1}, & \text{for } t \in [c,1] \\
   0,      & \text{for } t \in [0,c).
  \end{cases}
\]
\end{lemma}

\begin{proof}
From $(\phi (u'))'\leq 0$ on $[ 0,1] $, one has
that the function $\phi \circ u'$ $\ $is noincreasing on $[0,1]$.
Then, from $u'=\phi ^{-1}(\phi \circ u') $,
and $\phi ^{-1}$ increasing, we deduce that $u'$ is nonincreasing
on $[ 0,1] $. Thus $u$ is concave on $[ 0,1] $. On the
other hand, since the function $\phi \circ u'$ vanishes at $t=1$,
$\phi (u'(t) ) \geq 0$ for every $t\in [0,1] $. Then $u'\geq 0$ on
$[ 0,1]$, which shows
that $u$ is nondecreasing on $[0,1]$. If we have $u(0)<0$ then
$u'(0)=\frac{u(0)}{a}<0$ which is excluded by $u' \geq 0$ on $[0,1]$.
Hence $u(0) \geq 0$ and so $u$ is nonnegative, nondecreasing, concave
and $| u| _{\infty }=u(1) $. Inequality \ref{eq2} being obvious for
$t \in [0,c)$, it remains to prove it for $t \in [c,1]$.
If $\min_{t\in \lbrack c,1]}u(t) =0$, then the concavity of $u$ implies
$u=0$ on $[ 0,1] $, and so \eqref{eq2} holds.
If $\min_{t\in \lbrack c,1]}u(t) >0$, then we may assume without loss of
generality that $\min_{t\in \lbrack c,1]}u(t) =1$ (otherwise,
multiply \eqref{eq2} by a suitable positive constant). Then $u(c)=1$.
 Since $u$ is concave, its graph on $[c,1]$ is under the line containing
the points $(0,u(0))$ and $(c,1)$ and so at point $t=1$ we have
\[
u(1) \leq \frac{u(0)(c-1)+1}{c}.
\]
However, $u'(0)= \frac{u(0)}{a}$ and being the slope of the line,
\[
u'(0) \geq \frac{1-u(0)}{c}.
\]
Hence $u(0) \geq \frac{a}{c+a}$ and then $u(1) \leq \frac{a+1}{a+c}$.
Now, from $| u| _{\infty }=u(1)$ we have
\[
\frac{a+c}{a+1}|u|_{\infty} \leq 1.
\]
Finally, since $1\leq u(t) $ for $t\in \lbrack c,1]$, we obtain
\[
u(t) \geq \frac{a+c}{a+1}|u|_{\infty}, \quad\text{for all }t \in [c,1],
\]
as we wished.
Notice that a graphical representation would make more clear the above
reasoning.
\end{proof}

Our first result is the following theorem.

\begin{theorem} \label{t1}
Let {\rm (A1)} and {\rm (A2)} hold and assume that there exist
$\alpha ,\beta >0$ with $\alpha \neq \beta $ such that
\begin{gather}
\Phi (\alpha ):=a\phi ^{-1}\Big(\int_0^1f(s,\gamma(s)\alpha )\,ds\Big)
+ \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,\gamma(s)\alpha )\,ds\Big)
\,d\tau > \alpha ,  \label{c1} \\
\Psi (\beta ):=a\phi ^{-1}\Big(\int_0^1f(s,\beta )\,ds\Big)
+ \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,\beta )\,ds\Big) \,d\tau <\beta .
 \label{c2}
\end{gather}
Then \eqref{eq1} has at least one positive solution $u$
with $r\leq {|u|}_{\infty }\leq R$,
 where $r=\min \{\alpha ,\beta \}$ and $R=\max \{\alpha ,\beta \}$.
\end{theorem}

\begin{proof}
We shall apply Krasnosel'ski\u{\i}'s fixed point theorem in cones.
In our case, $X=C[0,1] $, the cone is
\begin{align*}
K= \big\{&u\in C([0,1];\mathbb{R}_{+}):u(0)-au'(0)=u'(1)=0 \text{ and} \\
&u(t)\geq \gamma(t)|u|_{\infty } \text{ for all }t\in \lbrack 0,1]\big\},
\end{align*}
and $N$ is the operator given by \eqref{op}.

Note that if $u,v\in C([0,1];\mathbb{R}_{+})$ and $v<u$, that is $u-v\in
K\setminus \left\{ 0\right\} $, then $(u-v) (1) \geq
\gamma(1) | u-v| _{\infty }>0$. Hence
\begin{equation}
| u| _{\infty }\geq u(1) >v(1) .
\label{aj}
\end{equation}

First we remark that $N(K) \subset K$. Indeed, if $u\in K$ and
$v:=N(u) $, then $-(\phi (v') )'=$ $f(t,u) $. We have
$f(t,u(t)) \geq 0$ for every $t\in [ 0,1] $, so
$(\phi (v') ) '\leq 0$ on $[ 0,1] $. Then
Lemma \ref{lem1} guarantees that $v(t) \geq \gamma(t)
| v| _{\infty }$ for $t\in [ 0,1] $, that is $v\in K$ as desired.

Next we prove that
\begin{equation}
u\ngtr N(u)\quad \text{for every }u\in K \text{ with }|
u| _{\infty }=\alpha .  \label{k1}
\end{equation}
To this end, assume the contrary, i.e. $u>N(u) $ for some
$u\in K $ with $| u| _{\infty }=\alpha $. Then using \eqref{aj},
the definition of $K$, and the monotonicity of $f$ and $\phi $, we deduce
\begin{align*}
\alpha
& =|u|_{\infty } \geq |N(u)|_{\infty } \geq N(u)(1) \\
& = a\phi ^{-1}\Big(\int_0^1 f(s,u(s))\,ds\Big)
 + \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,u(s))\,ds\Big)\,d\tau \\
& \geq a\phi ^{-1}\Big(\int_0^1f(s,\gamma(s)\alpha )\,ds\Big)
 + \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,\gamma(s)\alpha )\,ds\Big) \,d\tau ,
\end{align*}
which contradicts \eqref{c1}. Thus \eqref{k1} holds. The next step is to prove that
\begin{equation}
u\nless N(u)\quad \text{for every $u\in K$ with }|u|_{\infty}=\beta .  \label{k2}
\end{equation}
Assume the contrary, i.e. $u<N(u)$ for some $u\in K$ with
$|u| _{\infty }=\beta $. Then we would obtain
\begin{align*}
\beta
& =|u|_{\infty }\leq | N(u)| _{\infty }=N(u) (1) \\
& = a\phi ^{-1}\Big(\int_0^1 f(s,u(s))\,ds\Big)
 + \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,u(s))\,ds\Big)\,d\tau \\
& \leq a\phi ^{-1}\Big(\int_0^1f(s,\beta )\,ds\Big)
 + \int_0^1\phi ^{-1}\Big(\int_{\tau}^1f(s,\beta )\,ds\Big) \,d\tau ,
\end{align*}
which contradicts \eqref{c2}. Thus \eqref{k2} holds.
Now Krasnosel'ski\u{\i} theorem applies and yields the result.
\end{proof}

\begin{remark} \label{rmk1} \rm
The existence and localization result, Theorem \ref{t1}, immediately yields
multiplicity results for the problem \eqref{eq1}, in case that several
(finitely many or infinitely many) couples of distinct numbers
$\alpha,\beta $ satisfying \eqref{c1}, \eqref{c2} exist such any two of the
corresponding intervals $(\alpha ,\beta ) $\ are disjoint.
\end{remark}

The next theorems are about the existence of at least one pair
$\alpha,\beta $ satisfying the conditions \eqref{c1}, \eqref{c2},
and the existence of a sequence of positive solutions of  \eqref{eq1}, respectively.
Their proofs are as in \cite{hp}. However, for the readers convenience we
reproduce them.

\begin{theorem}\label{t2}
Let {\rm (A1) }and {\rm (A2)} hold and assume that one of the
following conditions is satisfied:
\begin{itemize}
\item[(i)] $\limsup_{\lambda \to \infty }\frac{\Phi (\lambda )}{
\lambda }>1\ $ and $\liminf_{\lambda \to 0}\frac{\Psi (\lambda )
}{\lambda }<1$;

\item[(ii)] $\limsup_{\lambda \to 0}\frac{\Phi (\lambda )}{
\lambda }>1\ $ and $\liminf_{\lambda \to \infty }\frac{\Psi
(\lambda )}{\lambda }<1$.

\end{itemize}
Then \eqref{eq1} has at least one positive solution.
\end{theorem}

\begin{proof}
To apply Theorem \ref{t1}, we look for two numbers $\alpha ,\beta
>0$, $\alpha \neq \beta $ with
\[
\Phi (\alpha ) > \alpha \quad\text{and}\quad \Psi (\beta ) <\beta .
\]
In case (i), one can chose $\alpha $ large enough and $\beta $ small enough;
while in case (ii), $\alpha $ is chosen small enough and $\beta $ is chosen
large enough.
\end{proof}

\begin{theorem}\label{t3}
Let {\rm (A1)} and {\rm (A2)} hold. If the condition
\begin{itemize}
\item[(iii)] $\limsup_{\lambda \to \infty }\frac{\Phi (\lambda )}{
\lambda }>1\ $ and $\liminf_{\lambda \to \infty }\frac{\Psi
(\lambda )}{\lambda }<1$
\end{itemize}
 holds, then \eqref{eq1} has a sequence of positive solutions
$(u_{n}) _{n\geq 1}$ such that $| u_{n}|_{\infty }\to \infty $ as $n\to \infty $.

If the condition
\begin{itemize}
\item[(iv)]  $\limsup_{\lambda \to 0}\frac{\Phi (\lambda )}{
\lambda }>1$ and $\lim \inf_{\lambda \to 0}\frac{\Psi (\lambda )
}{\lambda }<1$
\end{itemize}
holds, then \eqref{eq1} has a sequence of positive solutions
$(u_{n}) _{n\geq 1}$ such that $| u_{n}|_{\infty }\to 0$ as $n\to \infty $.
\end{theorem}

\begin{proof}
Clearly (iii) guarantees the existence of two sequences
$(\alpha_{n}) _{n\geq 1},(\beta _{n}) _{n\geq 1}$ such that
\begin{equation}
0<\alpha _{n}<\beta _{n}<\alpha _{n+1}\quad
\text{for every $n\geq 1$,
and $\alpha _{n}\to \infty$ as $n\to \infty$}.  \label{i'}
\end{equation}
For each $n$, Theorem \ref{t1} yields a positive solution $u_{n}$ with
 $\alpha _{n}\leq | u_{n}| _{\infty }\leq \beta _{n}$. The
condition \eqref{i'} implies that these solutions are distinct and that
$| u_{n}| _{\infty }\to \infty $ as $n\to \infty $. A similar reasoning
can be done in case (iv).
\end{proof}

Notice that the conditions (iii) and (iv) show that $f$ is oscillating
towards $\infty$ and zero, respectively.

\section{Positive solutions of $\phi $-Laplace systems}

In this section we extend the above results to the general case \eqref{ps}.
We shall allow the homeomorphisms $\phi _i$ have different ranges,
 namely $\phi _i:\mathbb{R} \to (-b_i,b_i)$,
$0<b_i\leq \infty $, and we shall assume that $\phi _i$ are
increasing with $\phi _i(0)=0$, and that
$f_i:[0,1]\times \mathbb{R}_{+}^n\to \mathbb{R}_{+}$ are continuous
functions $(i=1,2,\dots,n)$.
Under these assumptions problem \eqref{ps} is equivalent to the integral
system
\begin{align*}
u_i(t)=a_i \phi_i ^{-1}\Big(\int_0^1f_i(s,u(s))\,ds\Big)
+ \int_0^{t}\phi_i ^{-1}\Big(\int_{\tau}^1f_i(s,u(s))\,ds\Big)\,d\tau,
\end{align*}
for $i=1,2,\dots,n$ and $u=(u_1,u_2,\dots,u_{n})$.

By Lemma \ref{lem1}, for each $i$ and any constant $c_i\in (0,1) $, a weak Harnack
type inequality holds for the differential operator $L_iv:=-(\phi _i(v'))'$
and the boundary conditions $v(0)-av'(0) =v'(1) =0$. Based on this result we
define the cones
\begin{align*}
K_i =  \big\{&u_i\in C([0,1];\mathbb{R}_{+}):u_i(0)-au_i'(0) =u_i'(1) =0 \text{ and}\\
  & u_i(t)\geq\gamma_i(t)|u_i|_{\infty }, \text{for all }t\in \lbrack 0,1]\big\},
\end{align*}
 for $i=1,2,\dots,n$. We note that the functions $\gamma_i$ are given by
 Lemma \ref{lem1} for possibly different $c_i$ and $a_i$.
 Now we consider the product cone
$K:=K_1\times K_2\times \dots\times K_{n}$
 in $C([0,1],\mathbb{R}^n)$.

Let $N:C([0,1];\mathbb{R}_{+}^n)\to C([0,1];\mathbb{R}_{+}^n)$,
$N=(N_1,N_2,\dots,N_{n})$ be defined by
\[
N_i(u)(t)=a_i \phi_i ^{-1}\Big(\int_0^1f_i(s,u(s))\,ds\Big)
+ \int_0^{t}\phi_i ^{-1}\Big(\int_{\tau}^1f_i(s,u(s))\,ds\Big)\,d\tau ,
\]
for $i=1,2,\dots,n$.

If $u_{j}\in K_{j}$ for each $j$, then $f_i(s,u(s))\geq 0$ and from Lemma
\ref{lem1}, one has $N_i(u)\in K_i$. Thus the cone $K$ is invariant by $N$.
Moreover, the operator $N$ is completely continuous since, by standard
arguments, the components $N_i$ are completely continuous.

The following result is a generalization of Theorem \ref{t1} and guarantees
the existence of positive solutions to \eqref{ps} and their
component-wise localization. For any index $i\in \{ 1,2,\dots,n\} $,
we shall say that the homeomorphism $\phi _i:\mathbb{R}\to(-b_i,b_i)$
 satisfies (A1) if $\phi _i$ is increasing and $\phi_i(0)=0$, and that 
the continuous function $f_i:[0,1]\times \mathbb{R}_{+}^n\to \mathbb{R}_{+}$ 
satisfies (A2) if for each $t\in [0,1] $, $f_i(t,x_1,\dots,x_{n})$ 
is nondecreasing on $\mathbb{R}_{+}$ with respect to any variable 
$x_{j}$, $j=1,2,\dots,n$.

\begin{theorem}\label{t8}
Let $\phi _i$, $f_i$ satisfy {\rm (A1)} and {\rm (A2)} for $i=1,2,\dots,n$. 
Assume that there exist $c_i$, $\alpha _i$, $\beta _i>0$ with $c_i<1$ and 
$\alpha _i\neq \beta _i$ such that
\begin{gather*}
\begin{aligned}
\Phi _i(\alpha )  
:&= \ a_i\phi_i ^{-1}\Big(\int_0^1f_i(s,\gamma_1(s)\alpha_1,\dots,
\gamma_{n}(s)\alpha_{n})\,ds\Big) \\
&\quad + \int_0^1\phi_i ^{-1}\Big(\int_{\tau}^1f_i(s,\gamma_1(s)\alpha_1,
\dots,\gamma_{n}(s)\alpha_{n} )\,ds\Big) \,d\tau > \alpha _i,
\end{aligned}\\
\Psi _i(\beta ) :=  a_i\phi_i ^{-1}\Big(\int_0^1f_i(s,\beta )\,ds\Big) 
+ \int_0^1\phi_i ^{-1}\Big(\int_{\tau}^1f_i(s,\beta )\,ds\Big) \,d\tau  <\beta _i,
\end{gather*}
for $i=1,2,\dots,n$, where 
$\alpha =(\alpha _1,\alpha _2,\dots,\alpha _{n}) $ and 
$\beta =(\beta _1,\beta _2,\dots,\beta _{n})$.
Then \eqref{ps} has at least one positive solution 
$u=(u_1,u_2,\dots,u_{n}) $ with 
$r_i\leq {|u_i|}_{\infty }\leq R_i$, where 
$r_i=\min \{\alpha _i,\beta _i\}$, $R_i=\max \{\alpha_i,\beta _i\}$, $i=1,2,\dots,n$.
\end{theorem}

The above result is a consequence of the vectorial version of Krasnosel'ski\u{\i}
fixed point theorem in cones.

We shall say that for a given index $i$, the condition (i) from Theorem \ref{t2}
 holds if for every $\lambda _1,\lambda _2,\dots,\lambda _{i-1}>0$,
\[
\limsup_{\lambda _i\to \infty }\frac{\Phi _i(\lambda )}{
\lambda_i }>1\quad \text{and}\quad
\liminf_{\lambda _i\to 0}\frac{\Psi _i(\lambda )}{\lambda _i}<1,
\]
uniformly with respect to 
$\lambda _{i+1},\lambda _{i+2},\dots,\lambda _{n}\in(0,\infty ) $. 
We shall understand the condition (ii) in a
similar manner. Therefore, if for each $i$ the condition (i) or (ii) holds,
then we obtain pairs $(\alpha_i, \beta_i)$ satisfying the assumptions of
Theorem \ref{t8}.

Analogously, we say that (iii) from Theorem \ref{t3} holds for some index $i$, if
for every $\lambda _1,\lambda _2,\dots,\lambda _{i-1}>0$,
\[
\limsup_{\lambda _i\to \infty }\frac{\Phi _i(\lambda )}{
\lambda_i }>1\quad \text{and}\quad
\liminf_{\lambda _i\to \infty } \frac{\Psi _i(\lambda )}{\lambda_i }<1,
\]
uniformly with respect to $\lambda _{i+1},\lambda _{i+2},\dots,\lambda _{n}\in
(0,\infty )$. Condition (iv) is understood in a similar
manner. Under such type of conditions we obtain sequences of solutions 
for the system \eqref{ps}.

Finally, we note that \cite[Theorem 3.2]{hp} can be applied to our problem 
\eqref{ps} in order to guarantee the existence of multiple solutions.


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\section{Addendum posted on May 19, 2016}


The author would like to thank the anonymous reader for his valuable remarks
and suggestions, and to make accordingly the following amendments and
completions:

(I) Condition (A2) and Theorem 2.2 needs the additional hypothesis on $f$,
\[
f( t,x) <b
\]%
for all $t\in [ 0,1] $ and $x\in \mathbb{R}_{+}$. This assumption
guarantees that the integral operator $N$ is well defined on $C([0,1];\mathbb{R}_{+})$.

(II) Cases (i) and (iii) from Theorem 2.4 and Theorem 2.5 are possible
only if $b=\infty $. Indeed, if $b<\infty $, then
\[
\int_0^1f(s,\gamma (s)\alpha )\,ds=\int_c^1f \Big(s,\frac{a+c}{a+1}
\alpha \Big) \,ds\leq (1-c)b.
\]%
Also, for $0\leq \tau \leq c$, as above
\[
\int_{\tau }^1f(s,\gamma (s)\alpha )\,ds=\int_c^1f \Big(s,\frac{a+c}{a+1}
\alpha \Big)\,ds\leq (1-c)b,
\]
while for $c\leq \tau \leq 1$,
\[
\int_{\tau }^1f(s,\gamma (s)\alpha )\,ds=\int_{\tau }^1f\Big(s,\frac{a+c}{%
a+1}\alpha \Big)\,ds\leq (1-\tau )b\leq (1-c)b.
\]
These inequalities show that $\Phi (\lambda )$ given by (2.7) is bounded,
namely
\[
\Phi (\lambda )\leq (a+1)\Phi ^{-1}((1-c)b).
\]
Then
\[
\limsup_{\lambda \to \infty }\frac{\Phi (\lambda )}{\lambda }=0
\]%
and so (i) and (iii) can not be satisfied.

(III) The above remarks also apply to the systems considered in Section 3.

(IV) Concerning Theorems 2.4 and 2.5, we present the following four examples.

\begin{example}[in (ii), case $b<\infty $] \label{examp1} \rm
In (1.1) we let
\[
\phi (u)=\frac{u}{\sqrt{1+u^2}}\quad \text{and}\quad
f(t,x)=f(x)=\frac{x}{x+1}.
\]
In this case, $b=1$ and one can easily check that condition (A2),
particularly the inequality $f( t,x) <1$, holds. Direct
computations show that
\[
\Phi (\lambda )=A\Big( \frac{a+c}{\sqrt{1-A^2}}+\frac{1-c}{1+\sqrt{%
1-A^2}}\Big) ,\quad
 \Psi ( \lambda ) =B\Big( \frac{a}{\sqrt{%
1-B^2}}+\frac{1}{1+\sqrt{1-B^2}}\Big) ,
\]
where
\[
A=\frac{\lambda (a+c)(1-c)}{\lambda (a+c)+(a+1)}\quad \text{and}\quad
B=\frac{\lambda }{\lambda +1}.
\]
Now it is easy to see that
\[
\lim_{\lambda \to 0}\frac{\Phi (\lambda )}{\lambda }
=\frac{(a+c)(1-c)(2a+c+1)}{2(a+1)}\quad \text{and} \quad 
\lim_{\lambda \to \infty }\frac{\Psi (\lambda )}{\lambda }=0.
\]
Hence  condition (ii) from Theorem 2.4 is satisfied if
\[
C:=\frac{2(a+1)}{(a+c)(1-c)(2a+c+1)}<1,
\]
which holds for sufficiently large $a$. For example, we can choose $a=7$ and
$c=0.5$. For this case, Figure \ref{fig1} shows the behavior of the function $f$ with 
respect to the line $y=Cx$. 
\end{example}

\begin{figure}[ht] 
\begin{center}
\setlength{\unitlength}{0.8mm} %\scriptsize
\begin{picture}(84,64)(-4,-4)
\put(0,0){\vector(1,0){80}}
\put(0,0){\vector(0,1){60}}
\textcolor[rgb]{1,0,0}{\put(0,0){\line(4,3){80}}
\put(55,54){$y=Cx$}
}
\textcolor[rgb]{0,0,1}{
\qbezier(-1,0)(8,13)(20,15) 
\qbezier(20,15)(50,18)(80,18) 
\put(60,22){$y=f(x)$}
}
\multiput(15,-.5)(16,0){4}{\line(0,1){1}}
\put(14,-5){2} \put(30,-5){4}
\put(46,-5){6} \put(62,-5){8} \put(76,-5){10}
\put(77,2){$x$}

\multiput(-2,10)(0,10){5}{\line(1,0){1}}
\put(-6,18.5){1}  \put(-7,38.5){2} \put(-7,58.5){3}
\put(2,58.5){$y$}

\end{picture}
\end{center}
\caption{Behavior of $f$ with respect to the line $y=Cx$.}
\label{fig1} 
\end{figure}


\begin{example}[in (iii), case $b=\infty$] \label{examp2} \rm
If in (1.1) we let $\phi(u)=u$, then expressions (2.7) and (2.8) become
\[
\Phi (\lambda )=f\Big( \frac{a+c}{a+1}\lambda \Big) 
\Big( \frac{(1-c)(2a+c+1)}{2}\Big),\quad
\Psi (\lambda )=f(\lambda )\Big( \frac{2a+1}{2}\Big) .
\]
Consider
$f:[0,1]\times \mathbb{R}_{+}\to \mathbb{R}_{+}$, defined by
\[
f(t,x)=f(x)=mx+nx\sin (p\ln (x+1)).
\]
In this case $b=\infty $ and  condition (A2) holds if
\begin{equation}
m\geq n(p+1).  \label{cond1}
\end{equation}
Now it is easy to see that
\begin{gather*}
\limsup_{\lambda \to \infty }\frac{\Phi (\lambda )}{\lambda }
=(m+n)\frac{(a+c)(1-c)(2a+c+1)}{2(a+1)}, \\
\liminf_{\lambda \to \infty }\frac{\Psi (\lambda )}{\lambda }
=(m-n)\frac{2a+1}{2}.
\end{gather*}
Hence condition (iii) from Theorem 2.5 is satisfied if
\begin{equation}
m+n>A\quad \text{and}\quad m-n<B,  \label{cond2}
\end{equation}
where
\[
A=\frac{2(a+1)}{(a+c)(1-c)(2a+c+1)}\quad \text{and}\quad
B=\frac{2}{2a+1}.
\]
For example, conditions \eqref{cond1} and \eqref{cond2} hold for
\[
a=2.5,\quad c=0.3,\quad m=0.46,\quad n=0.15,\quad p=2.
\]
For this case, Figure \ref{fig2} shows the oscillatory behavior 
of the function $f$ with respect to the lines $y=Ax$ and $y=Bx$.
\end{example}

\begin{figure}[ht] 
\begin{center}
\setlength{\unitlength}{0.8mm} %\scriptsize
\begin{picture}(90,90)(-10,-10)
\put(0,0){\vector(1,0){80}}
\put(0,0){\vector(0,1){80}}
\textcolor[rgb]{1,0,0}{\put(0,0){\line(1,1){80}}
\put(58,74){$y=Bx$}
}
\textcolor[rgb]{0,0,1}{
\qbezier(-1.5,0)(2,1)(5,6)
\qbezier(5,6)(18,42)(48,48) 
\qbezier(48,48)(71,54)(80,74) 
\put(60,48){$y=f(x)$}
}
\textcolor[rgb]{0,1,0}{\put(-3,0){\line(2,3){54}}
\put(30,74){$y=Ax$}
}
\multiput(15,-.5)(16,0){4}{\line(0,1){1}}
\put(12,-5){0.2} \put(28,-5){0.4}
\put(44,-5){0.6} \put(60,-5){0.8} 
\put(65,-10){$\cdot 10^{12}$}
\put(77,-5){1}
\put(77,2){$x$}

\multiput(-3.4,16)(0,16){4}{\line(1,0){1}}
\put(-9,14.7){1} \put(-9,30.7){2} 
\put(-9,46.3){3}  \put(-9,62.3){4} \put(-9,78.3){5} 
\put(1,70){$\cdot 10^{11}$}   
\put(2,78.3){$y$}

\end{picture}
\end{center} 
\caption{Behavior $f$ with respect to the lines $y=Ax$ and $y=Bx$.}
\label{fig2}
\end{figure}


\begin{example}[in (iv), case $b=\infty $] \label{examp3} \rm
We consider $\phi (u)=u$ and the function
$f:[0,1]\times \mathbb{R}_{+}\to \mathbb{R}_{+}$ defined by
\[
f(t,x)=f(x)=mx+nx\sin ( p\ln \frac{1}{x}) ,
\]
for $x>0$ and $f(0)=0$. In this case $b=\infty $ and the condition (A2)
holds if
\begin{equation}
m\geq n(p+1).  \label{cond3}
\end{equation}%
Now it is easy to see that
\begin{gather*}
\limsup_{\lambda \to 0}\frac{\Phi (\lambda )}{\lambda }
=(m+n)\frac{(a+c)(1-c)(2a+c+1)}{2(a+1)}, \\
\lim \inf_{\lambda \to 0}\frac{\Psi (\lambda )}{\lambda }
=(m-n)\frac{2a+1}{2}.
\end{gather*}
Hence  condition (iv) from Theorem 2.5 is satisfied if
\begin{equation}
m+n>A\quad\text{and}\quad  m-n<B,  \label{cond4}
\end{equation}
where
\[
A=\frac{2(a+1)}{(a+c)(1-c)(2a+c+1)} \quad \text{and} \quad B=\frac{2}{2a+1}.
\]%
For example, conditions \eqref{cond3} and \eqref{cond4} hold for
\[
a=2, \quad c=0.2, \quad m=0.54,\quad n=0.16,\quad p=2.
\]
\end{example}


\begin{example}[in (iv), case $b<\infty $] \label{examp4} \rm
Consider $\phi (u)=u/\sqrt{1+u^2}$ and the function $f(t,x)=f(x)$ which on a small
interval $(0,\epsilon )$ is defined by
\[
f(x)=mx+nx\sin \big( p\ln \frac{1}{x}\big) .
\]
Here $\epsilon >0$ is chosen such that $f(x)<1$ on $(0,\epsilon )$. 
The function $f$ is increasing on $(0,\epsilon )$ if
\begin{equation}
m\geq n(p+1).  \label{cond5}
\end{equation}%
Here
\[
\Phi (\lambda )=(a+c)\phi ^{-1}\Big( (1-c)f\Big( \frac{a+c}{a+1}\lambda
\Big) \Big) +\int_c^1\phi ^{-1}\Big( (1-\tau )f\Big( \frac{a+c}{%
a+1}\lambda \Big) \Big) \,d\tau .
\]%
Since
\[
\int_c^1\phi ^{-1}\Big( (1-\tau )f\Big( \frac{a+c}{a+1}\lambda
\Big) \Big) \,d\tau \geq 0,
\]
a sufficient condition for $\Phi (\lambda )>\lambda $ to hold is
\[
\phi ^{-1}\Big( (1-c)f\Big( \frac{a+c}{a+1}\lambda \Big) \Big) >
\frac{\lambda }{a+c},
\]
or equivalently
\[
(1-c)f\Big( \frac{a+c}{a+1}\lambda \Big) >\phi \Big( \frac{\lambda }{a+c}\Big) .
\]
This gives the condition
\[
\frac{f\big( \frac{a+c}{a+1}\lambda \big) }
{\frac{a+c}{a+1}\lambda }>
\frac{a+1}{(a+c)^2(1-c)\sqrt{1+( \frac{\lambda }{a+c}) ^2}}.
\]
Letting $\lambda \to 0$ yields
\[
m+n>\frac{a+1}{(a+c)^2(1-c)}.
\]
Also
\[
\Psi (\lambda )=a\phi ^{-1}( f(\lambda )) +\int_0^1\phi
^{-1}( (1-\tau )f( \lambda ) ) \,d\tau ,
\]
and since
\[
\int_0^1\phi ^{-1}( (1-\tau )f( \lambda ) )
\,d\tau \leq \phi ^{-1}( f( \lambda ) ) ,
\]%
a sufficient condition for $\Psi (\lambda )<\lambda $ to hold is
\[
\phi ^{-1}( f( \lambda ) ) <\frac{\lambda }{a+1},
\]
or equivalently
\[
f( \lambda ) <\phi \Big( \frac{\lambda }{a+1}\Big) .
\]
This gives the condition
\[
\frac{f( \lambda ) }{\lambda }<\frac{1}{(a+1)\sqrt{1+(
\frac{\lambda }{a+1}) ^2}},
\]
which letting $\lambda \to 0$ yields
\[
m-n<\frac{1}{a+1}.
\]
Hence  condition (iv) from Theorem 2.5 is satisfied if
\begin{equation}
m+n>\frac{a+1}{(a+c)^2(1-c)}\quad \text{and}\quad m-n<\frac{1}{a+1}.
\label{cond6}
\end{equation}
For example, conditions \eqref{cond5}) and \eqref{cond6} hold for
\[
a=2.5,\quad c=0.1,\quad m=0.43,\quad n=0.17,\quad p=1.5.
\]
\end{example}

End of addendum.

\end{document}