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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 32, pp. 1--9.\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-2013/32\hfil Positive solutions]
{Positive solutions for anisotropic discrete boundary-value problems}

\author[M. Galewski, S. G\l \c{a}b, R. Wieteska \hfil EJDE-2013/32\hfilneg]
{Marek Galewski, Szymon G\l \c{a}b, Renata Wieteska}  % in alphabetical order

\address{Marek Galewski \newline
Institute of Mathematics,
Technical University of Lodz,
Wolczanska 215, 90-924 Lodz, Poland}
\email{marek.galewski@p.lodz.pl}

\address{Szymon G\l \c{a}b \newline
Institute of Mathematics,
Technical University of Lodz,
Wolczanska 215, 90-924 Lodz, Poland}
\email{szymon.glab@p.lodz.pl}

\address{Renata Wieteska \newline
Institute of Mathematics,
Technical University of Lodz,
Wolczanska 215, 90-924 Lodz, Poland}
\email{renata.wieteska@p.lodz.pl}

\thanks{Submitted August 12, 2012. Published January 30, 2013.}
\subjclass[2000]{39A10, 34B18, 58E30}
\keywords{Discrete boundary value problem; mountain pass theorem; 
\hfill\break\indent variational methods; Karush-Kuhn-Tucker Theorem; 
positive solution; anisotropic problem}

\begin{abstract}
 Using mountain pass arguments and the Karsuh-Kuhn-Tucker Theorem,
 we prove the existence of at least two positive solution for
 anisotropic discrete Dirichlet boundary-value problems.
 Our results generalized and improve those in \cite{TG}.
\end{abstract}

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


\section{Introduction}

In this note we consider an anisotropic difference equation with Dirichlet
type boundary condition on the form
\begin{equation}
\begin{gathered}
\Delta (|\Delta y(k-1)|^{p(k-1)-2}\Delta y(k-1)) +f(k,y(k))=0,
\quad k\in [1,T]  , \\
y(0)=y(T+1)=0,
\end{gathered} \label{zad}
\end{equation}
where $T\geq 2$ is a integer, $f:[1,T]\times\mathbb{R} \to (0,+\infty )$
 is a continuous function; $[1,T]$ is a discrete
interval $\{1,2,\dots ,T\}$, $\Delta y(k-1)=y(k) -y(k-1)$ is the
forward difference operator; $y(k) \in\mathbb{R}$ for all $k\in [1,T] $;
$p:[0,T+1] \to [ 2,+\infty )$.
Let $p^{-}=\min_{k\in [0,T+1] }p(k) $; $p^{+}=\max_{k\in [0,T+1] }p(k) $.

About the nonlinear term, we assume the following condition
\begin{itemize}
\item[(C1)] There exist a number $m>p^{+}$ and
functions $\varphi_1,\varphi_2:[1,T]\to (0,\infty )$,
$\psi_1,\psi_2:[1,T]\to (0,\infty ) $ such that
\[
\psi_1(k)+\varphi_1(k)|y|^{m-2}y\leq f(k,y)\leq \varphi
_2(k)|y|^{m-2}y+\psi_2(k)
\]
for all $y\geq 0$ and all $k\in [1,T]$.
\end{itemize}

Now, we show an example of a function that satisfies condition (C1).


\begin{example} \label{przyklad} \rm
Let $f:[1,T]\times\mathbb{R}\to (0,\infty )$ be given by
\[
 f(k,y)=| y|^{m-2}y\frac{2+\arctan(y)}{T^{2}k}+
\frac{\sin^{2}(k)e^{-| y| }+1}{T^{3}}
\]
for $(k,y) \in [ 1,T]\times\mathbb{R}$; here $m>p^{+}$. We see that
for $y\geq 0$ we have
\[
\frac{1}{T^{3}}+\frac{2}{T^{2}k}| y|^{m-2}y\leq
f(k,y)\leq \frac{4+\pi }{2T^{2}k}| y|^{m-2}y+\frac{2}{T^{3}}.
\]
Thus we may put
\[
\varphi_1(k)=\frac{2}{T^{2}k};\quad
\varphi_2(k)=\frac{4+\pi }{2T^{2}k};\quad
\psi_1(k)=\frac{1}{T^{3}};\quad \psi_2(k)=\frac{2}{T^{3}}.
\]
\end{example}

Solutions to \eqref{zad} will be investigated in a space
\[
Y=\{y:[0,T+1]\to \mathbb{R}:y(0)=y(T+1)=0\}
\]
with a norm
\[
\| y\| =\Big(\sum_{k=1}^{T+1}|\Delta y(k-1)|^{2}\Big)^{1/2}
\]
with which $Y$ becomes a Hilbert space. For $y\in Y$, let
\[
y_{+}=\max \{y,0\},\quad y_{-}=\max \{-y,0\}.
\]
Note that
$y_{+}\geq 0$,  $y_{-}\geq 0$, $y=y_{+}-y_{-}$, and $y_{+}\cdot y_{-}=0$.

In order to demonstrate that problem \eqref{zad} has at least two positive
solutions we assume additionally the  condition
\begin{itemize}
\item[(C2)]
$$
T^{\frac{p^{+}-2}{2}}\Big(\frac{1}{\sqrt{T+1
}}\Big)^{p^{+}}>\sum_{k=1}^{T}(\varphi_2(k)+\psi_2(k)).
$$
\end{itemize}

\begin{example} \label{examp2}\rm
We show that the function defined in Example \ref{przyklad} 
satisfies condition (C2), by taking $p^{+}=18$ and $T=200$:
\[
T^{\frac{p^{+}-2}{2}}\Big(\frac{1}{\sqrt{T+1}}\Big)
^{p^{+}}=0.009>0.002=\sum_{k=1}^{T}(\varphi_2(k)+\psi
_2(k)) .
\]
\end{example}

\begin{theorem}\label{maintheorem}
Suppose that assumptions {\rm (C1), (C2)} hold.
Then \eqref{zad} has at least two positive solutions.
\end{theorem}

Discrete boundary-value problems received some attention lately.
 Let us mention, far from being
exhaustive, the following recent papers on discrete BVPs investigated via
critical point theory,
\cite{agrawal,CIT,caiYu,Liu,sehlik,TianZeng,teraz,zhangcheng,nonzero}.
The tools employed cover the Morse theory, mountain pass
methodology, linking arguments; i.e. methods usually applied in continuous
problems.

Continuous versions of problems such as \eqref{zad} are known to be
mathematical models of various phenomena arising in the study of elastic
mechanics (see \cite{B}), electrorheological fluids (see \cite{A}) or image
restoration (see \cite{C}). Variational continuous anisotropic problems have
been started by Fan and Zhang in \cite{D} and later considered by many
methods and authors (see \cite{hasto} for an extensive survey of such
boundary value problems). The research concerning the discrete anisotropic
problems of type \eqref{zad} have only been started (see \cite{KoneOuro},
\cite{MRT} where known tools from the critical point theory are applied in
order to get the existence of solutions).

When compared with \cite{TG} we see that our problem is more general since
we consider variable exponent case instead of a constant one. While we do
not include term depending on $\Phi_{p^{-}}(y)=|y|^{p^{-}-2}y$ in the
nonlinear part as is the case in \cite{TG}, it is apparent that our results
would also hold should we have made our nonlinearity more complicated. We
note that term $\Phi_{p^{-}}(y)=|y|^{p^{-}-2}y$ does not influence the
growth of the nonlinearity.

\section{Auxiliary results}

We connect positive solutions to \eqref{zad} with critical points of
suitably chosen action functional. Let
\[
F(k,y)=\int_0^{y}f(k,s)ds\quad \text{for $y\in \mathbb{R}$  and
$k\in [1,T]$}.
\]
Let us define a functional $J:Y\to R$ by
\[
J(y)=\sum_{k=1}^{T+1}\frac{1}{p(k-1)}|\Delta
y(k-1)|^{p(k-1)}-\sum_{k=1}^{T}F(k,y_{+}(k)).
\]
Functional $J$ is slightly different from functionals applied in
investigating the existence of positive solutions, compare with
\cite{TianZeng}. Thus we indicate its properties. The functional $J$ is
continuously G\^{a}teaux differentiable and its derivative
at $y$ is
\begin{equation}
\begin{aligned}
\langle J'(y),v\rangle
&=\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta y(k-1)\Delta v(k-1)\\
&\quad -  \sum_{k=1}^{T}f(k,y_{+}(k))v(k)
\end{aligned}\label{functional1}
\end{equation}
for all $v\in Y$. Suppose that $y$ is a critical point to $J$; i.e.,
 $\langle J'(y),v\rangle =0$ for all $v\in Y$. Summing by parts and
taking boundary values into account, see \cite{GW}, we observe that
\[
0=-\sum_{k=1}^{T+1}\Delta (|\Delta y(k-1)|^{p(k-1)-2}\Delta
y(k-1))v(k)-  \sum_{k=1}^{T}f(k,y_{+}(k))v(k).
\]
Since $v\in Y$ is arbitrary, we see that $y$ satisfies \eqref{zad}.

Now, we recall some auxiliary material which we use later:
For (A1)-(A3) see \cite{MRT}, for (A4)-(A5) see \cite{GW},
for (A6) see \cite{TianZeng}.
\begin{itemize}
\item[(A1)]  For every $y\in Y$ with $\| y\| >1$, we have
\[
\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)}\geq T^{\frac{2-p^{-}}{2}}\|
y\|^{p^{-}}-T.
\]

\item[(A2)] For every $y\in Y$ with $\|y\| \leq 1$, we have
\[
\sum_{k=1}^{T+1}| \Delta y(k-1)|^{p(k-1)}\geq T^{
\frac{p^{+}-2}{2}}\| y\|^{p^{+}}.
\]

\item[(A3)] For every $y\in Y$ and any $m\geq 2$, we have
\[
(T+1)^{\frac{2-m}{2}}\| y\|^m\leq
\sum_{k=1}^{T+1}| \Delta y(k-1)|^m\leq
(T+1)\| y\|^m.
\]

\item[(A4)]  If $p^{+}\geq 2$, there exists 
$C_{p^{+}}>0$  such that for every  $y\in Y$,
\[
\sum_{k=1}^{T+1}| \Delta y(k-1)|^{p(k-1)}\leq
2^{p^{+}}(T+1)(C_{p^{+}}\| y\|^{p^{+}}+1).
\]

\item[(A5)] For every $y\in Y$ and  any $m\geq 2$, we have 
\[
\sum_{k=1}^{T+1}| \Delta y(k-1)|^m\leq
2^m\sum_{k=1}^{T}| y(k)|^m.
\]

\item[(A6)] For every $y\in Y$  and any $p,q>1$ such that 
$\frac{1}{p}+\frac{1}{q}=1$, we have 
\[
\| y\|_{C}=\max_{k\in [ 1,T]}|
y(k)| \leq (T+1)^{\frac{1}{q}}(\sum_{k=1}^{T+1}|\Delta
y(k-1)|^{p})^{1/p}. \quad
\]

\end{itemize}
Let $E$ be a real Banach space. We say that a functional $J:E\to
\mathbb{R}$ satisfies Palais-Smale condition if every sequence $(y_{n})$
such that $\{J(y_{n})\}$ is bounded and $J'(y_{n})\to 0$,
has a convergent subsequence.

\begin{lemma}[\cite{mp}] \label{lem2} 
Let $E$ be a Banach space and $J\in C^{1}(E,\mathbb{R}) $ satisfy 
Palais-Smale condition. Assume that there exist 
$x_0,x_1\in E $ and a bounded open neighborhood $\Omega $ of $x_0$ such that 
$ x_1\notin \overline{\Omega }$ and
\[
\max \{J(x_0),J(x_1)\}<\inf_{x\in \partial \Omega }J(x).
\]
Let
\begin{gather*}
\Gamma =\{h\in C([0,1],E):h(0)=x_0,h(1)=x_1\},\\
c=\inf_{h\in \Gamma }\max_{s\in [ 0,1]}J(h(s)).
\end{gather*}
Then $c$ is a critical value of $J$; that is, there exists $x^{\star }\in E$
such that $J'(x^{\star })=0$ and $J(x^{\star })=c$, where $c>\max
\{J(x_0),J(x_1)\}$.
\end{lemma}

Finally we recall the Karush-Kuhn-Tucker theorem with Slater qualification
conditions (for one constraint), see \cite{borwein}.

\begin{theorem}\label{KKT-THEO} 
Let $X$ be a finite-dimensional Euclidean space, 
$\eta,\mu:X\to\mathbb{R}$ be differentiable functions, with $\mu$ convex and 
$\inf_X \mu<0$, and $S=\{x\in X:\mu(x)\leq 0\}$. Moreover, let 
$\overline{x}\in S$ be such that $\eta(\overline{x})=\inf_S\eta$. 
Then, there exists $\sigma\geq 0$ such that
\[
\eta'(\overline{x})+\sigma\mu'(\overline{x})=0
\quad\text{and}\quad 
\sigma\mu(\overline{x})=0.
\]
\end{theorem}

We will provide now some results which are used in the proof of the Main
Theorem. The following lemma may be viewed as a kind of a discrete maximum
principle.

\begin{lemma}\label{lem4}
 Assume that $y\in Y$ is a solution of the equation
\begin{equation}
\begin{gathered}
\Delta (|\Delta y(k-1)|^{p(k-1)-2}\Delta y(k-1))
+f(k,y_{+}(k))=0,k\in [1,T] , \\
y(0)=y(T+1)=0,
\end{gathered}  \label{UKL2}
\end{equation}
then $y(k) >0$ for all $k\in [1,T] $ and moreover $y$
is a solution of \eqref{zad}.
\end{lemma}

\begin{proof}
We will show that
\[
\Delta y(k-1)\Delta y_{-}(k-1)\leq 0\quad \text{for every } 
 k\in [ 1,T+1].
\]
Indeed,
\begin{align*}
&\Delta y(k-1)\Delta y_{-}(k-1)\\
&=(y(k)-y(k-1))(y_{-}(k)-y_{-}(k-1))\\
&=[(y_{+}(k)-y_{+}(k-1)) -(y_{-}(k)-y_{-}(k-1))
] (y_{-}(k)-y_{-}(k-1))\\
&= (y_{+}(k)-y_{+}(k-1)) (y_{-}(k)-y_{-}(k-1))
-(y_{-}(k)-y_{-}(k-1))^{2}\\
&= y_{+}(k)y_{-}(k)-y_{+}(k)y_{-}(k-1)-y_{+}(k-1)y_{-}(k)\\
&\quad +y_{+}(k-1)y_{-}(k-1)-(y_{-}(k)-y_{-}(k-1))^{2}\\
&= -[y_{+}(k)y_{-}(k-1)+y_{+}(k-1)y_{-}(k)+(y_{-}(k)-y_{-}(k-1))^{2}
] \leq 0.
\end{align*}
Assume that $y\in Y$ is a solution of \eqref{UKL2}.
 Taking $v=y_{-}$ in \eqref{functional1} we obtain
\[
\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta y(k-1)\Delta
y_{-}(k-1)=\sum_{k=1}^{T}f(k,y_{+}(k))y_{-}(k).
\]
Since the term on the left is non-positive and the one on the right is
non-negative, so this equation holds true if {the both terms are equal zero,
which} leads to $y_{-}(k)=0$ for all $k\in [ 1,T]$. Then $y=y_{+}$.
Therefore, $y$ is a positive solution of \eqref{zad}. Arguing by
contradiction, assume that there exists $k\in [ 1,T]$ such that 
$y(k)=0 $, while we can assume $y(k-1)>0$. Then, by \eqref{UKL2} we have
\[
|y(k+1)|^{p(k)-2}y(k+1)=-y(k-1)^{p(k-1)-1}-f(k,0)<0,
\]
which implies $y(k+1)<0$, a contradiction. So $y(k)>0$ for all 
$k\in [1,T]$.
\end{proof}

Finally we prove that $J$ satisfies Palais-Smale condition.

\begin{lemma} \label{lem3} 
Assume that \textbf{(C1)} holds. Then the functional $J$
satisfies Palais-Smale condition.
\end{lemma}

\begin{proof}
Assume that $\{y_{n}\}$ is such that $\{J(y_{n})\}$ is bounded and 
$J'(y_{n})\to 0$. Since $Y$ is finitely dimensional, it is
sufficient to show that $\{y_{n}\}$ is bounded. Note that
\[
\Delta y_{+}(k)\Delta y_{-}(k)\leq 0\quad \text{for every }k\in [0,T].
\]
Using the above inequality we obtain
\begin{equation}
\begin{aligned}
&-\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta y(k-1)\Delta
y_{-}(k-1)\\
&= -\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta
(y_{+}(k-1)-y_{-}(k-1))\Delta y_{-}(k-1)\\
&= -\sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta y_{+}(k-1)\Delta
y_{-}(k-1) \\
&\quad + \sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}\Delta y_{-}(k-1)\Delta
y_{-}(k-1)\\
&\geq  \sum_{k=1}^{T+1}|\Delta y(k-1)|^{p(k-1)-2}(\Delta
y_{-}(k-1))^{2}\\
&\geq \sum_{k=1}^{T+1}|\Delta y_{-}(k-1)|^{p(k-1)}.
\end{aligned} \label{eq1}
\end{equation}

Since $y_{n}=(y_{n})_{+}-(y_{n})_{-}$, we will show
that $\{ (y_{n})_{-}\} $ and $\{ (y_{n})_{+}\} $ are
bounded. Suppose that $\{ (y_{n})_{-}\} $ is unbounded. Then we
may assume that there exists $N_0>0$ such that for $n\geq N_0$ we have 
$\| (y_{n})_{-}\| \geq T\geq 2$. Using \eqref{eq1} we obtain
\begin{align*}
&\langle J'(y_{n}),(y_{n})_{-}\rangle\\
&=\sum_{k=1}^{T+1}|\Delta y_{n}(k-1)|^{p(k-1)-2}\Delta
y_{n}(k-1)\Delta (y_{n})_{-}(k-1) \\
&\quad -\sum_{k=1}^{T}f(k,(y_{n})_{+}(k))(y_{n})_{-}(k)\\
&\leq -\sum_{k=1}^{T+1}|\Delta (y_{n})_{-}(k-1)|^{p(k-1)}.
\end{align*}
So by (A1) we obtain
\begin{align*}
T^{\frac{2-p^{-}}{2}}\| (y_{n})_{-}\|^{p^{-}}-T
&\leq \sum_{k=1}^{T+1}|\Delta (y_{n})_{-}(k-1)|^{p(k-1)}\\
&\leq \langle J'(y_{n}),-(y_{n})_{-}\rangle 
\leq \| J'(y_{n})\| \, \| (y_{n})_{-}\| .
\end{align*}
Next, we see that
\begin{align*}
T^{\frac{2-p^{-}}{2}}\| (y_{n})_{-}\|^{p^{-}}
&\leq \| J'(y_{n})\| \, \| (y_{n})_{-}\| +T\\
&\leq \| J'(y_{n})\| \, \| (y_{n})_{-}\| +\|
(y_{n})_{-}\| \\
&\leq (\| J'(y_{n})\| +1) \|(y_{n})_{-}\|
\end{align*}
and
\[
T^{\frac{2-p^{-}}{2}}\| (y_{n})_{-}\|^{p^{-}-1}\leq (\|J'(y_{n})\| +1) .
\]
Since, for a fixed $\varepsilon >0$, there exists some $N_1\geq N_0$ such
that $\| J'(y_{n})\| <\varepsilon $ for every $n\geq N_1$,
we obtain
\[
\| (y_{n})_{-}\|^{p^{-}-1}\leq \frac{(\varepsilon +1) }{
T^{\frac{2-p^{-}}{2}}}.
\]
This means that $\{ (y_{n})_{-}\} $ is bounded$. $

Now, we will show that $\{(y_{n})_{+}\}$ is bounded. Suppose that
 $\{(y_{n})_{+}\}$ is unbounded. We may assume that 
$\| (y_{n})_{+}\|\to \infty $. Since
\[
f(k,y)\geq \varphi_1(k)|y|^{m-2}y+\psi_1(k)\quad \text{for all }
 k\in [ 1,T],
\]
it follows that
\[
F(k,y)\geq \frac{\varphi_1(k)}{m}|y|^m+\psi_1(k)y.
\]
Thus by (A3) and (A5), we obtain
\[
\sum_{k=1}^{T}F(k,(y_{n})_{+}(k))\geq \frac{\varphi_1^{-}}{m}
\sum_{k=1}^{T}|(y_{n})_{+}(k)|^m\geq \frac{\varphi_1^{-}}{m}
2^{-m}(T+1)^{\frac{2-m}{2}}\| (y_{n})_{+}\|^m,
\]
where $\varphi_1^{-}=\min_{k\in [1,T] }\varphi_1(k) $. 
Therefore by (A4), we have
\begin{align*}
J(y_{n})
&=\sum_{k=1}^{T+1}[\frac{1}{p(k-1)}|\Delta
y_{n}(k-1)|^{p(k-1)}-F(k,(y_{n})_{+}(k))] \\
&\leq  2^{p^{+}}(T+1)(C_{p^{+}}\| \allowbreak (y_{n})
_{+}-(y_{n})_{-}\|^{p^{+}}+1) -\frac{\varphi
_1^{-}}{m}2^{-m}(T+1)^{\frac{2-m}{2}}\| (y_{n})_{+}\|
^m\\
&\leq  2^{p^{+}}(T+1)(C_{p^{+}}2^{p^{+}-1}(\| \allowbreak
(y_{n})_{+}\|^{p^{+}}+\| (y_{n})
_{-}\|^{p^{+}}) +1) \\
&\quad - \frac{\varphi_1^{-}}{m}2^{-m}(T+1)^{\frac{2-m}{2}}\|
(y_{n})_{+}\|^m .
\end{align*}
Since $p^{+}<m$ and $\{(y_{n})_{+}\}$ is unbounded and $\{(y_{n})_{-}\}$ is
bounded, so $J(y_{n})\to -\infty $. Thus we obtain a contradiction
with the assumption $\{J(y_{n})\}$ is bounded, so $\{(y_{n})_{+}\}$ is
bounded. It follows that $\{y_{n}\}$ is bounded.
\end{proof}

\section{Proof of the main result}

In this section we present the proof of Theorem \ref{maintheorem}.

\begin{proof}
Assume that $y_0\in Y$\ is a local minimizer of $J$ in
\[
B:=\{y\in Y:\mu (y) \leq 0\},
\]
where $\mu (y)=\frac{\| y\|^2}{2}-\frac{1}{2(T+1)}$. Note that for 
$y\in B$ by (A6) it follows that for all $k\in [ 1,T]$,
\[
| y(k) | \leq \max_{s\in [1,T]}| y(s) | 
\leq \sqrt{T+1}\| y\| 
\leq \frac{1}{\sqrt{T+1}}\sqrt{T+1}=1.
\]
We prove that $y_0\in Int B$, by contradiction. Thus suppose otherwise; i.e.,
we suppose that $y_0\in \partial B$. Then by Theorem \ref{KKT-THEO} there
exists $\sigma \geq 0$ such that for all $v\in Y$
\[
\langle J'(y_0),v\rangle +\sigma \langle y_0,v\rangle =0.
\]
Hence
\begin{align*}
&\sum_{k=1}^{T+1}|\Delta y_0(k-1)|^{p(k-1)-2}\Delta y_0(k-1)\Delta
v(k-1)\\
&- \sum_{k=1}^{T}f(k,(y_0)_{+}(k))v(k)+\sigma \sum_{k=1}^{T}\langle
y_0(k) ,v(k) \rangle =0.
\end{align*}
Taking $v=y_0$, we see that
\[
\sum_{k=1}^{T+1}|\Delta y_0(k-1)|^{p(k-1)}+\sigma \| y_0\|
^{2}=\sum_{k=1}^{T}f(k,(y_0)_{+}(k))y_0(k).
\]
Since $y_0\in \partial B$, we see that 
$\| y_0\| =\frac{1}{\sqrt{T+1}}$. Thus by (A2), we have
\[
\sum_{k=1}^{T+1}|\Delta y_0(k-1)|^{p(k-1)}+\sigma \| y_0\|
^{2}\geq \sum_{k=1}^{T+1}|\Delta y_0(k-1)|^{p(k-1)}\geq T^{\frac{
p^{+}-2}{2}}\Big(\frac{1}{\sqrt{T+1}}\Big)^{p^{+}}.
\]
On the other hand
\begin{align*}
&\sum_{k=1}^{T}f(k,(y_0)_{+}(k))y_0(k)\\
&= \sum_{k=1}^{T}f(k,(y_0)_{+}(k))(y_0)_{+}(k)-\sum
_{k=1}^{T}f(k,(y_0)_{+}(k))(y_0)_{-}(k)\\
&\leq  \sum_{k=1}^{T}\varphi_2(k)|(y_0)
_{+}(k)|^m+\sum_{k=1}^{T}\psi_2(k)|(y_0)_{+}(k)|\\
&\leq \sum_{k=1}^{T}\varphi_2(k)+\sum_{k=1}^{T}\psi_2(k).
\end{align*}
Thus,
\[
T^{\frac{p^{+}-2}{2}}(\frac{1}{\sqrt{T+1}})^{p^{+}}\leq
\sum_{k=1}^{T}(\varphi_2(k)+\psi_2(k)) .
\]
A contradiction with (C2). Hence $y_0\in IntB$
and $y_0$ is a local minimizer of $J$. 
Thus $J(y_0)<\min_{y\in \partial B}J(y)$.
 We will show that there exists $y_1$ such that 
$y_1\in Y\setminus B$ and $J(y_1)<\min_{y\in \partial B}J(y)$. 
Let $y_{\lambda}\in Y$ be define as follows: $y_{\lambda }(k)=\lambda $ 
for $k=1,\dots ,T$ and $y_{\lambda }(0)=y_{\lambda }(T+1)=0$. 
Then for $\lambda >1$ we have
\[
J(y_{\lambda })
\leq \frac{\lambda^{p(0)}}{p(0)}+\frac{\lambda^{p(T)}}{p(T)}
-\sum_{k=1}^{T}\frac{\varphi_1(k)\lambda^m}{m}
\leq \frac{\lambda ^{p^{+}}}{p(0)}+\frac{\lambda^{p^{+}}}{p(T)}
 -\frac{\varphi_1^{-}\lambda^m}{m}T-\psi_1^{-}\lambda T.
\]
Since $m>p^{+}$, then 
$\lim_{\lambda \to \infty }J(y_{\lambda })=-\infty $. 
Thus there exists $\lambda_0$ with
 $J(y_{\lambda _0})<\min_{y\in \partial B}J(y)$. By Lemma \ref{lem2} 
and Lemma \ref{lem3}
we obtain a critical value\ of the functional $J$ for some 
$y^{\star }\in Y\setminus \partial B$. Then $y_0$ and $y^{\star }$ 
are two different critical points of $J$ and therefore by Lemma \ref{lem4}
these are positive solutions of problem \eqref{zad}.
\end{proof}

\subsection*{Acknowledgements}
The authors would like to thank the anonymous referees for their suggestions 
which allowed us to improve both the results and their presentation.


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