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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 238, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/238\hfil Fourth-order discrete anisotropic BVPs]
{Fourth-order discrete anisotropic boundary-value problems}

\author[M. Leszczy\'nski \hfil EJDE-2015/238\hfilneg]
{Maciej Leszczy\'nski}

\address{Maciej Leszczy\'nski\newline
Institute of Mathematics, Technical University of Lodz, 
Wolczanska 215, 90-924 Lodz, Poland}
\email{165708@edu.p.lodz.pl}

\thanks{Submitted May 15, 2015. Published September 17, 2015.}
\subjclass[2010]{39A10}
\keywords{Discrete boundary value problem; critical point theory;
\hfill\break\indent discrete p-n-Laplacian; fourth-order}

\begin{abstract}
 In this article we consider the fourth-order discrete anisotropic boundary
 value problem with both advance and retardation. We apply the direct method
 of the calculus of variations and the mountain pass technique to prove
 the existence of at least one and at least two solutions. Non-existence of
 non-trivial solutions is also undertaken.
\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}

Below $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{R}$ denote the sets of all
natural numbers, integers and real numbers respectively. Let $k$ denote a
natural number. Let $a,b \in\mathbb{Z}$; we define $\mathbb{Z}(a) =
\{a,a+1,\dots\}$, and when $a<b$, $\mathbb{Z}(a,b) = \{a,a+1,\dots b\}$.

We consider the difference equation with both advance and
retardation,
\begin{equation}
\Delta ^2(\gamma _{n-1}\phi _{p_{n}}(\Delta
^2u_{n-2}))=f(n,u_{n+1},u_{n},u_{n-1}),\quad n\in \mathbb{Z}(1,k),  \label{ROW}
\end{equation}
with boundary values conditions
\begin{equation}
\Delta u_{-1} = \Delta u_0 = 0 , \quad u_{k+1} = u_{k+2} = 0 ,
\label{eq:warunki_brzegowe}
\end{equation}
where $\gamma_n$ is non-zero and real parameter for each 
$n\in \mathbb{Z}(0,k+1)$, $p_n$ is real valued and $2\leq p_n<\infty $ 
for all $n\in \mathbb{Z}(0,k)$, $\Delta$ is the forward difference operator 
$\Delta u_n = u_{n+1} - u_n$, $\Delta^i u_n = \Delta ( \Delta^{i-1} u_n)$ 
for $i\geq 2$, $\phi_{p_n}$ is the so called
$p_n$-Laplacian operator defined as
\[
\phi_{p_n} (s) = |s|^{p_n -2} s,
\]
and $f \in C(\mathbb{Z}(0,k) \times\mathbb{R}^3,\mathbb{R})$.

To determine whether boundary value problem (BVP) 
\eqref{ROW}-\eqref{eq:warunki_brzegowe} has any solutions, we  use the critical
point theory. Let us denote
\begin{gather*}
\overline{\gamma }=\max \{\gamma _{n}: n\in \mathbb{Z}(0,k+1)\},\quad
\underline{\gamma }=\min \{\gamma _{n}: n\in \mathbb{Z}(0,k+1)\},\\
\overline{p}=\max \{p_{n}: n\in \mathbb{Z}(0,k)\},\quad
\underline{p}=\min \{p_{n}: n\in \mathbb{Z}(0,k)\}.
\end{gather*}

Continuous versions of problem like\eqref{ROW}-\eqref{eq:warunki_brzegowe} 
are known to be mathematical models of various phenomena arising in the 
study of elastic mechanics (see \cite{B}),
electrorheological fluids (see \cite{A}), and image restoration 
(see \cite{C}). 
Variational continuous anisotropic problems were 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 fourth-order anisotropic problems have only
been started, see \cite{KoneOuro,MRT}, where known tools from the
critical point theory are applied prove the existence of
solutions. Concerning the investigation of discrete boundary value problems
we mention, far from being exhaustive, the following recent papers
that used critical point theory 
\cite{agrawal,caiYu,Liu,sehlik,TianZeng,teraz,zhangcheng,nonzero}. 
These papers employ in the discrete setting
the variational techniques already known for continuous problems of course
with necessary modifications. The tools employed cover the Morse theory,
mountain pass methodology, linking arguments. 
New critical point tool for fourth-order discrete BVPs
are considered in \cite{MBR}. In our setting, upon suitable changes, it seems
possible to obtain similar results.

Concerning the fourth-order problems we mainly follow \cite{LiuZhangShi}. We use
somehow simpler approach and consider a more complicated variable exponent case.

\section{Variational framework}

Let $X$ be a $k$-dimensional Euclidean space consisting of functions 
$x:\mathbb{Z}(-1,k+1)\to\mathbb{R}$ satisfying \eqref{eq:warunki_brzegowe} 
and equipped with a norm
\[
\| u\| :=\Big(\sum_{n=1}^{k}|\Delta ^2u_{n}|^2\Big)^{1/2}.
\]
For $r\in [ 1,\infty )$, we define the norm 
\[
\| u\| _{r}=\Big(\sum_{j=1}^{k}|u_{j}|^{r}\Big)^{1/r}.
\]
Since $\operatorname{dim}(X)=k$, all norms are equivalent, hence there exist 
constants $c_{1,r},c_{2,r}$ for $r\in [ 1,\infty )$ such that 
$c_{2,r}\geq c_{1,r}\geq 0$, and
\[
c_{1,r}\| u\| \leq \| u\| _{r}\leq c_{2,r}\| u\|\quad
\forall u\in X.
\]

\begin{remark} \rm
Following some ideas from \cite{GalewskiSmejda}, the values of the
above constants can be easily calculated as
\[
c_{1,2}=\frac{\sqrt{6}}{6},\quad c_{2,2}=\frac{k^2}{2}
\]
For any $u\in X$ we have
\[
|u_{m}|\leq \frac{k}{2} \sum_{n=1}^{k}|\Delta ^2u_{n}| \quad \forall
m\in \mathbb{Z}(1,k).
\]
Indeed, the following inequality is true
\[
\max_{i\in \mathbb{Z}(1,k)} |u(i)| \leq \frac{1}{2} \sum_{n=1}^{k} |\Delta u_n |,
\]
while the the boundary conditions $u_0 = u_{k+1} =0$ are satisfied. 
So, analogously we obtain
\[
\max_{i\in \mathbb{Z}(1,k)} |\Delta u(i)| 
\leq \frac{1}{2} \sum_{n=1}^{k} |\Delta^2 u_n |,
\]
with conditions $\Delta u_0 = \Delta u_{k+1} = 0$, which are satisfied 
for every $u\in X$.
Hence, for every $n\in\mathbb{Z}(1,k)$, we obtain
\[
|u_n| \leq \sum_{i=1}^k |\Delta u_i |
\leq k \max_{i\in  \mathbb{Z}(1,k)}|\Delta u_i | 
\leq \frac{k}{2} \sum_{i=1}^k |\Delta^2 u_n |.
\]
Thus, for $m\in \mathbb{Z}(1,k)$ we have
\begin{align*}
|u_{m}|^2
&\leq \frac{k^2}{4} \Big(\sum_{n=1}^{k}|\Delta ^2u_{n}|\Big)^2\\
&\leq  \frac{k^2}{4} \Big(\Big(\sum_{n=1}^{k}1^2\Big)^{1/2}
\Big(\sum_{n=1}^{k}|\Delta ^2u_{n}|^2\Big)^{1/2}\Big)^2\\
&=\frac{k^3}{4}\sum_{n=1}^{k}|\Delta ^2u_{n}|^2.
\end{align*}
Therefore,
\begin{align*}
&\sum_{n=1}^{k}|u_{n}|^2=|u_1|^2+|u_2|^2+\ldots +|u_{k}|^2\\
&\leq \frac{k^3}{4} \sum_{n=1}^{k}|\Delta ^2u_{n}|^2
+k\sum_{n=1}^{k}|\Delta ^2u_{n}|^2+\ldots 
+k\sum_{n=1}^{k}|\Delta ^2u_{n}|^2\\
&=\frac{k^4}{4}\sum_{n=1}^{k}|\Delta ^2u_{n}|^2.
\end{align*}
As a consequence, 
\[
\| u\| _2
=\Big(\sum_{n=1}^{k}|u_{n}|^2\Big)^{1/2}
\leq \Big(\frac{k^4}{4}\sum_{n=1}^{k}|\Delta ^2u_{n}|^2\Big)^{1/2}
=\frac{k^2}{2}\Big(\sum_{n=1}^{k}|\Delta^2u_{n}|^2\Big)^{1/2}
=\frac{k^2}{2}\| u\| .
\]
Additionally
\begin{align*}
\| u\|^2 
&=\sum_{n=1}^{k}|\Delta ^2u_{n}|^2
 =\sum_{n=1}^{k}|u_{n+2}-2u_{n+1}+u_{n}|^2\\
&\leq \sum_{n=1}^{k}|u_{n+2}|^2+4\sum_{n=1}^{k}|u_{n+1}|^2
 +\sum_{n=1}^{k}|u_{n}|^2 \\
&\leq \sum_{n=1}^{k}|u_{n}|^2
 +4\sum_{n=1}^{k}|u_{n}|^2+\sum_{n=1}^{k}|u_{n}|^2 \\
&=6\sum_{n=1}^{k}|u_{n}|^2=6\| u\|^2 _2.
\end{align*}
Hence, we have
\[
\| u\|^2 \leq 6\| u\|^2 _2\Leftrightarrow \frac{\sqrt{6}}{6}\| u\|
\leq \| u\| _2.
\]
To sum up, we obtain
\[
\frac{\sqrt{6}}{6}\| u\| \leq \| u\| _2\leq\frac{k^2}{2}\| u\|
\]
\end{remark}

Let us consider a functional $J$ defined on $X$ as follows
\begin{equation}
J(u)=\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
^2u_{n}|^{p_{n}}-\sum_{n=1}^{k}F(n,u_{n+1},u_{n}),  \label{eq:funkcjonal}
\end{equation}
where $F(n,x,y)$ is a function of three variables. 
We denote by $ F_x(n,v_1,v_2)$ the derivative of $F$ with respect to the 
second variable calculated at point $(n,v_1,v_2)$ and by $F_y(n,v_1,v_2)$ 
the derivative of $F$ with respect to the third one calculated at point 
$(n,v_1,v_2)$.
\[
F_x(n-1,v_2,v_3)+ F_y(n,v_1,v_2)=f(n,v_1,v_2,v_{3});
\]
It is easy to see that $J\in C^{1}(\mathbb{R}^{k},\mathbb{R})$ and for any 
$u\in \mathbb{R}^{k}$ exploiting boundary values 
\eqref{eq:warunki_brzegowe} we can calculate the partial derivative as
\[
\frac{\partial J}{\partial u_n} = \Delta^2 (\gamma_{n-1}
\phi_{p_n}(\Delta^2 u_{n-2})) - f(n,u_{n+1},u_{n},u_{n-1}),\quad \forall_{ n\in
\mathbb{Z}(1,k)}.
\]
As a consequence, $u$ is a critical point of $J$ on $\mathbb{R}^{k}$ if and
only if
\[
\Delta ^2(\gamma _{n-1}\phi _{p_{n}}(\Delta^2u_{n-2}))
=f(n,u_{n+1},u_{n},u_{n-1}),\forall {n\in \mathbb{Z}(1,k)}
\]
We redefine problem of proving the existence of solutions of
\eqref{ROW}-\eqref{eq:warunki_brzegowe} to the existence of critical points
of $J$ on $\mathbb{R}^{k}$. Thus, the functional $J$ is the variational
framework of our problem.

Now we recall some necessary background from \cite{Figueredo}. 
Let $E$ be a real Banach space, $J\in C^{1}(E,\mathbb{R})$ i.e, 
$J$ is a continuously Frechet-differentiable functional defined on $E$. 
$J$ is said to satisfy the Palais-Smale condition (P.S. condition for short) 
if every sequence 
$\{u^{(l)}\}\subset E$ for which $\{J(u^{(l)})\}$ is bounded and 
$J'(u^{(l)})\to 0$ (as $l\to \infty$) possesses a convergent
subsequence in $E$.

\begin{lemma}[Mountain Pass Lemma]
 Let $E$ be a real Banach space and $J\in C^1(E,\mathbb{R})$ satisfying the
 P.S. condition. If $J(0) = 0$ and
\begin{itemize}
\item[(J1)] there exist constants $\rho,a>0$ such that $J|_{\partial
B_{\rho}}\geq a$, and

\item[(J2)] there exists $e\in E \backslash B_{\rho} $ such that $J(e)
\leq 0$.
\end{itemize}
Then, $J$ possesses a critical value $c\geq a$ given by
\[
c= \inf_{g\in \Gamma} \max_{s\in [0,1]} J(g(s)),
\]
where
$\Gamma = \{g\in C([0,1],E); g(0)=0 \text{ and } g(1)=e \}$.
\end{lemma}

\section{Auxiliary results}

In this article we use following inequalities

(A1) For every $u \in X $ and for every $m\geq 1 $ we have
\[
\sum_{n=1}^{k} |\Delta^2 u_n|^m \leq k\|u\|^m .
\]
\begin{proof}
Of course we have
\begin{equation*}
|\Delta ^2u_{n}|^2\leq \sum_{i=1}^{k}|\Delta ^2u_{i}|^2,\quad
\text{ for every }n\in \mathbb{Z}(1,k),
\end{equation*}
hence we have
\begin{equation*}
|\Delta ^2u_{n}|^m\leq ((\sum_{i=1}^{k}|\Delta ^2u_{i}|^2)^{1/2})^m,
\quad \text{for every }n\in \mathbb{Z}(1,k),
\end{equation*}
Summing left hand side of the inequality from 1 to k we obtain
\begin{equation*}
\sum_{n=1}^{k}|\Delta ^2u_{n}|^m\leq k\Big(\Big(\sum_{n=1}^{k}|\Delta
^2u_{i}|^2\Big)^{1/2}\Big)^m.
\end{equation*}
which leads us to
\begin{equation*}
\sum_{n=1}^{k}|\Delta ^2u_{n}|^m\leq k\| u\| ^m,
\end{equation*}
\end{proof}

(A2) For every $u \in X $ and every $m> 2 $ we have
\[
\sum_{n=1}^{k} |\Delta^2 u_n|^m \geq k^{\frac{2}{m-2}}\|u\|^m .
\]
\begin{proof}
Using H\"older inequality for $m>2$ we obtain
\begin{equation*}
\sum_{n=1}^{k}|\Delta ^2u_{n}|^2
\leq \Big(\sum_{n=1}^{k}1^{\frac{m}{m-2}}\Big)^{\frac{m-2}{m}}
\Big(\sum_{n=1}^{k}(|\Delta ^2u_{n}|^2)^{m/2}\Big)^{2/m}
=k^{\frac{m-2}{m}}\sum_{n=1}^{k}(|\Delta ^2u_{n}|^m)^{2/m}.
\end{equation*}
Calculating further we obtain
\begin{equation*}
\| u\| =\Big(\sum_{n=1}^{k}|\Delta ^2u_{n}|^2\Big)^{1/2}
\leq k^{\frac{m-2}{2m}}\sum_{n=1}^{k}(|\Delta ^2u_{n}|^m)^{1/m}.
\end{equation*}
Thus, we see the thesis
\begin{equation*}
\| u\| ^m\leq k^{\frac{m-2}{2}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^m\Leftrightarrow k^{\frac{2}{m-2}}\| u\| ^m\leq
\sum_{n=1}^{k}|\Delta ^2u_{n}|^m.
\end{equation*}
\end{proof}

(A3) For every $u\in X$ such that $\| u\| \geq 1$ we have
\[
\sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}\geq k^{\frac{2}{\underline{p}-2}
}\| u\| ^{\underline{p}}-k.
\]
\begin{proof}
Let $u\in X$ be such that $\| u\| \geq 1$. We obtain
\begin{align*}
&\sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}\\
&=\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta ^2u_{n}|^{p_{n}}
 +\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|>1\}}|\Delta ^2u_{n}|^{p_{n}}\\
&\geq \sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta
^2u_{n}|^{\overline{p}}+\sum_{\{k\in \mathbb{Z}(1,k);|\Delta
^2u_{k}|>1\}}|\Delta ^2u_{n}|^{\underline{p}}\\
&=\sum_{n=1}^{k}|\Delta ^2u_{n}|^{\underline{p}}-\sum_{\{k\in \mathbb{Z}
(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta ^2u_{n}|^{\underline{p}
}+\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta
^2u_{n}|^{\overline{p}}\\
&\geq \sum_{n=1}^{k}|\Delta ^2u_{n}|^{\underline{p}}-\sum_{n=1}^{k}1\\
&= \sum_{n=1}^{k}|\Delta ^2u_{n}|^{\underline{p}}-k.
\end{align*}
Now we can use (A2) with $m:=\underline{p}$ to get
\begin{equation*}
\sum_{n=1}^{k}|\Delta ^2u_{n}|^{\underline{p}}-k\geq k^{\frac{2}{
\underline{p}-2}}\| u\| ^{\underline{p}}-k,
\end{equation*}
which is our assertion.
\end{proof}

(A4) For every $u\in X$ we have
\[
\sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}\leq k\| u\| ^{\overline{p}}+k.
\]
\begin{proof}
Let us decompose
\begin{align*}
&\sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}\\
&=\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta ^2u_{n}|^{p_{n}}
+\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|>1\}}|\Delta ^2u_{n}|^{p_{n}}\\
&\leq \sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta
^2u_{n}|^{\underline{p}}+\sum_{\{k\in \mathbb{Z}(1,k);|\Delta
^2u_{k}|>1\}}|\Delta ^2u_{n}|^{\overline{p}}\\
&=\sum_{n=1}^{k}|\Delta ^2u_{n}|^{\overline{p}}+\sum_{\{k\in \mathbb{Z}
(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta ^2u_{n}|^{\underline{p}
}-\sum_{\{k\in \mathbb{Z}(1,k);|\Delta ^2u_{k}|\leq 1\}}|\Delta
^2u_{n}|^{\overline{p}}\\
&\leq \sum_{n=1}^{k}|\Delta ^2u_{n}|^{\overline{p}}+\sum_{n=1}^{k}1\\
&=\sum_{n=1}^{k}|\Delta ^2u_{n}|^{\overline{p}}+k.
\end{align*}
Now, using (A1) we have
\begin{equation*}
\sum_{n=1}^{k}|\Delta ^2u_{n}|^{\overline{p}}+k\leq k\| u\| ^{\overline{p}}+k.
\end{equation*}
\end{proof}

(A5) For every $u\in X$ such that $\| u\| \leq 1$ we have
\[
\sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}\geq k^{-\frac{\overline{p}-2}{2}
}\| u\| ^{\overline{p}}.
\]
In this section we have used some ideas from \cite{GalewskiWieteska}.

\section{Existence of solutions}

This section gives theorems with sufficient conditions for the existence of
at least one solution to \eqref{ROW}-\eqref{eq:warunki_brzegowe}.

\begin{theorem} \label{thm3}
Assume that the following hypothesis are satisfied

\begin{itemize}
\item[(F0)] for any $n\in \mathbb{Z}(0,k+1)$, $\gamma_n <0$;

\item[(F1)] there exists a functional $F \in C^1 (\mathbb{Z} \times
\mathbb{R}^2 ,\mathbb{R})$, such that
\[
F_x(n-1,v_2,v_3)+ F_y(n,v_1,v_2)=f(n,v_1,v_2,v_{3});
\]

\item[(F2)] There exists $M_{0}>0$ such that, for all $
(n,v_1,v_2)\in \mathbb{Z}(1,k)\times \mathbb{R}^2$
\[
F_x(n,v_1,v_2)\leq M_{0},F_y(n,v_1,v_2)\leq M_{0}.
\]
Then \eqref{ROW}-\eqref{eq:warunki_brzegowe} possesses at least one
solution.
\end{itemize}
\end{theorem}

\begin{remark} \rm
Assumption (F2) implies that there exists a constant $M_1$ such that
\begin{itemize}
\item[(F2')]
$ |F(n,v_1,v_2)|\leq M_1+M_{0}(|v_1|+|v_2|)$
for all $(n,v_1,v_2)\in \mathbb{Z}(1,k)\times \mathbb{R}^2$.
\end{itemize}
Let us define a function $H:[0,1]\to \mathbb{R}$, 
$H(t)=F(n,tv_1,tv_2)$. Then $H$ is differentiable, and
\[
H'(t)=F_x(n,tv_1,tv_2)v_1+F_y(n,tv_1,tv_2)v_2.
\]
Using the mean value theorem on $[0,1]$ we obtain
\begin{align*}
F(n,v_1,v_2)-F(n,0,0)
&=H(1)-H(0)=H'(\theta )(1-0) \\
&=F_x(n,\theta v_1,\theta v_2)v_1+F_y(n,\theta v_1,\theta v_2)v_2
\end{align*}
for some $\theta \in [ 0,1]$. Now, using assumption (F2) we obtain
\begin{align*}
|F(n,v_1,v_2)-F(n,0,0)|
&=|F_x(n,\theta v_1,\theta v_2)v_1+F_y(n,\theta v_1,\theta v_2)v_2|\\
&\leq |F_x(n,\theta v_1,\theta v_2) 
||v_1|+|F_y(n,\theta v_1,\theta v_2) ||v_2|\\
&\leq M_{0}|v_1|+M_{0}|v_2|.
\end{align*}
On the other hand, using a well know inequality for absolute value we obtain
\[
|F(n,v_1,v_2)-F(n,0,0)|\geq ||F(n,v_1,v_2)|-|F(n,0,0)||,
\]
and combining both inequalities we produce the following statement
\[
||F(n,v_1,v_2)|-|F(n,0,0)||\leq M_{0}|v_1|+M_{0}|v_2|.
\]
By the definition of the absolute value, it is equivalent to
\[
-M_{0}(|v_1|+|v_2|)\leq |F(n,v_1,v_2)|-|F(n,0,0)|\leq
M_{0}|v_1|+M_{0}|v_2|,
\]
which leads us to thesis substituting 
$M_1:=\max_{n\in \mathbb{Z}(1,k)}\{|F(n,0,0)|\}$.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm3}]
By (F2'), for any $u\in \mathbb{R}^{k}$, we have
\begin{align*}
J(u)
&=\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta ^2u_{n}|^{p_{n}}
  -\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\leq \frac{\overline{\gamma }}{\overline{p}}
  \sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}
  -\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\leq \frac{\overline{\gamma }}{\overline{p}}
 \sum_{n=1}^{k}|\Delta^2u_{n}|^{p_{n}}+M_{0}\sum_{n=1}^{k}(|u_{n+1}|+|u_{n}|)+M_1k\\
& \leq \frac{\overline{\gamma }}{\overline{p}}
 \sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_{n}}+2M_{0}\sum_{n=1}^{k}|u_{n}|+M_1k.
\end{align*}
Now, using (A3) we obtain
\begin{align*}
&\frac{\overline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
  ^2u_{n}|^{p_{n}}+2M_{0}\sum_{n=1}^{k}|u_{n}|+M_1k\\
&\leq \frac{\overline{\gamma }}{\overline{p}}(k^{\frac{2}{\underline{p}-2}}
  \| u\| ^{\underline{p}}-k)+2M_{0}\sum_{n=1}^{k}|u_{n}|+M_1k\\
&\leq \frac{\overline{\gamma }}{\overline{p}}k^{\frac{2}{\underline{p}-2}}
 \| u\| ^{\underline{p}}-\frac{\overline{\gamma }}{\overline{p}}k
 +2M_{0}c_{2,1}\| u\| +M_1k\to -\infty \quad\text{as }\| u\| \to +\infty .
\end{align*}
Above inequality means that $J$ is anti coercive. With continuity of $J$, it
attains its maximum at some point. From necessity condition of extremal
point of differentiable functional, we acquire that 
$u_{0}:=\max \{J(u):u\in X\}$ is a critical point of $J$. 
This finishes the proof.
\end{proof}

\begin{theorem} \label{thm5}
Suppose that {\rm (F1)} and the following hypothesis are satisfied
\begin{itemize}
\item[(F0')] For every $n\in \mathbb{Z}(1,k+1), \gamma_n > 0 $;

\item[(F3)] There exist $R>0$, $1<\alpha <2 $ and constants $a_1,a_2 >0$
such that for $n\in \mathbb{Z}(1,k)$ and $\sqrt{(v_1^2 + v_2^2)} \geq R$;
\[
F(n,v_1,v_2) \leq a_1 (\sqrt{v_1^2 + v_2^2})^{\frac{\alpha}{2}\underline{p}}
-a_2\,.
\]
\end{itemize}
Then  \eqref{ROW}-\eqref{eq:warunki_brzegowe} possesses at least one
solution.
\end{theorem}

\begin{proof}
By (F3) for any $u\in \mathbb{R}^{k}$, we have
\begin{align*}
J(u)&=\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
^2u_{n}|^{p_n}-\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\geq \frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\geq \frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-a_1\sum_{n=1}^{k}(\sqrt{u_{n+1}^2+u_{n}^2})^{\frac{
\alpha }{2}\underline{p}}-a_2k\\
&\geq \frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-a_1\sum_{n=1}^{k}(\sqrt{\sum_{i=1}^k u_{i}^2})^{
\frac{\alpha }{2}\underline{p}}-a_2k\\
&\geq \frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-a_1 \sum_{n=1}^{k}\| u\|_2^{\frac{\alpha }{2}\underline{p}}-a_2k\\
&\geq \frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-a_1k\| u\|_2^{\frac{\alpha }{2}\underline{p}}-a_2k.
\end{align*}
Again we will use (A3). Indeed, we have
\begin{align*}
&\frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-a_1k\| u\|_2^{\alpha \underline{p}/2}-a_2k\\
&\geq \frac{\underline{\gamma }}{\overline{p}}(k^{\frac{2}{\underline{p}-2}}
 \| u\| ^{\underline{p}}-k)-a_1k\| u\|_2^{\frac{\alpha }{2}\underline{p}}-a_2k\\
&\geq \frac{\underline{\gamma }}{\overline{p}}k^{\frac{2}{\underline{p}-2}
}\| u\| ^{\underline{p}}-\frac{\underline{\gamma }}{\overline{p}}
k-a_1k\| u\|_2^{\frac{\alpha }{2}\underline{p}}-a_2k\to +\infty \quad
\text{as }\| u\| \to +\infty .
\end{align*}
This inequality implies that $J$ is coercive, and using similar reasoning,
we acquire that  \eqref{ROW}-\eqref{eq:warunki_brzegowe} 
possesses at least one solution.
\end{proof}

\section{Existence and multiplicity of solutions}

This section will give sufficient conditions to existing at least two
solutions to \eqref{ROW}-\eqref{eq:warunki_brzegowe}.

\begin{theorem} \label{thm6}
Suppose {\rm (F0'), (F1)} and the following conditions are satisfied:
\begin{itemize}

\item[(F4)] Functional $F$ satisfies
\[
\lim_{r\to 0} \frac{F(n,v_1,v_2)}{r^{\overline{p}}} = 0, \quad
 r=\sqrt{v_1^2+v_2^2};
\]

\item[(F5)] There exist $\beta > \overline{p}$ and $a_3>0$ such that
for $n\in \mathbb{Z}(1,k)$ and $\sqrt{(v_1^2 + v_2^2)} \geq R$
\[
F(n,v_1,v_2) > a_3(\sqrt{v_1^2 + v_2^2})^{\beta}.
\]
\end{itemize}
Then \eqref{ROW}-\eqref{eq:warunki_brzegowe} possesses at least two
nontrivial solutions.
\end{theorem}

\begin{proof}
To show that our functional satisfies the P.S. condition we use 
that any anti-coercive functional $ T:X \to \mathbb{R}$, 
where $\dim{X} < \infty $, satisfies the P.S. condition.
By (F5) we have
\begin{align*}
J(u)&=\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
^2u_{n}|^{p_{n}}-\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\leq \sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
^2u_{n}|^{p_{n}}-\sum_{n=1}^{k}a_{3}(\sqrt{u_{n+1}^2+u_{n}^2})^{\beta}.
\end{align*}
Now, using (A2) and (A3) we have
\begin{align*}
&\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
 ^2u_{n}|^{p_{n}}-\sum_{n=1}^{k}a_{3}(\sqrt{u_{n+1}^2+u_{n}^2})^{\beta}\\
&\leq \frac{\overline{\gamma }}{\underline{p}}k\| u\| ^{\overline{p}}
 +\frac{\overline{\gamma }}{\underline{p}}k-a_3
 \Big(\sum_{n=1}^{k}\sqrt{u_{n}^2}\Big)^{\beta }\\
&=\frac{\overline{\gamma }}{\underline{p}}k\| u\| ^{\overline{p}}
 +\frac{\overline{\gamma }}{\underline{p}}k-a_3\| u\| _1^{\beta }\\
&\leq \frac{\overline{\gamma }}{\underline{p}}k\| u\| ^{\overline{p}}
 +\frac{\overline{\gamma }}{\underline{p}}k-a_3 c_{2,1}^{\beta }
 \| u\| ^{\beta } \to -\infty \quad \text{as } \| u \| \to \infty,
\end{align*}
We proved that $J$ is anti coercive, thus, the P.S. condition is verified.

Now, we have to show that other conditions of Mountain Pass Lemma are
satisfied. By (F4), for any 
\[
\epsilon =k^{-\frac{\overline{p}-4}{2}}
\frac{\underline{\gamma }}{2\overline{p}(c_{2,2})^{\overline{p}}},
\]
there exists $\rho >0$ such that
\begin{equation*}
|F(n,v_1,v_2)|\leq \epsilon 
(v_1+v_2)^{\overline{p}/2}\quad  \forall _{n\in \mathbb{Z}(1,k)}
\end{equation*}
for $\sqrt{v_1^2+v_2^2}\leq 2\sqrt{\rho }$. Then
\begin{align*}
J(u)
&=\sum_{n=1}^{k}\frac{\gamma _{n+1}}{p_{n}}|\Delta
^2u_{n}|^{p_n}-\sum_{n=1}^{k}F(n,u_{n+1},u_{n})\\
&\geq \frac{\underline{\gamma }}{\overline{p}}
 \sum_{n=1}^{k}|\Delta ^2u_{n}|^{p_n}-\epsilon
\sum_{n=1}^{k}(u_{n+1}^2+u_{n}^2)^{\overline{p}/2}.
\end{align*}
Now, using (A5), we can estimate
\begin{align*}
&\frac{\underline{\gamma }}{\overline{p}}\sum_{n=1}^{k}|\Delta
^2u_{n}|^{p_{n}}-\epsilon \sum_{n=1}^{k}
 (u_{n+1}^2+u_{n}^2)^{\overline{p}/2}\\
&\geq \frac{\underline{\gamma }}{\overline{p}}
 k^{-\frac{\overline{p}-2}{2}}\| u\| ^{\overline{p}}
 -\epsilon \sum_{n=1}^{k}(\sqrt{\sum_{i=1}^k u_{i}^2})^{\overline{p}}\\\
&\geq \frac{\underline{\gamma }}{\overline{p}}k^{-\frac{\overline{p}-2}{2}}
 \| u\| ^{\overline{p}}-\epsilon \sum_{n=1}^{k}\| u\|_2^{\overline{p}}\\
&\geq \frac{\underline{\gamma }}{\overline{p}}
 k^{-\frac{\overline{p}-2}{2}}\| u\| ^{\overline{p}}
 -\epsilon k\| u\| _2^{\overline{p}}\\
&\geq \frac{\underline{\gamma }}{\overline{p}}k^{-\frac{\overline{p}-2}{2}}
 \| u\| ^{\overline{p}}-\epsilon kc_{2,2}^{\overline{p}}\| u\|^{\overline{p}}\\
&=\frac{\underline{\gamma }}{\overline{p}}k^{-\frac{\overline{p}-2}{2}}
 \| u\| ^{\overline{p}}-k^{-\frac{\overline{p}-2}{2}}
 \frac{\underline{\gamma }}{2\overline{p}(c_{2,2})^{\overline{p}}}(c_{2,2}
 )^{\overline{p}}\| u\| ^{\overline{p}}\\
&=\frac{\underline{\gamma }}{2\overline{p}}k^{-\frac{\overline{p}-2}{2}}\|
u\| ^{\overline{p}}
\end{align*}
Take $a=\frac{\underline{\gamma }}{2\overline{p}}k^{-\frac{\overline{p}-2}{2}}
\rho ^{\overline{p}}>0$. Therefore
\begin{equation*}
J(u)\geq a>0\quad \forall _{u\in \partial B}.
\end{equation*}
At the same time, we have also proved that there exist constants $a>0$ and $
\rho >0$ such that $J|_{\partial B}\geq a$. That is to say, J satisfies
(J1) of the Mountain Pass Lemma. 

For our setting, $J(0)=0$. To exploit the Mountain Pass Lemma in critical
point theory, we need to verify other conditions of the lemma. We have shown
that J satisfies the P.S. condition. So, it suffices to verify the condition
(J2). From the proof of the P.S. condition we know that
\begin{equation*}
J(u)\leq\frac{\overline{\gamma }}{\underline{p}}k\| u\| ^{\overline{p}}
+\frac{\overline{\gamma }}{\underline{p}}k-a_3 c_{2,1}^{\beta }\| u\| ^{\beta }.
\end{equation*}
Since $\beta >\overline{p}$, we can choose $u^{\ast }$ far enough to ensure
that $J(u^{\ast })<0$. By the Mountain Pass Lemma, $J$ possesses a critical
value $c\geq a>0$ where
\begin{gather*}
c=\inf_{h\in \Gamma }\sup_{s\in [ 0,1]}J(h(s)), \\
\Gamma =\{h\in C([0,1],\mathbb{R}^{k})|h(0)=0,h(1)=u^{\ast }\}.
\end{gather*}
Let $\overline{u}\in \mathbb{R}^{k}$ be a critical point associated to the
critical value c of J. Due to anti coercivity and continuity, we know
that there exists $\hat{u}$ such that
\begin{equation*}
J(\hat{u})=c_{\rm max}=\max_{s\in [ 0,1]}J(h(s)).
\end{equation*}
Clearly, $\hat{u}\neq 0$. If $\overline{u}\neq \hat{u}$ we reach the
assertion of the theorem. 

Suppose that $\overline{u}=\hat{u}$. Itimplies that
\begin{equation*}
J(\overline{u})=\inf_{h\in \Gamma }\sup_{s\in [ 0,1]}J(h(s)).
\end{equation*}
Hence for any function $h\in \Gamma $, $\max_{t\in [ 0,1]}J(h(t))=J(
\overline{u})$. Indeed, for any $h\in \Gamma $ we have
\begin{equation*}
J(\overline{u})\geq \max_{t\in [ 0,1]}J(h(t))
\end{equation*}
since
\begin{equation*}
J(\overline{p})=\max_{x\in X}J(x)\text{ and }J(\overline{p})\leq \max_{t\in
[ 0,1]}J(h(t))
\end{equation*}
by the definition of the minimum. Since $k>1$, the space $X\backslash \{
\overline{u}\}$ is path connected.
Then there exists a function $h_{0}\in \Gamma $ such that $h_{0}(t)\neq
\overline{u}$ for $t\in [ 0,1]$. Since $\max_{t\in [
0,1]}J(h_{0}(t))=J(\overline{u})$ it follows that there exists $t_{0}\in
(0,1)$ such that $J(h_{0}(t_{0}))=\max_{x\in X}J(x)$ and by assertion $
h_{0}(t_{0})\neq \overline{u}$. Thus $h_{0}(t_{0}$ is a critical point
different from $\overline{u}$. \newline
The above argumentation implies that  \eqref{ROW}-\eqref{eq:warunki_brzegowe} 
possesses at least two nontrivial solutions.
\end{proof}

\section{Nonexistence of solutions}

This section give  sufficient conditions for the  nonexistence of
nontrivial solutions to \eqref{ROW}-\eqref{eq:warunki_brzegowe}.

\begin{theorem} \label{thm7}
Let {\rm (F0), (F1)} and the following conditions be satisfied.
\begin{itemize}
\item[(F6)] For all $n\in \mathbb{Z}(1,k)$, $v_2\neq 0
 \Rightarrow v_2 f(n,v_1,v_2,v_3)>0$.
\end{itemize}
Then \eqref{ROW}-\eqref{eq:warunki_brzegowe} has no nontrivial solution.
\end{theorem}

\begin{proof}
Assume in the sake of contradiction that \eqref{ROW}-\eqref{eq:warunki_brzegowe} 
possesses a nontrivial solution. Then, functional $J$ has a nonzero critical point 
$u^{\ast }$. Since
\begin{equation*}
\frac{\partial J}{\partial u_{n}}=\Delta ^2(\gamma _{n-1}\phi
_{p_{n}}(\Delta ^2u_{n}))-f(n,u_{n+1},u_{n},u_{n-1}),
\end{equation*}
it follows that
\begin{align*}
&\sum_{n=1}^{k}f(n,u_{n+1}^{\ast },u_{n}^{\ast },u_{n-1}^{\ast })\cdot
u_{n}^{\ast }\\
&=\sum_{n=1}^{k}\Delta ^2(\gamma _{n-1}\phi _{p_{n}}(\Delta
^2u_{n-2}^{\ast }))\cdot u_{n}^{\ast }\\
&=\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n}^{\ast }-2\gamma _{n}|\Delta ^2u_{n-1}^{\ast
}|^{p_{n+1}-2}(\Delta ^2u_{n-1}^{\ast })u_{n}^{\ast }\\
&\quad +\gamma _{n-1}|\Delta ^2u_{n-2}^{\ast }|^{p_{n}-2}(\Delta
^2u_{n-2}^{\ast })u_{n}^{\ast }\\
&=\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n}^{\ast }-2\sum_{n=1}^{k}\gamma _{n}|\Delta
^2u_{n-1}^{\ast }|^{p_{n+1}-2}(\Delta ^2u_{n-1}^{\ast })u_{n}^{\ast }\\
&\quad +\sum_{n=1}^{k}\gamma _{n-1}|\Delta ^2u_{n-2}^{\ast }|^{p_{n}-2}(\Delta
^2u_{n-2}^{\ast })u_{n}^{\ast }\\
&=\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n}^{\ast }-2\sum_{n=0}^{k-1}\gamma _{n+1}|\Delta
^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta ^2u_{n}^{\ast })u_{n+1}^{\ast } \\
&\quad +\sum_{n=-1}^{k-2}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n+2}^{\ast } \\
&=\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n}^{\ast }-2\sum_{n=1}^{k}\gamma _{n+1}|\Delta
^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta ^2u_{n}^{\ast })u_{n+1}^{\ast } \\
&\quad +\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })u_{n+2}^{\ast }+[-2\gamma _1|\Delta ^2u_{0}^{\ast
}|^{p_2-2}(\Delta ^2u_{0}^{\ast })u_1^{\ast }\\
&\quad +2\gamma _{k+1}|\Delta ^2u_{k}^{\ast }|^{p_{k}+2-2}(\Delta ^2u_{k}^{\ast
})u_{k+1}^{\ast }+\gamma _{0}|\Delta ^2u_{-1}^{\ast }|^{p_1-2}(\Delta
^2u_{-1}^{\ast })u_1^{\ast } \\
&\quad +\gamma _1|\Delta ^2u_{0}^{\ast }|^{p_2-2}(\Delta ^2u_{0}^{\ast
})u_2^{\ast }-\gamma _{k}|\Delta ^2u_{k-1}^{\ast }|^{p_{k}+1-2}(\Delta
^2u_{k-1}^{\ast })u_{k+1}^{\ast } \\
&\quad -\gamma _{k+1}|\Delta ^2u_{k}^{\ast }|^{p_{k+2}-2}(\Delta ^2u_{k}^{\ast
})u_{k+2}^{\ast }].
\end{align*}
Using boundary values, it is easy to see that the expression in square
bracket is equal to zero. This implies 
\begin{align*}
&\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })(u_{n}^{\ast }-2u_{n+1}^{\ast }+u_{n+2}^{\ast}) \\
&=\sum_{n=1}^{k}\gamma _{n+1}|\Delta ^2u_{n}^{\ast }|^{p_{n+2}-2}(\Delta
^2u_{n}^{\ast })^2<0,
\end{align*}
which is a contradiction with assumption. 
Hence, the only critical point of $J$ is 0.
\end{proof}

\section{Final comments and examples}

Firstly note that the classical approach to the positive solutions do not
apply to the fourth-order problems. It is so because of the
inequality
\[
\Delta ^2u_{n}\cdot \Delta ^2u_{-n}\leq 0,
\]
where $u_{-}=\max\{-u,0\}$, is not satisfied for all $u\in X$. Indeed, take
\[
u_{n}=5,\quad u_{n+1}=1,\quad u_{n+2}=-2.
\]
Substituting symbols by numbers we obtain
\begin{align*}
\Delta ^2u_{n}\cdot \Delta^2u_{-n}
&=(u_{n+2}-2u_{n+1}+u_{n})(u_{-n+2}-2u_{-n+1}+u_{-n})\\
&=(-2-2+5)\cdot (2+0+0)=2>0.
\end{align*}

Now, we shown four examples to illustrate the main results.

\begin{example} \label{examp1} \rm
For $n\in \mathbb{Z}(1,k)$, assume that
\[
\Delta^2 (-2n \phi_{p_n}(\Delta^2 u_{n-2})) = \Phi(n-1)\cos{u_n}\cos{
u_{n-1}} -\Phi(n)\sin{u_{n+1}}\sin{u_n},
\]
with boundary value conditions \eqref{eq:warunki_brzegowe}, where 
$p_n: \mathbb{Z}(1,k)\to\mathbb{R}$, $\Phi (n) >0$,
$n\in \mathbb{Z}(1,k)$. 
We have
\begin{gather*}
\gamma_n = -2(n+1), \quad
 f(n,v_1,v_2,v_3) = \Phi(n-1)\cos{v_2}\cos{v_3}-\Phi(n)\sin{v_1}\sin{v_2},\\
F(n,v_1,v_2) = \Phi(n) \sin{v_1} \cos{v_2}.
\end{gather*}
It is easy to verify that all the assumptions of Theorem \ref{thm3} are satisfied,
thus our problem possesses at least one solution.
\end{example}

\begin{example} \label{examp2} \rm
For $n\in \mathbb{Z}(1,k)$, assume that
\[
\Delta^2 (6^{n-1} \phi_{p_n}(\Delta^2 u_{n-2})) 
= \alpha u_n [ \psi(n)(u^2_{n+1} + u_n^2)^{\frac{\alpha}{4} \underline{p} -1} +
\psi(n-1)(u^2_{n} + u_{n-1}^2)^{\frac{\alpha}{4} \underline{p} -1}],
\]
with boundary value conditions \eqref{eq:warunki_brzegowe}, where 
$p_n:\mathbb{Z}(1,k)\to\mathbb{R}$, $\psi (n) >0$,
$n\in \mathbb{Z}(1,k)$, $1<\alpha <2$. We have
\begin{gather*}
\gamma_n = 6^n, \quad
 f(n,v_1,v_2,v_3) = \alpha v_2 [ \psi(n)(v_1^2 +v_2^2)^{\frac{\alpha}{4} 
\underline{p} -1} + \psi(n-1)(v^2_2 +v_3^2)^{\frac{\alpha}{4} \underline{p} -1}],\\
F(n,v_1,v_2) = \psi(n) (v_1^2 + v_2^2)^{\frac{\alpha}{4} \underline{p}}.
\end{gather*}
We can easily check that all the assumptions of Theorem \ref{thm3} are satisfied,
hence our problem possesses at least one solution.
\end{example}

\begin{example} \label{examp3}\rm
For $n\in \mathbb{Z}(1,k)$, assume that
\[
\Delta^2 (\phi_{p_n}(\Delta^2 u_{n-2})) 
= \beta u_n [ \psi(n)(u^2_{n+1} + u_n^2)^{\frac{\beta}{2} -1} 
+ \psi(n-1)(u^2_{n} + u_{n-1}^2)^{\frac{\beta}{2}-1}]
\]
with boundary value conditions \eqref{eq:warunki_brzegowe}, where
 $p_n: \mathbb{Z}(1,k)\to\mathbb{R}$, $\psi (n) >0$,
$n\in \mathbb{Z}(1,k)$, $\beta > \overline{p} $. We have
\begin{gather*}
\gamma_n \equiv 1,\quad
 f(n,v_1,v_2,v_3) = \beta v_2 [ \psi(n)(v_1^2 +v_2^2)^{\frac{\beta}{2} -1} 
+ \psi(n-1)(v^2_2 +v_3^2)^{\frac{\beta}{2} -1}], \\
F(n,v_1,v_2) = \psi(n) (v_1^2 + v_2^2)^{\frac{\beta}{2} }.
\end{gather*}
Then, the assumptions of Theorem \ref{thm6} are satisfied, hence our problem
possesses at least two solutions.
\end{example}

\begin{example} \label{examp4} \rm
For $n\in \mathbb{Z}(1,k)$, assume that
\[
\Delta^2 (-\phi_{p_n}(\Delta^2 u_{n-2})) = 4 u_n [ (u^2_{n+1} +
u_n^2) + (u^2_{n} + u_{n-1}^2) ],
\]
with boundary value conditions \eqref{eq:warunki_brzegowe}, where 
$p_n:\mathbb{Z}(1,k)\to\mathbb{R}$.
We have
\begin{gather*}
\gamma _{n}\equiv -1,\;f(n,v_1,v_2,v_{3})
=4v_2[(v_1^2+v_2^2)+(v_2^2+v_{3}^2)], \\
F(n,v_1,v_2)=(v_1^2+v_2^2)^2.
\end{gather*}
Then, the assumptions of Theorem \ref{thm7} are satisfied, hence our problem 
has no nontrivial solutions.
\end{example}

\subsection*{Acknowledgements}
The Author thanks Professor Marek Galewski from Lodz University of
Technology for some suggestions on the topic of this research.


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\end{document}