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\title{\sc Spectral estimates for a nonhomogeneous difference problem\,\thanks{ Correspondence address: Vicen\c{t}iu
R\u{a}dulescu, Department of Mathematics, University of Craiova,
200585 Craiova, Romania. E-mail: {\tt
vicentiu.radulescu@imar.ro}}}
\author{\sc Alexandru Krist\'aly$\,^{a,b}$\qquad Mihai Mih\u ailescu$\,^{b,c}$\\ \sc  Vicen\c{t}iu
R\u{a}dulescu$\,^{c,d}$\qquad Stepan  Tersian$\,^{e}$\\ \small
$^a\,$Department of Economics, University of Babe\c s-Bolyai,
400591 Cluj-Napoca, Romania\\ \small $^b\,$Department of
Mathematics, Central European University,  1051 Budapest,
Hungary\\ \small $^c\,$Department of Mathematics, University of
Craiova, 200585 Craiova, Romania\\ \small $^d\,$Institute of
Mathematics ``Simion Stoilow" of the Romanian Academy,\\ \small
P.O. Box 1-764, 014700 Bucharest, Romania\\ \small
$^e\,$Department of Mathematical Analysis, University of Rousse,
7017 Rousse, Bulgaria\\ \small E-mail: {\tt
alexandrukristaly@yahoo.com}\quad {\tt mmihailes@yahoo.com}\\
\small {\tt vicentiu.radulescu@imar.ro}\quad {\tt
sterzian@ru.acad.bg}}

\begin{document}

\baselineskip16pt \maketitle\noindent{\small{\sc Abstract}. We
study an eigenvalue problem in the framework of difference
equations. We show that there exist two positive constants
$\lambda_0$ and $\lambda_1$ verifying $\lambda_0\leq\lambda_1$
such that any $\lambda\in(0,\lambda_0)$ is not an eigenvalue of
the problem, while any $\lambda\in[\lambda_1,\infty)$ is an
eigenvalue of the problem. Some estimates for $\lambda_0$ and
$\lambda_1$ are also given.\\ \\ \small{\bf 2000 Mathematics
Subject Classification:}  47A75, 35B38, 35P30, 34L05, 34L30. \\
\small{\bf Key words:}  eigenvalue problem, discrete boundary
value problem, critical point,  continuous spectrum.}


\section{Introduction and main results}

Discrete boundary value problems have been intensively studied in
the last decade. The modeling of certain nonlinear problems from
biological neural networks, economics, optimal control and other
areas of study have led to the rapid development
 of the theory of difference equations (see the
monographs of Agarwal \cite{A} and Kelley-Peterson \cite{KP} and
the papers of Agarwal-Perera-O'Regan \cite{APR,APR1},
Cabada-Iannizzotto-Tersian \cite{cit}, Cai-Yu \cite{CY}, Mih\u
ailescu-R\u adulescu-Tersian \cite{mrt}, Yu-Guo \cite{YG},
Zhang-Liu \cite{ZL} and the reference therein).

In view of developing a viable theory of discrete boundary
problems, special attention has been given to the study of the
{\it spectrum} of certain {\it eigenvalue problems}. A classical
result in the theory of eigenvalue problems involving difference
equations asserts that the spectrum of the problem
\begin{equation}\label{diri}
\left\{\begin{array}{lll} -\Delta(\Delta u(k-1))=\lambda u(k),
\quad k\in[1,T],\\ u(0)=u(T+1)=0\,,
\end{array}\right.
\end{equation}
where $T\geq 2$ is an integer, $[1,T]$ is the discrete interval
$\{1,2,...,T\}$ and $\Delta u(k)=u(k+1)-u(k)$ is the forward
difference operator, is {\it finite} and all the eigenvalues are
positive. Since $u\mapsto\Delta(\Delta u(\cdot))$ is a linear
operator, the above statement is well illustrated  if one
considers, instead of  (\ref{diri}) the algebraic equation
$cs=\lambda s,$ $s\in \mathbb R$, where $c>0$ is a fixed number.

On the other hand, some recent advances obtained in
\cite{APR,cit,mrt} show that for some eigenvalue problems
involving difference operators the spectrum contains a {\it
continuous family} of eigenvalues.

The goal of this paper is to complete the studies begun in the
above papers by presenting a new phenomenon concerning the
behaviour of eigenvalues of a nonhomogeneous
difference equation. %  that will be described in the sequel.
Using the above notations, this paper is concerned with the study
of the eigenvalue problem
\begin{equation}\label{1}
\left\{\begin{array}{lll} -\Delta(\Delta
u(k-1))+|u(k)|^{q-2}u(k)=\lambda g(k)|u(k)|^{r-2}u(k), \quad
k\in[1,T],\\ u(0)=u(T+1)=0\,,
\end{array}\right.
\end{equation}
where
%$\Delta u(k)=u(k+1)-u(k)$ is the forward difference
%operator,
$q$ and $r$ are two real numbers satisfying $2<r<q$ and
$g:[1,T]\rightarrow(0,\infty)$ is a given function.

{Similar to the previous case, we formally consider instead of
(\ref{1})
 the algebraic equation $$cs+|s|^{q-2}s=\lambda
|s|^{r-2}s,\ s\in \mathbb R.\eqno{(\ref{1}')}$$ Note that if we
define $f:(0,\infty)\to (0,\infty)$ by $f(t)=ct^{2-r}+t^{q-r}$ and
the positive number
$t_0=\left(\frac{c(r-2)}{q-r}\right)^\frac{1}{q-2}$, equation
$(\ref{1}')$ has no non-zero solutions for $0<\lambda<f(t_0)$,
while for any $\lambda \geq f(t_0)$, equation $(\ref{1}')$ has
non-zero solutions.}


{By finding solutions to problem  \eq{1}, we expect to obtain a
similar phenomenon  as the one we described for the algebraic
equation $(\ref{1}')$. Note, however, that the presence of the
function $g$ as well as the nonhomogeneous nature of problem
\eq{1} make this problem more difficult. We shall prove the
existence of {\it two} positive numbers $\lambda_0$ and
$\lambda_1$, with $\lambda_0\leq\lambda_1$ such that for
$\lambda\in(0,\lambda_0)$
 problem \eq{1} has no non-zero solutions while for any
$\lambda\in[\lambda_1,\infty)$  problem \eq{1} has non-zero
solutions in a specific function space. Moreover, useful estimates
will be also given for $\lambda_0$ and $\lambda_1$ with respect to
the initial data $q$, $r$, $T$ and $g$.}

{In order to describe our result in its full generality we first
define the function space} $$H=\{u:[0,T+1]\rightarrow\RR;\
u(0)=u(T+1)=0\} .$$ Clearly, $H$ is a $T$-dimensional Hilbert
space (see \cite{APR}) with the inner product
$$(u,v)=\sum_{k=1}^{T+1}\Delta u(k-1)\Delta
v(k-1),\;\;\;\forall\;u,v\in H\,.$$ The associated norm is defined
by $$\|u\|=\left(\sum_{k=1}^{T+1}|\Delta
u(k-1)|^2\right)^{1/2}\,.$$ We say that $\lambda\in\RR$ is an {\it
eigenvalue} of problem \eq{1} if there exists $u\in
H\setminus\{0\}$ such that $$\sum_{k=1}^{T+1}\Delta u(k-1)\Delta
v(k-1)+\sum_{k=1}^{T}|u(k)|^{q-2}u(k)v(k)
-\lambda\sum_{k=1}^{T}g(k)|u(k)|^{r-2}u(k)v(k)=0,\ \ \forall v\in
H.$$ The function $u$ in the above definition will be called an
{\it eigenvector} of problem \eq{1}. The set of all eigenvalues of
problem \eq{1} will be called the {\it spectrum} of problem
\eq{1}.

The following theorem represents the main result of our paper.

\begin{teo}\label{main}
Let $2<r<q$, $T\geq 2$ and $g:[1,T]\rightarrow(0,\infty)$ be a
given function. Then there exist two positive constants
$\lambda_0$ and $\lambda_1$ with $\lambda_0\leq\lambda_1$ such
that no $\lambda\in(0,\lambda_0)$ is an eigenvalue of problem
\eq{1} while any $\lambda\in[\lambda_1,\infty)$ is an eigenvalue
of problem \eq{1}. Moreover, we have
\begin{equation}\label{tetel-osszefugges}
  \lambda_1\leq \frac{r}{2}\lambda_0\ \ {\rm and}\ \
\frac{4}{(T+1)^2|g|_\infty}\leq \lambda_0\leq \lambda_1 \leq
\frac{r(q-2)}{(q-r)\sum_{k=1}^T
g(k)}\left(\frac{T(q-r)}{q(r-2)}\right)^{\frac{r-2}{q-2}},
\end{equation}
where $|g|_\infty=\max_{k\in[1,T]}g(k).$
%assuming that
%$$\min\left\{\frac{2}{(T+1)^2},\frac{1}{q}\right\}\cdot
%r\geq\max_{k\in[1,T]}{g(k)}\,,$$ then
%$\lambda_1([0,T+1])\leq\lambda_1$, where $\lambda_1([0,T+1])$
%denotes the first eigenvalue of problem \eq{diri}. Furthermore,
%assuming that $$\frac{4}{(T+1)^2}\geq\max_{k\in[1,T]}{g(k)}\,,$$
%then $\lambda_1([0,T+1])\leq\lambda_0$.
\end{teo}

\noindent {{\it Notation.} For any $a$ and $b$ integers satisfying
$a<b$ we denote by $[a,b]$ the discrete interval
$\{a,a+1,...,b\}$.}

\section{Some estimates of eigenvalues}
In this section we will point out certain remarks on how we can
estimate the positive eigenvalues corresponding to positive
eigenvectors for the problem
\begin{equation}\label{2}
\left\{\begin{array}{lll} -\Delta(\Delta u(k-1))=\lambda u(k),
\quad k\in[1,T],\\ u(0)=u(T+1)=0\,.
\end{array}\right.
\end{equation}
In this section, the main result is given by Theorem \ref{t1}
which is of interest in its own right as well:
\begin{teo}\label{t1}
Let $\lambda>0$ be an eigenvalue of problem \eq{2} with the
property that the corresponding eigenvector
$u=\{u(k):k\in[0,T+1]\}$ is positive, i.e. $u(k)>0$ for any
$k\in[1,T]$. Then we have the estimates
\begin{equation}\label{misto}\frac{4}{(T+1)^2}<\lambda\leq
\min\left\{1,\frac{1}{T}\cdot\left(1+\frac{\max\{u(1),u(T)\}}{\min\{u(1),u(T)\}}\right)\right\}\,.\end{equation}
\end{teo}
\proof First, we point out certain general remarks on the behavior
of  $\Delta u(k)$ for $k\in[0,T]$. Since $u(k)>0$ for $k\in[1,T]$
satisfies equation \eq{1} and $\lambda>0$ we have $$\Delta(\Delta
u(k-1))=-\lambda u(k)<0,\;\;\;\forall\;k\in[1,T]\,.$$ Thus, we
deduce that the sequence $(\Delta u(k))$ is decreasing for
$k\in[0,T]$.

Second, we show that the left inequality holds true. In order to
do that we start by defining $$m=\max\{s\in[1,T];\;\;\Delta
u(s-1)\geq 0,\;\Delta u(s)<0\}\,.$$ Undoubtedly, $m$ can be
defined as above since we have $u(T+1)=0$ and $\Delta
u(T)=u(T+1)-u(T)=-u(T)<0$. (Actually, $m$ is the largest local
maximum point of $u$ in $[1,T]$.)

On the other hand,  since $\Delta u(m)<0$ and $(\Delta u(k))$ is a
decreasing sequence for $k\in[0,T]$ we notice that $$\Delta
u(k)<0,\;\;\;\forall\;k\in[m,T]\,,$$ and thus,
$$u(k+1)<u(k),\;\;\;\forall\;k\in[m,T]\,,$$ i.e. the sequence
$(u(k))$ is strictly decreasing for $k\in[m,T]$. A similar
argument, based on the fact that $\Delta u(m-1)\geq 0$ implies
that $\Delta u(k)\geq 0$ for any $k\in[0,m-1]$, i.e. the sequence
$(u(k))$ is nondecreasing for $k\in[0,m]$.

Adding the identities $u(k)-u(k-1)=\Delta u(k-1)$ for $k\in
[m+1,T+1]$ we obtain $u(T+1)-u(m)=\sum_{k=m+1}^{T+1}\Delta u
(k-1)\geq (T+1-m)\Delta u(T),$ i.e.
\begin{equation}\label{stea}\frac{-u(m)}{T+1-m}\geq\Delta
u(T)\,.\end{equation} Since by equation \eq{2} we have that
$$\Delta(\Delta u(k-1))=-\lambda
u(k),\;\;\;\forall\;k\in[m,T]\,,$$ summing the above relations
with respect to $k\in[m,T]$ we obtain $$\Delta u(T)-\Delta
u(m-1)=-\lambda \sum_{i=m}^Tu(i)\,.$$ Taking into account that
$\Delta u(m-1)\geq 0$ the above equality implies $$\Delta u(T)\geq
-\lambda \sum_{i=m}^Tu(i)\,.$$ The above inequality, relation
\eq{stea} and the fact that the sequence $(u(k))$ is decreasing
for $k\in[m,T]$ yield $$ \frac{-u(m)}{T+1-m}\geq
-\lambda\sum_{i=m}^Tu(i)\geq-\lambda u(m)\sum_{i=m}^T1\, $$ or
\begin{equation}\label{prima}
\lambda\sum_{i=m}^T1\geq \frac{1}{T+1-m}.
\end{equation}
In order to go further, we add the identities $u(k)-u(k-1)=\Delta
u(k-1)$ for $k\in [1,m]$, obtaining that
$u(m)-u(0)=\sum_{k=1}^{m}\Delta u (k-1).$ This inequality and the
fact that the sequence $(\Delta u(k))$ is decreasing for
$k\in[0,T]$ imply
\begin{equation}\label{stea2} \Delta u(0)\geq\frac{u(m)}{m}\,.
\end{equation}
Since by equation \eq{2} we have that $$\Delta(\Delta
u(k-1))=-\lambda u(k),\;\;\;\forall\;k\in[1,m]\,,$$ summing the
above relations with respect to $k\in[1,m]$ we obtain $$\Delta
u(m)-\Delta u(0)=-\lambda \sum_{i=1}^mu(i)\,.$$ But $\Delta
u(m)<0$ and taking into account that relation \eq{stea2} holds
true, we infer by the above equality
$$\frac{u(m)}{m}<\lambda\sum_{i=1}^mu(i)\,.$$ Using the fact that
$(u(k))$ is nondecreasing for $k\in[0,m]$ we find
\begin{equation}\label{adoua}\frac{1}{m}<\lambda\sum_{i=1}^m1\end{equation}
Now, by \eq{prima} and \eq{adoua} we get
$$\lambda\sum_{i=m}^T1+\lambda\sum_{i=1}^m1>\frac{1}{T+1-m}+\frac{1}{m}\,.$$
Thus, we conclude that $$\lambda(T+1)>\frac{4}{T+1}\,,$$ or
$$\lambda>\frac{4}{(T+1)^2}\,.$$

Finally, we prove the second inequality. By equation \eq{2} we
have $$\Delta u(k)-\Delta u(k-1)=-\lambda
u(k),\;\;\;\forall\;k\in[1,T]\,.$$ Summing the above relations we
find $$u(T)+u(1)=\lambda\sum_{i=1}^Tu(i)\,.$$ Since $u(k)>0$  by using the above relation,
we find, on the one hand, that  $$u(T)+u(1)\geq\lambda
T\min_{k\in[1,T]}u(k)\,,$$ or
\begin{equation}\label{lalala}
\frac{u(T)+u(1)}{T\min_{k\in[1,T]}u(k)}\geq\lambda\,,
\end{equation}
and on the other hand, $$u(T)+u(1)\geq\lambda(u(1)+u(T))\,,$$ or
$$1\geq\lambda\,.$$ Furthermore, we notice that if
$u(k_0)=\min_{k\in[1,T]}u(k)$ then $k_0\in\{1,T\}$. Indeed, let us
assume by contradiction that $k_0\in[1,T]\setminus\{1,T\}$. Then,
since $$\Delta u(k_0)-\Delta u(k_0-1)=-\lambda u(k_0)\,,$$ or
$$0\leq u(k_{0}+1)-2u(k_0)+u(k_{0}-1)=-\lambda u(k_0)<0\,,$$ we obtain
a contradiction. Consequently, $k_0\in\{1,T\}$. That
fact and relation \eq{lalala} yield
$$\frac{1}{T}\cdot\left(1+\frac{\max\{u(1),u(T)\}}{\min\{u(1),u(T)\}}\right)\geq\lambda\,.$$
Theorem \ref{t1} is completely proved.  \endproof
\medskip

%%%%%%%%%%%%%%%%%%%%%%%%InsertRemark1%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent{\bf Remark 1.} We emphasize that for  the estimate in
the left-hand side of \eq{misto} we can give an alternative proof.
This idea is described in what follows. The eigenvalues of problem
\eq{2} can be calculated directly, solving the linear second-order
difference equation $$\Delta(\Delta u(k-1))+\lambda u(k)=0\,,$$
(see, e.g. \cite[Chapter 3]{KP}, \cite[pp.38]{bereanu}). The
eigenvalues of \eq{2} are
$$\lambda_k=2\left(1-\cos\left(\frac{k\pi}{T+1}\right)\right)=4\sin^2\left(\frac{k\pi}{2(T+1)}\right),\;\;\;k\in[1,T]\,,$$
and the corresponding eigenvectors are
$$\phi_k=\left\{0,\sin\left(\frac{k\pi}{T+1}\right),\sin\left(\frac{2k\pi}{T+1},\right),...,\sin\left(\frac{Tk\pi}{T+1},\right),0\right\}\,.$$
Note that $0<\lambda_k<4$ and the estimate from the left-hand side in
\eq{misto} implies
$$m(T):=\frac{4}{(T+1)^2}<\lambda_1=4\sin^2\left(\frac{\pi}{2(T+1)}\right)\,,$$
or, equivalently
$$\frac{1}{(T+1)}<\sin\left(\frac{\pi}{2(T+1)}\right)\,.$$ That
fact follows also from the elementary inequality
$$x<\sin\left(\frac{\pi x}{2}\right),\;\;\;\forall\;x\in(0,1)\,.$$
The last inequality is equivalent with the following fact
$$\frac{2}{\pi}x<\sin(x),\;\;\;\forall\;x\in\left(0,\frac{\pi}{2}\right)\,,$$
which geometrically means that the graph of $\sin(x)$ is above the
chord which joints the points $(0,0)$ and $(\pi/2,1)$.

%Since problem \eq{2} is homogeneous, we can assume that $u(1)=1$. Then the right-hand side estimate of \eq{misto} can be
%replaced by
%$$M(u,T):=\min\left\{1,\frac{1+u(T)}{Tu(T)},\frac{1+u(T)}{T}\right\}\,,$$
%and the corresponding eigenvector of $\lambda_k$ can be chosen as
%$$u_k=\left\{0,1,\frac{\sin(2k\pi/(T+1))}{\sin(k\pi/(T+1))},...,\frac{\sin(Tk\pi/(T+1))}{\sin(k\pi/(T+1))},0\right\}\,.$$
%Note that $\sin(T\pi/(T+1))=\sin(\pi/(T+1))$ which implies that $M(u_1,T)=2/T$ and
%$$\frac{4}{(T+1)^2}<\lambda_1<\frac{2}{T}\,.$$
%We also have $\sin(2T\pi/(T+1))=-\sin(2\pi/(T+1))$, which implies that $u_2(T)=-1$ and $M(u_2,T)=0$. In this case we
%obtain
%$$\lambda_2=4\sin^2\left(\frac{\pi}{T+1}\right)>0=M(u_2,T)\,,$$
%and $u_2$ is an oscillating eigenvector as well as $u_3$, ..., $u_T$.

%A numerical experiment for $T=7$ shows that $\lambda_1=0.15224093$, $m(7)=4/64=0.0625$ and
%$$u_1=\{0,1,1.8478,2.4142,2.6131,2.4142,1.8478,1,0\}\,,$$
%$M(u_1,7)=2/7=0.2857143$ and $m(7)<\lambda_1<M(u_1,7)$. We also have $\lambda_2=0.58578644$,
%$$u_2=\{0,1,1.4142,1,0,-1,-1.4142,-1,0\}\,,$$
%and $M(u_2,7)=0$. The list plots of $u_1$ and $u_2$ are presented on Figure 1.

%\begin{figure}
%    \centering
%    \includegraphics[width=15cm]{FIG1_EIG.eps}
%    \label{fig1}
%    \caption{List plots of $u_1$ and $u_2$ for $T=7$. }
%\end{figure}

\medskip

\noindent{\bf Remark 2.} We point out that for a problem of type
\eq{2} there always exists at least a positive eigenvalue with a
positive corresponding eigenfunction, namely, the first eigenvalue
(see, e.g., \cite{APR1} or \cite{A}). Thus, denoting by
$\lambda_1([0,T+1])$ the first eigenvalue of equation \eq{2}, by using
Theorem \ref{t1}, we deduce that
\begin{equation}\label{estimare}
\frac{4}{(T+1)^2}<\lambda_1([0,T+1])\leq 1\,.
\end{equation}
Moreover, we point out that the left-hand side inequality in
\eq{estimare} is a discrete variant of the celebrated  Faber-Krahn
inequality which is valid in the continuous case (see, e.g.,
\cite{faber,krahn,lieb}), since in the particular  case when $T=2$
a simple computation shows that $\lambda_1([0,3])=1$ (actually, in
this case $1$ is the only eigenvalue of the problem), and thus,
the left-hand side of inequality \eq{estimare} can be rewritten in
the following way
$$\frac{4}{(T+1)^2}\lambda_1([0,3])<\lambda_1([0,T+1]),\;\;\;\forall\;T\geq
2\,.$$
\medskip

\noindent{\bf Remark 3.} We notice that by a simple computation it
can be proved that in the degenerate case $T=1$ the only
eigenvalue of  problem \eq{2} is $\lambda_1([0,2])=2$ while in
the case $T=2$ the two eigenvalues of problem \eq{2} are equal
to $1$. Thus, under these conditions, we have  the equality case in
the right-hand side of inequality \eq{misto}.  In other words, the case when
there is equality can occur.
\medskip

We point out that with a similar proof the result of Theorem
\ref{t1} can be extended to the following:

\begin{teo}\label{t2}
Let $p>1$ be a fixed real number and let $a\geq 1$ and $b\geq a+2$
be two integers. Consider the problem
\begin{equation}\label{p2}
\left\{\begin{array}{lll} -\Delta(|\Delta u(k-1)|^{p-2}\Delta
u(k-1))=\lambda |u(k)|^{p-2}u(k), \quad k\in[a,b-1],\\
u(a-1)=u(b)=0\,.
\end{array}\right.
\end{equation}
Let $\lambda>0$ be an eigenvalue of problem \eq{p2} with the
property that the corresponding eigenvector $u, u(k)>0$ for any
$k\in[a,b-1]$. Then we have the estimates
\begin{equation}\label{pmisto}\frac{2^p}{(b-a+1)^p}<\lambda\leq
\min\left\{1,\frac{1}{b-a}\cdot\left(1+\frac{\max\{u(1)^{p-1},u(T)^{p-1}}
{\min\{u(1)^{p-1},u(T)^{p-1}}\right)\right\}\,.\end{equation}
\end{teo}
In the case when $p=2$, $a=1$ and $b=T+1$ in Theorem \ref{t2} we
obtain Theorem \ref{t1}.

Finally, we recall that in the hypotheses of Theorem \ref{t2} the
first eigenvalue, $\lambda_{1,p}([a-1,b])$, is defined from a variational point
of view by the so-called Rayleigh quotient, that is
\begin{equation}\label{ray}\lambda_{1,p}([a-1,b])=\inf_{u\not\equiv 0}\frac{\sum\limits_{k=a}^{b}|\Delta u(k-1)|^p}
{\sum\limits_{k=a}^{b-1}|u(k)|^p}\,.\end{equation} We note that
in the case $p=2$ we will use the notation $\lambda_1([a-1,b])$
instead of $\lambda_{1,2}([a-1,b])$. Theorem \ref{t2} shows that
relation \eq{estimare} can be extended thanks to the following relation
\begin{equation}\label{estimare2}
\frac{4}{(b-a+1)^2}<\lambda_1([a-1,b])\leq 1\,.
\end{equation}



\section{Proof of Theorem \ref{main}}
$\bullet$ First, we show the existence of $\lambda_0>0$ such that
any $\lambda\in(0,\lambda_0)$ is not an eigenvalue of problem
\eq{1}.

Define the Rayleigh type quotient
\begin{equation}\label{hahaha}
\lambda_0=\inf_{u\in
H\setminus\{0\}}\frac{\sum\limits_{k=1}^{T+1}|\Delta
u(k-1)|^2+\sum\limits_{k=1}^T|u(k)|^q}
{\sum\limits_{k=1}^Tg(k)|u(k)|^r}\,.\end{equation}

In a first instance we prove that $\lambda_0>0$. In order to show
that we start by pointing out that relations \eq{ray} and
\eq{estimare} imply
\begin{equation}\label{haha1}\sum\limits_{k=1}^{T+1}|\Delta u(k-1)|^2\geq\lambda_1([0,T+1])
\sum\limits_{k=1}^{T}|u(k)|^2\geq\frac{4}{(T+1)^2}\sum\limits_{k=1}^{T}|u(k)|^2,
\;\;\;\forall\;u\in H\,.\end{equation} Since we have $2<r<q$ we
deduce $$|u(k)|^2+|u(k)|^q\geq|u(k)|^r,\;\;\;\forall\;u\in
H,\;\forall\;k\in[1,T]\,.$$ Summing the above inequalities we
obtain
\begin{equation}\label{haha2}
\sum\limits_{k=1}^T|u(k)|^2+\sum\limits_{k=1}^T|u(k)|^q\geq
\sum\limits_{k=1}^T|u(k)|^r\geq\frac{1}{|g|_\infty}\sum\limits_{k=1}^Tg(k)|u(k)|^r,\;\;\;\forall\;u\in
H\,.
\end{equation}
 Combining relations \eq{haha1} and \eq{haha2} we infer
\begin{equation}\label{haha3}
\begin{array}{lll}
\di\sum\limits_{k=1}^{T+1}|\Delta
u(k-1)|^2+\sum\limits_{k=1}^T|u(k)|^q&\geq&
\min\left\{\di\frac{4}{(T+1)^2},1\right\}\di\frac{1}{|g|_\infty}\di\sum\limits_{k=1}^Tg(k)|u(k)|^r\\
&=&\di\frac{4}{(T+1)^2|g|_\infty}\di\sum\limits_{k=1}^Tg(k)|u(k)|^r,\;\;\;\forall\;u\in
H\,.
\end{array}
\end{equation}
The last inequality shows that
\begin{equation}\label{lambda0}
\lambda_0\geq\frac{4}{(T+1)^2|g|_\infty}>0\,.
\end{equation}

Let us now define, $J_1$, $I_1$, $J_0$, $I_0:H\rightarrow\RR$ by
$$J_1(u)=\frac{1}{2}\sum\limits_{k=1}^{T+1}|\Delta
u(k-1)|^2+\frac{1}{q}\sum\limits_{k=1}^T|u(k)|^q\;\;\;\;\;
I_1(u)=\frac{1}{r}\sum\limits_{k=1}^Tg(k)|u(k)|^r\,,$$ and
$$J_0(u)=\sum\limits_{k=1}^{T+1}|\Delta
u(k-1)|^2+\sum\limits_{k=1}^T|u(k)|^q\;\;\;\;\;
I_0(u)=\sum\limits_{k=1}^Tg(k)|u(k)|^r\,.$$ Standard arguments
imply  that $J_1$, $I_1\in C^1(H,\RR)$ with $$\langle
J_1^{'}(u),v\rangle=\sum_{k=1}^{T+1}\Delta u(k-1)\Delta
v(k-1)+\sum_{k=1}^{T}|u(k)|^{q-2}u(k)v(k)\,,$$ and $$\langle
I_1^{'}(u),v\rangle=\sum_{k=1}^{T}g(k)|u(k)|^{r-2}u(k)v(k)\,,$$
for any $u$, $v\in H$.
\begin{lemma}\label{l0}
Let $\lambda_0$ be defined by relation \eq{hahaha}. Then no
$\lambda\in(0,\lambda_0)$ is an eigenvalue of problem \eq{1}.
\end{lemma}
\proof Indeed, assuming by contradiction that there exists
$\lambda\in(0,\lambda_0)$ an eigenvalue of problem \eq{1}, it
follows that we can find $w_\lambda\in H\setminus\{0\}$ such that
$$\langle J_1^{'}(w_\lambda),v\rangle=\lambda\langle
I_1^{'}(w_\lambda),v\rangle,\ \ \forall v\in H.$$ Letting
$v=w_\lambda$ we deduce $\langle
J_1^{'}(w_\lambda),w_\lambda\rangle=\lambda\langle
I_1^{'}(w_\lambda),w_\lambda\rangle\,,$ or
$$J_0(w_\lambda)=\lambda I_0(w_\lambda)\,.$$ Since $w_\lambda\neq
0$ we have that $J_0(w_\lambda)>0$ and thus, $I_0(w_\lambda)>0$.
Combining that fact with the ideas that $\lambda\in(0,\lambda_0)$
and $\lambda_0=\inf_{u\in H\setminus\{0\}} \frac{J_0(u)}{I_0(u)}$
we infer $$J_0(w_\lambda)\geq\lambda_0 I_0(w_\lambda)>\lambda
I_0(w_\lambda)=J_0(w_\lambda),$$ which is a contradiction. The
proof of Lemma \ref{l0} is complete.
\endproof




$\bullet$ Secondly, we show that there exists $\lambda_1$ such that
any $\lambda\in(\lambda_1,\infty)$ is an eigenvalue of problem
\eq{1}.

For any $\lambda>0$ we define the functional
$S_\lambda:H\rightarrow\RR$ by $$S_\lambda(u)=J_1(u)-\lambda
I_1(u),\;\;\;\forall\;u\in H\,.$$ We notice that $S_\lambda\in
C^1(H,\RR)$ with the derivative given by $$\langle
S_\lambda^{'}(u),v\rangle=\langle
J_1^{'}(u),v\rangle-\lambda\langle I_1^{'}(u),v\rangle,\;\;\;
\forall\;u,\;v\in H\,.$$ Thus, $\lambda$ is an eigenvalue of
problem \eq{1} if and only if there exists $u_\lambda\in
H\setminus\{0\}$ a critical point of $S_\lambda$.

\begin{lemma}\label{l1}
For any $\lambda\in(0,\infty)$ the functional $S_\lambda$ is
coercive, i.e.
$\lim_{\|u\|\rightarrow\infty}S_\lambda(u)=\infty\,.$
\end{lemma}
\proof It is obvious that
$$S_\lambda(u)\geq\frac{1}{2}\|u\|^2+\frac{1}{q}\sum\limits_{k=1}^T|u(k)|^q-
\frac{|g|_\infty}{r}\sum\limits_{k=1}^T|u(k)|^r\,,$$ for any $u\in
H$. For any $m\geq 2$ let us denote
$$|u|_m=\left(\sum_{k=1}^{T}|u(k)|^m\right)^{1/m}\,.$$ It is not
difficult to notice that each $|\cdot|_m$, $m\geq 2$, is a norm on
$H$. Since $H$ is a finite dimensional Hilbert space we deduce
that for any $m_1$, $m_2\geq 2$ the norms $|\cdot|_{m_1}$,
$|\cdot|_{m_2}$ and $\|\cdot\|$ are equivalent.

The above pieces of information imply that there exist two
positive constants $C_1$ and $C_2$ such that
$$S_\lambda(u)\geq\frac{1}{2}\|u\|^2+C_1\|u\|^q-C_2\|u\|^r\,,$$
for any $u\in H$. Since $2<r<q$, the proof of Lemma \ref{l1} is
complete.
\endproof

Define
\begin{equation}\label{lambda1}
\lambda_1=\inf_{u\in
H\setminus\{0\}}\frac{\di\frac{1}{2}\sum\limits_{k=1}^{T+1}
|\Delta
u(k-1)|^2+\di\frac{1}{q}\sum\limits_{k=1}^T|u(k)|^q}{\di\frac{1}{r}\sum\limits_{k=1}^Tg(k)|u(k)|^r}\,.
\end{equation}
 Due to
\eq{hahaha}, a simple estimate shows that
$$r\min\left\{\frac{1}{2},\frac{1}{q}\right\}\lambda_0\leq
\lambda_1\leq
r\max\left\{\frac{1}{2},\frac{1}{q}\right\}\lambda_0.$$ Since
$2<r<q,$ we clearly have
\begin{equation}\label{lambda11}
\frac{r}{q}\lambda_0\leq \lambda_1\leq\frac{r}{2}\lambda_0.
\end{equation}
In particular, \eq{lambda0} and the left hand size of
\eq{lambda11} imply  $\lambda_1>0.$

%Using similar
%arguments as in the case of $\lambda_0$ we deduce
%\begin{eqnarray*}
%\frac{1}{2}\sum\limits_{k=1}^{T+1}|\Delta u(k-1)|^2+
%\frac{1}{q}\sum\limits_{k=1}^T|u(k)|^q&\geq&\frac{2}{(T+1)^2}\sum\limits_{k=1}^{T}|u(k)|^2+
%\frac{1}{q}\sum\limits_{k=1}^T|u(k)|^q\\
%&\geq&\min\left\{\frac{2}{(T+1)^2},\frac{1}{q}\right\}\sum\limits_{k=1}^T|u(k)|^r\\
%&\geq&\min\left\{\frac{2}{(T+1)^2},\frac{1}{q}\right\}\cdot\frac{1}{|g|_\infty}\cdot\sum\limits_{k=1}^Tg(k)|u(k)|^r\,,
%\end{eqnarray*}
%for any $u\in H$.
% Consequently, we find
%\begin{equation}\label{lambda11}
%\lambda_1\geq\min\left\{\frac{2}{(T+1)^2},\frac{1}{q}\right\}\cdot\frac{r}{|g|_\infty}>0\,.
%\end{equation}




\begin{lemma}\label{l2}
Any $\lambda\in(\lambda_1,\infty)$ is an eigenvalue of problem
\eq{1}.
\end{lemma}
\proof We fix $\lambda\in(\lambda_1,\infty)$. By Lemma \ref{l1} we
deduce that $S_\lambda$ is coercive. On the other hand, it is
clear that the functional $S_\lambda$ is weakly lower
semi-continuous. These two facts enable us to apply Theorem 1.2 in
\cite{S} in order to prove that there exists $u_\lambda\in H$ a
global minimum point of $S_\lambda$.

Next, we show  that $u_\lambda$ is not trivial. Indeed, since
$\lambda_1=\inf_{u\in H,\;u\neq 0} \frac{J_1(u)}{I_1(u)}$ and
$\lambda>\lambda_1$ it follows that there exists $v_\lambda\in H$
such that $$J_1(v_\lambda)<\lambda I_1(v_\lambda)\,,$$ or
$$S_\lambda(v_\lambda)<0\,.$$ In particular,
$\inf_{H}S_\lambda<0$, and we conclude that $u_\lambda\neq 0$.

Next, we show that $\lambda$ is an eigenvalue of problem \eq{1}.
Let $v\in H$ fixed. The above property of $u_\lambda$ gives that
 $$\frac{d}{d\epsilon}S_\lambda(u_\lambda+\epsilon
v)|_{\epsilon=0}=0\,,$$ or $$\langle
J_1^{'}(u_\lambda),v\rangle-\lambda\langle
I_1^{'}(u_\lambda),v\rangle=0,\;\;\;\forall\;v\in H\,,$$ that
means $\lambda$ is an eigenvalue of problem \eq{1}. The proof of
Lemma \ref{l2} is complete.  \endproof

$\bullet$ Next, we show that $\lambda_1$ is also an eigenvalue of
problem \eq{1}. In order to do that we first prove the following
result.
\begin{lemma}\label{l3}
$\lim_{\|u\|\rightarrow 0}\frac{J_0(u)}{I_0(u)}=
\lim_{\|u\|\rightarrow\infty}\frac{J_0(u)}{I_0(u)}=\infty\,.$
\end{lemma}
\proof Considering again the norms, $|\cdot|_m$, $m\geq 2$,
defined in Lemma \ref{l1} and recalling that they are equivalent
with the norm $\|\cdot\|$ we find that there exist two positive
constants $D_1$ and $D_2$ such that $$\frac{J_0(u)}{I_0(u)}\geq
\frac{\|u\|^2+D_1\|u\|^q}{D_2\|u\|^r}\,,\ \ \forall u\in
H\setminus \{0\}.$$ Now taking into account that $2<r<q$, the
conclusion of Lemma \ref{l3}  immediately holds.
\endproof

\begin{lemma}\label{l4}
The real number $\lambda_1$, given by relation \eq{lambda1}, is an
eigenvalue of problem \eq{1}.
\end{lemma}
\proof Let $(\lambda_n)$ be a sequence in $\RR$ such that
$\lambda_n\searrow\lambda_1$ as $n\rightarrow\infty$. By Lemma
\ref{l2} we deduce that for each $n$ there exists $u_n\in
H\setminus\{0\}$ such that
\begin{equation}\label{deni1}
\langle J_1^{'}(u_n),v\rangle-\lambda_n\langle
I_1^{'}(u_n),v\rangle=0,\;\;\;\forall\;v\in H\,.
\end{equation}
Taking $v=u_n$ in the above equality we find
\begin{equation}\label{deni2}
 J_0(u_n)=\lambda_nI_0(u_n),\;\;\;\forall\;n\,.
\end{equation}
The above equality  and Lemma \ref{l3} imply that $(u_n)$ is a
bounded sequence in $H$. Indeed, assuming by contradiction that
$(u_n)$ is not bounded in $H$ it follows that passing eventually
to a subsequence, still denoted by $(u_n)$ we have
$\|u_n\|\rightarrow\infty$. On the other hand, the fact that
$\lambda_n\searrow\lambda_1$ and relation \eq{deni2} imply that
for each $n$ large enough it holds true
$$\frac{J_0(u_n)}{I_0(u_n)}= \lambda_n\leq\lambda_1+1\,.$$ Lemma
\ref{l3} shows that the above inequality and the fact that
$\|u_n\|\rightarrow\infty$ lead to a contradiction. Consequently,
$(u_n)$ is bounded in $H$. We deduce the existence of $u\in H$
such that, passing eventually to a subsequence, $u_n$ converges to
$u$ in $H$. Passing to the limit as $n\rightarrow\infty$ in
\eq{deni1} we get $$\langle J_1^{'}(u),v\rangle-\lambda_1\langle
I_1^{'}(u),v\rangle=0,\;\;\;\forall\;v\in H\,,$$ i.e. $\lambda_1$
is an eigenvalue of problem  \eq{1} provided that $u\neq 0$.

Finally, we explain why  $u\neq 0$. Assuming by
contradiction that $u=0$ we deduce that $u_n$ converges to $0$ in
$H$. By relation \eq{deni2} we deduce that for any $n$ the following equality holds
$$\frac{J_0(u_n)}{I_0(u_n)}=\lambda_n\,.$$ Passing to
the limit as $n\rightarrow\infty$ and taking into account the
result of Lemma \ref{l3} and the fact that
$\lambda_n\searrow\lambda_1$ we obtain a contradiction. The proof
of Lemma \ref{l4} is complete.
\endproof

$\bullet$ Finally, we point out that the conclusion of Theorem
\ref{main} holds true.

%\begin{lemma}\label{l5}
%The numbers $\lambda_0$ and $\lambda_1$ defined by
%relations \eq{hahaha} and \eq{lambda1} verify
%$\lambda_0\leq\lambda_1$. Moreover, assuming that
%$$\min\left\{\frac{2}{(T+1)^2},\frac{1}{q}\right\}\cdot r\geq\max_{k\in[1,T]}{g(k)}\,,$$
%then $\lambda_1([0,T+1])\leq\lambda_1$. Furthermore, assuming that
%$$\frac{4}{(T+1)^2}\geq\max_{k\in[1,T]}{g(k)}\,,$$
%then $\lambda_1([0,T+1])\leq\lambda_0$.
%\end{lemma}
%\proof Undoubtedly, by Lemmas \ref{l0}, \ref{l2} and \ref{l4} we
%deduce that $\lambda_0\leq\lambda_1$.

%On the other hand estimates \eq{lambda0}, \eq{lambda11} and \eq{estimare} complete the proof of Lemma \ref{l5}.\endproof










\noindent{\sc Proof of Theorem \ref{main}.} In order to obtain the
first part, it is enough to combine  Lemmas \ref{l0}, \ref{l2} and
\ref{l4}; in particular, we clearly have $\lambda_0\leq\lambda_1$.
The first two inequalities of \eq{tetel-osszefugges} come from
\eq{lambda11} and \eq{lambda0}, respectively.

It remains to prove the right hand side inequality of
\eq{tetel-osszefugges}, i.e., $\lambda_1\leq A$, where we use the
notation $A=\frac{r(q-2)}{(q-r)\sum_{k=1}^T
g(k)}\left(\frac{T(q-r)}{q(r-2)}\right)^{\frac{r-2}{q-2}}.$ Fix
$\tilde u\in H\setminus \{0\}$ by $\tilde u(k)=s>0$, $k\in [1,T]$.
Due to  \eq{lambda1}, we have $$ \lambda_1\leq
\frac{\di\frac{1}{2}\sum\limits_{k=1}^{T+1} |\Delta \tilde
u(k-1)|^2+\di\frac{1}{q}\sum\limits_{k=1}^T| \tilde
u(k)|^q}{\di\frac{1}{r}\sum\limits_{k=1}^Tg(k)|\tilde
u(k)|^r}\,=\frac{r(s^2+\frac{T}{q}s^q)}{s^r\sum_{k=1}^T g(k)}. $$
Taking the function $h:(0,\infty)\to (0,\infty)$ defined by
$$h(s)=\frac{r(s^2+\frac{T}{q}s^q)}{s^r\sum_{k=1}^T g(k)},$$ one
can easily show that its minimum is attained at the point
$s_0=\left(\frac{q(r-2)}{T(q-r)}\right)^{\frac{1}{q-2}}$, the
minimum value being $h(s_0)=A.$ This concludes the proof.
\endproof
\smallskip

\noindent{\bf Remark 4.} We notice that the spectrum of problem
\eq{1} is not completely described by our paper. Although we have
estimates for $\lambda_0$ and $\lambda_1$, at this stage we are
not able to say if $\lambda_0=\lambda_1$ or $\lambda_0<\lambda_1$.
Note that $\lambda_0$ and $\lambda_1$ are very close to each other
whenever $r$ is close to $2$; in that sense, see the first inequality in
\eq{tetel-osszefugges}. Due to the nonhomogeneous nature of the
problem \eq{1}, we are strongly convinced that we usually have
$\lambda_0<\lambda_1$, i.e., there is a gap between $\lambda_0$
and $\lambda_1$. If so, the problem of the existence/nonexistence
of eigenvalues in the interval $[\lambda_0,\lambda_1)$ should be
elucidated. This problem will hopefully be considered in a
forthcoming paper.
 %For instance the
%existence of two Rayleigh type quotients $\lambda_0$ and
%$\lambda_1$ is a peculiarity of nonhomogenous equations. (For
%problem \eq{diri}, for instance, $\lambda_0$ and $\lambda_1$ are
%equals.

\medskip
{\bf Acknowledgments.} A. Krist\'aly is supported by Grant IDEI
527/2007 while  M. Mih\u ailescu  and V.~R\u adulescu have been
supported by Grant CNCSIS 79/2007 ``Degenerate and Singular
Nonlinear Processes". S. Tersian is supported by Grant
VU-MI-02/05. He is thankful to Department of mathematics and its
applications, CEU, Budapest where a part of this work was prepared
during his visit.



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