\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 59, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/59\hfil Existence of solutions]
{Existence of solutions to $p$-Laplacian
difference equations under barrier strips conditions}

\author[C. Gao\hfil EJDE-2007/59\hfilneg]
{Chenghua Gao}  % in alphabetical order

\address{Chenghua Gao \newline
College of Mathematics and Information Science,
Northwest Normal University \\
Lanzhou, Gansu 730070,   China}
\email{gaokuguo@163.com}

\thanks{Submitted January 24, 2007. Published April 22, 2007.}
\subjclass[2000]{39A10}
\keywords{Second-order p-Laplacian difference equation;
 barrier strips; \hfill\break\indent Leray-Schauder principle; existence}

\begin{abstract}
 We study the existence of solutions to the
 boundary-value problem
 \begin{gather*}
 \Delta(\phi_p(\Delta u(k-1)))=f(k,u(k),\Delta u(k)),\quad
 k\in \mathbb{T}_{[1, N]}, \\
 \Delta u(0)=A, \quad u(N+1)=B,
 \end{gather*}
 with barrier strips conditions,
 where $N>1$ is a fixed natural number, $\phi_p(s)=|s|^{p-2}s$,
 $p>1$.
\end{abstract}

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

\section{Introduction}

Given $a, b\in\mathbf{Z}$ and $a<b$, we employ
$\mathbb{T}_{[a, b]}$ to denote $\{a,a+1,a+2,\dots,b-1,b\}$. In
this paper, we are concerned with the following  $p$-Laplacian
difference equation
\begin{equation}
 \Delta(\phi_p(\Delta u(k-1)))=f(k,u(k),\Delta u(k)),\quad
k\in \mathbb{T}_{[1, N]},\label{e1.1}
\end{equation}
satisfying the boundary conditions
\begin{equation}
 \Delta u(0)=A, u(N+1)=B,\label{e1.2}
\end{equation}
where $N>1$ is a fixed natural number,
$f:\mathbb{T}_{[1, N]}\times \mathbb{R}^{2}\to\mathbb{R}$
is continuous,
$\phi_p(s)=|s|^{p-2}s, p>1, (\phi_p)^{-1}=\phi_q,
\frac{1}{p}+\frac{1}{q}=1$.

In recent years, $p$-Laplacian discrete boundary-value problems have
been investigated  in literature [1,2,4]. But, almost all of the
works discussed these problems when $f$ satisfies growth restriction
at $\infty$. Now, the question is: Is there still a solution to
those problems when $f$ is not restricted at $\infty$?

In 1994,  Kelevedjiev  [3] used Leray-Schauder principle to discuss
the solutions to the nonlinear differential boundary-value problem
\begin{gather}
x{''}(t)=f(t,x(t),x{'}(t)),\quad t\in[0,1],\label{e1.3} \\
x'(0)=A, x(1)=B.\label{e1.4}
\end{gather}
He established the following results:

\begin{theorem} \label{thmA}
Let $f:[0,1]\times \mathbb{R}^{2}\to\mathbb{R}$ be continuous. Suppose
there are constants $L_{i}$, $i=1,2,3,4$, such that
$L_{2}>L_{1}\geq A$,  $L_{3}<L_{4}\leq A$,
\begin{gather*}
f(t,x,p)\leq0,\quad (t,x,p)\in [0,1]\times \mathbb{R}\times [L_{1},L_{2}],
\\
f(t,x,p)\geq0,\quad (t,x,p)\in [0,1]\times \mathbb{R}\times [L_{3},L_{4}].
\end{gather*}
Then  \eqref{e1.3}-\eqref{e1.4} has at least one solution in
$C^2[0, 1]$, where $C^{2}[0,1]$ is the set of functions whose
second derivative is continuous on $[0,1]$.
\end{theorem}

Clearly, growth restrictions on $f$ are not imposed in Theorem \ref{thmA}.
So, we use the Leray-Schauder principle to discuss the existence of
solutions to boundary-value problem \eqref{e1.1}-\eqref{e1.2} when $f$ is not
restricted at $\infty$.


\section{Preliminaries}

Let $X:=\{u|u:\mathbb{T}_{[0, N+1]}\to\mathbb{R}\}$ be
equipped with the norm
$$
\|u\|_{X}=\max_{k\in\mathbb{T}_{[0,N+1]}}|u(k)|,
$$
 and
$Y:=\{u|u:\mathbb{T}_{[1, N]}\to\mathbb{R}\}$ with the norm
$$
\|u\|_{Y}=\max_{k\in\mathbb{T}_{[1, N]}}|u(k)|.
$$
 It is easy to see that $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ are Banach
spaces.

The main result of our work is based on the following special form
of Leray-Schauder principle.

\begin{theorem} \label{thmB}
 Let $f:\mathbb{T}_{[1, N]}\times \mathbb{R}^{2}\to \mathbb{R}$ be
continuous, $L:D(L)\subset X\to Y$  be a bijection, and
$L^{-1}$ be completely continuous. If there exists a
constant $M$ such that an  arbitrary solution of
the  boundary-value problem
$$
 Lu(k)=\lambda f(k,u(k),\Delta u(k)),\quad
 k\in \mathbb{T}_{[1, N]},\quad \lambda\in [0,1],\quad
 u\in D(L)
$$
satisfies $\|u\|_{X}<M$, then the boundary-value problem
$$
 Lu(k)=f(k,u(k),\Delta u(k)),\quad k\in \mathbb{T}_{[1,N]},\quad u\in D(L)
$$
has at least one solution in $X$.
\end{theorem}

Define the operator $L:D(L)\subset X\to Y$ by
$$
Lu(k)=\Delta(\phi_p(\Delta u(k-1))),
\quad u\in D(L),\; k\in \mathbb{T}_{[1,N]},
$$
where $D(L)=\{u|u\in X, \Delta u(0)=A, u(N+1)=B\}$.

\begin{lemma} \label{lem2.1}
Let $h\in Y$. Then the boundary-value problem
\begin{gather}
\Delta\phi_p(\Delta u(k-1))=h(k),\quad k\in \mathbb{T}_{[1,
N]},\label{e2.1} \\
\Delta u(0)=A, \quad u(N+1)=B\label{e2.2}
\end{gather}
has a unique solution
$$
u(k)=B-\sum_{s=k+1}^{N+1}\Big(\phi_q\Big(\sum_{l=1}^{s-1}h(l)
+\phi_p(A)\Big)\Big),\quad k\in \mathbb{T}_{[0, N+1]}.
$$
\end{lemma}

\begin{proof}
 Summing the equation \eqref{e2.1} from $s=1$ to $s=k-1$, we obtain
$$
\phi_p(\Delta u(k-1))=\phi_p(A)+\sum_{s=1}^{k-1}h(s).
$$
Applying $\phi_q$  on  both sides of the above equation, we
obtain
$$
\Delta u(k-1)=\phi_q(\phi_p(A)+\sum_{s=1}^{k-1}h(s)).
$$
Summing again from $s=k+1$ to $s=N+1$, we have
\begin{gather*}
B-u(k)=\sum_{s=k+1}^{N+1}(\phi_q(\sum_{l=1}^{s-1}h(l)+\phi_p(A))), \\
u(k)=B-\sum_{s=k+1}^{N+1}(\phi_q(\sum_{l=1}^{s-1}h(l)+\phi_p(A))),\quad
k\in \mathbb{T}_{[0, N+1]}.
\end{gather*}

Next, we show that there is only one solution to \eqref{e1.1}-\eqref{e1.2}.
Suppose that $u_1, u_2$ are solutions. Then
\begin{equation}
\Delta (\phi_p(\Delta u_1(k-1)))=\Delta (\phi_p(\Delta
u_2(k-1))),\quad k\in \mathbb{T}_{[1, N]},\label{e2.3}
\end{equation}
and $\Delta u_i(0)=A$, $u_i(N+1)=B$,   $i=1,2$.
Now, summing \eqref{e2.3} from $s=1$ to $s=k-1$, we get
$$
\phi_p(\Delta u_1(k-1))-\phi_p(\Delta u_2(k-1))=\phi_p(\Delta
u_1(0))-\phi_p(\Delta u_2(0)),
$$
furthermore, $\Delta u_i(0)=A, i=1,2$,
$$
\phi_p(\Delta u_1(k-1))=\phi_p(\Delta u_2(k-1)),
$$
and since $\phi_p$ is a bijection,
$$
\Delta u_1(k-1)=\Delta u_2(k-1).
$$
Summing the above equation from $s=k+1$ to $s=N+1$, we have
\begin{gather*}
\sum_{s=k+1}^{N+1}\Delta
u_1(k-1)=\sum_{s=k+1}^{N+1}\Delta u_2(k-1), \\
B-u_1(k)=B-u_2(k),
\end{gather*}
so
$u_1(k)=u_2(k)$, $k\in\mathbb{T}_{[1, N]}$,
and from the boundary conditions $\Delta u_i(0)=A$, $u_i(N+1)=B$, we
have
$$
u_1(k)=u_2(k),\quad k\in\mathbb{T}_{[0, N+1]}.
$$
\end{proof}

We remark that from  Lemma \ref{lem2.1}, it follows that $L$ is a bijection.

\begin{lemma} \label{lem2.2}
$L^{-1}:Y\to X$ is completely
continuous.
\end{lemma}

\begin{proof} Since the range of $L^{-1}$ has finite dimension, it is not
difficult to check that it is compact; and from the continuity of
$f$ and $\phi_q$, we can see that $L^{-1}$ is continuous.
\end{proof}

\section{Main results}

\begin{theorem} \label{thm3.1}
Let $f:\mathbb{T}_{[1, N]}\times \mathbb{R}^{2}\to\mathbb{R}$ be continuous.
Suppose there exist constants $L_{i}$, $i=1,2,3,4$ satisfying
$L_{2}>L_{1}\geq A$, $L_{3}<L_{4}\leq A$, such that
\begin{gather}
\label{H1}  f(k,u,p)\leq0,\quad (k,u,p)\in \mathbb{T}_{[1, N]}\times
\mathbb{R}\times [L_{1},L_{2}],
\\
\label{H2}  f(k,u,p)\geq0,\quad (k,u,p)\in
\mathbb{T}_{[1, N]}\times \mathbb{R}\times [L_{3},L_{4}].
\end{gather}
Then the boundary-value problem \eqref{e1.1}-\eqref{e1.2} has at least
one solution in $X$.
\end{theorem}

\begin{proof} Let us define the function
$\Phi:\mathbb{R}\to \mathbb{R}$ as follows.
$$
\Phi(v)=\begin{cases}
                L_2, & v>L_2,\\
                v,   & L_3\leq v\leq L_2,\\
                L_3, & v<L_3.
              \end{cases}
$$
Now, we consider the  problem
\begin{equation}
\Delta(\phi_p(\Delta u(k-1)))=f(k,u(k),\Phi(\Delta u(k))),\quad
k\in \mathbb{T}_{[1, N]}, u\in D(L).\label{e3.1}
\end{equation}
Suppose that $u\in D(L)$ is an arbitrary solution to the family of
problems
\begin{equation}
\Delta(\phi_p(\Delta u(k-1)))=\lambda f(k,u(k),\Phi(\Delta
u(k))),\quad k\in \mathbb{T}_{[1, N]}.\label{e3.2}
\end{equation}
To apply Theorem \ref{thmB}, we need a priori bounds for
$\|u\|_X$ independent of $\lambda\in[0, 1]$. First, let us examine
$\Delta u(k)$.
Now, we prove that the set
$$
S_{0}=\{k\in \mathbb{T}_{[0, N]}|\Delta u(k)>L_1\}
$$
is  empty.
Suppose it is not empty. Let $k_0\in S_0$ be fixed. Then
$\Delta u(k_0)>L_1$. From the construction of $\Phi$, we know
$$
L_1<\Phi(\Delta u(k_0))\leq L_2.
$$
From \eqref{H1} and $\Delta(\phi_p(\Delta u(k_0-1)))\leq 0$, we have
\begin{equation}
|\Delta u(k_0)|^{p-2}\Delta u(k_0)\leq |\Delta u(k_0-1)|^{p-2}\Delta
u(k_0-1).\label{e3.3}
\end{equation}
Now, we prove $k_0-1\in S_0$.
It will be discussed in three cases:


\noindent\textbf{Case 1:} $\Delta u(k_0)>0$. Then from \eqref{e3.3}, we know
$L_1<\Delta u(k_0)\leq \Delta u(k_0-1)$;

\noindent\textbf{Case 2:} $\Delta u(k_0)=0$. Then the result is obvious;

\noindent\textbf{Case 3:} $\Delta u(k_0)<0$. Then $\Delta u(k_0-1)$ will be discussed
under two cases.

\noindent\textbf{Case 3.1:}  $\Delta u(k_0-1)\geq0$. Then from \eqref{e3.3}, $\Delta
u(k_0-1)>L_1$;

\noindent\textbf{Case 3.2:}  $\Delta u(k_0-1)<0$. Then $p$ will be discussed under
different situations.

\noindent\textbf{Case 3.2.1:} $p$ is an odd number. Then
$(-\Delta u(k_0))^{p-2}=-(\Delta u(k_0))^{p-2}$. From \eqref{e3.3},
we know $-(\Delta u(k_0))^{p-1}\leq|\Delta u(k_0-1)|^{p-2}\Delta u(k_0-1)$.
Moreover, $\Delta u(k_0-1)<0$, we have $-(\Delta u(k_0))^{p-1}\leq-(\Delta
u(k_0-1))^{p-1}$. Since $p-1$ is an even number and $\Delta u(k_0),
\Delta u(k_0-1)<0$, it's not difficult to get
$$
L_1<\Delta u(k_0)\leq\Delta u(k_0-1);
$$

\noindent\textbf{Case 3.2.2:} $p$ is an even number. Then we have
$(\Delta u(k_0))^{p-1}\leq(\Delta u(k_0-1))^{p-1}$, and since $p-1$ is an
odd number, we know that
$$
L_1<\Delta u(k_0)\leq\Delta u(k_0-1);
$$
so, when $\Delta u(k_0)<0$, $\Delta u(k_0-1)<0$, there also exists
$$
L_1<\Delta u(k_0)\leq\Delta u(k_0-1).
$$


From Case 1, Case 2, Case 3, we  obtain
$$
L_1<\Delta u(k_0)\leq\Delta u(k_0-1),
$$
so $k_0-1\in S_0$.
If we continue the above process, we  get
$$
\Delta u(0)\geq\Delta u(1)>L_1,
$$
which contradicts with $\Delta u(0)=A$, so $S_0=\emptyset$.

Similarly, we can obtain that the set
$$
S_1=\{k\in\mathbb{T}_{[0, N]}|\Delta u(k)<L_4\}
$$
is also empty.

Then for $k\in\mathbb{T}_{[0, N]}$,
\begin{equation}
L_4\leq\Delta u(k)\leq L_1,\label{e3.4}
\end{equation}
i.e.,
\begin{equation}
\max_{k\in \mathbb{T}_{[0, N]}}|\Delta u(k)|\leq
C,\label{e3.5}
\end{equation}
where $C=\max\{|L_{1}|,|L_{4}|\}$.

On the other hand, for $k\in\mathbb{T}_{[0, N]}$, since $u(N+1)=B$,
we can construct $ u(k)=-\sum_{s=k}^{N}\Delta u(s)+B$. Thus
for $u\in D(L)$, we have
\begin{equation}
\max_{k\in \mathbb{T}_{[0, N+1]}}|u(k)|\leq C_{1},\label{e3.6}
\end{equation}
where $C_{1}=(N+1)\cdot C+|B|$.
 From \eqref{e3.6}, we can see that all of the solutions to
problems \eqref{e3.2} satisfy
$$
\|u\|_{X}\leq C_{1}.
$$
Then there exists at least one solution $u\in D(L)$ to problem
\eqref{e3.1}. And from \eqref{e3.4}, we know that
$$
L_3< L_4\leq\Phi(\Delta u(k))\leq L_1<L_2,\quad k\in
\mathbb{T}_{[1, N]},
$$
together with the definition of $\Phi$, the following conclusion
$$
\Phi(\Delta u(k))=\Delta u(k),\quad k\in \mathbb{T}_{[1, N]},
$$
can be obtained. Thus $u$ is also a solution to the problem
\eqref{e1.1}-\eqref{e1.2}.

\subsection*{Example} Consider the  problem
\begin{gather*}
\Delta(\phi_p(\Delta u(k-1)))=(\Delta u(k))^{4}-6(\Delta
u(k))^{3}+11(\Delta u(k))^{2}-6\Delta u(k),\quad
k\in\mathbb{T}_{[1, N]}, \\
\Delta u(0)=2, u(N+1)=B,
\end{gather*}
where $N>1$ is a fixed natural number, $B$ is an arbitrary number.
Let
$f(k,u,p)=p^{4}-6p^{3}+11p^{2}-6p$,
$L_{1}=\frac{5}{2}$, $L_{2}=3$, $L_{3}=1$, $L_{4}=\frac{3}{2}$,
$A=2$.
We can prove that $f(k,u,p)$ satisfies all conditions of
Theorem \ref{thm3.1}, so this problem has at least one solution.
\end{proof}

The next theorem can be proved by similar arguments.

\begin{theorem} \label{thm3.2}
Let $f:\mathbb{T}_{[1, N]}\times \mathbb{R}^{2}\to\mathbb{R}$ be continuous.
Suppose there are constants $L_{i}$, $i=1,2,3,4$ with
$L_{2}>L_{1}\geq B$, $L_{3}<L_{4}\leq B$, such that
\eqref{H1}, \eqref{H2} are satisfied.
 Then the boundary-value problem
\begin{gather*}
\Delta(\phi_p(\Delta u(k-1)))=f(k,u(k),\Delta u(k)),\quad k\in
\mathbb{T}_{[1, N]}, \\
 u(0)=A, \quad \Delta u(N)=B
\end{gather*}
has at least one solution in $X$.
\end{theorem}

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\bibitem{h1} Zhimin He;
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\bibitem{k1} P. Kelevedjiev;
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\bibitem{l1} Yongkun Li, Linghong Lu;
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\end{thebibliography}

\end{document}