\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 63, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/63\hfil Positive solutions for nonlinear difference equations] {Positive solutions for nonlinear difference equations involving the p-Laplacian with sign changing nonlinearity} \author[Y. Sang, H. Su \hfil EJDE-2009/63\hfilneg] {Yanbin Sang, Hua Su} % in alphabetical order \address{Yanbin Sang\newline Department of Mathematics, North University of China, Taiyuan, Shanxi, 030051, China} \email{syb6662004@163.com} \address{Hua Su \newline School of Statistics and Mathematics, Shandong University of Finance, Jinan, Shandong, 250014, China} \email{suhua02@163.com} \thanks{Submitted December 16, 2008. Published May 8, 2009.} \thanks{Project supported by the Youth Science Fund of Shanxi Province (2009021001-2)} \subjclass[2000]{39A10} \keywords{Difference equations; $p$-Laplacian; fixed point index; \hfill\break\indent sign changing nonlinearity} \begin{abstract} By means of fixed point index, we establish sufficient conditions for the existence of positive solutions to $p$-Laplacian difference equations. In particular, the nonlinear term is allowed to change sign. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} The aim of the paper is to prove the existence of positive solutions to the problem $$\begin{gathered} \Delta[\phi_{p}(\Delta u(t-1))]+a(t)f(u(t))=0,\quad t\in [1,T+1],\\ \Delta u(0)=u(T+2)=0, \end{gathered} \label{e1.1}$$ where $\phi_{p}$ is $p$-Laplacian operator, i.e. $\phi_{p}=|s|^{p-2}s$, $p>1$, $(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$, $T\geq 1$ is a fixed positive integer, $\Delta$ denotes the forward difference operator with step size 1, and $[a,b]=\{a,a+1,\dots ,b-1,b\}\subset \mathbb{Z}$ the set of integers. Our work focuses on the case when the nonlinear term $f(u)$ can change sign. By means of fixed point index, some new results are obtained for the existence of at least two positive solutions to the BVP \eqref{e1.1}, the method of this paper is motivated by \cite{j1,s1,z1}. Due to the wide application in many fields such as science, economics, neural network, ecology, cybernetics,etc., the theory of nonlinear difference equations has been widely studied since the 1970s: see, for example \cite{a1,a2,k1,l1}. At the same time, boundary value problem (BVP) of difference equations have received much attention from many author: see \cite{a1,a2,a3,a4,a5,c1,h1,h2,l3,m1,w1,w2,w3} and the reference therein. The approach is mainly based on fixed point theorem. For example, using the Guo-Krasnosel'skii fixed point theorem in cone and a fixed point index theorem, He \cite{h2} considered the existence of one or two positive solutions of \eqref{e1.1}. Li and Lu \cite{l3} studied \eqref{e1.1} and obtained at least two positive solutions by an application of a fixed point theorem due to Avery and Henderson. Motivated by \cite{h2,l3} Wang and Guan \cite{w1}, showed that \eqref{e1.1} has at least three positive solutions by applying the Avery Five Functionals Fixed Point Theorem. On the other hand, the application of critical point theory in difference equations has also been studied by Pasquale Candito \cite{c1} who considered the problem $$\begin{gathered} -\Delta[\phi_{p}(\Delta u(k-1))]=\lambda f(k,u(k)),\quad k\in [1,T],\\ u(0)=u(T+1)=0, \end{gathered}\label{e1.2}$$ he established the existence of at least three solutions and two positive solutions to \eqref{e1.2} using critical point theory. However, almost all of these works only considered the $p$-Laplacian equations with nonlinearity $f$ being nonnegative. Therefore, it is a natural problem to consider the existence of positive solution of $p$-Laplacian equations with sign changing nonlinearity. Throughout this paper, we assume that the following two conditions are satisfied: \begin{itemize} \item[(H1)] $a: [1,T+1]\to (0,+\infty)$; \item[(H2)] $f: [0,+\infty)\to \mathbb{R}$ is continuous. \end{itemize} \section{Preliminaries} Let $E=\{u: [0,T+2]\to \mathbb{R}:\Delta u(0)=u(T+2)=0\}$, with norm $\|u\|=\max_{t\in[0,T+2]}|u(t)|$, then $(E,\|\cdot\|)$ is a Banach space. We define two cones by \begin{gather*} P=\{u\in E: u(t)\geq0,\ t\in[0,T+2]\},\\ P'=\{u\in E: \text{$u$ is concave, nonnegative and decreasing on $[0,T+2]$}\}. \end{gather*} \begin{lemma}[\cite{h2,l3,w1}] \label{lem2.1} If $u\in P'$, then $u(t)\geq \frac{T+2-t}{T+2}\|u\|$ for $t\in[0,T+2]$. \end{lemma} Let \begin{align*} K&=\big\{u\in E: \text{$u$ is nonnegative and decreasing on $[0,T+2]$},\\ &\quad u(t)\geq \gamma\|u\|,\; t\in[0,l]\}, \end{align*} where $\gamma=\frac{T+2-l}{T+2}\gamma_{1}$, and for $l\in\mathbb{Z}$ with $l=T+1$, $$\gamma_{1}=\frac{(T+2-l)\phi_{q}[\sum_{i=1}^{l}a(i)]} {\sum_{s=0}^{T+1}\phi_{q}[\sum_{i=1}^{s}a(i)]}\,.$$ Note that $u$ is a solution of \eqref{e1.1} if and only if $$u(t)=\sum_{s=t}^{T+1}\phi_{q}\big[\sum_{i=1}^{s}a(i)f(u(i))\big],\quad t\in[0,T+2].$$ We define the operators $F: P\to E$ and $S:K\to E$ as follows \begin{gather} (Fu)(t)=\sum_{s=t}^{T+1}\phi_{q}\big[\sum_{i=1}^{s}a(i)f(u(i))\big],\quad t\in[0,T+2], \label{e2.1}\\ (Su)(t)=\sum_{s=t}^{T+1}\phi_{q}\big[\sum_{i=1}^{s}a(i)f^{+}(u(i))\big],\quad t\in[0,T+2],\label{e2.2} \end{gather} where $f^{+}(u(t))=\max\{f(u(t)),0\}$, $t\in[0,T+2]$. It is obvious that $S: K\to K$ is completely continuous (see \cite[Theorem 3.1]{h2}). \begin{lemma}[\cite{g1}] \label{lem2.2} Let $K$ be a cone in a Banach space $X$. Let $D$ be an open bounded subset of $X$ with $D_{K}=D\cap K\neq \phi$ and $\overline{D_{K}}\neq K$. Assume that $A:\overline{D_{K}}\to K$ is a completely continuous map such that $x\neq A x$ for $x\in \partial D_{K}$. Then the following results hold: \begin{enumerate} \item If $\|Ax\|\leq\|x\|$, $x\in \partial D_{K}$, then $i_{K}(A, D_{K})=1$; \item If there exists $x_{0}\in K\backslash\{\theta\}$ such that $x\neq Ax+\lambda x_{0}$, for all $x\in \partial D_{K}$ and all $\lambda>0$, then $i_{K}(A, D_{K})=0$; \item Let $U$ be an open set in $X$ such that $\overline{U}\subset D_{K}$. If $i_{K}(A, D_{K})=1$ and $i_{K}(A, U_{K})=0$, then $A$ has a fixed point in $D_{K}\backslash\overline{U}_{K}$. The same results holds, if $i_{K}(A, D_{K})=0$ and $i_{K}(A, U_{K})=1$. \end{enumerate} \end{lemma} \begin{lemma}[\cite{l2}] \label{lem2.3} Let $K_{\rho}=\{u(t)\in K:\|u\|<\rho\}$ and $\Omega_{\rho}=\{u(t)\in K:\min_{0\leq t\leq l}u(t)<\gamma\rho\}$. Then the following properties are satsified: \begin{itemize} \item[(a)] $K_{\gamma\rho}\subset \Omega_{\rho}\subset K_{\rho}$; \item[(b)] $\Omega_{\rho}$ is open relative to K; \item[(c)] $u\in\partial \Omega_{\rho}$ if and only if $\min_{0\leq t\leq l}u(t)=\gamma\rho$; \item[(d)] If $u\in\partial \Omega_{\rho}$, then $\gamma\rho\leq u(t)\leq \rho$ for $t\in [0, l]$. \end{itemize} \end{lemma} Let \begin{gather} m=\Big\{\sum_{s=0}^{T+1}\phi_{q}\big[\sum_{i=1}^{s}a(i)\big]\Big\}^{-1}, \label{e2.3}\\ M=\Big\{(T+2-l)\phi_{q}\big[\sum_{i=1}^{l}a(i)\big]\Big\}^{-1}.\label{e2.4} \end{gather} We remark that by (H1), 00. In fact, if it is not so, then there exist u_{0}\in \partial\Omega_{\rho}, and \lambda_{0}>0 such that u_{0} =Su_{0}+\lambda_{0}e. Then from \eqref{e2.2}, \eqref{e2.4} and \eqref{e2.6}, we obtain \begin{align*} u_{0}(t)&=(Su_{0})(t)+\lambda_{0}\\ &\geq (Su_{0})(l)+\lambda_{0}\\ &=\sum_{s=l}^{T+1}\phi_{q}\big[\sum_{i=1}^{s}a(i)f^{+}(u_{0}(i))\big] +\lambda_{0} \\ &\geq(T+2-l)\phi_{q}\big[\sum_{i=1}^{l}a(i)f^{+}(u_{0}(i))\big]+\lambda_{0} \\ &\geq(T+2-l)\phi_{q}\big[\sum_{i=1}^{l}a(i)\phi_{p}(M\gamma\rho)\big] +\lambda_{0} \\ &=(T+2-l)M\gamma\rho\phi_{q}\big[\sum_{i=1}^{l}a(i)\big]+\lambda_{0}\\ &=\gamma\rho+\lambda_{0}. \end{align*} This together with Lemma \ref{lem2.3}(c) implies that \gamma \rho\geq \gamma \rho+\lambda_{0}, a contradiction. Hence, it follows from Lemma \ref{lem2.2} (2) that i_{K}(S, \Omega_{\rho})=0. The proof is complete. \end{proof} \section{Existence of positive solutions} \begin{theorem} \label{thm3.1} Assume {\rm (H1), (H2)} and that one of the following two conditions holds: \begin{itemize} \item[(H3)] There exist \rho_{1}, \rho_{2} \in (0, +\infty) with \rho_{1}<\gamma \rho_{2} and \rho_{2}<\rho_{3} such that \begin{itemize} \item[(i)] f(u)\leq\phi_{p}(m\rho_{1}), for u\in [0,\rho_{1}]; \item[(ii)] f(u)\geq 0, for u\in[\gamma\rho_{1},\rho_{3}], moreover, f(u)\geq\phi_{p}(M\gamma\rho_{2}), for u\in[\gamma\rho_{2},\rho_{2}], x\neq Sx, x\in\partial\Omega_{\rho_{2}}; \item[(iii)] f(u)\leq \phi_{p}(m\rho_{3}), for u\in[0,\rho_{3}]. \end{itemize} \item[(H4)] There exist \rho_{1}, \rho_{2} and \rho_{3} \in (0, +\infty) with \rho_{1}<\rho_{2}<\gamma \rho_{3} such that \begin{itemize} \item[(i)] f(u)\geq\phi_{p}(M\gamma\rho_{1}), for u\in [\gamma^{2}\rho_{1},\rho_{2}]; \item[(ii)] f(u)\leq \phi_{p}(m\rho_{2}), for u\in[0,\rho_{2}], x\neq Sx, x\in\partial K_{\rho_{2}}; \item[(iii)] f(u)\geq 0, for u\in[\gamma\rho_{2},\rho_{3}], moreover, f(u)\geq\phi_{p}(M\gamma\rho_{3}), for u\in[\gamma\rho_{3},\rho_{3}]. \end{itemize} \end{itemize} Then \eqref{e1.1} has at least two positive solutions u_{1} and u_{2}. \end{theorem} \begin{proof} Assuming (H3), we show that S has a fixed point u_{1} either in \partial K_{\rho_{1}} or u_{1} in \Omega_{\rho_{2}}\setminus \overline{K_{\rho_{1}}}. If u\neq Su, u\in\partial K_{\rho_{1}}\cup\partial K_{\rho_{3}}, by Lemmas \ref{lem2.4} and \ref{lem2.5}, we have i_{K}(S,K_{\rho_{1}})=1,\quad i_{K}(S,\Omega_{\rho_{2}})=0,\quad i_{K}(S,K_{\rho_{3}})=1. $$By Lemma \ref{lem2.3}(b) and \rho_{1}<\gamma \rho_{2}, we have \overline{K}_{\rho_{1}}\subset K_{\gamma \rho_{2}}\subset \Omega_{\rho_{2}}. By Lemma \ref{lem2.2}(3), we have S has a fixed point u_{1}\in \Omega_{\rho_{2}}\backslash \overline{K_{\rho_{1}}}. Similarly, S has a fixed point u_{2}\in K_{\rho_{3}}\backslash \overline{\Omega_{\rho_{2}}}. Clearly,$$ \|u_{1}\|>\rho_{1},\quad \min_{t\in[0,l]}u_{1}(t)=u_{1}(l)\geq \gamma\|u_{1}\|>\gamma\rho_{1}. $$This implies \gamma\rho_{1}\leq u_{1}(t)\leq \rho_{2}, t\in[0,l]. By (H3)(ii), we have f(u_{1}(t))\geq0, t\in[0,l]; i.e., f^{+}(u_{1}(t))=f(u_{1}(t)). Combining with the fact that Su=Fu=0 if t=T+2, we can get Su_{1}=Fu_{1}. That means u_{1} is a fixed point of F. From u_{2}\in K_{\rho_{3}}\backslash \overline{\Omega_{\rho_{2}}}, \rho_{2}<\rho_{3} and Lemma \ref{lem2.3}(b) we have \overline{K}_{\gamma\rho_{2}}\subset \Omega_{\rho_{2}}\subset K_{\rho_{3}}. Obviously, \|u_{2}\|>\gamma\rho_{2}. This implies that$$ \min_{t\in[0,l]}u_{2}(t)=u_{2}(l)\geq \gamma\|u_{2}\|>\gamma^{2}\rho_{2}. $$Therefore,$$ \gamma^{2}\rho_{2}\leq u_{2}(t)\leq \rho_{3},\quad t\in[0,l]. $$By \rho_{1}<\gamma\rho_{2} and (H3)(ii), we have f(u_{2}(t))\geq0, t\in[0,l]; i.e., f^{+}(u_{2}(t))=f(u_{2}(t)). So u_{2} is another fixed point of F. Thus, we have proved that \eqref{e1.1} has at least two positive solutions u_{1} and u_{2}. The proof under assumption (H4) is similar to the case above. This completes the proof. \end{proof} \begin{theorem} \label{thm3.2} Assume {\rm (H1), (H2)} that one of the following two conditions holds: \begin{itemize} \item[(H5)] There exist \rho_{1}, \rho_{2} \in (0, +\infty) with \rho_{1}<\gamma \rho_{2} such that \begin{itemize} \item[(i)] f(u)\leq\phi_{p}(m\rho_{1}), for u\in [0,\rho_{1}]; \item[(ii)] f(u)\geq 0, for u\in[\gamma\rho_{1},\rho_{2}],\ moreover, f(u)\geq\phi_{p}(M\gamma\rho_{2}), for u\in[\gamma\rho_{2},\rho_{2}]. \end{itemize} \item[(H6)] There exist \rho_{1}, \rho_{2}\in (0, +\infty) with \rho_{1}<\rho_{2} such that \begin{itemize} \item[(i)] f(u)\geq\phi_{p}(M\gamma\rho_{1}), for u\in [\gamma^{2}\rho_{1},\rho_{2}]; \item[(ii)] f(u)\leq \phi_{p}(m\rho_{2}), for u\in[0,\rho_{2}]. \end{itemize} \end{itemize} Then \eqref{e1.1} has at least one positive solution. \end{theorem} \section{Example} In this section, we present a simple example to illustrate our results. Consider the following boundary-value problem $$\begin{gathered} \Delta[\phi_{p}(\Delta u(t-1))]+a(t)f(u(t))=0,\quad t\in [1,4],\\ \Delta u(0)=u(5)=0, \end{gathered} \label{e4.1}$$ where p=3/2, q=3, a(t)\equiv 1, T=3 and$$ f(u)=\begin{cases} (u-\frac{8}{75})^{11}, &u\in [0,\frac{8}{75}];\\ (\frac{1}{30})^{1/2}\sin(\frac{75}{67} \frac{\pi}{2}u-\frac{8}{67}\frac{\pi}{2}), & u\in [\frac{8}{75},1];\\ (\frac{1}{30})^{1/2}(\frac{8}{3}-\frac{5}{3}u) +(\frac{1}{10})^{1/2}(\frac{5}{3}u-\frac{5}{3}) & u\in [1,\frac{8}{5}];\\ (\frac{1}{10})^{1/2}+\frac{25}{52\times 67^{2}}(u-\frac{8}{5})^{2},& u\in [\frac{8}{5},15];\\ (\frac{1}{10})^{1/2}+\frac{25}{52\times 67^{2}}(15-\frac{8}{5})^{2}+[1+(u-15)(25-u)],& u\in [15,+\infty). \end{cases} It is easy to check that f: [0,+\infty)\to \mathbb{R} is continuous, l=4, it follows from a direct calculation that $\gamma_{1}=\frac{\phi_{q}\big[\sum_{i=1}^{4}a(i)\big]} {\sum_{s=0}^{4}\big[\sum_{i=1}^{s}a(i)\big]} =\frac{(1+1+1+1)^{2}}{\sum_{s=0}^{4}[a(1)+\dots +a(s)]^{2}}=\frac{8}{15},$ \begin{align*} m&=\Big\{\sum_{s=0}^{4}\phi_{q}\big[\sum_{i=1}^{s}a(i)\big]\Big\}^{-1} =\Big\{\sum_{s=0}^{4}[a(1)+a(2)+\dots +a(s)]^{2}\Big\}^{-1}\\ &=\{1^{2}+2^{2}+3^{2}+4^{2}\}^{-1}=\frac{1}{30}, \\ M&=\Big\{\phi_{q}\big[\sum_{i=1}^{4}a(i)\big]\Big\}^{-1} =\{\phi_{q} [a(1)+a(2)+a(3)+a(4)]\}^{-1}\\ &=\{4^{2}\}^{-1}=\frac{1}{16}, \end{align*} $\gamma=\frac{T+2-l}{T+2}\gamma_{1} =\frac{3+2-4}{3+2}\cdot\frac{8}{15}=\frac{1}{5}\cdot\frac{8}{15} =\frac{8}{75}.$ Choose \rho_{1}=1, \rho_{2}=15, \rho_{3}=25. Then \gamma\rho_{1}<\rho_{1}<\gamma \rho_{2}<\rho_{2}<\rho_{3}. In addition, by the definition of f, we have \begin{itemize} \item[(i)] f(u)\leq\phi_{p}(m\rho_{1})=(\frac{1}{30})^{1/2}, for u\in [0,1]; \item[(ii)] f(u)\geq0, for u\in[\frac{8}{75}\cdot1,25], moreover, f(u)\geq\phi_{p}(M\gamma\rho_{2})=(\frac{1}{16} \cdot\frac{8}{75}\cdot15)^{1/2} =(\frac{1}{10})^{1/2}, for u\in[\frac{8}{75}\cdot15,15]; \item[(iii)] f(u)\leq \phi_{p}(m\rho_{3})=(\frac{1}{30}\cdot25)^{1/2} =(\frac{5}{6})^{1/2}, for u\in[0,25]. \end{itemize} By \eqref{e2.2}, we have (Su)(t)=\sum_{s=t}^{4}\phi_{q}\big[\sum_{i=1}^{s}a(i)f^{+}(u(i))\big] =\sum_{s=t}^{4}[a(1)f^{+}(u(1))+\dots +a(s)f^{+}(u(s))]^{2}.$Since$f^{+}(u)\leq(\frac{1}{10})^{1/2}+\frac{25}{52\times 67^{2}}(15-\frac{8}{5})^{2}$,$u\in [0,15]$. For$u\in \partial\Omega_{15}, we have \begin{align*} \|Su\| &=Su(0)\\ &=\sum_{s=0}^{4}[a(1)f^{+}(u(1))+\dots +a(s)f^{+}(u(s))]^{2} \\ &=[f^{+}(u(1))]^{2}+[f^{+}(u(1))+f^{+}(u(2))]^{2}+[f^{+}(u(1)) +f^{+}(u(2))+f^{+}(u(3))]^{2}\\ &\quad +[f^{+}(u(1))+f^{+}(u(2))+f^{+}(u(3))+f^{+}(u(4))]^{2} \\ &\leq 30[f^{+}(u)]^{2}\\ & <15=\|u\|. \end{align*} This impliesSu\neq u$, for$u\in \partial\Omega_{15}$. Thus, (H3) of Theorem \ref{thm3.1} is satisfied. Then \eqref{e4.1} has two positive solutions$u_{1}, u_{2}$. \begin{thebibliography}{00} \bibitem{a1} R. P. Agarwal; \emph{Difference Equations and Inequalities}, Marcel Dekker, New York, 1992. \bibitem{a2} R. P. Agarwal, M. Bohner, P. J. Y. Wong; \emph{Eigenvalues and eigenfunctions of discrete conjugate boundary value problems}, Comput. Math. Appl. 38(3-4) (1999) 159-183. \bibitem{a3} R. P. Agarwal, J. Henderson; \emph{Positive solutions and nonlinear eigenvalue problems for third-order difference equations}, Comput. Math. 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