\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 99, pp. 1--18.\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 2003 Texas State University-San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/99\hfil Qualitative properties of solutions \dots] {Qualitative properties of solutions for quasi-linear elliptic equations} \author[Zhenyi Zhao\hfil EJDE--2003/99\hfilneg] {Zhenyi Zhao} \address{Department of Mathematical Sciences Tsinghua University, Beijing, 100084, China} \email{zhaozhenyi@tsinghua.org.cn} \date{} \thanks{Submitted May 29, 2003. Published September 25, 2003.} \subjclass[2000]{35J15, 35J25, 35J60} \keywords{Quasi-linear elliptic equations, comparison Principle, \hfill\break\indent boundary blow-up solutions, moving plane method, sliding method, symmetry of solution} \begin{abstract} For several classes of functions including the special case $f(u)=u^{p-1}-u^{m}$, $m>p-1>0$, we obtain Liouville type, boundedness and symmetry results for solutions of the non-linear $p$-Laplacian problem $-\Delta_{p}u=f(u)$ defined on the whole space $\mathbb{R}^n$. Suppose $u \in C^{2}(\mathbb{R}^{n})$ is a solution. We have that either (1) if $u$ doesn't change sign, then $u$ is a constant (hence, $u\equiv 1$ or $u\equiv0$ or $u\equiv-1$); or (2) if $u$ changes sign, then $u\in L^{\infty}(\mathbb{R}^{n})$, moreover $|u|<1$ on $\mathbb{R}^{n}$; or (3) if $|Du|>0$ on $\mathbb{R}^n$ and the level set $u^{-1}(0)$ lies on one side of a hyperplane and touches that hyperplane, i.e., there exists $\nu \in S^{n-1}$ and $x_{0}\in u^{-1}(0)$ such that $\nu \cdot (x-x_{0})\geq 0$ for all $x\in u^{-1}(0)$, then $u$ depends on one variable only (in the direction of $\nu$). \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \section{Introduction}\label{ev1} In this paper we consider the problem \begin{equation} \begin{gathered} -\Delta_{p}u=f(u) \quad\text{in $\Omega$} \\ u>0 \quad\text{in $\Omega$} \\ u=0 \quad\text{on $\partial \Omega$}, \end{gathered} \end{equation} where $\Delta_{p}$ denotes the $p$-Laplacian operator $\Delta_{p}=\mathop{\rm div}(|Du|^{p-2}Du)$, $p>1$, $\Omega=\mathbb{R}^{N},N\geq 2$, and $f(u)$ is locally Lipschitz continuous. In the case $p=2$, several results have been obtained starting with the famous paper by Gidas, Ni and Nirenberg \cite{GNN} where, among other things, it is proved that, if $\Omega$ is a ball and $p=2$, solutions of (1.1) are radially symmetric and strictly radially decreasing. This paper had a big impact not only in virtue of the several monotonicity and symmetry results that it contains, but also because it brought to attention the moving plane method which, since then, has been largely used in many different problems. This method, which is essentially based on maximum principles, goes back to Alexandrov \cite{A} and was first used by Serrin in \cite{S}. The moving plane method has been improved and simplified by Berestycki and Nirenberg in \cite{BN} with the aid of the maximum principle in small domains. Recently, In a series papers, Berestycki, Caffarelli and Nirenberg \cite{BCN1,BCN2,BCN3} began to study the qualitative properties of solutions when $\Omega$ is unbounded, for example slab, half plane, and $\mathbb{R}^{n}$. When $\Omega = \mathbb{R}^{n}$, it is related to the following conjecture of De Giorgi \cite{DG}: If $u$ is a solution of the scalar Ginzburg-Landau equation $$ \Delta{u}+u(1-u^{2})=0 \quad \text{on $\mathbb{R}^{n}$} $$ such that $|u|\leq 1$ and $\partial_{n}u>0$ on $\mathbb{R}^{n}$, and $$ \lim_{x_{n}\rightarrow \pm \infty} u(x',x_{n})=\pm1,\forall x\in \mathbb{R}^{n-1} $$ then all level sets of u are hyperplanes, at least for $n \leq 8$. Here $\partial_{n} u$ denotes the partial derivative of $u$ with respect to $x_{n}$, the last component of $x$, and $x'$ denotes the first $n-1$ components of $x$. When $n=2$, this conjecture was completely resolved by Ghoussoub and Gui \cite{GG}. When $n=3$, it was very recently proved by Ambrosio and Cabre \cite{AC}. Both solutions of the conjecture are based on a Liouville-type theorem due to Berestycki,Caffarelli and Nirenberg \cite{BCN3}. The first partial answer to the De Giorgi conjecture is from the work of 1980 by Modica and Mortola \cite{MM}. In 1985, Modica \cite{M} found a pointwise gradient bound for all bounded solutions. This estimate was further generalized by Caffarelli, Garofalo and Segala \cite{CGS} to more general nonlinear partial differential equations which include the p-Laplacian. Under more assumptions on the solutions, for example, if $u(x)=u(x',x_{n})\rightarrow \pm 1$ as $x_{n}\rightarrow \pm \infty$ holds uniformly for $x' \in \mathbb{R}^{n-1}$, the conclusion of this conjecture was confirmed in \cite{BBG, BHM, F} for any $n\geq 2$. The conjecture in its original form however, remains open for $n>3$. We refer to \cite{AAC} for a fuller account of the history and progress about this conjecture. Du and Ma \cite{DM1} recently removed the boundedness condition $|u|\leq 1$ in De Giorgi's conjecture. This point has already been observed by Farina \cite{F}, but his conclusion does not seem to include those nonlinearities covered by Du and Ma's result. Very little is known about the monotonicity and symmetry of solutions of (1.1) when $p \ne 2$. In this case the solutions can only be considered in a weak sense since, generally they belong to the space $C^{1,\alpha}(\Omega)$(See \cite{DB} and \cite{T}). Anyway this is not a difficulty because the moving plane method method can be adapted to weak solutions of strictly elliptic problems in divergence form(See \cite{D} and \cite{Da}). The real difficulty with problem (1.1), for $p\ne 2$, is that the $p$-Laplacian operator is degenerate in critical points of the solutions, so that comparison principle(which could substitute the maximum principles in order to use the moving plane and sliding method when the operator is not linear) are not available in the same as for $p=2$. Actually counterexamples both to validity of comparison principles and to the symmetry results are available(see \cite{Br})for any $p$ with different degrees of regularity of $f$. A first step towards extending the moving plane method to solutions of problems involving the $p$-Laplacian operator has been done in \cite{Da}. In that paper the author mainly proves some weak and strong comparison principles for solutions of differential inequalities involving the $p$-Laplacian. Using these principles he adapts the moving plane method to solutions of (1.1) getting some monotonicity and symmetry results in the case $1p-1>0$. \begin{thm}[Liouville Type Property] Suppose $u \in C^{2}(\mathbb{R}^{n})$ is a solution of (1.2). Furthermore $u$ doesn't change sign. Then $u$ is a constant (hence,$u\equiv1$,or $u\equiv0$,or $u\equiv-1$). \end{thm} \begin{thm}[Global Boundedness] Suppose $u\in C^{2}(\mathbb{R}^{n})$ is a changing-sign solution of (1.2). Then $u\in L^{\infty}(\mathbb{R}^{n})$, moreover $|u|<1$ on $\mathbb{R}^{n}$. \end{thm} \begin{thm}[One-dimensional Property] Suppose that $u\in C^{2}(\mathbb{R}^{n})$ solves (1.2) on $\mathbb{R}^{n}$ and $|Du|>0$. If $u^{-1}(0)$ lies on one side of a hyperplane and touches that hyperplane, i.e., there exists $\nu \in S^{n-1}$ and $x_{0}\in u^{-1}(0)$ such that $\nu \cdot (x-x_{0})\geq 0$ for all $x\in u^{-1}(0)$,then $u$ depends on one variable only (in the direction of $\nu$). \end{thm} Similar results to our Theorems 1.1 and 1.2 are obtained by Dancer and Du \cite{DD}, Du and Gu \cite{DGu} by different methods. Now we compare our results with the very interesting works of P.Pucci, J.Serrin, and H.Zou \cite{PS, PSZ, SZ}. In their papers \cite{PS, PSZ}, the aim is to find conditions which make the Maximum Principle to be true. So they have to assume the behavior at infinity or at some point of solutions. Since one of our aim is to get Liouville type result, we need only to use the Comparison Principles (see Theorem 2.1-2.4 below). In the paper \cite{SZ}, the authors consider the radial symmetry of the solutions with the assumption about the behavior at infinity of the solutions. But in our Theorem 1.3, we study the one dimensional property of solutions under different conditions of the solutions. Throughout this paper, for simplicity, we assume that $u\in C^{2}(\mathbb{R}^{N})$. The rest of this paper is organized as follows. Some preliminary results are given in section2. In section 3, we prove Theorem 1.1. Theorem 1.2 is proved in section 4. In section 5, we prove two lemmas which are needed in the proof of Theorem 1.3. Theorem 1.3 is proved in section 6. \section{Preliminary Results} \label{ev2} In this section, we collect the related weak and strong comparison principles. Let $\Omega$ be a domain in $\mathbb{R}^{N},N\geq 2$, and let $u,v\in C^{2}(\Omega)$ be solutions of \begin{equation} \begin{gathered} -\Delta_{p}u \leq f(u) \quad\text{in $\Omega$} \\ -\Delta_{p}v \geq f(v) \quad\text{in $\Omega$} \end{gathered} \end{equation} For a set $A \subseteq\Omega$ we define \begin{equation} \begin{aligned} M_{A}&=M_{A}(u,v)&=\sup_{A}(|Du|+|Dv|) \\ m_{A}&=m_{A}(u,v)&=\inf_{A}(|Du|+|Dv|) \end{aligned} \end{equation} Firstly we state the weak maximum principles. \begin{thm}[Weak Comparison Principle] Let $u,v$ be solutions of (2.1) in a bounded domain $\Omega$ and $f \in C[0,\infty),f(0)=0$ and $f$ is non-decreasing on some interval $[0,\delta]$. Suppose also that $u$ and $v$ are continuous in $D$, with $v<\delta$ in $\Omega$ and $u\geq v$ on $\partial \Omega$. Then $u\geq v$ in $\Omega$. \end{thm} \begin{thm}[Weak Comparison Principle] Let $u,v$ be respective solutions of (2.1) in D(maybe unbounded). Suppose that $u$ and $v$ are continuous in $D$,that $m_{D}>0$,and that $u\geq v$ on $\partial D$. Then $u\geq v$ in $D$ \end{thm} \begin{thm}[Weak Comparison Principle] Suppose that $10$, depending on $p$, $|\Omega|$, $M_{\Omega}$ and the $L^{\infty}$ norms of $u$ and $v$ such that: if an open set $\Omega ' \subseteq \Omega$ satisfies $\Omega '=A_{1}\cup A_{2},|A_{1}\cap A_{2}|=0,|A_{1}<\alpha|,M_{A_{2}}2$ and $m_{\Omega}>0$,there exist $\delta,m>0$ depending on $p,|\Omega|,m_{\Omega}$ such that the following holds: if $\Omega'=A_{1} \cup A_{2}$ with $|A_{1}\cap A_{2}|=0,|A_{1}<\delta|$ and $m_{A_{2}}>m$ then $u\leq v$ on $\partial \Omega '$ implies $u\leq v$ in $\Omega '$ \end{thm} For a proof, see \cite{Da} Now, we state a comparison principle in slab. \begin{lem} Let $w$ be a function satisfying $Lw\leq 0$ in $\Omega=\mathbb{R}^{n-1}\times (b,c)$, where $b,c \in R$ and where $$ Lw=\alpha_{ij}(x)\partial_{ij}w+\beta_{j}\partial _{j}u+\gamma (x) u\,. $$ Assume that the coefficients $\alpha_{ij}(x),\beta_{j}(x)$ are uniformly continuous in $\overline{\Omega}$ and that the $\alpha_{ij}$ satisfy $$ \exists c'_{0}\geq c_{0}>0, \forall x\in \mathbb{R}^{n},\forall \xi \in \mathbb{R}^{n},c_{0}|\xi|^{2}\leq \alpha_{ij}(x)\xi_{i}\xi_{j}\leq c'_{0}|\xi|^{2}. $$ Furthermore, assume that $$ -C\leq \gamma (x) \leq0 \quad \text{for all $x\in \Omega$} $$ for some positive real number $C$. The function $w$ is required to be continuous in $\overline{\Omega}$ and to satisfy $Lw \in L^{\infty}(\Omega)$ and $m\leq w\leq M$ in $\Omega$ for some $m,M\in R$. If $w\geq 0$ on $\partial \Omega$, then $w \geq 0$ in $\Omega$ \end{lem} For the proof of this lemma, we can refer to Lemma 3.1 of \cite{BHM}. From the maximum principle, we can get the following comparison result. \begin{thm} Let $f$ be a Lipschitz-continuous function, non-increasing on the intervals $[-1,-1+\delta]$ and $[1-\delta,1]$ for some $\delta>0$. Assume that $u_{1},u_{2}$ are solutions of $$ \Delta_{p}u_{i}+f(u_{i})=0 \quad \text{in $\Omega$} $$ and are such that $|u_{i}|\leq 1(i=1,2)$. Furthermore, assume that $$ u_{2}\geq u_{1} \quad \text{on $\partial \Omega$} $$ and that either $u_{2}\geq 1-\delta$ in $\Omega$ or $u_{1}\leq -1+\delta$ in $\Omega$, Where $\Omega=\mathbb{R}^{n-1} \times (b,c)$. Then $u_{2}\geq u_{1}$. \end{thm} Next we deal with a form of a strong comparison theorem.First we prove the following Harnack type comparison inequality \begin{lem}[Harnack type comparison inequality] Suppose $u,v$ satisfy \begin{equation} -\mathop{\rm div}A(x,Du)+\Lambda u \leq -\mathop{\rm div}(x,Dv)+\Lambda v,u\leq v \quad \text{in $\Omega$} \end{equation} where $\Lambda \in \mathbb{R}$ and $u,v\in W^{1,\infty}_{\rm loc}(\Omega)$ if $p\ne 2$;$u,v\in W^{1,2}_{\rm loc}(\Omega)$ if $p=2$. Suppose $\overline{}{B(x,5\delta)}\subseteq \Omega$ for some $\delta >0$ and,if $p\neq 2,\inf_{D}(|Du|+|Dv|)>0$. Then, for any positive number $s< \frac{n}{n-2}$ we have $$ \|v-u\|_{L^{s}(B(x,2\delta))}\leq c\delta^{N/2} \inf_{B(x,\delta)}(v-u) $$ where $c$ is a constant depending on $N,p,s,c_{2},\delta$ and, if $p\neq 2$, also on $m=\inf_{B(x,5\delta)}(|Du|+|Dv|)$ and $M_{B(x,5\delta)}$ \end{lem} This lemma implies the following strong comparison principle: \begin{thm} Let $u,v\in C^{2}(\Omega)$ be solutions of (2.1)with $11$ is a constant and $m>p-1$. Then $u$ must be a constant. \end{thm} The basic ingredients in the proof consist of the following three lemmas. For use in later sections and possible future applications,these lemmas are given in much more general form than what is required in the proof of Theorem 3.1. We consider the problem \begin{equation} \Delta_{p}u+\alpha(x) u^{p-1}-\beta(x)u^{m}=0 \quad \text{on $\mathbb{R}^{n}$} \end{equation} Here $p>1$ is a constant and $m>p-1$. \begin{lem}[Comparison Principle] Suppose that $\Omega$ is a bounded domain in $\mathbb{R}^{N}$, $\alpha(x)$ and $\beta(x)$ are continuous functions on $\Omega$ with $\|\alpha\|_{\infty}<\infty$ and $\beta(x)$ positive, $p>2$. Let $u_{1},u_{2}\in C^{2}(\Omega)$ be positive in $\Omega$ and satisfy \begin{equation} \Delta_{p}u_{1}+\alpha(x)u_{1}^{p-1}-\beta(x)u_{1}^{m}\leq 0 \leq \Delta_{p}u_{2}+\alpha(x)u_{2}^{p-1}-\beta(x)u_{2}^{m},x\in \Omega \end{equation} and $\limsup_{x\rightarrow \partial \Omega}(u_{2}-u_{1})\leq 0$, and $\alpha(x)\leq \beta(x)$. Then $u_{2}\leq u_{1}$ in $\Omega$ \end{lem} \begin{proof} Let $\varepsilon_{1}>\varepsilon_{2}>0$ and denote $w_{i}=(u_{i}+\varepsilon_{i})^{-1}((u_{2}+\varepsilon_{2})^{2}-(u_{1}+\varepsilon_{1})^{2})_{+}(i=1,2)$. Observe $w_{i}$ be $C^{2}$ nonnegative functions on $\Omega$ and vanishing near $\partial \Omega$. Using (3.2), applying integration by parts and subtracting, we obtain \begin{equation} \begin{aligned} &-\int_{\Omega}[|\nabla u_{2}|^{p-2}\nabla u_{2}\nabla w_{2}-|\nabla u_{1}|^{p-2}\nabla u_{1}\nabla w_{1}]dx \\ &\geq \int_{\Omega}\beta(x)[u_{2}^{m}w_{2}-u_{1}^{m}w_{2}]+\int_{\Omega}\alpha(x)(u_{1}^{p-1}w_{1}-u_{2}^{p-1}w_{2}) \end{aligned} \end{equation} Denote $\Omega_{+}(\varepsilon_{1},\varepsilon_{2})=\{x\in \Omega:u_{2}(x)+\varepsilon_{2}>u_{1}(x)+\varepsilon_{1}\}$ and note that the integrands in (3.3) vanishing outside this set. The left side of (3.3) equals \begin{equation} \begin{aligned} &-\int_{\Omega_{+}(\varepsilon_{1},\varepsilon_{2})}[|\nabla u_{1} |^{p-2}|\nabla u_{2}-\frac{u_{2}+\epsilon_{2}}{u_{1}+\varepsilon_{1}}\nabla u_{1}|^{2}+|\nabla u_{2}|^{p-2}|\nabla u_{1}-\frac{u_{2}+\epsilon_{2}}{u_{1}+\varepsilon_{1}}\nabla u_{2}|^{2}]\\ &-\int_{\Omega_{+}(\varepsilon_{1},\varepsilon_{2})}(|\nabla u_{2}|^{p-2}-|\nabla u_{1}|^{p-2})(\nabla u_{2}\nabla u_{2}-\nabla u_{1}\nabla u_{1})dx \end{aligned} \end{equation} Noting that $w_{1}>w_{2}$ in $\Omega_{+}(\varepsilon_{1},\varepsilon_{2})$. We conclude that the left side of (3.4) is not positive. On the other hand as $\varepsilon_{1}\rightarrow 0$ the right side of (3.3) converges to $$ \int_{\Omega_{+}(0,0)}[\beta(x)(u_{2}^{m-1}-u_{1}^{m-1})-\alpha(x)(u_{2}^{p-1}-u_{1}^{p-1})](u^{2}_{2}-u^{2}_{1}) $$ while last term in (3.3) converge to $0$. Unless $\Omega_{+}(0,0)$ is empty, the limiting value of the right side of (3.3) is positive. Since this leads to a contradiction we conclude that $u_{2}\leq u_{1}$in $\Omega$ \end{proof} \begin{lem}[Locally uniformly Boundedness] $u\in C^{2}$ is a positive solution of (3.1). Then we have the bound $$ \max_{G}u(x)\leq c_{0} $$ For every compact subset $G\subset \mathbb{R}^{n}$ and $c_{0}$ is a constant. \end{lem} \begin{proof} Suppose that $max_{G}u(x)=u_{x_{0}}$ for some $x_{0}\in G$. If $|Du(x_{0})|=0$, Then $u\leq \max_{G}(\alpha(x)/\beta(x))^{m-p+1}$. Otherwise, we may assume that there is a ball $B_{2r}:=B_{2r}(x_{0})\subset \mathbb{R}^{n}$with center $x_{0}\subset G$ such that $$ \max_{\bar{G}}=\max_{B_{r}}u(x):=M(r) \quad \mbox{and}\quad \min_{B_{2r}}|Du|>0 $$ Since,on $B_{2r}$, we have $$ \Delta_{p}u \geq -\alpha(x)u $$ Then as pointed out in \cite{DM1}, $u$ locally uniformly bounded. \end{proof} \begin{lem} Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with smooth boundary. Suppose $\alpha$ and $\beta$ are smooth positive functions on $\bar{\Omega}$, and let $\mu_{1}$ denote the first eigenvalue of $-\Delta_{p}u=\mu \alpha(x)u^{p-1}$ on $\Omega$ under Dirichlet boundary conditions on $\partial \Omega$. Then the problem $$ -\Delta_{p}u=\mu u[\alpha (x)u^{p-2}-\beta(x)u^{m-1}],u|_{\partial \Omega}=0 $$ has a unique positive solution for every $\mu>\mu_{1}$,and the unique positive solution $u_{\mu}$ satisfies $u_{\mu}\rightarrow [\alpha(x)/\beta(x)]^{1/(m-p+1)}$ \end{lem} \begin{proof} The existence from a simple upper and lower solution argument. clearly any constant greater that or equal to $M=\max_{\bar{\Omega}}[\alpha(x)/\beta(x)]^{1/(m-p+1)}$ is an upper solution. Let $\phi$ be a positive eigenfunction corresponding to $\mu_{1}$, then for each fixed $\mu>\mu_{1}$ and all small positive $\epsilon,\epsilon \phi0$ such that $\epsilon<\upsilon_{0}=[\alpha(x)/\beta(x)]^{1/(m-p+1)}$ on $\Omega$, we let $\upsilon_{\epsilon}=\upsilon_{0}+\epsilon$ , and find that $v_{\epsilon}(\alpha(x)v_{\epsilon}^{p-2}-\beta(x)\upsilon_{\epsilon}^{m-1})\leq -\delta$ on $\Omega$ for some positive constant $\delta=\delta(\epsilon)$ and $-\Delta_{p}\upsilon_{\epsilon}\geq -c$ on $\Omega$ for some positive constant $c=c(\epsilon)$. It follows that for all large $\mu$, $\upsilon_{\epsilon}$ is an upper solution of our problem. On the other hand, Let $\phi$ be the positive eigenfunction corresponding to $\mu_{1}$ with $\|\phi\|_{\infty}=1$. then we can find a small neighborhood of $\partial \Omega$ in $\Omega$, say $U$, such that $\phi$ is very small in $U$ so that for all $\mu>\mu_{1}+1,-\Delta_{p}\phi=\mu_{1}\alpha(x)\phi^{p-1} \leq \mu\phi(\alpha(x)\phi^{p-2}-\beta(x)\phi^{m-1})$ on $U$. By shrinking $U$ further if necessary, we can assume that $\bar{U}\cap K =\emptyset$ and $\phiw_{\epsilon}>0$ on the rest of $\Omega$. It is easily seen that such $w_{\epsilon}$ is a lower solution of our problem for all large $\mu$. since $w_{\epsilon}<\mu_{\epsilon}$, we deduce $w_{\epsilon}\leq u_{\mu}<\upsilon_{\epsilon}$ on $\Omega$ for all large $\mu$. In particular, $$ [\alpha(x)/\beta(x)]^{1/(m-p+1)}+\epsilon \geq u_{\mu}\geq [\alpha(x)/\beta(x)]^{1/(m-p+1)}-\epsilon $$ on $K$ for all large $\mu$. this is to say that $u_{\mu}\rightarrow (\alpha/\beta)^{1/(m-p+1)}$ as $\mu \rightarrow \infty$ uniformly on $K$, as required. \end{proof} \begin{lem} Let $\Omega$ be an arbitrary domain in $\mathbb{R}^{n}$ and suppose that there exists a large solution of the equation $\Delta_{p} u=u^{m}$ in $\Omega$. Let $\Xi$ be a compact subset of $\partial \Omega$ and let $P\in \Xi$. Suppose that,for every $\delta>0$,there exists an open,connected neighborhood of $P$,say $Q_{P}$ with $C^{2}$ boundary, such that, \begin{itemize} \item $\Omega_{P}=Q_{P}\cap \Omega$ is a simply connected domain. \item $Q_{P}\subset \Xi_{\sigma}=\{x:dist(x,\Xi)<\sigma \}$ and $\partial \Omega \cap \bar{\Omega}_{P}=\overline{\partial \Omega\cap Q_{P}}$ \end{itemize} Then there exists $\delta_{0}>0$(which depends on $\Xi$ but not on $P$) such that,if $\Omega_{P}$ is contained in $\Xi_{\delta_{0}}$,the following statements hold. \begin{itemize} \item[(a)] There exists a large solution of (3.1) in $\Omega_{P}$; \item[(b)] There exists a positive solution $v$ of (3.1) in $\Omega_{P}$ such that \begin{gather} v(x)\rightarrow\infty \quad \text{locally uniformly as $x\rightarrow \Gamma_{1}=\partial \Omega \cap Q_{P}$}\\ v\in C(\Omega_{P}\cup \Gamma_{2}) \quad \text{and $v=0$ on $\Gamma_{2}=\Omega\cap\partial Q_{P}$} \end{gather} \end{itemize} \end{lem} \begin{proof} (a) Let $b=2\sup_{\Omega}\beta(x)$ and let $c=\sup\{-\alpha(x)t^{p-1}-\frac{1}{2}bt^{m}:t>0,x\in \Omega\}$. Then, every positive solution $u$ of (3.1) satisfies $$ \Delta_{p}u\leq bu^{m}+c $$ Let $U$ be a large solution of $\Delta_{p}u=2bu^{m}$ in $\Omega$. This means that $u$ is a solution of $\Delta_{p}u=2bu^{m}$ with boundary value $u=+\infty$ on $\partial \Omega$. Let $M=\inf\{U(x):x\in \Omega\cap\Xi\}$ and choose $\delta_{0}$ sufficiently small so that $bM^{m}\geq c$. Then \begin{equation} \Delta_{p} U\geq bU^{m}+c \quad \text{in $\Omega_{P}$} \end{equation} Let $\{ \Theta_{n}\}$ be an increasing sequence of domains with $C^{2}$ boundary such that \begin{equation} \bar{\Theta}_{n}\subset \Theta_{n+1}\subset\Omega_{P}\quad \text{and}\quad \Theta_{n}\uparrow \Omega_{P}. \end{equation} Let $u_{n}$ and $V$ be large solutions of (3.1) in $\Theta_{n}$ and $Q_{P}$ respectively. By comparison principle $\{u_{n}\}$ is monotone decreasing and $u_{n}\geq V$ in $\Theta_{n}$. By the comparison principle, (3.7) and (3.8) $u_{n}\geq U$ in $\Theta_{n}$. Hence $\lim u_{n}$ is a large solution of (3.1) in $\Omega_{P}$ \noindent (b) For the proof of the second statement we may assume (in view of (a)) that there exists a large solution of (3.1) in $\Omega$. Now, Let $\{\Theta_{n}\}$ be an increasing sequence of domain with $C^{2}$ boundary such that, $$ \Theta_{n}\subset \Omega_{P},\Theta_{n}\uparrow \Omega_{P} \quad \text{and}\quad \Omega_{P}\setminus \Theta_{n} \subset K_{n}= \{x:dist(x,\Gamma_{1})<2^{-n}\}. $$ Denote $\Gamma_{1,n}=\partial\Theta_{n}\cap K_{n},\Gamma_{2,n}=\partial \Theta_{n}\cap (\bar{K_{n}})^{c}$. Thus $\Gamma_{2,n}\subset\Gamma_{2,n+1}\subset \Gamma_{2}$. We shall also assume that the sets $\Gamma_{1,n}$ are disjoint. For each $n$,consider a sequence of functions $\{\varphi_{n,k}\}_{k=1}^{\infty}$ on $\partial \Theta_{n}$ satisfying the following properties. \begin{itemize} \item $\varphi_{n,k}=k$ on $\Gamma_{1,n};\varphi_{n,k}=0$ for $x\in \Gamma_{2,n}$ such that $dist(x,\Gamma_{1,n})>2^{-n}$; \item $0\leq \varphi_{n,k}\leq k$ everywhere; $\varphi_{n,k}\in C^{2}(\partial \Theta_{n})$; \item $\varphi_{n,k}\geq \varphi_{n-1,k}$ on $\Gamma_{2,n}$ and $\varphi_{n,k}\leq\varphi_{n,k-1}$ on $\partial \Theta_{n}$ \end{itemize} Let $v_{n,k}$ be a solution of (3.1) in $\Theta_{n}$ in $\Theta_{n}$ such $v_{n,k}=\varphi_{n,k}$ on $\partial \Theta$. By comparison principle $\{v_{n,k}\}_{k=1}^{\infty}$ is monotone increasing and by Lemma 3.2 the sequence is locally bounded. Hence $v_{n}=\lim_{k\rightarrow\infty}v_{n,k}$ is a solution of (3.1) in $\Theta_{n}$ such that \begin{equation} \begin{gathered} v_{n}\rightarrow \infty \quad \text{as $x\rightarrow \Gamma_{1,n};v_{n} \in C(\Theta_{n}\cup \Gamma_{2,n})$}\\ v_{n}=0 \quad \text{on $\Gamma_{2,n}$} \end{gathered} \end{equation} Furthermore,by their construction,$v_{n,k}\geq v_{n+1,k}$ so that $\{ v_{n} \}$ is monotone decreasing. Consequently $v=\lim_{n\rightarrow \infty}v_{n}$ is a solution of (3.1) in $\Omega_{P}$. If $V$ is a large solution of (3.1) in $Q_{P},v_{n}+V$ is a supersolution of (3.1) in $\Theta_{n}$ which blows up on $\partial \Theta_{n}$. Hence $v_{n}+V\geq U$, where $U$ is a large solution of (3.1) in $\Omega$. Thus $v+V\geq U$ and this implies (3.5) Finally by (3.9), $v$ satisfies (3.6) \end{proof} \begin{lem} The problem \begin{equation} -\Delta_{p} u=\mu u[\alpha(x)^{p-2}-\beta(x)u^{m-1}],u|_{\partial \Omega}=\infty \end{equation} has a unique positive solution for each $\mu>0$, and the unique positive solution $u_{\mu}$ satisfies $u_{\mu}\rightarrow (\alpha/\beta)^{1/(m-p+1)}$ uniformly on any compact subset of $\Omega$ as $\mu \rightarrow \infty$ \end{lem} Here and throughout this paper, by $u|_{\partial \Omega}$, we mean $u(x)\rightarrow\infty$ as $d(x,\partial \Omega)\rightarrow 0$. We also write $x\rightarrow \partial \Omega$ when $d(x,\partial \Omega)\rightarrow 0$. \begin{proof} 1. Existence: The existence follows from a simple upper and lower solution argument. Suppose $\mu>0$. For any positive integer $n>M=\max_{\bar{\Omega}}(\alpha/\beta)^{1/(m-p+1)}$, the problem $$ -\Delta_{p}u=\mu u(\alpha u^{p-2}-\beta u^{m-1}),u|_{\partial \Omega}=n $$ has a unique positive solution. Indeed $u\equiv 0$ and $u\equiv n$ are lower and upper solution to this problem,and hence there is at least one positive solution. By comparison principle,there is at most one positive solution. Therefore there is a unique positive solution. Denoting this solution by $u_{n}$, we find, by comparison principle, that $u_{n}$ increases with $n$. By Lemma 3.3 we can find a uniform upper bound for $u_{n}$ on any compact subset of $\Omega$, then by a standard regularity argument,$u_{\mu}=lim_{n\rightarrow \infty}u_{n}$ would be a positive solution of (3.10). 2. Uniqueness: Suppose that $u$ is a large solution of (3.10). Note that for every $\epsilon>0$ there exists $\beta_{\epsilon}>0$ such that $$ k(1-\epsilon)u^{m}\leq \Delta_{p} u\leq k(1+\epsilon)u^{m} \quad \text{in $\{ x\in \Omega:\mathop{\rm dist}(x,\partial \Omega)>\beta_{\epsilon}\}$} $$ Let $P\in \partial \Omega$ and assume (as we may) that the set $Q_{p}$ mentioned above is an open,bounded spherical cylinder centered at $P$, with axis parallel to the $\xi_{n}$ axis. Thus, $$ Q_{p}=\{\eta:|\eta '|<\rho_{P},|\eta_{N}|<\tau_{P}\} $$ where $\eta=\xi-P$ and $\eta'=(\eta_{1},\dots,\eta_{N-1})$. By appropriately choosing $\sigma_{P}$ and $\tau_{P}$ we may also assume that $\partial \Omega$ is bounded away from the 'top' and 'bottom' of the cylinder $Q_{P}$ and that $\partial \Omega \cup \bar{Q_{P}}=\overline{\partial \Omega\bigcap Q_{P}}$. finally we assume that $\rho_{P}$ and $\tau_{P}$ are sufficiently small so that Lemma 3.5 can be applied to $Q_{P}$ and so that \begin{equation} \begin{gathered} k(P)(1-\epsilon)u^{m}(x)\leq\Delta_{p}u\leq k(P)(1+\epsilon)u^{m}(x) \\ \forall x \in \Theta =Q_{P}\cap \Omega. \end{gathered} \end{equation} Therefore there exists a solution $v$ of the problem \begin{align*} \Delta_{p}v=v^{m} \quad \text{and $v>0$ in $\Theta=Q_{P}\cap \Omega$} \\ v(x)\rightarrow \infty \quad \text{locally uniformly as $x\rightarrow Q_{P}\cap \partial \Omega$} \\ v(x)\rightarrow 0 \quad \text{locally uniformly as $x\rightarrow \partial Q_{P}\cap \Omega$} \end{align*} Next denote \begin{align*} v_{1}=(k(P)(1-\epsilon))^{-1/(m-1)}v \\ v_{2}=(k(P)(1+\epsilon))^{-1/(m-1)}v \end{align*} and let $w$ be the large solution of (3.10) in $Q_{P}$. We claim that \begin{equation} v_{2}0\}$. If $f$ is a function defined in $\Theta$, set $f_{\sigma}(x)=f(x+\sigma \xi)$ for $x\in \Theta_{\sigma}$. Assume that $\sigma$ is a sufficiently small positive number so that $\Theta_{\sigma}\subset\subset\Omega$. Then $v_{1,\sigma}+w_{\sigma}$ is a supersolution in $\Theta_{\sigma}$ and hence $v_{1,\sigma}+w_{\sigma}>u$ there. On the other hand, by(3.11), $v_{2,-\sigma}0$, $$ \lim_{x\rightarrow \partial \Omega}[(1+\epsilon)u_{1}-u_{2}]=\infty $$ As $(1+\epsilon)u_{1}$ is an upper solution to (3.10), we can apply comparison principle to conclude that $(1+\epsilon)u_{1}\geq u_{2}$ on $\Omega$. As $\epsilon >0$ is arbitrary, we deduce $u_{2}\geq u_{1}$. Thus $u_{1}=u_{2}$ on $\Omega$. This proves the uniqueness. \noindent 3. Asymptotic behavior. Now we know that the positive solution $u_{\mu}$ constructed above is the unique positive solution. Let $K$ be an arbitrary compact subset of $\Omega,v_{0}=(\alpha/\beta)^{1/(m-p+1)}$ and $\epsilon>0$ any small positive number satisfying $\epsilonw_{\epsilon}$. On the other hand,fix a $\mu_{0}>0$ then we can find a small neighborhood $U$ of $\partial \Omega$ in $\Omega$ such that $u_{0}=u_{\mu_{0}}>v_{0}+\epsilon$ on $U$. Therefore, $$ -\Delta_{p}u_{0}=\mu_{0}u_{0}(\alpha u_{0}^{p-2}-\beta u_{0}^{m-1})\geq \mu u_{0}(\alpha u_{0}^{p-2}-\beta u_{0}^{m-1}) $$ on $U$ for all $\mu >\mu_{0}$. Now let us choose a smooth function $v_{\epsilon}$ satisfying $v_{\epsilon}=u_{0}$ on $U$,$v_{\epsilon}=v_{0}+\epsilon$ on $K$ and $v_{\epsilon}=v_{0}+\epsilon /2$ on the rest of $\Omega$. Then it is easily checked that $v_{\epsilon}$ is an upper solution for the equation of $u_{n}$ provided that $\mu$ is large enough. As $w_{\epsilon}>v_{\epsilon}$ on $\Omega$,we must have $w_{\epsilon}\leq u_{n}\leq v_{\epsilon}$ on $\Omega$ for all large $\mu$ and every large $n$. It follows that $w_{\epsilon}\leq u_{\mu}\leq v_{\epsilon}$ on $\Omega$. This implies that $u_{\mu}\rightarrow v_{0}$ on $K$ as $\mu\rightarrow \infty$, as required. The proof of the lemma is now complete. \end{proof} \noindent\textbf{Remark.} In the above argument, we used the idea of \cite{BM}. However, our case is more complicated. We have to overcome this difficulty. \begin{proof}[Proof of Theorem 3.1] Let us first observe that a nonnegative entire solution of $\Delta_p(u)+\lambda u^{p-1}-u^{m-1}$ is either identically zero or positive everywhere,due to the Harnack inequality. therefore,we need only consider positive solutions. Set $\Omega=\{x:|Du(x)|=0\}$. if $\Omega=\mathbb{R}^{n}$,we are done. It is easy to see that $\Omega$ is closed. Let $x_{0}$ be an arbitrary point in $\mathbb{R}^{n}$, we will show that $u(x_{0})=\lambda ^{1/(m-p+1)}$,using only pointwise convergence of $v_{\alpha}$ and $w_{\alpha}$. For $\alpha>0$ let us define \begin{equation} u_{\alpha}(x)=u(x_{0}+\alpha(x-x_{0})) \end{equation} It easily checked that $u_{\alpha}$ satisfies $$ \Delta_{p}u+\alpha^{p}(\lambda u^{p-1}-u^{m}) $$ Let $B$ denote the a ball with center $x_{0}$ and $B \cap \Omega=\emptyset$. By Lemma 3.4, for large $\alpha$, the problem $$ \Delta_{p}v +\alpha^{p}(v^{p-1}-v^{m}),v|_{\partial B}=0 $$ has a unique positive solution $v _{\alpha}$ and as $\alpha \to \infty,v_{\alpha}\to \lambda^{1/(m-p+1)}$ at $x=x_{0}\in B$. Applying comparison principle we see that $u_{\alpha}\geq v_{\alpha}$ on $B$, and hence $$ u(x_{0})=u_{\alpha}(x_{0})\geq v_{\alpha}(x_{0}) $$ Letting $\alpha \rightarrow \infty$ in the above inequality we conclude that $u(x_{0})\geq \lambda ^{1/(m-p+1)}$. Let $w_{\alpha}$ be the unique positive solution of $$ \Delta_{p} w+\alpha^{p}(w^{p-1}-w^{m}),w|_{\partial B}=\infty $$ by Lemma 3.6 we know that as $\alpha\rightarrow \infty, w_{\alpha}(x) \rightarrow \lambda^{1/(m-p+1)}$ at $x=x_{0}\in B$. Applying comparison principle we can see that $u_{\alpha}\leq w_{\alpha}$ on B. Thus $$ u(x_{0})=u_{\alpha}(x_{0})\leq w_{\alpha}(x_{0}) $$ Letting $\alpha \rightarrow \infty$ we obtain $u(x_{0})\leq \lambda ^{1/(m-p+1)}$. Therefore $u(x_{0})=\lambda ^{1/(m-p+1)}$.As $x_{0}$ is arbitrary, we conclude that $u \equiv \lambda ^{1/(m-p+1)}$. \end{proof} \noindent\textbf{Remark:} We believe this result is also true when p-Laplacian is replaced by the MCO $\mathop{\rm div}(\frac{\nabla u}{\sqrt{1+|\nabla u|^{2}}})$. \section{Global Boundedness and Related Results} \label{ev4} In this section, we prove general result which contains Theorem 1.2 as a special case. Let us observe the following result for the ODE problem \begin{equation} u'=f(u),u(0)=u_{0}. \end{equation} \begin{lem} Suppose $f$ is $C^{1}$ and satisfies $$ f(0)=f(1)=0,\quad f(u)>0\; \forall u\in (0,1),\quad f(u)<0 \;\forall u>1 $$ Then for any $u_{0}>0$, the unique solution $u(t)$ of (4.1) satisfies $\lim _{t\rightarrow \infty}u(t)=1$. \end{lem} \begin{proof} If $u_{0}=1$, we have $u(t)\equiv 1$ and there is nothing to prove. If $01$ follows from a similar analysis, except that now $u(t)$ is decreasing. \end{proof} \begin{thm} Let $u\in C^{2}(\mathbb{R}^{n})$ be a solution of (1.2). Then the conclusions in Theorem 1.2 hold. \end{thm} \begin{proof} Let us first observe that it suffices to show $|u|\leq 1$ in $\mathbb{R}^{n}$. Indeed, if $|u(x_{0})|=1$ say $u(x_{0})=-1$,then, $w:=u+1$ satisfies $$ -\mathop{\rm div}(|\nabla w|^{p-2}\nabla u)=f(w-1),w\geq 0,w(x_{0})=0\,. $$ Hence, it follows from the strong maximum principle that $w\equiv 0$, contradicting our assumption that $u$ changes sign. We now prove $|u|\leq 1$ on $\mathbb{R}^{n}$. Set $D=\{x:|Du(x)|=0\}$ it is easily seen that $|u|\leq 1$ on $D$. On $\mathbb{R}^{n}\setminus D$, let $g(u)=-u^{r},p-1c$. Define $v(x)=v(x-x_{0})$. We find that the set $\{x\in B(x_{0}):u(x)>v(x)\}$ has a component $\Omega$ whose closure lies entirely in the open ball $B(x_{0})=\{x:x-x_{0}\in B\}$. On $\Omega$, we have $u(x)>v(x)\geq c>M$ where $M$ satisfies $-u^{r}(M)=u^{p-1}(M)-u^{m}(M)$ and $\Delta_{p}u+g(u) \geq 0=\Delta_{p}v+g(v)$. Moreover, $u=v$ on $\partial \Omega$. As $g(u)$ is decreasing for $u>M$, from comparison principle, we deduce that $u\equiv v$ in $\Omega$. This contradiction shows that we must have $u\leq c$ on $\mathbb{R}^{n}$. Applying the above argument to $w=-u$ which satisfies $$ \Delta_{p}w=g(w),g(w)=-f(-w), $$ we deduce that $u \geq -c$ on $\mathbb{R}^{n}$. Therefore, $$ -c\leq u(x),\forall x\in \mathbb{R}^{n}\setminus D. $$ Let $u_{c}$ and $u_{-c}$ denote the unique solution of $$ u'=f(u),\quad u(0)=u_{0} $$ with $u_{0}=c$ and $u_{0}=-c$, respectively. Then it follows from Lemma 4.1 that $u_{c}(t) \rightarrow 1$ and $u_{-c}(t)\rightarrow -1$ as $t \rightarrow + \infty$. One the other hand, $u,u_{c},u_{-c}$ are all bounded solutions of the parabolic problem $$ u_{t}-\mathop{\rm div}(|\nabla u|^{p-2}\nabla u)=f(u)\,. $$ Since $u_{c}(0)\geq u(x)\geq u_{-c}(0)$ on $\mathbb{R}^{n} \setminus D$, by the parabolic maximum principle and the boundedness of $u,u_{c},u_{-c}$, we conclude that $u_{-c}(t)\leq u(x)\leq u_{c}(t)$ for all $t>0$. Letting $t\rightarrow \infty$, we obtain $-1\leq u(x) \leq 1$, as required. This finishes our proof of Theorem 4.2. \end{proof} \section{Some Lemmas} \label{ev5} In this section, we prove two lemma which are needed in the proof of Theorem 1.3. Consider the solutions of the problem \begin{equation} \Delta_{p}u+f(u)=0 \quad \text{in $\mathbb{R}^{n}$} \end{equation} and that satisfy $|u|\leq 1$ together with the asymptotic conditions \begin{gather} u(x',x_{n})\rightarrow \pm 1 \quad \text{as $x_{n}\rightarrow \pm \infty$ uniformly in $x'=(x_{1},\dots,x_{n-1})$},\\ |Du|>0\,. \end{gather} Assume that the function $f=f(u)$ is Lipschitz-continuous on $[-1,1]$, and that there exists $\delta>0$ such that \begin{equation} f \quad \text{is non-increasing on $[-1,-1+\delta]$ and on $[1-\delta,1]$}. \end{equation} \begin{lem} Let $u$ be a solution of (5.1), (5.2) and (5.3) such that $|u|\leq 1$. Then $u(x',x_{n})=u_{0}(x_{n})$ \end{lem} The proof uses a sliding method and a version of comparison principle in slab; i.e., Theorem 2.6. \begin{proof} Let us now consider a solution of (5.1), (5.2) and (5.3) such that $|u|\leq 1$, and let $f$ satisfy (5.4). We are first going to prove that $u$ is increasing in any direction $v=(v_{1},\dots,v_{n})$ such that $v_{n}>0$. In order to do so, for any $t\in R$, we define the function $u^{t}$ by $u^{t}(x)=u(x+tv)$. From (5.2), there exists real $a>0$ such that $u(x',x_{n})\geq 1-\delta$ for all $x'\in \mathbb{R}^{n-1}$ and $x_{n}\geq a$ and $u(x',x_{n}) \leq -1+\delta$ for all $x' \in \mathbb{R}^{n-1}$ and $x_{n}\leq -a$. For any $t\geq 2a/v_{n}$, the functions $u$ and $u^{t}$ are such that \begin{gather*} u^{t}(x',x_{n})\geq 1-\delta \quad\text{for all $x'\in \mathbb{R}^{n-1}$ and for all $x_{n}\geq -a$}, \\ u(x',x_{n})\leq -1+\delta \quad\text{for all $x'\in \mathbb{R}^{n-1}$ and for all $x_{n}\leq -a$}, \\ u^{t}(x',-a)\geq u(x',-a) \quad\text{for all $x'\in \mathbb{R}^{n-1}$ and for all $x_{n}\leq -a$}, \end{gather*} We now apply comparison principle in slabs of the type $$ \Omega_{h}=\mathbb{R}^{n-1}\times (-a,h) $$ with $h>-a$. Due to (5.2), there exists a function $\varepsilon (h)\geq 0$ such that $u^{t}(x',h)-u(x',h)\geq -\varepsilon(h)$ for all $x'\in \mathbb{R}^{n-1}$ and $\varepsilon(h) \rightarrow 0$ as $h\rightarrow +\infty$. Choose any $h>-a$ and set $$ w=u^{t}(x)+\varepsilon (h)\,. $$ Then $w,u$ fulfill the assumption of Theorem 2.6. We have $w\geq u$ in $\Omega_{h}$. By passing to the limit $h\rightarrow \infty$, we conclude that $$ u^{t}(x',x_{n})\geq u(x',x_{n}) \quad \text{for all $x'\in \mathbb{R}^{n-1}$ and $x_{n}\geq -a$}. $$ Similarly, we could show that $$ u^{t} \geq u(x',x_{n}) \quad \text{for all $x'\in \mathbb{R}^{n-1}$ and $x_{n}\leq -a$} $$ whence $u^{t}\geq u$ in $\mathbb{R}^{n}$ Let us now decrease $t$. We claim that $u^{t} \geq u$ for all $t>0$. Indeed, define $\tau =\inf\{t>0,u^{t} \geq u \text{in $\mathbb{R}^{n}\}$}$. By continuity, we see that $u^{\tau}\geq u$ in $\mathbb{R}^{n}$. Let us now argue by contradiction and suppose that $\tau >0$. Two cases may occur. \noindent\textbf{Case 1.} Suppose that \begin{equation} \inf_{\mathbb{R}^{n-1}\times [-a,a]}(u^{\tau}-u)>0. \end{equation} From standard elliptic estimates, $u$ is globally Lipschitz-continuous. Hence, there exists a real $\eta_{0}$ small enough, which can be chosen smaller than $\tau$, such that for all $\tau \geq t>t-\eta_{0}$, we have $$ u^{t}(x',x_{n})-u(x',x_{n})>0 \quad \text{for all $x'\in \mathbb{R}^{n-1}$ and $x_{n}\in [-a,a]$} $$ Since $u\geq 1-\delta$ in $\mathbb{R}^{n-1}\times [a,+\infty)$ it follows that $$ u^{t}(x',x_{n})-u(x',x_{n})>0 \quad \text{for all $x'\in \mathbb{R}^{n-1}$ and $x_{n}\in [-a,a]$}. $$ We may now apply Theorem 2.6 in the two half-space $\Omega^{+}=\{x_{n}>a\}$ and $\Omega^{-}=\{x_{n}<-a\}$. We then infer that, for all $\eta \in [0,\eta_{0}]$, $u^{\tau-\eta}(x',x_{n})\geq u(x',x_{n})$ for all $x'\in \mathbb{R}^{n-1}$ and for all $x_{n}\in(-\infty,-a)\bigcup (a,+\infty)$ and so for all $x_{n}\in R$ owing to (5.2). This is contradiction with the minimality of $\tau$. Hence (5.5) is ruled out. \noindent\textbf{Case 2.} Suppose \begin{equation} \inf_{\mathbb{R}^{n-1}\times [-a,a]}(u^{\tau}-u)=0. \end{equation} Then there exists a sequence $x^{k}_{k\in N}\in \mathbb{R}^{n-1}\times [-a,a]$ such that $u^{\tau}(x^{k})-u(x^{k})\rightarrow 0$ as $k\rightarrow +\infty$. We normalize $u$ by translation on $\mathbb{R}^{n}$ by setting $u_{k}(x)=u(x+x_{k})$. Then by standard elliptic estimate we may assume that $u_{k}$ converges to a solution $u_{\infty}$ of (5.1) as $k \rightarrow \infty$. We have $u^{\tau}_{\infty}(0)=u_{\infty}(0)$ and $u^{\tau}_{\infty}\geq u_{\infty}$ because $u^{\tau}_{k}\geq u_{k}$ for any $k\in N$. We have \begin{gather*} \Delta_{p}u^({\tau}_{\infty})+f(u^{\tau}_{\infty})=\Delta_{p}u_{\infty}+f(u_{\infty}) \quad \text{in $\mathbb{R}^{n}$} \\ u^{\tau}_{\infty}\geq u_{\infty} \quad \text{in $\mathbb{R}^{n}$} \\ u^{\tau}_{\infty}(0)=u_{\infty}(0) \end{gather*} Strong Comparison Principle yields $u^{\tau}_{\infty}(x)\equiv u_{\infty}(x)$. This means that $u_{\infty}(x)\equiv u_{\infty}(x+\tau v)$. Letting $\xi=\tau v$, we see that $u_{\infty}$ is periodic with respect to the vector $\xi$. Recalling that $-a \leq x^{k}_{n}\leq a$, we see that the function $u_{\infty}$ also satisfies the uniform limiting conditions (5.2). hence, since $\xi _{n}>0$, the function $u_{\infty}$ cannot be $\xi -periodic$. So Case 2 is also ruled out. Therefore, we have proved that $\tau =0$. the function $u$ is then increasing in any direction$v=(v_{1},\dots,v_{n})$ such that $v_{n}>0$. From the continuity of $\nabla u$, we deduce that $\partial_{v}u\geq 0$ for any $v$ such that $v_{n}=0$. If $v_{n}=0$,by taking $v$ and $-v$, we find that $\partial_{v}u=0$. Since this is true for all $v$ with $v_{n}=0$. Since this is true for all $v$ with $v_{n}=0$, this implies that $u(x)=u(x_{n})$. Since the solutions of (5.1) are unique up to translations,it then follows that the solutions $u$ of (5.1), (5.2) such that $|u|\leq 1$ are unique up to translations of the origin. The proof is complete. \end{proof} \begin{lem} Let $f$ be a Lipschitz continuous function which is positive over (0,1), and satisfies $f(1)=0,f(t)\geq \delta_{0}t$ on $(0,t_{0})$ for some small $\delta_{0}>0$ and $t_{0}>0$. If $u$ is $C^{2}$ on the half plane $\Sigma_{M}:=\{x\in \mathbb{R}^{n}:x_{n}>M\}$ and satisfies $$ \Delta_{p}u+f(u)\leq 0,00$ in B} \\ z\leq 0 \quad \text{on $\partial B$} \end{gather*} Then, for any continuous one-parameter family of Euclidean motions(i.e., translations and rotations) $A(t)$ for $0\leq t\leq T$ with $A(0)=Id$ and $A(t)\overline{B}\subset D,\forall t$, we have for all $t\in [0,t]$: \begin{equation} z_{t}(x):=z(A(t)^{-1}x)0$ in $B_{t}$} \\ z_{t} \leq 0 \quad \text{on $\partial B$} \end{gather*} Thus in $B_{t},z(t),z$ satisfies $\Delta_{p}z_{t}+f(z_{t})\geq \Delta_{p}z+f(z) $ wherever $z_{t}>0$ in $B_{t}$ and \begin{equation} z_{t}0$. Consequently, by (5.8), any $\tilde{x}\in \partial G$ lies in $B_{t}$. Hence $z_{t}(\tilde{x})>0$ and $z_{t}(x)>0$ for $x$ near $\tilde{x}$, which shows that $\tilde{x}\in G$. We have reached a contradiction. Hence, for all $t\in [0,T]$, the graph of $z_{t}$ always lies below that of $u$ in $B_{t}$. \end{proof} \begin{lem} There exist $\epsilon_{1},R_{0}>0$ with $R_{0}$ depending only on $n$ and $\delta_{0}$ of Lemma 5.2 such that $$ u(x)>\epsilon_{1} \quad \text{if $\mathop{\rm dist}(x,\Gamma)>R_{0}$} $$ \end{lem} \begin{proof} Let $B_{R_{0}}$ be a ball with $R_{0}$ so large that the principal eigenvalue $\lambda_{1}=\lambda_{1}(B_{R_{0}})$ of $\Delta_{p}$ in $B_{R_{0}}$ under Dirichlet boundary conditions satisfies $$ \lambda_{1}=\lambda_{1}(B_{R_{0}})<\delta_{0}. $$ Let $\varphi_{1}$ be the eigenfunction of $-\Delta_{p}$ in $B_{R_{0}}$, i.e., \begin{gather*} \varphi_{1}>0, -\Delta_{p}\varphi_{1}=\lambda_{1}\varphi_{1}^{p-1} \quad \text{in $B_{R_{0}}$} \\ \varphi_{1}=0 \quad \text{on $\partial B_{R_{0}}$} \end{gather*} with $\max\varphi_{1}=1$. then for $0<\epsilon \leq s_{0}$ the function $z=\epsilon \varphi_{1}$ is a subsolution of our equation, i.e., \begin{gather*} \Delta_{p}z+f(z)\geq 0 \quad \text{in $B_{R_{0}}$} \\ z=0 \quad \text{on $\partial B_{R_{0}}$} \end{gather*} Let us choose $a=(0,a_{n})$ with $a_{n}$ large enough so that $\overline{B_{R_{0}}(a)}$ lies in $\Omega$. For $B=B_{R_{0}}(a)$, set $\epsilon_{0}=\min_{\bar{B}}u$ (clearly $\epsilon_{0}>0$), and set $\epsilon_{1}=\min(\epsilon_{0},s_{0})$. Since $\max_{\bar{B}}\varphi_{1}=1$, it follows that $$ \epsilon_{1}\varphi_{1}(x-a)\leq u(x) \quad \text{in $B_{R_{0}}(a)$}. $$ In view of Lemma 5.3, we find then that $\forall y \in \Omega$ with $\mathop{\rm dist}(y,\Gamma)>R_{0}$ $$ \epsilon_{1}\varphi_{1}(x-a)\epsilon_{1}$, thereby proving Lemma 5.4 \end{proof} Using Lemma 5.4, we prove a result that implies Lemma 5.2 \begin{lem} Let $y$ be a point with $dist(y,\Gamma)>R_{0}$. By Lemma 5.4, $\epsilon_{1}\leq u(y)$. Set $$ \delta=\delta(y)=\min\{ f(s):s\in [\epsilon_{1},u(y)]\} $$ There is a constant $C_{1}$ depending only on $n$ such that \begin{equation} C_{1}\delta \leq [\mathop{\rm dist}(y,\Gamma)-R_{0}]^{-2} \end{equation} \end{lem} \begin{proof} We first choose $C_{1}$. Let $v$ be the solution in $B_{1}(0)$ of \begin{gather*} \Delta_{p}v=-1 \quad \text{in $B_{1}(0)$} \\ v=0 \quad \text{on $\partial B_{1}(0)$} \end{gather*} We take $C_{1}=\max_{B_{1}(0)}v=v(x_{0})$. We assume that (5.9) does not hold and argue by contradiction;i.e., we assume $$ C_{1}\delta > [\mathop{\rm dist}(y,\Gamma)-R_{0}]^{-2} $$ Fix $R<\mathop{\rm dist}(y,\Gamma)-R_{0}$ such that $C_{1}\delta > \mathbb{R}^{-1}$. Since $\Delta_{p}u<0$ at $y,u$ cannot achieve a local minimum there. choose a point $y_{1}$ close to $y$ with $u(y_{1})R_{0}+R$. By Lemma 5.4, $u\geq \epsilon_{1}$ in $B_{R}(y_{1})=:B$. Let $z$ be the solution in $B$ of \begin{equation} \begin{gathered} \Delta_{p}z=-\delta \quad \text{in $B$} \\ z=0 \quad \text{on $\partial B$} \end{gathered} \end{equation} By scaling, one finds that $$ \max z=z(y_{1})=C_{1}\delta \mathbb{R}^{2} $$ For $0<\tau$ small, $\tau z\Delta_{p}u>0$ in this neighborhood $N$. this contradicts the fact that $\tau z-u$ has a local maximum at $x_{0}$ \end{proof} \begin{proof}[Proof of Lemma 5.2] Since $\mathop{\rm dist}(x,\Gamma)\rightarrow \infty$, from (5.9) follows that $\min_{[\epsilon_{1},u(x)]}f \rightarrow 0$ which implies that $u(x)\rightarrow 1$ uniformly in $\Omega$ \end{proof} \section{Odd Nonlinearity and Related Results}\label{ev6} In this section, we prove a generalization of Theorem 1.3. \begin{thm} Suppose $f$ is Lipschitz continuous and satisfies $$ f(-1)=f(0)=f(1)=0,\quad tf(t)>0 \quad \text{when $0<|t|<1$}, $$ and for small positive constants $\delta_{0},t_{0}$ and $\delta$ \begin{gather*} \frac{f(t)}{t}\geq \delta_{0} \quad \text{when $0<|t|0,\forall x'\in \mathbb{R}^{n-1},\forall x_{n}>0$; the other possibility that $u(x',x_{n})<0,\forall x^{'} \in \mathbb{R}^{n-1},\forall x_{n}>0$ can be handled analogously. For $\tau \geq 0$, let us define $$ u_{\tau}(x',x_{n})=-u(x',2\tau -x_{n}). $$ Since $f$ is odd, we easily see that $$ -\Delta_{p}u_{\tau}=f(u_{\tau}) $$ clearly $u|_{x_{n}=\tau}\geq 0 \geq u_{\tau}|_{x_{n}=\tau}$. We want to show that for every $\tau \geq 0$, $u\geq u_{\tau}$ on the half space$\{ x:x_{n}\geq \tau \}$. Since $u(x)>0$ when $x_{n}>0$, it follows from Lemma 5.2 that $u(x^{'},x_{n})\rightarrow 1$ as $x_{n}\rightarrow\infty$ uniformly in $x^{'}\in \mathbb{R}^{n-1}$. Therefore, for large $\tau$ we can apply comparison principle to $$ \Omega:=\{x:x_{n}>\tau\} $$ to conclude that $u\geq u_{\tau}$ on $\Omega$. Now define $$ \tau_{0}=\inf\{\tau \in [0,\infty):u(x{'},x_{n})\geq u_{\tau}(x^{'},x_{n}), \forall x^{'}\in \mathbb{R}^{n-1}, \forall x_{n}\geq \tau\}. $$ \noindent\textbf{Claim:} $\tau_{0}=0$. Otherwise,$\tau_{0}>0$ and $u(x)\geq u_{\tau_{0}}(x)$ on the set $\Omega_{0}:=\{x:x_{n}\geq \tau_{0}\}$. Clearly $u,u_{\tau_{0}}$ satisfied $$ \Delta_{p}u+f(u)=\Delta_{p}u_{\tau}+f(u_{\tau}) $$ Since $u>0>u_{\tau_{0}}$ on $\partial \Omega_{0}$, by the definition of $\tau_{0}$,we have two possibilities. \begin{itemize} \item[(a)] $u(x_{0})=u_{\tau}(x_{0})$ for some $x_{0}\in \Omega_{0}$, or \item[(b)] $u(x)>u_{\tau}(x)>0$ in $\Omega_{0}$ and $u(z_{k})-u_{\tau}(z_{k})\rightarrow 0$ for some $z_{k}\in \bar{\Omega}_{0}$ with $|z_{k}|\rightarrow \infty$. \end{itemize} If case(a) occurs, then the Harnack inequality forces $w\equiv 0$ on $\Omega_{0}$, which is impossible as $w>0$ on $\partial \Omega_{0}$. If (b) occurs, we set $u_{k}(x)=u(x+z_{k})$. By standard elliptic estimates, up to extraction of a subsequence, $u_{k}$ converges in $C^{2}_{\rm loc}(\mathbb{R}^{n})$ to a solution $u^{*}$ of (5.1) as $k\rightarrow \infty$. Moreover, $$ v:=u^{*}-u^{*}_{\tau_{0}} $$ satisfies $v(0)=0$ and $$ \Delta_{p}u^{*}+f(u^{*})=\Delta_{p}u_{\tau_{0}}^{*}+f(u_{\tau_{0}}^{*}),v\geq 0, \forall x\in \Omega ^{*} $$ where $\Omega^{*}=\{x:x_{n}>\tau^{*}\}$ with $\tau^{*}\in[-\infty,0]$ determined by (passing to a subsequence when necessary) $$ \tau^{*} =-\lim_{k\rightarrow \infty} d(z_{k},\partial \Omega_{0}). $$ If $0\in \Omega^{*}$ then we obtain from the Harnack inequality that $v\equiv 0$ on $\Omega^{*}$, i.e., $$ u^{*}(x^{'},x_{n})=-u^{*}(x{'},2\tau_{0}-x_{n}),\forall x^{'}\in \mathbb{R}^{n-1},\forall x_{n}>\tau^{*}\,. $$ Taking $x_{n}=\tau_{0}$ we deduce $u^{*}(x',\tau_{0})=0$. This implies that ${d(z_{k},\partial \Omega_{0})}$ is bounded, for otherwise, due to $u(x',x_{n})\rightarrow 1$ uniformly in $x'\in \mathbb{R}^{n-1}$ as $x_{n}\rightarrow +\infty$, we would have $u^{*}\equiv 1$. The boundedness of $\{d(z_{k},\partial \Omega_{0})\}$ and the fact that $u(x',x_{n})\rightarrow 1$ uniformly in $x'\in \mathbb{R}^{n-1}$ as $x_{n}\rightarrow +\infty$ imply $u^{*}(x',x_{n})\rightarrow 1$ uniformly in $x'\in \mathbb{R}^{n-1}$ as $x_{n}\rightarrow +\infty$. This together with comparison principle implies that $u^{*}(x',x_{n})\rightarrow -1$ uniformly in $x'\in \mathbb{R}^{n-1}$ as $x_{n}\rightarrow -\infty$. Hence we can use Lemma 5.1 to conclude that $u^{*}(x)=u^{*}(x_{n})$ and is increasing in $x_{n}$. On the other hand,since $u_{k}(0)=u(z_{k})>0$, we have $u^{*}(0)\geq 0$, a contradiction to the monotonicity of $u^{*}(x_{0})$ and $u^{*}(\tau_{0})=0$ If $0\in \partial \Omega^{*}$, we necessarily have $\{d(z_{k},\partial \Omega_{0})\}\rightarrow 0$ and hence $\tau^{*}=0,\Omega^{*}=\{ x:x_{n}>0 \}$. As before,this implies $u^{*}(x',x_{n})\rightarrow 1$ uniformly in $x'$ as $x_{n}\rightarrow \infty$. Moreover for any$\eta \geq -\tau_{0}$, since $u_{k}(x',\eta)=u((x',\eta)+z_{k})\geq 0$ , we deduce $$ u^{*}(x',\eta)\geq 0,\forall x' \in \mathbb{R}^{n-1}. $$ In particular, \begin{equation} u^{*}(0,x_{n})\geq 0,\forall x_{n}\geq\tau_{0} \end{equation} As $v(0)=0$, we have $u^{*}(0)=-u^{*}(0,2\tau_{0})$. Therefore we necessarily have $u^{*}(0)=u^{*}(0,2\tau_{0})=0$. In view of (6.1),the function $g(t):=u^{*}(0,t)$ has a local minimum at $t=0$ and at$t=2\tau_{0}$.Therefore,$g'(0)=g'(2\tau_{0})=0$.This implies that$\partial_{n}v(0)=0$.Since $v$ satisfies $$ \Delta_{p}u^{*}+f(u^{*})=\Delta_{p}u^{*}_{\tau}+f(u^{*}_{\tau})\geq 0,\forall x\in \Omega^{*},v(0)=0,0\in \partial \Omega * $$ an application of the strong comparison principle gives $v\equiv 0$, i.e., $u^{*}(x',x_{n})=-u^{*}(x',2\tau_{0}-x_{n})$ for all $x'\in \mathbb{R}^{n-1}$ and all $x_{n}\geq 0$. We can now argue as in the case that $0\in \Omega ^{*}$ to conclude that $u^{*}(x)=u^{*}(x_{n})$ and is increasing in $x_{n}$. But this is in contradiction with our earlier observation that $u^{*}(0)=u^{*}(2\tau_{0})$.This proves our claim. From $\tau_{0}=0$ we obtain $u(x^{'},x_{n})\geq -u(x^{'},-x_{n})$ for all $x^{'}\in \mathbb{R}^{n-1}$and $x_{n}\geq 0$. Hence we must have $u(x^{'},x_{n})=-u(x^{'},-x_{n})$ for all$x^{'}\in \mathbb{R}^{n-1}$ and all $x_{n}>0$. Recall that we have $u(x^{'},x_{n})\rightarrow -1$ uniformly in $x^{'}$ as $x_{n}\rightarrow -\infty$. Therefore we can use Lemma 5.1 and conclude. The proof of Theorem 1.3 is complete. \end{proof} \subsection*{Acknowledgments} This is my master thesis at the Tsinghua University. I am very grateful to my advisor, Professor Li Ma, for his insightful guidance, substantial support, constructive advice and encouragement. \begin{thebibliography}{02} \bibitem{A} A. D. Alexandrov, {\it A characteristic property of spheres}, Ann.Mat.Pure Appl. (4)58,1962, 303-315. \bibitem{AAC} G. Alberti, L. Ambrosio and X.Cabre, {\it On a long-standing conjecture of E. DE Giorgi:old and recent results}, to appear in Acta Applicandae Mathematicae. \bibitem{AC} L. Ambrosio and X. Cabre, {\it Entire solutions of semi-linear elliptic equations in $\mathbb{R}^{3}$ and a conjecture of De Giorgi}, Journal of Amer. Math. Soc., 13 (2000), 725-739. \bibitem{AW} D. G. Aronson and H. F. Weinberger, {\it Multiple nonlinear diffusion arising in population genetics}, Adv.Math., 30(1978), 33-76. \bibitem{BBG} M. T. Barlow, R. F. Bass and C. Gui, {\it The liouville property and a conjeccture of De Giorgi}, Comm.Pure Appl.Math., 53(2000), 1007-1038. \bibitem{BCN1} H. Berestycki, L. Caffarelli and L. Nirenberg, {\it Inequalities for second order elliptic equations with applications to unbounded domains I}. Duke Math.J., 81(1996), 467-494. \bibitem{BCN2} H. Berestycki, L. Caffarelli and L. Nirenberg, {\it Monotonicity for elliptic equations in an unbounded Lipschitz domain}, Comm.Pure Appl. Math., 50(1997), 1089-1111. \bibitem{BCN3} H. Berestycki, L. Caffarelli and L. Nirenberg, {\it Further properties for elliptic equations in unbounded domains}, Ann.Scuola Norm.Sup.Pisa Cl.Sci., (4)25(1997), 69-94. \bibitem{BHM} H. Berestycki, F. Hamel and R. Monneau, {\it One-dimensional symmetry of bounded entire solutions of some elliptic equations}, Duke Math. J., 103 (2000), 375-396. \bibitem{BM} C. Bandle and M. Marcus, {\it Asymptotic behaviour of solutions and their derivatives,for semilinear elliptic problems with blowup on the boundary}, Ann.Inst.Henri Poincare, 12, 1995, 155-171. \bibitem{BN} H. Berestycki and L. Nirenberg, {\it On the method of moving planes and the sliding method}, Bol.Soc.Brasil.Mat. 22(1991), 1-37. \bibitem{Br} F. Brock, {\it Continuous rearrangement and symmetry of solutions of elliptic problems}, Habilitation thesis, Cologne(1997). \bibitem{CGS} L. Caffarelli, N. Garofalo and F. Segala, {\it A gradient bound for entire solutions of quasi-linear equations and its consequences}, Comm.Pure Appl.Math., 47(1994), 1457-1473. \bibitem{D} E. N. Dancer, {\it Some remarks on the method of moving planes}, Differential Integral Equations, 11,3(1998), 493-501. \bibitem{DD} E. N. Dancer and Yihong Du, {\it some Remarks on liouville type results for quasilinear elliptic equations}, Proc.AMS, s 0002-9939(02)06733-3. \bibitem{Da} L. Damascelli, {\it Comparison theorems for some quasilinear degenerate elliptic operators and applications to symmetry and monotonicity results}, Ann.Inst.Henri Poincare, 15, 1998, 493-516. \bibitem{DP1} L. Damascelli and F. Pacela, {\it Monotonicity and Symmetry of Solutions of p-Laplace equations,$1