\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 182, 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 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/182\hfil A Liouville type theorem] {A Liouville type theorem for $p$-Laplace equations} \author[C. Enache \hfil EJDE-2015/182\hfilneg] {Cristian Enache} \address{Cristian Enache \newline Simion Stoilow Institute of Mathematics of the Romanian Academy, 010702,\newline Bucharest, Romania} \email{cenache23@yahoo.com} \thanks{Submitted October 18, 2014. Published July 1, 2015.} \subjclass[2010]{70H25} \keywords{p-Laplace equation; Liouville theorem; entire solutions} \begin{abstract} In this note we study solutions defined on the whole space $\mathbb{R}^N$ for the $p$-Laplace equation $$ \operatorname{div}(| \nabla u| ^{p-2}\nabla u)+f(u)=0. $$ Under an appropriate condition on the growth of $f$, which is weaker than conditions previously considered in McCoy \cite{McC07} and Cuccu-Mhammed-Porru \cite{CMP10}, we prove the non-existence of non-trivial positive solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} In this note we improve some Liouville type results previously obtained in McCoy \cite{McC07} and Cuccu-Mohammed-Porru \cite{CMP10} for solutions to the $p$-Laplace equation \begin{equation} \operatorname{div}( | \nabla u| ^{p-2}\nabla u) +f(u)=0\quad \text{in }\mathbb{R}^{N},\; p>1, \label{e1.1} \end{equation} where the nonlinearity $f$ is a real differentiable function. The classical Liouville Theorem states that \emph{any harmonic function on the whole Euclidian space} $\mathbb{R}^{N}$, $N\geq 2$, \emph{which is bounded from one side, must be identically constant}. Nowadays it is already known that this property is not anymore a prerogative of harmonic functions, since it is also shared by bounded (from below and/or above) entire solutions to many other differential equations (we refer the reader to the survey paper of Farina \cite{Fa07} for an overview on Liouville type theorems in PDEs). For instance, when $p=2$ in \eqref{e1.1}, McCoy \cite{McC07} has proved that \emph{if }$f$\emph{ is differentiable and satisfies} \begin{equation} f'(t) \leq \frac{N+1}{N-1}\frac{f(t) }{t}\quad \text{for all }t>0, \label{e1.2} \end{equation} \emph{then any positive solution of \eqref{e1.1} must be a constant}. Later, this result was extended to the more general case $p>1$ by Cuccu, Mohammed and Porru \cite{CMP10}, as follows: \emph{if} $f$ \emph{is differentiable and satisfies} \begin{equation} f'(t) \leq (p-1) \frac{N+1}{N-1}\frac{f(t) }{t}\quad \text{for all }t>0, \label{e1.3} \end{equation} \emph{then any positive solution of \eqref{e1.1} must be a constant.} Adapting the main idea from the above mentioned works, we are going to show that, under a weaker condition on the growth of $f$, the above Liouville type results still hold. More precisely, we have: \begin{theorem} \label{thm1.1} Assume that $u(\mathbf{x})>0$ satisfies \eqref{e1.1}. If $f$ is differentiable and satisfies \begin{equation} f'(t) \leq \beta (p-1) \frac{N+1}{N-1}\frac{f(t) }{t}\quad \text{for all }t>0, \label{e1.4} \end{equation} where \begin{equation} \beta \in \begin{cases} [ 1,\frac{N-1}{N-p}), &\text{when }11$ be a real number and $N\geq 2$. If $u( x) $ is a $C^2$ function and $u_{i}$ denotes partial differentiation with respect to $x_{i}$, then \begin{equation} (p-1) u_{11}^2+\sum_{i=2}^N u_{ii}^2\geq \frac{(p-1) ( N-1) +1}{N-1}u_{11}^2- \frac{2}{N-1}\Delta uu_{11}+\frac{1}{N-1}( \Delta u) ^2. \label{e2.1} \end{equation} \end{lemma} Let us now introduce the auxiliary function \begin{equation} P(u;\mathbf{x}):=\frac{| \nabla u(\mathbf{x})| ^2}{ u^{2\beta }(\mathbf{x})}. \label{e2.2} \end{equation} Let us also consider a point $\mathbf{x}^{\ast }$ where $| \nabla u| >0$. From a seminal work of Tolksdorf \cite{To84} we know that $u(\mathbf{x}) $ is smooth in $\omega :=\{ \mathbf{x} \in \Omega :| \nabla u| (\mathbf{x})>0\} $. Therefore, we may compute in $\omega $, successively, the following derivatives: \begin{gather} P_{k}=\frac{2}{u^{2\beta }}u_{ik}u_{i}-\frac{2\beta }{u^{2\beta +1}} | \nabla u| ^2u_{k} \label{e2.3} \\ \begin{aligned} P_{kl}&= \frac{2}{u^{2\beta }}u_{ik}u_{il}+\frac{2}{u^{2\beta }} u_{ikl}u_{i}-\frac{4\beta }{u^{2\beta +1}}u_{ik}u_{i}u_{l}-\frac{4\beta }{ u^{2\beta +1}}u_{il}u_{i}u_{k} \\ &\quad +\frac{2\beta ( 2\beta +1) }{u^{2\beta +2}}| \nabla u| ^2u_{k}u_{l}-\frac{2\beta }{u^{2\beta +1}}| \nabla u| ^2u_{kl}. \end{aligned} \label{e2.4} \end{gather} Now, performing eventually a translation and/or rotation if necessary, we choose the coordinate axes such that at $\mathbf{x^{\ast }}$ we have \begin{equation} | \nabla u| =u_{1}\,\quad u_{i}=0\quad \text{for }i=2,\dots ,N. \label{e2.5} \end{equation} Using \eqref{e2.5} in \eqref{e2.4} we find that \begin{equation} \begin{aligned} P_{11}&= \frac{2}{u^{2\beta }}u_{1i}u_{1i}+\frac{2}{u^{2\beta }} u_{111}u_{1}-\frac{8\beta }{u^{2\beta +1}}u_{11}u_{1}^2 \\ &\quad +\frac{2\beta ( 2\beta +1) }{u^{2\beta +2}}u_{1}^{4} -\frac{2\beta }{u^{2\beta +1}}u_{11}u_{1}^2, \end{aligned} \label{e2.6} \end{equation} respectively \begin{equation} \begin{aligned} \Delta P&= \frac{2}{u^{2\beta }}u_{ik}u_{ik}+\frac{2}{u^{2\beta }} (\Delta u) _{1}u_{1}-\frac{8\beta }{u^{2\beta +1}}u_{11}u_{1}^2 \\ &\quad +\frac{2\beta ( 2\beta +1) }{u^{2\beta +2}}u_{1}^{4}-\frac{ 2\beta }{u^{2\beta +1}}u_{1}^2\Delta u. \end{aligned} \label{e2.7} \end{equation} It then it follows that \begin{equation} \begin{aligned} &\Delta P+( p-2) P_{11} \\ & =\frac{2}{u^{2\beta }}[ ( \Delta u) _{1}+( p-2) u_{111}] u_{1}+\frac{2}{ u^{2\beta }}[ u_{ik}u_{ik}+( p-2) u_{1j}u_{1j}] \\ & \quad -\frac{2\beta }{u^{2\beta +1}}[ \Delta u+( 5p-6) u_{11}] u_{1}^2+\frac{2\beta ( 2\beta +1) }{u^{2\beta +2} }(p-1) u_{1}^{4}. \end{aligned} \label{e2.8} \end{equation} On the other hand, differentiating \eqref{e1.1} with respect to $x_{1}$ and evaluating the result making use of \eqref{e2.5}, we obtain that at $\mathbf{x}^{\ast }$ we have \begin{equation} \begin{aligned} &u_{1}^{p-1}[ ( \Delta u) _{1}+( p-2) u_{111}+2( p-2) \sum_{j=2}^N u_{1j}^2u_{1}^{-1}] \\ &\quad +( p-2) [ \Delta u+( p-2) u_{11} ] u_{11}u_{1}^{p-3}+f'u_{1}=0. \end{aligned} \label{e2.9} \end{equation} Also, evaluating equation \eqref{e1.1} at $\mathbf{x^{\ast }}$ we have \begin{equation} \Delta u+( p-2) u_{11}=-f( u) u_{1}^{2-p}. \label{e2.10} \end{equation} Inserting \eqref{e2.10} in \eqref{e2.9} we obtain \begin{equation} ( \Delta u) _{1}+( p-2) u_{111}=2( 2-p) \sum_{j=2}^N u_{1j}^2u_{1}^{-1}+( p-2) u_{11}fu_{1}^{1-p}-f'u_{1}^{3-p}. \label{e2.11} \end{equation} Making now use of Lemma \ref{lem2.1}, we also have \begin{equation} \begin{aligned} &u_{ij}u_{ij}+( p-2) u_{1j}u_{1j}\\ & \geq (p-1) u_{11}^2+\sum_{i=2}^N u_{ii}^2 +p\sum_{j=2}^N u_{1j}^2 \\ &\geq \frac{(p-1) ( N-1) +1}{N-1}u_{11}^2 -\frac{2}{N-1}\Delta uu_{11} +\frac{1}{N-1}( \Delta u) ^2+p\sum_{j=2}^N u_{1j}^2. \end{aligned} \label{e2.12} \end{equation} Therefore, inserting \eqref{e2.11} and \eqref{e2.12} in \eqref{e2.8} leads to \begin{equation} \begin{aligned} &\Delta P+( p-2) P_{11} \\ & \geq \frac{2}{u^{2\beta }}\Big\{( 4-p) \sum_{j=2}^N u_{1j}^2 +(p-2) u_{11}fu^{2-p}-f'u_{1}^{4-p}\\ &\quad +\frac{(p-1) ( N-1) +1}{N-1} u_{11}^2-\frac{2}{N-1}\Delta uu_{11} +\frac{1}{N-1}( \Delta u)^2 \\ &\quad-\beta [ \Delta u+( 5p-6) u_{11}] \frac{u_{1}^2}{u}+\beta ( 2\beta +1) (p-1) \frac{ u_{1}^{4}}{u^2}\Big\} . \end{aligned} \label{e2.13} \end{equation} From \eqref{e2.3} and \eqref{e2.5} we have \begin{equation} P_{i}=\frac{2}{u^{2\beta }}u_{i1}u_{1}\quad \text{for }i=2,\dots ,N. \label{e2.14} \end{equation} Therefore \begin{equation} \frac{4}{u^{4\beta }}u_{1}^2\sum_{i=2}^N u_{i1}u_{i1} =\sum_{i=2}^N P_{i}^2 \leq | \nabla P| ^2, \label{e2.15} \end{equation} so that \begin{equation} \frac{2}{u^{2\beta }}( 4-p) \sum_{j=2}^N u_{1j}^2 \geq -| 4-p| \frac{| \nabla P| ^2}{2P}. \label{e2.16} \end{equation} Finally, since \begin{equation} \begin{gathered} \frac{(p-1) ( N-1) +1+2( p-2) +(p-2) ^2}{N-1} =\frac{(p-1) ( p+N-2) }{N-1}, \\ \frac{2}{N-1}+\frac{2( p-2) }{N-1}=\frac{2(p-1) }{N-1}, \end{gathered} \label{e2.17} \end{equation} using \eqref{e2.10} and \eqref{e2.16} in \eqref{e2.13} we are lead to \begin{equation} \begin{aligned} &\Delta P+ ( p-2) P_{11}\\ &\geq -| 4-p| \frac{| \nabla P| ^2}{2P} +\frac{2}{u^{2\beta }}\Big\{-f'u_{1}^{4-p}+\frac{(p-1) ( p+N-2) }{N-1} u_{11}^2 \\ &\quad +\big( p-2+\frac{2(p-1) }{N-1}\big) fu_{1}^{2-p}u_{11}+\frac{1}{N-1}( fu_{1}^{2-p}) ^2 \\ &\quad -\beta [ 4(p-1) u_{11}-fu_{1}^{2-p} ] \frac{u_{1}^2}{u}+\beta ( 2\beta +1) (p-1) \frac{u_{1}^{4}}{u^2}\Big\} . \end{aligned} \label{e2.18} \end{equation} Now, evaluating \eqref{e2.3} at $\mathbf{x^{\ast }}$, by using \eqref{e2.5}, we have \begin{equation} P_{1}=\frac{2}{u^{2\beta }}u_{11}u_{1}-\frac{2\beta }{u^{2\beta +1}} u_{1}^{3}, \label{e2.19} \end{equation} so that \begin{equation} u_{11}=\frac{P_{1}u^{2\beta }}{2u_{1}}+\beta \frac{u_{1}^2}{u}. \label{e2.20} \end{equation} Inserting \eqref{e2.20} into \eqref{e2.18} we obtain \begin{equation} \begin{aligned} &\Delta P+ ( p-2) P_{11}\\ &\geq -| 4-p| \frac{| \nabla P| ^2}{2P} +\frac{2}{u^{2\beta }}\Big\{-f'u_{1}^{4-p} \\ &\quad +\frac{(p-1) ( p+N-2) }{N-1}\Big( \frac{ P_{1}u^{2\beta }}{2u_{1}}+\beta \frac{u_{1}^2}{u}\Big) ^2 \\ &\quad +\Big( p-2+\frac{2(p-1) }{N-1}\Big) fu_{1}^{2-p}\Big( \frac{P_{1}u^{2\beta }}{2u_{1}}+\beta \frac{u_{1}^2}{u }\Big) \\ &\quad +\frac{1}{N-1}( fu_{1}^{2-p}) ^2 -\beta \Big[ 4(p-1) \Big( \frac{P_{1}u^{2\beta }}{2u_{1}} +\beta \frac{u_{1}^2}{u} \Big) -fu_{1}^{2-p}\Big] \frac{u_{1}^2}{u} \\ &\quad +\beta ( 2\beta +1) (p-1) \frac{u_{1}^{4}}{u^2}\Big\} . \end{aligned} \label{e2.21} \end{equation} Next, using the restriction \eqref{e1.4} on $f$ we note that \begin{equation} \begin{aligned} &-f'u^{4-p}+\Big( p-2+\frac{2(p-1) }{N-1}\Big) \beta \frac{f}{u}u_{1}^{4-p}+\beta \frac{f}{u}u_{1}^{4-p} \\ & =u^{4-p}\big[ -f'+\beta (p-1) \frac{N+1}{N-1}\frac{f}{u}\big] \geq 0. \end{aligned} \label{e2.22} \end{equation} We also note that \begin{equation} \frac{(p-1) ( p+N-2) }{N-1}\Big( \frac{ P_{1}u^{2\beta }}{2u_{1}}\Big) ^2\geq 0. \label{e2.23} \end{equation} Therefore \begin{equation} \begin{aligned} \Delta P+( p-2) P_{11} &\geq -| 4-p| \frac{| \nabla P| ^2}{2P} \\ &\quad +\frac{(p-1) ( p+N-2) }{N-1}\Big( 2\beta \frac{P_{1}u_{1}}{u}+2\beta ^2\frac{u_{1}^{4}}{u^{2\beta +2}} \Big) \\ &\quad +\Big( p-2+\frac{2(p-1) }{N-1}\Big) P_{1}fu_{1}^{1-p}+\frac{2}{N-1}\frac{( fu_{1}^{2-p}) ^2}{ u^{2\beta }} \\ &\quad -4\beta (p-1) \frac{P_{1}u_{1}}{u}+2(p-1) ( \beta -2\beta ^2) \frac{u_{1}^{4}}{u^{2\beta +2}}\,. \end{aligned} \label{e2.24} \end{equation} Moreover, using the following two identities \begin{gather} \frac{2(p-1) ( p+N-2) }{N-1}-4(p-1) = \frac{2(p-1) ( p-N) }{N-1}, \label{e2.25} \\ \frac{2(p-1) ( p+N-2) }{N-1}\beta ^2+2(p-1) ( \beta -2\beta ^2) =\frac{p-1}{N-1}[ 2\beta ^2( p-N) +2\beta ( N-1) ] , \label{e2.26} \end{gather} one may easily see that \eqref{e2.24} becomes \begin{equation} \begin{aligned} \frac{\Delta P+( p-2) P_{11}}{P} & \geq -| 4-p| \frac{| \nabla P| ^2}{2P^2}+2\beta \frac{(p-1) ( p-N) }{N-1}\frac{P_{1}u_{1}}{Pu} \\ &\quad +\Big( p-2+\frac{2(p-1) }{N-1}\Big) \frac{P_{1}fu_{1}^{1-p}}{P}+\frac{2}{N-1}( fu_{1}^{1-p}) ^2 \\ &\quad +2\beta [ \beta ( p-N) +N-1] \frac{p-1}{N-1}\frac{u_{1}^2}{u^2}. \end{aligned}\label{e2.27} \end{equation} Next, let us consider a point $\mathbf{x_{0}}\in \mathbb{R}^{N}$ and define \begin{equation} J(\mathbf{x}) =( a^2-r^2) ^2P, \label{e2.28} \end{equation} where $a>0$ is a constant and $r:=| \mathbf{x}-\mathbf{x_{0}} | $. Let us denote by $B$ the ball centered at $\mathbf{x_{0}}$ and of radius $a$. Then we immediately notice that \begin{equation} J(\mathbf{x}) \geq 0\text{ in }B, \quad J( \mathbf{x }) =0 \text{ on }\partial B. \label{e2.29} \end{equation} Consequently, $J(\mathbf{x}) $ must attain its maximum at some (interior) point $\mathbf{x^{\ast }}$. Now, if $| \nabla u| ( \mathbf{x^{\ast }}) =0$, then $P\equiv 0$ in $B$. Since the ball was chosen arbitrarily, $ P\equiv 0$ in every ball, so that $\nabla u\equiv 0$ in $\mathbb{R}^{N}$ and our theorem follows. It thus remain to analyze the case $| \nabla u| ( \mathbf{x^{\ast }}) >0$. In such a case, we have the following complementary inequality at $x^{\ast }$ (see Cuccu-Mohammed-Porru \cite[p. 227]{CMP10} for the proof; they used a different auxiliary function $P$, but the proof is identical, since the form of $P$ does not really play a role in the proof): \begin{equation} \frac{\Delta P+( p-2) P_{11}}{P} \leq \frac{Ca^2}{(a^2-r^2) ^2},\quad C:=24+4N+28| p-2| . \label{e2.30} \end{equation} Combining \eqref{e2.27} and \eqref{e2.30} we obtain \begin{equation} \begin{aligned} \frac{Ca^2}{( a^2-r^2) ^2} & \geq -| 4-p| \frac{| \nabla P| ^2}{2P^2}+2\beta \frac{(p-1) ( p-N) }{N-1}\frac{P_{1}u_{1}}{Pu} \\ &\quad +\Big( p-2+\frac{2(p-1) }{N-1}\Big) \frac{ P_{1}fu_{1}^{1-p}}{P}+\frac{2}{N-1}( fu_{1}^{1-p}) ^2 \\ &\quad +2\beta [ \beta ( p-N) +N-1] \frac{p-1}{N-1}\frac{u_{1}^2}{u^2}. \end{aligned} \label{e2.31} \end{equation} On the other hand, differentiating \eqref{e2.28} we obtain that at $\mathbf{x^{\ast }}$ (the point of maximum for $J$ in $B$) we have \begin{equation} J_{i}=-2( a^2-r^2) ( r^2) _{i}P+(a^2-r^2) ^2P_{i}=0, \label{e2.32} \end{equation} so that \begin{equation} P_{1}=2\frac{( r^2) _{1}P}{a^2-r^2},\quad \nabla P=2\frac{\nabla r^2P}{a^2-r^2}. \label{e2.33} \end{equation} From \eqref{e2.5} and \eqref{e2.33} we then conclude that \begin{gather} \frac{P_{1}u_{1}}{P}=\frac{\nabla P\nabla u}{P}=2\frac{\nabla r^2\nabla u}{a^2-r^2}, \label{e2.34} \\ \frac{| \nabla P| }{P}=\frac{2| \nabla ( r^2) | }{a^2-r^2}=\frac{4r}{a^2-r^2}. \label{e2.35} \end{gather} Now using \eqref{e2.34} and \eqref{e2.35} in \eqref{e2.31} we obtain \begin{equation} \begin{aligned} \frac{Ca^2}{( a^2-r^2) ^2} & \geq -| 4-p| \frac{8r^2}{( a^2-r^2) ^2}+4\beta \frac{ (p-1) ( p-N) }{N-1}\frac{\nabla r^2\nabla u}{ ( a^2-r^2) u} \\ &\quad +2\Big( p-2+\frac{2(p-1) }{N-1}\Big) fu_{1}^{1-p} \frac{\nabla r^2\nabla u}{a^2-r^2}+\frac{2}{N-1}( fu_{1}^{1-p}) ^2 \\ &\quad +2\beta [ \beta ( p-N) +N-1] \frac{p-1}{ N-1}\frac{| \nabla u| ^2}{u^2}. \end{aligned} \label{e2.36} \end{equation} Moreover, by classical inequalities we have \begin{equation} \begin{aligned} &4\beta (p-1) ( p-N) \frac{\nabla r^2\nabla u}{( a^2-r^2) u}\\ &\geq -\beta ^2\gamma (p-1) ^2 \frac{| \nabla u| ^2}{u^2}-4( p-N) ^2 \frac{4r^2}{\gamma ( a^2-r^2) ^2}, \end{aligned} \label{e2.37} \end{equation} and \begin{equation} \begin{aligned} &2( p-2) +\frac{2(p-1) }{N-1}fu_{1}^{-p}\frac{\nabla r^2\nabla u}{a^2-r^2} \\ & \geq -4( | p-2| + \frac{2(p-1) }{N-1}) | f| u_{1}^{p-1} \frac{r}{a^2-r^2} \\ & \geq -\frac{2}{N-1}( fu_{1}^{1-p}) ^2-\widetilde{C}\frac{ r^2}{( a^2-r^2) ^2}, \end{aligned} \label{e2.38} \end{equation} with $\gamma >0$ to be chosen and $\widetilde{C}:=2[ ( N-1)( p-2) +2(p-1) ] ^2/( N-1) $. Inserting now estimates \eqref{e2.37} and \eqref{e2.38} into \eqref{e2.36} we find \begin{equation} \begin{aligned} \frac{Ca^2}{( a^2-r^2) ^2} & \geq -| 4-p| \frac{8r^2}{( a^2-r^2) ^2} -4(p-N) ^2\frac{4r^2}{\gamma ( a^2-r^2) ^2} - \widetilde{C}\frac{r^2}{( a^2-r^2) ^2} \\ & \quad+[ 2\beta ^2( p-N) +2\beta ( N-1) -\beta ^2\gamma (p-1) ] \frac{p-1}{N-1}\frac{| \nabla u| ^2}{u^2}. \end{aligned} \label{e2.39} \end{equation} Now let us analyze separately the following two cases: \smallskip \noindent\textbf{Case 1.} when $10. \label{e2.40} \end{equation} Therefore, we can choose $\gamma $ small enough so that we have \begin{equation} [ 2\beta ^2( p-N) +2\beta ( N-1) -\beta ^2\gamma (p-1) ] >0. \label{e2.41} \end{equation} smallskip \noindent\textbf{Case 2.} when $p\geq N$ and $\beta \in [ 1,+\infty ) $, as above, we can again choose a small enough $\gamma $ so that \eqref{e2.41} holds. In conclusion, for a well chosen $\gamma$, there exists a constant $K=K( N,p,\beta ,\gamma ) $ such that \begin{equation} \frac{| \nabla u| ^2}{u^2} \leq \frac{Ka^2}{( a^2-r^2) ^2}. \label{e2.42} \end{equation} Moreover, since $u(\mathbf{x}) $ is positive, there exists a constant $L>0$ such that $u^{2-2\beta }\leq L$. Therefore, at some point $\mathbf{x}^{\ast }$ we have \begin{equation} J( \mathbf{x}^{\ast }) =\frac{| \nabla u| ^2 }{u^2}\frac{1}{u^{2\beta -2}}( a^2-r^2) ^2\leq KLa^2. \label{e2.43} \end{equation} But $\mathbf{x}^{\ast }$ is a point of maximum for $J( \mathbf{x}) $ in $B$, so that we have \begin{equation} J( \mathbf{x}_{0}) =\frac{| \nabla u| ^2}{u^{2\beta }}a^{4}\leq KLa^2. \label{e2.44} \end{equation} It follows that at $\mathbf{x=x}_{0}$ we have \begin{equation} \frac{| \nabla u| ^2}{u^{2\beta }}\leq \frac{KL}{a^2}. \label{e2.45} \end{equation} Letting $a\to \infty $ we find that $\nabla u=0$ at $x_{0}$. Since $x_{0}$ is arbitrary, we must have $\nabla u=0$ in $\mathbb{R}^{N}$. The proof is thus achieved. \subsection*{Acknowledgments} This work was supported by a CNCSIS (Romania) research grant (PN-II-ID-PCE-2012-4-0021 : Variable Exponent Analysis: Partial Differential Equations and Calculus of Variations). \begin{thebibliography}{9} \bibitem{CMP10} F. Cuccu, A. Mohammed, G. Porru; \emph{Extensions of a theorem of Cauchy-Liouville}, J. Math. Anal. Appl., \textbf{369} (2011), 222--231. \bibitem{Fa07} A. Farina; \emph{Liouville-type theorems for elliptic problems}, Handbook of differential equations: stationary partial differential equations, \textbf{4} (2007), 61--116. \bibitem{McC07} J. A. McCoy; \emph{Bernstein properties of solutions to some higher order equations}, Diff. Int. Equ., \textbf{20} (2007), 1153--1166. \bibitem{To84} P. Tolksdorf; \emph{Regularity for a more general class of quasilinear elliptic equations}, J. Differential Equations, \textbf{51} (1984), 126--150. \end{thebibliography} \end{document}