\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 08, pp. 1--9.\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 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/08\hfil Asymptotic behavior of solutions] {Asymptotic behavior of solutions to nonlinear parabolic equation with nonlinear boundary conditions} \author[T. K. Boni, D. Nabongo\hfil EJDE-2008/08\hfilneg] {Th\'eodore K. Boni, Diabate Nabongo} % in alphabetical order \address{Th\'eodore K. Boni \newline Institut National Polytechnique Houphout-Boigny de Yamoussoukro, BP 1093 Yamoussoukro, C\^ote d'Ivoire} \email{theokboni@yahoo.fr} \address{Diabate Nabongo \newline Universit\'e d'Abobo-Adjam\'e, UFR-SFA, D\'epartement de Math\'ematiques et Informatiques, 16 BP 372 Abidjan 16, C\^ote d'Ivoire} \email{nabongo\_diabate@yahoo.fr} \thanks{Submitted December 4, 2007. Published January 17, 2008.} \subjclass[2000]{35B40, 35B50, 35K60} \keywords{Parabolic boundary value problem; asymptotic behavior; \hfill\break\indent nonlinear boundary conditions} \begin{abstract} We show that solutions of a nonlinear parabolic equation of second order with nonlinear boundary conditions approach zero as $t$ approaches infinity. Also, under additional assumptions, the solutions behave as a function determined here. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with smooth boundary $\partial\Omega$. Consider the boundary value problem \begin{gather} \frac{\partial\varphi(u)}{\partial t}-Lu+f(x,t,u)=0\quad\text{in } \Omega\times(0,\infty), \label{e1} \\ \frac{\partial u}{\partial N}+g(x,t,u)=0\quad\text{on } \partial\Omega\times(0,\infty), \label{e2} \\ u(x,0)=u_{0}(x)\quad\text{in } \overline{\Omega}, \label{e3} \end{gather} where \[ Lu=\sum^{n}_{i,j=1}\frac{\partial}{\partial x_{j}}(a_{ij}(x)\frac{\partial u}{\partial x_{i}})+\sum^{n}_{i=1}a_{i}(x)\frac{\partial u}{\partial x_{i}},\quad \frac{\partial u}{\partial N}=\sum^{n}_{i,j=1}\cos(\nu,x_{i})a_{ij}(x)\frac{\partial u}{\partial x_{j}}. \] Here the coefficients $a_{ij}(x)\in C(\Omega)$ satisfy the inequality \[ \sum^{n}_{i,j=1}a_{ij}(x)\xi_{i}\xi_{j}\geq C|\xi|^{2}\quad \text{for }\xi\in \mathbb{R}^{n},\; \xi\neq 0, \; C>0, \] $a_{ij}(x)=a_{ji}(x)$, $\nu$ is the exterior normal unit vector on $\partial\Omega$, $f_{x,t}(s)=f(x,t,s)$ and $g_{x,t}(s)=g(x,t,s)$ are positive, increasing and convex functions for $s\geq 0$ with $f_{x,t}(0)=f'_{x,t}(0)=g_{x,t}(0)=g'_{x,t}(0)=0$. For positive values of $s$, $\varphi(s)$ is a positive and concave function. Throughout this paper, we assume the following condition: \begin{itemize} \item[(H0)] There exist functions $f_{*}(s)$, $g_{*}(s)$ of class $C^{1}([0,\infty))$, positive for positive values of $s$ such that for any $\alpha(t)$ tending to zero as $t\to\infty$, \begin{gather*} \lim_{t\to \infty}\frac{f(x,t,\alpha(t))}{f_{*}(\alpha(t))}=a(x),\quad \lim_{t\to \infty}\frac{g(x,t,\alpha(t))}{g_{*}(\alpha(t))}=b(x),\\ \frac{f_{*}}{\varphi'}(0)=\frac{g_{*}}{\varphi'}(0) =(\frac{f_{*}}{\varphi'})'(0)=(\frac{g_{*}}{\varphi'})'(0)=0, \end{gather*} where $a(x)$ is a bounded nonnegative function in $\Omega$ and $b(x)$ is a bounded nonnegative function on $\partial\Omega$. \end{itemize} Existence of positive classical solutions, local in time, was proved by Ladyzenskaya, Solonnikov and Ural'ceva in \cite{p1}. In this paper, we are dealing with the asymptotic behavior as $t\to \infty$ of positive solutions of \eqref{e1}--\eqref{e3}. The asymptotic behavior of solutions for parabolic equations has been the subject of study of many authors (see, for instance \cite{a1,b1,b2,b3,k1,k2,v1}. In particular, Kondratiev and Oleinik \cite{k1} considered the problem \begin{gather} \frac{\partial u}{\partial t}-Lu+a|u|^{p-1}u=0\quad\text{in } \Omega\times(0,\infty), \label{e4}\\ \frac{\partial u}{\partial N}=0\quad\text{on } \partial\Omega\times(0,\infty), \label{e5} \\ u(x,0)=u_{0}(x)\quad\text{in } \overline{\Omega}, \label{e6} \end{gather} where $p>1$, and $a$ is a positive constant. They proved that if $u$ is a positive solution of Problem \eqref{e4}--\eqref{e6}, then \begin{equation} \lim_{t\to \infty}t^{\frac{1}{p-1}}u(x,t) =\Big(\frac{p-1}{|\Omega|}\int_{\Omega}av_{1}(x)dx\Big)^{\frac{-1}{p-1}} \end{equation} uniformly in $x\in\Omega$, where $v_{1}(x)$ is a positive solution of the boundary value problem \begin{equation} \label{e8} \begin{gathered} L^{*}(v)=0\quad\text{in } \Omega\\ \frac{\partial v}{\partial N}=\sum^{n}_{i=1}a_{i}(x)\cos(\nu,x_{i})v \quad\text{on } \partial\Omega, \end{gathered} \end{equation} with \[ L^{*}(v)=\sum^{n}_{i,j=1}\frac{\partial}{\partial x_{i}}(a_{ij}(x)\frac{\partial v}{\partial x_{j}})-\sum^{n}_{i,j=1}\frac{\partial}{\partial x_{i}}(a_{i}(x)v). \] Notice that Problem \eqref{e8} is the adjoint of the Neumann problem for the operator $L$. The same result with $v_{1}(x)=1$, $a=a(x)$ has been also obtained in \cite{b1} and \cite{k2} in the case where $a(x)$ is a bounded function in $\Omega$ and $a_{i}(x)=0$ $(i=1,\dots,n)$ (i.e. the operator $L$ is self-adjoint). In \cite{b3}, the second author has shown similar results about the asymptotic behavior of solutions for another particular case of Problem \eqref{e1}--\eqref{e3} which corresponds to this last for $a_{i}(x)=0$ $(i=1,\dots,n)$, $\varphi(u)=u$, $f(x,t,u)=a(x)f_{*}(u)$, $g(x,t,u)=b(x)g_{*}(u)$. Our aim in this paper is to generalize the above results, describing the asymptotic behavior of solutions for Problem \eqref{e1}--\eqref{e3}. Our paper is written in the following manner. Under some conditions, we obtain in the next section the asymptotic behavior of positive solutions for Problem \eqref{e1}--\eqref{e3}. Introduce the function class $Z_{p}$ defined as follows: $u\in Z_{p}$ if $u$ is continuous in $\overline{G}$, $\frac{\partial u}{\partial x_{i}}\in G'$ and $\frac{\partial u}{\partial t}$, $\frac{\partial^{2}u}{\partial x_{i}\partial x_{j}}\in G$, where $G=\Omega\times(0,\infty)$, $G'=\overline{\Omega}\times(0,\infty)$, and $\overline{G}$ is the closure of $G$. \section{Asymptotic behavior} In this section, we show that under some assumptions, any positive solution $u\in Z_{p}$ of Problem \eqref{e1}--\eqref{e3} tends to zero as $t\to \infty$ uniformly in $x\in \Omega$. We also describe its asymptotic behavior as $t\to \infty$. The following lemma will be useful later. \begin{lemma} \label{lem2.1} Let $u,v\in Z_{p}$ satisfying the following inequalities \begin{gather*} \frac{\partial\varphi(u)}{\partial t}-Lu+f(x,t,u)>\frac{\partial\varphi(v)}{\partial t}-Lv+f(x,t,v)\quad\text{in } \Omega\times(0,\infty), \\ \frac{\partial u}{\partial N}+g(x,t,u)>\frac{\partial v}{\partial N}+g(x,t,v)\quad\text{on } \partial\Omega\times(0,\infty), \\ u(x,0)>v(x,0)\quad\text{in } \overline{\Omega}. \end{gather*} Then we have $u(x,t)>v(x,t)$ in $\Omega\times(0,\infty)$. \end{lemma} \begin{proof} The function $w(x,t)=u(x,t)-v(x,t)$ is continuous in $\overline{\Omega}\times [0,\infty)$. Then its minimum value $m$ is attained at a point $(x_{0},t_{0})\in\overline{\Omega}\times [0,\infty]$. If $t_{0}=0$, then $m>0$. If $0v(x,t)$ for $0\leq t< t_{1}$ but $u(x_{1},t_{1})=v(x_{1},t_{1})$ for some $x_{1}\in\overline{\Omega}$.\\If $x_{1}\in\Omega$ then we have \[ \frac{\partial\varphi(u)-\varphi(v)}{\partial t}(x_{1},t_{1})\leq 0,\quad Lw(x_{1},t_{1})\geq 0,\quad f(u(x_{1},t_{1}))=f(v(x_{1},t_{1})). \] Consequently, we have a contradiction because \[ \frac{\partial\varphi(u)-\varphi(v)}{\partial t}(x_{1},t_{1})-Lw(x_{1},t_{1})+[f(x_{1},t_{1},u(x_{1},t_{1})) -f(x_{1},t_{1},v(x_{1},t_{1}))]>0. \] Finally if $x_{1}\in\partial\Omega$, then $\frac{\partial w}{\partial N}(x_{1},t_{1})\leq 0$. We have again an absurdity because of the fact that \[ \frac{\partial w}{\partial N}(x_{1},t_{1})+[g(x_{1},t_{1},u(x_{1},t_{1})) -g(x_{1},t_{1},v(x_{1},t_{1}))]>0. \] Therefore we have $m>0$. \end{proof} For the limit of $f_{*}(t)/g_{*}(t)$ as $t\to 0$, we have the following possibilities: \begin{itemize} \item[(P1)] $\lim_{t\to 0}\frac{f_{*}(t)}{g_{*}(t)}=0$, \item[(P2)] $\lim_{t\to 0}\frac{f_{*}(t)}{g_{*}(t)}=\infty$, \item[(P3)] $\lim_{t\to 0}\frac{f_{*}(t)}{g_{*}(t)}=C_{*}$, where $C_{*}$ is a positive constant. \end{itemize} Let $\varepsilon_{f}$ and $\varepsilon_{g}$ be such that: \begin{itemize} \item[(H1)] $\varepsilon_{f}=0$, $\varepsilon_{g}=1$ if (P1) is satisfied; \item[(H2)] $\varepsilon_{f}=1$, $\varepsilon_{g}=0$ if (P2) is satisfied; \item[(H3)] $\varepsilon_{f}=\sqrt{\frac{C_{*}}{1+C_{*}}}$, $\varepsilon_{g}=\sqrt{\frac{C_{*}}{1+C_{*}}}$ if (P3) is satisfied. \end{itemize} Assumption (P1) is always used with the coefficients $\varepsilon_{f}$, $\varepsilon_{g}$ defined in (H1)--(H3). The function \begin{equation} \label{e9} h(t)=\varepsilon_{f}f_{*}(t)+\varepsilon_{g}g_{*}(t) \end{equation} is crucial for the study of asymptotic behavior of solutions. Let \begin{equation} \label{e10} G(s)=\int^{1}_{s}\frac{\varphi'(t)dt}{h(t)} \end{equation} and let $H(s)$ be the inverse function of $G(s)$. In this notation the initial-value problem \begin{equation} \label{e11} \varphi'(\beta(t))\beta'(t)=-\lambda h(\beta(t)),\quad \beta(0)=1\quad (\lambda>0) \end{equation} has the unique solution $\beta(t)=H(\lambda t)$. It follows from $\frac{h}{\varphi'}(0)=(\frac{h}{\varphi'})'(0)=0$ that $0<\frac{h(t)}{\varphi'(t)}0)$ and hence \begin{equation} \label{e12} G(0)=\infty,\quad G(1)=0\quad and\quad H(0)=1,\quad H(\infty)=0, \end{equation} which implies that $\beta(\infty)=0.$ The function $\beta(t)$ will be used later in the construction of supersolutions and subsolutions of \eqref{e1}--\eqref{e3} to obtain the asymptotic behavior of solutions. \begin{remark} \label{rmk2.1} \rm If (P1)--(P3) are satisfied, then \begin{gather*} \lim_{t\to \infty}\{-\varepsilon_{f}a(x) +\frac{f(x,t,\beta(t))}{h(\beta(t))}\} =0, \\ \lim_{t\to \infty}\{-\varepsilon_{g}b(x) +\frac{g(x,t,\beta(t))}{h(\beta(t))}\}=0. \end{gather*} \end{remark} In the following theorems, we suppose that (P1) or (P2) or (P3) is satisfied. Consider the boundary-value problem \begin{equation} \label{e13} -\lambda-L\psi=-\varepsilon_{f}a(x)+\delta,\quad \frac{\partial \psi}{\partial N}=-\varepsilon_{g}b(x)+\delta. \end{equation} This problem has a solution if and only if \begin{equation} \label{e14} \delta\big(\int_{\Omega}v_{0}(x)dx+\int_{\partial\Omega}v_{0}(x)ds\big) =I(a,b)-\lambda\int_{\Omega}v_{0}(x)dx, \end{equation} where $v_{0}(x)$ is a solution of Problem \eqref{e8} and \begin{equation} \label{e15} I(a,b)=\varepsilon_{g}\int_{\partial\Omega}b(x)v_{0}(x)ds +\varepsilon_{f}\int_{\Omega}a(x)v_{0}(x)dx, \end{equation} (see, for instance \cite{k1}). Thus in this paper, for problem \eqref{e13}, we suppose that for given $\lambda>0$, $\delta$ satisfies \eqref{e14}, which implies that problem \eqref{e13} has a solution $\psi$. Without loss of generality, we may suppose that $\psi>0$. Indeed, when $\psi$ is a solution of \eqref{e13}, we see that $\psi+C$ is also a solution of \eqref{e13} for any constant $C>0$. The function $\psi$ will be used later to construct supersolutions and subsolutions of \eqref{e1}--\eqref{e3} for getting the asymptotic behavior of solutions. The function $v_{0}(x)$ does not change sign in $\Omega$. We shall suppose that $v_{0}(x)>0$ in $\Omega$. If $a_{i}(x)=0$, then the operator$L$ is self-adjoint and $v_{0}(x)=1$. \begin{theorem} \label{thm2.1} \begin{itemize} \item[(i)] Suppose that $I(a,b)>0$ and $\lim_{s\to 0}\frac{h(s)\varphi''(s)}{\varphi'(s)}=0$. If $u\in Z_{p}$ is a positive solution of \eqref{e1}--\eqref{e3}, then $$ \lim_{t\to \infty}u(x,t)=0 $$ uniformly in $x\in \overline{\Omega}$. \item[(ii)] Moreover if there exists a positive constant $c_{2}$ such that $$ \lim_{s\to \infty}\frac{sh(H(s))}{H(s)\varphi'(H(s))}\leq c_{2}, $$ we have $u(x,t)=H(c_{fg}t)(1+o(1))$ as $t\to\infty$, where $c_{fg}=\frac{I(a,b)}{\int_{\Omega}v_{0}(x)dx}$. \end{itemize} \end{theorem} \begin{proof} (i) Put $w(x,t)=\beta(t)+\psi(x)h(\beta(t))$, where $\beta(t)$ and $\psi(x)$ are solutions of \eqref{e11} and \eqref{e13} respectively for $\lambda\leq \frac{I(a,b)}{2\int_{\Omega}v_{0}(x)dx}$, which implies that $\delta>0$. A straightforward computation reveals that \begin{align*} &\frac{\partial\varphi(w)}{\partial t}-Lw+f(x,t,w)\\ &=h(\beta(t))(-\lambda-L\psi) -\lambda h(\beta(t))h'(\beta(t))\psi(x) +f(x,t,\beta(t))+\psi(x)h(\beta(t))f'_{x,t}(y)\\ &\quad -\lambda\psi(x)\frac{h^{2}(\beta(t))\varphi''(z)}{\varphi'(\beta(t))} -\lambda\psi^{2}(x)\frac{h^{2}(\beta(t))h'(\beta(t))\varphi''(z)}{\varphi' (\beta(t))}, \end{align*} $$ \frac{\partial w}{\partial N}+g(x,t,w) =h(\beta(t))\frac{\partial \psi}{\partial N}+g(x,t,\beta(t))+\psi(x)h(\beta(t))g'_{x,t}(l), $$ with $\{l,y,z\}\in[\beta(t),\beta(t)+\psi(x)h(\beta(t))]$. It follows from \eqref{e13} that \begin{align*} &\frac{\partial \varphi(w)}{\partial t}-Lw+f(x,t,w)\\ &=(\delta-\varepsilon_{f}a(x))h(\beta(t)) -\lambda h(\beta(t))h'(\beta(t))\psi(x)+f(x,t,\beta(t)) +\psi(x)h(\beta(t))f'_{x,t}(y)\\ &\quad -\lambda\psi(x)\frac{h^{2}(\beta(t))\varphi''(z)}{\varphi'(\beta(t))} -\lambda\psi^{2}(x)\frac{h^{2}(\beta(t))h'(\beta(t))\varphi''(z)}{\varphi' (\beta(t))}, \end{align*} $$ \frac{\partial w}{\partial N}+g(x,t,w) =(\delta-\varepsilon_{g}b(x))h(\beta(t)) +g(x,t,\beta(t))+\psi(x)h(\beta(t))g'_{x,t}(l). $$ Since $f'_{x,\infty}(0)=g'_{x,\infty}(0)=0$, $\lim_{s\to 0}\frac{h(s)\varphi''(s)}{\varphi'(s)}=0$, using Remark 2.1, there exists $t_{1}\geq 0$ such that \begin{gather*} \frac{\partial\varphi(w)}{\partial t}-Lw+f(x,t,w)>0\quad\text{in } \Omega\times(t_{1},\infty), \\ \frac{\partial w}{\partial N}+g(x,t,w)>0\quad\text{on } \partial\Omega\times(t_{1},\infty). \end{gather*} Let $k>1$ be large enough that $$ u(x,t_{1})0\quad\text{in } \Omega\times(t_{1},\infty), \\ \frac{\partial kw}{\partial N}+g(x,t,kw)>0\quad\text{on } \partial\Omega\times(t_{1},\infty). \end{gather*} It follows from Comparison Lemma \ref{lem2.1} that $$ u(x,t_{1}+t)0$ small enough, there exist $\tau$ and $T$ such that $$ u(x,t+\tau)\leq \beta_{1}(t+T)+\psi_{1}(x)h(\beta_{1}(t+T)), $$ where $\beta_{1}(t)$ and $\psi_{1}(x)>0$ are solutions of \eqref{e11} and \eqref{e13} respectively for $\lambda=c_{fg}-\frac{\varepsilon}{2}$. \end{lemma} \begin{proof} Put \[ w_{1}(x,t)=\beta_{1}(t)+\psi_{1}(x)h(\beta_{1}(t)). \] Since $c_{fg}=I(a,b)/\int_{\Omega}v_{0}(x)dx$, it follows that \[ \delta=\frac{\varepsilon\int_{\Omega}v_{0}(x)dx} {2(\int_{\Omega}v_{0}(x)dx+\int_{\partial\Omega}v_{0}(x)dx)}, \] which implies that for any $\varepsilon>0$ small enough $\delta>0$ and as in the proof of Theorem \ref{thm2.1} (i), there exists $T\geq 0$ such that \begin{gather*} \frac{\partial\varphi(w_{1})}{\partial t}-Lw_{1}+f(x,t,w_{1})>0\quad\text{in } \Omega\times(T,\infty), \\ \frac{\partial w_{1}}{\partial N}+g(x,t,w_{1})>0\quad\text{on } \partial\Omega\times(T,\infty). \end{gather*} Since $\lim_{t\to \infty}u(x,t)=0$ uniformly in $x\in\overline{\Omega}$, there exists a $\tau>T$ such that $$ u(x,\tau)u(x,\tau)\quad\text{in } \overline{\Omega},\\ \frac{\partial \varphi(z_{1})}{\partial t} =\frac{\partial \varphi(w_{1})}{\partial t}\quad\text{in } \Omega\times(\tau,\infty),\\ Lz_{1}=Lw_{1}\quad\text{in }\Omega\times(\tau,\infty),\\ \frac{\partial z_{1}}{\partial N}=\frac{\partial w_{1}}{\partial N} \quad\text{on } \partial\Omega\times(\tau,\infty). \end{gather*} Therefore, \begin{gather*} \frac{\partial \varphi(z_{1})}{\partial t}-Lz_{1}+f(x,t,z_{1})>0\quad\text{in } \Omega\times(\tau,\infty), \\ \frac{\partial z_{1}}{\partial N}+g(x,t,z_{1})>0\quad\text{on } \partial\Omega\times(\tau,\infty), \\ z_{1}(x,\tau)>u(x,\tau)\quad\text{in } \overline{\Omega}. \end{gather*} It follows from Comparison Lemma \ref{lem2.1} that $$ u(x,t+\tau)\leq w_{1}(x,t+T)=\beta_{1}(t+T)+\psi_{1}(x)h(\beta_{1}(t+T)), $$ which yields the result. \end{proof} \begin{lemma} \label{lem2.3} Under the hypotheses of Theorem \ref{thm2.1} (i), if $u\in Z_{p}$ is a positive solution of \eqref{e1}--\eqref{e3}, then for any $\varepsilon>0$ small enough, there exists $T_{2}$ such that $$ u(x,t+\tau)\geq \beta_{2}(t+T_{2})+\psi_{2}(x)h(\beta_{1}(t+T_{2})), $$ where $\beta_{2}(t)$ and $\psi_{2}(x)>0$ are solutions of \eqref{e11} and \eqref{e13} respectively for $\lambda=c_{fg}+\frac{\varepsilon}{2}$. \end{lemma} \begin{proof} Put $$ w_{2}(x,t)=\beta_{2}(t)+\psi_{1}(x)h(\beta_{2}(t)). $$ Since $c_{fg}=\frac{I(a,b)}{\int_{\Omega}v_{0}(x)dx}$, it follows that \[ \delta=\frac{-\varepsilon\int_{\Omega}v_{0}(x)dx}{2(\int_{\Omega}v_{0}(x)dx +\int_{\partial\Omega}v_{0}(x)dx)}, \] which implies that for any $\varepsilon>0$ small enough $\delta<0$. As in the proof of Theorem \ref{thm2.1} (i), $w_{2}$ satisfies \begin{align*} &\frac{\partial\varphi(w_{2})}{\partial t}-Lw_{2}+f(x,t,w_{2})\\ &=(\delta-\varepsilon_{f}a(x))h(\beta_{2}(t)) \\ &\quad -(c_{fg}+\frac{\varepsilon}{2})h(\beta_{2}(t))h'(\beta_{2}(t))\psi(x) +f(x,t,\beta_{2}(t))+\psi(x)h(\beta_{2}(t))f'_{x,t}(y_{2}),\\ &\quad -(c_{fg}+\frac{\varepsilon}{2})\psi(x)\frac{h^{2}(\beta(t))\varphi'' (z_{2})}{\varphi'(\beta(t))}-(c_{fg}+\frac{\varepsilon}{2})\psi^{2}(x) \frac{h^{2}(\beta(t))h'(\beta(t))\varphi''(z_{2})}{\varphi'(\beta(t))}, \end{align*} $$ \frac{\partial w_{2}}{\partial N}+g(x,t,w_{2}) =(\delta-\varepsilon_{g}b(x))h(\beta_{2}(t))+g(x,t,\beta_{2}(t)) +\psi(x)h(\beta_{2}(t))g'_{x,t}(l_{2}). $$ with $\{y_{2},z_{2},l_{2}\}\in[\beta_{2}(t),\beta_{2}(t) +\psi_{2}(x)h(\beta_{2}(t))]$. Since $f'_{x,\infty}(0)=g'_{x,\infty}(0)=0$, $\lim_{s\to 0}\frac{h(s)\varphi''(s)}{\varphi'(s)}=0$, using Remark 2.1, for any $\varepsilon>0$ small enough, there exists $T_{1}>0$ such that \begin{gather*} \frac{\partial\varphi(w_{2})}{\partial t}-Lw_{2}+f(x,t,w_{2})<0\quad\text{in } \Omega\times(T_{1},\infty), \\ \frac{\partial w_{2}}{\partial N}+g(x,t,w_{2})<0\quad\text{on } \partial\Omega\times(T_{1},\infty). \end{gather*} Since $\lim_{t\to \infty}w_{2}(x,t)=0$ uniformly for $x\in\overline{\Omega}$, then there exists a $T_{2}>T_{1}$ such that $$ u(x,\tau)>w_{2}(x,T_{2})\quad\text{in } \overline{\Omega}. $$ Set $$ z_{2}(x,t)=w_{2}(x,T_{2}-\tau+t)\quad\text{in } \overline{\Omega}\times(\tau,\infty). $$ We get \begin{gather*} z_{2}(x,\tau)=w_{2}(x,T_{2})0$, $$ \lim_{t\to\infty}\frac{\beta(t+\gamma,\lambda)}{\beta(t,\lambda)}=1\,. $$ \item[(ii)] if $\lim_{s\to\infty}\frac{sh(H(s))}{H(s)\varphi'(H(s))}\leq c_{2}$ and $\alpha>0$, then \begin{gather} 1\geq \lim_{t\to\infty}\sup\frac{\beta(t,\lambda+ \alpha)}{\beta(t,\lambda)}\geq \lim_{t\to\infty}\inf\frac{\beta(t,\lambda+\alpha)}{\beta(t,\lambda)}\geq 1-\frac{c_{2}\alpha}{\lambda}, \label{e16} \\ 1\leq \lim_{t\to\infty}\inf\frac{\beta(t,\lambda-\alpha)}{\beta(t,\lambda)} \leq \lim_{t\to\infty}\sup\frac{\beta(t,\lambda-\alpha)}{\beta(t,\lambda)} \leq 1+\frac{2c_{2}\alpha}{\lambda}, \label{e17} \end{gather} for $\alpha$ small enough. \end{itemize} \end{lemma} \begin{proof} (i) Since $\beta_{\lambda}(t)=\beta(t,\lambda)$ is decreasing and convex, $$ \beta(t,\lambda)-\gamma\lambda\frac{h(\beta(t,\lambda))}{\varphi' (\beta(t,\lambda))}\leq \beta(t+\gamma,\lambda)\leq \beta(t,\lambda), $$ which implies $\lim_{t\to\infty}\frac{\beta(\gamma+t,\lambda)}{\beta(t,\lambda)}=1$ because $\lim_{s\to 0}\frac{h(s)}{s\varphi'(s)}=0$. (ii) We have $$ 1\geq \frac{\beta(t,\lambda+ \alpha)}{\beta(t,\lambda)}=\frac{H(\lambda t+\alpha)}{H(\lambda t)}\geq \frac{H(\lambda t)-\alpha t\frac{h(H(\lambda t))}{\varphi'(H(\lambda t))}}{H(\lambda t)}. $$ Since $\lim_{s\to \infty}\frac{h(H(s))}{H(s)\varphi'(H(s))}\leq c_{2}$, we obtain \eqref{e16}. We also get by means of \eqref{e16} the following inequalities: $$ 1\leq \lim_{t\to\infty}\inf\frac{\beta(t,\lambda-\alpha)}{\beta(t,\lambda)}\leq \lim_{t\to\infty}\sup\frac{\beta(t,\lambda-\alpha)}{\beta(t,\lambda)}\leq \frac{1}{1-\frac{c_{2}\alpha}{\lambda-\alpha}}\leq 1+\frac{2c_{2}\alpha}{\lambda}, $$ which yields \eqref{e17}. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2.1} (ii)] From Lemmas \ref{lem2.2}, \ref{lem2.3} and \ref{lem2.4}, for any $\varepsilon>0$ small enough, we have $$ 1-k_{1}\varepsilon\leq \lim_{t\to\infty}\inf\frac{u(x,t)}{\beta(t)}\leq \lim_{t\to\infty}\sup\frac{u(x,t)}{\beta(t)}\leq 1+k_{2}\varepsilon $$ where $k_{1}$ and $k_{2}$ are two positive constants. Consequently $$ u(x,t)=\beta(t)(1+o(1))\quad as\quad t\to\infty, $$ which gives the result. \end{proof} \begin{remark} \label{rmk2.2} \rm Let $\varphi(u)=u^{m}$, $f(x,t,u)=a_{1}(x,t)u^{p}$, $g(x,t,u)=b_{1}(x,t)u^{q}$ with $01$. Assume that $\lim_{t\to\infty}a_{1}(x,t)=a(x)$, $\lim_{t\to\infty}b_{1}(x,t)=b(x)$, $$ \varepsilon_{q}\int_{\partial\Omega}b(x)ds +\varepsilon_{p}\int_{\Omega}a(x)dx>0\,, $$ where $\varepsilon_{p}=0$, $\varepsilon_{q}=1$ if $p>q$, $\varepsilon_{p}=1$, $\varepsilon_{q}=0$ if $p