\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small 2005-Oujda International Conference on Nonlinear Analysis. \newline {\em Electronic Journal of Differential Equations}, Conference 14, 2006, pp. 109--117.\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 2006 Texas State University - San Marcos.} \vspace{9mm}} \setcounter{page}{109} \begin{document} \title[\hfilneg EJDE/Conf/14 \hfil Nonresonance conditions] {Nonresonance conditions for a semilinear wave equation in one space dimension} \author[A. Anane, O. Chakrone, A. Zerouali \hfil EJDE/Conf/14 \hfilneg] {Aomar Anane, Omar Chakrone, Abdellah Zerouali} % in alphabetical order \address{Aomar Anane \newline D\'epartement de Math\'ematiques et Informatique \\ Facult\'e des Sciences \\ Universit\'e Mohammed 1er, Oujda, Maroc} \email{anane@sciences.univ-oujda.ac.ma} \address{Omar Chakrone \newline D\'epartement de Math\'ematiques et Informatique \\ Facult\'e des Sciences \\ Universit\'e Mohammed 1er, Oujda, Maroc} \email{chakrone@sciences.univ-oujda.ac.ma} \address{Abdellah Zerouali \newline D\'epartement de Math\'ematiques et Informatique \\ Facult\'e des Sciences \\ Universit\'e Mohammed 1er, Oujda, Maroc} \email{abdellahzerouali@hotmail.com} \date{} \thanks{Published September 20, 2006.} \subjclass[2000]{35J65, 35J25} \keywords{D'Alembertian; homotopy argument; eigenvalues} \begin{abstract} In this paper we study the existence of periodic weak solutions for semilinear wave equations in one space dimension in the case of nonresonance. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \section{Introduction} In this paper we consider the existence of periodic solutions for the wave equation \begin{equation} \label{P} \begin{gathered} \Box u = \alpha u + \beta u_{x}- \gamma u_{t} + g(x,t,u) + h(x,t) \quad \text{in }Q, \\ u(x,t + 2\pi) = u(x,t) \quad \text{in }]0,\pi[\times\mathbb{R}, \\ u(0,t) = u(\pi,t) = 0 \quad \forall t\in\mathbb{R}, \end{gathered} \end{equation} where $Q=]0,\pi[\times]0,2\pi[$, $\Box=\frac{\partial^{2}}{\partial t^{2}}-\frac{\partial^{2}}{\partial x^{2}}$ is the D'Alembertian, $(\alpha, \beta, \gamma)\in\mathbb{R}\times\mathbb{R}\times\mathbb{R}$, $h$ is a given function in $L^{2}(Q)$, and $g:]0,\pi[\times\mathbb{R}\times\mathbb{R} \to\mathbb{R}$ is $2\pi$-periodic in $t$ and a Carath\'eodory function (i.e. measurable in $(x,t)$ for each $s\in\mathbb{R}$ and continuous in $s$ for almost all $(x,t)\in Q$). We are interested in the nonresonance for the problem \eqref{P} (i.e. in the condition for the function $g$ such that there exist a solution $u\in L^{2}(Q)$ for any given $h\in L^{2}(Q))$. We will assume that g satisfies the following conditions: \begin{itemize} \item[(C1)] $g(x,t,s)$ is nondecreasing in $s$; \item[(C2)] for $s\neq r$, $(x,t)\in Q$, we have $$ e^{\beta\frac{x}{2}} \big(\frac{g(x,t,r)-g(x,t,s)}{r-s}\big)\geq \frac{\beta^{2}}{4}- \alpha; $$ \item[(C3)] for all $R>0$, there exists $\phi_{R}\in L^{2}(Q)$ such that a.e. $(x,t)\in Q$, $$ \max_{|s|\leq R}|g(x,t,s)|\leq \phi_{R}(x,t); $$ \item[(C4)] a.e. $(x,t)\in Q$, we have \begin{align*} \lambda_{k} - \frac{\gamma^{2}}{4}+\frac{\beta^{2}}{4}- \alpha &< l(x,t):= \liminf_{|s|\to + \infty}\frac{g(x,t,s)}{s} \\ &\leq\limsup_{|s|\to +\infty}\frac{g(x,t,s)}{s}:=k(x,t)\\ &<\lambda_{k+1}-\frac{\gamma^{2}}{4} + \frac{\beta^{2}}{4} - \alpha, \end{align*} where $\lambda_{k}$ and $\lambda_{k+1}$ are two consecutive eigenvalues of the D'Alembertian, and $\sigma(\Box)$ denotes the spectrum of the D'Alembertian. \end{itemize} Problem \eqref{P} has been studied with conditions of resonances by several authors mention in particular: In the case $\alpha = \beta = \gamma = 0$ and $h = 0$ Benaoum in \cite{b1,b2}, Mustonen and Berkovits in \cite{b3,b4,b5,b6}, and Br\'ezis and Nirenberg in \cite{b10}. The case $g(x,t,s) = g(s)$, has been studied by Mustonen and Berkovits in \cite{b7}. The case $\beta = \gamma = 0$ and $\alpha$ is a eigenvalue of the D'Alembertian operator $(\Box)$, has been studied in \cite{b5}. The case $\beta = 0 $ and $\alpha$ is a eigenvalue of the operator $T$ defined by $T u = \Box u + \gamma u_{t}$, where $u_{t}= \frac{\partial u}{\partial t}$, has been studied in \cite{b3,b10}. In the general case, Anane, Chakrone and Ghanim in \cite{a2}. In the case of nonresonance, the problem \eqref{P} has been studied by Mustonen and Berkovits in \cite{b4} and \cite{b8}, and by Brezis and Nirenberg in \cite{b10} but only in particular cases. The situation that we consider here is marked by the presence of one term of transportation $\beta \nabla u$, what constitutes an extension of the cases studied by Mustonen and Berkovits in \cite{b4,b8}. In our work, we show (see Corollary \ref{coro3.2}) while using homotopy argument given by Mustonen and Berkovits in \cite{b4}, and of analogous techniques developed by Anane and Chakrone in \cite{a1} for the Laplacian $(\Delta)$, that the problem \eqref{P} has at least a solution for all $ h\in L^{2}(Q)$. \section{Remarks and notation} Let $\delta, \mu\in\mathbb{R}$ such that $\delta < \mu$, we introduce the following general hypothesis %$(C_{\delta, \mu})$ For a.e. $(x,t)\in Q$, we have \begin{equation} \label{Cdm} \begin{aligned} \delta + \frac{\beta^{2}}{4}- \alpha &\leq\neq l(x,t):= \liminf_{|s|\to + \infty}\frac{g(x,t,s)}{s} \\ &\leq\limsup_{|s|\to +\infty}\frac{g(x,t,s)}{s}:=k(x,t)\\ &\leq\neq \mu +\frac{\beta^{2}}{4}- \alpha \end{aligned} \end{equation} The notation $\leq\neq$ means that one has an large inequality on $Q$ and strict on a set of measure different from zero. \begin{remark} \label{rmk2.1} \rm (1) We denote by $Tu = \Box u + \gamma u_{t}$. Then \begin{itemize} \item[(i)] $T$ is a densely defined closed linear operator with closed range. \item[(ii)] $\mathop{\rm Im}(T) = [\ker(T)]^{\bot}$. \item[(ii)] $\lambda$ is a eigenvalue of the D'Alembertian if and only if $ \lambda - \frac{\gamma^{2}}{4}$ is a eigenvalue of $T$ \item[(iii)] If $T_{0}$ is the restriction of the operator $T$ on $\mathop{\rm Im}(T)= T(D(T))$, with $D(T)$ is the domain of the operator $T$, then $T_{0}$ has compact inverse. \end{itemize} For the proof of the remarks (i)--(iii), see \cite{b2}. \noindent(2) We put $\tilde{g}(x,t,s) = (\alpha - \frac{\beta^{2}}{4})s + e^{\frac{\beta}{2}x}g(x,t,e^{-\frac{\beta}{2}x}s)$ and $\tilde{h}(x,t) = e^{\frac{\beta}{2}x}h(x,t)$. Let $N:L^{2}(Q)\to L^{2}(Q)$, $$ N(u)=\tilde{g}(x,t,u) $$ be the Nemytskii operator generated by the function $\tilde{g}$. For $r\in[0,1]$, consider the operator $ H_{r}:D(T) \subset L^{2}(Q)\to L^{2}(Q)$, $$ H_{r}(u)= T u - r(N(u)+\tilde{h})-(1-r)\lambda u, $$ where $\delta< \lambda< \mu$. \begin{equation} \label{I} \begin{aligned} &\text{If there exists $R>0$, for all $r\in[0,1]$ and all $u\in D(T)$,}\\ &\text{with $ \|u\|= \Big(\int_{Q}|u|^{2}\Big)^{1/2}= R$, then $H_{r}(u)\neq 0$.} \end{aligned} \end{equation} \noindent(3) If (C1) and (C2) are verified, then $\tilde{g}(x,t,s)$ is nondecreasing in $s$, thus the operator $N$ is monotone. This statement and the following are easy to prove. \noindent(4) Condition (C3) implies that for all $R>0$ there exists $\tilde{\phi_{R}}\in L^{2}(Q)$ such that for a.e. $(x,t)\in Q$ we have $$ \max_{|s|\leq R}|\tilde{g}(x,t,s)|\leq \tilde{\phi_{R}}(x,t). $$ \noindent(5) If (C4) is verified, then for a.e. $(x,t)\in Q$, we have $$ \lambda_{k}-\frac{\gamma^{2}}{4} < \tilde{l}(x,t):= \liminf_{|s|\to + \infty}\frac{\tilde{g}(x,t,s)}{s} \leq\limsup_{|s|\to +\infty}\frac{\tilde{g}(x,t,s)}{s}:=\tilde{k}(x,t)<\lambda_{k+1} -\frac{\gamma^{2}}{4} $$ (6) If \eqref{Cdm} is satisfied, then for a.e. $(x,t)\in Q$, we have $$ \delta \leq\neq \tilde{l}(x,t):= \liminf_{|s|\to + \infty}\frac{\tilde{g}(x,t,s)}{s} \leq\limsup_{|s|\to + \infty}\frac{\tilde{g}(x,t,s)}{s}:=\tilde{k}(x,t)\leq\neq\mu $$ i.e. for all $\varepsilon>0$ there exists $a_{\varepsilon}\in L^{2}(Q)$ such that for a.e. $(x,t)\in Q$, and all $s \in\mathbb{R}$, we have $$ (\tilde{l}(x,t)-\varepsilon)s^{2}-a_{\varepsilon}(x,t)|s|\leq s\tilde{g}(x,t,s)\leq(\tilde{k}(x,t)+\varepsilon)s^{2} + a_{\varepsilon}(x,t)|s|. $$ (7) Under hypothesis (C3) and \eqref{Cdm}, there exists $\theta>0$ and $\eta\in L^{2}(Q)$ such that a.e. $(x,t)\in Q$, and all $s \in\mathbb{R}$, we have \begin{equation} |\tilde{g}(x,t,s)|\leq \theta|s| + \eta(x,t). \label{II} \end{equation} \end{remark} \begin{proposition} \label{prop2.2} The problem \eqref{P} is equivalent to the problem \begin{equation} \label{tildeP} \begin{gathered} T v = \tilde{g}(x,t,v) + \tilde{h}(x,t) \quad \text{in }Q, \\ v(x,t + 2\pi)= v(x,t) \quad \text{in }]0,\pi[\times\mathbb{R}, \\ v(0,t) = v(\pi,t)= 0 \quad \forall t \in\mathbb{R}, \end{gathered} \end{equation} \end{proposition} \begin{proof} Assume that $u$ is a solution of the problem \eqref{P}. Let $v = e^{\frac{\beta}{2}x}u$, it is clear that $v$ is $2\pi$-periodic in $t$ and $v(0,t) = v(\pi,t)= 0$ $\forall t \in\mathbb{R}$. On the other hand, we have \begin{gather*} v_{x} =\frac{\partial v}{\partial x}= \frac{\beta}{2}e^{\frac{\beta}{2}x}u + e^{\frac{\beta}{2}x}u_{x}, \\ v_{xx} =\frac{\partial^{2} v}{\partial x^{2}} = \beta e^{\frac{\beta}{2}x}u_{x} + \frac{\beta^{2}}{4}e^{\frac{\beta}{2}x}u + e^{\frac{\beta}{2}x}u_{xx}, \\ v_{t}= \frac{\partial v}{\partial t} = e^{\frac{\beta}{2}x}u_{t}, \quad v_{tt} = \frac{\partial^{2} v}{\partial t^{2}} = e^{\frac{\beta}{2}x}u_{tt}; \end{gather*} thus \begin{align*} \Box v &= v_{tt} - v_{xx} \\ & = e^{\frac{\beta}{2}x}u_{tt} - \beta e^{\frac{\beta}{2}x}u_{x} - \frac{\beta^{2}}{4}e^{\frac{\beta}{2}x}u - e^{\frac{\beta}{2}x}u_{xx} \\ & = -\frac{\beta^{2}}{4}v + e^{\frac{\beta}{2}x}( \Box u - \beta u_{x} )\\ & = (\alpha - \frac{\beta^{2}}{4})v - \gamma v_{t} + e^{\frac{\beta}{2}x}(g(x,t,e^{-\frac{\beta}{2}x}v) + h(x,t)). \end{align*} Hence $ T v = \tilde{g}(x,t,v) + \tilde{h}(x,t)$, and $v$ is a solution of the problem \eqref{tildeP}. The reciprocal implication is demonstrated by an identical calculation. \end{proof} \section{Main results} \begin{theorem} \label{thm3.1} Assume (C1), (C2), (C3) and \eqref{Cdm}. If $(H_{r})$ does not satisfy \eqref{I}, then there exists $m(x,t)\in L^{\infty}(Q)$, $v\in L^{2}(Q)\setminus \{0\}$ and $(u_{n})\subset L^{2}(Q)$ such that $v$ is the nontrivial solution of the problem \begin{equation} \label{Pm} \begin{gathered} T u = m(x,t)u \quad \text{in }Q, \\ u(x,t + 2\pi) = u(x,t) \quad \text{in }]0,\pi[\times\mathbb{R}, \\ u(0,t) = u(\pi,t) = 0 \quad \forall t\in\mathbb{R} \end{gathered} \end{equation} and \begin{gather*} \|u_{n}\|\to +\infty, \quad \frac{u_{n}}{\|u_{n}\|}\to v \quad \text{in } L^{2}(Q),\\ \delta\leq\neq m(x,t)\leq\neq \mu \quad \text{a.e. in } Q. \end{gather*} \end{theorem} \begin{corollary} \label{coro3.2} Assume (C1), (C2), (C3) and \eqref{Cdm}. If there exist two consecutive eigenvalues of the D'Alembertian $\lambda_{k}$ and $\lambda_{k+1}$ such that $0\leq\lambda_{k} - \frac{\gamma^{2}}{4}<\delta<\mu<\lambda_{k+1} - \frac{\gamma^{2}}{4}$, then problem \eqref{P} has at least one solution for all $ h\in L^{2}(Q)$. \end{corollary} \subsection*{Proof of theorem \ref{thm3.1}} As the proof is relatively long, we organize it in several lemmas. Suppose that $(H_{r})$ does not satisfy the estimate \eqref{I}, then $ \forall n \in\mathbb{N}$, there exist $r_{n}\in [0,1]$, and $u_{n}\in D(T)$ with $\|u_{n}\|=n $ such that \begin{equation} \label{III} T u_{n} - r_{n}(N(u_{n})+\tilde{h})-(1-r_{_{n}})\lambda u_{n}=0 \end{equation} Let $$ v_{n}=\frac{u_{n}}{\|u_{n}\|},\quad g_{n}(x,t) = \frac{\tilde{g}(x,t,u_{n})}{\|u_{n}\|} \quad\text{a.e. in } Q. $$ The sequence $(v_{n})$ is bounded in $L^{2}(Q)$, then for subsequence $v_{n}\to v $ weakly in $L^{2}(Q)$. \begin{lemma} \label{lem3.3} Assume $\eqref{II}$ and \eqref{III}. (1) For a subsequence $g_{n}\to f$ weakly in $L^{2}(Q)$. (2) $v_{n} \to v$ strongly in $L^{2}(Q)$, in particular, $\|v\|=1$, thus $v \neq 0$. \end{lemma} \begin{proof} (1) Dividing \eqref{II} by $\|u_{n} \|$, we have $$ |g_{n}(x,t)|\leq \theta|v_{n}|+ \frac{\eta(x,t)}{n},$$ thus $$ \|g_{n}\|\leq \theta\|v_{n}\|+\frac{\|\eta\|}{n} \leq\theta+\frac{\|\eta\|}{n}, $$ hence $g_{n}$ is bounded in $L^{2}(Q)$, one deduces that for a subsequence $g_{n}\to f$ weakly in $L^{2}(Q)$. \noindent (2) Dividing by $\|u_{n}\|$ in \eqref{III}, we have $$ T v_{n} = r_{n}g_{n}+(1-r_{_{n}})\lambda v_{n}+ r_{n}\frac{\tilde{h}}{n}. $$ Which implies $$ v_{n}=(T^{-1}_{0})[r_{n}g_{n}+(1-r_{_{n}})\lambda v_{n}+ r_{n}\frac{\tilde{h}}{n}]. $$ Since $g_{n}\to f$ weakly in $L^{2}(Q)$ and $v_{n}\to v$ weakly in $L^{2}(Q)$, then $$ r_{n}g_{n}+(1-r_{_{n}})\lambda v_{n}+ r_{n}\frac{\tilde{h}}{n}\to rf+(1-r)\lambda v\quad\text{weakly in } L^{2}(Q), $$ where $r=\lim_{n} r_{n}$. The operator $T^{-1}_{0}$ is compact, thus $$ v_{n}=(T^{-1}_{0})[r_{n}g_{n}+(1-r_{_{n}})\lambda v_{n}+ r_{n}\frac{\tilde{h}}{n}]\to (T^{-1}_{0})[rf+(1-r)\lambda v]\quad\text{strongly in } L^{2}(Q). $$ Therefore, $ v_{n}\to (T^{-1}_{0})[r f+(1-r)\lambda v]=v$ strongly in $L^{2}(Q)$. \end{proof} \begin{lemma} \label{lem3.4} Assume \eqref{II} and (III). Then $ f(x,t)=0$ a.e. in $A=\{(x,t)\in Q : v(x,t)=0 \text{ a.e. in } Q\}$. \end{lemma} \begin{proof} Let $\psi$ be the function $$ \psi(x,t)= \mathop{\rm sign}(f(x,t))\chi_{A}(x,t)\text{ a.e. in } Q, $$ where $\chi_{A}$ is the indicatrice function. Since $ g_{n} \to f$ weakly in $L^{2}(Q)$, we have $\int_{Q}g_{n}\psi \to \int_{Q}f\psi = \int_{A}|f(x,t)|$. On the other hand, as $v_{n}\to v$, and using \eqref{II}, we have $$ \big|\int_{Q}g_{n}\psi \big|\leq\int_{Q}|g_{n}\psi| \leq\theta\int_{Q}|v_{n}\chi_{A}|+\int_{Q}\frac{\eta(x,t)\chi_{A}}{n} \to \theta \int_{Q}|v|\chi_{A} = 0, $$ thus $ \int_{A}|f(x,t)|=0$ and $f=0$ a.e. in $A$. \end{proof} We define the function $$ d(x,t)=\begin{cases} \frac{f(x,t)}{v(x,t)}&\text{ a.e. in } Q\setminus A, \\ \lambda & \text{ a.e. in } A. \end{cases} $$ \begin{lemma} \label{lem3.5} If one supposes \eqref{II} ,\eqref{III} and \eqref{Cdm}, then $\delta \leq d(x,t)\leq\mu $ a.e. in $Q$. \end{lemma} \begin{proof} We prove that $ \delta\leq d(x,t)$ a.e. in $Q$. We denote $B=\{(x,t)\in Q : \delta(v(x,t))^{2}>v(x,t)f(x,t) \text{ a.e.}\}$. It is sufficient to prove that $ \mathop{\rm meas}B=0$. Under condition \eqref{Cdm} (cf. Remark \ref{rmk2.1}. No. 5), we have $$ (\delta-\varepsilon)u_{n}^{2}-a_{\varepsilon}(x,t)|u_{_{n}}|\leq u_{n}\tilde{g}(x,t,u_{n}). $$ Dividing by $\|u_{n}\|^{2}$, we get $$ (\delta-\varepsilon)v_{n}^{2}- a_{\varepsilon}(x,t) \frac{|v_{n}|}{n}\leq v_{n}g_{n}(x,t). $$ Multiplying by $\chi_{B}$ and integrating, we get \begin{equation} \label{IV} (\delta-\varepsilon)\int_{Q}v_{n}^{2}\chi_{B} - \int_{_{Q}}\frac{a_{\varepsilon}(x,t)}{n}|v_{n}|\chi_{B} \leq\int_{Q}v_{n}\chi_{B}g_{n}(x,t). \end{equation} Under conditions \eqref{II} and \eqref{III}, $g_{n}\to f$ weakly in $L^{2}(Q)$ and $ v_{n}\to v$ strongly in $L^{2}(Q)$. Going to the limit in $(IV)$, we have $$ (\delta-\varepsilon)\int_{Q}|v(x,t)|^{2}\chi_{B} \leq\int_{Q}v(x,t)f(x,t)\chi_{B}. $$ Since $\varepsilon$ is arbitrary, one concludes that $$ \int_{Q}[v(x,t)f(x,t)-\delta|v(x,t)|^{2}]\chi_{B}\geq 0. $$ Therefore, by the definition of B, $\mathop{\rm meas}B=0$. By an analogous method, we prove that $ d(x,t)\leq\mu$ a.e. in $Q$. \end{proof} \begin{lemma} \label{lem3.6} If one supposes \eqref{II}, (III) and \eqref{Cdm}, then \begin{gather*} T v = m(x,t)v \quad \text{in }Q, \\ v(x,t + 2\pi) = v(x,t) \quad \text{in }]0,\pi[\times\mathbb{R}, \\ v(0,t) = v(\pi,t) = 0 \quad \forall t\in\mathbb{R}, \end{gather*} where $m(x,t)=rd(x,t)+(1-r)\lambda$ and $r=\lim_{n}r_{n}$. \end{lemma} \begin{remark} \label{rmk3.7} \rm It is easy to see that $m(x,t)$ is $2\pi$-periodic in $t$, and $\delta\leq m(x,t)\leq\mu $ a.e. in $Q$. \end{remark} \begin{proof} In the proof of the Lemma \ref{lem3.3}, we have $r f+(1-r)\lambda v = T v$. From the definition of the function $m$, we have $T v = m v $. \end{proof} It remains to prove only the following lemma. \begin{lemma} \label{lem3.8} If one supposes \eqref{II}, \eqref{III} and \eqref{Cdm}, then $$ \delta\leq\neq m(x,t)\leq\neq\mu \quad\text{a.e. in } Q . $$ \end{lemma} \begin{proof} We prove that $ m(x,t)\leq\neq\mu$ a.e. in $Q$. (By analogous method, we prove that $\delta\leq\neq m(x,t)$ a.e. in $Q$). Suppose by contradiction that $ m(x,t)=\mu $ a.e. in $Q$. Under assumption \eqref{Cdm}, we have \begin{equation} \label{V} v_{n}g_{n}\leq(\tilde{k}(x,t)+\varepsilon)v_{n}^{2}+ \frac{a_{\varepsilon}|v_{n}|}{n}, \end{equation} where $\tilde{k}(x,t)\in L^{\infty}(Q)$ such that $\tilde{k}(x,t)\leq\neq\mu$. By $(V)$, we have \begin{equation} \label{VI} \begin{aligned} &\int_{Q}r_{n}g_{n}v_{n}+(1-r_{n})\lambda v_{n}^{2}+ r_{n}\int_{Q}\frac{\tilde{h}v_{n}}{n} \\ &\leq \int_{Q}[r_{n}(\tilde{k}(x,t)+\varepsilon)+(1-r_{n})\lambda] v_{n}^{2}+ r_{n}\int_{Q}(\tilde{h}\frac{v_{n}}{n}+ a_{\varepsilon}\frac{|v_{n}|}{n}). \end{aligned} \end{equation} Under conditions \eqref{II} and \eqref{III}, $g_{n}\to f$ weakly in $L^{2}(Q)$ and $v_{n}\to v$ strongly in $L^{2}(Q)$. Going to the limit in $(V)$, we get $$ \int_{Q}[r(fv) + (1-r)\lambda v^{2}] \leq \int_{Q}[r(\tilde{k}(x,t)+\varepsilon) + (1-r)\lambda]v^{2}. $$ By the definition of $m$, we have $$ \int_{Q}[r(fv) + (1-r)\lambda v^{2}] = \int_{Q}m(x,t)v^{2}= \int_{Q}\mu v^{2}. $$ Thus $$ \int_{Q}\mu v^{2}\leq \int_{Q}[r(\tilde{k}(x,t) + \varepsilon)+(1-r)\lambda]v^{2}. $$ Since $\varepsilon$ is arbitrary, we have $$ \int_{Q}\mu v^{2}\leq \int_{Q}[r\tilde{k}(x,t)+(1-r)\lambda]v^{2}. $$ Hence $$ \int_{Q}[\mu-r\tilde{k}(x,t)-(1-r)\lambda]v^{2}\leq 0. $$ Since $\tilde{k}(x,t)\leq \mu$ a.e. in $Q$ and $\lambda < \mu$, we have $\mu - r\tilde{k}(x,t)-(1-r)\lambda\geq0$, then $$ \int_{Q}[\mu-r\tilde{k}(x,t)-(1-r)\lambda]v^{2}=0. $$ Therefore, $[\mu-r\tilde{k}(x,t)-(1-r)\lambda]v^{2}=0$ a.e. in $Q$. Since $m(x,t)=\mu$ a.e. in $Q$, by the definition of the function of $d$, $(d(x,t)\neq \lambda)$, we have $\mathop{\rm meas}A =0$ (i.e. $v(x,t)\neq0$ a.e. in $Q$). Thus, $\mu=r \tilde{k}(x,t)+(1-r)\lambda$ a.e. in $Q$, this contradiction completes the proof. \end{proof} For the proof of Corollary \ref{coro3.2} we will need the following two lemmas. \begin{lemma}[\cite{b4}] \label{lem3.9} Assume (C1), (C2), \eqref{II}, $\lambda\in\sigma(T)$ and $ \lambda\geq 0$. Let $\tilde{h}\in L^{2}(Q)$, if there exist $R>0$ such that $$ T u - r(N(u)+\tilde{h})-(1-r)\lambda u \neq 0, \quad \forall u\in D(T),\; \|u\|=R,\; 0\leq r\leq1, $$ then problem \eqref{tildeP} admits at least one solution $u\in D(T)$ with $\|u\|