\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 64, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/64\hfil Boussinesq equation] {Cauchy problem for the sixth-order damped multidimensional Boussinesq equation} \author[Y. Wang \hfil EJDE-2016/64\hfilneg] {Ying Wang} \address{Ying Wang \newline School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu 611731, china} \email{nadine\_1979@163.com} \thanks{Submitted January 20, 2016. Published March 10, 2016.} \subjclass[2010]{35L60, 35K55, 35Q80} \keywords{Damped Boussinesq equation; well-posedness; blow-up; \hfill\break\indent asymptotic behavior} \begin{abstract} In this article, we consider the Cauchy problem for sixth-order damped Boussinesq equation in $\mathbb{R}^n$. The well-posedness of global solutions and blow-up of solutions are obtained. The asymptotic behavior of the solution is established by the multiplier method. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} It is well-known that the generalized Boussinesq equation, in $\mathbb{R}$, \begin{equation} \label{e1.1} u_{tt}+ u_{xxxx}-u_{xx}=(f(u))_{xx}, \end{equation} is a very important and famous nonlinear evolution equation suggested for describing the motion of water with small amplitude and long wave. There have been many results on the local and global well-posedness of problem \eqref{e1.1} in \cite{Lin,L4,L5,Liu}. In \cite{A}, the authors studied a damped Boussinesq equation \begin{equation} \label{e1.2} u_{tt}-ku_{txx}-u_{xx}- u_{xxtt}=(f(u))_{xx}. \end{equation} Wang and Chen \cite{WC} considered the Cauchy problem for the generalized double dispersion equation \begin{equation} \label{e1.3} u_{tt}-ku_{txx}+ u_{xxxx}-u_{xx}- u_{xxtt}=(f(u))_{xx}, \end{equation} whose well-posedness of the local and global solutions and the blow-up of the solutions were established in $\mathbb{R}$. Polat \cite{NA,NE} generalized the results obtained in \cite{WC} and proved the existence of local and global, blow-up, and asymptotic behavior of solutions for the Cauchy problem of \eqref{e1.3} in $\mathbb{R}^n$. Schneider and Eugene \cite{SE} considered another class of Boussinesq equation which characterizes the water wave problem with surface tension as follows \begin{equation} \label{e1.4} u_{tt}=u_{xx}+u_{xxtt}+\mu u_{xxxx}-u_{xxxxtt}+(u^2)_{xx}, \end{equation} which can also be formally derived from the 2D water wave problem. For a degenerate case, they proved that the long wave limit can be described approximately by two decoupled Kawahara-equations. Wang and Mu \cite{W1,W2} studied the well-posedness of the local and global solutions, the blow-up of solutions and nonlinear scattering for small amplitude solutions to the Cauchy problem of \eqref{e1.4}. Piskin and Polat \cite{EN} considered the Cauchy problem of the multidimensional Boussinesq equation \begin{equation} \label{e1.5} u_{tt}=\Delta u+\Delta u_{tt}+\mu\Delta^2 u-\Delta^2 u_{tt}+\Delta f(u) +k\Delta u_{t}. \end{equation} The existence, both locally and globally in time, the global nonexistence, and the asymptotic behavior of solutions for the Cauchy problem of equation \eqref{e1.5} are established in $n$-dimensional space. Wang and Esfahani \cite{WE1, WE2} considered the Cauchy problem associated with the sixth-order Boussinesq equation with cubic nonlinearity \begin{equation} \label{e1.6} u_{tt}=u_{xx}+\beta u_{xxxx}+ u_{xxxxxx}+(u^2)_{xx}, \end{equation} where $\beta=\pm 1$, Equation \eqref{e1.6} arises as mathematical models for describing the bi-directional propagation of small amplitude and long capillary-gravity waves on the surface of shallow water for bond number (surface tension parameter) less than but very close to $\frac{1}{3}$ \cite{DD}. Equation \eqref{e1.6} has been also used as the model of nonlinear lattice dynamics in elastic crystals \cite{M}. In this article, we investigate the Cauchy problem of the sixth-order damped multidimensional Boussinesq equation \begin{gather} \label{e1.7} u_{tt}-\Delta u_{tt}-\Delta u+\Delta^2 u-\Delta^3u-r\Delta u_{t} =\Delta f(u), \quad (x,t)\in \mathbb{R}^n\times (0,+\infty),\\ u(x,0)=\phi(x),\quad u_t(x,0)=\psi(x),\quad x\in \mathbb{R}^n, \label{e1.8} \end{gather} where $u(x,t)$ denotes the unknown function, $f(s)$ is the given nonlinear function, $r$ is a constant, the subscript $t$ indicates the partial derivation with respect to $t$, and $\Delta$ denotes the Laplace operator in $\mathbb{R}^n$. Recently, the authors \cite{WYZ} proved the existence and asymptotic behavior of global solutions of \eqref{e1.7} for all space dimensions $n\geq 1$ provided that the initial value is suitably small. In \cite{W}, the authors obtained the global existence and asymptotic decay of solutions to the problem \eqref{e1.7}. For the initial boundary value problem of \eqref{e1.7} with $f(u)=u^2$, Zhang \cite{ZLL} and Lai \cite{SY,SYQ} established the well-posedness of strong solution and constructed the solution in the form of series in the small parameter present in the initial conditions. The long-time asymptotics was also obtained in the explicit form. The main purpose of this paper is to study the well-posedness of the global solution and the asymptotic behavior of the global solution for the Cauchy problem \eqref{e1.7}-\eqref{e1.8} in $\mathbb{R}^n$. Due to the sixth-order term $\Delta^3$, it seems difficult to construct the operator $\partial^2_t-\Delta$ which is similar to that in \cite{WC,NA} to solve the problem \eqref{e1.7}-\eqref{e1.8}. To overcome this difficulty, we transformed \eqref{e1.7} in another way and established the corresponding estimate. Throughout this article, we use $L_p$ to denote the space of $L^p$-function on $\mathbb{R}^n$ with the norm $\|f\|_p=\|f\|_{L^p}$. $H^s$ denotes the Sobolev space on $\mathbb{R}^n$ with norm $\|f\|_{H^s}=\|(I-\Delta)^{s/2}f\|_2$, where $1\leq p\leq \infty,s\in \mathbb{R}$. To prove the global well-posedness, we use the contraction mapping principle to the local-posedness of the problem \eqref{e1.7}-\eqref{e1.8}. \begin{theorem} \label{thm1.1} Assume that $s>\frac{n}{2},\phi\in H^s,\psi\in H^{s-2}$ and $f(s)\in C^{[s]+1}(R)$, then problem \eqref{e1.7}-\eqref{e1.8} admits a unique local solution $u(x,t)$ defined on a maximal time interval $[0, T_0)$ with $u(x,t)\in C([0,T_0), H^{s})\cap C^1( [0,T_0), H^{s-2})$. Moreover, if \begin{equation} \label{e1.9} \sup_{t\in[0,T_0)}(\|u(t)\|_{H^s}+\|u_t(t)\|_{H^{s-2}})<\infty, \end{equation} then $T_0=\infty$. \end{theorem} Now we arrive at the existence and uniqueness of global solutions for \eqref{e1.7}-\eqref{e1.8}. \begin{theorem} \label{thm1.2} Assume that $1\leq n\leq 4$, $s\geq\frac{n+1}{2},f(u)\in C^{[s]+1}(R)$, $F(u)=\int^u_0f(s)ds$ or $f'(u)$ is bounded below, i.e. there is a constant $A_0$ such that $f'(u)\geq A_0$ for any $u\in \mathbb{R}$, $|f'(u)|\leq A|u|^{\rho}+B$, $0<\rho\leq \infty$ for $2\leq n\leq 4, (-\Delta)^{-1/2}\psi\in L^2, \phi\in H^{s+1}$ and $\psi\in H^{s-1},F(\phi)\in L^1$. Then problem \eqref{e1.7}-\eqref{e1.8} admits a global solution $u(x,t)\in C([0,\infty),H^s)\cap C^1([0,\infty), H^{s-2})$ and $(-\Delta)^{-1/2}u_t\in L^2$. \end{theorem} In Lemma \ref{lem3.1} below we have the energy equality $E(t)=\|(-\Delta)^{-1/2}\psi\|^2_2+\|\psi\|^2_2 +\|\phi\|^2_2+\|\nabla \phi\|^2_2+\|\Delta \phi\|^2_2+2\int_{\mathbb{R}^n}F(u)dx$. Then we can obtain the blow-up results by the concavity method. \begin{theorem} \label{thm1.3} Assume that $r\geq 0, f(u)\in C(R)$, $\phi\in H^2,\psi\in L^2, (-\Delta)^{-1/2}\phi$, $(-\Delta)^{-1/2}\psi\in L^2, F(u)=\int^u_0f(s)ds$, $F(\phi)\in L^1$, and there exists a constant $\alpha>0$ such that \begin{equation} \label{e1.10} f(u)u\leq (\alpha+r+2)F(u)+\frac{\alpha}{2}u^2,\quad \forall u\in \mathbb{R}. \end{equation} Then the solution $u(x,t)$ of \eqref{e1.7}-\eqref{e1.8} will blow up in finite time if one of the following conditions hold: \begin{itemize} \item[(i)] $E(0)=\|(-\Delta)^{-1/2}\psi\|^2_2+\|\psi\|^2_2 +\|\phi\|^2_2+\|\nabla \phi\|^2_2+\|\Delta \phi\|^2_2 +2\int_{\mathbb{R}^n}F(\phi)dx<0$, \item[(ii)] $E(0)=0$ and $\big((-\Delta)^{-1/2}\phi, (-\Delta)^{-1/2}\psi\big)+(\phi,\psi)>0$, \item[(iii)] $E(0)>0$ and \[ \big((-\Delta)^{-1/2}\phi, (-\Delta)^{-1/2}\psi\big)+(\phi,\psi) >\sqrt{2\frac{4+2r+2\alpha}{\alpha+2} E(0)(\|(-\Delta)^{-1/2}\phi\|^2_2+\|\phi\|^2_2}). \] \end{itemize} \end{theorem} \begin{theorem} \label{thm1.4} Let $r> 0$ and assume that \[ 0\leq F(u)\leq f(u)u,\quad \forall u \in \mathbb{R}, \quad F(u)=\int^u_0f(s)ds. \] Then for the global solution of problem \eqref{e1.7}-\eqref{e1.8}, there exist positive constants $C$ and $\theta$ such that \begin{equation} \label{e1.11} E(t)\leq C E(0) e^{-\theta t},\quad 0\leq t\leq \infty, \end{equation} where \[ E(t)=\frac{1}{2}(\|(-\Delta) ^{-1/2}u_t\|^2_2 +\|u_t\|^2_2+\|u\|^2_2+\|\nabla u\|^2_2+\|\Delta u\|^2_2) +\int _{\mathbb{R}^n}F(u)dx. \] \end{theorem} The article is organized as follows. In the next section, we prove Theorem \ref{thm1.1} which is related to the local well-posedness for a general nonlinearity. In Section 3, we prove Theorem \ref{thm1.2}. The proof of the nonexistence of a global solution is given in Section 4. In the last section, the asymptotic behavior of the global solution is discussed. \section{Existence and uniqueness of the local solution} In this section, we prove the existence and the uniqueness of the local solution for \eqref{e1.7}-\eqref{e1.8} by contraction mapping principle. To do so, we construct the solution of the problem as a fixed point of the solution operator associated with related family of Cauchy problem for linear equation. For this purpose, we rewrite \eqref{e1.7} as follows: \begin{equation} \label{e2.1} u_{tt}+\Delta^2 u=\Gamma[f(u)+r u_{t}+u]. \end{equation} where $\Gamma=(I-\Delta)^{-1}\Delta$. Using the Fourier transform, it is easy to obtain \[ \Gamma f=\Delta(G\ast f)=G\ast f-f, \] where $G(x)=\frac{1}{2}e^{-|x|}$, and $u\ast v$ denotes the convolution of $u$ and $v$. We start with the linear equation. \begin{equation} \label{e2.2} u_{tt}+\Delta^2 u=q(x,t), \quad x\in \mathbb{R}^n,\; t>0, \end{equation} with the initial value condition \eqref{e1.8}. To prove Theorem \ref{thm1.1}, we need the following lemmas. \begin{lemma}[\cite{S}] \label{lem2.1} If $s>k+n/2$, where $k$ is a nonnegative integer, then \[ H^s(\mathbb{R}^n)\subset C^k(\mathbb{R}^n)\cap L^\infty(\mathbb{R}^n), \] where the inclusion is continuous. In fact, \[ \sum_{|\alpha|\leq k}\|\partial^\alpha u\|_{L^{\infty}}\leq C_s\|u\|_{H^{s}}, \] where $C_s$ is independent of $u$. \end{lemma} \begin{lemma}[\cite{H}] \label{lem2.2} Let $q\in[1,n]$ and $\frac{1}{p}=\frac{1}{q}-\frac{1}{n}$, then for any $u\in H^q_1(\mathbb{R}^n)$, \[ \|u\|_p\leq C(n,q)\|\nabla u\|_q, \] where $C(n,q)$ is a constant dependent on $n$ and $q$. \end{lemma} \begin{lemma}[\cite{WCH}] \label{lem2.3} Assume that $f(u)\in C^{k}(R), f(0)=0,u\in H^{s}\cap L^{\infty}$ and $k=[s]+1$, where $s\geq 0$. Then \[ \|f(u)\|_{H^s}\leq K_1(W)\|u\|_{H^{s}}, \] if $\|u\|_{\infty}\leq W$, where $K_1(W)$ is a constant dependent on $W$. \end{lemma} \begin{lemma}[\cite{WCH}] \label{lem2.4} Assume that $f(u)\in C^{k}(R), u,v \in H^{s}\cap L^{\infty} $ and $k=[s]+1$, where $s\geq 0$. Then \[ \|f(u)-f(v)\|_{H^s}\leq K_2(W)\|u-v\|_{H^s}, \] if $\|u\|_{\infty}\leq W,\|v\|_{\infty}\leq W$, where $K_2(W)$ is a constant dependent on $W$. \end{lemma} \begin{lemma}[\cite{F}] \label{lem2.5} If $1\leq p\leq \infty,u(x,t)\in L^{p}(\mathbb{R}^{n})$ for a.e. $t$ and the function $t\mapsto \|u(\cdot,t)\|_{p}$ is in $L^{1}(I)$, where $I\subset [0,\infty)$ is an interval, then \[ \|\int_{I}u(\cdot,t)\|_{p}\leq \int_{I}\|u(\cdot,t)\|_{p}dt. \] \end{lemma} \begin{lemma} \label{lem2.6} Let $s\in \mathbb{R}, \phi\in H^{s},\psi\in H^{s-2}$ and $q\in L^1([0,T]; H^{s-2})$. Then for every $T>0$, there is a unique solution $u\in C([0,T],H^s)\cap C^1([0,T],H^{s-2})$ of Cauchy problem \eqref{e2.2} and \eqref{e1.8}. Moreover, $u$ satisfies \begin{equation} \label{e2.3} \|u(t)\|_{H^s}+\|u_t(t)\|_{H^{s-2}} \leq C(1+T)(\|\phi\|_{H^s}+\|\psi\|_{H^{s-2}} +\int^t_0\|q(\tau)\|_{H^{s-2}}d\tau), \end{equation} for $0\leq t\leq T$, where $C$ dependeds only on $s$. \end{lemma} \begin{proof} The argument about the existence and uniqueness of the solution of the Cauchy problem for the linear problem \eqref{e2.2} and \eqref{e1.8} is similar to that in \cite{S}, we omit it. The solution of the linear equation is given in Fourier space by \[ \hat{u}(\xi,t)=\cos(t|\xi|^2)\hat{\phi}(\xi) +\frac{\sin(t|\xi|^2)}{|\xi|^2}\hat{|\psi|^2} +\int^t_0\frac{\sin((t-\tau)|\xi|^2)}{|\xi|^2}\hat{q}(\xi,\tau)d\tau, \] where $\hat{}$ denotes Fourier transform with respect to $x$. Since \[ \|(1+|\xi|^2)^{s/2}\cos (t|\xi|^2)\hat{\phi}(\xi)\| \leq \|(1+|\xi|^2)^{s/2}\hat{\phi}(\xi)\|=\|\phi\|_{H^s} \] and \begin{align*} &\|(1+|\xi|^2)^{s/2}\frac{\sin(t|\xi|^2)}{|\xi|^2}\hat{\psi}(\xi)\|^2\\ &=\int_{|\xi|<1}(1+|\xi|^2)^{s}\frac{\sin^2(t|\xi|^2)}{|\xi|^4} |\hat{\psi}(\xi)|^2d\xi +\int_{|\xi|\geq 1}(1+|\xi|^2)^{s}\frac{\sin^2(t|\xi|^2)}{|\xi|^4}| \hat{\psi}(\xi)|^2d\xi\\ &\leq t^2\int_{|\xi|<1}(1+|\xi|^2)^s|\hat{\psi}(\xi)|^2d\xi +\int_{|\xi|\geq 1}(1+|\xi|^2)^s\frac{1}{|\xi|^4}|\hat{\psi}(\xi)|^2d\xi\\ & \leq 4t^2\int_{|\xi|<1}(1+|\xi|^2)^{s-2}|\hat{\psi}(\xi)|^2d\xi +4\int_{|\xi|\geq 1}(1+|\xi|^2)^s\frac{1}{|\xi|^4}|\hat{\psi}(\xi)|^2d\xi\\ &\leq 4(1+t^2)\int_{\mathbb{R}^n}(1+|\xi|^2)^{s-2}|\hat{\psi}(\xi)|^2d\xi\\ &=4(1+t^2)\|\psi\|^2_{H^{s-2}}, \end{align*} we obtain \begin{gather*} \|u(t)\|_{H^s}\leq \|\phi\|_{H^s}+2(1+t)\|\psi\|_{H^{s-2}}+2(1+t) \int^t_0\|q(\tau)\|_{H^{s-2}}d\tau,\\ \|u_t(t)\|_{H^{s-2}}\leq \|\phi\|_{H^s} +\|\psi\|_{H^{s-2}}+\int^t_0\|q(\tau)\|_{H^{s-2}}d\tau. \end{gather*} Therefore \eqref{e2.3} holds. This completes the proof. \end{proof} \begin{lemma} \label{lem2.7} The operator $L$ is bounded on $H^{s}$ for all $s\geq 0$ and \[ \|\Gamma u\|_{H^s}\leq C \|u\|_{H^{s}}, \forall u\in H^{s}. \] \end{lemma} \begin{proof} For $u\in H^{s},s\geq 0$, we have \[ \|\Gamma u\|_{H^s}^2=\int_{\mathbb{R}^n}(1+|\xi|^2)^s \frac{|\xi|^4}{(1+|\xi|^2)^2}|\hat{u(|\xi|)}|^2d\xi\leq C\|u\|^2_{H^{s}}. \] \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] We will prove the theorem in four steps. \smallskip \noindent\textbf{Step 1.} Define the function space \[ X(T)=C([0,T],H^{s})\cap C^1([0,T],H^{s-2}), \] which is equipped with the norm \[ \|u\|_{X(T)}=\max_{0\leq t\leq T}(\|u\|_{H^{s}}+\|u_t\|_{H^{s-2}}),\quad \forall u\in X(T). \] It is easy to see that $X(T)$ is a Banach space. For $s>n/2$ and any initial values $\phi\in H^s,\psi\in H^{s-2}$, let $M=\|\phi\|_{H^{s}}+\|\psi\|_{H^{s-2}}$. Take the set \[ Y(M,T)=\{u\in X(T):\|u\|_{X(T)}\leq 2CM\}. \] Note that $Y(M,T)$ is a nonempty bounded closed convex subset of $X(T)$ for any fixed $M>0$ and $T>0$. From Lemma \ref{lem2.1}, $u\in C([0,T],L^{\infty})$ and $\|u\|_{L^{\infty}}\leq C_s \|u\|_{H^s}$, if $u\in X(T)$. For $v\in Y(M,T)$, we consider the linear equation \begin{equation} \label{e2.4} u_{tt}+\Delta^2 u=\Gamma [f(v)+r v_{t}+ v] \end{equation} and we let $S$ denote the map which carried $v$ into the unique solution of \eqref{e2.4} and \eqref{e1.8}. Our goal is to show that $S$ has a unique fixed point in $Y(M,T)$ for appropriately chosen $T$. To this end, we shall employ the contraction mapping principle and Lemma \ref{lem2.6}. \smallskip \noindent\textbf{Step 2.} We shall prove that $S$ maps $Y(M,T)$ into itself for $T$ small enough. Let $v\in Y(M,T)$ be given. Define $q(x,t)$ by \[ q(x,t)=\Gamma [f(v)+r v_t+ v]. \] Using lemmas \ref{lem2.3} and \ref{lem2.7}, it follows easily that \[ \|q(t)\|_{H^{s-2}}\leq C \|f(v)\|_{H^{s-2}}+|r|\|v_t\|_{H^{s-2}} +\|v\|_{H^{s-2}}\leq C_M\|v\|_{H^{s}}+|r|\|v_t\|_{H^{s-2}}, \] where $C_M$ is a constant dependent on $M$ and $s$. From the above inequality we conclude that $q(x,t)\in C^1([0,T], H^{s-2})$. From Lemma \ref{lem2.6}, the solution $u=Sv$ of problem \eqref{e2.2} and \eqref{e1.8} belongs to $C([0,T],H^{s})\cap C^1([0,T],H^{s-2})$ and \begin{align*} \|u(t)\|_{H^{s}}+\|u_t(t)\|_{H^{s-2}} &\leq C(1+T)(\|\phi\|_{H^s}+\|\psi\|_{H^{s-2}} +\int^t_0\|q(\tau)\|_{H^{s-2}}d\tau)\\ &\leq CM+C[1+2C((C_M)+|r|)(1+T)]M T. \end{align*} By choosing $T$ small enough, we have \begin{equation} \label{e2.5} [1+2C((C_M)+|r|)(1+T)]T\leq 1, \end{equation} then we obtain \begin{equation} \label{e2.6} \|Sv\|_{X(T)}\leq 2CM. \end{equation} Thus, if condition \eqref{e2.6} holds, then $S$ maps $Y(M,T)$ into $Y(M,T)$. \smallskip \noindent\textbf{Step 3.} We shall also claim that for $T$ small enough, $S$ is a strictly contractive map. Let $T>0$ and $v,\bar{v} \in Y(M,T)$ be given. Set $u=Sv, \bar{u}=S \bar{v}, U=u-\bar{u}, V=v-\bar{v}$ and note that $U$ satisfies \begin{gather} \label{e2.7} U_{tt}+\Delta^2 U=Q(x,t),(x,t)\in \mathbb{R}^n\times (0,+\infty),\\ U(x,0)=U_t(x,0)=0, \label{e2.8} \end{gather} where $Q(x,t)$ is defined by \begin{equation} \label{e2.9} Q(x,t)=\Gamma[f(v)-f(\bar{v})]+r \Gamma[V_t]+\Gamma[V]. \end{equation} Observed that $S$ has the smoothness required to apply Lemma \ref{lem2.6} to problem \eqref{e2.7} and \eqref{e2.8}. By Lemmas \ref{lem2.4}, \ref{lem2.6} and \ref{lem2.7}, from \eqref{e2.9} we obtain \begin{align*} &\|U(t)\|_{H^s}+\|U_t(t)\|_{H^{s-2}}\\ &\leq C(1+T)\int^t_0 [\|f(v(\tau))-f(\bar{v}(\tau))\|_{H^{s-2}} +|r|\|V_t\|_{H^{s-2}}+\|V\|_{H^{s-2}}]d\tau\\ &\leq C(1+T)[C_M\max_{0\leq t\leq T}\|V(t)\|_{H^s} +|r|\max_{0\leq t\leq T}\|V_t(t)\|_{H^{s-2}}]T. \end{align*} Hence, we obtain \[ \|U(t)\|_{X(T)}\leq C(1+T)[C_M+|r|+C]T\|V(t)\|_{X(T)}. \] By choosing $T$ so small that \eqref{e2.5} holds and \begin{equation} \label{e2.10} (1+T)[C_M+|r|+C]<1/C, \end{equation} then \[ \|S v-S\bar{v}\|_{X(T)}<\|v-\bar{v}\|_{X(T)}. \] This shows that $S:Y(M,T)\to Y(M,T)$ is strictly contractive. \smallskip \noindent\textbf{Step 4.} From the contraction mapping principle, it follows that for appropriately chosen $T>0$, $S$ has a unique fixed point $u(x,t)\in Y(M,T)$, which is a strong solution of problem \eqref{e1.7}-\eqref{e1.8}. Similarly to \cite{W2}, we can prove uniqueness and local Lipschitz dependence with respect to the initial data in the space $Y(M,T)$. Using uniqueness we can extend the result in the space $C([0,T],H^{s})\cap C^1([0,T],H^{s-2})$ by a standard technique. \end{proof} \section{Existence and uniqueness of a global solution} In this section, we prove the existence and the uniqueness of the global solution for problem \eqref{e1.7}-\eqref{e1.8}. For this purpose, we are going to make a priori estimates of the local solutions for problem \eqref{e1.7}-\eqref{e1.8}. \begin{lemma} \label{lem3.1} Suppose that $f(u)\in C(R),F(u)=\int^u_0f(s)ds,\phi\in H^2, (-\Delta)^{\frac{1}{2} }\psi\in L^2,\psi\in L^2$, and $F(\phi)\in L^1$. Then for the solution $u(x,t)$ of the problem \eqref{e1.7}-\eqref{e1.8}, it follows that \begin{equation} \label{e3.1} \begin{aligned} E(t)&=\|(-\Delta)^{-1/2}u_t\|^2_2+\|u_t\|^2_2+\|u\|^2_2 +\|\nabla u\|^2_2+\|\Delta u\|^2_2 \\ &\quad +2r\int^t_0\|u_\tau\|^2_2d\tau+2\int_{\mathbb{R}^n}F(u)dx=E(0). \end{aligned} \end{equation} Here and in the sequel $(-\Delta)^{-\alpha}u(x)=\mathcal{F}^{-1}[|x|^{-2\alpha} \mathcal{F}u(x)]$, $\mathcal{F}$ and $\mathcal{F}^{-1}$ denote Fourier transformation and inverse Fourier transformation in $\mathbb{R}^n$ respectively. \end{lemma} \begin{proof} Multiplying both sides of \eqref{e1.7} by $(-\Delta)^{-1} u_t$, integrating the product over $\mathbb{R}^n$ and integrating by parts, we obtain \begin{align*} &(u_{tt}-\Delta u-\Delta u_{tt}+\Delta^2u-\Delta^3 u-r \Delta u_t-\Delta f(u), (-\Delta)^{-1}u_t)\\ &= ((-\Delta)^{-1}u_{tt}+u+u_{tt}-\Delta u+\Delta^2 u+ r u_t+f(u), u_t)\\ &= ((-\Delta)^{-1/2}u_{tt},( -\Delta)^{-1/2}u_{t})+(u,u_t) +(u_{tt},u_t)+(\Delta^2 u,u_t)+(\Delta u,u_t)\\ &\quad +r(u_t,u_t)+(f(u),u_t) = 0. \end{align*} So, \begin{align*} &\frac{d}{dt}[\|(-\Delta)^{-1/2}u_t\|^2_2 +\|u_t\|^2_2+\|u\|^2_2+\|\Delta u\|^2_2+\|\nabla u\|^2_2\\ &+2r\int^t_0\|u_\tau\|^2_2d\tau+2\int_{\mathbb{R}^n}F(u)dx]=0. \end{align*} The lemma is proved. \end{proof} \begin{lemma} \label{lem3.2} Suppose that the assumptions of Lemma \ref{lem3.1} hold and $F(u)\geq 0$ or $f'(u)$ is bounded below, i.e there is a constant $A_0$ such that $f'(u)\geq A_0$ for any $u\in \mathbb{R}$, then the solution $u(x,t)$ of problem \eqref{e1.7}-\eqref{e1.8} has the estimate \begin{equation} \label{e3.2} E_1(t)=\|(-\Delta)^{-1/2}u_t\|^2_2+\|u_t\|^2_2+\|u\|^2_2 +\|\nabla u\|^2_2+\|\Delta u\|^2_2\leq M_1(T), \end{equation} for all $t\in [0,T]$. Here and in the sequel $M_i(T)(i=1,2, \dots)$ are constants dependent on $T$. \end{lemma} \begin{proof} If $F(u)\geq 0$, then from energy identity \eqref{e3.1}, we obtain \[ E_1(t)\leq E(0)+2|r|\int^t_0\|u_\tau\|^2_2d\tau. \] It follows from Gronwall's inequality and the above inequality that \begin{equation} \label{e3.3} E_1(t)\leq E(0)e^{2|r|T}. \end{equation} If $f'(u)$ is bounded below. Let $f_0(u)=f(u)-r_0u$, where $k_0=min\{A_0,0\}(\leq 0)$, then $f_0(0)=0, f'_0(u)=f'(u)-r_0\geq 0$ and $f_0(u)$ is a monotonically increasing function. Then $F_0(u)=\int^u_0 f_0(s)ds\geq 0$ and $F(u)=\int^u_0f(s)ds=\int^u_0(f_0(s)+r_0 s)ds=F_0(u)+\frac{r_0}{2}u^2$. From \eqref{e3.1}, we have \begin{align*} &E_1(t)+2\int_{\mathbb{R}^n}F_0(u)dx\\ &=E(0)-2r\int^t_0\|u_\tau\|^2_2d\tau-r_0\|u\|^2_2\\ &=E(0)-2r\int^t_0\|u_\tau\|^2_2d\tau-r_0\|u_0\|^2_2 +\int^t_0(r_0^2\|u\|^2_2+\|u_\tau\|^2_2)d\tau\\ &\leq E(0)-r_0\|u_0\|^2_2+(2|r|+1+r_0^2)\int^t_0(\|u\|^2_2+\|u_\tau\|^2_2)d\tau. \end{align*} It follows from Gronwall's inequality and the above inequality that \begin{equation} \label{e3.4} E_1(t)\leq (E(0)-r_0\|u_0\|^2_2)\exp[(2|r|+1+r_0^2)T]. \end{equation} We get \eqref{e3.2} from inequalities \eqref{e3.3} and \eqref{e3.4}. The lemma is proved. \end{proof} \begin{lemma} \label{lem3.3} Under the conditions of Lemma \ref{lem3.2}, assume that $1\leq n\leq 4, f(u)\in C^2(R)$ and $|f'(u)|\leq A|u|^{\rho}+B,0<\rho<\infty$ for $2\leq n\leq 4, \phi\in H^3$ and $\psi\in H^1$, then the solution $u(x,t)$ of problem \eqref{e1.7}-\eqref{e1.8} has the estimation \begin{equation} \label{e3.5} E_2(t)=\|u_t\|^2_2+\|\nabla u\|^2_2+\|\nabla u_t\|^2_2 +\|\Delta u\|^2_2+\|\nabla^3 u\|^2_2\leq M_2(T), \quad \forall t\in[0,T]. \end{equation} \end{lemma} \begin{proof} Multiplying \eqref{e1.7} by $u_t$ and integrating the product over $\mathbb{R}^n$, we obtain \begin{equation} \label{e3.6} \frac{d}{dt}E_2(t)+2r\|\nabla u_t\|^2_2+2(\nabla f(u), \nabla u_t)=0. \end{equation} When $n=1$, we conclude from Lemma \ref{lem2.1} and \ref{lem3.2} that $u\in L^{\infty}$. Therefore, from \eqref{e3.6}, H\"{o}lder inequality, Cauchy inequality, Lemma \ref{lem2.3} and \eqref{e3.2}, we obtain \begin{equation} \label{e3.7} \begin{aligned} \frac{d}{dt}E_2(t) &\leq 2|r|\|\nabla u_t\|^2_2+2|(\nabla f(u),\nabla u_t)| \\ &\leq 2|r|\|\nabla u_t\|^2_2+2\|\nabla f(u)\|_2\|\nabla u_t\|_2 \\ &\leq 2|r|\|\nabla u_t\|^2_2+2K_1(W)(\|u\|_{\infty})(\|u\|_2+\|\nabla u\|_2)\|\nabla u_t\|_2 \\ &\leq C_1(M_1(t))(\|\nabla u\|^2_2+\|\nabla u_t\|^2_2), \end{aligned} \end{equation} where and in the sequel $C_i(M_j(t))(i=1,2,\dots, j=1,2,\dots)$ are constants depending on $M_j(t)$. Integrating \eqref{e3.7} with respect to $t$ and using the Gronwall's inequality, we obtain \eqref{e3.5}. In the case $2\leq n\leq 4$, from H\"{o}lder inequality, Lemma \ref{lem2.2}, Cauchy inequality and \eqref{e3.2}, we have \begin{align*} \int_{\mathbb{R}^n}\nabla f(u)\nabla u_t dx &\leq A\|u^{\rho}\|_{\infty}\|\nabla u\|^2_2\|\nabla u_t\|_2 +B\|\nabla u\|_2\|\nabla u_t\|_2\\ &\leq \frac{A}{2}(C_2\|\Delta u\|^2_2\|\nabla u\|^2_2+\|\nabla u_t\|^2_2) +\frac{B}{2}(\|\nabla u\|^2_2+\|\nabla u_t\|^2_2)\\ &\leq \frac{A}{2}(C_2(M_1(t))\|\Delta u\|^2_2+\|\nabla u_t\|^2_2) +\frac{B}{2}(M_1(t)+\|\nabla u_t\|^2_2). \end{align*} Substitute the above inequality in \eqref{e3.6} to obtain \begin{equation} \label{e3.8} \begin{aligned} \frac{d}{dt}E_2(t) &\leq 2|r|\|\nabla u_t\|^2_2+2|(\nabla f(u),\nabla u_t)|\\ &\leq B M_1(t)+C_3 M_1(t)(\|\Delta u\|^2_2+\|\nabla u_t\|^2_2). \end{aligned} \end{equation} Integrating \eqref{e3.8} with respect to $t$ and using the Gronwall's inequality, we obtain \eqref{e3.5}. The lemma is proved. \end{proof} \begin{lemma} \label{lem3.4} Under the conditions of Lemma \ref{lem3.3}, assume that $s\geq 2, f(u)\in C^{[s]}(R), \phi\in H^{s+1},\psi\in H^{s-1}$, then the solution $u(x,t)$ of problem \eqref{e1.7}-\eqref{e1.8} has the estimate \begin{equation} \label{e3.9} \begin{aligned} E_3(t)&= \|\nabla ^{s-2}u_t\|^2_2+\|\nabla^{s-1}u\|^2_2 +\|\nabla^{s-1}u_t\|^2_2+\|\nabla^s u\|^2_2 +\|\nabla^{s+1} u\|^2_2 \\ &\leq M_3(T),\quad \forall t\in [0,T]. \end{aligned} \end{equation} \end{lemma} \begin{proof} Multiplying \eqref{e1.7} by $\Delta^{s-2}u_t$ and integrating the product over $\mathbb{R}^n$, we obtain \begin{equation} \label{e3.10} \frac{d}{dt}E_3(t)+2r\|\nabla ^{s-1}u_t\|^2_2+2(\nabla^{s-1}f(u), \nabla^{s-1}u_t)=0. \end{equation} From Lemmas \ref{lem2.2} and \ref{lem3.3}, we know that $u\in L^{\infty}$. From H\"{o} lder inequality, Cauchy inequality, Lemma \ref{lem2.3} and \eqref{e3.2} we obtain \begin{align*} \frac{d}{dt}E_3(t) &\leq 2|r|\|\nabla ^{s-1}u_t\|^2_2+2|(\nabla^{s-1}f(u), \nabla^{s-1}u_t)|\\ &\leq 2|r|\|\nabla ^{s-1}u_t\|^2_2 +2K_1(W)(\|u\|_{\infty})(\|u\|_2 +\|\nabla^{s-1}u\|_2)\|\nabla^{s-1}u_t\|_2\\ &\leq C_4(M_1(t))(\|\nabla^{s-1}u\|^2_2+\|\nabla^{s-1}u_t\|^2_2). \end{align*} Integrating the above inequality with respect to $t$ and using the Gronwall's inequality, we obtain \eqref{e3.9}. The lemma is proved. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] From Theorem \ref{thm1.1}, we need only to show that $$ \sup_{t\in [0, T_0]}(\|u(t)\|_{H^s}+\|u_t(t)\|_{H^{s-2}})<\infty. $$ From Lemmas \ref{lem3.2}--\ref{lem3.4}, we obtain \[ \|u(t)\|_{H^s}+\|u_t(t)\|_{H^{s-2}}0$ and $L_1,L_2$ are constants. If $I(0)>0$, $I'(0)>\gamma_2 \nu^{-1}I(0)$ and $L_1+L_2>0$, then $I(t)$ tends to infinity as \[ t\to t_1\leq t_2=\frac{1}{2\sqrt{L_1^2+\nu L_2}}\ln \frac{\gamma_1I(0)+\nu I'(0)}{\gamma_1I(0)+\nu I'(0)}, \] where $\gamma_{1,2}=-L_1\mp \sqrt{L_1^2+\nu L_2}$. If $I(0)>0$, $I'(0)>0$ and $L_1=L_2=0$, then $I(t)\to \infty$ as $t\to t_1\leq t_2=I(0)/\nu I'(0)$. \end{lemma} \begin{proof}[Proof of Theorem \ref{thm1.3}] Suppose $T=+\infty$, let \begin{equation} \label{e4.1} I(t)=\|(-\Delta)^{-1/2}u\|^2_2+\|u\|^2_2+\beta(t+\tau)^2, \end{equation} where $\beta,\tau\geq 0$ to be defined later. Then \begin{equation} \label{e4.2} I'(t)=2((-\Delta)^{-1/2}u_t,(-\Delta)^{-1/2}u )+2\beta(t+\tau)+2(u,u_t). \end{equation} So, \begin{equation} \label{e4.3} \begin{aligned} (I'(t))^2 &\leq 4[\|(-\Delta)^{-1/2}u\|^2_2+\|u\|^2_2 +\beta(t+\tau)^2][\|(-\Delta)^{-1/2}u_t\|^2_2+\|u_t\|^2_2+\beta] \\ &=4I(t)[\|(-\Delta)^{-1/2}u_t\|^2_2+\|u_t\|^2_2+\beta]. \end{aligned} \end{equation} By \eqref{e1.7}, we obtain \begin{equation} \label{e4.4} \begin{aligned} I''(t) &= 2\|(-\Delta)^{-1/2}u_t\|^2_2+2((-\Delta)^{-1/2}u,(-\Delta)^{-1/2}u_{tt}) +2\|u_t\|^2_2+2(u,u_{tt}) \\ &\quad +2\beta \\ &= 2\|(-\Delta)^{-1/2}u_t\|^2_2+2\|u_t\|^2_2+2\beta+2(u,(-\Delta)^{-1}u_{tt}+u_{tt}) \\ &= 2\|(-\Delta)^{-1/2}u_t\|^2_2+2\|u_t\|^2_2+2\beta -2(u,u-\Delta u+\Delta^2 u+r u_t+f(u) ) \\ &= 2\|(-\Delta)^{-1/2}u_t\|^2_2+2\|u_t\|^2_2+2\beta -2\|u\|^2_2-2\|\nabla u\|^2_2-2\|\Delta u\|^2_2\\ &-2r(u,u_t)-2\int_{\mathbb{R}^n}u f(u)dx. \end{aligned} \end{equation} With the aid of the Cauchy inequality we obtain \begin{equation} \label{e4.5} \begin{aligned} 2r(u,u_t)&\leq r(\|u\|^2_2+\|u_t\|^2_2) \\ &= r[E(0)-\|(-\Delta)^{-1/2}u_t\|^2_2-\|\nabla u\|^2_2-\|\Delta u\|^2_2 \\ &-2r\int^t_0\|u_\tau\|^2_2d\tau-2\int_{\mathbb{R}^n}F(u)dx]. \end{aligned} \end{equation} It follows from \eqref{e4.1}-\eqref{e4.5} that \begin{equation} \label{e4.6} \begin{aligned} &I(t)I''(t)-(1+\frac{\alpha}{4})(I'(t))^2 \\ &\geq I(t)I''(t)-(4+\alpha)I(t)[\|(-\Delta)^{-1/2}u_t\|^2_2 +\|\Delta u\|^2_2+\|u_t\|^2_2+\beta] \\ &\geq I(t)\{2\|(-\Delta)^{-1/2}u_t\|^2_2+2\|u_t\|^2_2 +2\beta-2\|\Delta u\|^2_2-2\|u\|^2_2-2\|\nabla u\|^2_2 \\ &\quad -2r(u,u_t)-2\int_{\mathbb{R}^n}u f(u)dx-(4+\alpha) [\|(-\Delta)^{-1/2}u_t\|^2_2+\|u_t\|^2_2+\beta]\} \\ &\geq I(t)\{(r-\alpha-2)\|(-\Delta)^{-1/2}u_t\|^2_2 +(-2-\alpha)\|u_t\|^2_2+(-4-\alpha)\beta \\ &\quad +(r-2)(\|\nabla u\|^2_2+\|\Delta u\|^2_2) +\int_{\mathbb{R}^n}[2rF(u)-2uf(u)-2u^2]dx \\ &\quad +2r^2\int^t_0\|u_{\tau}\|^2_2d\tau-r E(0)\}. \end{aligned} \end{equation} From \eqref{e3.1}, we obtain \begin{align*} &(r-\alpha-2)\|(-\Delta)^{-1/2}u_t\|^2_2+(-2-\alpha)\|u_t\|^2_2 +(r-2)(\|\nabla u\|^2_2+\|\Delta u\|^2_2) \\ &\geq (-\alpha-2)(\|(-\Delta)^{-1/2}u_t\|^2_2+\|\nabla u\|^2_2 +\|\Delta u\|^2_2+\| u_t\|^2_2)\\ &=(\alpha+2)(\|u\|^2_2+2r\int^t_0\|u_{\tau}\|^2_2d\tau +2\int_{\mathbb{R}^n}F(u)dx-E(0)). \end{align*} Thus, from the above inequality, \eqref{e1.10} and \eqref{e4.6}, we have \begin{equation} \label{e4.7} \begin{aligned} &I(t)I''(t)-(1+\frac{\alpha}{4})(I'(t))^2 \\ &\geq I(t)\Big\{-(4+\alpha)\beta-(2+\alpha+r)E(0) +\int_{\mathbb{R}^n}[2(2+\alpha+r)F(u)\\ &\quad +\alpha u^2-2u f(u)\big]dx +(2r(2+\alpha)+2r^2)\int^t_0\|u_\tau\|^2_2d\tau\Big\} \\ \geq &-\big[(4+\alpha)\beta+(2+\alpha+r)E(0)]I(t). \end{aligned} \end{equation} If $E(0)<0$, taking $\beta=-\frac{2+\alpha+r}{4+\alpha}E(0)>0$, then \[ I(t)I''(t)-(1+\frac{\alpha}{4})(I'(t))^2\geq 0. \] We may choose $\tau$ so large that $I'(t)>0$. From Lemma \ref{lem4.1} we know that $I(t)$ becomes infinite at a time $T_1$ at most equal to \[ T_1=\frac{4I(0)}{\alpha I'(t)}<\infty. \] If $E(0)=0$, taking $\beta=0$, from \eqref{e4.7}, we obtain \[ I(t)I''(t)-(1+\frac{\alpha}{4})(I'(t))^2\geq 0. \] Also $I'(t)>0$ by assumption (ii), Thus, we obtain from Lemma \ref{lem4.1} that $I(t)$ becomes infinite at a time $T_2$ at most equal to \[ T_2=\frac{4I(0)}{\alpha I'(t)}<\infty. \] If $E(0)>0$, then taking $\beta=0$, inequality \eqref{e4.7} becomes \begin{equation} \label{e4.8} I(t)I''(t)-(1+\frac{\alpha}{4})(I'(t))^2\geq -(2+\alpha+r)E(0)I(t). \end{equation} Define $J(t)=(I(t))^{-\lambda}$, where $\lambda=\alpha/4$. Then \begin{gather} J'(t)=-\lambda(I(t))^{-\lambda-1}I'(t), \nonumber\\ \begin{aligned} J''(t)&=-\lambda(I(t))^{-\lambda-2}[I(t)I''(t)-(1+\lambda)(I'(t))^2] \\ &\leq \lambda(2+r+4\lambda)E(0)(I(t))^{-\lambda-1}, \end{aligned} \label{e4.9} \end{gather} where inequality \eqref{e4.8} is used. Assumption (iii) implies $J'(0)<0$. Let \begin{equation} \label{e4.10} t^{*}=\sup\{t|J'(\tau)<0,\tau\in(0,t)\}. \end{equation} By the continuity of $J'(t)$, $t^{*}$ is positive. Multiplying \eqref{e4.9} by $2J'(t)$ yields \begin{equation} \label{e4.11} \begin{aligned}{} [(J'(t))^2]' &\geq -2\lambda^2(2+r+4\lambda)E(0)(I(t))^{-2\lambda-2}I'(t) \\ &=2\lambda^2\frac{2+r+4\lambda}{2\lambda+1}E(0)[I(t)^{-2\lambda-1}]'. \end{aligned} \end{equation} Integrate with respect to $t$ over $[0,t)$ to obtain \begin{align*} (J'(t))^2 &\geq 2\lambda^2\frac{2+r+4\lambda}{2\lambda+1}E(0)(I(t))^{-2\lambda-1}\\ &\quad +(J'(0))^2-2\lambda^2\frac{2+r+4\lambda}{2\lambda+1}E(0)(I(0))^{-2\lambda-1} \\ &\geq(J'(0))^2-2\lambda^2\frac{2+r+4\lambda}{2\lambda+1}E(0)(I(0))^{-2\lambda-1}. \end{align*} From assumption (iii), we obtain \[ (J'(0))^2-2\lambda^2\frac{r+2+4\lambda}{2\lambda+1}E(0)(I(0))^{-2\lambda-1}>0. \] Hence by continuity of $J'(t)$, we have \begin{equation} \label{e4.12} J'(t)\leq-[(J'(0))^2-2\lambda^2 \frac{2+r+4\lambda}{2\lambda+1}E(0)(I(0))^{-2\lambda-1}]^{1/2} \end{equation} for $0\leq t< t^{*}$. By the definition of $t^{*}$, it follows that \eqref{e4.12} holds for all $t\geq 0$. Therefore, \[ J(t)\leq J(0)-[(J'(0))^2-2\lambda^2\frac{2+r+4\lambda}{2\lambda+1} E(0)(I(0))^{-2\lambda-1}]^{1/2}t,\quad \forall t>0. \] So $J(T_1)=0$ for some $T_1$ and \[ 00$, we obtain \[ e^{kt}E(t)\leq2C_0E(0)+(2C_2k^2+2C_3k)\int^t_0e^{k\tau}E(\tau)d\tau, \] which together with the Gronwall inequality gives \begin{gather*} e^{kt}E(t)\leq2C_0E(0)e^{2C_2k^2t+2C_3kt}, \quad 0\leq t<\infty, \\ E(t)\leq 2C_0E(0)e^{-(k-2C_2k^2-2C_3k)t}, \quad 0\leq t\leq\infty. \end{gather*} Again taking $k$ satisfying $00$. The proof is complete. \end{proof} \subsection*{Acknowledgments} This work is partially supported by the Fundamental Research Funds for the Central Universities grant ZYGX2015J096. The author is very grateful to the referees for their helpful suggestions and comments. \begin{thebibliography}{00} \bibitem{A} E. Arevalo, Yu. Gaididei, F. G. 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