\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 167, pp. 1--26.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/167\hfil Kirchhoff-Love equation] {Existence, blow-up and exponential decay for Kirchhoff-Love equations with Dirichlet conditions} \author[N. A. Triet, V. T. T. Mai, L. T. P. Ngoc, N. T. Long \hfil EJDE-2018/167\hfilneg] {Nguyen Anh Triet, Vo Thi Tuyet Mai \\ Le Thi Phuong Ngoc, Nguyen Thanh Long} \address{Nguyen Anh Triet \newline Department of Mathematics, University of Architecture of Ho Chi Minh City, 196 Pasteur Str., Dist. 3, Ho Chi Minh City, Vietnam} \email{anhtriet1@gmail.com} \address{Vo Thi Tuyet Mai \newline University of Natural Resources and Environment of Ho Chi Minh City, 236B Le Van Sy Str., Ward 1, Tan Binh Dist., Ho Chi Minh City, Vietnam.\newline Department of Mathematics and Computer Science, VNUHCM - University of Science, 227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam} \email{vttmai@hcmunre.edu.vn} \address{Le Thi Phuong Ngoc \newline University of Khanh Hoa, 01 Nguyen Chanh Str., Nha Trang City, Vietnam} \email{ngoc1966@gmail.com} \address{Nguyen Thanh Long \newline Department of Mathematics and Computer Science, VNUHCM - University of Science, 227 Nguyen Van Cu Str., Dist. 5, Ho Chi Minh City, Vietnam} \email{longnt2@gmail.com} \dedicatory{Communicated by Dung Le} \thanks{Submitted May 21, 2018. Published October 4, 2018.} \subjclass[2010]{35L20, 35L70, 35Q74, 37B25} \keywords{Nonlinear Kirchhoff-Love equation; blow-up; exponential decay} \begin{abstract} The article concerns the initial boundary value problem for a nonlinear Kirchhoff-Love equation. First, by applying the Faedo-Galerkin, we prove existence and uniqueness of a solution. Next, by constructing Lyapunov functional, we prove a blow-up of the solution with a negative initial energy, and establish a sufficient condition for the exponential decay of weak solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In this article, we consider the initial boundary value problem with homogeneous Dirichlet boundary conditions \begin{gather} \begin{aligned} &u_{tt}-\frac{\partial }{\partial x}\big[ B\big( x,t,u,\| u\|^2,\| u_{x}\|^2,\| u_{t}\|^2,\| u_{xt}\|^2\big) \big(u_{x}+\lambda _1u_{xt}+u_{xtt}\big) \big] +\lambda u_{t} \\ &=F\big( x,t,u,u_{x},u_{t},u_{xt},\| u(t)\| ^2,\| u_{x}(t)\|^2,\| u_{t}(t)\| ^2,\| u_{xt}(t)\|^2\big) \\ &\quad-\frac{\partial }{\partial x}\big[ G\big( x,t,u,u_{x},u_{t},u_{xt},\| u(t)\|^2,\| u_{x}(t)\|^2,\| u_{t}(t)\|^2,\| u_{xt}(t)\|^2\big) \big] \\ &\quad +f(x,t),\quad x\in \Omega =(0,1),\; 00$, $\lambda _1>0$ are constants and $\tilde{u}_0, \tilde{u}_1\in H_0^1\cap H^2;$ $f$, $F$ and $G$ are given functions that assumptions stated later. This problem has the so called model of Kirchhoff-Love type because it connects Kirchhoff and Love equation, this type is also introduced in \cite{Tri}. More precisely \eqref{1} has its origin in the nonlinear vibration of an elastic string (Kirchhoff \cite{11m}), for which the associated equation is \begin{equation} \rho hu_{tt}=\Big( P_0+\frac{Eh}{2L}\int_0^{L}| \frac{ \partial u}{\partial y}(y,t)|^2dy\Big) u_{xx}, \label{4} \end{equation} here $u$ is the lateral deflection, $L$ is the length of the string, $h$ is the cross sectional area, $E$ is Young's modulus, $\rho $ is the mass density, and $P_0$ is the initial tension. On the other hand, \eqref{1} arises from the Love equation \begin{equation} u_{tt}-\frac{E}{\rho }u_{xx}-2\mu^2\omega^2u_{xxtt}=0, \label{5} \end{equation} presented by Radochov\'a \cite{Rad}. This equation describes the vertical oscillations of a rod, which was established from Euler's variational equation of an energy functional \begin{equation} \int_0^{T} \int_0^{L}\big[ \frac{1}{2}F\rho ( u_{t}^2+\mu^2\omega^2u_{tx}^2) -\frac{1}{2}F( Eu_{x}^2+\rho \mu^2\omega^2u_{x}u_{xtt}) \big]\, dx\,dt\,, \label{6} \end{equation} where $u$ is the displacement, $L$ is the length of the rod, $F$ is the area of cross-section, $\omega $ is the cross-section radius, $E$ is the Young modulus of the material and $\rho $ is the mass density. It is well known that the existence, global existence, decay properties and blow-up of solutions to the initial boundary value problem for Kirchhoff type models under different types of hypotheses in have been extensively studied by many authors, for example, we refer to \cite{TK2,TK1,TK6,TK5,TK4,TK7,TK3}, and references therein. In \cite{TK1}, the authors studied the existence of global solutions and exponential decay for a Kirchhoff-Carrier model with viscosity. In \cite{TK4}, the authors discussed the global well-posedness and uniform exponential stability for the Kirchhoff equation in $\mathbb{R}^{n}$. Here, the global solvability is proved when the initial data is taken small enough and the exponential decay of the energy is obtained in the strong topology $H^2(\mathbb{R}^{n})\times H^2(\mathbb{R}^{n})$. In \cite{TK5}, the author investigated the global existence, decay properties, and blow-up of solutions to the initial boundary value problem for the nonlinear Kirchhoff type. In \cite{TK7}, the viscoelastic equation of Kirchhoff type was considered and the authors established a new blow-up result for arbitrary positive initial energy, by using simple analysis techniques. The purpose of this paper is establishing the existence, blow up and exponential decay of weak solutions for\eqref{1}--\eqref{3}. To our knowledge, there is no decay or blow up result for equations of this type. However, the existence and exponential decay of solutions or blow up results for Love equation were studied in \cite{NL4}. Here, by combining the linearization method for the nonlinear term, the Faedo-Galerkin method and the weak compactness method, the existence of a unique weak solution of a Dirichlet problem for the nonlinear Love equation \begin{equation} \label{7} \begin{aligned} &u_{tt}-u_{xx}-u_{xxtt}-\lambda _1u_{xxt}+\lambda u_{t}\\ &= F(x,t,u,u_{x},u_{t},u_{xt}) -\frac{\partial }{\partial x}[ G(x,t,u,u_{x},u_{t},u_{xt})] +f(x,t), \end{aligned} \end{equation} for $0 < x<1$ and $t>0$, has been proved. When $F=F(u)=a| u|^{p-2}u$, $G=G(u_{x})=b| u_{x}|^{p-2}u_{x}$, $a$, $b\in\mathbb{R}$, $p>2$, the blow up and exponential decay of solutions were established. For details, in case of $a>0$, $b>0;$ $f(x,t)\equiv 0$, with negative initial energy, the solution of \eqref{7} blows up in finite time. In case of $a>0$, $b<0$, if $\| \tilde{u}_{0x}\|^2-a\| \tilde{u}_0\| _{L^p}^p>0$ and $f\in L^2((0,1)\times \mathbb{R}_{+})$, $\| f(t)\| \leq Ce^{-\gamma _0t}$, such that $f(t)$ decays exponentially\ as $t\to +\infty $, the energy of the solution decays exponentially\ as $t\to +\infty $. Finally, in case of $a<0$, $b<0$ and $\| f(t)\| $ is small enough, \eqref{7} has a unique global solution with energy decaying exponentially as $t\to +\infty $, without the initial data $(\tilde{u}_0,\tilde{u}_1) $ small enough. Our model was inspired in the above mentioned works and motivated by the results in \cite{NL4}, we study the existence, blow-up and exponential decay estimates for \eqref{1}--\eqref{3}. This article is organized as follows. Section 2 is devoted to preliminaries and an existence result for \eqref{1}--\eqref{3} in case $F$, $G\in C^1([0,1]\times [ 0,T]\times\mathbb{R}^4\times\mathbb{R}_{+}^4)$; $B\in C^1([0,1]\times [ 0,T]\times\mathbb{R}\times\mathbb{R}_{+}^4)$ with $B(x,t,y,z)\geq b_0>0$, $\forall (x,t)\in [0,1]\times [ 0,T]$, for all $y\in\mathbb{R}$, for all $z\in\mathbb{R}_{+}^4$. Since $f$, $G$, $B$ are arbitrary, we need to combine the linearization method, the Faedo-Galerkin method and the weak compactness method. In Sections 3, 4, Problem \eqref{1}--\eqref{3} is considered in the case $B=B(x,t)$ and $F=F(u,u_{x})$, $G=G(u,u_{x})$ such that $(F,G)=(\frac{\partial \mathcal{F} }{\partial u}$, $\frac{\partial \mathcal{F} }{\partial v})$. More details, in Section 3, with $f(x,t)\equiv 0$ and a negative initial energy,\ we prove that the solution of \eqref{1}--\eqref{3} blows up in finite time. In Section 4, we give a sufficient condition, in which the initial energy is positive and small, to guarantee the global existence and exponential decay of weak solutions. In the proof, a suitable Lyapunov functional is constructed. The results obtained here may be considered as the generalizations of those in \cite{L5,NL4,Tri}, based on the main tool in \cite{Tri} and the techniques in \cite{NL4}. \section{Existence of a weak solution} First, we set the preliminary as follows. Let $\langle \cdot ,\cdot \rangle $ be either the scalar product in $L^2$ or the dual pairing of a continuous linear functional and an element of a function space, $\| \cdot \| $ be the norm in $L^2$ and $\| \cdot \| _{X}$ be the norm in the Banach space $X$. Let $L^p(0,T;X)$, $1\leq p\leq \infty $ be the Banach space of the real functions $u:(0,T)\to X$ measurable, with \begin{equation*} \| u\| _{L^p(0,T;X)} =\Big(\int_0^{T}\| u(t)\| _{X}^pdt\Big) ^{1/p}<\infty \quad \text{for }1\leq p<\infty , \end{equation*} and \begin{equation*} \| u\| _{L^{\infty }(0,T;X)}=\operatorname{ess\,sup}_{00$. Without loss of generality, by the fact that $F$ contains the variable $u_{t}$ and $\lambda $ is arbitrary, we can suppose that $\lambda =0$. The weak formulation of \eqref{1}--\eqref{3} can be given in as follows: Find $u\in \widetilde{W}=\{u\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2) :u'$, $u''\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2) \}$, such that $u$ satisfies the variational equation \begin{equation} \begin{aligned} &\langle u''(t),w\rangle +\langle B[u](t)( u_{x}(t)+\lambda _1u_{x}'(t)+u_{x}''(t)) ,w_{x}\rangle \\ &=\langle f(t),w\rangle +\langle F[u](t),w\rangle +\langle G[u](t),w_{x}\rangle , \end{aligned} \label{b3} \end{equation} for all $w\in H_0^1$, a.e., $t\in (0,T)$, with the initial conditions \begin{equation} u(0)=\tilde{u}_0,\quad u_{t}(0)=\tilde{u}_1, \label{b4} \end{equation} where \begin{equation} \label{b5} \begin{gathered} B[u](x,t) = B\big( x,t,u(x,t),\| u(t)\|^2,\| u_{x}(t)\|^2,\| u'(t)\| ^2,\| u_{x}'(t)\|^2\big) , \\ \begin{aligned} F[u](x,t)& = F\Big( x,t,u(x,t),u_{x}(x,t),u'(x,t),u_{x}'(x,t), \| u(t)\|^2,\| u_{x}(t)\|^2, \\ &\quad \| u'(t)\|^2,\| u_{x}'(t)\|^2\Big) , \end{aligned} \\ \begin{aligned} G[u](x,t) &= G\Big( x,t,u(x,t),u_{x}(x,t),u'(x,t),u_{x}'(x,t), \| u(t)\|^2,\| u_{x}(t)\|^2, \\ &\quad \| u'(t)\|^2,\| u_{x}'(t)\|^2\Big) . \end{aligned} \end{gathered} \end{equation} We use the following assumptions: \begin{itemize} \item[(H1)] $\tilde{u}_0, \tilde{u}_1\in H_0^1\cap H^2$; \item[(H2)] $f, f'\in L^2(Q_{T})$, $Q_{T}=(0,1)\times (0,T)$; \item[(H3)] $B\in C^1([0,1]\times [ 0,T]\times\mathbb{R} \times\mathbb{R}_{+}^4)$ and there exists a constant $b_0>0$ such that $B(x,t,y,z)\geq b_0$, for all $(x,t)\in [ 0,1]\times [ 0,T]$, for all $y\in\mathbb{R}$, for all $z\in\mathbb{R}_{+}^4$; \item[(H4)] $F\in C^1([0,1]\times [ 0,T]\times\mathbb{R}^4 \times\mathbb{R}_{+}^4)$; \item[(H5)] $G\in C^1([0,1]\times [ 0,T]\times\mathbb{R}^4 \times\mathbb{R}_{+}^4)$. \end{itemize} \begin{theorem} \label{thm2.2} Let {\rm(H1)--(H5)} hold. Then Problem \eqref{1}--\eqref{3} has a unique local solution $u$ and \begin{equation} \begin{gathered} u\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2) ,\quad u'\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2) ,\\ \text{ }u''\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2), \end{gathered} \label{b6} \end{equation} for $T_{\ast }>0$ small enough. \end{theorem} \begin{remark} \label{rmk2.1} \rm Thanks to the regularity obtained by \eqref{b6}, Problem \eqref{1}--\eqref{3} has a unique strong solution \begin{equation} u\in C^1( [0,T_{\ast }];H_0^1\cap H^2) ,\quad u''\in L^{\infty }( 0,T_{\ast };H_0^1\cap H^2) . \label{b7} \end{equation} \end{remark} \begin{proof}[Proof of Theorem \ref{thm2.2}] We have two steps. Using linearization, step 1 constructs a linear recurrent sequence $\{u_{m}\}$. Step 2 shows that $\{u_{m}\}$ converges to $u$ and $u$ is exactly a unique local solution of \eqref{1}--\eqref{3}. \smallskip \noindent\textbf{Step 1.} Consider $T>0$ fixed, let $M>0$, we put \begin{equation} \label{b8} K_M(f) = ( \| f\| _{L^2(Q_{T})}^2+\|f'\| _{L^2(Q_{T})}^2)^{1/2}, \end{equation} \[ \| B\| _{C^0(\tilde{A}_M)} = \sup_{(x,t,y,z_1,\dots,z_4) \in \tilde{A}_M} | B( x,t,y,z_1,\dots,z_4) | , \] with \begin{gather*} \tilde{A}_M = [0,1]\times [ 0,T]\times [-M,M]\times [ 0,M^2]^4, \\ \bar{B}_M = \| B\| _{C^1(\tilde{A}_M)}=\| B\| _{C^0(\tilde{A}_M)}+\sum_{i=1}^7\| D_iB\| _{C^0(\tilde{A}_M)}, \end{gather*} \[ \| F\| _{C^0(A_M)} = \sup_x,t,y_1,\dots,y_4,z_1,\dots,z_4) \in A_M |F( x,t,y_1,\dots,y_4,z_1,\dots,z_4) | , \] with \begin{gather*} A_M = [0,1]\times [ 0,T]\times [ -M,M]^4\times [ 0,M^2]^4, \\ \bar{F}_M = \| F\| _{C^1(A_M)} =\| F\| _{C^0(A_M)}+\sum_{i=1}^{10}\|D_iF\| _{C^0(A_M)}, \\ \bar{G}_M = \| G\| _{C^1(A_M)} =\|G\| _{C^0(A_M)}+\sum_{i=1}^{10}\| D_iG\| _{C^0(A_M)}. \end{gather*} For each $T_{\ast }\in (0,T]$ and $M>0$, we put \begin{equation} \begin{gathered} \begin{aligned} W(M,T_{\ast })=\Big\{ & v\in L^{\infty }(0,T_{\ast };H_0^1\cap H^2):v'\in L^{\infty }(0,T_{\ast };H_0^1\cap H^2),\\ &v''\in L^{\infty }(0,T_{\ast };H_0^1), \text{ with } \| v\| _{L^{\infty }(0,T_{\ast };H_0^1\cap H^2)},\\ &\| v'\| _{L^{\infty }(0,T_{\ast };H_0^1\cap H^2)},\; \| v''\| _{L^{\infty}(0,T_{\ast };H_0^1)}\leq M\Big\}, \end{aligned} \\ W_1(M,T_{\ast })=\{v\in W(M,T_{\ast }):v''\in L^{\infty }(0,T_{\ast };H_0^1\cap H^2)\}, \end{gathered} \label{b9} \end{equation} where $Q_{T_{\ast }}=\Omega \times (0,T_{\ast })$. We establish the linear recurrent sequence $\{u_{m}\}$ as follows. We choose the first term $u_0\equiv 0$, suppose that \begin{equation} u_{m-1}\in W_1(M,T_{\ast }), \label{b10} \end{equation} and associate with problem \eqref{1}--\eqref{3} the following problem. Find $u_{m}\in W_1(M,T_{\ast })$ $(m\geq 1)$ which satisfies the linear variational problem \begin{equation} \begin{gathered} \begin{aligned} &\langle u_{m}''(t),w\rangle +\langle B_{m}(t)( u_{mx}(t)+\lambda _1u_{mx}'(t)+u_{mx}''(t)) ,w_{x}\rangle \\ &=\langle f(t),w\rangle +\langle F_{m}(t),w\rangle +\langle G_{m}(t),w_{x}\rangle , \quad \forall w\in H_0^1, \end{aligned} \\ u_{m}(0)=\tilde{u}_0,\quad u_{m}'(0)=\tilde{u}_1, \end{gathered} \label{b11} \end{equation} where \begin{gather} \begin{aligned} &B_{m}(x,t) = B[u_{m-1}](x,t) \\ &= B\Big( x,t,u_{m-1}(x,t),\| u_{m-1}(t)\|^2,\| \nabla u_{m-1}(t)\|^2,\| u_{m-1}'(t)\| ^2,\| \nabla u_{m-1}'(t)\|^2\Big) , \end{aligned} \nonumber \\ \begin{aligned} F_{m}(x,t) &= F[u_{m-1}](x,t) \\ &= F\Big( x,t,u_{m-1}(x,t),\nabla u_{m-1}(x,t),u_{m-1}'(x,t), \nabla u_{m-1}'(x,t), \\ &\quad \| u_{m-1}(t)\|^2,\| \nabla u_{m-1}(t)\|^2, \| u_{m-1}'(t)\|^2,\| \nabla u_{m-1}'(t)\|^2\Big) , \end{aligned} \label{b12}\\ \begin{aligned} G_{m}(x,t) &= G[u_{m-1}](x,t) \\ &= G\Big( x,t,u_{m-1}(x,t),\nabla u_{m-1}(x,t),u_{m-1}'(x,t), \nabla u_{m-1}'(x,t), \\ &\quad \| u_{m-1}(t)\|^2,\| \nabla u_{m-1}(t)\|^2,\| u_{m-1}'(t)\|^2, \| \nabla u_{m-1}'(t)\|^2\Big) . \end{aligned} \nonumber \end{gather} \begin{lemma} \label{lem2.3} Let {\rm (H1)--(H5)} hold. Then there exist positive constants $M$, $T_{\ast }>0$ such that, for $u_0\equiv 0$, there exists a recurrent sequence $\{u_{m}\}\subset W_1(M,T_{\ast })$ defined by \eqref{b10}--\eqref{b12}. \end{lemma} \begin{proof} The proof consists of several steps. \smallskip \noindent\textbf{(i)} The Faedo-Galerkin approximation (introduced by Lions \cite{Lio3}). Consider a special orthonormal basis $\{w_{j}\}$ on $H_0^1:w_{j}(x)=\sqrt{2}\sin (j\pi x)$, $j\in\mathbb{N}$, formed by the eigenfunctions of the Laplacian $-\Delta =-\frac{\partial ^2}{\partial x^2}$. It is clear to see that there exists $c_{mj}^{(k)}(t)$, $1\leq j\leq k$, on interval $[0,T]$ such that if we have expression in form \begin{equation} u_{m}^{(k)}(t)=\sum_{j=1}^{k}c_{mj}^{(k)}(t)w_{j}, \label{b13} \end{equation} then $u_{m}^{(k)}(t)\ $satisfies \begin{equation} \begin{gathered} \begin{aligned} &\langle \ddot{u}_{m}^{(k)}(t),w_{j}\rangle +\langle B_{m}(t)\big( u_{mx}^{(k)}(t)+\lambda _1\dot{u}_{mx}^{(k)}(t)+\ddot{u} _{mx}^{(k)}(t)\big) ,w_{jx}\rangle \\ &=\langle f(t),w_{j}\rangle +\langle F_{m}(t),w_{j}\rangle +\langle G_{m}(t),w_{jx}\rangle ,\quad 1\leq j\leq k, \end{aligned}\\ u_{m}^{(k)}(0)=\tilde{u}_{0k}, \quad \dot{u}_{m}^{(k)}(0)=\tilde{u}_{1k}, \end{gathered} \label{b14} \end{equation} in which \begin{equation} \begin{gathered} \tilde{u}_{0k}=\sum_{j=1}^{k}\alpha _{j}^{(k)}w_{j}\to \tilde{u}_0\quad\text{strongly in } H_0^1\cap H^2,\\ \tilde{u}_{1k}=\sum_{j=1}^{k}\beta _{j}^{(k)}w_{j}\to \tilde{u}_1\quad\text{strongly in } H_0^1\cap H^2. \end{gathered} \label{b15} \end{equation} Indeed, \eqref{b14} leads to an equivalent form of system \eqref{b14} as follows \begin{equation} \begin{gathered} \ddot{c}_{mi}^{(k)}(t)+\sum_{j=1}^{k}b_{ij}^{(m)}(t)( \ddot{c} _{mj}^{(k)}(t)+\lambda _1\dot{c}_{mj}^{(k)}(t)+c_{mj}^{(k)}(t)) =f_{mi}(t),\\ c_{mi}^{(k)}(0)=\alpha _i^{(k)}, \quad \dot{c}_{mi}^{(k)}(0)=\beta _i^{(k)}, \quad 1\leq i\leq k, \end{gathered} \label{b16} \end{equation} where \begin{equation} \begin{gathered} f_{mj}(t)=\langle f(t),w_{j}\rangle +\langle F_{m}(t),w_{j}\rangle +\langle G_{m}(t),w_{jx}\rangle , \\ b_{ij}^{(m)}(t)=\langle B_{m}(t)w_{ix},w_{jx}\rangle ,\quad 1\leq i, \; j\leq k. \end{gathered} \label{b17} \end{equation} System \eqref{b16}, \eqref{b17} has a unique solution $c_{mj}^{(k)}(t)$, $1\leq j\leq k$ on interval $[0,T]$, the proof is obtained through \eqref{b10} and normal argument (see \cite{Cod}). \smallskip \noindent\textbf{(ii)} A priori estimates. We shall give a priori estimates to show that there exist positive constants $M$, $T_{\ast}>0\ $such that $u_{m}^{(k)}\in W(M,T_{\ast })$, for all $m$ and $k$. Put \begin{equation} \label{b18} \begin{aligned} S_{m}^{(k)}(t) &= \| \sqrt{B_{m}(t)}u_{mx}^{(k)}(t)\|^2+\| \sqrt{B_{m}(t)}\Delta u_{m}^{(k)}(t)\|^2\\ &\quad +\| \dot{u}_{m}^{(k)}(t)\|^2+\| \dot{u} _{mx}^{(k)}(t)\|^2 \\ &\quad +2\| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\Delta \dot{u}_{m}^{(k)}(t)\|^2 \\ &\quad +\| \ddot{u}_{m}^{(k)}(t)\|^2+\| \sqrt{B_{m}(t)}\ddot{u}_{mx}^{(k)}(t)\|^2 \\ &\quad +2\lambda _1\int_0^{t}\Big[ \| \sqrt{B_{m}(s)}\dot{u}_{mx}^{(k)}(s)\|^2 +\| \sqrt{B_{m}(s)}\Delta \dot{u}_{m}^{(k)}(s)\|^2 \\ &\quad +\| \sqrt{B_{m}(s)}\ddot{u}_{mx}^{(k)}(s)\|^2\Big] \,ds. \end{aligned} \end{equation} It follows from \eqref{b14} and \eqref{b18} that \begin{equation} \label{b19} \begin{aligned} &S_{m}^{(k)}(t)\\ &= S_{m}^{(k)}(0) +2\int_0^{t}\langle f(s),\dot{u}_{m}^{(k)}(s)\rangle ds -2\int_0^{t}\langle f(s),\Delta \dot{u}_{m}^{(k)}(s)\rangle ds \\ &\quad +2\int_0^{t}\langle f'(s),\ddot{u}_{m}^{(k)}(s)\rangle\,ds +2\int_0^{t}\langle F_{m}(s),\dot{u}_{m}^{(k)}(s)\rangle\,ds \\ &\quad -2\int_0^{t}\langle F_{m}(s),\Delta \dot{u}_{m}^{(k)}(s)\rangle\,ds +2\int_0^{t}\langle G_{m}(s),\dot{u}_{mx}^{(k)}(s)\rangle\, ds \\ &\quad +2\int_0^{t}\langle F_{m}'(s),\ddot{u}_{m}^{(k)}(s)\rangle ds +2\int_0^{t}\langle G_{m}'(s),\ddot{u}_{mx}^{(k)}(s)\rangle \,ds\\ &\quad +2\int_0^{t}\langle G_{mx}(s),\triangle \dot{u}_{m}^{(k)}(s)\rangle ds +\int_0^{t}ds\int_0^1B_{m}'(x,s) \Big[| u_{mx}^{(k)}(x,s)|^2+| \Delta u_{m}^{(k)}(x,s)|^2 \\ &\quad +2| \dot{u}_{mx}^{(k)}(x,s)|^2 +| \Delta \dot{u}_{m}^{(k)}(x,s)|^2-| \ddot{u}_{mx}^{(k)}(x,s)|^2\Big]\, dx \\ &\quad -2\int_0^{t}\langle B_{m}'(s)( u_{mx}^{(k)}(s)+\lambda _1\dot{u}_{mx}^{(k)}(s)) ,\ddot{u} _{mx}^{(k)}(s)\rangle ds \\ &\quad -2\int_0^{t}\langle B_{mx}(s)( u_{mx}^{(k)}(s)+\lambda _1 \dot{u}_{mx}^{(k)}(s)+\ddot{u}_{mx}^{(k)}(s)) ,\Delta \dot{u} _{m}^{(k)}(s)\rangle ds \\ &= S_{m}^{(k)}(0)+\sum_{j=1}^{12}I_{j}. \end{aligned} \end{equation} First, we need to estimate $\xi _{m}^{(k)}=\| \ddot{u} _{m}^{(k)}(0)\|^2+\| \sqrt{B_{m}(0)}\ddot{u} _{mx}^{(k)}(0)\|^2$. Letting $t\to 0_{+}$ in \eqref{b14}$_1$, multiplying the result by $\ddot{c}_{mj}^{(k)}(0)$, it gives \begin{align*} % \label{b20} &\| \ddot{u}_{m}^{(k)}(0)\|^2+\| \sqrt{B_{m}(0)}\ddot{u}_{mx}^{(k)}(0)\|^2 \\ &\quad +\langle B_{m}(0)( \lambda _1\tilde{u}_{1kx}+\tilde{u} _{0kx}) ,\ddot{u}_{mx}^{(k)}(0)\rangle \\ &= \langle f(0),\ddot{u}_{m}^{(k)}(0)\rangle +\langle F_{m}(0),\ddot{u}_{m}^{(k)}(0)\rangle +\langle G_{m}(0),\ddot{u} _{mx}^{(k)}(0)\rangle . \end{align*} Then \begin{align*} %\label{b21} \xi _{m}^{(k)} &= \| \ddot{u}_{m}^{(k)}(0)\|^2+\| \sqrt{B_{m}(0)}\ddot{u}_{mx}^{(k)}(0)\|^2 \\ &\leq\big[ \lambda _1\| \sqrt{B_{m}(0)}\tilde{u}_{1kx}\| +\| \sqrt{B_{m}(0)}\tilde{u}_{0kx}\|\big] \| \sqrt{B_{m}(0)}\ddot{u}_{mx}^{(k)}(0)\| \\ &\quad +[ \| f(0)\| +\| F_{m}(0)\| \text{ } ] \| \ddot{u}_{m}^{(k)}(0)\| +\| G_{m}(0)\| \| \ddot{u}_{mx}^{(k)}(0)\| \\ &\leq[ \lambda _1\| \sqrt{B_{m}(0)}\tilde{u} _{1kx}\| +\| \sqrt{B_{m}(0)}\tilde{u}_{0kx}\| \text{ }] \sqrt{\xi _{m}^{(k)}} \\ &\quad +[ \| f(0)\| +\| F_{m}(0)\| ] \sqrt{\xi _{m}^{(k)}} +\| G_{m}(0)\| \sqrt{\frac{\xi _{m}^{(k)}}{b_0}} \\ &\leq[ \lambda _1\| \sqrt{B_{m}(0)}\tilde{u} _{1kx}\| +\| \sqrt{B_{m}(0)}\tilde{u}_{0kx}\| +\| f(0)\| +\| F_{m}(0)\| +\frac{1}{\sqrt{b_0}}\| G_{m}(0)\| ]^2. \end{align*} On the other hand, $B_{m}(x,0)=B( x,0,\tilde{u}_0,\| \tilde{u}_0\|^2,\| \tilde{u} _{0x}\|^2,\| \tilde{u}_1\|^2,\|\tilde{u}_{1x}\|^2) $ is independent of $m$ and the constant $\| F_{m}(0)\| +\|G_{m}(0)\|/\sqrt{b_0}$ is also independent of $m$, because \begin{align*} &\| F_{m}(0)\| +\frac{\| G_{m}(0)\| }{\sqrt{b_0}} \\ &= \| F\big( \cdot ,0,\tilde{u}_0,\tilde{u}_{0x}, \tilde{u}_1,\tilde{u}_{1x},\| \tilde{u}_0\| ^2,\| \tilde{u}_{0x}\|^2,\| \tilde{u} _1\|^2,\| \tilde{u}_{1x}\|^2\big)\| \\ &\quad +\frac{1}{\sqrt{b_0}}\| G\big( \cdot ,0,\tilde{u}_0,\tilde{u} _{0x},\tilde{u}_1,\tilde{u}_{1x},\| \tilde{u}_0\| ^2,\| \tilde{u}_{0x}\|^2,\| \tilde{u} _1\|^2,\| \tilde{u}_{1x}\|^2\big)\| . \end{align*} Therefore, \begin{equation} \xi _{m}^{(k)}\leq \bar{S}_0,\quad \text{for all }m,\; k, \label{b22} \end{equation} where $\bar{S}_0$ is a constant depending only on $f$, $\tilde{u}_0$, $\tilde{u}_1$, $B$, $F$, $G$ and $\lambda _1$. Equations \eqref{b15}, \eqref{b18} and \eqref{b22} imply that \begin{align*} \label{b23} S_{m}^{(k)}(0) &= \| \sqrt{B_{m}(0)}\tilde{u}_{0kx}\|^2 +\| \sqrt{B_{m}(0)}\Delta \tilde{u}_{0k}\|^2 +\| \tilde{u}_{1k}\|^2+\| \tilde{u}_{1kx}\|^2 \\ &\quad +2\| \sqrt{B_{m}(0)}\tilde{u}_{1kx}\|^2+\| \sqrt{B_{m}(0)}\Delta \tilde{u}_{1k}\|^2+\xi _{m}^{(k)} \\ &\leq S_0,\quad \text{for all }m,\; k\in\mathbb{N}, \end{align*} where $S_0$ is also a constant depending only on $f$, $\tilde{u}_0$, $\tilde{u}_1$, $B$, $F$, $G$ and $\lambda _1$. We estimate the terms $I_{j}$ of \eqref{b19}. By the Cauchy - Schwartz inequality, we obtain \begin{gather*} %\label{b24} I_1 = 2\int_0^{t}\langle f(s),\dot{u}_{m}^{(k)}(s)\rangle\, ds \leq \| f\|_{L^2(Q_{T})}^2+\int_0^{t}\| \dot{u}_{m}^{(k)}(s)\|^2ds; \\ I_2 = -2\int_0^{t}\langle f(s),\Delta \dot{u}_{m}^{(k)}(s)\rangle ds \leq \| f\|_{L^2(Q_{T})}^2+\int_0^{t}\| \Delta \dot{u}_{m}^{(k)}(s)\|^2ds; \\ I_3 = 2\int_0^{t}\langle f'(s),\ddot{u}_{m}^{(k)}(s)\rangle ds \leq \| f'\|_{L^2(Q_{T})}^2+\int_0^{t}\| \ddot{u}_{m}^{(k)}(s)\|^2ds. \end{gather*} Note that \begin{align*} S_{m}^{(k)}(t) &\geq \| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\Delta \dot{u}_{m}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\ddot{u}_{mx}^{(k)}(t)\|^2 \\ &\geq b_0\big( \| \dot{u}_{mx}^{(k)}(t)\| ^2+\| \Delta \dot{u}_{m}^{(k)}(t)\|^2+\| \ddot{ u}_{mx}^{(k)}(t)\|^2\big) , \end{align*} so \begin{equation} I_1+I_2+I_3\leq 2K_M^2(f)+\frac{1}{b_0}\int_0^{t}S_{m}^{(k)}(s)ds. \label{b25} \end{equation} Because \begin{equation} | F_{m}(x,t)| \leq \bar{F}_M,\text{ \ }| G_{m}(x,t)| \leq \bar{G}_M, \label{b26} \end{equation} we have \begin{gather*} %\label{b27} I_4 = 2\int_0^{t}\langle F_{m}(s),\dot{u}_{m}^{(k)}(s)\rangle ds \leq T_{\ast }\bar{F}_M^2+\int_0^{t}\| \dot{u}_{m}^{(k)}(s)\|^2ds; \\ I_{5} = -2\int_0^{t}\langle F_{m}(s),\Delta \dot{u} _{m}^{(k)}(s)\rangle ds\leq T_{\ast }\bar{F}_M^2+\int_0^{t} \| \Delta \dot{u}_{m}^{(k)}(s)\|^2ds; \\ I_{6} = 2\int_0^{t}\langle G_{m}(s),\dot{u}_{mx}^{(k)}(s)\rangle ds\leq T_{\ast }\bar{G}_M^2+\int_0^{t}\| \dot{u} _{mx}^{(k)}(s)\|^2ds. \end{gather*} By \begin{align*} S_{m}^{(k)}(t) &\geq 2\| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\Delta \dot{u}_{m}^{(k)}(t)\|^2 \\ &\geq b_0( \| \dot{u}_{m}^{(k)}(t)\| ^2+\| \dot{u}_{mx}^{(k)}(t)\|^2+\| \Delta \dot{u}_{m}^{(k)}(t)\|^2) , \end{align*} we have \begin{equation} I_4+I_{5}+I_{6}\leq 2T_{\ast }( \bar{F}_M^2+\bar{G} _M^2) +\frac{1}{b_0}\int_0^{t}S_{m}^{(k)}(s)ds. \label{b28} \end{equation} We remark that \begin{align*} \label{b29} F_{m}'(t) &= D_2F[u_{m-1}]+D_3F[u_{m-1}]u_{m-1}'+D_4F[u_{m-1}]\nabla u_{m-1}' \\ &\quad +D_{5}F[u_{m-1}]u_{m-1}''+D_{6}F[u_{m-1}]\nabla u_{m-1}'' \\ &\quad +2D_{7}F[u_{m-1}]\langle u_{m-1}(t),u_{m-1}'(t)\rangle +2D_{8}F[u_{m-1}]\langle \nabla u_{m-1}(t),\nabla u_{m-1}'(t)\rangle \\ &\quad +2D_{9}F[u_{m-1}]\langle u_{m-1}'(t),u_{m-1}''(t)\rangle +2D_{10}F[u_{m-1}] \langle \nabla u_{m-1}'(t),\nabla u_{m-1}''(t)\rangle \end{align*} yields \begin{equation} \| F_{m}'(t)\| \leq ( 1+4M+8M^2) \bar{ F}_M\equiv \tilde{F}_M. \label{b30} \end{equation} Thus \begin{equation} I_{7}=2\int_0^{t}\langle F_{m}'(s),\ddot{u} _{m}^{(k)}(s)\rangle ds\leq T_{\ast }\tilde{F}_M^2+\int_0^{t} \| \ddot{u}_{m}^{(k)}(s)\|^2ds. \label{b31} \end{equation} In a similar way, we obtain the estimate \begin{equation} I_{8}=2\int_0^{t}\langle G_{m}'(s),\ddot{u} _{mx}^{(k)}(s)\rangle ds\leq T_{\ast }\tilde{G}_M^2+\int _0^{t}\| \ddot{u}_{mx}^{(k)}(s)\|^2ds, \label{b32} \end{equation} with $\tilde{G}_M=( 1+4M+8M^2) \bar{G}_M$. From \begin{align} G_{mx}(t) &= D_1G[u_{m-1}]+D_3G[u_{m-1}]\nabla u_{m-1}+D_4G[u_{m-1}]\Delta u_{m-1} \label{b33} \\ &\quad +D_{5}G[u_{m-1}]\nabla u_{m-1}'+D_{6}G[u_{m-1}]\Delta u_{m-1}', \end{align} we obtain \begin{equation} \| G_{mx}(t)\| \leq ( 1+4M) \bar{G}_M\leq \tilde{G}_M. \label{b34} \end{equation} Hence \begin{equation} I_{9}=2\int_0^{t}\langle G_{mx}(s),\triangle \dot{u} _{m}^{(k)}(s)\rangle ds\leq T_{\ast }\tilde{G}_M^2+\int_0^{t} \| \triangle \dot{u}_{m}^{(k)}(s)\|^2ds. \label{b35} \end{equation} On the other hand \begin{align*} 2S_{m}^{(k)}(t) &\geq 2\| \sqrt{B_{m}(t)}\ddot{u} _{mx}^{(k)}(t)\|^2+2\| \sqrt{B_{m}(t)}\Delta \dot{u} _{m}^{(k)}(t)\|^2 \\ &\geq b_0( 2\| \ddot{u}_{mx}^{(k)}(t)\| ^2+2\| \Delta \dot{u}_{m}^{(k)}(t)\|^2) \\ &\geq b_0( \| \ddot{u}_{m}^{(k)}(t)\| ^2+\| \ddot{u}_{mx}^{(k)}(t)\|^2+\| \Delta \dot{u}_{m}^{(k)}(t)\|^2) . \end{align*} We have verified that \begin{equation} \label{b36} \begin{aligned} I_{7}+I_{8}+I_{9} &\leq2T_{\ast }( \tilde{F}_M^2+\tilde{G}_M^2) \\ &\quad +\int_0^{t}[ \| \ddot{u}_{m}^{(k)}(s)\| ^2+\| \ddot{u}_{mx}^{(k)}(s)\|^2+\| \triangle \dot{u}_{m}^{(k)}(s)\|^2] ds \\ &\leq2T_{\ast }( \tilde{F}_M^2+\tilde{G}_M^2) +\frac{2 }{b_0}\int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \end{equation} It is known that \begin{equation} \label{b37} \begin{aligned} B_{m}'(t) &= D_2B[u_{m-1}]+D_3B[u_{m-1}]u_{m-1}' \\ &\quad +2D_4B[u_{m-1}]\langle u_{m-1}(t),u_{m-1}'(t)\rangle \\ &\quad +2D_{5}B[u_{m-1}]\langle \nabla u_{m-1}(t),\nabla u_{m-1}'(t)\rangle \\ &\quad +2D_{6}B[u_{m-1}]\langle u_{m-1}'(t),u_{m-1}''(t)\rangle \\ &\quad +2D_{7}B[u_{m-1}]\langle \nabla u_{m-1}'(t),\nabla u_{m-1}''(t)\rangle , \end{aligned} \end{equation} so \begin{equation} | B_{m}'(x,t)| \leq ( 1+M+8M^2) \bar{B}_M\equiv \tilde{B}_M. \label{b38} \end{equation} We also have \begin{align*} S_{m}^{(k)}(t) &\geq \| \sqrt{B_{m}(t)}u_{mx}^{(k)}(t)\| ^2+\| \sqrt{B_{m}(t)}\Delta u_{m}^{(k)}(t)\| ^2+2\| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2 \\ &\quad +\| \sqrt{B_{m}(t)}\Delta \dot{u}_{m}^{(k)}(t)\| ^2+\| \sqrt{B_{m}(t)}\ddot{u}_{mx}^{(k)}(t)\|^2 \\ &\geq b_0[ \| u_{mx}^{(k)}(t)\|^2+\| \Delta u_{m}^{(k)}(t)\|^2+2\| \dot{u}_{mx}^{(k)}(t) \|^2+\| \Delta \dot{u}_{m}^{(k)}(t)\| ^2+\| \ddot{u}_{mx}^{(k)}(t)\|^2] , \end{align*} hence \begin{equation} \label{b39} \begin{aligned} | I_{10}| &= \Big|\int_0^{t}ds\int_0^1B_{m}'(x,s)[ | u_{mx}^{(k)}(x,s)|^2+| \Delta u_{m}^{(k)}(x,s)|^2+2| \dot{u}_{mx}^{(k)}(x,s)|^2 \\ &\quad +| \Delta \dot{u}_{m}^{(k)}(x,s)| ^2-| \ddot{u}_{mx}^{(k)}(x,s)|^2] dx\Big| \\ &\leq\tilde{B}_M\int_0^{t}\Big[ \| u_{mx}^{(k)}(s)\|^2+\| \Delta u_{m}^{(k)}(s)\| ^2+2\| \dot{u}_{mx}^{(k)}(s)\|^2+\| \Delta \dot{u}_{m}^{(k)}(s)\|^2 \\ &\quad +\| \ddot{u}_{mx}^{(k)}(s)\|^2\Big] ds \\ &\leq\frac{\tilde{B}_M}{b_0}\int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \end{equation} Note that \begin{align*} S_{m}^{(k)}(t) &\geq \| \sqrt{B_{m}(t)}u_{mx}^{(k)}(t)\|^2 +2\| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\ddot{u}_{mx}^{(k)}(t)\|^2 \\ &\geq b_0[ \| u_{mx}^{(k)}(t)\|^2+2\| \dot{ u}_{mx}^{(k)}(t)\|^2+\| \ddot{u}_{mx}^{(k)}(t)\|^2] , \end{align*} we deduce that \begin{equation} \label{b40} \begin{aligned} | I_{11}| &=\big| 2\int_0^{t}\langle B_{m}'(s)( u_{mx}^{(k)}(s)+\lambda _1\dot{u} _{mx}^{(k)}(s)) ,\ddot{u}_{mx}^{(k)}(s)\rangle ds\big| \\ &\leq2\tilde{B}_M\int_0^{t}( \| u_{mx}^{(k)}(s)\| +\lambda _1\| \dot{u} _{mx}^{(k)}(s)\| ) \| \ddot{u}_{mx}^{(k)}(s)\| ds \\ &\leq\frac{\tilde{B}_M}{b_0}( 2+\lambda _1) \int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \end{equation} Because of \begin{gather*} % \label{b41} B_{mx}(x,t) = D_1B[u_{m-1}]+D_3B[u_{m-1}]\nabla u_{m-1}, \\ | B_{mx}(x,t)| \leq\bar{B}_M( 1+2M) \equiv \hat{B}_M, \\ \begin{aligned} S_{m}^{(k)}(t) &\geq \| \sqrt{B_{m}(t)}u_{mx}^{(k)}(t)\| ^2+2\| \sqrt{B_{m}(t)}\dot{u}_{mx}^{(k)}(t)\|^2\\ &\quad +\| \sqrt{B_{m}(t)}\ddot{u}_{mx}^{(k)}(t)\|^2 +\| \sqrt{B_{m}(t)}\Delta \dot{u}_{m}^{(k)}(t)\|^2 \\ &\geq b_0\Big( \| u_{mx}^{(k)}(t)\|^2+2\| \dot{u}_{mx}^{(k)}(t)\|^2+\| \ddot{u}_{mx}^{(k)}(t) \|^2+\| \Delta \dot{u}_{m}^{(k)}(t)\|^2\Big) , \end{aligned} \end{gather*} we have the estimate \begin{equation} \label{b42} \begin{aligned} I_{12} &= 2\int_0^{t}\langle B_{mx}(s)( u_{mx}^{(k)}(s)+\lambda _1\dot{u}_{mx}^{(k)}(s)+\ddot{u} _{mx}^{(k)}(s)) ,\Delta \dot{u}_{m}^{(k)}(s)\rangle ds \\ &\leq2\hat{B}_M\int_0^{t}( \| u_{mx}^{(k)}(s)\| +\lambda _1\| \dot{u} _{mx}^{(k)}(s)\| +\| \ddot{u}_{mx}^{(k)}(s)\| ) \| \Delta \dot{u}_{m}^{(k)}(s)\| ds \\ &\leq\frac{\hat{B}_M}{b_0}( 4+\lambda _1) \int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \end{equation} Consequently, estimates \eqref{b19}, \eqref{b23}, \eqref{b25}, \eqref{b28}, \eqref{b36}, \eqref{b39}, \eqref{b40} and \eqref{b42} show that \begin{equation} \label{b43} \begin{aligned} S_{m}^{(k)}(t) &\leq S_0+2K_M^2(f)+4T_{\ast }( \bar{F}_M^2+\bar{G}_M^2) \\ &\quad +\frac{1}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] \int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \end{equation} We choose $M>0\ $sufficiently large such that \begin{equation} S_0+2K_M^2(f)\leq \frac{1}{2}M^2, \label{b44} \end{equation} and then choose $T_{\ast }\in (0,T]$ small enough such that \begin{equation} \big( \frac{1}{2}M^2+4T_{\ast }( \bar{F}_M^2+\bar{G} _M^2) \big) \exp [ \frac{T_{\ast }}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] ] \leq M^2, \label{b45} \end{equation} and \begin{equation} k_{T_{\ast }}=2\sqrt{\bar{D}_M}\sqrt{T_{\ast }}\exp [ T_{\ast }( 1+\frac{\tilde{B}_M}{2b_0}) ] <1, \label{b46} \end{equation} with \begin{equation*} \bar{D}_M=\frac{1}{b_0}[ 4(1+2M)^2( \bar{F}_M+\bar{G} _M)^2+(2+\lambda _1)^2(1+4M)^2M^2\bar{B}_M^2] . \end{equation*} From \eqref{b43}--\eqref{b45}, we have \begin{equation} \begin{aligned} S_{m}^{(k)}(t) &\leq \exp [ \frac{-T_{\ast }}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] ] M^2 \\ &\quad +\frac{1}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] \int_0^{t}S_{m}^{(k)}(s)ds. \end{aligned} \label{b47} \end{equation} Using Gronwall's Lemma, \eqref{b47} leads to \begin{equation} S_{m}^{(k)}(t)\leq \exp [ \frac{-T_{\ast }}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] ] M^2\exp [ \frac{ -t}{b_0}[ 4+( 7+2\lambda _1) \tilde{B}_M] ] \leq M^2, \label{b48} \end{equation} for all $t\in [ 0,T_{\ast }]$, for all $m$ and $k$, so \begin{equation} u_{m}^{(k)}\in W(M,T_{\ast }),\text{ for all }m\text{ and }k. \label{b49} \end{equation} \smallskip \noindent\textbf{(iii)} Limiting process. By \eqref{b48}, there exists a subsequence of $\{u_{m}^{(k)}\}$ with a same notation, such that \begin{equation} \begin{gathered} u_{m}^{(k)}\to u_{m} \quad \text{in } L^{\infty }(0,T_{\ast}; H_0^1\cap H^2)\text{ weakly*,} \\ \dot{u}_{m}^{(k)}\to u_{m}'\quad \text{in } L^{\infty }(0,T_{\ast };H_0^1\cap H^2)\text{ weakly*,} \\ \ddot{u}_{m}^{(k)}\to u_{m}'' \quad \text{in } L^{\infty }(0,T_{\ast };H_0^1)\text{ weakly*,}\\ u_{m}\in W(M,T_{\ast }). \end{gathered} \label{b50} \end{equation} Passing to limit in \eqref{b14}, \eqref{b15}, it is clear to see that $u_{m}$ is satisfying \eqref{b11}, \eqref{b12} in $L^2(0,T_{\ast })$. Furthermore, \eqref{b11}$_1\ $and \eqref{b50}$_4$ imply that \begin{align*} \label{b51} B_{m}(t)\Delta u_{m}''(t) &= -B_{m}(t)[ \Delta u_{m}(t)+\lambda _1\Delta u_{m}'(t)] -B_{mx}(t)\big( u_{mx}(t)+\lambda _1u_{mx}'(t)\\ &\quad +u_{mx}''(t)\big) +u_{m}''(t)-f(t)-F_{m}(t)+G_{mx}(t) \\ &\equiv \Psi _{m}\in L^{\infty }(0,T_{\ast };L^2). \end{align*} We have \begin{equation*} b_0\| \Delta u_{m}''(t)\| \leq \|B_{m}(t)\Delta u_{m}''(t)\| =\| \Psi_{m}(t)\| \leq \| \Psi _{m}\| _{L^{\infty }(0,T_{\ast };L^2)}. \end{equation*} Hence $u_{m}''\in L^{\infty }(0,T_{\ast };H_0^1\cap H^2)$, so we obtain $u_{m}\in W_1(M,T_{\ast })$, Lemma \ref{lem2.3} is proved. It means that step 1 is done. \end{proof} \textit{Step 2.} We state the following lemma. \begin{lemma} \label{lem2.4} Let {\rm (H1)--(H5)} hold. Then \begin{itemize} \item[(i)] Problem \eqref{1}--\eqref{3} has a unique weak solution $u\in W_1(M,T_{\ast })$, where $M>0$ and $T_{\ast }>0$ are chosen constants as in Lemma \ref{lem2.3}. \item[(ii)] The linear recurrent sequence $\{u_{m}\}$ defined by \eqref{b10}--\eqref{b12} converges to the solution $u$ of \eqref{1}--\eqref{3} strongly in the space \begin{equation} W_1(T_{\ast })=\{v\in L^{\infty }(0,T_{\ast };H_0^1):v'\in L^{\infty }(0,T_{\ast };H_0^1)\}. \label{b52} \end{equation} \end{itemize} \end{lemma} \begin{proof} We use the result obtained in Lemma \ref{lem2.3} and the compact imbedding theorems to prove Lemma \ref{lem2.4}. It means that the existence and uniqueness of a weak solution of Prob. \eqref{1}--\eqref{3} is proved. \smallskip \noindent\textbf{(i)} Existence. It is well known that $W_1(T_{\ast })$ is a Banach space (see Lions \cite{Lio3}), with respect to the norm \begin{equation} \| v\| _{W_1(T_{\ast })}=\| v\| _{L^{\infty }(0,T_{\ast };H_0^1)}+\| v'\| _{L^{\infty }(0,T_{\ast };H_0^1)}. \label{b53} \end{equation} It is clear that $\{u_{m}\}$ is a Cauchy sequence in $W_1(T_{\ast })$. Indeed, let $w_{m}=u_{m+1}-u_{m}$, we have \begin{equation} \begin{gathered} \begin{aligned} &\langle w_{m}''(t),w\rangle +\langle B_{m+1}(t)(w_{mx}(t) +\lambda _1w_{mx}'(t)+w_{mx}''(t)), w_{x}\rangle \\ &=\langle F_{m+1}(t)-F_{m}(t),w\rangle +\langle G_{m+1}(t)-G_{m}(t),w_{x}\rangle \\ &\quad -\langle [ B_{m+1}(t)-B_{m}(t)] ( u_{mx}(t)+\lambda _1u_{mx}'(t)+u_{mx}''(t)) ,w_{x}\rangle , \quad \forall w\in H_0^1, \end{aligned} \\ w_{m}(0)=w_{m}'(0)=0. \end{gathered} \label{b54} \end{equation} Consider \eqref{b54} with $w=w_{m}'$, and then integrating in $t$, we obtain \begin{equation} \label{b55} \begin{aligned} &Z_{m}(t) \\ &=2\int_0^{t}\langle F_{m+1}(s)-F_{m}(s),w_{m}'(s)\rangle ds+2\int_0^{t}\langle G_{m+1}(s)-G_{m}(s),w_{mx}'(s)\rangle ds \\ &\quad +\int_0^{t}ds\int_0^1B_{m+1}'(x,s)(w_{mx}^2(x,s)+| w_{mx}'(x,s)|^2) dx \\ &\quad -2\int_0^{t}\langle ( B_{m+1}(s)-B_{m}(s)) ( u_{mx}(s)+\lambda _1u_{mx}'(s)+u_{mx}''(s)) ,w_{mx}'(s)\rangle ds \\ &= J_1+J_2+J_3+J_4, \end{aligned} \end{equation} with \begin{align*} %\label{b56} Z_{m}(t) &= \| w_{m}'(t)\|^2+\| \sqrt{B_{m+1}(t)}w_{mx}'(t)\|^2 +\| \sqrt{B_{m+1}(t)}w_{mx}(t)\|^2 \\ &\quad +2\lambda _1\int_0^{t}\| \sqrt{B_{m+1}(s)} w_{mx}'(s)\|^2ds. \end{align*} From \begin{gather*} %\label{b57} \| F_{m+1}(s)-F_{m}(s)\| \leq2(1+2M)\bar{F} _M\| w_{m-1}\| _{W_1(T_{\ast })}, \\ \| G_{m+1}(s)-G_{m}(s)\| \leq2(1+2M)\bar{G} _M\| w_{m-1}\| _{W_1(T_{\ast })}, \\ | B_{m+1}'(x,s)| \leq( 1+M+8M^2) \bar{B}_M\equiv \tilde{B}_M, \\ | B_{m+1}(x,s)-B_{m}(x,s)| \leq(1+4M)\bar{B} _M\| w_{m-1}\| _{W_1(T_{\ast })}, \\ \| u_{mx}(s)+\lambda _1u_{mx}'(s)+u_{mx}''(s)\| \leq(2+\lambda _1)M, \end{gather*} we obtain the estimates \begin{equation} \label{b58} \begin{aligned} J_1+J_2 &= 2\int_0^{t}\langle F_{m+1}(s)-F_{m}(s),w_{m}'(s)\rangle ds \\ &\quad +2\int_0^{t}\langle G_{m+1}(s)-G_{m}(s),w_{mx}'(s)\rangle ds\\ &\leq\frac{4}{b_0}(1+2M)^2( \bar{F}_M+\bar{G}_M) ^2T_{\ast }\| w_{m-1}\| _{W_1(T_{\ast })}^2+\int_0^{t}Z_{m}(s)ds; \end{aligned} \end{equation} \begin{align*} J_3 &= \int_0^{t}ds\int_0^1B_{m+1}'(x,s)( w_{mx}^2(x,s)+| w_{mx}'(x,s)| ^2) dx \\ &\leq\tilde{B}_M\int_0^{t}( \| w_{mx}(s)\|^2+\| w_{mx}'(s)\| ^2) ds\leq \frac{\tilde{B}_M}{b_0}\int _0^{t}Z_{m}(s)ds; \end{align*} \begin{align*} J_4 &= -2\int_0^{t}\langle ( B_{m+1}(s)-B_{m}(s)) ( u_{mx}(s)+\lambda _1u_{mx}'(s) +u_{mx}''(s)) ,w_{mx}'(s)\rangle ds \\ &\leq2(2+\lambda _1)(1+4M)M\bar{B}_M\| w_{m-1}\| _{W_1(T_{\ast })}\int_0^{t}\| w_{mx}'(s)\| ds \\ &\leq\frac{1}{b_0}(2+\lambda _1)^2(1+4M)^2M^2\bar{B} _M^2T_{\ast }\| w_{m-1}\| _{W_1(T_{\ast })}^2+\int_0^{t}Z_{m}(s)ds. \end{align*} From \eqref{b55} and \eqref{b58} we have \begin{equation} Z_{m}(t)\leq T_{\ast }\bar{D}_M\| w_{m-1}\| _{W_1(T_{\ast })}^2+\big( 2+\frac{\tilde{B}_M}{b_0}\big) \int_0^{t}Z_{m}(s)ds, \label{b59} \end{equation} with \begin{equation} \bar{D}_M=\frac{1}{b_0}[ 4(1+2M)^2( \bar{F}_M+\bar{G} _M)^2+(2+\lambda _1)^2(1+4M)^2M^2\bar{B}_M^2] . \label{b60} \end{equation} Using Gronwall's Lemma, \eqref{b59} leads to \begin{equation} \| w_{m}\| _{W_1(T_{\ast })}\leq k_{T_{\ast }}\| w_{m-1}\| _{W_1(T_{\ast })}\quad \forall m\in\mathbb{N}, \label{b61} \end{equation} so \begin{equation} \| u_{m}-u_{m+p}\| _{W_1(T_{\ast })}\leq M(1-k_{T_{\ast }})^{-1}k_{T_{\ast }}^{m},\quad \forall m,p\in\mathbb{N}. \label{b62} \end{equation} It follows that $\{u_{m}\}$ is a Cauchy sequence in $W_1(T_{\ast })$, so there exists $u\in W_1(T_{\ast })$ such that \begin{equation} u_{m}\to u\text{ strongly in }W_1(T_{\ast }). \label{b63} \end{equation} Note that $u_{m}\in W_1(M,T_{\ast })$, so there exists a subsequence $\{u_{m_{j}}\}$ of $\{u_{m}\}$ such that \begin{equation} \begin{gathered} u_{m_{j}}\to u \quad \text{in } L^{\infty }(0,T_{\ast };H_0^1\cap H^2)\text{ weakly*,} \\ u_{m_{j}}'\to u'\quad \text{in } L^{\infty }(0,T_{\ast };H_0^1\cap H^2)\text{ weakly*,} \\ u_{m_{j}}''\to u''\quad \text{in } L^{\infty }(0,T_{\ast };H_0^1)\text{ weakly*,} \\ u\in W(M,T_{\ast }). \end{gathered} \label{b64} \end{equation} On the other hand, by \eqref{b8}, \eqref{b10}, \eqref{b12} and \eqref{b64}$_4$, we obtain \begin{equation} \label{b65} \begin{gathered} \| F_{m}(t)-F[u](t)\| \leq2(1+2M)\bar{F}_M\| u_{m-1}-u\| _{W_1(T_{\ast })}, \\ \| G_{m}(t)-G[u](t)\| \leq2(1+2M)\bar{G}_M\| u_{m-1}-u\| _{W_1(T_{\ast })}, \\ | B_{m+1}(x,t)-B[u](x,t)| \leq(1+4M)\bar{B} _M\| u_{m-1}-u\| _{W_1(T_{\ast })}. \end{gathered} \end{equation} Then \eqref{b63} and \eqref{b65} imply \begin{equation} \begin{gathered} F_{m}\to F[u]\quad \text{strongly in }L^{\infty }(0,T_{\ast };L^2),\\ G_{m}\to G[u]\quad \text{strongly in }L^{\infty }(0,T_{\ast };L^2),\\ B_{m}\to B[u]\quad \text{strongly in }L^{\infty }(Q_{T_{\ast }}). \end{gathered} \label{b66} \end{equation} Passing to limit in \eqref{b11}, \eqref{b12} as $m=m_{j}\to \infty $, by \eqref{b63}, \eqref{b64} and \eqref{b66}, there exists $u\in W(M,T_{\ast })$ satisfying \begin{equation} \label{b67} \begin{aligned} &\langle u''(t),w\rangle +\langle B[u](t)( u_{x}''(t)+\lambda _1u_{x}'(t)+u_{x}(t)),w_{x}\rangle \\ &= \langle f(t),w\rangle +\langle F[u](t),w\rangle +\langle G[u](t),w_{x}\rangle ,\quad \forall w\in H_0^1, \end{aligned} \end{equation} and satisfying the initial conditions \begin{equation} u(0)=\tilde{u}_0,\quad u'(0)=\tilde{u}_1. \label{b68} \end{equation} Furthermore, assumption (H2) implies, from \eqref{b64}$_4$ and \eqref{b67} that \begin{equation} \label{b69} \begin{aligned} B[u]\Delta u'' &= -B[u]( \Delta u+\lambda _1\Delta u') -\frac{\partial }{\partial x}( B[u]) (u_{x}+\lambda _1u_{x}'+u_{x}'') \\ &\quad +u''-F[u]+\frac{\partial }{\partial x}G[u]-f\equiv \Psi \in L^{\infty }(0,T_{\ast };L^2). \end{aligned} \end{equation} From \begin{equation*} b_0\| \Delta u''(t)\| \leq \| B[u](t)\Delta u''(t)\| =\| \Psi (t)\| \leq \| \Psi \| _{L^{\infty }(0,T_{\ast};L^2)}, \end{equation*} we obtain $u''\in L^{\infty }(0,T_{\ast };H_0^1\cap H^2)$, and so $u\in W_1(M,T_{\ast })$. The existence is proved. \smallskip \noindent\textbf{(ii)} Uniqueness. Let $u_1$, $u_2$ be two weak solutions of \eqref{1}--\eqref{3}, such that \begin{equation} u_i\in W_1(M,T_{\ast }),\quad i=1,2. \label{b70} \end{equation} Then $w=u_1-u_2$ satisfies \begin{equation} \begin{gathered} \begin{aligned} &\langle w''(t),w\rangle +\langle B_1(t)( w_{x}(t)+\lambda _1w_{x}'(t)+w_{x}''(t)),w_{x}\rangle \\ &=\langle F_1(t)-F_2(t),w\rangle +\langle G_1(t)-G_2(t),w_{x}\rangle \\ &\quad -\langle [ B_1(t)-B_2(t)] (u_{2x}(t) +\lambda _1u_{2x}'(t)+u_{2x}''(t)),w_{x}\rangle , \quad \forall w\in H_0^1, \end{aligned} \\ w(0)=w'(0)=0, \end{gathered} \label{b71} \end{equation} where \begin{equation} B_i=B[u_i],\quad F_i=F[u_i],\quad G_i=G[u_i],\quad i=1,2. \label{b72} \end{equation} Taking $v=w=u_1-u_2$ in \eqref{b71}$_1$ and integrating with respect to $t$, we obtain \begin{equation} \label{b73} \begin{aligned} \rho (t) &= 2\int_0^{t}\langle F_1(s)-F_2(s),w'(s)\rangle ds +2\int_0^{t}\langle G_1(s)-G_2(s),w_{x}'(s)\rangle ds \\ &\quad +\int_0^{t}ds\int_0^1B_1'(x,s)(w_{x}^2(x,s)+| w_{x}'(x,s)|^2) dx \\ &\quad +2\int_0^{t}\langle ( B_1(s)-B_2(s)) ( u_{2x}(s)+\lambda _1u_{2x}'(s)+u_{2x}''(s)) ,w_{x}'(s)\rangle ds, \end{aligned} \end{equation} where \begin{equation} \begin{aligned} \rho (t)&=\| w'(t)\|^2+\| \sqrt{B_1(t)} w_{x}'(t)\|^2+\| \sqrt{B_1(t)}w_{x}(t)\|^2 \\ &\quad +2\lambda _1\int_0^{t}\| \sqrt{ B_1(s)}w_{x}'(s)\|^2ds. \end{aligned} \label{b74} \end{equation} On the other hand, by (H3)--(H5), we deduce from \eqref{b8}, \eqref{b74}, that \begin{equation} \label{b75} \begin{gathered} | B_1'(x,s)| \leq( 1+M+8M^2) \bar{B}_M\equiv \tilde{B}_M, \\ | B_1(x,s)-B_2(x,s)| \leq\sqrt{\frac{2}{b_0}} (1+4M)\bar{B}_M\sqrt{\rho (s)}, \\ \| F_1(s)-F_2(s)\| \leq2\sqrt{\frac{2}{b_0}} (1+2M)\bar{F}_M\sqrt{\rho (s)}, \\ \| G_1(s)-G_2(s)\| \leq2\sqrt{\frac{2}{b_0}} (1+2M)\bar{G}_M\sqrt{\rho (s)}, \\ \| u_{2x}(s)+\lambda _1u_{2x}'(s)+u_{2x}''(s)\| \leq( 2+\lambda _1) M. \end{gathered} \end{equation} Combining \eqref{b73} and \eqref{b75} leads to \[ \rho (t)\leq [ 4\sqrt{\frac{2}{b_0}}(1+2M)( \bar{F}_M+\bar{G} _M) +\frac{\tilde{B}_M}{b_0}+\frac{2\sqrt{2}}{b_0}( 2+\lambda _1) (1+4M)M\bar{B}_M] \int_0^{t}\rho (s)ds. \] % \label{b76} By Gronwall's Lemma we have $\rho \equiv 0$, i.e., $u_1\equiv u_2$. This completes the proof. \end{proof} By proving Lemma \ref{lem2.4}, we complete the proof Theorem \ref{thm2.2}. \end{proof} \section{Blow up} In this section, we consider\eqref{1}--\eqref{3} with $\lambda $, $\lambda _1>0$, $B=B(x,t)\in C^1([0,1]\times [ 0,T])$, $B( x,t) \geq b_0>0$; $F=F(u,u_{x})-\lambda u_{t}$, $G=G(u,u_{x})$, $F$, $G\in C^1(\mathbb{R}^2;\mathbb{R})$ as follows \begin{equation} \begin{gathered} \begin{aligned} &u_{tt}-\frac{\partial }{\partial x}[ B( x,t) ( u_{x}+\lambda _1u_{xt}+u_{xtt}) ] +\lambda u_{t} \\ & =F(u,u_{x})-\frac{\partial }{\partial x}( G(u,u_{x})) +f(x,t), \quad 00$ small enough. Furthermore, if the following assumptions hold, then a blow up result is obtained. \begin{itemize} \item[(H2')] $f=0$; \item[(H3')] $B\in C^1([0,1]\times [ 0,T])$ and there exist the positive constants $b_0$, $\bar{b}_0$, $b_1$ such that \begin{itemize} \item[(i)] $b_0\leq B(x,t)\leq \bar{b}_0$, for all $(x,t)\in[ 0,1]\times [ 0,T]$, \item[(ii)] $-b_1\leq B'(x,t)\leq 0$, for all $(x,t)\in[ 0,1]\times [ 0,T]$; \end{itemize} \item[(H4')] There exist $\mathcal{F} \in C^2(\mathbb{R}^2;\mathbb{R})$ and the constants $p$, $q>2;$ $d_1$, $\bar{d}_1>0$, such that \begin{itemize} \item[(i)] $\frac{\partial \mathcal{F} }{\partial u}(u,v)=F(u,v)$, $\frac{\partial \mathcal{F} }{\partial v}(u,v)=G(u,v)$, \item[(ii)] $uF(u,v)+vG(u,v)\geq d_1\mathcal{F} (u,v)$, for all $(u,v)\in\mathbb{R}^2$, \item[(iii)] $\mathcal{F} (u,v)\geq \bar{d}_1( |v|^p+| u|^{q}) $, for all $(u,v)\in \mathbb{R}^2$; \end{itemize} \item[(H5')] $0<\lambda _1<\frac{b_1}{2b_0}$; \item[(H6')] $d_1>\max \big\{ 2+\frac{2\lambda _1b_1}{b_0}, \frac{b_1}{b_0\lambda _1}-2\big\}$ with $d_1$ as in (H4'). \end{itemize} \begin{example} \label{examp1} \rm The following functions satisfy (H4'): \begin{gather*} F(u,v) = \alpha \bar{\gamma}_2| u|^{\alpha -2}u| v|^{\beta }+q\bar{\gamma}_3| u|^{q-2}u, \\ G(u,v) = p\bar{\gamma}_1| v|^{p-2}v+\beta \bar{\gamma} _2| u|^{\alpha }| v|^{\beta -2}v, \end{gather*} where $\alpha $, $\beta $, $p$, $q>2;$ $\bar{\gamma}_1$, $\bar{\gamma} _2, $ $\bar{\gamma}_3>0$ are the constants, with \begin{equation*} \min \{p,q,\alpha +\beta \}>\max \{ 2+\frac{2\lambda _1b_1}{b_0}, \frac{b_1}{b_0\lambda _1}-2\} , \end{equation*} with $b_0$, $b_1$, $\lambda _1$ as in $(H3')$, $(H5')$. It is obvious that $(H4')$ holds, because there exists an $\mathcal{F} \in C^2(\mathbb{R}^2;\mathbb{R})$ defined by \begin{equation*} \mathcal{F} (u,v) =\bar{\gamma}_1| v|^p+\bar{\gamma}_2| u|^{\alpha }| v|^{\beta } +\bar{\gamma}_3| u|^{q}, \end{equation*} such that \begin{gather*} \frac{\partial \mathcal{F} }{\partial u}(u,v)=\alpha \bar{\gamma} _2| u|^{\alpha -2}u| v|^{\beta }+q \bar{\gamma}_3| u|^{q-2}u=F(u,v), \\ \frac{\partial \mathcal{F} }{\partial v}(u,v)=p\bar{\gamma}_1| v|^{p-2}v+\beta \bar{\gamma}_2| u|^{\alpha }| v|^{\beta -2}v=G(u,v), \\ uF(u,v)+vG(u,v)\geq d_1\mathcal{F} (u,v), \quad\text{for all } (u,v)\in \mathbb{R}^2, \end{gather*} in which $d_1=\min \{p,q,\alpha +\beta \}>\max \{ 2+\frac{2\lambda _1b_1}{b_0},\frac{b_1}{b_0\lambda _1}-2\}$, \begin{equation*} \mathcal{F} (u,v)\geq \bar{d}_1( | v| ^p+| u|^{q}) \quad \text{for all }(u,v)\in \mathbb{R}^2, \end{equation*} with $\bar{d}_1=\min \{\bar{\gamma}_1,\bar{\gamma}_3\}$. Let us put \[ H(0)=-\frac{1}{2}\| \tilde{u}_1\|^2-\frac{1}{2} \| \sqrt{B(0)}\tilde{u}_{1x}\|^2-\frac{1}{2}\| \sqrt{B(0)}\tilde{u}_{0x}\|^2+\int_0^1\mathcal{F} ( \tilde{u}_0(x),\tilde{u}_{0x}(x)) dx. \] % \label{z3} \end{example} \begin{theorem} \label{thm3.1} Let {\rm (H2')--(H6')} hold. Then, for any $\tilde{u}_0, \tilde{u}_1\in H_0^1\cap H^2$, such that $H(0)>0$, the weak solution $u=u(x,t)$ of \eqref{z1} blows up in finite time. \end{theorem} \begin{proof} It consists of two steps: the Lyapunov functional $L(t)$ is constructed in step1 and then the blow up is proved in step 2. \smallskip \noindent\textbf{Step 1.} We define the energy associated with \eqref{z1} as \begin{equation} \begin{aligned} E(t)&=\frac{1}{2}\| u'(t)\|^2+\frac{1}{2} \| \sqrt{B(t)}u_{x}'(t)\|^2+\frac{1}{2} \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad -\int_0^1\mathcal{F} ( u(x,t),u_{x}(x,t)) dx, \end{aligned}\label{z4} \end{equation} and we put $H(t)=-E(t)$, for all $t\in [ 0,T_{\ast })$. Multiplying \eqref{z1})$_1$ by $u'(x,t)$ and integrating the resulting equation over $[0,1]$, we have \begin{equation} \begin{aligned} H'(t)&=\lambda \| u'(t)\|^2+\lambda_1\| \sqrt{B(t)}u_{x}'(t)\|^2\\ &\quad -\frac{1}{2} \int_0^1B'(x,t)( u_{x}^2(x,t)+| u_{x}'(x,t)|^2) dx\geq 0. \end{aligned} \label{z5} \end{equation} This implies \begin{equation} 00$ such that \begin{equation} L'(t)\geq \bar{L}_1[ H(t)+\| u_{x}(t)\| _{L^p}^p+\| u(t)\| _{L^{q}}^{q} +\| u'(t)\|^2+\| u_{x}'(t)\|^2+\| u_{x}(t)\|^2] . \label{z11} \end{equation} Multiplying \eqref{z1}$_1$ by $u(x,t)$ and integrating over $[0,1]$ leads to \begin{equation} \begin{aligned} \Psi'(t) &= \| u'(t)\|^2+\|\sqrt{B(t)}u_{x}'(t)\|^2-\| \sqrt{B(t)} u_{x}(t)\|^2 \label{z12} \\ &\quad +\langle B'(t)u_{x}'(t),u_{x}(t)\rangle +\frac{\lambda _1}{2}\int_0^1B'(x,t)u_{x}^2(x,t)dx \\ &\quad +\langle F( u(t),u_{x}(t)) ,u(t)\rangle +\langle G( u(t),u_{x}(t)) ,u_{x}(t)\rangle . \end{aligned} \end{equation} Therefore, \begin{equation} L'(t)=(1-\eta )H^{-\eta }(t)H'(t)+\varepsilon \Psi '(t)\geq \varepsilon \Psi'(t). \label{z13} \end{equation} By $(H4')$, we obtain \begin{equation} \begin{gathered} \begin{aligned} &\langle F( u(t),u_{x}(t)) ,u(t)\rangle +\langle G( u(t),u_{x}(t)) ,u_{x}(t)\rangle \\ &\geq d_1\int_0^1\mathcal{F} ( u(x,t),u_{x}(x,t)) dx, \end{aligned}\\ \int_0^1\mathcal{F} ( u(x,t),u_{x}(x,t)) dx\geq \bar{d}_1( \| u_{x}(t)\| _{L^p}^p+\|u(t)\| _{L^{q}}^{q}). \end{gathered} \label{z14} \end{equation} On the other hand, by (H3'), we obtain \begin{equation} \label{z15} \begin{aligned} &\big| \langle B'(t)u_{x}'(t),u_{x}(t)\rangle +\frac{\lambda _1}{2}\int_0^1B'(x,t)u_{x}^2(x,t)dx\big| \\ &\leq\frac{b_1}{b_0}\| \sqrt{B(t)}u_{x}'(t)\| \| \sqrt{B(t)}u_{x}(t)\| +\frac{\lambda_1b_1}{2b_0}\| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\leq\frac{b_1}{2b_0}\Big( \frac{1}{\lambda _1}\| \sqrt{ B(t)}u_{x}'(t)\|^2+\lambda _1\| \sqrt{B(t)} u_{x}(t)\|^2\Big) +\frac{\lambda _1b_1}{2b_0}\|\sqrt{B(t)}u_{x}(t)\|^2 \\ &= \frac{b_1}{2b_0\lambda _1}\| \sqrt{B(t)}u_{x}'(t)\|^2 +\frac{\lambda _1b_1}{b_0}\| \sqrt{B(t)}u_{x}(t)\|^2. \end{aligned} \end{equation} From \eqref{z12}, \eqref{z14}, \eqref{z15} it follows that \begin{align*} %\label{z16} &\Psi'(t) \\ &\geq \| u'(t)\|^2+\| \sqrt{B(t)}u_{x}'(t)\|^2-\| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad -[ \frac{b_1}{2b_0\lambda _1}\| \sqrt{B(t)} u_{x}'(t)\|^2+\frac{\lambda _1b_1}{b_0}\| \sqrt{B(t)}u_{x}(t)\|^2] +d_1\int_0^1 \mathcal{F} ( u(x,t),u_{x}(x,t)) dx \\ &= \| u'(t)\|^2+( 1-\frac{b_1}{ 2b_0\lambda _1}) \| \sqrt{B(t)}u_{x}'(t)\|^2 -\big( 1+\frac{\lambda _1b_1}{b_0}\big) \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad +d_1\delta _1\int_0^1\mathcal{F} ( u(x,t),u_{x}(x,t)) dx \\ &\quad +d_1(1-\delta _1)[ H(t)+\frac{1}{2}\| u'(t)\|^2 +\frac{1}{2}\| \sqrt{B(t)}u_{x}'(t)\|^2+\frac{1}{2}\| \sqrt{B(t)}u_{x}(t)\| ^2] \\ &\geq d_1\delta _1\bar{d}_1( \| u_{x}(t)\| _{L^p}^p+\| u(t)\| _{L^{q}}^{q}) +d_1(1-\delta _1)H(t)+[ 1+\frac{d_1}{2}(1-\delta _1)] \| u'(t)\|^2 \\ &\quad +\frac{1}{2}[ 2+d_1-\frac{b_1}{b_0\lambda _1}-\delta _1d_1] \| \sqrt{B(t)}u_{x}'(t)\|^2 \\ &\quad +\frac{1}{2}[ d_1-2\big( 1+\frac{\lambda _1b_1}{b_0}\big) -\delta _1d_1] \| \sqrt{B(t)}u_{x}(t)\|^2, \end{align*} for all $\delta _1\in (0,1)$. From $d_1>\max \{ 2+\frac{2\lambda _1b_1}{b_0}, \frac{ b_1}{b_0\lambda _1}-2\} $, we have $d_1-2( 1+\frac{ \lambda _1b_1}{b_0}) >0$ and $2+d_1-\frac{b_1}{b_0\lambda_1}>0$, we can choose $\delta _1>0$ small enough such that \begin{equation} 2+d_1-\frac{b_1}{b_0\lambda _1}-\delta _1d_1>0\text{ and } d_1-2\big( 1+\frac{\lambda _1b_1}{b_0}\big) -\delta _1d_1>0, \label{z17} \end{equation} and then \eqref{z11} holds. From the formula of $L(t)$ and \eqref{z11}, we can choose $\varepsilon >0$ small enough such that \begin{equation} L(t)\geq L(0)>0,\quad \forall t\in [ 0,T_{\ast }). \label{z18} \end{equation} Using the inequality \begin{equation} \Big( \sum_{i=1}^{5}x_i\Big)^{r} \leq 5^{r-1}\sum_{i=1}^{5}x_i^{r},\quad \text{for all }r>1,\text{ and } x_1,\dots,x_{5}\geq 0, \label{z19} \end{equation} we deduce from \eqref{z8}--\eqref{z10} that \begin{equation} \label{z20} \begin{aligned} L^{1/(1-\eta )}(t) &\leq \text{const.} \big[ H(t) +| \langle u(t),u'(t)\rangle |^{1/(1-\eta )} +| \langle B(t)u_{x}'(t),u_{x}(t)\rangle |^{1/(1-\eta )} \\ &\quad +\| u(t)\|^{2/(1-\eta )}+\| \sqrt{B(t)}u_{x}(t)\|^{2/(1-\eta )}] . \end{aligned} \end{equation} Using Young's inequality, we have \begin{equation} \label{z21} \begin{aligned} | \langle u(t),u'(t)\rangle |^{1/(1-\eta )} &\leq\| u(t)\|^{1/(1-\eta )}\| u'(t)\|^{1/(1-\eta )} \\ &\leq\frac{1-2\eta }{2(1-\eta )}\| u(t)\|^{s}+\frac{1}{ 2(1-\eta )}\| u'(t)\|^2 \\ &\leq\text{const.}( \| u_{x}(t)\|^{s}+\| u'(t)\|^2) , \end{aligned} \end{equation} where $s=2/(1-2\eta )\leq \min \left\{ p,q\right\} $ as in \eqref{z10}. Similarly, we obtain \begin{equation} \label{z22} \begin{aligned} | \langle B(t)u_{x}'(t),u_{x}(t)\rangle |^{1/(1-\eta )} &\leq\bar{b}_0^{1/(1-\eta )}\|u_{x}(t)\|^{1/(1-\eta )}\| u_{x}'(t)\| ^{1/(1-\eta )} \\ &\leq \text{const.}\big \| u_{x}(t)\|^{s}+\|u_{x}'(t)\|^2\big) . \end{aligned} \end{equation} Combining \eqref{z20}--\eqref{z22}, we obtain \begin{equation} \begin{aligned} L^{1/(1-\eta )}(t) &\leq \text{const.}[ H(t)+\| u'(t)\| ^2+\| u_{x}'(t)\|^2 +\| u(t)\|^{2/(1-\eta )} \\ &\quad +\| u_{x}(t)\|^{2/(1-\eta )}+\| u_{x}(t)\|^{s}] . \end{aligned} \label{z23} \end{equation} \smallskip \noindent\textbf{Step 2.} We note that the following property for any $v\in H_0^1 $. \begin{lemma} \label{lem3.2} Let $2\leq r_1\leq q$, $2\leq r_2$, $r_3\leq p$. Then, for any $v\in H_0^1$, we have \begin{equation} \| v\|^{r_1}+\| v_{x}\|^{r_2}+\| v_{x}\|^{r_3} \leq 3\big( \|v\| _{L^{q}}^{q}+\| v_{x}\| _{L^p}^p+\| v_{x}\|^2\big) . \label{z24} \end{equation} \end{lemma} The proof of the above lemma is not difficult, so we omit it. Using \eqref{z23} and Lemma \ref{lem3.2} with $r_1=\frac{2}{1-\eta }$, $r_2=2/(1-\eta )$, $r_3=s$, we obtain \begin{equation} \label{z25} \begin{aligned} L^{1/(1-\eta )}(t) &\leq \text{const.}\big[ H(t)+\| u'(t)\|^2+\| u_{x}'(t)\|^2 +\|u_{x}(t)\|^2 \\ &\quad +\| u(t)\| _{L^{q}}^{q} +\|u_{x}(t)\| _{L^p}^p] ,\quad \forall t\in [0,T_{\ast }). \end{aligned} \end{equation} It follows from \eqref{z11} and \eqref{z25} that \begin{equation} L'(t)\geq \bar{L}_2L^{1/(1-\eta )}(t),\quad \forall t\in [ 0,T_{\ast }), \label{z26} \end{equation} where $\bar{L}_2$ is a positive constant. Integrating \eqref{z26} over $(0,t)$ leads to \begin{equation} L^{\eta /(1-\eta )}(t)\geq \frac{1}{L^{-\eta /(1-\eta )}(0)-\frac{\bar{L} _2\eta }{1-\eta }t},\quad 0\leq t<\frac{1}{\bar{L}_2\eta }(1-\eta)L^{-\eta /(1-\eta )}(0). \label{z27} \end{equation} Consequently, $L(t)$ blows up in a finite time given by $T_{\ast }=\frac{1}{\bar{L}_2\eta }(1-\eta )L^{-\eta /(1-\eta )}(0)$. The proof of Theorem \ref{thm3.1} is complete. \end{proof} \section{Exponential decay} In this section, we consider Problem \eqref{z1} under the following assumptions. \begin{itemize} \item[(H2'')] $f\in L^{\infty }( \mathbb{R}_{+};L^2) \cap L^1( \mathbb{R}_{+};L^2)$; \item[(H3'')] $B\in C^1([0,1]\times \mathbb{R}_{+})$ and there exist three positive constants $b_0$, $\bar{b}_0$, $b_1$ such that \begin{itemize} \item[(i)] $b_0\leq B(x,t)\leq \bar{b}_0$, for all $(x,t)\in [ 0,1]\times \mathbb{R}_{+}$, \item[(ii)] $-b_1\leq B'(x,t)\leq 0$, for all $(x,t)\in[ 0,1]\times \mathbb{R}_{+}$; \end{itemize} \item[(H4'')] There exist $\mathcal{F} \in C^2(\mathbb{R}^2;\mathbb{R})$ and the constants $p, q, \alpha, \beta >2$; $2<\alpha$, $\beta ,q\leq p$; $d_2, \tilde{d}_1, \bar{d}_2>0$, such that \begin{itemize} \item[(i)] $\frac{\partial \mathcal{F} }{\partial u}(u,v)=F(u,v)$, $\frac{\partial \mathcal{F} }{\partial v}(u,v)=G(u,v)$, for all $(u,v)\in\mathbb{R}^2$, \item[(ii)] $\mathcal{F} _1(u,v)\equiv \mathcal{F} (u,v)+\tilde{d} _1| v|^p\leq \bar{d}_2( |u|^{\alpha }| v|^{\beta }+|u|^{q}) $, for all $(u,v)\in\mathbb{R}^2$, \item[(iii)] $uF(u,v)+vG(u,v)\leq d_2\mathcal{F} (u,v)$, for all $(u,v)\in\mathbb{R}^2$; \end{itemize} \item[(H5'')] $d_22$; $\tilde{\gamma}_1, \tilde{\gamma}_2, \tilde{\gamma}_3>0$ are the constants, with $2<\alpha ,\beta ,q0$ small enough such that \begin{equation*} 0<\varepsilon 0, \] where\ $p>\max \{2,d_2\}$ with $d_2$ given in (H4'')(iii). Let $u=u(x,t)$ be a weak solution of \eqref{z1} satisfying \eqref{z2} as note in section 3. To obtain the decay result, we construct the functional \begin{equation} \mathcal{L} (t)=E(t)+\delta \Psi (t), \label{s1} \end{equation} with $\delta >0$; $E(t)$ and $\Psi (t)$ as definition in Section 3. We rewrite $E(t)$ as follows % \label{s2} \begin{align*} E(t) &= \frac{1}{2}\| u'(t)\|^2+\frac{1}{2} \| \sqrt{B(t)}u_{x}'(t)\|^2+\frac{1}{2} \| \sqrt{B(t)}u_{x}(t)\|^2+\tilde{d}_1\| u_{x}(t)\| _{L^p}^p \\ &\quad -\int_0^1\mathcal{F} _1( u(x,t),u_{x}(x,t)) dx \\ &= \frac{1}{2}\| u'(t)\|^2+\frac{1}{2} \| \sqrt{B(t)}u_{x}'(t)\|^2+( \frac{1}{2}- \frac{1}{p}) \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad +\tilde{d}_1\| u_{x}(t)\| _{L^p}^p +\frac{1}{p}I(t), \end{align*} where \begin{equation} I(t)=\| \sqrt{B(t)}u_{x}(t)\| ^2-p\int_0^1\mathcal{F} _1( u(x,t),u_{x}(x,t))dx. \label{s3} \end{equation} \end{example} \begin{theorem} \label{thm4.1} Assume that {\rm (H2'')-(H5'')} hold. Let $\tilde{u}_0, \tilde{u}_1\in H_0^1\cap H^2$ such that $I(0)>0$ and the initial energy $E(0)$ satisfy \begin{equation} \eta^{\ast }=b_0-p\bar{d}_2\Big[ \Big( \frac{2p}{(p-2)b_0}E_{\ast }\Big)^{\frac{\alpha -2}{2}}\Big( \frac{E_{\ast }}{\tilde{d}_1} \Big)^{\beta /p}+\Big( \frac{2p}{(p-2)b_0}E_{\ast }\Big)^{\frac{ q-2}{2}}\Big] >0, \label{s4} \end{equation} where \begin{equation} E_{\ast }=\Big( E(0)+\frac{1}{2}\| f\| _{L^1( \mathbb{R}_{+};L^2) }\Big) \exp \big( \| f\| _{L^1( \mathbb{R}_{+};L^2) }\big) . \label{s5} \end{equation} Assume that \begin{equation} \| f(t)\|^2\leq \bar{C}_1\exp (-\bar{\eta}_1t)\quad \text{for all } t\geq 0, \label{s6} \end{equation} where $\bar{C}_1$, $\bar{\eta}_1$ are two positive constants. Then, there exist positive constants $\bar{C}$, $\bar{\gamma}$ such that \begin{equation} \| u'(t)\|^2+\| u_{x}'(t)\|^2+\| u_{x}(t)\|^2+\| u_{x}(t)\| _{L^p}^p\leq \bar{C}\exp (-\bar{\gamma}t),\quad \text{for all }t\geq 0. \label{s7} \end{equation} \end{theorem} \begin{proof} It consists of three steps. \smallskip \noindent\textbf{Step 1.} An estimate of $E'(t)$. We have \begin{equation} \label{s8} \begin{gathered} E'( t) \leq \frac{1}{2}\| f(t)\| +\| f(t)\| \| u'(t)\|^2, \\ E'(t) \leq-( \lambda -\frac{\varepsilon _1}{2}) \| u'(t)\|^2-\lambda _1\| \sqrt{B(t)}u_{x}'(t)\|^2+\frac{1}{2\varepsilon _1}\| f(t)\|^2, \end{gathered} \end{equation} for all $\varepsilon _1>0$. Indeed, multiplying \eqref{z1}$_1$ by $u'(x,t)$ and integrating over $[0,1]$, we obtain \begin{equation} \label{s9} \begin{aligned} E'(t) &= -\lambda \| u'(t)\|^2-\lambda _1\| \sqrt{B(t)}u_{x}'(t)\|^2 \\ &\quad +\frac{1}{2}\int_0^1B'(x,t)( u_{x}^2(x,t)+| u_{x}'(x,t)|^2) dx+\langle f(t),u'(t)\rangle . \end{aligned} \end{equation} On the other hand \begin{equation} | \langle f(t),u'(t)\rangle | \leq \frac{1}{2} \| f(t)\| +\frac{1}{2}\| f(t)\| \| u'(t)\|^2. \label{s10} \end{equation} From $B'(x,t)\leq 0$, by \eqref{s9}, \eqref{s10}, it is easy to see that \eqref{s8}$_{(i)}$ holds. Similarly, \begin{equation} | \langle f(t),u'(t)\rangle | \leq \frac{1}{ 2\varepsilon _1}\| f(t)\| _0^2+\frac{\varepsilon _1 }{2}\| u'(t)\|^2,\quad \text{for all }\varepsilon_1>0. \label{s11} \end{equation} By $B'(x,t)\leq 0$, \eqref{s9} and \eqref{s11}, that \eqref{s8}$_{(ii)}$ is valid. \smallskip \noindent\textbf{Step 2.} An estimate of $I(t)$. By the continuity of $I(t)$ and $I(0)>0$, there exists $T_1>0$ such that \begin{equation} I(t)\geq 0,\quad \forall t\in [ 0,T_1], \label{s12} \end{equation} this implies that \begin{equation} E(t)\geq \frac{1}{2}\| u'(t)\|^2+( \frac{1 }{2}-\frac{1}{p}) \| \sqrt{B(t)}u_{x}(t)\|^2+ \tilde{d}_1\| u_{x}(t)\| _{L^p}^p,\text{ }\forall t\in [ 0,T_1]. \label{s13} \end{equation} Combining \eqref{s8}$_i$ with \eqref{s13} and using Gronwall's inequality we obtain \begin{equation} \frac{(p-2)b_0}{2p}\| u_{x}(t)\|^2+\tilde{d} _1\| u_{x}(t)\| _{L^p}^p\leq E(t)\leq E_{\ast },\quad \forall t\in [ 0,T_1]. \label{s14} \end{equation} Hence, it follows from $(\bar{H}_4,(iii))$, \eqref{s5}, \eqref{s14} that \begin{equation} \label{s15} \begin{aligned} &p\int_0^1\mathcal{F} _1( u(x,t),u_{x}(x,t)) dx \\ &\leq p\bar{d}_2\Big( \int_0^1| u(x,t)| ^{\alpha }| u_{x}(x,t)|^{\beta }dx+\int_0^1| u(x,t)|^{q}dx\Big) \\ &\leq p\bar{d}_2\Big( \| u_{x}(t)\|^{\alpha }\| u_{x}(t)\| _{L^{\beta }}^{\beta }+\|u_{x}(t)\|^{q}\Big) \\ &\leq p\bar{d}_2\Big( \| u_{x}(t)\|^{\alpha }\| u_{x}(t)\| _{L^p}^{\beta }+\| u_{x}(t)\|^{q}\Big)\\ &\leq p\bar{d}_2[ ( \frac{2p}{(p-2)b_0}E_{\ast })^{ \frac{\alpha -2}{2}}( \frac{E_{\ast }}{\tilde{d}_1})^{\beta /p}+( \frac{2p}{(p-2)b_0}E_{\ast })^{\frac{q-2}{2}}] \| u_{x}(t)\|^2, \end{aligned} \end{equation} for all $t\in [ 0,T_1]$. Consequently, $I(t)\geq \eta^{\ast }\| u_{x}(t)\|^2>0$, for all $t\in [ 0,T_1]$. Put $T_{\infty }=\sup \{ T>0:I(t)>0,\; t\in [0,T]\} $. If $T_{\infty }<+\infty $ then the continuity of $I(t)$ leads to $I(T_{\infty })\geq 0$. By the same arguments, there exists $T_{\infty }'>T_{\infty }$ such that $I(t)>0$, for all $t\in [0,T_{\infty }']$. Hence, we conclude that $I(t)>0$, for all $t\geq 0$. \smallskip \noindent\textbf{Step 3.} Decay result. First, we note that there exist the positive constants $\bar{\beta}_1$, $\bar{\beta}_2$ such that \begin{equation} \bar{\beta}_1E_1(t)\leq \mathcal{L} (t)\leq \bar{\beta}_2E_1(t), \quad \forall t\geq 0, \label{s16} \end{equation} for $\delta $ small enough, where \begin{equation} E_1(t)=\| u'(t)\|^2+\| \sqrt{B(t)} u_{x}'(t)\|^2+\| \sqrt{B(t)}u_{x}(t)\| ^2+\| u_{x}(t)\| _{L^p}^p+I(t). \label{s17} \end{equation} Indeed, we have \begin{equation} \begin{aligned} \label{s18} \mathcal{L} (t) &= \frac{1}{2}\| u'(t)\|^2+ \frac{1}{2}\| \sqrt{B(t)}u_{x}'(t)\|^2 +( \frac{1}{2}-\frac{1}{p}) \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad +\tilde{d}_1\| u_{x}(t)\| _{L^p}^p +\frac{1}{p}I(t)+\delta \Big[ \langle u'(t),u(t)\rangle \\ &\quad +\langle B(t)u_{x}'(t),u_{x}(t)\rangle +\frac{\lambda }{2}\| u(t)\|^2+\frac{\lambda _1}{2}\| \sqrt{B(t)} u_{x}(t)\|^2\Big] . \end{aligned} \end{equation} On the other hand, \begin{equation} \label{s19} \begin{gathered} \langle u(t),u'(t)\rangle \leq\frac{1}{2b_0}\| \sqrt{B(t)}u_{x}(t)\|^2+\frac{1}{2}\| u'(t)\|^2, \\ \langle B(t)u_{x}'(t),u_{x}(t)\rangle \leq\frac{1}{2}\| \sqrt{B(t)}u_{x}'(t)\|^2 +\frac{1}{2}\| \sqrt{B(t)}u_{x}(t)\|^2. \end{gathered} \end{equation} Then \begin{equation} \label{s20} \begin{aligned} \mathcal{L} (t) &\geq \frac{1}{2}( 1-\delta ) \|u'(t)\|^2+\frac{1}{2}( 1-\delta ) \|\sqrt{B(t)}u_{x}'(t)\|^2 \\ &\quad +( \frac{1}{2}-\frac{1}{p}-\frac{\delta }{2b_0}) \| \sqrt{B(t)}u_{x}(t)\|^2+\tilde{d}_1\| u_{x}(t)\| _{L^p}^p+\frac{1}{p}I(t) \\ &\geq \bar{\beta}_1E_1(t), \end{aligned} \end{equation} where $\delta $ is small enough, and \begin{equation} \bar{\beta}_1=\min \big\{ \frac{1-\delta }{2},\; \frac{1}{2}-\frac{1}{p}-\frac{\delta }{2b_0},\; \tilde{d}_1,\; \frac{1}{p}\big\} >0, \quad 0<\delta <\min \big\{1,\frac{(p-2)b_0}{p}\big\}. \label{s21} \end{equation} Similarly, \begin{equation} \begin{aligned} \mathcal{F} (t) &\leq\frac{1}{2}( 1+\delta ) \|u'(t)\|^2+\frac{1}{2}( 1+\delta ) \| \sqrt{B(t)}u_{x}'(t)\|^2 \label{s22} \\ &\quad +\tilde{d}_1\| u_{x}(t)\| _{L^p}^p+\frac{1}{p}I(t) \\ &\quad +[ \frac{1}{2}-\frac{1}{p}+\frac{\delta }{2}( 1+\frac{1}{b_0}+ \frac{\lambda }{b_0}+\lambda _1) ] \| \sqrt{B(t)} u_{x}(t)\|^2 \\ &\leq \bar{\beta}_2E_1(t), \end{aligned} \end{equation} where \begin{equation} \bar{\beta}_2=\max \big\{ \frac{1+\delta }{2},\text{ }\frac{1}{2}-\frac{1 }{p}+\frac{\delta }{2}( 1+\frac{1}{b_0}+\frac{\lambda }{b_0} +\lambda _1) ,\text{ }\tilde{d}_1\big\} >0. \label{s23} \end{equation} Next, we show that the functional $\Psi (t)$ satisfies \begin{equation} \label{s24} \begin{aligned} \Psi'(t) &\leq\| u'(t)\|^2+\Big( 1+ \frac{b_1^2}{2\varepsilon _2b_0}\Big) \| \sqrt{B(t)} u_{x}'(t)\|^2 \\ &\quad -\Big( 1-\frac{d_2}{p}-\frac{\varepsilon _2}{b_0}\Big) \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad -\frac{d_2}{p}I(t)-d_2\tilde{d}_1\| u_{x}(t)\| _{L^p}^p+\frac{1}{2\varepsilon _2}\| f(t)\|^2, \end{aligned} \end{equation} for all $\varepsilon _2>0$. The proof is as follows. Multiplying \eqref{z1}$_1$ by $u(x,t)$ and integrating over $[0,1]$, we obtain \begin{equation} \begin{aligned} \Psi'(t) &= \| u'(t)\|^2+\| \sqrt{B(t)}u_{x}'(t)\|^2-\| \sqrt{B(t)} u_{x}(t)\|^2 \label{s25} \\ &\quad +\langle B'(t)u_{x}'(t),u_{x}(t)\rangle +\frac{\lambda _1}{2}\int_0^1B'(x,t)u_{x}^2(x,t)dx \\ &\quad +\langle F( u(t),u_{x}(t)) ,u(t)\rangle +\langle G( u(t),u_{x}(t)) ,u_{x}(t)\rangle +\langle f(t),u(t)\rangle . \end{aligned} \end{equation} Furthermore, by (H4'')$_{(iii)}$, we obtain \begin{equation} \label{s26} \begin{aligned} &\langle F( u(t),u_{x}(t)) ,u(t)\rangle +\langle G( u(t),u_{x}(t)) ,u_{x}(t)\rangle \\ &\leq d_2\int_0^1\mathcal{F} ( u(x,t),u_{x}(x,t)) dx \\ &= d_2[ \int_0^1\mathcal{F} _1( u(x,t),u_{x}(x,t)) dx-\tilde{d}_1\| u_{x}(t)\|_{L^p}^p] \\ &= \frac{d_2}{p}( \| \sqrt{B(t)}u_{x}(t)\| ^2-I(t)) -d_2\tilde{d}_1\| u_{x}(t)\|_{L^p}^p. \end{aligned} \end{equation} We also have \begin{equation} \label{s27} \begin{gathered} \frac{\lambda _1}{2}\int_0^1B'(x,t)u_{x}^2(x,t)dx \leq0, \\ \langle B'(t)u_{x}'(t),u_{x}(t)\rangle \leq\frac{b_1^2}{2\varepsilon _2b_0}\| \sqrt{B(t)}u_{x}'(t)\|^2 +\frac{\varepsilon _2}{2b_0}\| \sqrt{B(t)} u_{x}(t)\|^2, \\ \langle f(t),u(t)\rangle \leq\frac{\varepsilon _2}{2b_0}\| \sqrt{B(t)}u_{x}(t)\|^2+\frac{1}{2\varepsilon _2}\|f(t)\|^2, \end{gathered} \end{equation} for all $\varepsilon _2>0$. Combining \eqref{s25}--\eqref{s27}, we obtain \eqref{s24}. The estimates \eqref{s8}$_{(ii)}$ and \eqref{s24} give \begin{equation} \label{s28} \begin{aligned} \mathcal{L}'( t) &\leq-\big( \lambda -\frac{ \varepsilon _1}{2}-\delta \big) \| u'(t)\|^2 \\ &\quad -\big[ \lambda _1-\delta \big( 1+\frac{b_1^2}{2\varepsilon _2b_0}\big) \big] \| \sqrt{B(t)}u_{x}'(t)\|^2 \\ &\quad -\delta ( 1-\frac{d_2}{p}-\frac{\varepsilon _2}{b_0}) \| \sqrt{B(t)}u_{x}(t)\|^2 \\ &\quad -\frac{\delta d_2}{p}I(t)-\delta d_2\tilde{d}_1\| u_{x}(t)\| _{L^p}^p +\frac{1}{2}\big( \frac{1}{\varepsilon _1 }+\frac{\delta }{\varepsilon _2}\big) \| f(t)\|^2, \end{aligned} \end{equation} for all $\delta $, $\varepsilon _1$, $\varepsilon _2>0$. Because $p>\max \{2,d_2\}\geq d_2$, we can choose $\varepsilon _2>0$ such that \begin{equation} \theta _1=1-\frac{d_2}{p}-\frac{\varepsilon _2}{b_0}>0. \label{s29} \end{equation} Then, for $\varepsilon _1$ small enough such that $0<\frac{\varepsilon _1}{2}<\lambda $ and if $\delta >0$ such that \begin{equation} \begin{gathered} \theta _2=\lambda -\frac{\varepsilon _1}{2}-\delta >0,\quad \theta_3=\lambda _1-\delta ( 1+\frac{b_1^2}{2\varepsilon _2b_0}) >0,\\ 0<\delta <\min \{1,\frac{(p-2)b_0}{p}\}. \end{gathered} \label{s30} \end{equation} By \eqref{s28}-\eqref{s30}, we obtain \begin{equation} \mathcal{L}'(t)\leq -\bar{\beta}_3E_1(t)+\tilde{C}_1e^{-\bar{ \eta}_1t}\leq -\frac{\bar{\beta}_3}{\bar{\beta}_2}\mathcal{L} (t)+ \tilde{C}_1e^{-\bar{\eta}_1t}\leq -\bar{\gamma}\mathcal{L} (t)+\tilde{C} _1e^{-\bar{\eta}_1t}, \label{s31} \end{equation} where $\bar{\beta}_3=\min \{\delta \theta _1$, $\theta _2$, $\theta _3$, $\frac{\delta d_2}{p}$, $\delta d_2\tilde{d}_1\}$, $0<\bar{\gamma}<\min \{\frac{\beta _{\ast }}{\bar{\beta}_2}$, $\bar{\eta}_1\}$, $\tilde{C}_1=\frac{1}{2}( \frac{1}{\varepsilon _1}+\frac{\delta }{ \varepsilon _2}) \bar{C}_1$. 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