A further study on the coupled Allen–Cahn/Cahn–Hilliard equations

Abstract

In this paper, we will show that solutions of the initial boundary value problem for the coupled system of Allen–Cahn/Cahn–Hilliard equations continuously depend on parameters of the system, and under some restrictions on the parameters all solutions of the initial boundary value problem for Allen–Cahn/Cahn–Hilliard equations tend to zero with an exponential rate as $t\rightarrow\infty$.

1 Introduction

In this paper, we consider the following Allen–Cahn/Cahn–Hilliard system:

$$\begin{aligned} &\partial_{t}A-\alpha A-\partial_{x}^{2}A+k \vert A \vert ^{2}A=Ah,\quad x\in(0,l), t>0, \end{aligned}$$

(1.1)

$$\begin{aligned} &\partial_{t}h+ \vert A \vert \partial _{x}h=m\partial_{x}^{2}\mu,\quad x\in(0,l), t>0, \end{aligned}$$

(1.2)

$$\begin{aligned} &A(x,0)=A_{0}(x),\qquad h(x,0)=h_{0}(x),\quad x\in (0,l), \end{aligned}$$

(1.3)

$$\begin{aligned} &A(0,t)=A(l,t)=\vec{0},\qquad \partial_{x}h(x,t)|_{x=0,l}= \partial_{x}^{3}h(x,t)|_{x=0,l}=0,\quad t>0, \end{aligned}$$

(1.4)

where $\mu=f'(h)-\gamma\partial_{x}^{2}h$, $f'(h)=h^{3}-h$, and $\vec {0}$ is a zero vector of $R^{N}$, $mk>\frac{5}{4}$, $m, k, \alpha>0$ are given numbers, $A(x,t)=(A_{1}(x,t),\ldots,A_{N}(x,t))$ is the unknown vector function, $h(x,t)$ is the unknown scalar function, $A_{0}(x)$ and $h_{0}(x)$ are given initial data.

System (1.1)–(1.4) was introduced to model simultaneous order-disorder and phase separation in binary alloys on a BCC lattice in the neighborhood of the triple point [1]. Here, h denotes the concentration of one of the components, while A is an order parameter. The Allen–Cahn equation and the Cahn–Hilliard equation have been intensively studied [2,3,4,5]. Miranville, Saoud, and Talhouk [5] studied the long time behavior, in terms of finite-dimensional attractors, of a coupled Allen–Cahn/Cahn–Hilliard system. In particular, they proved the existence of an exponential attractor and, as a consequence, the existence of a global attractor with finite fractal dimension. Çelebi and Kalantarov [6] proved the decay of solutions and structural stability for the coupled Kuramoto–Sivashinsky–Ginzburg–Landau equations.

The large time behavior and the structural stability of solutions are important for the study of a higher-order parabolic system. Many papers have already been published to study the decay and the structural stability of solutions [7,8,9]. In this paper, we consider the asymptotic behavior of solutions and the continuous dependence of solutions for system (1.1)–(1.4). We are going to show the continuous dependence when the coefficient changes, which helps us to know whether a coefficient in the system can cause a large change in the solution.

The following is the main result of the paper.

Theorem 1.1

If

$$ \alpha< \lambda_{1} \quad \textit{and}\quad \lambda_{1} \biggl( \gamma m-\frac{1}{2\lambda_{1}^{2}} \biggr)-\frac{3m}{2}>0, $$

(1.5)

then all solutions of problem (1.1)–(1.4) tend to zero with an exponential rate as $t\rightarrow\infty$.

Theorem 1.1 implies that the concentration of one of the components and the order parameter will tend to zero as $t\rightarrow \infty$. Hence one of the components will disappear and the system will become disorder in a background point of view.

To prove Theorem 1.1, the basic a priori estimates are the $L^{2}$ norm estimates on h and $\partial_{x} h$. The main difficulties are caused by the nonlinearity of both the diffusive and the convective factors in equation (1.2). To overcome such difficulty, we establish two new functionals $E_{1}(t)$ and $E_{2}(t)$. Our method is based on the global energy estimates and require some delicate local integral estimates.

This paper is arranged as follows. We first study a priori estimates in Sect. 2, and then establish the exponential decay of solution in Sect. 3. Subsequently, we discuss the continuous dependence results in Sect. 4.

2 A priori estimates

Similar to [10], we know that problem (1.1)–(1.4) has a unique global solution. The first step is to obtain a priori estimates of solutions of system (1.1), (1.2). Applying the operator $P^{2}$ to both sides of equation (1.2), here $P^{2}$ is the inverse operator of the operator $L=-\frac{d^{2}}{dx^{2}}$ with the domain of definition $D(L)=H^{2}(0,l)\cap H_{0}^{1}(0,l)$, we get the following problem:

$$\begin{aligned} &\partial_{t}A-\alpha A-\partial_{x}^{2}A+k \vert A \vert ^{2}A=Ah,\quad x\in(0,l), t>0, \end{aligned}$$

(2.1)

$$\begin{aligned} &P^{2}\partial_{t}h+P^{2}\bigl( \vert A \vert \partial_{x}h\bigr)=-m\mu,\quad x\in(0,l), t>0, \end{aligned}$$

(2.2)

$$\begin{aligned} &A(x,0)=A_{0}(x),\qquad h(x,0)=h_{0}(x),\quad x\in (0,l), \end{aligned}$$

(2.3)

$$\begin{aligned} &A(0,t)=A(l,t)=\vec{0},\qquad \partial_{x}h(0,t)=\partial _{x}h(l,t)=0,\quad t>0. \end{aligned}$$

(2.4)

Multiplying equation (2.1) by A and (2.2) by h shows

$$ \frac{1}{2}\frac{d}{dt} \bigl\Vert A(t) \bigr\Vert ^{2}-\alpha\bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx=\bigl( \bigl\vert A(t) \bigr\vert ^{2},h(t) \bigr) $$

(2.5)

and

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \int_{0}^{l}\bigl( \bigl\vert A(x,t) \bigr\vert \partial_{x}h(x,t)\bigr)P^{2}h(x,t)\,dx \\ &\quad =-m \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx+m \bigl\Vert h(t) \bigr\Vert ^{2}-\gamma m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}. \end{aligned}$$

(2.6)

Adding the two resulting equations together, we obtain

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\bigl( \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}\bigr)- \alpha\bigl\Vert A(t) \bigr\Vert ^{2} + \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2}+m \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx \\ &\qquad{}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\gamma m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad =m \bigl\Vert h(t) \bigr\Vert ^{2}+ \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}h(x,t)\,dx - \int_{0}^{l}\bigl( \bigl\vert A(x,t) \bigr\vert \partial_{x}h(x,t)\bigr)P^{2}h(x,t)\,dx \\ &\quad \leq m \bigl\Vert h(t) \bigr\Vert ^{2}+ \frac{1}{2m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\frac{m}{2} \bigl\Vert h(t) \bigr\Vert ^{2} +\frac{1}{8m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\qquad{}+\frac{m\lambda_{1}^{4}}{2} \int_{0}^{l} \bigl\vert P^{2}h(x,t) \bigr\vert ^{4}\,dx +\frac{1}{2\lambda_{1}^{2}} \int_{0}^{l} \bigl\vert \partial _{x}h(x,t) \bigr\vert ^{2}\,dx \\ &\quad\leq\frac{3m}{2} \bigl\Vert h(t) \bigr\Vert ^{2}+\frac{5}{8m} \int_{0}^{l} \bigl\vert A(x) \bigr\vert ^{4}\,dx +\frac{1}{2\lambda_{1}^{2}} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}+\frac{m}{2} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx, \end{aligned}$$

(2.7)

that is,

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\bigl( \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}\bigr)+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2} + \frac{m}{2} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx \\ &\qquad{}+\frac{k}{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\biggl(\gamma m-\frac{1}{2\lambda_{1}^{2}}\biggr) \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2} \\ &\quad \leq\alpha\bigl\Vert A(t) \bigr\Vert ^{2}+\frac{3}{2}m \bigl\Vert h(t) \bigr\Vert ^{2}. \end{aligned}$$

(2.8)

Due to the Cauchy–Schwarz inequality, we have

$$ \Vert h \Vert ^{2}=\bigl(P^{-1}h,Ph\bigr)\leq\bigl\Vert P^{-1}h \bigr\Vert \Vert Ph \Vert \leq\varepsilon \Vert \partial_{x}h \Vert ^{2} +C_{1}(\varepsilon) \Vert Ph \Vert ^{2}. $$

Choosing $\varepsilon=\frac{\gamma}{3}-\frac{1}{3m\lambda_{1}^{2}}$ and $C_{1}(\varepsilon)=\frac{1}{4\varepsilon}$ for the last inequality and using it in (2.8), we get

$$\begin{aligned} &\frac{d}{dt}\bigl( \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}\bigr)+2 \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2} +m \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx\\ &\qquad{}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx +\gamma m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad\leq2\alpha\bigl\Vert A(t) \bigr\Vert ^{2}+C_{2}( \gamma,m,\lambda_{1}) \bigl\Vert Ph(t) \bigr\Vert ^{2}. \end{aligned}$$

(2.9)

Thanks to (2.9), it is true that

$$\begin{aligned} \begin{aligned} & \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}\leq D_{1}(t),\quad \forall t\in R^{+}, \\ & \int_{0}^{t} \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2}\,ds, k \int_{0}^{t} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx\,ds\leq D_{2}(t), \end{aligned} \end{aligned}$$

(2.10)

and

$$\begin{aligned} \gamma m \int_{0}^{t} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2}\,ds, m \int_{0}^{t} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx\,ds \leq D_{2}(t),\quad \forall t\in R^{+}, \end{aligned}$$

(2.11)

where $D_{1}(t)=[\|A_{0}\|^{2}+\|Ph_{0}\|^{2}]e^{(2\alpha+C_{2})t}$, $D_{2}(t)=D_{1}(t)+\|A_{0}\|^{2}+\|Ph_{0}\|^{2}$.

Based on (2.9) and (2.10), we use the standard Faedo–Galerkin method to prove the existence of a global weak solution $[A,h]$ of problem (2.1)–(2.4) with the following properties:

$$\begin{aligned} & A\in C\bigl(0,T;L^{2}(0,l)\bigr)\cap L^{2} \bigl(0,T;H_{0}^{1}(0,l)\bigr), \end{aligned}$$

(2.12)

$$\begin{aligned} & h\in C\bigl(0,T;H^{-1}(0,l)\bigr)\cap L^{2} \bigl(0,T;H_{0}^{1}(0,l)\bigr). \end{aligned}$$

(2.13)

Multiplying (1.2) by h in $L^{2}(0,l)$, we have

$$\begin{aligned} \frac{1}{2}\frac{d}{dt} \bigl\Vert h(t) \bigr\Vert ^{2} ={}&{-}3m \int_{0}^{l}h^{2}(x,t)\partial _{x}h^{2}(x,t)\,dx+m \int_{0}^{l} \bigl\vert \partial _{x}h(x,t) \bigr\vert ^{2}\,dx \\ &{}+\gamma m \int_{0}^{l}\partial_{x}h(x,t)\partial _{x}^{3}h(x,t)\,dx - \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\partial_{x}h(x,t)\cdot h\,dx. \end{aligned}$$

By Young’s inequality, we see that

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert h(t) \bigr\Vert ^{2}+\gamma m \bigl\Vert \partial_{x}^{2}h(t) \bigr\Vert ^{2} \\ &\quad \leq m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}+\frac{1}{4m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx +\frac{1}{4m} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx +\frac{m}{2} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad \leq\frac{3m}{2} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}+\frac{1}{4m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx +\frac{1}{4m} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx. \end{aligned}$$

(2.14)

Applying (2.11), integrating inequality (2.14) in $(0,t)$ gives the following estimate:

$$\begin{aligned} &\frac{1}{2} \bigl\Vert h(t) \bigr\Vert ^{2}+ \gamma m \int_{0}^{t} \bigl\Vert \partial _{x}^{2}h(s) \bigr\Vert ^{2}\,ds \\ &\quad \leq \biggl(\frac{1}{4mk}+\frac{1}{4m^{2}} \biggr)D_{2}(t) +\frac{3m}{2} \int_{0}^{t} \bigl\Vert \partial _{x}h(s) \bigr\Vert ^{2}\,ds+ \frac{1}{2} \Vert h_{0} \Vert ^{2} \\ &\quad \leq D_{3}(t),\quad \forall t\in[0,T], \end{aligned}$$

(2.15)

where $D_{3}(t):= (\frac{1}{4mk}+\frac{1}{4m^{2}}+\frac{3}{2\gamma} )D_{2}(t) +\frac{1}{2}\|h_{0}\|^{2}$. Therefore, we deduce

$$ \bigl\Vert h(t) \bigr\Vert ^{2}, 2\gamma m \int_{0}^{t} \bigl\Vert \partial _{x}^{2}h(s) \bigr\Vert ^{2}\,ds\leq 2D_{3}(T), \quad \forall t\in[0,T]. $$

(2.16)

Multiplying equation (1.1) by $\partial_{t}A$ in $L^{2}(0,l)$, we see

$$\begin{aligned} & \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+\frac{d}{dt} \biggl[\frac{1}{2} \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2}- \frac{\mu}{2} \bigl\Vert A(t) \bigr\Vert ^{2}+\frac{k}{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \biggr] \\ &\quad = \bigl(\bigl\langle A(t),\partial_{t}A(t)\bigr\rangle ,h(t) \bigr). \end{aligned}$$

(2.17)

Depending on Sobolev’s imbedding theorem and estimates (2.11), (2.16), the term on the right-hand side of (2.17) can be estimated as follows:

$$\begin{aligned} \bigl\vert \bigl(\bigl\langle A(t),\partial_{t}A(t)\bigr\rangle ,h(t) \bigr) \bigr\vert \leq& \bigl\Vert A(t) \bigr\Vert _{L^{\infty}(0,l)} \bigl\Vert h(t) \bigr\Vert \bigl\Vert \partial _{t}A(t) \bigr\Vert \\ \leq&\frac{1}{2} \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert A(t) \bigr\Vert _{L^{\infty}(0,l)}^{2} \bigl\Vert h(t) \bigr\Vert ^{2} \\ \leq&\frac{1}{2} \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+\frac{l}{2} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2} \bigl\Vert h(t) \bigr\Vert ^{2} \\ \leq&\frac{1}{2} \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+lD_{3}(t) \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}, \quad \forall t\in[0,T]. \end{aligned}$$

Thus, according to (2.17),

$$\begin{aligned} & \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+ \frac{d}{dt} \biggl[ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}- \mu\bigl\Vert A(t) \bigr\Vert ^{2}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \biggr] \\ &\quad \leq 2lD_{3}(t) \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2},\quad \forall t\in[0,T]. \end{aligned}$$

It is easy to obtain the estimate

$$ \int_{0}^{t} \bigl\Vert \partial _{s}A(s) \bigr\Vert ^{2}\,ds, \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2}\leq D(T), \quad \forall t\in[0,T], $$

(2.18)

where

$$ D(T):=2lD_{3}(T)D_{2}(t)+\alpha D_{1}(T)+ \Vert \partial_{x}A_{0} \Vert ^{2}-\alpha \Vert A_{0} \Vert ^{2}+ k \int_{0}^{l} \bigl\vert A_{0}(x) \bigr\vert ^{4}\,dx. $$

Remark 2.1

If $\lambda_{1} (\gamma m-\frac{1}{2\lambda_{1}^{2}} )-\frac {3m}{2}=r_{0}>0$, the following uniform estimate holds true:

$$ \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2} \leq\bigl[ \Vert A_{0} \Vert ^{2}+ \Vert Ph_{0} \Vert ^{2} \bigr]e^{-\gamma_{1}t}+\frac{\alpha^{2}l}{2k\gamma_{1}}, $$

(2.19)

where $\gamma_{1}:=2\lambda_{1}\min{\{1,r_{0}\}}$.

Proof

We deduce from (2.6) the inequality

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert Ph(t) \bigr\Vert ^{2} \\ &\quad =-m \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx+m \bigl\Vert h(t) \bigr\Vert ^{2}-\gamma m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\qquad{}- \int_{0}^{l}\bigl( \bigl\vert A(x,t) \bigr\vert \partial_{x}h(x,t)\bigr)P^{2}h(x,t)\,dx \\ &\quad \leq -m \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx+m \bigl\Vert h(t) \bigr\Vert ^{2}-\gamma m \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} + \frac{1}{8m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\qquad{}+\frac{m\lambda_{1}^{4}}{2} \int_{0}^{l} \bigl\vert P^{2}h(x,t) \bigr\vert ^{4}\,dx +\frac{1}{2\lambda_{1}^{2}} \int_{0}^{l} \bigl\vert \partial _{x}h(x,t) \bigr\vert ^{2}\,dx \\ &\quad \leq-\frac{m}{2} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert ^{4}\,dx+m \bigl\Vert h(t) \bigr\Vert ^{2}- \biggl(\gamma m-\frac{1}{2\lambda_{1}^{2}} \biggr) \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\qquad{} +\frac{1}{8m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq m \bigl\Vert h(t) \bigr\Vert ^{2}- \biggl(\gamma m- \frac{1}{2\lambda_{1}^{2}} \biggr) \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} +\frac{1}{8m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx. \end{aligned}$$

From (2.5), we know that

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert A(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq \frac{1}{2m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\frac{m}{2} \bigl\Vert h(t) \bigr\Vert ^{2} +\frac{k}{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\frac{\alpha^{2}l}{2k}. \end{aligned}$$

Adding the above two inequalities, we derive

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr) + \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ & \quad \leq\frac{3m}{2} \bigl\Vert h(t) \bigr\Vert ^{2}- \biggl( \gamma m-\frac{1}{2\lambda_{1}^{2}} \biggr) \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2} +k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx +\frac{\alpha^{2}l}{2k}, \end{aligned}$$

that is,

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+ \biggl( \gamma m-\frac{1}{2\lambda_{1}^{2}} \biggr) \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2} \\ &\quad{}-\frac{3m}{2} \bigl\Vert h(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}\leq\frac{\alpha^{2}l}{2k}. \end{aligned}$$

Hence

$$ \frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+ r_{0} \bigl\Vert h(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}\leq\frac{\alpha^{2}l}{2k}. $$

Taking $\gamma_{1}=2\lambda_{1}\min{\{1,r_{0}\}}$, we have

$$ \frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+\gamma_{1} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr) \leq\frac{\alpha^{2}l}{2k}. $$

We get (2.19) by integrating the last inequality. □

3 Exponential decay of solution

In this section, we are going to prove the exponential decay of solution.

Proof of Theorem 1.1.

Multiplying in $L^{2}(0,l)$ (2.1) by A, (2.2) by h, and adding the obtained relations, we get

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2} +k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx-\alpha\bigl\Vert A(t) \bigr\Vert ^{2} \\ &\quad{}+ \biggl(\gamma m-\frac{1}{2\lambda_{1}^{2}} \biggr) \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} -\frac{3m}{2} \bigl\Vert h(t) \bigr\Vert ^{2}-\frac{5}{8m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx\leq0, \end{aligned}$$

(3.1)

that is,

$$ \frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+d_{0} \bigl\Vert A(t) \bigr\Vert ^{2} + \frac{k}{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+r_{0}\lambda_{1} \bigl\Vert Ph(t) \bigr\Vert ^{2}\leq0, $$

which implies

$$ \frac{1}{2}\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+ \gamma_{0}\bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2}\bigr) +\frac{k}{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx\leq0, $$

(3.2)

where $d_{0}:=\lambda_{1}-\alpha$, $r_{0}:=\lambda_{1} (\gamma m-\frac{1}{2\lambda_{1}^{2}} )-\frac{3m}{2}$, and $\gamma_{0}=\min{\{d_{0},r_{0}\lambda_{1}\}}$. Hence, we have

$$ \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2}\leq\bigl( \Vert Ph_{0} \Vert ^{2}+ \Vert A_{0} \Vert ^{2} \bigr)e^{-2\gamma_{0}t}. $$

(3.3)

We conclude from (3.1) that

$$\begin{aligned} &\frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr) +2d_{0}\lambda_{1}^{-1} \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2}+2r_{0}\lambda _{1}^{-1} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad{}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx\leq0. \end{aligned}$$

(3.4)

Integrating this inequality over the interval $(0,t)$ and employing estimate (3.3), we obtain

$$ \frac{d_{0}}{\lambda_{1}} \int_{0}^{t} \bigl\Vert \partial _{x}A(\tau) \bigr\Vert ^{2}\,d\tau+ \frac{r_{0}}{\lambda_{1}} \int_{0}^{t} \bigl\Vert \partial _{x}h(\tau) \bigr\Vert ^{2}\,d\tau\leq\frac{1}{2} \bigl( \Vert Ph_{0} \Vert ^{2}+ \Vert A_{0} \Vert ^{2} \bigr),\quad \forall t>0. $$

(3.5)

We can know that if $A_{0}, h_{0}\in H_{0}^{1}(0,l)$ then problem (1.1)–(1.4) has a unique weak solution such that

$$ A,h\in C\bigl(0,T;H_{0}^{1}(0,l)\bigr)\cap L^{2} \bigl(0,T;H^{2}(0,l)\bigr),\quad \forall T>0. $$

Taking the inner product of (2.2) with $-\partial _{x}^{2}h$, we have

$$ \frac{d}{dt} \bigl\Vert h(t) \bigr\Vert ^{2}+2 \frac{1}{\lambda_{1}}(\lambda_{1}\gamma m-m) \bigl\Vert \partial _{x}^{2}h(t) \bigr\Vert ^{2} \leq \frac{1}{2m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\,dx. $$

(3.6)

Besides, we know

$$ \frac{d}{dt} \bigl( \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+2d_{0} \bigl\Vert A(t) \bigr\Vert ^{2} +k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+2r_{0} \bigl\Vert h(t) \bigr\Vert ^{2}\leq0. $$

Let us multiply (3.6) by a positive parameter $\varepsilon_{1}$ and add it with the above inequality

$$\begin{aligned} &\frac{d}{dt} \bigl(\varepsilon_{1} \bigl\Vert h(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2} \bigr)+ \frac{2(\lambda_{1}\gamma m-m)\varepsilon_{1}}{\lambda_{1}} \bigl\Vert \partial_{x}^{2}h(t) \bigr\Vert ^{2} \\ &\quad{}+\biggl(2d_{0}-\frac{\varepsilon_{1}}{2m}\biggr) \bigl\Vert A(t) \bigr\Vert ^{2} +2r_{0} \bigl\Vert h(t) \bigr\Vert ^{2}+k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx\leq0. \end{aligned}$$

Choosing $\varepsilon_{1}=2md_{0}$, we obtain the inequality

$$\begin{aligned} &\frac{d}{dt}E_{1}(t)+\delta_{1}E_{1}(t)+ \frac{2(\lambda_{1}\gamma m-m)\varepsilon_{1}}{\lambda_{1}} \bigl\Vert \partial_{x}^{2}h(t) \bigr\Vert ^{2} +k \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\quad\leq0,\quad \forall t\geq0, \end{aligned}$$

(3.7)

where

$$ E_{1}(t):=\varepsilon_{1} \bigl\Vert h(t) \bigr\Vert ^{2}+ \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \bigl\Vert A(t) \bigr\Vert ^{2}, $$

and $\delta_{1}:=\min{\{d_{0},\lambda_{1}r_{0},\frac {r_{0}}{\varepsilon_{1}}\}}$. Then we obtain

$$ E_{1}(t)\leq E_{1}(0)e^{-\delta_{1}t},\quad \forall t\geq0. $$

(3.8)

Taking the inner product in $L^{2}(0,l)$ of (2.1) with $\partial_{t}A$, we get

$$\begin{aligned} & \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+ \frac{d}{dt} \biggl(-\frac{\alpha}{2} \bigl\Vert A(t) \bigr\Vert ^{2} +\frac{1}{2} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+\frac{k}{4} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \biggr) \\ &\quad \leq \bigl\Vert \partial_{t}A(t) \bigr\Vert ^{2}+ \frac{1}{4} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2} \bigl\vert h(x,t) \bigr\vert ^{2}\,dx. \end{aligned}$$

According to [6],

$$\begin{aligned} &\frac{d}{dt} \biggl(-\frac{\alpha}{2} \bigl\Vert A(t) \bigr\Vert ^{2} +\frac{1}{2} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+\frac{k}{4} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \biggr) \\ &\quad \leq\varepsilon_{2} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx+\varepsilon_{3} \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2} +\frac{1}{256\varepsilon_{2}^{2}\varepsilon _{3}} \bigl\Vert h(t) \bigr\Vert ^{6}. \end{aligned}$$

Multiplying (3.4) by $\frac{1}{2}$ and adding with the inequality, we obtain

$$\begin{aligned} &\frac{d}{dt}E_{2}(t)+\frac{d_{0}}{\lambda_{1}} \bigl\Vert \partial _{x}A(t) \bigr\Vert ^{2} +\biggl(\frac{r_{0}}{\lambda_{1}}- \varepsilon_{3}\biggr) \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} +\biggl(\frac{k}{2}-\varepsilon_{2} \biggr) \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq\frac{1}{256\varepsilon_{2}^{2}\varepsilon_{3}} \bigl\Vert h(t) \bigr\Vert ^{6}, \end{aligned}$$

where

$$ E_{2}(t):=\frac{(1-\alpha)}{2} \bigl\Vert A(t) \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert Ph(t) \bigr\Vert ^{2}+ \frac{1}{2} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}+\frac{k}{4} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx. $$

We choose in the last inequality $\varepsilon_{2}=\frac{k}{4}$, $\varepsilon_{3}=\frac{r_{0}}{2\lambda _{1}}$ and obtain the equality

$$ \frac{d}{dt}E_{2}(t)+\delta_{2}E_{2}(t)\leq A_{0} \bigl\Vert A(t) \bigr\Vert ^{2}+A_{1} \bigl\Vert h(t) \bigr\Vert ^{6}, \quad\forall t\geq0, $$

where

$$ \delta_{2}:=\min{\biggl\{ \frac{2d_{0}}{\lambda_{1}},1,2r_{0} \lambda_{1}\biggr\} },\qquad A_{0}=\frac{\delta_{2} \vert 1-\alpha \vert }{2},\qquad A_{1}=\frac{1}{256\varepsilon_{2}^{2}\varepsilon_{3}}. $$

Using estimate (3.8), we have

$$ \frac{d}{dt}E_{2}(t)+\delta_{2}E_{2}(t) \leq A_{2}e^{-\delta_{1}t}, $$

(3.9)

where $A_{2}=A_{1}E_{1}^{3}(0)+A_{0}E_{1}(0)$. Integrating (3.9) and by Gronwall’s inequality, we get

$$ E_{2}(t)\leq\frac{A_{2}}{\delta_{1}}e^{-\delta_{2}t}. $$

(3.10)

Combining with (3.3), we have

$$ \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2}\leq\frac{2A_{2}}{\delta_{1}}e^{-\delta_{2}t}+R_{0}e^{-2\gamma_{0}t}. $$

(3.11)

Multiplying (2.2) by $\partial_{t}h$, we obtain

$$\begin{aligned} & \bigl\Vert P\partial_{t}h(t) \bigr\Vert ^{2}+\frac{d}{dt} \biggl(-\frac{m}{2} \bigl\Vert h(t) \bigr\Vert ^{2}+ \frac{\gamma m}{2} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2} \biggr)\\ &\quad\leq-\frac{m}{4} \frac{d}{dt} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx \\ &\qquad{}+\frac{1}{4} \int_{0}^{l}\bigl[P\bigl( \bigl\vert A(x,t) \bigr\vert \partial_{x}h(x,t)\bigr)\bigr]^{2}\,dx+ \bigl\Vert P\partial_{t}h(t) \bigr\Vert ^{2}. \end{aligned}$$

(3.12)

Adding $\frac{m}{2}$ (3.6) to (3.12), we get

$$\begin{aligned} &\frac{d}{dt} \biggl(\frac{\gamma m}{2} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2}+ \frac{m}{4} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx \biggr) +2(\lambda_{1}\gamma m-m) \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad \leq\frac{1}{2m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\,dx+ \frac{l}{4\lambda_{1}} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{2} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}, \end{aligned}$$

thus

$$\begin{aligned} &\frac{d}{dt} \biggl(\frac{\gamma m}{2} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2}+ \frac{m}{4} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx \biggr) + \biggl[2(\lambda_{1}\gamma m-m)- \frac{l}{4\lambda_{1}} \biggr] \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2} \\ &\quad\leq\frac{1}{2m} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\,dx. \end{aligned}$$

Note that

$$ \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx\leq\bigl\Vert h(t) \bigr\Vert ^{2}_{\infty} \bigl\Vert h(t) \bigr\Vert ^{2} \leq l \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2} \bigl\Vert h(t) \bigr\Vert ^{2}\leq c \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}. $$

Hence, we know that

$$\begin{aligned} &\frac{d}{dt} \biggl(\frac{\gamma m}{2} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2}+ \frac{m}{4} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx \biggr) \\ &\quad{}+c \biggl(\frac{\gamma m}{2} \bigl\Vert \partial _{x}h(t) \bigr\Vert ^{2}+ \frac{m}{4} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx \biggr)\leq\frac{E_{2}(0)}{2m}e^{-\delta_{1}t}. \end{aligned}$$

(3.13)

Finally, we integrate (3.13) and get

$$ \frac{\gamma m}{2} \bigl\Vert \partial_{x}h(t) \bigr\Vert ^{2}+ \frac{m}{4} \int_{0}^{l} \bigl\vert h(x.t) \bigr\vert ^{4}\,dx\leq c_{1}e^{-ct}, $$

(3.14)

where $c_{1}=-\frac{E_{2}(0)}{2m\delta_{1}}e^{-\delta_{1}t}+\frac {m}{2}\|\partial_{x}h_{0}\|^{2}+ \frac{m}{4}\int_{0}^{l}|h_{0}|^{4}\,dx-\frac{E_{2}(0)}{2m\delta_{1}}$. The theorem is true. □

4 Continuous dependence results

Assume that $[\tilde{A},\tilde{h} ]$ is the weak solution of the problem

$$\begin{aligned} &\partial_{t}\tilde{A}-\alpha\tilde{A}-\partial _{x}^{2}\tilde{A}+\tilde{k} \vert \tilde{A} \vert ^{2}\tilde{A} =\tilde{A}\tilde{h},\quad x\in(0,l), t>0, \end{aligned}$$

(4.1)

$$\begin{aligned} &\partial_{t}\tilde{h}+ \vert \tilde{A} \vert \partial_{x}\tilde{h}=m\partial_{x}^{2}\tilde{ \mu},\quad x\in(0,l), t>0, \end{aligned}$$

(4.2)

$$\begin{aligned} &\tilde{A}(x,0)=A_{0}(x),\qquad \tilde{h}(x,0)=h_{0}(x),\quad x\in(0,l), \end{aligned}$$

(4.3)

$$\begin{aligned} &\tilde{A}(0,t)=\tilde{A}(l,t)=\vec{0},\qquad \partial_{x} \tilde{h}(0,t)=\partial_{x}^{3}\tilde{h}(0,t) =\partial _{x}\tilde{h}(l,t)=\partial_{x}^{3} \tilde{h}(l,t)=0,\quad t>0, \end{aligned}$$

(4.4)

where $\tilde{\mu}=f'(\tilde{h})-\gamma\partial_{x}^{2}\tilde{h}$, $f'(\tilde{h})=\tilde{h}^{3}-\tilde{h}$.

Theorem 4.1

Assume that $[A,h]$ is a solution of problem (1.1)–(1.4) and $[\tilde{A},\tilde{h} ]$ is a solution of problem (4.1)–(4.4). Let $[a,H]= [A-\tilde{A},h-\tilde{h} ]$, we have

$$ \bigl\Vert PH(t) \bigr\Vert ^{2}+ \bigl\Vert a(t) \bigr\Vert ^{2} \leq q_{0}e^{-\int_{0}^{t}R_{1}(s)\,ds} \int_{0}^{t} \bigl\Vert \partial _{x}A(s) \bigr\Vert ^{4}\,ds, $$

where $q_{0}=\frac{2k_{1}^{\frac{4}{3}}C_{0}^{4}}{(k_{2}b_{0})^{\frac{1}{3}}}+ \frac{C_{0}^{4}}{4m^{3}\lambda_{1}^{4}\gamma^{2}}$,

$$ R_{1}(t)=2C(\gamma,m)+2C_{1}(\gamma,m)+4l \bigl\Vert h(t) \bigr\Vert ^{2}+2\alpha+\frac{l}{\lambda_{1}^{2}m} \bigl\Vert \partial _{x}^{2}\tilde{h}(t) \bigr\Vert ^{2} +l \bigl\Vert \partial_{x}^{2}\tilde{A}(t) \bigr\Vert ^{2}. $$

Proof

Note that $[a,H]= [A-\tilde{A},h-\tilde{h} ]$ is a solution of the following problem:

$$\begin{aligned} &\partial_{t}a-\alpha a-\partial_{x}^{2}a+k \vert A \vert ^{2}A-\tilde{k} \vert \tilde{A} \vert ^{2}\tilde{A} =Ah-\tilde{A}\tilde{h},\quad x\in(0,l), t>0, \end{aligned}$$

(4.5)

$$\begin{aligned} &P^{2}\partial_{t}H-mH-\gamma m\partial _{x}^{2}H \\ &\quad =-m\bigl(h^{3}-\tilde{h}^{3} \bigr) +P^{2}\bigl( \vert \tilde{A} \vert \partial_{x} \tilde{h}\bigr) -P^{2}\bigl( \vert A \vert \partial _{x}h\bigr) ,\quad x\in(0,l), t>0, \\ &a(x,0)=\vec{0},\qquad H(x,0)=0, \quad x\in(0,l), \\ &a(0,t)=a(l,t)=\vec{0}, \qquad \partial_{x}H(0,t)=\partial _{x}H(0,t),\quad x\in(0,l), t>0. \end{aligned}$$

(4.6)

By

$$ Ah-\tilde{A}\tilde{h}=Ah-\tilde{A}h+\tilde{A}h-\tilde{A}\tilde {h}=ah+\tilde{A}H $$

and

$$\begin{aligned} P^{2}\bigl( \vert \tilde{A} \vert \partial_{x} \tilde{h}\bigr)-P^{2}\bigl( \vert A \vert \partial_{x}h \bigr) ={}&P^{2}\bigl( \vert \tilde{A} \vert \partial_{x} \tilde{h}- \vert A \vert \partial_{x}h\bigr) \\ ={}&P^{2}\bigl( \vert \tilde{A} \vert \partial_{x} \tilde{h}- \vert A \vert \partial_{x}\tilde{h} + \vert A \vert \partial_{x}\tilde{h}- \vert A \vert \partial_{x}h \bigr) \\ ={}&P^{2}\bigl(- \vert A \vert \partial_{x}H-\partial _{x}\tilde{h}\bigl( \vert A \vert - \vert \tilde{A} \vert \bigr)\bigr), \end{aligned}$$

we see that $[a,H]$ satisfies the following system:

$$\begin{aligned} &\partial_{t}a-\alpha a-\partial_{x}^{2}a+k_{2} \bigl( \vert A \vert ^{2}A- \vert \tilde{A} \vert ^{2} \tilde{A}\bigr) =k_{1} \vert A \vert ^{2}A+ah+\tilde{A}H, \end{aligned}$$

(4.7)

$$\begin{aligned} &P^{2}\partial_{t}H-mH-\gamma m\partial _{x}^{2}H \\ &\quad =-m\bigl(h^{3}-\tilde{h}^{3}\bigr) -P^{2}\bigl( \vert A \vert \partial_{x}H+\partial_{x}\tilde{h} \bigl( \vert A \vert - \vert \tilde{A} \vert \bigr)\bigr), \end{aligned}$$

(4.8)

where $k_{2}=\tilde{k}$, $k_{1}=\tilde{k}-k$.

On the other hand, we know that

$$ \bigl( \bigl\vert A(t) \bigr\vert ^{2}A(t)- \bigl\vert \tilde{A}(t) \bigr\vert ^{2}\tilde{A}(t),A(t)-\tilde{A}(t) \bigr) \geq b_{0} \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{4}\,dx. $$

(4.9)

Multiplying (4.7) by a and using inequality (4.9), we obtain

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert a(t) \bigr\Vert ^{2}-\alpha\bigl\Vert a(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}a(t) \bigr\Vert ^{2}+k_{2}b_{0} \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq k_{1} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\bigl\langle A(x,t),a(x,t)\bigr\rangle \,dx + \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{2}h(x,t)\,dx \\ &\qquad{}+ \int_{0}^{l}\bigl\langle \tilde{A}(x,t),a(x,t)\bigr\rangle H(x,t)\,dx. \end{aligned}$$

(4.10)

We are going to estimate the first integral on the right-hand side of (4.10) by the Nirenberg inequality as follows:

$$\begin{aligned} \biggl\vert \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{2}h(x,t)\,dx \biggr\vert &\leq\bigl\Vert a(t) \bigr\Vert _{\infty}^{2} \int_{0}^{l} \bigl\vert h(x,t) \bigr\vert \,dx \\ &\leq\sqrt{l} \bigl\Vert a(t) \bigr\Vert _{\infty}^{2} \bigl\Vert h(t) \bigr\Vert \\ &\leq2\sqrt{l} \bigl\Vert a(t) \bigr\Vert \bigl\Vert \partial _{x}a(t) \bigr\Vert \bigl\Vert h(t) \bigr\Vert \\ &\leq\frac{1}{2} \bigl\Vert \partial_{x}a(t) \bigr\Vert ^{2}+2l \bigl\Vert a(t) \bigr\Vert ^{2} \bigl\Vert h(t) \bigr\Vert ^{2}. \end{aligned}$$

(4.11)

We can infer from the Nirenberg inequality and the Friedrichs inequality that the following estimate of the second term on the right-hand side of (4.10) is true:

$$\begin{aligned} & \biggl\vert \int_{0}^{l}\bigl\langle \tilde{A}(x,t),a(x,t)\bigr\rangle H(x,t)\,dx \biggr\vert \\ &\quad \leq \int_{0}^{l} \bigl\vert \tilde{A}(x,t) \bigr\vert \bigl\vert a(x,t) \bigr\vert \bigl\vert H(x,t) \bigr\vert \,dx \\ &\quad \leq \bigl\Vert \tilde{A}(t) \bigr\Vert _{\infty} \bigl\Vert a(t) \bigr\Vert \bigl\Vert H(t) \bigr\Vert \\ &\quad \leq\frac{1}{2} \bigl\Vert \tilde{A}(t) \bigr\Vert _{\infty}^{2} \bigl\Vert a(t) \bigr\Vert ^{2}+ \frac{1}{2} \bigl\Vert H(t) \bigr\Vert ^{2} \\ &\quad \leq\frac{l}{2} \bigl\Vert \partial_{x} \tilde{A}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} +\varepsilon_{2} \bigl\Vert \partial _{x}H(t) \bigr\Vert ^{2}+C_{1}(\varepsilon _{2}) \bigl\Vert PH(t) \bigr\Vert ^{2}. \end{aligned}$$

(4.12)

Employing Young’s inequality and Sobolev’s inequality, we have

$$\begin{aligned} & \biggl\vert k_{1} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{2}\bigl\langle A(x,t),a(x,t)\bigr\rangle \,dx \biggr\vert \\ &\quad \leq k_{1} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{3} \bigl\vert a(x,t) \bigr\vert \,dx \\ &\quad \leq\frac{k_{1}^{\frac{4}{3}}}{(k_{2}b_{0})^{\frac{1}{3}}} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4} +k_{2}b_{0} \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq\frac{k_{1}^{\frac{4}{3}}C_{0}^{4}}{(k_{2}b_{0})^{\frac{1}{3}}} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{4}\,dx +k_{2}b_{0} \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{4}\,dx. \end{aligned}$$

Then employing the last inequality and (4.11), (4.12), we deduce the following inequality from (4.10):

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert a(t) \bigr\Vert ^{2}+ \bigl\Vert \partial_{x}a(t) \bigr\Vert ^{2}+k_{2}b_{0} \int_{0}^{l} \bigl\vert a(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq\alpha\bigl\Vert a(t) \bigr\Vert ^{2} + \frac{k_{1}^{\frac{4}{3}}C_{0}^{4}}{(k_{2}b_{0})^{\frac{1}{3}}} \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{4} +2l \bigl\Vert h(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2}+\frac{l}{2} \bigl\Vert \partial _{x}\tilde{A}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} \\ &\qquad{}+\varepsilon\bigl\Vert \partial_{x}H(t) \bigr\Vert ^{2}+C(\varepsilon) \bigl\Vert PH(t) \bigr\Vert ^{2}. \end{aligned}$$

(4.13)

Multiplying (4.8) in $L^{2}(0,l)$ by H, we get

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert PH(t) \bigr\Vert ^{2}-m \bigl\Vert H(t) \bigr\Vert ^{2}+ \gamma m \bigl\Vert \partial_{x}H(t) \bigr\Vert ^{2} \\ &\quad =-m\bigl(h^{3}-\tilde{h}^{3},H\bigr) +\bigl( \vert A \vert \partial_{x}H+\partial_{x}\tilde{h} \bigl( \vert A \vert - \vert \tilde{A} \vert \bigr),-P^{2}H \bigr) \end{aligned}$$

(4.14)

and

$$\begin{aligned} &\bigl( \vert A \vert \partial_{x}h+\partial_{x} \tilde{h}\bigl( \vert A \vert - \vert \tilde{A} \vert \bigr),-P^{2}H\bigr) \\ &\quad =- \int_{0}^{l} \vert A \vert \partial _{x}HP^{2}H\,dx - \int_{0}^{l}\partial_{x}\tilde{h}\bigl( \vert A \vert - \vert \tilde{A} \vert \bigr)P^{2}H\,dx \\ &\quad \leq\frac{1}{8m^{3}\lambda_{1}^{4}\gamma^{2}} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx +m\lambda_{1}^{4} \int_{0}^{l} \bigl\vert P^{2}H(t) \bigr\vert ^{4}\,dx +\frac{\gamma m}{2} \int_{0}^{l} \bigl\vert \partial _{x}H(t) \bigr\vert ^{2}\,dx \\ &\qquad{}+\frac{l}{2\lambda_{1}^{2}m} \bigl\Vert \partial_{x}^{2} \tilde{h}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} +\frac{\lambda_{1}^{2} m}{2} \int_{0}^{l} \bigl\vert P^{2}H(x,t) \bigr\vert ^{2}\,dx \\ &\quad \leq m \int_{0}^{l} \bigl\vert H(x,t) \bigr\vert ^{4}\,dx +\frac{l}{2\lambda_{1}^{2}m} \bigl\Vert \partial_{x}^{2} \tilde{h}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} +\frac{m}{2} \int_{0}^{l} \bigl\vert H(x,t) \bigr\vert ^{2}\,dx \\ &\qquad{}+\frac{\gamma m}{2} \int_{0}^{l} \bigl\vert \partial _{x}H(x,t) \bigr\vert ^{2}\,dx +\frac{1}{8m^{3}\lambda_{1}^{4}\gamma^{2}} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx \\ &\quad \leq m \int_{0}^{l} \bigl\vert H(x,t) \bigr\vert ^{4}\,dx +\frac{l}{2\lambda_{1}^{2}m} \bigl\Vert \partial_{x}^{2} \tilde{h}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} +\frac{5\gamma m}{8} \bigl\Vert \partial_{x}H(t) \bigr\Vert ^{2} \\ &\qquad{}+C_{1}(\gamma,m) \bigl\Vert PH(t) \bigr\Vert ^{2} + \frac{1}{8m^{3}\lambda_{1}^{4}\gamma^{2}} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx. \end{aligned}$$

Employing Sobolev’s imbedding theorem and the Nirenberg inequality, we get

$$\begin{aligned} & \bigl(h^{3}-\tilde{h}^{3},H\bigr)= \int_{0}^{l}H^{2}\bigl(h^{2}+h \tilde{h}+\tilde{h}^{2}\bigr)\,dx \geq \int_{0}^{l}H^{4}\,dx, \end{aligned}$$

(4.15)

$$\begin{aligned} &\vert m \vert \bigl\Vert H(t) \bigr\Vert ^{2} \leq\frac{\gamma m}{4} \bigl\Vert \partial_{x}H(t) \bigr\Vert ^{2}+C_{1}(m,\gamma) \bigl\Vert PH(t) \bigr\Vert ^{2}. \end{aligned}$$

(4.16)

Using (4.15), (4.16), we have

$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert PH(t) \bigr\Vert ^{2}+\frac{\gamma m}{8} \bigl\Vert \partial_{x}H(t) \bigr\Vert ^{2}\\ &\quad \leq C_{1}(\gamma,m) \bigl\Vert PH(t) \bigr\Vert ^{2} \\ &\qquad{}+\frac{l}{2\lambda_{1}^{2}m} \bigl\Vert \partial_{x}^{2} \tilde{h}(t) \bigr\Vert ^{2} \bigl\Vert a(t) \bigr\Vert ^{2} +\frac{1}{8m^{3}\lambda_{1}^{4}\gamma^{2}} \int_{0}^{l} \bigl\vert A(x,t) \bigr\vert ^{4}\,dx. \end{aligned}$$

(4.17)

Taking $\varepsilon=\frac{\gamma m}{8}$ in (4.13) and adding it to (4.17), we get

$$\begin{aligned} &\frac{d}{dt} \bigl( \bigl\Vert PH(t) \bigr\Vert ^{2}+ \bigl\Vert a(t) \bigr\Vert ^{2} \bigr) \\ &\quad \leq \biggl(\frac{2k_{1}^{\frac {4}{3}}C_{0}^{4}}{(k_{2}b_{0})^{\frac{1}{3}}}+ \frac {C_{0}^{4}}{4m^{3}\lambda_{1}^{4}\gamma^{2}} \biggr) \bigl\Vert \partial_{x}A(t) \bigr\Vert ^{4} +R_{1}(t) \bigl( \bigl\Vert PH(t) \bigr\Vert ^{2}+ \bigl\Vert a(t) \bigr\Vert ^{2} \bigr), \end{aligned}$$

(4.18)

where

$$ R_{1}(t)=2C(\gamma,m)+2C_{1}(\gamma,m)+4l \bigl\Vert h(t) \bigr\Vert ^{2}+2\alpha+\frac{l}{\lambda_{1}^{2}m} \bigl\Vert \partial _{x}^{2}\tilde{h}(t) \bigr\Vert ^{2} +l \bigl\Vert \partial_{x}^{2}\tilde{A}(t) \bigr\Vert ^{2}. $$

From (4.18) we derive the estimate

$$ \bigl\Vert PH(t) \bigr\Vert ^{2}+ \bigl\Vert a(t) \bigr\Vert ^{2} \leq q_{0}e^{-\int_{0}^{t}R_{1}(s)\,ds} \int_{0}^{t} \bigl\Vert \partial _{x}A(s) \bigr\Vert ^{4}\,ds, $$

where $q_{0}=\frac{2k_{1}^{\frac{4}{3}}C_{0}^{4}}{(k_{2}b_{0})^{\frac{1}{3}}}+ \frac{C_{0}^{4}}{4m^{3}\lambda_{1}^{4}\gamma^{2}}$. The proof is complete. □