Exponential attractors for the strongly damped wave equations with critical exponent

Abstract

In this paper, we prove the existence of global attractor and exponential attractor in some stronger spaces for the strongly damped nonlinear wave equation when the nonlinear term $f(u,u_{t})$ depends on $u_{t}$ and contains a critical exponent with respect to u and the external forcing term g merely belongs to the weak space $H^{-1}(\Omega)$.

1 Introduction

We study the following strongly damped nonlinear wave equation:

$$ \textstyle\begin{cases} u_{tt}-\Delta u_{t}-\Delta u+f(u,u_{t})=g &t>0, x\in\Omega,\\ u(x,t)=0 &t>0, x\in\partial\Omega,\\u(x,0)=u_{0}(x),\quad\quad u_{t}(x,0)=u_{1}(x) &t=0, x\in\Omega. \end{cases} $$

(1.1)

Here $u=u(x,t)$ is a real-valued function defined on $\Omega\times [0,\infty)$. Ω is an open bounded set of $\mathbb{R}^{3}$ with a smooth boundary ∂Ω. $f(u,v)\in C^{1}(\mathbb{R}\times \mathbb{R}, \mathbb{R})$, and $g\in H^{-1}(\Omega)$.

In the case that $f=f(u)\in C^{1}(\mathbb{R}, \mathbb{R})$ with $\liminf_{|r|\rightarrow\infty}\frac{f(r)}{r}>-\lambda_{1}$, where $\lambda_{1}$ is the first eigenvalue of −Δ on $H^{1}_{0}(\Omega)$, Webb first considered the asymptotic behavior of strongly damped wave equations in [1]. Then, in [2], Carvalho et al. showed the existence of the global attractor for wave equations with the critical nonlinearity. The regularity of solutions was also investigated via a bootstrapping technique in [3, 4], and we mention that a similar result has also been given by Pata et al. in [5, 6]. Recently, Sun and Yang in [7, 8] proved the existence of global attractor and exponential attractor for the same equation with the weaker external term $g\in H^{-1}(\Omega)$.

For another case, $f=f(u,u_{t})\in C^{1}(\mathbb{R}\times\mathbb{R}, \mathbb {R})$, Massatt [9] and Hale [10] proved the existence of global attractor when the continuous semigroup of the mapping $S(t):\{ u_{0},u_{1}\}\mapsto\{u,u_{t}\}$ is pointwise dissipative and a bounded map. Moreover, under the assumptions that $f(u,u_{t})$ is subcritical with respect to u and the external force term g belongs to $L^{2}(\Omega )$, the author in [11] proved the existence of global attractor in the space $\mathcal{H}=H^{1}_{0}(\Omega)\times L^{2}(\Omega)$.

In this paper, we investigate the latter case with the conditions given in [8, 11]. Compared with those in [11], the nonlinear term $f(u,u_{t})$ satisfies the critical exponent growth condition with respect to u (see (2.4)) and the external force $g\in H^{-1}(\Omega)$, which is weaker than the assumptions in [11]. We also remove the additional assumptions (4.26), (4.27) in [8]. Motivated by the key ideas in [8], by making a shifting on the semigroup $\{S(t)\}_{t\geqslant0}$ with a (proper) fixed point $\phi (x)$, we first show the global attractor $\mathcal{A}-\phi(x)$ is bounded in a stronger topology. More precisely, $\mathcal{A}-\phi(x)$ is bounded in the space $\mathcal{H}^{\sigma}=D((-\Delta)^{\frac {1+\sigma}{2}})\times D((-\Delta)^{\frac{\sigma}{2}})$, $\sigma\in [0,\frac{1}{2})$ (see Theorem 3.1). Then, by proving that the semigroup $\{S(t)\}_{t\geqslant0}$ is Fréchet differential with respect to the initial value, we apply our standard method established in [12] to obtain the exponential attractor for equation (1.1) without the restrictions (4.26), (4.27) in [8]. In addition, with the regularity of solutions as in [6], we establish the existence of exponential attractor in the stronger space $H^{1}_{0}(\Omega )\times H^{1}_{0}(\Omega)$.

In order to have a comparison, we organize this paper as follows. In Section 1, we briefly review some results. Section 2 is devoting to proving that the existence of global attractor in the space $\mathcal {H}^{\sigma}$. In Section 3, we obtain the exponential attractor in the space $H^{1}_{0}(\Omega)\times H^{1}_{0}(\Omega)$.

2 Preliminaries

Let

$$\begin{aligned}& (u,v)= \int_{\Omega}uv\,dx, \quad\quad \Vert u\Vert _{2}=(u,u)^{1/2},\quad \forall u,v \in L^{2}(\Omega), \\& \bigl((u,v)\bigr)= \int_{\Omega}\nabla u\nabla v\,dx,\quad\quad \Vert u\Vert _{H^{1}_{0}(\Omega )}= \bigl((u,v)\bigr)^{1/2},\quad \forall u,v\in H^{1}_{0}( \Omega), \\& \mathcal{H}=H^{1}_{0}(\Omega)\times L^{2}( \Omega), \\& \mathcal{H}^{\sigma}= \bigl(H^{1}_{0}(\Omega)\cap H^{1+\sigma} \bigr)\times H^{\sigma}(\Omega)=D\bigl((- \Delta)^{\frac{1+\sigma}{2}}\bigr)\times D\bigl((-\Delta )^{\frac{\sigma}{2}}\bigr), \quad\sigma \in\biggl[0,\frac{1}{2}\biggr), \end{aligned}$$

and

$$\begin{aligned}& (y_{1},y_{2})_{\mathcal{H}}=(y_{1},y_{2})_{H^{1}_{0}(\Omega),L^{2}(\Omega )}= \bigl((u_{1},u_{2})\bigr)+(v_{1},v_{2}),\quad\quad \Vert y\Vert _{H^{1}_{0}(\Omega)\times L^{2}(\Omega )}=(y,y)^{1/2}_{H^{1}_{0}(\Omega)\times L^{2}(\Omega)}, \\& \Vert y_{i}\Vert _{\sigma}=\Vert y_{i}\Vert _{\mathcal{H}^{\sigma}}=\bigl\Vert (u_{i},v_{i})^{T} \bigr\Vert _{H^{1+\sigma}(\Omega), H^{\sigma}(\Omega)}, \\& \forall y_{i}=(u_{i},v_{i})^{T},\quad\quad y=(u,v)^{T}\in H^{1}_{0}(\Omega)\times L^{2}(\Omega) \text{ or } H^{1+\sigma}(\Omega)\times H^{\sigma}(\Omega ), \quad i=1,2, \end{aligned}$$

denotes the usual inner products and norms in $L^{2}(\Omega)$, $H^{1}_{0}(\Omega)$, and $H^{1}_{0}(\Omega)\times L^{2}(\Omega)$, $H^{1+\sigma }(\Omega)\times H^{\sigma}(\Omega)$, respectively.

Let $u_{t}=v$, then equations (1.1) are equivalent to the following initial value problem in the space $\mathcal{H}$:

$$ \textstyle\begin{cases}\dot{Y}=\mathbb{L}Y+F(Y), & x\in\Omega, t>0,\\ Y(0)=Y_{0}=(u_{0}, u_{1})^{T}\in\mathcal{H}, & t=0, \end{cases} $$

(2.1)

where

$$ \begin{aligned} &Y= \begin{pmatrix} u \\ v \end{pmatrix},\quad\quad \mathbb{L} = \begin{pmatrix} 0 & I\\ -A & -A \end{pmatrix},\quad\quad F(Y)= \begin{pmatrix} 0 \\ -f(u,u_{t})+g \end{pmatrix}, \\ &D(\mathbb{L})=D(A)\times D(A),\quad\quad D(A)=D(-\Delta)=H^{2}(\Omega)\cap H^{1}_{0}(\Omega). \end{aligned} $$

(2.2)

Massatt in [9] proved that $\mathbb{L}$ defined in (2.2) is a sectorial operator on $\mathcal{H}$ and generates an analytic compact semigroup $e^{\mathbb{L}t}$ on $\mathcal{H}$ for $t>0$. By the appropriate assumptions on f and the external forcing term $g\in L^{2}(\Omega)$, they proved that there exists a unique function $Y(\cdot )=Y(\cdot, Y_{0})\in C(R_{+}, \mathcal{H})$ such that $Y(0,Y_{0})=Y_{0}$ and $Y(t)$ satisfies the integral equation

$$ Y(t,Y_{0})=e^{\mathbb{L}t}Y_{0}+ \int_{0}^{t}e^{\mathbb{L}(t-s)}F\bigl(Y(\tau)\bigr)\,d \tau, $$

which is also called a mild solution of equation (2.1).

The main purpose here is to study the case $g\in H^{-1}(\Omega)$ and to provide some weaker assumptions on $f(u,v)$ than the one in [8, 11], that is, the function $f(u,v)\in C^{2}(\mathbb{R}\times\mathbb{R}, \mathbb{R})$ with $f(0,0)=0$ satisfies the following condition:

$$ \liminf_{\vert s\vert \rightarrow+\infty}\frac{f(s,0)}{s}>- \lambda_{1} $$

(2.3)

and its partial derivatives $f_{1}^{\prime}(u,v)$, $f_{2}^{\prime}(u,v)$, $f_{11}^{\prime\prime}(u,v)$, $f_{12}^{\prime\prime}(u,v)$, $f_{22}^{\prime\prime}(u,v)$ satisfy

$$\begin{aligned}& \bigl\vert f_{1}^{\prime}(u,v)\bigr\vert \leqslant C\bigl(1+\vert u\vert ^{4}\bigr),\quad \forall u,v\in \mathbb{R}, \end{aligned}$$

(2.4)

$$\begin{aligned}& f_{1}^{\prime}(u,v)\geqslant-\ell,\quad \forall u,v\in \mathbb{R}, \end{aligned}$$

(2.5)

$$\begin{aligned}& f_{2}^{\prime}(u,v)\leqslant\delta\text{ (small enough)},\quad \forall u,v\in\mathbb{R}, \end{aligned}$$

(2.6)

$$\begin{aligned}& \bigl\vert f_{11}^{\prime\prime}(u,v)\bigr\vert , \bigl\vert f_{12}^{\prime\prime}(u,v)\bigr\vert , \bigl\vert f_{22}^{\prime\prime}(u,v)\bigr\vert \leqslant C\bigl(1+\vert u \vert ^{3}\bigr),\quad \forall u,v\in\mathbb{R}. \end{aligned}$$

(2.7)

Note again that in contrast to [8], here $f=f(u,u_{t})$ without the addition assumptions (4.26), (4.27) in [8], and in contrast to [11], here $f=f(u,u_{t})$ is critical with respect to u, and its partial derivatives $f'_{j}$, $f^{\prime\prime}_{ij}$ is weaker than assumptions (3), (4) in [11].

Obviously, such conditions are satisfied in particular for the nonlinearities $f(u,v)=u^{5}+\delta\sin v$ (in other words, a small perturbation of $u^{5}$), etc.

As is well known, if $g\in H^{-1}(\Omega)$, the solution of the elliptic equation ($\theta>\ell$)

$$ \textstyle\begin{cases}-\Delta u+f(u,0)+\theta u=g\in H^{-1}(\Omega),\\ u|_{\partial\Omega}=0, \end{cases} $$

(2.8)

only belongs to $H^{1}_{0}(\Omega)$. The regularity of the attractor (if it exists) is not higher than $\mathcal{H}$ in this case. However, by a decomposition as in [8], $u(t)=\hat{u}(t)+\phi(x)$ where $\phi(x)$ is the solution of equation (2.8) for some θ, and $\hat{u}(t)$ satisfies

$$ \textstyle\begin{cases} \hat{u}_{tt}-\Delta\hat{u}_{t}-\Delta\hat{u}+f(\hat{u}+\phi,\hat {u}_{t})-f(\phi,0)=\theta\phi,\\ \hat{u}|_{\partial\Omega}=0. \end{cases} $$

(2.9)

Next, we will get the regularity of the solution $\hat{u}(t)$.

3 Global attractor

We first present the following asymptotic regularity by the Galerkin approximate scheme (see [8, 13]).

Theorem 3.1

Let $f(u,v) \in C^{2}(\mathbb{R}\times\mathbb{R}, \mathbb{R})$ with $f(0,0)=0$ satisfying the above assumptions (2.3)-(2.7), $g\in H^{-1}$, and $\{S(t)\}_{t\geqslant0}$ be the semigroup generated by the weak solution of (1.1) in the space $H^{1}_{0}(\Omega)\times L^{2}(\Omega)$. Then, for each $0<\sigma<\frac{1}{2}$, there exist a subset $\mathcal{B}_{\sigma}$, a monotone increasing function $Q_{\sigma }(\cdot)$, and a positive constant ν (independent of σ) such that: for any bounded set $B\subset\mathcal{H}$,

$$ \operatorname{dist}_{\mathcal{H}} \bigl(S(t)B,\mathcal{B}_{\sigma} \bigr) \leqslant Q_{\sigma} \bigl(\Vert B\Vert _{\mathcal{H}} \bigr)e^{-\nu t}, \quad\textit{for all } t\geqslant0, $$

where $\mathcal{B}_{\sigma}$ satisfies, for some constant $\Lambda _{\sigma}>0$,

$$ \mathcal{B}_{\sigma}=\bigl\{ \varsigma\in\mathcal{H}:\bigl\Vert \varsigma-\bigl(\phi (x),0\bigr)\bigr\Vert _{H^{1+\sigma}(\Omega)\times H^{\sigma}(\Omega)}\leqslant \Lambda_{\sigma}< \infty\bigr\} , $$

and $\phi(x)$ is the unique solution of the above equation (2.8) by choosing $\theta=\eta_{0}$ large enough, that is,

$$ \textstyle\begin{cases}-\Delta\phi+f(\phi,0)+\eta_{0} \phi=g\in H^{-1}(\Omega), \quad\textit{in } \Omega,\\ \phi|_{\partial\Omega}=0. \end{cases} $$

(3.1)

Remark 3.1

From [8], we know that

• 1. for each θ (>ℓ), equation (2.8) has a unique solution $u_{\theta}(x)\in H^{1}_{0}(\Omega)$ satisfying

  $$ \Vert \nabla u_{\theta} \Vert ^{2}+2(\theta-\ell)\Vert u_{\theta} \Vert _{2}^{2}\leqslant \Vert g\Vert _{H^{-1}}^{2}; $$

• 2. $\Vert \nabla u_{\theta} \Vert \rightarrow0$, $\Vert u_{\theta} \Vert _{L^{p}}\rightarrow0$ as $\theta\rightarrow\infty$ for any fixed $p\in[2,6)$.

Now, denote $h_{\theta}(u, u_{t} )=f(u,u_{t})+\theta u$. From (2.4)-(2.6) and the mean value theorem, one has, for any $v\in C^{1} ((0,\infty), \mathcal{H} )$,

$$\begin{aligned}& \frac{1}{2}\Vert \nabla v\Vert ^{2}+ \frac{1}{2}\Vert v_{t}\Vert ^{2}+2\bigl\langle h_{\theta }(v+\phi, v_{t}+\phi_{t})-h_{\theta}( \phi,\phi_{t}),v\bigr\rangle -\bigl\langle h^{\prime}_{1\theta}( \phi,0)v,v\bigr\rangle \\& \quad= \frac{1}{2}\Vert \nabla v\Vert ^{2}+\frac{1}{2} \Vert v_{t}\Vert ^{2}+2\bigl\langle h_{\theta }(v+ \phi, v_{t})-h_{\theta}(\phi,0),v\bigr\rangle -\bigl\langle h^{\prime}_{1\theta}(\phi ,0)v,v\bigr\rangle \\& \quad= \frac{1}{2}\Vert \nabla v\Vert ^{2}+\frac{1}{2} \Vert v_{t}\Vert ^{2}+2\bigl\langle h_{\theta }(v+ \phi, v_{t})-h_{\theta}(\phi,v_{t})+h_{\theta}( \phi,v_{t})-h_{\theta}(\phi ,0),v\bigr\rangle \\& \quad\quad{}-\bigl\langle h^{\prime}_{1\theta}(\phi,0)v,v\bigr\rangle \\& \quad= \frac{1}{2}\Vert \nabla v\Vert ^{2}+\frac{1}{2} \Vert v_{t}\Vert ^{2}+2\bigl\langle h^{\prime}_{1\theta }( \vartheta_{1} v+\phi, v_{t})v,v\bigr\rangle +2\bigl\langle h^{\prime}_{2\theta}(\phi, \vartheta_{2} v_{t})v_{t},v\bigr\rangle -\bigl\langle h^{\prime}_{1\theta}( \phi,0)v,v\bigr\rangle \\& \quad\geqslant \frac{1}{2}\Vert \nabla v\Vert ^{2}+ \frac{1}{2}\Vert v_{t}\Vert ^{2}+2(\theta-\ell ) \Vert v\Vert ^{2}-\theta \Vert v\Vert ^{2}-2\delta \int_{\Omega} \vert v_{t}v\vert \,dx-C \int_{\Omega }\bigl(1+\vert \phi \vert ^{4}\bigr) \vert v\vert ^{2}\,dx \\& \quad\geqslant \frac{1}{2}\Vert \nabla v\Vert ^{2}+ \frac{1}{2}\Vert v_{t}\Vert ^{2}+(\theta-2\ell -C- \delta)\Vert v\Vert ^{2}-\delta \Vert v_{t}\Vert ^{2}-C\Vert \nabla\phi \Vert ^{4}\Vert \nabla v\Vert ^{2}, \end{aligned}$$

(3.2)

where the constants C, δ, and ℓ come from (2.4)-(2.6), respectively, and $\vartheta_{1}, \vartheta_{2}\in (0,1)$, ϕ is the solution of (3.1).

Hence, by choosing θ large enough in (3.2) with the assertion 2 in Remark 3.1, we know that

$$\begin{aligned}& \frac{1}{2}\Vert \nabla v\Vert ^{2}+ \frac{1}{2}\Vert v_{t}\Vert ^{2}+2\bigl\langle h_{\theta }(v+\phi, v_{t}+\phi_{t})-h_{\theta}( \phi,\phi_{t}),v\bigr\rangle -\bigl\langle h^{\prime}_{1\theta}( \phi,0)v,v\bigr\rangle \geqslant0, \\& \quad\text{for all } v\in C^{1} \bigl((0,\infty), \mathcal{H} \bigr). \end{aligned}$$

(3.3)

3.1 Decomposition of the equations

Let

$$ h(u,u_{t})=f(u,u_{t})+\eta_{0} u, $$

where the positive constant $\eta_{0}$ is large enough and such that (2.8) and (3.3) holds when $\theta=\eta_{0}$.

Now, we first decompose the solution $S(t)(u_{0},v_{0})=(u(t),u_{t}(t))$ into the sum

$$ \bigl(u(t),u_{t}(t)\bigr)=S(t)\xi_{u}(0)=K(t) \xi_{u}(0)+D(t)\xi _{u}(0)=\bigl(w(t),w_{t}(t) \bigr)+\bigl(z(t),z_{t}(t)\bigr), $$

where $K(t)\xi_{u}(0)=(w(t),w_{t}(t))$ and $D(t)\xi_{u}(0)=(z(t),z_{t}(t))$ solve the following equations, respectively:

$$ \textstyle\begin{cases} w_{tt}-\Delta w_{t}-\Delta w+f(u,u_{t})-f(z,z_{t})=\eta_{0} z \quad\text{in } \Omega\times\mathbb{R}^{+},\\ w \vert _{\partial\Omega}=0,\\ (w(x,0),w_{t}(x,0))=(0,0), \end{cases} $$

(3.4)

and

$$ \textstyle\begin{cases} z_{tt}-\Delta z_{t}-\Delta z+h(z,z_{t})=g(x) \quad\text{in } \Omega\times \mathbb{R}^{+},\\ z|_{\partial\Omega}=0,\\ (z(x,0),z_{t}(x,0))=\xi_{u}(0). \end{cases} $$

(3.5)

Then we decompose further the solution $z(x,t)$ of (3.5) as $z(x,t)=v(x,t)+\phi(x)$, where $\phi(x)$ is the unique solution of (2.8) and $v(x,t)$ solves the following equation:

$$ \textstyle\begin{cases} v_{tt}-\Delta v_{t}-\Delta v+h(z,z_{t})-h(\phi,0)=0 \quad\text{in } \Omega \times\mathbb{R}^{+},\\ v|_{\partial\Omega}=0,\\ (v(x,0),v_{t}(x,0))=\xi_{u}(0)- (\phi(x),0 ). \end{cases} $$

(3.6)

Hence,

$$\begin{aligned} \bigl(u(t),u_{t}(t)\bigr) =&\bigl(w(t),w_{t}(t) \bigr)+\bigl(z(t),z_{t}(t)\bigr) \\ =&\bigl(w(t),w_{t}(t)\bigr)+\bigl(v(t)+\phi,v_{t}(t)+ \phi_{t}\bigr) \\ =&\bigl(w(t),w_{t}(t)\bigr)+\bigl(v(t)+\phi,v_{t}(t)\bigr), \quad\text{due to } \phi_{t}=0. \end{aligned}$$

(3.7)

Hereafter, we always assume the assumptions in Theorem 3.1 hold and denote the unique solution of (2.8) by $\phi(x)$.

3.2 The prior estimates in spaces $\mathcal{H}, \mathcal {H}^{\sigma}(\sigma\in[0,\frac{1}{2}))$

Now, we will give the prior estimates in space $\mathcal{H}$ or regular space $\mathcal{H}^{\sigma}$ for the above decompositions of the solutions z, v, w, u, respectively.

First of all, we have the following estimate (e.g., see [5, 8]) for the solution z of (3.5).

Lemma 3.1

There exists an increasing function $Q_{1}(\cdot)$ such that, for any bounded set $B\subset\mathcal{H}$, one gets, for any $t\geqslant0$,

$$ \bigl\Vert \nabla z(t)\bigr\Vert ^{2}+ \int_{0}^{t}\bigl\Vert \nabla z_{t}(s) \bigr\Vert ^{2}\,dx\leqslant Q_{1}\bigl(\Vert B\Vert _{\mathcal{H}}+\Vert g\Vert _{H^{-1}}\bigr),\quad \forall \xi_{u}(0)\in B. $$

(3.8)

Proof

Indeed, we consider the functional (by choosing $\hat{\phi }(y)=f(y,0)+\eta_{0} y$ in [5])

$$ \mathcal{F}(t)=\mathcal{F}\bigl(z(t)\bigr)=2 \int_{\Omega} \int _{0}^{z(x,t)}\bigl(f(s,0)+\eta_{0} s \bigr)\,ds\,dx. $$

(3.9)

We set $\xi(t)=z_{t}+\epsilon z$ with $\epsilon\in(0,\epsilon_{0})$, for some $\epsilon_{0}\leqslant1$ to be determined later. Multiplying equation (3.5) by ξ yields

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt}E+\epsilon(1-\epsilon)\Vert \nabla z\Vert ^{2}+\Vert \nabla \xi \Vert ^{2} \\& \quad= \epsilon \Vert \xi \Vert ^{2}-\epsilon^{2}\langle z,\xi \rangle+\epsilon\langle g,z\rangle-\epsilon\bigl\langle f(z,0)+\eta_{0} z,z\bigr\rangle +\bigl\langle f(z,0)-f(z,z_{t}),z_{t}+\epsilon z\bigr\rangle , \end{aligned}$$

(3.10)

where the energy functional E is defined as

$$ E(t)=E\bigl(z(t)\bigr)=(1-\epsilon)\Vert \nabla z\Vert ^{2}+\bigl\Vert \xi(t)\bigr\Vert ^{2}+\mathcal {F}(t)-2 \langle g,z\rangle. $$

(3.11)

Obviously, from (2.4), we know that here the function $\hat{\phi }(y)=f(y,0)+\eta_{0} y$ satisfies the assumptions (8), (9), (11), (12) in [5], and due to the mean value theorem, we have

$$\begin{aligned} \bigl\langle f(z,z_{t})-f(z,0),z_{t}+\epsilon z\bigr\rangle =&\bigl\langle f_{2}^{\prime}(z,\vartheta z_{t})z_{t},z_{t}+\epsilon z\bigr\rangle \\ \leqslant&\delta \Vert z_{t}\Vert ^{2}+\delta\epsilon \int_{\Omega} \vert z_{t}z\vert \,dx, \end{aligned}$$

(3.12)

where $\vartheta\in(0,1)$.

As to the assumption (2.6), if δ is small enough, the term in (3.12) can be controlled by the left-hand side of (3.10). Therefore, with the application of the same argument as in [5], it is easy to get the inequality (3.8). It finishes the proof of Lemma 3.1. □

Then, for the solution v of (3.6), we have the following.

Lemma 3.2

There exist an increasing function $Q_{2}(\cdot)$ and some constant $k_{1}>0$, such that, for any bounded set $B\subset\mathcal{H}$,

$$ \bigl\Vert \bigl(v(x,t),v_{t}(x,t)\bigr)\bigr\Vert _{\mathcal{H}} \leqslant Q_{2}\bigl(\Vert B\Vert _{\mathcal {H}} \bigr)e^{-k_{1}t}, \quad\forall t\geqslant0, \xi_{v}(0)\in B, $$

that is,

$$ \bigl\Vert \bigl(z(x,t),z_{t}(x,t)\bigr)-\bigl(\phi(x),0\bigr)\bigr\Vert _{\mathcal{H}}\leqslant Q_{2}\bigl(\Vert B\Vert _{\mathcal{H}}\bigr)e^{-k_{1}t}, \quad\forall t\geqslant0, \xi_{v}(0) \in B. $$

Proof

As in [8, 14], for $\epsilon\in (0,1)$ to be determined later, we define the functional

$$ \Lambda(t)=\bigl\Vert \nabla v(t)\bigr\Vert ^{2}+\bigl\Vert v_{t}(t)\bigr\Vert ^{2}+\epsilon\bigl\Vert \nabla v(t) \bigr\Vert ^{2}+2\bigl\langle h(z,0)-h(\phi, 0),v\bigr\rangle +2 \epsilon\langle v_{t},v\rangle -\bigl\langle h^{\prime}_{1}( \phi,0)v,v\bigr\rangle . $$

Then, from (3.3) and by taking ϵ small enough, we have

$$ \Lambda(t)\geqslant\frac{1}{4}\bigl\Vert \xi_{v}(t)\bigr\Vert _{\mathcal{H}}^{2} \quad\text{for all } t\geqslant0, \xi_{0}\in B. $$

Multiplying (3.6) by $v_{t}+\epsilon v(t)$ we have (note that $z_{t}=v_{t}$ and $\phi_{t}=0$)

$$\begin{aligned}& \frac{d}{dt}\Lambda(t)+\epsilon\Lambda(t)+\Gamma+ \frac{\epsilon}{2}\bigl\Vert \nabla v(t)\bigr\Vert ^{2} \\& \quad=2\bigl\langle \bigl(h'_{1}(z,0)-h'_{1}( \phi,0)\bigr)z_{t}, v\bigr\rangle +2\bigl\langle \bigl(h(z,0)-h(z,z_{t}) \bigr), v_{t}+\epsilon v\bigr\rangle , \end{aligned}$$

(3.13)

where

$$ \Gamma=2\bigl\Vert \nabla v_{t}(t)\bigr\Vert ^{2}+ \frac{\epsilon}{2}\bigl\Vert \nabla v(t)\bigr\Vert ^{2}-3\epsilon \Vert v_{t}\Vert ^{2}-2\epsilon^{2}\langle v_{t},v\rangle-\epsilon \Vert \nabla v\Vert ^{2}+\epsilon \bigl\langle h'_{1}(\phi,0),v^{2}\bigr\rangle . $$

It is easy to see that $\Gamma\geqslant0$ as ϵ small enough, and from (2.7), we have

$$\begin{aligned} 2\bigl\langle \bigl(h'_{1}(z,0)-h'_{1}( \phi,0)\bigr)z_{t},v\bigr\rangle =&2\bigl\langle h^{\prime\prime}_{11} \bigl(rz+(1-r)\phi,0\bigr)z_{t}, v^{2}\bigr\rangle \\ \leqslant& C \int_{\Omega}\bigl(1+\vert z\vert ^{3}+\vert \phi \vert ^{3}\bigr)\vert z_{t}\vert \vert v\vert ^{2}\,dx \\ \leqslant& c_{2}\Vert \nabla z_{t}\Vert \Vert \nabla v \Vert ^{2}\leqslant\frac{\epsilon }{2}\Vert \nabla v\Vert ^{2}+\frac{c_{2}}{\epsilon} \Vert \nabla z_{t}\Vert ^{2}\Lambda, \end{aligned}$$

where $r\in(0,1)$ and the constant $c_{2}$ depends only on $\Vert B\Vert _{\mathcal{H}}+\Vert \nabla\phi \Vert$.

By the mean value theorem, for the last term in the right-hand side of (3.13), we get

$$\begin{aligned} 2\bigl\langle \bigl(h(z,0)-h(z,z_{t})\bigr), v_{t}+\epsilon v \bigr\rangle =&2\bigl\langle f(z,z_{t})-f(z,0),z_{t}+\epsilon z \bigr\rangle \\ =&\bigl\langle f_{2}^{\prime}(z,\vartheta z_{t})z_{t},z_{t}+ \epsilon v\bigr\rangle \\ \leqslant&\delta \Vert z_{t}\Vert ^{2}+\delta\epsilon \int_{\Omega} \vert z_{t}v\vert \, dx. \end{aligned}$$

Since δ is small enough, from Lemma 3.1 and by noticing $\Lambda(0)\leqslant Q(\Vert B\Vert _{\mathcal{H}}+\Vert \nabla\phi \Vert )$ and by applying Lemma 2.2 [15], we can finish the proof of Lemma 3.2. □

Second, for the solution $w(t)$ in (3.4), we have the following result.

Lemma 3.3

For each bounded subset $B\subset\mathcal{H}$ and any $\sigma\in [0,\frac{1}{2})$, there exists an increasing function $Q_{\sigma}(\cdot )$ such that

$$ \bigl\Vert K(t)\xi_{u}(0)\bigr\Vert _{\mathcal{H}^{\sigma}}=\bigl\Vert \bigl(w(t),w_{t}(t)\bigr)\bigr\Vert _{\mathcal {H}^{\sigma}}\leqslant Q_{\sigma}\bigl(\Vert B\Vert _{\mathcal{H}} \bigr)e^{\nu_{\sigma }t} \quad\forall t\geqslant0, \xi_{u}(0)\in B, $$

(3.14)

where the positive constant $\nu_{\sigma}$ depends only on $\Vert B\Vert _{\mathcal{H}}$ and σ.

Proof

Rewriting equation (1.1) as follows:

$$ \textstyle\begin{cases} u_{tt}-\Delta u_{t}-\Delta u+f(u,0)=g+f(u,0)-f(u,u_{t}) &t>0, x\in\Omega ,\\ u(x,t)=0 &t>0, x\in\partial\Omega,\\u(x,0)=u_{0}(x),\quad\quad u_{t}(x,0)=u_{1}(x) &t=0, x\in\Omega, \end{cases} $$

and applying the same argument as in the proof procedure of Lemma 3.1 with the assumptions (2.4)-(2.6), and combining with (3.8), it is easy to show that

$$ \bigl\Vert \nabla u(t)\bigr\Vert +\bigl\Vert \nabla z(t)\bigr\Vert \leqslant c\bigl(\Vert B\Vert _{\mathcal{H}}\bigr),\quad \forall t\geqslant0. $$

Now, rewrite equation (3.4) as follows:

$$ \textstyle\begin{cases} w_{tt}-\Delta w_{t}-\Delta w+f(u,0)+\eta_{0}u-(f(z,0)+\eta_{0}z),\\=\eta_{0} u+f(u,0)-f(u,u_{t})-(f(z,0)-f(z,z_{t})) \quad\text{in } \Omega\times\mathbb {R}^{+},\\ w \vert _{\partial\Omega}=0,\\ (w(x,0),w_{t}(x,0))=(0,0). \end{cases} $$

(3.15)

Denoting $\hat{\phi}(u)=f(u,0)+\eta_{0}u$, $\hat{\phi}(z)=f(z,0)+\eta_{0}z$ like the one in [5], and testing equation (3.15) with $A^{\sigma}w_{t}$, we are led to the identity (denote $\gamma(t)=(w(t),w_{t}(t))$)

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt}\bigl\Vert \gamma(t)\bigr\Vert _{\sigma}^{2}+\bigl\Vert A^{(1+\sigma)/2}w_{t} \bigr\Vert ^{2} \\& \quad=-\bigl\langle \hat{\phi}(u)-\hat{\phi}(z),A^{\sigma}w_{t} \bigr\rangle +\bigl\langle g,A^{\sigma}w_{t}\bigr\rangle \\& \quad\quad{}+\bigl\langle f(u,0)-f(u,u_{t})-\bigl(f(z,0)-f(z,z_{t}) \bigr),A^{\sigma}w_{t}\bigr\rangle . \end{aligned}$$

(3.16)

Due to (2.4), we get

$$\begin{aligned} -\bigl\langle \hat{\phi}(u)-\hat{\phi}(z),A^{\sigma}w_{t} \bigr\rangle \leqslant& c\bigl(1+\Vert u\Vert _{L^{6}}^{4}+ \Vert z\Vert _{L^{6}}^{4}\bigr)\Vert w\Vert _{L^{6/(1-2\sigma)}}\bigl\Vert A^{\sigma }w_{t}\bigr\Vert _{L^{6/(1+2\sigma)}} \\ \leqslant& c\bigl(1+\bigl\Vert A^{1/2}u\bigr\Vert ^{4}+ \bigl\Vert A^{1/2}v\bigr\Vert ^{4}\bigr)\bigl\Vert A^{(1+\sigma)/2}w\bigr\Vert \bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert \\ \leqslant& c\bigl\Vert \gamma(t)\bigr\Vert ^{2}_{\sigma}+ \frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert ^{2}. \end{aligned}$$

(3.17)

By virtue of (2.6), we have

$$\begin{aligned}& \bigl\langle f(u,0)-f(u,u_{t})-\bigl(f(z,0)-f(z,z_{t}) \bigr),A^{\sigma}w_{t}\bigr\rangle \\& \quad= \bigl\langle -f'_{2}(u,\vartheta_{2} u_{t})u_{t}+f'_{2}(z, \vartheta_{2} z_{t}) z_{t},A^{\sigma}w_{t} \bigr\rangle \\& \quad\leqslant \delta\bigl( \Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{\sigma}w_{t} \bigr\Vert _{L^{6/(1+2\sigma)}} \\& \quad\leqslant \delta\bigl(\Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{(1+\sigma)/2}w_{t} \bigr\Vert \\& \quad\leqslant c+\frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert ^{2}, \end{aligned}$$

(3.18)

where $\vartheta_{2}\in(0,1)$.

Additionally,

$$ \bigl\langle g, A^{\sigma}w_{t}\bigr\rangle \leqslant\bigl\Vert A^{-1/2}g\bigr\Vert \bigl\Vert A^{(1+\sigma )/2}w_{t}\bigr\Vert \leqslant c+\frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert ^{2}. $$

(3.19)

Plugging (3.17)-(3.19) into (3.16), we obtain

$$ \frac{d}{dt}\bigl\Vert \gamma(t)\bigr\Vert _{\sigma}^{2}\leqslant c\bigl\Vert \gamma(t)\bigr\Vert _{\sigma}^{2}+c, $$

(3.20)

and the Gronwall lemma entails

$$ \bigl\Vert \gamma(t)\bigr\Vert _{\sigma}^{2}\leqslant e^{kt}-1, $$

which concludes the proof. □

Now, based on Lemmas (3.2) and (3.3), one can also decompose the solution $u(t)$ as follows.

Lemma 3.4

For any $\epsilon>0$,

$$ u(t)=v_{1}(t)+w_{1}(t), \quad\textit{for all } t\geqslant0, $$

(3.21)

where $v_{1}(t)$ and $w_{1}(t)$ satisfy the following:

$$ \int_{s}^{t}\bigl\Vert \nabla v_{1}( \tau)\bigr\Vert ^{2}\,d\tau\leqslant\epsilon(t-s)+C_{\epsilon } \quad\textit{for all } t\geqslant s\geqslant0, $$

(3.22)

and

$$ \bigl\Vert A^{\frac{1+\sigma}{2}}w_{1}(t)\bigr\Vert ^{2}\leqslant K_{\epsilon} \quad\textit{for all } t\geqslant0, $$

(3.23)

with the constants $C_{\epsilon}$ and $K_{\epsilon}$ depending on ϵ, the initial value $\Vert \xi_{u}(0)\Vert _{\mathcal{H}}$ and $\Vert g\Vert _{H^{-1}}$.

Due to (3.7) and Lemma 4.5 in [8], one can easily deduce Lemma 3.4.

Next, we will show further that the estimate w in (3.14) can be chosen independent of the time t.

Lemma 3.5

For every $\sigma\in[0,\frac{1}{2})$, there exists a constant $J_{B,\sigma}$ which depends only on the $\mathcal{H}$-bound of B ($\subset\mathcal{H}$) and σ, such that

$$ \bigl\Vert K(t)\xi_{u}(0)\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}= \bigl\Vert \bigl(w(t),w_{t}(t)\bigr)\bigr\Vert _{\mathcal {H}^{\sigma}}^{2} \leqslant J_{B,\sigma} \quad\textit{for all } t\geqslant 0 \textit{ and } \xi_{u}(0)\in B. $$

Proof

The idea comes from [8, 16, 17] but with different details.

Multiplying (3.15) by $A^{\sigma}(w_{t}(t)+\epsilon w(t))$, we obtain

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\Omega}\bigl\vert A^{\frac{\sigma}{2}}(w_{t}+\epsilon w)\bigr\vert ^{2}-\bigl\langle \epsilon w_{t},A^{\sigma}(w_{t}+ \epsilon w)\bigr\rangle \\& \quad\quad{}-\bigl\langle Aw_{t}, A^{\sigma}(w_{t}+ \epsilon w)\bigr\rangle -\bigl\langle Aw, A^{\sigma }(w_{t}+\epsilon w)\bigr\rangle \\& \quad= -\bigl\langle f(u,0)-f(z,0), A^{\sigma}(w_{t}+\epsilon w)\bigr\rangle +\bigl\langle \eta _{0}z, A^{\sigma}(w_{t}+ \epsilon w)\bigr\rangle \\& \quad\quad{}+\bigl\langle f(u,0)-f(u,u_{t})-\bigl(f(z,0)-f(z,z_{t}) \bigr),A^{\sigma}w_{t}\bigr\rangle , \end{aligned}$$

where ϵ (>0) is small enough to be determined later.

We only need to deal with the right-hand side term, and the others can be estimated easily as those Lemma 4.4 in [18].

From (2.4), we first deal with the first dual product,

$$ \bigl\vert \bigl\langle f(u,0)-f(z,0), A^{\sigma}(w_{t}+ \epsilon w)\bigr\rangle \bigr\vert \leqslant C \int_{\Omega}\bigl(1+\vert u\vert ^{4}+\vert z \vert ^{4}\bigr)\vert w\vert \bigl\vert A^{\sigma}(w_{t}+ \epsilon w)\bigr\vert \,dx. $$

Applying Lemma 3.4, we have

$$ \int_{\Omega} \vert u\vert ^{4}\vert w\vert \bigl\vert A^{\sigma}w\bigr\vert \,dx\leqslant C \int_{\Omega }\bigl(\vert v_{1}\vert ^{4}+ \vert w_{1}\vert ^{4}\bigr)\bigl\vert w(t)\bigr\vert \bigl\vert A^{\sigma}w(t)\bigr\vert \,dx $$

(3.24)

and

$$\begin{aligned} \bigl\vert \bigl\langle f(u,0)-f(z,0), A^{\sigma} w)\bigr\rangle \bigr\vert \leqslant&c_{4}Q_{4}\bigl(\Vert B\Vert _{\mathcal{H}}\bigr)\bigl\Vert \nabla v_{1}(t)\bigr\Vert ^{2}\bigl\Vert A^{\frac{1+\sigma}{2}}w(t)\bigr\Vert ^{2} \\ &{}+c_{\sigma}\bigl(K_{\epsilon}+\Vert \phi \Vert _{H^{2}} \bigr)Q_{5}\bigl(\Vert B\Vert _{\mathcal {H}}\bigr)+C+ \frac{1}{4}\bigl\Vert A^{\frac{1+\sigma}{2}}w(t)\bigr\Vert ^{2}. \end{aligned}$$

Similarly,

$$\begin{aligned} \bigl\vert \bigl\langle f(u,0)-f(z,0), A^{\sigma}w_{t}\bigr\rangle \bigr\vert \leqslant&c_{4}Q_{4}\bigl(\Vert B\Vert _{\mathcal{H}}\bigr)\bigl\Vert \nabla v_{1}(t)\bigr\Vert ^{2}\bigl\Vert A^{\frac{1+\sigma}{2}}w(t)\bigr\Vert ^{2} \\ &{}+c_{\sigma}\bigl(K_{\epsilon}+\Vert \phi \Vert _{H^{2}} \bigr)Q_{5}\bigl(\Vert B\Vert _{\mathcal {H}}\bigr)+C+ \frac{1}{4}\bigl\Vert A^{\frac{1+\sigma}{2}}w_{t}(t)\bigr\Vert ^{2}. \end{aligned}$$

By the mean value theorem, similar to (3.18), we have

$$\begin{aligned}& \bigl\langle f(u,0)-f(u,u_{t})-\bigl(f(z,0)-f(z,z_{t}) \bigr),A^{\sigma}w_{t}\bigr\rangle \\& \quad= \bigl\langle -f'_{2}(u,\vartheta_{2} u_{t})u_{t}+f'_{2}(z, \vartheta_{2} z_{t})z_{t},A^{\sigma}w_{t} \bigr\rangle \\& \quad\leqslant \delta\bigl(\Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{\sigma}w_{t} \bigr\Vert _{L^{6/(1+2\sigma)}} \\& \quad\leqslant \delta\bigl(\Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{(1+\sigma)/2}w_{t} \bigr\Vert \\& \quad\leqslant c+\frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert ^{2} \end{aligned}$$

and

$$\begin{aligned}& \bigl\langle f(u,0)-f(u,u_{t})-\bigl(f(z,0)-f(z,z_{t}) \bigr),A^{\sigma}w\bigr\rangle \\& \quad= \bigl\langle -f'_{2}(u,\vartheta_{2} u_{t})u_{t}+f'_{2}(z, \vartheta_{2} z_{t})z_{t},A^{\sigma}w\bigr\rangle \\& \quad\leqslant \delta\bigl(\Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{\sigma}w\bigr\Vert _{L^{6/(1+2\sigma)}} \\& \quad\leqslant \delta\bigl(\Vert u_{t}\Vert _{L^{6/(5-2\sigma)}}+\Vert z_{t}\Vert _{L^{6/5-2\sigma }}\bigr)\bigl\Vert A^{(1+\sigma)/2}w\bigr\Vert \\& \quad\leqslant c+\frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w\bigr\Vert ^{2}. \end{aligned}$$

Therefore, we can finish the proof by using the Gronwall-type inequality as was done in [18], Lemma 4.4. □

Finally, for $u(t)$, the following decomposition is valid, which will be used later to construct an exponential attractor.

Lemma 3.6

For each $\sigma\in[0,\frac{1}{2})$ and for any bounded (in $\mathcal {H}^{\sigma}$) subset $B_{1}\subset\mathcal{H}^{\sigma}$, if the initial data $\xi_{u}(0)\in\phi(x)+B_{1}$, then

$$\begin{aligned}& \bigl\Vert S(t)\xi_{u}(0)-\bigl(\phi(x),0\bigr)\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}=\bigl\Vert \bigl(u(t),u_{t}(t)\bigr)- \bigl(\phi(x), 0\bigr)\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}\leqslant K_{B_{1},\sigma} \\& \quad \forall t\geqslant0, \xi_{u}(0)\in \phi(x)+B_{1}, \end{aligned}$$

where the constant $K_{B_{1},\sigma}$ depends only on the $\mathcal {H}^{\sigma}$-bound of $B_{1}$ and σ.

Proof

By taking the following decomposition: $u(t)=\hat{u}(t)+\phi(x)$, where $\phi(x)$ is the unique solution of (3.1) and $\hat{u}(t)$ solves the following equation:

$$ \textstyle\begin{cases} \hat{u}_{tt}-\Delta\hat{u}_{t}-\Delta\hat{u}+f(u,0)-f(\phi,0)=\eta_{0} \phi+f(u,0)-f(u,u_{t}) \quad\text{in } \Omega\times\mathbb{R}^{+},\\ \hat{u} \vert _{\partial\Omega}=0,\\ (\hat{u}(x,0),\hat{u}_{t}(x,0))=\xi_{u}(0)-(\phi,0), \end{cases} $$

by applying Lemma 3.4, we get similar estimates to those in Lemma 3.5. Noting that the initial value data $(\hat{u}(x,0),\hat {u}_{t}(x,0))=\xi_{u}(0)-(\phi,0)\in\mathcal{H}^{\sigma}$, the conclusion can be obtained. □

Hence, the proof of Theorem 3.1 follows from the above lemmas as in [8].

4 Exponential attractor

In this section, based on the asymptotic regularity obtained above, we will construct an exponential attractor by the abstract method devised in [12]. Here it is different from [8] to prove the asymptotic smooth property (as it was called by EMS 2000 in [19]) under the additional assumptions (4.26), (4.27) in that paper.

By our abstract method devised in [12], one defines here S as the map induced by Poincaré sections of a Lipschitz continuous semigroup $\{S(t)\}_{t\geqslant0}$ at the time $t=T^{*}$ for some $T^{*}>0$; that is, $S: =S(T^{*})$ and $S: B_{\epsilon_{0}}(\mathscr{A})\rightarrow B_{\epsilon_{0}}(\mathscr{A})$ is a $C^{1}$ map. $\mathscr{L}(X)=\{L|L:X\rightarrow X \text{ bounded linear maps}\}$, $\mathscr{L}_{\lambda}(X)=\{L|L\in \mathscr{L}(X)\mbox{ and }L=K+C \text{ with } K \text{ compact}, \Vert C\Vert <\lambda\}$. For the discrete semigroup $\{S^{n}\}^{\infty}_{n=1}$ generated by S, we have the following lemmas.

Lemma 4.1

(see Theorem 1.2 [12])

If there exists $\lambda\in(0,1)$ such that $D_{x}S(x)\in \mathscr{L}_{\lambda}(X)$ for all $x\in B_{\epsilon_{0}}(\mathscr{A})$ then $\{S^{n}\}^{\infty}_{n=1}$ possesses an exponential attractor $\mathscr{M}_{d}$.

Lemma 4.2

(see Theorem 1.4 [12])

Suppose that there is $T^{*}>0$ such that $S=S(T^{*})$ satisfies the condition of above lemma 4.1 and the map $F(x,t)=S(t)x$ is Lipschitz from $[0,T]\times X$ into X for any $T>0$. Then the flow $\{S(t)\}_{t\geqslant0}$ admits an exponential attractor $\mathscr{M}_{c}$.

As regards the Fréchet differential of semigroup, we have the following crucial lemma.

Lemma 4.3

Consider the linearized equation of (1.1),

$$ \textstyle\begin{cases} U_{tt}-\Delta U_{t}-\Delta U+f'_{1}(u,u_{t})U+f'_{2}(u,u_{t})U_{t}=0,\\ U(x,t)|_{\partial\Omega}=0, \\ (U(x,0),U_{t}(x,0))^{T}=(\xi,\eta)^{T}. \end{cases} $$

(4.1)

If the function $f(u,v)$ satisfies conditions (2.3)-(2.7), then (4.1) is a well-posed problem in E, the mapping $S(t)$ defined in (1.1) is Fréchet differentiable on E for any $t>0$, its differential at $\varphi_{0}=(u_{0},u_{1})^{T}$ is the linear operator on $E:(\xi,\eta)^{T}\mapsto(U(t),V(t))^{T}$, where U is the solution of (4.1) and $V=U_{t}$.

Proof

According to assumptions (2.4)-(2.6), (4.1) is a well-posed problem in $\mathcal{H}$.

In the sequel, we first consider the Lipschitz property of the semigroup $S(t)$ on the bounded sets B ($\subset\mathcal{H}$). Letting $\varphi_{0}=(u_{0},u_{1})^{T}\in D(\mathbb{L})$, $\tilde{\varphi }_{0}=\varphi_{0}+(\xi,\eta)^{T}=(u_{0}+\xi,u_{1}+\eta)^{T}\in D(\mathbb{L})$, it follows from the above estimate that the solutions $S(t)\varphi _{0}=\varphi(t)=(u(t),u_{t}(t))^{T}\in D(\mathbb{L})$, $S(t)\tilde{\varphi }_{0}=\tilde{\varphi}(t)=(\tilde{u}(t),\tilde{u}_{t}(t))^{T}\in D(\mathbb{L})$.

Obviously, the difference $\psi=\tilde{u}-u$ satisfies

$$ \psi_{tt}-\Delta\psi_{t}-\Delta\psi=-\bigl[f( \tilde{u},\tilde{u}_{t})-f(u,u_{t})\bigr]. $$

(4.2)

Taking the scalar product of (4.2) with $\psi_{t}=\tilde{u}_{t}-u_{t}$ in $L^{2}(\Omega)$ and by the mean value theorem, we have

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl(\Vert \psi_{t}\Vert ^{2}+\Vert \nabla\psi \Vert ^{2} \bigr)+\Vert \nabla\psi_{t}\Vert ^{2} \\& \quad=\bigl\langle - \bigl[f(\tilde{u},\tilde{u}_{t})-f(u,\tilde {u}_{t})\bigr]- \bigl[f(u,\tilde{u}_{t})-f(u,u_{t})\bigr],\psi_{t} \bigr\rangle \\& \quad= \bigl\langle -f'_{1}\bigl(u+\vartheta_{1}( \tilde{u}-u),u_{t}\bigr)\psi -f'_{2} \bigl(u,u_{t}+\vartheta_{2}(\tilde{u}_{t}-u_{t}) \bigr)\psi_{t}, \psi_{t}\bigr\rangle \\& \quad\bigl(\text{by (2.4), (2.6) and the Poincar\'{e} inequality}\bigr) \\& \quad\leqslant \int_{\Omega}C\bigl(1+\vert u\vert ^{4}+\vert \tilde{u}\vert ^{4}\bigr)\vert \psi \vert \vert \psi_{t}\vert \,dx+\delta \Vert \psi_{t}\Vert _{L^{2}(\Omega)}^{2} \\& \quad\leqslant C\bigl(1+\Vert u\Vert _{L^{6}}^{4}+\Vert \tilde{u}\Vert _{L^{6}}^{4}\bigr)\Vert \psi \Vert _{L^{6}}\Vert \psi_{t}\Vert _{L^{6}}+\delta \Vert \psi_{t}\Vert _{L^{2}(\Omega)}^{2} \\& \quad(\text{due to Lemma~3.6 and the Poincar\'{e} inequality}) \\& \quad\leqslant C(\delta)\Vert \nabla\psi \Vert _{L^{2}(\Omega)}^{2}+2 \delta \Vert \nabla \psi_{t}\Vert _{L^{2}(\Omega)}^{2}. \end{aligned}$$

(4.3)

Since δ is small enough, applying the Gronwall inequality to (4.3), it is easy to show the semigroup $\{S(t)\}_{t\geqslant0}$ is Lipschitz, i.e.,

$$\begin{aligned} \bigl\Vert \tilde{\psi}(t)-\psi(t)\bigr\Vert _{H^{1}_{0}\times L^{2}}^{2} =&\bigl\Vert \tilde{u}(t)-u(t)\bigr\Vert ^{2}+\bigl\Vert \nabla\tilde{u}(t)-\nabla u(t)\bigr\Vert ^{2} \\ \leqslant& e^{ct} \bigl(\Vert \eta \Vert ^{2}+\Vert \nabla\xi \Vert ^{2} \bigr),\quad \forall t\geqslant0. \end{aligned}$$

(4.4)

Integrating (4.3) in dτ on $[0, t]$, this, on account of (4.4), yields

$$ \int_{0}^{t} \Vert \nabla\psi \Vert ^{2}\, d\tau\leqslant e^{ct} \bigl(\Vert \eta \Vert ^{2}+\Vert \nabla\xi \Vert ^{2} \bigr),\quad \forall t \geqslant0. $$

(4.5)

Furthermore, applying the same argument as in [6] with the assumptions (2.4)-(2.6), we can obtain the same estimates for $\Vert \psi_{t}(t)\Vert$ and $\Vert \nabla\psi_{t}(t)\Vert$, that is,

$$\begin{aligned} \bigl\Vert \tilde{\psi}_{t}(t)-\psi_{t}(t) \bigr\Vert _{H^{1}_{0}\times L^{2}}^{2} =&\bigl\Vert \tilde {u}_{t}(t)-u_{t}(t)\bigr\Vert ^{2}+\bigl\Vert \nabla\tilde{u}_{t}(t)-\nabla u_{t}(t)\bigr\Vert ^{2} \\ \leqslant& e^{ct} \bigl(\Vert \eta \Vert ^{2}+\Vert \nabla\xi \Vert ^{2} \bigr), \quad\forall t\geqslant0. \end{aligned}$$

(4.6)

Next, consider the difference $\theta=\tilde{u}-u-U$, with U the solution of the linearized equation (4.1). Obviously,

$$ \theta(0)=\theta(0)=0, \quad\quad\theta_{t}(0)= \theta_{t}(0)=0; $$

(4.7)

and

$$ \theta_{tt}-\Delta\theta_{t}-\Delta\theta=- \bigl[f(\tilde{u}, \tilde {u}_{t})-f(u,u_{t})-f_{1}'(u,u_{t})U-f_{2}'(u,u_{t})U_{t} \bigr]=h, $$

(4.8)

where $h=- [f(\tilde{u}, \tilde {u}_{t})-f(u,u_{t})-f_{1}'(u,u_{t})U-f_{2}'(u,u_{t})U_{t} ]$.

By the mean value theorem, we have

$$\begin{aligned} h =&- \bigl[f_{1}'\bigl(u+ \vartheta_{3}(\tilde{u}-u),\tilde{u}_{t}\bigr)-f_{1}'(u, \tilde {u}_{t})+f_{1}'(u,\tilde{u}_{t})-f_{1}'(u,u_{t}) \bigr](\tilde{u}-u) \\ &{}- \bigl[f_{2}'\bigl(u,u_{t}+ \vartheta_{4}(\tilde{u}_{t}-u_{t}) \bigr)-f_{2}'(u,u_{t}) \bigr](\tilde {u}_{t}-u_{t}) \\ &{}+f_{1}'(u,u_{t})\theta+f_{2}'(u,u_{t}) \theta_{t}, \end{aligned}$$

(4.9)

where $\vartheta_{i}\in(0,1)$, $i=3,4$.

Taking the scalar product of each side of (4.8) with $\theta_{t}$ in $L^{2}(\Omega)$ and by (4.7), we find

$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl(\Vert \theta_{t}\Vert ^{2}+\Vert \nabla\theta \Vert ^{2}\bigr)+\Vert \nabla \theta_{t}\Vert ^{2} \\& \quad=(h,\theta_{t}) \\& \quad\bigl(\text{by assumptions (2.6), (2.7)}\bigr) \\& \quad\leqslant \int_{\Omega} \vert \theta_{t}\vert \bigl(C_{1}\bigl(1+\vert \tilde {u}\vert ^{3}+\vert u \vert ^{3}\bigr)\vartheta_{3}\vert \tilde{u}-u \vert ^{2}+C_{2}\bigl(1+\vert u\vert ^{3} \bigr) (\tilde {u}_{t}-u_{t}) (\tilde{u}-u) \\& \quad\quad{}+C_{3}\bigl(1+\vert u\vert ^{3}\bigr) \vartheta_{4}\vert \tilde{u}_{t}-u_{t}\vert ^{2}+C_{4}\bigl(1+\vert u\vert ^{4}\bigr) \vert \theta \vert +\delta \vert \theta_{t}\vert \bigr)\,dx, \end{aligned}$$

(4.10)

where $\vartheta_{3}, \vartheta_{4}\in(0,1)$.

We will deal with every term in the right-hand side of inequality (4.10); we have

$$\begin{aligned}& \int_{\Omega} \vert \theta_{t}\vert C_{1} \bigl(1+\vert \tilde{u}\vert ^{3}+\vert u\vert ^{3} \bigr) \vartheta_{3} \vert \tilde {u}-u\vert ^{2}\,dx \\& \quad\leqslant C \biggl( \int_{\Omega}\bigl(1+\vert \tilde{u}\vert +\vert u\vert ^{3}\bigr)^{2}\,dx \biggr)^{1/2} \biggl( \int_{\Omega}\bigl(\vert \tilde{u}-u\vert ^{2}\vert \theta_{t}\vert \bigr)^{2}\,dx \biggr)^{1/2} \\& \quad\leqslant C \biggl( \int_{\Omega} \vert \tilde{u}-u\vert ^{4}\vert \theta_{t}\vert ^{2}\,dx \biggr)^{1/2} \\ & \quad\leqslant C \biggl( \int_{\Omega}\bigl[\vert \tilde{u}-u\vert ^{4} \bigr]^{3/2}\,dx \biggr)^{1/3} \biggl( \int_{\Omega}\bigl(\vert \theta_{t}\vert ^{2}\bigr)^{3}\,dx \biggr)^{1/6} \\ & \quad\leqslant \frac{1}{4}\Vert \nabla\theta_{t}\Vert ^{2}+C\Vert \nabla\tilde{u}-\nabla u\Vert ^{4}; \end{aligned}$$

(4.11)

$$\begin{aligned}& \int_{\Omega} \bigl(C_{2}\bigl(1+\vert u\vert ^{3}\bigr) (\tilde{u}_{t}-u_{t}) (\tilde{u}-u) \vert \theta _{t}\vert \bigr)\,dx \\ & \quad\leqslant C \biggl( \int_{\Omega}\bigl(1+\vert u\vert ^{3} \bigr)^{2}\,dx \biggr)^{1/2} \biggl( \int_{\Omega }\bigl(\vert \tilde{u}_{t}-u_{t} \vert \vert \tilde{u}-u\vert \vert \theta_{t}\vert \bigr)^{2}\,dx \biggr)^{1/2} \\ & \quad\leqslant C \biggl( \int_{\Omega}\bigl(\vert \theta_{t}\vert ^{2}\bigr)^{3}\,dx \biggr)^{1/6} \biggl( \int _{\Omega}\bigl(\vert \tilde{u}_{t}-u_{t} \vert ^{2}\vert \tilde{u}-u\vert ^{2} \bigr)^{3/2}\,dx \biggr)^{1/3} \\ & \quad\leqslant C\Vert \theta_{t}\Vert _{L^{6}}^{2}+ \Vert \tilde{u}_{t}-u_{t}\Vert _{L^{6}}^{2} \Vert \tilde {u}-u\Vert _{L^{6}}^{2} \\ & \quad\leqslant \frac{1}{4}\Vert \nabla\theta_{t}\Vert ^{2}+C\Vert \nabla\tilde {u}_{t}-\nabla u_{t} \Vert ^{2}\Vert \nabla\tilde{u}-\nabla u\Vert ^{2}; \end{aligned}$$

(4.12)

$$\begin{aligned}& \int_{\Omega}\bigl(1+\vert u\vert ^{3}\bigr) \vartheta_{4}\vert \tilde{u}_{t}-u_{t}\vert ^{2}\vert \theta _{t}\vert \,dx \\ & \quad\leqslant C \biggl( \int_{\Omega}\bigl(1+\vert u\vert ^{3} \bigr)^{2}\,dx \biggr)^{1/2} \biggl( \int_{\Omega }\bigl(\vert \tilde{u}_{t}-u_{t} \vert ^{2}\vert \theta_{t}\vert \bigr)^{2}\,dx \biggr)^{1/2} \\ & \quad\leqslant C \biggl( \int_{\Omega} \vert \tilde{u}_{t}-u_{t}\vert ^{4}\vert \theta_{t}\vert ^{2}\,dx \biggr)^{1/2} \\ & \quad\leqslant C \biggl( \int_{\Omega}\bigl(\vert \tilde{u}_{t}-u_{t} \vert ^{4}\bigr)^{3/2}\,dx \biggr)^{1/3} \biggl( \int_{\Omega}\bigl(\vert \theta_{t}\vert ^{2}\bigr)^{3}\,dx \biggr)^{1/6} \\ & \quad\leqslant C\Vert \tilde{u}_{t}-u_{t}\Vert _{L^{6}}^{4}+\bigl\Vert \theta_{t}^{N} \bigr\Vert _{L^{6}}^{2} \\ & \quad\leqslant C\Vert \nabla\tilde{u}_{t}-\nabla u_{t}\Vert ^{4}+\frac{1}{4}\Vert \nabla \theta_{t}\Vert ^{2}; \end{aligned}$$

(4.13)

$$\begin{aligned}& \int_{\Omega}C_{4}\bigl(1+\vert u\vert ^{4} \bigr)\vert \theta \vert \vert \theta_{t}\vert \,dx \\ & \quad\leqslant C_{4} \biggl( \int_{\Omega}\bigl(1+\vert u\vert ^{4} \bigr)^{3/2}\,dx \biggr)^{2/3}\biggl( \int _{\Omega} \vert \theta \vert ^{3}\vert \theta_{t}\vert ^{3}\,dx\biggr)^{1/3} \\ & \quad\leqslant C\biggl( \int_{\Omega} \vert \theta \vert ^{6}\,dx \biggr)^{1/6}\biggl( \int_{\Omega} \vert \theta _{t}\vert ^{6}\,dx \biggr)^{1/6} \\ & \quad\leqslant C\frac{1}{4}\Vert \nabla\theta \Vert _{2}^{2}+ \frac{1}{4}\Vert \nabla\theta _{t}\Vert _{2}^{2}. \end{aligned}$$

(4.14)

Plugging (4.11)-(4.14) into (4.10), we have

$$\begin{aligned}& \frac{d}{dt} \bigl(\Vert \theta_{t}\Vert ^{2}+ \Vert \nabla\theta \Vert ^{2} \bigr) \\ & \quad\leqslant C_{1} \bigl(\Vert \theta_{t}\Vert ^{2}+\Vert \nabla\theta \Vert ^{2} \bigr) \\ & \quad\quad{}+C_{2} \bigl(\Vert \nabla\tilde{u}-\nabla u\Vert ^{4}+ \Vert \nabla\tilde{u}_{t}-\nabla u_{t}\Vert ^{2} \Vert \nabla\tilde{u}-\nabla u\Vert ^{2}+\Vert \nabla \tilde{u}_{t}-\nabla u_{t}\Vert ^{4} \bigr), \end{aligned}$$

where $C_{1}>0$, $C_{2}>0$. By the Gronwall inequality and the estimates (4.4), (4.5), (4.6), we obtain

$$\begin{aligned} \Vert \theta_{t}\Vert ^{2}+\Vert \nabla\theta \Vert ^{2} \leqslant& \frac{C_{2}}{C_{1}}\exp^{C_{1}t} \int_{0}^{t} \bigl(\bigl\Vert \nabla\tilde {u}(s)- \nabla u(s)\bigr\Vert ^{4} \\ &{}+\bigl\Vert \tilde{u}_{t}(s)- u_{t}(s)\bigr\Vert ^{2}\bigl\Vert \nabla\tilde{u}(s)-\nabla u(s)\bigr\Vert ^{2}+\bigl\Vert \nabla\tilde{u}_{t}(s)-\nabla u_{t}(s)\bigr\Vert ^{4} \bigr)\,ds \\ \leqslant& C_{3} \bigl(\vert \eta \vert ^{2}+\Vert \xi \Vert ^{2} \bigr)^{2}\cdot\exp^{C_{4}t},\quad \forall t \geqslant0, \end{aligned}$$

where $C_{3}>0$, $C_{4}>0$, that is,

$$ \bigl\Vert \tilde{\psi}(t)-\psi(t)-U(t)\bigr\Vert _{H^{1}_{0}\times L^{2}}^{2} \leqslant C_{3} \bigl(\bigl\Vert (\xi,\eta)^{T}\bigr\Vert _{H^{1}_{0}\times L^{2}}^{2} \bigr)^{2}\cdot\exp^{C_{4}t} \quad\forall t\geqslant0. $$

Therefore,

$$\begin{aligned}& \frac{\Vert \tilde{\psi}(t)-\psi(t)-U(t)\Vert _{H^{1}_{0}\times L^{2}}^{2}}{\Vert (\xi ,\eta)^{T}\Vert _{H^{1}_{0}\times L^{2}}^{2}} \\& \quad\leqslant C_{4}\bigl\Vert (\xi,\eta)^{T}\bigr\Vert ^{2}_{H^{1}_{0}\times L^{2}}\cdot\exp ^{C_{4}t} \\& \quad\rightarrow 0 \quad\text{as } (\xi, \eta)^{T}\rightarrow0 \text{ in } D( \mathbb{L}). \end{aligned}$$

(4.15)

Since $\mathcal{H}=H^{1}_{0}(\Omega)\times L^{2}(\Omega)$ is dense in $D(\mathbb{L})$, (4.15) is true for solutions $\tilde{\psi}(t)$, $\psi(t)$, $U(t)\in\mathcal{H}$.

Next, to prove the decomposition (4.1), one has the following.

Lemma 4.4

$L\cdot((\xi,\eta)^{T})=(U,U_{t})=(U_{1},U_{1t})+(U_{2},U_{2t})=C\cdot((\xi ,\eta)^{T})+K\cdot((\xi,\eta)^{T})$ (where the operator C is contractive and K is compact as in Lemma  4.1), separately, satisfying the following equations:

$$\begin{aligned}& \textstyle\begin{cases} U_{1tt}-\Delta U_{1t}-\Delta U_{1}=0,\\ U_{1}(x,t) \vert _{\partial\Omega}=0, \\ (U_{1}(x,0),U_{1t}(x,0))^{T}=(\xi,\eta)^{T}; \end{cases}\displaystyle \end{aligned}$$

(4.16)

$$\begin{aligned}& \textstyle\begin{cases} U_{2tt}-\Delta U_{2t}-\Delta U_{2}+f'_{1}(u,u_{t})U_{2}+f'_{2}(u,u_{t})U_{2t}=0,\\ U_{2}(x,t)|_{\partial\Omega}=0, \\ (U_{2}(x,0),U_{2t}(x,0))^{T}=(0,0)^{T}. \end{cases}\displaystyle \end{aligned}$$

(4.17)

Proof

For $(U_{1}, U_{1t})$, we set

$$ \zeta(t)=U_{1t}(t)+\epsilon U_{1}(t). $$

Here $\epsilon\in(0,\epsilon_{0})$, for some $\epsilon_{0}\leqslant1$ to be determined later. Testing equation (4.16) with ζ yields

$$ \frac{1}{2}\frac{d}{dt}E+\epsilon(1-\epsilon)\bigl\Vert A^{1/2}U_{1}\bigr\Vert ^{2}+\bigl\Vert A^{1/2}\zeta\bigr\Vert ^{2}=\epsilon \Vert \zeta \Vert ^{2}-\epsilon^{2}\langle U_{1},\zeta\rangle, $$

(4.18)

where the energy functional E is given as

$$ E=(1-\epsilon)\bigl\Vert A^{1/2}U_{1}(t)\bigr\Vert ^{2}+\bigl\Vert \zeta(t)\bigr\Vert ^{2}. $$

We have the inequality

$$ -\epsilon^{2}\langle U_{1},\zeta\rangle\leqslant \frac{\epsilon^{3}}{4}\bigl\Vert A^{1/2}U_{1}\bigr\Vert ^{2}+\epsilon \Vert \zeta \Vert ^{2}. $$

Inserting it into (4.18), one gets

$$ \frac{d}{dt}E+2\epsilon\biggl(1-\epsilon-\frac{\epsilon^{2}}{4}\biggr)\bigl\Vert A^{1/2}U_{1}\bigr\Vert ^{2}+(2 \lambda_{1}-4\epsilon)\Vert \zeta \Vert ^{2}\leqslant0, $$

(4.19)

so, for $\epsilon_{0}$ small enough,

$$ \frac{d}{dt}E+\epsilon\bigl\Vert A^{1/2}U_{1} \bigr\Vert ^{2}+(2\lambda_{1}-4\epsilon)\Vert \zeta \Vert ^{2}\leqslant0, $$

(4.20)

which implies that $(U_{1},U_{1t})=C\cdot((\xi,\eta)^{T})$ is contractive.

Furthermore, multiplying (4.16) by $A^{\sigma}U_{1t}+\epsilon A^{\sigma}U_{1}$ as in Lemma 3.5, we have

$$ \bigl\Vert C\cdot\bigl((\xi,\eta)^{T}\bigr)\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}=\bigl\Vert (U_{1},U_{1t}) \bigr\Vert _{\mathcal{H}^{\sigma}}^{2}\leqslant J_{B,\sigma}\quad \text{for all } t\geqslant0 \text{ and } \xi_{u}(0)\in B. $$

Similarly, multiplying (4.1) by $A^{\sigma}U_{t}+\epsilon A^{\sigma}U$, we have

$$ \bigl\Vert L\cdot(\xi,\eta)^{T}\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}= \bigl\Vert (U,U_{t})\bigr\Vert _{\mathcal {H}^{\sigma}}^{2} \leqslant J_{B,\sigma} \quad\text{for all } t\geqslant 0 \text{ and } \xi_{u}(0)\in B. $$

Thus,

$$\begin{aligned}& \bigl\Vert K\cdot(\xi,\eta)^{T}\bigr\Vert _{\mathcal{H}^{\sigma}}^{2}= \bigl\Vert (U_{2},U_{2t})\bigr\Vert _{\mathcal{H}^{\sigma}}^{2} = \bigl\Vert (U,U_{t})-(U_{1},U_{1t})\bigr\Vert _{\mathcal {H}^{\sigma}}^{2}\leqslant J_{B,\sigma} \\& \quad \text{for all } t\geqslant0 \text{ and } \xi_{u}(0)\in B, \end{aligned}$$

which implies that $K\cdot(\xi,\eta)^{T}=(U_{2},U_{2t})$ is compact and the proof of Lemma 4.4 is finished. □

We also need the following Lipschitz continuity of $\{S(t)\}$.

Lemma 4.5

The mapping $(t,\xi_{u}(0))\mapsto\xi_{u}(t)$ is Lipschitz continuous on $[0,t^{*}]\times\mathcal{B}_{\sigma}$, where the absorbing set $\mathcal {B}_{\sigma}$ is given in Theorem  3.1.

Proof

For any $\xi_{u_{i}}(0)\in\mathcal{B}_{\sigma}$, $t_{i}\in[0,t^{*}]$, $i=1,2$, we have

$$\begin{aligned}& \bigl\Vert S(t_{1})\xi_{u_{1}}(0)-S(t_{2}) \xi_{u_{2}}(0)\bigr\Vert _{\mathcal{H}} \\& \quad\leqslant \bigl\Vert S(t_{1})\xi_{u_{1}}(0)-S(t_{1}) \xi_{u_{2}}(0)\bigr\Vert _{\mathcal{H}}+\bigl\Vert S(t_{1}) \xi_{u_{2}}(0)-S(t_{2})\xi_{u_{2}}(0)\bigr\Vert _{\mathcal{H}}. \end{aligned}$$

The first term has been estimated in (4.4); for the second term, we have

$$\begin{aligned} \bigl\Vert S(t_{1})\xi_{u_{2}}(0)-S(t_{2}) \xi_{u_{2}}(0)\bigr\Vert _{\mathcal{H}} \leqslant& \biggl\vert \int_{t_{1}}^{t_{2}}\biggl\Vert \frac{d}{dt} \bigl(S(t)\xi_{u_{2}}(0)\bigr)\biggr\Vert _{\mathcal {H}} \biggr\vert \\ \leqslant&\biggl\Vert \frac{d}{dt}\bigl(S(t)\xi_{u_{2}}(0)\bigr) \biggr\Vert _{L^{\infty}(0,t^{*};\mathcal{H})}\cdot \vert t_{1}-t_{2}\vert \end{aligned}$$

and we note that $\Vert \frac{d}{dt}(S(t)\xi_{u_{2}}(0))\Vert _{L^{\infty }(0,t^{*};\mathcal{H})}$ can be estimated as in [6] with the assumptions (2.4)-(2.6). □

Therefore, applying the abstract results devised in [12] to Lemmas 4.4, 4.5, we obtain the exponential attractor for the original semigroup $\{S(t)\}_{t\geqslant0}$ in the space $\mathcal{H}$.

Also applying the same argument as in [6] with the assumptions (2.4)-(2.6), we can obtain the same estimates about $\Vert \nabla u_{t}(t)\Vert$ and $u_{tt}(t)$. Thus, similar to Theorem 4.13 in [8], we indeed have the following results (with a stronger attraction for the second component $u_{t}(t)$ of $(u(t),u_{t}(t))$).

Theorem 4.1

Let the assumptions of Theorem  3.1 hold, then there exists a set $\mathcal{E}$, such that

• (i) $\mathcal{E}$ is compact in $H^{1}_{0}(\Omega)\times H^{1}_{0}(\Omega)$ and positively invariant, i.e., $S(t)\mathcal{E}\subset\mathcal {E}$ for all $t\geqslant0$;

• (ii) $\dim_{F}(\mathcal{E},H^{1}_{0}(\Omega)\times H^{1}_{0}(\Omega) )<\infty$;

• (iii) there exist a constant $\alpha>0$ and an increasing function $Q:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for any subset $B\subset\mathcal{H}$ with $\Vert B\Vert _{\mathcal{H}}\leqslant R$,

  $$ \operatorname{dist}_{H^{1}_{0}(\Omega)\times H^{1}_{0}(\Omega)} \bigl(S(t)B,\mathcal {E} \bigr)\leqslant Q(R)\frac{1}{\sqrt{t}}e^{-\alpha t} \quad\textit{for all } t\geqslant0; $$

• (iv) $\mathcal{E}=(\phi(x),0)+\mathcal{E}_{\sigma}$, with $\mathcal {E}_{\sigma}$ bounded in $H^{1}_{0}(\Omega)\cap H^{1+\sigma}(\Omega)\times H^{1}_{0}(\Omega)$ ($\sigma<\frac{1}{2}$), where $\phi(x)$ is the unique solution of (3.1).