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# Exponential attractors for the strongly damped wave equations with critical exponent

Boundary Value Problems20162016:36

https://doi.org/10.1186/s13661-016-0543-5

• Accepted: 25 January 2016
• Published:

## 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 $$H^{-1}(\Omega)$$.

## Keywords

• wave equation
• critical nonlinearity
• exponential attractor

• 37L30
• 39A14

## 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 . Then, in , 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  and Hale  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  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 , 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 . We also remove the additional assumptions (4.26), (4.27) in . Motivated by the key ideas in , 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  to obtain the exponential attractor for equation (1.1) without the restrictions (4.26), (4.27) in . In addition, with the regularity of solutions as in , 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  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 , here $$f=f(u,u_{t})$$ without the addition assumptions (4.26), (4.27) in , and in contrast to , 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 .

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 , $$u(t)=\hat{u}(t)+\phi(x)$$ 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 , we know that
1. 1.
for each θ (>), 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. 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 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 )
$$\mathcal{F}(t)=\mathcal{F}\bigl(z(t)\bigr)=2 \int_{\Omega} \int _{0}^{z(x,t)}\bigl(f(s,0)+\eta_{0} s \bigr)\,ds\,dx.$$
(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 , 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 , 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 , 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 , 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 $$\vartheta_{2}\in(0,1)$$.
$$\bigl\langle g, A^{\sigma}w_{t}\bigr\rangle \leqslant\bigl\Vert A^{-1/2}g\bigr\Vert \bigl\Vert A^{(1+\sigma )/2}w_{t}\bigr\Vert \leqslant c+\frac{1}{3}\bigl\Vert A^{(1+\sigma)/2}w_{t}\bigr\Vert ^{2}.$$
(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 , 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 .

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 , 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 .

## 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 . Here it is different from  to prove the asymptotic smooth property (as it was called by EMS 2000 in ) under the additional assumptions (4.26), (4.27) in that paper.

By our abstract method devised in , 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 )

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 )

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 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  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  with the assumptions (2.4)-(2.6). □

Therefore, applying the abstract results devised in  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  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 , 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
1. (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 $$t\geqslant0$$;

2. (ii)

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

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

4. (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).

## Declarations

### Acknowledgements

Partially supported by NSFC Grant(11401100), the foundation of Fujian Education Department(JB14021), and the innovation foundation of Fujian Normal University (IRTL1206). 