Global attractor for the damped BBM equation in the sharp low regularity space

Abstract

The long-term behavior of low regularity solutions to the damped BBM equation with a distribution force on the torus is studied. Since the energy equation fails to hold for the low regularity solutions, the existence of a bounded absorbing set is not a trivial. This difficulty is overcome by splitting the solution into five parts, where some parts decay exponentially in gradually higher regularity spaces, the final remainder belongs the energy space and thus enjoys the dissipative effect. In this way, the existence of a global attractor is proved in the sharp low regularity space. Moreover, the attractor is shown to have a finite fractal dimension based on the quasi-stable estimate method.

1 Introduction and main results

The Benjamin–Bona–Mahony equation (BBM) equation reads that

$$ u_t-u_{txx}+u_x+uu_x=0, $$

where the solution u is a real-valued function, $u_t=\partial _t u$ and $u_{txx}=\partial _t\partial _x^2u$. The well known model is introduced in [1] to study the dynamics of small-amplitude surface water waves propagating unidirectionally. For the BBM equation, several important topics are discussed in references, such as the characterization of all conservation laws [21], (global) well-posedness in Sobolev spaces [2, 35], convergence to the entropy solutions [8, 9], stability properties of solitary waves [22].

If the BBM equation is endowed with a periodic boundary condition, and some damping effects are taken into account, then we are led to the following damped BBM on the torus $\mathbb {T}=[0,2\pi ]$

$$\begin{aligned} u_t-u_{txx}-u_{xx}+u_x+uu_x=f(x), \quad u(0,x)=u_0(x), \end{aligned}$$

(1.1)

where the force f is independent of time variable. In this paper, we are interested in the long time behavior of low regularity solutions to (1.1). A key concept to describe the long-term behavior of a system is the global attractor. The existence of global attractor has been studied for various damped BBM equations. We refer to [4, 16, 23, 28, 29, 32, 38, 39] for BBM equations on bounded domains, to [13, 24, 25, 31] for BBM equations on unboudned domains, to [5, 6, 17, 30] for BBM equations with noise, and to [10,11,12] for BBM equations with memory. In these works, the dissipative effect is exploited by the energy equation

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\int \limits _\mathbb {T}(|u|^2+|u_x|^2){\,\mathrm d}x +\int \limits _\mathbb {T}|u_x|^2{\,\mathrm d}x =\int \limits _\mathbb {T}f(x)u{\,\mathrm d}x \end{aligned}$$

(1.2)

or similar versions of (1.2). Indeed, (1.2) plays an important role in the proof of the existence of a bounded absorbing set in $H^1(\mathbb {T})$ under some regularity assumptions of f.

Observe that the identity (1.2) is valid for $H^1(\mathbb {T})$ (or smoother) solutions of (1.1), and does not make sense if the solutions merely belong to $H^s(\mathbb {T})$ for $s<1$. Thus it is not an easy task to find a bounded absorbing set for the system (1.1) in $H^s(\mathbb {T})$ with $s<1$. This is the reason that why we call $H^s(\mathbb {T})$ with $s<1$ low regularity spaces. The difficulty is overcome in [34] with the aid of I-method, which is introduced by I-team in [3] for the KdV equation. More precisely, choose an integral operator $I_N$ with a parameter $N>0$ such that $I_N: H^s(\mathbb {T})\mapsto H^1(\mathbb {T})$ and $I_N$ tends to the identity operator in some sense as $N\rightarrow \infty$. Next, one can establish an approximate energy equation for $I_Nu$, namely

$$\begin{aligned} \qquad \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\int \limits _\mathbb {T}(|I_Nu|^2+|(I_Nu)_x|^2){\,\mathrm d}x +\int \limits _\mathbb {T}|(I_Nu)_x|^2{\,\mathrm d}x =\int \limits _\mathbb {T}f(x)I_Nu{\,\mathrm d}x + \mathcal {O}(N^{-\frac{3}{2}})\Vert I_Nu\Vert ^3_{H^1(\mathbb {T})}.\qquad \end{aligned}$$

(1.3)

Clearly, (1.3) reduces to (1.2) as $N\rightarrow \infty$. Thus the term $\mathcal {O}(N^{-\frac{3}{2}})\Vert I_Nu\Vert ^3_{H^1(\mathbb {T})}$ is negligible if N is large. In this way, it is proved in [34] that (1.1) has a global attractor in $\dot{H}^s(\mathbb {T})$ for $0\le s\le 1$ and $f\in \dot{L}^2(\mathbb {T})$. A similar result also holds for the BBM equation on the real line [33].

Another interesting topic in the same spirit is the attractor theory for dissipative equations with irregular forces. This topic has been studied in several references. Indeed, when the force is a distribution, a global attractor is obtained for reaction diffusion equations in [27, 37], and for damped wave equations in [18, 19], respectively. Motivated by these works, the following question arises naturally, whether the regularity $f\in \dot{L}^2(\mathbb {T})$ can be relaxed to ensure the existence of a global attractor in $\dot{H}^s(\mathbb {T})$. In [36], it is shown that the system (1.1) has a global attractor in $\dot{H}^s(\mathbb {T})$ if $f\in \dot{H}^{s-2}(\mathbb {T})$ and $\frac{1}{4}\le s\le 1$. The main idea is a combination of an asymptotic regularity (in the terminology of [26]) and the I-method mentioned above. In this work, we extend the results in [36] to larger range of s. The main result reads as follows.

Theorem 1.1

Assume that $s\in [0,1]$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. The BBM equation (1.1) has a global attractor in $\dot{H}^s(\mathbb {T})$. Moreover, the attractor has a finite fractal dimension.

Some remarks are given in order. First, Theorem 1.1 also holds if $s>1$. In fact, in the case $s>1$, a similar result for the BBM equation on the real line has been proved in [15]. The argument used there is robust and also valid on the torus. Second, the assumption $s\ge 0$ is optimal. In fact, the BBM equation is ill-posed in some sense in $H^s$ with $s<0$, see e.g., [2, 35]. Thus in this case the global attractor theory is generally not expected. Third, the regularity assumption $f\in \dot{H}^{s-2}(\mathbb {T})$ is also sharp. Indeed, it is well known that the global attractor contains all equilibrium points, but the equilibrium point of (1.1) is bounded at most in $H^s(\mathbb {T})$ for general $f\in \dot{H}^{s-2}(\mathbb {T})$.

The proof of Theorem 1.1 is quite different from that in [36]. We give a sketch here for the reader’s convenience. The main idea is to introduce a new decomposition of the solution u to (1.1). To explain it more clear, we consider the most difficult case, namely $s=0$. Split

$$ u=Q+v, \quad v=v_1+v_2+v_3+v_4 $$

where $Q,v_1,v_2,v_3,v_4$ solve the following equations, respectively,

$$\begin{aligned} -Q_{xx}+Q_x+P^N(QQ_x)&=P^Nf, \end{aligned}$$

(1.4)

$$\begin{aligned} v_{1t}-v_{1txx}-v_{1xx}+v_{1x}+(v_1Q)_x&=0, \quad v_1(0,x)=v(0), \end{aligned}$$

(1.5)

$$\begin{aligned} v_{2t}-v_{2txx}-v_{2xx}+v_{2x}+(v_2Q)_x+v_1v_{1x}&=0, \quad v_2(0,x)=0, \end{aligned}$$

(1.6)

$$\begin{aligned} v_{3t}-v_{3txx}-v_{3xx}+v_{3x}+(v_3Q)_x+v_2v_{2x}+(v_1v_2)_x&=0, \quad v_3(0,x)=0, \end{aligned}$$

(1.7)

$$\begin{aligned} v_{4t}-v_{4txx}-v_{4xx}+v_{4x}+(v_4Q)_x+F(v_1,v_2,v_3,v_4)&=P_{N}(f-QQ_x), \;\; v_4(0,x)=0 \end{aligned}$$

(1.8)

where $P^N, P_N$ are Fourier projections on high and low frequencies, respectively, see Sect. 2 for a definition, and F is given by

$$ F(v_1,v_2,v_3,v_4)=\frac{1}{2}\partial _x\Big (v^2-(v_1+v_2)^2\Big ). $$

First, by the contraction mapping principle, we show that (1.4) has a unique solution $Q\in L^2(\mathbb {T})$ provided that $N>0$ is large enough. Moreover, $\Vert Q\Vert _{L^2(\mathbb {T})}$ tends to 0 as $N\rightarrow \infty$. Second, by choosing $N>0$ large enough, we prove that the solutions $v_1,v_2,v_3$ of (1.5)–(1.7) satisfy the exponential decay in a gradually stronger norm

$$ \Vert v_1(t)\Vert _{L^2(\mathbb {T})}+\Vert v_2(t)\Vert _{H^{\frac{1}{3}}(\mathbb {T})}+\Vert v_3(t)\Vert _{H^{\frac{2}{3}}(\mathbb {T})}\le C(\Vert u_0\Vert _{L^2},\Vert f\Vert _{H^{-2}})\textrm{e}^{-\frac{1}{4}t}, \quad \forall t>0. $$

Third, with the above estimates in hand, we find that $v_4(t)$ is uniformly bounded in $H^1(\mathbb {T})$ for all $t>0$. These imply the existence of an absorbing set of (1.1), and then a global attractor in $\dot{L}^2(\mathbb {T})$. We refer to Sect. 3 for details. Finally, in Sect. 4 we show that the global attractor has a finite fractal dimension in $H^s(\mathbb {T})$ by the quasi-stable estimate method [7]. Since Theorem 1.1 follows from Theorem (3.6) and Theorem 4.3 directly, we shall omit the proof of Theorem 1.1 in the sequel.

The notations used in this paper are collected in Sect. 2.

2 Local well-posedness

We recall some notations used in this paper. For every $s\in \mathbb {R}$, we use $H^s(\mathbb {T})$ to denote the Hilbert space endowed with the norm

$$\begin{aligned} \Vert u\Vert _{H^s(\mathbb {T})}=\Vert (1+n^2)^{\frac{s}{2}}\widehat{u}(n)\Vert _{l^2({\,\mathrm d}n)}=\left( \sum _{n\in {\mathbb {Z}}}(1+n^2)^{s}|\widehat{u}(n)|^2 \right) ^{1/2}, \end{aligned}$$

where $\widehat{u}(n)$ is the Fourier transform coefficient of u, namely

$$\begin{aligned} \widehat{u}(n)=\frac{1}{2\pi }\int \limits _\mathbb {T}\textrm{e}^{-\textrm{i}n x}u(x){\,\mathrm d}x. \end{aligned}$$

We use $\dot{H}^s(\mathbb {T})$ to denote the function space satisfying $\Vert u\Vert _{H^s(\mathbb {T})}<\infty$ and $\widehat{u}(0)=0$, or

$$\begin{aligned} \int \limits _\mathbb {T}u(x){\,\mathrm d}x =0. \end{aligned}$$

The space $\dot{H}^s(\mathbb {T})$ has the following equivalent norm

$$\begin{aligned} \Vert u\Vert _{\dot{H}^s(\mathbb {T})}\sim \left( \sum _{n\in {\mathbb {Z}}\backslash \{0\}}|n|^{2\,s}|\widehat{u}(n)|^2 \right) ^{1/2}. \end{aligned}$$

Here and below, $A\sim B$ means that both $A\lesssim B$ and $B\lesssim A$ hold, $A\lesssim B$ means $A\le CB$ for some constant $C>0$.

For every $N>0$, we define the frequency projection operator $P_N$ (resp. $P^N$) on low (resp. high) Fourier modes as

$$\begin{aligned} P_N u = \sum _{|n|\le N} \widehat{u}(n)\textrm{e}^{\textrm{i}nx}, \quad P^N u = \sum _{|n|> N} \widehat{u}(n)\textrm{e}^{\textrm{i}nx}. \end{aligned}$$

Clearly, we have the identity $u=P_N u+P^N u$.

In this section, we shall prove the local well-posedness of the BBM equation (1.1). The main difficulty comes from the low regularity force $f\in H^{s-2}(\mathbb {T})$. This is overcome by introducing a decomposition $u=v+Q$, where Q is the solution of

$$\begin{aligned} -Q_{xx}+Q_x+P^N(QQ_x)=P^Nf. \end{aligned}$$

(2.1)

If u solves the BBM equation (1.1), then $v=u-Q$ solves the equation

$$\begin{aligned} v_t-v_{txx}-v_{xx}+v_x+vv_x+(vQ)_x=g, \quad v(0,x)=u_0(x)-Q, \end{aligned}$$

(2.2)

where g is given by

$$\begin{aligned} g=P_Nf-P_N(QQ_x). \end{aligned}$$

(2.3)

One should show that the elliptic problem (2.1) has indeed a solution. To this end, we rewrite (2.1) as

$$\begin{aligned} Q=(-\partial _x^2+\partial _x)^{-1}(P^Nf-P^N(QQ_x)). \end{aligned}$$

(2.4)

We start with a lemma.

Lemma 2.1

Assume that $0\le s\le 1$. Then for all $u_1\in H^s(\mathbb {T})$ and all $u_2\in L^2(\mathbb {T})$

$$ \Vert \partial _x(1-\partial _x^2)^{-1}(u_1u_2)\Vert _{H^s(\mathbb {T})}\lesssim \Vert u_1\Vert _{H^s(\mathbb {T})}\Vert u_2\Vert _{L^2(\mathbb {T})}. $$

Proof

Since the nth Fourier coefficient of $\partial _x(1-\partial _x^2)^{-1}(u_1u_2)$ is

$$ \textrm{i}n (1+n^2)^{-1}\sum _{n=n_1+n_2}\widehat{u_1}(n_1)\widehat{u_2}(n_2), $$

by the definition of $H^s(\mathbb {T})$ norm and noting $|\textrm{i}n (1+n^2)^{-1}|\le (1+n^2)^{-\frac{1}{2}}$, we conclude Lemma 2.1 if one can show that

$$\begin{aligned} \Vert (1+n^2)^{\frac{s-1}{2}}\sum _{n=n_1+n_2}\widehat{u_1}(n_1)\widehat{u_2}(n_2)\Vert _{l^2({\,\mathrm d}n)}\lesssim \Vert (1+n^2)^{\frac{s}{2}}\widehat{u_1}(n)\Vert _{l^2({\,\mathrm d}n)}\Vert \widehat{u_2}(n)\Vert _{l^2({\,\mathrm d}n)}. \end{aligned}$$

(2.5)

Changing variable $(1+n^2)^{\frac{s}{2}}\widehat{u_1}(n)\mapsto \widehat{u_1}(n)$, we see that (2.5) is equivalent to

$$\begin{aligned} S\lesssim \Vert \widehat{u_1}(n)\Vert _{l^2({\,\mathrm d}n)}\Vert \widehat{u_2}(n)\Vert _{l^2({\,\mathrm d}n)}, \end{aligned}$$

(2.6)

where

$$ S=\Vert (1+n^2)^{\frac{s-1}{2}}\sum _{n=n_1+n_2}(1+n_1^2)^{-\frac{s}{2}}\widehat{u_1}(n_1)\widehat{u_2}(n_2)\Vert _{l^2({\,\mathrm d}n)}. $$

To prove (2.6), we first use the Hölder inequality to see that

$$\begin{aligned} S\lesssim {\left\{ \begin{array}{ll} \Vert \sum _{n=n_1+n_2}(1+n_1^2)^{-\frac{s}{2}}\widehat{u_1}(n_1)\widehat{u_2}(n_2)\Vert _{l^2({\,\mathrm d}n)}, \quad s\in (\frac{1}{2},1],\\ \Vert \sum _{n=n_1+n_2}(1+n_1^2)^{-\frac{s}{2}}\widehat{u_1}(n_1)\widehat{u_2}(n_2)\Vert _{l^\infty ({\,\mathrm d}n)}, \quad s\in [0,\frac{1}{2}),\\ \Vert \sum _{n=n_1+n_2}(1+n_1^2)^{-\frac{s}{2}}\widehat{u_1}(n_1)\widehat{u_2}(n_2)\Vert _{l^4({\,\mathrm d}n)}, \quad s=\frac{1}{2}. \end{array}\right. } \end{aligned}$$

(2.7)

Moreover, by the Young inequality of convolution integral, we deduce from (2.7) that

$$\begin{aligned} S\lesssim {\left\{ \begin{array}{ll} \Vert (1+n^2)^{-\frac{s}{2}}\widehat{u_1}(n)\Vert _{l^1({\,\mathrm d}n)}\Vert \widehat{u_2}(n)\Vert _{l^2({\,\mathrm d}n)}, \quad s\in (\frac{1}{2},1],\\ \Vert (1+n^2)^{-\frac{s}{2}}\widehat{u_1}(n)\Vert _{l^2({\,\mathrm d}n)}\Vert \widehat{u_2}(n)\Vert _{l^2({\,\mathrm d}n)}, \quad s\in [0,\frac{1}{2}),\\ \Vert (1+n^2)^{-\frac{s}{2}}\widehat{u_1}(n)\Vert _{l^{\frac{4}{3}}({\,\mathrm d}n)}\Vert \widehat{u_2}(n)\Vert _{l^2({\,\mathrm d}n)}, \quad s=\frac{1}{2}. \end{array}\right. } \end{aligned}$$

(2.8)

Applying the Hölder inequality to (2.8) gives (2.6). $\quad \square$

A direct consequence of Lemma 2.1 is

$$\begin{aligned} \Vert \partial _x(1-\partial _x^2)^{-1}(u_1u_2)\Vert _{H^s(\mathbb {T})}\lesssim \Vert u_1\Vert _{H^s(\mathbb {T})}\Vert u_2\Vert _{H^s(\mathbb {T})}. \end{aligned}$$

(2.9)

Thanks to (2.9) and the fact

$$ \lim _{N\rightarrow \infty } \Vert P^Nf\Vert _{H^{s-2}(\mathbb {T})}=0, $$

one can show the existence and uniqueness of solution to (2.4) by the contraction mapping principle, see e.g., [36]. More precisely, we have the following.

Proposition 2.2

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. Then there exists a constant $N_0>0$ such that, for all $N\ge N_0$, the elliptic equation (2.4) has a unique solution Q satisfying the bound

$$ \Vert Q\Vert _{H^s(\mathbb {T})}\lesssim \Vert P^N f\Vert _{H^{s-2}(\mathbb {T})}. $$

Now let $N\ge N_0$ and Q be the solution given by Proposition 2.2. We consider the well-posedness of Eq. (2.2). Rewrite (2.2) into an integral form as

$$\begin{aligned} v(t)=\textrm{e}^{-t\varphi (D)}v(0)+\int \limits _0^t \textrm{e}^{-(t-\tau )\varphi (D)}(1-\partial _x^2)^{-1}\Big (g-vv_x-(vQ)_x\Big ){\,\mathrm d}\tau , \end{aligned}$$

(2.10)

where $\varphi (D)=(-\partial _x^2+\partial _x)(1-\partial _x^2)^{-1}$, $\{\textrm{e}^{-t\varphi (D)}\}_{t\ge 0}$ is the $C_0$ semigroup on $H^s(\mathbb {T})$, which can be understood as a Fourier multiplier

$$ \widehat{\textrm{e}^{-t\varphi (D)} u }(n) = \textrm{e}^{-t\frac{n^2+\textrm{i}n}{1+n^2}}\widehat{u}(n), \quad n\in {\mathbb {Z}}. $$

Moreover, we have for all $s\in \mathbb {R}$

$$\begin{aligned} \Vert \textrm{e}^{-t\varphi (D)}u\Vert _{\dot{H}^s(\mathbb {T})}\le \sup _{n\in {\mathbb {Z}}\backslash \{0\}}|\textrm{e}^{-t\frac{n^2+\textrm{i}n}{1+n^2}}|\;\Vert u\Vert _{\dot{H}^s(\mathbb {T})}\le \textrm{e}^{-\frac{1}{2}t}\Vert u\Vert _{\dot{H}^s(\mathbb {T})}. \end{aligned}$$

(2.11)

It is convenient to consider (2.10) with a general data v(0) and force g.

Proposition 2.3

Assume that $0\le s \le 1$, $v(0)\in \dot{H}^s(\mathbb {T})$ and $g\in \dot{H}^{s-2}(\mathbb {T})$. Then the integral equation (2.10) has a unique solution $v\in C([0,T]; \dot{H}^s(\mathbb {T}))$ satisfying the bound

$$ \sup _{0\le t\le T}\Vert v(t)\Vert _{H^s(\mathbb {T})}\le 2(\Vert v(0)\Vert _{H^s(\mathbb {T})}+\Vert g\Vert _{H^{s-2}}), $$

where the life span $T>0$ depends on v(0), g and Q. Moreover, for every $t\in [0,T]$, the mapping $v(0)\mapsto v(t)$ is continuous from $H^s(\mathbb {T})$ to $H^s(\mathbb {T})$.

Proof

Fix $s\in [0,1]$. Consider the map

$$ \Gamma v = \textrm{e}^{-t\varphi (D)}v(0)+\int \limits _0^t \textrm{e}^{-(t-\tau )\varphi (D)}(1-\partial _x^2)^{-1}\Big (g-vv_x-(vQ)_x\Big ){\,\mathrm d}\tau $$

on the ball

$$ \mathcal {B}=\left\{ v\in L^\infty ([0,T];H^s): \sup _{t\in [0,T]}\Vert v(t)\Vert _{H^s(\mathbb {T})}\le 2(\Vert v(0)\Vert _{H^s(\mathbb {T})}+\Vert g\Vert _{H^{s-2}}) \right\} , $$

where $T>0$ is a constant determined later. Clearly, $\mathcal {B}$ is a complete metric space.

If $v\in \mathcal {B}$, then by (2.9) and (2.11)

$$\begin{aligned} \sup _{t\in [0,T]}\Vert \Gamma v\Vert _{H^s(\mathbb {T})} \le \Vert v(0)\Vert _{H^s(\mathbb {T})}+C_1T\sup _{t\in [0,T]}\Big (\Vert g\Vert _{H^{s-2}}+\Vert v\Vert ^2_{H^s(\mathbb {T})}+\Vert Q\Vert _{H^s(\mathbb {T})}\Vert v\Vert _{H^s(\mathbb {T})}\Big ). \end{aligned}$$

(2.12)

Moreover, if $v_1,v_2\in \mathcal {B}$, then

$$\begin{aligned} \sup _{t\in [0,T]}\Vert \Gamma v_1-\Gamma v_2\Vert _{H^s(\mathbb {T})} \le C_1T(\Vert v_1\Vert _{H^s(\mathbb {T})}+\Vert v_2\Vert _{H^s(\mathbb {T})}+\Vert Q\Vert _{H^s(\mathbb {T})})\sup _{t\in [0,T]}\Vert v_1-v_2\Vert _{H^s(\mathbb {T})}. \end{aligned}$$

(2.13)

Choosing T satisfying

$$ T\le \frac{1}{8C_1(1+\Vert Q\Vert _{H^s(\mathbb {T})}+\Vert v(0)\Vert _{H^s(\mathbb {T})}+\Vert g\Vert _{H^{s-2}(\mathbb {T})})}, $$

then we conclude from (2.12)–(2.13) that $\Gamma$ is a contraction mapping on $\mathcal {B}$. This shows the existence and uniqueness of solution $v\in L^\infty ([0,T];H^s(\mathbb {T}))$. Applying $(1-\partial _x^2)^{-1}$ to both sides of (2.2) we have

$$ v_t=(1-\partial _x^2)^{-1}\Big (-(-\partial _x^2+\partial _x)v-vv_x-(vQ)_x+g\Big ). $$

Then we deduce that $v_t\in L^\infty ([0,T];H^s(\mathbb {T}))$. Thus $v\in C([0,T];H^s(\mathbb {T}))$. If $\int \limits _\mathbb {T}v(0){\,\mathrm d}x=0$ and $\int \limits _\mathbb {T}g{\,\mathrm d}x=0$, then integrating (2.2) shows that $\int \limits _\mathbb {T}v(t,x){\,\mathrm d}x=0$ for all $t\in [0,T]$. So $v\in C([0,T];\dot{H}^s(\mathbb {T}))$. The continuity of $v(0)\mapsto v(t)$ on $H^s(\mathbb {T})$ follows from (2.13). $\quad \square$

3 Global attractor

The main new ingredients of this paper are contained in this section. It is well known that the first step to the global attractor theory is the existence of a bounded absorbing set. However, this is not an easy task since we are working in the low regularity space $H^s(\mathbb {T})$ ($0\le s<1$) and the force lies in the sharp irregular space $H^{s-2}(\mathbb {T})$. In the case $\frac{1}{4}\le s\le 1$, this is overcome by the I-method in [36]. Now we shall extend the theory to the full range $0\le s\le 1$. Our strategy starts with a new decomposition of the solution.

Let v be a solution of (2.2). Split $v=v_1+v_2+v_3+v_4$, where $v_1,v_2,v_3$ are solution of

$$\begin{aligned} v_{1t}-v_{1txx}-v_{1xx}+v_{1x}+(v_1Q)_x= & {} 0, \quad v_1(0,x)=v(0), \end{aligned}$$

(3.1)

$$\begin{aligned} v_{2t}-v_{2txx}-v_{2xx}+v_{2x}+(v_2Q)_x+v_1v_{1x}= & {} 0, \quad v_2(0,x)=0, \end{aligned}$$

(3.2)

$$\begin{aligned} v_{3t}-v_{3txx}-v_{3xx}+v_{3x}+(v_3Q)_x+v_2v_{2x}+(v_1v_2)_x= & {} 0, \quad v_3(0,x)=0, \end{aligned}$$

(3.3)

respectively,¹ while the remainder $v_4$ solves

$$\begin{aligned} v_{4t}-v_{4txx}-v_{4xx}+v_{4x}+(v_4Q)_x+F(v_1,v_2,v_3,v_4)=P_{N}(f-QQ_x), \quad v_4(0,x)=0 \end{aligned}$$

(3.4)

where F is given by

$$\begin{aligned} F(v_1,v_2,v_3,v_4)=\frac{1}{2}\partial _x\Big (v^2-(v_1+v_2)^2\Big ). \end{aligned}$$

(3.5)

Remark 3.1

The decomposition presented here looks a little unusual. But it is indeed useful to overcome the difficulty caused by the low regularity solutions. If one restricts to the attractor theory in a more regular space, say $\dot{H}^s(\mathbb {T})$ with $s\in (0,1]$, then a similar decomposition of v into three parts is sufficient.

In the sequel, we shall prove some global bounds of $v_1,v_2,v_3$ and $v_4$ for $N>0$ large enough. Note that both Eqs. (3.1) and (3.2) depend on N in an implicit way. In fact, by Proposition 2.2, if $f\in H^{s-2}(\mathbb {T})$ then

$$\begin{aligned} \Vert Q\Vert _{H^s(\mathbb {T})}\lesssim \Vert P^N f\Vert _{H^{s-2}(\mathbb {T})}\rightarrow 0, \quad \text{ as } N\rightarrow \infty . \end{aligned}$$

(3.6)

Lemma 3.2

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. There exists $N_1>N_0$ such that, if $N\ge N_1$ then (3.1) has a unique solution $v_1\in C([0,\infty ); \dot{H}^s(\mathbb {T}))$ satisfying

$$\begin{aligned} \Vert v_1(t)\Vert _{H^s(\mathbb {T})}\le 2\textrm{e}^{-\frac{1}{4}t}\Vert v(0)\Vert _{H^s(\mathbb {T})}, \quad \forall t\ge 0. \end{aligned}$$

(3.7)

Proof

The integral version of (3.1) reads

$$\begin{aligned} v_1(t)=\textrm{e}^{-t\varphi (D)}v(0)-\int \limits _0^t \textrm{e}^{-(t-\tau )\varphi (D)}(1-\partial _x^2)^{-1}(v_1Q)_x{\,\mathrm d}\tau , \end{aligned}$$

(3.8)

where $\varphi (D)=(-\partial _x^2+\partial _x)(1-\partial _x^2)^{-1}$ as before. Since $f\in \dot{H}^{s-2}(\mathbb {T})$, we have $v(0)\in \dot{H}^s(\mathbb {T})$. Then by the contraction mapping principle, (3.8) has a unique solution $v_1\in C([0,T]; \dot{H}^s(\mathbb {T}))$ for some $T>0$. It remains to show the global bound (3.7). Thanks to (2.11) and Lemma 2.1, we deduce from (3.8) that

$$\begin{aligned} \Vert v_1(t)\Vert _{H^s(\mathbb {T})}\le \textrm{e}^{-\frac{1}{2}t}\Vert v(0)\Vert _{H^s(\mathbb {T})}+C\int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_1(\tau )\Vert _{H^s(\mathbb {T})}{\,\mathrm d}\tau . \end{aligned}$$

(3.9)

In view of (3.6), one can choose $N_1>N_0$ such that $C\Vert Q\Vert _{L^2(\mathbb {T})}\le 1/4$ for all $N\ge N_1$. Then (3.9) implies that

$$\begin{aligned} \Vert v_1(t)\Vert _{H^s(\mathbb {T})}\le \textrm{e}^{-\frac{1}{2}t}\Vert v(0)\Vert _{H^s(\mathbb {T})}+\frac{1}{4}\int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_1(\tau )\Vert _{H^s(\mathbb {T})}{\,\mathrm d}\tau . \end{aligned}$$

(3.10)

Denote by

$$ y(t)=\int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_1(\tau )\Vert _{H^s(\mathbb {T})}{\,\mathrm d}\tau . $$

Then we deduce from (3.10) that

$$\begin{aligned} \frac{{\,\mathrm d}y}{{\,\mathrm d}t} + \frac{1}{4}y\le \textrm{e}^{-\frac{1}{2}t}\Vert v(0)\Vert _{H^s(\mathbb {T})}. \end{aligned}$$

(3.11)

Applying Gronwall’s lemma to (3.11) we see

$$ y(t)\le 4\textrm{e}^{-\frac{1}{4}t}\Vert v(0)\Vert _{H^s(\mathbb {T})}, \quad \forall t\ge 0. $$

This, together with (3.10), gives the desired bound (3.7). $\square$

Lemma 3.3

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. There exists $N_2>N_1$ such that, if $N\ge N_2$ then (3.2) has a unique solution $v_2\in L^\infty ([0,\infty ); \dot{H}^s(\mathbb {T})\bigcap H^{\frac{1}{3}}(\mathbb {T}))$ satisfying

$$\begin{aligned} \Vert v_2(t)\Vert _{H^s(\mathbb {T})}+\Vert v_2(t)\Vert _{H^{\frac{1}{3}}(\mathbb {T})}\lesssim \textrm{e}^{-\frac{1}{4}t}\Vert v(0)\Vert ^2_{H^s(\mathbb {T})}, \quad \forall t\ge 0. \end{aligned}$$

(3.12)

Proof

We rewrite (3.2) as an integral form

$$\begin{aligned} v_2(t)=-\int \limits _0^t \textrm{e}^{-(t-\tau )\varphi (D)}(1-\partial _x^2)^{-1}\Big ((v_2Q)_x+v_1v_{1x}\Big ){\,\mathrm d}\tau . \end{aligned}$$

(3.13)

By the bound in Lemma 3.2, using the contraction mapping, we can show that (3.13) has a unique solution $v_2\in C([0,T]; \dot{H}^s(\mathbb {T}))$ for some $T>0$. It remains to establish a global bound in $H^s(\mathbb {T})\bigcap H^{\frac{1}{3}}(\mathbb {T})$.

Taking $H^s(\mathbb {T})$ norm on both sides of (3.13), using (2.11) and Lemma 2.1, we deduce that

$$\begin{aligned} \Vert v_2(t)\Vert _{H^s(\mathbb {T})}\le C \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Big (\Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_2(\tau )\Vert _{H^s(\mathbb {T})}+\Vert v_1(\tau )\Vert ^2_{H^s(\mathbb {T})}\Big ){\,\mathrm d}\tau . \end{aligned}$$

(3.14)

To obtain an $H^{\frac{1}{3}}(\mathbb {T})$ version of (3.14), we claim that

$$\begin{aligned} \Vert \partial _x(1-\partial _x^2)^{-1}(h_1h_2)\Vert _{H^{\frac{1}{3}}(\mathbb {T}))}\lesssim \Vert h_1\Vert _{L^2(\mathbb {T})}\Vert h_2\Vert _{L^2(\mathbb {T})}. \end{aligned}$$

(3.15)

Indeed, by the definition of $H^s(\mathbb {T})$ norm, by the Hölder inequality we have

$$ \Vert \partial _x(1-\partial _x^2)^{-1}(h_1h_2)\Vert _{H^{\frac{1}{3}}(\mathbb {T})}=\Vert \textrm{i}n(1+n^2)^{-1+\frac{1}{6}}\widehat{h_1h_2}(n)\Vert _{l^2({\,\mathrm d}n)}\lesssim \Vert \widehat{h_1h_2}(n)\Vert _{l^\infty ({\,\mathrm d}n)}\lesssim \Vert h_1h_2\Vert _{L^1(\mathbb {T})}. $$

Using the Cauchy inequality $\Vert h_1h_2\Vert _{L^1(\mathbb {T})}\le \Vert h_1\Vert _{L^2(\mathbb {T})}\Vert h_2\Vert _{L^2(\mathbb {T})}$, we conclude (3.15). Similar to (3.14), but using (3.15) instead, we get

$$\begin{aligned} \Vert v_2(t)\Vert _{H^{\frac{1}{3}}(\mathbb {T})}\le C \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Big (\Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_2(\tau )\Vert _{H^{\frac{1}{3}}(\mathbb {T})}+\Vert v_1(\tau )\Vert ^2_{L^2(\mathbb {T})}\Big ){\,\mathrm d}\tau . \end{aligned}$$

(3.16)

Adding (3.14) and (3.16) together, noting $s\ge 0$, we infer that

$$\begin{aligned} \Vert v_2(t)\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}\le C \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Big (\Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_2(\tau )\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}+\Vert v_1(\tau )\Vert ^2_{H^s(\mathbb {T})}\Big ){\,\mathrm d}\tau , \end{aligned}$$

(3.17)

where $\Vert v_2(t)\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}=\Vert v_2(t)\Vert _{H^s(\mathbb {T})}+\Vert v_2(t)\Vert _{H^{\frac{1}{3}}(\mathbb {T})}$. Using (3.6) again, we can choose $N_2>N_1$ such that $C\Vert Q\Vert _{L^2(\mathbb {T})}\le \frac{1}{4}$ for all $N\ge N_2$. Then (3.17) becomes

$$\begin{aligned} \Vert v_2(t)\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}&\le \frac{1}{4} \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_2(\tau )\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}+C\int \limits _0^t\textrm{e}^{-\frac{1}{2}(t-\tau )}\Vert v_1(\tau )\Vert ^2_{H^s(\mathbb {T})} {\,\mathrm d}\tau \nonumber \\&\le \frac{1}{4} \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_2(\tau )\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})} + 4Ct\textrm{e}^{-\frac{t}{2}}\Vert v(0)\Vert ^2_{H^s(\mathbb {T})}, \end{aligned}$$

(3.18)

where in the last step we used Lemma 3.2.

Similar to (3.11), applying Gronwall’s lemma, we find

$$ \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_2(\tau )\Vert _{H^s\bigcap H^{\frac{1}{3}}(\mathbb {T})}\le \int \limits _0^t \textrm{e}^{-\frac{1}{4}(t-\tau )} 4C\tau \textrm{e}^{-\frac{\tau }{2}}\Vert v(0)\Vert ^2_{H^s(\mathbb {T})}\le 4^3C\textrm{e}^{-\frac{1}{4}t}\Vert v(0)\Vert ^2_{H^s(\mathbb {T})}. $$

This, together with (3.18), gives the desired bound. $\quad \square$

Lemma 3.4

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. There exists $N_3>N_2$ such that, if $N\ge N_3$ then (3.2) has a unique solution $v_3\in L^\infty ([0,\infty ); \dot{H}^s(\mathbb {T})\bigcap H^{\frac{2}{3}}(\mathbb {T}))$ satisfying

$$\begin{aligned} \Vert v_3(t)\Vert _{H^s(\mathbb {T})}+\Vert v_3(t)\Vert _{H^{\frac{2}{3}}(\mathbb {T})}\lesssim \textrm{e}^{-\frac{1}{4}t}(\Vert v(0)\Vert ^3_{H^s(\mathbb {T})}+\Vert v(0)\Vert ^4_{H^s(\mathbb {T})}), \quad \forall t\ge 0. \end{aligned}$$

(3.19)

Proof

The proof is the same to that of Lemma 3.3, but we shall use

$$\begin{aligned} \Vert \partial _x(1-\partial _x^2)^{-1}(h_1h_2)\Vert _{H^{\frac{2}{3}}(\mathbb {T}))}\lesssim \Vert h_1\Vert _{L^2(\mathbb {T})}\Vert h_2\Vert _{H^{\frac{1}{3}}(\mathbb {T})}. \end{aligned}$$

(3.20)

instead of (3.15). Indeed, by the Hölder inequality and the fact $\textrm{i}n(1+n^2)^{-1+\frac{1}{3}}\in l^6({\,\mathrm d}n)$ we have

$$\begin{aligned} \Vert \partial _x(1-\partial _x^2)^{-1}(h_1h_2)\Vert _{H^{\frac{2}{3}}(\mathbb {T})}&=\Vert \textrm{i}n(1+n^2)^{-1+\frac{1}{3}}\widehat{h_1h_2}(n)\Vert _{l^2({\,\mathrm d}n)}\lesssim \Vert \widehat{h_1h_2}(n)\Vert _{l^{3}({\,\mathrm d}n)}. \end{aligned}$$

Applying the Hausdorff-Young inequality and the Sobolev embedding $H^{\frac{1}{3}}(\mathbb {T})\hookrightarrow L^6(\mathbb {T})$, we infer that

$$ \Vert \widehat{h_1h_2}(n)\Vert _{l^{3}({\,\mathrm d}n)}\lesssim \Vert h_1h_2\Vert _{L^{\frac{3}{2}}(\mathbb {T})}\le \Vert h_1\Vert _{L^{2}(\mathbb {T})}\Vert h_2\Vert _{L^{6}(\mathbb {T})}\lesssim \Vert h_1\Vert _{L^2(\mathbb {T})}\Vert h_2\Vert _{H^{\frac{1}{3}}(\mathbb {T})}. $$

So (3.20) holds.

Rewrite (3.3) as an integral form

$$\begin{aligned} v_3(t)=-\int \limits _0^t \textrm{e}^{-(t-\tau )\varphi (D)}(1-\partial _x^2)^{-1}\Big ((v_3Q)_x+v_2v_{2x}+(v_1v_2)_x\Big ){\,\mathrm d}\tau . \end{aligned}$$

(3.21)

Similar to (3.17), we deduce from (3.21) that

$$\begin{aligned}{} & {} \Vert v_3(t)\Vert _{H^s\bigcap H^{\frac{2}{3}}(\mathbb {T})} \nonumber \\{} & {} \le C \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Big (\Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_3(\tau )\Vert _{H^s\bigcap H^{\frac{2}{3}}(\mathbb {T})}+(\Vert v_1(\tau )\Vert _{H^s(\mathbb {T})}+\Vert v_2(\tau )\Vert _{H^s(\mathbb {T})})\Vert v_2(\tau )\Vert _{H^{\frac{1}{3}}(\mathbb {T})}\Big ){\,\mathrm d}\tau , \nonumber \\ \end{aligned}$$

(3.22)

where $\Vert v_3(t)\Vert _{H^s\bigcap H^{\frac{2}{3}}(\mathbb {T})}=\Vert v_3(t)\Vert _{H^s(\mathbb {T})}+\Vert v_3(t)\Vert _{H^{\frac{2}{3}}(\mathbb {T})}$. Choosing $N_3>N_2$ such that $C\Vert Q\Vert _{L^2(\mathbb {T})}\le \frac{1}{4}$ for all $N\ge N_3$, by (3.22) we infer

$$\begin{aligned} \Vert v_3(t)\Vert _{H^s\bigcap H^{\frac{2}{3}}(\mathbb {T})}&\le \frac{1}{4} \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} \Vert v_3(\tau )\Vert _{H^s\bigcap H^{\frac{2}{3}}(\mathbb {T})}\\&\quad +C\int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )}(\Vert v_1(\tau )\Vert _{H^s(\mathbb {T})}+\Vert v_2(\tau )\Vert _{H^s(\mathbb {T})})\Vert v_2(\tau )\Vert _{H^{\frac{1}{3}}(\mathbb {T})}\Big ){\,\mathrm d}\tau . \end{aligned}$$

Combining the decay estimates of $\Vert v_1(t)\Vert _{H^s(\mathbb {T})}$ in Lemma 3.2 and that of $\Vert v_2(t)\Vert _{H^s(\mathbb {T})\bigcap H^{\frac{1}{3}}(\mathbb {T})}$ in Lemma 3.3, we conclude the desired bound. $\square$

Lemma 3.5

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. There exist $N_4>N_3$ such that, if $N\ge N_4$ then (3.4) has a global solution $v_4\in L^\infty ([0,\infty );H^1(\mathbb {T}))$ satisfying that for all $t\ge 0$

$$\begin{aligned} \Vert v_4(t)\Vert _{H^1(\mathbb {T})}^2\le \textrm{e}^{-\frac{1}{8}t}\frac{2}{C_1}\textrm{e}^{C_1T_0}\textrm{e}^{\frac{1}{8}T_0}\Big (C_0 + C_2\Big )+16\Big (C_0 + C_2 \textrm{e}^{-\frac{1}{8}t}\Big ), \end{aligned}$$

(3.23)

where $C_0=C_0(N)>0$ (which tends to infinity as $N\rightarrow \infty$), and the constants $T_0,C_1,C_2$ depend only on $\Vert v(0)\Vert _{H^s(\mathbb {T})}$.

Proof

The local existence of solution $v_4$ to (3.4) follows from the well-posedness of v in Proposition 2.3 and Lemma 3.2-Lemma 3.4. It remains to gives a global bound of $\Vert v_4(t)\Vert _{H^1(\mathbb {T})}$. Taking the $L^2(\mathbb {T})$ inner product of the equation in (3.4) with $v_4$, we obtain

$$ \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+\Vert v_{4x}\Vert ^2_{L^2(\mathbb {T})}+\Big ((Qv_4)_x, v_4\Big )+\Big (F(v_1,v_2,v_3,v_4),v_4\Big )=\Big (P_{N}(f-QQ_x), v_4\Big ). $$

This, together with the Poincaré inequality (noting $\int \limits _\mathbb {T}v_4(x){\,\mathrm d}x=0$)

$$ \Vert v_{4x}\Vert _{L^2(\mathbb {T})}\ge \Vert v_{4}\Vert _{L^2(\mathbb {T})}, $$

we deduce that

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+\frac{1}{2}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+\Big ((Qv_4)_x, v_4\Big )+\Big (F(v_1,v_2,v_3,v_4),v_4\Big ) \le \Big (P_{N}(f-QQ_x), v_4\Big ). \nonumber \\ \end{aligned}$$

(3.24)

We estimate the terms in (3.24) as follows. First, by the embedding $H^1(\mathbb {T})\hookrightarrow L^\infty (\mathbb {T})$,

$$\begin{aligned} \left| \Big ((Qv_4)_x, v_4\Big ) \right|&=\left| \Big (Qv_4, v_{4x}\Big ) \right| \le \Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_4\Vert _{L^\infty (\mathbb {T})}\Vert v_4\Vert _{H^1(\mathbb {T})}\nonumber \\&\le C \Vert Q\Vert _{L^2(\mathbb {T})}\Vert v_4\Vert _{H^1(\mathbb {T})}^2\le \frac{1}{8}\Vert v_4\Vert _{H^1(\mathbb {T})}^2 \end{aligned}$$

(3.25)

provided that we choose $N_4>N_3$ such that $C \Vert Q\Vert _{L^2(\mathbb {T})}\le 1/8$ for all $N\ge N_4$. Moreover,

$$\begin{aligned} \left| \Big (P_{N}(f-QQ_x), v_4\Big ) \right|&\le \Vert P_{N}(f-QQ_x)\Vert _{L^2(\mathbb {T})} \Vert v_4\Vert _{L^2(\mathbb {T})}\nonumber \\&\le \frac{1}{8}\Vert v_4\Vert _{H^1(\mathbb {T})}^2+2\Vert P_{N}(f-QQ_x)\Vert _{L^2(\mathbb {T})}^2\nonumber \\&\le \frac{1}{8}\Vert v_4\Vert _{H^1(\mathbb {T})}^2+C_0, \end{aligned}$$

(3.26)

where $C_0=C_0(N,\Vert f\Vert _{H^{s-2}(\mathbb {T})})$ and we used the fact that

$$\begin{aligned} 2\Vert P_{N}(f-QQ_x)\Vert _{L^2(\mathbb {T})}^2\lesssim (1+N^2)\Vert (f-QQ_x)\Vert _{H^{s-2}(\mathbb {T})}^2\lesssim (1+N^2)(\Vert f\Vert _{H^{s-2}(\mathbb {T})}+\Vert Q\Vert ^2_{H^s(\mathbb {T})}). \end{aligned}$$

It remains to estimate $(F(v_1,v_2,v_3,v_4),v_4)$. To this end, recalling the definition (3.5), we write

$$\begin{aligned} F(v_1,v_2,v_3,v_4)&=\frac{1}{2}\partial _x\Big ((v_3+v_4)^2+2(v_3+v_4)(v_1+v_2)\Big )\nonumber \\&=v_4v_{4x}+\Big (v_4(v_1+v_2+v_3)\Big )_x+v_3v_{3x}+\Big (v_3(v_1+v_2)\Big )_x. \end{aligned}$$

(3.27)

Using (3.27) and integration by parts, we find

$$\begin{aligned} -\Big (F(v_1,v_2,v_3,v_4), v_4\Big )&= \Big (v_4(v_1+v_2+v_3),v_{4x}\Big )+\Big (\frac{1}{2}v_3^2+v_3(v_1+v_2),v_{4x}\Big ) \end{aligned}$$

(3.28)

It follows from (3.28) that

$$\begin{aligned} \left| \Big (F(v_1,v_2,v_3,v_4), v_4\Big )\right|&\le \Vert v_4\Vert _{L^\infty (\mathbb {T})}\Vert v_1+v_2+v_3\Vert _{L^2(\mathbb {T})}\Vert v_{4x}\Vert _{L^2(\mathbb {T})} \nonumber \\&\quad + \Vert \frac{1}{2}v_3^2+v_3(v_1+v_2)\Vert _{L^2(\mathbb {T})} \Vert v_{4x}\Vert _{L^2(\mathbb {T})} \nonumber \\&\lesssim (\Vert v_1\Vert _{L^2(\mathbb {T})}+\Vert v_2\Vert _{L^2(\mathbb {T})}+\Vert v_3\Vert _{L^2(\mathbb {T})})\Vert v_4\Vert ^2_{H^1(\mathbb {T})}\nonumber \\&\quad + (\Vert v_3\Vert ^2_{L^4(\mathbb {T})}+\Vert v_3\Vert _{L^\infty (\mathbb {T})}(\Vert v_1\Vert _{L^2(\mathbb {T})}+\Vert v_2\Vert _{L^2(\mathbb {T})}))\Vert v_4\Vert _{H^1(\mathbb {T})}. \end{aligned}$$

(3.29)

Noting $s\ge 0$ and using the embedding $H^{\frac{2}{3}}(\mathbb {T})\hookrightarrow L^\infty (\mathbb {T}) \hookrightarrow L^4(\mathbb {T})$, we deduce from (3.29) that

$$\begin{aligned} \left| \Big (F(v_1,v_2,v_3,v_4), v_4\Big )\right|&\lesssim (\Vert v_1\Vert _{H^s(\mathbb {T})}+\Vert v_2\Vert _{H^s(\mathbb {T})}+\Vert v_3\Vert _{H^s(\mathbb {T})})\Vert v_4\Vert ^2_{H^1(\mathbb {T})}\nonumber \\&\quad + (\Vert v_3\Vert ^2_{H^{\frac{2}{3}}(\mathbb {T})}+\Vert v_3\Vert _{H^{\frac{2}{3}}(\mathbb {T})}(\Vert v_1\Vert _{H^s(\mathbb {T})}+\Vert v_2\Vert _{H^s(\mathbb {T})}))\Vert v_4\Vert _{H^1(\mathbb {T})}. \end{aligned}$$

(3.30)

Plugging the decay estimates in Lemma 3.2-Lemma 3.4 into (3.30) gives

$$\begin{aligned} \left| \Big (F(v_1,v_2,v_3,v_4), v_4\Big )\right| \le C_1\textrm{e}^{-\frac{1}{4}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+C_2 \textrm{e}^{-\frac{1}{4}t}+\frac{1}{8}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}, \end{aligned}$$

(3.31)

where $C_1,C_2>0$ depends only on $\Vert v(0)\Vert _{H^s(\mathbb {T})}$.

Now plugging (3.25), (3.26) and (3.31) into (3.24), we infer that

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+\frac{1}{8}\Vert v_4\Vert ^2_{H^1(\mathbb {T})} \le C_0 +C_1\textrm{e}^{-\frac{1}{4}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+C_2 \textrm{e}^{-\frac{1}{4}t} \end{aligned}$$

(3.32)

for all $t>0$. Rewrite (3.32) as

$$\begin{aligned} \frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+(\frac{1}{4}-C_1\textrm{e}^{-\frac{1}{4}t})\Vert v_4\Vert ^2_{H^1(\mathbb {T})} \le 2C_0 + 2C_2 \textrm{e}^{-\frac{1}{4}t}. \end{aligned}$$

(3.33)

Define a time

$$\begin{aligned} T_0=\max \left\{ t\ge 0: C_1\textrm{e}^{-\frac{1}{4}t}\ge \frac{1}{8}\right\} . \end{aligned}$$

(3.34)

Clearly, $T_0$ is well defined since $C_1$ is finite. We also remark that $T_0$ depends only on $\Vert v(0)\Vert _{H^s(\mathbb {T})}$ since $C_1$ is. We divide the analysis of (3.33) into two cases.

Case (1): $0\le t\le T_0$. In this case, (3.33) shows that

$$\begin{aligned} \frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}\le C_1\Vert v_4\Vert ^2_{H^1(\mathbb {T})} + 2C_0 + 2C_2. \end{aligned}$$

(3.35)

Applying Gronwall’s lemma to (3.35) and recalling $v_4(0)=0$, we get

$$\begin{aligned} \Vert v_4\Vert ^2_{H^1(\mathbb {T})}\le \frac{2}{C_1}\textrm{e}^{C_1T_0}\Big (C_0 + C_2\Big ), \quad 0\le t\le T_0. \end{aligned}$$

(3.36)

Case (2): $t> T_0$. In this case, using the definition (3.34), we deduce from (3.33) that

$$\begin{aligned} \frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert v_4\Vert ^2_{H^1(\mathbb {T})}+ \frac{1}{8}\Vert v_4\Vert ^2_{H^1(\mathbb {T})} \le 2C_0 + 2C_2 \textrm{e}^{-\frac{1}{4}t}. \end{aligned}$$

(3.37)

Applying Gronwall’s lemma to (3.37) gives

$$\begin{aligned} \Vert v_4(t)\Vert ^2_{H^1(\mathbb {T})}&\le \textrm{e}^{-\frac{1}{8}(t-T_0)}\Vert v_4(T_0)\Vert ^2_{H^1(\mathbb {T})}+ \int \limits _{T_0}^t \textrm{e}^{-\frac{1}{8}(t-\tau )} \Big (2C_0 + 2C_2 \textrm{e}^{-\frac{1}{4}\tau }\Big ){\,\mathrm d}\tau \nonumber \\&\le \frac{2}{C_1}\textrm{e}^{C_1T_0}\textrm{e}^{-\frac{1}{8}(t-T_0)}\Big (C_0 + C_2\Big )+16\Big (C_0 + C_2 \textrm{e}^{-\frac{1}{8}t}\Big ), \end{aligned}$$

(3.38)

where we used (3.36) with $t=T_0$. Noting (3.38) also holds if $t\in [0,T_0]$ (see (3.36)), we complete the proof. $\quad \square$

Theorem 3.6

Assume that $0\le s\le 1$ and $f\in \dot{H}^{s-2}(\mathbb {T})$. Then the BBM equation (1.1) is globally well-posed in $\dot{H}^s(\mathbb {T})$, and has a global attractor $\mathcal {A}$ in $\dot{H}^s(\mathbb {T})$.

Proof

Since the case $s=1$ is easier (see [36] and [15]), we assume now that $0\le s<1$. Fix $f\in \dot{H}^{s-2}(\mathbb {T})$. Let $N=N_4$ be given by Lemma 3.5 and let $u_0\in \dot{H}^s(\mathbb {T})$. Since $u=v+Q$, we find

$$\begin{aligned} \Vert v(0)\Vert _{H^s(\mathbb {T})}\lesssim \Vert u_0\Vert _{H^s(\mathbb {T})}+\Vert f\Vert _{H^{s-2}(\mathbb {T})}. \end{aligned}$$

(3.39)

Moreover, it follows from Lemma 3.2-Lemma 3.5 that

$$\begin{aligned} \Vert u(t)-Q\Vert _{H^s(\mathbb {T})}=\Vert v(t)\Vert _{H^s(\mathbb {T})}\le C'\textrm{e}^{-\frac{1}{8}t}+C'', \quad \forall t\ge 0, \end{aligned}$$

(3.40)

where $C'>0$ depends only on $\Vert v(0)\Vert _{H^s(\mathbb {T})}, \Vert f\Vert _{H^{s-2}(\mathbb {T})}$ and N, $C''>0$ depends only on $\Vert f\Vert _{H^{s-2}(\mathbb {T})}$. Combining (3.39) and (3.40) gives

$$\begin{aligned} \limsup _{t\rightarrow \infty } \sup _{u_0\in B}\Vert u(t)\Vert ^2_{H^s(\mathbb {T})}\le C''+\Vert Q\Vert _{H^s(\mathbb {T})}, \end{aligned}$$

(3.41)

where B is a bounded set in $H^s(\mathbb {T})$. This, together with the local well-posedenss in Proposition 2.3, gives the global well-posedness of the BBM equation in $\dot{H}^s(\mathbb {T})$. Recall that the solution of the elliptic equation (2.1) satisfies that $\Vert Q\Vert _{H^s(\mathbb {T})}\lesssim \Vert f\Vert _{H^{s-2}(\mathbb {T})}$. So (3.41) shows that the ball

$$\begin{aligned} \mathcal {B}=\left\{ u: \Vert u\Vert _{H^s(\mathbb {T})} \le \sqrt{C''+\Vert Q\Vert _{H^s(\mathbb {T})}} \right\} \end{aligned}$$

(3.42)

is a bounded absorbing set to the BBM equation (1.1).

Further, if $u_0\in \mathcal {B}$, then it follows from Lemmas 3.2–3.5 again that

$$\begin{aligned} \Vert v_4(t)\Vert _{H^1(\mathbb {T})}\lesssim 1, \quad \forall t\ge 0 \end{aligned}$$

(3.43)

and

$$\begin{aligned} \Vert v(t)-v_4(t)\Vert _{H^s(\mathbb {T})}\lesssim \textrm{e}^{-\frac{1}{8}t}, \quad \forall t\ge 0. \end{aligned}$$

(3.44)

Since the embedding $H^1(\mathbb {T})\hookrightarrow H^s(\mathbb {T})$ is compact (noting $s<1$), thanks to (3.43)–(3.44), we conclude that the solution mapping $v(0)\mapsto v(t)$ of (2.2) is $\omega$-limit compact in $H^s(\mathbb {T})$. In other words, for every bounded set $B\subset H^s(\mathbb {T})$,

$$ \lim _{T\rightarrow +\infty } \kappa _{H^s}\Big ( \big \{v(t): v(0)\in B, t\ge T \big \}\Big )=0, $$

where $\kappa _{H^s}(E)$ denotes the Kuratowski measure of non-compactness of E,

$$ \kappa _{H^s}(E)=\inf \Big \{\delta >0\Big |E \text { has a finite open cover of sets of diameter } <\delta \Big \}. $$

This, together with the relation $u=v+Q$, shows that the solution mapping $u_0\mapsto u(t)$ of (1.1) is also $\omega$-limit compact in $H^s(\mathbb {T})$. By [20, Theorem 3.8] or [14, Theorem 2.4.2], the existence of a global attractor $\mathcal {A}$ follows. $\quad \square$

As mentioned in the introduction, the global attractor $\mathcal {A}$ is bounded at most in $H^s(\mathbb {T})$ for general force $f\in H^{s-2}(\mathbb {T})$. But the proof in (3.6) shows that the attractor $\mathcal {A}$ has a further asymptotic regularity in the terminology of Sun [26].

Corollary 3.7

Let Q be the solution of the elliptic equation (2.1) with $N=N_4$ (given by Lemma 3.5). Let $\mathcal {A}$ be the global attractor obtained in Theorem 3.6. Then the set $\mathcal {A}-Q$ is bounded in $H^1(\mathbb {T})$, namely

$$ \sup _{u\in \mathcal {A}}\Vert u-Q\Vert _{H^1(\mathbb {T})}\lesssim 1. $$

Proof

The proof is standard. Let $u_0$ be an element in the attractor $\mathcal {A}$. Then there exists a bounded complete orbit $\{u(t)\}_{t\in \mathbb {R}}$ also contained in $\mathcal {A}$ such that $u(0)=u_0$ and

$$\begin{aligned} \sup _{t\in \mathbb {R}}\Vert u(t)\Vert _{H^s(\mathbb {T})}\lesssim 1. \end{aligned}$$

(3.45)

For every $t>0$, we can interpret u(0) as the solution of (1.1) with initial data $u(-t)$ at time t. Thus we deduce from Lemma 3.2-Lemma 3.4 that

$$\begin{aligned} \Vert u_0-Q-v_4(0)\Vert _{H^s(\mathbb {T})}\le C\textrm{e}^{-\frac{1}{4}t}, \end{aligned}$$

(3.46)

where C depends only on $\Vert f\Vert _{H^{s-2}(\mathbb {T})}$ and $\Vert u(-t)\Vert _{H^s(\mathbb {T})}$, thus by (3.45), C is bounded uniformly for all $t\in \mathbb {R}$. Taking the limit $t\rightarrow +\infty$ in (3.46) we obtain

$$ u_0-Q=v_4(0). $$

But by Lemma 3.5, $v_4(0)$ is bounded in $H^1(\mathbb {T})$, we conclude that $u_0-Q$ is also bounded in $H^1(\mathbb {T})$. Since $u_0$ can be chosen arbitrarily in $\mathcal {A}$, we complete the proof. $\quad \square$

4 Fractal dimension

We first recall the definition of the fractal dimension. Let $\mathcal {X}$ be a metric space.

Definition 4.1

Let $\mathcal {M}$ be a compact set in $\mathcal {X}$. The fractal dimension is defined by

$$ \textrm{dim} (\mathcal {M},\mathcal {X})=\limsup _{\varepsilon \rightarrow 0} \frac{\ln n(\mathcal {M}, \varepsilon )}{\ln (1 / \varepsilon )} $$

where $n(\mathcal {M}, \varepsilon )$ is the minimal number of closed balls of radius $\varepsilon$ which cover the set $\mathcal {M}.$

Clearly, the fractal dimension depends on the metric space $\mathcal {X}$. This explains the reason that we write the fractal dimension as $\textrm{dim} (\mathcal {M},\mathcal {X})$. The following criterion of Chueshov and Lasiecka [7, Theorem 2.15, p.23] is useful for proving the finite dimensionality of a set.

Proposition 4.2

Let $\mathcal {X}$ be a Banach space and $\mathcal {M}$ be a bounded closed set in $\mathcal {X}$. Assume that there exists a mapping $S: \mathcal {M} \mapsto \mathcal {X}$ such that $\mathcal {M}\subseteq S\mathcal {M}$ and

• (i) S is Lipschitz on $\mathcal {M}$, i.e., there exists $L > 0$ such that

  $$ \Vert Su_1 - Su_2\Vert _{\mathcal {X}} \le L\Vert u_1 - u_2\Vert _{\mathcal {X}}, \quad u_1, u_2 \in \mathcal {M}; $$

• (ii) There exists a compact semi-norm $\Vert \cdot \Vert _{\mathcal {Y}}$ with respect to $\Vert \cdot \Vert _{\mathcal {X}}$ (namely $\mathcal {X} \hookrightarrow \mathcal {Y}$ is compact) such that

  $$ \Vert Su_1 - Su_2\Vert _{\mathcal {X}} \le \eta \Vert u_1 - u_2\Vert _{\mathcal {X}} + K\big (\Vert u_1 - u_2\Vert _{\mathcal {Y}} + \Vert Su_1 - Su_2\Vert _{\mathcal {Y}}\big ) $$

  for any $u_1, u_2 \in \mathcal {M}$, where $0< \eta < 1$ and $K > 0$ are constants. Then $\textrm{dim} (\mathcal {M},\mathcal {X})<\infty$.

The following is a stronger version of Corollary 3.7, which says that the set $\mathcal {A}-Q$ is thinner than a compact set in $H^1(\mathbb {T})$.

Theorem 4.3

Under the same assumptions as that in Corollary 3.7, we have

$$ \textrm{dim} (\mathcal {A}-Q, H^1(\mathbb {T}))<\infty . $$

Proof

Let $u_1(t), u_2(t)$ be two complete orbits on the attractor $\mathcal {A}$. Then $v_i(t)=u_i(t)-Q (i=1,2)$ are complete orbits on the set $\mathcal {A}-Q$. Denote by

$$\begin{aligned} V(t)=v_1(t)-v_2(t), \quad t\in \mathbb {R}\end{aligned}$$

(4.1)

which, according to (2.2), satisfies the equation

$$\begin{aligned} V_t-V_{txx}-V_{xx}+V_x+(VQ)_x+(V\frac{v_1+v_2}{2})_x=0. \end{aligned}$$

(4.2)

We divide the proof into three steps.

Step 1: $H^1(\mathbb {T})$ estimate.

Taking the $L^2(\mathbb {T})$ inner product of (4.2) with V, we obtain

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert V\Vert ^2_{H^1}+\Vert V_x\Vert ^2_{L^2(\mathbb {R})}+(V_x,V)+((VQ)_x,V)+\Big ((V\frac{v_1+v_2}{2})_x,V\Big )=0. \end{aligned}$$

(4.3)

Using the Poincaré inequality and integration by parts, we deduce from (4.3) that

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert V\Vert ^2_{H^1}+\frac{1}{2}\Vert V\Vert ^2_{H^1(\mathbb {R})}\le (VQ,V_x)+\Big (\frac{v_1+v_2}{2},VV_x\Big ). \end{aligned}$$

(4.4)

On the one hand, by the Hölder inequality and the Sobolev embedding $H^1(\mathbb {T})\hookrightarrow L^\infty (\mathbb {T})$,

$$\begin{aligned} |(VQ,V_x)|\le \Vert Q\Vert _{L^2(\mathbb {T})}\Vert v\Vert _{L^\infty (\mathbb {T})}\Vert v_x\Vert _{L^2(\mathbb {T})}\le C\Vert Q\Vert _{L^2(\mathbb {T})}\Vert v\Vert ^2_{H^1(\mathbb {T})}\le \frac{1}{8}\Vert v\Vert ^2_{H^1(\mathbb {T})} \end{aligned}$$

(4.5)

provided that we choose $N=N_5>N_4$ such that $C\Vert Q\Vert _{L^2(\mathbb {T})}\le 1/8$.

On the other hand, similarly we have

$$\begin{aligned} \left| \Big (\frac{v_1+v_2}{2},VV_x\Big )\right| \lesssim \Vert V\Vert _{L^2(\mathbb {T})}\Vert v_1+v_2\Vert _{H^1(\mathbb {T})}\Vert V\Vert _{H^1(\mathbb {T})}. \end{aligned}$$

(4.6)

Thanks to Corollary 3.7, we find that $v_1(t),v_2(t)$ are uniformly bounded in $H^1(\mathbb {T})$, namely

$$ \sup _{t\in \mathbb {R}} \Vert v_1(t)\Vert _{H^1(\mathbb {T})}+\Vert v_2(t)\Vert _{H^1(\mathbb {T})}\lesssim 1. $$

From this, we infer from (4.6) that for some constant $C'>0$

$$\begin{aligned} \left| \Big (\frac{v_1+v_2}{2},VV_x\Big )\right| \le \frac{1}{8}\Vert V\Vert ^2_{H^1(\mathbb {T})}+C'\Vert V\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.7)

Plugging (4.5) and (4.7) into (4.4), we arrive at

$$\begin{aligned} \frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert V\Vert ^2_{H^1}+\frac{1}{2}\Vert V\Vert ^2_{H^1(\mathbb {R})}\le 2C'\Vert V\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.8)

Applying Gronwall’s lemma to (4.8), we get that for all $t\ge 0$

$$\begin{aligned} \Vert V(t)\Vert ^2_{H^1}\le \textrm{e}^{-\frac{1}{2}t}\Vert V(0)\Vert ^2_{H^1} + \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} 2C'\Vert V(\tau )\Vert ^2_{L^2(\mathbb {T})}{\,\mathrm d}\tau . \end{aligned}$$

(4.9)

Step 2: $L^2(\mathbb {T})$ estimate.

Acting $\mathcal {J}=(1-\partial _x^2)^{-1}$ on both sides of (4.2) we find

$$\begin{aligned} V_t-\partial _x^2\mathcal {J}V+\partial _x\mathcal {J}V+\partial _x\mathcal {J}(VQ)+\partial _x\mathcal {J}(V\frac{v_1+v_2}{2})=0. \end{aligned}$$

(4.10)

Taking the $L^2(\mathbb {T})$ inner product of (4.10) with V, and using

$$ (-\partial _x^2\mathcal {J}V,V)\ge 0, \quad (\partial _x\mathcal {J}V,V)=0 $$

we obtain that

$$\begin{aligned} \frac{1}{2}\frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert V\Vert ^2_{L^2(\mathbb {T})}+\Big (\partial _x\mathcal {J}(VQ),V\Big )+\Big (\partial _x\mathcal {J}(V\frac{v_1+v_2}{2}),V\Big )\le 0. \end{aligned}$$

(4.11)

By the Cauchy-Schwarz inequality and Lemma 2.1 (with $s=0$), and noting Q is bounded in $L^2$, we have

$$\begin{aligned} \left| \Big (\partial _x\mathcal {J}(VQ),V\Big )\right| \le \Vert \partial _x\mathcal {J}(VQ)\Vert _{L^2(\mathbb {T})}\Vert V\Vert _{L^2(\mathbb {T})}\lesssim \Vert V\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.12)

Similarly, Corollary 3.7 implies that

$$\begin{aligned} \left| \Big (\partial _x\mathcal {J}(V\frac{v_1+v_2}{2}),V\Big )\right| \lesssim \Vert V\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.13)

It follows from (4.11)–(4.13) that for some constant $C''>0$

$$\begin{aligned} \frac{{\,\mathrm d}}{{\,\mathrm d}t}\Vert V\Vert ^2_{L^2(\mathbb {T})}\le C''\Vert V\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.14)

Applying Gronwall’s lemma to (4.14), we get

$$\begin{aligned} \Vert V\Vert ^2_{L^2(\mathbb {T})}\le \textrm{e}^{C''t}\Vert V(0)\Vert ^2_{L^2(\mathbb {T})}, \quad \forall t\ge 0. \end{aligned}$$

(4.15)

Step 3: finish the proof.

Plugging (4.15) into (4.9) we find

$$\begin{aligned} \Vert V(t)\Vert ^2_{H^1}\le \textrm{e}^{-\frac{1}{2}t}\Vert V(0)\Vert ^2_{H^1} + \int \limits _0^t \textrm{e}^{-\frac{1}{2}(t-\tau )} 2C'\textrm{e}^{C''\tau }\Vert V(0)\Vert ^2_{L^2(\mathbb {T})}{\,\mathrm d}\tau . \end{aligned}$$

(4.16)

In particular, letting $t=1$ in (4.16) gives

$$\begin{aligned} \Vert V(1)\Vert ^2_{H^1}\le \textrm{e}^{-\frac{1}{2}}\Vert V(0)\Vert ^2_{H^1} + C'''\Vert V(0)\Vert ^2_{L^2(\mathbb {T})}. \end{aligned}$$

(4.17)

Let $S(t):v(0)\mapsto v(t)$ be the solution semigroup of (2.2). Since the embedding $H^1(\mathbb {T})\hookrightarrow L^2(\mathbb {T})$ is compact, (4.17) shows that the set $\mathcal {A}-Q$ satisfies the quasi-stable estimate (ii) of Proposition 4.2 with $S=S(1)$. Clearly, (i) of Proposition 4.2 also holds. Thus $\mathcal {A}-Q$ has a finite fractal dimension in $H^1(\mathbb {T})$. $\quad \square$

Corollary 4.4

Under the same assumption as that in Theorem 3.6, we have

$$ \textrm{dim} (\mathcal {A}, H^s(\mathbb {T}))<\infty . $$

Proof

For every $\varepsilon >0$ and $0\le s\le 1$, we define the closed ball

$$ B_\varepsilon (u, H^s(\mathbb {T}))=\Big \{v\in H^s(\mathbb {T}): \Vert v-u\Vert _{H^s(\mathbb {T})}\le \varepsilon \Big \}. $$

Since $\Vert \cdot \Vert _{H^s(\mathbb {T})}\le \Vert \cdot \Vert _{H^1(\mathbb {T})}$, if the set $\mathcal {A}-Q$ can be covered by the union of balls $B_\varepsilon (u_i, H^s(\mathbb {T}))$, $i=1,2,\ldots ,n$, then $\mathcal {A}$ can be covered by the union of balls $B_\varepsilon (u_i+Q, H^s(\mathbb {T}))$, $i=1,2,\ldots ,n$. The key point is that the number of balls covered $\mathcal {A}$ is the same to that of $\mathcal {A}-Q$. Thus

$$ \textrm{dim} (\mathcal {A}, H^s(\mathbb {T}))\le \textrm{dim} (\mathcal {A}-Q, H^1(\mathbb {T})). $$

Then the conclusion follows from Theorem 4.3. $\quad \square$