Global Attractors for the Generalized Wave Fronts in Chemical Reactions and Kuramoto-Sivashinsky Equations

1. Introduction

This paper is concerned with the initial-value problem (IVP) for the generalized equation of wave fronts in chemical reactions (gWFCR), given by

$$\left\{\begin{array}{c}{u}_t-\lambda {\partial }_x^{2}u+\beta {\partial }_x^{4}u+\gamma u-\mu {(1-{\partial }_x^2)}^{-1/2}u-\kappa {({\partial }_{x}u)}^2=f,x\in \mathbb{R},t\ge 0,\hfill \\ u(x,0)=\varphi (x),\hfill \end{array}\right.$$

(1.1)

where $\gamma >0$, $\beta ,\mu \ge 0$ and $\lambda ,\kappa \in \mathbb{R}$.

The last equation frequently appears in several applications, such as reaction-diffusion systems, flame-propagation, and viscous flow problems, where the time-independent function f represents the external excitation and $\gamma u$ is the damping term.

The above IVP, corresponds to a generalization of the initial value problem for the evolution equation of wave fronts in chemical reactions (WFCR)

$$\left\{\begin{array}{c}{u}_t-{\partial }_x^{2}u-\mu {(1-{\partial }_x^2)}^{-1/2}u-{\displaystyle \frac{1}{2}}{({\partial }_{x}u)}^2=0,x\in \mathbb{R},t\ge 0,\hfill \\ u(x,0)=\varphi (x),\hfill \end{array}\right.$$

(1.2)

where $\mu \ge 0$ is a constant and u is a real-valued function, see [10]. We see that the equation in (1.1) reduces to that in (1.2), when $\lambda =1$, $\beta =\gamma =0$, $\kappa ={\displaystyle \frac{1}{2}}$ and $f=0$. Next, we give some information about the physical phenomenon described by IVP (1.2). An initial value problem equivalent to (1.2) is given by

$$\left\{\begin{array}{c}{\displaystyle {\partial }_{t}H-{\mathcal{D}}_C{\partial }_x^{2}H-\frac{1}{2}{({\partial }_{x}H)}^2-{\displaystyle \frac{\delta G}{8\pi }}{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{e}^{ik(x-x^{\prime })}{\displaystyle \frac{H(x^{\prime },t)}{\sqrt{{\displaystyle \frac{1}{4}}+k^2}}}d{x}^{\prime }dk=0,}\hfill \\ H(x,0)=\varphi (x),\hfill \end{array}\right.$$

(1.3)

where ${\mathcal{D}}_C$ is dimensionless catalyst diffusivity, $\delta$ is relative density and G is dimensionless acceleration of gravity, was derived by G. I. Sivashinsky, et al ([23]), to describe vertical propagation of chemical waves fronts in the presence instability due to density gradients (possibly thermally induced). Above, H is the vertical position of the front. For further details, see [10], where a derivation of the equation in (1.2) is presented. See also [3].

Regarding the IVP (1.2), the existing results in the literature are as follows. For the local and global well-posedness in $H^s(\mathbb{R})$, for $s\ge 1$, see [2]. After that, in [10], the authors, using an application of the fixed point theorem in a suitable time-weighted function space, obtained the local and global well-posedness for the IVP (1.2), where $s>{\displaystyle \frac{1}{2}}$. More precisely, they proved the following theorems.

Theorem A. 

(Local well-posedness) Let $\mu >0$ and $s>1/2,$ then for all $\varphi \in H^s(\mathbb{R}),$ there exists $T=T(\parallel \varphi {\parallel }_{H^s})$, a space

$${\mathcal{X}}_T^s\hookrightarrow C([0,T];H^s(\mathbb{R}))$$

and a unique solution u of (1.2) in ${\mathcal{X}}_T^s$. In addition, the flow map data-solution

$$S:H^s(\mathbb{R})\to {\mathcal{X}}_T^s\cap C([0,T];H^s),\varphi \mapsto u$$

is smooth and

$$u\in C((0,T];H^{\infty }(\mathbb{R})).$$

Moreover, if $s^{\prime }>s$ then the solution with initial data $\varphi \in H^{s^{\prime }}(\mathbb{R})$ is defined in the same interval $[0,T]$, with $T=T(\parallel \varphi {\parallel }_{H^s})$.

Theorem B. 

(Global well-posedness) Let $\mu >0$ and $s>1/2$, then the initial value problem (1.2) is globally well-posed in $H^s(\mathbb{R})$.

Theorem C. 

(Ill-posedness) Let $s<1/2$, if there exists some $T>0$, such that the problem (1.2) is locally well-posed in $H^s(\mathbb{R})$, then the flow-map data solution

$$S:H^s(\mathbb{R})\to C([0,T];H^s(\mathbb{R})),\varphi \mapsto u,$$

is not $C^2$ at zero.

The results above are sharp in the sense that the flow-map data-solution is not $C^2$ at the origin. As a consequence, the Cauchy problem (1.2), for $s<1/2$, cannot be solved by a contraction argument on the integral equation (see [10] and references therein).

In this paper, as usual, we assume well-posedness in the sense of Kato, which includes existence, uniqueness, the persistence property, and the smoothness of the data-to-solution map (see [18]).

The IVP (1.1) is also a generalization of the Cauchy problem for the Kuramoto-Sivashinsky (KS) equation

$$\left\{\begin{array}{c}{u}_t+{\displaystyle \frac{1}{2}}{({\partial }_{x}u)}^2+\beta ({\partial }_x^{2}u+{\partial }_x^{4}u)=0,x\in \mathbb{R},t\ge 0,\hfill \\ u(x,0)=\varphi (x),\hfill \end{array}\right.$$

(1.4)

which is obtained from (1.1) by choosing $\lambda =-\beta$, $\mu =0$, $\kappa =-{\displaystyle \frac{1}{2}}$, $\beta >0$ and $f=0$. The IVP (1.4) has been extensively studied by numerous authors (see [15,22] and references therein). Furthermore, the IVP (1.2) is globally well-posed in $H^s(\mathbb{R})$, for $s>-1$, as established in [22], see also [9]. For insights into the dynamical behavior of the Kuramoto-Sivashinsky equation with periodic initial data, we refer to [16,21]. For the existence of a global attractor in Sobolev space $H^s$ for the KS equation, we refer to [4].

The main objective of this article is to prove the existence of global attractors in $H^s(\mathbb{R})$. More specifically, we establish a suitable relationship between the coefficients of the linear and nonlinear parts of (1.1), which enables us to conclude the existence of a global attractor (see Theorem 1.2 below). Here, the global attractor corresponds to a compact set that attracts every bounded set in $Z_{s,{\displaystyle \frac{1}{2}}}$ and, moreover, is maximal with respect to these properties. Next, we summarize the strategy to obtain it.

Throughout this work, $S(t)\varphi \equiv u(t)$ represents the solution of the IVP (1.1). Thus, we use the theory from [24], in which the global attractor $\mathcal{A}$ is the $\omega$-limit of an absorbing set in $H^s(\mathbb{R})$ if $S(t)$ is uniformly compact for large t. Since our domain is unbounded, to show the uniform compactness of the semigroup $S(t)$, we use the ideas described in [1], where a time-dependent weighted function (see (5.1)) is introduced. The ideas employed in this paper are also inspired by [5].

On the other hand, due to the presence of the term ${(1-{\partial }_x^2)}^{-1/2}u$ in (1.1), estimates involving such a weighted function are considered in the weighted Sobolev spaces $Z_{s,{\displaystyle \frac{1}{2}}}$. Now we state our main results.

Theorem 1.1. 

Let $s\ge 1$, $f\in L^2(\langle x\rangle dx)$, and $\kappa \in \mathbb{R}$. Then, the following statements are true.

a)

If $\beta ,\gamma >0$, $\mu \ge 0$ and $\lambda \in \mathbb{R}$, then the IVP (1.1) is globally well-posed in $Z_{s,{\displaystyle \frac{1}{2}}}$.

b)

If $\beta =\mu =0$ and $\lambda >0$, then the IVP (1.1) is globally well-posed in $Z_{s,{\displaystyle \frac{1}{2}}}$.

Theorem 1.2. 

Let $\varphi ,f\in Z_{s,{\displaystyle \frac{1}{2}}}$ and $\kappa \in \mathbb{R}$. Then the following statements are true.

a)

Assume that $s\ge 1$. If $\beta ,\gamma >0$, $\mu \ge 0$, $\lambda \in \mathbb{R}$ and $\gamma >{\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}+{\displaystyle \frac{2}{\beta }}{\lambda }^2+\beta +\mu$, then $\mathcal{A}=\omega (B_s)$ is the global attractor for the semigroup $S(t)$ associated with the Cauchy problem (1.1).

b)

Assume that $s=1$. If $\beta =\mu =0$ and $\lambda >0$, then $\mathcal{A}=\omega (B_s)$ is the global attractor for the semigroup $S(t)$ associated with the Cauchy problem (1.1).

This paper is organized as follows. In the next section, we present some preliminary estimates. Theorem 1.1 is proved in Section 3. In Section 4, we establish the existence of absorbing sets for (1.1). Finally, Section 5 is dedicated to proving Theorem 1.2.

2. Notation and Preliminaries Results

Here, we introduce the notations used throughout this paper. We use c to denote various positive constants that may appear in our arguments. A subscript indicates dependence on parameters; thus, for example, $c_{\alpha ,\beta ,\gamma }$ denotes a constant that depends on parameters $\alpha$, $\beta$, and $\gamma$. Additionally, for positive numbers a and b, we write $a\lesssim b$ if there exists a constant c such that $a\le cb$. Although our main theorems in this article are proved in one dimension, some definitions and results below are presented in a d-dimension setting. The Fourier transform of g is defined by

$$\widehat{g}(\xi )={\int }_{{\mathbb{R}}^d}g(x)e^{-i\xi ·x}dx,\xi \in {\mathbb{R}}^d.$$

Let any $s\in \mathbb{R}$ and a real function g. Then, the Bessel potential and Riesz potential are defined via their Fourier transforms, respectively as

$$\widehat{J^{s}g}(\xi )={(1+|\xi |}^2{)}^{s/2}\widehat{g}(\xi )and\widehat{D^{s}g}(\xi )={|\xi |}^s\widehat{g}(\xi ),\xi \in {\mathbb{R}}^d.$$

The $L^2$-norm is denoted by $\parallel ·\parallel$. Moreover, for $s\in \mathbb{R}$, we denote by $H^s:=H^s({\mathbb{R}}^d)$ the $L^2$-based Sobolev space, endowed with the norm ${\parallel ·\parallel }_s$, where ${\parallel g\parallel }_s:=\parallel J^{s}g\parallel$. For $r\ge 0$, we also set the weighted Sobolev space by

$$Z_{s,r}=H^s({\mathbb{R}}^d)\cap L^2({\langle x\rangle }^{2r}dx),$$

with norm given by ${\parallel g\parallel }_{s,r}^2:={\parallel g\parallel }_s^2+{\parallel {\langle x\rangle }^{r}g\parallel }^2$. Next, we introduce the Stein derivative of order b. For this we write $L_s^p:=J^{-s}{L}^p({\mathbb{R}}^d)$.

Theorem 2.1. 

Let $b\in (0,1)$ and $2d/(d+2b)<p<\infty .$ Then $g\in L_b^p({\mathbb{R}}^d)$ if and only if

a)

$g\in L^p({\mathbb{R}}^d),$

b)

${\mathcal{D}}^{b}g(x):={\displaystyle {\left({\int }_{{\mathbb{R}}^d}\frac{{|g(x)-g(y)|}^2}{{|x-y|}^{d+2b}}dy\right)}^{1/2}}\in L^p({\mathbb{R}}^d),$ with

$${\parallel g\parallel }_{b,p}:=\parallel J^b{g\parallel }_p\simeq {\parallel g\parallel }_p+\parallel D^b{g\parallel }_p\simeq {\parallel g\parallel }_p+{\parallel {\mathcal{D}}^{b}g\parallel }_p.$$

(2.1)

Proof. 

See [17] and references therein. □

The following lemma is useful as well.

Lemma 2.2. 

Let $b\in (0,1)$ and h be a measurable function on ${\mathbb{R}}^d$ such that $h,{\partial }_{x_j}h\in L^{\infty }({\mathbb{R}}^d)$. Then, for all $x\in {\mathbb{R}}^d$

$${\mathcal{D}}^{b}h(x)\lesssim {\parallel h\parallel }_{L^{\infty }({\mathbb{R}}^d)}+{\parallel h^{\prime }\parallel }_{L^{\infty }({\mathbb{R}}^d)}.$$

(2.2)

Moreover,

$$\parallel {\mathcal{D}}^b{(hg)\parallel }_{L^2({\mathbb{R}}^d)}\le \parallel {\mathcal{D}}^b{h\parallel }_{L^{\infty }({\mathbb{R}}^d)}{\parallel g\parallel }_{L^2({\mathbb{R}}^d)}+{\parallel h\parallel }_{L^{\infty }({\mathbb{R}}^d)}{\parallel {\mathcal{D}}^{b}g\parallel }_{L^2({\mathbb{R}}^d)}.$$

(2.3)

Proof. 

Inequalities (2.2) and (2.3) follow from [11] and references therein. □

We will also use the following interpolation estimate.

Lemma 2.3. 

Let $a,b>0.$ Assume that $J^{a}g\in L^2({\mathbb{R}}^d)$ and ${\langle x\rangle }^{b}g={(1+x^2)}^{{\displaystyle \frac{b}{2}}}g(x)\in L^2({\mathbb{R}}^d).$ Then, for any $\beta \in (0,1)$,

$$\parallel J^{a\beta }({\langle x\rangle }^{(1-\beta )b}g){\parallel }_{L^2({\mathbb{R}}^d)}{\lesssim \parallel \langle x\rangle }^b{g\parallel }_{L^2({\mathbb{R}}^d)}^{1-\beta }{\parallel J^{a}g\parallel }_{L^2({\mathbb{R}}^d)}^{\beta }.$$

(2.4)

Proof. 

See ([11], Lemma 2.9). □

Next, we introduce a sequence approximation of the function $\langle x\rangle :={(1+x^2)}^{{\displaystyle \frac{1}{2}}}$ as in [13,17]. Let $N\in {\mathbb{Z}}^{+}$; the truncated weights ${\langle x\rangle }_N:\mathbb{R}\to \mathbb{R}$ satisfy

$${\langle x\rangle }_N(x)=\left\{\begin{array}{cc}\hfill & \langle x\rangle ,\mathrm{if}|x|\le N,\hfill \\ \hfill & 2N,\mathrm{if}|x|\ge 3N,\hfill \end{array}\right.$$

(2.5)

where ${\langle x\rangle }_N$ is smooth and non-decreasing in $|x|$ with ${\partial }_x{\langle x\rangle }_N\le 1$, and there exists a constant c, independent of N, such that $|{\partial }_x{\langle x\rangle }_N|\le c|{\partial }_x\langle x\rangle |$.

Lemma 2.4. 

Let $\tau \ge 0$, $0<\alpha <2$ and $A\ge 0,$ then

$$-{\tau }^2+A{\tau }^{\alpha }\le c_{\alpha }{A}^{{\displaystyle \frac{2}{2-\alpha }}}.$$

Proof. 

The proof follows by an analysis of the function $\phi (x)=-x^2+A{x}^{\alpha }$. □

To formulate the next result, we denote the Sobolev seminorm ${\parallel ·\parallel }_{[s]}$ by

$${\parallel g\parallel }_{[s]}^2={\int }_{{\mathbb{R}}^d}{|\xi |}^{2s}{|\widehat{g}(\xi )|}^{2}d\xi ,s>-d/2.$$

Lemma 2.5. 

(Lemma A.1)

$$\parallel \widehat{g}{\parallel }_{L^1}\le {c\parallel g\parallel }_{[r]}^{(2s-d)/(2s-2r)}{\parallel g\parallel }_{[s]}^{(d-2r)/(2s-2r)},-d/2<r<d/2<s.$$

In particular,

$$\parallel \widehat{g}{\parallel }_{L^1}\le c{\parallel g\parallel }_s,s>d/2.$$

The proof of the above lemma can be found in [19].

3. Well-Posedness in ${\mathit{Z}}_{\mathit{s},{\displaystyle \frac{1}{2}}}$

First, observe that the semigroup associated with the linear part of the gWFCR equation is given, via Fourier transform, by

$$U(t)\varphi :={(e^{t\Phi (\xi )}\widehat{\varphi })}^{\vee },$$

(3.1)

where

$$\Phi (\xi )=-\lambda {\xi }^2-\beta {\xi }^4+\gamma +{\displaystyle \frac{\mu }{\langle \xi \rangle }}.$$

(3.2)

The global well-posedness for the IVP (1.1) in $H^s(\mathbb{R})$, for $s\ge 1$, follows from similar ideas presented in [10,22]. Hence, there exists $u\in C([0,T];H^s)$ that satisfies the following integral equation

$$u(t)=U(t)\varphi +{\int }_0^{t}U(t-\tau )\left(\kappa {({\partial }_{x}u)}^2+f\right)d\tau ,t\in [0,T],$$

(3.3)

for all $T>0$.

Thus, we focus only on the question of well-posedness in the weighted Sobolev spaces $Z_{s,{\displaystyle \frac{1}{2}}}$. To achieve this, we first attempt to apply the ideas presented in [11], where the IVP (1.1) and the truncated weights (2.5) are used to establish well-posedness by deriving an estimate for ${\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel$, where u is solution of (1.1). However, since we do not know how to handle the term ${\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{(1-{\partial }_x^2)}^{-1/2}u\parallel$, whether this strategy is viable remains an open question. Consequently, we conclude that a more effective approach is to analyze an estimate that incorporates the weights ${\langle x\rangle }^{{\displaystyle \frac{1}{2}}}$ and the semigroup (3.1), as detailed in Lemma 3.4 below.

The following results are the ingredients necessary to obtain the well-posedness in the weighted Sobolev spaces $Z_{s,{\displaystyle \frac{1}{2}}}$.

Lemma 3.1. 

Let there be given $\nu ,\mu \ge 0$, $\lambda ,\gamma >0$ and $\beta =0$. Then for all $t>0$ and $s\in \mathbb{R}$

$${\parallel U(t)\varphi \parallel }_{s+\nu }\lesssim \left[1+{(2t\lambda )}^{-{\displaystyle \frac{\nu }{2}}}\right]e^{t(\gamma +\mu )}{\parallel \varphi \parallel }_s.$$

(3.4)

Proof. 

It follows from arguments similar to those in [10]. □

Lemma 3.2. 

Let Φ as in (3.2), $\lambda \in \mathbb{R}$, $\beta ,\gamma >0$ and $\mu \ge 0$. Then, for all $\xi \in \mathbb{R}$ and $t\ge 0$

$$|e^{t\Phi (\xi )}|\le e^{t({\displaystyle \frac{{\lambda }^2}{\beta }}+\gamma +\mu )}$$

(3.5)

and

$$|{\partial }_{\xi }{e}^{t\Phi (\xi )}|\le t(2|\lambda |r+4\beta r^3+\mu )e^{t({\displaystyle \frac{{\lambda }^2}{\beta }}+\gamma +\mu )},$$

(3.6)

where

$$r={\left({\displaystyle \frac{|\lambda |}{2\beta }}+{\displaystyle \frac{2}{\sqrt{t\beta }}}\right)}^{{\displaystyle \frac{1}{2}}}.$$

(3.7)

Proof. 

First, observe that $-\lambda {\xi }^2-\beta {\xi }^4\le 0$, for $|\xi |\ge {({\displaystyle \frac{\lambda }{\beta }})}^{{\displaystyle \frac{1}{2}}}$. Consequently,

$$e^{-t(\lambda {\xi }^2+\beta {\xi }^4)}\le e^{{\displaystyle \frac{t{\lambda }^2}{\beta }}}.$$

The last inequality implies (3.5).

On the other hand, a simple computation provides us

$${\partial }_{\xi }{e}^{t\Phi (\xi )}=-(2\lambda \xi +4\beta {\xi }^3+\mu \xi {\langle \xi \rangle }^{-3})t{e}^{t\Phi (\xi )},$$

(3.8)

and by studying the maximum value of the function $y\in [0,\infty )\mapsto y^k{e}^{-t(\lambda y^2+\beta y^4)}$, we obtain

$$\underset{\xi \in \mathbb{R}}{sup}|{\xi }^k{e}^{-t(\lambda {\xi }^2+\beta {\xi }^4)}|\le r^k{e}^{t(\gamma +\mu )},k=1,3,$$

(3.9)

where r is given in (3.7).

Then, (3.8) and (3.9) imply (3.6).

This ends the proof. □

Proposition 3.3. 

Let there be given $\lambda \in \mathbb{R}$, $\nu ,\mu ,\ge 0$ and $\beta ,\gamma >0$. Then for all $t>0$ and $s\in \mathbb{R}$

$${\parallel U(t)\varphi \parallel }_{s+\nu }\lesssim {\Lambda }_t{\parallel \varphi \parallel }_s,$$

where

$${\Lambda }_t:={\left[{\displaystyle \frac{|\lambda |}{\beta }}+{\left({\displaystyle \frac{\nu +1}{t\beta }}\right)}^{{\displaystyle \frac{1}{4}}}\right]}^{\nu }{e}^{t({\displaystyle \frac{{\lambda }^2}{\beta }}+\gamma +\mu )}.$$

Proof. 

The proof follows from ideas analogous to those presented in [15]. □

Lemma 3.4. 

Let $\varphi \in L^2({\langle x\rangle }^{2}dx)$ and r as in (3.7), then

$${\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t)\varphi \parallel \lesssim G_t\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel ,for allt\ge 0,$$

where

$$G_t:=\left[3+t(2|\lambda |r+4\beta r^3+\mu )\right]e^{t({\displaystyle \frac{{\lambda }^2}{\beta }}+\gamma +\mu )}.$$

(3.10)

Proof. 

By using Plancherel’s identity, (2.1), and Lemma 2.2 we obtain

$$\begin{array}{cc}\hfill {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t)\varphi \parallel & \le \parallel U(t)\varphi \parallel +{\parallel |x|}^{{\displaystyle \frac{1}{2}}}U(t)\varphi \parallel \hfill \\ \hfill & \le \parallel e^{t\Phi (\xi )}\widehat{\varphi }\parallel +\parallel D_{\xi }^{{\displaystyle \frac{1}{2}}}(e^{t\Phi (\xi )}\widehat{\varphi })\parallel \hfill \\ \hfill & \lesssim \parallel e^{t\Phi (\xi )}{\parallel }_{L_{\xi }^{\infty }}\parallel \widehat{\varphi }\parallel +\parallel {\mathcal{D}}_{\xi }^{{\displaystyle \frac{1}{2}}}(e^{t\Phi (\xi )}){\parallel }_{L_{\xi }^{\infty }}\parallel \widehat{\varphi }\parallel +\parallel e^{t\Phi (\xi )}{\parallel }_{L_{\xi }^{\infty }}\parallel {\mathcal{D}}_{\xi }^{{\displaystyle \frac{1}{2}}}\widehat{\varphi }\parallel \hfill \\ \hfill & \lesssim \left(3\parallel e^{t\Phi (\xi )}{\parallel }_{L_{\xi }^{\infty }}+{\parallel {\partial }_{\xi }{e}^{t\Phi (\xi )}\parallel }_{L_{\xi }^{\infty }}\right)\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel .\hfill \end{array}$$

Hence, the proof follows from the above inequality combined with (3.5) and (3.6). □

In the following result, $\Gamma$ represents the standard Gamma function.

Proposition 3.5. 

Let $\beta ,\gamma >0$, $\lambda \in \mathbb{R}$ and $\mu \ge 0$. If $s\ge 0$, then

$${\parallel U(t)\eta \parallel }_s\le L_s(t){\parallel \eta \parallel }_{L^1(\mathbb{R})},$$

where

$$L_s(t)=c_{\lambda ,\beta ,s}\left(1+t^{-{\displaystyle \frac{1}{4}}}+t^{-{\displaystyle \frac{2s+1}{8}}}\right).$$

(3.11)

Proof. 

It follows from ideas similar to those in ([15], Proposition 2.2); however, for the sake of completeness, we will provide a sketch of the proof. Using $\parallel \widehat{\eta }{\parallel }_{L_{\xi }^{\infty }}\lesssim {\parallel \eta \parallel }_{L^1}$ and Hölder’s inequality we see that

$$\begin{array}{cc}\hfill {\parallel U(t)\eta \parallel }_s^2& \le {\parallel \eta \parallel }_{L^1}^2{e}^{2t(\gamma +\mu )}\underset{\mathcal{I}}{\underbrace{{\int }_{-\infty }^{\infty }{(1+{\xi }^2)}^s{e}^{-t(\lambda {\xi }^2+\beta {\xi }^4)}d\xi }}.\hfill \end{array}$$

(3.12)

Next, after the following change of variables ${\xi }^2=r$ and $\tau =2t\beta {(r+{\displaystyle \frac{\lambda }{2\beta }})}^2$ we get

$$\mathcal{I}\le 2e^{{\displaystyle \frac{{\lambda }^2}{4\beta }}}\left[\Gamma \left({\displaystyle \frac{3}{4}}\right)+{|\lambda |}^{s-{\displaystyle \frac{1}{2}}}{\beta }^{-1}\Gamma \left({\displaystyle \frac{1}{2}}\right)t^{-{\displaystyle \frac{1}{2}}}+{\beta }^{-{\displaystyle \frac{2s+1}{4}}}\Gamma \left({\displaystyle \frac{2s+1}{4}}\right)t^{-{\displaystyle \frac{2s+1}{4}}}\right].$$

(3.13)

Therefore, combining (3.12) and (3.13) we conclude the desired result. □

Lemma 3.6. 

If $u\in C([0,T];H^s)$ is a solution of (3.3), with $\varphi ,f\in L^1(\mathbb{R})$, then it follows that

$${\parallel u(t)\parallel }_{\sigma }\le L_{\sigma }(t)\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +\left(|\kappa |\underset{[0,T]}{sup}{\parallel u\parallel }_1^2+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right){\int }_0^t{L}_{\sigma }(t-\tau )d\tau ,$$

(3.14)

where $0\le \sigma <{\displaystyle \frac{7}{2}}$.

Proof. 

The proof is a direct application of Proposition 3.5 together with the embedding

$$L^2(\langle x\rangle dx)\hookrightarrow L^1(\mathbb{R}).$$

□

Next, the integral Equation (3.3) combined with the Banach fixed-point theorem is used to determine the existence of the solution.

Theorem 3.7. 

Let $s\ge 1$, then the IVP (1.1) is locally well-posed in $Z_{s,{\displaystyle \frac{1}{2}}}$.

Proof. 

Let $\varphi \in Z_{s,{\displaystyle \frac{1}{2}}}$, then putting $h_{\varphi }:={\parallel \varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel$ we define

$${\chi }_T=\left\{u\in C([0,T];Z_{s,{\displaystyle \frac{1}{2}}}){|\parallel u(t)-U(t)\varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}\le h_{\varphi },\forall t\in [0,T]\right\}.$$

(3.15)

Then, it is not difficult to see that ${\chi }_T$ is a complete metric space with

$$d(u,v):=\underset{[0,T]}{sup}{\parallel u(t)-v(t)\parallel }_{s,{\displaystyle \frac{1}{2}}}.$$

Our strategy is to show that there exists $0<T^{\prime }\le T$ for which the application

$$(\Psi u)(t):=U(t)\varphi +{\int }_0^{t}U(t-\tau )\left(\kappa {({\partial }_{x}u)}^2+f\right)d\tau ,t\in [0,T],$$

(3.16)

is a contraction in ${\chi }_{T^{\prime }}$.

To do this, Proposition 3.3 (with $\nu =0$) and Lemma 3.4 imply that if $u\in {\chi }_T$, then $\Psi u\in C([0,T];Z_{s,{\displaystyle \frac{1}{2}}})$.

Before starting our arguments, we need some auxiliary estimates. To simplify notation, we will omit the variable $\tau$ in the solution u. By (3.15), Lemma 3.4 and Proposition 3.3, we see that for all $u\in {\chi }_T$ and $\tau \in [0,T]$

$${\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel \le \parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}\left(u-U(\tau )\right){\varphi \parallel +\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(\tau )\varphi \parallel \le (1+G_{\tau })h_{\varphi },$$

(3.17)

and

$${\parallel u\parallel }_1\le {\parallel u-U(\tau )\varphi \parallel }_1+{\parallel U(\tau )\varphi \parallel }_1\le (1+{\Lambda }_{\tau })h_{\varphi }.$$

(3.18)

Here, we assume $0\le s<{\displaystyle \frac{7}{2}}$. The case $s\ge {\displaystyle \frac{7}{2}}$ can be treated similarly, as demonstrated in [15]. Thus, using Proposition 3.5 and (3.18) we obtain

$$\begin{array}{cc}\hfill {\parallel \Psi u(t)-U(t)\varphi \parallel }_s& \le {\int }_0^t\left(|\kappa |\parallel U(t-\tau ){({\partial }_{x}u)}^2{\parallel }_s+{\parallel U(t-\tau )f\parallel }_{H^s}\right)d\tau \hfill \\ \hfill & \le {\int }_0^t{L}_s(t-\tau )(|\kappa |\parallel {({\partial }_{x}u)}^2{\parallel }_{L^1}+{\parallel f\parallel }_{L^1})d\tau \hfill \\ \hfill & \le {\int }_0^t{L}_s{(t-\tau )(|\kappa |\parallel u\parallel }_1^2+{\parallel f\parallel }_{L^1})d\tau \hfill \\ \hfill & \le \left[|\kappa |{(1+{\Lambda }_T)}^2{h}_{\varphi }+1\right]h_{\varphi }{\int }_0^t{L}_s(t-\tau )d\tau ,\hfill \end{array}$$

(3.19)

where above, it is used ${\parallel f\parallel }_{L^1}\lesssim \parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \le h_{\varphi }$.

On the other hand, Proposition 3.5 and definitions above yields us

$$\begin{array}{cc}\hfill {|\kappa |}^{-1}{\parallel \Psi u(t)-\Psi v(t)\parallel }_s& \le {\int }_0^t\parallel U(t-\tau )({({\partial }_{x}u)}^2-{({\partial }_{x}v)}^2){\parallel }_{s}d\tau \hfill \\ \hfill & \le {\int }_0^t{L}_s(t-\tau ){\parallel {({\partial }_{x}u)}^2-{({\partial }_{x}v)}^2\parallel }_{L^1}d\tau \hfill \\ \hfill & \le {\int }_0^t{L}_s(t-\tau )\parallel {\partial }_x(u-v)\parallel \parallel {\partial }_x(u+v)\parallel d\tau \hfill \\ \hfill & \le {\int }_0^t{L}_s{(t-\tau )\parallel u-v\parallel }_1{\parallel u+v\parallel }_{1}d\tau \hfill \\ \hfill & \le 2(1+{\Lambda }_T)h_{\varphi }d(u,v){\int }_0^t{L}_s(t-\tau )d\tau .\hfill \end{array}$$

(3.20)

Assuming $3<\sigma <{\displaystyle \frac{7}{2}}$, we see that by (3.14)

$$\begin{array}{cc}\hfill {\parallel u\parallel }_{\sigma }& \lesssim L_{\sigma }(\tau )\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +\left(2|\kappa |h_{\varphi }^2+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right){\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime }\hfill \\ \hfill & \lesssim h_{\varphi }\left[L_{\sigma }(t)+(2|\kappa |h_{\varphi }+1){\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime }\right].\hfill \end{array}$$

(3.21)

Consequently, using that $H^{{\displaystyle \frac{\sigma }{2}}-1}(\mathbb{R})$ is a Banach algebra and from Lemma 2.3 (with $a=\sigma$, $\alpha =1/2$ and $b={\displaystyle \frac{1}{2}}$) we obtain

$$\begin{array}{cc}\hfill {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{({\partial }_{x}u)}^2\parallel =& \parallel {({\langle x\rangle }^{1/4}{\partial }_{x}u)}^2\parallel \hfill \\ \hfill \le & \parallel J^{{\displaystyle \frac{\sigma }{2}}-1}{\left({\langle x\rangle }^{1/4}{\partial }_{x}u\right)}^2\parallel \hfill \\ \hfill \le & \parallel J^{{\displaystyle \frac{\sigma }{2}}-1}\left({\langle x\rangle }^{1/4}{\partial }_{x}u\right){\parallel }^2\hfill \\ \hfill \le & \parallel J^{{\displaystyle \frac{\sigma }{2}}}({\langle x\rangle }^{1/4}u){\parallel }^2+{\parallel J^{{\displaystyle \frac{\sigma }{2}}-1}({\langle x\rangle }^{-{\displaystyle \frac{3}{4}}}u)\parallel }^2\hfill \\ \hfill \le & {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel +\parallel u\parallel }_{\sigma }+{\parallel u\parallel }_1^2\hfill \\ \hfill \lesssim & (1+G_{\tau })h_{\varphi }+h_{\varphi }\left[L_{\sigma }(t)+(2h_{\varphi }+1){\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime }\right]+{(1+{\Lambda }_{\tau })}^2{h}_{\varphi }^2:=\Theta (\tau ),\hfill \end{array}$$

(3.22)

where above its also used (3.17), (3.18) and (3.21).

Also, Lemma 3.4 and (3.17) gives us

$$\begin{array}{cc}\hfill {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}\left(\Psi u(t)-U(t)\varphi \right)\parallel & \le {\int }_0^t\left({|\kappa |\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t-\tau ){({\partial }_{x}u)}^2\parallel +\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t-\tau )f\parallel \right)d\tau \hfill \\ \hfill & \le G_T{\int }_0^t{(|\kappa |\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{({\partial }_{x}u)}^2\parallel +\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel )d\tau \hfill \\ \hfill & \le G_T{\int }_0^t\left[|\kappa |\left({\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel +\parallel u\parallel }_{\sigma }+{\parallel u\parallel }_1^2\right)+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right]d\tau \hfill \\ \hfill & \le G_T\left(|\kappa |{\int }_0^t\Theta (\tau )d\tau +T\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right).\hfill \end{array}$$

(3.23)

Next, we need an auxiliary estimate. From integral Equation (3.3), Proposition 3.5 and the definitions above we get

$$\begin{array}{cc}\hfill {|\kappa |}^{-1}{\parallel u-v\parallel }_{\sigma }& \le {\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime }){\parallel u-v\parallel }_{H^1}{\parallel u+v\parallel }_{H^1}d{t}^{\prime }\hfill \\ \hfill & \le 2d(u,v)\underset{{\rm Y}(\tau )}{\underbrace{{\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime })(1+{\Lambda }_{t^{\prime }})h_{\varphi }d{t}^{\prime }}}.\hfill \end{array}$$

(3.24)

In what follows, from Lemma 2.3 and since that $H^{{\displaystyle \frac{\sigma }{2}}-1}$ is a Banach algebra it follows that for all $u,v\in {\chi }_T$

$$\begin{array}{cc}\hfill {|\kappa |}^{-1}\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}(\Psi u(t)-\Psi v(t))\parallel \le & {\int }_0^t\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t-\tau )({(u_x)}^2-{(v_x)}^2)\parallel d\tau \hfill \\ \hfill \le & {\int }_0^{t}G(t-\tau )\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\left({(u_x)}^2-{(v_x)}^2\right)\parallel d\tau \hfill \\ \hfill \le & G_T{\int }_0^t\parallel {\langle x\rangle }^{1/4}{\partial }_x(u-v){\langle x\rangle }^{1/4}{\partial }_x(u+v)\parallel d\tau \hfill \\ \hfill \le & G_T{\int }_0^t\parallel J^{{\displaystyle \frac{\sigma }{2}}-1}\left({\langle x\rangle }^{1/4}{\partial }_x(u-v){\langle x\rangle }^{1/4}{\partial }_x(u+v)\right)\parallel d\tau \hfill \\ \hfill \le & G_T{\int }_0^t\parallel J^{{\displaystyle \frac{\sigma }{2}}-1}\left({\langle x\rangle }^{1/4}{\partial }_x(u-v)\right)\parallel \parallel J^{{\displaystyle \frac{\sigma }{2}}-1}\left({\langle x\rangle }^{1/4}{\partial }_x(u+v)\right)\parallel \hfill \\ \hfill \le & G_T{\int }_0^t\left({\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{(u-v)\parallel +\parallel u-v\parallel }_{\sigma }+{\parallel u-v\parallel }_1^2\right)\times \hfill \\ \hfill & \times \left({\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{(u+v)\parallel +\parallel u+v\parallel }_{\sigma }+{\parallel u+v\parallel }_1^2\right)d\tau \hfill \\ \hfill \le & 8G_{T}d(u,v){\int }_0^t\left(1+2|\kappa |{\rm Y}(\tau )+2(1+{\Lambda }_{\tau })h_{\varphi }\right)\Theta (\tau )d\tau ,\hfill \end{array}$$

(3.25)

where above we used (3.22), (3.18) and (3.24). Therefore, from (3.19)–(3.25) we conclude that there exists $0\le T^{\prime }\le T$ such that $\Psi :{\chi }_{T^{\prime }}\to {\chi }_{T^{\prime }}$ is a contraction. Consequently, the existence of a solution u follows by the Banach fixed point Theorem. This solution is unique, since we have uniqueness in $H^1$. The continuous dependence follows by an application of Gronwall’s Lemma.

This finishes the proof. □

Proof of Theorem 1.1 

Assume that $1\le s<{\displaystyle \frac{7}{2}}$. The case $s>{\displaystyle \frac{7}{2}}$ run as in [10,15]. Let be u the solution of (1.1) defined on the interval $[0,T]$, as given by Theorem 3.7.

First, we deal with the Case a). We need the following auxiliary estimate. Let $w:={\partial }_{x}u$. Then, from Equation (1.1), it follows that

$${\partial }_{t}w-\lambda {\partial }_x^{2}w+\beta {\partial }_x^{4}w+\gamma w-\mu {(1-{\partial }_x^2)}^{-1/2}w+2\kappa w{\partial }_{x}w={\partial }_{x}f.$$

(3.26)

Integrating by parts and using Lemma 2.4 (with $A=\parallel w\parallel$ and $\alpha =1$) we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel w(t)\parallel }^2& =(w,\lambda {\partial }_x^{2}w-\beta {\partial }_x^{4}w-\gamma w+\mu {(1-{\partial }_x^2)}^{-1/2}w-2\kappa w{\partial }_{x}w+{\partial }_{x}f)\hfill \\ \hfill & \le |\lambda |\parallel {\partial }_x{w\parallel }^2-\beta \parallel {\partial }_x^2{w\parallel }^2-{\gamma \parallel w\parallel }^2+{\mu \parallel w\parallel }^2+\parallel w\parallel \parallel {\partial }_{x}f\parallel \hfill \\ \hfill & \le \left({\displaystyle \frac{{\lambda }^2}{\beta }}+\mu +1\right){\parallel w\parallel }^2+{\parallel {\partial }_{x}f\parallel }^2,\hfill \end{array}$$

(3.27)

where it is also used that

$$(w,w{\partial }_{x}w)=0.$$

(3.28)

Hence, by Gronwall’s Lemma

$$\parallel {\partial }_x{u\parallel }^2\lesssim \left(\parallel {\partial }_x{\varphi \parallel }^2+{\parallel {\partial }_{x}f\parallel }^2\right)e^{({\displaystyle \frac{{\lambda }^2}{\beta }}+\mu +1)t},t\in [0,T].$$

(3.29)

Now, using (3.3) and Proposition 3.5 we conclude that

$$\begin{array}{cc}\hfill {\parallel u(t)\parallel }_s\le & {\parallel U(t)\varphi \parallel }_s+{\int }_0^t{\parallel U(t-\tau )\left(\kappa {({\partial }_{x}u)}^2+f\right)\parallel }_{s}d\tau \hfill \\ \hfill \le & {\parallel \varphi \parallel }_s+|\kappa |{\int }_0^t\parallel U(t-\tau ){({\partial }_{x}u)}^2\parallel d\tau +{\int }_0^t{\parallel U(t-\tau )f\parallel }_{s}d\tau \hfill \\ \hfill \le & {\parallel \varphi \parallel }_s+|\kappa |{\int }_0^t{L}_s(t-\tau )\parallel {({\partial }_{x}u)}^2{\parallel }_{L^1}d\tau +{\int }_0^t{L}_s(t-\tau ){\parallel f\parallel }_{L^1}d\tau \hfill \\ \hfill \le & {\parallel \varphi \parallel }_s+|\kappa |{\int }_0^t{L}_s(t-\tau )\parallel {\partial }_x{u\parallel }^{2}d\tau +\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel {\int }_0^t{L}_s(t-\tau )d\tau .\hfill \end{array}$$

(3.30)

Inequalities (3.29) and (3.30) yields us, for all $t\in [0,T]$

$${\parallel u(t)\parallel }_s\le \underset{C_s(\varphi ,f,T)}{\underbrace{{\parallel \varphi \parallel }_s+\left({\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel +|\kappa |\left(\parallel {\partial }_x{\varphi \parallel }^2+{\parallel {\partial }_{x}f\parallel }^2\right)e^{({\displaystyle \frac{{\lambda }^2}{\beta }}+\mu +1)T}\right){\int }_0^T{L}_s(t-\tau )d\tau }}.$$

(3.31)

On the other hand, let $\sigma$ be as in (3.21).

Then, from (3.14) and setting $C_1^2=C_1^2(\varphi ,f,T)$ it follows that

$$\begin{array}{cc}\hfill {\parallel u(\tau )\parallel }_{\sigma }\le & L_{\sigma }(\tau )\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +\left(|\kappa |\underset{[0,T]}{sup}{\parallel u\parallel }_1^2+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right){\int }_0^{\tau }{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime }\hfill \\ \hfill \le & L_{\sigma }(\tau )\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +\left(|\kappa |C_1^2+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right){\int }_0^T{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime },\tau \in [0,t].\hfill \end{array}$$

(3.32)

Also, in a similar way to (3.23) and using (3.32)

$$\begin{array}{cc}\hfill {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}u(t)\parallel \le & {\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t)\varphi \parallel +{\int }_0^t\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}U(t-\tau )\left(\kappa {({\partial }_{x}u)}^2+f\right)\parallel d\tau \hfill \\ \hfill \le & G_T\left[{\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +{\int }_0^t\left({|\kappa |\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{({\partial }_{x}u)}^2\parallel +\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right)d\tau \right]\hfill \\ \hfill \le & G_T\left[{\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel +{\int }_0^t\left[|\kappa |\left({\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel +\parallel u\parallel }_{\sigma }+{\parallel u\parallel }_1^2\right)+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right]d\tau \right]\hfill \\ \hfill \le & G_T{[\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{\varphi \parallel +|\kappa |\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}\varphi \parallel {\int }_0^T{L}_{\sigma }(\tau )d\tau +|\kappa |T\left(|\kappa |C_1^2+\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel \right){\int }_0^T{L}_{\sigma }(\tau -t^{\prime })d{t}^{\prime }\hfill \\ \hfill & +|\kappa |T{C}_1^2{+T\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}f\parallel +|\kappa |{\int }_0^t\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}u(\tau )\parallel d\tau ].\hfill \end{array}$$

(3.33)

As a consequence, the Gronwall’s Lemma and (3.33) imply that

$${\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}u(t)\parallel \le {\mathcal{C}}_1(\varphi ,f,T),t\in [0,T],$$

(3.34)

where ${\mathcal{C}}_1$ is a constant.

Therefore, the inequalities (3.31) and (3.34), combined with standard extension arguments, demonstrate the existence of a unique global solution in $Z_{s,{\displaystyle \frac{1}{2}}}$.

Now, we proceed to the proof of Case b). Here, the global well-posedness in $H^s(\mathbb{R})$ for $s\ge 1$ follows similarly to the arguments presented in [10,15]. Then, we omit the details of this proof. To conclude this case, it remains to prove that for an initial data $\varphi \in Z_{s,{\displaystyle \frac{1}{2}}}$, the corresponding solution $u\in C([0,T];H^s)$, defined for any $T>0$, satisfies $u\in C([0,T];L^2(\langle x\rangle dx))$. By multiplying Equation (1.1) (with $\beta =\mu =0$) by ${\langle x\rangle }_{N}u$ and integrating over the real line, we obtain

$${\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel }^2-\lambda \int {\langle x\rangle }_{N}u{\partial }_x^{2}u+\gamma \int {\langle x\rangle }_N{u}^2-\kappa \int {\langle x\rangle }_{N}u{({\partial }_{x}u)}^2=\int {\langle x\rangle }_{N}uf,x\in \mathbb{R},t\ge 0,$$

(3.35)

where we recall that ${\langle x\rangle }_N$ is given in (2.5).

All terms in (3.35) are estimated in the proof of Proposition 4.2 (see below for details). Thus, (4.25) and (4.26) imply

$$\int {\langle x\rangle }_{N}u{({\partial }_{x}u)}^2\lesssim {\displaystyle \frac{1}{6}}{\parallel u\parallel }_{\infty }{\parallel u\parallel }^2\lesssim C_1^3(\varphi ,f,T)$$

(3.36)

and

$$\int {\langle x\rangle }_{N}u{\partial }_x^{2}u\lesssim {\displaystyle \frac{1}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{1}{6}}{\parallel u\parallel }_{\infty }{\parallel u\parallel }^2\lesssim C_1^2(\varphi ,f,T)+C_1^3(\varphi ,f,T),$$

(3.37)

where above we also used Sobolev embedding and (3.31).

Additionally, by Hölder’s inequality, we get

$$\int {\langle x\rangle }_N{u}^2{=\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^{2}and\int {\langle x\rangle }_N{uf\le \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}f\parallel }^2.$$

(3.38)

Therefore, (3.35)–(3.38) yields us

$${\displaystyle \frac{d}{dt}}{\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2\lesssim C_1^2(\varphi ,f,T)+(1+|\kappa |)C_1^3{(\varphi ,f,T)+\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{f\parallel }^2+(\gamma +1){\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel }^2.$$

The last inequality and Gronwall’s Lemma give us

$${\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel \le \tilde{C}(\varphi ,f,T),t\in [0,T].$$

(3.39)

Setting $N\to \infty$ in (3.39), we conclude that $u\in L^{\infty }([0,T];L^2(\langle x\rangle dx))$, for all $T>0$. The continuity of $t\in [0,T]\mapsto u(t)\in L^2(\langle x\rangle dx)$ and the continuous dependence on the initial data follow from similar ideas present in [11,13]. This concludes the proof of Cases a) and b).

This ends the proof. □

Remark 3.8. 

The ideas presented in Theorems 3.7 and 1.1 can be applied to deduce global well-posedness for the IVP (1.1) with other weights. Indeed, it is possible to prove that Theorem 1.1 holds in $Z_{s,r}$, where $s\ge 1$ and $r\ge 0$. Further details on well-posedness in weighted spaces can be found in [6,7,8,12,14].

The following result enables us to conclude the regularity of solutions outside the origin.

Proposition 3.9. 

Let $\varphi \in H^s$, where $s\ge 1$. If u is a solution of (1.1) for Cases a) or b), in Theorem 1.1, then

$$u\in C((0,\infty );H^{\infty }(\mathbb{R})).$$

Proof. 

It is a consequence of Lemmas 3.3–3.1, and a bootstrapping argument presented in [10,15]. □

4. Absorbing Sets

Here, we establish the existence of absorbing sets for the IVP (1.1).

Firstly, we show the existence of absorbing sets for Case a) in Theorem 1.2, i.e., $\beta ,\gamma >0$, $\mu \ge 0$, $\lambda \in \mathbb{R}$.

Lemma 4.1. 

Let $\gamma >{\displaystyle \frac{{\lambda }^2}{\beta }}+\beta +\mu$ and ${\partial }_x\varphi ,{\partial }_{x}f\in L^2$. If u be a global solution of IVP (1.1), then there exists $\varrho =\varrho (\lambda ,\beta ,\gamma ,\mu ,\parallel {\partial }_{x}f\parallel )$ such that

$$\parallel {\partial }_{x}u(t)\parallel \le \varrho ,t\ge \tilde{T}:={\displaystyle \frac{1}{{\theta }_1}}ln\left({\displaystyle \frac{\beta {\theta }_1{\parallel {\partial }_x\varphi \parallel }^2}{\parallel {\partial }_x{f\parallel }^2}}\right),$$

where ${\theta }_1:=\gamma -{\displaystyle \frac{{\lambda }^2}{\beta }}-\beta -\mu$.

Proof. 

First, observe that from (3.27) and (3.28) we conclude that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel w(t)\parallel }^2& \le \lambda \parallel w\parallel \parallel {\partial }_x^{2}w\parallel -\beta \parallel {\partial }_x^2{w\parallel }^2-{\gamma \parallel w\parallel }^2+{\mu \parallel w\parallel }^2+\parallel w\parallel \parallel {\partial }_{x}f\parallel .\hfill \end{array}$$

(4.1)

Moreover, Lemma 2.4 gives us

$$\lambda \parallel w\parallel \parallel {\partial }_x^{2}w\parallel -\beta \parallel {\partial }_x^2{w\parallel }^2\lesssim {\displaystyle \frac{{\lambda }^2}{\beta }}{\parallel w\parallel }^2.$$

(4.2)

Thus, by Young’s inequality

$$\parallel w\parallel \parallel {\partial }_x{f\parallel \le \beta \parallel w\parallel }^2+{\displaystyle \frac{\parallel {\partial }_x{f\parallel }^2}{\beta }}.$$

(4.3)

Hence, (4.1)–(4.3) imply that

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\parallel w\parallel }^2+\left(\gamma -{\displaystyle \frac{{\lambda }^2}{\beta }}-\beta -\mu \right){\parallel w\parallel }^2& \lesssim {\displaystyle \frac{\parallel {\partial }_x{f\parallel }^2}{\beta }}\hfill \end{array}$$

(4.4)

which, after integration from 0 until t, yields

$${\parallel w(t)\parallel }^2\lesssim {\parallel {\partial }_x\varphi \parallel }^2{e}^{-t{\theta }_1}+{\displaystyle \frac{(1-e^{-{\theta }_{1}t}){\parallel {\partial }_{x}f\parallel }^2}{{\theta }_1\beta }}.$$

Therefore, setting

$${\varrho }^2={\displaystyle \frac{2\parallel {\partial }_x{f\parallel }^2}{{\theta }_1\beta }},$$

(4.5)

we obtain the desired result.

This ends the proof.

□

Proposition 4.2. 

Let $s\ge 0$, $\gamma >{\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}+{\displaystyle \frac{2}{\beta }}{\lambda }^2+\beta +\mu$ and $f\in Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R})$. Also, assume that ${\partial }_x\varphi ,{\partial }_{x}f\in L^2$, then, there exists a constant ${\rho }_{s,{\displaystyle \frac{1}{2}}}={\rho }_s{(\lambda ,\beta ,\gamma ,\parallel f\parallel }_{s,{\displaystyle \frac{1}{2}}})$ such that for all $R>0$ there exists $T_{s,{\displaystyle \frac{1}{2}}}(R)$ satisfying

$${\parallel S(t)\varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}\le {\rho }_{s,{\displaystyle \frac{1}{2}}},\forall \varphi \in Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R}),{\parallel \varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}\le Randt\ge T_{s,{\displaystyle \frac{1}{2}}}(R),$$

(4.6)

where $S(t)\varphi \equiv u(t)$ is a global solution of IVP (1.1).

Proof. 

We observe that, although Theorem 1.1 guarantees the existence of solutions only for $s\ge 1$, however, we emphasize that the argument presented here is valid for all $s\ge 0$. Our proof will be divided into two cases.

Case 1): $s=0$. From (1.1) we obtain

$${\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u(t)\parallel }^2=(u,\lambda {\partial }_x^{2}u-\beta {\partial }_x^{4}u-\gamma u+\mu {(1-{\partial }_x^2)}^{-1/2}u+\kappa {({\partial }_{x}u)}^2+f).$$

(4.7)

Lemma 2.5 (with $d=1$, $s=2$, $r=0$) and the inequality ${\parallel u\parallel }_{L^{\infty }}\lesssim {\parallel \widehat{u}\parallel }_{L^1}$ imply that

$${\parallel u\parallel }_{L^{\infty }}\lesssim {\parallel u\parallel }^{{\displaystyle \frac{3}{4}}}{\parallel {\partial }_x^{2}u\parallel }^{{\displaystyle \frac{1}{4}}}.$$

(4.8)

Hence

$$\begin{array}{cc}\hfill \left(u,{({\partial }_{x}u)}^2\right)& \le {\parallel u\parallel }_{L^{\infty }}\parallel {\partial }_x{u\parallel }^2\le {\parallel u\parallel }^{3/4}\parallel {\partial }_x^2{u\parallel }^{1/4}{\parallel {\partial }_{x}u\parallel }^2.\hfill \end{array}$$

(4.9)

Thus, by Cauchy-Schwartz inequality and (4.7)–(4.9)

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\parallel u\parallel }^2\lesssim & {|\kappa |\parallel u\parallel }^{3/4}\parallel {\partial }_x{u\parallel }^2\parallel {\partial }_x^2{u\parallel }^{1/4}+|\lambda |\parallel u\parallel \parallel {\partial }_x^{2}u\parallel -\beta \parallel {\partial }_x^2{u\parallel }^2-\gamma {\parallel u\parallel }^2\hfill \\ \hfill & +{\mu \parallel u\parallel }^2+\parallel u\parallel \parallel f\parallel \hfill \\ \hfill \lesssim & {|\kappa |\parallel u\parallel }^{3/4}\parallel {\partial }_x{u\parallel }^2\parallel {\partial }_x^2{u\parallel }^{1/4}-{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }^2+|\lambda |\parallel u\parallel \parallel {\partial }_x^{2}u\parallel -{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }^2-\gamma {\parallel u\parallel }^2\hfill \\ \hfill & +{\mu \parallel u\parallel }^2+\parallel u\parallel \parallel f\parallel .\hfill \end{array}$$

(4.10)

To handle the inequality above, we first establish some necessary estimates.

From Lemma 2.4 (with $\alpha ={\displaystyle \frac{1}{4}}$ and $A={\displaystyle \frac{2|\kappa |}{\beta }}{\parallel u\parallel }^{3/4}{\parallel {\partial }_{x}u\parallel }^2$) and Young’s inequality we see that

$${|\kappa |\parallel u\parallel }^{3/4}\parallel {\partial }_x{u\parallel }^2\parallel {\partial }_x^2{u\parallel }^{1/4}-{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }^2\lesssim {\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}{(\parallel u\parallel }^2+\parallel {\partial }_{x}u{\parallel }^4)$$

(4.11)

and

$$\lambda \parallel u\parallel \parallel {\partial }_x^{2}u\parallel -{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }^2\lesssim {\displaystyle \frac{2}{\beta }}{\lambda }^2{\parallel u\parallel }^2.$$

(4.12)

Also, again by Young’s inequality

$$\parallel u\parallel \parallel f\parallel \le {\beta \parallel u\parallel }^2+{\displaystyle \frac{{\parallel f\parallel }^2}{\beta }}.$$

(4.13)

Thus, (4.7)–(4.13) imply

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{\parallel u\parallel }^2+\underset{{\theta }_2}{\underbrace{\left(\gamma -{\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}-{\displaystyle \frac{2}{\beta }}{\lambda }^2-\beta -\mu \right)}}{\parallel u\parallel }^2\lesssim & {\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}{\parallel {\partial }_{x}u\parallel }^4+{\displaystyle \frac{{\parallel f\parallel }^2}{\beta }}.\hfill \end{array}$$

(4.14)

Therefore, by put

$${\rho }_0^2:={\displaystyle \frac{2}{{\theta }_2}}\left[{\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}{\varrho }^4+{\parallel f\parallel }^2\right],$$

(4.15)

and from similar way to Lemma 4.1 we see that

$$\parallel u(t)\parallel \lesssim {\rho }_0,fort\ge T_0:=max\left\{{\displaystyle \frac{1}{{\theta }_2}}ln{\displaystyle \frac{{\parallel \varphi \parallel }^2}{{\rho }_0^2}},\tilde{T}\right\},$$

where $\tilde{T}$ is given in Lemma 4.1.

Case 2): $s>0$. First, we need several auxiliary estimates. Using Lemma 2.5, once again, we obtain

$$\parallel {\partial }_x{u\parallel }_{\infty }\lesssim \parallel {\partial }_x{u\parallel }^{1/2}\parallel {\partial }_x^2{u\parallel }^{1/2}\lesssim \parallel {\partial }_x{u\parallel }^{1/2}{\parallel u\parallel }_2^{1/2}.$$

(4.16)

In the following, interpolation in Sobolev spaces allows us to conclude that

$${\parallel u\parallel }_s\le {\parallel u\parallel }^{{\displaystyle \frac{2}{s+2}}}{\parallel u\parallel }_{s+2}^{{\displaystyle \frac{s}{s+2}}}{,\parallel u\parallel }_{s+1}\le {\parallel u\parallel }^{{\displaystyle \frac{1}{s+2}}}{\parallel u\parallel }_{s+2}^{{\displaystyle \frac{s+1}{s+2}}}$$

(4.17)

and

$${\parallel u\parallel }_2^{1/2}\le {\parallel u\parallel }^{{\displaystyle \frac{s}{2(s+2)}}}{\parallel u\parallel }_{s+2}^{{\displaystyle \frac{1}{s+2}}}.$$

(4.18)

Then, from Kato-Ponce commutador (see [20]) and (4.16)–(4.18)

$$\begin{array}{cc}\hfill |{\left(u,{({\partial }_{x}u)}^2\right)}_s|& \lesssim {\parallel u\parallel }_s{\parallel {({\partial }_{x}u)}^2\parallel }_s\hfill \\ \hfill & \lesssim {\parallel u\parallel }_s\parallel {\partial }_x{u\parallel }_{\infty }{\parallel {\partial }_{x}u\parallel }_s\hfill \\ \hfill & \lesssim {\parallel u\parallel }_s\parallel {\partial }_x{u\parallel }^{1/2}{\parallel u\parallel }_2^{1/2}{\parallel u\parallel }_{s+1}\hfill \\ \hfill & \lesssim \left({\parallel u\parallel }^{{\displaystyle \frac{3s+8}{2(s+2)}}}+{\parallel {\partial }_x^{2}u\parallel }_s^{{\displaystyle \frac{s+1}{s+2}}}\right){\parallel {\partial }_{x}u\parallel }^{1/2}.\hfill \end{array}$$

(4.19)

On the other hand, an application of Lemma 2.4 (with $A={\displaystyle \frac{2|\kappa |}{\beta }}{\parallel {\partial }_{x}u\parallel }^{1/2}$) yields us

$$|\kappa |\parallel {\partial }_x^2{u\parallel }_s^{{\displaystyle \frac{s+1}{s+2}}}\parallel {\partial }_x{u\parallel }^{1/2}-{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }_s^2\lesssim {\left({\displaystyle \frac{2|\kappa |}{\beta }}\right)}^{{\displaystyle \frac{2(s+1)}{s+3}}}{\parallel {\partial }_{x}u\parallel }^{{\displaystyle \frac{s+2}{s+3}}}$$

(4.20)

and

$${\parallel u\parallel }_s\parallel {\partial }_x^2{u\parallel }_s-{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }_s^2\lesssim {\displaystyle \frac{2}{\beta }}{\lambda }^2{\parallel u\parallel }_s^2.$$

(4.21)

Therefore, by using (4.18)–(4.21) and Young’s inequality, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u(t)\parallel }_s^2=& {\left(u,\lambda {\partial }_x^{2}u-\beta {\partial }_x^{4}u-\gamma u+\mu {(1-{\partial }_x^2)}^{-1/2}u-\kappa {({\partial }_{x}u)}^2+f\right)}_s\hfill \\ \hfill \lesssim & {\lambda \parallel u\parallel }_s\parallel {\partial }_x^2{u\parallel }_s-\beta \parallel {\partial }_x^2{u\parallel }_s^2-{\gamma \parallel u\parallel }_s^2+{\mu \parallel u\parallel }_s^2+|\kappa {(u,{({\partial }_{x}u)}^2)}_s|+|{(u,f)}_s|\hfill \\ \hfill \lesssim & {\lambda \parallel u\parallel }_s\parallel {\partial }_x^2{u\parallel }_s-\beta \parallel {\partial }_x^2{u\parallel }_s^2-{\gamma \parallel u\parallel }_s^2+{\mu \parallel u\parallel }_s^2+\left({\parallel u\parallel }^{{\displaystyle \frac{3s+8}{2(s+2)}}}+{\parallel {\partial }_x^{2}u\parallel }_s^{{\displaystyle \frac{s+1}{s+2}}}\right){\parallel {\partial }_{x}u\parallel }^{1/2}\hfill \\ \hfill & +{\parallel u\parallel }_s{\parallel f\parallel }_s\hfill \\ \hfill \lesssim & {\lambda \parallel u\parallel }_s\parallel {\partial }_x^2{u\parallel }_s-{\displaystyle \frac{\beta }{2}}\parallel {\partial }_x^2{u\parallel }_s^2-{\gamma \parallel u\parallel }_s^2+{\mu \parallel u\parallel }_s^2+\parallel {\partial }_x^2{u\parallel }_s^{{\displaystyle \frac{s+1}{s+2}}}\parallel {\partial }_x{u\parallel }^{1/2}-{\displaystyle \frac{\beta }{2}}{\parallel {\partial }_x^{2}u\parallel }_s^2\hfill \\ \hfill & +{\parallel u\parallel }^{{\displaystyle \frac{3s+8}{2(s+2)}}}\parallel {\partial }_x{u\parallel }^{1/2}+\beta {\parallel u\parallel }_s^2+{\displaystyle \frac{{\parallel f\parallel }_s^2}{\beta }}\hfill \\ \hfill \lesssim & {\displaystyle \frac{2}{\beta }}{\lambda }^2{\parallel u\parallel }_s^2-{\gamma \parallel u\parallel }_s^2+{\mu \parallel u\parallel }_s^2+{\left({\displaystyle \frac{2|\kappa |}{\beta }}\right)}^{{\displaystyle \frac{2(s+1)}{s+3}}}\parallel {\partial }_x{u\parallel }^{{\displaystyle \frac{s+2}{s+3}}}+{|\kappa |\parallel u\parallel }^{{\displaystyle \frac{3s+8}{2(s+2)}}}{\parallel {\partial }_{x}u\parallel }^{1/2}\hfill \\ \hfill & +{\beta \parallel u\parallel }_s^2+{\displaystyle \frac{{\parallel f\parallel }_s^2}{\beta }}\hfill \end{array}$$

hence,

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u(t)\parallel }_s^2+\underset{{\theta }_3}{\underbrace{\left(\gamma -{\displaystyle \frac{2}{\beta }}{\lambda }^2-\beta -\mu \right)}}{\parallel u\parallel }_s^2\lesssim & {\left({\displaystyle \frac{2|\kappa |}{\beta }}\right)}^{{\displaystyle \frac{2(s+1)}{s+3}}}\parallel {\partial }_x{u\parallel }^{{\displaystyle \frac{s+2}{s+3}}}+{|\kappa |\parallel u\parallel }^{{\displaystyle \frac{3s+8}{2(s+2)}}}{\parallel {\partial }_{x}u\parallel }^{1/2}+{\displaystyle \frac{{\parallel f\parallel }_s^2}{\beta }}.\hfill \end{array}$$

Thus, recalling that $\varrho$ is is defined in (4.5), we choose

$$\begin{array}{c}\hfill {\rho }_s^2:={\displaystyle \frac{2}{{\theta }_3}}\left[{\left({\displaystyle \frac{2|\kappa |}{\beta }}\right)}^{{\displaystyle \frac{2(s+1)}{s+3}}}{\varrho }^{{\displaystyle \frac{s+2}{2(s+3)}}}+|\kappa |{\rho }_0^{{\displaystyle \frac{3s+8}{2(s+2)}}}{\varrho }^{{\displaystyle \frac{1}{4}}}+{\displaystyle \frac{{\parallel f\parallel }_s^2}{\beta }}\right]\end{array}$$

and

$$T_s(R):=max\left\{{\displaystyle \frac{1}{{\theta }_3}}ln{\displaystyle \frac{R}{{\rho }_s^2}},T_0\right\}.$$

Therefore, from Case 1) and proceeding analogously to Lemma 4.1, we obtain

$${\parallel S(t)\varphi \parallel }_s\le {\rho }_s,\forall \varphi \in H^s(\mathbb{R}),{\parallel \varphi \parallel }_s\le Randt\ge T_s(R).$$

(4.22)

Next, we deal with estimates necessary to show the existence of absorbing sets in the weighted Sobolev space $Z_{s,{\displaystyle \frac{1}{2}}}$.

Multiplying (1.1) by ${\langle x\rangle }_{N}u$ and integrating over the real line, we observe that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel }^2-\lambda \int {\langle x\rangle }_{N}u{\partial }_x^{2}u& +\beta \int {\langle x\rangle }_{N}u{\partial }_x^{4}u+\gamma \int {\langle x\rangle }_N{u}^2-\mu \int {\langle x\rangle }_{N}u{(1-{\partial }_x^2)}^{-1/2}u\hfill \\ \hfill & -\kappa \int {\langle x\rangle }_{N}u{({\partial }_{x}u)}^2=\int {\langle x\rangle }_{N}uf.\hfill \end{array}$$

(4.23)

Using integrating by parts we conclude

$$\begin{array}{cc}\hfill \int {\langle x\rangle }_{N}u{\partial }_x^{4}u& ={\displaystyle \frac{1}{2}}\int {\partial }_x^4{\langle x\rangle }_N{u}^2-\int \left({\displaystyle \frac{3}{2}}{\partial }_x^2{\langle x\rangle }_N+{\displaystyle \frac{1}{2}}{\partial }_x^3{\langle x\rangle }_N\right){({\partial }_{x}u)}^2+\int {\langle x\rangle }_N{({\partial }_x^{2}u)}^2\hfill \\ \hfill & \lesssim {\parallel u\parallel }^2+{\parallel {\partial }_{x}u\parallel }^2+\int {\langle x\rangle }_N{({\partial }_x^{2}u)}^2\hfill \end{array}$$

(4.24)

$$\int {\langle x\rangle }_{N}u{({\partial }_{x}u)}^2={\displaystyle \frac{1}{6}}\int {\partial }_x^2{\langle x\rangle }_N{u}^3\lesssim {\parallel u\parallel }_{\infty }{\parallel u\parallel }^2$$

(4.25)

and

$$\int {\langle x\rangle }_{N}u{\partial }_x^{2}u={\displaystyle \frac{1}{2}}\int {\partial }_x^2{\langle x\rangle }_N{u}^2-{\displaystyle \frac{1}{6}}\int {\partial }_x^2{\langle x\rangle }_N{u}^3\lesssim {\parallel u\parallel }^2{(1+\parallel u\parallel }_{\infty }).$$

(4.26)

In addition, using Sobolev embedding it is seen that

$${\parallel u\parallel }_{\infty }\lesssim \parallel u\parallel +\parallel {\partial }_{x}u\parallel .$$

(4.27)

An application of the Young’s inequality with $ϵ:={\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}+{\displaystyle \frac{2}{\beta }}{\lambda }^2+\beta +\mu$ gives us

$$\begin{array}{cc}\hfill \int {\langle x\rangle }_{N}uf& {\le \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{f\parallel \lesssim ϵ\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\displaystyle \frac{{\parallel \langle x\rangle }_N^{\frac{1}{2}}{f\parallel }^2}{ϵ}}.\hfill \end{array}$$

(4.28)

Also, Plancherel identity, inequalities (2.1), (2.2) and (2.3) (with $h={\langle \xi \rangle }^{-1}$) yields

$$\begin{array}{cc}\hfill \int {\langle x\rangle }_{N}u{(1-{\partial }_x^2)}^{-1/2}u& {\le \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{(1-{\partial }_x^2)}^{-1/2}u\parallel \hfill \\ \hfill & {\lesssim \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\parallel D_{\xi }^{{\displaystyle \frac{1}{2}}}\left({\langle \xi \rangle }^{-1}\widehat{u}\right)\parallel }^2\hfill \\ \hfill & {\lesssim \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2{+\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\parallel u\parallel }^2.\hfill \end{array}$$

(4.29)

Thus, the last inequality and (4.23)–(4.29) imply

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+\gamma {\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel }^2\lesssim & \left[\lambda +(\parallel u\parallel +\parallel {\partial }_{x}u\parallel )(\lambda +|\kappa |)+\mu +\beta \right]{\parallel u\parallel }^2\hfill \\ \hfill & +\beta \parallel {\partial }_x{u\parallel }^2{+\mu \parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+\mu {\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}u\parallel }^2\hfill \\ \hfill & {+ϵ\parallel \langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\displaystyle \frac{{\parallel \langle x\rangle }_N^{\frac{1}{2}}{f\parallel }^2}{ϵ}}.\hfill \end{array}$$

(4.30)

To conclude our result, we need to exchange the order of limit, when $N\to \infty$, with the derivative, with respect to t, on the left-hand side above. This is justified because, in view of (4.23), the convergence of ${\displaystyle \frac{d}{dt}}{\parallel {\langle x\rangle }_N^{{\displaystyle \frac{1}{2}}}u\parallel }^2$ is uniform in $[0,T]$, for all $T>0$. Hence, by setting $N\to \infty$ in (4.30) it follows that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+\underset{{\theta }_4}{\underbrace{(\gamma -ϵ-2\mu )}}{\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}u\parallel }^2\lesssim & \left[\lambda +(\parallel u\parallel +\parallel {\partial }_{x}u\parallel )(\lambda +|\kappa |)+\mu +\beta \right]{\parallel u\parallel }^2\hfill \\ \hfill & +\beta \parallel {\partial }_x{u\parallel }^2+{\displaystyle \frac{{\parallel \langle x\rangle }^{\frac{1}{2}}{f\parallel }^2}{ϵ}}.\hfill \end{array}$$

(4.31)

Observe that from our hypotheses, ${\theta }_4>0$. Thus, using (4.31), putting

$${\rho }_{s,{\displaystyle \frac{1}{2}}}^2:=max\left\{{\displaystyle \frac{2}{{\theta }_4}}\left[\left(\lambda +({\rho }_0+\varrho )(\lambda +|\kappa |)+\mu +\beta \right){\rho }_0^2+\beta \varrho +{\displaystyle \frac{{\parallel \langle x\rangle }^{\frac{1}{2}}{f\parallel }^2}{ϵ}}\right],{\rho }_s^2,{\rho }_0^2\right\}$$

(4.32)

and

$$T_{s,{\displaystyle \frac{1}{2}}}(R):=max\left\{{\displaystyle \frac{1}{{\theta }_4}}ln{\displaystyle \frac{R}{{\rho }_{s,\frac{1}{2}}^2}},T_s(R)\right\}$$

we obtain (4.6).

This ends the proof. □

In the following, we show the existence of absorbing sets for Case b) in Theorem 1.2, i.e., $\beta =\mu =0$ and $\lambda >0$. Since the following statements are slightly analogous to those in Case a), we will highlight only the passages that present new information compared to what was previously stated.

Lemma 4.3. 

Let $\gamma >\lambda$ and ${\partial }_x\varphi ,{\partial }_{x}f\in L^2$. If u be a global solution of IVP (1.1), then there exist $\varrho =(\gamma ,\lambda ,\parallel {\partial }_{x}f\parallel )$ such that

$$\parallel {\partial }_x{u(t)\parallel }^2\le \tilde{\varrho }:={\displaystyle \frac{2\parallel {\partial }_x{f\parallel }^2}{\tilde{\theta }\lambda }},t\ge \tilde{\tilde{T}}:={\displaystyle \frac{1}{\tilde{\theta }}}ln\left({\displaystyle \frac{\lambda \tilde{\theta }{\parallel {\partial }_x\varphi \parallel }^2}{\parallel {\partial }_x{f\parallel }^2}}\right),$$

where $\tilde{\theta }=\gamma -\lambda$.

Proof. 

Let $w={\partial }_{x}u$, then taking the derivative in Equation (1.1) it follows that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel w\parallel }^2& =(w,\lambda {\partial }_x^{2}w-\gamma w-2\kappa w{\partial }_{x}w+{\partial }_{x}f).\hfill \end{array}$$

(4.33)

On the other hand, using Young’s inequality

$$\int w{\partial }_{x}f\le \lambda {\parallel w\parallel }^2+{\displaystyle \frac{\parallel {\partial }_x{f\parallel }^2}{\lambda }}.$$

(4.34)

Therefore, by the inequalities above and (3.28)

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel w\parallel }^2+(\gamma -\lambda ){\parallel w\parallel }^2\le {\displaystyle \frac{\parallel {\partial }_x{f\parallel }^2}{\lambda }}.\end{array}$$

(4.35)

Hence, similarly to Lemma 4.1 we obtain the desired result.

This ends the proof. □

Proposition 4.4. 

Let $s=0$ or $s=1$, $\gamma >\lambda +|\kappa |$ and $f\in Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R})$. Then, there exists a constant ${\tilde{\rho }}_{s,{\displaystyle \frac{1}{2}}}={\tilde{\rho }}_s{(\lambda ,\beta ,\gamma ,\parallel f\parallel }_{s,{\displaystyle \frac{1}{2}}})$ such that for all $R>0$ there exists ${\tilde{T}}_{s,{\displaystyle \frac{1}{2}}}(R)$ satisfying

$${\parallel S(t)\varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}\le {\tilde{\rho }}_{s,{\displaystyle \frac{1}{2}}},\forall \varphi \in Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R}),{\parallel \varphi \parallel }_{s,{\displaystyle \frac{1}{2}}}\le Randt\ge {\tilde{T}}_{s,{\displaystyle \frac{1}{2}}}(R),$$

(4.36)

where $S(t)\varphi \equiv u(t)$ is a global solution of IVP (1.1).

Proof. 

First, we divide our proof in two cases.

Case 1): $s=0$. From (1.1) we obtain

$${\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u(t)\parallel }^2=(u,\lambda {\partial }_x^{2}u-\gamma u-\kappa {({\partial }_{x}u)}^2+f).$$

(4.37)

Hence, using Sobolev embedding

$$\begin{array}{cc}\hfill \left(u,{({\partial }_{x}u)}^2\right)& \le {\parallel u\parallel }_{L^{\infty }}{\parallel {\partial }_{x}u\parallel }^2\hfill \\ \hfill & \lesssim {\parallel u\parallel }_1{\parallel {\partial }_{x}u\parallel }^2\hfill \\ \hfill & \lesssim (\parallel u\parallel +\parallel {\partial }_{x}u\parallel )\parallel {\partial }_x{u\parallel }^2\hfill \\ \hfill & \lesssim {\parallel u\parallel }^2+\parallel {\partial }_x{u\parallel }^4+{\parallel {\partial }_{x}u\parallel }^3.\hfill \end{array}$$

(4.38)

Also,

$$(u,f)\lesssim {\lambda \parallel u\parallel }^2+{\displaystyle \frac{{\parallel f\parallel }^2}{\lambda }}.$$

(4.39)

Then by (4.37)–(4.39)

$${\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u(t)\parallel }^2+\left(\gamma -\lambda -|\kappa |\right){\parallel u\parallel }^2\lesssim |\kappa |\parallel {\partial }_x{u\parallel }^3+|\kappa |\parallel {\partial }_x{u\parallel }^4+{\displaystyle \frac{{\parallel f\parallel }^2}{\lambda }}.$$

(4.40)

The rest, run as in the proof of Proposition 4.2 (Case 1)).

Case 2): $s=1$.

In view of the inequality ${\parallel u\parallel }_1\lesssim \parallel u\parallel +\parallel {\partial }_{x}u\parallel$, the proof, in this case, follows from Proposition 4.3 and Case 1), above.

The estimates in the weighted Sobolev spaces $Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R})$ are similar to those in the proof of Proposition 4.2.

This concludes the proof. □

Remark 4.5. 

The strategy adopted to establish the existence of absorbing sets for the IVP (1.1) in Case a) of Theorem 1.2 cannot be applied to Case b). Indeed, the presence of the fourth-order derivative in (1.1) is necessary to derive inequalities (4.19)–(4.21). Consequently, we cannot determine whether Theorem 1.2 (Case b)) holds for $s>1$.

5. Global Attractors

By the theory developed in [24], the $\omega$-limite set of $B_s:=\{v\in Z_{s,{\displaystyle \frac{1}{2}}}{(\mathbb{R}):\parallel v\parallel }_{s,{\displaystyle \frac{1}{2}}}\le {\rho }_{s,{\displaystyle \frac{1}{2}}}\}$ (where ${\rho }_{s,{\displaystyle \frac{1}{2}}}$ is given in the Proposition 4.2),

$$\mathcal{A}=\omega (B_s)=\bigcap _{r\ge 0}\overline{{\bigcup }_{t\ge r}S(t)B_s},$$

is a global attractor if the semigroup operators $S(t)$ are uniformly compact for t large.

In the following, we proceed to prove uniform compactness for the semigroup operators $S(t)$. Due to the non compactness of the injection $Z_{\tilde{s},{\displaystyle \frac{1}{2}}}(\mathbb{R})$ into $Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R})$, the proof of uniform compactness for $S(t)$ in the case $Z_{s,{\displaystyle \frac{1}{2}}}(\mathbb{R})$, $s\ge 1$, is an application of the procedure described in [1], using weighted estimate time-dependent, introducing a weight function. Thus, let

$$\psi (x,t)={(1+x^2)}^{1/2}\left[1-e^{{\displaystyle \frac{-t}{{(1+x^2)}^{1/2}}}}\right],$$

(5.1)

where $\psi$ satisfies

i)

$\psi \in C^{\infty }({\mathbb{R}}^2)$, $\psi (x,t)\ge 0$, for all $(x,t)\in \Omega =\mathbb{R}\times [0,\infty )$ and $\psi (x,0)=0$.

ii)

all derivatives of order $\ge 1$ of $\psi$ are bounded functions in $\Omega$.

iii)

$\psi (x,t)\to +\infty$ when $(|x|,t)\to \infty$.

By defining ${\phi }^2(x,t)=\psi (x,t)$, we obtain the following.

Proposition 5.1. 

Assume $s\ge 1$. Let also, $\gamma ,\beta ,\mu ,\lambda ,\kappa$ and f as in statement of Theorem 1.2. If u is solution of (1.1), then

$$\underset{t\to \infty }{lim\; sup}\int u{(x,t)}^2\psi (x,t)dx\le C,$$

where C does not depend on the initial data ϕ.

Proof. 

First, assume that the coefficients of Equation (1.1) satisfy the hypotheses of Case a) in Theorem 1.2.

Let $R>0$ such that ${\parallel \varphi \parallel }_{1,{\displaystyle \frac{1}{2}}}\le R$, then Proposition 4.2 implies that there exist $T_{1,{\displaystyle \frac{1}{2}}}(R)>0$ and ${\rho }_{1,{\displaystyle \frac{1}{2}}}$ such that

$${\parallel u(t)\parallel }_{1,{\displaystyle \frac{1}{2}}}\le {\rho }_{1,{\displaystyle \frac{1}{2}}},t\ge T_{1,{\displaystyle \frac{1}{2}}}(R).$$

(5.2)

Next, we put $\rho :={\rho }_{1,{\displaystyle \frac{1}{2}}}$, $T_1:=T_{1,{\displaystyle \frac{1}{2}}}(R)$ and $\phi :={\psi }^{{\displaystyle \frac{1}{2}}}$. Using the Mean value theorem, observe that $\phi$ is bounded, for $t>0$. Hence, multiplying (1.1) by $u{\phi }^2$ we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}\int {(u\phi )}^2-\lambda \int u{\phi }^2{\partial }_x^{2}u+\beta \int u{\phi }^2{\partial }_x^{4}u+\gamma \int {(u\phi )}^2=& -\mu \int u{\phi }^2{(1-{\partial }_x^2)}^{-1/2}u\hfill \\ \hfill & +\int u{\phi }^2{\displaystyle \frac{1}{2}}{({\partial }_{x}u)}^2+\int u{\phi }^{2}f.\hfill \end{array}$$

(5.3)

In our estimates below, we assume $t\ge T_1$.

On the other hand, integrating by parts in a way similar to (4.24)–(4.26) we obtain

$$\begin{array}{cc}\hfill \int {\phi }^{2}u{\partial }_x^{4}u& ={\displaystyle \frac{1}{2}}\int {\partial }_x^4{\phi }^2{u}^2-\int \left({\displaystyle \frac{3}{2}}{\partial }_x^2{\phi }^2+{\displaystyle \frac{1}{2}}{\partial }_x^3{\phi }^2\right){({\partial }_{x}u)}^2+\int {\phi }^2{({\partial }_x^{2}u)}^2\hfill \\ \hfill & \lesssim {\parallel u\parallel }^2+{\parallel {\partial }_{x}u\parallel }^2+\int {\phi }^2{({\partial }_x^{2}u)}^2,\hfill \end{array}$$

(5.4)

$$\int {\phi }^{2}u{({\partial }_{x}u)}^2={\displaystyle \frac{1}{6}}\int {\partial }_x^2{\phi }^2{u}^3\lesssim {\parallel u\parallel }_{\infty }{\parallel u\parallel }^2$$

(5.5)

and

$$\int {\phi }^{2}u{\partial }_x^{2}u={\displaystyle \frac{1}{2}}\int {\partial }_x^2{\phi }^2{u}^2-{\displaystyle \frac{1}{6}}\int {\partial }_x^2{\phi }^2{u}^3\lesssim {\parallel u\parallel }^2{(1+\parallel u\parallel }_{\infty }).$$

(5.6)

We also have by Hölder’s inequality, Plancherel identity and (2.3)

$$\begin{array}{cc}\hfill \int u{\phi }^2{(1-{\partial }_x^2)}^{-1/2}u& \le \parallel \phi u\parallel \parallel \phi {(1-{\partial }_x^2)}^{-1/2}u\parallel \hfill \\ \hfill & \le \parallel \phi u\parallel \parallel D_{\xi }^{{\displaystyle \frac{1}{2}}}\left({(1+{\xi }^2)}^{-{\displaystyle \frac{1}{2}}}\widehat{u}\right)\parallel \hfill \\ \hfill & \lesssim {\parallel \phi u\parallel }^2{+\parallel \langle x\rangle }^{{\displaystyle \frac{1}{2}}}{u\parallel }^2+{\parallel u\parallel }^2.\hfill \end{array}$$

(5.7)

Finally, Hölder’s inequality and Young’s inequality, with ${ϵ}_1:={\displaystyle \frac{{(2\kappa )}^{\frac{8}{7}}}{{\beta }^{\frac{1}{7}}}}+{\displaystyle \frac{2}{\beta }}{\lambda }^2+\beta$ gives us

$$\begin{array}{cc}\hfill \int {\phi }^{2}uf& \le \parallel \phi u\parallel \parallel \phi f\parallel \lesssim {ϵ}_1{\parallel \phi u\parallel }^2+{\displaystyle \frac{{\parallel \phi f\parallel }^2}{{ϵ}_1}}.\hfill \end{array}$$

(5.8)

Therefore, using the above inequalities and (4.27) we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel \phi u\parallel }^2+{\parallel \phi u\parallel }^2\underset{{\theta }_5}{\underbrace{(\gamma -\mu -{ϵ}_1)}}\lesssim & \left[|\lambda |+(\parallel u\parallel +\parallel {\partial }_{x}u\parallel )(|\lambda |+|\kappa |)+\mu +\beta \right]{\parallel u\parallel }^2\hfill \\ \hfill & +\beta \parallel {\partial }_x{u\parallel }^2+\mu {\parallel {\langle x\rangle }^{{\displaystyle \frac{1}{2}}}u\parallel }^2+{\displaystyle \frac{{\parallel \phi f\parallel }^2}{{ϵ}_1}}.\hfill \end{array}$$

(5.9)

Thus, using inequality (5.9) together with (4.5), (4.15) and (4.32), and setting

$$k=\left(|\lambda |+({\rho }_0+\varrho )(|\lambda |+|\kappa |)+\mu +\beta \right){\rho }_0^2+\beta {\varrho }^2+\mu {\rho }_{s,{\displaystyle \frac{1}{2}}}^2+{\displaystyle \frac{{\parallel \langle x\rangle }^{\frac{1}{2}}{f\parallel }^2}{{ϵ}_1}}$$

we obtain

$${\displaystyle \frac{d}{dt}}{\parallel \phi u\parallel }^2+{\theta }_5{\parallel \phi u\parallel }^2\lesssim kfor allt\ge T_1.$$

Mltiplying the last inequality by $e^{t{\theta }_5}$ and integrating between $T_1$ and t we obtain

$${\parallel \phi u\parallel }^2\le {\displaystyle \frac{k}{{\theta }_5}}\left(1-e^{-{\theta }_5(t-T_1)}\right)+e^{-{\theta }_5(t-T_1)}{\parallel \phi (T_1)u(T_1)\parallel }^2,for anyt\ge T_1.$$

(5.10)

Consequently by setting $C={\displaystyle \frac{k}{{\theta }_5}}$, inequality (5.10) gives the desired result.

The proof for the case where the coefficients of Equation (1.1) satisfy the hypotheses of Case b) in Theorem 1.2 follows a similar approach, and thus we omit the details.

This concludes the proof. □

Based on the results obtained in Section 4 and Section 5, we are now in a position to address the proof of Theorem 1.2.

Proof of Theorem 1.2 

We will prove both Cases a) and b). By the Propositions 4.2 and 4.4, it follows that $B_s$ is an absorbing set. From Proposition 5.1 and [1] we conclude that the semigroup operator $S(t)$ is uniformly compact, for t large. Hence, the proof is a direct application of ([24] chapter I, Theorem 1.1).

This finishes the proof. □