$H^1$ Solutions for a Kuramoto–Velarde Type Equation

Abstract

Kuramoto–Velarde equation describes the spatiotemporal evolution of the morphology of steps on crystal surfaces, or the evolution of the spinoidal decomposition of phase separating systems in an external field. We prove the well-posedness of the classical solutions for the Cauchy problem, associated with this equation for each choice of the terminal time T.

1 Introduction

In this paper, we investigate the well-posedness of the following Cauchy problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu+\partial _x f(u)+\gamma \partial _{x}^2u+\alpha \partial _{x}^3u \\ \qquad \qquad +\kappa u\partial _{x}^2u+\delta (\partial _x u)^2 +\beta ^2\partial _{x}^4u=0, &{}\quad 0< t< T, \quad x\in \mathbb {R},\\ u(0,x)=u_0(x), &{}\quad x\in \mathbb {R}, \end{array}\right. }\nonumber \\ \end{aligned}$$

(1.1)

with

$$\begin{aligned} \gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,\beta \in \mathbb {R}, \quad \beta \ne 0. \end{aligned}$$

(1.2)

On the flux f), we assume

$$\begin{aligned} f\in C^1(\mathbb {R}), \quad \vert f'(u)\vert \le C_0\left( 1+\vert u\vert +\vert u\vert ^2\right) , \end{aligned}$$

(1.3)

for some positive constant $C_0$.

On the initial datum, we assume

$$\begin{aligned} u_0\in H^1(\mathbb {R}), \quad u_0\ne 0. \end{aligned}$$

(1.4)

Taking

$$\begin{aligned} f(u)=b_1u^2, \quad b_1\in \mathbb {R}\end{aligned}$$

(1.5)

Equation (1.1) reads

$$\begin{aligned} \partial _tu+b_1\partial _x u^2 +\gamma \partial _{x}^2u+\alpha \partial _{x}^3u+\kappa u\partial _{x}^2u+\delta (\partial _x u)^2 +\beta ^2\partial _{x}^4u=0. \end{aligned}$$

(1.6)

It is known as the Kuramoto–Velarde equation and it describes slow space-time variations of disturbances at interfaces, diffusion–reaction fronts and plasma instability fronts [26, 30, 31]. It also describes Benard–Marangoni cells that occur when there is large surface tension on the interface [38, 67, 71] in a microgravity environment. This situation arises in crystal growth experiments aboard an orbiting space station, although the free interface is metastable with respect to small perturbations. The nonlinearities $\kappa u\partial _{x}^2u$ and $\delta (\partial _x u)^2$ model pressure destabilization effects striving to rupture the interface. (1.6) is deduced in [66] to describe the long waves on a viscous fluid owing down an inclined plane, and in [25] to model the drift waves in a plasma.

From a mathematical point of view, in [40], the exact solutions for (1.6) are studied, while, in [60], the initial boundary problem is analyzed. In [7, 8], the authors prove the existence of the solitons for (1.6). Instead, in [56], the existence of traveling wave solutions for (1.6) is studied. In [41], the author analyzes the existence of the periodic solution for (1.6), under appropriate assumptions on $b_1,\,\gamma ,\,\alpha ,\,\kappa ,\,\delta$ and $\beta$. The well-posedness of the Cauchy problem for (1.6) is proven in [59], using the energy space technique and taking $b_1=0$. In [15], under assumption

$$\begin{aligned} u_0\in H^2(\mathbb {R}), \quad u_0\ne 0, \end{aligned}$$

(1.7)

and $\delta =2\kappa$, through a priori estimates together with an application of the Cauchy–Kovalevskaya Theorem, the well-posedness of the classical solutions of (1.6) is proven. In [11], under Assumption (1.7) and $\delta \ne 2\kappa$, the well-posedness of classical solutions is proven, under appropriate assumptions on $\beta$, T and $H^1$-norm of the initial datum. Finally, in [16], the well-posedness of classical solutions is proven, under Assumption (1.7) and under appropriate assumptions on $\beta$, T and $L^2$-norm of the initial datum.

Taking $\kappa =\delta =0$ in (1.6), we have

$$\begin{aligned} \partial _tu+b_1\partial _x u^2 +\gamma \partial _{x}^2u+\alpha \partial _{x}^3u+\beta ^2\partial _{x}^4u=0. \end{aligned}$$

(1.8)

It was also independently deduced by Kuramoto [45,46,47] to describe the phase turbulence in reaction–diffusion systems, and by Sivashinsky [64], to describe plane flame propagation, taking into account the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front.

Equation (1.8) can be used to study incipient instabilities in several physical and chemical systems [5, 37, 48]. Moreover, (1.8), which is also known as the Benney–Lin equation [3, 52], was derived by Kuramoto in the study of phase turbulence in Belousov–Zhabotinsky reactions [51].

From a mathematical point of view, the dynamical properties and the existence of exact solutions for (1.8) have been investigated in [29, 42, 44, 57, 58, 69]. The control problem for (1.8) are studied in [1, 4, 32]. In [6], the problem of global exponential stabilization of (1.8) with periodic boundary conditions is analyzed. In [39], it is proposed a generalization of optimal control theory for (1.8), while, in [55], the problem of global boundary control of (1.8) is considered. In [61], the existence of solitonic solutions for (1.8) is proven. In [2, 15, 17, 18, 65], the well-posedness of the Cauchy problem for (1.8) is proven, using the energy space technique, the fixed point method, a priori estimates together with an application of the Cauchy–Kovalevskaya Theorem and a priori estimates together with an application of the Aubin–Lions Lemma, respectively. Instead, in [19, 53, 54], the initial-boundary value problem for (1.8) is studied, using a priori estimates together with an application of the Cauchy–Kovalevskaya Theorem, and the energy space technique, respectively. Finally, following [20, 49, 62], in [21], the convergence of the solution of (1.8) to to the unique entropy one of the Burgers equation is proven.

Taking

$$\begin{aligned} f(u)=b_1u^2+b_2u^3,\quad b_1,\,b_2\in \mathbb {R}, \end{aligned}$$

(1.9)

(1.1) reads

$$\begin{aligned} \partial _tu+b_1\partial _x u^2 +b_2\partial _x u^3+\gamma \partial _{x}^2u+\alpha \partial _{x}^3u+\kappa u\partial _{x}^2u+\delta (\partial _x u)^2 +\beta ^2\partial _{x}^4u=0.\nonumber \\ \end{aligned}$$

(1.10)

It models the spinodal decomposition of phase separating systems in an external field [28, 50, 70], the spatiotemporal evolution of the morphology of steps on crystal surfaces [33, 43, 61] and the growth of thermodynamically unstable crystal surfaces with strongly anisotropic surface tension [34,35,36]. In the case of a growing crystal surface with strongly anisotropic surface tension, the function u represents the surface slope, while the constants $b_1$ and $b_1$ are the growth driving forces proportional to the difference between the bulk chemical potentials of the solid and fluid phases. Equation (1.10) is also deduced by Watson [68] as a small-slope approximation of the crystal growth model obtained in [27].

Here we complete the results of [11, 16]. Here we assume (1.4) on the initial condition and our arguments are based on the Aubin-Lions Lemma [13, 14, 23, 24, 63].

The main result of this paper is the following theorem.

Theorem 1.1

Assume (1.2), (1.3) and (1.4). Fixed $\gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,T$, there exists a unique solution u of (1.1), such that

$$\begin{aligned} \begin{aligned}&u \in L^{\infty }(0,T;H^1(\mathbb {R}))\cap L^4(0,\,T;W^{1,\,4}(\mathbb {R}))\cap L^6(0,\,T;W^{1,\,6}(\mathbb {R})),\\&\partial _{x}^3u\in L^2((0,T)\times \mathbb {R}). \end{aligned} \end{aligned}$$

(1.11)

Moreover, if $u_1$ and $u_2$ are two solutions of (1.1) corresponding to the initial data $u_{1,0}$ and $u_{2,0}$, we have that

$$\begin{aligned} \left\| u_1(t,\cdot )-u_2(t,\cdot ) \right\| _{L^2(\mathbb {R})}\le e^{C(T)t}\left\| u_{1,0}-u_{2,0} \right\| _{L^2(\mathbb {R})}, \end{aligned}$$

(1.12)

for some suitable $C(T)>0$, and every, $0\le t\le T$.

The paper is organized as follows. In Sect. 2, we prove several a priori estimates on a vanishing viscosity approximation of (1.1). Those play a key role in the proof of our main result, that is given in Sect. 3.

2 Vanishing Viscosity Approximation

Our existence argument is based on passing to the limit in a vanishing viscosity approximation of (1.1).

Fix a small number $0<\varepsilon <1$ and let $u_\varepsilon =u_\varepsilon (t,x)$ be the unique classical solution of the following problem [9,10,11]:

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _tu_\varepsilon +\partial _x f(u_\varepsilon )+\gamma \partial _{x}^2u_\varepsilon +\alpha \partial _{x}^3u_\varepsilon \\ \qquad \qquad +\kappa u_\varepsilon \partial _{x}^2u_\varepsilon +\delta (\partial _x u_\varepsilon )^2 +\beta ^2\partial _{x}^4u_\varepsilon =\varepsilon \partial _{x}^6u_\varepsilon , &{}\quad 0\le t\le T, \quad x\in \mathbb {R},\\ u_{\varepsilon }(0,x)=u_{\varepsilon ,\,0}(x), &{}\quad x\in \mathbb {R}, \end{array}\right. }\nonumber \\ \end{aligned}$$

(2.1)

where $u_{\varepsilon ,0}$ is a $C^{\infty }$ approximation of $u_0$, such that

$$\begin{aligned} u_{\varepsilon ,\,0}\ne 0, \quad \left\| u_{\varepsilon ,\,0} \right\| _{H^1(\mathbb {R})}\le \left\| u_0 \right\| _{H^1(\mathbb {R})}, \quad \sqrt{\varepsilon }\left\| \partial _{x}^2u_{\varepsilon ,\,0} \right\| _{L^2(\mathbb {R})}\le C_0, \end{aligned}$$

(2.2)

where $C_0$ is a positive constant, independent on $\varepsilon$.

Let us prove some a priori estimates on $u_\varepsilon$. We denote by $C_0$ the constants which depend only on the initial data, and by C(T), the constants which depend also on T.

Following [11, Lemma 2.1] and [22, Lemma 2.2] , we prove the following result.

Lemma 2.1

Fix $\gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,T$. There exists a constant $C(T)>0$, independent of $\varepsilon$, such that

$$\begin{aligned}{} & {} \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})},\,\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\le C(T),\end{aligned}$$

(2.3)

$$\begin{aligned}{} & {} \qquad \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\le C(T),\end{aligned}$$

(2.4)

$$\begin{aligned}{} & {} \int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot )^2 \right\| _{L^2(\mathbb {R})}\mathrm{{d}}s\le C(T),\end{aligned}$$

(2.5)

$$\begin{aligned}{} & {} \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})} \mathrm{{d}}s\le C(T),\end{aligned}$$

(2.6)

$$\begin{aligned}{} & {} \varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\le C(T), \end{aligned}$$

(2.7)

for every $0\le t\le T$.

Proof

We begin by proving that

$$\begin{aligned} \begin{aligned}&\frac{e^{\frac{(40AC_0+a^2_1)t}{2\beta ^2}}}{\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) ^2}+\frac{100AC_0+a^2_2}{A(40AC_0+a_1^2)} \left( e^{\frac{(40AC_0+a^2_1)t}{2\beta ^2}}-1\right) \\&\qquad \quad \ge \frac{1}{\left( \left\| u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}\right) ^2} \end{aligned}\nonumber \\ \end{aligned}$$

(2.8)

for every $0\le t\le T$, where A is a arbitrary positive constant, and

$$\begin{aligned} \begin{aligned} a^2_1:=28\gamma ^2+5(\delta -2\kappa )^2 +4\kappa ^2+4\ne 0, \quad a^2_2:=4\kappa ^2+(\delta -2\kappa )^2+16\ne 0. \end{aligned}\nonumber \\ \end{aligned}$$

(2.9)

Consider A a positive constant. Multiplying (2.1) by $2u_\varepsilon -2A\partial _{x}^2u_\varepsilon$, an integration on $\mathbb {R}$ gives

$$\begin{aligned}{} & {} \frac{\textrm{d}}{\textrm{d }t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \\{} & {} \quad =2\int _{\mathbb {R}}u_\varepsilon \partial _tu_\varepsilon \mathrm{{d}}x-2A\int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _tu_\varepsilon \mathrm{{d}}x\\{} & {} \quad =\underbrace{-2\int _{\mathbb {R}}u f'(u)\partial _x u \textrm{d}x}_{=0}+2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x\\{} & {} \qquad +2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} -2\alpha \int _{\mathbb {R}}u_\varepsilon \partial _{x}^3u_\varepsilon \textrm{d}x +2A\alpha \int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \textrm{d}x\\{} & {} \qquad -2\kappa \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \textrm{d}x+2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \textrm{d}x -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \textrm{d}x\\{} & {} \qquad +2A\delta \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\partial _{x}^2u_\varepsilon \textrm{d}x -2\beta ^2\int _{\mathbb {R}}u_\varepsilon \partial _{x}^4u_\varepsilon \textrm{d }x +2A\beta ^2\int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \textrm{d}x\\{} & {} \qquad +2\varepsilon \int _{\mathbb {R}}u_\varepsilon \partial _{x}^6u_\varepsilon \textrm{d}x -2A\varepsilon \int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^6u_\varepsilon \textrm{d}x \\{} & {} \quad =2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x+2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \qquad +2\alpha \int _{\mathbb {R}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x -2\kappa \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \textrm{d}x+2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \textrm{d}x\\{} & {} \qquad -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \textrm{d}x +2\beta ^2\int _{\mathbb {R}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \textrm{d}x -2A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \qquad -2\varepsilon \int _{\mathbb {R}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \textrm{d}x +2A\varepsilon \int _{\mathbb {R}}\partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon \textrm{d}x \\{} & {} \quad =2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x+2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \qquad -2\kappa \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \textrm{d}x +2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \textrm{d}x -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \textrm{d}x\\{} & {} \qquad -2\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}-2A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon \int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \textrm{d}x \\{} & {} \qquad -2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \quad =2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \textrm{d}x+2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \qquad -2\kappa \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \textrm{d}x +2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \textrm{d}x -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \textrm{d }x\\{} & {} \qquad -2\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}-2A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}-2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\{} & {} \qquad -2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Consequently, we have that

$$\begin{aligned}{} & {} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \quad +2\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad =2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x+2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad -2\kappa \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \mathrm{{d}}x +2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \mathrm{{d}}x -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \mathrm{{d}}x.\nonumber \\ \end{aligned}$$

(2.10)

Observe that

$$\begin{aligned} -2\delta \int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^2 \mathrm{{d}}x=-2\delta \int _{\mathbb {R}}u_\varepsilon \partial _x u_\varepsilon \partial _x u_\varepsilon \mathrm{{d}}x =\delta \int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \mathrm{{d}}x.\nonumber \\ \end{aligned}$$

(2.11)

Moreover,

$$\begin{aligned} 2A\kappa \int _{\mathbb {R}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 \mathrm{{d}}x= & {} 2A\kappa \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x=-2A\kappa \int _{\mathbb {R}}\partial _x u_\varepsilon \partial _x \left( u_\varepsilon \partial _{x}^2u_\varepsilon \right) \mathrm{{d}}x\nonumber \\= & {} -2A\kappa \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\partial _{x}^2u_\varepsilon \mathrm{{d}}x -2A\kappa \int _{\mathbb {R}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \mathrm{{d}}x\nonumber \\= & {} -2A\kappa \int _{\mathbb {R}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \mathrm{{d}}x. \end{aligned}$$

(2.12)

It follows from (2.10), (2.11) and (2.12) that

$$\begin{aligned}{} & {} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \qquad \quad +2\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad \quad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad =2A\int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x-2\gamma \int _{\mathbb {R}}u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x+2A\gamma \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad \quad +(\delta -2\kappa )\int _{\mathbb {R}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \mathrm{{d}}x -2A\kappa \int _{\mathbb {R}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \mathrm{{d}}x. \end{aligned}$$

(2.13)

Due to (1.3) and the Young inequality,

$$\begin{aligned}&2A\int _{\mathbb {R}}\vert f'(u_\varepsilon )\vert \vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\le 2AC_0\int _{\mathbb {R}}\left( 1+\vert u_\varepsilon \vert +\vert u_\varepsilon \vert ^2\right) \vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le 2AC_0\int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x +2AC_0\int _{\mathbb {R}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x \\&\qquad \quad +2AC_0\int _{\mathbb {R}}u_\varepsilon ^2\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le \frac{A^2C_0}{D_1\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{A^2C_0}{D_1\beta ^2}\int _{\mathbb {R}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2 \mathrm{{d}}x\\&\qquad \quad +\frac{A^2C_0}{D_1\beta ^2}\int _{\mathbb {R}}u_\varepsilon ^4(\partial _x u_\varepsilon )^2 \mathrm{{d}}x+3D_1\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad \le \frac{A^2C_0}{D_1\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +\frac{A^2C_0}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad \quad +\frac{A^2C_0}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +3D_1\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&2\vert \gamma \vert \int _{\mathbb {R}}\vert u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\le \frac{\gamma ^2}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+D_1\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&\vert \delta -2\kappa \vert \int _{\mathbb {R}}u_\varepsilon ^2\vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x=2\int _{\mathbb {R}}\left| \frac{(\delta -2\kappa )u_\varepsilon ^2}{2\beta \sqrt{D_1}}\right| \left| \beta \sqrt{D_1}\beta \partial _{x}^2u_\varepsilon \right| \mathrm{{d}}x\\&\qquad \le \frac{(\delta -2\kappa )^2}{4D_1\beta ^2}\int _{\mathbb {R}}u_\varepsilon ^4 \mathrm{{d}}x +D_1\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad \le \frac{(\delta -2\kappa )^2}{4D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +D_1\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&2A\vert \kappa \vert \int _{\mathbb {R}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert \vert \partial _{x}^3u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le \frac{A\kappa ^2}{\beta ^2}\int _{\mathbb {R}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2 \mathrm{{d}}x +A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad \le \frac{A\kappa ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} + A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}, \end{aligned}$$

where $D_1$ is a positive constant, which will be specified later. Therefore, by (2.13),

$$\begin{aligned}&\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \\&\qquad +\beta ^2(2-5D_1)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le \frac{A^2C_0}{D_1\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{A^2C_0}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +\frac{A^2C_0}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\gamma ^2}{D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +\frac{(\delta -2\kappa )^2}{4D_1\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +\frac{A\kappa ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\vert \gamma \vert \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Taking $D_1=\frac{1}{5}$, we have that

$$\begin{aligned}{} & {} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \qquad +\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\beta ^2\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad \le \frac{5A^2C_0}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{5A^2C_0}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{5A^2C_0}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{5\gamma ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{5(\delta -2\kappa )^2}{4\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\vert \gamma \vert \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{A\kappa ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

(2.14)

Observe that

$$\begin{aligned} 2A\vert \gamma \vert \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}=2A\vert \gamma \vert \int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x=-2A\vert \gamma \vert \int _{\mathbb {R}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \mathrm{{d}}x. \end{aligned}$$

Thanks to the Young inequality,

$$\begin{aligned} 2A\vert \gamma \vert \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le&2A\vert \gamma \vert \int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^3u_\varepsilon \vert \mathrm{{d}}x\\ \le&\frac{2A\gamma ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{A\beta ^2}{2}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

It follows from (2.14) that

$$\begin{aligned}{} & {} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \qquad +\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{A\beta ^2}{2}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2A\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad \le \frac{5A^2C_0}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{5A^2C_0}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{5A^2C_0}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{5\gamma ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{5(\delta -2\kappa )^2}{4\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{2A\gamma ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad +\frac{A\kappa ^2}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad \le \frac{7\gamma ^2+5AC_0+1}{\beta ^2}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \qquad +\frac{20AC_0+5(\delta -2\kappa )^2+4\kappa ^2}{4\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})} \times \nonumber \\{} & {} \qquad \times \left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \nonumber \\{} & {} \qquad +\frac{5AC_0+1}{\beta ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) .\nonumber \\ \end{aligned}$$

(2.15)

We define

$$\begin{aligned} \begin{aligned} X_{\varepsilon }(t)&:=\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}, \quad \ell _1=\frac{7\gamma ^2+5AC_0+1}{\beta ^2},\\ \ell _2&:=\frac{20AC_0+5(\delta -2\kappa )^2+4\kappa ^2}{4\beta ^2}, \quad \ell _3:=\frac{5AC_0+1}{\beta ^2}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.16)

It follows from (2.15) that

$$\begin{aligned} \frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}\le \ell _1 X_{\varepsilon }(t)+\ell _2\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}X_{\varepsilon }(t)+\ell _3\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}X_{\varepsilon }(t).\nonumber \\ \end{aligned}$$

(2.17)

Thanks to the Hölder inequality,

$$\begin{aligned} u_\varepsilon ^2(t,x)&=2\int _{-\infty }^{x}u_\varepsilon \partial _x u_\varepsilon \mathrm{{d}}y\le 2\int _{-\infty }^{x}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert \mathrm{{d}}y\le 2\int _{\mathbb {R}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert \mathrm{{d}}x\\&\le 2\left\| u_\varepsilon (t,\cdot ) \right\| _{L^{2}(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^{2}(\mathbb {R})}\\&=\frac{2}{\sqrt{A}}\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{2}(\mathbb {R})}}\sqrt{A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^{2}(\mathbb {R})}}\\&\le \frac{2}{\sqrt{A}}X_{\varepsilon }(t). \end{aligned}$$

Hence,

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\le \frac{2}{\sqrt{A}}X_{\varepsilon }(t). \end{aligned}$$

(2.18)

It follows from (2.17) and (2.18) that

$$\begin{aligned} \frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}\le \ell _1 X_{\varepsilon }(t)+\frac{2\ell _2}{\sqrt{A}}X^2_{\varepsilon }(t)+\frac{4\ell _3}{A}X^3_{\varepsilon }(t). \end{aligned}$$

(2.19)

Due to the Young inequality,

$$\begin{aligned} \frac{2}{\sqrt{A}}X^2_{\varepsilon }(t)=\frac{2}{\sqrt{A}}X^{\frac{1}{2}}_{\varepsilon }(t)X^{\frac{3}{2}}_{\varepsilon }(t)\le X_{\varepsilon }(t)+\frac{1}{A}X^3_{\varepsilon }(t). \end{aligned}$$

Consequently, by (2.19),

$$\begin{aligned} \frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}\le \left( \ell _1+\ell _2\right) X_{\varepsilon }(t)+\frac{4\ell _3+\ell _2}{A}X^3_{\varepsilon }(t). \end{aligned}$$

(2.20)

Defined

$$\begin{aligned} \ell _4:= \ell _1+\ell _2, \quad \ell _5:=4\ell _3+\ell _2, \end{aligned}$$

(2.21)

it follows from (2.20) that

$$\begin{aligned} \frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}\le \ell _4X_{\varepsilon }(t)+\frac{\ell _5}{A}X^3_{\varepsilon }(t), \end{aligned}$$

that is,

$$\begin{aligned} \frac{1}{X^3_{\varepsilon }(t)}\frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}\le \frac{\ell _4}{X^2_{\varepsilon }(t)}+\frac{\ell _5}{A}. \end{aligned}$$

(2.22)

Since

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \frac{1}{X^2_{\varepsilon }(t)}\right) =-\frac{2}{X^3_{\varepsilon }(t)}\frac{\mathrm{{d}}X_{\varepsilon }(t)}{\mathrm{{d}}t}, \end{aligned}$$

by (2.22), we have that

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \frac{1}{X^2_{\varepsilon }(t)}\right) \ge -\frac{2\ell _4}{X^2_{\varepsilon }(t)}-\frac{2\ell _5}{A}, \end{aligned}$$

which gives

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \frac{1}{X^2_{\varepsilon }(t)}\right) +\frac{2\ell _4}{X^2_{\varepsilon }(t)}\ge -\frac{2\ell _5}{A}. \end{aligned}$$

(2.23)

Multiplying (2.23) by $e^{2\ell _4t}$, we get

$$\begin{aligned} e^{2\ell _4t}\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \frac{1}{X^2_{\varepsilon }(t)}\right) +\frac{2\ell _4e^{2\ell _4t}}{X^2_{\varepsilon }(t)}\ge -\frac{2\ell _5e^{2\ell _4t}}{A}. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \frac{e^{2\ell _4t}}{X^2_{\varepsilon }(t)}\right) \ge -\frac{2\ell _5e^{2\ell _4t}}{A}. \end{aligned}$$

Integrating on (0, t), we have that

$$\begin{aligned} \frac{e^{2\ell _4t}}{X^2_{\varepsilon }(t)}-\frac{1}{X^2_{\varepsilon }(0)}\ge -\frac{\ell _5}{A\ell _4}\left( e^{2\ell _4t}-1\right) , \end{aligned}$$

that is,

$$\begin{aligned} \frac{e^{2\ell _4t}}{X^2_{\varepsilon }(t)}+\frac{\ell _5}{A\ell _4}\left( e^{2\ell _4t}-1\right) \ge \frac{1}{X^2_{\varepsilon }(0)}. \end{aligned}$$

(2.24)

Using (2.16) and (2.21) in (2.24), thanks to (2.9), we have (2.8).

We prove (2.3). We begin by observing that, by (2.2),

$$\begin{aligned} \left\| u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}\le Y_0+AY_1, \end{aligned}$$

(2.25)

where

$$\begin{aligned} Y_0:=\left\| u_0 \right\| ^2_{L^2(\mathbb {R})}, \quad Y_1:=\left\| \partial _x u_0 \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

(2.26)

Consequently, we have that

$$\begin{aligned} \left( \left\| u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}\right) ^2\le \left( Y_0+AY_1\right) ^2, \end{aligned}$$

which gives,

$$\begin{aligned} \frac{1}{\left( \left\| u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_{\varepsilon ,\,0} \right\| ^2_{L^2(\mathbb {R})}\right) ^2}\ge \frac{1}{\left( Y_0+AY_1\right) ^2}. \end{aligned}$$

(2.27)

Moreover,

$$\begin{aligned} e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}\ge e^{\frac{(40AC_0+a^2_1)t}{2\beta ^2}}. \end{aligned}$$

(2.28)

It follows from (2.8), (2.27) and (2.28) that

$$\begin{aligned}&\frac{e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}}{\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) ^2}+\frac{100AC_0+a^2_2}{A(40AC_0+a_1^2)} \!\left( e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}\!-1\right) \\&\quad \ge \frac{1}{\left( Y_0+AY_1\right) ^2}, \end{aligned}$$

that is

$$\begin{aligned} \begin{aligned}&\frac{e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}}{\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) ^2}\\&\qquad \quad \ge \frac{1}{\left( Y_0+AY_1\right) ^2}-\frac{100AC_0+a^2_2}{A(40AC_0+a_1^2)} \left( e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}-1\right) . \end{aligned}\nonumber \\ \end{aligned}$$

(2.29)

We search A such that

$$\begin{aligned} \frac{1}{\left( Y_0+AY_1\right) ^2}-\frac{100AC_0+a^2_2}{A(40AC_0+a_1^2)}\left( e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}-1\right) >0, \end{aligned}$$

(2.30)

that is,

$$\begin{aligned} e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}<1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}. \end{aligned}$$

Therefore,

$$\begin{aligned} 1<\frac{2\beta ^2}{(40AC_0+a^2_1)T}\log \left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) , \end{aligned}$$

which gives

$$\begin{aligned} e< \left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) ^{\frac{2\beta ^2}{(40AC_0+a^2_1)T}}. \end{aligned}$$

(2.31)

Taking $\beta$ as in

$$\begin{aligned} \vert \beta \vert =A^n, \quad n>\frac{3}{2}, \end{aligned}$$

(2.32)

(2.31) reads

$$\begin{aligned} e< \left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) ^{\frac{2A^{2n}}{(40AC_0+a^2_1)T}}. \end{aligned}$$

Thanks to (2.32), we have that

$$\begin{aligned} \lim _{A\rightarrow \infty }\left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) ^{\frac{2A^{2n}}{(40AC_0+a^2_1)T}}=\infty . \end{aligned}$$

(2.33)

$$\begin{aligned}&\lim _{A\rightarrow \infty }\left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) ^{\frac{2A^{2n}}{(40AC_0+a^2_1)T}}\\&\qquad =\lim _{A\rightarrow \infty }\left[ \left( 1+\frac{A(40AC_0+a_1^2)}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}\right) ^{\frac{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2}{(40AC_0+a^2_1)}}\right] ^{\frac{2A^{2n}}{(100AC_0+a^2_2)\left( Y_0+AY_1\right) ^2T}}\\&\qquad =e^{\infty }=\infty . \end{aligned}$$

Therefore, thanks to (2.33), (2.30), which is equivalent to (2.31), holds, taking A very big, and up to rescaling, we can have $\vert \beta \vert =A^n$, with n defined in (2.32).

Consequently, by (2.32), (2.29), (2.30), or (2.31), and (2.33), there exists a constant $C(T)>0$, independent of $\varepsilon$, such that

$$\begin{aligned} \frac{e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}}{\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) ^2}\ge C(T). \end{aligned}$$

Hence,

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+A\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le \sqrt{\frac{e^{\frac{(40AC_0+a^2_1)T}{2\beta ^2}}}{C(T)}} \end{aligned}$$

which gives (2.3).

We prove (2.4). Thanks to (2.3), (2.16) with $A=1$, and (2.18) with $A=1$, we have that

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\le 2\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \le C(T), \end{aligned}$$

which gives (2.4).

Finally, we prove (2.5), (2.6), (2.7). We begin by observing that, thanks to (2.3), (2.4) and (2.15) with $A=1$, we have that

$$\begin{aligned}&\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left( \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\right) \\&\quad +\beta ^2\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2}{2}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad +2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le C(T). \end{aligned}$$

Integrating on (0, t), by (2.2), we get

$$\begin{aligned}&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +\beta ^2\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s+\frac{\beta ^2}{2}\int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\\&\qquad +2\varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\\&\quad \le C_0+C(T)t\le C(T), \end{aligned}$$

which gives (2.5), (2.6), (2.7). $\square$

Lemma 2.2

Fix $\gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,T$. There exists a constant $C(T)>0$, independent of $\varepsilon$, such that

$$\begin{aligned}{} & {} \int _{0}^{t}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^4_{L^4(\mathbb {R})}\mathrm{{d}}s\le C(T),\end{aligned}$$

(2.34)

$$\begin{aligned}{} & {} \int _{0}^{t}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^6_{L^6(\mathbb {R})}\mathrm{{d}}s\le C(T), \end{aligned}$$

(2.35)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. We begin by proving (2.34). [12, Lemma 2.3] says that

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})}\le 9\left\| u \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} \end{aligned}$$

Thanks to (2.4), we have that

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})}\le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Integrating on (0, t), we have (2.34).

Finally, we prove (2.35). We begin by observing that

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^6(\mathbb {R})}=\int _{\mathbb {R}}(\partial _x u_\varepsilon )^5\partial _x u_\varepsilon \mathrm{{d}}x =-5\int _{\mathbb {R}}u_\varepsilon (\partial _x u_\varepsilon )^4\partial _{x}^2u_\varepsilon \mathrm{{d}}x.\nonumber \\ \end{aligned}$$

(2.36)

Due to the (2.4) and the Hölder and Young inequalities,

$$\begin{aligned}&5\int _{\mathbb {R}}\vert u_\varepsilon \vert (\partial _x u_\varepsilon )^4\vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le 5\left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\int _{\mathbb {R}}(\partial _x u_\varepsilon )^4\vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^2(\mathbb {R})}\int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\qquad \le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\\&\qquad \le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^{\infty }(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\\&\qquad \le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^{\infty }(\mathbb {R})} +C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

It follows from (2.36) that

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^6(\mathbb {R})}\le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^{\infty }(\mathbb {R})} +C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}.\nonumber \\ \end{aligned}$$

(2.37)

Due to the Young inequality,

$$\begin{aligned} C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^{\infty }(\mathbb {R})}&=C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\\&\le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4{L^{\infty }(\mathbb {R})}\!+C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^8_{L^{\infty }(\mathbb {R})}\!. \end{aligned}$$

Consequently, by (2.37), we have that

$$\begin{aligned} \begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^6(\mathbb {R})}&\le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}+C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^8_{L^{\infty }(\mathbb {R})}\\&\quad +C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.38)

Thanks to the Hölder inequality,

$$\begin{aligned} (\partial _x u_\varepsilon (t,x))^2&=2\int _{-\infty }^{x}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}y\le 2\int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\\&\le 2\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\le 4\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

(2.39)

Observe that

$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}=\int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x=-\int _{\mathbb {R}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \mathrm{{d}}x. \end{aligned}$$

Thanks to the Hölder inequality,

$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le \int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \mathrm{{d}}x\le \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

It follows from (2.39) that

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\le 4\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^2(\mathbb {R})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}, \end{aligned}$$

which gives

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^8_{L^{\infty }(\mathbb {R})}\le 16\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^2(\mathbb {R})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

(2.40)

It follows from (2.3), (2.39) and (2.40) that

$$\begin{aligned} \begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }(\mathbb {R})}\le&C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\ \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^8_{L^{\infty }(\mathbb {R})}\le&C(T)\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned} \end{aligned}$$

(2.41)

Consequently, by (2.38) and (2.41), we get

$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^6_{L^6(\mathbb {R})}\le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+C(T)\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

Integrating on (0, t), we have (2.35). $\square$

Lemma 2.3

Fix $\gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,T$. There exists a constant $C(T)>0$, independent of $\varepsilon$, such that

$$\begin{aligned}{} & {} \varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})} \mathrm{{d}}s \nonumber \\{} & {} \quad +2\varepsilon ^2\int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\le C(T), \end{aligned}$$

(2.42)

for every $0\le t\le T$.

Proof

Let $0\le t\le T$. Multiplying (2.1) by $2\varepsilon \partial _{x}^4u_\varepsilon$, an integration on $\mathbb {R}$ gives

$$\begin{aligned}&\varepsilon \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad =2\varepsilon \int _{\mathbb {R}}\partial _{x}^4u_\varepsilon \partial _tu_\varepsilon \mathrm{{d}}x\\&\quad =-2\varepsilon \int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x -2\gamma \varepsilon \int _{\mathbb {R}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x\\&\qquad -2\alpha \varepsilon \int _{\mathbb {R}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x-2\kappa \varepsilon \int _{0}^{t}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x\\&\qquad -2\delta \varepsilon \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\partial _{x}^4u_\varepsilon \mathrm{{d}}x -2\beta ^2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon ^2\int _{\mathbb {R}}\partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon \mathrm{{d}}x\\&\quad =-2\varepsilon \int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x+2\gamma \varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}-2\kappa \varepsilon \int _{0}^{t}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x\\&\qquad -2\delta \varepsilon \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\partial _{x}^4u_\varepsilon \mathrm{{d}}x-2\beta ^2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}-2\varepsilon ^2\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Therefore, we have that

$$\begin{aligned}{} & {} \varepsilon \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\beta ^2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon ^2\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad =-2\varepsilon \int _{\mathbb {R}}f'(u_\varepsilon )\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x+2\gamma \varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \qquad -2\kappa \varepsilon \int _{0}^{t}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \mathrm{{d}}x-2\delta \varepsilon \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\partial _{x}^4u_\varepsilon \mathrm{{d}}x. \end{aligned}$$

(2.43)

Since $0<\varepsilon <1$, thanks to (2.3), (2.4) and the Young inequality,

$$\begin{aligned}&2\varepsilon \int _{\mathbb {R}}\vert f'(u_\varepsilon )\vert \vert \partial _x u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x\\&\quad \le 2\varepsilon \left\| f' \right\| _{L^{\infty }(-C(T),\,C(T))}\int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x\\&\quad \le C(T)\varepsilon \int _{\mathbb {R}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x=\int _{\mathbb {R}}\left| \frac{C(T)\partial _x u_\varepsilon }{\beta }\right| \left| \beta \varepsilon \partial _{x}^4u_\varepsilon \right| \mathrm{{d}}x\\&\quad \le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +\frac{\beta ^2\varepsilon ^2}{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le C(T) +\frac{\beta ^2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&2\vert \kappa \vert \varepsilon \int _{0}^{t}\vert u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x\\&\quad \le 2\vert \kappa \vert \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\int _{\mathbb {R}}\vert \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x\\&\quad \le C(T)\varepsilon \int _{\mathbb {R}}\vert \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x=\int _{\mathbb {R}}\left| \frac{C(T)\partial _{x}^2u_\varepsilon }{\beta }\right| \left| \beta \varepsilon \partial _{x}^4u_\varepsilon \right| \mathrm{{d}}x\\&\quad \le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +\frac{\beta ^2\varepsilon ^2}{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2\varepsilon }{2} \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\&2\vert \delta \vert \varepsilon \int _{\mathbb {R}}(\partial _x u_\varepsilon )^2\vert \partial _{x}^4u_\varepsilon \vert \mathrm{{d}}x\\&\quad =\int _{\mathbb {R}}\left| \frac{2\delta (\partial _x u_\varepsilon )^2}{\beta }\right| \left| \beta \varepsilon \partial _{x}^4u_\varepsilon \right| \mathrm{{d}}x\\&\quad \le \frac{2\delta ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})} +\frac{\beta ^2\varepsilon ^2}{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le \frac{2\delta ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})} +\frac{\beta ^2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

It follows from (2.43) that

$$\begin{aligned}&\varepsilon \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\varepsilon ^2\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le C(T) +2\vert \gamma \vert \varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+ \frac{2\delta ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})}\\&\quad \le C(T) +2\vert \gamma \vert \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\qquad +C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+ \frac{2\delta ^2}{\beta ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^4(\mathbb {R})}.\ \end{aligned}$$

Integrating on (0, t), by (2.2) and (2.34), we get

$$\begin{aligned}&\varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s+2\varepsilon ^2\!\int _{0}^{t}\!\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\\&\quad \le C_0 +C(T)t + 2\vert \gamma \vert \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s\\&\qquad +C(T)\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}s+ \frac{2\delta ^2}{\beta ^2}\int _{0}^{t}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^4_{L^4(\mathbb {R})}\mathrm{{d}}s\\&\quad \le C(T), \end{aligned}$$

which gives (2.42). $\square$

3 Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

We begin by proving the following lemma.

Lemma 3.1

Fix $\gamma ,\,\alpha ,\,\kappa ,\,\delta ,\,T$. Then,

$$\begin{aligned} \hbox {the sequence}\, \, \{u_\varepsilon \}_{\varepsilon >0} \, \, \hbox {is compact in} \, \, L^2_\mathrm{{loc}}((0,\infty )\times \mathbb {R}). \end{aligned}$$

(3.1)

Consequently, there exists a subsequence $\{u_{\varepsilon _k}\}_{k\in \mathbb {N}}$ of $\{u_\varepsilon \}_{\varepsilon >0}$ and $u\in L^2_\mathrm{{loc}} ((0,\infty )\times \mathbb {R})$ such that, for each compact subset K of $(0,\infty )\times \mathbb {R})$,

$$\begin{aligned} u_{\varepsilon _k}\rightarrow u \, \, \text {in} \, \, L^2(K) \text {and a.e.} \end{aligned}$$

(3.2)

Moreover, u is a solution of (1.1), satisfying (1.11).

Proof

We begin by proving (3.1). To prove (3.1), we rely on the Aubin–Lions Lemma (see [13, 14, 23, 24, 63]). We recall that

$$\begin{aligned} H^1_\mathrm{{loc}}(\mathbb {R})\hookrightarrow \hookrightarrow L^2_\mathrm{{loc}}(\mathbb {R})\hookrightarrow H^{-1}_\mathrm{{loc}}(\mathbb {R}), \end{aligned}$$

where the first inclusion is compact and the second is continuous. Owing to the Aubin–Lions Lemma [63], to prove (3.1), it suffices to show that

$$\begin{aligned}{} & {} {\{u_\varepsilon \}_{\varepsilon >0}~\, \, \text {is uniformly bounded in} \, \, L^2(0,T;H^1_\mathrm{{loc}}(\mathbb {R})),}\end{aligned}$$

(3.3)

$$\begin{aligned}{} & {} {\{\partial _tu_\varepsilon \}_{\varepsilon >0} \, \, \text {is uniformly bounded in} \, \, L^2(0,T;H^{-1}_\mathrm{{loc}}(\mathbb {R})).} \end{aligned}$$

(3.4)

We prove (3.3). Thanks to Lemma 2.1,

$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{H^1(\mathbb {R})}=\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le C(T). \end{aligned}$$

Therefore,

$$\begin{aligned} \hbox {} \{u_\varepsilon \}_{\varepsilon >0} \, \, \hbox {~is uniformly bounded in} \, \, L^{\infty }(0,T;H^{1}(\mathbb {R})), \end{aligned}$$

which gives (3.3).

We prove (3.4). We begin by observing that

$$\begin{aligned} \partial _x \left( u_\varepsilon \partial _x u_\varepsilon \right) =(\partial _x u_\varepsilon )^2+u_\varepsilon \partial _{x}^2u_\varepsilon . \end{aligned}$$

(3.5)

Therefore, by (2.1) and (3.5), we have that

$$\begin{aligned} \partial _tu_\varepsilon =\partial _x \left( G(u_\varepsilon )\right) -f'(u_\varepsilon )\partial _x u_\varepsilon -\left( \delta +\kappa \right) (\partial _x u_\varepsilon )^2, \end{aligned}$$

where

$$\begin{aligned} G(u_\varepsilon )=\kappa u_\varepsilon \partial _x u_\varepsilon -\gamma \partial _x u_\varepsilon -\alpha \partial _{x}^2u_\varepsilon -\beta ^2\partial _{x}^3u_\varepsilon +\varepsilon \partial _{x}^5u_\varepsilon . \end{aligned}$$

(3.6)

We claim that

$$\begin{aligned} \kappa ^2\int _{0}^{T}\!\!\!\int _{\mathbb {R}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2\mathrm{{d}}t\mathrm{{d}}x\le C(T). \end{aligned}$$

(3.7)

Thanks to (2.3) and (2.4),

$$\begin{aligned} \kappa ^2\int _{0}^{T}\!\!\!\int _{\mathbb {R}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2\mathrm{{d}}t\mathrm{{d}}x\le \kappa ^2\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times \mathbb {R})}\int _{0}^{T}\!\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}t\le C(T). \end{aligned}$$

Moreover, thanks to (2.3) and (2.42),

$$\begin{aligned} \begin{aligned}&\gamma ^2\left\| \partial _x u_\varepsilon \right\| ^2_{L^2((0,T)\times \mathbb {R})},\, \alpha ^2\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^2((0,T)\times \mathbb {R})}\le C(T),\\&\beta ^4\left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^2((0,T)\times \mathbb {R})},\,\varepsilon ^2\left\| \partial _{x}^5u_\varepsilon \right\| ^2_{L^2((0,T)\times \mathbb {R})}\le C(T). \end{aligned} \end{aligned}$$

(3.8)

Therefore, by (3.6), (3.7) and (3.8), we have that

$$\begin{aligned} \{\partial _x \left( G(u_\varepsilon )\right) \}_{\varepsilon >0} \quad \hbox { is bounded in}\ H^1((0,T)\times \mathbb {R}). \end{aligned}$$

(3.9)

We have that

$$\begin{aligned} \int _{0}^{T}\!\!\!\int _{\mathbb {R}}(f'(u_\varepsilon ))^2(\partial _x u_\varepsilon )^2 \mathrm{{d}}t\mathrm{{d}}x\le C(T). \end{aligned}$$

(3.10)

Thanks to (2.3) and (2.4),

$$\begin{aligned} \int _{0}^{T}\!\!\!\int _{\mathbb {R}}(f'(u_\varepsilon ))^2(\partial _x u_\varepsilon )^2 \mathrm{{d}}t\mathrm{{d}}x{} & {} \le \left\| f' \right\| ^2_{L^{\infty }(-C(T),C(T))}\int _{0}^{T}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\mathrm{{d}}t\\{} & {} \le C(T). \end{aligned}$$

Moreover, thanks to (2.34),

$$\begin{aligned} \left( \delta +\kappa \right) ^2\int _{0}^{T}\!\!\!\int _{\mathbb {R}}(\partial _x u_\varepsilon )^4\mathrm{{d}}t\mathrm{{d}}x\le C(T). \end{aligned}$$

(3.11)

Consequently, (3.4) follows from (3.9), (3.10) and (3.11).

Thanks to the Aubin–Lions Lemma, (3.1) and (3.2) hold.

Therefore, arguing as in [14, Theorem 1.1], u is solution of (1.1) and, thanks to Lemmas 2.1 and 2.2, (1.11) holds. $\square$

Now, we prove Theorem 1.1.

Proof of Theorem 1.1

We begin by observing that, by (3.5), (1.1) reads:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _tu +\partial _x f(u)+\gamma \partial _{x}^2u+\alpha \partial _{x}^3u\\ \displaystyle \qquad \qquad -\kappa \partial _x \left( u\partial _x u\right) +\left( \delta +\kappa \right) (\partial _x u)^2+\beta ^2\partial _{x}^4u=0, &{}\quad 0\le t\le T, \quad x\in \mathbb {R},\\ u(0,x)=u_0(x), &{}\quad x\in \mathbb {R}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(3.12)

Lemma 3.1 gives the existence of a solution (3.12) satisfying (1.11). We prove (1.12). Let $u_1$ and $u_2$ two solutions of (3.12), which verify (1.11), that is

$$\begin{aligned}&{\left\{ \begin{array}{ll} \displaystyle \partial _tu_i+\partial _x f(u_i)+\gamma \partial _{x}^2u_i+\alpha \partial _{x}^3u_i\\ \displaystyle \qquad \qquad -\kappa \partial _x \left( u_i\partial _x u_i\right) +\left( \delta +\kappa \right) (\partial _x u_i)^2\\ \displaystyle \qquad \qquad +\beta ^2\partial _{x}^4u_i =0, &{}\quad 0\le t\le T, \quad x\in \mathbb {R},\\ \displaystyle u_i(0,x)=u_{i,\,0}(x),&{}\quad x\in \mathbb {R}, \end{array}\right. }&\quad i=1,2. \end{aligned}$$

Then, the function

$$\begin{aligned} \omega (t,x)=u_1(t,x)-u_2(t,x), \end{aligned}$$

(3.13)

is the solution of the following Cauchy problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \partial _t\omega +\partial _x \left( f(u_1)-f(u_2)\right) +\gamma \partial _{x}^2\omega +\alpha \partial _{x}^3\omega \\ \displaystyle \qquad \qquad -\kappa \partial _x \left( u_1\partial _x u_1-u_2\partial _x u_2\right) \\ \displaystyle \qquad \qquad +\left( \delta +\kappa \right) \left[ (\partial _x u_1)^2-(\partial _x u_2)\right] +\beta ^2\partial _{x}^4\omega =0, \quad &{} 0\le t\le T, \quad x\in \mathbb {R},\\ \displaystyle \omega (0,x)=u_{1,\,0}(x)-u_{2,\,0}(x),\quad &{}x\in \mathbb {R}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(3.14)

Fixed $T>0$, since $u_1,\, u_2\in H^1(\mathbb {R})$, for every $0\le t\le T$, we have that

$$\begin{aligned} \begin{aligned}&\left\| u_1 \right\| _{L^{\infty }((0,T)\times \mathbb {R})},\, \left\| u_2 \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\le C(T),\\&\left\| \partial _x u_1(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\, \left\| \partial _x u_1(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le C(T). \end{aligned}\nonumber \\ \end{aligned}$$

(3.15)

Since $f\in C^1(\mathbb {R})$, thanks to (3.13), there exists $\xi$ between $u_1$ and $u_2$, such that

$$\begin{aligned} f(u_1)-f(u_2)=f'(\xi )(u_1-u_2)=f'(\xi )\omega , \quad u_1<\xi<u_2, \text {or}, u_2<\xi <u_1.\nonumber \\ \end{aligned}$$

(3.16)

Moreover, by (2.38), we have that

$$\begin{aligned} \vert f'(\xi )\vert \le \left\| f' \right\| _{L^{\infty }(-C(T),\,C(T))}\le C(T). \end{aligned}$$

(3.17)

Observe that, thanks to (3.13)

$$\begin{aligned} \partial _x \left( u_1\partial _x u_1-u_2\partial _x u_2\right)= & {} \partial _x \left( u_1\partial _x u_1-u_2\partial _x u_1+u_2\partial _x u_1-u_2\partial _x u_2\right) \nonumber \\= & {} \partial _x \left( \partial _x u_1\omega \right) +\partial _x \left( u_2\partial _x \omega \right) , \nonumber \\ (\partial _x u_1)^2-(\partial _x u_2)^2= & {} (\partial _x u_1+\partial _x u_2)(\partial _x u_1-\partial _x u_2)=\partial _x u_1\partial _x \omega +\partial _x u_2\partial _x \omega .\nonumber \\ \end{aligned}$$

(3.18)

Thanks to (3.16) and (3.18), (3.14) reads

$$\begin{aligned} \begin{aligned} \partial _t\omega&= -\partial _x \left( f'(\xi )\omega \right) -\gamma \partial _{x}^2\omega -\alpha \partial _{x}^3\omega +\kappa \partial _x \left( \partial _x u_1\omega \right) \\&\quad +\kappa \partial _x \left( u_2\partial _x \omega \right) -(\delta +\kappa )\partial _x u_1\partial _x \omega -(\delta +\kappa )\partial _x u_2\partial _x \omega -\beta ^2\partial _{x}^4\omega . \end{aligned}\nonumber \\ \end{aligned}$$

(3.19)

Multiplying (3.19) by $2\omega$, an integration on $\mathbb {R}$ gives

$$\begin{aligned}&\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad =2\int _{\mathbb {R}}\omega \partial _t\omega \mathrm{{d}}x\\&\quad =-2\int _{\mathbb {R}}\omega \partial _x \left( f'(\xi )\omega \right) \mathrm{{d}}x-2\gamma \int _{\mathbb {R}}\omega \partial _{x}^2\omega \mathrm{{d}}x-2\alpha \int _{\mathbb {R}}\omega \partial _{x}^3\omega \\&\qquad +2\kappa \int _{\mathbb {R}}\omega \partial _x \left( \partial _x u_1\omega \right) \mathrm{{d}}x+2\kappa \int _{\mathbb {R}}\omega \partial _x \left( u_2\partial _x \omega \right) \mathrm{{d}}x\\&\qquad -2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_1\omega \partial _x \omega \mathrm{{d}}x-2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_2\omega \partial _x \omega \mathrm{{d}}x-2\beta ^2\int _{\mathbb {R}}\omega \partial _{x}^4\omega \mathrm{{d}}x\\&\quad =2\int _{\mathbb {R}}f'(\xi )\omega \partial _x \omega \mathrm{{d}}x -2\gamma \int _{\mathbb {R}}\omega \partial _{x}^2\omega \mathrm{{d}}x+2\alpha \int _{\mathbb {R}}\partial _x \omega \partial _{x}^2\omega \mathrm{{d}}x\\&\qquad -2\kappa \int _{\mathbb {R}}\partial _x u_1\omega \partial _x \omega \mathrm{{d}}x-2\kappa \int _{\mathbb {R}}u_2(\partial _x \omega )^2 \mathrm{{d}}x -2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_1\omega \partial _x \omega \mathrm{{d}}x\\&\qquad -2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_2\omega \partial _x \omega \mathrm{{d}}x+2\beta ^2\int _{\mathbb {R}}\partial _x \omega \partial _{x}^3\omega \mathrm{{d}}x\\&\quad =2\int _{\mathbb {R}}f'(\xi )\omega \partial _x \omega \mathrm{{d}}x -2\gamma \int _{\mathbb {R}}\omega \partial _{x}^2\omega \mathrm{{d}}x-2(\delta +2\kappa )\int _{\mathbb {R}}\partial _x u_1\omega \partial _x \omega \mathrm{{d}}x\\&\qquad -2\kappa \int _{\mathbb {R}}u_2(\partial _x \omega )^2 \mathrm{{d}}x -2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_2\omega \partial _x \omega \mathrm{{d}}x-2\beta ^2\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Consequently, we have that

$$\begin{aligned}{} & {} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+2\beta ^2\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\nonumber \\{} & {} \quad =2\int _{\mathbb {R}}f'(\xi )\omega \partial _x \omega \mathrm{{d}}x -2\gamma \int _{\mathbb {R}}\omega \partial _{x}^2\omega \mathrm{{d}}x\nonumber \\{} & {} \qquad -2\left( \delta +2\kappa \right) \int _{\mathbb {R}}\partial _x u_1\omega \partial _x \omega \mathrm{{d}}x -2\kappa \int _{\mathbb {R}}u_2(\partial _x \omega )^2 \mathrm{{d}}x \nonumber \\{} & {} \qquad -2(\delta +\kappa )\int _{\mathbb {R}}\partial _x u_2\omega \partial _x \omega \mathrm{{d}}x. \end{aligned}$$

(3.20)

Due to (3.15), (3.17) and the Young inequality,

$$\begin{aligned} 2\int _{\mathbb {R}}\vert f'(\xi )\vert \vert \omega \vert \vert \partial _x \omega \vert \mathrm{{d}}x&\le C(T)\int _{\mathbb {R}} \vert \omega \vert \vert \partial _x \omega \vert \mathrm{{d}}x\\&\le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +C(T)\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\ 2\vert \gamma \vert \int _{\mathbb {R}}\vert \omega \vert \vert \partial _{x}^2\omega \vert \mathrm{{d}}x)&=2\int _{\mathbb {R}}\left| \frac{\gamma \omega }{\beta }\right| \left| \beta \partial _{x}^2\omega \right| \mathrm{{d}}x\\&\le \frac{\gamma ^2}{\beta ^2}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +\beta ^2\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\ 2\left| \delta +2\kappa \right| \int _{\mathbb {R}}\vert \partial _x u_1\omega \vert \vert \partial _x \omega \vert \mathrm{{d}}x&\le (\delta +2\kappa )^2\int _{\mathbb {R}}\omega ^2(\partial _x u_1)^2 \mathrm{{d}}x+\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\le (\delta +2\kappa )^2\left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_1(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad +\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}+\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\ 2\vert \kappa \vert \int _{\mathbb {R}}\vert u_2\vert (\partial _x \omega )^2 \mathrm{{d}}x&\le 2\vert \kappa \vert \left\| u_2 \right\| _{L^{\infty }((0,T)\times \mathbb {R})}\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\le C(T) \left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})},\\ 2\vert \delta +\kappa \vert \int _{\mathbb {R}}\vert \partial _x u_2\omega \vert \vert \partial _x \omega \vert \mathrm{{d}}x&\le (\delta +\kappa )^2\int _{\mathbb {R}}\omega ^2(\partial _x u_2)^2 \mathrm{{d}}x +\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\le (\delta +\kappa )^2\left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\left\| \partial _x u_2(t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad +\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}+\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

It follows from (3.20) that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\beta ^2\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})} +C(T)\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.21)

Thanks to the Hölder inequality,

$$\begin{aligned} \omega ^2(t,x)&= 2\int _{-\infty }^{x}\omega \partial _x \omega \mathrm{{d}}y\le \int _{\mathbb {R}}\vert \omega \vert \vert \partial _x \omega \vert \mathrm{{d}}x\\&\le 2\left\| \omega (t,\cdot ) \right\| _{L^2(\mathbb {R})}\left\| \partial _x \omega (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\le 2\left\| \omega (t,\cdot ) \right\| _{L^2(\mathbb {R})}\left\| \partial _x \omega (t,\cdot ) \right\| _{L^2(\mathbb {R})}. \end{aligned}$$

Due to the Young inequality,

$$\begin{aligned} \left\| \omega (t,\cdot ) \right\| ^2_{L^{\infty }(\mathbb {R})}\le \left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Therefore, by (3.21), we have that

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\beta ^2\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\\&\quad \le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+C(T)\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.22)

Observe that

$$\begin{aligned} C(T)\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}=C(T)\int _{\mathbb {R}}\partial _x \omega \partial _x \omega \mathrm{{d}}x=-C(T)\int _{\mathbb {R}}\omega \partial _{x}^2\omega \mathrm{{d}}x. \end{aligned}$$

Hence, by the Young inequality,

$$\begin{aligned} C(T)\left\| \partial _x \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}&\le C(T)\int _{\mathbb {R}}\vert \omega \vert \vert \partial _{x}^2\omega \vert \mathrm{{d}}x=\int _{\mathbb {R}}\left| \frac{C(T)\omega }{\beta }\right| \left| \beta \partial _{x}^2\omega \right| \mathrm{{d}}x\\&\le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2}{2}\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

Consequently, by (3.22),

$$\begin{aligned} \frac{\mathrm{{d}}}{\mathrm{{d}}t}\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2}{2}\left\| \partial _{x}^2\omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}\le C(T)\left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}. \end{aligned}$$

The Gronwall Lemma and (3.14) give

$$\begin{aligned} \left\| \omega (t,\cdot ) \right\| ^2_{L^2(\mathbb {R})}+\frac{\beta ^2 e^{C(T)t}}{2}\!\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^2\omega (s,\cdot ) \right\| ^2_{L^2(\mathbb {R})} \mathrm{{d}}s\le e^{C(T)t}\left\| \omega _0 \right\| ^2_{L^2(\mathbb {R})}.\nonumber \\ \end{aligned}$$

(3.23)

(1.12) follows from (3.13) and (3.23).