A Note on the Solutions for a Higher-Order Convective Cahn–Hilliard-Type Equation

Abstract

The higher-order convective Cahn-Hilliard equation describes the evolution of crystal surfaces faceting through surface electromigration, the growing surface faceting, and the evolution of dynamics of phase transitions in ternary oil-water-surfactant systems. In this paper, we study the $H^3$ solutions of the Cauchy problem and prove, under different assumptions on the constants appearing in the equation and on the mean of the initial datum, that they are well-posed.

1. Introduction

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

$$\left\{\begin{array}{cc}{\partial }_{t}u+\kappa {\partial }_x{u}^2+\gamma {\partial }_x^{2}u-{\beta }^2{\partial }_x^{6}u+\alpha {\partial }_x^{4}u+{\delta }^2{\partial }_x^4\left(u^3\right)=0,\hfill & t>0,x\in \mathbb{R},\hfill \\ u(0,x)=u_0\left(x\right),\hfill & x\in \mathbb{R},\hfill \end{array}\right.$$

(1)

with

$$\begin{array}{cc}\hfill & \kappa ,\gamma ,\beta ,\alpha ,\delta \in \mathbb{R},\gamma <0,\alpha \ge 0,\kappa ,\beta ,\delta \ne 0,\mathrm{or},\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & \beta ,\alpha ,\delta \in \mathbb{R},\kappa =\gamma =0,\alpha \ge 0,\beta ,\delta \ne 0,\mathrm{or},\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill & \kappa ,\gamma ,\beta ,\alpha ,\delta \in \mathbb{R},\beta \ne 0,\delta =0,\mathrm{or},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill & \kappa ,\gamma ,\beta ,\alpha ,\delta \in \mathbb{R},\beta ,\delta \ne 0.\hfill \end{array}$$

(5)

On the initial datum, we assume

$$\begin{array}{cc}\hfill & u_0\in H^3\left(\mathbb{R}\right),\mathrm{or},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill & u_0\in H^3\left(\mathbb{R}\right)\cap L^1\left(\mathbb{R}\right),{\int }_{\mathbb{R}}{u}_0\left(x\right)dx=0.\hfill \end{array}$$

(7)

Inspired by [1,2,3,4,5,6,7,8,9,10,11,12], in light of (7), we define the following function:

$$P_0\left(x\right)={\int }_{-\infty }^x{u}_0\left(y\right)dy,$$

(8)

on which we assume

$${∥P_0∥}_{L^2\left(\mathbb{R}\right)}^2={\int }_{\mathbb{R}}{\left({\int }_{-\infty }^x{u}_0\left(y\right)dy\right)}^{2}dx<\infty .$$

(9)

The equation in (1) is derived in [13] to model the evolution of a crystalline surface with small slopes that undergoes faceting. The unknown u gives the surface slope, the constant $\kappa$ is proportional to the atomic flux deposition strength and the convective term $\kappa {\partial }_x{u}^2$ arises from the deposited atoms normal impingement. The sixth-order linear term ${\partial }_x^{6}u$ regularizes the equation, taking into account the surface curvature and the anisotropy of the surface energy under the surface diffusion.

From a mathematical point of view, the existence and uniqueness of weak solutions of (1) with periodic boundary conditions is proven in [14], under the assumptions $\kappa >0$ and $\gamma =0$. In the same setting, a similar result is proven in two space dimensions in [15]. In [16], the authors derived the stationary solutions of (1), again assuming $\kappa >0$ and $\gamma =0$. In [17], the existence of a global-in-time attractor is studied, while the well-posedness of the classical solutions of (1) is proven in [18], requiring (7)–(9), and $\gamma =0$. In this paper, we will prove that, if (2) or (3) hold, we have the well-posedness of the classical solutions of (1) assuming (6), while if (5) holds, we have the well-posedness of (1) assuming (7)–(9).

Taking $\kappa =0$, (1) becomes

$${\partial }_{t}u+\gamma {\partial }_x^{2}u-{\beta }^2{\partial }_x^{6}u+\alpha {\partial }_x^{4}u+{\delta }^2{\partial }_x^4\left(u^3\right)=0,$$

(10)

which is a Cahn-Hilliard type equation [19,20,21]. It was deduced in [22] to describe the evolution of crystal surfaces faceting through surface electromigration. It also describes the phase transition development in ternary oil-water-surfactant systems. One part of the surfactant is hydrophilic, and the other one (termed amphiphile) is lipophilic. Oil, water, and microemulsion (i.e., a homogeneous, isotropic mixture of oil and water) can coexist in equilibrium. The unknown u, in (10), gives the local difference between oil and water concentrations.

From a mathematical point of view, in [23] the initial-boundary-value problem for (10) is analyzed, under appropriate assumptions on $\gamma ,\beta ,\alpha ,\delta$. In [24], the authors analyze the existence of a global-in-time attractor. The existence of weak solutions for the initial-boundary-value problem for (10) is proven in the case of degenerate mobility in [25]. Finally, in [18], the well-posedness of the classical solution of the Cauchy problem of (10) is proven, assuming (7)–(9), with $\gamma =0$. In this paper, we will show that the classical solutions of the Cauchy problem of (10) are well-posed, assuming (6), if $\gamma \le 0$ and $\alpha >0$, while in the general case, we will prove the same result assuming (7)–(9).

Observe that in [13], it is proven that, as $\kappa \to \infty$, (1) reduces

$${\partial }_{t}u+{\partial }_x{u}^2+a_1{\partial }_x^{2}u-b_1^2{\partial }_x^{6}u+c_1{\partial }_x^{4}u=0,$$

(11)

which is known as the Kuramoto-Sivashinsky equation (see [26,27,28]). In Section 4, we will prove the well-posedness of the Cauchy problem for (11), assuming (6).

When $\beta =\delta =0$ and $\alpha =f^2\ne 0$, (1) reads

$${\partial }_{t}u+\kappa {\partial }_x{u}^2+\gamma {\partial }_x^{2}u+f^2{\partial }_x^{4}u=0.$$

(12)

(12) appears in several physical situations; for example, it models long waves on a viscous fluid flowing down an inclined plane [29] and drift waves in a plasma [30]. (12) was also independently deduced by Kuramoto [27,31,32] to describe the phase turbulence in reaction-diffusion systems, and by Sivashinsky [28] 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 (12) can be used to study incipient instabilities in several physical and chemical systems [33,34,35]. Moreover, (12) is also termed the Benney-Lin equation [36,37], and was deduced by Kuramoto as a model for phase turbulence in the Belousov-Zhabotinsky reaction [38].

The existence and the dynamical properties of the exact solutions for (12) can be found in [39,40,41,42,43,44]. The control problem for (12) with periodic boundary conditions, and on a bounded interval, are studied in [45,46,47]. The problem of global-in-time exponential stabilization of (12) with periodic boundary conditions is analyzed in [48]. A generalization of the optimal control theory for (12) is proposed in [49], while the f global boundary controllability of (12) is considered in [50]. The existence of solitonic solutions for (12) is proven in [51]. The well-posedness of the Cauchy problem for (12) is proven in [52,53,54], using the energy space technique, a priori estimates together with an application of the Cauchy-Kovalevskaya and the fixed point method, respectively. Instead, the initial-boundary value problem for (1) is studied, using a priori estimates together with an application of the Cauchy-Kovalevskaya, and the energy space technique in [55,56,57]. Inspired by [58,59,60], the convergence of the solution of (12) to the unique entropy one of the Burgers equation is proven in [61].

Finally, due to its general structure, we conjecture that (1) can have a possible application in machine learning (see [62,63]).

2. Results and Organization of the Paper

In this paper, we improve and complete the results of [14,16,17,18] working with $H^3$ initial data and having general assumptions on the constants appearing in (1). The main result of this paper is the following theorem. We prove the global-in-time existence, uniqueness, and stability of the solutions of the Cauchy problem (1).

Theorem 1.

Fix $T>0$. Assuming one of the following

(i)

(2) and (6),

($ii$)

(3) and (6),

($iii$)

(4) and (6),

($iv$)

(5) and (7),

and (9), there exists a unique solution u of (1) such that

$$u\in H^1((0,T)\times \mathbb{R})\cap L^{\infty }(0,T;H^3\left(\mathbb{R}\right)).$$

(13)

In particular, under the Assumptions (7) and (9), we have that

$${\int }_{\mathbb{R}}u(t,x)dx=0.$$

(14)

Moreover, if $u_1$ and $u_2$ are two solutions of (1), we have that

$${∥u_1(t,·)-u_2(t,·)∥}_{L^2\left(\mathbb{R}\right)}\le e^{C\left(T\right)t}{∥u_{1,0}-u_{2,0}∥}_{L^2\left(\mathbb{R}\right)},$$

(15)

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

The well-posedness of (1) is guaranteed for a short time by the Cauchy-Kowaleskaya Theorem [64]. The solutions are indeed global, thanks to suitable a priori estimates.

The paper is organized as follows. In Section 3 we prove Theorem 1, assuming (i) or ($ii$). In Section 4 we prove Theorem 1, assuming ($iii$). In Section 5 we prove Theorem 1, assuming ($iv$).

3. Proof of Theorem 1 Assuming ($\mathit{i}$) or ($\mathit{ii}$)

In this section, we prove Theorem 1, assuming (i) or ($ii$). For the sake of notational simplicity, define

$$\gamma =-a^2,\alpha =b^2,$$

(16)

and then (1) reads

$$\left\{\begin{array}{cc}{\partial }_{t}u+\kappa {\partial }_x{u}^2-a^2{\partial }_x^{2}u-{\beta }^2{\partial }_x^{6}u+b^2{\partial }_x^{4}u+{\delta }^2{\partial }_x^4\left(u^3\right)=0,\hfill & t>0,x\in \mathbb{R},\hfill \\ u(0,x)=u_0\left(x\right),\hfill & x\in \mathbb{R}.\hfill \end{array}\right.$$

(17)

Since the short time well-posedness of (17) is guaranteed by the Cauchy-Kowaleskaya Theorem [64], here we need to prove some suitable global a priori estimates.

From now on, we denote with $C_0$ the constants which depend only on the initial data, and with $C\left(T\right)$, the constants which depend also on T.

Following [65] (Lemma $2.2$), we begin with the following energy estimate in the space $L^{\infty }(0,\infty ;H^1\left(\mathbb{R}\right))\cap L^2(0,\infty ;H^2\left(\mathbb{R}\right)).$

Lemma 1.

Assuming (2), for each $t>0$, we have that

$$\begin{array}{cc}\hfill \frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +K_1^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+K_2^2{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +a^2{b}^2{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +{\int }_0^t{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dsdx\le C_0,\hfill \end{array}$$

(18)

where $K_1^2,K_2^2$ are two appropriate positive constants.

Assuming (3), for each $t>0$, we have that

$$\begin{array}{cc}\hfill \frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\int }_0^t{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dsdx\le C_0.\hfill \end{array}$$

(19)

Moreover, there exists $C_0>0$, independent on $\kappa ,a,b$, such that, for each $t\ge 0$,

$${∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}\le C_0.$$

(20)

Proof. 

We begin by proving (18). Assume (2). Multiplying (17) by $-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u$, we have that

$$\begin{array}{cc}\hfill & \left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_{t}u+2\kappa \left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right)u{\partial }_{x}u\hfill \\ \hfill & -a^2\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{2}u-{\beta }^2\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{6}u\hfill \\ \hfill & +b^2\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{4}u+{\delta }^2\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^4\left(u^3\right)=0.\hfill \end{array}$$

(21)

Performing some integration by parts, we gain

$$\begin{array}{cc}\hfill & {\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_{t}udx\hfill \\ \hfill & =\frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right),\hfill \\ \hfill & 2\kappa {\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right)u{\partial }_{x}udx=-2\kappa {\beta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx,\hfill \\ \hfill & -a^2{\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{2}udx\hfill \\ \hfill & =a^2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+3a^2{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+a^2{b}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & -{\beta }^2{\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{6}udx\hfill \\ \hfill & =-{\beta }^4{\int }_{\mathbb{R}}{\partial }_x^{3}u{\partial }_x^{5}udx+{\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x\left(u^3\right){\partial }_x^{5}udx+{\beta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{5}udx\hfill \\ \hfill & ={\beta }^4{\int }_{\mathbb{R}}{\left({\partial }_x^{4}u\right)}^{4}dx-{\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{4}udx-{\beta }^2{a}^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx,\hfill \\ \hfill & b^2{\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^{4}udx\hfill \\ \hfill & =-{\beta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx-{\delta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x\left(u^3\right){\partial }_x^{3}udx-b^4{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{3}udx\hfill \\ \hfill & =-{\beta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx+{\delta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{2}udx+b^4{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{2}dx,\hfill \\ \hfill & {\delta }^2{\int }_{\mathbb{R}}\left(-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u\right){\partial }_x^4\left(u^3\right)dx\hfill \\ \hfill & ={\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{3}u{\partial }_x^3\left(u^3\right)dx-{\delta }^4{\int }_{\mathbb{R}}{\partial }_x\left(u^3\right){\partial }_x^3\left(u^3\right)dx-b^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^3\left(u^3\right)dx\hfill \\ \hfill & =-{\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^2\left(u^3\right)dx+{\delta }^4{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx+b^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^2\left(u^3\right)dx.\hfill \end{array}$$

(22)

Therefore, thanks to (22), an integration of (21) on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +a^2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+3a^2{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+a^2{b}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^4{\int }_{\mathbb{R}}{\left({\partial }_x^{4}u\right)}^{4}dx-2{\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{4}udx-2{\beta }^2{a}^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx\hfill \\ \hfill & +2{\delta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{2}udx+b^4{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{2}dx+{\delta }^4{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx\hfill \\ \hfill & =2\kappa {\beta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx.\hfill \end{array}$$

(23)

Since

$$\begin{array}{cc}\hfill & b^4{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{2}dx+{\delta }^4{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx+{\beta }^4{\int }_{\mathbb{R}}{\left({\partial }_x^{4}u\right)}^{4}dx\hfill \\ \hfill & +2{\delta }^2{b}^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{2}udx-2{\beta }^2{a}^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx-2{\beta }^2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^2\left(u^3\right){\partial }_x^{4}udx\hfill \\ \hfill & ={\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dx,\hfill \end{array}$$

(23) becomes

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +a^2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+3a^2{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+a^2{b}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dx=2\kappa {\beta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx.\hfill \end{array}$$

(24)

Due to the Young inequality,

$$\begin{array}{cc}\hfill 2\left|\kappa \right|{\beta }^2{\int }_{\mathbb{R}}\left|u\right|{\partial }_{x}u\left|\right|{\partial }_x^{2}u|dx=& 2{\beta }^2{\int }_{\mathbb{R}}\left|\frac{\kappa u{\partial }_{x}u}{\sqrt{D_1}}\right|\left|\sqrt{D_1}{\partial }_x^{2}u\right|dx\hfill \\ \hfill \le & \frac{{\beta }^2{\kappa }^2}{D_1}{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_1{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

where $D_1$ is a positive constant, which will be specified later. Consequently, by (24),

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +{\beta }^2\left(a^2-D_1\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\left(3a^2{\delta }^2-\frac{{\beta }^2{\kappa }^2}{D_1}\right){∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +a^2{b}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dx\le 0.\hfill \end{array}$$

(25)

We search $D_1$, such that

$$a^2-D_1>0,3a^2{\delta }^2-\frac{{\beta }^2{\kappa }^2}{D_1}>0.$$

(26)

Since, after a rescaling, $\left|a\right|$ can be taken very big, $D_1$ does exist and (26) holds.

Therefore, by (25) and (26),

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +K_1^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+K_2^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+a^2{b}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dx\le 0,\hfill \end{array}$$

where $K_1^2,K_2^2$ are two appropriate positive constants. Integrating on $(0,t)$, by (6), we have that

$$\begin{array}{cc}\hfill \frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +K_1^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+K_2^2{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +a^2{b}^2{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +{\int }_0^t{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dsdx\le C_0,\hfill \end{array}$$

that is, (18).

We continue by proving (19). Assume (3). Since $\kappa =\gamma =0$, (17) reads

$${\partial }_{t}u-{\beta }^2{\partial }_x^{6}u+b^2{\partial }_x^{4}u+{\delta }^2{\partial }_x^4\left(u^3\right)=0.$$

(27)

Multiplying (27), by $-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+b^{2}u$, arguing as in the previous case, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{b^2}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\left[b^2{\partial }_x^{2}u+{\delta }^2{\partial }_x^2\left(u^3\right)-{\beta }^2{\partial }_x^{4}u\right]}^{2}dx=0.\hfill \end{array}$$

(6) and an integration on $(0,t)$ give (19).

Finally, we prove (20). Thanks to (18) or (19) and the Hölder inequality,

$$\begin{array}{cc}\hfill {\left|u(t,x)\right|}^3=& 3\left|{\int }_{-\infty }^x{u}^2{\partial }_{x}udy\right|\le 3{\int }_{\mathbb{R}}{u}^2\left|{\partial }_{x}u\right|dx\hfill \\ \hfill \le & 3{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0.\hfill \end{array}$$

Hence,

$${∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^3\le C_0,$$

which gives (20). □

We continue by proving an $L^2-$ estimate, which is independent on $\kappa ,a,b$.

Lemma 2.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$, such that

$$\begin{array}{cc}\hfill {∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{\int }_{\mathbb{R}}{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+{\beta }^2{\int }_{\mathbb{R}}{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \\ \hfill +2b^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \le C\left(T\right),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill {\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \le C\left(T\right),\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill {\int }_0^t{∥u(s,·){\left({\partial }_{x}u(s,·)\right)}^2∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \le C\left(T\right),\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill {\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds& \le C\left(T\right),\hfill \end{array}$$

(31)

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

Proof. 

Let $0\le t\le T$. We begin by observing that

$${\partial }_x{u}^3=3u^2{\partial }_{x}u,{\partial }_x^2\left(u^3\right)=6u{\left({\partial }_{x}u\right)}^2+3u^2{\partial }_x^{2}u.$$

(32)

Multiplying (17) by $2u$, an integration on $\mathbb{R}$ and several integrations by part give

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=& 2{\int }_{\mathbb{R}}u{\partial }_{t}udx\hfill \\ \hfill =& -4\kappa {\int }_{\mathbb{R}}{u}^2{\partial }_{x}dx+2a^2{\int }_{\mathbb{R}}u{\partial }_x^{2}udx+2{\beta }^2{\int }_{\mathbb{R}}u{\partial }_x^{6}udx\hfill \\ \hfill & -2b^2{\int }_{\mathbb{R}}u{\partial }_x^{4}udx-2{\delta }^2{\int }_{\mathbb{R}}u{\partial }_x^4\left(u^3\right)dx\hfill \\ \hfill =& -2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{5}udx+2b^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{3}udx\hfill \\ \hfill & +2{\delta }^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^3\left(u^3\right)dx\hfill \\ \hfill =& -2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx-2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^2\left(u^3\right)dx\hfill \\ \hfill =& -2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^2\left(u^3\right)dx,\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=-2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^2\left(u^3\right)dx.\hfill \end{array}$$

(33)

Due to the Young inequality, we can estimate the right-hand side of (33), as follows:

$$2{\delta }^2{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_x^2\left(u^3\right)|dx\le {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\delta }^4{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx.$$

Consequently, (33) becomes

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\delta }^4{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx.\hfill \end{array}$$

(34)

Observe that, by (32),

$$\begin{array}{cc}\hfill {\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx=& 36{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx+9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+36{\int }_{\mathbb{R}}{u}^3{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}udx\hfill \\ \hfill =& 36{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx+9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-36{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx\hfill \\ \hfill =& 9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=9{\int }_{\mathbb{R}}{u}^4{\left({\partial }_x^{2}u\right)}^{2}dx.\hfill \end{array}$$

(35)

Using (20),

$$\begin{array}{cc}\hfill {\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx\le & 9{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^4{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(36)

It follows from (34) and (36) that

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(37)

Since

$$C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=C_0{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{2}udx=-C_0{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{3}udx,$$

Lemma 1 and the Young inequality give

$$\begin{array}{cc}\hfill C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le & 2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_{x}u}{2\beta }\right|\left|\beta {\partial }_x^{3}u\right|dx\hfill \\ \hfill \le & C_0{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill \le & C_0+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(38)

Consequently, by (37),

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right).\hfill \end{array}$$

Integrating on $(0,t)$, by (6), we have that

$$\begin{array}{cc}\hfill {∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+{\beta }^2{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +2b^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C_0+C\left(T\right)t\le C\left(T\right),\hfill \end{array}$$

which gives (28).

(29) follows from (28), (38) and an integration on $(0,t)$.

We prove (30). We begin by observing that, by (35) and (36), we have that

$$\begin{array}{cc}\hfill 36{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx\le & C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -36{\int }_{\mathbb{R}}{u}^3{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}udx.\hfill \end{array}$$

(39)

Thanks to the Young inequality,

$$36{\int }_{\mathbb{R}}{\left|u\right|}^3{\left({\partial }_{x}u\right)}^2\left|{\partial }_x^{2}u\right|dx\le 18{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx+18{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Consequently, by (39),

$$18{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{4}dx\le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

(40)

Observe that, by (20),

$$9{∥u^2(t,·){\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=9{\int }_{\mathbb{R}}{u}^4{\left({\partial }_x^{2}u\right)}^{2}dx\le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Therefore, by (40),

$$18{∥u(t,·){\left({\partial }_{x}u(t,·)\right)}^2∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

(41)

Integrating (41) on $(0,t)$, by (29), we have (30).

Finally, we prove (31). We begin by observing that [66] (Lemma $2.3$) says that

$${∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\le 6\left({∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Therefore, by Lemma 1 and (28), we have that

$${∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\le C\left(T\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Integrating on $(0,t)$, by (29), we have (31). □

We continue with an a priori estimate in the space $L^2(0,\infty ;H^4\left(\mathbb{R}\right))$.

Lemma 3.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$, such that

$$\begin{array}{cc}\hfill {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+{\beta }^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +2b^2{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right)\hfill \end{array}$$

(42)

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

Proof. 

Let $0\le t\le T$. Multiplying (17) by $-2{\partial }_x^{2}u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=& -2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_{t}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx-2a^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{6}udx\hfill \\ \hfill & +2b^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx+2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^4\left(u^3\right)dx\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^{3}dx-2a^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{3}u{\partial }_x^{5}udx\hfill \\ \hfill & -2b^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{3}u{\partial }_x^3\left(u^3\right)dx\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^{3}dx-2a^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2b^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^2\left(u^3\right)dx,\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=2\kappa {\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^{3}dx+2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^2\left(u^3\right)dx.\hfill \end{array}$$

(43)

Due to Lemma 1, (36) and the Young inequality,

$$\begin{array}{cc}\hfill & 2\left|\kappa \right|{\int }_{\mathbb{R}}{\left|{\partial }_{x}u\right|}^{3}dx\le {\kappa }^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\hfill \\ \hfill & \le C_0+{∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4,\hfill \\ \hfill & 2{\delta }^2{\int }_{\mathbb{R}}|{\partial }_x^{4}u\left|\right|{\partial }_x^2\left(u^3\right)|dx=2{\int }_{\mathbb{R}}\left|\beta {\partial }_x^{4}u\right|\left|\frac{{\delta }^2{\partial }_x^2\left(u^3\right)}{\beta }\right|dx\hfill \\ \hfill & \le {\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^4}{{\beta }^2}{\int }_{\mathbb{R}}{\left[{\partial }_x^2\left(u^3\right)\right]}^{2}dx\hfill \\ \hfill & \le {\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Therefore, by (43),

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le {∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (6), (29) and (31),

$$\begin{array}{cc}\hfill {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+{\beta }^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +2b^2{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill \le & C_0+{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),\hfill \end{array}$$

which gives (42). □

We continue with an a priori estimate in the space $L^2(0,\infty ;H^5\left(\mathbb{R}\right))$.

Lemma 4.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$,

$$\begin{array}{cc}\hfill {∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}& \le C\left(T\right),\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \\ \hfill +{\beta }^2{\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+2b^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \le C\left(T\right),\hfill \end{array}$$

(45)

for every $0\le t\le T$, where $C\left(T\right)$ is independent on $\kappa ,a,b$.

Proof. 

Let $0\le t\le T$. We begin by observing that, by (32), we have that

$${\partial }_x^3\left(u^3\right)=6{\left({\partial }_{x}u\right)}^3+18u{\partial }_{x}u{\partial }_x^{2}u+3u^2{\partial }_x^{3}u.$$

(46)

Multiplying (17) by $2{\partial }_x^{4}u$, thanks to (46), an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=& 2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_{t}udx\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx+2a^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^{6}udx\hfill \\ \hfill & -2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^4\left(u^3\right)dx\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx-2a^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{5}u{\partial }_x^3\left(u^3\right)dx\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx-2a^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-12{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^3{\partial }_x^{5}udx-36{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}u{\partial }_x^{5}udx\hfill \\ \hfill & -6{\int }_{\mathbb{R}}{u}^2{\partial }_x^{3}u{\partial }_x^{5}udx,\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =-2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx-12{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^3{\partial }_x^{5}udx\hfill \\ \hfill & -36{\delta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}u{\partial }_x^{5}udx-6{\delta }^2{\int }_{\mathbb{R}}{u}^2{\partial }_x^{3}u{\partial }_x^{5}udx.\hfill \end{array}$$

(47)

Due to Lemma 1 and the Young inequality,

$$\begin{array}{cc}\hfill & 2\left|\kappa \right|{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_x^{4}u|dx\le {\kappa }^2{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{2}dx+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le {\kappa }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C_0{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 12{\delta }^2{\int }_{\mathbb{R}}|{\partial }_x{u|}^3\left|{\partial }_x^{5}u\right|dx=2{\int }_{\mathbb{R}}\left|\frac{6{\delta }^2{\left({\partial }_{x}u\right)}^3}{\beta \sqrt{D_2}}\right|\left|\beta \sqrt{D_2}{\partial }_x^{5}u\right|dx\hfill \\ \hfill & \le \frac{36{\delta }^4}{{\beta }^2{D}_2}{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^{6}dx+{\beta }^2{D}_2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{36{\delta }^4}{{\beta }^2{D}_2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+{\beta }^2{D}_2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 36{\delta }^2{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u{\partial }_x^{2}u\left|\right|{\partial }_x^{5}u|dx\le 36{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}|{\partial }_{x}u{\partial }_x^{2}u\left|\right|{\partial }_x^{5}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_{x}u{\partial }_x^{2}u\left|\right|{\partial }_x^{5}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_{x}u{\partial }_x^{2}u}{\beta \sqrt{D_2}}\right|\left|\beta \sqrt{D_2}{\partial }_x^{5}u\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_2}{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\left({\partial }_x^{2}u\right)}^{2}dx+{\beta }^2{D}_2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{C_0}{D_2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 6{\delta }^2{\int }_{\mathbb{R}}{u}^2|{\partial }_x^{3}u\left|\right|{\partial }_x^{5}u|dx\le 6{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^2{\int }_{\mathbb{R}}|{\partial }_x^{3}u\left|\right|{\partial }_x^{5}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_x^{3}u\left|\right|{\partial }_x^{5}u|dx\le 2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_x^{3}u}{\beta \sqrt{D_2}}\right|\left|\beta \sqrt{D_2}{\partial }_x^{5}u\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_2}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

where $D_2$ is a positive constant, which will be specified later. Therefore, by (47),

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^2\left(2-3D_2\right){∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C_0+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{36{\delta }^4}{{\beta }^2{D}_2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\hfill \\ \hfill & +\frac{C_0}{D_2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{C_0}{D_2}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

and taking $D_2=\frac{1}{3}$

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C_0+{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{108{\delta }^4}{{\beta }^2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\hfill \\ \hfill & +C_0{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (6), (29), (31) and (42), we obtain that

$$\begin{array}{cc}\hfill & {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{\int }_0^t{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^2{\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+2b^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C_{0}t+{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +\frac{108{\delta }^4}{{\beta }^2}{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds\hfill \\ \hfill & +C_0{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C\left(T\right)\left(1+{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2\right).\hfill \end{array}$$

(48)

We prove (44). Thanks to Lemma 1, (48) and the Hölder inequality,

$$\begin{array}{cc}\hfill {\left({\partial }_{x}u(t,x)\right)}^2=& 2{\int }_{-\infty }^x{\partial }_{x}u{\partial }_x^{2}dx\le 2{\int }_{\mathbb{R}}|{\partial }_x\left|\right|{\partial }_x^{2}u|dx\le 2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}\hfill \\ \hfill \le & C\left(T\right)\sqrt{\left(1+{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2\right)}.\hfill \end{array}$$

Hence,

$${∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^4-C\left(T\right){∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2-C\left(T\right)\le 0,$$

which gives (44).

Finally, (45) follows from (44) and (48). □

We continue with an a priori estimate in the space $L^4(0,\infty ;W^{2,4}\left(\mathbb{R}\right))$.

Lemma 5.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$, such that

$${\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds\le C\left(T\right),$$

(49)

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

Proof. 

Let $0\le t\le T$. We begin by observing that

$${\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{4}dx={\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^3{\partial }_x^{2}udx=-3{\int }_{\mathbb{R}}{\partial }_{x}u{\left({\partial }_x^{2}u\right)}^2{\partial }_x^{3}udx.$$

(50)

Due to the Young inequality,

$$3{\int }_{\mathbb{R}}\left|u\right|{\partial }_{x}u{\left({\partial }_x^{2}u\right)}^2\left|{\partial }_x^{3}u\right|dx\le \frac{9}{2}{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\left({\partial }_x^{3}u\right)}^{2}dx+\frac{1}{2}{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{4}dx.$$

It follows from (50) that

$${\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^{4}dx\le 9{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\left({\partial }_x^{3}u\right)}^{2}dx.$$

By (44), we have that

$${∥{\partial }_x^{2}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\le 9{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right){∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Integrating on $(0,t)$, by (28), we have that

$${\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds\le C\left(T\right){\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

which gives (49). □

We continue with an a priori estimate in the space $L^2(0,\infty ;H^6\left(\mathbb{R}\right))$.

Lemma 6.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$, such that

$$\begin{array}{cc}\hfill {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2a^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +{\beta }^2{\int }_0^t{∥{\partial }_x^{6}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+2b^2{\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right)\hfill \end{array}$$

(51)

In particular, we have that

$${∥{\partial }_x^{2}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}\le C\left(T\right),$$

(52)

where $C\left(T\right)$ is independent on $\kappa ,a,b$.

Proof. 

Let $0\le t\le T$. We begin by observing that, by (46),

$${\partial }_x^4\left(u^3\right)=36{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}u+18u{\left({\partial }_x^{2}u\right)}^2+24u{\partial }_{x}u{\partial }_x^{3}u+3u^2{\partial }_x^{4}u$$

(53)

Multiplying (17) by $-2{\partial }_x^{6}u$, thanks to (46), an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=& -2{\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_{t}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx-2a^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{6}udx-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2b^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^{6}udx+2{\delta }^2{\int }_{\mathbb{R}}{\partial }_x^{5}u{\partial }_x^4\left(u^3\right)dx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx+2a^2{\int }_{\mathbb{R}}{\partial }_x^{3}u{\partial }_x^{5}udx-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2b^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}u{\partial }_x^{6}udx\hfill \\ \hfill & +36{\delta }^2{\int }_{\mathbb{R}}u{\left({\partial }_x^{2}u\right)}^2{\partial }_x^{6}udx+48{\delta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{3}u{\partial }_x^{6}udx\hfill \\ \hfill & +6{\delta }^2{\int }_{\mathbb{R}}{u}^2{\partial }_x^{4}u{\partial }_x^{6}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx-2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & -2b^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}u{\partial }_x^{6}udx\hfill \\ \hfill & +36{\delta }^2{\int }_{\mathbb{R}}u{\left({\partial }_x^{2}u\right)}^2{\partial }_x^{6}udx+48{\delta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{3}u{\partial }_x^{6}udx\hfill \\ \hfill & +6{\delta }^2{\int }_{\mathbb{R}}{u}^2{\partial }_x^{4}u{\partial }_x^{6}udx.\hfill \end{array}$$

Therefore, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx+72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}u{\partial }_x^{6}udx\hfill \\ \hfill & +36{\delta }^2{\int }_{\mathbb{R}}u{\left({\partial }_x^{2}u\right)}^2{\partial }_x^{6}udx+48{\delta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{3}u{\partial }_x^{6}udx+6{\delta }^2{\int }_{\mathbb{R}}{u}^2{\partial }_x^{4}u{\partial }_x^{6}udx.\hfill \end{array}$$

(54)

Due to Lemma 1, (44), (45) and the Young inequality,

$$\begin{array}{cc}\hfill & 4\left|\kappa \right|{\int }_{\mathbb{R}}|u{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx\le 4|\kappa |{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_{x}u}{\beta \sqrt{D_3}}\right|\left|\beta \sqrt{D_3}{\partial }_x^{6}udx\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_3}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{C_0}{D_3}+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2|{\partial }_x^{2}u\left|\right|{\partial }_x^{6}u|dx\le 72{\delta }^2{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_x^{6}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_x^{2}u}{\beta \sqrt{D_3}}\right|\left|\beta \sqrt{D_3}{\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 36{\delta }^2{\int }_{\mathbb{R}}\left|u\right|{\left({\partial }_x^{2}u\right)}^2|{\partial }_x^{6}u|dx=36{\delta }^2{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^2\left|{\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^2\left|{\partial }_x^{6}u\right|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\left({\partial }_x^{2}u\right)}^2}{\beta \sqrt{D_3}}\right|\left|\beta \sqrt{D_3}{\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 48{\delta }^2{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_x^{3}u\left|\right|{\partial }_x^{6}u|dx\le 48{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{3}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{3}u\left|\right|{\partial }_x^{6}u|dx\le 2C_0{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_x^{3}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_x^{3}u\left|\right|{\partial }_x^{6}u|dx\le 2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_x^{3}u}{\beta \sqrt{D_3}}\right|\left|\beta \sqrt{D_3}{\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 6{\delta }^2{\int }_{\mathbb{R}}{u}^2|{\partial }_x^{4}u\left|\right|{\partial }_x^{6}u|dx\le 6{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^2{\int }_{\mathbb{R}}|{\partial }_x^{4}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_x^{4}u\left|\right|{\partial }_x^{6}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_x^{4}u}{\beta \sqrt{D_3}}\right|\left|\beta \sqrt{D_3}{\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_3}{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{D}_3{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

where $D_3$ is a positive constant, which will be specified later. It follows from (54) that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^2\left(2-5D_3\right){∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}+\frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{C\left(T\right)}{D_3}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{C_0}{D_3}{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Taking $D_3=\frac{1}{5}$, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2b^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){∥{\partial }_x^{2}u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+C\left(T\right){∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (6), (28), (42), (49), we have that

$$\begin{array}{cc}\hfill & {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2a^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +{\beta }^2{\int }_0^t{∥{\partial }_x^{6}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+2b^2{\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C\left(T\right){\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C\left(T\right){\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t∥{\partial }_x^{4}u(s,·)∥ds\hfill \\ \hfill & \le C\left(T\right),\hfill \end{array}$$

which gives (51).

Finally, we prove (52). Thanks to (45), (51) and the Hölder inequality,

$$\begin{array}{cc}\hfill {\left({\partial }_x^{2}u(t,x)\right)}^2=& 2{\int }_{-\infty }^x{\partial }_x^{2}u{\partial }_x^{3}udy\le 2{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_x^{3}u|dx\hfill \\ \hfill \le & 2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}\le C\left(T\right).\hfill \end{array}$$

Hence,

$${∥{\partial }_x^{2}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2\le C\left(T\right),$$

which gives (52). □

We continue with an a priori estimate in the space $H^1((0,\infty )\times \mathbb{R})$.

Lemma 7.

Fix $T>0$ and assume (2) or (3). There exists a constant $C\left(T\right)>0$, independent on $\kappa ,a,b$, such that

$$\begin{array}{cc}\hfill a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\int }_0^t{∥{\partial }_{t}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),\hfill \end{array}$$

(55)

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

Proof. 

Let $0\le t\le T$. Multiplying (17) by $2{\partial }_{t}u$, we have that

$$\begin{array}{cc}\hfill 2{\left({\partial }_{t}u\right)}^2& +4\kappa u{\partial }_{x}u{\partial }_{t}u-2a^2{\partial }_x^{2}u{\partial }_{t}u-2{\beta }^2{\partial }_x^{6}u{\partial }_{t}u+2b^2{\partial }_x^{4}u{\partial }_{t}u+2{\delta }^2{\partial }_{t}u{\partial }_x^4\left(u^3\right)=0.\hfill \end{array}$$

(56)

Since,

$$\begin{array}{cc}\hfill -2a^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_{t}u=& a^2\frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill -2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_{t}udx=& {\beta }^2\frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill 2b^2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_{t}udx=& b^2\frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

thanks to (53), an integration of (56) on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)+b^2\frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =-4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_{t}udx-72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2{\partial }_x^{2}u{\partial }_{t}udx\hfill \\ \hfill & +36{\delta }^2{\int }_{\mathbb{R}}u{\left({\partial }_x^{2}u\right)}^2{\partial }_{t}udx-48{\delta }^2{\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}u{\partial }_{t}udx-6{\delta }^2{\int }_{\mathbb{R}}{u}^2{\partial }_x^{4}u{\partial }_{t}udx.\hfill \end{array}$$

(57)

Due to Lemma 1, (44), (45), (52) and the Young inequality,

$$\begin{array}{cc}\hfill & 4\left|\kappa \right|{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_{t}u|dx\le 4|\kappa |{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_{t}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_{x}u}{\sqrt{D_3}}\right|\left|\sqrt{D_3}{\partial }_{t}u\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_3}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \frac{C_0}{D_3}+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 72{\delta }^2{\int }_{\mathbb{R}}{\left({\partial }_{x}u\right)}^2|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\le 72{\delta }^2{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_x^{2}u}{\sqrt{D_3}}\right|\left|\sqrt{D_3}{\partial }_{t}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \frac{C\left(T\right)}{D_3}+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 36{\delta }^2{\int }_{\mathbb{R}}\left|u\right|{\left({\partial }_x^{2}u\right)}^2|{\partial }_{t}u|dx\le 36{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^2\left|{\partial }_{t}u\right|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}{\left({\partial }_x^{2}u\right)}^2|{\partial }_{t}u|dx=2C_0{∥{\partial }_x^{2}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_x^{2}u}{\sqrt{D_3}}\right|\left|\sqrt{D_3}{\partial }_{t}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \frac{C\left(T\right)}{D_3}+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 48{\delta }^2{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\le 48{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\le 2C_0{∥{\partial }_{x}u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_x^{2}u\left|\right|{\partial }_{t}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_x^{2}u}{\sqrt{D_3}}\right|\left|\sqrt{D_3}{\partial }_x^{2}u\right|dx\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \frac{C\left(T\right)}{D_3}+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill & 6{\delta }^2{\int }_{\mathbb{R}}{u}^2|{\partial }_x^{4}u\left|\right|{\partial }_{t}u|dx=6{\delta }^2{∥u(t,·)∥}_{L^{\infty }\left(\mathbb{R}\right)}^2{\int }_{\mathbb{R}}|{\partial }_x^{4}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C_0{\int }_{\mathbb{R}}|{\partial }_x^{4}u\left|\right|{\partial }_{t}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_x^{4}u}{\sqrt{D_3}}\right|\left|\sqrt{D_3}{\partial }_{t}u\right|dx\hfill \\ \hfill & \le \frac{C_0}{D_3}{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_3{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

where $D_3$ is a positive constant, which will be specified later. As a consequence, (57) becomes

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +b^2\frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left(1-2D_3\right){∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le \frac{C\left(T\right)}{D_3}+\frac{C_0}{D_3}{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

and taking $D_3=\frac{1}{2}$,

$$\begin{array}{cc}\hfill \frac{d}{dt}\left(a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)& +b^2\frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill \le & C\left(T\right)+C_0{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (6) and (42), we have that

$$\begin{array}{cc}\hfill & a^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+b^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\int }_0^t{∥{\partial }_{t}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C\left(T\right)t+C_0{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),\hfill \end{array}$$

which gives (55). □

We are finally ready to prove Theorem 1, assuming (i) or ($ii$).

Proof of Theorem 1 assuming ($\mathit{i}$) or ($\mathit{ii}$).

The well-posedness of (1) is guaranteed for a short time by the Cauchy-Kowaleskaya Theorem [64]. Thanks to the a priori estimates proved in Lemmas 1–7, we have that the global-in-time existence of a is the solution of (1) that satisfies (13).

The stability estimates (15) can be proved using the same arguments of [18] (Theorem 1). □

4. Proof of Theorem 1 Assuming ($\mathit{iii}$)

In this section, we prove Theorem 1 assuming ($iii$). Due to (4), here, (1) becomes

$$\left\{\begin{array}{cc}{\partial }_{t}u+\kappa {\partial }_x{u}^2+\gamma {\partial }_x^{2}u-{\beta }^2{\partial }_x^{6}u+\alpha {\partial }_x^{4}u=0,\hfill & t>0,x\in \mathbb{R},\hfill \\ u(0,x)=u_0\left(x\right),\hfill & x\in \mathbb{R}.\hfill \end{array}\right.$$

(58)

The argument of this section is analogous to that of the previous one. We deduce the local-in-time well-posedness from the Cauchy-Kowaleskaya Theorem [64], and we improve the local-in-time existence to the global-in-time one, proving some suitable a priori estimates on u.

We begin with an energy estimate in the space $L_{loc}^{\infty }(0,\infty ;L^2\left(\mathbb{R}\right))\cap L_{loc}^2(0,\infty ;H^3\left(\mathbb{R}\right))$.

Lemma 8.

Fix $T>0$. There exists a constant $C\left(T\right)>0$, such that

$${∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{e}^{C_{0}t}{\int }_0^t{e}^{-C_s}{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

(59)

for every $0\le t\le T$. In particular, (29) holds. Moreover, we have that

$${\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

(60)

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

Proof. 

Let $0\le t\le T$. Multiplying (58) by $2u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=& 2{\int }_{\mathbb{R}}u{\partial }_{t}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}{u}^2{\partial }_{x}udx-2\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx+2{\beta }^2{\int }_{\mathbb{R}}u{\partial }_x^{6}udx-2\alpha {\int }_{\mathbb{R}}u{\partial }_x^{4}udx\hfill \\ \hfill =& -2\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx-2{\beta }^2{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{5}udx+2\alpha {\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{3}udx\hfill \\ \hfill =& -2\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx-2\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill =& -2\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx-2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

that is,

$$\frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=-2\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx-2\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

(61)

Since, using the Young inequality,

$$\begin{array}{c}\hfill 2\left|\gamma \right|{\int }_{\mathbb{R}}\left|u\right||{\partial }_x^{2}u|dx\le {\gamma }^2{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\end{array}$$

we can pass from (61) to

$$\frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

(62)

Observe that

$$C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=C_0{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{2}udx=-C_0{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_x^{3}udx.$$

Therefore, by the Young inequality,

$$\begin{array}{cc}\hfill C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le & 2{\int }_{\mathbb{R}}\left|\frac{C_0{\partial }_{x}u}{2\sqrt{D_4}}\right|\left|\sqrt{D_4}{\partial }_x^{3}u\right|dx\hfill \\ \hfill \le & \frac{C_0}{D_4}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_4{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

(63)

where $D_4$ is a positive constant, which will be specified later. Observe again that

$$\frac{C_0}{D_4}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=\frac{C_0}{D_4}{\int }_{\mathbb{R}}{\partial }_{x}u{\partial }_{x}udx=-\frac{C_0}{D_4}{\int }_{\mathbb{R}}u{\partial }_x^{2}udx.$$

Consequently, by the Young inequality,

$$\begin{array}{cc}\hfill \frac{C_0}{D_4}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le & 2{\int }_{\mathbb{R}}\left|\frac{C_0{D}_{4}u}{2\sqrt{D_5}}\right|\left|\sqrt{D_5}{\partial }_x^{2}u\right|dx\hfill \\ \hfill \le & \frac{C_0{D}_4^2}{D_5}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_5{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \end{array}$$

(64)

where $D_5$ is a positive constant, which will be specified later. Il follows from (63) and (64) that

$$\left(C_0-D_5\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \frac{C_0{D}_4^2}{D_5}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_4{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Choosing

$$D_5=\frac{C_0}{2},$$

(65)

we have that

$$\frac{C_0}{2}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le 2D_4^2{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+D_4{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,$$

that is

$$C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le 4D_4^2{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2D_4{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2$$

(66)

It follows from (62) and (66) that

$$\frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left({\beta }^2-D_4\right){∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le \left(C_0+4D_4^2\right){∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Choosing

$$D_4=\frac{{\beta }^2}{2},$$

(67)

we have that

$$\frac{d}{dt}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C_0{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

By the the Gronwall Lemma and (6), we get

$${∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{e}^{C_{0}t}{\int }_0^t{e}^{-C_{0}s}{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C_0{e}^{C_{0}t}\le C\left(T\right),$$

which gives (59).

We prove (29). Thanks to (59), (66) and (67),

$$C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right)+\frac{{\beta }^2}{2}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Integrating on $(0,t)$, by (59), we have that

$$C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right)t+\frac{{\beta }^2}{2}{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

which gives (29).

Finally, we prove (60). Thanks to (59), (64) and (67),

$$C_0{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right)+C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

Integrating on $(0,t)$, by (29), we have that

$$C_0{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right)t+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

which gives (60). □

We continue with an energy estimate in the space $L_{loc}^{\infty }(0,\infty ;H^1\left(\mathbb{R}\right))\cap L_{loc}^2(0,\infty ;H^4\left(\mathbb{R}\right))$.

Lemma 9.

Fix $T>0$. There exists a constant $C\left(T\right)>0$, such that

$$\begin{array}{cc}\hfill {∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}& \le C\left(T\right),\hfill \end{array}$$

(68)

$$\begin{array}{cc}\hfill {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds& \le C\left(T\right),\hfill \end{array}$$

(69)

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

Proof. 

Let $0\le t\le T$. Multiplying (58) by $-2{\partial }_x^{2}u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}& {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=-2{\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_{t}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx+2\gamma {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_x^{2}udx+2\alpha {\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_x^{2}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx+2\gamma {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{5}u{\partial }_x^{3}udx-2\alpha {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx+2\gamma {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Consequently, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx+2\gamma {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(70)

Due to the Young inequality,

$$\begin{array}{cc}\hfill 4\kappa {\int }_{\mathbb{R}}|u{\partial }_{x}u\left|\right|{\partial }_x^{2}u|dx\le & 2{\kappa }^2{\int }_{\mathbb{R}}{u}^2{\left({\partial }_{x}u\right)}^{2}dx+2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill \le & 2{\kappa }^2{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Therefore, by (70),

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le 2{\kappa }^2{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2(1+|\gamma \left|\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left|\alpha \right|{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C_0{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(6), (29), (59), (60) and an integration on $(0,t)$ give

$$\begin{array}{cc}\hfill & {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C_0{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C\left(T\right)\left(1+{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2\right).\hfill \end{array}$$

(71)

We prove (68). Thanks to (59), (69), and the Hölder inequality,

$$\begin{array}{cc}\hfill u^2(t,x)=& 2{\int }_{-\infty }^{x}u{\partial }_{x}udy\le 2{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u|dx\hfill \\ \hfill \le & {∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}\le C\left(T\right)\sqrt{\left(1+{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2\right)}.\hfill \end{array}$$

Hence,

$${∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^4-C\left(T\right){∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}^2-C\left(T\right)\le 0,$$

which gives (68).

Finally, (69) follows form (68) and (71). □

We continue with an energy estimate in the space $L_{loc}^{\infty }(0,\infty ;H^2\left(\mathbb{R}\right))\cap L_{loc}^2(0,\infty ;H^5\left(\mathbb{R}\right))$.

Lemma 10.

Fix $T>0$. There exists a constant $C\left(T\right)>0$, such that

$${∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{\int }_{\mathbb{R}}{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

(72)

for every $0\le t\le T$. In particular, (44) holds.

Proof. 

Let $0\le t\le T$. Multiplying (58) by $2{\partial }_x^{4}u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}& {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=2{\int }_{\mathbb{R}}{\partial }_x^{4}u{\partial }_{t}udx\hfill \\ \hfill =& -4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx-2\gamma {\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^{4}udx+2{\beta }^2{\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_x^{4}udx-2\alpha {∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill =& -4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx+2\gamma {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Therefore, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =-4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{4}udx+2\gamma {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(73)

Due to (68), (69) and the Young inequality,

$$\begin{array}{cc}\hfill & 4\left|\kappa \right|{\int }_{\mathbb{R}}|u{\partial }_{x}u\left|\right|{\partial }_x^{4}u|dx\le 4|\kappa |{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{4}u|dx\hfill \\ \hfill & \le C\left(T\right){\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{4}u|dx\le C\left(T\right){∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C\left(T\right){∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

It follows from (73) that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left|\gamma \right|{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left|\alpha \right|{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C\left(T\right){∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (6), (59) and (69), we have

$$\begin{array}{cc}\hfill & {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C\left(T\right)t+C\left(T\right){\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C\left(T\right){\int }_0^t{∥{\partial }_x^{3}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C\left(T\right),\hfill \end{array}$$

which gives (72).

Finally, by (69), (72) and the Hölder inequality, we have (44). □

We continue with an energy estimate in the space $L_{loc}^{\infty }(0,\infty ;H^3\left(\mathbb{R}\right))\cap L_{loc}^2(0,\infty ;H^6\left(\mathbb{R}\right))$.

Lemma 11.

Fix $T>0$. There exists a constant $C\left(T\right)>0$, such that

$${∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{\int }_0^t{∥{\partial }_x^{6}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right),$$

(74)

for every $0\le t\le T$. In particular, (52) holds.

Proof. 

Let $0\le t\le T$. Multiplying (58) by $-2{\partial }_x^{6}u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}& {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=-2{\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_{t}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx+2\gamma {\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_x^{2}udx-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\alpha {\int }_{\mathbb{R}}{\partial }_x^{6}u{\partial }_x^{4}udx\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx-2\gamma {\int }_{\mathbb{R}}{\partial }_x^{5}u{\partial }_x^{3}udx-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill =& 4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx+2\gamma {∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Therefore, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & =4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{6}udx+2\gamma {∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-2\alpha {∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(75)

Due to (68), (69) and the Young inequality,

$$\begin{array}{cc}\hfill & 4\left|\kappa \right|{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx\le 2|\kappa |{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_x^{6}u|dx=2{\int }_{\mathbb{R}}\left|\frac{C\left(T\right){\partial }_{x}u}{\beta }\right|\left|\beta {\partial }_x^{6}u\right|dx\hfill \\ \hfill & \le C\left(T\right){∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

It follows from (75) that

$$\begin{array}{cc}\hfill & \frac{d}{dt}{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{6}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+2\left|\gamma \right|{∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2\left|\alpha \right|{∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){∥{\partial }_x^{4}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C\left(T\right){∥{\partial }_x^{5}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

By (6), (69), (72), and an integration on $(0,t)$,

$$\begin{array}{cc}\hfill & {∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{\int }_0^t{∥{\partial }_x^{6}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C_0+C\left(T\right)t+C\left(T\right){\int }_0^t{∥{\partial }_x^{4}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C\left(T\right){\int }_0^t{∥{\partial }_x^{5}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C\left(T\right),\hfill \end{array}$$

which gives (74).

Finally, arguing as in Lemma 6, we have (52). □

We continue with an energy estimate in the space $L_{loc}^{\infty }(0,\infty ;H^3\left(\mathbb{R}\right))\cap H_{loc}^1((0,\infty )\times \mathbb{R})$.

Lemma 12.

Fix $T>0$. There exists a constant $C\left(T\right)>0$, such that

$${\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\int }_0^t{∥{\partial }_{t}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C\left(T\right).$$

(76)

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

Proof. 

Let $0\le t\le T$. Multiplying (58) by $2{\partial }_{t}u$, an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left({\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-\gamma {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & =-2{\beta }^2{\int }_{\mathbb{R}}{\partial }_{t}u{\partial }_x^{6}udx+2\alpha {\int }_{\mathbb{R}}{\partial }_{t}u{\partial }_x^{4}udx+2\gamma {\int }_{\mathbb{R}}{\partial }_t{\partial }_{x}u{\partial }_x^{2}udx\hfill \\ \hfill & =-2{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_{t}udx.\hfill \end{array}$$

Therefore, we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left({\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-\gamma {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +2{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=-4\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_{t}udx.\hfill \end{array}$$

(77)

Due to (68), (69), and the Young inequality,

$$\begin{array}{cc}\hfill & 4\left|\kappa \right|{\int }_{\mathbb{R}}\left|u\right||{\partial }_{x}u\left|\right|{\partial }_{t}u|dx\le 4|\kappa |{∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}{\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_{t}u|dx\hfill \\ \hfill & \le 2C\left(T\right){\int }_{\mathbb{R}}|{\partial }_{x}u\left|\right|{\partial }_{t}u|dx\le C\left(T\right){∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Consequently, by (77),

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left({\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-\gamma {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +{∥{\partial }_{t}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right).\hfill \end{array}$$

(6) and an integration on $(0,t)$ give

$$\begin{array}{cc}\hfill & {\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\alpha {∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-\gamma {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\int }_0^t{∥{\partial }_{t}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le C_0+C\left(T\right)t\le C\left(T\right).\hfill \end{array}$$

Therefore, by (69), (72), we have that

$$\begin{array}{cc}\hfill {\beta }^2{∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +{\int }_0^t{∥{\partial }_{t}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill \le & C\left(T\right)+\left|\alpha \right|{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\left|\gamma \right|{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le C\left(T\right),\hfill \end{array}$$

which gives (76). □

We are finally ready to prove Theorem 1 assuming ($iii$).

Proof of Theorem 1 assuming ($\mathit{iii}$). 

The well-posedness of (1) is guaranteed for a short time by the Cauchy-Kowaleskaya Theorem [64]. Thanks to the a priori estimates proved in Lemmas 8–12, we have that the global-in-time existence of a is solution of (1) that satisfies (13).

The stability estimates (15) can be proved using the same arguments of [18] (Theorem 1). □

5. Proof of Theorem 1 Assuming ($\mathit{iv}$)

In this final section, we prove Theorem 1 assuming ($iv$).

The argument is again analogous to the one of the previous sectionss. We deduce the local-in-time well-posedness from the Cauchy-Kowaleskaya Theorem [64], and we improve the local-in-time existence to the global-in-time one proving some suitable a priori estimates on u.

We begin with the zero mean estimate.

Lemma 13.

For each $t>0$, we have (14).

Proof. 

Integrating (1) on $\mathbb{R}$, we have that

$${\int }_{\mathbb{R}}{\partial }_{t}u(t,x)dx=\frac{d}{dt}{\int }_{\mathbb{R}}u(t,x)dx=0$$

(78)

(14) follows from (6) and (78). □

Remark 1.

In light of (14), we can consider the following equation:

$$P(t,x)={\int }_{-\infty }^{x}u(t,y)dy.$$

(79)

Moreover, again by (14), we have that

$$P(t,-\infty )=P(t,\infty )=0.$$

(80)

We continue by proving some energy estimates on the function P.

Lemma 14.

Let $T>0$. There exists a constant $C\left(T\right)>0$, such that

$$\begin{array}{cc}\hfill {∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2{\beta }^2{e}^{2t}{\int }_0^t{e}^{-2s}{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +2{\delta }^2{e}^t{\int }_0^{2t}{e}^{-2s}{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill \le & C\left(T\right)+C\left(T\right){\int }_0^t{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C\left(T\right){\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds,\hfill \end{array}$$

(81)

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

Proof. 

Let $0\le t\le T$. Integrating (1) on $(-\infty ,x)$, we have that

$${\int }_{-\infty }^x{\partial }_{t}udx+\kappa u^2+\gamma {\partial }_{x}u-{\beta }^2{\partial }_x^{5}u+\alpha {\partial }_x^{3}u+{\delta }^2{\partial }_x^3\left(u^3\right)=0.$$

(82)

Differentiating (79) with respect to t, we obtain that

$${\partial }_{t}P(t,x)=\frac{d}{dt}{\int }_{-\infty }^{x}u(t,y)dy={\int }_{-\infty }^x{\partial }_{t}u(t,y)dy.$$

(83)

It follows from (82) and (83) that

$${\partial }_{t}P+\kappa u^2+\gamma {\partial }_{x}u-{\beta }^2{\partial }_x^{5}u+\alpha {\partial }_x^{3}u+{\delta }^2{\partial }_x^3\left(u^3\right)=0.$$

(84)

Arguing as in [18] (Lemma 2), we have that

$$\begin{array}{cc}\hfill -2{\beta }^2{\int }_{\mathbb{R}}P{\partial }_x^{5}udx=& 2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill 2\alpha {\int }_{\mathbb{R}}P{\partial }_x^{3}udx=& 2\alpha {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill 2{\delta }^2{\int }_{\mathbb{R}}P{\partial }_x^3\left(u^3\right)dx=& 6{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(85)

Therefore, multiplying (84) by $2P$, thanks to (85), an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+6{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill =& -2\kappa {\int }_{\mathbb{R}}P{u}^{2}dx-2\gamma {\int }_{\mathbb{R}}P{\partial }_{x}udx-2\alpha {∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(86)

Due to the Young inequality,

$$\begin{array}{cc}\hfill 2\left|\kappa \right|{\int }_{\mathbb{R}}\left|P\right|u^{2}dx\le & {∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\kappa }^2{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4,\hfill \\ \hfill 2\left|\gamma \right|{\int }_{\mathbb{R}}\left|P\right||{\partial }_{x}u|dx\le & {∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\gamma }^2{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

It follows from (86) that

$$\begin{array}{cc}\hfill \frac{d}{dt}{∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+6{\delta }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill \le & 2{∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+C_0{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Therefore, by the Gronwall Lemma and (9), we have that

$$\begin{array}{cc}\hfill {∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2& +2{\beta }^2{e}^{2t}{\int }_0^t{e}^{-2s}{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & +6{\delta }^2{e}^{2t}{\int }_0^t{e}^{-2s}{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill \le & C_0+C_0{e}^{2t}{\int }_0^t{e}^{-2s}{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C_0{e}^{2t}{\int }_0^t{e}^{-2s}{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill \le & C\left(T\right)+C\left(T\right){\int }_0^t{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C\left(T\right){\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds,\hfill \end{array}$$

which gives (81). □

Lemma 15.

Let $T>0$. There exists a constant $C\left(T\right)>0$, such that

$$\frac{{\beta }^2}{6}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^2\le C\left(T\right),$$

(87)

for every $0\le t\le T$. In particular, we have (29), (31),

$$\begin{array}{cc}\hfill {∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}\le & C\left(T\right),\hfill \\ \hfill {∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}\le & C\left(T\right),\hfill \\ \hfill {\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\le & C\left(T\right),\hfill \\ \hfill {\int }_0^t{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^3\left(u^3\right)\right]}^{2}dsdx\le & C\left(T\right),\hfill \end{array}$$

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

$$\begin{array}{cc}\hfill {∥P∥}_{L^{\infty }((0,T)\times \mathbb{R})}\le & C\left(T\right)\hfill \\ \hfill {∥u∥}_{L^{\infty }((0,T)\times \mathbb{R})}\le & C\left(T\right).\hfill \end{array}$$

Proof. 

Let $0\le t\le T$. Consider an real constant A, which will be specified later. Observe that

$$\begin{array}{cc}\hfill {\delta }^2\gamma {\int }_{\mathbb{R}}{u}^3{\partial }_x^{2}udx=& -3{\delta }^2\gamma {∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2,\hfill \\ \hfill 2A\gamma {\int }_{\mathbb{R}}u{\partial }_x^{2}udx=& -2A{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(88)

Multiplying (1) by

$$-{\beta }^2{\partial }_x^{2}u+{\delta }^2{u}^3+Au,$$

thanks to (88) and arguing as in [18] (Lemma 3), an integration on $\mathbb{R}$ gives

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+\frac{A}{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dx\hfill \\ \hfill & =-{\beta }^2\left(A+\alpha \right){∥{\partial }_x^{3}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2-{\delta }^2\left(A+\alpha \right){\int }_{\mathbb{R}}{\partial }_x^{2}u{\partial }_x^2\left(u^3\right)dx\hfill \\ \hfill & -\left(A\alpha -\gamma {\beta }^2\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+3{\delta }^2\gamma {∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +2{\beta }^2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx.\hfill \end{array}$$

Taking

$$A=-\alpha ,$$

we have that

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4-\frac{\alpha }{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)\hfill \\ \hfill & +{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dx\hfill \\ \hfill & =\left({\alpha }^2+\gamma {\beta }^2\right){∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+3{\delta }^2\gamma {∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+2{\beta }^2\kappa {\int }_{\mathbb{R}}u{\partial }_{x}u{\partial }_x^{2}udx.\hfill \end{array}$$

(89)

Due to the Young inequality,

$$2{\beta }^2\left|\kappa \right|{\int }_{\mathbb{R}}|u{\partial }_{x}u\left|\right|{\partial }_x^{2}u|dx\le {\beta }^2{\kappa }^2{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+{\beta }^2{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.$$

It follows from (89) that

$$\begin{array}{cc}\hfill & \frac{d}{dt}\left(\frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4-\frac{\alpha }{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\right)+{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dx\hfill \\ \hfill & \le C_0{∥{\partial }_x^{2}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+C_0{∥u(t,·){\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

Integrating on $(0,t)$, by (7), we have that

$$\begin{array}{cc}\hfill & \frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4-\frac{\alpha }{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & +{\int }_0^t{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dsdx\hfill \\ \hfill & \le C_0+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds,\hfill \end{array}$$

that is

$$\begin{array}{cc}\hfill & \frac{{\beta }^2}{2}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+{\int }_0^t{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dsdx\hfill \\ \hfill & \le C_0+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+\frac{\alpha }{2}{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C_0+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(90)

Observe that, by (79) and (80),

$$C_0{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2=C_0{\int }_{\mathbb{R}}uudx=-C_0{\int }_{\mathbb{R}}P{\partial }_{x}udx.$$

Therefore, by the Young inequality,

$$\begin{array}{cc}\hfill C_0{∥u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\le & 2{\int }_{\mathbb{R}}\left|\frac{\sqrt{3}{C}_{0}P}{2\beta }\right|\left|\frac{\beta {\partial }_{x}u}{\sqrt{3}}\right|dx\hfill \\ \hfill \le & C_0{∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\beta }^2}{3}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2.\hfill \end{array}$$

(91)

It follows from (81), (90) and (91) that

$$\begin{array}{cc}\hfill & \frac{{\beta }^2}{6}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4+{\int }_0^t{\int }_{\mathbb{R}}{\left[{\beta }^2{\left({\partial }_x^{4}u\right)}^2-{\delta }^2{\partial }_x^2\left(u^3\right)\right]}^{2}dsdx\hfill \\ \hfill & \le C_0+C_0{\int }_0^t{∥{\partial }_x^{2}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{\int }_0^t{∥u(s,·){\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+C_0{∥P(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right){\int }_0^t{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds+C\left(T\right){\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right)\left(\frac{{\beta }^2}{6}{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+\frac{{\delta }^2}{4}{\int }_0^t{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds\right).\hfill \end{array}$$

Therefore, we have that

$$\begin{array}{cc}\hfill & \frac{{\beta }^2}{6}{∥{\partial }_{x}u(t,·)∥}_{L^2\left(\mathbb{R}\right)}^2+\frac{{\delta }^2}{4}{∥u(t,·)∥}_{L^4\left(\mathbb{R}\right)}^4\hfill \\ \hfill & \le C\left(T\right)+C\left(T\right)\left(\frac{{\beta }^2}{6}{\int }_0^t{∥{\partial }_{x}u(s,·)∥}_{L^2\left(\mathbb{R}\right)}^{2}ds+\frac{{\delta }^2}{4}{\int }_0^t{∥u(s,·)∥}_{L^4\left(\mathbb{R}\right)}^{4}ds\right).\hfill \end{array}$$

The Gronwall Lemma and (7) give (87).

Finally, arguing as in [18] (Lemma 3), the proof is concluded. □

Arguing as in [18] (Theorem 1), we have Theorem 1.

6. Conclusions

This paper is dedicated to the well-posedness of a solution to the Cauchy problem for a higher-order convective Cahn-Hilliard equation. Such an equation models the evolution of crystal surfaces faceting through surface electromigration, the growing surface faceting, and the evolution of dynamics of phase transitions in ternary oil-water-surfactant systems. The well-posedness of (1) is proved for a short time by the Cauchy-Kowaleskaya Theorem [64]. The global-in-time well-posedness is thus proved, proving several a priori estimates.