A review of pore-forming agents on the structures, porosities, and mechanical properties of porous ceramics

Mohamed Lokman Jalaluddin; Umar Al-Amani Azlan; Mohd Warikh Abd Rashid; Norfauzi Tamin; Mohamad Najmi Masri; Mohamed Lokman Jalaluddin; Umar Al-Amani Azlan; Mohd Warikh Abd Rashid; Norfauzi Tamin; Mohamad Najmi Masri

doi:10.3934/matersci.2024033

AIMS Materials Science

2024, Volume 11, Issue 4: 634-665. doi: 10.3934/matersci.2024033

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Review

A review of pore-forming agents on the structures, porosities, and mechanical properties of porous ceramics

1.
Faculty of Industrial and Manufacturing Technology and Engineering, Universiti Teknikal Malaysia Melaka, Malaysia
2.
Faculty of Technical and Vocational Education, Universiti Tun Hussein Onn Malaysia, Parit Raja, Johor, Malaysia
3.
Faculty of Bioengineering and Technology, Universiti Malaysia Kelantan, Campus Jeli, Kelantan, Malaysia

Received: 11 June 2024 Revised: 09 July 2024 Accepted: 23 July 2024 Published: 05 August 2024

This review article provided a thorough examination of porous ceramic materials, concentrating on production, characteristics, and the involvement of pore-forming agents. The primary objective of this research was to evaluate the effects of various ceramic materials and pore-forming agents on the structure, porosity, and mechanical characteristics of porous ceramics. The study's scope included a thorough investigation of key sources of literature, such as academic publications, review articles, and industry reports, to provide a comprehensive understanding of porous ceramic technology. According to the literature review, the selection of ceramic material and pore-forming agents has a significant influence on the pore size distribution, porosity, and mechanical strength of porous ceramics. Various manufacturing methods, including foaming, sintering, and sol-gel procedures, were explored in terms of their influence on porous ceramic microstructure and characteristics. Furthermore, the study emphasized the need to optimize processing settings and select pore-forming agents to obtain the necessary qualities in porous ceramic materials. Overall, this review is useful for researchers, engineers, and practitioners who desire to learn more about porous ceramic manufacturing, characteristics, and applications.

Keywords:

Citation: Mohamed Lokman Jalaluddin, Umar Al-Amani Azlan, Mohd Warikh Abd Rashid, Norfauzi Tamin, Mohamad Najmi Masri. A review of pore-forming agents on the structures, porosities, and mechanical properties of porous ceramics[J]. AIMS Materials Science, 2024, 11(4): 634-665. doi: 10.3934/matersci.2024033

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Abstract

1. Introduction

The equation:

$\begin{equation} \begin{cases} \partial_t u + \partial_x f(u)-\beta^2 \partial_{x}^2 u+\delta \partial_{x}^3 u+\kappa u +\gamma^2\vert u\vert u = 0, &\quad 0 < t < T, \quad x\in \mathbb{R},\\ u(0,x) = u_0(x), &\quad x\in \mathbb{R}, \end{cases} \end{equation}$

(1.1)

was originally derived in ^[14,17] with $f(u) = au^2$ focusing on microbubbles coated by viscoelastic shells. These structures are crucial in ultrasound diagnosis using contrast agents, and the dynamics of individual coated bubbles are explored, taking into account nonlinear competition and dissipation factors such as dispersion, thermal effects, and drag force.

The coefficients $\beta^2$ , $\delta$ , $\kappa$ , and $\gamma^2$ are related to the dissipation, the dispersion, the thermal conduction dissipation, and to the drag force, repsctively.

If $\kappa = \gamma = 0$ , we obtain the Kudryashov-Sinelshchikov ^[18] Korteweg-de Vries-Burgers ^[3,20] equation

$\begin{equation} \partial_t u +a \partial_x u^2-\beta^2 \partial_{x}^2 u+\delta \partial_{x}^3 u = 0, \end{equation}$

(1.2)

that models pressure waves in liquids with gas bubbles, taking into account heat transfer and viscosity. The mathematical results on Eq (1.2) are the following:

● analysis of exact solutions in ^[13],

● existence of the traveling waves in ^[2],

● well-posedness and asymptotic behavior in ^[7,11].

If $\beta = 0$ , we derive the Korteweg-de Vries equation:

$\begin{equation} \partial_t u +a \partial_x u^2+\delta \partial_{x}^3 u = 0, \end{equation}$

(1.3)

which describes surface waves of small amplitude and long wavelength in shallow water. Here, $u(t, x)$ represents the wave height above a flat bottom, $x$ corresponds to the distance in the propagation direction, and $t$ denotes the elapsed time. In ^{[4,6,10,12,15,16]}, the completele integrability of Eq (1.3) and the existence of solitary wave solutions are proved.

Through the manuscript, we will assume

● on the coefficients

$\begin{equation} \beta,\,\delta,\,\kappa,\,\gamma\in \mathbb{R}, \quad \beta,\,\delta,\,\gamma\ne0; \end{equation}$

(1.4)

● on the flux $f$ , one of the following conditions:

$\begin{align} &f(u) = a u^2+bu^3, \end{align}$

(1.5)

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

(1.6)

for some positive constant $C_0$ ;

● on the initial value

$\begin{equation} u_0\in H^1( \mathbb{R}). \end{equation}$

(1.7)

The main result of this paper is the following theorem.

Theorem 1.1. Assume Eqs (1.5)–(1.7). For fixed $T > 0$ , there exists a unique distributional solution $u$ of Eq (1.1), such that

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

(1.8)

Moreover, if $u_1$ and $u_2$ are solutions to Eq (1.1) corresponding to the initial conditions $u_{1, 0}$ and $u_{2, 0}$ , respectively, it holds that:

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

(1.9)

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

Observe that Theorem 1.1 gives the well-posedness of (1.1), without conditions on the constants. Moreover, the proof of Theorem 1.1 is based on the Aubin-Lions Lemma ^[5,21]. The analysis of Eq (1.1) is more delicate than the one of Eq (1.2) due to the presence of the nonlinear sources and the very general assumptions on the coefficients.

The structure of the paper is outlined as follows. Section 2 is dedicated to establishing several a priori estimates for a vanishing viscosity approximation of Eq (1.1). These estimates are crucial for proving our main result, which is presented in Section 3.

2. Vanishing viscosity approximation

To establish existence, we utilize a vanishing viscosity approximation of equation (1.1), as discussed in ^[19]. Let $0 < \varepsilon < 1$ be a small parameter, and denote by $u_\varepsilon\in C^\infty([0, T)\times \mathbb{R})$ the unique classical solution to the following problem ^[1,9]:

$\begin{equation} \begin{cases} \partial_t u_\varepsilon + \partial_x f( u_\varepsilon)-\beta^2 \partial_{x}^2 u_\varepsilon+\delta \partial_{x}^3 u_\varepsilon+\kappa u\\ \qquad\quad +\gamma^2\vert u\vert u = - \varepsilon \partial_{x}^4 u_\varepsilon, &\quad 0 < t < T, \quad x\in \mathbb{R},\\ u_{ \varepsilon}(0,x) = u_{ \varepsilon,\,0}(x), &\quad x\in \mathbb{R}, \end{cases} \end{equation}$

(2.1)

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

$\begin{equation} \left\| {u_{ \varepsilon,\,0}} \right\|_{H^1( \mathbb{R})}\le \left\| {u_0} \right\|_{H^1( \mathbb{R})}. \end{equation}$

(2.2)

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

We begin by proving the following lemma:

Lemma 2.1. Let $T > 0$ be fixed. There exists a constant $C(T) > 0$ , which does not depend on $\varepsilon$ , such that

$\begin{equation} \begin{split} \left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}&+2\gamma^2e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^{t}\!\!\displaystyle {\int}_{ \mathbb{R}}e^{-\vert\kappa\vert s} u_\varepsilon^2\vert u_\varepsilon\vert dsdx\\ &+2\beta^2e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^{t}e^{-\vert\kappa\vert s}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})} ds\\ &+2 \varepsilon e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^{t}e^{-\vert\kappa\vert s}\left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2(R)}\le C(T), \end{split} \end{equation}$

(2.3)

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

Proof. For $0\le t\le T$ . Multiplying equations (2.1) by $2 u_\varepsilon$ , and integrating over $\mathbb{R}$ yields

$\begin{align*} \frac{d}{dt}&\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} = 2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_t u_\varepsilon dx\\ = &\underbrace{-2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon f'( u_\varepsilon) \partial_x u_\varepsilon dx}_{ = 0}+2\beta^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_{x}^2 u_\varepsilon dx-2\delta\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_{x}^3 u_\varepsilon dx \\ &-\kappa\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^2 dx-2 \varepsilon\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_{x}^4 u_\varepsilon dx \\ = &-2\beta^2\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\delta\displaystyle {\int}_{ \mathbb{R}} \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx\\ &-\kappa\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^2 dx+2 \varepsilon\displaystyle {\int}_{ \mathbb{R}} \partial_x u_\varepsilon \partial_{x}^3 u_\varepsilon dx\\ = &-2\beta^2\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\kappa\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &-2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^2 dx-2 \varepsilon\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Thus, it follows that

$\begin{align*} \frac{d}{dt}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+&2\beta^2\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &+2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^2 dx+2 \varepsilon\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ = &\kappa\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\le \vert\kappa\vert \left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Therefore, applying the Gronwall's lemma and using Eq (2.2), we obtain

$\begin{align*} \left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}&+2\beta^2 e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^{t}e^{-\vert\kappa\vert s}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ &+ 2\gamma^2e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^t\!\!\displaystyle {\int}_{ \mathbb{R}}e^{-\vert\kappa\vert t}\vert u_\varepsilon\vert u_\varepsilon^2 dsdx+2 \varepsilon\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &+2 \varepsilon e^{\vert\kappa\vert t}\displaystyle {\int}_{0}^{t}e^{-\vert\kappa\vert s} \left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ \le& C_0e^{\vert\kappa\vert t}\le C(T), \end{align*}$

which gives Eq (2.3). □

Lemma 2.2. Fix $T > 0$ and assume (1.5). There exists a constant $C(T) > 0$ , independent of $\varepsilon$ , such that

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

(2.4)

$\begin{align} \left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\beta^2\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds& \end{align}$

(2.5)

$\begin{align} +2 \varepsilon\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^3 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds&\le C(T),\\ \displaystyle {\int}_{0}^{t}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^4_{L^4( \mathbb{R})} ds&\le C(T), \end{align}$

(2.6)

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

Proof. Let $0\le t\le T$ . Consider $A, \, B$ as two real constants, which will be specified later. Thanks to Eq (1.5), multiplying Eq (2.1) by

$\begin{equation*} -2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3, \end{equation*}$

we have that

$\begin{align} &\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_t u_\varepsilon+ 2a\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon \partial_x u_\varepsilon\\ &\qquad\quad+3b\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon^2 \partial_x u_\varepsilon-\beta^2\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^2 u_\varepsilon \\ &\qquad\quad+\delta\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^3 u_\varepsilon+\kappa\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon\\ &\qquad\quad+\gamma^2\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right)\vert u_\varepsilon\vert u_\varepsilon = - \varepsilon\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^4 u_\varepsilon. \end{align}$

(2.7)

Observe that

$\begin{align*} &\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_t u_\varepsilon dx\\ &\qquad = \frac{d}{dt}\left(\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\frac{A}{3}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx +\frac{B}{4}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\right),\\ &2a\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon \partial_x u_\varepsilon dx = -4a\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx,\\ &3b\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon^2 \partial_x u_\varepsilon dx = -6b\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2 \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx,\\ &{-\beta^2\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^2 u_\varepsilon dx}\\ &\qquad = 2\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2A\beta^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon( \partial_x u_\varepsilon)^2 dx +3B\beta^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2dx, \\ &\delta\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^3 u_\varepsilon dx = -2A\delta\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx -3B\delta\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2 \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx, \\ &\kappa\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) u_\varepsilon dx\\ &\qquad = {2\kappa\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+{A\kappa}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3 dx +{B\kappa}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx,}\\ &\gamma^2\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right)\vert u_\varepsilon\vert u_\varepsilon dx\\ &\qquad = -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx +A\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u\vert u_\varepsilon^3dx +B\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u^4 dx,\\ &- \varepsilon\displaystyle {\int}_{ \mathbb{R}}\left(-2 \partial_{x}^2 u_\varepsilon+A u_\varepsilon^2+B u_\varepsilon^3\right) \partial_{x}^4 u_\varepsilon dx\\ &\qquad = -2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2A \varepsilon\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_x u_\varepsilon{ \partial_{x}^3 u_\varepsilon} dx+3B \varepsilon\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2 \partial_x u_\varepsilon \partial_{x}^3 u_\varepsilon dx\\ &\qquad = -2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-A \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^3 dx -6B \varepsilon\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon( \partial_x u_\varepsilon)^2 \partial_{x}^2 u_\varepsilon dx\\ &\qquad\quad -3B \varepsilon\left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad = -2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-A \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^3 dx +2B \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4dx\\ &\qquad\quad -3B \varepsilon\left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Therefore, an integration on $\mathbb{R}$ gives

$\begin{align*} &\frac{d}{dt}\left(\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\frac{A}{3}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx +\frac{B}{4}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\right)\\ &\qquad\quad +\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad = -\left(4a+A\delta\right)\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx -3\left(2b+B\delta\right)\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2 \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx\\ &\qquad\quad -2A\beta^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon( \partial_x u_\varepsilon)^2 dx -3B\beta^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2dx\\ &\qquad\quad -\kappa\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\frac{A\kappa}{3}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3 dx -\frac{B\kappa}{4}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\\ &\qquad\quad +2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx {-A\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^3dx-B\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^4 dx}\\ &\qquad\quad -A \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^3 dx +2B \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4dx-3B \varepsilon\left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Taking

$\begin{equation*} \left(A,\,B\right) = \left(-\frac{4a}{\delta},\,-\frac{2b}{\delta}\right), \end{equation*}$

we get

$\begin{align} &\frac{d}{dt}\left(\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\frac{4a}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx -\frac{b}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\right)\\ &\qquad\quad +2\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad = \frac{8a\beta^2}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon( \partial_x u_\varepsilon)^2 dx +\frac{6b\beta^2}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2dx\\ &\qquad\quad -\kappa\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\frac{4a\kappa}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3 dx +\frac{b\kappa}{2}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\\ &\qquad\quad +2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx {+\frac{4a\gamma^2}{\delta}\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^3dx +\frac{2b\gamma^2}{\delta}\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon^4 dx} \\ &\qquad\quad +\frac{4a \varepsilon}{\delta}\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^3 dx-\frac{4b \varepsilon}{\delta}\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4dx+\frac{6b \varepsilon}{\delta}\left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align}$

(2.8)

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

$\begin{align*} &\frac{8a\beta^2}{\delta}\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert( \partial_x u_\varepsilon)^2 dx\\ &\qquad\le 4\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2 dx +\frac{4a^2\beta^4}{\delta^2}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le 4\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\frac{4a^2\beta^4}{\delta^2}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})},\\ &\left\vert\frac{6b\beta^2}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2dx\le \left\vert\frac{6b\beta^2}{\delta}\right\vert\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})},\\ &\vert\frac{4a\kappa}{3\delta}\vert\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert^3 dx\le \left\vert\frac{4a\kappa}{3\delta}\right\vert\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})},\\ &\left\vert\frac{b\kappa}{2}\right\vert\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\le \left\vert\frac{b\kappa}{2}\right\vert\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}, \\ &2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx{\le}2\displaystyle {\int}_{ \mathbb{R}}\left\vert\frac{\gamma^2\vert u_\varepsilon\vert u_\varepsilon}{\beta}\right\vert\left\vert\beta \partial_{x}^2 u_\varepsilon\right\vert dx\\ &\qquad\le\frac{\gamma^4}{\beta^2}\displaystyle {\int}_{ \mathbb{R}}{ u_\varepsilon}^4dx +\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad \le\frac{\gamma^4}{\beta^2}\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0.T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})},\\ &\left\vert\frac{4a\gamma^2}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}\vert { u_\varepsilon}\vert \vert u_\varepsilon\vert^3dx = \left\vert\frac{4a\gamma^2}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\\ &\qquad\le \left\vert\frac{4a\gamma^2}{\delta}\right\vert\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} \le C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}, \\ &\left\vert\frac{2b\gamma^2}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert { u_\varepsilon}^4 dx\le \left\vert\frac{2b\gamma^2}{\delta}\right\vert\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,\,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,\,T)\times \mathbb{R})},\\ &\left\vert\frac{4a \varepsilon}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}\vert \partial_x u_\varepsilon\vert^3 dx\le\left\vert\frac{4a \varepsilon}{\delta}\right\vert\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\left\vert\frac{4a \varepsilon}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4 dx\\ &\qquad \le \left\vert\frac{4a}{\delta}\right\vert\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\left\vert\frac{4a \varepsilon}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4 dx. \end{align*}$

It follows from Eq (2.8) that

$\begin{align} &\frac{d}{dt}\left(\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\frac{4a}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx -\frac{b}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\right)\\ &\qquad\quad +\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\\ &\qquad\quad +C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\\ &\qquad\quad+C_0 \varepsilon\displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4 dx +C_0 \varepsilon\left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\quad {+C_0\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}}. \end{align}$

(2.9)

[8, Lemma 2.3] says that

$\begin{equation} \displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^4 dx\le 9\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_{x}^2 u_\varepsilon)^2 dx\le 9\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,\,T)\times \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation}$

(2.10)

Moreover, we have that

$\begin{equation} \left\| { u_\varepsilon(t,\cdot) \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} = \displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_{x}^2 u_\varepsilon)^2 dx {\le}\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,\,T)\times \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation}$

(2.11)

Consequentially, by Eqs (2.9)–(2.11), we have that

$\begin{align*} &\frac{d}{dt}\left(\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\frac{4a}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx -\frac{b}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\right)\\ &\qquad\quad +\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\\ &\qquad\quad +C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\\ &\qquad\quad +C_0 \varepsilon\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,\,T)\times \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} {+C_0\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}.}\notag \end{align*}$

An integration on $(0, t)$ and Eqs (2.2) and (2.3) give

$\begin{align*} &\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-\frac{4a}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx -\frac{b}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx\\ &\qquad\quad +\beta^2\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds+2 \varepsilon\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^3 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\displaystyle {\int}_{0}^{t}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ &\qquad\quad+C(T)\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}t +C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}t+C(T)\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}t\\ &\qquad\quad +C_0 \varepsilon\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,\,T)\times \mathbb{R})}\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds {+C_0\displaystyle {\int}_{0}^{t}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds}\\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right). \end{align*}$

Therefore, by Eq (2.3),

$\begin{align} &\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\beta^2\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^2 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds+2 \varepsilon\displaystyle {\int}_{0}^{t}\left\| { \partial_{x}^3 u_\varepsilon(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right)\\ &\qquad\quad +\frac{4a}{3\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^3dx +\frac{b}{\delta}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx \\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right)\\ &\qquad\quad +\left\vert\frac{4a}{3\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert^3dx +\left\vert\frac{b}{\delta}\right\vert\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx \\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right)\\ &\qquad\quad +\left\vert\frac{4a}{3\delta}\right\vert \left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon({t},\cdot)} \right\|^2_{L^2( \mathbb{R})} +\left\vert\frac{b}{\delta}\right\vert\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right). \end{align}$

(2.12)

We prove Eq (2.4). Thanks to the Hölder inequality,

$\begin{equation*} u_\varepsilon^2(t,x) = 2\displaystyle {\int}_{-\infty}^{x} u_\varepsilon \partial_x u_\varepsilon dx\le 2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert\vert \partial_x u_\varepsilon\vert dx\le 2\left\| { u_\varepsilon(t,\cdot)} \right\|_{L^2( \mathbb{R})}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|_{L^2( \mathbb{R})}. \end{equation*}$

Hence, we have that

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

(2.13)

Thanks to Eqs (2.3) and (2.12), we have that

$\begin{equation} \begin{split} &\left\| { u_\varepsilon} \right\|^4_{L^{\infty}((0,T)\times \mathbb{R})}\\ &\qquad\le C(T)\left(1+\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})} +\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\right). \end{split} \end{equation}$

(2.14)

Due to the Young inequality,

$\begin{align*} C(T)\left\| { u_\varepsilon} \right\|^3_{L^{\infty}((0,T)\times \mathbb{R})}\le& \frac{1}{2}\left\| { u_\varepsilon} \right\|^4_{L^{\infty}((0,T)\times \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})},\\ C(T)\left\| { u_\varepsilon} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\le& C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+C(T). \end{align*}$

By Eq (2.14), we have that

$\begin{equation*} \frac{1}{2}\left\| { u_\varepsilon} \right\|^4_{L^{\infty}((0,T)\times \mathbb{R})}-C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}-C(T){\le0}, \end{equation*}$

which gives Eq (2.4).

Equation (2.5) follows from Eqs (2.4) and (2.12).

Finally, we prove Eq (2.6). We begin by observing that, from Eqs (2.4) and (2.10), we have

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

An integration on $(0, t)$ and Eqs (2.5) give Eq (2.6). □

Lemma 2.3. Fix $T > 0$ and assume (1.6). There exists a constant $C(T) > 0$ , independent of $\varepsilon$ , such that Eq (2.4) holds. Moreover, we have Eqs (2.5) and (2.6).

Proof. Let $0\le t\le T$ . Multiplying Eq (2.1) by $-2 \partial_{x}^2 u_\varepsilon$ , an integration on $\mathbb{R}$ gives

$\begin{align*} \frac{d}{dt}&\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} = -2\displaystyle {\int}_{ \mathbb{R}} \partial_{x}^2 u_\varepsilon \partial_t u_\varepsilon dx\\ = &-2\displaystyle {\int}_{ \mathbb{R}}f'( u_\varepsilon) \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx-2\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} -2\delta\displaystyle {\int}_{ \mathbb{R}} \partial_{x}^2 u_\varepsilon \partial_{x}^3 u_\varepsilon dx\\ &-2\kappa\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon \partial_{x}^2 u_\varepsilon dx-2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx+2 \varepsilon\displaystyle {\int}_{ \mathbb{R}} \partial_{x}^2 u_\varepsilon \partial_{x}^4 u_\varepsilon dx\\ = &-2\displaystyle {\int}_{ \mathbb{R}}f'( u_\varepsilon) \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx -2\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &+2\kappa\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx-2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Therefore, we have that

$\begin{equation} \begin{split} &\frac{d}{dt}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad = -2\displaystyle {\int}_{ \mathbb{R}}f'( u_\varepsilon) \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dx+2\kappa\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx. \end{split} \end{equation}$

(2.15)

Due Eqs (1.6) and (2.3) and the Young inequality,

$\begin{align*} &2\displaystyle {\int}_{ \mathbb{R}}\vert f'( u_\varepsilon)\vert\vert \partial_x u_\varepsilon\vert\vert \partial_{x}^2 u_\varepsilon\vert dx\le C_0\displaystyle {\int}_{ \mathbb{R}}\vert \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon\vert dx+C_0\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon \partial_x u_\varepsilon\vert\vert \partial_{x}^2 u_\varepsilon\vert dx\\ &\qquad = 2\displaystyle {\int}_{ \mathbb{R}}\left\vert\frac{C_0\sqrt{3} \partial_x u_\varepsilon}{2\beta}\right\vert\left\vert\frac{\beta \partial_{x}^2 u_\varepsilon}{\sqrt{3}}\right\vert dx+2\displaystyle {\int}_{ \mathbb{R}}\left\vert\frac{C_0\sqrt{3} u_\varepsilon \partial_x u_\varepsilon}{2\beta}\right\vert\left\vert\sqrt{3}\beta \partial_{x}^2 u_\varepsilon\right\vert dx\\ &\qquad\le C_0\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+C_0\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2( \partial_x u_\varepsilon)^2 dx +\frac{2\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+C_0\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\frac{2\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\frac{2\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})},\\ &2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_\varepsilon\vert u_\varepsilon \partial_{x}^2 u_\varepsilon dx\le 2\gamma^2\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^2\vert \partial_{x}^2 u_\varepsilon\vert dx = 2\displaystyle {\int}_{ \mathbb{R}}\left\vert\frac{\sqrt{3}\gamma^2 u_\varepsilon^2}{\beta}\right\vert\left\vert\frac{\beta \partial_{x}^2 u_\varepsilon}{\sqrt{3}}\right\vert dx\\ &\qquad\le \frac{3\gamma^4}{\beta^2}\displaystyle {\int}_{ \mathbb{R}} u_\varepsilon^4 dx + \frac{\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le\frac{3\gamma^4}{\beta^2}\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\left\| { u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} + \frac{\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}+\frac{\beta^2}{3}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

It follows from Eq (2.15) that

$\begin{align*} &\frac{d}{dt}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\beta^2\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2 \varepsilon\left\| { \partial_{x}^3 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ &\qquad\le C_0\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right)\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}. \end{align*}$

Integrating on $(0, t)$ , by Eq (2.3), we have that

(2.16)

Thanks to Eqs (2.3), (2.13), and (2.16), we have that

$\begin{equation*} \left\| { u_\varepsilon} \right\|^4_{L^{\infty}((0,T)\times \mathbb{R})}\le C(T)\left(1+\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\right). \end{equation*}$

Therefore,

$\begin{equation*} \left\| { u_\varepsilon} \right\|^4_{L^{\infty}((0,T)\times \mathbb{R})}-C(T)\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}-C(T)\le 0, \end{equation*}$

which gives (2.4).

Equation (2.5) follows from (2.4) and (2.16), while, arguing as in Lemma 2.2, we have Eq (2.6).

□

Lemma 2.4. Fix $T > 0$ . There exists a constant $C(T) > 0$ , independent of $\varepsilon$ , such that

$\begin{equation} \displaystyle {\int}_{0}^{t}\left\| { \partial_x u_\varepsilon(s,\cdot)} \right\|^6_{L^6( \mathbb{R})}ds\le C(T), \end{equation}$

(2.17)

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

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

$\begin{equation} \displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^6 dx\le \left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^4_{L^{\infty}( \mathbb{R})}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation}$

(2.18)

Thanks to the Hölder inequality,

$\begin{align*} ( \partial_x u_\varepsilon(t,x))^2 = &2\displaystyle {\int}_{-\infty}^{x} \partial_x u_\varepsilon \partial_{x}^2 u_\varepsilon dy \le 2\displaystyle {\int}_{ \mathbb{R}}\vert \partial_x u_\varepsilon\vert\vert \partial_{x}^2 u_\varepsilon\vert dx\\ \le&2\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|_{L^2( \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

Hence,

$\begin{equation*} \left\| {u(t,\cdot)} \right\|^4_{L^{\infty}( \mathbb{R})}\le 4\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation*}$

It follows from Eq (2.18) that

$\begin{equation*} \displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^6 dx\le 4\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^4_{L^2( \mathbb{R})}\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation*}$

Therefore, by Eq (2.5),

$\begin{equation*} \displaystyle {\int}_{ \mathbb{R}}( \partial_x u_\varepsilon)^6 dx\le C(T)\left\| { \partial_{x}^2 u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation*}$

An integration on $(0, t)$ and Eq (2.5) gives (2.17). □

3. Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

We begin by proving the following result.

Lemma 3.1. Fix $T > 0$ . Then,

$\begin{equation} {{the\ family\ }} \{ u_\varepsilon\}_{ \varepsilon > 0} {{ \ is\ compact\ in\ }} L^2_{loc}((0,T)\times \mathbb{R}) . \end{equation}$

(3.1)

Consequently, there exist a subsequence $\{ u_{\varepsilon_k} \}_{k\in \mathbb{N}}$ and $u\in L^2_{loc}((0, T)\times \mathbb{R})$ such that

$\begin{equation} u_{\varepsilon_k}\to u\ { in }\ L^2_{loc}((0,T)\times \mathbb{R}) {{\ and\ a.e.\ in\ }} (0,T)\times \mathbb{R} . \end{equation}$

(3.2)

Moreover, $u$ is a solution of Eq (1.1), satisfying Eq (1.8).

Proof. We begin by proving Eq (3.1). To prove Eq (3.1), we rely on the Aubin-Lions Lemma (see ^[5,21]). We recall that

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

where the first inclusion is compact and the second one is continuous. Owing to the Aubin-Lions Lemma ^[21], to prove Eq (3.1), it suffices to show that

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

(3.3)

$\begin{align} & \{ \partial_t u_\varepsilon\}_{ \varepsilon > 0} {\rm{\ is\ uniformly\ bounded\ in\ }} L^2(0,T;H^{-1}_{loc}( \mathbb{R})) . \end{align}$

(3.4)

We prove Eq (3.3). Thanks to Lemmas 2.1–2.3,

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

Therefore,

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

which gives Eq (3.3).

We prove Eq (3.4). Observe that, by Eq (2.1),

$\begin{equation*} \partial_t u_\varepsilon = - \partial_x\left( G( u_\varepsilon)\right)-f'( u_\varepsilon) \partial_x u_\varepsilon-\kappa u_\varepsilon -\gamma^2\vert u_\varepsilon\vert u_\varepsilon, \end{equation*}$

where

$\begin{equation} G( u_\varepsilon) = \beta^2 \partial_x u_\varepsilon-\delta \partial_{x}^2 u_\varepsilon- \varepsilon \partial_{x}^3 u_\varepsilon. \end{equation}$

(3.5)

Since $0 < \varepsilon < 1$ , thanks to Eq (2.5), we have that

$\begin{equation} \begin{split} &{\beta^2\left\| { \partial_x u_\varepsilon} \right\|^2_{L^2((0,T)\times \mathbb{R})}},\, \delta^2\left\| { \partial_{x}^2 u_\varepsilon} \right\|^2_{L^2((0,T)\times \mathbb{R})}\le C(T),\\ & \varepsilon^2\left\| { \partial_{x}^3 u_\varepsilon} \right\|^2_{L^2((0,T)\times \mathbb{R})}\le C(T). \end{split} \end{equation}$

(3.6)

Therefore, by Eqs (3.5) and (3.6), we have that

$\begin{equation} \{ \partial_x\left(G( u_\varepsilon)\right)\}_{ \varepsilon > 0} \quad {\rm{\ is\ bounded\ in}}\ { L^2(0, T ; H^{-1}( \mathbb{R})) }. \end{equation}$

(3.7)

We claim that

$\begin{equation} \displaystyle {\int}_{0}^{T}\!\!\!\displaystyle {\int}_{ \mathbb{R}}(f'( u_\varepsilon))^2( \partial_x u_\varepsilon)^2 dtdx\le C(T). \end{equation}$

(3.8)

Thanks to Eqs (2.4) and (2.5),

$\begin{equation*} \displaystyle {\int}_{0}^{T}\!\!\!\displaystyle {\int}_{ \mathbb{R}}(f'( u_\varepsilon))^2( \partial_x u_\varepsilon)^2 dtdx\le \left\| {f'} \right\|^2_{L^{\infty}(-C(T),C(T))}\displaystyle {\int}_{0}^{T}\left\| { \partial_x u_\varepsilon(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}dt\le C(T). \end{equation*}$

Moreover, thanks to Eq (2.3),

$\begin{equation} \vert\kappa\vert\displaystyle {\int}_{0}^{T}\!\!\!\displaystyle {\int}_{ \mathbb{R}}( u_\varepsilon)^2 dx \le C(T). \end{equation}$

(3.9)

We have that

$\begin{equation} \gamma^2\displaystyle {\int}_{0}^T\!\!\!\displaystyle {\int}_{ \mathbb{R}}(\vert u_\varepsilon\vert u_\varepsilon)^2 dsdx\le C(T). \end{equation}$

(3.10)

In fact, thanks to Eqs (2.3) and (2.4),

$\begin{align*} \gamma^2\displaystyle {\int}_{0}^T\!\!\!\displaystyle {\int}_{ \mathbb{R}}(\vert u_\varepsilon\vert u_\varepsilon)^2 dsdx\le& \gamma^2\left\| { u_\varepsilon} \right\|^2_{L^{\infty}((0,T)\times \mathbb{R})}\displaystyle {\int}_{0}^T\!\!\!\displaystyle {\int}_{ \mathbb{R}}( u_\varepsilon)^2 dsdx\\ \le&C(T)\displaystyle {\int}_{0}^T\!\!\!\displaystyle {\int}_{ \mathbb{R}}( u_\varepsilon)^2 dsdx\le C(T). \end{align*}$

Therefore, Eq (3.4) follows from Eqs (3.7)–(3.10).

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

Consequently, arguing as in [, Theorem 1.1], $u$ is solution of Eq (1.1) and, thanks to Lemmas 2.1–2.3 and Eqs (2.4), (1.8) holds. □

Proof of Theorem 1.1. Lemma 3.1 gives the existence of a solution of Eq (1.1).

We prove Eq (1.9). Let $u_1$ and $u_2$ be two solutions of Eq (1.1), which verify Eq (1.8), that is,

$\begin{align*} &\begin{cases} \partial_t u_i+ \partial_x f(u_i)-\beta^2 \partial_{x}^2 u_i+\delta \partial_{x}^3 u_i +\kappa u_i +\gamma^2\vert u_i\vert u_i = 0,&\quad 0 < t < T, \quad x\in \mathbb{R},\\ u_i(0,x) = u_{i,\,0}(x), &\quad x\in \mathbb{R}, \end{cases}\quad i = 1,2. \end{align*}$

Then, the function

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

(3.11)

is the solution of the following Cauchy problem:

$\begin{equation} \begin{cases} \partial_t\omega+ \partial_x\left( f(u_1)-f(u_2)\right)-\beta^2 \partial_{x}^2\omega+\delta \partial_{x}^2\omega \\ \qquad\qquad+\kappa\omega+\gamma^2\left(\vert u_1\vert u_1-\vert u_2\vert u_2\right) = 0, &\quad 0 < t < T, \quad x\in \mathbb{R},\\ \omega(0,x) = u_{1,\,0}(x)-u_{2,\,0}(x), &\quad x\in \mathbb{R}. \end{cases} \end{equation}$

(3.12)

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

$\begin{equation} \left\| {u_1} \right\|_{L^{\infty}((0,T)\times \mathbb{R})},\, \left\| {u_2} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\le C(T). \end{equation}$

(3.13)

We define

$\begin{equation} g = \frac{f(u_1)-f(u_2)}{\omega} \end{equation}$

(3.14)

and observe that, by Eq (3.13), we have that

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

(3.15)

Moreover, by Eq (3.11) we have that

$\begin{equation} \left\vert \vert u_1\vert-\vert u_2\vert\right\vert\le \vert u_1-u_2\vert = \vert\omega\vert. \end{equation}$

(3.16)

Observe that thanks to Eq (3.11),

$\begin{equation} \begin{split} \vert u_1\vert u_1-\vert u_2\vert u_2 = &\vert u_1\vert u_1-\vert u_1\vert u_2 +\vert u_1\vert u_2 - \vert u_2\vert u_2\\ = &\vert u_1\vert\omega +u_2\left(\vert u_1\vert- \vert u_2\vert\right). \end{split} \end{equation}$

(3.17)

Thanks to Eqs (3.14) and (3.17), Equation (3.12) is equivalent to the following one:

$\begin{equation} \partial_t\omega+ \partial_x ({g}\omega) -\beta^2 \partial_{x}^2\omega+\delta \partial_{x}^3\omega+\kappa\omega+\gamma^2\vert u_1\vert\omega+\gamma^2 u_2\left(\vert u_1\vert- \vert u_2\vert\right) = 0. \end{equation}$

(3.18)

Multiplying Eq (3.18) by $2\omega$ , an integration on $\mathbb{R}$ gives

$\begin{align*} \frac{dt}{dt}&\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} = 2\displaystyle {\int}_{ \mathbb{R}}\omega \partial_t\omega\\ = &-2\displaystyle {\int}_{ \mathbb{R}}\omega \partial_x ({g}\omega)dx+2\beta^2\displaystyle {\int}_{ \mathbb{R}}\omega \partial_{x}^2\omega dx-2\delta\displaystyle {\int}_{ \mathbb{R}}\omega \partial_{x}^3\omega dx \\ &-2\kappa\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_1\vert\omega^2 dx -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}u_2\left(\vert u_1\vert- \vert u_2\vert\right)\omega dx\\ & = 2\displaystyle {\int}_{ \mathbb{R}}{g}\omega \partial_x\omega dx -2\beta^2\left\| { \partial_x\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\delta\displaystyle {\int}_{ \mathbb{R}} \partial_x\omega \partial_{x}^2\omega dx\\ &-2\kappa\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_1\vert\omega^2 dx -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}u_2\left(\vert u_1\vert- \vert u_2\vert\right)\omega dx\\ = & 2\displaystyle {\int}_{ \mathbb{R}}{g}\omega \partial_x\omega dx -2\beta^2\left\| { \partial_x\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}\\ & -2\kappa\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}-2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_1\vert\omega^2 dx -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}u_2\left(\vert u_1\vert- \vert u_2\vert\right)\omega dx. \end{align*}$

Therefore, we have that

$\begin{equation} \begin{split} \left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}&+2\beta^2\left\| { \partial_x\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_1\vert\omega^2 dx\\ = & 2\displaystyle {\int}_{ \mathbb{R}}{g}\omega \partial_x\omega dx -\kappa\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} -2\gamma^2\displaystyle {\int}_{ \mathbb{R}}u_2\left(\vert u_1\vert- \vert u_2\vert\right)\omega dx. \end{split} \end{equation}$

(3.19)

Due to Eqs (3.13), (3.15) and (3.16) and the Young inequality,

$\begin{align*} &2\displaystyle {\int}_{ \mathbb{R}}\vert {g}\vert\vert\omega\vert\vert \partial_x\omega\vert dx\le 2C(T)\displaystyle {\int}_{ \mathbb{R}}\vert\omega\vert\vert \partial_x\omega\vert dx\\ &\qquad = 2\displaystyle {\int}_{ \mathbb{R}}\left\vert\frac{C(T)\omega}{\beta}\right\vert\left\vert\beta \partial_x\omega\right\vert dx\le C(T)\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})} +\beta^2\left\| { \partial_x\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})},\\ &2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_2\vert\left\vert\left(\vert u_1\vert- \vert u_2\vert\right)\right\vert\vert\omega\vert dx\le 2\gamma^2\left\| {u_2} \right\|_{L^{\infty}((0,T)\times \mathbb{R})}\displaystyle {\int}_{ \mathbb{R}}\left\vert\left(\vert u_1\vert- \vert u_2\vert\right)\right\vert\vert\omega\vert dx\\ &\qquad\le C(T)\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{align*}$

It follows from Eq (3.19) that

$\begin{equation*} \left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+\beta^2\left\| { \partial_x\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}+2\gamma^2\displaystyle {\int}_{ \mathbb{R}}\vert u_1\vert\omega^2 dx\le C(T)\left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}. \end{equation*}$

The Gronwall Lemma and Eq (3.12) give

$\begin{equation} \begin{split} \left\| {\omega(t,\cdot)} \right\|^2_{L^2( \mathbb{R})}&+\beta^2 e^{C(T)t}\displaystyle {\int}_{0}^{t}e^{-C(T)s}\left\| { \partial_x\omega(s,\cdot)} \right\|^2_{L^2( \mathbb{R})}ds\\ &+2\gamma^2 e^{C(T)t}\displaystyle {\int}_{0}^{t}\!\!\!\displaystyle {\int}_{ \mathbb{R}}e^{-C(T)s}\vert u_1\vert\omega^2 dsdx\le e^{C(T)t}\left\| {\omega_0} \right\|^2_{L^2( \mathbb{R})}. \end{split} \end{equation}$

(3.20)

Equation (1.9) follows from Eqs (3.11) and (3.20). □

Author contributions

Giuseppe Maria Coclite and Lorenzo Di Ruvo equally contributed to the methodologies, typesetting, and the development of the paper.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

Giuseppe Maria Coclite is an editorial boardmember for [Networks and Heterogeneous Media] and was not involved inthe editorial review or the decision to publish this article.

GMC is member of the Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). GMC has been partially supported by the Project funded under the National Recovery and Resilience Plan (NRRP), Mission 4 Component 2 Investment 1.4 -Call for tender No. 3138 of 16/12/2021 of Italian Ministry of University and Research funded by the European Union -NextGenerationEUoAward Number: CN000023, Concession Decree No. 1033 of 17/06/2022 adopted by the Italian Ministry of University and Research, CUP: D93C22000410001, Centro Nazionale per la Mobilità Sostenibile, the Italian Ministry of Education, University and Research under the Programme Department of Excellence Legge 232/2016 (Grant No. CUP - D93C23000100001), and the Research Project of National Relevance "Evolution problems involving interacting scales" granted by the Italian Ministry of Education, University and Research (MIUR Prin 2022, project code 2022M9BKBC, Grant No. CUP D53D23005880006). GMC expresses its gratitude to the HIAS - Hamburg Institute for Advanced Study for their warm hospitality.

Conflict of interest

The authors declare there is no conflict of interest.

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