Blockchain-based multi-authority revocable data sharing scheme in smart grid

Xiao-Dong Yang; Ze-Fan Liao; Bin Shu; Ai-Jia Chen; Xiao-Dong Yang; Ze-Fan Liao; Bin Shu; Ai-Jia Chen

doi:10.3934/mbe.2023531

Mathematical Biosciences and Engineering

2023, Volume 20, Issue 7: 11957-11977. doi: 10.3934/mbe.2023531

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Research article

Blockchain-based multi-authority revocable data sharing scheme in smart grid

1.
College of Computer Science and Engineering, Northwest Normal University, Lanzhou 730070, China
2.
China Telecom WanWei Information Technology Co., LTD, Lanzhou 730030, China

Academic Editor: Danilo Pelusi

Received: 26 February 2023 Revised: 09 April 2023 Accepted: 23 April 2023 Published: 12 May 2023

In view of the problems of inefficient data encryption, non-support of malicious user revocation and data integrity checking in current smart grid data sharing schemes, this paper proposes a blockchain-based multi-authority revocable data sharing scheme in the smart grid. Using online/offline encryption technology with hybrid encryption technology enhances the encryption performance for the data owner. The use of user binary tree technology enables the traceability and revocability of malicious users. The introduction of multiple attribute authorization authorities eliminates the threat of collusive attacks that exist in traditional data-sharing schemes. In addition, the semi-honest problem of third-party servers is solved by uploading data verification credentials to the blockchain. The security analysis results show that the scheme can resist selective plaintext attacks and collusion attacks. The performance analysis results show that the proposed scheme has lower computational overhead and better functionality than similar schemes, which is suitable for secure data sharing in smart grids.

Keywords:

Citation: Xiao-Dong Yang, Ze-Fan Liao, Bin Shu, Ai-Jia Chen. Blockchain-based multi-authority revocable data sharing scheme in smart grid[J]. Mathematical Biosciences and Engineering, 2023, 20(7): 11957-11977. doi: 10.3934/mbe.2023531

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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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