Wave propagation in dilatant granular materials

Abstract

The wave propagation in dilatant granular materials is described by a nonlinear evolution equation of the fifth order deduced by Giovine–Oliveri in (Meccanica 30(4) (1995) 341–357). In this paper, we study the well-posedness of the classical solutions for the Cauchy problem, associated with this equation.

Keywords

Existence Uniqueness Stability Wave propagation Dilatant granular materials Cauchy problem

1. Introduction

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

\begin{aligned} {\begin{cases} \partial_{t} u + κ u_{ε} \partial_{x} u_{ε} + δ \partial_{x}^{3} u - β^{2} \partial_{t} \partial_{x}^{2} u \\ + α \partial_{x}^{2} (u \partial_{x} u) + γ \partial_{x}^{5} u = 0, & 0 < t < T, x \in R, \\ u (0, x) = u_{0} (x), & x \in R, \end{cases} \end{aligned}

(1.1)

with

κ, δ, β, α, γ \in R

, such that

\begin{aligned} β, α \neq 0. \end{aligned}

(1.2)

\begin{aligned} γ \neq 0, \end{aligned}

(1.3)

on the initial datum, we assume

\begin{aligned} u_{0} \in H^{4} (R), u_{0} \neq 0. \end{aligned}

(1.4)

On the other hand, if

\begin{aligned} γ = 0, \end{aligned}

(1.5)

we assume

\begin{aligned} u_{0} \in H^{ℓ} (R), u_{0} \neq 0, ℓ \in {2, 3} . \end{aligned}

(1.6)

Equation (1.1) was deduced in [28] in order to describe the wave propagation in dilatant granular materials. The variable u is the bulk density, δ, α, γ are macro- and microlevel dispersion parameters, while β is a parameter involving the ratio of the grain size and the wavelength.

In [46], Equation (1.1) with $γ = 0$ was deduced to model the evolution of the non-linear wave propagation compatible with the non-diffusive version of the model of bubbly liquids introduced by Drumheller and Bedford [26].

The existence of the solitons for (1.1) is proven in [29,48], while numerical solutions are analyzed in [30].

Taking $β = α = 0$ , (1.1) reads

\begin{aligned} \partial_{t} u + κ u_{ε} \partial_{x} u_{ε} + δ \partial_{x}^{3} u + γ \partial_{x}^{5} u = 0, \end{aligned}

(1.7)

that was derived by Kawahara [37] to describe small-amplitude gravity capillary waves on water of a finite depth when the Weber number is close to

1 / 3

(see [45]). Moreover, in [37], the author deduced (1.7) to describe one-dimensional propagation of small-amplitude long waves in various problems of fluid dynamics and plasma physics. (1.7) is also known as the fifth-order Korteweg–de Vries equation or generalized Benney–Lin equation [7,41].

Mathematical properties of (1.7) were studied recently in many detail, including the local and global well-posedness in the Bourgain space [20,21,34,35,52], the local and global well-posedness in energy space [25,27,31,51,56,57], the existence of solitary wave solution [9,32], the stability of periodic traveling wave solutions see [1,2,44,50,58], the well-posedness of the initial-boundary value problem on a bounded domain [10,13,24,38], the initial-boundary value problem on the half-line [11,12], periodic solutions (see [8,47]), and numerical solutions see [5,6,23,33,42,53]. In particular, [25] shows that the Cauchy problem of (1.7) is ill-posed in some energy spaces.

To obtain the exact solutions for (1.7), a number of methods have been proposed in the literature. In particular, the Lie group analysis method is used in [4,36], the exp-function method is used in [3,39,43], the polygonal method is used in [22,54], the generalized F-expansion method is used in [55].

In [18], the well-posedness of the Cauchy problem for (1.7) is proven, assuming on the initial datum (1.4). Finally, following [14,15,40,49], the convergence of the solution of (1.7) as $α, β, γ, δ \to 0$ is studied in [16,17].

The main result of this paper are the following theorems.

Theorem 1.1.

Assume (1.2), (1.3) and (1.4). Given κ, δ, β, α, γ, T, there exists an unique solution u of (1.1) such that

\begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{4} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u \in L^{\infty} (0, T; L^{2} (R)) . \end{aligned}

(1.8)

Moreover, if

u_{1}

and

u_{2}

are two solutions of (1.1) in correspondence of the initial data

u_{1, 0}

and

u_{2, 0}

, we have that

\begin{aligned} | | u_{1} (t, \cdot) - u_{2} (t, \cdot) | |_{H^{1} (R)} ⩽ \frac{τ_{2}^{2} e^{C (T) t}}{τ_{1}^{2}} | | u_{1, 0} - u_{2, 0} | |_{H^{1} (R)}, \end{aligned}

(1.9)

for some suitable

C (T) > 0

, and every,

0 ⩽ t ⩽ T

, where

\begin{aligned} τ_{2}^{2} = min {1, β^{2}}, τ_{2}^{2} = max {1, β^{2}} . \end{aligned}

(1.10)

Theorem 1.2.

Assume (1.2), (1.5) and (1.6) with $ℓ = 2$ . Given κ, δ, β, α, T, there exists a solution u of (1.1) such that

\begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{2} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u \in L^{\infty} (0, T; L^{2} (R)) . \end{aligned}

(1.11)

Theorem 1.3.

Assume (1.2), (1.5) and (1.6) with $ℓ = 3$ . Given κ, δ, β, α, T, there exists an unique solution u of (1.1) such that

\begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{3} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u \in L^{\infty} (0, T; L^{2} (R)) . \end{aligned}

(1.12)

Moreover, (1.9) holds.

Using the Sobolev Immersion Theorem, Theorems 1.1, 1.2, 1.3 give the existence of a solution of (1.1), without additional assumption on the constants. Moreover, in Theorems 1.1 and 1.3 we have also the uniqueness and stability of the solution of (1.1).

The paper is organized as follows. In Sections 2, 3, 4, we prove Theorems 1.1, 1.2, 1.3, respectively.

2. Case:

γ \neq 0

In this section, we assume (1.3) and (1.4).

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

Fix a small number $0 < ε < 1$ and let $u_{ε} = u_{ε} (t, x)$ be the unique classical solution of the following problem:

\begin{aligned} {\begin{cases} \partial_{t} u_{ε} + κ u_{ε} \partial_{x} u_{ε} + δ \partial_{x}^{3} u_{ε} - β^{2} \partial_{t} \partial_{x}^{2} u_{ε} \\ + α \partial_{x}^{2} (u_{ε} \partial_{x} u_{ε}) + γ \partial_{x}^{5} u_{ε} = ε \partial_{x}^{6} u_{ε}, & 0 < t < T, x \in R, \\ u_{ε} (0, x) = u_{ε, 0} (x), & x \in R, \end{cases} \end{aligned}

(2.1)

where

u_{ε, 0}

is a

C^{\infty}

approximation of

u_{0}

, such that

\begin{aligned} | | u_{ε, 0} | |_{H^{4} (R)} ⩽ | | u_{0} | |_{H^{4} (R)}, \sqrt{ε} | | \partial_{x}^{5} u_{ε, 0} | |_{L^{2} (R)} ⩽ C_{0}, \end{aligned}

(2.2)

where

C_{0}

an a positive constant independent of

ε

Let us prove some a priori estimates on $u_{ε}$ , denoting with $C_{0}$ constants which depend only on the initial data, and with $C (T)$ , the constants which depend also on T.

Following [19, Lemma 2.2], we prove the following lemma

Lemma 2.1.

Given κ, δ, β, α, γ, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} | | u_{ε} (t, \cdot) | |_{L^{2} (R)}, | | \partial_{x} u_{ε} (t, \cdot) | |_{L^{2} (R)}, | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)} ⩽ C (T), \end{aligned}

(2.3)

\begin{aligned} | | \partial_{x} u_{ε} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \end{aligned}

(2.4)

\begin{aligned} | | u_{ε} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \end{aligned}

(2.5)

\begin{aligned} ε \int_{0}^{t} | | \partial_{x}^{3} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T), \end{aligned}

(2.6)

for every

0 ⩽ t ⩽ T

Proof.

We begin by proving the following inequality:

\begin{aligned} \frac{1}{\sqrt{| | u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + (β^{2} + A) | | \partial_{x} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + A β^{2} | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}}} \\ + \frac{2 \sqrt{2} | α - A κ | + 5 | α | t}{A^{\frac{1}{4}} | β |^{3}} \\ ⩾ \frac{1}{\sqrt{| | u_{ε, 0} | |_{L^{2} (R)}^{2} + (β^{2} + A) | | \partial_{x} u_{ε, 0} | |_{L^{2} (R)}^{2} + A β^{2} | | \partial_{x}^{2} u_{ε, 0} | |_{L^{2} (R)}^{2}}} \end{aligned}

(2.7)

for every

0 ⩽ t ⩽ T

, where A is an arbitrary positive constant.

We begin by observing that

\begin{aligned} \partial_{x} (u_{ε} \partial_{x} u_{ε}) = {(\partial_{x} u_{ε})}^{2} + u_{ε} \partial_{x}^{2} u_{ε}, \partial_{x}^{2} (u_{ε} \partial_{x} u_{ε}) = 3 \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} + u_{ε} \partial_{x}^{3} u_{ε} . \end{aligned}

(2.8)

Consider A a positive constant. Multiplying (2.1) by

2 u_{ε} - 2 A \partial_{x}^{2} u_{ε}

, thanks to (2.8), an integration on

R

gives

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 \int_{R} u_{ε} \partial_{t} u_{ε} d x - 2 (β^{2} + A) \int_{R} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 2 κ \int_{R} u_{ε}^{2} \partial_{x} u_{ε} d x + 2 A κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x - 2 δ \int_{R} u_{ε} \partial_{x}^{3} u_{ε} d x \\ + 2 A δ \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d x - 2 α \int_{R} u_{ε} \partial_{x}^{2} (u_{ε} \partial_{x} u_{ε}) + 6 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x \\ + 2 A α \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d x - 2 γ \int_{R} u_{ε} \partial_{x}^{5} u_{ε} d x + 2 A γ \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{5} u_{ε} d x \\ + 2 ε \int_{R} u_{ε} \partial_{x}^{6} u_{ε} d x - 2 A ε \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{6} u_{ε} d x \\ = - A κ \int_{R} (\partial_{x} u_{ε})^{3} d x + 2 δ \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x + 2 α \int_{R} \partial_{x} u_{ε} \partial_{x} (u_{ε} \partial_{x} u_{ε}) d x \\ + 5 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x - 2 ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{5} u_{ε} d x + 2 A ε \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{5} u_{ε} d x \\ = - A κ \int_{R} (\partial_{x} u_{ε})^{3} d x - 2 α \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} + 5 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x \\ + 2 ε \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} - 2 A ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = (α - A κ) \int_{R} (\partial_{x} u_{ε})^{3} d x + 5 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x \\ - 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} - 2 A ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 A ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = (α - A κ) \int_{R} (\partial_{x} u_{ε})^{3} d x + 5 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x . \end{aligned}

(2.9)

Observe that

\begin{aligned} | α - A κ | \int_{R} | \partial_{x} u_{ε} |^{3} d x & ⩽ | α - A κ | ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = \frac{| α - A κ |}{β^{2}} β^{2} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{| α - A κ |}{β^{2}} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 5 A | α | \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{2} u_{ε})^{2} d x & ⩽ 5 A | α | ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = 5 A β^{2} \frac{| α |}{β^{2}} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

It follows from (2.9) that

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 A ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{| α - A κ |}{β^{2}} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ + 5 A β^{2} \frac{| α |}{β^{2}} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{| α - A κ | + 5 | α |}{β^{2}} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{\infty} (R)} \times \\ \times ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) . \end{aligned}

(2.10)

We define

\begin{aligned} X_{ε} (t) = ‖ u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2} + (β^{2} + A) ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2} + A β^{2} ‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2} . \end{aligned}

(2.11)

Therefore, by (2.10), we have that

\begin{aligned} \frac{d X_{ε} (t)}{d t} ⩽ \frac{| α - A κ | + 5 | α |}{β^{2}} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{\infty} (R)} X_{ε} (t) . \end{aligned}

(2.12)

Thanks to (2.11) and the Hölder inequality,

\begin{aligned} (\partial_{x} u_{ε} (t, x))^{2} & = 2 \int_{- \infty}^{x} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d y ⩽ 2 \int_{R} | \partial_{x} u_{ε} | | \partial_{x}^{2} u_{ε} | d x \\ ⩽ 2 ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖_{L^{2} (R)} \\ = \frac{2}{β^{2} A^{\frac{1}{2}}} \sqrt{β^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}} \sqrt{A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}} \\ ⩽ \frac{2}{β^{2} A^{\frac{1}{2}}} \sqrt{(β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}} \sqrt{A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}} \\ ⩽ \frac{2}{β^{2} A^{\frac{1}{2}}} X_{ε} (t) . \end{aligned}

Hence,

\begin{aligned} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{\infty} (R)}^{2} ⩽ \frac{2}{β^{2} A^{\frac{1}{2}}} X_{ε} (t) . \end{aligned}

(2.13)

It follows from (2.12) that

\begin{aligned} \frac{d X_{ε} (t)}{d t} ⩽ \frac{\sqrt{2} | α - A κ | + 5 | α |}{A^{\frac{1}{4}} {| β |}^{3}} X_{ε}^{\frac{3}{2}} (t), \end{aligned}

that is,

\begin{aligned} X_{ε}^{- \frac{3}{2}} (t) \frac{d X_{ε} (t)}{d t} ⩽ \frac{\sqrt{2} | α - A κ | + 5 | α |}{A^{\frac{1}{4}} | β |^{3}} . \end{aligned}

(2.14)

An integration on

(0, t)

gives

\begin{aligned} \frac{1}{\sqrt{X_{ε} (t)}} - \frac{1}{\sqrt{X_{ε} (0)}} ⩾ - \frac{2 \sqrt{2} | α - A κ | + 5 | α |}{A^{\frac{1}{4}} | β |^{3}} . \end{aligned}

(2.15)

Using (2.11) in (2.15), we have (2.7).

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

\begin{aligned} ‖ u_{ε, 0} ‖_{L^{2} (R)}^{2} + (β^{2} + A) ‖ \partial_{x} u_{ε, 0} ‖_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε, 0} ‖}_{L^{2} (R)}^{2} ⩽ Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}, \end{aligned}

where

\begin{aligned} Y_{0} := ‖ u_{0} ‖_{L^{2} (R)}^{2}, Y_{1} := ‖ \partial_{x} u_{0} ‖_{L^{2} (R)}^{2}, Y_{2} := {‖ \partial_{x}^{2} u_{0} ‖}_{L^{2} (R)}^{2} . \end{aligned}

Consequentially,

\begin{aligned} \begin{array}{ll} \frac{1}{\sqrt{‖ u_{ε, 0} ‖_{L^{2} (R)}^{2} + (β^{2} + A) ‖ \partial_{x} u_{ε, 0} ‖_{L^{2} (R)}^{2} + A β^{2} ‖ \partial_{x}^{2} u_{ε, 0} ‖_{L^{2} (R)}^{2}}} \\ ⩾ \frac{1}{\sqrt{Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}}} . \end{array} \end{aligned}

(2.16)

Moreover

\begin{aligned} \frac{2 \sqrt{2} | α - A κ | + 5 | α | T}{A^{\frac{1}{4}} | β |^{3}} ⩾ \frac{2 \sqrt{2} | α - A κ | + 5 | α | t}{A^{\frac{1}{4}} | β |^{3}} . \end{aligned}

(2.17)

It follows from (2.7), (2.16) and (2.17) that

\begin{aligned} \frac{1}{\sqrt{| | u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + (β^{2} + A) | | \partial_{x} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + A β^{2} | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}}} \\ ⩾ \frac{1}{\sqrt{Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}}} - \frac{2 \sqrt{2} | α - A κ | + 5 | α | T}{A^{\frac{1}{4}} | β |^{3}} \\ = \frac{A^{\frac{1}{4}} | β |^{3} - (2 \sqrt{2} | α - A κ | + 5 | α | T) (\sqrt{Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}})}{A^{\frac{1}{4}} | β |^{3} (\sqrt{Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}})} . \end{aligned}

(2.18)

We search A such that

\begin{aligned} A^{\frac{1}{4}} | β |^{3} - (2 \sqrt{2} | α - A κ | + 5 | α | T) (\sqrt{Y_{0} + (β^{2} + A) Y_{1} + A β^{2} Y_{2}}) > 0. \end{aligned}

(2.19)

Taking

\begin{aligned} A = {| β |}^{n}, 0 < n < \frac{8}{5} \end{aligned}

(2.20)

(2.19) reads

\begin{aligned} | β |^{\frac{n + 12}{4}} - (2 \sqrt{2} | α - | β |^{n} κ | + 5 | α | T) (\sqrt{Y_{0} + (β^{2} + | β |^{n}) Y_{1} + | β |^{2 + n} Y_{2}}) > 0. \end{aligned}

(2.21)

Under Assumption (2.20), we have that

\begin{aligned} lim_{| β | \to \infty} | β |^{\frac{n + 12}{4}} - (2 \sqrt{2} | α - | β |^{n} κ | + 5 | α | T) (\sqrt{Y_{0} + (β^{2} + | β |^{n}) Y_{1} + | β |^{2 + n} Y_{2}}) = \infty . \end{aligned}

(2.22)

In fact, we observe that

\begin{aligned} \deg (\sqrt{Y_{0} + (β^{2} + | β |^{n}) Y_{1} + | β |^{2 + n} Y_{2}}) = \frac{2 + n}{2}, \\ \deg (2 \sqrt{2} | α - | β |^{n} κ | + 5 | α | T) = n . \end{aligned}

Therefore,

\begin{aligned} \deg (\sqrt{Y_{0} + (β^{2} + | β |^{n}) Y_{1} + | β |^{2 + n} Y_{2}}) \deg (2 \sqrt{2} | α - | β |^{n} κ | + 5 | α | T) = \frac{2 + 3 n}{2} . \end{aligned}

(2.23)

By (2.23), we have (2.22) if only if

\begin{aligned} \frac{n + 12}{4} > \frac{2 + 3 n}{2}, \end{aligned}

that is,

\begin{aligned} n < \frac{8}{5} . \end{aligned}

Therefore, thanks to (2.22), (2.21) holds, taking A very large and up to rescaling, we can have

A = | β |^{n}

, with n defined in (2.20).

Consequently, by (2.18), (2.19) and (2.22),

\begin{aligned} \frac{1}{\sqrt{‖ u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2} + (β^{2} + A) ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2} + A β^{2} ‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖_{L^{2} (R)}^{2}}} ⩾ C (T) . \end{aligned}

Therefore,

\begin{aligned} {‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \end{aligned}

which gives (2.3).

We prove (2.4). Thanks to (2.3), (2.11) with $A = 1$ and (2.13) with $A = 1$

\begin{aligned} ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)}^{2} \\ ⩽ \frac{2}{β^{2}} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) ⩽ C (T) . \end{aligned}

Hence,

\begin{aligned} ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)}^{2} ⩽ C (T), \end{aligned}

which gives (2.4).

We prove (2.5). Thanks to (2.3) and the Hölder inequality,

\begin{aligned} u_{ε}^{2} (t, x) & = 2 \int_{- \infty}^{x} u_{ε} \partial_{x} u_{ε} d y ⩽ 2 \int_{R} | u_{ε} | | \partial_{x} u_{ε} | d x \\ ⩽ 2 ‖ u_{ε} (t, \cdot) ‖_{L^{2} (R)} ‖ \partial_{x} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ⩽ C (T) . \end{aligned}

Hence,

\begin{aligned} ‖ u_{ε} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \end{aligned}

which gives (2.5).

We prove (2.6). We consider (2.9) with $A = 1$ , that is

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = (α - κ) \int_{R} (\partial_{x} u_{ε})^{3} d x + 5 α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x . \end{aligned}

(2.24)

Thanks to (2.3) and (2.4), we have that

\begin{aligned} | α - κ | \int_{R} | \partial_{x} u_{ε} |^{3} d x ⩽ α - κ | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \\ 5 | α | \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{2} u_{ε})^{2} d x ⩽ 5 | α | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) . \end{aligned}

It follows from (2.24) that

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) . \end{aligned}

Integrating on

(0, t)

, by (2.2), we get

\begin{aligned} {‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ + 2 ε \int_{0}^{t} {‖ \partial_{x}^{3} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s + 2 ε \int_{0}^{t} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C (T) t ⩽ C (T), \end{aligned}

which gives (2.6).

Lemma 2.2.

Given κ, δ, β, α, γ, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ + 2 ε e^{C (T) t} \int_{0}^{t} e^{- C (T) s} | | \partial_{x}^{4} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T), \end{aligned}

(2.25)

for every

0 ⩽ t ⩽ T

. In particular,

\begin{aligned} | | \partial_{x}^{2} u_{ε} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{aligned}

(2.26)

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (2.1) by $2 \partial_{x}^{4} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} \frac{d}{d t} ({‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} u_{ε} d x - 2 β^{2} \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 2 δ \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} \\ - 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x + 2 ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{x}^{6} u_{ε} d x \\ = 2 κ \int_{R} (\partial_{x} u_{ε})^{2} \partial_{x}^{3} u_{ε} + 2 κ \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d x + 7 α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x - 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 5 κ \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x + 7 α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x - 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} (| | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}) + 2 ε | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ = - 5 κ \int_{R} \partial_{x} u_{ε} {(\partial_{x}^{2} u_{ε})}^{2} d x + 7 α \int_{R} \partial_{x} u_{ε} {(\partial_{x}^{3} u_{ε})}^{2} d x . \end{aligned}

(2.27)

Thanks to (2.3) and (2.4),

\begin{aligned} 5 | κ | \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{2} u_{ε})^{2} d x ⩽ 5 | κ | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \\ 7 | α | \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{3} u_{ε})^{2} d x ⩽ 7 | α | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

It follows from (2.27) that

\begin{aligned} \frac{d}{d t} ({‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C (T) \\ ⩽ C (T) ({‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + C (T) . \end{aligned}

Therefore, by the Gronwall Lemma and (2.2), we obtain that

\begin{aligned} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε e^{C (T) t} \int_{0}^{t} e^{- C (T) s} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} e^{C (T) t} + C (T) e^{C (T) t} \int_{0}^{t} e^{- C (T) s} d s ⩽ C (T), \end{aligned}

which gives (2.25).

Finally, we prove (2.26). Due to (2.25) and the Hölder inequality,

\begin{aligned} (\partial_{x}^{2} u_{ε} (t, x))^{2} & = \int_{- \infty}^{x} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d y ⩽ 2 \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{3} u_{ε} | d x \\ ⩽ 2 ‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖_{L^{2} (R)} . \end{aligned}

Hence,

\begin{aligned} ‖ \partial_{x}^{2} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \end{aligned}

which gives (2.26).

Lemma 2.3.

Given κ, δ, β, α, γ, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ + 2 ε \int_{0}^{t} | | \partial_{x}^{6} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T) . \end{aligned}

(2.28)

for every

0 ⩽ t ⩽ T

. In particular,

\begin{aligned} | | \partial_{x}^{3} u_{ε} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T) \end{aligned}

(2.29)

Proof.

Let $0 ⩽ t ⩽$ . Multiplying (2.1) by $- 2 \partial_{x}^{6} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} \frac{d}{d t} ({‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = - 2 \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} u_{ε} d x + 2 β^{2} \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{6} u_{ε} d x + 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{6} u_{ε} d x + 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{6} u_{ε} d x \\ - 2 ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 2 κ \int_{R} (\partial_{x} u_{ε})^{2} \partial_{x}^{5} u_{ε} - 2 κ \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{5} u_{ε} d x - 6 α \int_{R} (\partial_{x}^{2} u_{ε})^{2} \partial_{x}^{5} u_{ε} d x \\ - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{5} u_{ε} d x - 2 ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = 6 κ \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x + 2 κ \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x + 12 α \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x \\ - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} - 2 ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 7 κ \int_{R} (\partial_{x} u_{ε}) (\partial_{x}^{3} u_{ε})^{2} d x + 18 α \int_{R} \partial_{x}^{2} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x - 2 ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} (| | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}) + 2 ε | | \partial_{x}^{6} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ = - 7 κ \int_{R} (\partial_{x} u_{ε}) {(\partial_{x}^{3} u_{ε})}^{2} d x + 18 α \int_{R} \partial_{x}^{2} u_{ε} {(\partial_{x}^{3} u_{ε})}^{2} d x . \end{aligned}

(2.30)

Thanks to (2.4), (2.25) and (2.26),

\begin{aligned} 7 | κ | \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{3} u_{ε})^{2} d x ⩽ 7 | κ | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \\ 18 | α | \int_{R} | \partial_{x}^{2} u_{ε} | (\partial_{x}^{3} u_{ε})^{2} d x ⩽ 18 | α | ‖ \partial_{x}^{2} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) . \end{aligned}

It follows from (2.30) that

\begin{aligned} \frac{d}{d t} ({‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) . \end{aligned}

Integrating on

(0, t)

, by (2.2), we have that

\begin{aligned} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε \int_{0}^{t} {‖ \partial_{x}^{6} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C (T) t ⩽ C (T), \end{aligned}

which gives (2.28).

Finally, we prove (2.29). Thanks to (2.28) and the Hölder inequality,

\begin{aligned} (\partial_{x}^{3} u_{ε} (t, x))^{2} & = 2 \int_{- \infty}^{x} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d y ⩽ 2 \int_{R} | \partial_{x}^{3} u_{ε} | | \partial_{x}^{4} u_{ε} | d x \\ ⩽ 2 ‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ⩽ C (T) . \end{aligned}

Hence,

\begin{aligned} {‖ \partial_{x}^{3} u_{ε} ‖}_{L^{\infty} ((0, T) \times R)}^{2} ⩽ C (T), \end{aligned}

which gives (2.29).

Lemma 2.4.

Given κ, δ, β, α, γ, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} ε | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} ε | | \partial_{x}^{5} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ + ε^{2} \int_{0}^{t} | | \partial_{x}^{7} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T) \end{aligned}

(2.31)

for every

0 ⩽ t ⩽ T

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (2.1) by $2 ε \partial_{x}^{8} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} \frac{d}{d t} (ε {‖ ε \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 ε \int_{R} \partial_{x}^{8} u_{ε} \partial_{t} u_{ε} d x + β^{2} ε \int_{R} \partial_{x}^{8} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = - 2 κ ε \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{8} u_{ε} d x - 6 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{8} u_{ε} d x - 2 α ε \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{8} u_{ε} d x \\ + 2 ε^{2} \int_{R} \partial_{x}^{6} u_{ε} \partial_{x}^{8} u_{ε} d x \\ = 2 κ ε \int_{R} (\partial_{x} u_{ε})^{2} \partial_{x}^{7} u_{ε} d x + 2 κ ε \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{7} u_{ε} d x + 6 α ε \int_{R} (\partial_{x}^{2} u_{ε})^{2} \partial_{x}^{7} u_{ε} d s \\ + 8 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{7} u_{ε} d x + 2 α ε \int_{R} u_{ε} \partial_{x}^{4} u_{ε} \partial_{x}^{7} u_{ε} d x - 2 ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} (ε {‖ ε \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = 2 κ ε \int_{R} (\partial_{x} u_{ε})^{2} \partial_{x}^{7} u_{ε} d x + 2 κ ε \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{7} u_{ε} d x \\ + 6 α ε \int_{R} (\partial_{x}^{2} u_{ε})^{2} \partial_{x}^{7} u_{ε} d s + 8 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{7} u_{ε} d x \\ + 2 α ε \int_{R} u_{ε} \partial_{x}^{4} u_{ε} \partial_{x}^{7} u_{ε} d x . \end{aligned}

(2.32)

Due to (2.3), (2.4), (2.5), (2.26) (2.28) and the Young inequality,

\begin{aligned} 2 | κ | ε \int_{R} (\partial_{x} u_{ε})^{2} | \partial_{x}^{7} u_{ε} | d x & ⩽ 2 | κ | ε ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ ⩽ 2 C (T) ε \int_{R} | u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ = 2 \int_{R} | \frac{C (T) \partial_{x} u_{ε}}{\sqrt{D_{1}}} | | ε \sqrt{D_{1}} \partial_{x}^{7} u_{ε} | \\ ⩽ \frac{C (T)}{D_{1}} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{1}} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | κ | ε \int_{R} | u_{ε} | | \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} u_{ε} | d x & ⩽ 2 | κ | ε ‖ u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ ⩽ 2 C (T) ε \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} | d x \\ = 2 \int_{R} | \frac{C (T) \partial_{x}^{2} u_{ε}}{\sqrt{D_{1}}} | | \sqrt{D_{1}} ε \partial_{x}^{7} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{1}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{1}} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{aligned}

\begin{aligned} 6 | α | ε \int_{R} (\partial_{x}^{2} u_{ε})^{2} | \partial_{x}^{7} u_{ε} | d s & ⩽ 6 | α | ε ‖ \partial_{x}^{2} u_{ε} ‖_{L^{\infty} ((0, T \times R)} \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ ⩽ 2 C (T) ε \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ = 2 \int_{R} | \frac{C (T) \partial_{x}^{2} u_{ε}}{\sqrt{D_{1}}} | | \sqrt{D_{1}} ε \partial_{x}^{7} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{1}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{1}} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 8 | α | ε \int_{R} | \partial_{x} u_{ε} | | \partial_{x}^{3} u_{ε} | | \partial_{x}^{7} u_{ε} | d x & ⩽ 8 | α | ε ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{2} | u_{ε} | \partial_{x}^{7} u_{ε} d x \\ ⩽ 2 C (T) ε \int_{R} | \partial_{x}^{2} | u_{ε} | \partial_{x}^{7} u_{ε} d x \\ = 2 \int_{R} | \frac{C (T) \partial_{x}^{2} u_{ε}}{\sqrt{D_{1}}} | | \sqrt{D_{1}} ε \partial_{x}^{7} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{1}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{1}} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | α | ε \int_{R} | u_{ε} | | \partial_{x}^{4} u_{ε} | | \partial_{x}^{7} u_{ε} | d x & ⩽ 2 | α | ε ‖ u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{4} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ ⩽ 2 C (T) ε \int_{R} | \partial_{x}^{4} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ = 2 \int_{R} | \frac{C (T) \partial_{x}^{4} u_{ε}}{\sqrt{D_{1}}} | | \sqrt{D_{1}} ε \partial_{x}^{7} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{1}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{1}} + D_{1} ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{aligned}

where

D_{1}

is a positive constant, which will be specified later. It follows from (2.32) that

\begin{aligned} \frac{d}{d t} (ε {‖ ε \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + (2 - 5 D_{1}) ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ \frac{C (T)}{D_{1}} . \end{aligned}

Choosing

D_{1} = \frac{1}{5}

, we have that

\begin{aligned} \frac{d}{d t} (ε {‖ ε \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) . \end{aligned}

Integrating on

(0, t)

, by (2.2), we have that

\begin{aligned} ε {‖ ε \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε^{2} \int_{0}^{t} {‖ \partial_{x}^{7} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C (T) t ⩽ C (T), \end{aligned}

which gives (2.31).

Lemma 2.5.

Given κ, δ, β, α, γ, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} | | \partial_{t} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{t} \partial_{x} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} ⩽ C (T), \end{aligned}

(2.33)

\begin{aligned} | | \partial_{t} u_{ε} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \end{aligned}

(2.34)

for every

0 ⩽ t ⩽ T

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (2.1) by $2 \partial_{t} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} 2 {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 β^{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = 2 \int_{R} (\partial_{t} u_{ε})^{2} d x - 2 β^{2} \int_{R} \partial_{t} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{t} u_{ε} d x - 2 δ \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} \partial_{x} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} \\ - 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} d x - 2 γ \int_{R} \partial_{x}^{5} u_{ε} \partial_{t} u_{ε} d x + 2 ε \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} u_{ε} d x \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{t} u_{ε} d x + 2 δ \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} d x \\ - 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} d x + 2 γ \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x} u_{ε} d x - 2 ε \int_{R} \partial_{x}^{5} u_{ε} \partial_{t} \partial_{x} u_{ε} d x . \end{aligned}

(2.35)

Since

0 < ε < 1

, thanks to (2.3), (2.4), (2.25), (2.28), (2.31) and the Young inequality,

\begin{aligned} 2 | κ | \int_{R} | u_{ε} | | \partial_{x} u_{ε} | | \partial_{t} u_{ε} | d x & ⩽ 2 | κ | ‖ u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x} u_{ε} | | \partial_{t} u_{ε} | d x \\ ⩽ 2 C (T) \int_{R} | \partial_{x} u_{ε} | | \partial_{t} u_{ε} | d x = 2 \int_{R} | \frac{C (T) \partial_{x} u_{ε}}{\sqrt{D_{2}}} | | \sqrt{D_{2}} t | d x \\ ⩽ \frac{C (T)}{D_{2}} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{1} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{2}} + D_{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | δ | \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x & = 2 \int_{R} | \frac{δ \partial_{x}^{2} u_{ε}}{β^{2} \sqrt{D_{3}}} | | β \sqrt{D_{3}} \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ \frac{δ^{2}}{β^{2} D_{3}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} D_{3} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{3}} + β^{2} D_{3} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{aligned}

\begin{aligned} 6 | α | \int_{R} | \partial_{x} u_{ε} | | \partial_{x}^{2} u_{ε} | | \partial_{t} u_{ε} | d x & ⩽ 6 | α | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{t} u_{ε} | d x \\ ⩽ 2 C (T) \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{t} u_{ε} | d x = 2 \int_{R} | \frac{C (T) \partial_{x}^{2} u_{ε}}{\sqrt{D_{2}}} | | \sqrt{D_{2}} \partial_{t} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{2}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{2}} + D_{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | α | \int_{R} | u_{ε} | | \partial_{x}^{3} u_{ε} | | \partial_{t} u_{ε} | d x & ⩽ 2 | α | ‖ u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{3} u_{ε} | | \partial_{t} u_{ε} | d x \\ ⩽ 2 C (T) \int_{R} | \partial_{x}^{3} u_{ε} | | \partial_{t} u_{ε} | d x = 2 \int_{R} | \frac{C (T) \partial_{x}^{3} u_{ε}}{\sqrt{D_{2}}} | | \partial_{t} u_{ε} | d x \\ ⩽ \frac{C (T)}{D_{2}} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + D_{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{2}} + D_{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | γ | \int_{R} | \partial_{x}^{4} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x & = 2 \int_{R} | \frac{γ \partial_{x}^{4} u_{ε}}{β \sqrt{D_{3}}} | | β \sqrt{D_{3}} \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ \frac{γ^{2}}{β^{2} D_{3}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} D_{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{3}} + β^{2} D_{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 ε \int_{R} | \partial_{x}^{5} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x & = 2 \int_{R} | \frac{ε \partial_{x}^{5} u_{ε}}{β \sqrt{D_{3}}} | | β \sqrt{D_{3}} \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ \frac{ε^{2}}{β^{2} D_{3}} {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} D_{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{ε}{β^{2} D_{3}} {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} D_{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C (T)}{D_{3}} + β^{2} D_{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{aligned}

where

D_{1}

D_{2}

are two positive constants, which will be specified later. It follows from (2.35) that

\begin{aligned} (1 - 3 D_{2}) {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} (2 - 3 D_{2}) {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ \frac{C (T)}{D_{2}} + \frac{C (T)}{D_{3}} . \end{aligned}

Taking

D_{2} = D_{3} = \frac{1}{3}

, we have that

\begin{aligned} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \end{aligned}

which gives (2.33).

Finally, we prove (2.34). Thanks to (2.33) and the Hölder inequality,

\begin{aligned} (\partial_{t} u_{ε} (t, x))^{2} & = 2 \int_{- \infty}^{x} \partial_{t} u_{ε} \partial_{t} \partial_{x} u_{ε} d y ⩽ 2 \int_{R} | \partial_{t} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ 2 ‖ \partial_{t} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖_{L^{2} (R)} ⩽ C (T) . \end{aligned}

Hence,

\begin{aligned} ‖ \partial_{t} u_{ε} ‖_{L^{\infty} ((0, T) \times R)}^{2} ⩽ C (T), \end{aligned}

which gives (2.34).

Using the Sobolev Immersion Theorem, we begin by proving the following result.

Lemma 2.6.

Fix $T > 0$ . There exist a subsequence ${u_{ε_{k}}}_{k \in N}$ of ${u_{ε}}_{ε > 0}$ and an a limit function u which satisfies (1.8) such that

\begin{aligned} u_{ε_{k}} \to u a . e . a n d i n L_{loc}^{p} ((0, T) \times R), 1 ⩽ p < \infty, \end{aligned}

(2.36)

Moreover, u is solution of (1.1).

Proof.

Let $0 ⩽ t ⩽ T$ . We begin by observing that, thanks to Lemmas 2.1, 2.2, 2.3 and 2.5,

\begin{aligned} {u_{ε}}_{ε > 0} is uniformly bounded in H^{1} ((0, T) \times R), \end{aligned}

(2.37)

which gives (2.36).

Observe that, thanks to Lemmas 2.1, 2.2 and 2.3

\begin{aligned} u \in L^{\infty} (0, T; H^{4} (R)), \end{aligned}

while by Lemma 2.5, we obtain that

\begin{aligned} u \in W^{1, \infty} ((0, T) \times R) . \end{aligned}

(2.38)

Moreover, again by Lemma 2.5, we get

\begin{aligned} \partial_{t} \partial_{x} u \in L^{\infty} (0, T; L^{2} (R)) . \end{aligned}

Therefore, u is solution of (1.1) and (1.8) holds.

We are ready for the proof of Theorem 1.1. Proof of Theorem 1.1. Proof of Theorem 1.1.

Lemma 2.6 gives the existence of a solution of (1.1), such that (1.8) holds.

We prove (1.9). Let $u_{1}$ and $u_{2}$ be two solutions of (1.1), which verify (1.8), that is

\begin{aligned} {\begin{cases} \partial_{t} u_{i} + κ u_{i} \partial_{x} u_{i} - β^{2} \partial_{t} \partial_{x}^{2} u_{i} + δ \partial_{x}^{3} u_{i} \\ + α \partial_{x}^{2} (u_{i} \partial_{x} u_{i}) + γ \partial_{x}^{5} u_{i} = 0, & 0 < t < T, x \in R, \\ u_{i} (0, x) = u_{i, 0} (x), & x \in R, \end{cases} i = 1, 2. \end{aligned}

Then, the function

\begin{aligned} ω (t, x) = u_{1} (t, x) - u_{2} (t, x), \end{aligned}

(2.39)

is the solution of the following Cauchy problem:

\begin{aligned} {\begin{cases} \partial_{t} ω + \partial_{x} (u_{1}^{2} - u_{2}^{2}) - β^{2} \partial_{t} \partial_{x}^{2} ω + δ \partial_{x}^{3} ω \\ + α \partial_{x}^{2} (u_{1} \partial_{x} u_{1} - u_{2} \partial_{x} u_{2}) + γ \partial_{x}^{5} ω = 0, & t > 0, x \in R, \\ ω (0, x) = u_{1, 0} (x) - u_{2, 0} (x), & x \in R . \end{cases} \end{aligned}

(2.40)

Observe that, thanks to (2.39), we have that

\begin{aligned} u_{1}^{2} - u_{2}^{2} = (u_{1} + u_{2}) ω . \end{aligned}

(2.41)

Moreover, again by (2.39),

\begin{aligned} u_{1} \partial_{x} u_{1} - u_{2} \partial_{x} u_{2} = u_{1} \partial_{x} u_{1} - u_{1} \partial_{x} u_{2} + u_{1} \partial_{x} u_{2} - u_{2} \partial_{x} u_{2} = u_{1} \partial_{x} ω + \partial_{x} u_{2} ω . \end{aligned}

(2.42)

Thanks to (2.41) and (2.42), Equation (2.40) is equivalent to

\begin{aligned} \partial_{t} ω + \partial_{x} ((u_{1} + u_{2}) ω) - β^{2} \partial_{t} \partial_{x}^{2} ω + δ \partial_{x}^{3} ω \\ + α \partial_{x}^{2} (u_{1} \partial_{x} ω) + α \partial_{x}^{2} (\partial_{x} u_{2} ω) + γ \partial_{x}^{5} ω = 0. \end{aligned}

(2.43)

Fixed

T > 0

, since

u_{1}, u_{2} \in H^{2} (R)

, for every

0 ⩽ t ⩽ T

, we have that

\begin{aligned} | | \partial_{x} u_{1} | |_{L^{\infty} ((0, T) \times R)}, | | \partial_{x} u_{2} | |_{L^{\infty} ((0, T) \times R)}, | | \partial_{x}^{2} u_{2} | |_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{aligned}

(2.44)

Therefore, by (2.44), we have that

\begin{aligned} | \partial_{x} u_{1} + \partial_{x} u_{2} | ⩽ | \partial_{x} u_{1} | + | \partial_{x} u_{2} | ⩽ C (T) . \end{aligned}

(2.45)

Multiplying (2.43) by

2 ω

, an integration on

R

gives

\begin{aligned} \frac{d}{d t} ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 \int_{R} ω \partial_{t} ω d x - 2 β^{2} \int_{R} ω \partial_{t} \partial_{x}^{2} ω d x \\ = - 2 \int_{R} \partial_{x} ((u_{1} + u_{2}) ω) ω d x - 2 δ \int_{R} ω \partial_{x}^{3} ω d x - 2 α \int_{R} ω \partial_{x}^{2} (u_{1} \partial_{x} ω) d x \\ - 2 α \int_{R} ω \partial_{x}^{2} (\partial_{x} u_{2} ω) d x - 2 γ \int_{R} ω \partial_{x}^{5} ω d x \\ = 2 \int_{R} ((u_{1} + u_{2}) ω \partial_{x} ω d x + 2 δ \int_{R} ω \partial_{x}^{2} ω d x + 2 α \int_{R} \partial_{x} ω \partial_{x} (u_{1} \partial_{x} ω) \\ + 2 α \int_{R} \partial_{x} ω \partial_{x} (\partial_{x} u_{2} ω) + 2 γ \int_{R} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x \\ = - 2 \int_{R} ((\partial_{x} u_{1} + \partial_{x} u_{2}) ω^{2} d x - 2 α \int_{R} u_{1} ω \partial_{x} ω d x - 2 α \int_{R} \partial_{x} u_{2} \partial_{x} ω \partial_{x}^{2} ω d x \\ - 2 γ \int_{R} \partial_{x}^{2} ω \partial_{x}^{3} ω d x \\ = 2 \int_{R} ((\partial_{x} u_{1} + \partial_{x} u_{2}) ω^{2} d x + α \int_{R} \partial_{x} u_{1} ω^{2} d x + α \int_{R} \partial_{x}^{2} u_{2} (\partial_{x} ω)^{2} d x . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} (| | ω (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x} ω (t, \cdot) | |_{L^{2} (R)}^{2}) \\ ⩽ \int_{R} (\partial_{x} u_{1} + \partial_{x} u_{2}) ω^{2} d x + α \int_{R} \partial_{x} u_{1} ω^{2} d x + α \int_{R} \partial_{x}^{2} u_{2} (\partial_{x} ω)^{2} d x . \end{aligned}

(2.46)

Thanks to (2.44) and (2.44),

\begin{aligned} \int_{R} | \partial_{x} u_{1} + \partial_{x} u_{2} | ω^{2} d x ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ | α | \int_{R} | \partial_{x} u_{1} | ω^{2} d x ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ | α | \int_{R} | \partial_{x}^{2} u_{2} | (\partial_{x} ω)^{2} d x ⩽ C (T) {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

It follows from (2.46) that

\begin{aligned} \frac{d}{d t} ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) & ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + C (T) {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) . \end{aligned}

The Gronwall Lemma and (2.40) give

\begin{aligned} | | ω (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x} ω (t, \cdot) | |_{L^{2} (R)}^{2} ⩽ e^{C (T) t} (| | ω_{0} | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x} ω_{0} | |_{L^{2} (R)}^{2}) . \end{aligned}

(2.47)

(1.9) follows from (1.10), (2.39) and (2.47).

3. Case:

γ = 0

u_{0} \in H^{2} (R)

In this section, we assume $γ = 0$ . Therefore, (1.1) reads

\begin{aligned} {\begin{cases} \partial_{t} u + κ u_{ε} \partial_{x} u_{ε} + δ \partial_{x}^{3} u - β^{2} \partial_{t} \partial_{x}^{2} u + α \partial_{x}^{2} (u \partial_{x} u) = 0, & 0 < t < T, x \in R, \\ u (0, x) = u_{0} (x), & x \in R, \end{cases} \end{aligned}

(3.1)

On the initial datum, we assume

\begin{aligned} u_{0} \in H^{2} (R) . \end{aligned}

(3.2)

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

Fix a small number $0 < ε < 1$ and let $u_{ε} = u_{ε} (t, x)$ be the unique classical solution of the following problem:

\begin{aligned} {\begin{cases} \partial_{t} u_{ε} + κ u_{ε} \partial_{x} u_{ε} + δ \partial_{x}^{3} u_{ε} - β^{2} \partial_{t} \partial_{x}^{2} u_{ε} \\ + α \partial_{x}^{2} (u_{ε} \partial_{x} u_{ε}) = - ε \partial_{x}^{4} u_{ε}, & 0 < t < T, x \in R, \\ u_{ε} (0, x) = u_{ε, 0} (x), & x \in R, \end{cases} \end{aligned}

(3.3)

where

u_{ε, 0}

is a

C^{\infty}

approximation of

u_{0}

, such that

\begin{aligned} | | u_{ε, 0} | |_{H^{2} (R)} ⩽ | | u_{0} | |_{H^{2} (R)}, \sqrt{ε} | | \partial_{x}^{3} u_{ε, 0} | |_{L^{2} (R)} ⩽ C_{0}, \end{aligned}

(3.4)

where

C_{0}

an a positive constant independent of

ε

We prove the following lemma

Lemma 3.1.

Given κ, δ, β, α, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that (2.3), (2.4), (2.5) hold. Moreover

\begin{aligned} ε \int_{0}^{t} | | \partial_{x}^{3} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T), \end{aligned}

(3.5)

for every

0 ⩽ t ⩽ T

Proof.

Consider A a positive constant. Multiplying (3.1) by $2 u_{ε} - 2 A \partial_{x}^{2} u_{ε}$ , thanks to (2.8), arguing as in Lemma 2.1, we have

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + A) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 A ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = (α - A κ) \int_{R} (\partial_{x} u_{ε})^{3} d x + 5 A α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{2} u_{ε})^{2} d x . \end{aligned}

(3.6)

Arguing as in Lemma 2.1, we have (2.3), (2.4) and (2.5).

Finally, we prove (3.5). Due to (2.3), (2.3) and (3.6) with $A = 1$ ,

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ | α - κ | \int_{R} | \partial_{x} u_{ε} |^{3} d x + 5 α \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{2} u_{ε})^{2} d x \\ ⩽ | α - κ | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ + 5 | α | ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) . \end{aligned}

Integrating on

(0, t)

, by (3.4), we get

\begin{aligned} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (β^{2} + 1) {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + 2 ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} + C (T) t ⩽ C (T), \end{aligned}

which gives (3.5).

Lemma 3.2.

Given κ, δ, β, α, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} ε | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} ε | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ + ε^{2} \int_{0}^{t} | | \partial_{x}^{4} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T) . \end{aligned}

(3.7)

for every

0 ⩽ t ⩽ T

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (3.3) by $2 ε \partial_{x}^{4} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} \frac{d}{d t} (ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} u_{ε} d x - 2 β^{2} ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = - 2 κ ε \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 2 δ ε \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x - 6 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x \\ - 2 α ε \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x - ε^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 2 κ ε \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 6 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x + 2 α ε \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x \\ - ε^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Therefore, we have that

\begin{aligned} \frac{d}{d t} (ε | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} ε | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}) + 2 ε^{2} | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ = - 2 κ ε \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 6 α ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x + 2 α ε \int_{R} \partial_{x} u_{ε} {(\partial_{x}^{3} u_{ε})}^{2} d x . \end{aligned}

(3.8)

Due to (2.3), (2.4) and the Young inequality

\begin{aligned} 2 | κ | ε \int_{R} | u_{ε} | | \partial_{x} u_{ε} | | \partial_{x}^{4} u_{ε} | d x & ⩽ 2 | κ | ε ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | u_{ε} | | \partial_{x}^{4} u_{ε} | d x \\ ⩽ C (T) ε \int_{R} | u_{ε} | | \partial_{x}^{4} u_{ε} | d x = \int_{R} | C (T) u_{ε} | | ε \partial_{x}^{4} u_{ε} | d x \\ ⩽ C (T) {‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) + \frac{ε^{2}}{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 6 | α | ε \int_{R} | \partial_{x} u_{ε} | | \partial_{x}^{2} u_{ε} | \partial_{x}^{4} u_{ε} | d x & ⩽ 6 | α | ε ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} \int_{R} | \partial_{x}^{2} u_{ε} | \partial_{x}^{4} u_{ε} | d x \\ ⩽ C (T) ε \int_{R} | \partial_{x}^{2} u_{ε} | \partial_{x}^{4} u_{ε} | d x = \int_{R} | C (T) \partial_{x}^{2} u_{ε} | | ε \partial_{x}^{4} u_{ε} | d x \\ ⩽ C (T) {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) + \frac{ε^{2}}{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 | α | ε \int_{R} | \partial_{x} u_{ε} | (\partial_{x}^{3} u_{ε})^{2} d x & ⩽ 2 | α | ε ‖ \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

It follows from (3.8) that

\begin{aligned} \frac{d}{d t} (ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + ε^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) + C (T) ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{aligned}

Integrating on

(0, t)

, by (3.4) and (3.5), we get

\begin{aligned} ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε^{2} \int_{0}^{t} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C (T) t + C (T) ε \int_{0}^{t} {‖ \partial_{x}^{3} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C (T), \end{aligned}

which gives (3.7).

Lemma 3.3.

Given κ, δ, β, α, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that (2.33) and (2.34) hold.

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (3.3) by $2 \partial_{t} u_{ε}$ , thanks to (2.8), an integration on $R$ gives

\begin{aligned} 2 {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 β^{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = 2 \int_{R} (\partial_{t} u_{ε})^{2} d x - 2 β^{2} \int_{R} \partial_{t} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{t} u_{ε} d x - 2 δ \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} d x \\ - 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} d x - 2 ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} u_{ε} d x \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{t} u_{ε} d x + 2 δ \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} d x \\ + 2 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} d x + 2 α \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x} u_{ε} d x + 2 ε \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} \partial_{x} u_{ε} d x . \end{aligned}

Thanks to (2.3), (2.4), (2.5) and (3.7), arguing as in Lemma 2.5, we have (2.33) and (2.33).

Proof of Theorem 1.2. Proof of Theorem 1.2.

Arguing as in Lemma 2.6, we have (1.2).

4. Case: $γ = 0$ , $u_{0} \in H^{3} (R)$

In this section, we consider (3.1), where on the initial datum we assume

\begin{aligned} u_{0} \in H^{3} (R) \end{aligned}

(4.1)

Consider Approximation (3.3), where

u_{ε, 0}

is a

C^{\infty}

approximation of

u_{0}

, such that

\begin{aligned} ‖ u_{ε, 0} ‖_{H^{3} (R)} ⩽ ‖ u_{0} ‖_{H^{3} (R)}, \end{aligned}

(4.2)

Observe that, since $H^{2} (R) \subset H^{3} (R)$ , Lemmas 3.1, 3.2 and Lemma 3.3 hold in this case.

We prove the following result:

Lemma 4.1.

Given κ, δ, β, α, T, there exists a constant $C (T) > 0$ , independent of $ε$ , such that

\begin{aligned} | | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ + ε e^{C (T) t} \int_{0}^{t} e^{- 2 C (t) s} | | \partial_{x}^{4} u_{ε} (s, \cdot) | |_{L^{2} (R)}^{2} d s ⩽ C (T) \end{aligned}

(4.3)

for every

0 ⩽ t ⩽ T

. Moreover, we have (2.26).

Proof.

Let $0 ⩽ t ⩽ T$ . Arguing as in Lemma 3.2, we have

\begin{aligned} \frac{d}{d t} (| | \partial_{x}^{2} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} + β^{2} | | \partial_{x}^{3} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2}) + 2 ε | | \partial_{x}^{4} u_{ε} (t, \cdot) | |_{L^{2} (R)}^{2} \\ = - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x + 2 α \int_{R} \partial_{x} u_{ε} {(\partial_{x}^{3} u_{ε})}^{2} d x \end{aligned}

(4.4)

Observe that

\begin{aligned} - 2 κ \int_{R} u_{ε} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x & = 2 \int_{R} (\partial_{x} u_{ε})^{2} \partial_{x}^{3} u_{ε} + 2 κ \int_{R} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{3} d x \\ = - 5 \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x, \\ - 6 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x & = 6 α \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε})^{2} d x . \end{aligned}

It follows from (4.4) and (2.4) that

\begin{aligned} \frac{d}{d t} ({‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 5 κ \int_{R} \partial_{x} u_{ε} (\partial_{x}^{3} u_{ε}) 2 d x + 8 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x \\ ⩽ C (T) {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) ({‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) . \end{aligned}

By the Gronwall Lemma and (4.1), we get

\begin{aligned} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε e^{C (T) t} \int_{0}^{t} e^{- C (T) s} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} e^{C (T) t} ⩽ C (T), \end{aligned}

which gives (4.3).

Finally, arguing as in Lemma 2.2, we have (2.26).

Proof of Theorem 1.3. Proof of Theorem 1.3.

Arguing as in Lemma 2.6, we have that u is solution of (3.1), such that (1.12). Since $u \in H^{3}$ , arguing as in Theorem 1.1, we have (1.9).

Footnotes

Acknowledgements

G. M. Coclite 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 Mobilitá 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.

The authors declare that they do not have any conflict of interest.

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Wave propagation in dilatant granular materials

Abstract

Keywords

1. Introduction

4. Case: γ = 0 , u 0 ∈ H 3 ( R )

Footnotes

Acknowledgements

References

4. Case: $γ = 0$ , $u_{0} \in H^{3} (R)$