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Article

On the Solutions for a Fifth Order Kudryashov–Sinelshchikov Type Equation

by
Giuseppe Maria Coclite
1,*,† and
Lorenzo di Ruvo
2,†
1
Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, 70125 Bari, Italy
2
Department of Mathematics, Università di Bari, 70125 Bari, Italy
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2022, 14(8), 1535; https://doi.org/10.3390/sym14081535
Submission received: 28 June 2022 / Revised: 17 July 2022 / Accepted: 25 July 2022 / Published: 27 July 2022
(This article belongs to the Section Mathematics)

Abstract

:
The fifth order Kudryashov–Sinelshchikov equation models the evolution of the nonlinear waves in a gas–liquid mixture, taking into account an interphase heat transfer, surface tension, and weak liquid compressibility simultaneously at the derivation of the equations for non-linear-waves. We prove the well-posedness of the solutions for the Cauchy problem associated with this equation for each choice of the terminal time T.

1. Introduction

In this paper, we investigate the well-posedness of the following Cauchy problem:
t u + x f ( u ) + ν x 2 u + δ x 3 u + α u x 3 u + κ x u x 2 u + q ( x u ) 2 + γ x 5 u + β 2 x 4 u = 0 , 0 < t < T , x R , u ( 0 , x ) = u 0 ( x ) , x R ,
with
ν , δ , α , κ , q , γ , β R , β , γ 0 .
On the flux f, we assume
f ( u ) C 1 ( R ) , | f ( u ) | C 0 1 + | u | 2 ,
for some positive constant C 0 .
On the initial datum, we assume
u 0 H 2 ( R ) , u 0 0 .
Taking
f ( u ) = b 1 u 2 + b 2 u 3 , b 1 , b 2 R ,
Equation (1) reads
t u + b 1 x u 2 + b 2 x u 3 + ν x 2 u + δ x 3 u + α u x 3 u + κ x u x 2 u + q ( x u ) 2 + γ x 5 u + β 2 x 4 u = 0 .
Equation (6) is deduced in [1,2] to model the evolution of the nonlinear-waves in a gas–liquid mixture, taking into account an interphase heat transfer, a surface tension, and a weak liquid compressibility simultaneously at the derivation of the equations for non-linear-waves. In particular, in [1], the authors show that (6) is obtained by a perturbation of the Burgers–Korteweg-de Vries equation, in correspondence with main influence of dispersion nonlinear waves propagation. Finally, Equation (6) is deduced in [3] in the context of ray tracing through a crystalline lens and pressure waves in mixtures liquid–gas bubbles under the consideration of heat transfer and viscosity.
From a mathematical point of view, in [4], special solutions of (6) are studied. In particular, some elliptic and simple periodic traveling waves solution are constructed. In [5], the authors proved that (6) does not belong to the class of integrable equations. Moreover, they also proved that (6) admits classical and non-classical symmetries. In [6], approximate invariant solutions for (6) are analyzed.
Equation (6) is a generalization of the following equation:
t u + b 1 x u 2 b 3 2 x 2 u + δ x 3 u + α u x 3 u + κ x u x 2 u = 0 ,
with is deduced in [7]. Equation (7) is also derived for water waves by Olver [8] (see also [9]) using Hamiltonian perturbation theory, with further generalization given by Craig and Groves [10].
Mathematical properties of (7) have been studied in significant detail, including the existence of the travelling wave solutions in [11,12,13,14,15], the solitary and periodic wave solutions [16,17], the periodic loop solutions [18], the soliton solutions [19], and the quasi-exact solutions [20]. Methods to find exact solutions are in [21,22,23,24,25], while, in [26], under appropriate assumptions of b 1 , α , κ , and
u 0 H ( R ) , { 1 , 2 } ,
the existence of the solutions for (7) is proven. Finally, following [27,28,29], in [30], the convergence of the solution of (7) to the unique entropy in one of the Burgers equations is proven.
Taking b 2 = b 3 = β = q = 0 , (7) reads
t u + b 1 x u 2 + δ x 3 u + α u x 3 u + κ x u x 2 u + γ x 5 u = 0 .
From a physical point of view, (7) was derived by Olver [8,31] in the context of water waves. In the case b 1 = 0 , (9) was derived by Benney [32] as a model to describe the interaction effects between short and long waves.
From a mathematical point of view, under suitable assumptions on b 1 , δ , α , κ , and γ , the existence of the travelling waves solutions for (9) is proven in [33,34], while a method to find exact solutions of (9) is given in [35]. In [36] the local well-posedness of the Cauchy problem of (9) is proven, while, in [37,38], under appropriate assumptions of b 1 , α , κ , the global well-posedness is showed.
Taking α = κ = q = 0 , (1) reads
t u + x f ( u ) + ν x 2 u + δ x 3 u + γ x 5 u + β 2 x 4 u = 0 .
Under Assumption (3), (10) was first introduced by Benney [39] and later by Lin [40] to describe the evolution of long waves in various problems in fluid dynamics (see also [41]).
In [42,43], under Assumptions f ( u ) = b 1 u 2 and appropriate assumptions on b 1 , ν , δ , γ , and β using the energy space technique, the local and global well-posedness of weak solutions for (10) is proven, while, in [44], the well-posedness is proven using the Bourgain bilinear estimate technique. In [45], the well-posedness of (10) is proven under Assumption (4) for every choice of β and T.
Taking b 2 = α = κ = q = γ = 0 , (6) reads
t u + b 1 x u 2 + ν x 2 u + δ x 3 u + β 2 x 4 u = 0 .
Equation (11) was derived independently by Kuramoto [46,47,48] as a model for phase turbulence in reaction–diffusion systems and by Sivashinsky [49] as a model for plane flame propagation, describing the combined influence of diffusion and thermal conduction of the gas on the stability of a plane flame front. It also describes incipient instabilities in a variety of physical and chemical systems [50,51,52]. It was derived by Kuramoto in the study of phase turbulence in the Belousov–Zhabotinsky reaction [53].
In [54,55,56] the well-posedness of the Cauchy problem for (11) is proven using the energy space technique the fixed point method, a priori estimates together with an application of the Cauchy–Kovalevskaya, and a priori estimates together with an application of the Aubin–Lions Lemma, respectively. In [57,58,59], the initial boundary value problem for (11) is studied using a priori estimates together with an application of the Cauchy–Kovalevskaya and the energy space technique. Finally, in [60], the convergence of the solution of (11) to the unique entropy in one of the Burgers equations is proven. Here, we extend some of those results considering the fifth order case.
Taking b 2 = α = κ = γ = 0 , in (6), we have the following equation:
t u + b 1 x u 2 + ν x 2 u + δ x 3 u + q ( x u ) 2 + β 2 x 4 u = 0 .
Equation (12), known as the Kuramoto–Velarde equation, describes slow space-time variations of disturbances at interfaces, diffusion–reaction fronts, and plasma instability fronts [61,62,63]. It also describes Benard–Marangoni cells that occur when there is large surface tension on the interface [64,65,66] in a microgravity environment.
In [67], the exact solutions for (12) are studied, while in [68], the initial boundary problem is analyzed. In [61,69], the existence of the solitons is proven, while in [70], the existence of traveling wave solutions for (12) is analyzed. In [71], the author analyzes the existence of the periodic solution for (12) under appropriate assumptions on b 1 , ν , δ , q , β . The local well-posedness of the Cauchy problem of (12) is proven in [72] using the energy space technique and assuming b 1 = 0 , while in [73], the well-posedness of classical solutions is proven under Assumption (4) and suitable assumptions on β , T , and u 0 . Finally, in [74], the authors prove the existence of appropriate rescalings, in which the well-posedness of the Cauchy problem of (12) holds for each choice of T, and under the assumption
u 0 H 1 ( R ) , u 0 0 .
Taking b 2 = ν = α = κ = q = γ = β = 0 , (6) reads
t u + b 1 x u 2 + δ x 3 u = 0 ,
known as the Korteweg–de Vries equation [75]. It has a very wide range of applications, such as magnetic fluid waves, ion sound waves, and longitudinal astigmatic waves.
In [76,77,78], the Cauchy problem for (14) is studied. In particular, in [76,77], the well-posedness of the Cauchy problem of (14) is proven under Assumption (4) and for each choice of T. In [79], the author reviewed the travelling wave solutions for (14), while in [28,29,80], the convergence of the solution of (14) to the unique entropy in one of the Burgers equations is proven.
Taking b 2 = ν = α = κ = q = β = 0 , we have the following equation:
t u + b 1 x u 2 + δ x 3 u + γ x 5 u = 0 .
This was derived by Kawahara [81] to describe small-amplitude gravity capillary waves on water of a finite depth when the Weber number is close to 1 / 3 (see [82]). Moreover, in [81], the author deduced (15) to describe one-dimensional propagation of small-amplitude long waves in various problems of fluid dynamics and plasma physics.
In [83,84,85,86,87], the local and global well-posedness in Bourgain space for (15) is proven, while in [88,89,90,91,92,93], the local and global well-posedness in energy space for (15) is studied. In [94,95,96], the well-posedness of the initial boundary value problem on a bounded domain is analyzed, while in [97] (see also [98]), the well-posedness of the classical solution of the Cauchy problem of (15) is proven for each choice of T. In [99], the authors prove that the solution of (15) converges to the solution of (14), while, in [100,101] (see also [102]), the convergence of the solution of (15) to the unique entropy in one of the Burgers equations is proven.
The main result of this paper is the following theorem.
Theorem 1.
Assume (2)–(4). Given ν , α , κ , q , T , there exists a unique solution u of (1), such that
u L ( 0 , T ; H 2 ( R ) ) L 4 ( 0 , T ; W 2 , 4 ( R ) ) L 6 ( 0 , T ; W 2 , 6 ( R ) ) , x 4 u L 2 ( ( 0 , T ) × R ) .
Moreover, if u 1 and u 2 are two solutions of (1) in correspondence with the initial data u 1 , 0 and u 2 , 0 , the following stability estimate holds:
u 1 ( t , · ) u 2 ( t , · ) L 2 ( R ) e C ( T ) t u 1 , 0 u 2 , 0 L 2 ( R ) ,
for some suitable C ( T ) > 0 and every, 0 t T .
The proof of Theorem 1 is based on the Aubin–Lions Lemma due to the functional setting [103,104,105,106]. Observe that using the Aubin–Lions Lemma under Assumption (4), Refs. [45,76,77] give the well-posedness of (10) and (14) for each choice of T. Therefore, thanks to Theorem 1, we find that the solution of (6) converges to the unique solutions of (10) and (14) under assumptions α = κ = q = γ = 0 and α = κ = q = γ = β = 0 , respectively.
The paper is organized as follows. In Section 2, we prove several a priori estimates on a vanishing viscosity approximation of (1). Those play a key role in the proof of our main result, which is given in Section 3.

2. Vanishing Viscosity Approximation

Our existence argument is based on passing to the limit in a vanishing viscosity approximation of (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 [73]:
t u ε + x f ( u ε ) + ν x 2 u ε + δ x 3 u ε + α u ε x 3 u ε + κ x u ε x 2 u ε + q ( x u ε ) 2 + γ x 5 u ε + β 2 x 4 u ε = ε x 6 u ε , 0 < t < T , x R , u ε ( 0 , x ) = u ε , 0 ( x ) , x R ,
where u ε , 0 is a C approximation of u 0 , such that
u ε , 0 H 2 ( R ) u 0 H 2 ( R ) , u ε , 0 0 .
Let us prove some a priori estimates on u ε . We denote with C 0 the constants which depend only on the initial data, and with C ( T ) the constants which depend also on T.
Following [73], Lemma 1 and [107], Lemma 2.2 , we prove the following result.
Lemma 1.
The following inequalities hold
u ε ( t , · ) L 2 ( R ) , x u ε ( t , · ) L 2 ( R ) C ( T ) ,
u ε L ( ( 0 , T ) × R ) C ( T ) ,
0 t x 2 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
0 t x 3 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
ε 0 t x 4 u ε ( t , · ) L 2 ( R ) 2 d s C ( T ) ,
0 t x u ε ( s , · ) L 4 ( R ) 4 d s C ( T ) ,
for every 0 t T .
Proof. 
We begin by proving that
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) t A β 2 u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 2 + 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) t A β 2 1 1 u ε , 0 L 2 ( R ) 2 + A x u ε , 0 L 2 ( R ) 2 2 ,
for every 0 t T , where A is a generic positive constant, and
a 1 2 = 12 ν 2 + 1 0 , a 2 2 = 3 ( C 0 + q 2 ) , a 3 2 : = 12 κ 2 + 6 α 2 , a 4 2 : = 3 ( 24 α 2 + 7 κ 2 ) 2 + 1 0 , a 5 2 : = a 1 2 + a 2 .
Consider A a positive constant. Multiplying (18) by 2 u ε 2 A x 2 u ε , an integration on R gives
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 = 2 R u ε t u ε d x 2 A R x 2 u ε t u ε d x = 2 R u f ( u ) x u d x = 0 + 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 2 δ R u ε x 3 u ε d x + 2 A δ R x 2 u ε x 3 u ε d x 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x + 2 A κ R x u ε ( x 2 u ε ) 2 d x 2 q R u ε ( x u ε ) 2 d x + 2 A q R ( x u ε ) 2 x 2 u ε d x 2 γ R u ε x 5 u ε d x + 2 A γ R x 2 u ε x 5 u ε d x 2 β 2 R u ε x 4 u ε d x + 2 A β 2 R x 2 u ε x 4 u ε d x + 2 ε R u ε x 6 u ε d x 2 A ε R x 2 u ε x 6 u ε d x = 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 δ R x u ε x 2 u ε d x 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x + 2 A κ R x u ε ( x 2 u ε ) 2 d x 2 q R u ε ( x u ε ) 2 d x + 2 γ R x u ε x 4 u ε d x 2 A γ R x 3 u ε x 4 u ε d x + 2 β 2 R x u ε x 3 u ε d x 2 A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 2 ε R x u ε x 5 u ε d x + 2 A ε R x 3 u ε x 5 u ε d x = 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x + 2 A κ R x u ε ( x 2 u ε ) 2 d x 2 q R u ε ( x u ε ) 2 d x 2 γ R x 2 u ε x 3 u ε d x 2 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 2 A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε R x 2 u ε x 4 u ε d x 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 = 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x + 2 A κ R x u ε ( x 2 u ε ) 2 d x 2 q R u ε ( x u ε ) 2 d x 2 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 2 A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 .
Therefore, we have that
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + 2 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 = 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x + 2 A κ R x u ε ( x 2 u ε ) 2 d x 2 q R u ε ( x u ε ) 2 d x .
Observe that
R x u ε ( x 2 u ε ) 2 d x = R x u ε x 2 u ε x 2 u ε d x = R x u ε x x u ε x 2 u ε d x = R x u ε ( x 2 u ε ) 2 d x R ( x u ε ) 2 x 3 u ε d x .
Consequently, we have that
2 R x u ε ( x 2 u ε ) 2 d x = R ( x u ε ) 2 x 3 u ε d x .
Moreover,
R u ε ( x u ε ) 2 d x = R u ε x x u ε d x = R u ε x u ε x u ε = R u ε ( x u ε ) 2 R u ε 2 x 2 u ε d x ,
which gives
2 R u ε ( x u ε ) 2 d x = R u ε 2 x 2 u ε d x .
Hence, by (28)–(30), we obtain
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + 2 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 = 2 A R f ( u ε ) x u ε x 2 u ε d x 2 ν R u ε x 2 u ε d x + 2 A ν x 2 u ε ( t , · ) L 2 ( R ) 2 2 α R u ε 2 x 3 u ε d x + 2 A α R u ε x 2 u ε x 3 u ε d x 2 κ R u ε x u ε x 2 u ε d x A κ R ( x u ε ) 2 x 3 u ε d x + q R u ε 2 x 2 u ε d x .
Due to (3) and the Young inequality,
2 A R | f ( u ε ) | | x u ε | | x 2 u ε | d x 2 A C 0 R | x u ε | | x 2 u ε | d x + 2 A C 0 R | u ε | | x u ε | | x 2 u ε | d x + 2 A C 0 R u ε 2 | x u ε | | | x 2 u ε | d x = 2 R A C 0 x u ε β D 1 D 1 β x 2 u ε d x + 2 R A C 0 u ε x u ε β D 1 D 1 β x 2 u ε d x + 2 R A C 0 u ε 2 x u ε β D 1 d x A 2 C 0 D 1 β 2 x u ε ( t , · ) L 2 ( R ) 2 + A 2 C 0 D 1 β 2 R u ε 2 ( x u ε ) 2 d x + A 2 C 0 D 1 β 2 R u ε 4 ( x u ε ) 2 d x + 3 D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 A 2 C 0 D 1 β 2 x u ε ( t , · ) L 2 ( R ) 2 + A 2 C 0 D 1 β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + A 2 C 0 D 1 β 2 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 3 D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 , 2 | ν | R | u ε | | x 2 u ε | d x = 2 R ν u ε β D 1 D 1 β x 2 u ε d x ν 2 β 2 D 1 u ε ( t , · ) L 2 ( R ) 2 + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 , 2 | α | R u ε 2 | x 3 u ε | d x = 2 R α u ε 2 A D 2 β A D 2 β x 3 u ε d x α 2 β 2 A D 2 R u ε 4 d x + A D 2 β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 α 2 β 2 A D 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + A D 2 β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 , 2 A | α | R | u ε | | x 2 u ε | | x 3 u ε | d x = 2 A R α u ε x 2 u ε β D 2 D 2 β x 3 u ε d x α 2 A D 2 β 2 R u ε 2 ( x 2 u ε ) 2 d x + A D 2 β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 α 2 A D 2 β 2 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + A D 2 β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 , 2 κ R | u ε x u ε | x 2 u ε | d x = 2 R κ u ε x u ε β D 1 D 1 β x 2 u ε d x κ 2 D 1 β 2 R u ε 2 ( x u ε ) 2 d x + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 κ 2 D 1 β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 = κ 2 A A D 1 β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 , A | κ | R ( x u ε ) 2 | x 3 u ε | d x = 2 A R κ ( x u ε ) 2 2 β D 2 D 2 β x 3 u ε d x κ 2 A 4 β 2 D 2 x u ε ( t , · ) L 4 ( R ) 4 + A D 2 β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 , | q | R u ε 2 | x 2 u ε | d x = 2 R q u ε 2 2 β D 1 D 1 x 2 u ε d x q 2 4 β 2 D 1 R u ε 4 d x + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 q 2 4 β 2 D 1 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + D 1 β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 ,
where D 1 , D 2 are two positive constants, which will be specified later. It follows from (31) that
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + 2 β 2 1 3 D 1 x 2 u ε ( t , · ) L 2 ( R ) 2 + A β 2 2 3 D 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 2 A | ν | x 2 u ε ( t , · ) L 2 ( R ) 2 + A 2 C 0 D 1 β 2 x u ε ( t , · ) L 2 ( R ) 2 + ν 2 β 2 D 1 u ε ( t , · ) L 2 ( R ) 2 + A 2 C 0 D 1 β 2 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + α 2 A D 2 β 2 + q 2 4 D 1 β 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + α 2 A D 1 β 2 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + κ 2 A 4 β 2 D 2 x u ε ( t , · ) L 4 ( R ) 4 + A ( A C 0 + κ 2 ) A β 2 D 1 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 .
Choosing D 1 = 1 6 and D 2 = 1 3 , we have that
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 2 A | ν | x 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A 2 C 0 β 2 x u ε ( t , · ) L 2 ( R ) 2 + 6 ν 2 β 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A 2 C 0 β 2 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 6 α 2 + 3 A q 2 2 A β 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + 6 α 2 A β 2 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 3 κ 2 A 4 β 2 x u ε ( t , · ) L 4 ( R ) 4 + 6 A ( A C 0 + κ 2 ) A β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 .
Following Lemma 2.3 in [108], we obtain
x u ε ( t , · ) L 4 ( R ) 4 9 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 .
Consequently, by (32),
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 2 A | ν | x 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A 2 C 0 β 2 x u ε ( t , · ) L 2 ( R ) 2 + 6 ν 2 β 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A 2 C 0 β 2 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 6 α 2 + 3 A q 2 2 A β 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + A ( 24 α 2 + 27 κ 2 ) β 2 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A ( A C 0 + κ 2 ) A β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 .
Observe that
x 2 u ε ( t , · ) L 2 ( R ) 2 = R x 2 u ε x 2 u ε d x = R x u ε x 3 u ε d x .
Therefore, by the Hölder inequality,
x 2 u ε ( t , · ) L 2 ( R ) 2 R | x u ε | | x 3 u ε | d x x u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) .
Hence, by (34), we obtain that
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + A β 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 2 A | ν | x u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) + 6 A 2 C 0 β 2 x u ε ( t , · ) L 2 ( R ) 2 + 6 ν 2 β 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A 2 C 0 β 2 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 6 α 2 + 3 A q 2 2 A β 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + A ( 24 α 2 + 27 κ 2 ) β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) + 6 A ( A C 0 + κ 2 ) A β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 .
Due to the Young inequality,
2 A | ν | x u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) = A 2 | ν | x u ε ( t , · ) L 2 ( R ) | β | | β | x 3 u ε ( t , · ) L 2 ( R ) 2 A ν 2 β 2 x u ε ( t , · ) L 2 ( R ) 2 + β 2 2 x 2 u ε ( t , · ) L 2 ( R ) , A ( 24 α 2 + 27 κ 2 ) β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) = 2 A 3 ( 24 α 2 + 27 κ 2 ) u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 | β | 3 | β | x 3 u ε ( t , · ) L 2 ( R ) 3 3 A ( 24 α 2 + 27 κ 2 ) 2 4 β 6 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + A β 2 3 x 3 u ε ( t , · ) L 2 ( R ) 2 .
It follows from (35) that
d d t u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 + β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + A β 2 6 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 A ε x 4 u ε ( t , · ) L 2 ( R ) 2 A ( 2 A C 0 + 6 ν 2 ) β 2 x u ε ( t , · ) L 2 ( R ) 2 + 6 ν 2 β 2 u ε ( t , · ) L 2 ( R ) 2 + A ( 24 A β 4 C 0 + 3 ( 24 α 2 + 27 κ 2 ) 2 ) 4 β 6 u ε ( t , · ) L ( R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 6 α 2 + 3 A q 2 2 A β 2 u ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + 6 A ( A C 0 + κ 2 ) A β 2 u ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 .
We define
X ε ( t ) : = u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 , 1 : = 2 A C 0 + 12 ν 2 + 1 β 2 , 2 : = 3 A ( C 0 + q 2 ) + 12 κ 2 + 6 α 2 A β 2 , 3 : = 24 A β 4 C 0 + 3 ( 24 α 2 + 27 κ 2 ) 2 + 1 4 β 6 .
It follows from (36) that
d X ε ( t ) d t 1 X ε ( t ) + 2 u ε ( t , · ) L ( R ) 2 X ε ( t ) + 3 u ε ( t , · ) L ( R ) 4 X ε ( t ) .
Due to (37) and the Hölder inequality,
u ε 2 ( t , x ) = 2 x u ε x u ε d y 2 R | u ε | | x u ε | d x 2 u ε ( t , · ) L 2 ( R ) x u ε ( t , · ) L 2 ( R ) = 2 A u ε ( t , · ) L 2 ( R ) 2 A x u ε ( t , · ) L 2 ( R ) 2 2 A X ε ( t ) .
Hence,
u ε ( t , · ) L ( R ) 2 2 A X ε ( t ) , for every 0 t T .
It follows from (38) and (39) that
d X ε ( t ) d t 1 X ε ( t ) + 2 2 A X ε 2 ( t ) + 4 3 A X ε 3 ( t ) .
Thanks to the Young inequality,
2 A X ε 2 ( t ) = 2 A X ε 1 2 ( t ) X ε 3 2 ( t ) X ε + 1 A X ε 3 ( t ) .
Consequently, by (40),
d X ε ( t ) d t 1 + 2 X ε ( t ) + 4 3 + 2 A X ε 3 ( t ) .
Define
4 : = 1 + 2 , 5 : = 4 3 + 2 .
It follows from (41) that
1 X ε 3 ( t ) d X ε ( t ) d t 4 X ε 2 ( t ) + 5 A .
Since
d d t 1 X ε 2 ( t ) = 2 X ε 3 ( t ) d X ε ( t ) d t ,
by (43), we have that
d d t 1 X ε 2 ( t ) 2 4 X ε 2 ( t ) 2 5 A ,
which gives
d d t 1 X ε 2 ( t ) + 2 4 X ε 2 ( t ) 2 5 A .
Multiplying (44) by e 2 4 t , we obtain
e 2 4 t d d t 1 X ε 2 ( t ) + 2 4 e 2 4 t X ε 2 ( t ) 2 5 e 2 4 t A .
Therefore,
d d t e 2 4 t X ε 2 ( t ) 2 5 e 2 4 t A .
Integrating on ( 0 , t ) , we have that
e 2 4 t X ε 2 ( t ) 1 X ε 2 ( 0 ) 5 A 4 e 2 4 t 1 ,
that is,
e 2 4 t X ε 2 ( t ) + 5 A 4 e 2 4 t 1 1 X ε 2 ( 0 ) .
Using (37) and (42) in (45), thanks to (27), we have (26).
We demonstrate (20). To this end, we begin by observing that, by (19),
u ε , 0 L 2 ( R ) 2 + A x u ε , 0 L 2 ( R ) 2 Y 0 + A Y 1 ,
where
Y 0 : = u 0 L 2 ( R ) 2 , Y 1 : = x u 0 L 2 ( R ) 2 .
Consequently, we have that
u ε , 0 L 2 ( R ) 2 + A x u ε , 0 L 2 ( R ) 2 2 Y 0 + A Y 1 2 ,
which gives,
1 u ε , 0 L 2 ( R ) 2 + A x u ε , 0 L 2 ( R ) 2 2 1 Y 0 + A Y 1 2 .
Moreover,
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) t A β 2 .
It follows from (26), (48), and (49) that
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 2 + 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 1 1 Y 0 + A Y 1 2 ,
that is
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 2 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 1 + 1 Y 0 + A Y 1 2 .
We search A such that
1 Y 0 + A Y 1 2 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 1 > 0 ,
that is,
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 < 1 + A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) Y 0 + A Y 1 2 ( 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A )
which gives
e < 1 + A β 4 ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) Y 0 + A Y 1 2 ( 24 C 0 A 2 β 4 + a 2 2 A β 4 + a 3 2 β 2 + a 4 2 A ) A β 2 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T .
Taking β as in
| β | = A n , n > 2 .
Equation (52) reads as follows
e < 1 + A 1 + 4 n ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) Y 0 + A Y 1 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) A 1 + 2 n 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T .
Thanks to (53), we have that
lim A 1 + A 1 + 4 n ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) Y 0 + A Y 1 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) A 1 + 2 n 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T = .
In fact,
lim A 1 + A 1 + 4 n ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) Y 0 + A Y 1 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) A 1 + 2 n 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T = lim A 1 + 1 a 7 2 ( A ) a 7 2 ( A ) a 8 2 ( A ) ,
where
a 7 2 ( A ) : = Y 0 + A Y 1 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) A 1 + 4 n ( 2 C 0 A 2 + a 5 2 A + a 3 2 ) , a 8 2 ( A ) : = A 6 n + 2 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) Y 0 + A Y 1 2 T .
Observe, thanks to (55),
lim A 1 + 1 a 7 2 ( A ) a 7 2 ( A ) = e ,
while, by (53) and (55),
lim A A 6 n + 2 2 ( 24 C 0 A 2 + 4 n + a 2 2 A 1 + 4 n + a 3 2 A 2 n + a 4 2 A ) Y 0 + A Y 1 2 T = .
Equations (55)–(57) give (54).
Therefore, thanks to (54), (51), which is equivalent to (52), holds; taking A very large and up to rescaling, we can have | β | = A n , with n defined in (53).
Consequently, by (53), (50), (51) or (52), and (54), there exists a constant C ( T ) > 0 , independent of ε , such that
e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 2 C ( T ) .
Hence,
u ε ( t , · ) L 2 ( R ) 2 + A x u ε ( t , · ) L 2 ( R ) 2 e 2 ( 2 A 2 C 0 + a 5 2 A + a 3 2 ) T A β 2 C ( T ) ,
which gives (20).
We prove (21). Thanks to (20) and (37) with A = 1 , and (39) with A = 1 ,
u ε ( t , · ) L ( R ) 2 2 u ε ( t , · ) L 2 ( R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
which gives (21).
We prove (22)–(24). Thanks to (20), (21), and (36) with A = 1 , we have that
d d t u ε ( t , · ) L 2 ( R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 + β 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + β 2 6 x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 C ( T ) .
Integrating on ( 0 , t ) by (19), we obtain
u ε ( t , · ) L 2 ( R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 + β 2 0 t x 2 u ε ( s , · ) L 2 ( R ) 2 d s + β 2 6 0 t x 3 u ε ( s , · ) L 2 ( R ) 2 d s + 2 ε 0 t x 3 u ε ( s , · ) L 2 ( R ) 2 d s + 2 ε 0 t x 4 u ε ( s , · ) L 2 ( R ) 2 d s C 0 + C ( T ) t C ( T ) ,
which gives (22)–(24).
Finally, we prove (25). Observe that, by (21) and (33), we have that
x u ε ( t , · ) L 4 ( R ) 4 C ( T ) x 2 u ε ( t , · ) L 2 ( R ) 2 .
Consequently, by an integration on R and (22)–(24), we have (25). □
Lemma 2.
The following inequalities hold
x u ε L ( R ) C ( T ) ,
x 2 u ε ( t , · ) L 2 ( R ) 2 + β 2 0 t x 4 u ε ( s , · ) L 2 ( R ) 2 d s + 2 ε 0 t x 5 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Multiplying (18) by 2 x 4 u ε , an integration on R gives
d d t x 2 u ε ( t , · ) L 2 ( R ) 2 = 2 R x 4 u ε t u ε d x = 2 R f ( u ε ) x u ε x 4 u ε d x 2 ν R x 2 u ε x 4 u ε d x 2 δ R x 3 u ε x 4 u ε d x 2 α R u ε x 3 u ε x 4 u ε d x 2 κ R x u ε x 2 u ε x 4 u ε d x 2 q R ( x u ε ) 2 x 4 u ε d x 2 γ R x 4 u ε x 5 u ε d x 2 β 2 x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 ε R x 4 u ε x 6 u ε d x = 2 R f ( u ε ) x u ε x 4 u ε d x + 2 ν x 3 u ε ( t , · ) L 2 ( R ) 2 + α + 2 κ R x u ε ( x 3 u ε ) 2 d x 2 q R ( x u ε ) 2 x 4 u ε d x 2 β 2 x 4 u ε ( t , · ) L 2 ( R ) 2 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 .
Therefore, we have that
d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 β 2 x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 = 2 R f ( u ε ) x u ε x 4 u ε d x + 2 ν x 3 u ε ( t , · ) L 2 ( R ) 2 + α + 2 κ R x u ε ( x 3 u ε ) 2 d x 2 q R ( x u ε ) 2 x 4 u ε d x 2 R | f ( u ε ) | | x u ε | | x 4 u ε | d x + 2 | q | R ( x u ε ) 2 | x 4 u ε | d x + 2 | ν | + | α + 2 κ | x u ε L ( ( 0 , T ) × R ) x 3 u ε ( t , · ) L 2 ( R ) 2 .
Due to (20), (21), and the Young inequality,
2 R | f ( u ε ) | | x u ε | | x 4 u ε | d x 2 f L ( C ( T ) , C ( T ) ) R | x u ε | | x 4 u ε | d x C ( T ) R | x u ε | | x 4 u ε | d x = R C ( T ) x u ε β β x 4 u ε d x C ( T ) x u ε ( t , · ) L 2 ( R ) 2 + β 2 2 x 4 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + β 2 2 x 4 u ε ( t , · ) L 2 ( R ) 2 , 2 | q | R ( x u ε ) 2 | x 4 u ε | d x = R 2 q ( x u ε ) 2 β β x 4 u ε d x 2 q 2 β 2 x u ε ( t , · ) L 4 ( R ) 4 + β 2 2 x 4 u ε ( t , · ) L 2 ( R ) 2 .
It follows from (60) that
d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + β 2 x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + 2 q 2 β 2 x u ε ( t , · ) L 4 ( R ) 4 + 2 | ν | + | α + 2 κ | x u ε L ( ( 0 , T ) × R ) x 3 u ε ( t , · ) L 2 ( R ) 2 .
Integrating on ( 0 , t ) , by (19), (22)–(25), we have that
x 2 u ε ( t , · ) L 2 ( R ) 2 + β 2 0 t x 4 u ε ( s , · ) L 2 ( R ) 2 d s + 2 ε 0 t x 5 u ε ( s , · ) L 2 ( R ) 2 d s C 0 + C ( T ) t + 2 q 2 β 2 0 t x u ε ( s , · ) L 4 ( R ) 4 d s + 2 | ν | + | α + 2 κ | x u ε L ( ( 0 , T ) × R ) 0 t x 3 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) .
We prove (58). Thanks to (20), (61), and the Hölder inequality,
( x u ε ( t , x ) ) 2 = 2 x x u ε x 2 u ε d y 2 R | x u ε | | x 2 u ε | d x 2 x u ε ( t , · ) L 2 ( R ) x 2 u ε ( t , · ) L 2 ( R ) C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) .
Hence,
x u ε L ( ( 0 , T ) × R ) 4 C ( T ) x u ε L ( ( 0 , T ) × R ) C ( T ) 0 .
Arguing as in [103], Lemma 2.3 or [104], Lemma 2.4 , we have (58).
Finally, (59) follows from (58) and (61). □
Lemma 3.
The following estimates hold
0 t x 2 u ε ( s , · ) L 4 ( R ) 4 d s C ( T ) ,
0 t x 2 u ε ( s , · ) L 6 ( R ) 6 d s C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . We begin by proving (62). We observe that
x 2 u ε ( t , · ) L 4 ( R ) 4 = R x 2 u ε ( x 2 u ε ) 3 d x = 3 R x u ε ( x 2 u ε ) 2 x 3 u ε d x .
Due to (58) and the Young inequality,
3 R | x u ε | ( x 2 u ε ) 2 | x 3 u ε | d x 3 x u ε L ( ( 0 , T ) × R ) R ( x 2 u ε ) 2 | x 3 u ε | d x C ( T ) R ( x 2 u ε ) 2 | x 3 u ε | d x = R ( x 2 u ε ) 2 | C ( T ) x 3 u ε | d x 1 2 x 2 u ε ( t , · ) L 4 ( R ) 4 + C ( T ) x 3 u ε ( t , · ) L 2 ( R ) 2 .
It follows from (64) that
1 2 x 2 u ε ( t , · ) L 4 ( R ) 4 C ( T ) x 3 u ε ( t , · ) L 2 ( R ) 2 .
Integrating on ( 0 , t ) , thanks to (22)–(24), we have (62).
Finally, we prove (63). We begin by observing that
x 2 u ε ( t , · ) L 6 ( R ) 6 = R x 2 u ε ( x 2 u ε ) 5 d x = 5 R x u ε ( x 2 u ε ) 4 x 3 u ε d x .
Due to (58), (59), and the Young inequality,
5 R | x u ε | | x 2 u ε | 4 | x 3 u ε | d x 5 x u ε L ( ( 0 , T ) × R ) R | x 2 u ε | 4 | x 3 u ε | d x C ( T ) R | x 2 u ε | 5 | x 3 u ε | d x = R | x 2 u ε | 3 | C ( T ) x 2 u ε x 3 u ε | d x 1 2 x 2 u ε ( t , · ) L 6 ( R ) 6 + C ( T ) R ( x 2 u ε ) 2 ( x 3 u ε ) 2 d x 1 2 x 2 u ε ( t , · ) L 6 ( R ) 6 + C ( T ) x 3 u ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 1 2 x 2 u ε ( t , · ) L 6 ( R ) 6 + C ( T ) x 3 u ε ( t , · ) L ( R ) 2 .
Therefore, by (65),
1 2 x 2 u ε ( t , · ) L 6 ( R ) 6 C ( T ) x 3 u ε ( t , · ) L ( R ) 2 .
Due to the Hölder inequality,
( x 3 u ε ( t , x ) ) 2 = 2 x x 3 u ε x 4 u ε d y 2 R | x 3 u ε | | x 4 u ε | d x 2 x 3 u ε ( t , · ) L 2 ( R ) x 4 u ε ( t , · ) L 2 ( R ) .
Hence,
x 3 u ε ( t , · ) L 2 ( R ) 2 2 x 3 u ε ( t , · ) L 2 ( R ) x 4 u ε ( t , · ) L 2 ( R ) .
Thanks to the Young inequality,
x 3 u ε ( t , · ) L 2 ( R ) 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + x 4 u ε ( t , · ) L 2 ( R ) 2 .
It follows from (66) that
1 2 x 2 u ε ( t , · ) L 6 ( R ) 6 C ( T ) x 3 u ε ( t , · ) L 2 ( R ) 2 + C ( T ) x 4 u ε ( t , · ) L 2 ( R ) 2 .
Integrating on ( 0 , t ) , by (22)–(24) and (59), we have (63). □

3. Theorem 1’s Proof Based on the Aubin–Lions Lemma

This section is devoted to the proof of Theorem 1.
Our compactness argument is based on the Aubin–Lions Lemma (see [105,106]).
Lemma 4
(Aubin–Lions). Let X , B , Y be Banach spaces such that
X B Y .
If 1 p , F is bounded in L p ( 0 , T ; X ) , and
τ h f f L p ( 0 , T h ; Y ) 0 as h 0 uniformly for f F ,
where τ h is the translation operator. Then, F is relatively compact in L p ( 0 , T ; B ) (and in C ( 0 , T ; B ) if p = ).
We begin by proving the following lemma.
Lemma 5.
Fix γ , α , κ , δ , T . Then,
the sequence { u ε } ε > 0 is compact in L loc 2 ( ( 0 , ) × R ) .
Consequently, there exists a subsequence { u ε k } k N of { u ε } ε > 0 and u L l o c 2 ( ( 0 , ) × R ) such that, for each compact subset K of ( 0 , ) × R ) ,
u ε k u in L 2 ( K ) and a . e .
Moreover, u is a solution of (1), satisfying (16).
Proof. 
We begin by proving (67). To prove (67), we rely on the Aubin–Lions Lemma (see Lemma 4). We recall that
H l o c 1 ( R ) L l o c 2 ( R ) H l o c 1 ( R ) ,
where the first inclusion is compact and the second is continuous. Owing to the Aubin–Lions Lemma (see Lemma 4), to prove (67), it suffices to show that
{ u ε } ε > 0 is uniformly bounded in L 2 ( 0 , T ; H loc 1 ( R ) ) ,
{ t u ε } ε > 0 is uniformly bounded in L 2 ( 0 , T ; H loc 1 ( R ) ) .
We prove (69). Thanks to Lemmas 1 and 2,
u ε ( t , · ) H 2 ( R ) 2 = u ε ( t , · ) L 2 ( R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 + x 2 u ε ( t , · ) L 2 ( R ) 2 C ( T ) .
Therefore,
{ u ε } ε > 0 is uniformly bounded in L ( 0 , T ; H 1 ( R ) ) ,
which gives (69).
We prove (70). We begin by observing that
x u ε x 2 u ε = 1 2 x ( x u ε ) 2 + u ε x 3 u ε .
Therefore, by (18) and (71), we have that
t u ε = x G ( u ε ) f ( u ε ) x u ε q ( x u ε ) 2 ,
where
G ( u ε ) = α κ 2 ( x u ε ) 2 α u ε x 2 u ε ν x u ε δ x 2 u ε γ x 4 u ε β 2 x 3 u ε + ε x 5 u ε .
We claim that
0 T R u ε 2 ( x 2 u ε ) 2 d t d x C ( T ) .
Thanks to (21)–(24)
κ 2 0 T R u ε 2 ( x u ε ) 2 d t d x κ 2 u ε L ( ( 0 , T ) × R ) 2 0 T x 2 u ε ( t , · ) L 2 ( R ) 2 d t C ( T ) .
Moreover, since 0 < ε < 1 , by (20), (22)–(25), and (59),
( α κ ) 2 4 x u ε L 4 ( ( 0 , T ) × R ) 4 , ν 2 x u ε L 2 ( ( 0 , T ) × R ) 2 , δ 2 x 2 u ε L 2 ( ( 0 , T ) × R ) C ( T ) , γ 2 x 4 u ε L 2 ( ( 0 , T ) × R ) 2 , β 4 x 3 u ε L 2 ( ( 0 , T ) × R ) 2 C ( T ) , ε 2 x 5 u ε L 2 ( ( 0 , T ) × R ) 2 C ( T ) .
Therefore, by (72)–(74), we have that
{ x G ( u ε ) } ε > 0 is bounded in H 1 ( ( 0 , T ) × R ) .
We have that
0 T R ( f ( u ε ) ) 2 ( x u ε ) 2 d t d x C ( T ) .
Thanks to (20) and (21),
0 T R ( f ( u ε ) ) 2 ( x u ε ) 2 d t d x f L ( C ( T ) , C ( T ) ) 2 0 T x u ε ( t , · ) L 2 ( R ) 2 d t C ( T ) .
Moreover, thanks to (25),
δ + κ 2 0 T R ( x u ε ) 4 d t d x C ( T ) .
Consequently, (70) follows from (75)–(77). Thanks to the Aubin–Lions Lemma, (67) and (68) hold. Therefore, u is solution of (1) and, thanks to Lemmas 1, 2, and 3, (16) holds. □
Now, we prove Theorem 1.
Proof of Theorem 1.
We begin by observing that, by (71), (1) reads:
t u + x f ( u ) + ν x 2 u + δ x 3 u + α x u x 2 u + κ α 2 x ( x u ) 2 + q ( x u ) 2 + γ x 5 u + β 2 x 4 u = 0 , 0 < t < T , x R , u ( 0 , x ) = u 0 ( x ) , x R .
Lemma 5 gives the existence of a solution (78) satisfying (16). We prove (17). Let u 1 and u 2 be two solutions of (78) that verify (16), that is
t u i + x f ( u i ) + ν x 2 u i + δ x 3 u i + α x u i x 2 u i + κ α 2 x ( x u i ) 2 + q ( x u i ) 2 + γ x 5 u i + β 2 x 4 u i = 0 , 0 < t < T , x R , u i ( 0 , x ) = u i , 0 ( x ) , x R , i = 1 , 2 .
Then, the function
ω ( t , x ) = u 1 ( t , x ) u 2 ( t , x ) ,
is the solution of the following Cauchy problem:
t ω + x f ( u 1 ) f ( u 2 ) + ν x 2 ω + δ x 3 ω + α x u 1 x 2 u 1 u 2 x 2 u 2 + κ α 2 x ( x u 1 ) 2 ( x u 2 ) 2 + q ( x u 1 ) 2 ( x u 2 ) 2 + γ x 5 ω + β 2 x 4 ω = 0 , 0 < t < T , x R , ω ( 0 , x ) = u 1 , 0 ( x ) u 2 , 0 ( x ) , x R .
Fixe T > 0 . Since u 1 , u 2 H 2 ( R ) , for every 0 t T , we have that
u 1 L ( ( 0 , T ) × R ) , u 2 L ( ( 0 , T ) × R ) C ( T ) , x u 1 L ( ( 0 , T ) × R ) , x u 2 L ( ( 0 , T ) × R ) C ( T ) , x u 2 ( t , · ) L 2 ( R ) 2 C ( T ) .
Since f C 1 ( R ) , thanks to (79), there exists ξ between u 1 and u 2 , such that
f ( u 1 ) f ( u 2 ) = f ( ξ ) ( u 1 u 2 ) = f ( ξ ) ω , u 1 < ξ < u 2 , or , u 2 < ξ < u 1 .
Moreover, by (81), we have that
| f ( ξ ) | f L ( C ( T ) , C ( T ) ) C ( T ) .
Observe that, thanks to (79),
u 1 x 2 u 1 u 2 x 2 u 2 = u 1 x 2 u 1 u 1 x 2 u 2 + u 1 x 2 u 2 u 2 x 2 u 2 = u 1 x 2 ω x 2 u 2 ω , ( x u 1 ) 2 ( x u 2 ) 2 = ( x u 1 + x u 2 ) ( x u 1 x u 2 ) = x u 1 x ω + x u 2 x ω .
Thanks to (82) and (84), (80) reads
t ω = x f ( ξ ) ω ν γ x 2 ω δ x 3 ω α x u 1 x 2 ω α x x 2 u 2 ω + α κ 2 x x u 1 x ω + α κ 2 x x u 2 x ω + q x u 1 x ω + q x u 2 x ω γ x 5 ω β 2 x 4 ω .
Multiplying (85), an integration on R gives
d d t ω ( t , · ) L 2 ( R ) 2 = 2 R ω t ω d x = 2 R ω x f ( ξ ) ω 2 ν R ω x 2 ω d x 2 δ R ω x 3 ω d x 2 α R ω x u 1 x 2 ω d x 2 α R ω x x 2 u 2 ω d x + ( α κ ) R ω x x u 1 x ω d x + ( α κ ) R ω x x u 2 x ω d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x 2 γ R ω x 5 ω d x 2 β 2 R ω x 4 ω d x = 2 R f ( ξ ) ω x ω d x 2 ν R ω x 2 ω d x + 2 δ R x ω x 2 ω d x + 2 α R u 1 ω x 2 ω d x + 2 α R ω x ω x 2 u 2 d x + ( κ α ) R x u 1 ( x ω ) 2 d x + ( κ α ) R x u 2 ( x ω ) 2 d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x + 2 γ R x ω x 4 ω d x + 2 β 2 R x ω x 3 ω d x = 2 R f ( ξ ) ω x ω d x 2 ν R ω x 2 ω d x 2 α R x u 1 ω x ω d x 2 α R u 1 ( x ω ) 2 d x + 2 α R ω x ω x 2 u 2 d x + ( κ α ) R x u 1 ( x ω ) 2 d x + ( κ α ) R x u 2 ( x ω ) 2 d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x 2 γ R x 2 ω x 3 ω d x 2 β 2 x 2 ω ( t , · ) L 2 ( R ) 2 = 2 R f ( ξ ) ω x ω d x 2 ν R ω x 2 ω d x + 2 δ R x ω x 2 ω d x + 2 α R u 1 ω x 2 ω d x + 2 α R ω x ω x 2 u 2 d x + ( κ α ) R x u 1 ( x ω ) 2 d x + ( κ α ) R x u 2 ( x ω ) 2 d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x + 2 γ R x ω x 4 ω d x + 2 β 2 R x ω x 3 ω d x = 2 R f ( ξ ) ω x ω d x 2 ν R ω x 2 ω d x 2 α R x u 1 ω x ω d x 2 α R u 1 ( x ω ) 2 d x + 2 α R ω x ω x 2 u 2 d x + ( κ α ) R x u 1 ( x ω ) 2 d x + ( κ α ) R x u 2 ( x ω ) 2 d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x 2 β 2 x 2 ω ( t , · ) L 2 ( R ) 2 .
Therefore, we have that
d d t ω ( t , · ) L 2 ( R ) 2 + 2 β 2 x 2 ω ( t , · ) L 2 ( R ) 2 = 2 R f ( ξ ) ω x ω d x 2 ν R ω x 2 ω d x 2 α R x u 1 ω x ω d x 2 α R u 1 ( x ω ) 2 d x + 2 α R ω x ω x 2 u 2 d x + ( κ α ) R x u 1 ( x ω ) 2 d x + ( κ α ) R x u 2 ( x ω ) 2 d x + 2 q R x u 1 ω x ω d x + 2 q R x u 2 ω x ω d x .
Due to (81), (83), and the Young inequality,
2 R | f ( ξ ) | | ω | | x ω | d x 2 f L ( C ( T ) , C ( T ) ) R | ω | | x ω | d x C ( T ) R | ω | | x ω | d x C ( T ) ω ( t , · ) L 2 ( R ) 2 + C ( T ) x ω ( t , · ) L 2 ( R ) 2 , 2 | ν | R | ω | | x 2 ω | d x = 2 R ν ω β β x 2 ω d x ν 2 β 2 ω ( t , · ) L 2 ( R ) 2 + β 2 x 2 ω ( t , · ) L 2 ( R ) 2 , 2 | α | R | x u 1 | ( x ω ) 2 d x 2 | α | x u 1 L ( ( 0 , T ) × R ) x ω ( t , · ) L 2 ( R ) 2 C ( T ) x ω ( t , · ) L 2 ( R ) 2 , 2 | α | R | u 1 | ( x ω ) 2 d x 2 | α | u 1 L ( ( 0 , T ) × R ) x ω ( t , · ) L 2 ( R ) 2 C ( T ) x ω ( t , · ) L 2 ( R ) 2 , 2 | α | R | ω | | x ω | | x 2 u 2 | d x x ω ( t , · ) L 2 ( R ) 2 + α 2 R ω 2 ( x 2 u 2 ) 2 d x x ω ( t , · ) L 2 ( R ) 2 + α 2 ω ( t , · ) L ( R ) 2 x 2 u 2 ( t , · ) L 2 ( R ) 2 x ω ( t , · ) L 2 ( R ) 2 + C ( T ) ω ( t , · ) L ( R ) 2 , | κ α | R | x u 1 | ( x ω ) 2 d x | κ α | x u 1 L ( ( 0 , T ) × R ) x ω ( t , · ) L 2 ( R ) 2 C ( T ) x ω ( t , · ) L 2 ( R ) 2 , | κ α | R | x u 2 | ( x ω ) 2 d x | κ α | x u 2 L ( ( 0 , T ) × R ) x ω ( t , · ) L 2 ( R ) 2 C ( T ) x ω ( t , · ) L 2 ( R ) 2 , 2 | q | R | x u 1 | | ω | | x ω | d x 2 | q | x u 1 L ( ( 0 , T ) × R ) R | ω | | x ω | d x C ( T ) R | ω | | x ω | d x C ( T ) ω ( t , · ) L 2 ( R ) 2 + C ( T ) x ω ( t , · ) L 2 ( R ) 2 , 2 | q | R | x u 2 | | ω | | x ω | d x 2 | q | x u 1 L ( ( 0 , T ) × R ) R | ω | | x ω | d x C ( T ) R | ω | | x ω | d x C ( T ) ω ( t , · ) L 2 ( R ) 2 + C ( T ) x ω ( t , · ) L 2 ( R ) 2 .
It follows from (86) that
d d t ω ( t , · ) L 2 ( R ) 2 + β 2 x 2 ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + C ( T ) x ω ( t , · ) L 2 ( R ) 2 + C ( T ) ω ( t , · ) L ( R ) 2 .
Thanks to the Hölder inequality,
ω 2 ( t , x ) = 2 x ω x ω d y R | ω | | x ω | d x 2 ω ( t , · ) L 2 ( R ) x ω ( t , · ) L 2 ( R ) .
Hence,
ω ( t , · ) L ( R ) 2 2 ω ( t , · ) L 2 ( R ) x ω ( t , · ) L 2 ( R ) .
Due to the Young inequality,
ω ( t , · ) L ( R ) 2 ω ( t , · ) L 2 ( R ) 2 + x ω ( t , · ) L 2 ( R ) 2 .
Therefore, by (87), we have that
d d t ω ( t , · ) L 2 ( R ) 2 + β 2 x 2 ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + C ( T ) x ω ( t , · ) L 2 ( R ) 2 .
Observe that
C ( T ) x ω ( t , · ) L 2 ( R ) 2 = C ( T ) R x ω x ω d x = C ( T ) R ω x 2 ω d x .
Hence, by the Young inequality,
C ( T ) x ω ( t , · ) L 2 ( R ) 2 C ( T ) R | ω | | x 2 ω | d x = R C ( T ) ω β β x 2 ω d x C ( T ) ω ( t , · ) L 2 ( R ) 2 + β 2 2 x 2 ω ( t , · ) L 2 ( R ) 2 .
Consequently, by (88),
d d t ω ( t , · ) L 2 ( R ) 2 + β 2 2 x 2 ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 .
The Gronwall Lemma and (80) give
ω ( t , · ) L 2 ( R ) 2 + β 2 e C ( T ) t 2 0 t e C ( T ) s x 2 ω ( s , · ) L 2 ( R ) 2 d s e C ( T ) t ω 0 L 2 ( R ) 2 .
Equation (17) follows from (79) and (89).

4. Discussion

In this paper we studied the Cauchy problem of the fifth order Kudryashov–Sinelshchikov equation. Our argument is based on several a priori estimates on a sixth order approximation of the equation. The compactness of such approximations is obtained using the Aubin–Lions Lemma. Finally, we obtained the uniqueness of solutions proving a stability estimate with respect to the initial conditions.

Author Contributions

Conceptualization, G.M.C. and L.d.R.; methodology, G.M.C. and L.d.R.; formal analysis, G.M.C. and L.d.R.; investigation, G.M.C. and L.d.R.; resources, G.M.C. and L.d.R.; writing—original draft preparation, G.M.C. and L.d.R.; writing—review and editing, G.M.C. and L.d.R. All authors have read and agreed to the published version of the manuscript.

Funding

G.M.C. has been partially supported by the Research Project of National Relevance “Multiscale Innovative Materials and Structures” granted by the Italian Ministry of Education, University and Research (MIUR Prin 2017, project code 2017J4EAYB and the Italian Ministry of Education, University and Research under the Programme Department of Excellence Legge 232/2016 (Grant No. CUP-D94I18000260001).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

G.M.C. 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).

Conflicts of Interest

The authors declare no conflict of interest.

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Coclite, G.M.; di Ruvo, L. On the Solutions for a Fifth Order Kudryashov–Sinelshchikov Type Equation. Symmetry 2022, 14, 1535. https://doi.org/10.3390/sym14081535

AMA Style

Coclite GM, di Ruvo L. On the Solutions for a Fifth Order Kudryashov–Sinelshchikov Type Equation. Symmetry. 2022; 14(8):1535. https://doi.org/10.3390/sym14081535

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Coclite, Giuseppe Maria, and Lorenzo di Ruvo. 2022. "On the Solutions for a Fifth Order Kudryashov–Sinelshchikov Type Equation" Symmetry 14, no. 8: 1535. https://doi.org/10.3390/sym14081535

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