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Article

Well-Posedness Results for the Continuum Spectrum Pulse Equation

by
Giuseppe Maria Coclite
1,* and
Lorenzo di Ruvo
2
1
Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, via E. Orabona 4, 70125 Bari, Italy
2
Dipartimento di Matematica, Università di Bari, via E. Orabona 4, 70125 Bari, Italy
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(11), 1006; https://doi.org/10.3390/math7111006
Submission received: 25 September 2019 / Revised: 18 October 2019 / Accepted: 21 October 2019 / Published: 23 October 2019
(This article belongs to the Special Issue The Application of Mathematics to Physics and Nonlinear Science)

Abstract

:
The continuum spectrum pulse equation is a third order nonlocal nonlinear evolutive equation related to the dynamics of the electrical field of linearly polarized continuum spectrum pulses in optical waveguides. In this paper, the well-posedness of the classical solutions to the Cauchy problem associated with this equation is proven.

1. Introduction

In this paper, we investigate the well-posedness of the classical solution of the following Cauchy problem:
t u + 3 g u 2 x u a x 3 u + q x ( u v ) = b P , t > 0 , x R , x P = u , t > 0 , x R , α x 2 v + β x v + γ v = κ u 2 , t > 0 , x R , P ( t , ) = 0 , t > 0 , u ( 0 , x ) = u 0 ( x ) , x R ,
where g , a , q , b , α , β , γ , κ R .
On the initial datum, we assume that
u 0 H 2 ( R ) L 1 ( R ) , R u 0 ( x ) d x = 0 .
Following [1,2,3,4,5,6], on the function
P 0 ( x ) = x u 0 ( y ) d y ,
we assume that
R P 0 ( x ) d x = R x u 0 ( y ) d y = 0 , P 0 L 2 ( R ) 2 = R x u 0 ( y ) d y 2 d x < .
In addition, we assume that
q β κ 0 , g 0 , a 0 , α 0 .
Observe that, since α cannot vanish, we can factorize it and deal with only three constants.
In the physical literature (1) is termed the continuum spectrum pulse equation (see [7,8,9,10,11,12,13,14]). It is used to describe the dynamics of the electrical field u of linearly polarized continuum spectrum pulses in optical waveguides, including fused-silica telecommunication-type or photonic-crystal fibers, as well as hollow capillaries filled with transparent gases or liquids.
The constants a , b , g , q , α , κ , β , γ , in (1), take into account the frequency dispersion of the effective linear refractive index and the nonlinear polarization response, the excitation efficiency of the vibrations, the frequency and the decay time (see [7,8,14]).
Moreover, (1) generalizes the following system:
t u + q x ( u v ) = b P , t > 0 , x R , x P = u , t > 0 , x R , α x 2 v + β x v + γ v = κ u 2 , t > 0 , x R , P ( t , ) = 0 , t > 0 , u ( 0 , x ) = u 0 ( x ) , x R ,
whose the well-posedness is studied in [15]. From a mathematical point of view, the presence of the term
3 g u 2 x u a x 3 u
in the first equation of (1) makes the analysis of such system more subtle than the one of (6).
Observe that, taking b = α = β = 0 , (1) becomes the modified Korteweg-de Vries equation (see [16,17,18,19,20])
t u + g + q κ γ x u 3 a x 3 u = 0 .
In [8,9,21,22,23,24], it is proven that (7) is a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons. In [6,18], the Cauchy problem for (7) is studied, while, in [16,19], the convergence of the solution of (7) to the unique entropy solution of the following scalar conservation law
t u + g + q κ γ x u 3 = 0
is proven.
On the other hand, taking α = β = 0 in (1), we have the following equations
t u + g + q κ γ x u 3 = b P , t > 0 , x R , x P = u , t > 0 , x R ,
that were deduced by Kozlov and Sazonov [12] for the description of the nonlinear propagation of optical pulses of a few oscillations duration in dielectric media and by Schäfer and Wayne [25] for the description of the propagation of ultra-short light pulses in silica optical fibers. Moreover, (9) is a non-slowly-varying envelope approximation model that describes the physics of few-cycle-pulse optical solitons (see [22,23,24,26,27,28]), a particular Rabelo equation which describes pseudospherical surfaces (see [29,30,31,32]), and a model for the descriptions of the short pulse propagation in nonlinear metamaterials characterized by a weak Kerr-type nonlinearity in their dielectric response (see [33]).
Finally, (9) was deduced in [34] in the context of plasma physic and that similar equations describe the dynamics of radiating gases [35,36], in [37,38,39,40] in the context of ultrafast pulse propagation in a mode-locked laser cavity in the few-femtosecond pulse regime and in [41] in the context of Maxwell equations.
The Cauchy problem for (9) was studied in [42,43,44] in the context of energy spaces, [4,5,45,46] in the context of entropy solutions. The homogeneous initial boundary value problem was studied in [47,48,49,50]. Nonlocal formulations of (9) were analyzed in [15,51] and the convergence of a finite difference scheme proved in [52].
Observe that, taking α = β = 0 , (1) reads
t u + g + q κ γ x u 3 a x 3 u = b P , t > 0 , x R , x P = u , t > 0 , x R .
It was derived by Costanzino, Manukian and Jones [53] in the context of the nonlinear Maxwell equations with high-frequency dispersion. Kozlov and Sazonov [12] show that (10) is an more general equation than (9) to describe the nonlinear propagation of optical pulses of a few oscillations duration in dielectric media.
Mathematical properties of (10) are studied in many different contexts, including the local and global well-posedness in energy spaces [43,53] and stability of solitary waves [53,54], while, in [6], the well-posedness of the classical solutions is proven.
Observe that letting a 0 in (10), we obtain (9). Hence, following [19,55,56], in [5,57], the convergence of the solution of (10) to the unique entropy solution of (9).
The main result of this paper is the following theorem.
Theorem 1.
Assume (2), (3), (4) and (5). Fix T > 0 , there exists an unique solution ( u , v , P ) of (1) such that
u L ( 0 , T ; H 2 ( R ) ) , v L ( 0 , T ; H 4 ( R ) ) , P L ( 0 , T ; H 3 ( R ) ) .
In particular, we have that
R u ( t , x ) d x = 0 , t 0 .
Moreover, if ( u 1 , v 1 , P 1 ) and ( u 2 , v 2 , P 2 ) are two solutions of (1), we have that
u 1 ( t , · ) u 2 ( t , · ) L 2 ( R ) e C ( T ) t u 1 , 0 u 2 , 0 L 2 ( R ) , v 1 ( t , · ) v 2 ( t , · ) H 2 ( R ) e C ( T ) t u 1 , 0 u 2 , 0 L 2 ( R ) , P 1 ( t , · ) P 2 ( t , · ) H 1 ( R ) e C ( T ) t P 1 , 0 P 2 , 0 H 1 ( R ) ,
where,
P 1 , 0 ( x ) = x u 1 , 0 ( y ) d y , P 2 , 0 ( x ) = x u 2 , 0 ( y ) d y ,
for some suitable C ( T ) > 0 , and every 0 t T .
The proof of Theorem 1 is based on the Aubin–Lions Lemma (see [58,59,60]).
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, that is given in Section 3. Appendix A is an appendix, where we prove the posedness of the classical solutions of (1), under the assumption
u 0 L 1 ( R ) H 3 ( R ) .

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 mixed problem [19,61,62]:
t u ε + 3 g u ε 2 x u ε a x 3 u ε + q x ( v ε u ε ) = b P ε ε x 4 u ε , t > 0 , x R , x P ε = u ε , t > 0 , x R , α x 2 v ε + β x v ε + γ v ε = κ u ε 2 , t > 0 , x R , P ε ( t , ) = 0 , t > 0 , 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 ) , R u ε , 0 d x = 0 , P ε , 0 L 2 ( R ) P 0 L 2 ( R ) , R P ε , 0 d x = 0 .
Let us prove some a priori estimates on u ε , P ε and v 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.
Lemma 1.
For each t 0 ,
P ε ( t , ) = 0 .
In particular, we have that
R u ε ( t , x ) d x = 0 .
Proof. 
We begin by proving (18). Thanks to the regularity of u ε and the first equation of (16), we have that
lim x t u ε + 3 g u ε 2 x u ε a x 3 u ε + q x ( v ε u ε ) ε x 5 u ε = b P ε ( t , ) = 0 ,
which gives (18).
Finally, we prove (19). Integrating the second equation of (16) on ( , x ) , by (16), we have that
P ε ( t , x ) = x u ε ( t , y ) d y .
Equation (19) follows from (18) and (20).  □
Arguing as in ([15], Lemma 2.2), we have the following result.
Lemma 2.
Assume (5). We have that
R u ε 2 x v ε d x = β κ x v ε ( t , · ) L 2 ( R ) 2 , i f   β 0 , 0 , i f   β = 0 .
Lemma 3.
Assume (5). If β 0 , then for each t 0 , there exists a constant C 0 > 0 , independent on ε, such that
u ε ( t , · ) L 2 ( R ) 2 + q β κ 0 t x v ε ( s , · ) L 2 ( R ) 2 d s + 2 ε 0 t x u ε ( t , · ) L 2 ( R ) 2 d s C 0 .
If β = 0 , then for each t 0 ,
u ε ( t , · ) L 2 ( R ) 2 + 2 ε 0 t x u ε ( s , · ) L 2 ( R ) 2 d s u 0 L 2 ( R ) 2 .
In particular, we have
x v ε ( t , · ) L ( R ) , x v ε ( t , · ) L 2 ( R ) C 0 , v ε ( t , · ) L ( R ) , v ε ( t , · ) L 2 ( R ) C 0 .
Moreover, fixed T > 0 , there exists a constant C ( T ) > 0 , independent on ε, such that
ε 0 t x u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) .
Proof. 
Multiplying by 2 u ε the first equation of (16), an integration on R gives
d d t u ε ( t , · ) L 2 ( R ) 2 = 2 R u ε t u ε d x = 6 g R u ε 3 x u ε d x 2 q R u ε x ( u ε v ε ) d x + 2 b R P ε u ε d x + 2 a R u ε x 3 u ε d x 2 ε R u ε x 4 u ε d x = 2 q R u ε x ( u ε v ε ) d x 2 a R x u ε x 2 u ε d x + 2 b R P ε u ε d x + 2 ε R x u ε x 3 u ε d x = 2 q R u ε x ( u ε v ε ) d x 2 a R x u ε x 2 u ε d x + 2 b R P ε u ε d x 2 ε x 2 u ε ( t , · ) L 2 ( R ) 2 .
Therefore,
d d t u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 2 u ε ( t , · ) L 2 ( R ) 2 = 2 b R P ε u ε d x 2 q R u ε x ( u ε v ε ) d x .
Arguing as in ([15], Lemma 2.2), we have (22), (23) and (24).
Finally, arguing as in ([6], Lemma 2.3), we have (25).  □
Lemma 4.
Assume (5). Fix T > 0 . There exists a constant C 0 > 0 , independent on ε, such that
x 2 v ε L ( ( 0 , T ) × R ) C 0 1 + u ε L ( ( 0 , T ) × R ) 2 .
Proof. 
Let 0 t T . Thanks to the third equation of (16), we have that
α x 2 v ε = κ u ε 2 β x v ε γ v ε .
Therefore, by (24),
| α | | x 2 v ε | = | κ u ε 2 β x v ε γ v ε | | κ | u ε 2 + | β | | x v ε | + | γ | | v ε | | κ | u ε L 2 ( ( 0 , T ) × R ) 2 + | β | x v ε ( t , · ) L ( R ) + | γ | v ε ( t , · ) L ( R ) | κ | u ε L 2 ( ( 0 , T ) × R ) 2 + C 0 C 0 1 + u ε L 2 ( ( 0 , T ) × R ) 2 ,
which gives (26).  □
Arguing as in ([6], Lemma 2.2), we have the following result.
Lemma 5.
For each t 0 , we have that
0 P ε ( t , x ) d x = A ε ( t ) ,
0 P ε ( t , x ) d x = A ε ( t ) ,
where
A ε ( t ) = 1 b t P ε ( t , 0 ) g b u ε 3 ( t , 0 ) a b x 2 u ε ( t , 0 ) q b u ε ( t , 0 ) v ε ( t , 0 ) + ε b x u ε ( t , 0 ) .
In particular, we have
R P ε ( t , x ) d x = 0 .
Lemma 6.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
P ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s x u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
for every 0 t T . In particular, we have that
P ε L ( ( 0 , T ) × R ) C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 4 .
Proof. 
Let 0 t T . We begin by observing that, by (28), we can consider the following function:
F ε ( t , x ) = x P ε ( t , y ) d y .
Integrating the second equation of (16) on ( , x ) , we have
P ε ( t , x ) = x u ε ( t , y ) d y .
Differentiating (34) with respect to t, we get
t P ε ( t , x ) = d d t x u ε ( t , y ) d y = x t u ε ( t , x ) d x .
Equation (33), (35) and an integration on ( , x ) of the first equation of (16) give
t P ε = b F ε ε x 3 u ε g u ε 3 + a x 2 u ε q v ε u ε .
Multiplying (36) by 2 P ε , an integration on R gives
d d t P ε ( t , · ) L 2 ( R ) 2 = 2 b R F ε P ε d x 2 ε R P ε x 3 u ε d x 2 g R P ε u ε 3 d x + 2 a R P ε x 2 u ε d x 2 q R P ε v ε u ε d x .
Observe that, by (18), (30), (33) and the second equation of (16),
2 b R F ε P ε d x = 2 b R F ε x F ε d x = b F ε 2 ( t , ) = b R P ε ( t , x ) d x 2 d x = 0 , 2 ε R P ε x 3 u ε d x = 2 ε R x P ε x 2 u ε d x = 2 ε R u ε x 2 u ε d x = 2 ε x u ε ( t , · ) L 2 ( R ) 2 , 2 a R P ε x 2 u ε d x = 2 a R x P ε x u ε d x = 2 a R u ε x u ε = 0 , 2 q R P ε v ε u ε d x = 2 q R P ε v ε x P ε d x = 2 q R x v ε P ε 2 d x .
Consequently, by (24) and (37),
d d t P ε ( t , · ) L 2 ( R ) 2 + 2 ε x u ε ( t , · ) L 2 ( R ) 2 = 2 q R x v ε P ε 2 d x 2 g R P ε u ε 3 d x 2 | q | R | x v ε | P ε 2 d x + 2 | g | R | P ε | | u ε | 3 d x 2 | q | x v ε ( t , · ) L ( R ) P ε ( t , · ) L 2 ( R ) 2 + 2 | g | R | P ε | | u ε | 3 d x C 0 P ε ( t , · ) L 2 ( R ) 2 + 2 | g | R | P ε | | u ε | 3 d x .
Due to Lemma 3 and the Young inequality,
2 | g | R | P ε | | u ε | 3 d x = R | 2 P ε u ε | | g u ε 2 | d x 2 R P ε 2 u ε 2 d x + g 2 2 R u ε 4 d x 2 P ε ( t , · ) L ( R ) 2 u ε ( t , · ) L 2 ( R ) 2 + g 2 2 u ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 C 0 P ε ( t , · ) L ( R ) 4 + C 0 u ε L ( ( 0 , T ) × R ) 2 P ε ( t , · ) L ( R ) 4 + C 0 + C 0 u ε L ( ( 0 , T ) × R ) 2 .
It follows from (38) that
d d t P ε ( t , · ) L 2 ( R ) 2 + 2 ε x u ε ( t , · ) L 2 ( R ) 2 C 0 P ε ( t , · ) L 2 ( R ) 2 + P ε ( t , · ) L ( R ) 4 + C 0 1 + u ε L ( ( 0 , T ) × R ) 2 .
Thanks to Lemma 3 and the Hölder inequality,
P ε 2 ( t , x ) = 2 x P ε u ε d y 2 R | P ε | | u ε | d x P ε ( t , · ) L 2 ( R ) u ε ( t , · ) L 2 ( R ) C 0 P ε ( t , · ) L 2 ( R ) .
Hence,
P ε ( t , · ) L ( R ) 4 C 0 P ε ( t , · ) L 2 ( R ) 2 .
It follows from (39) and (40) that
d d t P ε ( t , · ) L 2 ( R ) 2 + 2 ε u ε ( t , · ) L 2 ( R ) 2 C 0 P ε ( t , · ) L 2 ( R ) 2 + C 0 1 + u ε L ( ( 0 , T ) × R ) 2 .
The Gronwall Lemma and (17) give
P ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s u ε ( s , · ) L 2 ( R ) 2 d s C 0 e C 0 t + C 0 1 + u ε L ( ( 0 , T ) × R ) 2 e C 0 t 0 t e C 0 s d s C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 ,
which gives (31).
Finally, (32) follows from (31) and (40).  □
Following ([6], Lemma 2.5), we prove the following result.
Lemma 7.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
u ε L ( ( 0 , T ) × R ) C ( T ) .
In particular, we have
x u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s x 3 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
for every 0 t T . Moreover,
x 2 v ε L ( ( 0 , T ) × R ) C ( T ) , P ε ( t , · ) L 2 ( R ) C ( T ) , P ε L ( ( 0 , T ) × R ) C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Multiplying the first equation of (1) by 2 x 2 u ε + 2 g a u ε 3 , we have that
2 x 2 u ε + 2 g a u ε 3 t u ε + 3 g 2 x 2 u ε + 2 g a u ε 3 u ε 2 x u ε a 2 x 2 u ε + 2 g a u ε 3 x 3 u ε + q 2 x 2 u ε + 2 g a u ε 3 x ( v ε u ε ) = b 2 x 2 u ε + 2 g a u ε 3 P ε ε 2 x 2 u ε + 2 g a u ε 3 x 4 u ε .
Observe that, by (18) and the second equation of (16),
2 b R P ε x 2 u ε d x = 2 b R x P ε x u ε d x = 2 b R u ε x u ε d x = 0 .
Moreover,
R 2 x 2 u ε + 2 g a u ε 3 t u ε d x = d d t x u ε ( t , · ) L 2 ( R ) 2 + g 2 a R u ε 4 d x , 3 g R 2 x 2 u ε + 2 g a u ε 3 u ε 2 x u ε d x = 6 g R u ε 2 x u ε x 2 u ε d x , a R 2 x 2 u ε + 2 g a u ε 3 x 3 u ε d x = 6 g R u ε 2 x u ε x 2 u ε d x , ε R 2 x 2 u ε + 2 g a u ε 3 x 4 u ε d x = 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 + 6 g ε a R u ε 2 x u ε x 3 u ε d x .
Defined
G ( t ) : = x u ε ( t , · ) L 2 ( R ) 2 + g 2 a R u ε 4 d x ,
it follows from (45), (46) and an integration on R of (44) that
d G ( t ) d t + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 = 2 b g a R P ε u ε 3 d x + 6 g ε a R u ε 2 x u ε x 3 u ε d x + 2 q R x ( v ε u ε ) x 2 u ε d x 2 q g a R x ( v ε u ε ) u ε 3 d x .
Observe that
2 q R x ( v ε u ε ) x 2 u ε d x = 2 q R u ε x v ε x 2 u ε d x + 2 q R v ε x u ε x 2 u ε d x = 2 q R x 2 v ε u ε x u ε d x 3 q R x v ε ( x u ε ) 2 d x , 2 q g a R x ( v ε u ε ) u ε 3 d x = 2 q g a R x v ε u ε 4 d x 2 q g a R v ε x u ε u ε 3 d x 3 q g 2 a R x v ε u ε 4 d x .
Consequently, by (48),
d G ( t ) d t + 2 ε x 3 u ε ( t , · ) L 2 ( R ) 2 = 2 b g a R P ε u ε 3 d x + 6 g ε a R u ε 2 x u ε x 3 u ε d x 3 q R x v ε ( x u ε ) 2 d x 2 q R x 2 v ε u ε x u ε d x 3 q g 2 a R x v ε u ε 4 d x .
Due to (26), (32), Lemma 3 and the Young inequality,
2 b g a R P ε u ε 3 d x = R 2 P ε u ε b g u ε 2 a d x 2 R P ε 2 u ε 2 d x + b 2 g 2 2 a 2 R u ε 4 d x 2 P ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 + b 2 g 2 2 a 2 u ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 2 C 0 P ε L ( ( 0 , T ) × R ) 2 + C 0 u ε L ( ( 0 , T ) × R ) 2 P ε L ( ( 0 , T ) × R ) 4 + C 0 u ε L ( ( 0 , T ) × R ) 2 + C 0 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 , 3 | q | R | x v ε | ( x u ε ) 2 d x 3 | q | x v ε ( t , · ) L ( R ) x u ε ( t , · ) L 2 ( R ) 2 C 0 x u ε ( t , · ) L 2 ( R ) 2 , 2 | q | R x 2 v ε u ε x u ε d x = 2 R | q x 2 v ε u ε | x u ε | d x q 2 R ( x 2 v ε ) 2 u ε 2 d x + x u ε ( t , · ) L 2 ( R ) 2 q 2 x 2 v ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 C 0 x 2 v ε L ( ( 0 , T ) × R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 + x u ε ( t , · ) L 2 ( R ) 2 , 3 q g 2 a R | x v ε | u ε 4 d x 3 q g 2 a x v ε ( t , · ) L ( R ) R u ε 4 d x C 0 R u ε 4 d x C 0 u ε 2 L ( ( 0 , T ) × R ) u ε ( t , · ) L 2 ( R ) 2 C 0 u ε 2 L ( ( 0 , T ) × R ) , 6 g ε a R | u ε 2 x u ε | | x 3 u ε | d x = 2 ε R 3 g a u ε 2 x u ε x 3 u ε d x 9 g 2 ε a 2 R u ε 4 ( x u ε ) 2 d x + ε x 3 u ε ( t , · ) L 2 ( R ) 2 .
Consequently, by (49),
d G ( t ) d t + ε x 3 u ε ( t , · ) L 2 ( R ) 2 C 0 x u ε ( t , · ) L 2 ( R ) 2 + 9 g 2 ε a 2 R u ε 4 ( x u ε ) 2 d x + C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
Lemma 2.6 of [6] says that
R u ε 4 ( x u ε ) 2 d x 4 u ε ( t , · ) L 2 ( R ) 4 x 2 u ε ( t , · ) L 2 ( R ) 2 .
Hence, by Lemma 3, we have that
9 g 2 ε a 2 R u ε 4 ( x u ε ) 2 d x 36 g 2 ε a 2 u ε ( t , · ) L 2 ( R ) 4 x 2 u ε ( t , · ) L 2 ( R ) 2 C 0 ε x 2 u ε ( t , · ) L 2 ( R ) 2 .
Therefore, by (50),
. d G ( t ) d t + ε x 3 u ε ( t , · ) L 2 ( R ) 2 C 0 x u ε ( t , · ) L 2 ( R ) 2 + C 0 ε x 2 u ε ( t , · ) L 2 ( R ) 2 + C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
Observe that, by (47) and Lemma 3,
C 0 x u ε ( t , · ) L 2 ( R ) 2 = C 0 G ( t ) g C 0 2 a R u ε 4 d x C 0 G ( t ) + g C 0 2 a u ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 C 0 G ( t ) + C 0 u ε L ( ( 0 , T ) × R ) 2 .
It follows from (52) and (53) that
d G ( t ) d t + ε x 3 u ε ( t , · ) L 2 ( R ) 2 C 0 G ( t ) + C 0 ε x 2 u ε ( t , · ) L 2 ( R ) 2 + C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
The Gronwall Lemma, (17), (47) and Lemma 3 that
x u ε ( t , · ) L 2 ( R ) 2 + g 2 a R u ε 4 d x + ε e C 0 t 0 t e C 0 s x 3 u ε ( s , · ) L 2 ( R ) 2 d s C 0 e C 0 t + C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 e C 0 t 0 t e C 0 s d s + C 0 ε e C 0 t 0 t e C 0 s x 2 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 + C ( T ) ε 0 t x 2 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
Consequently, by Lemma 3,
x u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s x 3 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 g 2 a u ε ( t , · ) L 4 ( R ) 4 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 + g 2 a R u ε 4 d x C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 + g 2 a u ε L ( ( 0 , T ) × R ) 2 u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 + C 0 u ε L ( ( 0 , T ) × R ) 2 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
We prove (41). Thanks to (54), Lemma 3 and the Hölder inequality,
u ε 2 ( t , x ) = 2 x u ε x u ε d x R | u ε | | x u ε | d x u ε ( t , · ) L 2 ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + u ε L ( ( 0 , T ) × R ) 2 .
Hence,
u ε L ( ( 0 , T ) × R ) 4 C ( T ) u ε L ( ( 0 , T ) × R ) 2 C ( T ) 0 ,
which gives (41).
Finally, (42) follows from (41) and (54), while (26), (31), (32) and (41) give (43).  □
Arguing as in ([15], Lemmas 2.8 and 2.9), we have the following result.
Lemma 8.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
x 2 v ε ( t , · ) L 2 ( R ) , x 3 v ε ( t , · ) L 2 ( R ) C ( T ) ,
for every 0 t T .
Lemma 9.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
x u ε L ( ( 0 , T ) × R ) C ( T ) ,
In particular, we have that
x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s x 4 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Consider two real constants D , E which will be specified later Multiplying the first equation of (16) by
2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε ,
we have that
2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε t u ε + 3 g 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε u ε 2 x u ε a 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε x 3 u ε + q 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε u ε x v ε + q 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε v ε x u ε = b 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε P ε ε 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε x 4 u ε .
Observe that
R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε t u ε d x = a 2 d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + D a g R u ε ( x u ε ) 2 t u ε d x + E a g R u ε 2 x 2 u ε t u ε d x , 3 g R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε u ε 2 x u ε d x = 12 a 2 g R u ε ( x u ε ) 2 x 3 u ε d x 6 a 2 g R u ε 2 x 2 u ε x 3 u ε d x + 3 D 6 E a g 2 R u ε 3 ( x u ε ) 3 d x = 30 a 2 g R u ε x u ε ( x 2 u ε ) 2 d x + 3 D 6 E a g 2 R u ε 3 ( x u ε ) 3 d x , a R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε x 3 u ε d x = 2 D + E a 2 g R u ε x u ε ( x 2 u ε ) 2 d x , q R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε u ε x v ε d x = 2 a 2 q R x u ε x v ε x 3 u ε d x 2 a 2 q R u ε x 2 v ε x 3 u ε d x + D 3 E a g q R u ε 2 ( x u ε ) 2 x v ε d x a g q E R u ε 3 x u ε x 2 v ε d x = 2 a 2 q R x v ε ( x 2 u ε ) 2 d x + 4 a 2 q R x u ε x 2 v ε x 2 u ε d x + 2 a 2 q R u ε x 3 v ε x 2 u ε d x + a q D 3 E R u ε 2 ( x u ε ) 2 x v ε d x , q R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε v ε x u ε d x = 2 a 2 q R x v ε x u ε x 3 u ε d x 2 a 2 q R v ε x 2 u ε x 3 u ε d x + D E a g q R u ε v ε ( x u ε ) 3 d x E a g q 2 R u ε 2 ( x u ε ) 3 x 2 v ε d x = 2 a 2 q R x 2 v ε x u ε x 2 u ε d x + 3 a 2 q R x v ε ( x 2 u ε ) 2 d x + D E a g q R u ε v ε ( x u ε ) 3 d x E a g q 2 R u ε 2 ( x u ε ) 3 x 2 v ε d x , ε R 2 a 2 x 4 u ε + D a g u ε ( x u ε ) 2 + E a g u ε 2 x 2 u ε x 4 u ε d x = 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 + D a g ε R u ε ( x u ε ) 2 x 4 u ε d x + E a g ε R u ε 2 x 2 u ε x 4 u ε d x .
Consequently, an integration on R of (58) gives
a 2 d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + D a g R u ε ( x u ε ) 2 t u ε d x + E a g R u ε 2 x 2 u ε t u ε d x + 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 = a 2 g 30 + 2 D + E R u ε x u ε ( x 2 u ε ) 2 d x 3 D 6 E a g 2 R u ε 3 ( x u ε ) 3 d x 5 a 2 q R x v ε ( x 2 u ε ) 2 d x 6 a 2 q R x 2 v ε x u ε x 2 u ε d x 2 a 2 q R u ε x 3 v ε x 2 u ε d x a q D 3 E R u ε 2 ( x u ε ) 2 x v ε d x D E a g q R u ε v ε ( x u ε ) 3 d x + E a g q 2 R u ε 2 ( x u ε ) 3 x 2 v ε d x + 2 a 2 b R P ε x 4 u ε d x + D a g b R P ε u ε ( x u ε ) 2 d x + E a g b R P ε u ε 2 x 2 u ε d x D a g ε R u ε ( x u ε ) 2 x 4 u ε d x + E a g ε R u ε 2 x 2 u ε x 4 u ε d x .
Thanks to the second equation of (16) and (18), we have that
2 a 2 b R P ε x 4 u ε d x = 2 a 2 b R x P ε x 3 u ε = 2 a 2 b R u ε x 3 u ε d x = 2 a 2 b R x u ε x 2 u ε d x = 0 , E a g b R P ε u ε 2 x 2 u ε d x = E a g b R x P ε u ε 2 x u ε d x 2 E a g b R P ε u ε ( x u ε ) 2 d x = 2 E a g b R P ε u ε ( x u ε ) 2 d x E a g b R u ε 3 x u ε d x = 2 E a g b R P ε u ε ( x u ε ) 2 d x .
Therefore, by (59),
a 2 d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + D a g R u ε ( x u ε ) 2 t u ε d x + E a g R u ε 2 x 2 u ε t u ε d x + 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 = a 2 g 30 + 2 D + E R u ε x u ε ( x 2 u ε ) 2 d x 3 D 6 E a g 2 R u ε 3 ( x u ε ) 3 d x 5 a 2 q R x v ε ( x 2 u ε ) 2 d x 6 a 2 q R x 2 v ε x u ε x 2 u ε d x 2 a 2 q R u ε x 3 v ε x 2 u ε d x a q D 3 E R u ε 2 ( x u ε ) 2 x v ε d x D E a g q R u ε v ε ( x u ε ) 3 d x + E a g q 2 R u ε 2 ( x u ε ) 3 x 2 v ε d x + D 2 E a g b R P ε u ε ( x u ε ) 2 d x + D a g ε R u ε ( x u ε ) 2 x 4 u ε d x + E a g ε R u ε 2 x 2 u ε x 4 u ε d x .
Observe that
D a g R u ε ( x u ε ) 2 t u ε d x + E a g R u ε 2 x 2 u ε t u ε d x = D a g 2 R t ( u ε 2 ) ( x u ε ) 2 d x E a g R x u ε x ( u ε 2 t u ε ) d x = D a g 2 R t ( u ε 2 ) ( x u ε ) 2 d x 2 E a g R u ε ( x u ε ) 2 t u ε d x E a R u ε 2 x u ε t x 2 u ε d x a g D 2 E R t ( u ε 2 ) ( x u ε ) 2 d x E a g 2 R u ε 2 t ( ( x u ε ) 2 ) d x .
Consequently, by (60),
a 2 d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + a g D 2 E R t ( u ε 2 ) ( x u ε ) 2 d x E a g 2 R u ε 2 t ( ( x u ε ) 2 ) d x + 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 = a 2 g 30 + 2 D + E R u ε x u ε ( x 2 u ε ) 2 d x 3 D 6 E a g 2 R u ε 3 ( x u ε ) 3 d x 5 a 2 q R x v ε ( x 2 u ε ) 2 d x 6 a 2 q R x 2 v ε x u ε x 2 u ε d x 2 a 2 q R u ε x 3 v ε x 2 u ε d x a q D 3 E R u ε 2 ( x u ε ) 2 x v ε d x D E a g q R u ε v ε ( x u ε ) 3 d x + E a g q 2 R u ε 2 ( x u ε ) 3 x 2 v ε d x + D 2 E a g b R P ε u ε ( x u ε ) 2 d x + D a g ε R u ε ( x u ε ) 2 x 4 u ε d x + E a g ε R u ε 2 x 2 u ε x 4 u ε d x .
We search D , E such that
D 2 E = E 2 , 30 + 2 D + E = 0 ,
that is
D = E , 30 + 2 D + E = 0 .
Since D = E 10 is the unique solution of (62), it follows from (61) that
a 2 d d t x 2 u ε ( t , · ) L 2 ( R ) 2 + 5 a g R t ( u ε 2 ) ( x u ε ) 2 d x + 5 a g R u ε 2 t ( ( x u ε ) 2 ) d x + 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 = 30 a g 2 R u ε 3 ( x u ε ) 3 d x 5 a 2 q R x v ε ( x 2 u ε ) 2 d x 6 a 2 q R x 2 v ε x u ε x 2 u ε d x 2 a 2 q R u ε x 3 v ε x 2 u ε d x 20 a q R u ε 2 ( x u ε ) 2 x v ε d x 5 a g q R u ε 2 ( x u ε ) 3 x 2 v ε d x + 10 a g b R P ε u ε ( x u ε ) 2 d x 10 a g ε R u ε ( x u ε ) 2 x 4 u ε d x 10 a g ε R u ε 2 x 2 u ε x 4 u ε d x ,
that is
d d t a 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 5 a g R u ε 2 ( x u ε ) 2 d x + 2 a 2 ε x 4 u ε ( t , · ) L 2 ( R ) 2 = 30 a g 2 R u ε 3 ( x u ε ) 3 d x 5 a 2 q R x v ε ( x 2 u ε ) 2 d x 6 a 2 q R x 2 v ε x u ε x 2 u ε d x 2 a 2 q R u ε x 3 v ε x 2 u ε d x 20 a q R u ε 2 ( x u ε ) 2 x v ε d x 5 a g q R u ε 2 ( x u ε ) 3 x 2 v ε d x + 10 a g b R P ε u ε ( x u ε ) 2 d x 10 a g ε R u ε ( x u ε ) 2 x 4 u ε d x 10 a g ε R u ε 2 x 2 u ε x 4 u ε d x .
Due to (41), (42), (43), (55), Lemma 3 and the Young inequality,
| 30 a g 2 | R | u ε | 3 | x u ε | 3 d x | 30 a g 2 | u ε L ( ( 0 , T ) × R ) 3 R | x u ε | 3 d x C ( T ) R | x u ε | 3 d x C ( T ) x u ε ( t , · ) L 2 ( R ) 2 + C ( T ) R ( x u ε ) 4 d x C ( T ) + C ( T ) x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 , | 5 a 2 q | R | x v ε | ( x 2 u ε ) 2 d x | 5 a 2 q | x v ε ( t , · ) L ( R ) x 2 u ε ( t , · ) L 2 ( R ) 2 C 0 x 2 u ε ( t , · ) L 2 ( R ) 2 , | 6 a 2 q | R | x 2 v ε x u ε | | x 2 u ε | d x 3 a 4 q 2 R ( x 2 v ε ) 2 ( x u ε ) 2 d x + 3 x 2 u ε ( t , · ) L 2 ( R ) 2 3 a 4 q 2 x 2 v ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + 3 x 2 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + 3 x 2 u ε ( t , · ) L 2 ( R ) 2 , | 2 a 2 q | R | u ε x 3 v ε | | x 2 u ε | d x a 4 q 2 R u ε 2 ( x 3 v ε ) 2 d x + x 2 u ε ( t , · ) L 2 ( R ) 2 a 4 q 2 u ε L ( ( 0 , T ) × R ) 2 x 3 v ε ( t , · ) L 2 ( R ) 2 + x 2 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + x 2 u ε ( t , · ) L 2 ( R ) 2 , | 20 a q | R u ε 2 ( x u ε ) 2 | x v ε | d x | 20 a q | x v ε ( t , · ) L ( R ) u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C ( T ) x u ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
| 5 a g q | R u ε 2 | x u ε | 3 | x 2 v ε | d x | 5 a g q | u ε L ( ( 0 , T ) × R ) 2 x 2 v ε L ( ( 0 , T ) × R ) R | x u ε | 3 d x C ( T ) R | x u ε | 3 d x C ( T ) x u ε ( t , · ) L 2 ( R ) 2 + C ( T ) R ( x u ε ) 4 d x C ( T ) + C ( T ) x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 , | 10 a g b | R | P ε u ε | ( x u ε ) 2 d x | 10 a g b | P ε L ( ( 0 , T ) × R ) u ε L ( ( 0 , T ) × R ) x u ε ( t , · ) L 2 ( R ) 2 C ( T ) x u ε ( t , · ) L 2 ( R ) 2 C ( T ) , | 10 a g | ε R | u ε ( x u ε ) 2 | | x 4 u ε | d x = ε R | 10 g u ε ( x u ε ) 2 | | a x 4 u ε | d x 50 g 2 ε R u ε 2 ( x u ε ) 4 d x + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 50 g 2 ε u ε L ( ( 0 , T ) × R ) 2 R ( x u ε ) 4 d x + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 C ( T ) ε x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 , | 10 a g | ε R | u ε 2 x 2 u ε | | x 4 u ε | d x = ε R | 10 g u ε 2 x 2 u ε | | a x 4 u ε | d x 50 g 2 ε R u ε 4 ( x 2 u ε ) 2 d x + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 50 g 2 ε u ε L ( ( 0 , T ) × R ) 4 x 2 u ε ( t , · ) L 2 ( R ) 2 + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 C ( T ) ε x 2 u ε ( t , · ) L 2 ( R ) 2 + a 2 ε 2 x 4 u ε ( t , · ) L 2 ( R ) 2 .
Therefore, defining
G 1 ( t ) = a 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 5 a g R u ε 2 ( x u ε ) 2 d x ,
by (63) and (64), we have
d G 1 ( t ) d t + ε x 4 u ε ( t , · ) L 2 ( R ) 2 C 0 x 2 u ε ( t , · ) L 2 ( R ) 2 + C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 + C ( T ) ε x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + C ( T ) ε x 2 u ε ( t , · ) L 2 ( R ) 2 .
Observe that by (41), (42) and (64),
C 0 x 2 u ε ( t , · ) L 2 ( R ) 2 = C 0 a 2 a 2 x 2 u ε ( t , · ) L 2 ( R ) 2 = C 0 a 2 G 1 ( t ) 5 C 0 g a R u ε 2 ( x u ε ) 2 d x C 0 G 1 ( t ) + 5 C 0 g a R u ε 2 ( x u ε ) 2 d x C 0 G 1 ( t ) + 5 C 0 g a u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C 0 G 1 ( t ) + C ( T ) .
It follows from (65) and (66) that
d G 1 ( t ) d t + ε x 4 u ε ( t , · ) L 2 ( R ) 2 C 0 G 1 ( t ) + C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 + C ( T ) ε x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + C ( T ) ε x 2 u ε ( t , · ) L 2 ( R ) 2 .
The Gronwall Lemma, (17) and Lemma 3 give
x 2 u ε ( t , · ) L 2 ( R ) 2 + 5 a g R u ε 2 ( x u ε ) 2 d x + 2 ε e C 0 t 0 t e C 0 s x 4 u ε ( s , · ) L 2 ( R ) 2 d s C 0 e C 0 t + C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 e C 0 t 0 t e C 0 s d s + C ( T ) ε x u ε L ( ( 0 , T ) × R ) 2 e C 0 t 0 t e C 0 s x u ε ( s , · ) L 2 ( R ) 2 d s + C ( T ) ) ε e C 0 t 0 t e C 0 s x 2 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 + C ( T ) ε x u ε L ( ( 0 , T ) × R ) 2 0 t x u ε ( s , · ) L 2 ( R ) 2 d s + C ( T ) ) ε 0 t x 2 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 .
Therefore, thanks to (41) and (42),
x 2 u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C 0 t 0 t e C 0 s x 4 u ε ( s , · ) L 2 ( R ) 2 d s = C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 5 a g R u ε 2 ( x u ε ) 2 d x C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 + | 5 a g | R u ε 2 ( x u ε ) 2 d x C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 + | 5 a g | u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + x u ε L ( ( 0 , T ) × R ) 2 .
We prove (56). Due to (42), (67) and the Hölder inequality,
( x u ε ( t , x ) ) 2 = 2 x x u ε x 2 u ε d x 2 R | x u ε | | x 2 u ε | d x 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 ) 2 .
Therefore,
x u ε L ( ( 0 , T ) × R ) 4 C ( T ) x u ε L ( ( 0 , T ) × R ) 2 C ( T ) 0 ,
which gives (56).
Finally, (57) follows from (56) and (67).  □
Lemma 10.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
x 4 v ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
for every 0 t T . In particular, we have that
x 3 v ε L ( ( 0 , T ) × R ) C ( T ) ,
Proof. 
Let 0 t T . Differentiating the third equation of (16) twice with respect to x, we have
α x 4 v ε = 2 κ ( x u ε ) 2 + 2 κ u ε x 2 u ε β x 3 v ε γ x 2 v ε .
Since
u ε ( t , ± ) = x u ε ( t , ± ) = x 2 u ε ( t , ± ) = 0 ,
it follows from (24) and (55) that
x 4 v ε ( t , ± ) = 0 .
Multiplying (70) by 2 α x 4 v ε , an integration on R gives
2 α 2 x 4 v ε ( t , · ) L 2 ( R ) 2 = 2 κ α R ( x u ε ) 2 x 4 v ε d x + 2 κ α R u ε x 2 u ε x 4 v ε d x 2 β α R x 3 v ε x 4 v ε d x 2 γ α R x 2 v ε x 4 v ε d x .
Observe that, thanks to (24), (55) and (72),
2 β α R x 3 v ε x 4 v ε d x = 0 , 2 γ α R x 2 v ε x 4 v ε d x = 2 γ α x 3 v ε ( t , · ) L 2 ( R ) 2 .
Therefore, by (55) and (73),
2 α 2 x 4 v ε ( t , · ) L 2 ( R ) 2 2 | κ α | R ( x u ε ) 2 | x 4 v ε | d x + 2 | κ α | R | u ε x 2 u ε | | x 4 v ε | d x + 2 | γ α | x 3 v ε ( t , · ) L 2 ( R ) 2 2 | κ α | R ( x u ε ) 2 | x 4 v ε | d x + 2 | κ α | R | u ε x 2 u ε | | x 4 v ε | d x + C ( T ) .
Due to (41), (42), (56), (57) and the Young inequality,
2 | κ α | R ( x u ε ) 2 | x 4 v ε | d x κ 2 R ( x u ε ) 4 d x + α 2 x 4 v ε ( t , · ) L 2 ( R ) 2 κ 2 x u ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + α 2 x 4 v ε ( t , · ) L 2 ( R ) 2 C ( T ) + α 2 x 4 v ε ( t , · ) L 2 ( R ) 2 , 2 | κ α | R | u ε x 2 u ε | | x 4 v ε | = R | 2 κ u ε x 2 u ε | | α x 4 v ε | d x 2 κ 2 R u ε 2 ( x 2 u ε ) 2 d x + α 2 2 x 4 v ε ( t , · ) L 2 ( R ) 2 2 κ 2 u ε L ( ( 0 , T ) × R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + α 2 2 x 4 v ε ( t , · ) L 2 ( R ) 2 C ( T ) + α 2 2 x 4 v ε ( t , · ) L 2 ( R ) 2 .
Consequently, by (74),
α 2 2 x 4 v ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
which gives (68).
Finally, we prove (69). Due to (55), (68) and the Hölder inequality,
( x 3 v ε ( t , x ) ) 2 = 2 x x 3 v ε x 4 v ε d x 2 R | x 3 v ε | | x 4 v ε | d x x 3 v ε ( t , · ) L 2 ( R ) x 4 v ε ( t , · ) L 2 ( R ) C ( T ) .
Hence,
x 3 v ε L ( ( 0 , T ) × R ) 2 C ( T ) ,
which gives (69).  □

3. Proof of Theorem 1

This section is devoted to the proof of Theorem 1.
We begin by proving the following lemma.
Lemma 11.
Fix T > 0 . Then,
t h e   s e q u e n c e   { u ε } ε > 0   i s   c o m p a c t   i n   L l o c 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   i n   L 2 ( K )   a n d   a . e . ,
v ε k v   i n   H 1 ( ( 0 , T ) × R ) ,
P ε k P   i n   L 2 ( ( 0 , T ) × R ) .
Moreover, ( u , v , P ) is a solution of (1) satisfying (11) and (12).
Proof. 
We begin by proving (75). To prove (75), we rely on the Aubin–Lions Lemma (see [58,59,60]). 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 [60], to prove (75), it suffices to show that
{ u ε } ε > 0 is   uniformly   bounded   in   L 2 ( 0 , T ; H l o c 1 ( R ) ) ,
{ t u ε } ε > 0 is   uniformly   bounded   in   L 2 ( 0 , T ; H l o c 1 ( R ) ) .
We prove (79). Thanks to (42), (57) and Lemma 3,
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 2 ( R ) ) ,
which gives (79).
We prove (80). By the first equation of (16),
t u ε = x g u ε 3 + a x 2 u ε q v ε u ε ε x 3 u ε + b P ε .
We have that
u ε L 6 ( ( 0 , T ) × R ) C ( T ) .
Indeed, thanks to (41) and Lemma 3,
g 2 0 T R u ε 6 d t d x g 2 u ε L ( ( 0 , T ) × R ) 4 0 T R u ε 2 d t d x C ( T ) 0 T R u ε 2 d t d x C ( T ) .
We prove that
q 2 v ε u ε L 2 ( ( 0 , T ) × R ) 2 C ( T ) .
Due to Lemma 3,
q 2 0 T R v ε 2 u ε 2 d t d x q 2 v ε L ( ( 0 , T ) × R ) 2 0 T R u ε 2 d t d x C ( T ) 0 T R u ε 2 d t d x C ( T ) .
Observe that, since 0 < ε < 1 , thanks to (42) and (57),
ε x 3 u ε L 2 ( ( 0 , T ) × R ) 2 , β 2 x 2 u ε L 2 ( ( 0 , T ) × R ) 2 C ( T ) .
Therefore, by (82), (83) and (84),
x g u ε 3 + a x 2 u ε q v ε u ε ε x 3 u ε ε > 0 is   bounded   in H 1 ( ( 0 , T ) × R ) .
Moreover, by (43), we have that
b 2 P ε L 2 ( ( 0 , T ) × R ) 2 C ( T ) .
Equation (80) follows from (85) and (86).
Thanks to the Aubin–Lions Lemma, (75) and (76) hold.
Observe that, (77) follows from Lemma 3, while, by (43), we have (78). Consequently, ( u , v , P ) solves (1).
Observe again that, thanks to Lemmas 3, 7, 8, 9, (10) and the second equation of (16), we obtain (11).
Finally, we prove (12). Thanks to Lemmas 3 and 7, we have
u ε k u   in H 1 ( ( 0 , T ) × R ) .
Therefore, (12) follows from (19) and (87).  □
We are ready for the proof of Theorem 1.
Proof of Theorem 1.
Lemma 11 gives the existence of a solution of (1) satisfying (11) and (12). Let ( u 1 , P 1 ) and ( u 2 , P 2 ) be two solutions of (1) satisfying (11) and (12), namely
t u 1 + 3 g u 1 2 x u 1 a x 3 u 1 + q x ( u 1 v 1 ) = b P 1 , t > 0 , x R , x P 1 = u 1 , t > 0 , x R , α x 2 v 1 + β x v 1 + γ v 1 = κ u 1 2 , t > 0 , x R , P 1 ( t , ) = 0 , t > 0 , u 1 ( 0 , x ) = u 1 , 0 ( x ) , x R ,
t u 2 + 3 g u 2 2 x u 2 a x 3 u 2 + q x ( u 2 v 2 ) = b P 2 , t > 0 , x R , x P 2 = u 2 , t > 0 , x R , α x 2 v 2 + β x v 2 + γ v 2 = κ u 2 2 , t > 0 , x R , P 2 ( t , ) = 0 , t > 0 , u 2 ( 0 , x ) = u 2 , 0 ( x ) , x R .
Then, the triad ( ω , V , Ω ) defined by
ω ( t , x ) = u 1 ( t , x ) u 2 ( t , x ) , V ( t , x ) = v 1 ( t , x ) v 2 ( t , x ) , Ω ( t , x ) = x ω ( t , y ) d y = x u 1 ( t , y ) d y x u 2 ( t , y ) d y , Ω ( 0 , x ) = x ω ( 0 , y ) d y = x u 1 ( 0 , y ) d y x u 2 ( 0 , y ) d y .
is solution of the following Cauchy problem:
t ω + 3 g u 1 2 x u 1 u 2 2 x u 2 a x 3 ω + q x ( u 1 v 1 u 2 v 2 ) = b Ω , t > 0 , x R , x Ω = ω , t > 0 , x R , α x 2 V + β x V + γ V = κ ( u 1 2 u 2 2 ) , t > 0 , x R , Ω ( t , ) = 0 , t > 0 , ω ( 0 , x ) = u 1 , 0 u 2 , 0 ( x ) , x R .
Arguing as in ([15], Theorem 1.1), we have that
V ( t , · ) H 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 ,
V ( t , · ) L ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 ,
x V ( t , · ) L ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 .
Moreover, by (12) and (88),
Ω ( t , ) = R ω ( t , x ) d x = R u 1 ( t , y ) d y R u 2 ( t , x ) d x = 0 .
Observe that, by (88)
3 g u 1 2 x u 1 u 2 2 x u 2 = 3 g u 1 2 x u 1 u 2 2 x u 1 + u 2 2 x u 1 u 2 2 x u 2 = 3 g x u 1 u 1 2 u 2 2 + u 2 2 x ω = 3 g x u 1 u 1 + u 2 ω + u 2 2 x ω .
Moreover, arguing as in ([15], Theorem 1.1),
q x ( u 1 v 1 u 2 v 2 ) = q x ( u 1 v 1 u 2 v 1 + u 2 v 1 u 2 v 2 ) = q x ( v 1 ω ) + q x ( u 2 V ) .
Therefore, thanks to (94) and (95), the first equation of (89) is equivalent to the following one:
t ω = b Ω 3 g x u 1 u 1 + u 2 ω 3 g u 2 2 x ω + a x 3 ω q x ( v 1 ω ) q x ( u 2 V ) .
Multiplying (96) by 2 ω , an integration on R gives
d d t ω ( t , · ) L 2 ( R ) 2 = 2 b R Ω x ω d x 6 g R x u 1 u 1 + u 2 ω 2 d x 2 a R ω x 3 ω d x 6 g R u 2 2 ω x ω d x 2 q R x ( v 1 ω ) ω d x 2 q R x ( u 2 V ) ω d x .
Observe that, by (88) and (93),
2 b R Ω x ω d x = 2 b R Ω x Ω d x = b Ω 2 ( t , ) = 0 , 6 g R u 2 2 ω x ω d x = 6 g R u 2 x u 2 ω 2 d x , 2 a R ω x 3 ω d x = 2 a R x ω x 2 ω = 0 , 2 q R x ( v 1 ω ) ω d x = 2 q R v 1 ω x ω d x = q R x v 1 ω 2 d x , 2 q R x ( u 2 V ) ω d x = 2 q R x u 2 V ω d x 2 q R u 2 x V ω d x .
It follows from (97) and (98) that
d d t ω ( t , · ) L 2 ( R ) 2 = 6 g R x u 1 u 1 + u 2 ω 2 d x + 6 g R u 2 x u 2 ω 2 d x q R x v 1 ω 2 d x 2 q R x u 2 V ω d x 2 q R u 2 x V ω d x .
Since (11) holds, 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 v 1 L ( ( 0 , T ) × R ) , u 2 ( t , · ) L 2 ( R ) , x u 2 ( t , · ) L 2 ( R ) C ( T ) ,
for evert 0 t T . Consequently, by (91), (100) and the Hölder inequality,
| 6 g | R | x u 1 | | u 1 + u 2 | ω 2 d x | 6 g | x u 1 L ( ( 0 , T ) × R ) R | u 1 + u 2 | ω 2 d x C ( T ) u 1 L ( ( 0 , T ) × R ) + u 2 L ( ( 0 , T ) × R ) ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 , | 6 g | R | u 2 | | x u 2 | ω 2 d x | 6 g | u 1 L ( ( 0 , T ) × R ) R | x u 2 | ω 2 d x C ( T ) x u 2 L ( ( 0 , T ) × R ) ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 , | q | R | x v 1 | ω 2 d x | q | x v 1 L ( ( 0 , T ) × R ) ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 , | 2 q | R | x u 2 | | V | | ω | d x | 2 q | V ( t , · ) L ( R ) R | x u 2 | | ω | d x C ( T ) x u 2 ( t , · ) L 2 ( R ) ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 , | 2 q | R | u 2 | | x V | | ω | d x | 2 q | x V ( t , · ) L ( R ) R | u 2 | | | ω | d x C ( T ) u 2 ( t , · ) L 2 ( R ) ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 .
It follows from (99) that
d d t ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 .
The Gronwall Lemma and (89) give
ω ( t , · ) L 2 ( R ) 2 e C ( T ) t ω ( 0 , x ) L 2 ( R ) 2 .
Since (11) holds, by (88), arguing as in Lemma 5, Ω ( t , · ) is integrable at ± . Moreover, thanks to (93) and Lemma 5, we have that
R Ω ( t , x ) d x = 0 .
Consider the following function:
Ω 1 ( t , x ) = x Ω ( t , y ) d y ,
since, by the second equation of (89),
t Ω = d d t x ω ( t , y ) d y = x t ω ( t , y ) d y ,
integrating the first equation of (89) on ( , x ) , by (104) and (105), we have that
t Ω = b Ω 1 g u 1 3 u 2 3 + a x 2 ω q u 1 v 1 u 2 v 2 .
Observe that, by (88),
u 1 3 u 2 3 = u 1 2 + u 2 2 + u 1 u 2 ω , u 1 v 1 u 2 v 2 = v 1 ω + u 2 V .
Consequently, by (106),
t Ω = b Ω 1 g u 1 2 + u 2 2 + u 1 u 2 ω + a x 2 ω q v 1 ω q u 2 V .
It follows from (88), (93), (103) and (104) that
2 b R Ω 1 Ω d x = 2 b R Ω 1 x Ω 1 d x = b Ω 1 2 ( t , ) = b R Ω ( t , x ) d x 2 = 0 , 2 a R Ω x 2 ω d x = 2 a R x Ω x ω d x = 2 a R ω x ω d x = 0 .
Therefore, multiplying (107) by 2 Ω , an integration on R gives
d d t Ω ( t , · ) L 2 ( R ) 2 = 2 g R u 1 2 + u 2 2 + u 1 u 2 ω Ω d x 2 q R v 1 ω Ω d x 2 q R u 2 V Ω d x .
Due to (91), (100) and the Young inequality,
| 2 g | R u 1 2 + u 2 2 + u 1 u 2 | ω | | | Ω | d x g 2 R u 1 2 + u 2 2 + u 1 u 2 2 ω 2 d x + Ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 , | 2 q | R | v 1 ω | | Ω | d x q 2 R v 1 2 ω 2 d x + Ω ( t , · ) L 2 ( R ) 2 q 2 v 1 L ( ( 0 , T ) × R ) 2 ω ( t , · ) L 2 ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 , | 2 q | R | u 2 V | Ω d x q 2 R V 2 u 2 2 d x + Ω ( t , · ) L 2 ( R ) 2 q 2 V ( t , · ) L ( R ) 2 u 2 ( t , · ) L 2 ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 C ( T ) V ( t , · ) L ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + Ω ( t , · ) L 2 ( R ) 2 .
Therefore, by (108),
d d t Ω ( t , · ) L 2 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + 3 Ω ( t , · ) L 2 ( R ) 2 .
Adding (101) and (109), by (88) and the second equation of (89), we have that
d d t Ω ( t , · ) H 1 ( R ) 2 C ( T ) ω ( t , · ) L 2 ( R ) 2 + 3 Ω ( t , · ) L 2 ( R ) 2 C ( T ) Ω ( t , · ) H 1 ( R ) 2
and
Ω ( t , · ) H 1 ( R ) 2 e C ( T ) t Ω ( 0 , · ) H 1 ( R ) 2 .
Therefore, (13) follows (14), (88), (89), (90), (102) and (110).  □

4. Conclusions

In this paper we studied the Cauchy problem for the Spectrum Pulse equation. It is a third order nonlocal nonlinear evolutive equation related to the dynamics of the electrical field of linearly polarized continuum spectrum pulses in optical waveguides. Our existence analysis is based on on passing to the limit in a fourth order perturbation of the equation. If the initial datum belongs to H 2 ( R ) L 1 ( R ) and has zero mean we use the Aubin–Lions Lemma while if it belongs to H 3 ( R ) L 1 ( R ) and has zero mean we use the Sobolev Immersion Theorem. Finally, we directly prove a stability estimate that implies the uniqueness of the solution.

Author Contributions

Writing—original draft, G.M.C. and L.d.R.

Funding

This research received no external funding.

Acknowledgments

The authors are members 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.

Appendix A. u 0 H 3 ( R ) L 1 ( R )

In this appendix, we consider the Cauchy problem (1), where, on the initial datum, we assume
u 0 ( x ) H 3 ( R ) L 1 ( R ) , R u 0 ( x ) d x = 0 ,
while on the function P ( x ) , defined in (3), we assume (4). Moreover, we assume (5). The main result of this appendix is the following theorem.
Theorem A1.
Assume (3), (4), (5) and (A1). Fix T > 0 , there exists an unique solution ( u , v , P ) of (1) such that
u H 1 ( ( 0 , T ) × R ) L ( 0 , T ; H 3 ( R ) ) , v H 1 ( ( 0 , T ) × R ) L ( 0 , T ; H 5 ( R ) ) W 1 , ( ( 0 , T ) × R ) , t x 2 v L ( ( 0 , T ) × R ) L ( 0 , T ; L 2 ( R ) ) , P L ( 0 , T ; H 4 ( R ) ) .
Moreover, (12) and (13) hold.
To prove Theorem A1, we consider the approximation (16), where u ε , 0 is a C approximation of u 0 such that
u ε , 0 H 3 ( R ) u 0 H 3 ( R ) , R u ε , 0 d x = 0 , P ε , 0 L 2 ( R ) P 0 L 2 ( R ) , R P ε , 0 d x = 0 , ε x 4 u ε ( t , · ) L 2 ( R ) 2 C 0 ,
where C 0 is a positive constant, independent on ε .
Let us prove some a priori estimates on u ε , v ε and P ε .
Since H 2 ( R ) H 3 ( R ) , then Lemmas 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 are still valid.
We prove the following result.
Lemma A1.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that,
x 3 u ε L ( ( 0 , T ) × R ) C ( T ) .
In particular, we have that
x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C ( T ) t 0 t e C ( T ) s x 5 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Multiplying the first equation of (16) by 2 x 6 u ε , we have that
2 x 6 u ε t u ε = 2 b P ε x 6 u ε + 2 ε x 6 u ε x 4 u ε + 6 g u ε 2 x u ε x 6 u ε 2 a x 3 u ε x 6 u ε + 2 q u ε x v ε x 6 u ε + 2 q v ε x u ε x 6 u ε .
Observe that by (18) and the second equation of (16),
2 b R P ε x 6 u ε d x = 2 b R x P ε x 5 u ε d x = 2 b R u ε x 5 u ε d x = 2 b R x u ε x 4 u ε d x = 2 b R x 2 u ε x 3 u ε d x = 0 .
Moreover,
2 R x 6 u ε t u ε = d d t x 3 u ε ( t , · ) L 2 ( R ) 2 , 2 ε R x 6 u ε x 4 u ε d x = 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 , 2 a R x 3 u ε x 6 u ε d x = 2 a R x 4 u ε x 5 u ε d x = 0 .
It follows from (A7), (A8) and an integration of (A6) on R that
d d t x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 = 6 g R u ε 2 x u ε x 6 u ε d x + 2 q R u ε x v ε x 6 u ε d x + 2 q R v ε x u ε x 6 u ε d x .
Observe that
6 g R u ε 2 x u ε x 6 u ε d x = 12 g R u ε ( x u ε ) 2 x 5 u ε d x 6 g R u ε 2 x 2 u ε x 5 u ε d x = 12 g R ( x u ε ) 3 x 4 u ε d x + 36 g R u ε x u ε x 2 u ε x 4 u ε d x + 6 g R u ε 2 x 3 u ε x 4 u ε d x = 72 g R ( x u ε ) 2 x 2 u ε x 3 u ε d x 36 g R u ε ( x 2 u ε ) 2 x 3 u ε d x 42 g R u ε x u ε ( x 3 u ε ) 2 d x , 2 q R u ε x v ε x 6 u ε d x = 2 q R x u ε x v ε x 5 u ε d x 2 q R u ε x 2 v ε x 5 u ε d x = 2 q R x 2 u ε x v ε x 4 u ε d x + 4 q R x u ε x 2 v ε x 4 u ε d x + 2 q R u ε x 3 v ε x 4 u ε d x = 2 q R x v ε ( x 3 u ε ) 2 d x 6 q R x 2 u ε x 2 v ε x 3 u ε d x 6 q R x u ε x 3 v ε x 3 u ε d x 2 q R u ε x 4 v ε x 3 u ε d x = 2 q R x v ε ( x 3 u ε ) 2 d x + 3 q R x 3 v ε ( x 2 u ε ) 2 d x 6 q R x u ε x 3 v ε x 3 u ε d x 2 q R u ε x 4 v ε x 3 u ε d x , 2 q R v ε x u ε x 6 u ε d x = 2 q R x v ε x u ε x 5 u ε d x 2 q R v ε x 2 u ε x 5 u ε d x = 2 q R x 2 v ε x u ε x 4 u ε d x + 4 q R x v ε x 2 u ε x 4 u ε d x + 2 q R v ε x 3 u ε x 4 u ε d x = 2 q R x 3 v ε x u ε x 3 u ε d x 6 q R x 2 v ε x 2 u ε x 3 u ε d x 5 q R x v ε ( x 3 u ε ) 2 d x = 2 q R x 3 v ε x u ε x 3 u ε d x + 3 q R x 3 v ε ( x 2 u ε ) 2 d x 5 q R x v ε ( x 3 u ε ) 2 d x .
Consequently, by (A9),
d d t x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 = 72 g R ( x u ε ) 2 x 2 u ε x 3 u ε d x 36 g R u ε ( x 2 u ε ) 2 x 3 u ε d x 42 g R u ε x u ε ( x 3 u ε ) 2 d x 7 q R x v ε ( x 3 u ε ) 2 d x + 6 q R x 3 v ε ( x 2 u ε ) 2 d x 8 q R x 3 v ε x u ε x 3 u ε d x 2 q R u ε x 4 v ε x 3 u ε d x .
Due to (24), (41), (42), (56), (57), (68), (69) and the Young inequality,
| 72 g | R | ( x u ε ) 2 x 2 u ε | | x 3 u ε | d x 36 g 2 R ( x u ε ) 4 ( x 2 u ε ) 2 d x + 36 x 3 u ε ( t , · ) L 2 ( R ) 2 36 g 2 x u ε L ( ( 0 , T ) × R ) 4 x 2 u ε ( t , · ) L 2 ( R ) 2 + 36 x 3 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + 36 x 3 u ε ( t , · ) L 2 ( R ) 2 , | 36 g | R | u ε ( x 2 u ε ) 2 | | x 3 u ε | d x 18 g 2 R u ε 2 ( x 2 u ε ) 4 d x + 18 x 3 u ε ( t , · ) L 2 ( R ) 2 18 g 2 u ε L ( ( 0 , T ) × R ) 2 x 2 u ε L ( ( 0 , T ) × R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 18 x 3 u ε ( t , · ) L 2 ( R ) 2 C ( T ) x 2 u ε L ( ( 0 , T ) × R ) 2 + 18 x 3 u ε ( t , · ) L 2 ( R ) 2 , | 7 q | R | x v ε | ( x 3 u ε ) 2 d x | 7 q | x v ε ( t , · ) L ( R ) x 3 u ε ( t , · ) L 2 ( R ) 2 C 0 x 3 u ε ( t , · ) L 2 ( R ) 2 , | 6 q | R | x 3 v ε | ( x 2 u ε ) 2 d x | 6 q | x 3 v ε L ( ( 0 , T ) × R ) x 2 u ε ( t , · ) L 2 ( R ) 2 C ( T ) , | 8 q | R | x 3 v ε x u ε | | x 3 u ε | d x 4 q 2 R ( x 3 v ε ) 2 ( x u ε ) 2 d x + 4 x 3 u ε ( t , · ) L 2 ( R ) 2 4 q 2 x 3 v ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + 4 x 3 u ε ( t , · ) L 2 ( R ) 2 C ( T ) + 4 x 3 u ε ( t , · ) L 2 ( R ) 2 , | 2 q | R | u ε x 4 v ε | | x 3 u ε | d x q 2 R u ε 2 ( x 4 v ε ) 2 d x + x 3 u ε ( t , · ) L 2 ( R ) 2 , q 2 u ε L ( ( 0 , T ) × R ) 2 x 4 v ε ( t , · ) L 2 ( R ) 2 + x 3 u ε ( t , · ) L 2 ( R ) 2 .
It follows from (A10) that
d d t x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε x 5 u ε ( t , · ) L 2 ( R ) 2 C ( T ) x 3 u ε ( t , · ) L 2 ( R ) 2 + C ( T ) 1 + x 2 u ε L ( ( 0 , T ) × R ) 2 .
The Gronwall Lemma and (A3) give
x 3 u ε ( t , · ) L 2 ( R ) 2 + 2 ε e C ( T ) t 0 t e C ( T ) s x 5 u ε ( s , · ) L 2 ( R ) 2 d s C 0 e C ( T ) t + C ( T ) 1 + x 2 u ε L ( ( 0 , T ) × R ) 2 e C ( T ) t 0 t e C ( T ) s d s C ( T ) 1 + x 2 u ε L ( ( 0 , T ) × R ) 2 .
We prove (A4). Thanks to (57), (A11) and the Hölder inequality,
( x 2 u ε ( t , x ) ) 2 = 2 x x 2 u ε x 3 u ε d x 2 R | x 2 u ε | | x 3 u ε | d x x 2 u ε ( t , · ) L 2 ( R ) x 3 u ε ( t , · ) L 2 ( R ) 2 C ( T ) 1 + x 2 u ε L ( ( 0 , T ) × R ) 2 .
Hence,
x 2 u ε L ( ( 0 , T ) × R ) 4 C ( T ) x 2 u ε L ( ( 0 , T ) × R ) 2 C ( T ) 0 ,
which gives (A4).
Finally, (A5) follows from (A4) and (A11).  □
Lemma A2.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
x 5 v ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
for every 0 t T . In particular, we have that
x 4 v ε L ( ( 0 , T ) × R ) C ( T ) .
Proof. 
Let 0 t T . Differentiating (70) with respect to x, we have
α x 5 v ε = 6 κ x u ε x 2 u ε + 2 κ u ε x 3 u ε β x 4 v ε γ x 3 v ε .
Since x 3 u ε ( t , ± ) = 0 , by (55), (71) and (72), we have that
x 5 v ε ( t , ± ) = 0 .
Observe that
2 β α R x 4 v ε x 5 v ε d x = 0 , 2 α γ R x 3 v ε x 5 v ε d x = 2 α γ x 4 v ε ( t , · ) L 2 ( R ) 2 .
Consequently, multiplying (A14) by 2 α x 5 v ε , an integration on R gives
2 α 2 x 5 v ε ( t , · ) L 2 ( R ) 2 = 12 α κ R x u ε x 2 u ε x 5 v ε d x + 4 α κ R u ε x 3 u ε x 5 v ε d x + 2 α γ x 4 v ε ( t , · ) L 2 ( R ) 2 .
Due to (41), (56), (57), (A4) and the Young inequality,
12 | α κ | R | x u ε x 2 u ε | | x 5 v ε | d x = R | 12 κ x u ε x 2 u ε | | α x 5 v ε | d x 72 κ 2 R ( x u ε ) 2 ( x 2 u ε ) 2 d x + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 72 κ 2 x u ε L ( ( 0 , T ) × R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 C ( T ) + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 , | 4 α κ | R | u ε x 3 u ε | | x 5 v ε | d x = R | 4 κ u ε x 3 u ε | | | α x 5 v ε | d x 8 κ 2 R u ε 2 ( x 3 u ε ) 2 d x + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 8 κ 2 u ε L ( ( 0 , T ) × R ) 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 C ( T ) + α 2 2 x 5 v ε ( t , · ) L 2 ( R ) 2 .
It follows from (68) and (A16) that
α 2 x 5 v ε ( t , · ) L 2 ( R ) 2 C ( T ) + | 2 α γ | x 4 v ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
which gives (A12).
Finally, we prove (A13). Thanks to (68), (A12) and the Hölder inequality,
( x 4 v ε ( t , x ) ) 2 = 2 x x 4 v ε x 5 v ε d x 2 R | x 4 v ε | | x 5 v ε | d x x 4 v ε ( t , · ) L 2 ( R ) x 5 v ε ( t , · ) L 2 ( R ) C ( T ) .
Hence,
x 4 v ε L ( ( 0 , T ) × R ) 2 C ( T ) ,
which gives (A13).  □
Lemma A3.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that,
ε x 4 u ε ( t , · ) L 2 ( R ) 2 + ε 2 0 t x 6 u ε ( s , · ) L 2 ( R ) 2 d s C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Multiplying the first equation of (16) by 2 ε x 8 u ε , we have
2 ε x 8 u ε t u ε = 2 b ε P ε x 8 u ε 2 ε 2 x 4 u ε x 8 u ε 6 g ε u ε 2 x u ε x 8 u ε + 2 a ε x 3 u ε x 8 u ε 2 q ε u ε x v ε x 8 u ε 2 q ε v ε x u ε x 8 u ε .
Observe that by the second equation of (16) and (18),
2 b ε R P ε x 8 u ε d x = 2 b ε R x P ε x 7 u ε d x = 2 b ε R u ε x 7 u ε d x = 2 b ε R x u ε x 6 u ε d x = 2 b ε R x 2 u ε x 5 u ε d x = 2 b ε R x 3 u ε x 4 u ε d x = 0 .
Moreover,
2 ε R x 8 u ε t u ε d x = ε d d t x 4 u ε ( t , · ) L 2 ( R ) 2 , 2 ε 2 R x 4 u ε x 8 u ε d x = 2 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , 2 a ε R x 3 u ε x 8 u ε d x = 0 .
Therefore, (A19), (A20) and an integration of (A18) on R give
ε d d t x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 = 6 g ε R u ε 2 x u ε x 8 u ε d x 2 q ε R u ε x v ε x 8 u ε d x 2 q ε R v ε x u ε x 8 u ε d x .
Observe that
6 g ε R u ε 2 x u ε x 8 u ε d x = 12 g ε u ε ( x u ε ) 2 x 7 u ε d x + 6 g ε R u ε 2 x 2 u ε x 7 u ε d x = 12 g ε R ( x u ε ) 3 x 6 u ε d x 36 g ε R u ε x u ε x 2 u ε x 6 u ε d x 6 g ε R u ε 2 x 3 u ε x 6 u ε d x , 2 q ε R u ε x v ε x 8 u ε d x = 2 q ε R x u ε x v ε x 7 u ε d x + 2 q ε R u ε x 2 v ε x 7 u ε d x = 2 q ε R x 2 u ε x v ε x 6 u ε d x 4 q ε R x u ε x 2 v ε x 6 u ε d x 2 q ε R u ε x 3 v ε x 6 u ε d x , 2 q ε R v ε x u ε x 8 u ε d x = 2 q ε R x v ε x u ε x 6 u ε d x + 2 q ε R v ε x 2 u ε x 7 u ε d x = 2 q ε R x 2 v ε x u ε x 6 u ε d x 4 q ε R x v ε x 2 u ε x 6 u ε d x 2 q ε R v ε x 3 u ε x 6 u ε d x .
Consequently, by (A21),
ε d d t x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 = 12 g ε R ( x u ε ) 3 x 6 u ε d x 36 g ε R u ε x u ε x 2 u ε x 6 u ε d x 6 g ε R u ε 2 x 3 u ε x 6 u ε d x 6 q ε R x v ε x 2 u ε x 6 u ε d x 6 q ε R x 2 v ε x u ε x 6 u ε d x 2 q ε R u ε x 3 v ε x 6 u ε d x 2 q ε R v ε x 3 u ε x 6 u ε d x .
Due to (24), (41), (42), (43), (56), (57), (A5) and the Young inequality,
12 | g ε | R | x u ε | 3 | x 6 u ε | d x = 12 R g ( x u ε ) 3 D 1 ε D 1 x 6 u ε d x 6 g 2 D 1 R ( x u ε ) 6 d x + 6 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 6 g 2 D 1 x u ε L ( ( 0 , T ) × R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 6 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + 6 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , | 36 g ε | R | u ε x u ε x 2 u ε | | x 6 u ε | d x = 36 R g u ε x u ε x 2 u ε D 1 D 1 ε x 6 u ε d x 18 g 2 D 1 R u ε 2 ( x u ε ) 2 ( x 2 u ε ) 2 d x + 18 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 18 g 2 D 1 u ε L ( ( 0 , T ) × R ) 2 x u ε L ( ( 0 , T ) × R ) x 2 u ε ( t , · ) L 2 ( R ) 2 + 18 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + 18 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , | 6 g ε | R | u ε 2 x 3 u ε | | x 6 u ε | d x = 6 R g u ε 2 x 3 u ε D 1 D 1 ε x 6 u ε d x 3 g 2 D 1 R u ε 4 ( x 3 u ε ) 2 d x + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 3 g 2 D 1 u ε L ( ( 0 , T ) × R ) 4 x 3 u ε ( t , · ) L 2 ( R ) 2 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , | 6 q ε | R | x v ε x 2 u ε | | x 6 u ε | d x = 6 R q x v ε x 2 u ε D 1 D 1 ε x 6 u ε d x 3 q 2 D 1 R ( x v ε ) 2 ( x 2 u ε ) 2 d x + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 3 q 2 D 1 x v ε ( t , · ) L ( R ) 2 x 2 u ε ( t , · ) L 2 ( R ) 2 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 ,
| 6 q ε | R | x 2 v ε x u ε | | x 6 u ε | d x = 6 R q x 2 v ε x u ε D 1 D 1 ε x 6 u ε d x 3 q 2 D 1 R ( x 2 v ε ) 2 ( x u ε ) 2 d x + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 3 q 2 D 1 x 2 v ε L ( ( 0 , T ) × R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + 3 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , | 2 q ε | R | u ε x 3 v ε | | x 6 u ε | d x = 2 R q u ε x 3 v ε D 1 D 1 ε x 6 u ε d x q 2 D 1 R u ε 2 ( x 3 v ε ) 2 d x + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 q 2 D 1 u ε L ( ( 0 , T ) × R ) 2 x 3 v ε ( t , · ) L 2 ( R ) 2 + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 , | 2 q ε | R | v ε x 3 u ε | | x 6 u ε | d x = R q v ε x 3 u ε D 1 D 1 ε x 6 u ε d x q 2 D 1 R v ε 2 ( x 3 u ε ) 2 d x + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 q 2 D 1 v ε ( t , · ) L ( R ) 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 + D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 ,
where D 1 is a positive constant, which will be specified later. Consequently, by (A23),
ε d d t x 4 u ε ( t , · ) L 2 ( R ) 2 + 2 35 D 1 ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 1 .
Taking D 1 = 1 35 , we have that
ε d d t x 4 u ε ( t , · ) L 2 ( R ) 2 + ε 2 x 6 u ε ( t , · ) L 2 ( R ) 2 C ( T ) .
Equation (A3) and an integration on ( 0 , t ) give
ε x 4 u ε ( t , · ) L 2 ( R ) 2 + ε 2 0 t x 6 u ε ( s , · ) L 2 ( R ) 2 d s C 0 + C ( T ) t C ( T ) ,
that is (A17).  □
Lemma A4.
Assume (5). Fix T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that,
t u ε ( t , · ) L 2 ( R ) C ( T ) ,
for every 0 t T .
Proof. 
Let 0 t T . Multiplying the first equation of (16) by 2 t u ε , an integration on R gives
2 t u ε ( t , · ) L 2 ( R ) 2 = 2 b R P ε t u ε d x 2 ε R x 4 u ε t u ε d x 6 g R u ε 2 x u ε t u ε d x + 2 a R x 3 u ε t u ε d x 2 q R u ε x v ε t u ε d x 2 q R v ε x u ε t u ε d x .
Since 0 < ε < 1 , thanks to (24), (41), (42), (43), (A5), (A17) and the Young inequality,
| 2 b | R | P ε | | t u ε | d x = 2 R b P ε D 2 D 2 t u ε d x b 2 D 2 P ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 , 2 ε R | x 4 u ε | | t u ε | d x = 2 R ε x 4 u ε D 2 D 2 t u ε d x ε 2 D 2 x 4 u ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 ε D 2 x 4 u ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 , | 6 g | R | u ε 2 x u ε | | t u ε | d x = 6 R g u ε 2 x u ε D 2 D 2 t u ε d x 3 g 2 D 2 R u ε 4 ( x u ε ) 2 d x + 3 D 2 t u ε ( t , · ) L 2 ( R ) 2 3 g 2 D 2 u ε L ( ( 0 , T ) × R ) 4 x u ε ( t , · ) L 2 ( R ) 2 + 3 D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + 3 D 2 t u ε ( t , · ) L 2 ( R ) 2 , | 2 a | R | x 3 u ε | | t u ε | d x = 2 R a x 3 u ε D 2 D 2 t u ε d x a 2 D 2 x 3 u ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 , | 2 q | R | u ε x v ε | | t u ε | d x = 2 R q u ε x v ε D 2 D 2 t u ε d x q 2 D 2 R u ε 2 ( x v ε ) 2 d x + D 2 t u ε ( t , · ) L 2 ( R ) 2 q 2 D 2 u ε L ( ( 0 , T ) × R ) 2 x v ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 ,
| 2 q | R | v ε x u ε | | t u ε | d x = 2 R q v ε x u ε D 2 D 2 t u ε d x q 2 D 2 R v ε 2 ( x u ε ) 2 d x + D 2 t u ε ( t , · ) L 2 ( R ) 2 q 2 D 2 v ε ( t , · ) L ( R ) 2 x u ε ( t , · ) L 2 ( R ) 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 + D 2 t u ε ( t , · ) L 2 ( R ) 2 ,
where D 2 is a positive constant, which will be specified later. Therefore, by (A25),
2 1 4 D 2 t u ε ( t , · ) L 2 ( R ) 2 C ( T ) D 2 .
Choosing D 2 = 1 8 , we have that
t u ε ( t , · ) L 2 ( R ) 2 C ( T ) ,
which gives (A24).  □
Arguing as in ([15], Lemma 2.12), we have the following result.
Lemma A5.
Assume (5). Let T > 0 . There exists a constant C ( T ) > 0 , independent on ε, such that
t x 2 v ε ( t , · ) L ( R ) , t x 2 v ε ( t , · ) L 2 ( R ) C ( T ) , t v ε ( t , · ) L ( R ) , t v ε ( t , · ) L 2 ( R ) C ( T ) ,
for every 0 t T .
Using the Sobolev Immersion Theorem, we begin by proving the following result.
Lemma A6.
Fix T > 0 . There exist a subsequence { ( u ε k , v ε k , P ε k ) } k N of { ( u ε , v ε , P ε ) } ε > 0 and an a limit triplet ( u , v , P ) which satisfies (11) such that
u ε k u a . e . a n d i n L l o c p ( ( 0 , T ) × R ) , 1 p < , u ε k u i n H 1 ( ( 0 , T ) × R ) , v ε k v a . e . a n d i n L l o c p ( ( 0 , T ) × R ) , 1 p < , v ε k v i n H 1 ( ( 0 , T ) × R ) , P ε k P i n L 2 ( ( 0 , T ) × R ) .
Moreover, ( u , v , P ) is solution of (1) satisfying (12).
Proof. 
Let 0 t T . We begin by observing that, thanks to Lemmas 3, 7, 9, A1 and A4,
{ u ε } ε > 0 is   uniformly   bounded   in   H 1 ( ( 0 , T ) × R ) .
Lemmas 3 and A5 say that
{ v ε } ε > 0 is   uniformly   bounded   in   H 1 ( ( 0 , T ) × R ) .
Instead, by Lemma 7, we have that
{ P ε } ε > 0 is   uniformly   bounded   in   L 2 ( ( 0 , T ) × R ) .
Equation (A28), (A29) and (A30) give (A27).
Observe that, thanks to Lemmas 3, 7, 9, A1 and the second equation of (16), we have that
P L ( 0 , T ; H 4 ( R ) ) .
Lemmas 3, 7, 9, A1 say that
u L ( 0 , T ; H 3 ( R ) ) .
Instead, thanks to Lemmas 3, 8, 10, A2 and (A26), we get
v L ( 0 , T ; H 5 ( R ) ) W 1 , ( ( 0 , T ) × R ) .
Moreover, Lemmas A5 says also that
t x 2 v L 2 ( R ) L ( R ) ,
for every 0 t T . Therefore, (11) holds and ( u , v , P ) is solution of (1).
Finally, (12) follows from (19) and (A27).  □
Now, we prove Theorem A1.
Proof of Theorem A1.
Lemma A6 gives the existence of a solution of (1) such that (12) and (A27) hold. Arguing as in Theorem 1, we have (13).  □

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Coclite, G.M.; Ruvo, L.d. Well-Posedness Results for the Continuum Spectrum Pulse Equation. Mathematics 2019, 7, 1006. https://doi.org/10.3390/math7111006

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Coclite GM, Ruvo Ld. Well-Posedness Results for the Continuum Spectrum Pulse Equation. Mathematics. 2019; 7(11):1006. https://doi.org/10.3390/math7111006

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Coclite, Giuseppe Maria, and Lorenzo di Ruvo. 2019. "Well-Posedness Results for the Continuum Spectrum Pulse Equation" Mathematics 7, no. 11: 1006. https://doi.org/10.3390/math7111006

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Coclite, G. M., & Ruvo, L. d. (2019). Well-Posedness Results for the Continuum Spectrum Pulse Equation. Mathematics, 7(11), 1006. https://doi.org/10.3390/math7111006

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