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On H 2-solutions for a Camassa-Holm type equation

  • Giuseppe Maria Coclite EMAIL logo and Lorenzo di Ruvo
Published/Copyright: May 5, 2023
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Open Mathematics
From the journal Open Mathematics Volume 21 Issue 1

Abstract

Camassa-Holm type equations arise as models for the unidirectional propagation of shallow water waves over a flat bottom. They also describe finite length, small amplitude radial deformation waves in cylindrical compressible hyperelastic rods. Under appropriate assumption on the initial data, on the time T , and on the coefficients of such equation, we prove the well-posedness of the classical solutions for the Cauchy problem.

Keywords: existence; uniqueness; stability; Camassa-Holm type equation; Cauchy problem
MSC 2010: 35G25; 35K55

1 Introduction

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

(1.1) ∂ t u + δ ∂ x u + 2 κ ∂ x u 2 + a ∂ x 3 u − β 2 ∂ t ∂ x 2 u + α ∂ x u ∂ x 2 u + γ u ∂ x 3 u = 0 , 0 < t < T , x ∈ R , u ( 0 , x ) = u 0 ( x ) , x ∈ R ,

with δ , κ , a , β , α , γ ∈ R are constant, such that

(1.2) β ≠ 0 .

On the initial datum, we assume

(1.3) u 0 ∈ H 2 ( R ) , u 0 ≢ 0 ,

and one of the following:

(1.4) β 6 − 8 ( ∣ κ + α + 2 γ ∣ + ∣ 2 α − γ ∣ ) 2 [ A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 ] T 2 > 0 ,

(1.5) ∣ κ + α + 2 γ ∣ + ∣ 2 α − γ ∣ < ∣ β ∣ 3 2 2 T A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 ,

(1.6) ∣ β ∣ 3 log ( A 2 ( β ) ) > ( ∣ κ + α + 2 γ ∣ + ∣ 2 α − γ ∣ ) T 2 ,

(1.7) κ = 0 , γ = 2 α ,

(1.8) α = 2 γ , γ ≠ 0 , κ ≠ 0 , T < ( π − 2 arctan ( B 0 + β 2 E 0 ) ) ∣ β ∣ 5 2 τ 3 2 ( β 2 ) ,

where

(1.9) A 0 = ∥ u 0 ∥ L 2 ( R ) 2 , B 0 = ∥ ∂ x u 0 ∥ L 2 ( R ) 2 , E 0 = ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 , A 1 ≔ 8 ( ∣ κ + α + 2 γ ∣ + ∣ 2 α − γ ∣ ) 2 , τ 3 2 ( β 2 ) ≔ max { ∣ κ ∣ ( A 0 + β 2 B 0 ) , 3 ∣ γ ∣ } , A 2 ( β ) ≔ A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 + 1 A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

Observe that (1.4) is satisfied if one of the following holds

(1.10) ∣ β ∣ = T , T > A 1 ( B 0 + E 0 ) + A 1 2 ( B 0 + E 0 ) 2 + 4 A 1 ( A 0 + B 0 ) ,

(1.11) T = ∣ β ∣ 3 , A 1 < 1 A 0 + ( T 2 3 + 1 ) B 0 + 2 3 T E 0 .

Equation (1.1) arises as a model for the unidirectional propagation of shallow water waves over a flat bottom, where u ( t , x ) represents the water-free surface in nondimensional variables. It was first obtained in [1] as an abstract bi-Hamiltonian equation with infinitely many conservation laws [2–4], and subsequently from physical principles [5–8].

Equation (1.1) is also deduced in [9–11] as an equation describing finite length, small amplitude radial deformation waves in cylindrical compressible hyperelastic rods. Moreover, (1.1) is a re-expression of the geodesic flow in the group of compressible diffeomorphisms of the circle [12], just like the Euler equation that is an expression of the geodesic flow in the group of incompressible diffeomorphisms of the torus [13]. This geometric interpretation leads to a proof that equation (1.1) satisfies the least action principle [14]: a state of the system is transformed to another nearby state through a uniquely determined flow that minimizes the energy (see also [15]).

If (1.8) holds, (1.1) is known as the Camassa-Holm equation. Local well-posedness results are proved in [16–19]. It is also known that there exist global solutions for a certain class of initial data and solutions that blow up in finite time for a large class of initial data [16,20,21]. Existence and uniqueness results for global weak solutions are proven in [21–33]. The convergence of finite difference schemes is proved in [34,35], existence of the traveling wave solutions in [36], and the well-posedness of periodic solutions for the Cauchy problem in [37,38]. In [39–42], using a compensated compactness argument in the L p setting [43–46], the convergence of the solution of (1.1) to the unique entropy one of the Burgers equation is proven, and in [47], it was proved using a kinetic approach [48].

Equation (1.1) is also studied in [49,50]. By using the method of asymptotic integrability, the authors found that only three equations from this family were asymptotically integrable up to third order: the Camassa-Holm equation, the Korteweg-deVries equation, and one with

(1.12) κ = − 2 γ β 2 , α = γ .

The Korteweg-deVries equation ( β = α = γ = 0 ) models weakly nonlinear unidirectional long waves and arises in various physical contexts. For example, it models surface waves of small amplitude and long wavelength in shallow water. In this context, u ( t , x ) represents the wave height above a flat bottom, with x being proportional to distance in the propagation direction and t being proportional to the elapsed time. The Korteweg-deVries equation is completely integrable and possesses solitary wave solutions that are solitons. The Cauchy problem for the Korteweg-deVries equation is studied in [51,52], in [53], and the references cited therein. In particular, in [51,52], the authors prove that the well-posedness of the Korteweg-deVries equation, under Assumption (1.3) for any T .

Observe that, by using (1.12) in (1.1), and properly scaling, we gain

(1.13) ∂ t u + ∂ x u + 3 ∂ x u 2 + ∂ x 3 u − β 2 ∂ t ∂ x 2 u − 9 β 2 2 ∂ x u ∂ x 2 u − 3 β 2 2 u ∂ x 3 u = 0 .

By rescaling, shifting the dependent variable, and finally applying a Galilean boost, Equation (1.13) can be transformed into the following form [49,54]:

(1.14) ∂ t u − ∂ t ∂ x 2 u + 4 u ∂ x u = 3 ∂ x u ∂ x 2 u + u ∂ x 3 u .

From a mathematical point of view, the local and global well-posedness of (1.14) in energy spaces is proven in [55–58] and the references cited therein. In [59–61], the well-posedness of the entropy solution is proven, while, in [62], the well-posedness of the homogeneous initial boundary value problem is studied. In [63], the well-posedness of the periodic solution of (1.14) is analyzed, while, in [64], the convergence of some numerical schemes is proven. Possible estimates on the blow-up time T for (1.14) are given in [18,65,66] and the references cited therein.

In this article, we obtained the following estimates on the blow-up time T of the H 2 norm for (1.1). We can obtain some additional ones.

  1. Assuming (1.5) and choosing κ = 1 , β = 1 , α = 3 , and γ = 1 in (1.1), we obtain (1.14) and we can estimate T as follows:

    (1.15) T ≥ 1 18 2 ( A 0 + 2 B 0 + E 0 ) .

  2. By assuming (1.12) in (1.1), we have the following equation:

    (1.16) ∂ t u + δ ∂ x u − 2 γ β 2 ∂ x u 2 + a ∂ x 3 u − β 2 ∂ t ∂ x 2 u + γ ∂ x u ∂ x 2 u + γ u ∂ x 3 u = 0 ,

    and by assuming (1.5), we obtain

    (1.17) ∣ γ ∣ ( ∣ 3 β 2 − 2 ∣ + β 2 ) < ∣ β ∣ 5 2 2 T A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

    Therefore, fixed T and u 0 , we have that (1.16) admits an unique classical solution, when

    (1.18) ∣ γ ∣ < ∣ β ∣ 5 2 2 T ( ∣ 3 β 2 − 2 ∣ + β 2 ) A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

  3. By choosing β = 1 and γ ≠ 0 and assuming (1.17), we can estimate the blow-up time T for (1.16) as follows:

    (1.19) T ≥ 1 4 2 ∣ γ ∣ ( A 0 + 2 B 0 + E 0 ) .

One more interesting equation belonging to equation (1.1) is as follows:

(1.20) ∂ t u + δ ∂ x u + 3 u ∂ x u + a ∂ x 3 u − β 2 ∂ t ∂ x 2 u − 2 β 2 ∂ x u ∂ x 2 u − β 2 u ∂ x 3 u = 0 ,

which was deduced in [67], in the context of the integrable shallow water wave equation with linear and nonlinear dispersions. Here, we obtain the following on the blow-up time T for (1.20):

(1.21) T ≥ ∣ β ∣ 3 2 ( ∣ 3 − 8 β 2 ∣ + 6 β 2 ) A 0 + ( β 2 + 1 ) B 0 + β 2 E .

Taking κ = b + 1 2 , α = b β 2 , and γ = β 2 , (1.1) reads:

(1.22) ∂ t u + δ ∂ x u + ( b + 1 ) u ∂ x u + a ∂ x 3 u − β 2 ∂ t ∂ x 2 u + b β 2 ∂ x u ∂ x 2 u + β 2 u ∂ x 3 u ,

which is known as the b -equation and was deduced in [68–70] as the family of asymptotically equivalent shallow water wave equations that emerges at quadratic order accuracy, for each b ≠ − 1 , by an appropriate Kodama transformation. If b = − 1 , [68,69] show that the corresponding Kodama transformation is singular and the asymptotic ordering is violated.

The solution of (1.22) is studied numerically for various values of b in [70,71], under assumption δ = a = 0 . Using the energy space technique, the local and global well-posedness of the Cauchy problem for (1.22) is proven in [72,73]. Moreover, [72] gives an estimate on the blow-up time T . Here, by assuming (1.5), (1.2), and (1.3), we prove that (1.22) admits an unique classical solution, if the following inequality is verified:

(1.23) ( β 2 + 1 ) ∣ b + 1 ∣ + β 2 ∣ b ∣ < ∣ β ∣ 3 2 2 T A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 ,

which gives an estimate on the blow-up time T

(1.24) T ≥ ∣ β ∣ 3 2 2 ( ( β 2 + 1 ) ∣ b + 1 ∣ + β 2 ∣ b ∣ ) A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

Choosing

(1.25) δ = 0 , κ = 1 2 , a = 0 , α = θ − 1 , γ = − θ , 0 < θ < 1 ,

(1.1) becomes

(1.26) ∂ t u + u ∂ x u − β 2 ∂ t ∂ x 2 u + ( θ − 1 ) ∂ x u ∂ x 2 u − θ u ∂ x 3 u = 0 .

It models equations of some dispersive schemes [74]. In [75], by using the energy spaces technique, the local and global well-posedness of the Cauchy problem for (1.26) is proven. They provide an estimate of the blow-up time T . Here, by using (1.25), we show that (1.26) admits an unique classical solution, if

(1.27) ∣ 1 − 2 θ ∣ + 2 ∣ 3 θ − 2 ∣ < ∣ β ∣ 3 2 T A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 ,

that gives the following estimate on the blow-up time T

(1.28) T ≥ ∣ β ∣ 3 2 ( ∣ 1 − 2 θ ∣ + 2 ∣ 3 θ − 2 ∣ ) A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

Finally, observe that if δ = a = α = γ = 0 , (1.1) gives the following equation:

(1.29) ∂ t u + κ ∂ x u 2 − β 2 ∂ t ∂ x 2 u = 0 ,

which is known as the Benjamin-Bona-Mahony equation. In [76], the well-posedness of the classical solution of (1.29) is proven, for each choose of β and T , and under assumption,

(1.30) u 0 ∈ H 1 ( R ) .

Here, by assuming (1.5), we prove that (1.29) admits an H 2 solution, if

(1.31) ∣ κ ∣ < ∣ β ∣ 3 2 2 T A 0 + ( β 2 + 1 ) B 0 + β 2 E 0

holds.

The main result of this article is the following theorem.

Theorem 1.1

Assuming (1.2), (1.3), and one within (1.4), (1.10), (1.11), (1.5), (1.6), (1.7), and (1.8) hold, there exists a unique solution u of (1.1), such that

(1.32) u ∈ H 1 ( ( 0 , T ) × R ) ∩ L ∞ ( 0 , T ; H 2 ( R ) ) ∩ W 1 , ∞ ( ( 0 , T ) × R ) , ∂ t ∂ x u ∈ L ∞ ( 0 , T ; L 2 ( R ) ) .

Moreover, if u 1 and u 2 are two solutions of (1.1) in correspondence of the initial data u 1 , 0 and u 2 , 0 , we have that

(1.33) ∥ u 1 ( t , ⋅ ) − u 2 ( t , ⋅ ) ∥ H 1 ( R ) 2 ≤ τ 2 2 e C t τ 3 2 ∥ u 1 , 0 − u 2 , 0 ∥ H 1 ( R ) 2 ,

for some suitable C > 0 , and every 0 ≤ t ≤ T , where

(1.34) τ 1 2 = min { 1 , β 2 } , τ 2 2 = max { 1 , β 2 } .

Theorem 1.1 gives some conditions on κ , β , α , γ , T , and u 0 , to have classical solutions for (1.1). Moreover, it says that the solutions of (1.1) are classical, for each choice of κ , β , α , γ , T , and u 0 , if (1.7) holds (see Lemma 2.3), or, under the condition (Corollary 2.2):

(1.35) κ + α + 2 γ = 0 , 2 α − γ = 0 .

Observe that, if κ = α 2 , with α ≠ 0 , then (1.35) is equivalent

α 2 + 5 α = 0 , γ = 2 α .

Therefore, Theorem 1.1 holds also in the case

(1.36) ( κ , α , γ ) = ( 25 , − 5 , − 10 ) .

In general, thanks to [77, Lemma 2.2] [78], Theorem 1.1 holds also in the following cases:

(1.37) κ = α 2 n , α = − 5 1 2 n − 1 , n ≠ 1 2 , κ = ( α + q ) 2 n , q > 1 5 − 5 2 n 2 n 2 n − 1 + 5 2 n 2 n − 1 , n ≠ 0 .

Note that (1.36) and (1.37) do not imply (1.36).

This article is organized as follows. In Section 2, we prove several a priori estimates on a vanishing viscosity approximation of (1.1). 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 (1.1).

Fix a small number 0 < ε < 1 and let u ε = u ε ( t , x ) be the unique classical solution of the following problem [79–82]:

(2.1) ∂ t u ε + δ ∂ x u ε + 2 κ u ε ∂ x u ε + a ∂ x 3 u ε − β 2 ∂ t ∂ x 2 u ε + α ∂ x u ε ∂ x 2 u ε + γ u ε ∂ x 3 u ε = − ε ∂ x 4 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

(2.2) ∥ u ε , 0 ∥ H 2 ( R ) ≤ ∥ u 0 ∥ H 2 ( R ) , ε ∥ ∂ x 3 u ε , 0 ∥ L 2 ( R ) ≤ C 0 ,

where C 0 is a positive constant independent of ε .

Let us prove some a priori estimates on u ε .

Following [81, Lemma 1], we prove the following result.

Lemma 2.1

( H 2 estimate) We have that

(2.3) 1 A ε ( t ) + ( β 2 + 1 ) B ε ( t ) + β 2 E ε ( t ) + 2 2 ( ∣ κ + α + 2 γ ∣ + ∣ 2 α − γ ∣ ) t ∣ β ∣ 3 ≥ 1 A ε ( 0 ) + ( β 2 + 1 ) B ε ( 0 ) + β 2 E ε ( 0 )

for every 0 ≤ t ≤ T , where

(2.4) A ε ( t ) ≔ ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , B ε ( t ) ≔ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , E ε ( t ) ≔ ∥ ∂ x 2 u ( t , ⋅ ) ∥ L 2 ( R ) 2 .

In particular, under Assumption (1.3), there exists a constant C > 0 , dependent on β , u 0 , and T , but not ε , such that

(2.5) ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ,

(2.6) ∥ ∂ x u ε ∥ L ∞ ( ( 0 , T ) × R ) ≤ C ,

(2.7) ε ∫ 0 t ∥ ∂ x 2 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ C ,

(2.8) ε ∫ 0 t ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ C ,

(2.9) ∥ u ε ∥ L ∞ ( ( 0 , T ) × R ) ≤ C ,

for every 0 ≤ t ≤ T .

Proof

By multiplying (2.1) by 2 u ε − 2 ∂ x 2 u ε , an integration on R gives

d d t ( ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) = 2 ∫ R u ∂ t u d x − 2 β 2 ∫ R u ε ∂ t ∂ x 2 u ε d x − 2 ∫ R ∂ x 2 u ε ∂ t u ε d x + 2 β 2 ∫ R ∂ x 2 u ε ∂ t ∂ x 2 u ε d x = − 2 κ ∫ R u ε 2 ∂ x u ε d x + 2 κ ∫ R u ε ∂ x u ε ∂ x 2 u ε d x − 2 α ∫ R u ε ∂ x u ε ∂ x 2 u ε d x + 2 α ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x − 2 γ ∫ R u ε 2 ∂ x 3 u ε d x + 2 γ ∫ R u ε ∂ x 2 u ε ∂ x 3 u ε d x − 2 ε ∫ R u ε ∂ x 4 u ε d x + 2 ε ∫ R ∂ x 2 u ε ∂ x 4 u ε d x − 2 δ ∫ R u ε ∂ x u ε d x − 2 a ∫ R u ε ∂ x 3 u ε d x + 2 δ ∫ R ∂ x u ε ∂ x 2 u ε d x + 2 a ∫ R ∂ x 2 u ε ∂ x 3 u ε d x = − ( κ + α ) ∫ R ( ∂ x u ε ) 3 d x + ( 2 α − γ ) ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x + 4 γ ∫ R u ε ∂ x u ε ∂ x 2 u ε d x + 2 ε ∫ R ∂ x u ε ∂ x 3 u ε d x − 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 a ∫ R ∂ x u ε ∂ x 2 u ε d x = − ( κ + α + 2 γ ) ∫ R ( ∂ x u ε ) 3 d x + ( 2 α − γ ) ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x − 2 ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 − 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

Consequently, we have that

(2.10) d d t ( ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = − ( κ + α + 2 γ ) ∫ R ( ∂ x u ε ) 3 d x + ( 2 α − γ ) ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x ≤ ∣ κ + α + 2 γ ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ∣ 2 α − γ ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

We define

(2.11) X ε ( t ) ≔ ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , ℓ 1 ≔ ∣ κ + α + 2 γ ∣ , ℓ 2 ≔ ∣ 2 α − γ ∣ .

Observe that, thanks to (2.11),

(2.12) ℓ 1 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = ℓ 1 β 2 + 1 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ℓ 1 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ℓ 1 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) X ε ( t ) , ℓ 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = ℓ 2 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ℓ 2 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) X ε ( t ) .

Therefore, by (2.10)–(2.12),

(2.13) d X ε ( t ) d t ≤ ℓ 1 + ℓ 2 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) X ε ( t ) .

Thanks to (2.11) and the Hölder inequality,

( ∂ x u ε ( t , ⋅ ) ) 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 ) 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = 2 β 2 ∣ β ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) ∣ β ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) = 2 β 2 β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 2 β 2 ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 2 β 2 X ε ( t ) .

Hence,

(2.14) ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ≤ 2 ∣ β ∣ X ε 1 2 ( t ) .

By (2.11), (2.12), and (2.14), we obtain

(2.15) d X ε ( t ) d t ≤ 2 ( ℓ 1 + ℓ 2 ) ∣ β ∣ 3 X ε 3 2 ( t ) ,

that is,

X ε − 3 2 ( t ) d X ε ( t ) d t ≤ 2 ( ℓ 1 + ℓ 2 ) ∣ β ∣ 3 .

By integrating on ( 0 , t ) , we have that

(2.16) 1 X ε ( t ) ≥ 1 X ε ( 0 ) − 2 2 ( ℓ 1 + ℓ 2 ) t ∣ β ∣ 3 .

By using (2.11) in (2.16), thank to (2.4), we have (2.3).

Assume (1.4) and we prove (2.5). Thanks to (1.9), we can define

(2.17) X 0 ≔ ∥ u 0 ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 = A 0 + ( β 2 + 1 ) B 0 + β 2 E 0 .

Consequently, by (2.2) and (2.11),

(2.18) X ε ( 0 ) ≤ X 0 ⇒ 1 X ε ( 0 ) ≥ 1 X 0 .

Moreover,

(2.19) 2 2 ( ℓ 1 + ℓ 2 ) t ∣ β ∣ 3 ≤ 2 2 ( ℓ 1 + ℓ 2 ) T ∣ β ∣ 3 ⇒ − 2 2 ( ℓ 1 + ℓ 2 ) t ∣ β ∣ 3 ≥ − 2 2 ( ℓ 1 + ℓ 2 ) T ∣ β ∣ 3 .

It follows from (2.3), (2.11), (2.17), (2.18), and (2.19) that

1 X ε ( t ) ≥ 1 X 0 − 2 2 ( ℓ 1 + ℓ 2 ) T ∣ β ∣ 3 ,

which gives

1 X ε ( t ) ≥ ∣ β ∣ 3 − 2 2 ( ℓ 1 + ℓ 2 ) X 0 T ∣ β ∣ 3 X 0 .

Thanks to (1.4), (1.9), (2.11), and (2.17), there exists a constant C > 0 , dependent on β , u 0 , and T , but not ε , such that

1 X ε ( t ) ≥ C ∣ β ∣ 3 X 0 .

Therefore, by (2.17),

(2.20) X ε ( t ) ≤ ∣ β ∣ 3 X 0 C ≤ C ⇒ X ε ( t ) ≤ C .

By using (2.11) in (2.20), we have (2.5).

We prove (2.6). Thanks to (2.5), (2.11), and (2.14), we obtain

∥ ∂ x u ε ∥ L ∞ ( ( 0 , T ) × R ) 2 ≤ C ,

which gives (2.6).

Now, we prove (2.7) and (2.8). We begin by observing that, by (2.5), (2.6), and (2.10),

d d t ( ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C .

By integrating on ( 0 , t ) , by (2.2), (2.11), and (2.17), we obtain

∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ( β 2 + 1 ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∫ 0 t ∥ ∂ x 2 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s + 2 ε ∫ 0 t ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ X 0 + C t ≤ C ,

which gives (2.7) and (2.8).

Finally, we prove (2.9). Thanks to (2.5) and the Hölder inequality,

u ε 2 ( t , x ) = 2 ∫ − ∞ x u ε ∂ x u ε d y ≤ 2 ∫ R ∣ u ε ∣ ∣ ∂ x 2 u ε ∣ d x ≤ 2 ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) ≤ C .

Hence,

∥ u ε ∥ L ∞ ( ( 0 , T ) × R ) 2 ≤ C ,

which gives (2.9).□

Since (1.10) and (1.11) and (1.5) imply (1.4), we have the following corollary.

Corollary 2.1

If (1.10) or (1.11) or (1.5) hold, then we have (2.5)–(2.9).

Corollary 2.2

Fix T > 0 and assume (1.35). Then, (2.5)–(2.8) hold, for each choose of β , T , and u 0 .

Proof

Let 0 ≤ t ≤ T . We begin by observing that, thanks to (1.35), (1.4) reads

∣ β ∣ > 0 ,

which is verified by Assumption (1.2). Therefore, arguing as in Lemma 2.1, we have (2.5)–(2.8), for each choose of β , T , and u 0 .□

Lemma 2.2

We have that

(2.21) e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 A ε ( t ) + ( β 2 + 1 ) B ε ( t ) + β 2 E ε ( t ) + e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 − 1 ≥ 1 A 0 , ε + ( β 2 + 1 ) B 0 , ε + β 2 E 0 , ε ,

for every 0 ≤ t ≤ T , where A ε ( t ) , B ε ( t ) , E ε ( t ) , ℓ 1 , and ℓ 2 are defined in (2.4) and (2.11), respectively. In particular, under Assumption (1.6), we have (2.5). Moreover, (2.6)–(2.9) hold.

Proof

We begin observing that, arguing as in Lemma 2.1, we have (2.15). Thanks to the Young inequality,

(2.22) X ε 3 2 ( t ) ≤ 1 2 ( X ε + X ε 2 ) .

It follows from (2.15) and (2.22) that

d X ε ( t ) d t ≤ ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 ( X ε + X ε 2 ) ,

that is,

(2.23) 1 X ε 2 ( t ) d X ε ( t ) d t ≤ ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 X ε ( t ) + ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 .

Observe that

1 X ε 2 ( t ) d X ε ( t ) d t = − d d t 1 X ε ( t ) .

Consequently by (2.23), we have that

(2.24) d d t 1 X ε ( t ) + ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 X ε ( t ) ≥ − ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 .

By multiplying (2.24) by e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 , we obtain

d d t e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 X ε ( t ) ≥ − ( ℓ 1 + ℓ 2 ) 2 ∣ β ∣ 3 e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 .

It follows from an integration on ( 0 , t ) that

e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 X ε ( t ) − 1 X 0 , ε ≥ − e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 − 1 ,

that is,

(2.25) e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 X ε ( t ) + e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 − 1 ≥ 1 X 0 , ε .

By using (2.3) and (2.11), in (2.25), we have (2.21).

Assume (1.6) and we prove (2.5). We begin by observing that (2.4), (2.11), (2.21), and (2.25),

e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 + e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 − 1 X ε ( t ) X ε ( t ) ≥ 1 X 0 , ε .

Hence, by (2.2), (2.11), (2.17), and (2.18),

X ε ( t ) ≤ X 0 , ε e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 + X 0 , ε e ( ℓ 1 + ℓ 2 ) t 2 ∣ β ∣ 3 − 1 X ε ( t ) ≤ X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 + X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 − 1 X ε ( t ) .

Therefore,

1 − X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 − 1 X ε ( t ) ≤ X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 .

Thanks to (1.6), (1.9), and (2.11), there exists an constant C > 0 , dependent on β , u 0 , and T , but not ε , such that

X ε ( t ) C ≤ X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 .

Hence,

(2.26) X ε ( t ) ≤ C X 0 e ( ℓ 1 + ℓ 2 ) T 2 ∣ β ∣ 3 ≤ C .

By using (2.11) in (2.26), we have (2.5).

Finally, arguing as in Lemma 2.1, we have (2.6)–(2.9).□

Lemma 2.3

Fix T > 0 and assume (1.7). Then, (2.5)–(2.9) hold, for each choose of β , T , and u 0 .

Proof

Let 0 ≤ t ≤ T . We begin by observing that, by (1.7), (2.1) reads

(2.27) ∂ t u ε + δ ∂ x u ε − β 2 ∂ t ∂ x 2 u ε + γ ∂ x 3 u ε + α ∂ x u ε ∂ x 2 u ε + 2 α u ε ∂ x 3 u ε = − ε ∂ x 4 u ε .

We prove that

(2.28) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∫ R ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 .

Observe that

− 2 α ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x − 4 α ∫ R u ε ∂ x 2 u ε ∂ x 3 u ε d x = − 2 α ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x − 2 α ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x = 0 .

Consequently, multiplying (2.28) by − 2 ∂ x 2 u ε and arguing as in Lemma 2.1, and integration on R gives

d d t ( ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∫ 0 t ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s = 0 .

By integrating on ( 0 , t ) , by (2.2), we have (2.28).

We prove that

(2.29) ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ≤ 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 .

Thanks to (2.28) 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 ) ≤ 2 ( ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 ) .

Hence,

∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) 2 ≤ 2 ( ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 ) ,

which gives (2.29).

Now, we prove that

(2.30) ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∫ 0 t ∥ ∂ x 4 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ ∥ u 0 ∥ H 2 ( R ) 2 + 5 2 ∣ α ∣ ( ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 ) 3 2 T .

We begin by observing that

2 α ∫ R u ε ∂ x u ε ∂ x 2 u ε + 4 α ∫ R u ε 2 ∂ x 3 u ε d x = − α ∫ R ( ∂ x u ε ) 3 d x − 8 α ∫ R u ε ∂ x u ε ∂ x 2 u ε d x = 5 α ∫ R ( ∂ x u ε ) 3 d x .

Therefore, multiplying (2.27) by 2 u and arguing as in Lemma 2.1, integration on R gives

(2.31) d d t ( ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = − 5 α ∫ R ( ∂ x u ε ) 3 d x .

Thanks to (2.28) and (2.29),

5 ∣ α ∣ ∫ R ∣ ∂ x u ε ∣ 3 d x ≤ ∣ α ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 5 2 ∣ α ∣ ( ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 ) 3 2 .

It follows from (2.31) that

d d t ( ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 5 2 ∣ α ∣ ( ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 ) 3 2 .

By integrating on ( 0 , t ) , by (2.2), we have (2.28).

Finally, thanks to (2.30) and the Hölder inequality, we have that

(2.32) ∥ u ε ∥ L ∞ ( ( 0 , T ) × R ) ≤ C ( T ) .

Therefore, the proof is concluded.□

Lemma 2.4

Assume (1.8). Then, we have (2.5)–(2.9).

Proof

We begin by observing that, thanks to (1.8), (2.1) reads

(2.33) ∂ t u ε + δ ∂ x u ε − 2 κ u ε ∂ x u ε + a ∂ x 3 u ε − β 2 ∂ t ∂ x 2 u ε + 2 γ ∂ x u ε ∂ x 2 u ε + γ u ε ∂ x 3 u ε = − ε ∂ x 4 u ε .

We prove that

(2.34) ∥ u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε ∫ 0 t ∥ ∂ x 2 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ ∥ u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 .

Observe that,

4 γ ∫ R u ε ∂ x u ε ∂ x 2 u ε d x + 2 γ ∫ R u ε 2 ∂ x 3 u ε d x = 4 γ ∫ R u ε ∂ x u ε ∂ x 2 u ε d x − 4 γ ∫ R u ε ∂ x u ε ∂ x 2 u ε d x = 0 .

Therefore, by multiplying (2.33) by u ε and arguing as in Lemma 2.1, and integrating on R , we obtain (2.34).

Moreover, thanks to (2.34) and the Hölder inequality,

∥ u ε ( t , ⋅ ) ∥ L ∞ ( R ) ≤ 2 ( ∥ u 0 ∥ L 2 ( R ) + β 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 ) .

We prove that

(2.35) arctan ( B ε ( t ) + β 2 E ε ( t ) ) ≤ arctan ( B 0 + β 2 E 0 ) + τ 3 2 ( β 2 ) t 2 ∣ β ∣ 5 2 ,

where B 0 , E 0 , τ 3 2 ( β 2 ) , B ε ( t ) , and E ε ( t ) are defined in (1.9) and (2.4). Multiplying (2.33) by − 2 ∂ x 2 u ε and arguing as in Lemma 2.1, an integration on R gives

(2.36) d d t ( ∥ ∂ x u ε ( t , ⋅ ) ∥ 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = κ ∫ R ( ∂ x u ε ) 3 d x + 3 γ ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x ≤ ∣ κ ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 3 ∣ γ ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

Observe that, by (2.34),

∣ κ ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ∣ κ ∣ ( ∥ u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 ) β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) .

Therefore, by (2.36), we have

(2.37) d d t ( ∥ ∂ x u ε ( t , ⋅ ) ∥ 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ∣ κ ∣ ( ∥ u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 ) β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) + 3 ∣ γ ∣ β 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ( ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) .

Due to the Young inequality,

∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 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 ) = 2 ∣ β ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) ∣ β ∣ ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) = 2 ∣ β ∣ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 2 ∣ β ∣ ( ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) .

Hence,

(2.38) ∥ ∂ x u ε ( t , ⋅ ) ∥ L ∞ ( R ) ≤ 2 ∣ β ∣ ( ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) .

We define

(2.39) Y ε ( t ) ≔ ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

It follows from (2.37), (2.38), and (2.39) that

(2.40) d Y ε ( t ) d t + 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ∣ κ ∣ ( ∥ u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x u 0 ∥ L 2 ( R ) 2 ) ∣ β ∣ 5 2 Y ε 1 2 ( t ) + 3 ∣ γ ∣ ∣ β ∣ 5 2 Y ε 3 2 ( t ) ≤ τ 3 2 ( β 2 ) ∣ β ∣ 5 2 Y ε 1 2 ( t ) ( 1 + Y ε ( t ) ) ,

where τ 3 2 ( β 2 ) is defined in (1.9). Consequently, we have that

1 Y ε 1 2 ( t ) ( 1 + Y ε ( t ) ) d Y ε ( t ) d t ≤ τ 3 2 ( β 2 ) ∣ β ∣ 5 2 .

By integrating on ( 0 , t ) , by (2.2) and (2.39), we obtain that

(2.41) arctan Y ε 1 2 ( t ) ≤ arctan Y 0 , ε 1 2 + τ 3 2 ( β 2 ) t 2 ∣ β ∣ 5 2 ≤ arctan Y 0 1 2 + τ 3 2 ( β 2 ) t 2 ∣ β ∣ 5 2 ,

where Y 0 ≔ ∥ ∂ x u 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u 0 ∥ L 2 ( R ) 2 .

By using (1.9) and (2.4) in (2.41), we have (2.35).

We prove that under Assumption (1.8), there exists a constant C > 0 , dependent on β , u 0 and T , but not ε , such that

(2.42) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ,

for every 0 ≤ t ≤ T .

We begin by observing that, by (1.9), (2.4), and (2.35), we have that

(2.43) arctan ( B ε ( t ) + β 2 E ε ( t ) ) ≤ arctan ( B 0 + β 2 E 0 ) + τ 3 2 ( β 2 ) T 2 ∣ β ∣ 5 2 .

Moreover, by (1.8), we have that

(2.44) arctan ( B 0 + β 2 E 0 ) + τ 3 2 ( β 2 ) T 2 ∣ β ∣ 5 2 < π 2 .

Consequently, by (2.43) and (2.44), we obtain

(2.45) arctan ( B ε ( t ) + β 2 E ε ( t ) ) ≤ C , C ∈ 0 , π 2 .

(2.42) follows from (2.4) and (2.45).

Finally, arguing as in Lemma 2.1, we have (2.6) and (2.8). Therefore, the proof is concluded.□

Lemma 2.5

Assume that one within (1.4), (1.10), (1.11), and (1.5)–(1.8). There exists a constant C > 0 , dependent on β , T , and u 0 , but not ε , such that

(2.46) ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε 2 ∫ 0 t ∥ ∂ x 4 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ C ,

for every 0 ≤ t ≤ T .

Proof

By multiplying (2.1) by 2 ε ∂ x 4 u ε , an integration on R gives

d d t ( ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ε β 2 ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) = − 2 ε ∫ R ∂ x 4 u ε ∂ t u ε d x − 2 β 2 ε ∫ R ∂ x 4 u ε ∂ t ∂ x 2 u ε d x = − 2 κ ε ∫ R u ε ∂ x u ε ∂ x 4 u ε d x − 2 α ε ∫ R ∂ x u ε ∂ x 2 u ε ∂ x 4 u ε d x − 2 γ ε ∫ R u ε ∂ x 3 u ε ∂ x 4 u ε d x − 2 ε 2 ∥ ∂ x 4 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 − 2 δ ∫ R ∂ x u ε ∂ x 4 u ε d x − 2 a ∫ R ∂ x 3 u ε ∂ x 4 u ε d x = 2 κ ε ∫ R ( ∂ x u ε ) 2 ∂ x 3 u ε d x + 2 κ ε ∫ R u ε ∂ x 2 u ε ∂ x 3 u ε d x + ( 2 α − γ ) ε ∫ R ∂ x u ε ( ∂ x 3 u ε ) 2 d x − 2 ε 2 ∥ ∂ x 4 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 δ ∫ R ∂ x 2 u ε ∂ x 3 u ε d x = − 3 κ ε ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x + ( 2 α − γ ) ε ∫ R ∂ x u ε ( ∂ x 3 u ε ) 2 d x − 2 ε 2 ∥ ∂ x 4 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

Therefore, (2.6), we have that

d d t ( ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ε β 2 ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ) + 2 ε 2 ∥ ∂ x 4 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = − 3 κ ε ∫ R ∂ x u ε ( ∂ x 2 u ε ) 2 d x + ( 2 α − γ ) ε ∫ R ∂ x u ε ( ∂ x 3 u ε ) 2 d x ≤ 3 ∣ κ ∣ ε ∥ ∂ x u ε ∥ L ∞ ( ( 0 , T ) × R ) ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ∣ α − γ ∣ ε ∥ ∂ x u ε ∥ L ∞ ( ( 0 , T ) × R ) ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + C ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

By integrating on ( 0 , t ) , by (2.2), (2.7), and (2.8), we obtain

ε ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + ε β 2 ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ε 2 ∫ 0 t ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ C 0 + C ε ∫ 0 t ∥ ∂ x 2 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s + C ε ∫ 0 t ∥ ∂ x 3 u ε ( s , ⋅ ) ∥ L 2 ( R ) 2 d s ≤ C ,

which gives (2.46).□

Lemma 2.6

Assume that one within (1.4), (1.10), (1.11), and (1.5)–(1.8). There exists a constant C > 0 , dependent on β , T , and u 0 , but not ε , such that

(2.47) ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) + β 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ,

(2.48) ∥ ∂ t u ε ∥ L ∞ ( ( 0 , T ) × R ) ≤ C ,

for every 0 ≤ t ≤ T .

Proof

By multiplying (2.1) by 2 ∂ t u ε , an integration on R gives

2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = − 2 κ ∫ R u ε ∂ x u ε ∂ t u ε d x + 2 β 2 ∫ R ∂ t u ε ∂ x 2 u ε d x − 2 α ∫ R ∂ x u ε ∂ x 2 u ε ∂ t u ε d x − 2 γ ∫ R u ε ∂ x 3 u ε ∂ t u ε d x − 2 ε ∫ R ∂ x 4 u ε ∂ t u ε d x − 2 δ ∫ R ∂ x u ε ∂ t u ε d x − 2 a ∫ R ∂ x 3 u ε ∂ t u ε d x = − 2 κ ∫ R u ε ∂ x u ε ∂ t u ε d x − 2 β 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 ( γ − α ) ∫ R ∂ x u ε ∂ x 2 u ε ∂ t u ε d x + 2 γ ∫ R u ε ∂ x 2 u ε ∂ t ∂ x u ε d x + 2 ε ∫ R ∂ x 3 u ε ∂ t ∂ x u ε d x − 2 δ ∫ R ∂ x u ε ∂ t u ε d x + 2 a ∫ R ∂ x 2 u ε ∂ t ∂ x u ε d x .

Therefore, we have that

(2.49) 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 2 β 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 = − 2 κ ∫ R u ε ∂ x u ε ∂ t u ε d x + 2 ( γ − α ) ∫ R ∂ x u ε ∂ x 2 u ε ∂ t u ε d x + 2 γ ∫ R u ε ∂ x 2 u ε ∂ t ∂ x u ε d x + 2 ε ∫ R ∂ x 3 u ε ∂ t ∂ x u ε d x − 2 δ ∫ R ∂ x u ε ∂ t u ε d x + 2 a ∫ R ∂ x 2 u ε ∂ t ∂ x u ε d x .

Since, 0 < ε < 1 , due to (2.5), (2.6), (2.9), Lemma 2.5 and the Young inequality,

2 ∣ κ ∣ ∫ R ∣ u ε ∣ ∣ ∂ x u ε ∣ ∣ ∂ t u ε ∣ d x ≤ 2 ∣ κ ∣ ∥ u ∥ L ∞ ( ( 0 , T ) × R ) ∫ R ∣ ∂ x u ε ∣ ∣ ∂ t u ε ∣ d x ≤ C ∫ R ∣ ∂ x u ε ∣ ∣ ∂ t u ε ∣ d x = ∫ R ∣ C ∂ x u ε ∣ ∣ ∂ t u ε ∣ d x ≤ C ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , 2 ∣ γ − α ∣ ∫ R ∣ ∂ x u ε ∣ ∣ ∂ x 2 u ε ∣ ∣ ∂ t u ε ∣ d x ≤ 2 ∣ γ − α ∣ ∥ ∂ x u ε ∥ L ∞ ( ( 0 , T ) × R ) ∫ R ∣ ∂ x 2 u ε ∣ ∣ ∂ t u ε ∣ d x ≤ C ∫ R ∣ ∂ x 2 u ε ∣ ∣ ∂ t u ε ∣ d x = ∫ R ∣ C ∂ x 2 u ε ∣ ∣ ∂ t u ε ∣ d x ≤ C ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , 2 ∣ γ ∣ ∫ R ∣ u ε ∣ ∣ ∂ x 2 u ε ∣ ∣ ∂ t ∂ x u ε ∣ d x ≤ 2 ∣ γ ∣ ∥ u ε ∥ L ∞ ( ( 0 , T ) × R ) ∫ R ∣ ∂ x 2 u ε ∣ ∣ ∂ t ∂ x ∣ d x ≤ C ∫ R ∣ ∂ x 2 u ε ∣ ∣ ∂ t ∂ x u ε ∣ d x = ∫ R C ∂ x 2 u ε β ∣ β ∂ t ∂ x u ε ∣ d x ≤ C ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) ≤ C + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) , 2 ε ∫ R ∣ ∂ x 3 u ε ∣ ∣ ∂ t u ε ∣ d x = ∫ R 2 ε ∂ x 3 u ε β ∣ β ∂ t ∂ x u ε ∣ d x ≤ 2 ε 2 ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 2 ε ∥ ∂ x 3 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C 0 + + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , 2 ∣ δ ∣ ∫ R ∣ ∂ x u ε ∣ ∣ ∂ t u ε ∣ d x ≤ 4 δ 2 ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C + 1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 , 2 ∣ a ∣ ∫ R ∣ ∂ x 2 u ε ∣ ∣ ∂ t ∂ x u ε ∣ d x = ∫ R 2 a ∂ x 2 u ε β ∣ β ∂ t ∂ x u ε ∣ d x ≤ 2 a 2 β 2 ∥ ∂ x 2 u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C + β 2 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 .

Consequently, by (2.49), we obtain

1 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 + 1 2 ∥ ∂ t ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ,

which gives (2.47).

Finally, we prove (2.48). Thanks to (2.47) and the Hölder inequality,

( ∂ t u ε ( t , x ) ) 2 = 2 ∫ − ∞ x ∂ t u ε ∂ t ∂ x u ε d y ≤ 2 ∫ R ∣ ∂ t u ε ∣ ∣ ∂ t ∂ x u ε ∣ d x ≤ 2 ∥ ∂ t u ε ( t , ⋅ ) ∥ L 2 ( R ) ∥ ∂ x u ε ( t , ⋅ ) ∥ L 2 ( R ) ≤ C .

Hence,

∥ ∂ t u ε ∥ L ∞ ( ( 0 , T ) × R ) 2 ≤ C ,

which gives (2.48).□

3 Proof of Theorem 1.1

This section is devoted to the proof of theorem.

By using the Sobolev immersion theorem, we prove the following result.

Lemma 3.1

Assume that one within (1.4)–(1.8) and (1.10) holds. There exist a subsequence { u ε k } k ∈ N of { u ε } ε > 0 and a limit function u, which satisfies (1.32) such that

(3.1) u ε k → u a.e. and in L loc p ( ( 0 , T ) × R ) , 1 ≤ p < ∞ .

Moreover, u is the solution of (1.1).

Proof

Thanks to Lemma 2.1, Corollaries 2.1 and 2.2, Lemmas 2.2 and 2.3, (2.4), and, Lemma 2.6,

(3.2) { u ε } ε > 0 is uniformly bounded in H 1 ( ( 0 , T ) × R ) ,

which gives (3.1).

Observe that, thanks to Lemma 2.1, Corollaries 2.1 and 2.2, and Lemmas 2.2 and 2.3

u ∈ L ∞ ( 0 , T ; H 2 ( R ) ) ,

while, by Lemma 2.6,

u ∈ W 1 , ∞ ( ( 0 , T ) × R ) .

Moreover, by Lemma 2.6, we have that

∂ t ∂ x u ∈ L 2 ( ( 0 , T ) × R ) .

Therefore, (1.32) holds and u is the solution of (1.1).□

Following [83, Theorem 1.1], we prove Theorem 1.1.

Proof of Theorem 1.1

Lemma 3.1 gives the existence of a solution u of (1.1) such that (1.32) holds.

We prove (1.33). Let u 1 and u 2 be two solutions of (1.1), which satisfy (1.32), that is,

∂ t u 1 + δ ∂ x u 1 + κ ∂ x u 1 2 − β 2 ∂ t ∂ x 2 u 1 + a ∂ x 3 u 1 + α − γ 2 ∂ x ( ( ∂ x u 1 ) 2 ) − γ ∂ x ( u 1 ∂ x 2 u 1 ) , t > 0 , x ∈ R , u 1 ( 0 , x ) = u 1 , 0 ( x ) , x ∈ R ,

∂ t u 2 + δ ∂ x u 2 + κ ∂ x u 2 2 − β 2 ∂ t ∂ x 2 u 2 + a ∂ x 3 u 2 + α − γ 2 ∂ x ( ( ∂ x u 2 ) 2 ) − γ ∂ x ( u 2 ∂ x 2 u 2 ) , t > 0 , x ∈ R , u 2 ( 0 , x ) = u 2 , 0 ( x ) , x ∈ R .

Then, the function

(3.3) ω = u 1 − u 2

is the solution of the following Cauchy problem:

(3.4) ∂ t ω + ν ∂ x ω − β 2 ∂ t ∂ x 2 ω + κ ∂ x ( u 1 2 − u 2 2 ) + a ∂ x 3 ω + α − γ 2 ∂ x ( ( ∂ x u 1 ) 2 − ( ∂ x u 1 ) 2 ) − γ ∂ x ( u 1 ∂ x 2 u 1 − u 2 ∂ x 2 u 2 ) = 0 , t > 0 , x ∈ R , ω 0 ( x ) = u 1 , 0 ( x ) − u 2 , 0 ( x ) , x ∈ R .

Observe that, thanks to (3.3),

(3.5) ∂ x ( u 1 2 − u 2 2 ) = ∂ x ( ( u 1 + u 2 ) ( u 1 − u 2 ) ) = ∂ x ( ( u 1 + u 2 ) ω ) , ∂ x ( ( ∂ x u 1 ) 2 − ( ∂ x u 1 ) 2 ) = ∂ x ( ( ∂ x u 1 + ∂ x u 2 ) ( ∂ x u 1 − ∂ x u 2 ) ) = ∂ x ( ( ∂ x u 1 + ∂ x u 2 ) ω ) .

Moreover, again by (3.3),

(3.6) ∂ x ( u 1 ∂ x 2 u 1 − u 2 ∂ x 2 u 2 ) = ∂ x ( u 1 ∂ x 2 u 1 − u 1 ∂ x 2 u 2 + u 1 ∂ x 2 u 2 − u 2 ∂ x 2 u 2 ) = ∂ x ( u 1 ∂ x 2 ω − ∂ x 2 u 2 ω ) + ∂ x ( ∂ x 2 u 2 ω ) .

Consequently, by (3.5) and (3.6), equation (3.4) reads

(3.7) ∂ t ω + δ ∂ x ω − β 2 ∂ t ∂ x 2 ω + κ ∂ x ( ( u 1 + u 2 ) ω ) + a ∂ x 3 ω + α − γ 2 ∂ x ( ( ∂ x u 1 + ∂ x u 2 ) ω ) − γ ∂ x ( u 1 ∂ x 2 ω − ∂ x 2 u 2 ω ) = 0 .

Since u 1 , u 2 ∈ L ∞ ( 0 , T ; H 2 ( R ) ) , there exists a constant C > 0 , such that

(3.8) ∥ u 1 ∥ L ∞ ( ( 0 , T ) × R ) , ∥ u 2 ∥ L ∞ ( ( 0 , T ) × R ) ≤ C , ∥ ∂ x u 1 ∥ L ∞ ( ( 0 , T ) × R ) , ∥ ∂ x u 2 ∥ L ∞ ( ( 0 , T ) × R ) ≤ C , ∥ ∂ x 2 u 2 ( t , ⋅ ) ∥ L 2 ( R ) ≤ C .

Moreover, by (3.8), we have that

(3.9) ∣ u 1 + u 2 ∣ ≤ ∣ u 1 ∣ + ∣ u 2 ∣ ≤ C , ∣ ∂ x u 1 + ∂ x u 2 ∣ ≤ ∣ ∂ x u 1 + ∂ x u 2 ∣ ≤ C .

Observe that

2 ∫ R ω ( ∂ t ω − β 2 ∂ t ∂ x 2 ω ) d x = d d t ( ∥ ω ( t , ⋅ ) ∥ + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ) ,

(3.10) 2 δ ∫ R ω ∂ x ω d x = 0 ,

2 κ ∫ R ω ∂ x ( ( u 1 + u 2 ) ω ) d x = − 2 κ ∫ R ( u 1 + u 2 ) ω ∂ x ω d x , ( α − γ ) ∫ R ω ∂ x ( ( ∂ x u 1 + ∂ x u 2 ) ω ) d x = − ( α − γ ) ∫ R ( ∂ x u 1 + ∂ x u 2 ) ω ∂ x ω d x , 2 a ∫ R ω ∂ x 3 ω d x = − 2 a ∫ R ∂ x ω ∂ x 2 ω d x = 0 ,

(3.11) − 2 γ ∫ R ω ∂ x ( u 1 ∂ x 2 ω − ∂ x 2 u 2 ω ) d x = 2 γ ∫ R u 1 ∂ x ω ∂ x 2 ω d x − 2 γ ∫ R ∂ x 2 u 2 ω ∂ x ω d x = − γ ∫ R ∂ x u 1 ( ∂ x ω ) 2 d x − 2 γ ∫ R ∂ x 2 u 2 ω ∂ x ω d x .

By multiplying (3.7) by 2 ω , thanks to (3.10), we have that

(3.12) d d t ( ∥ ω ( t , ⋅ ) ∥ + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ) = 2 κ ∫ R ( u 1 + u 2 ) ω ∂ x ω d x + ( α − γ ) ∫ R ( ∂ x u 1 + ∂ x u 2 ) ω ∂ x ω d x + γ ∫ R ∂ x u 1 ( ∂ x ω ) 2 d x + 2 γ ∫ R ∂ x 2 u 2 ω ∂ x ω d x .

Due to (3.8), (3.9), and the Young inequality,

2 ∣ κ ∣ ∫ R ∣ u 1 + u 2 ∣ ∣ ω ∣ ∣ ∂ x ω ∣ d x ≤ C ∫ R ∣ ω ∣ ∣ ∂ x ω ∣ d x ≤ C ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + C ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 , ∣ α − γ ∣ ∫ R ∣ ∂ x u 1 + ∂ x u 2 ∣ ∣ ω ∣ ∣ ∂ x ω ∣ d x ≤ C ∫ R ∣ ω ∣ ∣ ∂ x ω ∣ d x ≤ C ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + C ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 , ∣ γ ∣ ∫ R ∣ ∂ x u 1 ∣ ( ∂ x ω ) 2 d x ≤ C ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 , 2 ∣ γ ∣ ∫ R ∣ ∂ x 2 u 2 ω ∣ ∣ ∂ x ω ∣ d x ≤ γ 2 ∫ R ω 2 ( ∂ x 2 u 2 ) 2 d x + ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ γ 2 ∥ ω ( t , ⋅ ) ∥ L ∞ ( R ) 2 ∥ ∂ x 2 u 2 ( t , ⋅ ) ∥ L 2 ( R ) 2 + ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ∥ ω ( t , ⋅ ) ∥ L ∞ ( R ) 2 + ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 .

It follows from (3.12) that

(3.13) d d t ( ∥ ω ( t , ⋅ ) ∥ + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ) ≤ C ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + C ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + C ∥ ω ( t , ⋅ ) ∥ L ∞ ( R ) 2 .

Thanks to the Hölder inequality,

∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ 2 ∫ R ∣ ω ∣ ∣ ∂ x ω ∣ d x ≤ 2 ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) .

Therefore, by the Young inequality,

∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 .

It follows from (3.13) that

d d t ( ∥ ω ( t , ⋅ ) ∥ + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ) ≤ C ∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ C ( ∥ ω ( t , ⋅ ) ∥ + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ) .

The Gronwall Lemma and (3.4) give

∥ ω ( t , ⋅ ) ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x ω ( t , ⋅ ) ∥ L 2 ( R ) 2 ≤ e C ( T ) t ( ∥ ω 0 ∥ L 2 ( R ) 2 + β 2 ∥ ∂ x ω 0 ∥ L 2 ( R ) 2 ) .

By (1.34), we have that

(3.14) τ 1 2 ∥ ω ∥ H 1 ( R ) 2 ≤ τ 2 2 e C ( T ) t ∥ ω 0 ∥ H 1 ( R ) 2 .

Therefore, (1.33) follows from (3.4) and (3.14).□

Acknowledgements

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). GMC has been partially supported by the Project funded under the National Recovery and Resilience Plan (NRRP), Mission 4 Component 2 Investment 1.4 -Call for tender No. 3138 of 16/12/2021of Italian Ministry of University and Research funded by the European Union -NextGenerationEUoAward Number: CN000023, Concession Decree No. 1033 of 17/06/2022 adopted by the Italian Ministry of University and Research, CUP: D93C22000410001, Centro Nazionale per la Mobilità Sostenibile and the Italian Ministry of Education, University and Research under the Programme Department of Excellence Legge 232/2016 (Grant No. CUP - D93C23000100001). GMC expresses its gratitude to HIAS - Hamburg Institute for Advanced Study for their warm hospitality.

  1. Conflict of interest: The authors state no conflict of interest.

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Received: 2022-11-09
Accepted: 2023-03-28
Published Online: 2023-05-05

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2022-0577
Keywords for this article
existence; uniqueness; stability; Camassa-Holm type equation; Cauchy problem
Creative Commons
BY 4.0

Articles in the same Issue

  1. Special Issue on Future Directions of Further Developments in Mathematics
  2. What will the mathematics of tomorrow look like?
  3. On H 2-solutions for a Camassa-Holm type equation
  4. Classical solutions to Cauchy problems for parabolic–elliptic systems of Keller-Segel type
  5. Control of multi-agent systems: Results, open problems, and applications
  6. Logical perspectives on the foundations of probability
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  9. Luenberger compensator theory for heat-Kelvin-Voigt-damped-structure interaction models with interface/boundary feedback controls
  10. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part II)
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  12. Analysis of solutions for the fractional differential equation with Hadamard-type
  13. Hilfer proportional nonlocal fractional integro-multipoint boundary value problems
  14. A comprehensive review on fractional-order optimal control problem and its solution
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  19. Regular Articles
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  85. Existence of solutions for semilinear retarded equations with non-instantaneous impulses, non-local conditions, and infinite delay
  86. On the quadratic residues and their distribution properties
  87. On average theta functions of certain quadratic forms as sums of Eisenstein series
  88. Connected component of positive solutions for one-dimensional p-Laplacian problem with a singular weight
  89. Some identities of degenerate harmonic and degenerate hyperharmonic numbers arising from umbral calculus
  90. Mean ergodic theorems for a sequence of nonexpansive mappings in complete CAT(0) spaces and its applications
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  94. On a class of stochastic differential equations driven by the generalized stochastic mixed variational inequalities
  95. Entire solutions of two certain Fermat-type ordinary differential equations
  96. Generalized Lie n-derivations on arbitrary triangular algebras
  97. Markov decision processes approximation with coupled dynamics via Markov deterministic control systems
  98. Notes on pseudodifferential operators commutators and Lipschitz functions
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  100. Strong limit of processes constructed from a renewal process
  101. Construction of analytical solutions to systems of two stochastic differential equations
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  103. Regularity and abundance on semigroups of partial transformations with invariant set
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  106. Properties of locally semi-compact Ir-topological groups
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  108. Ordering stability of Nash equilibria for a class of differential games
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  111. Ricci ϕ-invariance on almost cosymplectic three-manifolds
  112. Schur-power convexity of integral mean for convex functions on the coordinates
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  114. On Bohr's inequality for special subclasses of stable starlike harmonic mappings
  115. Properties of meromorphic solutions of first-order differential-difference equations
  116. A double-phase eigenvalue problem with large exponents
  117. On the number of perfect matchings in random polygonal chains
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  120. The 𝔪-WG° inverse in the Minkowski space
  121. Stability result for Lord Shulman swelling porous thermo-elastic soils with distributed delay term
  122. Approximate solvability method for nonlocal impulsive evolution equation
  123. Construction of a functional by a given second-order Ito stochastic equation
  124. Global well-posedness of initial-boundary value problem of fifth-order KdV equation posed on finite interval
  125. On pomonoid of partial transformations of a poset
  126. New fractional integral inequalities via Euler's beta function
  127. An efficient Legendre-Galerkin approximation for the fourth-order equation with singular potential and SSP boundary condition
  128. Eigenfunctions in Finsler Gaussian solitons
  129. On a blow-up criterion for solution of 3D fractional Navier-Stokes-Coriolis equations in Lei-Lin-Gevrey spaces
  130. Some estimates for commutators of sharp maximal function on the p-adic Lebesgue spaces
  131. A preconditioned iterative method for coupled fractional partial differential equation in European option pricing
  132. A digital Jordan surface theorem with respect to a graph connectedness
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