On the classical solutions for a Rosenau–Korteweg-deVries

Abstract

The Rosenau–Korteweg-deVries–Kawahara equation describes the dynamics of dense discrete systems or small-amplitude gravity capillary waves on water of a finite depth. In this paper, we prove the well-posedness of the classical solutions for the Cauchy problem.

Keywords

Existence uniqueness stability Rosenau–Korteweg-deVries–Kawahara type equation Cauchy problem

1. Introduction

In this study, we investigate the well-posedness of the classical solution of the following Cauchy problem: $\begin{array}{l} (1.1) & \{\begin{matrix} \partial_{t} u + \partial_{x} f (u) + α \partial_{x}^{3} u + γ \partial_{x}^{5} u + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = 0, & t > 0, x \in R, \\ u (0, x) = u_{0} (x), & x \in R, \end{matrix} \end{array}$ with $\begin{array}{l} (1.2) & α, γ, β \in R, β \neq 0, \end{array}$ and $\begin{array}{l} (1.3) & f \in C^{1} (R) . \end{array}$ If $γ \neq 0$ , on the initial datum, we assume $\begin{array}{l} (1.4) & u_{0} (x) \in H^{3} (R), \end{array}$ while, if $γ = 0$ , we assume $\begin{array}{l} (1.5) & u_{0} (x) \in H^{2} (R) \end{array}$ Observe that, if $f (u) = u^{2}$ and $γ = β = 0$ , (1.1) reads $\begin{array}{l} (1.6) & \partial_{t} u + 2 u \partial_{x} u + α \partial_{x}^{3} u = 0, \end{array}$ which is known as the Korteweg-deVries equation [30], which models a large number of physical phenomena. Rosenau observed that (1.6) is not able to describe the wave-wave and wave-wall interactions when study of compact discrete systems. Therefore, he proposed the following equation (see [36,37]): $\begin{array}{l} (1.7) & \partial_{t} u + 2 u \partial_{x} u + β^{2} \partial_{t} \partial_{x}^{4} u = 0 . \end{array}$ For further consideration of nonlinear waves, the term $α \partial_{x}^{3} u$ is included in (1.7) (see [20,26,36,37,39]). The resulting equation is termed the Rosenau–Korteweg-deVries equation: $\begin{array}{l} (1.8) & \partial_{t} u + 2 u \partial_{x} u + α \partial_{x}^{3} u + β^{2} \partial_{t} \partial_{x}^{4} u = 0 . \end{array}$ From a mathematical point of view, in [34,41], the existence and the uniqueness of the solution of (1.7) is proven, while, in [42], the initial-boundary value problem for (1.7) is studied. In [1–3,5–7,22,23,25,38], numerical schemes for (1.7) are studied. Instead, following [8,31,40], in [9,14], the convergence of the solution of (1.7) to the unique entropy of the Burgers equation is proven.

About (1.8), the existence of the solitary wave solutions is proven in [20,21], the existence of the soliton solutions is proven in [19,35,39], and the existence of the periodic solution is proven in [43]. Instead, in [26,27], numerical schemes for (1.8) are studied, while, in [13], the convergence of the solution of (1.8) to the unique entropy solution of the Burgers equation is proven.

If $f (u) = u^{3}$ and $γ = β = 0$ , (1.1) reads $\begin{array}{l} (1.9) & \partial_{t} u + 3 u^{2} \partial_{x} u + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = 0, \end{array}$ which is known as the modified Rosenau equation (see [35–37]). In [4], the authors investigated the solitary solution and the two invariance of (1.9), while, in [24], numerical schemes are studied. Finally, following [8,31,40], the convergence of the solution of (1.9) to the entropy solutions of the following scalar conservation law $\begin{array}{l} (1.10) & \partial_{t} + 3 u^{2} \partial_{x} u = 0, \end{array}$ is proven in [15].

Observe that, if $f (u) = u^{2}, u^{3}$ , $β = 0$ and $γ \neq 0$ , (1.1) reads $\begin{array}{l} (1.11) & \partial_{t} u + \partial_{x} f (u) + α \partial_{x}^{3} u + γ \partial_{x}^{5} u = 0, f (u) = u^{2}, u^{3}, \end{array}$ known as the Kawahara–Korteweg-de Vries equation. It was derived by Kawahara [28] to describe small-amplitude gravity capillary waves on water of a finite depth when the Weber number is close to $1 / 3$ (see [32]). It was also derived in the context of water waves by Olver [33] (see also [29]), using Hamiltonian perturbation theory, with further generalization given by Craig and Groves [18].

In [17], the well-posedness of the classical solutions of the Cauchy problem for (1.11) is proven, while the convergence of the solution of (1.11) to the Burges equation or (1.10) is proven in [10,11] and [12], respectively.

The main result of this paper is the following theorem.

Theorem 1.1.
Fix $T > 0$ . Assuming ( 1.3 ), ( 1.4 ) and $γ \neq 0$ , there exists an solution u of ( 1.1 ) such that $\begin{array}{l} (1.12) & \begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{3} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{2} u (t, \cdot) \in L^{2} (R), for every 0 ⩽ t ⩽ T . \end{aligned} \end{array}$ Assuming ( 1.3 ), ( 1.5 ) and $γ = 0$ , there exists an solution u of ( 1.1 ) such that $\begin{array}{l} (1.13) & \begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{2} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{2} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{3} u (t, \cdot) \in L^{2} (R), for every 0 ⩽ t ⩽ T . \end{aligned} \end{array}$ Moreover, if $f \in C^{2}$ , u is unique. Moreover, if $u_{1}$ and $u_{2}$ are two solutions of ( 1.1 ), we have that $\begin{array}{l} (1.14) & {‖ u_{1} (t, \cdot) - u_{2} (t, \cdot) ‖}_{H^{2} (R)}^{2} ⩽ \frac{3 τ_{2}^{2} e^{C (T) t}}{2 τ_{1}^{2}} ‖ u_{1, 0} - u_{2, 0} ‖_{H^{2} (R)}^{2}, \end{array}$ where $\begin{array}{l} (1.15) & τ_{1}^{2} = min {1, β^{2}}, τ_{2}^{2} = max {1, β^{2}}, \end{array}$ for some suitable $C (T) > 0$ , and every $0 ⩽ t ⩽ T$ .
Corollary 1.1.
Assume $f \in C^{2} (R)$ and ( 1.5 ). if $γ = α = 0$ , there exists an unique solution u of ( 1.1 ), such that $\begin{array}{l} (1.16) & \begin{aligned} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{2} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{2} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{3} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{4} u (t, \cdot) \in L^{2} (R), for every 0 ⩽ t ⩽ T . \end{aligned} \end{array}$ Moreover, ( 1.14 ) holds.

The paper 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, that is given in Section 3. In last section, we prove Corollary 1.1.
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: $\begin{array}{l} (2.1) & \{\begin{matrix} \partial_{t} u_{ε} + \partial_{x} f (u_{ε}) + α \partial_{x}^{3} u_{ε} + γ \partial_{x}^{5} u_{ε} + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = ε \partial_{x}^{6} u_{ε}, & t > 0, x \in R, \\ u_{ε} (0, x) = u_{ε, 0} (x), & x \in R, \end{matrix} \end{array}$ where $u_{ε, 0}$ is a $C^{\infty}$ approximation of $u_{0}$ , such that, under the Assumptions $γ \neq 0$ and (1.4), $\begin{array}{l} (2.2) & ‖ u_{ε, 0} ‖_{H^{3} (R)} ⩽ ‖ u_{0} ‖_{H^{3} (R)}, ε ({‖ \partial_{x}^{2} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{4} u_{ε, 0} ‖}_{L^{2} (R)}) ⩽ C_{0}, \end{array}$ with $C_{0}$ is a positive constant, independent on ε. Instead, under the Assumptions $γ = 0$ and (1.5), we take $\begin{array}{l} (2.3) & \begin{aligned} ‖ u_{ε, 0} ‖_{H^{2} (R)} ⩽ ‖ u_{0} ‖_{H^{2} (R)}, \\ ε ({‖ \partial_{x}^{2} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{4} u_{ε, 0} ‖}_{L^{2} (R)}) ⩽ C_{0}, \\ ε ({‖ \partial_{x}^{3} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{5} u_{ε, 0} ‖}_{L^{2} (R)}) ⩽ C_{0} . \end{aligned} \end{array}$ Let us prove some a priori estimates on $u_{ε}$ . We denote with $C_{0}$ the constants which depend only on the initial data, and with $C (T)$ the constants which depend also on T.

Lemma 2.1.
Assume $γ \neq 0$ or $γ = 0$ . For each $t ⩾ 0$ , $\begin{array}{l} (2.4) & {‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε \int_{0}^{t} {‖ \partial_{x}^{3} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C_{0} . \end{array}$ In particular, we have that $\begin{array}{l} (2.5) & {‖ u_{ε} (t, \cdot) ‖}_{L^{\infty} (R)}, {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}, {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{\infty} (R)} ⩽ C_{0} . \end{array}$
Proof.
Multiplying (2.1) by $2 u_{ε}$ , an integration on $R$ gives $\begin{array}{l} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = 2 \int_{R} u_{ε} \partial_{t} u_{ε} d x + 2 β^{2} \int_{R} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x \\ = \underset{= 0}{\underset{︸}{- 2 \int_{R} u_{ε} f^{'} (u_{ε}) \partial_{x} u_{ε} d x}} - 2 α \int_{R} u_{ε} \partial_{x}^{3} u_{ε} d x - 2 γ \int_{R} u_{ε} \partial_{x}^{5} u_{ε} d x + 2 ε \int_{R} u_{ε} \partial_{x}^{6} u_{ε} d x \\ = 2 α \int_{R} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x + 2 γ \int_{R} \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x - 2 ε \int_{R} \partial_{x} u_{ε} \partial_{x}^{5} u_{ε} d x \\ = - 2 γ \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d x + 2 ε \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x \\ = - 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Hence, $\begin{array}{l} \frac{d}{d t} ({‖ u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} = 0 . \end{array}$ (2.2) and an integration on $(0, t)$ give (2.4).

Finally, (2.5) follows from (2.4) and [17, Lemma 2.3]. □
Lemma 2.2.
Assume $γ \neq 0$ or $γ = 0$ . Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (2.6) & ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε^{2} e^{t} \int_{0}^{t} e^{- s} {‖ \partial_{x}^{5} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ .
Proof.
Let $0 ⩽ t ⩽ T$ . Multiplying (2.1) by $2 ε \partial_{x}^{4} u_{ε}$ , we have that $\begin{array}{l} 2 ε \partial_{t} u_{ε} \partial_{x}^{4} u_{ε} + 2 β^{2} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} = & - 2 ε f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} - 2 α \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} \\ (2.7) & - 2 γ \partial_{x}^{4} u_{ε} \partial_{x}^{5} u_{ε} + 2 ε \partial_{x}^{4} u_{ε} \partial_{x}^{6} u_{ε} . \end{array}$ Since $\begin{array}{l} 2 ε \int_{R} \partial_{t} u_{ε} \partial_{x}^{4} u_{ε} d x = ε \frac{d}{d t} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 β^{2} \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x = β^{2} ε \frac{d}{d t} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ - 2 α \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x = 0, \\ - 2 γ \int_{R} \partial_{x}^{4} u_{ε} \partial_{x}^{5} u_{ε} d x = 0, \\ 2 ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{x}^{6} u_{ε} d x = - 2 ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{array}$ an integration of (2.7) on $R$ gives $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (2.8) & = - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{4} u_{ε} d x . \end{array}$ Since $0 < ε < 1$ , thanks to (2.5) and the Young inequality, $\begin{array}{l} 2 ε \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{x}^{4} u_{ε} | d x \\ = 2 \int_{R} | \frac{f^{'} (u_{ε}) \partial_{x} u_{ε}}{β} | | ε β \partial_{x}^{4} u_{ε} (t, \cdot) | d x \\ ⩽ \frac{1}{β^{2}} \int_{R} {(f^{'} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{2} d x + ε^{2} β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{1}{β^{2}} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ It follows from (2.8) that $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ ε β^{2} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} \\ ⩽ ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} . \end{array}$ By the Gronwall Lemma and (2.2), we get $\begin{array}{l} ε {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε e^{t} \int_{0}^{t} e^{- s} {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} t ⩽ C (T), \end{array}$ which gives (2.6). □
Lemma 2.3.
Assume $γ \neq 0$ . Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (2.9) & {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε \int_{0}^{t} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ . In particular, we have $\begin{array}{l} (2.10) & {‖ \partial_{x}^{2} u_{ε} ‖}_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{array}$
Proof.
Let $0 ⩽ t ⩽ T$ . Multiplying (2.1), an integration on $R$ gives $\begin{array}{l} \frac{d}{d t} ({‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = - 2 \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} u_{ε} d x - 2 β^{2} \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x \\ = 2 \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x - 2 \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d x \\ - 2 γ \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{5} u_{ε} d x - 2 ε \int_{R} \partial_{x}^{2} u_{ε} \partial_{x}^{6} u_{ε} d x \\ = 2 \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x - 2 γ \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{4} u_{ε} d x \\ + 2 ε \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{5} u_{ε} d x \\ = 2 \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x - 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Therefore $\begin{array}{l} \frac{d}{d t} ({‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ (2.11) & + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} = 2 \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x . \end{array}$ Due to (2.4), (2.5) and the Young inequality, $\begin{array}{l} 2 \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{x}^{2} u_{ε} | d x \\ ⩽ \int_{R} {(f^{'} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{2} d x + {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C_{0} \\ ⩽ C_{0} (1 + {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2}) . \end{array}$ Consequently, by (2.11), $\begin{array}{l} \frac{d}{d t} ({‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} (1 + {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2}) . \end{array}$ It follows from (2.2) and an integration on $(0, t)$ that $\begin{array}{l} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + 2 ε \int_{0}^{t} {‖ \partial_{x}^{4} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C_{0} (1 + {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2}) t ⩽ C (T), \end{array}$ which gives (2.9).

Finally, we prove (2.10). Thanks to (2.4), (2.9) and the Hölder inequality, $\begin{array}{l} {(\partial_{x}^{2} u_{ε} (t, x))}^{2} = & 2 \int_{- \infty}^{x} \partial_{x}^{2} u_{ε} \partial_{x}^{3} u_{ε} d y ⩽ \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{x}^{3} u_{ε} | d x \\ ⩽ & {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) . \end{array}$ Hence, $\begin{array}{l} {‖ \partial_{x}^{2} u_{ε} ‖}_{L^{\infty} ((0, T) \times R)}^{2} ⩽ C (T), \end{array}$ which gives (2.10). □

Following [16, Lemma 2.1], we prove the following result.
Lemma 2.4.
Assume $γ \neq 0$ . Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (2.12) & \frac{1}{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{12} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ . In particular, $\begin{array}{l} (2.13) & {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)} ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ . Moreover, $\begin{array}{l} (2.14) & ‖ \partial_{t} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \\ (2.15) & ‖ \partial_{t} \partial_{x} u_{ε} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{array}$
Proof.
Let $0 ⩽ t ⩽ T$ . Multiplying (2.1) by $\partial_{t} u_{ε}$ , we have that $\begin{array}{l} (2.16) & {(\partial_{t} u_{ε})}^{2} = - f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{t} u_{ε} - α \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} - γ \partial_{x}^{5} u_{ε} \partial_{t} u_{ε} - β^{2} \partial_{t} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} + ε \partial_{x}^{6} u_{ε} \partial_{t} u_{ε} . \end{array}$ Since $\begin{array}{l} - α \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} u_{ε} d x = α \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x} u_{ε} d x, \\ - γ \int_{R} \partial_{x}^{5} u_{ε} \partial_{t} u_{ε} d x = - γ \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x, \\ - β^{2} \int_{R} \partial_{t} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x = - β^{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ ε \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} u_{ε} d x = ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x, \end{array}$ an integration on $R$ of (2.16) gives $\begin{array}{l} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{t} u_{ε} d x + α \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x} u_{ε} d x \\ (2.17) & - γ \int_{R} \partial_{x}^{3} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x + ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x . \end{array}$ Since $0 < ε < 1$ , due to (2.4), (2.5), (2.6) and the Young inequality, $\begin{array}{l} \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{t} u_{ε} | d x \\ ⩽ \frac{1}{2} \int_{R} {(f^{'} (u_{ε}) \partial_{x} u_{ε})}^{2} {(\partial_{x} u_{ε})}^{2} d x + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{1}{2} {‖ f^{'} ‖}_{(- C_{0}, C_{0})} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ | α | \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x \\ = 2 \int_{R} | \frac{α \partial_{x}^{2} u_{ε}}{2 \sqrt{A}} | | \sqrt{A} \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ \frac{α^{2}}{4 A} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + A {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{C_{0}}{A} + A {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ | γ | \int_{R} | \partial_{x}^{3} u_{ε} | | \partial_{t} \partial_{x}^{2} u_{ε} | d x \\ = \int_{R} | \frac{γ \partial_{x}^{3} u_{ε}}{β} | | β \partial_{t} \partial_{x}^{2} u_{ε} | d x \\ ⩽ \frac{γ^{2}}{2 β^{2}} {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C (T) + {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ ε \int_{R} | \partial_{x}^{4} u_{ε} | | \partial_{t} \partial_{x}^{2} u_{ε} | d x \\ = 2 \int_{R} | \frac{ε \sqrt{3} \partial_{x}^{4} u_{ε}}{2 β} | | \frac{β \partial_{t} \partial_{x}^{2} u_{ε}}{\sqrt{3}} | d x \\ ⩽ \frac{3 ε^{2}}{4 β^{2}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{3 ε}{4 β^{2}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{array}$ where A is a positive constant, which will specified later. It follows from (2.17) that $\begin{array}{l} \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{6} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (2.18) & ⩽ C (T) + \frac{C_{0}}{A} + A {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Observe that $\begin{array}{l} A {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} = A \int_{R} \partial_{t} \partial_{x} u_{ε} \partial_{t} \partial_{x} u_{ε} d x = - A \int_{R} \partial_{t} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x . \end{array}$ Consequently, by the Young inequality, $\begin{array}{l} A {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ & A \int_{R} | \partial_{t} u_{ε} | | \partial_{t} \partial_{x}^{2} u_{ε} | d x \\ = & 2 \int_{R} | \frac{\partial_{t} u_{ε}}{\sqrt{3}} | | \frac{A \sqrt{3} \partial_{t} \partial_{x}^{2} u_{ε}}{2} | d x \\ (2.19) & ⩽ & \frac{1}{3} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{3 A^{2}}{4} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Therefore, by (2.18), $\begin{array}{l} \frac{1}{6} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + (\frac{β^{2}}{6} - \frac{3 A^{2}}{4}) {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) + \frac{C_{0}}{A} . \end{array}$ Taking $\begin{array}{l} (2.20) & A = \frac{| β |}{3}, \end{array}$ we have (2.12).

(2.13) follows from (2.12), (2.19) and (2.20).

We prove (2.14). Due to (2.12), (2.13) and the Hölder inequality, $\begin{array}{l} {(\partial_{t} u_{ε} (t, x))}^{2} = & 2 \int_{- \infty}^{x} \partial_{t} u_{ε} \partial_{t} \partial_{x} u_{ε} d x ⩽ 2 \int_{R} | \partial_{t} u_{ε} | | \partial_{t} \partial_{x} u_{ε} | d x \\ ⩽ & 2 {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)} ⩽ C (T) . \end{array}$ Therefore, $\begin{array}{l} ‖ \partial_{t} u_{ε} ‖_{L^{\infty} ((0, T) \times R)}^{2} ⩽ C (T), \end{array}$ which gives (2.14).

Finally, (2.15) follows from (2.12), (2.13), the Hölder inequality and the following identity: $\begin{array}{l} {(\partial_{t} \partial_{x} u_{ε} (t, x))}^{2} = 2 \int_{- \infty}^{x} \partial_{t} \partial_{x} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x . \end{array}$ □
Lemma 2.5.
Assume $γ = 0$ . Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (2.21) & ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε \int_{0}^{t} {‖ \partial_{x}^{6} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ .
Proof.
Let $0 ⩽ t ⩽ T$ . We begin by observing that, since $γ = 0$ , (2.1) reads $\begin{array}{l} (2.22) & \partial_{t} u_{ε} + f^{'} (u_{ε}) \partial_{x} u_{ε} + α \partial_{x}^{3} u_{ε} + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = ε \partial_{x}^{6} u_{ε} . \end{array}$ Multiplying (2.22) by $- 2 ε \partial_{x}^{6} u_{ε}$ , an integration on $R$ gives $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ = - 2 ε \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} u_{ε} d x - 2 β^{2} ε \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x \\ = - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{6} u_{ε} d x - 2 α ε \int_{R} \partial_{x}^{3} u_{ε} \partial_{x}^{5} u_{ε} d x - 2 ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{6} u_{ε} d x + 2 α ε \int_{R} \partial_{x}^{4} u_{ε} \partial_{x}^{5} u_{ε} d x - 2 ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{6} u_{ε} d x - 2 ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Hence, $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ (2.23) & + 2 ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} = - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{6} u_{ε} d x . \end{array}$ Due to (2.5) and the Young inequality, $\begin{array}{l} 2 ε \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{x}^{6} u_{ε} | d x ⩽ & \int_{R} {(f^{'} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{2} d x + ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ & {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ & C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Consequently, by (2.23), $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ + ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} . \end{array}$ It follows from (2.3) and an integration on $(0, t)$ that $\begin{array}{l} ε {‖ \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε^{2} \int_{0}^{t} {‖ \partial_{x}^{6} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} t ⩽ C (T), \end{array}$ which gives (2.21). □

Following [13, Lemma 3.2], we prove the following result. Lemma 2.6.
Assume $γ = 0$ . Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{1}{4} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (2.24) & + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{4} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ . In particular, ( 2.14 ), ( 2.15 ), and $\begin{array}{l} (2.25) & {‖ \partial_{t} \partial_{x}^{2} u_{ε} ‖}_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{array}$ hold.
Proof.
Let $0 ⩽ t ⩽ T$ . Let B be a positive constant, which will specified later. Multiplying (2.22) by $\partial_{t} u_{ε} - B \partial_{t} \partial_{x}^{2} u_{ε}$ , arguing as in Lemma 2.4, we have that $\begin{array}{l} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + B {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ + β^{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + B β^{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ = - \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{t} u_{ε} d x + B \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{t} \partial_{x}^{2} u_{ε} d x \\ + α \int_{R} \partial_{t} \partial_{x} u_{ε} \partial_{x}^{2} u_{ε} d x - B α \int_{R} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x}^{3} u_{ε} d x \\ (2.26) & + ε \int_{R} \partial_{t} \partial_{x}^{2} u_{ε} \partial_{x}^{4} u_{ε} d x + B ε \int_{R} \partial_{t} \partial_{x}^{3} u_{ε} \partial_{x}^{5} u_{ε} d x . \end{array}$ Since $0 < ε < 1$ , thanks to (2.4), (2.5), (2.6), (2.21) and the Young inequality, $\begin{array}{l} \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{t} u_{ε} | d x ⩽ \frac{1}{2} \int_{R} {(f^{'} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{2} d x + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{1}{2} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ | α | \int_{R} | \partial_{t} \partial_{x} u_{ε} | | \partial_{x}^{2} u_{ε} | d x = \int_{R} | \sqrt{B} \partial_{t} \partial_{x} u_{ε} | | \frac{α \partial_{x}^{2} u_{ε}}{\sqrt{B}} | d x \\ ⩽ \frac{B}{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{α^{2}}{2 B} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{B}{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{C_{0}}{B}, \\ B | α | \int_{R} | \partial_{x}^{2} u_{ε} | | \partial_{t} \partial_{x}^{3} u_{ε} | d x = B \int_{R} | \frac{α \partial_{x}^{2} u_{ε}}{β} | | β \partial_{t} \partial_{x}^{3} u_{ε} | d x \\ ⩽ \frac{B α^{2}}{2 β^{2}} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{B β^{2}}{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} B + \frac{B β^{2}}{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ ε \int_{R} | \partial_{t} \partial_{x}^{2} u_{ε} | | \partial_{x}^{4} u_{ε} | d x = \int_{R} | β \partial_{t} \partial_{x}^{2} u_{ε} | | \frac{ε \partial_{x}^{4} u_{ε} (t, \cdot)}{β} | d x \\ ⩽ \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2 β^{2}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε}{2 β^{2}} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C (T), \\ B ε \int_{R} | \partial_{t} \partial_{x}^{3} u_{ε} | | \partial_{x}^{5} u_{ε} | d x = \int_{R} | B β \partial_{t} \partial_{x}^{3} u_{ε} | | \frac{ε \partial_{x}^{5} u_{ε}}{β} | d x \\ ⩽ \frac{B^{2} β^{2}}{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{B^{2} β^{2}}{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε}{2} {‖ \partial_{x}^{5} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{B^{2} β^{2}}{2} {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + C (T) . \end{array}$ It follows from (2.26) that $\begin{array}{l} \frac{1}{2} {‖ \partial_{t} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{B}{2} {‖ \partial_{t} \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{B β^{2}}{2} (1 - B) {‖ \partial_{t} \partial_{x}^{3} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (2.27) & ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + \frac{C_{0}}{B} + B C_{0} + C_{0} . \end{array}$ Choosing $B = \frac{1}{2}$ , (2.24) follows from (2.27).

Arguing as in Lemma 2.4, we have (2.14) and (2.15).

Finally, (2.25) follows from (2.24), the Hölder inequality and the following identity: $\begin{array}{l} {(\partial_{t} \partial_{x}^{2} u_{ε} (t, x))}^{2} = 2 \int_{- \infty}^{x} \partial_{t} \partial_{x}^{2} u_{ε} \partial_{t} \partial_{x}^{3} u_{ε} d y . \end{array}$ □

3. Proof of Theorem 1.1

This section is devoted to the proof of Theorem 1.1.

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

Lemma 3.1.
Fix $T > 0$ . Assume ( 1.3 ), ( 1.4 ) and $γ \neq 0$ . There exist a subsequence ${{(u_{ε_{k}}}}_{k \in N}$ of ${{(u_{ε}}}_{ε > 0}$ and an a limit function u which satisfies ( 1.12 ) such that $\begin{array}{l} (3.1) & u_{ε_{k}} \to u a.e. and in L_{loc}^{p} ((0, T) \times R), 1 ⩽ p < \infty, \end{array}$ Moreover, u is solution of ( 1.1 ).
Proof.
Let $0 ⩽ t ⩽ T$ . We begin by observing that, thanks to Lemmas 2.1, 2.3 and 2.4, $\begin{array}{l} (3.2) & {u_{ε}}_{ε > 0} is uniformly bounded in H^{1} ((0, T) \times R), \end{array}$ which gives (3.1).

Observe that, thanks to Lemmas 2.1 and 2.3, $\begin{array}{l} u \in L^{\infty} (0, T; H^{3} (R)), \end{array}$ while by Lemmas 2.1 and 2.4, we obtain that $\begin{array}{l} (3.3) & u \in W^{1, \infty} ((0, T) \times R) . \end{array}$ Moreover, by Lemma 2.4, we get $\begin{array}{l} \partial_{t} \partial_{x} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{2} u (t, \cdot) \in L^{2} (R), for every 0 ⩽ t ⩽ T . \end{array}$ Therefore, u is solution of (1.1) and (1.12) holds. □
Lemma 3.2.
Fix $T > 0$ . Assume ( 1.3 ), ( 1.5 ) and $γ = 0$ . There exist a subsequence ${u_{ε_{k}}}_{k \in N}$ of ${u_{ε}}_{ε > 0}$ and an a limit function u which satisfies ( 1.13 ) such that ( 3.1 ) holds. Moreover, u is solution of ( 1.1 ).
Proof.
Let $0 ⩽ t ⩽ T$ . We begin by observing that, thanks to Lemmas 2.1 and (2.6), we have (3.2) which gives (3.1).

Observe that, thanks to Lemma 2.1, $\begin{array}{l} u \in L^{\infty} (0, T; H^{2} (R)), \end{array}$ while by Lemmas 2.1 and 2.6, we obtain (3.3).

Moreover, by Lemma 2.6, we get $\begin{array}{l} u \in H^{1} ((0, T) \times R) \cap L^{\infty} (0, T; H^{2} (R)) \cap W^{1, \infty} ((0, T) \times R), \\ \partial_{t} \partial_{x} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{2} u (t, \cdot) \in L^{\infty} (R) \cap L^{2} (R), for every 0 ⩽ t ⩽ T, \\ \partial_{t} \partial_{x}^{3} u (t, \cdot) \in L^{2} (R), for every 0 ⩽ t ⩽ T . \end{array}$ Therefore, u is solution of (1.1) and (1.13) holds. □
Lemma 3.3.
If $f \in C^{2} (R)$ , then we have ( 1.14 ).
Proof.
Lemma 3.1 or 3.2 give the existence of a solution of (1.1), such that (1.12) or (1.13) hold. Let $u_{1}$ and $u_{2}$ be two solutions of (1.1), which verify (1.12) or (1.13), that is $\begin{array}{l} \{\begin{matrix} \partial_{t} u_{1} + f^{'} (u_{1}) \partial_{x} u_{1} + α \partial_{x}^{3} u_{1} + γ \partial_{x}^{5} u_{1} + β^{2} \partial_{t} \partial_{x}^{4} u_{1} = 0 & t > 0, x \in R, \\ u_{1} (0, x) = u_{1, 0} (x), & x \in R, \end{matrix} \\ \{\begin{matrix} \partial_{t} u_{2} + f^{'} (u_{2}) \partial_{x} u_{2} + α \partial_{x}^{3} u_{2} + γ \partial_{x}^{5} u_{2} + β^{2} \partial_{t} \partial_{x}^{4} u_{2} = 0 & t > 0, x \in R, \\ u_{2} (0, x) = u_{2, 0} (x), & x \in R . \end{matrix} \end{array}$ Then, the function $\begin{array}{l} (3.4) & ω (t, x) = u_{1} (t, x) - u_{2} (t, x), \end{array}$ is the solution of the following Cauchy problem: $\begin{array}{l} (3.5) & \{\begin{matrix} \partial_{t} ω + f^{'} (u_{1}) \partial_{x} u_{1} - f^{'} (u_{2}) \partial_{x} u_{2} + α \partial_{x}^{3} ω + γ \partial_{x}^{5} ω + β^{2} \partial_{t} \partial_{x}^{4} ω = 0, & t > 0, x \in R, \\ ω (0, x) = u_{1, 0} (x) - u_{2, 0} (x), & x \in R . \end{matrix} \end{array}$ Observe that, thanks to (3.4), $\begin{array}{l} f^{'} (u_{1}) \partial_{x} u_{1} - f^{'} (u_{2}) \partial_{x} u_{2} = & f^{'} (u_{1}) \partial_{x} u_{1} - f^{'} (u_{1}) \partial_{x} u_{2} + f^{'} (u_{1}) \partial_{x} u_{2} - f^{'} (u_{2}) \partial_{x} u_{2} \\ = & f^{'} (u_{1}) \partial_{x} ω + (f^{'} (u_{1}) - f^{'} (u_{2})) \partial_{x} u_{2} . \end{array}$ Therefore, (3.5) is equivalent to the following equation: $\begin{array}{l} (3.6) & \partial_{t} ω + f^{'} (u_{1}) \partial_{x} ω + (f^{'} (u_{1}) - f^{'} (u_{2})) \partial_{x} u_{2} + α \partial_{x}^{3} ω + γ \partial_{x}^{5} ω + β^{2} \partial_{t} \partial_{x}^{4} ω = 0 \end{array}$ Since $u_{1}, u_{2} \in L^{\infty} (0, T; H^{3} (R))$ or $u_{1}, u_{2} \in L^{\infty} (0, T; H^{2} (R))$ , there exists a constant $C (T)$ such that $\begin{array}{l} (3.7) & \begin{aligned} ‖ u_{1} ‖_{L^{\infty} ((0, T) \times R)} + ‖ u_{1} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T), \\ ‖ \partial_{x} u_{1} ‖_{L^{\infty} ((0, T) \times R)}, ‖ \partial_{x} u_{2} ‖_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{aligned} \end{array}$ Moreover, there exists ξ between $u_{1}$ and $u_{2}$ such that $\begin{array}{l} (3.8) & f^{'} (u_{1}) - f (u_{2}) = f^{″} (ξ) (u_{1} - u_{2}) = f^{″} (ξ) ω, u_{1} < ξ < u_{2}, or u_{2} < χ < u_{1} . \end{array}$ while, by (3.7), $\begin{array}{l} (3.9) & \begin{aligned} | f^{″} (ξ) | ⩽ {‖ f^{″} ‖}_{L^{\infty} (- C (T), C (T))} ⩽ C (T), \\ | f^{″} (u_{1}) | ⩽ {‖ f^{″} ‖}_{L^{\infty} (- C (T), C (T))} ⩽ C (T) . \end{aligned} \end{array}$ Since $\begin{array}{l} 2 \int_{R} f^{'} (u_{1}) ω \partial_{x} ω d x = - \int_{R} f^{″} (u_{1}) \partial_{x} u_{1} ω^{2} d x, \\ 2 α \int_{R} ω \partial_{x}^{3} ω d x = - 2 α \int_{R} \partial_{x} ω \partial_{x}^{2} ω d x = 0, \\ 2 γ \int_{R} ω \partial_{x}^{5} ω d x = - 2 γ \int_{R} \partial_{x} ω \partial_{x}^{4} ω d x = 2 γ \int_{R} \partial_{x}^{2} ω \partial_{x}^{3} ω = 0, \\ 2 β^{2} \int_{R} ω \partial_{t} \partial_{x}^{4} ω d x = - 2 β^{2} \int_{R} \partial_{x} ω \partial_{t} \partial_{x}^{3} ω d x = β^{2} \frac{d}{d t} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{array}$ multiplying (3.6) by $2 ω$ , (3.8) and an integration on $R$ give $\begin{array}{l} \frac{d}{d t} ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ (3.10) & = - \int_{R} f^{″} (u_{1}) \partial_{x} u_{1} ω^{2} d x + \int_{R} f^{″} (ξ) \partial_{x} u_{2} ω^{2} d x . \end{array}$ Due to (3.7) and (3.9), $\begin{array}{l} \int_{R} | f^{″} (u_{1}) | | \partial_{x} u_{1} | ω^{2} d x ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ \int_{R} | f^{″} (ξ) | | \partial_{x} u_{2} | ω^{2} d x ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ Hence, by (3.10), $\begin{array}{l} \frac{d}{d t} ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) \\ ⩽ C (T) {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T) ({‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2}) . \end{array}$ The Gronwall Lemma gives $\begin{array}{l} {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ e^{C (T) t} (‖ ω_{0} ‖_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{x}^{2} ω_{0} ‖}_{L^{2} (R)}^{2}) . \end{array}$ Thanks to (1.15), $\begin{array}{l} {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{τ_{2}^{2} e^{C (T) t}}{τ_{1}^{2}} (‖ ω_{0} ‖_{L^{2} (R)}^{2} + {‖ \partial_{x}^{2} ω_{0} ‖}_{L^{2} (R)}^{2}) \\ ⩽ \frac{τ_{2}^{2} e^{C (T) t}}{τ_{1}^{2}} (‖ ω_{0} ‖_{L^{2} (R)}^{2} + ‖ \partial_{x} ω_{0} ‖_{L^{2} (R)}^{2} + {‖ \partial_{x}^{2} ω_{0} ‖}_{L^{2} (R)}^{2}) \\ (3.11) & = \frac{τ_{2}^{2}}{τ_{1}^{2}} e^{C (T) t} ‖ ω_{0} ‖_{H^{2} (R)}^{2} . \end{array}$ Observe that $\begin{array}{l} {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} = \int_{R} \partial_{x} ω \partial_{x} ω d x = - \int_{R} ω \partial_{x}^{2} ω d x . \end{array}$ Therefore, by (3.11) and the Young inequality, $\begin{array}{l} {‖ \partial_{x} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ & \frac{1}{2} {‖ ω (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{1}{2} {‖ \partial_{x}^{2} ω (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (3.12) & ⩽ & \frac{τ_{2}^{2}}{2 τ_{1}^{2}} e^{C (T) t} ‖ ω_{0} ‖_{H^{2} (R)}^{2} . \end{array}$ Adding (3.11) and (3.12), we have $\begin{array}{l} (3.13) & {‖ ω (t, \cdot) ‖}_{H^{2} (R)}^{2} ⩽ \frac{3 τ_{2}^{2}}{2 τ_{1}^{2}} e^{C (T) t} ‖ ω_{0} ‖_{H^{2} (R)}^{2} \end{array}$ (3.4), (3.5) and (3.13) give (1.14). □
Proof of Theorem 1.1..
Theorem 1.1 follows from Lemmas 3.1, 3.2 and 3.3. □

4. Proof of Corollary 1.1

In this section, we prove Corollary 1.1. We begin by observing that, since $γ = α = 0$ , then (1.1) redas $\begin{array}{l} (4.1) & \{\begin{matrix} \partial_{t} u + \partial_{x} f (u) + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = 0, & t > 0, x \in R, \\ u (0, x) = u_{0} (x), & x \in R . \end{matrix} \end{array}$

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

Fix a small number $0 < ε < 1$ and let $u_{ε} = u_{ε} (t, x)$ be the unique classical solution of the following problem: $\begin{array}{l} (4.2) & \{\begin{matrix} \partial_{t} u_{ε} + \partial_{x} f (u_{ε}) + β^{2} \partial_{t} \partial_{x}^{4} u_{ε} = ε \partial_{x}^{6} u_{ε}, & t > 0, x \in R, \\ u_{ε} (0, x) = u_{ε, 0} (x), & x \in R . \end{matrix} \end{array}$ where $u_{ε, 0}$ is a $C^{\infty}$ approximation of $u_{0}$ , such that $\begin{array}{l} (4.3) & \begin{aligned} ‖ u_{ε, 0} ‖_{H^{2} (R)} ⩽ ‖ u_{0} ‖_{H^{2} (R)}, ε ({‖ \partial_{x}^{2} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{4} u_{ε, 0} ‖}_{L^{2} (R)}) ⩽ C_{0}, \\ ε ({‖ \partial_{x}^{3} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{5} u_{ε, 0} ‖}_{L^{2} (R)} + {‖ \partial_{x}^{6} u_{ε, 0} ‖}_{L^{2} (R)}) ⩽ C_{0} . \end{aligned} \end{array}$ with $C_{0}$ is a positive constant, independent on ε.

Let us prove some a priori estimates on $u_{ε}$ .

Observe that, since $f \in C^{2} (R)$ , then $f \in C^{1} (R)$ . Consequently, Lemma 2.1, 2.2, 2.5 and 2.6 are still valid.

We prove the following result.

Lemma 4.1.

Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (4.4) & ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} \int_{0}^{t} {‖ \partial_{x}^{7} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ .

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (4.2) by $2 ε \partial_{x}^{8} u_{ε}$ , we have that $\begin{array}{l} (4.5) & 2 ε \partial_{x}^{8} u_{ε} \partial_{t} u_{ε} + 2 β^{2} ε \partial_{x}^{8} u_{ε} \partial_{t x}^{5} u_{ε} = - 2 ε f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{8} u_{ε} + 2 ε^{2} \partial_{x}^{6} u_{ε} \partial_{x}^{8} u_{ε} . \end{array}$ Since $\begin{array}{l} 2 ε \int_{R} \partial_{x}^{8} u_{ε} \partial_{t} u_{ε} d x = ε \frac{d}{d t} {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 β^{2} ε \int_{R} \partial_{x}^{8} u_{ε} \partial_{t x}^{5} u_{ε} d x = β^{2} ε \frac{d}{d t} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ - 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{x}^{8} u_{ε} d x = 2 ε \int_{R} f^{″} (u_{ε}) {(\partial_{x} u_{ε})}^{2} \partial_{x}^{7} u_{ε} d x + 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x}^{2} u_{ε} \partial_{x}^{7} u_{ε} d x, \\ 2 ε^{2} \int_{R} \partial_{x}^{6} u_{ε} \partial_{x}^{8} u_{ε} d x = - 2 ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \end{array}$ an integration on $R$ of (4.5) gives $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + 2 ε^{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (4.6) & = 2 ε \int_{R} f^{″} (u_{ε}) {(\partial_{x} u_{ε})}^{2} \partial_{x}^{7} u_{ε} d x + 2 ε \int_{R} f^{'} (u_{ε}) \partial_{x}^{2} u_{ε} \partial_{x}^{7} u_{ε} d x . \end{array}$ Due to (2.4), (2.5) and the Young inequality, $\begin{array}{l} 2 ε \int_{R} | f^{″} (u_{ε}) {(\partial_{x} u_{ε})}^{2} | | \partial_{x}^{7} u_{ε} | d x \\ ⩽ \int_{R} {(f^{″} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{4} d x + ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ {‖ f^{″} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} \int_{R} {(\partial_{x} u_{ε})}^{4} d x + ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ {‖ f^{″} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{\infty} (R)}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{″} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + ε^{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ 2 ε \int_{R} | f^{'} (u_{ε}) \partial_{x}^{2} u_{ε} | | \partial_{x}^{7} u_{ε} | d x \\ = \int_{R} | 2 f^{'} (u_{ε}) \partial_{x}^{2} u_{ε} | | ε \partial_{x}^{7} u_{ε} | d x \\ ⩽ 2 \int_{R} {(f^{'} (u_{ε}))}^{2} {(\partial_{x}^{2} u_{ε})}^{2} d x + \frac{ε^{2}}{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ 2 {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + \frac{ε^{2}}{2} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} . \end{array}$ It follows from (4.6) that $\begin{array}{l} \frac{d}{d t} (ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}) + \frac{ε^{2}}{2} {‖ \partial_{x}^{7} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{″} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} . \end{array}$ Integrating on $(0, t)$ , by (4.3), we have that $\begin{array}{l} ε {‖ \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} ε {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{ε^{2}}{2} \int_{0}^{t} {‖ \partial_{x}^{7} u_{ε} (s, \cdot) ‖}_{L^{2} (R)}^{2} d s \\ ⩽ C_{0} + C_{0} {‖ f^{″} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} t + C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} t ⩽ C (T), \end{array}$ which gives (4.4). □

Lemma 4.2.

Fix $T > 0$ . There exists a constant $C (T) > 0$ , independent on ε, such that $\begin{array}{l} (4.7) & {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} ⩽ C (T), \end{array}$ for every $0 ⩽ t ⩽ T$ . In particular, $\begin{array}{l} (4.8) & {‖ \partial_{t} \partial_{x}^{3} u_{ε} ‖}_{L^{\infty} ((0, T) \times R)} ⩽ C (T) . \end{array}$

Proof.

Let $0 ⩽ t ⩽ T$ . Multiplying (4.2) by $\partial_{t} \partial_{x}^{4} u_{ε}$ , an integration on $R$ gives $\begin{array}{l} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + β^{2} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ (4.9) & = - \int_{R} f^{'} (u_{ε}) \partial_{x} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x + ε \int_{R} \partial_{x}^{6} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x . \end{array}$ Since $0 < ε < 1$ , thanks to (2.5) and (4.4), $\begin{array}{l} \int_{R} | f^{'} (u_{ε}) \partial_{x} u_{ε} | | \partial_{t} \partial_{x}^{4} u_{ε} | d x \\ = \int_{R} | \frac{f^{'} (u_{ε}) \partial_{x} u_{ε}}{β} | | β^{2} \partial_{t x}^{5} u_{ε} | d x \\ ⩽ \frac{1}{2 β^{2}} \int_{R} ({(f^{'} (u_{ε}))}^{2} {(\partial_{x} u_{ε})}^{2} d x + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ \frac{1}{2 β^{2}} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} {‖ \partial_{x} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + \frac{β^{2}}{2} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2}, \\ ε \int_{R} | \partial_{x}^{6} u_{ε} | | \partial_{t} \partial_{x}^{4} u_{ε} | d x \\ = 2 \int_{R} | \frac{\sqrt{3} ε \partial_{x}^{6} u_{ε}}{2 β} | | \frac{β \partial_{t} \partial_{x}^{4} u_{ε}}{\sqrt{3}} | d x \\ ⩽ \frac{3 ε^{2}}{4 β^{2}} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} d x \\ ⩽ \frac{3 ε}{4 β^{2}} {‖ \partial_{x}^{6} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} d x \\ ⩽ C (T) + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} d x . \end{array}$ Consequently, by (4.9), we have $\begin{array}{l} {‖ \partial_{t} \partial_{x}^{2} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} + \frac{β^{2}}{3} {‖ \partial_{t} \partial_{x}^{4} u_{ε} (t, \cdot) ‖}_{L^{2} (R)}^{2} \\ ⩽ C_{0} {‖ f^{'} ‖}_{L^{\infty} (- C_{0}, C_{0})}^{2} + C (T) ⩽ C (T), \end{array}$ which gives (4.7).

Finally, (4.8) follows from (2.24), (4.7), the Hölder inequality and the following identity: $\begin{array}{l} {(\partial_{t} \partial_{x}^{3} u_{ε} (t, x))}^{2} = 2 \int_{- \infty}^{x} \partial_{t} \partial_{x}^{3} u_{ε} \partial_{t} \partial_{x}^{4} u_{ε} d x . \end{array}$ □

Proof of Corollary 1.1. .

Thanks to Lemma 4.2, arguing as in Theorem 1.1, the proof is concluded. □

Footnotes

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

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On the classical solutions for a Rosenau–Korteweg-deVries–Kawahara type equation

Abstract

Keywords

1. Introduction

Footnotes

Acknowledgements

References