Abstract
The Olver–Benney equation is a nonlinear fifth-order equation, which describes the interaction effects between short and long waves. In this paper, we prove the global existence of solutions of the Cauchy problem associated with this equation.
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1 Introduction
In this study, we investigate the existence of solutions of the following Cauchy problem:
$$\begin{aligned} {\left\{ \begin{array}{ll} {\partial _t}u +\alpha \partial _x u\partial _{x}^2u +\beta u\partial _{x}^3u +\gamma \partial _{x}^5u =0, &{}\quad t>0, \quad x\in {\mathbb {R}},\\ u(0,x)=u_0(x),&{}\quad x\in {\mathbb {R}}, \end{array}\right. } \end{aligned}$$
(1.1)
with
$$\begin{aligned} \beta&=-2\alpha , \quad \gamma \ne 0, \quad {\text {or}} \end{aligned}$$
(1.2)
$$\begin{aligned} \alpha&=2\beta , \quad \gamma \ne 0, \quad {\text {or}} \end{aligned}$$
(1.3)
$$\begin{aligned} \beta&=2\alpha , \quad \gamma \ne 0. \end{aligned}$$
(1.4)
On the initial datum, we assume
$$\begin{aligned} u_0(x)\in H^4({\mathbb {R}}). \end{aligned}$$
(1.5)
From a physical point of view, (1.1) was derived in the context of water waves by Olver [16, 17] (see also [10, 18]), using Hamiltonian perturbation theory, with further generalization given by Craig and Groves [8], while, under Assumption (1.3), (1.1) was derived by Benney [1] as a model to describe the interaction effects between short and long waves.
[17] shows that (1.1) is a particular case of the following equation:
$$\begin{aligned} {\partial _t}u + \kappa u\partial _x u +q u^2\partial _x u +\tau \partial _{x}^3u +\alpha \partial _x u\partial _{x}^2u +\beta u\partial _{x}^3u +\gamma \partial _{x}^5u =0. \end{aligned}$$
(1.6)
If \(q=\alpha =\beta =\gamma =0\) in (1.6), (1.6) becomes the Korteweg–de Vries equation [11], whose the well-posedness is studied in [4].
Instead, if \(\alpha =\beta =0\), (1.6) becomes the Kawahara–Korteweg–de Vries type equation, which was derived by Kawahara [9] to describe small-amplitude gravity capillary waves on water of a finite depth when the Weber number is close to 1/3 (see [15]). In [2], the well-posedness of the Cauchy problem for the Kawahara–Korteweg–de Vries type equation is studied.
Moreover, assuming \(\kappa =q=\gamma =0\) and (1.2), (1.6) reads
$$\begin{aligned} {\partial _t}u +\tau \partial _{x}^3u_\varepsilon +\alpha \partial _x u\partial _{x}^2u -2\alpha u\partial _{x}^3u=0. \end{aligned}$$
(1.7)
It is a particular case of the Kudryashov–Sinelshchikov equation [5, 12], which describes pressure waves in liquids with gas bubbles taking into account heat transfer and viscosity. In [5], the existence of solutions of the Cauchy problem is proven.
From a mathematical point of view, under suitable assumptions on \(\kappa ,\,q,\,\tau ,\,\alpha ,\,\beta ,\,\gamma\), the existence of the travelling waves solutions for (1.6) is proven in [14, 20], while a method to find exact solutions of (1.6) is given in [13]. Instead, in [19], the local well-posedness of the Cauchy problem of (1.1) is proven.
The main result of this paper is the following theorem.
Theorem 1.1
Assume (1.2), or (1.3), or (1.4) and (1.5). There exists an unique solution u of (1.1) such that
$$\begin{aligned} u \in L^{\infty }(0,T;H^4({\mathbb {R}})),\quad T\ge 0. \end{aligned}$$
(1.8)
We remind that [19] the local in time well-posedness is \(H^s,\,s\ge 4\). Here, Theorem 1.1 gives the global well-posedness of the solution of the Cauchy problem of (1.1), under Assumptions (1.2), (1.3) and (1.4).
Since we are able to prove estimates only on the spatial derivatives of (1.1), the proof of Theorem 1.1 is based on the Aubin–Lions lemma (see [3, 6, 7, 21]), which requires only the \(H^{-1}\) boundedness of the time derivative.
One of the main point of our argument is the invariance of the energy space \(H^4\) (see Lemma 2.4). The key point in that direction is the \(H^2\) regularity. Assumptions (1.2), (1.3) and (1.4) are needed for that purpose. Indeed, assuming (1.2), (1.1) preserves (see Lemma 2.1):
$$\begin{aligned} t\rightarrow \int _{{\mathbb {R}}}(\partial _{x}^2u)^2 {\mathrm{d}}x, \end{aligned}$$
while assuming (1.3), the same equation preserves (see Lemma 3.1):
$$\begin{aligned} t\rightarrow \int _{{\mathbb {R}}} u^2 {\mathrm{d}}x, \end{aligned}$$
and if (1.4) holds, it preserves (see Lemma 4.1):
$$\begin{aligned} t\rightarrow \int _{{\mathbb {R}}} (\partial _x u)^2 {\mathrm{d}}x. \end{aligned}$$
So we can say that the assumptions on the constants are needed for the \(H^2\) regularity of u and only indirectly for the \(H^4\) one.
Observe again that, (1.5) is the same assumption to prove the well-posedness of the Cauchy problem for the Kawahara equation (see [2]). [2, Appendix A] shows that, for the Kawahara equation, it is possible also to assume on the initial datum also
$$\begin{aligned} u_0\in H^{5}({\mathbb {R}}). \end{aligned}$$
(1.9)
and obtain the well-posedness of the classical solution (see [2, Theorem A.1]). It is not possible to show a similar result for (1.1), under Assumption (1.9), due to the term \(\alpha \partial _x u\partial _{x}^2u +\beta u\partial _{x}^3u\).
The paper is organized as follows. In Sect. 2, we prove Theorem 1.1 under Assumption (1.2), while in Sects. 3 and 4, we prove Theorem 1.1, under Assumptions (1.3) and (1.4), respectively.
2 Proof of Theorem 1.1 under Assumption (1.2).
In section, we prove Theorem 1.1 under Assumption (1.2).
Our existence argument is based on passing to the limit in a vanishing viscosity approximation of (1.1).
Fix a small number \(0<\varepsilon <1\) and let \(u_\varepsilon =u_\varepsilon (t,x)\) be the unique classical solution of the following problem [2, 4, 19]:
$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle {\partial _t}u_\varepsilon +\alpha \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon +\beta u_\varepsilon \partial _{x}^3u_\varepsilon +\gamma \partial _{x}^5u_\varepsilon =\varepsilon \partial _{x}^6u_\varepsilon , \quad &{}t>0,\, x\in {\mathbb {R}},\\ u_{\varepsilon }(0,x)=u_{\varepsilon ,0}(x), \quad &{}x\in {\mathbb {R}}, \end{array}\right. } \end{aligned}$$
(2.1)
where \(u_{\varepsilon ,0}\) is a \(C^{\infty }\) approximation of \(u_0\) such that
$$\begin{aligned} \begin{aligned} \left\| u_{\varepsilon ,0} \right\| _{H^4({\mathbb {R}})}\le \left\| u_0 \right\| _{H^4({\mathbb {R}})}. \end{aligned} \end{aligned}$$
(2.2)
Let us prove some a priori estimates on \(u_\varepsilon\). 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 (1.2). For each \(t\ge 0\),
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C_0. \end{aligned}$$
(2.3)
Proof
Multiplying (2.1) by \(2\partial _{x}^4u_\varepsilon\), an integration on \({\mathbb {R}}\) gives
$$\begin{aligned} \frac{\mathrm{{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&2\int _{{\mathbb {R}}}\partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x \\ =\,&-2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -2\beta \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\\&-2\gamma \int _{{\mathbb {R}}}\partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+2\varepsilon \int _{{\mathbb {R}}}\partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\\ =\,&\left( 2\alpha +\beta \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x-2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Hence,
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\left( 2\alpha +\beta \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
Integrating on (0, t), thanks to (1.2) and (2.2), we have (2.3). \(\square\)
Lemma 2.2
Fix \(T>0\) and assume (1.2). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon e^{\frac{5t}{4}}\int _{0}^{t}e^{-\frac{5s}{4}}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(2.4)
for every \(0\le t\le T\). In particular, we have
$$\begin{aligned} \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})},\, \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})},\, \left\| \partial _x u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\le C(T). \end{aligned}$$
(2.5)
The proof of the previous lemma is based on the regularity of the functions \(u_\varepsilon\) and the following result.
Lemma 2.3
For each \(t\ge 0\), we have that
$$\begin{aligned} \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\le 2\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^7_{L^2({\mathbb {R}})}}. \end{aligned}$$
(2.6)
Proof
We begin by observing that, thanks to the regularity of \(u_\varepsilon\) and the Hölder inequality,
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=&\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _x u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\\ \le&\int _{{\mathbb {R}}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert {\mathrm{d}}x\le \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
Consequently,
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\le \sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}. \end{aligned}$$
(2.7)
Moreover, again by the regularity of \(u_\varepsilon\) and the Hölder inequality,
$$\begin{aligned} (\partial _x u_\varepsilon (t,x))^2=\,&2\int _{-\infty }^{x}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}y\le \, 2\int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert {\mathrm{d}}y\\ \le\,&2\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
Hence,
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le 2\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
(2.8)
It follows from (2.7) and (2.8) that
$$\begin{aligned} \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\le\,&\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&2\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}\\ \le\,&2\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^7_{L^2({\mathbb {R}})}}, \end{aligned}$$
which gives (2.6). \(\square\)
Proof of Lemma 2.2
Let \(0\le t\le T\). Multiplying (2.1) by 2u, an integration on \({\mathbb {R}}\) gives
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&2\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\\ =\,&-2\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x -2\beta \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _{x}^3u_\varepsilon {\mathrm{d}}x\\&-2\gamma \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x +2\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\\ =\,&2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x +2\gamma \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\\&-2\varepsilon \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\ =\,&2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x-2\gamma \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\\&+2\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\\ =\,&2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x-2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Hence,
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(2.9)
Due to (2.3), (2.6) and the Young inequality,
$$\begin{aligned} 2\left| \alpha -2\beta \right| \int _{{\mathbb {R}}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \vert {\mathrm{d}}x=\,&2\int _{{\mathbb {R}}}\vert u_\varepsilon \vert \left| (\alpha -2\beta )\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \right| {\mathrm{d}}x\\ \le\,&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+(\alpha -2\beta )^2\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\\ \le\,&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+ 2(\alpha -2\beta )^2\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^7_{L^2({\mathbb {R}})}}\\ \le\,&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C_0\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}+C_0\\ \le\,&\frac{5}{4}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C_0. \end{aligned}$$
Consequently, by (2.9),
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \frac{5}{4}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C_0. \end{aligned}$$
The Gronwall lemma and (2.2) give
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}&+2\varepsilon e^{\frac{5t}{4}}\int _{0}^{t}e^{-\frac{5s}{4}}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\ \le\,&C_0 e^{\frac{5t}{4}}+C_0 e^{\frac{5t}{4}}\int _{0}^{t}e^{-\frac{5t}{4}}{\mathrm{d}}s\le C(T). \end{aligned}$$
Therefore, (2.4) is proven.
Finally, we prove (2.5). Thanks to (2.4) and the Hölder inequality,
$$\begin{aligned} \begin{aligned} u_\varepsilon ^2(t,x)=&2\int _{-\infty }^{x}u_\varepsilon \partial _x u_\varepsilon {\mathrm{d}}y\le 2\int _{{\mathbb {R}}}\vert u_\varepsilon \vert \vert \partial _x u_\varepsilon \vert {\mathrm{d}}x\\ \le\,&2\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned} \end{aligned}$$
(2.10)
Hence, by (2.4),
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le C(T)\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
(2.11)
(2.5) follows from (2.3), (2.4), (2.7), (2.8) and (2.11). \(\square\)
Lemma 2.4
Fix \(T>0\) and assume (1.2). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\le C(T). \end{aligned}$$
(2.12)
In particular, we have
$$\begin{aligned} \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{{\mathbb {R}}}+\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^7u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(2.13)
for every \(0\le t\le T\). Moreover,
$$\begin{aligned} \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})},\, \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})},\, \left\| \partial _{x}^3u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\le C(T), \end{aligned}$$
(2.14)
for every \(0\le t\le T\).
The proof of the previous lemma is based on the regularity of the functions \(u_\varepsilon\) and the following result.
Lemma 2.5
For each \(t\ge 0\), we have that
$$\begin{aligned} \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\le&\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}},\end{aligned}$$
(2.15)
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\le&\sqrt{2}\root 4 \of {\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\root 4 \of {\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}},\end{aligned}$$
(2.16)
$$\begin{aligned} \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\le&\sqrt{2}\root 4 \of {\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\root 4 \of {\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}. \end{aligned}$$
(2.17)
In particular, we have
$$\begin{aligned} \int _{{\mathbb {R}}}\vert \partial _{x}^3u_\varepsilon \vert ^3 {\mathrm{d}}x \le \sqrt{2}\root 4 \of {\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^5_{L^2({\mathbb {R}})}}\root 4 \of {\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^7_{L^2({\mathbb {R}})}}. \end{aligned}$$
(2.18)
Moreover, fixed \(T>0\), there exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(2.19)
for every \(0\le t\le T\).
Proof
Arguing as in [2, Lemma 2.5], we have (2.15), (2.16) and (2.17).
Finally, we prove (2.19). Fix \(T>0\). Thanks to the regularity of \(u_\varepsilon\) and the Hölder inequality,
$$\begin{aligned} \varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&\varepsilon \int _{{\mathbb {R}}}\partial _{x}^4u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x=-\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\ \le\,&\varepsilon \int _{{\mathbb {R}}}\vert \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x \le \varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
Consequently, by the Young inequality,
$$\begin{aligned} \varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \frac{\varepsilon }{2}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Integrating on (0, t), by (2.3) and (2.4), we have that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le\,&\frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s +\frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\ \le\,&\frac{\varepsilon }{2}e^{\frac{5t}{4}}\int _{0}^{t}e^{-\frac{5s}{4}}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s +C_0\\ \le\,&C(T)+C_0\le C(T), \end{aligned}$$
that is (2.19). \(\square\)
Proof of Lemma 2.4
Let \(0\le t\le T\). Consider five real constants \(A,\,B,\,C,\,D,\,E,\), which will be specified later. Multiplying (2.1) by
$$\begin{aligned} 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2), \end{aligned}$$
thanks to (1.2), we have
$$\begin{aligned} \begin{aligned}&\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \right) {\partial _t}u_\varepsilon \\&\qquad \quad +\left( D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\right) {\partial _t}u_\varepsilon \\&\qquad \quad +\alpha \left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \\&\qquad \quad +\alpha \left( D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \\&\qquad \quad -2\alpha \left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \right) u_\varepsilon \partial _{x}^3u_\varepsilon \\&\qquad \quad -2\alpha \left( D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\right) u_\varepsilon \partial _{x}^3u_\varepsilon \\&\qquad \quad +\gamma \left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \right) \partial _{x}^5u_\varepsilon \\&\qquad \quad +\gamma \left( D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\right) \partial _{x}^5u_\varepsilon \\&\qquad =\varepsilon \left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \right) \partial _{x}^6u_\varepsilon \\&\qquad \quad +\varepsilon \left( D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon +E\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\right) \partial _{x}^6u_\varepsilon . \end{aligned} \end{aligned}$$
(2.20)
Observe that
$$\begin{aligned}&\int _{{\mathbb {R}}}\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \right) {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+A\int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x +B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^6u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +C\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+D\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x,\nonumber \\&\alpha \int _{{\mathbb {R}}}\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-2\alpha \int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2\partial _{x}^7u_\varepsilon {\mathrm{d}}x - 2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^7u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x -B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -B\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^5u_\varepsilon {\mathrm{d}}x -B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x -2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =6\alpha \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x +2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x+3B\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x+2B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x+\left( C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-6\alpha \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2\partial _{x}^5u_\varepsilon {\mathrm{d}}x -8\alpha \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x +A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x\nonumber \\&\qquad \quad +3B\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x+B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x-2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \end{aligned}$$
(2.21)
$$\begin{aligned}&=10\alpha \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x -2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x+3B\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x+3B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x+\left( C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x,\nonumber \\&-2\alpha \int _{{\mathbb {R}}}\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \right) u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =4\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^7u_\varepsilon {\mathrm{d}}x +4\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^7u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad -2A\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x +4B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2B\alpha \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x -2\left( C-D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +2D\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\\end{aligned}$$
$$\begin{aligned}&\qquad =-4\alpha \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x -8\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x -4\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^5u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad -2A\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x-4B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad -4B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -6B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad -2\left( C-D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2D\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad =4\alpha \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2\partial _{x}^5u_\varepsilon {\mathrm{d}}x +12\alpha \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x +10\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 dx\nonumber \\&\qquad \quad -2A\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x-4B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad -4B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -6B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad -2\left( C-D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2D\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad =-14\alpha \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +10\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -2A\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x-4B\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad -4B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -6B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad -2\left( C-D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x +2D\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2D\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x,\nonumber \\&\gamma \int _{{\mathbb {R}}}\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \right) \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\\end{aligned}$$
$$\begin{aligned}&\qquad =-\left( 2A\gamma +\frac{C\gamma }{2}\right) \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2{\mathrm{d}}x -\left( \frac{B\gamma }{2}+D\gamma \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\varepsilon \int _{{\mathbb {R}}}\left( 2\partial _{x}^8u_\varepsilon +A(\partial _{x}^3u_\varepsilon )^2+Bu_\varepsilon \partial _{x}^6u_\varepsilon +C\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon +D\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon \right) \partial _{x}^6u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-2A\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+B\varepsilon \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad +C\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x -\frac{D\varepsilon }{2}\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.21)
Moreover, since
$$\begin{aligned} \partial _x ((\partial _{x}^2u_\varepsilon )^2)=2\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon , \quad \partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)=2(\partial _{x}^3u_\varepsilon )^2+2\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon , \end{aligned}$$
we have that
$$\begin{aligned} E\alpha \int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x=\,&2E\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x \nonumber \\&+2E\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\ =&-2E\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x,\nonumber \\ -2E\alpha \int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x =&-4E\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x -4E\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\ =&-4E\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x+2E\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&+2E\alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x+2E\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x,\nonumber \\ E\gamma \int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\partial _{x}^5u_\varepsilon {\mathrm{d}}x =\,&2E\gamma \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2\partial _{x}^5u_\varepsilon {\mathrm{d}}x +2E\gamma \int _{{\mathbb {R}}}\partial _{x}^2\partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\ =\,&-5E\gamma \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x,\nonumber \\ E\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2)\partial _{x}^6u_\varepsilon {\mathrm{d}}x =\,&2E\varepsilon \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2\partial _{x}^6u_\varepsilon {\mathrm{d}}x +2E\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\nonumber \\ =\,&-6E\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x -2E\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.22)
It follows from (2.21), (2.22) and an integration of (2.20) that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+A\int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x +B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^6u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +C\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+D\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +E\int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2){\partial _t}u_\varepsilon {\mathrm{d}}x+2\varepsilon \left\| \partial _{x}^6u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 4\alpha +2A\gamma +\frac{C\gamma }{2}+5E\gamma \right) \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( \frac{B\gamma }{2}+D\gamma -8\alpha \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x -\left( 3B+C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2B-2D\right) \alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -3B\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 5B-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +\left( 2A-2E\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 4B+2C-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( 2A+6E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+C\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( \frac{D}{2}+2E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x+B\varepsilon \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.23)
Observe that
$$\begin{aligned} B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^6u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x =&-B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x -B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\ =&-B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon dx+B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\&+B\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}\partial _{x}^2u_\varepsilon {\mathrm{d}}x\\ =&-B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\&-B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon {\partial _t}\partial _{x}^2u_\varepsilon {\mathrm{d}}x -\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x\\ =&-2B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x -B\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\\&+B\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon {\partial _t}\partial _x u_\varepsilon +B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\&-\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x\\ =&-3B\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x -3B\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\\&-B\int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x -\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x. \end{aligned}$$
Consequently, by (2.23),
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left( A-B\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x \nonumber \\&\qquad \quad +\left( C-3B\right) \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+\left( D-3B\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +E\int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2){\partial _t}u_\varepsilon {\mathrm{d}}x+2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 4\alpha +2A\gamma +\frac{C\gamma }{2}+5E\gamma \right) \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( \frac{B\gamma }{2}+D\gamma -8\alpha \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x -\left( 3B+C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2B-2D\right) \alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +\left( B+2C-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 5B-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad +\left( 2A-2E\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x -\left( 2A+6E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +C\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-\left( \frac{D}{2}+2E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x. \nonumber \\&\qquad \quad +B\varepsilon \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.24)
Observe that
$$\begin{aligned} \left( C-3B\right) \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x=&-\left( C-3B\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x\\&-\left( C-3B\right) \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon {\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\ =&-\left( C-3B\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x\\&-\frac{C-3B}{2}\int _{{\mathbb {R}}}\partial _x ((\partial _{x}^2u_\varepsilon )^2){\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\ =&-\left( C-3B\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x\\&+\frac{C-3B}{2}\int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2){\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
Therefore, by (2.24),
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left( A+2B-C\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x \nonumber \\&\qquad \quad +\left( D-3B\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+\frac{2E+C-3B}{2}\int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2){\partial _t}u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad \quad +2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 4\alpha +2A\gamma +\frac{C\gamma }{2}+5E\gamma \right) \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( \frac{B\gamma }{2}+D\gamma -8\alpha \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x -\left( 3B+C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2B-2D\right) \alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad + \left( B+2C-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 5B-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +\left( 2A-2E\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( 2A+6E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+C\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( \frac{D}{2}+2E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x+B\varepsilon \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.25)
Observe that
$$\begin{aligned} \frac{2E+C-3B}{2}\int _{{\mathbb {R}}}\partial _{x}^2((\partial _{x}^2u_\varepsilon )^2){\partial _t}u_\varepsilon {\mathrm{d}}x=&-\frac{2E+C-3B}{2}\int _{{\mathbb {R}}}\partial _x ((\partial _{x}^2u_\varepsilon )^2){\partial _t}\partial _x u_\varepsilon {\mathrm{d}}x\\ =&\frac{2E+C-3B}{2}\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2{\partial _t}\partial _{x}^2u_\varepsilon {\mathrm{d}}x\\ =&\frac{2E+C-3B}{6}\frac{d}{dt}\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x. \end{aligned}$$
Consequently, by (2.25),
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left( A+2B-C\right) \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{B}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x \nonumber \\&\qquad \quad +\left( D-3B\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x+\frac{2E+C-3B}{6}\frac{{\mathrm{d}}}{{\mathrm{d}}t}\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&\qquad \quad +2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 4\alpha +2A\gamma +\frac{C\gamma }{2}+5E\gamma \right) \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( \frac{B\gamma }{2}+D\gamma -8\alpha \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -A\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x -\left( 3B+C-2D\right) \alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2B-2D\right) \alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad + \left( B+2C-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 5B-2D\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +\left( 2A-2E\right) \alpha \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( 2A+6E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+C\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\left( \frac{D}{2}+2E\right) \varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x+B\varepsilon \int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
(2.26)
We search \(A,\,B,\,C,\,D,\,E\) such that
$$\begin{aligned} \begin{array}{lll} &{}\displaystyle A+2B-C=-\frac{B}{2} ,\quad \displaystyle D-3B=0, &{}\quad \displaystyle E=-\frac{3C}{10}\\ &{}\displaystyle 4\alpha +2A\gamma +\frac{C\gamma }{2}+3E\gamma =0 , &{}\quad \displaystyle \frac{B\gamma }{2}+D\gamma -8\alpha =0, \end{array} \end{aligned}$$
that is
$$\begin{aligned} \begin{array}{lll} &{}\displaystyle 2A+5B-2C=0, \quad \displaystyle D=3B, &{} \quad \displaystyle E=-\frac{3C}{10},\\ &{} \displaystyle 4A\gamma +C\gamma +10E\gamma =-8\alpha , &{} \quad \displaystyle B\gamma +2D\gamma =16\alpha . \end{array} \end{aligned}$$
(2.27)
Since
$$\begin{aligned} (A,\,B,\,C,\,D,\,E)=\left( \frac{12\alpha }{7\gamma },\, \frac{16\alpha }{7\gamma },\, \frac{52\alpha }{7\gamma },\, \frac{48\alpha }{7\gamma },\,-\frac{78\alpha }{35\gamma }\right) , \end{aligned}$$
is the unique solution of (2.27), it follows from (2.26) that
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}&-\frac{8\alpha }{7\gamma }\int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{8\alpha }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _{x}^3u_\varepsilon )^2){\mathrm{d}}x+\ell _1\frac{{\mathrm{d}}}{{\mathrm{d}}t}\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x\\&+2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ =&\frac{12\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x+\frac{4\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\\&-\frac{64\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x-\frac{120\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\\&-\frac{16\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +\ell _2\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x\\&+\ell _3\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-\frac{20\alpha \varepsilon }{7\gamma } \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\&+\ell _4\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x+\frac{16\alpha \varepsilon }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x, \end{aligned}$$
where
$$\begin{aligned} \ell _1:= \frac{2E+C-3B}{6}, \quad \ell _2:= 2A-2E, \quad \ell _3:=2A+6E, \quad \ell _4:=\frac{D}{2}+2E. \end{aligned}$$
Consequently, we get
$$\begin{aligned} \frac{{\mathrm{d}}G(t)}{{\mathrm{d}}t}&+2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\ =&\frac{12\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x+\frac{4\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&-\frac{64\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x-\frac{120\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&-\frac{16\alpha ^2}{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x +\ell _2\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&+\ell _3\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-\frac{20\alpha \varepsilon }{7\gamma } \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&+\ell _4\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x+\frac{16\alpha \varepsilon }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x, \end{aligned}$$
(2.28)
where
$$\begin{aligned} G(t):=\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{8\alpha }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x+\ell _1\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x. \end{aligned}$$
(2.29)
Due to (2.3), (2.5), (2.15), (2.16), (2.17), (2.18), the Hölder inequality and the Young inequality,
$$\begin{aligned}&\left| \frac{12\alpha ^2}{7\gamma }\right| \int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \left| \frac{12\alpha ^2}{7\gamma }\right| \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \left| \frac{12\alpha ^2}{7\gamma }\right| \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le C(T)\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^{\frac{1}{2}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T),\\&\left| \frac{4\alpha ^2}{7\gamma }\right| \int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2\vert \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{16\alpha ^4}{49\gamma ^2}\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^4 {\mathrm{d}}x +\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{16\alpha ^4}{49\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T),\\&\left| \frac{64\alpha ^2}{7\gamma }\right| \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2\vert \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \left| \frac{64\alpha ^2}{7\gamma }\right| \left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{{\mathbb {R}}}\vert \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le C(T)\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le C(T)\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+ C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T),\end{aligned}$$
$$\begin{aligned}&\left| \frac{120\alpha ^2}{7\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{60\alpha ^4}{49\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _{x}^3u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x + \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{60\alpha ^4}{49\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} + \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}+ \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}+ \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}+ \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}+ \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T),\\&\left| \frac{16\alpha ^2}{7\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \partial _x u_\varepsilon \vert (\partial _{x}^4u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \left| \frac{16\alpha ^2}{7\gamma }\right| \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _x u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\vert \ell _2\vert \int _{{\mathbb {R}}}\vert u_\varepsilon \vert \vert \partial _{x}^3u_\varepsilon \vert ^3 {\mathrm{d}}x\\&\qquad \le \vert \ell _2\vert \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{{\mathbb {R}}}\vert \partial _{x}^3u_\varepsilon \vert ^3 {\mathrm{d}}x\\&\qquad \le C(T)\root 4 \of {\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^5_{L^2({\mathbb {R}})}}\root 4 \of {\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^7_{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\root 4 \of {\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T),\\&\vert \ell _3\vert \varepsilon \int _{{\mathbb {R}}}\vert \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad =2\int _{{\mathbb {R}}}\left| \frac{\ell _3\partial _{x}^4u_\varepsilon }{2}\right| \left| \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon \right| {\mathrm{d}}x\\&\qquad \le \frac{\ell _3^2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+ \varepsilon \int _{{\mathbb {R}}}(\partial _{x}^3u_\varepsilon )^2(\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \frac{\ell _3^2\varepsilon }{4}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\end{aligned}$$
$$\begin{aligned}&\left|\frac{20\alpha \varepsilon }{7\gamma }\right| \int _{{\mathbb {R}}}\vert \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^4u_\varepsilon \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{100\alpha ^2\varepsilon }{49\gamma ^2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2(\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \frac{100\alpha ^2\varepsilon }{49\gamma ^2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\vert \ell _4\vert \varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon (\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad =2\varepsilon \int _{{\mathbb {R}}}\left| \frac{\ell _4\partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon }{2}\right| \left| \partial _{x}^5u_\varepsilon \right| {\mathrm{d}}x\\&\qquad \le \frac{\ell _4^2\varepsilon }{4}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2(\partial _{x}^5u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \frac{\ell _4^2\varepsilon }{4}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\left| \frac{16\alpha \varepsilon }{7\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \vert (\partial _{x}^6u_\varepsilon )^2 {\mathrm{d}}x\\&\qquad \le \left| \frac{16\alpha \varepsilon }{7\gamma }\right| \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^6u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\varepsilon \left\| \partial _{x}^6u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
It follows from (2.28) that
$$\begin{aligned} \frac{{\mathrm{d}}G(t)}{{\mathrm{d}}t}&+2\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\ \le&C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)+C_0\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} \end{aligned}$$
(2.30)
$$\begin{aligned}&+\varepsilon \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\ell ^2_4\varepsilon }{4}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&+2\varepsilon \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\varepsilon \left\| \partial _{x}^6u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
(2.31)
Thanks to the Young inequality,
$$\begin{aligned} C(T)\varepsilon \left\| \partial _{x}^6u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&C(T)\varepsilon \int _{{\mathbb {R}}}\partial _{x}^6u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x=-C(T)\varepsilon \int _{{\mathbb {R}}}\partial _{x}^5u_\varepsilon \partial _{x}^7u_\varepsilon {\mathrm{d}}x\\ \le\,&C(T)\varepsilon \int _{{\mathbb {R}}}\vert \partial _{x}^5u_\varepsilon \vert \vert \partial _{x}^7u_\varepsilon {\mathrm{d}}x\\ \le\,&C(T)\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Consequently, by (2.30),
$$\begin{aligned} \frac{{\mathrm{d}}G(t)}{{\mathrm{d}}t}&+\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\ \le\,&C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)+C_0\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&+\varepsilon \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&+2\varepsilon \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
(2.32)
By (2.29), we have that
$$\begin{aligned} \begin{aligned} C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&C(T)G(t)+\frac{8C(T)\alpha }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^2{\mathrm{d}}x\\&+C(T)\ell _1\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x. \end{aligned} \end{aligned}$$
(2.33)
Moreover, by (2.3), (2.5), (2.15) and the Young inequality,
$$\begin{aligned}&\left| \frac{8C(T)\alpha }{7\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \vert (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \le \left| \frac{8C(T)\alpha }{7\gamma }\right| \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C(T)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))},\nonumber \\&\left| C(T)\ell _1\right| \int _{{\mathbb {R}}}\vert \partial _{x}^2u_\varepsilon \vert ^3 {\mathrm{d}}x\nonumber \\&\qquad \le \left| C(T)\ell _1\right| \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\nonumber \\&\qquad \le C(T)+ C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\nonumber \\&\qquad \le C(T)+C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }({\mathbb {R}})}\nonumber \\&\qquad \le C(T)+C(T)\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C(T) +C(T)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
(2.34)
Therefore, by (2.32), (2.33) and (2.34),
$$\begin{aligned} \frac{{\mathrm{d}}G(t)}{{\mathrm{d}}t}&+\varepsilon \left\| \partial _{x}^7u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&C(T)G(t) + C(T)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+C_0\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C(T)\\&+C(T)\left( 1+ \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right) \varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
The Gronwall lemma, (2.2), (2.3), (2.19) and (2.29) give
$$\begin{aligned}&\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{8\alpha }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x +\ell _1\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x\\&\qquad \quad +\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^7u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \le C_0e^{C(T)t} +C_0 e^{C(T)t}\varepsilon \int _{0}^{t}e^{-C_0s}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \quad +C(T)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}e^{C(T)t}\int _{0}^{t}e^{-C(T)s} ds+C(T)e^{C(T)t}\int _{0}^{t}e^{-C(T)s} {\mathrm{d}}s\\&\qquad \quad +C(T)\left( 1+ \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right) e^{C(T)t}\varepsilon \int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \quad +\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}e^{C(T)t}\varepsilon \int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \le C(T)\left( 1+\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) + C(T)\varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \quad +C(T)\left( 1+ \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right) \varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \le C(T)\left( 1+ \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
Consequently, by (2.34),
$$\begin{aligned} \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}&+\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^7u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\ \le\,&C(T)\left( 1+\left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right. \nonumber \\&\left. +\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&+\frac{8\alpha }{7\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x+\ell _1\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\ \le\,&C(T)\left( 1+\left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right. \nonumber \\&\left. +\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
(2.35)
Observe that, by (2.3), (2.16) and the Young inequality,
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le\,&2 \sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&C_0\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&C_0+C_0\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\ \le\,&C_0\left( 1+\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) ,\\ \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le\,&2\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\\ \le\,&C(T)\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\sqrt{\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&\frac{C(T)}{D_6}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} +C(T)D_6\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&\frac{C(T)}{D_6}\left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+C(T)D_6\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
where \(D_6\) is a positive constant, which will be specified later. Therefore,
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\le\,&C_0\left( 1+\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) ,\\ \left\| \partial _{x}^3u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\le\,&\frac{C(T)}{D_6}\left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+C(T)D_6\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
Consequently, by (2.35),
$$\begin{aligned} \begin{aligned}&\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon e^{C(T)t}\int _{0}^{t}e^{-C(T)s}\left\| \partial _{x}^7u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \le C(T)\left( 1 +\frac{C(T)}{D_6}\left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+(1+D_6)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned} \end{aligned}$$
(2.36)
Hence, by (2.36), we have
$$\begin{aligned} \left( 1-\frac{C(T)}{D_6}\right) \left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)(1+D_6)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)\le 0. \end{aligned}$$
Choosing
$$\begin{aligned} D_6=\frac{1}{2C(T)}, \end{aligned}$$
(2.37)
we have that
$$\begin{aligned} \frac{1}{2}\left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)\left\| \partial _{x}^4u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)\le 0, \end{aligned}$$
which give (2.12).
Finally, (2.13) follows from (2.12), (2.36) and (2.37), while (2.3), (2.13), (2.15), (2.16) and (2.17) give (2.14). \(\square\)
Now, we prove the following lemma.
Lemma 2.6
Fix \(T>0\). Then,
$$\begin{aligned} \{u_\varepsilon \}_{\varepsilon >0} {\text {is compact in}}\, L^2_{loc}((0,\infty )\times {\mathbb {R}}). \end{aligned}$$
(2.38)
Consequently, there exists a subsequence \(\{u_{\varepsilon _k}\}_{k\in {\mathbb {N}}}\) of \(\{u_\varepsilon \}_{\varepsilon >0}\) and \(u\in L^2_{loc}((0,\infty )\times {\mathbb {R}})\) such that, for each compact subset K of \((0,\infty )\times {\mathbb {R}})\),
$$\begin{aligned} u_{\varepsilon _k}\rightarrow u \, {\text {in}}\, L^2(K) {\text { and a.e.}} \end{aligned}$$
(2.39)
Moreover, u is a solution of (1.1) satisfying (1.8).
Proof
To prove (2.38), we rely on the Aubin–Lions lemma (see [3, 6, 7, 21]). We recall that
$$\begin{aligned} H^1_{loc}({\mathbb {R}})\hookrightarrow \hookrightarrow L^2_{loc}({\mathbb {R}})\hookrightarrow H^{-1}_{loc}({\mathbb {R}}), \end{aligned}$$
where the first inclusion is compact and the second one is continuous. Owing to the Aubin–Lions lemma [21], to prove (2.38), it suffices to show that
$$\begin{aligned}&\{u_\varepsilon \}_{\varepsilon >0}\, {\text {is uniformly bounded \, in}} \, {L}^2(0, \,{T}; {H}^1_{ {loc}}({\mathbb {R}})), \end{aligned}$$
(2.40)
$$\begin{aligned}&\{{\partial _t}u_\varepsilon \}_{\varepsilon >0} \,{\text {is uniformly bounded}} \, {\text {in}} \, {L}^2(0, {T}; {H}^{-1}_{ {loc}}({\mathbb {R}})). \end{aligned}$$
(2.41)
We prove (2.40). Thanks to Lemmas 2.1, 2.4 and 2.4,
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{H^4({\mathbb {R}})}=\,&\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&+\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C(T). \end{aligned}$$
Therefore,
$$\begin{aligned} \{u_\varepsilon \}_{\varepsilon >0} \, {\text {is uniformly bounded in}} \, {L}^{\infty }(0, {T}; {H}^{4}({\mathbb {R}})), \end{aligned}$$
which gives (2.40).
We prove (2.41). We begin by observing that
$$\begin{aligned} \beta u_\varepsilon \partial _{x}^3u_\varepsilon =\beta \partial _x (u_\varepsilon \partial _{x}^2u_\varepsilon )-\frac{\beta }{2}\partial _x ((\partial _x u_\varepsilon )^2). \end{aligned}$$
Therefore, by (2.1),
$$\begin{aligned} {\partial _t}u_\varepsilon =\partial _x \left( \frac{\beta -\alpha }{2}(\partial _x u_\varepsilon )^2-\beta u_\varepsilon \partial _{x}^2u_\varepsilon -\gamma \partial _{x}^4u_\varepsilon +\varepsilon \partial _{x}^5u_\varepsilon \right) . \end{aligned}$$
(2.42)
We have that
$$\begin{aligned} \frac{(\beta -\alpha )^2}{4}\left\| \partial _x u_\varepsilon \right\| ^4_{L^4((0,T)\times {\mathbb {R}})}\le C(T). \end{aligned}$$
(2.43)
Thanks to Lemma 2.5,
$$\begin{aligned} \frac{(\beta -\alpha )^2}{4}\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^4 {\mathrm{d}}t{\mathrm{d}}x\le\,&\frac{(\beta -\alpha )^2}{4}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2 {\mathrm{d}}t{\mathrm{d}}x\\ \le\,&C(T)\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2 {\mathrm{d}}t{\mathrm{d}}x\le C(T). \end{aligned}$$
We claim that
$$\begin{aligned} \beta ^2\left\| u_\varepsilon \partial _{x}^2u_\varepsilon \right\| ^2_{L^2((0,T)\times {\mathbb {R}})}\le C(T). \end{aligned}$$
(2.44)
$$\begin{aligned} \beta ^2\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}t{\mathrm{d}}x\le\,&\beta ^2\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}t{\mathrm{d}}x\\ \le\,&C(T)\int _{0}^{T}\!\!\!\int _{{\mathbb {R}}}(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}t{\mathrm{d}}x\le C(T). \end{aligned}$$
Moreover, since \(0<\varepsilon <1\), by Lemmas 2.3 and 2.4,
$$\begin{aligned} \gamma ^2\left\| \partial _{x}^4u_\varepsilon \right\| ^2_{L^2((0,T)\times {\mathbb {R}})},\, \varepsilon ^2\left\| \partial _{x}^5u_\varepsilon \right\| ^2_{L^2((0,T)\times {\mathbb {R}})}\le C(T). \end{aligned}$$
(2.45)
Therefore, by (2.43), (2.44) and (2.45),
$$\begin{aligned}&\left\{ \frac{\beta -\alpha }{2}(\partial _x u_\varepsilon )^2-\beta u_\varepsilon \partial _{x}^2u_\varepsilon -\gamma \partial _{x}^4u_\varepsilon +\varepsilon \partial _{x}^5u_\varepsilon \right\} _{\varepsilon >0} \\&\quad {\text { is bounded in}}\, \ L^2((0,T)\times {\mathbb {R}}). \end{aligned}$$
Thanks to the Aubin–Lions lemma, (2.38) and (2.39) hold.
Consequently, u is solution of (1.1) and (1.8) holds. \(\square\)
Proof of Theorem 1.1
Lemma (2.6) says that there exists a solution u of (1.1) such that (1.8) holds. Since \(H^2({\mathbb {R}})\subset H^4({\mathbb {R}})\), thanks to [19, Theorem 2.2], u is unique. \(\square\)
3 Proof of Theorem 1.1 under Assumption (1.3).
In section, we prove Theorem 1.1 under Assumption (1.3).
We consider approximation (2.1) of (1.1), where \(u_{\varepsilon ,0}\) is a \(C^{\infty }\) approximation of \(u_0\) such that (2.2) holds.
Let us prove some a priori estimates on \(u_\varepsilon\).
Lemma 3.1
Assume (1.3). For each \(t\ge 0\),
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C_0. \end{aligned}$$
(3.1)
Proof
Multiplying (2.1) by \(2u_\varepsilon\), an integration on \({\mathbb {R}}\) give
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&2\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\\ =&-2\alpha \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x -2\beta \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _{x}^3u_\varepsilon {\mathrm{d}}x\\&-2\gamma \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+2\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\\ =&-2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x \\&+2\gamma \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x-2\varepsilon \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\ =&-2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\\&-2\gamma \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x +2\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \\ =&-2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x-2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Hence,
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\left( \alpha -2\beta \right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x=0. \end{aligned}$$
(3.2)
Integrating on (0, t), by (1.2) and (2.2), we have (2.4). \(\square\)
Lemma 3.2
Assume (1.3) and fix \(T>0\). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(3.3)
for every \(0\le t\le T\). In particular, we have that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _x u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(3.4)
for every \(0\le t\le T\).
Proof
Let \(0\le t\le T\). We begin by observing that, thanks to the Hölder inequality,
$$\begin{aligned} \begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=&\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _x u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\\ \le&\int _{{\mathbb {R}}}\vert u_\varepsilon \vert \vert \partial _{x}^2u_\varepsilon \vert {\mathrm{d}}x\le \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned} \end{aligned}$$
(3.5)
Again by the Hölder inequality,
$$\begin{aligned} \begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=&\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\\ \le&\int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^3u_\varepsilon \vert {\mathrm{d}}x\le \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned} \end{aligned}$$
(3.6)
Therefore, by (3.5) and (3.6),
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
(3.7)
Due to (2.4) and the Young inequality,
$$\begin{aligned}&\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\&\qquad \le \frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{1}{8}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{1}{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C_0+ \frac{1}{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Consequently, by (3.7),
$$\begin{aligned} \frac{1}{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0+\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Since \(0<\varepsilon <1\), we have that
$$\begin{aligned} \frac{\varepsilon }{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \varepsilon C_0+\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0+\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
If follows from an integration on (0, t) and (2.4) that
$$\begin{aligned} \frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C_0t+\varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
which gives (3.3).
Finally, we prove (3.4). By (2.4), (3.5) and the Young inequality,
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \frac{1}{2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{1}{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0+\frac{1}{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Since \(0<\varepsilon <1\), we have that
$$\begin{aligned} \varepsilon \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \varepsilon C_0+\frac{\varepsilon }{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0+\frac{\varepsilon }{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Integrating on (0, t), by (3.3), we get
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0t+\frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
which gives (3.4). \(\square\)
Lemma 3.3
Assume (1.2) and fix \(T>0\). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\le C(T). \end{aligned}$$
(3.8)
In particular, we have
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{{\mathbb {R}}}+\varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(3.9)
for every \(0\le t\le T\). Moreover,
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})},\, \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})},\, \left\| \partial _x u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\le C(T), \end{aligned}$$
(3.10)
for every \(0\le t\le T\).
Proof
Let \(0\le t\le T\). Consider two real constants \(F,\,G\), which will be specified later. Multiplying (2.1) by
$$\begin{aligned} 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon , \end{aligned}$$
we have that
$$\begin{aligned}&\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) {\partial _t}u_\varepsilon \nonumber \\&\qquad \quad +\alpha \left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \nonumber \\&\qquad \quad +\beta \left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) u_\varepsilon \partial _{x}^3u_\varepsilon \nonumber \\&\qquad \quad +\gamma \left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _{x}^5u_\varepsilon \nonumber \\&\qquad =\varepsilon \left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _{x}^6u_\varepsilon . \end{aligned}$$
(3.11)
Observe that
$$\begin{aligned}&\int _{{\mathbb {R}}}\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+F\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x +G\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x,\nonumber \\&\alpha \int _{{\mathbb {R}}}\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 dx +\alpha G\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x, \nonumber \\&\beta \int _{{\mathbb {R}}}\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-\beta \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 dx -2\beta F\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x \nonumber \\&\qquad \quad -\beta G\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x, \nonumber \\&\gamma \int _{{\mathbb {R}}}\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-\gamma \left( 2F+G\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x -\gamma G\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x \nonumber \\&\qquad =\gamma \left( F+\frac{3G}{2}\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad =\gamma \left( F+\frac{3G}{2}\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x \nonumber ,\\&\varepsilon \int _{{\mathbb {R}}}\left( 2\partial _{x}^4u_\varepsilon +F(\partial _x u_\varepsilon )^2 +Gu_\varepsilon \partial _{x}^2u_\varepsilon \right) \partial _{x}^6u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\varepsilon \left( 2F+G\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -G\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(3.12)
It follows from (1.3), (3.12) and an integration on \({\mathbb {R}}\) of (3.11) that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+ F\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x+G\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad +2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 5\beta - F\gamma -\frac{3\gamma G}{2}\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2 F-G\right) \beta \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -\varepsilon \left( 2F+G\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-G\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(3.13)
Observe that
$$\begin{aligned} \begin{aligned}&F\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x+G\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^2u_\varepsilon {\partial _t}u_\varepsilon {\mathrm{d}}x\\&\qquad =\left( F-G\right) \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{G}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _x u_\varepsilon )^2){\mathrm{d}}x. \end{aligned} \end{aligned}$$
(3.14)
Consequently, by (3.13),
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\left( F-G\right) \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x-\frac{G}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _x u_\varepsilon )^2){\mathrm{d}}x\nonumber \\&\qquad \quad +2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 5\beta - F\gamma -\frac{3\gamma G}{2}\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( 2 F-G\right) \beta \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -\varepsilon \left( 2F+G\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-G\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(3.15)
We search \(F,\,G\) such that
$$\begin{aligned} F-G=-\frac{G}{2},\quad 5\beta - F\gamma -\frac{3\gamma G}{2} =0,\quad 2F-G=0, \end{aligned}$$
that is
$$\begin{aligned} F=\frac{G}{2}, \quad 5\beta - 2\gamma G =0. \end{aligned}$$
(3.16)
Since
$$\begin{aligned} (F,\,G)=\left( \frac{5\beta }{4\gamma },\,\frac{5\beta }{2\gamma }\right) . \end{aligned}$$
is the unique solution of (3.16), it follows from (3.15) that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{5\beta }{4\gamma } \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x\\&\qquad \quad -\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _x u_\varepsilon )^2){\mathrm{d}}x+2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad = -\frac{5\beta \varepsilon }{\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-\frac{5\beta \varepsilon }{2\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x, \end{aligned}$$
that is
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\right) +2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \qquad \qquad = -\frac{5\beta \varepsilon }{\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-\frac{5\beta \varepsilon }{2\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(3.17)
Due to the Young inequality,
$$\begin{aligned}&\left| \frac{5\beta \varepsilon }{\gamma }\right| \int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{25\beta ^2\varepsilon }{2\gamma ^2}\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 dx +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{25\beta ^2\varepsilon }{2\gamma ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} + \frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{25\beta ^2\varepsilon }{2\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} + \frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\left| \frac{5\beta \varepsilon }{2\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{25\beta ^2\varepsilon }{8\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x + \frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{25\beta ^2\varepsilon }{8\gamma ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{25\beta ^2\varepsilon }{8\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
It follows from (3.17) that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\right) \nonumber \\&\qquad \quad +\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le \frac{25\beta ^2\varepsilon }{2\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \quad + \frac{25\beta ^2\varepsilon }{8\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
(3.18)
Integration on (0, t), by (2.2), (3.1) and (3.3), we have that
$$\begin{aligned}&\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x+\varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \le C_0 + \frac{25\beta ^2\varepsilon }{2\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \quad + \frac{25\beta ^2\varepsilon }{8\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \le C_0\left( 1+\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\right) . \end{aligned}$$
(3.19)
Thanks to (3.1), [2, Lemma 2.5] and the Young inequality,
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le\,&\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&C_0 \sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&C_0 + C_0\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\\ \le\,&C_0+C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))},\\ \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le\,&\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}},\\ \le\,&C_0\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\\ =\,&C_0\frac{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}{\sqrt{D_1}}\sqrt{D_1\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&\frac{C_0}{D_1}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+D_1C_0\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&\frac{C_0}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+D_1C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}, \end{aligned}$$
where \(D_1\) is a positive constant, which will be specified later. Consequently,
$$\begin{aligned} \left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\le\,&C_0\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) ,\\ \left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\le\,&\frac{C_0}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+D_1C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
Thus, by (3.19),
$$\begin{aligned} \begin{aligned}&\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x+\varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\&\qquad \le C_0\left( 1+ \frac{1}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned} \end{aligned}$$
(3.20)
Thanks to (3.1), (3.5), [4, Lemma 2.6] and the Young inequality,
$$\begin{aligned}&\left| \frac{5\beta }{4\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \vert (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \le \frac{25\beta ^2}{32\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2 {\mathrm{d}}x +\frac{25\beta ^2}{32\gamma ^2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le \frac{25\beta ^2}{16\gamma ^2}\sqrt{\left\| u_\varepsilon (t,\cdot ) \right\| ^5_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\nonumber \\&\qquad \quad + \frac{25\beta ^2}{32\gamma ^2}\left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C_0\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}+C_0\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C_0\frac{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}{\sqrt{D_1}}\sqrt{D_1\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}} + C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\nonumber \\&\qquad \le \frac{C_0}{D_1}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} +D_1C_0\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})} + C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\nonumber \\&\qquad \le \frac{C_0}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+ C_0\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
(3.21)
Consequently, by (3.20) and (3.21),
$$\begin{aligned}&\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\varepsilon \int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \le C_0\left( 1+ \frac{1}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad +\frac{5\beta }{4\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \le C_0\left( 1+ \frac{1}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad +\left| \frac{5\beta }{4\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \vert (\partial _x u_\varepsilon )^2{\mathrm{d}}x\nonumber \\&\qquad \le C_0\left( 1+ \frac{1}{D_1}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
(3.22)
It follows from (3.22) that
$$\begin{aligned} \left( 1-\frac{C_0}{D_1}\right) \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C_0\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C_0\le 0. \end{aligned}$$
Choosing
$$\begin{aligned} D_1=2C_0, \end{aligned}$$
(3.23)
we obtain that
$$\begin{aligned} \frac{1}{2}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C_0\left( 1+D_1\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C_0\le 0, \end{aligned}$$
which gives (3.8).
Finally, thanks to (3.8), (3.22) and (3.23), we have (3.9), while (3.1), (3.9) and [2, Lemma 2.3] give (3.10). \(\square\)
Observe that, arguing as in Lemma 2.4, we have (2.13) and (2.14). Therefore, arguing as in Sect. 2, we have Theorem 1.1.
4 Proof of Theorem 1.1 under Assumption (1.4).
In section, we prove Theorem 1.1 under Assumption (1.4).
We consider approximation (2.1) of (1.1), where \(u_{\varepsilon ,0}\) is a \(C^{\infty }\) approximation of \(u_0\) such that (2.2) holds.
Let us prove some a priori estimates on \(u_\varepsilon\).
Lemma 4.1
Assume (1.4). For each \(t\ge 0\),
$$\begin{aligned} \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})} {\mathrm{d}}s \le C_0. \end{aligned}$$
(4.1)
Proof
Multiplying (2.1) by \(-2\partial _{x}^2u_\varepsilon\), an integration on \({\mathbb {R}}\) gives
$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\,&-2\int _{{\mathbb {R}}}{\partial _t}u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\\ =\,&2\alpha \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x +2\beta \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\\&+2\gamma \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x -2\varepsilon \int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^6u_\varepsilon {\mathrm{d}}x\\ =\,&\left( 2\alpha -\beta \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x-2\gamma \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\\&+2\varepsilon \int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\ =\,&\left( 2\alpha -\beta \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x-2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Therefore, we have that
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=\left( 2\alpha -\beta \right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x. \end{aligned}$$
Integrating on (0, t), thanks to (1.4) and (2.2), we get (4.1). \(\square\)
Lemma 4.2
Assume (1.4) and fix \(T>0\). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(4.2)
for every \(0\le t\le T\). In particular, we have that
$$\begin{aligned} \varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(4.3)
for every \(0\le t\le T\).
Proof
Let \(0\le t\le T\). We begin by observing that, thanks to the Hölder inequality,
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=&\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\\ \le&\int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert \vert \partial _{x}^3u_\varepsilon \vert {\mathrm{d}}x\le \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Hence, by (4.1),
$$\begin{aligned} \varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le \varepsilon \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\le C_0\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}. \end{aligned}$$
(4.4)
Moreover, by the Hölder inequality and the Young one,
$$\begin{aligned} \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=&\int _{{\mathbb {R}}}\partial _{x}^3u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}\partial _{x}^2u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\ \le&\int _{{\mathbb {R}}}\vert \partial _{x}^2u_\varepsilon \vert \partial _{x}^4u_\varepsilon \vert {\mathrm{d}}x\le \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\ =&\frac{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}{\sqrt{D_2}}\sqrt{D_2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\ \le&\frac{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}}{2D_2}+\frac{D_2}{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}, \end{aligned}$$
(4.5)
where \(D_2\) is positive constant, which will be specified later. Since \(0<\varepsilon <1\), it follows from (4.4), (4.5) and the Young inequality,
$$\begin{aligned} \varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le\,&C_0\varepsilon +C_0\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&C_0+\frac{C_0\varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}}{2D_2}+\frac{C_0D_2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Therefore,
$$\begin{aligned} \left( 1-\frac{C_0}{2D_2}\right) \varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0 + \frac{C_0D_2\varepsilon }{2}\left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Choosing
$$\begin{aligned} D_2=C_0, \end{aligned}$$
(4.6)
we have that
$$\begin{aligned} \frac{\varepsilon }{2}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0+C_0\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
It follows from an integration on (0, t) and (4.1) that
$$\begin{aligned} \frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le\,&C_0t+C_0\varepsilon \int _{0}^{t}\left\| \partial _{x}^4u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\ \le\,&C(T)+C_0\le C(T), \end{aligned}$$
which gives (4.2).
Finally, we prove (4.3). Thanks to (4.5) and (4.6),
$$\begin{aligned} \varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0\varepsilon \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C_0\varepsilon \left\| \partial _{x}^4u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}. \end{aligned}$$
Therefore, (4.3) follows from (4.1), (4.2) and an integration on (0, t). \(\square\)
Lemma 4.3
Assume (1.4) and fix \(T>0\). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T)\left( 1+\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}\right) , \end{aligned}$$
(4.7)
for every \(0\le t\le T\). In particular, we have that
$$\begin{aligned} \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\le\,&C(T)\sqrt{\left( 1+\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}\right) }, \end{aligned}$$
(4.8)
$$\begin{aligned} \left\| u_\varepsilon \right\| _{L^{\infty }(0,T;L^4({\mathbb {R}}))}\le\,&C(T)\root 4 \of {\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) }. \end{aligned}$$
(4.9)
Proof
Let \(0\le t\le T\). We begin by observing that
$$\begin{aligned} 2\int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x=-\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^3 {\mathrm{d}}x. \end{aligned}$$
(4.10)
Consequently, by (1.4) and (3.2), we have that
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}=-\alpha \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^3 {\mathrm{d}}x. \end{aligned}$$
(4.11)
$$\begin{aligned} \vert \alpha \vert \int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \vert ^3 {\mathrm{d}}x\le\,&\vert \alpha \vert \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| _{L^{\infty }({\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le\,&\vert \alpha \vert \sqrt{2}\sqrt{\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^3_{L^2({\mathbb {R}})}}\sqrt{\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})}}\\ \le\,&C_0\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}. \end{aligned}$$
Consequently, by (4.11),
$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+2\varepsilon \left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\le C_0\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}. \end{aligned}$$
It follows from (2.2) and an integration on \({\mathbb {R}}\) that
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}&+2\varepsilon \int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\\ \le\,&C_0+C_0\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}t\\ \le\,&C(T)\left( 1+\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}\right) , \end{aligned}$$
which gives (4.7).
We prove (4.8). Thanks to (2.10) and (4.1),
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\le C(T)\left( 1+\sqrt{\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}}\right) , \end{aligned}$$
which gives (4.8).
Finally, we prove (4.9). Thanks to (4.7) and (4.8),
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^4({\mathbb {R}})}=&\int _{{\mathbb {R}}}u_\varepsilon ^4{\mathrm{d}}x\le \left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^{\infty }({\mathbb {R}})}\left\| u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\ \le&C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
Consequently, we have that
$$\begin{aligned} \left\| u_\varepsilon \right\| ^4_{L^{\infty }(0,T;L^4({\mathbb {R}}))}\le C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) , \end{aligned}$$
which gives (4.9). \(\square\)
Lemma 4.4
Assume (1.4) and fix \(T>0\). There exists a constant \(C(T)>0\), independent on \(\varepsilon\), such that (3.8) holds. In particular, we have
$$\begin{aligned} \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\int _{0}^{t}\left\| \partial _{x}^5u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\le C(T), \end{aligned}$$
(4.12)
for every \(0\le t\le T\). Moreover,
$$\begin{aligned} \left\| u_\varepsilon (t,\cdot ) \right\| _{L^2({\mathbb {R}})},\, \left\| u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})},\, \left\| \partial _x u_\varepsilon \right\| _{L^{\infty }((0,T)\times {\mathbb {R}})}\le C(T). \end{aligned}$$
(4.13)
Proof
Let \(0\le t\le T\). Consider three real constants \(H,\,I,\,L\), which will be specified later. Multiplying (2.1) by
$$\begin{aligned} 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3, \end{aligned}$$
we have that
$$\begin{aligned}&\left( 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3\right) {\partial _t}u_\varepsilon \nonumber \\&\qquad \quad +\alpha \left( 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3\right) \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \nonumber \\&\qquad \quad +\beta \left( 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3\right) u_\varepsilon \partial _{x}^3u_\varepsilon \nonumber \\&\qquad \quad +\gamma \left( 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3\right) \partial _{x}^5u_\varepsilon \nonumber \\&\qquad =\varepsilon \left( 2\partial _{x}^4u_\varepsilon +H(\partial _x u_\varepsilon )^2+Iu_\varepsilon \partial _{x}^2u_\varepsilon +Lu_\varepsilon ^3\right) \partial _{x}^6u_\varepsilon . \end{aligned}$$
(4.14)
Observe that, thanks to (1.4),
$$\begin{aligned}&L\int _{{\mathbb {R}}}u_\varepsilon ^3{\partial _t}u_\varepsilon dx=\frac{L}{4}\frac{{\mathrm{d}}}{{\mathrm{d}}t}\int _{{\mathbb {R}}}u_\varepsilon ^4 {\mathrm{d}}x,\nonumber \\&\alpha L\int _{{\mathbb {R}}}u_\varepsilon ^3\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x +\beta L\int _{{\mathbb {R}}}u_\varepsilon ^4\partial _{x}^3u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-\frac{3\alpha L}{2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x -4\beta L\int _{{\mathbb {R}}}u_\varepsilon ^3\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-\frac{3\alpha L}{2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x+6\beta L\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x\nonumber \\&\qquad =\frac{21\alpha L}{2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x,\nonumber \\&L\gamma \int _{{\mathbb {R}}}u_\varepsilon ^3\partial _{x}^5u_\varepsilon {\mathrm{d}}x=-3L\gamma \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _x u_\varepsilon \partial _{x}^4u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =6L\gamma \int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2\partial _{x}^3u_\varepsilon {\mathrm{d}}x +3L\gamma \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _{x}^2u_\varepsilon \partial _{x}^3u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad =-15L\gamma \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\nonumber ,\\&L\varepsilon \int _{{\mathbb {R}}}u_\varepsilon ^3\partial _{x}^6u_\varepsilon {\mathrm{d}}x=-3L\varepsilon \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(4.15)
Consequently, by (1.4), (3.12) with \(H,\,I\) instead of \(F,\,G\), (3.14) with \(H,\,I\) instead of \(F,\,G\), (4.15) and an integration on \({\mathbb {R}}\) of (4.14), we have that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{L}{4}\int _{{\mathbb {R}}}u_\varepsilon ^4 {\mathrm{d}}x\right) +\left( H-I\right) \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2{\partial _t}u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\frac{I}{2}\int _{{\mathbb {R}}}u_\varepsilon {\partial _t}((\partial _x u_\varepsilon )^2) {\mathrm{d}}x +2\varepsilon \left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\left( 4\alpha -\gamma H -\frac{3I\gamma }{2}\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon (\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad +\left( \alpha I+4\alpha H+15\gamma L\right) \int _{{\mathbb {R}}}u_\varepsilon \partial _x u_\varepsilon (\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x\nonumber \\&\qquad \quad -\frac{21\alpha L}{2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x -\varepsilon \left( 2H+I\right) \int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -I\varepsilon \int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x-3L\varepsilon \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(4.16)
We search \(H,\,I,\,L\) such that
$$\begin{aligned} H-I=-\frac{I}{2},\quad \displaystyle 4\alpha -\gamma H -\frac{3I\gamma }{2} =0,\quad \alpha I+4\alpha H+15\gamma L=0, \end{aligned}$$
that is
$$\begin{aligned} 2H-2I=-I,\quad 8\alpha -2\gamma H -3I\gamma =0,\quad \alpha I+4\alpha H+15\gamma L=0. \end{aligned}$$
(4.17)
Since
$$\begin{aligned} (H,\,I,\,L)=\left( \frac{\alpha }{\gamma },\,\frac{2\alpha }{\gamma },\,-\frac{2\alpha ^2}{5\gamma ^2}\right) \end{aligned}$$
is the unique solution of (4.17), it follows from (4.16) that
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{\alpha ^2}{10\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^4 {\mathrm{d}}x-\frac{\alpha }{\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\right) +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad =\frac{21\alpha ^3 }{5\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^3 {\mathrm{d}}x -\frac{4\alpha \varepsilon }{\gamma }\int _{{\mathbb {R}}}\partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\nonumber \\&\qquad \quad -\frac{2\alpha \varepsilon }{\gamma }\int _{{\mathbb {R}}}u_\varepsilon \partial _{x}^3u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x+\frac{6\alpha ^2\varepsilon }{5\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x. \end{aligned}$$
(4.18)
Since \(0<\varepsilon <1\), due to (2.8), (4.1), (4.9) and the Young inequality,
$$\begin{aligned}&\left| \frac{21\alpha ^3 }{5\gamma ^2}\right| \int _{{\mathbb {R}}}u_\varepsilon ^2\vert \partial _x u_\varepsilon \vert ^3 {\mathrm{d}}x\\&\qquad =\left| \frac{21\alpha ^3 }{5\gamma ^2}\right| \int _{{\mathbb {R}}}\vert \sqrt{D_3}u_\varepsilon ^2\vert \left| \frac{(\partial _x u_\varepsilon )^3}{\sqrt{D_3}}\right| {\mathrm{d}}x\\&\qquad \le \left| \frac{21\alpha ^3D_3 }{10\gamma ^2}\right| \left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^4({\mathbb {R}})} +\left| \frac{21\alpha ^3 }{10\gamma ^2D_3}\right| \int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^6 {\mathrm{d}}x\\&\qquad \le \left| \frac{21\alpha ^3D_3 }{10\gamma ^2}\right| \left\| u_\varepsilon \right\| ^4_{L^{\infty }(0,T;L^4({\mathbb {R}}))}+\left| \frac{21\alpha ^3 }{10\gamma ^2D_3}\right| \left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^4_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)D_3\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) + \frac{C_0}{D_3}\left\| \partial _x u_\varepsilon \right\| ^4_{L^{\infty }((0,T)\times {\mathbb {R}})}\\&\qquad \le C(T)D_3\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) +\frac{C_0}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))},\\&\left| \frac{4\alpha \varepsilon }{\gamma }\right| \int _{{\mathbb {R}}}\vert \partial _x u_\varepsilon \partial _{x}^2u_\varepsilon \vert \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{8\alpha ^2\varepsilon }{\gamma ^2}\int _{{\mathbb {R}}}(\partial _x u_\varepsilon )^2(\partial _{x}^2u_\varepsilon )^2 {\mathrm{d}}x +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{8\alpha ^2\varepsilon }{\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\left| \frac{2\alpha \varepsilon }{\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon \partial _{x}^3u_\varepsilon \vert \vert \partial _{x}^5u_\varepsilon \vert {\mathrm{d}}x\\&\qquad \le \frac{4\alpha ^2\varepsilon }{\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _{x}^3u_\varepsilon )^2 {\mathrm{d}}x +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{4\alpha ^2\varepsilon }{\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+ \frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})},\\&\left| \frac{6\alpha ^2\varepsilon }{5\gamma ^2}\right| \int _{{\mathbb {R}}}u_\varepsilon ^2\partial _x u_\varepsilon \partial _{x}^5u_\varepsilon {\mathrm{d}}x\\&\qquad \le \frac{18\alpha ^4\varepsilon }{25\gamma ^4}\int _{{\mathbb {R}}}u_\varepsilon ^4(\partial _x u_\varepsilon )^2 {\mathrm{d}}x +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{18\alpha ^4\varepsilon }{25\gamma ^4}\left\| u_\varepsilon \right\| ^4_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C_0\left\| u_\varepsilon \right\| ^4_{L^{\infty }((0,T)\times {\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}, \end{aligned}$$
where \(D_3\) is a positive constant, which will be specified later. Therefore, by (4.18),
$$\begin{aligned}&\frac{{\mathrm{d}}}{{\mathrm{d}}t}\left( \left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{\alpha ^2}{10\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^4 dx-\frac{\alpha }{\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x\right) \\&\qquad \quad +\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le C(T)\left( 1+D_3+D_3\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\frac{1}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \\&\qquad \quad +\frac{8\alpha ^2\varepsilon }{\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \quad +\frac{4\alpha ^2\varepsilon }{\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _{x}^3u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \quad +C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
It follows from (2.2), (4.2), (4.3) and an integration on (0, t) that
$$\begin{aligned}&\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}-\frac{\alpha ^2}{10\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^4 {\mathrm{d}}x\nonumber \\&\qquad \quad -\frac{\alpha }{\gamma }\int _{{\mathbb {R}}}u_\varepsilon (\partial _x u_\varepsilon )^2 {\mathrm{d}}x+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C_0+ C(T)\left( 1+D_3+D_3\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\frac{1}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad + \frac{8\alpha ^2\varepsilon }{\gamma ^2}\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{t}\left\| \partial _{x}^2u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \quad +\frac{4\alpha ^2\varepsilon }{\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\int _{0}^{t}\left\| \partial _{x}^3u_\varepsilon (s,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}{\mathrm{d}}s\nonumber \\&\qquad \quad +C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) t\nonumber \\&\qquad \le C(T)\left( 1+D_3+D_3\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\frac{1}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad +C(T)\left( 1+\left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})} +\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
(4.19)
Due to (4.1), (4.8) and the Young inequality,
$$\begin{aligned}&\left| \frac{\alpha }{\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon (\partial _x u_\varepsilon )^2 \vert {\mathrm{d}}x\\&\qquad \le \frac{\alpha ^2}{2\gamma ^2}\int _{{\mathbb {R}}}u_\varepsilon ^2(\partial _x u_\varepsilon )^2 {\mathrm{d}}x +\frac{1}{2}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\\&\qquad \le \frac{\alpha ^2}{2\gamma ^2}\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\left\| \partial _x u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+C_0\\&\qquad \le C_0\left\| u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}+C_0\\&\qquad \le C_0\left\| u_\varepsilon \right\| ^4_{L^{\infty }((0,T)\times {\mathbb {R}})}+C_0\\&\qquad \le C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
$$\begin{aligned} \left\| \partial _x u_\varepsilon \right\| ^2_{L^{\infty }((0,T)\times {\mathbb {R}})}\le C_0\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}. \end{aligned}$$
Consequently, by (4.9) and (4.19),
$$\begin{aligned}&\left\| \partial _{x}^2u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}+\frac{\varepsilon }{2}\left\| \partial _{x}^5u_\varepsilon (t,\cdot ) \right\| ^2_{L^2({\mathbb {R}})}\nonumber \\&\qquad \le C(T)\left( 1+D_3+D_3\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\frac{1}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad +C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) +\frac{\alpha ^2}{10\gamma ^2}\left\| u_\varepsilon (t,\cdot ) \right\| ^4_{L^{4}({\mathbb {R}})}\nonumber \\&\qquad \quad +\left| \frac{\alpha }{\gamma }\right| \int _{{\mathbb {R}}}\vert u_\varepsilon (\partial _x u_\varepsilon )^2\vert {\mathrm{d}}x\nonumber \\&\qquad \le C(T)\left( 1+D_3+D_3\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}+\frac{1}{D_3}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) \nonumber \\&\qquad \quad +C(T)\left( 1+\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\right) . \end{aligned}$$
(4.20)
Therefore, by (4.20), we obtain that
$$\begin{aligned} \left( 1-\frac{C(T)}{D_3}\right) \left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}&-C(T)\left( 1+D_3\right) \left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}\\&-C(T)\left( 1+D_3\right) \le 0. \end{aligned}$$
Choosing
$$\begin{aligned} D_3=2C(T), \end{aligned}$$
(4.21)
we have that
$$\begin{aligned} \frac{1}{2}\left\| \partial _{x}^2u_\varepsilon \right\| ^2_{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)\left\| \partial _{x}^2u_\varepsilon \right\| _{L^{\infty }(0,T;L^2({\mathbb {R}}))}-C(T)\le 0, \end{aligned}$$
which give (3.8).
Finally, (4.12) follows from (3.8), (4.20) and (4.21), while (2.8), (3.8), (4.1), (4.7), (4.8) and (4.12) give (4.13). \(\square\)
Observe that, arguing as in Lemma 2.4, we have (2.13) and (2.14). Therefore, arguing as in Sect. 2, we have Theorem 1.1.
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Coclite, G.M., Ruvo, L.d. H4-Solutions for the Olver–Benney equation. Annali di Matematica 200, 1893–1933 (2021). https://doi.org/10.1007/s10231-020-01064-4
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DOI: https://doi.org/10.1007/s10231-020-01064-4