\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 71, pp. 1--24.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2004/71\hfil Gain of regularity]
{Gain of regularity for a Korteweg - de Vries - Kawahara type equation} 

\author[Octavio Paulo Vera Villagr\'{a}n\hfil EJDE-2004/71\hfilneg]
{Octavio Paulo Vera Villagr\'{a}n}  

\address{Facultad de Ingenier\'{\i}a \\
Universidad Cat\'{o}lica de la Sant\'{\i}sima Concepci\'{o}n \\
Paicav\'{\i} 3000, Concepci\'{o}n - Chile}
\email{overa@ucsc.cl \quad  Fax: (41)735300}


\date{}
\thanks{Submitted December 15, 2003. Published May 17, 2004.}
\thanks{Supported by MECESUP 9903, Universidad Cat\a'{o}lica de la
 Sant\a'{\i}sima Concepci\a'{o}n, Chile.}
\thanks{The author dedicates this work to Dr. Patricio Sierralta Standen 
in the Iquique Hospital.}
\subjclass[2000]{35Q53, 47J35}
\keywords{Evolution equations,  weighted Sobolev space}


\begin{abstract}
 We study the existence of local and global solutions, and the gain
 of regularity for the initial value problem associated to the
 Korteweg - de Vries - Kawahara (KdVK) equation perturbed by a
 dispersive term which appears in several fluids dynamics problems.
 The study of gain of regularity is motivated by the results obtained
 by Craig, Kappeler and Strauss \cite{c3}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section] % theorems numbered with section #
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In 1976, Saut and Temam [25] remarked that a solution $u$ of a
Korteweg-de Vries type equation cannot gain or lose
regularity. They showed that if $u(x, 0)=\varphi (x)
\in H^{s}(\mathbb{R})$ for $s\geq 2$, then $u( \cdot, t)\in H^{s}(\mathbb{R})$ for all $t>0$.
The same result was obtained independently by Bona and  Scott [3]
 through a different method.
For the Korteweg-de Vries equation on the line,  Kato [17]
motivated by work of  Cohen \cite{c2} showed that if
$u(x, 0)=\varphi (x)\in L_{b}^2\equiv
H^2(\mathbb{R} )\cap L^2(e^{bx}dx)$ $(b>0)$ then the solution
$u(x, t)$ of the KdV equation becomes
$C^{\infty }$ for all $t>0$. A main ingredient in the proof
was the fact that formally the semi-group
$S(t)=e^{-t \partial _{x}^{3}}$ in $L_{b}^2$ is equivalent
to $S_{b}(t)=e^{-t(\partial _{x} - b)^{3}}$ in $L^2$ when $t>0$.
One would be inclined to believe that this was a special
property of the KdV equation. This is not however the case. The
effect is due to the dispersive nature of the linear part of the
equation.  Kruzkov and  Faminskii \cite{k6} proved that for
$u(x, 0)=\varphi (x)\in L^2$ such that
$x^{\alpha }\varphi (x)\in L^2((0, +\infty ))$ the weak solution
of the KdV equation has $l$-continuous space derivatives for all $t>0$ if
$l<2\alpha $. The proof of this result is based on the
asymptotic behavior of the Airy function and its derivatives,
and on the smoothing effect of the KdV equation which was found
in \cite{k2,k6}. Similar work for some
special nonlinear Schr\"{o}dinger
equations was done by Hayashi et al.
\cite{h1,h2} and  Ponce \cite{p1}.
While the proof of Kato appears to
depend on special a priori estimates, some of its mystery has been
resolved by the result of local gain of finite regularity
for various others linear and nonlinear dispersive equations due
to  Constantin and Saut \cite{c6}, Sjolin \cite{s2},  Ginibre
and Velo \cite{g1} and others. However, all of them require growth
conditions on the nonlinear term.

 All the physically significant
dispersive equations and systems known to us have linear parts
displaying this local smoothing property. To mention only a few,
the KdV, Benjamin-Ono, intermediate long wave, various Boussinesq,
and Schr\"{o}dinger equations are included.
Continuing with the idea
of Craig, Kappeler and Strauss \cite{c5} we study a equation of Korteweg - de
Vries - Kawahara type (KdVK) which appears in fluids dynamics (see \cite{r1}
and references therein).
\begin{equation} \label{e1.1}
u_{{t}} + \eta  u_{xxxxx} + u_{xxx} + u u_{x}=0
\end{equation}
with $-\infty <x<+\infty$, $t>0$ and $\eta \in \mathbb{R}$.
It is shown that $C^{\infty }$
solutions $u(x, t)$ are obtained for all $t>0$ if the
initial data $u(x, 0)$ decays faster
than polynomially on $\mathbb{R} ^{+}=\{x\in \mathbb{R} : x>0\}$
and has certain initial Sobolev regularity.
In section three we prove the main inequality. In section 4 we prove an
important a priori estimate. In section 5 we prove a
basic local-in-time existence and uniqueness theorem. In section 6 we prove a
basic global existence theorem. In section
7 we develop a series of estimates for solutions of equation
\eqref{e1.1} in weighted Sobolev norms. These provide a starting point
for the a priori gain of regularity. In section 8 we prove the following
theorem.

\begin{theorem} \label{mainthm}
Let $T>0$ and $u(x, t)$
be a solution of \eqref{e1.1} in the region $\mathbb{R} \times [0, T]$ such that
\begin{equation} \label{e1.2}
u\in L^{\infty }([0, T]; H^{5}(W_{0L0}))
\end{equation}
for some $L\geq 2$ and all $\sigma >0$. Then
\[
u\in L^{\infty }([0, T]; H^{5 + l}(W_{{\sigma ,L - l,l}}))\bigcap
L^2([0, T]; H^{6 + l}(W_{{\sigma ,L - l,l}}) \cap H^{7 + l}
(W_{\sigma ,L - l - 1,l}))\]
for all $0\leq l\leq L - 1$.
\end{theorem}

\section{Preliminaries}
We consider the initial-value problem
\begin{equation} \label{e2.1}
u_{{t}} + \eta  u_{{xxxxx}} + u_{{xxx}} + u u_{{x}} =0
\end{equation}
with $-\infty <x<+\infty$, $t\in [0, T]$, $T$
an arbitrary positive time, and $\eta \in \mathbb{R} $ is a constant.

As a notation, we use
$ \partial =\partial /\partial x , \partial _{t}=\partial /\partial t$
and $u_{j}=\partial ^{j}u$,
$\partial _{j}=\partial /\partial u_{j}$.

\noindent{\bf Definition.} A function $\xi (x, t)$ belongs to the weight class
$W_{\sigma \,i\,k}$ if it is a positive $C^{\infty }$ function on
$\mathbb{R}\times [0, T], \xi _{x}>0$ and there are constant
$c_j, 0\leq j\leq 5$
such that
\begin{gather}
0<c_1\leq t^{- k} e^{- \sigma  x} \xi (x, t)
\leq c_2\quad \forall  x<-1,\; 0<t<T, \label{e2.2}\\
0<c_3\leq t^{- k} x^{- i} \xi (x, t)\leq c_4\quad \forall
 x>1,\; 0<t<T, \label{e2.3}\\
\left(t |  \xi _{t} | + |  \partial^{j}\xi  |  \right)/
\xi \leq c_5\quad \forall  (x, t)\in \mathbb{R}\times [0, T],  \forall  j\in \mathbb{Z}^{+}.
\label{e2.4}
\end{gather}
% Rmk 1
We remark that, we will always consider $\sigma \geq 0$, $i\geq 1$ and $k\geq 0$.
For example, let
\[
\xi (x)= \begin{cases}
1 + e^{-1/x} & \mbox {for }x>0   \\
1 & \mbox {for }x\leq 0\,;
\end{cases}
\]
then $\xi \in W_{{0 i 0}}$.

\noindent{\bf Definition.}
Let $s$ be a positive integer. We define the space
\[ H^{s}(W_{{\sigma  i k}})=
\{v\colon \mathbb{R} \to \mathbb{R} : \| v \|^2=
\sum _{j=0}^{s}\int_{-\infty }^{+\infty }|
\partial ^{j}v(x) |^2 \xi (x, \cdot  ) dx<+\infty \}
\]
with $\xi \in W_{{\sigma  i k}}$ fixed.
Note that $H^{s}(W_{{\sigma  i k}})$ depends on $t$ because
$\xi =\xi (x, t)$).

\begin{lemma}[\cite{c1}] \label{lm0}
For $\xi \in W_{{\sigma  i 0}}$ and $\sigma \geq 0$,  $i\geq 0$,
there exists a constant $c>0$ such that,
for $u\in H^{1}(W_{{\sigma  i 0}})$,
\[\sup _{x\in \mathbb{R}}| \xi  u^2|
\leq c\int _{-\infty }^{+\infty }\left( |  u | ^2 +
|  \partial u | ^2 \right) \xi  dx\]
\end{lemma}

\noindent {\bf Definition.} For fixed $\xi \in W_{{\sigma  i k}}$, we define
the spaces
\begin{gather*}
L^2([0, T]; H^{s}(W_{{\sigma  i k}}))  =  \{v(x, t):
||| v ||| ^2= \int_{0}^{T}\| v( \cdot  , t) \|^2 dt<+\infty \}
\\
L^{\infty }([0, T]; H^{s}(W_{{\sigma  i k}}))  =
\{v(x, t): |||v||| _{\infty }=
\mathop{\rm ess\,sup} _{t\in [0, T]} \| v( \cdot  , t) \|<+\infty \},
\end{gather*}
where $s$ is a positive integer.
Note that The usual Sobolev space $H^{s}(\mathbb{R})$ is
$H^{s}(W_{{0 0 0}})$, i.e., without weight.

We shall derive the a priori estimates assuming that the solution is
$C^{\infty}$, bounded as $x\to -\infty $, and rapidly decreasing
together with all of its derivatives as $x\to +\infty$.
We consider the following KdVK equation
\begin{equation} \label{e2.5}
u_{{t}} + \eta  u_5 + u_3 + u u_1 =0
\end{equation}
with $\eta \in \mathbb{R} $ constant. This equation will be studied for
$-\infty <x<+\infty $, $t\in[0, T]$
with $T$ an arbitrary positive time.

\section{Main Inequality}

\begin{lemma} \label{lm3.1}
Let  $u$ be a solution to \eqref{e2.5} with enough Sobolev regularity
(for instance, $u\in H^{N}(\mathbb{R}), N\geq \alpha + 5$), then
\[
\partial _{t}\int_{\mathbb{R}}\xi  u_{\alpha }^2 dx +
\int_{\mathbb{R}}{\mu _1} u_{\alpha + 1}^2 dx + \int_{\mathbb{R}}{\mu _2} u_{\alpha + 2}^2 dx +
\int_{\mathbb{R}}\theta  u_{\alpha }^2 dx +
\int_{\mathbb{R}}R_{\alpha }  dx\leq 0\]
with
\begin{gather*}
\mu _1  =   -c_5(5\eta + 3)\xi\quad \mbox {for }\quad\eta < -3/5
\quad (\mbox{Natural Condition}) \\
\mu _2   =   -5\eta \partial \xi \\
\theta   =  -\xi _{t} - \eta \partial ^{5}\xi -\partial ^{3}\xi - \partial (\xi u)\\
R_{\alpha }  =  O(u_{\alpha }, \dots)
\end{gather*}
\end{lemma}

\begin{proof} Taking $\alpha $-derivatives of \eqref{e2.5}
(for $\alpha \geq 3$) over  $x\in \mathbb{R} $
\begin{equation} \label{e3.1}
\partial _{{t}}u_{{\alpha }} + \eta  u_{{\alpha + 5}} + u_{{\alpha + 3}} +
u u_{{\alpha + 1}} + R_{{\alpha }}(u_{{\alpha }}, u_{{\alpha - 1}}, \dots)=0
\end{equation}
Multiply this equation by
$2\xi u_{\alpha }$ and, integrate over $x\in \mathbb{R} $ to have
\begin{align*}
2\int_{\mathbb{R}}\xi  u_{\alpha } \partial _{t}u_{\alpha } dx
+ 2 \eta \int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 5} dx
+ 2\int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 3} dx &\\
+2\int_{\mathbb{R}}\xi  u u_{\alpha } u_{\alpha + 1} dx +
2\int_{\mathbb{R}}\xi  u_{\alpha } R_{\alpha } dx &=0
\end{align*}
Integrating by parts,
\begin{align*}
\partial _{t}\int_{\mathbb{R}}\xi  u_{\alpha }^2 dx
+\int_{\mathbb{R}}( 5 \eta  \partial ^{3}\xi
+3 \partial \xi  ) u_{\alpha + 1}^2 dx &\\
+ \int_{\mathbb{R}}- 5 \eta  \partial \xi  u_{\alpha + 2}^2 dx +
\int_{\mathbb{R}}\theta  u_{\alpha }^2 dx + \int_{\mathbb{R}}R_{\alpha } dx &=0
\end{align*}
with $\theta = - \xi _{t} - \eta  \partial ^{5}\xi -
\partial^{3}\xi -
\partial (\xi  u)$.
Using \eqref{e2.4}, it follows for $c_5>0$ that
\[
\partial _{t}\int_{\mathbb{R}}\xi  u_{\alpha }^2 dx -
c_5( 5 \eta + 3 )\int_{\mathbb{R}}\xi  u_{\alpha + 1}^2 dx - 5 \eta  \int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx +
\int_{\mathbb{R}}\theta  u_{\alpha }^2 dx + \int_{\mathbb{R}}R_{\alpha } dx \leq 0\,,
\]
from where we obtain the {\em main inequality.}
\begin{equation} \label{e3.2}
\partial _{{t}}\int_{\mathbb{R}}\xi  u_{{\alpha }}^2 dx
+ \int_{\mathbb{R}}{\mu _1} u_{{\alpha + 1}}^2 dx
+ \int_{\mathbb{R}}{\mu _2} u_{{\alpha + 2}}^2 dx
+ \int_{\mathbb{R}}\theta  u_{{\alpha }}^2 dx
+ \int_{\mathbb{R}}R_{{\alpha }}  dx\leq 0
\end{equation}
with
\begin{gather*}
\mu _1  =   -c_5(5\eta + 3)\xi \quad \mbox {for }\eta < -3/5
\quad (\mbox {Natural Condition}) \\
\mu _2   =   -5\eta \partial \xi \\
\theta   =  -\xi _{{t}} - \eta \partial ^{5}\xi -
\partial ^{3}\xi - \partial (\xi  u)\\
R_{\alpha }  =  O(u_{{\alpha }}, \dots)
\end{gather*}
\end{proof}

\begin{lemma} \label{lm3.2}
 For $\mu _2 \in W_{{\sigma  i k}}$ an arbitrary weight
function and $\eta <-3/5$, there exists $\xi \in W_{{\sigma ,i + 1,k}}$
that satisfies
\begin{equation} \label{e3.3}
\mu _2 = -5\eta \partial \xi
\end{equation}
\end{lemma}

Indeed, we have
\begin{equation} \label{e3.4}
\xi = -\frac {1}{5 \eta } \int_{-\infty }^{x} \mu _2 (y, t) dy
\end{equation}

\begin{lemma} \label{lm3.3}
 The expression $R_{\alpha }$ in the inequality of Lemma \ref{lm3.1} is a
sum of terms of the form
\begin{equation} \label{e3.5}
\xi  u_{{\nu _1}} u_{{\nu _2}} u_{{\alpha }}
\end{equation}
where $1\leq \nu _1\leq \nu _2\leq \alpha $.
\begin{equation} \label{e3.6}
\nu _1 + \nu _2  =  \alpha + 1
\end{equation}
\end{lemma}

\begin{proof} Differentiating \eqref{e2.5} once with respect to $x$ and
multiplying by $2\xi u_{{\bf 1}}$ we have
\[
2 \xi  u_{{\bf 1}}\partial _{t}u_1 + 2 \eta  \xi  u_{{\bf 1}}u_6
+ 2 \xi  u_{{\bf 1}} u_{4} + 2 \xi  u_{{\bf 1}} u u_{2}
+ \xi  u_{{\bf 1}} u_1 u_1 =0\,,
\]
where
$R_1 \xi  u_{{\bf 1}} =\xi  u_1 u_1 u_{{\bf 1}}$.
Taking  2-$x$derivatives of the equation \eqref{e2.5}, and multiplying by
$2\xi u_{{\bf 2 }}$ we have
\[
2 \xi  u_{{\bf 2}} \partial _{t}u_{2} + 2 \eta  \xi  u_{{\bf 2}} u_7
+ 2 \xi  u_{{\bf 2}} u_5 + 2 \xi  u_{{\bf 2}} u u_{3}
+ 6 \xi  u_{{\bf 2}} u_1 u_{2} =0
\]
where $ R_{2} \xi  u_{{\bf 2}} = 6 \xi  u_1 u_{2} u_{{\bf 2}}$.
Taking  3-$x$derivatives of the equation \eqref{e2.5}, and multiplying by
$2\xi u_{{\bf 3}}$ we have
\[
2 \xi  u_{{\bf 3}} \partial _{t}u_{3} + 2 \eta  \xi  u_{{\bf 3}} u_8
+ 2 \xi  u_{{\bf 3}} u_6 + 2 \xi  u_{{\bf 3}} u u_{4}
+ 8 \xi  u_{{\bf 3}} u_1 u_{3} + 6 \xi  u_{{\bf 3}} u_{2} u_{2} =0
\]
where $R_{3} \xi  u_{{\bf 3}} = 8 \xi  u_1 u_{3} u_{{\bf 3}} +
6 \xi  u_{2} u_{2} u_{{\bf 3}}$.
Taking  4-$x$derivatives of \eqref{e2.5}, and multiplying by
$2\xi u_{{\bf 4 }}$ we have
\[
2 \xi  u_{{\bf 4}} \partial _{t}u_{4} + 2 \eta  \xi  u_{{\bf 4}} u_9
+ 2 \xi  u_{{\bf 4}} u_7 + 2 \xi  u_{{\bf 4}} u u_5
+ 10 \xi  u_{{\bf 4}} u_1 u_{4} + 20 \xi  u_{{\bf 4}} u_{2} u_{3} =0
\]
where $R_{4} \xi  u_{{\bf 4}}=10 \xi  u_1 u_{4} u_{{\bf 4}} + 20 \xi  u_{2} u_{3}
u_{{\bf 4}}$.
Taking  5-$x$derivatives of  \eqref{e2.5}, and multiplying by
$2\xi u_{{\bf 5 }}$ we have
\[
2 \xi  u_{{\bf 5}} \partial _{t}u_5 + 2 \eta  \xi  u_{{\bf 5}} u_{10}
+ 2 \xi  u_{{\bf 5}} u_8 + 2 \xi  u_{{\bf 5}} u u_6
+ 12 \xi  u_{{\bf 5}} u_1 u_5 + 30 \xi  u_{{\bf 5}} u_{2} u_{4}
+ 20 \xi  u_{{\bf 5}} u_{3} u_{3}=0
\]
where $R_5 \xi  u_{{\bf 5}}=
12 \xi  u_1 u_5 u_{{\bf 5}} +
30 \xi  u_{2} u_{4} u_{{\bf 5}} +
20 \xi  u_{3} u_{3} u_{{\bf 5}}$.
Throw away the first terms in each derivative and the result follows.
\end{proof}

\section{An a priori estimate}

We show a fundamental {\em a priori} estimate used for a
basic local-in-time existence  theorem. We construct a mapping
$Z:L^{\infty }([0, T]; H^{s}(\mathbb{R}))
\to L^{\infty }([0, T]; H^{s}(\mathbb{R}))$
with the following property:
Given $u^{(n)}=Z(u^{(n - 1)})$
and $\|u^{(n - 1)}\|_{{s}}\leq c_0$ then
$\| u^{(n)}\|_{{s}}\leq c_0$, where
$s$ and $c_0>0$ are constants. This
property tells us, in fact, that $Z:\mathbb{B}_{c_0}(0)
\to \mathbb{B}_{c_0}(0)$ where
$\mathbb{B}_{c_0}(0)=\{ v(x, t); \| v( \cdot  , t) \|_{{s}}\leq c_0 \}$
is a ball in
$L^{\infty }([0, T]; H^{s}(\mathbb{R}))$.
To guarantee this property, we will appeal to an a
priori estimate which is the main object of this section.
Differentiating \eqref{e2.5} four times leads to
\begin{equation} \label{e4.1}
\partial _{t}u_{4} + \eta  u_9 + u_7 + u u_5 +
5 u_1 u_{4} + 10 u_{2} u_{3}=0
\end{equation}
Let $u=\wedge v$ where $\wedge =(I - \partial ^{4})^{-1}$. Then
$\partial _{{t}}u_4= -v_{{t}} + u_{{t}}$ by
replacing in \eqref{e4.1} we have
\begin{equation} \label{e4.2}
-v_{t} + \eta {\wedge v}_9 + {\wedge v}_7 +
{\wedge v}{\wedge v}_5 +5{\wedge v}_1{\wedge v}_{4} +
10{\wedge v}_{2}{\wedge v}_{3}
   - [\eta {\wedge v}_5 +{\wedge v}_{3} -{\wedge v}{\wedge v}_1]=0
\end{equation}
The \eqref{e4.2} is linearized by substituting a new variable
$w$ in each coefficient;
\begin{equation} \label{e4.3}
-v_{t} + \eta {\wedge v}_9 + {\wedge v}_7 +
{\wedge w}{\wedge v}_5 +5{\wedge w}_1{\wedge v}_{4} +
10{\wedge w}_{2}{\wedge v}_{3}- [\eta {\wedge v}_5 +
{\wedge v}_{3} -{\wedge w}{\wedge v}_1]=0
\end{equation}
Equation \eqref{e4.3} is a linear equation at each iteration which can be
solved in any interval of time in which the coefficients are defined.
This equation has the form
\begin{equation} \label{e4.4}
\partial _{t}v=\eta  {\wedge v}_9^{(n)} + {\wedge v}_7^{(n)} +
b^{(1)} {\wedge v}_5^{(n)} + b^{(2)} {\wedge v}_{4}^{(n)} + b^{(3)}
\end{equation}
We consider the following lemma that  will help us setting
up the iteration scheme.

\begin{lemma} \label{lm4.1}
Let $\eta <-3/5$. Given initial data
$\varphi \in H^{\infty }(\mathbb{R})=\bigcap _{{N\geq 0}}H^{N}(\mathbb{R})$
there exists a unique solution of \eqref{e4.4}
where $b^{(1)}=b^{(1)}(\wedge w), b^{(2)}=b^{(2)}({\wedge w}_1)$ and $b^{(3)}=
b^{(3)}({\wedge w}_{3}, \dots, \wedge w)$ are smooth bounded
coefficients with $w\in H^{\infty }(\mathbb{R} )$.
The solution is defined in
any time interval in which the coefficients are defined.
\end{lemma}

\begin{proof} Let $T>0$ be arbitrary and $M>0$ a constant. Let
\[
\mathcal{L} = 2 \xi  (\partial _{t} - \eta  {\wedge \partial }^{9} -
{\wedge \partial }^{7} - b^{(1)} {\wedge \partial }^{5} - b^{(2)} {\wedge \partial }^{4})\]
where $0<c_6\leq \xi \leq c_7$.
We consider the bilinear form
$B\colon \mathcal{D}\times \mathcal{D}\mapsto \mathbb{R}$,
\[ B(u, v) =  \langle u, v\rangle
=\int_{0}^{T}\int_{\mathbb{R}}e^{- M t} u v  \,dx\, dt\, % *
\]
where $\mathcal{D}=\{u\in C_{0}^{\infty }(\mathbb{R}
\times [0, T]): u(x, 0)=0\}$.
We have
\begin{align*}
\int_{\mathbb{R}}\mathcal{L}u\cdot udx
& =  2\int_{\mathbb{R}}\xi uu_{t}dx - 2\eta \int_{\mathbb{R}}\xi u{\wedge u}_9dx
- 2\int_{\mathbb{R}}\xi u{\wedge u}_7dx \\
&\quad - 2\int_{\mathbb{R}}\xi b^{(1)}u{\wedge u}_5dx - 2\int_{\mathbb{R}}
\xi b^{(2)}u{\wedge u}_{4}dx
\end{align*}
Each term is treated separately. The first term yields
\[
2\int_{\mathbb{R} }\xi  u u_{t} dx  =
\partial _{t}\int _{\mathbb{R} }\xi  u^2 dx - \int_{\mathbb{R} }\xi _{t} u^2 dx
\]
In the second term, by integrating by parts we obtain
\begin{align*}
&-2\eta \int_{\mathbb{R} }\xi  u {\wedge u}_9 dx\\
& = -2 \eta \int_{\mathbb{R} }\xi  \wedge (I - \partial ^{4}) u {\wedge u}_9 dx
- 2 \eta \int_{\mathbb{R} }\xi  {\wedge u} {\wedge u}_9 dx +
2 \eta \int_{\mathbb{R} }\xi  {\wedge u}_{4} {\wedge u}_9 dx \\
& =  \eta \int_{\mathbb{R}}\partial ^{9}\xi  ( \wedge u )^2 dx - 9 \eta \int_{\mathbb{R}}\partial ^{7}\xi  ( \wedge u_1 )^2 dx + 27 \eta \int_{\mathbb{R}}\partial ^{5}\xi  (\wedge u_{2})^2 dx \\
&\quad  -  30\eta \int_{\mathbb{R}}\partial ^{3}\xi (\wedge u_{3})^2 dx - \eta \int_{\mathbb{R}}( \partial ^{5}\xi - 9 \partial \xi  ) ( \wedge u_{4} )^2 dx + 5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  ( \wedge u_5 )^2 dx \\
& \quad -  5\eta \int_{\mathbb{R}}\partial \xi  (\wedge u_6)^2 dx\,.
\end{align*}
All the others terms are calculated of the same way. We have
\begin{align*}
&\int_{\mathbb{R}}\mathcal{L}u \cdot  u  dx \\
& = \partial _{t}\int_{\mathbb{R}}\xi  u^2 dx -
\int_{\mathbb{R}}\xi _{t} u^2 dx +\eta \int_{\mathbb{R}}\partial ^{9}\xi  ( {\wedge u} )^2 dx -
9 \eta\int_{\mathbb{R}}\partial ^{7}\xi  ( {\wedge u}_1 )^2 dx  \\
&\quad +  27 \eta \int_{\mathbb{R}}\partial ^{5}\xi  ( {\wedge u}_{2} )^2 dx -
30 \eta \int_{\mathbb{R}}\partial ^{3} \xi  ( {\wedge u}_{3} )^2 dx -
\eta \int_{\mathbb{R}}( \partial ^{5}\xi - 9 \partial \xi  ) ( {\wedge u}_{4} )^2 dx  \\
&\quad +  5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  ({\wedge u}_5)^2 dx -
5 \eta \int_{\mathbb{R}}\partial
\xi  ({\wedge u}_6)^2 dx + \int_{\mathbb{R}}\partial ^{7}\xi  ({\wedge u})^2 dx - 7\int_{\mathbb{R}}\partial ^{5}\xi  ({\wedge u}_1)^2 dx \\
&\quad + 14\int_{\mathbb{R}}\partial ^{3}
\xi  ( {\wedge u}_{2} )^2 dx - 7\int_{\mathbb{R}}\partial \xi  ({\wedge u}_{3})^2 dx
- \int_{\mathbb{R}}\partial ^{3}\xi ({\wedge u}_{4})^2 dx + 3\int_{\mathbb{R}}\partial
\xi  ({\wedge u}_5)^2 dx \\
& \quad +  \int_{\mathbb{R}}\partial ^{5}( \xi  b^{(1)} )({\wedge u})^2 dx
-  5\int_{\mathbb{R}}\partial ^{3}( \xi  b^{(1)} ) ( {\wedge u}_1 )^2 dx +
5\int_{\mathbb{R}}\partial ( \xi  b^{(1)} ) ( {\wedge u}_{2} )^2 dx \\
&\quad - \int_{\mathbb{R}}\partial (\xi  b^{(1)} ) ( {\wedge u}_{4})^2 dx - \int_{\mathbb{R}}\partial ^{4}( \xi  b^{(2)} ) ( {\wedge u} )^2 dx +
4\int_{\mathbb{R}}\partial ^2
( \xi  b^{(2)} ) ( {\wedge u}_1 )^2 dx\\
& \quad -  \int_{\mathbb{R}}\xi  b^{(2)}  ( {\wedge u}_{4} )^2 dx
+ 2\int_{\mathbb{R}}\xi  b^{(2)} ({\wedge u}_{4})^2 dx\,.
\end{align*}
It follows that
\begin{align*}
&\int_{\mathbb{R} }\mathcal{L}u \cdot  u  dx \\
& = \partial _{t}\int_{\mathbb{R} }\xi  u^2 dx - 5 \eta \int_{\mathbb{R}}\partial \xi  ( {\wedge u}_6 )^2 dx +
\int_{\mathbb{R} }( 5 \eta  \partial ^{3}\xi + 3 \partial \xi  )
( {\wedge u}_5 )^2 dx   \\
& \quad +  \int_{\mathbb{R} }( - \eta  \partial ^{5}\xi -
\partial ^{3}\xi + 9 \eta  \partial \xi -
\partial ( \xi  b^{(1)} ) + 2 \xi  b^{(2)} ) ( {\wedge u}_{4} )^2 dx \\
&\quad +  \int_{\mathbb{R} }( - 30 \eta  \partial ^{3}\xi -
7 \partial \xi  ) ( {\wedge u}_{3} )^2 dx - \int_{\mathbb{R} }\xi _{t} u^2 dx \\
&\quad +  \int_{\mathbb{R} }( 27 \eta  \partial ^{5}\xi + 14 \partial ^{3}\xi +
5 \partial ^{3} ( \xi  b^{(1)} ) - 2 \xi  b^{(2)} ) ( {\wedge u}_{2} )^2 dx \\
&\quad  +  \int_{\mathbb{R} }( - 9 \eta  \partial ^{7}\xi - 7 \partial ^{5}\xi -
5 \partial ^{3}( \xi  b^{(1)} )
+ 4 \partial ^2( \xi  b^{(2)} ) ) ( {\wedge u} )^2 dx
\end{align*}
Using \eqref{e2.4}, ${\wedge u}_{n}=(I - (I - \partial ^{4})){\wedge u}_{{n - 4}}=
{\wedge u}_{{n - 4}} - u_{{n - 4}}$ for $n$ positive integer
and standard estimates it follows that
\begin{equation} \label{e4.5}
\int_{\mathbb{R}}\mathcal{L}u \cdot  u  dx\geq {\partial _{t}\int_{\mathbb{R}}}\xi  u^2 dx -
c\int_{\mathbb{R}}\xi  u^2 dx
\end{equation}
Multiply this equation by $e^{-M t}$, and integrate with respect to $t$ for $t\in [0, T]$ and $u\in \mathcal{D}$.
\begin{align*}
&\int_{0}^{T}\int_{\mathbb{R}}e^{- M t} \mathcal{L}u \cdot  u  dx dt \\
& \geq \int_{0}^{T}e^{- M t} \big( \partial _{t}\int_{\mathbb{R}}\xi  u^2 dx \big) dt -
c\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx dt \\
& =  e^{- M t}\int_{\mathbb{R}}\xi  u^2(x, t) dx |_{0}^{T} +
M\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx dt  \\
&\quad -  c\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{-M t} u^2 dx dt \\
& =  e^{- M T}\int_{\mathbb{R}}\xi (x, T) u^2(x, T) dx +
M\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx dt  \\
&\quad -  c\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx\, dt.
\end{align*}
Thus
\begin{align*}
&\int_{0}^{T}\int_{\mathbb{R}}e^{- M t} \mathcal{L}u \cdot  u  dx dt \\
& \geq e^{- M T}\int_{\mathbb{R}}\xi (x, T) u^2(x, T) dx +
(M - c)\int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx dt \\
& \geq  \int_{0}^{T}\int_{\mathbb{R}}\xi  e^{- M t} u^2 dx \,dt
\end{align*}
provided $M$ is chosen large enough. Then
$\langle \mathcal{L}u, u\rangle \geq \langle u, u\rangle$, for all
$u\in \mathcal{D}$.
Let
$\mathcal{L}^{*} = 2\xi (-\partial _{t} +
\eta {\wedge \partial }^{9} +{\wedge \partial }^{7} +
b^{(1)}{\wedge \partial }^{5} - b^{(2)}{\wedge \partial }^{4})$
be the formal adjoint of $\mathcal{L}$.
Let $\mathcal{D}^{*}=\{w\in C_{0}^{\infty }(\mathbb{R} \times [0, T]):w(x, L)=0\}$.
In the same way we prove that
\begin{equation} \label{e4.6}
\langle \mathcal{L}^{*}w, w\rangle \geq \langle w, w\rangle
\quad\forall w\in \mathcal{D}^{*}
\end{equation}
 From this equation, we have that $\mathcal{L}^{*}$ is one-one. Therefore
$\langle \mathcal{L}^{*}w, \mathcal{L}^{*}v\rangle $ is an inner product on
$\mathcal{D}^{*}$.
We denote by $X$ the completion of $\mathcal{D}^{*}$ with respect to this
inner product. By the Riesz Representation Theorem,
there exists a unique solution
 $V\in X$, such that for any $w\in \mathcal{D}^{*}$,
$\langle \xi b^{(3)}, w\rangle =\langle \mathcal{L}^{*}V, \mathcal{L}^{*}w\rangle $
where we use that $\xi b^{(3)}\in X$.
Then if $v = \mathcal{L}^{*}V$ we have
$\langle v, \mathcal{L}^{*}w\rangle =\langle \xi  b^{(3)}, w\rangle $
 or
$\langle \mathcal{L}^{*}w, v\rangle = \langle w, \xi  b^{(3)}\rangle $.
Hence $v=\mathcal{L}^{*}V$ is a weak solution of $\mathcal{L}v=\xi b^{(3)}$
with $v\in L^2(\mathbb{R} \times [0, T])\simeq
L^2([0, T]; L^2(\mathbb{R}))$.

\noindent\textbf{Remark} %rmk 5.
To obtain higher regularity of the solution, we repeat the proof
with higher derivatives. It is a standard approximation
procedure to obtain a result for general initial data.\smallskip

The next step is to estimate the corresponding solutions $v=v(x, t)$ of the
equation \eqref{e4.3} via
the coefficients of that equation.


\begin{lemma} \label{lm4.2}
Let  $v, w\in C^{k}([0,+\infty ); H^{N}(\mathbb{R}))$ for all $k, N$ which satisfy
\eqref{e4.3}. Let $0<c_8\leq \xi \leq c_9$ and $\eta <-3/5$.
For each integer $\alpha $ there exist
positive nondecreasing functions $G$ and $F$ such that for all $t\geq 0$,
\begin{equation} \label{e4.7}
 \partial _{t}\int_{\mathbb{R}}\xi  v_{\alpha }^2 dx \leq
G(\| w \|_{\lambda }) \| v \|_{\alpha }^2 + F(\| w \|_{\alpha })
\end{equation}
where $\| \cdot  \|_{\alpha }$ is the norm in
$H^{\alpha }(\mathbb{R})$ and $
\lambda =\mbox { max }\{1, \alpha \}$.
\end{lemma}


\begin{proof} Differentiating $\alpha $-times the equation \eqref{e4.3},
for some $\alpha \geq 0$, we obtain
\begin{equation} \label{e4.8}
-\partial _{{t}}v_{\alpha } + \eta {\wedge v}_{\alpha + 9} + {\wedge v}_{\alpha + 7} + \sum _{j=6}^{\alpha + 5}h^{(j)}{\wedge v}_{j} + c_{10}{\wedge v}_{3}{\wedge w}_{\alpha + 2} +
p(\wedge w_{\alpha + 1}, \dots)=0
\end{equation}
where $h^{(j)}$ is a smooth function depending on
$\wedge w_{i}, \dots$ with
$i=5 + \alpha - j$.
We multiply equation \eqref{e4.8} by $2\xi v_{\alpha }$, and integrate over
$\mathbb{R}$,
\begin{equation}  \label{e4.9}
\begin{aligned}
-2\int_{\mathbb{R}}\xi v_{\alpha }\partial _{t}v_{\alpha }dx
+ 2\eta \int_{\mathbb{R}}\xi v_{\alpha }{\wedge v}_{\alpha + 9}dx &\\
+ 2\int_{\mathbb{R}}\xi v_{\alpha }{\wedge v}_{\alpha + 7}dx
+ 2\sum _{j=6}^{\alpha + 5}\int_{\mathbb{R}}\xi h^{(j)}v_{\alpha }{\wedge v}_{j}dx &\\
+ 2c_{10}\int_{\mathbb{R}}\xi v_{\alpha }{\wedge v}_{3}{\wedge w}_{\alpha + 2}dx
+ 2\int_{\mathbb{R}}\xi v_{\alpha } p(\wedge w_{\alpha + 1}, \dots)dx & =0
\end{aligned}
\end{equation}
Each of these terms is treated separately.
The first term yields
\[
- 2\int_{\mathbb{R}}\xi  v_{\alpha } \partial _{t}v_{\alpha } dx =
 - \partial _{t}\int_{\mathbb{R}}\xi  v_{\alpha }^2 dx +
\int_{\mathbb{R}}\xi _{t} v_{\alpha }^2 dx
\]
In the second term we have, by integrating by parts
\begin{align*}
&2\eta \int_{\mathbb{R}}\xi  v_{\alpha } {\wedge v}_{\alpha + 9} dx \\
& = 2\eta \int_{\mathbb{R}}\xi  {\wedge (I - \partial ^{4})}
  v_{\alpha } {\wedge v}_{\alpha + 9} dx\\
&=  2\eta \int_{\mathbb{R}}\xi  {\wedge v}_{\alpha }{\wedge v}_{\alpha + 9} dx -
2 \eta \int_{\mathbb{R}}\xi  {\wedge v}_{\alpha + 4}{\wedge v}_{\alpha + 9} dx \\
& =  -\eta \int_{\mathbb{R}}\partial ^{9}\xi  ( {\wedge v}_{\alpha } )^2 dx +
9 \eta \int_{\mathbb{R}}\partial ^{7}\xi  ( {\wedge v}_{\alpha + 1} )^2 dx
- 27 \eta \int_{\mathbb{R}}\partial ^{5}\xi  ( {\wedge v}_{\alpha + 2} )^2 dx \\
&\quad + 30 \eta \int_{\mathbb{R}}\partial ^{3}\xi  ( {\wedge v}_{\alpha + 3} )^2 dx
+ \eta \int_{\mathbb{R}}( \partial ^{5}\xi
- 9 \partial \xi  )( {\wedge v}_{\alpha + 4} )^2 dx \\
&\quad - 5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  ( {\wedge v}_{\alpha + 5} )^2 dx
 +  5 \eta \int_{\mathbb{R}}\partial \xi  ( {\wedge v}_{\alpha + 6} )^2 dx
\end{align*}
The others terms are treated similarly.
Replacing the equations obtained, on \eqref{e4.9}, we have
\begin{align*}
& -\partial _{t}\int_{\mathbb{R}}\xi  v_{\alpha }^2 dx +
 \int_{\mathbb{R}}\xi _{t} v_{\alpha }^2 dx -
 \eta \int_{\mathbb{R}}\partial ^{9}\xi  ({\wedge v}_{\alpha })^2 dx +
 9 \eta \int_{\mathbb{R}}\partial ^{7}\xi  ({\wedge v}_{\alpha + 1})^2 dx  \\
&  -27\eta \int_{\mathbb{R}}\partial ^{5}\xi  ({\wedge v}_{\alpha + 2})^2 dx +
  30 \eta \int_{\mathbb{R}}\partial ^{3}\xi  ({\wedge v}_{\alpha + 2})^2 dx -
  3 \delta \int_{\mathbb{R}}\partial ^{3}\xi  ({\wedge v}_{\alpha + 3})^2 dx \\
& +\eta \int_{\mathbb{R}}(\partial ^{5}\xi - 9 \partial \xi ) ({\wedge v}_{\alpha  + 4})^2
  dx - 5 \eta \int_{\mathbb{R}}\partial ^{3}
  \xi  ({\wedge v}_{\alpha + 5})^2 dx + 5 \eta \int_{\mathbb{R}}\partial \xi
 ({\wedge v}_{\alpha + 6})^2 dx \\
& -\int_{\mathbb{R}}\partial ^{7}\xi  ({\wedge v}_{\alpha })^2 dx
 + 7\int_{\mathbb{R}}\partial \xi ({\wedge v}_{\alpha + 1})^2 dx
 - 14\int_{\mathbb{R}}\partial ^{3}\xi  ({\wedge v}_{\alpha + 2})^2 dx\\
& +7\int_{\mathbb{R}}\partial \xi  ({\wedge v}_{\alpha + 3})^2 dx  
 +\int_{\mathbb{R}}\partial ^{3}\xi  ({\wedge v}_{\alpha + 4})^2 dx
 -3\int_{\mathbb{R}}\partial \xi  ({\wedge v}_{\alpha + 5})^2 dx\\
&- \int_{\mathbb{R}}\partial ^{5}( \xi  h^{(\alpha + 5)} )
 ({\wedge v}_{\alpha })^2 dx 
 -\int_{\mathbb{R}}\partial ^{3}( \xi  h^{(\alpha + 5)} )
 ({\wedge v}_{\alpha + 1})^2 dx \\
& - 5\int_{\mathbb{R}}\partial ( \xi  h^{(\alpha + 5)} )
 ({\wedge v}_{\alpha + 2})^2 dx 
 + \int_{\mathbb{R}}\partial ( \xi  h^{(\alpha + 5)} )
 ({\wedge v}_{\alpha + 4})^2 dx \\
& +2\sum_{j=6}^{\alpha + 4}\int_{\mathbb{R}}\xi  h^{(j)} v_{\alpha }
 {\wedge v}_{j} dx 
+ 2 c_{10}\int_{\mathbb{R}}\xi  v_{\alpha }
 {\wedge v}_{3} {\wedge w}_{\alpha + 2} dx \\
&+ 2\int_{\mathbb{R}}\xi  v_{\alpha } p({\wedge w}_{\alpha + 1}, \dots) dx=0
\end{align*}
and
\begin{align*}
&\partial _{t}\int_{\mathbb{R}}\xi  v_{\alpha }^2 dx \\
& = 5 \eta \int_{\mathbb{R}}\partial \xi  ( {\wedge v}_{\alpha + 6} )^2 dx
- \int_{\mathbb{R}}( 5 \eta  \partial ^{3}\xi + 3 \partial \xi  ) ( {\wedge
v}_{\alpha + 5} )^2 dx + \int_{\mathbb{R}}\xi _{t} ( {\wedge v}_{\alpha } )^2 dx  \\
&\quad + \int_{\mathbb{R}}( \eta  \partial ^{5}\xi + \partial ^{3}\xi - 9 \eta  \partial \xi + \partial ( \xi  h^{(\alpha + 5)} ) ) ( {\wedge v}_{\alpha + 4} )^2 dx \\
&\quad + \int_{\mathbb{R}}(30\eta \partial ^{3}\xi 
+ 7\partial \xi )({\wedge v}_{\alpha + 3})^2dx \\
&\quad + \int_{\mathbb{R}}(-27\eta \partial ^{5}\xi - 14 \partial ^{3}\xi - 5\partial (\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha + 2})^2dx \\
&\quad + \int_{\mathbb{R}}(9\eta \partial ^{7}\xi + 7\partial \xi 
+ \partial ^{3}(\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha + 1})^2dx \\
&\quad + \int_{\mathbb{R}}(-\eta  \partial ^{9}\xi - \partial ^{7}\xi 
- \partial ^{5}(\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha })^2dx 
 + 2\sum _{j=6}^{\alpha + 4}\xi  h^{(j)} v_{\alpha } {\wedge v}_{j} dx \\
&\quad + 2 c_{10}\int_{\mathbb{R}}\xi  v_{\alpha } {\wedge v}_{3} {\wedge w}_{\alpha + 2} dx +
2\int_{\mathbb{R}}\xi  v_{\alpha } p( {\wedge w}_{\alpha + 1}, \dots) dx
\end{align*}
Using \eqref{e2.4} we have that the first and the second term in the right
hand side of the above expression are nonpositive. Hence
\begin{align*}
\partial _{t}\int_{\mathbb{R}}\xi  v_{\alpha }^2 dx 
& \leq \int_{\mathbb{R}}( \eta \partial ^{5}\xi + \partial ^{3}\xi
- 9\eta  \partial \xi + \partial ( \xi  h^{(\alpha + 5)} ) )
( {\wedge v}_{\alpha + 4} )^2 dx 
+ \int_{\mathbb{R}}\xi _{t}({\wedge v}_{\alpha })^2dx  \\
& \quad+ \int_{\mathbb{R}}(30\eta \partial ^{3}\xi 
+ 7\partial \xi )({\wedge v}_{\alpha + 3})^2dx \\
&\quad + \int_{\mathbb{R}}(-27\eta \partial ^{5}\xi - 14\partial ^{3}\xi 
- 5\partial (\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha + 2})^2dx \\
&\quad +  \int_{\mathbb{R}}(9\eta \partial ^{7}\xi + 7\partial \xi 
+ \partial ^{3}(\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha + 1})^2dx \\
&\quad + \int_{\mathbb{R}}(-\eta \partial ^{9}\xi - \partial ^{7}\xi 
- \partial ^{5}(\xi h^{(\alpha + 5)}))({\wedge v}_{\alpha })^2dx 
+ 2\sum _{j=6}^{\alpha + 4}\xi  h^{(j)} v_{\alpha } {\wedge v}_{j} dx \\
& \quad+ 2 c_{10}\int_{\mathbb{R}}\xi  v_{\alpha } {\wedge v}_{3} {\wedge w}_{\alpha + 2} dx 
+ 2\int_{\mathbb{R}}\xi  v_{\alpha } p({\wedge w}_{\alpha + 1}, \dots) dx
\end{align*}
Using that ${\wedge v}_{{n}}={\wedge v}_{{n - 4}} - v_{{n - 4}}$
and standard estimates, the Lemma follows.
\end{proof}

\section{Uniqueness and Existence of a Local Solution}

In this section, we study the uniqueness and the
existence of local strong solutions in the Sobolev space
$H^{N}(\mathbb{R})$ for $N\geq 5 $
for the problem \eqref{e2.5}.
To establish the  existence of strong solutions for
\eqref{e2.5} we use the a priori estimate together with an
approximation procedure.

\begin{theorem}[Uniqueness] \label{thm5.1}
Let $\eta <-3/5, \varphi \in H^{N}(\mathbb{R})$
with $N\geq 5$ and $0<T<+\infty $. Then there is
at most one strong solution $u\in L^{\infty }([0, T]; H^{N}(\mathbb{R}))$ of
\eqref{e2.5} with initial data $u(x, 0)=\varphi (x)$.
\end{theorem}

\begin{proof} Assume that $u, v\in L^{\infty }([0, T]; H^{N}(\mathbb{R}))$ are two
solutions of \eqref{e2.5} with $u_{t}, v_{t}\in L^{\infty }([0, T]; H^{N - 5}(\mathbb{R}))$
and with the same initial data. Then
\begin{equation} \label{e5.1}
(u - v)_{t} + \eta  (u - v)_5 + (u - v)_{3} + u u_1 - v v_1=0
\end{equation}
with $(u - v)(x, 0)=0$. By \eqref{e5.1},
\begin{equation} \label{e5.2}
(u - v)_{t} + \eta  (u - v)_5 + (u - v)_{3} + (u - v) u_1 + (u - v)_1 v =0\,.
\end{equation}
Multiplying \eqref{e5.2} by $2 \xi  (u - v)$ and
integrating with respect to $x$ over $\mathbb{R} $,
\begin{equation} \label{e5.3}
\begin{aligned}
& 2\int_{\mathbb{R}}\xi  (u - v) (u - v)_{t} dx
 + 2 \eta \int_{\mathbb{R}}\xi  (u - v) (u - v)_5 dx\\
&+ 2\int_{\mathbb{R}}\xi  (u - v) (u - v)_{3} dx
 + 2\int_{\mathbb{R}}\xi  u u_1 (u - v)^2 dx \\
&+ 2\int_{\mathbb{R}}\xi  v (u - v) (u - v)_1 dx = 0
\end{aligned}
\end{equation}
Each term is treated separately. In the first term
we obtain
\[
2\int_{\mathbb{R}}\xi  (u - v) (u - v)_{t} dx  =
\partial _{t}\int_{\mathbb{R}}\xi  (u - v)^2 dx -
\int_{\mathbb{R}}\xi _{t} (u - v)^2 dx
\]
In the others terms, we also integrate by parts,
\begin{gather*}
\begin{aligned}
2\eta \int_{\mathbb{R}}\xi  (u - v) (u - v)_5 dx
& = - \eta \int_{\mathbb{R}}\partial ^{5}\xi  (u - v)^2 dx +
5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  (u - v)_1^2 dx \\
& =  -5 \eta \int_{\mathbb{R}}\partial \xi  (u - v)_{2}^2 dx
\end{aligned}\\
2\int_{\mathbb{R}}\xi  (u - v) (u - v)_{3} dx  =
-\int_{\mathbb{R}}\partial ^{3}\xi  (u - v)^2 dx +
3\int_{\mathbb{R}}\partial \xi  (u - v)_1^2 dx \\
2\int_{\mathbb{R}}\xi  v (u - v) (u - v)_1 dx
= - \int_{\mathbb{R}}\partial ( \xi  v ) (u - v)^2 dx
\end{gather*}
Replacing these expression in \eqref{e5.3}, we have
\begin{align*}
& \partial _{t}\int_{\mathbb{R}}\xi  (u - v)^2 dx -
\int_{\mathbb{R}}\xi _{t} (u - v)^2 dx - \eta \int_{\mathbb{R}}\partial ^{5}\xi  (u - v)^2 dx +
5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  (u - v)_1^2 dx  \\
& -5\eta \int_{\mathbb{R}}\partial \xi  (u - v)_{2}^2 dx -
\int_{\mathbb{R}}\partial ^{3}\xi  (u - v) dx +
3\int_{\mathbb{R}}\partial \xi  (u - v)_1^2 dx  \\
& + 2\int_{\mathbb{R}}\xi  u_1 (u - v)^2 dx -
\int_{\mathbb{R}}\partial ( \xi  v ) (u - v)^2 dx =0
\end{align*}
then
\begin{align*}
&  \partial _{t}\int_{\mathbb{R}}\xi  (u - v)^2 dx +
\int_{\mathbb{R}}( 5 \eta  \partial ^{3}\xi +
3 \partial \xi  ) (u - v)_1^2 dx
- 5 \eta \int_{\mathbb{R}}\partial \xi  (u - v)_{2}^2 dx \\
&  + \int_{\mathbb{R}}( - \xi _{t} - \eta  \partial ^{5}\xi
- \partial ^{3}\xi + 2 \xi  u_1 - \partial ( \xi  v ) ) (u - v)^2 dx = 0
\end{align*}
By using \eqref{e2.4}, we obtain for $c_5>0$ and
$\eta <-3/5$ that
\begin{align*}
& \partial _{t}\int_{\mathbb{R}}\xi  (u - v)^2 dx -
c_5\int_{\mathbb{R}}( 5 \eta +3 ) \xi  (u - v)_1^2 dx
- 5 \eta  \int_{\mathbb{R}}\partial \xi  (u - v)_{2}^2 dx  \\
&  \leq  \int_{\mathbb{R}}( \xi _{t} + \eta  \partial ^{5}\xi + \partial ^{3}\xi
- 2 \xi  u_1 + \partial ( \xi  v ) ) (u - v)^2 dx
\end{align*}
and using Gagliardo-Nirenberg's inequality and standard estimates, we have
\[
\partial _{t}\int_{\mathbb{R}}\xi  (u - v)^2 dx\leq
c\int_{\mathbb{R}}\xi  (u - v)^2 dx
\]
By Gronwall's inequality and the fact that $(u - v)$ vanishes at $t=0$,
it follows that $u=v$. This proves the uniqueness of the solution.
\end{proof}

We construct the mapping $Z\colon L^{\infty }([0, T]; H^{s}(\mathbb{R}))
\to L^{\infty }([0, T]; H^{s}(\mathbb{R}))$
by
\begin{gather*}
u^{(0)}  =  \varphi (x) \\
u^{(n)}  =  Z(u^{(n - 1)})\quad n\geq 1,
\end{gather*}
where $u^{(n - 1)}$ is in place of $w$ in equation \eqref{e4.3} and $u^{(n)}$
is in place of $v$ which is the solution of equation \eqref{e4.3}.
By Lemma \ref{lm4.1}, $u^{(n)}$ exists and is unique in $C((0, +\infty ); H^{N}(\mathbb{R}))$.
A choice of $c_0$ and the use of the a priori estimate in \S 4 show that
$Z\colon \mathbb{B}_{c_0}(0)\to \mathbb{B}_{c_0}(0)$ where  $\mathbb{B}_{c_0}(0)$ is
a bounded ball in $L^{\infty}([0, T]; H^{s}(\mathbb{R}))$
\end{proof}

\begin{theorem}[Local solution] \label{thm5.2}
Let $\eta <-3/5$ and $N$ an integer $\geq 5$. If $\varphi \in H^{N}(\mathbb{R})$,
then there is $T>0$ and $u$ such that $u$
is a strong solution of \eqref{e2.5}, $u\in L^{\infty}
([0, T]; H^{N}(\mathbb{R}))$, and $u(x, 0)=\varphi (x)$
\end{theorem}

\begin{proof}  We prove that for
$\varphi \in H^{\infty }(\mathbb{R})=\bigcap _{{k\geq 0}}H^{k}(\mathbb{R})$
there exists a solution $u\in L^{\infty }([0, T]; H^{N}(\mathbb{R}))$ with
initial data $u(x, 0)=\varphi (x)$ which time of existence
$T>0$ only depends on the norm of $\varphi $.
We define a sequence of approximations to equation \eqref{e4.3} as
\begin{equation} \label{e5.4}
\begin{aligned}
&-v_{t}^{(n)} + \eta {\wedge v}_9^{(n)} +{\wedge v}_7^{(n)} +
{\wedge v}^{(n - 1)}{\wedge v}_5^{(n)} - \eta {\wedge v}_5^{(n)}\\
&+ 5{\wedge v}_1^{(n - 1)}{\wedge v}_{4}^{(n)}  + O({\wedge v}_{3}^{(n - 1)},
{\wedge v}_1^{(n - 1)},\dots)=0
\end{aligned}
\end{equation}
where the initial condition $v^{(n)}(x, 0)=\varphi (x) - \partial ^{4}\varphi (x)$.
The first approximation is given by
$v^{(0)}(x, 0)=\varphi (x) - \partial ^{4}\varphi (x)$.
Equation \eqref{e5.4} is a linear equation at each iteration which can be
solved in any interval of time in which the coefficients are defined.
This is shown in Lemma \ref{lm4.1}. By Lemma \ref{lm4.2}, it follows that
\begin{equation} \label{e5.5}
\partial _{t}\int_{\mathbb{R}}\xi  [ {v_{\alpha }^{(n)} ]^2} dx \leq
G( \| v^{(n - 1)} \|_{{\lambda }} ) \| v^{(n)} \|_{{\alpha }}^2 +
F( \| v^{(n - 1)} \|_{{\alpha }} )
\end{equation}
Choose $\alpha =1$ and let
$c\geq \|\varphi - \partial ^{4}\varphi \|_1 \geq \| \varphi  \|_5$.
For each iterate $n,
\| v^{(n)}( \cdot  , t) \|$ is continuous in
$t\in [0, T]$ and $\|v^{(n)}
( \cdot  , 0) \|\leq c$.
Define $c_0=\frac {c_9}{2c_8}c^2+ 1$.
Let $T_{0}^{(n)}$ be the maximum time such that
$\| v^{(k)}( \cdot  , t) \|_1\leq c_3$
for $0\leq t\leq T_0^{(n)}$, $0\leq k\leq n$.
Integrating \eqref{e5.5}
over $[0, t]$ we have for $0\leq t\leq T_0^{(n)}$ and
$j=0, 1$.
\[
\int_{0}^{t}\big( \partial _{s}\int_{\mathbb{R}} \xi  [ v_j^{(n)} ]^2 dx \big) ds
 \leq
\int_{0}^{t}G( \| v^{(n - 1)} \|_1 ) \| v^{(n)} \|_j^2 ds +
\int_{0}^{t}F( \| v^{(n - 1)} \|_j ) ds
\]
It follows that
\begin{align*}
&\int_{\mathbb{R}}\xi (x, t) [ v_j^{(n)}(x, t) ]^2 dx \\
& \leq \int_{\mathbb{R}}\xi (x, 0) [ v_j^{(n)}(x, 0) ]^2 dx +
\int_{0}^{t}G( \| v^{(n - 1)} \|_1 ) \| v^{(n)} \|_j^2 ds
 +  \int_{0}^{t}F( \| v^{(n - 1)} \|_j ) ds
\end{align*}
hence
\begin{align*}
c_8\int_{\mathbb{R}}[ v_j^{(n)} ]^2 dx
&\leq \int_{\mathbb{R}}\xi  [ v_j^{(n)} ]^2 dx \\
& \leq  \int_{\mathbb{R}}\xi (x, 0) [ v_j^{(n)}(x, 0) ]^2 dx +
\int_{0}^{t}G( \| v^{(n - 1)} \|_1 ) \| v^{(n)} \|_j^2 ds\\
&\quad +\int_{0}^{t}F( \| v^{(n - 1)} \|_j ) ds
\end{align*}
and
\[
\int_{\mathbb{R}}[ v_j^{(n)} ]^2 dx \leq \frac {c_9}{c_8}
\int_{\mathbb{R}}[ v_j^{(n)}(x, 0) ]^2 dx  +
\frac {G(c_3)}{c_8} c_3^2 t +\frac {F(c_3)}{c_8} t
\]
and we obtain for $j=0, 1$ that
\[
\| v^{(n)} \|_1\leq \frac {c_9}{c_8}c^2 +
 \frac {G(c_0)}{c_8} c_0^2 t +
 \frac {F(c_0)}{c_8} t
\]

\noindent\textbf{Claim: $T_0^{(n)}$ does not approach $0$}
\\
On the contrary, assume that $T_0^{(n)}\to 0$. 
Since $ \| v^{(n)}( \cdot  , t) \|$ is continuous for $t\geq 0$, 
there exists $\tau \in [0, T]$ such that
$\| v^{(k)}( \cdot  , \tau ) \|_1=c_0$
for $0\leq \tau \leq T_0^{(n)}, 0\leq k \leq n$. Then
\[
c_0^2\leq \frac {c_9}{c_8} c^2 +
 \frac {G(c_0)}{c_8} c_0^2 T_0^{(n)} +
 \frac {F(c_0)}{c_8} T_0^{(n)}.
\]
as $n\to +\infty $, we have
\[
\big(\frac {c_9}{2c_8}c^2 + 1 \big)^2
\leq  \frac {c_9}{c_8}c^2\quad \Longrightarrow \quad 
\frac {c_9^2}{4c_8^2}c^{4} + 1\leq 0
\]
which is a contradiction. Consequently $T_0^{(n)}\not \to 0$.
Choosing $T=T(c)$ sufficiently small, and $T$ not depending on
$n$, one concludes that
\begin{equation} \label{e5.6}
\| v^{(n)} \|_1\leq C
\end{equation}
for $0\leq t\leq T$. This shows that $T_0^{(n)}\geq T$.
Hence from \eqref{e5.6} we imply that there exists a subsequence 
$v^{(n_j)}:=v^{(n)}$ such that
\begin{equation} \label{e5.7}
v^{(n)}\stackrel {*}{\rightharpoonup }v\quad \mbox {weakly on}\quad
L^{\infty }([0, T]; H^{1}(\mathbb{R} ))
\end{equation}

\noindent\textbf{Claim: $u={\wedge v}$ is a solution.}
\\
In the linearized equation \eqref{e5.4} we have
\[
{\wedge v}_9^{(n)}  =
{\wedge (I - (I - \partial ^{4}))} v_5^{(n)}
 =  {\wedge v}_5^{(n)} - v_5^{(n)}
 =  \partial ^{4}
(\underbrace {{\wedge v}_1^{(n)}}_{\in L^2(\mathbb{R} )}) -
\underbrace {\partial ^{4}(v_1^{(n)})}_{\in H^{-4}(\mathbb{R} )}
\]
Since ${\wedge }= (I - \partial ^{4})^{-1}$ is bounded in
$H^{1}(\mathbb{R} )$
so ${\wedge v}_9^{(n)}$ belongs to $H^{-4}(\mathbb{R})$.
$v^{(n)}$ is still bounded in $
L^{\infty }([0, T]; H^{1}(\mathbb{R}))
\hookrightarrow L^2([0, T]; H^{1}(\mathbb{R}))$
and since ${\wedge }:L^2(\mathbb{R})\to H^{4}(\mathbb{R})$ is a
bounded operator,
\[
\|{\wedge v}_1^{(n)}\|_{{H^{4}(\mathbb{R})}}\leq c_{11}
\|v_1^{(n)}\|_{{L^2(\mathbb{R})}}\leq
c_{12}\|v_1^{(n)}\|_{{H^{1}(\mathbb{R})}}.
\]
Consequently ${\wedge v}_1^{(n)}$ is bounded in $
L^2([0, T]; H^{4}(\mathbb{R}))
\hookrightarrow L^2([0, T]; L^2(\mathbb{R}))$.
It follows that $\partial ^{4}({\wedge v}_1^{(n)})$ is bounded in $
L^2([0, T]; H^{-4}(\mathbb{R}))$,
and
\begin{equation} \label{e5.8}
{\wedge v}_9^{(n)}\quad \mbox {is bounded in}\quad
L^2([0, T]; H^{-4}(\mathbb{R}))
\end{equation}
Similarly, the other terms are bounded.
By  \eqref{e5.4}, $v_{t}^{(n)}$ is a sum of terms each of which
is the product of a coefficient, uniformly bounded on
$n$ and a function in $L^2([0, T]; H^{-4}(\mathbb{R}))$
uniformly bounded on
$n$ such that $v_{t}^{(n)}$ is bounded in $L^2([0, T]; H^{-4}(\mathbb{R}))$.
On the other hand, 
$H_{\rm loc}^{1}(\mathbb{R})\stackrel {c}{\hookrightarrow }
H_{\rm loc}^{1/2}(\mathbb{R})\hookrightarrow H^{-4}(\mathbb{R})$.
By Lions-Aubin's compactness Theorem  \cite{l1}
there is a subsequence
$v^{(n_j)}:=v^{(n)}$ such that $v^{(n)}\to v$ strongly on 
$L^2([0, T]; H_{\rm loc}^{1/2}(\mathbb{R}))$.
Hence, for a subsequence
$v^{(n_j)}:=v^{(n)}$, we have $v^{(n)}\to v$ a. e. in 
$L^2([0, T]; H_{\rm loc}^{1/2}(\mathbb{R}))$.
Moreover, from \eqref{e5.8}, ${\wedge v}_9^{(n)}\rightharpoonup
{\wedge v}_9$ weakly in $ L^2([0, T]; H^{-4}(\mathbb{R}))$.

Similarly, ${\wedge v}_5^{(n)}\rightharpoonup{\wedge v}_5$ weakly in 
$L^2([0, T]; H^{-4}(\mathbb{R}))$.
Since $\| {\wedge v}^{(n)} \|_{{H^{5}(\mathbb{R})}}
\leq c_{13} \| v^{(n)} \|_{{H^{1}(\mathbb{R})}}\leq c_{14}
\| v^{(n)} \|_{{H^{1/2}(\mathbb{R})}}$
and $v^{(n)}\to v$ strongly on $
L^2([0, T]; H_{\rm loc}^{1/2}(\mathbb{R}))$
then ${\wedge v}^{(n)}\to \wedge v$ strong in $
L^2([0, T]; H_{\rm loc}^{5}(\mathbb{R}))\hookrightarrow
L^2([0, T]; H_{\rm loc}^{4}(\mathbb{R}))$.
Thus the fourth term on the right hand side of \eqref{e5.4},
${\wedge v}^{(n - 1)}{\wedge v}_5^{(n)}\rightharpoonup
{\wedge v} {\wedge v}_5$ weakly in $
L^2([0, T]; L_{\rm loc}^{1}(\mathbb{R}))$
as ${\wedge v}_5^{(n)}\rightharpoonup {\wedge v}_{5}$ weakly in 
$L^2([0, T]; H^{-4}(\mathbb{R}))$ and ${\wedge v}^{(n - 1)}\to{\wedge v}$
strongly on $L^2([0, T]; H_{\rm loc}^{4}(\mathbb{R}))$.
Similarly, the other terms in \eqref{e5.4} converge to their
limits, implying $v_{t}^{(n)}\rightharpoonup v_{t}$ weakly in 
$L^2([0, T]; L_{\rm loc}^{1}(\mathbb{R}))$.
Passing to the limit
\begin{align*}
v_{t}  &=  \partial ^{4}(\eta {\wedge v}_5 +
{\wedge v}_{3} +{\wedge v}{\wedge v}_1) -(\eta {\wedge v}_5 
+ {\wedge v}_{3} +{\wedge v}{\wedge v}_1)\\
&= -(I - \partial ^{4})(\eta {\wedge v}_5 + {\wedge v}_{3} + {\wedge v}{\wedge v}_1)
\end{align*}
thus $v_{t} + (I - \partial ^{4})(\eta {\wedge v}_5 +
{\wedge v}_{3} + {\wedge v} {\wedge v}_1 ) =0$.
This way, we have that \eqref{e2.5} for $u={\wedge v}$.
Now, we prove that there exists a solution to \eqref{e2.5} with
$u\in L^{\infty }([0, T]; H^{N}(\mathbb{R}))$
and $N\geq 6$, where $T$ depends only on the norm of $\varphi $ in 
$H^{5}(\mathbb{R})$.
We already know that there is a solution 
$u\in L^{\infty }([0, T]; H^{5}(\mathbb{R}))$.
It is suffices to show that the approximating sequence  $v^{(n)}$ is
bounded in $L^{\infty }([0, T]; H^{N - 4}(\mathbb{R}))$.
Taking $\alpha = N - 2$ and considering \eqref{e5.5} for $\alpha \geq 2$,
we define $c_{{N - 5}}=\frac {c_9}{2 c_8} \| \varphi ( \cdot  ) \|_{{N}} + 1$.
Let $T_{{N - 5}}^{(n)}$ be the largest time such that
$\|v^{(k)}( \cdot  , t) \|_{{\alpha }}\leq c_{{N - 5}}$
for $0\leq t\leq T_{{N - 5}}^{(n)}, 0\leq k\leq n$.
Integrating \eqref{e5.5}
over $[0, t]$, for $0\leq t\leq T_{{N - 5}}^{(n)}$, we have
\[
\int_{0}^{t}\left(\partial _{s}\int_{\mathbb{R}}
\xi [v_{{\alpha }}^{(n)}]^2dx\right)ds
 \leq \int_{0}^{t}G( \| v^{(n - 1)} \|_{{\alpha }} ) \| v^{(n)} \|_{{\alpha }}^2 ds +
\int_{0}^{t}F( \| v^{(n - 1)} \|_{{\alpha }} ) ds.
\]
It follows that
\begin{align*}
&\int_{\mathbb{R}}\xi (x, t)[v_{{\alpha }}^{(n)}(x, t)]^2dx \\
& \leq \int_{\mathbb{R}}\xi (x, 0)[v_{{\alpha }}^{(n)}(x, 0)]^2dx +
\int_{0}^{t}G(\|v^{(n - 1)}\|_{{\alpha }})\|v^{(n)}\|_{{\alpha }}^2ds
 +  \int_{0}^{t}F(\|v^{(n - 1)}\|_{{\alpha }})ds
\end{align*}
hence
\begin{align*}
c_8\int_{\mathbb{R}}[ v_{{\alpha }}^{(n)} ]^2 dx
&\leq \int_{\mathbb{R}}\xi  [ v_{{\alpha }}^{(n)} ]^2 dx \\
& \leq \int_{\mathbb{R}}\xi (x, 0) [ v_{{\alpha }}^{(n)}(x, 0) ]^2 dx +
\int_{0}^{t}G( \| v^{(n - 1)} \|_{{\alpha }} ) \| v^{(n)} \|_{{\alpha }}^2 ds \\
&\quad +\int_{0}^{t}F( \| v^{(n - 1)} \|_{{\alpha }} ) ds.
\end{align*}
Then
\begin{align*}
\int_{\mathbb{R}}[ v_{{\alpha }}^{(n)} ]^2 dx 
& \leq  \frac {c_9}{c_8}\int_{\mathbb{R}}[ v_{{\alpha }}^{(n)}(x, 0) ]^2 dx  +
\frac {G(c_{{N - 5}})}{c_8} c_{{N - 5}}^2 t +
\frac {F(c_{{N - 5}})}{c_8} t  \\
& \leq  \frac {c_9}{c_8} \| v^{(n)}( \cdot  , 0) \|_{{\alpha }}^2  +
\frac {G(c_{{N - 5}})}{c_8 } c_{{N - 5}}^2 t +
\frac {F(c_{{N - 5}})}{c_8} t  \\
& \leq  \frac {c_9}{c_8}\| \varphi ( \cdot  , 0) \|_{{N}}^2  +
\frac {G(c_{{N - 5}})}{c_8} c_{{N - 5}}^2 t +
\frac {F(c_{{N - 5}})}{c_8} t
\end{align*}
and we obtain
\[
\| v^{(n)}( \cdot  , t) \|_{{\alpha }}^2\leq \frac {c_9}{c_8} 
\| \varphi ( \cdot  , 0) \|_{{N}}^2 +  \frac {G(c_3)}{c_8} c_3^2 t +
 \frac {F(c_3)}{c_8} t\,.
 \]
 
\noindent\textbf{Claim: $T_{{N - 5}}^{(n)}$ does not approach $0$.} 
\\
On the contrary, assume that $T_{{N - 5}}^{(n)}\to 0$. Since 
$\|v^{(n)}( \cdot  , t)\|$ is continuous for $t\geq 0$, there exists
$\tau \in [0, T_{{N - 5}}]$ such that
$\| v^{(k)}( \cdot  , \tau ) \|_{{\alpha }}=c_{{N - 5}}$
for $0\leq \tau \leq T^{(n)}$, $0\leq k \leq n$. Then
\[c_{{N - 5}}^2\leq \frac {c_9}{c_8} \| \varphi ( \cdot  , 0) \|_{{N}}^2 +
 \frac {G(c_{{N - 5}})}{c_8} c_{{N - 5}}^2 T_{{N - 5}}^{(n)} +
 \frac {F(c_{{N - 5}})}{c_8} T_{{N - 5}}^{(n)}.\]
as $n\to +\infty $ we have
\[
\big( \frac {c_9}{2 c_8} \| \varphi ( \cdot  , 0) \|_{{N}}^2 + 1  \big)^2
\leq  \frac {c_9}{c_8} \| \varphi ( \cdot  , 0) \|_{{N}}^2 \Longrightarrow  
\frac {c_9^2}{4 c_8^2} \| \varphi ( \cdot  , 0) \|_{{N}}^{4} + 1\leq 0
\]
which is a contradiction.
Then $T_{{N - 5}}^{(n)}\not \to 0$.
By choosing $T_{{N - 5}}=T_{{N - 5}}(\| \varphi ( \cdot  , 0) \|_{{N}}^2)$ 
sufficiently small, and $T_{{N - 5}}$ not depending on
$n$, we conclude that
\begin{equation}
\| v^{(n)}( \cdot  , t) \|_{{\alpha }}^2\leq c_{{N - 5}}^2\quad 
\mbox {for all}\quad  0\leq t\leq T_{{N - 5}}.
\end{equation}
This shows that $T_{{N - 5}}^{(n)}\geq T_{{N - 5}}$. Thus,
\[
v\in L^{\infty }([0, T_{{N - 5}}]; H^{\alpha }(\mathbb{R}))\equiv v\in L^{\infty }([0,
T_{{N - 5}}]; H^{N - 4}(\mathbb{R})).
\]
Now, denote by $0\leq T_{{N - 5}}^{*}\leq +\infty $ the maximal number
such that for all $0<t\leq T_{{N - 5}}^{*}$,
$u={\wedge v}\in L^{\infty }([0, t]; H^{N}(\mathbb{R}))$.
In particular $T_{{N - 5}}\leq T_{{N - 5}}^{*}$
for all $N\geq 6$. Thus,
$T$ can be chosen depending only on the norm of $\varphi $ in  $H^{5}(\mathbb{R})$.
Approximating $\varphi $
by $\{\varphi _j\}\in C_0^{\infty }(\mathbb{R})$ such that
$\| \varphi - \varphi _j \|_{{H^{N}(\mathbb{R})}}\to 0$ as
$j\to +\infty $.
Let $u_j $ be a solution of \eqref{e2.5} with
$u_j(x, 0)=\varphi _j(x)$.
According to the above argument, there exists $T$
which is independent on  $n$
but depending only on
$\sup_{j}\| \varphi _j \|$ such that $u_j$
exists on $[0, T]$ and a subsequence
$ u_j\stackrel {j\to +\infty }{\to }u$ in $
L^{\infty }([0, T]; H^{N}(\mathbb{R}))$.
\end{proof}

As a consequence of Theorems \ref{thm5.1} and \ref{thm5.2}
and its proof, one obtains the following result.

\begin{corollary} \label{coro5.3}
 Let $\varphi \in H^{N}(\mathbb{R})$ with $N\geq 5$ such
that $\varphi ^{(\gamma )}\to \varphi $ in $H^{N}(\mathbb{R})$.
Let $u$ and $u^{(\gamma )}$ be the corresponding unique
solutions given by Theorems \ref{thm5.1} and \ref{thm5.2} in $L^{\infty
}([0, T]; H^{N}(\mathbb{R}))$ with $T$ depending only on
$\sup_{\gamma }\| \varphi ^{(\gamma )} \|_{{H^{5}(\mathbb{R})}}$ then
\begin{gather*}
 u^{(\gamma )}\stackrel {*}{\rightharpoonup }u\quad \mbox{weakly on }
L^{\infty }([0, T]; H^{N}(\mathbb{R})), \\
 u^{(\gamma )}\to u\quad \mbox {strongly on } L^2([0, T];
H^{N + 1}(\mathbb{R})),\\
u^{(\gamma )}\to u\quad \mbox {strongly on } L^2([0, T];
H^{N + 2}(\mathbb{R}))
\end{gather*}
\end{corollary}

\section{Existence of Global Solutions}

Here, we will try to extend the local solution 
$u\in L^{\infty }([0, T]; H^{N}(W_{{0 i 0}}))$ of \eqref{e2.5} obtained in 
Theorem \ref{thm5.2} to $t\geq 0$. A standard way to obtain these extensions 
consists into deducing global estimations for the 
$H^{N}(W_{{0 i 0}})$-norm of $u$ in terms of the $H^{N}(W_{{0 i 0}})$-norm 
of $u(x, 0)=\varphi (x)$. These estimations are frecuently based on 
conservation laws which contain the $ L^2 $-norm of the solution and their 
spatial derivatives. It is not possible to do the same to give a solution 
of the problem of global existence because the difficulty here is that the 
weight depends on the variables $x$ and $t$ variables. To solve our problem 
we follow a different method using the Leibniz rule like in the proof of 
Theorem 3.1 of Bona and  Saut, cf. \cite{b5}.

\begin{theorem} \label{thm6.1}
 For $\eta <-3/5$  there exists a global solution to \eqref{e2.5} in the space
 $ H^{s}(\mathbb{R})\bigcap H^{N}(W_{{0 i 0}})$ with $N$ integer $\geq 5$ and 
 $ s\geq 2$.
\end{theorem}
 
\begin{proof}  
The first part was proved in \cite{b1}, with $N\geq 5$ and a nonegative 
integer $i$.
Taking $\partial ^{\alpha }$ derivatives of the equation \eqref{e2.5}
\begin{equation} \label{e6.1}
\partial _{{t}}u_{{\alpha }} + \eta  u_{{\alpha + 5}} + u_{{\alpha + 3}} 
+ ( u u_1 )_{{\alpha }} =0.
\end{equation}
We multiply \eqref{e6.1} by $2\xi u_{{\alpha }}$ and integrate over 
$\mathbb{R}$.
\begin{equation} \label{e6.2}
2\int_{\mathbb{R}}\xi  u_{\alpha }\partial _{t}u_{\alpha } dx 
+ 2 \eta \int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 5} dx 
+ 2\int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 3} dx 
+ 2\int_{\mathbb{R}}\xi  u_{\alpha } ( u u_1 )_{\alpha }dx =0.
\end{equation}
Each term is treated separately. The first term yields
\[
2\int_{\mathbb{R} }\xi  u_{\alpha } \partial _{t}u_{\alpha } dx  = 
\partial _{t}\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx -
\int_{\mathbb{R} }\xi _{t} u_{\alpha }^2 dx.
\]
In the second and third term, integrating by parts, we obtain
\begin{gather*}
2\eta \int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 5} dx  
=  -\eta \int_{\mathbb{R}}\partial ^{5}\xi  u_{\alpha }^2 dx + 5\eta 
\int_{\mathbb{R}}\partial ^{3}\xi  u_{\alpha + 1}^2 dx 
- 5\eta\int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx\,, \\
2\int_{\mathbb{R}}\xi  u_{\alpha } u_{\alpha + 3} dx  
=  -\int_{\mathbb{R}}\partial ^{3}\xi  u_{\alpha }^2 dx 
+ 3\int_{\mathbb{R}}\partial \xi  u_{\alpha + 1}^2 dx.
\end{gather*}
In the last term, using the Leibniz rule, we obtain
\begin{align*}
&2\int_{\mathbb{R}}\xi  u_{\alpha } ( u u_1 )_{\alpha } dx \\
& =  2\int_{\mathbb{R}}\xi  u u_{\alpha } u_{\alpha + 1} dx 
+ 2 \alpha \int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx 
+ 2 \frac {\alpha  (\alpha - 1)}{2}\int_{\mathbb{R}}\xi  u_{2} u_{\alpha - 1} u_{\alpha } dx \\
& \quad +  2\frac {\alpha !}{3!(\alpha - 3)!}\int_{\mathbb{R}}\xi u_{3}u_{\alpha - 2}
u_{\alpha }dx 
+ 2\frac {\alpha !}{4!(\alpha - 4)!}\int_{\mathbb{R}}\xi u_{4}u_{\alpha - 3}
u_{\alpha }dx \\
&\quad + \dots+ 2\int_{\mathbb{R}}\xi u_1u_{\alpha }^2dx.
\end{align*}
Integrating by parts it follows that
\begin{align*}
&2\int_{\mathbb{R}}\xi  u_{\alpha } ( u u_1 )_{\alpha } dx \\
& =  -\int_{\mathbb{R}}\partial ( \xi  u ) u_{\alpha }^2 dx 
+ 2 \alpha \int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx  
- \frac {\alpha  (\alpha - 1)}{2}\int_{\mathbb{R}}\partial ( \xi  u_{2} ) u_{\alpha - 1}^2 dx \\
&\quad +  2 \frac {\alpha !}{3! (\alpha - 3)!}\int_{\mathbb{R}}\xi  u_{3} u_{\alpha - 2} u_{\alpha } dx + 2 \frac {\alpha !}{4! (\alpha - 4)!}\int_{\mathbb{R}}\xi  u_{4} u_{\alpha - 3} u_{\alpha } dx \\
&\quad + \dots+ 2\int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx.
\end{align*}
Substituting in \eqref{e6.2}, we have
\begin{align*}
& \partial _{t}\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx -
\int_{\mathbb{R} }\xi _{t} u_{\alpha }^2 dx - \eta \int_{\mathbb{R}}\partial ^{5}\xi  u_{\alpha }^2 dx 
+ 5 \eta \int_{\mathbb{R}}\partial ^{3}\xi  u_{\alpha + 1}^2 dx\\
&- 5 \eta\int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx 
 - \int_{\mathbb{R}}\partial ^{3}\xi  u_{\alpha }^2 dx 
 + 3\int_{\mathbb{R}}\partial \xi  u_{\alpha + 1}^2 dx 
 -\int_{\mathbb{R}}\partial ( \xi  u ) u_{\alpha }^2 dx \\
&\quad + 2 \alpha \int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx  
- \frac {\alpha  (\alpha - 1)}{2}\int_{\mathbb{R}}\partial ( \xi  u_{2} ) 
u_{\alpha - 1}^2 dx 
+ 2 \frac {\alpha !}{3! (\alpha - 3)!}\int_{\mathbb{R}}\xi  u_{3} u_{\alpha - 2} 
u_{\alpha } dx \\
&\quad + 2 \frac {\alpha !}{4! (\alpha - 4)!}\int_{\mathbb{R}}\xi  u_{4} u_{\alpha - 3} u_{\alpha } dx
+ \dots+ 2\int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx=0
\end{align*}
hence
\begin{align*}
&  \partial _{t}\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx 
+ \int_{\mathbb{R}}( 5 \eta  \partial ^{3}\xi + 3 \partial \xi  ) u_{\alpha + 1}^2 dx 
- 5 \eta \int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx \\
& + \int_{\mathbb{R}}( - \xi _{t} - \eta  \partial ^{5}\xi - \partial ^{3}\xi 
- \partial ( \xi  u ) + 2 \alpha  \xi  u_1 ) u_{\alpha }^2 dx \\
& - \frac {\alpha  (\alpha - 1)}{2}\int_{\mathbb{R}}\partial ( \xi  u_{2} ) u_{\alpha - 1}^2 dx 
+ 2 \frac {\alpha !}{3! (\alpha - 3)!}\int_{\mathbb{R}}\xi  u_{3} u_{\alpha - 2}
u_{\alpha } dx \\
& + 2 \frac {\alpha !}{4! (\alpha - 4)!}\int_{\mathbb{R}}\xi  u_{4} u_{\alpha - 3} u_{\alpha } 
dx + \dots+ 2\int_{\mathbb{R}}\xi  u_1 u_{\alpha }^2 dx=0
\end{align*}
then using \eqref{e2.4}, Gagliardo - Nirenberg's inequality and standard estimates we get
\begin{equation} \label{e6.3}
\partial _{t}\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx + \int_{\mathbb{R}}( 5 \eta + 3 ) \xi  u_{\alpha 
+ 1}^2 dx - 5 \eta  \int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx\leq c\int _{\mathbb{R} }\xi  
u_{\alpha }^2 dx\,.
\end{equation}
Integrating \eqref{e6.3} in $t\in[0, T_{\mbox{max}}=T]$ we obtain
\begin{align*}
&\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx + \int_{0}^{t}\int_{\mathbb{R}}( 5 \eta + 3 ) 
\xi  u_{\alpha + 1}^2 dx ds 
- 5 \eta \int_{0}^{t}\int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx ds \\
&\leq \| \varphi  \|_{{\alpha }}^2 + \int_{0}^{t}
\big( c\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx \big) ds\,,
\end{align*}
where
\[
\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx\leq \| \varphi  \|_{{\alpha }}^2 
+ \int_{0}^{t}\big( c\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx \big) ds.
\]
Using Gronwall's inequality
\[
\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx\leq \| \varphi  \|_{{\alpha }}^2 e^{c t}
\leq \| \varphi  \|_{{\alpha }}^2 e^{c T}
\]
it follows that
\[
\int _{\mathbb{R} }\xi  u_{\alpha }^2 dx\leq C=C(T, \| \varphi  \|).
\]
Then for any $T=T_{{\mbox{max}}}>0$, there exists 
$C=C(T, \| \varphi  \|)$ such that
\[
\| u \|_{\alpha }^2 + \int_{0}^{t}\int_{\mathbb{R}}( 5 \eta + 3 ) \xi  u_{\alpha + 1}^2 dx ds 
- 5 \eta \int_{0}^{t}\int_{\mathbb{R}}\partial \xi  u_{\alpha + 2}^2 dx ds \leq C.
\]
This concludes the proof.
\end{proof}

\section{Persistence Theorem}

As a starting point for the a priori gain of regularity results that
will be discussed in the next section, we need to develop some estimates
for solutions of the equation \eqref{e2.5} in weighted Sobolev norms.
The existence
of these weighted estimates is often called the {\em persistence }
of a property
of the initial data $\varphi $. We show that if
$\varphi \in H^{5}(\mathbb{R})\bigcap H^{L}(W_{{0 i 0}})$ for $
L\geq 0, i\geq 1$ then the solution $u( \cdot  , t)$ evolves in
$H^{L}(W_{{0 i 0}})$ for $t\in [0, T]$. The time interval
of such persistence is at least as long as the interval
guaranteed by the existence Theorem \ref{thm5.2}.

\begin{theorem}[Persistence] \label{thm7.1}
 Let $i\geq 1$ and $L\geq 0$ be
non-negative integers, $0<T<+\infty $.
Assume that $u$ is the solution to \eqref{e2.5} in 
$L^{\infty}([0, T]; H^{5}(\mathbb{R}))$ with initial data 
$\varphi(x)=u(x, 0)\in H^{5}(\mathbb{R})$. If $\varphi (x)\in
H^{L}(W_{{0 i 0}})$ then
\begin{gather}
u\in L^{\infty }([0, T]; H^{5}(\mathbb{R})\bigcap H^{L}(W_{{0 i 0}}))
\label{e7.1}\\
\int_{0}^{T}\int_{\mathbb{R}}| \partial ^{L + 1}u(x, t) |^2\mu _1 dx dt<+\infty 
\label{e7.2}\\
\int_{0}^{T}\int_{\mathbb{R}}| \partial ^{L + 2}u(x, t) |^2\mu _2
 dx dt<+\infty \,, \label{e7.3}
\end{gather}
where $\sigma $ is arbitrary, $\mu _1 \in W_{{\sigma, i,  0}}$ and 
$\mu _2 \in W_{{\sigma , i - 1,0}}$ for $i\geq 1$.
\end{theorem}
 
\begin{proof} We use induction on $\alpha $. Let
\[
u\in L^{\infty }([0, T]; H^{5}(\mathbb{R})
\bigcap H^{\alpha }(W_{{0 i 0}}))\quad
\mbox {for}\quad 0\leq \alpha \leq L.
\]
We derive formally some a priori estimate for the
solution where the bound,
involves only the norms of  $u$ in $
L^{\infty }([0, T]; H^{5}(\mathbb{R}))$ and the norms of
$\varphi $ in $H^{5}(W_{{0 i 0}})$.
We do this by approximating $u(x, t)$ through smooth solutions, and the weight functions
by smooth bounded functions.
By Theorem \ref{thm5.2}, we have
\[
u(x, t)\in L^{\infty }([0, T]; H^{N}(\mathbb{R}))\quad\mbox {with}\quad
N=\mbox {max}\{L, 5\}.
\]
In particular
$u_{j}(x, t)\in L^{\infty }([0, T]\times \mathbb{R})$ for
$0\leq j\leq N - 1$.
To obtain \eqref{e7.1}-\eqref{e7.2} and \eqref{e7.3} there are two ways of 
approximation perform.
We approximate general solutions by smooth solutions, and
we approximate general weight functions by bounded weight functions.
The first of these procedures has already been discussed,
so we will concentrate
on the second.

Given a smooth weight function $\mu _2(x)\in W_{{\sigma ,i - 1,0}}$
with $\sigma >0$, we take a sequence $\mu
_2^{\beta }(x)$ of smooth bounded weight functions approximating
 $\mu _2(x)$ from below, uniformly on any half line
 $(-\infty , c)$.
Define the weight functions for the
$\alpha $-th induction step as
\[
\xi _{\beta}(x, t)=-\frac {1}{5 \eta }
\Big(1 + \int_{-\infty }^{x}\mu _2^{\beta }(y, t) dy\Big)
\]
then the $\xi _{\beta}$ are bounded weight functions
which approximate a desired
weight function
$\xi \in W_{{0 i 0}}$ from below, uniformly on a compact set.
For $\alpha =0$,
multiplying \eqref{e2.5} by $2\xi _{\beta}u$, and integrating over
$x\in \mathbb{R} $.
\begin{equation} \label{e7.4}
2\int_{\mathbb{R}}\xi _{\beta} u u_{t} dt +
2 \eta \int_{\mathbb{R}}\xi _{\beta} u u_5 dx +
2\int_{\mathbb{R}}\xi _{\beta} u u_{3} dx +
2\int_{\mathbb{R}}\xi _{\beta} u^2 u_1 dx =0.
\end{equation}
Each term is treated separately. In the first term we have
\[
2\int_{\mathbb{R}}\xi _{\beta} u u_{t} dx =
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx -
\int_{\mathbb{R}}\partial _{t}\xi _{\beta} u^2 dx.
\]
For the others terms, using integration by parts, we have
\begin{gather*}
2 \eta \int_{\mathbb{R}}\xi _{\beta} u u_5 dx  = 
- \eta \int_{\mathbb{R}}\partial ^{5}\xi _{\beta} u^2 dx 
+ 5 \eta \int_{\mathbb{R}}\partial ^{3}\xi _{\beta} u_1^2 dx 
- 5 \eta \int_{\mathbb{R}}\partial \xi _{\beta} u^2 dx.\\
2\int_{\mathbb{R}}\xi _{\beta} u u_{3} dx = 
-\int_{\mathbb{R}}\partial ^{3}\xi _{\beta} u^2 dx + 3\int_{\mathbb{R}}\partial \xi _{\beta}
u_1^2 dx\,,\\
2\int_{\mathbb{R}}\xi _{\beta} u^2 u_1 dx  
=  -\frac {2}{3}\int_{\mathbb{R}}\partial \xi _{\beta} u^{3} dx.
\end{gather*}
Replacing in \eqref{e7.4}, we obtain
\begin{align*}
&   \partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx -
\int_{\mathbb{R}}\partial _{t}\xi _{\beta} u^2 dx 
- \eta  \int_{\mathbb{R}}\partial ^{5}\xi _{\beta} u^2 dx 
+ 5 \eta \int_{\mathbb{R}}\partial ^{3}\xi _{\beta} u_1^2 dx \\
&   - 5 \eta \int_{\mathbb{R}}\partial \xi _{\beta} u^2 dx 
- \int_{\mathbb{R}}\partial ^{3}\xi _{\beta} u^2 dx + 3\int_{\mathbb{R}}\partial \xi _{\beta} u_1^2 dx 
- \frac {2}{3}\int_{\mathbb{R}}\partial \xi _{\beta} u^{3} dx =0
\end{align*}
then
\begin{align*}
&   \partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx 
+ \int_{\mathbb{R}}( 5 \eta    \partial ^{3}\xi _{\beta} 
+ 3 \partial \xi _{\beta} ) u_1^2 dx 
- 5 \eta \int_{\mathbb{R}}\partial \xi _{\beta} u_{2}^2 dx \\
&   + \int_{\mathbb{R}}( - \partial _{t}\xi _{\beta} - \eta  \partial ^{5}\xi _{\beta} 
- \partial ^{3}\xi _{\beta} - \frac {2}{3}\xi _{\beta} u ) u^2 dx =0\,.
\end{align*}
Using \eqref{e2.4}, for $c_5>0$ ($\eta <-3/5 $),
\begin{align*}
&  & \partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx - c_5 ( 5 \eta + 3 )\int_{\mathbb{R}}\xi _{\beta} u_1^2 dx - 5 \eta \int_{\mathbb{R}}\partial \xi _{\beta} u_{2}^2 dx \\
&  & + \int_{\mathbb{R}}( - \partial _{t}\xi _{\beta} 
- \eta  \partial ^{5}\xi _{\beta} - \partial ^{3}\xi _{\beta} 
- \frac {2}{3}\xi _{\beta} u ) u^2 dx \leq 0\,.
\end{align*}
Using again \eqref{e2.4} and Gagliardo-Nirenberg's inequality, we obtain
\[
 \partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx - c_5 ( 5 \eta + 3 )
 \int_{\mathbb{R}}\xi _{\beta} u_1^2 dx - 5 \eta \int_{\mathbb{R}}\partial \xi _{\beta} u_{2}^2 dx\leq 
 c\int_{\mathbb{R}}\xi _{\beta} u^2 dx
\]
thus
\[
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u^2 dx \leq c\int_{\mathbb{R}}\xi _{\beta} u^2 dx.
\]
We apply Gronwall's lemma to conclude
\begin{equation} \label{e7.5}
\int_{\mathbb{R}}\xi _{\beta} u^2 dx
\leq C=C( T, \| \varphi  \| )
\end{equation}
for $0\leq t\leq T$ and $c$ not depending on $\beta
>0$, the weighted estimate remains true for  $\beta
\to 0$. 

Now, we assume that the result is true for
$(\alpha - 1)$ and we prove that it is true for
 $\alpha $.
To prove this,  we start from the
main inequality \eqref{e3.2} with  $\mu _1, \mu _2$ and $\xi $
given by $\mu _1^{\beta }, \mu _2^{\beta }$ and $\xi _{\beta}$
respectively.
\[
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx +
\int_{\mathbb{R}}\mu _1^{\beta } u_{\alpha + 1}^2 dx + \int_{\mathbb{R}}\mu _2^{\beta } u_{\alpha + 2}^2 dx +
\int_{\mathbb{R}}\theta _{\beta} u_{\alpha }^2 dx +
\int_{\mathbb{R}}R_{{\alpha }}  dx\leq 0
\]
with
\begin{gather*}
\mu _1^{\beta }   =   -c_5 ( 5 \eta + 3 ) \xi _{\beta}\quad 
\mbox { for $\eta < - 3/5$ (Natural Condition)}\\
\mu _2^{\beta }   =   - 5 \eta  \partial \xi _{\beta} \\
\theta _{\beta}  =  -\partial _{t}\xi _{\beta} - \eta  \partial ^{5}\xi _{\beta} -
\partial
^{3}\xi _{\beta} - \partial ( \xi _{\beta} u )\\
R_{\alpha }  =  O(u_{\alpha }, \dots)
\end{gather*}
then
\begin{align*}
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx +
\int_{\mathbb{R}}\mu _1^{\beta } u_{\alpha + 1}^2 dx 
+ \int_{\mathbb{R}}\mu _2^{\beta } u_{\alpha + 2}^2 dx 
& \leq  -\int_{\mathbb{R}}\theta _{\beta} u_{\alpha }^2 dx -
\int_{\mathbb{R}}R_{\alpha }  dx \\
&  \leq  \big|-\int_{\mathbb{R}}\theta _{\beta} u_{\alpha }^2 dx -
\int_{\mathbb{R}}R_{\alpha }  dx\big| \\
&  \leq  \int_{\mathbb{R}}| \theta _{\beta} | u_{\alpha }^2 dx +
\int_{\mathbb{R}}| R_{\alpha } |  dx\,.
\end{align*}
Using \eqref{e2.4} and Gagliardo-Nirenberg in the first term of the right 
side we obtain
\[
\int_{\mathbb{R}}| \theta _{\beta} | dx \leq c\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx
\]
Thus
\[
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx +
\int_{\mathbb{R}}\mu _1^{\beta } u_{\alpha + 1}^2 dx 
+ \int_{\mathbb{R}}\mu _2^{\beta } u_{\alpha + 2}^2 dx 
 \leq  c\int_{\mathbb{R}}\xi _{\beta}u_{\alpha }^2 dx + \int_{\mathbb{R}}| R_{\alpha } |  dx.
\]
According to \eqref{e3.5}, $\int_{\mathbb{R}}R_{\alpha }dx$ contains a term of the form
\begin{equation} \label{e7.6}
\int_{\mathbb{R}}\xi _{\beta} u_{{\nu _1}} u_{{\nu _2}} u_{{\alpha }} dx.
\end{equation}
Let $\nu _2\leq \alpha - 2$. Integrating \eqref{e7.6} by parts and using
H\"{o}lder's inequality we obtain
\begin{equation} \label{e7.7}
c\Big[\Big(\int_{\mathbb{R}}\xi _{\beta}
u_{\nu _2 + 1}^2 dx\Big)^{1/2} +
\Big(\int_{\mathbb{R}}\xi _{\beta}u_{{\nu _2}}^2 dx\Big)^{1/2}\Big]
\Big(\int_{\mathbb{R}}\xi _{\beta}
u_{{\alpha - 1}}^2 dx\Big)^{1/2}
\end{equation}
where \eqref{e7.7} is bounded by hypothesis.
Now suppose that $\alpha - 1=\nu _1=\nu _2$, then in \eqref{e7.6} we obtain
\[
\big| \int_{\mathbb{R}}\xi _{\beta}
u_{{\alpha - 1}} u_{{\alpha - 1}} u_{{\alpha }} dx \big|
 \leq  \left\| u_{{\alpha - 1}} \right\|_{L^{\infty }(\mathbb{R})}
\Big(\int_{\mathbb{R}}\xi _{\beta}u_{\alpha - 1}^2 dx\Big)^{1/2}
\Big(\int_{\mathbb{R}}\xi _{\beta}u_{\alpha }^2 dx\Big)^{1/2}
\]
where $\|u_{\alpha - 1}\|_{{L^{\infty }(\mathbb{R})}}$
is bounded by hypothesis, and the estimate is complete.
Finally, for $\nu _1=\alpha - 2; \nu _2=\alpha - 1$ we have
\begin{align*}
\big| \int_{\mathbb{R}}\xi _{\beta} u_{{\alpha - 2}} u_{{\alpha - 1}} 
u_{{\alpha }} dx\big|
&= \big| \int_{\mathbb{R}}\sqrt {\xi _{\beta}}
 u_{{\alpha - 2}} u_{{\alpha - 1}} \sqrt {\xi _{\beta}} u_{{\alpha }} dx \big| \\
& \leq  \big\| \sqrt {\xi _{\beta}} u_{{\alpha - 2}} \big\|_{{L^{\infty }(\mathbb{R})}}
\big| \int_{\mathbb{R}}u_{{\alpha - 1}} \sqrt {\xi _{\beta}} u_{{\alpha }} dx \big| \\
 & \leq  \big\| \sqrt {\xi _{\beta}} u_{{\alpha - 2}} 
 \big\|_{L^{\infty }(\mathbb{R})}
 \big\| u_{\alpha - 1} \big\|_{L^2(\mathbb{R})}
 \Big(\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx\Big)^{1/2} \\
 & \leq  c \left\| u_{{\alpha - 1}} \right\|_{{L^2(\mathbb{R})}}
 \Big(\int_{\mathbb{R}}\xi _{\beta} u_{{\alpha }}^2 dx\Big)^{1/2}.
\end{align*}
Using these estimates in \eqref{e7.5},
and applying the Gronwall's  argument,
we obtain for  $0\leq t\leq T$,
\[
\partial _{t}\int_{\mathbb{R}}\xi _{\beta} u_{\alpha }^2 dx +
\int_{\mathbb{R}}\mu _1^{\beta } u_{\alpha + 1}^2 dx 
+ \int_{\mathbb{R}}\mu _2^{\beta } u_{\alpha + 2}^2 dx  \leq  c_0 e^{c_1 t}
\Big(\int_{\mathbb{R}}\xi _{\beta} \varphi _{{\alpha }}^2(x) dx + 1\Big)
\]
where $c_0$ and $c_1$ are independent $\beta $ such that
letting the parameter $\beta \to 0$ the desired
estimates \eqref{e7.2} and \eqref{e7.3} are obtained.
\end{proof}

 \section{Main Theorem}

In this section we state and prove our main Theorem,
which states that if the
initial data $u(x, 0)$ decays faster than polinomially on
$\mathbb{R}^{+}=\{x\in \mathbb{R} ; x>0\}$ and
possesses certain initial Sobolev regularity, then the solution
$u(x, t)\in C^{\infty }$ for all $t>0$.
For the main Theorem, we take $6\leq \alpha \leq L + 4$.
For $\alpha \leq L + 4$, we take 
\begin{gather} \label{e8.1}
\mu _1\in W_{{\sigma ,L -\alpha + 5,\alpha - 5}}
\quad\Longrightarrow\quad  \xi \in W_{{\sigma ,L - \alpha + 5,\alpha - 5}}\\
\mu _2\in W_{{\sigma ,L -\alpha + 4,\alpha - 5}}
\quad\Longrightarrow\quad \xi \in W_{{\sigma,L - \alpha + 5,\alpha - 5}}
\label{e8.2}
\end{gather}

\begin{lemma}[Estimate of error terms]   \label{lm8.1} 
Let $6\leq \alpha \leq L + 4$ and
the weight functions be chosen as in \eqref{e8.1}-\eqref{e8.2}, then
\begin{equation} \label{e8.3}
\big| \int_{0}^{T}\int_{\mathbb{R}}
\left(\theta  u_{\alpha }^2 + R_{\alpha } \right) dx dt\big| \leq c\,,
\end{equation}
where $c$ depends only on the norms of $u$ in
\begin{gather*}
L^{\infty }([0, T];
H^{\beta }(W_{{\sigma ,L - \beta + 5,\beta - 5}}))
\bigcap L^2([0, T]; H^{\beta + 1}
(W_{{\sigma ,L - \beta + 5,\beta - 5}})\\
\bigcap H^{\beta + 2}
(W_{{\sigma ,L - \beta + 4,\beta - 5}}))
\end{gather*}
for $5\leq \beta \leq \alpha - 1$, and the norms of $ u $
in $L^{\infty }([0, T]; H^{5}(W_{0L0}))$.
\end{lemma}
 
\begin{proof} 
We must estimate both $R_{\alpha }$ and $\theta$.
We begin with a term of $R_{\alpha }$ of the form
\begin{equation} \label{e8.4}
\xi  u_{{\nu _1}} u_{{\nu _2}} u_{{\alpha }}
\end{equation}
assuming that $\nu _1\leq \alpha - 2$.
By the induction hypothesis, $u$ is bounded in \\
$L^{\infty }([0, T]; H^{\beta }(W_{{\sigma ,L - (\beta -
5)^{+}, (\beta - 5)^{+}}}))$ for
$0\leq \beta \leq \alpha - 1$. By Lemma \ref{lm0},
\begin{equation} \label{e8.5}
\sup _{t>0} \sup _{x\in \mathbb{R}} \zeta  u_{\beta}^2< + \infty
\end{equation}
for $0\leq \beta \leq \alpha - 2$ and $\zeta \in W_{{\sigma
,L - (\beta - 4)^{+},(\beta - 4)^{+}}}$. We estimate $u_{{\nu _1}}$ using \eqref{e8.5}.
We estimate $u_{{\nu _2}}$ and $u_{{\alpha }}$
using the weighted
$ L^2 $ bounds
\begin{equation} \label{e8.6}
\int_{0}^{T}\int_{\mathbb{R}}\zeta  u_{\nu _2}^2 dx dt<+\infty
\quad \mbox {for }\zeta \in W_{{\sigma ,L - (\nu _2 -
5)^{+},(\nu _2 - 6)^{+}}}
\end{equation}
and the same with $\nu _2$ replaced by $\alpha $.
It suffices to check the powers of $t$, the powers of
$x$ as $x\to +\infty $ and
the exponential of $x$ as $x\to -\infty .$\\
For $x>1. $
In the term \eqref{e8.4}, the factor $\xi $ constributed
according to \eqref{e8.1}-\eqref{e8.2}
\[
\xi (x, t)  =  t^{(\alpha - 5)} x^{(L - \alpha + 5)}
t^{-(\alpha - 5)} x^{-(L - \alpha + 5)} \xi (x, t) \\
 \leq  c_2 t^{(\alpha - 5)} x^{(L - \alpha + 5)}
\]
by  \eqref{e2.3}.  Then 
$\xi   u_{{\nu _1}}u_{{\nu _2}} u_{{\alpha }}\leq
c_2 t^{(\alpha - 5)} x^{(L - \alpha + 5)} u_{{\nu _1}} u_{{\nu _2}} 
u_{{\alpha }}$.
Moreover
\begin{align*}
u_{{\nu _1}} u_{{\nu _2}} u_{{\alpha }} 
& =  t^{\frac {(\nu _1 -
4)^{+}}{2}} x^{\frac {L - (\nu _1 - 4)^{+}}{2}} t^{\frac {-(\nu _1
- 4)^{+}}{2}} x^{\frac {-(L - (\nu _1 - 4)^{+})}{2}} \\
&\quad\times u_{{\nu_1}} t^{\frac {(\nu _2 - 6)^{+}}{2}}
 x^{\frac {L - (\nu _2 - 5)^{+}}{2}} t^{\frac {-(\nu _2 -
6)^{+}}{2}} x^{\frac {-(L - (\nu _2 - 5)^{+})}{2}} u_{{\nu _2}} \\
&\quad\times t^{\frac {(\alpha -
6)^{+}}{2}} x^{\frac {L - (\alpha - 5)^{+}}{2}} t^{\frac {-(\alpha -
6)^{+}}{2}} x^{\frac {-(L - (\alpha - 5)^{+})}{2}} u_{\alpha }.
\end{align*}
It follows that
\begin{equation} \label{e8.7}
\begin{aligned}
&\xi  u_{{\nu_1}} u_{{\nu _2}} u_{{\alpha }}\\
&\leq  c_2 t^{M} x^{T}t^{\frac {(\nu _1 - 4)^{+}}{2}} x^{\frac {L - (\nu _1 -
4)^{+}}{2}} u_{{\nu _1}} t^{\frac {(\nu_2 - 6)^{+}}{2}} x^{\frac {L - (\nu _2 -
5)^{+}}{2}} u_{{\nu _2}} t^{\frac {(\alpha - 6)^{+}}{2}} x^{\frac
{L - (\alpha - 5)}{2}} u_{{\alpha }}
\end{aligned}
\end{equation}
where
$M  =  \alpha - 5 - \frac {1}{2}(\nu _1 - 4)^{+} -
\frac {1}{2}(\nu _2 - 6)^{+} -
\frac {1}{2}(\alpha - 6)^{+}$ and
\[
T  =  (T - \alpha + 5) -\frac {1}{2}(T - (\alpha -
5)^{+}) -
\frac {1}{2}(T - (\nu _2 - 5)^{+}) - \frac {1}{2}(T - (\nu _1 - 4)^{+}).
\]

\noindent\textbf{Claim $M\geq 0$ is large enough, that the extra power of  $t$
can be omitted.}
\begin{align*}
2M & =   2 \alpha - 10 - (\nu _1 - 4)^{+} -
(\nu _{2} - 6)^{+} - (\alpha - 6)^{+}\\
& =  \alpha - 4 - (\nu _1 - 4)^{+} -(\nu _2 - 6)^{+}\\
&=  \alpha - 4 - \nu _1 + 4 - \nu _2 + 6\\
& =  \alpha + 6 - (\nu _1 + \nu _2)\\
&=  \alpha + 6 - (\alpha + 1)
 =  5\geq 0.
\end{align*}

\noindent\textbf{Claim $T\leq 0$ is such that the extra power $x^{T}$
can be bounded as $x\to +\infty $.}
\[
T=L - \alpha + 5 - \frac {1}{2}(L - (\alpha - 5)^{+}) - \frac {1}{2}
(L - (\nu _2 - 5)^{+}) -\frac {1}{2}(L - (\nu _1 - 4)^{+}).
\]
Thus
\begin{align*}
2 T & =  2 L - 2 \alpha + 10 - (L - (\alpha - 5)^{+}) - L +
(\nu _2 - 5)^{+} - L + (\alpha - 4)^{+}\\
& =  - L - \alpha + \nu _1 + \nu _2 - 4\\
& =  - L - \alpha + \alpha + 1 - 4\\
& =  - (L + 3)\leq 0.
\end{align*}
Now, we study the behavior as $x \to -\infty $. 
Since each factor
$u_{{\nu _j}} (j=1, 2)$ must grow slower than an exponential
$e^{\sigma ^{^{,}} | x |}$ and $\xi $ decays as an
exponential $e^{-\sigma  | x |}$, we simply need to choose
the appropriate relationship between $\sigma $ and $
\sigma ^{^{,}}$ at each induction step.
The analysis of all the terms of
$R_{\alpha }$ will be completed with the case of
$\nu _1\geq \alpha - 1$. Then in \eqref{e3.6} if
$2(\alpha + 1)\leq \alpha + 1$,
$\alpha \leq 3$, but $\alpha \geq 5$ so this possibility is impossible.
For $x<1$ the estimate is similar,
except for an exponential weight. This completes the estimate of $R_{\alpha }$.

Now we estimate the term $\theta u_{{\alpha }}^2$ where $\theta $ is given in \eqref{e3.2}. We have that
$\theta $ involves derivatives of $u$ only up
to order one and hence $\theta u_{{\alpha }}^2$ is a sum of terms
of the same type which we have already encountered in $R_{\alpha }$. So,
 its integral can be bounded in the same manner.
Indeed \eqref{e3.2} shows that $\theta $ depends on $\xi _{t},
\partial ^{5}\xi $ and derivatives of lower order.
By using  \eqref{e3.3} we have the claim.
\end{proof}

\begin{theorem}[Main Theorem] \label{thm8.2.}
Let $T>0$ and $u(x, t)$ be a solution of \eqref{e2.5} in the region
 $\mathbb{R} \times [0, T]$ such that
\begin{equation} \label{e8.8}
u\in L^{\infty }([0, T]; H^{5}(W_{0L0}))
\end{equation}
for some $L\geq 2$ and all $\sigma >0$. Then
$u$ is in $L^{\infty }([0, T]; H^{5 + l}(W_{{\sigma ,L - l,l}}))\newline
\cap L^2([0, T]; H^{6 + l}(W_{{\sigma ,L - l,l}})
 \cap  H^{7 + l}(W_{{\sigma ,L - l - 1,l}}))
$ for all $0\leq l\leq L - 1$.
\end{theorem}

\begin{remark} \label{rmkl8.1} \rm
 If the assumption \eqref{e8.8} holds for all
$L\geq 2$, the solution is infinitely differentiable in the
$x$-variable. From \eqref{e2.5} we have that the solution is 
$C^{\infty }$ in both
variables.
\end{remark}
 
\begin{proof} We use induction on $\alpha $.
For $\alpha = 5$, let $u$ be a solution of \eqref{e2.5} satisfying \eqref{e8.8}.
Therefore, 
$u_{{t}}\in L^{\infty }([0, T]; L^2(W_{0L0}))$
where $
u\in L^{\infty }([0, T]; H^{5}(W_{0L0}))$ and
$u_{{t}}\in L^{\infty }([0, T]; L^2(W_{0L0}))$.
Then $
u\in C([0, T]; L^2(W_{0L0}))\bigcap
C_{w}([0, T]; H^{5}(W_{0L0}))$.
Hence $u\colon [ 0, T ]\mapsto H^{5}(W_{0L0})$ is a
weakly continuous function.
In particular, $u( \cdot  , t)\in H^{5}(W_{0L0})$
for all $t$.
Let $t_0\in (0, T)$ and $u( \cdot  , t_0)
\in H^{5}(W_{0L0})$, then there are
$\{ \varphi ^{(n)} \}\subset C_0^{\infty }(\mathbb{R})$ such that
$\varphi ^{(n)}( \cdot  )
\to u( \cdot  , t_0)$ in $H^{5}(W_{0L0})$.
Let $u^{(n)}(x, t)$ be a unique solution of \eqref{e2.5} with
$u^{(n)}(x, t_0)=\varphi ^{(n)}(x)$. Then by  Theorems \ref{thm5.1}
and \ref{thm5.2},  there exists in a time interval $[t_0, t_0 + \delta ]$ where $\delta >0$ does not
depend on $n$ and $u$ is a unique solution of \eqref{e2.5}
$u^{(n)}\in L^{\infty }([t_0, t_0 + \delta ]; H^{5}(W_{0L0}))$
with
$ u^{(n)}(x, t_0)\equiv \varphi ^{(n)}(x)
\to u(x, t_0)\equiv
\varphi (x)$ in $H^{5}(W_{0L0})$.
Now, by Theorem \ref{thm7.1}, we have
\[
u^{(n)}\in L^{\infty }([t_0, t_0 + \delta ]; H^{5}(W_{0L0}))
\bigcap L^2([t_0, t_0 + \delta ]; 
H^{6}(W_{{\sigma  L 0}})\cap H^{7}(W_{{\sigma ,L - 1,0}}))
\]
with a bound that depends only on the norm of
$\varphi ^{(n)}$ in $H^{5}(W_{0L0})$.
Furthermore, Theorem \ref{thm7.1} guarantees the non-uniform bounds
\[
\sup _{[t_0, t_0 + \delta ]} \sup _{x}
 (1 + | x_{+} | )^{k} | \partial ^{\alpha }u^{(n)}(x, t) |<+\infty
\]
for each $n, k$ and $\alpha $.
The main inequality \eqref{e3.2} and the estimate \eqref{e8.3} are therefore valid
for each $ u^{(n)} $ in the interval
$[t_0, t_0 + \delta ]$.
$\mu _2$ may be chosen arbitrarily in its weight class \eqref{e8.1} and then
$\xi $ is defined by \eqref{e3.4} and the constant $c_1, c_2, c_3,
c_4$
are independent of $n$.
 From \eqref{e3.2} and \eqref{e8.1}-\eqref{e8.2} we have
\begin{equation} \label{e8.9}
\sup _{[t_0, t_0 + \delta ]}
\int_{\mathbb{R}}\xi  [u_{\alpha }^{(n)}]^2 dx +
\int_{t_0}^{t_0 + \delta }\int_{\mathbb{R}}{\mu _1} [u_{\alpha + 1}^{(n)}]^2 dx dt 
+ \int_{t_0}^{t_0 + \delta }\int_{\mathbb{R}}{\mu _2} [u_{\alpha + 2}^{(n)}]^2 dx dt
\leq c
\end{equation}
where by \eqref{e8.3}, $c$ is independ of $n$.
Estimate \eqref{e8.9} is proved by induction for $\alpha =5, 6, \dots$
Thus $u^{(n)}$ is also bounded in
\begin{equation} \label{e8.10}
\begin{gathered}
L^{\infty }([t_0, t_0 + \delta ]; H^{\alpha }(W_{{\sigma ,L -
\alpha + 5,\alpha - 5}})) \bigcap 
L^2([t_0, t_0 + \delta ]; H^{\alpha + 1}(W_{{\sigma ,L -
\alpha + 5,\alpha - 5}}))  \\
 \bigcap  L^2([t_0, t_0 + \delta ]; H^{\alpha + 2}(W_{{\sigma ,L -
\alpha + 4,\alpha - 5}}))
\end{gathered}
\end{equation}
for $\alpha \geq 5$. Since
$u^{(n)}\longrightarrow u$ in $
L^{\infty }([t_0, t_0 + \delta ]; H^{5}(W_{0L0}))$.
By Corollary \ref{coro5.3} it follows that $u$ belongs to the space \eqref{e8.10}.
Since $\delta $ is fixed, this result is valid over the whole
interval  $[0, T]$. 
\end{proof}

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