# Asymptotics of two–point correlations in the multi–species $q$–TAZRP

Jeffrey Kuan and Zhengye Zhou

###### Abstract

A previous paper by the authors found explicit contour integral formulas for
certain joint moments of the multi–species $q$–TAZRP (totally asymmetric zero
range process), using algebraic methods. These contour integral formulas have
a “pseudo–factorized” form which makes asymptotic analysis simpler. In this
brief note, we use those contour integral formulas to find the asymptotics of
the two–point correlations. As expected the term arising from the
“shift–invariance” makes a non–trivial asymptotical contribution.

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##  Introduction

In interacting particle systems, control of the two-point correlations are
generally needed to determine the hydrostatic limit. For symmetric exclusion
processes, this has been done in [BMNS17] and [Tsu20]. For symmetric inclusion
process, this was done in [FGS22]. In asymmetric processes, calculating the
correlations is naturally harder, such as [UW05].

In more recent years, a certain type of interacting particle systems called
“multi–species” models have been well–studied, such as [BS15b, BS15a, BS18,
BGW, Gal, Kua17, Kua18a, KMMO16, Tak, Zho]. In this paper, we consider one
specific asymmetric multi–species model, called the multi–species $q$–TAZRP.
In previous work [FKZ], the authors and C. Franceschini found an explicit
expression for the two–point correlations of the height function of this
model. (That work originated with orthogonal polynomial duality (e.g. [ACR18,
ACR21, CFG21, CFGGR19, FG, Gro19, Zho]), but this current paper will not
reference duality). This expression is written as the sum of six terms, with
each term being a product of a simple $q$–Pochhammer (which only depends on
the number of initial particles) and a contour integral. Of these six terms,
five can be derived from the Markov projection (or color–blind) property, but
the sixth used a so–called “shift–invariance”.

The contour integral and product form of each term makes the asymptotics
suitable for analysis, using standard and well–known methods. In this brief
note, we carry out these calculations and make the observation that the term
arising from shift–invariance makes a non-trivial asymptotic contribution, as
expected.

Acknowledgements. The authors would like to thank Chiara Franceschini for
helpful discussions at the workshop “Recent Developments in Stochastic
Duality” in Eindhoven, and for support from EURANDOM and the NWO grant
613.001.753 “Duality for interacting particle systems”. The first author was
supported by National Science Foundation grant DMS-2000331.

##  Main Result

###  Definition of Model

The multi–species $q$–TAZRP will be defined informally here. It was first
introduced by Takeyama, generalizing the single–species version by Sasamoto
and Wadati. There are $n$ different species of particles, labeled $1,2,\ldots
n$. Arbitrary many particles of any number of species may occupy a lattice
site. The lattice is the one–dimensional lattice, labeled $\mathbf{Z}$, and
particles jump one step to the right at a time.

The jump rates depend only on the number of particles at a single site, as is
the case for any zero range process. Let $[k]_{q}=1-q^{k}$ denote the
$q$–integer in the below equations. If the number of species $k$ particles at
a site is denoted by $z_{k}$, then the jump rate for the species $k$ particle
is $q^{z_{1}+\cdots+z_{k-1}}[z_{k}]_{q}{/(1-q)}.$

Based on these jump rates, the multi–species $q$–TAZRP has a “Markov
projection” property: if one only ignores the species completely, one obtains
the (single–species) $q$–TAZRP. See the Figure 1 for an example of the jump
rates.

Figure 1: The jump rates for the $q$–TAZRP.

Recall the incomplete gamma function

$\Gamma(a,z)=\int_{z}^{\infty}t^{a-1}\mathrm{e}^{-t}\mathrm{~{}d}t$

and define its normalization

$Q(a,z)=\frac{\Gamma(a,z)}{\Gamma(a)}.$

Recall that for a continuous–time totally asymmetric simple random walk $X(t)$
starting at $0$, with jump rates $r$, then

$\mathbf{P}(X(t)\leq
n)=\sum_{k=0}^{n}e^{-rt}\frac{(rt)^{k}}{k!}=\frac{\Gamma(n+1,rt)}{n!}=Q(n+1,rt).$

We also have

${}_{1}\phi_{0}\left(\begin{array}[]{c}q^{-n}\\\
-\end{array};,z\right)=\left(zq^{-n};q\right)_{n},\quad n=0,1,2,\ldots$

where the $q$–Pochhammer is

$(a;q)_{n}=(1-a)(1-aq)\cdots\left(1-aq^{n-1}\right),\quad(a;q)_{0}:=1.$

In Theorem 3.5 of [FKZ], it was proved that

######  Theorem 1.

Suppose $\boldsymbol{\xi}$ has initial conditions with $n_{1}$ species 0
particles at site $x_{1}$ and $n_{2}$ species 1 particles at site $x_{2}$, or
in other words

$\boldsymbol{\xi}_{i}^{x}=\begin{cases}n_{1}\ \text{for}\ i=0,x=x_{1},\\\
n_{2}\ \text{for}\ i=1,x=x_{2},\\\ 0\ \ \text{else},\end{cases}$

and $\boldsymbol{\xi}$ evolves as a $q$–TAZRP with total asymmetry to the
right. Let $y_{1}$ and $y_{2}$ be some lattice sites, and additionally
assuming that $x_{1}<x_{2}$.

The dual process $\boldsymbol{\eta}$ has 1 species 0 particles at site
$y_{1}$, 1 species 1 particles at site $y_{2}$, so in other words

$\boldsymbol{\eta}_{i}^{y}=\begin{cases}1\ \text{for}\ i=0,y=y_{1},\\\ 1\
\text{for}\ i=1,y=y_{2},\\\ 0\ \ \text{else}.\end{cases}$

Then there is the explicit formula

$\mathbb{E}_{\boldsymbol{\xi}}[q^{1/2h(\boldsymbol{\xi}(t),\boldsymbol{\eta})}\prod_{i=0}^{1}\prod_{x=1}^{L}\
{}_{1}\phi_{0}\left(q^{-\xi_{i}^{x}(t)};q,q^{-\left(N_{x-1}^{-}(\boldsymbol{\xi}_{i}(t))+N_{x+1}^{+}(T(\boldsymbol{\eta})_{i+1})\right)+1/2}\right)]=\sum_{i=1}^{6}{\delta_{i}p_{i}},$

where

$h(\boldsymbol{\xi},\boldsymbol{\eta})=\sum_{x\in\Lambda_{L}}\sum_{i=0}^{n-1}\left(-\xi_{i}^{x}N_{x+1}^{+}\left(\boldsymbol{\eta}_{[0,n-2-i]}\right)+\eta_{i}^{x}N_{x}^{+}\left(\boldsymbol{\xi}_{[0,n-2-i]}\right)\right)$
$\displaystyle\delta_{1}:=q^{n_{1}/2}\left(q^{-n_{1}-1/2};q\right)_{n_{1}}\left(q^{-n_{2}+1/2};q\right)_{n_{2}},$
$\displaystyle\delta_{2}:=q^{n_{1}/2}\left(q^{-n_{1}+1/2};q\right)_{n_{1}}\left(q^{-n_{2}+1/2};q\right)_{n_{2}},$
$\displaystyle\delta_{3}:=q^{-n_{1}/2}\left(q^{-n_{1}-1/2};q\right)_{n_{1}}\left(q^{-n_{2}+1/2};q\right)_{n_{2}},$
$\displaystyle\delta_{4}:=q^{-n_{1}/2}\left(q^{-n_{1}+1/2};q\right)_{n_{1}}\left(q^{-n_{2}+1/2};q\right)_{n_{2}},$
$\displaystyle\delta_{5}:=q^{-n_{1}/2}\left(q^{-n_{1}-1/2};q\right)_{n_{1}}\left(q^{-n_{2}-1/2};q\right)_{n_{2}},$
$\displaystyle\delta_{6}:=q^{-n_{1}/2}\left(q^{-n_{1}+1/2};q\right)_{n_{1}}\left(q^{-n_{2}-1/2};q\right)_{n_{2}},$

and

$p_{1}=q_{1}-q_{4},\quad p_{2}=q_{4},\quad p_{3}=q_{2}+q_{3}-q_{5}-q_{6},\quad
p_{4}=-q_{3}+q_{5}+q_{6},\quad p_{5}=q_{5},\quad p_{6}=q_{3}-q_{5},$

with

$q_{1}={\delta_{y_{1}\leq x_{1}+}\delta_{y_{1}>x_{1}}}Q(y_{1}-x_{1},t),\quad
q_{3}=1-{\delta_{y_{1}\geq
x_{2}}Q(y_{1}-x_{2},t)}{-\delta_{y_{1}<x_{2}}},\quad q_{2}=1-q_{1}-q_{3}$

and

$\displaystyle q_{4}$ $\displaystyle=\frac{1}{(2\pi
i)^{2}}\int_{C}\frac{dw_{1}}{w_{1}}\int_{C}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{w_{1}-qw_{2}}\prod_{j=1}^{2}\left[(1-w_{j})^{-({y_{j}-x_{1}}+1)}e^{-w_{j}t}\right],$
$\displaystyle q_{5}$ $\displaystyle=\frac{q}{(2\pi
i)^{2}}\int_{\tilde{C}_{1}}\frac{dw_{1}}{w_{1}}\int_{\tilde{C}_{2}}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{w_{1}-qw_{2}}(1-w_{1})^{-({y_{1}-x_{2}}+1)}(1-w_{2})^{-({y_{2}-x_{1}}+1)}e^{-(w_{1}+w_{2})t},$
$\displaystyle q_{6}$ $\displaystyle=1-q_{1}-q\frac{1}{(2\pi
i)^{2}}\int_{\tilde{C}_{1}}\frac{dw_{1}}{w_{1}}\int_{\tilde{C}_{2}}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{w_{1}-qw_{2}}\prod_{j=1}^{2}\left[(1-w_{j})^{-({y_{j}-x_{1}}+1)}e^{-w_{j}t}\right],$

Here, the $C$ are “large contours” that are centered at $0$ and contain $1$,
and the $\tilde{C}$ are “small contours” where $\tilde{C_{2}}$ contains $1$
and not $0$, whereas $\tilde{C_{1}}$ contains $q\tilde{C_{2}},1$ and not $0$.

######  Remark 2.

Of these six terms, five of them (except $q_{5}$) can be derived from the two
Markov projections to single–species $q$–TAZRP. The term $q_{5}$ used the
“shift–invariance” property of [BGW, Gal, Kua21], and see also [He]; see the
fifth bullet point of section 4.9 of [FKZ]. Note that the $\delta_{j}$ terms
only depend on $n_{1},n_{2}$; while the $q_{j}$ terms do not depend on $n_{1}$
or $n_{2}$ at all. This makes asymptotic analysis significantly easier.

######  Remark 3.

There was a typo in Theorem 3.5 of [FKZ] (arXiv version), which has been
corrected here, and will be corrected when the refereed version of the paper
is updated on arXiv. The typo occured due to conflicting notation when citing
the results of [Kua21] in Lemmas 4.13 and 4.14 of [FKZ] . The correction has
been confirmed by all three authors of [FKZ].

###  Main Theorem

The main theorem of this paper is essentially an asymptotic analysis of the
contour integrals. Each term will be analyzed individually. Since $\delta_{j}$
only depend on $n_{j}$, those terms do not need to be analyzed.

######  Theorem 4.

Consider the limit as ${y_{j}-x_{i}}=L+c_{ij}L^{1/2}+O(1)$ and $t=L$. Then as
$L\rightarrow\infty$,

$\displaystyle q_{1}$
$\displaystyle\approx\frac{1}{2}\operatorname{erfc}\left(-\frac{c_{11}}{2^{\frac{1}{2}}}\right)+\frac{1}{\sqrt{2\pi
a_{1}}}\exp\left(-\frac{c_{11}^{2}}{2}\right)\sum_{n=0}^{\infty}\frac{C_{n}\left(-c_{11}\right)}{a_{1}^{n/2}},$
$\displaystyle q_{3}$ $\displaystyle\approx
1-\frac{1}{2}\operatorname{erfc}\left(-\frac{c_{21}}{2^{\frac{1}{2}}}\right)-\frac{1}{\sqrt{2\pi
a_{2}}}\exp\left(-\frac{c_{21}^{2}}{2}\right)\sum_{n=0}^{\infty}\frac{C_{n}\left(-c_{21}\right)}{a_{2}^{n/2}},$

where $\mathrm{erfc}$ denotes the complementary error function and
$C_{n}(\eta)$ are polynomials in $\eta$ of degree $3n+2$ and satisfy

$C_{0}(\eta)=\frac{1}{3}\eta^{2}-\frac{1}{3},$

and

$C_{n}(\eta)+\eta
C_{n}^{\prime}(\eta)-C_{n}^{\prime\prime}(\eta)=\eta\left(\eta^{2}-2\right)C_{n-1}(\eta)-\left(2\eta^{2}-1\right)C_{n-1}^{\prime}(\eta)+\eta
C_{n-1}^{\prime\prime}(\eta),$

and

$a_{i}=L+\frac{c_{i1}^{2}}{2}+c_{i1}\sqrt{L+\frac{c_{i1}^{2}}{4}}.$

Furthermore,

$\displaystyle q_{4}$
$\displaystyle\approx\frac{1}{2\pi}\int_{-\infty}^{c_{1j}}dc_{1j}\int_{-\infty}^{\infty}\exp\left(-\frac{u_{j}^{2}}{2}-c_{1j}iu_{j}\right)\left(1-\left(\frac{3iu_{j}}{2}+\frac{c_{1j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{j}$
$\displaystyle q_{5}$ $\displaystyle\approx q\cdot I(c_{21},c_{12})$
$\displaystyle q_{6}$ $\displaystyle\approx 1-q_{1}-q\cdot I(c_{11},c_{12})$

where

$\displaystyle I(\sigma_{1},\sigma_{2})=$ $\displaystyle-\frac{q}{(2\pi
i)^{2}}\int_{-\infty}^{\sigma_{1}}d\sigma_{1}\int_{-\infty}^{\sigma_{2}}d\sigma_{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{u_{1}-u_{2}}{u_{1}-qu_{2}}\prod_{j=1}^{2}\exp\left(-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}\right)$
$\displaystyle\quad\quad\quad\quad\times\left(1-\sum_{j=1}^{2}\left(\frac{3iu_{j}}{2}+\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{1}du_{2},$

where the $u_{1}$ and $u_{2}$ contours are deformed to avoid the pole at
$u_{1}-qu_{2}$.

##  Proof

###  Asymptotics of $q_{1},q_{2},q_{3}$

Because the terms $q_{1},q_{2},q_{3}$ are simply the incomplete gamma
function, an asymptotic analysis of the incomplete gamma function determines
the asymptotics of $q_{1},q_{2}$ and $q_{3}$. The full asymptotic expansion
was determined in a recent paper [ND19]. The result is that as
$a\rightarrow\infty$,

$Q\left(a,a+\tau
a^{\frac{1}{2}}\right)\sim\frac{1}{2}\operatorname{erfc}\left(2^{-\frac{1}{2}}\tau\right)+\frac{1}{\sqrt{2\pi
a}}\exp\left(-\frac{\tau^{2}}{2}\right)\sum_{n=0}^{\infty}\frac{C_{n}(\tau)}{a^{n/2}}.$

Somewhat awkwardly, the exponent of $1/2$ is in the second variable rather
than the first. So if $L=a+\tau a^{1/2}$ then

$a={-}\tau\sqrt{L+\tau^{2}/4}+L+\tau^{2}/2=L{-}\tau\sqrt{L}+O(1).$

Thus as $L\rightarrow\infty$,

$q_{1}=Q((L+c_{11}L^{1/2}+O(1),L)\approx\frac{1}{2}\operatorname{erfc}\left(-\frac{c_{11}}{2^{\frac{1}{2}}}\right)+\frac{1}{\sqrt{2\pi
a_{1}}}\exp\left(-\frac{c_{11}^{2}}{2}\right)\sum_{n=0}^{\infty}\frac{C_{n}\left(-c_{11}\right)}{a_{1}^{n/2}},$

where
$a_{1}=L+\frac{c_{11}^{2}}{2}+c_{11}\sqrt{L+\frac{c_{i1}^{2}}{4}}=L+c_{11}L^{1/2}+O(1)$.

###  Asymptotics of $q_{4},q_{5},q_{6}$

To analyze the asymptotics of $q_{5}$ and $q_{6}$, first denote

$q(\sigma_{1},\sigma_{2})=\frac{1}{(2\pi
i)^{2}}\int_{\tilde{C}_{1}}\frac{dw_{1}}{w_{1}}\int_{\tilde{C}_{2}}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{(w_{1}-qw_{2})(1-w_{1})(1-w_{2})}\\\
\times\prod_{j=1}^{2}\exp\left(LS\left(w_{j}\right)-L^{1/2}\sigma_{j}\log\left(1-w_{j}\right)\right)$

where $S(w)=-w-\log(1-w)$. Then, $q_{5}=q\cdot q(c_{21},c_{12})$ and
$q_{6}=1-q_{1}-q\cdot(c_{11},c_{12})$. As observed in section 5 of [Kua21],
$S(w)$ has a saddle point at $w=0$. Near $w=0$, we have the following Taylor
expansions:

$S(u)=\sum_{j=2}^{\infty}\frac{u^{j}}{j},\quad\quad\log(1-u)=-\sum_{j=1}^{\infty}\frac{u^{j}}{j}.$

Now change the contours to vertical lines through $0$, let
$w_{i}=-iL^{-1/2}u_{i}$

$\displaystyle LS\left(w_{j}\right)-L^{1/2}\sigma_{j}\log\left(1-w_{j}\right)$
$\displaystyle=-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}-\left(\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1}).$

To avoid the simple pole at $0$, we take partial derivatives with respect to
$\sigma_{j}$, thus by [Kua21] (The current arXiv version of [Kua21] omits the
comment about deforming the contours to avoid the pole, but it will be updated
in the journal version)

$\displaystyle\frac{\partial^{2}q(\sigma_{1},\sigma_{2})}{\partial\sigma_{1}\partial\sigma_{2}}$
$\displaystyle=\frac{L}{(2\pi
i)^{2}}\int\frac{\log(1-w_{1})}{w_{1}}dw_{1}\int\frac{\log(1-w_{2})}{w_{2}}dw_{2}\frac{w_{1}-w_{2}}{(w_{1}-qw_{2})(1-w_{1})(1-w_{2})}$
$\displaystyle\prod_{j=1}^{2}\exp\left(S\left(w_{j}\right)-L^{1/2}\sigma_{j}\log\left(1-w_{j}\right)\right)$
$\displaystyle=\frac{qL}{(2\pi
i)^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{u_{1}-u_{2}}{u_{1}-qu_{2}}\prod_{j=1}^{2}\frac{\log(1+iL^{-1/2}u_{j})}{u_{j}(1+iL^{-1/2}u_{j})}$
$\displaystyle\exp\left(-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}-\left(\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{j}$
$\displaystyle=-\frac{q}{(2\pi
i)^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{u_{1}-u_{2}}{u_{1}-qu_{2}}\prod_{j=1}^{2}\left(1-iL^{-1/2}u_{j}/2+O(L^{-1})\right)\left(1-iL^{-1/2}u_{j}+O(L^{-1})\right)$
$\displaystyle\exp\left(-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}\right)\left(1-\left(\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{j}$
$\displaystyle=-\frac{q}{(2\pi
i)^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{u_{1}-u_{2}}{u_{1}-qu_{2}}\prod_{j=1}^{2}\exp\left(-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}\right)$
$\displaystyle\left(1-\sum_{j=1}^{2}\left(\frac{3iu_{j}}{2}+\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{1}du_{2}.$

Thus

$\displaystyle q(\sigma_{1},\sigma_{2})\approx$ $\displaystyle-\frac{q}{(2\pi
i)^{2}}\int_{-\infty}^{\sigma_{1}}d\sigma_{1}\int_{-\infty}^{\sigma_{2}}d\sigma_{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{u_{1}-u_{2}}{u_{1}-qu_{2}}\prod_{j=1}^{2}\exp\left(-\frac{u_{j}^{2}}{2}-\sigma_{j}iu_{j}\right)$
$\displaystyle\left(1-\sum_{j=1}^{2}\left(\frac{3iu_{j}}{2}+\frac{\sigma_{j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{1}du_{2}.$

Next analyze $q_{4}$. Deform contours in $q_{4}$ so that $0$ is outside of new
contours:

$\displaystyle q_{4}=$ $\displaystyle\frac{1}{(2\pi
i)^{2}}\int_{C}\frac{dw_{1}}{w_{1}}\int_{C}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{w_{1}-qw_{2}}\prod_{j=1}^{2}\left[(1-w_{j})^{-(y_{j}-x_{1}+1)}e^{-w_{j}t}\right]$
$\displaystyle=$ $\displaystyle\frac{1}{(2\pi
i)^{2}}\int_{C}\frac{dw_{1}}{w_{1}}\int_{\tilde{C}_{2}}\frac{dw_{2}}{w_{2}}\frac{w_{1}-w_{2}}{w_{1}-qw_{2}}\prod_{j=1}^{2}\left[(1-w_{j})^{-(y_{j}-x_{1}+1)}e^{-w_{j}t}\right]$
$\displaystyle+\frac{1}{2\pi
i}\int_{C}(1-w_{1})^{-(y_{1}-x_{1}+1)}e^{-w_{1}t}\frac{dw_{1}}{w_{1}}$
$\displaystyle=$ $\displaystyle q(c_{11},c_{12})+1+\frac{1}{2\pi
i}\int_{\tilde{C}_{1}}(1-w_{1})^{-(y_{1}-x_{1}+1)}e^{-w_{1}t}\frac{dw_{1}}{w_{1}}$
$\displaystyle+\frac{1}{2\pi
i}\int_{\tilde{C}_{2}}(1-w_{2})^{-(y_{2}-x_{1}+1)}e^{-w_{2}t}\frac{dw_{2}}{w_{2}}.$

Last, we take the limit as $L\xrightarrow[]{}\infty$ of the following
integral:

$\displaystyle\frac{1}{2\pi
i}\int_{\tilde{C}_{j}}(1-w_{j})^{-(y_{j}-x_{1}+1)}e^{-w_{j}t}\frac{dw_{j}}{w_{j}(1-w_{j})}$
$\displaystyle=\frac{1}{2\pi
i}\int_{\tilde{C}_{j}}\exp\left(LS\left(w_{j}\right)-L^{1/2}c_{1j}\log\left(1-w_{j}\right)\right)\frac{dw_{j}}{w_{j}}$
$\displaystyle=\frac{-L^{1/2}}{2\pi
i}\int_{-\infty}^{c_{1j}}dc_{1j}\int_{\tilde{C}_{j}}\frac{\log(1-w_{j})}{w_{j}(1-w_{j})}\exp\left(LS\left(w_{j}\right)-L^{1/2}c_{1j}\log\left(1-w_{j}\right)\right)dw_{j}$
$\displaystyle\approx\frac{1}{2\pi}\int_{-\infty}^{c_{1j}}dc_{1j}\int_{-\infty}^{\infty}\exp\left(-\frac{u_{j}^{2}}{2}-c_{1j}iu_{j}\right)\left(1-\left(\frac{3iu_{j}}{2}+\frac{c_{1j}u_{j}^{2}}{2}-\frac{iu_{j}^{3}}{3}\right)L^{-1/2}+O(L^{-1})\right)du_{j}.$

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# Sharp global well-posedness for the higher order non-linear Schrödinger
equation on modulations spaces

X. Carvajal  ,  P. Gamboa  and  R. Santos<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We consider the initial value problem (IVP) associated to a higher order
nonlinear Schrödinger (h-NLS) equation

$\partial_{t}u+ia\partial^{2}_{x}u+b\partial^{3}_{x}u+ic_{1}|u|^{2}u+c_{2}|u|^{2}\partial_{x}u=0,\quad
x,t\in\mathbb{R},$

for given data in the modulation space $M_{s}^{2,p}(\mathbb{R})$. Using ideias
in Killip, Visan, Zhang [19], Oh, Wang [25] and [26], we prove that the IVP
associated to the h-NLS equation is globally well-posed in the modulation
spaces $M^{s,p}$ for $s\geq\frac{1}{4}$ and $p\geq 2$.

Key-words: Schrödinger equation, Korteweg-de Vries equation, Initial value
problem, Well-posedness, Sobolev spaces, Fourier-Lebesgue spaces, Modulation
spaces.

##  1\. Introduction

In this work we consider the initial value problem (IVP) associated to a
higher order nonlinear Schrödinger (h-NLS) equation

$\begin{cases}\partial_{t}u+ia\partial^{2}_{x}u+b\partial^{3}_{x}u+ic_{1}|u|^{2}u+c_{2}|u|^{2}\partial_{x}u=0,\quad
x,t\in\mathbb{R},\\\ u(x,0)=u_{0}(x),\end{cases}$ (1.1)

where $a,b,c_{1},c_{2}\in\mathbb{R}$ and $u=u(x,t)$ is a complex valued
function.

The main objective of this work is to investigate the global well-posedness
issues of the IVP (1.1) in the modulation spaces.

In recent time, well-posedness of the IVPs associated to the nonlinear
dispersive equations are being studied in some other scales of the function
spaces than the usual $L^{2}$ based Sobolev spaces $H^{s}(\mathbb{R})$ viz.,
the Fourier-Lebesgue spaces $\mathcal{F}L^{s,p}(\mathbb{R})$ with norm

$\|u\|_{\mathcal{F}L^{s,p}}=\|\langle\xi\rangle^{s}\widehat{u}(\xi)\|_{L^{p}},$

and modulation spaces $M_{s}^{r,p}(\mathbb{R})$ with norm given by (2.3)
(below). More specifically we mention the local well-posedness result for the
modified Korteweg-de Vries (mKdV) equation in $\mathcal{F}L^{s,p}(\mathbb{R})$
for $s\geq\frac{1}{2p}$ with $2\leq p<4$ obtained in [14] and its improvement
for the same range of $s$ with $2\leq p<\infty$ in [15]. For other discussion
about these results we refer the readers to [25] where the authors considered
the IVP associated to the complex-valued mKdV equation in the modulation
spaces $M_{s}^{2,p}(\mathbb{R})$ and proved local well-posedness for
$s\geq\frac{1}{4}$ with $2\leq p<\infty$. Quite recently, Oh and Wang [26]
introduced a new function space $H\\!M^{\theta,p}$ whose norm is given by the
$\ell^{p}$-sum of the modulated $H^{\theta}$-norm of a given function and
agrees with the modulation space $M^{2,p}(\mathbb{R})$ on the real line and
Fourier-Lebesgue space $\mathcal{F}L^{p}(\mathbb{T})$ on the circle. The
authors in [26] proved that the cubic NLS is globally well-posed in
$M^{2,p}(\mathbb{R})$ for any $p<\infty$ and the normalized cubic NLS is
globally well-posed in $\mathcal{F}L^{p}(\mathbb{T})$ for any $p<\infty$. As
far we we know, there are no known results about the global well-posedness
issues for the IVP (1.1) for given data in the modulation spaces.

Our interest here is in addressing the well-posedness issues for the h-NLS
equation (1.1) with given data in the modulation spaces. We obtain the global
well-posedness result for the IVP (1.1) in the same spirit to that for the
complex mKdV equation [25].

The h-NLS is a particular case of the more general equation (honse equation)

$\begin{cases}\partial_{t}u+ia\partial^{2}_{x}u+b\partial^{3}_{x}u+ic_{1}|u|^{2}u+c_{2}|u|^{2}\partial_{x}u+du^{2}\partial_{x}\overline{u}=0,\quad
x,t\in\mathbb{R},\\\ u(x,0)=u_{0}(x),\end{cases}$ (1.2)

when $d=0$. The honse equation (1.2), which is a mixed model of complex
Korteweg-de Vries (KdV) and Schrödinger type was proposed by Hasegawa and
Kodama in [16] and [20] to describe the nonlinear propagation of pulses in
optical fibers. The IVP (1.2) has also been studied by several authors in
recent years. Taking $a,b,c_{1},c_{2}$ and $d$ as real constants, Laurey [21]
proved that the IVP (1.2) for given data in $H^{s}(\mathbb{R})$ is locally
well-posed when $s>\frac{3}{4}$ and globally well-posed when $s\geq 1$. Later,
using the techniques developed by Kenig, Ponce and Vega [18], Staffilani [28]
improved the result in [21] by showing that the IVP (1.2) is locally well-
posed in $H^{s}(\mathbb{R}),\,s\geq\frac{1}{4}$. Using the method of almost
conserved quantities and the I-method introduced by Colliander et. al. [9],
Carvajal [6] proved the sharp global well-posedness of IVP associated to (1.2)
in $H^{s}(\mathbb{R})$ for $s\geq\frac{1}{4}$. The IVP (1.2) when $a$ and $b$
are functions of $t\in[-T_{0},T_{0}]$ for some $T_{0}>0$ and $b(t)\neq 0$ for
all $t\in[-T_{0},T_{0}]$ has also been a matter of study (see for instance
[21], [5] and [7]). In [8] the authors proved the local well-posedness to the
IVP (1.2) in the modulation spaces $M^{s,p}$ for $s\geq\frac{1}{4}$ and $p\geq
2$.

The IVP (1.2) posed on the circle $\mathbb{T}$ is also studied in the
literature, see for instance [29] and references therein.

As far as we know, no work is available in the literature that deals with the
global well-posedness issues of the IVP (1.2) for given data in the modulation
spaces. Motivated by the recent works in [25] and [8], our interest in this
work is to address this issue. In fact, we prove the global well-posedness
result for the IVP (1.2) for given data in the modulation space
$M_{s}^{2,p}(\mathbb{R})$, whenever $s\geq\frac{1}{4}$. This is the content
the following theorem which is the main result of this work.

######  Theorem 1.1.

For given $s\geq\frac{1}{4}$ and $2\leq p<\infty$, the IVP (1.1) is globally
well-posed in the modulation space $M_{s}^{2,p}(\mathbb{R})$.

We present the organization of this work. In Section 2 we introduce the
function spaces, their properties and record some preliminary results. Section
3 is devoted to properties of traces, multiplication of operators and we
derive the key conservation of the perturbation determinant $\alpha(k,u)$ (see
definition (3.42)) that is fundamental to prove the main result of this work.
In Section 4 we provide the proofs of an apriori estimate in modultion spaces
and the main result of this paper. We finish this section recording some
principal notations that will be used throughout this work.

Notations: We will use standard notations of the PDEs throughout this work. We
use $\widehat{f}$ to denote Fourier transform and is defined by
$\widehat{f}(\xi)=(2\pi)^{-1/2}\int e^{-ix\xi}f(x)\,dx.$ We write $A\lesssim
B$ if there exists a constant $c>0$ such that $A\leq cB$, we also write $A\sim
B$ if $A\lesssim B$ and $B\lesssim A$… etc.

##  2\. Function spaces and preliminary results

As described in the previous section, the best global well-posedness result
for the IVP (1.1) for given data in the Sobolev space $H^{s}(\mathbb{R})$,
$s\geq 1/4$ was obtained in [6], using the Fourier transform norm space
$Z^{s,b}$ defined for $s,b\in\mathbb{R}$, $Z^{s,b}$ is the Fourier transform
restriction norm space introduced by Bourgain [2] with norm

$\|u\|_{Z^{s,b}}:=\|\langle\xi\rangle^{s}\langle\tau-\phi(\xi)\rangle^{b}\widehat{u}(\xi,\tau)\|_{L^{2}_{\xi}L^{2}_{\tau}},$
(2.1)

where $\langle x\rangle:=1+|x|$ and $\phi(\xi)=\xi^{3}-\alpha\xi^{2}$ is the
phase function associated to the h-NLS equation (1.1). We note that, for
$b>\frac{1}{2}$ one has $Z^{s,b}\subset C(\mathbb{R};H^{s}(\mathbb{R}))$ and
these spaces play a very important role in obtaining the well-posedness
results for the IVP associated to the dispersive equations with low regularity
Sobolev data.

Now, we move on to introduce modulation spaces on which we are interested to
concentrate our work. For given $s\in\mathbb{R}$, $1\leq r,p\leq\infty$,
modulation spaces $M_{s}^{r,p}(\mathbb{R})$ are defined by [12, 13]

$M_{s}^{r,p}(\mathbb{R}):=\\{f\in\mathcal{S}^{\prime}(\mathbb{R}):\|f\|_{M_{s}^{r,p}}<\infty\\},$
(2.2)

where

$\|f\|_{M_{s}^{r,p}}:=\|\langle
n\rangle^{s}\|\psi_{n}(D)f\|_{L_{x}^{r}(\mathbb{R})}\|_{\ell_{n}^{p}(\mathbb{Z})},$
(2.3)

with $\psi\in\mathcal{S}(\mathbb{R})$ such that

${\mathrm{supp}}\,\psi\subset[-1,1],\qquad\sum_{k\in\mathbb{Z}}\psi(\xi-k)=1,$

and $\psi_{n}(D)$ is the Fourier multiplier operator with symbol

$\psi_{n}(\xi):=\psi(\xi-n).$

For given $n\geq 1$, let $P_{N}$ be the Littlewood-Paley projector on the
frequencies $\\{|\xi|\sim N\\}$.

For $n\in\mathbb{Z}$ we define

$\widehat{\Pi_{n}f}(\xi):=\psi_{n}(\xi)\widehat{f}(\xi).$ (2.4)

For any $1\leq q\leq p\leq\infty$, from Bernstein’s inequality we have the
followings

$\begin{split}&\|P_{N}f\|_{L_{x}^{p}}\lesssim
N^{\frac{1}{q}-\frac{1}{p}}\|f\|_{L_{x}^{q}},\\\
&\|\Pi_{n}f\|_{L_{x}^{p}}\lesssim\|f\|_{L_{x}^{q}}.\end{split}$ (2.5)

Now, we introduce the Bourgain’s type space $X_{p}^{s,b}$ adapted to the
modulation space $M_{s}^{2,p}(\mathbb{R})$ with norm given by

$\|f\|_{X_{p}^{s,b}}:=\Big{(}\sum_{n\in\mathbb{Z}}\langle
n\rangle^{sp}\|\tau-\xi^{3}\rangle^{b}\widehat{f}(\xi,\tau)\|^{p}_{L^{2}_{r,\xi}(\mathbb{R}\times[n,n+1])}\Big{)}^{\frac{1}{p}}\sim\|\|\Pi_{n}f\|_{X^{s,b}}\|_{\ell^{p}_{n}}.$
(2.6)

For $p=2$, the space $X_{p}^{s,b}$ simply reduces to the usual Bourgain’s
space $X^{s,b}$. Note that, for $b>\frac{1}{2}$, one has the following
inclusion

$X_{p}^{s,b}\subset C(\mathbb{R}:M_{s}^{2,p}(\mathbb{R})).$ (2.7)

Also the following estimates hold

$\|x_{n}\|_{\ell^{p}_{n}}\leq\|x_{n}\|_{\ell^{q}_{n}},\qquad p\geq q\geq 1,$
(2.8)

$\|u\|_{X_{p}^{s,b}}\leq\|u\|_{X_{q}^{s,b}},\qquad p\geq q\geq 1,$ (2.9)

$\|P_{N}u\|_{X_{q}^{s,b}}\lesssim
N^{\frac{1}{q}-\frac{1}{p}}\|P_{N}u\|_{X_{p}^{s,b}},\qquad p\geq q\geq 1.$
(2.10)

For a given time interval $I$, the local-in-time version $X^{s,b}_{p}(I)$ of
$X^{s,b}_{p}$ are defined with the norm

$\|f\|_{X^{s,b}_{p}(I)}:=\inf\\{\|g\|_{X^{s,b}_{p}}:g|_{I}=f\\}.$

In what follows we record some preliminary results. We start with the
estimates that the unitary group satisfies in the $X^{s,b}_{p}$ spaces from
[25].

######  Lemma 2.1.

Let $s,b\in\mathbb{R}$ and $1\leq p<\infty$. Then for any $0<T\leq 1$ the
following estimate holds

$\Big{\|}e^{-t\partial_{x}^{3}}f\Big{\|}_{X^{s,b}_{p}([0,T])}\lesssim\|f\|_{M^{2,p}_{s}}.$
(2.11)

######  Lemma 2.2.

Let $s\in\mathbb{R}$, $-\frac{1}{2}<b^{\prime}\leq 0\leq b\leq 1+b^{\prime}$
and $1\leq p<\infty$. Then for any $0<T\leq 1$ the following estimate holds

$\left\|\int_{0}^{t}e^{-(t-t^{\prime})\partial_{x}^{3}}F(t^{\prime})dt^{\prime}\right\|_{X^{s,b}_{p}([0,T])}\lesssim
T^{1+b^{\prime}-b}\|F\|_{X^{s,b^{\prime}}_{p}([0,T])}.$ (2.12)

##  3\. Operators and Traces

Let $f\in S(\mathbb{R})$, we define the linear operator $T$ com kernel
$K_{T}\in L^{2}(\mathbb{R}^{2})$,

$Tf(x):=\int_{\mathbb{R}}K_{T}(x,y)f(y)dy.$ (3.1)

Observe that $T:L^{2}\to L^{2}$, is a bounded linear operator, in fact using
Minkowsky and Cauchy-Schwartz inequalities holds $\|T\|_{L^{2}\to
L^{2}}\leq\|K_{T}\|_{L^{2}(\mathbb{R}^{2})}$. We define the trace of the
operator T as:

$\text{tr}(T)=\int_{\mathbb{R}}K_{T}(x,x)dx.$

Using Fubini’s Theorem in $\langle
Tf,g\rangle=\int_{\mathbb{R}}Tf(x)\overline{g(x)}dx$, $f,g\in S(\mathbb{R})$,
we obtain

$K_{T^{*}}(x,y)=\overline{K_{T}(y,x)}.$ (3.2)

Observe that if $T_{j}$ has kernel $K_{j}$, $j=1,2$, then

$T_{1}T_{2}f(x)=\int_{\mathbb{R}^{2}}K_{1}(x,y)K_{2}(y,z)f(z)dydz=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}K_{1}(x,y)K_{2}(y,z)dy\right)f(z)dz,$

thus $T_{1}T_{2}$ has kernel and trace

$K(x,z)=\int_{\mathbb{R}}K_{1}(x,y)K_{2}(y,z)dy,\quad\text{tr}(T_{1}T_{2})=\int_{\mathbb{R}^{2}}K_{1}(x,y)K_{2}(y,x)dydx.$

Using Fubini’s Theorem, we have

$\text{tr}(T_{1}T_{2})=\text{tr}(T_{2}T_{1}).$ (3.3)

We also set

$\displaystyle\|T\|^{2}=\text{tr}(TT^{*})=\int_{\mathbb{R}^{2}}|K_{T}(x,y)|^{2}dxdy.$
(3.4)

The operator (3.1) is a Hilbert–Schmidt operator and this norm (3.4) is the
Hilbert–Schmidt norm. In general if $T_{j}$ has kernel $K_{j}$,
$j=1,2,\dots,n$, then

$T_{1}T_{2}\dots
T_{n}f(x)=\int_{\mathbb{R}^{n}}K_{1}(x,x_{1})K_{2}(x_{1},x_{2})\dots
K_{n}(x_{n-1},x_{n})f(x_{n})dx_{1}dx_{2}\dots dx_{n},$ (3.5)

has kernel

$K(x,x_{n})=\int_{\mathbb{R}^{n-1}}K_{1}(x,x_{1})K_{2}(x_{1},x_{2})\dots
K_{n}(x_{n-1},x_{n})dx_{1}\dots dx_{n-1},$ (3.6)

and trace

$\text{tr}(T_{1}T_{2}\dots
T_{n})=\int_{\mathbb{R}^{n}}K_{1}(x,x_{1})K_{2}(x_{1},x_{2})\dots
K_{n}(x_{n-1},x)dxdx_{1}\dots dx_{n-1}.$ (3.7)

We also have

$|\text{tr}(T_{1}T_{2}\dots T_{n})|\leq\prod_{j=1}^{n}\|T_{j}\|.$ (3.8)

Let $m(\xi)$ a real function, we defined the multiplier operator $M$
associated to $m$, as

$\widehat{Mf}(\xi)=m(\xi)\widehat{f}(\xi).$

Let $u\in S(\mathbb{R})$ and $M$ a multiplier operator associated to $m$, we
define other operators $Mu$ and $uM$ as follow $(Mu)f(x):=M(uf)(x)$, and
$(uM)f(x)=u(x)(Mf)(x)$.

If $M_{j}$ is a multiplier operator with multiplicator $m_{j}$, $j=1,2$, since
that $M_{1}M_{2}=M_{2}M_{1}$ has multiplicator $m_{1}m_{2}=m_{2}m_{1}$, then

$M_{1}M_{2}u=M_{2}M_{1}u\quad\textrm{and}\quad uM_{1}M_{2}=uM_{2}M_{1}.$ (3.9)

We have the following examples

Operador Kernel, $K(x,\xi)=$ Trace M $m^{\vee}(x-\xi)$ Mu
$m^{\vee}(x-\xi)u(\xi)$ $\int m^{\vee}(0)u(x)dx=\left(\int
m(x)dx\right)\left(\int u(x)dx\right)$ uM $u(x)m^{\vee}(x-\xi)$ $\int
u(x)m^{\vee}(0)dx=\left(\int m(x)dx\right)\left(\int u(x)dx\right)$ uMv
$u(x)m^{\vee}(x-\xi)v(\xi)$ $\int u(x)m^{\vee}(0)v(x)dx=\left(\int
m(x)dx\right)\left(\int u(x)v(x)dx\right)$

Observe that using (3.5) and (3.6) we can define other operators as
$M_{1}uM_{2}=M_{1}(uM_{2})=(M_{1}u)M_{2}$ with kernel:

$K(x,z):=K(m_{1},u,m_{2})=\int_{\mathbb{R}}K_{1}(x,y)K_{2}(y,z)dy=\int_{\mathbb{R}}m_{1}^{\vee}(x-y)u(y)m_{2}^{\vee}(y-z)dy,$
(3.10)

and using (3.7), with trace:

$\quad\text{tr}(M_{1}uM_{2})=\int_{\mathbb{R}^{2}}K_{1}(x,y)K_{2}(y,x)dydx=\int_{\mathbb{R}^{2}}m_{1}^{\vee}(x-y)u(y)m_{2}^{\vee}(y-x)dydx.$
(3.11)

It is not difficult to see that if $m_{1},m_{2}\in L^{2}(\mathbb{R})$, then
$K(x,z)\in L^{2}(\mathbb{R}^{2})$.

######  Lemma 3.1.

For $k\neq 0$ and $u\in S(\mathbb{R})$,

$\|(k-\partial)^{-1/2}u(k+\partial)^{-1/2}\|^{2}\lesssim\int_{\mathbb{R}}\dfrac{|\widehat{u}(\xi)|^{2}}{|k|+|\xi|}d\xi\leq
C_{k}\|u\|^{2}_{H^{-1/2}},$ (3.12)

where $C_{k}=(\min\\{1,|k|\\})^{-1/2}$.

In order to prove (3.12), we need the following

######  Lemma 3.2.

If $a,b>0$ and $a+b>1$, we have

$\int_{\mathbb{R}}\dfrac{dx}{\langle x-\alpha\rangle^{a}\langle
x-\beta\rangle^{b}}\lesssim\dfrac{1}{\langle\alpha-\beta\rangle^{c}},\quad
c=\min\\{a,b,a+b-1\\}.$ (3.13)

###### Proof of Lemma 3.1.

From (3.10), using Plancherel’s identity and properties of the Fourier
transform we have

$\begin{split}\|K(x,z)\|_{L^{2}_{x}L^{2}_{z}}&=\|m_{1}^{\vee}*[u(\cdot)(m_{2}(\cdot)e^{-iz(\cdot)})^{\vee}](x)\|_{L^{2}_{x}L^{2}_{z}}\\\
&=\|m_{1}(\xi)\|[u(\cdot)(m_{2}(\cdot)e^{-iz(\cdot)})^{\vee}]^{\wedge}(\xi)\|_{L^{2}_{z}}\|_{L^{2}_{\xi}}\\\
&=\|m_{1}(\xi)\|[\widehat{u}*(m_{2}(\cdot)e^{-iz(\cdot)})(\xi)\|_{L^{2}_{z}}\|_{L^{2}_{\xi}}\\\
&=\|m_{1}(\xi)\|[(u(\cdot)e^{-i\xi(\cdot)})^{\vee}m_{2}(\cdot)]^{\wedge}(z)\|_{L^{2}_{z}}\|_{L^{2}_{\xi}}\\\
&=\|m_{1}(\xi)\|u^{\vee}(\xi-\eta)m_{2}(\eta)\|_{L^{2}_{\eta}}\|_{L^{2}_{x}}.\end{split}$
(3.14)

Thus making a change of variables $\xi=k\xi$ and $\eta=k\eta$ followed by
another change of variables $\xi-\eta=y$,

$\begin{split}\|K(x,z)\|_{L^{2}_{x}L^{2}_{z}}^{2}&=\int_{\mathbb{R}^{2}}\dfrac{|\widehat{u}(\xi-\eta)|^{2}}{\sqrt{k^{2}+\xi^{2}}\sqrt{k^{2}+\eta^{2}}}d\xi
d\eta\\\
&\sim\int_{\mathbb{R}^{2}}\dfrac{|\widehat{u}(k(\xi-\eta))|^{2}}{\langle\xi\rangle\langle\eta\rangle}d\xi
d\eta\\\ &=\int_{\mathbb{R}^{2}}\dfrac{|\widehat{u}(k(y))|^{2}}{\langle
y+\eta\rangle\langle\eta\rangle}dyd\eta\\\
&\lesssim\int_{\mathbb{R}}\dfrac{|\widehat{u}(k(y))|^{2}}{\langle
y\rangle}dy,\end{split}$ (3.15)

where in the last inequality was used Fubinni theorem and Lemma 3.2 with
$a=b=1$. ∎

If $T$ is a linear operator associated with the kernel $K$, we define the
operator $\overline{T}$ as

$\overline{T}f(x):=\int_{\mathbb{R}}\overline{K(x,y)}f(y)dy,$ (3.16)

where $\overline{K}$ denotes the complex conjugate of $K$. Thus if
$K(x,y)\in\mathbb{R}$, then $\overline{T}=T$. Observe that
$\overline{m^{\vee}(\eta)}=\widehat{\overline{m}}(\eta)=\overline{m}^{\vee}(-\eta)$,
thus

$\overline{Mu}=M^{-}\overline{u},$ (3.17)

where $M^{-}$ is the multiplier operator, associated to
$m^{-}(\xi)=\overline{m}(-\xi)$. In this way we get, for any $k$ real number

$\overline{(k-\partial)^{-1}u}=(k-\partial)^{-1}\overline{u},\quad\textrm{and}\quad\overline{(k+\partial)^{-1}u}=(k+\partial)^{-1}\overline{u},$
(3.18)

since that if $m(\xi)=(k\pm i\xi)^{-1}$, then $\overline{m}(-\xi)=m(\xi)$.

Let $M$ a multiplier operator associated to $m$, by (3.2)

$K_{M^{*}}(x,y)=\overline{K_{M}(y,x)}=\overline{m^{\vee}}(y-x)=\overline{\widehat{m}}(x-y)=\overline{m}^{\vee}(x-y)=K_{\overline{M}}(x,y),$

i.e.

$M^{*}=\overline{M}.$ (3.19)

Similarly using (3.2) and the above example, we have

$(Mu)^{*}=\overline{u}\overline{M}.$ (3.20)

The equalities (3.19) and (3.20) imply that

$(M_{1}uM_{2})^{*}=((M_{1}u)M_{2})^{*}=M_{2}^{*}(M_{1}u)^{*}=\overline{M_{2}}\overline{u}\overline{M_{1}}.$
(3.21)

###  3.1. Derivatives of the Multiplication Operator

Let $u\in S(\mathbb{R})$ we define $Pu:S(\mathbb{R})\to S(\mathbb{R})$ the
operator of multiplication associated to $u$ as:

$(Pu)f=uf,\quad f\in S(\mathbb{R}).$

We will use the notation $Pu:=u$. Let $n,l\in\mathbb{Z}^{+}$, we also define
the operators of multiplication $u\partial^{n}:S(\mathbb{R})\to S(\mathbb{R})$
by $(u\partial^{n})(f)=uf^{(n)}$ and $(\partial^{l}u):S(\mathbb{R})\to
S(\mathbb{R})$ by $(\partial^{l}u)(f):=(uf)^{(l)}$. Considering $n=l=1$, we
obtain $(\partial u)(f)=(uf)^{\prime}=u^{\prime}f+uf^{\prime}$ for all $f\in
S(\mathbb{R})$ or equivalently $\partial u=u^{\prime}+u\partial$ and thus
$u^{\prime}=\partial u-u\partial=[\partial,u]$. We can have combinations
between both operators such as

$\partial(u\partial)(f)=((u\partial)f)^{\prime}=(uf^{\prime})^{\prime}=u^{\prime}f^{\prime}+uf^{\prime\prime}\quad\iff\quad\partial(u\partial)=u^{\prime}\partial+u\partial^{2}.$
(3.22)

Similarly we can define other operators of multiplication, such as

$(\partial^{2}u)(f):=(uf)^{\prime\prime}=u^{\prime\prime}f+2u^{\prime}f^{\prime}+uf^{\prime\prime}$
(3.23)

which gives

$\partial^{2}u=u^{\prime\prime}+2u^{\prime}\partial+u\partial^{2}\quad\iff\quad
u^{\prime\prime}=\partial^{2}u-2u^{\prime}\partial-u\partial^{2},$ (3.24)

We define $(k\pm\partial)u:=ku\pm\partial u$, $(k\pm\partial)^{2}u:=k^{2}u\pm
2k\partial u+\partial^{2}u$ and by induction
$(k\pm\partial)^{n}u:=(k\pm\partial)^{n-1}(k\pm\partial)u$, $n=3,4,\dots$.

Using (3.22), adding and subtracting terms in the above equality, we have the
following identity

$\begin{split}u^{\prime\prime}=&u(\partial^{2}-2k\partial-k^{2})+(\partial^{2}+2k\partial-k^{2})u+2(k-\partial)u(k+\partial)\\\
=&u(k-\partial)^{2}+(k+\partial)^{2}u-4k^{2}u+2(k-\partial)u(k+\partial).\end{split}$
(3.25)

valid for all $k\in\mathbb{R}$. Also we have the operator of multiplication

$(\partial^{3}u)(f):=(uf)^{\prime\prime\prime}=u^{\prime\prime\prime}f+3u^{\prime\prime}f^{\prime}+3u^{\prime}f^{\prime\prime}+uf^{\prime\prime\prime\prime}\quad$
(3.26)

therefore

$\partial^{3}u=u^{\prime\prime\prime}+3u^{\prime\prime}\partial+3u^{\prime}\partial^{2}+u\partial^{3}\,\,\iff\,\,u^{\prime\prime\prime}=\partial^{3}u-3u^{\prime\prime}\partial-3u^{\prime}\partial^{2}-u\partial^{3}$
(3.27)

we also have the following identity

$u^{\prime\prime\prime}=\partial^{3}u-3\partial^{2}u\partial+3\partial
u\partial^{2}-u\partial^{3}=\sum_{j=0}^{3}(-1)^{j}\binom{3}{j}\partial^{3-j}u\partial^{j}.$

Using $u^{\prime}=\partial u-u\partial$, adding and subtracting terms in the
above equality, we obtain

$u^{\prime\prime\prime}=u(k-\partial)^{3}+(k+\partial)^{3}u-8k^{3}u+(k-\partial)(3u^{\prime}+6ku)(k+\partial)$
(3.28)

for any $k\in\mathbb{R}$. On the other hand
$\partial(|u|^{2}u)(f)=(|u|^{2}uf)^{\prime}=2|u|^{2}u^{\prime}f+u^{2}\overline{u}^{\prime}f+|u|^{2}uf^{\prime}$
thus

$\partial(|u|^{2}u)=2|u|^{2}u^{\prime}+u^{2}\overline{u}^{\prime}+|u|^{2}u\partial,$
(3.29)

adding and subtracting terms in the above equality, we obtain

$2|u|^{2}u^{\prime}=-(|u|^{2}u)(k+\partial)-(k-\partial)(|u|^{2}u)-u^{2}(\overline{u}^{\prime}-2k\overline{u}),$
(3.30)

###  3.2. Trace of Products of Multiplier Operators

Next, we will state some properties of the Products of Multiplier Operators.

######  Proposition 3.3.

Let $M_{j}$ the multipliers operator associated to $m_{j}$, $u_{j}\in
S(\mathbb{R})$, $j=1,\dots,n+1$ and $\sigma$ a shift permutation of the n-upla
$(1,2,\dots,n)$, we have

$\text{tr}\left(\Pi_{j=1}^{n}M_{j}u_{j}\right)=\text{tr}\left(\Pi_{j=1}^{n}M_{\sigma(j)}u_{\sigma(j)}\right)=\text{tr}\left(\left(\Pi_{j=1}^{n-1}u_{\sigma(j)}M_{\sigma(j+1)}\right)u_{\sigma(n)}M_{\sigma(1)}\right)$
(3.31)
$\text{tr}\left(\left(\Pi_{j=1}^{n}M_{j}u_{j}\right)M_{n+1}\right)=\text{tr}\left(M_{1}M_{n+1}u_{1}\left(\Pi_{j=2}^{n}M_{j}u_{j}\right)\right)$
(3.32)
$\text{tr}\left(\Pi_{j=1}^{n}u_{j}M_{j}\right)=\text{tr}\left(\left(\Pi_{j=1}^{n-1}M_{j}u_{j+1}\right)M_{n}u_{1}\right)$
(3.33)
$\text{tr}\left(\left(\Pi_{j=1}^{n}u_{j}M_{j}\right)u_{n+1}\right)=\text{tr}\left(\left(\Pi_{j=1}^{n-1}M_{j}u_{j+1}\right)M_{n}(u_{1}u_{n+1})\right)$
(3.34)

###### Proof.

Using (3.7) and example before, we have

$\text{tr}\left(\Pi_{j=1}^{n}M_{j}u_{j}\right)=\int_{\mathbb{R}^{n}}m_{1}^{\vee}(x-\xi_{1})u_{1}(\xi_{1})m_{2}^{\vee}(\xi_{1}-\xi_{2})u_{2}(\xi_{2})\cdots
m_{n}^{\vee}(\xi_{n-1}-x)u_{n}(x)dxd\xi_{1}\cdots d\xi_{n-1}$ (3.35)

applying Fubinni we get (3.31). Similarly

$\begin{split}&\text{tr}\left(\left(\Pi_{j=1}^{n}M_{j}u_{j}\right)M_{n+1}\right)=\\\
&\int_{\mathbb{R}^{n+1}}m_{1}^{\vee}(x-\xi_{1})u_{1}(\xi_{1})m_{2}^{\vee}(\xi_{1}-\xi_{2})u_{2}(\xi_{2})\cdots
m_{n}^{\vee}(\xi_{n-1}-\xi_{n})u_{n}(\xi_{n})m_{n+1}^{\vee}(\xi_{n}-x)\\\
=&\int_{\mathbb{R}^{n}}u_{1}(\xi_{1})m_{2}^{\vee}(\xi_{1}-\xi_{2})u_{2}(\xi_{2})\cdots
m_{n}^{\vee}(\xi_{n-1}-\xi_{n})u_{n}(\xi_{n})\int_{\mathbb{R}}m_{1}^{\vee}(x-\xi_{1})m_{n+1}^{\vee}(\xi_{n}-x)dx\\\
=&\int_{\mathbb{R}^{n}}u_{1}(\xi_{1})m_{2}^{\vee}(\xi_{1}-\xi_{2})u_{2}(\xi_{2})\cdots
m_{n}^{\vee}(\xi_{n-1}-\xi_{n})u_{n}(\xi_{n})(m_{1}m_{n+1})^{\vee}(\xi_{n}-\xi_{1})dx\end{split}$

and (3.32) follows. The proof of inequalities (3.33) and (3.34) are similar. ∎

Note that on the right side of (3.32), (3.33) and (3.34) we can still use the
(3.31) property. By property (3.33) we have

$\begin{split}\text{tr}\left(M_{1}u_{1}M_{2}u_{2}\right)=&\text{tr}\left(u_{2}M_{1}u_{1}M_{2}\right)=\int_{\mathbb{R}^{2}}m_{1}^{\vee}(x-\xi_{1})u_{1}(\xi_{1})m_{2}^{\vee}(\xi_{1}-x)u_{2}(x)d\xi_{1}dx\\\
=&\int_{\mathbb{R}}u_{1}(\xi_{1})\left\\{\left(\widehat{m_{1}}m_{2}^{\vee}\right)*u_{2}\right\\}(\xi_{1})d\xi_{1}.\end{split}$
(3.36)

######  Lemma 3.4.

For $k\neq 0$ and $u\in S(\mathbb{R})$,

$\text{Re\,tr}\left\\{(k-\partial)^{-1}u(k+\partial)^{-1}\overline{u}\right\\}=2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}(\xi)|^{2}}{4k^{2}+\xi^{2}}d\xi\sim_{k}\|u\|^{2}_{H^{-1}}.$
(3.37)

###### Proof.

Using (3.3), (3.36) with $u_{1}=\overline{u}$, $u_{2}=u$ and
$m_{1}(\xi)=\dfrac{1}{k+i\xi}$,
$m_{2}(\xi)=\dfrac{1}{k-i\xi}=\overline{m_{1}}$ and Plancherel identity, we
have $m_{2}^{\vee}=\overline{m_{1}}^{\vee}=\overline{\widehat{m_{1}}}$

$\begin{split}\text{Re\,tr}\left\\{(k-\partial)^{-1}u(k+\partial)^{-1}\overline{u}\right\\}=&\text{Re\,tr}\left\\{(k+\partial)^{-1}\overline{u}(k-\partial)^{-1}u\right\\}\\\
=&\text{Re\,}\int_{\mathbb{R}}\overline{\widehat{u}}(\xi)\widehat{u}(\xi)\widehat{\left(\widehat{m_{1}}m_{2}^{\vee}\right)}(\xi)d\xi\\\
=&\text{Re\,}\int_{\mathbb{R}}|\widehat{u}(\xi)|^{2}\widehat{|\widehat{m_{1}}|^{2}}(\xi)d\xi\\\
=&c\text{Re\,}\int_{\mathbb{R}}|\widehat{u}(\xi)|^{2}\dfrac{1}{2k-i\xi}d\xi.\end{split}$
(3.38)

∎

From the definition of norm of operator $M_{1}uM_{2}$, (see definition (3.4)),
(3.21) and property (3.32) holds

$\|M_{1}uM_{2}\|^{2}=\text{tr}(M_{1}uM_{2}\overline{M_{2}}\overline{u}\overline{M_{1}})=\text{tr}(\overline{M_{1}}M_{1}uM_{2}\overline{M_{2}}\overline{u}),$
(3.39)

similarly

$\|M_{2}\overline{u}M_{1}\|^{2}=\text{tr}(M_{2}\overline{u}M_{1}\overline{M_{1}}u\overline{M_{2}})=\text{tr}(\overline{M_{2}}M_{2}\overline{u}M_{1}\overline{M_{1}}u)$
(3.40)

the above equalities, (3.3) and (3.9) imply that

$\|M_{1}uM_{2}\|=\|M_{2}\overline{u}M_{1}\|.$ (3.41)

By analogy with has gone in the case of the NLS and mKdV, we define

$\alpha(u(t),k):=\text{Re}\sum_{l=1}^{\infty}\dfrac{(-1)^{l-1}}{l}\text{tr}\left\\{[(k-\partial)^{-1/2}u(k+\partial)^{-1}\overline{u}(k-\partial)^{-1/2}]^{l}\right\\}$
(3.42)

Let $R^{\pm}=(k\pm\partial)^{-1}$. Using (3.32) we deduce

$\text{tr}\left\\{[(k-\partial)^{-1/2}u(k+\partial)^{-1}\overline{u}(k-\partial)^{-1/2}]^{l}\right\\}=\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\right\\}.$
(3.43)

We consider $u$ a solution of h-NSE. For technical reasons, we will assume
that $c_{1}=-2a$ and $c_{2}=-6b$ i.e. $u$ satisfies the equation

$\partial_{t}u=-L(u)+F(u),\quad x,t\in\mathbb{R},$ (3.44)

where $L(u)=ia\partial^{2}_{x}u+b\partial^{3}_{x}u$ and
$F(u)=2ia|u|^{2}u+6b|u|^{2}\partial_{x}u$.

######  Theorem 3.5.

[Conservation of $\alpha$ for h-NSE] Let $u(x,t)$ denote a Schwartz-space
solution to HONSE. Then for any $k>0$ holds,

$\dfrac{d}{dt}\alpha(u(t),k)=0.$ (3.45)

###### Proof.

Using (3.43) and differentiating

$\begin{split}\dfrac{d}{dt}\alpha(u(t),k)&=\text{Re}\sum_{l=1}^{\infty}(-1)^{l-1}\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(\dfrac{d}{dt}u)R^{+}\overline{u}+R^{-}uR^{+}\dfrac{d}{dt}\overline{u}\right)\right\\}\\\
&=\text{Re}\sum_{l=1}^{\infty}(-1)^{l-1}\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(-Lu)R^{+}\overline{u}+R^{-}uR^{+}(-\overline{Lu})\right)\right\\}\\\
&+\text{Re}\sum_{l=1}^{\infty}(-1)^{l-1}\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(F(u))R^{+}\overline{u}+R^{-}uR^{+}(\overline{F(u)}\right)\right\\}\\\
&=\text{Re}\,\text{tr}\left\\{\left(R^{-}(-Lu)R^{+}\overline{u}+R^{-}uR^{+}(-\overline{Lu})\right)\right\\}\qquad(l=1)\\\
&+\text{Re}\sum_{l=1}^{\infty}(-1)^{l}\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\left(R^{-}(-Lu)R^{+}\overline{u}+R^{-}uR^{+}(-\overline{Lu})\right)\right\\}\quad(l:=l+1)\\\
&+\text{Re}\sum_{l=1}^{\infty}(-1)^{l-1}\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(F(u))R^{+}\overline{u}+R^{-}uR^{+}(\overline{F(u)}\right)\right\\}.\end{split}$
(3.46)

Thus (3.45) follows if

$\text{Re}\,\text{tr}\left\\{\left(R^{-}(Lu)R^{+}\overline{u}+R^{-}uR^{+}(\overline{Lu})\right)\right\\}=0,$
(3.47)

and

$\begin{split}\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\left(R^{-}(Lu)R^{+}\overline{u}+R^{-}uR^{+}(\overline{Lu})\right)\right\\}=&\\\
-\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(F(u))R^{+}\overline{u}+R^{-}uR^{+}(\overline{F(u)}\right)\right\\}.\end{split}$
(3.48)

Using (3.36) with $M_{1}=R^{-}$ and $M_{2}=R^{+}$, integrating three times we
get

$\text{tr}\left\\{\left(R^{-}(u_{xxx})R^{+}\overline{u}\right)\right\\}=\int_{\mathbb{R}}u_{xxx}(\xi_{1})\left(\widehat{m_{1}}m_{2}^{\vee}\right)*\overline{u}(\xi_{1})d\xi_{1}=-\int_{\mathbb{R}}u(\xi_{1})\left(\widehat{m_{1}}m_{2}^{\vee}\right)*\overline{u_{xxx}}(\xi_{1})d\xi_{1},$
(3.49)

similarly

$\text{tr}\left\\{\left(R^{-}(iu_{xx})R^{+}\overline{u}\right)\right\\}=\int_{\mathbb{R}}iu_{xx}(\xi_{1})\left(\widehat{m_{1}}m_{2}^{\vee}\right)*\overline{u}(\xi_{1})d\xi_{1}=-\int_{\mathbb{R}}u(\xi_{1})\left(\widehat{m_{1}}m_{2}^{\vee}\right)*\overline{iu_{xx}}(\xi_{1})d\xi_{1},$
(3.50)

using (3.49) and (3.50) we obtain (3.47). In order to prove (3.48),
considering k:=-k and taking complex conjugate in (3.25), (3.28) we obtain

$\overline{u_{xx}}=(k-\partial)^{2}\overline{u}+\overline{u}(k+\partial)^{2}-4k^{2}\overline{u}+2(k+\partial)\overline{u}(k-\partial).$
(3.51)

and

$\begin{split}\overline{u_{xxx}}=&\overline{u}(-k-\partial)^{3}+(-k+\partial)^{3}\overline{u}+8k^{3}\overline{u}+(-k-\partial)(3\overline{u_{x}}-6k\overline{u})(-k+\partial)\\\
=&-\overline{u}(k+\partial)^{3}-(k-\partial)^{3}\overline{u}+8k^{3}\overline{u}+(k+\partial)(3\overline{u_{x}}-6k\overline{u})(k-\partial).\end{split}$
(3.52)

Using, (3.28), (3.52), (3.25) and (3.51) it is not dificult to see that

$\begin{split}R^{-}(u_{xxx})R^{+}\overline{u}+R^{-}uR^{+}(\overline{u_{xxx}})=&R^{-}(k+\partial)^{3}uR^{+}\overline{u}+3u_{x}\overline{u}+6k|u|^{2}\\\
&-R^{-}uR^{+}\overline{u}(k+\partial)^{3}+R^{-}(3u\overline{u_{x}}-6k|u|^{2})(k-\partial),\end{split}$
(3.53)

and

$\begin{split}R^{-}(iu_{xx})R^{+}\overline{u}+R^{-}uR^{+}(\overline{iu_{xx}})=i\left(R^{-}\mathcal{A}uR^{+}\overline{u}+2|u|^{2}-R^{-}uR^{+}\overline{u}\mathcal{A}-2R^{-}(|u|^{2})(k-\partial)\right),\end{split}$
(3.54)

where $\mathcal{A}=(k+\partial)^{2}$.

From (3.53) and (3.54) we have that

$R^{-}L(u)R^{+}\overline{u}+R^{-}uR^{+}\overline{L(u)}=R^{-}\mathcal{B}uR^{+}\overline{u}-R^{-}uR^{+}\overline{u}\mathcal{B}+F_{1}(k)+R^{-}\overline{F_{1}(-k)}(k-\partial),$
(3.55)

where $F_{1}(k)=2ai|u|^{2}+3bu_{x}\overline{u}+6bk|u|^{2}$ and
$\mathcal{B}=ia\mathcal{A}+b(k+\partial)^{3}$, using Property (3.31) we have

$\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\left(R^{-}\mathcal{B}uR^{+}\overline{u}\right)\right\\}=\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}\mathcal{B}uR^{+}\overline{u}\right)\left(R^{-}uR^{+}\overline{u}\right)^{l}\right\\},$
(3.56)

and using Property (3.32) we get

$\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\left(R^{-}uR^{+}\overline{u}\mathcal{B}\right)\right\\}=\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}\mathcal{B}uR^{+}\overline{u}\right)\left(R^{-}uR^{+}\overline{u}\right)^{l}\right\\}$
(3.57)

consequently from (3.55), (3.56) and (3.57) it follows that

$\begin{split}\text{Re}\,&\text{tr}\\!\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l}\left(R^{-}L(u)R^{+}\overline{u}+R^{-}uR^{+}\overline{L(u)}\right)\right\\}\\!=\\!\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}R^{-}uR^{+}\overline{u}F_{1}(k)\right\\}\\\
&+\text{Re}\,\text{tr}\left\\{R^{-}uR^{+}\overline{u}\left(R^{-}uR^{+}\overline{u}\right)^{l-1}R^{-}\overline{F_{1}(-k)}(k-\partial)\right\\}\\\
=&\text{Re}\,\text{tr}\\!\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}R^{-}uR^{+}(\overline{u}F_{1}(k))\right\\}+\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}R^{-}u\left(\overline{F_{1}(-k)}\right)R^{+}\overline{u}\right\\}\end{split}$
(3.58)

in the last equality was used (3.31) (with the permutation $\sigma(n)=2$,
$\sigma(n-1)=1$ and $\sigma(j)=j+2,j=1,\dots n-2$, similar as in (3.56)). Now,
we will to consider the term with $|u|^{2}u_{x}$ of $F(u)$, considering k:=-k
and taking complex conjugate in (3.30), we obtain

$\begin{split}2|u|^{2}\overline{u_{x}}=&-(|u|^{2}\overline{u})(-k+\partial)-(-k-\partial)(|u|^{2}\overline{u})-\overline{u}^{2}(u_{x}+2ku)\\\
=&(k+\partial)|u|^{2}\overline{u}+(|u|^{2}\overline{u})(k-\partial)-\overline{u}^{2}(u_{x}+2ku),\end{split}$
(3.59)

By (3.30) and (3.59) we get

$\begin{split}R^{-}(2|u|^{2}u_{x})R^{+}\overline{u}&+R^{-}uR^{+}(2|u|^{2}\overline{u_{x}})=R^{-}(-|u|^{4})-|u|^{2}uR^{+}\overline{u}+R^{-}(-u^{2}\overline{u_{x}}+2k|u|^{2}u)R^{+}\overline{u}\\\
&+R^{-}(|u|^{4})+R^{-}uR^{+}(|u|^{2}\overline{u})(k-\partial)+R^{-}uR^{+}(-\overline{u}^{2}u_{x}-2k|u|^{2}\overline{u})\end{split}$
(3.60)

and using the properties (3.31) we have

$\begin{split}\text{Re}&\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(2|u|^{2}u_{x})R^{+}\overline{u}+R^{-}uR^{+}(2|u|^{2}\overline{u_{x}}\right)\right\\}=\\\
&\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(-u^{2}\overline{u_{x}}+2k|u|^{2}u)R^{+}\overline{u}+R^{-}uR^{+}(-\overline{u}^{2}u_{x}-2k|u|^{2}\overline{u}\right)\right\\}\end{split}$
(3.61)

where the second term in RHS(3.60) in the first line is canceled with the
second term in RHS(3.60) in the second line (using (3.31), similar as in
(3.56). Therefore

$\begin{split}&\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(F(u))R^{+}\overline{u}+R^{-}uR^{+}(\overline{F(u)}\right)\right\\}=\\\
&\text{Re}\,\text{tr}\left\\{\left(R^{-}uR^{+}\overline{u}\right)^{l-1}\left(R^{-}(-u\left(\overline{F_{1}(-k)}\right))R^{+}\overline{u}+R^{-}uR^{+}(-\overline{u}F_{1}(k)\right)\right\\},\end{split}$
(3.62)

where
$\overline{u}F_{1}(k)=2ia|u|^{2}\overline{u}+3b\overline{u}^{2}u_{x}+6kb|u|^{2}\overline{u}$.
This equality together with (3.58) prove equality (3.48) and therefore proves
the theorem

∎

######  Proposition 3.6.

Let $u$ solution of the IVP (1.2) and $v$ such that

$u(x,t):=v(x+d_{1}t,t)e^{i(d_{2}x+d_{3}t)}.$ (3.63)

where $d_{j}\in\mathbb{R}$, $j=1,2,3$. If $\dfrac{d\alpha(u(t),k)}{dt}=0$,
then $\dfrac{d\alpha(v(t),k)}{dt}=0$.

###### Proof.

By the definition of $\alpha$ and (3.43), holds

$\alpha(v(t),k):=\text{Re}\sum_{l=1}^{\infty}\dfrac{(-1)^{l-1}}{l}\text{tr}\left\\{\left(R^{-}ue^{-id_{2}x}R^{+}\overline{u}e^{id_{2}x}\right)^{l}\right\\}.$
(3.64)

and using Property (3.31) we have

$\alpha(u(t),k):=\text{Re}\sum_{l=1}^{\infty}\dfrac{(-1)^{l-1}}{l}\text{tr}\left\\{\left(\mathcal{R}^{-}u\mathcal{R}^{+}\overline{u}\right)^{l}\right\\}.$
(3.65)

where $\mathcal{R}^{\pm}=e^{\mp id_{2}x}R^{\pm}$. The rest of the proof is
equal to the proof of Theorem 3.5, replacing $R^{-}$ by $\mathcal{R}^{-}$ and
$R^{+}$ by $\mathcal{R}^{+}$. ∎

######  Remark 3.7.

We consider the following Gauge transform

$u(x,t):=v(x+d_{1}t,t)e^{i(d_{2}x+d_{3}t)}.$ (3.66)

Using this transformation the IVP (1.1) turns out to be

$\\!\\!\\!\\!\\!\\!\\!\begin{cases}\partial_{t}v+b\partial^{3}_{x}v+i(a+3bd_{2})\partial^{2}_{x}v+i(d_{3}-bd_{2}^{3}-ad_{2}^{2})v+(d_{1}-3bd_{2}^{2}-2ad_{2})\partial_{x}v\\\
+i(c_{1}+c_{2}d_{2})|v|^{2}v+c_{2}|v|^{2}v_{x}=0,\\\
v(x,0)=v_{0}(x):=u_{0}(x)e^{-id_{2}x}.\end{cases}$ (3.67)

If one chooses $d_{1}=-\frac{a^{2}}{3b}$, $d_{2}=\frac{-a}{3b}$,
$d_{3}=\frac{2a^{3}}{27b^{2}}$, the third, fourth, fifth terms in the first
equation in (3.67) vanish i.e.

$\\!\\!\\!\\!\\!\\!\\!\begin{cases}\partial_{t}v+b\partial^{3}_{x}v+ic_{3}|v|^{2}v+c_{2}|v|^{2}v_{x}=0,\\\
v(x,0)=v_{0}(x):=u_{0}(x)e^{iax/(3b)},\end{cases}$ (3.68)

where $c_{3}=c_{1}-\frac{ac_{2}}{3b}$. Also, we note that

$\|u_{0}\|_{H^{s}}\sim\|v_{0}\|_{H^{s}}\quad\mathrm{and}\quad\|u\|_{Z^{s,b}}\sim\|v\|_{X^{s,b}},$

where $X^{s,b}$ is the Fourier transform norm space with phase function
$\xi^{3}$, i.e., with norm

$\|u\|_{X^{s,b}}:=\|\langle\xi\rangle^{s}\langle\tau-\xi^{3}\rangle^{b}\widehat{u}(\xi,\tau)\|_{L^{2}_{\xi}L^{2}_{\tau}}.$

This observation allows one to consider the parameter $\alpha=0$.

##  4\. Apriori estimates

######  Proposition 4.1.

Let $2\leq p<\infty$ and $u\in S(\mathbb{R})$ a solution to (3.44). Then,
there exists $C=C(p)$ positive such that

$\|u(t)\|_{M^{2,p}(\mathbb{R})}\leq
C\left(1+\|u(0)\|_{M^{2,p}(\mathbb{R})}\right)^{p/2-1}\|u(0)\|_{M^{2,p}(\mathbb{R})},$
(4.1)

for any $t\in\mathbb{R}$.

###### Proof.

The proof is done in two steps. Initially we will to consider the initial data

$\|u_{0}\|_{M^{2,p}}<\epsilon\ll 1.$ (4.2)

Let $u$ solution of equation (3.44) and for any $n\in\mathbb{Z}$ we define
$u_{n}$ such that

$u(x,t)=e^{inx}e^{i(an^{2}+bn^{3})t}u_{n}(x+(2an+3bn^{2})t,t),$

then taking $v:=u_{n}$ where

$u_{n}(x,t)=e^{-inx}e^{i(an^{2}+2bn^{3})t}u(x-(2an+3bn^{2})t,t)$ (4.3)

is a solution of

$\begin{cases}\partial_{t}v+iA\partial^{2}_{x}v+b\partial^{3}_{x}v=2iA|v|^{2}v+6b|v|^{2}\partial_{x}v,\quad
x,t\in\mathbb{R},\\\ v(x,0)=e^{-inx}u(x,0),\end{cases}$ (4.4)

where $A=a+3bn$, Theorem 3.5 and Remark 3.6 imply that

$\dfrac{d\alpha(u_{n}(t))}{dt}=0,$ (4.5)

and also have
$\widehat{u}(\xi,t)=e^{i(an^{2}+bn^{3})t}e^{i(2an+3bn^{2})(\xi-n)t}\widehat{u_{n}}(\xi-n,t)$,
thus

$|\widehat{u_{n}}(\xi,t)|=|\widehat{u}(\xi+n,t)|$ (4.6)

for any $n\in\mathbb{Z}$ and $\xi,t\in\mathbb{R}$.

We denote $\alpha$ by $\alpha(u_{n}(t))=\alpha(u_{n}(t),k)$, $k>0$, from the
definition of $\alpha$ we get

$\alpha(u_{n}(t))=\text{Re}\sum_{l=1}^{\infty}\dfrac{(-1)^{l-1}}{l}\text{tr}\left\\{\left((k-\partial_{x})^{-\frac{1}{2}}u_{n}(k+\partial_{x})^{-1}\overline{u_{n}}(k-\partial_{x})^{-\frac{1}{2}}\right)^{l}\right\\}.$
(4.7)

but using (3.3)

$\text{tr}\left((k-\partial_{x})^{-\frac{1}{2}}u_{n}(k+\partial_{x})^{-1}\overline{u_{n}}(k-\partial_{x})^{-\frac{1}{2}}\right)=\text{tr}\left((k-\partial_{x})^{-1}u_{n}(k+\partial_{x})^{-1}\overline{u_{n}}\right),$

moreover of equation (3.37) and (4.7), we get

$\alpha(u_{n}(t))=2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi+\text{Re}\sum_{l=2}^{\infty}\dfrac{(-1)^{l-1}}{l}\text{tr}\left\\{\left[\,M^{-}u_{n}{M^{+}}M^{+}\overline{u}_{n}M^{-}\,\right]^{l}\,\right\\}$
(4.8)

where $M^{\mp}=(k\mp\partial_{x})^{-\frac{1}{2}}$, from (4.8) and (3.41), we
get

$\left|\,\alpha(u_{n}(t))-2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\,\right|\leq\sum_{l=2}^{\infty}\dfrac{1}{l}\,||M^{-}u_{n}M^{+}||^{2l}.$
(4.9)

However of Lemma 3.1,

$||M^{-}u_{n}M^{+}||^{2l}\,\lesssim\,\left[\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{|k|+|\xi|}\,d\xi\right]^{l}.$
(4.10)

then from (4.9) and (4.10), we get

$\left|\,\alpha(u_{n}(t))-2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\,\right|\lesssim\sum_{l=2}^{\infty}\left[\int_{\mathbb{R}}<\xi>^{-1}|\widehat{u}_{n}(\xi,t)|^{2}\,d\xi\right]^{l}.$
(4.11)

Using (4.6), making a change of variables in (4.11), we get

$\left|\,\alpha(u_{n}(t))-2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\,\right|\,\lesssim\,\sum_{l=2}^{\infty}\left[\int_{\mathbb{R}}<\xi-n>^{-1}|\widehat{u}(\xi,t)|^{2}\,d\xi\right]^{l}.$
(4.12)

For each $j\in\mathbb{Z}$ consider the interval

$I_{j}=[j-\frac{1}{2},j+\frac{1}{2}[$

Next, we will estimate the right side of (4.12), when time $t=0$. In this
sense, we olserve that

$\int_{\mathbb{R}}<\xi-n>^{-1}|\widehat{u}(\xi,0)|^{2}\,d\xi\;\sim\;\sum_{j\in\mathbb{Z}}<j-n>^{-1}\int_{I_{j}}\,|\widehat{u}(\xi,0)|^{2}\,d\xi.$
(4.13)

However, if $p\geq 2$ and $1<q\leq\infty$ such that
$\frac{2}{p}+\frac{1}{q}=1$, using Hölder’s inequality, we get

$\begin{split}\sum_{j\in\mathbb{Z}}<j-n>^{-1}\|\widehat{u}(\xi,0)\|_{L^{2}(I_{j})}^{2}&\lesssim\|<j-n>^{-1}\|_{\ell_{j}^{q}}\,\|\,\|\widehat{u}(\xi,0)\|_{L^{2}(I_{j})}^{2}\|_{\ell_{j}^{\frac{p}{2}}(\mathbb{Z})}\\\
&\lesssim\|\,\|\widehat{u}(\xi,0)\|_{L^{2}(I_{j})}\|_{\ell_{j}^{p}(\mathbb{Z})}^{2}\,\sim\,\|\,u(0)\|_{M^{2,p}}^{2}\end{split}$
(4.14)

uniformly in $n\in\mathbb{Z}$.

By continuity of the solution operator, there exists a neighborhood
$I_{\lambda}=]-\lambda,\lambda[$ around $t=0$ such that

$\|\,u(t)\,\|_{M^{2,p}}\,\leq\,\epsilon\ll 1$ (4.15)

Applying similar ideas to obtain (4.14) , we have

$r_{n}(t)\,=\,\int_{\mathbb{R}}<\xi-n>^{-1}|\widehat{u}(\xi,t)|^{2}\,d\xi\,\lesssim\,\|u(t)\|_{M^{2,p}}^{2}.$
(4.16)

Note that $\sum_{2}^{\infty}\,(r_{n}(t))^{\textit{l}}$ it is a geometric
series of ratio $0\leq r_{n}(t)\leq\epsilon\ll 1$ and $t\in I_{\lambda}$ fixed
any. Then

Of (4.12), (4.16), we have

$\left|\,\alpha(u_{n}(t))-2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\,\right|\,\lesssim\,\sum_{l=2}^{\infty}\,(r_{n}(t))^{\textit{l}}\,=\,\dfrac{(r_{n}(t))^{2}}{1-r_{n}(t)}\,\lesssim\,(r_{n}(t))^{2},$
(4.17)

for all $n\in\mathbb{Z},\;t\in I_{\lambda}$. Thus using Young’s inequality to
series, considering $\frac{1}{p}+1=\frac{1}{q}+\frac{2}{p}$
($q=\frac{p}{p-1}$), we have

$\begin{split}\left\|\alpha(u_{n}(t))-2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\,\right\|_{\ell_{n}^{p/2}}&\,\lesssim\,\left\|\int_{\mathbb{R}}<\xi-n>^{-1}|\widehat{u}(\xi,t)|^{2}\,d\xi\right\|_{\ell_{n}^{p}}^{2}\\\
&\,\lesssim\,\left\|\sum_{j\in\mathbb{Z}}<n-j>^{-1}\int_{I_{j}}\,|\widehat{u}(\xi,t)|^{2}\,d\xi.\right\|_{\ell_{n}^{p}}^{2}\\\
&\,\lesssim\,\|\,\|\widehat{u}(\xi,t)\|_{L_{\xi}^{2}(I_{j})}^{2}\,\|_{\ell_{j}^{p/2}}^{2}\\\
&\,\sim\,\|u(t)\|_{M^{2,p}}^{4}\end{split}$ (4.18)

evenly for all $t\in I_{\lambda}$.

On the other hand, note that

$||u(t)||_{MH^{-1,p}}^{2}\,=\,\left[\sum_{n\in\mathbb{Z}}||<\xi-n>^{-1}\widehat{u}(\xi,t)||_{L_{\xi}^{2}}^{p}\right]^{2/p}.$
(4.19)

Then, changing the variable $\mu=\xi-n$, we get

$\|u(t)\|_{MH^{-1,p}}^{2}\,=\,\left\|\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\mu,t)|^{2}}{1+\mu^{2}}\,d\mu\right\|_{\ell_{n}^{p/2}(\mathbb{Z})}.$
(4.20)

Of (4.18) and (4.20), we get

$\left\|2kc\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{4k^{2}+\xi^{2}}\,d\xi\right\|_{\ell_{n}^{p/2}}\,\sim_{k}\,\left\|\int_{\mathbb{R}}\dfrac{|\widehat{u}_{n}(\xi,t)|^{2}}{1+\xi^{2}}\,d\xi\right\|_{\ell_{n}^{p/2}}\,\lesssim\,\|\alpha(u_{n}(t))\|_{\ell_{n}^{p/2}}+\|u(t)\|_{M^{2,p}}^{4}.$
(4.21)

Furthermore

$||\alpha(u_{n}(t))||_{\ell_{n}^{p/2}}\,\lesssim\,||u(t)||_{M^{2,p}}^{4}+||u(t)||_{MH^{-1,p}}^{2}.$
(4.22)

Note that

$||u(t)||_{M^{2,p}}^{2}\,\sim\,||u(t)||_{MH^{-1,p}}^{2}\,\lesssim\,||\alpha(u_{n}(t))||_{\ell_{n}^{p/2}}+||u(t)||_{M^{2,p}}^{4}.$
(4.23)

Of (4.22), (4.23) and given that $\alpha$ is conservative, we get

$||u(t)||_{M^{2,p}}^{2}\,\lesssim\,||u(0)||_{M^{2,p}}^{4}+||u(0)||_{MH^{-1,p}}^{2}+||u(t)||_{M^{2,p}}^{4}.$
(4.24)

Then

$||u(t)||_{M^{2,p}}^{2}\,\lesssim\,||u(0)||_{M^{2,p}}^{2}+||u(0)||_{M^{2,p}}^{4}+||u(t)||_{M^{2,p}}^{4}.$
(4.25)

for all $t\in I_{\lambda}$.

Of (4.15) and the continuity, we get

$||u(t)||_{M^{2,p}}^{2}\,\lesssim\,||u(0)||_{M^{2,p}}^{2}.$ (4.26)

evenly for all $t\in\mathbb{R}$. Finally we will consider the general case,
let $u_{\lambda}(x,t)=\lambda^{-1}u(\lambda^{-1}x,\lambda^{-3}t)$, then
$v:=u_{\lambda}$ is a solution of

$v_{t}+ia\lambda^{-1}v_{xx}+bv_{xxx}=2ia\lambda^{-1}|v|^{2}v+6b|v|^{2}v_{x},$
(4.27)

using the inequality $(c_{1}+\cdots c_{n})^{a}\leq n^{a-1}(c_{1}^{a}+\cdots
c_{n}^{a})$, $c_{j}\geq 0$, $j=1,\dots n$, we have

$\begin{split}\|u_{\lambda}(\cdot,0)|\|_{M^{2,p}}\sim&\lambda^{-1/2}\|\,\|\widehat{u}(\cdot,0)\|_{L^{2}(J_{\lambda,n})}\,\|_{\ell^{p}_{n}},\qquad
J_{\lambda,n}=\lambda I_{n}=[\lambda
n-\frac{\lambda}{2},n\lambda+\frac{\lambda}{2}]\\\
=&\lambda^{-1/2}\left(\sum_{n\in\mathbb{Z}}\left(\sum_{j=0}^{\lambda-1}\|\widehat{u}(\cdot,0)\|^{2}_{L^{2}(J_{\lambda,n,j})}\right)^{p/2}\right)^{1/p}\\\
\leq&\lambda^{-1/p}\|u(\cdot,0)\|_{M^{2,p}}\end{split}$ (4.28)

where $J_{\lambda,n,j}=[\lambda
n-\frac{\lambda}{2}+j,n\lambda-\frac{\lambda}{2}+j+1]$,
$n,\lambda\in\mathbb{Z}^{+}$, $0\leq j\leq\lambda-1$, hence choosing
$\lambda\in Z^{+}$ such that

$\lambda\sim(1+\|u(\cdot,0)\|_{M^{2,p}})^{p}$

we obtain

$\|u_{\lambda}(\cdot,0)|\|_{M^{2,p}}<\epsilon\ll 1,$

by the small data case presented above follows that.

$\|u_{\lambda}(\cdot,t)|\|_{M^{2,p}}\lesssim\|u_{\lambda}(\cdot,0)\|_{M^{2,p}}$
(4.29)

for all $t\in\mathbb{R}$. By scaling, (4.29) and (4.28) holds

$\begin{split}\|u(t)\|_{M^{2,p}}\leq\lambda^{1/2}\|u_{\lambda}(\cdot,\lambda^{3}t)|\|_{M^{2,p}}\lesssim\lambda^{1/2}\|u_{\lambda}(\cdot,0)|\|_{M^{2,p}}\lesssim\lambda^{1/2-1/p}\|u(\cdot,0)\|_{M^{2,p}}.\end{split}$
(4.30)

∎

This inequality proves the proposition.

######  Proposition 4.2.

Let $2\leq p<\infty$, $0\leq s<1-\frac{1}{p}$ and $u\in S(\mathbb{R})$ a
solution to (3.44). Then, there exists $C=C(p)$ positive such that

$\|u(t)\|_{M^{2,p}_{s}(\mathbb{R})}\leq
C\left(1+\|u(0)\|_{M^{2,p}_{s}(\mathbb{R})}\right)^{p/2-1}\|u(0)\|_{M^{2,p}_{s}(\mathbb{R})},$
(4.31)

for any $t\in\mathbb{R}$.

###### Proof.

Is the same proof as the proof of Theorem B1 (i) in [26]. ∎

Proof of Theorem 1.1

The proof of Theorem 1.1 in the case $0\leq s\leq 1-\frac{1}{p}$, follows from
Propositions 4.1 and 4.2 and in the case $s>1-\frac{1}{p}$ the proof is
similar as in Section 3.6 of [25].

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